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http://openstudy.com/updates/55838e19e4b069a97a2f07d7 | 1,485,089,689,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281424.85/warc/CC-MAIN-20170116095121-00361-ip-10-171-10-70.ec2.internal.warc.gz | 206,529,340 | 88,187 | ## NY,NY one year ago Which trigonometric expression does not simplify to 1? (see link)
1. NY,NY
2. Vocaloid
well, let's try each answer one by one and see what we get, shall we?
3. NY,NY
ok.
4. Vocaloid
|dw:1434685599699:dw|
5. anonymous
help me please my question is at the top of the feed
6. Vocaloid
does that make sense?
7. NY,NY
I dont understand how cos^2/sin^2 became just cos^2.
8. Vocaloid
I distributed sin^2(x)
9. Vocaloid
|dw:1434685846593:dw|
10. Vocaloid
see how sin^2(x) cancels out?
11. Vocaloid
|dw:1434685945486:dw|
12. Vocaloid
|dw:1434685962311:dw|
13. NY,NY
Oh yes that makes more sense.
14. NY,NY
Thank you for explaining that.
15. Vocaloid
ok, so we know that #1 is not the right answer because we get 1 in the end
16. Vocaloid
let's move on to number 2
17. NY,NY
Ok.
18. Vocaloid
|dw:1434686053840:dw|
19. Vocaloid
|dw:1434686204867:dw|
20. Vocaloid
does this make sense so far?
21. NY,NY
yes, because you used the Pythagorean identity at the end, right?
22. Vocaloid
yup! ok, we know that #2 is out, because we ended up with 1, so let's move on to #3
23. NY,NY
Right.
24. Vocaloid
|dw:1434686335785:dw|
25. Vocaloid
now, I believe #3 is the right answer, but let's check #4 just in case
26. Vocaloid
|dw:1434686426089:dw|
27. Vocaloid
|dw:1434686523673:dw|
28. Vocaloid
whew, that took a while, but do you understand everything so far? :)
29. Vocaloid
the important thing to remember, is that when you have csc, sec, cot, or tan, you can re-write them in terms of cos or sin, which usually helps you cross things out and simplify :)
30. NY,NY
Im a little confused with the second step you took.
31. Vocaloid
which one? #3 or #4?
32. NY,NY
#4.
33. Vocaloid
can you tell me exactly which part is confusing?
34. NY,NY
the step that the arrow is pointing to |dw:1434687552340:dw|
35. NY,NY
Im confused how that concluded from the previous step, where you substituted the reciprocals.
36. Vocaloid
|dw:1434687726888:dw|
37. Vocaloid
which leaves me with|dw:1434687748727:dw|
38. Vocaloid
does it make it more clear? or is there something you're still unsure about?
39. NY,NY
Ok, yes that makes sense.
40. NY,NY
So that leaves us with choice 3.
41. Vocaloid
yup ~
42. NY,NY
Ok, thank you for helping. | 743 | 2,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-04 | longest | en | 0.896627 | ## NY,NY one year ago Which trigonometric expression does not simplify to 1? (see link). 1. NY,NY. 2. Vocaloid. well, let's try each answer one by one and see what we get, shall we?. 3. NY,NY. ok.. 4. Vocaloid. |dw:1434685599699:dw|. 5. anonymous. help me please my question is at the top of the feed. 6. Vocaloid. does that make sense?. 7. NY,NY. I dont understand how cos^2/sin^2 became just cos^2.. 8. Vocaloid. I distributed sin^2(x). 9. Vocaloid. |dw:1434685846593:dw|. 10. Vocaloid. see how sin^2(x) cancels out?. 11. Vocaloid. |dw:1434685945486:dw|. 12. Vocaloid. |dw:1434685962311:dw|. 13. NY,NY. Oh yes that makes more sense.. 14. NY,NY. Thank you for explaining that.. 15. Vocaloid. ok, so we know that #1 is not the right answer because we get 1 in the end. 16. Vocaloid. let's move on to number 2. 17. NY,NY. Ok.. 18. Vocaloid. |dw:1434686053840:dw|. 19. Vocaloid. |dw:1434686204867:dw|. 20. Vocaloid. does this make sense so far?. 21. NY,NY. yes, because you used the Pythagorean identity at the end, right?. 22. | Vocaloid. yup! ok, we know that #2 is out, because we ended up with 1, so let's move on to #3. 23. NY,NY. Right.. 24. Vocaloid. |dw:1434686335785:dw|. 25. Vocaloid. now, I believe #3 is the right answer, but let's check #4 just in case. 26. Vocaloid. |dw:1434686426089:dw|. 27. Vocaloid. |dw:1434686523673:dw|. 28. Vocaloid. whew, that took a while, but do you understand everything so far? :). 29. Vocaloid. the important thing to remember, is that when you have csc, sec, cot, or tan, you can re-write them in terms of cos or sin, which usually helps you cross things out and simplify :). 30. NY,NY. Im a little confused with the second step you took.. 31. Vocaloid. which one? #3 or #4?. 32. NY,NY. #4.. 33. Vocaloid. can you tell me exactly which part is confusing?. 34. NY,NY. the step that the arrow is pointing to |dw:1434687552340:dw|. 35. NY,NY. Im confused how that concluded from the previous step, where you substituted the reciprocals.. 36. Vocaloid. |dw:1434687726888:dw|. 37. Vocaloid. which leaves me with|dw:1434687748727:dw|. 38. Vocaloid. does it make it more clear? or is there something you're still unsure about?. 39. NY,NY. Ok, yes that makes sense.. 40. NY,NY. So that leaves us with choice 3.. 41. Vocaloid. yup ~. 42. NY,NY. Ok, thank you for helping. |
https://goprep.co/ex-12.3-q7-the-perimeter-of-a-triangular-field-is-420-m-and-i-1nkc90 | 1,620,991,895,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990449.41/warc/CC-MAIN-20210514091252-20210514121252-00081.warc.gz | 294,103,633 | 27,671 | Q. 74.5( 8 Votes )
# The perimeter of
Given that the sides are in ratio of 6:7:8
Let the sides of the triangle be 6x, 7x and 8x.
Perimeter(Δ) = 420
6x + 7x + 8x = 420
21x = 420
x = 20
Therefore the sides are 120m, 140m and 160m.
a = 120, b = 140, c = 160
s = (a + b + c)/2
s = (120 + 140 + 160)/2 = 420/2 = 210.
Area(Δ) = √s(s-a)(s-b)(s-c)
Area(Δ) = √210(210-120)(210-140)(210-160)
Area(Δ) = √210×90×70×50
Area(Δ) = 2100√15 m2
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Find the area of NCERT Mathematics Exemplar | 371 | 1,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-21 | latest | en | 0.736112 | Q. 74.5( 8 Votes ). # The perimeter of. Given that the sides are in ratio of 6:7:8. Let the sides of the triangle be 6x, 7x and 8x.. Perimeter(Δ) = 420. 6x + 7x + 8x = 420. 21x = 420. x = 20. Therefore the sides are 120m, 140m and 160m.. a = 120, b = 140, c = 160. s = (a + b + c)/2. s = (120 + 140 + 160)/2 = 420/2 = 210.. Area(Δ) = √s(s-a)(s-b)(s-c). Area(Δ) = √210(210-120)(210-140)(210-160). Area(Δ) = √210×90×70×50. Area(Δ) = 2100√15 m2. Rate this question :. | How useful is this solution?. We strive to provide quality solutions. Please rate us to serve you better.. Related Videos. Heron's Formula- Full Gyan46 mins. Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts. Dedicated counsellor for each student. 24X7 Doubt Resolution. Daily Report Card. Detailed Performance Evaluation. view all courses. RELATED QUESTIONS :. A piece of land iRS Aggarwal & V Aggarwal - Mathematics. The question consRS Aggarwal & V Aggarwal - Mathematics. The dimensions ofNCERT Mathematics Exemplar. A hand fan is madRD Sharma - Mathematics. Find the area of NCERT Mathematics Exemplar. |
http://docplayer.net/21763825-C-complex-numbers-1-complex-arithmetic.html | 1,550,457,506,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484020.33/warc/CC-MAIN-20190218013525-20190218035525-00541.warc.gz | 70,931,476 | 28,819 | # C. Complex Numbers. 1. Complex arithmetic.
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1 C. Complex Numbers. Complex arithmetic. Most people think that complex numbers arose from attempts to solve quadratic equations, but actually it was in connection with cubic equations they first appeared. Everyone knew that certain quadratic equations, like x 2 + = 0, or x 2 +2x+5 = 0, had no solutions. The problem was with certain cubic equations, for example x 6x+2 = 0. This equation was known to have three real roots, given by simple combinations of the expressions () A = + 7, B = 7; one of the roots for instance is A+B: it may not look like a real number, but it turns out to be one. What was to be made of the expressions A and B? They were viewed as some sort of imaginary numbers which had no meaning in themselves, but which were useful as intermediate steps in calculations that would ultimately lead to the real numbers you were looking for (such as A+B). This point of view persisted for several hundred years. But as more and more applications for these imaginary numbers were found, they gradually began to be accepted as valid numbers in their own right, even though they did not measure the length of any line segment. Nowadays we are fairly generous in the use of the word number : numbers of one sort or another don t have to measure anything, but to merit the name they must belong to a system in which some type of addition, subtraction, multiplication, and division is possible, and where these operations obey those laws of arithmetic one learns in elementary school and has usually forgotten by high school the commutative, associative, and distributive laws. To describe the complex numbers, we use a formal symbol i representing ; then a complex number is an expression of the form (2) a + bi, a, b real numbers. If a = 0 or b = 0, they are omitted (unless both are 0); thus we write a+0i = a, 0+bi = bi, 0+0i = 0. The definition of equality between two complex numbers is () a+bi = c+di a = c, b = d. This shows that the numbers a and b are uniquely determined once the complex number a+bi is given; we call them respectively the real and imaginary parts of a+bi. (It would be more logical to call bi the imaginary part, but this would be less convenient.) In symbols, (4) a = Re(a+bi), b = Im(a+bi)
2 2 8.0 NOTES Addition and multiplication of complex numbers are defined in the familiar way, making use of the fact that i 2 = : (5a) (5b) Addition Multiplication (a+bi)+(c+di) = (a+c)+(b+d)i (a+bi)(c+di) = (ac bd)+(ad+bc)i Division is a little more complicated; what is important is not so much the final formula but rather the procedure which produces it; assuming c+di 0, it is: (5c) Division a+bi c+di = a+bi c+di c di c di = ac+bd c 2 +d 2 + bc ad c 2 +d 2 i This division procedure made use of complex conjugation: if z = a + bi, we define the complex conjugate of z to be the complex number (6) z = a bi (note that z z = a 2 +b 2 ). The size of a complex number is measured by its absolute value, or modulus, defined by (7) z = a+bi = a 2 +b 2 ; (thus : z z = z 2 ). Remarks. For the sake of computers, which do not understand what a formal expression is, one can define a complex number to be just an ordered pair (a,b) of real numbers, and define the arithmetic operations accordingly; using (5b), multiplication is defined by (a,b)(c,d) = (ac bd, ad+bc). Then if we let i represent the ordered pair (0,), and a the ordered pair (a,0), it is easy to verify using the above definition of multiplication that i 2 = (0,)(0,) = (,0) = and (a,b) = (a,0)+(b,0)(0,) = a+bi, and we recover the human way of writing complex numbers. Since it is easily verified from the definition that multiplication of complex numbers is commutative: z z 2 = z 2 z, it does not matter whether the i comes before or after, i.e., whether we write z = x + yi or z = x + iy. The former is used when x and y are simple numbers because it looks better; the latter is more usual when x and y represent functions (orvaluesoffunctions), tomaketheistandoutclearlyortoavoidhavingtouseparentheses: 2. Polar representation. 2+i, 5 2πi; cos π 2 +i sin π 2, x(t)+iy(t). Complex numbers are represented geometrically by points in the plane: the number a+ib is represented by the point (a,b) in Cartesian coordinates. When the points of the plane represent complex numbers in this way, the plane is called the complex plane. By switching to polar coordinates, we can write any non-zero complex number in an alternative form. Letting as usual x = rcosθ, y = rsinθ, we get the polar form for a non-zero complex number: assuming x+iy 0, (8) x+iy = r(cosθ +isinθ). When the complex number is written in polar form, we see from (7) that r = x + iy. (absolute value, modulus)
3 C. COMPLEX NUMBERS We call θ the polar angle or the argument of x+iy. In symbols, one sometimes sees θ = arg (x+iy) (polar angle, argument). The absolute value is uniquely determined by x+iy, but the polar angle is not, since it can be increased by any integer multiple of 2π. (The complex number 0 has no polar angle.) To make θ unique, one can specify 0 θ < 2π principal value of the polar angle. This so-called principal value of the angle is sometimes indicated by writing Arg (x + iy). For example, Arg ( ) = π, arg ( ) = ±π,±π,±5π,.... Changing between Cartesian and polar representation of a complex number is essentially the same as changing between Cartesian and polar coordinates: the same equations are used. Example. Give the polar form for: i, +i, i, +i. Solution. i = i sin π 2 +i = 2(cos π 4 +i sin π 4 ) +i = 2(cos 2π +i sin 2π ) i = 2(cos π 4 +i sin π 4 ) The abbreviation cisθ is sometimes used for cosθ+isinθ; for students of science and engineering, however, it is important to get used to the exponential form for this expression: (9) e iθ = cosθ +isinθ Euler s formula. Equation (9) should be regarded as the definition of the exponential of an imaginary power. A good justification for it however is found in the infinite series e t = + t! + t2 2! + t! If we substitute iθ for t in the series, and collect the real and imaginary parts of the sum (remembering that i 2 =, i = i, i 4 =, i 5 = i,..., and so on, we get ) e iθ = ( θ2 2! + θ4 4!... + i = cosθ + isinθ, in view of the infinite series representations for cosθ and sinθ. ) (θ θ! + θ5 5!... Since we only know that the series expansion for e t is valid when t is a real number, the above argument is only suggestive it is not a proof of (9). What it shows is that Euler s formula (9) is formally compatible with the series expansions for the exponential, sine, and cosine functions. Using the complex exponential, the polar representation (8) is written (0) x+iy = re iθ The most important reason for polar representation is that multiplication and division of complex numbers is particularly simple when they are written in polar form. Indeed, by using Euler s formula (9) and the trigonometric addition formulas, it is not hard to show
4 4 8.0 NOTES () e iθ e iθ = e i(θ+θ ). This gives another justification for the definition (9) it makes the complex exponential follow the same exponential addition law as the real exponential. The law () leads to the simple rules for multiplying and dividing complex numbers written in polar form: (2a) multiplication rule re iθ r e iθ = rr e i(θ+θ ) ; to multiply two complex numbers, you multiply the absolute values and add the angles. (2b) reciprocal rule re iθ = r e iθ ; (2c) division rule r = r e iθ r e i(θ θ ) ; to divide by a complex number, divide by its absolute value and subtract its angle. re iθ The reciprocal rule (2b) follows from (2a), which shows that r e iθ re iθ =. The division rule follows by writing reiθ r e iθ = r e iθ re iθ and using (2b) and then (2a). Using (2a), we can raise x+iy to a positive integer power by first using x+iy = re iθ ; the special case when r = is called DeMoivre s formula: () (x+iy) n = r n e inθ ; DeMoivre s formula: (cosθ+isinθ) n = cosnθ+isinnθ. Example 2. Express a) (+i) 6 in Cartesian form; b) +i +i in polar form. Solution. a) Change to polar form, use (), then change back to Cartesian form: (+i) 6 = ( 2e iπ/4 ) 6 = ( 2) 6 e i6π/4 = 8e iπ/2 = 8i. +i b) Changing to polar form, = 2eiπ/ +i 2e = iπ/6 eiπ/6, using the division rule (2c). You can check the answer to (a) by applying the binomial theorem to (+i) 6 and collecting the real and imaginary parts; to (b) by doing the division in Cartesian form (5c), then converting the answer to polar form.. Complex exponentials Because of the importance of complex exponentials in differential equations, and in science and engineering generally, we go a little further with them. Euler s formula (9) defines the exponential to a pure imaginary power. The definition of an exponential to an arbitrary complex power is: (4) e a+ib = e a e ib = e a (cosb+isinb). We stress that the equation (4) is a definition, not a self-evident truth, since up to now no meaning has been assigned to the left-hand side. From (4) we see that (5) Re(e a+ib ) = e a cosb, Im(e a+ib ) = e a sinb.
5 The complex exponential obeys the usual law of exponents: (6) e z+z = e z e z, as is easily seen by combining (4) and (). C. COMPLEX NUMBERS 5 The complex exponential is expressed in terms of the sine and cosine by Euler s formula (9). Conversely, the sin and cos functions can be expressed in terms of complex exponentials. There are two important ways of doing this, both of which you should learn: (7) cosx = Re(e ix ), sinx = Im(e ix ) ; (8) cosx = 2 (eix +e ix ), sinx = 2i (eix e ix ). The equations in (8) follow easily from Euler s formula (9); their derivation is left for the exercises. Here are some examples of their use. Example. Express cos x in terms of the functions cosnx, for suitable n. Solution. We use (8) and the binomial theorem, then (8) again: cos x = 8 (eix +e ix ) = 8 (eix +e ix +e ix +e ix ) = 4 cosx+ 4 cosx. As a preliminary to the next example, we note that a function like e ix = cosx+isinx is a complex-valued function of the real variable x. Such a function may be written as u(x)+iv(x), u, v real-valued and its derivative and integral with respect to x are defined to be (9a,b) a) D(u+iv) = Du+iDv, b) (u+iv)dx = udx+i vdx. From this it follows by a calculation that (20) D(e (a+ib)x = (a+ib)e (a+ib)x, and therefore e (a+ib)x dx = a+ib e(a+ib)x. Example 4. Calculate e x cos2xdx by using complex exponentials. Solution. The usual method is a tricky use of two successive integration by parts. Using complex exponentials instead, the calculation is straightforward. We have ( e x cos2x = Re e (+2i)x), by (4) or (5); therefore ( ) e x cos2xdx = Re e (+2i)x dx, by (9b). Calculating the integral, e (+2i)x dx = +2i e(+2i)x by (20); ( = 5 2 ) (e 5 i x cos2x+ie x sin2x ), using (4) and complex division (5c). According to the second line above, we want the real part of this last expression. Multiply using (5b) and take the real part; you get 5 ex cos2x+ 2 5 ex sin2x.
6 6 8.0 NOTES In this differential equations course, we will make free use of complex exponentials in solving differential equations, and in doing formal calculations like the ones above. This is standard practice in science and engineering, and you need to get used to it. 4. Finding n-th roots. To solve linear differential equations with constant coefficients, you need to be able find the real and complex roots of polynomial equations. Though a lot of this is done today with calculators and computers, one still has to know how to do an important special case by hand: finding the roots of z n = α, where α is a complex number, i.e., finding the n-th roots of α. Polar representation will be a big help in this. Let s begin with a special case: the n-th roots of unity: the solutions to z n =. To solve this equation, we use polar representation for both sides, setting z = re iθ on the left, and using all possible polar angles on the right; using the exponential law to multiply, the above equation then becomes r n e inθ = e (2kπi), k = 0,±,±2,.... Equating the absolute values and the polar angles of the two sides gives from which we conclude that ( ) r =, θ = 2kπ n r n =, nθ = 2kπ, k = 0,±,±2,...,, k = 0,,...,n. In the above, we get only the value r =, since r must be real and non-negative. We don t need any integer values of k other than 0,...,n since they would not produce a complex number different from the above n numbers. That is, if we add an, an integer multiple of n, to k, we get the same complex number: θ = 2(k +an)π n = θ +2aπ; and e iθ = e iθ, since e 2aπi = (e 2πi ) a =. We conclude from ( ) therefore that (2) the n-th roots of are the numbers e 2kπi/n, k = 0,...,n. 2π i e πi e This shows there are n complex n-th roots of unity. They all lie on the unit circle in the complex plane, since they have absolute value ; they are evenly spaced around the unit circle, starting with ; the angle between two consecutive ones is 2π/n. These facts are illustrated on the right for the case n = 6. 4πi π i e e 5
7 C. COMPLEX NUMBERS 7 From (2), we get another notation for the roots of unity (ζ is the Greek letter zeta ): (22) the n-th roots of are,ζ,ζ 2,...,ζ n, where ζ = e 2πi/n. We now generalize the above to find the n-th roots of an arbitrary complex number w. We begin by writing w in polar form: w = re iθ ; θ = Arg w, 0 θ < 2π, i.e., θ is the principal value of the polar angle of w. Then the same reasoning as we used above shows that if z is an n-th root of w, then (2) z n = w = re iθ, so z = n re i(θ+2kπ)/n, k = 0,,...,n. Comparing this with (22), we see that these n roots can be written in the suggestive form (24) n w = z 0, z 0 ζ, z 0 ζ 2,..., z 0 ζ n, where z 0 = n re iθ/n. As a check, we see that all of the n complex numbers in (24) satisfy z n = w : (z 0 ζ i ) n = z0ζ n ni = z0 n i, since ζ n =, by (22); = w, by the definition (24) of z 0 and (2). Example 5. Find in Cartesian form all values of a) b) 4 i. Solution. a) According to (22), the cube roots of are,ω, and ω 2, where ω = e 2πi/ = cos 2π +isin 2π ω 2 = e 2πi/ = cos 2π +isin 2π = 2 +i 2 = 2 i 2. The greek letter ω ( omega ) is traditionally used for this cube root. Note that for the polar angle of ω 2 we used 2π/ rather than the equivalent angle 4π/, in order to take advantage of the identities cos( x) = cosx, sin( x) = sinx. Note that ω 2 = ω. Another way to do this problem would be to draw the position of ω 2 and ω on the unit circle, and use geometry to figure out their coordinates. b) To find 4 i, we can use (24). We know that 4 =,i,, i (either by drawing the unit circle picture, or by using (22)). Therefore by (24), we get 4 i = z 0, z 0 i, z 0, z 0 i, where z 0 = e πi/8 = cos π 8 +isin π 8 ; = a+ib, b+ia, a ib, b ia, where z 0 = a+ib = cos π 8 +isin π 8.
8 8 8.0 NOTES Example 6. Solve the equation x 6 2x +2 = 0. Solution. Treating this as a quadratic equation in x, we solve the quadratic by using the quadratic formula, the two roots are + i and i (check this!), so the roots of the original equation satisfy either x = +i, or x = i. This reduces the problem to finding the cube roots of the two complex numbers ± i. We begin by writing them in polar form: +i = 2e πi/4, i = 2e πi/4. (Once again, note the use of the negative polar angle for i, which is more convenient for calculations.) The three cube roots of the first of these are (by (2)), 6 2 e πi/2 = 6 ( 2 cos π 2 +isin π e πi/4 = 6 ( 2 cos π 4 +isin π e 7πi/2 = 6 ( 2 ) cos 7π 7π isin 2 2 ), since ), since The second cube root can also be written as 6 ( ) +i 2 2 π 2 + 2π = π 4 ; π 2 2π = 7π 2. = +i 2. This gives three of the cube roots. The other three are the cube roots of i, which may be found by replacing i by i everywhere above (i.e., taking the complex conjugate). The cube roots can also according to (24) be described as z, z ω, z ω 2 and z 2, z 2 ω, z 2 ω 2, where z = 6 2 e πi/2, z 2 = 6 2 e πi/2. Exercises: Section 2E
9 M.I.T. 8.0 Ordinary Differential Equations 8.0 Notes and Exercises c Arthur Mattuck and M.I.T. 988, 992, 996, 200, 2007, 20
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Most people think that complex numbers arose from attempts to solve quadratic equations, but actually it was in connection with cubic equations they first appeared. Everyone knew that certain quadratic equations, like x 2 + = 0, or x 2 +2x+5 = 0, had no solutions. The problem was with certain cubic equations, for example x 6x+2 = 0. This equation was known to have three real roots, given by simple combinations of the expressions () A = + 7, B = 7; one of the roots for instance is A+B: it may not look like a real number, but it turns out to be one. What was to be made of the expressions A and B? They were viewed as some sort of imaginary numbers which had no meaning in themselves, but which were useful as intermediate steps in calculations that would ultimately lead to the real numbers you were looking for (such as A+B). This point of view persisted for several hundred years. But as more and more applications for these imaginary numbers were found, they gradually began to be accepted as valid numbers in their own right, even though they did not measure the length of any line segment. Nowadays we are fairly generous in the use of the word number : numbers of one sort or another don t have to measure anything, but to merit the name they must belong to a system in which some type of addition, subtraction, multiplication, and division is possible, and where these operations obey those laws of arithmetic one learns in elementary school and has usually forgotten by high school the commutative, associative, and distributive laws. To describe the complex numbers, we use a formal symbol i representing ; then a complex number is an expression of the form (2) a + bi, a, b real numbers. If a = 0 or b = 0, they are omitted (unless both are 0); thus we write a+0i = a, 0+bi = bi, 0+0i = 0. The definition of equality between two complex numbers is () a+bi = c+di a = c, b = d. This shows that the numbers a and b are uniquely determined once the complex number a+bi is given; we call them respectively the real and imaginary parts of a+bi. (It would be more logical to call bi the imaginary part, but this would be less convenient.) In symbols, (4) a = Re(a+bi), b = Im(a+bi). 2 2 8.0 NOTES Addition and multiplication of complex numbers are defined in the familiar way, making use of the fact that i 2 = : (5a) (5b) Addition Multiplication (a+bi)+(c+di) = (a+c)+(b+d)i (a+bi)(c+di) = (ac bd)+(ad+bc)i Division is a little more complicated; what is important is not so much the final formula but rather the procedure which produces it; assuming c+di 0, it is: (5c) Division a+bi c+di = a+bi c+di c di c di = ac+bd c 2 +d 2 + bc ad c 2 +d 2 i This division procedure made use of complex conjugation: if z = a + bi, we define the complex conjugate of z to be the complex number (6) z = a bi (note that z z = a 2 +b 2 ). The size of a complex number is measured by its absolute value, or modulus, defined by (7) z = a+bi = a 2 +b 2 ; (thus : z z = z 2 ). Remarks. For the sake of computers, which do not understand what a formal expression is, one can define a complex number to be just an ordered pair (a,b) of real numbers, and define the arithmetic operations accordingly; using (5b), multiplication is defined by (a,b)(c,d) = (ac bd, ad+bc). Then if we let i represent the ordered pair (0,), and a the ordered pair (a,0), it is easy to verify using the above definition of multiplication that i 2 = (0,)(0,) = (,0) = and (a,b) = (a,0)+(b,0)(0,) = a+bi, and we recover the human way of writing complex numbers. Since it is easily verified from the definition that multiplication of complex numbers is commutative: z z 2 = z 2 z, it does not matter whether the i comes before or after, i.e., whether we write z = x + yi or z = x + iy. The former is used when x and y are simple numbers because it looks better; the latter is more usual when x and y represent functions (orvaluesoffunctions), tomaketheistandoutclearlyortoavoidhavingtouseparentheses: 2. Polar representation. 2+i, 5 2πi; cos π 2 +i sin π 2, x(t)+iy(t). Complex numbers are represented geometrically by points in the plane: the number a+ib is represented by the point (a,b) in Cartesian coordinates. When the points of the plane represent complex numbers in this way, the plane is called the complex plane. By switching to polar coordinates, we can write any non-zero complex number in an alternative form. Letting as usual x = rcosθ, y = rsinθ, we get the polar form for a non-zero complex number: assuming x+iy 0, (8) x+iy = r(cosθ +isinθ). When the complex number is written in polar form, we see from (7) that r = x + iy. (absolute value, modulus). 3 C. COMPLEX NUMBERS We call θ the polar angle or the argument of x+iy. In symbols, one sometimes sees θ = arg (x+iy) (polar angle, argument). The absolute value is uniquely determined by x+iy, but the polar angle is not, since it can be increased by any integer multiple of 2π. (The complex number 0 has no polar angle.) To make θ unique, one can specify 0 θ < 2π principal value of the polar angle. This so-called principal value of the angle is sometimes indicated by writing Arg (x + iy). For example, Arg ( ) = π, arg ( ) = ±π,±π,±5π,.... Changing between Cartesian and polar representation of a complex number is essentially the same as changing between Cartesian and polar coordinates: the same equations are used. Example. Give the polar form for: i, +i, i, +i. Solution. i = i sin π 2 +i = 2(cos π 4 +i sin π 4 ) +i = 2(cos 2π +i sin 2π ) i = 2(cos π 4 +i sin π 4 ) The abbreviation cisθ is sometimes used for cosθ+isinθ; for students of science and engineering, however, it is important to get used to the exponential form for this expression: (9) e iθ = cosθ +isinθ Euler s formula. Equation (9) should be regarded as the definition of the exponential of an imaginary power. A good justification for it however is found in the infinite series e t = + t! + t2 2! + t! If we substitute iθ for t in the series, and collect the real and imaginary parts of the sum (remembering that i 2 =, i = i, i 4 =, i 5 = i,..., and so on, we get ) e iθ = ( θ2 2! + θ4 4!... + i = cosθ + isinθ, in view of the infinite series representations for cosθ and sinθ. ) (θ θ! + θ5 5!... Since we only know that the series expansion for e t is valid when t is a real number, the above argument is only suggestive it is not a proof of (9). What it shows is that Euler s formula (9) is formally compatible with the series expansions for the exponential, sine, and cosine functions. Using the complex exponential, the polar representation (8) is written (0) x+iy = re iθ The most important reason for polar representation is that multiplication and division of complex numbers is particularly simple when they are written in polar form. Indeed, by using Euler s formula (9) and the trigonometric addition formulas, it is not hard to show. 4 4 8.0 NOTES () e iθ e iθ = e i(θ+θ ). This gives another justification for the definition (9) it makes the complex exponential follow the same exponential addition law as the real exponential. The law () leads to the simple rules for multiplying and dividing complex numbers written in polar form: (2a) multiplication rule re iθ r e iθ = rr e i(θ+θ ) ; to multiply two complex numbers, you multiply the absolute values and add the angles. (2b) reciprocal rule re iθ = r e iθ ; (2c) division rule r = r e iθ r e i(θ θ ) ; to divide by a complex number, divide by its absolute value and subtract its angle. re iθ The reciprocal rule (2b) follows from (2a), which shows that r e iθ re iθ =. The division rule follows by writing reiθ r e iθ = r e iθ re iθ and using (2b) and then (2a). Using (2a), we can raise x+iy to a positive integer power by first using x+iy = re iθ ; the special case when r = is called DeMoivre s formula: () (x+iy) n = r n e inθ ; DeMoivre s formula: (cosθ+isinθ) n = cosnθ+isinnθ. Example 2. Express a) (+i) 6 in Cartesian form; b) +i +i in polar form. Solution. a) Change to polar form, use (), then change back to Cartesian form: (+i) 6 = ( 2e iπ/4 ) 6 = ( 2) 6 e i6π/4 = 8e iπ/2 = 8i. +i b) Changing to polar form, = 2eiπ/ +i 2e = iπ/6 eiπ/6, using the division rule (2c). You can check the answer to (a) by applying the binomial theorem to (+i) 6 and collecting the real and imaginary parts; to (b) by doing the division in Cartesian form (5c), then converting the answer to polar form.. Complex exponentials Because of the importance of complex exponentials in differential equations, and in science and engineering generally, we go a little further with them. Euler s formula (9) defines the exponential to a pure imaginary power. The definition of an exponential to an arbitrary complex power is: (4) e a+ib = e a e ib = e a (cosb+isinb). We stress that the equation (4) is a definition, not a self-evident truth, since up to now no meaning has been assigned to the left-hand side. From (4) we see that (5) Re(e a+ib ) = e a cosb, Im(e a+ib ) = e a sinb.. 5 The complex exponential obeys the usual law of exponents: (6) e z+z = e z e z, as is easily seen by combining (4) and (). C. COMPLEX NUMBERS 5 The complex exponential is expressed in terms of the sine and cosine by Euler s formula (9). Conversely, the sin and cos functions can be expressed in terms of complex exponentials. There are two important ways of doing this, both of which you should learn: (7) cosx = Re(e ix ), sinx = Im(e ix ) ; (8) cosx = 2 (eix +e ix ), sinx = 2i (eix e ix ). The equations in (8) follow easily from Euler s formula (9); their derivation is left for the exercises. Here are some examples of their use. Example. Express cos x in terms of the functions cosnx, for suitable n. Solution. We use (8) and the binomial theorem, then (8) again: cos x = 8 (eix +e ix ) = 8 (eix +e ix +e ix +e ix ) = 4 cosx+ 4 cosx. As a preliminary to the next example, we note that a function like e ix = cosx+isinx is a complex-valued function of the real variable x. Such a function may be written as u(x)+iv(x), u, v real-valued and its derivative and integral with respect to x are defined to be (9a,b) a) D(u+iv) = Du+iDv, b) (u+iv)dx = udx+i vdx. From this it follows by a calculation that (20) D(e (a+ib)x = (a+ib)e (a+ib)x, and therefore e (a+ib)x dx = a+ib e(a+ib)x. Example 4. Calculate e x cos2xdx by using complex exponentials. Solution. The usual method is a tricky use of two successive integration by parts. Using complex exponentials instead, the calculation is straightforward. We have ( e x cos2x = Re e (+2i)x), by (4) or (5); therefore ( ) e x cos2xdx = Re e (+2i)x dx, by (9b). Calculating the integral, e (+2i)x dx = +2i e(+2i)x by (20); ( = 5 2 ) (e 5 i x cos2x+ie x sin2x ), using (4) and complex division (5c). According to the second line above, we want the real part of this last expression. Multiply using (5b) and take the real part; you get 5 ex cos2x+ 2 5 ex sin2x.. 6 6 8.0 NOTES In this differential equations course, we will make free use of complex exponentials in solving differential equations, and in doing formal calculations like the ones above. This is standard practice in science and engineering, and you need to get used to it. 4. Finding n-th roots. To solve linear differential equations with constant coefficients, you need to be able find the real and complex roots of polynomial equations. Though a lot of this is done today with calculators and computers, one still has to know how to do an important special case by hand: finding the roots of z n = α, where α is a complex number, i.e., finding the n-th roots of α. Polar representation will be a big help in this. Let s begin with a special case: the n-th roots of unity: the solutions to z n =. To solve this equation, we use polar representation for both sides, setting z = re iθ on the left, and using all possible polar angles on the right; using the exponential law to multiply, the above equation then becomes r n e inθ = e (2kπi), k = 0,±,±2,.... Equating the absolute values and the polar angles of the two sides gives from which we conclude that ( ) r =, θ = 2kπ n r n =, nθ = 2kπ, k = 0,±,±2,...,, k = 0,,...,n. In the above, we get only the value r =, since r must be real and non-negative. We don t need any integer values of k other than 0,...,n since they would not produce a complex number different from the above n numbers. That is, if we add an, an integer multiple of n, to k, we get the same complex number: θ = 2(k +an)π n = θ +2aπ; and e iθ = e iθ, since e 2aπi = (e 2πi ) a =. We conclude from ( ) therefore that (2) the n-th roots of are the numbers e 2kπi/n, k = 0,...,n. 2π i e πi e This shows there are n complex n-th roots of unity. They all lie on the unit circle in the complex plane, since they have absolute value ; they are evenly spaced around the unit circle, starting with ; the angle between two consecutive ones is 2π/n. These facts are illustrated on the right for the case n = 6. 4πi π i e e 5. 7 C. COMPLEX NUMBERS 7 From (2), we get another notation for the roots of unity (ζ is the Greek letter zeta ): (22) the n-th roots of are,ζ,ζ 2,...,ζ n, where ζ = e 2πi/n. We now generalize the above to find the n-th roots of an arbitrary complex number w. We begin by writing w in polar form: w = re iθ ; θ = Arg w, 0 θ < 2π, i.e., θ is the principal value of the polar angle of w. Then the same reasoning as we used above shows that if z is an n-th root of w, then (2) z n = w = re iθ, so z = n re i(θ+2kπ)/n, k = 0,,...,n. Comparing this with (22), we see that these n roots can be written in the suggestive form (24) n w = z 0, z 0 ζ, z 0 ζ 2,..., z 0 ζ n, where z 0 = n re iθ/n. As a check, we see that all of the n complex numbers in (24) satisfy z n = w : (z 0 ζ i ) n = z0ζ n ni = z0 n i, since ζ n =, by (22); = w, by the definition (24) of z 0 and (2). Example 5. Find in Cartesian form all values of a) b) 4 i. Solution. a) According to (22), the cube roots of are,ω, and ω 2, where ω = e 2πi/ = cos 2π +isin 2π ω 2 = e 2πi/ = cos 2π +isin 2π = 2 +i 2 = 2 i 2. The greek letter ω ( omega ) is traditionally used for this cube root. Note that for the polar angle of ω 2 we used 2π/ rather than the equivalent angle 4π/, in order to take advantage of the identities cos( x) = cosx, sin( x) = sinx. Note that ω 2 = ω. Another way to do this problem would be to draw the position of ω 2 and ω on the unit circle, and use geometry to figure out their coordinates. b) To find 4 i, we can use (24). We know that 4 =,i,, i (either by drawing the unit circle picture, or by using (22)). Therefore by (24), we get 4 i = z 0, z 0 i, z 0, z 0 i, where z 0 = e πi/8 = cos π 8 +isin π 8 ; = a+ib, b+ia, a ib, b ia, where z 0 = a+ib = cos π 8 +isin π 8.. 8 8 8.0 NOTES Example 6. Solve the equation x 6 2x +2 = 0. Solution. Treating this as a quadratic equation in x, we solve the quadratic by using the quadratic formula, the two roots are + i and i (check this!), so the roots of the original equation satisfy either x = +i, or x = i. This reduces the problem to finding the cube roots of the two complex numbers ± i. We begin by writing them in polar form: +i = 2e πi/4, i = 2e πi/4. (Once again, note the use of the negative polar angle for i, which is more convenient for calculations.) The three cube roots of the first of these are (by (2)), 6 2 e πi/2 = 6 ( 2 cos π 2 +isin π e πi/4 = 6 ( 2 cos π 4 +isin π e 7πi/2 = 6 ( 2 ) cos 7π 7π isin 2 2 ), since ), since The second cube root can also be written as 6 ( ) +i 2 2 π 2 + 2π = π 4 ; π 2 2π = 7π 2. = +i 2. This gives three of the cube roots. The other three are the cube roots of i, which may be found by replacing i by i everywhere above (i.e., taking the complex conjugate). The cube roots can also according to (24) be described as z, z ω, z ω 2 and z 2, z 2 ω, z 2 ω 2, where z = 6 2 e πi/2, z 2 = 6 2 e πi/2. Exercises: Section 2E. 9 M.I.T. 8.0 Ordinary Differential Equations 8.0 Notes and Exercises c Arthur Mattuck and M.I.T. 988, 992, 996, 200, 2007, 20. ### COMPLEX NUMBERS. a bi c di a c b d i. a bi c di a c b d i For instance, 1 i 4 7i 1 4 1 7 i 5 6i. COMPLEX NUMBERS _4+i _-i FIGURE Complex numbers as points in the Arg plane i _i +i -i A complex number can be represented by an expression of the form a bi, where a b are real numbers i is a symbol with. ### THE COMPLEX EXPONENTIAL FUNCTION. Math 307 THE COMPLEX EXPONENTIAL FUNCTION (These notes assume you are already familiar with the basic properties of complex numbers.) We make the following definition e iθ = cos θ + i sin θ. 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Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which. ### Student Performance Q&A:. Student Performance Q&A: AP Calculus AB and Calculus BC Free-Response Questions The following comments on the free-response questions for AP Calculus AB and Calculus BC were written by the Chief Reader,. ### MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education). MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education) Accurately add, subtract, multiply, and divide whole numbers, integers,. ### Precalculus with Geometry and Trigonometry. 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Fields 4 1.3. The field of complex numbers. 6 1.4. |
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# Tent map
Graph of tent map function
In mathematics, the tent map with parameter μ is the real-valued function fμ defined by
${\displaystyle f_{\mu }:=\mu \min\{x,\,1-x\},}$
the name being due to the tent-like shape of the graph of fμ. For the values of the parameter μ within 0 and 2, fμ maps the unit interval [0, 1] into itself, thus defining a discrete-time dynamical system on it (equivalently, a recurrence relation). In particular, iterating a point x0 in [0, 1] gives rise to a sequence ${\displaystyle x_{n}}$ :
${\displaystyle x_{n+1}=f_{\mu }(x_{n})={\begin{cases}\mu x_{n}&\mathrm {for} ~~x_{n}<{\frac {1}{2}}\\\\\mu (1-x_{n})&\mathrm {for} ~~{\frac {1}{2}}\leq x_{n}\end{cases}}}$
where μ is a positive real constant. Choosing for instance the parameter μ=2, the effect of the function fμ may be viewed as the result of the operation of folding the unit interval in two, then stretching the resulting interval [0,1/2] to get again the interval [0,1]. Iterating the procedure, any point x0 of the interval assumes new subsequent positions as described above, generating a sequence xn in [0,1].
The ${\displaystyle \mu =2}$ case of the tent map is a non-linear transformation of both the bit shift map and the r=4 case of the logistic map.
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• ✪ ChaosBook.org chapter Charting the state space - Tent map
• ✪ ChaosBook.org chapter Stretch, fold, prune - Tent map symbolic dynamics
• ✪ Lecture - 9 The Logistic Map and Period doubling
• ✪ [NJIT Math 491 - 695] Chaos - Lecture 4: Maps! What are they good for!?
• ✪ ChaosBook.org chapter Charting the state space - Bernoulli map
## Behaviour
Orbits of unit-height tent map
Bifurcation diagram for the tent map. Higher density indicates increased probability of the x variable acquiring that value for the given value of the μ parameter.
The tent map with parameter μ=2 and the logistic map with parameter r=4 are topologically conjugate,[1] and thus the behaviours of the two maps are in this sense identical under iteration.
Depending on the value of μ, the tent map demonstrates a range of dynamical behaviour ranging from predictable to chaotic.
• If μ is less than 1 the point x = 0 is an attractive fixed point of the system for all initial values of x i.e. the system will converge towards x = 0 from any initial value of x.
• If μ is 1 all values of x less than or equal to 1/2 are fixed points of the system.
• If μ is greater than 1 the system has two fixed points, one at 0, and the other at μ/(μ + 1). Both fixed points are unstable i.e. a value of x close to either fixed point will move away from it, rather than towards it. For example, when μ is 1.5 there is a fixed point at x = 0.6 (because 1.5(1 − 0.6) = 0.6) but starting at x = 0.61 we get
${\displaystyle 0.61\to 0.585\to 0.6225\to 0.56625\to 0.650625\ldots }$
• If μ is between 1 and the square root of 2 the system maps a set of intervals between μ − μ2/2 and μ/2 to themselves. This set of intervals is the Julia set of the map i.e. it is the smallest invariant sub-set of the real line under this map. If μ is greater than the square root of 2, these intervals merge, and the Julia set is the whole interval from μ − μ2/2 to μ/2 (see bifurcation diagram).
• If μ is between 1 and 2 the interval [μ − μ2/2, μ/2] contains both periodic and non-periodic points, although all of the orbits are unstable (i.e. nearby points move away from the orbits rather than towards them). Orbits with longer lengths appear as μ increases. For example:
${\displaystyle {\frac {\mu }{\mu ^{2}+1}}\to {\frac {\mu ^{2}}{\mu ^{2}+1}}\to {\frac {\mu }{\mu ^{2}+1}}{\mbox{ appears at }}\mu =1}$
${\displaystyle {\frac {\mu }{\mu ^{3}+1}}\to {\frac {\mu ^{2}}{\mu ^{3}+1}}\to {\frac {\mu ^{3}}{\mu ^{3}+1}}\to {\frac {\mu }{\mu ^{3}+1}}{\mbox{ appears at }}\mu ={\frac {1+{\sqrt {5}}}{2}}}$
${\displaystyle {\frac {\mu }{\mu ^{4}+1}}\to {\frac {\mu ^{2}}{\mu ^{4}+1}}\to {\frac {\mu ^{3}}{\mu ^{4}+1}}\to {\frac {\mu ^{4}}{\mu ^{4}+1}}\to {\frac {\mu }{\mu ^{4}+1}}{\mbox{ appears at }}\mu \approx 1.8393}$
• If μ equals 2 the system maps the interval [0,1] onto itself. There are now periodic points with every orbit length within this interval, as well as non-periodic points. The periodic points are dense in [0,1], so the map has become chaotic. In fact, the dynamics will be non-periodic if and only if ${\displaystyle x_{0}}$ is irrational. This can be seen by noting what the map does when ${\displaystyle x_{n}}$ is expressed in binary notation: It shifts the binary point one place to the right; then, if what appears to the left of the binary point is a "one" it changes all ones to zeroes and vice versa (with the exception of the final bit "one" in the case of a finite binary expansion); starting from an irrational number, this process goes on forever without repeating itself. The invariant measure for x is the uniform density over the unit interval.[2] The autocorrelation function for a sufficiently long sequence {${\displaystyle x_{n}}$} will show zero autocorrelation at all non-zero lags.[3] Thus ${\displaystyle {x_{n}}}$ cannot be distinguished from white noise using the autocorrelation function. Note that the r=4 case of the logistic map and the ${\displaystyle \mu =2}$ case of the tent map are homeomorphic to each other: Denoting the logistically evolving variable as ${\displaystyle y_{n}}$, the homeomorphism is
${\displaystyle x_{n}={\tfrac {2}{\pi }}\sin ^{-1}(y_{n}^{1/2}).}$
• If μ is greater than 2 the map's Julia set becomes disconnected, and breaks up into a Cantor set within the interval [0,1]. The Julia set still contains an infinite number of both non-periodic and periodic points (including orbits for any orbit length) but almost every point within [0,1] will now eventually diverge towards infinity. The canonical Cantor set (obtained by successively deleting middle thirds from subsets of the unit line) is the Julia set of the tent map for μ = 3.
### Numerical errors
Time series of the Tent map for the parameter m=2.0 which shows numerical error: "the plot of time series (plot of x variable with respect to number of iterations) stops fluctuating and no values are observed after n=50". Parameter m= 2.0, initial point is random.
## Magnifying the orbit diagram
Magnification near the tip shows more details.
• A closer look at the orbit diagram shows that there are 4 separated regions at μ ≈ 1. For further magnification, 2 reference lines (red) are drawn from the tip to suitable x at certain μ (e.g., 1.10) as shown.
Further magnification shows 8 separated regions.
• With distance measured from the corresponding reference lines, further detail appears in the upper and lower part of the map. (total 8 separated regions at some μ)
## Asymmetric tent map
The asymmetric tent map is essentially a distorted, but still piecewise linear, version of the ${\displaystyle \mu =2}$ case of the tent map. It is defined by
${\displaystyle v_{n+1}={\begin{cases}v_{n}/a&\mathrm {for} ~~v_{n}\in [0,a]\\\\(1-v_{n})/(1-a)&\mathrm {for} ~~v_{n}\in [a,1]\end{cases}}}$
for parameter ${\displaystyle a\in [0,1]}$. The ${\displaystyle \mu =2}$ case of the tent map is the present case of ${\displaystyle a={\tfrac {1}{2}}}$. A sequence {${\displaystyle v_{n}}$} will have the same autocorrelation function [3] as will data from the first-order autoregressive process ${\displaystyle w_{n+1}=(2a-1)w_{n}+u_{n+1}}$ with {${\displaystyle u_{n}}$} independently and identically distributed. Thus data from an asymmetric tent map cannot be distinguished, using the autocorrelation function, from data generated by a first-order autoregressive process. | 2,345 | 8,311 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 23, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-43 | latest | en | 0.749984 | To install click the Add extension button. That's it.. The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.. 4,5. Kelly Slayton. Congratulations on this excellent venture… what a great idea!. Alexander Grigorievskiy. I use WIKI 2 every day and almost forgot how the original Wikipedia looks like.. Live Statistics. English Articles. Improved in 24 Hours. Languages. Recent. Show all languages. What we do. Every page goes through several hundred of perfecting techniques; in live mode. Quite the same Wikipedia. Just better.. .. Leo. Newton. Brights. Milds. # Tent map. Graph of tent map function. In mathematics, the tent map with parameter μ is the real-valued function fμ defined by. ${\displaystyle f_{\mu }:=\mu \min\{x,\,1-x\},}$. the name being due to the tent-like shape of the graph of fμ. For the values of the parameter μ within 0 and 2, fμ maps the unit interval [0, 1] into itself, thus defining a discrete-time dynamical system on it (equivalently, a recurrence relation). In particular, iterating a point x0 in [0, 1] gives rise to a sequence ${\displaystyle x_{n}}$ :. ${\displaystyle x_{n+1}=f_{\mu }(x_{n})={\begin{cases}\mu x_{n}&\mathrm {for} ~~x_{n}<{\frac {1}{2}}\\\\\mu (1-x_{n})&\mathrm {for} ~~{\frac {1}{2}}\leq x_{n}\end{cases}}}$. where μ is a positive real constant. Choosing for instance the parameter μ=2, the effect of the function fμ may be viewed as the result of the operation of folding the unit interval in two, then stretching the resulting interval [0,1/2] to get again the interval [0,1]. Iterating the procedure, any point x0 of the interval assumes new subsequent positions as described above, generating a sequence xn in [0,1].. The ${\displaystyle \mu =2}$ case of the tent map is a non-linear transformation of both the bit shift map and the r=4 case of the logistic map.. • 1/5. Views:. 726. 489. 16 492. 416. 363. • ✪ ChaosBook.org chapter Charting the state space - Tent map. • ✪ ChaosBook.org chapter Stretch, fold, prune - Tent map symbolic dynamics. • ✪ Lecture - 9 The Logistic Map and Period doubling. • ✪ [NJIT Math 491 - 695] Chaos - Lecture 4: Maps! What are they good for!?. • ✪ ChaosBook.org chapter Charting the state space - Bernoulli map. ## Behaviour. Orbits of unit-height tent map. | Bifurcation diagram for the tent map. Higher density indicates increased probability of the x variable acquiring that value for the given value of the μ parameter.. The tent map with parameter μ=2 and the logistic map with parameter r=4 are topologically conjugate,[1] and thus the behaviours of the two maps are in this sense identical under iteration.. Depending on the value of μ, the tent map demonstrates a range of dynamical behaviour ranging from predictable to chaotic.. • If μ is less than 1 the point x = 0 is an attractive fixed point of the system for all initial values of x i.e. the system will converge towards x = 0 from any initial value of x.. • If μ is 1 all values of x less than or equal to 1/2 are fixed points of the system.. • If μ is greater than 1 the system has two fixed points, one at 0, and the other at μ/(μ + 1). Both fixed points are unstable i.e. a value of x close to either fixed point will move away from it, rather than towards it. For example, when μ is 1.5 there is a fixed point at x = 0.6 (because 1.5(1 − 0.6) = 0.6) but starting at x = 0.61 we get. ${\displaystyle 0.61\to 0.585\to 0.6225\to 0.56625\to 0.650625\ldots }$. • If μ is between 1 and the square root of 2 the system maps a set of intervals between μ − μ2/2 and μ/2 to themselves. This set of intervals is the Julia set of the map i.e. it is the smallest invariant sub-set of the real line under this map. If μ is greater than the square root of 2, these intervals merge, and the Julia set is the whole interval from μ − μ2/2 to μ/2 (see bifurcation diagram).. • If μ is between 1 and 2 the interval [μ − μ2/2, μ/2] contains both periodic and non-periodic points, although all of the orbits are unstable (i.e. nearby points move away from the orbits rather than towards them). Orbits with longer lengths appear as μ increases. For example:. ${\displaystyle {\frac {\mu }{\mu ^{2}+1}}\to {\frac {\mu ^{2}}{\mu ^{2}+1}}\to {\frac {\mu }{\mu ^{2}+1}}{\mbox{ appears at }}\mu =1}$. ${\displaystyle {\frac {\mu }{\mu ^{3}+1}}\to {\frac {\mu ^{2}}{\mu ^{3}+1}}\to {\frac {\mu ^{3}}{\mu ^{3}+1}}\to {\frac {\mu }{\mu ^{3}+1}}{\mbox{ appears at }}\mu ={\frac {1+{\sqrt {5}}}{2}}}$. ${\displaystyle {\frac {\mu }{\mu ^{4}+1}}\to {\frac {\mu ^{2}}{\mu ^{4}+1}}\to {\frac {\mu ^{3}}{\mu ^{4}+1}}\to {\frac {\mu ^{4}}{\mu ^{4}+1}}\to {\frac {\mu }{\mu ^{4}+1}}{\mbox{ appears at }}\mu \approx 1.8393}$. • If μ equals 2 the system maps the interval [0,1] onto itself. There are now periodic points with every orbit length within this interval, as well as non-periodic points. The periodic points are dense in [0,1], so the map has become chaotic. In fact, the dynamics will be non-periodic if and only if ${\displaystyle x_{0}}$ is irrational. This can be seen by noting what the map does when ${\displaystyle x_{n}}$ is expressed in binary notation: It shifts the binary point one place to the right; then, if what appears to the left of the binary point is a "one" it changes all ones to zeroes and vice versa (with the exception of the final bit "one" in the case of a finite binary expansion); starting from an irrational number, this process goes on forever without repeating itself. The invariant measure for x is the uniform density over the unit interval.[2] The autocorrelation function for a sufficiently long sequence {${\displaystyle x_{n}}$} will show zero autocorrelation at all non-zero lags.[3] Thus ${\displaystyle {x_{n}}}$ cannot be distinguished from white noise using the autocorrelation function. Note that the r=4 case of the logistic map and the ${\displaystyle \mu =2}$ case of the tent map are homeomorphic to each other: Denoting the logistically evolving variable as ${\displaystyle y_{n}}$, the homeomorphism is. ${\displaystyle x_{n}={\tfrac {2}{\pi }}\sin ^{-1}(y_{n}^{1/2}).}$. • If μ is greater than 2 the map's Julia set becomes disconnected, and breaks up into a Cantor set within the interval [0,1]. The Julia set still contains an infinite number of both non-periodic and periodic points (including orbits for any orbit length) but almost every point within [0,1] will now eventually diverge towards infinity. The canonical Cantor set (obtained by successively deleting middle thirds from subsets of the unit line) is the Julia set of the tent map for μ = 3.. ### Numerical errors. Time series of the Tent map for the parameter m=2.0 which shows numerical error: "the plot of time series (plot of x variable with respect to number of iterations) stops fluctuating and no values are observed after n=50". Parameter m= 2.0, initial point is random.. ## Magnifying the orbit diagram. Magnification near the tip shows more details.. • A closer look at the orbit diagram shows that there are 4 separated regions at μ ≈ 1. For further magnification, 2 reference lines (red) are drawn from the tip to suitable x at certain μ (e.g., 1.10) as shown.. Further magnification shows 8 separated regions.. • With distance measured from the corresponding reference lines, further detail appears in the upper and lower part of the map. (total 8 separated regions at some μ). ## Asymmetric tent map. The asymmetric tent map is essentially a distorted, but still piecewise linear, version of the ${\displaystyle \mu =2}$ case of the tent map. It is defined by. ${\displaystyle v_{n+1}={\begin{cases}v_{n}/a&\mathrm {for} ~~v_{n}\in [0,a]\\\\(1-v_{n})/(1-a)&\mathrm {for} ~~v_{n}\in [a,1]\end{cases}}}$. for parameter ${\displaystyle a\in [0,1]}$. The ${\displaystyle \mu =2}$ case of the tent map is the present case of ${\displaystyle a={\tfrac {1}{2}}}$. A sequence {${\displaystyle v_{n}}$} will have the same autocorrelation function [3] as will data from the first-order autoregressive process ${\displaystyle w_{n+1}=(2a-1)w_{n}+u_{n+1}}$ with {${\displaystyle u_{n}}$} independently and identically distributed. Thus data from an asymmetric tent map cannot be distinguished, using the autocorrelation function, from data generated by a first-order autoregressive process. |
http://mathhelpforum.com/pre-calculus/32393-help-problem-solving.html | 1,480,876,412,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541361.65/warc/CC-MAIN-20161202170901-00050-ip-10-31-129-80.ec2.internal.warc.gz | 187,062,592 | 10,449 | # Thread: Help with problem solving
1. ## Help with problem solving
A sculptor commissioned to design a monument for the long reach city council, has chosen a parabolic shape that stand 15m high with supports at 45° as shown in the diagram. For aesthetic reason the sculptor chooses the shape given by the function
Y = 15 -
π (pie)
ie. y equals 15 minus x² over π (pie)
Find the length of the support beam marked A. Provide justification for your answer.
2. do you mean
$E_1 \rightarrow y=\frac{15-x^2}{\pi}$
or
$E_2 \rightarrow y=15-\frac{x^2}{\pi}$
3. $\tan^{-1}(45^{\circ})=1$
so the equation of the line is $y=x$
setting $y=x$ equal to $y=15-\frac{x^2}{\pi}$
$x=15-\frac{x^2}{\pi} \iff x^2+\pi x=15 \pi$
so I will solve by completeing the square so we will add $\frac{\pi^2}{4}$ to both sides
$x^2+\pi x +\frac{\pi^2}{4}=15 \pi +\frac{\pi^2}{4} \iff (x-\frac{\pi}{2})^2=\frac{\pi(60+\pi)}{4}$
$x=\frac{\pi}{2} +\frac{\sqrt{\pi(60+\pi)}}{2}$
So this is half the distance from the diagram so if we multiply it by 2 we will get the length of the line.
$L=\pi+\sqrt{\pi(60+\pi)}$ | 371 | 1,097 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2016-50 | longest | en | 0.7832 | # Thread: Help with problem solving. 1. ## Help with problem solving. A sculptor commissioned to design a monument for the long reach city council, has chosen a parabolic shape that stand 15m high with supports at 45° as shown in the diagram. For aesthetic reason the sculptor chooses the shape given by the function. Y = 15 -. π (pie). ie. y equals 15 minus x² over π (pie). Find the length of the support beam marked A. Provide justification for your answer.. 2. do you mean. $E_1 \rightarrow y=\frac{15-x^2}{\pi}$. | or. $E_2 \rightarrow y=15-\frac{x^2}{\pi}$. 3. $\tan^{-1}(45^{\circ})=1$. so the equation of the line is $y=x$. setting $y=x$ equal to $y=15-\frac{x^2}{\pi}$. $x=15-\frac{x^2}{\pi} \iff x^2+\pi x=15 \pi$. so I will solve by completeing the square so we will add $\frac{\pi^2}{4}$ to both sides. $x^2+\pi x +\frac{\pi^2}{4}=15 \pi +\frac{\pi^2}{4} \iff (x-\frac{\pi}{2})^2=\frac{\pi(60+\pi)}{4}$. $x=\frac{\pi}{2} +\frac{\sqrt{\pi(60+\pi)}}{2}$. So this is half the distance from the diagram so if we multiply it by 2 we will get the length of the line.. $L=\pi+\sqrt{\pi(60+\pi)}$. |
https://www.eduzip.com/ask/question/displaystyle-int-frac-x-2-left-2-3x-2-right-52-dx-is-equal-to-581529 | 1,627,168,043,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00204.warc.gz | 751,723,943 | 8,707 | Mathematics
# $\displaystyle \int { \frac { { x }^{ 2 } }{ { \left( 2+{ 3x }^{ 2 } \right) }^{ 5/2 } } dx }$ is equal to
$\dfrac { 1 }{ 5 } { \left[ \dfrac { { x }^{ 2 } }{ 2+3{ x }^{ 2 } } \right] }^{ 3/2 }+C$
Its FREE, you're just one step away
Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105
#### Realted Questions
Q1 Subjective Medium
$-2 \int x^2 e^{-2x} dx = e^{-2x} (ax^2 + bx + c) + d$, then
1 Verified Answer | Published on 17th 09, 2020
Q2 Single Correct Medium
$\displaystyle\int \frac{\sqrt{2+x^{10}}}{x^{16}}dx$
• A. $\displaystyle \frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 3 }{ 2 } }$
• B. $\displaystyle { (\frac { 2 }{ { x }^{ 30 } } +1) }^{ \dfrac { 3 }{ 2 } }$
• C. $\displaystyle \frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 1 }{ 2 } }$
• D. $\displaystyle -\frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 3 }{ 2 } }$
1 Verified Answer | Published on 17th 09, 2020
Q3 Single Correct Hard
If $f(x) = \dfrac {x + 2}{2x + 3}$, then $\displaystyle \int \left (\dfrac {f(x)}{x^{2}}\right )^{1/2} dx = \dfrac {1}{\sqrt {2}}g \left (\dfrac {1 + \sqrt {2f(x)}}{1 - \sqrt {2f(x)}}\right ) - \sqrt {\dfrac {2}{3}}h \left (\dfrac {\sqrt {3f(x)} + \sqrt {2}}{\sqrt {3f(x)} - \sqrt {2}}\right ) + c$, where
• A. $g(x) = \log |x|, h(x) = \tan^{-1}x$
• B. $g(x) = h(x) = \tan^{-1}x$
• C. $g(x) = \log|x|, h(x) = \log |x|$
• D. $g(x) = \tan^{-1} x, h(x) = \log |x|$
1 Verified Answer | Published on 17th 09, 2020
Q4 Subjective Medium
Evaluate the following:
$\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{tan \,xdx}{1 + m^2 \,tan^2 x}$
1 Verified Answer | Published on 17th 09, 2020
Q5 Single Correct Medium
$\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \dfrac { 2x }{ 1-{ x }^{ 2 } } \right) dx } } =\dfrac{\pi}{a}-\ln a$. Find $a$.
• A. $1$
• B. $-1$
• C. None of these
• D. $2$ | 878 | 1,954 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-31 | latest | en | 0.372189 | Mathematics. # $\displaystyle \int { \frac { { x }^{ 2 } }{ { \left( 2+{ 3x }^{ 2 } \right) }^{ 5/2 } } dx }$ is equal to. $\dfrac { 1 }{ 5 } { \left[ \dfrac { { x }^{ 2 } }{ 2+3{ x }^{ 2 } } \right] }^{ 3/2 }+C$. Its FREE, you're just one step away. Single Correct Medium Published on 17th 09, 2020. Questions 203525. Subjects 9. Chapters 126. Enrolled Students 105. #### Realted Questions. Q1 Subjective Medium. $-2 \int x^2 e^{-2x} dx = e^{-2x} (ax^2 + bx + c) + d$, then. 1 Verified Answer | Published on 17th 09, 2020. Q2 Single Correct Medium. $\displaystyle\int \frac{\sqrt{2+x^{10}}}{x^{16}}dx$. • A. $\displaystyle \frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 3 }{ 2 } }$. • B. $\displaystyle { (\frac { 2 }{ { x }^{ 30 } } +1) }^{ \dfrac { 3 }{ 2 } }$. • C. $\displaystyle \frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 1 }{ 2 } }$. • D. $\displaystyle -\frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 3 }{ 2 } }$. 1 Verified Answer | Published on 17th 09, 2020. Q3 Single Correct Hard. | If $f(x) = \dfrac {x + 2}{2x + 3}$, then $\displaystyle \int \left (\dfrac {f(x)}{x^{2}}\right )^{1/2} dx = \dfrac {1}{\sqrt {2}}g \left (\dfrac {1 + \sqrt {2f(x)}}{1 - \sqrt {2f(x)}}\right ) - \sqrt {\dfrac {2}{3}}h \left (\dfrac {\sqrt {3f(x)} + \sqrt {2}}{\sqrt {3f(x)} - \sqrt {2}}\right ) + c$, where. • A. $g(x) = \log |x|, h(x) = \tan^{-1}x$. • B. $g(x) = h(x) = \tan^{-1}x$. • C. $g(x) = \log|x|, h(x) = \log |x|$. • D. $g(x) = \tan^{-1} x, h(x) = \log |x|$. 1 Verified Answer | Published on 17th 09, 2020. Q4 Subjective Medium. Evaluate the following:. $\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{tan \,xdx}{1 + m^2 \,tan^2 x}$. 1 Verified Answer | Published on 17th 09, 2020. Q5 Single Correct Medium. $\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \dfrac { 2x }{ 1-{ x }^{ 2 } } \right) dx } } =\dfrac{\pi}{a}-\ln a$. Find $a$.. • A. $1$. • B. $-1$. • C. None of these. • D. $2$. |
https://r-onetrading.com/wireless-connections/you-asked-how-many-volts-is-2-amps.html | 1,638,196,287,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358774.44/warc/CC-MAIN-20211129134323-20211129164323-00514.warc.gz | 561,700,772 | 18,288 | # You asked: How many volts is 2 amps?
Contents
Current Voltage Power
2 Amps 20 Volts 40 Watts
2 Amps 22.5 Volts 45 Watts
2 Amps 25 Volts 50 Watts
2 Amps 27.5 Volts 55 Watts
## How many amps are in a Volt?
A “volt” is a unit of electric potential, also known as electromotive force, and represents “the potential difference between two points of a conducting wire carrying a constant current of 1 ampere, when the power dissipated between these points is equal to 1 watt.” Stated another way, a potential of one volt appears …
## What does 2 amps mean?
The ampere, often shortened to “amp” or A, is the base unit of electric current in the International System of Units. … A current of 2 Amps can be written as 2A. The bigger the current the more electricity is flowing.
## How many amps is 11.1 volts?
11.1V Ohm to Amps
Assumed Ω = 20, the result of the conversion is 0.56 amps (rounded to two decimal digits).
## How many amps is 220 volts?
13.64 Amps, if you use 220 V.
## Is it better to charge a battery at 2 amps or 10 amps?
Slow charging using 10 amp chargers or less is generally accepted to be better. This is because repeatedly fast-charging a car battery may lead to overcharging and will severely reduce its performance. The short answer: 10 amp chargers are generally preferable than 2 amp chargers.
IT IS INTERESTING: Does xiaomi Redmi have Wi Fi calling?
## Is a 2 amp charger good?
Don’t worry about any problems using a 2-amp charger with a device designed to be used with a 1-amp charger. In some cases, it may speed up charging of the smaller device. In other cases, the smaller device will limit its power-sipping to 1 amp.
## How many amps are in a 12 volt?
Equivalent Volts and Amps Measurements
Voltage Current Power
12 Volts 0.8333 Amps 10 Watts
12 Volts 1.25 Amps 15 Watts
12 Volts 1.667 Amps 20 Watts
12 Volts 2.083 Amps 25 Watts
## How many volts is 70 amps?
The 70A ohm to volts formula is V = 70 × Ω. Replace Ω with your individual resistance in ohm. Assumed Ω = 8, the result of the conversion is 560 volts (rounded to two decimal places).
## How many volts are in 13 amps?
The 13A watts to volts formula is V = W / 13. Replace W with your actual power in watt. For example with W = 60, you obtain the result of 4.62 volts (rounded to two decimal digits). | 629 | 2,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-49 | latest | en | 0.915389 | # You asked: How many volts is 2 amps?. Contents. Current Voltage Power. 2 Amps 20 Volts 40 Watts. 2 Amps 22.5 Volts 45 Watts. 2 Amps 25 Volts 50 Watts. 2 Amps 27.5 Volts 55 Watts. ## How many amps are in a Volt?. A “volt” is a unit of electric potential, also known as electromotive force, and represents “the potential difference between two points of a conducting wire carrying a constant current of 1 ampere, when the power dissipated between these points is equal to 1 watt.” Stated another way, a potential of one volt appears …. ## What does 2 amps mean?. The ampere, often shortened to “amp” or A, is the base unit of electric current in the International System of Units. … A current of 2 Amps can be written as 2A. The bigger the current the more electricity is flowing.. ## How many amps is 11.1 volts?. 11.1V Ohm to Amps. Assumed Ω = 20, the result of the conversion is 0.56 amps (rounded to two decimal digits).. ## How many amps is 220 volts?. 13.64 Amps, if you use 220 V.. ## Is it better to charge a battery at 2 amps or 10 amps?. Slow charging using 10 amp chargers or less is generally accepted to be better. This is because repeatedly fast-charging a car battery may lead to overcharging and will severely reduce its performance. The short answer: 10 amp chargers are generally preferable than 2 amp chargers. | IT IS INTERESTING: Does xiaomi Redmi have Wi Fi calling?. ## Is a 2 amp charger good?. Don’t worry about any problems using a 2-amp charger with a device designed to be used with a 1-amp charger. In some cases, it may speed up charging of the smaller device. In other cases, the smaller device will limit its power-sipping to 1 amp.. ## How many amps are in a 12 volt?. Equivalent Volts and Amps Measurements. Voltage Current Power. 12 Volts 0.8333 Amps 10 Watts. 12 Volts 1.25 Amps 15 Watts. 12 Volts 1.667 Amps 20 Watts. 12 Volts 2.083 Amps 25 Watts. ## How many volts is 70 amps?. The 70A ohm to volts formula is V = 70 × Ω. Replace Ω with your individual resistance in ohm. Assumed Ω = 8, the result of the conversion is 560 volts (rounded to two decimal places).. ## How many volts are in 13 amps?. The 13A watts to volts formula is V = W / 13. Replace W with your actual power in watt. For example with W = 60, you obtain the result of 4.62 volts (rounded to two decimal digits). |
https://examshunt.com/corporation-bank-so-reasoning/number-ranking-and-time-sequence-test/exam/6/49 | 1,674,839,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495001.99/warc/CC-MAIN-20230127164242-20230127194242-00211.warc.gz | 255,674,346 | 11,385 | Top
# Number, Ranking and Time Sequence Test
1.
Sandhya's birthday falls on 15th August and Minu's birthday falls on 25th June. If Minu's birthday was on Wednesday, what was the day on Sandhya's birthday in the same year?
Total days from 25th June to 15th August = 5 + 31 + 15 = 51 days
51/7 = 7 weeks and 2 days
August 15 = Wednesday + 2 = Friday
Enter details here
2.
In the following question, select the odd number pair from the given alternatives.
No answer description available for this question.
Enter details here
3.
How many days will be there from 26th january 2008, to 15th may 2008(both days are included)?
Number of days = (6 + 29 + 31 + 30 + 15) = 111
2008 is a leap year, number of days in february = 29
Enter details here
4.
Three persons A, B and C are standing in a queue. There are five persons between A and B and eight persons between B and C. If there be three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue?
No answer description available for this question.
Enter details here
5.
How many numbers amogst the numbers 9 to 54 are there which are exactly divisible by 9 but not by 3 ?
Enter details here
6.
If the position of the first, sixth digits of the number 2796543018 are interchanged, similarly the positions of the the second and seventh digits are interchanged and so on, which of the following will be the left of seventh digit from the left end?
The new number formed is 4 3 0 1 8 2 7 9 6 5
The seventh digit from the left is 7, the 3rd digit from the left of 7 is 1
Enter details here
7.
If tuesday falls on 4th of month, then which day will fall three days after the 24th ?
The 4th day is tuesday, then 11, 18, 25th also tuesdays
Then 3 days after 24th is 27th, which is thursday
Enter details here
8.
If the seventh day of a month is three days earlier than Friday, what day will it be on the nineteenth day of the month?
No answer description available for this question.
Enter details here
9.
In a row of trees, one tree is fifth from either end of the row. How many trees are there in the row?
No answer description available for this question.
Enter details here
10.
If the day before yesterday was Saturday, what day will fall on the day after tomorrow? | 617 | 2,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-06 | latest | en | 0.932681 | Top. # Number, Ranking and Time Sequence Test. 1.. Sandhya's birthday falls on 15th August and Minu's birthday falls on 25th June. If Minu's birthday was on Wednesday, what was the day on Sandhya's birthday in the same year?. Total days from 25th June to 15th August = 5 + 31 + 15 = 51 days. 51/7 = 7 weeks and 2 days. August 15 = Wednesday + 2 = Friday. Enter details here. 2.. In the following question, select the odd number pair from the given alternatives.. No answer description available for this question.. Enter details here. 3.. How many days will be there from 26th january 2008, to 15th may 2008(both days are included)?. Number of days = (6 + 29 + 31 + 30 + 15) = 111. 2008 is a leap year, number of days in february = 29. Enter details here. 4.. Three persons A, B and C are standing in a queue. There are five persons between A and B and eight persons between B and C. If there be three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue?. No answer description available for this question.. Enter details here. 5. | How many numbers amogst the numbers 9 to 54 are there which are exactly divisible by 9 but not by 3 ?. Enter details here. 6.. If the position of the first, sixth digits of the number 2796543018 are interchanged, similarly the positions of the the second and seventh digits are interchanged and so on, which of the following will be the left of seventh digit from the left end?. The new number formed is 4 3 0 1 8 2 7 9 6 5. The seventh digit from the left is 7, the 3rd digit from the left of 7 is 1. Enter details here. 7.. If tuesday falls on 4th of month, then which day will fall three days after the 24th ?. The 4th day is tuesday, then 11, 18, 25th also tuesdays. Then 3 days after 24th is 27th, which is thursday. Enter details here. 8.. If the seventh day of a month is three days earlier than Friday, what day will it be on the nineteenth day of the month?. No answer description available for this question.. Enter details here. 9.. In a row of trees, one tree is fifth from either end of the row. How many trees are there in the row?. No answer description available for this question.. Enter details here. 10.. If the day before yesterday was Saturday, what day will fall on the day after tomorrow?. |
https://www.open.edu/openlearn/ocw/mod/oucontent/view.php?id=82662§ion=1.4 | 1,632,543,837,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00303.warc.gz | 915,884,600 | 19,986 | Teaching mathematics
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# 1.4 Using concrete resources
It is important to give learners of all ages a variety of concrete experiences connected with proportional reasoning to encourage them to compare, make conjectures, find relationships and generalise their learning.
Young learners may start with weighing and measuring objects in the classroom, and discussing comparisons between amounts.
## Activity 6 Sharing money
When asked to share out £24 between two cousins in the ratio of 3:1, learners can use counters to represent the pounds (Figure 7). How can you use counters to demonstrate the solution to this problem? Write responses in the box below.
Figure 7 Using counters to share out in a ratio
To use this interactive functionality a free OU account is required. Sign in or register.
Interactive feature not available in single page view (see it in standard view).
### Discussion
Grouping counters in groups of 1 and 3 is one way to work on this problem.
By grouping counters in this way, learners share out all the counters in the ratio of 1:3.
They can then look at how many of each colour there are altogether. Noticing that there are 6 groups of 3 and 6 groups of 1 would result in finding a total of 18 blue counters and 6 yellow counters, representing the £18 and £6 the cousins will each get.
Another way to approach this problem using counters is to create boxes (on paper or on mini-whiteboards) to represent the ratio. In this case, since the ratio is 3:1, you will need 3 boxes for the first cousin and 1 box for the second cousin.
The counters, which represent pounds, are then shared out equally between the 4 boxes. The first cousin is given 3 boxes, so receiving £18, and the second cousin is given 1 box, so receiving £6 (Figure 8).
Figure 8 Using ratio boxes with counters
This approach can also be used with pens on mini-whiteboards. Learners need to create the boxes to represent the ratio and then put a mark in every box until they get to the total, in this case 24 (Figure 9).
Figure 9 Using ratio boxes on mini-whiteboards
This imagery can then support learners in moving towards a more abstract concept of sharing amounts into a ratio. In other words, it can help them to understand that in a ratio of 3:1 there are 4 equal parts in total which the £24 needs to be divided equally between.
Note: When moving on to ratio problems involving fractions and decimals, using manipulative resources can become problematic. It is sensible to move onto more abstract methods with your learners before introducing these kinds of problems.
TM_1 | 593 | 2,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-39 | latest | en | 0.930884 | Teaching mathematics. Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.. Free course. # 1.4 Using concrete resources. It is important to give learners of all ages a variety of concrete experiences connected with proportional reasoning to encourage them to compare, make conjectures, find relationships and generalise their learning.. Young learners may start with weighing and measuring objects in the classroom, and discussing comparisons between amounts.. ## Activity 6 Sharing money. When asked to share out £24 between two cousins in the ratio of 3:1, learners can use counters to represent the pounds (Figure 7). How can you use counters to demonstrate the solution to this problem? Write responses in the box below.. Figure 7 Using counters to share out in a ratio. To use this interactive functionality a free OU account is required. Sign in or register.. Interactive feature not available in single page view (see it in standard view).. ### Discussion. Grouping counters in groups of 1 and 3 is one way to work on this problem.. By grouping counters in this way, learners share out all the counters in the ratio of 1:3. | They can then look at how many of each colour there are altogether. Noticing that there are 6 groups of 3 and 6 groups of 1 would result in finding a total of 18 blue counters and 6 yellow counters, representing the £18 and £6 the cousins will each get.. Another way to approach this problem using counters is to create boxes (on paper or on mini-whiteboards) to represent the ratio. In this case, since the ratio is 3:1, you will need 3 boxes for the first cousin and 1 box for the second cousin.. The counters, which represent pounds, are then shared out equally between the 4 boxes. The first cousin is given 3 boxes, so receiving £18, and the second cousin is given 1 box, so receiving £6 (Figure 8).. Figure 8 Using ratio boxes with counters. This approach can also be used with pens on mini-whiteboards. Learners need to create the boxes to represent the ratio and then put a mark in every box until they get to the total, in this case 24 (Figure 9).. Figure 9 Using ratio boxes on mini-whiteboards. This imagery can then support learners in moving towards a more abstract concept of sharing amounts into a ratio. In other words, it can help them to understand that in a ratio of 3:1 there are 4 equal parts in total which the £24 needs to be divided equally between.. Note: When moving on to ratio problems involving fractions and decimals, using manipulative resources can become problematic. It is sensible to move onto more abstract methods with your learners before introducing these kinds of problems.. TM_1. |
http://mathhelpforum.com/calculus/109147-integral-partial-fractions.html | 1,527,004,640,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864798.12/warc/CC-MAIN-20180522151159-20180522171159-00466.warc.gz | 187,072,334 | 11,354 | # Thread: Integral with partial fractions
1. ## Integral with partial fractions
so I got
A= -3 , B = 5 C = 5
and i got ∫-3/x-4 + 5x+5/x²+9 dx for my equation
my answer came out to be
I dont know why it is wrong.
2. I didn't check the details but one mistake is that it should be $\displaystyle -3\ln(x-4)$, not $\displaystyle -3\ln(x+4)$
3. Originally Posted by emurphy
so I got
A= -3 , B = 5 C = 5
and i got ∫-3/x-4 + 5x+5/x²+9 dx for my equation
my answer came out to be
I dont know why it is wrong.
As well as the mistake already mentioned (which may well be just a typo), your calculation of $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$ is very wrong. Please show all the details of your calculation so that it can be reviewed.
4. Oh yeah that is pretty bad. Use parentheses! You wrote $\displaystyle 3/x-4 + 5x+5/x^2 + 9$ and integrated it as $\displaystyle 3/(x-4) + 5x+5/(x^2 + 9)$ when in fact you should have written $\displaystyle 3/(x-4) + (5x+5)/(x^2 + 9)$. These expressions are completely different from one another.
5. For $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$
I separated the fraction into 2
∫5x+5 dx + ∫1/x²+9 dx
∫5x+5 dx = 5x²/2+5x
∫1/x²+9 dx = ∫1/(1/9)(x²+1) dx = 9 ∫ 1/x²+1
= 9tan^-1(x)
6. Originally Posted by emurphy
For $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$
I separated the fraction into 2
∫5x+5 dx + ∫1/x²+9 dx
∫5x+5 dx = 5x²/2+5x
∫1/x²+9 dx = ∫1/(1/9)(x²+1) dx = 9 ∫ 1/x²+1
= 9tan^-1(x)
NO! You cannot seperate the integrand like that!! Using this logic you would solve $\displaystyle \int \frac{x}{x} \, dx$ as $\displaystyle \int x \, dx + \int \frac{1}{x} \, dx$ which is clearly so wrong that I don't need to say anything more.
7. Originally Posted by mr fantastic
NO! You cannot seperate the integrand like that!! Using this logic you would solve $\displaystyle \int \frac{x}{x} \, dx$ as $\displaystyle \int x \, dx + \int \frac{1}{x} \, dx$ which is clearly so wrong that I don't need to say anything more.
Adding to this, as a diagnostic of where you're at could you please do the following two integrals:
$\displaystyle I_1 = \int \frac{2x}{x^2 + 9} \, dx$
$\displaystyle I_2 = \int \frac{5}{x^2 + 9} \, dx$
Edit: If you cannot do these there is little point in you attempting questions like the one you posted until you have thoroughly reviewed the basic integration techniques this questions assumes and have developed a more solid understanding of integration. | 850 | 2,454 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-22 | latest | en | 0.897838 | # Thread: Integral with partial fractions. 1. ## Integral with partial fractions. so I got. A= -3 , B = 5 C = 5. and i got ∫-3/x-4 + 5x+5/x²+9 dx for my equation. my answer came out to be. I dont know why it is wrong.. 2. I didn't check the details but one mistake is that it should be $\displaystyle -3\ln(x-4)$, not $\displaystyle -3\ln(x+4)$. 3. Originally Posted by emurphy. so I got. A= -3 , B = 5 C = 5. and i got ∫-3/x-4 + 5x+5/x²+9 dx for my equation. my answer came out to be. I dont know why it is wrong.. As well as the mistake already mentioned (which may well be just a typo), your calculation of $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$ is very wrong. Please show all the details of your calculation so that it can be reviewed.. 4. Oh yeah that is pretty bad. Use parentheses! You wrote $\displaystyle 3/x-4 + 5x+5/x^2 + 9$ and integrated it as $\displaystyle 3/(x-4) + 5x+5/(x^2 + 9)$ when in fact you should have written $\displaystyle 3/(x-4) + (5x+5)/(x^2 + 9)$. These expressions are completely different from one another.. 5. | For $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$. I separated the fraction into 2. ∫5x+5 dx + ∫1/x²+9 dx. ∫5x+5 dx = 5x²/2+5x. ∫1/x²+9 dx = ∫1/(1/9)(x²+1) dx = 9 ∫ 1/x²+1. = 9tan^-1(x). 6. Originally Posted by emurphy. For $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$. I separated the fraction into 2. ∫5x+5 dx + ∫1/x²+9 dx. ∫5x+5 dx = 5x²/2+5x. ∫1/x²+9 dx = ∫1/(1/9)(x²+1) dx = 9 ∫ 1/x²+1. = 9tan^-1(x). NO! You cannot seperate the integrand like that!! Using this logic you would solve $\displaystyle \int \frac{x}{x} \, dx$ as $\displaystyle \int x \, dx + \int \frac{1}{x} \, dx$ which is clearly so wrong that I don't need to say anything more.. 7. Originally Posted by mr fantastic. NO! You cannot seperate the integrand like that!! Using this logic you would solve $\displaystyle \int \frac{x}{x} \, dx$ as $\displaystyle \int x \, dx + \int \frac{1}{x} \, dx$ which is clearly so wrong that I don't need to say anything more.. Adding to this, as a diagnostic of where you're at could you please do the following two integrals:. $\displaystyle I_1 = \int \frac{2x}{x^2 + 9} \, dx$. $\displaystyle I_2 = \int \frac{5}{x^2 + 9} \, dx$. Edit: If you cannot do these there is little point in you attempting questions like the one you posted until you have thoroughly reviewed the basic integration techniques this questions assumes and have developed a more solid understanding of integration. |
https://learn.careers360.com/school/question-please-solve-rd-sharma-class-12-chapter-19-definite-integrals-exercise-19-point-5-question-3-maths-textbook-solution-2/?question_number=3.0 | 1,716,759,616,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00561.warc.gz | 302,235,309 | 39,880 | please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19.5 question 3 maths textbook solution
8
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{1}^{3}\left ( 3x-2 \right )dx$
Solution:
We have,
$\int_{1}^{3}\left ( 3x-2 \right )dx$
$\int_{a}^{b}\left ( fx\right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$
Where, $h=\frac{b-a}{n}$
Here,
$a=1,b=3,f(x)=(3x-2)\\ h=\frac{2}{n}$
Thus, we have
\begin{aligned} I &=\int_{1}^{3}(3 x-2) d x \\ I &=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \\ \end{aligned}
\begin{aligned} &=\lim _{h \rightarrow 0} h[1+(3(1+h)-2)+\{3(1+2 h)-2\}+\ldots\{3(1+(n-1) h)-2\}] \\ &=\lim _{h \rightarrow 0} h[n+3 h(1+2+3+\ldots(n-1))] \\ \end{aligned}\begin{aligned} &=\lim _{h \rightarrow 0} h\left[n+3 h\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[n+\frac{6}{n} \frac{n(n-1)}{2}\right] \\ \end{aligned} $\left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n} \\ n \rightarrow \infty \end{array}\right]$
\begin{aligned} &=\lim _{n \rightarrow \infty} 2+\frac{6}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right) \\ &=2+6=8 \end{aligned} | 541 | 1,265 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.352324 | please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19.5 question 3 maths textbook solution. 8. Hint:. To solve the given statement, we have to use the formula of addition limits.. Given:. $\int_{1}^{3}\left ( 3x-2 \right )dx$. Solution:. We have,. $\int_{1}^{3}\left ( 3x-2 \right )dx$. | $\int_{a}^{b}\left ( fx\right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$. Where, $h=\frac{b-a}{n}$. Here,. $a=1,b=3,f(x)=(3x-2)\\ h=\frac{2}{n}$. Thus, we have. \begin{aligned} I &=\int_{1}^{3}(3 x-2) d x \\ I &=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \\ \end{aligned}. \begin{aligned} &=\lim _{h \rightarrow 0} h[1+(3(1+h)-2)+\{3(1+2 h)-2\}+\ldots\{3(1+(n-1) h)-2\}] \\ &=\lim _{h \rightarrow 0} h[n+3 h(1+2+3+\ldots(n-1))] \\ \end{aligned}\begin{aligned} &=\lim _{h \rightarrow 0} h\left[n+3 h\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[n+\frac{6}{n} \frac{n(n-1)}{2}\right] \\ \end{aligned} $\left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n} \\ n \rightarrow \infty \end{array}\right]$. \begin{aligned} &=\lim _{n \rightarrow \infty} 2+\frac{6}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right) \\ &=2+6=8 \end{aligned}. |
http://gmatclub.com/forum/is-the-number-of-members-of-club-x-greater-than-the-number-10382.html#p60251 | 1,480,809,838,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541142.66/warc/CC-MAIN-20161202170901-00444-ip-10-31-129-80.ec2.internal.warc.gz | 113,877,594 | 41,830 | Is the number of members of Club X greater than the number : DS Archive
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# Is the number of members of Club X greater than the number
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Is the number of members of Club X greater than the number of members of Club Y ?
(1) Of the members of Club X, 20 percent are also members of Club Y.
(2) Of the members of Club Y, 30 percent are also members of Club X.
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03 Oct 2004, 03:35
C.
of course each one independently will not give anything.
for both looking at the intersection or members comon to X and Y
from (1) . intersection = 0.2 X
from (2). intersection = 0.3 Y
0.2X = 0.3 Y
this gives that memebres of X are greater than Y
Display posts from previous: Sort by | 510 | 1,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2016-50 | latest | en | 0.912711 | Is the number of members of Club X greater than the number : DS Archive. Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack. It is currently 03 Dec 2016, 16:03. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. # Events & Promotions. ###### Events & Promotions in June. Open Detailed Calendar. # Is the number of members of Club X greater than the number. Author Message. Intern. Joined: 25 Sep 2004. Posts: 21. Followers: 0. Kudos [?]: 98 [0], given: 0. Is the number of members of Club X greater than the number [#permalink]. ### Show Tags. 03 Oct 2004, 03:10. | 00:00. Difficulty:. (N/A). Question Stats:. 0% (00:00) correct 0% (00:00) wrong based on 0 sessions. ### HideShow timer Statistics. This topic is locked. If you want to discuss this question please re-post it in the respective forum.. Is the number of members of Club X greater than the number of members of Club Y ?. (1) Of the members of Club X, 20 percent are also members of Club Y.. (2) Of the members of Club Y, 30 percent are also members of Club X.. Manager. Joined: 07 Sep 2004. Posts: 60. Followers: 1. Kudos [?]: 1 [0], given: 0. ### Show Tags. 03 Oct 2004, 03:35. C.. of course each one independently will not give anything.. for both looking at the intersection or members comon to X and Y. from (1) . intersection = 0.2 X. from (2). intersection = 0.3 Y. 0.2X = 0.3 Y. this gives that memebres of X are greater than Y. Display posts from previous: Sort by. |
https://byjusexamprep.com/reasoning-booster-on-direction-test-for-teaching-exams-i-799d8b5a-8637-11e6-93de-05f4f33a0b1e | 1,675,509,616,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500126.0/warc/CC-MAIN-20230204110651-20230204140651-00458.warc.gz | 160,380,298 | 66,283 | # Reasoning Booster on Direction Test for Teaching Exams
By Ashish Kumar|Updated : December 31st, 2021
Verbal reasoning comprises of a very important concept that is - Direction. It is one of the easiest topics to score in most of the Teaching Exams. The Concept behind the Directions is the same that we use in our day-to-day life. This direction sense test is nothing but a precise of sensing the direction. To solve the direction sense test first you need to make a sketch of the data provided. We have clearly described the direction Sense test Tricks, tips and Shortcuts here.
Verbal reasoning comprises a very important concept which is - Direction. It is one of the easiest topics to score in most Teaching Exams. The Concept behind the Directions is the same that we use in our day-to-day life. This direction sense test is nothing but a precise of sensing the direction. To solve the direction sense test first you need to make a sketch of the data provided. We have clearly described the direction Sense test Tricks, tips and Shortcuts here.
### Short Tricks & Basic Concepts on Direction Test
Four main Directions - North, South, East, West
Four Cardinal Direction - North-East, North-West, South-East, South-West
• At the time of sunrise if a man stands facing the east, his shadow will be towards the west.
• At the time of sunset the shadow of an object is always in the east.
• If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.
• At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow.
The above is the Directional Diagram. The direction sense consists of four different types of problems. Let’s see the types of Direction problems with solutions.
1. A man starting from his home walks 5 km towards East, and then he turns left and goes 4 km. At last he turn to his left and walks 5 km. Now find the distance between the man and his home and also find at which direction he is facing?
Sol.
From the above diagram we can find he is 4 km from his house and facing the North Direction.
2. A man starting from his home moves 4 km towards East, then he turns right and moves 3 km. Now what will be the minimum distance covered by him to come back to his home?
Sol.
The minimum distance covered by him = √42 + 32 = 5 km
3. After Sunrise, Prakash while going to college suddenly met with Lokesh at a crossing point. Lokesh's Shadow was exactly to right of Prakash. If they were facing each other on which direction was Prakash facing?
Sol. Always Sun rises in the East Direction. So Shadow falls towards West
4. Prem started from his home and moved 5 km to reach the crossing point of the palace. In which direction was Prem going, if the road opposite to his direction goes to the hospital. The road to the right of Prem goes to the station. If the road which goes to station is just opposite to the road of the IT-Park, then in which direction is Prem which leads to the IT- Park?
Sol.
From the above diagram its shows that the road which goes towards the IT-Park is left of Prem.
That’s how Direction questions are solved easily. We hope these concepts have made the chapter easy for you all.
Thanks!
The most comprehensive exam prep app
#DreamStriveSucceed | 749 | 3,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-06 | latest | en | 0.953396 | # Reasoning Booster on Direction Test for Teaching Exams. By Ashish Kumar|Updated : December 31st, 2021. Verbal reasoning comprises of a very important concept that is - Direction. It is one of the easiest topics to score in most of the Teaching Exams. The Concept behind the Directions is the same that we use in our day-to-day life. This direction sense test is nothing but a precise of sensing the direction. To solve the direction sense test first you need to make a sketch of the data provided. We have clearly described the direction Sense test Tricks, tips and Shortcuts here.. Verbal reasoning comprises a very important concept which is - Direction. It is one of the easiest topics to score in most Teaching Exams. The Concept behind the Directions is the same that we use in our day-to-day life. This direction sense test is nothing but a precise of sensing the direction. To solve the direction sense test first you need to make a sketch of the data provided. We have clearly described the direction Sense test Tricks, tips and Shortcuts here.. ### Short Tricks & Basic Concepts on Direction Test. Four main Directions - North, South, East, West. Four Cardinal Direction - North-East, North-West, South-East, South-West. • At the time of sunrise if a man stands facing the east, his shadow will be towards the west.. • At the time of sunset the shadow of an object is always in the east.. • If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.. • At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow.. The above is the Directional Diagram. The direction sense consists of four different types of problems. Let’s see the types of Direction problems with solutions.. 1. A man starting from his home walks 5 km towards East, and then he turns left and goes 4 km. At last he turn to his left and walks 5 km. | Now find the distance between the man and his home and also find at which direction he is facing?. Sol.. From the above diagram we can find he is 4 km from his house and facing the North Direction.. 2. A man starting from his home moves 4 km towards East, then he turns right and moves 3 km. Now what will be the minimum distance covered by him to come back to his home?. Sol.. The minimum distance covered by him = √42 + 32 = 5 km. 3. After Sunrise, Prakash while going to college suddenly met with Lokesh at a crossing point. Lokesh's Shadow was exactly to right of Prakash. If they were facing each other on which direction was Prakash facing?. Sol. Always Sun rises in the East Direction. So Shadow falls towards West. 4. Prem started from his home and moved 5 km to reach the crossing point of the palace. In which direction was Prem going, if the road opposite to his direction goes to the hospital. The road to the right of Prem goes to the station. If the road which goes to station is just opposite to the road of the IT-Park, then in which direction is Prem which leads to the IT- Park?. Sol.. From the above diagram its shows that the road which goes towards the IT-Park is left of Prem.. That’s how Direction questions are solved easily. We hope these concepts have made the chapter easy for you all.. Thanks!. The most comprehensive exam prep app. #DreamStriveSucceed. |
https://math.stackexchange.com/questions/2190775/how-to-prove-gcd-with-prime-factor-decomposition | 1,643,392,331,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00206.warc.gz | 422,468,710 | 32,465 | How to prove $gcd$ with prime factor decomposition?
So we know that $n=\prod_{p\in\mathbb{P}} p^{v_{n}(p)}$ is formula of factor decomposition.
How to show that $gcd(m,n) = \prod_{p\in\mathbb{P}} p^{min(v_{n}(p),v_{n}(p))}$ using factor decomposition?
I know that for example when we have $gcd(36,360)$, we can write $36$ as
$36 = 2^{p_1}3^{p_2}5^{p_3}$ which is $36 = 2^{2}3^{2}5^{0}$
and $360$ as
$360 = 2^{q_1}3^{q_2}5^{q_3}$ which is $360 = 2^{3}3^{2}5^{1}$
To get $gcd$ we now need to take smallest $p$ and $q$ for every pair with same base. In this case these are 2,2,0, therefore our $gcd(36,360) = 36$ which is obvious.
So, I can show that this is valid $gcd(m,n) = \prod_{p\in\mathbb{P}} p^{min(v_{n}(p),v_{n}(p))}$ through an example with numbers, but I need proof, and I don't know how to prove that.
That's just because $m|n$ iff $\nu_m(p) \leq \nu_n(p)$ for all relevant $p$, and then use the definition of greatest common divisor. | 356 | 953 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2022-05 | latest | en | 0.912957 | How to prove $gcd$ with prime factor decomposition?. So we know that $n=\prod_{p\in\mathbb{P}} p^{v_{n}(p)}$ is formula of factor decomposition.. How to show that $gcd(m,n) = \prod_{p\in\mathbb{P}} p^{min(v_{n}(p),v_{n}(p))}$ using factor decomposition?. I know that for example when we have $gcd(36,360)$, we can write $36$ as. $36 = 2^{p_1}3^{p_2}5^{p_3}$ which is $36 = 2^{2}3^{2}5^{0}$. and $360$ as. | $360 = 2^{q_1}3^{q_2}5^{q_3}$ which is $360 = 2^{3}3^{2}5^{1}$. To get $gcd$ we now need to take smallest $p$ and $q$ for every pair with same base. In this case these are 2,2,0, therefore our $gcd(36,360) = 36$ which is obvious.. So, I can show that this is valid $gcd(m,n) = \prod_{p\in\mathbb{P}} p^{min(v_{n}(p),v_{n}(p))}$ through an example with numbers, but I need proof, and I don't know how to prove that.. That's just because $m|n$ iff $\nu_m(p) \leq \nu_n(p)$ for all relevant $p$, and then use the definition of greatest common divisor. |
http://www.math-math.com/2016/04/goldbach-conjecture-explained-in-5.html | 1,495,654,467,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607860.7/warc/CC-MAIN-20170524192431-20170524212431-00385.warc.gz | 567,396,203 | 9,039 | Goldbach Conjecture Explained in 5 Minutes
Goldbach Conjecture Explained in 5 Minutes
In 1742, the German mathematician Christian Goldbach, in a discussion with the mathematician Leonhard Euler, made a simple statement..
Every even integer greater than 2 can be written as the sum of two prime numbers.
Mathematicians have tried to prove this ever since. None have. It's a great example of how a simple statement in mathematics can be amazingly difficult to prove. Computers have checked billions of numbers and shown it to be true for every number tested, but that's not the same as a proof. A proof would show it to be true for all even numbers, period.
It's easy to prove that every integer can be written as a product of primes. This is called the prime decomposition of an integer. So for any integer m we have.. m=(p1^s1)*(p2^s2)*......*(pn^sn)
where p1,p2,...,pn are prime numbers and the s1,s2,s3,...,sn are just integer powers and represent the simple fact that primes may be repeated. By collecting like primes together and raising them to a power we make sure that p1,p2,p3,...,pn are distinct primes with no duplication.
The condition that m be even simply means that one of these primes must be 2. Since the order of multiplication does not matter we can make p1=2. So m now looks like this..
m=(2^s1)*(p2^s2)*...*(pn^sn)
Goldbach's Conjecture says that when m is even there exists two prime numbers, let's call them g1 and g2, such that..
m=(2^s1)*(p2^s2)*...*(pn^sn)=g1+g2
Are we on our way to a proof? No, but it's still fun to try!
Perhaps there's some deep clue in the fact that Goldbach's Conjecture only works if p1=2. In other words, if 2 does not appear in the prime decomposition of an integer then Goldbach's Conjecture does not work. What's so special about the number 2? | 467 | 1,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-22 | longest | en | 0.933723 | Goldbach Conjecture Explained in 5 Minutes. Goldbach Conjecture Explained in 5 Minutes. In 1742, the German mathematician Christian Goldbach, in a discussion with the mathematician Leonhard Euler, made a simple statement... Every even integer greater than 2 can be written as the sum of two prime numbers.. Mathematicians have tried to prove this ever since. None have. It's a great example of how a simple statement in mathematics can be amazingly difficult to prove. Computers have checked billions of numbers and shown it to be true for every number tested, but that's not the same as a proof. A proof would show it to be true for all even numbers, period.. It's easy to prove that every integer can be written as a product of primes. This is called the prime decomposition of an integer. So for any integer m we have.. m=(p1^s1)*(p2^s2)*......*(pn^sn). where p1,p2,...,pn are prime numbers and the s1,s2,s3,...,sn are just integer powers and represent the simple fact that primes may be repeated. | By collecting like primes together and raising them to a power we make sure that p1,p2,p3,...,pn are distinct primes with no duplication.. The condition that m be even simply means that one of these primes must be 2. Since the order of multiplication does not matter we can make p1=2. So m now looks like this... m=(2^s1)*(p2^s2)*...*(pn^sn). Goldbach's Conjecture says that when m is even there exists two prime numbers, let's call them g1 and g2, such that... m=(2^s1)*(p2^s2)*...*(pn^sn)=g1+g2. Are we on our way to a proof? No, but it's still fun to try!. Perhaps there's some deep clue in the fact that Goldbach's Conjecture only works if p1=2. In other words, if 2 does not appear in the prime decomposition of an integer then Goldbach's Conjecture does not work. What's so special about the number 2?. |
https://math.stackexchange.com/questions/2929894/determinants-of-block-tridiagonal-matrices-when-off-diagonal-blocks-are-not-mm | 1,566,616,552,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319470.94/warc/CC-MAIN-20190824020840-20190824042840-00103.warc.gz | 558,989,432 | 29,382 | Determinants of block tridiagonal matrices when off-diagonal blocks are not m*m matrices
We know that if we have the block-tridiagonal matrix $$M(z)$$ with blocks $$A_i$$, $$B_i$$ and $$C_{i−1}$$ (i =1,...,n) which are all $$m \times m$$ matrices $$M(z)=\begin{bmatrix} A_1 & B_1 & \cdots &\frac{1}{z}C_0\\ C_1 & \ddots & \ddots& \\ \vdots & \ddots & \ddots& B_{n-1}\\ zB_n & & C_{n-1}& A_n \end{bmatrix}$$ where the off-diagonal blocks are nonsingular, it can be associated with a transfer matrix, $$T=\begin{bmatrix} -B_n^{-1}A_n& -B_n^{-1}C_{n-1} \\ I_m & 0 \end{bmatrix} \cdots \begin{bmatrix} -B_1^{-1}A_1& -B_1^{-1}C_0 \\ I_m & 0 \end{bmatrix}$$ where $$I_m$$ is the $$m×m$$ unit matrix and $$\det(T)=\prod_{i=1}^\infty \det[B_i^{-1}C_{i-1}]$$
But what will the transfer matrix $$T$$ and the determinant $$\det(T)$$ be if $$B_i$$ and $$C_{i}$$ are $$m \times n$$ matrices ($$m$$ is not equal to $$n$$) instead of $$m\times m$$ matrices? What's the new formula for that?
Thank you!! | 379 | 990 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-35 | latest | en | 0.543278 | Determinants of block tridiagonal matrices when off-diagonal blocks are not m*m matrices. We know that if we have the block-tridiagonal matrix $$M(z)$$ with blocks $$A_i$$, $$B_i$$ and $$C_{i−1}$$ (i =1,...,n) which are all $$m \times m$$ matrices $$M(z)=\begin{bmatrix} A_1 & B_1 & \cdots &\frac{1}{z}C_0\\ C_1 & \ddots & \ddots& \\ \vdots & \ddots & \ddots& B_{n-1}\\ zB_n & & C_{n-1}& A_n \end{bmatrix}$$ where the off-diagonal blocks are nonsingular, it can be associated with a transfer matrix, $$T=\begin{bmatrix} -B_n^{-1}A_n& -B_n^{-1}C_{n-1} \\ I_m & 0 \end{bmatrix} \cdots \begin{bmatrix} -B_1^{-1}A_1& -B_1^{-1}C_0 \\ I_m & 0 \end{bmatrix}$$ where $$I_m$$ is the $$m×m$$ unit matrix and $$\det(T)=\prod_{i=1}^\infty \det[B_i^{-1}C_{i-1}]$$. But what will the transfer matrix $$T$$ and the determinant $$\det(T)$$ be if $$B_i$$ and $$C_{i}$$ are $$m \times n$$ matrices ($$m$$ is not equal to $$n$$) instead of $$m\times m$$ matrices? What's the new formula for that?. | Thank you!!. |
https://math.stackexchange.com/questions/1290948/special-representation-of-a-number | 1,555,835,288,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530505.30/warc/CC-MAIN-20190421080255-20190421102255-00370.warc.gz | 477,233,849 | 34,782 | # Special representation of a number
How can I check, if a number $n$ can be representated by
$$pq+rs$$
where $p,q,r,s$ are pairwise different prime numbers with the same number of digits.
For example,
$$105153899965560312960 = 3022993637\times 6003631993 + 9069920719\times 9592692301$$
has such a representation.
My questions :
• Is such a representation (if it exists), always unique ?
• How can I find the primes $p,q,r,s$ if a representation exists ?
• How can I check if a representation exists ?
• There is a very large number of counterexamples to your uniqueness conjecture, of which the smallest is $$11\cdot13 + 19\cdot 29 = 11\cdot 43 + 13\cdot 17$$ – MJD May 20 '15 at 13:12
• So, the first part is answered. Thank you. And I conjectured the uniqueness only for numbers large enough. It would be interesting, which is the largest counterexample (I think there is a largest). – Peter May 20 '15 at 13:14
You said in a comment:
I conjectured the uniqueness only for numbers large enough. It would be interesting, which is the largest counterexample (I think there is a largest).
I think your intuition is backwards here, and I think it will be instructive if I explain to you why; this is also why I was able to immediately guess that there would be counterexamples. Short summary: There so many 4-tuples compared with the number of possible sums that it is not possible for every 4-tuple to get its own sum.
Suppose our primes have $d$ digits and that there are $D$ primes with $d$ digits. The prime number theorem tells us that $D$ is around $O\left(\frac{10^d}{d}\right)$.
Then there are $O(D^4) = O\left(\frac{10^{4d}}{d^4}\right)$ quadruples $(p,q,r,s)$ of distinct $d$-digit primes. The quantity $n=pq+rs$ has around $2d$ digits and there are $10^{2d}$ of these numbers.
There is no reason to think that the quantities $pq+rs$ will be distributed among these $10^{2d}$ possibilities in any way other than randomly. (Additive properties of prime numbers are almost always random unless there is some obvious reason they cannot be. To take an example I made up on the spot, the mod-3 remainder of $p+q$ is equal to 0, 1, or 2 with probability close to $\frac12, \frac14, \frac14$ just as one would expect.)
So we should expect that what we are doing here is essentially throwing $O\left(\frac{10^{4d}}{d^4}\right)$ balls at random into only $10^{2d}$ bins . The number of balls greatly exceeds the number of bins when $d$ is large, so it is not at all surprising that some balls end up in the same bin; that is, that some tuples yield the same value for $pq+rs$. As $d$ increases, we should expect this to be vastly more likely, not less likely, and this argues strongly against the possibility of a maximal counterexample.
Indeed, one would expect the opposite to be true: let $N$ be given. Then we should expect to find many sets, each containing $N$ 4-tuples, for which all the 4-tuples in the set have the same $pq+rs$ value, no matter how large $N$ is large. If we want to find $n$ that can be represented as $n=pq+rs$ in one million different ways, we should expect that there will be many such.
(The tuples won't be distributed uniformly over the space of sums—for example, only a few tuples map to an odd sum—but this will tend to increase, not decrease, the number of collisions.)
• That is convincing. +1 for the well explanation. – Peter May 20 '15 at 13:37
• Experiments bear out this theory. There are 20 ways to represent 3150 in the form $n=pq+rs$, and this is the most ways for any $n$ under 10,000. But if we start looking at larger primes, there are over ten thousand such $n$ below 200,000; for example there are 20 or more ways to represent each of 170280, 170430, 170436, 170520, 170790, 170940, 170970, 171318, 171780, 171990, 172410… – MJD May 20 '15 at 13:49
(Just to elaborate on MJD's answer.)
Note that if $p, q, r, s$ are all $d$ digits long, which means that they all lie in the interval $[10^{d-1}, 10^d)$, then $pq + rs$ lies in the interval $[2\times10^{2d-2}, 2\times10^d)$. For numbers $N$ that lie close to the endpoints of this interval there will be few representations as $pq + rs$, while for numbers "well inside" this interval there ought to be many.
Let's dispose of the case of $1$-digit primes: in that case $\{p, q, r, s\} = \{2, 3, 5, 7\}$, and the only possible values of $pq + rs$ are $\{29, 31, 41\}$. For larger lengths, $p, q, r, s$ are all odd numbers, so $pq + rs$ is even.
I computed the number of representations as $pq + rs$ for $p, q, r, s$ being 3-digit and 4-digit primes, respectively. For 3-digit primes numbers with such a representation lie in the interval $[20000, 2000000]$ (actually $[22030,1948418]$), and for 4-digit primes they lie in the interval $[2000000, 200000000]$ (actually $[2062436,198303900]$). Plotting them gives:
Far from the representation as $pq+rs$ being unique,
• all even numbers in $[60198, 1217132]$ have at least $2$ representations,
• all even numbers in $[90376, 1038516]$ and in $[3667846, 161023932]$ have at least $10$ representations,
• all even numbers in $[12346780, 95078484]$ have at least $1000$ representations, etc.
There are even numbers with over $12000$ representations as you can see, and as you go to larger $N$ this will only increase.
I think there is nothing special about prime numbers here (once you start adding them): if you take merely "odd numbers" you'd probably see something similar.
• The remarkable resemblance of the two plots is interesting though, and I'm curious what the general shape (of the "upper" and "lower" curves) are. – ShreevatsaR Jun 28 '15 at 16:33 | 1,581 | 5,629 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-18 | latest | en | 0.921352 | # Special representation of a number. How can I check, if a number $n$ can be representated by. $$pq+rs$$. where $p,q,r,s$ are pairwise different prime numbers with the same number of digits.. For example,. $$105153899965560312960 = 3022993637\times 6003631993 + 9069920719\times 9592692301$$. has such a representation.. My questions :. • Is such a representation (if it exists), always unique ?. • How can I find the primes $p,q,r,s$ if a representation exists ?. • How can I check if a representation exists ?. • There is a very large number of counterexamples to your uniqueness conjecture, of which the smallest is $$11\cdot13 + 19\cdot 29 = 11\cdot 43 + 13\cdot 17$$ – MJD May 20 '15 at 13:12. • So, the first part is answered. Thank you. And I conjectured the uniqueness only for numbers large enough. It would be interesting, which is the largest counterexample (I think there is a largest). – Peter May 20 '15 at 13:14. You said in a comment:. I conjectured the uniqueness only for numbers large enough. It would be interesting, which is the largest counterexample (I think there is a largest).. I think your intuition is backwards here, and I think it will be instructive if I explain to you why; this is also why I was able to immediately guess that there would be counterexamples. Short summary: There so many 4-tuples compared with the number of possible sums that it is not possible for every 4-tuple to get its own sum.. Suppose our primes have $d$ digits and that there are $D$ primes with $d$ digits. The prime number theorem tells us that $D$ is around $O\left(\frac{10^d}{d}\right)$.. Then there are $O(D^4) = O\left(\frac{10^{4d}}{d^4}\right)$ quadruples $(p,q,r,s)$ of distinct $d$-digit primes. The quantity $n=pq+rs$ has around $2d$ digits and there are $10^{2d}$ of these numbers.. There is no reason to think that the quantities $pq+rs$ will be distributed among these $10^{2d}$ possibilities in any way other than randomly. (Additive properties of prime numbers are almost always random unless there is some obvious reason they cannot be. | To take an example I made up on the spot, the mod-3 remainder of $p+q$ is equal to 0, 1, or 2 with probability close to $\frac12, \frac14, \frac14$ just as one would expect.). So we should expect that what we are doing here is essentially throwing $O\left(\frac{10^{4d}}{d^4}\right)$ balls at random into only $10^{2d}$ bins . The number of balls greatly exceeds the number of bins when $d$ is large, so it is not at all surprising that some balls end up in the same bin; that is, that some tuples yield the same value for $pq+rs$. As $d$ increases, we should expect this to be vastly more likely, not less likely, and this argues strongly against the possibility of a maximal counterexample.. Indeed, one would expect the opposite to be true: let $N$ be given. Then we should expect to find many sets, each containing $N$ 4-tuples, for which all the 4-tuples in the set have the same $pq+rs$ value, no matter how large $N$ is large. If we want to find $n$ that can be represented as $n=pq+rs$ in one million different ways, we should expect that there will be many such.. (The tuples won't be distributed uniformly over the space of sums—for example, only a few tuples map to an odd sum—but this will tend to increase, not decrease, the number of collisions.). • That is convincing. +1 for the well explanation. – Peter May 20 '15 at 13:37. • Experiments bear out this theory. There are 20 ways to represent 3150 in the form $n=pq+rs$, and this is the most ways for any $n$ under 10,000. But if we start looking at larger primes, there are over ten thousand such $n$ below 200,000; for example there are 20 or more ways to represent each of 170280, 170430, 170436, 170520, 170790, 170940, 170970, 171318, 171780, 171990, 172410… – MJD May 20 '15 at 13:49. (Just to elaborate on MJD's answer.). Note that if $p, q, r, s$ are all $d$ digits long, which means that they all lie in the interval $[10^{d-1}, 10^d)$, then $pq + rs$ lies in the interval $[2\times10^{2d-2}, 2\times10^d)$. For numbers $N$ that lie close to the endpoints of this interval there will be few representations as $pq + rs$, while for numbers "well inside" this interval there ought to be many.. Let's dispose of the case of $1$-digit primes: in that case $\{p, q, r, s\} = \{2, 3, 5, 7\}$, and the only possible values of $pq + rs$ are $\{29, 31, 41\}$. For larger lengths, $p, q, r, s$ are all odd numbers, so $pq + rs$ is even.. I computed the number of representations as $pq + rs$ for $p, q, r, s$ being 3-digit and 4-digit primes, respectively. For 3-digit primes numbers with such a representation lie in the interval $[20000, 2000000]$ (actually $[22030,1948418]$), and for 4-digit primes they lie in the interval $[2000000, 200000000]$ (actually $[2062436,198303900]$). Plotting them gives:. Far from the representation as $pq+rs$ being unique,. • all even numbers in $[60198, 1217132]$ have at least $2$ representations,. • all even numbers in $[90376, 1038516]$ and in $[3667846, 161023932]$ have at least $10$ representations,. • all even numbers in $[12346780, 95078484]$ have at least $1000$ representations, etc.. There are even numbers with over $12000$ representations as you can see, and as you go to larger $N$ this will only increase.. I think there is nothing special about prime numbers here (once you start adding them): if you take merely "odd numbers" you'd probably see something similar.. • The remarkable resemblance of the two plots is interesting though, and I'm curious what the general shape (of the "upper" and "lower" curves) are. – ShreevatsaR Jun 28 '15 at 16:33. |
https://diffgeom.subwiki.org/w/index.php?title=Torus_in_3-space&oldid=1565 | 1,620,314,331,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988758.74/warc/CC-MAIN-20210506144716-20210506174716-00418.warc.gz | 226,665,543 | 7,802 | # Torus in 3-space
## Definition
The torus in $\R^3$ is constructed as follows:
• Take a base circle centered at a point which we shall call the origin. We shall call the radius of the base circle the average radius.
• Fix a length smaller than the average radius, called the tube radius.
• For each point on the torus, consider the circle centered at that point of radius equal to the tube radius, in the plane perpendicular to the plane of the base circle, and containing that center and the origin
• The union of all such circles is termed the torus
The torus can thus be thought of as the trace of a circle whose center is itself moving on a base circle, such that the plane of the circle always contains the origin of the base circle.
Some further terminology:
• The inner radius is the average radius minus the tube radius. It is the shortest possible distance between the origin and points on the torus. The set of points at this minimum distance forms a circle, called the inner rim or inner circle.
• The outer radius is the average radius plus the tube radius. It is the maximum distance between the origin and points on the torus. The set of points at this maixmum distance form a circle, called the outer rim or outer circle.
• Consider two planes parallel to the plane containing the base circle, with distance from it equal to the tube radius. Both these planes are tangent to the torus, meeting it at circles. These circles are termed the top circle and bottom circle (interchangeably). The radius of these circles is equal to the average radius.
## Equational descriptions
### Cartesian parametric equation
Let $r_1$ denote the average radius and $r_2$ denote the tube radius. Suppose the base circle is in the $xy$-plane and the origin of the torus is the origin. Then the parametric equations aer in terms of two angles, $\alpha$ and $\beta$, where:
• $x = r_1 \cos \alpha + r_2 \cos \alpha \cos \beta$
• $y = r_1 \sin \alpha + r_2 \sin \alpha \cos \beta$
• $z = r_2 \sin \beta$
Both $\alpha$ and $\beta$ are modulo $2\pi$.
## Abstract structure
Topologically, and even differentially, the torus is isomorphic to the direct product of circles, viz $S^1 \times S^1$. One coordinate describes the position of the center on the base circle, and the other coordinate describes the position of the point on that particular small circle. The Cartesian parametric equations given above make this explicit.
However, the metric structure on the torus is very different from that on $S^1 \times S^1$ with the direct product metric. The latter is actually the flat torus, which cannot be embedded in $\R^3$, but needs to be embedded in $\R^4$.
## Structure and symmetry
The isometry group of the torus is the semidirect product of the following two groups:
• The group $S^1$, which acts on the first coordinate by rotating the base circle.
• The group of order two which reflects the whole configuration about the $xy$-plane
In fact, this is precisely the group of isometries of $\R^3$ which fix the origin and the preserve the $z$-axis (though they may not preserve direction). | 729 | 3,103 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2021-21 | latest | en | 0.851063 | # Torus in 3-space. ## Definition. The torus in $\R^3$ is constructed as follows:. • Take a base circle centered at a point which we shall call the origin. We shall call the radius of the base circle the average radius.. • Fix a length smaller than the average radius, called the tube radius.. • For each point on the torus, consider the circle centered at that point of radius equal to the tube radius, in the plane perpendicular to the plane of the base circle, and containing that center and the origin. • The union of all such circles is termed the torus. The torus can thus be thought of as the trace of a circle whose center is itself moving on a base circle, such that the plane of the circle always contains the origin of the base circle.. Some further terminology:. • The inner radius is the average radius minus the tube radius. It is the shortest possible distance between the origin and points on the torus. The set of points at this minimum distance forms a circle, called the inner rim or inner circle.. • The outer radius is the average radius plus the tube radius. It is the maximum distance between the origin and points on the torus. The set of points at this maixmum distance form a circle, called the outer rim or outer circle.. • Consider two planes parallel to the plane containing the base circle, with distance from it equal to the tube radius. Both these planes are tangent to the torus, meeting it at circles. These circles are termed the top circle and bottom circle (interchangeably). The radius of these circles is equal to the average radius.. ## Equational descriptions. | ### Cartesian parametric equation. Let $r_1$ denote the average radius and $r_2$ denote the tube radius. Suppose the base circle is in the $xy$-plane and the origin of the torus is the origin. Then the parametric equations aer in terms of two angles, $\alpha$ and $\beta$, where:. • $x = r_1 \cos \alpha + r_2 \cos \alpha \cos \beta$. • $y = r_1 \sin \alpha + r_2 \sin \alpha \cos \beta$. • $z = r_2 \sin \beta$. Both $\alpha$ and $\beta$ are modulo $2\pi$.. ## Abstract structure. Topologically, and even differentially, the torus is isomorphic to the direct product of circles, viz $S^1 \times S^1$. One coordinate describes the position of the center on the base circle, and the other coordinate describes the position of the point on that particular small circle. The Cartesian parametric equations given above make this explicit.. However, the metric structure on the torus is very different from that on $S^1 \times S^1$ with the direct product metric. The latter is actually the flat torus, which cannot be embedded in $\R^3$, but needs to be embedded in $\R^4$.. ## Structure and symmetry. The isometry group of the torus is the semidirect product of the following two groups:. • The group $S^1$, which acts on the first coordinate by rotating the base circle.. • The group of order two which reflects the whole configuration about the $xy$-plane. In fact, this is precisely the group of isometries of $\R^3$ which fix the origin and the preserve the $z$-axis (though they may not preserve direction). |
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