Calculus + Factorial = ?

Recently, I've been messing around with factorials (mainly because I was just told that x!(xe)x×2πxx! \approx (\frac{x}{e})^x\times\sqrt{2\pi x}). I came across a couple calculus problems involving the function. Since I am very new to calculus, I figured I should ask the community on how to get the answers and what they are, not just by typing them into Wolfram Alpha and seeing what comes out.

Since x!x! increases so rapidly, I decided taking the xx root of x!x!. This gave me the function, f(x)=x!1x\color{#D61F06}{f(x)=x!^{\frac{1}{x}}}. That function is the red graph in the picture below.

I noticed that, even though f(0)f(0) is undefined (since10\frac{1}{0} is undefined), it still appears to have a value. That is my first problem: to find

limx0f(x)\lim_{x\rightarrow 0} \color{#D61F06}{f(x)}

Next, I noticed that f(x)\color{#D61F06}{f(x)} is not linear. However, it appeared to be linear, so I decided to graph the derivative, which gave me the function g(x)=ddx f(x)\color{#3D99F6}{g(x)=\dfrac{\text{d}}{\text{d}x}~f(x)}. This is the blue graph in the picture above. Since f(x)\color{#D61F06}{f(x)} approached linearity (?), I knew that g(x)\color{#3D99F6}{g(x)} must have a limit as it approached infinity. That leads me to my second problem: to find

limxg(x)\lim_{x\rightarrow \infty} \color{#3D99F6}{g(x)}

The graph is at this link.

NOTE: The domains of both functions stop before 171 because 171! is a massive number, to the point that most online calculators can't handle.

Note by Blan Morrison
1 year, 10 months ago

No vote yet
1 vote

  Easy Math Editor


When posting on Brilliant:


MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
\sqrt{2} 2 \sqrt{2}
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

  1. Stirling's formula is a good reference for the approximation. Note that this approximation holds for large values, and not necessarily for small values.

  2. According to Wolfram Alpha, limx0f(x) \lim_{x\rightarrow 0 } f(x) is an interesting value.
    It is not immediately apparant to me why this is true as yet.
    Note: γ\gamma is the Euler-Mascheroni constant.

  3. Note that f(x)xe f(x) \approx \frac{x}{e} for larget values. Hence, if there is any justice in the world (meaning that if the limit exists), it is most likely that \lim g(x) = \frac{1}[e} .

  4. However, there is a slight flaw in your logic. Namely, the following statement is not true: "If f(x)g(x)<ϵ | f(x) - g(x) | < \epsilon for all xx, then limg(x)=limf(x) \lim g'(x) = \lim f'(x) .".
    An extra condition will need to be added to arrive at "If f(x)g(x) f(x) \approx g(x) (in some manner), then limg(x)=limf(x) \lim g'(x) = \lim f'(x) ."

Calvin Lin Staff - 1 year, 10 months ago

Log in to reply

L=limn0(n!)1n=limn0(Γ(n+1))1n\displaystyle L=\lim_{n\to 0} (n!)^{\frac 1n}=\lim_{n\to 0} (\Gamma(n+1))^{\frac 1n} L=limn0eln(Γ(n+1))n=elimn0ln(Γ(n+1))n\displaystyle L=\lim_{n\to 0} e^{\frac {\ln (\Gamma(n+1))}{n}}=e^{\lim_{n\to 0} \frac {\ln(\Gamma(n+1))}{n}}

Using L'Hospital rule we get L=elimn0ψ(n+1)=eψ(1)\displaystyle L= e^{\lim_{n\to 0} \psi(n+1)}=e^{\psi(1)}

And using ψ(1)=γ\psi(1)=-\gamma we get L=eγL=e^{-\gamma}

Where Γ(.)\Gamma(.) is the Gamma function, ψ(z)\psi(z) is the Digamma function and γ\gamma is the Euler-Mascheroni constant

Rohan Shinde - 10 months, 1 week ago

Log in to reply

I did some experimenting with factorials, here are some interested facts:

  1. The limit as f(x) approaches 0 is equal to the negative root of the equation x!=2x!=2
  2. The limit as g(x) approaches infinity is equal to this limit: limx(x0.1ex)1x\large \lim_{x \to \infty}(x^{0.1}e^{-x})^{\frac{1}{x}} and also this limit: (x can be any postive real number) limn0(xnex)1x\large \lim_{n \to 0}(x^{n}e^{-x})^{\frac{1}{x}}

Those limits are derived from the gamma function, and the second one seem to evaluate to 1e\frac{1}{e}

Anthony Xu - 1 year, 9 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...