<s>[INST] Erroneously Finding the Lagrange Error Bound Consider $f(x) = \sin(5x + \pi/4)$ and let $P(x)$ be the third-degree Taylor polynomial for $f$ about $0$. I am asked to find the Lagrange error bound to show that $|(f(1/10) - P(1/10))| < 1/100$. Because $P(x)$ is a third-degree polynomial, I know the difference is in the fourth degree term. So I found the fourth derivative to be $f(x) = 625 \sin(\pi/4 + 5x)$. Then I substituted $1/10$ into the fourth derivative to find $M$. I substituted the $M$ I found and $x$ as $1/10$ into the formula for a Lagrange remainder: $(M(1/10)^4) / 4!$ This calculates to be $0.002498$. The answer is that the Lagrange error is $1 / 384$. Could someone please help me understand where I am going wrong?
 [/INST] We have:
$$f(x) = \sin\left(5x + \dfrac{\pi}{4}\right)$$
The third degree Taylor Polynomial is given by:
$$T_3(x) = -\frac{125 x^3}{6 \sqrt{2}}-\frac{25 x^2}{2 \sqrt{2}}+\frac{5 x}{\sqrt{2}}+\frac{1}{\sqrt{2}}$$
The error term is given by:
$$R_{n+1} = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \le \dfrac{M}{(n+1)!}(x-a)^{n+1}$$
We have $n= 3$, thus:
$$\dfrac{d^4}{dx^4} \left( \sin\left(5x + \dfrac{\pi}{4}\right)\right) = 625 \sin \left(5 x+\frac{\pi }{4}\right)$$
We have $a = 0, n= 3, x = \dfrac{1}{10}$ and the max of sine is $1$, so this yields:
$$R_{n+1} \le \dfrac{M}{(n+1)!}(x-a)^{n+1}= \dfrac{625}{4!} \left(\dfrac{1}{10}\right)^4 = \dfrac{1}{384}$$
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