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?


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:

$$R_{n+1} \le \dfrac{M}{(n+1)!}(x-a)^{n+1}= \dfrac{625}{4!} \left(\dfrac{1}{10}\right)^4 = \dfrac{1}{384}$$


You're not "going wrong". Your answer is slightly better than the 1/384, because you are choosing a lower max value on this interval than 1. (Although the max value of sine is 1, the max value on this interval is your choice for M.) However, IIRC, the AP problem to which you're referring didn't allow for a calculator, so you couldn't have obtained your M.


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