$$f(x) = \arctan(x^2)$$
- Decide the Taylor polynomial of the first degree around $x = 1$
Answer: $$ P(x)= \frac{\pi}{4}+ 1(x-1) $$
- Show that $${\frac{\pi}{4} + \frac 1 {10} - f(1.1)} < \frac 1 {50}$$
What I know: I'm supposed to show that the error of the approximation of $ f(1.1)$ which is $\dfrac{\pi}{4} + \dfrac 1 {10}$ is smaller than $\dfrac {1}{50}$.
The error is estimated with $$\frac{f''(c)}{3!}\left(\frac {1}{10}\right)^2$$ which gives me that the error is estimated by $$\frac{3c^4-1}{(1+c^4)^2}\left(\frac{1}{10}\right)^2$$ and that $1 \leq c \leq 1.1$. However, I don't really know where to go from there.
You have, since $1\leq c \leq1.1$, $$ \left|\frac{f''(c)}{6}\right|=\frac{c^4-1/3}{(1+c^4)^2}\leq1.1^4-1/3<2. $$ Then $$ \left|\frac{f''(c)}{6}\,\frac1{100}\right|<\frac2{100}=\frac1{50}. $$ With more care the first estimate can be reduced from $2$ to $0.19$, so the approximation is as good as $1/500$.