# Taylor polynomial remainder

$$f(x) = \arctan(x^2)$$

1. Decide the Taylor polynomial of the first degree around $x = 1$

Answer: $$P(x)= \frac{\pi}{4}+ 1(x-1)$$

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.

• Which value of $c$ maximizes the error? – Andrew Li Mar 10 '18 at 19:11
• I would guess that it's 1,1? However, "just" taking 1,1 as c doesn't fit the answer in my textbook so I kind of assumed that it was something more to it than that... – gbgult Mar 10 '18 at 19:25

## 1 Answer

You have, since $c>.9$, $$\left|\frac{f''(c)}{6}\right|=\frac{c^4-1/3}{(1+c^4)^2}\leq1.1^4-1/3<1.2.$$ Then $$\left|\frac{f''(c)}{6}\,\frac1{100}\right|<\frac2{100}=\frac1{50}.$$

• Isn't that just the first derivate of $arctan(x^2)$? – gbgult Mar 10 '18 at 19:21
• Never mind. I never saw the square. – Martin Argerami Mar 10 '18 at 19:24