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

  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.

  • $\begingroup$ Which value of $c$ maximizes the error? $\endgroup$ – Andrew Li Mar 10 '18 at 19:11
  • $\begingroup$ 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... $\endgroup$ – gbgult Mar 10 '18 at 19:25

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 $$ \left|\frac{f''(c)}{6}\,\frac1{100}\right|<\frac2{100}=\frac1{50}. $$

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

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