1
$\begingroup$

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

  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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.