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I have to approximate this integral with an error lesser than 0.1 using Taylor Series. This is the integral: $$\int_0^1 \arctan\left(\frac{1}{x^{10}}\right) \,\Bbb dx.$$ If I understood correctly, I have to determinate the Taylor Series expansion in terms of a series with alternating signs. I have to check if the series has a uniform convergence in [0,1] and then take the series sign outside the integral sign and integrate the $x$-dependent part. Once done, I have to estimate the error using Leibniz: $|S - S_{n+1}|<|a_n|$. To find it I just have to calculate $n$ terms and when I’ll find $a_n<0.1$ I can computer the series from 0 to $n-1$. Am I correct?

I think I figured out how to do this job but I really do not know how to determinate the Taylor series expansion… I tried to use the McLaurin’s $\arctan(x)$ where $x_0=0$ but the integrand function is not defined for $x=0$… how can I handle this kind of problem? Please, can someone explain me if everything I said is correct and explain me how to determinate the series expansion in this case?

Sorry for my bad English and thank you in advance for your help!

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    $\begingroup$ You might want to use $\arctan (1/u) = \pi/2 - \arctan u.$ $\endgroup$
    – zhw.
    May 25, 2016 at 17:04
  • $\begingroup$ Thank you so much, I didn't know about this relation. $\endgroup$
    – PeppeDAlterio
    May 25, 2016 at 17:30
  • $\begingroup$ Can you please explain me where did you take that relationship? I have more exercises like this (exp(1/x) for example) and I'd like to know how to handle these problems. Thank you in advance $\endgroup$
    – PeppeDAlterio
    May 27, 2016 at 20:18
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    $\begingroup$ You can't expect this nice relation to hold too often; $e^{1/x}$ is quite a different situation. $\endgroup$
    – zhw.
    May 27, 2016 at 20:55

1 Answer 1

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$$I=\int_{0}^{1}\arctan\left(\frac{1}{x^{10}}\right)\,dx = \frac{\pi}{2}-\int_{0}^{1}\arctan(x^{10})\,dx$$ hence:

$$ I = \frac{\pi}{2}-\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{0}^{1} x^{10+20n}\,dx = \color{red}{\frac{\pi}{2}-\frac{1}{11}+\frac{1}{93}}-\sum_{n\geq 2}\frac{(-1)^n}{(11+20n)(1+2n)}.$$

The last series is convergent by Leibniz' test and bounded by its first term.

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  • $\begingroup$ Thank you, I didn't know about this relation! Anyway , one more thing, I can conclude that the integral is PI/2 - 1/11 becase |1/93| is lesser than 0.1 and then say that the result is 1.4 . Can you please confirm me this? Thank you! $\endgroup$
    – PeppeDAlterio
    May 25, 2016 at 17:31
  • $\begingroup$ That's fine, but be careful to round correctly. $\pi/2 - 1/11 = 1.48$, so to one decimal place you should answer 1.5. $\endgroup$
    – Alex Meiburg
    Jul 8, 2016 at 0:59

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