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I want to find the Taylor series of $f(x)=\arctan x,\; x\in[-1,1],\;\xi=0$.

That's what I have tried do far:

$$f'(x)=\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}=\sum_{n=0}^{\infty} (-x^2)^n.$$

How can I continue?

Hakim
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evinda
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2 Answers2

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Notice that $\sum\limits_{n=0}^\infty (-1)^nx^{2n}$ is a power series with radius $R=1$ hence we can integrate term by term inside the interval of convergence $(-1,1)$ so for $|x|<1$ we have:

$$\arctan x=\int_0^x\frac{dt}{1+t^2}=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}$$

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    Why do we take these limits of integration ? – evinda May 26 '14 at 16:10
  • Do you know that $\arctan0=0$? –  May 26 '14 at 16:11
  • Yes,has it to do with that,that we take this limits of integration? :/ – evinda May 26 '14 at 16:11
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    To simplify we always take the limits of integration $0$ and $x$ but in this case we have $\arctan0=0$ and in the general case we write $$f(x)=f(0)+\int_0^x f'(t)dt$$ –  May 26 '14 at 16:15
  • A ok... But don't we have to check also what happens at the point $-1,1$ ? Or isn't it necessary? – evinda May 26 '14 at 16:21
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    Yes you can do it and it should be simple using the Leibniz theorem. –  May 26 '14 at 16:22
  • So,do I have to do this,to be right or can I just do it to verify the result? – evinda May 26 '14 at 16:26
  • It's simple to see that the founded series is convergent for $x=\pm1$ using as I said the Leibniz theorem so the interval of convergence is $[-1,1]$. –  May 26 '14 at 16:32
  • And something else... I saw now the solution in my notes,and there it is done like that:

    $$\frac{1}{1+t^2}=1-t^2+ \dots +(-1)^{n-1} t^{2n-2}+\frac{(-t^2)^n}{1+t^2}$$

    then they integrate and to show that it is the right Taylor series,they check if $\int_0^x \frac{(-t^2)^n}{1+t^2}dt \to 0$..

    Why do we take this equality: $\frac{1}{1+t^2}=1-t^2+ \dots +(-1)^{n-1} t^{2n-2}+\frac{(-t^2)^n}{1+t^2}$ ?

    Also,why do we check if $\int_0^x \frac{(-t^2)^n}{1+t^2}dt \to 0$,to check if it is the right Taylor series?

    Could you explain it to me? I haven't understood it.. :/

     – evinda May 26 '14 at 16:40
  • This method doesn't use the known result for power series. It just use the definition of a series $$\sum_{k=0}^\infty a_k=\sum_{k=0}^n a_k+R_n$$ where $R_n$ is called the remainder and it has in your case an integral representation and you should prove that this remainder tends to $0$ if $n$ tends to $\infty$. –  May 26 '14 at 16:47
  • So,if I want to find $R_n$ in that way,can't I find the Taylor series in the way that you explained me? – evinda May 26 '14 at 16:58
  • I can I use the powerseries that we have found? But then $R_n$ will not have an integral representation? Or am I wrong???? – evinda May 26 '14 at 17:07
  • $R_n=\int_0^x \frac{(-t^2)^n}{1+t^2}dt$ and in your notes there's a proof that $R_n\to0$ isn't it? that's all what you have to do! –  May 26 '14 at 17:09
  • Yes,but how do we know that $R_n=\int_0^x \frac{(-t^2)^n}{1+t^2}dt$,doing it in the way that you explained me? – evinda May 26 '14 at 17:13
  • Look carefully at the sum which you have wrote above and integrate it so if we should take $R_n$ what expression looks like a remainder? –  May 26 '14 at 17:19
  • From which sum do I have to find $R_n$ ? From this one: $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$ ? – evinda May 26 '14 at 17:24
  • No from the geometric sum $$\sum_{k=0}^{n-1}(-t^2)^k=\frac{1-(-t^2)^n}{1+t^2}$$ –  May 26 '14 at 17:27
  • Sorry evinda I have to go! Try to understand your notes and if you have more questions you can go to the chat room you'll find a very helpful guys there;-) –  May 26 '14 at 17:32
  • Sweet!${}{}{}{}$ – amWhy May 27 '14 at 12:18
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Thus, we have $$f(x)=\int f'(x)\,dx=\int\sum_{n\ge 0}(-1)^nx^{2n}\,dx\ =\\ =\ \sum_{n\ge 0}(-1)^n\frac{x^{2n+1}}{2n+1}\ +C$$ Then find $C$ by plugging in $x=0$.

What will you get if you plug in $x=1$?

Berci
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