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I know the Taylor series of $\arctan x$ at $x=0$.

But I have been stuck when $x=1$.

I computed $f'(1)=1/2$, $f''(1)=-1/2$, $f'''(1)=1/2$, $f''''(1)=0$, $f'''''(1)=-3$.

But.. I can't find any regularity.

What should I do to write regular Taylor series form..?

user26857
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Shineme
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    If you, instead of looking at the derivatives, look at the Taylor coefficients, the pattern is easier to spot: $\frac12, -\frac12, \frac14, 0, -\frac18, \frac18, -\frac1{16}, 0, \ldots$ – Arthur Dec 09 '19 at 11:02
  • Cf. https://math.stackexchange.com/q/3439994/96384, https://math.stackexchange.com/q/1858869/96384, https://math.stackexchange.com/q/1069869/96384. – Torsten Schoeneberg Jun 16 '23 at 19:10

1 Answers1

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Start from expanding the derivative of $\arctan(x)$, which is $\frac{1}{1+x^2}$.

To expand it around $x=1$ write it as

$$\begin{align}\frac{1}{1+x^2}&=\frac{i/2}{i+x}+\frac{-i/2}{-i+x}\\ &=\frac{i/2}{i+1+x-1}+\frac{-i/2}{-i+1+x-1}\\ &=\frac{i/2}{(i+1)}\frac{1}{1+(x-1)/(i+1)}+\frac{-i/2}{-i+1}\frac{1}{1+(x-1)/(1-i)}\\ &=\frac{i/2}{(i+1)}\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^n}{(1+i)^n}+\frac{-i/2}{-i+1}\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^n}{(1-i)}\\ &=\sum_{n=0}^{\infty}\left(\frac{(-1)^ni/2}{(i+1)^{n+1}}+\frac{(-1)^n(-i/2)}{(-i+1)^{n+1}}\right)(x-1)^{n}\end{align}$$

Now you can integrate term-by-term and add the constant $\arctan(1)$ to get

$$\arctan(x)=\arctan(1)+\sum_{n=0}^{\infty}\frac{\left(\frac{(-1)^ni/2}{(i+1)^{n+1}}+\frac{(-1)^n(-i/2)}{(-i+1)^{n+1}}\right)}{n+1}(x-1)^{n+1}$$

Note that all coefficients are real numbers. You can simplify them further if you like.

topeik
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