I am trying to find a Maclaurin Series for $\arctan(x)$ up to the term with the fifth power of x and I have to use the method of successive differentiation. I know (from an example in my notes) the series found by integration of the $\dfrac{1}{1+x^2}$ series is $$x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}$$ When I try with successive differentiation, I cannot get this same series. Are both series correct, or is it a methodical error on my part? I checked the derivatives using WolframAlpha, so I can't see it being computational.
BY SUCCESSIVE DIFFERENTIATION:
To find coefficients of the power series, evaluate the function and it's successive derivatives at $x=0$:
$\arctan(0) =0$
$\arctan^{(1)}(x) = (1+x^2)^{-1}$ so $\arctan^{(1)}(0)= 1$
$\arctan^{(2)}(x) = -2x(1+x^2)^{-2}$ so $\arctan^{(2)}(0)= 0$
$\arctan^{(3)}(x) = (6x^2-2)(1+x^2)^{-3}$ so $\arctan^{(3)}(0) = -2$
$\arctan^{(4)}(x) = -24x(x^2-1)((1+x^2)^{-4}$ so $\arctan^{(4)}(0) = 0$
$\arctan^{(5)}(x) = 24(5x^4-10x^2+1)(1+x^2)^{-5}$ so $\arctan^{(5)}(0) = 24$
Which leads to the series
$$x-2x^3+24x^5-\cdots$$
which captures the alternation of the series but the coefficients are all off. If this is correct and there are indeed the two series, an explanation of this oddity would be much appreciated, though I don't see how this can be the case.
Best Answer
You're simply forgetting to divide by the factorials, as far as I can see. That is, you have $-2$ when you should be getting $$\frac{-2}{3!}=-\frac 1 3$$
You have $24=4!$ when you should be getting $$\frac{4!}{5!}=\frac 1 5$$
and so on.
Recall that the coefficients are given by $$\frac{f^{(n)}(0)}{n!}$$
There is a nice method to get the derivatives of all orders of $\arctan$: write $$\frac{1}{1+x^2}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)$$ and blindly differentiate as if $i\in\Bbb R$. Of course this can be made rigorous.
ADD A slick solution is the following:
$$\frac{1}{1+x^2}=\sum_{k=0}^\infty (-1)^k x^{2k}$$
It follows (by uniqueness of representation) that if $f(x)=\dfrac{1}{1+x^2}$, $$f^{(2k+1)}(0)=0\; ; \;k\geq 0$$ while $$f^{(2k)}(0)=(-1)^k (2k)!\; ;\; k\geq 0$$
but these are $n$th derivatives of $(\tan^{-1}(x))'$, so are $(n+1)$th derivatives of $\tan^{-1}(x)$.