Vrftjkry

I encounter the problem when I try to get the Taylor series of $arctan x$ at $x=1$. It seems that the method for expanding it at $x=0$ does not work anymore (which is obtained by observing that $arctan' x=frac{1}{1+x^2}=sum_{n=0}^infty (-1)^n x^{2n}$, and then integration over the convergence domain).

So I try to compute the derivatives at $x=1$: Note that if we set $y(x)=arctan x$, then
$$y'=frac{1}{1+x^2},quad y''=frac{-2x}{(1+x^2)^2}implies
(1+x^2)y''=-2x y'.$$
By Leibniz rule, taking derivative of order $n$ at both side, we have
$$
y^{(n+2)}(1)+(n+1)y^{(n+1)}(1)+frac{n(n+1)}{2} y^{(n)}(1)=0.
$$
It is easy to show
$$
y(1)=pi/4,quad y'(1)=1/2,quad y''(1)=-1/2,quad y'''(1)=1/2,quad y''''(1)=0.
$$
I don't know how to get a general formula from the above recursive equation, any ideas?

In fact, I am also searching for a general theory about recursive equations, any reference there?

If you simply want to find the Taylor series about $x=1$ of $text{arctan}(x)$, then you can use $$frac{text{d}}{text{d}x},text{arctan}(x)=frac{1}{1+x^2}=frac{1}{1+(z+1)^2}=frac{1}{2text{i}}left(frac{1}{1-text{i}+z}-frac{1}{1+text{i}+z}right),,$$
where $z:=x-1$. This gives
$$frac{text{d}}{text{d}x},text{arctan}(x)=frac{1}{2text{i}},left(frac{left(1+frac{z}{1-text{i}}right)^{-1}}{1-text{i}}-frac{left(1+frac{z}{1+text{i}}right)^{-1}}{1+text{i}}right),.$$
Therefore, if $|z|<sqrt{2}$, then
$$frac{text{d}}{text{d}x},text{arctan}(x)=frac{1}{2text{i}},left(frac{sumlimits_{k=0}^infty,(-1)^k,left(frac{z}{1-text{i}}right)^{k}}{1-text{i}}-frac{sumlimits_{k=0}^infty,(-1)^k,left(frac{z}{1+text{i}}right)^{k}}{1+text{i}}right),.$$
Simplifying this, we have
$$frac{text{d}}{text{d}x},text{arctan}(x)=sum_{k=0}^infty,(-1)^k,left(frac{frac1{(1-text{i})^{k+1}}-frac{1}{(1+text{i})^{k+1}}}{2text{i}}right),(x-1)^k$$
for $xinmathbb{C}$ such that $|x-1|<sqrt{2}$.

Using $1pmtext{i}=sqrt{2},expleft(pmdfrac{text{i}pi}{4}right)$, we conclude that
$$frac{text{d}}{text{d}x},text{arctan}(x)=sum_{k=0}^infty,frac{(-1)^k}{2^{frac{k+1}{2}}},sinleft(frac{(k+1)pi}{4}right),(x-1)^k$$
for all complex numbers $x$ with $|x-1|<sqrt{2}$.
Integrating the series above, we have
$$text{arctan}(x)=frac{pi}{4}+sum_{k=1}^{infty},frac{(-1)^{k-1}}{2^{frac{k}{2}},k},sinleft(frac{kpi}{4}right),(x-1)^k,.$$
for every $xinmathbb{C}$ such that $|x-1|<sqrt{2}$. In this way, it follows that
$$y^{(k)}(1)=begin{cases}frac{pi}{4}&text{if }k=0,,\
frac{(-1)^{k-1},(k-1)!}{2^{frac{k}{2}}},sinleft(frac{kpi}{4}right)&text{if }k=1,2,3,ldots,,
end{cases}$$
where $y(x):=text{arctan}(x)$.

However, if you really want to solve the recursion without the expansion above, then let $a_k:=dfrac{y^{(k)}(1)}{(k-1)!}$ for $k=1,2,3,ldots$. From
$$y^{(k+2)}(1)+(k+1),y^{(k+1)}(1)+frac{k(k+1)}{2},y^{(k)}(1)=0text{ for }k=1,2,3,ldots,,$$
we divide both sides by $(k+1)!$ to get
$$a_{k+2}+a_{k+1}+frac{1}{2},a_k=0text{ for }k=1,2,3,ldots,.$$
The characteristic polynomial of the recursion above is $lambda^2+lambda+dfrac{1}{2}$, whose roots are $dfrac{-1pmtext{i}}{2}$. Therefore, for $k=1,2,3,ldots$,
$$a_k=p,left(frac{-1+text{i}}{2}right)^k+q,left(frac{-1-text{i}}{2}right)^k$$
for some fixed $p,qinmathbb{C}$. Since $a_1=dfrac{1}{2}$ and $a_2=-dfrac{1}{2}$, we get $$p=+frac{text{i}}{2}text{ and }q=-frac{text{i}}{2},,$$
whence
$$y^{(k)}(1)=frac{left(frac{-1+text{i}}{2}right)^k-,left(frac{-1-text{i}}{2}right)^k}{2text{i}},(k-1)!=frac{(-1)^{k-1},(k-1)!}{2^{frac{k}{2}}},sinleft(frac{kpi}{4}right)$$
for $k=1,2,3,ldots$.