Taylor expansion of $\frac{1}{1+x^{2}}$ at $0.$ I am trying to find the Taylor expansion for the function $$f(x) = \frac{1}{1+x^{2}}$$ at $a=0.$ I have looked up the Taylor expansion and concluded that it would be sufficient to show that
\begin{eqnarray} f^{(2k)}(0) & = & (-1)^{k}(2k)! \\ f^{(2k+1)}(0) & = & 0 \end{eqnarray}
The problem I am having is that the derivatives of $f$ are horrid - even proving them to be of a particular form seems hopeless. For example, I have tried without success to show that $$ f^{(2k)}(x) = \frac{(2k)!\left( (-1)^{k} + p(x) \right)}{(x^{2}+1)^{2k+1}} $$ where $p$ is a polynomial satisfying $p(0) = p^{\prime}(0) = 0$ and similar for $f^{(2k+1)}.$
I am sure that there is some trick I am missing. The book I am reading hasn't yet covered the binomial expansion for non-integer exponents and so it would seem circular to have to use the binomial expansion for $(1+x^{2})^{-1}.$
How do I derive the Taylor polynomial of order $2n$ at $0$ for $f$?
 A: It is easier to use a geometric series to find the Taylor series. Note that $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$
Now substitute $-x^2$ in for $x$ to find:
$$\sum_{n=0}^\infty (-1)^n x^{2n} = \frac{1}{1+x^2}$$
Note that the radius of convergence of the geometric series is 1. That is is works provided $|x|<1$. This also means that the new series is valid provided $|-x^2| < 1$, which means $|x|<1$. Hence it has the same radius of convergence.
A: Direct Taylor expansion can be managed.
Let us rewrite the relation $y=\frac1{1+x^2}$ as
$$(1+x^2)y=1.$$
Then by successive derivations,
$$2xy+(1+x^2)y'=0,$$
$$2y+4xy'+(1+x^2)y''=0,$$
$$6y'+6xy''+(1+x^2)y'''=0,$$
$$12y''+8xy'''+(1+x^2)y''''=0,$$
$$20y'''+10xy''''+(1+x^2)y'''''=0,$$
$$30y''''+12xy'''''+(1+x^2)y''''''=0,$$
$$...$$
$$n(n-1)y^{(n-2)}+2nxy^{(n-1)}+(1+x^2)y^{(n)}=0.$$
(from a row to the next, the coefficient of the middle term increases by $2$, and the coefficient of the left term is the sum of the coefficients of the middle terms.)
When setting $x=0$, this reduces to
$$y=1,$$
$$y'=0,$$
$$2y+y''=0,$$
$$6y'+y'''=0,$$
$$12y''+y''''=0,$$
$$20y'''+y'''''=0,$$
$$30y''''+y''''''=0,$$
$$...$$
$$n(n-1)y^{(n-2)}+y^{(n)}=0.$$
Then by recurrence, $$y^{(2k)}=(-1)^k(2k)!y=(-1)^k(2k)!,\\y^{(2k+1)}=(-1)^k(2k+1)!y'=0.$$
A: Note that for $|t| < 1$ the Taylor expansion of $\frac{1}{1+t}$ is
$$ \frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 + ... $$
Now insert $t = x^2$ to get
$$ \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + x^8 + ... $$
and you see there are no odd power terms in $x$.
A: Let us generalize the problem. We want expand $\dfrac1{1+x^2}$ about $a$ providing we know Taylor series of
$$
\frac{1}{1-y}=\sum_{n=0}^\infty y^n\qquad;\qquad\text{for $|y|<1$}.\tag1
$$
Rewrite
\begin{align}
\frac1{1+x^2}&=\frac1{1+a-(a-x^2)}\\
&=\frac1{a+1}\left[\frac1{1-\left(\dfrac{a-x^2}{a+1}\right)}\right].\tag2
\end{align}
Setting $y=\dfrac{a-x^2}{a+1}$ and using $(1)$, then $(2)$ becomes
$$
\color{blue}{\frac1{1+x^2}=\frac1{a+1}\sum_{n=0}^\infty \left(\frac{a-x^2}{a+1}\right)^n}\qquad;\qquad\text{for $\color{red}{\left|\dfrac{a-x^2}{a+1}\right|<1}$}.\tag3
$$
The last step is setting $a=0$.
A: Since
$$\frac{1}{1+x}=1-x+x^2-x^3+\cdots+(-1)^nx^n+o(x^n),$$
we have
$$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\cdots+(-1)^nx^{2n}+o(x^{2n}).$$
