Taylor expansion of $\frac{1}{1+x^{2}}$ at $0.$

Question

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 $f$?

Answer 1

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.

Answer 2

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 $x$.

Answer 3

Since $$\frac{1}{1+x}=1-x+x^2-x^3+\cdots+(-1)^nx^n+o(x^n),$$ we have $$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\cdots+(-1)^nx^{2n}+o(x^{2n}).$$

Answer 4

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, $$y^{(2k)}=(-1)^k(2k)!y=(-1)^k(2k)!,\\y^{(2k+1)}=(-1)^k(2k+1)!y'=0.$$

Answer 5

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 $a=0$.