$f(x) = \displaystyle \frac{1}{1+x^{2}}$

Compute

$f(x)$

$=$

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$f(0)$

$=$

$f^\prime(x)$

$=$

$\hskip 30pt$

$f^\prime(0)$

$=$

$f^{\prime\prime}(x)$

$=$

$\hskip 30pt$

$f^{\prime\prime}(0)$

$=$

$f^{\prime\prime\prime}(x)$

$=$

$\hskip 30pt$

$f^{\prime\prime\prime}(0)$

$=$

$f^{(iv)}(x)$

$=$

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$f^{(iv)}(0)$

$=$

$f^{(v)}(x)$

$=$

$\hskip 30pt$

$f^{(v)}(0)$

$=$

We see that for the odd terms $f^{(2k+1)}(0)$ =
and we also see that for the even derivatives $f^{(2k)}(0)$ = .
Hence the Taylor series for $f$ centered at $0$ is given by

$f(x) = \displaystyle \sum_{k=0}^\infty$ $x^{2k}$.