CVPL5

Write the Taylor expansion of $f(z)=z^{10}\,$ at $z=2\,$ .

In general, the Taylor expansion around $z_0\,$ is $f(z) = \sum_{k=0}^\infty \frac{f^{(k)}(z_0)}{k!}(z-z_0)^k\,$ .

In this case,

$z^{10} = \sum_{k=0}^{10} \frac{\frac{d^k}{dz^k}z^{10}\big|_{z=2}}{k!}(z-2)^k\,$

$= \sum_{k=0}^{10} \frac{ \frac{10!}{(10-k)!}z^{10-k}\big|_{z=2}}{k!}(z-2)^k\,$

$= \sum_{k=0}^{10} \frac{10!}{(10-k)!k!} 2^{10-k}(z-2)^k\,$

$= 2^{10}\sum_{k=0}^{10} {10 \choose k}\left(\frac{1}{2}\right)^k (z-2)^k\,$

$= 2^{10}\sum_{k=0}^{10} {10 \choose k} \left(\frac{z}{2}-1\right)^k\,$

Note that the answer may also be written down straight away, as an application of the Binomial Theorem:

$z^{10} = (2+z-2)^{10} = \sum_{k=0}^{10}{10 \choose k} 2^{10-k}(z-2)^k\,$