CVPL5

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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}\,


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