CVPL6

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Write the Taylor expansion of f(z)=(z-1)(z-2)^{3}\, 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,

f(z)=(z-1)(z-2)^{3}=z^{4}-7z^{3}+18z^{2}-20z+8\,

f'(z)=4z^{3}-21z^{2}+36z-20\,

f''(z)=12z^{2}-42z+36\,

f'''(z)=24z-42\,

f^{{(4)}}(z)=24\,

Finally,

f(z)=(16-56+72-40+8)+(32-84+72-20)(z-2)+{\frac  {1}{2}}(48-84+36)(z-2)^{2}\,

+{\frac  {1}{6}}(48-42)(z-2)^{3}+{\frac  {24}{24}}(z-2)^{4}\,

f(z)=(z-2)^{3}+(z-2)^{4}\,

Of course this is much more easily done by noting that f(z)=(z-2+1)(z-2)^{3}=(z-2)^{4}+(z-2)^{3}

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