CVPL6

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 $f (z) = (z - 2 + 1)(z - 2) 3 = (z - 2) 4 + (z - 2) 3$