This one seems a lot harder...

Find the power series about the origin of:

z^2/(4-z)^2

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- Apr 10th 2010, 10:28 PMjzelltAnother Power Series
This one seems a lot harder...

Find the power series about the origin of:

z^2/(4-z)^2 - Apr 11th 2010, 12:16 AMmr fantastic
You only need to bother calculating the series for 1/(4 - z)^2 (why?).

Did you bother to read the link I gave in my reply here: http://www.mathhelpforum.com/math-he...er-series.html

Now look for the pattern in $\displaystyle f^{n}(0)$ where n = 0, 1, 2, 3, ....

If you need more help, please say where you are stuck. - Apr 11th 2010, 02:39 AMFailure
If you know the power series of $\displaystyle \frac{1}{(1-z)^2}$ you are essentially done. Consider:

$\displaystyle \frac{1}{(1-z)^2}=\frac{d}{dz}\frac{1}{1-z}=\frac{d}{dz}\sum_{n=0}^\infty z^n=\sum_{n=1}^\infty n z^{n-1}=\sum_{n=0}^\infty (n+1)z^n$

To know this is perhaps generally useful, but now let's solve your original problem:

$\displaystyle \frac{z^2}{(4-z)^2}=\left(\frac{z}{4}\right)^2\cdot\frac{1}{\lef t(1-\left(\frac{z}{4}\right)\right)^2}=\left(\frac{z}{ 4}\right)^2\cdot\sum_{n=0}^\infty(n+1)\left(\frac{ z}{4}\right)^n=\ldots$