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- #1

- Thread starter StefanM
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- #1

- Feb 9, 2012

- 118

This could work: $\dfrac{1}{{1 + z}} = \dfrac{1}{{6 + (z - 5)}} = \dfrac16\cdot\dfrac{1}{{1 + \frac{{z - 5}}{6}}}$

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- #3

- Feb 12, 2012

- 26

With a $\frac 16$ factorThis could work: $\dfrac{1}{{1 + z}} = \dfrac{1}{{6 + (z - 5)}} = \dfrac{1}{{1 + \frac{{z - 5}}{6}}}$

- Feb 9, 2012

- 118

Stefan because given a power series $\displaystyle\sum_{k=0}^\infty b_k(z-a)^k,$ you have $a$ is the center, so what I'm doing there is to invoke the geometric series centered at $z=5.$

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- #6

- Feb 9, 2012

- 118

I think I get what you mean, so yes.

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- #8

- Feb 9, 2012

- 118

Oh I'm sorry, I misread the question, yes, that's it.

I thought you were asking it for $z=5.$

I thought you were asking it for $z=5.$

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- #10

Just to summarise what has been written...so the power series is ......Sum 1/(4^n+1)*(z+5)....with R=-5.....?

\[ \displaystyle \begin{align*} \frac{1}{1 + z} &= \frac{1}{-4 + (z + 5)} \\ &= -\frac{1}{4}\left[\frac{1}{1 - \frac{z + 5}{4}}\right] \end{align*} \]

and keeping in mind that $\displaystyle \sum_{n = 0}^{\infty}a\,r^n = \frac{a}{1 - r} $ for $ \displaystyle |r| < 1 $, and taking note that $\displaystyle \frac{1}{1 - \frac{z + 5}{4}} $ is of the form $ \displaystyle \frac{a}{1 - r} $, that means

\[ \displaystyle \begin{align*} -\frac{1}{4}\left[\frac{1}{1 - \frac{z + 5}{4}}\right] &= -\frac{1}{4} \sum_{n = 0}^{\infty} \left(\frac{z + 5}{4}\right)^n \end{align*} \]

and this series is convergent where

\[ \displaystyle \begin{align*} \left|\frac{z + 5}{4}\right| &< 1 \\ \frac{|z + 5|}{|4|} &< 1 \\ \frac{|z + 5|}{4} &< 1 \\ |z + 5| &< 4 \end{align*} \]

in other words, the series is convergent inside but not on the circle of radius 4 units centred at z = -5.

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- #12

- Feb 7, 2012

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