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Power series expansion

StefanM

New member
Jan 30, 2012
28
I'm trying to find the power series expansion of 1/1+z at z=-5 and the radius of convergence.How should I think and solve this problem?
I'm looking for a step by step explanation because I want to understand the mechanics behind it.Thank you.
 

Krizalid

Active member
Feb 9, 2012
118
This could work: $\dfrac{1}{{1 + z}} = \dfrac{1}{{6 + (z - 5)}} = \dfrac16\cdot\dfrac{1}{{1 + \frac{{z - 5}}{6}}}$
 
Last edited:

StefanM

New member
Jan 30, 2012
28
Thanks but how did you come up with that....and why in this form?
 

Moo

New member
Feb 12, 2012
26
This could work: $\dfrac{1}{{1 + z}} = \dfrac{1}{{6 + (z - 5)}} = \dfrac{1}{{1 + \frac{{z - 5}}{6}}}$
With a $\frac 16$ factor :p
 

Krizalid

Active member
Feb 9, 2012
118
Haha, yes, Moo, I forgot to add it.

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 $z=5.$
 

StefanM

New member
Jan 30, 2012
28
again same question also is z=-5 meaning it should be z+5 (below) if it was 5 =5 the it should be .....z-5(below),right?
 

Krizalid

Active member
Feb 9, 2012
118
I think I get what you mean, so yes.
 

StefanM

New member
Jan 30, 2012
28
so below should be -4+(z+5) maybe?
 

Krizalid

Active member
Feb 9, 2012
118
Oh I'm sorry, I misread the question, yes, that's it.
I thought you were asking it for $z=5.$
 

StefanM

New member
Jan 30, 2012
28
so the power series is ......Sum 1/(4^n+1)*(z+5)....with R=-5.....?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
so the power series is ......Sum 1/(4^n+1)*(z+5)....with R=-5.....?
Just to summarise what has been written...

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

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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,680
A useful check in problems like this is that the radius of convergence is always the distance from the centre of expansion to the nearest singularity. In this case the expansion is centred at $z=-5$, and the function $\frac1{1+z}$ has a singularity at $z=-1$. The distance from –5 to –1 is 4, so that is the radius of convergence (confirming that the answer in the previous comment is correct).