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[SOLVED] powers of z-1

dwsmith

Well-known member
Feb 1, 2012
1,673
How do I expand $\displaystyle f(z) =\frac{2z + 3}{z + 1}$ in powers of $z-1$?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Use the substitution $u=z-1$ .
 

dwsmith

Well-known member
Feb 1, 2012
1,673

pickslides

Member
Feb 1, 2012
57
z=u+1, now find f(z)= f(u+1)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
So $\displaystyle f(z) = f(u + 1) = \frac{2u + 5}{u + 2}$.

I am a little confused. How do I expand this to find the radius of convergence? By a Taylor series?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
That would be one method. Perhaps simpler is to use the fact that u+2= 2((u/2)+ 1)= 2(1- (-u/2)) and $\frac{1}{1- z}= \sum z^n$ (sum of a geometric series). So $\frac{1}{u+ 2}= \frac{1}{2}\frac{1}{1- (-u/2)}= \frac{1}{2}\sum (-u/2)^n$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
That would be one method. Perhaps simpler is to use the fact that u+2= 2((u/2)+ 1)= 2(1- (-u/2)) and $\frac{1}{1- z}= \sum z^n$ (sum of a geometric series). So $\frac{1}{u+ 2}= \frac{1}{2}\frac{1}{1- (-u/2)}= \frac{1}{2}\sum (-u/2)^n$
So the $R = 2$ then correct?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
So the $R = 2$ then correct?
First series expansion

$f(z)=2+\dfrac{1}{u+2}=2+\dfrac{1}{2}\dfrac{1}{1+u/2}=2+\dfrac{1}{2}\displaystyle\sum_{n=0}^{\infty}(-u/2)^n=2+\dfrac{1}{2}\displaystyle\sum_{n=0}^{\infty}(-(z-1)/2)^n$ ( if 0<|z-1|<2 )

Second series expansion

$f(z)=2+\dfrac{1}{u+2}=2+\dfrac{1}{u}\dfrac{1}{1+2/u}=2+\dfrac{1}{u}\displaystyle\sum_{n=0}^{\infty}(-2/u)^n=2+\displaystyle\sum_{n=0}^{\infty}(-2)^n/(z-1)^{n+1}$ ( if 2<|z-1|<$+\infty$ )
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$\displaystyle\frac{1}{2}\sum_n^{\infty}(2z + 3)\left(-\frac{u}{2}\right)^n$

By ratio test,

$\displaystyle \frac{1}{2} \lim_{n\to\infty}\left|\frac{(2z+3)\left(-\frac{u}{2}\right)^{n+1}}{(2z+3)\left(-\frac{u}{2}\right)^n}\right|$
$\displaystyle =\frac{1}{2}\lim_{n\to\infty}\left|\left(-\frac{u}{2}\right)\right|=\frac{1}{4}\left|u\right|<1\Rightarrow\left| u\right| < 4$

Is this not correct?
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
No, that is not correct. You should not have that "1/2" in front of the limit.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
No, that is not correct. You should not have that "1/2" in front of the limit.
$\displaystyle \lim_{n\to\infty}\left|\frac{(2z+3)\left(-\frac{u}{2}\right)^{n+1}}{(2z+3)\left(-\frac{u}{2}\right)^n}\right|$
$\displaystyle =\lim_{n\to\infty}\left|\left(-\frac{u}{2}\right)\right|=\frac{1}{2}\left|u\right|<1\Rightarrow\left| u\right| < 2$

Is this correct then?

Also, if the 1/2 was part of the factorization, why was it neglected?
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
I think I have it.

$u = z - 1$

$|z - 1|<2\Rightarrow |x - 1 + yi|<2\Rightarrow \sqrt{(x-1)^2+y^2}<2$

$\Rightarrow (x-1)^2+y^2<4\Rightarrow -1<x<3$ and $-2<y<2$.

Now is this correct?

Assuming this is correct, what would I say the Radius of convergence is?
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
The radius of convergence is 2 right?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661