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- Thread starter dwsmith
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- #1

- Jan 29, 2012

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Use the substitution $u=z-1$ .

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Where does the $u$ go?Use the substitution $u=z-1$ .

- Feb 1, 2012

- 57

z=u+1, now find f(z)= f(u+1)

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- Jan 29, 2012

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- #7

So the $R = 2$ then correct?

- Jan 29, 2012

- 661

So the $R = 2$ then correct?

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

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

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- #9

$\displaystyle\frac{1}{2}\sum_n^{\infty}(2z + 3)\left(-\frac{u}{2}\right)^n$

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?

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:

- Jan 29, 2012

- 1,151

No, that is not correct. You should not have that "1/2" in front of the limit.

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- #11

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

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

Is this correct then?

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

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- #12

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?

$u = z - 1$

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

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

Now is this correct?

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

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- #13

- Jan 29, 2012

- 661

Right.The radius of convergence is 2 right?