anyone can? Find the Taylor series expansion of a function of f(z)=z exp(2z) about z = -1 and find the region of convergence.,

- anonymous

- schrodinger

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- anonymous

you help me? @Jemurray3

- anonymous

Let u = x+1 -> x = u-1
Substitute that in first, and what do you get?

- anonymous

why u = x+1 ??

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## More answers

- anonymous

Because then you expand what you have left around u = 0

- anonymous

substitute to z?

- anonymous

What?

- anonymous

Oh, yeah. Z instead of x, whatever.
u = z+1

- anonymous

ah ok..., then f(u-1)=(u-1) exp(2(u-1)) ?

- anonymous

Yep. Now just expand that in a power series around u = 0. I'll get you started.
\[f(u) = e^{-2}\cdot \left( (u-1) \cdot e^{2u}\right) \]
\[ =e^{-2} \cdot(u-1) \cdot \sum \frac{(2u)^n}{n!} \]

- anonymous

i dont know how to expand

- anonymous

Why are you having him make the substitution? That makes the problem no easier and weakens the intuition of what is happening.

- anonymous

have you idea @dfyodor ??

- anonymous

Well, we might as well take it from jemurray3's last equation. What is thee next part of the problem?

- anonymous

the next is find the region of convergence...,

- anonymous

Right, what does that mean?

- anonymous

As in, the definition with equations.

- anonymous

For a power series \[\sum c_n z^n\] converges if and only if \[|z|

- anonymous

We know the root test, that is, that \[\sum c_n\] converges if and only if \[\sqrt{c_n}<1\] as n goes to infinity. We also know (presumably) that \[\frac{a^n}{n!}\] converges to zero for any a greater than 0.

- anonymous

I don't feel comfortable giving much more at this point, as I think that should be plenty enough to work out the solution to the problem.

- anonymous

then \the region of convergence |z-1| < 1 ??

- anonymous

@dfyodor I made the substitution because I'd rather that than carry around a whole bunch of (z+1)'s instead. It doesn't weaken anything. It certainly is algebraically easier to expand around zero.

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