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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.,

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you help me? @Jemurray3
Let u = x+1 -> x = u-1 Substitute that in first, and what do you get?
why u = x+1 ??

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Because then you expand what you have left around u = 0
substitute to z?
What?
Oh, yeah. Z instead of x, whatever. u = z+1
ah ok..., then f(u-1)=(u-1) exp(2(u-1)) ?
Yep. Now just expand that in a power series around u = 0. I'll get you started. \[f(u) = e^{-2}\cdot \left( (u-1) \cdot e^{2u}\right) \] \[ =e^{-2} \cdot(u-1) \cdot \sum \frac{(2u)^n}{n!} \]
i dont know how to expand
Why are you having him make the substitution? That makes the problem no easier and weakens the intuition of what is happening.
have you idea @dfyodor ??
Well, we might as well take it from jemurray3's last equation. What is thee next part of the problem?
the next is find the region of convergence...,
Right, what does that mean?
As in, the definition with equations.
For a power series \[\sum c_n z^n\] converges if and only if \[|z|
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.
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.
then \the region of convergence |z-1| < 1 ??
@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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