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

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