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gerryliyana
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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.,
 one year ago
 one year ago
gerryliyana Group Title
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.,
 one year ago
 one year ago

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gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
you help me? @Jemurray3
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Let u = x+1 > x = u1 Substitute that in first, and what do you get?
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
why u = x+1 ??
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Because then you expand what you have left around u = 0
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
substitute to z?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Oh, yeah. Z instead of x, whatever. u = z+1
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
ah ok..., then f(u1)=(u1) exp(2(u1)) ?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Yep. Now just expand that in a power series around u = 0. I'll get you started. \[f(u) = e^{2}\cdot \left( (u1) \cdot e^{2u}\right) \] \[ =e^{2} \cdot(u1) \cdot \sum \frac{(2u)^n}{n!} \]
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
i dont know how to expand
 one year ago

dfyodor Group TitleBest ResponseYou've already chosen the best response.0
Why are you having him make the substitution? That makes the problem no easier and weakens the intuition of what is happening.
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
have you idea @dfyodor ??
 one year ago

dfyodor Group TitleBest ResponseYou've already chosen the best response.0
Well, we might as well take it from jemurray3's last equation. What is thee next part of the problem?
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
the next is find the region of convergence...,
 one year ago

dfyodor Group TitleBest ResponseYou've already chosen the best response.0
Right, what does that mean?
 one year ago

dfyodor Group TitleBest ResponseYou've already chosen the best response.0
As in, the definition with equations.
 one year ago

dfyodor Group TitleBest ResponseYou've already chosen the best response.0
For a power series \[\sum c_n z^n\] converges if and only if \[z<R\], and R is called the radius of convergeance.
 one year ago

dfyodor Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

dfyodor Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.0
then \the region of convergence z1 < 1 ??
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago
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