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gerryliyana
 2 years ago
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.,
gerryliyana
 2 years ago
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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gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0you help me? @Jemurray3

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

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1Because then you expand what you have left around u = 0

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.1Oh, yeah. Z instead of x, whatever. u = z+1

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

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

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0i dont know how to expand

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

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0have you idea @dfyodor ??

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

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0the next is find the region of convergence...,

dfyodor
 2 years ago
Best ResponseYou've already chosen the best response.0Right, what does that mean?

dfyodor
 2 years ago
Best ResponseYou've already chosen the best response.0As in, the definition with equations.

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

dfyodor
 2 years ago
Best ResponseYou've already chosen the best response.0We 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.

dfyodor
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0then \the region of convergence z1 < 1 ??

Jemurray3
 2 years ago
Best 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.
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