Question

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.,

Answer 1

you help me? @Jemurray3

Answer 2

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

Answer 3

why u = x+1 ??

Answer 4

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

Answer 5

substitute to z?

Answer 6

What?

Answer 7

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

Answer 8

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

Answer 9

Yep. Now just expand that in a power series around u = 0. I'll get you started.

\[f(u) = e^{-2}\cdot \left( (u-1) \cdot e^{2u}\right) \]
\[ =e^{-2} \cdot(u-1) \cdot \sum \frac{(2u)^n}{n!} \]

Answer 10

i dont know how to expand

Answer 11

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

Answer 12

have you idea @dfyodor ??

Answer 13

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

Answer 14

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

Answer 15

Right, what does that mean?

Answer 16

As in, the definition with equations.

Answer 17

For a power series \[\sum c_n z^n\] converges if and only if \[|z|<R\], and R is called the radius of convergeance.

Answer 18

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.

Answer 19

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.

Answer 20

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

Answer 21

@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.