anonymous 3 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.,

1. anonymous

you help me? @Jemurray3

2. anonymous

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

3. anonymous

why u = x+1 ??

4. anonymous

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

5. anonymous

substitute to z?

6. anonymous

What?

7. anonymous

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

8. anonymous

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

9. anonymous

Yep. Now just expand that in a power series around u = 0. I'll get you started. $f(u) = e^{-2}\cdot \left( (u-1) \cdot e^{2u}\right)$ $=e^{-2} \cdot(u-1) \cdot \sum \frac{(2u)^n}{n!}$

10. anonymous

i dont know how to expand

11. anonymous

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

12. anonymous

have you idea @dfyodor ??

13. anonymous

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

14. anonymous

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

15. anonymous

Right, what does that mean?

16. anonymous

As in, the definition with equations.

17. anonymous

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

18. 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.

19. 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.

20. anonymous

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

21. 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.