## 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.,

1. gerryliyana

you help me? @Jemurray3

2. Jemurray3

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

3. gerryliyana

why u = x+1 ??

4. Jemurray3

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

5. gerryliyana

substitute to z?

6. Jemurray3

What?

7. Jemurray3

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

8. gerryliyana

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

9. Jemurray3

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. gerryliyana

i dont know how to expand

11. dfyodor

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

12. gerryliyana

have you idea @dfyodor ??

13. dfyodor

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

14. gerryliyana

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

15. dfyodor

Right, what does that mean?

16. dfyodor

As in, the definition with equations.

17. dfyodor

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. dfyodor

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. dfyodor

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. gerryliyana

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

21. Jemurray3

@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.