## 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

1. gerryliyana Group Title

you help me? @Jemurray3

2. Jemurray3 Group Title

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

3. gerryliyana Group Title

why u = x+1 ??

4. Jemurray3 Group Title

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

5. gerryliyana Group Title

substitute to z?

6. Jemurray3 Group Title

What?

7. Jemurray3 Group Title

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

8. gerryliyana Group Title

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

9. Jemurray3 Group Title

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 Group Title

i dont know how to expand

11. dfyodor Group Title

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

12. gerryliyana Group Title

have you idea @dfyodor ??

13. dfyodor Group Title

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

14. gerryliyana Group Title

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

15. dfyodor Group Title

Right, what does that mean?

16. dfyodor Group Title

As in, the definition with equations.

17. dfyodor Group Title

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 Group Title

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 Group Title

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 Group Title

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

21. Jemurray3 Group Title

@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.