## 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

2. anonymous

Gimme a second, I'm not that great at using the equation pad.

3. anonymous

The Taylor summation of e^x is as follows $e ^{x} = \Sigma \frac{ x^n }{ n! }$ So then you let x = 2z to get $\sum_{0}^{\infty}\frac{ (2z)^n }{ n! }$ Which can be split up as follows $\sum_{0}^{\infty}\frac{ 2^n*z^n }{ n! }$ And then we multiply this by z to get $\sum_{0}^{\infty}\frac{2^n*z ^{n+1} }{ n! }$ and then the last step is to put the summation around z = -1, which makes the summation look as follows $\sum_{0}^{\infty}\frac{2^n*(z + 1) ^{n+1} }{ n! }$

4. anonymous

To find the radius of convergence you use the equation $\left| \frac{ f(x)_{n+1} }{ f(x)_{n} } \right| < 1$ where f(x) is everything inside the summation from the previous part

5. anonymous

ah ok.., and then?

6. anonymous

so you find what the absolute value of x is and that's your radius of convergence

7. anonymous

is it |z-1| < 0 ???

8. anonymous

sorry, it's R = f(x) of n / f(x) of n+1

9. anonymous

i dont understand first one, why u multiplied by z ??

10. anonymous

@gerryliyana we multiply through by $$z$$ because our function is $$z\exp(2z)$$ not just $$\exp(2z)$$ :-)

11. anonymous

ah yea

12. anonymous

We started with something we knew was true by definition, that $$\exp(z)$$ has the following (Taylor) power series:$$\exp(z)=\sum\frac1{n!}z^n$$... just like its real analog. Now, since our function has $$\exp(2z)$$, we merely substitute for $$z$$ as follows.$$\exp(2z)=\sum\frac1{n!}(2z)^n=\sum\frac{2^n}{n!}z^n$$We're almost there. Recall that since our desired function is $$z\exp(2z)$$, we merely need to multiply both sides by $$z$$ (wow... power series are neat :-)$$z\exp(2z)=\sum \frac{2^n}{n!}z^{n+1}$$

13. anonymous

Now, for determining the radius of convergence for our series, why not try something like the ratio test?$$C=\lim_{n\to\infty}\left|\frac{2^{n+1}/(n+1)!\,z^{n+2}}{2^n/n!\,z^{n+1}}\right|=\lim_{n\to\infty}\left|\frac{2z}{n+1}\right|=\lim_{n\to\infty}\frac2{n+1}|z|=0<1$$... thus we converge for any $$z$$ and our radius of convergence is $$\infty$$.

14. anonymous

is it region of convergence??

15. anonymous

@oldrin.bataku

16. anonymous

@gerryliyana the region of convergence is all of $$\mathbb{C}$$.

17. anonymous

@oldrin.bataku how about z = -1?? the Taylor series expansion of a function of f(z)=z exp(2z) about z = -1, is that already fulfilled??

18. anonymous

let me try., $\lim_{n \rightarrow \infty} \left| \frac{ (2^{n+1}/(n+1)!)(z+1)^{n+2} }{ (2^{n}/(n)!)(z+1)^{n+1} } \right|=\lim_{n \rightarrow \infty} \left| \frac{ 2(z+1) }{ n+1 } \right|=\lim_{n \rightarrow \infty} \left| \frac{ 2 }{ n+1 } \right| \left| z+1 \right| = 0 < 1$ then the region convergence is |z+1|<1

19. anonymous

@gerryliyana it is not $$(z+1)^{n+2}$$ but merely $$z^{n+2}$$, and you'll notice that $$\frac2{n+1}\to0$$ regardless of $$z$$.

20. anonymous

@oldrin.bataku then the region of convergence is $$|z+1| < \infty$$ ???

21. anonymous

@gerryliyana oops I forgot to make it centered at $$z=-1$$ :o