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

have idea @CanadianAsian ???

Answer 2

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

Answer 3

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

\[\sum_{0}^{\infty}\frac{2^n*(z + 1) ^{n+1} }{ n! }\]

Answer 4

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

Answer 5

ah ok.., and then?

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

ah yea

Answer 12

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 :-)$$z\exp(2z)=\sum \frac{2^n}{n!}z^{n+1}$$

Answer 13

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 \(\infty\).

Answer 14

is it region of convergence??

Answer 15

@oldrin.bataku

Answer 16

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

Answer 17

@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??

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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