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gerryliyana
 one year 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.,
gerryliyana
 one year 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.,

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gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0have idea @CanadianAsian ???

CanadianAsian
 one year ago
Best ResponseYou've already chosen the best response.3Gimme a second, I'm not that great at using the equation pad.

CanadianAsian
 one year ago
Best ResponseYou've already chosen the best response.3The 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 \[\sum_{0}^{\infty}\frac{2^n*(z + 1) ^{n+1} }{ n! }\]

CanadianAsian
 one year ago
Best ResponseYou've already chosen the best response.3To 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

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0ah ok.., and then?

CanadianAsian
 one year ago
Best ResponseYou've already chosen the best response.3so you find what the absolute value of x is and that's your radius of convergence

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0is it z1 < 0 ???

CanadianAsian
 one year ago
Best ResponseYou've already chosen the best response.3sorry, it's R = f(x) of n / f(x) of n+1

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0i dont understand first one, why u multiplied by z ??

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0@gerryliyana we multiply through by \(z\) because our function is \(z\exp(2z)\) not just \(\exp(2z)\) :)

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0We 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 :)$$z\exp(2z)=\sum \frac{2^n}{n!}z^{n+1}$$

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0Now, 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 \(\infty\).

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0is it region of convergence??

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0@gerryliyana the region of convergence is all of \(\mathbb{C}\).

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0@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??

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0let 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

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0@gerryliyana it is not \((z+1)^{n+2}\) but merely \(z^{n+2}\), and you'll notice that \(\frac2{n+1}\to0\) regardless of \(z\).

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku then the region of convergence is \(z+1 < \infty \) ???

oldrin.bataku
 one year ago
Best ResponseYou've already chosen the best response.0@gerryliyana oops I forgot to make it centered at \(z=1\) :o
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