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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
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

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

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

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

CanadianAsianBest ResponseYou've already chosen the best response.3
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
 one year ago

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

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

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

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

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

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

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

oldrin.batakuBest ResponseYou've already chosen the best response.0
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 \(\infty\).
 one year ago

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

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

gerryliyanaBest 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??
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
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
 one year ago

oldrin.batakuBest 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 \(z\).
 one year ago

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

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