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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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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1I wish I had known how to do this. It wouldn't take anything more than the definition of the Taylor Series. I mightn't know how to apply it here though. :

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0ok.., no problem parthkohli...,

CarlosGP
 one year ago
Best ResponseYou've already chosen the best response.2The expression of a Taylor Series of f(z) around point "a" is:\[f(z)=f(a)+\sum_{n=1}^{\infty}f^{(n}(a)(za)/n!\] where\[f^{(n}(a)\]stands for the value of the nth derivative of f(z) in z=a, which in our case is a=1. Thus the expression for our function would be: \[f(z)=f(1)+\sum_{n=1}^{\infty}f^{(n}(1)(z+1)^n/n!\] \[f(1)=e^{2}\] \[f^{(1}(z)=e^{2z}(2z+1)\]\[f^{(2}(z)=2e^{2z}(2z+2)\]\[f^{(3}(z)=4e^{2z}(2z+3)\]\[f^{(4}(z)=8e^{2z}(2z+4)\] and so on. If you observe this four derivatives we can obtain a general expression like this:\[f^{(n}(z)=2^{(n1)}e^{2z}(2z+n)\rightarrow f^{(n}(1)=2^{(n1)}e^{2}(n2)\] If you replace this in Taylor´s expression, you get:\[f(z)=e^{2}+e^{2}\sum_{n=1}^{\infty}2^{(n1)}(n2)(z+1)^n/n!\]

CarlosGP
 one year ago
Best ResponseYou've already chosen the best response.2In order to get the region of convergence, apply a criterium such as Alembert's and see how the series converges for any value of z, which means the region of convergence is plus/minus infinite

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0@CarlosGP how about the region of convergence ??

niksva
 one year ago
Best ResponseYou've already chosen the best response.0in order to find the region of convergence we need to apply ration test as suggested by @CarlosGP

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0i thought the region convergence is \[z+1 < \infty \] hbu ??

niksva
 one year ago
Best ResponseYou've already chosen the best response.0how do u come about this region of convergence?

niksva
 one year ago
Best ResponseYou've already chosen the best response.0i have tried to apply alembert ratio test look at the steps \[a _{n}= 2^{n1}\frac{ (n2) (z+1)^{n} }{ n! }\] \[a _{n+1} = 2^{n}\frac{ (n1) (z+1)^{n+1} }{ (n+1)! }\] now \[\lim_{n \rightarrow infinity} \frac{ a _{n+1} }{ a _{n} } = \lim_{n \rightarrow infinity} \frac{ 2(n1)(z+1) }{ (n+1)(n2) }= \frac{ 1 }{ R }\] where R is region of convergence

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

niksva
 one year ago
Best ResponseYou've already chosen the best response.0yeah it will come out to be infinity after applying the limits hope u understand the region of convergence

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0ah yea..., it's ok now.,
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