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gerryliyana
 2 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.,
gerryliyana
 2 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.,

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ParthKohli
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0ok.., no problem parthkohli...,

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

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

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

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

niksva
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0ah yea..., it's ok now.,
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