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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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ParthKohliBest ResponseYou've already chosen the best response.1
I 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. :
 one year ago

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

CarlosGPBest ResponseYou've already chosen the best response.2
The 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:\[f(z)=e^{2}+e^{2}\sum_{n=1}^{\infty}2^{(n1)}(n2)(z+1)^n/n!\]
 one year ago

CarlosGPBest ResponseYou've already chosen the best response.2
In 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
 one year ago

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

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

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

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

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

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

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

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