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.,

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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.,

Statistics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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. :-|
ok.., no problem parthkohli...,

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

The expression of a Taylor Series of f(z) around point "a" is:\[f(z)=f(a)+\sum_{n=1}^{\infty}f^{(n}(a)(z-a)/n!\] where\[f^{(n}(a)\]stands for the value of the n-th 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:\[f^{(n}(z)=2^{(n-1)}e^{2z}(2z+n)\rightarrow f^{(n}(-1)=2^{(n-1)}e^{-2}(n-2)\] If you replace this in Taylor´s expression, you get:\[f(z)=-e^{-2}+e^{-2}\sum_{n=1}^{\infty}2^{(n-1)}(n-2)(z+1)^n/n!\]
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
@CarlosGP how about the region of convergence ??
in order to find the region of convergence we need to apply ration test as suggested by @CarlosGP
*ratio
i thought the region convergence is \[|z+1| < \infty \] hbu ??
how do u come about this region of convergence?
i have tried to apply alembert ratio test look at the steps \[a _{n}= 2^{n-1}\frac{ (n-2) (z+1)^{n} }{ n! }\] \[a _{n+1} = 2^{n}\frac{ (n-1) (z+1)^{n+1} }{ (n+1)! }\] now \[\lim_{n \rightarrow infinity} \frac{ a _{n+1} }{ a _{n} } = \lim_{n \rightarrow infinity} \frac{ 2(n-1)(z+1) }{ (n+1)(n-2) }= \frac{ 1 }{ R }\] where R is region of convergence
@gerryliyana is this ok?
and R = infnty
yeah it will come out to be infinity after applying the limits hope u understand the region of convergence
ah yea..., it's ok now.,

Not the answer you are looking for?

Search for more explanations.

Ask your own question