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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.,

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  1. gerryliyana
    • one year ago
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    @ParthKohli

  2. ParthKohli
    • one year ago
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    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. :-|

  3. gerryliyana
    • one year ago
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    ok.., no problem parthkohli...,

  4. CarlosGP
    • one year ago
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    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:\[f(z)=-e^{-2}+e^{-2}\sum_{n=1}^{\infty}2^{(n-1)}(n-2)(z+1)^n/n!\]

  5. CarlosGP
    • one year ago
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    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

  6. gerryliyana
    • one year ago
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    @CarlosGP how about the region of convergence ??

  7. niksva
    • one year ago
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    in order to find the region of convergence we need to apply ration test as suggested by @CarlosGP

  8. niksva
    • one year ago
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    *ratio

  9. gerryliyana
    • one year ago
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    i thought the region convergence is \[|z+1| < \infty \] hbu ??

  10. niksva
    • one year ago
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    how do u come about this region of convergence?

  11. niksva
    • one year ago
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    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

  12. niksva
    • one year ago
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    @gerryliyana is this ok?

  13. gerryliyana
    • one year ago
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    and R = infnty

  14. niksva
    • one year ago
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    yeah it will come out to be infinity after applying the limits hope u understand the region of convergence

  15. gerryliyana
    • one year ago
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    ah yea..., it's ok now.,

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