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gerryliyana

  • 3 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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  1. gerryliyana
    • 3 years ago
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    you help me? @Jemurray3

  2. Jemurray3
    • 3 years ago
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    Let u = x+1 -> x = u-1 Substitute that in first, and what do you get?

  3. gerryliyana
    • 3 years ago
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    why u = x+1 ??

  4. Jemurray3
    • 3 years ago
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    Because then you expand what you have left around u = 0

  5. gerryliyana
    • 3 years ago
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    substitute to z?

  6. Jemurray3
    • 3 years ago
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    What?

  7. Jemurray3
    • 3 years ago
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    Oh, yeah. Z instead of x, whatever. u = z+1

  8. gerryliyana
    • 3 years ago
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    ah ok..., then f(u-1)=(u-1) exp(2(u-1)) ?

  9. Jemurray3
    • 3 years ago
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    Yep. Now just expand that in a power series around u = 0. I'll get you started. \[f(u) = e^{-2}\cdot \left( (u-1) \cdot e^{2u}\right) \] \[ =e^{-2} \cdot(u-1) \cdot \sum \frac{(2u)^n}{n!} \]

  10. gerryliyana
    • 3 years ago
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    i dont know how to expand

  11. dfyodor
    • 3 years ago
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    Why are you having him make the substitution? That makes the problem no easier and weakens the intuition of what is happening.

  12. gerryliyana
    • 3 years ago
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    have you idea @dfyodor ??

  13. dfyodor
    • 3 years ago
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    Well, we might as well take it from jemurray3's last equation. What is thee next part of the problem?

  14. gerryliyana
    • 3 years ago
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    the next is find the region of convergence...,

  15. dfyodor
    • 3 years ago
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    Right, what does that mean?

  16. dfyodor
    • 3 years ago
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    As in, the definition with equations.

  17. dfyodor
    • 3 years ago
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    For a power series \[\sum c_n z^n\] converges if and only if \[|z|<R\], and R is called the radius of convergeance.

  18. dfyodor
    • 3 years ago
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    We know the root test, that is, that \[\sum c_n\] converges if and only if \[\sqrt{c_n}<1\] as n goes to infinity. We also know (presumably) that \[\frac{a^n}{n!}\] converges to zero for any a greater than 0.

  19. dfyodor
    • 3 years ago
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    I don't feel comfortable giving much more at this point, as I think that should be plenty enough to work out the solution to the problem.

  20. gerryliyana
    • 3 years ago
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    then \the region of convergence |z-1| < 1 ??

  21. Jemurray3
    • 3 years ago
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    @dfyodor I made the substitution because I'd rather that than carry around a whole bunch of (z+1)'s instead. It doesn't weaken anything. It certainly is algebraically easier to expand around zero.

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