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stgreen

  • 3 years ago

Show that the singular point of the following function is a pole. Determine the order m of that pole and the corresponding residue B. f(z)=(1-coshz)/(z^3)

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  1. malevolence19
    • 3 years ago
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    I really wish my university had an applied complex variables class :/

  2. anonymous
    • 3 years ago
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    i think we can at least start this if i recall \(\cosh(z)=\cos(iz)\) if so, we can expand in a taylor series

  3. anonymous
    • 3 years ago
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    we can try \[\cosh(z)=\sum_{k=0}^{\infty}\frac{z^{2k}}{(2k)!}\]

  4. anonymous
    • 3 years ago
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    this should give \[1-\cosh(z)=-\sum_{k=1}^{\infty}\frac{z^{2k}}{(2k)!}\]

  5. stgreen
    • 3 years ago
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    yeah got it

  6. malevolence19
    • 3 years ago
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    Which would then give: \[-\sum_{k=1}^{\infty}\frac{z^{2k-3}}{(2k)!}\] right?

  7. anonymous
    • 3 years ago
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    divide by \(z^3\) first term is \(\frac{1}{2z}\) and all the others have \(z\) in the numerator

  8. anonymous
    • 3 years ago
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    @malevolence19 yes, that is what i get too. or shift the index either way.

  9. malevolence19
    • 3 years ago
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    So the the residue is just 1/2?

  10. stgreen
    • 3 years ago
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    so b1 comes out to be -1/2.yeah?

  11. anonymous
    • 3 years ago
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    yes if my memory serves me in this context the residue is the coefficient of the \(z^{-1}\) term

  12. stgreen
    • 3 years ago
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    ^right

  13. malevolence19
    • 3 years ago
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    Yeah, -1/2 i forgot about that in front.

  14. anonymous
    • 3 years ago
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    i would not place much money on this, because it has been a while, but i am pretty sure it is correct

  15. stgreen
    • 3 years ago
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    and pole is??i don't have any idea what the heck is it

  16. malevolence19
    • 3 years ago
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    It would just be zero right?

  17. malevolence19
    • 3 years ago
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    It it was like: \[\frac{B}{(z-\lambda)}\] then the pole would be lambda? I think, but I'm going off very little knowledge.

  18. stgreen
    • 3 years ago
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    lambda is a singular point..not sure about pole

  19. anonymous
    • 3 years ago
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    yes, the pole is at 0, because that is where the denominator is undefined. it is a simple pole because it is a zero of order one

  20. stgreen
    • 3 years ago
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    ^the denominator becomes undefined at singular points...so every singular point is a pole??

  21. stgreen
    • 3 years ago
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    i think its pole is 1

  22. anonymous
    • 3 years ago
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    a "pole" is a kind of singular point think of a simple example from pre calc class \[\frac{x^2-4}{x-2}\] vs \[\frac{x^2}{x-2}\] in the first example, 2

  23. stgreen
    • 3 years ago
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    this thing implies m=1 at b1

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  24. anonymous
    • 3 years ago
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    in both examples, 2 is a singular point because the function is not defined at 2 but in the first example, 2 is a removable singularity, because you can remove it

  25. anonymous
    • 3 years ago
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    in the second example, 2 is a pole, in pre calc a "vertical asypmtote"

  26. anonymous
    • 3 years ago
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    however it is spelled

  27. stgreen
    • 3 years ago
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    isolated singular point.what is it??

  28. anonymous
    • 3 years ago
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    ok hold the phone

  29. anonymous
    • 3 years ago
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    what you posted was about the order of the pole, not the pole itself

  30. anonymous
    • 3 years ago
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    in your case above, we had \(\frac{1}{2z}\) plus stuff with positive exponents, means you have a pole of ORDER 1, not a pole at \(z=1\)

  31. anonymous
    • 3 years ago
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    here is a nice succinct explanation (very short) with example http://www.solitaryroad.com/c612.html

  32. stgreen
    • 3 years ago
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    implies order of pole=k if bk is the co-efficient of z^-1 term.right?

  33. anonymous
    • 3 years ago
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    the order of the pole is the multplicty of the zero on the denominator, to put it in pre calculus terms, example \[\frac{1}{(z-5)^3}\] has a pole of order 3 at \(z=5\)

  34. anonymous
    • 3 years ago
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    \[\frac{1}{z-2}\] has a pole of order 1 at \(z=2\) a pole of order one is called a simple pole

  35. stgreen
    • 3 years ago
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    oh *shouts got it*

  36. anonymous
    • 3 years ago
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    good!

  37. stgreen
    • 3 years ago
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    thanks bro

  38. anonymous
    • 3 years ago
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    yw

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