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Math Help - why its not removable singular point??

  1. #1
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    why its not removable singular point??

    \frac{z-\sin z}{z^4}
    my singular point is 0
    if i will do lhopital 4 times on the lmit i will get -sin z/6 =0
    which is singular point because the result is constant

    so why its a first order pole?
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    \frac{z-\sin z}{z^4}
    my singular point is 0
    if i will do lhopital 4 times on the lmit i will get -sin z/6 =0
    which is singular point because the result is constant

    so why its a first order pole?
    Expand z-\sin (z) in its Taylor series and you'll find it has an order 3 zero at 0

    Your reasoning is wrong since applying L'hopital gives you \lim_{z\rightarrow 0} \frac{\cos (z)}{24z} =\infty
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  3. #3
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    ok now i tried to determine the power of the pole
    i did
    g(x)=1/f(x)
    where x=0
    and i got that g'(0)=0/0 which means that its not a first order pole
    actualyy i should get zero but i got 0/0
    ??
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    ok now i tried to determine the power of the pole
    i did
    g(x)=1/f(x)
    where x=0
    and i got that g'(0)=0/0 which means that its not a first order pole
    actualyy i should get zero but i got 0/0
    ??
    Post #2 has given you a big hint.

    The singularity of \displaystyle f(z) = \frac{z - \sin z}{z^4} at z = 0 is a simple pole. Do what post #2 said to see why.
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  5. #5
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    i want to solve it by the derivative way
    not by developing into a series
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    i want to solve it by the derivative way
    not by developing into a series
    Well then, where is your trouble in using l'Hopital's Rule a couple of times to find that \displaystyle {\lim_{z \to 0} \frac{z(z - \sin z)}{z^4} = {\lim_{z \to 0} \frac{z - \sin z}{z^3}} exists and is finite?
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