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Math Help - how to calculate residium at infinity?

  1. #1
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    how to calculate residium at infinity?

    i have such a function
    <br />
z^3 \sin \frac{1}{3}<br />

    i need to calculate its residium at z=infinity

    if i substitue infinity instead of a"" into the formal formula
    res(f(x),a)=\lim_{x->a}(f(x)(x-a))

    i get infinity
    am i correct?
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    i have such a function
    <br />
z^3 \sin \frac{1}{3}<br />

    i need to calculate its residium at z=infinity

    if i substitue infinity instead of a"" into the formal formula
    res(f(x),a)=\lim_{x->a}(f(x)(x-a))

    i get infinity
    am i correct?
    The best way to find the residue at infinity is to replace "z" by "1/z" and find the residue at z=0.

    Here, " sin(\frac{1}{3})" is a constant so replacing z by 1/z gives \frac{sin(\frac{1}{3})}{z^3} which has a pole of order 3 at z=0. That has residue 0 at z= 0 so the residue of sin(\frac{1}{3})z^3 at infinity is also 0.
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  3. #3
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    i am sorry i wrote the function in a wrong way
    its
    <br />
z^3 \sin \frac{1}{z}<br />
    so by folowing your method
    <br />
\frac{1}{z^3} \sin z<br />
    so the third derivative of sine is -cos z
    and divide by 3!
    and put 0 instead of z
    and in the end i get
    -1/6
    correct?
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    i am sorry i wrote the function in a wrong way
    its
    <br />
z^3 \sin \frac{1}{z}<br />
    so by folowing your method
    <br />
\frac{1}{z^3} \sin z<br />
    so the third derivative of sine is -cos z
    and divide by 3!
    and put 0 instead of z
    and in the end i get
    -1/6
    correct?
    Note that \frac{\sin z}{z^3} has a double pole at z = 0.

    But in fact it's trivial to get the residue just by writing out the first few terms of the Laurent series around z = 0. Then it's crystal clear that the residue is .....
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