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According to Fractional Calculus, the power rule can be written as

(d

For example,

(d

To find the residue of f(z) = f(z)/(z-z

For example, when I use the formula for the residue of 1/(z-1)

1/(1/2-1)! d

= 1/(-1/2)! (2/√π) z

Actually, the formula for any power residue can also be computed.

Res(z→z

= 1/(m/2-1)! (d

= 1/[(m/2-1)!(1-m/2)!] z

Using this formula, I can construct a table:

m=1, Res = 2/π √z

m=3, Res = 2/π 1/√z

m=5, Res = -2/(3π) 1/z

m=7, Res = 2/(5π) 1/z

m=9, Res = -2/(7π) 1/z

m=11, Res = 2/(9π) 1/z

.

.

.

m, Res = (-1)

http://www.wolframalpha.com/input/?i=1/[(m/2-1)!+(1-m/2)!]+z^(1-m/2),+m=[1,15]

It's interesting to note that for m=2, Res=1, and every other even integer above that gives a 0 value. This is consistent with the correct results.

What is the meaning of such a residue? Every source I can find on the internet says there shouldn't be a residue, and Wolfram says the residue is 0.

(d

^{m}/dz^{m}) z^{n}= n!/(n-m)! z^{n-m}For example,

(d

^{1/2}/dz^{1/2}) z^{1/2}= (1/2)!/(1/2-1/2)! z^{0}= (1/2)√πTo find the residue of f(z) = f(z)/(z-z

_{0})^{m}at z→z_{0}, the formula is Res(z→z_{0}) f(z) = 1/(m-1)! d^{m-1}/dz^{m-1}f(z).For example, when I use the formula for the residue of 1/(z-1)

^{1/2}at z→1, I get the following:1/(1/2-1)! d

^{1/2-1}/dz^{1/2-1}1= 1/(-1/2)! (2/√π) z

^{1/2}= 2/πActually, the formula for any power residue can also be computed.

Res(z→z

_{0}) f(z)= 1/(m/2-1)! (d

^{m/2-1}/dz^{m/2-1}) z^{0}= (1/(m/2-1)!)(0!/(1-m/2)!) z^{1-m/2}= 1/[(m/2-1)!(1-m/2)!] z

^{1-m/2}, where m is an odd integer.Using this formula, I can construct a table:

m=1, Res = 2/π √z

_{0}m=3, Res = 2/π 1/√z

_{0}m=5, Res = -2/(3π) 1/z

_{0}^{3/2}m=7, Res = 2/(5π) 1/z

_{0}^{5/2}m=9, Res = -2/(7π) 1/z

_{0}^{7/2}m=11, Res = 2/(9π) 1/z

_{0}^{9/2}.

.

.

m, Res = (-1)

^{(m+1)/2}2/[(m-2)π] 1/z^{m/2-1}http://www.wolframalpha.com/input/?i=1/[(m/2-1)!+(1-m/2)!]+z^(1-m/2),+m=[1,15]

It's interesting to note that for m=2, Res=1, and every other even integer above that gives a 0 value. This is consistent with the correct results.

What is the meaning of such a residue? Every source I can find on the internet says there shouldn't be a residue, and Wolfram says the residue is 0.

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