Interesting value for W, h involving phi,Omega?
#5
ThanksSmile

I had mistake in the sign of n=4 for I*pi/2 so correctly:

W(-n*(pi/2) - I*n*ln(n) ) = ln(n) - i*pi/2 for n>=2

for n=1 the sign at i(pi/2) is +..

The resulting formula for n>=2 can also be rewritten as:

W(-n*(pi/2)-I*n*ln(n))= ln(n/I)

but -n(pi/2)-I*n*ln(n) =-n*ln(I)/I-I*n*ln(n)= +I*n*(ln(I)-ln(n))

So W(I*n*ln(I/n)) = ln(n/I) and

W(ln((I/n)^(I*n)))=ln(n/I) which makes calculations for h simpler.

It seems to hold for all x>1; at x=1 sign changes for some reason.

W(ln((I/x)^(I*x)))=ln(x/I)

h((I/x)^(x/I))= (I/x)
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Messages In This Thread
RE: Interesting value for W, h involving phi? - by Ivars - 01/19/2008, 07:40 PM

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