Interesting value for W, h involving phi,Omega?

I tried to find h(z) = h((I/2)^(2/I)) = h( exp(pi)/exp(-I*2*ln2)

ln ( exp*pi/exp(-I*2*ln2)) = pi+ I*2*ln2

-ln(z) = - pi- I*2*ln2

so as W(-pi-I*2*ln2) = ln2 - I*(pi/2)

Than h(z) = (ln2-I*(pi/2))/ (-pi - i*2*ln2)

This can be braught to :

h(z) = 0.5*I ( i guess with plus sign, not sure)

now, since I/2 = sin (I*ln(phi)) than

h((i/2)^(2/i)) = sin(I*ln(phi))

so Arcsin(h(I/2^(2/I)) = I*ln(phi)

phi= -I*e^Arcsin(h(I/2)^(2/I))

so taking a self root of I/2 can be related to phi in this way. Without an additional partitioning, 2 in any side of the formula involving h(i^(1/i)), it is not possible to get from I to phi.

I think it is possible to continue with I/3, I/4 etc in the same way to find interesting things.

is this well known, if there is no mistake?
I tried to find h(z) = h((I/3)^(3/I)) = h( exp(3pi/2)/exp(-I*3*ln3))

ln ( exp(3pi/2)/exp(-I*2*ln2)) = 3pi/2+ I*3*ln3

-ln(z) = - 3pi/2- I*3*ln3

so as W(-3pi/2-I*3*ln3) = ln3 - I*(pi/2)

Than h(z) = (-ln3+I*(pi/2))/ (-3pi/2 - i*3*ln3)

This can be brought to :

h(z) = I/3 ( i guess with plus sign, not sure)

For h(i/4)^(4/i)) we have :

W(-2pi-I*4*ln4) = ln4 -I*(pi/2)

So most likely:

W(-n*pi/2 -I*n*ln n) = ln n -n* I*(pi/2) n>1
Ivars Wrote:W(-pi-I*2*ln2) = ln2 - I*(pi/2)

True (numerically verified).

Ivars Wrote:W(-3pi/2-I*3*ln3) = ln3 - I*(pi/2)

True (numerically verified).

Ivars Wrote:W(-2pi-I*4*ln4) = ln4 + I*(pi/2)
True (numerically verified).

Ivars Wrote:H((I/2)^(I/2)^(-1)) = I/2
True (numerically verified), plus this is practically the definition of H.

Ivars Wrote:H((I/3)^(I/3)^(-1)) = I/3

(not numerically verified), this is practically the definition of H, so it should be true, but for some reason my CAS gives a different value. These are the values I'm getting:
H((i/3)^{1/(i/3)}) \approx -0.008442966 & +\ 0.343894471 i \\
i/3 = 0 & +\ 0.333333333 i
although, this is bothering me, since it is the definition of H that this should be true. I'm guessing that that large value \( |(i/3)^{(3/i)}|\approx 111.318 \) might be why. Maybe my CAS never tested the LambertW for this large a value, and maybe it really does give the wrong value...

About your claim for all n, since you have found 3 instances of it being true, it is probably true, but that does not constitute a proof. Since your logic doesn't require n to be constant, you could probably re-work the math to start with n rather than 2 or 3. Then it would be a proof. Smile

Andrew Robbins
This is the result of my computation:

\( z=\left(\frac{i}{n}\right)^{\frac{n}{i}}=\left(\frac{i}{n}\right)^{-ni}=\frac{e^{i\frac{\pi}{2}(-ni)}}{n^{-ni}}=e^{\frac{\pi}{2}n} n^{ni}=e^{\frac{\pi}{2}n} e^{n\ln(n)i}=e^{\frac{\pi}{2}n+n\ln(n)i} \)
\( y:=-\ln(z)=-\frac{\pi}{2}n-n\ln(n)i \)

If we now set \( x:=\ln(n)-\frac{\pi}{2}i \) then
\( xe^x=(\ln(n)-\frac{\pi}{2}i)e^{\ln(n)-\frac{\pi}{2}i}=(\ln(n)-\frac{\pi}{2}i)n(-i)=-\ln(n)ni-\frac{\pi}{2}n=y \)

And thatswhy \( W(y)=x \) and \( h(z)=\frac{W(y)}{y}=\frac{x}{y} \).
We see above that \( y=xn(-i) \) and hence \( h(z)=\frac{i}{n} \).

So except some minor differences (and not considering branches) this confirms your conjecture.
Well done Ivars!

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.


h((I/x)^(x/I))= (I/x)
If we use substitution


so that x= +- sqrt(y) we can get complex conjugate values if we use both roots separately:

W(-(+sqrt(y)*(pi/2) - I*(+sgrt(y)*ln(+sqrt(y)) = ln(+sqrt(y)) - i*pi/2 for y>1 but

W(-(+sqrt(y)*(pi/2) - I*(-sgrt(y)*ln(+sqrt(y)) = ln(+sqrt(y)) + i*pi/2 for y>1

than the change of sign and existance of 2 values -i,+i in h(i^-i) does not cause any questions-we just use both square roots of 1, however, not symmetrically...

Alternatively, we can decide to use only one value, but I^3 instead of I*(-1).

W(-(+sqrt(y)*(pi/2) - I^3*(+sgrt(y)*ln(+sqrt(y)) = ln(+sqrt(y)) + i*pi/2 for y>1

So we can get both branches of W by switching from i to - i in argument of W instead of taking negative square roots. From the point of view of the result of W, these are equivalent approaches.

By involving other root of y we will get h((I/x)^(x/I))= (1/I*x)

However it seemed as if couldnt see the wood because all those trees:
Of course \( h(x) \) is the inverse function of \( x^{1/x} \).
And thatswhy it is clear that \( h(x^{1/x})=x \),
especially for \( x=I/n \).
bo198214 Wrote:However it seemed as if couldnt see the wood because all those trees:
Of course \( h(x) \) is the inverse function of \( x^{1/x} \).
And thatswhy it is clear that \( h(x^{1/x})=x \),
especially for \( x=I/n \).

Also for x= I/y, y -real>1?

Ivars Wrote:Also for x= I/y, y -real>1?

I would say so.
Good. One score. Of course, it is a tetration definition, nice that it works also for I/x , but values for Lambert function of such arguments like are not so obvious (were not initially).

W(-x*(pi/2) - I*x*ln(x) ) = ln(x) - I*(pi/2)=ln(x)-lnI=ln(x/I) for x>1

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