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+--- Thread: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) (/showthread.php?tid=54)
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/06/2007
Ivars Wrote:Im (W) = +- ipi/2
then branches k=-1 and k=+1 are accordingly
-i(pi/2 - 2pi) = - 5/2pi i ; +i(pi/2+2pi) = + 5/2 pi*i and so on for all branches.
Again, no! Try for example Maple (LambertW) or Mathematica (ProductLog), you can chose the branch there, for example
Code:
LambertW(-3,-Pi/2)=-2.198342630-13.98120831*I
LambertW(-2,-Pi/2)=-1.604290913-7.647192276*I
LambertW(-1,-Pi/2)=-1.570796327*I
LambertW(0,-Pi/2)=1.570796327*I
LambertW(1,-Pi/2)=-1.604290913+7.647192276*I
LambertW(2,-Pi/2)=-2.198342630+13.98120831*IThe numbering of the branches is of course convention, but seems to match we the one given in your article (which seems to be a quite good article).
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/06/2007
Hej BO,
bo198214 Wrote:Again, no! Try for example Maple (LambertW) or Mathematica (ProductLog), you can chose the branch there, for example
where the first argument denotes the branch.Code:LambertW(-3,-Pi/2)=-2.198342630-13.98120831*I
LambertW(-2,-Pi/2)=-1.604290913-7.647192276*I
LambertW(-1,-Pi/2)=-1.570796327*I
LambertW(0,-Pi/2)=1.570796327*I
LambertW(1,-Pi/2)=-1.604290913+7.647192276*I
LambertW(2,-Pi/2)=-2.198342630+13.98120831*I
I see, thanks. I appreciate a lot the lessons I get from You.
I wonder if there is analytical way to derive all branch values for W(-pi/2) as function of k and hence h(e^pi/2). What does Mathematica says about the values of h(e^pi/2) and h(e^-pi/2) along few first branches? (I have no access to such instruments...). Any kind of pattern?
From paper I have I see these formulas for Quadratix of Hippias but can not yet understand how these could be used to derive values of W(-pi/2) on other branches than 0 and -1.
I still think there are actually 4 values h(e^pi/2) and may be even in some more general divergent cases of real number tetration: 2 corresponding to -i, 2 to +i , but I do not know yet how to find out if that is true and what are these values. The only case when there are only 2 distinquishable values ar +-i bacause the real parts here are 0..
I also make a conjecture that as number of branch k-> infinity,
h(e^pi/2) = -1 independent of way it is reached.
Also, I think there must be some relation between all h(e^pi/2) and
h(e^-pi/2) branch values, possibly type of Mobius transform. Again , just a conjecture, to be proved wrong to make progress
Best regards,
Ivars
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/07/2007
Ivars Wrote:But, if I understand correctly ( probably not) the picture 4. in the article http://www.cs.uwaterloo.ca/research/tr/1993/03/W.pdf if
nice coincidence. I found a formula in this paper, which agrees perfectly with my fixpoints-for-real-bases b>e^(1/e) formula. In that article they discuss boundaries for the lambert-w-function such that
x = eta ctg(eta) + i * eta
and similar (page 15, formula 4.1 - 4.5)
For the complex fixpoints for real b>e^(1/e) I had the same type of formula; such that
u = beta cos(beta)/sin(beta) + I * beta
t = exp(u)
b = exp(u/t) is real
and this last expression is a branch-enabled version of
b = t^(1/t)
Nice...
Gottfried
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/07/2007
bo198214 Wrote:Gottfried Wrote:there is an article of L Euler, which seems to deal with it.
Where did you dig out this article?!
That was *not* complicated (as I erroneously recalled). In fact, the scan is in the Euler-archive
E489
I even had the link in my July-thread in sci.math. (Where is my brain currently? )
Gottfried
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/07/2007
Gottfried Wrote:For the complex fixpoints for real b>e^(1/e) I had the same type of formula; such that
u = beta cos(beta)/sin(beta) + I * beta
t = exp(u)
b = exp(u/t) is real
and this last expression is a branch-enabled version of
b = t^(1/t)
So how do you finally compute the \( k \)th branch of \( h \) or \( W \) with this formula?
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/07/2007
Hej Gotfried,
Congratulations! There are no accidents.
Quote:So how do you finally compute the \( k \)th branch of \( h \) or \( W \) with this formula?
Yes, how would You? And particularly, in case
h(e^pi/2) and h(e^-pi/2)?
Waiting impatiently even
Ivars
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Gottfried - 11/07/2007
Ivars Wrote:Hej Gotfried,
Congratulations! There are no accidents.
Quote:So how do you finally compute the \( k \)th branch of \( h \) or \( W \) with this formula?
Yes, how would You? And particularly, in case
h(e^pi/2) and h(e^-pi/2)?
Waiting impatiently even
Ivars
No chance.... ;-)
Yes, there is the coincidence; but I did not compute the W-function but the function, which gives real bases b (or, to avoid confusion with the parameter-notation in the article:bases s) for exp(u/t) where s=t^(1/t) is the principal branch, given u (or more precisely: given the imaginary part of u), and get real bases s>1 .
I set imag(u) = beta = any real value -pi < beta < pi
compute real(u) according to the above formula thus having u completed,
then compute t = exp(u)
and then compute s = exp(u/t), which is then surely real.
The formula for this is in the file fixpoints.pdf
In this article I've not yet included the extension of the range for beta>pi; but I've done some computations with this and found further fixpoints for the real s in the regions 2*k*pi< beta < 2*(k+1)*pi. However, there is no exact periodicity, the consecutive fixpoints for the same s approach the lower bound 2*k*pi with the index k.
I can find the inverse, to compute a fixpoint t by a given s, only by numerical approximation, since s=f(u) is monotonic, using a binary iterative process.
If I *had* the branch-enabled Lambert-W-function, this would be easier, but Pari/GP doesn't have it and I still have only the python-example from wikipedia for the real-valued region, and have not yet invested in programming it myself.
Gottfried
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/07/2007
It is not that difficult to compute the branches of \( h \).
The branches of \( h(b) \) are simply the fixed points of \( b^z \). In this way you can compute the branches \( k=0,\dots,\infty \) via
\( h_k(b)=\lim_{n\to\infty}\log_{k,b}^{\circ n}(-1) \), where \( \log_{k,b}(z)=\frac{\log(z)+2\pi i k}{\log(b)} \).
This is however valid only for \( b\notin [e^{-e},e^{1/e}] \) as the numbering of the branches in this interval may be a bit different. For \( b=e^{1/e} \) the branch 0 and -1 (which are non-real) join and for slightly smaller \( b \) it splits again into two real branches.
The negative branches are simply the conjugates of the positive branches:\( h_{-k}(b)=\overline{h_{k-1}(b)} \).
Edit:
As it is so easy to compute \( h \) it is perhaps more appropriate to compute \( W \) from \( h \) by \( W(z)=z\cdot h(e^{-z}) \) instead of computing \( h \) from \( W \).
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - Ivars - 11/07/2007
hej BO,
PLEASE post at least one numerical example how to use this limit formula e.f for k=-2, h(e^(pi/2)).
I am not familiar (entirely my fault) with the notations You use - what is n, how to take the limit, what is z, what is b (I assume b=e^(pi/2) in my case.
Thank You in advance,
Ivars
RE: Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) - bo198214 - 11/07/2007
Ivars Wrote:PLEASE post at least one numerical example how to use this limit formula e.f for k=-2, h(e^(pi/2)).
I am not familiar (entirely my fault) with the notations You use - what is n, how to take the limit, what is z, what is b (I assume b=e^(pi/2) in my case.
Yes the notation, mainly you have to know what \( f^{\circ n}(x) \) means, it is simply the \( n \) times repetition of \( f \) applied to \( x \), for example \( f^{\circ 3}(x)=f(f(f(x))) \).
In our example is then \( b=e^{\frac{\pi}{2}} \). Because \( k=-2 \) we compute \( h_{-2}(b)=\overline{h_{2-1}(b) \) which means the complex conjugate of \( h_1(b) \). To compute this we consider \( \log_{1,b}(z) = \frac{\log(z)+2\pi i}{\pi/2}=\frac{2}{\pi}\log(z)+4i \) and then compute \( h_1(b)=\lim_{n\to\infty} \log_{1,b}(z)^{\circ n}(-1) \). This can of course only approximately computed. Have a look at some sample values:
\( \log_{1,b}(-1)=6*I \)
\( \log_{1,b}^{\circ 2}(-1)=\log_{1,b}(\log_{1,b}(-1))=1.140669505+5.000000000*I \)
\( \log_{1,b}^{\circ 10}(-1)=1.021323305+4.868353812*I \)
\( \log_{1,b}^{\circ 100}(-1)=1.021323316+4.868353806*I \)
But you can see that the value converges, \( h_1(b)\approx 1.021323+4.8683538*I \) hence \( h_{-2}(b)\approx 1.021323-4.8683538*I \).
This is indeed the same value, you get when computing
\( \text{LambertW}(-2,-\pi/2)/(-\pi/2) \) with Maple.