A small, and largely useless hint:
A friend of mine researched this, and apparently the only way he found to do it involved an odd function that neither of us had heard of before.
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A friend of mine researched this, and apparently the only way he found to do it involved an odd function that neither of us had heard of before.
2 f( +/- ln(2) / 2, n)
x = - ------------------------
ln(2)
m f( - zeta ln(2) / 2, n)
x = - ---------------------------
ln(m)
Treatid basically wrote:widdout elephants deh be no starting points. deh be no ZFC.
Darth Eru wrote:A small, and largely useless hint:A friend of mine researched this, and apparently the only way he found to do it involved an odd function that neither of us had heard of before.
That "odd function" is the Lambert W function, right?
gmalivuk wrote:Let f(z, k) be the kth x that solves z = x e^x. (k=0 is the principal solution)
Cosmologicon wrote:gmalivuk wrote:Let f(z, k) be the kth x that solves z = x e^x. (k=0 is the principal solution)
I'm not familiar enough with complex functions to know this. Is it easier to solve for the kth root of x exp(x) - z than it is to solve for the kth root of x / ln(x) - z? If so, how do you go about it?
If not, you can just as easily say let y(z,k) be the kth root of x / ln(x) - z, and then solutions are of the form y(2/ln(2), n).
Treatid basically wrote:widdout elephants deh be no starting points. deh be no ZFC.
shill wrote:Darth Eru wrote:A small, and largely useless hint:A friend of mine researched this, and apparently the only way he found to do it involved an odd function that neither of us had heard of before.That "odd function" is the Lambert W function, right?
Yeah, that sounds right.
shill wrote:Darth Eru wrote:A small, and largely useless hint:A friend of mine researched this, and apparently the only way he found to do it involved an odd function that neither of us had heard of before.That "odd function" is the Lambert W function, right?
What?Ulianov wrote:that means f auxousa for x belongs to (o,e]
and f fthinousa for x belongs to [e,+~)
Treatid basically wrote:widdout elephants deh be no starting points. deh be no ZFC.
gmalivuk wrote:What?Ulianov wrote:that means f auxousa for x belongs to (o,e]
and f fthinousa for x belongs to [e,+~)
Aniviller wrote:4) For x^y=y^x, the only integer solutions are (2,2), (2,4), (4,2) and (4,4).
rauni wrote:This is my single favourite equation!
Did I miss something or is there something else cool hidden in this problem that I did not notice?
EvanED wrote:be aware that when most people say "regular expression" they really mean "something that is almost, but not quite, entirely unlike a regular expression"
MHD wrote:2^x = x^2
Substitute x? (is this legal?)
2^2 = 2^2
Fairly obvious...
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
jestingrabbit wrote:... but its exactly the "by observation" that you're being told not to use by the thread title.