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Topic: x^x=5 (Read 5630 times) |
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temporary
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The title says it all. Try x^x=i if you have already solved the previous one. Hintsqrt(n)^sqrt(n)^sqrt(n)...=n when done from up to down
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| « Last Edit: Jan 23rd, 2008, 6:40pm by temporary » |
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Obob
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Re: x^x=5
« Reply #1 on: Jan 23rd, 2008, 11:39pm » |
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Please don't post riddles when you have no idea what you are talking about.
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Michael Dagg
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Re: x^x=5
« Reply #2 on: Jan 24th, 2008, 12:20am » |
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Thinking sensibly, you can contrive an exponential
function for this equation that is a Lambert function
-- not really difficult either.
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Michael Dagg
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Icarus
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Re: x^x=5
« Reply #4 on: Jan 24th, 2008, 4:56pm » |
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The "hint" is easily demonstrated to be false. If it were true, then ( n)n = n. But this is not true for most values of n.
That ii = i is a fun little result. iz is normally a multi-valued function. But all the various values collapse at z = i to just one.
As for xx = 5, I get x = 2.129372483...
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ThudnBlunder
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Re: x^x=5
« Reply #5 on: Jan 24th, 2008, 6:20pm » |
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on Jan 24th, 2008, 4:56pm, Icarus wrote:
That ii = i is a fun little result.
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I would call that funny. 
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Obob
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Re: x^x=5
« Reply #6 on: Jan 24th, 2008, 6:25pm » |
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Unfortunately ii=i also isn't true... It is true that ii isn't multi-valued, but the value is e- pi/2.
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Icarus
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Re: x^x=5
« Reply #7 on: Jan 24th, 2008, 6:35pm » |
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&$#*(&!! 
Dropped an i and didn't notice it!
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I wonder: Which is larger
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temporary
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Re: x^x=5
« Reply #8 on: Jan 24th, 2008, 7:52pm » |
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I don't know where you get i^i=i, but i^i=e^-pi/2 +-2pi(k) where k is any integer.
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| « Last Edit: Jan 24th, 2008, 7:53pm by temporary » |
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Icarus
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Re: x^x=5
« Reply #9 on: Jan 24th, 2008, 8:29pm » |
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You might consider reading the three posts following that one, where this little mystery is solved.
Of course, I have the small satisfaction of partially confusing Obob with my gibberish. You are correct that ii is multivalued.
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"Pi goes on and on and on ...
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I wonder: Which is larger
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temporary
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Re: x^x=5
« Reply #10 on: Jan 24th, 2008, 9:14pm » |
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Especially since ^i involves natural logarithm, which is endlessly multivalued.
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Obob
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Re: x^x=5
« Reply #11 on: Jan 24th, 2008, 10:31pm » |
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Ah yes, ii is multivalued. I must have fallen victim to the same mistake Icarus did. In fact zw is always multivalued unless either z=0 or w=0, even if say z and w are integers, so that we know what zw means in the normal sense.
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Icarus
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Re: x^x=5
« Reply #12 on: Jan 25th, 2008, 6:10pm » |
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It wasn't the multi-valuedness of the logarithm that was my downfall. I dropped an i from my calculation and so ended up with ei /2 + 2ki = i for all values of k.
Of course, it was silly not to have noticed such a farcical result even while posting it.
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Eigenray
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Re: x^x=5
« Reply #13 on: Jan 25th, 2008, 8:00pm » |
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on Jan 24th, 2008, 10:31pm, Obob wrote:| In fact zw is always multivalued unless either z=0 or w=0, even if say z and w are integers, so that we know what zw means in the normal sense. |
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Even if w is an integer? 
on Jan 25th, 2008, 6:10pm, Icarus wrote:| It wasn't the multi-valuedness of the logarithm that was my downfall. I dropped an i from my calculation and so ended up with ei\subpi/2 + 2k\subpii = i for all values of k. |
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So what you really showed is that i1 = i, which is a bit more plausible. 
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Icarus
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Re: x^x=5
« Reply #14 on: Jan 25th, 2008, 8:05pm » |
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Indeed. I showed the full value of my 11 1/2 years of college mathematics to produce this amazing and unexpected result!
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"Pi goes on and on and on ...
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I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Obob
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Re: x^x=5
« Reply #15 on: Jan 25th, 2008, 10:29pm » |
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Wow, I've been making a lot of mistakes lately. You are of course right Eigenray. For some reason in my back-of-the-envelope calculation I deduced that e2 pi i w = 1 implies w = 0...
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temporary
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Re: x^x=5
« Reply #16 on: Jan 26th, 2008, 11:08am » |
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Or any other integer.
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Icarus
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Re: x^x=5
« Reply #17 on: Jan 26th, 2008, 8:10pm » |
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I think I see the idea behind the ill-formed hint:
If xx = z, then x = z1/x = z1/z^(1/x) = z1/z^(1/(z^(1/x))) = ...
If you make an initial guess x0 for x and form the sequence xn = z^(1/xn-1), and if the sequence converges, it will converge to x.
If the sequence fails to converge, then try xn = ln(z)/ln(xn-1).
The first series converges for both z = 5 and z = i.
For z = i, the value is about 1.36062487029112 + i 1.11943916624235 (using the principle branch of ln(z) when calculating all values).
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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temporary
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Re: x^x=5
« Reply #18 on: Jan 27th, 2008, 11:25am » |
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Good, now here is another one. x^x=0. Except this time, I know the answer.
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Icarus
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Re: x^x=5
« Reply #19 on: Jan 27th, 2008, 11:34am » |
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So does everyone here: there is no solution (not in  anyway).
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| « Last Edit: Jan 27th, 2008, 11:36am by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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temporary
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Re: x^x=5
« Reply #20 on: Jan 27th, 2008, 3:04pm » |
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There is a solution. Srn347 might have known it.
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Obob
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Re: x^x=5
« Reply #21 on: Jan 27th, 2008, 5:49pm » |
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I've told this to srn347 as well:
Take your crappy nonsense out of Putnam. This is not the place for it.
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temporary
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Re: x^x=5
« Reply #22 on: Jan 27th, 2008, 7:34pm » |
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Take your ****** language and attitude out of my thread. This is not a place for it.
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My goal is to find what my goal is, once I find what my goal is, my goal will be complete.
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Icarus
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Re: x^x=5
« Reply #23 on: Jan 27th, 2008, 8:53pm » |
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There is NO solution in . 0^0 = 1, and if x  0, then xy 0 either, for any y, and any acceptable definition of exponentiation.
I'm guessing that you are thinking something like (- )- . But infinities don't work that way. (-)- is an undefined form, and for very good reason.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Icarus
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Boldly going where even angels fear to tread.
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Re: x^x=5
« Reply #24 on: Jan 27th, 2008, 9:10pm » |
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on Jan 27th, 2008, 5:49pm, Obob wrote:I've told this to srn347 as well:
Take your crappy nonsense out of Putnam. This is not the place for it. |
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Obob, it really isn't your call what goes here. The x^x = 5 and x^x = i puzzles are entirely appropriate to this forum, even if they are not as abstract as most of the stuff here. There are a number of other problems in this forum that they match quite well with. Admittedly, temporary doesn't understand the issues he brought up, but there is interesting stuff here non-the-less. When I was in college, my text and reference books were mostly written in the 1970s before people started paying attention to the Lambert W function, and since I wasn't involved in this area of study, I never knew anything about it, until I encountered it here.
Temporary's smug misunderstandings can be annoying at times, but he and his idol are not the first to match this description. And there is value here.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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