wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> putnam exam (pure math) >> x^x=5
        
(Message started by: temporary on Jan 23^rd, 2008, 6:40pm)

Title: x^x=5
Post by temporary on Jan 23^rd, 2008, 6:40pm

The title says it all. Try x^x=i if you have already solved the previous one. Hint[hide]sqrt(n)^sqrt(n)^sqrt(n)...=n when done from up to down[/hide]

Title: Re: x^x=5
Post by Obob on Jan 23^rd, 2008, 11:39pm

Please don't post riddles when you have no idea what you are talking about.

Title: Re: x^x=5
Post by Michael_Dagg on Jan 24^th, 2008, 12:20am

Thinking sensibly, you can contrive an exponential
function for this equation that is a Lambert function
-- not really difficult either.

Title: Re: x^x=5
Post by towr on Jan 24^th, 2008, 5:02am

It's pretty much example 2 on http://en.wikipedia.org/wiki/Lambert_function

So [hide]ln(5)/W(ln(5))[/hide]
Not a very elucidating result, mind you.

Title: Re: x^x=5
Post by Icarus on Jan 24^th, 2008, 4:56pm

The "hint" is easily demonstrated to be false. If it were true, then (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifn)ⁿ = n. But this is not true for most values of n.

That iⁱ = i is a fun little result. i^z is normally a multi-valued function. But all the various values collapse at z = i to just one.

As for x^x = 5, I get x = 2.129372483...

Title: Re: x^x=5
Post by ThudanBlunder on Jan 24^th, 2008, 6:20pm

on 01/24/08 at 16:56:47, Icarus wrote:

That iⁱ = i is a fun little result.

I would call that funny. ::)

Title: Re: x^x=5
Post by Obob on Jan 24^th, 2008, 6:25pm

Unfortunately iⁱ=i also isn't true... It is true that iⁱ isn't multi-valued, but the value is e^{- pi/2}.

Title: Re: x^x=5
Post by Icarus on Jan 24^th, 2008, 6:35pm

&$#*(&!! :-[ :-[

Dropped an i and didn't notice it!

Title: Re: x^x=5
Post by temporary on Jan 24^th, 2008, 7:52pm

I don't know where you get i^i=i, but i^i=e^-pi/2 +-2pi(k) where k is any integer.

Title: Re: x^x=5
Post by Icarus on Jan 24^th, 2008, 8:29pm

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.

Title: Re: x^x=5
Post by temporary on Jan 24^th, 2008, 9:14pm

Especially since ^i involves natural logarithm, which is endlessly multivalued.

Title: Re: x^x=5
Post by Obob on Jan 24^th, 2008, 10:31pm

Ah yes, iⁱ is multivalued. I must have fallen victim to the same mistake Icarus did. In fact z^w is always multivalued unless either z=0 or w=0, even if say z and w are integers, so that we know what z^w means in the normal sense.

Title: Re: x^x=5
Post by Icarus on Jan 25^th, 2008, 6:10pm

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 e^{ihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subpi.gif/2 + 2khttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subpi.gifi} = i for all values of k.

Of course, it was silly not to have noticed such a farcical result even while posting it.

Title: Re: x^x=5
Post by Eigenray on Jan 25^th, 2008, 8:00pm

on 01/24/08 at 22:31:19, Obob wrote:

In fact z^w is always multivalued unless either z=0 or w=0, even if say z and w are integers, so that we know what z^w means in the normal sense.

Even if w is an integer? ???

on 01/25/08 at 18:10:20, 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 e^{i\subpi/2 + 2k\subpii} = i for all values of k.

So what you really showed is that i¹ = i, which is a bit more plausible. ;)

Title: Re: x^x=5
Post by Icarus on Jan 25^th, 2008, 8:05pm

Indeed. I showed the full value of my 11 1/2 years of college mathematics to produce this amazing and unexpected result!

Title: Re: x^x=5
Post by Obob on Jan 25^th, 2008, 10:29pm

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 e^{2 pi i w} = 1 implies w = 0...

Title: Re: x^x=5
Post by temporary on Jan 26^th, 2008, 11:08am

Or any other integer.

Title: Re: x^x=5
Post by Icarus on Jan 26^th, 2008, 8:10pm

I think I see the idea behind the ill-formed hint:

If x^x = z, then x = z^1/x = z^{1/z^(1/x)} = z^{1/z^(1/(z^(1/x)))} = ...

If you make an initial guess x₀ for x and form the sequence x_n = z^(1/x_n-1), and if the sequence converges, it will converge to x.

If the sequence fails to converge, then try x_n = ln(z)/ln(x_n-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).

Title: Re: x^x=5
Post by temporary on Jan 27^th, 2008, 11:25am

Good, now here is another one. x^x=0. Except this time, I know the answer.

Title: Re: x^x=5
Post by Icarus on Jan 27^th, 2008, 11:34am

So does everyone here: there is no solution (not in http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbc.gifanyway).

Title: Re: x^x=5
Post by temporary on Jan 27^th, 2008, 3:04pm

There is a solution. Srn347 might have known it.

Title: Re: x^x=5
Post by Obob on Jan 27^th, 2008, 5:49pm

I've told this to srn347 as well:

Take your crappy nonsense out of Putnam. This is not the place for it.

Title: Re: x^x=5
Post by temporary on Jan 27^th, 2008, 7:34pm

Take your ****** language and attitude out of my thread. This is not a place for it.

Title: Re: x^x=5
Post by Icarus on Jan 27^th, 2008, 8:53pm

There is NO solution in http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbc.gif. 0^0 = 1, and if x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif0, then x^y http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif0 either, for any y, and any acceptable definition of exponentiation.

I'm guessing that you are thinking something like (-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif)^{-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif}. But infinities don't work that way. (-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif)^{-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif} is an undefined form, and for very good reason.

Title: Re: x^x=5
Post by Icarus on Jan 27^th, 2008, 9:10pm

on 01/27/08 at 17:49:41, Obob wrote:

I've told this to srn347 as well:

Take your crappy nonsense out of Putnam. This is not the place for it.

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.

Title: Re: x^x=5
Post by Obob on Jan 27^th, 2008, 10:44pm

I respect your opinion Icarus, but you weren't around much when srn347 turned this forum into a wasteland of complete nonsense. I agree I may have been a bit hasty in my denouncing "temporary"'s posts, but the amount of abuse that I and other regulars on the forum endured from srn has made me very impatient to what is no doubt his reincarnation.

Sure there is the Lambert W function, but that is nearly a cop out as far as this "riddle" is concerned: its practically like saying, if you want to solve the equation f(x) = c, then the correct answer is "the inverse function of f, evaluated at c." The actual content to the solution is debatable at best.

And my initial reaction in the second post of the thread was to his hint, which is completely false and meaningless. In fact, so long as n is at least 4+epsilon there isn't even a chance that it could be true, since the tower of exponents on sqrt(n) will be bigger than 2, rendering the sqrt(n)^(tower of exponents) bigger than n.

I do admit that this thread has for the most part turned into a decent discussion, but putnam is most emphatically not the place for riddles based on mathematical misunderstandings to be discussed.

Title: Re: x^x=5
Post by Obob on Jan 27^th, 2008, 11:30pm

Also, just to clarify, my statement to "take your nonsense elsewhere" was in particular directed at his insistence that x^x=0 has a solution, exactly the kind of thing that srn used to wildly profess, despite any arguments to the contrary.

Srn had this way of receiving replies to his threads or opinions and dismiss them, even when all he did was spout nonsense. I am convinced he didn't learn a single thing from all his time here; previously I thought maybe he learned that people don't have to put up with him if they didn't want to, and that they have ways of dealing with him without disrupting the entire community. It appears that he doesn't even respect that.

Title: Re: x^x=5
Post by temporary on Mar 15^th, 2008, 10:02pm

on 01/27/08 at 20:53:52, Icarus wrote:

(-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif)^{-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif} is correct, which I should have noticed you saying sooner. It is not undefined, although which zero it will become(negative zero, positive zero, regular zero) is undefined, it is also irrelevant.

Title: Re: x^x=5
Post by Obob on Mar 16^th, 2008, 5:59am

on 03/15/08 at 22:02:56, temporary wrote:

Wow.

Title: Re: x^x=5
Post by Icarus on Mar 16^th, 2008, 11:50am

Alright, Obob. You've got me on this one! :P

temporary:

First of all, +0 = -0 = 0. They are the same number.

Second, it is easy to get non-zero numbers out of (-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif)^{-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif} forms, so to say that it is 0 in any way is wrong. There is no accepted convention that defines it to be 0.

Title: Re: x^x=5
Post by temporary on Mar 25^th, 2008, 6:24pm

(-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif)^{-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif} can be rewritten as 1/(-infinity)^infinity. Since (-infinity)^infinity is +-infinity, it's reciprocal must be zero.