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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ie, mainipulate the equation in some way to show the answer.

A small, and largely useless hint:

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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- Yakk
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Solving for x in (0, infinity):

x ln 2 = 2 ln x

x/ln x = 2/ln 2

Let f(Z) = Z/ln Z

On (1, infinity) this is monotonicly increasing and analytical (C-infinity), and as such has an inverse with the same properties. Similarly, on (0,1).

Let g = (f|(1, infinity))^(-1)| and h = (f|(0,1))^(-1)

(t|d is the function t restricted to the domain d).

Both inverses are constructable from the original function.

The original function (f) is constructable.

So both g and h is construcable.

Then x = g(2/ln 2) or x = h(2/ln 2). (2/ln 2 may lie outside of the domain of g or h, in which case there is only 1 or 0 solutions).

QED

This leaves the imaginary/negative/etc solutions unsolved.

x ln 2 = 2 ln x

x/ln x = 2/ln 2

Let f(Z) = Z/ln Z

On (1, infinity) this is monotonicly increasing and analytical (C-infinity), and as such has an inverse with the same properties. Similarly, on (0,1).

Let g = (f|(1, infinity))^(-1)| and h = (f|(0,1))^(-1)

(t|d is the function t restricted to the domain d).

Both inverses are constructable from the original function.

The original function (f) is constructable.

So both g and h is construcable.

Then x = g(2/ln 2) or x = h(2/ln 2). (2/ln 2 may lie outside of the domain of g or h, in which case there is only 1 or 0 solutions).

QED

This leaves the imaginary/negative/etc solutions unsolved.

- gmalivuk
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Let f(z, k) be the kth x that solves z = x e^x. (k=0 is the principal solution)

Then the solutions to x^2 = 2^x are of the form

Where n is an integer.

Notably, these include x=2 and x=4.

I believe that, in general the solutions to x^m = m^x (m an integer > 1) are of the form

Where, again, n is an integer, and zeta is an mth root of unity.

Then the solutions to x^2 = 2^x are of the form

Code: Select all

` 2 f( +/- ln(2) / 2, n)`

x = - ------------------------

ln(2)

Where n is an integer.

Notably, these include x=2 and x=4.

I believe that, in general the solutions to x^m = m^x (m an integer > 1) are of the form

Code: Select all

` m f( - zeta ln(2) / 2, n)`

x = - ---------------------------

ln(m)

Where, again, n is an integer, and zeta is an mth root of unity.

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?

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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).

- gmalivuk
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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).

Dunno. The point is that my f is the specific function that shill mentioned. There are probably plenty of other functions you can define to get the answer, but the one I used (without naming it) already has a presence in math.

shill wrote:Darth Eru wrote:A small, and largely useless hint:That "odd function" is the Lambert W function, right?

Yeah, that sounds right.

^^Not actually a spoiler, I just didn't want to break the trend.

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Dvorak > QWERTY

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shill wrote:Darth Eru wrote:A small, and largely useless hint:That "odd function" is the Lambert W function, right?

This is my second favourite function (after the Wright function - which is the inverse Laplace transform of the Mittag-Leffler function (another fav (btw I love nested parenthesis))). It's awesome because if A is a nxn matrix and it has p < n distinct eigenvalues, then Xexp(X) = A has an uncountable number of solutions (if p=n then it has a countable number (if p > n then you need to redo linear algebra))

2^x=X^2

Ln2^x=Lnx^2 , x>0

xLn2=2lnx

Lnx/x=Ln2/2

fx=lnx/x , x>0

f'x=1-lnx/x^2 , x>0

f'x>0 for x<e

that means f auxousa for x belongs to (o,e]

and f fthinousa for x belongs to [e,+~)

As a result f has to possible solution one in (0,e] and the other in [e,+~)

f have the possible solution of x=2 and x=4 As a result 2^x-X^2 has to solutions x=2 and x=4

Ln2^x=Lnx^2 , x>0

xLn2=2lnx

Lnx/x=Ln2/2

fx=lnx/x , x>0

f'x=1-lnx/x^2 , x>0

f'x>0 for x<e

that means f auxousa for x belongs to (o,e]

and f fthinousa for x belongs to [e,+~)

As a result f has to possible solution one in (0,e] and the other in [e,+~)

f have the possible solution of x=2 and x=4 As a result 2^x-X^2 has to solutions x=2 and x=4

- gmalivuk
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What?Ulianov wrote:that means f auxousa for x belongs to (o,e]

and f fthinousa for x belongs to [e,+~)

gmalivuk wrote:What?Ulianov wrote:that means f auxousa for x belongs to (o,e]

and f fthinousa for x belongs to [e,+~)

I tried to figure it out from context, and I am pretty sure it involves taking the derivative of C'thulu.

Transliterated Greek. "Ascending" and "descending".

All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.

- Yakk
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And ~ is being used as infinity!

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

This is my single favourite equation!

Lets look at the general problem [imath]x^y=y^x[/imath], where [imath]x,y \in (0,\infty)[/imath].

Lets bring in new variable: [imath]y=a*x[/imath].

Now [imath]x^{(a*x)}=(a*x)^x[/imath].

[imath](x^a)^x=(a*x)^x[/imath]

Lets take both sides to power (1/x): [imath]x^a=a*x[/imath]

Lets divide both sides with x: [imath]x^{(a-1)}=a[/imath]

Lets take both sides to power (1/(a-1)): [imath]x=a^{(1/(a-1))}[/imath]

Because [imath]y=a*x[/imath], [imath]y=a*a^{(1/(a-1))}=a^{(1/(a-1)+1)}=a^{(a/(a-1))}[/imath].

Notice that [imath]a \in (0,\infty)[/imath].

Let [imath]f(a)=a^{(1/(a-1))}[/imath].

It is easy to see that f(2)=2, so x=2 and y=4.

Also f(1/2)=4, so x=4 and y=2.

f is decreasing. (I tried to prove, but the derivatives are bit nasty. I will try again later, but I am sure someone will beat me to it. )

My understanding is that this function f gives all positive real pairs of solutions (that are non-equal). If I missed something, please tell me

So, what do You think? Did I miss something or is there something else cool hidden in this problem that I did not notice?

PS. Generally, f(z)*z=f(1/z). (prove Yourself)

PPS. Interestingly, f(1)=e (prove Yourself). Of course, if x=y, then x^y=y^x always, but this also shows that if x[imath]= e - \epsilon[/imath], then y~[imath]= e+ \epsilon[/imath].

Yours,

Rauni

Lets look at the general problem [imath]x^y=y^x[/imath], where [imath]x,y \in (0,\infty)[/imath].

Lets bring in new variable: [imath]y=a*x[/imath].

Now [imath]x^{(a*x)}=(a*x)^x[/imath].

[imath](x^a)^x=(a*x)^x[/imath]

Lets take both sides to power (1/x): [imath]x^a=a*x[/imath]

Lets divide both sides with x: [imath]x^{(a-1)}=a[/imath]

Lets take both sides to power (1/(a-1)): [imath]x=a^{(1/(a-1))}[/imath]

Because [imath]y=a*x[/imath], [imath]y=a*a^{(1/(a-1))}=a^{(1/(a-1)+1)}=a^{(a/(a-1))}[/imath].

Notice that [imath]a \in (0,\infty)[/imath].

Let [imath]f(a)=a^{(1/(a-1))}[/imath].

It is easy to see that f(2)=2, so x=2 and y=4.

Also f(1/2)=4, so x=4 and y=2.

f is decreasing. (I tried to prove, but the derivatives are bit nasty. I will try again later, but I am sure someone will beat me to it. )

My understanding is that this function f gives all positive real pairs of solutions (that are non-equal). If I missed something, please tell me

So, what do You think? Did I miss something or is there something else cool hidden in this problem that I did not notice?

PS. Generally, f(z)*z=f(1/z). (prove Yourself)

PPS. Interestingly, f(1)=e (prove Yourself). Of course, if x=y, then x^y=y^x always, but this also shows that if x[imath]= e - \epsilon[/imath], then y~[imath]= e+ \epsilon[/imath].

Yours,

Rauni

Splendid equation, I deeply respect it since elementary school.

1) It is solved routinely using special "Lambert W" function, which is conveniently enough defined as a solution of a similar equation y=x exp(x). Reachable through mathematica as ProductLog.

2) x^y=y^x in positive has two solutions: x=y and another solution that looks something like (y-1)=1/(x-1) (just as a visual aid). The solutions cross at e (I guessed it at first but it turned out to be correct). So e^x=x^e only has one solution.

3) The negative (the interesting one) solution of 2^x=x^2 is also expressible as a power tower:

- (sqrt(2)/2^sqrt(2)/2^sqrt(2)/2^... )

the powers have to be evaluated from right to left. This makes it easy to get the solution with a handheld calculator.

For y^x=x^y at even y's, there are negative solutions of x. They decrease towards -1 with increasing y (but not monotonously at first).

4) For x^y=y^x, the only integer solutions are (2,2), (2,4), (4,2) and (4,4).

1) It is solved routinely using special "Lambert W" function, which is conveniently enough defined as a solution of a similar equation y=x exp(x). Reachable through mathematica as ProductLog.

2) x^y=y^x in positive has two solutions: x=y and another solution that looks something like (y-1)=1/(x-1) (just as a visual aid). The solutions cross at e (I guessed it at first but it turned out to be correct). So e^x=x^e only has one solution.

3) The negative (the interesting one) solution of 2^x=x^2 is also expressible as a power tower:

- (sqrt(2)/2^sqrt(2)/2^sqrt(2)/2^... )

the powers have to be evaluated from right to left. This makes it easy to get the solution with a handheld calculator.

For y^x=x^y at even y's, there are negative solutions of x. They decrease towards -1 with increasing y (but not monotonously at first).

4) For x^y=y^x, the only integer solutions are (2,2), (2,4), (4,2) and (4,4).

Aniviller wrote:4) For x^y=y^x, the only integer solutions are (2,2), (2,4), (4,2) and (4,4).

(n,n) is a solution for all integer n. (except possibly at 0, depending on whether you define 0^0 or not.)

Also, there is (-2,-4) and (-4,-2)

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?

Yes, the general form is rather pretty. I think you covered most of the main points (apart from the stuff about Lambert's omega function). But you could've been a bit more explicit about the connection with e.

Since e is bounded by a

With a change of variable, we can put this into a more familiar form.

Let a = 1 + 1/n, so n = 1/(a - 1)

For n>0,

(1 + 1/n)

or, more conveniently for calculation,

e ~= (1 + 1/(m - 1/2))

If we let m be a power of 2, the preceding expression can easily be calculated by repeated squaring.

Eg, using bc:

scale=50; n=25; m=2^n; p=1+1/(m-.5); for(i=0;i<n;i++)p*=p; p

2.71828182845904543655355902971283842309033434109472

We can make it converge even faster by playing with that -1/2 in the denominator.

[imath]2^x = x^2[/imath]

Substitute x? (is this legal?)

[imath]2^2 = 2^2[/imath]

Fairly obvious...

Substitute x? (is this legal?)

[imath]2^2 = 2^2[/imath]

Fairly obvious...

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"

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MHD wrote:[imath]2^x = x^2[/imath]

Substitute x? (is this legal?)

[imath]2^2 = 2^2[/imath]

Fairly obvious...

Yeah, sure, its legal to substitute a value for x, but how do you know that you end up with an equality? In general you don't (see what happens for 3 for instance). You guessed that 2 was good, it was, but its exactly the "by observation" that you're being told not to use by the thread title.

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.

No no no. They are using "Fairly obvious" we just need to avoid "by observation" So I think this one stands.

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