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## #1 2021-10-14 21:18:55

Relentless
Member
Registered: 2015-12-15
Posts: 631

### x ° y = y ° x

Hi everyone,

It sure has been a while.

I was recently pursuing this line of questioning:

and
are obviously always true (commutativity).

is pretty clearly true when
or

is true whenever
or
(multiply both sides by x*y).

What would be involved, however, in demonstrating when

?

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## #2 2021-10-14 22:57:32

Jai Ganesh Registered: 2005-06-28
Posts: 43,555

### Re: x ° y = y ° x

Hi Relentless,

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #3 2021-10-14 23:45:45

Relentless
Member
Registered: 2015-12-15
Posts: 631

### Re: x ° y = y ° x

I'm going to list some solutions you can compute manually.
(1,1)
(2,4)
(2,-0.767...)
(3,2.478...)
(4,-0.767...)
(5,1.765...)
(6,-0.79...)
(6,1.624...)

I lack the knowledge to explain what is going on here in general terms. It apparently has something to do with the "product log function" and complex numbers.

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## #4 2021-10-15 09:54:22

Bob Registered: 2010-06-20
Posts: 9,719

### Re: x ° y = y ° x

hi Relentless,

Hhmm. Interesting!

I'm still working on this so I may have more to offer.  Sometimes a graph will help. MIF has two graph plotters; the function grapher will work with equations of the form y = f(x).  The equation grapher only requires an equation connecting x and y.  Here's what I got: I didn't get any negative points.  I think that is just the way the plotter has been set up as we know that negative values are possible.  You can see (2,4) there.  The point where x = y looks like it might be (e,e) but I have yet to prove it.

It would seem that there are values of y for every x and values of x for every y, possibly except where x or y is zero.

I'm hoping to discover more.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob Offline

## #5 2021-10-15 20:23:04

Bob Registered: 2010-06-20
Posts: 9,719

### Re: x ° y = y ° x

Proof that the curve is cut by y = x at (e,e)

differentiate wrt x

If x and y are swapped the equation is unchanged. This means that the function is symmetrical about the line y = x, which means the gradient at the point where the line cuts the curve must be -1

So I can substitute in the above y = x and dy/dx = -1

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob Offline

## #6 2021-10-15 20:54:06

Relentless
Member
Registered: 2015-12-15
Posts: 631

### Re: x ° y = y ° x

Hi guys,

If you plot the graphs for x^y and y^x for any particular y you can see some patterns.

0^x=x^0 has no solutions because x^0 is always 1 and 0^x is 0, x>0.

x=1 is a unique solution of 1^x=x^1 (you must set x^1=1).

For natural numbers higher than one we can find there are two solutions for odd numbers 3^x=x^3 etc. and three solutions, including a negative one, for even numbers 2^x=x^2 etc. The negative solution for x is able to produce a positive result for both sides in the case of even numbers (e.g. x^2 can be positive even if x is negative), but in the case of odd numbers a negative x will produce a negative RHS and positive LHS (e.g. x^3 is negative for negative x, but 3^x is always positive).

More generally, for positive y-values:
There are two or three solutions for x if y>1, depending on whether x^y can be positive if x is negative.
There are one or two solutions for x if 0<y≤1, again depending on whether x^y can be positive if x is negative.

For negative y-values the graph of y^x is only continuous in the complex plane. It looks like if x is an integer we get real values. Also every integer y will have solution x=y.

This is about as far as I would get with a plotter, with nothing to tie the different results together. If we can algebraically derive some of the non-obvious solutions that would be an improvement.

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## #7 2021-10-15 21:02:37

Relentless
Member
Registered: 2015-12-15
Posts: 631

### Re: x ° y = y ° x

Bob wrote:

Proof that the curve is cut by y = x at (e,e)

differentiate wrt x

If x and y are swapped the equation is unchanged. This means that the function is symmetrical about the line y = x, which means the gradient at the point where the line cuts the curve must be -1

So I can substitute in the above y = x and dy/dx = -1

Bob

Hi Bob,

I'm afraid I don't understand this result, where it comes from or how it is significant.

Edit: Oh. x^e=e^x has exactly one solution, namely, x=e. It is the place where two solutions normally distinct coincide.
Doesn't this tell us that for a general form y=f(x), f(e)=e?

Last edited by Relentless (2021-10-15 22:02:10)

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## #8 2021-10-15 22:11:46

zetafunc
Moderator Registered: 2014-05-21
Posts: 2,419
Website

### Re: x ° y = y ° x

Hi Relentless,

I'm not sure what maths background you have but I'll try to explain what's going on as best I can -- please let me know if anything isn't clear (happy to talk through it in more detail if needed). You'll find the formulae you need to generate the non-trivial solutions in this post and in the first half of post #12.

The standard method for generating such solutions is as follows. Consider the equation
Taking logs of both sides gives us:

i.e.

Now suppose that
for some value
This gives us:

i.e.

and since
then we also have

This is called a parametrisation (also called a 'change of variables'). Now to generate some solutions to your equation we just need to vary
Clearly setting
gives us
so let's try a different value, say,
Simplifying, we get:

and similarly you can generate (infinitely) many similar solutions by plugging in some additional values of
(By the way -- these types of solutions are called 'algebraic numbers'.) OK, so how about negative solutions? Well, let's try setting
This gives us:

Note that the cube root of -2 has three solutions: one is real, the other are complex (and conjugate) roots -- so you could generate complex solutions this way, too.

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## #9 2021-10-15 22:21:16

zetafunc
Moderator Registered: 2014-05-21
Posts: 2,419
Website

### Re: x ° y = y ° x

Relentless wrote:

Hi Bob,

I'm afraid I don't understand this result, where it comes from or how it is significant.

Bob's result can also be seen in another way: as can be seen from the post above, one such 'solution set' is given by the following:

Clearly this doesn't make much sense when
We can, however, consider the limit as
so let's do that instead. Setting
gives us:

One way of interpreting this result is that
is exactly the point at which the trivial solutions (that's everything on the line y = x) and the non-trivial solutions (that's the curved bit of Bob's graph) meet.

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## #10 2021-10-15 23:05:46

zetafunc
Moderator Registered: 2014-05-21
Posts: 2,419
Website

### Re: x ° y = y ° x

You also mentioned the product-log function (also called the Lambert W-function). I'll try to explain where this comes from (and why WolframAlpha tends to spit up that kind of result!). First: what is it?

The definition of Lambert's W-function is that it's the inverse of this function below:

So in other words, if
then the W-function tells us which values of
we need, i.e.
OK great, but how does this relate to our equation? Let's go back to this line in post #8:

let's try a different substitution, say...
This gives us:

But from the definition of the W-function, the value of
which satisfies the above must be given by:

Re-expressing all of this in terms of
and
only, we get:

So there's another way we can generate some solutions. The problem is that the W-function isn't expressible (generally) in terms of elementary functions -- so given an arbitrary value of
which satisfies
might end up being (irreducibly) some infinite series!

The other issue is that I haven't defined the W-function particularly well. Does
even have an explicit inverse? Well, no, not really -- as it stands, our W-function would be multi-valued, so that's no good. We instead need to do what's called a 'branch cut' in complex analysis. (This basically means you define the possible inputs for W such that it only spits out unique values.) So the full solution set would look something like:

where, for each
the function
represents a different branch of the W-function for each of the ranges of possible values of

I haven't done a full video on branch cuts but I do motivate them to some extent in this video if you want to have a look. I've also got a few videos on contour integration which -- while not relevant here -- do involve choosing branch cuts, so I can link to those if you want them too.

By the way, one way you can visualise the connection of the W-function to the graphs Bob and ganesh have shown is by thinking of the straight line as the trivial solution (i.e. y = x) and the curved component as being the equation involving the W-function. Indeed, if you plug in x = e into the equation above then we find that x = y = e also.

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## #11 2021-10-16 00:05:37

Jai Ganesh Registered: 2005-06-28
Posts: 43,555

### Re: x ° y = y ° x

Hi,

This image may be of help. It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #12 2021-10-16 04:26:23

zetafunc
Moderator Registered: 2014-05-21
Posts: 2,419
Website

### Re: x ° y = y ° x

I should mention that the parametrisation
isn't the only choice we could have used in post #8 -- there are others we could use to generate solutions to your equation. (By the way, in mathematics we call this kind of 'guess' at a solution an ansatz. This is very common for solving things like difference equations where we don't know what the solution is, but we hazard a guess at what we think it 'should' look like -- and then prove it's a solution.) For example, let's consider something a bit more different. (This one is a lot more beautiful I think!)

As before, suppose that
satisfy
Let's instead consider an alternative mapping
given by:

where
and
(These are called polar co-ordinates.) Then:

-- i.e. raising both sides to the power of
.

Expanding the brackets on both sides (because we want
on one side and
on the other), we get:

Pulling 'like' terms across the equality:

This gives us another tool for generating solutions in
and
since we can now just choose a particular
which will determine
and hence
and
For example, you can show that if we set

then it follows that
and hence

which is the classical solution to

Setting
gives us the interesting solution:

In particular, we again see Bob's result showing up here too. Note that our expression for
in terms of
wasn't defined when
i.e. when
However, if we consider the limit instead, we get:

So as
then
and so we have:

i.e.
and
both converge to
as
approaches

By the way, the proof of the first line in this limit is as follows. Let's split the function up into two parts, i.e. let

so that

We'll try to calculate the limits of these functions as
separately. If both exist, then the product of these limits exists (and is equal to the limit of the product). We clearly have:

For the first function, we have:

The function inside the limit satisfies the conditions of L'Hopital's rule, so this is just equal to:

and so

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## #13 2021-10-17 00:27:16

zetafunc
Moderator Registered: 2014-05-21
Posts: 2,419
Website

### Re: x ° y = y ° x

So what does this mean graphically? It turns out that, if we plot the graph of this function (in terms of x and y):

then we get this graph, which should be identical to Bob's graph except without the trivial line y = x: https://www.desmos.com/calculator/xsgqovhxul

For
we trace out everything to the right of the line
which you can see here: https://www.desmos.com/calculator/ntzw4opu6x

...and similarly for
we trace out everything to the left of the line
which you can see here: https://www.desmos.com/calculator/ehl1zmc5nq

You'll notice that if you try to plot values between, say,
and
then it doesn't work -- that's because you get complex solutions.

For example, taking
gives us the following valid solution to
:

but both of these are complex numbers which won't show up in the (real) Cartesian plane.

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## #14 2021-10-17 09:12:55

zetafunc
Moderator Registered: 2014-05-21
Posts: 2,419
Website

### Re: x ° y = y ° x

As a final thought (because I think I've flooded your thread enough with my ramblings!), here's one more parametrisation you could think about -- but it's pretty volatile, so use with caution.

Let's consider the standard parametric equations for a hyperbola: let
satisfy
such that for some
we have:

Then after some algebra analogous to that which was detailed in post #12, we find that

which produces this curve: https://www.desmos.com/calculator/aggjlrrwlk

This also generates solutions to
For example, taking
yields:

from which we find that one solution pair is:

Similarly we can obtain the classical solution
by setting

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## #15 2021-10-23 10:11:23

Relentless
Member
Registered: 2015-12-15
Posts: 631

### Re: x ° y = y ° x

Thank you zetafunc, I do appreciate the various solutions in case as they may come in handy

To be sure I understand I'm going to apply what you said first to the related case of

Taking logs of both sides we get:

Rearranging:

If we now make the y=x^t substitution:

And since y=x^t:

We now have a way to generate solutions. When I try to use this method to find those more complicated answers I began listing earlier, however, I still don't know how, other than by graphing it (which is very difficult for negative answers). For example, graphically it seems there must be some t the real part of which is around -2.56 for which

, yielding the solution to the original problem y=2, x=-0.767... . Is this what happens when you consider t over all the complex numbers? I can see now the solution space for positive t is rather simple (for both equations), but it's still tricky to assess when t is negative. I suppose I am still uninformed, and will come back in a little bit. Offline