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

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 (add x+y to both sides). is true whenever or (multiply both sides by x*y).What would be involved, however, in demonstrating when

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 46,646

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

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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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 10,464

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

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 10,464

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

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

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

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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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.**

i.e.

Now suppose that for some value This gives us:i.e.

and since then we also haveThis 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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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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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:Now instead of putting 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 your corresponding 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 ofI 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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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 46,646

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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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:

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

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 setthen it follows that and hencewhich 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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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/ehl1zmc5nqYou'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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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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**Relentless****Member**- Registered: 2015-12-15
- Posts: 631

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

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