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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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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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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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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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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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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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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.
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 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: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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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:
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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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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.
Online
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: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:
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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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