Ferdinando Posted December 3, 2006 Share Posted December 3, 2006 Could you find all positive integer solutions to ... x^y - y^x = x + y Link to comment Share on other sites More sharing options...

the tree Posted December 3, 2006 Share Posted December 3, 2006 x=1,y=1 is one of them, I'm afraid I can't see a methodical aproach to this. Besides, I smell homework. Link to comment Share on other sites More sharing options...

timo Posted December 3, 2006 Share Posted December 3, 2006 x=1,y=1 is one of them, ... Not really. Link to comment Share on other sites More sharing options...

5614 Posted December 3, 2006 Share Posted December 3, 2006 Yer, tree, too much homework in one day? x=0, y=0 is one solution. x=1, y=0 is another solution. I don't believe there are any other real solutions to this equation. Link to comment Share on other sites More sharing options...

the tree Posted December 3, 2006 Share Posted December 3, 2006 Gah, I read a minus as a plus or something. And that was seriously early in morning as well. 5614, isn't 0^{0} undefined? Link to comment Share on other sites More sharing options...

D H Posted December 3, 2006 Share Posted December 3, 2006 x=0, y=0 is one solution.x=1, y=0 is another solution. I don't believe there are any other real solutions to this equation. Neither x=0, y=0 nor x=1, y=1 are solutions. The OP asked for positive integral solutions. Define [math]f(x,y) = x^y - (y^x + x + y)[/math]. Then [math]f(1,y) = -2y\ \forall y \ne 0[/math] [math]f(y,y) = -2y \ \forall y \ne 0[/math] [math]f(x,y) < 0 \ \forall (x,y)\in (\mathbb I \times \mathbb I), x > y >= 2[/math] [math]f(y-1,y) > 0 \ \forall y\in \mathbb I, y >= 4[/math] There are a two positive real solutions [math]x[/math] to [math]f(x,y)=0[/math] for each integer [math]y\in(4,5,...)[/math]. (Actually, there are two positive real solutions [math]x[/math] to [math]f(x,y)=0[/math] for each [math]y>3.6475[/math] or so.) There is at least one integral solution (i.e., I found one). Link to comment Share on other sites More sharing options...

Ferdinando Posted December 4, 2006 Author Share Posted December 4, 2006 Hi D H, 0^0 = 1, if we substitute, then the eq. would be valid And, if x = 1 and y = 0, it will work. Link to comment Share on other sites More sharing options...

D H Posted December 5, 2006 Share Posted December 5, 2006 Hi D H, 0^0 = 1, if we substitute, then the eq. would be valid And, if x = 1 and y = 0, it will work. Neither is a solution. [math]0^0[/math] is undefined, but more importantly, Could you find all positive integer solutions to x^y - y^x = x + y This is your stinking problem, and you have violated your own constraints. Link to comment Share on other sites More sharing options...

Ferdinando Posted December 5, 2006 Author Share Posted December 5, 2006 Sorry, I was thinking of something else. I got distracted. 0^0 is undefined (Of course !) Link to comment Share on other sites More sharing options...

D H Posted December 5, 2006 Share Posted December 5, 2006 0^0 is undefined (Of course !) Many mathematicians define 0^0 as 1. Even with that definition in hand, x=0,y=0 is not a solution because 0 is not a positive integer. If this is homework, the deadline is undoubtedly long past. The only solution I could find was x=2,y=5. I cannot prove this is the only solution. Link to comment Share on other sites More sharing options...

Ferdinando Posted December 5, 2006 Author Share Posted December 5, 2006 Yes (2,5) is what i found too. I am trying to resolve this equation formally. Thanks for your input. Link to comment Share on other sites More sharing options...

5614 Posted December 5, 2006 Share Posted December 5, 2006 I was aware of the issue of 0^0 being undefined when I posted the solution, however as it's: x^y - y^x = x + y you would get (0^0 - 0^0) and can you not argue that that is equal to 0? Link to comment Share on other sites More sharing options...

D H Posted December 5, 2006 Share Posted December 5, 2006 Is zero a positive integer? Link to comment Share on other sites More sharing options...

Ferdinando Posted December 8, 2006 Author Share Posted December 8, 2006 According to some Calculus textbooks, 0^0 is an "indeterminate form." What mathematicians mean by "indeterminate form" is that in some cases we think about it as having one value, and in other cases we think about it as having another. When evaluating a limit of the form 0^0, you need to know that limits of that form are "indeterminate forms," and that you need to use a special technique such as L'Hopital's rule to evaluate them. For instance, when evaluating the limit Sin[x]^x (which is 1 as x goes to 0), we say it is equal to x^x (since Sin[x] and x go to 0 at the same rate, i.e. limit as x->0 of Sin[x]/x is 1). Then we can see from the graph of x^x that its limit is 1. Other than the times when we want it to be indeterminate, 0^0 = 1 seems to be the most useful choice for 0^0 . This convention allows us to extend definitions in different areas of mathematics that would otherwise require treating 0 as a special case. Notice that 0^0 is a discontinuity of the function f(x,y) = x^y, because no matter what number you assign to 0^0, you can't make x^y continuous at (0,0), since the limit along the line x=0 is 0, and the limit along the line y=0 is 1. This means that depending on the context where 0^0 occurs, you might wish to substitute it with 1, indeterminate or undefined/nonexistent. Some people feel that giving a value to a function with an essential discontinuity at a point, such as x^y at (0,0), is an inelegant patch and should not be done. Others point out correctly that in mathematics, usefulness and consistency are very important, and that under these parameters 0^0 = 1 is the natural choice. Link to comment Share on other sites More sharing options...

Ferdinando Posted December 8, 2006 Author Share Posted December 8, 2006 Typically, 0 is considered to be neither positive or negative. When you want to include 0, you typically use the phrase "nonnegative" or "nonpositive" instead. However, it is important to keep in mind that many people do use the word "positive" to include 0, even though it isn't precisely correct. Link to comment Share on other sites More sharing options...

Ragib Posted December 9, 2006 Share Posted December 9, 2006 0^0 is formally undefined, but in certain fields such as measure theory, it is convenient to assume it is 1. And i'm very surprised no one has noticed that the equation you describe is transcendental, meaning there are not polynomials with rational coeffients of which the solutions are roots of. Essentially this means there is no analytical way of getting solutions, merely approximations. Link to comment Share on other sites More sharing options...

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