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Thought of 2 Difficult Math Problems

  1. Oct 16, 2007 #1
    I was futzing around with some equations and i came across 2 problems that I'm having trouble solving.
    In the first one, you must solve for y as a function of x:
    [tex]y^x = x^y + 1[/tex]

    For the second one, I know that [tex]\sum_{ x=1}^\infty \frac{1}{x^2} = \frac{ \pi^2}{6}[/tex], but then what's the exact value of [tex]\sum_{ x=1}^\infty \frac{1}{x^3}[/tex]?
     
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  3. Oct 16, 2007 #2

    matt grime

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    I will define FOO(x) as the function such that FOO(X)^x=x^FOO(X)+1. Assuming this even exists.

    The second one is hard, I think (unknown?). Somewhat surprising, really, is that even exponents are 'easy' and odd ones are 'hard'.
     
  4. Oct 16, 2007 #3
    All you did here was change the representation of a variable. That doesn't make it any easier... You'd still have to solve it algebraically.
     
  5. Oct 16, 2007 #4

    arildno

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    Why?
    If matt, or somebody else takes the the trouble of proving the existence of Foo(x), the case is over and done with.
     
  6. Oct 16, 2007 #5
    But you haven't solved for y, so you don't know what FOO(x) actually is.
    For this problem, I was trying to put it in the form of y=blah, where "blah" is some function of x. Let me give you an example:
    If I say "solve this equation for y as a function of x: 2x+y=3," I mean that i want you to put it in the form: y=3-2x. Or, if I said "solve for y: y^x=3," I want it in the form: y=3^(1/x).
     
    Last edited: Oct 16, 2007
  7. Oct 16, 2007 #6

    arildno

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    Set
    [tex]g(x,y)=y^{x}-x^{y}-1[/tex]
    We see that g(1,2)=0, furthermore, we have:
    [tex]\frac{\partial{g}}{\partial{y}}=xy^{x-1}-x^{y}\ln(x)[/tex],
    whose value at (1,2) equals 1.

    Thus, foo(x) exists in a region about (1,2), and we have:
    [tex]\frac{dFoo(x)}{dx}=\frac{(Foo(x))^{x}\ln(Foo(x))-(Foo(x))x^{Foo(x)-1}}{x(Foo(x))^{x-1}-x^{(Foo(x))}\ln(x)}[/tex]
    This is a differential equation for Foo(x).
     
  8. Oct 16, 2007 #7
    Well, that helps, but it still isn't the answer, because I still don't know what FOO(x) is. (I only know how to do the most basic of differential equations.)
     
  9. Oct 16, 2007 #8

    arildno

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    So?
    Foo is perfectly well-defined now, through the initial condition Foo(1)=2.
     
  10. Oct 16, 2007 #9
    But I didn't want to define Foo, i wanted the equation to be put in the form y=blah, with only x's on the "blah" side. That's what i was asking.
     
  11. Oct 16, 2007 #10

    arildno

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    Note, by the way, that the line x=0 is a different solution to the equation as well..
     
  12. Oct 16, 2007 #11

    arildno

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    Well, in a neighbourhood of (1,2), Foo exists, and within that neighbourhood, you may determine the function value of Foo at any point to any desirable degree of accuracy using numerical techniques.

    What more do you want?
     
  13. Oct 16, 2007 #12
    Yeah, but if you put domain restrictions on it, you can still put it in the form of y=blah.
     
  14. Oct 16, 2007 #13

    arildno

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    With the initial condition Foo(1)=2, this is enough.
    To be nice to you, I'll show you how to find the best linear approximation to Foo close to x=1, i.e, compute the 2-term Taylor series about x=1:

    Now, we readily see that [tex]\frac{dFoo}{dx}\mid_{x=1}=\frac{2\ln(2)-2}{1}=2\ln(\frac{2}{e})[/tex]

    Thus, we have:
    [tex]y=Foo(x)\approx{2}+2\ln(\frac{2}{e})(x-1)[/tex]

    EDIT:
    I made the wrong sign on the derivative of Foo, meaning that our 2-term Taylor expansion of Foo should be:
    [tex]y=Foo(x)\approx{2}+2\ln(\frac{e}{2})(x-1)[/tex]
     
    Last edited: Oct 16, 2007
  15. Oct 16, 2007 #14

    CRGreathouse

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    The irrational number zeta(3) = 1.2020569... is Apéry's constant. No closed form is known, but the first 2 billion (at least) decimal places have been calculated by Howard Cheng, Guillaume Hanrot, Emmanuel Thomé, Eugene Zima, and Paul Zimmermann.
     
  16. Oct 16, 2007 #15
  17. Oct 17, 2007 #16

    matt grime

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    And you don't know what sin(3) is, but I suspect you think that that y=sin(x) is a proper type of solution, don't you?

    http://www.dpmms.cam.ac.uk/~wtg10/equations.html
     
  18. Oct 17, 2007 #17

    CompuChip

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    For some functions, this is just not possible. For example, the error function is defined through an integral equation, and we can describe it's properties, but we cannot write it in closed form.
    Same goes for the Bessel function, which is defined as a (specific) solution to some differential equation, and countless others (of which some have special names, like [itex]J, B_i, A_n[/itex] or [itex]\operatorname{Foo}(x)[/itex]).
     
  19. Oct 17, 2007 #18

    HallsofIvy

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    What do you mean by "actually is"? What "actually is" the Lambert W function, W(x), which is defined as the inverse of the function f(x)= xex? What "actually is" the Bessel function of the first order, which is defined as one of the two independent solutions to Bessel's differential equation? For that matter, what "actually is" sin(x) or cos(x) which can be defined in a variety of ways, none of whichy give you an easy formula in which you can just "plug in" a value for x?

    That's exactly what matt grime did: it is now in the form y= Foo(x), where Foo is some function of x.

    Do you understand that the great majority of functions ("almost all" functions) CANNOT be written in such a way? That they are NOT "rational functions" nor "radical functions" nor even "transcendental functions"?
     
  20. Oct 17, 2007 #19
    Yes, because it is in the form of y=blah. I just wanted it in a form where I could plug any number in for x and instantly know what y is, and the only way I know how to do that is to put it in the form y=blah, where there are only x's on the blah side. Also, I do kind of know what sin(3) is, because I can approximate it with an infinite sum. I did not know how to approximate FOO(x) when it was in the form originally given to me.

    If that is the case, then it should at least be proved that the equation I gave you can't be put into the form y=blah (where blah is some function I can approximate on a calculator or using some infinite sum) before you resort to defining "blah" as "FOO(x)."
     
    Last edited: Oct 17, 2007
  21. Oct 17, 2007 #20

    matt grime

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    That really makes no sense. Being transcedental, or not expressible in terms of basic functions has nothing to do with being able to approximate f(x) for any given value of x. I can do that with the FOO function too, or with ERF, or the Lambert function. You really have the wrong idea here.
     
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