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An interesting(or very stupid) doubt that came to mind

  1. Jun 26, 2013 #1
    I was dealing with the derivation of Stefan's law from Planck's law. In this derivation one gets to the following equation


    At this point most places I looked provide an aproximate numerical solution to this transcendental equation and carry on. But I wwas thinking about this. Suppose there is x that satisfies the equation. Then,




    The numerical solution they provide is less than 5, so we can use the geometrical series sum.


    Now, if I write the exponential series, and using that two power series must have the same coefficients i arrive at my problem. I thought a bit about it and since I can neither figure it out nor know exactly how to look for it I decided to ask it here. Maybe it is a stupid doubt and I am forgeting something. Either way, I would apreciate some help. Thanks.
  2. jcsd
  3. Jun 26, 2013 #2


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    They would have the same coefficients if that equation was true for every value of x. But it's only true for one value of x, so the coefficients can go crazy. As a stupid example, if you solve sin(x) = 0 you wouldn't expect the coefficients on both sides to be the same
  4. Jun 26, 2013 #3


    Staff: Mentor

    Office_Shredder makes a good point about the difference between an equation that is true only for specific values (a conditional equation) and one that is true for all values of the variable (an identity).

    When you solve a quadratic equation such as x2 - 4x = -3 you are solving for the two values of x that make the equation a true statement.

    OTOH, when you do partial fraction decomposition to turn 1/(x2 - 4) into the sum of two rational expressions of the form A/(x - 2) + B/(x + 2), you are finding the values of the constants A and B so that the equation 1/(x2 - 4) = A/(x - 2) + B/(x + 2) is true for all values of x (excepting, of course, 2 and -2).

    The equation you started with, ##\frac{xe^{x}}{e^{x}-1}=5##, is not identically true, so equating ex with a geometric series as you did is invalid.
  5. Jun 27, 2013 #4
    Hi !
    the method proposed by assed is not stupid at all if we consider the problem on the practical viewpoint (i.e. approximate numerical method, but not exact analytical method)
    But, as shown in attachment, it will require a very large limited expansion of the function to be accurate enough : That is the drawback.

    Attached Files:

  6. Jun 27, 2013 #5


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    First off I get x~4.965114231744276303698759131322893944056 so 4<x<5
    The exact solution can be expressed using the Lambert W function
    x = W(-5/e^5)+5
    you can use http://www.wolframalpha.com/input/?i=x+e^x/(e^x-1)=5

    As far as the proposed method I would describe the problem differently
    we desire to solve
    it is totally valid to write f as a power series and solve it for x
    the question is how to do so as in general I do not think much is gained
    As was mentioned above if you truncate the series the zeros move.
    You would need to read an approximation book about methods that preserve zeroes
    usual fixed point methods (like Newton's method and its variations) are probably a better approach
    Last edited: Jun 27, 2013
  7. Jun 27, 2013 #6
    Of course, I agree with lurflurf's remark.
  8. Jun 27, 2013 #7
    Thanks for the replies. As I expected, there was something I was missing. The first two answers showed the (rather stupid) mistake I was making, forgeting the x in my ratiocination was a specific number and not a variable and therefore there was no power series at all. Maybe I was sleepy or something. :smile:
    Anyway, thanks again for the answers.
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