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Details about the function n/ln(n)

  1. May 17, 2012 #1
    From the file attached I would like to know the following.
    if Y=n/ln(n), Is there a way of explicitly expressing n in terms of Y.


    Relations I found are:

    There are 2 values of n for every Y. Except at Y=e , the two values converge to n=e.
    If n1 and n2 are the values of n
    then
    n1^n2=n2^n1.

    So is there a way of finding n1, given n2?

    What could be the possible type of functions involved?
    From the graph it is seen that |n1-e| and |n2-e| are related inversely.
    What could be this relation?
     

    Attached Files:

  2. jcsd
  3. May 18, 2012 #2

    haruspex

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    No, there's no way to express n as a function of y using any standard functions. Of course, it's always possible to invent a new one for the purpose.
    Mixtures functions of different types (polynomial, exponential, logarithmic, trigonometric..) are nearly always impossible to invert. E.g. y = x.exp(x), y = sin(x)/x, ...
     
  4. May 20, 2012 #3

    Mute

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    There is a special function called the Lambert-W function (aka the Product-Log) which you can use to write n in terms of y. The Lambert-W function is the function w=W(x) such that
    x = w exp(w).

    If you invert your equation so that 1/y = ln(n)/n, and then let n = exp(a), this gives

    [tex]\frac{1}{y} = a e^{-a}[/tex]

    We see that if we multiply both sides by -1 this will be in Lambert-W form, giving a = W(-1/y). Inverting n = exp(a), this gives

    [tex]n = \ln W_k\left(-\frac{1}{y}\right).[/tex]

    Some very important notes: The Lambert-W function has two real-valued branches, corresponding to k = 0 and k = -1. Usually the k=0 branch corresponds to the desired solution. The other branches gives complex values for W, so inverting ln n = W(-1/y) is more complicated if want solutions for these others branches. However, since you seem mostly interested in real values you don't need to worry about this.
     
  5. May 22, 2012 #4
    Thanks 'Mute'. Got my way through.

    tuhinrao
     
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