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Deriving Planck's law with Taylor series

  1. Sep 12, 2009 #1
    Expanding exp(hc / lambda*k_b * T) by Taylor series
    = 1 + hc /lambda*k_B * T +...

    But don't you take the derivative with respect to lambda? So I don't get how it would be this.
     
  2. jcsd
  3. Sep 12, 2009 #2
    [tex]e^x = 1 + x + x^2/2! + ... + x^n/n! + ...
    [/tex]
    if x<<1, then ignoring [tex] O(x^2) [/tex] and higher,

    [tex]e^x= 1 + x [/tex]

    there is no derivative involved.
     
  4. Sep 12, 2009 #3
    No, there is a derivative: the f^(n) term in the Taylor series evaluated at a point a.

    So wouldn't you need to take the derivative of the exponential function with respect to one of the variables, namely lambda?

    i.e. f' = - hcexp(hc / lambda k_B T) / lambda^2 k_B T => f'(0) = -hc/lambda^2 k_B T
     
  5. Sep 12, 2009 #4
    taylor expansion of [tex] f(x-a) = \sum_{n=0} ^ {\infin } \frac {f^{(n)}(a)}{n!} \, (x-a)^{n} [/tex]

    take a=0 and x=hc/lambda*K_b*T

    its the same.
     
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