Deriving Planck's law with Taylor series

jaejoon89
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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.
 
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e^x = 1 + x + x^2/2! + ... + x^n/n! + ... <br />
if x<<1, then ignoring O(x^2) and higher,

e^x= 1 + x

there is no derivative involved.
 
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
 
taylor expansion of f(x-a) = \sum_{n=0} ^ {\infin } \frac {f^{(n)}(a)}{n!} \, (x-a)^{n}

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

its the same.
 
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