Deriving Planck's law with Taylor series

1. Sep 12, 2009

jaejoon89

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. Sep 12, 2009

elduderino

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

$$e^x= 1 + x$$

there is no derivative involved.

3. Sep 12, 2009

jaejoon89

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

4. Sep 12, 2009

elduderino

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.