Deriving Planck's law with Taylor series

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Homework Help Overview

The discussion revolves around deriving Planck's law using Taylor series expansions, specifically focusing on the exponential function involved in the expression.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the Taylor series expansion of the exponential function and its application to the problem. Questions arise regarding the necessity of taking derivatives with respect to the variable lambda and the implications of ignoring higher-order terms.

Discussion Status

The discussion is active, with participants presenting different interpretations of the Taylor series and its application. Some provide insights into the role of derivatives in the expansion, while others challenge these interpretations, leading to a productive exchange of ideas.

Contextual Notes

There appears to be some confusion regarding the treatment of the variable lambda in the context of the Taylor series expansion, as well as the assumptions made about the terms being neglected.

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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