Taylor series representation of Planck's law

Question

I am to find the Taylor series for Planck's law, $$\frac{8\pi \:hc\lambda ^{-5}}{e^{\frac{hc}{\lambda \:kt}}-1}$$ T is temperature, h, c, and k are constants, and $$\lambda $$ is wavelength. I am supposed to show that Planck's law gives an approximation of Raleigh-Jeans law at large wavelengths. first derivative is $$\frac{40\pi \:kt\lambda \:^{-4}}{e^{\frac{hc}{kt\lambda \:}}}$$ (as far as I can tell) The next, I think, is $$\frac{160\pi \:k^2t^2}{che^{\frac{ch}{kt\lambda }}\lambda ^3}$$ But I am unsure of what value to set $$ a $$ to, in the formula $$ f(x)=f(a)+f'(a)/1!*(x-a)+f''(a)/2!*(x-a)^2... $$ Do I use a Maclaurin series? would that mean setting $$\lambda$$ to 0? Is $$ \lambda$$ equivalent to a?

Answer 1

Here is a summary of the comments:

1. Large wavelengths means that $\lambda$ is much larger than $0$. You need to expand in the limit $\lambda\rightarrow\infty$.

2. Obviously the MacLaurin expansion will have issues in this case, since you will have terms as powers of$(x-\infty)$. If you need to go along this route, your variable should be $1/\lambda$, which in the case of long wavelengths goes to $0$. You have then $x=1/\lambda$ and $$f(x)=\frac{8\pi h c x^5}{e^{\frac{hc}{kT} x}-1}$$

3. You need to calculate the derivatives correctly

The simplest way to solve this problem is to note that for small $x$, $e^{\frac{hc}{kT} x}\approx 1+\frac{hc}{kT} x$