A note on Taylor series and the Wien law

Just a brief note with a comment on a mistake I saw in a number of blog posts on deriving the Wien displacement law from the blackbody distribution.  Many of you used a Taylor series for \(e^u-1\) to solve the transcendental equation
$$5=\frac{ue^u}{e^u-1}$$.

You argued that by Taylor series, \(e^u-1\approx (1+u)-1=u\) so that the equation becomes, expanding both top and bottom,
$$5=\frac{u(1+u)}{(1+u-1)}=1+u\to u=4.$$

However, this isn't right.  A Taylor series is only valid inside the "radius of convergence":  fancy language for "small displacements from the point you are expanding about."  This should be no surprise: for any function, if you approximate it by a line, that will only be a good approximation locally (unless the function is just a line!)

More formally, a Taylor series says
$$f(x+\Delta x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!}(\Delta x)^n$$

and the expansion for \(e^u-1\) above takes only the n=0 and n=1 terms.  But consider \(\Delta x >1\):   then the powers of \(\Delta x\) above grow with n, and the higher n terms contribute more.  In short, the series does not converge: you have gone outside the radius of convergence.  This is all to say that you cannot use a Taylor expansion of \(e^u-1\) about u=0 for a value of u near 4!

Indeed, the correct logic was to recongnize that for u larger than, say, 2, \(e^u\gg 1\) so the denominator can be written as approximately \(e^u\), canceled with part of the numerator, and the correct answer,
$$u=5,$$

determined.