According to Zettili, we can derive Wien's displacement law from Planck's energy density
$$
\tilde{u}(\lambda, T) = \frac{8\pi hc}{\lambda^{5}}\frac{1}{e^{hc/kT\lambda}-1}
$$
where $\lambda$ is the wavelength and $T$ is the temperature.
Taking the derivative of this, setting it equal to zero, and rearranging terms, one gets
$$
\frac{\alpha}{\lambda} = 5(1-e^{-\alpha / \lambda})
$$
where $\alpha = hc/(kT)$. To solve this, Zettili goes on to say that we can write
$$
\frac{\alpha}{\lambda} = 5-\epsilon
$$
so that the previously mentioned equation becomes
$$
5-\epsilon = 5-5e^{-5+\epsilon}
$$
Apparently, $\epsilon \approx 5e^{-5}$ is a good answer. My first question is why do we use $5-\epsilon$ to rewrite the equation? What method is this? I've never seen that before and other resources just say that they got the values by solving numerically. My second question is why $\epsilon \approx 5e^{-5}$ when there is an extra factor of $e^{\epsilon}$ in the equation? Why can we ignore that factor?
Best Answer
Well this is sort of a Taylor expansion. Let us assume we do not take 5 from the beginning, what is the method behind? We want to solve something like: $$x=5(1-\mathrm e ^{-x})$$ Ok then, let us assume the solution is close to some value $x_0$ that we might guess, but only differs by a small $\epsilon\ll1$, i.e. $x=x_0-\epsilon$. We then have: $$x_0-\epsilon=5(1-\mathrm e ^{-x_0+\epsilon})$$ We then obviously have $$\epsilon=x_0-5+5\mathrm e ^{-x_0}\mathrm e^ \epsilon$$ with $\mathrm{e}^\epsilon\approx 1$. We need $\epsilon$ to be small, because that was our assumption in the first place.The value of $e ^{-x_0}$ is probably a small number, but what about $x_0-5$? Well this is small if $x_0=5$, as in this case this is zero. The result then must provide self consistently that $\epsilon$ is small. Otherwise the ansatz was wrong, but this is OK here. So it works.
Note that this requires a large factor for the parenthesis. 5 is OK, it would have been less precise for, e.g., 3 and would completely fail for 1.
To continue the idea let's see what we can do in case of 3. The True numerical solution for $x=3(1-\mathrm e^{-x})$ is $x=2.82144...$ but the simple idea from above gives $\epsilon=0.149...$, i.e. $x=3-0.149...=2.8506...$ so we are notably off. Our mistake is the assumption that $\mathrm e^\epsilon =1$ but with $0.149$ this is $1.16...$, i.e. $16\%$ off. So lets make a Taylor expansion of this. $\mathrm e ^\epsilon\approx 1+\epsilon$. Plugging this into the above for a factor of 3 we get
$$x_0-\epsilon=3-3\mathrm e ^{-x_0}(1+\epsilon)$$ as before (but now with 3) setting $x_0=3$ and solving for $\epsilon$ we get $$\epsilon=\frac{3 \mathrm e^ {-3}}{1-3 \mathrm e^{-3}}= 0.175...$$ and $3-\epsilon=2.8244$, reducing the error from $1.2\%$ to $0.4\%$
And just because it is sort of fun, let me even improve. Let's approximate $\mathrm{e}^{\epsilon}\approx 1+\epsilon+\epsilon^2/2$. This results in a quadratic equation for $\epsilon$ which gives us:
$$\epsilon= \frac 1 a -1-\sqrt{(\frac 1 a -1)^2-2}; \quad a=3\mathrm{e}^{-3}$$
The above result is actually an approximation of this as (just to practice Taylor series a little bit) $$\begin{eqnarray} \epsilon&= &(\frac 1 a -1)-\sqrt{(\frac 1 a -1)^2-2}\\ &= &(\frac 1 a -1)-(\frac 1 a -1)\sqrt{1-\frac{2}{(\frac 1 a -1)^2}}\\ &= &(\frac 1 a -1)\left(1-\sqrt{1-\frac{2}{(\frac 1 a -1)^2}}\right)\\ &\approx &(\frac 1 a -1)\left(1-\left(1-\frac{1}{(\frac 1 a -1)^2}\right)\right)\\ &=&\frac{1}{(\frac 1 a -1)}=\frac{a}{1-a} \end{eqnarray}$$ In any case, plugging in the value of $3$, the quadratic approximation gives $3-\epsilon=2.82162$, which is $0.006\%$
If you would have done this with the factor $5$ you'd get for the exact solution $ x= 4.96511423...$ and for the approximation $x\approx 4.96511448...$ providing a relative error of $5\cdot 10^{-8}$