What uses do products of power series have?

Aug 23, 2015

One example that I find useful is the use and manipulation of the products of power series to derive ${e}^{i x} = \cos x + i \sin x$, which is an identity used many, many times to solve the Schroedinger Equation in Physical Chemistry, by substituting $i$ for various different constants.

What is accepted by Physical Chemists is that you can write out the general solution to the Equation as:

$y \left(x\right) = {c}_{1} {e}^{\alpha x} + {c}_{2} {e}^{- \alpha x}$

where $\alpha = i \omega t$ and, after various proofs and empirical tests, it is agreed that we can use ${e}^{\alpha x}$ as a working "trial function" when we guess the form of the overall solution in terms of a finite addition of these $c \cdot {e}^{\alpha x}$ functions so that we can predict molecular properties:

$\psi \left(x\right) = {\sum}_{i = 1}^{N} {c}_{i} {\phi}_{i} \left(x\right)$
where each $\phi$ could, for example, represent an atomic orbital, and $\psi \left(x\right)$ would in that case be the molecular orbital.

A common example of solving the time-dependent Schroedinger equation is (example 2-4 in Physical Chemistry: A Molecular Approach):

$\frac{{d}^{2} x \left(t\right)}{{\mathrm{dt}}^{2}} + {\omega}^{2} x \left(t\right) = 0$

subject to the boundary conditions $x \left(0\right) = A$ and $\frac{\mathrm{dx} \left(0\right)}{\mathrm{dt}} = 0$. These boundary conditions define the fact that a stationary transverse wave with one antinode has two endpoints, and these are at $x = 0$ and $x = l$, half of the wavelength.

To solve this one, one would have to use identity written at the top, with $\alpha$ substituted for $i$ like so:

${c}_{1} {e}^{\alpha x} + {c}_{2} {e}^{- \alpha x}$

$= {c}_{1} \left(\cos x + \alpha \sin x\right) + {c}_{2} \left(\cos x - \alpha \sin x\right)$

$= {c}_{1} \cos x + {c}_{1} \alpha \sin x + {c}_{2} \cos x - {c}_{2} \alpha \sin x$

$= \left({c}_{1} + {c}_{2}\right) \cos x + \left({c}_{1} \alpha - {c}_{2} \alpha\right) \sin x$

and it is generally written out by absorbing the arbitrary constants ${c}_{1}$, $\alpha$, and ${c}_{2}$ into new arbitrary constants ${c}_{3}$ and ${c}_{4}$, with ${c}_{1} + {c}_{2} = {c}_{3}$ and ${c}_{1} \alpha - {c}_{2} \alpha = {c}_{4}$:

$= {c}_{3} \cos x + {c}_{4} \sin x$

Then, substituting $\omega t$ for $x$, we get:

${c}_{3} \cos \left(\omega t\right) + {c}_{4} \sin \left(\omega t\right)$

for the solution to the so-called common example.

Looking at the boundary condition $\frac{\mathrm{dx} \left(0\right)}{\mathrm{dt}} = 0$, we get:

$= {c}_{3} \left(- \sin \left(\omega t\right)\right) \cdot \omega + {c}_{4} \cos \left(\omega t\right) \cdot \omega$

$= {\cancel{{c}_{3} \left(- \sin \left(\omega \left(0\right)\right)\right) \cdot \omega}}^{0} + {c}_{4} \cos \left(\omega \left(0\right)\right) \cdot \omega$

$= {c}_{4} \omega$

But we know that at $t = 0$, $\omega = 0$ because time has not passed yet.

$\implies {c}_{4} \omega = 0$, thus satisfying the condition $\frac{\mathrm{dx} \left(0\right)}{\mathrm{dt}} = 0$.

Using the $x \left(0\right) = A$ boundary condition we get:

$x \left(0\right) = {c}_{3} \cos \left(\omega \left(0\right)\right) + {\cancel{{c}_{4} \sin \left(\omega \left(0\right)\right)}}^{0}$

$= {c}_{3}$

with ${c}_{3}$ taken as $A$---which is the amplitude of an initialized stationary wave---since it is the only contributor to the wave. Thus, since ${c}_{3} = A$, we have satisfied the condition $x \left(0\right) = A$, and we just have:

$\textcolor{b l u e}{x \left(t\right) = A \cos \left(w t\right)}$

which is the familiar physics equation for a transverse wave, as depicted in the image above! :)