Product of Power Series

Key Questions

What uses do products of power series have?

One example that I find useful is the use and manipulation of the products of power series to derive $e^(ix) = cosx + isinx$ , 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:

$y(x) = c_1 e^(alphax) + c_2 e^(-alphax)$

where $alpha = iomegat$ and, after various proofs and empirical tests, it is agreed that we can use $e^(alphax)$ as a working "trial function" when we guess the form of the overall solution in terms of a finite addition of these $c*e^(alphax)$ functions so that we can predict molecular properties:

$psi(x) = sum_(i=1)^N c_i phi_i(x)$
where each $phi$ could, for example, represent an atomic orbital, and $psi(x)$ 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):

$(d^2x(t))/(dt^2) + omega^2x(t) = 0$

subject to the boundary conditions $x(0) = A$ and $(dx(0))/(dt) = 0$ . 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:

$c_1e^(alphax) + c_2e^(-alphax)$

$= c_1(cosx + alphasinx) + c_2(cosx - alphasinx)$

$= c_1cosx + c_1alphasinx + c_2cosx - c_2alphasinx$

$= (c_1 + c_2)cosx + (c_1alpha - c_2alpha)sinx$

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 $c_1alpha - c_2alpha = c_4$ :

$= c_3cosx + c_4sinx$

Then, substituting $omegat$ for $x$ , we get:

$c_3cos(omegat) + c_4sin(omegat)$

for the solution to the so-called common example.

Looking at the boundary condition $(dx(0))/(dt) = 0$ , we get:

$= c_3(-sin(omegat))*omega + c_4cos(omegat)*omega$

$= cancel(c_3(-sin(omega(0)))*omega)^(0) + c_4cos(omega(0))*omega$

$= c_4omega$

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

$=> c_4omega = 0$ , thus satisfying the condition $(dx(0))/(dt) = 0$ .

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

$x(0) = c_3cos(omega(0)) + cancel(c_4sin(omega(0)))^(0)$

$= c_3$

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 $x(0) = A$ , and we just have:

$color(blue)(x(t) = Acos(wt))$

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

Truong-Son N. · 1 · Aug 23 2015
How do you find the product of power series?

$(sum_{n=0}^infty a_nx^n)cdot(sum_{n=0}^infty b_nx^n)=sum_{n=0}^infty(sum_{k=0}^na_kb_{n-k})x^n$

Let us look at some details.

$(sum_{n=0}^infty a_nx^n)cdot(sum_{n=0}^infty b_nx^n)$

by writing out the first few terms,

$=(a_0+a_1x+a_2x^2+cdots)cdot(b_0+b_1x+b_2x^2+cdots)$

by collecting the like terms,

$=a_0b_0x^0+(a_0b_1+a_1b_0)x^1+(a_0b_2+a_1b_1+a_2b_0)x^2+cdots$

by using sigma notation,

$=sum_{n=0}^infty(sum_{k=0}^na_kb_{n-k})x^n$

Wataru · · Sep 25 2014

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Power Series

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Product of Power Series

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Power Series