What uses do products of power series have?

1 Answer
Aug 23, 2015

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.

http://www.physicsclassroom.com/

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! :)