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.

    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! :)

    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

Questions

Power Series