Product of Power Series
Key Questions

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 timedependent Schroedinger equation is (example 24 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 socalled 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 wavesince 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! :)

#(sum_{n=0}^infty a_nx^n)cdot(sum_{n=0}^infty b_nx^n)=sum_{n=0}^infty(sum_{k=0}^na_kb_{nk})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_{nk})x^n#
Questions
Power Series

Introduction to Power Series

Differentiating and Integrating Power Series

Constructing a Taylor Series

Constructing a Maclaurin Series

Lagrange Form of the Remainder Term in a Taylor Series

Determining the Radius and Interval of Convergence for a Power Series

Applications of Power Series

Power Series Representations of Functions

Power Series and Exact Values of Numerical Series

Power Series and Estimation of Integrals

Power Series and Limits

Product of Power Series

Binomial Series

Power Series Solutions of Differential Equations