- 11-Sep-2013 --- Original version.

- Apply the standard Frobenius method, but now you need to expand the potential in a series as well. You only need a finite number of terms, so there's no need to try to determine a general expression: just work your way up until you have three nonzero terms. The condition on the wave function should lead you to a particular choice of the possible solutions.
**Fourier series for***f(x)=x*- For an expansion from 0 to L, equations (4.13), (4.14),
and (4.15) in the Lea excerpt give the expansion coefficients.
Mathematica makes quick work of the integrals! But
you need to use simplification (as in the fourier_series.nb
notebook on the 7701 Mathematica webpage)
to get the coefficients in simplest form. For example:
will give you the
`f[x_] := x`

an[n_] = FullSimplify[

2 Integrate[f[x]* Sin[2 Pi n x], {x, 0, 1}],

Assumptions -> {Element[{n}, Integers]}]

*a*coefficients._{n} - Now you want the range -L < x < L but you need to decide if you want the odd or even expansion, which takes you to equation (4.25) or (4.26)-(4.27). Be careful: those equations already take into account your choice (unlike those in part a) so that they can switch the integral back to the 0 to L interval. (You can also use Lea (4.16) and get the same answer.) If you use Mathematica (recommended!), be careful to use Sin[Pi n x] rather than Sin[2 Pi n x], etc.
- The Mathematica notebook fourier_series.nb has examples
of such plots.
It's efficient to define a function to make your plot if you've
defined the coefficients as above (also define b0 and the bn
coefficients) :
For similarities and differences, observe which expansion does better in different regions (e.g., at the origin), whether they exhibit the Gibbs overshoot, what values they have at a discontinuity, whether there is a constant term or not, and anything else you can think of! Try extending the range of your plots to -2 to 2. This shows you the periodic extension of each function.
`fs[x_, nmax_] := b0 + Sum[an[n]*Sin[n*2 Pi x] + bn[n]*Cos[n*2 Pi x], {n, 1, nmax}]`

Plot[{f[x], fs[x, 1], fs[x, 2], fs[x, 3]}, {x, 0, 1}]

- For an expansion from 0 to L, equations (4.13), (4.14),
and (4.15) in the Lea excerpt give the expansion coefficients.
Mathematica makes quick work of the integrals! But
you need to use simplification (as in the fourier_series.nb
notebook on the 7701 Mathematica webpage)
to get the coefficients in simplest form. For example:
**Complex exponential Fourier series.**Again, nothing tricky here if you use the correct formulas for expansion coefficients. Note that you are to use the exponential (complex) expansion intially, then afterward convert to sines and cosines. Mathematica has no trouble with the complex series, just use`I`for*i*.**Damped, perioically driven, harmonic oscillator.**The first step is to analyze*f(t)*as a Fourier series. Then the coefficients of the*x(t)*expansion follow algebraically. What type of expansion (exponential or sines and cosines) is best if you have both first and second derivatives? Can you show that your answer for*x(t)*is real?**Guitar string physics.**- You can follow the example in Lea's text, only
replace the initial shape of the string. Note that
the Fourier series can still be taken as just a sine
series because of the boundary conditions at
*x=0*and*x=L*. - Are there zero coefficients in the expansion?
How about when the string is plucked in the middle?
(Think about even and odd harmonics about the center
of the string.) Do the coefficients vanish for any
values of
*n*? (The Mathematica`Table`function is useful here to look at a range of*n*.) - See the fourier_series.nb notebook once again! You can take reasonable values (like unity) for variables like L and v.

- You can follow the example in Lea's text, only
replace the initial shape of the string. Note that
the Fourier series can still be taken as just a sine
series because of the boundary conditions at
**Fourier series for**. This is a straightforward application. You might try the complex expansion, but you can also do this just as in problem 1. If you used Mathematica as in the example code above, it's just a matter of changing the function! Example plots in fourier_series.nb.*f(x)=x*^{2}- One solution here should be apparent by inspection because
there are just derivatives
of
*y*with no term proportional to*y*. This should also come out from your analysis. I'm not actually sure how the other solution is supposed to work, since Frobenius seems to give identically zero if you use the recurrence relations! However, to get the point, just look at the recurrence relation for general*m*and*m+1*and apply a convergence test for the series (try section 2.3.1 of Lea for possible tests). What kind of singularity doesn't converge for any value of*x*when expanded about*x=0*? Mathematica can tell you the general solution (use DSolve). -
**Constraints on complex coefficients.**We already gave the answer to this in class, but the problem is to carry out the proof. It's straightforward: just take the complex conjugate and use the fact that*n*is a dummy summation variable.

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