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Cell[CellGroupData[{
Cell[TextData[{
 StyleBox["Phys 263",
  FontSlant->"Italic"],
 " Notebook for 20 May 09"
}], "Title",
 CellChangeTimes->{{3.451810900551*^9, 3.451810934407*^9}}],

Cell[TextData[{
 "The goal for the day is to work through as much of the homework as we can.\n\
Your job is to evaluate each cell, think through why you expected the answer\n\
",
 StyleBox["Mathematica",
  FontSlant->"Italic"],
 " gave (or why it didn't do what you expected) and typically to play\nwith \
the cell, e.g. changing parameters.\nYou may find it useful to create your \
own notebook, copying cells out this one.\nAnd of course you can come back \
later and finish reading through as you do the homework."
}], "Text",
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Cell[CellGroupData[{

Cell["Morin 4.35", "Section",
 CellChangeTimes->{{3.451749363558*^9, 3.451749380788*^9}, {
  3.4517533232539997`*^9, 3.451753323612*^9}, {3.451811033073*^9, 
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Cell["\<\
asks about the stadard issue 3 spring / 2 mass problem where the masses are \
driven
by an external force which just happens to go at 2\[Omega] with twice as much \
force on one of them\
\>", "Text",
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  Axes->True,
  AxesOrigin->{0, 0},
  PlotRange->{{0, 12}, {-1, 1}},
  PlotRangeClipping->True,
  PlotRangePadding->{Automatic, Automatic}]], "Input"],

Cell["Here is the drawing function which created the above:", "Text",
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Cell["\<\
Here are the equations of motion, including damping and forcing:\
\>", "Text",
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Cell["And ask for the solution:", "Text",
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Cell["\<\
Maybe it will look simpler when we pick boundary conditions, say starting at \
zero at rest:\
\>", "Text",
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Cell["\<\
Which part of this solution solution is the \"particular\" solution, i.e. the \
piece which comes from the driving
force and has to be there in every solution?  And which piece is the \
\"homogeneous\" solution, i.e. the modes
that would be there even without a driving force?
Your task: find boundary conditions which excite only the \"particular\" \
solution.
For the homework you want to work \"by hand\", taking as an ansatz a motion \
which wiggles
with frequency 2\[Omega] and seeing what works.
Can you think of a way to solve this system for a general force?  Try solving \
for the motion
when the force is the *same* on both particles, and when the forces are \
*opposite*.
What happens when the driving frequency is more general?
To explore, you may find it useful to just solve things numerically, as \
below:\
\>", "Text",
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Cell["\<\
on page 276 asks about the quadratic form (i.e. quadratic function of the \
variables)\
\>", "Text",
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Cell["To find the critical point (there is only one):", "Text",
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Cell["\<\
Note that means we can find the matrix of second derivatives at this point \
(actually, the matrix 
of second derivatives is the *same* for any point):\
\>", "Text",
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Cell["\<\
The character of the critical point is determined by the Hessian:\
\>", "Text",
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Cell["\<\
Are the eigenvalues both positive (in which case we have a minimum), both \
negative
(max) or opposite in sign (saddlepoint)?\
\>", "Text",
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Cell["So now you won't be surprise to see the following:", "Text",
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Cell["\<\
The character of such matrices is quickly determined by taking the trace and \
the determinant,
once we know that the trace is the sum of eigenvalues and the determinant is \
the product.
The job is to verify this statement \"by hand\".  Here let's let the machine \
do the algebra.
Generalizing to the class of Hermitian matrices (which includes real \
symmetric matrices):\
\>", "Text",
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Cell["Check out the sum and product", "Text",
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Cell["\<\
Thus if our Hessian has negative determinant we know the product of \
eigenvalues is negative,
i.e. one is positive and one is negative, and we have a saddle point.  To \
distinguish between
maxima and minima (both of which have positive det) we have to look at the \
trace: positive
trace means both eigenvalues are positive, i.e. a local minimum, and negative \
trace means max.\
\>", "Text",
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}, Open  ]],

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Cell["BTM 9.6.6", "Section",
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Cell["\<\
ask you to use the insight of BTM 9.6.6 as you find and analyze the critical \
points of the functions\
\>", "Text",
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Cell["\<\
To get you going, here is the function in Figure 9.4 (page 276):\
\>", "Text",
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Cell["We can look at this in 3D as well.", "Text",
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Cell["To find the critical points (now there are several)", "Text",
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Cell["Make a list of the points and then plot them as red dots:", "Text",
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Cell["\<\
To add infcrmation on the eigendirections, we construct the Hessian at each \
point:\
\>", "Text",
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Cell["The direction vectors at one of the critical points are:", "Text",
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Cell["To draw the vectors we need more functions:", "Text",
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Cell["\<\
Your job: add p4 and p5, i.e. vectors for the other two points,
and then produce analagous plots for the other target functions.\
\>", "Text",
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}, Open  ]],

Cell[CellGroupData[{

Cell["BTM 9.7.1", "Section",
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Cell["\<\
asks you to check that these functions are properly normalized, i.e. their \
absolute squares
integrate to 1 over the domain (here the interval from 0 to L).\
\>", "Text",
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Cell["\<\
The problem only asks about n=1 and n=2, but we can be general (and so can \
you, by
writing the sin(\[Theta]) as (1/2I)(Exp[I \[Theta]]-Exp[-I \[Theta]]) ):\
\>", "Text",
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Cell["Likewise we check orthogonality:", "Text",
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Cell["\<\
but more often than not we choose the basis of periodic functions:\
\>", "Text",
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Cell["Check that we normalized them correctly:", "Text"],

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Cell["\<\
Your task: verify they are orthogonal for general integers m and n.\
\>", "Text",
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Cell["Choose an interval, e.g. -1 to 1:", "Text"],

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Cell["Define normalized periodic basis functions:", "Text"],

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Cell["Check that they are orthogonal etc.:", "Text"],

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Cell["Pick a function (evaluate one or the other of these)", "Text",
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Cell["Compute the Fourier series:", "Text"],

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Cell["Note that the n=0 mode has to be treated as a limit:", "Text",
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Cell["\<\
Define the n-th approximation by summing the modes from -n to n:\
\>", "Text",
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Cell["The value of at x=1 is1 :", "Text",
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Cell["\<\
On the other hand the Fourier series evaluated at that point is the sum of \
terms:
\
\>", "Text",
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Cell["\<\
Note we are double counting the n = 0, but we can fix it with:\
\>", "Text",
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Cell["\<\
The sum of all these numbers is f(1), i.e. 1, or equivalently, the sum of the \
following is \[Pi]^2/4:\
\>", "Text",
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Cell["\<\
Since \[Pi]^2/4 - \[Pi]^2/6 = \[Pi]^2/12, we've just computed the following \
sum:\
\>", "Text",
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Cell["BTM 9.7.3 etc.", "Section",
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Cell["Asks for the same game to be applied to:", "Text",
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Cell["\<\
Before plotting, take a look at f(x) in your mind's eye: what does it look \
like for x<L/2?
what is the value f(L/2)?  what does it look like for x>L/2?   Now peek:\
\>", "Text",
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Note we can simplify this further when m's are known to be integers:\
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Note that this expression still isn't right for m=0, so we fill in the gap by \
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And here is the approximation with just the three contributions from \
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Cell["\<\
Finally, noting that f[L/2] = h, we learn how to sum another infinite series, \
just along
the lines of Sum[1/n^2] above.\
\>", "Text",
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Cell["BTM 9.7.5", "Section"],

Cell["Drive an LRC circuit with this v[t]:", "Text"],

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Cell["\<\
And take a look.  q[t] itself looks kind of normal, but the current q'[t] has \
an
interesting hiccup at the time the applied voltage jumps:\
\>", "Text",
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To solve by Fourier series, resolve the applied voltage in a sum of exp()'s.
Then recall that for one pure frequency \[Omega], i.e. v=vn Exp[I \[Omega]n \
t]  the solution for q(t) is:\
\>", "Text",
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Cell["\<\
Your task: resolve v[t] in to a sum Sum[vn Exp[I \[Omega]n t]], i.e. find the \
Fourier coeffiecients vn.
Then q[t] can be found as Sum[qn[t]].  Taking one time derivative we then get \
the current,
which according to the back of BTM should be:\
\>", "Text",
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Cell["\<\
Note that after a dozen or so modes even that little kink starts to become \
apparent:\
\>", "Text",
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Cell["And ask for the coeffients for our v[t]= v0 t/T:", "Text",
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This is the general expression, but has a problem at m=0, where we want\
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Pausing to take a look at how the series does for our target function:\
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Notice how our approximants have to be continuous, smooth and periodic, so \
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Notice this says the average charge is c V0/2, as appropriate for an applied \
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Taking a time derivative, the current is then a sum over all modes except m=0 \
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