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# [Phys-L] qualitative reasoning +- parametric methods

• From: "John S. Denker" <jsd@AV8N.COM>
• Date: Sun, 06 Mar 2005 22:23:40 -0500

Hi --

Here's a little trick that is not as well known as it ought to
be. It is useful for quickly getting a qualitative feel for
a problem, and often almost as quickly getting an exact answer.

First I will illustrate the method on a simple math problem.
Then I will apply it to a physics problem that came up on the
list last week.

1) Suppose you are asked to find the sum of the first N
multiples of three.
N=1 S=3
N=2 S=3+6
N=3 S=3+6+9

Now presumably most of you can answer this in less time
than it takes to ask the question. But suppose you were a
student who didn't know the trick for this particular series,
or suppose I had chosen a more-challenging example ... the
point here is the method, not the result.

Anyway, the method goes like this. We know the sum is bounded
above by the integral of a straight line. That is, draw the
graph that has N on one axis and piecewise-flat steps 3N high
in the other direction. Draw a line that passes through the
left end of each tread (i.e. the top of each riser). The area
under the step-function is the quantity we seek, and it is
bounded above by the area under the line.

We also know the sum (S) is bounded below by the area under another
line, i.e. the line passing through the right end of each tread
(i.e. the bottom of each riser).

For both lines, we know the area (as a function of N) is going to
be some polynomial of degree 2. Since S is bounded above and
below by such things, you don't have to be a rocket scientist to
guess that S itself is some polynomial of degree 2. (You can
actually prove that, but I won't bother.)

So we write
S = a + b N + c N^2
and all we have to do is determine the parameters (a, b, and c).

The parameter (a) is a gimme, since S=0 when N=0.

Since S=3 when N=1, we have
3 = b + c

And since S=9 when N=2, we have
9 = 2 b + 4 c

At this point we have two linear (!) equations in two unknowns.

Now ... as a second task, consider the sum of the _squares_ of
the first N multiples of 3. There is not (so far as I know) a
handy trick for doing this sum (as there is for arithmetic series),
so the parametric method is (as far as I know) easiest way to
attack this problem. You get a cubic, with four parameters to
be determined. Again one is trivial. You are left with three
linear equations in three unknowns.

This approach (slightly generalized) plays a huge role in other
fields including pattern recognition and machine learning.

Computer-math programs (like Mathematica, Macsyma, Maple ....)
use parametric methods for doing indefinite integrals.

=====================

Now you may be wondering what this has got to do with physics.

In the context of a ball of mass m entering a box of mass M
and bouncing around, on 02/26/05 23:20, Carl Mungan wrote:

In the above example, the box is
stationary half the time *regardless* of the values of m and M.

This makes a tasty exercise in qualitative reasoning.

At first glance, the result is highly nontrivial. In the
general case, the two halves of the motion involve unequal
velocities and unequal distances. So it might seem like
quite a coincidence if the two times come out equal.

The first question is, without writing down any equations, can
you come up with a good argument why "half the time" is the

Now, to make things interesting: can you come up with *two*
independent qualitative arguments?

1) Switch to the center-of-mass frame. The CoM is stationary. The
world line of the ball is then completely symmetric; in effect it
just bounces off the CoM, so its reflected velocity equals its
incident velocity. Ditto for the box. And nonrelativistically
speaking it's clear that timing in the CoM frame is the same as in
the lab frame.

2) You can figure this out, qualitatively, without switching to the
CoM frame. Observe that "half" is the right answer in the limit
where m << M. The ball just bounces back in forth in the essentialy
stationary box. The same line of reasoning applies in the limit
where m >> M. The box just bounces back and forth, bracketing the
essentially stationary ball.

Interestingly, the same thing happens when m is just the same as
M. This result should be familiar from playing caroms or playing
with the Newton's cradle machine. Half the time m is moving, and
half the time M is moving, with the same velocity in each case.

Now let's think about the parametric method. We needn't assume
the behavior of the system is governed by a polynomial ... indeed
the velocities etc. are given by rational functions, not polynomials.
But there are limits to how complicated the function can be.

We are looking for a function that has the value "half" when m/(m+M)
goes to zero, has the value "half" when m/(m+M) goes to 1, and has
the value "half" in the middle, where m/(m+M)=1/2.

We conclude that either the function we are looking for is insanely
complicated, or it is a constant, equal to "half".

Now for the zinger: I know of a third qualitative argument. It is
arguably not 100% independent of the previous two, but it is simpler
than the first and more convincing than the second.

Pedagogical note: Students find it really hard to find multiple
solutions to the same problem. For some reason, knowledge of the
first solution puts them into a rut that interferes with finding
the second solution. Doing the exercise interactively, in class,
makes the process seem easier than it is, because different
people come up with different solutions. At the other extreme,
assigning it as a do-your-own-work exercise is problematic ...
the trick is to motivate them to not give up. Don't give up!
Don't give up! Don't give up! Finding the second solution is
hard, but worthwhile.

For one thing, knowing how to get yourself out of a rut is
more-or-less a prerequisite for any kind of original creative
thinking.

For another thing, the harder a problem, the more important it
is to work the problem in two independent ways, as a cross-check
to increase reliability. Suppose you are doing a reeeally hard
problem, and lives depend on getting it right. You want to do
it at least twice.

The good part is that when you have (finally!) taught them to do
some out-of-the-box thinking, many of them come to enjoy it.
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