Chronology |
Current Month |
Current Thread |
Current Date |

[Year List] [Month List (current year)] | [Date Index] [Thread Index] | [Thread Prev] [Thread Next] | [Date Prev] [Date Next] |

*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.

You can solve this in your head.

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

right answer, regardless of mass?

Now, to make things interesting: can you come up with *two*

independent qualitative arguments?

Here are my answers:

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.

_______________________________________________

Phys-L mailing list

Phys-L@electron.physics.buffalo.edu

https://www.physics.buffalo.edu/mailman/listinfo/phys-l

- Prev by Date:
**[Phys-L] Re: suggestions for 1-semester textbook** - Next by Date:
**[Phys-L] LabRats** - Previous by thread:
**[Phys-L] Re: suggestions for 1-semester textbook** - Next by thread:
**[Phys-L] LabRats** - Index(es):