Finding a Rule to Fit Given Data

Date: 11/14/2005 at 22:26:36
From: Rhonda
Subject: solve for j (what math process will you use:add,subtract,etc

Given that 1J1 = 2, 3J5 = 34, 6J9 = 117, and 10J14 = 296, conjecture a
value for 3J8.  Justify your answer.

For 1J1, the obvious answer is addition but that doesn't fit the rest
of the problem.  I tried to multiply and divide and square it and cube 
it and I cannot come up with the answer.

Date: 11/15/2005 at 09:42:44
From: Doctor Peterson
Subject: Re: solve for j (what math process will you use:add,subtract,etc

Hi, Rhonda.

Without knowing the context of the problem (what you have been 
learning about, other similar problems you have seen), I'll guess that 
J represents an unknown binary operation that involves combining its 
operands in some simple way using the basic operations.

Rather than look at one case, I would look at how changing the 
operands affects the result. I'll start by making a table:

   a   b   aJb
  ---+---+----
   1 | 1 |   2
   3 | 5 |  34
   6 | 9 | 117
  10 |14 | 296

It would be nice if one operand were the same between two cases, so we 
could see how changing one variable affected the result. But I notice 
that going from 1,1 to 3,5, we've multiplied the two variables by 3 
and 5 respectively, and multiplied the result by 17, which is close to 
3*5. So I suspect that the product ab is involved in the definition of 
J. Let's add a column to the table for ab:

   a   b   ab   aJb
  ---+---+----+----
   1 | 1 |  1 |   2
   3 | 5 | 15 |  34
   6 | 9 | 54 | 117
  10 |14 |140 | 296

Hmmm... aJb seems to be close to twice ab; let's look at the 
difference between the two:

   a   b  2ab   aJb  aJb-2ab
  ---+---+----+-----+-------
   1 | 1 |  2 |   2 |   0
   3 | 5 | 30 |  34 |   4
   6 | 9 |108 | 117 |   9
  10 |14 |280 | 296 |  16

Interesting! Can you take it from here, and think of a way to 
calculate aJb from a and b?

Notice my strategy: I am trying to approach the answer step by step, 
by making a guess and comparing the results with the goal, then 
adjusting my guess to take the error into account.

If you need more help, please write back and show me how far you got.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 11/15/2005 at 22:00:19
From: Rhonda
Subject: solve for j (what math process will you use:add,subtract,etc

Ok, I have a little more of an understanding.  I see that in the table
a increases by aJb - 2ab (by adding the 1 and 3, then the 3 and 
6, then the 6 and 10).  I also notice that aJb - 2ab increases by the 
next square number (25 would come next in the table).  After this I 
have tried to do 3ab, look for square roots (ab times 2 plus square 
root of 4).  I still cannot find a value for aJb and especially can't 
see how to fit 3J8 into the table.

Thank you so much!

Rhonda

Date: 11/15/2005 at 22:38:04
From: Doctor Peterson
Subject: Re: solve for j (what math process will you use:add,subtract,etc

Hi, Rhonda.

I'm not sure what you have in mind when you talk about how something
"increases"; you should be thinking of each row by itself, not of
connections between rows, because the values they've chosen for a and
b are meant only to be sample values, not a specific sequence.  I can
see how the numbers tempt one to think that way, but you'll want to
avoid the temptation!  We have to be able to take ANY two numbers a 
and b, and calculate something called aJb.

What I had in mind was that aJb - 2ab is a square--the square of what? 
Can you see some way to get that number from the other numbers in the 
same row?  Once you do, you can say that the formula for aJb is 2ab 
plus that new expression.

Once you have that formula, you can use it to find 3J8.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 11/15/2005 at 22:49:58
From: Rhonda
Subject: solve for j (what math process will you use:add,subtract,etc

You are fabulous!  I think I get it now.  aJb is: 2ab + the square of
(b-a)?  So 3J8 would be 2*3*8 + (8-3)^2 = 48 + 25 = 73?

Wow!  Thank you so much!

Date: 11/15/2005 at 23:35:52
From: Doctor Peterson
Subject: Re: solve for j (what math process will you use:add,subtract,etc

Hi, Rhonda.

That's what I got:

  aJb = 2ab + (b-a)^2

But now we can simplify that algebraically; it turns out to be
surprisingly simple!  I'm sure there are other ways we could have come
across it more directly.

It's worth noting, also, that the problem was worded just right: all
we can really do is to make a conjecture, based on the data we have;
the simplicity of the answer convinces us that it is what a teacher
would have come up with, but there are many more complicated formulas
that would also make the same results.  What we're doing here is more
like science than math: we look at some data, work out what formula
might lie behind it, and then design an experiment to see if it's
right.  Our experiment is to submit the conjecture that 3J8 = 73, and
see whether we get a good grade.


- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/