Finding a Rule to Fit Given DataDate: 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/ |
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