Defining Quadratic Formula
Date: 01/21/2002 at 14:31:51 From: Monica Kerai Subject: Defining quadratic formula Here's what I understand so far: a+b+c: 5 8 12 17 23 ... 3a+b: 3 4 5 6 ... 2b: 1 1 1 ... You figure out what a, b, and c are, and then you put them into this equation: an^2 + bn + c. Where do I get the (a+b+c) (3a +b) (2b) part from? Please help me.
Date: 01/22/2002 at 14:08:41 From: Doctor Peterson Subject: Re: Defining quadratic formula Hi, Monica. I think you meant 2a, not 2b. As you'll see, you're more likely to remember it right if you see where it comes from. What you are doing is taking successive differences in a sequence. When they all turn out to be zero, you know that the sequence is a polynomial of that degree; in this case, the third difference will be zero, so the sequence represents a quadratic equation. Suppose you have an arbitrary quadratic sequence x[n] = an^2 + bn + c Then your differences will look like this: n=1 n=2 n=3 a+b+c 4a+2b+c 9a+3b+c 3a+b 5a+b 2a Those (on the left) are the expressions you were told to use. You can simplify the work a bit by separating out each power, since if you add two sequences together, the differences will be the sum of the differences of the two sequences. Look at these cases: an^2 + bn + c a 4a 9a 16a b 2b 3b 4b c c c c 3a 5a 7a b b b 0 0 0 2a 2a 0 0 0 So the first difference row will start with 3a+b, and the second difference row with 2a. The constant term (c) has no effect on any differences; the linear term (b) drops out after the first difference, and so on. That helps in remembering or reconstructing the formula. See what you get if you add on a cubic term. We have some discussions of the method of finite differences in our archives, which may help clarify the process; start here: Method of Finite Differences http://mathforum.org/dr.math/problems/gillett.10.12.00.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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