hi, i helped a neighbour's son with a difficult question the other day.

this is the gist of the question.. i have figured out it myself, half youtube'd it.

and the question was a polynomial with 3 degrees, not the 2 here.

so here is x and y

x = 0 1 2 3..... y = 23, 24, 29 38

first change in y = 1 5 9

2nd change in y = 4 4.......

therefore 2nd deriv of A*x^2 = 2*A... therefore 2*A = 4... A = 2...

i could do it again for the x..........

so the polynomial = 2* x^2 - x +23............

that seems like a really nifty trick to come up with the 2nd derivative two different ways and then solve for A.....

thanks in advance........ just look for what this is called.. now that i think about, probably solving a polynomial (but i confuse that with roots of a polynomial)

and there's another really painful "by hand" way to solve this........

You might be looking for the term "finite calculus" or "the calculus of finite differences."

Aaron, thx so much

surprised this is high school or AP.... of course, i'm not actually sure what year the person i helped is in.

This problem can be done as a system of linear equations. Let f(x) = ax^2 + bx + c:

f(0) = c = 23
f(1) = a + b + c = 24
f(2) = 4a + 2b + c = 29

Substitute the first value into the second two equations:

a + b = 1
4a + 2b = 6

And it's just a two variable linear system that you can solve by inspection and conclude that a = 2 and b = -1.

I don't see the relationship between the derivatives and the differences shown.


PairTheBoard

I suspect that it was a made up example and that he didn't actually do the work for the specific numbers he gave when he said he worked out the x part. But that doesn't mean that it doesn't actually work.

Let D(f) = f(x+1) - f(x). Notice that this calculate the difference between two consecutive terms in a sequence defined by f. We can quickly calculate that D(f + g) = D(f) + D(g) and D(kf) = kD(f) (where k is a constant), and also calculate D(k) = 0, D(x) = 1, and D(x^2) = 2x + 1. (The +1 is the thing that I'm not sure whether he was doing correctly.)

Consider the function defined by the following points (and assume that the solution is a quadratic polynomial).

We can compute the forward differences:

By assumption, we have that f(x) = ax^2 + bx + c for some integers a, b, and c.

Notice that D(ax^2 + bx + c) = a(2x + 1) + b, so that D^2(ax^2 + bx + c) = 2a. Since D^2(f)(x) = 4 for all x, we get that a = 2.

We can now substitute back into the equation to get D(f) = 4x + 2 + b. When x = 0, we get D(f)(0) = 1 = 4(0) + 2 + b. This implies b = -1.

Lastly, we see that f(0) = c, which implies that c = 23.

Excellent.

PTB

aaron and PTB, thx for the responses.

i did realize you could just solve it as a bunch of basic linear equations.. but it was x^3 in the example i actually did (not posted here). and it can get very messy very fast.

what i had figured out my own is this:

say it is 5x^2+x+2..... hypothetical

so x and f(x) = 0,1,2,3,4.... and 2, 8, 24, 50, 86...

first difference in f(x) = 6,16,26,36...

second difference = 10, 10, 10

what i didn't figure out is that the second difference is crudely the same as the second derivative of Ax^2... first derivative = 2Ax.. 2nd derivative = 2A..

so A = 5.......... you haven't proven that every point between all the x's follows this relationship. but you are just trying to fit those 5 points



so plug that in and come up with new f(x). and then solve the same way again. i think that's described correctly...... you can probably do it visually at this level.