## Tuesday, September 11, 2012

### NBI Week 4 -- Finding a Polynomial

Here's a classic problem: Find a polynomial that passes through the points $(1, 12)$ and $(12, 9230)$.

I realized today that for years I've been thinking "I like math because there is only one answer to every problem" — but that's entirely untrue! For one thing, I actually like ambiguities a lot of the time, and enjoy hearing how other people interpret things differently from me. Secondly, a whole host of problems have way more than one answer! Sure, you could say "the one solution" to this question consists of the infinitely many polynomials going through these points, but then you end up saying something like "Math problems have one answer, except sometimes they have multiple answers, but then those are just one answer" (real quote I have probably said myself).

And could that be any more confusing to a new math learner?

Perhaps a better view is to say every response to a math problem is decidedly either correct or incorrect — of course, this isn't strictly true either, but for most high school math it will be. (Aside: What is definitely true is that if you have a collection of statements, there is a way to do mathematics where all the statements are true [if any of them contradict each other, this follows since you can prove anything from an inconsistent system; if none contradict each other, we can create a maximally consistent set of formulae which includes all these, then factor out cleverly to create a Henkin Model where these statements all hold]).

Back to the problem at hand, we can narrow down number of solutions by adding a few extra restrictions. For instance, if we further say the polynomial must be of degree 1, there is exactly one solution. What is that solution? Let's program it: I can use the factor $(x-1)$ to 'turn off' parts of my equation — whenever $x=1$, this term will evaluate to $0$, so anything I choose to multiply it by won't contribute to the overall equation. So, using $(x-1)(X) + 12$ for any (equation) $X$ will get me the first point of $(1, 12)$. Similarly, $(x-12)(X) + 9230$ will get me a point of $(12, 9230)$. I can thread those together to get $(x-1)(9230) + (x-12)(12)$, but that's not quite it, since now when $x=1$, the overall equation has the value $(1-12)(12)$, which is $(1-12)$ times what I wanted! But this can easily be fixed by scaling the term appropriately: use $(x-1)(9230) + \frac{1}{(1-12)}(x-12)(12)$ and the output is again $12$ when $x=1$; we can fix the output for $x=12$ in a similar fashion, arriving at the final solution $p(x) = \frac{9230}{11}(x-1)-\frac{12}{11}(x-12)$.

I love this method for finding polynomials passing through certain points. Notice how easy it is to see there are infinitely many quadratics passing through those two points: the term $(x-1)(x-12)(X)$ will allow $X$ to contribute to the output only when $x$ is neither $1$ nor $12$. So we can just add in this term to the linear equation we had! Since $X$ can be absolutely anything, there are infinitely many quadratics through these points!

What's that? You wanted to see how many $n^{th}$ degree polynomials go through the points $(1, 12)$ and $(12, 9230)$? Just use our linear model $\frac{9230}{11}(x-1)-\frac{12}{11}(x-12)$ to make sure the points are hit, then add in the 'off switch' multiplied by $ax^{n-2}$. That is, $p(x) = \frac{9230}{11}(x-1)-\frac{12}{11}(x-12) + (x-1)(x-12)(ax^{n-2})$. We can let $a$ be arbitrary, so again we see there are infinitely many solutions!

But I digress.

Restricting the degree of the polynomial is one easy way to whittle down the number of solutions. Another way to force just one solution is to restrict the types of coefficients. If we allow only positive integer coefficients...

I'll leave it for you to explore the 'why', but let's write $9230$ in base $12$. $(9236)_{10} = (5412)_{12}$, and wouldn't you know it? With $p(x) = 5x^3 + 4x^2 + x + 2$, $p(1) = 12$ and $p(12) = 9230$.