I used polynomials and specifically special symmetric polynomials extensively during my Phd research (mathematical physics).

For instance symmetric polynomials (x_1^2 + x_2^2 + ...) can describe the state of a system of particles where exchanging any 2 particles does not change the system at all (exchange x_1 and x_2 in the previous expression, same polynomial = same system & state).

If you have a system of equation that you can solve exactly with special polynomials, you can approximate real world system governed by similar equations by using your special polynomials as a starting point and adding a correction to your solution.

There's so much to say about polynomials, but I'll leave you with a basic example that shows how multiplying infinite polynomials allow you to count the number of ways there are to hand you back your change at the till:

The basic units of change are 0.01$, 0.05$, 0.10$, 0.25$, 1$, 5$, 10$, ... For example, you can always give exact change back with only 0.01$.

So the set of all the different amount of change you can produce with 0.01$ is given by the exponents in the following

sum_n>=0 (q^(0.01))^n = q^0 + q^0.01 + q^0.02 + ... q^348.47 + ...

Now we can get all the amounts you can generate with 5 cents:

sum_k>=0 (q^(0.05))^k = q^0 + q^0.05 + q^0.10 + .... and so on for 25cents, 1$, ...

Notice now that given the multiplication properties of polynomials, that multiplying:

(sum_n (q^0.01)^n) * (sum_k (q^0.05)^k) * (sum_l (q^0.10)^l) Will give you all the different amounts you can generate with 0.01, 0.05 and 0.10.

For instance with n=5, k=1, l=0 you get 0.10$

q^(0.01 ^ 5) * q^(0.05 * 1)

You can get 0.10$ with n=10

q^(0.01 * 10)

You can get 0.10$ with l=1

q^(0.10 * 1)

Finally you can get 0.10$ with k=2 q^(0.05 * 2)

So when you multiply

(sum_n (q^0.01)^n) * (sum_k (q^0.05)^k) * (sum_l (q^0.10)^l)

altogether, you get

1 + ... + q^(0.01 * 5) * q^(0.05 * 1) + q^(0.01 * 10) + q^(0.10 * 1) + q^(0.05 * 2) + ... = ... + 4 q ^ (0.10)

There are thus 4 ways of handing back exactly 10 cents.

So for any amount, you take the following: product(c in (0.01, 0.05, 0.10, 0.25,...) (sum_n (q^c)^n) = sum_(a >= 0.01) [Number Of Way To Give Back Change for amount `a`] * q^(a)

So that would be the "generating series" of the number of ways to hand back change.

In this context, polynomials bridge the gaps between combinatorics and analytical computation.

user:	lav42
created:	May 9, 2024
karma:	6