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.
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.