[Discrete Mathematics] Coefficient Extraction and Extended Binomial Theorem

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Recommends Makina · 2018-10 · Mathematics Stack Exchange · on “How many solutions are there to the equation $x_1+x_2+x_3+x_4+x_5=30$ with restrictions using generating functions?”

The theory is rather ... computationaly extensive. Check out these 2 videos, they should give a full explanation [LINK] [this video] being even means, that the generating function's part related to this x will be something like $(x^0 + x^2 + x^4 + x^6 +...)$ For odd - similar scenario: $x^1 + x^3 + x^5 + x^7 + ... $ the generating function for (a) would be correct if you removed the denominator. the generating function for (b) should look like this I believe: $(1+x+x^2+x^3+...)^3(x+x^3+x^5+x^7+...)(1+x^2+x^4+x^6+...) = x(1-x)^{-3}(1-x^2)^{-2}$

Recommends Aji Wibowo · 2021-10 · Mathematics Stack Exchange · on “A simple and more generalised way to define this coefficient”

I used this video reference [this video] The video said when you encounter with that, you must make all possible way of product of the binomial coefficient to reach x^n

Cites Antoni Parellada · 2017-01 · Mathematics Stack Exchange · on “How to extract coefficients of a generating function using a computer?”

The Cauchy product in R is handled with the discrete convolution [LINK] function: > (l= lapply(seq_len(4)-1, function(i) rep(c(1, integer(i)), length.out=6))) [[1]] [1] 1 1 1 1 1 1 [[2]] [1] 1 0 1 0 1 0 [[3]] [1] 1 0 0 1 0 0 [[4]] [1] 1 0 0 0 1 0 > round(Reduce(f = function(x, y) convolve(x, y, type = "open"), x = l)) $[1] \text{0 0 0 0 1 1 2 3 5 6 6 8 8 8 6 6}\bullet \color{blue}{5}\bullet \text{3 2 1 1}$ This is consistent with the results in Wolfram Alpha [LINK] : Input: $$\small (1+x+x^2+x^3+x^4+x^5) \times (1 + x^2 + x^4) \times (1 + x^3) \times (1 + x^4)$$ Output: $$\small x^{16}+x^{15}+2 x^{14}+3 x^{13}+5 x^{12}+6 x^{11}+6 x^{10}+8 x^9+8 x^8+8 x^7+6 x^6+6 x^5+\large{\color{blue}{5}}\s

Cites cosmicbrownie · 2018-10 · Mathematics Stack Exchange · on “How many solutions are there to the equation $x_1+x_2+x_3+x_4+x_5=30$ with restrictions using generating functions?”

How many solutions are there to the equation $x_1+x_2+x_3+x_4+x_5=30$ where $a) 2 \leq x_1 \leq 4$ , and $3 \leq x_i \leq 8$ , for all $2 \leq i \leq 5$ . $b) 0 \leq x_i$ for all $1 \leq i \leq 5$ , with $x_2$ even and $x_3$ odd. Attempt 1: So far with part $a$ I have this: $\frac{(x^2+x^3+x^4)(x^3+x^4+x^5+x^6+x^7+x^8)^4}{(1-x)}$ I'm struggling with how to go further from where I am at by using generating functions. I don't know how I would deal with the case where $x_2$ is even and $x_3$ is odd. Attempt 2: I've followed the videos posted before and here's where I'm at now: Part a: $[x^{30}](x^2+x^3+x^4)(x^3+x^4+x^5+x^6+x^7+x^8)^4=[x^{30}]x^2 \frac{1-x^3}{1-x} x^{12} (\frac{1-x^6}{1-x})^4 =