Hey.

Question

Hey.

parthkohli · Answer

Why is the number of solutions to the following equation:$$x_1 + x_2 + \cdots + x_k=n$$given $a_i <x_i < b_i $ equal to the coefficient of $x^{n}$ in the expansion of$$\prod_{i=1}^{n} \sum_{k=a_i}^{b_i}x^{k}$$

imqwerty · Answer

number of solutions =$$x_{i} $$?

parthkohli · Answer

Number of solutions = ordered pairs.

parthkohli · Answer

@ganeshie8

parthkohli · Answer

yeah

parthkohli · Answer

ah, kinda makes sense to me now.

imqwerty · Answer

ahhgh that given expression is so bad..

parthkohli · Answer

I understood why. Now as an example if we want to calculate solutions of$$x_1 + x_2  +x_3=20$$subject to $x_i \ge -3$ then$$(x^{-3} + x^{-2} + \cdots + x^{26})^3$$coefficient of $x^{20}$

parthkohli · Answer

$$\left(x^{-3}\cdot \left(\frac{1 - x^{30}}{1 - x}\right)\right)^{3}$$

parthkohli · Answer

^ how do we calculate the coefficient of $x^{20}$ in that?

imqwerty · Answer

i didn't get that 2nd step? 
(x^-3 +x^-2..x^26)^3

parthkohli · Answer

ah, the lower restriction is given to us: $x_1, x_2, x_3 \ge -3$ since their sum is $20$, the max. value any of them can take is $26$ (if and only if the other two are -3)
thus $-3 \le x_i \le 26$

parthkohli · Answer

now refer to the above identity

imqwerty · Answer

:o ok now i get the ques :)

parthkohli · Answer

but how do we calculate the coefficient? :(

parthkohli · Answer

ah, since $1 -x^{30}$ cannot generate $x^{20}$ we can remove that and find the coefficient of $x^{20}$ in $x^{-9} \cdot \dfrac{1}{(1-x)^3}$

parthkohli · Answer

I'm not sure about the above step... can you confirm it?

imqwerty · Answer

wait what was that equation where did it go? http://prntscr.com/8nkh3q

parthkohli · Answer

maybe there are displaying problems... refresh?

ganeshie8 · Answer

generating functions