Question

Continuing what I left off yesterday.

Answer 1

@ganeshie8 urgent...

Answer 2

\[x_1 + x_2 + x_3 = 20\]where\[-3 \le x_1 , x_2 , x_3 \]

Answer 3

That reduces to the coefficient of \(x^{20}\) in\[(x^{-3} + x^{-2} + \cdots + x^{20})^3\]

Answer 4

How do I calculate the above?

Answer 5

@dan815 use elementary techniques please

Answer 6

When it comes to these, it really depends on what exactly you want to do with it. There's a few different ways I can play around. Be more precise

Answer 7

\[=\left( x^{-3}\cdot\dfrac{1-x^{30}}{1-x}\right)^3\]

Answer 8

\[= x^{-9}(1-x^{30})^3(1-x)^{-3 }\]

Answer 9

\[x^{-9} (1 - x^{90} - 3x^{30} + 3x^{60})(1-x)^{-3}\]

Answer 10

\[= (x^{-9} - x^{81}-3x^{21} + 3x^{51} )(1-x)^{-3}\]

Answer 11

now the coefficient of \(x^r\) in \((1 - x)^{-n}\) is\[\binom{n+r-1}{r-1}\]

Answer 12

you could also simply use stars and bars

Answer 13

\[x_1 + x_2 + x_3 = 20\]where\[-3 \le x_1 , x_2 , x_3 \]

is same as solving
\[(y_1+3) + (y_2+3) + (y_3+3) = 20\]where\[0 \le y_1 , y_2 , y_3 \]

Answer 14

hmm, yeah. is it \(y_1 + 3\) or \(y_1 - 3\)?

Answer 15

\(y_i=x_i+3\)
so yeah it should be \(y_i-3\)

Answer 16

ok, that's nice.\[\binom{n-1}{k-1}\]

Answer 17

watcha doing?

Answer 18

u want int solutions?

Answer 19

\[\binom{19}{2}\]

Answer 20

I really wasn't trying to solve the problem ... I just want to know the approach to calculate the coefficient of \(x^k\) in huge expressions like the above one.

Answer 21

id just use stars and bars

Answer 22

but sure ill bite, how does it come out to be the coeff of x^k again

Answer 23

oh sorry, I missed all the threes.

Answer 24

\[y_1 + y_2 + y_3 = 29\]

Answer 25

should the number of nonnegative solutions be\[\binom{29-1}{3-1}\]

Answer 26

oh nvm i get whwy its the coeff of x^20 in that expansion kinda neat trick

Answer 27

yeah, it is.

Answer 28

you would still use stars and bars to solve for the coeff i guess now, since theres no real nice way of expanding that

Answer 29

the way to think about this problem is the same as
x1+x2+x3=k
x1,x2,x3 >= 0

Answer 30

do you know how to solve that problem

Answer 31

allowed -3 what u basicaaly did is just allow more 1s in each place

Answer 32

by introducing -3 you just shifted your number index

Answer 33

ik, ik. there are various tricks to find the coefficient.

Answer 34

ok so watcha looking for now

Answer 35

ya x1+x2+x3=29 is right

Answer 36

for x1,x2,x3 non negative solutions

Answer 37

ok, so nonnegative solutions