Continuing what I left off yesterday.

- ParthKohli

Continuing what I left off yesterday.

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- schrodinger

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- ParthKohli

@ganeshie8 urgent...

- ParthKohli

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

- ParthKohli

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

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## More answers

- ParthKohli

How do I calculate the above?

- ParthKohli

@dan815 use elementary techniques please

- Compassionate

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

- ParthKohli

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

- ParthKohli

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

- ParthKohli

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

- ParthKohli

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

- ParthKohli

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

- ganeshie8

you could also simply use stars and bars

- ganeshie8

\[x_1 + x_2 + x_3 = 20\]where\[-3 \le x_1 , x_2 , x_3 \]
is same as solving
\[(y_1+3) + (y_2+3) + (y_3+3) = 20\]where\[0 \le y_1 , y_2 , y_3 \]

- ParthKohli

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

- ganeshie8

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

- ParthKohli

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

- dan815

watcha doing?

- dan815

u want int solutions?

- ParthKohli

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

- ParthKohli

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.

- dan815

id just use stars and bars

- dan815

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

- ParthKohli

oh sorry, I missed all the threes.

- ParthKohli

\[y_1 + y_2 + y_3 = 29\]

- ParthKohli

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

- dan815

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

- ParthKohli

yeah, it is.

- dan815

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

- dan815

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

- dan815

do you know how to solve that problem

- dan815

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

- dan815

by introducing -3 you just shifted your number index

- ParthKohli

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

- dan815

ok so watcha looking for now

- dan815

ya x1+x2+x3=29 is right

- dan815

for x1,x2,x3 non negative solutions

- ParthKohli

ok, so nonnegative solutions|dw:1444026999220:dw|

- dan815

hahahaha

- ParthKohli

eh, that thing was just meant to illustrate stars-and-bars for x1 + x2 + x3 = 4

- dan815

ya i know

- ParthKohli

hey shut up

- dan815

but yes parth

- dan815

heres a very easy way to work it out

- ParthKohli

\[\binom{29+3-1}{3-1}\]

- dan815

suppose
x1,x2,x3 = 20
but u allow no 0 x1,x2,x3>=1

- dan815

what does this really mean?

- dan815

that means we must start with 1 in x1 x2 x3

- dan815

1 +k1+ 1 + k2 +1 +k3 = 20
so your new zero beginning equation is now
k1+k2+k3=17

- dan815

now u see how we went from the 1 case to 0 case

- dan815

same logic from 0 to -3 case

- dan815

just number indices here they are just symbols

- ParthKohli

i know what's happening here jesus

- dan815

dont sweat

- dan815

swear

- ParthKohli

ok so by stars and bars, it turns out to be 465
however http://www.wolframalpha.com/input/?i=%28x%5E%28-3%29+%2B+x%5E%28-2%29+%2B+...+%2B+x%5E%2820%29%29%5E3 says x^20's coefficient is 402

- ParthKohli

ohhh I got where I went wrong lol

- Astrophysics

the n?

- Astrophysics

Oh wait..

- Astrophysics

\[\binom{29+3-1}{3-1} \] this should be fixed no?

- ParthKohli

i need \((x^{-3} + x^{-2} + \cdots + x^{26})^3 \)

- ParthKohli

yay 465 is right baby

- ParthKohli

http://www.wolframalpha.com/input/?i=%28x%5E%28-3%29+%2B+x%5E%28-2%29+%2B+...+%2B+x%5E%2826%29%29%5E3+expansion

- Astrophysics

Yeah see it should be \[\binom{29+3-1}{29}\] ye?

- ParthKohli

well both of them are equal so it doesn't even matter

- Astrophysics

Ah I see, you're too good Mr. Kohli!

- ParthKohli

how many solutions to \(x+y+z+t=29\) when \(x\ge 1, y\ge 2, z\ge 3, t\ge0 \)

- dan815

same thing

- ParthKohli

again we transform in the following manner\[x=a+1\]\[y=b+2\]\[z=c+3\]

- dan815

yea

- ParthKohli

how many solutions to \(x_1 + x_2 + x_3 + 4x_4 = 20\)

- ParthKohli

coefficient of \(x^{20}\) in\[(1 + x + \cdots + x^{20})^3(1 + x^{4}+x^8+\cdots + x^{20}) \]

- ParthKohli

looks like you're back

- ParthKohli

I'm looking at a sample problem where they removed \(1-x^{21}\) because it doesn't contribute to \(x^{20}\).

- ParthKohli

so can you use that trick here? I haven't got the real feel for it.

- ParthKohli

\[\frac{\left(1 - x^{21}\right)^3 }{(1-x)^{3}} \cdot \frac{1- x^{20}}{1-x^{4} }\]

- ParthKohli

now can we remove \((1-x^{21})^3\) because it doesn't contribute to \(x^{20}\)?

- ParthKohli

\[(1-x)^{-3} \left(1 + x^4 +x^8 + x^{12}+x^{16}+x^{20}\right)\]

- ParthKohli

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

- ParthKohli

or is it the opposite... coefficient of \(x^n\) in \((1-x)^{-r}\) is \(\binom{n+r-1}{r-1}\)

- ParthKohli

\[\frac{1}{(1-x)^s}=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^k \]

- ParthKohli

\[ \frac{1}{(1-x)^s}=\sum_{k=0}^{\infty} \binom{s+k-1}{k }x^k \]

- ganeshie8

\[\dbinom{-n}{r} = (-1)^r\dbinom{n+r-1}{r-1}\]

- ParthKohli

that's nice

- ParthKohli

i'm appearing for an all-india test tomorrow and i'm really nervous :(
so many expectations!
how do I get the best out of myself?

- ganeshie8

\[(1-x)^{-s}=\sum_{k=0}^{\infty} (-1)^k\binom{s+k-1}{s-1}x^k \]
right ?

- ParthKohli

hmm, but that \((-1)^k\) shouldn't be there because there's another \((-1)^k\) attached to \(x^k\) so they cancel out in a way

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