ParthKohli
  • ParthKohli
Continuing what I left off yesterday.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
@ganeshie8 urgent...
ParthKohli
  • ParthKohli
\[x_1 + x_2 + x_3 = 20\]where\[-3 \le x_1 , x_2 , x_3 \]
ParthKohli
  • 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
  • ParthKohli
How do I calculate the above?
ParthKohli
  • ParthKohli
@dan815 use elementary techniques please
Compassionate
  • 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
  • ParthKohli
\[=\left( x^{-3}\cdot\dfrac{1-x^{30}}{1-x}\right)^3\]
ParthKohli
  • ParthKohli
\[= x^{-9}(1-x^{30})^3(1-x)^{-3 }\]
ParthKohli
  • ParthKohli
\[x^{-9} (1 - x^{90} - 3x^{30} + 3x^{60})(1-x)^{-3}\]
ParthKohli
  • ParthKohli
\[= (x^{-9} - x^{81}-3x^{21} + 3x^{51} )(1-x)^{-3}\]
ParthKohli
  • ParthKohli
now the coefficient of \(x^r\) in \((1 - x)^{-n}\) is\[\binom{n+r-1}{r-1}\]
ganeshie8
  • ganeshie8
you could also simply use stars and bars
ganeshie8
  • 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
  • ParthKohli
hmm, yeah. is it \(y_1 + 3\) or \(y_1 - 3\)?
ganeshie8
  • ganeshie8
\(y_i=x_i+3\) so yeah it should be \(y_i-3\)
ParthKohli
  • ParthKohli
ok, that's nice.\[\binom{n-1}{k-1}\]
dan815
  • dan815
watcha doing?
dan815
  • dan815
u want int solutions?
ParthKohli
  • ParthKohli
\[\binom{19}{2}\]
ParthKohli
  • 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
  • dan815
id just use stars and bars
dan815
  • dan815
but sure ill bite, how does it come out to be the coeff of x^k again
ParthKohli
  • ParthKohli
oh sorry, I missed all the threes.
ParthKohli
  • ParthKohli
\[y_1 + y_2 + y_3 = 29\]
ParthKohli
  • ParthKohli
should the number of nonnegative solutions be\[\binom{29-1}{3-1}\]
dan815
  • dan815
oh nvm i get whwy its the coeff of x^20 in that expansion kinda neat trick
ParthKohli
  • ParthKohli
yeah, it is.
dan815
  • 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
  • dan815
the way to think about this problem is the same as x1+x2+x3=k x1,x2,x3 >= 0
dan815
  • dan815
do you know how to solve that problem
dan815
  • dan815
allowed -3 what u basicaaly did is just allow more 1s in each place
dan815
  • dan815
by introducing -3 you just shifted your number index
ParthKohli
  • ParthKohli
ik, ik. there are various tricks to find the coefficient.
dan815
  • dan815
ok so watcha looking for now
dan815
  • dan815
ya x1+x2+x3=29 is right
dan815
  • dan815
for x1,x2,x3 non negative solutions
ParthKohli
  • ParthKohli
ok, so nonnegative solutions|dw:1444026999220:dw|
dan815
  • dan815
hahahaha
ParthKohli
  • ParthKohli
eh, that thing was just meant to illustrate stars-and-bars for x1 + x2 + x3 = 4
dan815
  • dan815
ya i know
ParthKohli
  • ParthKohli
hey shut up
dan815
  • dan815
but yes parth
dan815
  • dan815
heres a very easy way to work it out
ParthKohli
  • ParthKohli
\[\binom{29+3-1}{3-1}\]
dan815
  • dan815
suppose x1,x2,x3 = 20 but u allow no 0 x1,x2,x3>=1
dan815
  • dan815
what does this really mean?
dan815
  • dan815
that means we must start with 1 in x1 x2 x3
dan815
  • dan815
1 +k1+ 1 + k2 +1 +k3 = 20 so your new zero beginning equation is now k1+k2+k3=17
dan815
  • dan815
now u see how we went from the 1 case to 0 case
dan815
  • dan815
same logic from 0 to -3 case
dan815
  • dan815
just number indices here they are just symbols
ParthKohli
  • ParthKohli
i know what's happening here jesus
dan815
  • dan815
dont sweat
dan815
  • dan815
swear
ParthKohli
  • 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
  • ParthKohli
ohhh I got where I went wrong lol
Astrophysics
  • Astrophysics
the n?
Astrophysics
  • Astrophysics
Oh wait..
Astrophysics
  • Astrophysics
\[\binom{29+3-1}{3-1} \] this should be fixed no?
ParthKohli
  • ParthKohli
i need \((x^{-3} + x^{-2} + \cdots + x^{26})^3 \)
ParthKohli
  • ParthKohli
yay 465 is right baby
ParthKohli
  • 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
  • Astrophysics
Yeah see it should be \[\binom{29+3-1}{29}\] ye?
ParthKohli
  • ParthKohli
well both of them are equal so it doesn't even matter
Astrophysics
  • Astrophysics
Ah I see, you're too good Mr. Kohli!
ParthKohli
  • ParthKohli
how many solutions to \(x+y+z+t=29\) when \(x\ge 1, y\ge 2, z\ge 3, t\ge0 \)
dan815
  • dan815
same thing
ParthKohli
  • ParthKohli
again we transform in the following manner\[x=a+1\]\[y=b+2\]\[z=c+3\]
dan815
  • dan815
yea
ParthKohli
  • ParthKohli
how many solutions to \(x_1 + x_2 + x_3 + 4x_4 = 20\)
ParthKohli
  • ParthKohli
coefficient of \(x^{20}\) in\[(1 + x + \cdots + x^{20})^3(1 + x^{4}+x^8+\cdots + x^{20}) \]
ParthKohli
  • ParthKohli
looks like you're back
ParthKohli
  • ParthKohli
I'm looking at a sample problem where they removed \(1-x^{21}\) because it doesn't contribute to \(x^{20}\).
ParthKohli
  • ParthKohli
so can you use that trick here? I haven't got the real feel for it.
ParthKohli
  • ParthKohli
\[\frac{\left(1 - x^{21}\right)^3 }{(1-x)^{3}} \cdot \frac{1- x^{20}}{1-x^{4} }\]
ParthKohli
  • ParthKohli
now can we remove \((1-x^{21})^3\) because it doesn't contribute to \(x^{20}\)?
ParthKohli
  • ParthKohli
\[(1-x)^{-3} \left(1 + x^4 +x^8 + x^{12}+x^{16}+x^{20}\right)\]
ParthKohli
  • ParthKohli
now the coefficient of \(x^r\) in \((1-x)^{-n}\) is\[\binom{n+r-1}{r-1}\]is that right?
ParthKohli
  • ParthKohli
or is it the opposite... coefficient of \(x^n\) in \((1-x)^{-r}\) is \(\binom{n+r-1}{r-1}\)
ParthKohli
  • ParthKohli
\[\frac{1}{(1-x)^s}=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^k \]
ParthKohli
  • ParthKohli
\[ \frac{1}{(1-x)^s}=\sum_{k=0}^{\infty} \binom{s+k-1}{k }x^k \]
ganeshie8
  • ganeshie8
\[\dbinom{-n}{r} = (-1)^r\dbinom{n+r-1}{r-1}\]
ParthKohli
  • ParthKohli
that's nice
ParthKohli
  • 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
  • ganeshie8
\[(1-x)^{-s}=\sum_{k=0}^{\infty} (-1)^k\binom{s+k-1}{s-1}x^k \] right ?
ParthKohli
  • 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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