Continuing what I left off yesterday.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Continuing what I left off yesterday.

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

@ganeshie8 urgent...
\[x_1 + x_2 + x_3 = 20\]where\[-3 \le x_1 , x_2 , x_3 \]
That reduces to the coefficient of \(x^{20}\) in\[(x^{-3} + x^{-2} + \cdots + x^{20})^3\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question