How can i find coefficient of $x^{51}$ in expansion of $$(x-1)(x^2-2)(x^3-3)\ldots (x^{10}-10)?$$ I tried all methods and formulae but couldn't find the right answer.
Asked
Active
Viewed 1,235 times
1
-
1is the third term $(x^3-3)$? – Eleven-Eleven Dec 15 '13 at 14:04
-
Pari states -1, but you surely want to have a formula ... – Peter Dec 15 '13 at 14:11
-
? f=prod(j=1,10,x^j-j);print(polcoeff(f,51)) – Peter Dec 15 '13 at 14:14
-
What about finding all partitions of 51 with the numbers 1 to 10, only available once ? This would give the necessary terms. – Peter Dec 15 '13 at 14:16
-
Peter how did u get it???whats pari? – The Champ Dec 15 '13 at 14:19
-
PARI is a computational tool. The comment commencing with ? is the input line. – Peter Dec 15 '13 at 14:20
-
@user115908 Pari/GP – Pece Dec 15 '13 at 14:20
-
2The title is misleading (the answer to that question would be: compute the binomial coefficient); in fact this question has nothing to do with the binomial theorem. – Marc van Leeuwen Dec 15 '13 at 14:26
2 Answers
5
Note that $1+2+\cdots+10=55$.
When you multiply out this polynomial, you will get a sum of terms; each term is the result of choosing either one term or the other from each binomial, and multiplying all of them together.
So, you get an $x^{51}$ by choosing $x^{m}$ terms whose exponents add up to $51$, and using the constant terms for the others.
Now, because $1+2+\cdots+10=55$, you can equivalently consider this as choosing to use constant terms in place of $x^m$'s where the $m$'s add up to $4$. How many different ways can you choose a subset of $\{1,2,3,\ldots,10\}$ that add up to $4$? (Not very many!)
Nick Peterson
- 32,430
-
-
@user115908 Well, then it gets harder! It is likely not a coincidence that you weren't asked that. If the expression was of a nicer variety, you could use generating function methods; you could also take derivatives (probably using logarithmic differentiation) and use them to find the Taylor series for this function about $x=0$. – Nick Peterson Dec 15 '13 at 14:31
-
@NicholasR.Peterson. I also think that Taylor series is really the best for the more general problem. – Claude Leibovici Dec 15 '13 at 14:33
-
-
@user115908 It's possible, but also significantly more complicated! You need to compute the $51$st derivative of that product at $x=0$. – Nick Peterson Dec 15 '13 at 14:40
2
To expand on the answer by Nicholas R. Peterson, you can perform the following operations: put $y=x^{-1}$, multiply each factor $x^d-d$ by$~y^d$ (giving $1-dy^d$) and now the term in the expansion of the product of degree $55-k$ in $x$ has become a term of degree $k$ in$~y$. So concretely you are asking for the coefficient of $y^4$ in the product $\prod_{d\geq1}(1-dy^d)$; this is easy to compute.
I have made the product infinite; this does not change the coefficients in degrees${}\leq10$ (so it still gives an answer to your question), because the infinite product is mathematically more interesting. Nevertheless, it does not seem that interesting; OEIS knows nothing about its sequence of coefficients $1,-1,-2,-1,-1,5,1,13,4,2,-8,-61,-31,13,-156,21,11,223,\ldots$
Marc van Leeuwen
- 115,048