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Say we have something like $$ (x+2+y)^{23} $$

How does one go about finding the co efficient of say $$ x^6y^7 $$

DillPixel
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4 Answers4

5

Consider the entire expansion $(x+2+y)(x+2+y) \ldots (x+2+y)$. You want the term $x^6 y^7$, which means that you have to pick 6 $x$'s and 7 $y$'s. Thus, you need to pick $23-6-7$ 2's.

Hence, the term is given by

$$ {23 \choose 6} {23-6 \choose 7} { 23-6-7 \choose 23-6-7} \times x^6 \times y^7 \times (2)^{23-6-7}$$

It get's slightly more interesting when the base has multiple powers of $x$. You can try this problem of determining the coefficient of $x^{16}$ in $ (x^3+x+1)^8$.

Calvin Lin
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  • I don't understand the final part with the 2's. Could you elaborate on this? Why do we need to pick (23 - 6 - 7) 2's? – DillPixel Jan 15 '13 at 03:17
  • @DillPixel We need to pick 1 term from each bracket. Since we're only interested in $x^6 y^7$, it means that we need to pick 6 $x$'s and 7 $y$'s. The rest can be neither $x$ not $y$, hence have to be 2's. Since there are 23 brackets, it means that the remainder ($23-6-7$) must be 2's. – Calvin Lin Jan 15 '13 at 03:31
3

You can also use the multinomial theorem. Your problem is then equivalent to compute

$$\binom{23}{6,10,7}2^{10}x^6y^7=\dfrac{23!}{6!\ 10!\ 7!}2^{10}x^6y^7=2010334470144 x^6 y^7$$

ՃՃՃ
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2

You can also use the binomial theorem twice over:

$$ (x+y+2)^{23} = \sum_{k=0}^{23} \binom{23}{k} 2^{23-k} (x+y)^k$$

$$ = \sum_{k=0}^{23} \binom{23}{k} 2^{23-k} \sum_{\ell=0}^k \binom{k}{\ell} x^{\ell} y^{k-\ell}$$

Seen this way, $k$ represents the net order of the monomial of interest, which in this case is 13. It then follows that $\ell = 6$ and the coefficient of $x^6 y^7$ is

$$2^{10} \binom{23}{13} \binom{13}{6}$$

Ron Gordon
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1

Related techniques: (I), (II). We will use the Taylor series in two variables

$$ \sum_{k=0}^{\infty}\sum_{m=0}^{\infty}\frac{x^m y^k}{m! k!}\frac{\partial^{m+k}}{\partial x^m \partial y^k }f(x,y)|_{x=0,y=0} ,$$

to get the desired result. In your case, we are looking for the coefficient of $x^6 y^7$, so we need to find

$$ \frac{1}{6!\,7!}\frac{\partial^{13}}{\partial x^6 \partial y^7 }(x+y+2)^{23}|_{x=0,y=0} = 2010334470144 $$

Note: We used the formula

$$ D^n x^m = \frac{\Gamma(m+1)}{\Gamma(m-n+1)}x^{m-n} $$

to find

$$ \frac{\partial^{13}}{\partial x^6 \partial y^7 }(x+y+2)^{23}. $$

$$ \frac{\partial^{6}}{\partial x^6 }\frac{\Gamma(23+1)}{\Gamma(23-7+1)}(x+y+2)^{16} = \frac{\Gamma(23+1)}{\Gamma(23-7+1)}\frac{\Gamma(16+1)}{\Gamma(16-6+1)}(x+y+2)^{10} $$

$$ = \frac{(23)!}{(16)!}\frac{(16)!}{(10)!}(x+y+2)^{10} .$$

Community
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Mhenni Benghorbal
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