Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
add comment

4 Answers

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$.

share|improve this answer
    
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 3:17
    
@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 3:31
add comment

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

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

share|improve this answer
add comment

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

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

share|improve this answer
add comment

A related technique. 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

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

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.