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It's been a while since I have been playing with these so excuse me if it is too obvious.

How can I represent

$$ f(x) = \frac{(1-x^{10} ) ^6}{(1-x)^6} $$

as sum of powers of $x$

I am especially interested in the coefficient in front of $x^{27} $ in that sum.

The book I am reading gives this coefficient as obviously being

$${32 \choose 5}- {6 \choose 1} {22 \choose 5}+{6 \choose 2}{12 \choose 5}$$

but I don't know where this comes from.

Many thanks in advance.

Mike Earnest
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peter.petrov
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  • First take off the powers of $6$ from the whole fraction. Then see if the formula $\frac{1-r^n}{1-r}$ looks at all familiar… – Jam Aug 30 '18 at 22:29
  • @Jam Yeah it looks familiar, I know what it is. I am intentionally giving this short representation here. But when I take it to power 6... how do I know which number I get in front of $x^{27}$?! Or in front of any $x^k$? – peter.petrov Aug 30 '18 at 22:32
  • how can you get $27$ from adding $6$ non-negative integers together? – Jam Aug 30 '18 at 22:36
  • @Jam Not sure what you mean here. Sorry. What 6 non-negative numbers? – peter.petrov Aug 30 '18 at 22:48
  • @Jam Oh, I see. Well I tried that too. Leads me to 2 Diophantine equations which I don't find too obvious either. But it is not 6 numbers, it is 10 numbers. – peter.petrov Aug 30 '18 at 22:53

3 Answers3

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First from the geometric series $$ \frac{1}{1-x}=\sum_{n=0}^\infty x^n $$ combined with repeated differentiation yields that $$ \frac{1}{(1-x)^6}=\sum_{n=0}^\infty\binom{n+5}{5} x^n\tag{1} $$ (for $|x|<1$, if you don't want to deal with formal series). The binomial theorem implies that $$ (1-x^{10})^6=1-6x^{10}+\binom{6}{2}x^{20}-\binom{6}{3}x^{30}+\binom{6}{4}x^{40}-\binom{6}{5}x^{50}+x^{60}\tag{2}. $$ We want the coefficient of $x^{27}$ in the product of (1) and (2). We get an $x^{27}$ term from the products $1\times x^{27}$, $x^{10}\times x^{17}$ and $x^{20}\times x^{7}$. In particular the coefficient is then $$ \binom{27+5}{5}-6\binom{17+5}{5}+\binom{6}{2}\binom{7+5}{5}. $$

Sri-Amirthan Theivendran
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4

As an alternative recall that

$$(1-x^{10} )=(1-x)(1+x+x^2+\ldots+x^9)$$

therefore

$$f(x) = \frac{(1-x^{10} ) ^6}{(1-x)^6}=(1+x+x^2+\ldots+x^9)^6$$

then we can evaluate the coefficient for $x^{27}$ by the stars and bars method by the equivalent problem, discussed here, of the numbers of way to distribute $b=27$ balls in $c=6$ container with no more than $n=9$ balls in any container that is

$$N(b,c,n)=\sum_i(-1)^i\binom{c}i\binom{b+c-1-i(n+1)}{c-1}=$$

$$=\binom{27+6-1}{5}-\binom{6}{1}\binom{27+6-1-(9+1)}{5}+\binom{6}{2}\binom{27+6-1-2(9+1)}{5}=$$

$$={32 \choose 5}- {6 \choose 1} {22 \choose 5}+{6 \choose 2}{12 \choose 5}$$

user
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4

Expand out the numerator with the binomial theorem: $$ (1-x^{10})^6=1-\binom{6}1x^{10}+\binom{6}2x^{20}-\dots $$ We could do more terms, but since you want the $x^{27}$ coefficient this is all we will need.

The numerator is being multiplied by $(1-x)^{-6}$, which can be expanded using the generalized binomial theorem, sometimes called Newton's binomial theorem: $$ (1-x)^{-6}=\sum_{k\ge 0}\binom{-6}k(-x)^k=\sum_{k\ge 0}\binom{6+k-1}kx^k $$ Finally, since we are multiplying these, we have to find all ways of choosing monomials from both so their powers add up to $27$. We can only get powers of $0, 10$ or $20$ from the first factor, so the second factor must contribute either $27,17$ or $7$. Adding up these three contributions gives $$ 1\cdot \binom{6+27-1}{27}-\binom{6}1\binom{6+17-1}{17}+\binom{6}2\binom{6+7-1}{7} $$

Mike Earnest
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