Re: coefficients

In article <20070214.084753@xxxxxxxx>,
rob@xxxxxxxxxxxxxx (Rob Johnson) wrote:

In article <1171470176.060012.170740@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
se16@xxxxxxxxxxxxxx wrote:
On 14 Feb, 16:17, r...@xxxxxxxxxxxxxx (Rob Johnson) wrote:
In article <1171464223.131431.98...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

"Toxician" <fridgechemis...@xxxxxxxxx> wrote:
F(x)= (x^2+x^3+x^4+x^5)(x+x^2+....+x^7)(1+x+...+x^15)

What is the coefficient of x^15 ? I try to multiply all of them but it
becomes too long.

F(x) = x^3 + 3x^4 + 6x^5 + 10x^6 + 14x^7 + 18x^8 + 22x^9 + 25x^10
+ 27x^11 + 28x^12 + 28x^13 + 28x^14 + 28x^15 + 28x^16 + 28x^17
+ 28x^18 + 27x^19 + 25x^20 + 22x^21 + 18x^22 + 14x^23 + 10x^24
+ 6x^25 + 3x^26 + x^27

So the answer is 28.

Rob Johnson <r...@xxxxxxxxxxxxxx>
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Indeed, but since only x^15 is needed you could just look at
0+0+0+1+2+3+4+4+4+4+3+2+1+0+0+0 (and just slide left and right for
the other sums)

Yes, I did resort to the easy way and used program to multiply the
polynomials. However, to explain your sum to the OP, we first
multiply the first two polynomials, which is pretty simple:

x^3 + 2x^4 + 3x^5 + 4x^6 + 4x^7 + 4x^8 + 4x^9 + 3x^10 + 2x^11 + x^12

Then, to compute the coefficient of x^15, we match the terms in this
product with those of (1+x+...+x^15) that make x^15. We start with
the x^3 term in (1+x+...+x^15) and end with the x^12 term:
1 + 2 + 3 + 4 + 4 + 4 + 4 + 3 + 2 + 1 = 28.

As others have pointed out, it's easier than that. Multiplying
the first two polys together with no simplification yields the
sum of 4*7 = 28 terms, each of the form x^k, k = 3, ..., 12. Each
of those terms can be paired with exactly one term in the last
polynomial to give x^15. The coefficient of x^15 is therefore 28.
.

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