Re: coefficients

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.

Rob Johnson <rob@xxxxxxxxxxxxxx>
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