Expand (x-1)(x-2)...(x-n)

The solution to a problem is  P(x) = (x-1)(x-2)...(x-n)  (for some integer n).

However, I want to expand this polynomial, so as to find the coefficient of the ith power.

It's quite clear that:  a_n=1,  and  a_0=n!(-1)^n, but I've yet to figure a pattern for the others...

How would I approach this?


Thanks
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anas_aliraqiCommented:
you can use Pascal's triangle to find any cof.
see this link
http://en.wikipedia.org/wiki/Pascal's_triangle
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TalmashCommented:
a_1 = sigma ( -i ) for i=1..n  {or -(n+1)*n/2 }
a_n-1 = sigma ( PI {-i} for i=1..j-1 X PI {-i} for i=j+1..n ) for j=1..n
   you need a script/algo to calculate this even for a_n-1

a_n-k = sigma (choose subset k out of n) PI (k elements)

you need some functions:
1. subset_k_n (input n input k input start input prev_subset output next_subset output last)
2. PI_subset (input subset_array output subset_product)
3. SIGMA
  3.1 calls subset_k_n with start=1
  3.2 subset_k_n return with the 1st subset (1..k) = prev_subset
  3.3 calls PI_subset with (1..k)
  3.4 PI_subset return with k(k+1)/2
  3.5 SIGMA set this value as a_n-k
  3.6 call subset_k_n with prev_subset
  3.7 call PI_subset with the new subset
  3.8 add a_n-k the returned value from PI_subset
  3.9 goto 3.6 until subset_k_n->stop = 1

tal



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TalmashCommented:
wow! didn't knew about these stirling triangles!
anyway, no equation compute them stright forward.
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Infinity08Commented:
I see ozo posted already, but I found this myself :

        a(n,n)   = 1

and for j = 1..n :

        a(n,n-j) = Sum(i=j..n) [ a(i-1,i-j) * (-i) ]

where a(x,y) is the y-th coëfficient for the x-th order polynomial.



Or, worked out :

a(0,0) = 1
    => 1 = 1

a(1,1) = 1
a(1,0) = Sum(i=1..1) [ a(i-1,i-1) * (-i) ] = -1
    => (x - 1) = x - 1

a(2,2) = 1
a(2,1) = Sum(i=1..2) [ a(i-1,i-1) * (-i) ] = (-1) + (-2) = -3
a(2,0) = Sum(i=2..2) [ a(i-1,i-2) * (-i) ] = 2
    => (x - 1)(x - 2) = x^2 - 3x + 2

a(3,3) = 1
a(3,2) = Sum(i=1..3) [ a(i-1,i-1) * (-i) ] = (-1) + (-2) + (-3) = -6
a(3,1) = Sum(i=2..3) [ a(i-1,i-2) * (-i) ] = 2 + 9 = 11
a(3,0) = Sum(i=3..3) [ a(i-1,i-3) * (-i) ] = -6
    => (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6

a(4,4) = 1
a(4,3) = Sum(i=1..4) [ a(i-1,i-1) * (-i) ] = (-1) + (-2) + (-3) + (-4) = -10
a(4,2) = Sum(i=2..4) [ a(i-1,i-2) * (-i) ] = 2 + 9 + 24 = 35
a(4,1) = Sum(i=3..4) [ a(i-1,i-3) * (-i) ] = (-6) + (-44) = -50
a(4,0) = Sum(i=4..4) [ a(i-1,i-4) * (-i) ] = 24
    => (x - 1)(x - 2)(x - 3)(x - 4) = x^4 - 10x^3 + 35x^2 - 50x + 24


etc.
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ozoCommented:
#!/usr/bin/perl
$s[0][0]=1;
for$n( 0..20 ){
    print $s[$n+1][$_] = $s[$n][$_-1]-$n*$s[$n][$_],"x^",$_-1," " for reverse 1..$n+1;
    print "\n";
}
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InteractiveMindAuthor Commented:
Wow
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