a small error, though
f(x) = x^i should be x^k instead, same to the following
in order to know sum of i^k, we need to do know k-1 sums of i^l ,for l=1,2,...k-1
but to me, sum of squares is enough to derive Spearman's Coefficient
There's also and binomial proof, which is more usable and universal, but it's harder too.
]]> 3(1²+2²+3²+...+n²)
3(1 +2 +3 +... +n)
+) 1 +1 +1 +... +1
___________________
(-1³+2³)+(-2³+3³)+(-3³+4³)+...+(-n³+(n+1)³)
=-1+(n+1)³=n³+3n²+3n
3x + 3n(n+1)/2+n=n³+3n²+3n
3x=n³+3/2 n²+1/2 n= 1/2 n(2n²+3n+1)
x= 1/6 n (2n+1) (n+1)
]]>I'll try to find the link now. Hang on.
Edit: That was easier than I thought! Here it is.
]]>it's a huge challenge!! Cuz Google doesn't provide a derivation!!:P
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