[EM] EM] Satisfaction Approval Voting - A Better Proportional Representation Electoral Method

[fsimmons at pcc.edu]

Yes, that's right.

Actually it would have been simpler to use the sum

1+1/3+...+1/(2n-1), 

i.e. with a minus in the denominator of the last term instead of a plus,

so that the n tells how many terms in the sum.

Then the integrand would have been simply

(1-t^(2n))/(1-t^2).

Forest

----- Original Message -----
From: Kristofer Munsterhjelm
Date: Tuesday, May 25, 2010 12:42 pm

> fsimmons at pcc.edu wrote:
> > Note that the sum
> >
> > 1+1/3+…+1/(2n+1)
> >
> > is the integral (with respect to t) from zero to one of the sum
> >
> > 1+t^2+…+t^(2n),
> >
> > and that this integrand is a finite geometric sum with closed form
> >
> > (1-t^(2n+1))/(1-t^2) .
> >
> > So this is the appropriate integrand for a Sainte-Lague
> version that
> > allows fractional values of n, i.e. that works with any kind
> of range
> > ballot, not just approval.
>
> For the list, I'll note that the 2n+1 in the integrand appears
> to really
> be 2(n+1).
>
> This gives
>
> f(x) = H(x + 0.5)/2 + ln(2),
>
> where H is the harmonic number function, so an approximation to
> H(x) can
> be used here as well.
>
> (If the term had been 2n+1, then we'd have got f(x) = H(x)/2 +
> ln(2),
> which is not right.)
>