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From: TideMan <mulgor@gmail.com>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Problem with the Limits
Date: Mon, 10 May 2010 16:45:57 -0700 (PDT)
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On May 11, 10:46 am, "Roger Stafford"
<ellieandrogerxy...@mindspring.com.invalid> wrote:
> TideMan <mul...@gmail.com> wrote in message <50885f83-7fdc-4db8-8a6e-7644a4332...@a2g2000prd.googlegroups.com>...
> > You're right, but I only got there after a page of algebra.
> > How could I have gotten there without the algebra?
> > BTW, the answer I got for the limit is 6/30, correct?
>
> - - - - - - - -
>   Yes, that's the correct limit, Tideman.  (Shashishekar will be grateful to you for spilling the beans.  :-)  )
>
>   Whether it takes a lot of algebra or not depends on whether you already know the fourth identity, which is in fact:
>
>  1^4 + 2^4 + 3^4 + 4^4 + ... + (n-1)^4 = (n-1)*n*(2*n-1)*(3*n^2-3*n-1)/30
>
> After that it is all smooth sailing, because the summation satisfies
>
>  sum((1-(0:n-1)/(n-1)).^4) = sum(((n-1-(0:n-1))/(n-1)).^4) =
>  sum((n-1:-1:0).^4)/(n-1)^4 = sum((0:n-1).^4)/(n-1)^4 ,
>
> this last equality being obtained by doing a 'fliplr' which doesn't alter the sum.  So, hardly any algebra is needed after that.
>
>   My ancient "CRC Standard Mathematical Tables" only gave the summation of powers of i up to cubes, so I'll have to confess it took some algebra on my part to work it out for the fourth power, but I am sure some other tables are floating around which contain that formula.
>
> Roger Stafford

Oh, I got the 4th power OK from my 40-year old Gradshteyn & Ryzhik
Table of Integrals Series and Products.
What I wanted to know is how you knew that the first 4 terms cancelled
to just leave just the 4th power sum.