Path: news.mathworks.com!newsfeed-00.mathworks.com!newsfeed2.dallas1.level3.net!news.level3.com!postnews.google.com!j36g2000prj.googlegroups.com!not-for-mail 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) Organization: http://groups.google.com Lines: 37 Message-ID: <d27c8072-9547-429f-bdfd-9eb7e032c95c@j36g2000prj.googlegroups.com> References: <hs6hln$old$1@fred.mathworks.com> <hs8jbl$1ft$1@fred.mathworks.com> <50885f83-7fdc-4db8-8a6e-7644a4332835@a2g2000prd.googlegroups.com> <hsa2bd$lst$1@fred.mathworks.com> NNTP-Posting-Host: 202.78.152.105 Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable X-Trace: posting.google.com 1273535158 10007 127.0.0.1 (10 May 2010 23:45:58 GMT) X-Complaints-To: groups-abuse@google.com NNTP-Posting-Date: Mon, 10 May 2010 23:45:58 +0000 (UTC) Complaints-To: groups-abuse@google.com Injection-Info: j36g2000prj.googlegroups.com; posting-host=202.78.152.105; posting-account=qPexFwkAAABOl8VUndE6Jm-9Z5z_fSpR User-Agent: G2/1.0 X-HTTP-UserAgent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.9.2.3) Gecko/20100401 Firefox/3.6.3,gzip(gfe) Xref: news.mathworks.com comp.soft-sys.matlab:634423 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.