Problem with the Limits
in [Matlab]
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From: Roger Stafford on 10 May 2010 18:46 TideMan <mulgor(a)gmail.com> wrote in message <50885f83-7fdc-4db8-8a6e-7644a4332835(a)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
From: TideMan on 10 May 2010 19:45 On May 11, 10:46 am, "Roger Stafford" <ellieandrogerxy...(a)mindspring.com.invalid> wrote: > TideMan <mul...(a)gmail.com> wrote in message <50885f83-7fdc-4db8-8a6e-7644a4332...(a)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.
From: Shashishekar on 12 May 2010 04:43
The problem has been resolved now, It would not have been possible without the help of this forum. Thank you very much for the support |