S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Thu, 23 Oct 2014 17:28:20 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 23 posts ]  Go to page 1, 2  Next
Author Message
 Post subject: Calculus (13)
PostPosted: Sat, 21 Jun 2008 21:36:59 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

Joined: Sat, 18 Mar 2006 08:42:24 UTC
Posts: 834
Evaluate: $\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(n-k)\int_{\frac{k}{n}}^{\frac{k+1}{n}}\frac{dx}{x^4 + 1}.
.......................................................................................................
Remark: The limit is a special case of a general result. Have fun! :)


Top
 Profile  
 
 Post subject:
PostPosted: Tue, 16 Sep 2008 05:30:54 UTC 
Offline
Senior Member

Joined: Thu, 11 Sep 2008 01:59:14 UTC
Posts: 73
I know the result of that integral and how to calculate the limit but... how do I put all those symbols that you used :?


Top
 Profile  
 
 Post subject:
PostPosted: Tue, 16 Sep 2008 05:40:11 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame
User avatar

Joined: Wed, 1 Oct 2003 04:45:43 UTC
Posts: 9834
illusion: if you click the quote button above commutative's post, then you can see how he wrote all those symbols.


Top
 Profile  
 
 Post subject:
PostPosted: Tue, 16 Sep 2008 23:02:32 UTC 
Offline
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 14149
Location: Austin, TX
I also recommend this link illusion.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject:
PostPosted: Mon, 22 Sep 2008 16:11:33 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame
User avatar

Joined: Sat, 20 Oct 2007 18:26:04 UTC
Posts: 385
Haven't solved it, but am I right in thinking it can be 'simplified' to:

$\lim_{n\to \infty} \left[ \left(\frac{n^{2}-n+1}{n}\int_{0}^{1} \frac{dx}{x^{4}+1}\right) - \left(\frac{1}{n}\sum_{k=1}^{n-1} \int_{0}^{kn^{-1}} \frac{dx}{x^{4}+1}}\right)\right]$


Top
 Profile  
 
 Post subject: Re: Calculus (13)
PostPosted: Tue, 23 Sep 2008 00:43:37 UTC 
Offline
Senior Member
User avatar

Joined: Fri, 9 Nov 2007 00:04:38 UTC
Posts: 128
I think I have it
Spoiler:
$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(n-k)\int_{\frac{k}{n}}^{\frac{k+1}{n}}\frac{dx}{x^4 + 1}
Make the substitution $ \alpha = nx, \; dx = \frac{d \alpha}{n} so it is transformed into $ \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(n-k)
\frac{1}{n} \int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}}
The integral converges to 1 as n approaches infinity, since $ \lim_{n \to \infty}\int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} = \int_{k}^{k+1} \frac{d \alpha}{1 + 0} = k+1-k = 1
So now we have $ \lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=0}^{n-1}(n-k) \times 1
The sum is $ \sum_{k=0}^{n-1}(n-k) = (\sum_{k=0}^{n-1} n) - (\sum_{k=0}^{n-1} k) = (n \sum_{k=0}^{n-1} 1) - \frac{1}{2} (n-1+1)(n-1) = nn - \frac{1}{2} n(n-1) = \frac{1}{2}(nn+n)
Dividing this by nn and taking the limit as n approaches infinity gives the answer as 1/2.

_________________
"The reason that every major university maintains a department of mathematics is that it is cheaper to do this than to institutionalize all those people."

Some people live their lives by the KISS principle (Keep It Simple Stupid), but I prefer a slightly different one:
Make It Fun And Complicate It Further, or MIFACIF. Ever try and find the volume of a cube in spherical coordinates?


Top
 Profile  
 
 Post subject: Re: Calculus (13)
PostPosted: Tue, 23 Sep 2008 00:53:41 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame
User avatar

Joined: Fri, 4 Jun 2004 16:55:13 UTC
Posts: 577
Location: New Junk City
simplethinker wrote:
I think I have it
...
The integral converges to 1 as n approaches infinity, since $ \lim_{n \to \infty}\int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} = \int_{k}^{k+1} \frac{d \alpha}{1 + 0} = k+1-k = 1



Doesn't k depend on n?

_________________
"I'm proud to live in a country where anyone, regardless of species, can buy a college degree!"


Top
 Profile  
 
 Post subject: Re: Calculus (13)
PostPosted: Tue, 23 Sep 2008 01:27:59 UTC 
Offline
Senior Member
User avatar

Joined: Fri, 9 Nov 2007 00:04:38 UTC
Posts: 128
Zone Ranger wrote:
simplethinker wrote:
I think I have it
...
The integral converges to 1 as n approaches infinity, since $ \lim_{n \to \infty}\int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} = \int_{k}^{k+1} \frac{d \alpha}{1 + 0} = k+1-k = 1


Doesn't k depend on n?

k itself doesn't depend on n, only the limit of the sum. I split the equation into two parts with the basic property of limits that if $ \lim_{n \to a} p(n) and \lim_{n \to a} q(n) exist then \lim_{n \to a} p(n)q(n) = (\lim_{n \to a} p(n))(\lim_{n \to a} q(n))

_________________
"The reason that every major university maintains a department of mathematics is that it is cheaper to do this than to institutionalize all those people."

Some people live their lives by the KISS principle (Keep It Simple Stupid), but I prefer a slightly different one:
Make It Fun And Complicate It Further, or MIFACIF. Ever try and find the volume of a cube in spherical coordinates?


Top
 Profile  
 
 Post subject: Re: Calculus (13)
PostPosted: Tue, 23 Sep 2008 02:10:24 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame
User avatar

Joined: Fri, 4 Jun 2004 16:55:13 UTC
Posts: 577
Location: New Junk City
simplethinker wrote:
Zone Ranger wrote:
simplethinker wrote:
I think I have it
...
The integral converges to 1 as n approaches infinity, since $ \lim_{n \to \infty}\int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} = \int_{k}^{k+1} \frac{d \alpha}{1 + 0} = k+1-k = 1


Doesn't k depend on n?

k itself doesn't depend on n, only the limit of the sum. I split the equation into two parts with the basic property of limits that if $ \lim_{n \to a} p(n) and \lim_{n \to a} q(n) exist then \lim_{n \to a} p(n)q(n) = (\lim_{n \to a} p(n))(\lim_{n \to a} q(n))



How did you get the limit next to the integral?

_________________
"I'm proud to live in a country where anyone, regardless of species, can buy a college degree!"


Top
 Profile  
 
 Post subject: Re: Calculus (13)
PostPosted: Tue, 23 Sep 2008 03:22:25 UTC 
Offline
Senior Member
User avatar

Joined: Fri, 9 Nov 2007 00:04:38 UTC
Posts: 128
I'm not quite sure what you mean by "the limit next to the integral". If you mean how did I get that particular limit as part of the equation:
$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(n-k)
\frac{1}{n} \int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} = \lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=0}^{n-1}(n-k) \int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} =
$ ( \lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=0}^{n-1}(n-k) ) \times ( \lim_{n \to \infty} \int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} )

_________________
"The reason that every major university maintains a department of mathematics is that it is cheaper to do this than to institutionalize all those people."

Some people live their lives by the KISS principle (Keep It Simple Stupid), but I prefer a slightly different one:
Make It Fun And Complicate It Further, or MIFACIF. Ever try and find the volume of a cube in spherical coordinates?


Top
 Profile  
 
 Post subject: Re: Calculus (13)
PostPosted: Tue, 23 Sep 2008 03:24:14 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame
User avatar

Joined: Fri, 4 Jun 2004 16:55:13 UTC
Posts: 577
Location: New Junk City
simplethinker wrote:
I'm not quite sure what you mean by "the limit next to the integral". If you mean how did I get that particular limit as part of the equation:
$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(n-k)
\frac{1}{n} \int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} = \lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=0}^{n-1}(n-k) \int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} =
$ ( \lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=0}^{n-1}(n-k) ) \times ( \lim_{n \to \infty} \int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} )


you can't do that...it doesn't work that way..also the answer isn't 1/2

_________________
"I'm proud to live in a country where anyone, regardless of species, can buy a college degree!"


Top
 Profile  
 
 Post subject:
PostPosted: Tue, 23 Sep 2008 03:37:52 UTC 
Offline
Senior Member
User avatar

Joined: Fri, 9 Nov 2007 00:04:38 UTC
Posts: 128
It doesn't work that way? It says it right here (on the SOS math site) as well as mathworld, wikipedia and one of my old calculus books. Since both limits exist it's valid. Granted my answer might not be correct, but I don't think it's that step where I messed up.

_________________
"The reason that every major university maintains a department of mathematics is that it is cheaper to do this than to institutionalize all those people."

Some people live their lives by the KISS principle (Keep It Simple Stupid), but I prefer a slightly different one:
Make It Fun And Complicate It Further, or MIFACIF. Ever try and find the volume of a cube in spherical coordinates?


Top
 Profile  
 
 Post subject:
PostPosted: Tue, 23 Sep 2008 03:40:38 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame
User avatar

Joined: Fri, 4 Jun 2004 16:55:13 UTC
Posts: 577
Location: New Junk City
simplethinker wrote:
It doesn't work that way? It says it right here (on the SOS math site) as well as mathworld, wikipedia and one of my old calculus books. Since both limits exist it's valid. Granted my answer might not be correct, but I don't think it's that step where I messed up.


simplethinker wrote:
...\lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=0}^{n-1}(n-k) \int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} =
$ ( \lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=0}^{n-1}(n-k) ) \times ( \lim_{n \to \infty} \int_{k}^{k+1} \frac{d \alpha}{1 + (\frac{\alpha}{n})^{4}} )



you cant pass the limit through this sum...that is the rule you are breaking.

_________________
"I'm proud to live in a country where anyone, regardless of species, can buy a college degree!"


Top
 Profile  
 
 Post subject:
PostPosted: Tue, 23 Sep 2008 03:54:38 UTC 
Offline
Senior Member
User avatar

Joined: Fri, 9 Nov 2007 00:04:38 UTC
Posts: 128
Ah, that makes sense. I knew it was too easy :D
Thanks for catching it. I was playing around with the different parts and didn't pay attention to the rest of the equation. Now that I think about it, since the integral depends on 'k' I couldn't have split it up into a product.
I still have this feeling that taking the limit makes the integral equal one.

_________________
"The reason that every major university maintains a department of mathematics is that it is cheaper to do this than to institutionalize all those people."

Some people live their lives by the KISS principle (Keep It Simple Stupid), but I prefer a slightly different one:
Make It Fun And Complicate It Further, or MIFACIF. Ever try and find the volume of a cube in spherical coordinates?


Top
 Profile  
 
 Post subject:
PostPosted: Tue, 23 Sep 2008 03:56:57 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame
User avatar

Joined: Fri, 4 Jun 2004 16:55:13 UTC
Posts: 577
Location: New Junk City
simplethinker wrote:
Ah, that makes sense. I knew it was too easy :D
Thanks for catching it. I was playing around with the different parts and didn't pay attention to the rest of the equation. Now that I think about it, since the integral depends on 'k' I couldn't have split it up into a product.
I still have this feeling that taking the limit makes the integral equal one.



the answer is close to 1/2...

try finding upper and lower bounds for the integral part

_________________
"I'm proud to live in a country where anyone, regardless of species, can buy a college degree!"


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 23 posts ]  Go to page 1, 2  Next

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA