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  <math> E \le \frac{M(b-a)^2}{N}. </math>
 
  <math> E \le \frac{M(b-a)^2}{N}. </math>
  
*So what does this equation "E < M(b-a)^2/N" mean.  This reads that the error is less than the Maximum value of the derivative of the function of x multiplied by the interval squared from x=a to x=b all divided by the total number of subintervals N.
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(I moved all the discussion that used to be here to the discussion tab. I still think it would be nice to put a polished proof of the estimate here.  --[[User:Bell|Bell]] 14:49, 14 October 2008 (UTC))
 
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*I don't understand why this must be true.  Maybe I'm wrong, but if f(x) were a horizontal line, wouldn't E=0 and M(b-a)^2/N also be =0.  That would mean it is a false statement that E < M(b-a)^2/N.  Are we to assume that E <= M(b-a)^2/N?
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*Yeah, he said in class today (Wed.) to assume that, right? --[[User:Chumbert|Chumbert]]
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*'''Bell''' - Oops!  Sorry about that.  You're right.  It needs to be <math>\le</math>.  (I can show that the ''only'' time it is actually equal is when the function <math>f(x)</math> is a constant function.)
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*Ctuchek - I do remember him saying that we will need to use the Mean Value Theorem.
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*Logically, I think I got it, but I'm not entirely sure how to prove it mathematically: The <math>M(b-a)</math> gives the height of one section(slope=(y/x), so slope*x=y), where <math>\frac{(b-a)}{N}</math> gives the width, and when multiplied together, they give you a rectangle which, if you remember from class, is the error--take the R-sum, then stack the extra blocks on to of each other. Does anyone else remember that? Or should I explain it better? --[[User:Chumbert|Chumbert]]
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*It's a bit easier to follow the discussion when one puts a signature after a comment. Just push the signature button in the edit page, or type two dashes followed by four ~, i.e. <nowiki>--~~~~</nowiki>, and your signature with the date will appear. --[[User:Mboutin|Mboutin]] 17:42, 19 September 2008 (UTC)
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*Dryg - I get what you're saying [[User:Chumbert|Chumbert]].. Yeah I remember the explanation of the stacking of blocks of the error from each sum. Unfortunately, I don't know how to prove it mathematically either  --[[User:Idryg|Idryg]] 14:30, 22 September 2008 (UTC)
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*Somebody ought to be able to find the argument in their [http://www.math.purdue.edu/~bell/MA181/basicestimate.mht class notes].  I sketched the argument in class one day.  --[[User:Bell|Bell]] 12:07, 23 September 2008 (UTC)
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*[http://kiwi.ecn.purdue.edu/MA181Fall2008bell/images/6/6e/Gbrizends_error_notes.pdf Here are the Error Notes.]--[[User:Gbrizend|Gbrizend]]
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*So is this problem officially solved now that we have the notes of both Professor Bell and Gbrizend?  I wish I had had more time to actually contribute to this.  Or have we not actually proven anything yet?  It sure looks like the solution is in the notes. --[[User:Jhunsber|Jhunsber]]
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*I think so... With said notes and the connections made, I think we're golden =]. Of course, my opinion doesn't really count for anything.. so we'll wait for Professor Bell's approval. We could write out in logical steps how it connects, with fancy LaTeX and everything. Anyone want to volunteer? =P --[[User:Chumbert|Chumbert]]
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*Well, since I haven't actually contributed anything yet, and therefore feel that I am not eligible for the extra credit, I definitely volunteer.  I'll have time this week, I'm sure.  It should be up by Saturday.[[User:Jhunsber|Jhunsber]]
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*I can't get Prof. Bell's class notes to work properly. I believe the link should be basicestimate.htm not basicestimate.mht --[[User:Ctuchek|Ctuchek]] 15:04, 1 October 2008 (UTC)
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*The .mht files work, but you cannot view them properly in firefox until you download the plugin.  They work in explorer without any downloads. --[[User:Dparkos|Dparkos]]
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Latest revision as of 10:49, 14 October 2008

Suppose that $ f(x) $ is continuously differentiable on the interval [a,b].

  • Let N be a positive integer
  • Let $ M = Max \{ |f'(x)| : a \leq x \leq b \} $
  • Let $ h = \frac{(b-a)}{N} $
  • Let $ R_N $ denote the "right endpoint"

Riemann Sum for the integral

$  I = \int_a^b f(x) dx . $

In other words,

$  R_N = \sum_{n=1}^N f(a + n h) h . $

Explain why the error, $ E = | R_N - I | $, satisfies

$  E \le \frac{M(b-a)^2}{N}.  $

(I moved all the discussion that used to be here to the discussion tab. I still think it would be nice to put a polished proof of the estimate here. --Bell 14:49, 14 October 2008 (UTC))

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