Difference between revisions of "ExtraCreditRiemannSum MA181Fall2008bell" - Rhea

Latest revision as of 09:49, 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))

@@ Line 14: / Line 14: @@
   <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.
+(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))
-*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?
-*Chumbert - Yeah, he said in class today (Wed.) to assume that, right?
-*'''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.)
-*Ctuchek - I do remember him saying that we will need to use the Mean Value Theorem.
-*Chumbert - 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?
-*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)
-*Dryg - I get what you're saying 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)
-*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)
-*[[Error_notes_MA181Fall2008bell| Here are the Error Notes.]]--[[User:Gbrizend|Gbrizend]]
-*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.

Difference between revisions of "ExtraCreditRiemannSum MA181Fall2008bell" - Rhea

Latest revision as of 09:49, 14 October 2008

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