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 [[RiemannSumErrorProof_MA181Fall2008bell| ought to be able to find the argument in their class notes]].  I sketched the argument in class one day.  --[[User:Bell|Bell]] 12:07, 23 September 2008 (UTC)

Difference between revisions of "ExtraCreditRiemannSum MA181Fall2008bell" - Rhea

Latest revision as of 09:49, 14 October 2008

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