Hi !

Let increasing.

How can I prove that the sequence defined by (Riemann sum) is increasing ?

Thanks (Wink)

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- Apr 17th 2010, 02:57 AMInfophileRiemann sum increasing
Hi !

Let increasing.

How can I prove that the sequence defined by (Riemann sum) is increasing ?

Thanks (Wink) - Apr 17th 2010, 04:45 AMLaurent
I think (I'm even almost sure) this is wrong. Are you sure this is what the exercise is asking? This could be true if the Riemann sum was defined with respect to a sequence of "increasing" subdivisions, i.e. such that the subdivision for is finer than that for ; for instance, at the points .

- Apr 17th 2010, 05:03 AMInfophile
It's the case no ? The step is 1/n for u_n and 1/(n+1) for u_(n+1) (finer subdivision than for u_n).

In my exercice I have f(x)=xexp(x), but I suppose this is true for all increasing function.

I verify with Maple for this f. - Apr 17th 2010, 05:35 AMLaurent
Finer means that the subdivision from one step to the next is obtained by subdividing it: we only add points in the subdivision.

Consider the following example: if , if (yes, I know, it is not continuous). Then for the Riemann sum is . And for it is . To get a continuous example, just consider going quickly from 0 to 1 just before .

In your example, f is convex; this could help. - Apr 17th 2010, 06:35 AMLaurent
Sufficient conditions are and is convex on .

Proof sketch: for , we have and the convex combination . Using convexity and value at 0, we get

.

Furthermore, , hence . QED. - Apr 17th 2010, 09:30 AMInfophile
Merci Laurent ! (je n'avais pas vu que tu étais Français)

But I don't understand this equality , it seems false. - Apr 17th 2010, 11:11 AMLaurent
Does it seem false because you don't understand it or for more sound reasons? If you look at the sum globally, each term appears twice (once for index and once for index ), except for which appear once. The coefficient in front of is the sum of the two corresponding coefficients. I can't ensure my computation is correct but I let you check it.

- Apr 17th 2010, 11:46 AMInfophile
Yes I saw that each terms appears twice, but I think the coefficient in front of is rather .

Do you agree ? - Apr 17th 2010, 12:13 PMLaurent
- Apr 17th 2010, 12:26 PMInfophile
Indeed, sorry !

Thanks !