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Let a partition $\{t_0,\ldots,t_n\}$ of the interval $[a,b]$ and let $f$ an integrable function. (we may also assume that $f$ is differentiable on $[a,b]$)

We know that the Right Riemann sum is

$$R(f)=\sum_{i=1}^n f(t_{i})(t_i-t_{i-1})$$

What's the relation ( in order sense terms) between $R(f)$ and the integral $\displaystyle \int_a^b f(x)dx$?

Can I say that there exists a positive constant $C>0$ such that $\displaystyle \int_a^b f(x)dx\leq C\cdot R(f)$? I know that if $f$ is increasing function, this is correct, but in the general case, i may say that?

EDIT: Assume that $f(t)\geq 0$ on $[a,b]$

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    $\begingroup$ You are probably assuming that $f(t)\ge 0$ for all $t\in [a, b]$, right? $\endgroup$ – Giuseppe Negro Mar 4 '14 at 21:58
  • $\begingroup$ yes, I assuming that. Thanks for your observation $\endgroup$ – Jorge Alexis Castillo Seplveda Mar 4 '14 at 21:59
  • $\begingroup$ There's no relationship whatsoever, right? The value of $f$ can be anything at the $t_i$s, since there's only finitely many of them. $\endgroup$ – Mark McClure Mar 4 '14 at 22:03
  • $\begingroup$ From wikipedia: en.wikipedia.org/wiki/Riemann_sum , I get the following information $\displaystyle \left|\int_a^bf(x)dx-R(f)\right|\leq \dfrac{M_1(b-a)^2}{2n}$ With $M_1$ the maxiu value of $|f'|$ along the interval $\endgroup$ – Jorge Alexis Castillo Seplveda Mar 4 '14 at 22:06
  • $\begingroup$ Then, $\displaystyle\int_a^b f(x)dx\leq \dfrac{M_1(b-a)^2}{2n}+R(f)$, but i'd like a constant multiplying $R(f)$, not adding $\endgroup$ – Jorge Alexis Castillo Seplveda Mar 4 '14 at 22:09
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As Michael E2 points out, the answer is definitely "No" in general. Let me suggest the counterexample: (fix some positive integer $N$) $$\int_0^1 \sin^2(N!\,\pi x)\,dx$$ The integrand is a non-negative function which hits zero when $x$ is any integer multiple of $\frac{1}{N!}$. Because of this any right hand approximation with up to $N$ rectangles will be identically zero (the partition points for the $n$-th approximation would be $x_i=0+i\Delta x = \frac{i}{n} = \frac{i \cdot 1 \cdot 2 \cdots (n-1)(n+1)\cdots N}{N!}$ which is an integer multiple of $1/N!$). Since the integral itself is positive, we can't make your inequality work.

However, when the right hand sums are positive, such a constant always exists. But this isn't anything special about right hand approximations. It's a fact about any positive sequence.

Let $\{ a_i \}_{i=1}^\infty$ be a sequence of positive real numbers which converges to $A$. If $A=0$, then $A=0 \leq Ca_i$ is trivially true for any constant $C \geq 0$. So let's consider the case when $A \not= 0$. In such a case, it is not hard to show that $A>0$.

Let $\epsilon = A/2 > 0$. There exists some $N>0$ such that $|a_n-A|<\epsilon=A/2$ for all $n \geq N$. Thus $A/2 = A-A/2 = A -\epsilon < a_n$ for all $n \geq N$. So that $A < 2a_n$ for all $n \geq N$.

Let $M = \mathrm{min} \{a_1,\dots,a_N \}$ (note $M>0$ since we assumed our sequence was positive). Then $A = \frac{A}{M}\cdot M \leq \frac{A}{M} a_i$ for $i=1,\dots,N$.

Finally, let $C = \mathrm{max} \{ \frac{A}{M}, 2\}$ and we have $a_i \leq C \cdot A$ for all $i$.

You can see that the argument breaks down when a term of the sequence is allowed to be 0 (in this case $M=0$ and $A/M$ is undefined).

Now back to your question. Letting $a_i$ be the $i$-th right hand approximation of $A = \int_a^b f(x)\,dx$, we have $A \leq Ca_i$ for all $i$ such that $a_i$ is not zero.

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  • $\begingroup$ Oops! I forgot to add in the trivial case when the limit is zero. :) Thanks for pointing that out...fixed. And, yes, I'm only assuming $f$ is integrable and nonnegative (so that the right hand sums are nonnegative and their limit -- the integral -- exists). $\endgroup$ – Bill Cook Mar 17 '14 at 0:53
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Basic question

I think this question boils down to something simpler than what has been suggested: if $f$ is bounded and integrable, and the Riemann sum is nonzero, there will always be such a $c$. This is because both the Riemann sum and the integral are constants constants. For any two nonzero constants this is true (think: if $c>\frac{b}{a}$, the exist infinitely many constants $C$ such that $b<C\cdot a$).

More generally

The much more interesting question is when you get equality. Equality absolutely holds with a monotonically increasing function, as you suggested. But in fact these are not the only functions for which this is true. This is true, in the precise way you formulated it, for any function and partition (this example IS dependent on the partition) such that for each $i\leq n$, $\max_{x\in\left[t_i,t_{i-1}\right]}{f(x)=f(t_i)}$. I think it is pretty straightforward from here, but for thoroughness' sake, I will show this in another way. Let $g(x)=\min\left \{t_i:t_i>x\right\}$. Then $g\geq f$ for all $x$, so $\int_a^b{gdx}\geq\int_a^b{fdx}$. However, $\int_a^b{gdx}$ is precisely equal to the Riemann sum $\sum_{i=1}^n{f(t_i)(t_i-t_{i-1})}$.

Even more generally

If we slightly modify the Riemann sum, this will always be true. Instead of taking $f(t_i)$ as the height of the rectangle, we can take the maximum of $f(x)$ on the interval $\left[t_{i-1},t_i\right]$, and the inequality is almost always true. We still have to worry about the case that $f$ is bounded and integrable, and it is still a problem if the sum is $0$, since $f$ can be negative. If $f(x)>0$ for all $x$ or we take $|f|$ we don't have to worry about the sum anymore. Speaking of changes of the height of the rectangle, herein lies a cool application of the mean value theorem. If $f$ is continuous on $\left[a,b\right]$, given a partition $P=\left\{t_0,t_1,\cdots,t_n\right\}$ there exists a sequence $\left\{a_i\right\}_{i=1}^n$ where $\left\{t_0<a_1<t_1<a_2<t_3,\cdots,t_{n-1},a_n,t_n\right\}$ such that the Riemann sum $\sum_{i=1}^n{f(a_i)(t_i-t_{i-1})}$ is precisely equal to $\int_a^b{f(x)dx}$.

Difference Bound (not quite)

Unfortunately the bound you linked to in the comments is only true if $f$ is monotonic. Even then, the purpose of the bound is not to bound the integral by the Riemann sum. The purpose of that bound is just to bound the difference between the Riemann sum and the integral; it is speaking to the accuracy of Riemann sums as an approximation for the integral. In fact, we can see that as we change the partition so $n\rightarrow\infty$, $\frac{M_1(b-a)^2}{2n}\rightarrow 0$, and the Riemann sum is equal to the integral (hence the definition of the Riemann integral).

Does a definitive bound make sense at all?

Not really. Say we're taking the right-handed Riemann sum of a function $f$, but function $f$ has jump discontinuities at each $t_i$ (but assume it is still bounded, and therefore integrable). Then we can show the Riemann sum can be made arbitrarily close to $0$. Consider a partition $P=\left\{t_i\right\}=\left\{a, a+\frac{b}{n}, a+\frac{2b}{n}, \cdots, b\right\}$. Let $f(t_i)=\frac{\varepsilon}{b-a}$ for all $t_i$ for some $\varepsilon>0$. Then the Riemann sum is \begin{equation} R(f)=\sum_{i=1}^n{f(t_i)(t_i-t_{i-1})}=\sum_{i=1}^n{\frac{\varepsilon}{b-a}\frac{b-a}{n}}=\sum_{i=1}^n\frac{\varepsilon}{n}=\varepsilon \end{equation} but since $\varepsilon$ can be arbitrarily small, so can the Riemann sum. Even in the continuous case, if a function dips at each $t_i$ the same phenomena can happen. This is the exact reason we use limit processes for integrals and the exact same reason Riemann integration can become unwieldy at times (consider $\frac{1}{x}\sin\left(\frac{1}{x^3}\right)$ -- it's not even Riemann integrable). Riemann sums themselves are remarkably inaccurate, at least when partition size is small. For these reasons, I really don't think it makes sense to have such a bound of a function by a Riemann sum.

I hope I've helped answer your question, and given you a bit of insight into how this machinery works.

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  • $\begingroup$ Great answer, really helps a lot for beginners like me. $\endgroup$ – Nhat Apr 16 '14 at 23:55
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From Wikipedia I get the following formula

$$\left|\int_a^b f(t)dt-R(f)\right|\leq \dfrac{M_1(a-b)^2}{2n}$$

Then,

$$\int_a^b f(t)dt\leq \dfrac{M_1(a-b)^2}{2n}+R(f)\leq \max\left\{\dfrac{M_1(a-b)^2}{2n},1\right\}(1+R(f))$$

I'd like a constant multiplying $R(f)$.

Could this be improved?

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