Riemann sum / difference quotient

[noslen]

How does the difference quotient undo what the Riemann sum does or vice versa. In terms of the two formulas?

I would assume that working a difference quotient backwards would be similar to working a Riemann sum forward, but in reality as the operations go this couldn't be further from the truth.

any takers?

 

[Data]

Let's take a look. First note that the Riemann sum deals with definite integrals, so it makes the most sense to look at this from the perspective of the second part of the Fundamental Theorem of Calculus, which states (for sufficiently well behaved functions f and F),

$$F(x) = \int_a^x f(t) \ dt \Longrightarrow \frac{dF}{dx} = f(x).$$

We'll see if we can prove this from the definitions of the Riemann integral and the derivative. I'll use one of the more simple definitions of the Riemann integral to make this easy on me (which of course means that this is only valid in a limited range of circumstances), namely, letting $x_i^* \in \left[a + (b-a)(i-1)/n, \ a + (b-a)i/n\right]$, I define

$$\int_a^b f(t) \ dt = \lim_{n \rightarrow \infty} \sum_{i = 1}^n f(x_i^*) \frac{b-a}{n},$$

for any $f(t)$ continuous on [a, b].

Also, recall the definition of the derivative,

$$\frac{df}{dx} = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}.$$

We'll approach this directly. Let

$$F(x) = \int_a^x f(t) \ dt$$

be a differentiable function. Then using the definition above we have

$$\frac{dF}{dx} = \lim_{h \rightarrow 0} \frac{\int_a^{x+h} f(t) \ dt - \int_a^x f(t) \ dt}{h}$$

$$= \lim_{h \rightarrow 0} \frac{\int_x^{x+h} f(t) \ dt}{h}.$$

I now substitute in using the definition for the Riemann integral that I mentioned above, to get (for $x_i^* \in \left[ x + \frac{(x+h-x)(i-1)}{n}, \ x+\frac{(x+h-x)i}{n} \right] = \left[ x+\frac{h(i-1)}{n}, \ x+\frac{hi}{n}\right]$),

$$\frac{dF}{dx} = \lim_{h \rightarrow 0} \frac{\lim_{n\rightarrow \infty} \sum_{i=1}^n f(x_i^*)\frac{(x+h)-x}{n}}{h} = \lim_{h \rightarrow 0}\frac{\lim_{n \rightarrow \infty}\sum_{i=1}^n f(x_i^*) \frac{h}{n}}{h}$$

and using the properties of limits and series I can bring the h in the denominator inside the sum to get

$$\frac{dF}{dx} = \lim_{h\rightarrow 0}\lim_{ n \rightarrow \infty} \sum_{i=1}^n f(x_i^*)\frac{1}{n}.$$

Here the only thing depending on $h$ is $x_i^*$, and $\lim_{h \rightarrow 0} x_i^* = x$, so we just get

$$\frac{dF}{dx} = \lim_{n \rightarrow \infty} \sum_{i=1}^n f(x)\frac{1}{n} = \lim_{n \rightarrow \infty} f(x) \sum_{i=1}^n \frac{1}{n} = f(x)\lim_{n \rightarrow \infty} \sum_{i=1}^n \frac{1}{n}$$

$$= f(x)\lim_{n \rightarrow \infty} 1 = f(x),$$

as claimed.

 

[Data]

And by the way, I should note that the FTC is not an obvious result. As you say, at first glance you wouldn't think, from the definitions, that integration and differentiation would have any sort of "special" relationship.

 

[noslen]

thanks man. sweet proof

 

[marteinson]

Actually I think we can also prove that there are infinitely many prime pairs also using a related phenomenon. But I'm very rusty on formal proofs, so maybe one of you experts could formalize on what I'm saying will work.

There are three types of primes: (a) 2 and 3, (b) those which operated on by mod(6) = 5, and (c) those which under mod (6) = 1.

In other words, every multiple of six, 6n, has a pair of potential primes at 6n +/- 1, as noticed by eratosthenes.

However, no one seems to have used modular arithmetic as I suggest in my paper
http://www.chass.utoronto.ca/french...article7en.html
to separate the primes above 3 into two series, equalling 1 and 5 in mod6, or, you could think of them as equalling 7 and 5 in mod6. There is no interdependency between the primeness of the terms of the two series, 6n+1 and 6n-1, and both series display the only candidates for primeness, and contain all primes, and all their members -are- primes unless factorizable by an inferior member of the same series.

See the new "modulus 6 clock spiral" which I propose to replace Ulam's spiral, in the article, and you'll see what I mean.

Peter