Riemann sum / difference quotient

In summary, the conversation discusses the relationship between the difference quotient and the Riemann sum in terms of the Fundamental Theorem of Calculus. The participants also mention using modular arithmetic to prove the existence of infinitely many prime pairs. The conversation ultimately concludes that there is a special relationship between integration and differentiation.
  • #1
noslen
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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?
 
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  • #2
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),

[tex]F(x) = \int_a^x f(t) \ dt \Longrightarrow \frac{dF}{dx} = f(x).[/tex]

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 [itex]x_i^* \in \left[a + (b-a)(i-1)/n, \ a + (b-a)i/n\right][/itex], I define

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

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

Also, recall the definition of the derivative,

[tex]\frac{df}{dx} = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}.[/tex]

We'll approach this directly. Let

[tex]F(x) = \int_a^x f(t) \ dt[/tex]

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

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

[tex] = \lim_{h \rightarrow 0} \frac{\int_x^{x+h} f(t) \ dt}{h}.[/tex]

I now substitute in using the definition for the Riemann integral that I mentioned above, to get (for [itex]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][/itex]),

[tex]\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}[/tex]

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

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

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

[tex]\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}[/tex]

[tex]= f(x)\lim_{n \rightarrow \infty} 1 = f(x),[/tex]

as claimed.
 
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  • #3
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.
 
  • #4
thanks man. sweet proof
 
  • #5
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
 

Related to Riemann sum / difference quotient

What is a Riemann sum?

A Riemann sum is a method of approximating the area under a curve by dividing the area into smaller rectangles and calculating the sum of their individual areas.

What is the difference between a left, right, and midpoint Riemann sum?

A left Riemann sum uses the left endpoint of each subinterval to calculate the height of the rectangle, a right Riemann sum uses the right endpoint, and a midpoint Riemann sum uses the midpoint of each subinterval.

What is the purpose of a Riemann sum?

The purpose of a Riemann sum is to estimate the area under a curve when it is not possible to find the exact area using traditional methods.

What is the formula for a Riemann sum?

The formula for a Riemann sum is:
S = ∑ f(xi)Δx
where f(xi) is the height of the rectangle at each subinterval and Δx is the width of the subinterval.

What is the connection between Riemann sums and the definite integral?

Riemann sums are closely related to definite integrals. As the number of rectangles in a Riemann sum increases, the approximation becomes more accurate and approaches the exact value of the definite integral.

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