A Numerical Insight for the Fundamental Theorem of Calculus - Comments

[Delta2]

Greg Bernhardt submitted a new blog post

A Numerical Insight for the Fundamental Theorem of Calculus

Continue reading the Original Blog Post.

 

[Svein]

Well - it is correct for a restricted class of functions. Saying \int f'(x)dx =f(x) presupposes that f(x) is differentiable (otherwise the expression is meaningless). The other way around (\frac{d}{dx}\int f(x)dx) allows for a larger class of functions (the Riemann-integrable functions).

Going to an even larger class of functions, we have the following theorem of Lebesgue:
If φ(x) is a summable function, its indefinite integral F(x)=\int_{a}^{x}\phi(t)dt is a continuous function of bounded variation and it has almost everywhere a derivative equal to φ(x).

Observe the "almost everywhere" clause which is typical for all integrals based on measure theory. Lebesgue also proved a theorem about the other direction:
The derivative φ(x) of an absolutely continuous function F(x) defined on the closed interval [a, b] is summable and for every x \int_{a}^{x}\phi(t)dt = F(x)-F(a).

Observe the restriction on φ(x)!

 

[jim mcnamara]

I think the point of the insight is that for someone unfamiliar with the Fundamental Theorem of Calculus, that reader would find the discussion useful.

Mathematicians often do exactly what @Svein did - generalize or expand the scope. Which is not an incorrect position in any way. Simply put: Sometimes knowing when to limit scope can be instructive, too.

 

[lavinia]

These approximations will always be true for any function. For instance

$Σ((f(x_{i}+Δx)-f(x_{i})/Δx)Δx$ will always give $f(x_{n})-f(x_{0})$ no matter what the function. So in order for the argument to be true one needs a limiting argument. One might ask "Why if this formula is true for any function at all, does it illustrate the theorem in the case that $f$ is differentiable?".

On the other hand, the picture is right as @Svein says for differentiable functions and is certainly helpful.

 

[Svein]

Σ((f(xi+Δx)−f(xi)/Δx)ΔxΣ((f(x_{i}+Δx)-f(x_{i})/Δx)Δx will always give f(xn)−f(x0)f(x_{n})-f(x_{0}) no matter what the function.

As usual, someone has to come up with a pathological function. Not the Dirichlet function this time, but \int_{0}^{1}\sin(\frac{1}{x})dx...

 

[stevendaryl]

The "telescoping sum" argument as to why $\int_A^B f'(x) dx = f(A) - f(B)$ is a simplified case of the argument leading to Stoke's theorem. You get perfect cancellations everywhere except the boundary.

 

[Delta2]

Sorry for the late reply, its because only now I am being able to see the comments and reply (due to some conflict with my account and the insights forum login subsystem)

Well about as @jim mcnamara said, this insight was meant for students that are now introduced to calculus, for high school students or for students of technical schools that aren't taught calculus with mathematical rigor but possibly they want to gain some intuitive simple insight on calculus.

When I was writing the insight I had in mind the Rieman integrable functions but @Svein is right. But then again Lavinia is more right cause here the derivative is not the real derivative, it is just an approximation of the real derivative, so the approximations just hold for any class of functions.

 

[atyy]

Does this view relate to the one from physics without calculus?

derivative = velocity = distance / time

integral = distance = velocity X time

 

[Delta2]

Does this view relate to the one from physics without calculus?

derivative = velocity = distance / time

integral = distance = velocity X time

It is not exactly like this but one could say that this is the basic idea.

More precisely what I do is that I remove the limit operator (if I can call it that way) from the definition of integral and derivative , and I just take as integral the sum of $f(x_i)\Delta x_i$ for small enough $\Delta x_i$ and as derivative the ratio of difference for small enough $\Delta x$. Without the limit operator everything becomes numerical or algebraic.