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20201212, 07:52 AM (ISO 8601)
 Join Date
 Jan 2007
Re: Approximating a diagonal slope through 9 degree steps. Why does it not work?
I think I understand the problem a bit better then before.
Aside from proofs that not every iterative approximation works I think it would be nice to see, which ones do and which ones do not. The simplest method to distinguish good and bad approximations is to look at the derivative of a given function.
All bad approximations have one thing in common: the derivative does not have a proper limit at all. For example, if we consider a rightangled zigzag function and how well it resembles a straight, horizontal line, the derivative of the zigzag would be either 1 or 1 depending on the particular place, while for the horizontal line it will obviously be 0 everywhere. If we make the steps in the zigzag smaller, the derivative will still be either 1 or 1, but the intervals are getting shorter. In the limit of infinitely small zigzag steps we cannot tell anymore whether the derivative should be 1 or 1 at any given point. Derivative becomes undefined.
All the good approximations have a well defined derivative in the limit (as long as the curve we want to approximate is also differentiable). What does it mean? that if you zoom in enough at any point, those curves look like a straight line. Thanks to that, even when we take the limit, the length will be consistent with the curve we want to approximate.In a war it doesn't matter who's right, only who's left.

20201214, 05:21 AM (ISO 8601)
 Join Date
 Feb 2012
 Location
 Boston, MA
Re: Approximating a diagonal slope through 9 degree steps. Why does it not work?
This is known as the Staircase Paradox. It's the same logic used to prove π = 4.
It does, but that's not what's happening here.
An arc length can be approximated by dividing the curve into triangles and adding up all the hypotenuses:
S ≈ ∑ √(Δx_{i}^{2} + Δy_{i}^{2})
If we take a Riemann sum, the limit as the number of triangles approaches infinity lets us get rid of that approximately qualifier, and we get the actual arc length. The integral looks a little different, but is ultimately derived from this.
S = _{a}∫^{b} √[1 + f'(x)^{2}] dx
The difference is that in the Staircase Paradox, you are summing the sides and not the hypotenuse. It doesn't matter if you do this an infinite amount of times, because you are summing something different from what you want which converges on a different value.
Another way to think about it intuitively is that what you are summing is always longer than the line. If you do this an infinite amount of times, you are still going to have something longer than the line you want.

20201214, 06:00 AM (ISO 8601)
 Join Date
 Feb 2015

20201215, 06:39 AM (ISO 8601)
 Join Date
 Feb 2012
 Location
 Boston, MA
Re: Approximating a diagonal slope through 9 degree steps. Why does it not work?
Well yes, and you can sum 1 + 2 + 3 ... to be 1/12 if you want. With infinity, any sort of intuition eventually breaks down.
But I think in this case, it's clear to see that if you are adding line segments that are always longer than the line segment you are trying to measure—if they weren't, you wouldn't have a triangle—then there is no intuitive reason for that to converge on the correct answer. Conveniently, there is also no mathematical reason.