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I'm having trouble deriving arc length of functions. I'm using a vector-oriented approach, if that's alright. I'm doing this because I don't understand the [math]1+(f')^2[/math] part of the standard formula, and I don't like using formulae I don't understand.
So far I have this:
Let [math]d(\vec{u},\vec{v})[/math] be the Euclidean distance function between two vectors [math]\vec{u}[/math] and [math]\vec{v}[/math], defined by [eqn]d(\vec{u},\vec{v})=\left \| \vec{v}-\vec{u} \right \|[/eqn]. Let [math]f(x)[/math] be any "reasonable" function. (By "reasonable", I mean continuously differentiable on the closed interval [math][a,b][/math], with no singularities, crazy fractal stuff, etc. I think the proof works as long as it's continuous, actually. I could be wrong.)
To obtain the arc length of [math]f(x)[/math] on the interval [math][a,b][/math], we have to rectify the curve generated by [math]f(x)[/math]. We add up the distances between infinitesimal parts of [math]f[/math]:
[eqn]\int_{a}^{b}d(f(x),f(x+h))dx[/eqn]
We are working in 2D, so here's the explicit formula for [math]d(\vec{u},\vec{v})[/math].
Let [math]\vec{u}=\left \langle x_1,y_1 \right \rangle[/math] and [math]\vec{v}=\left \langle x_2,y_2 \right \rangle[/math]
[eqn]\vec{v}-\vec{u}=\left \langle x_2-x_1,y_2-y_1 \right \rangle[/eqn]
[eqn]d(\vec{u},\vec{v})=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2)}[/eqn]
Now, let's substitute [math]\vec{u}=\left \langle a,f(a) \right \rangle[/math] and [math]\vec{v}=\left \langle a+h,f(a+h) \right \rangle[/math] where [math]a[/math] is some number on [math]f[/math] and [math]h[/math] is a small number (infinitesimal, you know?).
[eqn]d(\vec{u},\vec{v})=\lim_{h \to 0}\sqrt{((a+h)-(a))^2+(f(a+h)-f(a))^2}[/eqn]
That's where I hit a hitch. I'm not sure how to get rid of that h and replace it with a derivative or something more practical. Oh,[math]((a+h)-a)^2=h^2[/math], right? [math]h[/math] is infinitesimal. Is squaring an issue? Or does it become so small as to be considered nothing?
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Why don't the eqn tags work?
[eqn]test[/eqn]
[math]test[/math]
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Curves are stupid
join the straight-line-and-rational-numbers-only master race
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>>7754029
>he proof works as long as it's continuous, actually. I could be wrong.
f needs to be of bounded variation
>I'm not sure how to get rid of that h and replace it with a derivative
Pull h out of the square root. Also you start with a sum, not with an integral. So you obtain the integral in the limit.
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You take n amount points in arc and make a approximation by drawing straight lines by the points. Now when n approaches infinity we get the desired lenght
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You have a curve [math](t,f(t)): [a,b] \to \mathbb{R^2} [/math].
You take a sequence of equidistant partitions [math] (t^{(n)}_i) [/math] of the interval [math] [a,b] [/math],
[math]a = t^{(n)}_0 < t^{(n)}_1 < \ldots < t^{(n)}_n = b [/math]
then the arc length is
[math] \lim_{n \to \infty} \sum_{i=0}^{n-1} d(c(t^{(n)}_{i+1}), c(t^{(n)}_i)) = \lim_{n \to \infty} \sum_{i=0}^{n-1} \sqrt{(t^{(n)}_{i+1} - t^{(n)}_i)^2 + (f(t^{(n)}_{i+1}) - f(t^{(n)}_i))^2}[/math]
Now thanks to Taylor we know that [math]f((t^{(n)}_{i+1}) - f(t^{(n)}_i)) \approx f'(t^{(n)}_i) (t^{(n)}_{i+1} - t^{(n)}_i) [/math]
So
[math] \lim_{n \to \infty} \sum_{i=0}^{n-1} \sqrt{(t^{(n)}_{i+1} - t^{(n)}_i)^2 + (f(t^{(n)}_{i+1}) - f(t^{(n)}_i))^2} = \lim_{n \to \infty}\sum_{i=0}^{n-1} \sqrt{1 + (f'(t^{(n)}_i))^2 } (t^{(n)}_{i+1} - t^{(n)}_i) = \int_a^b \sqrt{1 + (f'(t))^2} dt[/math]
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>>7754034
>yfw any straight line contains more points that have an irrational ordinate than points that have only rational ordinates
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>>7754034
*tips berdora*
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>>7754525
and yet you cannot show us any of those "irrational ordinate" because they are ill defined.
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>>7754649
>they're not defined the way I want them to be
>therefore they do not exist
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>>7754731
sue me
present me with any construction of "irrational numbers" and I will debunk it right here.
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>>7754737
Cauchy sequences
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>>7754768
what about them?
Q is not complete
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>>7754779
Cauchy sequences are used to construct the reals by completing Q. If you don't understand the construction how are you going to "debunk" it?
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>>7754737
what the hell is your problem with Dedekind cuts to construct the reals and then irrationals are just numbers which aren't rational?
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>>7754737
The completion of Q
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[math]\frac{255}{\pi}\left(sin^{-1}\left(sin\left(\frac{\pi}{2}x\right)\right)+cos^{-1}\left(cos\left(\frac{\pi}{2}x\right)\right)\right)[/math]
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