Average Arc Length Question?

Question

Consider the set S of all functions f(x) such that each f(x) is strictly
increasing, continuous and differentiable on the interval [0,1] and such that
f(0) = 0 and f(1) = 1. What is the maximum arc length from (0,0) to (1,1) from
amongst the functions in S and is it possible to compute the average arc length
from (0,0) to (1,1) for all functions in S? Clearly the shortest arc length is sqrt(2),
and I'm thinking that the longest approaches 2, but I'm not sure how to prove that.
As for the average, there are some symmetries to work with but I'm not sure if
the question is even possible to answer.

If the condition that f'(x) was also strictly increasing, continuous and differentiable
on [0,1] for all functions f(x) in S was also applied, would that make my questions
more tractable? I realize that in both cases S is uncountably infinite in size, so an
average may not be possible because of that fact. I'm in a bit over my depth here,
but it seemed like an interesting question to ask and a good way of learning
something new. :)

Answer 1

The assumption that the functions be differentiable is actually unnecessary, as any continuous increasing function f on [0,1] is rectifiable, since it is of bounded variation (and in fact has total variation equal to f(1) - f(0)).



Recall, for a metric space (X,d) and a path α : [0,1] → X,

the arc-length L[α] of α, if it exists, is defined by

Sup_P {Σ_[t(k)∈P] d(α(t(k)),α(t(k-1))},

for partitions P = {0 = t(0) < ... < t(n) = 1} of [0,1].



For f ∈ S this becomes

L[f] = Sup_P {Σ_[t(k)∈P] √[(t(k) - t(k-1))² + (f(t(k)) - f(t(k-1)))²]}

≤ Sup_P {Σ_[t(k)∈P] |t(k) - t(k-1)| + |f(t(k)) - f(t(k-1))|}

= Sup_P {Σ_[t(k)∈P] |t(k) - t(k-1)| + Σ_[t(k)∈P] |f(t(k) - f(t(k-1))|}

= Sup_P {2} (since Σ_[t(k)∈P] |t(k) - t(k-1)| = Σ_[t(k)∈P] |f(t(k)) - f(t(k-1))| = 1 for any P)

= 2.



Anyway, as messy as the above proof looks, the thinking behind it is fairly straightforward from the definition of arc-length, the fact that √(x² + y²) ≤ |x| + |y|, and the "telescoping" of the |t(k) - t(k-1)|'s and |f(t(k)) - f(t(k-1))|'s. What's more interesting would be a proof that *no* element of S has arc-length 2. There are examples of continuous increasing functions f on [0,1] with f(0) = 0 and f(1) = 1 with L[f] = 2, (such as the Cantor function) but they're likely all very non-differentiable.







As for the "average" of L, S ⊂ C₀[0,1] has Wiener measure zero (I think), so I don't know of any standard way to give S a probability measure.

Answer 2

I'll give you part of it. You stated that you don't know the maximum arc length. The maximum arc length is not achievable. At most, you can approach the maximum as a limit, thus a least upper bound or supremum.



Consider the curve that follows two edges of the unit square from (0,0) to (1,1). Since it is not a function allowed by the conditions, it is not admissible, but it has an arc length of 2.



Also, we can see that no curve that satisfies the requirements of continuity, differentiability and monotonicity can have a longer arc length. For example, any monotonic stair-step path between those points has the same arc length of 2.



Does there exist a set of curves with arc length that approaches 2 as a limit? Consider the function y = x^n, with integer n. Clearly each member of this family is continuous, monotone and infinitely differentiable over the interval. Integer n ensures the curves are differentiable at zero. As n grows large, the curves also clearly approximate a step function arbitrarily closely, so there is some member of that family that has arc length at least 2-epsilon, for any non-zero epsilon. Just make n large enough.



Therefore the maximum arc length is 2, but it cannot be achieved by any member of the set you have defined.

Answer 3

Your average question needs to be made more precise since there is not a standard measure on your space of functions.



For all pratical purposes one can work with continuous piecewise affine functions.



For instance you could decide to take (n - 1) random points between 0 and 1, put them in increasing order y_1 < y_2 etc



and define f(1/n) = y_1, f(2/n) = y_2,................., f(1-1/n) = y_(n-1).



One joins these points to get a piecewise affine function. One can measure the average length of the graph for n fixed, and this average has a limit at infinity, giving you a reasonable average.



However if you choose the x_i's and y_i's independently and again put them in increasing order, join the points, calculate the average length and let n tend to infinity, you'll get another average but larger than the previous one. So there is a choice to be made somewhere..



edit: Here are the values of the averages



m_1 = int [0,+infty] (1+x^2)^1/2 * e^(-x) dx = 1.5388662



m_2 = 2*int[0,pi/2] dx/(sin x + cos x)^3 = 1+ln(1+sqrt(2)) / sqrt(2) = 1.6232252









edit for some reason I can't edit my answer to your other q,



https://answersrip.com/question/index?qid=20120730213439AATvELR



so I post it here:



Here are a few facts about your paths. It is convenient to use (0,0) and (n,n) as the end points of your paths. With these notations one has



....n =1.....2......3........4.........5......…



P(n) =3....13....63.....321.....1683.....8989



A(n) = 5...44...321...2184...14325... 91860...



A general formula for P(n) is sum [k<=n] (n+k)! / [(n-k)!k!k!]



This enables you to get exact values with Wolfram for small values of n



http://www.wolframalpha.com/input/?i=sum…



Similarly P(n) is sum [k<=n](n+k)*(n+k)! / [(n-k)!k!k!]



http://www.wolframalpha.com/input/?i=sum…



Using stirling's formula one can find asymptotics both for A(n) and P(n). One gets



A(n) ~ C*(1+sqrt(2))^2n /sqrt(n) where C = (2+sqrt(2))*2^(1/4) / 4*sqrt(pi)



P(n) ~ (1+sqrt(2)/2)*n*A(n)



From which it follows that A(n)*P(n) ~ (1+sqrt(2))^(4n+3) / 8pi.



A consequence of these asymptotics is that the probabilities of going north, east or north-east are not 1/3,1/3,1/3 but sqrt(2)/4, sqrt(2)/4, 1 - sqrt(2)/2, at least for n large.