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I started to learn calculus by myself. First chapter of my textbook is about the limit of the sequence. I did all exercises in my textbook except one problem. There is some special problems in the end of this chapter: you need to find a mistake in the given definition. The last one is very weird. I don't understand how to strictly mathematically prove why this definition is wrong:

$L(a_n)$ - length of the curve $a_n$.

$D(a_n(P),S_{AB})$ - distance between point $P \in a_n$ and segment $S_{AB}$ (perpendicular from the point P to the segment $S_{AB}$).

Definition: Sequence of smooth continuous curves $a_n$ is called an approximation for segment $S_{AB}$ if:

  1. All curves $a_n$ begins at point A and ends at B.

  2. For any $m<n, \{m,n\} \in \mathbb{N}, \ L(a_m) \geq L(a_n)$.

  3. For each $\epsilon >0$ there exists a natural number $N$ such that, for every $n\geq N$, for every points $P \in a_n$ we have $D(a_n(P),S_{AB})<\epsilon$.

  4. If (1-3) true then the sequence of smooth continuous curves $a_n$ is an approximation for a segment $S_{AB}$, their length tends to the limit L, which is length of a segment $S_{AB}$.


It is definitely wrong. With this definition we can prove that $5=4$.

May be we should change in 2) that $L(a_m) > L(a_n)$? Or this is unfixable?

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A definition does not assert anything; it does not have a truth value. –  Qiaochu Yuan Aug 27 '12 at 5:00
    
You have to find out that part of the definition that is not consistent with the conclusion we can make using it and some other facts we already know are true. For example, in your case, the second condition of the definition is falsely assumed I think. –  Jayesh Badwaik Aug 27 '12 at 5:02
    
What do you mean by proving that $5=4$? What is $5$? –  Thomas E. Aug 27 '12 at 6:28
    
@Thomas "the approximating curve $a_n$ can be near the given curve in the sense of (3), but can have length much greater than the length of the curve" –  Mike Aug 27 '12 at 6:42

1 Answer 1

up vote 4 down vote accepted

The definition is in principle fixable (after all, one could fix it by giving the standard definition), but there is a fundamental flaw: the approximating curve $a_n$ can be near the given curve in the sense of $(3)$, but can have length much greater than the length of the curve.

The definition of length needs to give the "right answer" for simple cases where we have a good intuition, in particular for straight line segments. However, if we take the diagonal of a $1\times 1$ square, we can find a sequence of zig-zag straight line segments, with each straight line segment parallel to a side of the square, which is close to the diagonal in the sense of $(3)$, and always has total length $2$.

True, these zig-zag paths have the property that $L(a_n)$ is constant for all $n$. But replacing $L(a_m)\ge L(a_n)$ for $m\le n$ by $L(a_m)\gt L(a_n)$ will not help much. A small modification of the zig-zag paths can make their lengths strictly decreasing, with limit any number in the interval $[\sqrt{2},2)$.

Note that not all of the "definition" is a definition. Assertion $(4)$ is in fact a theorem statement. That "theorem" happens to be false. The limit is very dependent on the sequence of "approximating curves" that we choose.

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Look how I think about this: We can approximate the length of a circle by computing the perimeters of circumscribed and inscribed polygons. This polygons form two sequences, which satisfy to this definition. (We can smooth angles of polygon if you wish). In another hand, i can construct an example, as you said "approximating curve an can be near the given curve in the sense of (3), but can have length much greater than the length of the curve." So, just to be clear, this example shows that "theorem" (4) and all definition is false? This is strict prove? –  Mike Aug 27 '12 at 5:48
    
If the definition had said something like "if all $a_n$ give the same limit, that limit is called the length of the curve," then the definition would be inadequate but not false. (It would be inadequate because even in the simple case of a straight line, not all sequences satisfying $(1)$, $(2)$, and $(3)$ give the same limit. So if they all give the same limit never holds.) The assertion in $(4)$ that they all give the same limit is outright false. –  André Nicolas Aug 27 '12 at 5:54
    
I see! It is my mistake when I translate it. In original, last 4) says that every next $a_n$ tends to the length of AB. And when $n \rightarrow \infty$ its limit is equal to the length of AB. Not that all this curves give the same limit. –  Mike Aug 27 '12 at 6:02
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@Mike: For any particular sequence $(a_n)$ of approximating curves, the limit does exist (non-increasing sequence, bounded below). However, that limit is very dependent on the sequence $(a_n)$ chosen, so cannot serve as the definition of length: a definition should, at least in nice cases, give a definite answer. If $L$ is the actual length, by appropriate choice of $(a_n)$ satisfying $(1,2,3)$ we can get any limit in $[L,\infty)$. –  André Nicolas Aug 27 '12 at 6:15
    
Thank you for your patience. –  Mike Aug 27 '12 at 6:22

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