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Suppose I write down the Manhattan distance from the origin to a point (x,y) in terms of a series of n steps of length x/n in the x direction, alternated with m steps of length y/m in the y direction: $$d_{Manhattan} = \sum_{i=1}^n \frac{x}{n} + \sum_{i=1}^m \frac{y}{m}$$

Imagine walking an approximately diagonal line towards (x,y), zig-zagging parallel to the x and y axes.

Now, if we take the limits as $n\rightarrow \infty$ and $m\rightarrow \infty$ our path should approach the straight line connecting the origin to (x,y), suggesting that in the limit the Manhattan distance should equal $\sqrt{x^2+y^2}$. Why is this not the case? Is there a way to correctly arrive at Pythagoras by taking a limit using infinitesimal steps along the axis directions?

Robert Sim
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    "our path should approach the straight line connecting" Why do you think this? Although these seems intuitive there is not mathematical basis for it. The path is not a straight line and the path, no matter how small intervals you take, never "approaches" being a straight line. – fleablood Mar 24 '20 at 05:35
  • This is a good question. I believe it has to do with the nature of limits and spaces, but I don't know enough to write an answer. I will, however, pose a question of my own - why would you expect the Manhattan/taxicab distance to approach the Euclidean distance? They're different metrics, with wildly different properties. For instance, there is a single unique path that connects two points to give a shortest Euclidean distance, but many paths can give the shortest taxicab distance between two points. – Deepak Mar 24 '20 at 05:35
  • "Is there a way to correctly arrive at Pythagoras by taking a limit using infinitesimal steps along the axis directions?" No. – fleablood Mar 24 '20 at 05:37
  • Related (duplicate?): "If calculus is just based on minute approximations, can't it be wrong?". The body of the question, and the answers, discuss the matter in the above question. I'm sure there are more questions like this, but that's the first I found. (Related discussions about the circumference of a circle have also appeared here.) – Blue Mar 24 '20 at 06:21
  • Represent the length of a path $P$ by $l(P)$. Then the problem with your argument is that you are trying to determine $l(P)$ (with $P$ being the straight path to $(x,y)$) by inventing a series of zigzag paths $P_1, P_2,\dots$, and considering $l(P_1), l(P_2),\dots$. But it is invalid to argue that $\lim l(P_i)$ is $l(\lim P_i)$. Even if $\lim P_i$ were a meaningful concept. This is in addition to fleablood's point. – Rosie F Mar 24 '20 at 06:50
  • @fleablood why do you think it does not approach the line? Pointwise it converges to the straight line (both in euclidean and manhattan metric), doesn’t it? Just because every number in a sequence is rational (hence far from being irrational), its limit can be irrational, so I don’t really understand your point... – Jonas Linssen Mar 24 '20 at 09:30
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    One point is, as you decrease the increment size (I.e. increase $m$ and $n$ for fixed $x$ and $y$), the Manhattan distance from $x$ to $y$ doesn’t change, so the fact that the squiggly path stays near the line is irrelevant for arc length — same length as First going all the way East, then all the way North. That why the arc length differential in calculus isn’t $dx + dy$. – Ned Mar 24 '20 at 11:06
  • A Line is a geometrical figure. Not an numerical value. Pointwise the figure goes toward a line but our method of measuring it does not change. Because we can $\lim_{x\to 0} 2x = \lim_{x\to 0}\sqrt 2 x$ our method of measuring need not and does not change. To assume it would would be to assume that $\lim_{x\to 0}2x = \lim_{x\to 0}\sqrt 2x \implies 2=\sqrt 2$ which of course it does not. – fleablood Mar 24 '20 at 17:21
  • In essence we measure length by #of steps $\times$ length of steps. Although a zig-zaggy path and a line path my both converge to appear to be the same figure, the two quantuum sized ants wearing fitbits who are counting their steps end up walking at different speeds even though they end up both walking the same path. – fleablood Mar 24 '20 at 17:26
  • Related: The staircase paradox, or why $≠4$ – epimorphic Apr 17 '20 at 05:41

1 Answers1

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You can't add up an uncountably many number of immeasurably small points and expect to get the length as a sum.

Yes, the shape and resulting figures both end up being the same line segment but our methods of measuring the length is different. It's a bit like one person measuring the line with a ruler marked in inches saying it is $7$ and another measuring it with a ruler marked in centimeters and saying it is $17.78$.

And lest you think I am being glib, consider we have rulers that measures number of steps. Yours measures number of centimeter steps a tiny gnome would take and you measure that it is $17.78$ centimeter straight steps. Mine measures the number of steps the gnome takes if he goes $1$ centimeter in one direction then turns 90 degrees and goes $1$ centimeter in another. I measure it is $25.15$ centimeter steps. Your ruler is marked of in fractions of steps to infinite precision. So is mine but each of my units is $\frac 1{\sqrt 2}$ as long as each of your units.

We both have the formula. Length = number of steps $\times$ length of steps and an but my number of steps is $\sqrt {2}$ times your number of steps. And for both of us, as length of steps $\to 0$ then number of steps $\to \infty$ and the length of the line as each of us measure it will be $\lim_{\text{length of steps}\to 0}$ number of steps $\times$ length of steps. But for you: number of steps $=\frac {17.78}{\text{length of step}}$ and for me: number of steps $=\frac {\sqrt 2\cdot 17.78}{\text{length of step}}$

If we measure our tiny gnomes with a timer we will notice.... mine is walking slower than yours.

Now notice. Since we can't have measure length as length $=\infty\cdot 0$. But we also have speed = length/time. And we can have instantaneous speed by taking limits. If speed = length/time then length = speed$\cdot$ time. .... And my gnome is slower than yours.

But it's not that my gnome is always slower than yours. Our gnomes live with different metrics. Your gnomes speed is constant at any angle. Mine is not. My gnomes speed matches yours at $0, 90, 180, 270$ degree angles but otherwise is slower than yours depending on the angle with my slowest speed and $45,135, 225, 315$ angles where my gnomes speed is $\frac 1{\sqrt 2}$ of yours.

J. W. Tanner
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fleablood
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