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Diagonal Lines

  1. Jun 25, 2008 #1
    Alright, I have a question that I know is very dumb, but despite my best efforts I've been unable to come up with an answer that satisfies me. Here it is:

    We know the shortest distance b/n two points is a straight line. So, if we have a right triangle with legs of unit length, the shortest distance between the endpoints is the hypotenuse (which will have length [tex]\sqrt{2}[/tex]).

    Okay, now let's walk the distance b/n the two points in several different ways:

    1.) Walk 1 unit East, then 1 unit North. Total distance = 1 + 1 = 2.

    2.) Walk 1/2 unit East, then 1/2 unit North, then 1/2 unit East, then 1/2 unit North. Total Distance = 1/2 + 1/2 + 1/2 + 1/2 = 2.

    3.) Walk 1/3 unit East, then 1/3 unit North, 1/3 unit East, then 1/3 unit North, 1/3 unit East, then 1/3 unit North. Total distance = 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 = 2.

    I'm sure you see where I'm going by now. So, my question is, why doesn't the sum of this "stair step" method approach [tex]\sqrt{2}[/tex] as your step size approaches 0? Is there something fundamentally different about a diagonal line? I just can't see what the difference is b/n a "stair step" of infinitesimal step size and a diagonal line...

    This has been bugging me for a while, so someone please set me straight (pun intended).
  2. jcsd
  3. Jun 25, 2008 #2
    What do you mean "why"? As you have pointed out, the total distance traveled remains 2. For a "stair", if you look at horizontal segments alone, you clearly see that they add up to the horizontal distance between the two bases, independently of how many segments (or stairs, same number) there are. Same goes with the vertical segments.
  4. Jun 26, 2008 #3
    Yes, of course I see that, but it doesn't answer my actual question: What is the difference b/n a stair step of infinitesimally small step size and an actual diagonal line? So, to apply the question to the example at hand: Why would a stair step of infinitesimally small step size from start to end yield a total distance of 2, whereas a diagonal line would yield a total distance of [tex]\sqrt{2}[/tex]?
  5. Jun 26, 2008 #4
  6. Jun 26, 2008 #5
    Thanks for the great article. It's nice to realize that I'm not the only one to have been confused by this. I was actually getting a little depressed (I just commented to my wife that I felt like I needed to return my recently-earned physics degree...)

    But, despite the great points in the article, it doesn't really satisfy me. Basically, it just says that you have to be careful not to assume that because some limiting case has some of the properties of entity it's approaching that it has all the same properties. That's a great point, but it doesn't actually get into the why of things.

    So, why does this limiting scenario (the stair step) not possess all the properties of the entity it's approaching (the diagonal line)? I'm not seeing, fundamentally, what the difference is.
  7. Jun 26, 2008 #6
    It seems as though the source of your questions stems from your concept of "infinitely small". The "infinitely small" is a nebulous notion, no longer regarded as meaningful by modern mathematicians. One must speak of the passage to the limit (2), which has a clear definition, instead of obstructing the matter with unclear terms (i.e. "infinitely small step sizes).
  8. Jun 26, 2008 #7


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    You made an assumption about how things behave -- that it didn't matter in which order you applied the steps "compute the length" and "take the limit". You looked at an actual example. The example demonstrated your assumption was flawed. I'm not sure what you find unsatisfactory about this state of affairs.

    Maybe simply restating what you have shown will help? In whatever topological space you're using to make sense of that limit, you've proven that the following cannot both be true:
    (1) That sequence of stairsteps converges to that line segment
    (2) The length function is continuous

    In particular, your work says (almost literally) that if (1) holds, then that line segment is a discontinuity of the length function.
    Last edited: Jun 26, 2008
  9. Jun 26, 2008 #8
    If you draw a picture of your stairs, and make the step size really small, the picture will "look like" a diagonal line, but it is not! Take just one of your stair "steps" and pake the picture bigger - you will see that the diagonal length is still the hypotenuse and its length is still sqrt(x^2 + y^2). No matter how small the step size, this picture of just one step is always the same. When you add up all those little hypotenuses, you get sqrt(2); you do not get 1+1 = 2.

    Does that help?
  10. Jun 27, 2008 #9

    Progress, progress, progress. Thanks to the generous people on this forum (generous with their time), you know know that your question is very very far from being a dumb question.

    Notice your criteria for an answer, however. You want "an answer that satisfies me."

    (Somehow, "You want 'an answer that satisfies you.'" seems like a more natural way to put that last sentence.)

    That's not a criteria that anybody can really work to in a logical manner except you. (More accurately, in a "deductive" manner.)

    I applaud you for holding out for an answer that satisfies your criteria.

    Here's my contribution to the mix.

    Let f:[0,1]->[0,1] be the function defined by f(x) = x. the derivative of f(x) is one (f'(x)=1).

    Let f_n be the nth staircase function as you describe it.

    Then, for f_n converges uniformly to f (as n-->infinity).

    ("Converges uniformly" means converges pointwise AND for every epsilon in the delta-epsilon definition of a limit, the same N_0 works for all points in the unit interval [1,1].)

    (f_n converges pointwise to f as n goes to infinity) iff (for all real numbers x)(for all epsilon greater than zero)(there exists a natural number N_0 such that)(n>N_0 ==> |f_n(x)-f(x)|<epsilon).

    However, the limit as n-->infinity of the derivatives of f_n (f_n') do not converge to any function. In fact, loosely speaking, the sequence f_n'(x) is equal to zero an infinite number of times and it is equal to infinity an infinite number of times.

    BTW, a question that I have forgotten the answer to, and maybe one of you gurus know off the top of your head is, is the following statement true when (f and f_n are differentiable functions mapping R into R)?

    If f_n -> f uniformly and f_n'->g pointwise as n-->infinity, then f = g?

    If not, what if f and f_n are continuously differentiable?

    What if f_n'->g uniformly?

    Are there other cases that are interesting?

    Last edited: Jun 27, 2008
  11. Jun 28, 2008 #10


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    Yes, if fn goes to f uniformly then it also goes to f pointwise and the limit is unique.
  12. Jun 28, 2008 #11
    Thanks, HallsOfIvy! DJ
  13. Jul 22, 2009 #12
    This is an interesting problem, intuitively I would imagine that one can draw a stair case with a blunt pencil, there is a limit where the thickness of the line would cause the staircase to look like a diagonal. This is the point at which definitions of length would be ambiguous because they are the same order of magnitude as the width of the line drawn. No matter how the pencil is sharpened, this point will always be reached. This would indicate that there is not a gradual approximation from a staircase to a diagonal and that the two are fundamentally different objects when viewed at the appropriate resolution. Not too sure how that translates into mathematical terms though.
  14. Jul 22, 2009 #13


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    Did you notice that this thread was a year old?

    In any case, the "mathematical terms" for what you are saying is what I said before: that the convergence is pointwise but not uniform.
  15. Jul 23, 2009 #14
    Thanks for the clarification (it helped me understand some of the more technical aspects that were put forward earlier) I did notice the time stamp of a year ago but as I was thinking about the same problem just recently, I decided to venture a comment anyway.
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