Deriving the diagonal with limens

[bedaone]

I haven't made any real calculations on this, just thought a bit in my head...

Assume a square with side 1, meaning the diagonal would be 1.41.
The distance between 2 corners if moving along the edges is of course 2.

Now starting in upper left corner let's move
0.5 in x
-0.5 in y
0.5 in x
-0.5 in y
forming a staircase to lower right. Still total distance 2, right?

Dividing this into smaller steps, i.e. lim x,y -> 0, the distance would always be 2, even when the steps are in fact forming a diagonal.

Obvioulsy I am making some error in my thoughts, but i can't figure out what...

 

[mathman]

You have observed a case of a general situation in limit theory, i.e. if a sequence of functions converges to a specific function, it doesn't follow the derivatives of the sequence of function converge to the derivative of the limit. To see it a little more clearly, rotate your figure by 45 deg. so that the diagonal from the upper left to the lower right is horizontal. The staircase now looks like a saw tooth. The length is still 2, while the horizontal line is sqrt(2). The derivatives of the sawtooth segments are +1 and -1 alternatively, but the sequence doesn't converge at all, while the derivative of the limit is 0.

 

[bedaone]

Hmm... I am not sure i understand fully...
Even infinitesmall sawtooths will have a derivative of +-1, while the horizontal line of course has a derivative of 0. Because of this, deriving the diagonal in this way is not possible as i understand it.

However, you mention that in a sequence converging to a specific function the derivatives does not converge. Does that mean that the sawtooth sequence is indeed converging towards c^2 = sqrt(a^2 + b^2) in some meaning, but the derivatives are not?

 

[mathman]

Sequence converges - derivative does not. To further clarify, the length of the function is given by the integral of (y'²+1)^1/2), where y' is the derivative of the function. The sawtooth square is 1, while the limit has a derivative of 0.

 

[mathwonk]

however if the derivatives of the fn are continuous and do converge uniformly to something, that something, interestingly, is the derivative of the uniform limit of the fn.

 

[mathman]

however if the derivatives of the fn are continuous and do converge uniformly to something, that something, interestingly, is the derivative of the uniform limit of the fn.

Unfortunately, in the original problem, the derivative sequence doesn't converge at all.