4
$\begingroup$

Suppose one is given the following visual proof that

$$\lim\limits_{n \to \infty} \sum_{k=1}^n \frac{1}{2^k} = 1$$

which is the following construction over $[0,1]\times[0,1]$

enter image description here

What this is suggesting is that the sum of the areas of the triangles, which is $1/2,1/4,1/8,\dots$ indeed covers the unitary square with area 1.

What I want to know if it is possible to give any sense to this construction, or if it is possible to give a geometric/analytic proof, by showing the sequence of points converges to $(1,1)$. The sequence is defined as, $x(P)$ and $y(P)$ being the $x$ and $y$ entry of the point $P$:

$$P_1 = (1,0)$$ $$P_2 = \left(\frac{x(P_1)+y(P_1)}{2},\frac{x(P_1)+y(P_1)}{2} \right)$$ $$P_3 = (1, x(P_2))$$ $$P_{2n} = \left(\frac{x(P_{2n-1})+y(P_{2n-1})}{2},\frac{x(P_{2n-1})+y(P_{2n-1})}{2} \right) $$ $$P_{2n+1} = \left(1,x(P_{2n}) \right) $$

this is to say, we take the arithmetic mean of the $x,y$ entries, and then move the point to the line $x=1$. The idea is that since the limit/fixed point is $1$ then it must be the case that the square is covered by the infinitely many triangles produced. I'm not sure what other mathematical notion is needed to interpret this, so I hope you can understand what I'm striving for and help me out.

$\endgroup$
  • $\begingroup$ I don't understand, we can find a simple explicit expression for the coordinates of the $P_i$. $\endgroup$ – André Nicolas Apr 16 '12 at 17:56
  • $\begingroup$ @AndréNicolas What is it that you don't understand? $\endgroup$ – Pedro Tamaroff Apr 16 '12 at 17:59
  • $\begingroup$ Since one can find explicit expression, the limit of the $P_i$ is easy to compute. $\endgroup$ – André Nicolas Apr 16 '12 at 18:01
  • $\begingroup$ Maybe that would help ? en.wikipedia.org/wiki/Newton's_method $\endgroup$ – Student Apr 16 '12 at 18:06
  • $\begingroup$ @AndréNicolas What I want to know is what is needed to give sense to the proof. I have no formal interpretation of the fact the triangle "covers" the unit square, which I thing are pertinent to measures, topology, or something of the kind. $\endgroup$ – Pedro Tamaroff Apr 16 '12 at 18:06
6
$\begingroup$

It's a two-dimensional visualization of the fact that $$\Bigl[0,{1\over2}\Bigl[\ \cup\ \Bigl[{1\over2},{3\over4}\Bigl[ \ \cup\ \Bigl[{3\over4},{7\over8}\Bigl[\ \cup\ \ldots=\ [0,1[$$ where the successive intervals have lengths ${1\over2}$, ${1\over4}$, $\ldots\ $. So it has to do with $\sigma$-additivity of measure. If you are uneasy with this it is simpler to consider the one-dimensional situation.

$\endgroup$
  • $\begingroup$ Very nice ! (the best I ever seen ) $\endgroup$ – Khosrotash Mar 25 '16 at 9:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.