Total length of graph

**Mhmh96** · July 8th 2012, 01:43 PM

How to solve it?

**emakarov** · July 8th 2012, 03:55 PM

**Mhmh96** · July 8th 2012, 04:02 PM

I've drawn that , It's a series of triangles compressed along the x-axis,and..?

**emakarov** · July 8th 2012, 04:09 PM

Guess the number of triangles in the graph of $f^n(x)$ as a function of n. ( $f^n(x)$ has $n$ nested occurrences of $f$ .) When you know the number of triangles, you know all the characteristics of each triangle (first the altitude and the base length, from which you can find the side lengths since the triangles are isosceles).

**Mhmh96** · July 8th 2012, 04:16 PM

Originally Posted by emakarov

Guess the number of triangles in the graph of $f^n(x)$ as a function of n. ( $f^n(x)$ has $n$ nested occurrences of $f$ .) When you know the number of triangles, you know all the characteristics of each triangle (first the altitude and the base length, from which you can find the side lengths since the triangles are isosceles).

But how to know what is $f^n(x)$ ,and how to know the number of triangles ?

**emakarov** · July 8th 2012, 04:20 PM

Originally Posted by Mhmh96

But to know what is $f^n(x)$ ?

The graphs in post #2 are graphs of $f(x)$ , $f^2(x)$ , $f^3(x)$ and $f^4(x)$ , and you need to find the length of the graph of $f^{2012}(x)$ .

**Mhmh96** · July 8th 2012, 04:24 PM

Originally Posted by emakarov

The graphs in post #2 are graphs of $f(x)$ , $f^2(x)$ , $f^3(x)$ and $f^4(x)$ , and you need to find the length of the graph of $f^{2012}(x)$ .

and then I have to do that another 2,008 times?

**emakarov** · July 8th 2012, 04:26 PM

Originally Posted by Mhmh96

how to know the number of triangles ?

Based on the graphs of $f^n(x)$ for small values of n, make a hypothesis about the relationship between n and the number of triangles in the graph of $f^n(x)$ . I am not sure how strictly you need to prove this hypothesis, but going from the graph of $f(x)$ to that of $f^2(x)$ should give you the idea of the proof of the hypothesis.

**Soroban** · July 8th 2012, 04:31 PM

Hello, Mhmh96!

What a strange problem . . . bizarre!

I did a lot of sketching and muttering
. . and I think I've understood the composite functions.
If I'm wrong, my apologies for wasting your time . . .

$\text{Let }f\text{ be the function such that: }\:f(x) \;=\;\begin{Bmatrix}2x && \text{if }x \le \frac{1}{2} \\ 2-2x && \text{if }x > \frac{1}{2}\end{array}$

$\text{What is the total length of }\:\underbrace{f(f(\hdots f}_{2012\:f's} (\hdots)))\:\text{ from }x = 0\text{ to }x = 1\,?$

$y_1 \,=\,f(x)$ has this graph.

Code:

      |
     1+ - - - - - - - *
      |             *   *
      |           *       *
      |         *           *
      |       *               *
      |     *                   *
      |   *                       *
      | *                           *
  - - * - - - - - - - : - - - - - - - * - -
      0              1/2              1

It is an isosceles triangle with base 1 and height 1.
The slanted side has length $\sqrt{1^2+\left(\tfrac{1}{2}\right)^2} \:=\:\tfrac{\sqrt{5}}{2}$ . . . and there two of them.
Hence: . $L_1 \:=\:\sqrt{5}$

$y_2 \:=\:f(f(x))$ has this graph.

Code:

      |
     1+ - - - o - - - - - - - o
      |      o o             o o
      |     o   o           o   o
      |    o     o         o     o
      |   o       o       o       o
      |  o         o     o         o
      | o           o   o           o
      |o             o o             o
  - - o - - - : - - - o - - - : - - - o - -
      0              1/2              1

It has two isosceles triangles with base 1/2 and height 1.
The slanted side has length: $\sqrt{1^2 + \left(\tfrac{1}{4}\right)^2} \:=\:\frac{\sqrt{17}}{4}$ . . . and there are four of them.
Hence: . $L_2 \:=\:\sqrt{17}$

$y_3 \:=\:f(f(f(x)))$ has this graph.

Code:

      |
     1+ - * - - - * - - - * - - - *
      |
      |  * *     * *     * *     * *
      |
      | *   *   *   *   *   *   *   *
      |
      |*     * *     * *     * *     *
      |
  - - * - - - * - - - * - - - * - - - * - -
      0              1/2              1

It has four isosceles triangles with base 1/4 and height 1.
The slanted side has length $\sqrt{1^2 + \left(\tfrac{1}{8}\right)^2} \:=\:\frac{\sqrt{65}}{8}$ . . . and there are eight of them.
Hence: . $L_3 \:=\:\sqrt{65}$

I conjecture that: . $L_n \:=\:\sqrt{2^{2n}+1}}$

And therefore: . $L_{2012} \;=\;\sqrt{2^{4024}+1}$

**Mhmh96** · July 8th 2012, 04:32 PM

Originally Posted by emakarov

Based on the graphs of $f^n(x)$ for small values of n, make a hypothesis about the relationship between n and the number of triangles in the graph of $f^n(x)$ . I am not sure how strictly you need to prove this hypothesis, but going from the graph of $f(x)$ to that of $f^2(x)$ should give you the idea of the proof of the hypothesis.

But what is the "number of triangles" ?its just look equal to n

**Mhmh96** · July 8th 2012, 04:38 PM

Originally Posted by Soroban

Hello, Mhmh96!

What a strange problem . . . bizarre!

I did a lot of sketching and muttering
. . and I think I've understood the composite functions.
If I'm wrong, my apologies for wasting your time . . .

$y_1 \,=\,f(x)$ has this graph.

Code:

      |
     1+ - - - - - - - *
      |             *   *
      |           *       *
      |         *           *
      |       *               *
      |     *                   *
      |   *                       *
      | *                           *
  - - * - - - - - - - : - - - - - - - * - -
      0              1/2              1

It is an isosceles triangle with base 1 and height 1.
The slanted side has length $\sqrt{1^2+\left(\tfrac{1}{2}\right)^2} \:=\:\tfrac{\sqrt{5}}{2}$ . . . and there two of them.
Hence: . $L_1 \:=\:\sqrt{5}$

$y_2 \:=\:f(f(x))$ has this graph.

Code:

      |
     1+ - - - o - - - - - - - o
      |      o o             o o
      |     o   o           o   o
      |    o     o         o     o
      |   o       o       o       o
      |  o         o     o         o
      | o           o   o           o
      |o             o o             o
  - - o - - - : - - - o - - - : - - - o - -
      0              1/2              1

It has two isosceles triangles with base 1/2 and height 1.
The slanted side has length: $\sqrt{1^2 + \left(\tfrac{1}{4}\right)^2} \:=\:\frac{\sqrt{17}}{4}$ . . . and there are four of them.
Hence: . $L_2 \:=\:\sqrt{17}$

$y_3 \:=\:f(f(f(x)))$ has this graph.

Code:

      |
     1+ - * - - - * - - - * - - - *
      |
      |  * *     * *     * *     * *
      |
      | *   *   *   *   *   *   *   *
      |
      |*     * *     * *     * *     *
      |
  - - * - - - * - - - * - - - * - - - * - -
      0              1/2              1

It has four isosceles triangles with base 1/4 and height 1.
The slanted side has length $\sqrt{1^2 + \left(\tfrac{1}{8}\right)^2} \:=\:\frac{\sqrt{65}}{8}$ . . . and there are eight of them.
Hence: . $L_3 \:=\:\sqrt{65}$

I conjecture that: . $L_n \:=\:\sqrt{2^{2n}+1}}$

And therefore: . $L_{2012} \;=\;\sqrt{2^{4024}+1}$

No , your conjecture is right ,nice job indeed.
And there is a typo in the final answer

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