Composing the function $f(\rho,\theta)=(\frac{2 \rho}{\rho +1}\cos(\theta+\frac\pi4),\frac{2 \rho}{\rho +1}\sin(\theta+\frac\pi4))$ with itself

Question

I would like to compose $k$-times this function $$f(\rho,\theta)=\left(\dfrac{2 \rho}{\rho +1}\cos\left(\theta+\dfrac{\pi}{4}\right),\dfrac{2 \rho}{\rho +1}\sin\left(\theta+\dfrac{\pi}{4}\right)\right)$$ I know that the answer is : $$f^{k}(\rho,\theta)=\left(\dfrac{2^{k} \rho}{(2^{k}-1)\rho +1}\cos\left(\theta+\dfrac{k\pi}{4}\right),\dfrac{2^{k} \rho}{(2^{k}-1)\rho +1}\sin\left(\theta+\dfrac{k\pi}{4}\right)\right)$$ but I can't compute that.

Answer 1

If we agree to use square brackets $[\rho, \theta]$ for polar coordinates, in order to distinguish them from the usual Cartesian coordinates $(\rho \cos \theta, \rho \sin \theta)$, then your function is $f([\rho, \theta]) = [\frac {2 \rho} {\rho + 1}, \theta + \frac \pi 4]$. Let us compute a few of those compositions in order to get a feeling about how they behave:

• $$(f \circ f) ([\rho, \theta]) = f \left( \left[ \frac {2 \rho} {\rho + 1}, \theta + \frac \pi 4 \right] \right) = \left[ \frac {2 \frac {2 \rho} {\rho + 1}} {\frac {2 \rho} {\rho + 1} + 1}, \left( \theta + \frac \pi 4 \right) + \frac \pi 4 \right] = \left[ \frac {2^2 \rho} {3 \rho + 1}, \theta + \frac {2\pi} 4 \right] ,$$

• $$(f \circ f \circ f) ([\rho, \theta]) = f \left( \left[ \frac {2^2 \rho} {3 \rho + 1}, \theta + \frac {2\pi} 4 \right] \right) = \left[ \frac {2 \frac {2^2 \rho} {3 \rho + 1}} {\frac {2^2 \rho} {3 \rho + 1} + 1}, \left( \theta + \frac {2\pi} 4 \right) + \frac \pi 4 \right] = \\ \left[ \frac {2^3 \rho} {7 \rho + 1}, \theta + \frac {3\pi} 4 \right] .$$

The above two are enough to guess that

$$\underbrace{(f \circ \dots \circ f)} _{k \text{ times}} ([\rho, \theta]) = \left[ \frac {2^k \rho} {a_k \rho + 1}, \theta + \frac {k\pi} 4 \right]$$

with $a_k$ unknown for the moment and $a_1 = 1$. Reasoning by induction will give us $a_k$ and also verify the rest of the formula.

For $k=1$ things are clear. For $k+1$ we have

$$\underbrace{(f \circ \dots \circ f)} _{k+1 \text{ times}} ([\rho, \theta]) = f \left( \left[ \frac {2^k \rho} {a_k \rho + 1}, \theta + \frac {k\pi} 4 \right] \right) = \left[ \frac {2 \frac {2^k \rho} {a_k \rho + 1}} {\frac {2^k \rho} {a_k \rho + 1} + 1}, \left(\theta + \frac {k \pi} 4 \right) + \frac \pi 4 \right] = \\ \left[ \frac {2^{k+1} \rho} {(2^k + a_k) \rho + 1}, \theta + \frac {(k+1) \pi} 4 \right] ,$$

whence we deduce that our guessing is correct and that

$$a_{k+1} = a_k + 2^k = a_{k-1} + 2^{k-1} + 2^k = \dots = a_1 + 2^1 + 2^2 + \dots + 2^k = \sum _{i=0} ^k 2^i = 2^{k+1} - 1 ,$$

so, indeed, the formula becomes

$$\underbrace{(f \circ \dots \circ f)} _{k \text{ times}} ([\rho, \theta]) = \left[ \frac {2^k \rho} {(2^k - 1) \rho + 1}, \theta + \frac {k \pi} 4 \right] = \\ \left( \frac {2^k \rho} {(2^k - 1) \rho + 1} \cos \left( \theta + \frac {k \pi} 4 \right), \frac {2^k \rho} {(2^k - 1) \rho + 1} \sin \left( \theta + \frac {k \pi} 4 \right) \right) .$$