1. Let $r(\theta) = \frac{1}{1-\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?
2. If $f(a) = \frac{1}{1-a}$, find the product $f^{-1}(a) \times a \times f(a)$. (Assume $a \neq 0$ and $a \neq 1$.)
1. \(r(\theta) = \frac{1}{1-\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)? \)
r(30) = 1 / (1 -30) = 1/-29 = -1/29
r(r(30)) = 1 / (1 - (- 1/29)) = 1 / (29/29 + 1/29) = 1 / (30/29) = 29/30
r (r((r(30))) = 1 / ( 1 - 29/30) = 1 / ( 30/30 - 29/30) = 1 / (1/30) = 30
r (r(r(r(30)))) = 1 / (1 - ( 30) ) = 1/ (-29) = -1/29
r (r (r(r(r(30))))) = 1 / ( 1 - (-1/29) ) = ( 29/29 + 1/29) = 1 / (30/29) = 29/30
r (r (r (r(r(r(30)))))) = 1 / (1 - (29/30) ) = 1 / ( 30/30 - 29/30) = 1/(1/30) = 30
\(f(a) = \frac{1}{1-a}$, find the product $f^{-1}(a) \times a \times f(a)$. (Assume $a \neq 0$ and $a \neq 1$.)\)
Write y = 1 / ( 1 - a) swap a and y and we want to get y by itself
a = 1 / ( 1 - y) multiply both sides by ( 1 - y)
a ( 1 - y) = 1 divide both sides by a
1 - y = 1 /a rearrange as
1 - 1 / a = y get a common denominator on the left side
(a - 1) / a = y = f-1(a) and this is the inverse
So
f-1 (a) * a = (a - 1) / a * a = (a -1)
And this product multiplied by f(a) = [ (a -1) ] * [ 1 / (1 - a) ] = [ (a -1) ] * [ - 1 (a - 1) ] = -1
f
1. Let $r(\theta) = \frac{1}{1-\theta}$. What is $r(r(r(r(r(r(30))))))$ (where $r$ is applied $6$ times)?
\(\begin{array}{|rcll|} \hline r(\theta) &=& \dfrac{1}{1-\theta} \\ \hline r(r(\theta)) &=& \dfrac{1}{1-\dfrac{1}{1-\theta}} \\\\ &=& \dfrac{1-\theta}{1-\theta-1 } \\\\ &=& \dfrac{1-\theta}{ -\theta } \\\\ &=& \dfrac{\theta-1}{ \theta } = 1-\dfrac{1}{\theta} \\\\ \hline r(r(r(\theta))) &=& \dfrac{1}{1- (1-\dfrac{1}{\theta}) } \\\\ &=& \dfrac{1}{1- 1 + \dfrac{1}{\theta} }\\\\ &=& \dfrac{1}{ \dfrac{1}{\theta} } \\\\ &=& \theta \\ \hline r(r(r(r(\theta)))) &=& \dfrac{1}{1-\theta} \\ \hline \end{array}\)
This is a cycle:
\(\begin{array}{|r|r|c|} \hline \text{cycle} & & \ldots r(r(\theta) \\ \hline 1 & \text{once} & \color{red}\dfrac{1}{1-\theta} \\ \hline 1 & \text{twice} & \color{green}\dfrac{\theta-1}{ \theta } \\ \hline 1 & 3\text{ times} & \color{blue}\theta \\ \hline\hline 2 & 4\text{ times} & \color{red}\dfrac{1}{1-\theta} \\ \hline 2 & 5\text{ times} & \color{green}\dfrac{\theta-1}{ \theta } \\ \hline 2 & 6\text{ times} & \color{blue}\theta \\ \hline \ldots & \ldots& \ldots \\ \hline \end{array} \)
\(\text{So $r(r(r(r(r(r(\theta)))))) = \theta $ and $ r(r(r(r(r(r(30)))))) = \mathbf{30} $ } \)
2. If $f(a) = \frac{1}{1-a}$, find the product $f^{-1}(a) \times a \times f(a)$. (Assume $a \neq 0$ and $a \neq 1$.)
\(\begin{array}{|rcll|} \hline f{\color{red}(}f^{-1}(a){\color{red})} &=& a \\\\ f {\color{red}\Big(}\dfrac{a-1}{a}{\color{red}\Big)} &=& a \quad & | \quad \text{see table above } r\left(\dfrac{\theta-1}{\theta} \right) = \theta \\\\ \text{so } \\\\ f^{-1}(a) &=& \dfrac{a-1}{a} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline && f^{-1}(a) \times a \times f(a) \\\\ &=& \dfrac{a-1}{a}\times a \times \dfrac{1}{1-a} \\\\ &=& \dfrac{a-1}{1-a} \\\\ &=& -\dfrac{1-a}{1-a} \\\\ &=& -1 \\ \hline \end{array}\)