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2 functuon questions

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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 $a \neq 1$.)

Jul 17, 2018

#1
+101871
+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

Jul 17, 2018
#2
+101871
+1

$$f(a) = \frac{1}{1-a}, find the product f^{-1}(a) \times a \times f(a). (Assume a \neq 0 and a \neq 1.)$$

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

Jul 17, 2018
#3
+22550
0

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 $a \neq 1$.)

$$\begin{array}{|rcll|} \hline f{\color{red}(}f^{-1}(a){\color{red})} &=& a \\\\ f {\color{red}\Big(}\dfrac{a-1}{a}{\color{red}\Big)} &=& a \quad & | \quad \text{see table above } r\left(\dfrac{\theta-1}{\theta} \right) = \theta \\\\ \text{so } \\\\ f^{-1}(a) &=& \dfrac{a-1}{a} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && f^{-1}(a) \times a \times f(a) \\\\ &=& \dfrac{a-1}{a}\times a \times \dfrac{1}{1-a} \\\\ &=& \dfrac{a-1}{1-a} \\\\ &=& -\dfrac{1-a}{1-a} \\\\ &=& -1 \\ \hline \end{array}$$

Jul 18, 2018
edited by heureka  Jul 18, 2018