0
$\begingroup$

$$c = x \left( 1 - \left( 1- \frac{1}{x} \right)^a \right)$$

where $c$ and $a$ are constants. How can one solve this equation?

$\endgroup$
  • 1
    $\begingroup$ Is $a$ an integer? Or more specifically a natural number? $\endgroup$ – user546996 Mar 29 '18 at 17:26
  • $\begingroup$ Yes, $a$ is a natural number. $\endgroup$ – potrto Mar 29 '18 at 17:51
0
$\begingroup$

Obviously $x\ne 0.$ If $a$ is not an integer, then we should assume that $1-\frac{1}{x}>0$, so $\frac{1}{x}<1$ and $x\in(-\infty,0)\cup(1,\infty).$ Uur equation reads as $$\frac{c}{x}=1-\left(1-\frac{1}{x}\right)^a.$$ Next, if $c=0$, then $$1-\frac{1}{x}=1,$$ which is impossible. After next rearrangement we arrive at $$\frac{1}{c}\left[1-\left(1-\frac{1}{x}\right)^a\right]=\frac{1}{x}.$$ Put $C=\frac{1}{c}$ and $t=\frac{1}{x}.$ Then $t$ is a fixed point of $f(t)=C(1-(1-t)^a).$

$\endgroup$
  • $\begingroup$ I'd like to numerically calculate the fixed point of your function. Should I use fixed-point iteration method? $\endgroup$ – potrto Mar 29 '18 at 17:59
  • $\begingroup$ You could use this method if $f$ is a contraction. $\endgroup$ – szw1710 Mar 29 '18 at 18:27
0
$\begingroup$

Note that

$$c = x \left( 1 - \left( 1- \frac{1}{x} \right)^a \right)\iff \frac c x=1 - \left( 1- \frac{1}{x} \right)^a\iff1- \frac{c}{x}=\left( 1- \frac{1}{x} \right)^a\\\iff c\left( 1- \frac{1}{x} \right)-c+1=\left( 1- \frac{1}{x} \right)^a\iff\frac{\left( 1- \frac{1}{x} \right)^a-1}{\left( 1- \frac{1}{x} \right)-1}=c$$

and by $y= 1- \frac{1}{x}$

$$f(y)=\frac{y^a-1}{y-1}=c$$

$\endgroup$
  • $\begingroup$ I'm sorry, I'm not a mathematician. How can this help me to calculate $y$? $\endgroup$ – potrto Mar 29 '18 at 17:56
0
$\begingroup$

Put $$1-\dfrac{1}{x}=t\iff x=\frac{1}{1-t}\Rightarrow c(1-t)=1-t^a$$ Therefore your equation is equivalent to a trinomial $$t^a-ct+(c-1)=0$$ from which the difficulty is clear for arbitrary constant $a$. You do have to apply numerical techniques when $a\notin\{0,1,2\}$ .

$\endgroup$

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.