10
$\begingroup$

Solve: $$x = \left(x-\frac{1}{x}\right) ^ {1/9} + \left(1-\frac{1}{x}\right)^{1/9}$$

Simplifying, $$x^{10/9} = (x^2-1)^{1/9}+(x-1)^{1/9}$$
I don't know how to start. Any hint will be helpful.

$\endgroup$
11
  • $\begingroup$ Hint: you can write $x^2-1$ as $(x+1)(x-1)$ $\endgroup$ Feb 2, 2016 at 14:48
  • 1
    $\begingroup$ I think if it is $1/2$ instead of $1/9$ then it is easy to solve. Then we get a very well known equation $x^2-x-1=0$ Which has solution $x=\dfrac{1+\sqrt(5)}{2}$ $\endgroup$ Feb 2, 2016 at 14:59
  • 1
    $\begingroup$ @RezwanArefin, are you sure there is a 9 here? Also, are you sure there is a symbolic solution? Maybe you just need a numerical or approximate solution? $\endgroup$
    – Yuriy S
    Feb 2, 2016 at 15:12
  • $\begingroup$ There are two roots, one at $x = 1.001391$ and one at $x = 1.966965$. $\endgroup$ Feb 2, 2016 at 15:12
  • 1
    $\begingroup$ @HansEngler I also got those two using Wolfram Alpha. But how to solve in hand? $\endgroup$ Feb 2, 2016 at 15:17

1 Answer 1

1
$\begingroup$

Seeing as OP was not satisfied with the numerical solution generated by Wolframalpha, there are ways to generate it with calculator alone.

As I see it, the problem is well posed for fixed-point iteration method. Let's start with $x \approx 2$ (it is easy to guess, as I will show later).

So we take $x_0=2$ for the first root. Then:

$$ x_1=\left(x_0−\frac{1}{x_0}\right)^{1/9}+\left(1−\frac{1}{x_0}\right)^{1/9}= $$

$$ =\left(\frac{3}{2}\right)^{1/9}+\left(\frac{1}{2}\right)^{1/9}=1.97196... $$

So we take $x_1=1.97196...$ and find $x_2$:

$$ x_2=\left(x_1−\frac{1}{x_1}\right)^{1/9}+\left(1−\frac{1}{x_1}\right)^{1/9}=1.9677... $$

We are already very close to the solution (which is $1.966965...$) so the number of iterations we use depends on the precision we need.

The second root is tricky because it's too close to $1$. We can't take $x_0=1$ because then we'll get $x_1=0$, so we need to carefully choose the starting point. I will not elaborate further.


On the other hand, there is another way to approximate the second root. Since $x$ is really close to $1$ in this case, we can take $x=1+a$ and use the Taylor expansion:

$$ x = \left(1+a−\frac{1}{1+a}\right)^{1/9}+\left(1−\frac{1}{1+a}\right)^{1/9} $$

$$ 1+a \approx \left(2a\right)^{1/9}+\left(a\right)^{1/9} $$

$$ a \approx \frac{1}{(2^{1/9}+1)^9}=0.00137... $$

This answer is again very close to the accurate solution $1.00139...$. We can make it better by keeping 2 order terms in Taylor expansions.


As for $x \approx 2$ it is easy to show without calculator. We know that $a^{1/n} \approx 1$ for large $n$, unless $a$ is too small or too large. Let's eliminate these possibilities:

If $x \gg 1$ then from taking limits in the original equation we have: $$ x \approx x^{1/9} \Rightarrow x \approx 1 $$ We arrived at a contradiction so $x$ can't be very large.

If $x \ll 1$ then from taking limits in the original equation we have: $$ x \approx -2 \left(\frac{1}{x}\right)^{1/9} $$ Which doesn't even have real roots, so $x$ can't be very small.

From these we can show that both brackets have the same order, so if we take them both to be of the same order as $1$, we have:

$$ x \approx 1+1=2 $$

Note, that 'of the same order as $1$' doesn't mean they have to be especially close, for example $8^{1/9}=1.2599...$, $0.3^{1/9}=0.8748...$. Both values are close enough to $1$ after taking the root.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .