I need help to solve this equation, please.
$$\left(\sqrt{\frac{x-1}{x}}\right)^{x^2}=\left(\frac{1}{x}\right)^{x+1}$$
I know that the solution is $x=\varphi$ (the golden ratio). I got this result by equating the bases and the exponents. I want to know if there are another way to obtain this.
Thank you.
On
Consider that you lookfor the zero of function $$f(x)=\left(\sqrt{\frac{x-1}{x}}\right)^{x^2}-\left(\frac{1}{x}\right)^{x+1}$$ By inspection or plotting, you know that the solution is between $1$ and $2$ and probably closer to $2$ that to $1$ since $f(1)=-1$ and $f(2)=\frac 18$.
However, as already said in comments, if you plot the functio $$g(x)=\log\Bigg[\left(\sqrt{\frac{x-1}{x}}\right)^{x^2}\Bigg]-\log\Bigg[\left(\frac{1}{x}\right)^{x+1}\Bigg] $$ you should notice that it is close to linearity when $x \geq 2$. Simplifying, we have $$g(x)=\frac 12 x^2 \log \left(\frac{x-1}{x}\right)+(x+1)\log(x)$$ Expand it as a Taylor series around $x=2$, you have $$g(x)=\log(2)+\frac{5-2\log (2)}{2}(x-2)+\frac{3-4 \log (2)}{8} (x-2)^2-$$ $$\sum_{n=3}^\infty (-1)^n \frac{ \left(2^n+2\right) (n-5) n+2^{n+3}+4}{2^{n+1}(n-2) (n-1) n} (x-2)^n$$
Truncate to some low order and use series reversion and obtain the estimate $$x=2+t+\frac{ (3-4 \log (2))}{4 (2 \log (2)-5)}t^2+\frac{ \left(7+48 \log ^2(2)-64 \log (2)\right)}{24 (2 \log (2)-5)^2}t^3-\frac{5 \left(189+192 \log ^3(2)-344 \log ^2(2)-4 \log (2)\right)}{192 (2 \log (2)-5)^3}t^4+O\left(t^5\right)$$ where $$t=\frac{\log (4)-2 g(x)}{\log (4)-5}$$
Make $f(x)=0$ as desired; the decimal representation of the result is $x=1.618019$ while the golden ratio is $x=1.618034$
Notice that truncating the series to $O\left(t^2\right)$ (this would correspond to the first iteration of Newton method) already gives $$x_1=\frac{10-6 \log (2)}{5-2 \log (2)}=1.61638$$
Now, for sure, you can use Newton-like methods. The iterates would be $$\left( \begin{array}{cccc} n & \text{Newton} & \text{Halley} & \text{Householder} \\ 0 & 2.0000000000000000000 & 2.0000000000000000000 & 2.0000000000000000000 \\ 1 & 1.6163787259801535786 & 1.6140493818327460672 & 1.6166687509536829242 \\ 2 & 1.6180336777746660995 & 1.6180340035848826741 & 1.6180339887489227152 \\ 3 & 1.6180339887498839464 & 1.6180339887498948482 & 1.6180339887498948482 \\ 4 & 1.6180339887498948482 & 1.6180339887498948482 & 1.6180339887498948482 \end{array} \right)$$
There is no solution in $N$, $Z$, $Q$ or else. But in the reals there is a numerical solution.
Solution is
$x=1.61803$
This can be done by simple numerical methods or the graphical methods:
It is not worth attempting to simplify this.
This sequence so how fast the convergence is:
$x -> 1.61803, 1.6, 1.61762, 1.61803, 1.61803, 1.61803$
Here is graphics of that:
Fixpoint iterations are the methods of choice. There are many out there for that problem type.
As can be seen in the graph the curve are almost linear and can be approximated by Taylor series around the crossing point.
This search gives a nice overview here at math.stackexchange.com: iteration methods.
This tedious and long and takes time to cope with. Try make a good guess or a rough first table of function values and than refine.
Why does this not simplify?
Both functions under consideration are potential functions with variable potence, exponent and basis. There is a common coordinate system possible for the basis. But this is not a good reduction point for the solution start. The different basis and the different exponents, potences do the block for simplification. This is a real composition of not any more simplifable operations. It is therefore a case for numerical or graphical solution.
The continuity and the values in $x=0$ and fo very large $x$ suggest that there is a real crossing.
Apply the series expansion direct to the two given function is faster.
The series expansion around the crossing are
For simplicity.
-0.44091 + 1.34852 x - 0.793004 x^2 + 0.146074 x^3, 4.36948 - 5.42705 x + 2.3944 x^2 - 0.37139 x^3
Hope this helps and convinces that this is all that has to be done.