how can i solve the following equation?
The equation can be written in more convenient form as...
$\displaystyle \displaystyle x^{2}\ ln x - \ln 2 = 0$ (1)
... i.e in the form $\displaystyle f(x)=0$, and it can be solved iteratively using the Newton-Raphson method...
$\displaystyle \displaystyle x_{n+1}= x_{n} - \frac{f(x_{n})}{f^{'} (x_{n})}$ (2)
... starting from an arbitrary $\displaystyle x_{0}$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
can i solve it without using any numerical method? also, when you solve using this method, -sqrt(2) (which is a solution) will be eliminated.The equation can be written in more convenient form as...
that's true.I am positive that there are only two real solutions:
x_1=sqrt(2)
x_2=-sqrt(2)
The Newton's sequence for solving the equation...
$\displaystyle \displaystyle x^{2}\ \ln x - \ln 2 =0 $ (1)
... is expressed as...
$\displaystyle \displaystyle \Delta_{n} = x_{n+1} - x_{n} = \frac{\ln 2 - x_{n}^{2}\ \ln x_{n}}{x_{n}\ (1+ 2\ \ln x_{n})} = \varphi (x_{n})$ (2)
The function $\displaystyle \varphi(x)$ is represented here...
If x has to be consider positive real, there is only one attractive fixed point at $\displaystyle x=\sqrt{2}$ and for any $\displaystyle x_{0} > \frac{1}{\sqrt{2}}$ the sequence converges to the solution. For x negative or in general complex the task is much more complex and the 'solution' $\displaystyle x=- \sqrt{2}$ found by Zarathustra, though perfectly correct, is symply among the complex solutions the only that has the imaginary part equal to 0...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
$\displaystyle 2=\left(\sqrt{2}\right)^2=\left(\sqrt{2}\right)^{\ left(\sqrt{2}\right)^2}$
$\displaystyle 2=\left(-\sqrt{2}\right)^2=\left(-\sqrt{2}\right)^\left(-\sqrt{2}\right)^2$
$\displaystyle x^{x^2}=2\ \Rightarrow\ x=\pm\sqrt{2}$
In fact, you can continue stacking that index to infinity
$\displaystyle x^{x^2} = 2 \Rightarrow x^{x^{x^{x^2}}} = x^{x^2} = 2$ and we can continue with as many
times as we want for $\displaystyle x$ to appear in those powers (an even number of times, though).
Taking limit we have $\displaystyle u = x^{x^{...}}} = 2$. Therefore $\displaystyle 2 = u = x^{x^{...}}} = x^{x^{x^{...}}}} = x^u = x^2$ so $\displaystyle x=\pm\sqrt{2}$
Generalize: $\displaystyle x^{x^{...}^{x^\beta}} = \beta\ ,\ \ \beta>0$ for a constant number of times x appears in the power.
In case you want it more rigorously:
Let $\displaystyle a_n$ be defined by $\displaystyle a_n = x^{x^{...^{x^\beta}}$, where $\displaystyle x$ appears $\displaystyle rn$ times in those powers
(where in the given equation we have $\displaystyle x$ appear $\displaystyle r$ times in the powers).
It's easy to show that $\displaystyle a_n = \beta$ for all $\displaystyle n\in\mathbb{N}$.
$\displaystyle a_n$ is a sub-sequence of $\displaystyle b_n$ which is defined just like $\displaystyle a_n$ but for $\displaystyle r=1$.
$\displaystyle \lim_{n\rightarrow \infty} b_n =b_1 = x^\beta$ so it is also the limit of its sub-sequence $\displaystyle \beta = \lim_{n\rightarrow \infty} \beta= \lim_{n\rightarrow \infty} a_n = x^\beta$.
Therefore $\displaystyle x$ is the solution to $\displaystyle x^\beta = \beta$, which would be $\displaystyle \beta^\frac{1}{\beta}$ and possibly the negative of it, if it can be substituted in the original equation.
Note an interesting point here. From what we proved it follows that if we're talking in the domain $\displaystyle \mathbb{R}^+$ then $\displaystyle x^{x^{...}}$
converges only if $\displaystyle x$ is a solution to $\displaystyle \beta^\frac{1}{\beta}$ for some $\displaystyle \beta\in\mathbb{R}^+$.
But using the usual invetigation methods of functions it follows that the maximal value of $\displaystyle \beta^\frac{1}{\beta}$ in $\displaystyle \mathbb{R}^+$
is when $\displaystyle \beta = e$. That is, the maximal value is $\displaystyle e^{e^{-1}}$, so $\displaystyle x^{x^{...}}$ converges for $\displaystyle x\in\mathbb{R}^+$ only if $\displaystyle x\le e^{e^{-1}}$
We see that $\displaystyle e^{e^{-1}}$ is about $\displaystyle 1.44466786 > \sqrt{2}$ as we would expect, as a result of the original problem.