solve?

**eq123** · July 18th 2010, 10:24 PM

how can i solve the following equation?

**chisigma** · July 18th 2010, 11:22 PM

The equation can be written in more convenient form as...

$\displaystyle x^{2}\ ln x - \ln 2 = 0$ (1)

... i.e in the form $f(x)=0$ , and it can be solved iteratively using the Newton-Raphson method...

$\displaystyle x_{n+1}= x_{n} - \frac{f(x_{n})}{f^{'} (x_{n})}$ (2)

... starting from an arbitrary $x_{0}$ ...

Kind regards

$\chi$ $\sigma$

**Also sprach Zarathustra** · July 18th 2010, 11:53 PM

I am positive that there are only two real solutions:
x_1=sqrt(2)
x_2=-sqrt(2)

**eq123** · July 19th 2010, 12:04 AM

The equation can be written in more convenient form as...

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.

I am positive that there are only two real solutions:
x_1=sqrt(2)
x_2=-sqrt(2)

that's true.

**Also sprach Zarathustra** · July 19th 2010, 12:20 AM

can i solve it without using any numerical method?

Yes, first of all by guessing one root... then investigate the function f(x)=x^x^2-2 (What you need to prove is that: the function is crossing only one time the positive x-axis)

**chisigma** · July 19th 2010, 01:13 AM

The Newton's sequence for solving the equation...

$\displaystyle x^{2}\ \ln x - \ln 2 =0$ (1)

... is expressed as...

$\displaystyle \Delta_{n} = x_{n+1} - x_{n} = \frac{\ln 2 - x_{n}^{2}\ \ln x_{n}}{x_{n}\ (1+ 2\ \ln x_{n})} = \varphi (x_{n})$ (2)

The function $\varphi(x)$ is represented here...

If x has to be consider positive real, there is only one attractive fixed point at $x=\sqrt{2}$ and for any $x_{0} > \frac{1}{\sqrt{2}}$ the sequence converges to the solution. For x negative or in general complex the task is much more complex and the 'solution' $x=- \sqrt{2}$ found by Zarathustra, though perfectly correct, is symply among the complex solutions the only that has the imaginary part equal to 0...

Kind regards

$\chi$ $\sigma$

**eq123** · July 21st 2010, 11:23 AM

does that means that we cannot solve this equation with some sort of algebraic maneuver?

**mr fantastic** · July 21st 2010, 01:08 PM

Originally Posted by eq123

does that means that we cannot solve this equation with some sort of algebraic maneuver?

Did the original question suggest that it could? Please post all the details about where this equation has come from.

**Archie Meade** · July 21st 2010, 01:36 PM

Originally Posted by eq123

how can i solve the following equation?

$2=\left(\sqrt{2}\right)^2=\left(\sqrt{2}\right)^{\ left(\sqrt{2}\right)^2}$

$2=\left(-\sqrt{2}\right)^2=\left(-\sqrt{2}\right)^\left(-\sqrt{2}\right)^2$

$x^{x^2}=2\ \Rightarrow\ x=\pm\sqrt{2}$

In fact, you can continue stacking that index to infinity

**Unbeatable0** · July 21st 2010, 01:37 PM

$x^{x^2} = 2 \Rightarrow x^{x^{x^{x^2}}} = x^{x^2} = 2$ and we can continue with as many

times as we want for $x$ to appear in those powers (an even number of times, though).

Taking limit we have $u = x^{x^{...}}} = 2$ . Therefore $2 = u = x^{x^{...}}} = x^{x^{x^{...}}}} = x^u = x^2$ so $x=\pm\sqrt{2}$

Generalize: $x^{x^{...}^{x^\beta}} = \beta\ ,\ \ \beta>0$ for a constant number of times x appears in the power.

In case you want it more rigorously:

Let $a_n$ be defined by $a_n = x^{x^{...^{x^\beta}}$ , where $x$ appears $rn$ times in those powers

(where in the given equation we have $x$ appear $r$ times in the powers).

It's easy to show that $a_n = \beta$ for all $n\in\mathbb{N}$ .

$a_n$ is a sub-sequence of $b_n$ which is defined just like $a_n$ but for $r=1$ .

$\lim_{n\rightarrow \infty} b_n =b_1 = x^\beta$ so it is also the limit of its sub-sequence $\beta = \lim_{n\rightarrow \infty} \beta= \lim_{n\rightarrow \infty} a_n = x^\beta$ .

Therefore $x$ is the solution to $x^\beta = \beta$ , which would be $\beta^\frac{1}{\beta}$ 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 $\mathbb{R}^+$ then $x^{x^{...}}$

converges only if $x$ is a solution to $\beta^\frac{1}{\beta}$ for some $\beta\in\mathbb{R}^+$ .

But using the usual invetigation methods of functions it follows that the maximal value of $\beta^\frac{1}{\beta}$ in $\mathbb{R}^+$

is when $\beta = e$ . That is, the maximal value is $e^{e^{-1}}$ , so $x^{x^{...}}$ converges for $x\in\mathbb{R}^+$ only if $x\le e^{e^{-1}}$

We see that $e^{e^{-1}}$ is about $1.44466786 > \sqrt{2}$ as we would expect, as a result of the original problem.

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