Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How could we solve $$\sqrt{x} - \ln(x) -1 = 0$$ without using Mathematica? Obviously a solution is $x = 1$, but what are the other exact solutions? This question is inspired by my first question How can we solve: $\sqrt{x} + \ln(x) -1 = 0$?. Here the situation is not clear. Any ideas?

share|improve this question

2 Answers 2

up vote 11 down vote accepted

The other real solution, according to Maple, is $4 \text{LambertW}(-1, -e^{-1/2}/2)^2$, which is approximately $12.34020237$. Note that if $f(x) = \sqrt{x} - \ln(x)$, $f'(x) = \dfrac{1}{2\sqrt{x}} - \dfrac{1}{x} = \dfrac{\sqrt{x}-2}{x}$, so $f(x)$ is decreasing on $(0,4]$ and increasing on $[4,\infty)$. Since $f(4) = 2 - 2 \ln(2) < 0$ while $f(x) \to +\infty$ as $x \to 0+$ and as $x \to +\infty$, there are two real solutions, one in $(0,4)$ and one in $(4,\infty)$.

To explain that LambertW solution: if $x = 4 t^2$ with $t > 0$, the equation says $$4 t^2 = x = e^{\sqrt{x}-1} = e^{2 t-1}$$ and thus $$ -t e^{-t} = -e^{-1/2}/2 $$ Now $\text{LambertW}(s)$ is defined to be a solution $w$ of $w e^w = s$. If $e^{-1} < s < 0$, as in this case $s = -e^{-1/2}/2$, there are two real solutions, the "principal branch" (which in this case gives $t=1/2$ and thus $x=1$), and the "$-1$ branch". There are infinitely many other branches, which are all complex.

share|improve this answer
    
thank you for the clear explanation! if I understand correctly - after researching the LambertW functions, we will resort to Newton's method to find an approximation to our second solution? –  bigollo Jun 6 '12 at 10:44
    
Newton's method is one handy way of calculating LambertW, yes. It's not the only way. For example, you might use a Taylor series $${\it LambertW} \left( -1, \left( t-2 \right) {{\rm e}^{-2}} \right) = -2-t-{\frac {1}{6}}{t}^{3}+{\frac {1}{12}}{t}^{4}-{\frac {13}{120}}{t} ^{5}+{\frac {37}{360}}{t}^{6}-{\frac {593}{5040}}{t}^{7}+{\frac {883}{ 6720}}{t}^{8}-{\frac {55781}{362880}}{t}^{9}+{\frac {47123}{259200}}{t }^{10}-{\frac {1749037}{7983360}}{t}^{11}+\ldots $$ –  Robert Israel Jun 6 '12 at 17:28

You could let $x=e^t$, and deal with the equation $f(t) = e^{\frac{t}{2}}-t-1 = 0$ instead. It is immediate that $f''(t)> 0$ (ie, strictly convex), $f(0) = 0$, $f'(0) <0$ and $\lim_{t\to\infty} f(t) = \infty$. Hence the equation has exactly two solutions.

Since $f(4)>0$ and $f$ is convex, using Newton's method (ie, $t_{new} = t-\frac{f(t)}{f'(t)}$) will produce a decreasing sequence that converges rapidly (quadratically) to the other solution. Starting from $4$, in 7 iterations the sequence hits the noise floor of my machine with $t \approx 2.51286241725234$. This corresponds to $x \approx 12.3402023625440$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.