Who can solve ln x = 1/x? I can get an approximate solution to any degree of accuracy, but is there an exact solution?

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- Dec 31st 2008, 01:49 PMbillyatesNeed solution to mixed equation.
Who can solve ln x = 1/x? I can get an approximate solution to any degree of accuracy, but is there an exact solution?

- Dec 31st 2008, 02:15 PMgalactus
Let's just use Newton's method. It's always a good standby.

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

$\displaystyle f(x)=ln(x)-\frac{1}{x}$

$\displaystyle f'(x)=\frac{1}{x}+\frac{1}{x^{2}}$

Let's make an initial guess of 1.5:

$\displaystyle x_{n+1}=1.5-\frac{ln(1.5)-\frac{1}{1.5}}{\frac{1}{1.5}+\frac{1}{(1.5)^{2}}}\ approx 1.7350814027$

Iterate:

$\displaystyle 1.7350814027-\frac{ln(1.7350814027)-\frac{1}{1.7350814027}}{\frac{1}{1.7350814027}+\fr ac{1}{(1.7350814027)^{2}}}\approx 1.76291539065$

Keep going until you reach the desired accuracy.

The solution is 1.76291539065....

As far as an exact solution, I do not think so.

$\displaystyle \frac{9ln(\frac{2}{3})}{10}+\frac{21}{10}=\frac{3( 3ln(2/3)+7)}{10}$

is pretty close of an approximation. - Dec 31st 2008, 03:39 PMmr fantastic
The exact solution can be written in terms of the Lambert W-function:

$\displaystyle \ln x = \frac{1}{x} \Rightarrow x = e^{1/x} \Rightarrow 1 = \frac{1}{x} \cdot e^{1/x}$.

Therefore $\displaystyle \frac{1}{x} = W(1) \Rightarrow x = \frac{1}{W(1)}$ where W is the Lambert W-function.

W(1) is the Omega constant: http://en.wikipedia.org/wiki/Omega_constant.