I was messing around the other day and I noticed this:

and I wondered if there is a pattern, i.e.
$x=y-1=z-2,\space y\neq1,\space y\neq3$
Then solve for $y$, and repeat with larger integers in place of the 3 until I can find a pattern.

The only way I know to solve this is with iterative equations. The problem is, I can't find an iterative equation and starting value combination that give the right answer.

Also, I have no idea what to tag this question.

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    $\begingroup$ $y \approx 2.478$ $\endgroup$ Commented Jul 1, 2014 at 10:30

3 Answers 3


Graphs are useful in such situations, if you would like to reduce your calculation efforts. I plot the graph, it looks absolutely wavy and it intersects the x axis at 2 points like below

enter image description here

I tried zooming in and calculating the value of the first intersect it comes out $y\approx2.4781$


Let us suppose that you do not know Lambert function and that you want to solve $$f(y)=3^y-y^3=0$$

If you plot the function, you notice that there is a root between $2$ and $3$. So, use Newton method, which, starting with a "reasonable" guess $y_0$, will update it according to $$y_{n+1}=y_n-\frac{f(y_n)}{f'(y_n)}$$ So, let us start with $y_0=2$. Newton iterates will then be successively : $2.47338$, $2.47803$, $2.47805$ which is the solution for six significant figures.

Please notice that any equation of the form $$A+Bx+C\log(D+Ex)=0$$ has a closed form solution using Lambert function.

Added later to this answer

If we look at the solution of the more general equation $$f(y)=a^y-y^a=0$$ the solution is given by $$y=-\frac{a W\left(-\frac{\log (a)}{a}\right)}{\log (a)}$$ and, then, deside the trivial solution $y=a$, a second root exist if and only if $a>e$. This second solution is a decreasing function of $a$, starting at $e$ and decreasing asymptotically to $1$ if $a$ goes to infinity.

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    $\begingroup$ You missed out on a $ and I can't submit edit! :( $\endgroup$
    – MonK
    Commented Jul 1, 2014 at 11:12
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    $\begingroup$ @Sid. Thank you ! Cheers :) $\endgroup$ Commented Jul 1, 2014 at 11:21


The Lambert W function $W(x)$ is multivaluated in the range $-\frac{1}{e}<x<0$

$x=-\frac{\ln(3)}{3}$ The two roots are $y=3$ and $y=2.47805...$


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