I was studying a statistics paper the other day in which the author said to solve

*t* log( 1 + *n*/*t* ) = *k*

for *t* as part of an algorithm. Assume 0 < *k* < *n*.

## Is this well posed?

First of all, *can* this equation be solved for *t*? Second, if there is a solution, is it unique?

It’s not hard to show that as a function of *t*, the left side approaches 0 as *t* approaches 0, and it approaches *n* as *t* goes to infinity. So there is a solution for all *k* between 0 and *n*. The restriction on *k* is necessary since the left side cannot exceed *n*.

With a little more work one can show that the derivative is always positive, so the left side is a monotonically increasing function, and so the solution given each value of *k* is unique.

## Analytic solution

Now if we fix *n*, we can think of the equation above as defining *t* as a function of *k*. Can we solve for *t* exactly? I suspected that there might be a solution in terms of the Lambert *W* function because if you divide by *t* and exponentiate, you get an equation that smells like the equation

*z* = *w* exp(*w*)

defining the function *W*(*z*). It turns out this is indeed the case.

If we ask Mathematica

Solve[t Log[1 + n/t] == k, t]

we get

Great! There’s a closed-form solution, if you accept using the *W* function as being closed form.

## Problems with the solution

I found the solution in Mathematica, but I tested it in Python to make sure both platforms define *W* the same way.

from numpy import log, exp from scipy.special import lambertw def f(t, n): return t*log(1 + n/t) def g(k, n): r = k/n return -k/(lambertw(-r*exp(-r)) + r) n, k = 10, 8 t = g(k, n) print(f(t, n))

This should print *k*, so it prints 8, right? No, it prints 10.

What’s up with that?

If we look back at the equation for *t* above, we see that the *W* function is being evaluated at *x* exp(*x*) where *x* = –*k*/*n*, so we should get –*k*/*n* back since *W*(*x* exp(*x*)) = *x* by definition. But that means our denominator is zero, and so the equation doesn’t make sense!

Things are getting worse. At first we had a wrong value, but at least it was finite!

The problem is not a difference between Mathematica and Python.

## Resolution

The problem is we’ve glossed over a branch cut of the *W* function. To make a long story short, we were using the principle branch of the *W* function, but we should have used a different branch.

Let’s go back to where I asked Mathematica

Solve[t Log[1 + n/t] == k, t]

I ran the solution through `TeXForm`

to get the TeX code that produced the image for the solution equation. I made a few aesthetic changes to the TeX, but it was essentially Mathematica’s output.

Without the `TeXForm`

, Mathematica’s solution was in terms of `ProductLog`

, not in terms of *W*; the `TeXForm`

function turned `ProductLog`

into *W*. If you look up `ProductLog`

, it will tell you

`ProductLog[z]`

gives the principal solution for *w* in *z* = *w**e*^{w}.

The *principle* solution. So we should be on the alert for difficulties with branches. There are two real solutions to *z* = *w**e*^{w} for some values of *z*, and we have to choose the right one. For example, if z = -0.1, the *w* could be -0.1118 or -3.5772.

Mathematica gave me the wrong branch. But to be fair, it did try to warn me.

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

The solution is to use the -1 branch. In Mathematica’s notation, the branch comes before the argument. In SciPy, the branch comes second. So to fix the code above, we need to change

lambertw(-r*exp(-r))

to

lambertw(-r*exp(-r), -1)

and then the code will be correct.

If *x* is negative, and we use the -1 branch of *W*, then

*W*_{-1}(*x* exp(*x*)) ≠ *x*

and so we’re not dividing by zero in our solution.

Another way to see that there is a solution is by setting t/n=x. The equation can then be rearranged as log(1+x)=r x, with 0 < r = k/n 0.

(repost without inequality symbols)

Another way to see that there is a solution is by setting t/n=x. The equation can then be rearranged as log(1+x)=r x, with r = k/n between 0 and 1.

On a graph it is easy to see that r x always intersects log(1+x) at x=0 and some other positive x.

> There’s a closed-form solution, if you accept using the W function as being closed form.

Dunno if your aware of the research below to calculate the inverse of W, or do you mean something else by the above? Lambert W stuff is v neat though sometimes the math that uses it is right at the level to make my brain hurt a bit

https://www.hindawi.com/journals/tswj/2015/909231/