I'm an electrical engineer and I recently came across an unforeseen issue in my masters thesis because I lack a deeper mathematical education.

I want to know for which positive real $x$ the following inequality is an equality:

$$ n \log(1 + \tfrac{x}{n}) - \log(1 + x) \leq a\, , \;\;\;\; (\ast) $$ i.e. $$ n \log(1 + \tfrac{x}{n}) - \log(1 + x) - a = 0\, , $$ where $x \in \mathbb{R} \geq 0$, $n\in \mathbb{N}\gg 1$ and $a \in \mathbb{R} > 0$.

This is equivalent to $$ (1+\tfrac{x}{n})^n = \mathrm{e}^a\cdot(1 + x) \, $$ which looks basically not so hard.

My questions is: Is there a "closed form" solution of this equation for the unknown $x$ and I am just too dumb to get it? By closed form solution I mean any solution that I can nicely write like $$ x \leq \ldots $$ to solve $(\ast)$.

If there is no "closed form" solution, I would be interested why and how I could have seen this. Unfortunately I'm not really familiar enough with Transcendence Theory or Galois Theory to see this on my own.


Assuming that $(\ast)$ has no "closed form" solution, I also thought about a workaround.

Since $\lim\limits_{n \rightarrow \infty} (1 + \tfrac{x}{n})^n = \mathrm{e}^x$, I could write write $$ \mathrm{e}^x - \mathrm{e}^a\cdot(1 + x) = 0 \, $$ for the case that $n \rightarrow \infty$. I thought about this as an approximation for finite $n$. Unfortunately for finite $n$, $(1 + \tfrac{x}{n})^n < \mathrm{e}^x$ yields and this approximation would clash my basic inequality $(\ast)$. If I could show that $$ \lim\limits_{n \rightarrow \infty} (1 + \tfrac{x}{n})^{n+2} = \mathrm{e}^x $$ and $$ (1 + \tfrac{x}{n})^{n+2} > \mathrm{e}^x $$ for finite $n$, it might be possible to solve $$ \mathrm{e}^x\cdot(1 + \tfrac{x}{n})^{-2} = \mathrm{e}^a\cdot(1 + x) $$ for an approximate solution of $(\ast)$. Here, the Lambert W function might be helpful but I didn't succeed on this problem so far as well.

Any thoughts on this? Thanks a lot!

  • 2
    $\begingroup$ What you have is basically an $n$th degree polynomial equation in $x$. When $n \ge 5$, there is no general formula, and I suspect your polynomial can be shown at least for some $a$ to be such that no "nice" solution exists. $\endgroup$
    – J. J.
    May 6, 2013 at 14:57
  • $\begingroup$ In my problem setting, $a$ is very small or arbitrary small. Might this open opportunities to open backdoor cuts? $\endgroup$
    – ArminB
    May 6, 2013 at 15:08
  • $\begingroup$ I used Lambert W function in a similar case using the approximation of $(1-\frac{x}{n})^n$ and then doing some algebra $\endgroup$
    – Alex
    May 6, 2013 at 15:28
  • $\begingroup$ @Alex: By now I didn't find a way to utilize the Lambert W function, mainly because of the $\mathrm{e}^a$-factor. I would be happy for any hints! $\endgroup$
    – ArminB
    May 6, 2013 at 15:34

3 Answers 3


Consider the equation $\left(1+\frac{x}{n}\right)^n=e^a(1+x)$. Choose $c\in\mathbb{C}$ such that $c^{n-1} = -\frac{1}{n}e^{-a}$ and define $y = c\left(1+\frac{x}{n}\right)$, then the equation becomes $$c^{-n}y^n = e^a\left(\frac{n}{c}y+1-n\right)$$ and after multiplying with $c^n$ $$ y^n = e^a\left(nc^{n-1} y+(1-n)c^n\right).$$ Using $c^n = cc^{n-1} = -\frac{c}{n}e^{-a}$ this may finally be rewritten to $$ y^n +y= c\left(1-\frac{1}{n}\right).$$

The solution of this equation can in general not be written in terms of roots and elementary arithmetic operations, but it can in terms of an ultraradical:

$$y = \mathrm{ur}\left(c\left(1-\frac{1}{n}\right)\right)$$ and hence $$x = \frac{n}{c}\mathrm{ur}\left(c\left(1-\frac{1}{n}\right)\right)-n.$$

  • $\begingroup$ WOW!! Thanks a lot! Now I need some minutes to go through this in detail and understand that. I'll be back soon :-) $\endgroup$
    – ArminB
    May 6, 2013 at 15:42
  • $\begingroup$ How is the ultraradical conventionally defined? There are five solutions to the defining quintic equation. Is there some sort of 'positive root', like how the square-root is defined? $\endgroup$
    – QuantumDot
    May 6, 2013 at 16:22
  • $\begingroup$ @QuantumDot The ultraradical of a positive real number is defined to be a positive real number. Then you can extend this continuously to $\mathbb{C}\backslash \mathbb{R}^-$. $\endgroup$
    – Abel
    May 6, 2013 at 16:27
  • $\begingroup$ @Abel: This looks very elegant, especially because in my problem $x \geq 0$ is defined anyways. But can you give me some more backup on the ultraradical $ur(\ldots)$? Unfortunately, the info on MathWorld is not really extensive and google does not provide much help either. Is there a way to compute the ultraradical in MATLAB? Thanks a lot! $\endgroup$
    – ArminB
    May 8, 2013 at 15:37

I think you should continue like this ($\ t=1+x$): $$ e^x=e^a(1+x)\\ \frac{e^x}{(1+x)}=e^a\\ -te^{-t}=-e^{-(a+1)}\\ W(-e^{-(a+1)})=-t\\ t=-W(-e^{-(a+1)})\\ x=1-W(-e^{-(a+1)}) $$

  • 1
    $\begingroup$ Thank you very much for showing me how to use the Lambert W function in this situation. For me it would have taken a very long time to find this :-) $\endgroup$
    – ArminB
    May 8, 2013 at 15:27
  • $\begingroup$ You are welcome $\endgroup$
    – Alex
    May 8, 2013 at 15:43

As noted, your equation is the same as $$\left(1+\frac{x}{n}\right)^n=e^a(1+x)$$ Take a look at the graphs of $y=\left(1+\frac{x}{n}\right)^n$ and $y=e^a(1+x)$. The first is a polynomial with a single negative repeated root. The other is a line with slope $e^a$ and root at $-1$. When $n$ is even, there are two solutions, one positive and one negative (but larger than $-1$). If $n$ is odd, there is still a positive solution, but a negative solution only exists if $a$ is large enough to give that line steep slope.

For $n$ larger than 4, there is no general solution by radicals for such a polynomial equation. I would recommend arguing that solutions exist as I have done (only more formally) and noting that when needed, they can be found to arbitrarily high precision using any one of a number of methods, including Newton's method.


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