# Solving $x=W(x)e^{W(x)}$ for $W(x)$

Suppose you're given the functional equation $x = W(x)e^{W(x)}$ and asked to solve it. At this point, it seems pretty obvious that this won't have an analytical solution. This is the step where most people decide it's time to turn to a series solution for help. Its inverse function, $w(x) = xe^x$, has a very simple taylor series from which you can invert it to find a taylor series for $W(x)$, but it has a very small radius of convergence ($1/e$). You deserve better than that. We all deserve better than that.

What I'm about to do may seem very hand-wavy, but be assured there is a rigorous theory behind it; such that covering too much of it would just lead me off on tangents and make this derivation too painful to read.

Let's start with $x = W(x)e^{W(x)}$. The approach here is to approximate $W(x)$ at large values of $x$ and work our way down to smaller and smaller orders. This is exactly the opposite approach by using Taylor series, where you start with a local approximation at a given point and move outwards. It'll become clear in a minute how we're going to do this. For now, take the natural logarithm of both sides.

$ln(x) = ln(W) + W$

This is where some theory comes in: For any term, we can assign a magnitude to it. Qualitatively, it denotes a function's growth at extremely large values. A function such as $f(x) + x^2 + x$ has a magnitude $x^2$, and this can be written as $f(x) \asymp x^2$. This holds because $x^2$ will outgrow $x$ enough to make the latter term insignificant. Rigorously, you can say that two functions $f(x)$ and $g(x)$ have equal magnitudes if $\displaystyle\lim_{x \to \infty} \frac{f(x)}{g(x)} = c$

For some constant $c$. Magnitudes are well-ordered, so we can determine whether a given term's magnitude is less than or greater than any other term's. We have special symbols to represent this: $\succ, \prec,$ and $\asymp$. As an example, $ln(x) \prec x \asymp ax \prec e^x$. It's a good exercise to check that for yourself. Notice that this is NOT the same as $<$, as both $x and $x>e^x$ are true for certain values of $x$, while $x \prec e^x$ is true no matter what.

Magnitudes tell us about the growth of a term at high values of $x$. Thus, $x \prec e^x$ holds because $e^x$ terms will always outgrow $x$ terms, even if it may take a while to do so (compare $1000000x$ with $0.000001e^x$, the latter will eventually reach a point where it overcomes the former in value). Let's get back on topic. We know that $ln(W) \prec W$, so we can write

$W \simeq ln(x)$ $W = ln(x) + Q \qquad Q \prec ln(x)$

As according to the exposition above, $W$ will outgrow $ln(W)$, so at large values of $x$ the latter term may be ignored. However, we want to go deeper. In order to sharpen our asymptotic approximation, we introduce a new term $Q$ with a lower magnitude than $ln(x)$. This holds because $ln(x)$ is already the highest magnitude term in $W(x)$, as told by the functional equation. To find $Q$, let's plug $W$ back into the functional equation.

$x = We^W = (ln(x) + Q)e^{ln(x) + Q}$ $e^{-Q} = ln(x) + Q$

Taking the natural log of both sides

$Q = -ln(ln(x) + Q) \simeq -ln(ln(x)) \simeq -ln(ln(x)) + R \qquad R \prec ln(ln(x))$

The second step holds because $ln(x)$ outgrows $Q$ (remember $Q \prec ln(x)$). Putting this back into $W$ gives

$W = ln\left(\frac{x}{ln(x)}\right) + R \qquad R\prec ln\left(\frac{x}{ln(x)}\right)$

Again, to improve our approximation we need to find $R$. Plugging this back into the functional equation leads to:

$x = \left(ln\left(\frac{x}{ln(x)}\right) + R \right)e^{ln\left(\frac{x}{ln(x)}\right)+R}$ $e^{-R} = \left(ln \left( \frac{x}{ln(x)} \right) +R \right) \frac{1}{ln(x)}$ $R \simeq -ln \frac{ln \frac{x}{ln(x)}}{ln(x)}$

Putting this back into $W$...

$W \simeq ln\left(\frac{x}{ln(x)}\right) -ln \frac{ln \frac{x}{ln(x)}}{ln(x)} = ln \frac{x}{ln \frac{x}{ln(x)}}$

This can be continued indefinitely, but I'll stop here. There is a very clear pattern appearing, however. Should you decide to continue, you'll get the identity

$\Large W(x) = ln \frac{x}{ln \frac{x}{ln \frac{x}{ln \frac{x}{ln\frac{x}{ln(\cdots)}}}}}$

Which solves $x = W(x)e^{W(x)}$ as long as $W(x)>1$.

In a way, this can be seen as an "inverse" Taylor Series. It's typical to approximate functions locally, to within a (typically small) neighborhood of values around a known finite point of interest. This method takes the opposite approach, approximating a function globally around an "infinite" point where the function is well-behaved under the ordering I showed above.

Note by Levi Walker
2 years ago

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@Levi Adam Walker Check this out......... https://en.wikipedia.org/wiki/LambertWfunction

- 2 years ago

I saw that already :) what inspired me to write this was actually a paper on transseries.

- 2 years ago

Ohh, I see.....!!! But, thanks for sharing the link!!!!! :)

- 2 years ago

You mean "@Levi Adam Walker Check this out.

- 2 years ago

- 2 years ago

Sorry if my derivation isn't clear - if you'd like you can read this paper which gives a more formal exposition on what method I'm using.

- 2 years ago

ok

- 2 years ago

that is good book for beginners right

- 2 years ago

You didn't write correct. This is how it's right "I don't understand your writing."

- 2 years ago

my keyboard piece is gone.

- 2 years ago

Which one? The shift button or the dot. And writhing mean isn't the same as writing.

- 2 years ago

AAA! It's too hard for me. I wish I will be a mathematical genius.

- 1 year, 3 months ago