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. 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, \(ln(x) \prec x \asymp ax \prec e^x\). Notice that this is NOT the same as \(<\), as both \(x<e^x\) 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 \(W(x)>1\).

Note by Levi Adam Walker
3 weeks, 6 days ago

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

Aaghaz Mahajan - 3 weeks, 6 days ago

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You mean "@Levi Adam Walker Check this out.

B D - 2 weeks, 1 day ago

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

Levi Adam Walker - 3 weeks, 5 days ago

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

Aaghaz Mahajan - 3 weeks, 5 days ago

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i dont understand your writhing

Nahom Assefs - 2 weeks, 4 days ago

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

B D - 2 weeks, 1 day ago

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my keyboard piece is gone.

Nahom Assefs - 2 weeks, 1 day ago

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

B D - 1 week, 6 days ago

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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.

Levi Adam Walker - 2 weeks, 4 days ago

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that is good book for beginners right

Nahom Assefs - 2 weeks, 4 days ago

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ok

Nahom Assefs - 2 weeks, 4 days ago

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