Solving x=W(x)eW(x)x=W(x)e^{W(x)} for W(x)W(x)

Suppose you're given the functional equation x=W(x)eW(x)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)=xexw(x) = xe^x, has a very simple taylor series from which you can invert it to find a taylor series for W(x)W(x), but it has a very small radius of convergence (1/e1/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)eW(x)x = W(x)e^{W(x)} . The approach here is to approximate W(x)W(x) at large values of xx 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 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)+x2+xf(x) + x^2 + x has a magnitude x2x^2, and this can be written as f(x)x2f(x) \asymp x^2. This holds because x2x^2 will outgrow xx enough to make the latter term insignificant. Rigorously, you can say that two functions f(x)f(x) and g(x)g(x) have equal magnitudes if limxf(x)g(x)=c \displaystyle\lim_{x \to \infty} \frac{f(x)}{g(x)} = c

For some constant cc. 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)xaxexln(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<exx<e^x and x>exx>e^x are true for certain values of xx, while xexx \prec e^x is true no matter what.

Magnitudes tell us about the growth of a term at high values of xx. Thus, xexx \prec e^x holds because exe^x terms will always outgrow xx terms, even if it may take a while to do so (compare 1000000x1000000x with 0.000001ex0.000001e^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)Wln(W) \prec W, so we can write

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

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

x=WeW=(ln(x)+Q)eln(x)+Q x = We^W = (ln(x) + Q)e^{ln(x) + Q} eQ=ln(x)+Q e^{-Q} = ln(x) + Q

Taking the natural log of both sides

Q=ln(ln(x)+Q)ln(ln(x))ln(ln(x))+RRln(ln(x)) 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)ln(x) outgrows QQ (remember Qln(x)Q \prec ln(x)). Putting this back into WW gives

W=ln(xln(x))+RRln(xln(x)) 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 RR. Plugging this back into the functional equation leads to:

x=(ln(xln(x))+R)eln(xln(x))+R x = \left(ln\left(\frac{x}{ln(x)}\right) + R \right)e^{ln\left(\frac{x}{ln(x)}\right)+R} eR=(ln(xln(x))+R)1ln(x) e^{-R} = \left(ln \left( \frac{x}{ln(x)} \right) +R \right) \frac{1}{ln(x)} Rlnlnxln(x)ln(x) R \simeq -ln \frac{ln \frac{x}{ln(x)}}{ln(x)}

Putting this back into WW...

Wln(xln(x))lnlnxln(x)ln(x)=lnxlnxln(x) 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

W(x)=lnxlnxlnxlnxlnxln() \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)eW(x)x = W(x)e^{W(x)} as long as W(x)>1W(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
11 months ago

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

Aaghaz Mahajan - 11 months ago

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

Levi Walker - 11 months ago

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

Aaghaz Mahajan - 11 months ago

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You mean "@Levi Adam Walker Check this out.

B D - 10 months, 2 weeks ago

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i dont understand your writhing

Nahom Assefs - 10 months, 3 weeks 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 Walker - 10 months, 3 weeks ago

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ok

Nahom Assefs - 10 months, 3 weeks ago

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that is good book for beginners right

Nahom Assefs - 10 months, 3 weeks ago

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

B D - 10 months, 2 weeks ago

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my keyboard piece is gone.

Nahom Assefs - 10 months, 2 weeks 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 - 10 months, 2 weeks ago

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

Veronica Pauli - 2 months ago

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