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As the title states, I want to differentiate the identity of the Lambert W function. (I have a tendency to use brackets)

Identity:

$\frac{x}{W(x)}=e^{W(x)}$

If you don't know what the Lambert W function is, it is as follows:

$W(x)=y$ and $x=ye^y$, which, using substitution and dividing by $W(x)$, produces the above identity.

Since $W(x)=f^{-1}(x)$

when $f(x)=xe^x$,

we can find $W'(x)$.

$W'(x)=\frac{1}{f'[W(x)]}=\frac{1}{f[W(x)]+e^{W(x)}}=\frac{1}{x+e^{W(x)}}$

which using the identity...

$\frac{1}{x+e^{W(x)}}=\frac{1}{x+\frac{x}{W(x)}}=\frac{W(x)}{x[W(x)+1]}$

By then using chain rule, I attempted to differentiate each part of the identity separately.

$\frac{x}{W(x)}dx=\frac{1}{W(x)}-\frac{x}{W^2(x)}$

$e^{W(x)}dx=W'(x)e^{W(x)}=\frac{W(x)e^{W(x)}}{x[W(x)+1]}=\frac{x}{x[W(x)+1]}=\frac{1}{W(x)+1}$

By the identity, these two solutions should be equivalent to each other.

$\frac{1}{W(x)}-\frac{x}{W^2(x)}=\frac{1}{[W(x)+1]}$

I finish this off by solving for $W(x)$.

$\frac{1}{W(x)}-\frac{x}{W^2(x)}=\frac{1}{[W(x)+1]}$

I multiply by the GCF.

$[W(x)+1][W(x)-x]=W^2(x)$

Multiply.

$W^2(x)+(-x+1)W(x)-x=W^2(x)$

Subtract $W^2(x)$ and add $x$

$(-x+1)W(x)=x$

Divide by $-x+1$.

$W(x)=\frac{x}{-x+1}$

Then I realized this was completely untrue and that apparently, the Lambert $W$ function can't be expressed in terms of elementary functions, according to Wikipedia.

I was, in fact, trying to prove that the Lambert $W$ function might be solvable in terms of elementary functions with this method, but it appears to fail me.

Did I do something wrong or is this just plain weird? If I did everything right, is this just a mystery of the Lambert $W$ function or can someone explain it to me. No sources that I have found on the internet have ever even attempted this sort of an approach, so I am unsure if everything I have posted here. I am confident in my calculus skills, but I could have made a mistake. If I did, point out my mistake and solve for $W(x)$.

Thank you.

P.S. If this becomes famous or revolutionary, I call the credit for thinking of this unless someone can prove otherwise. If someone else solves this, they can have they credit of solving it.

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  • $\begingroup$ I didn't read through your entire post, but I can already tell you you made a mistake, because $W(x)=\frac{x}{-x+1}$ is not true, and you can check that by proving that $x\neq \frac{x}{-x+1}\cdot e^{\frac{x}{-x+1}}$... $\endgroup$ – 5xum Oct 29 '15 at 0:33
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$\frac{x}{W(x)}dx=\frac{1}{W(x)}-\frac{x}{W^2(x)}$

Nope.

$$\frac{d}{dx} (\frac{x}{W(x)}) = \frac{W(x) - xW'(x)}{W^2(x)} = \frac{1}{W(x)} - \frac{W'(x) x}{W^2(x)} \neq \frac{1}{W(x)}-\frac{x}{W^2(x)}$$

So your hunch was correct, you did make a mistake at some point. There was a lot of derivatives in your calculation, so it was bound to happen :)


Also, a piece of advice: there is a lot of very very smart people doing mathematics, and they work hard to produce theorems that are proven up to the highest standard. It is very highly unlikely that any theorem (i.e. important and proven statement) you learn or hear about in your studies up to graduate level is false.

If you get a result that is both simple and clearly invalidates a well known mathematical theorem, your intuition should not say "Did I just discover something revolutionary!"... Your intuition should be "Damn, I obviously made a mistake".

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