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I have this term: $-\frac{1}{p}W(-\frac{p}{e})\bigg[\log\bigg(-\frac{1}{p}W(-\frac{p}{e})\bigg)-1\bigg]$.

In order to reduce complexity I've taken $-\frac{p}{e}=x$ which gives $f(x)=\frac{1}{xe}W(x)\bigg[\log\bigg(\frac{1}{xe}W(x)\bigg)-1\bigg]$.

I need to do a power series expansion for this term at $x=-1/e$.

From the basics of Taylor expansion: $f(x)=f(-1/e)+\frac{df}{dx}(-1/e) (x+1/e)^1+\frac{d^2f}{dx^2}(-1/e)(x+1/e)^2+.....$

The first term $f(-1/e)=-1$ but the second term $\frac{df}{dx}=-\frac{[W(x)]^2\bigg[log\bigg(\frac{W(x)}{x}\bigg)-1\bigg]}{ex^2[w(x)+1]}$ is not defined at $x=-1/e$ since $W(-1/e)=-1$ making the denominator to be $0$ also the numerator is $0$ as well. If the second term is undefined then the Taylor expansion does not work (to my understanding).

I'm not sure where I'm going wrong with this approach or if there's any other way to expand this term as a power series at $x=-1/e$. Any suggestions/help appreciated. Thanks!

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According to R. M. Corless et al. if $p=+\sqrt{2(ez+1)}$ and $\mid p \mid<\sqrt{2}$ then $W(z)=-1+p-\frac{1}{3}p^2+\frac{11}{72}p^3+....$

Also according to the definition of Lambert-W function $We^W=z$, taking logarithm on both sides of this equation yields: $\ln W=\ln z-W$ gives us a relatively easier way to break down the log term in our function.

Now getting back to the function we have, $-\frac{1}{y}W(-\frac{y}{e})\bigg[\ln\bigg(-\frac{1}{y}W(-\frac{y}{e})\bigg)-1\bigg]$. Let's take $z=-\frac{y}{e}$ and so we get $p=\sqrt{2(1-y)}$. Now we are able to do the power series expansion for our function:

$W(z)=-1+\sqrt{2}(1-y)^{1/2}-\frac{2}{3}(1-y)+\frac{11 \times 2^{3/2}}{72}(1-y)^{3/2}+....$

Now, using this expansion and after some very tedious multiplication the final expression for the expansion can be derived.

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