# How can I calculate $W(-x)$ using $W(x)$?

The question is with regards to Lambert W Function:

Given $$W(x)$$, I need to calculate $$W(-x)$$.

Is there any way to do that?

I've searched through the function identities, but couldn't find anything useful.

Thank you.

• Can you provide some context on the question? Where does it come from, why do you want to do this? May 19 '20 at 11:06
• @GEdgar I think you mean: if $a = W(x)$ and $b=W(-x)$, can we write $a$ in terms of $b$? This is equivalent to: if $ae^a =x$ and $be^b=-x$, can we write $a$ in terms of $b$? May 19 '20 at 11:46
• @JensRenders You are right. Then this amounts to: solve $ae^a = -be^b$ for $b$. May 19 '20 at 11:57
• @GEdgar Ineed, but without evaluating Lambert $W$ in $-ae^a$ explicitly, because $-ae^a$ is just $-x$. May 19 '20 at 12:02
• Do you want the real-valued W function? In order for both $x$ and $-x$ to belong to the domain, you need $-1/e < x < 1/e$. May 19 '20 at 12:07

This is just the start of what I am working on (hoping to get something better later).

Since we are restricted to the range $$-\frac 1 e \leq x\leq \frac 1 e$$, we could consider, around $$x=0$$ the expansion $$W(-x)\, W(x)=-x^2\sum_{n=0}^\infty a_n \,x^{2n}$$ where the $$a_n$$'s are the sequence $$\left\{1,2,\frac{22}{3},\frac{168}{5},\frac{18142}{105},\frac{2702944}{2835},\frac{ 286762684}{51975},\frac{66973259776}{2027025},\frac{9972382736698}{49116375},\cdots\right\}$$ which is not recognized by $$OEIS$$ which lead to quite good results except very close to $$x=\frac 1 e$$.

Similarly, we could write $$W(-x)+ W(x)=-2x^2\sum_{n=0}^\infty b_n \,x^{2n}$$ where the $$b_n$$'s are the sequence $$\left\{1,\frac{8}{3},\frac{54}{5},\frac{16384}{315},\frac{156250}{567},\frac{298598 4}{1925},\frac{7909306972}{868725},\frac{35184372088832}{638512875},\frac{508373 1656658}{14889875},\cdots\right\}$$

If required, these series expansions could be made much more compact transforming them is Padé approximants.

Similarly, we could write $$\frac{W(-x)\, W(x)}{W(-x)+W(x)}=\frac 12-\frac {x^2}3\sum_{n=0}^\infty c_n \,x^{2n}$$ where the $$c_n$$'s are the sequence $$\left\{1,\frac{38}{15},\frac{634}{63},\frac{75568}{1575},\frac{39393118}{155925}, \frac{301414530016}{212837625},\frac{1763848931644}{212837625},\frac{2716085710533 376}{54273594375},\cdots\right\}$$

Edit

We could do better using Winitzki's approximation $$W(x)\sim\frac{e x}{1+\frac{1}{a+\frac{1}{\sqrt{2 e x+2}}}}\qquad \text{where}\qquad a=\frac{1}{e-1}-\frac{1}{\sqrt{2}}$$ which is very good for the range of interest. So, $$\color{blue}{W(-x)=-W(x)\,\,\frac{1+\frac{1}{a+\frac{1}{\sqrt{2 e x+2}}}}{1+\frac{1}{a+\frac{1}{\sqrt{2-2 e x}}}}}$$ Since this is an approximation, we can use $$a=-\frac 18$$ and even retain $$W(-x)=-W(x)\frac{\left(4 \sqrt{2}-\sqrt{1-e x}\right) \left(7 \sqrt{1+e x}+4 \sqrt{2}\right)}{\left(4 \sqrt{2}-\sqrt{1+e x}\right)\left(7 \sqrt{1-e x}+4 \sqrt{2}\right) }$$

A few values for comparison $$\left( \begin{array}{ccc} x & \text{approximation} & \text{exact} \\ -0.35 & +0.261884 & +0.267777 \\ -0.30 & +0.231621 & +0.236755 \\ -0.25 & +0.199919 & +0.203888 \\ -0.20 & +0.166157 & +0.168916 \\ -0.15 & +0.129852 & +0.131515 \\ -0.10 & +0.090490 & +0.091277 \\ -0.05 & +0.047464 & +0.047672 \\ +0.00 & +0.000000 & +0.000000 \\ +0.05 & -0.052937 & -0.052706 \\ +0.10 & -0.112804 & -0.111833 \\ +0.15 & -0.181790 & -0.179491 \\ +0.20 & -0.263475 & -0.259171 \\ +0.25 & -0.364499 & -0.357403 \\ +0.30 & -0.500250 & -0.489402 \\ +0.35 & -0.732766 & -0.716639 \end{array} \right)$$

Update

Speaking about Padé approximants, we could also write (have a look here) $$W(x) \sim \log(1+x)\,P_{n,n}(x)$$ and then write $$W(-x)\sim W(x)\,\, \frac{\log(1-x) } {\log(1+x) }\,\, \frac{P_{n,n}(-x) } {P_{n,n}(x) }$$ Since the linked page gives the coefficients for $$n=2$$, I give you the next $$P_{3,3}(x)=\frac{1+\frac{26213 }{5194}x+\frac{82391 }{10388}x^2+\frac{85291 }{22260}x^3 } {1+\frac{14405 }{2597}x+\frac{76259 }{7791}x^2+\frac{564353 }{103880}x^3 }$$

In the linked answer, the answerer wrote that "probably one might do better with a minimax rational approximation". This is what I did for $$-\frac 1e \leq x \leq \frac 1e$$ writing $$W(x)\sim Q_{3,3}(x)= x\,\frac{a_0+a_1 x+a_2x^2}{b_0+b_1 x+b_2x^2+b_3 x^3}$$ the optimum parameters being $$a_0=1+\frac{13489}{3916800 e^6}+\frac{13489}{326400 e^4}-\frac{121401}{343400 e^2}$$ $$a_1=\frac{228}{85}-\frac{13489}{326400 e^4}-\frac{94423}{343400 e^2}\qquad a_2=\frac{451}{340}+\frac{13489}{137360 e^2}$$ $$b_0=1+\frac{13489}{326400 e^4}-\frac{121401}{343400 e^2}$$ $$b_1=\frac{313}{85}-\frac{26978}{42925 e^2}\qquad b_2=\frac{1193}{340}\qquad b_3=\frac{133}{204}$$

So, the proposed approximation is $$W(-x) \sim W(x)\,\, \frac{Q_{3,3}(-x) } {Q_{3,3}(x) }$$

A new table for comparison with the previous one $$\left( \begin{array}{ccc} -0.35 & +0.274095 & +0.267777 \\ -0.30 & +0.236981 & +0.236755 \\ -0.25 & +0.203889 & +0.203888 \\ -0.20 & +0.168917 & +0.168916 \\ -0.15 & +0.131517 & +0.131515 \\ -0.10 & +0.091277 & +0.091277 \\ -0.05 & +0.047672 & +0.047672 \\ +0.00 & +0.000000 & +0.000000 \\ +0.05 & -0.052706 & -0.052706 \\ +0.10 & -0.111832 & -0.111833 \\ +0.15 & -0.179489 & -0.179491 \\ +0.20 & -0.259170 & -0.259171 \\ +0.25 & -0.357401 & -0.357403 \\ +0.30 & -0.488936 & -0.489402 \\ +0.35 & -0.700120 & -0.716639 \end{array} \right)$$ which seems to be much better.

Solve $$ae^a=-be^b$$ for $$b$$.
I found nothing explicit, but the Taylor series near $$a=0$$ is $$b=-a-2\,{a}^{2}-4\,{a}^{3}-{\frac{28}{3}}{a}^{4}-24\,{a}^{5}-{\frac{ 328}{5}}{a}^{6}-{\frac{8416}{45}}{a}^{7}-{\frac{173216}{315}}{a}^{8}-{ \frac{104288}{63}}{a}^{9}-{\frac{2878496}{567}}{a}^{10}+O \left( {a}^{11 } \right)$$

Looking in the Online Enyclopedia of Integer Sequences, I conjecture that this is $$b = -\sum_{j=1}^\infty \frac{S_{j-1} 2^{j-1}}{(j-1)!}\;a^j$$ where $$(S_j)_{j\ge 0}$$ is the sequence A007889. But it turns out that is not surprising. One of the formulas for the exponential generating function of A007889 involves the Lambert W function.