# Why is it that the Lambert W relation cannot be expressed in terms of elementary functions?

According to this Wikipedia page, the Lambert W relation cannot be expressed in terms of elementary functions.

However, it does not explain why this is the case.

An elementary function is "a function of one variable which is the composition of a finite number of arithmetic operations (+ – × ÷), exponentials, logarithms, constants, and solutions of algebraic equations (a generalization of nth roots)," according to Wikipedia.

Note

I know that I should not always refer to the same source for information, but I believe that this is an accurate definition of the term.

Questions

Does there exist a proof that the Lambert W relation cannot be expressed in terms of elementary functions?

Why is it that the Lambert W relation cannot be expressed in terms of elementary functions?

• @ZainPatel The OP is asking why there isn't some clever formula like $W(t) = \ln t/\ln\ln t$. – Stephen Montgomery-Smith Jul 15 '15 at 21:51
• @Taylor But you have the same issue with the exponential and trig functions. – Stephen Montgomery-Smith Jul 15 '15 at 21:53
• @Taylor : what do you means by "you cannot input a value, and have it return an exact value". If I understand what you mean, even with $x \mapsto \sqrt{x}$ "you cannot input a value, and have it return an exact value" – Tryss Jul 15 '15 at 21:54
• What's the difference between $\ln(3)$ and $W(3)$? Can you calculate exactly $\ln(3)$? – Tryss Jul 15 '15 at 21:58
• @Taylor, here you go: mathoverflow.net/questions/135911/… – Zain Patel Jul 15 '15 at 22:02

The branches of Lambert W are the local inverses of the Elementary function $f$ with $f(z)=ze^z$, $z \in \mathbb{C}$.

The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 answers which kinds of Elementary functions can have an inverse which is an Elementary function.

The non-elementarity of LambertW was already shown by Joseph Liouville. It is also treated in

Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22

and in

But Ritt's theorem shows that no antiderivatives, no differentiation and no differential fields are needed for defining the Elementary functions.

I would recommend that you could possibly look into hyper-operations. They are quite interesting and related. For quick reference, a hyper-operation of n is a repetition of they {n-1}th hyper-operation.$$5*4=5+5+5+5$$Here, multiplication was turned into a repetition of addition. Similarly, exponentiation is a repetition of multiplication. And so on...

But more importantly, all of this can be represented in terms of elementary functions. Whether it be addition, multiplication, or exponentiation.

Furthermore, the inverses of each individual hyper-operation is considered elementary. The opposites of addition, multiplication, and exponentiation are subtraction, division, and logarithms or roots, respectively.

However, a combination of different levels of hyper-operations (except addition) cannot be inverted. For example:$$f(x)=xe^x$$$$f^{-1}=?$$For this example, we assign the Lambert W function as the solution.$$f^{-1}=W(x)$$But the problem is that we cannot turn this into something involving only addition, multiplication, exponentiation, and their inverses (or higher hyper-operations like tetrations and such).

More specifically, we cannot turn this into a $finite$ amount of terms being added or multiplied and such.

To answer your first question, the proof is simply that the definition of the Lambert W function cannot be solved with elementary functions.

As to why, it is because, as far as I can explain, two hyper-operations, multiplication and exponentiation, were combined. In general, combining different hyper-operations results in unsolvable inverses (or a manipulation of the Lambert W function). Try solving (without the Lambert W function): $$f(x)=x^a+bx$$$$f^{-1}(x)=?$$Now try solving it with $a=3,2,1,0$. Much easier? To solve for any known $a$ is easy, but solving for all $a$'s is more difficult.

The reason why you were able to solve for the above values of $a$ were most likely because of factoring. However, you cannot factor with exponents like you can with polynomials. Envision the following:$$a^{(ax)^{(ax^2)^{..^{..^{..}}}}}$$Compare it to$$a+ax+ax^2+...$$$$and$$$$a*ax*ax^2*ax^3*...$$The last two are simplifiable, but the exponential one was not. This is why we stop being able to find inverses of functions when they are exponential, tetrational, or higher.

• And my answer may not to be the best, FYI. – Simply Beautiful Art Dec 7 '15 at 23:30
• What you wrote does not answer any question, as far as i can tell... The sentence «To answer your first question, the proof is simply that the definition of the Lambert W function cannot be solved with elementary functions.» is akin to saying «The sky is blue because it is blue». – Mariano Suárez-Álvarez Feb 25 '17 at 17:15
• @MarianoSuárez-Álvarez Haha, that's one of my old answers.... I suppose I'll look it through :-) – Simply Beautiful Art Feb 25 '17 at 17:18