Non-elementary functions

This is a big subject. What I have in mind at present is xˣ, xˣ⁻¹ and W(x), the Lambert W-function. (To be used at other places.)


The Lambert W-funtion

This has two (real) branches W₀ and W₋₁ defined thus:
(1) If f(x) = x·eˣ, x ≥ -1 then W₀(x) ≡ f⁻¹(x), x ≥ -1/e.
(2) If f(x) = x·eˣ, x ≤ -1 then W₋₁(x) ≡ f⁻¹(x), -1/e ≤ 0.

So W(x·eˣ) = x and W(x)·e^(W(x)) = x with suitable branch.
Example: Find the negative solution of the equation x² = 2ˣ.

Solution: Manipulate, to an equation of the form g(x)·e^g(x) = a.

x² = 2ˣ, x² = e^(ln 2ˣ), x² = e^(x·ln 2), -x = e^(x·(ln 2)/2) (x < 0),
e^(-x·(ln 2)/2) = -1/x, (-x)·e^(-x·(ln 2)/2) = 1, (-x·(ln 2)/2)·e^(-x·(ln 2)/2) = (ln 2)/2,
-x·(ln 2)/2 = W₀((ln 2)/2) ((ln 2)/2 > 0), x = -(2/ln 2)W₀((ln 2)/2)

Differentiate W(x)·e^(W(x)) = x and use the chain rule.
W'(x)·e^W(x) + W(x)·e^W(x)·W'(x) = 1, so W'(x) = 1/(e^W(x)(1 + W(x))), or
W'(x) = W(x)/(x(1 + W(x))).


More about this function and applications:


The function f(x) = xˣ, x > 0 (or x ≥ 0 if 0⁰ is defined as 1). lim [x→0⁺] xˣ = 1.

f'(x) = xˣ(1 + ln x)
The decreases strictly in ]0,1/e] taking all values in [e^(-1/e),1[, increases strictly in [1/e,∞[, and xˣ → ∞ when x → ∞.
We will concentrate on the inverse in the intervals ]0,1/e] and and [1/e,∞[.

Let y = xˣ, ln y = x·ln x = e^(ln x)·ln x. Apply W on both side; this gives
W(ln y) = ln x, x = exp(W(ln y)), so f⁻¹(x) = exp(W(ln x)). However this must be made more precise since xˣ is not invertible in the entire interval x > 0 and W has two branches.

If e^(-1/e) ≤ x < 1, -1/e ≤ ln x ≤ 0. Then if 0 < f⁻¹(x) ≤ 1/e, f⁻¹(x) = exp(W₋₁(ln x)).

If x ≥ e^(-1/e) and f⁻¹(x) ≥ 1/e, then ln x ≥ -1/e and f⁻¹(x) = exp(W₀(ln x)).
Example 1: The equation xˣ = 2 has the real solution x = f⁻¹(2) = exp(W₀(ln 2)) ≈ 1.559610441.

Example 2: The equation xˣ = ln 2 (> e^(-1/e)) has two real solutions,
x₁ = exp(W₋₁(ln ln 2)) ≈ 0.3366297865 and x₂ = exp(W₀(ln ln 2)) ≈ 0.40004026.

Black curve: y = x^x
Blue curve: y = exp(W₀(lnx))
Red curve: y = exp(W₋₁(ln x))


Black curve: y = x·e^x
Blue curve: y = W₀(x)
Red curve: W₋₁(x)


The function f(x) = xˣ⁻¹, x > 0.

f'(x) = xˣ⁻¹(1 + ln x – 1/x)
The function decreases strictly in ]0,1], increases strictly in [1,∞[, and xˣ → ∞ when x → ∞ and when x → 0⁺. So xˣ⁻¹ ≥ 1 for all real x with equality iff x = 1. This will be of use in later posts.
Much like xˣ the function has an inverse in the intervals ]0,1] and and [1,∞[. I will not go into details but to give a closed expression for f⁻¹(x) a ”cousin” of the W-function can be used, defined as the inverse of x·eˣ – x.



Elliptic functions (and more generally Abelian functions) use integrals as well as inverses. For example the inverse (in suitable interval) of the integral of 1/√((1 – t²)(1 – k²t²)), t = 0 to x, defines the so called sn function, one of the elliptic functions of Jacobi type.

In particular, for parameter value k = 0, this gives first arcsin x then sin x. So the theory of trigonimetric functions can be built from the algebraic function 1/√(1 – x²). For general k the theory of (Jacobi type) elliptic functions are built up.


6 Responses to Non-elementary functions

  1. Heriberto skriver:

    I’ll immediately grab your rss feed as I can not find your
    e-mail subscription link or e-newsletter service.
    Do you’ve any? Kindly let me know so that I could subscribe. Thanks.

  2. fitness skriver:

    Pretty great post. I just stumbled upon your weblog
    and wished to say that I have really enjoyed surfing around your blog posts.
    In any case I will be subscribing to your rss feed and I hope you write
    again very soon!

  3. game skriver:

    Ahaa, its pleasant discussion about this post here
    at this blog, I have read all that, so at this time me also commenting here.

  4. whoah this weblog is excellent i love reading your posts.
    Stay up the great work! You recognize, a lot of persons are looking round for this info, you can
    aid them greatly.

  5. Clash of Clans skriver:

    Your method of describing all in this piece of writing is truly
    fastidious, every one be able to easily know it, Thanks a lot.

  6. yoga skriver:

    It is not my first time to pay a quick visit this website, i am browsing this website dailly and take pleasant facts from here all the time.


Fyll i dina uppgifter nedan eller klicka på en ikon för att logga in: Logo

Du kommenterar med ditt Logga ut /  Ändra )


Du kommenterar med ditt Google+-konto. Logga ut /  Ändra )


Du kommenterar med ditt Twitter-konto. Logga ut /  Ändra )


Du kommenterar med ditt Facebook-konto. Logga ut /  Ändra )

Ansluter till %s

%d bloggare gillar detta: