## 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.)

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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.
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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)
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Differentiation

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))).

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http://www.orcca.on.ca/LambertW/
http://en.wikipedia.org/wiki/Lambert_W_function
http://2000clicks.com/MathHelp/BasicSimplifyingLambertWFunction.aspx

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

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 → ∞.
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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)).
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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))

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

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The function f(x) = xˣ⁻¹, x > 0.
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Differentiation

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.
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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.

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(Appendix)

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.

Annonser

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