Inverse function -- need help

Question

I'm a senior software developer but my math lessons are a bit rusty. I know the name of what I want, but not anymore how to compute it ;)

I've found (by myself with Grapher.app) a simple easing function (for transitions), that's a bijection from $[0,1]$ to $[0,1]$:

$$ f(x) = \frac{27}{4} \cdot \left( \left( \frac{2x}3 \right) ^2 - \left( \frac{2x}3 \right) ^3 \right) $$

Edit: in fact, after simplifying, it's as simple as:

$$ f(x) = 3 x^2 - 2 x^3 $$

The derivatives at $0$ and $1$ are both zero, so it's a perfect ease-in-out function that's easy to compute.

In order to start my transitions from anywhere between $0$ and $1$, I need $f^{-1}$, the inverse of this function, but I'm completely lost as to where to start... Would someone shed some light on my oh-gosh-it-was-so-long-ago math course?

It's simpler and easier to understand if you write $f(x)=3x^2-2x^3$. — user1551, Jan 31 '13 at 19:54
Oh. My. Didn't even think of simplifying it. When I told you I'm rusty... — Cyrille, Jan 31 '13 at 20:48

score 5 · Accepted Answer · answered Jan 31 '13 at 20:17

5

Take your equation $3 x^2 - 2 x^3 = y$, substitute $x = X + 1/2$, multiply by $2$ and expand to get $-4X^3 + 3 X = 2 y - 1$. Now note that $\sin(3t) = - 4 \sin(t)^3 + 3 \sin(t)$. Thus a solution is $X = \sin(t)$ where $\sin(3t) = 2y-1$, i.e. $$ x = \sin\left(\frac{\arcsin(2y-1)}{3}\right) + 1/2 $$

Note that $0 \le x \le 1$ when $0 \le y \le 1$.

answered Jan 31 '13 at 20:17

Robert Israel

448,999

Wait wait wait... sinuses? – Cyrille Jan 31 '13 at 20:51
Okay, after re-reading it like ten times, it makes sense. – Cyrille Jan 31 '13 at 21:02
1

What I've always wondered (and that's where the beauty of maths comes from) is: where did you take the "substitute x = X + 1/2" from? I mean, did you just have to say "shazam" to find this? – Cyrille Jan 31 '13 at 21:11
Take $x = X + a$ and see what $a$ has to be to eliminate the $X^2$ term. – Robert Israel Jan 31 '13 at 23:35

score 0 · Answer 2 · answered Jan 31 '13 at 19:53

The inverse function $g(y)$ satisfies $f[g(y)] = y$. In this case:

$$\frac{27}{4} \cdot \left( \left( \frac{2g(y)}3 \right) ^2 - \left( \frac{2g(y)}3 \right) ^3 \right) = y$$

Solve for $g(y)$ algebraically. Might be tough because it's a cubic, but it is possible.

Emanuele Paolini · Answer 3 · 2013-02-01T07:05:31.013

0

You can ask to wolfram alpha:

1/2 (1+1/(1-2 x+2 sqrt(-x+x^2))^(1/3)+(1-2 x+2 sqrt(-x+x^2))^(1/3))

Or either you could implement the bisection method to solve $f(x) = y$. One should check, but it is possible that in this case the bisection method is faster than using the explicit formula for inverse.

edited Feb 01 '13 at 07:05

answered Jan 31 '13 at 20:55

Emanuele Paolini

21,447

Didn't even knew you could ask this to Wolfram. Thanks! – Cyrille Jan 31 '13 at 21:05
But notice that when $x \in (0,1)$, you're taking the square root of a negative number. This is no accident: see http://en.wikipedia.org/wiki/Casus_irreducibilis – Robert Israel Jan 31 '13 at 23:39

Inverse function -- need help

3 Answers3