Blending two functions, where one is inverse

Let me first explain the idea. The actual math question is below the screenshots. For musical purpose I am building a groove algorithm where event positions are translated by a mathematical function F(X). The positions are normalized inside the groove range, so I am basically dealing with values between zero and one (which makes shaping groove curves way easier-the only limitation is x'>=0). This groove algorithm accepts any event position and also work by filtering static notes from a data-structure like a timeline note-track. For filtering e开发者_开发技巧vents in a certain range (audio block-size) I need the inverse groove-function to locate the notes in the track and transform them into the groove space. So far so good. It works!

In short: I use an inverse function for the fact that it is mirrored to (y=x). So I can plug in a value x and get a y. This y can obviously plugged into the inverse function to get first x again.

Problem: I now want to be able to blend the groove into another, but the usual linear (hint hint) blending code does not behave like I expected it. To make it easier, I first tried to blend to y=x.

B(x)=alpha*F(x) + (1-alpha)*x;
iB(x)=alpha*iF(x) + (1-alpha)*x;

For alpha=1 we get the full curve. For alpha=0 we get the straight line. But for alpha between 0 and 1 B(x) and iB(x) are not mirrored anymore (close, but not enough), F(x) and iF(x) are still mirrored.

Is there a solution for that (besides quantizing the curve into line segments)? Any subject I should throw an eye on?

you are combining two functions, f(x) and g(x), so that y = a f(x) + (1-a) g(x). and given some y, a, f and g, you want to find x. at least, that is what i understand.

i don't see how to do this generally (although i haven't tried very hard - i mean, it would be worth asking someone else), but i suspect that for "nice" shaped functions, like you seem to be using, newton's method would be fairly quick.

you want to find x such that y = a f(x) + (1-a) g(x). in other words, when 0 = a f(x) + (1-a) g(x) - y.

so let's define r(x) = a f(x) + (1-a) g(x) - y and find the "zero" of that. start with a guess in the middle, x_0 = 0.5. calculate x_1 = x_0 - r(x_0) / r'(x_0). repeat. if you are lucky this will rapidly converge (if not, you might consider defining the functions relative to y=x, which you already seem to be doing, and trying it again).

see wikipedia

This problem can't be solved algebraically, in general.

Consider for instance

y = 2e^x (inverse x = log 0.5y)

and

y = 2x (inverse x = 0.5y).

Blending these together with weight 0.5 gives y = e^x+x, and it is well-known that it is not possible to solve for x here using only elementary functions, even though the inverse of each piece was easy to find.

You will want to use a numerical method to approximate the inverse, as discussed by andrew above.