digitalmars.D.learn - D and math, can you isolate this ?
- Basile B. (21/21) Sep 20 2016 I've recently started an easing/interpolation family of function
- Basile B. (13/27) Sep 20 2016 If you don't understand, these function have a control point, for
- H. S. Teoh via Digitalmars-d-learn (14/24) Sep 20 2016 I couldn't manage to solve it. Nested exponentials are very nasty to
- Basile B. (3/26) Sep 20 2016 Thanks for trying, you're not the first to tell me about the
- H. S. Teoh via Digitalmars-d-learn (19/30) Sep 20 2016 [...]
- Nicholas Wilson (9/30) Sep 20 2016 So if we rearrange and take the logs of both sides and divide by
- Basile B. (5/42) Sep 21 2016 Y is a floating point value. I think I'm gonna make a LUT for
- Nicholas Wilson (8/26) Sep 21 2016 What does Y being float have to do with this? LUT is a good idea,
- Basile B. (5/12) Sep 21 2016 Forget to say yesterday that actually the original equation is
I've recently started an easing/interpolation family of function in my D user library. It's based on something I know well since I've already used them in 2012 in a VST plugin called GrainPlot (RIP). However for one of the function, I can't manage to get the inverse. A function that's fully implemented: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L598 - f(x,c) = x*x*x - x*x*c + x*c; - c(f(0.5)) = 4 * (y - 0.125)); Another: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L749 - f(x,c) = pow(x, c); - c(f(0.5)) = log(y) / log(0.5)); The problem is here: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L849 - f(x,c) = 1.0 - pow(1.0 - pow(x, 2.0/c), c * 0.5); - c(f0.5)) = ? Which means that I ask you if you can isolate c for y = 1.0 - pow(1.0 - pow(0.5, 2.0/c), c * 0.5); y is always f(0.5,c)
Sep 20 2016
Basile B. <b2.temp gmx.com> On Tuesday, 20 September 2016 at 12:35:18 UTC, Basile B. wrote:I've recently started an easing/interpolation family of function in my D user library. It's based on something I know well since I've already used them in 2012 in a VST plugin called GrainPlot (RIP). However for one of the function, I can't manage to get the inverse. [...] The problem is here: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L849 - f(x,c) = 1.0 - pow(1.0 - pow(x, 2.0/c), c * 0.5); - c(f0.5)) = ? Which means that I ask you if you can isolate c for y = 1.0 - pow(1.0 - pow(0.5, 2.0/c), c * 0.5); y is always f(0.5,c)If you don't understand, these function have a control point, for "parabol" and "pow" it's easy to get the c Coefficient that manages the slope. But for the ellipse (aka the super ellipse) it's a math nightmare !!!!) For example is use the three functions in the same order (parabol, pow, ellipse): http://sendvid.com/ygti5jmr for the ellipse you can see that the mouse position is not in sync with the control point at the middle...it's the problem. I need to isolate c when "y = 1.0 - pow(1.0 - pow(0.5, 2.0/c), c * 0.5)". I know it's hard...otherwise I wouldn't ask ;]
Sep 20 2016
On Tue, Sep 20, 2016 at 12:35:18PM +0000, Basile B. via Digitalmars-d-learn wrote: [...]The problem is here: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L849 - f(x,c) = 1.0 - pow(1.0 - pow(x, 2.0/c), c * 0.5); - c(f0.5)) = ? Which means that I ask you if you can isolate c for y = 1.0 - pow(1.0 - pow(0.5, 2.0/c), c * 0.5); y is always f(0.5,c)I couldn't manage to solve it. Nested exponentials are very nasty to invert. :-( At first, I thought it might be solvable in terms of the Lambert W function (aka ProductLog) but I couldn't manage to get the equation into the right form. Then I checked on Wolfram Alpha and it says "no result found in terms of standard mathematical functions". That probably means the inverse cannot be expressed in terms of elementary functions. Probably the only thing you can do is to use some kind of numerical approximation, like some form of Newton's method or some such, to find the value of c. T -- Questions are the beginning of intelligence, but the fear of God is the beginning of wisdom.
Sep 20 2016
On Tuesday, 20 September 2016 at 16:22:19 UTC, H. S. Teoh wrote:On Tue, Sep 20, 2016 at 12:35:18PM +0000, Basile B. via Digitalmars-d-learn wrote: [...]Thanks for trying, you're not the first to tell me about the Newton's method...The problem is here: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L849 - f(x,c) = 1.0 - pow(1.0 - pow(x, 2.0/c), c * 0.5); - c(f0.5)) = ? Which means that I ask you if you can isolate c for y = 1.0 - pow(1.0 - pow(0.5, 2.0/c), c * 0.5); y is always f(0.5,c)I couldn't manage to solve it. Nested exponentials are very nasty to invert. :-( At first, I thought it might be solvable in terms of the Lambert W function (aka ProductLog) but I couldn't manage to get the equation into the right form. Then I checked on Wolfram Alpha and it says "no result found in terms of standard mathematical functions". That probably means the inverse cannot be expressed in terms of elementary functions. Probably the only thing you can do is to use some kind of numerical approximation, like some form of Newton's method or some such, to find the value of c. T
Sep 20 2016
On Tue, Sep 20, 2016 at 09:22:19AM -0700, H. S. Teoh via Digitalmars-d-learn wrote:On Tue, Sep 20, 2016 at 12:35:18PM +0000, Basile B. via Digitalmars-d-learn wrote: [...][...][...]Which means that I ask you if you can isolate c for y = 1.0 - pow(1.0 - pow(0.5, 2.0/c), c * 0.5); y is always f(0.5,c)That probably means the inverse cannot be expressed in terms of elementary functions. Probably the only thing you can do is to use some kind of numerical approximation, like some form of Newton's method or some such, to find the value of c.[...] It may be analytically very hard to solve this equation, but it's probably not so hard to solve numerically. Based on the graph of the equation produced by Wolfram Alpha, it seems that y must always lie between 0 and 1, and that it has a horizontal asymptote at y=1. At around c=6 or thereabouts, y becomes very close to 1. The value of c for y=0.5 is approximately 2, so that seems like a good initial guess for an iterative method. So if y<0 or y>1, return NaN. If y=1, return +inf. Otherwise, use an iterative method with a starting value of c=2. Because of the horizontal asymptote at y=1, though, values of c much greater than 6 will probably be quite inaccurate, so hopefully your application doesn't depend on the exact value in that case! T -- Freedom of speech: the whole world has no right *not* to hear my spouting off!
Sep 20 2016
On Tuesday, 20 September 2016 at 12:35:18 UTC, Basile B. wrote:I've recently started an easing/interpolation family of function in my D user library. It's based on something I know well since I've already used them in 2012 in a VST plugin called GrainPlot (RIP). However for one of the function, I can't manage to get the inverse. A function that's fully implemented: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L598 - f(x,c) = x*x*x - x*x*c + x*c; - c(f(0.5)) = 4 * (y - 0.125)); Another: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L749 - f(x,c) = pow(x, c); - c(f(0.5)) = log(y) / log(0.5)); The problem is here: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L849 - f(x,c) = 1.0 - pow(1.0 - pow(x, 2.0/c), c * 0.5); - c(f0.5)) = ? Which means that I ask you if you can isolate c for y = 1.0 - pow(1.0 - pow(0.5, 2.0/c), c * 0.5); y is always f(0.5,c)So if we rearrange and take the logs of both sides and divide by c we get 2*log(1-y)/c = log(1-2^(-2/c)) and then that we have one occurrence of c on each side do an iterative back substitution to find the intersection given that you know for y=0.5 ,c = 2. We used this method for finding voltages and currents in circuits with semiconductors.
Sep 20 2016
On Wednesday, 21 September 2016 at 01:34:06 UTC, Nicholas Wilson wrote:On Tuesday, 20 September 2016 at 12:35:18 UTC, Basile B. wrote:Y is a floating point value. I think I'm gonna make a LUT for let's say 100 values to find the initial range where the result stands.I've recently started an easing/interpolation family of function in my D user library. It's based on something I know well since I've already used them in 2012 in a VST plugin called GrainPlot (RIP). However for one of the function, I can't manage to get the inverse. A function that's fully implemented: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L598 - f(x,c) = x*x*x - x*x*c + x*c; - c(f(0.5)) = 4 * (y - 0.125)); Another: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L749 - f(x,c) = pow(x, c); - c(f(0.5)) = log(y) / log(0.5)); The problem is here: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L849 - f(x,c) = 1.0 - pow(1.0 - pow(x, 2.0/c), c * 0.5); - c(f0.5)) = ? Which means that I ask you if you can isolate c for y = 1.0 - pow(1.0 - pow(0.5, 2.0/c), c * 0.5); y is always f(0.5,c)So if we rearrange and take the logs of both sides and divide by c we get 2*log(1-y)/c = log(1-2^(-2/c)) and then that we have one occurrence of c on each side do an iterative back substitution to find the intersection given that you know for y=0.5 ,c = 2. We used this method for finding voltages and currents in circuits with semiconductors.
Sep 21 2016
On Wednesday, 21 September 2016 at 08:21:29 UTC, Basile B. wrote:On Wednesday, 21 September 2016 at 01:34:06 UTC, Nicholas Wilson wrote:What does Y being float have to do with this? LUT is a good idea, a round number like 64 or 128 (or even 32) is probably better. then do g = 2*log(1-y);//constant c(n+1) = g/log(1-2^(-2/c(n))) where c(1) is a guess from the LUT. the iteration should converge very fast.On Tuesday, 20 September 2016 at 12:35:18 UTC, Basile B. wrote:Y is a floating point value. I think I'm gonna make a LUT for let's say 100 values to find the initial range where the result stands.[...]So if we rearrange and take the logs of both sides and divide by c we get 2*log(1-y)/c = log(1-2^(-2/c)) and then that we have one occurrence of c on each side do an iterative back substitution to find the intersection given that you know for y=0.5 ,c = 2. We used this method for finding voltages and currents in circuits with semiconductors.
Sep 21 2016
On Tuesday, 20 September 2016 at 12:35:18 UTC, Basile B. wrote:The problem is here: https://github.com/BBasile/iz/blob/master/import/iz/math.d#L849 - f(x,c) = 1.0 - pow(1.0 - pow(x, 2.0/c), c * 0.5); - c(f0.5)) = ? Which means that I ask you if you can isolate c for y = 1.0 - pow(1.0 - pow(0.5, 2.0/c), c * 0.5); y is always f(0.5,c)Forget to say yesterday that actually the original equation is x = pow(cos(angle), 0.25*c); y = pow(sin(angle), 0.25*c); But I couldn't use this form.
Sep 21 2016