Matlab: Help understanding sinusoidal curve fit - c
I have an unknown sine wave with some noise that I am trying to reconstruct. The ultimate goal is to come up with a C algorithm to find the amplitude, dc offset, phase, and frequency of a sine wave but I am prototyping in Matlab (Octave actually) first. The sine wave is of the form
y = a + b*sin(c + 2*pi*d*t)
a = dc offset
b = amplitude
c = phase shift (rad)
d = frequency
I have found this example and in the comments John D'Errico presents a method for using Least Squares to fit a sine wave to data. It is a neat little algorithm and works remarkably well but I am having difficulties understanding one aspect. The algorithm is as follows:
Algorithm
Suppose you have a sine wave of the form:
(1) y = a + b*sin(c+d*x)
Using the identity
(2) sin(u+v) = sin(u)*cos(v) + cos(u)*sin(v)
We can rewrite (1) as
(3) y = a + b*sin(c)*cos(d*x) + b*cos(c)*sin(d*x)
Since b*sin(c) and b*cos(c) are constants, these can be wrapped into constants b1 and b2.
(4) y = a + b1*cos(d*x) + b2*sin(d*x)
This is the equation that is used to fit the sine wave. A function is created to generate regression coefficients and a sum-of-squares residual error.
(5) cfun = #(d) [ones(size(x)), sin(d*x), cos(d*x)] \ y;
(6) sumerr2 = #(d) sum((y - [ones(size(x)), sin(d*x), cos(d*x)] * cfun(d)) .^ 2);
Next, sumerr2 is minimized for the frequency d using fminbnd with lower limit l1 and upper limit l2.
(7) dopt = fminbnd(sumerr2, l1, l2);
Now a, b, and c can be computed. The coefficients to compute a, b, and c are given from (4) at dopt
(8) abb = cfun(dopt);
The dc offset is simply the first value
(9) a = abb(1);
A trig identity is used to find b
(10) sin(u)^2 + cos(u)^2 = 1
(11) b = sqrt(b1^2 + b2^2)
(12) b = norm(abb([2 3]));
Finally the phase offset is found
(13) b1 = b*cos(c)
(14) c = acos(b1 / b);
(15) c = acos(abb(2) / b);
Question
What is going on in (5) and (6)? Can someone break down what is happening in pseudo-code or perhaps perform the same function in a more explicit way?
(5) cfun = #(d) [ones(size(x)), sin(d*x), cos(d*x)] \ y;
(6) sumerr2 = #(d) sum((y - [ones(size(x)), sin(d*x), cos(d*x)] * cfun(d)) .^ 2);
Also, given (4) shouldn't it be:
[ones(size(x)), cos(d*x), sin(d*x)]
Code
Here is the Matlab code in full. Blue line is the actual signal. Green line is the reconstructed signal.
close all
clear all
y = [111,140,172,207,243,283,319,350,383,414,443,463,483,497,505,508,503,495,479,463,439,412,381,347,311,275,241,206,168,136,108,83,63,54,45,43,41,45,51,63,87,109,137,168,204,239,279,317,348,382,412,439,463,479,496,505,508,505,495,483,463,441,414,383,350,314,278,245,209,175,140,140,110,85,63,51,45,41,41,44,49,63,82,105,135,166,200,236,277,313,345,379,409,438,463,479,495,503,508,503,498,485,467,444,415,383,351,318,281,247,211,174,141,111,87,67,52,45,42,41,45,50,62,79,104,131,163,199,233,273,310,345,377,407,435,460,479,494,503,508,505,499,486,467,445,419,387,355,319,284,249,215,177,143,113,87,67,55,46,43,41,44,48,63,79,102,127,159,191,232,271,307,343,373,404,437,457,478,492,503,508,505,499,488,470,447,420,391,360,323,287,254,215,182,147,116,92,70,55,46,43,42,43,49,60,76,99,127,159,191,227,268,303,339,371,401,431,456,476,492,502,507,507,500,488,471,447,424,392,361,326,287,287,255,220,185,149,119,92,72,55,47,42,41,43,47,57,76,95,124,156,189,223,258,302,337,367,399,428,456,476,492,502,508,508,501,489,471,451,425,396,364,328,294,259,223,188,151,119,95,72,57,46,43,44,43,47,57,73,95,124,153,187,222,255,297,335,366,398,426,451,471,494,502,507,508,502,489,474,453,428,398,367,332,296,262,227,191,154,124,95,75,60,47,43,41,41,46,55,72,94,119,150,183,215,255,295,331,361,396,424,447,471,489,500,508,508,502,492,475,454,430,401,369,335,299,265,228,191,157,126,99,76,59,49,44,41,41,46,55,72,92,118,147,179,215,252,291,328,360,392,422,447,471,488,499,507,508,503,493,477,456,431,403]';
fs = 100e3;
N = length(y);
t = (0:1/fs:N/fs-1/fs)';
cfun = #(d) [ones(size(t)), sin(2*pi*d*t), cos(2*pi*d*t)]\y;
sumerr2 = #(d) sum((y - [ones(size(t)), sin(2*pi*d*t), cos(2*pi*d*t)] * cfun(d)) .^ 2);
dopt = fminbnd(sumerr2, 2300, 2500);
abb = cfun(dopt);
a = abb(1);
b = norm(abb([2 3]));
c = acos(abb(2) / b);
d = dopt;
y_reconstructed = a + b*sin(2*pi*d*t - c);
figure(1)
hold on
title('Signal Reconstruction')
grid on
plot(t*1000, y, 'b')
plot(t*1000, y_reconstructed, 'g')
ylim = get(gca, 'ylim');
xlim = get(gca, 'xlim');
text(xlim(1), ylim(2) - 15, [num2str(b) ' cos(2\pi * ' num2str(d) 't - ' ...
num2str(c * 180/pi) ') + ' num2str(a)]);
hold off
(5) and (6) are defining anonymous functions that can be used within the optimisation code. cfun returns an array that is a function of t, y and the parameter d (that is the optimisation parameter that will be varied). Similarly, sumerr2 is another anonymous function, with the same arguments, this time returning a scalar. That scalar will be the error that is to be minimised by fminbnd.
Related
Erroneous result using inverse Vincenty's formula in C
I have written a C script to implement the inverse Vincenty's formula to calculate the distance between two sets of GPS coordinates based on the equations shown at https://en.wikipedia.org/wiki/Vincenty%27s_formulae
However, my results are different to the results given by this online calculator https://www.cqsrg.org/tools/GCDistance/ and Google maps. My results are consistently around 1.18 times the result of the online calculator.
My function is below, any tips on where I could be going wrong would be very much appreciated!
double get_distance(double lat1, double lon1, double lat2, double lon2)
{
double rad_eq = 6378137.0; //Radius at equator
double flattening = 1 / 298.257223563; //flattenig of earth
double rad_pol = (1 - flattening) * rad_eq; //Radius at poles
double U1,U2,L,lambda,old_lambda,sigma,sin_sig,cos_sig,alpha,cos2sigmam,A,B,C,u_sq,delta_s,dis;
//Convert to radians
lat1=M_PI*lat1/180.0;
lat2=M_PI*lat2/180.0;
lon1=M_PI*lon1/180.0;
lon2=M_PI*lon2/180.0;
//Calculate U1 and U2
U1=atan((1-flattening)*tan(lat1));
U2=atan((1-flattening)*tan(lat2));
L=lon2-lon1;
lambda=L;
double tolerance=pow(10.,-12.);//iteration tollerance should give 0.6mm
double diff=1.;
while (abs(diff)>tolerance)
{
sin_sig=sqrt(pow(cos(U2)*sin(lambda),2.)+pow(cos(U1)*sin(U2)-(sin(U1)*cos(U2)*cos(lambda)),2.));
cos_sig=sin(U1)*cos(U2)+cos(U1)*cos(U2)*cos(lambda);
sigma=atan(sin_sig/cos_sig);
alpha=asin((cos(U1)*cos(U2)*sin(lambda))/(sin_sig));
cos2sigmam=cos(sigma)-(2*sin(U1)*sin(U2))/((pow(cos(alpha),2.)));
C=(flattening/16)*pow(cos(alpha),2.)*(4+(flattening*(4-(3*pow(cos(alpha),2.)))));
old_lambda=lambda;
lambda=L+(1-C)*flattening*sin(alpha)*(sigma+C*sin_sig*(cos2sigmam+C*cos_sig*(-1+2*pow(cos2sigmam,2.))));
diff=abs(old_lambda-lambda);
}
u_sq=pow(cos(alpha),2.)*((pow(rad_eq,2.)-pow(rad_pol,2.))/(pow(rad_pol,2.)));
A=1+(u_sq/16384)*(4096+(u_sq*(-768+(u_sq*(320-(175*u_sq))))));
B=(u_sq/1024)*(256+(u_sq*(-128+(u_sq*(74-(47*u_sq))))));
delta_s=B*sin_sig*(cos2sigmam+(B/4)*(cos_sig*(-1+(2*pow(cos2sigmam,2.)))-(B/6)*cos2sigmam*(-3+(4*pow(sin_sig,2.)))*(-3+(4*pow(cos2sigmam,2.)))));
dis=rad_pol*A*(sigma-delta_s);
//Returns distance in metres
return dis;
}
This formula is not symmetric:
cos_sig = sin(U1)*cos(U2)
+ cos(U1)*cos(U2) * cos(lambda);
And turns out to be wrong, a sin is missing.
Another style of formatting (one including some whitespace) could also help.
Besides the fabs for abs and one sin for that cos I also changed the loop; there were two abs()-calls and diff had to be preset with the while-loop.
I inserted a printf to see how the value progresses.
Some parentheses can be left out. These formulas are really difficult to realize. Some more helper variables could be useful in this jungle of nested math operations.
do {
sin_sig = sqrt(pow( cos(U2) * sin(lambda), 2)
+ pow(cos(U1)*sin(U2)
- (sin(U1)*cos(U2) * cos(lambda))
, 2)
);
cos_sig = sin(U1) * sin(U2)
+ cos(U1) * cos(U2) * cos(lambda);
sigma = atan2(sin_sig, cos_sig);
alpha = asin(cos(U1) * cos(U2) * sin(lambda)
/ sin_sig
);
double cos2alpha = cos(alpha)*cos(alpha); // helper var.
cos2sigmam = cos(sigma) - 2*sin(U1)*sin(U2) / cos2alpha;
C = (flat/16) * cos2alpha * (4 + flat * (4 - 3*cos2alpha));
old_lambda = lambda;
lambda = L + (1-C) * flat * sin(alpha)
*(sigma + C*sin_sig
*(cos2sigmam + C*cos_sig
*(2 * pow(cos2sigmam, 2) - 1)
)
);
diff = fabs(old_lambda - lambda);
printf("%.12f\n", diff);
} while (diff > tolerance);
For 80,80, 0,0 the output is (in km):
0.000885870048
0.000000221352
0.000000000055
0.000000000000
9809.479224
which corresponds to the millimeter with WGS-84.
Is there a way to improve this pygame colour filter algorithm [duplicate]
I've made a function to find a color within a image, and return x, y. Now I need to add a new function, where I can find a color with a given tolerence. Should be easy? Code to find color in image, and return x, y: def FindColorIn(r,g,b, xmin, xmax, ymin, ymax): image = ImageGrab.grab() for x in range(xmin, xmax): for y in range(ymin,ymax): px = image.getpixel((x, y)) if px[0] == r and px[1] == g and px[2] == b: return x, y def FindColor(r,g,b): image = ImageGrab.grab() size = image.size pos = FindColorIn(r,g,b, 1, size[0], 1, size[1]) return pos Outcome: Taken from the answers the normal methods of comparing two colors are in Euclidean distance, or Chebyshev distance. I decided to mostly use (squared) euclidean distance, and multiple different color-spaces. LAB, deltaE (LCH), XYZ, HSL, and RGB. In my code, most color-spaces use squared euclidean distance to compute the difference. For example with LAB, RGB and XYZ a simple squared euc. distance does the trick: if ((X-X1)^2 + (Y-Y1)^2 + (Z-Z1)^2) <= (Tol^2) then ... LCH, and HSL is a little more complicated as both have a cylindrical hue, but some piece of math solves that, then it's on to using squared eucl. here as well. In most these cases I've added "separate parameters" for tolerance for each channel (using 1 global tolerance, and alternative "modifiers" HueTol := Tolerance * hueMod or LightTol := Tolerance * LightMod). It seems like colorspaces built on top of XYZ (LAB, LCH) does perform best in many of my scenarios. Tho HSL yields very good results in some cases, and it's much cheaper to convert to from RGB, RGB is also great tho, and fills most of my needs.
Computing distances between RGB colours, in a way that's meaningful to the eye, isn't as easy a just taking the Euclidian distance between the two RGB vectors.
There is an interesting article about this here: http://www.compuphase.com/cmetric.htm
The example implementation in C is this:
typedef struct {
unsigned char r, g, b;
} RGB;
double ColourDistance(RGB e1, RGB e2)
{
long rmean = ( (long)e1.r + (long)e2.r ) / 2;
long r = (long)e1.r - (long)e2.r;
long g = (long)e1.g - (long)e2.g;
long b = (long)e1.b - (long)e2.b;
return sqrt((((512+rmean)*r*r)>>8) + 4*g*g + (((767-rmean)*b*b)>>8));
}
It shouldn't be too difficult to port to Python.
EDIT:
Alternatively, as suggested in this answer, you could use HLS and HSV. The colorsys module seems to have functions to make the conversion from RGB. Its documentation also links to these pages, which are worth reading to understand why RGB Euclidian distance doesn't really work:
http://www.poynton.com/ColorFAQ.html
http://www.cambridgeincolour.com/tutorials/color-space-conversion.htm
EDIT 2:
According to this answer, this library should be useful: http://code.google.com/p/python-colormath/
Here is an optimized Python version adapted from Bruno's asnwer:
def ColorDistance(rgb1,rgb2):
'''d = {} distance between two colors(3)'''
rm = 0.5*(rgb1[0]+rgb2[0])
d = sum((2+rm,4,3-rm)*(rgb1-rgb2)**2)**0.5
return d
usage:
>>> import numpy
>>> rgb1 = numpy.array([1,1,0])
>>> rgb2 = numpy.array([0,0,0])
>>> ColorDistance(rgb1,rgb2)
2.5495097567963922
Instead of this: if px[0] == r and px[1] == g and px[2] == b: Try this: if max(map(lambda a,b: abs(a-b), px, (r,g,b))) < tolerance: Where tolerance is the maximum difference you're willing to accept in any of the color channels. What it does is to subtract each channel from your target values, take the absolute values, then the max of those.
Assuming that rtol, gtol, and btol are the tolerances for r,g, and b respectively, why not do: if abs(px[0]- r) <= rtol and \ abs(px[1]- g) <= gtol and \ abs(px[2]- b) <= btol: return x, y
Here's a vectorised Python (numpy) version of Bruno and Developer's answers (i.e. an implementation of the approximation derived here) that accepts a pair of numpy arrays of shape (x, 3) where individual rows are in [R, G, B] order and individual colour values ∈[0, 1]. You can reduce it two a two-liner at the expense of readability. I'm not entirely sure whether it's the most optimised version possible, but it should be good enough. def colour_dist(fst, snd): rm = 0.5 * (fst[:, 0] + snd[:, 0]) drgb = (fst - snd) ** 2 t = np.array([2 + rm, 4 + 0 * rm, 3 - rm]).T return np.sqrt(np.sum(t * drgb, 1)) It was evaluated against Developer's per-element version above, and produces the same results (save for floating precision errors in two cases out of one thousand).
A cleaner python implementation of the function stated here, the function takes 2 image paths, reads them using cv.imread and the outputs a matrix with each matrix cell having difference of colors. you can change it to just match 2 colors easily import numpy as np import cv2 as cv def col_diff(img1, img2): img_bgr1 = cv.imread(img1) # since opencv reads as B, G, R img_bgr2 = cv.imread(img2) r_m = 0.5 * (img_bgr1[:, :, 2] + img_bgr2[:, :, 2]) delta_rgb = np.square(img_bgr1- img_bgr2) cols_diffs = delta_rgb[:, :, 2] * (2 + r_m / 256) + delta_rgb[:, :, 1] * (4) + delta_rgb[:, :, 0] * (2 + (255 - r_m) / 256) cols_diffs = np.sqrt(cols_diffs) # lets normalized the values to range [0 , 1] cols_diffs_min = np.min(cols_diffs) cols_diffs_max = np.max(cols_diffs) cols_diffs_normalized = (cols_diffs - cols_diffs_min) / (cols_diffs_max - cols_diffs_min) return np.sqrt(cols_diffs_normalized)
Simple: def eq_with_tolerance(a, b, t): return a-t <= b <= a+t def FindColorIn(r,g,b, xmin, xmax, ymin, ymax, tolerance=0): image = ImageGrab.grab() for x in range(xmin, xmax): for y in range(ymin,ymax): px = image.getpixel((x, y)) if eq_with_tolerance(r, px[0], tolerance) and eq_with_tolerance(g, px[1], tolerance) and eq_with_tolerance(b, px[2], tolerance): return x, y
from pyautogui source code def pixelMatchesColor(x, y, expectedRGBColor, tolerance=0): r, g, b = screenshot().getpixel((x, y)) exR, exG, exB = expectedRGBColor return (abs(r - exR) <= tolerance) and (abs(g - exG) <= tolerance) and (abs(b - exB) <= tolerance) you just need a little fix and you're ready to go.
Here is a simple function that does not require any libraries: def color_distance(rgb1, rgb2): rm = 0.5 * (rgb1[0] + rgb2[0]) rd = ((2 + rm) * (rgb1[0] - rgb2[0])) ** 2 gd = (4 * (rgb1[1] - rgb2[1])) ** 2 bd = ((3 - rm) * (rgb1[2] - rgb2[2])) ** 2 return (rd + gd + bd) ** 0.5 assuming that rgb1 and rgb2 are RBG tuples
Number of recursive calls in gcd() function
Recently I have been given a gcd() function, written in C programming language which takes two arguments n and m and compute the GCD of these two numbers using recursion.I have been asked that "How many recursive calls are made by the function if n>=m?" Can any one provide the solution with explanation to my problem as I am unable to figure it out.
Here is the source code of the function :
int gcd(int n, int m)
{
if (n%m==0)
return m;
else
n=n%m;
return gcd(m, n);
}
Euclidean algorithm gives #steps = T(a, b) = 1 + T(b, r0) = 2 + T(r0, r1) = … = N + T(rN - 2, rN - 1) = N + 1 where a and b are the inputs, and r_i the remainder. We used that T(x, 0) = 0 Running an example in paper would help you get a better grasp of the aforementioned equation: gcd(1071, 462) is calculated from the equivalent gcd(462, 1071 mod 462) = gcd(462, 147). The latter GCD is calculated from the gcd(147, 462 mod 147) = gcd(147, 21), which in turn is calculated from the gcd(21, 147 mod 21) = gcd(21, 0) = 21 So a = 1071 and b = 462, and we have: T(a, b) = 1 + T(b, a % b) = 1 + T(b, r_0) = (1) 2 + T(r_0, b % r_0) = 2 + T(r_0, r_1) = 3 + T(r_1, r_0 % r_1) = 3 + T(r_1, r_2) = (2) 3 + T(r_1, 0) = 3 + 0 = 3 which says that we needed to take 3 steps to compute gcd(1071, 462). (1): notice that the 1 is the step already done before, i.e. T(a, b) (2): r_2 is equal to 0 in this example You could run a plethora of examples in paper, and see how this unfolds, and eventually you will be able to see the pattern, if you don't see it already. Note: While #Ian'Abott's comments are also correct, I decided to present this approach, since it's more generic, and can be applied to any similar recursive method.
How do I use Metropolis Sampling in MATLAB to calculate an integral?
I am trying to write a matlab function to solve a test integral using the Metropolis Method. My function is listed below. The integral is from 0 to infinity of x*e(-x^2), divided by the integral from 0 to infinity of e(-x^2) This function converges to ~0.5 (notably, it does fluctuate about this answer a little) however analytically the solution is ~0.5642 or 1/sqrt(pi). The code I use to run the function is also below. What I have done wrong? How do I use metropolis to correct solve this test function? % Metropolis Method for Integration % Written by John Furness - Computational Physics, KTH function [I S1 S2] = metropolis(f,a,b,n,sig) % This function calculates an integral using Metropolis Mathod % Only takes input as a function, f, on an interval between a and b, % where n is the number of points. %Defining burnin %burnin = n/20; burnin = 0; % Finding maximum point x = linspace(a,b,1000); f1 = f(x); max1 = max(f1); %Setting Up x-vector and mu x(1) = rand(1); mu=0; % Generating Random Poins for x with Gaussian distribution. % Proposal Distribution will be the normal distribution strg = 'exp(-1*((x-mu)/sig).^2)'; norm = inline(strg,'x','mu','sig'); for i = 2:n % This loop generates a new state from the proposal distribution. y = x(i-1) + sig*randn(1); % generate a uniform for comparison u = rand(1); % Alpha is the acceptance probability alpha = min([1, (f(y))/((f(x(i-1))))]); if u <= alpha x(i) = y; else x(i) = x(i-1); end end %Discarding Burnin %x(1:burnin) = []; %I = ((inside)/length(x))*max1*(b-a); I = (1/length(f(x)))*((sum(f(x))))/sum(norm(x,mu,sig)); %My investigation variables to see what's happening %S1 = sum(f(x)); %S2 = sum(norm1(x,mu,sig)); S1 = min(x); S2 = max(x); end Code used to run the above function: % Code for Running Metropolis Method % Written by John Furness - Computational Physics % Clearing Workspace clear all close all clc % Equation 1 % Changing Parameters for Equation 1 a1 = 0; b1 = 10; n1 = 10000; sig = 2; N1 = #(x)(x.*exp(-x.^2)); D1 = #(x)(exp(-x.^2)); denom = metropolis(D1,a1,b1,n1,sig); numer = metropolis(N1,a1,b1,n1,sig); solI1 = numer/denom
Calculate (x exponent 0.19029) with low memory using lookup table?
I'm writing a C program for a PIC micro-controller which needs to do a very specific exponential function. I need to calculate the following: A = k . (1 - (p/p0)^0.19029) k and p0 are constant, so it's all pretty simple apart from finding x^0.19029 (p/p0) ratio would always be in the range 0-1. It works well if I add in math.h and use the power function, except that uses up all of the available 16 kB of program memory. Talk about bloatware! (Rest of program without power function = ~20% flash memory usage; add math.h and power function, =100%). I'd like the program to do some other things as well. I was wondering if I can write a special case implementation for x^0.19029, maybe involving iteration and some kind of lookup table. My idea is to generate a look-up table for the function x^0.19029, with perhaps 10-100 values of x in the range 0-1. The code would find a close match, then (somehow) iteratively refine it by re-scaling the lookup table values. However, this is where I get lost because my tiny brain can't visualise the maths involved. Could this approach work? Alternatively, I've looked at using Exp(x) and Ln(x), which can be implemented with a Taylor expansion. b^x can the be found with: b^x = (e^(ln b))^x = e^(x.ln(b)) (See: Wikipedia - Powers via Logarithms) This looks a bit tricky and complicated to me, though. Am I likely to get the implementation smaller then the compiler's math library, and can I simplify it for my special case (i.e. base = 0-1, exponent always 0.19029)? Note that RAM usage is OK at the moment, but I've run low on Flash (used for code storage). Speed is not critical. Somebody has already suggested that I use a bigger micro with more flash memory, but that sounds like profligate wastefulness! [EDIT] I was being lazy when I said "(p/p0) ratio would always be in the range 0-1". Actually it will never reach 0, and I did some calculations last night and decided that in fact a range of 0.3 - 1 would be quite adequate! This mean that some of the simpler solutions below should be suitable. Also, the "k" in the above is 44330, and I'd like the error in the final result to be less than 0.1. I guess that means an error in the (p/p0)^0.19029 needs to be less than 1/443300 or 2.256e-6
Use splines. The relevant part of the function is shown in the figure below. It varies approximately like the 5th root, so the problematic zone is close to p / p0 = 0. There is mathematical theory how to optimally place the knots of splines to minimize the error (see Carl de Boor: A Practical Guide to Splines). Usually one constructs the spline in B form ahead of time (using toolboxes such as Matlab's spline toolbox - also written by C. de Boor), then converts to Piecewise Polynomial representation for fast evaluation.
In C. de Boor, PGS, the function g(x) = sqrt(x + 1) is actually taken as an example (Chapter 12, Example II). This is exactly what you need here. The book comes back to this case a few times, since it is admittedly a hard problem for any interpolation scheme due to the infinite derivatives at x = -1. All software from PGS is available for free as PPPACK in netlib, and most of it is also part of SLATEC (also from netlib).
Edit (Removed)
(Multiplying by x once does not significantly help, since it only regularizes the first derivative, while all other derivatives at x = 0 are still infinite.)
Edit 2
My feeling is that optimally constructed splines (following de Boor) will be best (and fastest) for relatively low accuracy requirements. If the accuracy requirements are high (say 1e-8), one may be forced to get back to the algorithms that mathematicians have been researching for centuries. At this point, it may be best to simply download the sources of glibc and copy (provided GPL is acceptable) whatever is in
glibc-2.19/sysdeps/ieee754/dbl-64/e_pow.c
Since we don't have to include the whole math.h, there shouldn't be a problem with memory, but we will only marginally profit from having a fixed exponent.
Edit 3
Here is an adapted version of e_pow.c from netlib, as found by #Joni. This seems to be the grandfather of glibc's more modern implementation mentioned above. The old version has two advantages: (1) It is public domain, and (2) it uses a limited number of constants, which is beneficial if memory is a tight resource (glibc's version defines over 10000 lines of constants!). The following is completely standalone code, which calculates x^0.19029 for 0 <= x <= 1 to double precision (I tested it against Python's power function and found that at most 2 bits differed):
#define __LITTLE_ENDIAN
#ifdef __LITTLE_ENDIAN
#define __HI(x) *(1+(int*)&x)
#define __LO(x) *(int*)&x
#else
#define __HI(x) *(int*)&x
#define __LO(x) *(1+(int*)&x)
#endif
static const double
bp[] = {1.0, 1.5,},
dp_h[] = { 0.0, 5.84962487220764160156e-01,}, /* 0x3FE2B803, 0x40000000 */
dp_l[] = { 0.0, 1.35003920212974897128e-08,}, /* 0x3E4CFDEB, 0x43CFD006 */
zero = 0.0,
one = 1.0,
two = 2.0,
two53 = 9007199254740992.0, /* 0x43400000, 0x00000000 */
/* poly coefs for (3/2)*(log(x)-2s-2/3*s**3 */
L1 = 5.99999999999994648725e-01, /* 0x3FE33333, 0x33333303 */
L2 = 4.28571428578550184252e-01, /* 0x3FDB6DB6, 0xDB6FABFF */
L3 = 3.33333329818377432918e-01, /* 0x3FD55555, 0x518F264D */
L4 = 2.72728123808534006489e-01, /* 0x3FD17460, 0xA91D4101 */
L5 = 2.30660745775561754067e-01, /* 0x3FCD864A, 0x93C9DB65 */
L6 = 2.06975017800338417784e-01, /* 0x3FCA7E28, 0x4A454EEF */
P1 = 1.66666666666666019037e-01, /* 0x3FC55555, 0x5555553E */
P2 = -2.77777777770155933842e-03, /* 0xBF66C16C, 0x16BEBD93 */
P3 = 6.61375632143793436117e-05, /* 0x3F11566A, 0xAF25DE2C */
P4 = -1.65339022054652515390e-06, /* 0xBEBBBD41, 0xC5D26BF1 */
P5 = 4.13813679705723846039e-08, /* 0x3E663769, 0x72BEA4D0 */
lg2 = 6.93147180559945286227e-01, /* 0x3FE62E42, 0xFEFA39EF */
lg2_h = 6.93147182464599609375e-01, /* 0x3FE62E43, 0x00000000 */
lg2_l = -1.90465429995776804525e-09, /* 0xBE205C61, 0x0CA86C39 */
ovt = 8.0085662595372944372e-0017, /* -(1024-log2(ovfl+.5ulp)) */
cp = 9.61796693925975554329e-01, /* 0x3FEEC709, 0xDC3A03FD =2/(3ln2) */
cp_h = 9.61796700954437255859e-01, /* 0x3FEEC709, 0xE0000000 =(float)cp */
cp_l = -7.02846165095275826516e-09, /* 0xBE3E2FE0, 0x145B01F5 =tail of cp_h*/
ivln2 = 1.44269504088896338700e+00, /* 0x3FF71547, 0x652B82FE =1/ln2 */
ivln2_h = 1.44269502162933349609e+00, /* 0x3FF71547, 0x60000000 =24b 1/ln2*/
ivln2_l = 1.92596299112661746887e-08; /* 0x3E54AE0B, 0xF85DDF44 =1/ln2 tail*/
double pow0p19029(double x)
{
double y = 0.19029e+00;
double z,ax,z_h,z_l,p_h,p_l;
double y1,t1,t2,r,s,t,u,v,w;
int i,j,k,n;
int hx,hy,ix,iy;
unsigned lx,ly;
hx = __HI(x); lx = __LO(x);
hy = __HI(y); ly = __LO(y);
ix = hx&0x7fffffff; iy = hy&0x7fffffff;
ax = x;
/* special value of x */
if(lx==0) {
if(ix==0x7ff00000||ix==0||ix==0x3ff00000){
z = ax; /*x is +-0,+-inf,+-1*/
return z;
}
}
s = one; /* s (sign of result -ve**odd) = -1 else = 1 */
double ss,s2,s_h,s_l,t_h,t_l;
n = ((ix)>>20)-0x3ff;
j = ix&0x000fffff;
/* determine interval */
ix = j|0x3ff00000; /* normalize ix */
if(j<=0x3988E) k=0; /* |x|<sqrt(3/2) */
else if(j<0xBB67A) k=1; /* |x|<sqrt(3) */
else {k=0;n+=1;ix -= 0x00100000;}
__HI(ax) = ix;
/* compute ss = s_h+s_l = (x-1)/(x+1) or (x-1.5)/(x+1.5) */
u = ax-bp[k]; /* bp[0]=1.0, bp[1]=1.5 */
v = one/(ax+bp[k]);
ss = u*v;
s_h = ss;
__LO(s_h) = 0;
/* t_h=ax+bp[k] High */
t_h = zero;
__HI(t_h)=((ix>>1)|0x20000000)+0x00080000+(k<<18);
t_l = ax - (t_h-bp[k]);
s_l = v*((u-s_h*t_h)-s_h*t_l);
/* compute log(ax) */
s2 = ss*ss;
r = s2*s2*(L1+s2*(L2+s2*(L3+s2*(L4+s2*(L5+s2*L6)))));
r += s_l*(s_h+ss);
s2 = s_h*s_h;
t_h = 3.0+s2+r;
__LO(t_h) = 0;
t_l = r-((t_h-3.0)-s2);
/* u+v = ss*(1+...) */
u = s_h*t_h;
v = s_l*t_h+t_l*ss;
/* 2/(3log2)*(ss+...) */
p_h = u+v;
__LO(p_h) = 0;
p_l = v-(p_h-u);
z_h = cp_h*p_h; /* cp_h+cp_l = 2/(3*log2) */
z_l = cp_l*p_h+p_l*cp+dp_l[k];
/* log2(ax) = (ss+..)*2/(3*log2) = n + dp_h + z_h + z_l */
t = (double)n;
t1 = (((z_h+z_l)+dp_h[k])+t);
__LO(t1) = 0;
t2 = z_l-(((t1-t)-dp_h[k])-z_h);
/* split up y into y1+y2 and compute (y1+y2)*(t1+t2) */
y1 = y;
__LO(y1) = 0;
p_l = (y-y1)*t1+y*t2;
p_h = y1*t1;
z = p_l+p_h;
j = __HI(z);
i = __LO(z);
/*
* compute 2**(p_h+p_l)
*/
i = j&0x7fffffff;
k = (i>>20)-0x3ff;
n = 0;
if(i>0x3fe00000) { /* if |z| > 0.5, set n = [z+0.5] */
n = j+(0x00100000>>(k+1));
k = ((n&0x7fffffff)>>20)-0x3ff; /* new k for n */
t = zero;
__HI(t) = (n&~(0x000fffff>>k));
n = ((n&0x000fffff)|0x00100000)>>(20-k);
if(j<0) n = -n;
p_h -= t;
}
t = p_l+p_h;
__LO(t) = 0;
u = t*lg2_h;
v = (p_l-(t-p_h))*lg2+t*lg2_l;
z = u+v;
w = v-(z-u);
t = z*z;
t1 = z - t*(P1+t*(P2+t*(P3+t*(P4+t*P5))));
r = (z*t1)/(t1-two)-(w+z*w);
z = one-(r-z);
__HI(z) += (n<<20);
return s*z;
}
Clearly, 50+ years of research have gone into this, so it's probably very hard to do any better. (One has to appreciate that there are 0 loops, only 2 divisions, and only 6 if statements in the whole algorithm!) The reason for this is, again, the behavior at x = 0, where all derivatives diverge, which makes it extremely hard to keep the error under control: I once had a spline representation with 18 knots that was good up to x = 1e-4, with absolute and relative errors < 5e-4 everywhere, but going to x = 1e-5 ruined everything again.
So, unless the requirement to go arbitrarily close to zero is relaxed, I recommend using the adapted version of e_pow.c given above.
Edit 4
Now that we know that the domain 0.3 <= x <= 1 is sufficient, and that we have very low accuracy requirements, Edit 3 is clearly overkill. As #MvG has demonstrated, the function is so well behaved that a polynomial of degree 7 is sufficient to satisfy the accuracy requirements, which can be considered a single spline segment. #MvG's solution minimizes the integral error, which already looks very good.
The question arises as to how much better we can still do? It would be interesting to find the polynomial of a given degree that minimizes the maximum error in the interval of interest. The answer is the minimax
polynomial, which can be found using Remez' algorithm, which is implemented in the Boost library. I like #MvG's idea to clamp the value at x = 1 to 1, which I will do as well. Here is minimax.cpp:
#include <ostream>
#define TARG_PREC 64
#define WORK_PREC (TARG_PREC*2)
#include <boost/multiprecision/cpp_dec_float.hpp>
typedef boost::multiprecision::number<boost::multiprecision::cpp_dec_float<WORK_PREC> > dtype;
using boost::math::pow;
#include <boost/math/tools/remez.hpp>
boost::shared_ptr<boost::math::tools::remez_minimax<dtype> > p_remez;
dtype f(const dtype& x) {
static const dtype one(1), y(0.19029);
return one - pow(one - x, y);
}
void out(const char *descr, const dtype& x, const char *sep="") {
std::cout << descr << boost::math::tools::real_cast<double>(x) << sep << std::endl;
}
int main() {
dtype a(0), b(0.7); // range to optimise over
bool rel_error(false), pin(true);
int orderN(7), orderD(0), skew(0), brake(50);
int prec = 2 + (TARG_PREC * 3010LL)/10000;
std::cout << std::scientific << std::setprecision(prec);
p_remez.reset(new boost::math::tools::remez_minimax<dtype>(
&f, orderN, orderD, a, b, pin, rel_error, skew, WORK_PREC));
out("Max error in interpolated form: ", p_remez->max_error());
p_remez->set_brake(brake);
unsigned i, count(50);
for (i = 0; i < count; ++i) {
std::cout << "Stepping..." << std::endl;
dtype r = p_remez->iterate();
out("Maximum Deviation Found: ", p_remez->max_error());
out("Expected Error Term: ", p_remez->error_term());
out("Maximum Relative Change in Control Points: ", r);
}
boost::math::tools::polynomial<dtype> n = p_remez->numerator();
for(i = n.size(); i--; ) {
out("", n[i], ",");
}
}
Since all parts of boost that we use are header-only, simply build with:
c++ -O3 -I<path/to/boost/headers> minimax.cpp -o minimax
We finally get the coefficients, which are after multiplication by 44330:
24538.3409, -42811.1497, 34300.7501, -11284.1276, 4564.5847, 3186.7541, 8442.5236, 0.
The following error plot demonstrates that this is really the best possible degree-7 polynomial approximation, since all extrema are of equal magnitude (0.06659):
Should the requirements ever change (while still keeping well away from 0!), the C++ program above can be simply adapted to spit out the new optimal polynomial approximation.
Instead of a lookup table, I'd use a polynomial approximation:
1 - x0.19029 ≈ - 1073365.91783x15 + 8354695.40833x14 - 29422576.6529x13 + 61993794.537x12 - 87079891.4988x11 + 86005723.842x10 - 61389954.7459x9 + 32053170.1149x8 - 12253383.4372x7 + 3399819.97536x6 - 672003.142815x5 + 91817.6782072x4 - 8299.75873768x3 + 469.530204564x2 - 16.6572179869x + 0.722044145701
Or in code:
double f(double x) {
double fx;
fx = - 1073365.91783;
fx = fx*x + 8354695.40833;
fx = fx*x - 29422576.6529;
fx = fx*x + 61993794.537;
fx = fx*x - 87079891.4988;
fx = fx*x + 86005723.842;
fx = fx*x - 61389954.7459;
fx = fx*x + 32053170.1149;
fx = fx*x - 12253383.4372;
fx = fx*x + 3399819.97536;
fx = fx*x - 672003.142815;
fx = fx*x + 91817.6782072;
fx = fx*x - 8299.75873768;
fx = fx*x + 469.530204564;
fx = fx*x - 16.6572179869;
fx = fx*x + 0.722044145701;
return fx;
}
I computed this in sage using the least squares approach:
f(x) = 1-x^(19029/100000) # your function
d = 16 # number of terms, i.e. degree + 1
A = matrix(d, d, lambda r, c: integrate(x^r*x^c, (x, 0, 1)))
b = vector([integrate(x^r*f(x), (x, 0, 1)) for r in range(d)])
A.solve_right(b).change_ring(RDF)
Here is a plot of the error this will entail:
Blue is the error from my 16 term polynomial, while red is the error you'd get from piecewise linear interpolation with 16 equidistant values. As you can see, both errors are quite small for most parts of the range, but will become really huge close to x=0. I actually clipped the plot there. If you can somehow narrow the range of possible values, you could use that as the domain for the integration, and obtain an even better fit for the relevant range. At the cost of worse fit outside, of course. You could also increase the number of terms to obtain a closer fit, although that might also lead to higher oscillations.
I guess you can also combine this approach with the one Stefan posted: use his to split the domain into several parts, then use mine to find a close low degree polynomial for each part.
Update
Since you updated the specification of your question, with regard to both the domain and the error, here is a minimal solution to fit those requirements:
44330(1 - x0.19029) ≈ + 23024.9160933(1-x)7 - 39408.6473636(1-x)6 + 31379.9086193(1-x)5 - 10098.7031260(1-x)4 + 4339.44098317(1-x)3 + 3202.85705860(1-x)2 + 8442.42528906(1-x)
double f(double x) {
double fx, x1 = 1. - x;
fx = + 23024.9160933;
fx = fx*x1 - 39408.6473636;
fx = fx*x1 + 31379.9086193;
fx = fx*x1 - 10098.7031260;
fx = fx*x1 + 4339.44098317;
fx = fx*x1 + 3202.85705860;
fx = fx*x1 + 8442.42528906;
fx = fx*x1;
return fx;
}
I integrated x from 0.293 to 1 or equivalently 1 - x from 0 to 0.707 to keep the worst oscillations outside the relevant domain. I also omitted the constant term, to ensure an exact result at x=1. The maximal error for the range [0.3, 1] now occurs at x=0.3260 and amounts to 0.0972 < 0.1. Here is an error plot, which of course has bigger absolute errors than the one above due to the scale factor k=44330 which has been included here.
I can also state that the first three derivatives of the function will have constant sign over the range in question, so the function is monotonic, convex, and in general pretty well-behaved.
Not meant to answer the question, but it illustrates the Road Not To Go, and thus may be helpful:
This quick-and-dirty C code calculates pow(i, 0.19029) for 0.000 to 1.000 in steps of 0.01. The first half displays the error, in percents, when stored as 1/65536ths (as that theoretically provides slightly over 4 decimals of precision). The second half shows both interpolated and calculated values in steps of 0.001, and the difference between these two.
It kind of looks okay if you read from the bottom up, all 100s and 99.99s there, but about the first 20 values from 0.001 to 0.020 are worthless.
#include <stdio.h>
#include <math.h>
float powers[102];
int main (void)
{
int i, as_int;
double as_real, low, high, delta, approx, calcd, diff;
printf ("calculating and storing:\n");
for (i=0; i<=101; i++)
{
as_real = pow(i/100.0, 0.19029);
as_int = (int)round(65536*as_real);
powers[i] = as_real;
diff = 100*as_real/(as_int/65536.0);
printf ("%.5f %.5f %.5f ~ %.3f\n", i/100.0, as_real, as_int/65536.0, diff);
}
printf ("\n");
printf ("-- interpolating in 1/10ths:\n");
for (i=0; i<1000; i++)
{
as_real = i/1000.0;
low = powers[i/10];
high = powers[1+i/10];
delta = (high-low)/10.0;
approx = low + (i%10)*delta;
calcd = pow(as_real, 0.19029);
diff = 100.0*approx/calcd;
printf ("%.5f ~ %.5f = %.5f +/- %.5f%%\n", as_real, approx, calcd, diff);
}
return 0;
}
You can find a complete, correct standalone implementation of pow in fdlibm. It's about 200 lines of code, about half of which deal with special cases. If you remove the code that deals with special cases you're not interested in I doubt you'll have problems including it in your program.
LutzL's answer is a really good one: Calculate your power as (x^1.52232)^(1/8), computing the inner power by spline interpolation or another method. The eighth root deals with the pathological non-differentiable behavior near zero. I took the liberty of mocking up an implementation this way. The below, however, only does a linear interpolation to do x^1.52232, and you'd need to get the full coefficients using your favorite numerical mathematics tools. You'll adding scarcely 40 lines of code to get your needed power, plus however many knots you choose to use for your spline, as dicated by your required accuracy.
Don't be scared by the #include <math.h>; it's just for benchmarking the code.
#include <stdio.h>
#include <math.h>
double my_sqrt(double x) {
/* Newton's method for a square root. */
int i = 0;
double res = 1.0;
if (x > 0) {
for (i = 0; i < 10; i++) {
res = 0.5 * (res + x / res);
}
} else {
res = 0.0;
}
return res;
}
double my_152232(double x) {
/* Cubic spline interpolation for x ** 1.52232. */
int i = 0;
double res = 0.0;
/* coefs[i] will give the cubic polynomial coefficients between x =
i and x = i+1. Out of laziness, the below numbers give only a
linear interpolation. You'll need to do some work and research
to get the spline coefficients. */
double coefs[3][4] = {{0.0, 1.0, 0.0, 0.0},
{-0.872526, 1.872526, 0.0, 0.0},
{-2.032706, 2.452616, 0.0, 0.0}};
if ((x >= 0) && (x < 3.0)) {
i = (int) x;
/* Horner's method cubic. */
res = (((coefs[i][3] * x + coefs[i][2]) * x) + coefs[i][1] * x)
+ coefs[i][0];
} else if (x >= 3.0) {
/* Scaled x ** 1.5 once you go off the spline. */
res = 1.024824 * my_sqrt(x * x * x);
}
return res;
}
double my_019029(double x) {
return my_sqrt(my_sqrt(my_sqrt(my_152232(x))));
}
int main() {
int i;
double x = 0.0;
for (i = 0; i < 1000; i++) {
x = 1e-2 * i;
printf("%f %f %f \n", x, my_019029(x), pow(x, 0.19029));
}
return 0;
}
EDIT: If you're just interested in a small region like [0,1], even simpler is to peel off one sqrt(x) and compute x^1.02232, which is quite well behaved, using a Taylor series:
double my_152232(double x) {
double part_050000 = my_sqrt(x);
double part_102232 = 1.02232 * x + 0.0114091 * x * x - 3.718147e-3 * x * x * x;
return part_102232 * part_050000;
}
This gets you within 1% of the exact power for approximately [0.1,6], though getting the singularity exactly right is always a challenge. Even so, this three-term Taylor series gets you within 2.3% for x = 0.001.