Calculate the cosine of a sequence

Question

Hi,

I have to calculate the following:

float2 y = CONSTANT;
for (int i = 0; i < totalN; i++)
   h[i] = cos(y*i);

totalN is a large number, so I would like to make this in a more efficient way. Is there any way to improve this? I suspect there is, because, after all, we know what's the result of cos(n), for n=1..N, so maybe there's some theorem that allows me to compute this in a faster way. I would really appreciate any hint.

Thanks in advance,

Federico

Answer 1

You can do this using complex numbers.

if you define x = sin(y) + i cos(y), cos(y*i) will be the real part of x^i.

You can compute for all i iteratively. Complex multiply is 2 multiplies plus two adds.

Answer 2

Knowing cos(n) doesn't help -- your math library already does these kind of trivial things for you.

Knowing that cos((i+1)y)=cos(i*y+y)=cos(i*y)*cos(y)-sin(i*y)*sin(y) can help, if you precompute cos(y) and sin(y), and keep track of both cos(i*y) and sin(i*y) along the way. It may result in some loss of precision, though - you'll have to check.

Answer 3

Here's a method, but it uses a little bit of memory for the sin. It uses the trig identities:

cos(a + b) = cos(a)cos(b)-sin(a)sin(b)
sin(a + b) = sin(a)cos(b)+cos(a)sin(b)

Then here's the code:

h[0] = 1.0;
double g1 = sin(y);
double glast = g1;
h[1] = cos(y);
for (int i = 2; i < totalN; i++){
    h[i] = h[i-1]*h[1]-glast*g1;
    glast = glast*h[1]+h[i-1]*g1;

}

If I didn't make any errors then that should do it. Of course there could be round-off problems so be aware of that. I implemented this in Python and it is quite accurate.

Answer 4

Using one of the most beautiful formulas of mathematics, Euler's formula
exp(i*x) = cos(x) + i*sin(x),

substituting x := n * phi:

cos(n*phi) = Re( exp(i*n*phi) )
sin(n*phi) = Im( exp(i*n*phi) )

exp(i*n*phi) = exp(i*phi) ^ n

Power ^n is n repeated multiplications. Therefore you can calculate cos(n*phi) and simultaneously sin(n*phi) by repeated complex multiplication by exp(i*phi) starting with (1+i*0).

Code examples:

Python:

from math import *

DEG2RAD = pi/180.0 # conversion factor degrees --> radians
phi = 10*DEG2RAD # constant e.g. 10 degrees

c = cos(phi)+1j*sin(phi) # = exp(1j*phi)
h=1+0j
for i in range(1,10):
  h = h*c
  print "%d %8.3f"%(i,h.real)

or C:

#include <stdio.h>
#include <math.h>

// numer of values to calculate:
#define N 10

// conversion factor degrees --> radians:
#define DEG2RAD (3.14159265/180.0)

// e.g. constant is 10 degrees:
#define PHI (10*DEG2RAD)

typedef struct
{
  double re,im;
} complex_t;


int main(int argc, char **argv)
{
  complex_t c;
  complex_t h[N];
  int index;

  c.re=cos(PHI);
  c.im=sin(PHI);

  h[0].re=1.0;   
  h[0].im=0.0;
  for(index=1; index<N; index++)
  {
    // complex multiplication h[index] = h[index-1] * c;
    h[index].re=h[index-1].re*c.re - h[index-1].im*c.im; 
    h[index].im=h[index-1].re*c.im + h[index-1].im*c.re; 
    printf("%d: %8.3f\n",index,h[index].re);
  }
} 

Answer 5

I'm not sure what kind of accuracy vs. performance compromises you're willing to make, but there are extensive discussions of various sinusoid approximation techniques at these links:

Fun with Sinusoids - http://www.audiomulch.com/~rossb/code/sinusoids/
Fast and accurate sine/cosine - http://www.devmaster.net/forums/showthread.php?t=5784

Edit (I think this is the "Don Cross" link that's broken on the "Fun with Sinusoids" page):

Optimizing Trig Calculations - http://groovit.disjunkt.com/analog/time-domain/fasttrig.html

Answer 6

How accurate do you need the resulting cos(x) to be? If you can live with some, you could create a lookup table, sampling the unit circle at 2*PI/N intervals and then interpolate between two adjacent points. N would be chosen to achieve some desired level of accuracy.

What I don't know is whether an interpolation is actually less costly than computing a cosine. Since its usually done in microcode in modern CPUs, it may not be.

Answer 7

Maybe the simplest formula is

cos(n+y) = 2cos(n)cos(y) - cos(n-y).

If you precompute the constant 2*cos(y) then each value cos(n+y) can be computed from the previous 2 values with one single multiplication and one subtraction. I.e., in pseudocode

h[0] = 1.0
h[1] = cos(y)
m = 2*h[1]
for (int i = 2; i < totalN; ++i)
  h[i] = m*h[i-1] - h[i-2]

Answer 8

There are some good answers here but they are all recursive. Recursive calculation will not work for cosine function when using floating point arithmetic; you will invariably get rounding errors which quickly compound.

Consider calculation y = 45 degrees, totalN 10 000. You won't end up with 1 as the final result.

Answer 9

To address Kirk's concerns: all of the solutions based on the recurrence for cos and sin boil down to computing

x(k) = R x(k - 1),

where R is the matrix that rotates by y and x(0) is the unit vector (1, 0). If the true result for k - 1 is x'(k - 1) and the true result for k is x'(k), then the error goes from e(k - 1) = x(k - 1) - x'(k - 1) to e(k) = R x(k - 1) - R x'(k - 1) = R e(k - 1) by linearity. Since R is what's called an orthogonal matrix, R e(k - 1) has the same norm as e(k - 1), and the error grows very slowly. (The reason it grows at all is due to round-off; the computer representation of R is in general almost, but not quite orthogonal, so it will be necessary to restart the recurrence using the trig operations from time to time depending on the accuracy required. This is still much, much faster than using the trig ops to compute each value.)