First, is this the correct C++ representation of the pdf gaussian function ?
float pdf_gaussian = ( 1 / ( s * sqrt(2*M_PI) ) ) * exp( -0.5 * pow( (x-m)/s, 2.0 ) );
Second, does it make sense of we do something like this ?
if(pdf_gaussian < uniform_random())
do something
else
do other thing
EDIT: An example of what exactly are you trying to achieve:
Say I have a data called Y1. Then a new data called Xi arrive. I want to see if I should associated Xi to Y1 or if I should keep Xi as a new data data that will be called Y2. This is based on the distance between the new data Xi and the existing data Y1. If Xi is "far" from Y1 then Xi will not be associated to Y1, otherwise if it is "not far", it will be associated to Y1. Now I want to model this "far" or "not far" using a gaussian probability based on the mean and stdeviation of distances between Y and the data that where already associated to Y in the past.
Technically,
float pdf_gaussian = ( 1 / ( s * sqrt(2*M_PI) ) ) * exp( -0.5 * pow( (x-m)/s, 2.0 ) );
is not incorrect, but can be improved.
First, 1 / sqrt(2 Pi) can be precomputed, and using pow with integers is not a good idea: it may use exp(2 * log x) or a routine specialized for floating point exponents instead of simply x * x.
Example better code:
float normal_pdf(float x, float m, float s)
{
static const float inv_sqrt_2pi = 0.3989422804014327;
float a = (x - m) / s;
return inv_sqrt_2pi / s * std::exp(-0.5f * a * a);
}
You may want to make this a template instead of using float:
template <typename T>
T normal_pdf(T x, T m, T s)
{
static const T inv_sqrt_2pi = 0.3989422804014327;
T a = (x - m) / s;
return inv_sqrt_2pi / s * std::exp(-T(0.5) * a * a);
}
this allows you to use normal_pdf on double arguments also (it is not that much more generic though). There are caveats with the last code, namely that you have to beware not using it with integers (there are workarounds, but this makes the routine more verbose).
yes. boost::random has gaussian distribution.
See, for example, this question: How to use boost normal distribution classes?
As an alternative, there's a standard way of converting two uniformly distributed random numbers into two normally distributed numbers.
See, e.g. this question: Generate random numbers following a normal distribution in C/C++
In response to your last edit (note that the question is completely different as edited, hence my answer to an original one is irrelevant). I think you'd better off first formulating for yourself what exactly do you mean to mean by "modelling far or not far using a gaussian distribution". Then reformulate that understanding in math terms and only then start programming. As it stands, I think the problem is underspecified.
Use Box-Muller transform. This creates values with a normal/gaussian distribution.
http://en.wikipedia.org/wiki/Normal_distribution#Generating_values_from_normal_distribution
http://en.wikipedia.org/wiki/Box_Muller_transform
It's not very complex to code using math libraries.
eg.
Generate 2 uniform numbers, use them to get two normally distributed numbers. Then Return one and save the other so that you have it for your 'next' request of a random number.
Since C++ 11, std::normal_distribution defined in the standard header random can be used to generate Gaussian random samples. More information can be found herein.
Related
I am trying to write a short C++ routine to calculate the following function F(i,j,z) for given integers j > i (typically they lie between 0 and 100) and complex number z (bounded by |z| < 100), where L are the associated Laguerre Polynomials:
The issue is that I want this function to be callable from within a CUDA kernel (i.e. with a __device__ attribute). Standard library/Boost/etc functions are therefore out of the questions, unless they are simple enough to re-implement on my own - this especially relates to the Laguerre polynomials which exist in Boost and C++17. Regardless if I manage to wrap any standard function for Laguerre polynomials, I still have a similar pre-factor to calculate of the form (z^j/j!).
Question: How can I do a relatively simple implementation of such a function, without introducing significant numerical instability?
My idea so far is to calculate L and its pre-factor independently. The pre-factor I will calculate by first looping from 0 to j-i and calculate (z^1 * z^2/2 * ... * z^(j-1)/(j-i)!). I will then calculate the remaining factor exp(-|z|^2/2) *(j-i)! * sqrt(i!/j!) (either in a similar way, or through the Gamma-function, which is implemented in CUDA math). The idea is then to find a minimal algorithm to calculate the associated Laguerre polynomial, unless I manage to wrap an implementation from e.g. Boost or GNU C++.
Edit/side note: The expression for F actually blows up numerically for some values of i/j. It was derived wrong in the source where I got it, and the indices of the associated Laguerre polynomials should instead be L_i^(j-i). That does not invalidate the approaches suggested in the answers/comments.
I recommend finding a recurrence relation for the coefficients of the Laguerre Polynomial:
C(k+1) = g(k)C(k)
g(k) = C(k+1) / C(k)
g(k) = -z * (j - k) / ((j - i + k + 1) * (k + 1)) //Verify this yourself :)
This allows you to avoid most of factorials in computing the polynomial.
After that I would follow Severin's idea of doing the calculations in logarithms
so as to not overload the double floating point range:
log(F) = log(sqrt(i!/j!)) - |z|^2 + (j-i) * log(-z) + log(L(|z|^2))
log(L) = log((2*j - i)!) + log(sum) // where the summation is computed using the recurrence relation above
and using the fact that:
log(a!) = sum(k=1..a, log(k))
and also:
log(z) = log(|z|) + I * arg(z) for complex z
log(-z) = log(|z|) + I * arg(-z)
log(-z) = log(|z|) - I * arg(z)
for the log(sqrt(i!/j!)) part I would do (assuming that j >= i):
log(sqrt(i!/j!))
= 0.5 * (log(i!) - log(j!))
= -0.5 * sum(k==i+1..j, log(k))
I haven't tried this out so there could definitely be little mistakes here and there. This answer is more about the technique rather than a copy-paste-ready answer
Well, what you should do is to logarithm it
Assuming natural logarithm,
q = log(z^j/j!) = log(z^j) - log(j!) = j*log(z) - log(Gamma(j+1))
First term is simple, second term is standard C++ function lgamma(x) (or you could use GSL).
compute value of q and return cexp(q)
You could fold exponent in this method as well
I browsed through the recently released Doom 3 BFG source code, when I came upon something that does not appear to make any sense. Doom 3 wraps mathematical functions in the idMath class. Some of the functions just foward to the corresponding functions from math.h, but some are reimplementations (e.g. idMath::exp16()) that I assume have a higher performance than their math.h counterparts (maybe at the expense of precision).
What baffles me, however, is the way they have implemented the float idMath::Sqrt(float x) function:
ID_INLINE float idMath::InvSqrt( float x ) {
return ( x > FLT_SMALLEST_NON_DENORMAL ) ? sqrtf( 1.0f / x ) : INFINITY;
}
ID_INLINE float idMath::Sqrt( float x ) {
return ( x >= 0.0f ) ? x * InvSqrt( x ) : 0.0f;
}
This appears to perform two unnecessary floating point operations: First a division and then a multiplication.
It is interesting to note that the original Doom 3 source code also implemented the square root function in this way, but the inverse square root uses the fast inverse square root algorithm.
ID_INLINE float idMath::InvSqrt( float x ) {
dword a = ((union _flint*)(&x))->i;
union _flint seed;
assert( initialized );
double y = x * 0.5f;
seed.i = (( ( (3*EXP_BIAS-1) - ( (a >> EXP_POS) & 0xFF) ) >> 1)<<EXP_POS) | iSqrt[(a >> (EXP_POS-LOOKUP_BITS)) & LOOKUP_MASK];
double r = seed.f;
r = r * ( 1.5f - r * r * y );
r = r * ( 1.5f - r * r * y );
return (float) r;
}
ID_INLINE float idMath::Sqrt( float x ) {
return x * InvSqrt( x );
}
Do you see any advantage in calculating Sqrt(x) as x * InvSqrt(x) if InvSqrt(x) internally just calls math.h's fsqrt(1.f/x)? Am I maybe missing something important about denormalized floating point numbers here or is this just sloppiness on id software's part?
I can see two reasons for doing it this way: firstly, the "fast invSqrt" method (really Newton Raphson) is now the method used in a lot of hardware, so this approach leaves open the possibility of taking advantage of such hardware (and doing potentially four or more such operations at once). This article discusses it a little bit:
How slow (how many cycles) is calculating a square root?
The second reason is for compatibility. If you change the code path for calculating square roots, you may get different results (especially for zeroes, NaNs, etc.), and lose compatibility with code that depended on the old system.
As far as I know, the InvSqrt is used to compute colors in the sense that color depends on the angle from which light bounces off a surface, which gives you some function using the inverse of the square root.
In their case, they don't need huge precision when computing these numbers, so the engineers behind Doom 3's code (originally from Quake III) came up with a very very very fast method of computing an approximation for InvSqrt using only several Newton-Raphson's iterations.
This is why they use InvSqrt in all their code, instead of using built-in (slower) functions. I guess the use of x * InvSqrt(x) is there to avoid multiplying work by two (by having two very efficient functions, one for InvSqrt and another for Sqrt).
You should read this article, it might shed some light on this issue.
When code has been modified by multiple people, it becomes hard to answer questions about why it has its current form, especially without revision history.
However, given a third of a century of programming experience, this code fits the pattern others have mentioned: At one time, InvSqrt was fast, and it made sense to use it to compute the square root. Then InvSqrt changed, and nobody updated Sqrt.
It is also possible that they came across a relatively naive version of sqrtf which was notably slower for bigger numbers.
I am wondering if there is a C/C++ library or Matlab code technique to determine real and complex numbers using a minimization solver. Here is a code snippet showing what I would like to do. For example, suppose that I know Utilde, but not x and U variables. I want to use optimization (fminsearch) to determine x and U, given Utilde. Note that Utilde is a complex number.
x = 1.5;
U = 50 + 1i*25;
x0 = [1 20]; % starting values
Utilde = U * (1 / exp(2 * x)) * exp( 1i * 2 * x);
xout = fminsearch(#(v)optim(v, Utilde), x0);
function diff = optim(v, Utilde)
x = v(1);
U = v(2);
diff = abs( -(Utilde/U) + (1 / exp(2 * x)) * exp( 1i * 2 * x ) );
The code above does not converge to the proper values, and xout = 1.7318 88.8760. However, if U = 50, which is not a complex number, then xout = 1.5000 50.0000, which are the proper values.
Is there a way in Matlab or C/C++ to ensure proper convergence, given Utilde as a complex number? Maybe I have to change the code above?
If there isn't a way to do this natively in Matlab, then perhaps one
gist of the question is this: Is there a multivariate (i.e.
Nelder-Mead or similar algorithm) optimization library that is able
to work with real and complex inputs and outputs?
Yet another question is whether the function is convergent or not. I
don't know if it is the algorithm or the function. Might I need to change something in the Utilde = U * (1 / exp(2 * x)) * exp( 1i * 2 * x) expression to make it convergent?
The main problem here is that there is no unique solution to this optimization or parameter fitting problem. For example, looking at the expected and actual results above, Utilde is equivalent (ignoring round-off differences) for the two (x, U) pairs, i.e.
Utilde(x = 1.5, U = 50 + 25i) = Utilde(x = 1.7318, U = 88.8760)
Although I have not examined it in depth, I even suspect that for any value of x, you can find an U that computes to Utilde(x, U) = Utilde(x = 1.5, U = 50 + 25i).
The solution here would thus be to further constrain the parameter fitting problem so that the solver yields any solution that can be considered acceptable. Alternatively, reformulate Utilde to have a unique value for any (x, U) pair.
UPDATE, AUG 1
Given reasonable starting values, it actually seems like it is sufficient to restrict x to be real-valued. Performing unconstrained non-linear optimization using the diff function formulated above, I get the following result:
x = 1.50462926953244
U = 50.6977768845879 + 24.7676554234729i
diff = 3.18731710515855E-06
However, changing the starting guess to values more distant from the desired values does yield different solutions, so restricting x to be real-values does not alone provide a unique solution to the problem.
I have implemented this in C#, using the BOBYQA optimizer, but the numerics should be the same as above. If you want to try outside of Matlab, it should also be relatively simple to turn the C# code below into C++ code using the std::complex class and an (unconstrained) nonlinear C++ optimizer of your own choice. You could find some C++ compatible codes that do not require gradient computation here, and there is also various implementations available in Numerical Recipes. For example, you could access the C version of NR online here.
For reference, here are the relevant parts of my C# code:
class Program
{
private static readonly Complex Coeff = new Complex(-2.0, 2.0);
private static readonly Complex UTilde0 = GetUTilde(1.5, new Complex(50.0, 25.0));
static void Main(string[] args)
{
double[] vars = new[] {1.0, 25.0, 0.0}; // xstart = 1.0, Ustart = 25.0
BobyqaExitStatus status = Bobyqa.FindMinimum(GetObjfnValue, vars.Length, vars);
}
public static Complex GetUTilde(double x, Complex U)
{
return U * Complex.Exp(Coeff * x);
}
public static double GetObjfnValue(int n, double[] vars)
{
double x = vars[0];
Complex U = new Complex(vars[1], vars[2]);
return Complex.Abs(-UTilde0 / U + Complex.Exp(Coeff * x));
}
}
The documentation for fminsearch says how to deal with complex numbers in the limitations section:
fminsearch only minimizes over the real numbers, that is, x must only consist of real numbers and f(x) must only return real numbers. When x has complex variables, they must be split into real and imaginary parts.
You can use the functions real and imag to extract the real and imaginary parts, respectively.
It appears that there is no easy way to do this, even if both x and U are real numbers. The equation for Utilde is not well-posed for an optimization problem, and so it must be modified.
I've tried to code up my own version of the Nelder-Mead optimization algorithm, as well as tried Powell's method. Neither seem to work well for this problem, even when I attempted to modify these methods.
I'm using bits of Perry Cook's Synthesis Toolkit (STK) to generate saw and square waves. STK includes this BLIT-based sawtooth oscillator:
inline STKFloat BlitSaw::tick( void ) {
StkFloat tmp, denominator = sin( phase_ );
if ( fabs(denominator) <= std::numeric_limits<StkFloat>::epsilon() )
tmp = a_;
else {
tmp = sin( m_ * phase_ );
tmp /= p_ * denominator;
}
tmp += state_ - C2_;
state_ = tmp * 0.995;
phase_ += rate_;
if ( phase_ >= PI )
phase_ -= PI;
lastFrame_[0] = tmp;
return lastFrame_[0];
}
The square wave oscillator is broadly similar. At the top, there's this comment:
// A fully optimized version of this code would replace the two sin
// calls with a pair of fast sin oscillators, for which stable fast
// two-multiply algorithms are well known.
I don't know where to start looking for these "fast two-multiply algorithms" and I'd appreciate some pointers. I could use a lookup table instead, but I'm keen to learn what these 'fast sin oscillators' are. I could also use an abbreviated Taylor series, but thats way more than two multiplies. Searching hasn't turned up anything much, although I did find this approximation:
#define AD_SIN(n) (n*(2.f- fabs(n)))
Plotting it out shows that it's not really a close approximation outside the range of -1 to 1, so I don't think I can use it when phase_ is in the range -pi to pi:
Here, Sine is the blue line and the purple line is the approximation.
Profiling my code reveals that the calls to sin() are far and away the most time-consuming calls, so I really would like to optimise this piece.
Thanks
EDIT Thanks for the detailed and varied answers. I will explore these and accept one at the weekend.
EDIT 2 Would the anonymous close voter please kindly explain their vote in the comments? Thank you.
Essentially the sinusoidal oscilator is one (or more) variables that change with each DSP step, rather than getting recalculated from scratch.
The simplest are based on the following trig identities: (where d is constant, and thus so is cos(d) and sin(d) )
sin(x+d) = sin(x) cos(d) + cos(x) sin(d)
cos(x+d) = cos(x) cos(d) - sin(x) sin(d)
However this requires two variables (one for sin and one for cos) and 4 multiplications to update. However this will still be far faster than calculating a full sine at each step.
The solution by Oli Charlesworth is based on solutions to this general equation
A_{n+1} = a A_{n} + A_{n-1}
Where looking for a solution of the form A_n = k e^(i theta n) gives an equation for theta.
e^(i theta (n+1) ) = a e^(i theta n ) + b e^(i theta (n-1) )
Which simplifies to
e^(i theta) - e^(-i theta ) = a
2 cos(theta) = a
Giving
A_{n+1} = 2 cos(theta) A_{n} + A_{n-1}
Whichever approach you use you'll either need to use one or two of these oscillators for each frequency, or use another trig identity to derive the higher or lower frequencies.
How accurate do you need this?
This function, f(x)=0.398x*(3.1076-|x|), does a reasonably good job for x between -pi and pi.
Edit
An even better approximation is f(x)=0.38981969947653056*(pi-|x|), which keeps the absolute error to 0.038158444604 or less for x between -pi and pi.
A least squares minimization will yield a slightly different function.
It's not possible to generate one-off sin calls with just two multiplies (well, not a useful approximation, at any rate). But it is possible to generate an oscillator with low complexity, i.e. where each value is calculated in terms of the preceding ones.
For instance, consider that the following difference equation will give you a sinusoid:
y[n] = 2*cos(phi)*y[n-1] - y[n-2]
(where cos(phi) is a constant)
(From the original author of the VST BLT code).
As a matter of fact, I was porting the VST BLT oscillators to C#, so I was googling for good sin oscillators. Here's what I came up with. Translation to C++ is straightforward. See the notes at the end about accuumulated round-off errors.
public class FastOscillator
{
private double b1;
private double y1, y2;
private double fScale;
public void Initialize(int sampleRate)
{
fScale = AudioMath.TwoPi / sampleRate;
}
// frequency in Hz. phase in radians.
public void Start(float frequency, double phase)
{
double w = frequency * fScale;
b1 = 2.0 * Math.Cos(w);
y1 = Math.Sin(phase - w);
y2 = Math.Sin(phase - w * 2);
}
public double Tick()
{
double y0 = b1 * y1 - y2;
y2 = y1;
y1 = y0;
return y0;
}
}
Note that this particular oscillator implementation will drift over time, so it needs to be re-initialzed periodically. In this particular implementation, the magnitude of the sin wave decays over time. The original comments in the STK code suggested a two-multiply oscillator. There are, in fact, two-multiply oscillators that are reasonably stable over time. But in retrospect, the need to keep the sin(phase), and sin(m*phase) oscillators tightly in synch probably means that they have to be resynched anyway. Round-off errors between phase and m*phase mean that even if the oscillators were stable, they would drift eventually, running a significant risk of producing large spikes in values near the zeros of the BLT functions. May as well use a one-multiply oscillator.
These particular oscillators should probably be re-initialized every 30 to 100 cycles (or so). My C# implementation is frame based (i.e. it calculates an float[] array of results in a void Tick(int count, float[] result) method. The oscillators are re-synched at the end of each Tick call. Something like this:
void Tick(int count, float[] result)
{
for (int i = 0; i < count; ++i)
{
...
result[i] = bltResult;
}
// re-initialize the oscillators to avoid accumulated drift.
this.phase = (this.phase + this.dPhase*count) % AudioMath.TwoPi;
this.sinOsc.Initialize(frequency,this.phase);
this.mSinOsc.Initialize(frequency*m,this.phase*m);
}
Probably missing from the STK code. You might want to investigate this. The original code provided to the STK did this. Gary Scavone tweaked the code a bit, and I think the optimization was lost. I do know that the STK implementations suffer from DC drift, which can be almost entirely eliminated when implemented properly.
There's a peculiar hack that prevents DC drift of the oscillators, even when sweeping the frequency of the oscillators. The trick is that the oscillators should be started with an initial phase adjustment of dPhase/2. That just so happens to start the oscillators off with zero DC drift, without having to figure out wat the correct initial state for various integrators in each of the BLT oscillators.
Strangely, if the adjustment is re-adjusted whenever the frequency of the oscillator changes, then this also prevents wild DC drift of the output when sweeping the frequency of the oscillator. Whenever the frequency changes, subtract dPhase/2 from the previous phase value, recalculate dPhase for the new frequency, and then add dPhase/2.I rather suspect this could be formally proven; but I have not been able to so. All I know is that It Just Works.
For a block implementation, the oscillators should actually be initialized as follows, instead of carrying the phase adjustment in the current this.phase value.
this.sinOsc.Initialize(frequency,phase+dPhase*0.5);
this.mSinOsc.Initialize(frequency*m,(phase+dPhase*0.5)*m);
You might want to take a look here:
http://devmaster.net/forums/topic/4648-fast-and-accurate-sinecosine/
There's some sample code that calculates a very good appoximation of sin/cos using only multiplies, additions and the abs() function. Quite fast too. The comments are also a good read.
It essentiall boils down to this:
float sine(float x)
{
const float B = 4/pi;
const float C = -4/(pi*pi);
const float P = 0.225;
float y = B * x + C * x * abs(x);
return P * (y * abs(y) - y) + y;
}
and works for a range of -PI to PI
If you can, you should consider memorization based techniques. Essentially store sin(x) and cos(x) values for a bunch values. To calculate sin(y), find a and b for which precomputed values exist such that a<=y<=b. Now using sin(a), sin(b), cos(a), cos(b), y-a and y-b approximately calculate sin(y).
The general idea of getting periodically sampled results from the sine or cosine function is to use a trig recursion or an initialized (barely) stable IIR filter (which can end up being pretty much the same computations). There are bunches of these in the DSP literature, of varying accuracy and stability. Choose carefully.
I updated the code.
What i am trying to do is to hold every lagrange's coefficient values in pointer d.(for example for L1(x) d[0] would be "x-x2/x1-x2" ,d1 would be (x-x2/x1-x2)*(x-x3/x1-x3) etc.
My problem is
1) how to initialize d ( i did d[0]=(z-x[i])/(x[k]-x[i]) but i think it's not right the "d[0]"
2) how to initialize L_coeff. ( i am using L_coeff=new double[0] but am not sure if it's right.
The exercise is:
Find Lagrange's polynomial approximation for y(x)=cos(π x), x ∈−1,1 using 5 points
(x = -1, -0.5, 0, 0.5, and 1).
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;
const double pi=3.14159265358979323846264338327950288;
// my function
double f(double x){
return (cos(pi*x));
}
//function to compute lagrange polynomial
double lagrange_polynomial(int N,double *x){
//N = degree of polynomial
double z,y;
double *L_coeff=new double [0];//L_coefficients of every Lagrange L_coefficient
double *d;//hold the polynomials values for every Lagrange coefficient
int k,i;
//computations for finding lagrange polynomial
//double sum=0;
for (k=0;k<N+1;k++){
for ( i=0;i<N+1;i++){
if (i==0) continue;
d[0]=(z-x[i])/(x[k]-x[i]);//initialization
if (i==k) L_coeff[k]=1.0;
else if (i!=k){
L_coeff[k]*=d[i];
}
}
cout <<"\nL("<<k<<") = "<<d[i]<<"\t\t\tf(x)= "<<f(x[k])<<endl;
}
}
int main()
{
double deg,result;
double *x;
cout <<"Give the degree of the polynomial :"<<endl;
cin >>deg;
for (int i=0;i<deg+1;i++){
cout <<"\nGive the points of interpolation : "<<endl;
cin >> x[i];
}
cout <<"\nThe Lagrange L_coefficients are: "<<endl;
result=lagrange_polynomial(deg,x);
return 0;
}
Here is an example of lagrange polynomial
As this seems to be homework, I am not going to give you an exhaustive answer, but rather try to send you on the right track.
How do you represent polynomials in a computer software? The intuitive version you want to archive as a symbolic expression like 3x^3+5x^2-4 is very unpractical for further computations.
The polynomial is defined fully by saving (and outputting) it's coefficients.
What you are doing above is hoping that C++ does some algebraic manipulations for you and simplify your product with a symbolic variable. This is nothing C++ can do without quite a lot of effort.
You have two options:
Either use a proper computer algebra system that can do symbolic manipulations (Maple or Mathematica are some examples)
If you are bound to C++ you have to think a bit more how the single coefficients of the polynomial can be computed. You programs output can only be a list of numbers (which you could, of course, format as a nice looking string according to a symbolic expression).
Hope this gives you some ideas how to start.
Edit 1
You still have an undefined expression in your code, as you never set any value to y. This leaves prod*=(y-x[i])/(x[k]-x[i]) as an expression that will not return meaningful data. C++ can only work with numbers, and y is no number for you right now, but you think of it as symbol.
You could evaluate the lagrange approximation at, say the value 1, if you would set y=1 in your code. This would give you the (as far as I can see right now) correct function value, but no description of the function itself.
Maybe you should take a pen and a piece of paper first and try to write down the expression as precise Math. Try to get a real grip on what you want to compute. If you did that, maybe you come back here and tell us your thoughts. This should help you to understand what is going on in there.
And always remember: C++ needs numbers, not symbols. Whenever you have a symbol in an expression on your piece of paper that you do not know the value of you can either find a way how to compute the value out of the known values or you have to eliminate the need to compute using this symbol.
P.S.: It is not considered good style to post identical questions in multiple discussion boards at once...
Edit 2
Now you evaluate the function at point y=0.3. This is the way to go if you want to evaluate the polynomial. However, as you stated, you want all coefficients of the polynomial.
Again, I still feel you did not understand the math behind the problem. Maybe I will give you a small example. I am going to use the notation as it is used in the wikipedia article.
Suppose we had k=2 and x=-1, 1. Furthermore, let my just name your cos-Function f, for simplicity. (The notation will get rather ugly without latex...) Then the lagrangian polynomial is defined as
f(x_0) * l_0(x) + f(x_1)*l_1(x)
where (by doing the simplifications again symbolically)
l_0(x)= (x - x_1)/(x_0 - x_1) = -1/2 * (x-1) = -1/2 *x + 1/2
l_1(x)= (x - x_0)/(x_1 - x_0) = 1/2 * (x+1) = 1/2 * x + 1/2
So, you lagrangian polynomial is
f(x_0) * (-1/2 *x + 1/2) + f(x_1) * 1/2 * x + 1/2
= 1/2 * (f(x_1) - f(x_0)) * x + 1/2 * (f(x_0) + f(x_1))
So, the coefficients you want to compute would be 1/2 * (f(x_1) - f(x_0)) and 1/2 * (f(x_0) + f(x_1)).
Your task is now to find an algorithm that does the simplification I did, but without using symbols. If you know how to compute the coefficients of the l_j, you are basically done, as you then just can add up those multiplied with the corresponding value of f.
So, even further broken down, you have to find a way to multiply the quotients in the l_j with each other on a component-by-component basis. Figure out how this is done and you are a nearly done.
Edit 3
Okay, lets get a little bit less vague.
We first want to compute the L_i(x). Those are just products of linear functions. As said before, we have to represent each polynomial as an array of coefficients. For good style, I will use std::vector instead of this array. Then, we could define the data structure holding the coefficients of L_1(x) like this:
std::vector L1 = std::vector(5);
// Lets assume our polynomial would then have the form
// L1[0] + L2[1]*x^1 + L2[2]*x^2 + L2[3]*x^3 + L2[4]*x^4
Now we want to fill this polynomial with values.
// First we have start with the polynomial 1 (which is of degree 0)
// Therefore set L1 accordingly:
L1[0] = 1;
L1[1] = 0; L1[2] = 0; L1[3] = 0; L1[4] = 0;
// Of course you could do this more elegant (using std::vectors constructor, for example)
for (int i = 0; i < N+1; ++i) {
if (i==0) continue; /// For i=0, there will be no polynomial multiplication
// Otherwise, we have to multiply L1 with the polynomial
// (x - x[i]) / (x[0] - x[i])
// First, note that (x[0] - x[i]) ist just a scalar; we will save it:
double c = (x[0] - x[i]);
// Now we multiply L_1 first with (x-x[1]). How does this multiplication change our
// coefficients? Easy enough: The coefficient of x^1 for example is just
// L1[0] - L1[1] * x[1]. Other coefficients are done similary. Futhermore, we have
// to divide by c, which leaves our coefficient as
// (L1[0] - L1[1] * x[1])/c. Let's apply this to the vector:
L1[4] = (L1[3] - L1[4] * x[1])/c;
L1[3] = (L1[2] - L1[3] * x[1])/c;
L1[2] = (L1[1] - L1[2] * x[1])/c;
L1[1] = (L1[0] - L1[1] * x[1])/c;
L1[0] = ( - L1[0] * x[1])/c;
// There we are, polynomial updated.
}
This, of course, has to be done for all L_i Afterwards, the L_i have to be added and multiplied with the function. That is for you to figure out. (Note that I made quite a lot of inefficient stuff up there, but I hope this helps you understanding the details better.)
Hopefully this gives you some idea how you could proceed.
The variable y is actually not a variable in your code but represents the variable P(y) of your lagrange approximation.
Thus, you have to understand the calculations prod*=(y-x[i])/(x[k]-x[i]) and sum+=prod*f not directly but symbolically.
You may get around this by defining your approximation by a series
c[0] * y^0 + c[1] * y^1 + ...
represented by an array c[] within the code. Then you can e.g. implement multiplication
d = c * (y-x[i])/(x[k]-x[i])
coefficient-wise like
d[i] = -c[i]*x[i]/(x[k]-x[i]) + c[i-1]/(x[k]-x[i])
The same way you have to implement addition and assignments on a component basis.
The result will then always be the coefficients of your series representation in the variable y.
Just a few comments in addition to the existing responses.
The exercise is: Find Lagrange's polynomial approximation for y(x)=cos(π x), x ∈ [-1,1] using 5 points (x = -1, -0.5, 0, 0.5, and 1).
The first thing that your main() does is to ask for the degree of the polynomial. You should not be doing that. The degree of the polynomial is fully specified by the number of control points. In this case you should be constructing the unique fourth-order Lagrange polynomial that passes through the five points (xi, cos(π xi)), where the xi values are those five specified points.
const double pi=3.1415;
This value is not good for a float, let alone a double. You should be using something like const double pi=3.14159265358979323846264338327950288;
Or better yet, don't use pi at all. You should know exactly what the y values are that correspond to the given x values. What are cos(-π), cos(-π/2), cos(0), cos(π/2), and cos(π)?