I have implemented a MCMC algorithm in C++ using the Eigen library. The main part of the algorithm is a loop in which first some some matrix calculations are performed after which the determinant of the resulting matrix is obtained and added to the output. E.g.:
MatrixXd delta0;
NumericVector out(3);
out[0] = 0;
out[1] = 0;
for (int i = 0; i < s; i++) {
...
delta0 = V*(A.cast<double>()-(A+B).cast<double>()*theta.asDiagonal());
...
I = delta0.determinant()
out[1] += I;
out[2] += std::sqrt(I);
}
return out;
Now on certain matrices I unfortunately observe a numerical underflow so that the determinant is outputted as zero (which it actually isn't).
How can I avoid this underflow?
One solution would be to obtain, instead of the determinant, the log of the determinant. However,
I do not know how to do this;
how could I then add up these logs?
Any help is greatly appreciated.
There are 2 main options that come to my mind:
The product of eigenvalues of square matrix is the determinant of this matrix, therefore a sum of logarithms of each eigenvalue is a logarithm of the determinant of this matrix. Assume det(A) = a and det(B) = b for compact notation. After applying aforementioned for 2 matrices A and B, we end up with log(a) and log(b), then actually the following is true:
log(a + b) = log(a) + log(1 + e ^ (log(b) - log(a)))
Yes, we get a logarithm of the sum. What would you do with it next? I don't know, depends on what you have to. If you have to remove logarithm by e ^ log(a + b) = a + b, then you might be lucky that the value of a + b does not underflow now, but in some cases it can still underflow as well.
Perform clever preconditioning; there might be tons of options here, and you better read about them from some trusted sources as this is a serious topic. The simplest (and probably the cheapest ever) example of preconditioning for this particular problem could be to recall that det(c * A) = (c ^ n) * det(A), where A is n by n matrix, and to premultiply your matrix with some c, compute the determinant, and then to divide it by c ^ n to get the actual one.
Update
I thought about one more option. If on the last stages of #1 or #2 you still experience underflow too frequently, then it might be a good idea to increase precision specifically for these last operations, for example, by utilizing GNU MPFR.
You can use Householder elimination to get the QR decomposition of delta0. Then the determinant of the Q part is +/-1 (depending on whether you did an even or odd number of reflections) and the determinant of the R part is the product of the diagonal elements. Both of these are easy to compute without running into underflow hell---and you might not even care about the first.
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 asked this question yesterday: Simulating matlab's mrdivide with 2 square matrices
And thats got mrdivide working. However now I'm having problems with mldivide, which is currently implemented as follows:
cv::Mat mldivide(const cv::Mat& A, const cv::Mat& B )
{
//return b * A.inv();
cv::Mat a;
cv::Mat b;
A.convertTo( a, CV_64FC1 );
B.convertTo( b, CV_64FC1 );
cv::Mat ret;
cv::solve( a, b, ret, cv::DECOMP_NORMAL );
cv::Mat ret2;
ret.convertTo( ret2, A.type() );
return ret2;
}
By my understanding the fact that mrdivide is working should mean that mldivide is working but I can't get it to give me the same results as matlab. Again the results are nothing alike.
Its worth noting I am trying to do a [19x19] \ [19x200] so not square matrices this time.
Like I've previously mentioned in your other question, I am using MATLAB along with mexopencv, that way I can easily compare the output of both MATLAB and OpenCV.
That said, I can't reproduce your problem: I generated randomly matrices, and repeated the comparison N=100 times. I'm running MATLAB R2015a with mexopencv compiled against OpenCV 3.0.0:
N = 100;
r = zeros(N,2);
d = zeros(N,1);
for i=1:N
% double precision, i.e CV_64F
A = randn(19,19);
B = randn(19,200);
x1 = A\B;
x2 = cv.solve(A,B); % this a MEX function that calls cv::solve
r(i,:) = [norm(A*x1-B), norm(A*x2-B)];
d(i) = norm(x1-x2);
end
All results agreed and the errors were very small in the order of 1e-11:
>> mean(r)
ans =
1.0e-12 *
0.2282 0.2698
>> mean(d)
ans =
6.5457e-12
(btw I also tried x2 = cv.solve(A,B, 'IsNormal',true); which sets the cv::DECOMP_NORMAL flag, and the results were not that different either).
This leads me to believe that either your matrices happen to accentuate some edge case in the OpenCV solver, where it failed to give a proper solution, or more likely you have a bug somewhere else in your code.
I'd start by double checking how you load your data, and especially watch out for how the matrices are laid out (obviously MATLAB is column-major, while OpenCV is row-major)...
Also you never told us anything about your matrices; do they exhibit a certain characteristic, are there any symmetries, are they mostly zeros, their rank, etc..
In OpenCV, the default solver method is LU factorization, and you have to explicitly change it yourself if appropriate. MATLAB on the hand will automatically choose a method that best suits the matrix A, and LU is just one of the possible decompositions.
EDIT (response to comments)
When using SVD decompositition in MATLAB, the sign of the left and right eigenvectors U and V is arbitrary (this really comes from the DGESVD LAPACK routine). In order to get consistent results, one convention is to require that the first element of each eigenvector be a certain sign, and multiplying each vector by +1 or -1 to flip the sign as appropriate. I would also suggest checking out eigenshuffle.
One more time, here is a test I did to confirm that I get similar results for SVD in MATLAB and OpenCV:
N = 100;
r = zeros(N,2);
d = zeros(N,3);
for i=1:N
% double precision, i.e CV_64F
A = rand(19);
% compute SVD in MATLAB, and apply sign convention
[U1,S1,V1] = svd(A);
sn = sign(U1(1,:));
U1 = bsxfun(#times, sn, U1);
V1 = bsxfun(#times, sn, V1);
r(i,1) = norm(U1*S1*V1' - A);
% compute SVD in OpenCV, and apply sign convention
[S2,U2,V2] = cv.SVD.Compute(A);
S2 = diag(S2);
sn = sign(U2(1,:));
U2 = bsxfun(#times, sn, U2);
V2 = bsxfun(#times, sn', V2)'; % Note: V2 was transposed w.r.t V1
r(i,2) = norm(U2*S2*V2' - A);
% compare
d(i,:) = [norm(V1-V2), norm(U1-U2), norm(S1-S2)];
end
Again, all results were very similar and the errors close to machine epsilon and negligible:
>> mean(r)
ans =
1.0e-13 *
0.3381 0.1215
>> mean(d)
ans =
1.0e-13 *
0.3113 0.3009 0.0578
One thing I'm not sure about in OpenCV, but MATLAB's svd function returns the singular values sorted in decreasing order (unlike the eig function), with the columns of the eigenvectors in corresponding order.
Now if the singular values in OpenCV are not guaranteed to be sorted for some reason, you have to do it manually as well if you want to compare the results against MATLAB, as in:
% not needed in MATLAB
[U,S,V] = svd(A);
[S, ord] = sort(diag(S), 'descend');
S = diag(S);
U = U(:,ord)
V = V(:,ord);
I have a scientific code that uses both sine and cosine of the same argument (I basically need the complex exponential of that argument). I was wondering if it were possible to do this faster than calling sine and cosine functions separately.
Also I only need about 0.1% precision. So is there any way I can find the default trig functions and truncate the power series for speed?
One other thing I have in mind is, is there any way to perform the remainder operation such that the result is always positive? In my own algorithm I used x=fmod(x,2*pi); but then I would need to add 2pi if x is negative (smaller domain means I can use a shorter power series)
EDIT: LUT turned out to be the best approach for this, however I am glad I learned about other approximation techniques. I will also advise using an explicit midpoint approximation. This is what I ended up doing:
const int N = 10000;//about 3e-4 error for 1000//3e-5 for 10 000//3e-6 for 100 000
double *cs = new double[N];
double *sn = new double[N];
for(int i =0;i<N;i++){
double A= (i+0.5)*2*pi/N;
cs[i]=cos(A);
sn[i]=sin(A);
}
The following part approximates (midpoint) sincos(2*pi*(wc2+t[j]*(cotp*t[j]-wc)))
double A=(wc2+t[j]*(cotp*t[j]-wc));
int B =(int)N*(A-floor(A));
re += cs[B]*f[j];
im += sn[B]*f[j];
Another approach could have been using the chebyshev decomposition. You can use the orthogonality property to find the coefficients. Optimized for exponential, it looks like this:
double fastsin(double x){
x=x-floor(x/2/pi)*2*pi-pi;//this line can be improved, both inside this
//function and before you input it into the function
double x2 = x*x;
return (((0.00015025063885163012*x2-
0.008034350857376128)*x2+ 0.1659789684145034)*x2-0.9995812174943602)*x;} //7th order chebyshev approx
If you seek fast evaluation with good (but not high) accuracy with powerseries you should use an expansion in Chebyshev polynomials: tabulate the coefficients (you'll need VERY few for 0.1% accuracy) and evaluate the expansion with the recursion relations for these polynomials (it's really very easy).
References:
Tabulated coefficients: http://www.ams.org/mcom/1980-34-149/S0025-5718-1980-0551302-5/S0025-5718-1980-0551302-5.pdf
Evaluation of chebyshev expansion: https://en.wikipedia.org/wiki/Chebyshev_polynomials
You'll need to (a) get the "reduced" argument in the range -pi/2..+pi/2 and consequently then (b) handle the sign in your results when the argument actually should have been in the "other" half of the full elementary interval -pi..+pi. These aspects should not pose a major problem:
determine (and "remember" as an integer 1 or -1) the sign in the original angle and proceed with the absolute value.
use a modulo function to reduce to the interval 0..2PI
Determine (and "remember" as an integer 1 or -1) whether it is in the "second" half and, if so, subtract pi*3/2, otherwise subtract pi/2. Note: this effectively interchanges sine and cosine (apart from signs); take this into account in the final evaluation.
This completes the step to get an angle in -pi/2..+pi/2
After evaluating sine and cosine with the Cheb-expansions, apply the "flags" of steps 1 and 3 above to get the right signs in the values.
Just create a lookup table. The following will let you lookup the sin and cos of any radian value between -2PI and 2PI.
// LOOK UP TABLE
var LUT_SIN_COS = [];
var N = 14400;
var HALF_N = N >> 1;
var STEP = 4 * Math.PI / N;
var INV_STEP = 1 / STEP;
// BUILD LUT
for(var i=0, r = -2*Math.PI; i < N; i++, r += STEP) {
LUT_SIN_COS[2*i] = Math.sin(r);
LUT_SIN_COS[2*i + 1] = Math.cos(r);
}
You index into the lookup table by:
var index = ((r * INV_STEP) + HALF_N) << 1;
var sin = LUT_SIN_COS[index];
var cos = LUT_SIN_COS[index + 1];
Here's a fiddle that displays the % error you can expect from different sized LUTS http://jsfiddle.net/77h6tvhj/
EDIT Here's an ideone (c++) with a ~benchmark~ vs the float sin and cos. http://ideone.com/SGrFVG For whatever a benchmark on ideone.com is worth the LUT is 5 times faster.
One way to go would be to learn how to implement the CORDIC algorithm. It is not difficult and pretty interesting intelectually. This gives you both the cosine and the sine. Wikipedia gives a MATLAB example that should be easy to adapt in C++.
Note that you can augment speed and reduce precision simply by lowering the parameter n.
About your second question, it has already been asked here (in C). It seems that there is no simple way.
You can also calculate sine using a square root, given the angle and the cosine.
The example below assumes the angle ranges from 0 to 2π:
double c = cos(angle);
double s = sqrt(1.0-c*c);
if(angle>pi)s=-s;
For single-precision floats, Microsoft uses 11-degree polynomial approximation for sine, 10-degree for cosine: XMScalarSinCos.
They also have faster version, XMScalarSinCosEst, that uses lower-degree polynomials.
If you aren’t on Windows, you’ll find same code + coefficients on geometrictools.com under Boost license.
Quick method to quickly compute Fibonacci, using Matrix property
Divide_Conquer_Fib(n) {
i = h = 1;
j = k = 0;
while (n > 0) {
if (n%2 == 1) { // if n is odd
t = j*h;
j = i*h + j*k + t;
i = i*k + t;
}
t = h*h;
h = 2*k*h + t;
k = k*k + t;
n = (int) n/2;
}
return j;
}
How do i understand this code? What would your strategy be? Would you put lots of print statements to see how states of variables change?
It is important to see how various developers' minds would go about understanding this code.
I would start off by running it against a few vales of n to check that it actually appears to give the correct answers. Then I'd read up on the mathematical theory to understand how it is likely to be working, and finally use that knowledge to take it to bits…
The Wikipedia entry section on the Matrix form explains the basis for this algorithm.
Well, the proper way to look at this code is to know what it does: Fibonacci numbers are coming up as an interesting exercise frequently, plus there is quite a bit of context saying what it does: it uses a matrix property together with divide and conquer. It turns out that you can compute the vector (Fibn, Fibn-1) as a product of some matrix and (Fibn-1, Fibn-2). Let's assume two rows in the code below are just two rows of the same matrix:
(Fib[n] ) (1 1) (Fib[n-1])
( ) = ( ) * ( )
(Fib[n-1]) (1 0) (Fib[n-2])
Now, matrix multiplication of quadratic matrices is associative, i.e., if the matrix above is M you can compute Fibn as Mn times (1, 0).
The next step is to compute Mn using divide and conquer. The basic trick here is that Mn can be decomposed according to the bits of n: Instead of computing the power by n multiplication you decompose the computation into computing squares and multiplying an extra term if the value is odd.
This is the basic underlying approach. The computation of the powers is done in the other direction, however, which works - I think - because the matrix is symmetric. I don't think you can derive the algorithm from the code easily if you are unaware of the basic approach.
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)?