Optimization to find complex number as input - c++

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.

Related

Performance bottlenecks in fast evaluation of trig functions using Eigen and MEX

In a project using Matlab's C++ MEX API, I have to compute the value exp(j * 2pi * x) for over 100,000 values of x where x is always a positive double. I've written some helper functions that breakdown the computation into sin/cos using euler's formula. I then apply the method of range reduction to reduce my values to their corresponding points in the domain [0,T/4] where T is the period of the exponential I'm computing. I keep track of which quadrant in [0, T] the original value would have fallen into for later. I can then compute the trig function using a taylor series polynomial in horner form and apply the appropriate shift depending on which quadrant the original value was in. For further information on some of the concepts in this technique, check out this answer. Here is the code for this function:
Eigen::VectorXcd calcRot2(const Eigen::Ref<const Eigen::VectorXd>& idxt) {
Eigen::VectorXd vidxt = idxt.array() - idxt.array().floor();
Eigen::VectorXd quadrant = (vidxt.array()*2+0.5).floor();
vidxt.array() -= (quadrant.array()*0.5);
vidxt.array() *= 2*3.14159265358979;
const Eigen::VectorXd sq = vidxt.array()*vidxt.array();
Eigen::VectorXcd M(vidxt.size());
M.real() = fastCos2(sq);
M.imag() = fastSin2(vidxt,sq);
M = (quadrant.array() == 1).select(-M,M);
return M;
}
I profiled the code segment in which this function is called using std::chrono and averaged over 500 calls to the function (where each call to the mex function processes all 100,000+ values by calling calcRot2 in a loop. Each iteration passes about 200 values to calcRot2). I find the following average runtimes:
runtime with calcRot2: 75.4694 ms
runtime with fastSin/Cos commented out: 50.2409 ms
runtime with calcRot2 commented out: 30.2547 ms
Looking at the difference between the two extreme cases, it seems like calcRot has a large contribution to the runtime. However, only a portion of that comes from the sin/cos calculation. I would assume Eigen's implicit vectorization and the compiler would make the runtime of the other operations in the function effectively negligible. (floor shouldn't be a problem!) Where exactly is the performance bottleneck here?
This is the compilation command I'm performing (It uses MinGW64 which I think is the same as gcc):
mex(ipath,'CFLAGS="$CFLAGS -O3 -fno-math-errno -ffast-math -fopenmp -mavx2"','LDFLAGS="$LDFLAGS -fopenmp"','DAS.cpp','DAShelper.cpp')
Reference Code
For reference, here is the code segment in the main mex function where the timer is called, and the helper function that calls calcRot2():
MEX function call:
chk1 = std::chrono::steady_clock::now();
// Calculate beamformed signal at each point
Eigen::MatrixXcd bfVec(p.nPoints,1);
#pragma omp parallel for
for (int i = 0; i < p.nPoints; i++) {
calcPoint(idxt.col(i),SIG,p,bfVec(i));
}
chk2 = std::chrono::steady_clock::now();
auto diff3 = chk2 - chk1;
calcPoint:
void calcPoint(const Eigen::Ref<const Eigen::VectorXd>& idxt,
const Eigen::Ref<const Eigen::MatrixXcd>& SIG,
Parameters& p, std::complex<double>& bfVal) {
Eigen::VectorXcd pRot = calcRot2(idxt*p.fc/p.fs);
int j = 0;
for (auto x : idxt) {
if(x >= 0) {
int vIDX = static_cast<int>(x);
bfVal += (SIG(vIDX,j)*(vIDX + 1 - x) + SIG(vIDX+1,j)*(x - vIDX))*pRot(j);
}
j++;
}
}
Clarification
To clarify, the line
(vidxt.array()*2+0.5).floor()
is meant to yield:
0 if vidxt is between [0,0.25]
1 if vidxt is between [0.25,0.75]
2 if vidxt is between [0.75,1]
The idea here is that when vidxt is in the second interval (quadrants 2 and 3 on the unit circle for functions with period 2pi), then the value needs to map to its negative value. Otherwise, the range reduction maps the values to the correct values.
The benefits of Eigen's vectorization are outweighed because you evaluate your expressions into temporary vectors. Allocating, deallocating, filling and reading these vectors has cost that seems significant. This is especially so because the expressions themselves are relatively simple (just a few scalar operations).
Expression objects
What usually helps here is aggregating into fewer expressions. For example line 3 and 4 can be collapsed into one:
vidxt.array() = 2*3.14159265358979 * (vidxt.array() - quadrant.array()*0.5);
(BTW: Note that that math.h contains a constant M_PI with pi in double precision).
Beyond that, Eigen expressions can be combined and reused. Something like this:
auto vidxt0 = idxt.array() - idxt.array().floor();
auto quadrant = (vidxt0*2+0.5).floor();
auto vidxt = 2*3.14159265358979 * (vidxt0 - quadrant.array()*0.5);
auto sq = vidxt.array().square();
Eigen::VectorXcd M(vidxt.size());
M.real() = fastCos2(sq);
M.imag() = fastSin2(vidxt,sq);
M = (quadrant.array() == 1).select(-M,M);
Note that none of the auto values are vectors. They are expression objects that behave like arrays and can be evaluated into vectors or arrays.
You can pass these on to your fastCos2 and fastSin2 function by declaring them as templates. The typical Eigen pattern would be something like
template<Derived>
void fastCos2(const Eigen::ArrayBase<Derived>& sq);
The idea here is that ultimately, everything compiles into one huge loop that gets executed when you evaluate the expression into a vector or array. If you reference the same sub-expression multiple times, the compiler may be able to eliminate the redundant computations.
Unfortunately, I could not get any better performance out of this particular code, so it is no real help here but it is still something worth exploring in these kind of cases.
fastSin/Cos return value
Speaking of temporary vectors: You didn't include the code for your fastSin/Cos functions but it looks a lot like you return a temporary vector which is then copied into the real and imaginary parts or the actual return value. This is another temporary that you may want to avoid. Something like this:
template<class Derived1, class Derived2>
void fastCos2(const Eigen::MatrixBase<Derived1>& M, const Eigen::MatrixBase<Derived2>& sq)
{
Eigen::MatrixBase<Derived1>& M_mut = const_cast<Eigen::MatrixBase<Derived1>&>(M);
M_mut = sq...;
}
fastCos2(M.real(), sq);
Please refer to Eigen's documentation on the topic of function arguments.
The downside of this approach in this particular case is that now the output is not consecutive (real and imaginary parts are interleaved). This may affect vectorization negatively. You may be able to work around this by combining the sin and cos functions into one expression for both. Benchmarking is required.
Using a plain loop
As others have pointed out, using a loop may be easier in this particular case. You noted that this was slower. I have a theory why: You did not specify -DNDEBUG in your compile options. If you don't, all array indices in Eigen vectors are range-checked with an assertion. These cost time and prevent vectorization. If you include this compile flag, I find my code significantly faster than using Eigen expressions.
Alternatively, you can use raw C pointers to the input and output vector. Something like this:
std::ptrdiff_t n = idxt.size();
Eigen::VectorXcd M(n);
const double* iidxt = idxt.data();
std::complex<double>* iM = M.data();
for(std::ptrdiff_t j = 0; j < n; ++j) {
double ival = iidxt[j];
double vidxt = ival - std::floor(ival);
double quadrant = std::floor(vidxt * 2. + 0.5);
vidxt = (vidxt - quadrant * 0.5) * (2. * 3.14159265358979);
double sq = vidxt * vidxt;
// stand-in for sincos
std::complex<double> jval(sq, vidxt + sq);
iM[j] = quadrant == 1. ? -jval : jval;
}
Fixed sized arrays
To avoid the cost of memory allocation and make it easier for the compiler to avoid memory operations in the first place, it can help to run the computation on blocks of fixed size. Something like this:
std::ptrdiff_t n = idxt.size();
Eigen::VectorXcd M(n);
std::ptrdiff_t i;
for(i = 0; i + 4 <= n; i += 4) {
Eigen::Array4d idxt_i = idxt.segment<4>(i);
...
M.segment<4>(i) = ...;
}
if(i + 2 <= n) {
Eigen::Array2D idxt_i = idxt.segment<2>(i);
...
M.segment<2>(i) = ...;
i += 2;
}
if(i < n) {
// last index scalar
}
This kind of stuff needs careful tuning to ensure that vectorized code is generated and there are no unnecessary temporary values on the stack. If you can read assembler, Godbolt is very helpful.
Other remarks
Eigen includes vectorized versions of sin and cos. Have you compared your code to these instead of e.g. Eigen's complex exp function?
Depending on your math library, there is also an explicit sincos function to compute sine and cosine in one function. It is not vectorized but still saves time on range reduction. You can (usually) access it through std::polar. Try this:
Eigen::VectorXd scale = ...;
Eigen::VectorXd phase = ...;
// M = scale * exp(-2 pi j phase)
Eigen::VectorXd M = scale.binaryExpr(-2. * M_PI * phase,
[](double s, double p) noexcept -> std::complex<double> {
return std::polar(s, p);
});
If your goal is an approximation instead of a precise result, shouldn't your first step be to cast to single precision? Maybe after the range reduction to avoid losing too many decimal places. At the very least it will double the work done per clock cycle. Also, regular sine and cosine implementations take less time in float.
Edit
I had to correct myself on the cast to int64 instead of int. There is no vectorized conversion to int64_t until AVX512
The line (vidxt.array()*2+0.5).floor() bugs me slightly. This is meant to round down to negative infinity for [0, 0.5) and up to positive infinity for [0.5, 1), correct? vidxt is never negative. Therefore this line should be equivalent to (vidxt.array()*2).round(). With AVX2 and -ffast-math that saves one instruction. With SSE2 none of these actually vectorize, as can be seen on Godbolt

Problems when substituting a matrix in a polynomial

Example: let
M = Matrix([[1,2],[3,4]]) # and
p = Poly(x**3 + x + 1) # then
p.subs(x,M).expand()
gives the error :
TypeError: cannot add <class'sympy.matrices.immutable.ImmutableDenseMatrix'> and <class 'sympy.core.numbers.One'>
which is very plausible since the two first terms become matrices but the last term (the constant term) is not a matrix but a scalar. To remediate to this situation I changed the polynomial to
p = Poly(x**3 + x + x**0) # then
the same error persists. Am I obliged to type the expression by hand, replacing x by M? In this example the polynomial has only three terms but in reality I encounter (multivariate polynomials with) dozens of terms.
So I think the question is mainly revolving around the concept of Matrix polynomial:
(where P is a polynomial, and A is a matrix)
I think this is saying that the free term is a number, and it cannot be added with the rest which is a matrix, effectively the addition operation is undefined between those two types.
TypeError: cannot add <class'sympy.matrices.immutable.ImmutableDenseMatrix'> and <class 'sympy.core.numbers.One'>
However, this can be circumvented by defining a function that evaluates the matrix polynomial for a specific matrix. The difference here is that we're using matrix exponentiation, so we correctly compute the free term of the matrix polynomial a_0 * I where I=A^0 is the identity matrix of the required shape:
from sympy import *
x = symbols('x')
M = Matrix([[1,2],[3,4]])
p = Poly(x**3 + x + 1)
def eval_poly_matrix(P,A):
res = zeros(*A.shape)
for t in enumerate(P.all_coeffs()[::-1]):
i, a_i = t
res += a_i * (A**i)
return res
eval_poly_matrix(p,M)
Output:
In this example the polynomial has only three terms but in reality I encounter (multivariate polynomials with) dozens of terms.
The function eval_poly_matrix above can be extended to work for multivariate polynomials by using the .monoms() method to extract monomials with nonzero coefficients, like so:
from sympy import *
x,y = symbols('x y')
M = Matrix([[1,2],[3,4]])
p = poly( x**3 * y + x * y**2 + y )
def eval_poly_matrix(P,*M):
res = zeros(*M[0].shape)
for m in P.monoms():
term = eye(*M[0].shape)
for j in enumerate(m):
i,e = j
term *= M[i]**e
res += term
return res
eval_poly_matrix(p,M,eye(M.rows))
Note: Some sanity checks, edge cases handling and optimizations are possible:
The number of variables present in the polynomial relates to the number of matrices passed as parameters (the former should never be greater than the latter, and if it's lower than some logic needs to be present to handle that, I've only handled the case when the two are equal)
All matrices need to be square as per the definition of the matrix polynomial
A discussion about a multivariate version of the Horner's rule features in the comments of this question. This might be useful to minimize the number of matrix multiplications.
Handle the fact that in a Matrix polynomial x*y is different from y*x because matrix multiplication is non-commutative . Apparently poly functions in sympy do not support non-commutative variables, but you can define symbols with commutative=False and there seems to be a way forward there
About the 4th point above, there is support for Matrix expressions in SymPy, and that can help here:
from sympy import *
from sympy.matrices import MatrixSymbol
A = Matrix([[1,2],[3,4]])
B = Matrix([[2,3],[3,4]])
X = MatrixSymbol('X',2,2)
Y = MatrixSymbol('Y',2,2)
I = eye(X.rows)
p = X**2 * Y + Y * X ** 2 + X ** 3 - I
display(p)
p = p.subs({X: A, Y: B}).doit()
display(p)
Output:
For more developments on this feature follow #18555

Numerically calculate combinations of factorials and polynomials

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

Using the gaussian probability density function in C++

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.

lagrange approximation -c++

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)?