问题
Given two points in the x, y plane:
x, f(x)
1, 3
2, 5
I can interpolate them using Lagrange and find f(1.5), which result in 4. Thinking a little I managed to find a way to discover the coefficients of the equation:
void l1Coefficients(const vector<double> &x, const vector<double> &y) {
double a0 = y[0]/(x[0]-x[1]);
double a1 = y[1]/(x[1]-x[0]);
double b0 = (-x[1]*y[0])/(x[0]-x[1]);
double b1 = (-x[0]*y[1])/(x[1]-x[0]);
double a = a0 + a1;
double b = b0 + b1;
cout << "P1(x) = " << a << "x +" << b << endl;
}
That gives me P1(x) = 2x +1.
Thinking a little more I was able to extend that to 2nd order equations. So, given the points:
1, 1
2, 4
3, 9
I found the equation P2(x) = 1x^2 +0x +0 with the following:
void l2Coefficients(const vector<double> &x, const vector<double> &y) {
double a0 = y[0] / ((x[0]-x[1])*(x[0]-x[2]));
double a1 = y[1] / ((x[1]-x[0])*(x[1]-x[2]));
double a2 = y[2] / ((x[2]-x[0])*(x[2]-x[1]));
double b0 = -(x[1]+x[2])*y[0] / ((x[0]-x[1])*(x[0]-x[2]));
double b1 = -(x[0]+x[2])*y[1] / ((x[1]-x[0])*(x[1]-x[2]));
double b2 = -(x[0]+x[1])*y[2] / ((x[2]-x[0])*(x[2]-x[1]));
double c0 = (x[1]*x[2])*y[0] / ((x[0]-x[1])*(x[0]-x[2]));
double c1 = (x[0]*x[2])*y[1] / ((x[1]-x[0])*(x[1]-x[2]));
double c2 = (x[0]*x[1])*y[2] / ((x[2]-x[0])*(x[2]-x[1]));
double a = a0 + a1 + a2;
double b = b0 + b1 + b2;
double c = c0 + c1 + c2;
cout << "P2(x) = " << a << "x^2 +" << b << "x +" << c << endl;
}
Working hard I actually was able to find the coefficients for equations of order up to 4th.
How to find the coefficients of order n equations? Where
Pn(x) = c_2x^2 + c_1x^1 + c_0x^0 + ...
回答1:
It's a simple linear algebra problem.
We have a set of N samples of the form xk -> f(xk) and we know the general form of function f(x), which is:
f(x) = c0x0 + c1x1 + ... + cN-1xN-1
We want to find the coefficients c0 ... cN-1. To achieve that, we build a system of N equations of the form:
c0xk0 + c1xk1 + ... + cN-1xkN-1 = f(xk)
where k is the sample number. Since xk and f(xk) are constants rather than variables, we have a linear system of equations.
Expressed in terms of linear algebra, we have to solve:
Ac = b
where A is a Vandermonde matrix of powers of x and b is a vector of f(xk) values.
To solve such a system, you need a linear algebra library, such as Eigen. See here for example code.
The only thing that can go wrong with such an approach is the system of linear equations being under-determined, which will happen if your N samples can be fit with with a polynomial of degree less than N-1. In such a case you can still solve this system with Moore-Penrose pseudo inverse like this:
c = pinv(A)*b
Unfortunately, Eigen doesn't have a pinv() implementation, though it's pretty easy to code it by yourself in terms of Singular Value Decomposition (SVD).
回答2:
I created a naive implementation of the matrix solution:
#include <iostream>
#include <vector>
#include <stdexcept>
class Matrix
{
private:
class RowIterator
{
public:
RowIterator(Matrix* mat, int rowNum) :_mat(mat), _rowNum(rowNum) {}
double& operator[] (int colNum) { return _mat->_data[_rowNum*_mat->_sizeX + colNum]; }
private:
Matrix* _mat;
int _rowNum;
};
int _sizeY, _sizeX;
std::vector<double> _data;
public:
Matrix(int sizeY, int sizeX) : _sizeY(sizeY), _sizeX(sizeX), _data(_sizeY*_sizeX){}
Matrix(std::vector<std::vector<double> > initList) : _sizeY(initList.size()), _sizeX(_sizeY>0 ? initList.begin()->size() : 0), _data()
{
_data.reserve(_sizeY*_sizeX);
for (const std::vector<double>& list : initList)
{
_data.insert(_data.end(), list.begin(), list.end());
}
}
RowIterator operator[] (int rowNum) { return RowIterator(this, rowNum); }
int getSize() { return _sizeX*_sizeY; }
int getSizeX() { return _sizeX; }
int getSizeY() { return _sizeY; }
Matrix reduce(int rowNum, int colNum)
{
Matrix mat(_sizeY-1, _sizeX-1);
int rowRem = 0;
for (int y = 0; y < _sizeY; y++)
{
if (rowNum == y)
{
rowRem = 1;
continue;
}
int colRem = 0;
for (int x = 0; x < _sizeX; x++)
{
if (colNum == x)
{
colRem = 1;
continue;
}
mat[y - rowRem][x - colRem] = (*this)[y][x];
}
}
return mat;
}
Matrix replaceCol(int colNum, std::vector<double> newCol)
{
Matrix mat = *this;
for (int y = 0; y < _sizeY; y++)
{
mat[y][colNum] = newCol[y];
}
return mat;
}
};
double solveMatrix(Matrix mat)
{
if (mat.getSizeX() != mat.getSizeY()) throw std::invalid_argument("Not square matrix");
if (mat.getSize() > 1)
{
double sum = 0.0;
int sign = 1;
for (int x = 0; x < mat.getSizeX(); x++)
{
sum += sign * mat[0][x] * solveMatrix(mat.reduce(0, x));
sign = -sign;
}
return sum;
}
return mat[0][0];
}
std::vector<double> solveEq(std::vector< std::pair<double, double> > points)
{
std::vector<std::vector<double> > xes(points.size());
for (int i = 0; i<points.size(); i++)
{
xes[i].push_back(1);
for (int j = 1; j<points.size(); j++)
{
xes[i].push_back(xes[i].back() * points[i].first);
}
}
Matrix mat(xes);
std::vector<double> ys(points.size());
for (int i = 0; i < points.size(); i++)
{
ys[i] = points[i].second;
}
double w = solveMatrix(mat);
std::vector<double> result(points.size(), 0.0);
if(w!=0)
for (int i = 0; i < ys.size(); i++)
{
result[i] = solveMatrix(mat.replaceCol(i, ys));
result[i] /= w;
}
return result;
}
void printCoe(std::vector<double> coe)
{
std::cout << "f(x)=";
bool notFirstSign = false;
for (int i = coe.size() - 1; i >= 0; i--)
{
if (coe[i] != 0.0)
{
if (coe[i] >= 0.0 && notFirstSign)
std::cout << "+";
notFirstSign = true;
if (coe[i] != 1.0)
if (coe[i] == -1.0)
std::cout << "-";
else
std::cout << coe[i];
if (i == 1)
std::cout << "x";
if (i>1)
std::cout << "x^" << i;
}
}
std::cout << std::endl;
}
int main()
{
std::vector< std::pair<double, double> > points1 = { {3,31}, {6,94}, {4,48}, {0,4} };
std::vector<double> coe = solveEq(points1);
printCoe(coe);
std::vector< std::pair<double, double> > points2 = { { 0,0 },{ 1,-1 },{ 2,-16 },{ 3,-81 },{ 4,-256 } };
printCoe(solveEq(points2));
printCoe(solveEq({ { 0,0 },{ 1,1 },{ 2,8 },{ 3,27 } }));
std::cin.ignore();
return 0;
}
来源：https://stackoverflow.com/questions/43566648/how-to-find-coefficients-of-polynomial-equation