Related
I spent quiet some time looking on the internet to find a solution to this, maybe it's out there but nothing of what I saw helped me.
I have a function !
double integrand(double r, double phi, double theta)
That I want to integrate with some given definite bounds over the three dimensions. I found multiple lines of code on the internet that implement single variable definite integrals numerical schemes. I was thinking to myself "well, I'll just integrate along one dimension after the other".
Algorithmically speaking what I wanted to do was :
double firstIntegral(double r, double phi) {
double result = integrationFunction(integrand,lower_bound,upper_bound);
return result;
}
And simply do it again two more times. This works easily in languages like Matlab where I can create functions handler anywhere but I don't know how to do it in C++. I would have to first define a function that some r and phi will calculate integrand(r, phi, theta) for any theta and make it in C++ a function of one variable only but I don't know how to do that.
How can I compute the triple integral of my three-variables function in C++ using a one -dimensional integration routine (or anything else really...) ?
This is a very slow and inexact version for integrals over cartesian coordinates, which should work with C++11.
It is using std::function and lambdas to implement the numerical integration. No steps have been taken to optimize this.
A template based solution could be much faster (by several orders of magnitude) than this, because it may allow the compiler to inline and simplify some of the code.
#include<functional>
#include<iostream>
static double integrand(double /*x*/, double y, double /*z*/)
{
return y;
}
double integrate_1d(std::function<double(double)> const &func, double lower, double upper)
{
static const double increment = 0.001;
double integral = 0.0;
for(double x = lower; x < upper; x+=increment) {
integral += func(x) * increment;
}
return integral;
}
double integrate_2d(std::function<double(double, double)> const &func, double lower1, double upper1, double lower2, double upper2)
{
static const double increment = 0.001;
double integral = 0.0;
for(double x = lower2; x < upper2; x+=increment) {
auto func_x = [=](double y){ return func(x, y);};
integral += integrate_1d(func_x, lower1, upper1) * increment;
}
return integral;
}
double integrate_3d(std::function<double(double, double, double)> const &func,
double lower1, double upper1,
double lower2, double upper2,
double lower3, double upper3)
{
static const double increment = 0.001;
double integral = 0.0;
for(double x = lower3; x < upper3; x+=increment) {
auto func_x = [=](double y, double z){ return func(x, y, z);};
integral += integrate_2d(func_x, lower1, upper1, lower2, upper2) * increment;
}
return integral;
}
int main()
{
double integral = integrate_3d(integrand, 0.0, 1.0, 0.0, 1.0, 0.0, 1.0);
std::cout << "Triple integral: " << integral << std::endl;
return 0;
}
You can use functors
#include <iostream>
struct MyFunctorMultiply
{
double m_coeff;
MyFunctorMultiply(double coeff)
{
m_coeff = coeff;
}
double operator()(double value)
{
return m_coeff * value;
}
};
struct MyFunctorAdd
{
double m_a;
MyFunctorAdd(double a)
{
m_a = a;
}
double operator()(double value)
{
return m_a + value;
}
};
template<class t_functor>
double calculate(t_functor functor, double value, double other_param)
{
return functor(value) - other_param;
}
int main()
{
MyFunctorMultiply multiply2(2.);
MyFunctorAdd add3(3.);
double result_a = calculate(multiply2, 4, 1); // should obtain 4 * 2 - 1 = 7
double result_b = calculate(add3, 5, 6); // should obtain 5 + 3 - 6 = 2
std::cout << result_a << std::endl;
std::cout << result_b << std::endl;
}
If your concern is just about getting the right prototype to pass to the integration function, you can very well use alternative data passing mechanisms, the simpler of which is using global variables.
Assuming that the order of integration is on theta, then phi, then r, write three functions of a single argument:
It(theta) computes the integrand from the argument theta passed explicitly and the global phi and r.
Ip(phi) computes the bounds on theta from the argument phi passed explicitly and the global r; it also copies the phi argument to the global variable and invokes integrationFunction(It, lower_t, upper_t).
Ir(r) computes the bounds on phi from the argument r passed explicitly; it also copies the r argument to the global variable and invokes integrationFunction(Ip, lower_p, upper_p).
Now you are ready to call integrationFunction(Ir, lower_r, upper_r).
It may also be that integrationFunction supports a "context" argument where you can store what you want.
I've been asked to solve this differential equation:
(x,y,vx,vy)'=(vx,vy,vy,-vx)
which should return a circular motion with a 2*pi period.
I implemented the function:
class FunzioneBase
{
public:
virtual VettoreLineare Eval(double t, const VettoreLineare& v) const = 0;
};
class Circonferenza: public FunzioneBase
{
private:
double _alpha;
public:
Circonferenza(double alpha) { _alpha = alpha; };
void SetAlpha(double alpha) { _alpha = alpha; };
virtual VettoreLineare Eval(double t, const VettoreLineare& v) const;
};
VettoreLineare Circonferenza::Eval(double t, const VettoreLineare& v) const
{
VettoreLineare y(4);
if (v.GetN() != 4)
{
std::cout << "errore" << std::endl;
return 0;
};
y.SetComponent(0, v.GetComponent(2));
y.SetComponent(1, v.GetComponent(3));
y.SetComponent(2, pow(pow(v.GetComponent(0), 2.) + pow(v.GetComponent(1), 2.), _alpha) * v.GetComponent(3));
y.SetComponent(3, - pow(pow(v.GetComponent(0), 2.) + pow(v.GetComponent(1), 2.), _alpha)) * v.GetComponent(2));
return y;
};
where _alpha equals to 0.
Now, this works just fine with Euler's method: if I integrate this ODE for 2 * pi * 10, given the initial condition (1, 0, 0, -1), with a 0.003 precision, I get that the position is comparable to (1, 0) within a range of 1 ± 0.1, as we should expect. But if I integrate the same ODE with Runge Kutta's Method (with a 0.003 precision, for 2 * pi * 10 seconds) implemented as follows:
class EqDifferenzialeBase
{
public:
virtual VettoreLineare Passo (double t, VettoreLineare& x, double h, FunzioneBase* f) const = 0;
};
class Runge_Kutta: public EqDifferenzialeBase
{
public:
virtual VettoreLineare Passo(double t, VettoreLineare& v, double h, FunzioneBase* f) const;
};
VettoreLineare Runge_Kutta::Passo(double t, VettoreLineare& v, double h, FunzioneBase* _f) const
{
VettoreLineare k1 = _f->Eval(t, v);
VettoreLineare k2 = _f->Eval(t + h / 2., v + k1 *(h / 2.));
VettoreLineare k3 = _f->Eval(t + h / 2., v + k2 * (h / 2.));
VettoreLineare k4 = _f->Eval(t + h, v + k3 * h);
VettoreLineare y = v + (k1 + k2 * 2. + k3 * 2. + k4) * (h / 6.);
return y;
}
the program returns an x position which equals to 0.39 aproximately, when the precision should theorically be, for a 4th order Runge Kutta's method, around 1E-6. I checked and found that the period, with Runge_Kutta's, seems to almost quadruplicate (since in a 2 * pi lapse, x gets from 1 to 0.48), but I don't understand why. This is the content of my main:
VettoreLineare DatiIniziali (4);
Circonferenza* myCirc = new Circonferenza(0);
DatiIniziali.SetComponent(0, 1.);
DatiIniziali.SetComponent(1, 0.);
DatiIniziali.SetComponent(2, 0.);
DatiIniziali.SetComponent(3, -1.);
double passo = 0.003;
Runge_Kutta myKutta;
for(int i = 0; i <= 2. * M_PI / passo; i++)
{
DatiIniziali = myKutta.Passo(0, DatiIniziali, passo, myCirc);
cout << DatiIniziali.GetComponent(0) << endl;
};
cout << 1 - DatiIniziali.GetComponent(0) << endl;
Thank you in advance.
Update: One error identified: Always compile with the -Wall option to catch all warnings and automatic code corrections of the compiler. Then you would have found
fff: In member function ‘virtual VettoreLineare Circonferenza::Eval(double, const VettoreLineare&) const’:
fff:xxx:114: error: invalid operands of types ‘void’ and ‘double’ to binary ‘operator*’
y.SetComponent(3, - pow(pow(v.GetComponent(0), 2.) + pow(v.GetComponent(1), 2.), _alpha)) * v.GetComponent(2));
^
where you are closing too early after _alpha so that the void of SetComponent gets to be a factor.
Update II: second error identified: In another post of yours the code of the linear vector class is given. There, in contrast to the addition (operator+), the scalar-vector product (operator*(double)) is modifying the calling instance. Thus in computing k2 the components of k1 get multiplied with h/2. But then this modified k1 (and also modified k2 and k3) are used in assembling the result y resulting in some almost completely useless state update.
Original rapid prototyping: I can tell you that a stripped down bare-bones implementation in python works flawlessly
import numpy as np
def odeint(f,t0,y0,tf,h):
'''Classical RK4 with fixed step size, modify h to fit
the full interval'''
N = np.ceil( (tf-t0)/h )
h = (tf-t0)/N
t = t0
y = np.array(y0)
for k in range(int(N)):
k1 = h*np.array(f(t ,y ))
k2 = h*np.array(f(t+0.5*h,y+0.5*k1))
k3 = h*np.array(f(t+0.5*h,y+0.5*k2))
k4 = h*np.array(f(t+ h,y+ k3))
y = y + (k1+2*(k2+k3)+k4)/6
t = t + h
return t, y
def odefunc(t,y):
x,y,vx,vy = y
return [ vx,vx,vy,-vx ]
pi = 4*np.arctan(1);
print odeint(odefunc, 0, [1,0,0,-1], 2*pi, 0.003)
ends with
(t, [ x,y,vx,vy]) = (6.2831853071794184,
[ 1.00000000e+00, -6.76088739e-15, 4.23436503e-12,
-1.00000000e+00])
as expected. You will need a debugger or intermediate output to find where the computation goes wrong.
I'm writing a sine function that has to be recursive. I have written a sine function but am not really sure how to do it recursively. Could someone explain how to get started on this?
This is what I have so far:
/*--------------------------------------------------------------
Name: sine( double X );
Return: Function "sine" will return the
sine of X, where X is measured in radians.
--------------------------------------------------------------*/
double sine(double X)
{
double result = 0;
double term;
int k;
double lim;
k = 0;
lim = power(10, -8);
term = power(-1, k)*power(X, ((2*k) + 1)) / (factorial((2*k)+1));
result = term;
while (absolute(term) > lim)
{
k += 1;
term = power(-1, k)*power(X, ((2*k) + 1)) / (factorial((2*k)+1));
result += term;
}
return result;
}
EDIT: I used a wrapper function to solve this. Basically created another function called
double sine_rec(double X, double k)
and changed around the current code to fit in with that.
The way I would approach this would be to have another function sine(double X, int n) which takes another integer parameter - the number of terms to include in the power series approximation. Then this function could return something like [nth term in series] + sine(X, n - 1) (just remember a prior if statement to deal with n = 1).
You can eliminate the while loop by recursion in following way:
double sine(double X, int k = 0)
{
double result = 0;
double term;
double lim;
lim = power(10, -8);
term = power(-1, k)*power(X, ((2*k) + 1)) / (factorial((2*k)+1));
if (absolute(term) > lim)
{
return sine(X, k+1) + term;
}
else
{
return term;
}
}
But I cannot recommend doing this at all. (There are better solutions even to this recursion, but find them on your own)
Is there a way to calculate mean and standard deviation for a vector containing samples using Boost?
Or do I have to create an accumulator and feed the vector into it?
I don't know if Boost has more specific functions, but you can do it with the standard library.
Given std::vector<double> v, this is the naive way:
#include <numeric>
double sum = std::accumulate(v.begin(), v.end(), 0.0);
double mean = sum / v.size();
double sq_sum = std::inner_product(v.begin(), v.end(), v.begin(), 0.0);
double stdev = std::sqrt(sq_sum / v.size() - mean * mean);
This is susceptible to overflow or underflow for huge or tiny values. A slightly better way to calculate the standard deviation is:
double sum = std::accumulate(v.begin(), v.end(), 0.0);
double mean = sum / v.size();
std::vector<double> diff(v.size());
std::transform(v.begin(), v.end(), diff.begin(),
std::bind2nd(std::minus<double>(), mean));
double sq_sum = std::inner_product(diff.begin(), diff.end(), diff.begin(), 0.0);
double stdev = std::sqrt(sq_sum / v.size());
UPDATE for C++11:
The call to std::transform can be written using a lambda function instead of std::minus and std::bind2nd(now deprecated):
std::transform(v.begin(), v.end(), diff.begin(), [mean](double x) { return x - mean; });
If performance is important to you, and your compiler supports lambdas, the stdev calculation can be made faster and simpler: In tests with VS 2012 I've found that the following code is over 10 X quicker than the Boost code given in the chosen answer; it's also 5 X quicker than the safer version of the answer using standard libraries given by musiphil.
Note I'm using sample standard deviation, so the below code gives slightly different results (Why there is a Minus One in Standard Deviations)
double sum = std::accumulate(std::begin(v), std::end(v), 0.0);
double m = sum / v.size();
double accum = 0.0;
std::for_each (std::begin(v), std::end(v), [&](const double d) {
accum += (d - m) * (d - m);
});
double stdev = sqrt(accum / (v.size()-1));
Using accumulators is the way to compute means and standard deviations in Boost.
accumulator_set<double, stats<tag::variance> > acc;
for_each(a_vec.begin(), a_vec.end(), bind<void>(ref(acc), _1));
cout << mean(acc) << endl;
cout << sqrt(variance(acc)) << endl;
Improving on the answer by musiphil, you can write a standard deviation function without the temporary vector diff, just using a single inner_product call with the C++11 lambda capabilities:
double stddev(std::vector<double> const & func)
{
double mean = std::accumulate(func.begin(), func.end(), 0.0) / func.size();
double sq_sum = std::inner_product(func.begin(), func.end(), func.begin(), 0.0,
[](double const & x, double const & y) { return x + y; },
[mean](double const & x, double const & y) { return (x - mean)*(y - mean); });
return std::sqrt(sq_sum / func.size());
}
I suspect doing the subtraction multiple times is cheaper than using up additional intermediate storage, and I think it is more readable, but I haven't tested the performance yet.
It seems the following elegant recursive solution has not been mentioned, although it has been around for a long time. Referring to Knuth's Art of Computer Programming,
mean_1 = x_1, variance_1 = 0; //initial conditions; edge case;
//for k >= 2,
mean_k = mean_k-1 + (x_k - mean_k-1) / k;
variance_k = variance_k-1 + (x_k - mean_k-1) * (x_k - mean_k);
then for a list of n>=2 values, the estimate of the standard deviation is:
stddev = std::sqrt(variance_n / (n-1)).
Hope this helps!
My answer is similar as Josh Greifer but generalised to sample covariance. Sample variance is just sample covariance but with the two inputs identical. This includes Bessel's correlation.
template <class Iter> typename Iter::value_type cov(const Iter &x, const Iter &y)
{
double sum_x = std::accumulate(std::begin(x), std::end(x), 0.0);
double sum_y = std::accumulate(std::begin(y), std::end(y), 0.0);
double mx = sum_x / x.size();
double my = sum_y / y.size();
double accum = 0.0;
for (auto i = 0; i < x.size(); i++)
{
accum += (x.at(i) - mx) * (y.at(i) - my);
}
return accum / (x.size() - 1);
}
2x faster than the versions before mentioned - mostly because transform() and inner_product() loops are joined.
Sorry about my shortcut/typedefs/macro: Flo = float. CR const ref. VFlo - vector. Tested in VS2010
#define fe(EL, CONTAINER) for each (auto EL in CONTAINER) //VS2010
Flo stdDev(VFlo CR crVec) {
SZ n = crVec.size(); if (n < 2) return 0.0f;
Flo fSqSum = 0.0f, fSum = 0.0f;
fe(f, crVec) fSqSum += f * f; // EDIT: was Cit(VFlo, crVec) {
fe(f, crVec) fSum += f;
Flo fSumSq = fSum * fSum;
Flo fSumSqDivN = fSumSq / n;
Flo fSubSqSum = fSqSum - fSumSqDivN;
Flo fPreSqrt = fSubSqSum / (n - 1);
return sqrt(fPreSqrt);
}
In order to calculate the sample mean with a better presicion the following r-step recursion can be used:
mean_k=1/k*[(k-r)*mean_(k-r) + sum_over_i_from_(n-r+1)_to_n(x_i)],
where r is chosen to make summation components closer to each other.
Create your own container:
template <class T>
class statList : public std::list<T>
{
public:
statList() : std::list<T>::list() {}
~statList() {}
T mean() {
return accumulate(begin(),end(),0.0)/size();
}
T stddev() {
T diff_sum = 0;
T m = mean();
for(iterator it= begin(); it != end(); ++it)
diff_sum += ((*it - m)*(*it -m));
return diff_sum/size();
}
};
It does have some limitations, but it works beautifully when you know what you are doing.
//means deviation in c++
/A deviation that is a difference between an observed value and the true value of a quantity of interest (such as a population mean) is an error and a deviation that is the difference between the observed value and an estimate of the true value (such an estimate may be a sample mean) is a residual. These concepts are applicable for data at the interval and ratio levels of measurement./
#include <iostream>
#include <conio.h>
using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
int main(int argc, char** argv)
{
int i,cnt;
cout<<"please inter count:\t";
cin>>cnt;
float *num=new float [cnt];
float *s=new float [cnt];
float sum=0,ave,M,M_D;
for(i=0;i<cnt;i++)
{
cin>>num[i];
sum+=num[i];
}
ave=sum/cnt;
for(i=0;i<cnt;i++)
{
s[i]=ave-num[i];
if(s[i]<0)
{
s[i]=s[i]*(-1);
}
cout<<"\n|ave - number| = "<<s[i];
M+=s[i];
}
M_D=M/cnt;
cout<<"\n\n Average: "<<ave;
cout<<"\n M.D(Mean Deviation): "<<M_D;
getch();
return 0;
}
I have written a program to approximate the solutions to ordinary differential equations using Adam's Method.
Running the program with gdb gives me:
Program received signal EXC_BAD_ACCESS, Could not access memory.
Reason: KERN_PROTECTION_FAILURE at address: 0x00007fff5f3ffff8
0x0000000100003977 in std::vector<double, std::allocator<double> >::push_back (this=0x100005420, __x=#0x100005310) at stl_vector.h:604
604 this->_M_impl.construct(this->_M_impl._M_finish, __x);
Clearly something is wrong with my treatment of vector.push_back, but I do not know where to begin looking. I can not think of a case where modifying a vector is illegal.
Call differentiate() to begin. Mathematics is done in step(). Adaptive time advancing across an interval with advance(). Check the chosen time step with checkdt() before running step() again.
Sorry for the huge code dump. I am sure many improvements can be made purely from the standpoint of C++, without knowledge of the mathematics:
//============================================================================
// Description : An Adam's Method Ordinary Differential Equation Approximation
// Disclaimer : Posted to StackOverflow for help on a segmentation fault
//============================================================================
#include <iostream> //IO
#include <vector> //std::vector
#include <cmath> //abs, pow, sqrt
#include <numeric> //accumulate
using namespace std;
/* Terminology:
* f(t, y) = the ordinary differential equation that will be solved
* y(t) = solution of f at point t.
* told = the previous point at which f was evaluated and solved
* tnow = the current point at which f is evaluated and solved
* tnew = the new (interesting) point at which f will be evaluated and solved
*
* Yold = the corrected solution of the differential equation at told
* Ynow = the corrected solution of the differential equation at tnow
* Ynew = the corrected solution of the differential equation at tnew
*
* fold = the value of the given differential equation at told
= f(told, Yold)
* fnow = the value of the given differential equation at tnow
= f(tnow, Ynow)
* fnew = the value of the given differential equation at tnew
= f(tnew, Ynew)
*
* Pnew = prediction for the value of Ynew
* dt = abs(tnew - tnow)
* dtold = abs(tnow - told)
*/
//Information storage
struct simTime {
double told;
double tnow;
double tnew;
double dt;
double dtold;
double tol;
double agrow;
double ashrink;
double dtmin;
double dtmax;
double endTime;
double fold;
double fnow;
double fnew;
double Yold;
double Ynow;
double Ynew;
double Pnew;
int stepsSinceRejection;
int stepsRejected;
int stepsAccepted;
} test;
//Define global variables
std::vector<double> errorIndicators(0);
std::vector<double> solutions(0);
std::vector<double> differencesDDY(0);
std::vector<double> differencesDDYSquared(0);
std::vector<double> timesTNew(0);
//Function declarations
void step(double fnow, double fold, double Ynow, double tnew, double tnow,
double dtold, double dt, double(*f)(double t, double y), double told);
void advance();
void shiftvariables();
void printvector(std::vector<double>& vec);
void differentiate(double(*f)(double t, double y), double y0, double a,
double b);
double f(double t, double y);
void checkdt();
int main() {
differentiate(f, 0, 1, 5);
cout << "Time values:" << endl;
printvector(timesTNew);
cout << "Solutions:" << endl;
printvector(solutions);
cout << "Differences between Prediction and Solution:" << endl;
printvector(differencesDDY);
return 0;
}
//Shift back all the variables to make way for the new values
void shiftvariables() {
test.tnow = test.tnew;
test.dtold = test.dt;
test.Yold = test.Ynow;
test.Ynow = test.Ynew;
test.fold = test.fnow;
test.fnow = test.fnew;
advance();
}
//Ordinary differential equation to be solved
double f(double t, double y) {
return pow(t, 2);
}
//Calculate the predicted and corrected solution at a chosen tnew
void step(double fnow, double fold, double Ynow, double tnew, double tnow,
double dtold, double dt, double(*f)(double t, double y), double told) {
//The calculation for Ynew requires integration. I first thought I would need to
// use my project 1 code to calculate the integration, but now I see in class we
// solved it analytically such that integration is not required:
//Linear prediction of Ynew using Ynow and fnow
double Pnew = Ynow + (dt * fnow) + (dt * dt / (2 * dtold)) * (fnow - fold);
test.Pnew = Pnew;
//Predict the value of f at tnew using Pnew
double fnew = f(tnew, Pnew);
test.fnew = fnew;
//Calculate the corrected solution at tnew
double interpolationFactor = fnew - (fnow + dt * (fnow - fold) / dtold);
double integration = (dt / 6) * (2 * dt + 3 * dtold) / (dt + dtold);
double Ynew = Pnew + interpolationFactor * integration;
test.Ynew = Ynew;
//Update the variables for the next round
shiftvariables();
}
//Check the previous solution and choose a new dt to continue evaluation
void advance() {
//The error indicator is the l2-norm of the prediction minus the correction
double err_ind = sqrt(
std::accumulate(differencesDDYSquared.begin(),
differencesDDYSquared.end(), 0));
errorIndicators.push_back(err_ind);
// Case where I reject the step and retry
if (err_ind > test.tol && test.dt > test.dtmin) {
++test.stepsRejected;
test.stepsSinceRejection = 0;
test.dt = test.dt * 0.5;
test.tnew = test.tnow + test.dt;
checkdt();
}
// Cases where I accept the step and move forward
else {
++test.stepsAccepted;
++test.stepsSinceRejection;
solutions.push_back(test.Ynew);
differencesDDY.push_back(abs(test.Pnew - test.Ynew));
differencesDDYSquared.push_back(pow((test.Pnew - test.Ynew), 2));
//Decrease dt
if (err_ind >= 0.75 * test.tol) {
test.dtold = test.dt;
test.dt = (test.dt * test.ashrink);
test.tnew = test.tnow + test.dt;
checkdt();
}
//Increase dt
else if (err_ind <= test.tol / 4) {
if ((test.stepsRejected != 0) && (test.stepsSinceRejection >= 2)) {
test.dt = (test.dt * test.agrow);
test.tnew = test.tnow + test.dt;
checkdt();
} else if (test.stepsRejected == 0) {
test.dt = (test.dt * test.agrow);
test.tnew = test.tnow + test.dt;
checkdt();
}
}
}
}
//Check that the dt chosen by advance is acceptable
void checkdt() {
if ((test.tnew < test.endTime) && (test.endTime - test.tnew < test.dtmin)) {
cout << "Reached endTime." << endl;
} else if (test.dt < test.dtmin) {
test.dt = test.dtmin;
test.tnew = test.tnow + test.dt;
timesTNew.push_back(test.tnew);
step(test.fnow, test.fold, test.Ynow, test.tnew, test.tnow, test.dtold,
test.dt, f, test.told);
} else if (test.dt > test.dtmax) {
test.dt = test.dtmax;
test.tnew = test.tnow + test.dt;
timesTNew.push_back(test.tnew);
step(test.fnow, test.fold, test.Ynow, test.tnew, test.tnow, test.dtold,
test.dt, f, test.told);
} else if ((test.tnew + test.dt) > test.endTime) {
test.dt = test.endTime - test.tnew;
test.tnew = test.tnow + test.dt;
checkdt();
} else if (((test.tnew + test.dt) < test.endTime)
&& ((test.tnew + 2 * test.dt) > test.endTime)) {
test.dt = (test.endTime - test.tnew) / 2;
test.tnew = test.tnow + test.dt;
checkdt();
}
//If none of the above are satisfied, then the chosen dt
// is ok and proceed with it
else {
timesTNew.push_back(test.tnew);
step(test.fnow, test.fold, test.Ynow, test.tnew, test.tnow, test.dtold,
test.dt, f, test.told);
}
}
//meta function to solve a differential equation, called only once
void differentiate(double(*f)(double t, double y), double y0, double a,
double b) {
//Set the starting conditions for the solving of the differential equation
test.fnow = f(a, y0);
test.endTime = b;
test.Ynow = y0;
//solutions.push_back(y0);
timesTNew.push_back(a);
//Set the constants
test.ashrink = 0.8;
test.agrow = 1.25;
test.dtmin = 0.05;
test.dtmax = 0.5;
test.tol = 0.1;
//Set fold = fnow for the first step
test.fold = test.fnow;
test.tnow = a;
test.told = a - test.dtmin;
test.dtold = abs(test.tnow - test.told);
//Create the first prediction, which will then lead to correcting it with step
advance();
}
// Takes a vector as its only parameters and prints it to stdout
void printvector(std::vector<double>& vec) {
for (vector<double>::iterator it = vec.begin(); it != vec.end(); ++it) {
cout << *it << ", ";
}
cout << "\n";
}
Thank you.
Since you're using recursion, could it be possible that you are running out of stack memory, thus causing the segfault? This could happen if either your app recurses too many times or if some bug causes it to recurse infinitely.
Note, as sth suggests in a comment, a debugger may help you decide whether or not this is the case.