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How can you define a function to calculate the value of a definite integral in C++? For example to solve the integral of the function x^2 * cos(x)?
Interestingly enough, I ran across this article a little while ago explaining one method for calculating numerical integrals using function pointers.
https://helloacm.com/c-function-to-compute-numerical-integral-using-function-pointers/
For something like x^2 * cos(x):
You would need an overloaded integral function:
double integral(double(*f)(double x), double(*g)(double x, double y), double a, double b, int n)
{
double step = (b - a)/n; // width of rectangle
double area = 0.0;
double y = 0; // height of rectangle
for(int i = 0; i < n; ++i)
{
y = f(a + (i + 0.5) * step) * g(a + (i + 0.5) * step, y);
area += y * step // find the area of the rectangle and add it to the previous area. Effectively summing up the area under the curve.
}
return area;
}
To call:
int main()
{
int x = 3;
int low_end = 0;
int high_end = 2 * M_PI;
int steps = 100;
cout << integral(std::powf, std::cosf, low_end, high_end, steps);
return 0;
}
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I know that it may seem a duplicate question, but I could not find my answer in previous questions.
I mean how to write a log base 10 function by simple loops and not using built in log function in c++.
The easiest way is to calculate the natural logarithm (ln) with a Taylor series. Once you have found the natural logarithm, just divide it by ln(10) and you get the base-10 log.
The Taylor series is quite simple to implement in C. If z is the number for which you are seeking the log, you just have to loop a few iterations multiplying an accumulator by (z-1) each time. To a limit, the more iterations you run, the more accurate your result will be. Check it a few times against the libC log10() version until you are happy with the precision.
This is a "numeric approach". There are other numeric solutions to finding the logarithm of a number which can give more accurate results. Some of them can be found in that Wikipedia link I gave you.
Assuming by "log base 10" you mean "the number of times n can be divided by 10 before resulting in a value < 10":
log = 0;
// Assume n has initial value N
while ( n >= 10 ) {
// Invariant: N = n * 10^log
n /= 10;
log += 1;
}
You'll get faster convergence with Newton's Method. Use something like this (hand written not compiled or tested uses f(r) = 2**r - x to compute log2(x) ):
double next(double r, double x) {
static double one_over_ln2 = 1.4426950408889634;
return r - one_over_ln2 * (1 - x / (1 << static_cast<int>(r)));
double log2(double x) {
static double epsilon = 0.000000001; // change this to change accuracy
double r = x / 2;. // better first guesses converge faster
double r2 = next(r, x);
double delta = r - r2;
while (delta * delta > epsilon) {
r = r2;
r2 = next(r, x);
delta = r - r2
}
return r2;
}
double log10(double x) {
static double log2_10 = log2(10);
return log2(x) / log2_10;
}
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Using boost c++ odeint library, how to solve the following ordinary differential equations of motion,
z'' = -n²·z.
The analytical solution of the aforementioned ODE is
z(t) = (z0'/n)·sin(n·t) + z0·cos(n·t).
Solution attempt as per comment (Jan 17 at 23:01) by the OP in answer
My program is:
void ode( const state_type &z , state_type &dzdt , double t ) {
dzdt[0] = z[1];
dzdt[1] = -1 * z[0] * w * w;
}
void write_ode( const state_type &z , const double t ) {
cout << t << '\t' << z[0] << '\t' << z[1] << endl;
}
int main { ...
integrate( ode , z , t , 1000 , 0.1 , write_ode );
}
But the integrate function return only the values of z0, and z0'. I need to find the values of z(t).
You can use the standard way of transforming an N th-order differential equation of a first order ODE of dimension N. In your case this is:
x = z
y = dz/dt
dx/dt = y
dy/dt = n^2 x
You can easily put this into any solver of odeint. But, your ODE seems to be Hamiltonian - you might also think about a symplectic solver. They conserve the phase space volume and the energy "only" oscillates about the initial energy. Odeint has implemented two symplectic solvers.
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Can anyone tell me how to find the centroid of an object in point cloud?.
I haven't tried any code yet because I have no slight idea as to how to go about it.
If you have the point locations, you should be able to just average the x and y positions.
int x = 0; y = 0;
for ( i = 0; i < num_pts; i++ )
{
x += pt[i].x;
y += pt[i].y;
}
centroid.x = x / num_pts;
centroid.y = y / num_pts;
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I'm having the following code and the number of my channels are 3
IplImage* img_crop_mat = cvLoadImage("....", 1);
...
int b = 0;
uchar* rgb = (uchar*) img_crop_mat->imageData;
I would like to have R, G and B matrices in a loop, skimming the entire image:
for (int y = b; y < height - b; y++)
{
???
for (int x = b; x < width - b; x++)
{
????
}
}
The previous forums regarding my question deal with CvMat but not with pointers as my code.
What are the indexes that I must take into account?
You can use the following macro to access an arbitrary pixel of a 3-channel, 8U-image:
CV_IMAGE_ELEM(myImage, unsigned char, y, x*3 + ChannelOfInterest)
This is an lvalue so you can take and use its value, or you can change the pixel's value.
By default,
ChannelOfInterest = 0, blue
ChannelOfInterest = 1, green
ChannelOfInterest = 2, red
The actual data structure is pretty straightforward, look up the definition of CV_IMAGE_ELEM.
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can anyone tell me how to generate a 2d gaussian filter kernel using the gaussian filter equation? how does the x and y value vary?
ref: http://en.wikipedia.org/wiki/Gaussian_function
To generate the kernel is quite simple. If your problem is in applying the kernel, you need to update the question.
The kernel is simply a square matrix of values, generally an odd number size so that there's a clearly defined center. To fill it, the x and y values go from -(n-1)/2 to (n-1)/2 where n is the size of the matrix.
double half_n = (n - 1) / 2.0;
for (i = 0; i < n; ++i)
{
double x = i - half_n;
for (j = 0; j < n; ++j)
{
double y = j - half_n;
kernel[i][j] = // use formula with x and y here
}
}