How to find the intersection between a line and a mathematical function?

Question

I have a problem where I need to find the intersection of two mathematical functions. I have the formula of the first function and two coordinates, where ab and cd are my points. To pass these coordinates to a function, I use the line equation by two points, and this is my problem.

The first function is represented by f(x) = |x| - cos(3x), who create this drawing:

Since the formula is already ready, I wrote it in C# with:

static Vector2 CossineFormula(int x)
{
     double y = Math.Sqrt(Math.Pow(Math.Abs(x), 2)) - Math.Cos(x * 3);
     return new Vector2(x, y);
}

So far, so good. Now, I have two points that I mentioned, and I can’t go on with the two-point line equation because I can’t define a function dynamically, or at least I can’t find how.

The intersection is given by the continuation of the points ab and cd until you create the interception Ef:

I can’t find that intersection because I can’t pass the two points to mathematical function.

I tried to use the formula of the ramp (Slope) as it is in the above mentioned tutorial, but I stopped in this part (in the tutorial is Step 2):

static Vector2 CossineIntersection(Vector2 a, Vector2 b)
{
    double distance = Math.Sqrt(Math.Pow(b.X - a.X, 2) + Math.Pow(b.Y - a.Y, 2));
    double slope = (b.Y - a.Y) / (b.X - a.X);
    double intercept = (2 * slope); // onde continuo com b?
}

Slope was calculated, but at this stage:

Substitute the Slope for’m' in the Slope Intercept form of the Equation y = mx + b

I went to y = (2 * slope) but I didn’t have to continue b.

I also asked about how to find this intersection in Mathematics.

All the ways of doing intersection, I can’t perform in programming. I know it’s possible, but I can’t find anywhere anything like.

Note: is not an intersection between two lines, but between a line and a function.

Answer 1

You can use the bisection method. You’re basically trying to find the root of this equation:

y = |x| - cos(3x) - (mx + b)

The root of this equation is where its two lines meet.

Like Vector2 works with floats, so let’s treat all the numbers as floats.

First, the linear function:

static float LinearFormula(float x, Vector2 a, Vector2 b)
{
     float slope = (b.Y - a.Y) / (b.X - a.X);
     float dx = x - a.X;
     float dy = slope * dx; 
     return dy + a.Y;
}

Its function with cosine:

static float CossineFormula(float x)
{
     return Math.Abs(x) - (float) Math.Cos(x * 3);
}

Note that I simplified the Math.Sqrt(Math.Pow(Math.Abs(x), 2)) for Math.Abs(x). There is no reason to square just to take out the square root later, even more that you are already applying the Abs.

Its function:

static float MyFormula(float x, Vector2 a, Vector2 b)
{
     return CossineFormula(x) - LinearFormula(x, a, b);
}

And then we look for the root for consecutive bisections:

static Vector2 FindIntersection(Vector2 a, Vector2 b)
{
    float x1 = 2;
    float x2 = 3;

    for (int i = 0; i < 100; i++)
    {
        float d = (x2 + x1) / 2;
        if (d == x1 || d == x2) break;
        float v3 = MyFormula(d, a, b);
        if (v3 > 0)
        {
            x2 = d;
        }
        else
        {
            x1 = d;
        }
    }
    return new Vector2(x1, CossineFormula(x1));
}

In this function, 2 and 3 are the initial values of the interval where we will look for the root of the function. The choice of these values is arbitrary, but I did it knowing that (a) for 2 we have a negative value, (b) for 3 a positive and (c) in this interval there is only one single root for the function. Therefore, by dividing this interval into smaller and smaller parts, we can find where the root is.

The for should rotate the number of times enough to find the root, and could even be a while (true). But I put a counter to limit the number of iterations, in case something goes wrong.

Finally, to test:

public static void Main()
{
    Console.WriteLine(FindIntersection(new Vector2(0, 4), new Vector2(1, 3)));
}

The result is this:

<2.354912, 1.645088>

See here working on ideone.