 An important problem in mathematics. Given a set of data values, find a function that produces those data values. We can generally view our data values as being a set of points, so what we generally want is to find a curve that passes through a given set of points. This is known as the problem of curve-fitting. It's useful to consider the differentiability class of a function. A function is Cn if its nth derivative is continuous. We'll regard the 0th derivative as a function itself, so a C0 function is a continuous function. And if a function has derivatives of all orders, we say it's a C infinity function. Now if a function is Ck, then its kth derivative is continuous. So the only thing preventing it from being a Ck plus 1 is a point of non-differentiability. On a continuous function, this is usually because the graph has a cusp or a corner. So intuitively, a smooth function has no corners. Analytically, this means that all derivatives are continuous, and so a smooth function is a C infinity function. For example, let's find the differentiability class of the absolute value function. Now remember the absolute value function is continuous. However, it has this corner at x equals 0, and so it's non-differentiable at x equals 0, and so its derivative is not continuous. And this means it's in differentiability class, C0. So let's think about this. Since polynomials have derivatives of all orders, they are smooth functions. And so the question to ask yourself is, self, can we find a polynomial that passes through a set of points? And the answer is, yes, but it may take a lot of work. So suppose you have n plus 1 points, with coordinates x0, y0, x1, yn, and so on. You can try to find an nth degree polynomial, f of x, passing through all the points, where here the unknowns are the coefficients a0 through an. Now since the curve passes through x0, y0, we must have f of x0 equals y0, and so this gives us one equation with n plus 1 unknowns. Similarly, since the curve also passes through x1, y1, we must also have f of x1 equals y1. And so again, this gives us a second equation with n plus 1 unknowns. But wait, there's more. The curve also passes through x2, y2, and so we have f of x2 equals y2. And this gives us a third equation with n plus 1 unknowns. And the remaining points give us more equations. And what this means is that we can use the given points to produce and solve a system of n plus 1 equations in n plus 1 unknowns. So now let's say we want to find a quadratic curve that passes through three given points. So our curve has equation f of x equals a0 plus a1x plus a2x squared, where our unknowns are the coefficients a0, a1, a2. Using our given points gives us a system of equations. So the point 1, 3 on y equals f of x gives us the equation where if x is equal to 1, the function value is equal to 3. And this gives us our first equation. Our second point 4 negative 5 on y equals f of x, if x is equal to 4, then our function value is equal to negative 5. And this gives us a second equation. And finally the point 6, 4 on y equals f of x gives us another equation. And that gives us a system of three equations in three unknowns. Solving this equation gives us, and so our quadratic function is going to be. If we graph it, we see that it does actually go through our three points.