#### Introduction

• Given a set of points:
• We define two functions f1 and f2 that will be our quadratic splines:

#### Solving the unknowns

• We can solve the unknowns in these equations.
• First we plug in our known values:
• This gives us a system of linear equations:
• Which simplifies to:
• An extra contraint we can add to this system is that the gradient of the two splines should be equal at the point they meet:
• First we find the derivitives of our two functions:
• And then solve them at x = 0.5:
• And finally set them equal to each other and re-arrange
• We now have 5 equations, but 6 unknowns
• So it's not quite solvable yet
• However we can rewrite each of the terms with respect to a1:

#### Solving a1

• There are several different options for solving a1
• But in order to compare them, let's start with a more complex problem:
• Given a new set of points:
• We can perform the same steps as above to produce a set of coefficients for our 4 splines:
##### a1 = 0
• One simple solution is to set a1 to 0:
• This system is now solvable, giving:
• Which produces splines of:
• Setting a1 to 0 makes it easy and efficient to compute the splines.
• However the first spline is a straight line.
• And all other splines have large gradients.
##### a1 = a2
• Another simple solution would be to set a1 to a2
• The system is now solvable, giving:
• Which produces splines of:
• The first two splines are equal
• The average gradient of the splines is less than for a1 = 0
##### Minimizing the gradients of the splines
• An ideal interpolation minimizes the amount that each spline curves
• We can describe this mathematically as minimizing the mean squared error of the a terms:
• First we get all the a terms with respect to a1:
• Then the error function contains a single term:
• We can find its minimum by finding the derivitive:
• And solving for 0:
• If we now use a1 = 5.7 as an extra equation in our system, it solves as:
• Which produces splines of:
• The average gradient of the splines is minimized
• However solving a1 was complicated and required finding the derivative of the error function

#### Comparison

• The value we choose for a1 makes a big difference for subsequent splines
• using mean square error (mse) to minimize the average gradient produces splines that have the least deviation against a linear interpolation
• However it is by far the most complex option
• If we increase the number of points we are interpolating, the changes get even more pronounced: