Interpolation Algorithm: Laplace Interpolation

Let's take a look at the derivation of the formula and some key points. [two-dimensional interpolation is used here, and so on]

 

1. interpolation problem: a pile of discrete data is often obtained during the experiment. Now I want to use a mathematical formula to simulate this pile of discrete data. What should we do? mathematicians raised the interpolation question. The interpolation problem is such that a pile of data points (x0, y0), (x1, Y1), (X2, Y2) are given )... (Xn, yn), requires a Function Y = f (x), requires the function to pass through all the above data points.

 

2. Polynomial Interpolation and its uniqueness: among all functions, polynomial functions are the simplest functions, so as long as they are human beings, they will think of using polynomial functions as interpolation functions, given n + 1 points above, a n times polynomial y = An * x ^ N +... a1 * x + a0, let them pass through the n + 1 points. Through the Van Dimon and the Klein law, we can determine that if the X values of the n + 1 points are different, then this polynomial is unique. The result is unique, but it is not easy to use the direct method. Now, we can try another way. This is: the Laplace polynomial.

 

3. The structure of the Laplace polynomial, with four points as an Example

 

Since the function goes through four points (x0, y0), (x1, Y1), (X2, Y2), (X3, Y3), you can set the function:

 

F (x) = b0 (x) * y0 + B1 (x) * Y1 + B2 (x) * y2 + B3 (x) * Y3

 

Note: B0 (x),..., B3 (x) are all three polynomials of X, called the Laplace Interpolation Basis Function.

 

Because f (x) = y0 is required when X is x0, the simplest way is to make B0 (x0) = 1, B1 (x0) = B2 (x0) = B3 (x0) = 0;

 

Likewise, on points x1, x2, and X3, the value structure of the interpolation base function is as follows:

 

B0 (x) B1 (x) B2 (x) B3 (X)

 

X = x0 1 0 0 0

 

X = x1 0 1 0 0

 

X = x2 0 0 1 0

 

X = X3 0 0 0 1

 

Question 1: Determine the B0 (x) expression based on these values,

 

Because B0 (X1) = b0 (X2) = b0 (X3) = 0, so x1, x2, X3 is the zero point of B0 (x), because B0 (X) it is a cubic polynomial.

 

B0 (x) = C0 * (x-x1) * (x-x2) * (x-x3)

 

Because B0 (x0) = 1, so 1 = C0 * (x0-x1) * (x0-x2) * (x0-x3) Get C0 = 1/[(x0-x1) (x0-x2) (x0-x3)]

 

So: B0 (x) = (x-x1) * (x-x2) * (x-x3)/[(x0-x1) * (x0-x2) * (x0-x3)]

 

Similarly, B1 (x), B2 (x), and

 

Question 2: according to the preceding table, describe a property of the Interpolation Basis Function: regardless of the X value, their sum is 1. [This is called a Harmonic Function]

 

The following example shows how to add each row in the preceding table. The result is g (x) = b0 (x) + B1 (x) + B2 (X) + The values of B3 (x) in x0, x1, x2, and X3 are 1.

 

B0 (x) + B1 (x) + B2 (x) + B3 (X)

 

X = x0 1 + 0 + 0 + 0 = 1

 

X = x1 0 + 1 + 0 + 0 = 1

 

X = x2 0 + 0 + 1 + 0 = 1

 

X = X3 0 + 0 + 0 + 1 = 1

 

Therefore, the equation g (x)-1 = 0 should have four x0, x1, x2, X3; however, due to B0 (x), B1 (X), B2 (x), B3 (x) are all three polynomial, so g (x) is also a maximum of three polynomial, at most three root, so the equation: G (x) = 1 should be an equality. .

 

Question 3: basic functions: B0 (x), B1 (x), B2 (x), and B3 (x) are linearly independent.

 

Set: Numbers T0, T1, T2, and T3: t0 * B0 (x) + T1 * B1 (x) + T2 * B2 (x) + T3 * B3 (X) = 0

 

X = x0: 0 = T0 * B0 (x0) + T1 * B1 (x0) + T2 * B2 (x0) + T3 * B3 (x0) = T0 * 1 + T1 * 0 + T2 * 0 + T3 * 0: T0 = 0;

 

Likewise, T1 = t2 = T3 = 0, as defined (all coefficients are 0 ). Therefore, interpolation functions are linearly independent.