 # Natural Cubic Spline Interpolation

by admin in on June 18, 2019

Cubic spline interpolation is a special case for Spline interpolation that is used very often to avoid the problem of Runge’s phenomenon. This method gives an interpolating polynomial that is smoother and has smaller error than some other interpolating polynomials such as Lagrange polynomial and Newton polynomial.

### Example On Using This Code

#### Input

xf = [0 1 2 3];     % X Values
yf = [exp(xf(1)) exp(xf(2)) exp(xf(3)) exp(xf(4))];       % Y Values
xx = 1.5;           % Value we interest to interpolate.

#### Output  The numbers x(0), ..., x(n) are:
0.0000   1.0000   2.0000   3.0000

The coefficients of the spline on the subintervals are:
a(i)          b(i)          c(i)          d(i)
1.00000000    1.46599761    0.00000000    0.25228421
2.71828183    2.22285026    0.75685264    1.69107137
7.38905610    8.80976965    5.83006675   -1.94335558

Interpolated value at 1.5 is equal to = 4.2303

All Splines:
1.465997*conj(x) + 0.25228421*conj(x)^3 + 1.0
5.7823590*conj(x) - 4.3163614*conj(x)^2 + 1.69107137*conj(x)^3 - 0.4387871
17.4902*conj(x)^2 - 37.8307*conj(x) - 1.943355*conj(x)^3 + 28.6366

### Contents

• Definition
• Boundary Conditions
• Methods
• Example
• Exercise
• References

### Definition

Given a set of n + 1 data points (xi,yi) where no two xi are the same and a = x_(0) < x_(1) < … < x_(n) = b, the spline S(x) is a function satisfying:

1. S(x) in C^(2) [a,b];
2. On each subinterval [x_(i-1), x_(i)], S(x) is a polynomial of degree 3, where i = 1,… ,n.
3. S(x_(i)) = y_(i), for all i = 0, 1,… ,n.

Let us assume that where each is a cubic function, i = 1,… ,n.

### Boundary Conditions

To determine this cubic spline S(x), we need to determine a_(i), b_(i), c_(i) and d_(i) for each i by:

• C_(i)(x_(i-1)) = y_(i-1) and C_(i)(x_(i)) = y_(i), i = 1,… ,n.
• C’_(i)(x_(i)) = C’_(i+1)(x_(i)), i = 1,… ,n-1.
• C”_(i)(x_(i)) = C”_(i+1)(x_(i)), i = 1,… ,n-1.

We can see that there are n + n + (n-1) + (n-1) = 4n-2 conditions, but we need to determine 4n coefficients, so usually we add two boundary conditions to solve this problem. There are three types of common boundary conditions:

I. First derivatives at the endpoints are known: This is called clamped boundary conditions.

II. Second derivatives at the endpoints are known: .

The special case is called natural or simple boundary conditions.

III. When the exact function f(x) is a periodic function with period x_(n) – x_(0)S(x) is a periodic function with period x_(n) – x_(0) too. Thus .

The spline functions S(x) satisfying this type of boundary condition are called periodic splines.

### Methods

There are several methods that can be used to find the spline function S(x) according to its corresponding conditions. Since there are 4n coefficients to determine with 4nconditions, we can easily plug the values we know into the 4n conditions and then solve the system of equations. Note that all the equations are linear with respect to the coefficients, so this is workable and computers can do it quite well.

The algorithm given in w:Spline interpolation is also a method by solving the system of equations to obtain the cubic function in the symmetrical form.

The other method used quite often is w:Cubic Hermite spline, this gives us the spline in w:Hermite form.

Here, we discuss another method using second derivatives S”(x_(i)) = M_(i)(i = 0,1,… ,n) to find the expression for spline S(x).

Let h_(i) = x_(i) – x_(i-1), i = 1,… ,n, S”(xi) = C”_(i)(xi) = C”_(i+1) (xi) = Mi (i = 1,… ,n-1) and S”(x0) = C”_(1)(x0) = M0, and S”(xn) = C”_(n)(xn) = Mn. Note that M_{i} are unknown (except for type II boundary condition, M_{0} and M_{n} are given).

Since each C_(i) is a cubic polynomial, C”_(i) is linear.

By w:Lagrange interpolation, we can interpolate each C”_(i) on [x_(i-1), x_(i)] since C”_(i)(x_(i-1)) = M_(i-1) and C”_(i)(xi) = Mi, the Lagrange form of this interpolating polynomial is: for x in [x_{i-1},x_{i}].

Integrating the above equation twice and using the condition that C_{i}(x_{i-1}) = y_{i-1} and C_{i}(x_{i}) = y_{i} to determine the constants of integration, we have (1)

This expression gives us the cubic spline S(x) if M_{i},i=0,1,…. ,n can be determined.

For i=1,… ,n-1 when x in [x_{i}, x_{i+1}], we can calculate that Therefore, Similarly, when x in [x_{i-1}, x_{i}], we can shift the index to obtain (2)

Thus, Since , we can derive (3)

where (4)

and f[x_{i-1}, x_{i} ,x_{i+1}] is a divided difference. According to different boundary conditions, we can solve the system of equations above to obtain the values of M_{i}‘s.

I. For type I boundary condition, we are given C’_{1}(x_{0}) = f’_{0} and C’_{n}(x_{n}) = f’_{n}. According to equation (2 ), we can obtain   (5.1)

Similarly, simplifying we will have (5.2)

Therefore, let and , combine (3 ), (5.1 ) and (5.2 ) together, so the system of equations that we need to solve is (6)

II. For type II boundary condition, we are given and (7)

directly, so let lambda _{0} = mu _{n} = 0, d_{0} = 2f”_{0}, and d_{n} = 2f”_{n}, and we need to solve the system of equations in the same form as (6).

### Example

For points (0,0), (1,0.5), (2,2) and (3,1.5), find the interpolating cubic spline S(x) satisfying S'(0) = 0.2 and S'(3) = -1.

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June 18, 2019

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June 19, 2019