UNIT 2 • UNEQUAL INTERVAL INTERPOLATION

Cubic Spline
Natural & Clamped

A cubic spline joins the data points with a separate cubic on each interval, matched so the curve, its slope and its curvature are continuous at every point. The unknowns are the second derivatives $M_i = S''(x_i)$ .

THE SPLINE EQUATIONSWORKS FOR UNEQUAL STEPS

h_{i-1}M_{i-1} + 2(h_{i-1} + h_i)M_i + h_iM_{i+1} = 6\left[\frac{y_{i+1} - y_i}{h_i} - \frac{y_i - y_{i-1}}{h_{i-1}}\right]

1. One Equation per Inner PointWrite the equation above for each interior

x_i

2. End ConditionsNatural:

M_0 = M_n = 0

. Clamped: two extra equations from the end slopes.

3. Solve, then EvaluateSolve for the M's, then use the piece for the interval containing x.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Fit a natural cubic spline through (1, 2), (2, 3), (3, 5) and estimate y at x = 1.5.

STEP 01 • THE EQUATION FOR THE MIDDLE POINT

Here h = 1 and natural means $M_0 = M_2 = 0$ :

M_0 + 4M_1 + M_2 = 6\left[(5 - 3) - (3 - 2)\right] = 6 \;\Rightarrow\; M_1 = 1.5

STEP 02 • THE PIECE ON [1, 2]

S_0(x) = 2 + 0.75(x - 1) + 0.25(x - 1)^3

Check: $S_0(1) = 2$ , $S_0(2) = 2 + 0.75 + 0.25 = 3$ ✓

STEP 03 • EVALUATE AT 1.5

S_0(1.5) = 2 + 0.75(0.5) + 0.25(0.5)^3 = 2 + 0.375 + 0.03125 = 2.40625

FINAL CONCLUSION

y(1.5) ≈ 2.40625

S(1.5) = 2.40625

The calculator below opens on these points, shows the system and every piece, and draws the spline.

LABORATORY

Interactive Cubic Spline Calculator

LABORATORY

Cubic Spline Engine

x values (increasing)y valuesFind y at x = (optional)

End conditions

Common Questions

Why use a cubic spline instead of one big polynomial?

A single high-degree polynomial through many points can swing wildly between them (Runge's phenomenon). A spline uses a separate cubic on each interval, joined smoothly, so it stays close to the data.

What is the difference between a natural and a clamped spline?

A natural spline sets the second derivative to 0 at both ends, so the curve straightens out there. A clamped spline fixes the slope at each end to values you give, which is better when you know them.

Why is the system tridiagonal?

Each equation links only

M_{i-1}

M_i

and

M_{i+1}

, so the matrix has entries only on three diagonals. That is why it can be solved quickly by the Thomas algorithm.

PreviousLagrange's Interpolation Formula

Unit 2•17 of 63

NextNumerical Differentiation

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 2 • UNEQUAL INTERVAL INTERPOLATION

Cubic Spline
Natural & Clamped

THE SPLINE EQUATIONSWORKS FOR UNEQUAL STEPS

h_{i-1}M_{i-1} + 2(h_{i-1} + h_i)M_i + h_iM_{i+1} = 6\left[\frac{y_{i+1} - y_i}{h_i} - \frac{y_i - y_{i-1}}{h_{i-1}}\right]

1. One Equation per Inner PointWrite the equation above for each interior

x_i

2. End ConditionsNatural:

M_0 = M_n = 0

. Clamped: two extra equations from the end slopes.

3. Solve, then EvaluateSolve for the M's, then use the piece for the interval containing x.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Fit a natural cubic spline through (1, 2), (2, 3), (3, 5) and estimate y at x = 1.5.

STEP 01 • THE EQUATION FOR THE MIDDLE POINT

Here h = 1 and natural means $M_0 = M_2 = 0$ :

M_0 + 4M_1 + M_2 = 6\left[(5 - 3) - (3 - 2)\right] = 6 \;\Rightarrow\; M_1 = 1.5

STEP 02 • THE PIECE ON [1, 2]

S_0(x) = 2 + 0.75(x - 1) + 0.25(x - 1)^3

Check: $S_0(1) = 2$ , $S_0(2) = 2 + 0.75 + 0.25 = 3$ ✓

STEP 03 • EVALUATE AT 1.5

S_0(1.5) = 2 + 0.75(0.5) + 0.25(0.5)^3 = 2 + 0.375 + 0.03125 = 2.40625

FINAL CONCLUSION

y(1.5) ≈ 2.40625

S(1.5) = 2.40625

The calculator below opens on these points, shows the system and every piece, and draws the spline.

LABORATORY

Interactive Cubic Spline Calculator

LABORATORY

Cubic Spline Engine

x values (increasing)y valuesFind y at x = (optional)

End conditions

Common Questions

Why use a cubic spline instead of one big polynomial?

What is the difference between a natural and a clamped spline?

A natural spline sets the second derivative to 0 at both ends, so the curve straightens out there. A clamped spline fixes the slope at each end to values you give, which is better when you know them.

Why is the system tridiagonal?

Each equation links only

M_{i-1}

M_i

and

M_{i+1}

, so the matrix has entries only on three diagonals. That is why it can be solved quickly by the Thomas algorithm.

PreviousLagrange's Interpolation Formula

Unit 2•17 of 63

NextNumerical Differentiation

Cubic Spline Natural & Clamped

Step-by-Step Worked Example

y(1.5) ≈ 2.40625

Interactive Cubic Spline Calculator

Cubic Spline Engine

Common Questions

NETZ Engine

Cubic Spline Natural & Clamped

Step-by-Step Worked Example

y(1.5) ≈ 2.40625

Interactive Cubic Spline Calculator

Cubic Spline Engine

Common Questions

Cubic Spline
Natural & Clamped

Cubic Spline
Natural & Clamped