UNIT 2 • EQUAL INTERVAL INTERPOLATION

Stirling's Formula
Central Difference Interpolation

Stirling's formula interpolates near the middle of an equally spaced table, using central differences around the nearest point $x_0$ . It is the average of Gauss's forward and backward formulas.

STIRLING'S FORMULABEST FOR |p| ≤ ¼

\begin{gathered} y = y_0 + p\,\frac{\Delta y_{-1} + \Delta y_0}{2} + \frac{p^2}{2!}\Delta^2 y_{-1} \\ + \frac{p(p^2 - 1)}{3!}\cdot\frac{\Delta^3 y_{-2} + \Delta^3 y_{-1}}{2} \\ + \frac{p^2(p^2 - 1)}{4!}\Delta^4 y_{-2} + \cdots \\ p = \frac{x - x_0}{h} \end{gathered}

1. Pick x₀Choose the tabulated x nearest to the one you want; it becomes index 0.

2. Find p

p = (x - x_0)/h

, a small number between −½ and ½.

3. Middle RowUse the differences straddling

x_0

: odd orders averaged, even orders on the row.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

From x = 1, 2, 3, 4, 5 and y = 1, 8, 27, 64, 125, estimate y at x = 3.2.

STEP 01 • CHOOSE x₀ AND p

x_0 = 3, \quad h = 1, \quad p = \frac{3.2 - 3}{1} = 0.2

The differences around x₀: $\Delta y_{-1} = 19$ , $\Delta y_0 = 37$ , $\Delta^2 y_{-1} = 18$ , $\Delta^3 y_{-2} = \Delta^3 y_{-1} = 6$ .

STEP 02 • THE TERMS

y_0 = 27

p\,\frac{19 + 37}{2} = 0.2 \times 28 = 5.6

\frac{p^2}{2}\,\Delta^2 y_{-1} = \frac{0.04}{2} \times 18 = 0.36

\frac{p(p^2 - 1)}{6}\cdot\frac{6 + 6}{2} = \frac{0.2 \times (-0.96)}{6} \times 6 = -0.192

STEP 03 • ADD THEM UP

y(3.2) \approx 27 + 5.6 + 0.36 - 0.192 = 32.768

The data come from $y = x^3$ , and $3.2^3 = 32.768$ exactly: with a cubic, the formula stops at Δ³ and is exact.

FINAL CONCLUSION

y(3.2) ≈ 32.768

y(3.2) = 32.768

The calculator below opens on this table and tints the differences it used.

LABORATORY

Interactive Stirling Calculator

LABORATORY

Stirling Engine

x values (equally spaced)y values

Find y at x =

Common Questions

When should you use Stirling's formula?

When the x you want is near the middle of the table, close to a tabulated point: it works best for

-\tfrac{1}{4} \le p \le \tfrac{1}{4}

. Near the start use Newton forward, near the end Newton backward, and halfway between two middle points use Bessel.

How is Stirling's formula related to Gauss's formulas?

It is the average of Gauss's forward and backward formulas. That is why its odd terms use the mean of two differences.

What is p?

p = \frac{x - x_0}{h}

: how many steps x is from the chosen middle point

x_0

PreviousFinite Difference Table

Unit 2•13 of 63

NextBessel's Formula

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 2 • EQUAL INTERVAL INTERPOLATION

Stirling's Formula
Central Difference Interpolation

Stirling's formula interpolates near the middle of an equally spaced table, using central differences around the nearest point $x_0$ . It is the average of Gauss's forward and backward formulas.

STIRLING'S FORMULABEST FOR |p| ≤ ¼

\begin{gathered} y = y_0 + p\,\frac{\Delta y_{-1} + \Delta y_0}{2} + \frac{p^2}{2!}\Delta^2 y_{-1} \\ + \frac{p(p^2 - 1)}{3!}\cdot\frac{\Delta^3 y_{-2} + \Delta^3 y_{-1}}{2} \\ + \frac{p^2(p^2 - 1)}{4!}\Delta^4 y_{-2} + \cdots \\ p = \frac{x - x_0}{h} \end{gathered}

1. Pick x₀Choose the tabulated x nearest to the one you want; it becomes index 0.

2. Find p

p = (x - x_0)/h

, a small number between −½ and ½.

3. Middle RowUse the differences straddling

x_0

: odd orders averaged, even orders on the row.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

From x = 1, 2, 3, 4, 5 and y = 1, 8, 27, 64, 125, estimate y at x = 3.2.

STEP 01 • CHOOSE x₀ AND p

x_0 = 3, \quad h = 1, \quad p = \frac{3.2 - 3}{1} = 0.2

The differences around x₀: $\Delta y_{-1} = 19$ , $\Delta y_0 = 37$ , $\Delta^2 y_{-1} = 18$ , $\Delta^3 y_{-2} = \Delta^3 y_{-1} = 6$ .

STEP 02 • THE TERMS

y_0 = 27

p\,\frac{19 + 37}{2} = 0.2 \times 28 = 5.6

\frac{p^2}{2}\,\Delta^2 y_{-1} = \frac{0.04}{2} \times 18 = 0.36

\frac{p(p^2 - 1)}{6}\cdot\frac{6 + 6}{2} = \frac{0.2 \times (-0.96)}{6} \times 6 = -0.192

STEP 03 • ADD THEM UP

y(3.2) \approx 27 + 5.6 + 0.36 - 0.192 = 32.768

The data come from $y = x^3$ , and $3.2^3 = 32.768$ exactly: with a cubic, the formula stops at Δ³ and is exact.

FINAL CONCLUSION

y(3.2) ≈ 32.768

y(3.2) = 32.768

The calculator below opens on this table and tints the differences it used.

LABORATORY

Interactive Stirling Calculator

LABORATORY

Stirling Engine

x values (equally spaced)y values

Find y at x =

Common Questions

When should you use Stirling's formula?

When the x you want is near the middle of the table, close to a tabulated point: it works best for

-\tfrac{1}{4} \le p \le \tfrac{1}{4}

. Near the start use Newton forward, near the end Newton backward, and halfway between two middle points use Bessel.

How is Stirling's formula related to Gauss's formulas?

It is the average of Gauss's forward and backward formulas. That is why its odd terms use the mean of two differences.

What is p?

p = \frac{x - x_0}{h}

: how many steps x is from the chosen middle point

x_0

PreviousFinite Difference Table

Unit 2•13 of 63

NextBessel's Formula

Stirling's Formula Central Difference Interpolation

Step-by-Step Worked Example

y(3.2) ≈ 32.768

Interactive Stirling Calculator

Stirling Engine

Common Questions

NETZ Engine

Stirling's Formula Central Difference Interpolation

Step-by-Step Worked Example

y(3.2) ≈ 32.768

Interactive Stirling Calculator

Stirling Engine

Common Questions

Stirling's Formula
Central Difference Interpolation

Stirling's Formula
Central Difference Interpolation