- 2071 Bhadra · 4 marks
Difference table (3.0 to 3.8)
Short answer
With h=0.2, forward differences: Δy₀=0.0280, Δ²y₀=−0.0016, Δ³y₀=0.0000, Δ⁴y₀=0.0003. Third and fourth differences are nearly zero, indicating nearly quadratic behavior.
Create difference table from following data.
| X | 3.0 | 3.2 | 3.4 | 3.6 | 3.8 |
|---|---|---|---|---|---|
| Y | 0.4771 | 0.5051 | 0.5315 | 0.5563 | 0.5798 |
The values are equally spaced with h = 0.2. Each entry of a column is the difference of the two neighbouring entries of the previous column: Δyᵢ = yᵢ₊₁ − yᵢ, Δ²yᵢ = Δyᵢ₊₁ − Δyᵢ, and so on.
Forward difference table
| x | y | Δy | Δ²y | Δ³y | Δ⁴y |
|---|---|---|---|---|---|
| 3 | 0.4771 | 0.0280 | -0.0016 | 0.0000 | 0.0003 |
| 3.2 | 0.5051 | 0.0264 | -0.0016 | 0.0003 | |
| 3.4 | 0.5315 | 0.0248 | -0.0013 | ||
| 3.6 | 0.5563 | 0.0235 | |||
| 3.8 | 0.5798 |
Working for the first row:
Δy₀ = 0.5051 − 0.4771 = 0.0280
Δ²y₀ = 0.0264 − 0.0280 = −0.0016
Δ³y₀ = −0.0016 − (−0.0016) = 0.0000
Δ⁴y₀ = 0.0003 − 0.0000 = 0.0003
The third and fourth differences are almost zero, which shows the data behave like a smooth, nearly quadratic function (these are log₁₀ x values). The first row (y₀, Δy₀, Δ²y₀, Δ³y₀, Δ⁴y₀) is what Newton's forward formula uses, and the last diagonal is what the backward formula uses.
✓ Answer checked by Claude Opus 5.5 (high effort)
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