- 2073 Bhadra · 8 marks
Stirling formula: u28
Short answer
For x = 28, take x0 = 30 and h = 5 so p = -0.4. Stirling's central difference formula using differences around x0 gives u28 = 47691.8256, approximately 47691.83.
Using Stirling formula find u28, given; u20 = 49225, u25 = 48316, u30 = 47236, u35 = 45926, u40 = 44306
x = 28 lies near the middle of the table, so Stirling's central difference formula is suitable. Take x₀ = 30, h = 5, so p = (28 − 30)/5 = -0.4.
Difference table
| x | y | Δy | Δ²y | Δ³y | Δ⁴y |
|---|---|---|---|---|---|
| 20 | 49225 | -909 | -171 | -59 | -21 |
| 25 | 48316 | -1080 | -230 | -80 | |
| 30 | 47236 | -1310 | -310 | ||
| 35 | 45926 | -1620 | |||
| 40 | 44306 |
Central differences about x₀ = 30: Δu₋₁ = −1080, Δu₀ = −1310; Δ²u₋₁ = −230; Δ³u₋₂ = −59, Δ³u₋₁ = −80; Δ⁴u₋₂ = −21.
Stirling's formula:
u(x₀ + ph) = u₀ + p·(Δu₋₁ + Δu₀)/2 + p²/2!·Δ²u₋₁ + p(p²−1)/3!·(Δ³u₋₂ + Δ³u₋₁)/2 + p²(p²−1)/4!·Δ⁴u₋₂
| Term | Coefficient | Difference | Value |
|---|---|---|---|
| u₀ | 1 | 47236 | 47236.0000 |
| p·(Δu₋₁+Δu₀)/2 | -0.4000 | -1195.0 | 478.0000 |
| p²/2!·Δ²u₋₁ | 0.0800 | -230 | -18.4000 |
| p(p²−1)/3!·(Δ³u₋₂+Δ³u₋₁)/2 | 0.0560 | -69.5 | -3.8920 |
| p²(p²−1)/4!·Δ⁴u₋₂ | -0.0056 | -21 | 0.1176 |
u₂₈ = 47691.8256
Answer: u₂₈ ≈ 47691.83
✓ Answer checked by Claude Opus 5.5 (high effort)
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