Binomial calibration of a sensor
A calibration correction for a pressure sensor is modelled by (1 + kx)^n for small normalized error x, with k and n positive constants and n an integer. Bench tests give the coefficient of x as 10 and the coefficient of x² as 40. A laboratory assistant assumes n = 10 and k = 1 because the first coefficient is 10, but the second coefficient then does not match. The manufacturer needs the exact integer n and constant k, the coefficient of x³, and a numerical check at x = 0.02. The acceptance rule is that the correction estimate may differ from the exact model by at most 0.001. The engineer asks whether a linear truncation is acceptable and whether a quadratic truncation passes; all coefficients must be justified from the test data, not guessed.
1. Determine the exact integer n and constant k, and find the coefficient of x³.
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The x coefficient is nk = 10. The x² coefficient is C(n,2)k² = n(n-1)k²/2 = 40. Since nk=10, n(n-1)k² = (nk)((n-1)k) = 10((n-1)k), so 10((n-1)k)/2 = 5k(n-1) = 40, giving k(n-1)=8. Subtracting from nk=10 gives k=2, hence n=5. The expansion is (1+2x)^5 = 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵. The x³ coefficient is 80. The assistant's n=10, k=1 would give x² coefficient C(10,2)=45, not 40.
2. At x = 0.02, compare the linear and quadratic truncations with the exact value. Decide which truncation meets the 0.001 tolerance.
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At x=0.02, exact = (1.04)^5 = 1.2166529024. Linear truncation = 1 + 10(0.02) = 1.2; error = 0.01665 > 0.001, so fails. Quadratic truncation = 1 + 10(0.02) + 40(0.02²) = 1 + 0.2 + 0.016 = 1.216; error = 1.2166529024 - 1.216 = 0.0006529024 < 0.001, so passes. The quadratic estimate is acceptable under the stated rule; the linear one is not.