- 2069 Bhadra · 8 marks
Fit y = a e^(bx)
Short answer
Linearize via ln y = ln a + bx. Least squares on the transformed data gives A=1.0301, B=0.4613, so a=2.8015, b=0.4613. The curve is y = 2.8015 e^(0.4613x).
Fit the following set of data to a curve of the form y = a e^(bx) from the following observation by least square method.
| x | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| y | 5.5 | 6.5 | 9.4 | 15.2 | 30.6 | 49.8 |
Take natural logarithms: y = a·e^(bx) → ln y = ln a + b·x. Put Y = ln y, A = ln a, so Y = A + bx is a straight line.
| x | y | Y = ln y | x² | xY |
|---|---|---|---|---|
| 1 | 5.5 | 1.7047 | 1.0000 | 1.7047 |
| 2 | 6.5 | 1.8718 | 4.0000 | 3.7436 |
| 3 | 9.4 | 2.2407 | 9.0000 | 6.7221 |
| 4 | 15.2 | 2.7213 | 16.0000 | 10.8852 |
| 5 | 30.6 | 3.4210 | 25.0000 | 17.1050 |
| 6 | 49.8 | 3.9080 | 36.0000 | 23.4481 |
| Σx = 21 | 15.8676 | 91.0000 | 63.6088 |
Normal equations (b = B)
ΣY = n·A + B·Σx → 15.8676 = 6A + 21.0000B
ΣxY = A·Σx + B·Σx² → 63.6088 = 21.0000A + 91.0000B
Solving: B = (nΣxY − ΣxΣY)/(nΣx² − (Σx)²) = 0.4613, A = (ΣY − BΣx)/n = 1.0301.
a = e^A = 2.8015, b = 0.4613.
Answer: y = 2.8015 e^(0.4613x) (at x = 6 it gives 44.60 against 49.8).
✓ Answer checked by Claude Opus 5.5 (high effort)
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