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Question
Find f(0.5) if f(– 1) = 202, f(0) = 175, f(1) = 82 and f(2) = 55
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Solution
From the given data
| x | – 1 | 0 | 1 | 2 |
| f(x) | 202 | 175 | 82 | 55 |
Here we have to apply Newton’s forward interpolation formula,
Since the value of f(x) is required near the beginning of the table.
`y_((x = x_0 + "nh")) = "f"(x_0) + "n"/(1!) Delta"f"(x_0) + ("n"("n" - 1))/(2!) Delta^2"f"(x_0) + ("n"("n" - 1)("n" - 2))/(3!) Delta^3"f"(x_0) + .......`
Given:
x = 0.5 and h = 1
x0 + nh = x
– 1 + n(1) = 0.5
n = 1 + 0.5
∴ n = 1.5
| x | f(x) | `Delta"f"(x)` | `Delta^2"f"(x)` | `Delta^3"f"(x)` |
| – 1 | 202 | |||
| – 27 | ||||
| 0 | 175 | – 66 | ||
| – 93 | 132 | |||
| 1 | 82 | 66 | ||
| – 27 | ||||
| 2 | 55 |
`"y"_((0.5)) = 202 + 1.5/(1!) (-27) + ((1.5)(1.5 - 1))/(2!) (- 66) + ((1.5)(1.5 - 1)(1.5 - 2))/(3!) (132)`
= `202 + 1.5(- 27) + ((1.5)(0.5)(- 66))/2 + ((1.5)(0.5)(- 0.5)(132))/6`
= 202 – 40.5 – 24.75 – 8.25
= 202 – 73.5
f(0.5) = 128.5
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