numerical

1. proof of Simpson’s rule∗
drini†
2013-03-21 18:20:43
We want to derive Simpson’s rule for
b
a
f(x) dx.
We will use Newton and Cotes formulas for n = 2. In this case, x0 = a,
x2 = b and x1 = (a + b)/2. We use Lagrange’s interpolation formula to get a
polynomial p(x) such that p(xj) = f(xj) for j = 0, 1, 2.
The corresponding interpolating polynomial is
p(x) = f(x1)
(x − x2)(x − x3)
(x1 − x2)(x1 − x3)
+f(x2)
(x − x1)(x − x3)
(x2 − x1)(x2 − x3)
+f(x3)
(x − x1)(x − x2)
(x3 − x1)(x3 − x2)
.
and thus
b
a
f(x) dx ≈
b
a
f(x1)
(x − x2)(x − x3)
(x1 − x2)(x1 − x3)
+f(x2)
(x − x1)(x − x3)
(x2 − x1)(x2 − x3)
+f(x3)
(x − x1)(x − x2)
(x3 − x1)(x3 − x2)
dx.
Since integration is linear, we are concerned only with integrating each term
in the sum. Now, taking xj = a + hj where j = 0, 1, 2 and h = |b − a|/2, we can
rewrite the quotients on the last integral as
b
a
p(x) dx = hf(x0)
2
0
(t − 1)(t − 2)
(0 − 1)(0 − 2)
dt+hf(x1)
2
0
(t − 0)(t − 2)
(1 − 0)(1 − 2)
dt+hf(x2)
2
0
(t − 0)(t − 1)
(2 − 0)(2 − 1)
dt.
and if we calculate the integrals on the last expression we get
b
a
p(x) dx = hf(x0)
1
3
+ hf(x1)
4
3
+ hf(x2)
1
3
,
which is Simpson’s rule:
b
a
f(x) dx ≈
h
3
(f(x0) + 4f(x1) + f(x2)).
∗ ProofOfSimpsonsRule created: 2013-03-21 by: drini version: 36509 Privacy
setting: 1 Proof 65D32 41A55 26A06 28-00
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