(c) We want to find a three-point formula, [*^* ƒ (x)dx = w₁ f(x1) + w2f (x2) + wzf(xx) + E, -h similar to the two-point formula in part ((b)). X1 < x2 < X3, (3) Definite integrals of f(x) and f(x) on a symmetric interval should be equal, i.e. L f(x)dx = L f(x)dx. -h (4) The quadrature rule should also possess such a property of integrals. Using this ar- gument of symmetry, show that x2 = 0 and find relations between x₁ and x3; and between w₁ and w3. Then, calculate the weights and points that make this rule ex- act for polynomials of the highest degree possible. Find the error term and write the quadrature formula (3).

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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(c) We want to find a three-point formula,
[*^* ƒ (x)dx = w₁ f(x1) + w2f (x2) + wzf(xx) + E,
-h
similar to the two-point formula in part ((b)).
X1 < x2 < X3,
(3)
Definite integrals of f(x) and f(x) on a symmetric interval should be equal, i.e.
L
f(x)dx =
L
f(x)dx.
-h
(4)
The quadrature rule should also possess such a property of integrals. Using this ar-
gument of symmetry, show that x2 = 0 and find relations between x₁ and x3; and
between w₁ and w3. Then, calculate the weights and points that make this rule ex-
act for polynomials of the highest degree possible. Find the error term and write the
quadrature formula (3).
Transcribed Image Text:(c) We want to find a three-point formula, [*^* ƒ (x)dx = w₁ f(x1) + w2f (x2) + wzf(xx) + E, -h similar to the two-point formula in part ((b)). X1 < x2 < X3, (3) Definite integrals of f(x) and f(x) on a symmetric interval should be equal, i.e. L f(x)dx = L f(x)dx. -h (4) The quadrature rule should also possess such a property of integrals. Using this ar- gument of symmetry, show that x2 = 0 and find relations between x₁ and x3; and between w₁ and w3. Then, calculate the weights and points that make this rule ex- act for polynomials of the highest degree possible. Find the error term and write the quadrature formula (3).
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