[2] [Derivation of Simpon's Rule] In this exercise will derive Simpson's rule using Taylor's theorem. Suppose ƒ € C¹¹ ([a, b]) and consider three equispaced points {xo, x₁, x2} with xo = a, x₁ = a + h and £₂ b, where h = = (b-a)/2. (a) Use Taylor's theorem to write f(x) as a third order polynomial plus a remainder term using the point of expansion ₁. (b) Insert the expression for f(x) from (a) into the expression ² f(x)dx and explicitly compute the integral of each term in the Taylor polynomial to derive Cx2 x2 [*"* ƒ (x) dx = 2hƒ(x1) + ²/²_ƒ″(x1) + 12/17 [** ƒ(¹) ({(x)) (x − x1) ª da 3 24 XO (hint: two of the integrals will vanish identically because of parity, i.e. be- cause they are antisymmetric functions) (c) Justify that the Weighted Mean Value Theorem can be used for the remainder term. Then, apply the theorem to obtain: x2 1 27/4 * * ƒ (¹) (§ (x)) (x − x₁) ªdx - хо f(4) (c) h5 60
[2] [Derivation of Simpon's Rule] In this exercise will derive Simpson's rule using Taylor's theorem. Suppose ƒ € C¹¹ ([a, b]) and consider three equispaced points {xo, x₁, x2} with xo = a, x₁ = a + h and £₂ b, where h = = (b-a)/2. (a) Use Taylor's theorem to write f(x) as a third order polynomial plus a remainder term using the point of expansion ₁. (b) Insert the expression for f(x) from (a) into the expression ² f(x)dx and explicitly compute the integral of each term in the Taylor polynomial to derive Cx2 x2 [*"* ƒ (x) dx = 2hƒ(x1) + ²/²_ƒ″(x1) + 12/17 [** ƒ(¹) ({(x)) (x − x1) ª da 3 24 XO (hint: two of the integrals will vanish identically because of parity, i.e. be- cause they are antisymmetric functions) (c) Justify that the Weighted Mean Value Theorem can be used for the remainder term. Then, apply the theorem to obtain: x2 1 27/4 * * ƒ (¹) (§ (x)) (x − x₁) ªdx - хо f(4) (c) h5 60
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Taykor
![[2] [Derivation of Simpon's Rule]
In this exercise will derive Simpson's rule using Taylor's theorem. Suppose
ƒ € C¹([a, b]) and consider three equispaced points {xo, x₁, x2} with x。 = a,
X1 = a + h and x₂ = b, where h = (b − a)/2.
(a) Use Taylor's theorem to write f(x) as a third order polynomial plus
a remainder term using the point of expansion ₁.
(b) Insert the expression for f(x) from (a) into the expression f(x)dx
and explicitly compute the integral of each term in the Taylor polynomial to
derive
Cx2
X2
["^ f(x)dx = 2hf(x) + ²/³f²(xs) + 21 [ * * ƒ^9 (E (x)} (x − xi) ºdz
(x₁)
ƒ(4) (§(x)) x₁) ¹dx
-
3
24
XO
so
(hint: two of the integrals will vanish identically because of parity, i.e. be-
cause they are antisymmetric functions)
(c) Justify that the Weighted Mean Value Theorem
can be used for the remainder term. Then, apply the theorem to obtain:
Cx2
1/4 * * =
ƒ(¹) (§(x)) (x − x₁)¹ dx
24
for some c = (a, b).
ƒ(4) (c) h³
5
60](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7fd47556-f3ce-4f39-818d-be563d9523c8%2F497fb1fa-642c-4dd3-a0cc-7d8959cea334%2Ffnxfqvi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[2] [Derivation of Simpon's Rule]
In this exercise will derive Simpson's rule using Taylor's theorem. Suppose
ƒ € C¹([a, b]) and consider three equispaced points {xo, x₁, x2} with x。 = a,
X1 = a + h and x₂ = b, where h = (b − a)/2.
(a) Use Taylor's theorem to write f(x) as a third order polynomial plus
a remainder term using the point of expansion ₁.
(b) Insert the expression for f(x) from (a) into the expression f(x)dx
and explicitly compute the integral of each term in the Taylor polynomial to
derive
Cx2
X2
["^ f(x)dx = 2hf(x) + ²/³f²(xs) + 21 [ * * ƒ^9 (E (x)} (x − xi) ºdz
(x₁)
ƒ(4) (§(x)) x₁) ¹dx
-
3
24
XO
so
(hint: two of the integrals will vanish identically because of parity, i.e. be-
cause they are antisymmetric functions)
(c) Justify that the Weighted Mean Value Theorem
can be used for the remainder term. Then, apply the theorem to obtain:
Cx2
1/4 * * =
ƒ(¹) (§(x)) (x − x₁)¹ dx
24
for some c = (a, b).
ƒ(4) (c) h³
5
60
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