What is the weight w₁ of the Newton-Cotes formula with nodes O a. W₁=1/3 O b. w₁ 1/2 O c. W₁=1/√3 O d. W₁=1 Oe. W₁=√3 Of. W₁=2 Og. W₁=3 Q[ƒ; −1, 1] = woƒ(-1) + w₁ƒ(√√ ;-) + w₂f(1) Xo = -1, X₁ = 3 x₂ = 1?
What is the weight w₁ of the Newton-Cotes formula with nodes O a. W₁=1/3 O b. w₁ 1/2 O c. W₁=1/√3 O d. W₁=1 Oe. W₁=√3 Of. W₁=2 Og. W₁=3 Q[ƒ; −1, 1] = woƒ(-1) + w₁ƒ(√√ ;-) + w₂f(1) Xo = -1, X₁ = 3 x₂ = 1?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![What is the weight w₁ of the Newton-Cotes formula
with nodes
a. w₁=1/3
O b. w₁=1/2
O c. W₁=1/√3
O d. W₁=1
e. W₁=√3
Of. W₁=2
g. W₁=3
Q[ƒ; −1, 1] = wof(−1) + w₁ƒ(√√3/3) + + w₂f(1)
Xo = -1,
x₁ =
- اس
X₂ = 1?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6fe1fc35-672a-49fd-831b-9642c77888ed%2Fdd8194b2-7a58-41a5-af38-09d8bcfb4301%2Fvxmsfbt_processed.png&w=3840&q=75)
Transcribed Image Text:What is the weight w₁ of the Newton-Cotes formula
with nodes
a. w₁=1/3
O b. w₁=1/2
O c. W₁=1/√3
O d. W₁=1
e. W₁=√3
Of. W₁=2
g. W₁=3
Q[ƒ; −1, 1] = wof(−1) + w₁ƒ(√√3/3) + + w₂f(1)
Xo = -1,
x₁ =
- اس
X₂ = 1?
Expert Solution
Step 1 formula
The newton cotes formula is valid for all polynomials up-to degree n if the n=(N-1) is odd and
it is n+1 if n=(N-1) is even. Where N is the number of nodes.
Here number of nodes N=3 .
So n=N-1=2 , even number.
Therefore the formula is valid for polynomials of degree at-most 3.
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