Consider the nonlinear system: • Fcxx₂y³ = *4* xy + x² = y ²³²-1=0 g(x₂₁47= x+2y-x₁³²³² - 2 = 0 (9) For X₂=0 and 4₁ = 2 we have F₁₂(04₂₁, y₁) = Fy (x₁, 4₁) = 9x (04₁,4₁ ) = 94 (x₁, y₁) = b) The system of nonlinear equations above has a solution near the paint (O.₁, 4₁) = (0,2), Talking the initial approximations X₁₂ = 0 and 1₁=2, and applying one iteration of Newton's method, gives the improved appel approximation Here 4ə (9) (AT₁) + Ax₁ = 44₁= JG₂ = N₂ = Ya
Consider the nonlinear system: • Fcxx₂y³ = *4* xy + x² = y ²³²-1=0 g(x₂₁47= x+2y-x₁³²³² - 2 = 0 (9) For X₂=0 and 4₁ = 2 we have F₁₂(04₂₁, y₁) = Fy (x₁, 4₁) = 9x (04₁,4₁ ) = 94 (x₁, y₁) = b) The system of nonlinear equations above has a solution near the paint (O.₁, 4₁) = (0,2), Talking the initial approximations X₁₂ = 0 and 1₁=2, and applying one iteration of Newton's method, gives the improved appel approximation Here 4ə (9) (AT₁) + Ax₁ = 44₁= JG₂ = N₂ = Ya
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
![Consider the nonlinear system:
• Fcxx₂y³ = *4* xy + x² = y ²³²-1=0
g(x,y)= x+2y-x₁²²³² - 2 = 0
a) For X₂=0 and 4₁₁ = 2 we have
F₂(x₂₁, y₁) =
Fy (x₁, 4₁) =
gr (04₁,4₁ ) =
94 (X₁, Y₁ ) =
b) The system of nonlinear equations above has a solution
near the paint (XC₁₁, 4₁) = (0,2). Telking the initial
approximations X1₁=0 and 4₁=2, and applying one
iteration of Newton's method, gives the improved appel
approximation :
450
хо
X
Here
(x₂)
Y₁
n
#+
Ax₁ = A4, =
[
36₂ = Y = [
Ax
Ay](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06089eb2-ff0d-4c09-8152-ab59813eefd5%2F1d65f566-3acd-46cb-9f33-f6d2b20bd9e7%2Fx5ld4zo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the nonlinear system:
• Fcxx₂y³ = *4* xy + x² = y ²³²-1=0
g(x,y)= x+2y-x₁²²³² - 2 = 0
a) For X₂=0 and 4₁₁ = 2 we have
F₂(x₂₁, y₁) =
Fy (x₁, 4₁) =
gr (04₁,4₁ ) =
94 (X₁, Y₁ ) =
b) The system of nonlinear equations above has a solution
near the paint (XC₁₁, 4₁) = (0,2). Telking the initial
approximations X1₁=0 and 4₁=2, and applying one
iteration of Newton's method, gives the improved appel
approximation :
450
хо
X
Here
(x₂)
Y₁
n
#+
Ax₁ = A4, =
[
36₂ = Y = [
Ax
Ay
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