i need the answer for these questions
Newton’s Method for Vector-Valued Functions
Our system of equations is,
f1(x,y,z) = 0
f2(x,y,z) = 0
f3(x,y,z) = 0
f1(x,y,z) = xyz – x^2 + y^2 – 1.34
f2(x,y,z) = xy –z^2 – 0.09
f3(x,y,z) = e^x + e^y + z – 0.41
we can think of (x,y,z) as a vector x and (f1,f2,f3) as a vector-valued function f. With this notation, we can write the system of equations as,
f(x) = 0
i.e. we wish to find a vector x that makes the vector function f equal to the zero vector.
Linear Approximation for Vector Functions
In the single variable case, Newton’s method was derived by considering the linear approximation of the function f at the initial guess x0. From Calculus, the following is the linear approximation of f at x0, for vectors and vector-valued functions:
f(x) ≈ f(x0) + Df(x0)(x − x0).
Here Df(x0) is a 3 × 3 matrix whose entries are the various partial derivative of the components of f. Speciﬁcally,
∂f1/ ∂x (x0)
∂f1/ ∂y (x0)
∂f1/ ∂z (x0)
∂f2/ ∂x (x0)
∂f2/ ∂y (x0)
∂f2/ ∂z (x0)
∂f3/ ∂x (x0)
∂f3/ ∂y (x0)
∂f3/ ∂z (x0)
We wish to find x that makes f equal to the zero vector, so let’s choose x = x1 so that
f(x0) + Df(x0)(x1 − x0) = f(x) = 0.
Since Df(x0)) is a square matrix, we can solve this equation by
x1 = x0 − (Df(x0))^(−1)f(x0),
provided that the inverse exists. The formula is the vector equivalent of the Newton’s method formula for single variable functions. However, in practice we never use the inverse of a matrix for computations, so we cannot use this formula directly. Rather, we can do the following. First solve the equation
Df(x0)∆x = −f(x0)
Since Df(x0) is a known matrix and −f(x0) is a known vector, this equation is just a system of linear equations, which can be solved efficiently and accurately. Once we have the solution vector ∆x, we can obtain our improved estimate x1 by x1 = x0 + ∆x.
For subsequent steps, we have the following process:
• Solve Df(xi)∆x = −f(xi).
• Let xi+1 = xi + ∆x Read More: ...