Using Newton’s method, solve the system is

xyz-x^2+y^2=1.34

x y-z^2=0.09

e^x+e^y+z=0.41

by newton method to obtain a solution near (1,1,1) use three iteration only

recategorized

Part 1

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

with,

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) Df(x0) = ∂f2/ ∂x (x0) ∂f2/ ∂y (x0) ∂f2/ ∂z (x0) ∂f3/ ∂x (x0) ∂f3/ ∂y (x0) ∂f3/ ∂z (x0)

Newton’s Method

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)(x1x0) = 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

by Level 11 User (81.5k points)
edited by

Part 2

Using our equations, we have

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

and

 ∂f1/ ∂x (x) ∂f1/ ∂y (x) ∂f1/ ∂z (x) Df(x) = ∂f2/ ∂x (x) ∂f2/ ∂y (x) ∂f2/ ∂z (x) ∂f3/ ∂x (x) ∂f3/ ∂y (x) ∂f3/ ∂z (x)

i.e.

 yz – 2x xz + 2y xy Df(x) = y x -2z e^x e^y 1

1st Iteration

For x0 = [x0,y0,z0] = [1,1,1],

 – 1 3 1 Df(x0) = 1 1 -2 e e 1

f1 = -0.34

f2 = – 0.09

f3 = 2e + 0.59

the system of equations, in matrix format, is Df(x0)∆x = −f(x0).

 – 1 3 1 ∆x = 0.34 1 1 -2 ∆y = 0.09 e e 1 ∆z = -2e - 0.59

or,

-∆x + 3∆y + ∆z = 0.34

∆x + ∆y – 2∆z = 0.09

2.7183∆x + 2.7183∆y + ∆z = -6.0266

Solving this system of equations gives us,

∆x = -1.72254, ∆y = -0.13608, ∆z = -0.97431

From which,

x1 = x0 + ∆x, y1 = y0 + ∆y, z1 = z0 + ∆z,

x1 = 1 - 1.72254, y1 = 1 - 0.13608, z1 = 1 - 0.97431,

x1 = -0.72254, y1 = 0.86392, z1 = 0.02569,

x1 = [-0.72254, 0.86392, 0.02569]

Part 3

2nd Iteration

For x1 = [x1,y1,z1] = [-0.72254, 0.86392, 0.02569]

 1.46728 1.70928 -0.624221 Df(x0) = 0.863922 -0.722543 -0.0513797 0.485516 2.37245 1

f1 = -1.131742062

f2 = - 0.7148808356

f3 = 2.473654395

the system of equations, in matrix format, is Df(x0)∆x = −f(x0).

 1.46728 1.70928 -0.624221 ∆x = 1.13174 0.863922 -0.722543 -0.0513797 ∆y = 0.714881 0.485516 2.37245 1 ∆z = -2.47365

or,

1.467279449∆x + 1.709282902∆y – 0.6242208∆z = 1.131742062

0.8639224733∆x – 0.72254268∆y – 0.0513797∆z = 0.7148808356

0.4855161696∆x + 2.372448332∆y + ∆z = -2.473654395

Solving this system of equations gives us,

∆x = 0.4174047065, ∆y = -0.360889953, ∆z = -1.820118362

From which,

x2 = x1 + ∆x, y2 = y1 + ∆y, z2 = z1 + ∆z,

x2 = -0.72254 + 0.4174047, y1 = 0.86392 - 0.36089, z2 = 0.02569 - 1.82012,

x2 = -0.305135, y2 = 0.50303, z2 = -1.79443,

x2 = [-0.305135, 0.50303, -1.79443]

3rd Iteration

For x2 = [x2,y2,z2] = [-0.305135, y2 = 0.50303, z2 = -1.79443]

 -0.29238 1.55361 -0.153494 Df(x0) = 0.503033 -0.305138 3.58886 0.737022 1.65373 1

f1 = -0.9046327596

f2 = -3.463468076

f3 = 0. 186321774

the system of equations, in matrix format, is Df(x0)∆x = −f(x0).

 -0.29238 1.55361 -0.153494 ∆x = 0.904633 0.503033 -0.305138 3.58886 ∆y = 3.46347 0.737022 1.65373 1 ∆z = -0.186322

or,

-0.292379716∆x + 1.553613689∆y – 0.1534944∆z = 0.9046327596

0.5030325661∆x – 0.30513813∆y + 3.58885701∆z = 3.463468076

0.737021563∆x + 1.653728716∆y + ∆z = -0.186321774

Solving this system of equations gives us,

∆x = -2.592141381, ∆y = 0.2276073932, ∆z = 1.347741436

From which,

x3 = x2 + ∆x, y3 = y2 + ∆y, z3 = z2 + ∆z,

x3 = -0.305135 - 2.592141381, y3 = 0.50303 + 0.2276073932, z3 = -1.79443 + 1.347741436,

x3 = -2.897279, y3 = 0.730637, z3 = -0.446688,

x3 = [-2.897279, 0.730637, -0.446688]

This 3d iteration is far from accurate, although it is converging. 10 iterations will give 4 d.p. accuracy with the solution:

X10 = [-0.3287, -1.3027. -0.5816]