Multivariable Calculus (1st order approximation)

[plzen90]

Homework Statement

Define f:R²→R³ by
f(x,y,z)=(xy+z)
...(x²-yz)

let p = (1,1,1)^T and h=(δ,ε,θ)

a)what are n and m? evaluate f(p) and f(p+h)
b)Calculate the Jacobian Matrix Df(x,y,z) and evaluate Df(p)
c) Calculate the error e(h) in the first order approximation to f(p+h)
d) show clearly that

lim h→0 |e(h)| =0
......|h|


Explain why this is what you expect

Homework Equations

The Attempt at a Solution

a)
n=3
m=2

f(p)=(1,1,1) = (2)
......(0)

f(p+h) = f(1+δ, 1+ε, 1+θ)

=(2+δε+δ+ε+θ)
(δ²+2δ-εθ-ε-θ)

b)
jac= (y x 1)
...(2x -z -y)

Df(p)=(1 1 1)
...(2 -1 -1)

c)
f(p+h)≈f(p)+Df(p)h

only calculation of Df(p)h needed to work out error.

=(y+x+1)h
(2x-z-y)

=(ε+δ+1)
(2δ-θ-ε)

e(h)=f(p+h)-(f(p) + Df(p)h)

=(δε+θ-ε-1)
(δ²-εθ)

(not confident on this)

d)not attempted yet/ don't know how to

 

[mighty2000]

show it clearly.

The first part of your solution looks correct. The Jacobian matrix is a 3x2 matrix, and the derivative at point p is a 1x2 matrix.

For part c), you are correct in saying that the error is equal to f(p+h) - (f(p) + Df(p)h). However, the calculation of Df(p)h is incorrect. It should be:

Df(p)h = (y+x+1)δ + (2x-z-y)ε
(y+x+1)ε + (2x-z-y)θ

Substituting in the values for p and h, we get:

Df(p)h = (2+1)δ + (2-1-1)ε
(2+1)ε + (2-1-1)θ

= 3δ + 0ε
3ε + 0θ

= (3δ, 3ε)

Therefore, the error e(h) becomes:

e(h) = (δε+θ-ε)
(δ2-εθ)

- (2+0+0)
(0+0-0)

= (δε+θ-ε-2)
(δ2-εθ)

For part d), you can show that the limit as h goes to 0 of e(h)/|h| is equal to 0 by simplifying the expression:

lim h→0 |e(h)|/|h| = lim h→0 ((δε+θ-ε-2)^2 + (δ2-εθ)^2)/(|δ|+|ε|+|θ|)

As h goes to 0, all the terms with δ, ε, and θ go to 0 as well. Therefore, the numerator becomes 0, and the denominator becomes 0+0+0 = 0. This means that the limit is equal to 0/0, which is an indeterminate form. To evaluate this limit, we can use L'Hopital's rule, which states that if we have a limit of the form 0/0, we can take the derivative of the numerator and denominator separately and evaluate the limit again. Applying this rule, we get:

lim h→0 |e(h)|/|h| = lim h→0 (2