Is This Function a Contraction Map?

[simo1]

how do i prove that this function is a contraction map?

f(x)=⟨(1/9)cos(x1+sin(x2)),(1/6)arctan(x1+x2)⟩;

x1=⟨0,−1⟩.i wantd to use the matrix form of the jacobian
i said
x(1)' = -1÷9sin(x_1 +sinx_2) ∙ (x'_1 + cos_2)

x(2)' = 1÷(6(1+ (x_1 + x_2)^2) ∙ x'_1 + x'_2

I don't know how to put this function in a matrix form and move on(Sadface)

 

[jvicens]

Firstly, to prove that a function is a contraction map, we need to show that it satisfies the definition of a contraction map. A contraction map is a function that maps elements from a metric space to itself, and it must satisfy the condition that the distance between two images of any two points in the metric space is always less than the distance between the original points.

In this case, the metric space is R^2, and the distance between two points (x1, x2) and (y1, y2) is given by d((x1, x2), (y1, y2)) = √((x1-y1)^2 + (x2-y2)^2).

Now, to show that the given function is a contraction map, we need to prove that for any two points (x1, x2) and (y1, y2), the distance between their images under the function f is always less than the distance between the original points.

Let's consider the points x = (0, -1) and y = (y1, y2). The distance between these points is given by d(x, y) = √(y1^2 + (y2+1)^2).

Now, let's find the images of these points under the given function f:

f(x) = ((1/9)cos(0+sin(-1)), (1/6)arctan(0+(-1))) = ((1/9)cos(-1), (1/6)arctan(-1)) = (0.123, -0.165)

f(y) = ((1/9)cos(y1+sin(y2)), (1/6)arctan(y1+y2))

The distance between these two images is given by d(f(x), f(y)) = √((0.123-y1)^2 + (-0.165-(y2+1))^2)

Now, let's simplify this expression using the triangle inequality:

d(f(x), f(y)) = √((0.123-y1)^2 + (-0.165-(y2+1))^2)
≤ √((y1^2 + (y2+1)^2) + (0.123-y1)^2 + (-0.165-(y2+1))^2)
= √(2y1^2 +