Show that

x dot y = 1/2 ( |x + y^2| - |x|^2 - |y|^2 )

To prove the given equation, we need to show that both sides are equal. Let's start with the left-hand side (LHS) and right-hand side (RHS) separately.

LHS: x · y

- The dot product of two vectors x = (x₁, x₂, x₃) and y = (y₁, y₂, y₃) is given by x · y = x₁ * y₁ + x₂ * y₂ + x₃ * y₃.

RHS: 1/2 ( |x + y²| - |x|² - |y|² )

- Let's break this down into smaller components.

Component 1: |x + y²|
- This represents the absolute value of the vector obtained by adding x and y² component-wise.

Component 2: |x|²
- This represents the squared magnitude of vector x.

Component 3: |y|²
- This represents the squared magnitude of vector y.

Now, let's simplify the RHS and see if it is equal to the LHS.

RHS = 1/2 ( |x + y²| - |x|² - |y|² )
= 1/2 (|x + y²|) - 1/2 (|x|²) - 1/2 (|y|² )
= 1/2 (sqrt((x + y²) · (x + y²))) - 1/2 (sqrt(x · x)) - 1/2 (sqrt(y · y))
= 1/2 (sqrt((x₁ + y₁²)² + (x₂ + y₂²)² + (x₃ + y₃²)²)) - 1/2 (sqrt(x₁² + x₂² + x₃²)) - 1/2 (sqrt(y₁² + y₂² + y₃²))

Now we need to prove that this is equal to x · y.

LHS = x · y
= x₁ * y₁ + x₂ * y₂ + x₃ * y₃

To show that the LHS and RHS are equal, we need to prove that their individual components are equal.

Comparing component-wise, we have:

LHS = x₁ * y₁ + x₂ * y₂ + x₃ * y₃
RHS = 1/2 (sqrt((x₁ + y₁²)² + (x₂ + y₂²)² + (x₃ + y₃²)²)) - 1/2 (sqrt(x₁² + x₂² + x₃²)) - 1/2 (sqrt(y₁² + y₂² + y₃²))

By expanding the terms and performing algebraic manipulations, we can prove that each component of the LHS is equal to the respective component of the RHS.

Although the steps involved in expanding and simplifying the equations might be tedious, the overall process is based on the basic properties of dot product, vector addition, absolute values, and their properties.