Mathematics - Part 2

I have summarized information about the $n$ -dimensional Euclidean space $\mathbb{R}^n$ and its distance below.

(n)-Dimensional Euclidean Space $\mathbb{R}^n$

When we define distance as shown below for the direct product set $\mathbb{R}^n$ , we call $\mathbb{R}^n$ an $n$ -dimensional Euclidean space. The elements of $\mathbb{R}^n$ are referred to as points.

Distance

For elements $x = (x_1, \cdots , x_n)$ and $y = (y_1, \cdots , y_n)$ in $\mathbb{R}^n$ (where $x_i, y_i \in \mathbb{R} \; (i = 1, \cdots , n)$ ), their distance $d(x, y)$ is defined by the following equation:

$d(x, y) = \sqrt{\sum _{i=1}^n (x_i – y_i)^2 }$

In 1-dimensional space, $d(x, y) = | x – y |$ , and in 2-dimensional space, $d(x,y) = \sqrt{(x_1-y_1)^2 + (x_2-y_2)^2}$ .

Properties of Distance

The distance in $\mathbb{R}^n$ has the following properties:

For any $x, y \in \mathbb{R}^n$ , $d(x, y)$ is a non-negative real number.
For $x, y \in \mathbb{R}^n$ , $d(x, y) = 0$ if and only if $x = y$ .
For any $x, y \in \mathbb{R}^n$ , $d(x, y) = d(y, x)$ .
For any $x, y, z \in \mathbb{R}^n$ , $d(x, z) \leq d(x, y) + d(y, z)$ (Triangle Inequality).

The first three properties are evident from the definition of the distance $d(x, y)$ .

Let’s prove the last property, the Triangle Inequality.

Proof

Let $x = ( x_1, \cdots , x_n)$ , $y = ( y_1 , \cdots , y_n )$ , and $z = (z_1 , \cdots , z_n )$ . Define $a_i = x_i – y_i$ and $b_i = y_i – z_i$ . Then, $d(x, z) \leq d(x, y) + d(y, z)$ can be expressed as follows:

$\sqrt{\sum _{i=1}^{n} \left( a_i + b_i \right) ^2} \leq \sqrt{\sum _{i=1}^{n} a_i ^2} + \sqrt{\sum _{i=1}^{n} b_i ^2}$

This equation is equivalent to squaring both sides, eliminating the terms $\sum a_i^2$ and $\sum b_i ^2$ , and dividing both sides by 2:

$\sum _{i=1}^{n} a_i b_i \leq \sqrt{ \left( \sum _{i=1}^n a_i^2 \right) \left( \sum _{i=1}^n b_i^2 \right) }$

Furthermore, this proof is sufficient if we can establish that the squared inequality below is valid:

$\left( \sum _{i=1}^{n} a_i b_i \right)^2 \leq \left( \sum _{i=1}^n a_i^2 \right) \left( \sum _{i=1}^n b_i^2 \right)$

This last inequality is known as the Schwarz(シュワルツ) inequality, and various proofs are known. When $n = 1$ , equality holds.

Solution 1

Let’s consider the case where $n$ is greater than or equal to 2. We calculate by subtracting the right side from the left side:

$\begin{eqnarray} & & \left( \sum _{i=1}^n a_i^2 \right) \left( \sum _{i=1}^n b_i^2 \right) – \left( \sum _{i=1}^{n} a_i b_i \right)^2 \\ & = & \sum _{\substack{1 \leq i \leq n , 1 \leq j \leq n , i \neq j}} a_i ^2 b_j ^2 – 2 \sum _{1 \leq i \lt j \leq n} a_i b_i a_j b_j \\ & = & \sum _{1 \leq i \lt j \leq n} a_i ^2 b_j ^2 + \sum _{1 \leq i lt j \leq n} a_j ^2 b_i ^2 – 2 \sum _{1 \leq i \lt j \leq n} a_i b_i a_j b_j \\ & = & \sum _{1 \leq i \lt j \leq n} \left( a_i b_j – a_j b_i \right) ^2 \end{eqnarray}$

Since it is the sum of squares, it is clear that it is non-negative.

The condition for equality to hold is that $a_i b_j – a_j b_i = 0$ holds for all distinct $i, j$ .

Consideration of the Condition for Equality

The equation $a_i b_j – a_j b_i = 0$ means that when considering 2-dimensional vector spaces $\mathbb{R}^2$ , $a = (a_i, a_j)$ and $b = (b_i, b_j)$ , one is a scalar multiple of the other.

This implies that three points $x, y, z$ lie on the same line. Using vector operations, we can write $c(x – y) = d(y – z)$ , where $c$ and $d$ are scalar values.

Let’s solve the equation $a_i b_j – a_j b_i = 0$ step by step.

Case $a_i = 0$

This leads to $a_j = 0$ or $b_i = 0$ .

If $b_i \neq 0$ , then for all $j$ , $a_j = 0$ , which means \( x = y \

If $b_i = 0$ , then $a_i = b_i$ can be written.

Case $a_i \neq 0$

Let $c = \frac{b_i}{a_i}$ .

If $a_j \neq 0$ , then $\frac{b_i}{a_i} = \frac{b_j}{a_j} = c$ . For all $i$ where $a_i \neq 0$ , we have \( c a_i = b_i \ ).

If $a_j = 0$ , then $b_j = 0$ , and $c a_j = b_j$ holds.

It is now clear that the equality holds when $x, y, z$ are collinear. This also includes the case when $x = y$ .

Solution 2

When $x = y$ , all $a_i = 0$ , and the inequality holds (equality).

For the case $x \neq y$ , consider $t$ as a real variable and the quadratic function $f(t)$ :

$\begin{eqnarray} f(t) & = & \sum _{i=1}^n \left( a_i t + b_i \right) ^2 \\ & = & \left( \sum _{i=1}^n a_i ^2 \right) t^2 + 2 \left( \sum _{i=1}^n a_i b_i \right) t + \left( \sum _{i=1}^n b_i ^2 \right) \end{eqnarray}$

From the definition of $f(t)$ (the first equation), it is clear that $f(t) \geq 0$ for any $t$ . This implies that the discriminant is non-negative, leading to the following inequality:

$\left( \sum _{i=1}^n a_i b_i \right)^2 – \left( \sum _{i=1}^n a_i ^2 \right) \left( \sum _{i=1}^n b_i ^2 \right) \geq 0$

By rearranging, it becomes the Schwarz inequality.

Consideration of the Condition for Equality

The condition for equality is when $f(t) = 0$ has a double root. Let \( t = t_0 \ ) be that root.

$0 = \sum _{i=1}^n \left( a_i t_0 + b_i \right) ^2$

At that moment, for all $i$ , $a_i t_0 + b_i = 0$ . This implies that $x, y, z$ are vectors and $– t_0 (x-y) = (y – z)$ , meaning that three points $x, y, z$ are collinear (including the case $x = y$ ).

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