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Category Archives: Mathematics
(日本語) 二項分布とその平均・分散
Example of Geometric Distribution, Negative Binomial Distribution, and Central Limit Theorem.
Let’s introduce examples of the geometric distribution and the negative binomial distribution.
Continue reading Example of Geometric Distribution, Negative Binomial Distribution, and Central Limit Theorem.Proof: Triangle Inequality
Let’s prove \(|PR| \leq |PQ| + |QR| \) for the destances among 3 points \(P, Q, R \in \mathbb{R}^n\).
Continue reading Proof: Triangle InequalityProperties of n-dimensional Euclidean Space \(\mathbb{R}^n\) and Distance
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 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 \( x = y \)).