description/proof of that finite composition of motions is motion
Topics
About: vectors space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of motion.
Target Context
- The reader will have a description and a proof of the proposition that any finite composition of motions is a motion.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\{F_1, ..., F_n\}\): \(F_j \in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(\{F'_1, ..., F'_n\}\): \(F'_j \in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure, such that \(\forall j \in \{1, ..., n - 1\} (F'_j = F_{j + 1})\)
\(\{V_1, ..., V_n\}\): \(V_j \in \{\text{ the normed } F_j \text{ vectors spaces }\}\), with the norm, \(\Vert \bullet \Vert_j\)
\(\{V'_1, ..., V'_n\}\): \(V'_j \in \{\text{ the normed } F'_j \text{ vectors spaces }\}\) such that \(\forall j \in \{1, ..., n - 1\} (V'_j \subseteq V_{j + 1})\), with the norm, \(\Vert \bullet \Vert_{j + 1}\)
\(\{f_1, ..., f_n\}\): \(f_j: V_j \to V'_j\), \(\in \{\text{ the motions }\}\)
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Statements:
\(f_n \circ ... \circ f_1: V_1 \to V'_n \in \{\text{ the motions }\}\)
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2: Natural Language Description
For any \(\{F_1, ..., F_n\}\) where \(F_j \in \{\mathbb{R}, \mathbb{C}\}\) with the canonical field structure, any \(\{F'_1, ..., F'_n\}\) where \(F'_j \in \{\mathbb{R}, \mathbb{C}\}\) with the canonical field structure, such that for each \(j \in \{1, ..., n - 1\}\), \(F'_j = F_{j + 1}\), any \(\{V_1, ..., V_n\}\) where \(V_j \in \{\text{ the normed } F_j \text{ vectors spaces }\}\) with the norm, \(\Vert \bullet \Vert_j\), any \(\{V'_1, ..., V'_n\}\) where \(V'_j \in \{\text{ the normed } F'_j \text{ vectors spaces }\}\) such that for each \(j \in \{1, ..., n - 1\}\), \(V'_j \subseteq V_{j + 1}\), with the norm, \(\Vert \bullet \Vert_{j + 1}\), and any \(\{f_1, ..., f_n\}\) where \(f_j: V_j \to V'_j\) is any motion, \(f_n \circ ... \circ f_1: V_1 \to V'_n\) is a motion.
3: Proof
Whole Strategy: prove it inductively with respect to \(n\); Step 1: prove it for the \(n = 1\) case; Step 2: prove it for the \(n = 2\) case; Step 3: suppose it for the \(n = 1, ..., n' - 1\) cases, and prove it for the \(n = n'\) case.
Step 1:
Let us suppose that \(n = 1\).
\(f_1\) is a motion.
Step 2:
Let us suppose that \(n = 2\).
For each \(v, v' \in V_1\), \(\Vert v - v' \Vert_1 = \Vert f_1 (v) - f_1 (v') \Vert_2 = \Vert f_2 \circ f_1 (v) - f_2 \circ f_1 (v') \Vert_3\). So, \(f_2 \circ f_1\) is a motion.
Step 3:
Let us suppose that the proposition holds for the \(n = 1, ..., n' - 1\) cases.
Let us suppose that \(n = n'\).
\(f_{n'} \circ ... \circ f_1 = f_{n'} \circ (f_{n' - 1} \circ ... \circ f_1)\). \(f_{n' - 1} \circ ... \circ f_1\) is a motion, by the induction hypothesis for the \(n = n' - 1\) case. \(f_{n'} \circ (f_{n' - 1} \circ ... \circ f_1)\) is a motion, by the induction hypothesis for the \(n = 2\) case.
