Nonlinear Dynamics and Chaos
2nd Edition
ISBN: 9780813349107
Author: Steven H. Strogatz
Publisher: PERSEUS D
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Chapter 11.4, Problem 1E
Interpretation Introduction
Interpretation:
To find the dimension of the von Koch snowflake.
Concept Introduction:
The Koch curve starts with a straight line that is divided up into three equal parts. Using the middle segment as a base, an equilateral triangle is created. Finally, the base of the triangle is removed, leaving us with the first iteration of the Koch curve.
From the Koch curve comes the Koch Snowflake. Instead of one line, the snowflake begins with an equilateral triangle. The steps in creating the Koch Curve are then repeatedly applied to each side of the equilateral triangle, creating a snowflake shape.
The Koch snowflake is an example of a figure that is self-similar, meaning it looks the same on any scale.
The formula for the number of sides in the Koch snowflake is:
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Let A be a vector space with basis 1, a, b. Which (if any) of the following rules
turn A into an algebra? (You may assume that 1 is a unit.)
(i) a² = a, b² = ab = ba = 0.
(ii) a²=b, b² = ab = ba = 0.
(iii) a²=b, b² = b, ab = ba = 0.
No chatgpt pls will upvote
= 1. Show
(a) Let G = Z/nZ be a cyclic group, so G = {1, 9, 92,...,g" } with g":
that the group algebra KG has a presentation KG = K(X)/(X” — 1).
(b) Let A = K[X] be the algebra of polynomials in X. Let V be the A-module
with vector space K2 and where the action of X is given by the matrix
Compute End(V) in the cases
(i) x = p,
(ii) xμl.
(67) ·
(c) If M and N are submodules of a module L, prove that there is an isomorphism
M/MON (M+N)/N.
(The Second Isomorphism Theorem for modules.)
You may assume that MON is a submodule of M, M + N is a submodule of L
and the First Isomorphism Theorem for modules.
Chapter 11 Solutions
Nonlinear Dynamics and Chaos
Ch. 11.1 - Prob. 1ECh. 11.1 - Prob. 2ECh. 11.1 - Prob. 3ECh. 11.1 - Prob. 4ECh. 11.1 - Prob. 5ECh. 11.1 - Prob. 6ECh. 11.1 - Prob. 7ECh. 11.2 - Prob. 1ECh. 11.2 - Prob. 2ECh. 11.2 - Prob. 3E
Ch. 11.2 - Prob. 4ECh. 11.2 - Prob. 5ECh. 11.2 - Prob. 6ECh. 11.3 - Prob. 1ECh. 11.3 - Prob. 2ECh. 11.3 - Prob. 3ECh. 11.3 - Prob. 4ECh. 11.3 - Prob. 5ECh. 11.3 - Prob. 6ECh. 11.3 - Prob. 7ECh. 11.3 - Prob. 8ECh. 11.3 - Prob. 9ECh. 11.3 - Prob. 10ECh. 11.4 - Prob. 1ECh. 11.4 - Prob. 2ECh. 11.4 - Prob. 3ECh. 11.4 - Prob. 4ECh. 11.4 - Prob. 5ECh. 11.4 - Prob. 6ECh. 11.4 - Prob. 7ECh. 11.4 - Prob. 8ECh. 11.4 - Prob. 9ECh. 11.4 - Prob. 10ECh. 11.5 - Prob. 1E
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