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April 7, 2020, 11:52 p.m.

Manipulating Group Elements

By Maurice Ticas

Tags:

Algebra

Group Theory

Proof

What is there to say when given a group and a few of its elements? With very little information, group behavior of elements can reveal simple relationships. We summarize some simple relationships when manipulatiing group elements.

Consider the group \(<G,\cdot>\) and let \(a_1,a_2,a_3 \in G\). We then have that \[a_{1}^{3} = e \Rightarrow a_1 = (a_{1}^{-1})^2\] \[a_{1}^{2}=e \Rightarrow a_{1}=(a_{1}^{-1})^3\] \[\text{If }(a_{1})^{-1} = a_{2}^{3} \text{ for some } a_2 \in G \text{, then } a_1 = (a_2^{-1})^3\] \[a_{1}^{2}a_{2}a_{1} = a_{2}^{-1} \Rightarrow a_2 = \left [ (a_{1}a_{2}a_{1})^{-1} \right]^{3} \] \[a_{1}a_{2}a_{1} = a_{3} \Rightarrow a_{2}a_{3}=(a_{2}a_{1})^{2}\]

You can read and study the proofs. They just use the properties of a group. Arguments will not be more involved than a typical usage of induction.

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March 25, 2020, 9:32 a.m.

Groups, Subgroups and Algebraic Structure

By Maurice Ticas

Tags:

Algebra

Group Theory

Here we'll give a very high level overview of the subject to summarize the material from our Secolinsky group theory publication. Anyone wanting to better understand groups is invited to read its beginning theoretic development. The publication will help you understand the details and concrete nature of the rich subject.

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Nov. 27, 2019, 5:01 p.m.

The Trapezoid Geometric Method of Finding Area

By Maurice Ticas

Tags:

Continuity

Geometry

Integrals

Consder a real-valued continuous function from points a, b of its domain where a < b. We will use the formula for the area of a trapezoid to find an approximation of the area \(\int_{a}^{b} f(x) \, dx\). Let's begin.

The formula for the area of a trapezoid is \[(c + d)/2 \times h \] where \(c, d\) are the lengths of the trapezoid basis, respectively, and \(h\) is the trapezoid height. After dividing our closed interval \([a,b]\) into n smaller subintervals of width \(\frac{b-a}{n}\), we then have that \(\int_{a}^{b} f(x) \, dx \) is approximately equal to \[ \frac{f(x_0) + f(x_1)}{2} \times \frac{b-a}{n} + \frac{f(x_1) + f(x_2)}{2} \times \frac{b-a}{n} + \dots + \frac{f(x_n-1) + f(x_n)}{2} \times \frac{b-a}{n} \\ = \frac{b-a}{2n} ( f(x_0) + 2 \, f(x_1) + \cdots + 2 \, f(x_{n-1}) + f(x_n)) \]

Our geometric intuition would allow us to be convinced that when \(f\) is concave up on \([a, b]\), then the trapezoid method is an over-approximation. When \(f\) is concave down on \([a, b]\), the method is an under-approximation.

Now consider applying the method for \(\int_{-\pi}^{\pi} \sin(x) \, dx\) for when \(n = 3\) and allow your imagination to give you the geometric intuition to see that the approximation gives the correct answer of \(0\). In this case, the trapezoids become triangles and what your are left with are two pairs of triangles with areas that cancel each other out to zero.

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