Continuous Democracy System

The future of participatory direct democracy, as advocated by Bernie Sander, lies in information systems of coordination that allow deep public opinion to be integrated within a whole reflexive administrative state.  The ideal of a fully adaptive and sensitive autonomous governance system can be called a continuous democracy system since it samples the population opinions and recompiles these in its inner communication system on a continuous basis, thereby adapting as circumstances change and as public opinion shifts by developing on an issue of public importance.

The basic operation of a direct democracy system is simply voting on a referendum or a candidate for an office.  To perform these operations on a more continuous basis is to change the public vote on a policy or candidate, perhaps before the official duration has been completed or at least within shorter election-cycles.  Such would crowdsource administrative decisions of reflection to the wider public, and could even include modifications.  Yet, there is still the problem of who sets the agenda, itself the function of electoral politics.  The general problem with the presently practiced populist and yearly cycle of electoral democracy is often that the electoral system is not sensitive enough to the preferences of the total population or to its changes with time.  This can be solved through including more frequent votes with lesser weight within government functioning and more depth in the voting as through partially ranked choices rather than single choices to generate changing public systems from the wisdom of the crowds.  The choices for each deep vote are established through gaining a threshold of petitions.  Clearly the official use of available virtual technologies can significantly improve populist democracy to allow temporal and spatial sensitivity without changing its underlying process structures.

One may also begin to think of a more complex component to continuous democracy systems by conceiving of it as an Artificial-Intelligence system that samples the population in order to both represent the population’s aggregate deep opinion, as a psuedo-law, but also functionally coordinate the society through this functional communication system, an emergent-economy.  Clearly, as an autonomous system, participation (and thus coordination) would only be optional and so real economic transactions unlikely, rendering the functions as communicative, rather than directive towards peoples’ actual behaviors as with population control or commercial enforcement by the state.  Yet, integrations between this complex system and the real economic-commerce and state-administration can be made.

In this Complex Democracy system, the ideals of frequent voting and deep opinion can be realized to a further level since it has less official validity and therefore real institutional administrative checks that consume much human resources.  The underlying public opinion can be considered to be a quantum system, and hence a random variable with an underlying distribution.  While with the real component to complex democracy a single conclusive vote is the output, for the imaginary component of complex democracy, the underlying distribution (a complex function) is the sought-after solution.  In order to properly use this Social Information AI system to solve real problems, it is important to recognize that the crowdsourcing of research to the population, as distributed cognitive loads, performs these underlying quantum-computing operations.  Its functional-system ‘code of operation’ is itself a pseudo or emergent legal-economy, interpreted both by the humans – for their quantum operations of cognition and communication – and the main digital AI computer system that learns and evolves with each iteration of population sampling and recompiling.

I am presently developing an experimental virtual continuous complex democracy system with the migrant population in Honduras, in partnership with Foundation ALMA (www.foundationalam.org) to help them reintegrate into the places they fled by helping them organize to solve the normative disputes in their communities and society that have caused such high local violence and national systems of violence.

Statistics as the Logic of Science

\documentclass{article} \usepackage[utf8]{inputenc} % !BIB TS-program = biber \usepackage[backend=biber,style=numeric, citestyle=authoryear]{biblatex} \addbibresource{blog.bib} \usepackage{amssymb} \usepackage{dirtytalk} \usepackage{csquotes} \usepackage{amsmath} \usepackage{calc} \usepackage{textcomp} \usepackage{mathtools} \usepackage[english]{babel} \usepackage{fancyhdr} \usepackage{url} \def\UrlBreaks{\do\/\do-} \usepackage{breakurl} \usepackage[breaklinks]{hyperref} \usepackage{graphicx} \graphicspath{ {images/} } \usepackage{wrapfig} \usepackage{float} \usepackage[T1]{fontenc} \usepackage{outlines} \usepackage{enumitem} \setenumerate[1]{label=\null} \setenumerate[2]{label=\null} \setenumerate[3]{label=\roman*.} \setenumerate[4]{label=\alph*.} \newcommand{\midtilde}{\raise.17ex\hbox{$\scriptstyle\mathtt{\sim}$}} \usepackage{CJKutf8} \pagestyle{fancy} \fancyhf{} \title{Statistical Analysis by Communicative Functionals: \\ Lecture 2 – Statistics as The Logic of Science} \author{Justin Petrillo} \begin{document} \maketitle The question of $science \ itself$ has never been its particular object of inquiry but the existential nature, in its possibility and thereby the nature of its actuality. Science is power, and thus abstracts itself as the desired meta-good, although it is always itself about particularities as an ever-finer branching process. Although a philosophic question, the \textit{question of science} is inherently a political one, as it is the highest good desired by the society, its population, and its government. To make sense of science mathematically-numerically, as statistics claims, it is the scientific process itself that must be understood through probability theory as \say{The Logic of Science.} \footcite{QTSTATS} \section{Linguistic Analysis of the Invariants of Science: The Laws of Nature} The theory of science, as the proof of its validity in universality, must consider the practice of science, as the negating particularity. The symbolic language of science, within which its practice and results are embedded, necessarily negates its own particularity as well, as thus to represent a structure universally. Science, in the strict sense of having achieved already the goal of universality, is $de-linguistified$. While mathematics, in its extra-linguistic nature, often has the illusion of universal de-linguistification, such is only a semblance and often an illusion. The numbers of mathematics always can refer to things, and in the particular basis of their conceptual context always do. The non-numeric symbols of mathematics too represented words before short-hand gave them a distilled symbolic life. The de-linguistified nature of the extra-linguistic property of mathematics is that to count as mathematics, the symbols must themselves represent universal things. Thus, all true mathematical statements may represent scientific phenomena, but the context and work of this referencing is not trivial and sometimes the entirety of the scientific labor. The tense of science, as the time-space of the activity of its being, is the $tensor$, which is the extra-linguistic meta-grammar of null-time, and thus any and all times. \section{The Event Horizon of Discovery: The Dynamics between an Observer \& a Black Hole} The consciousness who writes or reads science, and thereby reports or performs the described tensor as an action of experimentation or validation, is the transcendental consciousness. Although science is real, it is only a horizon. The question is thus of its nature and existence at this horizon. What is knowable of science is thereby known as \say{the event horizon}, as that which has appeared already, beyond which is mere a \say{black hole} as what has not yet revealed itself – always there is a not-yet to temporality and so such a black hole can always be at least found as all that of science that has not and cannot be revealed since within the very notion of science is a negation of withdrawal (non-appearance) as the condition of its own universality (negating its particularity). Beginning here with the null-space of black-holes, the physical universe – at least the negative gravitational entities – have a natural extra-language – at least for the negative linguistic operation of signification whereby what is not known is the \say{object} of reference. In this cosmological interpretation of subjectivity within the objectivity of physical space-time, we thus come to the result of General Relativity that the existence of a black-hole is not independent of the observer, and in fact is only an element in the Null-Set, or negation, of the observer. To ‘observer’ a black-hole is to point to and outline something of which one does not know. If one ‘knew’ what it was positively then it would be not ‘black’ in the sense of not-emitting light \textit{within the reference frame (space-time curvature) of the observer}. That one $cannot$ see something, as receive photons reflecting space-time measurements, is not a property of the object but rather of the observer in his or her subjective activity of observation since to be at all must mean there is some perspective from which it can be seen. As the Negation of the objectivity of an observer, subjectivity is the $negative \ gravitational \ anti-substance$ of blackholes. Subjectivity, as what is not known by consciousness, observes the boundaries of an aspect (a negative element) of itself in the physical measurement of an ‘event horizon.’ These invariants of nature, as the conditions of its space-time, are the laws of dynamics in natural science. At the limit of observation we find the basis of the conditionality of the observation and thus its existence as an observer. From the perspective of absolute science, within the horizon of universality (i.e. the itself as not-itself of the black-hole or Pure Subjectivity), the space-time of the activity of observation (i.e. the labor of science) is a time-space as the hyperbolic negative geometry of conditioning (the itself of an unconditionality). What is a positive element of the bio-physical contextual condition of life, from which science takes place, for the observer is a negative aspect from the perspective of transcendental consciousness (i.e. science) as the limitation of the observation. Within Husserlian Phenomenology and Hilbertian Geometry of the early 20th century in Germany, from which Einstein’s theory arose, a Black-Hole is therefore a Transcendental Ego as the absolute measurement point. Our Solar System is conditioned in its space-time geometry by the MilkyWay galaxy it is within, which is conditioned by the blackhole Sagittarius A* ($SgrA*$). Therefore, the unconditionality of our solar space-time (hence the bio-kinetic features) is an unknown of space-time possibilities, enveloped in the event horizon of $SgrA*$. What is the inverse to our place (i.e. space-time) of observation will naturally only exist as a negativity, what cannot be seen. \section{Classical Origins of The Random Variable as The Unknown: Levels of Analysis} Strictly speaking, within Chinese Cosmological Algebra of 4-variables ($\mu$, X,Y,Z), this first variable of primary Unknowing, is represented by $X$, or Tiān (\begin{CJK*}{UTF8}{gbsn}天\end{CJK*}), for ‘sky’ as that which conditions the arc of the sky, i.e. “the heavens” or the space of our temporal dwelling ‘in the earth.’ We can say thus that $X=SgrA*$ is the largest and most relevant primary unknown for solarized galactic life. While of course X may represent anything, in the total cosmological nature of science, i.e. all that Humanity doesn’t know yet is conditioned by, it appears most relevantly and wholistically as $SgrA*$. It can be said thus that all unknowns ($x$) in our space-time of observation are within \say{\textit{the great unknown}} ($X$) of $SgrA*$, as thus $x \in X$ or $x \mathcal{A} X$ for the negative aspectual ($\mathcal{A}$) relationship \say{x is an aspect of X}. These are the relevant, and most general (i.e. universal) invariants to our existence of observation. They are the relative absolutes of, from, and for science. Within more practical scientific judgements from a cosmological perspective, the relevant aspects of variable unknowns are the planets within our solar system as conditioning the solar life of Earth. The Earthly unknowns are the second variable Y, or Di (\begin{CJK*}{UTF8}{gbsn}地\end{CJK*}) for “earth.” They are the unknowns that condition the Earth, or life, as determining the changes in climate through their cyclical dynamics. Finally, the last unknown of conditionals, Z, refers to people, Ren (\begin{CJK*}{UTF8}{gbsn}人\end{CJK*}) for ‘men,’ as what conditions their actions. X is the macro unknown (conditionality) of the gravity of ‘the heavens,’ Y the meso unknown of biological life in and on Earth, and Z the micro unknown of psychology as quantum phenomena. These unknowns are the subjective conditions of observation. Finally, the 4th variable is the “object”, or Wu (\begin{CJK*}{UTF8}{gbsn}物\end{CJK*}), $\mu$ of measurement. This last quality is the only $real$ value in the sense of an objective measurement of reality, while the others are imaginary in the sense that their real values aren’t known, and can’t be within the reference of observation since they are its own conditions of measurement within \say{the heavens, the earth, and the person}. \footcite[p.~82]{CHIN-MATH} In the quaternion tradition of Hamilton, ($\mu$, X,Y,Z) are the quaternions, ($\mu$, i,j,k). Since the real-values of X,Y,Z in the scientific sense can’t be known truly and thus must always be themselves unknowns, they are treated as imaginary numbers ($i=\sqrt{-1}$) with their ‘values’ merely coefficients to the quaternions $i,j,k$. These quaternions are derived as quotients of vectors, as thus the unit orientations of measurement’s subjectivity, themselves representing the space-time. We often approximate this with the Cartesian X,Y,Z of 3 independent directions as vectors, yet such is to assume Euclidean Geometry as independence. \printbibliography \end{document}
Cardinality and Countably Infinite Sets

Cardinality and Countably Infinite Sets

Cardinality is a term used to describe the size of sets. Set A has the same cardinality as set B if a bijection exists between the two sets. We write this as |A| = |B|. One important type of cardinality is called “countably infinite.” A set A is considered to be countably infinite if a bijection exists between A and the natural numbers ℕ. Countably infinite sets are said to have a cardinality of אo (pronounced “aleph naught”).

Remember that a function f is a bijection if the following condition are met:

1. It is injective (“1 to 1”): f(x)=f(y)⟹x=y

2. It is surjective (“onto”): for all b in B there is some a in A such that f(a)=b.

A set is a bijection if it is both a surjection and in injection.

A set is a bijection if it is both a surjection and an injection.

Example 1. Show that the set of integers ℤ is countably infinite.

To show that ℤ is countably infinite, we must find a bijection between ℕ and ℤ, i.e. we need to find a way to match up each element of ℕ to a unique element of ℤ, and this function must cover each element in ℤ.

We can start by writing out a pattern. One pattern we can use is to count down starting at 0, then going back and “picking up” each positive integer. This follows the pattern {0, -1, 1, -2, 2, -3, 3…}. We match to ℕ to ℤ as follows:

0 1 2 3 4
0 -1 1 -2 2

Notice that each even natural number is matched up to it’s half. It follows the function f(n) = n/2.

The odd numbers follow the function f(n) = -(n+1)/2

We can write this as a piecewise function as:

$f(n) =

\begin{cases}

n/2, & \text{if n is even} \\

-(n+1)/2, & \text{if n is odd}

\end{cases} $

Now we need to check if our function is a bijection.

Injectivity: Suppose the function is not injective. Then there exists some natural numbers x and y such that f(x)=f(y) but x≠y.

For even integers, x/2 = y/2 ⟹ x=y

For odd integers, Then (-x+1)/2 = -(y+1)/2 ⟹ x=y

These are contradictions, so the function is injective.

Surjectivity: Suppose the function is not surjective. Then there is some integer k such that there is no n in ℕ for which f(n) = k.

For k≥0, k=n/2 ⟹2k=n ⟹ n is an even number in ℕ

For k<0, k=-(n+1)/2 ⇒ -2k-1 = n ⇒ n is an odd number in ℕ

These are contradictions, so the function is surjective.

Since f is both injective and surjective, it is a bijection. Therefore, |ℤ| = |N| =אo. ∎

This result is often surprising to students because the set ℕ is contained in the set ℤ.

Example 2. For A = {2n | n is a number in ℕ}, show that |A| = |ℤ| (the set of integers has the same cardinality as the set of even natural numbers).

We can either find a bijection between the two sets or find a bijection from each set to the natural numbers. Since we already found a bijection from ℤ to ℕ in the previous example, we will now find a bijection from A to ℕ.

One function that will work is f(n) = n/2. Checking that it is a bijection is very similar to Example 1.

Since |A| = |ℕ| and |ℤ| = |ℕ|, then |A| = |ℤ| = אo.∎

There are many sets that are countably infinite, ℕ, ℤ, 2ℤ, 3ℤ, nℤ, and ℚ. All of the sets have the same cardinality as the natural numbers ℕ. Some sets that are not countable include ℝ, the set of real numbers between 0 and 1, and ℂ.

Georg Cantor was a pioneer in the field of set theory and was the first to explore countably infinite sets

Georg Cantor was a pioneer in the field of set theory and was the first to explore countably infinite sets

 

Completing the Square

Completing the Square

by: HT Goodwill

What Is Completing the Square?

Completing the Square is a technique in algebra that allows us to rewrite a quadratic expression that was in standard form:

$$ax^2 + bx + c$$

into vertex form:

$$a(x-h)^2 + k$$

Why Do We Complete the Square?

It should be noted that many times, students think (are taught?) that completing the square is just a way to solve a quadratic equation that has the form:

$$ax^2 + bx + c = 0$$.

However, in other areas of mathematics, we sometimes need to express a quadratic in the vertex form.

Leading up to How To

Suppose we have a perfect square binomial that we expand using FOIL:

$$\begin{align*}
(x + B)^2 & = (x + B)(x + B)\\
& = x^2 + Bx + Bx + B^2\\
& = x^2 + 2Bx + B^2
\end{align*}$$

If we reverse this string of equations, we see that any quadratic that has this pattern: \(x^2 + 2Bx + B^2\) which is a perfect square and we can factor it as \((x + B)^2\).

To Complete the Square, we adjust a quadratic expression so that it exhibits the perfect-square pattern. Here are some examples:

Introductory Examples

Example 1: Complete the square on \(x^2 + 14x + 40\)

Match up the quadratic expression with the perfect square pattern, starting with the \(x^2\) term.

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad x^2 + 14x + 40
\end{align*}$$

Step 1: Matching The Quadratic Terms

The quadratic terms match since they have the same coefficient.

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad x^2 + 14x + 40\\
\mbox{Match?} & \quad
\end{align*}$$

Step 2: Matching the Linear Terms

The value of \(2B\) is assigned be us and is whatever we need it to be so we can match the perfect-square pattern. In this example, they match, if we assign \(2B = 14\).

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + { 2B}x + B^2\\
\mbox{Our Quadratic:} & \quad x^2 + {14}x + 40\\
\mbox{Match?} & \quad \quad\quad
\end{align*}$$

Step 3: Matching the Constant Term

In order to match the perfect square pattern, the constant term has to be equal to \(B^2\). Since we defined \(2B = 14\) in Step 2, we know that \(B = 7\) which implies \(B^2 = 7^2 = 49\).

To adjust our constant term, we add zero (which won’t change the value of anything) and choose to rewrite the zero as \(B^2 – B^2\).

$$\begin{align*}
x^2 + 14x + 40 & = x^2 + 14x + 0 + 40 &&\mbox{Adding } 0.\\
& = x^2 + 14x + { (49 – 49)} + 40 &&\mbox{Since } B^2 – B^2 = 49-49 = 0\\
& = x^2 + 14x + 49 – 49 + 40\\
& = (x^2 + 14x + 49) – 49 + 40&&\mbox{Regrouping}
\end{align*}$$

Matching up to the pattern:

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad (x^2 + 14x + 49) – 49 + 40\\
\mbox{Match?} & \quad \quad\quad \quad\quad
\end{align*}$$

Since the first three terms (the ones in the parentheses) match the perfect-square pattern, we know those three terms will factor into \((x+B)^2\).

$$\begin{align*}
(x^2 + 14x + 49)- 49 + 40
& = (x+7)^2 – 49 + 40\\
& = (x+7)^2 – 9
\end{align*}$$

Answer: \(x^2 + 14x + 40 = (x+7)^2 – 9\)

Example 2: Complete the square on \(x^2 – 9x + 3\).

As before, we match up our quadratic with the perfect-square pattern. We’ll do this one a bit more quickly.

Step 1: Quadratic and Linear Terms

Both quadratic terms have a coefficient of \(1\) so they already match. The linear terms will match if we assign \(2B = -9\). So without any real work or effort, we match the first two terms already.

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad x^2-9x + 3\\
\mbox{Match?} & \quad \quad\,\,
\end{align*}$$

Step 2: Constant Term

Since we assigned \(2B = -9\), we divide by 2 to get \(B = -\frac{9}{2}\). And we know our constant term needs to be \(B^2 = \left(-\frac{9}{2}\right)^2 = \frac{81}{4}\). and we introduce it the same way we did before, by adding zero.

$$\begin{align*}
x^2 – 9x + 3
& = x^2 – 9x + 0 + 3 && \mbox{Adding }0\\
& = x^2 – 9x + \left(\frac{81}{4} – \frac{81}{4}\right) + 3 && \mbox{Since } B^2 – B^2 = \frac{81}{4}-\frac{81}{4} = 0\\
& = x^2 – 9x + \frac{81}{4} – \frac{81}{4} + 3\\
& = \left(x^2 – 9x + \frac{81}{4}\right) – \frac{81}{4} + 3 && \mbox{Regrouping}
\end{align*}$$

Making sure we match the pattern, we have

$$\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad\quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad \left(x^2 + 14x + \frac{81}{4}\right) – \frac{81}{4} + 40\\
\mbox{Match?} & \quad\,\,\,\quad\quad\quad\quad
\end{align*}$$

Since the three terms in the parentheses match the perfect-square pattern, we know those three terms factor as a perfect square.

$$\begin{align*}
\overbrace{\left(x^2 – 9x + \frac{81}{4}\right)}^{\mbox{Perfect Square}\frac{81}{4} + 3
& = \overbrace{\left(x – \frac 9 2\right)^2}^{\mbox{Factored}} – \frac{81}{4} + \frac{12}{4}\\
& = \left(x – \frac{9}{2}\right)^2  – \frac{69}{4}
\end{align*}$$

Answer: \(x^2 – 9x + 3 = \left(x – \frac{9}{2}\right)^2 – \frac{69}{4}\)

Example 3: Dealing With Leading Coefficients. Complete the square on \(2x^2 + 12x – 5\).

Solution In order to complete the square, we first need to reduce the leading coefficient to 1.

Step 1: Factor out the leading coefficient out of the first two terms.

$$2x^2 + 12x – 5 = 2[x^2 + 6x] – 5$$

Â

Step 2: Complete the Square \textit{Inside the Braces}\vskip 2mm

Â

\qquad Quadratic Term: \textit{Inside} the braces, our quadratic term has a coefficient of 1.\vskip 2mm

Â

\qquad Linear Term: \textit{Inside} the braces, we set \({\color{blue} 2B = 6}\).\vskip 2mm

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A quick check and we see that we are matching the perfect-square pattern \textit{inside the braces}.

Â

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad\,\,\,\, x^2 + {\color{blue}2B}x + B^2\\
\mbox{Our Quadratic (Inside Braces):} & \quad 2[x^2 + {\color{blue}6}x] – 5\\
\mbox{Match?}&\quad\quad \checkmark\quad\,\,\,\checkmark
\end{align*}

Â

Let’s move on to the constant term.\vskip 2mm

Â

\qquad Constant Term: \textit{Inside} the braces we have \({\color{blue} 2B = 6}\) which means \(B = 3\) and so \({\color{red} B^2 = 3^2 = 9}\). As before, adjust our constant term by adding {\color{red} 0}, but now it is \textit{inside the braces}.

Â

\begin{align*}
2[x^2 + {\color{blue} 6}x] – 5
& = 2[x^2 + {\color{blue} 6}x + {\color{red} 0}] – 5 && \mbox{Adding } {\color{red} 0} \mbox{ \textit{inside} the braces}\\
& = 2[x^2 + {\color{blue} 6}x + {\color{red} (9 – 9)}] – 5 && \mbox{Since } B^2 – B^2 = {\color{red} 9 – 9 = 0}\\
& = 2[x^2 + 6x + 9 – 9] – 5\\
& = 2[(x^2 + 6x + 9) – 9] – 5 && \mbox{Regrouping \textit{inside} the braces}
\end{align*}
\vskip 2mm

Â

Still keeping our attention focused \textit{inside} the braces, we check to see if we match the perfect-square pattern.

Â

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic (Inside Braces):} & \quad 2[(x^2 + 6x + 9) – 9] – 5\\
\mbox{Match?}&\quad\quad \checkmark\quad\,\,\,\checkmark\quad\,\,\checkmark
\end{align*}

Â

Since the three terms inside the parentheses match the pattern, we know those three terms form a perfect square, so they will easily factor into \((x + B)^2\).

Â

\begin{align*}
2\big[{\color{blue}\overbrace{\left(x^2 + 6x + 9\right)}^{\mbox{Perfect Square}}} – 9\big] – 5
& = 2\big[{\color{blue}\overbrace{\left(x + 3\right)^2}^{\mbox{Factored}}} – 9\big] – 5
\end{align*}
\vskip 5mm

Â

Step 3: Adjusting the Final Form\vskip 2mm

Â

Start by multiplying the leading coefficient \textit{through the square braces} (NOT the parentheses!), and simplify.

Â

\begin{align*}%
{\color{blue} 2}\big[\left(x + 3\right)^2 – 9\big] – 5
& = \big[{\color{blue} 2}(x+3)^2 – {\color{blue} (2)} 9\big] – 5 && \mbox{Multiply through braces}\\
& = \big[2(x + 3)^2 – 18\big] – 5\\
& = 2(x + 3)^2 – 18 – 5 && \mbox{Braces no longer needed}\\
& = 2(x+3)^2 – 23 && \mbox{Combine the constants}
\end{align*}
\vskip 5mm

Â

\textbf{Answer:} \(2x^2 + 12x – 5 = 2(x+3)^2 – 23\)
\vskip 1.5cm

Â

\textbf{\large Quick Example}\vskip 2mm

Â

\textbf{Example 4: A Quick Example}\vskip 2mm

Â

Complete the Square on \(3x^2 – 4x + 8\).\vskip 5mm

Â

\textbf{Solution:}\vskip 2mm

Â

\begin{align*}%
3x^2 – 4x + 8 & = 3\left[x^2 – \frac 4 3 x\right] + 8 && \mbox{Factor out leading coefficient}\\[6pt]
& = 3\left[x^2 {\color{blue}- \frac 4 3} x\right] + 8 && \mbox{Identify } {\color{blue} 2B = -\frac 4 3} \longrightarrow {\color{red} B^2 = \left(-\frac 2 3\right)^2 = \frac 4 9}\\[6pt]
& = 3\left[x^2 {\color{blue}- \frac 4 3} x + {\color{red} \left(\frac 4 9 – \frac 4 9\right)}\right] + 8 && \mbox{Add } {\color{red} 0}\\[6pt]
& = 3\left[\left(x^2 {\color{blue}- \frac 4 3} x + {\color{red}\frac 4 9}\right)\,\,{\color{red}- \frac 4 9}\right] + 8 && \mbox{Regrouping}\\[6pt]
& = 3\left[\left(x – \frac 2 3\right)^2 – \frac 4 9\right] + 8 && \mbox{Factor}\\[6pt]
& = 3\left(x – \frac 2 3\right)^2 – \frac 4 3 + 8 && \mbox{Distribute}\\[6pt]
& = 3\left(x – \frac 2 3\right)^2 – \frac 4 3 + \frac{24} 3\\[6pt]
& = 3\left(x – \frac 2 3\right)^2 + \frac{20} 3 && \mbox{Combine like terms}
\end{align*}
\vskip 5mm

Â

\textbf{Answer:} \(3x^2 – 4x + 8 = 3\left(x – \frac 2 3\right)^2 + \frac{20} 3\)
\vskip 1.5cm

Â

\textbf{\large The Quick Method}\vskip 2mm

Â

The method shown in the examples can be made more efficient if we recognize that the pattern is always the same. For a simple quadratic with a leading coefficient of \(1\), the completed square form looks like this:

Â

\begin{center}
\begin{tikzpicture}
% Help Lines
% \draw [step=0.25cm,help lines] (-2,-1) grid (2,1);
% \draw (-2,-1) grid (2,1);
% Equation
\node {%
\(%
x^2 + {\color{blue} 2B}x + c = (x + {\color{blue} B})^2 – {\color{blue} B^2} + c
\)};
% Arrows
\node (linear) [inner sep=0pt] at (-1.625,-0.15){};
\node (BLow) [inner sep=0pt] at (0.75, -0.15){};
\node (BUp) [inner sep=0pt] at (0.75, 0.2){};
\node (B2) [inner sep=0pt] at (1.75,0.2){};
\draw [-latex] (linear.south) to [out=270,in=270] (BLow.south);
\draw [-latex] (BUp.north) to [out=90,in=90] (B2.north);
% Labels
\node at (-0.375, -0.625) {\(\div 2\)};
\node at (1.5, 0.75) {\scriptsize Minus the Square};
\node at (3,-1) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}

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Inside the final parentheses we always end up with \(x + B\), where \(B\) is half of the coefficient of the original \(x\) term.\vskip 2mm

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Next, we subtract \(B^2\) \textit{outside} the parentheses. Let’s try it with one of our previous examples to see it in action.\vskip 5mm

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\textbf{Example 5: Simple Quick Method}\vskip 2mm

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Use the quick version of Completing the Square on \(x^2 + 14x + 40\).

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\begin{center}
\begin{tikzpicture}
% Help Lines
% \draw [step=0.25cm,help lines] (0,-1) grid (5,1);
% \draw (0,-1) grid (5,1);
% Equation
\node at (1.2,-0.25){%
\(%
x^2 + {\color{blue} 14}x + 40 = (x + {\color{blue} 7})^2 \underbrace{- {\color{blue} 7^2} + 40}_{\mbox{\scriptsize Add Together}} = (x + 7)^2 – 9
\)};
% Arrows
\node (linear) [inner sep=0pt] at (-1.625,-0.15){};
\node (BLow) [inner sep=0pt] at (0.75, -0.15){};
\node (BUp) [inner sep=0pt] at (0.75, 0.2){};
\node (B2) [inner sep=0pt] at (1.75,0.2){};
\draw [-latex] (linear.south) to [out=270,in=270] (BLow.south);
\draw [-latex] (BUp.north) to [out=90,in=90] (B2.north);
\draw [-latex] (2.85,-0.575) .. controls (3,-0.625) and (5,-0.75)..(5,-0.25);
% Labels
\node at (-0.375, -0.625) {\(\div 2\)};
\node at (1.5, 0.75) {\scriptsize Minus the Square};
\node at (5,-1) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}
\vskip 5mm

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\textbf{Answer:} \(x^2 + 14x + 40 = (x + 7)^2 – 9\)

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Note: This is the same answer we got in Example 1.\vskip 1.5cm

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\textbf{Example 6: Quick Method with Leading Coefficient}\vskip 2mm

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The quick version can also be used when the leading coefficient isn’t \(1\). Like before, we just factor out the leading coefficient first, and then work inside the braces.\vskip 2mm

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Let’s complete the square on \(5x^2 + 80x + 13\).

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\noindent\textbf{Solution}\vskip 2mm

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Step 1) Factor out the leading coefficient.

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\[%
5x^2 + 80x + 13 = 5[x^2 + 16x] + 13
\]

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Step 2) Now use the quick method to complete the square on the \textit{inside} of the braces.

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\begin{center}
\begin{tikzpicture}
% Help Lines
% \draw [step=0.25cm,help lines] (-2,-2) grid (4,2);
% \draw (-2,-1) grid (4,1);
% \draw [red, step = 2] (0,-1) grid (3,1);
% Equation
\node at (1.2,0){%
\begin{minipage}{\textwidth}
\begin{align*}%
5[x^2 + {\color{blue} 16}x] + 13
& = 5[x^2 + {\color{blue} 16}x] + 13\\[5mm]
& = 5[(x + {\color{blue} 8})^2 – {\color{blue} 64}] + 13\\[6mm]
& = 5(x + {\color{blue} 8})^2 – 320 + 13\\
& = 5(x + {\color{blue} 8})^2 – 307%
\end{align*}%
\end{minipage}
};
% Arrows
\node (linear) [inner sep=0pt] at (2.375,0.95){};
\node (BLow) [inner sep=0pt] at (2.23, -0.05){};
\node (BUp) [inner sep=0pt] at (2.23, 0.25){};
\node (B2) [inner sep=0pt] at (3.23,-0.05){};
\draw [-latex] (linear.south) to [out=270,in=90] (BUp.south);
\draw [-latex] (BLow.south) to [out=270,in=270] (B2.south);
% Labels
\node at (2.625, 0.625) {\scriptsize \(\div 2\)};
\node at (3, -0.55) {\scriptsize Minus the Square};
\node at (4.5,-2) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}
\vskip 5mm

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\textbf{Answer:} \(5x^2 + 80x + 13 = 5(x + 8)^2 – 307\)

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\vskip 1.5cm

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\textbf{\large Solving Equations}\vskip 2mm

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When solving a quadratic equation, I find the technique of completing the square is not very efficient, except in the simple cases where the leading coefficient is \(1\) and the linear coefficient is even.\vskip 2mm

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But that doesn’t mean we shouldn’t know \textit{how} to solve more complicated quadratic equations with this technique.\vskip 5mm

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\textbf{Two Approaches}\vskip 2mm

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When solving an equation through completing the square, there are two basic approaches: (1) Complete the square first, \textit{then} solve the equation, or (2) Solve the equation \textit{as} you complete the square.\vskip 5mm

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\textbf{Example 7: Solving a Quadratic Equation by Completing the Square}\vskip 2mm

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Use Completing the Square to solve \(4x^2 + 3x – 10 = 0\).\vskip 5mm

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\textbf{Solution (Method 1)}\vskip 2mm

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For this approach, we’ll first complete the square, then solve the equation.\vskip 2mm

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Step 1) Complete the Square on \(4x^2 + 3x – 10\).\vskip 2mm

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Using the techniques we’ve discussed above, we get the following.

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\[%
4x^2 + 3x – 10 = 4\left(x + \frac 3 8\right)^2 – \frac{169}{16}
\]
\vskip 5mm

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Step 2) Solve the equation.

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\begin{align*}
4x^2 + 3x – 10 & = 0\\[6pt]
4\left(x + \frac 3 8\right)^2 – \frac{169}{16} & = 0 && \mbox{Complete the Square}\\[6pt]
4\left(x + \frac 3 8\right)^2 & = \frac{169}{16} && \mbox{Add the constant}\\[6pt]
\left(x + \frac 3 8\right)^2 & = \frac{169}{64} && \mbox{Divide by \(4\)}\\[6pt]
x + \frac 3 8 & = \pm\sqrt{\frac{169}{64}} && \mbox{Take the square root}\\[6pt]
x + \frac 3 8 & = \pm\frac{13} 8 && \mbox{Simplify}\\[6pt]
x & = – \frac 3 8 \pm\frac{13} 8 && \mbox{Subtract}
\end{align*}

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The two solutions are

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\[%
\begin{array}{rl}
x & = \displaystyle – \frac 3 8 + \frac {13} 8 = \frac{10} 8 = \frac 5 4\\[12pt]
x & = \displaystyle – \frac 3 8 – \frac{13} 8 = – \frac{16} 8 = – 2
\end{array}
\]
\vskip 5mm

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\textbf{Solution (Method 2)}\vskip 2mm

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This time, we’ll complete the square as we solve the equation. It should be noted that this approach is what many students think \textit{is} Completing the Square.\vskip 2mm

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\begin{align*}
4x^2 + 3x – 10 & = 0\\[6pt]
4x^2 + 3x & = 10 && \mbox{Add the constant}\\[6pt]
x^2 + \frac 3 4 x & = \frac{10} 4 && \mbox{Divide by 4}\\[6pt]
x^2 + \frac 3 4 x + \frac 9 {64} & = \frac{10} 4 + \frac 9 {64} && \mbox{Add \(B^2\)}\\[6pt]
\left(x + \frac 3 8\right)^2 & = \frac{160}{64} + \frac 9 {64} && \mbox{Factor the left side}\\[6pt]
x + \frac 3 8 & = \pm \sqrt{\frac{169}{64}} && \mbox{Take square root}\\[6pt]
x & = -\frac 3 8 \pm \frac{13} 8 && \mbox{Subtract}
\end{align*}

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Again, we see the two answers are \(x =-\frac 3 8 + \frac{13} 8 = \frac 5 4\) and \(x = -\frac 3 8 – \frac{13} 8 = -2\).
\vskip 5mm

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\textbf{Answer:} \(x = \frac 5 4\) or \(x = -2\)
\vskip 1.5cm

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\textbf{Conclusion}\vskip 2mm

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Admittedly, most high school student will Complete the Square in the context of solving quadratic equations. However, it is important that they understand that the technique is more than just an equation-solving technique. Outside of high school math, much of its use is in changing the form of a quadratic function so that it can be used in higher mathematical techniques.

10 Ways Tutoring Can Help You

10 Ways Tutoring Can Help You

Finding a good math tutor can be essential to your success in school. Tutoring can do much more than just improving math grades. Here are 10 ways in which you can benefit from tutoring.

1. Improve your confidence

When working one-on-one outside of the classroom setting, you might find it easier to focus on the subject material and return to math class with increased confidence.

2. Improve your grades 

Supplementing your coursework with tutoring can help you understand the material more thoroughly so you can do better in school.

3. Work at a pace you are comfortable with 

Your teacher might be moving through the material at a speed that is too fast for you, but when you are studying with a tutor they will structure their sessions at a pace that makes you feel comfortable. Alternatively, if you feel that your class is moving too slow, your tutor can work ahead with you to prepare for future material.

4. Build mathematical foundation

It is essential that you understand the basics before you can succeed in more difficult math. While your teacher must follow a strict curriculum,a tutor will work with you to review the material that you didn’t understand before. Building a strong mathematical foundation now will allow you to tackle the more complicated topics later.

5. Provide supplemental work

If you are confused about a certain topic, your tutor can find you extra material and work through it with you.

6. Tailor lessons to your needs 

In a classroom setting, students work at different paces and understand things differently. In one-on-one tutoring, your lessons can be perfectly tailored to your own learning style, which can make learning math easier and more enjoyable.

7. Find a new challenge 

Is your math class too easy for you? Tutoring might be the perfect way to push your own boundaries. A tutor can expand and enrich upon the material you’re learning in school, or work on interesting material that’s not even offered at your school. Your tutor can also help you find local math competitions to challenge you in a fun way.

8. Improve SAT/ACT scores 

Tutoring is a great way to make sure you ace your college entrance exam. Tutors can help you learn test taking strategies, as well as helping you learn what kind of questions to expect.

9. Bring a fresh new perspective 

If you’re feeling stuck in a mathematical rut, working with a tutor can help. Your tutor can discuss math from a new point of view that might make more sense to you than the way your teacher approaches the material.

10. Make math more interesting 

Tutors are passionate about math and would love to share that passion with you. If there is a certain topic of interest to you, your tutor can build some lessons around that to keep you interested.

Modular Arithmetic & Fermat’s Little Theorem

Modular Arithmetic & Fermat’s Little Theorem

Modular Arithmetic and Fermat’s Little Theorem

Modular arithmetic is a way of counting in which the numbers wrap around after reaching a certain value. The clock is often used as an analogy. While time always progresses forward, the 12-hour clock “resets” to 1 after passing 12 (13 o’clock is equivalent to 1 o’clock). If we replace 12 with 0, we have the set known as ℤ12 which is made up of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}. If we consider midnight as t0=0, then 5 hours later is t5=5. 17 hours later will also be t17=5. 84 hours later will be t84 = 0. 85 hours later will be t85 = 1. There is a pattern here. The value a in the set ℤ12 corresponding to any other integer b is the remainder when b is divided by 12. We can write this mathematically as b ≡ a (mod 12), and we say “b is congruent to a modulo 12”

A clock is a good analogy for modular arithmetic in Z12

A clock represents modular arithmetic in ℤ12

More generally, we can write the set of integers for any modulus n as ℤn = {0, 1, 2, 3, … n-1} . Each element a in ℤn represents not only a itself, but all of the b’s for which b ≡ a (mod n). These elements are called congruence classes, and the set ℤn is a set made up of these classes. Sometimes, it’s more convenient to write the relationship as b-a ≡ 0 (mod n), i.e. each b is assigned to a congruence class depending on how much needs to be subtracted to be divisible by n. Addition and multiplication can be done using modular arithmetic if you are using the same modulus. If b ≡ a (mod n) and d ≡ c (mod n), then

  1. (b+d) ≡ (a+c) (mod n)
  2. (b*d) ≡ (a*c) (mod n)

Example 1. What is 84,872 (mod 5)? A simple way to approach this problem is to find the closest multiple of 5 to 84,872. That is 84,870. Therefore, we can write 84,872 (mod 5) ≡ 84,870 (mod 5) + 2 (mod 5) ≡ 0 (mod 5) + 2 (mod 5) ≡ 2 (mod 5). Example 2. In modulo 10, what is 19,374 · 3,172? One way to attempt this problem is to multiply out these numbers and then find the remainder when dividing by 10. However, since we want our answer in modulo 10, we can instead multiply just the representatives from their congruence classes. The remainder when 19,374 is divided by 10 is 4, so 19,374 ≡ 4 (mod 10) and 3,172 ≡ 2 (mod 10), so (19,374 · 3,172) ≡ (2·4) (mod 10) ≡ 8 (mod 10). Example 3. In modulo 4, what is 180,700,247 + 64,987,422? Again we first find the congruence class for each number modulo 4. We can simplify each by breaking it into the sum of smaller multiples of 4. One possible way is to write 180,700,247 ≡ 180,700,200 (mod 4) + 44 (mod 4) + 3 (mod 4) ≡ 0 (mod 4) + 0 (mod 4) + 3 (mod 4) ≡ 3 (mod 4) Similarly, 64,987,422 ≡ 2 (mod 4). Therefore, (180,700,247+ 64,987,422) ≡ (3 + 2) (mod 4) ≡ 5 (mod 4). However, 5 is not in the set ℤ4 = {0, 1, 2, 3}, so we must reduce 5 (mod 4) to something in the set. We can simplify by noticing that 5 (mod 4) ≡ 4 (mod 4) + 1 (mod 4) ≡ 0 (mod 4) + 1 (mod 4) ≡ 1 (mod 4).

Fermat’s Little Theorem

One important application for modular arithmetic is Fermat’s Little Theorem which states that if p is a prime number and a is not divisible by p, then ap-1 ≡ 1 (mod p). This theorem is useful because allows you to find a remainder when dividing a really big number by a prime number. Example 4. What is the remainder when 2784 is divided by 11? 2784 is too big to put in most calculators and too time-consuming to multiply by hand. We can apply Fermat’s Little Theorem to make it easier to solve. From the theorem we know that 210 ≡ 1 (mod 11). And we also know that we can multiply congruence classes with the same modulus. Therefore, one strategy we can apply is to write 2784 as the product of as many 210’s as possible. We can simplify as follows: 2784= (210)78·(24) ≡ 178 · 24 (mod 11) ≡ 24 (mod 11) ≡16 (mod 11) ≡ 5 (mod 11). The remainder is 5.

"Pierre