The post Cardinality and Countably Infinite Sets appeared first on Math Academy.
]]>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 א_{o} (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.
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:
ℕ | 0 | 1 | 2 | 3 | 4 | … |
ℤ | 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| =א_{o}. ∎
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| = א_{o}.∎
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.
The post Cardinality and Countably Infinite Sets appeared first on Math Academy.
]]>The post Cryptography: How the Internet We Love Almost Never Existed appeared first on Math Academy.
]]>The post Cryptography: How the Internet We Love Almost Never Existed appeared first on Math Academy.
]]>The post Completing the Square appeared first on Math Academy.
]]>A brief discussion about what completing the square is and what we use it for. Focuses on using the technique for other reasons than solving equations.
Development of the patterns we use for completing the square. This part is important since it introduces the notation we will be using!
The first three examples step through the process very slowly with a great deal of explanation for each step. These are intended for people who are unfamiliar with the process and (perhaps) need to take the algebra slowly.
Note that the three examples increase in difficulty.
Example 4 is similar in difficulty to Example 3, but worked through quickly and with less explanation.
With a solid grounding in the ideas and methods of completing the square, this discussion shows how we can cut out most of the background steps and basically just do things in one step (at least for the easy ones).
Demonstrating the Quick Method.
An example and discussion of solving a quadratic equation by Completing the Square
What Is Completing the Square?
Completing the Square is a technique in algebra that allows us to rewrite a quadratic expression that is in standard form in vertex form. That is
\[%
ax^2 + bx + c\quad \underset{\mbox{\scriptsize the Square}}{\xrightarrow{\mbox{\scriptsize Complete}}}\quad a(x – h)^2 + k
\]
so that \(ax^2 + bx + c = a(x-h)^2 + k\).
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. The following are two examples of this.
Please note, you don’t have to understand these examples! That’s not the point of them!
\begin{itemize}
\item \textit{Integral Calculus:}
\begin{center}
\begin{tikzpicture}
\node [align=center] at (-1,0.75) {\scriptsize Cannot\\[-2mm]\scriptsize Integrate};
\node [align=center] at (1,0.75) {\scriptsize Can Now\\[-2mm] \scriptsize Integrate};
\node at (0,0) {\(\int \frac{dx}{x^2 + 4x + 5} = \int \frac{dx}{(x+2)^2 + 1}\)};
\draw [-latex] (-1,-0.3) arc [start angle=-180, end angle=0, x radius=1cm, y radius=0.5cm];
\node at (0,-1) [anchor=north,inner sep=0pt,align=center] {\scriptsize Complete the Square\\[-1mm] \scriptsize in Denominator};
\node at (3,-2) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}
\item \textit{Differential Equations (Laplace Transforms):}
\begin{center}
\begin{tikzpicture}
% \draw[help lines,step=5mm] (-2,-1) grid (2.5,1);
\node [align = center] at (-1.25,1) {\scriptsize Cannot Find\\[-2mm] \scriptsize Transform};
\node [align=center] at (1.25,1) {\scriptsize Can Now Find\\[-2mm] \scriptsize Transform};
\node at (0,0) {\(\mathcal{L}^{-1}\left\{\frac{5}{s^2 + 4s + 5}\right\} = \mathcal{L}^{-1}\left\{\frac{5}{(s+2)^2 + 1}\right\}\)};
\draw [-latex] (-1.25,-0.3) arc [start angle=-180, end angle=0, x radius=1.5cm, y radius=0.5cm];
\node at (0,-1) [anchor=north,inner sep=0pt,align=center] {\scriptsize Complete the Square\\[-1mm] \scriptsize in Denominator};
\node at (3,-2) [anchor=east] {\tiny www.mathtutoringacademy.com};
\end{tikzpicture}
\end{center}
\end{itemize}
The two examples above are \textit{only} to show you that completing the square is used for \textit{other reasons} than to solve equations.\vskip 1.5cm
\textbf{\large Background (Leading up to \textit{How To\ldots})}
\vskip 2mm
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: \(x^2 + 2Bx + B^2\) is a perfect square and we can factor it as \((x + B)^2\).
\[%
\underbrace{x^2 + 2Bx + B^2}_{\mbox{\parbox{3cm}{\centering \scriptsize The Perfect-Square\\Pattern\ldots}}} = \underbrace{(x+B)^2}_{\mbox{\parbox{3cm}{\centering\scriptsize \ldots always factors\\this way.}}}
\]\vskip 2mm
To Complete the Square, we adjust a quadratic expression so that it exhibits the perfect-square pattern. Here is an overview of the rest of the article so you can focus on what you need most.\vskip 1.5cm
\textbf{\large Introductory Examples}\vskip 2mm
\textbf{Example 1 How To Complete the Square}\vskip 2mm
Complete the square on \(x^2 + 14x + 40\).
\vskip 5mm
\textbf{Solution}\vskip 2mm
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*}
\vskip 5mm
Step 1: Matching The Quadratic Terms\vskip 2mm
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 \checkmark
\end{align*}
\vskip 5mm
Step 2: Matching the Linear Terms\vskip 2mm
The \textit{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 \({\color{blue} 2B = 14}\).
\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + {\color{blue} 2B}x + B^2\\
\mbox{Our Quadratic:} & \quad x^2 + {\color{blue}14}x + 40\\
\mbox{Match?} & \quad \checkmark\quad\quad \checkmark
\end{align*}
\vskip 5mm
Step 3: Matching the Constant Term\vskip 2mm
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 \({\color{red} B^2 = 7^2 = 49}\).
\vskip 2mm
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 + {\color{red} 0} + 40 &&\mbox{Adding }{\color{red} 0}.\\
& = x^2 + 14x + {\color{red} (49 – 49)} + 40 &&\mbox{Since } B^2 – B^2 = {\color{red} 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 \checkmark\quad\quad \checkmark\quad\quad \checkmark
\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*}
{\color{blue}\overbrace{(x^2 + 14x + 49)}^{\mbox{Perfect Square}}} – 49 + 40
& = {\color{blue}\overbrace{(x+7)^2}^{\mbox{Factored}}} {\color{red}- 49 + 40}\\
& = (x+7)^2 {\color{red}- 9}
\end{align*}
\textbf{Answer:} \(x^2 + 14x + 40 = (x+7)^2 – 9\)
\vskip 1.5cm
\textbf{Example 2 (A \textit{Little} More Complicated)}\vskip 2mm
Complete the square on \(x^2 – 9x + 3\).
\vskip 5mm
\textbf{Solution} As before, we match up our quadratic with the perfect-square pattern. We’ll do this one a bit more quickly.\vskip 2mm
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 \({\color{blue}2B = -9}\). So without any real work or effort we match the first two terms already.
\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + {\color{blue} 2B}x + B^2\\
\mbox{Our Quadratic:} & \quad x^2\,{\color{blue}-\,9}x + 3\\
\mbox{Match?} & \quad \checkmark \quad\,\, \checkmark
\end{align*}
Step 2: Constant Term
Since we assigned \({\color{blue} 2B = -9}\), we divide by 2 to get \(B = -\frac 9 2\). And we know our constant term needs to be
\[%
{\color{red}B^2 = \left(-\frac 9 2\right)^2 = \frac{81} 4},
\]
and we introduce it the same way we did before, by adding zero.
\begin{align*}
x^2 – 9x + 3
& = x^2 – 9x + {\color{red} 0} + 3 && \mbox{Adding }{\color{red} 0}\\
& = x^2 – 9x + {\color{red}\left(\frac{81} 4 – \frac{81} 4\right)} + 3 && \mbox{Since } B^2 – B^2 = {\color{red} \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\,\,\, \checkmark\quad\quad \checkmark\quad\quad \checkmark
\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*}
{\color{blue}\overbrace{\left(x^2 – 9x + \frac{81} 4\right)}^{\mbox{Perfect Square}}} {\color{red}- \frac{81} 4 + 3}
& = {\color{blue} \overbrace{\left(x – \frac 9 2\right)^2}^{\mbox{Factored}}} {\color{red} – \frac{81} 4 + \frac{12} 4}\\
& = \left(x – \frac 9 2\right)^2 {\color{red} – \frac{69} 4}
\end{align*}
\vskip 5mm
\textbf{Answer:} \(x^2 – 9x + 3 = \left(x – \frac 9 2\right)^2 – \frac{69} 4\)
\vskip 1.5cm
\textbf{Example 3: Dealing With Leading Coefficients}\vskip 2mm
Complete the square on \(2x^2 + 12x – 5\).
\vskip 5mm
\textbf{Solution} In order to complete the square, we first need to reduce the leading coefficient to 1.
\vskip 2mm
Step 1: Factor out the leading coefficient out of the first two terms.
\[%
{\color{blue} 2}x^2 + 12x – 5 = {\color{blue} 2}[x^2 + 6x] – 5
\]
\vskip 5mm
Step 2: Complete the Square \textit{Inside the Braces}\vskip 2mm
\qquad Quadratic Term: \textit{Inside} the braces, our quadratic term has a coefficient of 1.\vskip 2mm
\qquad Linear Term: \textit{Inside} the braces, we set \({\color{blue} 2B = 6}\).\vskip 2mm
A quick check and we see that we are matching the perfect-square pattern \textit{inside the braces}.
\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*}
Let’s move on to the constant term.\vskip 2mm
\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}.
\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
Still keeping our attention focused \textit{inside} the braces, we check to see if we match the perfect-square pattern.
\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*}
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 \((x + B)^2\).
\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
Step 3: Adjusting the Final Form\vskip 2mm
Start by multiplying the leading coefficient \textit{through the square braces} (NOT the parentheses!), and simplify.
\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
\textbf{Answer:} \(2x^2 + 12x – 5 = 2(x+3)^2 – 23\)
\vskip 1.5cm
\textbf{\large Quick Example}\vskip 2mm
\textbf{Example 4: A Quick Example}\vskip 2mm
Complete the Square on \(3x^2 – 4x + 8\).\vskip 5mm
\textbf{Solution:}\vskip 2mm
\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
\textbf{Answer:} \(3x^2 – 4x + 8 = 3\left(x – \frac 2 3\right)^2 + \frac{20} 3\)
\vskip 1.5cm
\textbf{\large The Quick Method}\vskip 2mm
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:
\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}
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
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
\textbf{Example 5: Simple Quick Method}\vskip 2mm
Use the quick version of Completing the Square on \(x^2 + 14x + 40\).
\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
\textbf{Answer:} \(x^2 + 14x + 40 = (x + 7)^2 – 9\)
Note: This is the same answer we got in Example 1.\vskip 1.5cm
\textbf{Example 6: Quick Method with Leading Coefficient}\vskip 2mm
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
Let’s complete the square on \(5x^2 + 80x + 13\).
\noindent\textbf{Solution}\vskip 2mm
Step 1) Factor out the leading coefficient.
\[%
5x^2 + 80x + 13 = 5[x^2 + 16x] + 13
\]
Step 2) Now use the quick method to complete the square on the \textit{inside} of the braces.
\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
\textbf{Answer:} \(5x^2 + 80x + 13 = 5(x + 8)^2 – 307\)
\vskip 1.5cm
\textbf{\large Solving Equations}\vskip 2mm
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
But that doesn’t mean we shouldn’t know \textit{how} to solve more complicated quadratic equations with this technique.\vskip 5mm
\textbf{Two Approaches}\vskip 2mm
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
\textbf{Example 7: Solving a Quadratic Equation by Completing the Square}\vskip 2mm
Use Completing the Square to solve \(4x^2 + 3x – 10 = 0\).\vskip 5mm
\textbf{Solution (Method 1)}\vskip 2mm
For this approach, we’ll first complete the square, then solve the equation.\vskip 2mm
Step 1) Complete the Square on \(4x^2 + 3x – 10\).\vskip 2mm
Using the techniques we’ve discussed above, we get the following.
\[%
4x^2 + 3x – 10 = 4\left(x + \frac 3 8\right)^2 – \frac{169}{16}
\]
\vskip 5mm
Step 2) Solve the equation.
\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*}
The two solutions are
\[%
\begin{array}{rl}
x & = \displaystyle – \frac 3 8 + \frac {13} 8 = \frac{10} 8 = \frac 5 4\\[12pt]
x & = \displaystyle – \frac 3 8 – \frac{13} 8 = – \frac{16} 8 = – 2
\end{array}
\]
\vskip 5mm
\textbf{Solution (Method 2)}\vskip 2mm
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
\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*}
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
\textbf{Answer:} \(x = \frac 5 4\) or \(x = -2\)
\vskip 1.5cm
\textbf{Conclusion}\vskip 2mm
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.
The post Completing the Square appeared first on Math Academy.
]]>The post Find the Vertex by Using the Quadratic Formula appeared first on Math Academy.
]]>The Quadratic Formula is primarily used to identify the roots (\(x\)-intercepts) of a quadratic function. What many people don’t know is that you can also easily find the vertex of the function by simply looking at the Quadratic Formula!
When graphing quadratic functions by hand, we will often use the quadratic formula to obtain the roots (or \(x\)-intercepts) of the function. Then separately, we will work on finding the vertex of the function since this point defines the lowest (or highest) point on the graph, as well as the line of symmetry.
For the general quadratic function, \(f(x) = ax^2 + bx + c\), the quadratic formula identifies the roots at:
\[%
x = \frac{-b \pm\sqrt{b^2 – 4ac}}{2a}
\]
while the the vertex is found at \(\left(-\frac b {2a}, f\left(-\frac b {2a}\right)\right)\).
The \(x\)-value of the vertex, \(x = -\frac b {2a}\), is often referred to in algebra textbooks as The Vertex Formula.
Many people notice (or are taught) that the Vertex Formula can be found nestled in the Quadratic Formula. Specifically,
\[%
\mbox{Quadratic Formula: } x = \frac{\color{blue}{-b} \pm\sqrt{b^2 – 4ac}}{\color{blue}{2a}}\longrightarrow \color{blue}{\frac{-b}{2a}} = x\mbox{-value of the vertex.}
\]
Less well known is the fact that the \(y\)-coordinate of the vertex can also be obtained easily from the quadratic formula. The vertex of a quadratic function can be written as
\[%
\mbox{Vertex: } \left(\color{blue}{-\frac b {2a}}, \color{red}{-\frac D {4a}}\right)
\]
where \(D\) is the discriminant of the quadratic function. Recall that the discriminant of a quadratic function is \(\color{red}{D = b^2 – 4ac}\) and it appears under the radical in the quadratic formula:
\[%
\mbox{Quadratic Formula: } x = \frac{-b \pm\sqrt{\color{red}{b^2 – 4ac}}}{2a} = \frac{-b \pm\sqrt{\color{red} D}}{2a}.
\]
Also, notice that the denominator of the \(x\)-coordinate and \(y\)-coordinate of the vertex are very similar. The \(y\)-coordinate’s denominator is just twice as big as the \(x\)-coordinate’s!
Now, let’s try out this idea and find the vertex of a quadratic from just the quadratic formula.
Suppose \(f(x) = 5x^2 + 4x – 3\). Use the quadratic formula to identify the entire vertex of the function, and then verify the answer using the standard method.
Solution: The quadratic formula for this function is
\[%
x = \frac{-4 \pm \sqrt{4^2 – 4(5)(-3)}}{2(5)} = \frac{-4\pm \sqrt{76}}{10}.
\]
The discriminant for this function is \(\color{red}{D = 76}\). So, the vertex should be at
\[%
\mbox{Vertex: } \left(\color{blue}{-\frac b {2a}}, \color{red}{-\frac D {4a}}\right) = \left(-\frac 4 {10}, -\frac{\color{red}{76}}{20}\right) = \left(-\frac 2 5, -\frac{19} 5\right).
\]
Let’s double check our \(y\)-coordinate by using the standard method (plugging \(x = -\frac b {2a} = -\frac 2 5\) into the original function).
\begin{align*}
f\left(\color{blue}{-\frac 2 5}\right)
& = 5\left(\color{blue}{-\frac 2 5}\right)^2 + 4\left(\color{blue}{-\frac 2 5}\right) – 3 && \mbox{Plugging in the \(x\)-value.}\\[6pt]
& = 5\left(\frac 4 {25}\right) – \frac 8 5 – \frac 3 1 && \mbox{Simplifying.}\\[6pt]
& = \frac 4 5 – \frac 8 5 – \frac{15} 5\\[6pt]
& = \frac{-19} 5 && \mbox{\(y\)-value of the vertex.}
\end{align*}
Let’s try another example. Suppose \(f(x) = -2x^2 +12x -25\). Identify the vertex using the quadratic formula, and verify the answer with the standard method.
Solution: The quadratic formula for this function is
\[%
x = \frac{-12 \pm\sqrt{(-12)^2-4(-2)(-25)}}{2(-2)} = \frac{\color{blue}{-12} \pm\sqrt{\color{red}{-56}}}{-4}
\]
The vertex should be \(\displaystyle \left(\frac{\color{blue}{-12}}{-4}, -\frac{\color{red}{-56}}{-8}\right) = \left(3, -7\right)\). Let’s verify the result. Using the standard method, we evaluate \(f(3)\).
\begin{align*}
f(\color{blue} 3) & = -2(\color{blue} 3)^2 + 12(\color{blue} 3) – 25\\[6pt]
& = -2(9) + 36 – 25\\
& = -18 + 11\\
& = -7
\end{align*}
Again, we see the method works.
However, the savvy math student will immediately ask, “Does it work every time? Or was there something `special’ or `unusual’ about these particular quadratic functions?”
This is an important question. In order to claim something is valid in mathematics, we have to do more than look at a few examples. We have to prove its validity. So, without further ado, here is the proof of this technique.
Given: The \(x\)-coordinate of the vertex of a quadratic function is \(x = -\frac b {2a}\).
Show: The \(y\)-coordinate of the vertex of a quadratic function is \(y = -\frac D {2a}\), where \(D\) is the discriminant of the function.
Proof: Simply evaluate \(f(x) = ax^2 + bx + c\) at \(x = -\frac b {2a}\).
\begin{align*}
f\left(\color{blue}{-\frac b {2a}}\right)
& = a\left(\color{blue}{-\frac b {2a}}\right)^2 + b\left(\color{blue}{-\frac b {2a}}\right) + c\\[6pt]
& = a\left(\frac{b^2}{4a^2}\right) -\frac{b^2}{2a} + c && \mbox{Simplify a little.}\\[6pt]
& = \frac{b^2}{4a} -\frac{b^2}{2a} + \frac c 1 && \mbox{Cancel the common factor.}\\[6pt]
& = \frac{b^2}{4a} -\frac{2b^2}{4a} + \frac{4ac}{4a} && \mbox{Common denominator.}\\[6pt]
& = \frac{-b^2 + 4ac}{4a} && \mbox{Combine numerators.}\\[6pt]
& = -\frac{\color{red}{b^2 – 4ac}}{4a} && \mbox{Factor out a negative 1.}\\[6pt]
& = -\frac{\color{red}{D}}{4a}
\end{align*}
Well, how about that. It seems like any time we evaluate a quadratic at \(x = -\frac b {2a}\), we will end up with \(-\frac D {4a}\).
So, what does this do for us in practical terms? Suppose we were asked to graph (by hand) the function \(f(x) = 3x^2 -6x + 8\). A good first step is to find the roots, and a good method for that is to use the quadratic formula.
\begin{align*}
x & = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\\[6pt]
& = \frac{6 \pm \sqrt{(-6)^2 – 4(3)(8)}}{2(3)}\\[6pt]
& = \frac{6 \pm \sqrt{-60}} 6
\end{align*}
Oops. This quadratic function doesn’t have real-valued \(x\)-intercepts (Yes, it has complex roots, but that won’t help us graph, will it?).
But the function has a graph, we just have to find it. If only we had the vertex and one other point! Then we could graph the quadratic! But wait! We already have written out the quadratic formula, so that should tell us the vertex:
\[%
\mbox{Vertex: } \left(-\frac b {2a}, -\frac D {4a}\right) = \left(-\frac 6 6, -\frac{-60}{12}\right) = (1, 5).
\]
Now, together with the \(y\)-intercept at \((0,8)\), we can easily plot the function.
In conclusion, the more we understand about the tools we have at our disposal, the easier it becomes to work through the mathematics that crop up in school and careers. And the more we learn, the more we become intrigued by the idea of other possibilities. I mean, doesn’t this insight into the quadratic formula just beg the question, “What else can we learn from the Quadratic Formula? Or the discriminant?”
© HT Goodwill and www.mathacademytutoring.com, 2016. Unauthorized use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to HT Goodwill and mathacademytutoring.com with appropriate and specific direction to the original content.
The post Find the Vertex by Using the Quadratic Formula appeared first on Math Academy.
]]>The post 10 Ways Tutoring Can Help You appeared first on Math Academy.
]]>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.
The post 10 Ways Tutoring Can Help You appeared first on Math Academy.
]]>The post Modular Arithmetic & Fermat’s Little Theorem appeared first on Math Academy.
]]>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 ℤ_{12} which is made up of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}. If we consider midnight as t_{0}=0, then 5 hours later is t_{5}=5. 17 hours later will also be t_{17}=5. 84 hours later will be t_{84} = 0. 85 hours later will be t_{85} = 1. There is a pattern here. The value a in the set ℤ_{12} 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”
More generally, we can write the set of integers for any modulus n as ℤ_{n} = {0, 1, 2, 3, … n-1} . Each element a in ℤ_{n} 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 ℤ_{n} 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
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 ℤ_{4} = {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).
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 a^{p-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 2^{784} is divided by 11? 2^{784} 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 2^{10} ≡ 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 2^{784} as the product of as many 2^{10}’s as possible. We can simplify as follows: 2^{784}= (2^{10})^{78}·(2^{4}) ≡ 1^{78} · 2^{4} (mod 11) ≡ 2^{4} (mod 11) ≡16 (mod 11) ≡ 5 (mod 11). The remainder is 5.
The post Modular Arithmetic & Fermat’s Little Theorem appeared first on Math Academy.
]]>The post Algebraic Rings appeared first on Math Academy.
]]>An algebraic ring is one of the most fundamental algebraic structures. It builds off of the idea of algebraic groups by adding a second operation (For more information please review our article on groups). For rings we often use the notation of addition and multiplication because the integers are a good analogy for a ring, but the operations “+” and “⋅” do not necessarily represent addition and multiplication in the ring. A ring is defined as a set S combined with two operations (“+” and “⋅”) that has the following properties: I. The set S is a group under one operation (we will call this operation “+”): a. It is closed under this operation: For all elements a and b in the set S, a+b is also in S. b. It contains an additive identity element (which we will call “0”, although it does not necessarily represent the integer 0): There is some element 0 in the set S such that for every element a in S, a+0 = 0+a = a. c. The operation is associative: For all a, b, and c in the set S, (a+b)+c = a+(b+c). d. Inverses exist: For every element a in the set S, there is a (-a) in S such that a+(-a)= (-a)+a = 0. II. The group operation is commutative: For all a and b in the set S, a + b = b + a. III. The second operation (written as “⋅”) follows these rules: a. It is closed under this operation. b. The set contains a multiplicative identity element (which we call “1”, although it does not necessarily represent the integer 1): such that for all elements a in the set S, a⋅1 = 1⋅a = a. c. It is associative: For all a, b, and c in the set S, (a⋅b)⋅c = a⋅(b⋅c). d. It is distributive over “+” on both the right and left side: For all elements a, b, and c in the set S, a⋅(b+c) = a⋅b + a⋅c and (b+c)⋅a = b⋅a + c⋅a. Note that inverses are not required for the “⋅” operation. If inverses do exist under “⋅”, we have a special kind of ring called a field. We will now look at some examples of rings and sets that aren’t rings. Example 1. The integers under addition and multiplication. I. The integers are a group under addition (see our group article). I. Addition can be carried out in any order so it is commutative. III. Multiplication follows the additional rules laid out for rings. The integers are closed under multiplication. The identity element is 1. Multiplication is associative and it is also distributive. Therefore, the integers are a ring under addition and multiplication. ∎ This is the reason addition, multiplication, 0, and 1 show up in our notation. If a set “acts like” the integers by following the ring axioms, it is a ring. Example 2. The set of 2x2 matrices under addition and multiplication. I. The set of 2x2 matrices form a group under addition. a. For any 2x2 matrices A and B, A+B is also a 2x2 matrix. b. The additive identity “0” is the zero matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$. c. Matrix addition is associative. d. The additive inverse of a matrix A is just -A. Observe $\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} + \begin{bmatrix} -a & -b \\ -c & -d \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$. II. Addition of 2x2 matrices is commutative, A+B = B+A. III. Multiplication of 2x2 matrices satisfies the following conditions: a. It is closed under the operation. The multiplication of two 2x2 matrices results in another 2x2 matrix. b. The identity matrix is I = $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$ and for any 2x2 matrix A, A⋅I = I⋅A = A. c. Matrix multiplication is associative. d. Matrix multiplication distributes over addition on the left and right side (This can be easily proven, but to save space it will not be proven here). Therefore, the set of 2x2 matrices is a ring. ∎ Example 3. The set of 2x2 matrices with non-negative entries under addition and multiplication. This set satisfies all of the requirements for the multiplication operation, but it is not a group under addition. Without negative entries, most matrices in this set do not have additive inverses. Therefore, this set is not a ring. ∎ Like groups, rings do not have to be composed of real numbers or matrices. Some examples include sets of polynomials, the complex numbers, and integers modulo n. The word “ring” comes from the German word “Zahlring,” which means a collection of numbers that circle back on themselves. However, now that the field has been studied more extensively, we know that cyclic rings are just one type of rings. The symbol ℤ for the integers is also named after a German word for numbers “Zahlen.”
The post Algebraic Rings appeared first on Math Academy.
]]>The post Algebraic Groups appeared first on Math Academy.
]]>One of the most fundamental algebraic structures in mathematics is the group. A group is a set of elements paired with an operation that satisfies the following four conditions: I. It is closed under an operation (represented here by “+”, although it does not necessarily mean addition): For all elements a and b in the set S, a+b is also in S. II. It contains an identity element (often written as “e”): There is some element e in the set S such that for every element a in S, a+e = e+a = a. III. The operation is associative: For all a, b, and c in the set S, (a+b)+c = a+(b+c). IV. Inverses exist: For every element a in the set S, there is an a^{-1} in S such that a+a^{-1} = a^{-1}+a = e We will look at some examples of groups and sets that aren’t groups. Example 1. Integers under addition, (ℤ, +). I. The group operation is addition. For any two integers n and m, n+m is also an integer. II. The identity element is 0 because for any integer n, n+0 = 0+n = n. III. Addition is associative. IV. Inverses exist. The inverse of any integer n is -n, and n + (-n) = (-n) + n = 0. Therefore, (ℤ, +) is a group. ∎ Example 2. Integers under multiplication, (ℤ, ⋅) I. The group is closed under multiplication, because for any two integers n and m, n⋅m is also an integer. II. The identity element is 1. For any integer n, n⋅1 = 1⋅n = n. III. Multiplication is associative. IV. Inverses do NOT exist. For any integer n, 1/n is not an integer except when n=1. Since the the set does not meet all four criteria, (ℤ, ⋅) is not a group. ∎ Example 3. The set of rational numbers not including zero, under multiplication (ℚ – 0, ⋅) I. The group is closed under multiplication. II. 1 is the identity element. III. Multiplication is associative. IV. Inverses exist in this group. If q is a rational number, 1/q is also a rational number. And q⋅ (1/q) = (1/q) ⋅ q = 1. Notice that if 0 were in this set, it would not be a group because 0 has no inverse. Therefore, ℚ – 0 is a group. ∎ Example 4. The set {0, 1, 2} under addition I. This set is not closed under addition because 1 + 2 = 3, and 3 is not part of the set. Therefore, the set cannot be a group. ∎ Example 5. The trivial group {e} This group consists only of the identity element. We don’t need to specify the operation here because it works for both multiplication and addition. For addition, e=0. 0+0=0 so the group is closed under addition, has an identity element, and is closed under inverses (and addition is associative). For multiplication, e=1 and similarly, 1⋅1 = 1. ∎ The examples here involved only numbers, but there are many different types of groups. For example, the ways to transform a triangle is called a Dihedral Group, with the group operation being the act of rotating or reflecting the shape around an axis. The possible permutations of a Rubik’s Cube are also a group, with the operation being a sequence of moves. It’s powerful to know that a set is a group because it gives you an understanding of how the elements will always behave. Groups have many other properties that are useful to mathematicians, and there is a whole field of study built off of this knowledge called Group Theory.
The post Algebraic Groups appeared first on Math Academy.
]]>The post L’Hospital’s Rule appeared first on Math Academy.
]]>L’Hospital’s Rule is a useful way to evaluate tricky limits. It is most often used for limits of indeterminate form. The rule is as follows:
If $f(x)$ and $g(x)$ are differentiable on some interval around the number $a$ (or if $a=∞$, $f(x)$ and $g(x)$ are differentiable for all $x>ε$ for some $ε$), and ${\lim\limits_{x \to a} g(x)} ≠ 0$, then
$$\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$$
We can take the derivatives of the numerator and the denominator and then take the limit and it will be the same as the limit of the original ratio. Indeterminate form includes limits of the forms $\frac{0}{0}$, $\frac{∞}{∞}$, $0\cdot∞$, $∞-∞$, $0^0$, $1^∞$ and $∞^0$.
Example 1. $\lim\limits_{x \to 0} \frac{\sin x}{x}$
In this example, we see that taking the limit of the numerator and denominator results in $\frac{0}{0}$, which is an indeterminate form. And since there is not an obvious way to simplify the fraction in order to make the limit easier to compute, it is a good candidate for L’Hospital’s Rule. We take the limit of the numerator and denominator as follows:
$$\lim\limits_{x \to 0} \frac{\sin x}{x} = \lim\limits_{x \to 0} \frac{\frac d{dx} (\sin x)}{\frac d{dx} (x)} = \lim\limits_{x \to 0} \frac{\cos x}{1} = cos(0) = 1 $$
Example 2. $\lim\limits_{x \to ∞} \frac{e^x}{x^2+x+4}$
Here we see that attempting to take the limit as is results in $\frac{∞}{∞}$. So again we use L’Hospital’s Rule:
$$\lim\limits_{x \to ∞} \frac{e^x}{x^2+x+4} = \lim\limits_{x \to ∞} \frac{\frac d{dx} (e^x)}{\frac d{dx} (x^2+x+4)} = \lim\limits_{x \to ∞} \frac{e^x}{2x+1}$$
Taking the limit at this point still results in $\frac{∞}{∞}$, so we apply L’Hospital’s Rule again.
$$\lim\limits_{x \to ∞} \frac{e^x}{2x+1} = \lim\limits_{x \to ∞} \frac{\frac d{dx} (e^x)}{\frac d{dx} (2x+1)} = \lim\limits_{x \to ∞} \frac{e^x}{2} = ∞$$
$∞$ is not an indeterminate form and so that is our final answer.
Example 3. $\lim\limits_{x \to 0} (x \cdot \ln x)$
If we plug zero into the limit, we get the indeterminant form $0 \cdot (-∞)$. We can’t apply L’Hospital’s Rule right away because we do not have a ratio. Instead, we must manipulate the function so that it becomes a ratio. We can move $x$ to the denominator by rewriting as $ \frac {1}{x}$.
$$ \lim\limits_{x \to 0} (x \cdot \ln x) = \lim\limits_{x \to 0} \frac{\ln x}{\frac {1}{x}}$$
Now we have a ratio and the indeterminate form $\frac {-∞}{∞}$. Applying L’Hospital’s Rule,
$$ \lim\limits_{x \to 0} \frac{\ln x}{\frac {1}{x}} = \lim\limits_{x \to 0} \frac{\frac d{dx} (\ln x)}{\frac d{dx} (\frac {1}{x})} = \lim\limits_{x \to 0} \frac{ \frac {1}{x}}{- \frac {1}{x^2}}$$
We still have an indeterminate form $\frac {∞}{-∞}$, but before applying L’Hospital’s Rule again, we should notice that we can cancel the $\frac {1}{x}$ in the top and the bottom. It is always best to simplify after applying L’Hospital’s Rule if possible.
$$ \require{cancel} \lim\limits_{x \to 0} \frac{ \frac {1}{\cancel {x}}}{- \frac {1}{x^\cancel{2}}} = \lim\limits_{x \to 0} \frac {1}{- \frac{1}{x}} = \lim\limits_{x \to 0} (-x) = 0 $$
If we had tried to use L’Hospital’s Rule before canceling, our ratio would become more complicated with each application of the rule.
Example 4. $\lim\limits_{x \to ∞} \frac {e^x – e^{-x}}{e^x + e^{-x}}$
Although L’Hospital’s Rule can always be applied to functions that meet the criteria of being differentiable around an interval, it does not always yield a useful answer, even if we see indeterminate forms. Here we will try to use L’Hospital’s Rule.
$$\lim\limits_{x \to ∞} \frac {e^x – e^{-x}}{e^x + e^{-x}} = \lim\limits_{x \to ∞} \frac {\frac d{dx} (e^x – e^{-x})}{\frac d{dx}(e^x + e^{-x})} = \lim\limits_{x \to ∞} \frac {e^x + e^{-x}}{e^x – e^{-x}} $$
Noticing that this ratio isn’t any simpler, we can try to apply L’Hospital’s Rule again.
$$ \lim\limits_{x \to ∞} \frac {e^x + e^{-x}}{e^x – e^{-x}} = \lim\limits_{x \to ∞} \frac {\frac d{dx} (e^x + e^{-x})}{\frac d{dx}(e^x – e^{-x})} = \lim\limits_{x \to ∞} \frac {e^x – e^{-x}}{e^x + e^{-x}}$$
This is the original limit we started with, so we are not getting anywhere using L’Hospital’s Rule. Instead, we should try to simplify our problem by dividing everything by the element with the highest power, in this case $e^x$.
$$\lim\limits_{x \to ∞} \frac {e^x – e^{-x}}{e^x + e^{-x}} = \lim\limits_{x \to ∞} \frac {\frac {e^x}{e^x} – \frac {e^{-x}}{e^x}}{\frac {e^x}{e^x} + \frac{e^{-x}}{e^x}} = \lim\limits_{x \to ∞} \frac {1 – e^{-2x}}{1 + e^{-2x}} $$
Noticing that $e^{-2x}$ approaches $0$ as $x$ approaches infinity, we can now take the limit
$$\lim\limits_{x \to ∞} \frac {1 – e^{-2x}}{1 + e^{-2x}} = \frac {1-0}{1+0} = 1 $$
L’Hospital’s Rule is often useful to evaluate limits of indeterminate form. It is most useful when it makes the numerator and denominator simpler than they were before. Remember to always simplify after applying the rule. L’Hospital’s Rule is applicable to all limits of ratios of differentiable functions, but is not guaranteed to make the limit easier to find.
The post L’Hospital’s Rule appeared first on Math Academy.
]]>The post Absolute Value and Logarithms appeared first on Math Academy.
]]>Absolute values often turn up unexpectedly in problems involving logarithms. That’s because you can’t take the log of a negative number.
Let’s first review the definition of the logarithm function:
Log_{b }x = y ⇔ b^{y} = x
(The double arrow is a bi-conditional, which means that one side is true if and only if the other side is true).
There are several restrictions on this definition when using real numbers (those not involving i = √-1 ).
If any of these conditions are not met, we risk introducing imaginary solutions or an infinite number of solutions. To prevent this from happening, we must follow these rules when working with real numbers, and to do so we often need to introduce the absolute value.
Example 1. Expand the function f(x,y) = log(xy).
We must always have a positive number inside the log function, and since x and y are being multiplied together, they must then always have the same sign. Therefore, the domain of our function is xy>0 which means that f(x,y) lies in Quadrants I and III on the x-y plane.
To expand, we must follow the above rules, with emphasis in this case on Rule 3. To make sure that the values inside the log are positive, we introduce the absolute value.
log(xy) = log|x| + log|y|
If we had not included the absolute values, log(x) + log(y) would only be defined when x and y are both positive (Quadrant I). Since the domain is different than that of log(xy), they cannot be equivalent functions.
Example 2. Simplify the function g(x,y) = log|x| + log|y|
It is important to notice that the above solution is not true in the reverse direction. g(x,y) = log|x| + log|y| is defined for all real, nonzero values of x and y. Therefore, to simplify this function, we must make sure the domain remains all real nonzero numbers. To do so, we use the absolute value. Simplifying,
log|x| + log|y| = log|xy|
Example 3. Evaluate the indefinite integral ∫ 1⁄x dx
This common integral can be found in many integral tables. It is equal to ln|x| + C. The absolute value is important because this is an indefinite integral, which means x might range through the entire real number line (There is a singularity at x=0, but log(0) is undefined too). We introduce the absolute value into the log to ensure that the antiderivative is defined everywhere the integral is.
When working in the real numbers with log functions, remember to always follow the three rules listed above. For many expressions involving logs, it is often necessary to introduce absolute values to ensure that these rules are met.
The post Absolute Value and Logarithms appeared first on Math Academy.
]]>