Monthly Archives: March 2010

Proof of Mathematical Induction

Well-Ordering Property of the Natural Numbers (axiom)
Every non-empty subset of natural numbers has a least element.
Principle of Mathematical Induction (theorem)
Let S be a subset of \( \mathbb{N} \). If S possess the following two properties:
(i) \( 1 \in S \)
(ii) \( \forall j \in \mathbb N, \left(j \in S \right) \rightarrow \left(j+1 \in S \right) \)

then \( S= \mathbb N \)

Proof: (Argue by contradiction by assuming \( S \neq \mathbb N \).)
If \( S \neq \mathbb N \) then \( \mathbb N \setminus S \) is nonempty by definition of complement.
\( \mathbb N \setminus S \) has a least element, k , by the Well-Ordering Principle of the Natural Numbers.
\( k \notin S \) by definition of the complement \( \mathbb N \setminus S \)
If \( 1 \in S \) then \( k>1 \) from the definition of \( \mathbb N \setminus S \).
\( k-1 \in S \).
By hypothesis (ii) if \( j=k-1 \in S \) then \( j+1= \left(k-1 \right) + 1 \in S \) which contradicts the assumption.

Consider a specific example
Assume \( S \neq \mathbb N \) and possess properties (i) and (ii)
Since hypothesis (i) demands that \( 1 \in S \) let \( S = \left \{1, 2, 3, 4, 5, 6 \right \} \)
\( \mathbb N \setminus S = \left \{7, 8, 9, 10, \dots , n, \dots \right \} \) by the definition of complement.
\( \mathbb N \setminus S \) has a least element \( k=7 \) by the Well-Ordering property of the natural numbers.
\( 7 \notin S \) by the definition of complement
\( 7 > 1 \) and since 7 is the least element of \( \mathbb N \setminus S \) then \( 7-1=6 \in S \), again, by the definition of the complement.
By hypothesis (ii) if \( 6 \in S \) then \( 6+1=7 \in S \) but this contradicts \( 7 \notin S \) which implies that the assumption \( S \neq \mathbb N \) is false.
Therefore, \( S= \mathbb N \)

Proof of the Quadratic Formula

The following is a derivation of the quadratic formula. Assume \(a, b, c \in \mathbb{R}\)

\( ax^2 + bx + c = 0 \)

The first step will be to complete the square. Factor out \(a\).

\( a \left( x^2 + \frac{b}{a}x + \frac{c}{a} \right) =0 \)

Now add and subtract the square of half of the coefficient of the \(x\) term.

\( a \left( x^2 + \frac{b}{a}x + \frac{c}{a} + \left( \frac{b}{2a} \right)^2 – \left( \frac{b}{2a} \right)^2 \right) = 0 \)

Now complete the square.

\( a \left[ \left( x + \frac{b}{2a} \right)^2 + \frac{c}{a} – \left( \frac{b}{2a} \right)^2 \right] = 0 \)

Now solve for \(x\).

Either \( a=0 \) or \( \left( x + \frac{b}{2a} \right)^2 + \frac{c}{a} – \left( \frac{b}{2a} \right)^2 = 0 \)

Isolate \(x\)

\( \left( x + \frac{b}{2a} \right)^2 = \frac{b^2}{4a^2} – \frac{c}{a} \)

Take the square root of both sides

\( \left| x + \frac{b}{2a} \right| = \frac{ \sqrt {b^2 – 4ac}}{2a} \)

Apply the definition of absolute value to determine intervals.

\( \left| x + \frac{b}{2a} \right| = \begin{cases} \ \ x + \frac{b}{2a}, & x + \frac{b}{2a} \ge 0 \\ -\left( x + \frac{b}{2a} \right), & x + \frac{b}{2a} < 0 \end{cases} \)

For \(x \ge -\frac{b}{2a}\)

\( x + \frac{b}{2a} = \frac{ \sqrt {b^2 – 4ac}}{2a}\)

\( x = – \frac{b}{2a} + \frac{ \sqrt {b^2 – 4ac}}{2a}\)

For \(x < -\frac{b}{2a}\)

\( -\left( x + \frac{b}{2a} \right) = \frac{ \sqrt {b^2 – 4ac}}{2a} \)

\( x = – \frac{b}{2a} – \frac{ \sqrt {b^2 – 4ac}}{2a}\)

Therefore, the formula for the quadratic formula is written as

\( x = \frac{ -b \pm \sqrt {b^2 – 4ac}}{2a}\)

Review the specific example below

\( 3x^2 + 8x + 2 = 0 \)

\( 3 \left( x^2 + \frac{8}{3}x + \frac{2}{3} \right) =0 \)

\( 3 \left( x^2 + \frac{8}{3}x + \frac{2}{3} + \left( \frac{8}{2\cdot3} \right)^2 – \left( \frac{8}{2\cdot3} \right)^2 \right) = 0 \)

\( 3 \left[ \left( x + \frac{8}{6} \right)^2 + \frac{2}{3} – \left( \frac{8}{6} \right)^2 \right] = 0 \)

\( a \neq 0 \) we only need to consider \( \left( x + \frac{4}{3} \right)^2 + \frac{3}{2} – \left( \frac{4}{3} \right)^2 = 0 \)

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

\( \left| x + \frac{4}{3} \right| = \frac{ \sqrt {64 – 24}}{6} \)

\( \left| x + \frac{4}{3} \right| = \frac{ \sqrt {40}}{6} \)

\( \left| x + \frac{4}{3} \right| = \frac{ \sqrt{4} \sqrt {10}}{6} \)

\( \left| x + \frac{4}{3} \right| = \frac{ 2 \sqrt {10}}{6} \)

\( \left| x + \frac{4}{3} \right| = \frac{ \sqrt {10}}{3} \)

\( \left| x + \frac{4}{3} \right| = \begin{cases} \ \ x + \frac{4}{3}, & x + \frac{4}{3} \ge 0 \\ -\left( x + \frac{4}{3} \right), & x + \frac{4}{3} < 0 \end{cases} \)

\( x + \frac{4}{3} = \frac{ \sqrt {10}}{3}\)

\( x = \frac{ – 4 + \sqrt {10}}{3}\)

\( -\left( x + \frac{4}{3} \right) = \frac{ \sqrt {10}}{3}\)

\( x = \frac{ -4 – \sqrt {10}}{3}\)

\( \sin x \) centered about \( \frac{ \pi }3 \)

The Taylor Series for \( \sin x \) centered at \( x=\frac{\pi}3 \)

\( g(x)=\sin x \) \( g^{\prime}\left(\frac{\pi}3 \right)=\frac{1}2 \)
\( g^{\prime}(x)=\cos x \) \( g^{\prime}\left(\frac{\pi}3 \right)=\frac{1}2 \)
\( g^{\prime\prime}(x)=-\sin x \) \( g^{\prime\prime}\left(\frac{\pi}3 \right)=\frac{-\sqrt 3}2 \)
\( g^{\prime\prime\prime}(x)=-\cos x \) \( g^{\prime\prime\prime}\left(\frac{\pi}3 \right)=\frac{-1}2 \)
\( g^{(4)} (x)=\sin x \) \( g^{(4)}\left(\frac{\pi}3 \right)=\frac{\sqrt 3}2 \)
\(\sin x =\frac {\sqrt 3}{2 \cdot 0!}\left(x-\frac{\pi}3 \right)^{0}+\frac {1}{2\cdot 1!}\left(x-\frac {\pi}3 \right)^{1}-\frac {\sqrt 3}{2 \cdot 2!}\left(x-\frac {\pi}3 \right)^{2}-\frac {1}{2 \cdot 3!} \left(x- \frac{\pi}3 \right)^{3}+\frac {\sqrt 3}{2 \cdot 4!} \left(x- \frac{\pi}3 \right)^{4}+ \cdots\)
\(\sin x =\frac{\sqrt 3}{2} \left(1-\frac{1}{2!} \left(x-\frac{\pi}3 \right)^{2}+\frac{1}{4!} \left(x-\frac{\pi}3 \right)^{4}- \cdots \right)+ \frac{1}2 \left( \left( x – \frac{\pi}3 \right) – \frac{1}{3!} \left( x – \frac{\pi}3 \right)^{3} + \frac{1}{5!} \left( x – \frac{\pi}3 \right)^{5} + \cdots \right)\)
\(\sin x=\frac {\sqrt 3}2 \sum_{n=0}^{\infty} \frac{ \left(-1 \right)^n}{ \left(2n \right)!} \left( x – \frac {\pi}3 \right)^{2n}+ \frac {1}2 \sum_{n=0}^{\infty} \frac { \left(-1 \right)^{n}}{ \left(2n+1 \right)!} \left( x – \frac{\pi}3 \right)^{2n+1}\)