Category Archives: Mathematics

Logistic Differential Equation

general solution

$$\frac{dy}{dt} = ky \left( M – y \right)$$
$$\frac{dy}{y \left( M – y \right)} = k \ dt$$
$$\int \! \frac{dy}{y \left( M – y \right)} = \int \! k \ dt = kt + C_0$$

In order to integrate the right side of the equation the integrand must be decomposed and integrated by the technique commonly known as “integration by partial fractions.”

$$\frac{1}{y \left( M – y \right)} = \frac{A}{y} + \frac{B}{ \left( M – y \right)}$$
$$1 = A \left( M – y \right) + By = \left( B – A \right)y + AM$$

or

$$0y + 1 = \left( B – A \right)y + AM$$

For this equation to be true the coefficients must be equal. Therefore,

$$B – A = 0$$ and $$AM=1$$

$$A=B$$
$$A = \frac{1}{M}$$

Therefore, the decomposition is

$$\frac{1}{y \left( M – y \right)} = \frac{1}{My} + \frac{1}{ M\left( M – y \right)}$$

Replacing the integrand with its decomposition and splitting the integral yields

$$\int \! \frac{dy}{y \left( M – y \right)} = \frac{1}{M} \left( \int \! \frac{dy}{y} + \int \! \frac{dy}{M – y} \right)$$

From above it follows

$$\frac{1}{M} \left( \int \! \frac{dy}{y} + \int \! \frac{dy}{M – y} \right) = kt + C_0$$

Now integrating and solving for \( y \)

$$\frac{1}{M} \left( \ln \left| y \right| – \ln \left| M – y \right| + C_1 \right) = kt + C_0$$
$$\ln \left|y \right| – \ln \left| M – y \right| = Mkt + C_2$$
$$\ln \left| \frac{y}{M-y} \right| = Mkt + C_2$$

exponentiate both sides

$$\left| \frac{y}{M-y} \right| = e^{Mkt + C_2} = C_3 e^{Mkt}$$

The absolute value yields two cases.

$$\left| \frac{y}{M-y} \right| = \begin{cases} \ \ \frac{y}{M-y}, & \frac{y}{M-y} \ge 0 \\ -\left( \frac{y}{M-y} \right), & \frac{y}{M-y} < 0 \end{cases}$$

Solving the case for \(y \ge 0\) and assuming that \(y \left( 0 \right) = y_0\)

$$\frac{y_0}{M-y_0} = C_3 e^0 = C_3$$
$$\frac{y}{M-y} = \frac{y_0 e^{Mkt}}{M-y_0}$$
$$y = \frac{y_0 e^{Mkt}}{M-y_0} \left( M – y \right)$$
$$y \left( 1 + \frac{y_0 e^{Mkt}}{M-y_0} \right) = \frac{y_0 M e^{Mkt}}{M-y_0}$$
$$y \left( \frac{M – y_0 + y_0 e^{Mkt}}{M-y_0} \right) = \frac{y_0 M e^{Mkt}}{M-y_0}$$
$$y = \frac{y_0 M e^{Mkt}}{M – y_0 + y_0 e^{Mkt}}$$
$$y = \frac{y_0 M}{ y_0 + \left( M – y_0 \right) e^{-Mkt}}$$

Derivatives of Some Trig Functions

In order to find the derivatives of the trigonometric functions the definition of the derivative must be used. Each of the following cases uses the form

$$ f’\left(x \right) = \lim_{h \rightarrow 0} \frac{f \left( x + h \right) – f \left( x \right)}{h} $$

Let \( f \left(x \right) = \sin \left( x \right) \)

$$ \lim_{h \rightarrow 0} \frac{ \sin \left( x + h \right) – \sin \left( x \right)}{h} $$

$$ \lim_{h \rightarrow 0} \frac{ \sin \left( x \right) \cos \left( h \right) + \sin \left( h \right) \cos \left( x \right) – \sin \left( x \right)}{h} $$

$$ \lim_{h \rightarrow 0} \frac{ \sin \left( x \right) \left( \cos h – 1 \right) + \cos \left( x \right) \sin \left( h \right)}{h} $$

$$ \lim_{h \rightarrow 0} \frac{ \sin \left( x \right) \left( \cos h – 1 \right)}{h} + \frac{\cos \left( x \right) \sin \left( h \right)}{h} = \cos \left( x \right) $$

Let \( f \left(x \right) = \cos x \)

$$ \lim_{h \rightarrow 0} \frac{ \cos \left( x + h \right) – \cos x}{h} $$

$$ \lim_{h \rightarrow 0} \frac{ \cos x \cos h – \sin h \sin x – \cos x}{h} $$

$$ \lim_{h \rightarrow 0} \frac{ \cos x \left( \cos h – 1 \right) – \sin h \sin x}{h} $$

$$ \lim_{h \rightarrow 0} \frac{ \cos x \left( \cos h – 1 \right)}{h} – \frac{\sin x \sin h}{h} = – \sin x $$

Let \( f \left(x \right) = \tan x \)

$$\lim_{h \rightarrow 0} \frac{ \tan \left( x + h \right) – \tan x}{h}$$

$$\lim_{h \rightarrow 0} \frac{ \frac{\sin \left(x + h \right)}{\cos \left(x + h \right)} – \frac{ \sin x}{\cos x}}{h}$$

$$ \lim_{h \rightarrow 0} \frac{\sin \left(x + h \right) \cos x – \sin x \left( \cos \left( x + h \right) \right)}{h \cos \left(x + h \right) \cos x} $$

$$ \lim_{h \rightarrow 0} \frac{\left( \sin x \cos h + \sin h \cos x \right) \cos x – \sin x \left( \cos x \cos h – \sin x \sin h \right)}{h \cos \left(x + h \right) \cos x} $$

$$ \lim_{h \rightarrow 0} \frac{ \sin h \cos^2 x}{h \cos \left(x + h \right) \cos x} + \frac{\sin^2 x \sin h}{h \cos \left(x + h \right) \cos x}= 1 + \tan^2 x = \sec^2 x $$

Let \( f \left(x \right) = \csc x \)

$$ \lim_{h \rightarrow 0} \frac{ \csc \left( x + h \right) – \csc x}{h} $$

$$ \lim_{h \rightarrow 0} \frac{ \frac{1}{\sin \left(x + h \right)} – \frac{1}{\sin x}}{h} $$

$$ \lim_{h \rightarrow 0} \frac{\sin x – \sin \left( x + h \right)}{h \sin \left(x + h \right) \sin x} $$

$$ \lim_{h \rightarrow 0} \frac{ \sin x – \sin x \cos h – \sin h \cos x}{h \sin \left(x + h \right) \sin x} $$

$$ \lim_{h \rightarrow 0} \frac{ \sin x \left( 1 – \cos h \right)}{h \sin \left(x + h \right) \sin x} – \frac{\sin h \cos x}{h \sin \left(x + h \right) \sin x}= – \cot x \csc x $$

Let \( f \left(x \right) = \sec x \)

$$\lim_{h \rightarrow 0} \frac{ \sec \left( x + h \right) – \sec x}{h}$$

$$\lim_{h \rightarrow 0} \frac{ \frac{1}{\cos \left(x + h \right)} – \frac{1}{\cos x}}{h}$$

$$\lim_{h \rightarrow 0} \frac{\cos x – \cos \left( x + h \right)}{h \cos \left(x + h \right) \cos x}$$

$$\lim_{h \rightarrow 0} \frac{ \cos x – \cos x \cos h + \sin h \sin x}{h \cos \left(x + h \right) \cos x}$$

$$\lim_{h \rightarrow 0} \frac{ \cos x \left( 1 – \cos h \right)}{h \cos \left(x + h \right) \cos x} + \frac{\sin h \sin x}{h \cos \left(x + h \right) \cos x}= \tan x \sec x$$

Mathematical Induction Proofs

In this section are completed proofs using induction. As time progresses I will slowly begin to improve the English explanation included in each proof. Below are the proof statements.

I. Bernoulli’s Inequality: If \( x > -1 \) then \( \forall n \in \mathbb{N} \) it is the case that \( \left( 1 + x \right)^n \ge 1 + nx \)

II. Prove that \( n^3 + 5n \) is divisible by \( 6 \ \forall n \in \mathbb{N} \)

III. Prove \( 1^2 – 2^2 +3^2 + \ldots + \left(-1 \right)^{n+1}n^2 = \left(-1 \right)^{n+1} \frac{n \left(n + 1 \right)}2 \ \forall n \in \mathbb{N} \)

IV. Prove using induction $$ E_n \left( x \right) = 1 + x + \frac{ x^2 }{2} + \frac{ x^3 }{3!}+ \cdots + \frac{ x^n }{n!} $$

I. Bernoulli’s Inequality:
If \( x > -1 \) then \( \forall n \in \mathbb{N} \) it is the case that \( \left( 1 + x \right)^n \ge 1 + nx \) \( \left( 1 + x \right)^1 \ge 1 + \left(1 \right) x \) By the inductive hypothesis \( \forall k \in \mathbb{N} \) it is the case that \( \left( 1 + x \right)^k \ge 1 + kx \). Since \( \left( 1 + x \right) > 0 \) it is true that \( \left( 1 + x \right)^k \left( 1 + x \right) \ge \left( 1 + kx \right) \left( 1 + x \right)\) $$ \left( 1 + x \right)^{k + 1} \ge 1 + x + kx + kx^2 = 1 + \left(k + 1 \right)x + kx^2 \ge 1 + \left(k + 1 \right)x $$. Therefore, by mathematical induction Bernoulli’s equation is correct.
II. Prove that \( n^3 + 5n \) is divisible by \( 6 \ \forall n \in \mathbb{N} \)
By definition \( 6 \vert \left(n^3 + 5n \right) \) if and only if there exists some \( m \in \mathbb{Z} \) such that \( n^3 + 5n = 6m \).
\( \left(1 \right)^3 + 5\left(1\right) = 6m \)
By the inductive hypothesis it is true that \( k^3 + 5k = 6m \) for all \( k \in \mathbb{N} \) and for some \( m \in \mathbb{Z} \)
\( k^3 + 5k +3k \left( 3k + 1 \right) + 6= 6m + 3k \left( 3k + 1 \right) + 6 \)
\( \left(k + 1 \right)^3 + 5 \left(k + 1 \right)= 6m + 3k \left( 3k + 1 \right) + 6 \)
If the quantity \( 3k \left( 3k + 1 \right) \) contains a factor of six then the proof is complete.
Since \( k \in \mathbb{Z}\) and since all integers are either even or odd then
i. if \( k \) is even then by definition \( k = 2p \) which implies \( 3k \left(3k + 1 \right) = 6p \left(6p + 1 \right)=6\left(6p + p \right) \)
ii. if \( k \) is odd then by definition \( k = 2p +1 \) which implies \( 3k \left(3k + 1 \right) = \left( 6p +3 \right) \left(6p + 4 \right)=6 \left( 6p^2 +7p + 2 \right) \)
III. Prove \( 1^2 – 2^2 +3^2 + \ldots + \left(-1 \right)^{n+1}n^2 = \left(-1 \right)^{n+1} \frac{n \left(n + 1 \right)}2 \ \forall n \in \mathbb{N} \)
\( \left(-1 \right)^2 1^2 = \left(-1 \right)^2 \frac{1 \left( 1 + 1 \right)}2 \).
By the inductive hypothesis \( \forall k \in \mathbb{N} \) it is the case that \( 1^2 – 2^2 +3^2 + \ldots + \left(-1 \right)^{k+1}k^2 = \left(-1 \right)^{k+1} \frac{k \left(k + 1 \right)}2 \). \( \sum_{l=1}^k \left(-1 \right)^{l+1}l^2 + \left(-1 \right)^{k+2} \left(k+1 \right)^2 = \left(-1 \right)^{k+1} \frac{k \left(k + 1 \right)}2 + \left(-1 \right)^{k+2} \left(k+1 \right)^2 \).
Simplifying the left-hand side and creating a common denominator on the right yields \( \sum_{l=1}^{k+1} \left(-1 \right)^{l+1}l^2 = \left(-1 \right)^{k+1} \frac{k \left(k + 1 \right)}2 + \frac{2 \left(-1 \right)^{k+2} \left(k+1 \right)^2}2 \).
\( \sum_{l=1}^{k+1} \left(-1 \right)^{l+1}l^2 = \left(-1 \right)^{k+1} \left(k+1 \right) \frac{k – 2 \left(k+1 \right)}2 = \left(-1 \right)^{k+1} \left(k+1 \right) \frac{\left(-1 \right) \left(k+2 \right)}2 = \left(-1 \right)^{k+2} \frac{\left(k + 1 \right) \left(k+2 \right)}2 \)
IV. Let \(E_n\) be a sequence of continuous functions defined as $$ E_1 \left( x \right) = 1 + x $$ $$ E_{ n + 1 } = 1 + \int_0^x E_n \left( x \right) \ dt $$ Prove using induction $$ E_n \left( x \right) = 1 + x + \frac{ x^2 }{2} + \frac{ x^3 }{3!}+ \cdots + \frac{ x^n }{n!} $$ \( \forall n \in \mathbb{N} \) and \( \forall x \in \mathbb{R} \)
First show that \( E_1 \) is true.$$ E_1 = 1 + x $$ $$E_2 = 1 + \int_0^x E_1 \left( x \right) \ dt = 1 + x + \frac{ x^2 }{2} $$
Since \(E_1\) is true then by the induction hypothesis $$ E_k \left( x \right) = 1 + \int_0^x E_{k – 1} \left( x \right) \ dt = 1 + x + \frac{ x^2 }{2} + \frac{ x^3 }{3!}+ \cdots + \frac{ x^k }{k!} $$ $$E_k + \frac{ x^{ k + 1 }}{ \left( k + 1 \right)! } = 1 + x + \frac{ x^2 }{2} + \frac{ x^3 }{3!}+ \cdots + \frac{ x^k }{k!} + \frac{ x^{ k + 1 }}{ \left( k + 1 \right)! } = E_{ k + 1 } $$
therefore by mathematical induction
$$ E_n \left( x \right) = 1 + x + \frac{ x^2 }{2} + \frac{ x^3 }{3!}+ \cdots + \frac{ x^n }{n!} $$

Congruence Theorems I-IX

Congruence
Let \(a,b,n \in \mathbb{Z}\) and \(n>0\). \(a\) and \(b\) are congruent modulo \(n\) if and only if \(n \vert \left(a-b \right)\) and it is written
$$a \equiv b\; (\text{mod }n) $$
Algebraic Properties
I. Reflexive: \( n \vert 0 \Rightarrow a \equiv a\; ( \text{mod }n) \)
Proof: By definition \(n \vert 0 \) iff \( \exists k \in \mathbb{Z} \ni 0=nk\) and since \( 0=a-a=nk\) then by definition \(a \equiv a \; ( \text{mod }n) \)
II. Symmetric: \( a \equiv b\; ( \text{mod }n)\ \Rightarrow b \equiv a \; ( \text{mod }n) \)
Proof: By definition \( \left(a-b \right)=nk \) and \( -\left(b-a \right)=n \left(-k\right) \) which implies \(b \equiv a \; ( \text{mod }n) \)
III. Transitive: If \( a \equiv b\; ( \text{mod }n) \) and \( b \equiv c \; ( \text{mod }n) \) then \( a \equiv c \; ( \text{mod }n) \)
Proof: By definition \( \left(a-b \right)=nk\) and \( -\left(b-c \right)=nl\). Also, \(a=nk+b\) and \(c=b-nl\) and \(a-c=nk+b-b+nl=nk+nl=n \left(k+l \right)\) which implies \(a \equiv c\; ( \text{mod }n) \)
IV. If \(a \equiv b\; ( \text{mod }n) \) and \(c \equiv d\; ( \text{mod }n) \) then \(a+c \equiv b+d\; ( \text{mod }n) \)
Proof: By definition \( \left(a-b \right)=nk\) and \( \left(c-d \right)=nl\). Also, \(a=nk+b\) and \(c=d+nl\) and \(a+c=nk+b+d+nl=b+d+nk+nl=\left(b+d \right) + n\left(k+l\right)\) which implies \(a + c \equiv b + d \; ( \text{mod }n) \) since \(\left(a+c \right) – \left(b+d \right) = n\left(k+l\right).\)
V. If \(a \equiv b\; ( \text{mod }n) \) and \(c \equiv d\; ( \text{mod }n) \) then \(a-c \equiv b-d\; ( \text{mod }n) \)
Proof: By definition \( \left(a-b \right)=nk\) and \( \left(c-d \right)=nl\). Also, \(a=nk+b\) and \(c=d+nl\) and \(a-c=nk+b-d-nl=b-d+nk-nl=\left(b-d \right) + n\left(k-l\right)\) which implies \(a – c \equiv b – d\; ( \text{mod }n) \) since \( \left(a-c \right) – \left(b-d \right) = n\left(k-l\right).\)
VI. If \(a \equiv b\; ( \text{mod }n) \) and \(c \equiv d\; ( \text{mod }n) \) then \(ac \equiv bd\; ( \text{mod }n) \)
Proof: By definition \( \left(a-b \right)=nk\) and \( \left(c-d \right)=nl\). Also, \(a=nk+b\) and \(c=d+nl\) and \(ac=\left(nk+b\right)\left(d+nl\right)=dnk+n^2kl+bd+nbl=bd+n\left(dk+nkl+bl\right)\) which implies \(ac \equiv bd \; ( \text{mod }n) \) since \(ac – bd =n \left(dk+nk+bl\right)\)
VII. If \(a \equiv b\; ( \text{mod }n) \) then \(a^k \equiv b^k\; ( \text{mod }n) \)
Proof: By mathematical induction note that \( a^1 \equiv b^1\; ( \text{mod }n) \) is true by definition therefore, by the inductive hypothesis we assume \(a^t \equiv b^t \; ( \text{mod }n) \) to be true for all \(t\) contained in the natural numbers. It must be shown that \(a^{t+1} \equiv b^{t+1}\; ( \text{mod }n) \) is true. Notice \( aa^t=a^{t+1}\) and \( bb^t=b^{t+1}\). Since \( a \equiv b \; ( \text{mod }n)\ \) is true by definition and \(a^t \equiv b^t \; ( \text{mod }n) \) is true by the inductive hypothesis then by theorem VI \(a^{t+1} \equiv b^{t+1}\; ( \text{mod }n) \) is true.
VIII. If \(n\) is a natural number expressed in base \(10\) as \( n= a_k a_{k-1} \cdots a_1 a_0 \) and \( m= a_k + a_{k-1} + \ldots + a_1 + a_0 \) then \( n \equiv m\; ( \text{mod }3) \)
Proof: By definition \( n – m = 10^k a_k + 10^{k-1} a_{k-1} + \ldots + 10 a_1 + a_0 – a_k – a_{k-1} – \ldots – a_1 – a_0\). Simplifying and collecting like terms \(n – m = a_k \left( 10^k – 1 \right) + a_{k-1} \left( 10^{k-1} – 1 \right) + \ldots + a_1 \left ( 10 -1 \right)\) If \( 10^k \equiv 1^k \; (\text{mod }3) \) then \(n – m = a_k \left( 3t_k \right) + a_{k-1} \left( 3t_{k-1} \right) + \ldots + a_1 \left( 3t_1 \right)=3t \) where \( t = \left( a_kt_k + a_{k-1} t_{k-1} + \ldots + a_1 t_1 \right) \in \mathbb{Z}\) therefore, by definition \( n \equiv m \; ( \text{mod }3)\ \)
IX. If \(n\) is a natural number expressed in base \(10\) as \( n= a_k a_{k-1} \cdots a_1 a_0 \) and \( m= a_k + a_{k-1} + \ldots + a_1 + a_0 \) then \( n \equiv m \; ( \text{mod }9) \)
Proof: By definition \( n – m = 10^k a_k + 10^{k-1} a_{k-1} + \ldots + 10 a_1 + a_0 – a_k – a_{k-1} – \ldots – a_1 – a_0\). Simplifying and collecting like terms \(n – m = a_k \left( 10^k – 1 \right) + a_{k-1} \left( 10^{k-1} – 1 \right) + \ldots + a_1 \left ( 10 -1 \right)\) If \( 10^k \equiv 1^k \; ( \text{mod }9) \) then \(n – m = a_k \left( 9t_k \right) + a_{k-1} \left( 9t_{k-1} \right) + \ldots + a_1 \left ( 9t_1 \right)=9t\) where \( t = \left( a_kt_k + a_{k-1} t_{k-1} + \ldots + a_1 t_1 \right) \in \mathbb{Z}\) therefore, by definition \( n \equiv m \; ( \text{mod }9) \)

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 \( x=\frac{\pi}3 \)

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