Matemathics Explanations of Eulers Identity

Math Explanation CODE relations

From Andraelity, how CODE using math.

In thy following lines we are going to verify or create the connexion that would allow us to unite code with mathematics

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EULERS IDENTITY We are going to detail information and ideas that could help us clarify from which place all this properties came from in a more cohesive and organized way to validate that every single proof that we set is presented in an order that could help us understand the reasons of equality that could elevate in the future to a order of probabilities, the orders we are structuring right now are base entirely on very well define amounts, that we could follow in a relationship of quantity, a quantity of numbers that we could start to count, and keep counting till we are able to reason about more inmense and overlimited set of items.

Index Webpages Redirect

Eulers Identity

Factorial Operator

$$n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1$$

$Factorial$	$Calculation$	$Result$
$4!$	$4 \times 3 \times 2 \times 1$	$24$
$3!$	$24 / 4$	$6$
$2!$	$6 / 3$	$2$
$1!$	$2 / 2$	$1$
$0!$	$1 / 1$	$1$

The next formula describe the behavior and the proof of the factorial operator

n! = \frac{(n + 1)!}{n + 1}

Permutational Formula

How many different ways can I pick a group of k items from a total set of n items,

if the order matters?

P (n, k) = \frac{n!}{(n - k)!}

P (10, 3) = \frac{10!}{(10 - 3)!} = \frac{10!}{7!} = \frac{10 \times 9 \times 8 \times \dots}{7 \times 6 \times \dots} = 10 \times 9 \times 8 = 720

\begin{array}{l} P (n, k) = \frac{n!}{(n - k)!} \\ P (3, 2) = \frac{3!}{(3 - 2)!} \\ = \frac{3!}{1!} \\ = \frac{3 \times 2 \times 1}{1} = 6 \end{array}

Binomial Coefficient / Combinations Formula / "n choose k"

How many different ways can I pick a group of k items from a total set of n items,

if the order doesn't matter?

C (n, k) = (\binom{n}{k}) = \frac{n!}{k! (n - k)!}

(\binom{5}{3}) = \frac{5!}{3! (5 - 3)!} = \frac{120}{6 \times 2} = 10

Binomial Theorem

(a + b)^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) a^{n - k} b^{k}

Example #1,

(x + 2)^{4}

(x + 2)^{4} = (x + 2) (x + 2) (x + 2) (x + 2)

(x + 2)^{4} = (x^{2} + 4 x + 4) (x^{2} + 4 x + 4)

(x + 2)^{4} = x^{4} + (4 x^{3} + 4 x^{3}) + (4 x^{2} + 16 x^{2} + 4 x^{2}) + (16 x + 16 x) + 16

(x + 2)^{4} = x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16

(x + 2)^{4} = \sum_{k = 0}^{4} (\binom{4}{k}) (x)^{4 - k} (2)^{k}

\begin{matrix} 1 (x^{4}) (2^{0}) = 1 \cdot x^{4} \cdot 1 = x^{4} \\ 4 (x^{3}) (2^{1}) = 4 \cdot x^{3} \cdot 2 = 8 x^{3} \\ 6 (x^{2}) (2^{2}) = 6 \cdot x^{2} \cdot 4 = 24 x^{2} \\ 4 (x^{1}) (2^{3}) = 4 \cdot x \cdot 8 = 32 x \\ 1 (x^{0}) (2^{4}) = 1 \cdot 1 \cdot 16 = 16 \end{matrix}

(x + 2)^{4} = x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16

Example #2,

{(2 x - 3 y)}^{3}

{(2 x - 3 y)}^{3} = (2 x - 3 y) (2 x - 3 y) (2 x - 3 y)

(2 x - 3 y) (2 x - 3 y) = 4 x^{2} - 12 x y + 9 y^{2}

{(2 x - 3 y)}^{3} = (4 x^{2} - 12 x y + 9 y^{2}) (2 x - 3 y)

{(2 x - 3 y)}^{3} = 8 x^{3} - 36 x^{2} y + 54 x y^{2} - 27 y^{3}

(a - b)^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) a^{n - k} (- b)^{k} = \sum_{k = 0}^{n} (- 1)^{k} (\binom{n}{k}) a^{n - k} b^{k}

{(2 x - 3 y)}^{3} = \sum_{k = 0}^{3} (- 1)^{k} (\binom{3}{k}) (2 x)^{3 - k} (3 y)^{k}

1 {(2 x)}^{3} {(- 3 y)}^{0} = 1 (8 x^{3}) (1) = 8 x^{3}

3 {(2 x)}^{2} {(- 3 y)}^{1} = 3 (4 x^{2}) (- 3 y) = - 36 x^{2 y}

3 {(2 x)}^{1} {(- 3 y)}^{2} = 3 (2 x) (9 y^{2}) = 54 x y^{2}

1 {(2 x)}^{0} {(- 3 y)}^{3} = 1 (1) (- 27 y^{3}) = - 27 y^{3}

{(2 x - 3 y)}^{3} = 8 x^{3} - 36 x^{2} y + 54 x y^{2} - 27 y^{3}

Euler's Number Existence

First proof of obtention

e = {(1 + \frac{1}{n})}^{n}

e = lim_{n \to \infty} {(1 + \frac{1}{n})}^{n}

e = {(1 + \frac{1}{10})}^{10} = \sum_{k = 0}^{10} (\binom{10}{k}) {(\frac{1}{10})}^{k}

= (\binom{10}{0}) {(\frac{1}{10})}^{0} + (\binom{10}{1}) {(\frac{1}{10})}^{1} + (\binom{10}{2}) {(\frac{1}{10})}^{2} + (\binom{10}{3}) {(\frac{1}{10})}^{3} + (\binom{10}{4}) {(\frac{1}{10})}^{4}

+ (\binom{10}{5}) {(\frac{1}{10})}^{5} + (\binom{10}{6}) {(\frac{1}{10})}^{6} + (\binom{10}{7}) {(\frac{1}{10})}^{7} + (\binom{10}{8}) {(\frac{1}{10})}^{8} + (\binom{10}{9}) {(\frac{1}{10})}^{9} + (\binom{10}{10}) {(\frac{1}{10})}^{10}

e = {(1 + \frac{1}{10})}^{10} = 1 + \frac{10}{10} + \frac{45}{100} + \frac{120}{1000} + \frac{210}{10000} + \frac{252}{100000}

+ \frac{210}{1000000} + \frac{120}{10000000} + \frac{45}{100000000} + \frac{10}{1000000000} + \frac{1}{10000000000} \approx 2.59374

$Target Accuracy$	$Required Value of n$	$Resulting Value$
$1 Decimal Place (2.7)$	$n = 14$	$2.701 \dots$
$2 Decimal Places (2.71)$	$n = 135$	$2.710 \dots$
$3 Decimal Places (2.718)$	$n = 1,354$	$2.717 \dots$
$4 Decimal Places (2.7182)$	$n = 13,534$	$2.7181 \dots$

Second proof of obtention

(a + b)^{n} = {(1 + \frac{1}{n})}^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) a^{n - k} b^{k} = (\binom{n}{0}) {(1)}^{n} {(\frac{1}{n})}^{0} + (\binom{n}{1}) {(1)}^{n - 1} {(\frac{1}{n})}^{1} + (\binom{n}{2}) {(1)}^{n - 2} {(\frac{1}{n})}^{2} + \dots

if k = 0, Binomial Theorem

(\binom{n}{0}) {(1)}^{n} {(\frac{1}{n})}^{0} = \frac{n!}{0! (n - 0)!} {(1)}^{n} {(\frac{1}{n})}^{0}

$$\text{Total} = \frac{\color{red}{\cancel{n!}}}{1 \cdot \color{red}{\cancel{n!}}} \cdot (1)^n \cdot \left( \frac{1}{n} \right)^0 = 1 \cdot 1 \cdot 1 = 1 $$

if k = 1, Binomial Theorem

(\binom{n}{1}) {(1)}^{n - 1} {(\frac{1}{n})}^{1} = \frac{n!}{1! (n - 1)!} {(1)}^{n - 1} {(\frac{1}{n})}^{1}

$$\text{Total} = \left( \frac{n \cdot (n - 1)!}{1 \cdot (n - 1)!} \right) \cdot (1)^{n - 1} \cdot \left(\frac{1}{n}\right)^1 $$ $$\\[5ex]$$ $$\text{Total} =\left( \frac{\color{blue}{\cancel{n}} \cdot \color{red}{\cancel{(n - 1)!}}}{1 \cdot \color{red}{\cancel{(n - 1)!}}} \right) \cdot (1)^{n - 1} \cdot \left(\frac{1}{\color{blue}{\cancel{n}}}\right)^1 = 1$$

if k = 2, Binomial Theorem

$$ \binom{n}{2} (1)^{n-2} \left(\frac{1}{n}\right)^2 = \left( \frac{n!}{2!(n-2)!} \right) \cdot (1)^{n-2} \cdot \left( \frac{1}{n} \right)^2$$ $$\\[5ex]$$ $$\text{Total} = \left( \frac{n!}{2!(n-2)!} \right) \cdot (1)^{n-2} \cdot \left( \frac{1}{n} \right)^2 = \frac{n(n-1)(n-2)!}{2 \cdot (n-2)!} \cdot 1 \cdot \frac{1}{n^2} $$ $$\\[5ex]$$ $$\text{Total} = \frac{\color{blue}{\cancel{n}}\color{white}{(n-1)}\color{red}{\cancel{(n-2)!}}} {2 \cdot \color{red}{\cancel{(n-2)!}}} \cdot 1 \cdot \frac{1}{ n \cdot \color{blue}{\cancel{n}} }$$ $$\\[5ex]$$ $$\text{Total} = \frac{n-1}{2} \cdot 1 \cdot \frac{1}{n}$$ $$\\[5ex]$$ $$\text{Total} = \frac{n-1}{2n}$$

if k = 3, Binomial Theorem

$$\binom{n}{3} (1)^{n-3} \left(\frac{1}{n}\right)^3 = \left( \frac{n!}{3!(n-3)!} \right) \cdot (1)^{n-3} \cdot \left( \frac{1}{n} \right)^3$$ $$\\[5ex]$$ $$\text{Total} =\left( \frac{n!}{3!(n-3)!} \right) \cdot (1)^{n-3} \cdot \left( \frac{1}{n} \right)^3 = \frac{n(n-1)(n-2)(n-3)!}{6 \cdot (n-3)!} \cdot (1)^{n-3} \cdot \frac{1}{n \cdot n^2}$$ $$\\[5ex]$$ $$\text{Total} = \frac{\color{blue}{\cancel{n}}\color{white}{(n-1)(n-2)}\color{red}{\cancel{(n-3)!}}}{6 \cdot \color{red}{\cancel{(n-3)!}}} \cdot \color{white}{(1)^{n-3}} \cdot \frac{1}{\color{blue}{\cancel{n}}\color{white}{\cdot n^2}}$$ $$\\[5ex]$$ $$\text{Total} = \frac{(n-1)(n-2)}{6} \cdot 1 \cdot \frac{1}{n^2}$$ $$\\[5ex]$$ $$\text{Total} = \frac{(n-1)(n-2)}{6n^2}$$

if k = k, Binomial Theorem

$$\\[5ex]$$ $$\binom{n}{k} (1)^{n-k} \left(\frac{1}{n}\right)^k = \left( \frac{n!}{k!(n-k)!} \right) \cdot (1)^{n-k} \cdot \left( \frac{1}{n} \right)^k$$ $$\\[5ex]$$ $$\text{Total} = \frac{n(n-1)(n-2)\cdots(n-(k-1))(n-k)!}{k! \cdot (n-k)!} \cdot (1)^{n-k} \cdot \frac{1}{n \cdot n^{k-1}}$$ $$\\[5ex]$$ $$\text{Total} = \frac{\color{blue}{\cancel{n}}\color{white}{(n-1)(n-2)\cdots(n-(k-1))}\color{red}{\cancel{(n-k)!}}}{\color{white}{k!} \cdot \color{red}{\cancel{(n-k)!}}} \cdot \color{white}{1} \cdot \frac{1}{\color{blue}{\cancel{n}}\color{white}{\cdot n^{k-1}}}$$ $$\\[5ex]$$ $$\text{Total} = \frac{(n-1)(n-2)\cdots(n-(k-1))}{k!} \cdot 1 \cdot \frac{1}{n^{k-1}}$$ $$\\[5ex]$$ $$\text{Total} = \frac{n-1}{n} \cdot \frac{n-2}{n} \dots \frac{n-(k-1)}{n} \cdot \frac{1}{k!}$$ $$\\[5ex]$$ $$\text{Total} = \left(1 - \frac{1}{n}\right) \cdot \left(1 - \frac{2}{n}\right) \dots \left(1 - \frac{k-1}{n}\right) \cdot \frac{1}{k!}$$ $$\\[5ex]$$

General Expression for Binomial

$$\\[5ex]$$ $$ e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n$$ $$\\[5ex]$$ $$(a + b)^n = \left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k = \sum_{k=0}^{n} \left(1 - \frac{1}{n}\right) \cdot \left(1 - \frac{2}{n}\right) \dots \left(1 - \frac{k-1}{n}\right) \cdot \frac{1}{k!}$$ $$\\[5ex]$$ $$\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = \lim_{n \to \infty} \sum_{k=0}^{n} \left(1 - \frac{1}{n}\right) \cdot \left(1 - \frac{2}{n}\right) \dots \left(1 - \frac{k-1}{n}\right) \cdot \frac{1}{k!}$$ $$\\[5ex]$$ $$\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^{n} \lim_{n \to \infty} \left(1 - \frac{1}{n}\right) \cdot \left(1 - \frac{2}{n}\right) \dots \left(1 - \frac{k-1}{n}\right) \cdot \frac{1}{k!}$$ $$\\[5ex]$$ $$\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^{n} (1 - 0) \cdot (1 - 0) \dots (1 - 0) \cdot \frac{1}{k!}$$ $$\\[5ex]$$ $$\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^{n} 1 \cdot 1 \dots 1 \cdot \frac{1}{k!} = \sum_{k=0}^{n} \frac{1}{k!}$$ $$\\[5ex]$$ $$ e = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^{n} \frac{1}{k!}$$

e = \sum_{n = 0}^{\infty} \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots

$Terms (n)$	$Calculation$	$Sum (e ≈)$	$Accuracy$
$1 term$	$1$	$1.0$	$Off by 63%$
$2 terms$	$1 + 1$	$2.0$	$Off by 26%$
$3 terms$	$2 + 0.5$	$2.5$	$Off by 8%$
$4 terms$	$2.5 + 0.1666...$	$2.6666...$	$Off by 1.9%$
$7 terms$	$\dots + 1 / 720$	$2.71805...$	$99.99% accurate$
$10 terms$	$\dots + 1 / 362,880$	$2.7182815...$	$6 decimal places$
$13 terms$	$\dots + 1 / 479,001,600$	$2.718281828...$	$10 decimal places$

Third proof of obtention

if n > 0

and if a_{0} = 1

and if a_{n} = \frac{a_{n - 1}}{n}

e = a_{0} + \sum_{n = 0}^{\infty} \frac{a_{0 + n}}{n + 1}

$Term (a_{n})$	$Contribution$	$Accuracy Margin (\frac{a_{n - 1}}{n})$	$Running Total$
$a_{0}$	$1$	$Undefined$	$1.0$
$a_{1}$	$1$	$\frac{a_{0}}{1}$	$2.0$
$a_{2}$	$0.5$	$\frac{a_{1}}{2}$	$2.5$
$a_{3}$	$0.166666$	$\frac{a_{2}}{3}$	$2.666666$
$a_{4}$	$0.041666$	$\frac{a_{3}}{4}$	$2.708333$
$a_{5}$	$0.008333$	$\frac{a_{4}}{5}$	$2.716666$
$a_{6}$	$0.001388$	$\frac{a_{5}}{6}$	$2.718055$
$a_{7}$	$0.000198$	$\frac{a_{6}}{7}$	$2.718253$
$a_{8}$	$0.000024$	$\frac{a_{7}}{8}$	$2.718278$
$a_{9}$	$0.000002$	$\frac{a_{8}}{9}$	$2.718281$

Euler's Power to X

e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots

e^{4} = \sum_{n = 0}^{\infty} \frac{4^{n}}{n!} = 1 + 4 + \frac{4^{2}}{2!} + \frac{4^{3}}{3!} + \dots \approx 54.5981500331 \dots

e^{4} \approx 54.5981500331 \dots

$Term (n)$	$Value of Term (\frac{4^{n}}{n!})$	$Value of Term % of Total$	$Accuracy$	$Running Total$
$0$	$1$	$1.83 %$	$1.8 %$	$1$
$1$	$4$	$7.33 %$	$9.2 %$	$5$
$2$	$8$	$14.65 %$	$23.8 %$	$13$
$3$	$10.667$	$19.54 %$	$43.3 %$	$23.667$
$4$	$10.667$	$19.54 %$	$62.9 %$	$34.333$
$5$	$8.533$	$15.63 %$	$78.5 %$	$42.867$
$6$	$5.689$	$10.42 %$	$88.9 %$	$48.556$
$7$	$3.251$	$5.95 %$	$94.9 %$	$51.806$
$8$	$1.625$	$2.98 %$	$97.9 %$	$53.432$
$9$	$0.722$	$1.32 %$	$99.2 %$	$54.154$
$15$	$0.00045$	$0.0008 %$	$99.99 %$	$54.598$

Imaginary Number $i$

$$i = \sqrt{-1} \quad \text{or} \quad i^2 = -1$$ $$\\[5ex]$$ $$i^1 = \mathbf{i}$$ $$\\[5ex]$$ $$i^2 = \mathbf{-1}$$ $$\\[5ex]$$ $$i^3 = -i$$ $$\\[5ex]$$ $$i^4 = \mathbf{1}$$ $$\\[5ex]$$ $$i^5 = i \text{(The cycle starts over!)}$$ $$\\[5ex]$$

Sine and Cosine

$$\\[5ex]$$ $$\text{if Point } A = (x,y) = (1,0), \text{ then } \cos(A) = 1 \\[2ex]$$ $$\\[5ex]$$ $$\text{if Point } A = (x,y) = (1,0), \text{ then } \sin(A) = 0 \\[2ex]$$ $$\\[5ex]$$ $$\text{if Point } B = (x,y) = (0,1), \text{ then } \sin(B) = 1 \\[2ex]$$ $$\\[5ex]$$ $$\text{if Point } B = (x,y) = (0,1), \text{ then } \cos(B) = 0 \\[2ex]$$ $$\\[5ex]$$

$$\\[5ex]$$ $$\text{If } \theta = 45^\circ = \frac{\pi}{4},\hspace{5pt} C = (0.707106811865, 0.707106811865)$$ $$\\[3ex]$$ $$\text{If } C = (0.707106811865, 0.707106811865),\hspace{5pt} C, sin = C.y = 0.707106811865$$ $$\\[3ex]$$ $$\text{If } C = (0.707106811865, 0.707106811865),\hspace{5pt} C, \cos = C.x = 0.707106811865$$ $$\\[5ex]$$ $$\text{If } \theta = 45^\circ = \frac{\pi}{4} ,\hspace{5pt} \sin(\theta) = 0.707106811865$$ $$\\[5ex]$$ $$\text{If } \theta = 45^\circ = \frac{\pi}{4},\hspace{5pt} \cos(\theta) = 0.707106811865$$ $$\\[5ex]$$ $$|\vec{C}| = \sqrt{x^2 + y^2}$$ $$\\[3ex]$$ $$\text{hipotenuse} = \hspace{3pt} h \hspace{3pt} = \sqrt{\text{base}^2 + \text{height}^2} \implies |\vec{C}| = \sqrt{x^2 + y^2}$$ $$\\[5ex]$$ $$|\vec{C}| = \sqrt{x^2 + y^2} = \hspace{3pt} 1 \hspace{3pt} = \sqrt{(\cos 45^\circ)^2 + (\sin 45^\circ)^2} = \sqrt{\left(\cos \left(\frac{\pi}{4}\right)\right)^2 + \left(\sin \left(\frac{\pi}{4}\right)\right)^2}$$

MORE PROOFS AND CLARIFICATION

Derivative

$$\text{General Formula for Derivative}$$ $$\\[5ex]$$ $$\large\boxed{f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}}$$ $$\\[5ex]$$ $$\large \text{Example \#1}$$ $$\\[5ex]$$ $$\large \boxed {\quad f(x) = x^2 \quad}$$ $$\\[5ex]$$ $$f(x+h) = (x+h)^2$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$$ $$\\[5ex]$$ $$(x+h)^2 = x^2 + 2xh + h^2$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{\color{red}{\cancel{x^2}} \color{white}{+ 2xh + h^2} \color{red}{\cancel{- x^2}}}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{\color{red}{\cancel{h}} \color{white}{(2x + h)}}{\color{red}{\cancel{h}}}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} (2x + h)$$ $$\\[5ex]$$ $$f'(x) = 2x + 0$$ $$\\[5ex]$$ $$\large \boxed {\quad f'(x) = 2x \quad}$$ $$\\[5ex]$$

$$\large \text{Theorem \#1 }$$ $$\\[5ex]$$ $$\large \boxed {\quad f(x) = x^n \quad}$$ $$\\[5ex]$$ $$f(x) = x^n \implies f'(x) = nx^{n-1}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{(x + h)^n - x^n}{h}$$ $$\\[5ex]$$ $$\text{Binomial Theorem}$$ $$(x + h)^n = x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots + h^n$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{\left( \color{red}{\cancel{x^n}} \color{white}{+ nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots + h^n} \right) \color{red}{\cancel{- x^n}}}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots + h^n}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{h \left( nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \dots + h^{n-1} \right)}{h}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \frac{\color{red}{\cancel{h}} \color{white}{\left( nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \dots + h^{n-1} \right)}}{\color{red}{\cancel{h}}}$$ $$\\[5ex]$$ $$f'(x) = \lim_{h \to 0} \left( nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \dots + h^{n-1} \right)$$ $$\\[5ex]$$ $$f'(x) = nx^{n-1} + 0 + 0 + \dots + 0$$ $$\\[5ex]$$ $$\large \boxed {\quad f'(x) = nx^{n-1}\quad}$$ $$\\[5ex]$$

MORE PROOFS AND CLARIFICATION

Taylor Series

$$\text{General Formula for Taylor Series}$$ $$\\[5ex]$$ $$\large\boxed{f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n}$$ $$\\[5ex]$$ $$\text{where } a \in \mathbb{R} \text{ is the center point}$$ $$\\[5ex]$$ $$\circ \quad \text{Find } c_0$$ $$f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \dots$$ $$\\[5ex]$$ $$\text{if } x = a \implies x - a = 0$$ $$\\[5ex]$$ $$f(x) = c_0 + c_1(0) + c_2(0)^2 + c_3(0)^3 + \dots$$ $$\\[5ex]$$ $$f(x) = c_0$$ $$\\[5ex]$$ $$c_0 = f(a)$$ $$\\[5ex]$$ $$\circ \quad \text{Find } c_1$$ $$\\[5ex]$$ $$f'(x) = 0 + c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + \dots$$ $$\\[5ex]$$ $$\text{if } x = a \implies x - a = 0$$ $$\\[5ex]$$ $$f'(a) = 0 + c_1 + 2c_2(0) + 3c_3(0)^2 + \dots$$ $$\\[5ex]$$ $$f'(a) = c_1 + 0 + 0 \dots$$ $$\\[5ex]$$ $$f'(a) = c_1 $$ $$\\[5ex]$$ $$c_1 = f'(a)$$ $$\\[5ex]$$ $$\circ \quad \text{Find } c_2$$ $$\\[5ex]$$ $$f''(x) = 2c_2 + 6c_3(x-a) + 12c_4(x-a)^2 + 20c_5(x-a)^3 + 30c_6(x-a)^4 + \dots$$ $$\\[5ex]$$ $$\text{if } x = a \implies x - a = 0$$ $$\\[5ex]$$ $$f''(a) = 2c_2 + 6c_3(0) + 12c_4(0)^2 + 20c_5(0)^3 + 30c_6(0)^4 + \dots$$ $$\\[5ex]$$ $$f''(a) = 2c_2 $$ $$\\[5ex]$$ $$c_2 = \frac{f''(a)}{2}\quad\mid \quad\frac{f''(a)}{2!}$$ $$\\[5ex]$$ $$f^{(n)}(a) = n! \cdot c_n$$ $$\\[5ex]$$ $$\text{So... } $$ $$\\[5ex]$$ $$f^{(n)}(a) = n! \cdot c_n$$ $$\\[5ex]$$ $$c_n = \frac{f^{(n)}(a)}{n!}$$ $$\\[10ex]$$ $$\large \text{Example \#1}$$ $$\\[5ex]$$ $$\text{if } f(x) = e^x $$ $$\\[5ex]$$ $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \implies e^x = \sum_{n=0}^{\infty} \frac{e^a}{n!}(x-a)^n$$ $$\\[5ex]$$ $$\text{if, \quad } a = 0 $$ $$\\[5ex]$$ $$e^x = \sum_{n=0}^{\infty} \frac{e^a}{n!}(x)^n$$ $$\\[5ex]$$ $$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots$$ $$\\[5ex]$$ $$e^x = e^a + e^a(x-a) + \frac{e^a}{2!}(x-a)^2 + \frac{e^a}{3!}(x-a)^3 + \frac{e^a}{4!}(x-a)^4 + \dots$$ $$\\[5ex]$$ $$a=0,\quad then \quad e^a = e^0 = 1$$ $$\\[5ex]$$ $$e^x = 1 + (x-0) + \frac{1}{2!}(x-0)^2 + \frac{1}{3!}(x-0)^3 + \frac{1}{4!}(x-0)^4 + \dots$$ $$\\[5ex]$$ $$\large e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$$ $$\\[5ex]$$ $$\large \boxed{e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots}$$ $$\\[10ex]$$ $$\large \text{Example \#2}$$ $$\\[5ex]$$ $$\text{if } f(x) = \cos(x) $$ $$\\[5ex]$$ $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \implies \cos(x) = \sum_{n=0}^{\infty} \frac{\frac{d^n}{dx^n}[\cos(x)]_{}}{n!}(x-a)^n$$ $$\\[5ex]$$ $$\text{if, \quad } a = 0 $$ $$\\[5ex]$$ $$\cos(x) = \sum_{n=0}^{\infty} \frac{\frac{d^n}{dx^n}[\cos(x)]_{}}{n!}(x)^n$$ $$\\[5ex]$$ $$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots$$ $$\\[5ex]$$ $$\begin{aligned} f(a) &= \cos(a) \\ f'(a) &= -\sin(a) \\ f''(a) &= -\cos(a) \\ f'''(a) &= \sin(a) \\ f^{(4)}(a) &= \cos(a) \end{aligned}$$ $$\\[5ex]$$ $$\cos(x) = \cos(a) - \sin(a)(x-a) - \frac{\cos(a)}{2!}(x-a)^2 + \frac{\sin(a)}{3!}(x-a)^3 + \frac{\cos(a)}{4!}(x-a)^4 - \dots$$ $$\\[5ex]$$ $$\begin{aligned} f(0) = \cos(0) &= 1 \\ f'(0) = -\sin(0) &= 0 \\ f''(0) = -\cos(0) &= -1 \\ f'''(0) = \sin(0) &= 0 \\ f^{(4)}(0) = \cos(0) &= 1 \end{aligned}$$ $$\\[5ex]$$ $$\cos(x) = \cos(0) - \sin(0)(x) - \frac{\cos(0)}{2!}x^2 + \frac{\sin(0)}{3!}x^3 + \frac{\cos(0)}{4!}x^4 - \dots$$ $$\\[5ex]$$ $$\cos(x) = 1 - (0)x - \frac{1}{2!}x^2 + \frac{0}{3!}x^3 + \frac{1}{4!}x^4 - \dots$$ $$\\[5ex]$$ $$\large\boxed{\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \dots}$$ $$\\[15 ex]$$ $$\large \text{Example \#3}$$ $$\\[5ex]$$ $$\text{if } f(x) = \sin(x) $$ $$\\[5ex]$$ $$\\[5ex]$$ $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \implies \sin(x) = \sum_{n=0}^{\infty} \frac{\frac{d^n}{dx^n}[\sin(x)]_{}}{n!}(x-a)^n$$ $$\\[5ex]$$ $$\text{if, \quad } a = 0 $$ $$\\[5ex]$$ $$\sin(x) = \sum_{n=0}^{\infty} \frac{\frac{d^n}{dx^n}[\sin(x)]_{}}{n!}(x)^n$$ $$\\[5ex]$$ $$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots$$ $$\\[5ex]$$ $$\begin{aligned} f(a) &= \sin(a) \\ f'(a) &= \cos(a) \\ f''(a) &= -\sin(a) \\ f'''(a) &= -\cos(a) \\ f^{(4)}(a) &= \sin(a) \end{aligned}$$ $$\\[5ex]$$ $$\sin(x) = \sin(a) + \cos(a)(x-a) - \frac{\sin(a)}{2!}(x-a)^2 - \frac{\cos(a)}{3!}(x-a)^3 + \frac{\sin(a)}{4!}(x-a)^4 + \dots$$ $$\\[5ex]$$ $$\begin{aligned} f(0) = \sin(0) &= 0 \\ f'(0) = \cos(0) &= 1 \\ f''(0) = -\sin(0) &= 0 \\ f'''(0) = -\cos(0) &= -1 \\ f^{(4)}(0) = \sin(0) &= 0 \end{aligned}$$ $$\\[5ex]$$ $$\sin(x) = \sin(0) + \cos(0)(x) - \frac{\sin(0)}{2!}x^2 - \frac{\cos(0)}{3!}x^3 + \frac{\sin(0)}{4!}x^4 + \dots$$$$\\[5ex]$$$$\sin(x) = 0 + (1)x - \frac{0}{2!}x^2 - \frac{1}{3!}x^3 + \frac{0}{4!}x^4 + \frac{1}{5!}x^5 - \dots$$$$\\[5ex]$$$$\large\boxed{\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \dots}$$

Eulers Identity

$$\\[5ex]$$ $$\text{ Exponential}$$ $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots$$ $$\\[5ex]$$ $$\text{ Cosine}$$ $$\\[5ex]$$ $$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$$ $$\\[5ex]$$ $$\text{ Sine}$$ $$\\[5ex]$$ $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots $$ $$\\[5ex]$$ $$\text{If, }\quad x = ix \quad \text{,in}\quad e^x \implies e^ix $$ $$\\[5ex]$$ $$e^{ix} = 1 + (ix) + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \dots$$ $$\\[5ex]$$ $$\begin{aligned} \text{Since } i^2 &= -1, \\ (ix)^1 &= ix \\ (ix)^2 &= i^2 x^2 = -x^2 \\ (ix)^3 &= i^3 x^3 = -ix^3 \\ (ix)^4 &= i^4 x^4 = x^4 \\ (ix)^5 &= i^5 x^5 = ix^5 \end{aligned}$$ $$\\[5ex]$$ $$e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \dots$$ $$\\[5ex]$$ $$e^{ix} = \underbrace{\left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \right)}_{\text{This is } \cos(x)} + i \underbrace{\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right)}_{\text{This is } \sin(x)}$$ $$\\[5ex]$$ $$e^{ix} = \cos(x) + i\sin(x)$$ $$\text{If,} \quad x = \pi $$ $$\cos(\pi) = -1$$ $$\sin(\pi) = 0$$ $$\\[5ex]$$ $$e^{i\pi} = -1 + i(0)$$$$e^{i\pi} = -1$$ $$\\[5ex]$$ $$\large\boxed{e^{i\pi} + 1 = 0}$$ $$\\[10ex]$$

Matemathics Explanations of Eulers Identity

Math Explanation CODE relations

Eulers Identity

Factorial Operator

Permutational Formula

Binomial Coefficient / Combinations Formula / "n choose k"

Binomial Theorem

Euler's Number Existence

Euler's Power to X

Imaginary Number $i$

Sine and Cosine

Derivative

Taylor Series

Eulers Identity

End

Mathematics-Code_INDEX

Matemathics Explanations of Eulers Identity

Math Explanation CODE relations

Eulers Identity

Factorial Operator

Permutational Formula

Binomial Coefficient / Combinations Formula / "n choose k"

Binomial Theorem

Euler's Number Existence

Euler's Power to X

Imaginary Number i

Sine and Cosine

Derivative

Taylor Series

Eulers Identity

End

Mathematics-Code_INDEX

Imaginary Number $i$