unsorted dependencies

exponentiation

Suppose

b \in R

, and

n \in N_{1}

, then

b^{n} := \underset{n times}{\underset{⏟}{b \cdot b \cdot \dots \cdot b \cdot b}}

, note that

b^{n} : N_{1} \to R

exponentiation of a positive number is positive

Suppose that

x \in R^{> 0}

, then

x^{n} \in R^{> 0}

product of two reals of the same sign is positive

Given

x, y \in R

, if

x, y < 0

x, y > 0

, then

x \cdot y = 0

multiplicative inverse

Suppose that

x \in R^{\neq 0}

, then if

w \in R

satisfies

x \cdot w = 1

, then it is said to be the reciprocal or multiplicative inverse of

x

and we write

w = \frac{1}{x}

Suppose that

n \in N_{1}

and

f : R^{> 0} \to R

is defined as

f (x) = x^{n}

, then

f

is strictly increasing.

Suppose that

n \in N_{1}

and

x \in R^{> 0}

, then

x^{n} > 0

(TODO: product of two numbers of the same sign is positive and induction.)

Suppose that

a, b, c \in R

, with

a < b

and

c > 0

, then

a c < b c

Suppose that

a_{n}, b_{n}

are two real sequences and that

\sum_{n = 0}^{\infty} a_{n}

and

\sum_{n = 0}^{\infty} b_{n}

are convergent series, if for each

n \in N, a_{n} < b_{n}

, then

\sum_{n = 1}^{\infty}

Suppose

x, y \in R

with the same sign. Then

x < y

if and only if

\frac{1}{y} \leq \frac{1}{x}

binomial

Suppose that

x, y \in R

such that

x + y \geq 0

and that

n \in N_{0}

, then

{(x + y)}^{n} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} y^{k}

TODO

exponential function

exponential

e^{x} := \sum_{k = 0}^{\infty} \frac{x^{k}}{k!}

, note that

e^{x} : R \to R

and that

x^{k}

is exponentiation. We say that

e^{x}

is the exponential.

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

TODO

\frac{d}{d x} [e^{x}] = e^{x}

$\frac{d}{d x} = lim_{h \to 0} \frac{e^{x + h} - e^{x}}{h} = lim_{h \to \infty} \frac{lim_{n \to \infty} {(1 + \frac{x + h}{n})}^{n} - lim_{n \to \infty} {(1 + \frac{x}{n})}^{n}}{h} = lim_{h \to 0} (lim_{n \to \infty} \frac{{(1 + \frac{x + h}{n})}^{n} - {(1 + \frac{x}{n})}^{n}}{h})$

Therefore by the binomial theorem...

exponential sum product equality

Suppose

x, y \in R

, then

e^{x + y} = e^{x} \cdot e^{y}

TODO

reciprocal of exponential

\frac{1}{e^{x}} = e^{- x}

$e^{0} = 1$ as if we sub in $x = 0$ $e^{x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$ , and we know that $0^{0}$ is defined to be $0$ , then $e^{0} = 1$

But at the same time $e^{0} = e^{x - x}$ which equals $e^{x} \cdot e^{- x} = 1$ , so thus $\frac{1}{e^{x}} = e^{- x}$ as needed.

x < 0

,iff

e^{x} < 1

since

e^{x}

is increasing it's fine (TODO finish)

0 < x < 1

iff

\ln (x) < 0

TODO

exponential is always positive

\forall x \in R, e^{x} > 0

If $x > 0$ , then by one of the unordered dependencies we have $x^{n} > 0$ , then by one of the unorder dependencies about real numbes since $\frac{1}{n!} > 0$ we have $\frac{x^{n}}{n!} > \frac{0}{n!} = 0$ and using $\frac{x^{n}}{n!} = \sum_{n = 0}^{\infty} a_{n}$ with $a_{i} = 0$ for $i \in {0, \dots, n - 1}$ and $a_{n} = \frac{x^{n}}{n!}$ and $b_{i} = \frac{x^{i}}{i!}$ the comparison test tells us that $0 < e^{x}$

If $x < 0$ , then $- x > 0$ , and thus by the above argumnet $0 < e^{- x} = \frac{1}{e^{x}}$ , recall that $e^{- x} \cdot e^{x} = 1$ , and we know that $e^{- x}$ and $1$ are positive so therefore $e^{x} > 0$ as well, as needed.

If $x = 0$ , then $e^{x} = e^{0} = 1$

logarithm

Suppose that

a, b \in R

, then

m = \log_{b} (a)

is a number such that

b^{m} = a

natural logarithm

We define

\ln (x) := \log_{e} (x)

and call it the natural logarithm.

\forall r \in R, r^{x} := e^{\ln (r) \cdot x}

exponentiation inverse

given

r \in R

, then the multiplicative inverse of

r^{x}

r^{- x}

TODO

Suppose that

0 < r < 1

and

x \in R^{> 0}

, then

0 < r^{x} < 1

TODO

Let

a, b \in R^{> 0}

and assume that

a < b

, then if

0 < r < 1

, then

r^{a} > r^{b}

Since

a < b

, then

0 < b - a

, and thus