Math formulas proofs

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Euler's formula

Let $e^{ix} = f + ig$. (Eq. 1)

Differentiate both sides with respect to $x$:

$(e^{ix})' = (f + ig)'$

$ie^{ix} = f' + ig'$ (Eq. 2)

Substitute Eq. 1 into the left-hand side:

$i(f + ig) = f' + ig'$

$-g + if = f' + ig'$

Equate the real and imaginary parts of both sides:

$-g = f'$

$f = g'$

Which can be rewritten as:

$f'' = -f$

$g'' = -g$

These are second-order differential equations. By substituting $x = 0$ into Eqs. 1 and 2, we obtain the initial conditions $f(0) = 1$, $f'(0) = 0$, $g(0) = 0$, and $g'(0) = 1$. The unique solutions are:

$f = \cos x$

$g = \sin x$

Algebraic identities

Quadratic formula

Exponent rules

Let $a$ and $b$ be real numbers, and let $m$ and $n$ be positive integers.

Definition (Exponentiation):

$a^1 = a$, (Def. 1)

$a^{n + 1} = a^n a$ for $n \ge 1$. (Def. 2)

By induction on $n$:

Rule 1

$a^m a^n = a^{m + n}$

Base case ($n = 1$):

$a^m a^1 = a^m a$ (by Def. 1)

$= a^{m + 1}$. (by Def. 2)

Inductive step: Assume $a^m a^n = a^{m + n}$. Then:

$a^m a^{n + 1} = a^m (a^n a)$ (by Def. 2)

$= (a^m a^n) a$ (by associativity)

$= a^{m + n} a$ (by the induction hypothesis)

$= a^{(m + n) + 1}$ (by Def. 2)

$= a^{m + (n + 1)}$. (by associativity)

Rule 2

$(a^m)^n = a^{mn}$

Base case ($n = 1$):

$(a^m)^1 = a^m$ (by Def. 1)

$= a^{m \cdot 1}$. (by identity)

Inductive step: Assume $(a^m)^n = a^{mn}$. Then:

$(a^m)^{n + 1} = (a^m)^n a^m$ (by Def. 2)

$= a^{mn} a^m$ (by the induction hypothesis)

$= a^{mn + m}$ (by Rule 1)

$= a^{m(n + 1)}$. (by distributivity)

Rule 3

$(ab)^n = a^n b^n$

Base case ($n = 1$):

$(ab)^1 = ab$ (by Def. 1)

$= a^1 b^1$. (by Def. 1)

Inductive step: Assume $(ab)^n = a^n b^n$. Then:

$(ab)^{n + 1} = (ab)^n (ab)$ (by Def. 2)

$= a^n b^n (ab)$ (by the induction hypothesis)

$= (a^n a) (b^n b)$ (by associativity and commutativity)

$= a^{n + 1} b^{n + 1}$. (by Def. 2)

Logarithm rules

Let $x, y > 0$ and $k$ be real numbers, and let $b, c > 0$ be real numbers with $b, c \ne 1$. Since a logarithm is the inverse of exponentiation, we have:

$b^{\log_b x} = \log_b (b^x) = x$,

just like $f(f^{-1}(x)) = f^{-1}(f(x)) = x$. Using this:

Rule 1

$\log_b (xy) = \log_b (b^{\log_b x} b^{\log_b y})$

$= \log_b (b^{\log_b x + \log_b y})$

$= \log_b x + \log_b y$.

Rule 2

$\log_b (x^k) = \log_b ((b^{\log_b x})^k)$

$= \log_b (b^{k \log_b x})$

$= k \log_b x$.

Rule 3

$\log_b x = \log_b x \cdot \dfrac{\log_c b}{\log_c b}$

$= \dfrac{\log_c (b^{\log_b x})}{\log_c b}$

$= \dfrac{\log_c x}{\log_c b}$.

Trigonometric identities

Pythagorean identity

Addition formulas

$\cos(\alpha + \beta) + i \sin(\alpha + \beta) = e^{i(\alpha + \beta)}$

$= e^{i \alpha} e^{i \beta}$

$= (\cos \alpha + i \sin \alpha)(\cos \beta + i \sin \beta)$

$= (\cos \alpha \cos \beta - \sin \alpha \sin \beta) + i(\sin \alpha \cos \beta + \cos \alpha \sin \beta)$

Double-angle formulas

$\sin 2 \theta = \sin(\theta + \theta)$

$= \sin \theta \cos \theta + \cos \theta \sin \theta$

$= 2 \sin \theta \cos \theta$

$\cos 2 \theta = \cos(\theta + \theta)$

$= \cos \theta \cos \theta - \sin \theta \sin \theta$

$= \cos^2 \theta - \sin^2 \theta$

$\cos^2 \theta - \sin^2 \theta = (\cos^2 \theta + \sin^2 \theta) - 2 \sin^2 \theta$

$= 1 - 2 \sin^2 \theta$

$\cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - (\cos^2 \theta + \sin^2 \theta)$

$= 2 \cos^2 \theta - 1$

Multiple-angle formulas

$\cos n \theta + i \sin n \theta = (\cos \theta + i \sin \theta)^n$

$= \displaystyle \sum_{k = 0}^n \binom{n}{k} \cos^{n - k} \theta i^k \sin^k \theta$

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