Proofs of ultimate math formulas - yuuki7/yuukinder GitHub Wiki

< Ultimate math formulas

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 $n$:

Product of powers

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

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)

Power of a power

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

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 product of powers)

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

Power of a product

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

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)

Let $a, b, M, N, x &gt; 0$ and $n$ be real numbers, with $a, b \ne 1$. Since a logarithm is an inverse of exponentiation, we have

$\log_a(a^x) = a^{\log_a x} = x$,

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

$\log_a(MN) = \log_a(a^{\log_a M} a^{\log_a N}) = \log_a(a^{\log_a M + \log_a N}) = \log_a M + \log_a N$

$\log_a(M^n) = \log_a((a^{\log_a M})^n) = \log_a(a^{n \log_a M}) = n \log_a M$

$\log_a M = \displaystyle \frac{\log_a M \cdot \log_b a}{\log_b a} = \frac{\log_b(a^{\log_a M})}{\log_b a} = \frac{\log_b M}{\log_b a}$

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

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

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

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

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

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

$\cos^2 \theta - \sin^2 \theta = (1 - \sin^2 \theta) - \sin^2 \theta = 1 - 2 \sin^2 \theta$

$\cos^2 \theta - \sin^2 \theta = \cos^2 \theta - (1 - \cos^2 \theta) = 2 \cos^2 \theta - 1$

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

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 with initial conditions $f(0) = 1$, $f'(0) = 0$, $g(0) = 0$, and $g'(0) = 1$, obtained by substituting $x = 0$ into Eqs. 1 and 2. The unique solutions are:

$f = \cos x$

$g = \sin x$

#math