Proofs of ultimate math formulas
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Expansion and factorization
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:
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)
Logarithm rules
Let a, b, M, N, x > 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}
Fractions
Systems of linear equations
Sequences
Derivatives
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 = (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
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
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 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