User:Ans/math


 * $$a \iff a \land a \iff a \lor a \iff \neg\neg a \iff a \land T \iff a \lor F$$
 * $$a \land \neg a \iff F \iff a \land F$$
 * $$a \lor \neg a \iff T \iff a \lor T$$
 * $$a \land \neg b \lor b \iff (a \lor b) \land (\neg b \lor b) \iff (a \lor b) \land T \iff a \lor b$$
 * $$(a \lor \neg b) \land b \iff a \land b \lor \neg b \land b \iff a \land b \lor F \iff a \land b$$
 * $$(a \lor \neg b) \land b \iff  (a \lor \neg b) \land (b \lor \neg(a\lor\neg b))  \iff  (a\lor\neg b) \land (b \lor \neg a\land b)  \iff  (a\lor\neg b) \land (b\lor\neg a) \land (b\lor b)  \iff$$
 * $$\exists ... \iff  \forall x\ f(x)$$
 * $$[a \implies a \lor b] \iff \neg a \lor a \lor b  \iff  T \lor b  \iff  T$$
 * $$[a \land c \implies (a \lor b) \land c] \iff \neg (a\land c) \lor (a \lor b) \land c  \iff  \neg a \lor \neg c \lor (a \lor b) \land c  \iff  \neg a \lor \neg c \lor a \lor b  \iff  T \lor \neg c \lor b  \iff  T$$
 * $$[\forall x\ f(x) \implies \forall x (f(x) \lor g(x))] \iff \neg\forall x\ f(x) \lor \forall x (f(x) \lor g(x))  \iff  \exists x\ \neg f(x) \lor$$
 * $$T \iff  [\neg a \land \neg c \implies (\neg a \lor \neg b) \land \neg c]  \iff  [\neg (a \lor c) \implies \neg (a\land b \lor c)]  \iff  [a\land b \lor c \implies a \lor c]$$
 * $$[a \land b \implies a] \iff \neg(a \land b) \lor a  \iff \neg a \lor \neg b \lor a  \iff  T \lor \neg b  \iff  T$$
 * $$a \land b \implies a \implies a \lor b$$
 * $$[a \implies b] \implies [a \lor b \implies b \lor b \iff b] \land [b \implies a \lor b] \iff [b \iff a \lor b]$$
 * $$[a \implies b] \implies [a \iff a \land a \implies a \land b] \land [a \land b \implies a] \iff [a \iff a \land b]$$
 * $$[a \implies b] \implies [f(a) \iff f(a \land b)]$$
 * $$f(a \land [a \implies b]) \iff f(a \land [\neg a \lor b]) \iff f(a \land \neg a \lor a \land b) \iff f(F \lor a \land b) \iff f(a \land b)$$
 * $$[(a \implies b) \implies (a \land c \implies b)] \iff \neg(\neg a \lor b) \lor (\neg(a \land c) \lor b)  \iff  a \land \neg b \lor \neg a \lor \neg c \lor b  \iff  a \lor b \lor \neg a \lor \neg c  \iff  T \lor b \lor \neg c  \iff  T$$
 * $$[a \implies b] \iff \neg a \lor b \implies \neg a \lor b \lor \neg c \iff \neg(a \land c) \lor b \iff [a\land c \implies b]$$
 * $$(a \implies b) \land (c \implies d) \iff (\neg a \lor b) \land (\neg c \lor d) \implies  (\neg a \lor b \lor \neg c) \land (\neg c \lor d \lor \neg a)  \iff  (\neg(a\land c) \lor b) \land (\neg(a\land c) \lor d)  \iff  \neg(a\land c) \lor b\land d  \iff  [a\land c \implies b\land d]$$


 * $$(a \lor b) \land (b \implies c) \iff  (a \lor b) \land (\neg b \lor c)  \iff  a\land\neg b \lor a\land c \lor b\land\neg b \lor b\land c  \iff  a\land\neg b \lor a\land c \lor b\land c  \iff  (a\lor a) \land (a\lor c) \land (\neg b\lor a) \land (\neg b\lor c) \lor b\land c$$
 * $$(a \lor b) \land (b \implies c) \iff  (a\lor b) \land (a\lor c\lor b) \land (\neg b\lor a\lor b) \land (\neg b\lor c\lor b) \land (a\lor c) \land (a\lor c\lor c) \land (\neg b\lor a\lor c) \land (\neg b\lor c\lor c)  \implies  a\lor c$$
 * $$(b \implies c) \implies (a \lor b \implies a \lor c)$$
 * $$[a \land b \implies c] \iff \neg(a \land b) \lor c \iff  \neg a \lor \neg b \lor c  \iff [a \implies (b \implies c)]$$
 * $$(a \land b) \land (b \implies c) \iff  a \land b \land (\neg b \lor c)  \iff  a\land b\land\neg b \lor a\land b\land c  \iff  F \lor a\land b\land c  \iff  a\land b\land c  \implies  a\land c$$
 * $$(b \implies c) \implies (a \land b \implies a \land c)$$
 * $$a \land (a \implies b) \iff a \land (\neg a \lor b) \iff a\land\neg a \lor a\land b \iff F \lor a\land b \iff a\land b \implies b$$


 * $$f(a) \iff a \land f(T) \lor \neg a \land f(F)$$ (need prove)
 * $$a \land f(a) \iff a \land f(T)$$ (need prove)
 * $$a \lor f(a) \iff \neg(\neg a \land \neg f(\neg\neg a)) \iff \neg(\neg a \land \neg f(\neg T)) \iff a \lor f(F)$$
 * $$[a \implies f(a)] \iff \neg a \lor f(\neg\neg a) \iff \neg a \lor f(\neg F) \iff [a \implies f(T)]$$
 * $$\forall x \in \mathbb{A} \ f(x)$$ $$\iff$$ $$\forall x\ [x \in \mathbb{A} \implies f(x)]$$ $$\iff$$ $$\forall a\ [a \in \mathbb{A} \implies f(a)]$$ definition? $$\implies$$ $$[c \in \mathbb{A} \implies f(c)]$$
 * $$\forall x \in \mathbb{A} \ f(x)$$ $$\cancel\iff$$ $$[x \in \mathbb{A} \implies f(x)]$$, since the second form has one more non-local variable $$x$$ which not exist in the first form
 * counter example: $$\neg\forall x\in\mathbb{A}\ f(x) \cancel\iff  \neg(x\in\mathbb{A} \implies f(x))  \iff  \neg(\neg x\in\mathbb{A} \lor f(x))  \iff  x\in\mathbb{A} \land \neg f(x)$$
 * $$\forall a\in\mathbb{A}\ f(a) \lor \forall b\in\mathbb{B}\ f(b) \cancel\iff [a\in\mathbb{A} \implies f(a) \lor b\in\mathbb{B} \implies f(b)]  \iff  \neg a\in\mathbb{A} \lor f(a) \lor \neg b\in\mathbb{B} \lor f(b)$$- (bad example)
 * $$\forall a\in\mathbb{A}\ f(a) \lor \forall b\in\mathbb{B}\ f(b) \cancel\iff \neg (a\in\mathbb{A} \land b\in\mathbb{B}) \lor f(a) \lor f(b)  \iff  [a\in\mathbb{A} \land b\in\mathbb{B} \implies f(a) \lor f(b)]$$- (bad example)
 * $$\forall x\in\mathbb{A}\ f(x) \iff  \forall a\in\mathbb{A}\ f(a)$$ need prove?
 * $$a\in\mathbb{A} \land \forall x\in\mathbb{A}\ f(x) \implies f(a)$$
 * $$\forall x\in\mathbb{A}\ f(x\in\mathbb{A}) \iff \forall x\ [x\in\mathbb{A} \implies f(x\in\mathbb{A})] \iff \forall x\ [x\in\mathbb{A} \implies f(T)] \iff \forall x\in\mathbb{A}\ f(T)$$
 * $$\exists x\in\mathbb{A}\ f(x\in\mathbb{A}) \iff \neg\forall x\in\mathbb{A}\ \neg f(x\in\mathbb{A}) \iff \neg\forall x\in\mathbb{A}\ \neg f(T) \iff \exists x\in\mathbb{A}\ f(T)$$
 * $$a \land \forall x\in\mathbb{A}\ f(x) \iff  ...  \forall x\in\mathbb{A} (a \land f(x))$$
 * $$\forall x\in\mathbb{A}\ f(x) \iff  \forall x\in\mathbb{A}\ f(x) \land \forall x\in\mathbb{A}\ f(x)  \iff  \forall x\in\mathbb{A}\ f(x) \land \forall a\in\mathbb{A}\ f(a)  \iff  $$
 * $$\forall x [f(x) \land g(x)] \iff \forall x\ f(x) \land \forall x\ g(x)$$
 * $$\forall x\ f(x) \lor \forall x\ g(x) \implies \forall x\ [f(x)\lor g(x)] \lor \forall x\ [g(x)\lor f(x)] \iff \forall x [f(x) \lor g(x)]$$
 * $$\neg \neg (\forall x\ f(x) \lor \forall x\ g(x)) \implies \neg \neg \forall x [f(x) \lor g(x)]$$
 * $$\neg (\exists x\ \neg f(x) \land \exists x\ \neg g(x)) \implies \neg \exists x [\neg f(x) \land \neg g(x)]$$
 * $$\neg (\exists x\ f(x) \land \exists x\ g(x)) \implies \neg \exists x [f(x) \land g(x)]$$
 * $$\exists x [f(x) \land g(x)] \iff \exists x [f(x) \land g(x)] \land \exists x [f(x) \land g(x)] \implies \exists x\ f(x) \land \exists x\ g(x)$$
 * $$\exists x [f(x) \lor g(x)] \iff \exists x\ f(x) \lor \exists x\ g(x)$$
 * $$\forall x\ \neg f(x) \iff \neg \exists x\ f(x)$$
 * $$\forall x\ f(x) \implies \exists x\ f(x)$$; only apply for non-empty set (need prove)
 * $$\forall x (f(x) \land g(x)) \iff \forall x\ f(x) \land \forall x\ g(x) \implies \forall x\ f(x)$$
 * $$\forall x\ f(x) \implies \forall x (f(x) \lor g(x)) $$ (need prove)
 * $$[\forall x (f(x) \implies g(x)) \implies (\forall x\ f(x) \implies \forall x\ g(x))] \iff \neg\forall x (\neg f(x) \lor g(x)) \lor (\neg\forall x\ f(x) \lor \forall x\ g(x))$$
 * $$[\forall x (f(x) \implies g(x)) \implies (\forall x\ f(x) \implies \forall x\ g(x))] \iff \exists x (f(x) \land \neg g(x)) \lor \exists x\ \neg f(x) \lor \forall x\ g(x)$$
 * $$[\forall x (f(x) \implies g(x)) \implies (\forall x\ f(x) \implies \forall x\ g(x))] \iff \exists x (f(x) \land \neg g(x) \lor \neg f(x)) \lor \forall x\ g(x) \iff \exists x (\neg g(x) \lor \neg f(x)) \lor \forall x\ g(x)$$
 * $$[\forall x (f(x) \implies g(x)) \implies (\forall x\ f(x) \implies \forall x\ g(x))] \iff \exists x\ \neg g(x) \lor \exists x\ \neg f(x) \lor \neg\exists x\ \neg g(x) \iff  T \lor \exists x\ \neg f(x)  \iff  T$$
 * TODO: prove by "\implies"
 * $$\forall x (f(x) \implies g(x)) \iff  \forall x (\neg f(x) \lor g(x))  \implies  \forall x (\neg f(x) \lor g(x) \lor \neg\forall x\ f(x))???$$
 * $$\forall x [f(x) \iff g(x)] \iff \forall x [f(x) \land g(x) \lor \neg f(x) \land \neg g(x)] \Longleftarrow \forall x [f(x) \land g(x)] \lor \forall x [\neg f(x) \land \neg g(x)] \iff \forall x\ f(x) \land \forall x\ g(x) \lor \forall x \neg f(x) \land \forall x \neg g(x)$$
 * $$\forall x [f(x) \iff g(x)] \iff \forall x [f(x) \land g(x) \lor \neg f(x) \land \neg g(x)] \implies \exists x [f(x) \land g(x) \lor \neg f(x) \land \neg g(x)] \iff \exists x [f(x) \land g(x)] \lor \exists x[\neg f(x) \land \neg g(x)] \implies \exists x\ f(x) \land \exists x\ g(x) \lor \exists x\ \neg f(x) \land \exists x\ \neg g(x) \iff \exists x\ f(x) \land \exists x\ g(x) \lor \neg\forall x\ f(x) \land \neg\forall x\ g(x)$$
 * $$\neg \forall x [f(x) \iff g(x)] \iff \exists x [f(x) \iff \neg g(x)] \iff \exists x [f(x) \land \neg g(x) \lor \neg f(x) \land g(x)] \iff \exists x [f(x) \land \neg g(x)] \lor \exists x [\neg f(x) \land g(x)] \implies \exists x\ f(x) \land \exists x\ \neg g(x) \lor \exists x\ \neg f(x) \land \exists x\ g(x) \iff \exists x\ f(x) \land \neg\forall x\ g(x) \lor \neg\forall x\ f(x) \land \exists x\ g(x)$$
 * $$\forall x [f(x) \iff g(x)] \iff \forall x [f(x) \land g(x) \lor \neg f(x) \land \neg g(x)] \iff \forall x [(f(x) \lor \neg g(x)) \land (\neg f(x) \lor g(x))] \iff \forall x [f(x) \lor \neg g(x)] \land \forall [\neg f(x) \lor g(x)]$$


 * 1) $$(x - \varepsilon)^2 << x^2 + a << (x + \varepsilon)^2;\ x >> \varepsilon > 0$$
 * 2) $$x - \varepsilon < \sqrt{x^2+a} < x + \varepsilon;\ x >> \varepsilon > 0$$
 * 3) $$x - \varepsilon-x < \sqrt{x^2+a}-x < x + \varepsilon-x;\ x >> \varepsilon > 0$$
 * 4) $$-\varepsilon < \sqrt{x^2+a}-x < \varepsilon;\ x >> \varepsilon > 0$$
 * 5) $$\lim_{x\to\infty}(\sqrt{x^2+a}-x) = 0$$
 * 6) $$\lim_{x\to\infty}(\sqrt{(x+b)^2+a}-(x+b)) = 0$$
 * 7) $$\lim_{x\to\infty}(\sqrt{(x+b)^2+a}-x) = b$$

$$ \lim_{x \to c}f(x) = L $$ $$ \lim_{x \to c^+}f(x) = L$$ $$ \lim_{x \to 0^+}f(x) = L = \lim_{1/x \to \infty}f(1/(1/x)) = \lim_{x \to \infty}f(1/x) $$
 * $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x (0 < |x - c| < \delta \ \implies \ |f(x) - L| < \varepsilon)$$
 * "0 < |x - c|" means that "x <> c"
 * $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x (0 < x - c < \delta \ \implies \ |f(x) - L| < \varepsilon)$$
 * 1) $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x (0 < x < \delta \ \implies \ |f(x) - L| < \varepsilon)$$
 * 2) $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x>0 (0 < x < \delta \ \implies \ |f(x) - L| < \varepsilon)$$
 * 3) $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x>0 (\infty > 1/x > 1/\delta \ \implies \ |f(1/(1/x)) - L| < \varepsilon)$$
 * 4) $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x>0 (x > 1/\delta \ \implies \ |f(1/x) - L| < \varepsilon)$$
 * 5) $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x>0 (x > \delta \ \implies \ |f(1/x) - L| < \varepsilon)$$


 * $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x (0 < |x - c| < \delta \ \implies \ |f(x) - L| < \varepsilon \ \implies \ \Big| |f(x)| - |L| \Big| < |f(x) - L| < \varepsilon).$$
 * $$ \lim_{x \to c}f(x) = L \implies \lim_{x \to c}\left|f(x)\right| = |L| = \left| \lim_{x \to c}f(x) \right|$$
 * $$\left| f'(x) \right| = \left| \frac{df(x)}{dx} \right| = \left| \lim_{dx \to 0}\frac{f(x+dx)-f(x)}{dx} \right| = \lim_{dx \to 0}\left| \frac{f(x+dx)-f(x)}{dx} \right| = \lim_{dx \to 0} \frac{|f(x+dx)-f(x)|}{|dx|} = \lim_{dx \to 0} \frac{dx(f(x+dx)-f(x))|f(x+dx)-f(x)|}{dx(f(x+dx)-f(x))|dx|}$$
 * $$\lim_{dx \to 0} \frac{dx(f(x+dx)-f(x))|f(x+dx)-f(x)|}{dx(f(x+dx)-f(x))|dx|} = \lim_{dx \to 0} \frac{f(x+dx)-f(x)}{dx} \lim_{dx \to 0} \frac{dx|f(x+dx)-f(x)|}{|dx|(f(x+dx)-f(x))} = \lim_{dx \to 0} \frac{f(x+dx)-f(x)}{dx} \lim_{dx \to 0} \frac{|dx(f(x+dx)-f(x))|}{dx(f(x+dx)-f(x))} = f'(x) ... ???$$
 * $$\frac{x}{|x|} = \frac{x|x|^2}{|x|x^2} = \frac{|x|}{x}$$


 * $$ \exists\delta>0 \ \forall x ( 0 < |x - c| < \delta \implies f(x) = g(x) ) $$
 * $$ \forall x ( 0 < |x - c| < \delta_g \implies f(x) = g(x) ); \delta_g>0 $$
 * $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x (0 < |x - c| < \delta \ \implies \ |f(x) - L| < \varepsilon) $$
 * $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x (0 < |x - c| < \min(\delta,\delta_g) \ \implies \ |f(x) - L|=|g(x) - L| < \varepsilon) $$
 * $$ \forall \varepsilon > 0 \ \ \exists \min(\delta,\delta_g) > 0 \ \ \forall x (0 < |x - c| < \min(\delta,\delta_g) \ \implies \ |g(x) - L| < \varepsilon) $$
 * $$ \lim_{x \to c}f(x) = L = \lim_{x \to c}g(x) $$


 * $$ \exists\delta>0 \ \forall x ( 0 < x - c < \delta \implies f(x) = g(x) ) $$
 * $$ \forall x ( 0 < x - c < \delta_g \implies f(x) = g(x) ); \delta_g>0 $$
 * $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x (0 < x - c < \delta \ \implies \ |f(x) - L| < \varepsilon) $$
 * $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x (0 < x - c < \min(\delta,\delta_g) \ \implies \ |f(x) - L|=|g(x) - L| < \varepsilon) $$
 * $$ \forall \varepsilon > 0 \ \ \exists \min(\delta,\delta_g) > 0 \ \ \forall x (0 < x - c < \min(\delta,\delta_g) \ \implies \ |g(x) - L| < \varepsilon) $$
 * $$ \lim_{x \to c^+}f(x) = L = \lim_{x \to c^+}g(x) $$
 * $$ \exists\delta>0 \ \forall x ( 0 < x - c < \delta \implies f(x) = f(\min(x,c+k)) ); k>0 $$
 * $$ \forall x ( 0 < x-c < k \iff c < x < c+k \implies x=\min(x,c+k) \implies f(x) = f(\min(x,c+k)) ); k>0 $$
 * $$ \forall k>0\ \forall x ( 0 < x-c < k \implies f(x) = f(\min(x,c+k)) )$$
 * $$ \forall \delta>0\ \forall x ( 0 < x-c < \delta \implies f(x) = f(\min(x,c+\delta)) )$$
 * $$ \exists \delta>0\ \forall x ( 0 < x-c < \delta \implies f(x) = f(\min(x,c+\delta)) )$$; cannot apply in this prove
 * $$ \lim_{x \to c^+}f(x) = L = \lim_{x \to c^+}f(\min(x,c+k)); k>0 $$
 * $$ \forall x ( 0 < x-c < k \implies x=\min(x,c+k) \implies f(x,x) = f(\min(x,c+k),x) ); k>0 $$
 * $$ \lim_{x \to c^+}f(x,x) = L = \lim_{x \to c^+}f(\min(x,c+k),x); k>0 $$
 * $$ \forall x ( 0 < x-c < k \implies x=\max(x,c) \implies f(x,x) = f(\max(x,c),x) ); k>0 $$
 * $$ \lim_{x \to c^+}f(x,x) = L = \lim_{x \to c^+}f(\max(x,c),x); k>0 $$
 * $$ \exists\delta>0 \ \forall x ( 0 < c-x < \delta \implies f(x) = g(x) ) $$
 * $$ \forall x ( 0 < c-x < \delta_g \implies f(x) = g(x) ); \delta_g>0 $$
 * $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x (0 < c-x < \delta \ \implies \ |f(x) - L| < \varepsilon) $$
 * $$ \forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ \forall x (0 < c-x < \min(\delta,\delta_g) \ \implies \ |f(x) - L|=|g(x) - L| < \varepsilon) $$
 * $$ \forall \varepsilon > 0 \ \ \exists \min(\delta,\delta_g) > 0 \ \ \forall x (0 < c-x < \min(\delta,\delta_g) \ \implies \ |g(x) - L| < \varepsilon) $$
 * $$ \lim_{x \to c^-}f(x) = L = \lim_{x \to c^-}g(x) $$


 * 1) $$y' = 1/x; y = \ln x + c$$
 * 2) * $$y = \ln x + \ln 1 = \ln x + 0; e^y = x --> y = \ln x + 2n\pi j$$
 * 3) * $$y = \ln -x = \ln x + \ln -1 = \ln x + \pi j; e^{y-\pi j} = x --> y = \ln x + (2n+1)\pi j$$
 * 4) * $$0 = \ln (1) = \ln (-1 \cdot -1) = \ln (-1) + \ln (-1) = \pi j + \pi j = 2 \pi j !!!$$
 * 5) * $$1 = \sqrt 1 = \sqrt {-1 \cdot -1} = \sqrt {-1} \sqrt {-1} = j \cdot j = -1 !!!$$

$$ \int_{a}^{a+t} f(x) dx = \lim_{d_x \to 0^+} \sum_{i=1}^{\lfloor t/d_x \rfloor-1} f(a+id_x) d_x = F(t) ; t>0$$
 * $$ \int_{a}^{a+t} f'(x) dx ???$$ the function to be integrated may not be a differential of any function
 * $$ \int_{a}^{a+t} f'(x) dx = f(a+t)-f(a)$$
 * $$ \frac{d(\int_{a}^{a+t} f(x) dx)}{dt} = \frac{d(F(a+t)-F(a))}{dt} = \frac{dF(a+t)}{dt}-\frac{dF(a)}{dt} = f(a+t)-0$$
 * 1) $$ \frac{dF(t)}{dt} = \lim_{d_t \to 0} \frac{F(t+d_t)-F(t)}{d_t} = \lim_{d_t \to 0} \frac{ \lim_{d_x \to 0^+} \sum_{i=1}^{\lfloor (t+d_t)/d_x \rfloor-1} f(a+id_x) d_x - \lim_{d_x \to 0^+} \sum_{i=1}^{\lfloor t/d_x \rfloor-1} f(a+id_x) d_x }{d_t} ; t>0$$
 * 2) $$ \forall d_t ( 0 < |d_t - 0| < t \implies t+d_t>0 ) ; t>0$$
 * 3) $$ \frac{dF(t)}{dt} = \frac{d(\int_{a}^{a+t} f(x) dx)}{dt} $$
 * 4) $$ \frac{dF(t)}{dt} = \lim_{d_t \to 0} \frac{ \lim_{d_x \to 0^+} \left( \sum_{i=1}^{\lfloor (t+d_t)/d_x \rfloor-1} f(a+id_x) d_x - \sum_{i=1}^{\lfloor t/d_x \rfloor-1} f(a+id_x) d_x \right) }{d_t} $$
 * 5) * $$ \sum_{i=1}^{n+m} f(i) = \sum_{i=1}^n f(i) + \sum_{i=n+1}^{n+m} f(i) $$
 * $$ F(t) = \lim_{d_x \to 0^+} \sum_{i=1}^{\lfloor t/d_x \rfloor-1} f(a+id_x) d_x = \lim_{d_x \to 0^+} \sum_{i=1}^{\lfloor t/\min(d_x,0+t/2) \rfloor-1} f(a+id_x) d_x; t/2>0$$
 * $$ \lim_{x \to c^+}f(x,x) = L = \lim_{x \to c^+}f(\min(x,c+k),x); k>0 $$

paradox http://m.facebook.com/story.php?story_fbid=4357574176677&id=1207419880&comment_id=4544749&_rdr
 * "ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ"
 * "ประโยค a เป็นเท็จ" and ประโยค a == "ประโยค a เป็นเท็จ"
 * conclusion: [ประโยค a == "ประโยค a เป็นเท็จ"] เป็นเท็จ
 * "ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ" <==> "ประโยค a เป็นเท็จ" and ประโยค a == "ประโยค a เป็นเท็จ" ???
 * not "ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ" <==> not "ประโยค a เป็นเท็จ" or not ประโยค a == "ประโยค a เป็นเท็จ" ???
 * not "ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ" <==> "ประโยค a เป็นจริง" or ประโยค a != "ประโยค a เป็นเท็จ" ???
 * $$a = b \iff c = a \land c = b$$??? (rather be $$\exists c\ c = a \land c = b$$??? --Ans 13:42, 16 June 2017 (UTC))
 * $$c = a \land c = b \implies a = b$$
 * $$a = b \implies [c = a \iff c = b]$$
 * $$[c = a \iff c = b] \implies a = b$$??? can easily disprove
 * "ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ" <== "ประโยค a เป็นเท็จ" and ประโยค a == "ประโยค a เป็นเท็จ"
 * not "ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ" ==> not "ประโยค a เป็นเท็จ" or not ประโยค a == "ประโยค a เป็นเท็จ"
 * not "ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ" ==> "ประโยค a เป็นจริง" or ประโยค a != "ประโยค a เป็นเท็จ" ???
 * all a ["ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ" <== "ประโยค a เป็นเท็จ" and ประโยค a == "ประโยค a เป็นเท็จ"]
 * all a [not "ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ" ==> not "ประโยค a เป็นเท็จ" or not ประโยค a == "ประโยค a เป็นเท็จ"]
 * all a [not "ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ" ==> "ประโยค a เป็นจริง" or ประโยค a != "ประโยค a เป็นเท็จ" ???]
 * a = !a, ถามว่า a เป็น T (true) หรือ F (false)?
 * ตอบ a ไม่สามารถเป็นได้ทั้ง T หรือ F
 * จึงสรุปว่า "a = !a" เป็นเท็จ, เมื่อมันเป็นเท็จ นั่นก็หมายความว่า "a != !a"?
 * ถ้ากำหนดให้, ประโยค a == "ประโยค a เป็นเท็จ", ถามว่า ประโยคที่ว่า "ประโยค a เป็นเท็จ" เป็นจริงหรือเท็จ?
 * "ประโยค a เป็นเท็จ" ไม่สามารถเป็นได้ทั้ง จริง หรือ เท็จ
 * ก็เหมือนกับคำถามว่า กำหนดให้ a == !a ถามว่า a เป็นจริงหรือเท็จหละ? คำตอบก็คือไม่สามารถเป็นได้ทั้งจริงและเท็จ นั่นก็คือไม่มีคำตอบ, หรือ ไม่มีค่า a ที่ทำให้ a == !a เป็นจริง, ซึ่งก็หมายความว่า a == !a ไม่มีทางเป็นจริง.  เมื่อ a == !a ไม่มีทางเป็นจริง ก็ต้องกลับไปบอกว่า มันผิดตั้งแต่ตอนที่กำหนดให้ a == !a แล้ว, คือผิดตั้งแต่ตั้งคำถามแล้ว.
 * ด้วยตรรกะแบบเดียวกัน จึงไม่มีทางที่จะมี ประโยค a ที่ว่า ประโยค a == "ประโยค a เป็นเท็จ", นั่นก็หมายความว่า มันผิดตั้งแต่ตอนกำหนดให้ ประโยค a == "ประโยค a เป็นเท็จ" แล้ว.
 * "ประโยค a == ประโยคที่จะพูดถัดจากนี้". "ประโยค a เป็นเท็จ"
 * "ประโยคที่ผมกำลังพูดอยู่นี้เป็นเท็จ", ถามว่าประโยคข้างหน้านี้เป็นจริงหรือเท็จ?
 * สมมุติว่ามันเป็นจริง, ถ้ามันเป็นจริง ถ้างั้นก็หมายความว่ามันเป็นเท็จนะสิ, ก็ในเมื่อเชื่อว่ามันเป็นจริง ถ้าเชื่อมัน ก็มันบอกว่ามันเองเป็นเท็จก็ต้องเชื่อมัน ดังนั้นก็ต้องเชื่อตามมันว่ามันเป็นเท็จสิ. ซึ่งขัดแย้งกันจึงสรุปได้ว่าเป็นความเชื่อที่ผิดตั้งแต่ต้น เป็น assumption ที่ผิด จึงเป็นไปไม่ได้ที่มันจะเป็นจริง.  (จริงๆ ตรงที่เชื่อตามมันว่ามันเป็นเท็จ ถ้าเชื่อว่ามันเป็นเท็จ ก็แสดงว่าจริงๆ แล้วไม่ได้เชื่อมันนะสิ, ขัดแย้งกันอีกละ ตกลงว่าจะเชื่อหรือไม่เชื่อมันกันแน่)
 * ถ้าเป็นไปไม่ได้ที่มันจะเป็นจริง ถ้างั้นก็หมายความว่ามันเป็นเท็จนะสิ. ถ้ามันเป็นเท็จ ก็ต้องไม่เชื่อมันใช่มั้ย, ในเมื่อไม่เชื่อมัน ดังนั้นที่มันบอกว่ามันเป็นเท็จก็ต้องไม่เชื่อมัน, เมื่อไม่เชื่อที่มันบอกว่ามันเป็นเท็จ ถ้างั้นมันก็ต้องเป็นจริงนะสิ.  อ้าวขัดแย้งกับ assumption อีกละ ก็เป็นไปไม่ได้อีกเหมือนกันที่มันจะเป็นเท็จ.
 * อ้าว จริงก็ไม่ใช่ เท็จก็ไม่ใช่ แล้วตกลงค่าความจริงของมันเป็นอะไรหละ? ถ้างั้นสรุปได้ไหมว่า มันไม่ใช่ทั้งจริงและเท็จ?  จะสรุปอย่างนั้นมันก็แปลกๆ อยู่นา ประโยคที่พูดมาถ้ามันไม่จริง มันก็ต้องเท็จ ถ้ามันไม่เท็จ มันก็ต้องจริง มันเป็นได้แค่สองอย่างนี้เท่านั้น, มันจะเป็นอย่างอื่นนอกเหนือจากนี้ได้ยังไง?  ถ้ามันไม่ใช่ทั้งจริงและเท็จ แล้วมันจะหมายความว่ายังไงหละ มันเป็นสถานะที่แปลกๆ อยู่นา.
 * ยังบอกไม่ได้ว่าจริงหรือเท็จ เนื่องจากประโยคติดตัวแปร หรือเรียกตัวเอง (recursive) แบบไม่สิ้นสุด?
 * $$f(a) \iff \forall x\in\{a\} f(x) \iff \forall x (x=a \implies f(x)) \iff \exists x (x=a \land f(x))$$
 * $$f(a(x)) \cancel\iff \forall x\in\{x|x=a(x)\} f(x) \iff \forall x (x=a(x) \implies f(x)) \cancel\iff \exists x (x=a(x) \land f(x))$$

Given that f(x) is a boolean function,


 * 1) $$g(x) \in \mathbb{R} \implies f(g(x))$$
 * 2) $$x \in \mathbb{R} \implies f(x)$$

Is #1 equivalent to #2?
 * School of Mathematics Help Desk


 * 1) $$\forall x \in \mathbb{A} \ f(x)$$ (for any f)
 * 2) * $$f(x_1) \land f(x_2) \land f(x_3) \land \dots \land f(x_n)$$
 * 3) $$x \in \mathbb{A} \implies f(x)$$ (for any x and f)
 * 4) $$\forall x \in \mathbb{A} \ f(x)$$ (for any f) $$\iff$$ $$x \in \mathbb{A} \implies f(x)$$ (for any x and f); ???
 * 5) $$\forall x \in \mathbb{A} \ f(x)$$ $$\iff$$ $$x \in \mathbb{A} \implies f(x)$$ (for any x); for any f???


 * 1) $$\exists x \in \mathbb{A} \ f(x)$$
 * 2) $$\neg\neg(\ \exists x \in \mathbb{A} \ f(x)\ )$$
 * 3) $$\neg(\ \forall x \in \mathbb{A} \ \neg f(x)\ )$$
 * 4) $$\neg(\ x \in \mathbb{A} \implies \neg f(x)\ )$$
 * 5) $$\neg(\ x \notin \mathbb{A} \lor \neg f(x)\ )$$
 * 6) $$x \in \mathbb{A} \land f(x)$$


 * 1) $$x \in \mathbb{A} \implies \neg f(x)$$
 * 2) $$x \notin \mathbb{A} \lor \neg f(x)$$
 * 3) $$\neg\neg(\ x \notin \mathbb{A} \lor \neg f(x)\ )$$
 * 4) $$\neg(\ x \in \mathbb{A} \land f(x)\ )$$


 * 1) $$x \in \mathbb{A} \implies f(x)$$
 * 2) $$x \notin \mathbb{A} \lor f(x)$$
 * 3) $$\neg\neg(\ x \notin \mathbb{A} \lor f(x)\ )$$
 * 4) $$\neg(\ x \in \mathbb{A} \land \neg f(x)\ )$$


 * 1) $$\neg(\ \exists x \in \mathbb{A} \ f(x)\ )$$
 * 2) $$\forall x \in \mathbb{A} \ \neg f(x)$$
 * 3) $$x \in \mathbb{A} \implies \neg f(x)$$
 * 4) $$x \notin \mathbb{A} \lor \neg f(x)$$
 * 5) $$\neg(\ x \in \mathbb{A} \land f(x)\ )$$


 * 1) $$\exists x \in \mathbb{A} \ f(x)$$
 * 2) $$\neg\neg(\ \exists x \in \mathbb{A} \ f(x)\ )$$
 * 3) $$\neg(\ \forall x \in \mathbb{A} \ \neg f(x)\ )$$
 * 4) $$\neg(\ x \in \mathbb{A} \implies \neg f(x);\ for\ any\ x\ )$$
 * 5) $$\neg(\ x \notin \mathbb{A} \lor \neg f(x);\ for\ any\ x\ )$$
 * 6) $$\neg(\ x \notin \mathbb{A} \lor \neg f(x)\ )$$; there's x
 * 7) $$x \in \mathbb{A} \land f(x)$$; there's x

$$\forall x \in \mathbb{R} \implies 2x \in \mathbb{R}$$

$$(\ \forall x \in \mathbb{R} \implies f(x)\ ) \implies (\ \forall x \in \mathbb{R} \implies f(2x)\ )$$

^^^ need $$\forall$$?

$$x \in \mathbb{R} \implies 2x \in \mathbb{R}$$

$$(\ x \in \mathbb{R} \implies f(x)\ ) \implies (\ x \in \mathbb{R} \implies f(2x)\ )$$

ellipse motion

 * $$\left[\left[a\right]b\right] \left[(a)b\right] \left[\left((a)\right)b\right] \left(f\right) \left(f'\right) \left(f^2\right) \left(f'^2\right)$$
 * $$\vec s(t) = \left[a\cos f(t), b\sin f(t)\right]$$
 * $$\vec s\,'(t) = \left[-a f'(t)\sin f(t), b f'(t)\cos f(t)\right]$$
 * $$\vec s\,(t) = \left[-a \left( f(t)\sin f(t) + f'^2(t)\cos f(t) \right), b \left( f''(t)\cos f(t)-f'^2(t)\sin f(t) \right)\right] \propto \vec s(t) + [k, 0]$$
 * $$\left| \vec s\,(t) \right|^2 = a^2 \left( f(t)\sin f(t) + f'^2(t)\cos f(t) \right)^2 + b^2 \left( f''(t)\cos f(t)-f'^2(t)\sin f(t) \right)^2$$
 * $$\left| \vec s\,(t) \right|^2 = f^2(t) \left( a^2 \sin^2 f(t) + b^2 \cos^2 f(t) \right) + f'^4(t) \left( a^2 \cos^2 f(t) + b^2 \sin^2 f(t) \right) + 2 f''(t)f'^2(t) \sin f(t) \cos f(t) \left( a^2 - b^2 \right)$$
 * $$\left| \vec s\,(t) \right|^2 = f^2(t) \left( a^2 + (b^2-a^2) \cos^2 f(t) \right) + f'^4(t) \left( (a^2-b^2) \cos^2 f(t) + b^2 \right) + f''(t)f'^2(t) \sin 2f(t) \left( a^2 - b^2 \right)$$
 * $$\frac {a\cos f(t)} {b\sin f(t)} = \frac { -a \left( f(t)\sin f(t) + f'^2(t)\cos f(t)\right) } { b \left( f(t)\cos f(t)-f'^2(t)\sin f(t) \right) }$$
 * $$\frac {a\cos f(t)+k} {b\sin f(t)} = \frac { -a \left( f(t)\sin f(t) + f'^2(t)\cos f(t)\right) } { b \left( f(t)\cos f(t)-f'^2(t)\sin f(t) \right) }$$
 * $$\frac {a\cos f(t)+k} {\sin f(t)} = \frac { -a \left( f(t)\sin f(t) + f'^2(t)\cos f(t)\right) } { \left( f(t)\cos f(t)-f'^2(t)\sin f(t) \right) }$$
 * $$a \left( f(t)\cos^2 f(t)-f'^2(t)\sin f(t)\cos f(t) \right) + k \left( f(t)\cos f(t)-f'^2(t)\sin f(t) \right) = -a \left( f''(t)\sin^2 f(t) + f'^2(t)\sin f(t)\cos f(t) \right)$$
 * $$a f(t)\left(\cos^2 f(t)+\sin^2 f(t)\right) + k \left( f(t)\cos f(t)-f'^2(t)\sin f(t) \right) = 0$$
 * $$a f(t) + k f(t) \cos f(t) - k f'^2(t) \sin f(t) = 0$$
 * $$\left( a + k \cos f(t) \right) f''(t) = k \sin f(t) f'^2(t)$$
 * $$f''(t) = \frac{ k \sin f(t) f'^2(t) }{ a + k \cos f(t) }$$
 * $$\frac{f''(t)}{f'^2(t)} = \frac{df'(t)}{f'^2(t)dt} = \frac{ k \sin f(t) }{ a + k \cos f(t) }$$
 * $$\frac{-d1/f'(t)}{dt} = \frac{ k \sin f(t) }{ a + k \cos f(t) }$$
 * $$f''(t) = \frac{ k \sin f(t) \frac{df(t)}{dt} f'(t) }{ a + k \cos f(t) } = \frac{ -k \frac{d\cos f(t)}{dt} f'(t) }{ a + k \cos f(t) } = \frac{ - \frac{d( a+k\cos f(t) )}{dt} f'(t) }{ a + k \cos f(t) }$$
 * $$f''(t) (a + k \cos f(t)) = k \sin f(t) \frac{df(t)}{dt} f'(t) = -k \frac{d\cos f(t)}{dt} f'(t) = - \frac{d( a+k\cos f(t) )}{dt} f'(t)$$
 * $$f''(t) = \frac{df'(t)}{dt} = \frac{ - d\ln| a+k\cos f(t) |}{dt} f'(t)$$
 * $$\frac{d\ln| f'(t) |}{dt} = \frac{ - d\ln| a+k\cos f(t) |}{dt}$$
 * $$\frac{d\ln| f'(t) |}{dt} + \frac{ d\ln| a+k\cos f(t) |}{dt} = 0$$
 * $$\frac{d\left( \ln| f'(t) | + \ln| a+k\cos f(t) | \right)}{dt} = 0$$
 * $$\ln| f'(t) | + \ln| a+k\cos f(t) | = c$$
 * $$e^{\ln| f'(t) | + \ln| a+k\cos f(t) |} = e^c$$
 * $$| f'(t) | \cdot | a+k\cos f(t) | = e^c$$
 * $$| f'(t) \left( a+k\cos f(t) \right) | = e^c$$
 * $$\left| \frac{df(t)}{dt} \left( a+k\cos f(t) \right) \right| = e^c$$
 * $$\left| k \sin f(t) \frac{df(t)}{dt} \left( a+k \cos f(t) \right) \right| = |k\sin f(t)|e^c$$
 * $$\left| -\frac{d(a+k\cos f(t))}{dt} \left( a+k\cos f(t) \right) \right| = |k\sin f(t)|e^c$$
 * $$\left| \frac{d(a+k\cos f(t))^2}{dt} \right| = 2|k\sin f(t)|e^c$$
 * $$\left| \frac{d \left( a f(t) + k \sin f(t) \right)}{dt} \right| = e^c = e^c \frac{dt}{dt} = \frac{de^c t}{dt} = \frac{d \left( a f(t) + k \sin f(t) \right)}{dt} \text{abs}' \frac{d \left( a f(t) + k \sin f(t) \right)}{dt}$$
 * $$f'(x) \text{abs}' f(x) = g'(x) \text{abs}' g(x)$$; ???
 * $$f'(x) d|f(x)|/df(x) = f'(x) |f(x)|/f(x) = f'(x) \frac{d|f(x)|/df(x)}{d|x|/dx} |x|/x = f'(x) \frac{d|f(x)|/d|x|}{f'(x)} |x|/x = \frac{d|f(x)|}{d|x|} |x|/x$$
 * $$\frac{d|f(x)|}{d|x|} |x|/x = \frac{df(x)|f(x)|/f(x)}{dx|x|/x} |x|/x = \frac{|f(x)|/f(x)}{|x|/x} \frac{df(x)}{dx} |x|/x = \frac{d|f(x)|/df(x)}{d|x|/dx} \frac{df(x)}{dx} \cdot |x|/x = \frac{d|f(x)|/d|x|}{df(x)/dx} f'(x) |x|/x$$
 * $$\text{abs}' f(x) df(x) = \text{abs}' g(x) dg(x)$$
 * $$d |f(x)| = d |g(x)|$$
 * $$dx|x|/dx = |x|dx/dx+xd|x|/dx = |x|+|x| = 2|x|$$; where x!=0 (d|x|/dx = x/|x| = |x|/x)
 * $$\frac{(0+dx)|0+dx|-0|0|}{dx} = dx|dx|/dx = |dx| => 0$$
 * $$d|x|/dx = |x|/x; d|x|/dx \cdot dx/d|x| = d|x|/dx \cdot x/|x| = 1 = |dx/dx|$$
 * $$d|f(x)|/dx = d|f(x)|/d|x| \cdot d|x|/dx$$
 * $$f(x)|f(x)| = g(x)|g(x)|$$
 * $$f'(x) 2|f(x)| = g'(x) 2|g(x)|$$
 * $$|f(x)| df(x) = |g(x)| dg(x)$$
 * $$|f'(x)| = |g'(x)| = f'(x)|f'(x)|/f'(x) ??? = f'(x)d|f'(x)|/df'(x) = \frac{f'(x)|df(x)/dx|}{df(x)/dx}$$
 * $$|df(x)/dt| = |dg(x)/dt|$$
 * $$a f(t) + k \sin f(t) = \pm e^c t + c_2 = c_1 t + c_2$$
 * $$f'(t) (a + k \cos f(t)) = \pm e^c = c_1$$
 * $$f''(t) (a + k \cos f(t)) - f'^2(t) k \sin f(t) = 0$$
 * $$f''(t) (a + k \cos f(t)) = f'^2(t) k \sin f(t)$$
 * $$f''(t) (a + k \cos f(t))^3 = f'^2(t) (a + k \cos f(t))^2 k \sin f(t)$$
 * $$f''(t) (a + k \cos f(t))^3 = c_1^2 k \sin f(t)$$
 * $$\vec s\,'(t) (a + k \cos f(t)) = c_1 \left[-a \sin f(t), b \cos f(t)\right]$$
 * $$\vec s\,''(t) (a + k \cos f(t))^3 = c_1^2 \left[-a \left( k \sin^2 f(t) + (a + k \cos f(t)) \cos f(t) \right), b \left( k \sin f(t) \cos f(t) - (a + k \cos f(t)) \sin f(t) \right)\right]$$
 * $$\vec s\,''(t) (a + k \cos f(t))^3 = c_1^2 \left[-a \left( k + a\cos f(t) \right), - ab \sin f(t) \right]$$
 * $$\vec s\,''(t) (a + k \cos f(t))^3 = -a c_1^2 \left[k + a\cos f(t), b \sin f(t) \right]$$
 * $$\vec s\,''(t) (a + k \cos f(t))^3 = -a c_1^2 (\vec s(t) + [k, 0])$$
 * $$\left| \vec s\,''(t) \right|^2 (a + k \cos f(t))^6 = \left(a c_1^2\right)^2 \left| \vec s(t) + [k, 0] \right|^2 = a^2 c_1^4 R^2$$
 * $$\vec s(t) + [k, 0] = \left[a\cos f(t) + k, b\sin f(t)\right]$$
 * $$R^2 = \left| \vec s(t) + [k, 0] \right|^2 = (a\cos f(t) + k)^2 + b^2 \sin^2 f(t) = k^2 + 2ak\cos f(t) + a^2 \cos^2 f(t) + b^2 \sin^2 f(t)$$
 * If k is focus point, then $$k^2 = a^2 - b^2$$
 * $$R^2 = k^2 + 2ak\cos f(t) + a^2 \cos^2 f(t) + \left(a^2-k^2\right) \sin^2 f(t)$$
 * $$R^2 = k^2 \left( 1 - \sin^2 f(t) \right) + 2ak\cos f(t) + a^2 = k^2 \cos^2 f(t) + 2ak\cos f(t) + a^2 = \left( a + k \cos f(t) \right)^2$$
 * $$\left| \vec s\,''(t) \right|^2 R^6 = a^2 c_1^4 R^2$$
 * $$\left| \vec s\,''(t) \right| = \frac{a c_1^2}{R^2}$$
 * $$\left| \vec s\,(t) \right|^2 = f^2(t) \left( a^2 \sin^2 f(t) + \left(a^2-k^2\right) \cos^2 f(t) \right) + f'^4(t) \left( a^2 \cos^2 f(t) + \left(a^2-k^2\right) \sin^2 f(t) \right) + 2 f''(t)f'^2(t) \left( a^2 \sin f(t) \cos f(t) - \left(a^2-k^2\right) \sin f(t) \cos f(t) \right)$$
 * $$\left| \vec s\,(t) \right|^2 = f^2(t) \left( a^2 - k^2 \cos^2 f(t) \right) + f'^4(t) \left( a^2 - k^2 \sin^2 f(t) \right) + 2 f''(t)f'^2(t) k^2 \sin f(t) \cos f(t)$$
 * $$\left| \vec s\,(t) \right|^2 = f^2(t) \left( a^2 - k^2 \cos^2 f(t) \right) + f'^4(t) \left( k^2 \cos^2 f(t) + a^2 - k^2 \right) + f''(t)f'^2(t) k^2 \sin 2f(t)$$
 * $$R^4 = a^4 + 4 a^3 k \cos f(t) + 6 a^2 k^2 \cos^2 f(t) + 4 a k^3 \cos^3 f(t) + k^4 \cos^4 f(t)$$
 * $$\left( a^2 - k^2 \sin^2 f(t) \right) R^4 = a^6 + 4 a^5 k \cos f(t) + 6 a^4 k^2 \cos^2 f(t) + 4 a^3 k^3 \cos^3 f(t) + k^4 a^2 \cos^4 f(t) - \left( a^4 k^2 \sin^2 f(t) + 4 a^3 k^3 \cos f(t) \sin^2 f(t) + 6 a^2 k^4 \cos^2 f(t) \sin^2 f(t) + 4 a k^5 \cos^3 f(t) \sin^2 f(t) + k^6 \cos^4 f(t) \sin^2 f(t) \right)$$
 * $$a^2 R^4 = a^6 + 4 a^5 k \cos f(t) + 6 a^4 k^2 \cos^2 f(t) + 4 a^3 k^3 \cos^3 f(t) + a^2 k^4 \cos^4 f(t)$$
 * $$-k^2 R^4 = -\left( a^4 k^2 + 4 a^3 k^3 \cos f(t) + 6 a^2 k^4 \cos^2 f(t) + 4 a k^5 \cos^3 f(t) + k^6 \cos^4 f(t) \right)$$
 * $$k^2 \cos^2 f(t) R^4 = a^4 k^2 \cos^2 f(t) + 4 a^3 k^3 \cos^3 f(t) + 6 a^2 k^4 \cos^4 f(t) + 4 a k^5 \cos^5 f(t) + k^6 \cos^6 f(t)$$
 * $$(a^6 - a^4 k^2) + (4 a^5 k - 4 a^3 k^3) \cos f(t) + (6 a^4 k^2 - 6 a^2 k^4 + a^4 k^2) \cos^2 f(t) + (4 a^3 k^3 - 4 a k^5 + 4 a^3 k^3) \cos^3 f(t) + (a^2 k^4 - k^6 + 6 a^2 k^4) \cos^4 f(t) + 4 a k^5 \cos^5 f(t) + k^6 \cos^6 f(t)$$
 * $$(a^6 - a^4 k^2) + (4 a^5 k - 4 a^3 k^3) \cos f(t) + (7 a^4 k^2 - 6 a^2 k^4) \cos^2 f(t) + (8 a^3 k^3 - 4 a k^5) \cos^3 f(t) + (7 a^2 k^4 - k^6) \cos^4 f(t) + 4 a k^5 \cos^5 f(t) + k^6 \cos^6 f(t)$$
 * $$f'(t) f(t) = c \iff \frac{df^2(t)}{dt} = 2c \iff f^2(t) = 2ct + c_2$$
 * $$f'(t) f(t) = 0 \iff \frac{df^2(t)}{dt} = 0 \iff f^2(t) = c_2$$
 * $$\forall t f'(t) f(t) = 0 \iff??? \forall f'(t)=0 \lor \forall f(t)=0 \iff??? \forall f(t)=c \lor f(t)=0$$


 * 1) $$\binom n i + \binom n {i-1} = \frac {n!} {i!(n-i)!} + \frac {n!} {(i-1)!(n-i+1)!} = \frac {n!} {(i-1)!(n-i)!} \left( \frac 1 i + \frac 1 {n-i+1} \right) ;\ 0 < i \le n$$
 * 2) $$= \frac {n!} {(i-1)!(n-i)!} \frac {n-i+1 + i} {i(n-i+1)} = \frac {n!} {(i-1)!(n-i)!} \frac {n+1} {i(n-i+1)} = \frac {(n+1)!} {i!(n-i+1)!} = \frac {(n+1)!} {i!(n+1-i)!} = \binom {n+1} i ;\ 0 < i \le n$$
 * 3) $$\binom n i + \binom n {i-1} = \binom n 0 + \binom n {-1} = 1 + 0 = \binom {n+1} 0 = \binom {n+1}i ;\ 0=i \le n$$
 * 4) $$\binom n i + \binom n {i-1} = 0 + 0 = 0 = \binom {n+1}i ;\ i<0 \le n$$
 * 5) $$\binom n i + \binom n {i-1} = \binom n {n+1} + \binom n {n+1-1} = 0 + \binom n n = 1 = \binom {n+1}{n+1} = \binom {n+1}i ;\ 0 \le n, n+1=i$$
 * 6) $$\binom n i + \binom n {i-1} = 0 + 0 = 0 = \binom {n+1}i ;\ 0 \le n, n<n+1<i, n<i-1$$
 * 7) $$\binom n i + \binom n {i-1} = \binom {n+1}i ;\ 0 \le n$$


 * 1) $$P(n)\!:\ \ \forall 0 \le k \le n \sum_{i=0}^n\binom n i \prod_{j=1}^k(a_j-i) (-1)^i = n! \cdot 0^{n-k} = n! \binom k n$$
 * 2) $$P(0)\!:\ \ \forall 0 \le k \le 0 \sum_{i=0}^0\binom 0 i \prod_{j=1}^k(a_j-i) (-1)^i = 0! \binom k 0$$
 * 3) $$\forall 0 \le k \le 0 \binom 0 0 \prod_{j=1}^k(a_j-0) (-1)^0 = 1 \cdot \binom k 0$$
 * 4) $$\forall 0 \le k \le 0\ 1 \cdot \prod_{j=1}^k a_j = \binom k 0$$
 * 5) $$\forall 0 \le k \le 0\ \prod_{j=1}^k a_j = \binom k 0$$
 * 6) $$\prod_{j=1}^0 a_j = \binom 0 0$$
 * 7) $$1 = 1$$
 * 8) $$P(n+1)\!:\ \ \forall 0 \le k \le n+1 \sum_{i=0}^{n+1} \binom {n+1} i \prod_{j=1}^k(a_j-i) (-1)^i = (n+1)! \binom k {n+1} ;\ n\ge0$$
 * 9) $$\forall 0 \le k \le n+1 \sum_{i=0}^{n+1} \left( \binom n i + \binom n {i-1} \right) (a-i)^k (-1)^i = (n+1)! \binom k {n+1} ;\ n\ge1$$
 * 10) $$\forall 0 \le k \le n+1 \sum_{i=0}^{n+1} \binom n i (a-i)^k (-1)^i + \sum_{i=0}^{n+1} \binom n {i-1} (a-i)^k (-1)^i = (n+1)! \binom k {n+1} ;\ n\ge1$$
 * 11) $$\forall 0 \le k \le n+1 \sum_{i=0}^n \binom n i (a-i)^k (-1)^i + \binom n {n+1} (a-n-1)^k (-1)^{n+1} + \sum_{i=-1}^n \binom n i (a-i-1)^k (-1)^{i+1} = (n+1)! \binom k {n+1} ;\ n\ge1$$
 * 12) $$\forall 0 \le k \le n+1 \sum_{i=0}^n \binom n i (a-i)^k (-1)^i + 0 \cdot (a-n-1)^k (-1)^{n+1} - \sum_{i=-1}^n \binom n i (a-1-i)^k (-1)^i = (n+1)! \binom k {n+1} ;\ n\ge1$$
 * 13) $$\forall 0 \le k \le n+1 \sum_{i=0}^n \binom n i (a-i)^k (-1)^i + 0 - \binom n {-1} (a-1+1)^k (-1)^{-1} - \sum_{i=0}^n \binom n i (a-1-i)^k (-1)^i = (n+1)! \binom k {n+1} ;\ n\ge1$$
 * 14) $$\forall 0 \le k \le n+1 \sum_{i=0}^n \binom n i (a-i)^k (-1)^i - 0 \cdot (a-1+1)^k (-1)^{-1} - \sum_{i=0}^n \binom n i (a-1-i)^k (-1)^i = (n+1)! \binom k {n+1} ;\ n\ge1$$
 * 15) $$\forall 0 \le k \le n+1 \sum_{i=0}^n \binom n i (a-i)^k (-1)^i - \sum_{i=0}^n \binom n i (a-1-i)^k (-1)^i = (n+1)! \binom k {n+1} ;\ n\ge1$$
 * 16) $$\forall 0 \le k \le n \sum_{i=0}^n \binom n i (a-i)^k (-1)^i - \sum_{i=0}^n \binom n i (a-1-i)^k (-1)^i = (n+1)! \binom k {n+1} ;\ n\ge1$$
 * 17) $$\forall 0 \le k \le n\ n! \binom k n - n! \binom k n = (n+1)! \cdot 0 ;\ n\ge1$$
 * 18) $$\forall 0 \le k \le n\ 0 = 0 ;\ n\ge1$$
 * 19) $$\forall k = n+1 \sum_{i=0}^n \binom n i (a-i)^k (-1)^i - \sum_{i=0}^n \binom n i (a-1-i)^k (-1)^i = (n+1)! \binom k {n+1} ;\ n\ge1$$
 * 20) $$\sum_{i=0}^n \binom n i (a-i)^{n+1} (-1)^i - \sum_{i=0}^n \binom n i (a-1-i)^{n+1} (-1)^i = (n+1)! \binom {n+1} {n+1} ;\ n\ge1$$
 * 21) $$\sum_{i=0}^n \binom n i (a-i)^{n+1} (-1)^i - \sum_{i=0}^n \binom n i (a-i-1)^{n+1} (-1)^i = (n+1)! \cdot 1 ;\ n\ge1$$
 * 22) $$\sum_{i=0}^n \binom n i (a-i)^{n+1} (-1)^i - \sum_{i=0}^n \binom n i \sum_{k=0}^{n+1} \binom {n+1} k (a-i)^k (-1)^{n+1-k} (-1)^i = (n+1)! ;\ n\ge1$$
 * 23) $$\sum_{i=0}^n \binom n i (a-i)^{n+1} (-1)^i - \sum_{i=0}^n \sum_{k=0}^{n+1} \binom n i \binom {n+1} k (a-i)^k (-1)^{n+1-k} (-1)^i = (n+1)! ;\ n\ge1$$
 * 24) $$\sum_{i=0}^n \binom n i (a-i)^{n+1} (-1)^i - \sum_{k=0}^{n+1} \sum_{i=0}^n \binom n i \binom {n+1} k (a-i)^k (-1)^{n+1-k} (-1)^i = (n+1)! ;\ n\ge1$$
 * 25) $$\sum_{i=0}^n \binom n i (a-i)^{n+1} (-1)^i - \sum_{i=0}^n \binom n i \binom {n+1} {n+1} (a-i)^{n+1} (-1)^{n+1-(n+1)} (-1)^i - \sum_{k=0}^{n} \sum_{i=0}^n \binom n i \binom {n+1} k (a-i)^k (-1)^{n+1-k} (-1)^i = (n+1)! ;\ n\ge1$$
 * 26) $$\sum_{i=0}^n \binom n i (a-i)^{n+1} (-1)^i - \sum_{i=0}^n \binom n i \cdot 1 \cdot (a-i)^{n+1} (-1)^0 (-1)^i - \sum_{k=0}^{n} \binom {n+1} k (-1)^{n+1-k} \sum_{i=0}^n \binom n i (a-i)^k (-1)^i = (n+1)! ;\ n\ge1$$
 * 27) $$\sum_{i=0}^n \binom n i (a-i)^{n+1} (-1)^i - \sum_{i=0}^n \binom n i (a-i)^{n+1} (-1)^i - \sum_{k=0}^{n} \binom {n+1} k (-1)^{n+1-k} n! \binom k n = (n+1)! ;\ n\ge1$$
 * 28) $$0 - \sum_{k=0}^{n-1} \binom {n+1} k (-1)^{n+1-k} n! \binom k n - \binom {n+1} n (-1)^{n+1-n} n! \binom n n = (n+1)! ;\ n\ge1$$
 * 29) $$- \sum_{k=0}^{n-1} \binom {n+1} k (-1)^{n+1-k} n! \cdot 0 - (n+1) (-1)^1 n! \cdot 1 = (n+1)! ;\ n\ge1$$
 * 30) $$- \sum_{k=0}^{n-1} 0 + (n+1) n! = (n+1)! ;\ n\ge1$$
 * 31) $$- 0 + (n+1)! = (n+1)! ;\ n\ge1$$
 * 32) (below is canceled unfinished proof)
 * 33) $$\forall 0 \le k < n+1 \sum_{i=0}^n \frac{(n+1)!}{i!(n+1-i)!} (-1)^i \sum_{j=0}^k \binom k j (n-i)^j 1^{k-j} + \binom {n+1} {n+1}(n+1-(n+1))^k (-1)^{n+1} = 0 ;\ n\ge1$$
 * 34) $$\forall 0 \le k < n+1 \sum_{i=0}^n \frac{(n+1)n!}{i!(n+1-i)(n-i)!} (-1)^i \sum_{j=0}^k \binom k j (n-i)^j + 1\cdot 0^k (-1)^{n+1} = 0 ;\ n\ge1$$
 * 35) $$\forall 0 \le k < n+1 \sum_{i=0}^n \frac{n+1}{n+1-i} \binom n i (-1)^i \sum_{j=0}^k \binom k j (n-i)^j + 0^k (-1)^{n+1} = 0 ;\ n\ge1$$
 * 36) $$\forall 0 \le k < n+1 \sum_{i=0}^n \frac{n+1-i+i}{n+1-i} \binom n i (-1)^i \left( \sum_{j=0}^{k-1} \binom k j (n-i)^j + \binom k k (n-i)^k \right) + 0^k (-1)^{n+1} = 0 ;\ n\ge1$$
 * 37) $$\forall 0 \le k < n+1 \sum_{i=0}^n \left(1+\frac{i}{n+1-i}\right) \binom n i (-1)^i \left( \sum_{j=0}^{k-1} \binom k j (n-i)^j + (n-i)^k \right) + 0^k (-1)^{n+1} = 0 ;\ n\ge1$$
 * 38) $$\forall 0 \le k < n+1 \ Q(k) ;\ n\ge1$$
 * 39) $$Q(k):\ \ \sum_{i=0}^n \left(1+\frac{i}{n+1-i}\right) \binom n i (-1)^i \left( \sum_{j=0}^{k-1} \binom k j (n-i)^j + (n-i)^k \right) + 0^k (-1)^{n+1} = 0 ;\ n\ge1, n+1>k$$
 * 40) $$Q(0):\ \ \sum_{i=0}^n \left(1+\frac{i}{n+1-i}\right) \binom n i (-1)^i \left( \sum_{j=0}^{0-1} \binom 0 j (n-i)^j + (n-i)^0 \right) + 0^0 (-1)^{n+1} = 0 ;\ n\ge1, n+1>0$$
 * 41) $$\sum_{i=0}^n \left(1+\frac{i}{n+1-i}\right) \binom n i (-1)^i \left( 0 + 1 \right) + 1\cdot(-1)^{n+1} = 0 ;\ n\ge1$$
 * 42) $$\sum_{i=0}^n \left(1+\frac{i}{n+1-i}\right) \binom n i (-1)^i + (-1)^{n+1} = 0 ;\ n\ge1$$
 * 43) $$\sum_{i=0}^n \left( (n-i)^0 \binom n i (-1)^i + \frac{i}{n+1-i} \binom n i (-1)^i \right) + (-1)^{n+1} = 0 ;\ n\ge1$$
 * 44) $$\sum_{i=0}^n \left( 0 + \frac{i}{n+1-i} \binom n i (-1)^i \right) + (-1)^{n+1} = 0 ;\ n\ge1$$
 * 45) $$\sum_{i=0}^n \frac{i}{n+1-i} \binom n i (-1)^i + (-1)^{n+1} = 0 ;\ n\ge1$$
 * 46) $$R(n):\ \ \sum_{i=0}^n \frac{i}{n+1-i} \binom n i (-1)^i + (-1)^{n+1} = 0 ;\ n\ge1$$
 * 47) $$R(1):\ \ \sum_{i=0}^1 \frac{i}{1+1-i} \binom 1 i (-1)^i + (-1)^{1+1} = 0 ;\ 1\ge1$$
 * 48) $$\frac{0}{1+1-0} \binom 1 0 (-1)^0 + \frac{1}{1+1-1} \binom 1 1 (-1)^1 + 1 = 0$$
 * 49) $$0\cdot 1 + 1\cdot 1\cdot -1 + 1 = 0$$
 * 50) $$R(n+1):\ \ \sum_{i=0}^{n+1} \frac{i}{n+1+1-i} \binom {n+1} i (-1)^i + (-1)^{n+1+1} = 0 ;\ n\ge1$$
 * 51) $$\sum_{i=0}^n \frac{i(n+1)}{(n+1+1-i)(n+1-i)} \binom n i (-1)^i + \frac{n+1}{n+1+1-(n+1)} \binom {n+1} {n+1} (-1)^{n+1} - (-1)^{n+1} = 0 ;\ n\ge1$$
 * 52) $$\sum_{i=0}^n \frac{(n+1)i}{(n+1+1-i)(n+1-i)} \binom n i (-1)^i + \frac{n+1}{1} \cdot 1 \cdot (-1)^{n+1} - (-1)^{n+1} = 0 ;\ n\ge1$$
 * 53) $$\sum_{i=0}^n \frac{(n+1-(n+1+1-i)+(n+1+1-i))i}{(n+1+1-i)(n+1-i)} \binom n i (-1)^i + n (-1)^{n+1} + (-1)^{n+1} - (-1)^{n+1} = 0 ;\ n\ge1$$
 * 54) $$\sum_{i=0}^n \frac{(i-1 + (n+1+1-i))i}{(n+1+1-i)(n+1-i)} \binom n i (-1)^i + (-1)^{n+1} + (n-1) (-1)^{n+1} = 0 ;\ n\ge1$$
 * 1) $$\sum_{i=0}^n \frac{(i-1 + (n+1+1-i))i}{(n+1+1-i)(n+1-i)} \binom n i (-1)^i + (-1)^{n+1} + (n-1) (-1)^{n+1} = 0 ;\ n\ge1$$

(canceled unfinished proof)
 * 1) $$P(k)\!:\ \ \forall n>k \sum_{i=0}^n\binom n i(n-i)^k (-1)^i = 0$$
 * 2) $$P(0)\!:\ \ \forall n>0 \sum_{i=0}^n\binom n i(n-i)^0 (-1)^i = 0$$
 * 3) $$\forall n>0 \sum_{i=0}^n\binom n i 1 (-1)^i = 0$$
 * 4) $$\forall n>0 \sum_{i=0}^n\binom n i 1^{n-i} (-1)^i = 0$$
 * 5) $$\forall n>0\ (1-1)^n = 0$$
 * 6) $$P(k+1)\!:\ \ \forall n>k+1 \sum_{i=0}^n\binom n i(n-i)^{k+1} (-1)^i = 0;\ k\ge0$$