text:
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) $$P(k+1)\!:\ \ \forall n>k+1 \sum_{i=0}^n\binom n i(n-i)^{k+1} (-1)^i = 0;\ k\ge0$$

title: User:Ans/math


text:
User:Egm6322.s09/Report table/Best of report R3

 Under construction; not final (but close to being stable after a couple of weeks). The intention here is to document the best features in any report for the readers (including you); if you see any excellent features in any of the reports (including your team's) that I may have missed noticing, don't hesitate to let me know. I don't have time to read all reports in detail. Due to time constraint, I would only selectively looked at a few important features. I also added some annotations (e.g., related to errors, improvements, etc.) where I looked through for the benefit of everyone. Egm6322.s09 12:52, 23 February 2009 (UTC)

Latest archived version (possibly with comments): Team Mafia

Current version (possibly with comments): Team Mafia

 * classnotes (including simple HW problems), graded over 100%

There are redundancies; continue to edit, and resubmit another updated version together with Report R4.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Artificial Consciousness/Neural Correlates/Synaptic Models/Synaptogenesis

Synaptogenesis: How Synapses are Formed
While Synaptogenesis might be still somewhat controversial, at least after the growth stage of the Neuron, There is some indirect evidence that the percentage of Permease proteins goes up by proportion of weight of neuron puree after learning, suggesting that more synapses might be formed as part of the learning process. Since this is Neurochemistry, many scientists are not sure exactly how to interpret this, in terms of learning processes.

Some scientists note that we do not yet understand the functions of early DNA involved with the cascade reactions formed during LTP, and therefore that there might be a Synaptogenesis component to the LTP cascade reactions, that proliferates synapses as part of the conversion from short term to long term memory. If this synaptogenesis component exists, then it is thought that it might be responsive to synaptic activity, favoring more active synapses over less active synapses.

One possible interpretation is that during membrane replacement, some synapses get cloned according to a protein tag, that survives membrane replacement and indicates active synapses. Thus the more active a synapse is, the more likely that it will be cloned, and a new synapse will form near it. To keep the volume of synapses down, there would probably be a synapse absorption mechanism that absorbs synapses that become too inactive. In simulations we simply remove all synapses with 0 weights.--Graeme E. Smith 17:11, 23 February 2009 (UTC)


text:
User:Wolfe~enwikiversity

{| border="1" cellpadding="5" cellspacing="0" ! ||  [[ talk] ||
 * rowspan="2"|
 * rowspan="2"|

 * colspan="2" align="left"| Bio:

Blog: Wolfe's Blog

Wikipedia:USERPAGE

Contributions

MY SANDBOX click here

Wiki Tags

 * }

title: User:Wolfe~enwikiversity


text:
User talk:Wolfe~enwikiversity

 Hello Wolfe, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --AFriedman 17:20, 23 February 2009 (UTC)

 Hello , we are the Baah and Bahh twins and we'd both like to welcome you to Wikiversity. While you sit back and explore Wikiversity, enjoy some of our fresh goat milk. You can help us learn by creating new learning resources or help us by improving the guided tours and other exploration tools. We look forward to seeing you around. Baaaahbye for now.

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Wolfe. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Wolfe~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:48, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:25, 19 April 2015 (UTC)

title: User talk:Wolfe~enwikiversity


text:
User talk:ChrisSlade

 Hello ChrisSlade, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --AFriedman 17:21, 23 February 2009 (UTC)

 Woof! My name is Jack(Russell). I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

title: User talk:ChrisSlade


text:
User:Wolvesdread

WOLVESDREAD (aka Fletcher)
Just your average primate whose happy as long as he has his bananas and friends to enjoy them with.

title: User:Wolvesdread


text:
User talk:Wolvesdread

not sure what to do with this yet - so just putting some placeholder text here as an initial entry. thx.

title: User talk:Wolvesdread


text: Category:Taoism

title: Category:Taoism


text: Category:Teachers

title: Category:Teachers


text: Category:Copyright law

title: Category:Copyright law


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Talk:Perl/Installing on Win32 Systems

The topic of this page is generic enough that I think there should be some serious content added regarding Strawberry perl (another Win32 distribution that mimics unix-like syntax and use a little better) and the pros and cons of using one version versus the other version. I could start a section discussing this, but I honestly don't know enough about both distributions to develop it too terribly in depth. - BJ Covert Action 19:28, 23 February 2009 (UTC)

Okay, I added some Strawberry Perl installation instructions, but they are far from complete. Its been awhile since I actually put Strawberry Perl on my machine and, frankly, I might have done it wrong. If anyone can improve or expand upon the Strawberry Perl section of this page it would be greatly appreciated. - BJ Covert Action 20:29, 23 February 2009 (UTC)

title: Talk:Perl/Installing on Win32 Systems


text: File:Strawberry Perl 1.png

title: File:Strawberry Perl 1.png


text: File:Strawberry Perl 2.png

title: File:Strawberry Perl 2.png


text: File:Strawberry Perl 3.png

title: File:Strawberry Perl 3.png


title: User:Mattr54321


title: User:Mattr54321/Spanish


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Talk:Perl/Modules on CPAN

Thanks for the page lkluft, its very informative. - BJ Covert Action 20:35, 23 February 2009 (UTC)

title: Talk:Perl/Modules on CPAN


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Charles V

School of History, Department of European History Charles V sought to become leader of a universal empire. His imperial dreams were encouraged by M. A. di Gattinara, whose influence replaced that of Charles's Flemish advisers. The chief problems Charles faced were the Protestant Reformation in Germany; the dynastic conflict with King Francis I of France, particularly for supremacy in Italy; and the advance of the Ottoman Turks.

Shortly after his election Charles began his lifelong struggle with France (see Italian Wars), which required immense expenditures. In 1520 he signed the Treaty of Gravelines with King Henry VIII of England, and in 1521 he invaded N Italy, then controlled by France. The fiscal onus for the war rested on Spain and provoked violent reaction, particularly in Castile, which resented Charles's high-handedness in obtaining funds from the Castilian Cortes. Toledo, Segovia, and other Castilian cities revolted in the brief war (1520–21) of the comuneros. Initially aimed at limiting the royal power, the uprising was later marked by violent class warfare. It was put down at the battle of Villalar; Juan de Padilla and other leaders were executed. Charles later won the loyalty of his Spanish subjects.

In Germany, at the fateful Diet of Worms (see Worms, Diet of) in 1521, Charles secured a satisfactory compromise regarding the Reichsregiment but unyieldingly opposed the doctrines of Martin Luther. In his written opinion, Charles declared himself ready to stake his dominions, friends, blood, life, and soul on the extinction of heresy. Late in May, 1521, he signed the Edict of Worms, outlawing Luther and his followers. However, Charles's preoccupation with the war with France prevented him from checking the spread of Luther's doctrines. Also, Charles was not always supported by the popes, who were concerned with the threat to their temporal power and independence posed by imperial domination of Italy.

After the French defeat at Pavia (1525) and the capture of Francis I, Charles seemed triumphant in Italy; Francis signed (1526) the humiliating Treaty of Madrid, by which he renounced his Italian claims and ceded Burgundy to Charles. On his release, however, Francis repudiated the treaty and organized the anti-imperial League of Cognac. The pope, Venice, Milan, and Florence joined the league. Charles sent an imperial army to Italy composed mostly of German Lutherans. Led first by Georg von Frundsberg and then by Charles de Bourbon, the army defeated the league and then marched on Rome, where the force sacked (May 1527) the city and besieged Pope Clement VII. Although the “German Fury” was disavowed by Charles, he profited from the outrage, extorting large sums of money from the pope.

The Treaty of Cambrai (see Cambrai, Treaty of) with France and the Peace of Barcelona with the pope (both 1529) confirmed Charles's position in Italy and secured his coronation as Holy Roman emperor at Bologna (1530). Charles was the last German emperor to be crowned by the pope. His brother Ferdinand, king of Bohemia and Hungary (later Holy Roman Emperor Ferdinand I), was elected king of the Romans, or German king, in 1531. Charles, who had awarded Ferdinand the Austrian duchies in 1521, delegated increasing authority to him in Germany, which was then torn by religious and social struggles. The rebellion (1522–23) of Franz von Sickingen was followed by the more serious Peasants' War (1524–26), and the Swabian League in 1531 made way for the Lutheran Schmalkaldic League. The Reformation progressed, and the breach between Catholics and Protestants widened.

Before dealing with the religious problem, Charles had to make peace abroad. Ottoman assaults in Austria and Hungary and along the Mediterranean coast posed a serious threat to the Hapsburg lands. In 1535, Charles launched a successful expedition against Tunis. In E Europe, Ferdinand attempted to hold back the Ottomans. In 1536, war broke out with Francis I over the succession to Milan. Intent on recouping in Italy, Francis allied himself with the Ottoman sultan, Sulayman I. Although a truce ended the fighting with Francis in 1538, the Ottomans continued their assaults on the Italian coast. A second expedition by Charles, this time to Algiers, was unsuccessful (1541). In 1542, Francis, again allied with Sulayman, renewed warfare. Charles joined (1543) with Henry VIII and in 1544 forced Francis to make peace at Crépy.

A subsequent truce with the Ottomans, however humiliating, gave Charles and Ferdinand some respite. At last the way opened for the Counter Reformation, ardently desired by Charles and forwarded by St. Ignatius of Loyola, when the Council of Trent (see Trent, Council of) convened in 1545. Turning on the Protestant princes of Germany, Charles split their ranks by winning over Maurice of Saxony and others, attacked the Schmalkaldic League in 1546, defeated (1547) John Frederick I of Saxony at Mühlberg, and imprisoned Philip of Hesse. At the Diet of Augsburg (1547) he secured the incorporation of the Netherlands into the Hapsburg hereditary possessions and forced through the Augsburg Interim (1548), a compromise profession of doctrine that he then tried to impose on the Protestants with the help of Spanish troops. In 1552, Maurice of Saxony changed sides again, called in Henry II of France, Francis's successor, and even attempted to capture Charles at Innsbruck.

Balked in his efforts to recapture Metz, which had been seized by Henry II, and realizing the necessity of compromising with Protestantism, Charles preferred to empower Ferdinand to treat, and he left Germany, never to return. Ferdinand negotiated the religious Peace of Augsburg (see Augsburg, Peace of), but war with France continued. It ended after Charles's death, with the Treaty of Cateau-Cambrésis (1559), a triumph for Spain.

In his remaining years Charles made a series of abdications that left the Hapsburg dominions divided between Austria and Spain. In 1554 he gave Naples and Milan to his son Philip, whom he married to Queen Mary I of England; in 1555 he turned over the Netherlands to Philip, and in 1556 he made him king of Spain and Sicily as Philip II. In 1556 also, he practically surrendered the empire to Ferdinand, and in 1558 he formally abdicated as emperor.

It is worth noting that Charles V abdicated the same year that Elizabeth I took the English throne.

title: Charles V


text:
User talk:Mattr54321

 Hello Mattr54321, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --AFriedman 22:14, 23 February 2009 (UTC)

 Hi! My name is WikiversityJack. I am a pumpkin and a Wikiversity mascot. My aunt told me that fun + learning = Wikiversity. Let me know if I can help you out.

title: User talk:Mattr54321


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Artificial Consciousness/Neural Correlates/Synaptic Models/Pre-Synaptic Bud Model

Pre-Synaptic Bud Model
A Synapse, consists of a Pre-Synaptic Bud, a Post-Synaptic sensitive patch, and a gap containing extra-cellular fluid. The pre-synaptic bud, is usually at the end of an Axion process, but some synapses in dendrites have been found to be pre-synaptic buds suggesting that Axons are not the only location for pre-synaptic buds.

Essentially, how the pre-synaptic bud is thought to work, is that as the voltage at the end of the process increases, with the Action potential of the cell, at some point it promotes the secretion of Neuro-Transmitters, this usually happens as a result of vessicles attaching to the cellular membrane at the end of the process. The vessicles release the Neuro-Transmitters into the extra-cellular fluid, where it diffuses across the synaptic gap and can be detected by the post-synaptic sensitive patch.

To complicate this model however, we have to take into account the fact that the voltage, and the attachment of the vessicles to the cellular membrane seem somehow related to the level of Calcium that can be found at the pre-synaptic bud. A condition called Habituation has been discovered where the strength of the connection of the synapse gradually gets weaker, as the calcium levels in the pre-synaptic bud drop, and fewer vessicles attach to the cellular membrane.

This condition is reversed however, with the opening of Synapses like the S synapse or NMDA synapse that pump more calcium into the cell, so that it can be transported to the end of the process. The change in habituation that this causes is called Facilitation. The S type synapse causes short term facilitation, the NMDA type synapse causes Long-Term Facilitation.


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Talk:Boundary Value Problems/Introduction to BVPs

Comment on this material.


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User talk:Shnur

Japanese
Hi Justin,

I just noticed your name on Topic:Foreign_Language_Learning/Participant_Coordination and figured I'd drop you a quick note. I'm not involved in the Japanese course as such, but have worked on the Japanese textbook over at Wikibooks. Feel free to trot on over if you're interested in joining that project. --Swift 11:10, 24 February 2009 (UTC)

title: User talk:Shnur


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Portal talk:FLOSS4Science/Featured/more

Structure and Purpose
The purpose was originally simply to highlight prominent FLOSS4Science for people looking for established options for their field of interest.

Consider expanding it to include articles on the software, such as reviews, case studies illustrating real world use of the software, learning resources using the software, etc. Then structure the page to look better and be more useful.

Thoughts/ideas? K 12:56, 11 September 2009 (UTC)

 * Include software license and any dependencies (e.g. on proprietary software). - K 17:27, 14 August 2010 (UTC)

title: Portal talk:FLOSS4Science/Featured/more


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User:Egm6322.s09.Three.ge/MyReport4

=Laplace Equation= The Laplace equation is the heat conduction equation with constant thermal conductivity and no heat generation.

The Laplace equation, symbolically:

$$div(grad \ u)=\triangledown \cdot(\triangledown u)=\triangledown^{2}u$$

Taking the Laplace equation in polar coordinates gives the following.

$$div(grad \ u)=0=u_{rr}+\frac {1} {r} u_{r}+ \frac {1} {r^{2}} u_{\theta \theta}$$

Axisymmetric Solution

For an axis-symmetric problem, the theta terms drop out.

$$u_{\theta}=u_{\theta\theta}=\cdots =0$$

The equation then reduces to,

$$u_{rr}+\frac {1} {r} u_{r}=0$$

which is an ordinary differential equation (ODE).

In order to solve this ODE one should note that it may be rearranged.

$$u_{rr}+\frac {1} {r} u_{r}=\frac{1}{r}\frac{d}{dr}(r\frac{du}{dr})=0$$

Separating variables and integrating gives the solution:

$$u(r)=A_{0}ln \ r+B_{0}$$

where Ao and Bo are constants.

It should be noted that if the domain $$\omega$$ includes the origin (where r=0) then Ao must be zero for a finite solution.

Separation of Variables solution

If the problem is not axis-symmetric, the Laplace equation may be solved using separation of variables.

Multiplying the Laplace equation by r2 gives:

$$\begin{matrix} r^{2} \cdot\left \{u_{rr}+\frac {1} {r} u_{r}+ \frac {1} {r^{2}} u_{\theta \theta}=0 \right \}\\

\\ =r^{2}(u_{rr}+\frac {1} {r} u_{r}) +u_{\theta\theta} \end{matrix}$$ ... (a)

Observing that the equation has a portion that depends on r only and a part that depends on theta only, one may assume a solution of the form:

$$u(r,\theta)=F(r)\cdot G(\theta)$$

Thus the solution is a product of 2 functions: one which depends only on r [$$F(r)$$] and the other which only depends on theta [$$G(\theta)$$].

Plugging in this solution into (a) produces:

$$r^{2}G(\theta)\left [\frac{d^{2}F(r)}{dr^{2}}+ \frac{1}{r}\frac{dF(r)}{dr} \right ]+F(r)\frac{d^{2}G}{d\theta^{2}}=0$$

Dividing by $$F(r)G(\theta)$$ and rearranging gives:

$$\frac{1}{F(r)}\left (r^{2}\frac{d^{2}F(r)}{dr^{2}}+ r\frac{dF(r)}{dr} \right )= \frac{-1}{G(\theta)}\frac{d^{2}G}{d\theta^{2}}=n^{2}$$

HW why a positive n2

For the two differential equations to be equal to each other the must each be equal to a constant, n2. This effectively separates the variables and provides 2 uncoupled ODE's.

$$ \begin{matrix} r^{2}\frac{d^{2}F(r)}{dr^{2}}+ r\frac{dF(r)}{dr}-n^{2}F(r) =0\\ \\ \frac{d^{2}G}{d\theta^{2}}-n^{2}G(\theta)=0 \end{matrix}$$

HW explanation of how the answer was got

And results in the following solution for $$n\neq 0$$.

$$\begin{matrix} F(r)=Ar^{n}+\frac{B}{r^{n}}\\ \\ G(\theta)=C cos(n\theta)+D sin(n\theta) \end{matrix}$$

If n=0:

$$\begin{matrix} F(r)=A_{0}ln \ r +B_{0}\\ \\ G(\theta)=C_{0}\theta+D_{0} \end{matrix}$$

HW why is Laplace solution called harmonic

In general, we wand the solution to be periodic, such that:

$$ \begin{matrix} k=interger=1,2,3,\cdots\\ u(r,\theta+k2\pi)=u(r,\theta) \end{matrix}$$

The general form of the Laplace equation in polar coordinates takes the form that follows.

$$u(r,\theta)=A_{0}ln \ r+\sum_{n=1}^{\infty }r^{n}\left [A_{n}cos(n\theta)+B_{n}sin(n\theta) \right ]+\sum_{n=1}^{\infty }\frac{}{r^{n}}\left [C_{n}cos(n\theta)+D_{n}sin(n\theta) \right ]+C_{0}$$

HW why it is periodic

R4 Edits
Homework:Derive LP p.14 (1.2.13) $$\begin{matrix} \overline{a}=a \phi_x^2 + 2b \phi_x \phi_y +c \phi_y^2\\ \\ \overline{b}=a \phi_x \psi_x + b( \phi_x \psi_y +\phi_y \psi_x ) + c \phi_y \psi_y \\ \\ \overline{c}=a \psi_x^2 + 2b \psi_x \psi_y +c \psi_y^2 \end{matrix}$$

Substituting into the equation yields:

$$\overline{ac}-\overline{b}^2 =(a \phi_x^2 + 2b \phi_x \phi_y +c \phi_y^2)(a \psi_x^2 + 2b \psi_x \psi_y +c \psi_y^2)$$ - $$[a \phi_x \psi_x + b( \phi_x \psi_y +\phi_y \psi_x ) + c \phi_y \psi_y ]^2 $$

Multiplying out the equations and rearranging terms gives:

=$$a^2\phi_x^2\psi_y^2 + 2ab\phi_x^2\psi_x\psi_y  + 2bc\phi_y^2\psi_x\psi_y  +  2bc\phi_x\phi_y\psi_y^2  +   c^2\phi_y^2\psi_y^2  +  2ab\phi_x\phi_y\psi_y^2  +  ac\phi_y^2\psi_y^2  +    4b^2\phi_x\phi_y\psi_x\psi_y  +    ac\phi_x^2\psi_y^2$$

-

$$a^2\phi_x^2\psi_x^2 +  2ab\phi_x^2\psi_x\psi_y  +    2bc\phi_y^2\psi_x\psi_y  +  2bc\phi_x\phi_y\psi_y^2  +  c^2\phi_y^2\psi_y^2  +  b^2\phi_x^2\psi_y^2  +  b^2\phi_y^2\psi_x^2  +  2b^2\phi_x\phi_y\psi_x\psi_y   +    2ab\phi_x\phi_y\psi_x^2  +  2ac\phi_x\phi_y\psi_x\psi_y $$

One can see clearly that the first five terms cancel. This leaves the following equation:

$$\begin{matrix}2ab\phi_x\phi_y\psi_y^2 +  ac\phi_y^2\psi_y^2  +    4b^2\phi_x\phi_y\psi_x\psi_y  +    ac\phi_x^2\psi_y^2\\ -\\ b^2\phi_x^2\psi_y^2 +  b^2\phi_y^2\psi_x^2  +  2b^2\phi_x\phi_y\psi_x\psi_y   +    2ab\phi_x\phi_y\psi_x^2  +  2ac\phi_x\phi_y\psi_x\psi_y \end{matrix}$$

Which, after some manipulation, results in:

$$(ac-b^2)(\phi_x \psi_y - \phi_y \psi_x)^2$$

Showed intermediate steps to make it easier to follow. Changed the notation of a,b and c. Still need to correct the answer and cite "LP".

Homework:problem 5.12 in Selvadurai (2000)

An unloaded weightless membrane roof over an annular enclosure. The outer circular boundary is r=b, and the inner circular boundary, r=a, is subject to the following displacement:

$$\Delta_{0} + \Delta_{1}sin \theta$$

where $$\Delta_{0}$$ and $$\Delta_{1}$$ are constants.

i) Formulate the Boundary value problem:

The boundary conditions are:
 * 1) $$w(r,\theta)=w(r,\theta+2\pi)$$
 * 2) $$w(b,\theta)=0$$
 * 3) $$w(a,\theta)=\Delta_{0} + \Delta_{1}sin \theta$$
 * 4) $$\frac{\partial w}{\partial r}_{r=b} =0$$

ii) Develop an expression for the membrane, given that:

$$w=A \ ln \ r +B\theta ln \ r +C\theta +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n\theta +D_{n}cos \ n\theta)$$

where $$A, B, C, D, A_{n}, B_{n}, C_{n}, and \ D_{n}$$ are constants.

Solution: Using boundary condition 1, the equation becomes:

$$\begin{matrix} A \ ln \ r +B\theta ln \ r +C\theta +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n\theta +D_{n}cos \ n\theta)\\

=\\ A \ ln \ r +B(\theta+2\pi) ln \ r +C(\theta+2\pi) +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n(\theta+2\pi) +D_{n}cos \ n(\theta+2\pi)) \end{matrix}$$

Clearly,

$$(C_{n}sin \ n\theta +D_{n}cos \ n\theta)=(C_{n}sin \ n(\theta+2\pi) +D_{n}cos \ n(\theta+2\pi))$$

and,

$$\begin{matrix} A \ ln \ r +B\theta ln \ r +C\theta +D =A \ ln \ r +B(\theta+2\pi) ln \ r +C(\theta+2\pi) +D\\ \\ \therefore B=C=0 \end{matrix}$$

Thus the equation simplifies to:

$$w(r,\theta)=A \ ln \ r +D + \sum_{n=1}^{\infty }\left (A_{n}r^{n}+\frac{B_{n}}{r^{n}} \right )(C_{n}sin \ n\theta +D_{n}cos \ n\theta)$$

iii) Calculate the resultant force and moment necessary to maintain the inner rigid disk shaped region in the displaced position.

title: User:Egm6322.s09.Three.ge/MyReport4


text:
Empathy Models/Empathy Model/Predatory economics

Email with Ellen Brown

Deflation vs. predatory economics (Humbolt squid)

Hi Ellen, I liked the article very much. It is so refreshing to see someone with communication skills serve-up economic reality.

You wrote: "However, the scenario most feared today is actually deflation – a lack of available dollars to fuel the economy. Asset values have collapsed, and savings have collapsed along with them."

I feel that is nothing to fear! But only if the cash supply dries up because people are keeping their money in their accounts, and not giving it to the government or corporations, both of which are pyramid structures, and hence ponzi cultures.

The last idea, ponzi culture, is what I am successfully describing as universally made up of mental defectives, specifically predators, through my study of empathic neurons in animals. Think of the say "The more people I meet, the more I like my dogs."

The people running the show are not just mildly retarded, they are so deficient in functioning empathic neurons that they are below any mammal in terms of conceptual or collaborative ability. All they can do is kill to take which explains our endless war situation. And they can create complicated ponzi schemes; they are intelligent and cunning, just lacking the key evolutionary mental advances.

John Ellen Brown Hi John, that's an interesting take! I think we've all got those genes though... Feb 23 (1 day ago) Ellen BrownLoading...   Feb 23 (1 day ago) Ellen Brown to me show details Feb 23 (1 day ago) Reply Hi John, that's an interesting take! I think we've all got those genes though, and we're sort of forced to resort to them in the system we're stuck in. You HAVE to have your money making money, so you have something to live on in old age when you can't work. But if money is making money, it's not producing anything; and it's necessarily taking money away from those who are producing. We're all in on the ponzi scheme. The system itself has to be changed, not just the mindset. Best, Ellen - Show quoted text -

On Mon, Feb 23, 2009 at 6:48 AM, John wrote:

Hi Ellen, I liked the article very much. It is so refreshing to see someone with communication skills serve-up economic reality.

You wrote: "However, the scenario most feared today is actually deflation – a lack of available dollars to fuel the economy. Asset values have collapsed, and savings have collapsed along with them."

I feel that is nothing to fear! But only if the cash supply dries up because people are keeping their money in their accounts, and not giving it to the government or corporations, both of which are pyramid structures, and hence ponzi cultures.

The last idea, ponzi culture, is what I am successfully describing as universally made up of mental defectives, specifically predators, through my study of empathic neurons in animals. Think of the say "The more people I meet, the more I like my dogs." The people running the show are not just mildly retarded, they are so deficient in functioning empathic neurons that they are below any mammal in terms of conceptual or collaborative ability. All they can do is kill to take which explains our endless war situation. And they can create complicated ponzi schemes; they are intelligent and cunning, just lacking the key evolutionary mental advances. John

John van V. to Ellen show details Feb 23 (1 day ago) Reply Yes that is all correct. I avoid the real definitions online, because I have learned that there are defective-advocates who uses google to stalk anyone mentioning empathic defects.

The diseases are:
 * aspergers
 * bi-polar
 * cocaine/methadrine addiction

The last two can be treated, but the first is genetic.

Unfortunately, lack or loss of morals is the main symptom of an unempathic disorder, so lying is easy for the "sufferers." In fact, "faking" human nature is the accepted Asperger's syndrome therapy.

What we are learning from the Humbolt giant squid, is that empathic neurons are in no way necessary for predatory cooperation; actually neither is intelligence. All that is necessary is rudimentary binary communication. And they consume each other, much the organ-harvesters of China and India.

What is the solution? Going back to nature. So far I think the best text is Kropotkin's Mutual Aid. Here in my New England city of Torrington, CT, I suggest simply re-opening the factories -- just open them. Let the workers and community figure out what to do from there to start manufacturing.

Probably 100% recycling is a good place to start, but I know rural people: they like fast cars and pickups. Cars are terrible as they are, ask any mechanic--its an open market. We can use the Interent to design and distribute component parts. In fact we have no need for the economic super-struture.

Hugs, John

PS, I am going to try to re-work your text a little to start a wikiversity.org page on the topic. - Show quoted text -

John van V. to Ellen show details Feb 23 (1 day ago) Reply Oh, a misconception that I am proving w/ my cat. She has dropped the blood model she used to survive before I found her, and now "catches and releases" mice since she has become a house cat.

You wrote: " I think we've all got those genes > though, and we're sort of forced to resort to them in the system we're stuck > in."

You are talking about "selfish genes." There is no such thing, what you are thinking of is "digital" as opposed to "emotional" communication and thinking (but mostly communication). The genes are emotional communication and mathematical. Since we live in an "asperger's culture," then we are forced to shut down the emotional communication facilities and become mathematical or robotic. We are told this is viciousness, but in fact it is simply devoid of emotion. That is what is called anthropomorphism.

Since there is no literature to support my model directly, I am constructing it from the environment around me, and finding that psychology is nearly pure lie. When using a model system rather than a hypothesis, the only criteria is success. If the model works for me and others try it, then it is a success and can move on to the theory stage of the scientific method.


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Neurosociety Media Centre/Projects

FP6
ERA-Net Neuron a strategically operating group of relevant research funding organisations in Europe and a coordinated implementation of research programmes in the area of research into brain diseases. Quote from the website: "In order to disseminate the results of its activities, the ERA-Net NEURON publishes on a regular basis flyers, newsletters and reports about scientific workshops... Eventually this will lead to a structured European health research policy in this area. Its dissemination products can be found here".

COST
Electric Neuronal Oscillations and Cognition (ENOC) to study complex cognitive functions by analysis of temporal and spatial synchronisation of frequency-specific oscillations; the application of the concept frequency-specific synchronised oscillations may lead to a new scientific paradigm for understanding brain-mind processes. ENOC MoU pdf

Neural regeneration and plasticity (NEREPLAS) to increase the knowledge on the neuronal processes underlying functional recovery following traumatic, ischemic or degenerative damage of the nervous system, and their relationship with those neural processes involved in motor and cognitive learning. NEREPLAS MoU (pdf)

Advanced Methods For The Estimation Of Human Brain Activity And Connectivity (NEUROMATH) to increase the knowledge on the mathematical methods able to estimate the cortical activity and connectivity in the human brain from non invasive neuroelectric and hemodynamic measurements.Neuromath MoU

Consciousness: A Transdisciplinary, Integrated Approach to increase the understanding of (1) the defining features, (2) the behavioural markers, (3) the computational principles, and (4) the neural mechanisms associated with conscious experience in humans and animals, and to identify the clinical, societal, and ethical implications of such findings. Because the proposed Action also has societal relevance, it is also targeted towards the educational and mental health care communities. MoU pdf

COST Action BM0601, “Advanced Methods For The Estimation Of Human Brain Activity and Connectivity”

COST Action A28 on ‘Human Rights, Peace and Security in EU Foreign Policy’

COST Action IS0702, ‘The Role of the EU in UN Human Rights Reform’) will

EUROCORES programmes
Consciousness in a Natural and Cultural Context (2006-2009) The leading idea behind the CNCC programme is that consciousness is both a cultural and a biological phenomenon. This implies overcoming the debate between dualism - according to which consciousness cannot be reduced to physical phenomena - and neurophysiological reductionism - maintaining that the central nervous system is the sole physical cause of conscious experience. There is more to consciousness than just brain activity, but this does not make consciousness something apart from physical reality. The aim of CNCC is to create a powerful platform for international and interdisciplinary research on the nature, origins, and dynamics of consciousness.

Research Networking Programmes
European Neuroscience and Society Network to establish a multidisciplinary forum for timely and necessary engagement with these issues, through the development of research strategies, conferences and workshops that will bring together leading European neuroscientists and social scientists for sustained discussions and cross-disciplinary exchanges about the present and future impact of advances in the neurosciences on our lives. See ENSN website

European Social Cognition Network 2 (2009-2014) Bringing together researchers from neurosciences, developmental psychology, behavioural economics, anthropology, linguistics, etc to create a platform that facilitates European research collaboration in the field of social cognition.

Evolution of Social Cognition: Comparisons and integration across a wide range of human and non-human animal species (2008-2013) aims to provide an evolution-based approach by presenting a broader view of cognition in diverse animal species. It brings together researchers (and laboratories) which have a different scientific background and work on different species. Providing a new forum to exchange views and discuss both long-debated and newly-emerging problems of comparative cognition and a unified system to collate research methods and results across various species.

Exploratory Workshops
Modularity For Versatile Motor Learning: From Neuroscience To Robotics And Back (April 2009) Experts in neurophysiology, computational neuroscience, robotics, and neural networks will explore how modularity, the partitioning of the control process into sensorimotor primitives that can be flexibly combined, may enable versatile motor learning in biological and artificial systems, and whether constructing a modular robotic controller might bring new insights to neuroscience.

Social, Cognitive And Affective Dimensions Of Collaborative Learning Interactions: Towards An Integrated Analysis (May 2009) This workshop will explore theoretical and methodological foundations for an approach to analysing communicative interactions in collaborative learning situations, integrating social, cognitive and affective dimensions of activity. Specialists in educational sciences, linguistics and psychology will elaborate the common approach by confronting analyses along different dimensions, on a common interaction corpus. The workshop will give rise to a common book publication, and response to the HERA JRP call, 2009.

The Dreaming Mind-Brain, Consciousness And Psychosis: Bridging The Gap From The Phenomenology Of Mentation To Neurones (May 2009) The aim is to bring together neuroscientists, psychiatrists, neurologists, psychologists and philosophers to discuss the validity of this hypothesis, its implications for research in severe mental conditions such as schizophrenia and our understanding of consciousness and the mind-brain relationship

Music And The Brain: New Perspectives For Stimulating Cognitive And Sensory Processes (July 2009) Recent findings in the neuroscience of music urge scientists to create a bridge between basic research and applied research aimed at rehabilitating brain functions (e.g., auditory processing, language, and motor processes). Interdisciplinary research involving clinical and cognitive neuroscientists is needed to capitalize on existing knowledge about the brain in order to devise effective rehabilitation programs.

Neuroesthetics: When Art And The Brain Collide (September 2009) Art and neuroscience are both crucial in the process of interbreeding in the seminal field of neuroesthetics. Art provides a unique window on mental processes and conversely neuroscience expands our understanding of the elusive aspects of aesthetic experience. The workshop aims at gathering the most skilled and innovative researchers in this new field of research. The meeting will allow the consolidation of a growing academic community which is now scattered in several separate departments, also due to the highly interdisciplinary nature of neuroesthetics.

[http://www.esf.org/activities/exploratory-workshops/workshops-list/workshops-detail.html?ew=7824 The Adolescent Brain. New Insights From Clinical And Animal Models (October 2009)] Adolescence, the transitional period between childhood and adulthood, is often characterized by extreme-oriented behaviours, possibly mediated by discontinuities in brain development. Discussion of recent evidence from humans and animal models may point to new directions in research so as to better understand neurobiological basis of normative adolescence, and, in turn, the neural and genetic factors underlying the increased vulnerability of adolescents to psychopathology, including drug abuse.

Qualities In Perception Science (November 2009) Perceptively speaking, quality is not reducible to quantity. However, when dealing with qualitative phenomena, many sciences read then only in quantitative terms. As far as cognitive science is concerned, this means treating qualities only as stimuli. The conference aims to suggest that before developing any model of perceptive data, one should develop an unbiased and detailed description of observables. Formal model will have to be developed only subsequently, and should possibly maintain a link to the carefully described observables.

Communication Of Deaf-Blind People As A Model For Exploring Language Modality, Social Communication, And Neural Plasticity (November 2009) The aim is to bring together researchers studying functional imaging of tactile communication, the role of visual and haptic feedback in gesture and sign language, the role of the face, and representations of space in language, in order to jointly develop an agenda for research using deaf-blind communication as a model system for exploring cognition, language and interaction.

Research Conferences
Law and Neuroscience: our growing understanding of the Human Brain and its impact on our Legal System (October 2009) Series on Cognitive Sciences and Technologies (from 2011 onwards). The aim is to establish a dialogue between neuroscientists, legal practitioners, researchers in sociolegal studies and social scientists, to further mutual understanding and make some realistic evaluations of the potential developments at the intersection of neuroscience and law.

Other
DISCOS Disorders and Coherence of the Embodied Self Marie Curie Research Training Network. An interdisciplinary project of both the biomedical sciences and the humanities to investigate the fundaments of the human person. The major breakthrough to be expected from DISCOS will be an integrative framework for research and interventions on the embodied human self that will overcome the fragmentation of approaches and disciplines in the field.

title: Neurosociety Media Centre/Projects


text:
Conservation Entrepreneurship/Kayapo

In the early 1990s, The Body Shop met a Kayapo village leader at a rally for indigenous peoples of Brazil. The Kayapo had a history of concessions to miners and loggers, but those lucrative endeavors were on rapid decline, so the Kayapo were in need of a new profit-making scheme (since, by now, they were accustomed to some modern accomodations, such as televisions, clothing, etc.). The Body Shop’s owners were committed to social, economic and ecological sustainability, and as part of their business, were dedicated to working with people in need. So, they conducted research for several months to determine if there were products the Kayapo could harvest sustainably. A scientist commissioned to conduct the survey determined that Brazil nut was a viable product for the area. Attempting to capture maximum profits for the community, The Body Shop discovered that the Kayapo could get more money for Brazil nut oil rather than the unshelled nuts. The Body Shop, then, put US $80,000 investment into the first year of production, which included the manual press the Kayapo would need to extract the oil from the nuts. They continued to front money for operations, since production didn’t result in any profits. They even introduced other endeavors, such as bracelet making, largely for women. This project failed, and it is unsure that the Brazil nut oil will succeed.

title: Conservation Entrepreneurship/Kayapo


text:
Deuteronomist History

From Prof. Creach's Handout

DtrH
 * division of this material is Theological not literary

 * under this division Deut. belongs with Joshua, Judges, Samuel, & Kings

 * edited according to religious and national purity expressed in Deut.

 * formed in 2 stages -
 * 1. The Deuteronomist (Dtr) incorporates older material into an editorial framework
 * Ark narrative of 1 Sam 4-6
 * The succession narrative referred to as Dtr 1
 * This work probably occurred ca. 622 BCE, during reign of Josiah of Judah
 * Highlights Southern Kings Hezekiah & Josiah who showed great interest in the kind of religio-political structures outlined in Deut.
 * At the same time Dtr1 condemns the Northern King Jereboam,
 * for establishing cultic centers,
 * for sponsoring a system of worship that included 2 golden calves representing YHWH(1 Kings 12.25-33)

 * 2. Second Phase (Dtr2) incorporates older material into an editorial framework
 * This work probably occurred ca. 550 BCE, during Bbaylonian exile
 * Central to the exilic interpolations are
 * the condemnation of King Manasseh of Judah blaming him for the defeat at Babylon's hands (2Kings 21.12-15)
 * an account of exiled King Jeoiachin being released from prison

 * Deut-Kings is meant to be read together as an extended account of how ancient Israel was given polity to guide its social, political and religious life and how the nation failed to uphold the ideals expressed by God to Moses on Mt. Sinai.

 * Other Material

The Deuteronomist (D) that treats the texts of Scripture as products of human intellect, working in time.
 * Martin Noth argued that there was an underlying unity in language and cultural content of the books from Deuteronomy to 2 Kings (Noth 1943).
 * Noth presented the persona of "The Deuteronomist" as a single author who was using pre-Exilic material but was editing and writing in the age of Babylonian exile, the mid-sixth century BCE.
 * Others suggest that "the Deuteronomist" is a close-knit group of Temple scholars rather than a sole individual.
 * Some suggest that the same source may also have written the account of Jeremiah.
 * Since Noth's work, some scholars attribute two separate stages to the text, a first (referred to as Dtr1) and second (referred to as Dtr2) edition of the text,
 * most still consider that both editions were the result of the same author.

First edition
King Hezekiah centralised the religion and destroyed places and objects of worship that were outside of the control of the Jerusalem Temple and its priesthood. The Assyrian empire invaded Judah shortly after Hezekiah died, and gained suzerainty. Subsequent kings of Judah, owing allegiance to the Assyrians, restored the places and objects of worship outside the temple. However, Hezekiah's great grandson Josiah instituted a new reform.

According to the narratives of 2 Kings and 2 Chronicles, in 622/621 BCE, Josiah's high priest Hilkiah found part of the Torah in the Temple, a mainly spartan and empty building. In reaction to the text, King Josiah again centralised the religion, and destroyed places and objects of worship which were neither the Jerusalem Temple nor specified to be housed in it. Since before the 5th century scholars (such as Jerome) have insisted that the text found by Hilkiah was the law code of Deuteronomy. Scholars allege that the text was written at Josiah's instigation and "found" to justify his actions.

According to the documentary hypothesis, the priests of Shiloh wrote the law code to support their views. The code was written to support the king, a centralised religion, Levites generally rather than just Aaronids, and a balance on the king's power (for example by supporting a militia rather than an organised army) due to the way in which kings had previously treated them.

D then created, according to the hypothesis, a history of rulers, judging them by their actions according to the code, culminating in Josiah. D inserted the law code at the start, framed as Moses' last words since D was not trying to change the pre-existing JE account. The purpose of this was to show that Josiah's rule was an act of God, Josiah being the hero to save Israel - a Moshiach. Josiah was the only person described as being comparable to Moses. The story of Josiah reflected the wording given by Moses in D, terms such as "do not turn, right or left", "and none arose like him", and "Love god with all your heart, all your soul, and all your might", are only ascribed to or of Moses and to or of Josiah in the whole of the Deuteronomical history. Parallels are not described between Moses and other kings.

D's history was composed by redacting earlier originally independent sources, and a few lines of text to provide a more continuous narrative, which included

 * The Deuteronomistic law code
 * The story of Joshua
 * The story of Jericho
 * The story of the conquest of the land
 * The story of Deborah
 * The story of Gideon
 * The story of Samuel
 * The story of Saul
 * The story of David
 * The Court History of David, a text composing most of 2 Samuel
 * The Davidic Covenant, a tradition concerning the perpetuity of the reign of King David's descendants.
 * The history of the Kings of Israel
 * The history of the Kings of Judah

The book referred to as Joshua was redacted together by associating each event with Joshua; the book of Judges by comparing each protagonist with the law code between the stories; the book of Kings by alternating between a king of Israel and that of Judah (both were originally covered in separate texts). At each alternation in the book of Kings, a description of the king's parentage, and an accession date compared to the reign of the king ruling in the other nation, is given.

Second edition
Josiah died when he traveled to Megiddo to fight the Egyptian army (in a battle so famed among the Jews that it established Meggido as the traditional location for the eschatological final battle between good and evil) which was passing through Judah to support the Assyrians in their conflict with Babylon. The next kings, the first a son of Josiah, reversed the reform of Josiah, once more restoring the non-centralised holy places.

Egypt conquered Judah and exiled the king, replacing him with another of Josiah's sons, who was subsequently killed in a Babylonian attack and replaced by his son. Babylon eventually gained control of Judah, appointing a new king, who was a third son of Josiah. The king, after 10 years, rebelled against the Babylonian emperor Nebudchadnezzar, resulting in Nebudchadnezzar destroying Jerusalem, killing the whole of the king of Judah's family, and blinding and exiling the king himself.

Babylon decided to appoint a governor who would be favourable to it, choosing an anti-Assyrian from Judah. However, those connected to the family of King David, the royal dynasty that Nebudchadnezzar had extinguished, were so offended by this appointment that they assassinated the governor. The denizens of Judah, afraid of Nebudchadnezzar's response, almost entirely chose to become refugees, initially in Egypt. Nebudchadnezzar chose to burn Jerusalem to the ground, destroy the temple, and exile the remaining nobility and officials to Babylon as slaves. Judah no longer existed.

This posed some problems for the first edition of D. In its histories, it had implied that the dynasty of king David had been promised by God that it would rule Judah forever. In its text, it describes certain things as still existing which Nebudchadnezzar had destroyed. It describes Josiah as the saviour of Israel, but he had been killed, and his reforms undone.

D was edited. D couldn't be completely rewritten because the text was relatively well known. The text was added to by scattering references to the threat of dispersion of Judah amongst the nations should they disobey the law code. The promise of the survival of the royal family's reign was also amended to imply that, although it was true, it would be irrelevant if Judah did not exist to be ruled over. A curse was added threatening to send Judah back to Egypt, as had happened.

Another set of additions were references to the command against worshipping other Gods, including a description of God's last words to Moses, which describes the future destruction of Judah for this very crime. Additions were made to the histories emphasising the consequences of the undoing of Hezekiah's reforms that had occurred amongst his descendants.

Finally, a few additional sources appear to have been added into the text, and expanded on by other additions, specifically

 * The Song of Moses, a poem existing in Deuteronomy 32:1 - 43
 * The Blessing of Moses, in Deuteronomy 33:2 - 27.

title: Deuteronomist History


text:
Anathoth

Jeremiah was a descendant from a family of priests in Anathoth. Later on Jeremiah visited there to negotiate the purchase of some field. He was arrested for desertion and thrown in prison. He was visited by the king and he gave him the same message as before. After this he was transferred to a better prisoner quarters. His uncle Shalum visited which was who he bought the field from (the field was an act of affirmation of faith for the future of the people and their devastated land). Anathoth was not where Jeremiah got his support – "which is the place he condemned – but from certain families at the court and from the ‘elders of the land,’ who, to judge by their intervention in his trial were more in line with the Judean Micah than with the Ephraimite Hosea" (Blenkinsopp p145).

title: Anathoth


text:
Talk:Conservation Entrepreneurship/Community Projects

Hey everyone, I added the case studies I knew about but I didn't write down who was doing what so someone else will have to add that (Brian did you record it?). There is a space for the summarizer and discussants for each case study. I think I missed one, so add it if you are leading the discussion. You can post your paragraph summary under the list of summarizer/discussants. Thanks! - Kifaro50 20:32, 24 February 2009 (UTC)


text:
Artificial Consciousness/Neural Correlates/Synaptic Models/Post-Syanptic (Sensitive Patch) Model

Post Synaptic (Sensitive Patch) Model
As you probably know by now, all synapses consist of a pre-synaptic bud, on the sending neuron, a Sensitive patch on the receiving neuron, and a gap filled with extracellular fluid called the Synaptic gap. The Neuro-Transmitters are secreted by the pre-synaptic bud, diffuse across the synaptic gap and are detected by the sensitive patch.

Just, as it is the Activation Potential of the sending neuron, that determines at least partially how much neuro-transmitter is secreted, The purpose of the sensitive patch, is to convert the chemical signals back into an electrical charge and thus Action Potential in the receiving Neuron.

To do this, the sensitive patch has to do two things, it has to detect the presence of small amounts of neuro-transmitter, and somehow convert those detections into an electrical charge. There are two separate mechanisms that the neuron uses to achieve these effects, Permease Molecules that detect and transmit the detection of Neuro-Transmitters into the cell where they can influence the cell, and Ion Channels that pump specific ions into or out of the cell and into or out of the extracellular fluid across the cell membrane.

An interesting note, is that these Sensitive Patches persist despite the fact that the Cellular Membrane that they interpenetrate is replaced periodically as part of the survival mechanism of the cell. Another interesting fact, is that depending on which ion is transferred and in which direction, the electrical characteristics of the neuron change. Some ions when pumped into the neuron increase the Activation Potential, Some Ions when pumped out of the neuron increase the Activation potential,

Some ions when pumped into the cell, decrease habituation, and some ions when pumped out of the cell increase the Potentiation of the Cell, While other Ions pumped into the cell act to shunt the cell keeping it from firing.

Each Synaptic Sensitive Patch, is associated with specific Ion channels and specific permease molecules that act like keyholes that only certain neuro-transmitters will activate, as a result, certain Neuro-Transmitters are associated most often with certain ion channels in a specific type of neuron. The combination of different types of Sensitive Patches with different phenotypes of neurons produce upwards of 150 different types of neurons in the Mammalian brain. Different types of Neurons tend to specialize in different types of processing probably because their phenotypes and sensitive patches are tuned to that type of processing. DNA probably determines which types of Neurons can be found in which parts of the Nervous System.