幻想境界「醉生夢死」 - 兔 @ ウィキ

temp3

2014-01-18T14:04:26+09:00

$$ S = \int \frac{z}{z-5}dz $$ $$ = \int \frac{x+iy}{(x-5)+iy} (dx+idy)$$ $$ \int_{-1}^{1} \frac{iy+1}{iy-4} idy$$ $$ = \int_{-1}^{1} \frac{(y^2-5iy-4)}{16+y^2}(idy)$$ $$ = \int_{-1}^{1} \frac{-20}{16+y^2}(idy) + 2i + \int_{-1}^{1} \frac{-5iy}{16+y^2}(idy)$$ $$ = -10i*arctan\frac{1}{4} + 2i$$ $$ \int_{1}^{-1} \frac{iy-1}{iy-6}(idy)$$ $$ = \int \frac{y^2-7iy-6}{y^2+36}(idy) $$ $$ = -2i -42i\int_{1}^{-1} \frac{dy}{y^2+36} +7 \int_{1}^{-1}\frac{ydy}{y^2+36} $$ $$ = -2i + (14i)arctan\frac{1}{4} $$ $$

temp2

2012-01-27T18:50:12+09:00

$$ \verb|Assumptions: | $$ $$ P \equiv \verb|chance of winning| $$ $$ \verb|Every win gives you b*A golds, while every lose gives you nothing.|$$ $$ \verb|where A is the investment, and b is an arbitrary constant.| $$ $$ \verb|Analysis of Snake's Gambling method :| $$ $$ \verb|1st round(A invested): we have a P chance of getting bA. |$$ $$ \verb|If not, we invest c*A in the next round, where c is an arbitrary constant.| $$ $$ \verb|Following this procedure, by wining at n-th round we have:| $$ $$ \verb|Total investment = | \sum^{n-1}_{k=0} A*c^{k} = \frac{A(c{^n}-1)}{c-1} \verb|(assuming c>1)|$$ $$ \verb|Total income = | bc^{n-1}A $$ $$ \verb|Worth it? Total income - Total investment = | \frac{bc^{n-1}A(c-1)-A(c^n-1)}{c-1} $$ $$ = \frac{A}{c-1}(c^n(b-1)-bc^{n-1}+1) $$ $$ \verb|Chance of getting this result is | P(1-P)^{n-1} $$ $$ \verb|Expectation value | \equiv \equiv \sum_{i} X_iP(X_i) $$ $$ = \sum^{\infty}_{k=0} P(1-P)^{k}\frac{A}{c-1}(c^k(b-1)-bc^{k-1}+1)$$ $$ = \frac{PA}{c-1}((b-1)\sum^{\infty}_{k=0}c^k(1-P)^k - \frac{b}{c}\sum^{\infty}_{k=0}c^k(1-P)^k + \sum^{\infty}_{k=0}(1-P)^k)$$ $$ \verb|(assuming | \sum^{\infty}_{k=0}c^k(1-P)^k \verb| exists)| $$ $$ = \frac{PA}{c-1}(\frac{(b-1)(c-1)-1}{c}\frac{1}{1-c(1-P)}+\frac{1}{1-(1-P)}) $$ $$ = \frac{PA}{c-1}(\frac{(b-1)(c-1)-1}{c(1-c(1-P))} + \frac{1}{P}) $$ $$ = \frac{PA}{c-1}\frac{ P[(b-1)(c-1)-1]-c[1-c(1-P)] }{Pc(1-c(1-P))} $$ $$ = \frac{A}{c(c-1)(1-c+cP)}(P(bc-b-c)-c(1-c+cP)) $$ $$ = \frac{A}{c(c-1)(1-c+cP)}(Pbc-Pb-Pc-c+c^2-Pc^2) $$ $$ \verb|Reference: expectation value = | PbA - A = A(Pb-1)$$

temp

2011-09-26T01:21:45+09:00

$$\verb|Q:Given that |A|a\rangle = a|a\rangle$$ $$\verb|, where A is an arbitrary operator and |$$ $$|a\rangle \verb|is an eigenvector of A,|$$ $$a \verb| is the correspondent eigenvalue to ||a\rangle,$$ $$\verb|is it always true that | f(A)|a\rangle = f(a)|a\rangle$$ $$ \verb| for any arbitrary function f?|$$ $$\verb|My attempt:|$$ $$\verb|If f is analytic,then|$$ $$ f(A)= \sum_{i}c_{i}A^{i} $$ $$ f(A)|a\rangle = \sum_{i}c_iA^i|a\rangle$$ $$ = \sum_{i}c_ia^i|a\rangle = f(a)|a\rangle $$ $$\verb|for any arbitrary function f.|$$ $$\verb|But what about non-analytic functions?|$$

An attempt to derive the MMR system

2010-11-02T01:23:28+09:00

gradient, curl, divergence in arbitrary coordinates

2010-08-20T16:24:17+09:00

physics (INYU)

2010-07-27T16:11:41+09:00

娘娘HW2

2010-07-21T17:54:42+09:00

1.求f在所予閉區間上的絕對極大值與絕對極小值 $$ f(x) = x^{3} - 6x^{2} + 9x + 2; \verb|[0,2]| $$ sol: $$ f'(x)= 3x^{2} - 12x + 9 $$ $$ \verb|Set | f'(x) = 0 \verb| => | 3(x-1)(x-3)=0 $$ Thus we have a maximum at x=1, and a minimum at x = 3. Because the slope of this function is positive in [0,1) and negative in (1,2], f(1) = 6 is the absolute max in [0,2]. f(0) = 2, f(2) = 4, so f(0) = 2 is the absolute min in [0,2] 2.求函數的遞增區間與遞減區間 $$ f(x) = x^{4} - 4x^{3} - 8x^{2} +3 $$ sol: $$ f'(x) = 4x^{3} - 12x^{2} - 16x = 4x(x-4)(x+1) $$ $$\verb|Ths slope of f(x) (namely f'(x)) is negative in |$$ $$\verb|(|\infty\verb|,-1) and (0,4), positive in (-1,0) and| $$ $$\verb|(4,|\infty\verb|), so it's decreasing in |$$ $$\verb|(|\infty\verb|,-1) and (0,4), increasing in (-1,0) and |$$ $$\verb|(4, |\infty\verb|).|$$ 3.討論函數圖形的凹性並找出反曲點 $$ f(x) = \frac{1}{x^{2}+1}$$ sol: $$ f'(x) = \frac{-2x}{(x^{2}+1)^{2}} $$ $$ f''(x) = \frac{-2(x^{2}+1)^{2} + 4x(x^{2}+1)(2x)}{(x^{2}+1)^{4}} $$ $$ = \frac{6x^{2}-2}{(x^{2}+1)^{3}}$$ $$\verb|Turning points at |x = \pm\frac{1}{\sqrt{3}} $$ 開口向上 - 反曲點 - 開口向下 - 反曲點 - 開口向上 4.試作函數的圖形 $$ y = f(x) = \frac{2x^{2}}{x^{2}-1} $$ sol: $$ f'(x) = \frac{4x(x^{2}-1) - 2x^{2}(2x)}{(x^{2}-1)^{2}} $$ $$ = \frac{-4x}{(x^{2}-1)^{2}} $$ $$ f''(x) = \frac{-4(x^{2}-1)^{2} + 8x(x^{2}-1)(2x)}{(x^{2}-1)^{4}} $$ $$ = \frac{12x^{2} + 4}{(x^{2}-1)^{3}} $$ f(x) has a local maximum at x=0, and two singularities at x = +1, -1. Note that f(x) gets closer and closer to (yet still larger than) y=2 while x goes to infinity at the two ends of the x-axis. Plot the function according to the conditions above and watch out for the singularities. To better draw the graph, you may start with drawing the asymptotes(漸進線) first: y=2, x=1, x=-1. #ref(graph.gif) 5.求 $$ \lim_{x \to 1} \frac{sin(x-1)}{x^{2}+x-2}$$ sol: $$ \verb|By L'Hospital's Rule, | $$ $$ \lim_{x \to 1} \frac{sin(x-1)}{x^{2}+x-2} $$ $$ = \lim_{x \to 1} \frac{cos(x-1)}{2x+1} $$ $$ = \frac{1}{3} $$ #comment()

這是首頁

2010-07-21T17:16:53+09:00

兔子曰：人生之中最美好的事情莫過於醉生夢死了。(茶) #comment()

娘娘HW1

2010-07-21T03:58:14+09:00

$$4.\verb| |y = \frac{ln(x)}{2+ln(x)} $$ $$\frac{dy}{dx} = \frac{(2+ln(x))\frac{d}{dx}ln(x) - ln(x)\frac{d}{dx}(2+ln(x))}{(2 + ln(x))^{2}} $$ $$ = \frac{(2+ln(x))\frac{1}{x} - ln(x)\frac{1}{x}}{(2+ln(x))^{2}} $$ $$ = \frac{2}{x(2+ln(x))^{2}} $$ $$5.\verb| | ln(x^{2} + y^{2}) = x + y $$ $$ \verb|Set g(x) | = ln(x^{2} + y^{2}) = x + y $$ $$ \frac{dg(x)}{dx} = \frac{2x + 2y\frac{dy}{dx}}{x^{2} + y^{2}} = 1 + \frac{dy}{dx} $$ $$ \frac{dy}{dx} = \frac{x^{2}+y^{2}-2x}{2y - x^{2} - y^{2}}$$ $$\verb|An alternative approach:| $$ $$\verb|Set g(x)| = x^{2} + y^{2} = e^{x+y} $$ $$\frac{dg}{dx} = 2x + 2y\frac{dy}{dx} = e^{x+y}(1+\frac{dy}{dx})$$ $$\frac{dy}{dx} = \frac{e^{x+y}-2x}{2y - e^{x+y}} $$ $$6.\verb| | y = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} $$ $$ \frac{dy}{dx} = \frac{(e^{x} + e^{-x})\frac{d}{dx}(e^{x} - e^{-x}) - (e^{x} - e^{-x})\frac{d}{dx}(e^{x} + e^{-x})}{(e^{x} + e^{-x})^{2}} $$ $$ = \frac{(e^{x} + e^{-x})^{2} - (e^{x} - e^{-x})^{2}}{(e^{x} + e^{-x})^{2}} $$ $$ = \frac{4e^{x}e^{-x}}{(e^{x} + e^{-x})^{2}} $$ $$ = \frac{4}{(e^{x} + e^{-x})^{2}} $$ $$7.\verb| | y = x^{ln(x)} $$ $$ ln(y) = (ln(x))^{2} $$ $$ y = e^{(ln(x))^{2}} $$ $$ \frac{dy}{dx} = e^{(ln(x))^{2}}\frac{d}{dx}(ln(x)^{2})$$ $$ = e^{(ln(x))^{2}}\frac{(2ln(x))}{x}$$ $$ = \frac{2ln(x)e^{(ln(x))^{2}}}{x}$$ $$\verb|Appendix A: derivative of ln(x)|$$ $$y=ln(x)$$ $$x=e^{y}$$ $$\frac{dx}{dy} = e^{y} = x$$ $$\frac{dy}{dx} = \frac{1}{x} $$ $$\verb|derivative of ln(f(x))|$$ $$y=ln(f)$$ $$\frac{dy}{dx} = \frac{dln(f)}{df}\frac{df}{dx} = \frac{f'}{f}\verb|by chain rule|$$

娘娘吉祥

2010-07-21T03:56:48+09:00

[[娘娘HW1]] [[娘娘HW2]]