设 $f:\bbR\to\bbR$ 二阶可微, 适合 $f(0)=1$, $f'(0)=0$, 并且 $$\bex f''(x)-5f'(x)+6f(x)\geq 0. \eex$$ 试证: $$\bex f(x)\geq 3e^{2x}-2e^{3x},\quad \forall\ x\in [0,\infty). \eex$$…

设 $A,B,C$ 是同阶方阵, 试证: $$\bex (A-B)C=BA^{-1}\ra C(A-B)=A^{-1}B. \eex$$…

设 $f:\bbR\to\bbR$ 三阶可微, 试证: 存在 $\xi\in (-1,,1)$, 使得 $$\bex \frac{f'''(\xi)}{6}=\frac{f(1)-f(-1)}{2}-f'(0). \eex$$…

设 $$\bex t=\tan \frac{x}{2}, \eex$$ 则 $$\bex \cos x=\frac{1-t^2}{1+t^2},\quad \sin x=\frac{2t}{1+t^2}, \eex$$ 经过化简有 $$\bex (\cos x+2)(\sin x+1)=\frac{(t+1)^2(t^2+3)}{(t^2+1)^2}\equiv f(t). \eex$$ 求导有 $$\bex f'(t)=-\frac{2(t+1)(t^3+t^2+5t-3)}{(t^2+1)^…

Quote Of The Day: “Everyday is an Opportunity to Learn and Grow, Don’t Waste Your Opportunity.” – Alan THE END…

A very hard mathematic problem Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4282 Description Haoren is very good at solving mathematic problems. Today he is working a problem like this: Find three positive in…

Design principles: Conceptual models Feedback Constraints Affordances All are important. This is what makes design such a challenge and rewarding descipline: it grapples with the need to accommodate apparently conflicting requirements. Much of our ev…

We see three varied examples of where natural transformations come in handy. const Right = x => ({ chain : f => f(x), ap : other => other.map(x), traverse : (of, f) => f(x).map(Right), map : f => Right(f(x)), fold : (f, g) => g(x), conca…

题意:问方程X^Z + Y^Z + XYZ = K (X<Y,Z>1)有多少个正整数解 (K<2^31) 解法:看K不大,而且不难看出 Z<=30, X<=sqrt(K), 可以枚举X和Z,然后二分找Y,这样的话不把pow函数用数组存起来的话好像会T,可以先预处理出1~47000的2~30次幂,这样就不会T了. 但是还可以简化,当Z=2时,X^2+Y^2+2XY = (X+Y)^2 = K, 可以特判下Z= 2的情况,即判断K是否为平方数,然后Z就可以从3开始了,这样的话X^…

证明: $$\bex \frac{2}{\pi}\int_0^\infty \frac{1-\cos 1\cos \lm-\lm \sin 1\sin \lm}{1-\lm^2}\cos \lm x\rd \lm =\sedd{\ba{ll} |\sin x|,&-1<x<1,\\ \frac{1}{2}|\sin x|,&|x|=1,\\ 0,&|x|>1. \ea} \eex$$…