\( \newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\P}{\mathcal P} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathbb{F}} \newcommand{\E}{\mathcal E} \newcommand{\brac}[1]{\left(#1\right)} \newcommand{\abs}[1]{\left|#1\right|} \newcommand{\matrixx}[1]{\begin{bmatrix}#1\end {bmatrix}} \newcommand{\vmatrixx}[1]{\begin{vmatrix} #1\end{vmatrix}} \newcommand{\lims}{\mathop{\overline{\lim}}} \newcommand{\limi}{\mathop{\underline{\lim}}} \newcommand{\limn}{\lim_{n\to\infty}} \newcommand{\limsn}{\lims_{n\to\infty}} \newcommand{\limin}{\limi_{n\to\infty}} \newcommand{\nul}{\mathop{\mathrm{Nul}}} \newcommand{\col}{\mathop{\mathrm{Col}}} \newcommand{\rank}{\mathop{\mathrm{Rank}}} \newcommand{\dis}{\displaystyle} \newcommand{\spann}{\mathop{\mathrm{span}}} \newcommand{\range}{\mathop{\mathrm{range}}} \newcommand{\inner}[1]{\langle #1 \rangle} \newcommand{\innerr}[1]{\left\langle #1 \right \rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\toto}{\rightrightarrows} \newcommand{\upto}{\nearrow} \newcommand{\downto}{\searrow} \newcommand{\qed}{\quad \blacksquare} \newcommand{\tr}{\mathop{\mathrm{tr}}} \newcommand{\bm}{\boldsymbol} \newcommand{\cupp}{\bigcup} \newcommand{\capp}{\bigcap} \newcommand{\sqcupp}{\bigsqcup} \newcommand{\re}{\mathop{\mathrm{Re}}} \newcommand{\im}{\mathop{\mathrm{Im}}} \newcommand{\comma}{\text{,}} \newcommand{\foot}{\text{。}} \)

Sunday, March 21, 2010

屋企真係一個寶庫

無喇喇搵到一本關於 ODE ge 書 (學校 textbook 太貴買唔起),內有 500 多條 solved problem = =,又有 brief review of 相關 material,正好補返我冇上堂聽 ge 不足,睇 powerpoint 真係睇到好攰 (我一直都係電腦睇)。

今日返學校去拎返份 math 190 功課,TA 語重心長咁俾左一個 comment 我。
645644
在 latex 裏打 $\sum$ 會有 $\sum$,打 $\displaystyle \sum$ 會有 $\displaystyle \sum$,這就是 displaystyle。

Problem. Let  $ f:[0,\infty)\to\mathbb{R}$ with $ f(0)=-1$ be a differentiable function so that $ |f(x)-f'(x)|<1,\forall x\ge 0$.

a) Prove that $ f$ does have a limit that is infinite.
b) Give an example of such a function.

Problem. Evaluate $\displaystyle\lim_{n\to\infty }n\left(\frac{1^{\alpha }+2^{\alpha }+...+n^{\alpha }}{n^{\alpha+1 }}-\frac{1}{\alpha+1 }\right)$, $ \alpha > 1 $.

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