煩

Problem. Verify that $  z=z(x,y)$ which is implicitly defined by $  \displaystyle x^2+y^2+z^2=yf\left(\frac{z}{y}\right)$ satisfies the partial diiferential equation
$  \displaystyle (x^2-y^2-z^2)\frac{\partial z}{\partial x}+2xy\frac{\partial z}{\partial y}=4xz$.

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要證 2 個 statement 等價，只須證明 $  (1)\implies (2)\implies (1)$, 但去到 4, 5 或更多個 statement，順序證明有可能令我們讚進死胡同。打個比方，若要證明 4 個 statement 等價，先證 $  (1)\implies (2)\implies (3)$ 卻發現自己 $  (3)\implies (4)$ 怎樣也想不到，可是 $  (2)\implies (4)$ 卻十分簡單，不妨先完成 $  (1)\implies (2)\implies (3)\implies (1)$，再從中把 $  (2)$ 抽出來證明 $  (2)\iff (4)$。這裏我們需要多做一步，但相對地可把問題變得簡單。

Math202 那份關於 integrability 的 notes，最後的第二條，有關證明四個命題等價的問題，若發現由第 3 到第 4 出現困難的話，不妨蹺一條長一點，但較平坦的路。

最後，當大家大至上認為自己對 integrability 有一定的認識，可嘗試 09 spring math203 final 有關 integrability 的題目：

Problem. Suppose $  f(x)$ and $  g(x)$ are integrable on $  [a,b]$. Prove that for any $  \epsilon > 0$, there is $  \delta>0$, such that for any partition $  P$ satisfying $  \|P\|<\delta$ and choices $  x_i^*,x_i^{**}\in [x_{i-1},x_i]$, we have \[
\left|\sum f(x_i^*)g(x_i^{**})\Delta x_i-\int_a^bf(x)g(x)\,\mathrm{d} x\right|<\epsilon.
\] 這大至上證明了，就算 $  x_i^*$ 和 $  x_i^{**}$ 所取的值不同，同樣有和相同選擇時的結果。

痾清嘔清一身輕

過度進食科大酸辣粉，嘔返出黎。

科大獸醫

一位我們學校某學生的經歷 = =。

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###!!!@@@ceev 說: (18小時前)
我比科大獸醫玩死

###!!!@@@@ceev 說: (18小時前)
上個星期去看獸醫 諗住傷風 easy job啦
開左三日藥比我 點知食完之後無好之餘 仲要大獲左
今日頂唔順去看醫生 點知話個鼻發炎
真係吹漲 獸醫可以完全唔知我個鼻發炎 我一早d鼻涕就green

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把 pdf 合併

同學以前問我懂不懂把數個 pdf 合併，嗯... 我不懂。今天心血來潮在網上找找，比較易找的都是 freeware，要錢的。用 ``Merge pdf" 找找看，發現一個不錯的網頁 (link)。

不用下載任何工具，上傳數個想要合併的 pdf，按 merge，完成！最後下載回來，方便快捷簡單易用。

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Problem. 設函數 $f$ 在 $latex x=0$ 處連續，如果 $ \displaystyle \lim_{x\to 0}\frac{f(2x)-f(x)}{x}=m$，求證 $ f'(0)=m$.

做這題時可能還須要用到以下結論。

Prerequisite. 設 $ \lim_{x\to 0}f(x)=0$，且 $ \displaystyle f(x)-f\left(\frac{x}{2}\right)=o(x)$ ($x\to 0$)，求證：$\displaystyle f(x)=o(x)$ ($latex x\to 0$)。

Well...

睇黎我要背鬼左呢兩條 inequality 佢=.=，起碼要由右手邊 sense 到左手邊舊野。

• $  a^3+b^3+c^3+3abc \geq a^2 b+a^2 c+b^2 a+b^2 c+c^2 a+c^2 b$.
• $  a^4+b^4+c^4+abc(a+b+c) \geq a^3 b+a^3 c+b^3 a+b^3 c+c^3 a+c^3 b$.

熱死人

呢幾日都好翳焗，熱死人了……。

看着嚴民教授那 inverse function theorem 的三頁證明，目前還抱着監賞的目光。

Problems of MATH202 # 17 of the last week

Since I am busy doing my work in another course, this weekly post is delayed til today :(.
Of the problems, only 5, 6, 7 are of our interest.

Problem 5. Let $  f:\mathbb{R}\to\mathbb{R}$, be a three times differentiable function. If $  f(x)$ and $  f'''(x)$ are bounded functions on $  \mathbb{R}$, show that $  f'$ and $  f''$ are also bounded functions on $  \mathbb{R}$.

My way.
Just make good use of the expansion $  f(x+h)=f(x)+f'(x)h+\frac{1}{2}f''(x)h^2+\frac{1}{3!}f'''(x+\theta h)h^3$, for some $  \theta \in (0,1)$.

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Problem 6. If $  f(x)$ and $  g(x)$ are $  n$ times differentiable and $  f^{(n-1)}(x),g^{(n-1)}(x)$ are both continuous in $  [a,b$. Then there exists a number $  c \in(a,b)$ such that \[ \frac{\displaystyle f(b)-f(a)-\sum_{k=1}^{n-1}\frac{(b-a)^k}{k!}f^{(k)}(a)}{\displaystyle g(b)-g(a)-\sum_{k=1}^{m-1}\frac{(b-a)^k}{k!}g^{(k)}(a)}=\frac{(m-1)!}{(n-1)!}(b-c)^{n-m}\left(\frac{f^{(n)}(c)}{g^{(m)}(c)}\right).\]

NO IDEA, I am just able to prove the case when $  m=n$.

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Problem 7. Let $  f$ be $  p$ times differentiable on $  \mathbb{R}$ and let $  M_k=\sup\{|f^{(k)}(x)|:x\in\mathbb{R}\}<\infty$, $  k=0,1,2,\dots,p$ and $  p\ge 2$. Prove that $  M_1\leq \sqrt{2M_0M_2}$ and \[M_k\leq 2^{\frac{k(p-k)}{2}}M_0^{1-\frac{k}{p}}M_p^{\frac{k}{p}}$, for $  k=1,2,\dots,p-1.\]
My way.
(1) Same as the case of problem 5, replace $  h$ by $  -h$ to construct another equation (remember to choose different $  \theta$), subtract two equation, observe that $  0\leq 2M_0 +2M_1h+M_2h^2$ for any $  h$, while discriminant $  \leq 0$, we are done.
(2) In exactly the same manner as (1), we conclude that $  M_{j+1}\leq \sqrt{2M_jM_{j+2}}$ for all $  j\leq p-2$. Now we take product $  \prod_{j=m}^{n}$ on both sides, having \[ \sqrt{M_{n+1}M_{m+1}}\leq (\sqrt{2})^{n-m+1}\sqrt{M_{n+2}M_m}.
\] Before we proceed, we first consider two cases. If $  M_k=0$, then the inequality we are asked to prove obviously holds since right hand side is always non-negative. In case if $  M_k>0$, then we take the product $  \prod_{m=0}^{k-1}$ on both sides of $  \sqrt{M_{n+1}M_{m+1}}\leq (\sqrt{2})^{n-m+1}\sqrt{M_{n+2}M_m}$, it results in \[ M_n^k\leq \left(\frac{M_0}{M_k}2^{k(2n-k+1)/2}\right)M_{n+1}^k.\]
(I remember I have replaced $  n$ by $  n-1$ to make the inequality seem better), we are interested in this because it is a beautiful (in the sense of solving the problem) reccurence relation, we have a direct consequence \begin{align*}
 M_n^k&\leq \left(\frac{M_0}{M_k}\right)^{p-k}2^{\sum_{j=n}^{n+p-k-1}k(2j-k+1)/2}M_{n+p-k}^k\\
&=\left(\frac{M_0}{M_k}\right)^{p-k}2^{k(p-k)(2n-2k+p)/2}M_{n+p-k}^k.
\end{align*} Finally, we take $  n=k$, $  M_k^p\leq M_0^{p-k}2^{\frac{k(p-k)p}{2}}M_p^{k}$.