# Ngô Quốc Anh

## December 28, 2007

### Đề thi giữa kỳ Giải tích II K52-A1T

Filed under: Giải Tích 2, Đề Thi — Ngô Quốc Anh @ 0:48

Mới thi hôm qua.

ktgk_k52-a1t_gt2.pdf

## December 19, 2007

### Quen quen :-?

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 2:34

Let $f \in C^1$ and ${\lim_{x \to \infty} {f'^2}(x) + {f^3}(x) = 0$. Prove that ${\lim_{x \to \infty} f(x) = 0$.

Proof. Let $h(x) = \{ f'(x) \}^2 + \{ f(x) \}^3$. For any $\epsilon > 0$, there is $x_0$ such that for $x>x_0$, $|h(x)| < \epsilon$. Consider the set $U = \{ x>x_0 \, : \, |f(x)| > \sqrt{2\epsilon} \}$. If $U$ is empty, then the result immediately follows. So assume $U$ is nonempty and choose any $a \in U$. If $f(a) > 0$, then $\{ f(a) \}^3 > 2\epsilon$, so $\{ f'(a) \}^2 + 2\epsilon < h(a) < \epsilon,$ which is a contradiction. Thus $f(a) < 0$ and $\{ f(a) \}^3 < -2\epsilon$. Then $-\epsilon < \{ f'(a) \}^2 + \{ f(a) \}^3 < \{ f'(a) \}^2 - 2\epsilon,$

or $\epsilon < \{ f'(a) \}^2$. Since $f'$ is continuous, this inequality implies $f'(a)$ does not vanish at any point in $U$.

Now assume there is a point $b \in U$ and $x_0 < a < b$ such that $f(b) < f(a)$, Applying intermediate value theorem, we may assume $a$ is in the neighborhood of $b$ in $U$. By mean value theorem, there is $c \in U$ such that $f'(c) < 0$. Since $f'$ cannot change its sign on $U$, $f$ is strictly decreasing on $U$. Especially, $I = [a, \infty) \subset U$.

Now let $f(x) = - u^{-2}$, where $u = u(x) > 0$ on $I$. Then $h(x) = \{ f'(x) \}^2 + \{ f(x) \}^3 = 4 u'^2 u^{-6} - u^{-6}$,

or $2u' = - \sqrt{1 + u^6 h(x)}$.

Since $\epsilon < \{ f'(x) \}^2$ and $f'(x) < 0$, we have $-\sqrt{\epsilon} > f'(x) = 2u'u^{-3}$ and $u$ is strictly decreasing. So $u$ must converge and we have $\lim_{x\to\infty} 2u' = - \lim_{x\to\infty} \sqrt{1 + u^6 h(x)} = -1,$

which contradicts that $u$ converges.

So, for all $b \in U$ and $x_0 < a < b$, $f(b) \geq f(a)$. This means $f$ is monotone increasing on $U$. If $f(x) < -\sqrt{2\epsilon}$ for all $x > x_0$, then $f(x)$ must converge to some $-L$ where $L \geq \sqrt{2\epsilon}$. But this implies $\lim_{x\to\infty} f'(x) = \sqrt{L} > 0$, leading to a contradiction. Hence $|f(x_1)| \leq \sqrt{2\epsilon}$ for some sufficient large $x_1$, and by our preceeding argument, $|f(x)| \leq \sqrt{2\epsilon}$ for all $x \geq x_1$. So $f(x) \to 0$ as $x \to \infty$.

## December 13, 2007

### BĐT liên quan đến đạo hàm và quy tắc l’Hospital

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 0:51

Giả sử $f\in C^{2}[0,\infty)$ sao cho $\displaystyle |f''(x)+2xf'(x)+(x^2+1)f(x)|\leq 1$

với mọi $x\in [0,\infty)$. Chứng minh rằng $\lim_{x\rightarrow\infty}f(x)=0$.

Lời giải. Đặt $\displaystyle g(x)=e^{\frac{x^2}2}(xf(x)+f'(x))$.

Khi đó ta có $\displaystyle g'(x)=e^{\frac{x^2}2}\left(f''(x)+2xf'(x)+(x^2+1)f(x) \right)$.

Vì vậy $\displaystyle\lim_{x \to +\infty}(xf(x)+f'(x))=\lim_{x \to +\infty}\frac{g(x)}{e^{\frac{x^2}2}}=\lim_{x \to +\infty}\frac{g'(x)}{xe^{\frac{x^2}2}}=0$.

Tiếp tục đặt $\displaystyle h(x)=e^{\frac{x^2}2}f(x)$.

Khi đó $\displaystyle h'(x)=e^{\frac{x^2}2}(xf(x)+f'(x))$.

Và vì vậy $\displaystyle \lim_{x \to +\infty}f(x)=\lim_{x \to +\infty}\frac{h(x)}{e^{\frac{x^2}2}}=\lim_{x \to +\infty}\frac{h'(x)}{xe^{\frac{x^2}2}}=\lim_{x \to +\infty}\frac{xf(x)+f'(x)}{x}=0$.

Điều phải chứng minh.

## December 6, 2007

### Surface integral: The symmetric property

Filed under: Các Bài Tập Nhỏ, Giải Tích 5 — Ngô Quốc Anh @ 4:59

Why is $\iint_S(x^2+y^2-2z^2)dA=0$ for $S$ being the unit sphere. $\displaystyle \int_S x^2\,dA=\int_S y^2\,dA=\int_S z^2\,dA=C$,

by rotational symmetry of the sphere. The integral is $\displaystyle C+C-2C=0$.

And the easiest way to find $C$ (should you want to) is again by symmetry $\displaystyle 3C=\iint_S(x^2+y^2+z^2)\,dS=\iint_S 1\,dS=4\pi$,

hence $C=\frac{4\pi}{3}$.

## December 2, 2007

### Đề thi giữa kỳ GT5 của K51 Toán Tin A2

Filed under: Giải Tích 5, Đề Thi — Ngô Quốc Anh @ 23:04