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[3]{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[3]{2\epsilon} for all x > x_0, then f(x) must converge to some -L where L \geq \sqrt[3]{2\epsilon}. But this implies \lim_{x\to\infty} f'(x) = \sqrt{L} > 0, leading to a contradiction. Hence |f(x_1)| \leq \sqrt[3]{2\epsilon} for some sufficient large x_1, and by our preceeding argument, |f(x)| \leq \sqrt[3]{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

ktgk_k51-a2_gt5_2.pdf

ktgk_k51-a2_gt5_2_dapan.jpg

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