Ngô Quốc Anh

September 23, 2007

“Hàm số” vs. “Số phức”

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 18:27

1) Let z_{1},z_{2}\in\mathbb{C} such as z_{1}= f(a)+if(b), z_{2}= f(b)-f(a)i and also \parallel z_{1}|-|z_{2}\parallel = |z_{1}|+|z_{2}|, where f is a differentiable function at [a,b]. Prove that exists \xi_{1},\xi_{2}\in(a,b) such as f'(\xi_{1})+f'(\xi_{2}) = 0.

2) Let f continuous function to [a,b] such as f(x)\neq0 \forall x\in[a,b]. Also let z\in\mathbb{C} such as z+\frac{1}{z}= f(a) and z^{2}+\frac{1}{z^{2}}= f^{2}(b). Prove that i)|f(a)| > |f(b)| and ii) the equation f(b)+x^{3}f(a) = 0 has at least one real root to (-1,1).

3) Let f differentiable function to [a,b]. Also let z,w\in\mathbb{C} such as |z-iw|^{2}= |z|^{2}+|iw|^{2} and f(b)\neq0 . Prove that exists \xi\in (a,b) such as \xi f'(\xi) = f(\xi).

Một bài thi OLP’ SV

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 14:55

Chứng minh rằng \forall a\in\mathbb{R},

\int_{0}^{\infty}\frac{dx}{(1+x^{2})(1+x^{a})}=\frac{\pi}{4}.

Chứng minh.

Using x=\tan (\theta )

\int_{0}^{\infty }\frac{1}{\left(x^{2}+1\right)\left(x^{\alpha }+1\right)}\, dx=\int_{0}^{\frac{\pi }{2}}\frac{1}{\tan^{\alpha }(\theta )+1}\, d\theta

P_{1}=\int_{0}^{\frac{\pi }{2}}\frac{1}{\tan^{\alpha }(\theta )+1}\, d\theta =\int_{0}^{\frac{\pi }{2}}\frac{\cos^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta

Putting \theta\to\frac{\pi }{2}-\theta

P_{2}=\int_{0}^{\frac{\pi }{2}}\frac{1}{\cot^{\alpha }(\theta )+1}\, d\theta =\int_{0}^{\frac{\pi }{2}}\frac{\sin^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta

Of course P_{1}=P_{2}

P_{1}+P_{2}=\int_{0}^{\frac{\pi }{2}}\frac{\cos^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta+\int_{0}^{\frac{\pi }{2}}\frac{\sin^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta =\int_{0}^{\frac{\pi }{2}}\frac{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}{\cos^{\alpha }(\theta )+\sin^{\alpha }(\theta )}\, d\theta 
P_{1}+P_{2}=\int_{0}^{\frac{\pi }{2}}1\, d\theta =\frac{\pi }{2}=2 P_{1}=2 P_{2} 
And we are done.

Tính 2 tích phân

Filed under: Giải Tích 2 — Ngô Quốc Anh @ 14:51

Tính các tích phân sau

1){\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x+\cos x+1}dx}
2){\int_{0}^{\frac{\pi}{2}}\frac{x^{2}}{\sin x+\cos x+1}dx}

Create a free website or blog at WordPress.com.