Dãy cho bởi phương trình


Trong de thi HSGQG cua Vietnam chung ta thuong gap bai toan sau: Cho cac phuong trinh f_n(x) = 0 (n\in\mathbb{N}) , chung minh moi phuong trinh do co nghiem thuc duy nhat tren mot mien D nao do, goi nghiem do la x_n, tim \lim_{n\to\infty} x_n.

Topic nay danh de trao doi ve nhung bai toan do, xin moi cac ban post cac loi giai va cung cap cac vi du.

14 thoughts on “Dãy cho bởi phương trình”

  1. Theo toi thi chung ta can luu y den cac diem sau khi lam viec voi cac bai Toan nay:
    1)Thanh thao ki thuat chung minh phuong trinh co nghiem duy nhat.
    2)Boi vi thong thuong se khong co cong thuc tuong minh cho x_n nen chung ta thuong dung dinh li sau de tim gioi han cua day so do:

    Dinh li. Cho ba day (u_n),(v_n),(w_n) thoa man:
    a)Ke tu luc nao do u_n\leq v_n\leq w_n.
    b)\lim_{n\to\infty}u_n=\lim_{n\to\infty}w_n=l.
    Khi do day (v_n) hoi tu va gioi han cua no cung bang l.

  2. Bai 4. (Viet nam MO 2007)
    Cho a>2 la mot so thuc va f_n(x)=a^{10}x^{n+10}+x^n+\cdots+x+1. Chung minh rang voi moi so nguyen duong n, phuong trinh f_n(x)=a co duy nhat nghiem duong. Ki hieu x_n la nghiem duong noi tren, chung minh rang day (x_n) la day hoi tu.

  3. Bai 9. Chung minh rang voi moi so nguyen duong n phuong trinh x^{2n+1}=x+1 co duy nhat nghiem thuc. Ki hieu nghiem cua phuong trinh do voi moi n la x_n. Tim \lim_{n\to\infty} x_n.

  4. Bai 10. Xet phuong trinh x^n=x+n.
    a) Chung minh rang voi moi so nguyen duong n>1, phuong trinh tren co duy nhat nghiem duong, ki hieu x_n.
    b) Chung minh rang \lim x_n=1.
    c) Tinh \lim\dfrac{n(x_n-1)}{\ln n}.

  5. Bai 11. Xet ham so f_n(x)=e^{-x}\sum_{m=0}^n\dfrac{x^m}{m!} trong do x la so thuc duong va n la 1 so nguyen duong .

    1, CMR voi moi so thuc 0<k<1 ta co phuong trinh f_n(x)=k co nghiem duy nhat \alpha_n .

    2, Tim \lim_{n\to\infty}\dfrac{1}{\alpha_n}

  6. Problem 12. Prove that for any n\in N,\ n\ge 3, the equation x^n = nx - 1 has
    two positive roots 0 < x_n < 1 < y_n  , the equation \sqrt {x + 1} - \sqrt x = \dfrac{1}{n}, x\in [0,1] has one zero x_n\in (0,1) and x_n\rightarrow \dfrac{1}{e - 1}.

    Problem 13. Prove that for any n\in N^*, the equation x^n + nx = 1
    has one positive root x_n, x_n\searrow 0 and \lim_{n\to\infty} (1 + x_n)^n = e.

    Problem 14. Prove that for the fixed p\in N^* and for any n\in N^*, the equation
    x^n + px = 1 has one positive root x_n and x_n\nearrow \dfrac{1}{p}.

  7. Problem 15. Let x_n be the zero of the equation \ln (n + x) + E = \sum\limits_{k = 1}^n\dfrac{1}{k}, where E
    is the Euler’s constant, i.e.
    \lim_{n\to\infty} \left(\sum_{k = 1}^n \dfrac{1}{k} - \ln n\right) = E. Prove that x_n\rightarrow \dfrac{1}{2}.

    Problem 16. Prove that for any n\in\mathbb{N}, the equation x^n + \arctan x = 1
    has one positive zero x_n and x_n\rightarrow 1.

    Problem 17. Prove that for any n\in\mathbb{N}, the equation x^n + \ln x = 0 has
    one zero x_n\in (0,1) and x_n\nearrow 1.

    From Mathlinks.ro😀

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s