Ian Chen 發問時間: 科學數學 · 8 年前

我有數學的問題~急x)

1.(a) Prove that the sequence defined by x1= 3 and Xn+1 =1/4−xn

converges.

(b) Now that we know limxn exists, explain why limxn+1 must also exist

and equal the same value.

(c) Take the limit of each side of the recursive equation in part (a) of this

exercise to explicitly compute limxn.

2.Let x1= 2, and definexn+1 =1/2(xn+2/xn)

.

(a) Show that x上2下n is always greater than 2, and then use this to prove thatxn−xn+1 ≥ 0. Conclude that limXn=√2.

(b) Modify the sequence (xn) so that it converges to √c.

已更新項目:

is 1/(4- xn)

2 個解答

評分
  • 8 年前
    最佳解答

    (1/4)-(xn) or 1/(4- xn) ?

    2013-10-06 22:50:39 補充:

    1. (a)

    (1)Show that xn in [1,3] by M.I.

    n=1, obviously

    Assume that xk in [1,3], then 4-xk in [1,3], x(k+1)=1/(4-xk) in [1/3, 1]

    so, x(k+1) in [1,3]

    (2)Show that xn is decreasing.

    x(n+1)-x(n)= 1/(4-xn) - xn = (xn^2-4xn+1)/(4-xn)

    =[(xn-2)^2 -3]/(4-xn) < 0 ( since xn in [1,3] )

    From (1),(2) then the sequence {xn} is decreasing and bounded,

    so that {xn} is convergents.

    1. (b)

    Set lim(n-> ∞) xn =L, then for all ε > 0, there exists N>0, such that

    when n > N, then |x(n)-L| < ε , so |x(n+1) - L| < ε, ie. lim(n-> ∞) x(n+1)= L

    1. (c)

    lim(n-> ∞) x(n+1) = lim(n-> ∞) 1/(4-xn), then L= 1/(4-L)

    L^2- 4L +1=0, L=2-√3 (2+√3 is rejected since xn in [1,3] )

    2. x1=2, x(n+1)= (xn+ 2/xn)/2, then xn > 0

    (a) x(n+1)= (xn+ 2/xn)/2 > = √2 (since AP > = GP), so (xn)^2 > = 2

    x(n+1)-xn= (2/xn - xn)/2 = (2-xn^2)/(2xn) < = 0

    so {xn} is bounded and decreasing, then {xn} is convergent.

    Let {xn} converge to L, then

    lim x(n+1)= lim (xn+ 2/xn)/2 ( n -> inf is omitted)

    L= (L + 2/L)/2 , so L=√2

    (b) x1=c+1, x(n+1)= (xn + c/xn)/2, c > 0, then lim (xn)= √c

  • 8 年前

    看不懂啦!

    不會寫國語嗎?

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