http://i.imgur.com/67Dcq2O.png I found the first two to be pretty easy, just kin

Question

http://i.imgur.com/67Dcq2O.png

I found the first two to be pretty easy, just kinda basic computation (please answer them though, just as to confirm). The last one I'm thinking that you need to replace the terms with en+1, en and q and use that en+1/en=q(1+delta n) as delta n ->0 but I don't see the answer. Any help would be appreciated.

I found the first two to be pretty easy, just kinda basic computation (please answer them though, just as to confirm). The last one I'm thinking that you need to replace the terms with en+1, en and q and use that en+1/en=q(1+delta n) as delta n ->0 but I don't see the answer. Any help would be appreciated.

Explanation / Answer

(xn+1 - l)/(xn - l) = (xn+2 - l)/(xn+1 - l) = q

Then, cross-multiplying, (xn+1 - l)^2 = (xn - l)(xn+2 - l), or

xn+1^2 - 2lxn+1 + l^2 = xnxn+2 - l(xn + xn+2) + l^2, or

xn+1^2 - 2lxn+1 = xnxn+2 - 2l(xn + xn+2)

xn+1^2 - xnxn+2 = l(2xn+1 - xn - xn+2), or

l = (xn+1^2 - xnxn+2)/(2xn+1 - xn - xn+2)

If we multiply through by -1, we get

l = (xnxn+2 - xn+1^2)/(xn+xn+2-2xn+1)

delta xn = xn+1 - xn

delta^2 xn = xn+2 + xn - 2xn+1

Then,

vn = xn - (delta xn)^2/(delta^2 xn) =

xn - (xn+1 - xn)^2/(xn+2 + xn - 2xn+1) = getting a common denominator

xn(xn+2 + xn - 2xn+1) - (xn+1 - xn)^2)/(xn+2 + xn - 2xn+1) = multiplying out

(xnxn+2 + xn^2 - 2xnxn+1 -(xn+1^2 + xn^2 - 2xnxn+1))/(xn+2 + xn - 2xn+1) =

(xnxn+2 + xn^2 - 2xnxn+1 -xn+1^2 - xn^2 + 2xnxn+1))/(xn+2 + xn - 2xn+1) =

(xnxn+2 -xn+1^2 ))/(xn+2 + xn - 2xn+1) =

vn

Thus, the equation works.

xn+1 is approximately l + q(xn - l) and xn+2 is approximately l + q^2(xn-l)

Let xn = l + c

Then, xn+1 = l + qc+ epsilon1 c, and xn+2 = l + (q+epsilon2) (xn+1-l )= l + (q+epsilon2)(q+epsilon1)c

Then, xn - (delta xn)^2/(delta^2 xn) = l + c - (1 - q - epsilon1)^2c^2/(l+c + l + (q+epsilon2)(q+epsilon1)c - 2(l+(q+epsilon1)c)) =

l + c - (1 - q - epsilon1)^2 c/(1 - 2q + q^2 - 2 epsilon1 + q epsilon1 + q epsilon2 + epsilon1epsilon2) =

l + ((1 - 2q + q^2 - 2 epsilon1 + q epsilon1 + q epsilon2 + epsilon1epsilon2) - (1 - q - epsilon1)^2)c/

(1 - 2q + q^2 - 2 epsilon1 + q epsilon1 + q epsilon2 + epsilon1epsilon2) =

l + ( -q epsilon1 +q epsilon2 - epsilon1^2 + epsilon1 epsilon2)c

/(1 - 2q + q^2 - 2 epsilon1 + q epsilon1 + q epsilon2 + epsilon1epsilon2)

Then, (vn-l)/(xn-l) = (-q epsilon1 +q epsilon2 - epsilon1^2 + epsilon1 epsilon2)c

/(1 - 2q + q^2 - 2 epsilon1 + q epsilon1 + q epsilon2 + epsilon1epsilon2) /c =

(-q epsilon1 + q epsilon2 - epsilon1^2 + epsilon1 epsilon2)

/(1 - 2q + q^2 - 2 epsilon1 + q epsilon1 + q epsilon2 + epsilon1epsilon2)

Thus, as epsilon1 and epsilon2 go to 0, we have (1-q)^2 in the denominator along with epsilon terms, but only epsilon terms in the numerator, so

lim n -> infinity (vn - l)/(xn-l) = 0

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