Find lt n->(infinity) f(n)

[chaoseverlasting]

Homework Statement

Given:
$$f(n+1)=\frac{1}{2}(f(n)+\frac{9}{f(n)}$$

Find lt n->(infinity) f(n)

Homework Equations

The Attempt at a Solution

No idea.

$$f(n+1)-f(n)=0.5(\frac{9}{f(n)} -f(n))$$
Replacing n by 1/n, dividing by 1/n and taking limit 1/n->0 which becomes something similar to the f'(n) except 1 should be h. Even still, you can't apply L'hospitals here. I am totally lost.

 

[AKG]

In your equation f(n+1) = 0.5(f(n) - 9/f(n)), the left side equals the right side, therefore the limit as n goes to infinity of the right side is the limit as n goes to infinity of the right side. Using the above fact, let L stand for the limit of f(n) as n goes to infinity, and solve for L.

 

[HallsofIvy]

Your orginal equation was
$$f(n+1)= \frac{1}{2}a(f(n)+ 9/f(n)$$
I think your subtracting f(n) from both sides confused AKG (not to mention me)!

The sequence {f(n+1)} is exactly the same as {f(n)} but with the index changed. If L= lim f(n) then L= lim f(n+1) also. AKG's suggestion is that you take the limit, as n goes to infinity, on both sides of the equation and so get a single equation involving L. Solve that equation.

 

[chaoseverlasting]

So, the equation should be something like L=0.5(L+9/L)?
Solving this gives you L=+3 and L=-3. What should my answer be?

 

[AKG]

Well if you've posted the question exactly as it's given to you, then both 3 and -3 are possible. Observe that if f is a sequence that satisfies the given relation, then so is -f, so if the limit of f is 3, then the limit of -f is -3.

 

[HallsofIvy]

In other words, whether the limit is 3 or -3 depends upon f(0).

If, for example, you know that f(0)= 1, then f(1)= (1/2)(1+ 9/1)= 5, f(2)= (1/2)(5+ 9/5)= 6.8/2= 3.4, f(3)= (1/2)(3.4+ 9/3.4)= 3.0234.., converging to 3.

If, instead, f(0)= -1, then f(1)= (1/2)(-1+9/-1)= -5, etc. converging to -3.