What is the value of f^{n}(x) when n approaches infinity?

[Bipolarity]

Consider the function defined by
f(x) = 0.6x + 2100

Suppose this function is iterated n times.

Express f^{n}(x) as an elementary function of x.

The problem is actually a simplified version of a finance problem I am trying to solve. I am not sure there is a solution so if a solution does not exist, please help me prove it.

BiP

 

[slider142]

This is straightforward if you iterate a few times.
f2(x) = 0.6(0.6x + 2100) + 2100 = (0.6^2)x + 0.6*2100 + 2100
f3(x) = 0.6((0.6^2)x + 0.6*2100 + 2100) + 2100 = (0.6^3)x + (0.6^2)2100 + 0.6*2100 + 2100
This should be enough to see the general pattern for fn(x).

 

[Bipolarity]

Would this be correct?

f^{n}(x) = .6^{n}x + 3500(.6^{n-1} + .6^{n-2} + .6^{n-3} + ... + .6^{0}) for n>1

BiP

 

[slider142]

Yes, assuming you meant 2100 instead of 3500. It is a linear function of x, which satisfies the elementary function requirement. The constants are just rather large polynomials of 0.6.

 

[Bipolarity]

Yes, assuming you meant 2100 instead of 3500. It is a linear function of x, which satisfies the elementary function requirement. The constants are just rather large polynomials of 0.6.

Interesting.
What if we make n → ∞ ?

Can we simplify in that case?

My guess is that the answer will be 3500*(\frac{.6}{1-.6}) = 5250.

BiP

 

[slider142]

Interesting.
What if we make n → ∞ ?

Can we simplify in that case?

My guess is that the answer will be 3500*(\frac{.6}{1-.6}) = 5250.

BiP

It should be 3500*(1/(1-0.6)) = 35000/4 = 8750 for the sum of the geometric series. Is 3500 the correct value? You have 2100 in its place in the first post.