Anyone able to help me find the solution for f(n) in terms of f(0) for the recurrence relation:
$$f\left(0 \right) = a$$$$f\left(n \right) = f\left(n-1\right)^2+b$$?

Question

amistre64 · Answer

write it out for starters, see if theres an apparent pattern to spot

anonymous · Answer

Tried it. It gets rather complicated.

amistre64 · Answer

f0 = a
f1= a^2 + b
f2=(a^2+b)^2 + b
f3=((a^2+b)^2 + b)^2 + b

it does doesnt it lol

amistre64 · Answer

it might be one of those; homogenous part and nonhomogenous part type thing

anonymous · Answer

I couldn't find anything on this type though.. I'll carry on looking.

anonymous · Answer

Oops, Mathworld seems to have the exact form. Hopefully it'll help me..

amistre64 · Answer

good luck :)

jamesj · Answer

*bookmark.  I'm curious to see the answer

amistre64 · Answer

an=(an-1)^2; a0=n

a0=n
a1=n^2
a2=n^4
a3=n^8
a4=n^16

an=n^(2^n) would be the homogenous part; then we would need to construct that particular part.

amistre64 · Answer

spose the "n" choice for the unknown was a bad option :)

amistre64 · Answer

$$f(n) = a^{2^n}$$
might look more better

anonymous · Answer

Right, after searching a little bit I found out that it's called a Quadratic Map. (See http://mathworld.wolfram.com/QuadraticMap.html) Unfortunately: "The well-known recurrence $$x_{n+1}={x_n}^2+c$$that is often called "the" quadratic map is not in general solvable in closed form." At least it's been cleared up. Had no idea that it was pretty much the Mandelbrot set.. Thanks for the help, and sorry to waste your time.