How to find a function $f$ that satisfies the recurrence relation:
$f(1)=k$
$f(x)=af(x-1)+b(f(x-1))^2$ for $x\ge 2$.
Where $x$ is a member of the set of positive integers, $f$ is a member of the set of real numbers and $a=0.5, b=0.25, k=0.25$
There’s probably more than one function $f$ that satisfies the equation. So is there like a general form for all $f$ that do so?
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COMMENT.-$$f(2)=ak+bk^2\\f(x-1)=\frac{-a\pm\sqrt{a^2+4bf(x)}}{2b}\\f(1)=\frac{-a\pm\sqrt{a^2+4b(ak+bk^2)}}{2b}=\frac{-a\pm(2bk+a)}{2b}=k$$ We must use negative sign (positive one gives nothing) so we have $$\boxed{a=-2bk}$$ It follows $$f(2)=\frac{-a^2}{2b}\\f(3)= \frac{a^4-2a^2}{2b}\\.......\\.......$$
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This is iteration of the quadratic function $\phi(z) = az+bz^2$. Then $$ f(n) = \phi\big(\phi\big(\dots \phi(k)\big)\big) $$ where the number of $\phi$s is $n$.
Mandelbrot (among others) studied this question. Any such iteration is equivalent to another iteration of the form $\psi(z) = z^2+c$ for some $c$. A "closed form" is known for the answer only when $c=0$ or $c=-2$.
In your case, $a=1/2, b=1/4$, we compute $c=3/16$, so there is no simple closed form.
Alright, first let's sort out the terminology confusion here. A function is a mathematical object which associates each value in a specified domain to a value in a codomain.
For instance, $g(x)=x^2$ on the domain $\mathbb{R}$, all real numbers, is a function. So is the function $h$ on the domain of positive integers described by the process "if the input is less than $10$, add $8$ to it; otherwise triple the number and subtract $154$".
The function you have given has domain "positive integers" and codomain "real numbers". We can write that as $f:\mathbb{Z}^+\rightarrow \mathbb{R}$, if we like.
You have then given a description of how the function maps each input to an output. The recurrence relation uniquely defines each input-output pair: for instance
$$ f(1)=k=0.25\\ f(2)=af(1)+bf(1)^2=0.5\times 0.25+0.25\times 0.25^2=0.140625\\f(3)=af(2)+bf(2)^2=0.07525634765625$$
and so on. The method of calculating $f(100)$ - by calculating $f(n)$ from $n=1$ upwards - is clear. In fact, it's also clear that it doesn't matter what the values of $a,b,k$ are: we could delay the choice of values for them and find expressions for $f(n)$ in terms of $a,b,k$: $f(2)=bk^2+ak$, $f(3)=b^3k^4+2ab^2k^3+a(a+1)bk^2+a^2k$.
What I gather you are asking for is a simpler way to calculate the values of the function, and moreover a way to calculate the sum of the values.
Unfortunately, the particular type of recurrence you have is a quadratic map and it is not, in general, easy to find a closed-form expression for it.