Writing the recursive as explicit This question was first asked elsewhere on stackexchange (it has since been deleted$^1$). It quickly was put on hold and I didn't think it got a fair shot. I am rewriting it to see if I can enliven it. 
First we will define a function recursively and we will then we can ask some questions about it. Let $f$ be a function which is $0$ on the negative numbers and $1$ on $[0,1)$. For all other values (that is, $x>1$) we let our function $f(x)$ simply record the area underneath itself on the interval $(x-2,x-1)$. Symbolically we can write this: 
We have a function $f:\mathbb R \to \mathbb R$ with such properties: 
$$f(x)=\begin{cases}0 & x<0 \\ 1 & 0 \le x < 1 \\ \\ \displaystyle \int^{x-1}_{x-2}{f(t)\ dt}& x\ge 1\end{cases}$$
Dawid Bucki (the author of the linked post)  computed $f$ for small $x$ and it (if he is to be believed) looks like this:
$$f(x)=\begin{cases} 1 & 0 \le x <1 \\ (x-1) & 1 \le x \le 2 \\ \frac 12 (x-2)^2-(x-2)+1 & 2 \le x \le 3 \\ \frac 16 (x-3)^3-(x-3)^2+(x-3)+\frac 12 & 3 \le x \le 4 \end{cases}$$
Dawid also claims that it can be shown to be bounded by exploiting induction and the mean value theorem. 
I think this function is kind of interesting and speaks to a question which I enjoy: "When can recursively defined things be made explicit?"
Question$^2$: Can this function be written explicitly? 
Question$^3$: What can we say about $$\lim_{x\to\infty}f(x)$$
It seems like the function limits to $0$ to me. Can we prove/disprove that? 
Footnote 1: Which seems unfortunate the answers/comments seemed valuable to me.
Footnote 2: This is my question and I get to write the questions in the order I want to because I was granted permission to rewrite this question.
Footnote 3: This is Dawid's question. And he's probably very angry at my ordering of the questions. 
 A: This is a longish comment. A simple example of a linear Delay differential equation with discrete delays is $\;f'(x) = f(x-1) - f(x-2).\;$ The usual ansatz is $\;f(x) = e^{cx}\;$ where $\;c\;$ satisfies $\;c = e^c + e^{2c}.\;$ The obvious solution is $\;c=0\;$ and the other is $\;c = -.51272\dots - 4.02555\dots i\;$ and its conjugate. The solution with $\;c=0\;$ is $\;f(x)=1,\;$ a constant function, and the other is a simple exponential decay function. These are analytic solutions. Something similar happens with the simpler one delay equation $\;f'(x) = f(x-1),\;$ which is MSE question 2245492 "Continuous recursive iteration".
The complication arises where we specify initial values for $\;f(x)\;$ on an interval $\;[0,1]\;$ as in this question. On each interval $\;[n,n+1]\;$ the function is a polynomial $\;p_n(x)\;$ of degree $\;n.\;$ Numeric computations suggest that  as $\;x\to\infty\;$ the function $\;f(x)\;$ approaches some linear combination $\;g(x):=a + b_1e^{cx} + b_2e^{\bar cx},\;$ and since we have exponential decay terms, $\;f(x)\to a\;$ as $\;x\to\infty.\;$ One interesting feature is that $\;f(x)-a\;$ has almost regularly spaced zeros. The polynomial zeros of $\;p_n(x)-a\;$ are all positive reals except for two conjugate pairs early on and one or two of them closely match $\;f(x)-a\;$ zeros.
Note carefully the difference between the nice $\;g(x)\;$ and the piecewise polynomial $\;f(x).\;$ The polynomial sequence $\;p_n(x)\;$ is an interesting one with several unexplored properties.
A: This is a partial answer only.Since I have not seen the previous post, some of the comments might have been made before.
It is easy to show that the function $f(x)$ is formed by piecewise polynomial expressions of degree $n$ on intervals $x \in [n,n+1]$. The description on the interval $x \in [0,4]$ is indeed correct and it is straightforward to calculate the function on consecutive intervals, although using the recursive relation with some program like Mathematica or otherwise will definitely help preventing making mistakes. Whether there is a "simple" general expression, I do not know.
Furthermore the function $f(x)$ is continuous for $x > 1$, differentiable for $x>3$, twice differentiable for $x>5$, etc..
What is also easy to show is that 
$$\lim_{x \rightarrow \infty} f(x) \neq 0$$
Hereto consider the following simple argument. On the interval $t \in [x-2,x]$ for all $x$ we have 
$$\min_{s \in [x-2,x]} f(s) \leq f(t) \leq  \max_{s \in [x-2,x]} f(s)$$
Therefor from the recurrence relation for $x \geq 1$ we get
$$\min_{s \in [x-2,x]} f(s) \leq f(x)=\int_{x-2}^{x-1}  f(t) d t \leq  \max_{s \in [x-2,x]} f(s)$$ 
Given the definition for $f(x)$ for $x<1$ we can also extend it to the weaker form
$$\min_{s \leq x} f(s) \leq f(x) \leq  \max_{s \leq x} f(s)$$
or extend it to
$$\min_{s \in [y-2,y]} f(s) \leq f(x \geq y \geq 1) \leq  \max_{s \in [y-2,y]} f(s)$$
In particular for $y=2$ we find $0 \leq f(x \geq 2) \leq 1$. The function $f(x)=$ for $2 \leq x \leq 4$ already gives $\frac{1}{2} \leq f(x>4) \leq 1$. For the interval $x \in [9,11]$ we get $0.6598 < f(x) < 0.6713$. So the limiting value of the function, if it exists, will be in that interval as well and 2/3 seems like a good candidate. This is of course not demonstrated here and this approach cannot exclude the possibility of a limiting oscillating function.
Note that we can use the fact that $f(x)$ is continuous for $x>1$ to show that the inequalities become strict for $x>2$.
