solution of recurrence relation I need help in solving recurrence relation:
$$x_{n+1} = \exp (-x_n) + 2; x_1 = 1$$
I suppose that I should, assuming the limit exists, solve: $a=e^{-a} + 2$. But how?
 A: Let $f(x)=2+e^{-x}$. Our recurrence solves the equation $f(x)=x$ by Fixed Point Iteration. 
After the initial (poor) starting point $1$, the terms of the sequence are always above $2$.Because $|f'(x)|$ is quite a bit under $1$ in this region, the fixed point iteration converges. 
A: The equation $a=e^{-a}+2$ cannot be solved algebraically. A way to find the solution will be to plot a graph of $e^{-a}$ and $a-2$ and see where they intersect. Geogebra gives me the value $a\approx2.12$.
But this is all assuming the limit exists. I do not know how to prove that.
A: I assume you're studing Edexcel's Core 3 A2 module? This recurrence relation gives you a numerical method to solve equations that you could not otherwise solve using algebra. If the recurrence relation generates a sequence that tends towards a limit, i.e. gets closer and closer to a number, then that limit will be a solution to the equation $x = \operatorname{e}^{-x}+2$.
Recurrence relations don't have a "solution". They give you a rule which you use to general a sequence, i.e. a long list of numbers.
In your case, $x_1 = 1$, and so $x_2 = \operatorname{e}^{-x_1}+2 = \operatorname{e}^{-1}+2 \approx 2.368$ (3dp).
Likewise, $x_2 \approx 2.368$ and so $x_3 = \operatorname{e}^{-x_2}+2 = \operatorname{e}^{-2.368}+2 \approx 2.094$ (3dp).
We can continue this process, and we see that
\begin{array}{ccc}
x_4 &\approx& 2.123 \\
x_5 &\approx& 2.120 \\
x_6 &\approx& 2.120
\end{array}
It's tempting to think that $x_n \to 2.120$ (3dp) as $n \to \infty$. 
If this recurrence relation heads towards a fixed value, then when it gets there it'll stay there. In other words, if the number $x_k$ get closer and closer to some number, say $a$, then $a$ will be a solution to $a = \operatorname{e}^{-a}+2$. In your example, you need to keep repeating the iteration process until your answers settle down to a fixed level of accuracy. I get
$$x_{12} \approx 2.120028239$$.
This is the solution to $x = \operatorname{e}^{-x}+2$, to 9 decimal places.
