foxjwill said:
So,
a=\frac{1+ \sqrt{4x+1}}{2}.
I threw out
a = \frac{1- \sqrt{4x+1}}{2},
which is negative, because \forall x \in \mathbb{R}, \sqrt{x+a}\geq 0.
Is this correct?
EDIT:I don't really understand why we have to show this. Well, actually, I don't think I really undertstand
what we're trying to show. Could you possibly give a more detailed explanation?
dear foxjwill
your solution is correct( more on this at the end) . Why can't we stop there? we know that that if a_1=a then we will get a constant sequence. we do not know if it will converge to a if we start at any other number! we will have to prove it!
Let's consider an example:
a_{n+1} = 1 + x a_n
solving a = 1 + x a we get a = \frac{1}{1-x}. Let's consider the the case where x=2 . a quick calculation will show that a=-1. now let's consider a_1=10. Then a_2=21 and so on. so althought a=-1 is a solution you will only reach it if you start with a_1=-1! Try a_1=-10
So it is not trivial that the sequence a_{n+1} = \sqrt{x+a_n} will converge if you don't start with a_1=a!
How can you prove that a sequence converges? well if a sequence is monoton increasing and bounded from above then it is convergent. (if a sequence is monoton decreasing and bounded from below it is also convergent). if you don't know that theorem then try to prove it! it is an important one!
How do we know that our sequence will be monoton? Well I will give you a visual prove! Without loss of generality let us asume that x=2 . we will study a_{n+1} = \sqrt{2+a_n}. What is the value if a?
Get a piece of paper with a plot of the functions
f(x)=\sqrt{2+x}
g(x) = x
h(x)=2
on it. Note that f(a)=f(2)=g(2)=2 so the intersection of f and g is the Point P=(2,2).
let's see what happens when we start with a_1 = 1. Make a mark at the point P_1= (1,1). draw a line through P_1 which is parallel to the y-axis.
This line will intersect the the graph of f in the point Q_1 = (1,\sqrt{3}). Draw a line through Q_1 which is parallel to the x-axis. This line will intersect the graph of g in the point P_2= (\sqrt{3},\sqrt{3}) = (a_2,a_2). Repeat that instruction and find P_3,P_4,.... Those points will be on g and the will get closer to P from below. Why will we get closer to P? Because the graph of f is above the graph of g in (-2,2)! Why can't we get to the right of P? Because the graph of f is below the graph of h in (-2,2). Why is it impossible that we stop at a Point B=(b,b) with b < 2? If start at a_1=b-\epsilon then a_2=\sqrt{2+b-\epsilon} >b for \epsilon sufficently small.
What will happen when you start with a_1=10? your sequence will be decreasing and I will leave the details to you.
Was our assumption x=2 special? No the only thing that changes is that the graph of f will move to the right or to the left. ( Well as long as x>0 nothing changes. If x=0 then you will get 2 solutions for a and the case where x<0 is quite complicated because the series may not be defined for all choices of a_1 and I don't think you are expected to consider that case)
An algebraic prove of convergence will look quite similar to our visual prove but you may not get the same insight.
