assuming convergence

Question

anonymous · Answer

$$x = \sqrt{3 + \sqrt{3 + \sqrt{3 + . . .}}}$$

i am not very familiar with convergence or how knowing this converges makes the problem any easier to solve.  i am open to ideas.  i assume i am looking for some kind of symmetry.

like i know if i multiplied both sides by x^2 i get

x^2 = 3 + .... 

but x = ... so

x^2 = 3 + x

x^2 - x - 3 = 0

Is the positive solution to this the answer i am looking for ?

anonymous · Answer

$$x = \sqrt{3 + \sqrt{3 + \sqrt{3 + . . .}}}$$
$$x^{2} = 3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + . . .}}}$$
$$x^{2} = 3 + x$$

so i assume the solution to this quadratic is the solution to the series.

zehanz · Answer

This seems the right way to do this, because you get results...
You could describe this with a recursively defined sequence:

Suppose$$u_0=\sqrt{3}$$and$$u_{n+1}=\sqrt{3+u_n}$$then$$\lim_{n \rightarrow \infty}u_n=u$$(if it exists).
Because u is the limit, the following should be true:$$u=\sqrt{3+u}$$so:$$u^2=3+u \Leftrightarrow u^2-u-3=0$$It is clear that if u exists, it is positive, so$$u=\frac{ 1+\sqrt{13} }{ 2 }$$