Evaluating Continued Fraction: \langle 1, 2, 1, 2, \ldots \rangle

  • Thread starter Thread starter math_grl
  • Start date Start date
  • Tags Tags
    Fractions
math_grl
Messages
46
Reaction score
0
Ok I need to know which is the right answer for evaluating the continued fraction \langle 1, 2, 1, 2, \ldots \rangle?

Here's my work:
x = 1 + \frac{1}{2+x} \Rightarrow x^2 + x - 3 = 0 and by quadratic formula, we get x = \frac{-1 \pm \sqrt{13}}{2} but we only want the positive root so I get x = \frac{-1 + \sqrt{13}}{2} for my answer but the answer given was x = \frac{1 + \sqrt{3}}{2}, so I'm confused at which it is...

Moreover, I can't seem to find any other example except for \langle 1, 1, 1, \ldots \rangle to see if I'm doing my computation right. Please help.
 
Physics news on Phys.org
Your equation for x is incorrect. It should be

1 + \frac{1}{2 + \frac{1}{x}}
 
:blushing:
that's embarassing.
 
You'll do better next time!
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top