+0  
 
0
168
7
avatar

Find the value of \(x= 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\)

 Nov 25, 2022
 #2
avatar
+2

Sorry, it  wasn't right

 Nov 25, 2022
edited by Guest  Nov 25, 2022
 #3
avatar+288 
+1

Please be aware that the Guest who 'answered' your question did miss a lot of steps. And there is no such thing as "theory of continued fractions"???

I would recommend you checking this out. https://web2.0calc.com/questions/deliberately-misguiding-answers#r2

 

Denote y = x-1

y = 1/(2 + 1/....

 

1/y = 2 + 1(2 + 1/(2..

 

1/y - 2 = y 

 

y^2 = -2y + 1

 

y^2 + 2y - 1 = 0

 

By Quadratic Formula,

y = -1 +- sqrt(2)

Because y has to be nonnegative, y = -1 + sqrt(2)

 

y = x - 1

-1 + sqrt(2) = x -1

 

x = sqrt(2) 

 Nov 25, 2022
edited by Voldemort  Nov 25, 2022
 #6
avatar+288 
-2

I am not quite sure if I understand you.

First off, I put 3  question marks to make it apparent that I wasn't actually 100% sure. Instead of thinking that I am assuming things, you shouldn't assume about things yourself.

2. The link I sent about "misguiding answers" is not a suggestion that Guest is doing it all wrong. He did miss a lot of steps and as you know, there have been anynomous peolpe who have been deliberately sending wrong answers. 

3. AIME 2019? Interesting which one, I or II? Both of them absolutely don't require such "continued fractions"? If I am mistaken, again, let me know. 

4. There is a difference between a theorem and a theory. A theory is something big, like the "BIg Bang Theory". I am pretty sure not anyone would name  such a small thing in math, a theory.

5. So you said something about derivatives? Could you explain this "theorem" to me because it looks like it requires calculus and requires more understanding for the person who asked the question.

 Nov 26, 2022

3 Online Users

avatar
avatar