A community for students.
Here's the question you clicked on:
 0 viewing
geerky42
 one year ago
Suspicious Brilliant.org question:
https://brilliant.org/problems/xisanegativenumberreally/?group=cwa197UMb6fO
\(\large{\dfrac{x}{x + \dfrac{x}{x + \dfrac{x}{x + \dfrac{x}{x + ...}}}} = 2}\)
Find the value of \(x\) that satisfy the equation above
I *think* limit doesn't exist. But I cannot see how to show that.
geerky42
 one year ago
Suspicious Brilliant.org question: https://brilliant.org/problems/xisanegativenumberreally/?group=cwa197UMb6fO \(\large{\dfrac{x}{x + \dfrac{x}{x + \dfrac{x}{x + \dfrac{x}{x + ...}}}} = 2}\) Find the value of \(x\) that satisfy the equation above I *think* limit doesn't exist. But I cannot see how to show that.

This Question is Closed

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0Let us define \(f_n(x) = \dfrac{x}{x+f_{n1}(x)}\), \(n\in\mathbb N\). Not sure about \(f_1(x)\), but for this problem, I think it is irrelevant. So we have statement "If \(\displaystyle \lim_{n\to\infty}f_n(x)=2\), then \(x=4\)." How can we prove or disprove it?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Hmm... this one is fun to think about. If the limit exists, then x must equal 4 as stated. But, I'm still thinking through with you how to actually prove whether the limit DOES exist in the first place...

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0Any idea? @dan815 @oldrin.bataku

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Not sure how rigorous this is, but we can assume that x = 4 is a candidate for our answer. Then, notice that \[2 = \frac{4}{4+2}\] Since 2 appears again in the denominator, we can repeat this over and over to infinity. Thus, \[2 = \frac{4}{4+\frac{4}{4+\cdots}}\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1This is almost "true by definition". It would seem that the limit does exists since 2 exists and all we are doing is replacing 2 with the equivalent 4/(4+2) each step.

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.0but "assume that x = 4 is a candidate for our answer" is the same as assuming the limit indeed exists. so the subsequent operations neither prove nor disprove the existence of the limit

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1I don't think they are necessarily the same. There is nothing magical about choosing x = 4. I could just as easily have chosen x = 1. The difference is that x = 1 leads to a dead end as I can't seem to get the form 1/(1 + 1/(1+1)) to cooperate and equal 2. I just got "lucky" by picking x = 4, which algebraically is equal to 2 and has the necessary form: 2 = 4/(4 + 2) = 4/(4 + 4/(4 + ...)))

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0Let's use what I have defined. We can see that if we have base \(f_1(x)=2\), then \(\displaystyle \lim_{n\to\infty}f_n(x)\) does exist, and we have \(\displaystyle \lim_{n\to\infty}f_n(x)=2\quad\Longrightarrow\quad x=4\). But what if \(f_1(x)\neq 2\) ? Would limit still exist? Well, for \(f_\infty(x)\), base \(f_1(x)\) technically isn't there.

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.0@jtvatsim OK I understand n agree w what u said :) Just for the fun of it, u can make a similar substitution...

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(x)=\frac{x}{x+f(x)} \\ \text{ we want } f(x)=2 \\ 2=\frac{x}{x+2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[2(x+2)=x \\ 2x+4=x \\x=4\]

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.01 = (1/2)/((1/2)1) noting 1 in the bottom, substituting 1 = (1/2)/((1/2)((1/2)/((1/2)1))) sorry Latex not working for me so i may be missing some () somewhere...

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Yeah @sdfgsdfgs , I like that one :)

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0It's just that I am having hard time see how \( \dfrac{x}{x + \dfrac{x}{x + \dfrac{x}{x + \dfrac{x}{x + ...}}}} \) still have value for \(x<0\) Maybe I am making things too hard lol.

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.0\[1 = \frac{ \frac{ 1 }{ 2 } }{ \frac{ 1 }{ 2 }  \frac{ \frac{ 1 }{ 2 } }{ \frac{ 1 }{ 2 }  1 } }\] n so on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u help me?The length of the hypotenuse of a right triangle is 24 inches. If the length of one leg is 8 inches, what is the approximate length of the other leg?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0@LivingK1ng Try to use Pythagorean Theorem? ( \(a^2+b^2=c^2\) ) Also it's kind of not nice posting your question on another user's question.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle x=2 }\) \(\Large\color{black}{ \displaystyle \frac{x}{x+\frac{x}{x}}=2 }\) (x=2) \(\LARGE\color{black}{ \displaystyle \frac{x}{x+\frac{x}{x+\frac{x}{x}}}=2 }\) (x=3) \(\LARGE\color{black}{ \displaystyle \frac{x}{x+\frac{x}{x+\frac{x}{x+\frac{x}{x}}}}=2 }\) (x=2√2) (x≈3.41) \(\huge\color{black}{ \displaystyle \frac{x}{x+\frac{x}{x+\frac{x}{x+\frac{x}{x+\frac{x}{x}}}}}=2 }\) (x≈3.61) so it seems to end somewhere......

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Oh my third line (x=2√2) is the exact value

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.0@geerky42 why not? @jtvatsim 's construct will work perfectly to generate the expression as posted by the prob.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1So it seems to converge to 4 (the sequence convergence of course)

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0Guess this is something I have to force myself to accept. Thanks for efforts though.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, sometimes, I guess it is helpful to use this kind of a technique. I have sequences that can not be give the \(a_n\) starting from \(a_1\) (like simple arithm. geom. sequences)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I did a similar one for my project some time ago, but that was a while ago.... did all that I can right now:) :(

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.0@geerky42 thanx 4 a fun one :) this gives an easy way to construct some very interesting expressions....basically start w/

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437098012080:dw

sdfgsdfgs
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ x }{ x + y } = y\] select y then solve for x by substituting the y expression back into the left hand side, u generate an infinite series rightaway!

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you can also figure out what this approaches: \[\frac{1}{3+\frac{1}{3+\frac{1}{3+\cdots }}} \\ \text{ Let } x=\frac{1}{3+\frac{1}{3+\frac{1}{3+\cdots }}} \\ \text{ so that means we can also substitute that other thing for } x \\ x=\frac{1}{3+x} \\ 3x+x^2=1 \\ x^2+3x1=0 \\ x=\frac{3 \pm \sqrt{3^24(1)(1)}}{2} \\ x=\frac{3 \pm \sqrt{9+4}}{2} \\ x=\frac{3 \pm \sqrt{13}}{2} \\ \text{ but we did have } x>0 \text{ so } x=\frac{3 + \sqrt{13}}{2} \\ \text{ so we have concluded that } \\ \frac{1}{3+\frac{1}{3+\frac{1}{3+\cdots }}} \rightarrow \frac{3 + \sqrt{13}}{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3http://webserv.jcu.edu/math//vignettes/continued.htm I bet that one with the e thing is harder to derive

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Consider the sequence \(\{a_n\}\) given by $$a_1=4\\a_n=\frac{4}{4+a_{n+1}}\implies a_{n+1}=4\left(\frac1{a_n}1\right)$$ now suppose \(a_n=b_{n+1}/b_n\) so $$a_{n+1}=4\left(\frac1{a_n}1\right)\left(\frac1{a_n}1\right)\\\frac{b_{n+2}}{b_{n+1}}=x\left(\frac{b_n}{b_n+1}1\right)\\b_{n+2} =4(b_nb_{n+1})\\b_{n+2}4b_{n+1}+4b_n=0$$ so suppose \(b_n=r^n\) giving $$r^24r+4=0\\ (r2)^2=0\implies r=2$$ so we get $$b_n=(A+Bn)\cdot2^n$$ meaning that $$a_n=\frac{(A+B(n+1))\cdot 2^{n+1}}{(A+Bn)\cdot 2^n}=2\cdot \frac{(A+B)+Bn}{A+Bn}$$ so then we have \(a_n\to 2\) as \(n\to\infty\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, that should read $$a_{n+1}=4\left(\frac1{a_n}1\right)$$ on the line under \(a_n=b_{n+1}/b_n\)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.