geerky42
  • geerky42
Suspicious Brilliant.org question: https://brilliant.org/problems/x-is-a-negative-number-really/?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.
Mathematics
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geerky42
  • geerky42
Suspicious Brilliant.org question: https://brilliant.org/problems/x-is-a-negative-number-really/?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.
Mathematics
katieb
  • katieb
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geerky42
  • geerky42
Let us define \(f_n(x) = \dfrac{x}{x+f_{n-1}(x)}\), \(n\in\mathbb N\). 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
  • jtvatsim
Hmm... 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
  • geerky42
Any idea? @dan815 @oldrin.bataku

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jtvatsim
  • jtvatsim
Not 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
  • jtvatsim
This 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
  • sdfgsdfgs
but "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
  • jtvatsim
I 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
  • geerky42
Let'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
  • sdfgsdfgs
@jtvatsim OK I understand n agree w what u said :) Just for the fun of it, u can make a similar substitution...
freckles
  • freckles
\[f(x)=\frac{x}{x+f(x)} \\ \text{ we want } f(x)=2 \\ 2=\frac{x}{x+2}\]
freckles
  • freckles
\[2(x+2)=x \\ 2x+4=x \\x=-4\]
sdfgsdfgs
  • sdfgsdfgs
-1 = (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
  • jtvatsim
Yeah @sdfgsdfgs , I like that one :)
geerky42
  • geerky42
It'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
  • sdfgsdfgs
\[-1 = \frac{ \frac{ 1 }{ 2 } }{ \frac{ 1 }{ 2 } - \frac{ \frac{ 1 }{ 2 } }{ \frac{ 1 }{ 2 } - 1 } }\] n so on
anonymous
  • anonymous
can 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
  • geerky42
@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
  • SolomonZelman
\(\large\color{black}{ \displaystyle x=2 }\) \(\Large\color{black}{ \displaystyle \frac{x}{x+\frac{x}{x}}=2 }\) (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
  • SolomonZelman
Oh my third line (x=-2-√2) is the exact value
sdfgsdfgs
  • sdfgsdfgs
@geerky42 why not? @jtvatsim 's construct will work perfectly to generate the expression as posted by the prob.
SolomonZelman
  • SolomonZelman
So it seems to converge to 4 (the sequence convergence of course)
SolomonZelman
  • SolomonZelman
oh, -4. xD
geerky42
  • geerky42
Guess this is something I have to force myself to accept. Thanks for efforts though.
SolomonZelman
  • SolomonZelman
Yeah, 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
  • SolomonZelman
I did a similar one for my project some time ago, but that was a while ago.... did all that I can right now:) -:(
sdfgsdfgs
  • sdfgsdfgs
@geerky42 thanx 4 a fun one :) this gives an easy way to construct some very interesting expressions....basically start w/
SolomonZelman
  • SolomonZelman
|dw:1437098012080:dw|
sdfgsdfgs
  • sdfgsdfgs
\[\frac{ x }{ x + y } = y\] select y then solve for x by substituting the y expression back into the left hand side, u generate an infinite series rightaway!
freckles
  • freckles
you can also figure out what this approaches: \[\frac{1}{3+\frac{1}{3+\frac{1}{3+\cdots }}} \\ \text{ Let } x=\frac{1}{3+\frac{1}{3+\frac{1}{3+\cdots }}} \\ \text{ so that means we can also substitute that other thing for } x \\ x=\frac{1}{3+x} \\ 3x+x^2=1 \\ x^2+3x-1=0 \\ x=\frac{-3 \pm \sqrt{3^2-4(1)(-1)}}{2} \\ x=\frac{-3 \pm \sqrt{9+4}}{2} \\ x=\frac{-3 \pm \sqrt{13}}{2} \\ \text{ but we did have } x>0 \text{ so } x=\frac{-3 + \sqrt{13}}{2} \\ \text{ so we have concluded that } \\ \frac{1}{3+\frac{1}{3+\frac{1}{3+\cdots }}} \rightarrow \frac{-3 + \sqrt{13}}{2}\]
freckles
  • freckles
http://webserv.jcu.edu/math//vignettes/continued.htm I bet that one with the e thing is harder to derive
anonymous
  • anonymous
Consider 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 $$a_{n+1}=-4\left(\frac1{a_n}-1\right)\left(\frac1{a_n}-1\right)\\\frac{b_{n+2}}{b_{n+1}}=x\left(\frac{b_n}{b_n+1}-1\right)\\b_{n+2} =-4(b_n-b_{n+1})\\b_{n+2}-4b_{n+1}+4b_n=0$$ so suppose \(b_n=r^n\) giving $$r^2-4r+4=0\\ (r-2)^2=0\implies r=2$$ 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
  • anonymous
oops, that should read $$a_{n+1}=-4\left(\frac1{a_n}-1\right)$$ on the line under \(a_n=b_{n+1}/b_n\)

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