Question

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.

Answer 1

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?

Answer 2

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...

Answer 3

Any idea? @dan815 @oldrin.bataku

Answer 4

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,
\[2 = \frac{-4}{-4+\frac{-4}{-4+\cdots}}\]

Answer 5

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.

Answer 6

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

Answer 7

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 + ...)))

Answer 8

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.

Answer 9

@jtvatsim OK I understand n agree w what u said :)

Just for the fun of it, u can make a similar substitution...

Answer 10

\[f(x)=\frac{x}{x+f(x)} \\ \text{ we want } f(x)=2 \\ 2=\frac{x}{x+2}\]

Answer 11

\[2(x+2)=x \\ 2x+4=x \\x=-4\]

Answer 12

-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...

Answer 13

Yeah @sdfgsdfgs , I like that one :)

Answer 14

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.

Answer 15

\[-1 = \frac{ \frac{ 1 }{ 2 } }{ \frac{ 1 }{ 2 } - \frac{ \frac{ 1 }{ 2 } }{ \frac{ 1 }{ 2 } - 1 } }\]

n so on

Answer 16

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?

Answer 17

@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.

Answer 18

\(\large\color{black}{ \displaystyle x=2 }\)

\(\Large\color{black}{ \displaystyle \frac{x}{x+\frac{x}{x}}=2 }\) (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......

Answer 19

Oh my third line (x=-2-√2) is the exact value

Answer 20

@geerky42 why not? @jtvatsim 's construct will work perfectly to generate the expression as posted by the prob.

Answer 21

So it seems to converge to 4 (the sequence convergence of course)

Answer 22

oh, -4. xD

Answer 23

Guess this is something I have to force myself to accept. Thanks for efforts though.

Answer 24

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)

Answer 25

I did a similar one for my project some time ago, but that was a while ago.... did all that I can right now:) -:(

Answer 26

@geerky42 thanx 4 a fun one :)

this gives an easy way to construct some very interesting expressions....basically start w/