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.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- geerky42

- katieb

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

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...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

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

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

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

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

@jtvatsim OK I understand n agree w what u said :)
Just for the fun of it, u can make a similar substitution...

- freckles

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

- freckles

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

- 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

Yeah @sdfgsdfgs , I like that one :)

- 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

\[-1 = \frac{ \frac{ 1 }{ 2 } }{ \frac{ 1 }{ 2 } - \frac{ \frac{ 1 }{ 2 } }{ \frac{ 1 }{ 2 } - 1 } }\]
n so on

- 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

@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

\(\large\color{black}{ \displaystyle x=2 }\)
\(\Large\color{black}{ \displaystyle \frac{x}{x+\frac{x}{x}}=2 }\) (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-âˆš2) (xâ‰ˆ-3.41)
\(\huge\color{black}{ \displaystyle \frac{x}{x+\frac{x}{x+\frac{x}{x+\frac{x}{x+\frac{x}{x}}}}}=2 }\) (xâ‰ˆ-3.61)
so it seems to end somewhere......

- SolomonZelman

Oh my third line (x=-2-âˆš2) is the exact value

- SolomonZelman

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

- SolomonZelman

oh, -4. xD

- geerky42

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

- 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

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

@geerky42 thanx 4 a fun one :)
this gives an easy way to construct some very interesting expressions....basically start w/

- SolomonZelman

|dw:1437098012080:dw|

- 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

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

http://webserv.jcu.edu/math//vignettes/continued.htm
I bet that one with the e thing is harder to derive

- 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

oops, that should read $$a_{n+1}=-4\left(\frac1{a_n}-1\right)$$ on the line under \(a_n=b_{n+1}/b_n\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.