One One One One One One Forever

Calculus Level 3

$\Large { \frac{1}{ \frac{1}{\frac{1}{\ldots}+1+\frac{1}{\ldots}} +1+ \frac{1}{\frac{1}{\ldots}+1+\frac{1}{\ldots}} }}$

What is the value of the nested functions above?

This problem is a part of set nested radicals .

\frac{1}{\sqrt{2}}

\frac{1}{2}

1

\frac{1}{4}

3 solutions

Ankith A Das
Mar 7, 2015

$let\quad x=\dfrac { 1 }{ \dfrac { 1 }{ \dfrac { 1 }{ .... } +1+\dfrac { 1 }{ ... } } +1+\dfrac { 1 }{ \dfrac { 1 }{ ... } +1+\dfrac { 1 }{ ... } } } \\ then\\ x=\dfrac { 1 }{ x+1+x } \\ x=\dfrac { 1 }{ 2x+1 } \\ 2{ x }^{ 2 }+x=1\\ 2{ x }^{ 2 }+x-1=0\\ 2{ x }^{ 2 }+2x-x-1=0\\ 2x\left( x+1 \right) -1\left( x+1 \right) =0\\ \left( 2x-1 \right) \left( x+1 \right) =0\\ x=\dfrac { 1 }{ 2 } ;\quad x=-1\\ But\quad x\quad cannot\quad be\quad negative\quad \\ \therefore \quad x=\dfrac { 1 }{ 2 } \\$

How can you conclude that x can't be negative.

X=-1 perfectly fits the fraction even though its counterintuitive

Aditya Mishra - 6 years, 3 months ago

Think about it. What conditions do you need for a fraction $\dfrac{a}{b}$ to be negative? Does the given fraction satisfy those conditions?

Hint: Even though the denominator has continued fraction components, the convergents of the continued fraction(s) are positive, so we have both the numerator and denominator of the given original fraction as positive.

Prasun Biswas - 6 years, 3 months ago

do you know that sum of all natural number can be written as -1/12 ? even physicists have used that value.

sum of all binary exponents 1+2+4+8+.. = -1. because if you multiply LHS or RHS by 2, value decreases by one.

I admit that i don't understand these results very well, but all i am saying is that infinite sum and radicals behave much differently than we would expect.

If you go beyond your personal bias that infinite positives will result in positive, you will see that if you assume the whole given sum to be -1, and put x=-1 on the RHS ,

$\frac{1}{-1+1-1}=-1$

which is equal to x and hence, is consistent solution.

Aditya Mishra - 6 years, 3 months ago

@Aditya Mishra – Each of the instances you mentioned has a separate story behind it.

For example., let's take the case of the sum of the binary exponents. I still remember this to be a problem made by Finn Hulse and there was a huge discussion on it. The sum is actually divergent. It isn't $(-1)$ . Let me explain it a bit more clearly.

Take $S=1+2+4+8+\ldots$

Now, what YOU proposed was this:

$S=1+2S\implies S=(-1)$

Isn't it? Let me tell you that this result is completely WRONG mathematically. Here's why! When are you allowed to perform arithmetic operations on numbers? Is the following statement: $\infty-\infty=0$ true? Of course not! Same is the case with the problem at hand. When you are performing the operation $2S-S$ , you are subtracting two divergent sums from one another. Arithmetic operations on sequence sums are only valid if they converge. The divergence of this sequence sum can easily be verified using ratio test. As such, the sum of the binary exponents upto infinite terms is divergent and has no particular value.

The summation of all natural numbers has a different story to it. The value $\left(-\dfrac{1}{12}\right)$ isn't the value of the sum. That sum is also divergent. You can read more about it here .

Prasun Biswas - 6 years, 3 months ago

@Prasun Biswas – I almost completely agree with you :)

Sorry for not phrasing it properly, but i do not claim that those sums are equal to -1 and $\frac{-1}{12}$ .

All i want to say is that under certain conditions, they share some properties. For example, if we multiply all terms of binary exponents by 2, a one seems to be missing from original terms. Also, if you multiply -1 by 2, it reduces by one. I know very well that the procedure is meant for converging series but the result is somewhat interesting.

Similarly, in given question, -1 shares the same self-similarity as $\frac{1}{2}$ . Both -1 and $\frac{1}{2}$ result in themselves if you multiply them by 2, add 1 to it and then get the reciprocal.

Instead of setting hard and fast restrictions about what can be done and what can't be done, it's more fun to bend those concepts to see if some interesting results occur.

If you do not reject complex numbers as solutions of polynomials with real coefficients, why reject -1 as a candidate without even thinking about what it means and why it was produced as a solution using standard algebraic methods.

Aditya Mishra - 6 years, 3 months ago

@Aditya Mishra – Because nobody says $\sqrt{1+\sqrt{1+\ldots}}$ is negative. If I iterate $\sqrt{1+x}$ , it is always positive for $x = 1$ . Since we're taking the limiting value of the nested radicals, we always end up with a positive value. In the same vein, $-1$ is a solution to $x = \frac{1}{2x+1}$ but NOT to the continued fraction.

If you study complex analysis, you'll understand more about why such divergent sums can be expressed as finite values. Fun stuff. It's great to have an open mindset like yours, but the result $\zeta(-1) = -\frac{1}{12}$ has its motivations, rather than just "oooh let's be a bit cheeky here". (Probably.)

Jake Lai - 6 years, 3 months ago

@Jake Lai – Let's start where we both agree. You agree that $-1$ is a solution to the following,

$x=\frac{1}{x+1+x}$

now, what if i substitute the value of x on RHS,

$x=\frac{1}{\frac{1}{x+1+x}+1+\frac{1}{x+1+x}}$

Repeating this process infinitely will result in our fraction. What i infer from this process is that the equation and the fraction are representing the exact same thing, even though the equation is finite and fraction is not.

That's why we could replace the fraction by the equation in the first place.So anything satisfying the given equation for x would fit as the solution of fraction, and $-1$ does so.

Also, in the nested radical you presented as example, you completely ignored the fact the Square root is a multivalued function in order to restrict the solution while if you allow it to be negative (which it naturally is) you will see that both $\frac{1+\sqrt{5}}{2} and \frac{1-\sqrt{5}}{2}$ have equal capability to satisfy the infinite radical.

Though in geometric problems, we would ignore the negative lengths, pure algebra poses no such restrictions to values.

And thanks, i would certainly study more about zeta function and complex analysis.

Aditya Mishra - 6 years, 3 months ago

@Aditya Mishra – The way you're constructing the "infinitely" continued fraction is analogous to building from the ground up to the sky; it can go in both directions, certainly, because it will always be a finite iteration. The presented continued fraction is an already constructed tower, and it remains for us to resolve it. It can only, then, go in one direction, dictated by the closedness of the positive reals under addition.

Jake Lai - 6 years, 3 months ago

@Jake Lai – Is there a rigorous proof that positive reals will be closed under infinite additions and divisions? If there is, then my argument is invalid.

One problem with that is if you add 1 to 0 an infinite number of times, you don't get positive real number, you get infinity.

Otherwise, The only way to resolve it is to assume that at any level, the value of fraction from that level on-wards is equal to the value of overall solution. If you find such a value, tower collapses to one consistent final value.

Also, you are missing the point of my construction. The recipe of creating the fraction tower is that equation. Even though i can't iterate my construction to produce exact fraction, It doesn't mean that equation doesn't have capability to represent it exactly.

Aditya Mishra - 6 years, 3 months ago

@Aditya Mishra – i liked your argument saying that the sum of all natural numbers turns out to be -1/12. i too would've said the same thing. but I didn't like the fact that you backed off from it thinking that the sum is not implied by reality and is just a mathematical anomaly. In fact this result including many others is just a silent indicator that our mathematics is not fully matured yet.... I mean Euler derived that sum and then it was also recognised to be zeta(-1), i.e, the Reimann zeta function with argument -1. Ironically, nobody has yet been able to give a proof of Reimann's hypothesis which says that all non trivial zeroes of this function lie on the line r=1/2 of the complex plane.... And although I'm no mathematician but still I think that the root of this problem lies in our ignorance of the concept of infinity which has deluded us since the dawn of mathematics itself!!!

For eg: take Euler's proof of the sum -1/12.... In it he takes the infinte series S=1-2+3-4+5...shifts it one place and then adds it with itself claiming it is

2S=1-1+1-1+.....now this is the part where we err or probably make the correct guess (no doubt blindly) that shifting and adding the numbers would have no effect upon the sum b'cuz its already infinity!!! I say, why even bother to make it 2S since it'd be the same as S, b'cuz S is infinity!!! These contradictions arise due to a single fact; we don't comprehend infinity well enough!!! So in my personal opinion, although I agree to the answer 1/2 for the infinite sum above mentioned, I still find it difficult to dismiss the possibility that -1 could also be a solution to the problem. In fact, until and unless a thorough understanding of the concept of infinity has been developed, it would be highly erroneous on our part to genuinely claim the absolute correctness of a single solution!!!

Once Bertrand Russell visited a town in England to give a lecture on planetary motion or some such stuff. In the hall, there was an old lady who couldn't digest Russell's arguments and claimed that the earth stood on the back of a giant tortoise. Amused, Russell asked in a superior tone - " well whats the tortoise standing on?"... To this the lady replied - "well its tortoises all the way down!!!". Now why did Russell think he knew better?

a 110 years back, before Einstein's paper on special relativity, nobody would've believed you if you told someone that time wasn't absolute. It would've been dismissed as another old woman's imagination!!! In the same way, maybe the integers don't act the way they seem to us. maybe they cycle back on the negative side after reaching positive infinity which could explain the negative sum...

maybe the number line isn't a line at all, but something like a number cycle!!!

This is just one of my highly speculative arguments that neither have any mathematical backing nor any practical confirmation but they still can't be ruled out as a possibility no matter how unlikely ;)

Somesh Singh - 6 years, 3 months ago

@Somesh Singh – i certainly agree with you that infinity is probably the most elusive concept of all :)

But I backed off from using the claim regarding those series because i simply do not understand them well enough.

Also, the number circle is an interesting concept. I also thought the exact same thing when i looked at the graph of $tan(x)$ in 8th grade. It's good to know that others also share these wild imaginations.

Aditya Mishra - 6 years, 3 months ago

@Aditya Mishra – sure man...and these wild imaginations are the key to great mathematical and scientific discoveries, so never give up the good work and keep imagining ;)

Somesh Singh - 6 years, 2 months ago

Curtis Clement
Mar 7, 2015

Rewrite the infinite fraction as: $x = \cfrac{1}{1 + \cfrac{2}{1 + \cfrac{2}{1+ \cfrac{2}{1...} } } }$ Now by writing the reciprocal of the infinite fraction (by moving it up by one) and subtracting 1 we can obtain: $\frac{1}{x} - 1 = 2x \Rightarrow\ 2x^2 +x-1 = (2x-1)(x+1) = 0$ Now all the terms are positive so ${x}$ > 0.

$\huge\color{#69047E}{\therefore\ x = \boxed{\frac{1}{2}} }$

Do we need to think about the thorny issue of convergence?

Otto Bretscher - 5 years, 2 months ago

Savira Utami
Mar 9, 2015

One One One One One One Forever

This problem is a part of set nested radicals .

3 solutions

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