Four Questions about this weird sequence

The sequence is defined as the following

\[a_1 = 2c^2 \quad (0<c<1) \]

an+1=(1Sn)c2+c(1Sn)2c2+Sn(2Sn) a_{n+1} = (1-S_n) c^2 + c \sqrt{(1-S_n)^2 c^2 + S_n (2-S_n)}

Sn=k=1nak S_n = \sum_{k=1}^n a_k

Q1. What is limnan \lim \limits_{n\to\infty} a_n?
Q2. What is limnSn \lim \limits_{n\to\infty} S_n?

Q3. Let c=cosθ,c = \cos \theta, where 0<θ<π20<\theta < \frac\pi2.

For n=2,3,n=2,3,\ldots, define bnb_n such that the following non-zero limit exists.

cn=limθ0+anθbn c_n = \lim_{\theta\to0^+} \dfrac{a_n}{\theta^{b_n}}

What is limncn\lim \limits_{n\to\infty} c_n?

Q4. Find n=1an2 \displaystyle\sum_{n=1}^\infty a_n^2

Q5. What is limc0+n=1an2c \lim \limits_{c\to0^+} \dfrac{\displaystyle\sum_{n=1}^\infty a_n^2}{c}?


(Note: For Q1 my guess is 0)
(Note:For Q2 my guess is 2)
(Note: For Q3. I have calculated c2=c3=c4=2c_{2} = c_{3} = c_{4} = 2

Note by Inquisitor Math
2 weeks, 4 days ago

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Let's start with Q2. As discussed above, we can formulate the recurrence purely in terms of SnS_n as Sn+1=Sn+(1Sn)c2+c(1Sn)2c2+Sn(2Sn)S_{n+1}=S_n+\left(1-S_n \right) c^2+c \sqrt{\left(1-S_n \right)^2 c^2 + S_n \left(2-S_n \right)}

Consider the recurrence as an iterated function scheme. Does it have any fixed points? Say xx is a fixed point; then

x=x+(1x)c2+c(1x)2c2+x(2x)(1x)c2=c(1x)2c2+x(2x)(1x)c=(1x)2c2+x(2x)(1x)2c2=(1x)2c2+x(2x)x(2x)=0\begin{aligned} x&=x+(1-x) c^2+c \sqrt{(1-x)^2 c^2 + x (2-x)} \\ (1-x) c^2 &= -c \sqrt{(1-x)^2 c^2 + x (2-x)} \\ (1-x) c &= - \sqrt{(1-x)^2 c^2 + x (2-x)} \\ (1-x)^2 c^2 &= (1-x)^2 c^2 + x (2-x) \\ x(2-x)&=0 \end{aligned}

In other words, the system has two fixed points, x=0x=0 and x=2x=2. However, plugging in x=0x=0 to the original expression, we find it is not a fixed point (the reason for this apparent contradiction is we lost some information when we squared both sides of the expression.)

It's easy to check that x=2x=2 is a fixed point.

Let f(x)=x+(1x)c2+c(1x)2c2+x(2x)f(x)=x+(1-x) c^2+c \sqrt{(1-x)^2 c^2 + x (2-x)}

Differentiating, we find f(2)=0f'(2)=0

so that x=2x=2 is in fact a stable fixed point, and the iterated function system does indeed converge to Sn=2S_n=2.


Now Q1 is simple; since Sn2S_n \to 2, SnSn1=an0S_{n}-S_{n-1}=a_n \to 0


For q3, we can use small angle approximations. The thing to keep track of is Tn=2SnT_n=2-S_n. As above, we get Tn+1=Tn+(1Tn)c2c(1Tn)2c2+Tn(2Tn)T_{n+1}=T_n+\left(1-T_n \right) c^2-c \sqrt{\left(1-T_n \right)^2 c^2 + T_n \left(2-T_n \right)}

With c=cosθ112θ2c=\cos\theta \approx 1-\frac12 \theta^2, neglecting terms in θ3\theta^3 and above, this becomes Tn+1Tn+(1Tn)(1θ2)(112θ2)(1Tn)2(1θ2)+Tn(2Tn)=Tn+(1Tn)(1θ2)(112θ2)1(1Tn)2θ2\begin{aligned} T_{n+1} &\approx T_n+\left(1-T_n \right) (1-\theta^2)-\left(1-\frac12 \theta^2 \right) \sqrt{\left(1-T_n \right)^2 (1-\theta^2) + T_n \left(2-T_n \right)} \\ &=T_n+\left(1-T_n \right) (1-\theta^2)-\left(1-\frac12 \theta^2 \right) \sqrt{1-\left(1-T_n \right)^2 \theta^2} \end{aligned}

Since both θ\theta and 1Tn1-T_n are small, we can also approximate the square root: Tn+1Tn+(1Tn)(1θ2)(112θ2)(112(1Tn)2θ2)T_{n+1} \approx T_n+\left(1-T_n \right) (1-\theta^2)-\left(1-\frac12 \theta^2 \right) \left(1-\frac12 \left(1-T_n \right)^2 \theta^2 \right)

After tidying up, this (amazingly) becomes Tn+112θ2Tn2T_{n+1} \approx \frac12 \theta^2 T_n^2

I'd appreciate some double-checking here, but this seems pretty accurate. If we take θ=0.2\theta=0.2 (which is not even that small; it's about 11.511.5^\circ), and compare the exact and approximated TT values, the errors are very small.

So, this approximate TT has a much more simple recurrence, and it's solvable; since T0=2T_0=2, we get Tn=2θ2n+12T_n=2\theta^{2^{n+1}-2}

This still looks a bit crazy, but hopefully you'll recognise the exponents:

nn2n+122^{n+1}-2
0000
1122
2266
331414
443030

Chris Lewis - 6 days, 7 hours ago

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These are wonderful! Thank you for these amazing proofs! (I forgot to mention that earlier.)

I have a question about fixed points; if x is a fixed point is 1<f(x)<1-1<f'(x)<1 a sufficient condition of showing that that fixed point is attractive. Or are these two equivalent? (I haven't learned about these yet so this might be a silly question. I'd be grateful if you could give me some insight.)

Inquisitor Math - 3 days, 4 hours ago

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Perhaps you can write the problem on a sheet and upload a snapshot. That way, it will be readable.

Karan Chatrath - 2 weeks, 4 days ago

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Since it is readable now, can you help me?

Inquisitor Math - 2 weeks, 3 days ago

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Yes, it is readable. I will try this problem and if I can solve it, I will reply on this thread.

Karan Chatrath - 2 weeks, 3 days ago

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Have you made any progress yet?

Inquisitor Math - 2 weeks, 2 days ago

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I think that your guesses for Q 1 and Q 2 are correct. I have had no success in solving part 3.

Karan Chatrath - 2 weeks, 2 days ago

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@Karan Chatrath If you have proved q1&2, would you kindly share it too?

Inquisitor Math - 2 weeks, 1 day ago

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What's the derivation of this question? Did you find this recurrence from another calculation? If so that might be useful.

No solution yet but some observations:

  • The form of the recurrence looks a lot like the solution of a quadratic; in fact, an+1a_{n+1} is one of the roots of x2c22(1Sn)xSn(2Sn)=0\frac{x^2}{c^2} - 2(1-S_n)x - S_n (2-S_n) = 0

  • we can reformulate this purely in terms of SnS_n; we have S1=2c2S_1=2c^2 and Sn+1=Sn+(1Sn)c2+c(1Sn)2c2+Sn(2Sn)S_{n+1}=S_n+\left(1-S_n \right) c^2+c \sqrt{\left(1-S_n \right)^2 c^2 + S_n \left(2-S_n \right)}

  • if we let Tn=2SnT_n=2-S_n and substitute in (with the idea of proving that Tn0T_n \to 0 in order to answer Q1 and Q2), we get Tn+1=Tn+(1Tn)c2c(1Tn)2c2+Tn(2Tn)T_{n+1}=T_n+\left(1-T_n \right) c^2-c \sqrt{\left(1-T_n \right)^2 c^2 + T_n \left(2-T_n \right)} which is really similar to the relation for SnS_n

  • alternatively, letting Rn=1SnR_n=1-S_n and s=1c2s=\sqrt{1-c^2}, we find Rn+1=s2Rnc1s2Rn2R_{n+1}=s^2 R_n - c\sqrt{1-s^2 R_n^2}

which looks a little like some sort of elliptic function. (That choice of ss is to make it equal to sinθ\sin \theta).

Chris Lewis - 2 weeks, 3 days ago

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Wow. Well, I came up with this problem to reinforce (prove) my intuition. It’s easy. Think of a semi circle whose original circle’s radius is 1 and from one end point draw a line which has a slope of tanθ. From the point that the line and circle intersects, draw a line that is perpendicular to the diameter. From the point that this line and diameter intersects repeat this process(starting from drawing a line whose slope is tanθ). Define the distance of the points on the diameter (ie from endpoint to the first point, first point to the second point) as an{a_n}. If you followed through, the recurrence that I posted can be found easily. And it is easy to see that the infinite sum of an {a_n} will be the length of the diameter ‘2’. I hope this can help :)

Inquisitor Math - 2 weeks, 3 days ago

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OK, cool, that makes sense. Just to be completely sure, we're looking at this diagram, right?

I'm wondering if there's a different way to write your q3. It looks like you're after the general behaviour of the sequence {an}\{a_n\} for small θ\theta, ie for c=cosθc=\cos \theta close to 11 (by the way, it's a bit confusing to use "cc" in two places the way you have - it might be better to use something other than cnc_n.)

If that's right, there might be some small-angle approximations you can use. The diagram suggests that these should be pretty good for small θ\theta.

Before I go on too much, could you let me know if I've understood your questions correctly? And if you could rephrase or discuss q3 that might be helpful too.

Chris Lewis - 2 weeks, 2 days ago

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@Chris Lewis Q3 is asked cuz I am just curious if {c_n}=2 for n=2,3,4... It may seem a little preposterous but I’m curious that’s all.

My calculations are the following for the two sequences: b2=2, b3=6, b4=14 and c2=c3=c4=2 From that point on, it is out of my limits. The questions seem crystal clear now and I am willing to know if there has been any progress made from now on. Thx again!

Inquisitor Math - 2 weeks, 2 days ago

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@Inquisitor Math So, the small angle idea does work. The thing to keep track of is Tn=2SnT_n=2-S_n. As above, we get Tn+1=Tn+(1Tn)c2c(1Tn)2c2+Tn(2Tn)T_{n+1}=T_n+\left(1-T_n \right) c^2-c \sqrt{\left(1-T_n \right)^2 c^2 + T_n \left(2-T_n \right)}

With c=cosθ112θ2c=\cos\theta \approx 1-\frac12 \theta^2, neglecting terms in θ3\theta^3 and above, this becomes Tn+1Tn+(1Tn)(1θ2)(112θ2)(1Tn)2(1θ2)+Tn(2Tn)=Tn+(1Tn)(1θ2)(112θ2)1(1Tn)2θ2\begin{aligned} T_{n+1} &\approx T_n+\left(1-T_n \right) (1-\theta^2)-\left(1-\frac12 \theta^2 \right) \sqrt{\left(1-T_n \right)^2 (1-\theta^2) + T_n \left(2-T_n \right)} \\ &=T_n+\left(1-T_n \right) (1-\theta^2)-\left(1-\frac12 \theta^2 \right) \sqrt{1-\left(1-T_n \right)^2 \theta^2} \end{aligned}

Since both θ\theta and 1Tn1-T_n are small, we can also approximate the square root: Tn+1Tn+(1Tn)(1θ2)(112θ2)(112(1Tn)2θ2)T_{n+1} \approx T_n+\left(1-T_n \right) (1-\theta^2)-\left(1-\frac12 \theta^2 \right) \left(1-\frac12 \left(1-T_n \right)^2 \theta^2 \right)

After tidying up, this (amazingly) becomes Tn+112θ2Tn2T_{n+1} \approx \frac12 \theta^2 T_n^2

I'd appreciate some double-checking here, but this seems pretty accurate. If we take θ=0.2\theta=0.2 (which is not even that small; it's about 11.511.5^\circ), and compare the exact and approximated TT values, the errors are very small.

So, this approximate TT has a much more simple recurrence, and it's solvable; since T0=2T_0=2, we get Tn=2θ2n+12T_n=2\theta^{2^{n+1}-2}

This still looks a bit crazy, but hopefully you'll recognise the exponents:

nn2n+122^{n+1}-2
0000
1122
2266
331414
443030

which I think answers your q3!

Chris Lewis - 2 weeks, 2 days ago

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@Chris Lewis I have one question tho. Is it valid to find Tn from that approximation that gives the recurrence of Tn?

Inquisitor Math - 2 weeks, 2 days ago

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@Inquisitor Math It gives you the first term in the expansion in powers of θ\theta; that's what you need for q3 (you were looking for cnc_n, which is the coefficient of this term, and is always 22; and bnb_n is the exponent). I realise it looks pretty rough, but you can either (a) do some δ,ϵ\delta,\epsilon style bounding (the analysis approach) or (b) just go with it (the fluid dynamics approach ;-)). The point with (b) is that the approximations get better as θ\theta gets smaller, and that's exactly what we're interested in.

As for q1, q2, I hadn't been looking at them - but I think they follow directly from the geometry, no?

Although...now I've typed that, I'm irritated with myself, because I tested for cc values outside the range (0,1)(0,1), and it still seems to work. So maybe something more "pure algebra" would be better.

I managed to get somewhere with q3 by using small angle approximations, but of course they don't help for the general case. I'm back to looking at q2 now (because that does solve q1).

Chris Lewis - 2 weeks, 2 days ago

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@Chris Lewis The reason why I did not mention the geometry in the first place was cuz I want to see it proven just by using the recurrence itself, that is pure algebra.(it would be very weird to think the opposite way ie going to that geometric approach, right?)

Inquisitor Math - 2 weeks, 1 day ago

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@Inquisitor Math Fair enough - I agree an algebraic proof would be interesting. But I'd challenge the idea that solving a non-geometry question with geometry is weird; the first proof I saw of the law of quadratic reciprocity used trigonometry; a whole load of arguments in dynamical systems are geometric. I'm sure there are other examples too.

Chris Lewis - 2 weeks, 1 day ago

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@Chris Lewis Yeah I think so too. My bad for misuse of language. Maybe I should change it to “difficult to think about for beginners”

Inquisitor Math - 2 weeks ago

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@Chris Lewis Just asking, is there any field in math or any mathematical tool that can be useful for discussing about ‘bouncing’(you get the expression right?this line kinda bounces between the curve and the line)? Writing this down, this seems like a stupid question!

Inquisitor Math - 2 weeks, 1 day ago

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@Inquisitor Math I can only think of coordinate geometry, really, but you've already been using that. I guess with these problems I quite often move between geometry/trigonometry, coordinate geometry and properties of polynomials. So for instance when you're looking at two curves being tangent to each other, it's often useful to think in terms of a double root; then you've got all the machinery of discriminants and some differential calculus you can use.

If this question had been about something like a light-ray, with different reflection properties, then vectors would have been more useful.

I think - especially with the kind of problem that you come up with yourself (so, not a textbook problem, which fits a certain topic, or an olympiad problem, which you at least know has an answer), it's worth trying as many ideas as you can. Sometimes it doesn't go anywhere, sometimes it takes ages to think of the right approach, and then the problem's easy, but sometimes you find something new and interesting. It also helps a lot in developing an intuition for problem solving.

Chris Lewis - 2 weeks, 1 day ago

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@Chris Lewis I hope this problem does not end up being nothing!

Inquisitor Math - 2 weeks, 1 day ago

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@Inquisitor Math Well, if it's any consolation, I've enjoyed it - it was cool to see everything work out with a small-angle approximation.

Chris Lewis - 2 weeks, 1 day ago

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@Chris Lewis Me too! It was mesmerizing and fantastic to see how that worked out! By the way are you going to tackle the problem in an algebraic way for the remaining problems?

Inquisitor Math - 2 weeks, 1 day ago

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I would like to add a problem Q4 which just came up to my mind. Take a look at it. (Since {a_n} was the length, in this case it is the area.)

Just saying en passant, approaching this geometrically I think this does not necessarily have to be a circle, instead maybe some arbitrary function that satisfies some conditions for this scenario. Dunno this will have any meaning tho

Inquisitor Math - 2 weeks ago

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Have you got any preliminary results or conjectures for your q4? I've done some numerical calculations, but nothing obvious is standing out. I'm not quite convinced there would be a "nice" expression for this in general.

Chris Lewis - 6 days, 8 hours ago

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@Chris Lewis I've started a separate comment for "final" results (since this thread is getting complicated). I've put in an algebraic proof for q1 and q2, and the proof from above for q3. Let me know your thoughts on q4.

Chris Lewis - 6 days, 7 hours ago

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@Chris Lewis How about showing(proving) that Q4 approaches π/2 as θ approaches π/2 first? That seems like a meaningful question, and maybe help for the general case :)

Inquisitor Math - 4 days, 2 hours ago

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@Inquisitor Math That's actually not what I'm finding; I seem to get n=1an2cπ2\sum_{n=1}^\infty a_n^2 \approx \frac{c \pi}{2}

as c0c \to 0.

Chris Lewis - 3 days, 4 hours ago

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@Chris Lewis What do you mean by seem? Do you have some calculations supporting that? And does that mean the sum approaches 0?

Inquisitor Math - 3 days, 4 hours ago

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@Inquisitor Math I calculated the sum (up to n=2000n=2000) for a few small values of cc:

ccan2\sum a_n^2cπ2c \frac{\pi}{2}
0.020.020.0314230.0314230.0314160.031416
0.010.010.0157090.0157090.0157080.015708
0.0050.0050.0078540.0078540.0078540.007854
0.00250.00250.0039270.0039270.0039270.003927

The calculations were based on the original recurrence you posted. They could be subject to rounding errors (I'm just using Excel). If you're getting different results could you post them for comparison?

If you check your recurrence, when c=0c=0, you just get an=0a_n=0 for all nn, so the limit does make sense.

Chris Lewis - 3 days, 4 hours ago

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@Chris Lewis If it is possible to approximate like that is the answer π/2? (By the way where did π/2 come from? Did you approach the problem in a algebraic way or some other way?)

Inquisitor Math - 3 days, 3 hours ago

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@Inquisitor Math It was just a numerical observation based on playing with the data. It's interesting to now look at c0c \approx 0 (where before we were looking at c1c \approx 1). I've tried simply discarding terms in c2c^2 from the recurrence but this doesn't quite work.

Chris Lewis - 1 day, 5 hours ago

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@Chris Lewis I have posted the question asked as Q5. Please let me know if you solved it

Inquisitor Math - 1 day, 8 hours ago

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@Chris Lewis Any progress on Q5? And do you mind looking at my question about fixed points? It is posted under your proof

Inquisitor Math - an hour ago

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