Hello all! I have been trying the following question for quite some time and haven't been able to reach the answer. I hope you math geniuses could shed some light on this and help me to get through this. :)

Let \( x_k=k \) for \(k \leq 31\) and \(x_{k+1}=\frac{x_1+x_2+.........+x_k}{k} \) for \(k \geq 31 \). Also let \(y_k=x_k \) for \(k \leq 31 \) and \(y_{k+1}=\frac{y_k+y_{k-1}.......y_{k-30}}{31}\) for \(k \geq 31 \). Now if \(z_k=y_k-x_k \) for all \(k \in N \). Find \( \lim_{n \rightarrow \infty} z_n \).

I was only able to figure out \( x_{k+1}=x_{k+2}=x_{k+3}.....=16\) for \( k \geq 31\). The trouble is finding \( \lim_{n \rightarrow \infty} y_n \). I have no idea about how would I go on evaluating this.

Any help is appreciated. Thanks!

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TopNewestrepresent the limit of the sequence based on 31 consecutive terms by f(y

1,y2...y31), and observe that every change in any of the initial yi affects the limit linearly (this needs to be shown!)In other words, f(y

1,y2,y3...,y31) = (m1)(y1) + (m2)(y2) + ... + (m31)(y31) for some constants mi. Furthermore, observe that sum(mi) = 1, since decreasing all of them by x uniformly obviously reduces f by x.Also observe that f(y

2,y3,y4...,y32) = f(y1,y2,...y_31), that is,(Note: m

i is always the same regardless of the values of yi. Thus, its worth it to substitute more simple values - try f(1,0,0,0,0,0,0,....,0), f(0,1,0,0,0,0...) and so on.)f(1,0,0,...0) = f(0,0,0,....1/31) implying m

31 = 31m1f(0,1,0,0,0....,0) = f(1,0,0....,0,1/31) implying m

2 = m1 + (1/31) (m31) = 2m1f(0,0,1,0....,0) = f(0,1,...,0,1/31) implying m

3 = m2 + (1/31)(m31) = 3m1Inductively, we get what we want: m

1:m2:m3:...:m31 = 1:2:3...:31Since the sum of them is 1. m

1 = 1/(496), m2 = 1/248 ... m_31 = 1/16Substituting them in will give the desired answer. – Gabriel Wong · 3 years, 11 months ago

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1,y2...y31)? Can't we simply is say that limit is infinite as the terms yk go on increasing (I found this by substituting a few values for k).Thank you once again. :) – Pranav Arora · 3 years, 11 months ago

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1...y31) represents the limit of the sequence, given that it has 31 consecutive terms y1, y2 ..., y_31, as the limit of the sequence will not be dependent on previous elements.You do in fact need to show the sequence converges, and that the function f is linear with respect to all y

i. There's a theorem that says an infinite bounded set has a limit point - and the set of yi is clearly bounded between 1 and 31 - the theorem isnt sufficient, but it does suggest the sequence converges. – Gabriel Wong · 3 years, 11 months agoLog in to reply

Anyways, I have the solution too. The solution simply evaluates \( \displaystyle \lim_{n \rightarrow \infty} y_n \) as \( \displaystyle \frac{\sum_{n=1}^{31} n^2}{\sum_{n=1}^{31} n} \) and it was stated without proof so I had to post this here. – Pranav Arora · 3 years, 11 months ago

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(1/496) + (2)(2/496)+(3)(3/496) ... = (1^2+2^2+3^2+...+31^2)/(496)

= (sum n^2)/(sum n)

If we dispense with the need to prove the sequence converges to a fixed point, and assume it does, as we can now use the function notation and prove linearity very quickly:

f(y

1,y2,....y31) - f(y1,y2....,y31+k) = f(y1,y2+....,y31+k) - f(y1,y2+....,y31 +2k) = ...g(x) = f(y

1,y2...y30,x) for fixed y1 to y_30g(x+kd) = k(g(x+d) - g(x)), implying linearity over rationals.

(you can do the same for the other y_i)

Which you can see fairly easily. Because the function is clearly strictly increasing, and you can quickly prove it is linear for all rationals, it is also linear for reals.

Edit: in fact, if you don't need to prove convergence, you can also use the invariant.

For any 31 consecutive terms a

1, a2, ...a31, sum( (i)(ai) ) is constant.Substituting the initial values (1,2...,31) and comparing with (L,L,L...L) will quickly yield the value of limit L as well. – Gabriel Wong · 3 years, 11 months ago

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I have got most of it but not completely. I am still trying to understand your posts. Can you tell me to which part of mathematics does this belong? I have never seen this kind of problem before. I will post if I get any doubts.

Thanks! – Pranav Arora · 3 years, 11 months ago

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– Gabriel Wong · 3 years, 11 months ago

Frankly I'm not sure where it belongs - since I approached it like an Olympiad problem, I'm not even sure if my method used is 'standard' to begin with. Hopefully Calvin or someone more familiar with analysis can shed light on thisLog in to reply

I think the limit of y

n is a case of infinite continued fraction series. If you solve y{k+1} it forms a continued fraction. – Shubham Srivastava · 3 years, 11 months agoLog in to reply

How many determinants of order 3by 3 can be formed , in which each element is either a or b (both non zero) and value of determinant is zero – Sachin Arora · 3 years, 11 months ago

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