Just an "Average" Proof

I honestly have no idea how this problem came to mind. Oh well; here it goes.

Start with a sequence an=an1+an22a_n=\frac{a_{n-1}+a_{n-2}}{2}, where the each term is the average of the previous terms. Define a function ff that chooses the base terms, a1a_1 and a2a_2,: f(x, y)={x=a1y=a2an=an1+an22limnanf(x,~y)=\begin{cases} x=a_1 \\ y=a_2 \\ a_n=\frac{a_{n-1}+a_{n-2}}{2} \\ \displaystyle\lim_{n\rightarrow \infty}a_n \end{cases}

After coding up the function in Python and testing a few cases, I conjectured (i.)(i.):

If f(x, y)={x=a1y=a2an=an1+an22limnanf(x,~y)=\begin{cases} x=a_1 \\ y=a_2 \\ a_n=\frac{a_{n-1}+a_{n-2}}{2} \\ \displaystyle\lim_{n\rightarrow \infty}a_n \end{cases}, then f(x, y)=x+2y3f(x,~y)=\frac{x+2y}{3}.

It is interesting to note that the proposed function is the average of the numbers x, y,x,~y, and yy. While it seems pretty obvious that this is true, nothing is true in mathematics until it is proven. A proof by induction probably seems like the most logical method, but I don't really know how to perform such a proof, much less with 2 variables. So, I tried deriving the formula.


Start with the first few terms of the sequence: x, y, x+y2, x+y2+y2, x+y2+x+2y4, x+y2+2x+5y8,x,~y,~\frac{x+y}{2},~\frac{\frac{x+y}{2}+y}{2},~\frac{\frac{x+y}{2}+x+2y}{4},~\frac{\frac{x+y}{2}+2x+5y}{8},\dots A pattern is most definitely emerging:

  1. The third term (x+y2)\left(\frac{x+y}{2}\right) is always present in the sequence.

  2. The denominator seems to be increasing in powers of 2 (the algebra shows why).

  3. The yy-coefficient in the nnth term is the xx-coefficient in the n1n-1th term (looking at the Python program shows why this is true).

Next, find the recursive formula for the yy-coefficient of the nnth term. Work: x+y2+x+2y4+x+y2+2x+5y82=x+y+2x+4y+x+y2+2x+5y82=x+y2+5x+10y16\dfrac{\dfrac{\frac{x+y}{2}+x+2y}{4}+\dfrac{\frac{x+y}{2}+2x+5y}{8}}{2}=\dfrac{\dfrac{x+y+2x+4y+\frac{x+y}{2}+2x+5y}{8}}{2}=\dfrac{\frac{x+y}{2}+5x+10y}{16}

Firstly, multiply first term by 22\frac{2}{2}, and then combine like terms. The new yy-coefficient is the sum of the previous coefficient, double the coefficient before that, and 1. In other words, cn=cn1+2cn2+1c_n=c_{n-1}+2c_{n-2}+1, with c1=c2=0c_1=c_2=0.

Since the denominator of the nnth term is 2n32^{n-3}, that means (ii.) f(x, y)=x+2y3    limncn2n3=23(ii.)~f(x,~y)=\frac{x+2y}{3}\iff \displaystyle\lim_{n\rightarrow \infty}\frac{c_n}{2^{n-3}}=\frac{2}{3} In other words, in order for the function to be true, the yy-coefficient in the algebraic expansion needs to be two-thirds of the denominator as the order of terms gets larger. (This is the link that can prove the function!)


Firstly, define limncn2n3=L\displaystyle\lim_{n\rightarrow \infty} \frac{c_n}{2^{n-3}}=L. Secondly, it must be acknowledged that the xx-coefficient is, as established earlier, cn1c_{n-1}. Therefore, the xx-coefficient limit is limncn12n3=limncn2n2=limncn2n32=L2\displaystyle\lim_{n\rightarrow \infty} \frac{c_{n-1}}{2^{n-3}}=\displaystyle\lim_{n\rightarrow \infty} \frac{c_n}{2^{n-2}}=\displaystyle\lim_{n\rightarrow \infty} \frac{c_n}{2^{n-3}\cdot 2}=\frac{L}{2}

Next, find the sum of these two limits; the yy-limit is the same as ignoring the xx value of the function and plugging in 1 for yy (since 1 is the multiplicative identity); the opposite is true for the xx-limit. Therefore, the sum of these to limits is the output of the function when x=y=1x=y=1:

    L+L2=f(1, 1)\implies L+\frac{L}{2}=f(1,~1)

Recall that ff was the limit of a recurring sequence that averaged 2 terms. Consider f(b, b)f(b,~b): taking the average of 2 identical numbers outputs that number. If the process is repeated infinitely many times, it will make no difference. Therefore, L+L2=f(1, 1)=1L+\frac{L}{2}=f(1,~1)=1 32L=1\frac{3}{2}L=1 L=23L=\frac{2}{3} Therefore, by (ii.)(ii.), the original conjecture is true!

β\huge{\beta_{\lceil \mid \rceil}}


Please feel free to critique this proof; I am here to learn! I feel that it is too unprofessional, but I am still proud of it; after writing the proof, I googled "recursive sequence limits" and found a much shorter proof on StackExchange, so I know this isn't the best possible proof. I feel like such an amateur after writing this because it is so informal, and I wonder if this what being a "real" mathematician is like. Of course, it probably isn't and I'm just being a kid with excessively high hopes.

Note by Blan Morrison
1 year, 1 month ago

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A really interesting note! Have you thought about generalising what happens with this sequence for an nthn^{\text{th}}-order linear recurrence relations? Would be interesting to see what happens...

A Former Brilliant Member - 1 year, 1 month ago

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I was just thinking about that this morning! I'll definitely make an update about that sometime in the next 12 hours.

Blan Morrison - 1 year, 1 month ago

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Interesting! If you add in a little linear algebra, you would be able to generalise to more linear recurrence relations of higher order, and even know whether it converges at all:

You can split your recurrence relation into multiple degree 1 recurrence relations. For your recurrence this goes from an=an1+an22\displaystyle a_n=\frac{a_{n-1}+a_{n-2}}{2} to:

bn=an1an=an1+bn12\begin{aligned} b_n&=a_{n-1} \\ a_n&=\frac{a_{n-1}+b_{n-1}}{2} \end{aligned}

The reason for doing this is so that you can express the nth term of the recurrence relation as a matrix product:

(a0b0)=(yx)(anbn)=(0.50.510)n(yx)\begin{aligned} \begin{pmatrix} a_0 \\ b_0 \end{pmatrix} &= \begin{pmatrix} y \\ x \end{pmatrix} \\ \begin{pmatrix} a_n \\ b_n\end{pmatrix} &= \begin{pmatrix} 0.5 & 0.5 \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix} y \\ x \end{pmatrix} \end{aligned}

Let's make (0.50.510)=A\displaystyle \begin{pmatrix} 0.5 & 0.5 \\ 1 & 0 \end{pmatrix} = A and (yx)=x\displaystyle \begin{pmatrix} y \\ x \end{pmatrix} = \vec x for notation sake. If we want to find ana_n as nn \rightarrow \infty, then we want to find limnAnx \displaystyle \lim_{n \rightarrow \infty} A^n \vec x

We can diagonalise the matrix AA using its eigenvalues and eigenvectors. For this question,

A=PQP1=(10.511)(1000.5)(10.511)1A = PQP^{-1} = \begin{pmatrix} 1 & -0.5 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -0.5 \end{pmatrix} \begin{pmatrix} 1 & -0.5 \\ 1 & 1 \end{pmatrix}^{-1}

And the reason why we diagonalise it is because we can easily calculate AnA^n:

An=PQnP1=(10.511)(1n00(0.5)n)(10.511)1A^n = PQ^nP^{-1} = \begin{pmatrix} 1 & -0.5 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1^n & 0 \\ 0 & (-0.5)^n \end{pmatrix} \begin{pmatrix} 1 & -0.5 \\ 1 & 1 \end{pmatrix}^{-1}

It is here we can see if ana_n diverges. If it does, the terms in QnQ^n would blow up, making AnxA^n \vec x blow up as well. Luckily, (and expectedly), for this value of AA, the terms in Q converge, making:

limnAn=(10.511)(1000)(10.511)1\lim_{n\rightarrow \infty} A^n = \begin{pmatrix} 1 & -0.5 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & -0.5 \\ 1 & 1 \end{pmatrix}^{-1}

And so the coveted value we wanna find can be calculated by:

(anbn)=(110.51)(1000)(110.51)1(yx)=13(2x+y2x+y)\begin{aligned}\begin{pmatrix} a_n \\ b_n\end{pmatrix} &= \begin{pmatrix} 1 & 1 \\ -0.5 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ -0.5 & 1 \end{pmatrix}^{-1} \begin{pmatrix} y \\ x \end{pmatrix} \\ &= \frac{1}{3} \begin{pmatrix} 2x+y \\ 2x+y \end{pmatrix} \end{aligned}

Interestingly, in general, ana_n is likely to diverge or go to 0. This is because in order for ana_n to converge to a non-zero value, QQ must have an entry of value 11, or QnQ^n diverges or becomes the null matrix. You are very lucky to have chanced upon this recurrence relation.

Oh and as for the LaTeX\LaTeX, you can use \begin{align} <ur eqns> \end{align} to line them up neatly.

Julian Poon - 1 year, 1 month ago

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I'm confused when you say "generalize." Do you mean the average of the previous nn terms instead of 2? It looked to me as if you did something different, but I honestly have little to no comprehension of linear algebra (yet).

Blan Morrison - 1 year, 1 month ago

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By generalise I mean any recurrence relation of the form an=i=1kciani \displaystyle a_n = \sum_{i=1}^{k} c_i a_{n-i}

Julian Poon - 1 year, 1 month ago

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@Julian Poon Ah, okay.

Blan Morrison - 1 year, 1 month ago

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Have a look into linear recurrence relations and how to solve them in general.

A Former Brilliant Member - 1 year, 1 month ago

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@A Former Brilliant Member I saw that wiki, but thank you for pointing that out. Also, thanks for helping with Julian's generalization.

Blan Morrison - 1 year, 1 month ago

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To add to the last point: You'd want the eigenvalues of QQ to all be between 1-1 and 11 (inclusive, I think?), otherwise the limit of the SVD approaches infinity.

On a more sour note: Can you check your calculations again? The matrix AA should read

(010.50.5) \begin{pmatrix} 0 & 1 \\ 0.5 & 0.5 \end{pmatrix}

if you define the recurrence as such. If we modify Blan's initial terms to a0a_0 and a1a_1 instead of a1a_1 and a2a_2 then you would want the matrix equation to start at n=1n = 1, so that your initial terms are given by b1b_1 and a1a_1, not b0b_0 and a0a_0. Aside from that, maybe you could check your SVD again, but it shouldn't change the QQ matrix if you modify the PP matrix. I'd also put a lim\lim sign in front of the bnb_n, ana_n vector at the end, to be precise.

A Former Brilliant Member - 1 year, 1 month ago

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My bad, this can be easily remedied by switching the positions of ana_n and bnb_n, a typo on my part. If we want ana_n to converge to a finite non zero value, all the entries of QQ have to be between -1 and 1, inclusive of 11 but not inclusive of 1-1. Furthermore, at least 1 entry of QQ has to be 11. With regards to the undex starting from 00, it is essentially the same, but i chose to start from 0 so that I don't have a n+1n+1 power in my QQ, which is kinda ugly

Julian Poon - 1 year, 1 month ago

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@Julian Poon No worries. Just wanted to make sure your notation is consistent with Blan’s.

A Former Brilliant Member - 1 year, 1 month ago

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Very interessting practice!

Anton Goetze - 1 year, 1 month ago

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One afterthought: generalization.

Others have considered the same idea, but I'm specifically generalizing where ana_n is the average of the previous kk terms and ff has kk inputs. I edited my Python code I used in the proof to fit k=3k=3, and I successfully predicted multiple outputs with the function f(x, y, z)=x+2y+3z6f(x,~y,~z)=\frac{x+2y+3z}{6}.

I currently don't have the time (and quite possibly the ability, but I'll try later) to prove the general case for kk, but I conjecture

If ana_n is the average of the previous kk terms and ff has kk inputs, then f(x1, x2,  xk1, xk)=i=1kixii=1ki=2i=1kixik2+kf(x_1,~x_2,~\dots ~x_{k-1},~x_k)=\frac{\displaystyle\sum_{i=1}^{k}ix_i}{\displaystyle\sum_{i=1}^{k}i}=2\cdot \dfrac{\displaystyle\sum_{i=1}^{k}ix_i}{k^2+k}

Blan Morrison - 1 year, 1 month ago

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