# Arithmetic, geometric and other sequences

You all probably know what an arithmetic sequence is. It's a sequence of numbers such that consecutive terms always have the same difference. But, there is another, more general definition: An arithmetic sequence is a sequence of real numbers such that every term in the sequence is the arithmetic mean of the preceding and the following term.

But what if we take some other mean instead of the arithmetic mean?

Other well known means are the geometric mean, the harmonic mean and the quadratic mean.

Geometric mean

The geometric mean $M_g$ of two numbers $a$ and $b$ is $M_g(a,b) \stackrel{\text{def}}{=} \sqrt{ab}$. Applied to our definition for a (now geometric) sequence $\{ a_n \}$, this means

\begin{aligned} & a_n = \sqrt{a_{n-1}a_{n+1}} \\ \Leftrightarrow & a_n^2 = a_{n-1}a_{n+1} \\ \Leftrightarrow & a_{n+1} = \frac {a_n^2}{a_{n-1}} \end{aligned}

If we define $r \stackrel{\text{def}}{=} \frac {a_n}{a_{n-1}}$, we see $a_{n+1} = \frac {a_n^2}{a_{n-1}} = ra_n$, which is another known definition of a geometric series.

Harmonic mean

The harmonic mean $M_h$ of two numbers $a$ and $b$ is defined as $M_h(a,b) \stackrel{\text{def}}{=} \frac 2{\frac 1a + \frac 1b}$. To get a recursive formula,

\begin{aligned} & a_n = \frac 2{\frac 1{a_{n-1}} + \frac 1{a_{n+1}}} \\ \Leftrightarrow & \frac 1{a_n} = \frac 1{a_{n-1}} + \frac 1{a_{n+1}} \\ \Leftrightarrow & \frac 1{a_{n+1}} = \frac 1{a_n} - \frac 1{a_{n-1}} \\ \Leftrightarrow & a_{n+1} = \frac 1{\frac 1{a_n} - \frac 1{a_{n-1}}} \end{aligned}

But this is not what I am aiming for. These means are too restrictive. However, there js a generization that involves all of these means as special cases. This is called the Hölder mean and it generalizes by introducing a new parameter $p$. In the case of 2 numbers $a$ and $b$, their Hölder mean with parameter $p$ is defined as

$M_p(a,b) \stackrel{\text{def}}{=} \sqrt[p]{\frac {a^p+b^p}{2}}$

With these means, we can also generalize our sequences to general "p-mean sequences".

In such a sequence, the equation $a_n = M_p(a_{n-1},a_{n+1})$ has to hold for all $n$.

We can rearrange this equation to get a recursive equation for any p-mean sequence

\begin{aligned} & a_n = \sqrt[p]{\frac {a_{n-1}^p + a_{n+1}^p}{2}} \\ \Leftrightarrow & 2a_n^p = a_{n-1}^p + a_{n+1}^p \\ \Leftrightarrow & a_{n+1}^p = 2a_n^p - a_{n-1}^p \\ \Leftrightarrow & \boxed{a_{n+1} = \sqrt[p]{2a_n^p - a_{n-1}^p}} \end{aligned}

To get a feeling of these sequences, let's look at some examples. For simplicity, we will always take $a_1 = 1$ and $a_2 = 2$.

$\begin{array}{cc} p & a_1, a_2, \ldots, a_6 \\ 1 & 1, 2, 3, 4, 5, 6 \\ 2 & 1, 2, \sqrt{7}, \sqrt{10}, \sqrt{13}, 4 \\ 3 & 1, 2, \sqrt{15}, \sqrt{22}, \sqrt{29}, \sqrt{36} \end{array}$

Since these numbers aren't that "nice" (because they involve roots), let's define another sequence $\{ b_n \} \stackrel{\text{def}}{=} \left\{ a_n^p \right\}$ to get rid of these roots.

We can adopt the definition of $\{ a_n \}$ to get a recursive equation for $\{ b_n \}$

$a_{n+1} = \sqrt[p]{2a_n^p - a_{n-1}^p}$

$b_{n+1} = 2b_n - b_{n-1}$

The starting values are $b_1 = a_1^p = 1$ and $b_2 = a_2^p = 2^p$.

Let's look at the table again, this time also with $\{ b_n \}$.

$\begin{array}{cc} p & a_1, a_2, \ldots, a_6 & b_1, \ldots, b_6 \\ 1 & 1, 2, 3, 4, 5, 6 & 1, 2, 3, 4, 5, 6 \\ 2 & 1, 2, \sqrt{7}, \sqrt{10}, \sqrt{13}, 4 & 1, 4, 7, 10, 13 \\ 3 & 1, 2, \sqrt{15}, \sqrt{22}, \sqrt{29}, \sqrt{36} & 1, 8, 15, 22, 29, 36 \\ 4 & 1, 2, \sqrt{31}, \sqrt{46}, \sqrt{61}, \sqrt{76} & 1, 16, 31, 46, 61, 76 \end{array}$

We observe something interesting; $\{ b_n \}$ is always an arithmetic sequence. To get its general formula, let's find the formula for each $p$ individually

$\begin{array}{cc} p & \text{formula} \\ 1 & n \\ 2 & 3n-2 \\ 3 & 7n-6 \\ 4 & 15n-14 \end{array}$

It looks like the general equation is $b_n = \left( 2^p-1 \right) n - \left( 2^p-2 \right)$.

Let's prove this by induction

Base case 1: $n=1$

\begin{aligned} b_1 &= \left( 2^p-1 \right) \cdot 1 - \left( 2^p-2 \right) \\ &= 2^p-1-2^p+2 \\ &= 1 \end{aligned}

Base case 2: $n=2$

\begin{aligned} b_2 &= \left( 2^p-1 \right) \cdot 2 - \left( 2^p-2 \right) \\ &= 2\cdot 2^p - 2\cdot 1 -2^p+2 \\ &= 2 \cdot 2^p - 2^p \\ &= 2^p \end{aligned}

Induction step

Suppose $b_k = \left( 2^p-1 \right) k - \left( 2^p-2 \right)$ holds true for some $k$. Then, by the induction hypothesis (i.h.), it must also be true for $b_{k+1}$.

\begin{aligned} b_{k+1} &\stackrel{\text{formula for }b_n}{=} 2 b_k - b_{k-1} \\ &\stackrel{\text{i.h.}}{=} 2 \left[ \left( 2^p-1 \right) k - \left( 2^p-2 \right) \right] - \left[ \left( 2^p-1 \right) (k-1) - \left( 2^p-2 \right) \right] \\ &= 2 \left[ 2^pk -k -2^p +2 \right] - \left[ 2^pk -2^p -k+1 - 2^p +2 \right] \\ &= 2^{p+1}k-2k-2^{p+1}+4 - 2^pk+2^p+k-1+2^p -2 \\ &= 2^pk-k +1 \\ &= \left(2^pk-k+2^p-1\right) -\left( 2^p-2 \right) \\ &= \left( 2^p-1 \right) (k+1) - \left( 2^p-2 \right) \\ &= b_{k+1} \end{aligned}

Since we end up with $b_{k+1}$ in the end and used the formula for $b_n$ and the definition of a p-mean sequence, this completes our proof for the formula of such a p-mean sequence

$\large\boxed{b_n = \left( 2^p-1 \right) n - \left( 2^p-2 \right)}$

$\large\boxed{a_n = \sqrt[p]{\left( 2^p-1 \right) n - \left( 2^p-2 \right)}}$ Note by Henry U
1 year, 8 months ago

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I still want to get the formula for a p-mean sequence with arbitrary starting values $a$ and $b$ and I think it won't be that difficult since I only have to linearly transform the formula for $1,2$ (at least I think so).

Then, maybe I will look at the asymptotic growth of such sequences.

- 1 year, 8 months ago

Whoah........!!! This is nice!!!!! Never thought about generalizing the sequences.......!!! And yup........the p mean sequence with arbitrary starting values isn't that difficult..........You can derive that easily.......... Meanwhile, I am gonna proceed with the asymptotic growth while waiting for your next update!!!

- 1 year, 8 months ago

For starting values $a$ and $b$ I get $b_n = (b^p - a^p)n - (b^p - 2a^p)$ and $a_n = \sqrt[p]{(b^p - a^p)n - (b^p - 2a^p)}$

- 1 year, 8 months ago

That looks reasonable! Thanks for your help; I will add this soon.

I think this also shows what one can conjecture before getting the exact formula, since larger $p$ values increase the influence of the larger term, which is the following term because the sequences are increasing, the diffetrence between consecutive terms should decrease more and more, and this matches the $\sqrt[p]{\cdot}$.

- 1 year, 8 months ago

.

- 1 year, 7 months ago