The Oldest Math Book?

Sometime in the 17 century BCE, some people living in the Fertile Crescent wrote on a clay tablet how to calculate the sum of numbers following some predictable patterns. They may have explored 1 + 2 + 3 + 4 + 5 + ... + 100, for example. Shall we?

Let's say S = 1 + 2 + 3 + 4 + 5 + ... + 100. We could certainly Google the answer or mindlessly keep adding until we get there, but what's the point in that? The point is to develop new skills and be the cleverest person in the room at parties. Let's look for patterns in the partial sums. What are partial sums? We may define the nth partial sum of S by

$S_{n} = 1 + 2 + 3 + 4 + \cdots + n.$

Then

$S_{1} = 1.$
$S_{2} = 3.$
$S_{3} = 6.$
$S_{4} = 10.$
$S_{5} = 15.$
$S_{6} = 21.$
$\vdots$
$S_{100} = \text{ ?}$

Are the partial sums revealing a way to calculate the 100th sum without rote? I don't think so either. Let's try breaking the problem into simpler parts. Notice that the sum we want is two smaller sums together: the odds + the evens. That is

$1 + 2 + 3 + 4 + \cdots + 100 = (1 + 3 + 5 + 7 + \cdots + 99) + (2 + 4 + 6 + 8 + \cdots + 100).$

Now, let's look at the partial sums of

$O = 1 + 3 + 5 + 7 + \cdots + 99 \text{ and}$
$E = 2 + 4 + 6 + 8 + \cdots + 100.$

Denoting the nth partial sums of O and E by O_n and E_n, we may observe that

$O_{n} = n^{2} \text{ and } E_{n} = n(n + 1).$

Thus,

$S_{100} = O_{50} + E_{50} = 5050.$

It must be noted that our reasoning so far is empirical. That is to say our final calculation is based on the prediction that things will keep happening as they have happened so far as we know. Thus, there is an unproved hypothesis afoot. However, our hypothesis suggests that

$S_{2n} = O_{n} + E_{n} = n^{2} + n(n + 1)$

We still don't have an nth term formula for the case where the number of terms is odd. We could easily write it, but there is a better way to get at it. Consider that 2S_n = E_n. Therefore,

$S_{n} = \frac{n(n + 1)}{2}$

If we take a different perspective on this result, we gain further insight. Writing it as

$S_{n} = n \left ( \frac{1 + n}{2} \right )$

reveals to us that 1 + 2 + 3 + ... + n is the average of the first and last terms, multiplied by the number of terms. This is a common theme we will make use of later.

Sequences and Series

How to Analyze a Sequence

Suppose we have the sequence 7, 11, 15, 19, 23, 27, 31, 35, 39, ... and we want to know how the rest of the sequence will play out. We can analyze its structure in various ways. This is the opposite of producing a number that answers a question. What is called for is to dig and dig into the past until we find artifacts from the buried structure; we may also apply the analogy of chemical reagents to separate substances in a solution. The most important thing is to keep sleuthing until the mystery is solved.

To solve any problem, we must collect data and then look at it through various lenses.

Tabulation

n	1	2	3	4	5	6	7	8	9	10
a_n	7	11	15	19	23	27	31	35	39	43

Numerical/Graphing

Colloquial Description

Complication/Unsimplifying/Structural Analysis/Reverse Engineering

Reformulation/Recursive Formulation/Explicit Formulation

Use the Scientific Method (or "Get Einstein on that bitch!}

Let's begin by supposing that there is a simple rule which generates the numbers in the sequence, a hypothesis if you will (hypothesize, experiment, observe, reason, question, repeat). Experiment 1: look at differences in consecutive terms: 4, 4, 4, 4, 4, .... Observations: Each number is 4 more than the previous. Conclusion 1: The sequence can be extended by simply adding 4 to the last number: 7, 7 + 4, 7 + 4 + 4, 7 + 4 + 4 + 4, .... Conclusion 2: Let a_n denote the nth term of the sequence; a_n = 7 + (n - 1)4.

Generally, when a sequence is generated by a common difference d, we have

$a_{2} - a_{1} = d$ $a_{3} - a_{2} = d$ $a_{4} - a_{3} = d$ $a_{5} - a_{4} = d$ $\vdots$ $a_{n} - a_{n - 1} = d$

which gives us

$a_{2} = a_{1} + d$ $a_{3} = a_{2} + d$ $a_{4} = a_{3} + d$ $a_{5} = a_{4} + d$ $\vdots$ $a_{n} = a_{n - 1} + d$

If we substitute the expression on the right side of each equation into the equation below it, then we see that

$a_{n} = a_{1} + (n - 1)d$

A sequence in which the forward differences Δa_i = a_{i + 1} − a_i are constant for all i, such as this one, we describe as arithmetic. When we want to add all the terms of sequence, we talk of a the series whose addends are the terms of the sequence. The sum of an arithmetic series is given by

$\sum_{i = 1}^{n}[a_{1} + (i - 1)d] = na_{1} + d\sum_{i = 1}^{n}(i - 1)$
$\text{...........................} = na_{1} + \frac{dn(n - 1)}{2}$
$\text{...........................} = \frac{n(a_{1} + [a_{1} + (n - 1)d])}{2}$
$\text{...........................} = \frac{n(a_{1} + a_{n})}{2}$

This result is celebrated to be one of the oldest and most important theorems, here. The idea is to take the average of the first and last terms, then multiply the result by the number of terms. The Babylonians are credited with its first appearance, but they did not prove it. Having had no formal deductive edifice upon which to build proofs (that we know of), they probably did something like this (without the modern symbolism, of course).

If S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10, then 2S = 10(1 + 10). Therefore, S = 10(1 + 10) / 2.

If S = 7 + 11 + 15 + 19 + 23 + 27 + 31 + 35 + 39, then 2S = 9(7 + 39). Therefore, S = 9(11 + 39) / 2.

One wonders why the Babylonians concerned themselves with what seems like a very impractical problem. It certainly suggests there was an intellectual culture developing around mathematical calculations of the time. It is possible that this sort of puzzle found its way into parlor games, gambling and perhaps ultimatlely became child's play, which would explain how the formula would come to be a rule of thumb in a culture that did not concern itself with formal mathematical proof.

They may have even used counters or geometric representations to see the relations more clearly. Such would be a great way to guide students to reconstructing the formula for themselves.

Let's turn our attention to different sequence. The forward differences in 3, 6, 12, 24, 48, 96, 192, 384, ... are not constant. In fact, as we see, Δa_i = a_i for all i. We will have to look at a different property: forward quotients. Let's define the forward quotient as follows. The forward quotient between two consecutives terms of a sequence is

$\xi a_{i} = \frac{a_{i + 1}}{a_{i}}$

Notice that ξa_i = 2 for all i. This tells us that a_n = 3·2^{n − 1}. We describe by geometric any sequence whose forward quotients are constant. More technically, a sequence of the form {arⁿ} is called geometric. Let's consider the sum of a geometric series.

$\sum_{i = 0}^{n}ar^{i}$

We begin with the simplest geometric series: 1 + 2 + 4 + 8 + 16 + 32 + ⋯ + 2ⁿ. If S = 1 + 2 + 4 + 8 + 16 + 32 + ⋯ + 2ⁿ, then 2S = 2 + 4 + 8 + 16 + 32 + 64 + ⋯ + 2^{n + 1}. We see that S − 1 = 2S − 2^{n + 1}.

$\text{It follows that } \sum_{i = 0}^{n}2^{i} = 2^{n + 1} - 1. \text{ Similarly, } \sum_{i = 0}^{n}3^{i} = \frac{3^{n + 1} - 1}{2}. \text{ More generally, for } a \neq 0 \text{ and } r > 1 \text{, } \sum_{i = 0}^{n}ar^{i} = \frac{ar^{n + 1} - a}{r - 1}.$
$\text{Proof: Let } S_{n} = \sum_{i = 0}^{n}r^{i} \text{ for all positive integers n and } r > 1 \text{, and suppose there is an integer } k \text{ such that } S_{k} = \frac{r^{k + 1} - 1}{r - 1}. \text{ Then } S_{k + 1} = S_{k} + r^{k + 1} = \frac{r^{k + 1} - 1}{r - 1} + r^{k + 1} = \frac{r^{k + 2} - 1}{r - 1}. \text{ Since } \sum_{i = 0}^{n}r^{i} = \frac{r^{n + 1} - 1}{r - 1} \text{ works for n = 1 and it works for n = k + 1 whenever it works for n = k, therefore } \sum_{i = 0}^{n}ar^{i} = \frac{ar^{n + 1} - a}{r - 1} \text{ works for all positive integers n.}$

We have seen what happens when the forward differences of a sequence are either constant or reproduce the sequence itself. What if the forward differences of a sequence are arithmetic with common difference d? That is, suppose

$\Delta_{n} = \Delta_{1} + (n - 1)d, \text { where } \Delta_{i} = \Delta a_{i}.$

By referring to those as first forward differences, we may define second forward differences in the following way.

$\Delta__{i}^{2} = \Delta_{i + 1} - \Delta_{i}.$

We see that

$\Delta__{i}^{2} = d \text{ for all } i. \text{ For the nth term, we observe } a_{n} = a_{n - 1} + \Delta_{n - 1} \text{ gives us}$
$a_{n} = a_{1} + \sum_{i = 1}^{n - 1}\Delta_{i} \text{ by means of iterated substitution. Making use of our initial assumption shows us that }$
$a_{n} = a_{1} + \sum_{i = 1}^{n - 1}[\Delta_{1} + (i - 1)d]. \text{ It is easy to show that this is a quadratic function in n, using only the first two terms of the sequence and the constant second forward difference. In particular,}$
$a_{n} = \frac{d}{2} n^{2} + \left ( a_{2} - a_{1} - \frac{3}{2}d \right )n + 2a_{1} - a_{2} + d.$

Let d be a constant. We have seen that

$\Delta_{i} = d \text{ for i = 1, ..., n if and only if } a_{n} \text{ is linear in n, and}$
$\Delta_{i}^{2} = d \text{ for i = 1, ..., n if and only if } a_{n} \text{ is quadratic in n.}$
$\text{Could it be that } \Delta_{i}^{k} = d \text{ for i = 1, ..., n if and only if } a_{n} \text{ is a polynomial of degree k in n?}$

Let's take a look at the triangular numbers.

n	1	2	3	4	5	6	7	8	9	10
a_n	1	3	6	10	15	21	28	36	45	55
Δ_n	2	3	4	5	6	7	8	9	10	---
Δ²_n	1	1	1	1	1	1	1	1	---	---

Theorem: For a sequence {a_n}, a_n is a quadratic in n if and only if Δ²_i is constant for all i from 1 to n.

How to Analyze a Series

Partial Sums/Convergence/Special Functions

matHUMANtics Series

The Oldest Math Book?

What is a series?

What is a sequence?

What are the next ten terms of the sequence?

What is the 100th term of the sequence?

What is a sequence?

What is a sequence?

What is a sequence?

Sequences and Series

How to Analyze a Sequence

Tabulation

Numerical/Graphing

Colloquial Description

Complication/Unsimplifying/Structural Analysis/Reverse Engineering

Reformulation/Recursive Formulation/Explicit Formulation

Use the Scientific Method (or "Get Einstein on that bitch!}

How to Analyze a Series