NTIC Average of Tau

Subsection 20.2.1 Beginnings

Let's begin by observing Figure 20.2.1, which plots the average for

τ

up to

n = 100 .

Average of $\tau\text{,}$ number of divisors — 🔗
Figure 20.2.1. Average of $τ,$ number of divisors

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Sage note 20.2.2. Try to be efficient.

Observe the following two cells. The first cell records the successive sums of $τ$ in a variable out (for ‘output’), so that we don't have to recalculate the entire sum each time we compute the average value for a different input value. We record the actual averages sequentially in a separate list ls.

Then the interactive cell is very simple indeed. Try being efficient in your programming!

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def L(n):
    ls = []
    out = 0
    for i in range(1,n+1):
        out += sigma(i,0)
        ls.append((i,out/i))
    return ls

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@interact
def _(n=100):
    P = line(L(n))
    show(P)

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These graphics shows how the average value of

τ

up to

n

changes as we let

n

get bigger. This isn't enough data to tell whether there is a limiting value for the average value of

τ (n),

even if you look out to the first 1000 integers, but it's suggestive. Part of the unpredictability is from primes; every prime number contributes just 2 to the total (and so reduces the average value)!

🔗

Nonetheless, thinking about this might lead us to look a little deeper. For example, the ‘trend’ is concave down. So let's look at comparing it with various concave down functions. (The following interact supports multiplied constants with them as well.)

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@interact
def _(n=100,C=.5,f=[x^(1/2), x, x^(1/3), x^(1/4), log(x), log(log(x)), x^(1.5), x^2]):
    f(x) = f
    P = line(L(n),legend_label=r'average of $\tau$')
    P += plot(C*f,(x,1,n), color='black', linestyle='--', legend_label='$%s%s$'%(RDF(C),latex(f(x))))
    show(P)

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At the very least I can estimate that the average value is Big Oh of a certain function. But how does it go on?

Average of $\tau$ to one million — 🔗
Figure 20.2.3. Average of $τ$ to one million

🔗

In Figure 20.2.3 we have our graph of averages of

τ (n)

versus

n,

out to one million. Certainly this looks akin to some fractional exponent function. On the other hand, it seems to grow more slowly than

\sqrt{x} = x^{1 / 2},

our initial estimate in the interact, so if it is, the exponent must be pretty small. (If you are familiar with semilog or log-log plots and are willing to look up how to do them in Sage, see Exercise 20.6.7 and then try to plot this on those axes.)

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Subsection 20.2.2 Heuristics for tau

🔗

We'll start with a heuristic, going right back to the sieve of Eratosthenes.

🔗

In that algorithm (6.2.3), we proved that in order to test whether

n

is prime, you just have to check all numbers up through

\sqrt{n} .

This is because any divisor

\sqrt{n} < d < n

implies the existence of a divisor

\frac{n}{d}

such that

1 = \frac{n}{n} < \frac{n}{d} < \frac{n}{\sqrt{n}} = \sqrt{n} .

🔗

So the absolute most number of divisors possible (for a given

n

) is if every number

d

less than

\sqrt{n}

was a divisor, and then all the

\frac{n}{d} > \sqrt{n}

you get were also divisors.

🔗

This is a silly idea beyond such small

n,

but let's go with it anyway. Even if all those divisors were there, you would have

τ (n) = 2 ⌊ \sqrt{n} ⌋ \leq 2 \sqrt{n}

so that

τ (n)

is

O (\sqrt{n}) .

🔗

Example 20.2.4.

For $n = 24$ this idea is actually true. We can line these up in pairs as $(1, 24),$ $(2, 12),$ $(3, 8),$ $(4, 6),$ and that gives $2 \cdot ⌊ \sqrt{24} ⌋ = 8$ total divisors.

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That estimate is very important! It means we can get a sense of a first bound on the average value of

τ .

At the very least we have that

\frac{1}{n} \sum_{k = 1}^{n} τ (k) \leq \frac{1}{n} \sum_{k = 1}^{n} 2 \sqrt{k} .

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Subsection 20.2.3 Using sums to get closer

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Let's rewrite this inequality in a more suggestive form by noting

k = n (k / n) :

\frac{1}{n} \sum_{k = 1}^{n} τ (k) \leq \sum_{k = 1}^{n} \frac{1}{n} 2 \sqrt{n (k / n)} .

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This form looks an awful lot like a Riemann sum with

x = k / n

and

Δ x = \frac{1}{n} .

To review, recall writing a Riemann sum for

\int_{0}^{1} x^{2} d x

in the form

\frac{1}{n} {(\frac{1}{n})}^{2} + \frac{1}{n} {(\frac{2}{n})}^{2} + \dots + \frac{1}{n} {(\frac{n}{n})}^{2} .

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(If you need a calculus refresher, there are several great free calculus texts in the American Institute of Mathematics list of approved textbooks.)

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Doing the same type of summation for the function

2 \sqrt{n x}

would give

\sum_{k = 1}^{n} \frac{1}{n} 2 \sqrt{n (k / n)} \approx \int_{0}^{1} 2 \sqrt{n x} d x = 2 \sqrt{n} \int_{0}^{1} \sqrt{x} d x = \frac{4}{3} \sqrt{n} .

🔗

That certainly suggests that the average of

τ

might be

O (\sqrt{n})

with

C = 4 / 3 .

🔗

To make this rigorous, we will need to make a slight change of point of view in order to ensure it will be viewed as a left-hand sum of an increasing function (and hence the Riemann sum is less than the actual value of the integral).

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Namely, consider that

\frac{1}{n} \sum_{k = 1}^{n} 2 \sqrt{k} = \sum_{k = 0}^{n - 1} (\frac{1}{n}) 2 \sqrt{k + 1} = \sum_{k = 0}^{n - 1} (\frac{1}{n}) 2 \sqrt{n (k / n) + 1} \leq \int_{0}^{1} 2 \sqrt{n x + 1} d x

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This integral evaluates to

\frac{4}{3} \sqrt{n} [{(1 + \frac{1}{n})}^{3 / 2} - {(\frac{1}{n})}^{3 / 2}] .

🔗

The big extra factor on the right can be shown to be decreasing as a function of

n

(using derivatives), and hence is always less than

2

for positive integers (plug in

n = 1

to see), so the entire expression will always be less than

\frac{8}{3} \sqrt{n} .

🔗

Thus one can write

\frac{1}{n} \sum_{k = 1}^{n} τ (k) \leq \frac{1}{n} \sum_{k = 1}^{n} 2 \sqrt{k} \leq \frac{8}{3} \sqrt{n}

🔗

so that the average value is bounded by a constant times

\sqrt{n}

and is hence

O (\sqrt{n}) .

This implies, perhaps, that the average number of divisors goes steadily up! (If so, it guarantees that the trend is, on the whole, concave down.)

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Subsection 20.2.4 But Big-Oh isn't enough

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However, we might also want to know what the average value of

τ

is. The preceding subsections only tell us what it's less than! In the next interact, it seems that it's hard to find the “right” value of

C

so that the average value would be the same order as

\sqrt{n} .

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def L(n):
    ls = []
    out = 0
    for i in range(1,n+1):
        out += sigma(i,0)
        ls.append((i,out/i))
    return ls
​
P = line(L(1000000))
​
@interact
def _(a=.02,n=2):
    show(P + plot(a*x^(1/n), (x,1,10^6), color='red',linestyle='--'))
    pretty_print(html(r"Blue is the average value of $\tau$"))
    pretty_print(html("Red is $%sx^{1/%s}$"%(a,n)))

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Try

x^{1 / 3}

in the interact; it doesn't seem to make matters any better.

🔗

In fact, one can show that

τ (n) = O (\sqrt[3]{n})

as well. Here are the steps one might take. We make fleshing out the details Exercise 20.6.10 (adapted from [E.4.5, Section 3.5]):

First, note that $τ$ is multiplicative.
For a given prime $p,$ note that $τ (p^{x}) = x + 1$ grows much more slowly than ${(p^{x})}^{1 / 3} = p^{x / 3},$ which is exponential in $x .$
- What value do each of these have at $x = 0 ?$
- Take derivatives of both functions at $x = 0$ to show that the growth statement is definitely true for $p \geq 23 .$
- Show that for each prime $p$ less than $23$ there is an $x_{p}$ such that the growth statement is true after $x_{p} .$
Put these pieces of information together to show that $τ$ is $O (x^{1 / 3}) .$

Number Theory: In Context and Interactive

Section 20.2 Average of Tau