Lecture 16: Mu and Convolution


Today we began class by discussing some of the relevant details for the upcoming Group Projects. We then talked a little more about odd perfect numbers. The majority of the class was focused on defining convolution of arithmetic functions, viewing some of the results we already know in the light of these convolutions, and then using an important convolution identity to state and prove the Mőbius Inversion Formula (MIF).

Group Projects

I started by assigning each of you to a group and then discussed the "ground rules" for the group projects. You can find more information here or by checking out relevant deadlines on the course calendar.

More on Odd Perfect Numbers

We'll wrap up last class period's discussion of odd perfect numbers by giving a theorem which says that an odd perfect number needs to have "lots" of prime factors. In class "lots" will mean "at least 3," but this can be jacked up quite a bit with more sophisticated means.

In order to make progress in this direction, we'll need the following

Lemma: For any integer n,

$\displaystyle \frac{\sigma(n)}{n} < \prod_{p\mid n}\frac{p}{p-1}.$

Proof: To see this is true, notice that if $n = p_1^{a_1}\cdots p_k^{a_k}$, then we have

\begin{align} \begin{split} \frac{\sigma(n)}{n} &= \frac{\prod_{i=1}^k\sigma(p_i^{a_i})}{p_i^{a_i}} = \frac{\prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p_i-1}}{\prod_{i=1}^k p^{a_i}}\\ &= \prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p^{a_i}(p_i-1)}= \prod_{i=1}^k \frac{p_i-\left(\frac{1}{p_i}\right)^{a_i}}{p_i-1}< \prod_{i=1}^k \frac{p_i}{p_i-1}. \end{split} \end{align}


Now we're ready to prove our

Theorem: If n is an odd perfect number, then n has at least 3 distinct prime factors.

Proof: Suppose first that n is an odd perfect number so that $n = p^a$, where p is prime. Then we get

\begin{align} \frac{\sigma(n)}{n} = 2 \end{align}

since n is perfect. But by our theorem we also have

\begin{align} \frac{\sigma(n)}{n} < \frac{p}{p-1}. \end{align}

Now the function $f(x) = \frac{x}{x-1}$ is decreasing on the interval $(1,\infty)$ (you can graph the function or take a derivative, if you like), and so the quantity on the right hand side of this last equation is made largest when the prime p is as small as possible. Since $p=3$ is the smallest possible prime (this is an odd number, after all), we have

\begin{align} \frac{\sigma(n)}{n} < \frac{p}{p-1} \leq \frac{3}{3-1} = 1.5 <2. \end{align}

This is a clear contradiction.

Suppose, then, that $n = p^aq^b$ for distinct primes p,q. Then Equation (2) still holds (since n is still perfect). But our previous lemma now says

\begin{align} \frac{\sigma(n)}{n} < \frac{p}{p-1}\frac{q}{q-1}. \end{align}

Again, the quantity on the right hand side is maximized when p and q are chosen as small as possible. Hence we have

\begin{align} \frac{\sigma(n)}{n} < \frac{p}{p-1}\frac{q}{q-1} \leq \frac{3}{3-1}\frac{5}{5-1} = \frac{15}{8} < 2. \end{align}

Again we are left with a contradiction.

We see that both of these scenarios are impossible, and so our odd number cannot have exactly 1 or 2 prime divisors. This means that if an odd perfect number exists, it has at least 3 distinct prime divisors. $\square$

The Mu Function

We're going to change gears now and talk more about arithmetic functions. In particular we're going to define a new operation on arithmetic functions which provides a new way of combining two arithmetic functions into another arithmetic function. To start, though, we're going to establish some notation for functions we've already been talking about.

Example: Some Arithmetic Functions

One of the functions we're already talked about is the function which raises an integer to a (fixed) power. We'll now give this function a name: $P_k(n)$ is the function defined by

\begin{equation} P_k(n) = n^k. \end{equation}

Notice that when $k=0$ this is the function is the constant function 1.

Another important arithmetic function is called the identity function, which is defined as

\begin{align} I(n) = \left\{\begin{array}{ll}1 &, \mbox{ if }n=1\\0 &, \mbox{ if }n>1.\end{array}\right. \end{align}

You can verify that this function is not just arithmetic, but is also multiplicative.

Finally, we define the function $\mu$ as

\begin{align} \mu(n) = \left\{\begin{array}{ll}1 &,\mbox{ if }n=1,\\0 &,\mbox{ if }p^2 \mid n \mbox{ for some prime }p,\\(-1)^r &, \mbox{ if }n=p_1\cdots p_r \mbox{ with each }p_i \mbox{ a distinct prime}.\end{array}\right. \end{align}


Even though it seems really strange, this $\mu$ function is actually quite important. We'll be investigating many properties of this function, but let's start by noting that

Lemma: The $\mu$ function is multiplicative.

Proof: We need to show that if $(m,n) = 1$ then $\mu(mn) = \mu(m)\mu(n)$. For this, suppose first that $mn=1$, so that $m=n=1$ as well. Then we have

\begin{align} \mu(mn) = 1 = 1\cdot 1=\mu(m)\mu(n). \end{align}

Otherwise we have $mn>1$, in which case mn has a prime factorization. Notice that if $m = p_1^{a_1}\cdots p_k^{a_k}$ and $n = q_1^{b_1}\cdots q_s^{b_s}$ then the factorization for mn is given by $mn = p_1^{a_1}\cdots p_k^{a_k} q_1^{b_1}\cdots q_s^{b_s}$, and that moreover since $(m,n) = 1$ none of the primes $p_i$ overlap with any of the primes $q_j$.

Suppose now that some prime p has $p^2 \mid mn$, then either $p^2 \mid m$ or $p^2 \mid m$ (since this prime p will correspond to either one of the $p_i$'s or one of the $q_j$'s). If this is the case, then we have one of $\mu(m)$ or $\mu(n)$ is zero. If $mu(m)=0$, though, we then have

\begin{align} \mu(mn) = 0 = 0\cdot \mu(n) = \mu(m)\mu(n), \end{align}

so that our desired equality holds. If instead $\mu(n)=0$, then a similar equality again shows $\mu(mn) = \mu(m)\mu(n)$.

The only case we have left is if all the exponents in the prime factorization for mn are 1. But in this case, we see that $\mu(mn) = (-1)^{k+s}$, since the number of prime factors of mn is $k+s$. But we also know that $\mu(n) = k$ and $\mu(m) = s$, so that

\begin{align} \mu(m)\mu(n) = (-1)^s(-1)^k = (-1)^{k+s} = \mu(mn). \end{align}


To get a little practice with the $\mu$ function, we'll prove the following

Lemma: For any integer n,

$\sum_{d \mid n} \mu(d) = I(n)$.

Proof: To see this is true, notice first that it holds true for $n=1$:

\begin{align} \sum_{d \mid 1} \mu(d) = \mu(1) = 1 = I(1). \end{align}

Since $\mu$ is a multiplicative function, then, we can verify this identity for $n>1$ by showing that $\mu(p^a) = I(p^a)$ for any prime p and any positive exponent a. Since $I(p^a) = 0$, this means we need to verify that $\mu(p^a) = 0$.

Now for this, notice that

\begin{split} \sum_{d\mid p^a} \mu(d) = \sum_{i=0}^a \mu(p^i) &= \mu(1)+\mu(p) + \mu(p^2)+\mu(p^3)+\cdots+\mu(p^a)\\&= 1-1+0+0+\cdots+0 = 0. \end{split}


Example: The Identity $I(n) = \sum_{d \mid n} \mu(d)$ in Action

Let's just compute $\sum_{d \mid n} \mu(d)$ for a specific n to see how the above lemma "really" works. We'll pick $n=12$, which has divisors $\{1,2,3,4,6,12\}$. With this information, we get

\begin{align} \sum_{d\mid 12}\mu(d) = \mu(1)+\mu(2)+\mu(3)+\mu(4)+\mu(6)+\mu(12). \end{align}

Now we know that $\mu(1) = 1$ by definition, and we know that if $p^2 \mid d$ for some prime $p$ then $\mu(d) = 0$. In particular, we know $\mu(4) = \mu(12) = 0$, since both 4 and 12 are divisible by 22. Finally, since 2 and 3 both have a single prime appearing in their factorization, this means $\mu(2) = \mu(3) = (-1)^1 = -1$, whereas because 6 has two primes in its factorization we get $\mu(6) = (-1)^2 = 1$. Putting this all together, we get

\begin{align} \sum_{d \mid 12} \mu(d) = \mu(1)+\mu(2)+\mu(3)+\mu(4)+\mu(6)+\mu(12) = 1-1-1+0+1-0 = 0. \end{align}



Now that we've met a few more arithmetic functions, we're going to talk about a new way of combining arithmetic functions.

Definition: For two arithmetic functions f and g, the convolution of f and g, written $f*g$, is defined as the function

$\displaystyle (f*g)(n) = \sum_{d\mid n}f\left(\frac{n}{d}\right)g(d).$

Example: Convolution in Practice

Though we didn't use this language before, it turns out we've spent a lot of time thinking about convolutions so far. For instance, we know that

\begin{align} \nu(n) = \sum_{d \mid n} 1. \end{align}

But notice that

\begin{align} \sum_{d\mid n}1 = \sum_{d \mid n}1\cdot 1 = \sum_{d \mid n}\left(\frac{n}{d}\right)^0 \left(d\right)^0 = \sum_{d\mid n} P_0\left(\frac{n}{d}\right)P_0(d) = (P_0 * P_0)(n). \end{align}

Hence we have $\nu = P_0 * P_0$.

Similarly we know that

\begin{align} \sigma(n) = \sum_{d \mid n} d. \end{align}

But this can be rewritten as

\begin{align} \sum_{d\mid n} d = \sum_{d \mid n}\left(\frac{n}{d}\right)^0\left(d\right)^1 = \sum_{d \mid n}P_0\left(\frac{n}{d}\right)P_1(d) = (P_0 * P_1)(n), \end{align}

and so $\sigma = P_0 * P_1$. $\square$

These are both special cases of the following

Lemma: For any integer k, $\sigma_k = P_0 * P_k$.

Proof: We can rewrite the formula

\begin{align} \sigma_k(n) = \sum_{d \mid n} d^k. \end{align}


\begin{align} \sum_{d\mid n} d^k = \sum_{d \mid n}\left(\frac{n}{d}\right)^0\left(d\right)^k = \sum_{d \mid n}P_0\left(\frac{n}{d}\right)P_k(d) = (P_0 * P_k)(n). \end{align}

This gives $\sigma_k = P_0 * P_k$.

Some Important Convolution Identities

One of the important facts about convolution is that the identity function we defined at the beginning of class acts as "the identity for convolution." The meaning of this phrase is captured in the next

Lemma: For any arithmetic function f, we have $f * I = f$.

Proof: To verify this identity, we'll just try to compute

\begin{align} (f*I)(n) = \sum_{d \mid n} f\left(\frac{n}{d}\right)I(d). \end{align}

Now notice that whenever $d>1$ we have $I(d) = 0$, and hence the sum on the right hand side of the previous equation has only one non-zero term (the term corresponding to $d=1$). Hence we have

\begin{align} (f*I)(n) = \sum_{d \mid n} f\left(\frac{n}{d}\right)I(d) = f\left(\frac{n}{1}\right)I(1) = f(n). \end{align}


This last lemma is how the identity function earned its name: it is the unique function which leaves all other functions fixed after convolution. So much in the way that 1 is the multiplicative identity because the product of 1 and any given number x is just $1\cdot x = x$, the identity function is the identity under convolution because for any arithmetic function f, the convolution of f and I is just f.

There are some other properties about convolution that are important. We didn't get a chance to prove them in class, but they are summarized in the following

Lemma: For any arithmetic functions f,g and h, we have

(1) $\displaystyle f*g = g*f$(commutativity of convolution)
(2) $\displaystyle (f*g)*h = f*(g*h)$ (associativity of convolution)

Also, if f and g are both multiplicative, so too is $f*g$.

These rules mean that we can do some fancy "algebra" with convolutions that makes some otherwise tricky identities easy to prove. The key to these kinds of tricks is the the following

Lemma: $I(n) = (P_0 * \mu)(n)$

Proof: We have already seen that $I(n) = \sum_{d \mid n}\mu(d)$. But notice that

\begin{align} \sum_{d \mid n}\mu(d) = \sum_{d \mid n}1\cdot \mu(d) = \sum_{d\mid n}\left(\frac{n}{d}\right)^0\mu(d) = \sum_{d\mid n}P_0\left(\frac{n}{d}\right)\mu(d) = (P_0 * \mu)(n). \end{align}

This is the desired equality. $\square$

Though it looks bizarre, this identity is the key to the so-called Mobius Inversin Formula, a result that gives us new ways to express identities amongst arithmetic functions.

We'll state and prove the Mobius Inversion Formula next class period, but here's what the theorem says

Theorem (Mobius Inversion Formula): For arithmetic functions f and g,

$\displayestyle f(n) = \sum_{d \mid n}g(d) \quad \mbox{ if and only if } \quad g(n) = \sum_{d \mid n} \mu\left(\frac{n}{d}\right) f(d).$

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