Lecture 16 - Mu and Convolution


We started today's class by introducing a few new arithmetic functions. When then spent the balance of the class period 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 the MÅ‘bius Inversion Formula (MIF).

Some new arithmetic functions

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

We're going to define some new functions. The first 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}



Notice that $\mu(6) = \mu(2\cdot 3) = (-1)^2 = 1$, since each prime factor of 6 appears with exponent 1.

On the other hand, $\mu(12) = \mu(2^2\cdot 3) = 0$ since the prime factor 2 appears to the exponent 2 (and $2>1$). $\square$

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}


Notice that since $\mu(n)$ is multiplicative, we know that

\begin{align} F(n) = \sum_{d \mid n} \mu(d) \end{align}

is also multiplicative.

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

Let's compute the values of this function for a few numbers:

\begin{split} F(5) &= \sum_{d \mid 5} \mu(d) = \mu(1)+\mu(5) = 1 + (-1) = 0\\ F(12) &= \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\\ F(15) &= \sum_{d \mid 15}\mu(d) = \mu(1)+\mu(3)+\mu(5)+\mu(15) = 1-1-1+1 =0. \end{split}

Notice that it seems we're getting mostly 0 out of this function F. Aviva noticed, however, that we don't always get 0: $F(1) = 1$.

Our computation hints at the fact

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 $F(p^a) = I(p^e)$ for any prime p and any positive exponent e. Once we have this, then for $n = p_1^{e_1}\cdots p_k^{e_k}$, we'll have

\begin{align} F(n) = F(p_1^{e_1}) \cdots F(p_k^{e_k}) = I(p_1^{e_1})\cdots I(p_k^{e_k}) = I(p_1^{e_1}\cdots p_k^{e_k}) = I(n). \end{align}

Notice that since $I(p^a) = 0$, this means we need to verify that $\mu(p^a) = 0$.

To verify this, let's compute:

\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}



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: A specific convolution

Let's evaluation $(\sigma * \phi)(6)$. By definition, we have

\begin{align} (\sigma*\phi)(6) = \sum_{d \mid 6}\sigma\left(\frac{6}{d}\right) \phi(d). \end{align}

Of course the divisors of 6 are just 1, 2, 3 and 6, so we have

\begin{split} (\sigma*\phi)(6) &= \sigma\left(\frac{6}{1}\right)\phi(1)+\sigma\left(\frac{6}{2}\right)\phi(2)+\sigma\left(\frac{6}{3}\right)\phi(3)+\sigma\left(\frac{6}{6}\right)\phi(6)\\ &= 12\cdot 1 + 4\cdot 1 + 3\cdot 2 + 1\cdot 2 = 24. \end{split}


As it turns out, we've seen convolutions in other places before; we just didn't realize they were convolutions.

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

Perhaps the most important convolution identity to remember is the following

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

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.

Some Algebraic Facts about Convolution

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.

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 $f*g$.

Mobius Inversion

These rules mean that we can do some fancy "algebra" with convolutions that makes some otherwise tricky identities easy to prove. We'll see that these rules let us prove the next result

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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