Proofs involving the totient function

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This page provides proofs for identities involving the totient function $\varphi(k)$ and the Möbius function $μ(k)$ .

1 Sum of integers relatively prime to and less than or equal to n
2 Proofs of totient identities involving the floor function
3 Average order of the totient
- 3.1 Average order of φ(n)/n
4 Inequalities
5 External links

[edit] Sum of integers relatively prime to and less than or equal to n

Claim:

$\sum_{1\le k\le n \atop {\gcd(k,n)=1}} k = \frac{1}{2} \, \varphi(n) \, n \quad \text{ for integers } n \ge 2.$

Since $\gcd(n,n)\ne 1$ and $gcd(k, n) = gcd(n - k, n)$ for all integers $k$ , a change of index yields

$\sum_{1\le k\le n \atop {\gcd(k,n)=1}} k \,= \sum_{1\le k\le n \atop {\gcd(k,n)=1}} (n-k)$ .

Therefore

$2\sum_{1\le k\le n \atop {\gcd(k,n)=1}} k \,= \sum_{1\le k\le n \atop {\gcd(k,n)=1}} (k+n-k) = \varphi(n)\,n$ .

[edit] Proofs of totient identities involving the floor function

The proof of the identity

$\sum_{k=1}^n\frac{\varphi(k)}{k} = \sum_{k=1}^n\frac{\mu(k)}{k}\left\lfloor\frac{n}{k}\right\rfloor$

is by mathematical induction on $n$ . The base case is $n = 1$ and we see that the claim holds:

$\varphi(1)/1 = 1 = \frac{\mu(1)}{1} \left\lfloor 1\right\rfloor.$

For the induction step we need to prove that

$\frac{\varphi(n+1)}{n+1} = \sum_{k=1}^n\frac{\mu(k)}{k} \left(\left\lfloor\frac{n+1}{k}\right\rfloor - \left\lfloor\frac{n}{k}\right\rfloor\right) + \frac{\mu(n+1)}{n+1}.$

The key observation is that

$\left\lfloor\frac{n+1}{k}\right\rfloor - \left\lfloor\frac{n}{k}\right\rfloor \; = \; \begin{cases} 1, & \mbox{if }k|(n+1) \\ 0, & \mbox{otherwise, } \end{cases}$

so that the sum is

$\sum_{k|n+1,\; k<n+1} \frac{\mu(k)}{k} + \frac{\mu(n+1)}{n+1} = \sum_{k|n+1} \frac{\mu(k)}{k}.$

Now the fact that

$\sum_{k|n+1} \frac{\mu(k)}{k} = \frac{\varphi(n+1)}{n+1}$

is a basic totient identity. To see that it holds, let $p_1^{v_1} p_2^{v_2} \ldots p_q^{v_q}$ be the prime factorization of n+1. Then

$\frac{\varphi(n+1)}{n+1} = \prod_{l=1}^q \left( 1 - \frac{1}{p_l} \right) = \sum_{k|n+1} \frac{\mu(k)}{k}$

by definition of $μ(k).$ This concludes the proof.^{[citation needed]}

An alternate proof proceeds by substituting $\frac{\varphi(k)}{k} = \sum_{d|k}\frac{\mu(d)}{d}$ directly into the left side of the identity, giving $\sum_{k=1}^n \sum_{d|k}\frac{\mu(d)}{d}.$

Now we ask how often the term $\begin{matrix}\frac{\mu(d)}{d}\end{matrix}$ occurs in the double sum. The answer is that it occurs for every multiple $k$ of $d$ , but there are precisely $\begin{matrix}\left\lfloor\frac{n}{d}\right\rfloor\end{matrix}$ such multiples, which means that the sum is

$\sum_{d=1}^n\frac{\mu(d)}{d}\left\lfloor\frac{n}{d}\right\rfloor$

as claimed.^{[citation needed]}

The trick where zero values of^{[citation needed]} $\begin{matrix} \left\lfloor\frac{n+1}{k}\right\rfloor - \left\lfloor\frac{n}{k}\right\rfloor \end{matrix}$ are filtered out may also be used to prove the identity

$\sum_{k=1}^n\varphi(k) = \frac{1}{2}\left(1+ \sum_{k=1}^n \mu(k)\left\lfloor\frac{n}{k}\right\rfloor^2\right).$

The base case is $n = 1$ and we have

$\varphi(1) = 1 = \frac{1}{2} \left(1+ \mu(1) \left\lfloor\frac{1}{1}\right\rfloor^2\right)$

and it holds. The induction step requires us to show that

$\varphi(n+1) = \frac{1}{2} \sum_{k=1}^n \mu(k) \left( \left\lfloor\frac{n+1}{k}\right\rfloor^2 - \left\lfloor\frac{n}{k}\right\rfloor^2 \right) \; + \; \frac{1}{2} \; \mu(n+1) \; \left\lfloor\frac{n+1}{n+1}\right\rfloor^2 .$

Next observe that^{[citation needed]}

$\left\lfloor\frac{n+1}{k}\right\rfloor^2 - \left\lfloor\frac{n}{k}\right\rfloor^2 \; = \; \begin{cases} 2\frac{n+1}{k} - 1, & \mbox{if }k|(n+1) \\ 0, & \mbox{otherwise.} \end{cases}$

This gives the following for the sum

$\frac{1}{2} \sum_{k|n+1, \; k<n+1} \mu(k) \left( 2\frac{n+1}{k} - 1 \right) \; + \; \frac{1}{2} \; \mu(n+1) = \frac{1}{2} \sum_{k|n+1} \mu(k) \left( 2 \; \frac{n+1}{k} - 1 \right).$

Treating the two inner terms separately, we get

$(n+1) \sum_{k|n+1} \frac{\mu(k)}{k} - \frac{1}{2} \sum_{k|n+1} \mu(k).$

The first of these two is precisely $\varphi(n+1)$ as we saw earlier, and the second is zero, by a basic property of the Möbius function (using the same factorization of $n + 1$ as above, we have $\begin{matrix} \sum_{k|n+1} \mu(k) = \prod_{l=1}^q (1-1) = 0 \end{matrix}$ .) This concludes the proof.

This result may also be proved by inclusion-exclusion. Rewrite the identity as

$-1 + 2\sum_{k=1}^n\varphi(k) = \sum_{k=1}^n \mu(k)\left\lfloor\frac{n}{k}\right\rfloor^2.$

Now we see that the left side counts the number of lattice points (a, b) in [1,n] × [1,n] where a and b are relatively prime to each other. Using the sets $A p$ where p is a prime less than or equal to n to denote the set of points where both coordinates are divisible by p we have

$\left| \bigcup_p A_p \right| = \sum_p \left| A_p \right| \; - \; \sum_{p<q} \left| A_p \cap A_q \right| \; + \; \sum_{p<q<r} \left| A_p \cap A_q \cap A_r \right| \; - \; \cdots \; \pm \; \left| A_p \cap \; \cdots \; \cap A_s \right|.$

This formula counts the number of pairs where a and b are not relatively prime to each other. The cardinalities are as follows:

$\left| A_p \right| = \left\lfloor \frac{n}{p} \right\rfloor^2, \; \left| A_p \cap A_q \right| = \left\lfloor \frac{n}{pq} \right\rfloor^2, \; \left| A_p \cap A_q \cap A_r \right| = \left\lfloor \frac{n}{pqr} \right\rfloor^2, \; \ldots$

and the signs are $\begin{matrix}-\mu(pqr\cdots)\end{matrix}$ , hence the number of points with relatively prime coordinates is

$\mu(1)\, n^2 \; + \; \sum_p \mu(p) \left\lfloor \frac{n}{p} \right\rfloor^2 \; + \; \sum_{p<q} \mu(p q) \left\lfloor \frac{n}{pq} \right\rfloor^2 \; + \; \sum_{p<q<r} \mu(p q r) \left\lfloor \frac{n}{pqr} \right\rfloor^2 \; + \; \cdots$

but this is precisely $\sum_{k=1}^n \mu(k) \left\lfloor \frac{n}{k} \right\rfloor^2$ and we have the claim.

[edit] Average order of the totient

We will use the last formula of the preceding section to prove the following result:

$\frac{1}{n^2} \sum_{k=1}^n \varphi(k)= \frac{3}{\pi^2} + \mathcal{O}\left(\frac{\log n }{n}\right)$

Using $x-1 < \lfloor x \rfloor \le x$ we have the upper bound

$\frac{1}{2 n^2} \left(1+ \sum_{k=1}^n \mu(k)\frac{n^2}{k^2}\right) = \frac{1}{2 n^2} + \frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2}$

and the lower bound

$\frac{1}{2 n^2} \left(1+ \sum_{k=1}^n \mu(k)\left(\frac{n^2}{k^2} - 2\frac{n}{k} + 1\right)\right)$

which is

$\frac{1}{2 n^2} + \frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2} - \frac{1}{n} \sum_{k=1}^n \frac{\mu(k)}{k} + \frac{1}{2 n^2} \sum_{k=1}^n \mu(k)$

Working with the last two terms and using the asymptotic expansion of the nth harmonic number we have

$- \frac{1}{n} \sum_{k=1}^n \frac{\mu(k)}{k} > - \frac{1}{n} \sum_{k=1}^n \frac{1}{k} = - \frac{1}{n} H_n > -\frac{1}{n} (\log n +1)$

and

$\frac{1}{2 n^2} \sum_{k=1}^n \mu(k) > - \frac{1}{2 n}.$

Now we check the order of the terms in the upper and lower bound. The term $\begin{matrix}\sum_{k=1}^n \frac{\mu(k)}{k^2}\end{matrix}$ is $\mathcal{O}(1)$ by comparison with $ζ(2)$ , where $ζ(s)$ is the Riemann zeta function. The next largest term is the logarithmic term from the lower bound.

So far we have shown that

$\frac{1}{n^2} \sum_{k=1}^n \varphi(k)= \frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2} + \mathcal{O}\left(\frac{\log n }{n}\right)$

It remains to evaluate $\begin{matrix}\sum_{k=1}^n \frac{\mu(k)}{k^2}\end{matrix}$ asymptotically, which we have seen converges. The Euler product for the Riemann zeta function is

$\zeta(s) = \prod_p \left(1 - \frac{1}{p^s} \right)^{-1} \mbox{ for } \Re(s)>1.$

Now it follows immediately from the definition of the Möbius function that

$\frac{1}{\zeta(s)} = \prod_p \left(1 - \frac{1}{p^s} \right) = \sum_{n \ge 1} \frac{\mu(n)}{n^s}.$

This means that

$\frac{1}{2} \sum_{k=1}^n \frac{\mu(k)}{k^2} = \frac{1}{2} \frac{1}{\zeta(2)} + \mathcal{O}\left(\frac{1}{n}\right)$

where the integral $\begin{matrix} \int_{n+1}^\infty \frac{1}{t^2} dt\end{matrix}$ was used to estimate $\begin{matrix} \sum_{k>n} \frac{\mu(k)}{k^2}\end{matrix}.$ But $\begin{matrix}\frac{1}{2} \frac{1}{\zeta(2)} = \frac{3}{\pi^2}\end{matrix}$ and we have established the claim.

[edit] Average order of φ(n)/n

The material of the preceding section, together with the identity

$\sum_{k=1}^n\frac{\varphi(k)}{k} = \sum_{k=1}^n\frac{\mu(k)}{k}\left\lfloor\frac{n}{k}\right\rfloor$

also yields a proof that

$\frac{1}{n} \sum_{k=1}^n \frac{\varphi(k)}{k} = \frac{6}{\pi^2} + \mathcal{O}\left(\frac{\log n }{n}\right).$

Reasoning as before, we have the upper bound

$\frac{1}{n} \sum_{k=1}^n\frac{\mu(k)}{k}\frac{n}{k} = \sum_{k=1}^n \frac{\mu(k)}{k^2}$

and the lower bound

$-\frac{1}{n} \sum_{k=1}^n\frac{\mu(k)}{k} + \sum_{k=1}^n \frac{\mu(k)}{k^2}.$

Now apply the estimates from the preceding section to obtain the result.

[edit] Inequalities

We first show that

$\lim \inf \frac{\varphi (n)}{n}=0 \mbox{ and } \lim \sup \frac{\varphi (n)}{n}=1.$

The latter holds because when n is a power of a prime p, we have

$\frac{\varphi (n)}{n} = 1 - \frac{1}{p},$

which gets arbitrarily close to 1 for p large enough (and we can take p as large as we please since there are infinitely many primes).

To see the former, let n_k be the product of the first k primes, call them $p 1, p 2,..., p k$ . Let

$r_k = \frac{\varphi (n_k)}{n_k} = \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right).$

Then

$\frac{1}{r_k} = \prod_{i=1}^k \left( 1 - \frac{1}{p_i} \right)^{-1} > \sum_{m=1}^{p_k} \frac{1}{m} = H_{p_k},$

a harmonic number. Hence, by the well-known bound $H n > log n$ , we have

$\frac{1}{r_k} > \log p_k.$

Since the logarithm is unbounded, taking k arbitrarily large ensures that r_k achieves values arbitrarily close to zero.

We use the same factorization of n as in the first section to prove that

$\frac {6 n^2} {\pi^2} < \varphi(n) \sigma(n) < n^2$ .

Note that

$\varphi(n) \sigma(n) = n \prod_{l=1}^q \left(1 - \frac{1}{p_l}\right) \prod_{l=1}^q \frac{p_l^{v_l+1}-1}{p_l-1} = n \prod_{l=1}^q \frac{p_l-1}{p_l} \; \frac{p_l^{v_l+1}-1}{p_l-1}$