Multiplicative Functions : Number Theory #Multiplicative #Functions #Number #Theory

Multiplicative functions :

Definition : ( Multiplicative function )

An arithmetical function f is called multiplicative if f is not identically zero and if f (mn) = f(m)f(n) whenever (m,n) = 1.

Definition : ( completely multiplicative function )

A multiplicative function f is called completely multiplicative if f(mn) = f(m)f(n) ∀ m,n irrespective of (m,n) = 1.

Examples:

· Let f_m(n) = n^m, when m is a fixed real or complex number. This fm is completely multiplicative . This function fm(n) = n^m is called power function.

· The identity function I(n) = [1/n] is completely multiplicative.

· The Euler totient function Ჶ(n) is multiplicative.

But this Euler totient function is not completely |∵ Ჶ( 4 )=2 where as Ჶ(2)Ჶ(1) = 1|

· Since ⴷ (1) = 0, the Mangold’s function ⴷ(n) is not multiplicative.

Theorem 1 :

If f is multiplicative then f(1) = 1.

Proof :

Suppose f is multiplicative.

We have f(n) = f(1.n)

1.f(n) = f(1) f(n)

1 = f(1) | ∵ by right cancellation |

Hence f(1) = 1.

Theorem 2 :

If both g and f*g are multiplicative then f is also multiplicative .

Proof :

Suppose g and f*g are multiplicative.

Therefore g(1) = (f*g)(1) = 1 …………> I

Now we prove that f is multiplicative.

Assume that f is not multiplicative.

Let h = f*g.

Since f is not multiplicative , there exists positive integers m and n with (m,n) = 1 such that f(mn) ≠ f(m)f(n).

Case 1 . let mn = 1.

Since mn=1, (m,n) = 1 , we have m=n=1.

f(1) = f(mn)

≠ f(m)f(n)

≠ f(1)f(1)

1.f(1) ≠ f(1)f(1)

f(1) ≠ 1 | ∵ by right cancellation |

Now h(1) = (f *g) (1)

= f(1)g(1)

= f(1) 1 | ∵ g(1) = 1 |

= f(1)

≠ 1

Hence h(1) ≠ 1

i.e. (f*g) (1) ≠ 1

this is a contradiction to f*g is multiplicative.

Our assumption is false.

Hence f is multiplicative , if mn = 1 .

Case 2 .

Let mn > 1.

We have f(ab) = f(a)f(b) ∀ positive integers a,b with (a,b) = 1 and ab < mn.

Now h(mn) = (f*g) (mn)

= 𝜮 f(ab) g(mn/ab) + f(mn) g(1)

= 𝜮 f(a)f(b) g(m/a) g(n/b) + f(mn) | ∵ g(1) = 1 |

= 𝜮 f(a) g(m/a) 𝜮f(b) g(n/b) – f(m)f(n) + f(mn)

= h(m)h(n) – f(m)f(n) + f(mn)

≠ h(m)h(n) | ∵ f(mn) ≠ f(m)f(n) |

∴ h is not multiplicative i.e. f*g is not multiplicative.

This is again a contradiction.

∴ our assumption is false.

∴ f is multiplicative if mn > 1.

Hence in any case , if g and f*g are multiplicative then f is also multiplicative.

Hence the proof.

Theorem 3 :

If f and g are multiplicative then their Dirichlet product f*g is also multiplicative

The Dirichlet product of two multiplicative functions is also multiplicative.

Proof:

Suppose f and g are multiplicative.

Now we product their Dirichlet product f*g is also multiplicative.

Let h = f*g and choose m,n ,two positive integers with (m,n) = 1.

Now h(mn) = (f*g) (mn)

= 𝜮f ( c) g(mn/c) for c/mn

Since c/mn , we have c=ab where a/m and b/n.

More over , (a,b) = 1

∴ (m/a, n/b ) = 1.

Also there is a 1-1 correspondence between the set of products ab and the divisors of mn.

Hence h(mn) = 𝜮 f(ab) g(mn/ab)

= 𝜮 f(a) f(b) g(m/a)g(n/b)

= 𝜮 f(a)g(m/a) 𝜮 f(b)g(n/b)

= h(mh(n)

Therefore h(mn) = h(m)h(n) ∀ m,n in Z⁺.

i.e (f*g) (mn) = (f*g) (m).(f*g) (n) ∀ m,n in Z⁺.

∴ f*g is multiplicative.

Hence the proof.

Note :

The Dirichlet product of two completely multiplicative functions need not be completely multiplicative but the Dirichlet product of multiplicative functions is multiplicative.

* Dirichlet Multiplication

* Euclidean Algorithm

* Fundamental Theorem of Arithmetic

* Properties of Numbers

* Historical Introduction to Number Theory

* GCD of morethan 2 numbers

* The Mobius Function 𝝻 ( n ) .

* The Euler Totient Function