Difference between revisions of "Euler's totient function"

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'''Euler's totient function''', <math>\phi(n)</math>, is defined as the number of positive integers less than or equal to a given positive integer that are [[relatively prime]] to that integer.
 
'''Euler's totient function''', <math>\phi(n)</math>, is defined as the number of positive integers less than or equal to a given positive integer that are [[relatively prime]] to that integer.
  
=== Formulas ===
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== Formulas ==
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To derive the formula, let us first define the prime factorization of <math> n </math> as <math> n = p_1^{e_1}p_2^{e_2}\cdot p_n^{e_n} </math> where the <math>\displaystyle p_i </math> are primes.  Now, we can use a [[PIE]] argument to count the number of numbers less than or equal to  <math> n </math> are are relatively prime to it.
  
The formal definition is <math>\phi(n):=\# \left\{ a \in \mathbb{Z}: 1 \leq a \leq n , \gcd(a,n)=1 \right\} </math>.
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First, let's count the complement of what we want (i.e. all the numbers less than <math> n </math> that share a common factor with it).  There are <math> p_1^{e_1-1}p_2^{e_2}\cdots p_n^{e_n} </math> numbers less than <math> n </math> that are divisible by <math> p_1 </math>.  If we do the same for each <math> p_k </math> and add these up, we get
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<center><math> p_1{e_1-1}p_2^{e_2}\cdots p_n{e_n} + p_1^{e_1}p_2^{e_2-1}\cdots p_n^{e_n} + \cdots + p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n - 1}.</math></center>
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We can factor out though:
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<center><math> p_1^{e_1-1}p_2^{e_2-1}\cdots p_n^{e_n-1}(p_1+p_2+\cdots + p_n).</math></center>
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But we are obviously overcounting.  So we then subtract out those divisible by two of the <math> p_1 </math>.  We continue with this PIE argument to figure out that the number of elements in the complement of what we want is
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<center><math>p_1^{e_1-1}p_2^{e_2-1}\cdots p_n^{e_n-1}[(p_1+p_2+\cdots+p_n)-(p_1p_2+p_1p_3+\cdots+(-1)^np_{n-1}p_n)+\cdots+p_1p_2\cdots p_n]</math></center>
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which we can factor further as
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<center><math>p_1^{e_1-1}p_2^{e_2-1}\cdots p_n^{e_n-1}(p_1-1)(p_2-1)\cdots(p_n-1).</math></center>
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Making one small adjustment, we write this as
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<center><math> \phi(n) = n\left(1-\frac 1{p_1}\right)\left(1-\frac 1{p_2}\right)\cdots\left(1-\frac 1{p_n}\right).</math></center>
  
 
Given the general [[prime factorization]] of <math>{n} = {p}_1^{e_1}{p}_2^{e_2} \cdots {p}_m^{e_m}</math>, one can compute <math>\phi(n)</math> using the formula <math> \phi(n) = n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right) \cdots \left(1-\frac{1}{p_m}\right) </math>.
 
Given the general [[prime factorization]] of <math>{n} = {p}_1^{e_1}{p}_2^{e_2} \cdots {p}_m^{e_m}</math>, one can compute <math>\phi(n)</math> using the formula <math> \phi(n) = n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right) \cdots \left(1-\frac{1}{p_m}\right) </math>.
  
=== Identities ===
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== Identities ==
  
 
For [[prime]] p, <math>\phi(p)=p-1</math>, because all numbers less than <math>{p}</math> are relatively prime to it.
 
For [[prime]] p, <math>\phi(p)=p-1</math>, because all numbers less than <math>{p}</math> are relatively prime to it.

Revision as of 20:45, 25 June 2006

Euler's totient function, $\phi(n)$, is defined as the number of positive integers less than or equal to a given positive integer that are relatively prime to that integer.

Formulas

To derive the formula, let us first define the prime factorization of $n$ as $n = p_1^{e_1}p_2^{e_2}\cdot p_n^{e_n}$ where the $\displaystyle p_i$ are primes. Now, we can use a PIE argument to count the number of numbers less than or equal to $n$ are are relatively prime to it.

First, let's count the complement of what we want (i.e. all the numbers less than $n$ that share a common factor with it). There are $p_1^{e_1-1}p_2^{e_2}\cdots p_n^{e_n}$ numbers less than $n$ that are divisible by $p_1$. If we do the same for each $p_k$ and add these up, we get

$p_1{e_1-1}p_2^{e_2}\cdots p_n{e_n} + p_1^{e_1}p_2^{e_2-1}\cdots p_n^{e_n} + \cdots + p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n - 1}.$

We can factor out though:

$p_1^{e_1-1}p_2^{e_2-1}\cdots p_n^{e_n-1}(p_1+p_2+\cdots + p_n).$

But we are obviously overcounting. So we then subtract out those divisible by two of the $p_1$. We continue with this PIE argument to figure out that the number of elements in the complement of what we want is

$p_1^{e_1-1}p_2^{e_2-1}\cdots p_n^{e_n-1}[(p_1+p_2+\cdots+p_n)-(p_1p_2+p_1p_3+\cdots+(-1)^np_{n-1}p_n)+\cdots+p_1p_2\cdots p_n]$

which we can factor further as

$p_1^{e_1-1}p_2^{e_2-1}\cdots p_n^{e_n-1}(p_1-1)(p_2-1)\cdots(p_n-1).$

Making one small adjustment, we write this as

$\phi(n) = n\left(1-\frac 1{p_1}\right)\left(1-\frac 1{p_2}\right)\cdots\left(1-\frac 1{p_n}\right).$

Given the general prime factorization of ${n} = {p}_1^{e_1}{p}_2^{e_2} \cdots {p}_m^{e_m}$, one can compute $\phi(n)$ using the formula $\phi(n) = n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right) \cdots \left(1-\frac{1}{p_m}\right)$.

Identities

For prime p, $\phi(p)=p-1$, because all numbers less than ${p}$ are relatively prime to it.

For relatively prime ${a}, {b}$, $\phi{(a)}\phi{(b)} = \phi{(ab)}$.

In fact, we also have for any ${a}, {b}$ that $\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})$.

For any $n$, we have $\sum_{d|n}\phi(d)=n$ where the sum is taken over all divisors d of $n$.

See also

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