Difference between revisions of "Euler's totient function"

Latest revision as of 14:49, 27 July 2024

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

Video

https://youtu.be/T7INnve59JM

Formula

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

Derivation

To derive the formula, let us first define the prime factorization of $n$ as $n =\prod_{i=1}^{m}p_i^{e_i} =p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$ where the $p_i$ are distinct prime numbers. Now, we can use a PIE argument to count the number of numbers less than or equal to $n$ that are relatively prime to it.

First, let's count the complement of what we want (i.e. all the numbers less than or equal to $n$ that share a common factor with it). There are $\frac{n}{p_1}$ positive integers less than or equal to $n$ that are divisible by $p_1$ . If we do the same for each $p_i$ and add these up, we get

$\frac{n}{p_1} + \frac{n}{p_2} + \cdots + \frac{n}{p_m} = \sum^m_{i=1}\frac{n}{p_i}.$

But we are obviously overcounting. We then subtract out those divisible by two of the $p_i$ . There are $\sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}$ such numbers. We continue with this PIE argument to figure out that the number of elements in the complement of what we want is

$\sum_{1 \le i \le m}\frac{n}{p_i} - \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}.$

This sum represents the number of numbers less than $n$ sharing a common factor with $n$ , so

$\phi(n) = n - \left(\sum_{1 \le i \le m}\frac{n}{p_i}- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}\right)$

$\phi(n)= n\left(1 - \sum_{1 \le i \le m}\frac{1}{p_i} + \sum_{1 \le i_1 < i_2 \le m}\frac{1}{p_{i_1}p_{i_2}} - \cdots + (-1)^{m}\frac{1}{p_1p_2\ldots p_m}\right)$

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

Note: Another way to find the closed form for $\phi(n)$ is to show that the function is multiplicative, and then breaking up $\phi(n)$ into its prime factorization.

Identities

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

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

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

(d) If $p$ is prime and $n\ge{1},$ then $\phi(p^n)=p^n-p^{n-1}$

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

Proof. Split the set $\{1,2,\ldots,n\}$ into disjoint sets $A_d$ where for all $d\mid n$ we have $A_d=\{x:1\leq x\leq n\quad\text{and}\quad \operatorname{syt}(x,n)=d \}.$ Now $\operatorname{gcd}(dx,n)=d$ if and only if $\operatorname{gcd}(x,n/d)=1$ . Furthermore, $1\leq dx\leq n$ if and only if $1\leq x\leq n/d$ . Now one can see that the number of elements of $A_d$ equals the number of elements of $A_d^\prime=\{x:1\leq x \leq n/d\quad\text{and}\quad \operatorname{gcd}(x,n/d)=1 \}.$ Thus by the definition of Euler's phi we have that $|A_d^\prime|=\phi (n/d)$ . As every integer $i$ which satisfies $1\leq i\leq n$ belongs in exactly one of the sets $A_d$ , we have that $n=\sum_{d \mid n}\phi \left (\frac{n}{d} \right )=\sum_{d \mid n}\phi (d).$

(f) Another interesting thing to note is that $\phi(n)\leq n - 1$ . This does seem very obvious but is helpful in solving many problems.

Graph

Notation

Sometimes, instead of $\phi$ , $\varphi$ is used. This variation of the Greek letter phi is common in textbooks, and is standard usage on the English Wikipedia

@@ Line 1: / Line 1: @@
 '''Euler's totient function''' <math>\phi(n)</math> applied to a [[positive integer]] <math>n</math> is defined to be the number of positive integers less than or equal to <math>n</math> that are [[relatively prime]] to <math>n</math>.  <math>\phi(n)</math> is read "phi of n."
-== Formulas ==
+==Video==
+https://youtu.be/T7INnve59JM
+== Formula ==
+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 <cmath>\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).</cmath>
+== Derivation ==
 To derive the formula, let us first define the [[prime factorization]] of <math> n </math> as <math> n =\prod_{i=1}^{m}p_i^{e_i} =p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m} </math> where the <math>p_i </math> are distinct [[prime number]]s.  Now, we can use a [[PIE]] argument to count the number of numbers less than or equal to  <math> n </math> that are relatively prime to it.
-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> \frac{n}{p_1} </math> numbers less than <math> n </math> that are divisible by <math> p_1 </math>.  If we do the same for each <math> p_i </math> and add these up, we get
+First, let's count the complement of what we want (i.e. all the numbers less than or equal to <math> n </math> that share a common factor with it).  There are <math> \frac{n}{p_1} </math> positive integers less than or equal to <math> n </math> that are divisible by <math> p_1 </math>.  If we do the same for each <math> p_i </math> and add these up, we get
-<cmath> \frac{n}{p_1} + \frac{n}{p_2} + \cdots + \frac{n}{p_m} = \sum^m_{i=1}\frac{n}{p_i}.</cmath>
+<center><cmath> \frac{n}{p_1} + \frac{n}{p_2} + \cdots + \frac{n}{p_m} = \sum^m_{i=1}\frac{n}{p_i}.</cmath></center>
 But we are obviously overcounting.  We then subtract out those divisible by two of the <math> p_i </math>.  There are <math>\sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}</math> such numbers. We continue with this PIE argument to figure out that the number of elements in the complement of what we want is
-\[ \sum_{1 \le i \le m}\frac{n}{p_i}
+<center><cmath> \sum_{1 \le i \le m}\frac{n}{p_i}
-- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}\] \[
+- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}
-+ \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}.\]
++ \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}. </cmath></center>
 This sum represents the number of numbers less than <math>n</math> sharing a common factor with <math>n</math>, so
-<math>\phi(n) = n - (\sum_{1 \le i \le m}\frac{n}{p_i}</math> <math>- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}
+<math>\phi(n) = n - \left(\sum_{1 \le i \le m}\frac{n}{p_i}- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}
-+ \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m})</math>
++ \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}\right)</math>
+<math>\phi(n)= n\left(1 - \sum_{1 \le i \le m}\frac{1}{p_i}
++ \sum_{1 \le i_1 < i_2 \le m}\frac{1}{p_{i_1}p_{i_2}} - \cdots + (-1)^{m}\frac{1}{p_1p_2\ldots p_m}\right)</math>
-<math>\phi(n)= n(1 - \sum_{1 \le i \le m}\frac{1}{p_i}
+<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>
-+ \sum_{1 \le i_1 < i_2 \le m}\frac{1}{p_{i_1}p_{i_2}}</math> <math>- \cdots + (-1)^{m}\frac{1}{p_1p_2\ldots p_m})</math>
-<math>\phi(n)= n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\cdots(1-\frac{1}{p_m}).</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 <cmath>\phi(n) = n(1-\frac{1}{p_1})(1-\frac{1}{p_2}) \cdots (1-\frac{1}{p_m}).</cmath>
+*Note: Another way to find the closed form for <math>\phi(n)</math> is to show that the function is multiplicative, and then breaking up <math>\phi(n)</math> into its prime factorization.
 == Identities ==
-For [[prime]] p, <math>\phi(p)=p-1</math>, because all numbers less than <math>{p}</math> are relatively prime to it.
+(a) For [[prime]] <math>p</math>, <math>\phi(p)=p-1</math>, because all numbers less than <math>{p}</math> are relatively prime to it.
+(b) For relatively prime <math>{a}, {b}</math>, <math> \phi{(a)}\phi{(b)} = \phi{(ab)} </math>.
-For relatively prime <math>{a}, {b}</math>, <math> \phi{(a)}\phi{(b)} = \phi{(ab)} </math>.
+(c) In fact, we also have for any <math>{a}, {b}</math> that <math>\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})</math>.
-In fact, we also have for any <math>{a}, {b}</math> that <math>\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})</math>.
+(d) If <math>p</math> is prime and <math>n\ge{1},</math> then <math>\phi(p^n)=p^n-p^{n-1}</math>
-For any <math>n</math>, we have <math>\sum_{d|n}\phi(d)=n</math> where the sum is taken over all divisors d of <math> n </math>.
+(e) For any <math>n</math>, we have <math>\sum_{d|n}\phi(d)=n</math> where the sum is taken over all divisors <math>d</math> of <math>n</math>.
 Proof. Split the set <math>\{1,2,\ldots,n\}</math> into disjoint sets <math>A_d</math> where for all <math>d\mid n</math> we have
@@ Line 41: / Line 52: @@
 <cmath>A_d^\prime=\{x:1\leq x \leq n/d\quad\text{and}\quad \operatorname{gcd}(x,n/d)=1 \}.</cmath>
 Thus by the definition of Euler's phi we have that <math>|A_d^\prime|=\phi (n/d)</math>. As every integer <math>i</math> which satisfies <math>1\leq i\leq n</math> belongs in exactly one of the sets <math>A_d</math>, we have that
-<cmath>n=\sum_{d \mid n}\varphi \left (\frac{n}{d} \right )=\sum_{d \mid n}\phi (d).</cmath>
+<cmath>n=\sum_{d \mid n}\phi \left (\frac{n}{d} \right )=\sum_{d \mid n}\phi (d).</cmath>
+(f) Another interesting thing to note is that <math>\phi(n)\leq n - 1</math>. This does seem very obvious but is helpful in solving many problems.
+==Graph==
+[[Image:Euler_totient_50000.png]]
 ==Notation==

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