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Totient Function and phi(n)==phi(2n) for odd nby timo1023
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#1
Jul608, 09:47 PM

P: 8

Hello,
Last year in number theory I discovered something with Euler's Totient Function that I couldn't explain. I asked my professor and he couldn't figure it out either. Here is what I discovered: As a premise, it should be clear that phi(n)==phi(2n) where n is an odd, positive, integer. Using the formula for Euler's totient function, this is clear. However, looking at the numbers, it becomes more complex. For example, let n = 9. We know that phi(9)=phi(18)=6. So we have the pos. integers from 1 to 9 ( n in red, coprime integers in blue, noncoprime integers in orange). Note that there are 4 primes in this range (2,3,5,7): [itex]\definecolor{orange}{rgb}{1,0.5,0} {\color{blue}1}, {\color{blue}2}, {\color{orange}3}, {\color{blue}4}, {\color{blue}5}, {\color{orange}6}, {\color{blue}7}, {\color{blue}8}, {\color{red}9} [/itex] Next, the integers from 1 to 18 for n=18. Note there are only 3 new primes from 918 [itex]\definecolor{orange}{rgb}{1,0.5,0} {\color{blue}1}, {\color{orange}2}, {\color{orange}3}, {\color{orange}4}, {\color{blue}5}, {\color{orange}6}, {\color{blue}7}, {\color{orange}8}, {\color{orange}9}, {\color{orange}10}, \color{blue}11}, \color{orange}12}, \color{blue}13}, \color{orange}14}, \color{orange}15}, \color{orange}16}, \color{blue}17}, \color{red}18}[/itex] I'm having a hard time phrasing this, but I think what I mean to say is this: How can the phi function know to compensate for the number of prime numbers between n and 2n? Wouldn't this be related to the density of primes? Also, could this be used to prove that there exist infinite prime numbers? I'm sorry If my wording is hard to understand. Once I gain some more understanding, I would like to write it up properly and show it to my math teacher or something. Thanks in advance. 


#3
Jul908, 01:20 AM

P: 122

You ask "How can the phi function know to compensate for the number of prime numbers between n and 2n?" But, does the data indicate that it really does that as opposed to compensating for the number of coprime numbers between n and 2n? Nothing more than a (hopefully) seminal question, but, I really would like it if you could give me your best shot at an answer. DJ 


#4
Jul908, 09:19 AM

P: 8

Totient Function and phi(n)==phi(2n) for odd n
Upon further investigation, it seems as though you are right. Here is what I came up with (please forgive my lack of precision in stating x is a positive integer, etc.):
[itex] \phi (2n)=\left{}\{xx\leq 2n1\}\{xx \textrm{ is even}\}  \{i\cdot x  x \textrm{ odd}, xn, i \textrm{ odd}\}\right{} [/itex] So, for [itex]n=15[/itex], we have: [itex] \phi (30) = \{1,\dots, 29\}\{\textrm{evens}\}\{3,9,15,21,27,5,25\} [/itex] [itex] \phi (30) =  29147  = 8 [/itex] As we already know, [itex]\phi (30)=\phi (15)=8[/itex], and thus this method works. So it would seem that it has nothing to do with the number of primes between 1 and 2n. Please give me your thoughts. 


#5
Feb2010, 01:46 AM

P: 25

The formula for euler's function ϕ(n) is given by
ϕ(n) = n (11/p1) (1 – 1/p2) … (1 – 1/pk) where n is prime factorized into p1a1p2a2...pkak and each pk is a prime. Assume n is an odd number. None of the pk will be an even number. When n is multiplied by 2, the prime factorization of n will b (p1^a1)(p2^a2)...(pk^ak) * 2. The formula for ϕ(n) for 2n will be ϕ(2n) = 2n ((11/p1) (1 – 1/p2) … (1 – 1/pk) (1  ½) = n ((11/p1) (1 – 1/p2) … (1 – 1/pk) which equals ϕ(n) Assume n is an even number. Multiplying n by 2 will increase the power of 2 by 1. ϕ(2n) = 2n ((11/p1) (1 – 1/p2) … (1 – 1/pk) = 2ϕ(n) You're welcome. 


#6
Feb2010, 02:14 AM

P: 688

n=17 could be another interesting example. In the upper half of the coprimes to 34, you find not only new primes (19,23,29,31), but also powers of smaller primes (25,27) as well as other composite numbers (21,33).
As n gets bigger, the panorama gets only more varied. 


#7
Feb2010, 10:42 AM

P: 105

Doesn't ϕ(mn) = ϕ(m)ϕ(n)?
And doesn't ϕ(2) = 1? What's the mystery? 


#8
Feb2110, 04:47 PM

P: 25

@Mensanator
ϕ(2m) = 2ϕ(m) ONLY IF m is even thats where's the mystery. And i think I did a pretty good job explaining it in post #5 if anyone is interested... 


#9
Feb2110, 08:46 PM

P: 105

Have you noticed that ϕ(3m) is twice ϕ(m) if m is not divisible by 3? Spooky! 


#10
Feb2110, 08:49 PM

P: 25




#11
Feb2110, 08:53 PM

P: 25

If the OP's question is why phi(2n) = phi(n) when n is odd, I think my explanation is clear enough,.... see post #5.
please feel free to ask questions and dont say stuff like even numbers have no factor of two. 


#12
Feb2110, 11:24 PM

P: 105




#13
Feb2210, 12:16 AM

P: 2

[QUOTE=timo1023;1793133]
Also, could this be used to prove that there exist infinite prime numbers? QUOTE] Not alone it can't. Phi(n) gives no information about the number of prime factors of the numbers less than and coprime to n, just the quantity of them. Phi(n) does give an upper bound on the number of primes. (Its a truly awful upper bound though, as n goes to infinity Phi(n) goes to c*n for some constant, c, I can't remember. The number of primes less than n goes to n/ln(n), a MUCH smaller number for large n). But you could say that phi(n) is related to the infinite product (11/p1)*(11/p2)*(11/p3)...over all primes. This product is equal to the sum 1/k for every positive integer k. Since that sum diverges there must be an infinite number of primes (Euler). 


#14
Feb2210, 07:57 AM

Sci Advisor
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P: 3,684




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