My investigation is about the Phi Function ?. I am investigating the different ways on how to find the Phi Functions of different numbers and finding easier ways of finding the Phi Functions of large numbers. I will go through four parts for this coursework.
I will start from the simplest cases of numbers and will go to more complicated. For any positive integer n, the Phi function ?(n) is defined as the number of positive integers less than n which have no factor (other than 1) in common (are co-prime) with n:So ?(12)=4, because the positive integers less than 10 which have no factors other than 1, in common with 12 are 1, 5, 7, 11 i.e. 4 of them. These four numbers are not factors of 12.
Also ?(6)=2, because the positive integers less than 6 which have no factors other than 1, in common with 6 are 1, 5 i.e. 2 of them. These two numbers are not factors of two.For the first part I will find the Phi Functions of many simple numbers and I will try to find a pattern on the Phi Functions of different types of numbers e.g. Odd numbers, even numbers, prime numbers, squared numbers, triangular numbers and so on. I will start from the numbers I obtained from the coursework sheet.
1. ?(3)=1, 2. Two of the numbers are not the factors of 3.So ?(3)=22. ?(8)=1, 2, 3, 4, 5, 6, 7. Four of the numbers are not factors of 8.So ?(8)=43.
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?(11)=1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Ten of the numbers are not factors of 11.So ?(11)=104. ?(24)=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23.I will now obtain the Phi Functions from numbers of my own choice and I will try to find patterns, if any. I will try to find patterns from even numbers, odd numbers, prime numbers, squared numbers, and triangular numbers.EVEN NUMBERS:?(2)=1?(4)=2?(6)=2?(8)=4?(10)=4?(12)=4?(14)=6?(16)=8?(18)=6?(20)=8ODD NUMBERS:?(3)=2?(5)=4?(7)=6?(9)=6?(11)=10?(13)=12?(15)=8?(17)=16?(19)=18?(21)=12PRIME NUMBERS:?(2)=1?(3)=2?(5)=4?(7)=6?(11)=10?(13)=12?(17)=16?(19)=18?(23)=22?(29)=28SQUARED NUMBERS:?(4)=2?(9)=6?(16)=8?(25)=20?(36)=12?(49)=42?(64)=32?(81)=54?(100)=20TRIANGULAR NUMBERS:?(3)=2?(6)=2?(10)=4?(15)=8TRIANGULAR NUMBERS CONT’D:?(21)=12?(28)=12?(36)=14?(45)=24?(55)=40?(66)=20I couldn’t find any patterns from any of the types of the numbers. But I have found out that there is an easier way of finding the Phi Functions of prime numbers.
I have found a formula for finding the Phi Function for prime numbers, which is “P-1”. “P” stands for Prime number.I the second part I am going to check whether ?(n�m)=?(n)�?(m) or ?(n�m)=?(n)�?(m). I am going to check the first two from the sheet.1.
?(7�4)=?(7)�?(4)?(28)=12. There are 12 numbers which are not factors of 28.?(7)=6. There are 6 numbers which are not factors of 7.?(4)=2. There are 2 numbers which are not factors of4.So ?(7�4)=?(7)�?(4)2. ?(6�4)=?(6)�(4)?(24)=8?(6)=2?(4)=2So ?(6�4)=?(6)�?(4)I am now going to check whether or not ?(nxm)=?(n)x?(m) for at least two separate choices of my own of n and m.
1. ?(2�8)=?(2)�?(8)?(16)=8. There are 8 positive numbers less than 16 which are not factors of 16.?(2)=1.
There is 1 positive number less than 2 which is not a factor of 2.?(8)=4. There are 4 positive numbers less than 8 which are not factors of 8.So ?(2�8)=?(2)�?(8)2.
?(7�3)=?(7)�?(3)?(21)=12?(7)=6?(3)=2So ?(7�3)=?(7)�?(3)For the third part I am investigating why ?(n�m)=?(n)�?(m) whilst in other cases this is not so. I have found out that it depends on the types of numbers used as n and m. Now I will check whether or not ?(n�m)=?(n)�?(m) for different types of choices of n and m.
1. Even numbers of n and m.?(6�4)=?(6)�?(4)?(24)=8. There are 8 positive numbers less than 24 which are not factors of 24.
?(6)=2. There are 2 positive numbers less than 6 which are not factors of 6.?(4)=2. There are 2 positive numbers less than 4 which are not factors of 4.So ?(6�4)=?(6)�?(4)2.
Odd numbers of n and m.?(9�9)=?(9)�?(9)?(81)=54?(9)=6?(9)=6So ?(9�9)=?(9)�?(9)3. Co-prime numbers of n and m.?(5×3)=?(5)x?(3)?(15)=9. There are 9 positive numbers less than 15 which are not factors of 15.?(5)=4.
There are 4 positive numbers less than 5 which are not factors of 5.?(3)=2. There are 2 positive numbers less than 2 which are not factors of 3.So ?(5�3)=?(5)�?(3)4. n(odd) and m(even).?(9×4)=?(9)x?(4)?(36)=14?(9)=6?(4)=2So ?(9�4)=?(9)�?(4)5. n(prime) and m(even).
?(7×4)=?(7)x?(4)?(28)=12. There are 12 positive numbers which are not factors of 28.?(7)=6. There are 6 positive integers which are not factors of 7.?(4)=2. There are 2 positive integers which are not factors of 4.So ?(7�4)=?(7)�?(4)I have found out that there are two ways of having ?(nxm)=?(n)x?(m).
The first way is that n and m should be co-prime. The other way is that either n or m should be prime and even. The rest of the number types result ?(n�m)=?(n)�?(m).I have to investigate “?(Pn Q )” if P and Q are prime in the fourth part. I have found two formulas for finding the Phi Functions of large numbers.
One of the formulas is “(Pn��)(P-1)”. The other formula is “Pn(1-1/P)”.FIRST FORMULA:(Pn��)(P-1)To find ?(81):1. Firstly the 81 has to be split up into prime numbers.
8132739333181 converted into prime numbers=3So the formula can now be used to find ?(81):2. (3 ��)(3-1)=3. 3��24.
54So ?(81)=54SECOND FORMULA: Pn(1-1/P)To find ?(81):1. Firstly 81 has to be split into prime numbers.8132739333181 converted into prime numbers=3So the formula can now be used to find ?(81):2. 3 (1-1/3)=3. 81�2/3=4. 54So ?(81)=54The two formulas give out the same results for the ?(81).
The “P” on the formulas stands for prime number. The formulas are faster at finding the Phi rather than writing all the numbers out and cancelling or circling them. There are four steps to work out the Phi’s from the formulas and that is why I have numbered my working out.
