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  • Level: GCSE
  • Subject: Maths
  • Word count: 3837

Identify and explain the rules and equations associated with the Phi function.

Extracts from this document...

Introduction

Graham Pollock 5P                

Introduction

The Phi is described as the number of positive integers less than n (positive integer) which have no factor, other than 1, in common, co-prime, with n.

In my investigation I aim to identify and explain the rules and equations associated with the Phi function.  To go about this I will investigate Φ (n).  This will be further touched upon and will help me investigate and

...read more.

Middle

27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42, 43,44

24

Φ46

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42, 43,44,45

22

Φ47

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42, 43,44,45,46

46

Φ48

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42, 43,44,45,46,47

16

Φ49

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42, 43,44,45,46,47,48

42

Φ50

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42, 43,44,45,46,47,48,49

20

Φ51

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40, 41,42,43,44,45,46,47,48,49,50

32

Φ52

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40, 41,42,43,44,45,46,47,48,49,50,51

24

Φ53

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40, 41,42,43,44,45,46,47,48,49,50,51,52

52

Φ54

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40, 41,42,43,44,45,46,47,48,49,50,51,52,53

18

Φ55

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22, 23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40, 41,42,43,44,45,46,47,48,49,50,51,52,53,54

40

In my coursework there are some examples which range up to 60 these were obtained from a reliable website and checked over to ensure they were correct.


Investigation

From the Phi table shown previously we can see that some clear patterns emerged.  For example I noticed that for all the prime numbers the Phi is one less than the prime number.

Φ2=1=1                                Φ11=1 2 3 4 5 6 7 8 9 10=10

Φ3=1 2=2                                Φ13=1 2 3 4 5 6 7 8 9 10 11 12=12

Φ7=1 2 3 4 5 6=6                        Φ17=1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16=16

This is because a prime number only has two numbers which divide into it, 1 and itself.  The actual number is not included in our counting while the 1 is hence why the number is one less than the prime.

The Phi of 2 is the only odd value while the rest all have even values.  This is because only one number is counted in

This can be arranged into an equation like so.

(p is a prime number)

Φ p = p – 1

Phi values of 2

I also looked at some other patterns in order to see if there was a formula which could solve the Phi of any number.  Beginning with the number 2.

Φ2=1                        Φ18=6        

Φ4=2                        Φ20=8

Φ6=2                        Φ22=10

Φ8=4                        Φ24=8

Φ10=4                        Φ26=12

Φ12=4                        Φ28=12

Φ14=6                        Φ30=8

Φ16=8                        Φ32=16

From this there seems to be no apparent pattern between all the multiples of two however I did spot a pattern where all the squares of 2 were always half of the Phi number.  I will further investigate this by placing them in a table.

x

2x

Φ(2

...read more.

Conclusion

Therefore this means Φ(c x d)= Φ(c) x Φ(d) (where c and d are co-primes)

Φ(e x f) = Φ(e) x Φ(f) (where e and f have more than 1 common factors other than 1)


Conclusions

Throughout this coursework I have found at wide range of equations allowing us to work out different aspects of Phi.  At the beginning I found the basic formulas, these being

  • Φ p = p – 1

A number of minor formulas helped me amalgamate and form a formula for all prime numbers which in turn helped me later. These formulas were:

  • Φ2x=½ (2x)
  • Φ3x=2/3(3x)
  • Φ3x=3x - 3x-1
  • Φ5x=4/5(5x)
  • Φ7x=6/7(7x)
  • Φ px=px-px-1(where p is prime)

This particular one helped me support, form and follow the equation:

  • Φ(x)= Φ(y)x Φ(z)

(Where y and z are the breakdowns of x into its lowest prime factors)

In part two I have found that (a x b) = (a) x (b) only in the cases where a and b are co-primes.  These examples can be solved using the equations shown previously.  When the numbers are not co-prime and have more than one common factor other than 1 they do not balance this equation may be used.

Φ(m x n)=½ Φ(m x n) (where m and n are even)

These formulas allow me to work out the Phi of any number providing it can be split into its prime factors.

Graham Pollock 5P                Phi function Coursework

...read more.

This student written piece of work is one of many that can be found in our GCSE Phi Function section.

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