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• Level: GCSE
• Subject: Maths
• Word count: 4988

# The Phi-Function.

Extracts from this document...

Introduction

Maths Coursework

by Yasir Al-Wakeel

The Phi-Function

A Function in mathematics is the term used to indicate the

relationship or correspondence between two or more quantities. It was

first used in 1637 by a French mathematician by the name of Rene

Descartes to designate a power xn of a variable x. Later, Gottfried Wilhelm

Leibniz in 1694 applied the term to various aspects of a curve, such as its

slope. The most widely used meaning until quite recently was defined in

1829 by the German mathematician Peter Dirichlet. Dirichlet conceived  a

function as a variable y, called the dependent variable, having its values

fixed or determined in some definite manner by the values assigned to the

independent variable x, or to several independent variables x1, x2, …, xk.

The Phi-Function is a means of breaking down numbers.  It is

defined as the number of positive integers less than n, where n is a positive

integer, which are co-prime with n.  Zero is neither a positive integer nor a

negative integer, it is on the boundary and so does not come under the

notation of n.  The phi function of a positive inter, n, is expressed as ?(n).

Two terms are co-prime when they have no factor in common other than

one. For example 3 and 4 are co-prime or 5, 7, and 8 are all co-prime.

When numbers are co-prime they can be written: (n , m)=1 such as

(5,7)=1.

Therefore to find the phi-function of a number several steps may be

required.  If we take ?(6) as an example, we can start by listing all the

positive integers that are smaller the than 6; 1, 2, 3,4, and 5.  We may then

list the factors of 6 which are 2 and3.  We also remove any multiples of

the factors of 6 i.e.4.  Therefore 2,3 and 4 cannot be co-prime with 6, by

Middle

? ?(3 x 6) ? ?(3) x ?(6)

n=4

If m =4, we must check whether or not  ?(4 x 4) = ?(4) x ?(4),

n                     16

Co prime   1,3,5,7,9,11,13,15,

with n

?(n)                    8

?(4) x ?(4) = 2 x 2 =4,  ?(4 x 4) = ?(16)=8,  8?4

? ?(4 x 4) ? ?(4) x ?(4)

If m =5, we must check whether or not  ?(4 x 5) = ?(4) x ?(5),

?(4) x ?(5) = 2 x 4 =8,  ?(4 x 5) = ?(20)=8,  8=8

? ?(4 x 5) = ?(4) x ?(5)

If m =6, we must check whether or not  ?(4 x 6) = ?(4) x ?(6),

?(4) x ?(6) = 2 x 2 =4,  ?(4 x 6) = ?(24)=8,  8?4

? ?(4 x 6) ? ?(4) x ?(6)

After completing a series of examples we may illustrate our results graphically(- means as

above):

Values of      Specific series of                  Does m=nr ( where r is an          Are n and m    Does ?(n x m)

n and m      numbers investigated                 integer)? (or n=mr if n?m)        co-prime?   Equal ?(n) x?(m)?

2,2                 when n=m                             yes                        no                no

3,3                   -                                yes                        no                no

4,4                   -                                yes                        no                no

2,4        when n and m are even            yes                        no                no

4,6                              -                            yes                        no                no

2,3              When m-n or n-m=1                     no                        yes                yes

3,4                      -                            no                        yes                yes

4,5                      -                            no                        yes                yes

2,5            When n and m are prime             no                        yes                yes

2,7                      -                            no                        yes                yes

3,5                      -                            no                        yes                yes

m=nr is derived from   n   by the use of cross-multiplication:

m

i.e. If n = 2 and m=4.   2    = 0.5, when cross-multiplying we get 2= 4 x 0.5

4

==>  4 =  2     , the reciprocal of 0.5 =   1    =  2      ==>  4= 2 x 2 ? (in this case) m=nr

0.5                                   0.5

However the limitations of the formula are that the denominator must be greater than the

numerator for r to be an integer, so if m is larger than n we place m as the numerator and

thus obtain the formula n=mr.

It may be suffice to say that when n divided by m gives us  m=rn(or m divided by n,

if n is greater than m, gives us m=nr), where r is an integer, than since m and n are not co-

Conclusion

-  sm pn+1 qm+1 + pn+1 qm sm + sm pn qm+1  - pn  qm sm) / pqs

Note that Equation 2a ? Equation 2b

Equation 1a can also be written differently if we express q and s in terms of the base p:

Equation 3a:

?(pn  qm sm)= pn + m log q/log p + m log s/log p  - pn + (m-1 x log q/log p)+m log

s/log p

- pn-1 + m log q/log p + m log s/log p + pn-1 + (m-1 x log q/log p) + m log s/log p - pn

+ m log q/log p + (m-1 x log s/log p) + pn + (m-1 x log q/log p)+(m-1 x log s/log p) + pn-1

+ m log q/log p + m-1 log s/log p - pn-1 + (m-1 x log q/log p) + (m-1 x log s/log p)

(note the above is only an approximation)

Using Equation 1a:

?(24 x 52 x 72)= (400 - 80 - 200 + 40) x (49 - 7) = 160 x 42 = 6720

Using Equation 1b:

?(24 x 52 x 72)=  (24 x 52 x 72) - (72 x 24 x 5) - (72 x 23 x 52) + (72 x 23 x 5)

- (7 x 52 x 24) + (7 x 24 x 5) +  (7 x 23 x 52) - (7 x 23 x 5) = 19600 - 3920 -

9800 +1960 - 2800 + 560 + 1400 - 280 = 6720

Using Equation 2a :

?(24 x 52 x 72)= 25 x 53 - 25 x 52- 24 x53 + 24 x 52        x     73-72

2 x 5                                7

=  160 x  42 = 6720

Using Equation 2b :

?(24 x 52 x 72)= (73 x 25 x53 - 73 x 25 x 52 - 73 x24 x53 + 73 x24 x 52  - 72 x

25 x53 + 72 x 25 x 52  + 72 x 24 x53- 72 x24 x 52) / 2 x 5 x 7 = 470400 / 70

= 6720

Using Equation 3a :

?(24 x 52 x 72)= 24 + 2 x log 5/log 2 + 2 x log 7/log 2 - 24 + (1 x log 5/log 2)+2 log 7/log 2   - 23 + 2  x

log 5/log 4 + 2 log 7/log 2+ 23 + (1 x log 5/log 2) + 2 log 7/log 2 - 24 + 2 x log 5/log 2 + (1 x log 7/log 2) + 24 +

(1 x log 5/log 2)+(1 x log 7/log 2) + 23 + 2 log 5/log 2 + 1 x log 7/log 2 - 23 + (1 x log 5/log 2) + (1 x log 7/log 2)

is approximately equal to 19600 - 3920 - 9800 +1960 - 2800 + 560 + 1400

- 280 = 6720

?  ?(19600) = 6720

"Function," Microsoft (R) Encarta. Copyright (c) 1994 Microsoft Corporation. Copyright (c) 1994 Funk

& Wagnall's Corporation.

"Logarithm," Microsoft (R) Encarta. Copyright (c) 1994 Microsoft Corporation. Copyright (c) 1994

Funk & Wagnall's Corporation.

1

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