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• Level: GCSE
• Subject: Maths
• Word count: 2123

# Transforming numbers

Extracts from this document...

Introduction

TRANSFORMING NUMBERS

Name: Waseem Ahmed

Course: PGCE Mathematics

Question

a → a + 2b   where a,b are whole numbers

b      a + b

1 → 3 → 7 → 17 →…….

1      2      5      12

Investigate transformations of this kind

Problem Statement

This problem involves fractions and the aim is to investigate how these numbers can be transformed to the next number in the sequence. How will I go about investigating this problem? First I will like to know where this sequence leads me.  From this I will get a better idea of approaching the problem.

Approach

Using excel spreadsheet, starting with numerator and denominator being equal, ie a=1 and b=1. I found that the sequence of the transformation eventually converging to the square root of 2.

a equal to b (a=b=1)

Sequence      Sequence

Table 1                                of                  of

a=b=1                       Numerator   Denominator     Result

 1 1 1 3 2 1.5 7 5 1.4 17 12 1.41667 41 29 1.41338 99 70 1.41429 239 169 1.4142 577 408 1.41421 1393 985 1.41421 3363 2378 1.41421

From table 1, it was noticed that the sequence converges towards √2. I wanted to investigate what happens if a and b have different values and are not equal to each other. Again I used excel to develop the transformation. So my next step was to investigate what happens when a is greater than b.

Table 2

a=2, b=1

a greater than b by 1 ( a>b)

Sequence       Sequence

of                  of

Numerator    Denominator     Result

 2 1 2 4 3 1.33333 10 7 1.42857 24 17 1.41176 58 41 1.41463 140 99 1.41414

Again I noticed the transformation converges to √2. To do a thorough investigation, I decided to use excel spreadsheet with a > b, by 2,3,4,and so on.

Middle

11584

6688

1.732049

As I suspected the result converges towards √3. Now my question is why does it converge to the square root of n?  I am now in a stuck moment of how should I go about proving that it goes to √n.

What I am now going to do is investigate the pattern being produced within the transformations of a and b. Hopefully this might help me to understand why it tends to √n.

Investigating pattern of a and b in the formula  a/b → a + 2b

a + b

We are given the sequence

1 → 3 →717→….

1      2     5    12

We now have to solve the next sequence of the algebra, Numerator and Denominator separately.

NUMERATOR

Adding the coefficients of a and b in the 2 previous terms, gives us the next term of the numerator.

1  +  2    =   3a

↓      ↓           ↓

a          a + 2b      3a + 4b

↑             ↑     ↑

1    +       1 +  2     =  4b

This gives us the formula to find the numerator of the next term in the sequence, as shown below:

Un= 2Un-1 + Un-2

DENOMINATOR

Adding the coefficients of a and b in the 2 previous terms, gives us the next term in the denominator.

1 + 1    =    2a

↓     ↓          ↓

b         a + b       2a + 3b

↑            ↑     ↑                  ↑

1    +      1 + 1   =             3b

This gives us the formula to find the denominator of the next term in the sequence, as shown below:

Un = 2Un-1 + Un-2

Using the formula developed to find the next term in the sequence of a+2b

a + b

 1  1 3     2 7    5 17             12 41     29 99    70 239    169 577   408 1393         985 3363     2378 a        b

Conclusion

2. For example, if X1=1 and 2/X1 =2/1=2, then X1 is less than 2 and 2 is greater than 2.

Hopefully using the average mean of X1 and 2/X1 will give us a better approximation to 2 than X1 does.

If given X1>0, then to find the next term X2 in this particular sequence is:

n ≥ 1. Xn+1 = 1 (Xn + 2/Xn)   for

2

Therefore, xnis converged to a particular value, then we have a limit of:

lim Xn+1 = lim (Xn/2 + 1/Xn)

Therefore, this property of limit, L must satisfy the condition   L=L/2 + 1/L  .

2L2=L2 + 2

2L2-L2= 2

L2= 2

From this we get L2= 2.          If  Xn >0, then xn+1 will give the average of two positive numbers. Therefore, when X1>0 this leads to positive limit, giving the positive square root of 2.

We need to show that the sequence has a limit, for positive initial prediction.

If  x1= 1, then using the following formula

Xn+1 = 1 (Xn + 2/Xn)……….(1)

2

So the next term of this sequence is:

X2 = 1 (1 + 2/1)

2

X2 = 3    Putting the value of X2 into the above equation 1, we get the next term of X3 = 17  and so on.

1. 12

However we notice the initial term of X1=1 is less than 2, the next term X2=3/2 is greater than 2. From this step, the sequence starts to decrease and is bounded below 2. Therefore the Monotone Convergence Theorem implicates the existing of the limit.

So we have proved that if X1 > 0, then at X2 it starts to monotone decrease and is bounded below by 2.

This means the transformation converges.

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

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