Sequence Sequence
of of
Numerator Denominator Result
It is seen from table 3 that, even when a is less than b, it still converges to √2. On the excel spreadsheet I investigated various values of a and b, keeping a less than b. But it always gave me the same result, converging to √2. This is very interesting, could it to be anything to do with the coefficient of b. Since the coefficient of b is 2, and the transformation converges to √2. I will now investigate what happens if the coefficient of b to 3.
Investigating when the coefficient of b is changed to 3
a → a + 3b , where a,b are whole numbers
b a + b
To investigate this phenomena of changing the coefficient of b to 3. I decided to use excel spreadsheet to see what number the sequence would converge to. My expectation was that it may converge to 3. The results which came from excel spreadsheet are shown in the below.
Table 4
a=b=1 a equal to b (a=b)
Sequence Sequence
of of
Numerator Denominator Results
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 →7→17→….
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
OBSERVATION
By observation the sequence looked as if it is related Fibonacci sequences. I remember with the number cells you were given the first two terms, and then you added the two terms to give the next term. The sequence continued by keeps adding the last two terms to get the next term. But there is a difference with this sequence, because the last term is multiplied by 2. I did further research into the equation we derived earlier and came with PELL NUMBERS. This gave the following sequence:
1,2,5,12,29.70,169,408….. and its equation was PK= 2PK-1 + PK-2, and its associated numbers are 1,3,17.41,99,….
This is the equation I had derived earlier. Our transformation also produced the sequences.
Now I am in stuck mode again, because I still haven’t proved why the transformation tends to √2.
I have also noticed that the coefficient of b in the numerator is twice the coefficient of a in the denominator.
Also coefficient of a in the numerator is the same as coefficient of b in the denominator. Again, the coefficient of a in the denominator produce PELL numbers and whilst the coefficient of b produce its associated numbers. In the numerator only the coefficient of a produced the PELL numbers.
I will now investigate the pattern of a and b developed for the formula a+3b .
a+b
Investigating pattern of a and b in the formula a/b → a + 3b
a + b
NUMERATOR
Adding the coefficients of a and b in the 2 previous terms, gives us the next term of the numerator.
1 + 3 = 4a
↓ ↓ ↓
a a + 3b 4a + 6b
↑ ↑ ↑ ↑
2*1 + 1 + 3 = 6b
This gives us the formula to find the numerator of the next term in the sequence, as shown below:
Un= 2Un-1 +2Un-2
DENOMINATOR
Adding the coefficients of a and b in the 2 previous terms, gives us the next term of the denominator.
1 + 1 = 2a
↓ ↓ ↓
b a + b 2a + 4b
↑ ↑ ↑ ↑
2*1 + 1 + 1 = 4b
This gives us the formula to find the denominator of the next term in the sequence, as shown below:
Un= 2Un-1 +2Un-2
Using the formula developed to find the next term in the sequence of a+3b
a + b
OBSERVATION
I have noticed a similar pattern recurring, the same as for formula (a+2b)/(a+b). The coefficient of b in the numerator is three times the coefficient of a in the denominator. As before it was two times greater.
Also coefficient of a in the numerator is the same as coefficient of b in the denominator. However, I cannot see any PELL number sequence in this transformation. So that theory has ‘gone out of the window’. Now I am in a stuck moment. AHA!!! What kind of transformation converges to a certain number? Well it’s the Golden Ratio, which converges to 1.6. Now I have to prove that our sequence converges to √n.
GOLDEN RATIO GRAPH
2 °
1.8
1.6 _______________________°_______________________
°
1.4
1.2
1 ° fib(i)__
fib(i-1)
0.8
0.6
0.4
0.2
0 2 4 6 8 10
Prove that a → a+2b converges to √2.
b a+b
Dividing throughout by b
a/b → a/b+2b/b
b/b a/b+b/b
a → a/b+2
b a/b+1
Replace a/b with x
X = X+2
X+1
X(X+1)=X+2
X2 + X =X +2
X2 = 2
X =√2
Prove that a → a+3b converges to √3
b a+b
Dividing throughout by b
a/b → a/b+3b/b
b/b a/b+b/b
a → a/b+3
b a/b+1
Replace a/b by X
X = X+3
X+1
X2 = 3X = √3
Prove that our transformation converges to √n
a → a+nb
b a+b
Dividing throughout by b
a/b → a/b+nb/b
b/b a/b+b/b
a → a/b+n
b a/b+1
Let X= a/b
X = X+n
X+1
X2+X = X+n
X2 = n
X =√n
Prove transformation converges to √n+1
a → a+(n+1)b
b a+b
Dividing throughout by b
a/b → a/b+(n+1)b/b
b/b a/b+b/b
a → a/b+(n+1)
b a/b+1
Let X= a/b
X = X+(n+1)
X+1
X2+X = X+(n+1)
X2 = n+1
X =√n+1 We have now proved that the transformation converges to √n+1. But we need to know why does
it converge to √n.
Why does our transformation converge to √n
We have proved that X2= 2, this is the same as X*X=2. Therefore the value of X=2/X. However if we begin with a positive number X1, then either X1 or 2/X1 will be greater than √2 and the other will be smaller than √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, xn is 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.
- 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.
