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Introduction

Anousha Manji

Maths Portfolio

Continued Fractions This ‘infinite fraction’ can be considered as a sequence of terms, tn:    A general formula for tn+1in terms of tncan now be determined. It can be seen that tn+1is 1added to 1 divided by the previous term.

i.e.  Decimal equivalents of each term can be computed. Here are the values for the first ten terms (correct to 5d.p as we can then see how the numbers differ):

t1=1
t2=1.50000
t3= 1.66667
t4= 1.60000
t5= 1.62500
t6= 1.61538
t7= 1.61905
t8= 1.61765
t9= 1.61818
t10= 1.61798 From looking at the graph we can see that for the first few terms the values fluctuate, but eventually the values fluctuate less and become very close together. I.e. the values become closer together as the value of n increases.

We can then conclude that as n increases tn≈tn+1.

Middle (We know that  must be a positive value because all the numbers in the continued fraction are positive.)

If we look at the list of values for the first 10 terms we can see that they are very close to 1.61, which is the value we obtained from our above formula.

Here is another continued fraction:    General formula:  Values of the first 10 terms (5d.p):

t1= 3.00000
t2= 2.33333
t
3= 2.42857
t
4= 2.41176
t
5= 2.41463
t
6= 2.41414
t
7= 2.41423
t
8= 2.41421
t9= 2.41421
t
10= 2.41421 Again we can see a similar pattern; as n increases, the values for each term become closer together.
What we can conclude:

• eventually tn≈tn+1
• if tn≈tn+1, then tn-tn+1=0

Finding the value for the nth term can be found using the general formula. However, like first example it takes a long time and

Conclusion

k will be used to be 1.5 and the same method used for values for k used above (1, 2 and 3) will be used.   General formula:    Values of the first 10 terms (5d.p)

t1= 2.50000

t2= 1.90000

t3= 2.02632

t4= 1.99351

t5= 2.00163

t6= 1.99959

t7= 2.00010

t8= 1.99997

t9= 2.00001

t10= 2.00000 This graph is similar to those we have seen on previous pages and we could conclude that a fraction with a value above 1 (numbers below 1 may show a different result) follows the same formula as the values of n that are integers.

Here is the continued fraction for when the value of k is a number below 1. General formula:  t1= 1.30000

t2= 1.06923

t3= 1.23525

t4= 1.10955

t5= 1.20127

t6= 1.13246

t7= 1.18304

t8= 1.14528

t9= 1.17315

t10= 1.15241 From looking at the graph we can see that the same general trend is present here, but for tn=tn-1the values of n must be larger. If we extend this graph we can see that the values come closer together nearer to when n=16 This student written piece of work is one of many that can be found in our GCSE Consecutive Numbers section.

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