• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9

Continued Fractions

Extracts from this document...

Introduction

        Anousha Manji

        Maths Portfolio

Continued Fractions

image00.png

This ‘infinite fraction’ can be considered as a sequence of terms, tn:

image01.pngimage01.png

image18.png

image27.png
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. image36.pngimage36.png

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

image44.png

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.

...read more.

Middle

image07.png

 (We know that image08.pngimage08.pngmust 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:

image09.png

image10.png

image11.png

image12.png

General formula: image13.pngimage13.png

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

image14.png

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

...read more.

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.

image33.png

image34.pngimage34.pngGeneral formula: image35.pngimage35.png

image37.png

image38.png

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

image39.png

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.

image40.png

General formula: image41.pngimage41.png

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

image42.png

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

image43.png

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Consecutive Numbers essays

  1. GCSE Maths Coursework - Maxi Product

    2x6=12 (3,5)= 8 --> 3+5 --> 3x5=15 (4,4)= 8 --> 4+4 --> 4x4=16 I have found that 16 is the highest number so far that can be retrieved from 4 and 4 when the number is 8, in whole numbers.

  2. Investigate the Maxi Product of numbers

    Results of Numbers Number Two numbers used Two numbers added Two numbers multiplied Maxi Product 12 5 and 7 5+7 5x7 35 12 6 and 6 6+6 6x6 36- Maxi Product 12 4 and 8 4+8 4x8 32 12 6.5 and 5.5 6.5+5.5 6.5x5.5 35.75 12 6.7 and 5.3 6.7+5.3

  1. Portfolio: Continued Fractions

    but also because the difference between the 200th term and the 199th term is so small that it is hard, even with a computer, to find a value of the 200th term which has enough significant figures to differ from the 199th term.

  2. In this investigation I will explore the relationship between a series of straight, non-parallel, ...

    is the number of lines in the diagram (where n can be any natural number) then By inspection, CR(1) = 0 CR(2) = 0 CR(3) = 1 CR(4) = 3 CR(5) = 6 CR(6) = 10 This sequence can be written as: Terms (n)

  1. Nth Term Investigation

    x 2 8 404 398 99 Cuboids 2 Another set of cuboids. Here is a table showing my results: n x n x n 1 x 3 x 3 8 16 8 0 2 x 3 x 3 8 20 16 4 3 x 3 x 3 8 24 24

  2. Investigate calendars, and look for any patterns.

    9 6 11 7 12 As you can see from the table above, the formula is quite simple: n = n + 5 Ex 3.3 The third row in June: n 1 13 2 14 3 15 4 16 5 17 6 18 7 19 The above table gives another

  1. To investigate consecutive sums. Try to find a pattern, devise a formulae and establish ...

    20 = 74 18 + 19 + 20 + 21 = 78 19 + 20 + 21 + 22 = 82 0 + 1 + 2 + 3 + 4 = 10 1 + 2 + 3 + 4 + 5 = 15 2

  2. About Triangular Square Numbers

    55420693056, that it is the square of 235416, and that it is the 332928th triangular number. Also, which is closer to = 1.41421356237... My conjecture is that this algorithm generates a countable infinity of triangular squares, and that any and all triangular squares which exist will be generated by this algorithm.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work