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# Infinite Surds Coursework

Extracts from this document...

Introduction

Petri Alexia

12T1

Standard Maths

Infinite Surds Coursework

The following expression is an example of an infinite surd. Find the formula for an+1 in terms of a

a1 = a2 = a2 = a3 = a3 = an+1 = an= Calculate the decimal values of the first ten terms of the sequence

a1 = 1.414213562373100

a2 = 1.553773974030040

a3 = 1.598053182478620

a4 = 1.611847754125250

a5 = 1.616121206508120

a6 = 1.617442798527390

a7 = 1.617851290609670

a8 = 1.617977530934740

a9 = 1.618016542231490

a10 = 1.618028597470230

Using technology, plot the relation between n and an. Describe what you notice.

 21 1.61803 22 1.61803 23 1.61803 24 1.61803 25 1.61803 26 1.61803 27 1.61803 28 1.61803 29 1.61803 30 1.61803 31 1.61803 32 1.61803 33 1.61803 34 1.61803 35 1.61803 36 1.61803 37 1.61803 38 1.61803 39 1.61803 40 1.61803
 n an 1 1.414213562373100 2 1.553773974030040 3 1.598053182478620 4 1.611847754125250 5 1.616121206508120 6 1.617442798527390 7 1.617851290609670 8 1.617977530934740 9 1.618016542231490 10 1.618028597470230 11 1.618032322752000 12 1.618033473928150 13 1.618033829661220 14 1.618033939588790 15 1.618033973558280 16 1.618033984055430 17 1.618033987299220 18 1.618033988301610 19 1.618033988611370 20 1.618033988707090

By plotting the relation between n and an, one notices that as n increases, anincreases. However this increase is not proportional to the increase of n, an

Middle

an. Describe what you notice.
 21 2.0000000000 22 2.0000000000 23 2.0000000000 24 2.0000000000 25 2.0000000000 26 2.0000000000 27 2.0000000000 28 2.0000000000 29 2.0000000000 30 2.0000000000 31 2.0000000000 32 2.0000000000 33 2.0000000000 34 2.0000000000 35 2.0000000000 36 2.0000000000 37 2.0000000000 38 2.0000000000 39 2.0000000000 40 2.0000000000 n an 1 1.847759065 2 1.9615705608 3 1.9903694533 4 1.9975909124 5 1.9993976374 6 1.9998494037 7 1.9999623506 8 1.9999905876 9 1.9999976469 10 1.9999994117 11 1.9999998529 12 1.9999999632 13 1.9999999908 14 1.9999999977 15 1.9999999994 16 1.9999999999 17 2.0000000000 18 2.0000000000 19 2.0000000000 20 2.0000000000 By plotting the relation between n and an, one notices that as n increases, anincreases. However this increase is not proportional to the increase of n, anseems to be increasing towards 2.

Once n reaches 17 anceases to increase, remaining stable at 2.

This suggests that as n becomes very large an – an+1 = 0

As such, we can conclude that the exact value for this infinite surd is 2.

Consider the general infinite surd : Find the formula for an+1 in terms of a

a1 = a2 = a2 = a3 = a3 =  x

Conclusion

>20

19

1368

342

6

120

30

20

1520

380

7

168

42

21

1680

420

8

224

56

22

1848

462

9

288

72

23

2024

506

10

360

90

24

2208

552

11

440

110

25

2400

600

12

528

132

26

2600

650

13

624

156

27

2808

702

14

728

182

28

3024

756

For to be an integer, has to be an odd perfect square.

 Values of odd perfect squares 57 Values of 3249 1 59 1 3481 3 61 9 3721 5 63 25 3969 7 65 49 4225 9 67 81 4489 11 69 121 4761 13 71 169 5041 15 73 225 5329 17 75 289 5625 19 77 361 5929 21 79 441 6241 23 81 529 6561 25 83 625 6889 27 85 729 7225 29 87 841 7569 31 89 961 7921 33 91 1089 8281 35 93 1225 8649 37 95 1369 9025 39 97 1521 9409 41 99 1681 9801 43 101 1849 10201 45 103 2025 10609 47 105 2209 11025 49 107 2401 11449 51 109 2601 11881 53 111 2809 12321 55 113 3025 12769

One can notice that the value of the odd perfect square is a series with  Therefore: y+1 is always an integer

The general statement that represents all the values of k for which the expression is an integer is with n any integer.

The general statement is valid for all integers and therefore has no limitations.

You can see from the steps above the process I took to find the general statement.

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