- Level: International Baccalaureate
- Subject: Maths
- Word count: 945
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 | 1.618033988736670 |
22 | 1.618033988745810 |
23 | 1.618033988748630 |
24 | 1.618033988749500 |
25 | 1.618033988749770 |
26 | 1.618033988749860 |
27 | 1.618033988749880 |
28 | 1.618033988749890 |
29 | 1.618033988749890 |
30 | 1.618033988749890 |
31 | 1.618033988749890 |
32 | 1.618033988749890 |
33 | 1.618033988749890 |
34 | 1.618033988749890 |
35 | 1.618033988749890 |
36 | 1.618033988749890 |
37 | 1.618033988749890 |
38 | 1.618033988749890 |
39 | 1.618033988749890 |
40 | 1.618033988749890 |
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
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
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.
Page of
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
Found what you're looking for?
- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month