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Infinite surds Maths Portifolio

Extracts from this document...

Introduction

Mathematics SL

Portifolie

Infinite Surds

An infinite surd is a never-ending positive irrational number. It is a number that can only be expressed exactly using the root sign . This sequence above is known as an infinite surd and can be expressed in the terms ofan:

a1 = = 1.414213

a2 = = 1.553773

a3= = 1.598053

a4 = = 1.611847

a5 = = 1.616121

a6= = 1.617442

a7= = 1.617851

a8= = 1.617977

a9= = 1.618016 etc.

This is the first ten terms and the formula for these sequences is: because if we use the term a2 as an example, this could be proven as: a1+1 = a2 = a2 = 1.553773

By plotting a graph of the ten first term of this sequence the relationship between n and an could be shown: As can be seen from this graph is that the values increase, but then flattens out. The values of an moves towards the value of 1.618 approximately, but will actually never reach it. This can be understood by:

an - an+1

as n gets very large.

lim(

Middle

= 1.990369453

b4 = = 1.997590912

b5 = = 1.999397637

b6 = = 1.999849404

b7 = = 1.999962351

b8 = = 1.999990588

b9 = = 1.999997647 By replacing an by bn the formula for this sequence is:

bn+1 = because if we use b2 as an example, the answer will be:

b1+1 = b2 = b2 = 1.961570561

By plotting a graph, the relationship between n and bn can be shown:  As can be seen from this graph is that the value of bn increases as n gets larger. After a while the curve flattens out, but continue to rise. The values of bn approaches 2, but will actually never reach it. This can be understood by:

bn – bn+1

As n becomes very large, the value approaches 0 because:

lim(bn – bn+1) → 0

When n becomes very large and reaches infinity, the value approaches 0. This is because the difference between the two values is so small that the difference becomes insignificant.

The exact value of this infinite surd is:

bn+1 can also be written as bn so:

bn = Conclusion

k = 2.1

an = an = an = 2.03297

Ex. 6

k = 2.5

an = an = an = 2.15831

As can be seen from example number 1 is that if k is a negative number we get square root of a negative number and therefore there will be no solution. Because of this, k cannot be a negative number.

From example 2 and 4 this can be found: There are few numbers that gives the answer of an integer in this sequence expressed with k. Some of the values are the numbers 0, 2, 6, 12 and 20 etc. From this we can see an increasing trend were:

0 = 2 2 = 2 6 = 2 12 = 2 20 = 2 We can see that from 0 to 2 the number we multiply by the constant, 2, increases by 1, and from 2 to 6 it increases by 2. From 6 to 12 it increases by 3, and from 12 to 20 it increases by 4.

If we use other numbers from 0 to 20 than those above, we get decimal numbers. This can be seen from example 3, 5 and 6. As k increases, the value of an increases as well.

The scope is 0< n > , because n cannot be a negative number.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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