- Level: International Baccalaureate
- Subject: Maths
- Word count: 911
Infinite surds Maths Portifolio
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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
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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