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# Infinite Surds

Extracts from this document...

Introduction

Infinite Surds

An infinite surd is a never ending irrational number.  Its exact value would be left in square root form.

The following example is the infinite surd of 1.  This is the first infinite surd being investigated:

When we see this, we can generate a pattern.

a1=

a2=

a3 =

So using this pattern, we can find the next 10 consecutive terms.

a1 1.414213562

a2 1.553773974

a3 1.598053182

a4 1.611847754

a5 1.616121206

a61.617442798

a7 1.617851290

a8 1.617977531

a9 1.618016542

a101.618028597

We start to see a pattern form.  We see that for each consecutive surd there is 1+ added to the previous.  This can be shown using an+1in terms of an:.  This pattern can also be clarified in a graph:

In this graph, one can see the pattern as well.  One sees that the graph is already approaching its asymptote, suggesting that even if an is greater than a10, the numbers will still be very close to 1.618.  One can already see its horizontal trend.

To find what an is, one must perceive an

Middle

The asymptote is approximately 1.618.  This makes sense when one looks at the data.  Since the graph only shows numbers above 0, there are no negative numbers.  Therefore, one must only look for the positive answer.  One can also see that there is a limit for our “a” value.  One can take this process when solving any other infinite surd, example:

When one sees this, one can generate a pattern

b1 =

b2 =

b3 =

Using this pattern, one can find the next 10 consecutive terms.

a1 1.847759065

a2 1.961570561

a3 1.990369453

a4 1.997590912

a5 1.999397637

a6 1.999849404

a71.999984940

a81.999990588

a9 1.999997647

a10 1.999999412

One can also observe that this graph also reaches an asymptote.  This means that an approaches 2, but never quite reaches it.  The formula that one can see is an+1 = n.  Again, one must find the asymptote:

an+1 = Get rid of the subscripts.

(a = 2

Conclusion

is odd if 4k+1 is a perfect square.  Now we can substitute integers and see what new patterns can be generated.

4 = 8 = 1 +  7 =49 = 1+4k 48 = 4k k = 12

3 = 6 = 1 +  5=25 = 1+4k 24 = 4k k = 6

5 = 10 = 1 +  9 =81 = 1+4k 80 = 4k k = 20

A general statement evolved from this is to use the value of k (2) and multiply it by the surd than add it to the “k” value found and you will find the next consecutive term.

Example:

(4(2)) + 12 = 20

(3(2)) + 6 = 12

Therefore we can predict that

(5(2)) + 20 = 30

When 6 is the infinite surd, k will be 30.

This statement however does have a limitation, it being that this statement needs a value of k to find the next consecutive term.  However if using the formula, we can get k and then apply this generalization. This generalization is just a pattern formulated as more examples were being created and developed.  This is why when solving such a complex mathematical problem, one must always work it out and break it down where necessary.

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

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