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Infinite Surds

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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: image00.pngimage00.png

When we see this, we can generate a pattern.  

a1= image07.pngimage07.png

a2= image15.pngimage15.png

a3 =image00.pngimage00.png

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

a1image01.pngimage01.png 1.414213562

a2image01.pngimage01.png 1.553773974

a3image01.pngimage01.png 1.598053182

a4image01.pngimage01.png 1.611847754

a5image01.pngimage01.png 1.616121206

a6image01.pngimage01.png1.617442798

a7image01.pngimage01.png 1.617851290

a8image01.pngimage01.png 1.617977531

a9image01.pngimage01.png 1.618016542

a10image01.pngimage01.png1.618028597

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:image08.png.  This pattern can also be clarified in a graph:

image09.png

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

...read more.

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:

image17.pngimage17.png

When one sees this, one can generate a pattern

b1 = image18.pngimage18.png

b2 = image19.pngimage19.png

b3 = image17.pngimage17.png

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

a1image01.pngimage01.png 1.847759065

a2image01.pngimage01.png 1.961570561

a3image01.pngimage01.png 1.990369453

a4image01.pngimage01.png 1.997590912

a5image01.pngimage01.png 1.999397637

a6image01.pngimage01.png 1.999849404

a7image01.pngimage01.png1.999984940

a8image01.pngimage01.png1.999990588

a9image01.pngimage01.png 1.999997647

a10image01.pngimage01.png 1.999999412

image20.png

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 = image21.pngimage21.pngn.  Again, one must find the asymptote:

an+1 =image21.pngimage21.png Get rid of the subscripts.

(a = image22.pngimage22.png2

...read more.

Conclusion

image03.pngimage03.png is odd if 4k+1 is a perfect square.  Now we can substitute integers and see what new patterns can be generated.

4 = image04.pngimage04.png8 = 1 + image05.pngimage05.png 7 =image06.pngimage06.png49 = 1+4k 48 = 4k k = 12

3 = image04.pngimage04.png6 = 1 + image05.pngimage05.png 5=image06.pngimage06.png25 = 1+4k 24 = 4k k = 6

5 = image04.pngimage04.png10 = 1 + image05.pngimage05.png 9 =image06.pngimage06.png81 = 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.

...read more.

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

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