# IB Math SL portfolio

IB Math SL Portfolio Type 1

This work was done in accordance with the NA honor Code x_______________________

The following is as example of and infinite surd of 1:

1+     1+      1+       1+ …

The progression from one surd to another, can be used to formulate a pattern.

a1 =    1  +     1

a2 =      1 +     1 +       1

a3 =        1 +    1 +     1 +       1

A pattern appears to form. After each consecutive term there is an additional     1 +…

added onto the previous term in the sequence.  This pattern can be utilized to invent a formula for an+1 in terms of an.

an+1 =    1 + an

Now, using a TI–83+, the first ten terms of this sequence are as follows:

a1 ≈ 1.4142135623731

a2 ≈ 1.55377397403

a3 ≈ 1.5980531824786

a4 ≈ 1.6118477541252

a5  ≈ 1.6161212065081

a6  ≈ 1.6174427985274

a7  ≈ 1.6178512906097

a8  ≈ 1.6179775309347

a9  ≈ 1.6180165422315

a10≈ 1.6180285974702

The graph of these decimal values appears as:

The data begins to increase by smaller increments at each consecutive n, suggesting that the data is approaching an asymptote.  As these values grow larger, they will in all likelihood not increase above the value of a10, because there already appears to be a horizontal trend by the tenth term. Analyzing the graph, one can determine that the asymptote is between the value of 6 and seven, although finding the exact value of said asymptote requires a different approach.

Given the general statement found above:

an+1 =    1 + an

By treating the values of an+1 and an as one variable (a), an exact answer can be calculated.

( a=   1 + a    )2

a2 = 1 + a

a2 – a – 1 = 0

This simplified equation has the appearance of a quadratic regression equation, or a set of data with a “best fit” of a parabolic like shape.

The equation can be simplified even more by removing the variable from the rest of the equation, thus allowing for the use of the quadratic formula to solve the equation.

a = 1

b = -1

c = -1

a =  1±    (-1)2–4(1)(-1)

2(1)

a =  1 ±   5

2

Looking back at the data, it appears that the answer cannot equal a negative number, an obvious indicator being that the graph shows all values above zero, and there that there are no negatives in the initial equation.  Therefore the answer is:

a =  1 +   5

2

In decimal form:

a ≈ 1.618033989

This number completes the criterion of the graph, as it is between 1.6 and 1.7, directly where the asymptote should be.  Also, the decimal value of a10≈ 1.6180285974702, which is shown by the graph, is very close to the asymptote.  The difference between the approximate value for the infinite surd and a10 is only 5.3912797∙10-6, or 0.0000053912797, which is very precise and means that the asymptote is very near a10, like the graph predicts. If this ...