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Infinite Surds portfolio

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Internal Assessment number 1

Nazha AlFaraj

Ms. Leana Ackerman

IB Mathematics SL (year 2)

Sunday, February 19, 2012

Infinite Surds

This following expression is known as an infinite surd.


The previous infinite surd can be changed into the following sequence:

a1= √1+√1= 1,414213

a2= √1+√1+√1= 1,553773

a3= √1+√1+√1+√1= 1,598053

a4= √1+√1+√1+√1+√1= 1,611847

a5= √1+√1+√1+√1+√1+√1= 1,616121

a6= √1+√1+√1+√1+√1+√1+√1= 1,617442

a7= √1+√1+√1+√1+√1+√1+√1+√1= 1,617851

a8= √1+√1+√1+√1+√1+√1+√1+√1+√1= 1,617977

a9=√1+√1+√1+√1+√1+√1+√1+√1+√1+√1= 1,618016

a10= √1+√1+√1+√1+√1+√1+√1+√1+√1+√1= 1,618028

The first 10 terms can be represented by:

an+1= √1 + an

If we

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The data begins to increase by a smaller amount about each consecutive n, suggesting
that the data may be approaching as asymptote. As these values get very large, they willprobably not get much higher than the value of a10, because there already appears to bealmost horizontal trend. The data also suggests that the asymptote is between the value of 6 and seven, although to find the exact value requires a different approach

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x²= √k+√k+√k…²

x²= k+ √k+√k+√k…

Because we are working with an infinite surd we can deduce that:

x² = k + x

0= k + x – x²

0 = (x+k)(x-k)

The null factor law can be used to portray that any value of k represents an integer.

(x + 4) (x – 4) = 0

→ x² - 4x + 4x – 16 = 0

→ x² - 16 = 0

→ x² = 16

→ x = 4

 As we compare this result to the general statement we provided we can easily establish that our general statement is valid.

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