Infinite Surds portfolio
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Introduction
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.
√1+√1+√1+√1+…
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
Middle
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
Conclusion
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.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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