Megan Griffith ID#: 0000007-‐013 December 16, 2011 Ms. Griffith IB Mathematics SL, 2nd period
Infinite Surds IB Mathematics SL Portfolio, Type I
I. Introduction The following is an example of an infinite surd:
1+ 1+ 1+ 1+ 1... Infinite surds are radicals within radicals within radicals that go on forever. We can express some infinite surds as real numbers, and even some as integers. In this portfolio, I will show the steps I took to find which values of k (the number inside this type of infinite surd) will yield an integer value for the infinite surd. Here are the steps I took:
1. Tested two values of k (k = 1 and k = 2) to find the values of the infinite surd. 2. Developed an expression for the general case (k) based on the pattern I recognized in
the previous two cases. 3. Developed a general statement for the possible values of k that would give an integer
value for the entire surd. 4. Tested my general statement. 5. Found the scope and limitations of my general statement.
II. Data Collection
A. Infinite Surd, k=1 We can also consider the above infinite surd as a sequence of terms an , where:
a1 = 1+ 1
a2 = 1+ 1+ 1
a3 = 1+ 1+ 1+ 1 etc. I found that I was able to substitute part of each term with the previous term. For example:
a2 = 1+ a1 a3 = 1+ a2 etc.
This led me to a formula to describe an+1 in terms of an : an+1 = 1+ an
After finding the formula for an+1 , I calculated the decimal values of the first ten terms of the sequence. I have included them in the table below. The decimal values have been included to five decimal places in order to show best how the terms change in value. n an an simplified an decimal 1 1+ 1 1+ 1 1.41421
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2 1+ 1+ 1 1+ a1 1.55377
3 1+ 1+ 1+ 1
1+ a2 1.59805
4 1+ 1+ 1+ 1+ 1
1+ a3 1.61185
5 1+ 1+ 1+ 1+ 1+ 1
1+ a4 1.61612
6 1+ 1+ 1+ 1+ 1+ 1+ 1
1+ a5 1.61744
7
1+ 1+ 1+ 1+ 1+ 1+ 1+ 1 1+ a6 1.61785
8
1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
1+ a7 1.61798
9
1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
1+ a8 1.61801
10
1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
1+ a9 1.61802
As n is getting the larger, the value of an is increasing more and more slowly. This can also be seen in a plot of the relation between n and an below. The plot was developed using a TI-‐84 calculator/MathCracker.com’s online scatterplot tool, etc.
When graphed, the relation looks similar to the square root parent function y = x , which makes sense considering we are dealing with surds. The graph also appears to increase more quickly for smaller values of n, and it appears to be leveling off or approaching a limit as n gets
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larger. That means the difference between consecutive terms is getting smaller. We can confirm that in the following table, by calculating the values of an ! an!1 for each value of n we have found so far: n an an ! an+1 1 1.41421 -‐ 0.13956 2 1.55377 -‐ 0.04428 3 1.59805 -‐ 0.01380 4 1.61185 -‐ 0.00427 5 1.61612 -‐ 0.00132 6 1.61744 -‐ 0.00041 7 1.61785 -‐ 0.00013 8 1.61798 -‐ 0.00003 9 1.61801 -‐ 0.00001 10 1.61802 -‐-‐ Clearly, as n is getting larger, the difference between an and an ! an+1 is getting smaller. In fact, we should expect that as n approaches infinity, the difference between an and an ! an+1 approaches zero. We can use that knowledge to set up the following equation to find an expression for an as n approaches infinity:
an ! an+1 = 0 By substituting an+1 = 1+ an , we can rearrange to see an :
an ! an+1 = 0
an ! 1+ an = 0
an = 1+ anan2 =1+ an
an2 ! an !1= 0
This is a quadratic, which can be solved using the quadratic formula: a =1,!b = !1,!c = !1
an =!b± b2 ! 4ac
2a
an =!(!1)± (!1)2 ! 4(1)(!1)
2(!1)
an =1± 52
The quadratic formula yields two answers, 1.618… and -‐0.618…. Only the first (positive) answer is valid for our investigation, since a radical with no sign in front is implied to be
positive. Therefore, we can express the infinite surd, an , as 1+ 52
. This is the Golden Ratio.
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However, so far we only know this is the case for an infinite surd using 1 (as in,
1+ 1+ 1+ 1+ 1... ). We will follow a similar investigation for the infinite surd using 2 (
2+ 2+ 2+ 2+ 2... ) below.
B. Infinite Surd, k=2 Similar to how we wrote out the initial infinite surd as a sequence of terms, we can do the same with our new infinite surd. To distinguish the two, we will write out our new sequence as bn :
b1 = 2+ 2
b2 = 2+ 2+ 2
b3 = 2+ 2+ 2+ 2 etc. We can also use the same method as with the first infinite surd to express each term using the previous term: b2 = 2+ b1 b3 = 2+ b2 etc. This led me to a formula to describe bn+1 in terms of bn : bn+1 = 2+ bn After finding the formula for bn+1 , I calculated the decimal values of the first ten terms of the sequence. I have included them in the table below. The decimal values have been included to six decimal places in order to show best how the terms change in value. n bn bn simplified bn decimal 1 2+ 2 2+ 2 1.847759
2 2+ 2+ 2 2+ b1 1.961571
3 2+ 2+ 2+ 2 2+ b2 1.990369
4 2+ 2+ 2+ 2+ 2
2+ b3 1.997591
5 2+ 2+ 2+ 2+ 2+ 2
2+ b4 1.999398
6 2+ 2+ 2+ 2+ 2+ 2+ 2
2+ b5 1.999849
7
2+ 2+ 2+ 2+ 2+ 2+ 2+ 2 2+ b6 1.999962
8
2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2
2+ b7 1.999991
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9
2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2
2+ b8 1.999997
10
2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2
2+ b9 1.999999
As n is getting the larger, the value of bn is increasing more and more slowly (It seems to be approaching 2.). This can also be seen in a plot of the relation between n and bn below. The plot was developed using a MathCracker.com’s online scatterplot tool.
When graphed, the relation still looks similar to the square root parent function y = x . The graph also appears to increase more quickly for smaller values of n, and it appears to be leveling off or approaching a limit as n gets larger. That means the difference between consecutive terms is getting smaller. We can confirm that in the following table, by calculating the values of bn ! bn!1 for each value of n we have found so far, just as we did with the first infinite surd. n bn bn ! bn+1 1 1.847759 -‐ 0.113812 2 1.961571 -‐ 0.028798 3 1.990369 -‐ 0.007222 4 1.997591 -‐ 0.001807 5 1.999398 -‐ 0.000451 6 1.999849 -‐ 0.000113 7 1.999962 -‐ 0.000029 8 1.999991 -‐ 0.000006 9 1.999997 -‐ 0.000002 10 1.999999 -‐-‐
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Clearly, as n is getting larger, the difference between bn and bn ! bn+1 is getting smaller. We would expect that as n approaches infinity, the difference between bn and bn ! bn+1 approaches zero. In fact, we had to show more decimal places for bn and bn ! bn+1 than with an and an ! an+1 , as it starts to level off more quickly. Just as we did with the initial infinite surd, we can use the knowledge that the difference between consecutive terms is approaching zero to set up the following equation to find an expression for bn as n approaches infinity:
bn ! bn+1 = 0 By substituting bn+1 = 2+ bn , we can rearrange to see bn :
bn ! bn+1 = 0
bn ! 2+ bn = 0
bn = 2+ bnbn2 = 2+ bn
bn2 ! bn ! 2 = 0
This still yields a quadratic, which can be solved using the quadratic formula: a =1,!b = !1,!c = !2
bn =!b± b2 ! 4ac
2a
bn =!(!1)± (!1)2 ! 4(1)(!2)
2(!1)
bn =1± 92
=1±32
The quadratic formula yields two answers, 2 and -‐1. Only the first (positive) answer is valid for our investigation, since a radical with no sign in front is implied to be positive. Therefore, we can express the infinite surd, an , as 2 . This is not the Golden Ratio. Instead, it is the number repeated inside the infinite surd.
III. General Statement
A. Infinite Surd, k= any number It would be worth our while to find a way to express all infinite surds in the form
k + k + k + k + k... to avoid going through the same long process for each value k. From our previous investigations, we know that an ! an+1 = 0 and bn ! bn+1 = 0 . It follows that the same would be the case for this infinite surd, which we will express as cn . Therefore:
an ! 1+ an = 0
bn ! 2+ bn = 0
cn ! k + cn = 0
We can use this to solve for cn using the quadratic formula, just as we did for the other two infinite surds:
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cn ! k + cn = 0
cn = k + cncn2 = k + cn
cn2 ! cn ! k = 0
a =1,!b = !1,!c = !k
cn =!b± b2 ! 4ac
2a
cn =!(!1)± (!1)2 ! 4(1)(!k)
2(1)
cn =1± 1+ 4k
2
We saw with the first two surds that only the positive solution is valid (1+ 1+ 4k2
).
Before we move any further with the general statement we have found, we need to test it again with the data we used previously.
For k = 1 (the first infinite surd), we get the following: 1+ 1+ 4(1)2
=1+ 52
For k = 2 (the second infinite surd), we get: 1+ 1+ 4(2)2
=1+ 92
=1+32
= 2
This matches what we found during our data collection, so we can now move on to judging the scope and limitations of our expression for the infinite surd with k.
B. Numbers of k that give an integer answer Even though this formula yielded an integer value (2) for the second infinite surd, it yielded an irrational number (granted, the Golden Ratio) for the first infinite surd. We can find all the values for which cn is an integer by first setting up an equation where cn equals an integer, Z. 1+ 1+ 4k
2= Z
By solving for k, we can see the possible values of k that will yield integer values for the infinite surd. 1+ 1+ 4k
2= Z
1+ 1+ 4k = 2Z
1+ 4k = 2Z !11+ 4k = (2Z !1)2
4k = (2Z !1)2 !1
k = (2Z !1)2 !1
4
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This means that for the infinite surd to equal an integer, k must equal (2Z !1)2 !1
4 where Z is
an integer. To test this, we will use a variety of real numbers.
IV. Testing General Statement
A. Test Values Z=-‐4, 6, 0, 0.67, 3 Although we cannot yet prove our general statement, we can test it with various values to see whether it works for all of those. In each case, we need to test a value of Z (some of which are integers and some of which are not), find the corresponding value of k, and find the value of the infinite surd to see whether it will yield an integer value: Z k Infinite Surd Infinite Surd -‐4
k = (2(!4)!1)2 !1
4= 20 1+ 1+ 4(20)
2
5
0 k = (2(0)!1)
2 !14
= 0 1+ 1+ 4(0)2
1
0.67 k = (2(0.67)!1)
2 !14
= !0.2211 1+ 1+ 4(!0.2211)2
0.67
3 k = (2 3 !1)2 !14
"1.27 1+ 1+ 4(1.27)2
1.73… ( 3 )
6 k = (2(6)!1)
2 !14
= 30 1+ 1+ 4(30)2
6
B. Judging the validity of the general statement My general statement held true for all of my test values, which means that it is still valid. Although I have not conclusively proven it, I am satisfied that it is a good working general statement because it worked for a variety of test values.
V. Scope and Limitations
A. Domain of the infinite surd
We know that to obtain an integer for the infinite surd, k must equal (2Z !1)2 !1
4 where Z is an
integer. However, there is a broader restriction on the value of k, since the value of the
infinite surd in general is 1+ 1+ 4k2
. Since we want to find real values for the infinite surd,
the value inside the radical (1 + 4k) must be greater than or equal to zero: 1+ 4k ! 04k ! "1
k ! " 14
Thus, k must always be at least -‐0.25 for the infinite surd to be a real number.
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B. Range of the infinite surd As long as k ≥ -‐0.25, the infinite surd will give us a positive real number because the radical in 1+ 1+ 4k
2 will always give zero or a positive, the 1 in the numerator is positive, and the 2 in
the denominator is also positive. This means the smallest real value for the infinite surd is ½, but there is no limit on how large it can be.
VI. Conclusion Infinite surds can sometimes be expressed as integers. I was able to find which infinite surds can be simplified to integers by breaking up the infinite surd into a sequence of terms. I could have improved my investigation by testing more even and odd values of k to see if there was a pattern connected with those, or if prime numbers k gave a different pattern than composite values for k.