Post on 26-Mar-2015
Call the Feds!
We’ve Got Nested Radicals!
Alan Craig
F. Lane Hardy Seminar
11-3-08
What are Nested Radicals?
2222
222
22
Examples:
We could keep this up forever!
222222
If we did, what would we get?
222222 = ?
Let’s work up to it.What are the values of these expressions?
? 22222
? 2222
? 222
? 22
? 2
Let’s work up to it.What are the values of these expressions?
? 22222
? 2222
? 222
? 22
1.414 2
What are the Values?
? 22222
? 2222
? 222
1.848 22
1.414 2
What are the Values?
? 22222
? 2222
1.962 222
1.848 22
1.414 2
What are the Values?
? 22222
1.990 2222
1.962 222
1.848 22
1.414 2
What are the Values?
1.998 22222
1.990 2222
1.962 222
1.848 22
1.414 2
What value is this sequence of numbers approaching?
1.998
1.990
1.962
1.848
1.414
Now what do you think the value of this infinite nested radical is?
? 222222
1.998 22222
1.990 2222
1.962 222
1.848 22
1.414 2
You’re Right!
2 222222
Let’s see an example of where an infinite nested radical could arise.
Warning: Brief Excursion into Trigonometry!
Trigonometry
Half-Angle Formula
• We will use the half-angle formula for cosine to take another look at this sequence and its limit.
2
cos1
2cos
Let’s use the formula to find .
so ,2
2
4cos Now,
24
cos1
8cos
8cos
Let’s use the formula to find .
222
1
24
cos1
8cos
8cos
Let’s rationalize the last expression by multiplying numerator and denominator by 2.
Let’s use the formula to find .
4
22
22
222
1
222
1
24
cos1
8cos
8cos
Let’s use the formula to find .
2
22
4
22
222
1
24
cos1
8cos
8cos
Let’s use the formula to find .
2
22
4
22
222
1
24
cos1
8cos
8cos
Now multiply both sides by 2.
Let’s use the formula to find .
228
cos2
2
22
4
22
222
1
24
cos1
8cos
8cos
Repeatedly using the ½ angle formula:
32cos22222
16cos2222
8cos222
Repeatedly using the ½ angle formula:
32cos22222
16cos2222
8cos222
As the angle gets smaller and smaller approaching 0, what value is the cos() approaching?
Repeatedly using the ½ angle formula:
32cos22222
16cos2222
8cos222
Recall cos(0) = 1, so
2 cos() is approaching 2 as approaches 0.
Repeatedly using the ½ angle formula:
2120cos22222
0 as 1cos
That is,
That’s all the trigonometry for this session.
We have shown in two different ways
that the equation ‘ought’ to be true
2 222222
To Recap:
Now let’s ‘prove’ it.
2 222222
Set x equal to the expression.
222222x
Square both sides.
2222222x
Subtract the original equation from the squared equation.
222222
2222222
x
x
2
222222
222222
2
2
xx
x
x
Subtract the original equation from the squared equation.
Now solve the equation.
22 xx
Solve the equation.
02
22
2
xx
xx
Solve the equation.
0)1)(2(
02
22
2
xx
xx
xx
Solve the equation.
2
0)1)(2(
02
22
2
x
xx
xx
xx
Why did we not use x = -1?
So
2 222222
What about?
333333
Does
333333
= 3 ???
Using the same process as before, we get
033
333333
333333
22
2
xxxx
x
x
Recall the Quadratic Formula
a
acbbxcbxax
2
40
22
032 xx• We have
• So a = 1, b = -1, and c = -3 and
2
131
)1(2
)3)(1(4)1()1( 2
x
x
So, No, we do not get 3
3.2333333
so ,3.22
131032
xxxx
Let’s ask a slightly different question.
• Is there a positive integer a, such that if we replace 3 under the nested radical with a, the nested radical will equal 3?
Let’s ask a slightly different question.
• That is, is there an a that makes the equation below true?
? 3 aaaaaa
Let’s ask a slightly different question.
• That is, is there an a that makes the equation below true?
• Yes! And we are going to find it.
? 3 aaaaaa
Subtract the original equation from the squared equation.
axx
aaaaaax
aaaaaax
2
2
Finding a
2
41102 a
xaxx
(Using the quadratic formula)
Finding a
We want x = 3, so
2
411 ax
32
411
a
Finding a
641132
411a
a
Finding a
2541541
641132
411
aa
aa
Finding a
6
2541541
641132
411
a
aa
aa
So we have shown that
3666666
Now let’s generalize our result.
• ‘Prove’ that for any integer k > 1, there is a unique positive integer a, such that
kaaaaaa
Note: The following is not a true mathematical proof of this theorem (which would use limits of bounded, monotonically increasing sequences) but does suggest the core reasoning and result of such a proof.
Finding a
12412
411
kak
a
Finding a
14441
12412
411
2
kka
kaka
Finding a
14441
12412
411
22
kkakka
kaka
Finding a
)1(
14441
12412
411
22
kka
kkakka
kaka
We have shown that
For any integer k > 1, there is exactly one integer a = k (k - 1), such that
kaaaaaa
We have shown that
For any integer k > 1, there is exactly one integer a = k (k - 1), such that
kaaaaaa
That is, every integer can be represented as an infinite nested radical!
Example: k = 4
4121212121212
1234)1( kka
5202020202020
2045)1( kka
Example: k = 5
Alternatively, we might have noticed that we need to solve
in such a way that we get two numbers that multiply to make a and subtract to make 1. Further, one of the numbers must be k. (Why?) Thus, the other number must be k - 1 and a must be k (k - 1).
Another Way
02 axx
That is
)1(
1
,1 and
:other the and numbers theof one be Let
2
2
kkkka
kakk
ak
k
ah
hkahk
hk
The END?
The END?
No!
This is way too much fun!
Let’s Kick it Up a Notch!
abababababa
abababababa
Note that what we did before was a special case of this expression with b = 1.
Let’s Kick it Up a Notch!
kabababababa
For each integer k > 1, there are exactly k - 1 pairs of integers a and b, 0 < b < k, that satisfy this equation. Further, ).( bkka
Let’s Kick it Up a Notch!
As before, square the equation.
abababababab
ax
abababababax
2
2
But before we subtract the original equation from the squared equation, we must isolate the radical (so that it will subtract away).
Now subtract.
02
2
xb
ax
abababababax
abababababab
ax
Now subtract.
00 22
2
abxxxb
ax
abababababax
abababababab
ax
We will solve this by factoring now but keep it in mind for later.
For integer solutions of
we need two integers that multiply to make a and have a difference of b. One of the numbers must be k, so the other is k - b. Thus,
Factor
02 abxx
)( bkka
There are exactly k – 1 such pairs a and b:
(k – 1) Pairs
1 1
2 2
3 )3(
2 )2(
1 )1(
)(
kk
kk
kk
kk
kk
bbkka
(difference)
Recall t
hat 0 < b < k
If k = 4, the k – 1 = 3 pairs a and b are:
Example: k = 4
3 414
2 824
1 1234
ba
Example: k = 4
482828282828
4121212121212
443434343434
One Last Thought
ba
b
ab
ab
ab
Consider this continued fraction:
ba
b
ab
ab
abx
Suppose it converges to x, then
ba
b
ab
ab
abx
Notice the shaded area is also x
Rewriting the continued fraction
x
abx
ba
b
ab
ab
abx
See what we get!
02
abxx
x
abx
ba
b
ab
ab
abx
Does this look familiar?
Yes, these are equal!!!
ababababa
ba
b
ab
ab
ab
In particular, set a = b = 1.
???111111111
11
1
11
11
11
The Golden Ratio
2
51111111111
11
1
11
11
11
(But that’s another F. Lane Hardy talk.)
EndEndEndEndEnd
?Reference
Zimmerman, S., & Ho, C. (2008). On infinitely nested radicals. Mathematics Magazine, 81(1), 3-15.