Call the Feds Weve Got Nested Radicals

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Call the Feds!Weve Got
Nested Radicals!

• Alan Craig
• F. Lane Hardy Seminar
• 11-3-08

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What are Nested Radicals?
Examples
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We could keep this up forever!
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If we did, what would we get?
?
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Lets work up to it.What are the values of these
expressions?
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Lets work up to it.What are the values of these
expressions?
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What are the Values?
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What are the Values?
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What are the Values?
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What are the Values?
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What value is this sequence of numbers
approaching?
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Now what do you think the value of this infinite
nested radical is?
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Youre Right!
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Lets see an example of where an infinite nested
radical could arise.Warning Brief Excursion
into Trigonometry!
Trigonometry
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Half-Angle Formula

• We will use the half-angle formula for cosine to
  take another look at this sequence and its limit.

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Lets use the formula to find .
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Lets use the formula to find .
Lets rationalize the last expression by
multiplying numerator and denominator by 2.
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Lets use the formula to find .
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Lets use the formula to find .
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Lets use the formula to find .
Now multiply both sides by 2.
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Lets use the formula to find .
Wow! A nested radical.
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Repeatedly using the ½ angle formula
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Repeatedly using the ½ angle formula
As the angle a gets smaller and smaller
approaching 0, what value is the cos(a)
approaching?
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Repeatedly using the ½ angle formula
Recall cos(0) 1, so 2 cos(a) is approaching 2
as a approaches 0.
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Repeatedly using the ½ angle formula
That is,
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Thats all the trigonometry for this session.
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We have shown in two different ways that the
equation ought to be true
To Recap
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Now lets prove it.
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Set x equal to the expression.
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Square both sides.
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Subtract the original equation from the squared
equation.
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Subtract the original equation from the squared
equation.
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Now solve the equation.
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Solve the equation.
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Solve the equation.
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Solve the equation.
Why did we not use x -1?
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So
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What about?
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Does
3 ???
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Using the same process as before, we get
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Recall the Quadratic Formula

• We have
• So a 1, b -1, and c -3 and

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So, No, we do not get 3
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Lets 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?

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Lets ask a slightly different question.

• That is, is there an a that makes the equation
  below true?

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Lets 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.

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Subtract the original equation from the squared
equation.
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Finding a
(Using the quadratic formula)
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Finding a

• We want x 3, so

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Finding a
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Finding a
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Finding a
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So we have shown that
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Now lets generalize our result.

• Prove that for any integer k gt 1, there is a
  unique positive integer a, such that

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.
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Finding a
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Finding a
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Finding a
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Finding a
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We have shown that

• For any integer k gt 1, there is exactly one
  integer a k (k - 1), such that

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We have shown that

• For any integer k gt 1, there is exactly one
  integer a k (k - 1), such that

• That is, every integer can be represented as an
  infinite nested radical!

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Example k 4
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Example k 5
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Another Way
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).
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That is
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The END?
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The END?

• No!
• This is way too much fun!

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Lets Kick it Up a Notch!
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Lets Kick it Up a Notch!
Note that what we did before was a special case
of this expression with b 1.
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Lets Kick it Up a Notch!
For each integer k gt 1, there are exactly k - 1
pairs of integers a and b, 0 lt b lt k, that
satisfy this equation. Further,
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As before, square the equation.
But before we subtract the original equation from
the squared equation, we must isolate the radical
(so that it will subtract away).
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Now subtract.
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Now subtract.
We will solve this by factoring now but keep it
in mind for later.
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Factor

• 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,

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(k 1) Pairs

• There are exactly k 1 such pairs a and b

(difference)
Recall that 0 lt b lt k
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Example k 4

• If k 4, the k 1 3 pairs a and b are

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Example k 4
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One Last Thought
Consider this continued fraction
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Suppose it converges to x, then
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Notice the shaded area is also x
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Rewriting the continued fraction
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See what we get!
Does this look familiar?
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Yes, these are equal!!!
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In particular, set a b 1.
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The Golden Ratio f
(But thats another F. Lane Hardy talk.)
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?
Reference Zimmerman, S., Ho, C. (2008). On
infinitely nested radicals. Mathematics Magazine,
81(1), 3-15.