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Call the Feds! We’ve Got Nested Radicals! Alan Craig
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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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Let’s work up to it. What are the values of these expressions?
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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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You’re Right!
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Let’s 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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Let’s use the formula to find.
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Let’s rationalize the last expression by multiply numerator and denominator by 2.
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Let’s use the formula to find.
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Now multiply both sides by 2.
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Let’s use the formula to find.
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Repeatedly using the ½ angle formula:
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As the angle gets smaller and smaller, what value is the cos( ) approaching?
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Repeatedly using the ½ angle formula: Recall cos(0) = 1, so 2 cos( ) is approaching 2 as approaches 0.
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That’s all the trigonometry for this session.
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Now let’s ‘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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Now solve the equation.
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Solve the equation.
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Why did we not use x = -1 ?
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So
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What about? = 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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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?
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Let’s ask a slightly different question. That is, is there an a that makes the equation below true?
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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.
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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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So we have shown that
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Now let’s generalize our result. ‘Prove’ that for any integer k > 1, there is a 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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We have shown that For any integer k > 1, there is exactly one integer a = k (k - 1), such that
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Examples
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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. Thus, the other number must be k - 1 and a must be k ( k - 1). Another Way
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That is
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The END?
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No! This is way too much fun!
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Let’s Kick it Up a Notch!
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Note that what we did before was a special case of this expression with b = 1.
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Let’s Kick it Up a Notch! For each integer k > 1, there are exactly k - 1 pairs of integers a and b, 0 < b < 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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Note one solution of this quadratic equation: Just keep this in mind for now. We will prove our assertion by factoring and come back to this later.
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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
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There are exactly k – 1 such pairs a and b : (k – 1) Pairs (difference)
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If k = 4, the k – 1 = 3 pairs a and b are : Example: k = 4
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One Last Thought Consider this continued fraction:
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Suppose it converges to x, then Does this look familiar?
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So these are equal!
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In particular, set a = b = 1. The Golden Ratio
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?
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