Let R 1 = 1 6 2 1 + 1 6 2 1 + 1 6 2 1 + 1 6 2 1 + 1 6 2 1 + . . . R 2 = 1 6 2 1 − 1 6 2 1 − 1 6 2 1 − 1 6 2 1 − 1 6 2 1 − . . . R 3 = 1 6 2 1 + 1 6 2 1 − 1 6 2 1 + 1 6 2 1 − 1 6 2 1 + . . . R 4 = 1 6 2 1 − 1 6 2 1 + 1 6 2 1 − 1 6 2 1 + 1 6 2 1 − . . . .
Given that the values of the four infinitely nested radical expressions above can be expressed
R 1 = q p 1 , R 2 = q p 2 , R 3 = q p 3 , R 4 = q p 4
where p 1 , p 2 , p 3 , p 4 , q are positive integers with g cd ( p i , q ) = 1 for 1 ≤ i ≤ 4 , find p 1 + p 2 + p 3 + p 4 .
Clarification: as a sequence, R 4 is defined in the following way: R 4 1 : = 1 6 2 1 , R 4 n : = 1 6 2 1 − 1 6 2 1 + R 4 n − 1 , where R 4 n is n -th term of the sequence. R 4 : = n → ∞ lim R 4 n .
R 3 is defined analogously.
Warning: in R 4 you may need to extract square root from a negative number. This root is always principal (i.e − 2 = i 2 , not − i 2 ).
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Finding the fixed points is the easy part... the challenge is to verify convergence!
Algebric ways.Not calculus...
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Convergence involves calculus, by definition ;)
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We see that R 1 = 1 6 2 1 + R 1 R 2 = 1 6 2 1 − R 2 Solving and we get R 1 = 4 7 ; R 2 = 4 3 .
Now for R 3 and R 4 , we create a system of equation ⎩ ⎨ ⎧ R 3 = 1 6 2 1 + R 4 R 4 = 1 6 2 1 − R 3 Solving it and we get { R 3 = 4 5 R 4 = 4 1 .So now ⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ p 1 = 7 p 2 = 3 p 3 = 5 p 4 = 1 ⇒ p 1 + p 2 + p 3 + p 4 = 1 6