Recurring Styles #11 - Smallest non-integral term of a Sequence!

Algebra Level 5

y n + 1 = 1 2 ( 3 y n + 5 y n 2 − 4 ) \large{y_{n+1} = \dfrac12 \left( 3y_n + \sqrt{5 y_n^2 - 4} \right)}

Define a sequence ( y n ) n ≥ 0 (y_n)_{n \geq 0} by the above recurrence relation such that y 0 = 1 , n ∈ Z y_0 = 1, \ n \in \mathbb Z . Find the smallest value of n n such that y n y_n is not an integer.

10 There exists no such smallest value of n n as every term of the sequence ( y n ) (y_n) is an integer. 17 12 4 15 3 5

Satyajit Mohanty by Satyajit Mohanty , 24, India

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1 solution

Surya Prakash
Aug 26, 2015

It can actually be proved by induction that y n ∈ Z ∀ n ∈ N y_{n} \in \mathbb{Z} \forall n \in \mathbb{N} . But we shall try to prove that in actual way.

y n + 1 = 1 2 ( 3 y n + 5 y n 2 − 4 ) y_{n+1} = \dfrac{1}{2} \left(3y_{n} + \sqrt{5y_{n} ^{2} - 4} \right) ( 2 y n + 1 − 3 y n ) 2 = 5 y n 2 − 4 (2y_{n+1} - 3y_{n})^{2} = 5y_{n} ^{2} - 4 y n + 1 2 − 3 y n y n + 1 + y n 2 = − 1 y_{n+1}^{2} - 3y_{n} y_{n+1} +y_{n} ^{2} = -1

Replace n n with n − 1 n-1 .

y n 2 − 3 y n − 1 y n + y n − 1 2 = − 1 y_{n} ^{2} - 3y_{n-1} y_{n} + y_{n-1} ^{2} = -1

Subtracting last two equations,

y n + 1 2 − 3 y n y n + 1 + 3 y n − 1 y n − y n − 1 2 = 0 y_{n+1}^{2} - 3y_{n} y_{n+1} + 3y_{n-1} y_{n} - y_{n-1} ^{2} = 0

Simplifying above yields two equations, either y n + 1 = y n y_{n+1} = y_{n} or y n + 1 + y n − 1 = 3 y n y_{n+1} + y_{n-1} = 3 y_{n} .

If y n + 1 = y n y_{n+1} = y_{n} , we get y 1 = y 0 = 1 y_{1} = y_{0} = 1 . But from the given reccurence we get y 1 = 2 y_{1} = 2 . So, y n + 1 ≠ y n y_{n+1} \neq y_{n} .

So, it implies that y n + 1 + y n − 1 = 3 y n y_{n+1} + y_{n-1} = 3 y_{n} . But, we have y 0 = 1 y_{0}=1 and from the given reccurence y 1 = 2 y_{1} = 2 . So, the reccurence y n + 1 = 3 y n − y n − 1 y_{n+1} = 3 y_{n} -y_{n-1} always produces integers because the coefficients of the reccurence relation are integers and the initial values are also integers.

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Moderator note:

This solution would be much more motivated if we first calculated the series of y n y_n , and then observed that it satisfied the linear recurrence y n + 2 − 3 y n + 1 + y n = 0 y_{n+2} - 3 y_{n+1} + y_n = 0 .

Note that the last paragraph of your solution still has a gap. What you have concluded is that "for all n n , we must either have y n + 1 = y n y_{n+1} = y_n or y n + 1 + y n − 1 = 3 y n y_{n+1} + y_{n-1} = 3 y_n ". What you have only shown, is that for n = 0 n = 0 , the first equation does not hold. You have to explain why "for all n n , the first equation does not hold".

What's this? Please recheck your solution. I don't agree to it. I'm obtaining y 1 = 2 , y 2 = 5 y_1 = 2 , \ y_2 = 5

Satyajit Mohanty - 5 years, 9 months ago

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Sorry, somewhere gone wrong. corrected it.

Surya Prakash - 5 years, 9 months ago

This solution would be much more motivated if we first calculated the series of y n y_n , and then observed that it satisfied the linear recurrence y n + 2 − 3 y n + 1 + y n = 0 y_{n+2} - 3 y_{n+1} + y_n = 0 .

Note that the last paragraph of your solution still has a gap. What you have concluded is that "for all n n , we must either have y n + 1 = y n y_{n+1} = y_n or y n + 1 + y n − 1 = 3 y n y_{n+1} + y_{n-1} = 3 y_n ". What you have only shown, is that for n = 0 n = 0 , the first equation does not hold. You have to explain why "for all n n , the first equation does not hold".

Calvin Lin Staff - 5 years, 9 months ago

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