Recurring Theme
Consider a recurrence relation, a n + 3 = 2 a n + 2 + a n + 1 − a n for n ≥ 0 and a 0 = 1 , find the sum of all possible values of n → ∞ lim a n a n + 1 , rounded to three significant figures. Here, a 1 and a 2 are arbitrary real numbers.
If you reach the conclusion that the limit fails to exist in some cases, enter 0.666
The answer is 2.00.
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1 solution
Are you saying that it works this way for all linear recurrences? What about something like a n + 3 = 3 a n + 2 − 3 a n + 1 + a n ?
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I was able to get that a n = f x n + g y n + h z n where f + g + h = 1 and x, y and z are roots of t 3 = 2 t 2 + t − 1 . How do I proceed?
Edit Got it now.
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@Srijit Mukherjee and @Deeparaj Bhat : This method works only if (a) The roots are distinct and (b) The roots have distinct moduli.
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@Otto Bretscher – Sir, I did a method similar to the ones above. Could you please tell how the conditions that you have mentioned are necessary.
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@Vignesh S – Consider the two examples I gave in this post, one with a multiple root and one with distinct roots of equal modulus.
@Vignesh S – Check out the wiki! Just added :)
@Otto Bretscher – If the roots are repeated, like in your comment above, I know a way out. But what to do if the moduli aren't distinct?
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@A Former Brilliant Member – Then the limit may not exist, as in a 1 = 1 , a 2 = 2 and a n + 2 = a n
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@Otto Bretscher – I meant with respect to solving the recurrence relation...
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@A Former Brilliant Member – The solution of the linear recurrence will be the same if the moduli aren't distinct as long as the roots are distinct.
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@Otto Bretscher – Thanks for that! Just one more thing, if the roots were repeated, even then the limits would've existed in all the cases and the answer would've been equal to the sum of all the roots, counted with multiplicities, right?
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@A Former Brilliant Member – Well, you don't count limits with multiplicities, so the sum may be different in that case... it's the sum of the distinct roots (again, as long as the moduli are distinct as well).
we are using a(n+1)/a(n)=L=a(n+2)/a(n+1)=a(n+3)/a(n+2) as n---->infinity using this we get a cubic equation the required solution is the negative of the coefficient of the L^2.