Use iteration to guess an explicit formula

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Homework Help Overview

The discussion revolves around finding an explicit formula for the recurrence relation defined by u_k = u_{k−2} * u_{k−1}, with initial conditions u_0 = u_1 = 2. The original poster attempts to express the solution in terms of the Fibonacci sequence.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the values of u_k for various k and explore the relationship to the Fibonacci sequence. There are questions about the validity of the derived expressions and the formatting of mathematical notation.

Discussion Status

Some participants affirm the original poster's reasoning and suggest that they may have reached a satisfactory point in their exploration. There is mention of needing to prove the conjectured formula, indicating ongoing engagement with the problem.

Contextual Notes

There is uncertainty regarding the proper formatting for mathematical expressions in the forum, which may affect the clarity of communication. The original poster expresses some hesitation about the completeness of their solution.

Arew
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Homework Statement



Use iteration to guess an explicit formula for u_k = u_{k−2} * u_{k−1}, for all integers k ≥ 2, u_0 = u_1 = 2 and prove it .

Homework Equations



Hint: Express the answer using the Fibonacci sequence.

The Attempt at a Solution



u_k = u_{k−2} * u_{k−1} and u_0 = u_1 = 2, so

u_2 = 2^2

u_3 = 2^3

u_4 = 2^5

u_5 = 2^8

u_6 = 2^13

u_7 = 2^21

Then (I think) in general we have ,

2^{F_k} = 2^{F_k-1 + F_k-2}

Not finished yet... But does it make sense so far?

BTW, I tried putting $ signs around expressions, but mathjax(?) doesn't seem to render in preview. [/B]
 
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Arew said:
But does it make sense so far?
Yes.
Arew said:
BTW, I tried putting $ signs around expressions, but mathjax(?) doesn't seem to render in preview.
That was dollar signs in pairs, right? $$latex$$. Not sure if mathjax is same as LaTeX.
 
haruspex said:
Yes.

That was dollar signs in pairs, right? $$latex$$. Not sure if mathjax is same as LaTeX.

Oh, I never knew you had to put the dollar signs in pairs. Thanks.$$2^{F_k} = 2^{F_{k-1} + F_{k-2}}$$ then $$2^{F_n} = 2^ {\frac{\frac {1 + \sqrt 5}{2} - \frac {1 + \sqrt 5}{2}}{\sqrt 5}}$$ which implies $$\log_22^{F_n} = \log_22^ {\frac{\frac {1 + \sqrt 5}{2} - \frac {1 + \sqrt 5}{2}}{\sqrt 5}}$$ meaning $$F_n = {\frac{\frac {1 + \sqrt 5}{2} - \frac {1 + \sqrt 5}{2}}{\sqrt 5}}$$

Can I claim this to be the solution to the given recurrence relation?
 
Arew said:

Homework Statement



Use iteration to guess an explicit formula for u_k = u_{k−2} * u_{k−1}, for all integers k ≥ 2, u_0 = u_1 = 2 and prove it .

Homework Equations



Hint: Express the answer using the Fibonacci sequence.

The Attempt at a Solution



u_k = u_{k−2} * u_{k−1} and u_0 = u_1 = 2, so

u_2 = 2^2

u_3 = 2^3

u_4 = 2^5

u_5 = 2^8

u_6 = 2^13

u_7 = 2^21

Then (I think) in general we have ,

2^{F_k} = 2^{F_k-1 + F_k-2}

Not finished yet... But does it make sense so far?

BTW, I tried putting $ signs around expressions, but mathjax(?) doesn't seem to render in preview. [/B]
Suggest you google Fibonacci series. I think you might have already solved this one as much as you can in post #1. (You still need to prove that your guess is correct.)
 
Last edited:
Charles Link said:
Suggest you google Fibonacci series. I think you might have already solved this one as much as you can in post #1. (You still need to prove that your guess is correct.)

I was unsure. Usually when we find an explicit formula for a recurrence, we get rid of indices in the recurrence relation so that we do at most one substitution into the formula. But if you say I am done - I am happy to take a break :) As for the proof, I think induction should take care of it. Thanks.
 

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