Hello everyone
There are 2n inequations:
and
where
Proof that
(3)
i could really use a hand here, I dont know how to solve it
I thought , but and therefor is possible.
Help is much appreciated
Thank you,
Rapha
Hello Rapha-
I haven't got a complete proof in an elegant form, but I think it's all to do with the terms of the Fibonacci sequence. If the term of the Fibonacci sequence is , then you can prove that
Write the inequalities and on alternate lines, and, in the case of the , write the term before the term:
(Note that this final inequality does not have the term in a, because
We now have 2n inequalities. Then, for to multiply the inequality by the term in the Fibonacci sequence, , and the final inequality by . Now add all the resulting inequalites. You'll then find that all the coefficients of the 's and 's are equal to , and the RHS is .
For instance, when n = 2, multiply the inequalities by 1, 1, 2 and 1 respectively, and add. The result is
and when n = 3, multiply by 1, 1, 2, 3, 5, 3 to get
Dividing both sides by completes the proof.
I hope you can improve on this. I'd like to see a more formal proof.
Grandad
Hello Rapha -
Thanks for your kind comment. I found this a most interesting problem, and I have now written out a formal proof. I assume that you have come up with one as well, but here's mine, if you're interested:
First, two lemmas relating to the Fibonacci sequence.
Let be the term of the Fibonacci sequence. So:
and
LEMMA 1
PROOF
Define the propositional function
Then from the recurrence relation for the sequence.
So
Now , which is true.
Hence
LEMMA 2
PROOF
Define the propositional function
Then from the recurrence relation for the sequence.
So
Now , which is true.
Hence
GIVEN
where
TO PROVE
PROOF
For , multiply each by :
For , multiply each by :
Multiply by :
Add , and :
Now consider the coefficient of :
For , coefficient of , from LEMMA 1
For , coefficient of
, from LEMMA 1
, from LEMMA 1
Now consider the coefficient of in :
For , coefficient of
, from LEMMA 2
, from LEMMA 2
, from the recurrence relation.
For , the coefficient of , from the recurrence relation
Finally, using LEMMA 1 and LEMMA 2 on the RHS, becomes:
, from the recurrence relation
QED
I would be interested to see if you have found any significant way of shortening this proof.
Grandad
Hi
I found this a most interesting problem, and I have now written out a formal proof. I assume that you have come up with one as well,To tell the truth I tried… and failed…but here's mine, if you're interested:Of course I’m interested, I would love to see it.......QEDWow. That's awesome. I really appreciate your formal proof. Thanks for all the help.I would be interested to see if you have found any significant way of shortening this proof.I did not, but in like three months my teacher will make a proposal for a solution.Kind regards,Rapha