The function
Of course it's integrable for MX1. But I am talking about the sketching part. Prehaps Zeebobby would like to alter the question.The functionis integrable using methods taught in Extension 1.
Perhaps the sketching part is inappropriate for MX1. Everything else seems okay.
Just realised that the value was half that of the golden ratio.
really cool fact (kinda unrelated but alright...)
I'll let somebody else give it a try, but does this proof happen to use the following properties:really cool fact (kinda unrelated but alright...)
you know every positive integer can be written as sum of distinct integer powers of phi?
e.g.
(where)
pretty hard question: prove it.
, i got it from the cambridge 3u(yr11) though it was on the extension part, probz been in others too alrite stuff that q, lets just moveee on
Uses the first property only - don't need anything to do with fibonacci numbers. Try it, its a fun questionI'll let somebody else give it a try, but does this proof happen to use the following properties:
I have a proof in mind, but it uses two of the aforementioned identities. I'll give it a try once I get back from work.Uses the first property only - don't need anything to do with fibonacci numbers. Try it, its a fun question
You've answered a slightly different question - I said *sum* of distinct integer powers of phi - you can't have differences. But if you could, yeah that still works ^^ And I believe that theorem is actually pretty easy to prove (you can induct with a pretty simple bounding argument I think). But yeah the question without differences you can do without any knowlegde of the fibonacci sequence - and with barely any algebra actually.Larg, regarding your question with the powers of phi.
Did you mean...
...instead of...
I managed to prove the following:
If n is odd, then:
If n is even, then:
Using a combination of the above, we can acquire the integers 1, 3, 4, 7, 11, 18 ...
For example:
This is a relationship using the same recursive formula as the Fibbonaci Sequence:
But with the conditions:
As opposed to:
Zeckendorf's Theorem (http://en.wikipedia.org/wiki/Zeckendorf's_theorem) states that all positive integers can be expressed as the sum of distinct terms of the Fibbonaci sequence.
Not quite sure about the validity of this, but since we are utilising the same recursive definition (but different starting conditions), all integers can be expressed as distinct sums (and subtractions) of powers of phi (though usually a positive power is paired with its equivalent negative power).
For example:
 + ({\phi ^{4}} + {\phi ^{-4}}) + ({\phi ^{5}} - {\phi ^{-5})} + ({\phi ^{6}} - {\phi ^{-6}}) \end{align*} )
You dont even need to use the centre. here's a pretty simple proof:
wahhhh... A, B and D are always going to be concyclic so long as <ADB is not zero (i.e. A,D,B collinear) since any three noncollinear points are concyclic. If you meant A, B, D are on a circle centre C, thats not true; although A, B, C and the centre of the circle through A, B and D are themselves concyclic
so long as... ? I don't see anything lol.
