Can somebody please point me into a really helpful website / video to help me get my head around these, I under stand the basics but getting confused when you start adding

(fn2+3) + (fn)2 and (fn2+3) + fn2 N = 1,2,... N-3 etc

thanks, Matt

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- Mar 21st 2011, 07:21 AMentrepreneurforum.co.ukLinear second order recurrence sequence
Can somebody please point me into a really helpful website / video to help me get my head around these, I under stand the basics but getting confused when you start adding

(fn2+3) + (fn)2 and (fn2+3) + fn2 N = 1,2,... N-3 etc

thanks, Matt - Mar 24th 2011, 10:03 PMChris11
Sorry, I can't understand your notation

- Mar 25th 2011, 01:03 AMAckbeet
Note that if you go into the Advanced Editing mode, you can use the TeX button to enclose math tags around LaTeX code. Also note that subscripts can be written as a_{n}, which displays as $\displaystyle a_{n}.$

As Chris11 pointed out, it's unclear exactly what values are in the "subscript", or function argument. You also have n's and N's. Mathematics is case-sensitive, so you should standardize on one or the other. Typically, capital N's are used to denote a fixed integer value, whereas little n's are more used to denote a variable integer value. - Mar 25th 2011, 02:55 AMentrepreneurforum.co.uk
What i'm looking for is help on understanding this and the cassini identity and how i can get from a fibonnaci expression such as;

$\displaystyle Un+1Un+2 - Un+1(Un)-Un(Un+2)$

using the cassini identity;

$\displaystyle UnUn+2 = U2n+1 + (-1)^n+1$

to show that

$\displaystyle Un+1Un+2-UnUn+3 = (-1)^n$

where;

$\displaystyle Un+1Un+2 - Un+1(Un)-Un(Un+2) = Un+1Un+2-UnUn+3 = (-1)^n$ - Mar 25th 2011, 04:46 AMAckbeet
I'm going to take a stab at re-writing your whole post, using my guesses for what you mean (you still need a lot of cleaning-up of notation). Please correct any errors.

What i'm looking for is help on understanding this and the cassini identity and how i can get from a fibonnaci expression such as;

$\displaystyle U_{n+1}U_{n+2} - U_{n+1}U_{n}-U_{n}U_{n+2}$

using the cassini identity;

$\displaystyle U_{n}U_{n+2} = U_{2n+1} + (-1)^{n+1}$

to show that

$\displaystyle U_{n+1}U_{n+2}-U_{n}U_{n+3} = (-1)^{n}$

where;

$\displaystyle U_{n+1}U_{n+2} - U_{n+1}U_{n}-U_{n}U_{n+2}=U_{n+1}U_{n+2}-U_{n}U_{n+3} = (-1)^{n}$.

How is that for the mathematical notation? As for your English, that needs cleaning up as well, since you have very convoluted syntax, if it's even a sentence at all. - Mar 25th 2011, 07:33 AMentrepreneurforum.co.uk
Actbeet, yes thats what i meant, I'm just not as familiar to TEX as you might be, but you were able to understand what I had written, thanks.

- Mar 28th 2011, 11:15 AMJester
Let's see if we (or should I say I) get this right.

Given $\displaystyle U_n +U_{n+1} = U_{n+2}$ (the Fibinacci sequence recursion relation)

and Cassini's identity

$\displaystyle U_{n-1}U_{n+1} - U_n^2 = (-1)^n$

prove

$\displaystyle U_{n+1}U_{n+2} - U_n U_{n+3} = (-1)^n$

Is this it? - Mar 28th 2011, 11:35 AMAckbeet
Reply to Danny.

Perhaps, but it looks more to me like this:

Assume

$\displaystyle U_{n}U_{n+2}=U_{2n+1}+(-1)^{n+1}\quad(\text{Cassini}),$ and

$\displaystyle U_{n+1}U_{n+2}-U_{n+1}U_{n}-U_{n}U_{n+2}=(-1)^{n}.$ Show that

$\displaystyle U_{n+1}U_{n+2}-U_{n}U_{n+3}=(-1)^{n}.$

The fact that post # 4 uses the word "Fibonacci" may or may not imply that the OP means $\displaystyle U_{n}$ to be the Fibonacci sequence. He never actually posts that relation. That's my impression, but I could be wrong. - Mar 28th 2011, 11:47 AMJester
When I wikipedia'd Cassini it gave what I gave in post #7. I think we need the OP to give some clarification before anyone wastes their time working on an answer to what we "think they meant."

Just my two cents. - Mar 28th 2011, 12:46 PMAckbeet