Linear second order recurrence sequence

**entrepreneurforum.co.uk** · March 21st 2011, 07:21 AM

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

**Chris11** · March 24th 2011, 10:03 PM

Sorry, I can't understand your notation

**Ackbeet** · March 25th 2011, 01:03 AM

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 $a_{n}.$

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.

**entrepreneurforum.co.uk** · March 25th 2011, 02:55 AM

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;

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

using the cassini identity;

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

to show that

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

where;

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

**Ackbeet** · March 25th 2011, 04:46 AM

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;

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

using the cassini identity;

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

to show that

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

where;

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

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.

**entrepreneurforum.co.uk** · March 25th 2011, 07:33 AM

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.

**Danny** · March 28th 2011, 11:15 AM

Let's see if we (or should I say I) get this right.

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

and Cassini's identity

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

prove

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

Is this it?

**Ackbeet** · March 28th 2011, 11:35 AM

Reply to Danny.

Perhaps, but it looks more to me like this:

Assume

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

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

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

The fact that post # 4 uses the word "Fibonacci" may or may not imply that the OP means $U_{n}$ to be the Fibonacci sequence. He never actually posts that relation. That's my impression, but I could be wrong.

**Danny** · March 28th 2011, 11:47 AM

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.

**Ackbeet** · March 28th 2011, 12:46 PM

Originally Posted by Danny

When I wikipedia'd Cassini it gave what I gave in post #8. 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.

Agreed.

Thread: Linear second order recurrence sequence

LinkBack

Thread Tools

Display

Linear second order recurrence sequence

Similar Math Help Forum Discussions

Logistsic Recurrence Sequence

2nd-order recurrence sequence (simultaneous equation mind block)

Linear recurrence sequence, closed form.

recurrence sequence

Linear order of a recurrence relation

Search Tags