MHB Recursive sequences and finding their expressions

delc1
Messages
9
Reaction score
0
Hi all,

I don't understand what is being asked by this question?

View attachment 2445

If anyone knows could they please describe the process, that would be greatly appreciated.
 

Attachments

  • Untitled.jpg
    Untitled.jpg
    11.5 KB · Views: 117
Physics news on Phys.org
delc1 said:
Hi all,

I don't understand what is being asked by this question?

View attachment 2445

If anyone knows could they please describe the process, that would be greatly appreciated.
The procedure for solving this type of linear second order difference equations is illustrated in...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-ii-860-post4544.html#post4544

Kind regards

$\chi$ $\sigma$
 
delc1 said:
Hi all,

I don't understand what is being asked by this question?

View attachment 2445

If anyone knows could they please describe the process, that would be greatly appreciated.

You are being asked to find the closed-form for the given linear homogeneous recurrence. The first step is to find the roots of the associated characteristic equation. Can you state this equation and its roots?
 
Hmmmm I tried doing this equation myself but am also stuck.

I tried subbing in n=2, 3 and 4 into the equation and have found that:
S2= -8
S3= -36
S4= -112
S5= -304

So the pattern that I have found is that there is a difference of -28, -76 and -192 but this doesn't lead me to an easily findable equation.
 
Writing the difference equation in the form...

$\displaystyle s_{n+2} - 4\ s_{n+1} + 4\ s_{n} = 0,\ s_{0}=3,\ s_{1}=1\ (1)$

... the associated characteristic equation is...

$\displaystyle x^{2} -4\ x +4 = 0\ (2)$

... the solution of which is x=2 with multiplicity 2. That means that the general solution of (1) is...

$\displaystyle s_{n} = (c_{1} + c_{2}\ n)\ 2^{n}\ (3)$

The constants$c_{1}$ and $c_{2}$ cn befound from the initial conditions, so that is...

$\displaystyle s_{n} = (3 - \frac{5}{2}\ n)\ 2^{n}\ (4)$

Kind regards$\chi$ $\sigma$
 
chisigma said:
Writing the difference equation in the form...

$\displaystyle s_{n+2} - 4\ s_{n+1} + 4\ s_{n} = 0,\ s_{0}=3,\ s_{1}=1\ (1)$

... the associated characteristic equation is...

$\displaystyle x^{2} -4\ x +4 = 0\ (2)$

... the solution of which is x=2 with multiplicity 2. That means that the general solution of (1) is...

$\displaystyle s_{n} = (c_{1} + c_{2}\ n)\ 2^{n}\ (3)$

The constants$c_{1}$ and $c_{2}$ cn befound from the initial conditions, so that is...

$\displaystyle s_{n} = (3 - \frac{5}{2}\ n)\ 2^{n}\ (4)$

Kind regards$\chi$ $\sigma$

Cheers, that's a much simpler way of thinking of solving the problem, I guess it was the fact that 2 was a double root that confused me.
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top