Recursive sequences and finding their expressions

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Discussion Overview

The discussion revolves around understanding and solving a linear homogeneous recurrence relation, specifically focusing on finding closed-form expressions for recursive sequences. Participants explore the process of deriving solutions from given initial conditions and characteristic equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion about the question and seek clarification on the process of solving linear second-order difference equations.
  • One participant shares their attempts to find values for the sequence and notes the differences between them, indicating a struggle to identify a pattern or formula.
  • Another participant provides a detailed breakdown of the difference equation, including the associated characteristic equation and its roots, leading to a general solution involving constants derived from initial conditions.
  • There is a reiteration of the solution process, emphasizing the significance of the double root in the characteristic equation and its impact on the general solution.

Areas of Agreement / Disagreement

Participants generally agree on the steps involved in solving the recurrence relation, but there is no consensus on the initial understanding of the problem, as some express confusion while others provide clarifications.

Contextual Notes

Some participants may have missing assumptions regarding the initial conditions or the specific form of the recurrence relation, which could affect their understanding and approach to the problem.

delc1
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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.
 

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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.
 

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