Kyra's question at Yahoo Answers regarding linear difference equations

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The discussion centers on Kyra's question regarding linear difference equations, specifically identifying the nature of a function based on its second differences. It is established that if the second differences are constant, the function is quadratic, leading to the conclusion that option B is correct. The derivation involves inhomogeneous linear recurrence relations and results in a characteristic equation with a root of multiplicity three, confirming the quadratic form of the solution.

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Here is the question:

Math hmk help asap?

help.,, Homework .,, thank you :))

If the second differences are the same, then the function is:
A. linear
B. quadratic
C. exponential
D. neither

Here is a link to the question:

Math hmk help asap? - Yahoo! Answers

I have posted a link there to this topic so the Op can find my response.
 
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Re: Kyra's question at Yahoo! Answers regardin linear difference equations

Hello Kyra,

Let the $n$th term of the sequence be given by $A_n$. If the second difference is constant, then we may state:

$$\left(A_{n}-A_{n-1} \right)-\left(A_{n-1}-A_{n-2} \right)=k$$ where $$0\ne k\in\mathbb{R}$$

Combining like terms, we may arrange this as the inhomogeneous linear recurrence:

(1) $$A_{n}=2A_{n-1}-A_{n-2}+k$$

We may increase the indices by 1, to prepare for symbolic differencing:

(2) $$A_{n+1}=2A_{n}-A_{n-1}+k$$

Subtracting (1) from (2), we obtain the homogeneous linear recurrence:

$$A_{n+1}=3A_{n}-3A_{n-1}+A_{n-2}$$

The characteristic equation is then:

$$r^3-3r^2+3r-1=(r-1)^3=0$$

Since the root $r=1$ is of multiplcity 3, we know the closed form will be:

$$A_n=k_1+k_2n+k_3n^2$$

We see then that the closed form is quadratic, hence B is the answer.

To Kyra and any other guest viewing this topic, I invite and encourage you to post other difference equation problems in our http://www.mathhelpboards.com/f15/ forum.

Best Regards,

Mark.
 

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