Help with this difference equation

[eljose]

Let be the equation:

$$A(2n)-A(2n-1)=3$$

i,m a bit stuck..i don,t know how to solve it it it weren't for the 3 term i would try $$A(n)=r^n$$ where r is an unknown number..however the 3 factor spoils all..also we could try the identity:

$$\sum_{n=1}^{k}A(n)-A(n-1)=3k=A(k)-A(1)$$ but it seems not to work..i,m really messed up with this nasty equation..

 

[arildno]

Try
$$A(n)=an^{2}+bn+c$$
Determine what a,b,c must be
EDIT:
See latter post.

 

[BoTemp]

Is n required to be an integer? If so, I don't think you have enough information. Well, for starters you'd need to know what A(0) is, or at least what A(n) is for some n, otherwise the solution will have a constant at the end.

But if n is required to be an integer, you only know the difference between an even and the odd directly below it. You'd need another equation to find the difference between an odd and the even below it, or between adjacent odds or evens.

 

[arildno]

"between adjacent odds or evens" Eeh?

eljose:
Your system of equations is DECOUPLED; in each equation, only two unknowns appear, neither of which appears anywhere else.

What follows from this?

 

[HallsofIvy]

This is a "non-homogeous, linear difference equation with constant coefficients". It can be treated exactly like you would a similar differential equation: find the general solution to the associated homogeneous equation, A(2n)- A(2n-1)= 0, then add anyone solution to the entire equation. The general solution is A(n)= C+ 3n where C is any constant.

 

[arildno]

No, HallsofIvy!
You have too few equations here, each unknown member of the sequence appears (in pairs) in only one equation.

Thus, the general solution is:
$$C_{2n-1}=b_{n}, C_{2n}=3+C_{2n-1}, n=1,2\cdots,$$
where $b_{n}$ is an arbitrary sequence.

 

[HallsofIvy]

So you are saying that my solution, A_n= C+ 3n is does satisfy the equation but is not the general solution?

If we are given b_n= 1, 3, 7 , 5, 9, ... then
A₁= 1, A₂= 3+ 1= 4, A₃= 3, A₄= 3+ 3= 6, A₅= 7, A_{6= 3+ 7= 10, ...
Your point, then, is that we don't need to worry about the fact that
A5- A4= 7- 6 is not 3, because the equation
A2n- A2n-1 requires that the first index always be even. Okay, that's clear. Thanks.}

 

[arildno]

It was BoTemp who was the first to point this out; I stepped into the same "trap" of indicating a linear solution as the gen. solution.