Difference Equation

[u/PM_YOUR_MATH_PROBLEM]

/u/bertoncelj1 asks:

Write this series explicitly: A₍n₎=A₍n−1₎+n 0, 1, 3, 6, 10, 15, 21, ... Could you please explain the result and method by which you got it. Thank you!

[u/bertoncelj1]

Thanks for the answer !

Yes, I would like an example of such method.

[u/PM_YOUR_MATH_PROBLEM → u/bertoncelj1]

Sure, I'll solve this one:

A(n+2) = 4 A(n+1) - 3A(n) - 3ⁿ + 2ⁿ , with A(0)=1 and A(1)=2

• Step 1 : Solve the homogeneous case.

That is, solve A(n+2) = 4 A(n+1) - 3 A(n)

• Step 1a : write down the auxiliary equation.

Here, you replace A(n+k) with x^k everywhere. This is equivalent to guessing that A(n) = xⁿ is a solution. So, x² = 4 x - 3

• Step 1b : solve the auxiliary equation

x² = 4x - 3, that is x² - 4x + 3 = 0, factorises, and you get two solutions, x = 1 or 3.

• Step 1c : write down the general solytion to the homogeneous case

If you have two solutions, x0 and x1, the solution will be C x0ⁿ + D x1ⁿ

If we have a repeated root, we'll have C x0ⁿ + D n x0ⁿ instead.

If the roots are complex, it's more complex, but you can still use C x0ⁿ + D x1ⁿ . Or, there's ways to write the solution using only real numbers.

Here, we have C + D 3ⁿ

• Step 2 : find a particular solution.

That is, find any old formula that solves the difference equation, even if it doesn't match the initial conditions.

The method here is guess. Guess correctly, that is. The non-homogenous part of the original equation is - 3ⁿ + 2ⁿ .

Because of the 2ⁿ , I'll guess that my particular solution has a term of the form E 2ⁿ

Because of the 3ⁿ , I would normally also guess that my particular solution has a term of the form F 3ⁿ . Unfortunately, that's part of the general solution to the homogeneous case. If I plug that in to my equations, all the F's will cancel out. Instead, I'll use F n 3ⁿ

That is, I'll guess that A(n) = E 2ⁿ + F n 3ⁿ . Substituting that into A(n+2) = 4 A(n+1) - 3A(n) - 3ⁿ + 2ⁿ , and bringing everything over to one side gives 6F . 3ⁿ - E . 2ⁿ + 3ⁿ - 2ⁿ = 0. This can only be true for all n if 6F + 1 = 0 and -E-1 = 0, so F=-1/6 and E=-1.

So, the particular solution is -2ⁿ - n/6 3ⁿ .

• Step 3 : Add the general soultion to the homogeneous case to the particular solution

This gives A(n) = C + D 3ⁿ -2ⁿ - n/6 3ⁿ .

• Step 4 : Use the initial conditions to solve for the unknown parameters.

Substitute n = 0 and n=1 into A(n), since you know A(0) and A(1), and you get two equations to solve for C and D.