Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$a_{0}=0$$ $$a_{1}=0$$ $$a_{2}=-1$$ $$a_{n+3}-6a_{n+2}+12a_{n+1}-8a_{n}=n$$

It's just that...I don't know what to do if there are $a_{n+1}$ instead of $a_{n-1}$, I don't know what to do with that $n$...How can I solve this?

share|improve this question
1  
If you want negative indices, transform $n+3 \to n$ and you'll get $a_n - 6a_{n-1} + 12a_{n-2} - 8a_{n-3} = n-3$, $n \ge 3$. –  Daniel R Aug 12 '13 at 11:49
    
Try increasing n by 1 and taking the difference of the resulting equations, you get the recurrence relation: $$a_{n+4}-7a_{n+3}+18a_{n+2}-20a_{n+1}+8a_n=1.$$ Now try finding a homogenous solution and a particular solution and add those together to get a general solution. –  walcher Aug 12 '13 at 12:13
    
@walcher, in the first place, I think that's the hard way, and in the second place, I don't think it engages with OP's discomfort with $a_{n+1}$. –  Gerry Myerson Aug 12 '13 at 12:28
    
I think I do not understand the exact nature of the discomfort @GerryMyerson is mentioning. OP: you might want to clarify "what to do if there are $a_{n+1}$ instead of $a_{n−1}$", perhaps by giving a case you can solve. –  Did Aug 12 '13 at 13:52
1  
@Did, my impression is that OP has a recipe for solving equations like $a_n-6a_{n-1}+12a_{n-2}-8a_{n-3}=n$ and doesn't understand how to twiddle the recipe to deal with the equation actually given. –  Gerry Myerson Aug 13 '13 at 0:10

2 Answers 2

We are given $$a_{n+3}-6a_{n+2}+12a_{n+1}-8a_n=n.$$ Increasing $n$ by one gives us $$a_{n+4}-6a_{n+3}+12a_{n+2}-8a_{n+1}=n+1.$$ Subtracting these two equations yields $$a_{n+4}-7a_{n+3}+18a_{n+2}-20a_{n+1}+8a_n=1,$$ so we will focus on solving this recurrence relation. First we find the general homogenous solution, i.e. the general solution to $$a_{n+4}-7a_{n+3}+18a_{n+2}-20a_{n+1}+8a_n=0.$$ This is a linear recurrence with characteristic equation $$x^4-7x^3+18x^2-20x+8=(x-2)^3(x-1)=0,$$ which we know to have the general solution $a_n=2^n(n^2+an+b)+c$. To find a particular solution we first note that $n-6n+12n-8n=-n$, so make the Ansatz $a_n=-n+d$. Substituting into $$a_{n+3}-6a_{n+2}+12a_{n+1}-8a_n=n$$ gives an equation for $d$, which is solved by $d=-3$, so our general solution is $$a_n=a_n=2^n(n^2+an+b)+c-n-3.$$ Substitute the three initial conditions and solve for $a$, $b$ and $c$ to get the solution.

share|improve this answer
1  
This does not address the question: it deals neither with the specific issue (what to do with indices greater than $n$) nor with the desired method of solution (via generating functions). –  Brian M. Scott Aug 12 '13 at 13:24
    
Moreover, the method of finding general and particular solutions can be applied to the original equation, without the increasing-$n$-and-subtracting step. Or, if you're going to do that step, why not do it twice, and get to a homogeneous equation, and forget about particular solutions? –  Gerry Myerson Aug 13 '13 at 0:13

Let $f(x) =\sum_{n=0}^{\infty} a_n x^n$. Usually relations like

$$a_{n+3}-6a_{n+2}+12a_{n+1}-8a_{n}=n$$

come with a restriction on n. I'm going to assume the equation holds for all $n \ge 0$.

Multiply by $x^n$:

$$a_{n+3} x^n-6a_{n+2}x^n+12a_{n+1}x^n-8a_{n}x^n=n x^n$$

Sum over $n= 0, 1, 2, \dots$:

$$\sum_{n=0}^{\infty} a_{n+3} x^n-6\sum_{n=0}^{\infty} a_{n+2}x^n+12\sum_{n=0}^{\infty} a_{n+1}x^n-8\sum_{n=0}^{\infty} a_{n}x^n=\sum_{n=0}^{\infty} n x^n$$

To deal with a sum like $\sum_{n=0}^{\infty} a_{n+1} x^n$, for example, observe that

$$x \sum_{n=0}^{\infty} a_{n+1} x^n = \sum_{n=0}^{\infty} a_{n+1} x^{n+1}= \sum_{n=1}^{\infty} a_n x^n = f(x) - a_0$$ so $$\sum_{n=0}^{\infty} a_{n+1} x^n = \frac{f(x)-a_0}{x}$$

Maybe you can handle the rest?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.