Let $r>4$ be a positive integer. Let us consider this difference equation: $$v_{n+1}-r^{2n+1}v_{n}=w$$ where $w$ is real number.

How I can solve this equation with respect to $v_{n}$. I am not able to find the good idea.

  • 1
    $\begingroup$ This is basically an interest problem with deposits and a growing interest rate. I do not know a closed for for such problems however. $\endgroup$ – Ian Oct 2 '15 at 10:05
  • $\begingroup$ @Ian: Did you mean that it is impossible to solve this equation. $\endgroup$ – DER Oct 2 '15 at 10:15
  • $\begingroup$ No, I just do not know whether a closed form exists. If you unfold a little you will see it looks a bit like a geometric series, so maybe you can exploit that. $\endgroup$ – Ian Oct 2 '15 at 10:19

Let's find an "integrating factor" where the quotient is $r^{2n+1}$. As $(n+1)^2-n^2=2n+1$, we get that $$ \frac{v_{n+1}}{r^{(n+1)^2}}-\frac{v_n}{r^{n^2}}=\frac{w}{r^{(n+1)^2}} $$ The summation over the square-degree powers of $r^{-1}$ however has no elementary simplification, so that $$ \frac{v_n}{r^{n^2}}=\frac{w}{r^{n^2}}+\frac{w}{r^{(n-1)^2}}+…+\frac{w}{r^{1}}+\frac{v_0}{r^0} $$ is as simple as it gets.


We have $v_{n+1}=w + r^{2n+1} v_n$, hence $$ v_{n+2} = w + r^{2n+3} v_{n+1} = w + r^{2n+3} w + r^{4n+4} v_n $$ and by induction:

$$ v_{n+k} = w\sum_{j=1}^{k}r^{2jn+k^2-(k+1-j)^2}+v_n r^{2kn+k^2}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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