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I'm not sure if this equation is a linear homogeneous recurrence relation because I didn't learn math in english. what is the explicit formula for $f$?

$$f(x) = f(x-2)-{n^{x-1}\over NW-1} f(x-1)+n^{x-1}$$

$n$ and $NW$ are constants. $n = 3$ and $NW = 7000000000 = $7 billion I'm having a hard time because of the $n^{x-1}$ please help me.

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    $\begingroup$ I'm having hard time reading it. Please, use $\LaTeX$. $\endgroup$
    – Kaster
    Jul 27, 2015 at 16:57
  • $\begingroup$ Welcome to MSE! You should read [this](meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) before posting questions $\endgroup$
    – Michael Galuza
    Jul 27, 2015 at 16:57
  • $\begingroup$ Please review and edit your post. Is $m$ supposed to be the same as $n$? What is NW-? is the $*nf(d-1)$ supposed to be in the denominator or numerator? $\endgroup$
    – Ross Millikan
    Jul 27, 2015 at 16:58
  • $\begingroup$ sorry, I didn't know how to use it, and i got many miss spellings. $\endgroup$
    – urw7RSeeh8FR
    Jul 27, 2015 at 17:12
  • $\begingroup$ Is the second term on the right hand side really $-[f(x-1)]^2/(NW-1)$? $\endgroup$
    – Intelligenti pauca
    Jul 27, 2015 at 18:12

1 Answer 1

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This is a linear second order recurrence with variable coefficients. Just as in the case of differential equations, there are few such recurrences with closed form solutions.

As a first approximation (for smallish $x$) you can disregard the term in $f(x - 1)$, and get a first order linear recurrence (of step two) in $f(x)$:

$$ f(x) - f(x - 2) = n^{x - 1} $$

This has solution:

$$ f(2 x) = f(0) + \sum_{0 \le k \le x} n^{k - 1} = f(0) + \frac{n^{x + 1} - 1}{n (n - 1)} $$

This might be enough for your purposes, or use this as a starting point to construct a closer approximation.

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