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  "body": "\n## Definition\n\nLet $c \\in \\C \\setminus \\set{0}$ and let $n \\ge 1$.\nLet $F$ be a polynomial or a rational function in its last $n + 1$ arguments, with coefficients that are meromorphic functions of $z$.\nA *complex difference equation* for an unknown [[meromorphic function|0ZZ7ZYDR8W]] $f$ on $\\C$ is an equation\n$$F\\big(z, f(z), f(z + c), \\dots, f(z + nc)\\big) = 0.$$\nFor such a solution $f$, write $T(r, f)$ for its [[Nevanlinna characteristic|YT6Z0GWS0W]] and $m(r, f)$ for its proximity function.\nA quantity $S(r, f)$ is *small* with respect to $f$ if $S(r, f) = o(T(r, f))$ as $r \\to \\infty$ outside a set of $r$ of finite logarithmic measure.\n*Nevanlinna theory of difference equations* is the study of the meromorphic solutions of such difference equations through $T$, $m$ and the counting function $N$.\nIts basic tools are estimates of the type $m\\big(r, f(z + c)/f(z)\\big) = S(r, f)$.\n\n## Remarks\n\nExample: $f(z) = e^{z}$ solves $f(z + 1) = e f(z)$.\nHere $T(r, f) = r/\\pi$ and $f(z + 1)/f(z) = e$ is constant, so $m\\big(r, f(z+1)/f(z)\\big) = 1$, which is small compared with $T(r, f)$.\n\nNon-example: for $f(z) = e^{e^z}$, which has infinite order, the quotient $f(z + 1)/f(z) = e^{(e - 1)e^z}$ has proximity function comparable to $T(r, f)$, so the small-term estimate fails; this is why the difference lemma requires finite order.\n\nThe basic estimate holds for every meromorphic $f$ of finite order and every $c \\in \\C$, outside an exceptional set of finite logarithmic measure.\n",
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