{
  "id": "math/0408096",
  "version": "v1",
  "published": "2004-08-07T14:03:30.000Z",
  "updated": "2004-08-07T14:03:30.000Z",
  "title": "Differentiating the absolutely continuous invariant measure of an interval map f with respect to f",
  "authors": [
    "David Ruelle"
  ],
  "comment": "10 pages, plain TeX",
  "doi": "10.1007/s00220-004-1267-4",
  "categories": [
    "math.DS"
  ],
  "abstract": "Let the map $f:[-1,1]\\to[-1,1]$ have a.c.i.m. $\\rho$ (absolutely continuous $f$-invariant measure with respect to Lebesgue). Let $\\delta\\rho$ be the change of $\\rho$ corresponding to a perturbation $X=\\delta f\\circ f^{-1}$ of $f$. Formally we have, for differentiable $A$, $$ \\delta\\rho(A)=\\sum_{n=0}^\\infty\\int\\rho(dx) X(x){d\\over dx}A(f^nx) $$ but this expression does not converge in general. For $f$ real-analytic and Markovian in the sense of covering $(-1,1)$ $m$ times, and assuming an {\\it analytic expanding} condition, we show that $$\\lambda\\mapsto\\Psi(\\lambda)=\\sum_{n=0}^\\infty\\lambda^n \\int\\rho(dx) X(x){d\\over dx}A(f^nx) $$ is meromorphic in ${\\bf C}$, and has no pole at $\\lambda=1$. We can thus formally write $\\delta\\rho(A)=\\Psi(1)$.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2004-08-07T14:03:30.000Z"
    }
  ],
  "analyses": {
    "keywords": [
      "absolutely continuous invariant measure",
      "interval map",
      "differentiating",
      "perturbation",
      "expression"
    ],
    "tags": [
      "journal article"
    ],
    "publication": {
      "publisher": "Springer",
      "journal": "Commun. Math. Phys."
    },
    "note": {
      "typesetting": "Plain TeX",
      "pages": 10,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}