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  "body": "\n## Definition\n\nLet $K$ be a field and let $f(x) = \\sum_{k \\ge 0} f_k x^k$ be a [[formal power series|033WGN00BF]] over $K$.\nFor integers $j, k \\ge 0$, let $M[f]_{jk}$ be the coefficient of $x^k$ in the power $f(x)^j$, with $f(x)^0 = 1$.\nThe *Carleman matrix* of $f$ is the infinite [[matrix|0VD9ZSWWJG]] $M[f] = (M[f]_{jk})_{j, k \\ge 0}$.\nSo\n$$f(x)^j = \\sum_{k \\ge 0} M[f]_{jk} \\, x^k \\quad \\text{for every } j \\ge 0.$$\n\n## Remarks\n\nExample: for $f(x) = ax$ with $a \\in K$, one has $f(x)^j = a^j x^j$, so $M[f]$ is diagonal with entries $1, a, a^2, \\ldots$.\nFor $f(x) = x$ it is the identity matrix.\n\nNon-example: the diagonal matrix with entries $1, 2, 3, \\ldots$ is not a Carleman matrix.\nIts row $1$ would force $f(x) = 2x$, and then row $2$ would have to list the coefficients of $4x^2$, not of $3x^2$.\n\nIf $g(0) = 0$, then $M[f \\circ g] = M[f] \\, M[g]$, a product of infinite matrices in which each entry is a finite sum.\nCheck: $f(g(x))^j = \\sum_k M[f]_{jk} \\, g(x)^k = \\sum_k M[f]_{jk} \\sum_l M[g]_{kl} x^l$, and $M[g]_{kl} = 0$ for $k \u003e l$ because $g(x)^k$ starts at $x^k$.\nIn particular, if $f(0) = 0$, then $M[f^n] = M[f]^n$ for the [[iterates|AYVTR2ES5B]] of $f$, and $M[f]$ is upper triangular with diagonal entries $f_1^j$.\nThe transpose of $M[f]$ is the Bell matrix of $f$, which reverses the order of products.\n",
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