A product approximation to the Taylor series of the exponential - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T00:03:09Z http://mathoverflow.net/feeds/question/81472 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81472/a-product-approximation-to-the-taylor-series-of-the-exponential A product approximation to the Taylor series of the exponential Charles Rezk 2011-11-21T01:18:55Z 2011-12-06T05:54:45Z <p>I recently came across the following in something I'm working on, and I'd never seen it before. Consider \begin{align*} f_1(x) &amp;= (1+x)^{1/1} \\ f_2(x) &amp;= (1+x)^{2/1} (1+2x)^{-1/2} \\ f_3(x) &amp;= (1+x)^{3/1} (1+2x)^{-3/2} (1+3x)^{1/3} \\ f_4(x) &amp;= (1+x)^{4/1} (1+2x)^{-6/2} (1+3x)^{4/3} (1+4x)^{-1/4} \\ &amp; \cdots \\ f_n(x) &amp;= \prod_{k=1}^n (1+kx)^{(-1)^{k-1}\binom{n}{k}/k}. \end{align*} Think of these as formal power series with coefficients in $\mathbb{Q}$. It turns out that these series approximate the Taylor expansion for the exponential, in the sense that $$f_n(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} +O(x^{n+1}).$$</p> <p>(This isn't too hard to prove, using a logarithmic derivative: \begin{align*} 1-f_n'(x)/f_n(x) &amp;= \sum_{k=0}^n (-1)^k\binom{n}{k}(1+kx)^{-1} \\ &amp;= \sum_j \biggl(\sum_k (-1)^k\binom{n}{k}k^j\biggr) (-x)^j. \end{align*} The coefficient of $x^j$ calculates the number of surjections <code>$\{1,\dots,j\}\to \{1,\dots,n\}$</code> (up to a sign), and so is $0$ if <code>$j&lt;n$</code>.)</p> <p>My questions include (but are not limited to): Do these things have a name? Is there some kind of combinatorial interpretation of the $f_n(x)$ (e.g., as a generating function of some kind)? Is there a more direct proof that $f_n(x)$ agrees with $e^x$ to order $n+1$?</p> http://mathoverflow.net/questions/81472/a-product-approximation-to-the-taylor-series-of-the-exponential/81598#81598 Answer by Gjergji Zaimi for A product approximation to the Taylor series of the exponential Gjergji Zaimi 2011-11-22T11:41:02Z 2011-11-22T11:41:02Z <p>I'm going to switch a couple of signs and focus on $g_n(x)=\frac{1}{f_n(-x)}$ for a moment. It is an exponential generating function for the number of properly $n$-colored rooted monotonic planar forests of depth at most $1$.</p> <p>I should clarify that by "$n$-colored" I mean that each edge is assigned one of given $n$ colors, and by "properly $n$-colored" I mean $n$-colored with the property that or any pair $i,j$, any edge with color $i$ is adjacent to some edge of color $j$. A monotonic rooted tree is a labelled tree for which the label of a vertex is greater than the labels of its children. It is now clear that $g_n(x)$ agrees with $e^x$ up to order $n$, since the only $g$-structures of order up to $n$ are graphs with no edges (if there were any edges, "properly $n$ colored" needs at least $n+1$ vertices).</p> <p>It remains to see why the generating function of these structures is what it is. Combinatorial species give a nice framework for interpreting combinatorially what different operations on the generating function mean at the level of objects being counted. Here is a sketch:</p> <blockquote> <p>Exponential Generating function for labelled monotonic planar rooted trees of depth at most 1 is $\log(1-x)$.</p> <p>Therefore the EGF for $k$-colored labelled monotonic planar rooted trees of depth at most 1 is $\frac{1}{k}\log(1-kx)$.</p> <p>The <a href="http://en.wikipedia.org/wiki/Binomial_transform" rel="nofollow">binomial tranform</a> of order $n$ changes "$k$-colored" to "properly $n$-colored".</p> <p>The <a href="http://en.wikipedia.org/wiki/Exponential_formula" rel="nofollow">exponential formula</a> changes "tree" to "forest".</p> <p>So this all gives $$g_n(x)=\exp\left(\sum_{k=1}^n (-1)^k\binom{n}{k}\frac{1}{k}\log(1-kx)\right)$$</p> </blockquote> http://mathoverflow.net/questions/81472/a-product-approximation-to-the-taylor-series-of-the-exponential/82483#82483 Answer by Tom Copeland for A product approximation to the Taylor series of the exponential Tom Copeland 2011-12-02T16:31:40Z 2011-12-06T05:54:45Z <p>See <a href="http://oeis.org/A019538" rel="nofollow">A019538</a>, <a href="http://oeis.org/A049019" rel="nofollow">A049019</a>, and <a href="http://oeis.org/A133314" rel="nofollow">A133314</a> for extensive formula for the coefficients and relation to face vectors of <a href="http://en.wikipedia.org/wiki/Permutohedron" rel="nofollow">permutahedra/permutohedra</a>. Your proof is a nice one in that A133314 clearly shows why the lower order coefficients as given by the finite differences must vanish if $1/f_n(x)$ truly is an approximation of $exp(-x)$. </p>