User a e charman - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:30:50Z http://mathoverflow.net/feeds/user/9829 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41226/pade-approximant-to-exponential-function Pade approximant to exponential function A E Charman 2010-10-06T02:30:30Z 2010-10-06T02:47:58Z <p>Suppose:</p> <p>a) \$p(z)\$ is an even degree polynomial (of degree \$k = 2j\$) with real coefficients;</p> <p>b) \$p(0) = 1\$;</p> <p>c) \$p(z)\$ and \$p(-z)\$ have no roots in common anywhere in the complex plane;</p> <p>d) \$f(z) = p(z)/p(-z)\$ is a Pade approximant to \$\exp(z) = e^z\$, such that the Taylor expansion of \$f(z)\$ agrees with that of \$\exp(z)\$ up to \$(2k)\$th order.</p> <p>Are there simple proofs of the following conjectures:</p> <p>i) the coefficients of \$p(z)\$ are all positive</p> <p>ii) \$f(x) \le \exp(x)\$ for all nonnegative \$x\$</p> <p>iii) \$p(x) p(-x) \exp(ax)\$ has all positive coefficients in its Taylor expansion for any \$a \ge 1\$</p> <p>iv) \$p(x) p(-x)\$ has no real roots.</p> <p>Comments: By Descartes rule of signs, (i) implies \$p(z)\$ has no positive roots. By a theorem of Laguerre (Ouvres, Tome 1), (iii) would imply (iv)</p>