Integral representation of higher order derivatives - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:38:38Z http://mathoverflow.net/feeds/question/65264 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65264/integral-representation-of-higher-order-derivatives Integral representation of higher order derivatives Pietro Majer 2011-05-17T18:40:50Z 2013-03-25T19:37:30Z <p>I'm quite curious about the following phenomena, that still puzzle me although I have a proof, and I'd be really glad if someone may shred some light, showing an interpretation or a generalization. I will sketch the computations at request, which consist on a manipulation of the integral formula of the remainder of the Taylor expansion.</p> <p><strong>1.</strong> Let $v\in C^\infty(\mathbb{R})$, vanishing at $0$ with some order $p\in\mathbb{N} _ +$ . In other words, the formal Taylor series of $v$ at $0$ belongs to the ideal $x^p\mathbb{R}[[x]]$ . Then, the function $w(x):=v(x)/x^p$ is also (extensible to) a $C^\infty(\mathbb{R})$ function, and we can express the $k-$ order derivative of $w$ at $x$ (say $x\ge0$) in terms of the derivatives of order $k+p$ of $v$ on $[0,x]$ as follows:</p> <p>$$\frac{w^{(k)}(x)}{k!}=\frac{\int_0^x (x-s)^{p-1}s^k\ \frac{v^{(k+p)}(s)}{(k+p)!}ds}{\int_0^x (x-s)^{p-1}s^k\ ds} \ .$$ In other terms, the $k-$th Taylor coefficient of $w$ at $x$ is an integral mean of the $(k+p)-$th Taylor coefficients of $v$, weighted with a Beta distribution on $[0,x]$. This is quite clear if $x=0$ and $v$ is analytic there, and not immediately obvious in general, but has it a special meaning, or is it an instance of a more general principle?</p> <p><strong>2.</strong> Let $u\in C^\infty(\mathbb{R})$, and assume that the formal Taylor series of $u$ at $0$ belongs to $\mathbb{R}[[x^2]]$ . Then, the function $w(x):=u( \sqrt { x } )$ is also (extendible to) a $C^\infty(\mathbb{R})$ function, and we can express the $k-$ order derivative of $w$ at $x^2$ in terms of the $2k-$ order derivatives of $v$ on $[0,x]$ as follows:</p> <p>$$w^{(k)}(x^2)=\frac{\ (2x)^{-2k+1}}{(k-1)!\ }\ \int_0^x (x^2-t^2)^{k-1}u^{(2k)}(t) dt\$$ (this may also be written as an equality relating Taylor coefficients by means of an integral mean).</p> <p><strong>3.</strong> There is also a more general statement for a function $w(x):=u(x^{1/p})$ for $p\in\mathbb{N} _ +$, assuming that the formal Taylor series of $u$ is in $\mathbb{R}[[x^p]]$; the $k-$th Taylor coefficient of $w$ at $x^p$ is then an integral mean of the $kp-$th Taylor coefficients of $v$, supported on $[0,x]$, with certain densities depending on $p$ and $n$ recursively defined. Is there a more general statement connecting analogously operations in $\mathbb{R}[[x]]$ and $C^\infty$ functions via integral means of their Taylor formal series?</p> http://mathoverflow.net/questions/65264/integral-representation-of-higher-order-derivatives/74548#74548 Answer by Bill Cook for Integral representation of higher order derivatives Bill Cook 2011-09-05T01:45:46Z 2011-09-05T22:28:25Z <p>Hello,</p> <p>I happened upon your question and noticed that your integrals look a lot like fractional derivatives. I think they provide the <i>generalization</i> you're looking for. </p> <p><a href="http://en.wikipedia.org/wiki/Fractional_derivative" rel="nofollow">http://en.wikipedia.org/wiki/Fractional_derivative</a></p> <p>I know this doesn't fully answer your question, but should be a good place to start looking. Good luck!</p> http://mathoverflow.net/questions/65264/integral-representation-of-higher-order-derivatives/125558#125558 Answer by Peter Michor for Integral representation of higher order derivatives Peter Michor 2013-03-25T19:37:30Z 2013-03-25T19:37:30Z <ol> <li><p>This is follows directly from the Taylor development to order $p-1$ with integral remainder: $$f(x) = 0+ \dots + 0+ x^p \int_0^1\frac{(1-t)^{p-1}}{(p-1)!}f^{(p)}(tx)dt.$$ Your formulas are differentiated versions of this.</p></li> <li><p>By Whitney (Duke Math J. 10, 1943), if $f$ is invariant under the $\mathbb Z/(2)$-action $x\mapsto -x$ on $\mathbb R$, then $f(x)=g(x^2)$ for some smooth $g$. Gleaser (Ann. Math. 77, 1963) extended this as follows: If $f\in C^\infty(\mathbb R^n)$ is invariant under all permutations of the coordinates, then $f(x)=g(\sigma_1(x),\dots,\sigma_n(x))$ for a smooth function $g$, in the elementary symmetric functions $\sigma_i$. G. Schwartz (Topology 14, 1975) extended this for any representation of a compact Lie group $G$ on $\mathbb R^n$, for any generating system $\rho_1,\dots,\rho_k$ of the algebra of $G$-invariant polynomials: $\rho^\star: C^\infty(\mathbb R^k)\to C^\infty(\mathbb R^n)^G$ is surjective. Mather (Topology 16, 1977) reproved this and showed that there is a linear continuous section of $\rho^\star$. Luna (Ann Inst Fourier 26, 1976) extended this to reductive Lie groups. <BR> The relation to your question is follows: You require only that the Taylor series of $f$ at zero is invariant under $x\mapsto -x$. So let $\tilde f(x) = \frac12(f(x)+f(-x))$. Then $\tilde f - f$ is infinitely flat at 0, so $(\tilde f- f)(\sqrt x)$ for $x\ge 0$ makes sense and is smooth and can be extended to a smooth function $g$ on $\mathbb R$ evenly, oddly, or by 0. $\tilde f(x) = \tilde g(x^2)$ for some smooth $\tilde g$. Then $\tilde g -g$ is what you looked for. This recipe works for all representations of compact groups by integrating over the group.</p></li> <li><p>Consider $F(z) = f(Re(z))$ and let $\mathbb Z/(p)$ act on $\mathbb C$ by $z\mapsto e^{2\pi i k/p}.z$ where $k=0,1,\dots p-1$. Then we are in the situation of 2. The invariant real polynomials are generated by $Re(z^p)$ and $Im(z^p)$. Integrate over $\mathbb Z/(p)$ and use flat functions as in 2.</p></li> </ol>