Re: Trouble : simple stuffs for beginner

*To*: mathgroup at smc.vnet.net*Subject*: [mg48279] Re: Trouble : simple stuffs for beginner*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Fri, 21 May 2004 03:54:28 -0400 (EDT)*Organization*: The University of Western Australia*References*: <c8f236$d3l$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

In article <c8f236$d3l$1 at smc.vnet.net>, Julien MARY <julien.mary at free.fr> wrote: > I am defining : > > g[x_] := Sum[ > Integrate[(-1)^k(2k - t)(2k - 1 < x <= 2k + 1)E[-(t - x)^2/2], > t], {k, -Infinity, Infinity}] Mathematica is not yet smart enough to understand such a syntax for piecewise functions. > To be clear, it the convolution of arcsin(sin(x))/2Pi with exp(-x²/2) I obtain ArcSin[Sin[t]] = (-1)^k (t - k Pi) for (k - 1/2) Pi <= t <= (k + 1/2) Pi. Hence the piecewise integral is int[k_][x_] = Integrate[(-1)^k (t - k Pi) Exp[-(t - x)^2 / 2], {t, (k - 1/2) Pi, (k + 1/2) Pi}] which Mathematica can compute in closed-form. Then, you need to sum over all the pieces. Here is the (symmetric) partial sum: g[n_][x_] := Sum[int[k][x], {k, -n, n}] Mathematica cannot compute the infinite sum in closed-form (and I suspect no simple closed-form exists). > The I want to plot g[x] with Plot[g[x],{x,-10,10}] and Mathematica is > insulting me. Is it really insulting you? Anyway, Plot[Evaluate[g[15][x]], {x,-10,10}] does what you want (you will see little change as you increase the number of terms in the partial sum). Note the use of Evaluate to compute the sum _once_ prior to plotting. Also, observe that the result is "very close" to 1.94 Sin[x] (i.e., other Fourier components are very small). An alternative derivation uses the Fourier Sin expansion of ArcSin[Sin[t]]: ArcSin[Sin[t]] == Sum[(-1)^(k+1) Sin[(2k-1) t]/(k-1/2)^2,{k,1,Infinity}]/Pi Then the convolution integral can be obtained by computing SetOptions[Integrate, GenerateConditions -> False]; Integrate[Exp[I (2k-1) t] Exp[-(t-x)^2 / 2], {t,-Infinity,Infinity}] and taking the imaginary part of the result Simplify[ComplexExpand[Im[%]], k > 0] obtaining Sqrt[2 Pi] Sin[(2k-1) x] Exp[-(2k-1)^2 / 2] Hence the (partial sums of the) convolution are h[n_][x_] := Sqrt[2/Pi] Sum[(-1)^(k+1)/(k-1/2)^2 Sin[(2k-1) x] * Exp[-(2k-1)^2 / 2],{k,1,n}] which we plot as Plot[Evaluate[h[15][x]], {x,-10,10}] Cheers, Paul -- Paul Abbott Phone: +61 8 9380 2734 School of Physics, M013 Fax: +61 8 9380 1014 The University of Western Australia (CRICOS Provider No 00126G) 35 Stirling Highway Crawley WA 6009 mailto:paul at physics.uwa.edu.au AUSTRALIA http://physics.uwa.edu.au/~paul