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

```

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