Re: calculation error in series
- To: mathgroup at smc.vnet.net
 - Subject: [mg125646] Re: calculation error in series
 - From: Bob Hanlon <hanlonr357 at gmail.com>
 - Date: Sun, 25 Mar 2012 00:15:41 -0500 (EST)
 - Delivered-to: l-mathgroup@mail-archive0.wolfram.com
 - References: <201203240702.CAA04167@smc.vnet.net>
 
expr = Sum[1/2^(n + 1)
   Sum[(-1)^k  Binomial[n, k] E^(-k/2) Cos[a k],
    {k, 0, n}], {n, 0, Infinity}]
(Sqrt[E] + 2*E^(1 + I*a) +
      E^(1/2 + 2*I*a))/
   (2*(1 + E^(1/2 + I*a))*
      (Sqrt[E] + E^(I*a)))
Note that the internal algorithms produce a representation that
includes I. Numerical evaluation with this representation can produce
a complex artifact.
expr /. a -> 14.134725141734695
0.7305593181773564 -
   5.551115123125783*^-17*I
The simplest approach is just to use Chop to eliminate the artifact
% // Chop
0.7305593181773564
Alternatively,
expr // ExpToTrig // FullSimplify
(E + Sqrt[E]*Cos[a])/(1 + E +
      2*Sqrt[E]*Cos[a])
% /. a -> 14.134725141734695
0.7305593181773565
Bob Hanlon
On Sat, Mar 24, 2012 at 3:02 AM, Maurice Coderre
<mauricecoderre at gmail.com> wrote:
> In[52]:= \!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
> FractionBox[\(1\),
> SuperscriptBox[\(2\), \((n + 1)\)]] \(
> \*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(n\)]\((
> SuperscriptBox[\((\(-1\))\), \(k\)] \((\((
> \*FractionBox[\(n!\), \(\(\((n - k)\)!\) \(k!\)\)])\)
> \*SuperscriptBox[\(E\), \(-
> \*FractionBox[\(k\), \(2\)]\)]\ )\) Cos[14.134725141734695  k])\)\)\)
> \)
>
> Out[52]= 0.730559318177 + 5.55111512313*10^-17 I
>
> In[53]:= \!\(
> \*UnderoverscriptBox[\(\[Sum]\), \(n = 0\), \(\[Infinity]\)]\(
> FractionBox[\(1\),
> SuperscriptBox[\(2\), \((n + 1)\)]] \(
> \*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(n\)]\((
> SuperscriptBox[\((\(-1\))\), \(k\)] \((\((
> \*FractionBox[\(n!\), \(\(\((n - k)\)!\) \(k!\)\)])\)
> \*SuperscriptBox[\(E\), \(-
> \*FractionBox[\(k\), \(2\)]\)]\ )\))\)\)\)\)
>
> Out[53]= Sqrt[E]/(1 + Sqrt[E])
>
> Why does the insertion of a purely real trigonometric function in a
> purely real infinit series, as shown above, give a complex result? Is
> it the result of an accumulated imprecision in the numerical
> evaluation?
>
- References:
- calculation error in series
- From: Maurice Coderre <mauricecoderre@gmail.com>
 
 
 - calculation error in series