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