       Re: calculation error in series

• To: mathgroup at smc.vnet.net
• Subject: [mg125706] Re: calculation error in series
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Thu, 29 Mar 2012 03:03:27 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201203280958.EAA11087@smc.vnet.net>

```Close the sum before using the values for w1 or w2 and the results are
what you expect.

equ1 = 1/2^(n + 1);

equ2 = (-1)^k*Binomial[n, k]/E^(k/2);

w1 = 14.134725141734695;

w2 = Rationalize[w1, 0];

equ3 = Sum[equ1*equ2*Cos[a*k], {n, 0, Infinity}, {k, 0, n}]

(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)))

equ4 = equ3 // ExpToTrig // FullSimplify

(E + Sqrt[E]*Cos[a])/
(1 + E + 2*Sqrt[E]*Cos[a])

{equ3, equ4} /. a -> w1 // Chop

{0.730559, 0.730559}

{equ3, equ4} /. a -> w2 // N // Chop

{0.730559, 0.730559}

Bob Hanlon

On Wed, Mar 28, 2012 at 5:58 AM, Dana DeLouis <dana01 at me.com> wrote:
>>  ... Is it the result of an accumulated imprecision ...
>
> Hi.  Just for fun, here's an opposite view, where we think we have precise input.
> I've never understood the algorithm, but it appears that if the input has a large numerator or denominator, then this situation appears to change the algorithm.
> Here are 2 examples.
>
> You had 2 sum functions, so just to be different, I'll use one.
>
> equ1=Divide[1,Power[2,(n+1)]];
> equ2=(Power[-1,k](Binomial[n,k]E^(-(k/2))));
>
> Sum[equ1*equ2,{n,0,Infinity},{k,0,n}]
>
> Sqrt[E]/(1+Sqrt[E])
>
> The above checks with what you have.
>
> Here are 2 numbers, that are considered (almost) equal.
>
> w1=14.134725141734695;
> w2=Rationalize[w1,0];
>
> w1==w2
> True
>
> One might think w2 would produce the more accurate answer.  However, the algorithm gives different answers.
>
> Sum[equ1*equ2*Cos[w1*k],{n,0,Infinity},{k,0,n}]  //Chop
> 0.730559
>
> Sum[equ1*equ2*Cos[w2*k],{n,0,Infinity},{k,0,n}]  //N
> 0.36528 +0.221225 I
>
> Here are 2 numbers that are as close together as possible.
> Yet, they give different answers.
>
> w1=(2/3)-\$MachineEpsilon;
> w2=Rationalize[w1,0];
>
> w1==w2
> True
>
> Sum[equ1*equ2*Cos[w1*k],{n,0,Infinity},{k,0,n}]  //Chop
> 0.636162
>
> Sum[equ1*equ2*Cos[w2*k],{n,0,Infinity},{k,0,n}]  //N//Chop
> 0.318081 +0.0807899 I
>
> = = = = = = = = = = = =
> I just find it interesting
> HTH  :>)
> Dana DeLouis
> Mac & Math 8
> = = = = = = = = = = = =
>
>
> On Mar 24, 3:04 am, Maurice Coderre <mauricecode... at gmail.com> wrote:
>> In:= \!\(
>> \*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= 0.730559318177 + 5.55111512313*10^-17 I
>>
>> In:= \!\(
>> \*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= 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?

```

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