Mathematica 9 is now available
Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2009

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Incongruence? hmm...


 Bug? Oversight? I don't know how to classify it. Actually, Mathematica
makes a bit of a meal of sums of this sort: compare its results for

Sum[Cos[m x]/m^2, {m, 1, Infinity}, GenerateConditions -> True]
Sum[Sin[m x]/m^3, {m, 1, Infinity}, GenerateConditions -> True]
Sum[Sin[m x]/m^5, {m, 1, Infinity}, GenerateConditions -> True]
Sum[Cos[m x]/m^2, {m, 1, Infinity}]
Series[%, {x, 0, 5}]
Sum[Sin[m x]/m^3, {m, 1, Infinity}]
Series[%, {x, 0, 5}]
Sum[Sin[m x]/m^5, {m, 1, Infinity}]
Series[%, {x, 0, 5}]

 with Gradshteyn and Ryzhik's results under item 1.443. I find it
particularly strange that including GenerateConditions -> True seems to
wrong-foot Mathematica on the first of these.

  Tony Harker

]-> -----Original Message-----
]-> From: Filippo Miatto [mailto:miatto at gmail.com] 
]-> Sent: 23 August 2009 10:34
]-> To: mathgroup at smc.vnet.net
]-> Subject: [mg102657] Re: [mg102734] Re: [mg102717] Re: 
]-> [mg102710] Incongruence? hmm...
]-> 
]-> I see. So, since the solution that Mathematica finds is 
]-> valid between 0 and 2Pi, what if I take the result
]-> 
]-> (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi]x^3 - 15 x^4)/720
]-> 
]-> and substitute Mod[x,2Pi] in place of all x? That would 
]-> make it valid for every x.
]-> I don't get why Mathematica doesn't find the right answer 
]-> for x in [-2pi,2pi], though.. Either using Abs[x] for the 
]-> 3rd power term, or at least using the right sign in the 
]-> interval [-2pi,0].
]-> is it a bug?
]-> 
]-> 2009/8/22, DrMajorBob <btreat1 at austin.rr.com>:
]-> > Yes, Cos is an even function, so the two results should 
]-> agree at 3 and 
]-> > -3, for instance:
]-> >
]-> > one = Sum[Cos[m x]/m^4, {m, 1, \[Infinity]},
]-> >    Assumptions -> {0 < x < 2 Pi}]
]-> > one3 = one /. x -> 3;
]-> >
]-> > 1/720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi] x^3 - 15 x^4)
]-> >
]-> > two = Sum[Cos[m x]/m^4, {m, 1, \[Infinity]},
]-> >    Assumptions -> {-2 Pi < x < 0}]
]-> > two3 = two /. x -> -3;
]-> >
]-> > 1/720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi] x^3 - 15 x^4)
]-> >
]-> > two3 - one3 // Expand
]-> >
]-> > -((9 \[Pi])/2)
]-> >
]-> > But they're not the same, so one of those results is 
]-> wrong. As Tony 
]-> > explained, the sum MUST be between + and - Pi^4/40 for 
]-> ALL x, and the 
]-> > second result fails that test:
]-> >
]-> > limit = Pi^4/90.
]-> >
]-> > 1.08232
]-> >
]-> > two /. x -> -Pi // N
]-> >
]-> > -17.1819
]-> >
]-> > Plot[{limit, -limit, two}, {x, -Pi, 0}, PlotRange -> All]
]-> >
]-> > An expression that's correct (or even and properly 
]-> bounded, at least) 
]-> > over both intervals is
]-> >
]-> > three = 1/
]-> >     720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi] Abs[x]^3 - 
]-> 15 x^4); 
]-> > Plot[{limit, -limit, three}, {x, -2 Pi, 2 Pi}]
]-> >
]-> > This expression agrees with the first result on 0 to 2Pi, as well.
]-> >
]-> > Plot[{one, three}, {x, 0, 2 Pi}, PlotRange -> All]
]-> >
]-> > Bobby
]-> >
]-> > On Sat, 22 Aug 2009 06:41:33 -0500, Filippo Miatto 
]-> <miatto at gmail.com>
]-> > wrote:
]-> >
]-> >> But the cosine is an even function, why should the 
]-> result not be even 
]-> >> in x?
]-> >> I mean, the assumption 0<x<2pi should be equivalent to 
]-> the assumption 
]-> >> -2pi<x<0 since every term of the sum does not change value.
]-> >> F
]-> >>
]-> >> 2009/8/22, DrMajorBob <btreat1 at austin.rr.com>:
]-> >>> So, of these, Mathematica gets the second one wrong:
]-> >>>
]-> >>> Sum[Cos[m x]/m^4, {m, 1, \[Infinity]},
]-> >>>    Assumptions -> {0 < x < 2 Pi}]
]-> >>>
]-> >>> 1/720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi] x^3 - 15 x^4)
]-> >>>
]-> >>> Sum[Cos[m x]/m^4, {m, 1, \[Infinity]},
]-> >>>    Assumptions -> {-Pi < x < 0}]
]-> >>>
]-> >>> 1/720 (8 \[Pi]^4 - 60 \[Pi]^2 x^2 + 60 \[Pi] x^3 - 15 x^4)
]-> >>>
]-> >>> Bobby
]-> >>>
]-> >>> On Fri, 21 Aug 2009 03:43:01 -0500, Tony Harker 
]-> <a.harker at ucl.ac.uk>
]-> >>> wrote:
]-> >>>
]-> >>>>
]-> >>>>  I don't see how either expression can be correct for 
]-> all x. The 
]-> >>>> 'direct'
]-> >>>> form is clearly wrong, as every term in the summation 
]-> is an even 
]-> >>>> function of x. However, the modulus of each term is 
]-> less than or 
]-> >>>> equal to 1/m^4, so the sum must be bounded above and 
]-> below by plus 
]-> >>>> and minus Pi^4/90, which neither result is.
]-> >>>>
]-> >>>>   The indirect approach obviously has a limited radius of 
]-> >>>> convergence for the sum over m, and 
]-> GenerateConditions->True will 
]-> >>>> show that, so that's what goes wrong there. The direct 
]-> approach 
]-> >>>> does not generate any conditions, so seems to be just 
]-> plain wrong.
]-> >>>>
]-> >>>>   The key to sorting this out is to note that the original 
]-> >>>> expression is a Fourier series, so any polynomial form 
]-> can only be 
]-> >>>> valid over an interval such as -Pi to Pi, and must then repeat 
]-> >>>> periodically.
]-> >>>>
]-> >>>>   In fact, the result from the direct sum is correct 
]-> for 0<x<Pi, 
]-> >>>> and for -Pi<x<0 the expression is similar, but the sign of the 
]-> >>>> coefficient of
]-> >>>> x^3 is
]-> >>>> changed. Basically, the original sum is a polynomial in (Pi-x) 
]-> >>>> which has been made symmetrical about x=0.
]-> >>>>
]-> >>>>   Tony Harker
]-> >>>>
]-> >>>>
]-> >>>> ]-> To: mathgroup at smc.vnet.net
]-> >>>> ]-> Subject: [mg102710] Incongruence? hmm...
]-> >>>> ]->
]-> >>>> ]-> Dear all,
]-> >>>> ]-> I'm calculating the sum
]-> >>>> ]->
]-> >>>> ]-> Sum[Cos[m x]/m^4, {m, 1, \[Infinity]}] ]-> ]-> in 
]-> two different 
]-> >>>> ways that do not coincide in result.
]-> >>>> ]-> If i expand the cosine in power series ]-> ]-> ((m x)^(2n) 
]-> >>>> (-1)^n)/((2n)!m^4) ]-> ]-> and sum first on m i obtain ]-> ]-> 
]-> >>>> ((-1)^n x^(2n) Zeta[4-2n])/(2n)!
]-> >>>> ]->
]-> >>>> ]-> then I have to sum this result on n from 0 to 
]-> infinity, but ]-> 
]-> >>>> Zeta[4-2n] is different from 0 only for n=0,1,2 and 
]-> the result is 
]-> >>>> ]-> ]-> \[Pi]^4/90 - (\[Pi]^2 x^2)/12 - x^4/48 ]-> ]-> 
]-> Three terms, 
]-> >>>> one independent on x, with x^2, one with x^4.
]-> >>>> ]->
]-> >>>> ]-> however if I perform the sum straightforwardly 
]-> (specifying that 
]-> >>>> ]-> 0<x<2pi) the result that Mathematica gives me is ]-> ]-> 
]-> >>>> \[Pi]^4/90 - (\[Pi]^2 x^2)/12 + (\[Pi] x^3)/12 - 
]-> x^4/48 ]-> ]-> 
]-> >>>> with the extra term (\[Pi] x^3)/12. Any idea on where 
]-> it ]-> comes 
]-> >>>> from??
]-> >>>> ]-> Thank you in advance,
]-> >>>> ]-> Filippo
]-> >>>> ]->
]-> >>>> ]->
]-> >>>>
]-> >>>>
]-> >>>
]-> >>>
]-> >>>
]-> >>> --
]-> >>> DrMajorBob at bigfoot.com
]-> >>>
]-> >>>
]-> >
]-> >
]-> >
]-> > --
]-> > DrMajorBob at bigfoot.com
]-> >
]-> 
]-> 



  • Prev by Date: Re: Incongruence? hmm...
  • Next by Date: Re: Viewing packages in mathematica
  • Previous by thread: Re: Re: Re: Incongruence? hmm...
  • Next by thread: Re: Incongruence? hmm...