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Re: Re: Re: Incongruence? hmm...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg102656] Re: [mg102734] Re: [mg102717] Re: [mg102710] Incongruence? hmm...
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Sun, 23 Aug 2009 05:33:32 -0400 (EDT)
  • References: <200908200856.EAA05738@smc.vnet.net>
  • Reply-to: drmajorbob at bigfoot.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


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