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Re: Bug in Sum?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg119978] Re: Bug in Sum?
*From*: Dana DeLouis <dana01 at me.com>
*Date*: Sun, 3 Jul 2011 04:10:51 -0400 (EDT)
Hi. If x had an assumed value of 1, then the equation would error with
Div by zero. However, no combination of assumptions, or Methods, appear
to overcome this.
If we change (1-x) to a single variable, then it appears to work.
equ=1/(n (1+n)) 1/((x)^4);
Sum[equ,{n,m}]
m/((1+m) x^4)
% /. x->(1-x)
m / ((1+m) (1-x)^4)
I'm not sure why x works, and not 1-x.
Even Wolfram Alpha has a hard time with this.
WolframAlpha["Sum (1/(n*(1 + n)*(x)^4)), n=1 to m=94]
<< Nice Output with Graph...>>
However, (1-x) causes a problem...
WolframAlpha["Sum (1/(n*(1 + n)*(1-x)^4)), n=1 to m=94]
.. Wolfram Alpha doesn't know how to interpret your input
Wow. I don't see the problem. Changing (x) to (1-x), and even Wolfram Alpha doesn't even know how to interpret this.
Not sure, but it sounds like a logic bug somewhere, unless there's a
mathematical reason for this behavior.
Anyone??
= = = = = = = = = =
HTH : >)
Dana DeLouis
$Version
8.0 for Mac OS X x86 (64-bit) (November 6, 2010)
On Jun 29, 5:42 pm, Dario <dario.benede... at aei.mpg.de> wrote:
> I am very puzzled by the fact that Mathematica does not evaluate the following Sum (it keeps running endlessly):
>
> Sum[1/(n (1 + n)) 1/((1 - x)^4) , {n, 1, M}]
>
> where x and M are some undefined variables,
>
> but it does evaluate
>
> Sum[1/(n (1 + n)) 1/((1 - x)^2) , {n, 1, M}]
>
> where I have only changed the power of (1-x), which should not matter as it must factor out!
>
> Any help on this please?
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