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Re: A curious result?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg91511] Re: A curious result?
*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>
*Date*: Sun, 24 Aug 2008 07:08:02 -0400 (EDT)
*References*: <g8o89u$l8q$1@smc.vnet.net>
amzoti <amzoti at gmail.com> wrote:
> a. Sum[1/n^1.001, {n, 1 , Infinity}] = 1000.58
>
> b. Integrate[1/n^(1.001), {n, 1, Infinity}] = 1000
I don't see anything curious about either result. Copied below is my
response to what you had posted earlier in sci.math.
David
-------------------------------------------------
David W. Cantrell <DWCantrell at sigmaxi.net> wrote:
> amzoti <amzoti at gmail.com> wrote:
> > Hi All,
> >
> > I just finished reading this nice write-up on precision that be
> > required reading for all students.
> >
> > www.stewartcalculus.com/data/default/upfiles/LiesCalcAndCompTold.pdf
> >
> > Which one of these is correct?
>
> Both.
>
> > a. Sum[1/n^1.001, {n, 1, Infinity}] = 1000.58
>
> You appear to be using Mathematica. Note that it is not a good idea,
> typically, to combine an "approximate" number (in this case, 1.001) with
> a symbolic operation (in this case, Sum). Much better would be to use,
> say,
>
> Sum[1/n^(1001/1000), {n, 1, Infinity}]
>
> which returns the precise result
>
> Zeta[1001/1000]
>
> and that can then be approximated, if desired. Rounded to two decimal
> places, 1000.58 is correct.
>
> > b. Integrate[1/n^(1.001), {n, 1, Infinity}] = 1000
>
> Well, actually that's not what Mathematica gives for the result. Since
> the input involved an "approximate" number, the result is 1000. instead.
>
> But, as for part a., it would have been much better to have asked for
> something which is exact:
>
> Integrate[1/n^(1001/1000), {n, 1, Infinity}]
>
> yields 1000 precisely.
>
> David
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