Re: Need help with infinite series
- From: "M.A.Fajjal" <h2maf@xxxxxxxxx>
- Date: Thu, 05 Jan 2006 08:08:57 EST
> >
> >The summation of 10^9 terms is 0.604882832034161
> >
> >So, 0.604914454810747 < s < 0.604882832034161
> >
> >The average is 0.604898643422454
>
> The expected error in the sum of 10^9 terms would be
> the last 4 or 5
> digits. Your computation seems to be off in the last
> 4 places.
>
> Using the asymptotic expansion from another thread in
> this topic, we
> can derive the asymptotic expansion for the
> alternating series:
>
> 2n
> --- k-1 1
> > (-1) -------
> --- sqrt(k)
> k=1
>
> 1 1
> 1 1 5
> ~ (1 - sqrt(2)) zeta(1/2) + sqrt(2n) ( - -- +
> -- + ----- - -------
> 4n
> 4n 32n^2
> 4n 32n^2 2048n^4
>
> 63 7293
> + -------- - ---------- + ... )
> 65536n^6 8388608n^8
>
> We also have both of the following:
>
> 1 1
> -------- = sqrt(2n) ( -- )
> sqrt(2n) 2n
>
> 1 1 1 3 5
> ---------- ~ sqrt(2n) ( -- - ---- + ----- -
> -- - ------ + ... )
> sqrt(2n+1) 2n 8n^2 64n^3
> 4n^3 256n^4
>
> Thus, if we average two adjacent partial sums, either
> 2n-1 and 2n or
> 2n and 2n+1, the error is about
>
> 1 -3/2
> - (2n)
> 8
>
> which is about 4e-15 when 2n = 10^9. So your answer,
> to the number of
> decimal places you have, should be off in only the
> last place, but it
> is off in the last 4.
>
> This means that the major source of error in your
> computation is the
> rounding error coming from adding 10^9 terms.
10^9 terms was determined using simple loop for summations. Is there any way to increase accuracy of soft wares like MATLAP for such cases
.
- References:
- Re: Need help with infinite series
- From: Rob Johnson
- Re: Need help with infinite series
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