anonymous
  • anonymous
Approximate the sum of the series (Sigma) 1/(n^3) by using the first 5 terms. Which is just: 1+1/8+1/27+1/64+1/125 = 1.18543.... How do I estimate the error involved in this approximation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
the eror is never greater than the next term, or something along those lines
anonymous
  • anonymous
So I would be comparing the error of 1.18543 and 1.18543 + 1/(6^3)?
amistre64
  • amistre64
wish i could tell, i never really got that far with it :)

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amistre64
  • amistre64
http://ecademy.agnesscott.edu/~lriddle/apcalculus/approxSeries.pdf might be useful
amistre64
  • amistre64
total sum = partial + remainder \[s=s_n+R_n\] \[R_n=s-s_n\] \[R_{max}=\int_{n+1}^{inf} f(x)dx\] \[R_{min}=\int_{n}^{inf} f(x)dx\]
amistre64
  • amistre64
i think i might have those backwards tho
amistre64
  • amistre64
anyhoos \[\int_{6}^{inf}x^{-3}dx=-\frac{1}{2(inf)^2}+\frac{1}{2(6)^2}\] \[\int_{5}^{inf}x^{-3}dx=-\frac{1}{2(inf)^2}+\frac{1}{2(5)^2}\] so it looks like the total sum rests between sn + 1/72 and sn+1/50
amistre64
  • amistre64
well, sn being the 5th partial sum in this case

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