Partial Sum with in 10^-10

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  • #1
mattmannmf
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determine the smallest positive integer N such that the partial sum is within 10^-10 of the series sum.

{Sigma} 2(-1/4)^(n-1) < 10^-10

ok divide the 2 out

(-1/4)^(n-1) < 5^-10


I know i have to take the natural log like so:
(n-1) ln|-1/4| < nl|5^-10|

But theres the negative 1/4, how do i get it to be positive? i get the right answer when its positive but im not sure how to do it.
 

Answers and Replies

  • #2
Dick
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It's an alternating series sum{a_n}. You have an error estimate for the partial sum up to n in terms of |a_(n+1)|. Note the absolute value.
 
  • #3
HallsofIvy
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determine the smallest positive integer N such that the partial sum is within 10^-10 of the series sum.

{Sigma} 2(-1/4)^(n-1) < 10^-10
Is [itex]\sum 2 (-1/4)^{n-1}[/itex] the sum? You have stated this wrong- you don't want the partial sum, which is what you have here, to be less than [itex]10^{-10}[/itex], you want its difference from the actual sum to be less than [itex]10^{-10}[/itex] but have then done that.

One way to do that is to find the actual sum. That is not difficult because this is a "geometric" series and there is a formula for the sum.

But, better, as Dick says, because this is alternating + and -, the "error" is always less than the next term. You just want to find n so that the absolute value of the next term is less than [itex]10^{-10}[/itex]

ok divide the 2 out

(-1/4)^(n-1) < 5^-10


I know i have to take the natural log like so:
(n-1) ln|-1/4| < nl|5^-10|

But theres the negative 1/4, how do i get it to be positive? i get the right answer when its positive but im not sure how to do it.
[tex]\left|\left(-\frac{1}{4}\right)^{n-1}\right|= \left(\frac{1}{4}\right)^{n-1}< 5^{-10}[/tex]
 
  • #4
mattmannmf
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so all i have to do is take the absolute value, THEN take the natural log? and solve from there pretty much
 
  • #5
Dick
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so all i have to do is take the absolute value, THEN take the natural log? and solve from there pretty much

Pretty much, yes. Provided you understand why that works.
 

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