Partial Sum with in 10^-10

In summary, the conversation discusses finding the smallest positive integer N for which the partial sum of a geometric series is within 10^-10 of the series sum. It is suggested to find the actual sum or to determine the smallest N such that the absolute value of the next term is less than 10^-10. Taking the natural log and solving from there is also recommended.
  • #1
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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 there's the negative 1/4, how do i get it to be positive? i get the right answer when its positive but I am not sure how to do it.
 
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  • #2
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
mattmannmf said:
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 there's the negative 1/4, how do i get it to be positive? i get the right answer when its positive but I am 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
so all i have to do is take the absolute value, THEN take the natural log? and solve from there pretty much
 
  • #5
mattmannmf said:
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.
 

1. What is "Partial Sum with in 10^-10"?

"Partial Sum with in 10^-10" refers to a mathematical concept where a series of numbers is added together until the difference between the partial sum and the actual sum is less than or equal to 10^-10.

2. Why is "Partial Sum with in 10^-10" important?

This concept is important in mathematics as it allows for a more precise calculation of sums in situations where an infinite series cannot be easily calculated. It is also useful in calculating limits and approximations in various fields such as physics and engineering.

3. How is "Partial Sum with in 10^-10" calculated?

The calculation of "Partial Sum with in 10^-10" involves adding up a series of numbers and checking the difference between the partial sum and the actual sum after each addition. The process continues until the difference is less than or equal to 10^-10.

4. What is the significance of 10^-10 in "Partial Sum with in 10^-10"?

The value of 10^-10 is used as a measure of precision in the calculation of partial sums. It represents a very small number, and using it ensures that the partial sum is accurate to a high degree.

5. Can "Partial Sum with in 10^-10" be used for any series of numbers?

Yes, "Partial Sum with in 10^-10" can be applied to any series of numbers, as long as the series converges (the sum of the series approaches a finite value). However, the accuracy of the partial sum may vary depending on the properties of the series.

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