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Am I correct or is the book's answer correct

  • #1

Homework Statement


find the sum of (-1)^(n-1)*(n^2)/(10^n) correct to 4 decimal places


Homework Equations





The Attempt at a Solution


I got s5 since s6 is 36/1000000<0.0001
According to Wolframalpha, s5 is 0.06765 and s6 is 0.067614. Am i missing something here, or reading the problem wrong? The book claims it's s6 since s7 is 0.0000049 and that is the first value accurate to the 4th decimal place
 

Answers and Replies

  • #2
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5,192

Homework Statement


find the sum of (-1)^(n-1)*(n^2)/(10^n) correct to 4 decimal places


Homework Equations





The Attempt at a Solution


I got s5 since s6 is 36/1000000<0.0001
According to Wolframalpha, s5 is 0.06765 and s6 is 0.067614. Am i missing something here, or reading the problem wrong? The book claims it's s6 since s7 is 0.0000049 and that is the first value accurate to the 4th decimal place
To be accurate to 4 decimal places you want the first value in the sum that is smaller than 1/2 X 10-4. In other words, by adding this amount, it won't change the first four decimal places. 0.0001 is a bit too large, and could cause a change in the 4th decimal place.
 
  • #3
D H
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find the sum of (-1)^(n-1)*(n^2)/(10^n) correct to 4 decimal places
This is an alternating convergent series, so the error is bounded by the first omitted term. The sum is 0.06761833... To be accurate to four decimal places, you need to find where the first omitted term is less than this times 0.5*10-4, or 3.4*10-6. That's your s6. s5 is just a tad shy of delivering the required accuracy.
 
  • #4
1.s5=.06765 and s6=.067614. The fourth digit after the decimal is 6, and it did not change between s5 and s6. Thus isn't s5 correct?

2.And why is it (1/2)*10^-4 instead of 1*10^-4? 0.0001 seems right to me since the biggest possible values smaller than that would be in the 0.00009... range and that's 4 zeros between the decimal point and a nonzero digit. More specifically, why the (1/2) in front? I would think it would be 1*10^-4 or 1*10^-5 or something like that. I think there's some concept that I'm completely missing or not understanding right now. Can someone explain this to me and give me examples and counterexamples?
 
  • #5
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freshman2013 said:
According to Wolframalpha, s5 is 0.06765 and s6 is 0.067614. Am i missing something here, or reading the problem wrong? The book claims it's s6 since s7 is 0.0000049 and that is the first value accurate to the 4th decimal place
You are mixing up the partial sums with the individual terms in the series.
S5 means the sum of the first 5 terms. What you call s7 would be better labeled as a7, the 7th term in the series. a6 is the first term that is smaller than .0001.

For your first question, I'll need to double check the values you got.
For the second question, if the first unused term is near .0001, adding it to the partial sum is likely to change the digit in the fourth decimal place. Adding a number that is smaller than .00005 (= .5 * 10-4) won't change what's in the fourth decimal place.
 
  • #6
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As for your first question, you are correct in saying that s5 gives the right answer to four decimal places. No doubt your textbook is using a theorem called something like Alternating Series Estimation Theorem, which is what D H cited. That theorem gives an upper bound on the error in truncating such a series. Since this is an upper bound, the actual error could be less.
 
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  • #7
Ray Vickson
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Homework Statement


find the sum of (-1)^(n-1)*(n^2)/(10^n) correct to 4 decimal places


Homework Equations





The Attempt at a Solution


I got s5 since s6 is 36/1000000<0.0001
According to Wolframalpha, s5 is 0.06765 and s6 is 0.067614. Am i missing something here, or reading the problem wrong? The book claims it's s6 since s7 is 0.0000049 and that is the first value accurate to the 4th decimal place
You can use standard tricks to evaluate the infinite summation exactly; Maple gets
[tex] \sum_{n=1}^{\infty} (-1)^{n-1} \frac{n^2}{10^n} = \frac{90}{1331} \doteq 0.06761833208[/tex]
Furthermore, you can also get the finite sums in closed form, but numerical methods are, of course, easier.

Here are s1..s10 as computed by Maple:
.1000000000, .6000000000e-1, .6900000000e-1, .6740000000e-1, .6765000000e-1, .6761400000e-1, .6761890000e-1, .6761826000e-1, .6761834100e-1, .6761833100e-1

Note that there is a difference between 4-decimal-place accuracy and 4 significant-figure accuracy; maybe the book means "significant figures" instead of "decimal places".
 
  • #8
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Note that there is a difference between 4-decimal-place accuracy and 4 significant-figure accuracy; maybe the book means "significant figures" instead of "decimal places".
That thought occurred to me as well.
 

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