Why does this trick work? Screenshot inside

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In summary, the conversation discusses a method to evaluate an error in a series using a given formula. The person is confused about why a large number of terms is needed to get an answer with an error < 0.005, but the formula provided allows for an answer with an error < 0.005 using only ~130 terms. The difference lies in how the series is summed, with the alternative method allowing for a faster decrease in error.
  • #1
tamtam402
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Why does this "trick" work? Screenshot inside

I managed to do the computations but unfortunately I don't understand why the trick works in part b). It is said that N has to be very very big to find the answer within 0.005, yet in part b I find N ≈ 130 (which I know is right). That means the solution can now be found with error < 0.005 with only 130 terms. Is that right? Could someone explain to me what's going on brievely? This is a scan from Mary L Boas mathematical methods book which I'm self-learning from, and unfortunately this is the only information found on this trick to evaluate the error.
 

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Thanks for replying. Unfortunately the linked website wasn't much help, I already figured out how to use the given expressions to bind a series, but I was hoping someone would explain to me why that method works. I'm mostly confused by the 2 steps listed in my book: first it is stated that a very big number of terms is necessary to get an answer with an error < 0.005, yet immediately after that I'm given the formula to get the answer with an error < 0.005 using only ~130 terms. What's going on?
 
  • #4


The difference lies in how you sum the series, which is what this example is trying to get at. Summing the terms one by one requires many, many terms to ensure the error falls into place, but this alternative method (whatever that is; "Figures 6.1 and 6.2" are not in the picture) seems to make it clear that you can massage the error to decrease much faster.
 

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