SOLUTION: First the problem explains the how series in caculus can be shown by the sigma sign form (such as on the main page of the series section on this site) The problem is : Approxima

Algebra ->  Sequences-and-series -> SOLUTION: First the problem explains the how series in caculus can be shown by the sigma sign form (such as on the main page of the series section on this site) The problem is : Approxima      Log On


   

Question 127883This question is from textbook College Algebra with Trigonometry
: First the problem explains the how series in caculus can be shown by the sigma sign form (such as on the main page of the series section on this site)
The problem is : Approximate e^-.5 using the first five terms of the series. Compare this approximation with your caculator evaluation of e^-.5 This question is from textbook College Algebra with Trigonometry

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
I'm presuming that the series to which you refer is e%5Ex=sum%28%28x%5En%29%2Fn%21%2Cn=0%2Cinfinity%29

The first 5 terms:

1%2Bx%2B%28x%5E2%2F2%21%29%2B%28x%5E3%2F3%21%29%2B%28x%5E4%2F4%21%29

1+%2B+%28-0.5%29+%2B+%280.25%2F2%29+%2B+%28-0.125%2F6%29+%2B+%280.0625%2F24%29

1+-+0.5+%2B+0.125+-+0.0208+%2B+0.0026+=+.6068 (roughly)

Calculator result: 0.6065 Certainly close enough for government work.