document.write( "Question 127883This question is from textbook College Algebra with Trigonometry
\n" ); document.write( ": 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)\r
\n" ); document.write( "\n" ); document.write( "The problem is : Approximate e^-.5 using the first five terms of the series. Compare this approximation with your caculator evaluation of e^-.5
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Algebra.Com's Answer #93746 by solver91311(24713)\"\" \"About 
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\"\r
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\n" ); document.write( "\n" ); document.write( "The first 5 terms:\r
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\n" ); document.write( "\n" ); document.write( "\"1%2Bx%2B%28x%5E2%2F2%21%29%2B%28x%5E3%2F3%21%29%2B%28x%5E4%2F4%21%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"1+%2B+%28-0.5%29+%2B+%280.25%2F2%29+%2B+%28-0.125%2F6%29+%2B+%280.0625%2F24%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"1+-+0.5+%2B+0.125+-+0.0208+%2B+0.0026+=+.6068\" (roughly)\r
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\n" ); document.write( "\n" ); document.write( "Calculator result: \"0.6065\" Certainly close enough for government work.
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