Question: 3.6 Evaluate e-5 using two approaches x2 x3 e* = 1-x+ 2 3! and 1 e x2 1 + x + 2 x3 + +

3.6 Evaluate e-5 using two approaches x2 x3 e* = 1-x+ 2 3! and 1 e x2 1 + x + 2 x3 + + 3! and compare with the true value of 6.737947 x 10-3. Use 20 terms to evaluate each series and compute true and approximate relative errors as terms are added

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