Evaluate
e^{-5}
e
−5
using two approaches
e^{-x} = 1 – x + \frac{x^2}{2} - \frac{x^3}{3!} + …
e
−x
=1–x+
2
x
2
−
3!
x
3
+…
and
e^{-x} = \frac{1}{e^x} = \frac{1}{1+ x + \frac{x^2}{2} + \frac{x^3}{3!}+…}
e
−x
=
e
x
1
=
1+x+
2
x
2
+
3!
x
3
+…
1
and compare with the true value of
6.737947 \times 10^{-3}
6.737947×10
−3
. Use 20 terms to evaluate each series and compute true and approximate relative errors as terms are added
