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3.6 Evaluate \( e^{-5} \) using two approaches
\[
e^{-x}=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{3 !}+\cdots
\]
and
\[
e^{-x}=\frac{

using exel using exel using exel using exel using exel using exel

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


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Given that series of e^-x We have e^-x=1-x+x^2/2!-x^3/3! +x^4/4! -x^5/5! +...
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