Originally Posted by

**namelessguy** $\displaystyle e^{-5}$ $\displaystyle \approx$ $\displaystyle \sum$ $\displaystyle \frac{(-5)^i}{i!}$ = $\displaystyle \sum$ $\displaystyle \frac{(-1)^i 5^i}{i!}$

i runs from 0 to 9

$\displaystyle e^{-5}$= $\displaystyle \frac{1}{e^5}$ $\displaystyle \approx$ $\displaystyle \frac{1}{\sum\frac{5^i}{i!}}$

i runs from 0 to 9

Sorry, I don't know how to put the indices on the summation. The question is which formula gives the most accuracy comparing to the actual value of $\displaystyle e^{-5}$ correct to three digits, which is 6.74 x $\displaystyle 10^{-3}$.

I computed using both formulas and figured out that the second formula gives the most accuracy, but I don't have a solid answer for why it is more accurate than the other.

I guess it's is because in the second formula we write down the exact formula for $\displaystyle e^{-5}$, then use the approximation after. On the first formula, we approximate first.

Anyone can give me some thought on this. Thank you.