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Number $e$ is defined as $$ e=\sum^\infty_{n=0} \frac{1}{n!}.$$ Use four-digit chopping arithmatic to compute the following

  1. Approximations to $e$ $$ e\approx \sum^5_{n=0} \frac{1}{n!}$$

Attempt: \begin{align} \sum^5_{n=0} \frac{1}{n!} &= \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!} \\ &= \frac{1}{1}+\frac{1}{2}+\frac{1}{3 \cdot 2 \cdot 1}+\frac{1}{4 \cdot 3 \cdot 2 \cdot 1}+\frac{1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \\ &= 1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120} \end{align}

to decimals

$$=1+.5000+0.16666666666+.04166666667+.0083333333333333$$

and I am not sure how $4$-digit chopping works. I think that doing $4$-digit chopping is

$$=1+.5000+0.1666+.04166+.008333=1.716593=1.716$$

I am getting that it is wrong according to webwork wich would never ever be wrong.

Rócherz
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Tiger Blood
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1 Answers1

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That is quite easy: you forgot the $1/ 0!$ in your calculations. Then you get something like $\mathrm e \approx2.716$. Also rounding in the end of your calculation might be the best.