How can we calculate $e^{-0.4}, ~e^{-1.67}, ~e^{-5}, ~e^{0.6},~e^{1.23}, ~\text{and} ~e^{7}$ by hand (without using a calculator)?
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$\begingroup$It depends a lot on what you know and how accurate you want to be. For small values of $x$ you can use the Taylor series $e^x \approx 1+x+\frac {x^2}2$. Any more terms than that is too hard. This will work pretty well for $-0.4, 0.6$ The error for $0.6$ is about $0.036$ on $1.78$ or $2\%$, which I would call very good. Then for $1.23$ you can square $1.78$ and multiply by $1.03$. For $-0.4$ I get $0.68$ from the Taylor series.
For $7$, maybe you know that $\log 10 \approx 2.30$, which gives us that $e^{6.9} \approx 1000$ and $e^7\approx 1.1\cdot 1000 =1100$. The correct value is about $1096.63$
For $-5$ I would say $e^{-4.6} \approx 0.01$, and so $e^{-5}=e^{-4.6}e^{-.4}\approx 0.0068$
For $1.23$ I would probably use $e^{1.23}=\sqrt{e^{2.3}}e^{0.08}\approx 1.08\sqrt{10}\approx 1.08 \cdot 3.162$
For $-1.67$ (if I had already done the above) I would probably invert the $1.23$ result and multiply by $e^{-0.4}\approx 0.68$. If not, I would probably use $\log 2 \approx 0.69$ anddo $e^{-1.67}=e^{-1.38}e^{-.35}e^{0.04}\approx 0.25\cdot 0.707\cdot 1.04$
I played fair, though I checked the numbers with a calculator, I did these by hand.
$\endgroup$ $\begingroup$As for "how we can calculate $e^x$ for given values of $x$:"
See the Wikipedia entry: Taylor Series, for common functions, first listing the expansion of $e^x$:
$$e^{x} = \sum^{\infty}_{n=0} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\quad\text{ for all } x\!$$
- It the section immediately following that, there is an outline for Calculation of Taylor Series, using specific examples.
As to actually calculating the specific values you list: I'll leave that work to you. Knowing this, you have all the tools you need linked/listed above. And you can minimize your work by noting $$\;e^{x+y} = e^xe^y\;\text{ and}\;\;e^{x-y} = \dfrac{e^x}{e^y},\;$$ so once compute, e.g., $\;e^{0.6},\;$ then you can easily compute $\;e^{-0.4} = e^{0.6 - 1} = \dfrac{e^{0.6}}{e}$.
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