Dear Uncle Colin,

How would you work out $e^{-0.56}$ without a calculator?

- Tell All Your Logic Or Reasoning

Hi, TAYLOR, and thanks for your message!

I think most people would suggest a Taylor((Hey! That’s your name!)) series here – either centred around 0 (which converges fairly fast) or around some sensible nearby number – for example, since $\ln(2) \approx 0.693$, you can work out that $e^{-0.693}$ is about 0.5.

So, $e^{-0.56} \approx e^{-0.693} \times e^{0.133}$, and the series for $e^{0.133}$ converges much more quickly than the one for $e^{-0.56}$ (because the power is smaller). Let’s do that, for form’s sake:

  • $e^{x} \approx \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!}$
  • $e^{0.133} \approx 1 + 0.133 + 0.018$ (I’m kind of guessing at the third decimal place).
  • $e^{0.133} \approx 1.151$

And that means $e^{-0.56}$ is about half of that, or 0.576.


I should have expected that.

$\ln(3) \approx 1.099$, and $-\frac{1}{2}\ln{3} \approx 0.549$. That means $e^{-0.549} \approx 3^{-1/2}$, which is 0.577.

That’s pretty close, sensei.

We can do better. The discrepancy in the power is 0.01, so we need to lose about 1%. Call it 0.006 and we end up with 0.571.

I… understand that that’s not a bad approximation.

Hope that helps!

- Uncle Colin