Physics Asked on January 5, 2021

From Braddick’s *The Physics of Experimental Method* (1963):"It should be remembered that the number of significant figures in a number $y$, and one derived from it, say $e^y$, are not always the same. Thus if $y=1.32times 10^{-2}$ (three significant figures) then $e^y=1.1034$ (five significant figures)".

What is the rationale behind the number of significant figures in the above example ?

Following does not address the query: Number of significant figures

If $y=1.32times 10^{-2}$, then this means that exact value of $y$ can be anywhere in the interval $[1.315times 10^{-2}, 1.325times 10^{-2}].$ If you now exponentiate the endpoints, then, since the exponential function is monotonic, you'll get the interval where $exp(y)$ can be: $[1.01323684148769,1.01333817023819].$ Now what's the difference between these endpoints? It's about $0.0001$. This is the absolute error of $exp(y)$. You can now easily see that we know this exponent to (a bit less than) 5 significant figures.

Correct answer by Ruslan on January 5, 2021

Remember that one representation for the exponential function is

$$ e^y = 1 + y + frac{y^2}{2!} + frac{y^3}{3!} + cdots $$

Suppose I know $y$ to only one significant figure, $y = 2times10^{-12}$. I immediately know $e^y$ to *twelve* significant figures:

begin{align} exp({2times10^{-12}}) &= 1 + 2times10^{-12} + frac{2^2}{2!}times10^{-24} + cdots & approx 1.000,000,000,002 end{align}

Answered by rob on January 5, 2021

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