# Number of significant figures on exponentiation

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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