# Significant figures for special functions (square roots)

• B
• yucheng
In summary, the rule of thumb for significant figures is not applicable in physics. Instead, you should report the uncertainty explicitly and use the propagation of errors formula to calculate the error after taking the square root. This will result in a more accurate and precise measurement.

#### yucheng

TL;DR Summary
I am trying to decide which scheme to follow....
I am using square roots, however, I am confused over how many significant figures (s.f.) to keep.
Suppose I have ##\sqrt{3.0}##, which has 2 s.f.

From three different sources, I'll put a summary in brackets:
(if 2 s.f. in the data, keep 3 s.f. in the result; this also appeared in the answers for Kleppner's Introduction to Mechanics)
http://cda.morris.umn.edu/~mcintogc/classes/modern/sigfig.htm
(if 2 s.f., keep 2s.f.)
(if 2 s.f., keep 2s.f.)

So, which is correct (not for high school, but for Physics in general)? Thanks in advance.

For physics in general none of the approaches is correct. Significant figures is a rule of thumb for students.

In a scientific paper you would report your uncertainty explicitly. So if you measured something to be 3.0 with a standard uncertainty of 0.2 then you would report it as ##3.0 \pm 0.2## or more concisely ##3.0(2)##.

You would use the propagation of errors formula to report the error after the square root.

berkeman, sophiecentaur, yucheng and 1 other person
yucheng said:
I am using square roots, however, I am confused over how many significant figures (s.f.) to keep.
Suppose I have ##\sqrt{3.0}##, which has 2 s.f.
If you use the propagation of errors formula, you will find that taking the square root cuts the relative error in half. This buys you 3 dB -- 30% of one significant digit. [If you know x to plus or minus 1 percent, you know ##\sqrt{x}## to plus or minus half a percent].

Conversely, squaring a number worsens the relative error by a factor of two. This costs you 30% of a significant digit. [If you know know x to plus or minus 1 percent, you only know ##x^2## to plus or minus 2 percent].

yucheng, sophiecentaur and berkeman