I'm trying to find solution of two exercises from Demidowicz book ,,Problems in mathematical analysis''.
I can find there definition when number has $n$ significant digits.
,,Number has $n$ significant digits, when absolute error of this number equals no more than half of order of magnitude expressed by $n$-th significant digit'' – That's my translation, I can't find English version of similar definition. I don't know what does it mean ,,half of order of magnitude''. In Polish also. This definition is not clear for me.
Ex. 32: Figure out how many significant digits has number
$$x=2.3752$$
if relative error of this number equals $1\%$.
My try: Relative error $\delta=\frac{\Delta}{|x|}$, where $\Delta$– absolute error
$$\frac{\Delta}{x}=\frac{1}{100}\Rightarrow\Delta=\frac{x}{100}\Rightarrow \Delta=0.023752$$
Second place after dot is less than $5$, so $x$ has three significant digits $2.37$. But answer in book is equals two digits. What is wrong with my solution?
Ex. 33: Number
$$x=12.125$$
has three significant digits. Find relative error of this number.
My try: If number has three significant numbers, then last significant digit is $1$ after dot. So the bigger possible value of absolute error equals $0.4999…$. From the other side the value of absolute error has to be bigger than $0.099…$. So relative error:
$$\delta_1\approx\frac{0.4999}{12.125}\approx 0.04123\approx 4\%$$
$$\delta_2\approx\frac{0.0999}{12.125}\approx 0.00823\approx 0.8\%$$
Answer in book: $\delta\le 0.41\%$. So i'm again wrong.
I think I don't understand a way to calculate relative error or how to relate single value with error/significant digits. Can you guide me how to approach to similar exercises? What about definition when number has $n$ significant digits?
Best Answer
Significant digits essentially means the digits you know are correct after rounding. To take the second exercise first, $x=12.125$ with three significant digits really means that that $x$ is known to round to the three digit number $12.1$, so can be any number in the range $12.05\le x < 12.15$. This is an absolute error of at most $0.05$ to either side of $12.1$, which is a relative error of $\frac{0.05}{12.1}\times 100\% = 0.41\%$.
Note that the absolute error does not depend on what the digits of $x$ actually are, only on where the last significant digit is relative to the decimal point. The absolute error is half the value of what a $1$ in that last digit position is worth, i.e. half the value of the power of ten (order of magnitude) that corresponds to that last digit position.
Also, leading zeroes do not count as significant figures, so for example $0.03210$ has $4$ significant figures.
For the first exercise, $x=2.3752$, and you calculated from the relative error of $1\%$ that the absolute error is $\Delta=0.023752$. That means that the actual value could be anything in the range $2.3752-0.023752$ to $2.3752+0.023752$, i.e. $2.351448\le x<2.398952$. If we round this to 3 figures (2 decimals), we get different last digits ($2.35$ versus $2.40$), so we cannot say it has three significant figures. If we round this to 2 figures (1 decimal), we get $2.4$ in either case, so it does have two significant figures.
This last example shows how crude using significant figures can be compared to using a specified absolute/relative error, since you are making it less accurate ($2.4$ with absolute error $\pm0.05$), but it is less complicated to communicate.