[Math] Correct to four decimal places

calculussequences-and-series

Suppose we have an alternating series $s:=\sum (-1)^{n+1} a_n$ where $a_n$ is positive and decreasing to zero.

If I want to find an approximating partial sum to be correct to four decimal places, I think what I am looking for is
$$|s – s_n|\leq a_{n+1} < 0.0001.$$
However the solution is choosing the term $a_{n+1}$ to be less than $0.00005$, which is like saying
$$-0.00005 < |s-s_n| < 0.00005.$$

So when we say correct to four decimal places, what do we really mean?

Best Answer

The first issue to settle is how to make a finite decimal approximation of a known real number. Ask yourself: what's the best 4 decimal approximation to $0.123401$? By your rule, $0.1235$ would work. But $0.1234$ is much better, since $|0.123401 - 0.1234| = .000001$ is $99$ times smaller than $|0.123401 - 0.1234| = .000099$.

The problem is that requiring the difference only to be $<0.0001$ allows two different choices of four decimal place approximations, whereas the stronger requirement that the difference be $<0.00005$ allows a unique four decimal place approximation. The reason $0.00005$ is used is that it is exactly half the distance between any two consecutive four decimal place numbers such as $0.1234$ and $0.1235$.

Now let me turn to your actual question. You have an unknown real number $x$, and you have a method for finding arbitrarily small open intervals $a < x < b$ (using alternating series, for example).

Your question amounts to this: How small should one take $|b-a|$ in order to obtain a well-defined 4 decimal place approximation of $x$?

The quick answer is: it may not be possible. For instance, perhaps you obtain the interval $a = .123449 < x < .123451 = b$. If you round $a$ to the nearest four decimals you get $.1234$, whereas if you round $b$ to the nearest 4 decimals you instead get $.1235$. There's no way to choose between these two approximations, because all you know about $x$ is that it is between $a$ and $b$: you have no further information about whether $x$ is closer to $a$ than it is to $b$.

A longer answer is this: there are two sources of error in the approximation you are trying to make. The first source of error is the mathematical approximation $a<x<b$; you don't know anything else about $x$ other than that it is between $a$ and $b$. The second source of error is rounding to the nearest 4 decimal place number. Combining errors in this fashion does not always produce a well-defined approximation, as my example shows.