Is there a way to quickly estimate the reciprocal of a number?

Question

Is there a way to estimate the reciprocal of a decimal number? e.g. 1/0.388. Are there any formulas for this sort or thing, or do I just have to calculate it by hand?

Answer 1

$ 1/0.388 = 1/(388/1000)=1000/388=250/97$. Now do it by hand, as precise as you need.

Answer 2

It depends what facts you have in your head. Probably you know that 1/2=0.5, 1/3=0.3333, 1/4=0.25, 1/5=0.2. It is less common to know 1/6=0.16666, 1/7=0.142857, 1/8=0.125, 1/9=0.11111. Leading zeros just make a factor of 10. You can just "look down the list" to see 1/.388 is between 2 and 3. Then 0.388 is about 15% bigger than 0.333, so 1/0.388 is about 15% less than 3, or about 2.55. Here we are using that $(1+x)^{-1}\approx 1-x$ for $x$ small compared with 1. This is only 1% off

Answer 3

Older books are often a good resource for such things, e.g. quoting (with slight editing) from the 1910 book of Edward M. Langley, A treatise on computation: an account of the chief methods for contracting and abbreviating arithmetical calculations, p.71:

From the algebraical result, $$\frac1{x-a}=\frac1x + \frac a{x^2} + \frac{a^2}{x^3}+\frac{a^3}{x^4}+\cdots$$ many reciprocals can be easily written down to a great many decimal places. Thus, since 98 = 100 - 2, $$\frac1{98}\approx0.010204081632.$$

This is a geometric series, an it can be got directly from one's preferred standard form of the geometric series. Langley's reciprocal expansion is easily remembered from the dimensional homogeneity of its terms. It can be interpreted as a base approximation, the first term, corrected by a series of ever-smaller subsequent terms. Convergence of the series requires only that $|a|<|x|$.

To apply the expansion to the given problem, we may take $\frac1{0.4}=2.5$ as the base approximation, and truncate after only one correction term. This gives:

$$\begin{eqnarray}\frac1{0.388}&=&\frac1{0.4-0.012}=\frac1{0.4}+\frac{0.012}{0.4^2}+\frac{0.012^2}{0.4^3}+\cdots,\\ &\approx&\frac1{0.4}+\frac{0.012}{0.4^2},\\ &=&2.5+0.075=2.575.\end{eqnarray}$$

This already compares well $\frac1{0.388}=2.57731958\ldots$, and indeed, if 0.388 is a measured value reported to three significant digits, than 2.58 is already the reciprocal given to the same number of significant digits.

Of course if more exactness is warranted, more terms of the expansion can be taken.

Answer 4

I very quickly estimated $\frac 1{0.388}$ as follows:

$\frac 1{0.388} \approx \frac 1{0.3 + 0.\overline 8} = \frac 1{\frac3{10} + \frac 8{90}} = \frac 1{\frac{35}{90}}= \frac {18}7$, which has only $-0.23 \%$ error or so.

While this is an answer to your specific question, a more general answer would simply be: become comfortable enough with spotting patterns in numbers that you can very quickly exploit familiar ones.

Answer 5

To me, by far the most natural way to estimate $\frac{1}{0.388} $ is: $\frac{1}{0.388} \approx \frac{1}{0.4} = \frac{1}{\frac{2}{5}} = \frac{5}{2} = 2.5.$