[Math] How to get the amount of digits after the decimal point

arithmeticdecimal-expansion

How can I easily find the amount of digits after the decimal point?
For example:

0.2 would result 1 (1 digit after the point)
0.09 would result 2 (2 digits after the point)
0.641 would result 3 (3 digits after the point)

I think this is fairly simple, but I just can't get my head around it right now…

Thanks!

Best Answer

The answer depends on just how you store your number.

If you are storing it as a rational number or as a fraction $\frac ab$ in lowest terms, or if you have a routine to convert your number to this format, here is how you can do it.

Find the prime factorization of $b$. Let's say it is

$$b=2^r5^sp^t\ldots$$

If $b$ has a prime factor other than $2$ or $5$, the number of decimal places is infinite. Otherwise, the number of decimal places in $\frac ab$ is

$$\max(r,s)$$

Do you need an explanation of why this works?

If your number $x$ is a float or a real number, find the smallest $t$ such that

$$10^t\cdot x=\lfloor 10^t\cdot x \rfloor$$

(The brackets are the greatest integer function.) Then $t$ is the number of decimal places. If no such $t$ exists, the number of decimal places is infinite.

Related Solutions

Algorithms – How to Calculate the Number of Decimal Digits for a Binary Number?

I would post this as a comment if I could, because it seems too simple for an answer. You already said that you can find the highest bit with a single instruction (if the number fits into a register). For every result $r$ you get (that is, your number is between $2^r$ and $2^{r+1}-1$), there are only two possible numbers $k$ and $k+1$ of decimal digits. So you could just use $r$ as an index into a lookup table that stores both $k$ and $10^k$, and output $k$ if your number is less than $10^k$, or $k+1$ otherwise.

[Math] How to evalulate a fraction (example 1/998001) accurately to 100 decimal places and then display it in maxima/wxmaxima?

$$ \frac{1}{(1000-1)^2} =\frac{10^{-6}}{(1-\frac{1}{1000})^2} =\frac{10^{-6}}{(1-x)^2} =10^{-6}\left(\sum_{n=0}^{\infty}x^n\right)^2 =10^{-6}\sum_{n=1}^{\infty}nx^{n-1} $$ for $x=.001$, which has no carries in its decimal expansion for $n<1000$. Thus the first $1000$ triplets ($3000$ digits after the decimal place) will be $0.000\;001\;002\;003\;\dots\;996\;997\;999\;000$, which brings us up to the first digit which receives a carry. If you want to see it without the scientific notation, add one before formatting. In sage:

(1+(10^3-1)^(-2)).n(digits=3001) # ignore the leading one in the output!

or in scientific notation:

((10^3-1)^(-2)).n(digits=3001) # note the e-6 at the end, meaning times 10^{-6}

As to the formula above, it can be derived by differentiation: $$ \left(\sum_{n=0}^{\infty}x^n\right)^2 =(1-x)^{-2} =\frac{d}{dx}(1-x)^{-1} =\frac{d}{dx}\left(\sum_{n=0}^{\infty}x^n\right) =\sum_{n=1}^{\infty}nx^{n-1} \qquad\text{for} \qquad|x|<1 $$

Best Answer

Related Solutions

Algorithms – How to Calculate the Number of Decimal Digits for a Binary Number?

[Math] How to evalulate a fraction (example 1/998001) accurately to 100 decimal places and then display it in maxima/wxmaxima?

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