Calculating the Number of Digits in x

[soandos]

is there a way to determine (does not need to be exact) the number of digits in x! ?

sorry if this is kind of pointless

 

[mathman]

For large n, use Stirling's formula:

n! ~ (2[pi]n)^1/2 (n/e)ⁿ

 

[CRGreathouse]

is there a way to determine (does not need to be exact) the number of digits in x! ?

O(x log x).

To be more precise, x log x - x log e + O(log x), where the base of the logarithms is the base you want to express the number.

 

[Gib Z]

Don't forget to then apply

$$D = 1 + \lfloor \log_{10} \lfloor n! \rfloor \rfloor$$

after that for the number of digits.

 

[disregardthat]

Don't forget to then apply

$$D = 1 + \lfloor \log_{10} \lfloor n! \rfloor \rfloor$$

after that for the number of digits.

If $$n!$$ is hard to calculate, the series

$$D = 1 + \lfloor \sum^n_{i=1} \log_{10}i \rfloor$$

should be easier to calculate - for integer $$n$$.

 

[Gib Z]

If $$n!$$ is hard to calculate, the series

$$D = 1 + \lfloor \sum^n_{i=1} \log_{10}i \rfloor$$

should be easier to calculate - for integer $$n$$.

Depends how large n is. If it's quite large, calculating logs for each term up to it may not be easier than using Stirlings Formula which simplifies into something that doesn't look that bad after a log anyway. But that definitely is a good idea =]

 

[soandos]

Anyone know of a way to get an answer just to a certain precision in mathematica?
I do no mean truncating the answer, but more the way a person would, so for example,
1/"insert incredibly long hard to calculate mess here" as zero, or instead of getting the inside of a natural log to hundreds of decimal points, just get it to 10 or so.
is there a way to do that?