[Haskell] for large x, log (x::Integer) :: Double

[Howard Oakley]

On 30/6/04 14:47, Hal Daume III wrote:

> i'm looking for an accurate way to take the log of a very large integer,
> for example:
> 
>   let x :: Integer =
> 130142727215188116061276556022688196621810140343691778718485630367238262325689
> 845541676397895906730024965277394371574303273329260262483498476173923323279461
> 919361195473572028476105814689924661111323670085360089179896892077534449168518
> 590692259602654391532136757777452291231593014452347270238624064599385936823085
> 594101937144705866411597403257188107243160465138552039367484067881179355426659
> 501377394743411557958891296796968015047325823672783086783214986710043714270547
> 671666903964025267795520158937805183611280026836733145529671590438773283635061
> 353921824995082955541839719790928834530340719498354530821282866299962327922291
> 308021419628714011758281176918848669320822757025713685194594340820628167255555
> 828946025686701689606333414064007570808358186629749461083454555486484630638301
> 4549439540479675828018496049574066533167553894586573246931377586176000
> 
> this # is approximately 10^850.

Hal,

Based on the old manual way of doing this, I would first obtain the log10
value. This is easiest if you can convert the number to a string:
- if the original number is an integer, the string length gives you the
exponent, which in log10 forms the integer part of the result
- then take the first few digits of the string, concat '0.' in front,
convert to a double, and obtain the log10 of that, which forms the decimal
part of the result.

And given the log10, if you want loge or ln then all you have to do is
multiply by loge(10), if I remember correctly.

Howard.

Dr Howard Oakley
The Works columnist for MacUser magazine (UK)
http://www.macuser.co.uk/