Re: finding ln(x)
- From: "alainverghote@xxxxxxxx" <alainverghote@xxxxxxxx>
- Date: 19 Feb 2007 01:56:39 -0800
On 19 fév, 08:27, "MQ" <michaelquinli...@xxxxxxxxx> wrote:
Hi all***************************
I'm just wondering how a calculator finds logarithms. My idea is the
following:
If x > e, then find x/e, which is the integer portion of the result
Take the remainder and put this into the infinite series definition
for ln (ie. x - x^2/2 + x^3/3 - ....)
For logarithms of any other base b, it would just calculate ln(x)/
ln(b). Am I right or is there some more efficient algorithm?
Bonjour,
Pour les calculettes il existe des procédés genre CORDIC
rapidement convergent:suppose connus:
Ln(1,1) ; Ln(1,01) ; Ln(1,001) ...
Soit x ton nombre
x dans l'intervalle { (1,1)^n ; (1,1)^(n+1) }
alors on considère y = (x/(1,1)^n , placé par essais
dans l'intervalle {(1,01)^p ;(1,01)^(p+1) }
etc
Amicalement, Alain
.
- References:
- finding ln(x)
- From: MQ
- finding ln(x)
- Prev by Date: Re: "Regression" average VS the "Arithematic" Average
- Next by Date: Re: finding ln(x)
- Previous by thread: finding ln(x)
- Next by thread: Re: finding ln(x)
- Index(es):
Relevant Pages
|
