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• *********************************************************************************** what legendre probably did is this : he knew that Pi(x) is approximately
Message 1 of 2 , Feb 17, 2005
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what legendre probably did is this :
he knew that Pi(x) is approximately " x/ln(x)"
so :
d(Pi(x)) 1 1
---------- = ------- - ------------
d(x) ln(x) ln(x)^2

know if we integrate again --->>> Pi(x)= Li(x) - int(1/ln(x)^2)

he found that the second part is not valuable.so Li(x) was discoverd.

********************************************************************************

Jean-Pierre Pratali <jpierre.pratali@...> wrote:

----- Original Message -----
From: "Jud McCranie" <j.mccranie@...>
To: "Jean-Pierre Pratali" <jpierre.pratali@...>
Sent: Sunday, February 13, 2005 3:31 AM
function Li(x) ?

> I use formula (6) and (7) at
> http://mathworld.wolfram.com/LogarithmicIntegral.html
>
> Note: according to that reference, li(x) is the logarithmic integral,
Li(x)
> is the
> Polylogarithm.
>
> We would like to see it - that could be important.
______________________________________________________________
Thanks. I now understand that formula (6) (Nielsen-Ramanujan) of MathWorld
is the one used to compute li(x) (a short time ago it was written Li(x)).

I'll try to set out briefly how I established my own formula:
1. Taking li(x), I had the idea to integrate "by parts" :
li(x)=x/logx+(Integral of 1/logx^2), and then to reiterate that at each time
on the remaining integral :
2. After n times : li(x)=x/logx(Sum of k!/logx^k, k from 0 to n-1)
+n!(Integral of 1/logx^(n+1))
3. I saw that, first that I could not get rid of an integral, and secondly
that the series k!/logx^k is (very) divergent!
4. So I had the thought to truncate it at its least term : using the
logarithmic derivate of k!/logx^k and the simplified Stirling formula
k(logk-1) for logk!, it's easy to show that this derivate is nil for k=logx.
5. Finally I built the sum x/logx(Sum of k!/logx^k, k from 0 to |logx|),
|logx| meaning "integer part" of logx (preferably "floor").
That's all.
Using Mathematica :
n = ...;
L := N[LogIntegral[n], 100]; N1 := Floor[L];
S := 0;
p = N[n/Log[n], 100];
For [k = 0, k < Floor[Log[n]], k++, S = S + N[k!/Log[n]^k, 100]];
S = p*S; N2 := Floor[S];
Print[N1, " ", N2, " ", N2 - N1]

it's easy to note that there is a nearly perfect sticking between N1 and N2.
The absolute value of the difference is never greater than 1 (and probably
for the little numbers only).

Thanks.

Jean-Pierre Pratali

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