Logarithms of fractions

The lazy integer application 

Let's apply formula

log(w)  =  2 * Sum( ((w-1)/(w+1))^(2*n-1) / (2*n-1)  : n = 1 2 3 ...)

for every real w > 0; see
https://whmth.blogspot.com/2024/05/the-logarithmic-function.html

EXAMPLES

  • For  w := 2  we obtain

            log(2)  =  2 * (1/3  +  1/(3*3^3)  +  1/(5*3^5)  +  1/(7*3^7)  + ... )

  • For  w:=3  we obtain

            log(3)  =  2 *     (1/2  +  1/(3*2^3)  +  1/(5*2^5)  +  1/(7*2^7)  +  ... )

  • For  w:=4  it's more efficient to apply
    log(4) = 2*log(2)

    The same goes for all composite integers:
    log(6) = log(2)+log(3)        log(8) = 3*log(2)        log(9) = 2*log(3)
    etc.

  • Thus, let's concentrate on primes. We obtain

    log(5)  =  2 * (2/3  +  (2/3)^3 / 3  +  (2/3)^5 / 5  +  (2/3)^7 / 7  +  ...)
    log(7)  =  2 * (3/4  +  (3/4)^3 / 3  +  (3/4)^5 / 5  +  (3/4)^7 / 7  +  ...)
    log(11)  =  2*(5/6  +  (5/6)^3 / 3  +  (5/6)^5 / 5  +  (5/6)^7 / 7  +  ...)
    etc.

Fractions, near 1 fractions, and near 1 odd fractions

Let  a b  be two positive real numbers. Then for  w:=a/b  we obtain

log(a/b)  =  2 * Sum( ((a/b-1)/(a/b+1))^(2*n-1) / (2*n-1)  : n = 1 2 3 ...)

hence
log(a/b)  =  2 * Sum( ((a-b)/(a+b))^(2*n-1) / (2*n-1)  : n = 1 2 3 ...)

Two special cases seem to be the simplest, the near 1 fractions  (t+1)/t,  and the near 1 odd fractions (2*t+1)/(2*t-1)  where  t  is an arbitrary natural number (1 2 3 ...). For these cases we obtain:

log((t+1)/t)  =  2 * Sum( (1/(2*t+1)^(2*n-1) / (2*n-1)  : n = 1 2 3 ...)

and
log((2*t+1)/(2*t-1))  =  2 * Sum( (1/(2*t)^(2*n-1) / (2*n-1)  : n = 1 2 3 ...)

Looking at the right hand side of the above two equations, the two cases are not dramatically different. The first series converges a bit faster always, and only a little bit faster when t is large.

EXAMPLES

  • Since  3 = 3/1 (a near 1 odd fraction), we let  t:=1 so that  3=(2*t+1)/(2*t-1)=(2*1+1)/(2*1-1), and we again obtain

            log(3)  =  2 * (1/2  +  1/(3 * 2^3)  +  1/(5 * 2^5)  +  1/(7 * 2^7)  + ... )

    just as in the above lazy application.

  • Since  2 = 2/1 (a near 1 fraction),  we let  t:=1 so that  2=(t+1)/t=(1+1)/1, and we again obtain

            log(2)  =  2 * (1/3  +  1/(3 * 3^3)  +  1/(5 * 3^5)  +  1/(7 * 3^7)  + ... )

    just as in the above lazy application.
  • Consider 5/3 (a near 1 odd fraction) so let  t:=2. Thus,

    log(5/3)  =  2 * (1/4  +  1/(3 * 4^3)  +  1/(5 * 4^5)  +  1/(7 * 4^7)  + ... )

  • Consider 3/2 (a near 1 odd fraction) so let  t:=2. Thus,

    log(3/2)  =  2 * (1/5  +  1/(3 * 5^3)  +  1/(5 * 5^5)  +  1/(7 * 5^7)  + ... )

  • Consider 7/5 (a near 1 odd fraction), so let  t:=3. Thus,

    log(7/5)  =  2 * (1/6  +  1/(3 * 6^3)  +  1/(5 * 6^5)  +  1/(7 * 6^7)  + ... )

  • Consider 4/3 (a near fraction), so let  t:=3. Thus

    log(4/3)  =  2 * (1/7  +  1/(3 * 7^3)  +  1/(5 * 7^5)  +  1/(7 * 7^7)  + ... )

  • Consider 9/7 (a near 1 odd fraction), so let  t:=4. Thus,

    log(9/7)  =  2 * (1/8  +  1/(3 * 8^3)  +  1/(5 * 8^5)  +  1/(7 * 8^7)  + ... )


  • Consider 5/4 (a near fraction), so let  t:=4. Thus

    log(5/4)  =  2 * (1/9  +  1/(3 * 9^3)  +  1/(5 * 9^5)  +  1/(7 * 9^7)  + ... )

  • etc.



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