The logarithmic function

The shifted logarithm, and the logarithmic function

 Let  s t  be two arbitrary positive real numbers. The shifted logarithm from  s  to  t, symbolically LOG(s t),  is the (oriented) area of a curvilinear  trapezoid in plane xy that is bounded by the graph of function  y=1/x, and by the x-axis, and which stretches between parallel ("vertical") lines  x=s and x=t -- "oriented" means that this area is the customary geometric area when  s < t, and it is the negative of that geometric are when  s > t. Of course,  LOG(s t) = 0  when s=t. Thus,

LOG(t s)   =   - LOG(s t)

Now we define the logarithmic function

log(t) := LOG(1 t)

for arbitrary positive real number  t.  In particular:

log(1) = 0

Obviously,

LOG(q  s) + LOG(s  t)  =  LOG(q  t)

for arbitrary positive real numbers  q s t.

Logarithm as a translation from multiplication to addition

Let lengths of the sides of a rectangle be  w  and h. where  w h  be two arbitrary positive real numbers. Next, lets squeeze/stretch side  v  by a factor of  u>0, and let's stretch/squeeze side  h  by 1/u. Then the modified rectangle will have area = (v*u)*(h/u) = v*h  which is equal to the area of the original rectangle.

Something similar holds for the trapezoids related to logarithm:

LOG(u*s   u*t)  =  LOG(s  t)

-- the original trapezoid spanned between  x=s  and  x=t  got stretched/squeezed by a factor of  u  along the x-axis, and it got squeezed/stretched by a factor of  1/u in the y-axis direction.

Now,

log(s*t)  =  LOG(1  s*t)  =  LOG(1  s) + (LOG(s  s*t)

=  LOG(1  s) + LOG(s*1  s*t)  =  LOG(1  s) + LOG(1  t)

=  log(s) + log(t)

i.e.

log(s*t)  =  log(s) + log(t)

for arbitrary positive real numbers  s t.

In particular,  log(s) + log(1/s) = log(s * 1/s) = log(1) = 0  hence

log(1/s) = -log(s)            for every real  s>0

and
log(t/s)  =  log(t) - log(s)
for arbitrary positive real  s t.


The classical logarithmic Taylor series, and more.

Watching how the trapezoid area of  log(t)  changes when  t varies, and applying the definition of log(t), we see that the derivative of log is simply

log'(t) = 1/t

hence

log'(1+u) = 1/(1+u)            for  u > -1

But

1/(1+u)   =   1 - u + u^2 - u^3 + ... 

               =   Sum( (-u)^n  :  n=0 1 2 ...)            for  |u| < 1

hence

log(1+u)  =  u - u^2/2 + u^3/3 - u^4/4 + ...                                                  

      =  Sum( (-1)^(n-1)*u^n / n  :  n-1 2 3 ...)            for |u| < 1

and this is the classical logarithmic Taylor series. Now, lets substitute  -u  for  u -- then:

 log(1-u)  =  - Sum(u^n/n : n=0 1 2 ...)            for |u| < 1

Thus,

    log((1+u)/(1-u))   =   log(1+u) -  log(1-u)  =  2 * Sum(u^(2*n-1)/(2*n-1))            for |u| < 1

i.e.

 log((1+u)/(1-u))   =   2 * Sum(u^(2*n-1)/(2*n-1))            for |u| < 1

Let's substitute  w = (1+u)/(1-u)  so that  u = (w-1)/(w+1).  Then, when w > 0  then  |u|<1  hence

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

for every w > 0.


REMARK  The above formula describes a Riemann analytical extension of logarithm over all complex numbers  w  such that Re(w) > 0, where. Re(w) is the real component of complex  w.


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