An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

~ 47. 48.

Logarithmic series.

83

+ 3- X R (~+ X3 X5 x2k -1
~ - 48 Lxix + s + +* 2sk+ 1<
Now substitute:
1+- Z +-a and so x a a 0,
X1- 2 z + a
then: 0 < Z < oo
(+ a) (z) + 2( a ( )3+ ( a )5+ )
1(~+ a~=I(~s)+2( 2x + + a 3 ~Z+ )a - 2z +....
For instance, = 1, a =- 1:
1(2)= 2 (1 + ()3 + - (1) + - (~7 +...);
- 2, a = 1:
1(3) -= (2) + 2 ( ) + I () + (i)5 + *.)
To pass from natural logarithms, with the base e, to common
logarithms with the base 10, since '0loga == elog a elog 10 we have
to calculate the number:
1(10) = 1(2) + (5) = 2.3025850929...,
then we must multiply all values by -() =0.4342944819...
48. Circular series. y= tan-lx.
An independent expression of the nth derivate of the circular
function f(x) = tan-lx was given in ~ 36:
fn(x) =- In-i cosny sin n (y + tr) L —. sin n(tan- x + r).
(1 + 2)
Now since for x = 0, y is also = 0, it follows that for this value:
f"(0) 0, fV (0) = 0, (0) = 0 f( k. (0) == (
f' (0) 1, f"'(0) — 12, f (0)- =. f 2k+l (0) - (- 1)k 2k.
The remainder B is by the first formula:
n
_-l' n siln (tan-'x 0 r+ Z) ( )sino(tanx x+2
L(1I +022)2
This first factor converges to zero, the third has a finite value. The
middle factor does not become infinite for n = oc when the quotient
within brackets is equal to or less than 1 for all values of 0, i. e. when
x2 < 1. We have thus:
x3 x 5 2 k+1i
tan- 1 +X x(- 1 < X < +1
1-' + i -'' -l 1.=The value of x being. any proper fraction, this series presents the
corresponding angle between - tC and + -rt.
The angle whose tangent has the value + 1 is equal to tzc; this
6*