(i) ln(x) is undefined for x=0, but is defined ln(1+X), where X=0, so if we work out the series for X we can replace X with x-1, because x=1+X. Let y=a0+a1X+a2X2+a3X3+... represented by ∑anXn.
When X=0, y=ln(1)=0, so a0=0.
y(1)=1/(1+X), y(2)=-1/(1+X)2, y(3)=2/(1+X)3, y(4)=-6/(1+X)4, ..., y(n)=(-1)n+1n!/(1+X)n+...
Also, y(1)(0)=1=a1, so a1=1; y(2)(0)=-1=2a2, so a2=-½; y(3)(0)=2=6a3, so a3=⅓; y(4)(0)=-6=4!a4, so a4=-¼; etc. Therefore, an=(-1)n+1/n for n>0. The series is therefore:
ln(X)=X-X2/2+X3/3-X4/4+... ln(1+x)=x-x2/2+x3/3-x4/4+...
ln(x)=x-1-(x-1)2/2+(x-1)3/3-(x-1)4/4+... When x=0 ln(0)=-1-1/2-1/3-1/4-... which must be (and is) divergent.
(ii) Let y=cos(x), then:
y(1)=-sin(x); y(2)=-cos(x); y(3)=sin(x); y(4)=cos(x)=y; y(5)=-sin(x)=y(1), ...
In particular, y(0)(0)=y(0)=1, y(1)(0)=0, y(2)(0)=-1, y(3)(0)=0. The pattern (1,0-1,0) repeats indefinitely.
If y=∑anxn as a series, then a0=1 because y(0)=1=a0.
All odd powers of x have a zero coefficient, so a2k+1=0, where k is an integer≥0.
a4k=1/(4k)! and a4k+2=-1/(4k+2)! making the series 1-x2/2!+x4/4!-x6/6!+...