(1) f(x) = ln(x) implies f(1)=0f(x)=ln(x)⇒f(1)=0
f^'(x) = 1/x implies f^'(1) = 1
f''(x) = -1/x^2 implies f''(1) = -1
f'''(x) = ((-1)(-2))/x^3 implies f'''(1) = 2
f^((4))(x) = ((-1)(-2)(-3))/x^4 implies f^((4))(1) = -6
The general pattern should be clear from this:
f^((n))(x) = ((-1)(-2)...(-n+1))/x^n implies
f^((n))(1) =(-1)^(n-1)(n-1)!
(2) So, the Taylor series for ln(x) centered at x=1 is given by
ln (x) = sum_{n=0}^oo f^((n))(1)/(n!) (x-1)^n
qquadqquad = sum_{n=0}^oo (-1)^(n-1)((n-1)!)/(n!) (x-1)^n
qquad qquad = sum_{n=0}^oo (-1)^{n-1} (x-1)^n/n
(3) The n-th term in the Taylor series above is given by
t_n = (-1)^(n-1)(x-1)^n/n
and thus
t_{n+1}/t_n = -n/(n+1)(x-1)
For the series to converge, we must have
lim_{n to oo}|t_{n+1}/t_n|<1 implies
lim_{n to oo}|-n/(n+1)(x-1)|<1 implies
color(red)(|x-1|<1)
which is the domain of convergence.
(4)
ln(4/3) = sum_{n=0}^oo (-1)^{n-1} (4/3-1)^n/n
qquad qquad = sum_{n=0}^oo (-1)^{n-1} 1/(n3^n)
ln(2/3) = sum_{n=0}^oo (-1)^{n-1} (2/3-1)^n/n
qquad qquad = sum_{n=0}^oo (-1)^{n-1} (-1)^n/(n3^n)
Hence
ln (2) = ln(4/3)-ln(2/3)
qquad = sum_{n=0}^oo (-1)^{n-1} (1-(-1)^n)/(n3^n)
In this sum, all the even terms cancel, so that we get
ln (2) = sum_{n=0}^oo (-1)^{2n+1-1} 2/((2n+1)3^(2n+1))
and hence
ln(sqrt2) = 1/2 ln(2) = sum_{n=0}^oo 1/((2n+1)3^(2n+1))
