Michael K. answered • 04/22/19

PhD professional for Math, Physics, and Computer Tutoring

Using Taylor's Theorem, we can express the natural logarithm over a specific domain of applicability.

(z-3)(z-1) = z^{2} -4z + 3

ln(z^{2} -4z + 3) = f(z)

f(z) = sum_[k=0]_[k=infinity] a_{k}(z-z_{0})^{k}

We define x_{0} as the center point and a_{k} as the Taylor coefficient at the kth derivative.

a_{k} = f^{k}(z_{0})/k!

d/dz [ ln(z^{2} -4z + 3) ] = (2z - 4)/(z^{2} -4z + 3)

d^{2}/dz^{2 }[ ln(z^{2} -4z + 3) ] = d/dz [ (2z - 4)/(z^{2} -4z + 3) ] = -2(z^{2} -4z + 5)/(z^{2} -4z + 3)^{2}

d^{3}/dz^{3 }[ ln(z^{2} -4z + 3) ] = d^{2}/dz^{2 }[ (2z - 4)/(z^{2} -4z + 3) ] = d/dx [ -2(z^{2} -4z + 5)/(z^{2} -4z + 3)^{2} ] --> 4(z-2)*(z^{2} -4z + 7)/(z^{2}-4z+3)^{3}

d^{4}/dz^{4 }[ ln(z^{2} -4z + 3) ] = ...

Evaluating all derivatives at z_{0} = 0...

a_{0} = f^{0}(0)/0! = f(0) = ln(3)

a_{1} = f^{1}(0)/1! = f'(0) = -4/3

a_{2} = f^{2}(0)/2! = f''(0)/2 = -10/18

a_{3} = f^{3}(0)/3! = f'''(0)/6 = -56/162

...

So the Taylor series looks like --> ln(z-3)(z-1) ≈ ln(3) - (4/3)x - (10/18)x^{2} - (56/162)x^{3} + ...

We want to compare the nth term to the (n+1)th term and look at the limit as n goes to infinity

We see that the the coefficients continue to decrease and if we wrote a pattern expression for the nth derivative we would find that the radius of convergence approaches infinity (nth term / (n+1)th term) which means it would convergence for all z about the center of z_{0} = 0.

The domain of the function are this values of z allowed. Since the ln function cannot be negative we are interested in any regions where (z-3)(z-1) is less than or equal to zero...

We see that z ≥ 1 and z ≤ 3 would be in the region where the parabola is negative and not-allowed...

Domain f(x) --> (-∞,1) U (3,∞)

Range of f(x) --> all reals ( y ∈ ℜ )