What is Taylor series of ln(z-3)(z-1)?

Question

Around 0? And what is the radius of convergence? (And the domain of the function?)

Answer 1

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

(z-3)(z-1) = z² -4z + 3

ln(z² -4z + 3) = f(z)

f(z) = sum_[k=0]_[k=infinity] a_k(z-z₀)^k

We define x₀ as the center point and a_k as the Taylor coefficient at the kth derivative.

a_k = f^k(z₀)/k!

d/dz [ ln(z² -4z + 3) ] = (2z - 4)/(z² -4z + 3)

d²/dz² [ ln(z² -4z + 3) ] = d/dz [ (2z - 4)/(z² -4z + 3) ] = -2(z² -4z + 5)/(z² -4z + 3)²

d³/dz³ [ ln(z² -4z + 3) ] = d²/dz² [ (2z - 4)/(z² -4z + 3) ] = d/dx [ -2(z² -4z + 5)/(z² -4z + 3)² ] --> 4(z-2)*(z² -4z + 7)/(z²-4z+3)³

d⁴/dz⁴ [ ln(z² -4z + 3) ] = ...

Evaluating all derivatives at z₀ = 0...

a₀ = f⁰(0)/0! = f(0) = ln(3)

a₁ = f¹(0)/1! = f'(0) = -4/3

a₂ = f²(0)/2! = f''(0)/2 = -10/18

a₃ = f³(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)x² - (56/162)x³ + ...

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 z₀ = 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 )