How do you find the Taylor series for #ln(x)# about the value x=1?

Question

William Abdalla · Accepted Answer

firstly we look at the formula for the Taylor series, which is:

#f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n #

which equals:

#f(a) + f'(a)(x-a) + (f''(a)(x-a)^2)/(2!) + (f'''(a)(x-a)^3)/(3!) + ... #

So you would like to solve for #f(x) = ln(x)# at #x=1# which I assume mean centered at #1# of which you would make #a=1#

To solve:

#f(x) = ln(x)# and #f(1) = ln(1) = 0#

#f'(x) = 1/x# and #f'(1) = 1/1 = 1#

#f''(x) = -1/x^2# and #f''(1) = -1/(1)^2 = -1#

#f^((3))(x) = 2/x^3# and #f^((3))(1) = 2/(1)^3 = 2#

#f^((4))(x) = -((2)(3))/x^4# and #f^((4))(1) = -((2)(3))/(1)^4 = -(2)(3)#

Where now we can already start to see a pattern forming, so we starting using our formula(2):

#0 + 1(x-1) - (1(x-1)^2)/(2!) + (2(x-1)^3)/(3!) - ((2)(3)(x-1)^4)/(4!) .....#

and now try try see how we can write this as a series, which we get: (we start will n=1 as our first term is 0)

#f(x) = ln(x) = sum_(n=1)^oo (-1)^(n-1) (((n-1)!)(x-1)^n)/(n!) #

Which can then simplify to:

#f(x) = ln(x) = sum_(n=1)^oo (-1)^(n-1) (x-1)^n/n#

Scarlett Allison · Answer

undefined To find the Taylor series for ln(x) about the value x=1, you can start by finding the derivatives of ln(x) and evaluating them at x=1 to obtain the coefficients of the Taylor series. The Taylor series for ln(x) about x=1 is given by:

ln(x) = ln(1) + (x - 1) * ln'(1) / 1! + (x - 1)^2 * ln''(1) / 2! + (x - 1)^3 * ln'''(1) / 3! + ...

To find the coefficients, you'll need to find the derivatives of ln(x) and evaluate them at x=1:

ln'(x) = 1/x
ln''(x) = -1/x^2
ln'''(x) = 2/x^3

Evaluate these derivatives at x=1:

ln'(1) = 1
ln''(1) = -1
ln'''(1) = 2

Substitute these values into the Taylor series formula:

ln(x) = 0 + (x - 1) * 1 / 1! + (x - 1)^2 * (-1) / 2! + (x - 1)^3 * 2 / 3! + ...

Simplify:

ln(x) = (x - 1) - (x - 1)^2 / 2 + (x - 1)^3 / 3 - ...

This is the Taylor series for ln(x) about the value x=1.