How do you find the Maclaurin series for #ln(2-3x) #?

#-1<=X<1#,

#ln(2-3x)=ln 2 + ln (1-1.5 x)#

#=ln 2 - 1.5 x -(1.5 x)^2/2-1.5 x)^3/3-...-(1.5 x)^n/n-...#,

#-1<=1.5x<1#, or explicitly, #-2/3<=x<2/3#.,

In order to find such Maclaurin series, which is just a special case of a Taylor series centered at #x=0#, we could also calculate a few derivatives and construct a series representation.

Remember that a Maclaurin series can be expressed in the following way:

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

# = sum_(n=0)^(∞) f^(n)(0)/(n!) x^n#

Let #f(x) = ln(2-3x) -> f(0) = ln2#

Calculating our derivatives:

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

# = -3(2-3x)^(-1) #

#f^(2)(x) = 3(2-3x)^(-2)(-3)#

#= -9(2-3x)^(-2)#

#f^(3)(x) =18(2-3x)^(-3)(-3)#

#= -54(2-3x)^(-3)#

#f^(4)(x) =162(2-3x)^(-4)(-3)#

#= -486(2-3x)^(-4)#

Since a Maclaurin series is centered at #x=0#, we evaluate each derivative at that point.

#f(0) = 0#

#f^(1)(0) = -3/2 #

#f^(2)(0) = -9/4#

#f^(3)(0) = -54/8 #

#f^(4)(0) = -486/16 #

Now that we have our derivatives, we can simply substitute them into our infinite sum:

#f(x) = ln 2 - (3/2)^(1)/(1!) x - (3/2)^(2)/(2!) x^2 - (3/2)^(3)/(3!) x^3 - (3/2)^(4)/(4!) x^4 - ...#

#=ln2 - sum_(n=1)^(∞) (3/2)^(n)/(n!) x^n#

Notice that our infinite sum does not start at #n = 0# but #n =1#.

To get an idea for how this can be visualized - here are the graphs of each function (the original and the approximation).

Graph of #ln(2-3x)#

graph{ln(2-3x) [-10, 10, -5, 5]}

Graph of # ln 2 - (3/2)^(1)/(1!) x - (3/2)^(2)/(2!) x^2 - (3/2)^(3)/(3!) x^3 - (3/2)^(4)/(4!) x^4#

graph{ln 2 - (3/2)^(1)/(1!) x - (3/2)^(2)/(2!) x^2 - (3/2)^(3)/(3!) x^3 - (3/2)^(4)/(4!) x^4 [-10, 10, -5, 5]}

When both overlap, we can see that the approximation looks to be correct.