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

Using the Maclaurin series ln (1-X)=-X-X^2/2-X^3/3-...-X^n/n-...,#

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Let

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

In order to find such Maclaurin series, which is just a special case of a Taylor series centered at

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

# = sum_(n=0)^(∞) f^(n)(0)/(n!) x^n#

Let

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.

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