Question

What is the Maclaurin series for the function `ln(x+e^x)`?

Answer 1

`ln(x+e^x) = 2x - 3/2 x^2 + 11/6 x^3 - ...`

Explanation:

The Maclaurin series is given by

`f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n + ...`

We start with the function

`f^((0))(x) = f(x) = ln(x+e^x)`

Then, we compute the first few derivatives:

`f^((1))(x) = 1/(x+e^x)(1+e^x)`
`\ \ \ \ \ \ \ \ \ \ \ \ = (1+e^x)/(x+e^x)`

`f^((2))(x) = ((x+e^x)(e^x)-(1+e^x)(1+e^x))/(x+e^x)^2`
`\ \ \ \ \ \ \ \ \ \ \ \ = (xe^x+e^(2x)-1-2e^x-e^(2x))/(x+e^x)^2`
`\ \ \ \ \ \ \ \ \ \ \ \ = (xe^x-1-2e^x)/(x+e^x)^2`

`f^((3))(x) = ((x-2)e^x+e^x)/(e^x+x)^2-(2(e^x+1)((x-2)e^x-1))/(e^x+x)^3`
`\ \ \ \ \ \ \ \ \ \ \ \ = -((x-3)e^(2x)+(-x^2+3x-6)e^x-2)/(e^x+x)^3`

`vdots`

Now we have the derivatives, we can compute their values when `x=0`

# {: (f^((0))(x), = ln(0+1), =0), (f^((1))(x), = (1+1)/(1), =2), (f^((3))(x), = (0-1-2)/(1)^2, =-3), (f^((4))(x), = -((-3)+(-6)-2)/(1)^3, =11) :} #

`vdots`

Which permits us to form the Maclaurin serie:

`f(x) = (0) + (2)/(1)x + (-3)/(2)x^2 + (11)/(6)x^3 + ... (f^((n))(0))/(n!)x^n + ...`

`\ \ \ \ \ \ \ = 2x -3/2 x^2 + (11)/(6)x^3 + ...`