(a) Find the Maclaurin series for ln(1-- x), and also find it\'s interval of con

Question

(a) Find the Maclaurin series for ln(1-- x), and also find it's interval of convergence.

(b) Find the Maclaurin series for ln(1 + x), and also find it's interval of convergence.

(c) Find the Maclaurin series for ln[(1 + x)/(1-- x)], and also find it's interval of convergence.

(d) Notice that any real number, y, can be written as y =(1 + x)/(1-- x) for some real number x such that --1 < x < 1. With this in mind, use your result from the previous step combined with the alternating series test to approximate ln(7) accurately to 6 decimal places.

Explanation / Answer

2. y = ln (1 + x)

dy/dx = 1/ (1 + x)

d2y/dx2 = -(1 + x)^-2

d3y/dx3 = 2(1 + x)^-3

d4y/dx4 = -6(1 + x)^-4

d5y/dx5 = 24 (1 +x)^-5

therefore,
f(0) = ln(1+0) = ln1 = 0
f'(0) = 1/(1 + 0) = 1
f''(0) = -(1+0)^-2 = -1
f'''(0) = 2(1 + 0)^-3 = 2
f^4(0) = -6(1 + 0)^-4 = -6
f^5(0) = 24 (1 + 0)^-5 = 24

maclaurin's series for ln (1+x)
= 0 + x - 1/2x^2 + 2/3!x^3 - 6/4!x^4 + 24/5!x^5
= x - 1/2x^2 + 2/6x^3 - 6/24x^4 + 24/120x^5
= x - 1/2x^2 + 1/3x^3 - 1/4x^4 + 1/5x^5

1. in the similar way u can find out the series for ln(1-x) which comes out to be eqaul to

ln(1 - x) = -x - (x^2)/2 - (x^3)/3 - (x^4)/4 . . .
valid if -1 < x < 1

3.

The expansion for log(1+x) is not ??n=0(?1)n+1xnn, but

Consequently

and

because

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