How do you find the maclaurin series expansion of #ln(1+x)#?

Answer 1

Elijah Abram

See the explanation.

#f(x) = ln(1+x)#

#f'(x) = 1/(1+x) = (1+x)^-1#

#f''(x) =-(1+x)^-2#

#f'"'(x) =2(1+x)^-3#

#f^((4))(x) =-6(1+x)^-4#

#f^((5))(x) =30(1+x)^-5#

#f^((n))(x) =(-1)^(n-1) (n-1)! (1+x)^-n#

Utilize the Maclaurin Series formula now.

Note #f^((n))(0) =(-1)^(n-1) (n-1)!#

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Answer 2

Adam Adkins

To find the Maclaurin series expansion of ln(1+x), you can start by finding the derivatives of ln(1+x) with respect to x, and then evaluating them at x=0 to obtain the coefficients of the series. The Maclaurin series expansion of ln(1+x) is:

ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ...

This series converges for |x| < 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.