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zagonek34

Answered question

2021-12-13

How do you find the Maclaurin series of

f (x) = \ln (1 + x) ?

Fasaniu

Beginner2021-12-14Added 46 answers

The Maclaurin series of

f (x) = \ln (1 + x) ?

is:

f (x) = \sum_{{n = 0}}^{\infty} {(- 1)}^{n} \frac{x^{n + 1}}{n + 1}

where

| x | < 1 .

First, let us find the Maclaurin series for

f^{'} (x) = \frac{1}{1 + x} = \frac{1}{1 - (- x)}

Remember that

\frac{1}{1 - x} = \sum_{{n = 0}}^{\infty} x^{n} if | x | < 1

(Note: This can be justified by viewing it as a geometric series.)
By replacing x by − x ,

f^{'} (x) = \frac{1}{1 - (- x)} = \sum_{{n = 0}}^{\infty} {(- x)}^{n} = \sum_{{n = 0}}^{\infty} {(- 1)}^{n} x^{n}

By integrating using Power Rule,

f (x) = \int \sum_{{n = 0}}^{\infty} {(- 1)}^{n} x^{n} d x = \sum_{{n = 0}}^{\infty} {(- 1)}^{n} \frac{x^{n + 1}}{n + 1} + C

(Note: integration can be done term by term.)
Since

f (0) = \ln [1 + (0)] = 0,

f (0) = \sum_{{n = 0}}^{\infty} {(- 1)}^{n} \frac{{(0)}^{n + 1}}{n + 1} + C = C = 0

Hence,

f (x) = \sum_{{n = 0}}^{\infty} {(- 1)}^{n} \frac{x^{n + 1}}{n + 1}

scoollato7o

Beginner2021-12-15Added 26 answers

Oh my god, how difficult it is, thanks for helping me understand and solve

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