Question #0973a

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Question #0973a

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Answer 1 · 2017-02-06T16:00:49

$ln (2 + x) = ln 2 + \infty \sum n = 0 {(- 1)}^{n} \frac{x^{n + 1}}{2^{n + 1} (n + 1)}$ $ln(2+x) = ln2 + sum_(n=0)^oo (-1)^n x^(n+1)/(2^(n+1)(n+1))$

Explanation:

Write the function as:

$ln (2 + x) = ln (2 (1 + \frac{x}{2})) = ln 2 + ln (1 + \frac{x}{2})$ $ln(2+x) = ln (2 (1+x/2)) = ln2 +ln(1+x/2)$

Now use the MacLaurin development of $ln (1 + x)$ $ln (1+x)$ , substituting $\frac{x}{2}$ $x/2$ to $x$ $x$ :

$ln (1 + \frac{x}{2}) = \infty \sum n = 0 {(- 1)}^{n} \frac{{(\frac{x}{2})}^{n + 1}}{n + 1} = \infty \sum n = 0 {(- 1)}^{n} \frac{x^{n + 1}}{2^{n + 1} (n + 1)}$ $ln(1+x/2) = sum_(n=0)^oo (-1)^n (x/2)^(n+1)/(n+1) = sum_(n=0)^oo (-1)^n x^(n+1)/(2^(n+1)(n+1))$

In conclusion:

$ln (2 + x) = ln 2 + \infty \sum n = 0 {(- 1)}^{n} \frac{x^{n + 1}}{2^{n + 1} (n + 1)}$ $ln(2+x) = ln2 + sum_(n=0)^oo (-1)^n x^(n+1)/(2^(n+1)(n+1))$

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Question #0973a

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