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Question Number 62882 by mathmax by abdo last updated on 26/Jun/19

l e t f (x) = l n ∣ \frac{x - 1}{x + 1} ∣

1) d e t e r m i n e D_{f}

2) {c a l c u l a t e f}^{(n)} (x) a n d f^{(n)} (0)

3) d e v e l o p p f a t i n t e g r s e r i e

4) c a l c u l a t e \int_{- \frac{1}{2}}^{\frac{1}{2}} f (x) d x .

Commented by mathmax by abdo last updated on 27/Jun/19

4) \int_{- \frac{1}{2}}^{\frac{1}{2}} l n ∣ \frac{x - 1}{x + 1} ∣ d x = \int_{- \frac{1}{2}}^{\frac{1}{2}} l n ∣ 1 - x ∣ d x - \int_{- \frac{1}{2}}^{\frac{1}{2}} l n ∣ 1 + x ∣ d x

= \int_{- \frac{1}{2}}^{\frac{1}{2}} l n (1 - x) d x - \int_{- \frac{1}{2}}^{\frac{1}{2}} l n (1 + x) d x = H - K

H =_{1 - x = t} \int_{\frac{3}{2}}^{\frac{1}{2}} l n (t) (- d t) = \int_{\frac{1}{2}}^{\frac{3}{2}} l n (t) d t = {[t l n (t) - t]}_{\frac{1}{2}}^{\frac{3}{2}}

= \frac{3}{2} l n (\frac{3}{2}) - \frac{3}{2} - \frac{1}{2} l n (\frac{1}{2}) + \frac{1}{2}

K = \int_{- \frac{1}{2}}^{\frac{1}{2}} l n (1 + x) d x =_{1 + x = t} \int_{\frac{1}{2}}^{\frac{3}{2}} l n (t) d t = {[t l n (t) - t]}_{\frac{1}{2}}^{\frac{3}{2}}

= \frac{3}{2} l n (\frac{3}{2}) - \frac{3}{2} - \frac{1}{2} l n (\frac{1}{2}) + \frac{1}{2} \Rightarrow

A = \frac{3}{2} l n (\frac{3}{2}) - \frac{3}{2} - \frac{1}{2} l n (\frac{1}{2}) + \frac{1}{2} - \frac{3}{2} l n (\frac{3}{2}) + \frac{3}{2} + \frac{1}{2} l n (\frac{1}{2}) - \frac{1}{2} \Rightarrow

A = 0

a n o t h e r [w a y l e t p r o v e t h a t f i s o d d w e h a v e

f (- x) = l n ∣ \frac{- x - 1}{- x + 1} ∣ = l n ∣ \frac{x + 1}{x - 1} ∣ = - l n ∣ \frac{x - 1}{x + 1} ∣= - f (x) \Rightarrow \int_{- \frac{1}{2}}^{\frac{1}{2}} f (x) d x = 0 .

Commented by mathmax by abdo last updated on 27/Jun/19

1) D_{f} = R - {- 1, 1}

2) w e h a v e f (x) = l n ∣ x - 1 ∣ - l n ∣ x + 1 ∣ \Rightarrow

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

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

f^{(n)} (x) = {(- 1)}^{n - 1} (n - 1)! \frac{{(x + 1)}^{n} - {(x - 1)}^{n}}{{(x^{2} - 1)}^{n}} w i t h x ⩾ 1

f^{(n)} (0) = {(- 1)}^{n - 1} (n - 1)! \frac{1 - {(- 1)}^{n}}{{(- 1)}^{n}} = - (n - 1)! (1 - {(- 1)}^{n}) = {{(- 1)}^{n} - 1} (n - 1)!

\Rightarrow f^{(2 p)} (0) = 0 a n d f^{(2 p + 1)} (0) = - 2 (2 p)!

3) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = \sum_{n = 1}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} = \sum_{p = 0}^{\infty} \frac{- 2 (2 p)!}{(2 p + 1)!} x^{2 p + 1} \Rightarrow

f (x) = - 2 \sum_{p = 0}^{\infty} \frac{x^{2 p + 1}}{2 p + 1}

Commented by mathmax by abdo last updated on 27/Jun/19

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