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Question Number 95786 by abdomathmax last updated on 27/May/20

let f (x) = \frac{1}{x} \ln (1 + 2 x)

1) calculate f^{(n)} (x) and f^{(n)} (1)

2) developp f at integr serie at x_{0} = 1

3) developp f at integr serie at x_{0} = 0

Commented by Rio Michael last updated on 28/May/20

sir when you talk about integer series

do you mean taylors series ?

Commented by mathmax by abdo last updated on 29/May/20

yes

Answered by mathmax by abdo last updated on 02/Jun/20

1) f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(\ln (2 x + 1))}^{(k)} {(\frac{1}{x})}^{(n - k)}

= C_{n}^{0} \ln (2 x + 1) \frac{{(- 1)}^{n} n!}{x^{n + 1}} + \sum_{k = 1}^{n} C_{n}^{k} {(\ln (2 x + 1))}^{(k)} \times \frac{{(- 1)}^{n - k} (n - k)!}{x^{n - k + 1}}

we have \frac{d}{dx} \ln (2 x + 1) = \frac{2}{2 x + 1} = \frac{1}{x + \frac{1}{2}} \Rightarrow \frac{d^{k}}{{dx}^{k}} \ln (2 x + 1) = {(\frac{1}{x + \frac{1}{2}})}^{(k - 1)}

= \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + \frac{1}{2})}^{k}} \Rightarrow

f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{x^{n + 1}} \ln (2 x + 1) + \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + \frac{1}{2})}^{k}} \times \frac{{(- 1)}^{n - k} (n - k)!}{x^{n - k + 1}}

= \frac{{(- 1)}^{n} n!}{x^{n + 1}} \ln (2 x + 1) + \sum_{k = 1}^{n} {(- 1)}^{n - 1} \frac{n!}{k! (n - k)!} \times \frac{(k - 1)! (n - k)!}{x^{n - k + 1} {(x + \frac{1}{2})}^{k}}

f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{x^{n + 1}} \ln (2 x + 1) + n! {(- 1)}^{n - 1} \sum_{k = 1}^{n} \frac{1}{k x^{n - k + 1} {(x + \frac{1}{2})}^{k}}

Commented by mathmax by abdo last updated on 02/Jun/20

f^{(n)} (1) = n! {(- 1)}^{n} \ln (3) + n! {(- 1)}^{n - 1} \sum_{k = 1}^{n} \frac{1}{k {(\frac{3}{2})}^{k}}

= n! {(- 1)}^{n} \ln (3) + n! {(- 1)}^{n - 1} \sum_{k = 1}^{n} \frac{1}{k} {(\frac{2}{3})}^{k}