let-f-x-ln-2-x-3-if-f-x-a-n-x-n-find-a-n-

Question Number 135384 by Bird last updated on 12/Mar/21

l e t f (x) = l n (2 + x^{3})

i f f (x) = Σ a_{n} x^{n}

f i n d a_{n}

Answered by Dwaipayan Shikari last updated on 12/Mar/21

f (x) = \underset{n = 0}{\sum^{\infty}} \frac{f^{n} (0)}{n!} x^{n}

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

f (x) = l o g (α^{3} + x^{3}) = l o g (α + x) + l o g (α^{2} - α x + x^{2})

= l o g (α + x) + l o g (x - β) + l o g (x - γ) β = α e^{\frac{2 π i}{3}} γ = α e^{\frac{- 2 π i}{3}}

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

f^{″} (x) = \frac{- 1}{{(α + x)}^{2}} + \frac{- 1}{{(x - β)}^{2}} - \frac{1}{{(x - γ)}^{2}}

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

= {(- 1)}^{n + 1} (n - 1)! (\frac{1}{{(α + x)}^{n}} + \frac{{(x - β)}^{n} + {(x - γ)}^{n}}{{(α^{2} - α x + x^{2})}^{n}})

a_{n} = \frac{{(- 1)}^{n + 1}}{n} (\frac{1}{α^{n}} + \frac{{(- 1)}^{n} (e^{\frac{2 π i n}{3}} + e^{- \frac{2 π i n}{3}})}{{(α^{2})}^{n}}) α = \sqrt[3]{2}

a_{n} = \frac{{(- 1)}^{n + 1}}{n} (\frac{1}{2^{\frac{n}{3}}} + \frac{2 {(- 1)}^{n} c o s (\frac{2 π n}{3})}{{(\sqrt[3]{4})}^{n}})

Commented by mathmax by abdo last updated on 12/Mar/21

thank you sir

Answered by Olaf last updated on 12/Mar/21

f (x) = \ln (2 + x^{3})

f^{'} (x) = \frac{3 x^{2}}{2 + x^{3}} = \frac{3}{2} . \frac{x^{2}}{1 + {(\frac{x}{2^{1 / 3}})}^{3}}

f^{'} (x) = \frac{3}{2} x^{2} \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} {(\frac{x}{2^{1 / 3}})}^{n}

f^{'} (x) = \frac{3}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2^{n / 3}} x^{n + 2}

f (x) = \frac{3}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(n + 3) 2^{n / 3}} x^{n + 3} + C

f (0) = \ln (2 + 0) = \ln 2 \Rightarrow C = \ln 2

f (x) = 3 \underset{n = 3}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n 2^{n}} x^{n} + \ln 2

a_{0} = \ln 2, a_{1} = a_{2} = 0 and a_{n} = \frac{3 {(- 1)}^{n}}{n 2^{n}}, n ⩾ 3

Commented by mathmax by abdo last updated on 12/Mar/21

thank you sir

Answered by mathmax by abdo last updated on 14/Mar/21

a_{n} = \frac{f^{) n)} (0)}{n!} let determine f^{) n)} (0) we have f^{^{'}} (x) = \frac{{3 x}^{2}}{x^{3} + 2}

let α =^{3} \sqrt{2} \Rightarrow f^{^{'}} (x) = \frac{{3 x}^{2}}{x^{3} + α^{3}} = \frac{{3 x}^{2}}{(x + α) (x^{2} - α x + α^{2})}

= \frac{a}{x + α} + \frac{bx + c}{x^{2} - α x + α^{2}} we have a = \frac{3 α^{2}}{3 α^{2}} = \frac{3}{2}

\lim_{x \to + \infty} {xf}^{^{'}} (x) = 3 = a + b \Rightarrow b = 3 - \frac{3}{2} = \frac{3}{2}

f^{^{'}} (0) = 0 = \frac{a}{α} + \frac{c}{α^{2}} \Rightarrow o = α a + c \Rightarrow c = - α . \frac{3}{2} \Rightarrow

f^{^{'}} (x) = \frac{3}{2 (x + α)} + \frac{\frac{3}{2} x - \frac{3 α}{2}}{x^{2} - α x + α^{2}} = \frac{3}{2 (x + α)} + \frac{3}{2} \frac{x - α}{x^{2} - α x + α^{2}}

let decompose inside C (x) the fraction F (x) = \frac{x - α}{x^{2} - α x + α^{2}}

Δ = α^{2} - 4 α^{2} = - 3 α^{2} \Rightarrow z_{1} = \frac{α + i α \sqrt{3}}{2} = α (\frac{1}{2} + i \frac{\sqrt{3}}{2}) = α e^{\frac{i π}{3}}

z_{2} = α e^{- \frac{i π}{3}} \Rightarrow F (x) = \frac{a}{x - z_{1}} + \frac{b}{x - z_{2}} (= \frac{x - α}{(x - z_{1}) (x - z_{2})})

a = \frac{z_{1} - α}{z_{1} - z_{2}} = \frac{z_{1} - α}{2 i α \frac{\sqrt{3}}{2}} = \frac{1}{i α \sqrt{3}} (z_{1} - α)

b = \frac{z_{2} - α}{z_{2} - z_{1}} = - \frac{1}{i α \sqrt{3}} (z_{2} - α) \Rightarrow

f^{^{'}} (x) = \frac{3}{2 (x + α)} + \frac{3 (z_{1} - α)}{2 i α \sqrt{3} (x - z_{1})} - \frac{3 (z_{2} - α)}{2 i α \sqrt{3} (x - z_{2})} \Rightarrow

f^{(n)} (x) = \frac{3}{2} {\frac{1}{x + α}}^{(n - 1)} + \frac{3 (z_{1} - α)}{2 i \sqrt{3}} {\frac{1}{x - z_{1}}}^{(n - 1)} - \frac{3 (z_{2} - α)}{2 i \sqrt{3}} {\frac{1}{x - z_{2}}}^{(n - 1)}

= \frac{3}{2} \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + α)}^{n}} + \frac{3 (z_{1} - α)}{2 i \sqrt{3}} \times \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - z_{1})}^{n}}

- \frac{3 (z_{2} - α)}{2 i \sqrt{3}} \times \frac{{(- 1)}^{n} (n - 1)!}{{(x - z_{2})}^{n}} \Rightarrow

a_{n} = \frac{1}{n!} f^{(n)} (0) = \frac{1}{n!} {\frac{3}{2} α^{- n} {(- 1)}^{n - 1} (n - 1)! + \frac{3 (z_{1} - α)}{2 i \sqrt{3}} {(- z_{1})}^{- n} {(- 1)}^{n - 1} (n - 1)!

- \frac{3 (z_{2} - α)}{2 i \sqrt{3}} {(- z_{2})}^{- n} {(- 1)}^{n} (n - 1)!}