let-f-x-x-2-3x-arctan-2x-1-1-determine-f-n-x-and-f-n-0-2-developp-f-at-integr-serie-3-calculate-0-1-f-x-dx-

Question Number 68238 by mathmax by abdo last updated on 07/Sep/19

l e t f (x) = (x^{2} - 3 x) a r c t a n (2 x + 1)

1) d e t e r m i n e f^{(n)} (x) a n d f^{(n)} (0)

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

3) c a l c u l a t e \int_{0}^{1} f (x) d x

Commented by mathmax by abdo last updated on 08/Sep/19

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

1) f (x) = (x^{2} - 3 x) a r c t a n (2 x + 1) l e i b n i z f o r m u l a g i v e

f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(x^{2} - 3 x)}^{(k)} {(a r c t a n (2 x + 1))}^{(n - k)}

= (x^{2} - 3 x) {(a r c t a n (2 x + 1))}^{(n)} + n (2 x - 3) {(a r c t a n (2 x + 1))}^{(n - 1)}

+ n (n - 1) {(a r c t a n (2 x + 1))}^{(n - 2)} l e t A (x) = a r c t a n (2 x + 1) \Rightarrow

A^{^{'}} (x) = \frac{2}{1 + {(2 x + 1)}^{2}} = \frac{2}{1 + 4 x^{2} + 4 x + 1} = \frac{2}{4 x^{2} + 4 x + 2} = \frac{1}{2 x^{2} + 2 x + 1}

\Rightarrow A^{(n)} (x) = {(\frac{1}{2 x^{2} + 2 x + 1})}^{(n - 1)}

2 x^{2} + 2 x + 1 = 0 \to Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow x_{1} = \frac{- 1 + i}{2} = - \frac{1 - i}{2} = - \frac{\sqrt{2}}{2} e^{- \frac{i π}{4}}

x_{2} = \frac{- 1 - i}{2} = - \frac{\sqrt{2}}{2} e^{\frac{i π}{4}} \Rightarrow

\frac{1}{2 x^{2} + 2 x + 1} = \frac{1}{2 (x + \frac{\sqrt{2}}{2} e^{- \frac{i π}{4}}) (x + \frac{\sqrt{2}}{2} e^{\frac{i π}{4}})}

= \frac{1}{2 i} {\frac{1}{x + \frac{\sqrt{2}}{2} e^{- \frac{i π}{4}}} - \frac{1}{x + \frac{\sqrt{2}}{2} e^{\frac{i π}{4}}}} \Rightarrow

A^{(n)} = \frac{1}{2 i} {\frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + \frac{\sqrt{2}}{2} e^{- \frac{i π}{4}})}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + \frac{\sqrt{2}}{2} e^{\frac{i π}{4}})}^{n}}}

= \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{{(x + \frac{\sqrt{2}}{2} e^{\frac{i π}{4}})}^{n} - {(x + \frac{\sqrt{2}}{2} e^{- \frac{i π}{4}})}^{n}}{{(x^{2} + x + \frac{1}{2})}^{n}}}

= \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} \times \frac{2 i I m ({(x + \frac{\sqrt{2}}{2} e^{\frac{i π}{4}})}^{n})}{{(x^{2} + x + \frac{1}{2})}^{n}} \Rightarrow

A^{(n)} (x) = \frac{{(- 1)}^{n - 1} (n - 1)! I m ({(x + \frac{\sqrt{2}}{2} e^{\frac{i π}{4}})}^{n})}{{(x^{2} + x + \frac{1}{2})}^{n}} \Rightarrow

f^{(n)} (x) = (x^{2} - 3 x) A^{(n)} (x) + n (2 x - 3) A^{(n - 1)} (x) + n (n - 1) A^{(n - 2)} (x)

\Rightarrow f^{(n)} (0) = - 3 n A^{(n - 1)} (0) + n (n - 1) A^{(n - 2)} (0)

A^{(n)} (0) = 2^{n} {(- 1)}^{n - 1} (n - 1)! I m {(\frac{\sqrt{2}}{2} e^{\frac{i π}{4}})}^{n}

= 2^{n} {(- 1)}^{n - 1} (n - 1)! \frac{2^{\frac{n}{2}}}{2^{n}} s i n (n \frac{π}{4}) = 2^{\frac{n}{2}} {(- 1)}^{n - 1} (n - 1)! s i n (\frac{n π}{4}) \Rightarrow

f^{(n)} (0) = - 3 n 2^{\frac{n - 1}{2}} {(- 1)}^{n - 2} (n - 2)! s i n (\frac{(n - 1) π}{4})

+ n (n - 1) 2^{\frac{n - 2}{2}} {(- 1)}^{n - 3} (n - 3)! s i n (\frac{(n - 2) π}{4})

Commented by mathmax by abdo last updated on 08/Sep/19

2) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

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

= f^{^{'}} (0) x + \frac{f^{(2)} (0)}{2} x^{2}

+ \sum_{n = 3}^{\infty} (\frac{- 3 n 2^{\frac{n - 1}{2}} {(- 1)}^{\boldsymbol n - 2} (\boldsymbol n - 2)! s i n (\frac{(n - 1) π}{4}) + n (n - 1) 2^{\frac{n - 2}{2}} {(- 1)}^{\boldsymbol n - 3} (\boldsymbol n - 3)! s i n (\frac{(n - 2) π}{4})}{n!}) x^{n}