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Question Number 143381 by Mathspace last updated on 13/Jun/21

l e t f (x) = a r c t a n (\sqrt{2} x^{2})

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

2) i f f (x) = Σ a_{n} x^{n} f i n d t h e

s e q u e n c e a_{n}

Answered by TheHoneyCat last updated on 13/Jun/21

not in the right order, to make it more simple

\forall x \in] - 1, 1 [

\frac{1}{1 - x} = \underset{n = 0}{\sum^{\infty}} x^{n}

t h u s : \frac{1}{1 + x^{2}} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{2 n}

k n o w i n g t h a t \arctan^{'} = x \to \frac{1}{1 + x^{2}} a n d t h a t \arctan (0) = 0

\arctan (x) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 1}

a n d t h e r e f o r :

\arctan (\sqrt{2} x^{2}) = 2 \sqrt{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} x^{4 n + 2}

s o \forall n \in N if n \equiv 2 [4] a_{n} = \frac{{(- 1)}^{\frac{n - 2}{4}}}{\frac{n - 2}{2} + 1}

otherwise a_{n} = 0

and \forall n \in N f^{(n)} (0) = a_{n}

a s f o r f^{(n)} (x) I a m q u i t e n o t c o u r a g e o u s e n o u g h t

f^{(0)} (x) = \frac{2 \sqrt{2} x}{1 + 2 x^{4}}

f r o m t h a t p o i n t o n, o n e m u s t u s e {L e i b n i e t z}^{'} s f o r m u l a f o r t h e d e r i v a t i o n o f a p r o d u c t

{(f g)}^{(n)} = \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) f^{(n - k)} g^{(k)}

w i t h f : x \to 2 \sqrt{2} x a n d g : x \to \frac{1}{1 + 2 x^{2}}

it obviously simplifies due to f^{″} = x \to 0

but like I said, I am not courrageous enougth

maybe there is a faster method

(for instance, if you guess a solution, you only have to verify it with a recurence)

good luck to you with that last bit .

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

1) f (x) = \arctan (\sqrt{2} x^{2}) \Rightarrow f^{^{'}} (x) = \frac{2 \sqrt{2} x}{1 + {2 x}^{4}} = \frac{2 \sqrt{2} x}{(\sqrt{2} x^{2} - i) (\sqrt{2} x^{2} + i)}

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

(λ = \frac{1}{\sqrt{\sqrt{2}}}) = \frac{a}{x - λ e^{\frac{i π}{4}}} + \frac{b}{x + λ e^{\frac{i π}{4}}} + \frac{c}{x - λ e^{- \frac{i π}{4}}} + \frac{d}{x + λ e^{- \frac{i π}{4}}}

a = \frac{λ \sqrt{2} e^{\frac{i π}{4}}}{2 λ e^{\frac{i π}{4}} (λ^{2} i + \frac{i}{\sqrt{2}})} = \frac{1}{\sqrt{2} (\frac{2 i}{\sqrt{2}})} = \frac{1}{2 i} its eazy to find other coefficient

\Rightarrow f^{(n)} (x) = \frac{a {(- 1)}^{n - 1} (n - 1)!}{{(x - λ e^{\frac{i π}{4}})}^{n}} + \frac{b {(- 1)}^{n - 1} (n - 1)!}{(x + λ e^{\frac{i π}{4}})}

+ \frac{c {(- 1)}^{n - 1} (n - 1)!}{{(x - λ e^{- \frac{i π}{4}})}^{n}} + \frac{d {(- 1)}^{n - 1} (n - 1)!}{(x + λ e^{- \frac{i π}{4}})}

= {(- 1)}^{n - 1} (n - 1)! {\frac{a}{{(x - λ e^{\frac{i π}{4}})}^{n}} + \frac{b}{{(x + λ e^{\frac{i π}{4}})}^{n}} + \frac{c}{{(x - λ e^{- \frac{i π}{4}})}^{n}}

+ \frac{d}{{(x + λ e^{- \frac{i π}{4}})}^{n}}} \Rightarrow

f^{(n)} (0) = {(- 1)}^{n - 1} (n - 1)! {\frac{{ae}^{- \frac{in π}{4}}}{{(- λ)}^{n}} + \frac{{be}^{- \frac{in π}{4}}}{λ^{n}} + \frac{{ce}^{\frac{in π}{4}}}{{(- λ)}^{n}} + \frac{{de}^{\frac{in π}{4}}}{λ^{n}}} (λ = \frac{1}{(^{4} \sqrt{2})})

2) if f (x) = Σ a_{n} x^{n} \Rightarrow a_{n} = \frac{f^{(n)} (0)}{n!}