let-f-x-1-x-2-2-cos-x-1-1-calculate-f-n-x-2-find-0-f-x-dx-3-find-0-f-n-x-dx-

Question Number 81165 by abdomathmax last updated on 09/Feb/20

l e t f (x) = \frac{1}{x^{2} - 2 (c o s θ) x + 1}

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

2) f i n d \int_{0}^{\infty} f (x) d x

3) f i n d \int_{0}^{\infty} f^{(n)} (x) d x

Commented by mathmax by abdo last updated on 10/Feb/20

1) l e t s o l v e x^{2} - 2 c o s θ) x + 1 = 0 \to Δ^{^{'}} = {c o s}^{2} x - 1 = - {s i n}^{2} x \Rightarrow

x_{1} = c o s θ + i s i n θ = e^{i θ} a n d x_{2} = c o s θ - i s i n θ = e^{- i θ}

f (x) = \frac{1}{(x - e^{i θ}) (x - e^{- i θ})} = \frac{1}{2 i s i n θ} (\frac{1}{x - e^{i θ}} - \frac{1}{x - e^{- i θ}}) (i f θ \neq k π)

\Rightarrow f^{(n)} (x) = \frac{1}{2 i s i n θ} {\frac{{(- 1)}^{n} n!}{{(x - e^{i θ})}^{n + 1}} - \frac{{(- 1)}^{n} n!}{{(x - e^{- i θ})}^{n + 1}}}

= \frac{{(- 1)}^{n} n!}{2 i s i n θ} {\frac{{(x - e^{- i θ})}^{n + 1} - {(x - e^{i θ})}^{n + 1}}{{(x^{2} - 2 c o s θ + 1)}^{n + 1}}}

= \frac{{(- 1)}^{n + 1} n!}{2 i s i n θ} {\frac{2 i I m ({(x - e^{i θ})}^{n + 1})}{{(x^{2} - 2 c o s θ + 1)}^{n + 1}}} \Rightarrow

f^{(n)} (x) = \frac{{(- 1)}^{n + 1} n! \times I m ({(x - e^{i θ})}^{n + 1})}{s i n θ {(x^{2} - 2 c o s θ + 1)}^{n + 1}}

Commented by mathmax by abdo last updated on 10/Feb/20

2) i f θ = k π \Rightarrow f (x) = \frac{1}{{(x \bar{+} 1)}^{2}} \Rightarrow \int f (x) d x = - \frac{1}{x \bar{+} 1} + c

i f θ \neq k π \Rightarrow \int_{0}^{\infty} f (x) d x = \int_{0}^{\infty} \frac{d x}{x^{2} - 2 c o s θ x + {c o s}^{2} θ + {s i n}^{2} θ}

Commented by mathmax by abdo last updated on 10/Feb/20

\int_{0}^{\infty} f (x) d x = \int_{0}^{\infty} \frac{d x}{{(x - c o s θ)}^{2} + {s i n}^{2} θ}

=_{x - c o s θ =∣ s i n θ ∣ u} \int_{- \frac{c o s θ}{∣ s i n θ ∣}}^{+ \infty} \frac{1}{{s i n}^{2} θ (1 + u^{2})} ∣ s i n θ ∣ d u

= \frac{1}{∣ s i n θ ∣} {[a r c t a n u]}_{- \frac{c o s θ}{∣ s i n θ ∣}}^{+ \infty} = \frac{1}{∣ s i n θ ∣} {\frac{π}{2} + a r c t a n (\frac{c o s θ}{∣ s i n θ ∣})} \Rightarrow

i f s i n θ > 0 \Rightarrow \int_{0}^{\infty} f (x) d x = \frac{1}{s i n θ} {\frac{π}{2} + a r c t a n (\frac{1}{t a n θ})}

= \frac{1}{s i n θ} {π - θ}

i f s i n θ < 0 \Rightarrow \int_{0}^{\infty} f (x d x = - \frac{1}{s i n θ} {\frac{π}{2} - a r c t a n (\frac{1}{t a n θ})}

= - \frac{1}{s i n θ} {\frac{π}{2} - (- \frac{π}{2} - θ)} = - \frac{1}{s i n θ} {π + θ}

Answered by mind is power last updated on 09/Feb/20

x^{2} - 2 c o s (θ) x + 1 = (x - e^{i θ}) (x - e^{- i θ})

f (x) = \frac{1}{2 i s i n (θ)} . \frac{1}{x - e^{i θ}} - \frac{1}{2 i s i n (θ)} . \frac{1}{x - e^{- i θ}}

f (x) = 2 R e (\frac{1}{2 i . s i n (θ) (x - e^{i θ})})

g (x) = \frac{1}{x - e^{i θ}} . \frac{1}{2 i s i n (θ)}

f^{n} = 2 R e (g^{n} (x))

g^{n} (x) = \frac{{(- 1)}^{n} n! .}{{(x - e^{i θ})}^{n + 1}} . \frac{1}{2 i s i n (θ)}

f (x) = 2 R e (\frac{{(- 1)}^{n} n!}{{(x - e^{i θ})}^{n + 1}} . \frac{1}{2 i s i n (θ)})

= {(- 1)}^{n} n! R e {\frac{{(x - e^{- i θ})}^{n + 1}}{{(x^{2} - 2 c o s (θ) x + 1)}^{n + 1}} . \frac{1}{2 i s i n (θ)}}

x - e^{- i θ}

= x - c o s (θ) + i s i n (θ)

= \sqrt{x^{2} - 2 c o s (θ) x + 1} . (\frac{x - c o s (θ)}{\sqrt{x^{2} - 2 x c o s (θ) + 1}} + \frac{i s i n (θ)}{\sqrt{x^{2} - 2 x c o s (θ) + 1}})

= \sqrt{x^{2} - 2 c o s (θ) x + 1} . e^{i (k π + a r c t a n (\frac{s i n (θ)}{x - c o s (θ)}))}, k \in {- 1, 0, 1}

= {(- 1)}^{n} n! \frac{s i n ((n + 1) {k π + a r c t a n (\frac{s i n (θ)}{x - s i n (θ)})})}{{(x^{2} - 2 x c o s (θ) + 1)}^{\frac{n + 1}{2}} s i n (θ)}

m a y b e e e r r o r o f c a l c u l u s

2) (x^{2} - 2 c o s (θ) x + 1) = {(x - c o s (θ))}^{2} + {s i n}^{2} (θ)

= {s i n}^{2} (θ) ({(\frac{x}{s i n (θ)} - c o t (θ))}^{2} + 1)

\int \frac{d x}{(x^{2} - 2 c o s (θ) x + 1)} = \int \frac{d x}{{s i n}^{2} (θ) ({(\frac{x}{s i n (θ)} - c o t (θ))}^{2} + 1)}

\forall θ \in] 0, 2 π [∣ s i n (θ) \neq 0

i f s i n (θ) = 0 \Rightarrow c o s (θ) \in {- 1, 1}

s i n c e f (- x) = x^{2} - 2 x c o s (θ) + 1 \Rightarrow c o s (θ) = 1 i s e n n o u g h

t o f i n d f (x),

\Rightarrow f (x) = \frac{1}{{(x - 1)}^{2}} n o t i n t e g r a b l e o v e r {I R}_{+}

\Rightarrow θ \neq 0

θ = π \Rightarrow f (x) = \frac{1}{{(x + 1)}^{2}}

\Rightarrow \int_{0}^{+ \infty} \frac{d x}{{(1 + x)}^{2}} = 1

\forall θ \in [0, 2 π [- {0, π}

f (x) = \frac{1}{{s i n}^{2} (θ) {{(\frac{x}{s i n (θ)} - c o t (θ))}^{2} + 1}}

\int_{0}^{+ \infty} f (x) d x = \frac{1}{s i n (θ)} {[a r c t a n (\frac{x}{s i n (θ)} - c o t (θ))]}_{0}^{+ \infty}

i f θ \in] 0, π [

f (x) = \frac{1}{s i n (θ)} {\frac{π}{2} + a r c t a n (c o t (θ))}

= \frac{1}{s i n (θ)} {π - θ}

i f θ \in] π, 2 π [

f (x) = \frac{1}{s i n (θ)} {- \frac{π}{2} + a r c t a n (c o t (θ))}

= \frac{1}{s i n (θ)} {- \frac{π}{2} + a r c t a n (c o t (π + (θ - π))}

= \frac{1}{s i n (θ)} {- \frac{π}{2} + a r c t a n (c o t (θ - π))} = \frac{1}{s i n (θ)} {- \frac{π}{2} + a r c t a n (t a n (\frac{3 π}{2} - θ))

= \frac{1}{s i n (θ)} {- \frac{π}{2} + \frac{3 π}{2} - θ} = \frac{1}{s i n (θ)} (π - θ)

\Rightarrow \int_{0}^{+ \infty} f (x) d x = \frac{π - θ}{s i n (θ)}, θ \in [0, 2 π] - {0, π}

θ = π \Rightarrow f (x) = 1 w e c a n s e e t h a t

\lim_{x \to π} \frac{π - θ}{s i n (θ)} = - \frac{1}{c o s (π)} = 1

3 \forall n ⩾ 2 = \lim_{x \to \infty} f^{n - 1} (x) - f^{n - 1} (0)

Commented by mathmax by abdo last updated on 10/Feb/20

t h a n k y o u s i r .

Commented by mind is power last updated on 10/Feb/20

w i t h e p l e a s u r s i r