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Question Number 139708 by qaz last updated on 30/Apr/21

P r o v e : 1 + 2 \underset{n = 1}{\sum^{\infty}} a^{n} \cos (n x) = \frac{1 - a^{2}}{1 - 2 a \cos x + a^{2}}, (∣ a ∣< 1)

A n d c a l c u l a t e :: \int_{0}^{π} \frac{x^{2}}{1 + 8 \sin^{2} x} d x = \frac{5 π^{3}}{36} - \frac{π}{6} {l n}^{2} 2

Answered by Dwaipayan Shikari last updated on 30/Apr/21

1 + \underset{n = 1}{\sum^{\infty}} a^{n} e^{i n x} + a^{n} e^{- i n x} = 1 + \underset{n = 1}{\sum^{\infty}} {({a e}^{i x})}^{n} + {({a e}^{- i x})}^{n}

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

= 1 + \frac{2 a c o s (x) - 2 a^{2}}{1 - 2 a c o s x + a^{2}} = \frac{1 - a^{2}}{1 - 2 a c o s (x) + a^{2}}

Commented by qaz last updated on 30/Apr/21

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

Answered by mindispower last updated on 30/Apr/21

::

= \int_{0}^{π} \frac{x^{2}}{1 + 8 (1 - {c o s}^{2} (x))} d x

= \int_{0}^{π} \frac{x^{2}}{(3 - 2 \sqrt{2} c o s (x)) (3 + 2 \sqrt{2} c o s (x))}

= \frac{1}{6} (\int_{0}^{π} \frac{x^{2}}{3 - 2 \sqrt{2} c o s (x)} d x + \int_{0}^{π} \frac{x^{2}}{3 + 2 \sqrt{2} c o s (x)} d x)

= \frac{1}{6} (A + B)

6 A = \int_{0}^{π} \frac{1 - {(\frac{1}{\sqrt{2}})}^{2}}{1 + \frac{1}{{(\sqrt{2})}^{2}} - 2 . \frac{1}{\sqrt{2}} c o s (x)} x^{2} d x

6 B = \int_{0}^{π} \frac{1 - {(- \frac{1}{\sqrt{2}})}^{2}}{1 + {(- \frac{1}{\sqrt{2}})}^{2} + 2 . - \frac{1}{\sqrt{2}} c o s (x)} x^{2} d x

A = \frac{1}{6} \int_{0}^{π} (x^{2} + 2 \sum_{n ⩾ 1} {(\frac{1}{\sqrt{2}})}^{n} x^{2} c o s (n x))

B = \frac{1}{6} \int_{0}^{π} (x^{2} + 2 \sum_{n ⩾ 1} {(- \sqrt{2})}^{n} x^{2} c o s (n x))

\int_{0}^{π} x^{2} c o s (n x) d x = - \frac{2}{n} \int_{0}^{π} x s i n (n x) d x

= \frac{2}{n^{2}} [π {(- 1)}^{n}]

A = \frac{π^{3}}{18} + \frac{1}{3} \sum_{n ⩾ 1} {(\frac{1}{\sqrt{2}})}^{n} . \frac{2 π}{n^{2}} ({(- 1)}^{n})

B = \frac{π^{3}}{18} + \frac{1}{3} \sum_{n ⩾ 1} {(- \frac{1}{\sqrt{2}})}^{n} . \frac{2}{n^{2}} ({(- 1)}^{n})

A + B = \frac{π^{3}}{9} + \frac{1}{3} \sum_{n ⩾ 1} {(\frac{1}{2})}^{n} . \frac{π}{n^{2}}

{l i}_{2} (z) = \sum_{n ⩾ 1} \frac{z^{n}}{n^{2}}; \forall ∣ z ∣⩽ 1

= \frac{π^{3}}{9} + \frac{π}{3} {L i}_{2} (\frac{1}{2}), {l i}_{2} (\frac{1}{2}) = \frac{π^{2}}{12} - \frac{{l n}^{2} (2)}{2}

= \frac{π^{3}}{9} + \frac{π}{3} (\frac{π^{2}}{12} - \frac{{l n}^{2} (2)}{2}) = \frac{5 π^{3}}{36} - \frac{π {l n}^{2} (2)}{6}

Commented by mnjuly1970 last updated on 30/Apr/21

b r a v o b r a v o s i r p o w e r \dots

Commented by mindispower last updated on 01/May/21

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

Answered by mathmax by abdo last updated on 01/May/21

\sum_{n = 1}^{\infty} a^{n} \cos (nx) = Re (\sum_{n = 0}^{\infty} a^{n} e^{inx} - 1)

\sum_{n = 0}^{\infty} {({ae}^{ix})}^{n} = \frac{1}{1 - {ae}^{ix}} = \frac{1}{1 - acosx - iasinx} = \frac{1 - acosx + iasinx}{{(1 - acosx)}^{2} + a^{2} \sin^{2} x}

\Rightarrow \sum_{n = 1}^{\infty} a^{n} \cos (nx) = \frac{1 - acosx}{1 - 2 acosx + a^{2}} - 1 \Rightarrow

1 + 2 \sum_{n = 1}^{\infty} a^{n} \cos (nx) = 1 + \frac{2 - 2 acosx}{1 - 2 acosx + a^{2}} - 2

= \frac{2 - 2 acosx}{1 - 2 acosx + a^{2}} - 1 = \frac{2 - 2 acosx - 1 + 2 acosx - a^{2}}{a^{2} - 2 acosx + 1} = \frac{1 - a^{2}}{a^{2} - 2 acosx + 1}