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Question Number 101500 by mathmax by abdo last updated on 03/Jul/20

let f (x) = cosx . \cos (2 x) . \cos (3 x)

1) calculate f^{(n)} (x) and f^{(n)} (0)

2) developp f at integr serie

3 . calculate \int_{0}^{\frac{π}{2}} f (x) dx

Commented by bemath last updated on 03/Jul/20

(3) \underset{0}{\int^{π / 2}} f (x) dx = [\frac{1}{4} x + \frac{1}{24} \sin 6 x +

{\frac{1}{8} \sin 2 x + \frac{1}{16} \sin 4 x]}_{0}^{π / 2}

= \frac{π}{8}

\cos (3 x) \cos (x) \cos (2 x) =

\frac{1}{2} (\cos 4 x + \cos 2 x) \cos 2 x =

\frac{1}{2} (\cos 4 xcos 2 x + \cos^{2} 2 x) =

\frac{1}{2} (\frac{1}{2} (\cos 6 x + \cos 2 x) + \frac{1}{2} + \frac{1}{2} \cos 4 x) =

\frac{1}{4} \cos 6 x + \frac{1}{4} \cos 2 x + \frac{1}{4} + \frac{1}{4} \cos 4 x

Answered by Ramajunan last updated on 03/Jul/20

Nice question

Expressing f (x) as the sum of cosines angles

cosxcos (2 x) \cos (3 x)

\Rightarrow \frac{1}{4} (\cos (5 x) + cosx) cosx

\Rightarrow \frac{1}{2} \cos (5 x) cosx + \frac{1}{2} \cos^{2} x

\Rightarrow f (x) = \frac{1}{4} [\cos (6 x) + \cos (4 x) + \cos (2 x) + 1]

f^{^{'}} (x) = \frac{1}{4} [- 6 \sin (6 x) - 4 \sin (4 x) - 2 \sin (2 x)]

f^{^{″}} (x) = \frac{1}{4} (- 6^{2} \cos (6 x) - 4^{2} \cos (4 x) - 2^{2} \cos (2 x)]

f^{(3)} (x) = \frac{1}{4} (6^{3} \sin (6 x) + 4^{3} \sin (x) + 2^{3} \sin (2 x)

f^{(n)} (x) = \frac{1}{4} (6^{n} \cos (6 x + \frac{n π}{2}) + 4^{n} \cos (4 x + \frac{n π}{2}) + 2^{n} c o s (2 x + \frac{n π}{2})] for \lor n

f^{n} (0) = \frac{1}{4} (6^{n} \cos (\frac{n π}{2}) + 4^{n} \cos (\frac{n π}{2}) + 2^{n} \cos (\frac{n π}{2}))

= \frac{1}{4} (6^{n} + 4^{n} + 2^{n}) \cos (\frac{n π}{2})

2) {D o n}^{'} t understand number 2

Number 3

\int_{0}^{\frac{π}{2}} c o s x c o s (2 x) c o s (3 x) d x

\int_{0}^{\frac{π}{2}} f (x) = \frac{1}{4} \int_{0}^{\frac{π}{2}} [\cos (6 x) + \cos (4 x) + \cos (2 x) + 1]

{[\frac{1}{4} (\frac{\sin (6 x)}{6} + \frac{\cos (4 x)}{4} + \frac{\sin (2 x)}{2} + x)]}_{0}^{\frac{π}{2}}

Commented by abdomathmax last updated on 03/Jul/20

thanks sir .

Answered by mathmax by abdo last updated on 04/Jul/20

1) let transform f (x) to sum we have

f (x) = cosx \cos (2 x) . \cos (3 x) = \frac{1}{2} {\cos (3 x) + cosx} \cos (3 x)

= \frac{1}{2} \cos^{2} (3 x) + \frac{1}{2} cosx \cos (3 x)

= \frac{1}{4} (1 + \cos (6 x)) + \frac{1}{4} {\cos (4 x) + \cos (2 x)}

= \frac{1}{4} (1 + \cos (2 x) + \cos (4 x) + \cos (6 x)} so for n ⩾ 1 we get

f^{(n)} (x) = \frac{1}{4} {\cos^{(n)} (2 x) + \cos^{(n)} (4 x) + \cos^{(n)} (6 x)}

= \frac{1}{4} {{(\frac{e^{2 ix} + e^{- 2 ix}}{2})}^{(n)} + {(\frac{e^{4 ix} + e^{- 4 ix}}{2})}^{(n)} + {(\frac{e^{6 ix} + e^{- 6 ix}}{2})}^{(n)}}

= \frac{1}{8} {{(2 i)}^{n} e^{2 ix} + {(- 2 i)}^{n} e^{- 2 ix} + {(4 i)}^{n} e^{4 ix} + {(- 4 i)}^{n} e^{- 4 ix} + {(6 i)}^{n} e^{6 ix} + {(- 6 i)}^{n} e^{- 6 ix}}

\Rightarrow f^{(n)} (0) = \frac{1}{8} {{(2 i)}^{n} + {(- 2 i)}^{n} + {(4 i)}^{n} + {(- 4 i)}^{n} + {(6 i)}^{n} + {(- 6 i)}^{n}}

= \frac{1}{8} {2^{n} (i^{n} + {(- i)}^{n}) + 4^{n} (i^{n} + {(- i)}^{n}) + 6^{n} (i^{n} + {(- i)}^{n})}

= \frac{1}{8} {2^{n} (2 \cos (\frac{n π}{2})) + 4^{n} (2 \cos (\frac{n π}{2})) + 6^{n} (2 \cos (\frac{n π}{2})}

f^{(n)} (0) = \frac{1}{4} \cos (\frac{n π}{2}) {2^{n} + 4^{n} + 6^{n}}

Commented by mathmax by abdo last updated on 04/Jul/20

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}

f (x) = 1 + \frac{1}{4} \sum_{n = 1}^{\infty} \frac{2^{n} + 4^{n} + 6^{n}}{n!} \cos (\frac{n π}{2}) x^{n}

Commented by mathmax by abdo last updated on 04/Jul/20

3) we have f (x) = \frac{1}{4} {1 + \cos (2 x) + \cos (4 x) + \cos (6 x)} \Rightarrow

\int_{0}^{\frac{π}{2}} f (x) dx = \frac{π}{8} + \frac{1}{4} \int_{0}^{\frac{π}{2}} \cos (2 x) dx + \frac{1}{4} \int_{0}^{\frac{π}{2}} \cos (4 x) dx + \frac{1}{4} \int_{0}^{\frac{π}{2}} \cos (6 x) dx

= \frac{π}{8} + \frac{1}{8} {[\sin (2 x)]}_{0}^{\frac{π}{2}} + \frac{1}{16} {[\sin (4 x)]}_{0}^{\frac{π}{2}} + \frac{1}{24} {[\sin (6 x)]}_{0}^{\frac{π}{2}}

= \frac{π}{8} + 0 + 0 + 0 = \frac{π}{8} \Rightarrow \int_{0}^{\frac{π}{2}} f (x) dx = \frac{π}{8}