Задача 6 (Розклад функції в тригонометричний ряд Фур’є)

Розкласти функцію в ряд Фур’є  f(x)=\frac{\pi +x}{2}  на проміжку х ∈ (0; π)

♦  f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}{\left(a_{n}cos\frac{\pi nx}{l}+b_{n}sin\frac{\pi nx}{l} \right)}

 l=\frac{\pi }{2}

 a_{0}=\frac{1}{l}\int_{a}^{b}{f(x)dx};

 a_{n}=\frac{1}{l}\int_{a}^{b}{f(x)\cdot cos\frac{\pi nx}{l}dx};

 b_{n}=\frac{1}{l}\int_{a}^{b}{f(x)sin\frac{\pi nx}{l}dx}.

 a_{0}=\frac{2}{\pi }\int_{0}^{\pi }{\frac{\pi +x}{2}dx}=\frac{2}{\pi }\int_{0}^{\pi }{\left(\frac{\pi }{2}+\frac{x}{2} \right)}dx=

 =\frac{2}{\pi }\left(\frac{\pi }{2}x+\frac{x^{2}}{4} \right)|_{0}^{\pi }=\frac{2}{\pi }\left(\frac{\pi ^{2}}{2}+\frac{\pi ^{2}}{4}-0-0 \right)=

 =\frac{2}{\pi }\cdot \frac{3\pi ^{2}}{4}=\frac{3\pi }{2};

 a_{n}=\frac{2}{\pi }\int_{0}^{\pi }{\frac{\pi +x}{2}\cdot cos\frac{2\pi nx}{\pi }dx}=

 =\frac{2}{\pi }\int_{0}^{\pi }{\frac{\pi +x}{2}\cdot cos\left(2nx \right)dx}=

 =\frac{2}{\pi }\left(\int_{0}^{\pi }{\frac{\pi }{2}cos\left(2nx \right)dx}+\int_{0}^{\pi }{\frac{x}{2}cos\left(2nx \right)dx} \right)=

 =\frac{2}{\pi }\left(\frac{\pi }{2}\cdot \frac{sin\left(2nx \right)}{2n}|_{0}^{\pi }+\frac{1}{2}\int_{0}^{\pi }{x\cdot cos\left(2nx \right)dx} \right)=

 \int_{0}^{\pi }{x\cdot cos\left(2nx \right)dx}=\begin{vmatrix} U=x, &dV=cos\left(2nx \right) \\ dU=dx, & V=\frac{sin\left(2nx \right)}{2n}dx \end{vmatrix}=

 =x\cdot \frac{sin\left(2nx \right)}{2n}|_{0}^{\pi }-\int_{0}^{\pi }{\frac{sin\left(2nx \right)}{2n}dx}=

 =\pi \cdot \frac{sin\left( 2\pi n\right)}{2n}-0+\frac{cos\left(2nx \right)}{4n^{2}}|_{0}^{\pi }=

 =0+\frac{1-1}{4n^{2}}=0

 =\frac{2}{\pi }\left(\frac{\pi }{2}\cdot 0+\frac{1}{2}\cdot 0 \right)=0\Rightarrow a_{n}=0

 b_{n}=\frac{2}{\pi }\int_{0}^{\pi }{\frac{\pi +x}{2}\cdot sin\frac{2\pi nx}{\pi }dx}=

 =\frac{2}{\pi }\int_{0}^{\pi }{\left(\frac{\pi }{2}\cdot sin\left(2nx \right) \right)+\frac{x}{2}sin\left(2nx \right)dx}=

 =\frac{2}{\pi }\left(\frac{\pi }{2}\cdot \frac{-cos\left(2nx \right)}{2n}|_{0}^{\pi } +\frac{1}{2}\int_{0}^{\pi }{x\cdot sin\left(2nx \right)dx}\right)=

 \int_{0}^{\pi }{x\cdot sin\left(2nx \right)dx}=\begin{vmatrix}  U=x, &dV=sin\left(2nx \right)dx \\ dU=dx, & V=-\frac{cos\left(2nx \right)}{2n} \end{vmatrix}=

 =-\frac{x\cdot cos\left(2nx \right)}{2n}|_{0}^{\pi }+\int_{0}^{\pi }{\frac{cos\left(2nx \right)}{2n}dx}=

 =\frac{-\pi \cdot 1}{2n}-0-\frac{1}{4n^{2}sin\left(2nx \right)|_{0}^{\pi }}=

 =-\frac{\pi }{2n}-0-0=-\frac{\pi }{2n}

 =\frac{2}{\pi }\left(\frac{\pi }{2}\left(-\frac{1}{2n}+\frac{1}{2n} \right)+\left(-\frac{\pi }{2n} \right) \right)=

 =\frac{2}{\pi }\left(0-\frac{\pi }{4n} \right)=\frac{2}{\pi }\left(-\frac{\pi }{4n} \right)=-\frac{1}{2n}\Rightarrow

 \Rightarrow b_{n}=-\frac{1}{2n}

Отже,  \frac{\pi +x}{2}=\frac{3\pi }{4}+\sum_{n=1}^{\infty}{-\frac{1}{2n}sin\left(2nx \right)}=

 =\frac{3\pi }{4}-\sum_{n=1}^{\infty}{\frac{sin\left(2nx \right)}{2n}} .

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