Math, asked by faizanahmed21092000, 1 month ago

Find the Fourier cosine series of the function

ƒ( in ) x) = π − x (0, π )​

Answers

Answered by jnkook9492
2

Answer:

(1)

The Fourier series of f1(x) is called the Fourier Sine series of the function f(x), and is given by

\begin{displaymath}f(x) \sim \sum_{n=1}^{\infty} b_n\sin(nx).\end{displaymath}

where

\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin(nx)dx.\end{displaymath}

(2)

The Fourier series of f2(x) is called the Fourier Cosine series of the function f(x), and is given by

\begin{displaymath}f(x) \sim a_0 + \sum_{n=1}^{\infty} a_n\cos(nx).\end{displaymath}

where

\begin{displaymath}a_0 = \frac{1}{\pi} \int_{0}^{\pi} f(x) \cos(nx)dx,\;\;\mbox{...

...}{\pi} \int_{0}^{\pi} f(x) \cos(nx)dx\;\;\mbox{for $n \geq 1$}.\end{displaymath}

Example. Find the Fourier Cosine series of f(x) = x for $ x \in [0,\pi]$.

Answer. We have

\begin{displaymath}a_0 = \frac{1}{\pi} \int_{0}^{\pi} xdx = \frac{\pi}{2},\end{displaymath}

and

\begin{displaymath}a_n = \frac{2}{\pi} \int_{0}^{\pi} x\cos(nx)dx =

\frac{2}{n^2\pi} (\cos(n\pi) - 1) = \frac{2}{n^2\pi} ((-1)^n - 1).\end{displaymath}

Therefore, we have

\begin{displaymath}f(x) \sim \frac{\pi}{2} - \frac{4}{\pi}\left(\cos(x) +

\frac{1}{9}\cos(3x) + \frac{1}{25}\cos(5x)+\ldots\right).\end{displaymath}

Example. Find the Fourier Sine series of the function f(x) = 1 for $ x \in [0,\pi]$.

Answer. We have

\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi} \sin(nx)dx =

\frac{2}{n\pi}(-\cos(n\pi) + 1) = \frac{2}{n\pi}(1-(-1)^n).\end{displaymath}

Hence

\begin{displaymath}f(x) \sim \frac{4}{\pi}\left(\sin(x) + \frac{1}{3}\sin(3x) +

\frac{1}{5}\sin(5x)...\right).\end{displaymath}

Example. Find the Fourier Sine series of the function $f(x) = \cos(x)$ for $ x \in [0,\pi]$.

Answer. We have

\begin{displaymath}b_n = \frac{2}{\pi} \int_{0}^{\pi}\cos(x)\sin(nx)dx\end{displaymath}

which gives b1 = 0 and for n > 1, we obtain

\begin{displaymath}b_n = \frac{2n}{\pi}\left(\frac{1 + (-1)^n}{n^2 -1}\right).\end{displaymath}

Hence

\begin{displaymath}\cos(x) \sim \frac{8}{\pi} \sum_{n=1}^{\infty}

\frac{n\sin(2nx)}{4n^2-1}.\end{displaymath}

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