Find the Fourier series of f on the given interval. f(x)=\left\{\begin{matrix} 0,\;\;\; -\pi...


Find the Fourier series of {eq}f {/eq} on the given interval.

{eq}f(x)=\left\{\begin{matrix} 0,\;\;\; -\pi < x<0 & \\ sin\;x, 0 \leq x <\pi & \end{matrix}\right. {/eq}

{eq}f(x) = \;\rule{20mm}{1pt} \;\;+ \sum_{N=2}^{\infty} \left [ (\rule{20mm}{1pt} \right ) \cos \;nx + (\rule{20mm}{1pt}) \sin nx] {/eq}

Give the number {eq}F_0 {/eq} to which the Fourier series converges at a point {eq}x_0 {/eq} of discontinuity of {eq}f {/eq}.

{eq}(x_0, F_0) = \;\rule{20mm}{.5pt} {/eq}

Fourier Series

To compute the Fourier series of function f(x) on the interval {eq}[-L,L] {/eq}, we apply the general definition of Fourier series:

{eq}f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos(\frac{n\pi x}{L})+b_n\sin(\frac{n\pi x}{L})) {/eq}


{eq}a_0=\frac{1}{L}\int_{-L}^{L}f(x)dx {/eq}

{eq}a_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos(\frac{n\pi x}{L}) {/eq}

{eq}b_n=\frac{1}{L}\int_{-L}^{L}f(x)\sin(\frac{n\pi x}{L})dx {/eq}

Answer and Explanation:

a) We have {eq}f(x)=\left\{\begin{matrix} 0,\;\;\; -\pi < x<0 & \\ sin\;x, 0 \leq x <\pi & \end{matrix}\right. {/eq}

Step 1) Now ...

See full answer below.

Become a member to unlock this answer! Create your account

View this answer

Learn more about this topic:

Summation Notation and Mathematical Series

from Math 101: College Algebra

Chapter 12 / Lesson 4

Related to this Question

Explore our homework questions and answers library

福建福彩网 天水市 永康市 成都市 商洛市 大庆市 镇江市 临夏市 阜新市 巴中市 萍乡市 崇州市 邓州市 平度市 河津市 台中市 衡水市 明光市 凤城市 吉林省 石首市 龙海市 黄石市 叶城市 都匀市 武穴市 朝阳市 青岛市 凤城市 葫芦岛市 仙桃市 合肥市 孝感市 邢台市 兴城市 平度市 利川市 洮南市 信阳市 常州市 宁国市 南阳市 徐州市 北宁市 邢台市 鹿泉市 池州市 北宁市 台中市 华阴市 延吉市 铁力市 兴城市 淮安市 汉川市 东阳市 焦作市 西安市 佛山市 潍坊市 甘肃省