# Mathematics-I (3110014)

BE | Semester-1   Winter-2019 | 17-01-2020

## Q2) (c)

#### Find the fourier series of $\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{\pi }-\mathrm{x}}{2}$.

$⟹\mathrm{L}=\mathrm{\pi }$.

${{\mathrm{a}}}_{{0}}=\frac{1}{2\mathrm{\pi }}{\left[\mathrm{\pi x}-\frac{{\mathrm{x}}^{2}}{2}\right]}_{0}^{2\mathrm{\pi }}$

${{\mathrm{a}}}_{{0}}=\frac{1}{2\mathrm{\pi }}\left[\left\{\mathrm{\pi }\left(2\mathrm{\pi }\right)-\frac{{\left(2\mathrm{\pi }\right)}^{2}}{2}\right\}-\left\{\mathrm{\pi }\left(0\right)-\frac{{\left(0\right)}^{2}}{2}\right\}\right]$

${{\mathrm{a}}}_{{0}}=\frac{1}{2\mathrm{\pi }}\left[\left\{2{\mathrm{\pi }}^{2}-\frac{4{\mathrm{\pi }}^{2}}{2}\right\}-\left\{0-0\right\}\right]$

${{\mathrm{a}}}_{{0}}=\frac{1}{2\mathrm{\pi }}\left[\left\{2{\mathrm{\pi }}^{2}-2{\mathrm{\pi }}^{2}\right\}-\left\{0-0\right\}\right]$

$⇒{\mathrm{a}}_{0}=0$

${{\mathrm{a}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}{\left[\left(\mathrm{\pi }-\mathrm{x}\right)\left(\frac{\mathrm{sin}\left(\mathrm{nx}\right)}{\mathrm{n}}\right)-\left(0-1\right)\left(-\frac{\mathrm{cos}\left(\mathrm{nx}\right)}{{\mathrm{n}}^{2}}\right)\right]}_{0}^{2\mathrm{\pi }}$

${{\mathrm{a}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}\left[\left\{\left(\mathrm{\pi }-2\mathrm{\pi }\right)\left(\frac{\mathrm{sin}\left(2\mathrm{n\pi }\right)}{\mathrm{n}}\right)-\left(0-1\right)\left(-\frac{\mathrm{cos}\left(2\mathrm{n\pi }\right)}{{\mathrm{n}}^{2}}\right)\right\}-\left\{\left(\mathrm{\pi }-0\right)\left(\frac{\mathrm{sin}\left(0\right)}{\mathrm{n}}\right)-\left(0-1\right)\left(-\frac{\mathrm{cos}\left(0\right)}{{\mathrm{n}}^{2}}\right)\right\}\right]$

${{\mathrm{a}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}\left[\left\{\left(-\mathrm{\pi }\right)\left(0\right)-\left(-1\right)\left(-\frac{1}{{\mathrm{n}}^{2}}\right)\right\}-\left\{\left(\mathrm{\pi }\right)\left(\frac{0}{\mathrm{n}}\right)-\left(-1\right)\left(-\frac{1}{{\mathrm{n}}^{2}}\right)\right\}\right]$

${{\mathrm{a}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}\left[\left\{0-\frac{1}{{\mathrm{n}}^{2}}\right\}-\left\{0-\frac{1}{{\mathrm{n}}^{2}}\right\}\right]$

${{\mathrm{a}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}\left[-\frac{1}{{\mathrm{n}}^{2}}+\frac{1}{{\mathrm{n}}^{2}}\right]$

$⇒{\mathrm{a}}_{n}=0$

${{\mathrm{b}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}{\left[\left(\mathrm{\pi }-\mathrm{x}\right)\left(-\frac{\mathrm{cos}\left(\mathrm{nx}\right)}{\mathrm{n}}\right)-\left(0-1\right)\left(-\frac{\mathrm{sin}\left(\mathrm{nx}\right)}{{\mathrm{n}}^{2}}\right)\right]}_{0}^{2\mathrm{\pi }}$

${{\mathrm{b}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}\left[\left\{\left(\mathrm{\pi }-2\mathrm{\pi }\right)\left(-\frac{\mathrm{cos}\left(2\mathrm{n\pi }\right)}{\mathrm{n}}\right)-\left(0-1\right)\left(-\frac{\mathrm{sin}\left(2\mathrm{n\pi }\right)}{{\mathrm{n}}^{2}}\right)\right\}-\left\{\left(\mathrm{\pi }-0\right)\left(-\frac{\mathrm{cos}\left(0\right)}{\mathrm{n}}\right)-\left(0-1\right)\left(-\frac{\mathrm{sin}\left(0\right)}{{\mathrm{n}}^{2}}\right)\right\}\right]$

${{\mathrm{b}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}\left[\left\{\left(-\mathrm{\pi }\right)\left(-\frac{1}{\mathrm{n}}\right)-\left(-1\right)\left(0\right)\right\}-\left\{\left(\mathrm{\pi }\right)\left(-\frac{1}{\mathrm{n}}\right)-\left(-1\right)\left(0\right)\right\}\right]$

${{\mathrm{b}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}\left[\left\{\frac{\mathrm{\pi }}{\mathrm{n}}+0\right\}-\left\{-\frac{\mathrm{\pi }}{\mathrm{n}}+0\right\}\right]$

${{\mathrm{b}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}\left[\frac{\mathrm{\pi }}{\mathrm{n}}+\frac{\mathrm{\pi }}{\mathrm{n}}\right]$

${{\mathrm{b}}}_{{\mathrm{n}}}=\frac{1}{2\mathrm{\pi }}\left[\frac{2\mathrm{\pi }}{\mathrm{n}}\right]$

$⇒{\mathrm{b}}_{\mathrm{n}}=\frac{1}{\mathrm{n}}$

$⇒\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\infty }\frac{\mathrm{sin}\left(\mathrm{nx}\right)}{\mathrm{n}}$