Mathematics-I (3110014)

$\mathrm{Here}, \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{\pi }-\mathrm{x}}{2}$

$\mathrm{Also},(\mathrm{c},\mathrm{c}+2\mathrm{L})= (0,2\mathrm{\pi })$

$⟹\mathrm{c}=0 \& \mathrm{c}+2\mathrm{L}=2\mathrm{\pi }$

$⟹ 2\mathrm{L}=2\mathrm{\pi }$

$⟹\mathrm{L}=\mathrm{\pi }$.

${\mathrm{a}}_0=\frac{1}{\mathrm{L}}{\int }_{\mathrm{c}}^{\mathrm{c}+2\mathrm{L}}\mathrm{f}\left(\mathrm{x}\right) d\mathrm{x}$

${\mathrm{a}}_0=\frac{1}{\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}\left(\frac{\mathrm{\pi }-\mathrm{x}}{2}\right) d\mathrm{x}$

${\mathrm{a}}_0=\frac{1}{2\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}\left(\mathrm{\pi }-\mathrm{x}\right) d\mathrm{x}$

${\mathrm{a}}_0=\frac{1}{2\mathrm{\pi }}{\left[\mathrm{\pi x}-\frac{{\mathrm{x}}^2}{2}\right]}_0^{2\mathrm{\pi }}$

${\mathrm{a}}_0=\frac{1}{2\mathrm{\pi }}\left[\left\{\mathrm{\pi }\left(2\mathrm{\pi }\right)-\frac{{\left(2\mathrm{\pi }\right)}^2}{2}\right\}-\left\{\mathrm{\pi }\left(0\right)-\frac{{\left(0\right)}^2}{2}\right\}\right]$

${\mathrm{a}}_0=\frac{1}{2\mathrm{\pi }}\left[\left\{2{\mathrm{\pi }}^2-\frac{4{\mathrm{\pi }}^2}{2}\right\}-\left\{0-0\right\}\right]$

${\mathrm{a}}_0=\frac{1}{2\mathrm{\pi }}\left[\left\{2{\mathrm{\pi }}^2-2{\mathrm{\pi }}^2\right\}-\left\{0-0\right\}\right]$

$\Rightarrow {\mathrm{a}}_0=0$

${\mathrm{a}}_{\mathrm{n}}=\frac{1}{\mathrm{L}}{\int }_{\mathrm{c}}^{\mathrm{c}+2\mathrm{L}}\mathrm{f}\left(\mathrm{x}\right)\cos \left(\frac{\mathrm{n\pi x}}{\mathrm{L}}\right) d\mathrm{x}$

${\mathrm{a}}_{\mathrm{n}}=\frac{1}{\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}\left(\frac{\mathrm{\pi }-\mathrm{x}}{2}\right) · \cos \left(\mathrm{nx}\right) d\mathrm{x}$

${\mathrm{a}}_{\mathrm{n}}=\frac{1}{2\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}\left(\mathrm{\pi }-\mathrm{x}\right) · \cos \left(\mathrm{nx}\right) d\mathrm{x}$

${\mathrm{a}}_{\mathrm{n}}=\frac{1}{2\mathrm{\pi }}{\left[\left(\mathrm{\pi }-\mathrm{x}\right)\left(\frac{\sin \left(\mathrm{nx}\right)}{\mathrm{n}}\right)-\left(0-1\right)\left(-\frac{\cos \left(\mathrm{nx}\right)}{{\mathrm{n}}^2}\right)\right]}_0^{2\mathrm{\pi }}$

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

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

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

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

$\Rightarrow {\mathrm{a}}_{\mathrm{n}}=0$

${\mathrm{b}}_{\mathrm{n}}=\frac{1}{\mathrm{L}}{\int }_{\mathrm{c}}^{\mathrm{c}+2\mathrm{L}}\mathrm{f}\left(\mathrm{x}\right)\sin \left(\frac{\mathrm{n\pi x}}{\mathrm{L}}\right) d\mathrm{x}$

${\mathrm{b}}_{\mathrm{n}}=\frac{1}{\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}\left(\frac{\mathrm{\pi }-\mathrm{x}}{2}\right) · \sin \left(\mathrm{nx}\right) d\mathrm{x}$

${\mathrm{a}}_{\mathrm{n}}=\frac{1}{2\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}\left(\mathrm{\pi }-\mathrm{x}\right) · \sin \left(\mathrm{nx}\right) d\mathrm{x}$

${\mathrm{b}}_{\mathrm{n}}=\frac{1}{2\mathrm{\pi }}{\left[\left(\mathrm{\pi }-\mathrm{x}\right)\left(-\frac{\cos \left(\mathrm{nx}\right)}{\mathrm{n}}\right)-\left(0-1\right)\left(-\frac{\sin \left(\mathrm{nx}\right)}{{\mathrm{n}}^2}\right)\right]}_0^{2\mathrm{\pi }}$

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

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

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

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

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

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

$\mathrm{Now}, \mathrm{Fourier} \mathrm{series} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is},$

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}_0+\sum _{\mathrm{n}=1}^{\infty }\left[{\mathrm{a}}_{\mathrm{n}} \cos \left(\frac{\mathrm{n\pi x}}{\mathrm{L}}\right) + {\mathrm{b}}_{\mathrm{n}} \sin \left(\frac{\mathrm{n\pi x}}{\mathrm{L}}\right) \right]$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=0+\sum _{\mathrm{n}=1}^{\infty }\left[\left(0\right) \cos \left(\mathrm{nx}\right) +\left(\frac{1}{\mathrm{n}}\right) \sin \left(\mathrm{nx}\right)\right]$

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