Question

find the finite cosine transform of (1-x/π)^2

Answer 1

SOLUTION

TO DETERMINE

The finite cosine transform of

$\displaystyle\sf{ { \bigg( 1 - \frac{x}{\pi} \bigg)}^{2} }$

EVALUATION

Here the given function is

$\displaystyle\sf{ f(x) = { \bigg( 1 - \frac{x}{\pi} \bigg)}^{2}}$

Hence the required finite Fourier cosine transform is

$\displaystyle \sf{F_c(n) = \int\limits_{0}^{\pi} f(x) \cos \: \frac{n\pi x}{\pi} \, dx }$

$\displaystyle \sf{ = \int\limits_{0}^{\pi} f(x) \cos \: nx \, dx }$

$\displaystyle \sf{ = \int\limits_{0}^{\pi} { \bigg( 1 - \frac{x}{\pi} \bigg)}^{2} \cos \: nx \, dx }$

$\displaystyle \sf{ = \frac{1}{ {\pi}^{2} } \int\limits_{0}^{\pi} { \big( \pi - x \big)}^{2} \cos \: nx \, dx }$

$\displaystyle \sf{ = \frac{1}{ {\pi}^{2} } \int\limits_{0}^{\pi} { \big( \pi - \pi + x \big)}^{2} \cos \: n(\pi - x) \, dx }$

$\displaystyle \sf{ = \frac{1}{ {\pi}^{2} } \int\limits_{0}^{\pi} {x}^{2} \cos \: (n\pi - nx) \, dx }$

$\displaystyle \sf{ = \frac{1}{ {\pi}^{2} } \int\limits_{0}^{\pi} {x}^{2} {( -1 )}^{n} \cos \: nx \, dx }$

$\displaystyle \sf{ = \frac{{( -1 )}^{n}}{ {\pi}^{2} } \int\limits_{0}^{\pi} {x}^{2} \cos \: nx \, dx }$

$\displaystyle \sf{ = \frac{{( -1 )}^{n}}{ {\pi}^{2} } \bigg[ \frac{ {x}^{2} \sin nx }{n} \bigg]_{0}^{\pi} - \frac{{( -1 )}^{n}}{n {\pi}^{2} } \int\limits_{0}^{\pi} 2x \sin \: nx \, dx }$

$\displaystyle \sf{ = \frac{{( -1 )}^{n}}{ {\pi}^{2} } \bigg[ \frac{ {\pi}^{2} \sin n\pi }{n} - 0 \bigg]_{0}^{\pi} - \frac{{( -1 )}^{n}}{n {\pi}^{2} } \int\limits_{0}^{\pi} 2x \sin \: nx \, dx }$

$\displaystyle \sf{ = - \frac{{( -1 )}^{n}}{n {\pi}^{2} } \int\limits_{0}^{\pi} 2x \sin \: nx \, dx }$

$\displaystyle \sf{ = - \frac{{( -1 )}^{n}}{n {\pi}^{2} } \int\limits_{0}^{\pi} 2x \sin \: nx \, dx }$

$\displaystyle \sf{= - \frac{{( -1 )}^{n}}{n {\pi}^{2} } \bigg[ - \frac{ 2x \cos nx }{n} \bigg]_{0}^{\pi} - \frac{{( -1 )}^{n}}{ {n}^{2} {\pi}^{2} } \int\limits_{0}^{\pi} \cos \: nx \, dx}$

$\displaystyle \sf{= - \frac{{( -1 )}^{n}}{n {\pi}^{2} } \bigg[ - \frac{ 2\pi \cos n\pi }{n} + 0\bigg]_{0}^{\pi} - \frac{{( -1 )}^{n}}{ {n}^{2} {\pi}^{2} } \bigg[ \frac{ \sin nx }{n} \bigg]_{0}^{\pi} }$

$\displaystyle \sf{= - \frac{{( -1 )}^{n}}{n {\pi}^{2} } \bigg[ - \frac{ 2\pi . {( - 1)}^{n} }{n} \bigg] - \frac{{( -1 )}^{n}}{ {n}^{2} {\pi}^{2} } \bigg[ \frac{ \sin n\pi }{n} - 0 \bigg] }$

$\displaystyle \sf{= \frac{{( -1 )}^{2n} \times 2\pi}{ {n}^{2} {\pi}^{2} } }$

$\displaystyle \sf{= \frac{ 2\pi}{ {n}^{2} {\pi}^{2} } }$

$\displaystyle \sf{= \frac{ 2}{ {n}^{2} {\pi}^{} } }$

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