Explanation Given data: The given function is, f ( t ) = { t 2 , − π < t < π Calculation: The timer period T for the given function is, T = 2 π The expression for ω 0 is given by, ω 0 = 2 π T Substitute 2 π for T in the above expression. ω 0 = 2 π 2 π = 1 rad / s The value of a 0 is given by, a 0 = 1 T ∫ T f ( t ) d t Substitute t 2 for f ( t ) , 2 π for T and range of the function is − π < t < π in the above expression. a 0 = 1 2 π ∫ − π π t 2 d t = 1 2 π [ t 3 3 ] − π π = 1 2 π [ π 3 3 − ( − π ) 3 3 ] = 1 2 π [ π 3 3 + π 3 3 ] Further solve as, a 0 = 1 2 π [ 2 π 3 3 ] = π 2 3 The value of a n is given by, a n = 2 T ∫ T f ( t ) cos ( n ω 0 t ) d t Substitute t 2 for f ( t ) , 2 π for T , 1 for ω 0 and range of the function is − π < t < π in the above expression. a n = 2 2 π ∫ T t 2 cos ( n ( 1 ) t ) d t = 1 π ∫ − π π t 2 cos n t d t = 1 π [ t 2 sin n t n + 2 t cos n t n 2 − 2 sin n t n 3 ] − π π = 1 π [ ( π 2 sin n π n − ( − π ) 2 sin n ( − π ) n ) + ( 2 π cos n π n 2 − 2 ( − π ) cos n ( − π ) n 2 ) − ( 2 sin n π n 3 − 2 sin n ( − π ) n 3 ) ] Solve further as, a n = 1 π [ ( π 2 sin n π n + π 2 sin n π n ) + ( 2 π cos n π n 2 + 2 π cos n π n 2 ) − ( 2 sin n π n 3 + 2 sin n π n 3 ) ] (1) Substitute 0 for sin n π in equation (1). a n = 1 π [ ( 0 n + 0 n ) + ( 2 π cos n π n 2 + 2 π cos n π n 2 ) − ( 0 n 3 + 0 n 3 ) ] = 1 π [ 2 π cos n π n 2 + 2 π cos n π n 2 ] = 1 π [ 4 π cos n π n 2 ] = 4 n 2 [ cos n π ] (2) The value of b n , b n = 2 T ∫ T f ( t ) sin ( n ω 0 t ) d t Substitute t 2 for f ( t ) , 2 π for T , 1 for ω 0 and range of the function is − π < t < π in the above expression

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Most of the incidents studied in the field of Engineering and Science as a function of time. For example, current and voltage are in the alternating current circuit. These occasional activities can be analyzed in the activity components (fundamental freq…

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The type of symmetry function exhibits and the Fourier coefficients for n=1,2 and 3.

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