And finally, we plug the expressions of ak and bk in the required condition and simplify By matching the the expression of a (t)with the synthesis equation, we can conclude that P=∑ak2=∑2=2∑a2-1(: a is real) (:0<la|< P=∑|2=∑向ak2, where N is the largest integer, such that N≤W/o k=-0 2 k=0 1 1--, for any complex B≠0 Plugging in(* N+1 1>R 2-2a 2R R+1 1-a2 2R+(1-)( 2-2 ≥R+1-(1-Ra2( ≥19-0.1a2( pluggin ar2+< 0.05+0.05a2(simplifying a bit) (2N+2)log(a)≤log(0.05+0.05a2) N+1> log(0.05+0.05a2) 2 log(a) (recall that a <1- log(a)< 0 log(0.05+0.05a2) 2 log(a) After choosing an integer n that satisfies the inequality above, W can be chosen such that�� �� �� � � � � � � � � � And finally, we plug the expressions of ak and bk in the required condition and simplify. By matching the the expression of x(t) with the synthesis equation, we can conclude that 0 = � and ak = �|k| 4 Px = |ak| 2 = |�|k| | 2 = 2 �2k − 1 (� � is real) k=−� k=−� k=0 2 = − 1 (� 0 < |�| < 1) 1 − �2 � N Py = |bk| 2 = |ak| 2 , where N is the largest integer, such that N ← W/�0 k=−� k=−N N N = |�|k| | 2 = 2 �2k − 1 (� � is real) k=−N k=0 1 − (�2) N+1 M � � 1 − �M+1 n = 2 − 1 � = , for any complex � ≤= 0 1 − �2 1 − � n=0 Plugging in (⇒): 1 − (�2) N+1 � � 2 2 − 1 → R − 1 1 − �2 1 − �2 2 − 2�2N+2 2R → − R + 1 1 − �2 1 − �2 2 − 2�2N+2 2R + (1 − R)(1 − �2) → 1 − �2 1 − �2 2 − 2�2N+2 → R + 1 − (1 − R)�2 (� 1 − �2 > 0) 2 − 2�2N+2 � → 1.9 − 0.1�2 (plugging in R=0.9) 2N+2 ← 0.05 + 0.05�2 (simplifying a bit) (2N + 2)log(�) ← log(0.05 + 0.05�2 ) log(0.05 + 0.05�2) N + 1 → (recall that � < 1 � log(�) < 0) 2 log(�) log(0.05 + 0.05�2) N → − 1 2 log(�) After choosing an integer N that satisfies the inequality above, W can be chosen such that W → N�0. 2