Global Journal of Science Frontier Research, F: Mathematics and Decision Science, Volume 21 Issue 4

Now if we take = 1 0 , we obtain 2 − − 2 (1 − 0 ) = 0 (6.8) and if we take = −1 0 , we obtain 2 (1 − 0 ) 2 + − 1 = 0 (6.9) Thus, the solution to equation (6.8) and equation (6.9) is given by = � ±� 2 −4 2 ( 0 −1) 2 ±� 2 −4 2 ( 0 −1) 2 2 ( 0 −1) (6.10) Where � 2 � 2 > 0 − 1 . Solutions relating to equation (6.9) have been given by virtually all the authors mentioned above for specific value of 0 and . Using equation (6.8), we have that = 1 0 . Now, by equation (6.4) it follows that 0 = = . Thus, 2 − 0 − 2 (1 − 0 ) = 0 = 0 1 0 And 2 (1 − 0 ) 2 + 0 − 1 = 0 = 0 −1 0 If we put 0 = 0 1 0 and 0 = 0 −1 0 , then we have ( ) = 0; ( ) = 0 2 − 0 0 + 2 ( 0 − 1) And ( ) = 0; ( ) = − 2 ( 0 − 1) 0 2 + 0 0 − 1 This reduces to solving for the zero of the functions ( ) and ( ) . For ( ) , this implies that given 0 ≤ 1 < 2 , if we take = � 0 0 − 0 2 − 1 0 −1 and = � 0 0 − 0 2 + 2 0 −1 , then �� 0 0 − 0 2 − 1 0 −1 � = − 1 ≤ 0 and �� 0 0 − 0 2 + 2 0 −1 � = 2 > 0 It follows that �� 0 0 − 0 2 − 1 0 − 1 � �� 0 0 − 0 2 + 2 0 − 1 � = − 1 2 < 0 1 ≠ 0 This implies that there exists a sequence � � =1 ∞ ⊂ ( , ) and at least one point 0 ∈ ( , ) such that the � � =1 ∞ converges to 0 ∈ ( , ) (i.e. ⟶ 0 ⟶ ∞ ) and ( 0 ) = 0 On the Generalized Power Transformation of Left Truncated Normal Distribution © 2021 Global Journals 1 Global Journal of Science Frontier Research Volume XXI Issue IV Year 2021 24 ( F ) Version I N otes