To Prove: (-i)! = (i/PI) & (i)! = i*PI;
Proof: If (-i)! = (i/PI), Then, (i/i)*(i/PI) = (-1)/(i*PI) = (-i)!;
If (-i)! = (-1)/(i*PI), &, (i)! = i*PI; then, -- ((-i)!)/(i*PI) = -1 & ((i)!/(i*PI) = 1;
If e^(i*PI) = e^((i)!) = (((i)!)/(i*PI))*(((-i)!)/(i*PI)) = 1*(-1) = -1;
Consequently, -- ((-i)!/(i*PI) MUST = -1 & ((i)!/(i*PI) = 1;
So -- If ((-i)! = ((-1)/(i*PI), Then,
((-i)!/(i*PI)) = (((-1/(i*PI)/(i*PI) = (-i*PI)/(i*PI) = -1; & ;
(((i)!/(i*PI)) = ((i*PI)/(i*PI)) = 1.
Therefore, -- (-i)! = (i/(PI)), &, (i)! = i*PI.
To Be Continued! Sincerely, Samuel David Roach.
No comments:
Post a Comment