where is this problem discontinuous?
2007-09-23 14:30:46 · 5 answers · asked by Connor O 1 in Science & Mathematics ➔ Mathematics
yes, the graph is periodic and discontinuous...
asymptotic at all odd multiples of 1/2 ... 1/2,3/2,5/2,7/2,... ; -1/2,-3/2,-5/2,-7/2,...
it is continuous from -1/2 to 1/2. the graph is then repeated (thus periodic)... §
2007-09-23 14:45:02 · answer #1 · answered by Alam Ko Iyan 7 · 2⤊ 0⤋
If pi was a function, it's be discontinuous everywhere. Just look at its graph.
2007-09-23 14:40:28 · answer #2 · answered by Anonymous · 0⤊ 0⤋
if it is 1/2 tan(pi X)
piX= (2n+1)pi/2 so X= (2n+1)/2 n integer
If it is tan(piX/2)
piX/2 = (2n+1)pi/2 so X= (2n+1)
2007-09-23 14:41:04 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋
pi is not a function.
What you have written is nonsense.
2007-09-23 14:36:09 · answer #4 · answered by AnswerMan 4 · 0⤊ 2⤋
what is that?
2007-09-23 14:35:09 · answer #5 · answered by Anonymous · 0⤊ 1⤋