- #1
Tspyros
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I am giving the following input to Ti-89 Titanium:
limit(S((2 x cos(100 x Pi x t)+2 x sin(10 x Pi x t))2 ,t,-t1,t1)/(2 x t1),t1,inf)
where S is the integral symbol, Pi is π = 3.14, x is the multiplication symbol and inf is the infinity symbol
The TI-89 answer is undef
If I do it by hand it is 4 and Wolfram Mathematica with the following input agrees:
Limit[Integrate[(2Sin[10Pit] + 2Cos[100Pit])2, t, -T, T}]/(2*T), T -> Infinity]
Do you see any errors in my TI89 input?
(I have made the same question in the Ti89 subreddit but it doesn't seem to be used by more than 5 people.)
Thank you!
Some more info:
For the integral alone the Ti89 gives:
-20 π T-10 sin(10 π T) cos(10 π T)+sin(100 π T) cos(100 π T))/((25 π)
whereas mathematica gives:
(400 π T - 10 Sin[20 π T] + Sin[200 π T])/(50 π) with the limit again 4 in Mathematica
If I limit the Ti-89 result with mathematica, it returns -2/5
limit(S((2 x cos(100 x Pi x t)+2 x sin(10 x Pi x t))2 ,t,-t1,t1)/(2 x t1),t1,inf)
where S is the integral symbol, Pi is π = 3.14, x is the multiplication symbol and inf is the infinity symbol
The TI-89 answer is undef
If I do it by hand it is 4 and Wolfram Mathematica with the following input agrees:
Limit[Integrate[(2Sin[10Pit] + 2Cos[100Pit])2, t, -T, T}]/(2*T), T -> Infinity]
Do you see any errors in my TI89 input?
(I have made the same question in the Ti89 subreddit but it doesn't seem to be used by more than 5 people.)
Thank you!
Some more info:
For the integral alone the Ti89 gives:
-20 π T-10 sin(10 π T) cos(10 π T)+sin(100 π T) cos(100 π T))/((25 π)
whereas mathematica gives:
(400 π T - 10 Sin[20 π T] + Sin[200 π T])/(50 π) with the limit again 4 in Mathematica
If I limit the Ti-89 result with mathematica, it returns -2/5