hi all,
I input \( \int_0^{\pi}{\sqrt{1-cos(t)^{2}} dt} \) and I get 0, after this message: "searching int of sqrt(-taylorx2)*abs(-taylorx2+1)/(2*taylorx2^2) where taylorx2 is on the unit circle, using residues...
But the integral should be equal to \int(\sin(t),0,π), then 2...
What am I missing?
Thank you for help!
salvo
EDIT: I used "exact" in CAS...
What the difference for normal and "rigorous" use?, please...
Inside Xcas int(sqrt(1-cos(t)^2),t,0,pi) returns 2, I'll check on the emulator.
(02-04-2015 06:57 AM)parisse Wrote: [ -> ]Inside Xcas int(sqrt(1-cos(t)^2),t,0,pi) returns 2, I'll check on the emulator.
here in Home it give 2, but with "T" (upper case).
In CAS with "exact result" set (and with "t") it give 0, without the setting it give 2. (approx result)...
Tried with my latest version of the emulator, I get 2 with default CAS settings (hence exact). Perhaps a good consequence of another bug fix since the last release.
(02-04-2015 09:31 AM)parisse Wrote: [ -> ]Tried with my latest version of the emulator, I get 2 with default CAS settings (hence exact). Perhaps a good consequence of another bug fix since the last release.
then good!
In 6975 there's still the bug...
Let's hope a new release soon!
How do you use the residue function or the RESIDUE function
seems this function is a ghost function for the HP Prime..
complex mode? program only?...little help here...my goodness..
The help menu is not a help for this...
J
ok, if you use residue or RESIDUE..make sure that the variable is not used by
the home directory..or something like this..or you will get error messages..
I have two primes and found the error by comparing calculations with the residue function...pretty neat that it works..
Also, the residue function works with both residue(func,z=x) and
residue(func,z,x)...Im pretty sure..the help menu as std is limited..
J