Problems with sin, cos, tan and cot
4 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
why there is a problem with tan, cot, sin and cos of Pi,0, or Pi/2
some of these should give zero but it gives a very small number.
0 Kommentare
Antworten (2)
Ameer Hamza
am 4 Nov. 2020
Bearbeitet: Ameer Hamza
am 4 Nov. 2020
This is caused by the finite-precision of floating-point datatypes and numerical algorithm to calculate the values of these functions. If you want an exact answer for any input to the trigonometric functions, then you need to use symbolic mathematics.
>> cos(pi/2)
ans =
6.1232e-17
>> cos(sym(pi)/2)
ans =
0
4 Kommentare
Stephen23
am 4 Nov. 2020
Bearbeitet: Stephen23
am 4 Nov. 2020
"But such a software should not be working like this."
Excel:
COS(PI()/2)
6.1257422745431E-17
Haskell:
main = do
print(cos(pi/2))
6.123233995736766e-17
Java:
public class Main {
public static void main(String args[]) {
double x = Math.PI / 2;
System.out.println(Math.cos(x));
}
}
6.123233995736766E-17
Julia:
print(cos(pi/2))
6.123233995736766e-17
Lua:
io.write( math.cos(math.pi / 2) )
6.1232339957368e-17
Octave:
>> cos(pi/2)
ans = 6.123031769111886e-17
Python:
import math
math.cos(math.pi/2)
Out[7]: 6.123233995736766e-17
R:
cos(pi/2)
[1] 6.123234e-17
Ruby:
puts Math.cos(Math::PI/2)
6.123233995736766e-17
Scala:
object HelloWorld {
def main(args: Array[String]) {
println(math.cos(math.Pi/2))
}
}
$scala HelloWorld
6.123233995736766E-17
Scilab:
cos(%pi/2)
ans =
6.123D-17
etc. etc.
"A simple calculator does it much better."
Ameer Hamza
am 4 Nov. 2020
The numerical errors in using finite-precision are not limited to MATLAB and are fundamental because of the way they are defined. As Stephen already mentioned, symbolic computation will be much slower than floating-point operations. It is a compromise between speed and accuracy. You can try to reduce it, but never completely avoid it.
Steven Lord
am 4 Nov. 2020
If you're computing sin, cos, etc. of either angles in degrees or angles that a multiple of pi radians there are other ways to compute than the straightforward sin(x*180/pi) or sin(x*pi).
format
A = 0:45:360
sineInDegrees = sind(A)
isSindOf180Exactly0 = sineInDegrees(5) == 0
sineOfMultiplesOfPi = sinpi(A/180)
isSinpiOfPiExactly0 = sineOfMultiplesOfPi(5) == 0
0 Kommentare
Siehe auch
Kategorien
Mehr zu Startup and Shutdown finden Sie in Help Center und File Exchange
Produkte
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!