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multiplying vectors (easy one!)

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Paul Rogers
Paul Rogers el 11 de Jun. de 2020
Cerrada: MATLAB Answer Bot el 20 de Ag. de 2021
Sorry guys, an easy question:
how do I evaluete this function in the interval
x=[-0.2,0.8]
the function is f = 2.2785*T_0(x) - 0.2142*T_1(x) - 0.3595*T_2(x) - 0.1793*T_3(x) + 0.1810*T_4(x)
where T_k(x)=cos*(k*cos^-1(x))
c =
2.2785
-0.2142
-0.3595
-0.1793
0.1810

Respuestas (1)

KSSV
KSSV el 11 de Jun. de 2020
m = 100 ;
x = linspace(-0.2,0.8,m) ;
Evaluate the function for every value of x. Read about element by element operations.
  3 comentarios
KSSV
KSSV el 11 de Jun. de 2020
Yes you can use that too.....two ways
  1. Fix the number of discretozations use linspace
  2. Fix the step size and ise : .
Paul Rogers
Paul Rogers el 11 de Jun. de 2020
thanks, but how do I write the rest?
f(x=0)=2.2785*T_0(x) - 0.2142*T_1(x) - 0.3595*T_2(x) - 0.1793*T_3(x) + 0.1810*T_4(x)
where for x=0 (i.e)
T_0(x=0)=cos*(0*cos^-1(x))
T_1(x=0)=cos*(1*cos^-1(x))
T_2(x=0)=cos*(2*cos^-1(x))
T_3(x=0)=cos*(3*cos^-1(x))
T_4(x=0)=cos*(4*cos^-1(x))
or x=-0.2
T_0(x=-0.2)=cos*(0*cos^-1(x))
T_1(x=-0.2)=cos*(1*cos^-1(x))
T_2(x=-0.2)=cos*(2*cos^-1(x))
T_3(x=-0.2)=cos*(3*cos^-1(x))
T_4(x=-0.2)=cos*(4*cos^-1(x))
c =
2.2785
-0.2142
-0.3595
-0.1793
0.1810
and
x = [-0.2:0.01:0.8];

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