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multiplying vectors (easy one!)

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Paul Rogers
Paul Rogers on 11 Jun 2020
Commented: Paul Rogers on 11 Jun 2020
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

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Answers (1)

KSSV
KSSV on 11 Jun 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 Comments

Paul Rogers
Paul Rogers on 11 Jun 2020
x = [-0.2:0.01:0.8];
KSSV
KSSV on 11 Jun 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 on 11 Jun 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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