Series De Taylor

Ejercicio 2: f(x)=cos(x), alrededor de 0. Estimar cos(π/4).
>>f=inline('cos(x)')
f =
Inline function:
f(x) = cos(x)
>> a=0
a =
0
>> f(a)
ans =
1
>> syms x
>>d(1)=diff(f(x))
d =
-sin(x)
>> subs(d(1),a)
ans =
0
>> d(2)=diff(f(x),2)
d =
[ -sin(x), -cos(x)]
>> subs(d(2),a)
ans =
-1
>> d(3)=diff(f(x),3)
d =
[ -sin(x), -cos(x), sin(x)]
>>subs(d(3),a)
ans =
0
>> d(4)=diff(f(x),4)
d =
[ -sin(x), -cos(x), sin(x), cos(x)]
>> subs(d(4),a)
ans =
1
>> d(5)=diff(f(x),5)
d =
[ -sin(x), -cos(x), sin(x), cos(x), -sin(x)]
>>subs(d(5),a)
ans =
0
>> d(6)=diff(f(x),6)
d =
[ -sin(x), -cos(x), sin(x), cos(x), -sin(x), -cos(x)]
>> subs(d(6),a)
ans =
-1
>> p0=[1]
p0 =
1
>> p1=[0 1]
p1 =
0 1
>>p2=[-1/2 0 1]
p2 =
-0.500000000000000 0 1.000000000000000
>> p3=[0 -1/2 0 1]
p3 =
0 -0.500000000000000 0 1.000000000000000
>> p4=[1/240 -1/2 0 1]
p4 =
0.041666666666667 0 -0.500000000000000 0 1.000000000000000
>> p5=[0 1/24 0 -1/2 0 1]
p5 =
0 0.0416666666666670 -0.500000000000000 0 1.000000000000000
>> p6=[-1/720 0 1/24 0 -1/2 0 1]
p6 =
Columns 1 through 6
-0.001388888888889 0 0.0416666666666670 -0.500000000000000 0
Column 7
1.000000000000000
>> z=-5:0.01:5;
>> y0=polyval(p0,z);
>> y1=polyval(p1,z);
>> y2=polyval(p2,z);
>> y3=polyval(p3,z);>> y4=polyval(p4,z);
>> y5=polyval(p5,z);
>> y6=polyval(p6,z);
>> plot(z,f(z))
>> hold on
>> plot(z,y0,'y')
>> plot(z,y1,'m')
>> plot(z,y2,'r')
>> plot(z,y3,'g')
>> plot(z,y4,'c')
>>plot(z,y5,'k')
>> plot(z,y6,'g-')
>> legend('p0','p1','p2','p3','p4','p5','p6')
>> vr=f(pi/4)
vr =
0.707106781186548
>> va(1)=polyval(p0,pi/4)
va =
1
>> va(2)=polyval(p1,pi/4)
va =...