find the 6th degree Taylor polynomial for f(x)=cosx centered at pi/2
To find the 6th degree Taylor polynomial for f(x) = cosx centered at pi/2, we can use the formula for the nth degree Taylor polynomial:
P_n(x) = f(c) + f'(c)(x - c) + (f''(c)/2!)(x - c)^2 + (f'''(c)/3!)(x - c)^3 + ... + (f^n(c)/n!)(x - c)^n,
where c is the center of the Taylor polynomial.
First, let's calculate the values of f(x), f'(x), f''(x), f'''(x), f''''(x), f'''''(x), and f''''''(x) for the given function:
f(x) = cosx
f'(x) = -sinx
f''(x) = -cosx
f'''(x) = sinx
f''''(x) = cosx
f'''''(x) = -sinx
f''''''(x) = -cosx
Next, we need to evaluate these derivatives at the center c = pi/2:
f(pi/2) = cos(pi/2) = 0
f'(pi/2) = -sin(pi/2) = -1
f''(pi/2) = -cos(pi/2) = -0
f'''(pi/2) = sin(pi/2) = 1
f''''(pi/2) = cos(pi/2) = 0
f'''''(pi/2) = -sin(pi/2) = -1
f''''''(pi/2) = -cos(pi/2) = -0
Now, we can substitute these values into the formula for the 6th degree Taylor polynomial:
P_6(x) = f(pi/2) + f'(pi/2)(x - pi/2) + (f''(pi/2)/2!)(x - pi/2)^2 + (f'''(pi/2)/3!)(x - pi/2)^3 + (f''''(pi/2)/4!)(x - pi/2)^4 + (f'''''(pi/2)/5!)(x - pi/2)^5 + (f''''''(pi/2)/6!)(x - pi/2)^6
P_6(x) = 0 + (-1)(x - pi/2) + (-0/2!)(x - pi/2)^2 + (1/3!)(x - pi/2)^3 + (0/4!)(x - pi/2)^4 + (-1/5!)(x - pi/2)^5 + (-0/6!)(x - pi/2)^6
Simplifying, we get:
P_6(x) = -1(x - pi/2) + (1/6)(x - pi/2)^3 - (1/120)(x - pi/2)^5
Thus, the 6th degree Taylor polynomial for f(x) = cosx centered at pi/2 is:
P_6(x) = -1(x - pi/2) + (1/6)(x - pi/2)^3 - (1/120)(x - pi/2)^5