Final answer:
Using the Taylor polynomial formula, the quadratic approximation can be expressed as: P(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)^2/2
Step-by-step explanation:
To compute the quadratic Taylor polynomial approximation to sin(x) expanded about the point x=π/4, we need to find the first three derivatives of sin(x) and evaluate them at x=π/4. The first derivative of sin(x) is cos(x), the second derivative is -sin(x), and the third derivative is -cos(x).
Using the Taylor polynomial formula, the quadratic approximation can be expressed as:
P(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)^2/2
Plugging in the values for sin(x), cos(x), -sin(x), a=π/4, and x-a, we get:
P(x) = sin(π/4) + cos(π/4)(x - π/4) - sin(π/4)(x - π/4)^2/2
The student asked for the calculation of the quadratic Taylor polynomial approximation to sin(x) expanded about the point x=π/4. To find this, we need to calculate the first few derivatives of sin(x) at x=π/4 and construct the polynomial.
The first derivative of sin(x) is cos(x), and at x=π/4, cos(π/4) = √2/2. The second derivative is -sin(x), so at x=π/4, -sin(π/4) = -√2/2. We don't need the third derivative since we're constructing a quadratic polynomial.
The Taylor polynomial is given by:
T_2(x) = sin(π/4) + (x-π/4)cos(π/4) - ½(x-π/4)^2sin(π/4)
Plugging in the values we get:
T_2(x) = √2/2 + (x-π/4)√2/2 - ½(x-π/4)^2(√2/2)