Questions: Calculate the Taylor polynomials T2(x) and T3(x) centered at x=a for f(x)=26 sin (x), a=π/2. (Express numbers in exact form. Use symbolic notation and fractions where needed.) T2(x)= T3(x)=

$Calculate the Taylor polynomials T2(x) and T3(x) centered at x=a for f(x)=26 sin (x), a=π/2. (Express numbers in exact form. Use symbolic notation and fractions where needed.) T2(x)= T3(x)=$

Transcript text: Calculate the Taylor polynomials $T_{2}(x)$ and $T_{3}(x)$ centered at $x=a$ for $f(x)=26 \sin (x), a=\frac{\pi}{2}$. (Express numbers in exact form. Use symbolic notation and fractions where needed.) \[ T_{2}(x)= \] \[ T_{3}(x)= \]

Solution

Solution Steps

Step 1: Define the Function and Center

We are given the function $ f(x) = 26 \sin(x) $ and the center $ a = \frac{\pi}{2} $.

Step 2: Calculate the Derivatives

We need to find the first, second, and third derivatives of $ f(x) $: \[ f'(x) = 26 \cos(x) \] \[ f''(x) = -26 \sin(x) \] \[ f'''(x) = -26 \cos(x) \]

Step 3: Evaluate the Function and Derivatives at $ x = a $

Evaluate $ f(x) $ and its derivatives at $ x = \frac{\pi}{2} $: \[ f\left(\frac{\pi}{2}\right) = 26 \sin\left(\frac{\pi}{2}\right) = 26 \] \[ f'\left(\frac{\pi}{2}\right) = 26 \cos\left(\frac{\pi}{2}\right) = 0 \] \[ f''\left(\frac{\pi}{2}\right) = -26 \sin\left(\frac{\pi}{2}\right) = -26 \] \[ f'''\left(\frac{\pi}{2}\right) = -26 \cos\left(\frac{\pi}{2}\right) = 0 \]

Step 4: Construct the Taylor Polynomials

Using the Taylor series formula, we construct the polynomials $ T_2(x) $ and $ T_3(x) $.

For $ T_2(x) $: \[ T_2(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 \] Substitute the values: \[ T_2(x) = 26 + 0 \cdot (x - \frac{\pi}{2}) + \frac{-26}{2}(x - \frac{\pi}{2})^2 \] Simplify: \[ T_2(x) = 26 - 13(x - \frac{\pi}{2})^2 \]

For $ T_3(x) $: \[ T_3(x) = T_2(x) + \frac{f'''(a)}{3!}(x - a)^3 \] Since $ f'''(a) = 0 $: \[ T_3(x) = T_2(x) = 26 - 13(x - \frac{\pi}{2})^2 \]