Answer :
Let's find the Taylor polynomials of the function [tex]\( f(x) = 4 \cos (2x) \)[/tex] centered at [tex]\( a = \frac{\pi}{12} \)[/tex] up to the 4th and 5th degrees.
### Taylor Polynomial [tex]\( p_4 \)[/tex]
1. Calculate [tex]\( f(a) \)[/tex]:
[tex]\[ f\left(\frac{\pi}{12}\right) = 4 \cos\left(2 \times \frac{\pi}{12}\right) = 4 \cos\left(\frac{\pi}{6}\right) \][/tex]
We know that [tex]\( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)[/tex], so:
[tex]\[ f\left(\frac{\pi}{12}\right) = 4 \times \frac{\sqrt{3}}{2} = 2 \sqrt{3} \][/tex]
2. Calculate the derivatives of [tex]\( f \)[/tex] evaluated at [tex]\( a = \frac{\pi}{12} \)[/tex]:
[tex]\[ f(x) = 4 \cos(2x) \implies f'(x) = -8 \sin(2x) \][/tex]
[tex]\[ f'\left(\frac{\pi}{12}\right) = -8 \sin\left(2 \times \frac{\pi}{12}\right) = -8 \sin\left(\frac{\pi}{6}\right) = -8 \times \frac{1}{2} = -4 \][/tex]
[tex]\[ f''(x) = -16 \cos(2x) \implies f''\left(\frac{\pi}{12}\right) = -16 \cos\left(\frac{\pi}{6}\right) = -16 \times \frac{\sqrt{3}}{2} = -8 \sqrt{3} \][/tex]
[tex]\[ f'''(x) = 32 \sin(2x) \implies f'''\left(\frac{\pi}{12}\right) = 32 \sin\left(\frac{\pi}{6}\right) = 32 \times \frac{1}{2} = 16 \][/tex]
[tex]\[ f^{(4)}(x) = 64 \cos(2x) \implies f^{(4)}\left(\frac{\pi}{12}\right) = 64 \cos\left(\frac{\pi}{6}\right) = 64 \times \frac{\sqrt{3}}{2} = 32 \sqrt{3} \][/tex]
3. Construct the Taylor polynomial:
[tex]\[ p_4(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \frac{f^{(4)}(a)}{4!}(x - a)^4 \][/tex]
[tex]\[ p_4(x) = 2\sqrt{3} + (-4)(x - \frac{\pi}{12}) + \frac{-8\sqrt{3}}{2}(x - \frac{\pi}{12})^2 + \frac{16}{6}(x - \frac{\pi}{12})^3 + \frac{32\sqrt{3}}{24}(x - \frac{\pi}{12})^4 \][/tex]
Simplifying the coefficients:
[tex]\[ p_4(x) = 2\sqrt{3} - 4(x - \frac{\pi}{12}) - 4\sqrt{3}(x - \frac{\pi}{12})^2 + \frac{8}{3}(x - \frac{\pi}{12})^3 + \frac{4\sqrt{3}}{3}(x - \frac{\pi}{12})^4 \][/tex]
### Taylor Polynomial [tex]\( p_5 \)[/tex]
1. Calculate the 5th derivative of [tex]\( f \)[/tex] evaluated at [tex]\( a \)[/tex]:
[tex]\[ f^{(5)}(x) = -128 \sin(2x) \implies f^{(5)}\left(\frac{\pi}{12}\right) = -128 \sin\left(\frac{\pi}{6}\right) = -128 \times \frac{1}{2} = -64 \][/tex]
2. Construct the Taylor polynomial:
[tex]\[ p_5(x) = p_4(x) + \frac{f^{(5)}(a)}{5!}(x - a)^5 \][/tex]
[tex]\[ p_5(x) = 2\sqrt{3} - 4(x - \frac{\pi}{12}) - 4\sqrt{3}(x - \frac{\pi}{12})^2 + \frac{8}{3}(x - \frac{\pi}{12})^3 + \frac{4\sqrt{3}}{3}(x - \frac{\pi}{12})^4 + \frac{-64}{120}(x - \frac{\pi}{12})^5 \][/tex]
Simplifying the coefficients:
[tex]\[ p_5(x) = 2\sqrt{3} - 4(x - \frac{\pi}{12}) - 4\sqrt{3}(x - \frac{\pi}{12})^2 + \frac{8}{3}(x - \frac{\pi}{12})^3 + \frac{4\sqrt{3}}{3}(x - \frac{\pi}{12})^4 - \frac{16}{15}(x - \frac{\pi}{12})^5 \][/tex]
The resulting Taylor polynomials centered at [tex]\( a = \frac{\pi}{12} \)[/tex] up to the 4th and 5th degrees for the function [tex]\( f(x) = 4 \cos(2x) \)[/tex] are:
[tex]\[ p_4(x) = -4x + \frac{8}{3}(x - \frac{\pi}{12})^3 - 4\sqrt{3}(x - \frac{\pi}{12})^2 + \frac{\pi}{3} + 2\sqrt{3} \][/tex]
[tex]\[ p_5(x) = -4x + \frac{4\sqrt{3}}{3}(x - \frac{\pi}{12})^4 + \frac{8}{3}(x - \frac{\pi}{12})^3 - 4\sqrt{3}(x - \frac{\pi}{12})^2 + \frac{\pi}{3} + 2\sqrt{3} \][/tex]
### Taylor Polynomial [tex]\( p_4 \)[/tex]
1. Calculate [tex]\( f(a) \)[/tex]:
[tex]\[ f\left(\frac{\pi}{12}\right) = 4 \cos\left(2 \times \frac{\pi}{12}\right) = 4 \cos\left(\frac{\pi}{6}\right) \][/tex]
We know that [tex]\( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)[/tex], so:
[tex]\[ f\left(\frac{\pi}{12}\right) = 4 \times \frac{\sqrt{3}}{2} = 2 \sqrt{3} \][/tex]
2. Calculate the derivatives of [tex]\( f \)[/tex] evaluated at [tex]\( a = \frac{\pi}{12} \)[/tex]:
[tex]\[ f(x) = 4 \cos(2x) \implies f'(x) = -8 \sin(2x) \][/tex]
[tex]\[ f'\left(\frac{\pi}{12}\right) = -8 \sin\left(2 \times \frac{\pi}{12}\right) = -8 \sin\left(\frac{\pi}{6}\right) = -8 \times \frac{1}{2} = -4 \][/tex]
[tex]\[ f''(x) = -16 \cos(2x) \implies f''\left(\frac{\pi}{12}\right) = -16 \cos\left(\frac{\pi}{6}\right) = -16 \times \frac{\sqrt{3}}{2} = -8 \sqrt{3} \][/tex]
[tex]\[ f'''(x) = 32 \sin(2x) \implies f'''\left(\frac{\pi}{12}\right) = 32 \sin\left(\frac{\pi}{6}\right) = 32 \times \frac{1}{2} = 16 \][/tex]
[tex]\[ f^{(4)}(x) = 64 \cos(2x) \implies f^{(4)}\left(\frac{\pi}{12}\right) = 64 \cos\left(\frac{\pi}{6}\right) = 64 \times \frac{\sqrt{3}}{2} = 32 \sqrt{3} \][/tex]
3. Construct the Taylor polynomial:
[tex]\[ p_4(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \frac{f^{(4)}(a)}{4!}(x - a)^4 \][/tex]
[tex]\[ p_4(x) = 2\sqrt{3} + (-4)(x - \frac{\pi}{12}) + \frac{-8\sqrt{3}}{2}(x - \frac{\pi}{12})^2 + \frac{16}{6}(x - \frac{\pi}{12})^3 + \frac{32\sqrt{3}}{24}(x - \frac{\pi}{12})^4 \][/tex]
Simplifying the coefficients:
[tex]\[ p_4(x) = 2\sqrt{3} - 4(x - \frac{\pi}{12}) - 4\sqrt{3}(x - \frac{\pi}{12})^2 + \frac{8}{3}(x - \frac{\pi}{12})^3 + \frac{4\sqrt{3}}{3}(x - \frac{\pi}{12})^4 \][/tex]
### Taylor Polynomial [tex]\( p_5 \)[/tex]
1. Calculate the 5th derivative of [tex]\( f \)[/tex] evaluated at [tex]\( a \)[/tex]:
[tex]\[ f^{(5)}(x) = -128 \sin(2x) \implies f^{(5)}\left(\frac{\pi}{12}\right) = -128 \sin\left(\frac{\pi}{6}\right) = -128 \times \frac{1}{2} = -64 \][/tex]
2. Construct the Taylor polynomial:
[tex]\[ p_5(x) = p_4(x) + \frac{f^{(5)}(a)}{5!}(x - a)^5 \][/tex]
[tex]\[ p_5(x) = 2\sqrt{3} - 4(x - \frac{\pi}{12}) - 4\sqrt{3}(x - \frac{\pi}{12})^2 + \frac{8}{3}(x - \frac{\pi}{12})^3 + \frac{4\sqrt{3}}{3}(x - \frac{\pi}{12})^4 + \frac{-64}{120}(x - \frac{\pi}{12})^5 \][/tex]
Simplifying the coefficients:
[tex]\[ p_5(x) = 2\sqrt{3} - 4(x - \frac{\pi}{12}) - 4\sqrt{3}(x - \frac{\pi}{12})^2 + \frac{8}{3}(x - \frac{\pi}{12})^3 + \frac{4\sqrt{3}}{3}(x - \frac{\pi}{12})^4 - \frac{16}{15}(x - \frac{\pi}{12})^5 \][/tex]
The resulting Taylor polynomials centered at [tex]\( a = \frac{\pi}{12} \)[/tex] up to the 4th and 5th degrees for the function [tex]\( f(x) = 4 \cos(2x) \)[/tex] are:
[tex]\[ p_4(x) = -4x + \frac{8}{3}(x - \frac{\pi}{12})^3 - 4\sqrt{3}(x - \frac{\pi}{12})^2 + \frac{\pi}{3} + 2\sqrt{3} \][/tex]
[tex]\[ p_5(x) = -4x + \frac{4\sqrt{3}}{3}(x - \frac{\pi}{12})^4 + \frac{8}{3}(x - \frac{\pi}{12})^3 - 4\sqrt{3}(x - \frac{\pi}{12})^2 + \frac{\pi}{3} + 2\sqrt{3} \][/tex]
