1. **Problem statement:** Given the function $f(x) = 2x \cos(2x) - (x - 2)^2$ and $x_0 = 0$, find the third Taylor polynomial $P_3(x)$ around $x_0=0$ and use it to approximate $f(0.4)$. Then, find an upper bound for the error $|f(0.4) - P_3(0.4)|$ using Taylor's theorem and compute the actual error. 2. **Taylor polynomial formula:** The Taylor polynomial of degree 3 centered at $x_0$ is $$ P_3(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \frac{f'''(x_0)}{3!}(x - x_0)^3 $$ 3. **Calculate derivatives:** - $f(x) = 2x \cos(2x) - (x - 2)^2$ - First derivative: $$ f'(x) = 2\cos(2x) + 2x(-2\sin(2x)) - 2(x - 2) = 2\cos(2x) - 4x\sin(2x) - 2x + 4 $$ - Second derivative: $$ f''(x) = -4\sin(2x) - 4\sin(2x) - 8x\cos(2x) - 2 = -8\sin(2x) - 8x\cos(2x) - 2 $$ - Third derivative: $$ f'''(x) = -16\cos(2x) - 8\cos(2x) + 16x\sin(2x) = -24\cos(2x) + 16x\sin(2x) $$ 4. **Evaluate derivatives at $x_0=0$:** - $f(0) = 2\cdot0\cdot\cos(0) - (0 - 2)^2 = 0 - 4 = -4$ - $f'(0) = 2\cos(0) - 4\cdot0\sin(0) - 2\cdot0 + 4 = 2 + 0 + 0 + 4 = 6$ - $f''(0) = -8\sin(0) - 8\cdot0\cos(0) - 2 = 0 - 0 - 2 = -2$ - $f'''(0) = -24\cos(0) + 16\cdot0\sin(0) = -24 + 0 = -24$ 5. **Form the Taylor polynomial $P_3(x)$:** $$ P_3(x) = -4 + 6x + \frac{-2}{2}x^2 + \frac{-24}{6}x^3 = -4 + 6x - x^2 - 4x^3 $$ 6. **Approximate $f(0.4)$ using $P_3(0.4)$:** $$ P_3(0.4) = -4 + 6(0.4) - (0.4)^2 - 4(0.4)^3 = -4 + 2.4 - 0.16 - 4(0.064) = -4 + 2.4 - 0.16 - 0.256 = -2.016 $$ 7. **Error bound using Taylor's theorem:** The remainder term for degree 3 is $$ R_3(x) = \frac{f^{(4)}(c)}{4!} (x - x_0)^4 $$ for some $c$ between 0 and 0.4. 8. **Find $f^{(4)}(x)$:** $$ f^{(4)}(x) = \frac{d}{dx}f'''(x) = \frac{d}{dx}(-24\cos(2x) + 16x\sin(2x)) = 48\sin(2x) + 16\sin(2x) + 32x\cos(2x) = 64\sin(2x) + 32x\cos(2x) $$ 9. **Estimate maximum of $|f^{(4)}(c)|$ for $c \in [0,0.4]$:** - $|\sin(2c)| \leq 1$ - $|\cos(2c)| \leq 1$ - $c \leq 0.4$ So, $$ |f^{(4)}(c)| \leq 64 \cdot 1 + 32 \cdot 0.4 \cdot 1 = 64 + 12.8 = 76.8 $$ 10. **Calculate error bound:** $$ |R_3(0.4)| \leq \frac{76.8}{4!} (0.4)^4 = \frac{76.8}{24} (0.0256) = 3.2 \times 0.0256 = 0.08192 $$ 11. **Calculate actual error:** - Compute $f(0.4)$ exactly: $$ f(0.4) = 2(0.4)\cos(0.8) - (0.4 - 2)^2 = 0.8 \cos(0.8) - ( -1.6)^2 = 0.8 \cos(0.8) - 2.56 $$ - Using $\cos(0.8) \approx 0.6967$, $$ f(0.4) \approx 0.8 \times 0.6967 - 2.56 = 0.55736 - 2.56 = -2.00264 $$ - Actual error: $$ |f(0.4) - P_3(0.4)| = |-2.00264 - (-2.016)| = |0.01336| = 0.01336 $$ **Final answers:** - Third Taylor polynomial: $$ P_3(x) = -4 + 6x - x^2 - 4x^3 $$ - Approximation: $$ P_3(0.4) = -2.016 $$ - Error bound: $$ |R_3(0.4)| \leq 0.08192 $$ - Actual error: $$ 0.01336 $$