1. Problem (a): Find Taylor polynomials $p_0, p_1, p_2, p_3$ for $f(x)=\ln x$ at $x=1$. - We use Taylor series expansion about $a=1$: $$p_n(x) = \sum_{k=0}^n \frac{f^{(k)}(1)}{k!}(x-1)^k$$ - First find derivatives: $$f(x) = \ln x,\quad f'(x) = \frac{1}{x},\quad f''(x) = -\frac{1}{x^2},\quad f^{(3)}(x) = \frac{2}{x^3}$$ - Evaluate derivatives at $x=1$: $$f(1) = 0, \quad f'(1)=1, \quad f''(1)=-1, \quad f^{(3)}(1)=2$$ - Construct polynomials: $$p_0(x) = 0$$ $$p_1(x)= 0 + 1(x-1)= x-1$$ $$p_2(x)= x-1 - \frac{1}{2}(x-1)^2$$ $$p_3(x)= x-1 - \frac{1}{2}(x-1)^2 + \frac{2}{6}(x-1)^3 = x-1 - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3$$ 2. Problem (b): Expand $\sin\left(\frac{\pi}{6} + h\right)$ in ascending powers of $h$ up to $h^4$ using Taylor series about $h=0$. - Define $g(h) = \sin\left(\frac{\pi}{6}+h\right)$. - Derivatives at $h=0$: $$g(0) = \sin\frac{\pi}{6} = \frac{1}{2}$$ $$g'(h) = \cos\left(\frac{\pi}{6}+h\right),\quad g'(0)= \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$$ $$g''(h) = -\sin\left(\frac{\pi}{6}+h\right),\quad g''(0) = -\frac{1}{2}$$ $$g^{(3)}(h) = -\cos\left(\frac{\pi}{6}+h\right),\quad g^{(3)}(0)= -\frac{\sqrt{3}}{2}$$ $$g^{(4)}(h) = \sin\left(\frac{\pi}{6}+h\right),\quad g^{(4)}(0) = \frac{1}{2}$$ - Taylor expansion: $$g(h) = g(0) + g'(0)h + \frac{g''(0)}{2}h^2 + \frac{g^{(3)}(0)}{6}h^3 + \frac{g^{(4)}(0)}{24}h^4 + \cdots$$ - Substitute values: $$\sin(\frac{\pi}{6}+h) = \frac{1}{2} + \frac{\sqrt{3}}{2}h - \frac{1}{4}h^2 - \frac{\sqrt{3}}{12}h^3 + \frac{1}{48}h^4$$ 3. Problem (c): Find power series of $\cos^2(2x)$ up to $x^6$ term. - Use identity: $$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$$ - Expand $\cos(4x)$ using Taylor series about $x=0$: $$\cos(4x) = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \frac{(4x)^6}{6!} + \cdots = 1 - 8x^2 + \frac{128}{24}x^4 - \frac{4096}{720}x^6 + \cdots$$ - Simplify coefficients: $$\frac{128}{24} = \frac{16}{3}, \quad \frac{4096}{720} = \frac{256}{45}$$ - Substitute back: $$\cos^2(2x) = \frac{1}{2} + \frac{1}{2}\left(1 - 8x^2 + \frac{16}{3}x^4 - \frac{256}{45}x^6\right) = 1 - 4x^2 + \frac{8}{3}x^4 - \frac{128}{45}x^6$$ Final answers: - (a) $$p_0=0,$$ $$p_1 = x-1,$$ $$p_2 = x-1 - \frac{1}{2}(x-1)^2,$$ $$p_3 = x-1 - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3$$ - (b) $$\sin\left(\frac{\pi}{6}+h\right) = \frac{1}{2} + \frac{\sqrt{3}}{2}h - \frac{1}{4}h^2 - \frac{\sqrt{3}}{12}h^3 + \frac{1}{48}h^4$$ - (c) $$\cos^2(2x) = 1 - 4x^2 + \frac{8}{3}x^4 - \frac{128}{45}x^6$$ Graphs for $f(x) = \ln x$ and polynomials $p_0,p_1,p_2,p_3$ show increasing approximation accuracy near $x=1$.