1.2

1. \(f(x) = \cos(x^7)\) and \(T_{4, \cos(x), 0}(x)\) (Taylor polynomial of \(\cos x\) at \(a = 0\))

   1. Show that \(g(x) = T_{4, \cos(x), 0}(x^7)\) is the Taylor polynomial of order 28 at \(x = 0\) of \(f\).

      \(T_{4, \cos(x), 0}(x) = \sum_{i=0}^2 \frac{(-1)^i}{(2i)!} x^{2i}\) \(= 1 - \frac{1}{2}x^2 + \frac{1}{4!}x^4\)

      \(g(x) = T_{4, \cos(x), 0}(x^7)\) \(= 1 - \frac{1}{2}(x^7)^2 - \frac{1}{4!}(x^7)^4\) \(=1-\frac12x^{14}-\frac{1}{4!}x^{28}\overset{\text{prove}}{=}T_{28,\cos(x^7),0}(x)\)

      Prove \(T_{28,\cos(x^7),0}(x)=1-\frac12x^{14}-\frac{1}{4!}x^{28}\) is unique since this

      \(\lim_{x \to 0} \frac{f(x) - g(x)}{(x - 0)^{28}} = \lim_{x \to 0} \frac{\cos(x^7) - (1 - \frac{1}{2}x^{14} - \frac{1}{4!}x^{28})}{x^{28}}\).

      Let \(\mu = x^7, \mu^4 = (x^7)^4 = x^{28}\), then \(=\lim_{x\to0}\frac{\cos(\mu)-(1-\frac12\mu^2-\frac{1}{4!}\mu^4)}{\mu^4}=0\) by theorem

      \(g(x)=T_{28,\cos(x^7),0}(x)\)​ 2. Compute \(f^{(14)}(0)\)

      \(a_{14} = \frac{f^{(14)}(0)}{14!} = -\frac{1}{2} \implies f^{(14)}(0) = -\frac{14!}{2}\)​

      \(T_{28, \cos(x^7), 0}(x) = 1 - \frac{1}{2}x^{14} + \frac{1}{4!}x^{28}\)

2. Suppose \(a_i, b_i\) are coefficients of Taylor polynomials at \(x = a\) of \(f\) and \(g\), respectively. Find the coefficients \(c_i\) of the Taylor polynomial at \(x = a\) of the following functions:

   \(T_{n, f, a}(x) = a_0 + a_1(x - a) + a_2(x - a)^2 + \dots + a_n(x - a)^n\)

   \(T_{n, g, a}(x) = b_0 + b_1(x - a) + b_2(x - a)^2 + \dots + b_n(x - a)^n\)

   1. \(f + g\)

      \(T_{n, f+g, a}(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + \dots + c_n(x-a)^n = T_{n, f, a}(x) + T_{n, g, a}(x)\)

      \((f+g)' = f' + g', \quad (f+g)'' = f'' + g'', \quad (f+g)^{(n)} = f^{(n)} + g^{(n)}\)

      \(c_i = \frac{(f+g)^{(i)}(a)}{i!} = \frac{f^{(i)}(a)}{i!} + \frac{g^{(i)}(a)}{i!} = a_i + b_i\) 2. \(f \cdot g\) \((f \cdot g)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(n-k)} g^{(k)}\) 3. \(f'\)

      \(h(x)=f^{\prime}(x)\) \(h'(x) = f''(x), \quad h''(x) = f^{(3)}(x), \quad h^{(n)}(x) = f^{(n+1)}(x)\)

      \(c_{i}=\frac{h^{(i)}(a)}{i!}=\frac{f^{(i+1)}(a)}{i!}=a_{i+1}\cdot\frac{(i+1)}{i!}=a_{i+1}\cdot(i+1)\)

      \(a_{i+1} = \frac{f^{(i+1)}(a)}{(i+1)!}, \quad c_i = a_{i+1} \cdot (i+1)\)

      \(T_{n, h, a}(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + \dots + c_n(x-a)^n\)

      \(= a_1 + a_2 \cdot 2(x-a) + a_3 \cdot 3(x-a)^2 + \dots + a_{n+1}(n+1)(x-a)^n\)

      \(= \sum_{i=0}^{n} a_{i+1}(i+1)(x-a)^i\)

3. Determine Taylor Polynomial of order 5 of:

   1. \(g(x) = \sin(x) + \frac{1}{1+x^2}\) at \(a = 0\)

      \(\sin(x) \rightarrow T_{5, \sin(x), 0}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}\)

      \(f(x) = \arctan(x), \quad f'(x) = \frac{1}{1+x^2} \rightarrow T_{5, \frac{1}{1+x^2}, 0}(x)\)

      Using 4(c): \(T_{5,\frac{1}{1+x^2},0}(x)=1\cdot x^0+0+\left(-\frac13\right)\cdot3x^2+0+\frac15\cdot x^4+0\cdot x^5\)

      \(T_{5, \sin(x) + \frac{1}{1+x^2}, 0}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + 1 - x^2 + x^4\) \(= 1 + x - x^2 - \frac{1}{6}x^3 + x^4 + \frac{1}{120}x^5\) 2. \(g(x) = \frac{\sin(x)}{1+x^2}\) at \(x = 0\)​

      \(g(x) = \sin(x) \cdot \frac{1}{1+x^2}\)

      \(T_{5, \sin(x), 0}(x) = \left(x - \frac{x^3}{3!} + \frac{x^5}{5!}\right), \quad T_{5, \frac{1}{1+x^2}, 0}(x) = \left(1 - x^2 + x^4\right)\)

      Since we know \(T_{5,\sin(x)\cdot\frac{1}{1+x^2},0}(x)=T_{5,\sin(x),0}(x)\cdot T_{5,\sin(x),0}(x)\) and \(g(x) = \left(x - \frac{x^3}{6} + \frac{x^5}{120}\right) \cdot \left(1 - x^2 + x^4\right)\)

      Then \(T_{5, g, 0}(x) = x - \left(1 + \frac{1}{3!}\right)x^3 + \left(1 + \frac{1}{3!} + \frac{1}{5!}\right)x^5\)

4. \(f(x) = \sin(x)\)

   1. \(T_{4, \sin(x), 0}(x) = x - \frac{x^3}{3!}\)

   2. Determine an upper bound for the remainder term when \(-0.3 \leq x \leq 0.3\), i.e., \(|x| \leq 0.3\).
      ​\(R_{4,0}(x) = \left| \frac{f^{(5)}(c)}{5!}(x - 0)^5 \right| = \left| \frac{f^{(5)}(0)}{5!} |x|^5 \right|\)

      For \(|x| \leq 0.3\): \(R_{4,0}(x)\leq\frac{|1|\cdot0.3}{5!}=\frac{3}{5!\cdot10}\)​ 3. Use this approximation to compute \(\sin(1)\).

      In points close to \(a = 0\): \(\sin(x) \approx T_{4, \sin(x), 0}(x) = x - \frac{x^3}{3!}\)

      \(\sin(1) \approx 1 - \frac{1}{6} = \frac{5}{6}\)​

5. Approximate \(\sqrt[3]{7}\) using Taylor polynomials of degree 2 at \(a = 8\) for the function \(f(x) = \sqrt[3]{x} = x^{\frac{1}{3}}\).

   Determine \(R_{2,8}\):

   \(f'(x) = \frac{1}{3} x^{\frac{1}{3} - 1} = \frac{1}{3} x^{-\frac{2}{3}}\) \(f'(8) = \frac{1}{3 \cdot 8^{\frac{2}{3}}} = \frac{1}{3 \cdot 4} = \frac{1}{12}\) \(a_1 = \frac{1}{12 \cdot 1!}\)

   \(f''(x) = \frac{-2}{9} x^{-\frac{5}{3}} = -\frac{2}{9} x^{-\frac{5}{3}}\) \(f''(8) = \frac{-2}{9} \cdot \frac{1}{8^{\frac{5}{3}}} = \frac{-2}{9 \cdot 8^2 \cdot 4} = \frac{-1}{16 \cdot 9}\) \(a_2 = \frac{-1}{16 \cdot 9 \cdot 2!} = \frac{-1}{16 \cdot 9 \cdot 2}\)

   \(f'''(x)=\frac29\cdot\frac53x^{-\frac53-1}=\frac{10}{27}x^{-\frac83}\)​

   \(T_{2,f,8}(x) = 2 + \frac{1}{12}(x - 8) - \frac{1}{288}(x - 8)^2\)

   \(\sqrt[3]{7} \approx T_{2,f,8}(7) = 2 + \frac{1}{12}(7 - 8) - \frac{1}{288}(7 - 8)^2\) \(= 2 - \frac{1}{12} - \frac{1}{288}\)

   \(R_{2,8}(x) = \frac{f^{(3)}(c)}{3!}(x - 8)^3 \quad \text{for some } c \text{ between 8 and } x\)

   \(R_{2,8}(x) = \frac{10}{27} \cdot c^{-8/3} \cdot \frac{1}{6} \cdot (x - 8)^3 = \frac{10}{162 \cdot c^{8/3}}(x - 8)^3\)

   \(R_{2,8}(7) = \frac{-10}{162 \cdot c^{8/3}}\)

8.

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