1.3

1. Find the Taylor polynomial of order 3 at \(a = 0\) for the function \(f(x)=e^{e^{x}}\)

\(T_{3,0}(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3\)

Given \(f(x) = e^{e^x}\), we calculate derivatives step-by-step:

\(f(0)=e^{e^0}=e\)
\(f'(x) = e^{e^x} \cdot e^x\), \(f^{\prime}(0)=e^{e}\cdot e^0=e\)
\(f''(x)=\left(e^{e^{x}}\cdot e^{x}\right)^{\prime}=e^{e^{x}}\cdot e^{x}\cdot e^{x}+e^{e^{x}}\cdot e^{x}\)

\(f''(0) = e + e = 2e\)
\(f^{(3)}(x)=\left(e^{e^{x}}\cdot e^{x}+e^{e^{x}}\cdot e^{x}\right)^{\prime}=e^{e^{x}}\cdot e^{x}\cdot e^{2x}+e^{e^{x}}\cdot e^{2x}\cdot2+e^{e^{x}}\cdot e^{x}\cdot e^{x}+e^{e^{x}}\cdot e^{x}\)

\(f^{(3)}(0)=e+2e+e+e=5e\)

Thus, the Taylor polynomial is \(T_{3,0}(x)=e+ex+ex^2+\frac{5e}{6}x^3\)

2. Let \(f(x) = \sqrt{1 + x}\). Using the Taylor polynomial of order 3 at \(a = 0\), compute an approximate value for \(\sqrt{2}\) and \(\sqrt{9}\). Determine the remainder term.

\(T_{3,0}(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3\)

\(f(0) = \sqrt{1 + 9} = 1, \quad f(x) = \sqrt{1+x}, \quad f'(x) = \frac{1}{2\sqrt{1+x}}, \quad f''(x) = -\frac{1}{4(1+x)^{3/2}}, \quad f^{(3)}(x) = \frac{3}{8}(1+x)^{-5/2}\)

\(f'(0) = \frac{1}{2}, \quad f''(0) = -\frac{1}{4}, \quad f^{(3)}(0) = \frac{3}{8}\)

\(T_{3,0}(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3\)

\(\sqrt{2} \approx T_{3,0}(1) = 1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} = \frac{16 + 8 - 2 + 1}{16} = \frac{23}{16}\)

\(\sqrt{9} \approx T_{3,0}(8) = 1 + \frac{8}{2} - \frac{64}{8} + \frac{512}{16} = 1 + 4 - 8 + 32 = 29\)

\(R_{3,0} = \frac{f^{(4)}(c)}{4!}x^4\)

\(f^{(4)}(x) = \frac{3}{8} \cdot -\frac{5}{3}(1+x)^{-7/2} = -\frac{15}{16}(1+x)^{-7/2}\)

\(R_{3,0} = -\frac{15}{16 \cdot 24 (1+c)^{7/2}}x^4 \quad \text{for some } c \in [0, x]\)

3.
(a) For which values of \(x\) can \(\sin(x)\) be approximated by \(x - \frac{x^3}{3!}\) with a remainder less than \(10^{-4}\)?

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

Then, \(\frac{\sin(c)}{24} x^4 < 10^{-4}\)

So, \(x^4 < \frac{10^{-4} \cdot 24}{\sin(c)}\). For \(c \in [0, x]\), \(x^4 < \frac{24}{10000 \cdot \sin(c)}\), then \(x < \sqrt[4]{\frac{24}{10000}}\)

(b) Compute the remainder of approximating \(\sin(x)\) by \(x - \frac{x^3}{3!} + \frac{x^5}{5!}\) for \(|x| < 0.2\).

\(-\frac{\left|x\right|^6}{6!}\leq R_{5,0}=-\frac{\sin(c)}{6!}x^6\leq\frac{\left|x\right|^6}{6!}\), since \(|x| < 0.2\), then \(|R_{5,0}|\leq\frac{|x|^6}{6!}=\frac{0.2^6}{6!}\)

4. Use Taylor Theorem to compute the following numbers with the indicated accuracy:
(a) \(e\) with a remainder less than \(10^{-3}\).

\(f(x) = e^x, \, R_{N,0} = \frac{e^c}{(N+1)!} \leq 10^{-3}, \, c \in [0,1]\)

Since \(\frac{e^{c}}{(N+1)!}\leq\frac{e}{(N+1)!}<\frac{3}{(N+1)!}\leq10^{-3}\), then \((N+1)!\geq\frac{3}{10^{-3}}=3000\)

Thus \(N = 6\)

We have \(f(x) - T_{6,0}(N) = \frac{e^c}{7!}\), then \(e^x - T_{6,0}(x) = \frac{e^c}{7!}\)

\(T_{6,0}(N) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \frac{x^6}{720}\)

\(T_{6,0}(1) = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + \frac{1}{720}\)

\(=2+\frac{360}{720}+\frac{120}{720}+\frac{30}{720}+\frac{6}{720}+\frac{1}{720}=2+\frac{517}{720}=2.718056\)

(b) \(\cos(0.1)\) with a remainder less than \(10^{-4}\).

\(f(x) = \cos x\) \(R_{N,0} = \frac{f^{(N+1)}(c)}{(N+1)!}(x)^{N+1}\)

Then \(R_{N,0}\leq10^{-4}\implies\frac{f^{\left(N+1\right)}\left(c\right)}{\left(N+1\right)!}\left(0.1^{N+1}\right)\leq\frac{0.1^{N+1}}{\left(N+1\right)!\left.\right.}\leq10^{-4}\)

When \(N = 2\), \(\frac{0.1^3}{4!}=\frac{0.1^3}{24}=\frac{0.1^4}{2.4}\leq10^{-4}\)

Then \(f(x) - T_{2,0}(x) = R_{2,0}\)

\(T_{2,0}(x)=f(0)+\frac{f^{\prime}(0)}{1!}x+\frac{f^{(2)}(0)}{2!}x^2=1-\frac{\cos0}{2}x^2=1-\frac12x^2\)

Then \(T_{2,0}(0.1) = 1 - \frac{0.01}{2} = 1 - 0.005 = 0.9950\)

5. Determine the following limits using a suitable Taylor polynomial. Justify using the remainder:
(a) \(\lim_{x \to 0} \frac{\sin(x) - x}{x^3}.\)
(b) \(\lim_{x \to 0} \frac{x - \sin(x)}{x \left(1 - \cos(3x)\right)}.\)

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