Taylor Polynomial Approximations - Calculus 2

1. The third-degree Taylor polynomial for $f(x) = e^{2x}$ centered at $x = 0$ is used to approximate $f(0.1)$. If the actual value is $e^{0.2} \approx 1.2214$, what is the absolute error in this approximation?
   Answer: Approximately 0.0014 The third-degree Taylor polynomial for $e^{2x}$ at $x = 0$ is $P_3(x) = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} = 1 + 2x + 2x^2 + \frac{4x^3}{3}$. At $x = 0.1$: $P_3(0.1) = 1 + 0.2 + 0.02 + \frac{0.004}{3} \approx 1.2200$. The absolute error is $|1.2214 - 1.2200| \approx 0.0014$. Choice A uses only terms up to $x^2$, choice C incorrectly computes the third derivative term, and choice D subtracts the linear approximation from the actual value.
2. If $P_2(x)$ is the second-degree Taylor polynomial for $f(x) = \ln(1 + x)$ centered at $x = 0$, and $Q_2(x)$ is the second-degree Taylor polynomial for $g(x) = \ln(1 + 2x)$ centered at $x = 0$, which relationship between $P_2(x)$ and $Q_2(x)$ is correct?
   Answer: $Q_2(x) = P_2(2x)$ for all values of $x$ For $\ln(1+x)$: $P_2(x) = x - \frac{x^2}{2}$. For $\ln(1+2x)$: the derivatives at $x=0$ are $f'(0) = 2$, $f''(0) = -4$, so $Q_2(x) = 2x - \frac{4x^2}{2} = 2x - 2x^2$. Note that $P_2(2x) = 2x - \frac{(2x)^2}{2} = 2x - 2x^2 = Q_2(x)$. Choice A gives $2P_2(x) = 2x - x^2 \neq Q_2(x)$. Choice C is the same as A. Choice D adds an incorrect correction term.
3. Consider the function $h(x) = e^x \cos(x)$. If the second-degree Taylor polynomial for $h(x)$ centered at $x = 0$ is $P_2(x) = a_0 + a_1 x + a_2 x^2$, what is the value of $a_2$?
   Answer: $a_2 = 0$ from the second derivative evaluation at $x = 0$ Using the product rule: $h'(x) = e^x \cos(x) - e^x \sin(x)$ and $h''(x) = e^x \cos(x) - e^x \sin(x) - e^x \sin(x) - e^x \cos(x) = -2e^x \sin(x)$. At $x = 0$: $h(0) = 1$, $h'(0) = 1$, and $h''(0) = -2e^0 \sin(0) = 0$. Therefore $a_2 = \frac{h''(0)}{2!} = 0$. Choice A gives the right answer but wrong reasoning. Choice B incorrectly applies the product rule. Choice C assumes $h''(0) = -1$ which is incorrect.
4. The third-degree Taylor polynomial for $f(x) = \sqrt{1 + x}$ centered at $x = 0$ is used to approximate $\sqrt{1.08}$. Which calculation setup is correct for finding this approximation?
   Answer: $P_3(0.08) = 1 + \frac{1}{2}(0.08) - \frac{1}{8}(0.08)^2 + \frac{1}{16}(0.08)^3$ For $f(x) = (1+x)^{1/2}$, we need: $f(0) = 1$, $f'(x) = \frac{1}{2}(1+x)^{-1/2} \Rightarrow f'(0) = \frac{1}{2}$, $f''(x) = -\frac{1}{4}(1+x)^{-3/2} \Rightarrow f''(0) = -\frac{1}{4}$, $f'''(x) = \frac{3}{8}(1+x)^{-5/2} \Rightarrow f'''(0) = \frac{3}{8}$. So $P_3(x) = 1 + \frac{1}{2}x + \frac{-1/4}{2!}x^2 + \frac{3/8}{3!}x^3 = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3$. Choice B has wrong sign on the cubic term. Choice C uses wrong coefficients. Choice D has wrong signs on the quadratic and cubic terms.
5. Two functions $f(x)$ and $g(x)$ have the same second-degree Taylor polynomial $P_2(x) = 2 + 3x - x^2$ centered at $x = 0$. If $f(0.1) = 2.29$ and the second-degree Taylor polynomial approximation gives $P_2(0.1) = 2.29$, what can be concluded about $f'''(0)$?
   Answer: $f'''(0)$ cannot be determined from the given information The fact that $f(0.1) = P_2(0.1)$ is a coincidence at this particular point and doesn't imply anything about $f'''(0)$. The Taylor remainder theorem shows that $f(0.1) - P_2(0.1) = \frac{f'''(c)}{3!}(0.1)^3$ for some $c$ between 0 and 0.1. Since this equals zero, we know $f'''(c) = 0$ for some $c$, but this doesn't determine $f'''(0)$. Choice A incorrectly assumes the derivative at 0 equals the derivative at the intermediate point. Choice C assumes $f'''(0) \neq 0$ without justification. Choice D incorrectly concludes the function must be polynomial.
6. If $P_n(x)$ represents the $n$th-degree Taylor polynomial for $\sin(x)$ centered at $x = 0$, which statement about $P_4(x)$ and $P_5(x)$ is correct?
   Answer: $P_5(x) = P_4(x) + \frac{x^5}{5!}$ with the additional term being positive The Taylor series for $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$. Since there is no $x^4$ term, $P_4(x) = x - \frac{x^3}{6}$. The $x^5$ term has coefficient $+\frac{1}{5!}$ (positive in the alternating series), so $P_5(x) = x - \frac{x^3}{6} + \frac{x^5}{120} = P_4(x) + \frac{x^5}{5!}$. Choice A incorrectly identifies which term is missing. Choice B is wrong about both the reasoning and which polynomial has the missing term. Choice D has the wrong sign for the $x^5$ coefficient.
7. The function $k(x) = \ln(\cos(x))$ is approximated by its second-degree Taylor polynomial centered at $x = 0$. Which expression represents this polynomial?
   Answer: $P_2(x) = -\frac{x^2}{2}$ since $\ln(1) = 0$ and odd derivatives vanish We have $k(0) = \ln(\cos(0)) = \ln(1) = 0$. For the derivatives: $k'(x) = \frac{-\sin(x)}{\cos(x)} = -\tan(x)$, so $k'(0) = 0$. $k''(x) = -\sec^2(x)$, so $k''(0) = -1$. Therefore $P_2(x) = 0 + 0 \cdot x + \frac{-1}{2!}x^2 = -\frac{x^2}{2}$. Choice B incorrectly starts with 1 instead of 0. Choice C uses wrong derivative calculations. Choice D includes a fourth-degree term which exceeds the requested degree, and the coefficient is incorrect anyway.

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