Homework 3

HW 3.pdf

(20 points) Find the first four terms of the Taylor series for the following functions:
1. \(x^4 + x - 2\) centered at \(a = 1\).
  
  \(T_{4,1} = f(1) + \frac{f'(1)}{1!} (x - 1) + \frac{f''(1)}{2!} (x - 1)^2 + \frac{f'''(1)}{3!} (x - 1)^3 + \frac{f''''(1)}{4!} (x - 1)^4\)
  
  Let \(f(x) = x^4 + x - 2, \quad f'(x) = 4x^3 + 1, \quad f''(x) = 12x^2, \quad f'''(x) = 24x, \quad f''''(x) = 24\)
  
  Then \(T_{4,1}=5\left(x-1)+6(x-1\right)^{2}+4(x-1)^{3}+(x-1)^{4}\) 2. \(\frac{x}{\sqrt{1 - x^2}}\) centered at \(0\).
  
  \(T_{4,0} = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f''''(0)}{4!} x^4\)
  
  Let \(f(x) = \frac{x}{\sqrt{1 - x^2}}\)
  \(f'(x) = (1 - x^2)^{-\frac{1}{2}} + x^2(1 - x^2)^{-\frac{3}{2}}\)
  
  \(f''(x) = 3x(1 - x^{2})^{-\frac{3}{2}}+ 3x^3(1-x^{2})^{-\frac{5}{2}}\)
  
  \(f'''(x) = 3(1 - x^{2})^{-\frac{3}{2}}+ 18x^{2}(1 - x^{2})^{-\frac{5}{2}}+ 15x^{4} (1 - x^{2})^{-\frac{7}{2}}\)
  
  \(f''''(x)=9x(1-x^{2})^{-\frac{5}{2}}+[18x^{2}(1 - x^{2})^{-\frac{5}{2}}+ 15x^{4} (1 - x^{2})^{-\frac{7}{2}}]'\)
  
  Then \(f(0) = 0, \quad f'(0) = 1, \quad f''(0) = 0, \quad f'''(0) = 3, \quad f''''(0) = 0\)
  
  Then \(T_{4,0}=x+\frac{1}{2}x^{3}\) 3. \(x e^{-x}\) centered at \(0\).
  
  \(T_{4,0} = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \frac{f''''(0)}{4!} x^4\)
  
  Let \(f(x) = x e^{-x}\), then \(f'(x) = e^{-x} - x e^{-x}\), \(f''(x) = -2 e^{-x} + x e^{-x}\)
  
  \(f'''(x) = 3 e^{-x} - x e^{-x}\), \(f''''(x) = -4 e^{-x} + x e^{-x}\)
  
  Then \(f(0) = 0, \quad f'(0) = 1, \quad f''(0) = -2, \quad f'''(0) = 3, \quad f''''(0) = -4\)
  
  Then \(T_{4,0} = x - x^2 + \frac{x^3}{2} - \frac{x^4}{6}\)
(30 points)
1. Estimate the error in approximating \(\cos(x)\) by \(1 - \frac{x^2}{2}\) on the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\).
  
  We know \(1-\frac{x^{2}}{2}=T_{3,\cos x,0}\), then by remainder theorem \(E_{3}(x)=\frac{f^{(4)}(c)}{4!}x^{5}\quad\text{for some }c\text{ between }0\text{ and }x\)
  Then \(\left|E_{3}(x)\right|=\left|\frac{\cos^{(4)}(c)}{4!}x^{4}\right|\leq\left|\frac{1}{4!} x^{4}\right|\leq\frac{1}{4!}\cdot\frac{1}{16}=\frac{1}{384}\) 2. Compute the value of \(\sqrt[3]{7}\) with an error smaller than \(10^{-4}\).
  
  Since \(1 < \sqrt[3]{7} < 2\), then consider \(f(x) = \sqrt[3]{x}\) and Taylor series around \(a = 8\).
  
  Consider \(\left|E(x)\right|=\left|\frac{f^{(n+1)}(c)}{(n+1)!}(x-8)^{n+1}\right|<10^{-4}\)
  When \(n = 3\), \(\left|E(x)\right|=\left|\frac{f^{(4)}(c)}{4!}(x-8)^{4}\right|\leq\left|\frac{80}{24\times81} c^{-\frac{11}{3}}\right|\leq\frac{10}{243}\left(\frac{9^{3}}{5^{3}}\right)^{-\frac{11}{3}} \leq10^{-4}\) since \(c\) between \(7\) and \(8\)
  
  Thus \(T_3,8 = f(8) + \frac{f'(8)}{1!} (x - 8) + \frac{f''(8)}{2!} (x - 8)^2 + \frac{f'''(8)}{3!} (x - 8)^3\)
  \(= 2 + \frac{1}{12} (x - 8) - \frac{1}{288} (x - 8)^2 + \frac{5}{20736} (x - 8)^3\)
  
  Then \(f(7)=\sqrt[3]{7}\approx T_{3,8}(7)\approx1.9130\)
(20 points) Solve the following limits using Taylor Series:
1. \(\lim_{x \to 0} \frac{\ln(1 - x^2)}{x^2}\).
  
  \(\lim_{x\to0}\frac{\ln(1-x^{2})}{x^{2}}=\lim_{x\to0}\frac{T_{2,0}\left(x\right)+R_{2,0}}{x^{2}} =\lim_{x\to0}\frac{\frac{2x^2}{x^2-1}-\frac{x^2\left(x^2-1\right)-2x^4}{\left(x^2-1\right)^2}+Ax^{3}}{x^{2}} =\lim_{x\to0}\frac{\frac{2x^{2}}{x^{2}-1}}{x^{2}}-\lim_{x\to0}\frac{\frac{x^{2}\left(x^{2}-1\right)-2x^{4}}{\left(x^{2}-1\right)^{2}}}{x^{2}} =-2-\left(-1\right)=-1\). 2. \(\lim_{x \to 0} \frac{1 - \cos(\sin(x))}{x^2}\).
  
  \(\lim_{x\to0}\frac{1-\cos(\sin(x))}{x^{2}}=\lim_{x\to0}\frac{1-\left(T_{2,0}\left(x\right)+R_{2,0}\left(x\right)\right)}{x^{2}} =\lim_{x\to0}\frac{1-\left(1-\sin\left(\sin x\right)\cdot\cos\left(x\right)\cdot x+\frac{\cos\left(\sin x\right)\cdot\cos^2(x)-\sin(\sin x)\cdot\sin(x)}{2}x^{2}+Ax^{3}\right)}{x^{2}} =\)
  
  \(=\lim_{x\to0}\frac{\frac{1}{2x^{2}}}{x^{2}}=\frac{1}{2}\)
(30 points) Prove if the following functions are Riemann integrable or not. If they are, give the value for the integral.
1. \(f(x) = c\), for \(c\) constant, on \([a, b]\).
  
  Take \(P_n\) as \(\left\lbrace x_{0},\ldots,x_{n}\right\rbrace\), \(m_{i}(f,P)=c\), \(M_{i}(f,P)=c\)
  
  \(L(f,P_{n})=\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)c=c\left(b-a\right)\) and \(U(f,P_{n})=\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)c=c\left(b-a\right)\)
  
  Then \(L(f,P_{n})=U(f,P_{n})\). Letting \(n \to \infty\), \(L(f)=U(f)=c\left(b-a\right)\). 2. \(f(x) = \begin{cases} \frac{1}{x}, & 0 < x \leq 1 \\ 0, & x = 0 \end{cases} \quad \text{on } [0,1].\)
  
  Take \(P_n\) as \(\left\lbrace0,\frac{1}{n},\frac{2}{n},\dots,\frac{n-1}{n},1\right\rbrace\), \(m_{i}(f,P)=\frac{n}{i}/0\), \(M_{i}(f,P)=\frac{n}{i-1}\)
  
  \(L(f,P_{n})=\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)m_{i}\left(f,P\right)=\frac{1}{n} \frac{n}{2}+\frac{1}{n}\frac{n}{3}+\cdots+\frac{1}{n}=\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}\)
  
  Similarly, \(U(f,P_{n})=1+\frac{1}{2}+\ldots+\frac{1}{n-1}\). Letting \(n \to \infty\), \(L(f),U(f)=\infty\) since \(\sum \frac1n\) is not bounded
  
  Thus it's not Riemann integrable 3. \(f(x) = \begin{cases} 0, & 0 < x \leq 1 \\ 1, & x = 0 \end{cases} \quad \text{on } [0,1].\)
  
  Take \(P_n\) as \(\left\lbrace0,\frac{1}{n},\frac{2}{n},\dots,\frac{n-1}{n},1\right\rbrace\), \(m_{i}(f,P)=0\), \(M_{i}(f,P)=1/0\)
  
  \(L(f,P_{n})=\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)0=0\) and \(U(f,P_{n})=\frac{1}{n}\)
  
  Letting \(n \to \infty\), \(L(f)=U(f)=0\).

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