Key Idea 8.8.1 gives the \(n^th\) term of the Taylor series of common functions. In Exercises 3 – 6, verify the formula given in the Key Idea by finding the first few terms of the Taylor series of the given function and identifying a pattern.
\(f(x)=e^x\) , \(c=0\)
\(f(x)=sinx\) , \(c=0\)
\(f(x)= \frac{1}{1-x}\) , \(c=0\)
\(f(x)=tan^{-1} x\) , \(c=0\)
Since \(c=0\) on all functions above, then we can find the Maclaurin Series of these functions:
\[ \begin{align} \sum_{n=0}^\infty \frac{f^{(n)} (c)}{n!} x^n\\ \end{align} \]
\[ \begin{align} f(x) &= e^x \Rightarrow f(0) =1 \\ f'(x) &= e^x \Rightarrow f'(0) =1 \\ f''(x) &= e^x \Rightarrow f''(0) =1 \\ f'''(x) &= e^x \Rightarrow f'''(0)=1 \\ f^{(4)} (x) &= e^x \Rightarrow f^{(4)} (0) = 1\\ ... \end{align} \]
So the Taylor series expansion of \(f(x)=e^x\) where \(c=0\) is:
\[ \begin{align} e^x &= f(0) + f'(0)x + f''(0) \frac{x^2}{2!} + f'''(0) \frac{x^3}{3!} + f^{(4)} (0) \frac{x^4}{4!} + f^{(5)} (0)\frac{x^5}{5!}+...\\ &= \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + ...\\ &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + ...\\ e^x &= \sum_{n=0} ^{\infty} \frac{x^n}{n!} \end{align} \]
\[ \begin{align} f(x)&=sin x \Rightarrow f(0) = 0\\ f'(x) &= cos x \Rightarrow f'(0) = 1\\ f''(x) &= -sinx \Rightarrow f''(0) = 0\\ f'''(x) &=-cosx \Rightarrow f'''(0) = -1 \\ f^{(4)} (x) &= sinx \Rightarrow f^{(4)}(0) = 0\\ f^{(5)} (x) &= cosx \Rightarrow f^{(5)} (0) = 1\\ f^{(6)} (x) &= -sinx \Rightarrow f^{(6)} (0) = 0\\ f^{(7)} (x) &= -cosx \Rightarrow f^{(7)} (0) = -1\\ ... \end{align} \]
So Taylor series expansion of \(f(x)=sinx\) where \(c=0\) is:
\[ \begin{align} sinx &= f(0) + f'(0)x + f''(0) \frac{x^2}{2!} + f'''(0) \frac{x^3}{3!} + f^{(4)} (0) \frac{x^4}{4!} + f^{(5)}(0) \frac{x^5}{5!}+...\\ &= 0 + (-1)^0 x^1 + 0 + (-1)^1 \frac{x^3}{3!} + 0 +(-1)^2 \frac{x^5}{5!} + 0 + (-1)^3 \frac{x^7}{7!}+...\\ &= 0 + (-1)^0 \frac{x^{(2*0+1)}}{2*0+1} + 0 + (-1)^1 \frac{x^{(2*1+1)}}{(2*1+1)!} + 0 +(-1)^2 \frac{x^{(2*2+1)}}{(2*2+1)!} + 0 + (-1)^3 \frac{x^{(2*3+1)}}{(2*3+1)!}+...\\ sinx &= \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} \end{align} \]
To find the derivative of this function, we can use the quotient rule: \((\frac{u}{v})' = \frac{u'v-uv'}{v^2}\) where \(u=1\) and \(v=1-x\) for the first derivative:
\[ \begin{align} f(x)&=\frac{1}{1-x} \quad \Rightarrow f(0) = 1\\ f'(x) &= -\frac{-1}{(1-x)^2} = \frac{1}{(1-x)^2} \quad \Rightarrow f'(0) = 1\\ f''(x) &= \frac{-2(1-x)(-1)}{(1-x)^4} = \frac{2}{(1-x)^3} \quad \Rightarrow f''(0) = 2\\ f'''(x) &=\frac{(-3)(2)(-1)(1-x)^2}{(1-x)^6} = \frac{6}{(1-x)^4} \quad \Rightarrow f'''(0) = 6 \\ f^{(4)} (x) &= \frac{(-6)(4)(-1)(1-x)^3}{(1-x)^8}=\frac{24}{(1-x)^5} \quad \Rightarrow f^{(4)}(0) = 24\\ f^{(5)} (x) &= \frac{(-24)(5)(-1)(1-x)^4}{(1-x)^10} = \frac{120}{(1-x)^6} \quad \Rightarrow f^{(5)} (0) = 120\\ ... \end{align} \]
So Taylor series expansion for \(f(x)= \frac{1}{1-x}\) where \(c=0\) is:
\[ \begin{align} \frac{1}{1-x} &= f(0) + f'(0)x + f''(0) \frac{x^2}{2!} + f'''(0) \frac{x^3}{3!} + f^{(4)} (0) \frac{x^4}{4!} + f^{(5)} (0)\frac{x^5}{5!}+...\\ &= 1 + x + 2 \frac{x^2}{2!} + 6 \frac{x^3}{3!} + 24 \frac{x^4}{4!} + 120 \frac{x^5}{5!} + ...\\ &= x^0 + x^1 + x^2 + x^3 + x^4 + x^5 + ...\\ \frac{1}{1-x} &= \sum_{n=0}^{\infty} x^n \end{align} \]
\[ \begin{align} f(x)&=tan^{-1} x \quad \Rightarrow f(0) = 0\\ f'(x) &= \frac{1}{1+x^2} \quad \Rightarrow f'(0) = 1\\ f''(x) &= \frac{-2x}{(1+x^2)^2} \quad \Rightarrow f''(0) = 0\\ f'''(x) &=\frac{(-2)(1+x^2)^2 - (-2x)(2)(1+x^2)(2x)}{(1+x^2)^4} = \frac{2(3x^2-1 )}{(1+x^2)^3} \quad \Rightarrow f'''(0) = -2 \\ f^{(4)} (x) &= \frac{-8x(2x^2-3)}{(1+x^2)^4} \quad \Rightarrow f^{(4)}(0) = 0\\ f^{(5)} (x) &= \frac{8(1-10x^2+5x^4)}{(1+x^2)^5} \quad \Rightarrow f^{(5)} (0) = 8\\ ... \end{align} \]
So the Taylor series expansion of \(f(x)=tan^{-1} x\) where \(c=0\) is:
\[ \begin{align} tan^{-1} x &= f(0) + f'(0)x + f''(0) \frac{x^2}{2!} + f'''(0) \frac{x^3}{3!} + f^{(4)} (0) \frac{x^4}{4!} + f^{(5)} (0)\frac{x^5}{5!}+...\\ &= 0 + x + 0 + -2 \frac{x^3}{3!} + 0 + 8 \frac{x^5}{5!} + ...\\ &= x + (-1)^1\frac{x^3}{3} + (-1)^2 \frac{x^5}{5} +...\\ tan^{-1} x &= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} \end{align} \]
Note: the third term on the Taylor series above; \((-1)^2 \frac{x^5}{5}\) , I still don’t know how it was simplified to be that I just write it that way so I can reach the desired sereis on the key idea 8.8.1. Explanation is below:
\[ \begin{align} f^{(5)} (0)\frac{x^5}{5!} &= 8 \frac{x^5}{5!} = 8 \frac{x^5}{120} = \frac{x^5}{15} \neq \frac{x^5}{5} ?? \end{align} \] I still cannot figure out my mistake??