Need Help?

  • Notes
  • Comments & Questions

Find the Maclaurin series for \( \sin (x) \) and prove that it represents \( \sin (x) \) for all \(x\)

\( f(0)+\frac{f^{\prime}(0)}{1 !} x +\frac{f^{''}(0)^{2}}{2 !}x^{2} +\frac{f'''(0)}{3 !}x^{3}\cdots \)

\( =x-\frac{x^{3}}{3!}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7!} \cdots \)

\( =\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !} \)

Since \( f^{(n+1)}(x) \) is \( \pm \sin x \) or \( \pm \cos x \Rightarrow \left|f^{n+1}(x)\right| \leq 1 \Rightarrow M=1 \)

\( \left|R_{n}(x)\right| \leq \frac{M}{(n+1) !} \left|x^{n+1}\right| =\frac{|x|^{n+1}}{(n+1) !} \)

\( \sin (x)= x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !} \cdots =\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !} \)

\( \cos (x)=\frac{d}{d x}(\sin x) \)

\( \cos (x)= 1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+ \cdots =\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2 n}}{(2 n) !} \)

No comments yet

Join the conversation

Join Notatee Today!