Answer to Question #137236 in Calculus for Usman

Question

1- F(x)=sin(x^4) About x=0 find taylor series.
2- f(x)=6x^2 cos(7x^5) About x=0

Answer 1

$(1)\\ \textsf{The Taylor series of}\hspace{0.1cm}\sin{x} \hspace{0.1cm} \textsf{about}\hspace{0.1cm} x = 0 \hspace{0.1cm} \textsf{is}\\ \displaystyle\sin{x} = \sum_{k = 0}^\infty\frac{x^{2k + 1} (-1)^k}{(2k + 1)!} \\ \displaystyle\sin{x} = x^4 - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} +...\\ \displaystyle\sin(x^4) = x^4 - \frac{(x^4)^3}{3!} + \frac{(x^4)^5}{5!} - \frac{(x^4)^7}{7!} + \frac{(x^4)^9}{9!} - \frac{(x^4)^{11}}{11!} +...\\ \displaystyle\sin(x^4) = x^4 - \frac{x^{12}}{3!} + \frac{x^{20}}{5!} - \frac{x^{28}}{7!} + \frac{x^{36}}{9!} - \frac{x^{44}}{11!} +...\\ (2)\\ \textsf{The Taylor series of}\hspace{0.1cm}\cos{x} \hspace{0.1cm} \textsf{about}\hspace{0.1cm} x = 0 \hspace{0.1cm} \textsf{is}\\ \displaystyle\cos{x} = \sum_{k = 0}^\infty\frac{x^{2k} (-1)^k}{(2k)!} \\ \displaystyle\cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} +...\\ \displaystyle\cos(7x^5) = 1 - \frac{(7x^5)^2}{2!} + \frac{(7x^5)^4}{4!} - \frac{(7x^5)^6}{6!} + \frac{(7x^5)^8}{8!} - \frac{(7x^5)^{10}}{10!} +...\\ \displaystyle 6x^2\cos(7x^5) = 6x^2 - \frac{6.x^{12}.7^2}{2!} + \frac{6.x^{22}.7^4}{4!} - \frac{6.x^{32}.7^6}{6!} + \frac{6.x^{42}.7^8}{8!} - \frac{6.x^{52}.7^{10}}{10!} +...\\$