Find the Maclaurin series for $f(x)=\sin (5 x)$ .
A) $5 x-\frac{5^{3} x^{3}}{3 !}+\frac{5^{5} x^{5}}{5 !}-$
B) $1-\frac{5^{2} x^{2}}{2 !}+\frac{5^{4} x^{4}}{4 !}-\cdots$
C) $x-\frac{5^{2} x^{3}}{3 !}+\frac{5^{4} x^{5}}{5 !}-\cdots$
D) $5 x-\frac{5^{2} x^{2}}{2 !}+\frac{5^{3} x^{3}}{3 !}-\cdots$

Question

Quizplus · Accepted Answer

The Maclaurin series for $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$. For $\sin(5x)$, replace $x$ with $5x$ in the series: $5x - \frac{(5x)^3}{3!} + \frac{(5x)^5}{5!} - \cdots$, which simplifies to $5x - \frac{5^3 x^3}{3!} + \frac{5^5 x^5}{5!} - \cdots$.

Find the Maclaurin Series For f(x)=sin⁡(5x)f(x)=\sin (5 x)f(x)=sin(5x) A) 5x−53x33!+55x55!−5 x-\frac{5^{3} x^{3}}{3 !}+\frac{5^{5} x^{5}}{5 !}-5x−3!53x3​+5!55x5​−

Find the Maclaurin Series For $f(x)=\sin (5 x)$
A) $5 x-\frac{5^{3} x^{3}}{3 !}+\frac{5^{5} x^{5}}{5 !}-$