use a maclaurin series in this table to obtain the maclaurin series for the given function. f(x)=11x…

use a maclaurin series in this table to obtain the maclaurin series for the given function. f(x)=11x cos(1/15 x^2) step 1 we are asked to find the maclaurin series for a function involving cos(x). recall the maclaurin series for cos(x). cos(x)=sum(n = 0 to infinity) (-1)^n x^(2n)/(2n)! the same equality would be true for any variable, and in particular for u = 1/15 x^2. therefore, the maclaurin series for cos(1/15 x^2) is sum(n = 0 to infinity) (-1)^n ( )^(2n)/(2n)!=sum(n = 0 to infinity) (-1)^n x^(4n)/ (2n)!.

use a maclaurin series in this table to obtain the maclaurin series for the given function. f(x)=11x cos(1/15 x^2) step 1 we are asked to find the maclaurin series for a function involving cos(x). recall the maclaurin series for cos(x). cos(x)=sum(n = 0 to infinity) (-1)^n x^(2n)/(2n)! the same equality would be true for any variable, and in particular for u = 1/15 x^2. therefore, the maclaurin series for cos(1/15 x^2) is sum(n = 0 to infinity) (-1)^n ( )^(2n)/(2n)!=sum(n = 0 to infinity) (-1)^n x^(4n)/ (2n)!.

Answer

Explanation:

Step1: Substitute $u = \frac{1}{15}x^{2}$ into $\cos(u)$ series

We know $\cos(u)=\sum_{n = 0}^{\infty}(-1)^{n}\frac{u^{2n}}{(2n)!}$. Substituting $u=\frac{1}{15}x^{2}$, we get $\sum_{n = 0}^{\infty}(-1)^{n}\frac{(\frac{1}{15}x^{2})^{2n}}{(2n)!}=\sum_{n = 0}^{\infty}(-1)^{n}\frac{x^{4n}}{15^{2n}(2n)!}$.

Step2: Multiply by $11x$

Since $f(x)=11x\cos(\frac{1}{15}x^{2})$, we multiply the series of $\cos(\frac{1}{15}x^{2})$ by $11x$. So $f(x)=11x\sum_{n = 0}^{\infty}(-1)^{n}\frac{x^{4n}}{15^{2n}(2n)!}=\sum_{n = 0}^{\infty}(-1)^{n}\frac{11x^{4n + 1}}{15^{2n}(2n)!}$.

Answer:

$\sum_{n = 0}^{\infty}(-1)^{n}\frac{11x^{4n+1}}{15^{2n}(2n)!}$