find the taylor polynomial p₅(x) centered at x = 0 for the function f(x)=2x cos(4x²)

Explanation:

Step1: Recall Taylor - series formula

The Taylor series of $\cos t$ centered at $t = 0$ is $\cos t=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}t^{2n}=1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\frac{t^{6}}{6!}+\cdots$.

Step2: Substitute $t = 4x^{2}$ into $\cos t$ series

We get $\cos(4x^{2})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}(4x^{2})^{2n}=1-\frac{(4x^{2})^{2}}{2!}+\frac{(4x^{2})^{4}}{4!}-\cdots=1 - 8x^{4}+\frac{256}{24}x^{8}-\cdots$.

Step3: Multiply by $2x$

$f(x)=2x\cos(4x^{2})=2x\left(1 - 8x^{4}+\frac{256}{24}x^{8}-\cdots\right)=2x-16x^{5}+\frac{256}{12}x^{9}-\cdots$.

Step4: Determine $P_5(x)$

The Taylor polynomial $P_5(x)$ of degree 5 is the sum of the terms with degree less than or equal to 5. So $P_5(x)=2x - 16x^{5}$.

Answer:

$P_5(x)=2x - 16x^{5}$