calculus - calculator allowed\n1. what is the coefficient of x^4 in the taylor series for the function…

Explanation:

Step1: Recall Taylor - series formulas

The Taylor series of (e^{x}=\sum_{n = 0}^{\infty}\frac{x^{n}}{n!}=1 + x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\cdots) and (\cos x=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}x^{2n}}{(2n)!}=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots).

Step2: Multiply the two series

[ \begin{align*} e^{x}\cos x&=\left(1 + x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\cdots\right)\left(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots\right)\ \end{align*} ] To find the coefficient of (x^{4}), we consider the following products:

1. (1\times\frac{x^{4}}{4!}=\frac{x^{4}}{24})
2. (\frac{x^{2}}{2!}\times\left(-\frac{x^{2}}{2!}\right)=\frac{x^{2}}{2}\times\left(-\frac{x^{2}}{2}\right)=-\frac{x^{4}}{4})
3. (\frac{x^{4}}{4!}\times1=\frac{x^{4}}{24})

Step3: Sum up the contributions

[ \begin{align*} \text{Coefficient of }x^{4}&=\frac{1}{24}-\frac{1}{4}+\frac{1}{24}\ &=\frac{1 - 6+1}{24}\ &=-\frac{4}{24}\ &=-\frac{1}{6} \end{align*} ]

Answer:

(-\frac{1}{6})