write the first four non - zero terms of the maclaurin series for each function. 3. f(x)=x sin x 4…

Explanation:

Step1: Recall Maclaurin series of $\sin x$

The Maclaurin series of $\sin x=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{(2n+1)!}x^{2n + 1}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots$

Step2: Multiply by $x$ for $f(x)=x\sin x$

$f(x)=x\sin x=x\left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}\right)=x^{2}-\frac{x^{4}}{3!}+\frac{x^{6}}{5!}-\frac{x^{8}}{7!}$ The first four non - zero terms are $x^{2},-\frac{x^{4}}{6},\frac{x^{6}}{120},-\frac{x^{8}}{5040}$

Step3: Recall Maclaurin series of $e^{x}$

The Maclaurin series of $e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1 + x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots$

Step4: Substitute $x$ with $3x$ for $f(x)=e^{3x}$

$f(x)=e^{3x}=\sum_{n = 0}^{\infty}\frac{(3x)^{n}}{n!}=1+3x+\frac{(3x)^{2}}{2!}+\frac{(3x)^{3}}{3!}=1 + 3x+\frac{9x^{2}}{2}+\frac{27x^{3}}{6}=1+3x+\frac{9x^{2}}{2}+\frac{9x^{3}}{2}$ The first four non - zero terms are $1,3x,\frac{9x^{2}}{2},\frac{9x^{3}}{2}$

Step5: Recall Maclaurin series of $\cos x$

The Maclaurin series of $\cos x=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{2n}=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots$

Step6: Find $f(x)=x - 1+\cos x$

$f(x)=x-1+\left(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}\right)=x-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}$ The first four non - zero terms are $x,-\frac{x^{2}}{2},\frac{x^{4}}{24},-\frac{x^{6}}{720}$

Answer:

For $f(x)=x\sin x$: $x^{2},-\frac{x^{4}}{6},\frac{x^{6}}{120},-\frac{x^{8}}{5040}$ For $f(x)=e^{3x}$: $1,3x,\frac{9x^{2}}{2},\frac{9x^{3}}{2}$ For $f(x)=x - 1+\cos x$: $x,-\frac{x^{2}}{2},\frac{x^{4}}{24},-\frac{x^{6}}{720}$