2. if the function f is defined by f(x)=∑n = 0∞xnn!, then f(x)=? write the first four nonzero terms and the…

Explanation:

Step1: Recall the power - series derivative rule

If $f(x)=\sum_{n = 0}^{\infty}a_nx^n$, then $f^{\prime}(x)=\sum_{n = 1}^{\infty}na_nx^{n - 1}$. Given $f(x)=\sum_{n=0}^{\infty}\frac{x^{4n}}{n!}$, we apply the derivative rule.

Step2: Differentiate the power - series

$f^{\prime}(x)=\sum_{n = 1}^{\infty}\frac{4n\cdot x^{4n-1}}{n!}=\sum_{n = 1}^{\infty}\frac{4x^{4n - 1}}{(n - 1)!}$.

Step3: Find the first four non - zero terms

When $n = 1$: $a_1=\frac{4x^{4\times1-1}}{(1 - 1)!}=4x^{3}$ When $n = 2$: $a_2=\frac{4x^{4\times2-1}}{(2 - 1)!}=4x^{7}$ When $n = 3$: $a_3=\frac{4x^{4\times3-1}}{(3 - 1)!}=\frac{4x^{11}}{2}=2x^{11}$ When $n = 4$: $a_4=\frac{4x^{4\times4-1}}{(4 - 1)!}=\frac{4x^{15}}{6}=\frac{2}{3}x^{15}$

Step4: Find the general term

The general term of the Taylor series for $f^{\prime}(x)$ about $x = 0$ is $a_n=\frac{4x^{4n-1}}{(n - 1)!}$, for $n\geq1$.

Answer:

The first four non - zero terms of the Taylor series of $f^{\prime}(x)$ about $x = 0$ are $4x^{3},4x^{7},2x^{11},\frac{2}{3}x^{15}$. The general term is $\frac{4x^{4n-1}}{(n - 1)!}$, for $n\geq1$.