calculator allowed what is the coefficient of x^6 in the taylor series about x = 0 for the function, f(x) =…

Explanation:

Step1: Recall Taylor - Series formula

The Taylor - series of a function (f(x)) about (x = a) is given by (f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^n). When (a = 0), it is a Maclaurin series: (f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n). First, we know that the Maclaurin series of (e^t=\sum_{n = 0}^{\infty}\frac{t^n}{n!}=1 + t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots).

Step2: Rewrite the given function

We have (f(x)=\frac{e^{2x}}{4}). Let (t = 2x), then (e^{2x}=\sum_{n = 0}^{\infty}\frac{(2x)^n}{n!}=\sum_{n = 0}^{\infty}\frac{2^n x^n}{n!}).

Step3: Find the coefficient of (x^6)

So, (f(x)=\frac{e^{2x}}{4}=\frac{1}{4}\sum_{n = 0}^{\infty}\frac{2^n x^n}{n!}). We want to find the coefficient of (x^6). When (n = 6), the term in the series is (\frac{1}{4}\cdot\frac{2^6x^6}{6!}). Calculate (\frac{1}{4}\cdot\frac{2^6}{6!}=\frac{1}{4}\cdot\frac{64}{720}=\frac{64}{2880}=\frac{2}{90}=\frac{1}{45}).

Answer:

(\frac{1}{45})