calculus - calculator allowed\n1. what is the coefficient of x^4 in the taylor series about x = 0 for the…

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). The Maclaurin series of the exponential function (e^t=\sum_{n = 0}^{\infty}\frac{t^n}{n!}=1 + t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots).

Step2: Substitute (t=\frac{x^2}{2}) into the (e^t) series

Let (t=\frac{x^2}{2}). Then (e^{\frac{x^2}{2}}=\sum_{n = 0}^{\infty}\frac{(\frac{x^2}{2})^n}{n!}=\sum_{n = 0}^{\infty}\frac{x^{2n}}{2^n n!}=1+\frac{x^2}{2}+\frac{x^4}{2^2\times2!}+\frac{x^6}{2^3\times3!}+\cdots).

Step3: Identify the coefficient of (x^4)

We want to find the coefficient of (x^4). When (2n = 4) (i.e., (n = 2)) in the series (\sum_{n = 0}^{\infty}\frac{x^{2n}}{2^n n!}), the term is (\frac{x^4}{2^2\times2!}). Since (2^2\times2!=4\times2 = 8), the coefficient of (x^4) is (\frac{1}{8}).

Answer:

(\frac{1}{8})