Step-by-step solution for the Taylor series expansion of e^x about x = 0

Here is the complete derivation. **Step 1: Derivatives at zero.** For $f(x) = e^x$, we have $f'(x) = e^x$, $f''(x) = e^x$, and in general $f^{(n)}(x) = e^x$ for all $n$. Evaluating at $x = 0$: $$f(0) = e^0 = 1, \quad f'(0) = 1, \quad f''(0) = 1, \quad \ldots, \quad f^{(n)}(0) = 1$$ **Step 2: Apply the Maclaurin formula.** $$e^x = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = \sum_{n=0}^{\infty} \frac{1}{n!} x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$ **Step 3: Radius of convergence via ratio test.** $$\lim_{n \ o \infty} \left| \frac{a_{n+1}}{a_n} \ ight| = \lim_{n \ o \infty} \left| \frac{x^{n+1}/(n+1)!}{x^n/n!} \ ight| = \lim_{n \ o \infty} \frac{|x|}{n+1} = 0$$ Since $0 < 1$ for all $x$, the series converges for all real $x$. Thus the radius of convergence is $R = \infty$.