video example find the maclaurin series of the function f(x) = e^2x and its radius of convergence. solution…

Explanation:

Step1: Find n - th derivative of $f(x)$

If $f(x)=e^{2x}$, by the chain - rule, $f^{(n)}(x)=2^{n}e^{2x}$. Evaluating at $x = 0$, we get $f^{(n)}(0)=2^{n}e^{0}=2^{n}$.

Step2: Write the Maclaurin series

The Maclaurin series of a function $f(x)$ is given by $\sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}$. Substituting $f^{(n)}(0)=2^{n}$, we have $\sum_{n = 0}^{\infty}\frac{2^{n}x^{n}}{n!}=1+\frac{2x}{1!}+\frac{4x^{2}}{2!}+\frac{8x^{3}}{3!}+\cdots$.

Step3: Find the radius of convergence using the Ratio Test

Let $a_{n}=\frac{2^{n}x^{n}}{n!}$. Then $a_{n + 1}=\frac{2^{n+1}x^{n + 1}}{(n + 1)!}$. Calculate $\left|\frac{a_{n+1}}{a_{n}}\right|=\left|\frac{\frac{2^{n + 1}x^{n+1}}{(n + 1)!}}{\frac{2^{n}x^{n}}{n!}}\right|=\left|\frac{2^{n+1}x^{n+1}n!}{2^{n}x^{n}(n + 1)!}\right|=\frac{2|x|}{n + 1}$. As $n\to\infty$, $\lim_{n\rightarrow\infty}\frac{2|x|}{n + 1}=0$. Since $0<1$ for all $x\in R$, the series converges for all $x$.

Answer:

The Maclaurin series of $f(x)=e^{2x}$ is $\sum_{n = 0}^{\infty}\frac{2^{n}x^{n}}{n!}=1 + 2x+\frac{4x^{2}}{2!}+\frac{8x^{3}}{3!}+\cdots$, and the radius of convergence $R=\infty$.