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

example video example find the maclaurin series of the function f(x) = e^2x and its radius of convergence. solution if f(x) = e^2x, then f^(n)(x) = , so f^(n)(0) = 2^n*e^0 = for all n. therefore the taylor series for f at 0 (that is, the maclaurin series) is given. sum(n = 0 to infinity) f^(n)(0)/n! * x^n = sum(n = 0 to infinity) 2^n*x^n/n! = 1 + 2x/1! + + 8x^3/3! + ... to find the radius of convergence we let a_n = 2^n*x^n/n!. then we have the following. |a_n + 1/a_n| = * n!/2^n*x^n = 2|x|/n + 1 -> < 1 so, by the ratio test, the series converges for all x and the radius of convergence is r = infinity. need help? read it submit answer

example video example find the maclaurin series of the function f(x) = e^2x and its radius of convergence. solution if f(x) = e^2x, then f^(n)(x) = , so f^(n)(0) = 2^n*e^0 = for all n. therefore the taylor series for f at 0 (that is, the maclaurin series) is given. sum(n = 0 to infinity) f^(n)(0)/n! * x^n = sum(n = 0 to infinity) 2^n*x^n/n! = 1 + 2x/1! + + 8x^3/3! + ... to find the radius of convergence we let a_n = 2^n*x^n/n!. then we have the following. |a_n + 1/a_n| = * n!/2^n*x^n = 2|x|/n + 1 -> < 1 so, by the ratio test, the series converges for all x and the radius of convergence is r = infinity. need help? read it submit answer

Answer

Explanation:

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

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

Step2: Write out the Maclaurin series

The Maclaurin series $\sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}=\sum_{n = 0}^{\infty}\frac{2^{n}x^{n}}{n!}=1+\frac{2x}{1!}+\frac{4x^{2}}{2!}+\frac{8x^{3}}{3!}+\cdots$.

Step3: Apply the Ratio Test for radius of convergence

Let $a_{n}=\frac{2^{n}x^{n}}{n!}$, then $\left|\frac{a_{n + 1}}{a_{n}}\right|=\left|\frac{\frac{2^{n+1}x^{n + 1}}{(n + 1)!}}{\frac{2^{n}x^{n}}{n!}}\right|=\left|\frac{2^{n+1}x^{n + 1}n!}{2^{n}x^{n}(n + 1)!}\right|=\frac{2|x|}{n + 1}$. As $n\to\infty$, $\lim_{n\rightarrow\infty}\frac{2|x|}{n + 1}=0<1$ for all $x$.

Answer:

$f^{(n)}(x)=2^{n}e^{2x}$; $f^{(n)}(0)=2^{n}$; $\frac{4x^{2}}{2!}$; $\left|\frac{2^{n+1}x^{n + 1}n!}{2^{n}x^{n}(n + 1)!}\right|$; $0$