If f⁽ⁿ⁾(0) = (n+1)! for n = 0,1,2,., find the Maclaurin series for f and its radius of convergence.

Solution:

f⁽ⁿ⁾(0) = (n+1)! for n = 0,1,2,..., (Given)

We should determine the Maclaurin series for f and its radius of convergence.

We know that the Maclaurin series for the function f is

\(f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n} \)

\(=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2}+....\)

To find the Maclaurin series for f

\(f(x)=\sum_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}x^{n} \)

\(=\sum_{n=0}^{\infty}\frac{(n+1)!}{n!}x^{n} \)

\(=\sum_{n=0}^{\infty }\frac{(n+1)n!}{n!}x^{n} \)

\(=\sum_{n=0}^{\infty}(n+1)x^{n} \)

Therefore, the Maclaurin series for the function f is

\(f(x)=\sum_{n=0}^{\infty }(n+1)x^{n} \)

Ratio test:

The series \(\sum_{n=0}^{\infty}a_{n} \) converges if \(\lim_{n\rightarrow\infty}\left | \frac{a_{n+1}}{a_{n}} \right | \) < 1 and the series diverges if \(\lim_{n\rightarrow \varpi }\left | \frac{a_{n+1}}{a_{n}} \right | \) > 1

For the series \(\sum_{n=0}^{\infty } c_{n}(x-a)^{n}\), the radius of convergence is

The series diverges if |x - a| > R,

The series converges if |x - a| < R

The radius of convergence for \(\sum_{n=0}^{\infty } (n+1)x^{n}\) is

a_n = (n + 1) xⁿ

By applying the ratio test

\(\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_{n}}|=\lim_{n\rightarrow \infty}|\frac{(n+2)x^{n+1}}{(n+1)x^{n}}|\)

\(=|x|\lim_{n\rightarrow \infty}\frac{(n+2)}{(n+1)}\)

\(=|x|\lim_{n\rightarrow\infty}\frac{(1+2/n)}{(1+1/n)}\)

= |x|

From the ratio test, |x| < 1 , the series converges if |x - a| < R.

Therefore, the radius of convergence is 1.