Question

Let f(x)= (Summation) 2 ^2n n!/(2n+1)! x^2n+1
at n=0.. =x+2/3x^3+4/3*5x^5+8/3*5*7+...

a) Find the radius of convergence of this power series
b)Show that f'(x)=1+2x*f(x)
c) What is d/dx[(e^-x^2 f(x)]
d) Express f(x) in terms of an integral

Answer 1

\[f (x)\sum_{0}^{\infty}\frac{ 2^{2n}n! }{ (2n+1) } x ^{2n+1}\]

\[=x+\frac{ 2 }{ 3}x^3+\frac{ 4 }{ 3\times5 }x^5+\frac{ 8 }{ 3\times5\times7 }x^7+...\]

Answer 2

a) Find the radius of convergence of this power series
b)Show that f'(x)=1+2x*f(x)
c) What is d/dx[(e^-x^2 f(x)]
d) Express f(x) in terms of an integral

Answer 3

\[f(x)=\sum_{n=0}^\infty \frac{2^{2n}n!}{(2n+1)!}x^{2n+1}\]
You can use the ratio test to determine the radius of convergence:
\[\begin{align*}\lim_{n\to\infty} \left|\frac{\dfrac{2^{2(n+1)}~(n+1)!}{(2(n+1)+1)!}x^{2(n+1)+1}}{\dfrac{2^{2n}~n!}{(2n+1)!}x^{2n+1}}\right|&=\lim_{n\to\infty} \left|\frac{\dfrac{4\cdot2^{2n}~(n+1)!}{(2n+3)!}x^2\cdot x^{2n+1}}{\dfrac{2^{2n}~n!}{(2n+1)!}x^{2n+1}}\right|\\\\\\
&=\lim_{n\to\infty} \left|\frac{\dfrac{4(n+1)~n!}{(2n+3)~(2n+2)~(2n+1)!}x^2}{\dfrac{n!}{(2n+1)!}}\right|\\\\\\
&=\lim_{n\to\infty} \left|\frac{4(n+1)}{(2n+3)~(2n+2)}x^2\right|\\\\\\
&=\cdots
\end{align*}\]
The ratio test tells us that the series diverges if the limit is greater than 1, converges if less than 1, and fails if equal to 1.