Question

(a) Show that the function
f(x) = \[\sum_{k=0}^{infinity}\] ((-1)^k * x^2k) / (2^2k (k!)^2)
satisfies the differential equation
x^2 f'' (x) + xf'(x) + x^2 f(x) = 0
(b) Evaluate integral from 0 to infinity f(x) dx correct to three decimal places.

Answer 1

@mathaddict @math111

Answer 2

Take the derivative of the summation:
\[\begin{align*}f(x)&=\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{2^{2k}(k!)^2}\\
f'(x)&=\sum_{k=1}^\infty \frac{(-1)^k2kx^{2k-1}}{2^{2k}(k!)^2}\\
&=\sum_{k=1}^\infty \frac{(-1)^kkx^{2k-1}}{2^{2k-1}(k!)^2}\\
f''(x)&=\sum_{k=2}^\infty \frac{(-1)^kk(2k-1)x^{2k-2}}{2^{2k-1}(k!)^2}
\end{align*}\]
Substitute into the differential equation:
\[\begin{align*}x^2\color{red}{f''(x)}+x\color{blue}{f'(x)}+x^2\color{green}{f(x)}&=0\\
x^2\color{red}{\sum_{k=2}^\infty \frac{(-1)^kk(2k-1)x^{2k-2}}{2^{2k-1}(k!)^2}}+x\color{blue}{\sum_{k=1}^\infty \frac{(-1)^kkx^{2k-1}}{2^{2k-1}(k!)^2}}+x^2\color{green}{\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{2^{2k}(k!)^2}}&=0\\
\sum_{k=2}^\infty \frac{(-1)^kk(2k-1)x^{2k}}{2^{2k-1}(k!)^2}+\sum_{k=1}^\infty \frac{(-1)^kkx^{2k}}{2^{2k-1}(k!)^2}+\sum_{k=0}^\infty \frac{(-1)^kx^{2k+2}}{2^{2k}(k!)^2}&=0
\end{align*}\]
Now, I think the next step is to write out a few terms of each series and somehow show that they all cancel each other out and sum to 0.