I have a question which asks me to use termwise differentation on the series $$\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(n!)^22^{2n}}$$ to show that it satisfies the differential equation $$x^2y''+xy'+x^2y=0$$
I dont understand what this question is asking me to do. I have found the interval and radius of convergence and the first 3 terms of this in previous questions if that is relevant at all? Can someone explain this to me or the method etc so that I know how to do complete it?
On
Excellent question. If you have a power series, you are allowed to differentiate it termwise (within the radius of convergence), just like a polynomial. I.e. $$ \frac{d}{dx}\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty na_nx^{n-1} $$ Now, you can just differentiate the given series and check that it satisfies the differential equation as you would any other function.
Let $$y(x):=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{n!^22^{2n}}.$$ Since a power series can be differentiated term-by-term inside its radius of convergence, we have $$ \begin{align*} y'(x)&=\sum_{n=1}^\infty\frac{(-1)^n2nx^{2n-1}}{n!^22^{2n}}\\ y''(x)&=\sum_{n=1}^\infty\frac{(-1)^n2n(2n-1)x^{2n-2}}{n!^22^{2n}} \end{align*} $$ and they converges absolutely on the same region. So $$ \begin{align*} &x^2y''(x)+xy'(x)+x^2y\\ &=\sum_{n=1}^\infty\frac{(-1)^n2n(2n-1)x^{2n}}{n!^22^{2n}} +\sum_{n=1}^\infty\frac{(-1)^n2nx^{2n}}{n!^22^{2n}} +\sum_{n=0}^\infty\frac{(-1)^nx^{2n+2}}{n!^22^{2n}}\\ &=\sum_{n=1}^\infty\frac{(-1)^n2n(2n-1)x^{2n}}{n!^22^{2n}} +\sum_{n=1}^\infty\frac{(-1)^n2nx^{2n}}{n!^22^{2n}} +\sum_{n=1}^\infty\frac{(-1)^{n-1}x^{2n}}{(n-1)!^22^{2(n-1)}}\\ &=\sum_{n=1}^\infty\frac{(-1)^n}{n!^2 2^{2n}}[2n(2n-1) +2n-4n^2]x^{2n}\\ \end{align*} $$ which is equal to $0$.