Given the differential equation": $$\frac{\partial^2 }{\partial x^2}y(x) + p(x)\frac{\partial }{\partial x}y(x) + r(x) y(x)=0$$

if the solution is assumed to be a power series $y(x)=\psi(x)=\sum_{n=0}^{\infty } a_n(x-x_0)^n$

Then the coefficient $a_n$ can found in terms of $a_0$ and $a_1$ from the differential equation using the fact that:

$$\frac{\partial^n }{\partial x^n}\psi(x_0)=(n!)a_n$$

So it seems to be that we can find a series solution to the equation as long as $p(x)$ and $q(x)$ are infinitely differentiable. However, according to my book this is a weak condition, and $p(x)$ and $q(x)$ not only need to be infinitely differentiable but also analytic at $x_0$

Why is that so? should not the infinitely differentiable condition for $p(x)$ and $q(x)$ guarntee we can find $a_n$ for n = 0 ,1 ,2 ,3 ,4 .... and so on?


Because every analytic function is infinitely differentiable,but the converse need not to be true. In fact Some infinitely differentiable functions don't have a power series representation around some certain points in their domain.

  • $\begingroup$ So why wouldn't infinite differentiability be the only condition since all coefficient can be found using it? $\endgroup$ – NegativeTension Apr 17 '18 at 20:34
  • $\begingroup$ consider the function $ [f\left( x \right)=\left\{ \begin{align} & {{e}^{\frac{-1}{{{x}^{2}}}}},\quad x\ne 0 \\ & 0\quad ,\quad x\ne 0 \\ \end{align} \right.\ $ which is infinitely differentiable. The Taylor series has a positive radius of convergence but does not converge to the original function $\endgroup$ – user547564 Apr 18 '18 at 12:50

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