I'm reading over solving linear differential equations with analytic coefficients, and finding the solutions that are near regular singular points. In the earlier section on solving similar equations near ordinary points, when you assume the Taylor series of a solution and compute a derivative, sensibly, the index of the sum shifts. For instance, if $\displaystyle y = \sum_{n=0}^{\infty}a_{n}x^{x}$ then $\displaystyle y' = \sum_{n=1}^{\infty}a_{n}x^{n-1}$.
However, when in the context of finding a solution near a regular singular point, we assume $\displaystyle y = x^{r}\sum_{n=0}^{\infty}a_{x}x^{n}$ and after pushing the $x^{r}$ inside the sum, in the book I'm reading, the derivative is given by $\displaystyle y' = \sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r-1}$. Why is it that the index doesn't also shift here?
My guess about the answer to this question is: Well, if $r$ is large, then by taking a derivative you don't necessarily lose a constant term, so the index should remain where it was.
But then my question to that is: Ok, so long as $r>0$ then that's true for the first derivative. But the book also doesn't shift index for the second derivative, and there if $r=1$ then by taking two derivatives you will lose a constant term. Shouldn't this be a consideration in solving these equations?
In fact, an example given in my book had $r=1$ as one of the solutions for $r$ and yet the index wasn't shifted. Is this legitimate?
When the exponent on $x$ is zero, the derivative with respect to $x$ is zero. Therefore the sum
$$\sum_{n=0}^\infty a_nx^n$$
has derivative
$$\sum_{n=1}^\infty na_nx^{n-1}$$
as the $a_0x^0$ term becomes $0$ after deriving.
In contrast, for $r\gt 0$ our sum
$$x^r\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_nx^{n+r}$$ derives as $$\sum_{n=0}^\infty (n+r)a_nx^{n+r-1}\tag 1$$
As you correctly guessed, if $r=0$ then this reduces to the previous form, where the $n=0$ term vanishes.
For the second derivative, assuming that $r\ge1$ then we have a $0$ in the first term. This is ok, and really it would be ok to leave the indexing unshifted in the first derivative too since only a zero term is removed. It would be similar to including $0!$ in a multiplicative definition of the factorial function; we need only multiply down to $1$ (or $2$ really), but we define $0!=1$ and so nothing is gained or lost by including the extra term or removing it.