# Different indexes in derivatives of summations

I have a question about derivatives of summations while working with ODEs. From what I know (and have seen up until now), to take the derivative of a summation, such as a power series, the index of the sum has to increase by 1 to account for the loss of the constant in the original sum:

$$(\sum_{n=0}^{\infty} c_n x^n)'=\sum_{n=1}^{\infty} n c_n x^{n-1}$$

However, while working on the Frobenius' method for solving ODEs about singular points, I have found that the derivatives of indexes do not change, such that, if $$y=\sum_{n=0}^{\infty} c_n x^{n+r}$$, then

$$y'=\sum_{n=0}^{\infty} (n+r) c_n x^{n+r-1}$$.

And similarly for $$y''$$. I have no idea why this "exception" is the case. Could anyone help with this? Many thanks!

• if r≠0 in the second case you loos no constant.So to be valid r>=1 Commented Jul 22, 2022 at 19:14
• Note that the sum is still valid starting at $n = 0$. $0 x^{-1} = 0$, and adding $0$ to a sum doesn't change it. Commented Jul 22, 2022 at 19:16

Consider $$y(x)=\sum_{n=0}^{\infty} c_n x^{n+r},$$ then $$\frac{dy}{dx}=\sum_{n=0}^{\infty} (n+r) c_n x^{n+r}.$$
For $$r\neq 0$$, that sum stays as it is whereas for $$r=0$$ we have
$$\frac{dy}{dx}=\sum_{n=0}^{\infty} (n+r) c_n x^{n+r}=\sum_{n=0}^{\infty} (n) c_n x^{n}=\sum_{n=1}^{\infty} n c_n x^{n},$$
because the $$n=0$$ term equates to $$0\times c_0 \times x^0 =0$$.