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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!

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  • $\begingroup$ if r≠0 in the second case you loos no constant.So to be valid r>=1 $\endgroup$
    – trula
    Commented Jul 22, 2022 at 19:14
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    $\begingroup$ 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. $\endgroup$ Commented Jul 22, 2022 at 19:16

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As eyeballfrog's comment suggests, your first sum is the exception, whereas the second is the rule.

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 $0\times c_0 \times x^0 =0$.

Note that when computing the derivative of the formal power series, we interchange summation and differentiation, which is possible only under certain conditions.

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