Integrating a Function as a Power Series
Integrating a Function as a Power Series

Why there is no need to decrease n by 1 in the case of integration, while it increases 1 for n in the case of differentiation in this theorem?

Does using integration for representing power series requires n-1 as the starting point?

• $\begingroup$ Can you clarify your question? When integrating or differentiating a power series, you do so term by term following the usual rules. The result is the power series expansion for $\int f$ or $f'(x)$, respectively. $\endgroup$– gabeNov 22, 2016 at 5:05
• $\begingroup$ Look at the derivative and integral of $x^n$. That might help. Although you are a CS student, I would suggest learn calculus nicely…it helps a lot. $\endgroup$ Nov 22, 2016 at 5:10

$$f'(x) = \sum_{n=0}^{\infty} \frac{d}{dx} \left[ c_n (x-a)^n \right]$$
For $n=0$, we have $c_n (x-a)^n = c_0 (x-a)^0 = x_0 \cdot 1 =c_0$. Thus $\frac{d}{dx} c_0 =0$ because the derivative of a constant is always zero. Therefore, we can rewrite the above as: $$f'(x) = \sum_{n=0}^{\infty} \frac{d}{dx} \left[ c_n (x-a)^n \right] =\frac{d}{dx}\left[c_0 \right]+ \sum_{n=1}^{\infty} \frac{d}{dx}\left[c_n (x-a)^n \right] = 0+ \sum_{n=1}^{\infty} n\cdot c_n(x-a)^{n-1}.$$
In other words, the book is being lazy to some extent and just ignoring the $0$th term in the power series corresponding to the derivative because it is equal to $0$. There isn’t any “need” to “increase the dummy index by $1$” for differentiation.