So, essentially, I want to start with a summation:
$$s = \sum_{k=a}^b{ f(k,x) }$$
and differentiate with respect to $x$, inside the summation. My question is: When does the following equality hold:
$$\frac{ds}{dx} = \sum_{k=a}^b{ \left( f(k,x) \frac{d}{dx} \right) }$$
IMPORTANT NOTE
I am especially interested in the case of infinite sums. Any information on differentiation inside infinite sums will be extremely helpful.
This question is similar to this one.
EXAMPLE
For example, Suppose I have the following sum:
$$s = \sum_{k=a}^b{ \cos(k x) }$$
If I now differentiate with respect to $x$ inside the summation, I get, on the right-hand side:
$$\sum_{k=a}^b{ -k \sin(k x) }$$
I'm wondering when I'm allowed to take the derivative, with respect to $x$, on the left-hand side, to get the same result.
Please note: This is just one example. I'm wondering when I can do this in general, for any $f(k,x)$.
Best Answer
For finite sums where each term is differentiable, the derivative of the sum is always the sum of the derivatives, as is seen inductively from $(f+g)' = f' + g'$.
For infinite sums or series, to be able to differentiate termwise, one needs good enough convergence of the differentiated series. A sufficient condition is locally uniform convergence of the differentiated series, together with pointwise convergence of the original series. If the differentiated series converges locally uniformly, we may interchange integration and summation in
$$\begin{align} \int_a^b \sum_{n=1}^\infty \frac{df_n}{dx}(x)\,dx &= \sum_{n=1}^\infty \int_a^b \frac{df_n}{dx}(x)\,dx\\ &= \sum_{n=1}^\infty \left(f_n(b) - f_n(a)\right)\\ &= \left(\sum_{n=1}^\infty f_n(b)\right) - \left(\sum_{n=1}^\infty f_n(a)\right). \end{align}$$
The argument shows that any kind of convergence that allows the interchange of integration and summation is sufficient. For example, if the derivatives are all non-negative, and the sum function is locally integrable, the monotone convergence theorem yields the result.