I am currently looking through a proof in a book and am failing to understand one particular step. The step is this:
$\sum_{n=0}^{\infty} x(\frac{d}{dx})x^n= x(\frac{d}{dx})\sum_{n=0}^{\infty}x^n$
How is it that you can treat the: "$x(\frac{d}{dx})$" term as a constant here? Does the derivative or $x$ not vary?
Thanks a lot
Lewis
you don't treat is a constant, you just use the fact that
$(f + g)' = f' + g'$
so
$\sum x \frac{d}{dx} x^n = x \frac{d}{dx} x + x \frac{d}{dx} x^2 + ... $
$ = x (\frac{d}{dx} x + \frac{d}{dx} x^2 + ...) = $
$ x \frac{d}{dx} (x + x^2 + ...)$
As noted by others, uniform convergence is required to extend this to infinite sum: Can the Sum Rule for derivatives be extended to infinite series?
The $x$ is indeed a constant for the series, so can be taken out.
The other "taking out" is saying that the derivative of (the series as a function in $x$) equals the series of derivatives of the terms. This is not always the case (it needs things like absolute convergence) for it to hold as functions on the reals, say. Often in a formal treatment of series (not regarding convergence etc.) the fact that this can be done is more like an axiom for a derivative operator. I don't know the context of the question, so it's hard to say which is meant here.
Hint: you don't treat it as a constant. Just try evaluating both sides using the normal differentiation rules and you will see that they come out the same. But it is not because we took the $x\frac{d}{dx}$ out as a constant.
You are not treating the derivative as a constant, you are actually noticing that it is a linear operator, so (under certain conditions) you can commute it with the summation sign. This always works for finite sums, to pass to the limit you have to ask that your series converges uniformly in a neighborhood of the point where you are differentiating (if I recall correctly).
Also, for the $x$ term, if you are looking at the series at a point, then $x$ will not vary and you can certainly treat it as a constant.
