Are these funcitons linearly independent? Let $a,b,c,\ldots$ be a finite set of distinct positive real numbers.
Are the functions
$(a+x)^{-r}$, where $r$ is a positive real number, linearly independent function on $[0,\infty)$? Are there any references for this? Would the answer depend on r?
 A: Let $0 < a_1 < \dots < a_n$ and assume that $\left( (a_i + x)^{-r} \right)_{i=1}^n$ are linearly dependent over $[0,\infty)$ and so we can find $b_1,\dots,b_n \in \mathbb{R}$ with
$$ \sum_{i=1}^n \frac{b_i}{(a_i + x)^r} = \sum_{i=1}^n \frac{b_i}{e^{r \ln(a_i + x)}}=  0 $$
for all $x \in [0,\infty)$. Note that the function $\sum_{i=1}^n \frac{b_i}{(a_i + x)^r}$ is in fact defined on the interval $(-a_1,\infty)$. In fact, it can be extended to a complex analytic function $z \mapsto \sum_{i=1}^n \frac{b_i}{e^{r \ln(a_i + z)}}$ on the set $\{ z \in \mathbb{C} \, | \, \Re(z) > -a_1 \}$ and by assumption, it vanishes on $[0,\infty)$ and hence it must actually vanish on $\{ z \in \mathbb{C} \, | \, \Re(z) > -a_1 \}$. In particular, we must have
$$ 0 = \lim_{x \to -a_1} \sum_{i=1}^n \frac{b_i}{(a_i + x)^r} = \sum_{i=2}^n \frac{b_i}{(a_i - a_1)^r} + \lim_{x \to -a_1} \frac{b_1}{(x + a_1)^r}$$
which shows that $b_1 = 0$. But then $\sum_{i=1}^n \frac{b_i}{(a_i + x)^r} = \sum_{i=2}^n \frac{b_i}{(a_i + x)^r}$ is actually well-defined on $(-a_2,\infty)$ and repeating the argument we see that $b_2 = 0$, etc.
A: I'll use the same notation as in levap's answer for the setup.  If you have $\sum\limits_{k=1}^n\dfrac{b_k}{(a_k+x)^r}=0$ for $n$ distinct positive $a_k$s and $n$ real $b_k$s, where the equation holds for all $x\geq 0$, then you want to show this implies all $b_k$s are $0$.  We may assume $a_1<a_2<\ldots<a_n$.
If you take the derivative $m$ times on both sides of the equation, you get 
$$\sum\limits_{k=1}^n(-1)^m r(r+1)(r+2)\cdots(r+m-1)\dfrac{b_k}{(a_k+x)^{r+m}}=0,$$ and dividing by the common constant factor yields $\sum\limits_{k=1}^n\dfrac{b_k}{(a_k+x)^{r+m}}=0$, which holds for all $m\in\mathbb N$ and $x>0$.  By continuity it will also hold when $x=0$, so you have $\sum\limits_{k=1}^n\dfrac{b_k}{a_k^{r+m}}=0$ for all $m\in\mathbb N$.  By factoring out $\dfrac{1}{a_1^{r+m}}$ from each term, this implies $$b_1=-\sum_{k=2}^n\dfrac{b_k}{(a_k/a_1)^{r+m}}.$$  Taking the limit as $m\to \infty$, this implies that $b_1=0$.  Then repeat with $$b_2=-\sum_{k=3}^n\dfrac{b_k}{(a_k/a_2)^{r+m}}$$ to get $b_2=0$, and so on.  All $b_k$s are $0$, proving that $\{(a_k+x)^{-r}:k\in\{1,2,\ldots,n\}\}$ is linearly independent.
