Linear independence of $x^{a_1}, x^{a_2}, ... , x^{a_n}, x^{a_n} \ln (x)$ I am trying to prove linear independence of $x^{a_1}, ... , x^{a_n}, x^{a_n}\cdot \ln x, x > 0$.
I understand how to prove linear independence of the first $n$ functions but how would one expand it to the power functions with natural log?
Any ideas are greatly appreciated.
 A: EDIT/Revision: Let $E={x\partial\over \partial x}$ be the "one-dimensional Euler operator". Then $Ex^a=a\cdot x^a$, and $(E-b)x^a=(a-b)\cdot x^a$. Thus, $E-a$ annihilates $x^a$ and multiplies $x^b$ for $b\not=a$ by a non-zero constant. Also, $(E-a)(x^a\log x)=x^a$.
Thus, given a linear relation $cx^{a_n}+\sum_i c_i x^{a_i}=0$ for all $x$ in some non-empty interval, application of $(E-a_1)\ldots(E-a_n)$ annihilates the multiples of $x^{a_1},\ldots,x^{a_n}$, and multiplies the log term by a non-zero constant. This implies that the coefficient of the log term is zero.
For the rest, we can say we're doing an induction on the size of the smallest non-trivial relation, apply $E-a_j$ for some $a_j$ appearing non-trivially, and obtain a shorter relation, contradiction.
A: Suppose $$c_1x^{a_1}+\cdots +c_nx^{a_n}+c_{n+1}x^{a_{n+1}}\ln x=0 \text{ for all }x>0.$$
As $c\to 0^+$, all summands but the last one tend to $0$, whereas the last is unbounded - unless $c_{n+1}=0$. After that, you are left with the cas you already know.
A: It is require that all the $a_j$'s are distinct, otherwise the statement is false.
Suppose then $a_j$'s are distinct. We consider the $x\in(0,\infty)$. For finite subintervals $(a, b)\subset(0,\infty)$ see the note at the end.
Suppose
$$ \sum^n_{j=1}c_jx^{a_j} + \log x c_{n+1} x^{a_n}\equiv0$$

*

*If  $c_{n+1}=0$ then all other $c_n's$ are $0$ since $\{x^{a_1},\ldots,x^{a_n}\}$ is a collection of linearly independent functions.



*

*We prove that $c_{n+1}=0$: Suppose $c_{n+1}\neq0$. Then not all the of other $c_j$'s are zero (otherwise $\log x\equiv0$ for all $x>0$ which is nonsense). Then
$$
\begin{align}
\log x=\sum^n_{j=1}c_jx^{(a_j-a_n)}\tag{1}\label{one}
\end{align}
$$
Let $a_k-a_n=\max_{1\leq j\leq n}\{a_j-a_n:c_j\neq0\}$ (this is well defined since $\{j:1\leq j\leq n, \, c_j\neq0\}\neq\emptyset$).
Then
$$ x^{-(a_k-a_n)}\log x=\sum^n_{j=1}c_jx^{(a_j-a_k)}$$
Taking the limit as $x\rightarrow\infty$ gives $0=c_k$ (recall that $x^{-\alpha}\log x\xrightarrow{x\rightarrow\infty}0$ for $\alpha>0$, and $x^{-\beta}\xrightarrow{x\rightarrow\infty}0$ for $\beta>0$). This is a contradiction to the choice of $a_k-a_n$ ($c_k\neq0)$. The contradiction is derived by assuming that $c_{n+1}\neq0$.



*

*Hence
$c_{n+1}=0$, in which case all other $c_j$'s are zero


*Linearly independence of $\{x^{a_1},\ldots, x^{a_n},x^{a_n}\log x\}$ follows.

Edit: The result for $x\in (a, b)\subset(0,\infty)$ follows from the result in $(0,\infty)$. To see this notice that the functions $f(z)=z^{a_n-a_k}\log(z)$ and $g(z)=\sum^n_{j=1}c_jz^{a_j-a_k}$ are analytic on $D=(0,\infty)\times\mathbb{R}$ (take the standard bath of $\log$). Then, if $f(z)=g(z)$ for all $z\in (a, b)\times\{0\}$, then $f\equiv g$ in$D$. This follows from the following well known result in complex analysis:
Theorem: Assume $f$ is analytic in an open connected set $D\subset\mathbb{C}$. Let $T\subset D$ and assume $T$ has an accumulation point $z_0\in D$. If $f(z)=0$ on $T$, then $f\equiv 0$ in $D$.
(This can be found in many textbooks in analysis or complex variables. For example, Apostol's Mathematical Analysis, 2nd edition, pp. 452)
A: If there were a nontrivial linear dependence with the coefficient of $x^{a_n}\ln x$ nonzero, then dividing by $x^{a_n}$ and rearranging, we could write
$$\ln(x)=\sum k_i x^{\alpha_i}$$
for some scalars $k_i,\alpha_i\in \mathbb R$.  Since $\lim_{c\to \infty} \ln(x)=\infty$, at least one   $\alpha_j$ is positive (and the coefficient on the largest one is positive).  Let $\epsilon\in \mathbb R$ such that $0<\epsilon < \alpha_j$.  Then
$$\lim_{x\to\ inffty} \ln(x)/x^{\epsilon} =0.$$
However, $\sum k_i x^{\alpha_i -\epsilon}$ contains a term with positive exponent, so
$$\lim_{x\to \infty} \sum k_i x^{\alpha_i -\epsilon}=\infty.$$
This is a contradiction, so $\ln(x)$ cannot be written in this form, and hence there cannot be a linear dependence.
