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I apologize in advance if this question has been asked before, but I am unable to find it. How would one prove $\{e^{ix} x^j\}_{i,j\in \mathbb Z},\,x\in \mathbf R$ is linearly independent?

Hans
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ketsi
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1 Answers1

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We use induction in the number of distinct terms in the liner combination $$c_1e^{n_1t}t^{m_1}+\ldots + c_ke^{n_kt}t^{m_k}=0$$ For $k=1$ the statement holds trivially.

Suppose the any $1\leq j<k$ distinct terms of the form $t^me^{nt}$ are linearly independent. Consider the sum $$c_1t^{m_1}e^{n_1t}+c_2e^{n_2t}t^{m_2}+\ldots+c_ke^{n_kt}t^{m_k}=0$$ Suppose the terms in the sum above have been ordered in such a way that $m_1\leq m_2\leq \ldots \leq m_k$. Dividing by $t^{m_1}$ we get that $$\begin{align} c_1e^{n_1t}+c_2e^{n_2t}t^{m_1-m_2}+\ldots+c_ke^{n_kt}t^{m_k-m_1}=0\tag{1}\label{one} \end{align}$$

  • If all $n_j$, $1\leq j\leq k$, are the same, then the $m_j$ are all distinct and $$c_1+c_2t^{m_2-m_2}+\ldots + t^{m_k-m_1}=0$$ for all $t$, which means that $c_1=\ldots=c_k=0$ by the fundamental theorem of algebra.

  • If not all the $n_j$'s are the same, then we can rearrange the terms in \eqref{one} as $$p_1(t)e^{n'_1t}+\ldots+ p_\ell(t)e^{n'_\ell t}=0$$ so that $p_1,\ldots, p_\ell$ are polynomials and $n'_1<\ldots< n'_\ell$. Hence $$p_1(t)e^{(n'_1-n'_\ell)t}+\ldots + p_{\ell-1}(t)e^{(n'_{\ell-1}-n_\ell)t}+ p_\ell(t)=0$$ for all $t$. Letting $t\rightarrow\infty$ yields $$0=\lim_{t\rightarrow\infty}|p_1(t)e^{(n'_1-n'_\ell)t}+\ldots + p_{\ell-1}(t)e^{(n'_{\ell-1}-n_\ell)t}+ p_\ell(t)|=\lim_{t\rightarrow\infty}|p_\ell(t)|$$ Hence $p_\ell(t)=0$ (otherwise $0=\lim_{t\rightarrow\infty}|p_\ell(t)|>0$ which is absurd. This reduced the terms number of terms in \eqref{one} and we can then apply the induction hypothesis.

Mittens
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  • This makes sense, I figured out it will be related to limits since for example $e^{3x}$ dominates any $c\cdot e^x$ in the long run ($x\to\infty$). Thanks! – ketsi Sep 21 '21 at 04:02
  • +1. My answer with the same approach (but over $\mathbb{C}$, not $\mathbb{R}$). – metamorphy Sep 21 '21 at 04:32
  • Good answer. But what if the domain of $t$ is bounded? – Hans Sep 21 '21 at 05:04
  • Limiting your domain to a bounded region would make the problem more difficult. Notice that in the course if the solution (before taking notable values for $t$ ($0$ or $\pm\infty$) it is enough to consider elements of the set of functions in your OP of the form $t^me^{nt}$ with $m\geq0$ and $n\geq0$. The functions involve then are defined all over the real line (they are even entire functions, that is analytic over $\mathbb{C}$). One should take advantage of that. – Mittens Sep 21 '21 at 08:32
  • I am fully aware that the bounded domain made the question harder. That is precisely why I pose the question. – Hans Sep 22 '21 at 20:38
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    Not that harder. For example if you are in an interval $I$ with nonempty interior, the result follows from the linear independence in $\mathbb{R}$. The reason is that, as I mentioned in my comment, it is enough to consider terms of the form $c_1t^{m_1}e^{n_1t}+\ldots +c_k t^{m_k}e^{n_kt}=f(t)\equiv0$ with $m_k\geq0$ and $n_k\geq0$. with $t\in I$. $f$ admits a unique entire extension to all of $\mathbb{C}$, and in particular, the restriction to $\mathbb{R}$ will imply that all the $c_j$'s equal to $0$. In fact, as long as $I$ admits accumulation points, the conclusion will hold ). – Mittens Sep 22 '21 at 20:53
  • Recall that if $f$ is analytic in a domain $D\subset\mathbb{C}$ and $f=0$ on a set $A\subset D$ that has accumulations points in $D$, then $f\equiv0$. – Mittens Sep 22 '21 at 20:54
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    @Hans: In case you insist on "real" methods, a possible approach is via power series (each of the functions we're considering here has a power series expansion converging everywhere; a power series representing $0$ on any set with nonempty interior is the only one, meaning identically zero). – metamorphy Sep 22 '21 at 21:02
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    I just realized the problem could be resolved through analytical continuation and came back to say as much. Good answer +1 – Hans Sep 22 '21 at 21:11
  • @metamorphy: Agree. – Hans Sep 22 '21 at 21:12