Examine linear dependence of two general systems 1^° 1, x…

Examine linear dependence of two general systems $1^\circ\quad 1,\ x,\ x^2,\ \ldots,\ x^n;\qquad\qquad 2^\circ\quad e^{k_1x},\ e^{k_2x},\ \ldots,\ e^{k_nx},$ where in $2^\circ$ the exponents are pairwise distinct: $k_i\neq k_j$ for $i\neq j$.

Step 1: System $1^\circ$: proof by definition

Suppose there exist constants $c_0,c_1,\ldots,c_n$ with

\[ c_0+c_1x+c_2x^2+\cdots+c_nx^n\equiv0\quad\text{for all }x. \tag{1} \]

The left side is a polynomial of degree at most $n$ that vanishes identically; a polynomial identically equal to zero has every coefficient equal to zero. Hence $c_0=c_1=\cdots=c_n=0$, so only the trivial combination gives the zero function.

Alternative: differentiating (1) successively and evaluating at $x=0$ gives $c_0=0$, then $c_1=0$, then $k!\,c_k=0\Rightarrow c_k=0$ at each stage – the same conclusion by a different route.

Step 2: System $1^\circ$: Wronskian proof

The $r$-th derivative of $x^c$ vanishes whenever $r>c$, so in the matrix of derivatives $0,1,\ldots,n$ (rows) applied to $1,x,\ldots,x^n$ (columns), every entry below the diagonal is zero: the Wronskian is upper triangular. Its diagonal entries are the $k$-th derivative of $x^k$ at each $k$, namely $k!$:

\[ W(x)=\begin{vmatrix} 1&x&x^2&\cdots&x^n\\ 0&1&2x&\cdots&nx^{n-1}\\ 0&0&2&\cdots&n(n-1)x^{n-2}\\ \vdots& & &\ddots&\vdots\\ 0&0&0&\cdots&n!\end{vmatrix}=\prod_{k=0}^{n}k!\;=\;0!\,1!\,2!\cdots n!. \]

This is a nonzero constant, independent of $x$, so the system is independent.

$1,\,x,\,x^2,\,\ldots,\,x^n:\quad W(x)=\prod_{k=0$.

Step 3: System $2^\circ$: proof by induction on $n$

Base case $n=1$: a single nonzero function $e^{k_1x}$ is trivially independent.

Inductive step: assume the claim for any $n-1$ exponentials with distinct exponents. Suppose

\[ c_1e^{k_1x}+c_2e^{k_2x}+\cdots+c_ne^{k_nx}\equiv0. \tag{2} \]

Divide by $e^{k_nx}>0$, then differentiate to eliminate the resulting constant term: \[c_1(k_1-k_n)e^{(k_1-k_n)x}+\cdots+c_{n-1}(k_{n-1}-k_n)e^{(k_{n-1}-k_n)x}\equiv0. \tag{3}\]

The $n-1$ exponents $k_i-k_n$ $(i

Step 4: System $2^\circ$: Wronskian proof (Vandermonde)

Since the $j$-th derivative of $e^{k_ix}$ is $k_i^{\,j}e^{k_ix}$, factoring $e^{k_ix}$ out of column $i$ leaves exactly the Vandermonde matrix in $k_1,\ldots,k_n$: \[W(x)=e^{k_1x}e^{k_2x}\cdots e^{k_nx} \begin{vmatrix}1&1&\cdots&1\\ k_1&k_2&\cdots&k_n\\ \vdots& &&\vdots\\ k_1^{n-1}&k_2^{n-1}&\cdots&k_n^{n-1}\end{vmatrix} = e^{(k_1+\cdots+k_n)x}\prod_{1\le i

Every factor $k_j-k_i\neq0$ because the exponents are distinct, so the Vandermonde product is nonzero and $W(x)\neq0$ for every $x$.

$e^{k_1x$.

Step 5: Summary