This following proof is from Apostol's Calculus Volume 2.

Example 7
If $a_1,...,a_n$ are distinct real numbers, the $n$ exponential functions
are independent. We can prove this by induction on $n$. The result holds trivially when $n=1$. Therefore, assume it is true for $n-1$ exponential functions and consider scalars $c_1,...,c_n$ such that
$\sum_{k=1}^{n}c_ke^{a_kx}=0$.     (1.2)
Let $a_M$ be the largest of the $n$ numbers $a_1,...,a_n$. Multiplying both members of (1.2) by $e^{-a_Mx}$, we obtain
$\sum_{k=1}^{n}c_ke^{(a_k-a_M)x}=0$.     (1.3)
If $k \not= M$, the number $a_k-a_M$ is negative. Therefore, when $x \rightarrow +\infty$ in Equation (1.3), each term with $k \not= M$ tends to zero and we find that $c_M=0$. Deleting the $M$th term from (1.2) and applying the induction hypothesis, we find that each of the remaining $n-1$ coefficients $c_k$ is zero.

I understand what is happening up to and including when he states that "If $k \not= M$, the number $a_k-a_M$ is negative." However, then I am confused as to why he is considering the limit as $x$ tends to infinity. it seems arbitrary. And also, how does applying the induction hypothesis make sense? From what I have learned, the inductive step is proving that if $k$ is true $k+1$ is true, but I don't see how he is doing that.


2 Answers 2


Our assumption is that $\sum_{k=1}^{n}c_ke^{a_kx}=0$ for all $x\in \Bbb R $. So in particular, it should hold as $x $ tends to $\infty $. When $x\to\infty $, $e^{(a_k-a_M)x}\to 0$ for all $k\neq M $. Then the only term left on the LHS is $c_M $, which is equal to zero on the RHS. So we have shown that $c_M =0$. Now we have only $n-1$ terms on the LHS. By induction assumption, we already know that $n-1$ terms of $e^{a_kx} $ are linearly independent. Hence $c_k=0$ for all $1\leq k\leq n $.

  • $\begingroup$ But then is he not proving it only for $x \rightarrow \infty$ instead of for all $x \in R$? $\endgroup$
    – John Arg
    Jun 23, 2019 at 2:26
  • $\begingroup$ @JohnArg Our assumption is that $\sum_{k=1}^{n}c_ke^{a_kx}=0$ for all $x\in \Bbb R $. Note that the coefficients are independent of $x$. So the same coefficients should be there for $x $ tending to infinity also. $\endgroup$
    – cqfd
    Jun 23, 2019 at 2:31

Note that $u_1(x),u_2(x),...,u_n(x)$ are functions with domains $(-\infty,\infty)$ so it makes perfect sense to fond the limit as $x\to \infty $

Regarding your induction question note that once one of the coefficients becomes zero the case on $n$ coefficients goes back to the case of $n-1$ coefficients hence the induction hypothesis kicks in.


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