Prove that $\{\cos x, \sin x, e^x, e^{-x}\}$ is a linearly independent subset of $C^\infty (\mathbb R)$ I am going to prove $\cos x, \sin x, e^x$ and $e^{-x}$ is a linearly independent subset of $C^\infty (\mathbb R)$, which is smooth functions.
first we have $a\cos x+b\sin x+ce^x+de^{-x}=0$, WTS that $a=b=c=d=0$.
Suppose $c \ne 0$, then for all $x$ in $\mathbb R$, $ce^x$ is sometimes much larger than other 3 terms which is contradiction. So $c=0$.
So we have $a\cos x+b\sin x+de^{-x}=0$, if $d \ne 0$, then by the same logic in $c$, for all $x$ in $\mathbb R$, $x$ is also sometimes much larger than other 2 terms which is contradiction. So $d=0$.
Here it becomes $a\cos x+b\sin x=0$,
when $x=0, a*0+b*1=0$
when $x=\pi/2, a*1+b*0=0$ there is no such situation that $\cos x=\sin x=0$.
So $a=b=0$.
Above is my proof, we have not learned Wronskian or det, so I could only prove it by definition. While since it is going to prove subset of $C^\infty(\mathbb R)$, is there any correction or improvement for the above? Or is there a more clear way to prove this?
 A: A bit different solution using derivatives.
Assume $a\cos x + b\sin x +ce^x + de^{-x} = 0$. Taking the derivative twice gives
$$-a\cos x - b\sin x +ce^x + de^{-x} = 0$$ 
so adding and subtracting the two equations yields
$$a\cos x + b\sin x = 0, \quad ce^x+de^{-x} = 0$$
You already established that the first equation implies $a = b  =0$. For the second one take the derivative to obtain
$$ce^x - de^{-x} = 0$$
Again adding and subtracting the two equations gives $c = d = 0$.
A: The “much larger” idea is good and you can formalize it rigorously, avoiding handwaving.
The idea is that $\lim_{x\to\infty}0=0$, so
$$
0=\lim_{x\to\infty}\frac{a\sin x+b\cos x+ce^x+de^{-x}}{e^x}=c
$$
because
$$
\lim_{x\to\infty}\frac{\sin x}{e^x}=0,\qquad
\lim_{x\to\infty}\frac{\cos x}{e^x}=0,\qquad
\lim_{x\to\infty}\frac{e^{-x}}{e^x}=0
$$
Similarly you conclude that $d=0$, by considering the limit at $-\infty$.
For $a\sin x+b\cos x$ you can indeed reason by substituting special values, or remember that, when $a^2+b^2\ne0$,
$$
a\sin x+b\cos x=A\sin(x+\varphi)
$$
where
$$
A=\sqrt{a^2+b^2},\quad \cos\varphi=\frac{a}{A},\quad \sin\varphi=\frac{b}{A}
$$
and this function is not constant.
A: "Sometimes much larger" isn't a precise term, although it can be phrased more rigorously. For example, I suggest considering limits $\lim_{x\rightarrow\pm\infty}$. If $$ a \cos x + b\sin x+ c e^x +  d e^{-x} =0$$
for all $x$, then $$ \lim_{x\rightarrow\pm\infty} (a \cos x + b\sin x+ c e^x +  d e^{-x}) = 0$$
and you can use that to show that $c=0=d$. 
A: You just need to say "much" larger a bit more rigorously.
If $c>0$, let $x=\ln(\frac{a+b+d+\epsilon}{c}),\epsilon>0$ Then
$$
a \cos x + b \sin x + c \exp x + d \exp (-x)\geq -a-b-d+a+b+d+\epsilon>0,
$$
So $c=0$.
Similarly $d=0$.
