Prove that $\{\cos x, \sin x, e^x, e^{-x}\}$ is a linearly independent subset of $C^\infty (\mathbb R)$

Question

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?

Answer 1

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 $c = d = 0$.

Answer 2

"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 $c=0=d$.

Answer 3

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 $d=0$.

Answer 4

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.