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?

  • 2
    $\begingroup$ Your proof is fine. Well done. $\endgroup$ – Crostul Jul 7 '19 at 22:41
  • 1
    $\begingroup$ Welcome to MSE! You might want to check how to write the equations in TeX to help other readers understand your question in the future. You can find the details in the link here math.meta.stackexchange.com/questions/5020/… $\endgroup$ – BigbearZzz Jul 7 '19 at 22:44
  • $\begingroup$ What does WTS mean, please? I've never seen it before. $\endgroup$ – TonyK Jul 7 '19 at 22:55
  • $\begingroup$ I think your third paragraph is mistyped. Do you mean "$e^{-x}$ is also sometimes much larger..."? $\endgroup$ – TonyK Jul 7 '19 at 22:56
  • $\begingroup$ WTS means Want To Show, in third paragraph, I mean when x is a real number less than 0, then it is also a condition that e^-x is much larger than sinx and cosx. $\endgroup$ – WaterBro Jul 7 '19 at 23:15

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 $c = d = 0$.


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.


"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 $c=0=d$.

  • $\begingroup$ I think "sometimes much larger" is fine, and easier to understand than a more formal description. $\endgroup$ – TonyK Jul 7 '19 at 22:58
  • $\begingroup$ Especially since $\cos$ and $\sin$ don't have limits in $\pm\infty$. $\endgroup$ – mechanodroid Jul 7 '19 at 23:04
  • $\begingroup$ @mechanodroid You can actualy use that to prove that also $a=b=0$, otherwise the limit, as you say, wouldn't exist, and it has to be $0$. $\endgroup$ – Adam Latosiński Jul 7 '19 at 23:19
  • $\begingroup$ Ok, perhaps. Assume $\lim_{x\to\infty}(a\cos x + b\sin x + ce^x + de^{-x}) = 0$. Since $e^{-x} \xrightarrow{x\to\infty} 0$ we have $$0 = \lim_{x\to\infty}(a\cos x + b\sin x + ce^x ) \ge \lim_{x\to\infty}(-a -b+ ce^x ) = +\infty$$ if $c> 0$. Similarly we get $0 \le -\infty$ if $c < 0$. Therefore $c = 0$ so $\lim_{x\to\infty}(a\cos x + b\sin x) = 0$. Now we can consider points of the form $x_n = n\pi$ to get $0 = \lim_{n\to\infty} a(-1)^n$ so $a = 0$. Then $b = 0$. $\endgroup$ – mechanodroid Jul 7 '19 at 23:27

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 $d=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.