Linear independence of $C^\infty $ functions Prove that the set $\mathscr B=\{e^x,\cos x, \sin x\}$ cannot span $C^\infty(\mathbb R)$.
I tried to prove $x$ is linearly independent to the set $\mathscr B$, but how? Or how to prove the condition with Wronskian?
 A: Any linear combination $f$ satisfies $f^{(4)}=f$, but the function $x$ does not. 
A: You could try sampling the function at a few points.

Theory.
Let $y_1(x),y_2(x),\ldots,y_n(x)$ be the functions under consideration. In theory, if there exists a point $x_0$ in $I$, such that the vectors 
$(y_i(x_0),y_i'(x_0),\ldots,y_i^{(n-1)}(x_0))$ for all $i=1,2,\ldots,n$ 
are linearly independent, then the functions are linearly independent in $C(I)$.
For simplicity, I assume $n=3$. We require
$\begin{align}
c_1 y_1(x) + c_2y_2(x) + c_3 y_3(x)  = 0\\
c_1 y_1'(x) + c_2y_2'(x) +c_3 y_3'(x) = 0\\
c_1 y_1''(x) + c_2y_2''(x) +c_3 y_3''(x) = 0\\
\end{align}$
to hold for all $x \in I$. If $x=x_0$ is a point in the interval $I$, we must have:
$\begin{align}
c_1 y_1(x_0) + c_2y_2(x_0) + c_3 y_3(x_0)  = 0\\
c_1 y_1'(x_0) + c_2y_2'(x_0) +c_3 y_3'(x_0) = 0\\
c_1 y_1''(x_0) + c_2y_2''(x_0) +c_3 y_3''(x_0) = 0\\
\end{align}$
The above system of homogeneous equations has a non-trivial solution, if and only if  the coefficient matrix 
$W=\begin{bmatrix}
y_1(x_0) & y_2(x_0) & y_3(x_0) \\
y_1'(x_0) & y_2'(x_0) & y_3'(x_0) \\
y_1''(x_0) & y_2''(x_0) & y_3''(x_0) 
\end{bmatrix}$
does not have full rank. 
To sum up, if the $det(W)$ is non-zero, or columns of the matrix $W$ are linearly independent, then the system has only the trivial solution $c_1=c_2=c_3=0$. But that means, the functions are linearly independent. 

So, pick a point $x_0$ and find these vectors in $R^n$. Their linear independence implies, the functions are linearly independent.
Clearly, if we take $x_0=0$, we get the vectors 
$(y_1(0),y_1'(0),y_1''(0))=(1,1,1)$
$(y_2(0),y_2'(0),y_2''(0))=(1,0,-1)$
$(y_3(0),y_3'(0),y_3''(0))=(0,1,0)$
are linearly independent. Hence, the functions in question are linearly independent.
A: $e^{2x}$ does not belong to the span: if $e^{2x}=ae^{x}+b\cos x+c \sin x$ then we can divide by $e^{2x}$ and let $x \to \infty$ to get the contradiction $1=0$.
Also the Wronskian of $\{x, e^{x}, \cos x,\sin x\}$ is easy to compute and you get $2e^{x}$. Since this is never $0$ we get linear independence of $\{x, e^{x}, \cos x,\sin x\}$
A: Call $S$ the vector space generated by $e^x$, $\cos x$ and $\sin x$. You may show that the are no constant function in $S$ different to zero. How?
Take the constant function $1$, that is $f(x)=1$ for every $x\in\Bbb R$ and write
$$1=f(x)=\alpha e^x+\beta\cos x+\gamma\sin x \qquad \text{ for every } x\in\Bbb R$$
Pick $x=0$, then $$1=f(0)=\alpha+\beta$$
Next, pick $x=2\pi$ and then
$$1=f(2\pi)=\alpha e^{2\pi}+\beta$$
These equations tell you that $\alpha=0$ and $\beta=1$, necessarily!!
Hence: $$1=f(x)=\cos x+\gamma\sin x$$
Again, for $x=\frac{\pi}{2}$, you have $\cos \frac{\pi}{2}=0$ and $\sin\frac{\pi}{2}=1$, so $\gamma=1$. However, when $x=\frac{3\pi}{2}$, you have $\cos \frac{3\pi}{2}=0$ and $\sin\frac{3\pi}{2}=-1$, so $\gamma=-1$. A contradiction.
