I am supposed to prove that ${x,e^x,\sin(x)}$ is a linearly independent set.
I know that if $\{x,e^x,\sin(x)\}$ is linearly independent, then we would have $ax+be^x+c\sin(x)=0$, for all $x \in \mathbb{R}$, and $a,b,c \in \mathbb{R}$, where a=b=c=0.
However, I'm skeptical how to proceed.
My gut tells me to begin taking derivatives, since I'm given a set of real function, and I may be able to to show that $a$, $b$, and $c$ must be equal to zero.
\begin{align*}
(ax+be^x+c\sin(x))'&= a1+be^x+c\cdot cos(x)=0\\
(ax+be^x+c\sin(x))''&=a0+be^x-c\cdot sin(x)=0\\
(ax+be^x+c\sin(x))'''&=a0+be^x-c\cdot cos(x)=0\\
(ax+be^x+c\sin(x))''''&=a0+be^x+c\cdot sin(x)=0\\
\end{align*}
Now subtracting $c\cdot \sin(x)$, we have
\begin{equation*}
be^x=-c(\sin x)
\end{equation*}
From this I can't determine much, I know that $e^x >0$ for all $x\in\mathbb{R}$.
Can someone give me a push?
Best Answer
$ax +be^x +c\sin(x) =0$, $x=0$ implies $b=0$, $ax+c\sin(x) = 0$, $x=\pi$, $a=0$ thus $c=0$