Prove that $\{ 1, e^x, e^{2x}, e^{3x}, e^{4x} \} \;$ is a linearly independent set in space of infinitely differentiable functions, $C ^\infty (R)$
What I tried is next:
To prove that $\{ 1, e^x, e^{2x}, e^{3x}, e^{4x} \} \;$ is linearly independent then I have to prove that the only one solution to the next linear combination:
$\alpha_1*1+ \alpha_2 *e^x+ \alpha_3* e^2x+ \alpha_4* e^3x+ \alpha_5* e^4x = 0 \quad $ is: $\quad \alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = \alpha_5 = 0 $
Step 1: I found the derivative of both sides of the equation and the result is:
$ \alpha_2 *e^x+ 2\alpha_3* e^{2x}+ 3\alpha_4* e^{3x}+ 4\alpha_5* e^{4x} = 0 $
Step 2: Then I divided both sides by $e^x$ and the result is:
$ \alpha_2 + 2\alpha_3* e^x+ 3\alpha_4* e^{2x}+ 4\alpha_5* e^{3x} = 0 $
Then I repeat some times more steps 1 and 2, and I got: $24\alpha_5 = 0 \to \alpha_5 = 0 $
So I can replace that result in the equation and do it again with $\alpha_4, \alpha_3, \alpha_2 \; $and$ \; \alpha_1$
and get that $ \alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = \alpha_5 = 0 $
But I'm not sure if that's right.
Best Answer
hint
Assume there exist $ (a,b,c,d,f)\in \Bbb R ) $ such that
$$(\forall x\in \Bbb R) \;\; F(x)=$$ $$a+be^x+ce^{2x}+de^{3x}+fe^{4x}=0$$
then $$\lim_{x\to-\infty}F(x)=a=0$$ and
$$F(x)=e^x(b+ce^x+de^{2x}+fe^{3x})$$ $$=e^xG(x)=0$$ So $$(\forall x\in\Bbb R)\;\; G(x)=0$$ thus
$$\lim_{x\to-\infty}G(x)=b=0$$ and so on.