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Good day ! I don't understand the following problem:

"Prove that the three functions $x^2,\cos{x},e^x$ are linearly independent"

So I think so I have to prove that the linear combination:

$a x^2+b\cos(x)+c e^x =0 \Rightarrow a=b=c=0 $

Does it right?

Also I know that $e^x \neq 0$

Happy Christmas and every help is welcomed.

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Note that the equality $\;ax^2+b\cos x+ c\,e^x=0\;$ is an equality of functions, so that it remains for any value of $\;x\;$ you choose, so for example:

$$x=0\implies b+c=0$$

$$\begin{align}&x=\frac\pi2\implies \frac{a\pi^2}4+ce^{\pi/2}=0\\{}\\&x=-\frac\pi2\implies \frac{a\pi^2}4+ce^{-\pi2}\end{align}$$

Substracting both equations above:

$$c(e^{\pi/2}-e^{-\pi/2})=0\implies c=0$$

and etc.

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Right, you have to show that if $x \mapsto ax^2 + b \cos(x) + ce^x$ is the zero function (i.e., $ax^2 + b \cos(x) + ce^x = 0$ for every real number $x$), then $a = b = c = 0$.

Hint: One possible approach is to think about the asymptotic growth rates of the functions. What happens as $x \to \infty$?

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See here. Take enough derivatives of each function to form a $nxn$ matrix (you have three functions, so take 2nd derivative). Form a matrix with the functions and derivatives, take determinant of that matrix.

\begin{bmatrix} f_{1}(x) & f_{2}(x) & f_{3}(x)\\ {f_{1}'(x)} & {f_{2}'(x)} & {f_{3}'(x)} \\ {f_{1}''(x)} & {f_{2}''(x)} & {f_{3}''(x)} \\ \end{bmatrix}

If the determinant is not $0$, your functions are linear independent.

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