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I assume by $C(\mathbb{R})$ the question means the vector space of continuous real functions, but I'm not completely sure of that.

How might I go about formally proving this? Obviously I could say that there's no way you can get the graph of any of these functions through linear combinations of the other two, but that's not very formal. Writing out $$c_1x +c_2cosx+c_3\frac{x^2}{1+x^2} =0$$ isn't very helpful either as far as I can see. Any help is appreciated!

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The equation you wrote must hold for any $x\in\mathbb{R}$. Put $x=0$ into the equation and you will get $c_2=0$. Then put $x=1$ and $x=-1$ to show that $c_1=c_3=0$.

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  • $\begingroup$ I see, that makes sense thank you very much! $\endgroup$ – James Ronald Apr 22 at 15:02
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Alternative approach: $\frac{x^2}{1+x^2}$ cannot be a linear combination of $x$ and $\cos x$ since the latter are entire functions, while $\frac{x^2}{1+x^2}$ is not. $x$ and $\cos x$ are obviously linearly independent since the former vanishes at a single point, while the latter is a periodic function.

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  • $\begingroup$ Thank you! Could you explain what you mean when you say $x$ vanishes at a single point? $\endgroup$ – James Ronald Apr 22 at 15:05
  • $\begingroup$ @JamesRonald: the only solution of $x=0$ is $x=0$, while the solutions of $\cos x=0$ form the set $\frac{\pi}{2}+\pi\mathbb{Z}$. $\endgroup$ – Jack D'Aurizio Apr 22 at 15:06

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