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How can I prove that the function $f(x)= 1-x^2$ will be greater than $g(x) = \cos(\pi x)$ on the interval $[-1,1]$? I feel like it should be pretty basic but it seems so hard.

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    $\begingroup$ Try considering the Taylor expansion of $\cos(x)$. $\endgroup$ – Lukas Kofler Sep 14 '18 at 20:45
  • $\begingroup$ $\cos(\pi x)$ is bound between $[-1,1]$ $\endgroup$ – Shogun Sep 14 '18 at 20:57
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Since both $f(x)$ and $g(x)$ are even it suffices to prove that for $x\in[0,1]$, and since for $x>\frac12$ $\cos(\pi x)< 0$ it suffices to consider $x\in[0,1/2]$.

Then we need to show that

$$h(x)=f(x)-g(x)=1-x^2-\cos (\pi x)> 0$$

for $x\in[0,1/2]$.

Firstly observe that $h(0)=0$ and therefore the inequality doesn't hold for $x=0$, then note that for $x\in[0,1/2]$

$$h'(x)=\pi\sin(\pi x)-2x\ge 0 \quad h'(x)=0 \iff x=0\tag{1}$$

therefore $h(x)$ is strictly increasing on the interval and the inequality holds for $x\in(0,1/2]$.

To show $(1)$ just observe that

  • $\pi\sin(\pi x)$ is concave on that interval
  • at $x=0$ we have $[\pi\sin(\pi x)]_{x=0}=[2x]_{x=0} \implies h'(0)=0$
  • at $x=1/2$ we have $[\pi\sin(\pi x)]_{x=1/2}>[2x]_{x=1/2}$
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  • $\begingroup$ Oops. I meant it to be $cos(\frac{\pi x}{2})$. I tried all of the tricks you used for the previous one but I can't seem to get it to work. Thanks for the original answer tho. $\endgroup$ – Opyau Sep 15 '18 at 3:10
  • $\begingroup$ @Opyau In that case we need some variation but it can be proved in a similar way. Since this OP is interesting you may let this OP in that way and ask a different question for the other problem. $\endgroup$ – gimusi Sep 15 '18 at 7:28

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