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I was solving the following problem:

"Prove the following inequality for all $x \geq 0$: $\arctan(x) \geq x-\frac{x^3}{3}$"

I am aware this is able to be solved using the regular MVT, but when I attempted to solve it using a Taylor expansion at x=0 and the mean value form of the remainder, I found that for all odd powers of the expansion, Taylor's theorem did not hold. That is, there was no intermediate point c for which the remainder was equal (for example): $\frac{(\arctan(c))^{(4)}}{4!}x^4$ for any $x>0$ (as the remainder is negative for c>1, and checking against desmos not every positive $x$ has an intermediate point such that the Taylor expansion equals $\arctan (x)$.

Even though supposedly all the conditions for the theorem hold ($\arctan(x)$ is infinitely differentiable on $\mathbb{R}$ and in particular around the point $x=0$).

My question is, what makes Taylor's theorem fail for odd powers here?

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    $\begingroup$ Huh? $\arctan^{(4)}(c)=\dfrac{24c(1-c^2)}{(1+c^2)^4}$ is positive for $\lvert c\rvert<1$. $\endgroup$ Feb 3, 2021 at 13:45
  • $\begingroup$ If Taylor's theorem appears to fail it's because you made an error somewhere. We can't help you find the error unless you show us what you did - exactly how do you imagine the theorem fails??? $\endgroup$ Feb 3, 2021 at 13:53
  • $\begingroup$ David C. Ullrich I realised my mistake. I was considering that c had to grow larger as x does, which is not the case. Thank you anyways! $\endgroup$
    – ShiNyyf
    Feb 3, 2021 at 14:05

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I realised my mistake. I was assuming c had to grow larger as x does, which is not the case, in which case taylor's theorem definitely holds. Thanks anyways!

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