# Why does the Mean Value form of the remainder of the Taylor expansion not work here?

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?

• Huh? $\arctan^{(4)}(c)=\dfrac{24c(1-c^2)}{(1+c^2)^4}$ is positive for $\lvert c\rvert<1$. Feb 3, 2021 at 13:45
• 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??? Feb 3, 2021 at 13:53
• 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! Feb 3, 2021 at 14:05