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Is the following always true? $$ \arctan(f(x))\leq f(x),\ \ \text{where}\ \ |f(x)|<1\ \ \forall x\in\mathbb{R} $$

For example, is it correct to say that $$ \arctan\left(\frac{\sin x}{\sqrt{x+1}}\right)\le\frac{\sin x}{\sqrt{x+1}}\ \ \forall x\in\mathbb{R}, $$ since $\arctan t=t-\frac{t^3}{3}+\frac{t^5}{5}-\dots$, where each term is not greater than the one before?

I'm confused because in the example above $\arctan$'s argument can be negative (i.e., $\sin x$ can be negative).

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3 Answers 3

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If $0\leq f(x) <1$, yes, of course. Since $\frac{d}{dy}\arctan y=\frac{1}{1+y^2}\leq1$ (and $\arctan(0)=0$), you'll have $$ \arctan y \leq y, \quad\text{for }\ y\geq0. $$ Clearly, the situation is reversed if $y<0$, because both functions are negative. You can say that $|\arctan f(x)|\leq|f(x)|$ with no restrictions on $f(x)$, though.

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The claim is false, since whenever $\xi<0$, $\arctan(\xi)>\xi$. On the other hand, if we include absolute values, the inequality $$(\forall \xi\in[-1,1])\quad|\arctan(\xi)|\leq|\xi|$$ is true.

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If you derive the function $x\mapsto \text{Arctan} (x) -x$, you get that:

  • if $x>0$ then $\text{Arctan} (x) < x$
  • but if $x<0$ then $\text{Arctan} (x) > x$.
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