$|\arctan(x)-\arctan(y)| $for all reals $x,y$ is it bounded?
for all $x,y\in \mathbb{R}$ is $$|\arctan(x)-\arctan(y)| $$ bounded?
plotted $\arctan(x)$ in wolfram
graphically $|\arctan(x)| < 3 $ so $$ |\arctan(x)-\arctan(y)| < |\arctan(x)|+|\arctan(y)|< 6$$
I do know the highest slope of unit circle is at angle $\pi /2 $ which seems to be the highest. It is not elegeant using 6. Not sure how to say it elegangly that it is bounded Is it just from definion of arctan?
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3 Answers
$\begingroup$Hint: The usual definition of $\arctan x$ says that $-\frac {\pi }2 \lt \arctan x \lt \frac {\pi}2$, so it never gets close to $3$ in absolute value. Now you can use the triangle inequality to get a bound on $|\arctan x - \arctan y|$ Your statement that it is bounded by $6$ is correct, but you can get a lower upper bound.
$\endgroup$ 5 $\begingroup$As $\;\lvert\arctan x\rvert<\dfrac\pi2$, the triangle inequality results in $$\lvert\arctan x+\arctan y\rvert\le\lvert\arctan x\rvert+\lvert\arctan y\rvert<\pi.$$
$\endgroup$ 4 $\begingroup$It is better than bounded: arctan is $1$-Lipschitz, i.e $|\arctan x - \arctan y| \le |x-y|$ for all $x,y$. To see this, show that the absolute-value of the derivative of arctan is uniformly bounded by $1$.
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