Arctan absolute value problem, calculus 1

[timon]

Homework Statement

prove

$$|arctan(x)-arctan(y)| \leq |x-y|$$

The Attempt at a Solution

I tried separating cases:
1. both x and y are in the same quadrant;
2. x and y are not in the same quadrant;
3. x and y are zero (trivial);
4. either x or y is zero.

Now, i can see why each of these is true from drawing arctan (the asymptotic behavior and |arctan(x)| =< |x| being important), but I'm having trouble producing a mathematical argument.. I tried doing something with the derivative but i just get confused. Also, proving the equality |arctan(x)| =< |x| is the _next_ question, so i doubt I'm supposed to use it here.

 

[awkward]

Try applying the Mean Value Theorem to arctan(x).

 

[timon]

i can get $$arctan(x) - arctan(y) \leq x-y$$, but how do i get that to say something about the absolute value?

 

[timon]

is this correct?
$$x-y > 0 \Rightarrow x - y = |x-y|, arctan(x)-arctan(y) = |arctan(x)-arctan(y)|$$
$$\Rightarrow |x-y| \geq |arctan(x)-arctan(y)|$$

$$x-y < 0 \Rightarrow x - y = -|x-y|, arctan(x)-arctan(y) = -|arctan(x)-arctan(y)|$$
$$\Rightarrow -|x-y| \geq -|arctan(x)-arctan(y)| \rightarrow |x-y| \geq |arctan(x)-arctan(y)|$$

$$x-y=0 \Rightarrow x=y=0$$

 

[awkward]

is this correct?
$$x-y > 0 \Rightarrow x - y = |x-y|, arctan(x)-arctan(y) = |arctan(x)-arctan(y)|$$
$$\Rightarrow |x-y| \geq |arctan(x)-arctan(y)|$$

True, but how do you know $$|x-y| \geq |arctan(x)-arctan(y)|$$? You have to justify it.

$$x-y < 0 \Rightarrow x - y = -|x-y|, arctan(x)-arctan(y) = -|arctan(x)-arctan(y)|$$
$$\Rightarrow -|x-y| \geq -|arctan(x)-arctan(y)| \rightarrow |x-y| \geq |arctan(x)-arctan(y)|$$

The step
$$\Rightarrow -|x-y| \geq -|arctan(x)-arctan(y)| \rightarrow |x-y| \geq |arctan(x)-arctan(y)|$$
is wrong.

$$x-y=0 \Rightarrow x=y=0$$

No, you can't conclude x=y=0.

Have you considered the original suggestion? What does the Mean Value Theorem tell you about arctan(x)? (I'm assuming you have studied the theorem.)

You have the right idea about breaking the proof into three cases, though: consider x > y, x < y, and x = y.

 

[timon]

True, but how do you know $$|x-y| \geq |arctan(x)-arctan(y)|$$? You have to justify it.

i used the mean value theorem, as you suggested:

if

$$f(x) = arctan(x)$$

then

$$\exists$$ $$c$$

such that

$$f(c)' = \frac{f(y)-f(x)}{y-x}$$

$$\Leftrightarrow f(c)'(y-x)=f(y)-f(x)$$

for any $$x, y \in \Re$$ where x < y, since arctan is continuous and differentiable everywhere. also,

$$f(x)'= \frac{d}{dx}[arctan(x)] = \frac{1}{1+x^2} \leq 1$$

thus

$$y-x \geq arctan(y)-arctan(x).$$

the same process for x > y gives

$$x-y \geq arctan(x)-arctan(y).$$

I think this should give me the wanted result, using the method i used for the first case in my previous post, and realizing $$|x-y| = |y-x|$$. I think i didn't split the cases soon enough when i did it before. I find all this splitting of cases rather confusing..

No, you can't conclude x=y=0.

Why not? could there be any other place where x=y?

 

[awkward]

You are OK up to the point x=y=0.

Can you think of a number other than 0?

 

[timon]

o wait i see what I've been doing. x and y are both on the x axis, so x=y doesn't imply x=y=0, in fact x and y could be any real number.
but the equality is still true, and trivially so. I guess I'm done? if so, thanks a lot for the assistance!

 

[awkward]

Right!