Proving Injectivity $x + \sin(x)$

Question

I'm trying to prove injectivity of a particular function (without calculus), but I've come across a bit of a problem. The function is:

$$f(x) = x+\sin(x)$$

I started by (abiding by common standards) assuming the following:

• For any arbitrary $x_1$ and $x_2$, if $f(x_1) = f(x_2)$, $x_1=x_2$

So we begin:

1. $\quad x_1 + \sin(x_1) = x_2 + \sin(x_2)$

2. $\quad(x_1 - x_2)+(\sin(x_1)-\sin(x_2))=0$

3. $\quad x_1-x_2 =-2\cos\left(\dfrac{x_1+x_2}{2}\right)\sin\left(\dfrac{x_1-x_2}{2}\right)$

4. $\quad\dfrac{x_1-x_2}{2}=-\cos\left(\dfrac{x_1+x_2}{2}\right)\sin\left(\dfrac{x_1-x_2}{2}\right)$

I'm stuck at step 4. I know that, of course, if $x_1=x_2$, we get a true statement for all possible $x_1$ and $x_2$, but I'm trying to get to that fact (not just assume it).

Is there any way to break this (step 4) down further, or should I go down a completely different route? I'm really just requesting a bit of guidance.

P.s: I also tried assuming that there existed an inverse function of the form

$$f^{-1}(x) = c_0 + c_1x + c_2x^2 + \cdots,$$

plugging that in, and finding proper coefficients, but that didn't work out so well.

Answer 1

If you know that $0<\sin t<t$ for $t>0$, or more generally $|\sin t|<|t|$ for $t\ne 0$, then from $4$ we have (for $x_1\ne x_2$) $$ \left|\frac{x_1-x_2}{2}\right |=\left|\cos\frac{x_1+x_2}{2}\right|\left|\sin \frac{x_1-x_2}{2}\right |\le \left|\sin \frac{x_1-x_2}{2}\right |<\left|\frac{x_1-x_2}{2}\right |.$$

Answer 2

My approach is to use Calculus, more specifically the first- and second- derivative tests to show that the function is injective. More specifically, we show that the function is monotone and non-decreasing. From this it follows ( for a Real-valued function of a Real variable) that the function must be injective:

Using calculus, you see that all the critical values x are those for which $f'(x)=cosx+1 =0 $ at $x= \pi/2+ 2k\pi$ . Using the second derivative test, you see that, for all the critical values $x$ we get that $f''(x)=-sinx =-sin(\pi/2+ 2k\pi)<0$, you get that the values at the critical points are all local maxima, so your function is strictly non-decreasing, i.e., it is monotone non-increasing, so it is injective.