Is It Injective?

Take the derivative of x+sinx. Which is 1+cosx, note that this is always greater than or equal to zero. Hence the function is monotonously increasing and also, it is continuous. Hence, it is injective.

I'm not exactly sure how to solve this without some very basic calculus. But here is some intuition: note that the smallest possible slope of a sine wave is $-1$ . This cancels out neatly with the slope of $x$ , which is $1$ ; thus, the lowest possible slope of $x+\sin x$ is $0$ . That means it is injective.

It seems that all of the comments so far either use calculus, or don't show what needs to be shown. (My apologies if I missed someone's comment.) However, it isn't terribly difficult.

Taking any $\alpha<\beta$ , start at the point $A=e^{i\alpha}$ on the unit circle, and then move along the unit circle, in the positive direction, $(\beta-\alpha)$ radians, thus ending up at the point $B=e^{i\beta}$ . (Complex representation is not essential to this solution, but I think it makes it easier to follow.) Since the shortest distance between two points is a straight line:

$\beta-\alpha> |A-B| \ge Im(A-B) = sin \alpha - sin \beta$

(note the first inequality is strict). In other words $\beta + sin \beta > \alpha + sin \alpha$ , which is to say, $f$ is a strictly increasing function, and therefore injective.

Let $x,y\in\mathbb{R}$ be distinct. Then $f(x)=f(y) \implies x+\sin x=y+\sin y \implies x-y=-(\sin x-\sin y)=-\left[\left(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\right)-\left(y-\dfrac{y^3}{3!}+\dfrac{y^5}{5!}-\cdots\right)\right]=-(x-y)+\left(\dfrac{x^3}{3!}-\dfrac{y^3}{3!}\right)+\cdots$ ,which is true only for $x=y.$ Thus $f(x)=f(y) \implies x=y.$ i.e. $f$ is injective.

Perhaps the question should specify whether x is required to be real or complex

For injection if f(x1)=f(x2) then x1=x2 Now,x+sinx=x+x-x^3+... which for small angles is approx x+x=~2x So f(x)=2x f(x1)=f(x2) 2 x1=2 x2 $o x1=x2 $o injective

(\f(x) = x+ sin x ) is clearly continuous and differentiable at all points. Also, $f'(x) = 1 + cos x \geq 0$ , with equality when $x= 2n\pi + \pi ,$ , where $n$ is any integer. So $f'(x) = 0$ only at Countably Infinite Points , and $f(x)$ is, therefore, a strictly increasing function ( and hence injective).

An Injective function is a one-to-one function (i.e. for every input value there is a unique output value).

Since the sine function is a many-to-one function (e.g. sin(Pi/2) = sin(5Pi/2)=1), the question is asking whether, when added to x, we change the function from a many-to-one function to a one-to-one function.

When we differentiate y=x + sin(x), we get dy/dx = 1 + cos(x). Since the minimum of cos(x) is -1, the minimum gradient is 0. This means that the function is increasing (the graph never goes down), i.e. there are no two y-values for each x value therefore injective.

Assume by way of contradiction that there exists some $k>0$ such that $x+\sin x=(x+k)+\sin(x+k)$ for some $x$ . We can rewrite this as $\sin x=k+\sin x\cos k+\sin k\cos x$ . After squaring, we get a quadratic in $\sin x$ :

$0=2\sin^2x(1-\cos k)-2k\sin x(1-\cos k)+k^2-\sin^2k$

It suffices to show that the discriminant of this quadratic equation is non-positive, or that

$(2k(1-\cos k))^2\le8(1-\cos k)(k^2-\sin^2k)$

Rewriting, this is

$0\le(1-\cos k)(2\cos^2k+k^2\cos k+k^2-2)$

But since $\cos k\le1$ , we only need to show that $2\cos^2k+k^2\cos k+k^2-2\ge0$ , which is rewritten as

$\boxed{2\le2\cos^2k+k^2(1+\cos k)}$

Can anyone finish from here without using calculus? #1variableinequalities I know equality of the original inequality occurs where $\cos k=1$ and then all the points where $2\cos^2k+k^2(1+\cos k)=2$ .

What I was working up to was that since the discriminant is non-positive and some contradiction gives $k=0$ , implying that the function is injective. We can get this contradiction by examining the equality cases.