# Inverse of sin(x)+x

1. Sep 27, 2014

### ellipsis

The normal answer is "no, there is no inverse function in terms of the normal operators and trigonometric functions."

That is to say, given a value of x, you cannot find the value of y of the input function reflected over the x-y axis using standard functions.

"Standard functions" is what is getting me here. All functions we can run on a computer is eventually compiled down to complex loops of additions, multiplications, subtractions, divisions, bit shifts, reads, and writes.

I wish there was a reflect( <expression> ) operator, that just returned the value you would get if you flipped <expression> over the xy axis first.

More to the point, sin(x) + x does not even fail any horizontal line tests. How can you determine that there is no inverse?

2. Sep 27, 2014

### pwsnafu

That is not the answer! The answer is that "the inverse cannot be expressed in a closed form expression of elementary functions".
[/PLAIN] [Broken]
If you allow infinite series then see the Lagrange inverse formula.

Last edited by a moderator: May 7, 2017
3. Sep 27, 2014

### qspeechc

How do you know sin(x) + x doesn't 'fail any horizontal line tests'? I'm just curious.
And as you point out, the inverse does exist, it just cannot be expressed in any of the standard functions we use. All we can do is say if f(x)=sinx + x, then the inverse is f-1(x)

4. Sep 27, 2014

### ellipsis

qspeechc: I used a graphing program. sin(x)+x was an arbitrary choice; the simplest function I knew without an inverse. I have since realized that the horizontal line test does not tell you if it's inverse is expressible in a closed form expression of elementary functions, only if it's inverse is a function in the sense that is has one output for every input.

I suspect determining whether or not the inverse is solvable in such a way is part of abstract algebra, Galois theory maybe.

5. Sep 27, 2014

### qspeechc

:/
On the one hand, you state that sinx+x doesn't fail any horizontal line tests, and on the other hand you claim it doesn't have an inverse. Please enlighten me. I thought the 'horizontal line test' was what they taught you in school to determine if a function has an inverse or not.

Let's be clear, f(x)=sinx+x is invertible. That is, there exists a function g(x), and that function is the inverse of f(x). The only thing is that g(x) is not expressible in simple or elementary functions.

6. Sep 27, 2014

### ellipsis

That's what I thought too. What it does is subtle: It tells you if the inverse is a function or not. The inverse of y = x^2 is not a function, because it has two solutions for every positive x. We take the principal square root, of course.

For sin(x)+x, it's inverse /is/ a function, and hence exists. But it is not expressible without invoking Lagrange's inversion theorem or something similar.

My interest in this is so: I have found a closed form expression for "Time" in terms of "Height" for one-dimensional gravity under the inverse square law. But I cannot solve for height as a function of time. A closed-form solution for this and similar problems has immense application in the accurate and efficient design of physics engines.

7. Sep 27, 2014

### qspeechc

An inverse is a function by definition. y=x2 fails the horizontal line test, does it not? Therefore according to the test it does not have an inverse. Am I missing something?

The problem is functions and inverses are done poorly in school. What we usually do for y=x2 is that we only consider non-negative values of x. Therefore in the graph we would only draw from x=0 onwards. If we restrict the function in this way, you will see that it now passes the horizontal line test, and therefore has an inverse, which is sqrt(x).

8. Sep 27, 2014

### ellipsis

You are not. y = x^2 does not have an inverse, by the normal definition of "function". I was always skeptical of that myself, though! It seems arbitrary to throw away the negative solution.

9. Sep 27, 2014

### qspeechc

I think we are responding to one another before the other person edits their comment :p

I'm struggling to understand the difficulties you are having. Let's be concrete, what is the function you are trying to invert, and what problems are you having with it?

For y=x2, we could restrict x to non-positive values, and then the inverse would be -sqrt(x).

The root of the problem is that x2 defined on the reals is a different function from x2 defined for non-negative numbers, and both are different from x2 defined for non-positive numbers. Those are all different functions, let's say f(x), g(x), and h(x) respectively. To define a function, we need the 'rule', in this case 'x2', and the input values for which it is 'defined', in other words, which input values we decide we are going to use. For f(x) the input values are all real numbers, for g(x) it is all the non-negative numbers, etc. So f(x), g(x), and h(x) are different functions because although the rule is the same, we have decided they will use different input values. If you look at it this way, you will see f(x) has no inverse, but g(x) and h(x) do, which are different (To see this, draw the graphs for each. In fact, I have simplified it here: we also need to define output values for functions, but don't worry about that).

Last edited: Sep 27, 2014
10. Sep 27, 2014

### ellipsis

Well, it's nasty. I want to transfer the following from parametric form, to explicit form:
$$\begin{cases} x = \sqrt{\frac{x_i}{2}}\Bigg(\sqrt{Tx_i-T^2} - x_i\sin^{-1}{\Big(\sqrt{\frac{T}{x_i}}\Big)} + \frac{x_i\pi}{2}\Bigg)\\ y = T \end{cases}$$

I've already started simplifying, and analyzing it a bit. The first term inside the parentheses is a semicircle, and the second is an arcsin shifting over and stretched. I can find the inverse of the function's component parts, but I cannot find the inverse of the function itself.

I should note: I have found very close approximations to height as a function of time. I might use those.

Last edited: Sep 27, 2014
11. Sep 27, 2014

### qspeechc

I doubt very much you'll find an inverse to that in simple functions, according to wolfram alpha it doesn't exist. And by the way, the first part isn't a semi-circle, look closely at it. Anyway, what is your problem exactly?