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I'm only curious why the inverse of $f(x)$ can not be determined algebraically. Is it because the inverse of $\sin(x)$ cannot be converted into a formula?

Paramanand Singh
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JJW22
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    Well, even $\sin(x)$ has no such inverse. And even $\sin(x)$ is not such an "elementary function". – leonbloy May 15 '16 at 05:09
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    @leonbloy actually most formal definitions of elementary functions include both sin and asin – ASKASK May 15 '16 at 05:33
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    Agree with @ASKASK, elementary functions consists of all algebraic functions, trigonometric functions (direct as well as inverse), exponential and logarithmic functions and any function which can be obtained via finite number of compositions of the functions mentioned earlier. – Paramanand Singh May 15 '16 at 05:36
  • @ASKASK The previous version of this question (before the editing) spoke of "elementary functions" and did not include sin/asin among them. – leonbloy May 15 '16 at 14:25
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    @leonbloy the previous definition I gave was a copy from wikipedia, but it's commonsense we should include trigonometric functions in it – JJW22 May 15 '16 at 15:28
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    @JanetWang In math, commonsense takes precedence only in absence of strict definitions :-) Furthermore, to add to the confusion, you are using the word "algebraically" (in title and question body); now, in case you imagine that it's common sense that trigonometric functions are algebraic functions... well, they are not, see Paramanand Singh answer. – leonbloy May 15 '16 at 15:38
  • see https://math.stackexchange.com/questions/652277/inverse-of-fx-sinxx/4495992#4495992, https://math.stackexchange.com/questions/1053472/how-to-solve-keplers-equation-m-e-varepsilon-sin-e-for-e/4732665#4732665 – IV_ Jul 08 '23 at 16:55

2 Answers2

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The equation $x + \sin(x) = y$ is Kepler's equation in the case $e = -1$. This has been well-studied. It just happens that the inverse function is not one of the special functions to which standard names have been attached.

Robert Israel
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Note that the function $f(x) = x + \sin x$ is transcendental function. Also we have a theorem which says that inverse of algebraic functions are also algebraic, therefore it follows that the inverse of $f$ is not algebraic.

That said I think you want to know something more. The function $f$ is transcendental but elementary (meaning that value of $f(x)$ can be obtained from $x$ via a finite number of algebraic operations together with composition of algebraic functions, trigonometric functions (direct and inverse), exponential and logarithmic functions). The inverse function is also transcendental but not elementary (it is difficult to prove in general that a function is non-elementary). Hence it is not possible to expect inverse of $f$ to be expressed in terms of usual functions we see in analysis.

Paramanand Singh
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