# Multiple function values for a single x-value

I'm curious if it's possible to define a function that would have more than two functionvalues for one single x-value. I know that it's possible to get two y-values by using the root (one positive, one negative: $\sqrt{4} = -2 ; +2$).

Is it possible to get three or more function values for one single x-value?

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Please, refrain from writing $\sqrt{4} = -2$. Unless you have reached a certain point in complex analysis where you get to "choose" a different definition with branch cuts, the only value of $\sqrt{4}$ is $2$. –  TMM Mar 3 '12 at 20:06

## 4 Answers

The concept of function is usually defined as a mapping from one $x$ to one $f(x)$.

There are such things as multiple-valued functions, though they're not strictly speaking functions.

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The usual mathematical concept of "function" includes a requirement that there is exactly one function value for each argument. So if you have something that produces two different output for the same input, that something is -- by definition -- not a function. It can still be an interesting mathematical object to study; we just don't call it a "function". One can view it as a function whose values are sets of numbers rather than individual numbers, however.

But it appears that what you're interested in here is not really the function concept but more "naively" just things that can be expressed as arithmetic expressions. In that case, you could consider something like $$\sqrt{\sqrt{x}+2}$$ where for $x=2$ one could get four different values by considering different signs of the square roots. But that's not how the square root sign is usually taken to work.

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$\sqrt{\sqrt{x} + 2}$ is a well-defined function. For $x = 2$ you will get exactly one answer, $y \approx 1.85$. And the square roots don't have "different signs". –  TMM Mar 3 '12 at 20:15
I'm supposing the same somewhat nonstandard multi-valued meaning of the square root as the OP used in considering $\sqrt 4$ to mean $\pm 2$. –  Henning Makholm Mar 3 '12 at 20:51

I am not sure I understand your question. There are two tones in your question:

1. Is there a function such that, for a single $y$-value, there are many $x$-values? ;
2. Is there a function such that for a single $x$-value, there are many $y$-values?

Answer to 1:

Let $S$ be a finite set. Consider the function $f: S \to \{1\}$. Clearly, there is a unique map that takes every element to $1$. So, by choosing the cardinality of $S$ as you please, you can have a function that maps $|S|$ values of $x$ to a single $y$-value, $1$.

Answer to 2:

By the definition of function, every $x$ has a unique image in the range (co-domain). So, the answer is there are no dunctions that satisfy this requirement.

That is, a function maps every $x$ to a unique $y$. This is often said as: When you apply a function $f$ to any $x$, $x$ must know a unique place where it would like to go.

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Let's consider some multivalued functions (not 'functions' since these are one to one by definition) :

$y=x^n$ has $n$ different solutions $\sqrt[n]{y}\cdot e^{2\pi i \frac kn}$ (no more than two will be real)

The inverse of periodic functions will be multivalued (arcsine, arccosine and so on...) with an infinity of branches (the principal branch will usually be considered and the principal value returned).

The logarithm is interesting too (every branch gives an additional $2\pi i$).

$i^i$ has an infinity of real values since $i^i=(e^{\pi i/2+2k \pi i})^i=e^{-\pi/2-2k \pi}$ (replace one of the $i$ by $ix$ to get a multivalued function).

The Lambert W function has two rather different branches $W_0$ and $W_{-1}$

and so on...

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I hate being a nitpick, but saying that $\sqrt{.}$ or $\sqrt[n]{.}$ is multivalued is just bad. The equation $z^n = x$ has $n$ solutions for $z$, and, depending on the principal branch, one of those is defined to be $\sqrt[n]{x}$. –  TMM Mar 3 '12 at 20:45
@TMM: you are right of course but this was an answer in 'the spirit' of the question (I'll edit it anyway) –  Raymond Manzoni Mar 3 '12 at 20:59