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I'm reading a book on trigonometry and the author shown that some of the examples contain one abcissa for one ordinate and he showed that a parabola has two abscissas for every positive ordinates, I got curious:

  • Is it possible to have more than two abscissas for one and only one ordinate?
  • Is it possible to have more than two abscissas for every element of a specific set of numbers?
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If we do not make continuity assumptions, anything can happen. With continuity, the Intermediate Value Theorem severely restricts the possibilities. –  André Nicolas Aug 15 '12 at 22:12

2 Answers 2

up vote 1 down vote accepted

The answer to both questions is easily yes if you don’t require the function to be continuous. Let

$$f(x)=\begin{cases} 2,&\text{if }x\in\{0,1,2\}\\ \tan^{-1}x,&\text{otherwise}\; \end{cases}$$

The function $\tan^{-1}x$ is one-to-one and takes values between $-\frac{\pi}2$ and $\frac{\pi}2$, so every ordinate except $1$ corresponds to at most one abscissa, while $1$ corresponds to three abscissae.

The same basic idea can be used to provide a positive answer to the second question. Let $Y$ be the set of ordinates that are to have more than two abscissae. For each $y\in Y$ pick a set $X_y$ of three real numbers, making sure that the sets $X_y$ are pairwise disjoint and that $\left|\Bbb R\setminus\bigcup_{y\in Y}X_y\right|=|\Bbb R\setminus Y|$. (This is always possible.) Let $$g:\Bbb R\setminus\bigcup_{y\in Y}X_y\to\Bbb R\setminus Y$$ be any bijection, and let

$$f(x)=\begin{cases} y,&\text{if }x\in X_y\\ g(x),&\text{otherwise}\;; \end{cases}$$

this function $f$ has the desired properties.

The answer to the first question is still yes even if you require the function to be continuous; for instance, take

$$f(x)=\begin{cases} x,&\text{if }x\le 0\\ 0,&\text{if }0\le x\le 1\\ x-1,&\text{if }x\ge 1\;. \end{cases}$$

Apart from the trivial example of a constant function, however, you cannot get such an example if $f$ is required to be a polynomial.

If $F$ is any finite set of real numbers, you can easily modify the last example to get a continuous function that has a whole interval of abscissae for each ordinate in $F$ and is one-to-one everywhere else. It's also possible to modify it to get examples for countably infinite sets $F$, but that’s just a bit tricky.

If you want $F$ to be a closed interval, you can do it with a polynomial of degree $5$. For instance, to get $F=[a,b]$ you can use a fifth degree polynomial with two local minima at height $a$ and two local maxima at height $b$. More generally, with a polynomial of degree $4n+1$ you can get $F$ to be the union of $n$ pairwise disjoint closed intervals.

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What artifacts could be used to make a non continuous function? The concept of piecewise defined functions? –  Jesus Christ Aug 16 '12 at 14:42

The sine function keeps repeating. It's a wave. Therefore you have infinitely many abscissas for each ordinate that actually occurs. (If the ordinate is more than $1$ or less than $-1$, then it doesn't "actually occur" (unless you bring in imaginary numbers).) Look at it's graph and you'll see this.

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Oh, I got curious with one thing: It seems that a graph such as the graph of$f(x)=x^2$ have a relation with $x$ and $y$. Functions like trigonometric functions seems to have a relation with a circle and/or right triangles, as you can see here and here, is this true? –  Jesus Christ Aug 16 '12 at 14:29
It may be a naive question, but I feel the objects for these relations are different somehow, they end up being translated into a graph(which makes them similar). I'm trying to understand this distinction: The object that generates $n$ values and it's representation in the graph. –  Jesus Christ Aug 16 '12 at 14:37
Here is part of the graph: en.wikipedia.org/wiki/File:Sinus.svg –  Michael Hardy Aug 16 '12 at 16:18
Here's a simpler example: Let $f(x)=x(x-1)(x+1)$. Then the ordinate $0$ occurs when the abscissa is $1$, $0$, or $-1$. –  Michael Hardy Aug 16 '12 at 16:19
Here's a better version of the graph of the sine function: themathpage.com/atrig/trig_IMG/sine.gif –  Michael Hardy Aug 16 '12 at 16:31

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