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I'm reading this term (odd function) in my numerical analysis book, but I have never heard of this. What does it mean that an function is odd ?

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    $\begingroup$ Go see Wikipedia. $\endgroup$ – user37238 Sep 19 '13 at 15:00
  • $\begingroup$ Three possible definitions include: $f(-x)=-f(x)$ (the most likely), for integer $x$, $f(x)$ is odd (unlikely), degree$(f)=2k+1$ (unlikely). $\endgroup$ – abiessu Sep 19 '13 at 15:00
  • $\begingroup$ Answers in here may be worth a look math.stackexchange.com/questions/463139/… $\endgroup$ – Jean-Sébastien Sep 19 '13 at 15:37
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It means $\forall x\in \operatorname{D}_f:f(-x)=-f(x)$. There's also something called even function that is defined as $f(x)=f(-x)$.

The terminology is because these functions show some properties that are common with odd and even integers.

EDIT: I think the terminology isn't very bad. I think the terminology is good when you compose functions.

The composition of odd/even functions behave exactly like multiplying odd and even integers. So, suppose $f,h$ are odd functions and $g,k$ are even functions.

$\operatorname{f \circ h}(-x)=\operatorname{f}(\operatorname{h}(-x))=\operatorname{f}(-\operatorname{h}(x))=-\operatorname{f}(\operatorname{h}(x))=-\operatorname{f \circ h}(x)$. That means $\operatorname{f \circ h}$ is odd. $\operatorname{g \circ k}(-x)=\operatorname{g}(\operatorname{k}(-x))=\operatorname{g}(\operatorname{k}(x))=\operatorname{g \circ k}(x)$. That means $\operatorname{g \circ k}$ is even. $\operatorname{g \circ f}(-x)=\operatorname{g}(\operatorname{f}(-x))=\operatorname{g}(-\operatorname{f}(x))=\operatorname{g}(\operatorname{f}(x))=\operatorname{g \circ f}(x)$. That means $\operatorname{g \circ f}$ is even. $\operatorname{h \circ k}(-x)=\operatorname{h}(\operatorname{k}(-x))=\operatorname{h}(\operatorname{k}(x))=\operatorname{h \circ k}(x)$. That means $\operatorname{h \circ k}$ is even.

It's easy to construct as many functions as you want that are neither odd nor even. So, most functions are neither odd nor even. However, the only function that is both odd and even is $\operatorname{f}(x)=0$. Because if $\operatorname{f}$ is both odd and even then we have $\operatorname{f}(-x)=-\operatorname{f}(x)$ because it's odd and we have $\operatorname{f}(-x)=\operatorname{f}(x)$ because it's even. So we must have $\operatorname{f}(x)=-\operatorname{f}(x)$ which gives $2\operatorname{f}(x)=0$, i.e. $\operatorname{f}(x)=0$

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  • $\begingroup$ The name is a really bad one. It arises because if $f$ is a polynomial with only odd-degree terms, then it is an odd function. Other than that, odd functions have very few properties in common with odd integers. For example, the sum of two odd functions is not in general an even function; it is an odd function; and the product of two odd functions is not in general odd. $\endgroup$ – MJD Sep 19 '13 at 15:09
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    $\begingroup$ @MJD: I disagree. It's true that they don't behave exactly like integers. But there are some nice properties too. For example if $f$ is odd/even, its derivative is even/odd respectively. The composition of two even functions is even, the composition of two odd functions is odd, and the composition of an even with an odd function is even. This gives us a homomorphism from the set of functions under composition as a monoid to $\mathbb{Z}_2$ under multiplication. It's not as bad as you say, even though it's far from perfect either. $\endgroup$ – user66733 Sep 19 '13 at 15:18
  • $\begingroup$ \circ will make the composition ymbol in $f\circ g$. $\endgroup$ – MJD Sep 19 '13 at 15:35
  • $\begingroup$ @MJD: Thanks. I didn't remember that code. Is there a shorter code than \operatorname{ } to force LaTeX to write $\operatorname{f}$ instead of $f$ which looks Italic? $\endgroup$ – user66733 Sep 19 '13 at 15:39
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$f$ is odd if $f(-x) = -f(x)$. The graph is symmetrical with respect to a 180 degree rotation about the origin. For example $f(x) = x^3$ is odd.

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An odd function is a function where: $$f(-x)=-f(x)$$

If you graph an odd function, you will find that the graph has rotational symmetry around the point $(0,0)$.

Also worth noting is that odd functions are made up of polynomials where $x$ is raised to an odd power

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A function is said to be odd if changing the sign of the variable changes the sign of the function (keeping the absolute value the same). It is even if changing the sign of the variable does not change the function. We express this mathematically as:

Odd: $f(-x)=-f(x)$

Even: $f(-x)=f(x)$

It is sometimes useful to know that any function can be expressed as the sum of an even function and an odd function $f(x)=\left(\frac{f(x)+f(-x)}{2}\right)+\left(\frac{f(x)-f(-x)}{2}\right)$

The "odd"ness of a function is therefore an antisymmetric property, while evenness is a symmetric property. There are occasions in other contexts where it is useful to split something into symmetric and antisymmetric parts. So although the terminology may seem a little odd, it actually expresses something which is quite deep, and introduces an idea which can be fruitful. For example determinants are antisymmetric functions.

If the symmetric integral of an odd function exists, it is equal to zero, which sometimes saves a lot of work if you spot it - i.e. $\int_{-a}^{a}f(x)dx=0$ if $f(x)$ is an odd function.

It is useful to know that the function $\sin x$ is odd, and $\cos x$ is even. The hyperbolic functions $\sinh x$ and $\cosh x$ are the decomposition of the exponential function $e^x$ into its odd and even parts.

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