7

I found this definition in this PDF file on page 17: http://math.byu.edu/home/sites/default/files/u107/proofs_crash_course.pdf

I couldn't quite grasp it. I tried substituting x for a value in both f(x) and f(-x) but the result wasn't equal. So then, how was this proven? Is it correct?

Please understand that I'm requesting an explanation of the definition : f(x) is even if f(-x) = f(x). I'm not asking for an answer to the question asked in the PDF reference.

JohnD
  • 14,392
Anfernee
  • 243
  • 11
    It's a definition. – ploosu2 Dec 04 '14 at 00:32
  • I know that. I'm requesting proof of that definition – Anfernee Dec 04 '14 at 00:44
  • The question you've asked doesn't make sense. Are you, by chance, trying to ask different questions like "How do I prove the function $g$ defined by $g(x) = x^2$ is even?" or "How do I show that $\sin$ is not an even function?" –  Dec 04 '14 at 00:49
  • he wants to know how to the statement "If $f(x)$ is even Then $f(-x)=f(x)$" – Fmonkey2001 Dec 04 '14 at 00:56
  • 1
    Definitions do not require proof. It's like baptizing. I baptized myself ypercube. Therefore ypercube is me. We baptized even functions all those functions f that satisfy f(x) = f(-x) for all x that f is defined. – ypercubeᵀᴹ Dec 04 '14 at 01:02
  • Well then, how was that definition created? – Anfernee Dec 04 '14 at 01:05
  • 1
    For example: You define x^2 as the function that multiplies x by itself. You are asking why x^2=x*x. – Teoc Dec 04 '14 at 01:22
  • @AjPennster Saying something is "definition" means it's not something we need to prove as a consequence of something else. What it is instead is a statement we are saying is true. For example, I can define a new operation $\oplus$ such that for any real numbers $x$ and $y$, $x \oplus y = 2x + 3y$. Since I defined it this way (i.e. I said that this is going to be true), I don't have to prove that $\oplus$ behaves this way -- that is the way it is defined. Now, if instead I said let $\oplus$ be defined such that $x \oplus y = 2x + 2y$, and I claimed that under $\oplus$, the sum of any... – layman Dec 04 '14 at 01:25
  • ...two real numbers is even, this would have to be proven, because I'm not defining that the sum of the two functions is even. I'm defining that $x \oplus y = 2x + 2y$, and I'm saying that a consequence of this definition is that any sum is even. Then you would have to prove that if $\oplus$ is defined as it is above, then the sum of any two numbers is even. – layman Dec 04 '14 at 01:27
  • 1
    Ah I understand. – Anfernee Dec 04 '14 at 01:27
  • Also you might be confused on this: An even FUNCTION does not relate to even NUMBERS. – Teoc Dec 04 '14 at 01:31

6 Answers6

13

There's a chance you might be getting confused with the word "even". It doesn't mean that the function equals 2,4,6, etc. We use the word "even" for that too (when we are talking about numbers), but when people say that a function is even, they mean that it is mirrored over the y-axis.

For example, $f(x)=x^{2}$ is "even", not because it equals 2 or 4 or 6, but because it is symmetric across the y-axis. This mirror property is a result of the fact that $f(x)=f(-x)$. Why is it that way? Well, if you plug in 1, it's the same as if you had plugged in -1. Same with 2 and -2. Thus $f(x)$ is the same for positive and negative x; in other words, $f(x)=f(-x)$. $cos(x)$ has the property too, and tons of other functions. I'll show you some pics for clarity, credit to wolfram alpha:

x^2

cosx

cosx/|x|

If you're asking proof that $f(x)$ is even if $f(x)=f(-x)$, it's not something that you can "prove", at least not in a mathematical way. Let me say a bit more: It's not like we have a concept of what "even" means in our head, and we have a concept of what it means that $f(x)=f(-x)$ as well, and we can show that those are equivalent. Instead, "even" is just a label that we put on functions for which $f(x)=f(-x)$, rather than saying that outright each time. We could never use the word even, but it's just easier. We observed a trait and then we gave it a name.

An analogous example: you can't "prove" that we call those round edible fruits with stems "apples", that's just the name we gave them. The apples came first, and then came the name. I guess the only "proof" that those things are called apples would be to take a survey and see what percent of people actually DO call them apples. You could do that for this too, and I'm confident that you'd find that when it came to mathematicians, in English, at least 95% would call such a function even. Though this wouldn't be a proof, it would just be strong evidence if you had a large sample size. Proofs are, in theory, reasoning which makes it IMPOSSIBLE for the statement to be false.

doublefelix
  • 890
  • 1
    Thank you for a clear and descriptive answer even though my question was not well constructed. The diagrams helped a lot. – Anfernee Dec 04 '14 at 01:31
6

You also might be asking why we would call a function like $f(x) = x^2$ "even" - that is, why would we choose that word. Well, it turns out that $f(x) = x^k$ is an even function when $k$ is an even number, and is an odd function when $k$ is an odd number. So although we could have chosen other words for these types of functions (for example, "symmetric" and "antisymmetric"), choosing "even" and "odd" as the names has some mnemonic value.

Michael Lugo
  • 22,354
  • 1
    I was going to make the same point, as the terminology certainly isn't a coincidence. I'll note that the "Earliest Known Uses of Some Words of Mathematics" site attributes the term "even function" to Euler himself in a 1727 presentation. I suspect, as you assert, that "even" & "odd" were chosen to evoke the even-power and odd-power functions as archetypal examples. This suspicion could be confirmed (or refuted) by anyone with access to the text of Problematis traiectoriarum reciprocarum solutio (and a degree in Latin) ... but, sadly, I am not that anyone. – Blue Dec 04 '14 at 02:04
  • 2
    I found a translation of Problematis traiectroiarum. On "even": In the first place functions are to be noted that I call even, of which this is the property: that they remain unchanged if $-x$ is put in place of $x$. All the powers of $x$ with even exponents are of this kind [...]. On "odd": In the second place, I take heed of odd functions, which in short produce the negative of these, if $x$ is changed to $-x$. All the powers of $x$ with odd exponents, such as $x$ itself, $x^3$, $x^5$, etc. are of this kind [...]. – Blue Dec 04 '14 at 02:17
  • 4
    Kinda relevant: For numbers, $a\times b$ is odd iff $a,b$ are both odd. For functions, $f\circ g$ is odd iff $f,g$ are both odd. (The notation $\circ$ means composition. For example, $(f\circ g)(x)=f(g(x))$.) – Akiva Weinberger Dec 04 '14 at 03:05
  • @columbus8myhw And also, $a + b$ is odd if and only if one of $a,b$ is even and the other is odd; similarly, $f\times g$ is odd if and only if exactly one of $f, g$ is even and the other is odd. –  Dec 05 '14 at 03:59
  • 1
    @Strants So, we're replacing $+,\times$ with $\times,\circ$ respectively. This may seem random, but note that if $f_n(x):=x^n$, we have $f_a\times f_b=f_{a+b}$, and $f_a\circ f_b=f_{a\times b}$. This makes sense, because $f_a$ is even iff $a$ is, and similarly for odd functions/numbers. – Akiva Weinberger Dec 05 '14 at 04:36
2

A definition is a definition, you are not allowed to request a proof, or to question it, you are expected to memorize it and be able to use it.

But, you may ask for an intuitive justification and a motivation for a definition. Consider the function $g(x)=3x^4-6x^2-5$. Notice things like
$g(-1)=3(-1)^4-6(-1)^2-5 = 3 (1)^4-6(1)^2-5 = g(1)$, and
$g(-4)=3(-4)^4-6(-4)^2-5 = 3 (4)^4-6(4)^2-5 = g(4)$, and more generally
$g(-x)=3(-x)^4-6(-x)^2-5 = 3 (x)^4-6(x)^2-5 = g(x)$ for all $x$.
This function has the property that $g(-x)=g(x)$ for all $x$. It seems this has something to do with the powers of $x$ that appear in it: $x^4$, $x^2$, and, you may say, $x^0$ (like $5x^0=5$). It happens more generally that if all powers of $x$ in a function are even, then the values of that function at $-x$ and at $x$ are the same. But, this also happens for some other functions, like $\cos(-x)=\cos(x)$ for all $x$ even if you see no powers of $x$ here. It is convenient to give a name to functions that satisfy this property. Since the easiest examples involved even powers, we call such functions even functions. But, we do not want to be restricted to only consider functions in which the powers of $x$ are even: It turns out to be convenient to use this same name whenever a function $f$ has the property that $f(-x)=f(x)$ for all $x$. All such functions are by definition called even (this is just a convention, nothing to prove here, like you need no proof to justify why the color red is called red), but once this convention, or definition, if accepted you could use it to prove that one function or another is even, i.e. that it satisfied the defining property of even functions, that $f(-x)=f(x)$ for all $x$. For example the function $f(x)=|x|+5$ is even (even if it looks like $x$ appears to the first power which is odd), since $f(-x)=|-x|+5=|x|+5=f(x)$ for all $x$.

Another example $h(x)=x^3-4x$. Here if you start with $h(-x)$ we get
$h(-x)=(-x)^3-4(-x)= -(x^3)+4x= -\Bigl(x^3-4x\Bigl)=-h(x)$.
So $h(-x)=-h(x)$ for all $x$. It is convenient to call all functions with this property odd functions. That is $f$ is an odd function if, by definition (convention, agreement) $f(-x)=-f(x)$ for all $x$ (in the domain of $f$).
There are many examples of such functions where all powers of $x$ are odd, and this justifies the name, but there are other examples too, like $\sin(-x)=-\sin(x)$ for all $x$, so $\sin$ is an odd function.

The graph of each even function is symmetric about the $y$-axis. This is something one could prove, using the definition. Indeed, if the point $(x,f(x))$ is on the graph, the so is the point $(-x,f(x))=(-x,f(-x))$,
but $(x,f(x))$ and $(-x,f(x))$ are symmetric about the $y$-axis.
Like, the graph of $y=x^2$ is symmetric about the $y$ axis, and so is the graph of $y=\cos x$.

The graph of each odd function is symmetric about the origin. This is something one could prove, using the definition. Indeed, if the point $(x,f(x))$ is on the graph, the so is the point $(-x,-f(x))=(-x,f(-x))$,
but $(x,f(x))$ and $(-x,-f(x))$ are symmetric about the origin.
Like, the graph of $y=x^3$ is symmetric about the origin, and so is the graph of $y=\sin x$.

Finally, there are functions that are neither even nor odd. Usually that happens when one mixes (additively) even and odd powers of $x$, like in $p(x)=x^3-5x^2$. Here $p(1)=-4$ and $p(-1)=-6$. So, $p(-1)\not=p(1)$ and this function $p$ is not even. But we also have $-p(1)=-(-4)=4$ and $4\not=-6$, that is $p(-1)\not=-p(1)$ so this function is not odd either. So, it is neither. One alternative way to see this, if to start with $p(-x)$ and see if one obtains $p(x)$ (in that case $p$ would be even), or one obtains $-p(x)$ (in that case $p$ would be odd), or one obtains neither $p(x)$ nor $-p(x)$. In this example $p(-x)=(-x)^3-5(-x)^2 = -x^3-5x^2$ which is different from both
$p(x)=x^3-5x^2$ and from $-p(x)=-x^3+5x^2$, so $p$ is neither an even nor an odd function.

Other examples of functions that are neither even not odd are $2-\sin(x)$ and $x+\cos(x)$. On the other hand $\dfrac{x^2-1}{x^3}$ is odd, regardless that we mix even and odd powers, because this time we do not add/subtract them, but instead miltiply/divide them (the numerator has only even powers, and the denominator has only odd powers). You may use the definition to verify (that is, prove) that $\dfrac{x^2-1}{x^3}$ is indeed an odd function. You may also try to find out things like whether the product of two odd functions is odd or even (it is even), and whether the product of an odd function and an even function is even or odd.

Mirko
  • 13,445
0

The definition of an even function is that it satisfies f(x)=f(-x). Likewise, an odd function satisfies f(-x)=-f(x). An example of an even function is cos(x).

Teoc
  • 8,700
0

I'm trying to upload some images, but my computer is going very slow. Consider: $$f(x)=\sin(x)$$ If you were to do $$f\big({\pi\over2}\big)=\sin\big({\pi\over2}\big)$$ and also were to do $$f\big(-{\pi\over2}\big)=\sin\big(-{\pi\over2}\big)$$ you could see that $$f\big({\pi\over2}\big)=1\ \text{and} \ f\big(-{\pi\over2}\big)=-1$$ So $$f({\pi\over2})=-f(-{\pi\over2})$$ making it an odd function. You could do the same argument for even functions using $$f(x)=\cos(x)$$ Does that help any?

Fmonkey2001
  • 1,240
  • So what is really the even/odd thing here? The result of the function? The function itself? – Anfernee Dec 04 '14 at 00:48
  • Even: means that if you fold the function over the $Y$-axis the same $y$ value touches two different $x$ values such that $f(-x)=f(x)$

    Odd: means that if you fold the function over the $Y$-axis and the $X$-axis the same $y$ value touches two different $x$ values such that $f(-x)=-f(x)$

    or at least that's my understanding of it.

     – Fmonkey2001 Dec 04 '14 at 00:52
  • 3
    The even/odd thing is the function itself. – Fmonkey2001 Dec 04 '14 at 00:53
0

You want to proof this: If $f$is even, then $f(x)$ is not one-to-one. You can proof this by deduction. Let be any $x$, by definition, since $f$ is even, then $f(x)=f(-x)$, therefore $f(x)$ is not unique, then $f(x)$ is not one-to-one.

You must remember this: $f$ is one-to-one if $f(x)=f(y)$ then $x=y$.

Hugo Rodrigo Mas Ku
  • 333
  • I'm asking for proof of the definition stated in the title, not the example given in the reference. – Anfernee Dec 04 '14 at 00:47
  • You can not prove that because it is a definition, you must understand what is the diference beetween one definition and one proof. For example, if I define a prime number: A prime number p is a number with only 2 divisors. So I can not ask for a proof like: Prove that if a number p has only 2 divisors, then it is prime. There is no proof for this because it is a definition. – Hugo Rodrigo Mas Ku Dec 05 '14 at 06:36