How many even functions are there from $\{-n, \dots,n \}$ to itself?

Question

If $A=\{-n,-n+1, \dots, n-1,n \}$, how many functions $A \to A$ are there,that are even,so they satisfy the condition $f(-x)=f(x), \forall x \in A$?

Is it maybe $(\frac{|A|}{2})^{|A|}$ ?

Note that $|A|=2n+1$ is odd, so your formula cannot be right (its value is non-integer) — Marc van Leeuwen, Apr 24 '14 at 21:12
What you do is just "count" them. You look like you are on the right track, by multiplying out all the points that can be "chosen" and excluding the ones that are fixed by symmetry. Now you just need to notice that all of 0 to n can be chosen, and all of -n to -1 are fixed. How many is that? How many different choices are available at each point? — ex0du5, Apr 24 '14 at 21:22
@evinda: Switch base and exponent, and round the latter up to an integer. — Marc van Leeuwen, Apr 24 '14 at 21:22

Marc van Leeuwen · Accepted Answer · 2014-04-25T04:34:15.660

5

The requirement of being even means your function is determined by its values on the non-negative integers, and these values can be arbitrary (within the specified range): one can put $f(-i)=f(i)$ for $i=1,\ldots,n$ to complete to an even function. So the answer is the number of maps from the $n+1$ element set $\{0,1,\ldots,n\}$ to the $2n+1$ element set $A$, which number is $(2n+1)^{n+1}$.

edited Apr 25 '14 at 04:34

answered Apr 24 '14 at 21:21

Marc van Leeuwen

115,048

Could you explain me further the sentence: $\text{ So the answer is the number of maps from a } n+1 \text{ element set to a } 2n+1 \text{ element set, which number is } (2n+1)^{n+1}.$ – evinda Apr 24 '14 at 21:29
I've detailed which precise sets are meant. Once a map as indicated in te final sentence is fixed, apply the procedure at the end of my first sentence to construct an even function from $A$ to itself. – Marc van Leeuwen Apr 25 '14 at 04:35

drhab · Answer 2 · 2014-04-24T21:22:10.553

2

The values on negative integers is determined by the value on the positive integers and nonnegative integers can be sent to every element of $A$ giving $(2n+1)^{n+1}$ possibilities.

edited Apr 24 '14 at 21:22

answered Apr 24 '14 at 21:12

drhab

151,093

1

Edit this to read $(2n+1)^n$ and I'll retract my downvote :) – Hagen von Eitzen Apr 24 '14 at 21:14
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It is not necessary that $f(0)=0$. $f(0)=f(-0)$ automatically. – Ben Whitney Apr 24 '14 at 21:14
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There are even functions that aren't $0$ at $0$. For example constant functions. – kahen Apr 24 '14 at 21:15
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Thank you. I feel a bit stupid now. – drhab Apr 24 '14 at 21:16
I haven't understood why it is $(2n+1)^n$. :/ Could you explain it to me? – evinda Apr 24 '14 at 21:16
@kahen Ah yes, $(2n+1)^n$ counts odd functions. We need yet another factor $(2n+1)$ to get the even ones. – Hagen von Eitzen Apr 24 '14 at 21:16
So,what is now the right answer..?I am confused right now. – evinda Apr 24 '14 at 21:18
@evinda For each $k\in{0,1,\ldots , n}$ you have the choice among $2n+1$ values, namely ${-n,\ldots,n}$. Thus the count is $\underbrace{(2n+1)\cdots (2n+1)}_{n+1}=(2n+1)^{n+1}$. – Hagen von Eitzen Apr 24 '14 at 21:18
I am confused myself now. I will delete my answer and will (hopefully) come up with a new one. – drhab Apr 24 '14 at 21:19
@evinda Sorry for my mistake. I hope it is all clear now. – drhab Apr 24 '14 at 21:24

Gopi · Answer 3 · 2014-04-25T05:16:08.870

1

Suppose n=1.Then A={-1,0,1} are elements of your set A. Now define x goes to x^2n where n€N. That is f from A to A is defined as f(x) = x^2n (x power 2n). Now thing how many even functions are there. There are countably infinite even function.

edited Apr 25 '14 at 05:16

answered Apr 24 '14 at 21:45

Gopi

21

1

The codomain of the function is $A$, not $\mathbb{N}$. There are a finite number of such functions, so there are a finite number of even ones. – Henry Swanson Apr 24 '14 at 22:03

How many even functions are there from $\{-n, \dots,n \}$ to itself?

3 Answers3