/sci/ - Science & Math

>>11942141
factorial are (n + 1 * n * n-1....*1)/n+1, 0! = 1/1 = 1

0 not equal to 1 is true

>>11942141
>!=

Number of bijections from the empty set to itself is 1.
The number of bijections from a set with n elements to itself is n!

>>11942141
0! results in the empty product. The result of multiplying no factors. This is by convention the multiplicative identity, which is 1 (x*1=x). Just as 0 is the identity of summation (x+0=x). You have your heart in the right place but lack the knowledge which means you need to read a bit more about the fundamentals of mathematics.

Remember that 90% of these things aren't out there in nature (which is why you have a certain feel for mathematic operations) they are rather conventions that we agreed upon to do more complex math. You could define 0! as 0 but then you would have to change a lot of things about the operation itself which would be inconvenient to use it properly.

>>11942141
4! = 5! / 5
3! = 4! / 4
2! = 3! / 3
1! = 2! / 2
0! = 1! / 1

>>11942141
>What is the gamma function?

>>11942310
-1! = 0!/0 = 1/0
-2! = -1!/0 = 1/(0^2)
-3! etc.

why hasn't any progress been made on the nature, magnitudes, and orders of infinity?

>>11942343
1/0 is undefined my friend
Meaning negative factorials are all undefined

>>11942353
what a weakass copout. I get that they aren't currently defined (outside of complex analysis), but what's stopping the progress?

http://mathforum.org/library/drmath/view/57128.html

>>11942141
how many times can you arrange 0? once

>>11942343
> /0
your iq

>>11942400
>how many times can you arrange 8? once

>>11942400
bs, I just did it 300 times

>>11942374
What's stopping it is defining x/0 using axioms that don't make the axiomatic system inconsistent

>>11942435
non-faggotism

>>11942404
th-thanks?
>>11942435
your sentence is incoherent

>>11942141
The sum over the empty set is the neutral element with respect to addition, no one would ever argue that the result of adding nothing is not 0.
So it makes sense to have the product over the empty set be the neutral element with respect to multiplication. 1.
What is supposed to be bullshit about that?
We do the same for [math]n^0[/math]. The real bullshit is [math]0^0[/math], but that remains undefined after all.

What does the "!" after "0" indicate?

>>11942709
shouting

>>11942709
.9...=0!!!

>>11942746
Math is beautiful.

Think about it this way.

I have two rooms of people. The first room has n people and the second room has m people. Say we wanted to line up the people in the first room in a row. Then there are n! ways of doing this. What if we do this to the second room of people? Then there are m! ways of doing this. So how many total ways are there to order both rooms. Well for each of the n! ways to permute the people in the first room, there are m! ways to permute the people in the second room. So the answer is n! * m!. But what if m = 0? Then there are still n! ways to permute the first room so it only makes sense that m! = 0! is defined to be 1.

>>11942141
yes it is
factorials are [math]\mathit{defined}[/math] as [eqn]
0 ! = 1 \\
n ! = n (n - 1)!
[/eqn] for all natural numbers [math]n[/math].

Same reason that 1 + 2 + 3 + 4 + ... = - 1/12

>>11942310
fake news
4!= 4*3*2*1
3!= 3*2*1
2!=2*1
1!=1*1
0!=0*1

>>11942823
thats obviously not true either. how does so much bunk catch on in the field of mathematics?

>>11942141
yeah it is, 0 != 1; simple programming

>>11942823
That only works out by assuming that 1+2+3+... converges.
It trivially does not, and you should be ashamed of yourself.

>>11942858
>That only works out by assuming that 1+2+3+... converges.

>>11942781
>n!=n(n−1)!
>something is defined as itself plus some extra stuff

technically true but this is a bit silly

>>11942852
brainlets ask dumb sounding questions and mathfags can't be arsed to teach
ignorance, IFLS, and garbage youtube videos ensue

>>11942982
wew lad
ever heard of recursion?

>>11942830
>0!=0*1
read definition again

>>11942141
it is a programming language statement that evaluates to true

>>11942982
I seriously hope this is bait, holy shit

>>11942141
1=0!

makes you think

>>11942991
>>11943019
maybe i misunderstood something or my wording was off
but imagine there was a mathematical operation notated like n& or something, then the definition couldnt contain that & sign right? it would be like a circular argument
as far as i know "proper" definition of the factorial doesnt rely on the factorial being an established operation, thats why i found it silly to explain factorials using factorials

>>11942435
https://en.wikipedia.org/wiki/Wheel_theory

>>11943277
this is called recursion and it's a completely normal and common thing. I'm not even gonna explain, just read the wikipedia page or something.

>>11943233
It seems if you shout 0 loud enough it become 1, have ethnic studies and feminism taken over math?

>>11942995
the definition is arbitrary. if something is arbitrary it is astrology not mathematics

>>11943296
i understand recursion (i think), my point was more about using recursion as a fundamental definition
n!=n(n−1)! makes sense, but could it be the only definition for a mathematical operation? yes it describes the factorial but the actual definition is all integers from 1 to n multiplied, not n!=n(n−1)!

>>11943521
>but the actual definition is
says who

>>11942830
>by the logic you apply to the last expression consider the following

4! = 4*3*2*1*0 = 0 you fag
3! = 3*2*1*0 = 0 you fag
2! = 2*1*0 = 0 you fag
1! = 1*0 = 0 you fag
0! = 0 kek

>>11942830
Means the same thing as >>11942310
If you divide by 4 to remove the 4 from 4!, you get the same term as 3!
4!=4*3*2*1
(4*3*2*1)/4 = (4*3*2*1)/4
3*2*1 = (4*3*2*1)/4
3! = 4!/4

If you absolutely want a non-recursive definition:
[math]n! := \prod_{j \in ([1,n] \cap \mathbb{N})}j[/math]

So consequently,
[math]0! = \prod_{j \in ([1,0] \cap \mathbb{N})}j = \prod_{j \in \emptyset}j = 1[/math]

1! = 1
0! = 1
therefore
1! + 0! = 2

>>11943722
This, what is the number of bijections from the empty set to itself brainlets?

>>11943730
1! + 0! = 2!

>>11943741

>>11943521
This is real late, but: anon, this is a perfectly valid way to define the function (assuming you can show it's defined for the relevant inputs, etc), with the caveat you need a terminating condition. So:

n! = n * (n-1)! for all positive integer n,
0! = 1

...is a complete definition of a function over the set of nonnegative integers. In the case of the factorial, you can reeatedly apply the first rule to itself to get a 'simpler' expression, aka, n*(n-1)*...*2*1. Either is valid.
However, some functions are more usefully defined using a recursive definition -- breaking it down like above is difficult, impossible, or just not helpful for understanding the function. A recursive definition is perfectly valid, as long as you can show it's self-consistent and defined for all values in its domain (usually shown by induction).

>>11943277
Read Analysis by Terrence Tao and unironically you will know why the recursive definition works

>>11943734
Can it be defined as the card of the family of functions of a given finite set to itself such that the said function is bijective and be proven equivalent to the recursive definition?

>>11943722
What is your definition of the capital pi notation? I know a definition but it is itself recursive.

>>11942404
>infinity
c'mon

>>11942982
>>11943277
>this is a bit silly
>it's circular
>proper definition of the factorial doesn't rely on the factorial being an established operation
create a recursive factorial function in your favorite programming language and think about why it works

>>11943970
>Can it be defined as the card of the family of functions of a given finite set to itself such that the said function is bijective and be proven equivalent to the recursive definition?
Yes

>>11943970
>What is your definition of the capital pi notation? I know a definition but it is itself recursive.
https://ncatlab.org/nlab/show/cartesian+product

Why is everyone yelling?

[math] N_n = \{1, \ldots, n\} [/math] with [math] N_0 = \emptyset [/math] in particular. We define for each [math] n \geq 0 [/math] the set
[eqn] S_n = \{f : N_n \to N_n : f(x) = f(y) \rightarrow x = y\} [/eqn]
With this, take [math] n! = |S_n| [/math], where we recall that such functions are subsets of [math] N_n \times N_n [/math]. Therefore, since
[eqn] N_0 \times N_0 = \emptyset \times \emptyset = \emptyset [/eqn]
we see that there is exactly one function (subset) in the space [math] N_0 \times N_0 [/math], and furthermore, this function satisfies the criteria to be counted in [math] S_0 [/math], proving that [math] 0! = 1 [/math].