/sci/ - Science & Math

[Anonymous No.14744996]

Bullshit. Utter nonsense

[Anonymous No.14745012]

>>14744996
0! = 1! / 1 = 1 / 1 = 1
0! = 1
There's exactly 1 way to arrange elements of an empty set.

[Anonymous No.14745020]

>>14744996
I know right. How can a zero shouting be the same as a lonely one.

[Anonymous No.14745029]

n! = (n+1)! / n

[Anonymous No.14745040]

Javascript disagrees.

[Anonymous No.14745046]

Brainlet thread
OP proven to be a retard again

[Anonymous No.14745097]

>>14744996
It's a convention that becomes particularly convenient when writing summations/sequences in general. A lot of results would require ugly special casing if 0! weren't defined this way.
Think about the Taylor series for example.

[Anonymous No.14745127]

>>14745012
>>14745040
>>14745046
>>14745097
but 0 is not equal to 1, it doesnt matter how you say it. you could say
0! = 1!
0? = 1
0 = 1.
0> = 1>
0% = 1
youd be wrong each time, the grammar of the equation doesnt change its meaning

[Anonymous No.14745142]

>>14745127
Why are you assuming that the factorial function is injective?
[math] sin(0)=sin(2\pi) \implies 0=2\pi?[/math]

[Anonymous No.14745150]

>>14745142
but sin(0) is not equal to sin(2housesymbol) because 0 is not equal. it doesnt matter what letters and brackets you put in the equation.

[Anonymous No.14745157]

>>14745150
it does, they're both equal to zero, this is basic trigonometry
I think you're confusing equality with equivalence

[Anonymous No.14745160]

>>14744996
Is cos 0 not 1, either?

[Anonymous No.14745161]

>>14745150
How do you explain then why your mom is such a dirty whore?

[Anonymous No.14745193]

[eqn]n!=1\cdot2\cdot3\cdots n[/eqn]
[eqn]n!=n(n-1)(n-2)(n-3)\cdots3\cdot2\cdot1[/eqn]
[eqn]n!=n(n-1)![/eqn]
[eqn]n!=n(n-1)(n-1-1)!=n(n-1)(n-2)!=n(n-1)(n-2)(n-2-1)![/eqn]
[eqn]n!=n(n-1)(n-2)(n-3)\cdots(n-n)! [/eqn]
Meaning that [math]0![/math] is in every factorial operation [math]n![/math]. The value of [math]0![/math] is not immediately apparent, so I assume that 0!=1. Can anyone provide a formal proof?

[Anonymous No.14745254]

>>14745193
n!=n(n-1)!
--> (n-1)! = n! / n
n=1 --> 0! = 1! / 1 = 1

[Anonymous No.14745267]

If that's utter nonsense explain this
[math]\frac{1}{2} = \sqrt{\frac{π}{2}}[/math]
How can you arrange half an object [math]0.88[/math] times?

[Anonymous No.14745275]

>>14745267
*
\frac{1}{2}! = \sqrt{\frac{\pi}{2}}

[Anonymous No.14745287]

>>14745193
>Can anyone provide a formal proof
There is no proof. That's just how it is defined. There is one way of arranging 0 objects, therefore 0! is 1. Similarly, factorial of a non natural number excluding 0, is 0, since there is no way of choosing objects in such fashion.

[Anonymous No.14745316]

>>14745012
>There's exactly 1 way to arrange elements of an empty set

>>14745287
>. There is one way of arranging 0 objects

I was on team [math] 0! = 1 [/math] until I read these, how does one arrange that which is not there, you got nothing to arrange

[Anonymous No.14745346]

>>14745316
Duh! Is a set containing the empty set.

[Anonymous No.14745523]

>>14745012
>>14745316
>>14745316
There is exactly one way to arrange zero objects, but somehow there are infinite ways to distribute n candies to zero children.

[Anonymous No.14745594]

>>14745040
Underrated

[Anonymous No.14745622]

How do you define the factorial? Let's use
[math]n! =\begin{cases} 1, &n=0 \\ n\cdot (n-1)!, &n>0\end{cases}[/math]
and it should be obvious why [math]0!=1[/math]

[Anonymous No.14745648]

There are [math]\binom{n}{k} = \frac{n!}{k! (n-k)!}[/math] ways to choose k elements from a set of n elements. How many ways are there to choose n elements from a set of n elements? 1 obviously. Therefore, [math]\tbinom{n}{n} = \tfrac{n!}{n!0!} = 1[/math]

[Anonymous No.14745681]

>>14745523
[math] _n \mathrm P_0 = 1[/math]

[Anonymous No.14745699]

>>14745523
>There is exactly one way to arrange zero objects

how does a function act on an input that doesn't exist?

if you tried to arrange 0 apples on a table, how do you grab(apple) ? that function wont return a success bool as apple isnt in memory. Perhaps you say apple = null . but than thats taking space in memory, that is something describing nothing, a quintessential illogicality

to highligh, I'm perfectly fine with the expression in >>14745648
but to interpret it as arranging nothingness reveals to me to search Wildbergers persepective on this

[Anonymous No.14745710]

>>14745127
stupid cuck nigger
if 2^2 = 4, is 2 equal to 4? STUPID KEKNIGGER STUPID RETARD

[Anonymous No.14745733]

>>14745699
It is mathematically true tho. There is exactly one way to arrange a set of zero apples and if you try to give N>0 apples to zero children, well, that's either undefined or infinite.
Also, proving 0!=1 using the binomial coefficient like in >>14745648 feels a bit like cheating because you are using a formula, instead of the definition/meaning of factorial.

[Anonymous No.14745795]

>>14745733
I've been told 0.999... = 1 is mathematically true as well, same for a zero the equation p(x) = x^5 - 2x + 3 but inspection reveals substandard handling of the edge cases with hand wavy ill precise definitions.

Ergo, if its mathematically true but not mirroring the logical wheelwork of nature, than the math is faulty, or at least to loosely bound

[Anonymous No.14746028]

>>14744996
[eqn]0!=\Gamma(1)\\
=\int_{0}^{\infty}{e^{-x}x^{1-1}dx}\\
=\int_{0}^{\infty}{e^{-x}dx}\\
=1
[/eqn]

[Anonymous No.14746034]

>>14745699
Arrangements is defined as number of tuples containing elements of a set. There is only one tuple for the null set.

[Anonymous No.14746041]

>>14745733
>proving 0!=1 using the binomial coefficient like in >>14745648 feels a bit like cheating because you are using a formula, instead of the definition/meaning of factorial
But the definition of the factorial already contains 0!=1 as in >>14745622

[Anonymous No.14746055]

>>14746041
Saying "it's true because they defined it that way" leads to the question "why didn't they define it in this different way?". That's why I wrote definition/meaning.
If you think of n! = n * (n-1) * (n-2) * ... * 1 then it is still questionable whether or not 0! should be 1 or 0, so its combinatorial definition might be more interesting.
Then you have to understand why >>14745733
>There is exactly one way to arrange a set of zero apples and if you try to give N>0 apples to zero children, well, that's either undefined or infinite.

[Anonymous No.14746079]

>>14744996
[math]\lim_{x\;\ll\;\infty} x! = 1 - \gamma x + \frac{1}{12}\left(6 \gamma^2 + \pi^2\right) x^2 + ...[/math]
Therefore
[math]\lim_{x \to 0} x! = 1[/math]
To within a third order correction or less.

[Anonymous No.14746094]

>>14746034
> one tuple for the null set.

this is easier to swallow, null takes space in memory. It defines a space by which the object can be determined to be present or not. Null takes entropy when taking space in memory. By definition the nothingness is just that, nothing. There is no physical process by which one can measure it, by definition lest it not be nothing.

these other definitions however, proclaim 0 = nothing

[Anonymous No.14746102]

>>14745699
>>14746094
You smoked too much crap. You don't know what you're talking about.

[Anonymous No.14746118]

Everyone here who thinks this needs proving is a fucking moron.

It's a definition.

[Anonymous No.14746121]

>>14744996
0 isn't ever a factor.
3! = 3+2+1
2! = 2+1
0 is never included in these so why would it be included in 0!

[Anonymous No.14746122]

>>14746102
I'm fine with all maths presented here

just the abstractions to language from the maths i find troublesome. To interpret this equation as one as 'there is only one was to arrange nothingness' as an assumption upon the continuum that is ill thought. Nothingness cannot be arranged. Pointers to null in memory can be however

[Anonymous No.14746143]

>>14746122
It's okay anon, inability to comprehend abstract thought is a common symptom of autism, it shouldn't affect your ability to learn to code at all

[Anonymous No.14746172]

>>14746122
Nothing (the empty set) can be arranged (in one way), mathematically speaking. NULL is a programming concept that is implemented in several way depending on what language you're considering: it can be a pointer to address 0 (C), it can be a special kind of object (python), it can be a special data type that is not even an object (python), it can be the empty list (lisp) and maybe it's a concept that does not exists at all (ML). You are comparing apples to oranges, it doesn't make sense.

[Anonymous No.14746197]

>>14746143
>abstract thought

as we see in Wildbergers critique on set theory, is that arithmetic with natural numbers, with comp sci forms the basic of combinatorics, of which set theory forms a part. A general theme is to stop taking variable instantiate for granted. You will never be able to arrange nothingness, coping out saying mathematically speaking is a self-reflexive non sequitor, whats the point of math in the first place if not to model reality?

When arithemtic with natural numbers with com sci informs the nature of the continuum instead of set theory
>>14746172
the difference between zero, nothingness, false and null become first order objects, not language games

[Anonymous No.14746240]

>>14745699
Make a list of k white balls and n-k black balls, numbered 1 through n so you can tell them all apart. There are n! ways of arranging the list. We can partition the arrangements into sets of arrangements that can be changed into one another without swapping any white balls with black balls. Each partition contains k! * (n-k)! arrangements, since this is the number of ways you can arrange the white and black balls separately. Each partition corresponds with a choice of which k out of the n balls to make white. The number of partitions is [math]\frac{n!}{k! (n-k)!}[/math]. Does this sort of argument make sense to you when k is 0?

[Anonymous No.14746242]

>>14744996
>0 is not equal to 1
Idk OP that makes sense to me

[Anonymous No.14746294]

>>1474499
0! = 1
1! = 1
0 = 1

[Anonymous No.14746319]

>>14746121
greetings from Kazakhstan

[Anonymous No.14746345]

10 * 0! = 10
10 * 1! = 10 * 1
10 * 2! = 10 * 1 * 2
10 * 3! = 10 * 1 * 2 * 3

[Anonymous No.14746382]

>>14746121
she got everything she dreamt of, eg shoes, Levi's pants, yellow t-shirts. She should be sucking Putins dick till eternity for killing her brother.

[Anonymous No.14746419]

>>14746240
here is a paste to the code i got started to help understand your query to help align any miscommunication

https://pastebin.com/PgbqMhzE

Assuming your logic is correct (turning into code i wasn't sure on some details) i believe it does, just with alot of wasted space on the stack machine. When k is zero we are performing the run without any white balls. though it is clear we are instantiating alot of space in memory to accomodate the edge case.

[Anonymous No.14746486]

>>14746419
Not sure how what the code is doing is relevant. I'll try writing some code myself to illustrate what I mean.

[Anonymous No.14746495]

>>14744996
0! is an expression which evaluates to a number, it is not a number per se

[Anonymous No.14746617]

>>14746419
>>14746486
https://pastebin.com/4fqM7692
What I'm saying is that getPermutations(s1+s2) should return the same set of results as getPermutations2(s1, s2), not necessarily in the same order. Since getPermutations(s).length is factorial(s.length) and getPermutations2(s1, s2).length is getCombinations((s1+s2).length, s1.length).length * getPermutations(s1).length * getPermutations(s2).length, we have getCombinations(i+j, i).length = factorial(i+j) / factorial(i) / factorial(j).

[Anonymous No.14746889]

>>14746617
thanks for that, i had a console.log party on that code and its much clearer to me what your saying now

> We can partition the arrangements into sets of arrangements that can be changed into one another without swapping any white balls with black balls.

the partition piece has me arranging subsets of s1 and s2 that were different numbers but same colour, but from the code its quite clear exactly what you mean.

I believe i came across this formula years ago and just accepted it as fact, but now I feel I have an understanding on how combinations and permutations are related

[Anonymous No.14746929]

>>14744996
Gamma function disagrees.

[Anonymous No.14746942]

>>14744996
Yes, you are

[Anonymous No.14747661]

>>14745795
Wildberger believes 0.999... = 1 though
For the standard meaning of ... as 9 repeats forever.

[Anonymous No.14747868]

>>14746055
It's not undefined or infinite, what the fuck are you talking about? NC0 and NP0 both are 1.

[Anonymous No.14747905]

>>14747868
>if you try to give N>0 apples to zero children, well, that's either undefined or infinite.
You could rephrase this as n / 0, and division by zero is undefined. The limit n/x of x->0 is infinity.

[Anonymous No.14747912]

>>14747868
> nCk is the number of all subsets of k elements of a given set of n elements.
> nPk is the number of all ordered subsets of k elements of a given set of n elements.
That's obviously zero for k=0. There's only one subset like that, and only one ordered subset (both are the empty set).

[Anonymous No.14747927]

>>14747661
not under all solutions

[Anonymous No.14747947]

>>14744996
Why is this one so much more universally accepted or at least less disputed than 0^0=1?

[Anonymous No.14748486]

>>14747947
most people dont know what a factorial is

[Anonymous No.14748490]

>>14747947
The argument
(n-1)! = n! / n
0! = 1! / 1
doesn't work for 0^0:
a^(b-1) = a^b / a
0^0 = 0^1 / 0 [undefined!]

[Anonymous No.14748494]

>>14748486
[math]\Gamma|_\mathbb{N}[/math] ?

[Anonymous No.14748496]

>>14747947
also x^y is discontinuous at x=0, y=0

[Anonymous No.14748645]

>>14747912
>that's obviously 0
>that's obviously 1
What?

>>14747905
It's literally 1. NC0/NP0 is the number of un/ordered subsets of size 0. There's only one subset, the null set. There's exactly one way of putting N chocolates in hands of 0 children, i.e., not putting anything at all. For similar reasons 0PN/0CN = 0, if N>0. None of it is undefined for nonnegative integers.

[Anonymous No.14748905]

>>14748496
>discontinuous
what is discontinuous? i don't see a function anywhere in your post

[Anonymous No.14749097]

>>14748905
[eqn] f : \mathbb R \times [0,\infty) \to \mathbb R\\
\phantom{f:} (x,y) \mapsto x^y [/eqn]
There! Is that good enough for you faggot?

[Anonymous No.14749123]

>>14744996
This is an early example of analytic continuation
ANALYTIC CONTINUATION

[Anonymous No.14749149]

>>14748905
[eqn] f : \mathbb R \times [0,\infty) \to \mathbb C\\
\phantom{f:} (x,y) \mapsto x^y [/eqn]
There! Is that good enough for you faggot?

[Anonymous No.14749185]

>>14749149
well of course that's discontinuous at 0, 0 is literally an endpoint

[Anonymous No.14749198]

>>14749185
Counter example: [math] f : \{ 0 \} \to \{ 1 \} [/math], where [math] f(0) = 1[/math]. 0 is an endpoint but it is continuous.

[Anonymous No.14749253]

>>14749198
not an endpoint

[Anonymous No.14749600]

>>14746122
You earlier invoked "0 apples". Well, this was your undoing, because "0 apples" distinctly is not the same as "nothingness", as you qualify that quantity of 0 with "apples" (as you should). There is exactly one figurative bucket of zero apples in the world; and this bucket that contains "apple-nothingness" corresponds to exactly one way to arrange 0 apples.
Now mentally exchange these apples for the concept of integers (or permutations) itself.

[Anonymous No.14750398]

>>14749253
Give an example of a set with endpoint

[Anonymous No.14750497]

>>14745040
> trusting Javascript for anything

[Anonymous No.14750944]

>>14746028
based

[Anonymous No.14750969]

>>14750398
>>14749149
>[0,∞)

[Anonymous No.14750984]

>>14750969
Counterexample:
[eqn] f : [0,\infty) \to [0,\infty) \\ \phantom{f : } x \mapsto x [/eqn]

[Anonymous No.14751092]

>>14750984
but that's discontinuous at 0 it literally just stops

[Anonymous No.14751190]

>>14751092
If it is discontinuous, surely you could find an [math] x \in [0,\infty) [/math], such that: [math] | x - 0 | < \epsilon /2 \land |f(x) - f(0)| \geq \epsilon [/math]

[Anonymous No.14751200]

>>14744996
We know it doesn't make sense. That's why 0! := 1

[Anonymous No.14751552]

when n is 1 n! = 1
the function described for n! is n! = 1*2....*n-1*n
therefor (n-1)! = n!/n
from this we can say (1-1)! = 1!/1
0! = 1/1
0! = 1

[Anonymous No.14751745]

>>14744996
I wrote a paper on Factorials and the Gamma function. This might explain it.

If the factorial calculates the number of ways you can order n objects, or the number of states n objects can exist in, then it become obvious why 0!=1. Nothingness can only ever exist as nothing. It has one state. This at least gives a more intuitive proof.

[Anonymous No.14752841]

>>14751745
>How many states can nothingness exists in?
>If I have nothing, how many ways I can arrange it?
>The answer is clearly 1
It's debatable whether or not this is should be accepted as an obvious statement.
For example, is there anything that I can arrange in zero ways? Is there something that is stateless? Maybe nothingness doesn't need a state. Motivate your answer.

[Anonymous No.14753454]

>>14744996
[math]\displaystyle z!=\Gamma (z+1)=\int_{0}^{\infty }e^{-t}t^{z}dt[/math]
so [math]\displaystyle 0!=\Gamma (1)=\int_{0}^{\infty }e^{-t}dt=1[/math]
EZ-PZ

[Anonymous No.14753567]

{}=1
{1}=1
{1,2},{2,1}=2
{1,2,3},{1,3,2},{2,1,3},{2,3,1},{3,1,2},{3,2,1}=6

Simple as.

[Anonymous No.14753702]

>>14745040
/thread

[Anonymous No.14753801]

>>14753702

[Anonymous No.14754040]

>>14745699
the nothing in itself by just being is the only arrangement of itself it can be, therefore 0!=1

[Anonymous No.14754141]

>>14754040
Please arrange this 'the nothing' for me

place it on a table
strike it so it makes a sound
determine its fragrance
qualify its taste
show me where this nothing is so i can see it

oh wait

[Anonymous No.14754208]

>>14754141
if you could do any of that it wouldn't be "nothing" now would it, silly

[Anonymous No.14754666]

>>14754208
So the factorial function should only be defined for integers greater than zero and 0! = undefined because you cannot arrange "nothing".

[Anonymous No.14754784]

>>14754666
>>14745254

[Anonymous No.14754812]

>>14754141
the nothing was on the table already

[Anonymous No.14754820]

>>14754666
nothing to do with arranging, 0! and 0^0 both equal 1 for the same reason. ! and ^ are both multiplying operations but no multiplication is happening so all that's there is the background identity 1

[Anonymous No.14754822]

>>14745254
>n!=n(n-1)!
>--> (n-1)! = n! / n
n=-1 --> -2! = -1! / (-1) = -1/-1 1!/1 = 1
-2! = 1

[Anonymous No.14755279]

>>14754822
>>-1! = -1
>source: my ass

[Anonymous No.14755575]

>this b8 thread is still up
Never change /sci/

[Anonymous No.14756423]

>>14754822
Retard. You could have at least came up with some nonsense a somewhat more correct way.
[eqn] 0! = 0 (-1)! \implies 0! = 0[/eqn]

[Anonymous No.14759674]

>>14754820
So multiplying nothing always yields something?

[Anonymous No.14759702]

>>14759674
>why is there something instead of nothing
that's how the universe got started

[Anonymous No.14759704]

>>14759702
Nothing is something, though, it even has one specific way to be arranged.

[Jealous No.14759712]

[eqn]\frac{n!}{(n-1)!}=n[/eqn]
Choose n=1
[eqn]\frac{1!}{0!}=1\quad\implies\quad0!=1[/eqn]

[Anonymous No.14759846]

>>14759712
[math]\frac{n!}{(n-1)!} = n \implies \frac{0!}{(-1)!} = 1 \implies (-1)! = 1[/math]

[Anonymous No.14759879]

>>14759846
>>14759846
0!/(-1)!=0 retard

[Anonymous No.14759925]

>>14745040
lol

[Anonymous No.14760024]

Numbers are an irrational concept and purely a human creation, all nature cares about is if something is equal to or a fraction of a variable whole, depending on the situation, and is not necessarily tied to any dimension but if some defining characteristic is fractionally less or more than another.

[Anonymous No.14760040]

>>14760024
Especially the modern number system that expects us to believe that the origin number is its own negation/opposite number.

[Anonymous No.14760045]

>>14745040
>>14753702
>>14753801
i dont know if this math operator is correspond to boolean logic, if so OP is retarded

[Anonymous No.14760080]

>>14746121
Empathetic people, everyone.

[Anonymous No.14760093]

>>14746121
>0 is never included
It is always included 3+2+1=3+2+1+0, 2+1=2+1+0

[Anonymous No.14761187]

>>14744996
>He's never heard of the gamma function.