/sci/ - Science & Math

and ?

Ha, I was going to make a thread about this.

I just used this in a proof today actually

>>7624407
[eqn] x^x=1 \implies x \ln(x)=0 \implies x=0 [/eqn]

Proof me mistaken, fgts.

>>7624427
not ln

>>7624416
That's like saying 0 * 0 = 1

doesn't make cents

>>7624459
0 * 0 = 0^2, not 0^0

Empty product is 1.

>>7624509
Show workings :^)

>>7624427
1^1 = 1 you retard

>>7624459
>Cents

OP is wrong, [math]0^0\neq0[/math]
since [math]0^0=0^{x-x}=\frac{0^x}{0^x}=\frac{0}{0}[/math]
it's undefined as shit

>>7624641
[math]0^0=0^{x-x}=\frac{0^x}{0^x}=\frac{0}{0}[/math]

>>7624645
[math] 0^0 = 0^{x-x} = \frac{0^x}{0^x} = \frac{0}{0} [/math]

>>7624645
[math]0^0=0^{x-x} = \frac{0^x}{0^x} = \frac{0}{0}[/math]

3^5 = 3^6 ÷ 3 = 243
3^4 = 3^5 ÷ 3 = 81
3^3 = 3^4 ÷ 3 = 27
3^2 = 3^3 ÷ 3 = 9
3^1 = 3^2 ÷ 3 = 3

Then logically 3^0 = 3^1 ÷ 3 = 3 ÷ 3 = 1.

Your welcome Fgts

>>7624651
what you're meaning to say is
[math] n^0 = n^{x-x} = \frac{n^x}{n^x} = \frac{n}{n} = 1[/math]
but you can't do that for zero, as i stated

0^0 = 1 is the kind of thing a computer science retard would say

It's not even undefined you fucking retards. 0^0 is of an indeterminate form.
/thread

>>7624663
don't really know the difference, i'm mostly copying from wikipedia here

>>7624427

lol

xln(x) = 0

x =0
ln(x) = 0
x = 1

check mate

>>7624659
I mean, if you want to call people like Euler "like a computer scientist," you won't get any complaints from me.

:^)

>>7624658
Why not?

But 0! = 1 is okay?
Faggots
0^(-1) = 1/0
0^0 = (1/0) . 0
X . 0 = 0
0^0 = 0
So 0^3 = 3/0 = 2,5

>>7624688
Ln has an asymptote you dimwits you can't just ln everything you see when obviously the domain of ln doesn't reach the domain of your "x". Learn to math.

>>7624622
>>7624506
gotcha fams

There is one function from the empty set to the empty set, the one that takes no argument and returns no value. So 0^0 = 1.

How the fuck do you do that fancy math text?

>>7624407
>It's one of those /sci/ threads

>>7625059
>Hom(null, null) is nonempty

>>7625110
There is no reason why that image should be a png

>>7624709
Unlike you Euler was not a retard who did not understand basic fucking calculus (precalculus really) so no , he would not say that.

>>7624407
i always wonder about this.
>0^0=1
How something appeared from the nothing?

>>7624459
>0x0 = 0^1
>yes i majored in womens studies

>>7624650
This is not necessarily true for general x.

>>7624407
This is just one of those things that the public like to make a big deal of, when in fact what we do is just take 0^0 = 1 by convention (some fields take this to be 0).
There is no value in making a big philosophical fuss over it.
Similarly, there are such arguments for whether 0 is a natural number because "it's not a counting number" etc etc, when in actual fact we take 0 to be a natural number in some fields, and 0 to not be a natural number in other fields.

>>7625116
Would you prefer:

0^0 is the number of ways to place 0 objects in 0 places. Hence, 1 (do nothing).

>>7625173
I majored in underwater basket weaving actually.

>>7624659
Doubt a computational mathematician would say that.

>>7624902
Have you ever heard of proof by contradiction?

>>7624666
Indeterminate forms are forms of numbers that can have more than one value.
For example, 0/0, infinity/infinity, 0 * infinity, and 0^0.
To see that they can have any value, let k be any real number. Then 0/0 = (k*0)/0 = k*(sin(0)/0) = k.

>>7625087
It's called LaTeX for real but I don't have any idea how to use it

>>7625163
0 = 1-1
Omg something appeared from the nothing

>>7625312
Why have we involved sin? And why such a convoluted way?
Try this:
Assume that 0/0 has a value, say k.
Then, 0/0 = k iff 0 = 0k.
Observe that k can be any number (not necessarily real).
It's called indeterminate because we can't determine which value we should take it to be.

>>7625177
If you want to stop talking now that would be okay. Define a system of natural numbers to be a set N, an element 0∈N, and an injection s:N->N satisfying the following:
-for all n∈N, s(n)∈N
-there exists no element n∈N s.t. s(n)=0
Define the operation +:N->N, m+s(n) := s(m+n), m+0:=m.
> fuck the Dedekind/Peano axioms tho, 0 isn't a counting number.
Only idiots say zero isn't a natural number.

>>7625320
I included it because most calc 1 students have to show or at least observe lim(x->0){sin(x)/x}=1, so it would be more familiar to them. I have no idea how much they know.

>>7625334
Did you read the whole post before being so defensive?
I gave an example of a situation where needless arguments are had by the general public, when in actualy fact we take 0 to be or not be in the set of natural numbers depending on the convention that we choose to take.
>0 isn't a counting number
Note the " " marks.
i.e. something that other people that are not me like to say when arguing irrationally
Some courses in set and number theory do not include 0 in the natural numbers.

>>7625341
Thanks for the clarification.

>>7624641
this

0 can't be a successor in a peeano system tbh

>>7625126
>wanting extra artifacting

>>7625531
>9 times the file size to preserve the quality of a screenshot of a video still
>can't even tell which is jpg or png without switching directly between them and being told which is which

[math] \int {x^n} = x^{n+1} / {n+1} + c [/math]

>>7624407
epic meme faggot

IM INVINCIBLE

>>7625295
Why are you calling computational mathematicians retards?

>>7624649
0^x is not always 0, therefore false

>>7625180

I'm not anywhere near you guys on this in understanding but, wouldn't the fact that if there are 0 objects in 0 places, there are 0 functions as (doing nothing) isn't a function?

Would like to learn more thanks

There is exactly one function from the empty set to itself, namely the empty set. That's all you need. 0^0 = 1

>>7624407
10/10

>>7624427

x*ln(x) is undefined for x = 0.

x = 1 => 1*ln(1) = 0.

>>7624427
Pretty sure ln 0 goes to minus infinity m8

>>7627561
But it is.

Zero, zero times is still 1.

Zero goes into all real numbers once.

>>7624586
both

https://www.google.com/#q=y%3Dx%5Ex

>>7628158
>not having an adblocker installed

>>7628235
Are you retarded

>>7627998
>>7628014
>What is a limit

For all [math] x > 0 [/math] then [eqn] -x \leq \ln (x) \leq x [/eqn] thus [eqn] \ln(x) \to 0 [/eqn] by the squeeze theorem.

>>7628262
You know you have adblockers for your entire phone, right? I never see ads in apps anymore and have not so for years.

>>7628235
Cool story bro, but i dont care about the ads one bit

>>7624407

>>7628276
>squeeze theorem.
Shhh, don't use big words, you'll confuse them.

>>7625110
Hi Friend,
I noticed your PNG was unoptimized. I've taken the liberty of optimizing it for you.

Can't you just accept that
>x^1 = x
>x^2 = x * x
>x^3 = x * x * x
etc
and that
>x^0 = [the absence of x and any other number at all]
thus meaning
>x^0 = 0

>>7624407
except it isn't

that's like saying that 0/0 is 1

>>7628819
>being this much of a pedantic faggot

>>7628290
link plz?

>>7624427

false proof:

ln(0) doesn't exists or is minus infinity

when x=1 or x=-1, you have x^x==1

you have to prove x=0 => x^x=1
and not x^x=1 => x=0

ru retard?

0^0 is an undefined case.

This means that 0^0 cannot directly be evaluated.

HOWEVER, the context of the result can be used to find a limit in which a finite value likely exists.

The limit of x^0 as x approaches 0 is 1

But the limit of 0^x as x approaches 0 is 0

You can manipulate 0^0 into a 0/0 or a 0*inf case pretty easily, which are both undefined forms

>>7629243
>when x=1 or x=-1, you have x^x==1
>ru retard?

When x = - 1, x^x is - 1

>>7629250
yes, sorry

>>7625140
But he did say that anon. It sounds a bit like you're projecting.

>>7629243
>x=-1, you have x^x==1
Y R U so retard?

>>7627173
>no parentheses around n+1 in the denominator

TRIGGERED

>>7624649
>0/0
is it defined?
divide by a number x=multiply by x^-1
but 0^-1 doens't exist or is infinity

0^0 = number of possible functions from the empty set to itself = 1

>>7624407
So silly of you.

[math]\lim_{x \to 0^{+}} x^{x} = \lim_{x \to 0^{+}} \exp(\log(x^{x}))

= \lim_{x \to 0^{+}} \exp(x \log(x))

= \exp( \lim_{x \to 0^{+} } x \log(x) )

= \exp( \lim_{x \to 0^{+} } \frac{\log(x)}{ x^{-1} } )

= \exp( \lim_{x \to 0^{+} } \frac{ \frac{d}{dx} \log(x) }{ \frac{d}{dx} x^{-1} } )

= \exp( \lim_{x \to 0^{+} } \frac{x^{-1}}{- x^{-2}} )

= \exp( \lim_{x \to 0^{+} } -x )

= \exp( 0)

= 1

So, since \lim_{x \to 0^{+}} x^{x} = 1, that means that 0^{0} = 1.[/math]

>>7624407
0.999... =/=1

>>7624407
[math]0+0=0[/math]
Because sum of two empty sets is an empty set.

[math]0*0=0[/math]
Because product of two empty sets is an empty set.

[math]0^{0}=0[/math]
Because number of functions between two empty sets is one, an empty function.

[math]0+0=0[/math]
Because sum of two empty sets is an empty set.

[math]0*0=0[/math]
Because product of two empty sets is an empty set.

[math]0^{0}=1[/math]
Because number of functions between two empty sets is one, an empty function.

>>7628368
That's why you block them.

>>7629246
/thread tbqhwyf

>>7633672
Ayyyyy lmaoooo :^)