What if x=0?

If images are not shown try to refresh the page. If you like this website, please disable any AdBlock software!

You are currently reading a thread in /sci/ - Science & Math

You are currently reading a thread in /sci/ - Science & Math

Thread images: 7

What if x=0?

>>

Still equals 1, just like 0! = 1.

>>

not defined

>>

>>7795550

[eqn] x^x \approx 1+ x\log(x) + \frac{1}{2} x^2 \log^2(x) + \cdots [/eqn] Now let [math] b = \log(x) [/math] then [eqn] x^x \approx = 1+xb + \frac{1}{2} x^2 b^2 + \cdots [/eqn] Now setting x=0 we get [eqn] 0^0 = 1 + 0 + 0 + \cdots = 1 [/eqn]

>>

>>7795562

kek

>>

[math] (x, y) \mapsto x^y[/math] is discontinous at (0, 0), so there is no definition that statisfies all needs.

>>

>>

>>7795562

ayy lmao

>>

http://www.askamathematician.com/2010/12/q-what-does-00-zero-raised-to-the-zeroth-power-equal-why-do-mathematicians-and-high-school-teachers-disagree/

wew

>>

>>7795718

Highschool teachers are so dumb

>>

Its 1, this is a bait thread, everyone abort

>>

>>7795587

It's undefined so mathematicians just went and gave it their own definition.

>>

>>7795550

>multiply 1 by 0, 0 times.

still 1 m8.

>>

>>7795550

1, 'cause if you take lim of x^x when x -> 0- and x -> 0+ you get 1

>>

>>7795558

It is because assuming we have a function [math]f:x \longmapsto x^0 \,=\, 1[/math], we would naturally extend it by continuity, thus having [math]f(0) \,=\, 1[/math].

>>

>>7795550

The limit is 1

>>

>>7795961

and if we have a function mappin x to 0^x=0, we could extend it by continuity, thus having 0^0=0.

Analysis doesn't give a satisfying result to this question, but if you look at it from a set theory perspective, you can define n^m as the number of functions between an m-element set and an n-element set.

Since there exists exactly one function between the empty set and itself, we can set 0^0 = 1

>>

>>7795961

It is because assuming we have a function [math]f:x \longmapsto 0^x \,=\, 0[/math], we would naturally extend it by continuity, thus having [math]f(0) \,=\, 0[/math]

>>

>>7795550

0 is not a number

>>

>>7795550

its undefined, but for certain cases (some limits, for example) you can get 1 as answer

>>

[math]\lim _ {x \rightarrow 0} x^x = 1[/math]

[math]\lim _ {x \rightarrow 0} 0^x = 0[/math]

The limit doesn't exist. When we do sums and series, however, we generally define it as 0^0=1.

Here's why:

Series for e^x = [math]\sum_ {n=0} ^{\infty} \frac {x^n} {n!} [/math]

would look a lot worse as a piece-wise function, as well as binomial theorem, trig functions, or really any function defined as a Taylor polynomial.

It's the equivalent of asking why 1 isn't prime. The reason behind that is so whenever you're defining something that involves primes or factors you don't have to explicitly exclude 1.

>>

>>7795550

0^0 is undefined, let's say you have

f(x)^g(x)

= exp[g(x) * ln(f(x))]

= exp[ln(f(x) / g(x)^-1]

you apply L'Hopital's rule to ln(f(x)) / g(x)^-1 and plug the number back in the exponential

>>

>these many posts

>only a few mentioning l'hopital's rule

is /sci/ actually clinically retarded?

>>

I think everyone misread the question. OP asked to consider [math]x^0=1[/math], and what happens when x = 0. So the standard method, if using calculus, is to consider the value as x approaches 0.

Some people here utilised multivariable calculus. Here, we consider the function [math]f(x,y) = x^y[/math] and what happens as (x,y) approaches (0,0). There are multiple paths which give different limits. But OP was asking us to consider only one path, and that is the path where y = 0. So the fact that other paths like x = 0 give different limits doesn't matter in this case.

Same thing if you're single variable calculus; the correct function to use is [math]f(x) = x^0[/math] and not [math]f(x) = x^x[/math] (the latter is equivalent to considering the path y = x in the above multivariable function). In this case the result is the same since the limit for both as x approaches 0 is 1, but the method is wrong. It's like using [math]f(x) = 0^x[/math] instead, which would indeed give a different limit, as the path considered in the above example is x = 0.

The limit is 1. The answer is 1.

>>7796605

I guess amidst all the dickwaving, no one thought of using it.

>>

>>7796625

There's not 10 words in the OP

Is this bait?

>>

>>7796647

>The limit is 1. The answer is 1.

This doesn't follow though. Not unless you simply want to define 0^0 as the limit of x^0 as x goes to zero.

How do you know x^0 is continuous at zero? You can't, unless you choose 0^0 to be one beforehand.

It is equal to one, but there's no way to prove that. It's just defined that way because it works out a bit better than saying it's zero.

>>

>>7795562

either bait or the autism is strong with this one.

>>

>>7796652

Clearly it's not bait, there's not 10 words in that post.

>>

>>7795562

Prove your premise

>>

>>7795550

It's still 1. It's the number of maps from the empty set to the empty set. There's only the identity map. So the answer is 1.

Duh.

>>

>>7795562

tha resta you faggots know shit

>>

>>7795606

it is

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

>>

File: Screen Shot 2016-01-20 at 3.06.05 AM.png (108 KB, 1102x694)
Image search:
[iqdb]
[SauceNao]
[Google]

108 KB, 1102x694

Where is your god now /sci/ ?

>>

Just 1 bcz its mathematically like that u cant change this shit !!!

>>

Lol check out the first two digits of the minimum of the real valued solutions of x^x

>>

>>7795550

X^0 is same as 1 quantity of x's. There for x=1

>>

Empty products are always 1, just like empty sums being 0.

>>

0 isn't a number.

You can have 3 sheep, or 2 sheep, or perhaps 56 sheep.

1 isn't really a number either. You don't count your sheep if you have a sheep. There's just a sheep.

Likewise, you have no sheep. You don't have 0 sheep. You don't start counting sheep if no sheep are to be seen.

0 and 1 are more like concepts just like infinity.

>>

>>7796697

It could still be bait, where he's just playing dumb.

>>

>>7797207

> 1 is not a number

> you can't have 1 sheep

Wat

>>

>>7797341

He is right though, if we had only one thing we wouldn't need to invent the concept of numbers including 1.

>>

>>7797434

>implying you wouldn't need to be able to distinguish between 1 and 0

>>

>>7797446

not if was an everpresent thing

it's like if you had no point of refernce you couldn't tell whether somethign is moving in the same direction as you at the same velocity or growing at the same rate as you

>>

>>7797448

>if you had no point of reference

You're retarded.

>>

>>7797450

may be you should start looking up relativity to save some face

>>

>>7795562

Check your symbols: 0^0 ≈ 1 not 0^0 = 1

>>

>>7797455

>no reference frame

Nope

>>

>>7795562

>log(0)

>>

rollinnnng

>>

0^0=1 because Cardinal Numbers>Real Numbers.

>>

Thread images: 7

Thread DB ID: 445051

All trademarks and copyrights on this page are owned by their respective parties. Images uploaded are the responsibility of the Poster. Comments are owned by the Poster.

This is a 4chan archive - all of the shown content originated from that site. This means that 4Archive shows their content, archived. If you need information for a Poster - contact them.

If a post contains personal/copyrighted/illegal content, then use the post's