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Math

Name: Anonymous 2008-05-29 16:45

Whats y´ of y=x^2/x+1

Name: Anonymous 2008-06-17 19:05

>>40
h(x)=x^2/x

Name: Anonymous 2008-06-17 19:55

>>40
So you're saying x^2 / x != x?
Like >>37 said, the singularity is removable, so there is no discontinuity.

Name: Anonymous 2008-06-17 22:14

>>42 Yes, you can't divide by zero.
So what if the "singularity" is "removable" (Please define both), f=x^2/x is still not defined in zero, thus is not continuous in 0.
In conclusion. If g(x)=x, then f!=g. BUT f ∪ {(0,0)} = g

Name: Anonymous 2008-06-17 22:37

>>43
Why aren't you canceling the x before substituting? If you can cancel out the singularity, you are certainly allowed to do so. In fact, it would be foolish not to. I honestly don't see the rationale behind this lack of simplification.

So then, following your reasoning x^2 is discontinuous at 0 also?

Also, if you want to be pedantic with definitions, functions aren't sets. The range of a function is a set, as is the domain. A function is simply a rule that assigns elements of one set(domain) to elements of another set(range).

If you don't even know this, I think you're in a bit over your head.

Name: Anonymous 2008-06-17 22:49

>>44
¿do you know whats the meaning of cancelation? Is to multiply by the inverse of a number, and YES, 0 does not have multiplicative inverse. Your arguments are making that the proof of 0=1 is valid.

And, functions between sets ARE sets.

If you don't even know this, I think you're in a bit over your head.

Name: Anonymous 2008-06-18 1:51

>>45
a function is a set of what?  I thought a function was a mapping

Name: Anonymous 2008-06-18 4:48

x^2/x = x, except at x = 0, where it is discontinuous. You aren't allowed to just ``cancel out the singularity''.
This is middle school level stuff.

That's really all that needs to be said about it. The rest is trolling.

Name: Anonymous 2008-06-18 13:26

>>46, a function F between sets is a set of ordered pairs (a,b) with certain properties ( for example if (a,b) is in F and (a,c) is in F => b=c)

also, an ordered pair is a set (a,b) = {{a},{a,b}}

Name: Anonymous 2008-06-18 14:15

>>48
No. That's like, a high school definition of functions. Not all functions are bijective, and not all functions can be represented by ordered pairs.

Name: Anonymous 2008-06-18 16:32

x^2/x is not defined at zero and thus neither is (x^2/x)'. Did you not learn a single goddamn thing in high school? You can cancel x^2/x to x for purposes of making things easier, but if something is outside the domain of the original function, guess the fuck what, you can't use it in the canceled version. Christ. As >>47 says, this is fucking middle school level. Remember when they told you that when you wanted to graph a rational function, say, f(x)=(x+2)(x-1)/(x+2), you could just graph (x-1) since it cancels BUT the function is still not defined for x=-2?

Name: Anonymous 2008-06-18 21:36

>>49
All functions between sets ARE sets of ordered pairs. And, obviously, not all functions are bijective, but that doesn't concern this discussion.

Name: Anonymous 2008-06-18 22:19

>>43

Everyone from this post onwards seems to be talking a lot of stupid shit.

If you don't know what a removable singularity is, or analytic continuation, then you shouldn't really be fucking talking.

There is no singularity.

You could argue that the function is not defined at 0, but it's a removable singularity, so it's basically well defined there if you're not an idiot.

Name: Anonymous 2008-06-18 22:25

>>52
ITT someone read something about removable singularities and doesn't know when and where to apply them.

Name: Anonymous 2008-06-18 22:33

>>53


Um, no. That's just not right.

Just consider the function x^2/x as a function defined on the complex plane.

Obviously x^2/x is holomorphic on the plane without {0}, and thus analytic on the real line without {0}

Now, you could argue it two ways, you could say that x^2/x = x and therefore f(0) = 0 and thus the function is holomorphic on the entire plane. This sort of cancellation is pretty trivially allowed in any sort of construction of a theory of functions.

And even if you don't see the cancellation at trivial, the function is only not defined at a singleton point, and at that point there is a removable singularity. Therefore the function can be holmorphically extended to an analytic function on the complex plane and thus the real line. And it's pretty fucking trivial again that this analytic continuation is x^2/x = x.

ITT shut the fuck up.

Name: Anonymous 2008-06-18 22:45

>>54
Extended functions are not the same as the original function, I really dont understand why you extended the function to the complex plane.
In conclusion, you really don't know what are you talking about, or you are just trolling.

Name: Anonymous 2008-06-18 23:37

>>54
What can you do with the original function that you can't do with the extended function?

And R is a subset of C, so the treatment as a function on C is valid.

Name: Anonymous 2008-06-18 23:51

>>56

Sorry, meant to quote >>55 here

Name: Anonymous 2008-06-19 0:26

>>56
It is valid, but it won't help. The original and the extended function are different sets, and have different properties. For example: The domain and image sets are different, the extended function is a field isomorphism, while the original isn't, the extended function is not essential(a function is essential if it is not homotopic to a constant function, I'm not sure if that term is used in English language) while the original is essential.

Name: Anonymous 2008-06-19 6:30

>>58

I extended to the complex plane since I don't want to quote results about singularities that I've only ever seen defined in terms of complex functions with respect to a real function. It was more to make things clearer for someone trying to follow what I was saying, as obviously it applies to real functions.

Why would you ever not consider the extended function? It's pointless to consider a function which is undefined on points at which the limit of the function exists there. I cannot think of one reason not to.

Obviously they're not the same function, but for intents and purposes they are, and the extended function is much more useful.

Name: Anonymous 2008-06-19 8:15

>>58
What does being essential(sorry, not familiar with the term) give us?

Name: Anonymous 2008-06-19 9:17

>>60

Not >>58 here, but being homotopic to a constant function is also often called null-homotopic.

Basically a function f is homotopic to a function g

f,g:X -> Y (spaces X and Y)

if there exists a continuous function H: X x [0,1] -> Y

s.t H(x,0) = f(x)  H(x,1) = g(x)

I've not used this definition at all, but I'd have thought any continuous function on the real line would be null-homotopic :-/

given f cts. let H(x,t) = f(x) - t.(f(x)-1)

H(x,t) pretty trivially cts and H(x,0) = f(x)  H(x,1) = 1

But this contradicts >>58 since he said the extended function is not essential, whereas in fact the original function would not be essential, being undefined at a point. Whereas the extended function would be essential.

Odd.

Name: Anonymous 2008-06-19 12:02

I am extremly dumb in maths.

Should i kill myself?

I mean, i am so slow in understanding that it is useless.

Name: Anonymous 2008-06-19 13:28

OK which one of you faggots brought complex analysis into this first

Name: Anonymous 2008-06-19 15:04

>>61 maybe I didn't explained correctly, or maybe I didn't understood you correctly. What I meat to say is:
The original function is essential, that means that it is not homotopic to a constant function, while the extended function is not essential, that is, is homotopic to a constant function (the homotopy is very simple).

Name: Anonymous 2008-06-19 15:09

\latex

Name: Anonymous 2008-06-19 16:04

>>65
\LaTeX

Name: Anonymous 2008-06-19 16:52

>>66
Only \TeX works. It seems like math support makes post previews really important since it's easy to mistype.

\overbrace{ 1+2+\cdots+100 }^{5050}

Name: Anonymous 2008-06-19 16:58

\sqrt{hax my anus}

Name: Anonymous 2008-06-19 17:31

>>69
If you want spaces to show up you have to escape them.

\sqrt{hax\ my\ anus}

Name: Anonymous 2008-06-19 19:03

>>63

Me. Only bought it in as a way of defining singularities though.

Never really seen a rigorous treatment of singularities on real functions, although that's likely because it's done when constructing a theory of complex functions, so there's no point.

Name: Anonymous 2008-06-19 19:39

>>64
So why do we want the function to be essential? What can we get from essentiality(or lack thereof)? Are there some results or something that depend on this?

Name: Anonymous 2008-06-19 20:32

>>72
The essenciality discussion came just to show that both functions are different and they have different properties.
Not applicable to show the differentiability of a function, I believe. With essenciality you can prove interesting things, like the fundamental theorem of algebra, or that there exist one place in earth that has the same temperature as the opossite place in earth, nothing that a lot people in this thread are interested in.

Name: Anonymous 2008-06-19 21:20

>>73

The essential thing is a bit odd anyway. Why not just claim one was path connected and the other not, that's a simple enough reason why they're different. Still doesn't change the point.

Not to be a broken record, but complex analysis > essentiality to prove fundamental theorem of algebra. Liouville's theorem all the fucking way. Or Rouche's, whatever you like.

Hairy ball theorem is quite nice though.

Name: 4tran 2008-06-20 4:57

>>68
ZOMG when was \TEX support introduced to 4chan?
\frac{\sqrt{169}}{3.1415926}

Name: Anonymous 2008-06-20 12:22

>>75
It's [math], for all your \TeX needs.

Name: Anonymous 2008-06-20 16:41

\LaTeX

Name: Anonymous 2008-06-20 17:45

\text{hello world}

Name: Anonymous 2008-06-20 17:56

\int_0^3 \frac{\left(x^3\left(3-x\right)\right)^{1/4}}{5-x} \mathrm{d}x = \frac{\pi}{2\sqrt{2}}\left(17-40^{3/4}\right)

dus latex srsly work?

Name: Anonymous 2008-06-20 17:57

>>79

HOLY FUCKSHIT IT DOES, FORGIVE ME FOR BEING A NEWFAG BUT THIS IS AWESOME
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