You take a function of x and you call it y
pick any x-naught that you care to try
Make a little change and call it delta-x
the corresponding change in y is what we find next
Then we take the quotient and now, carefully,
send delta-x to zero and I think you'll see
that what this gives us if our work all checks
is what we call dy/dx, it's just dy/dx!
>>26
in fact, you CAN do that. Product rule + chain rule implies quotient rule.
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CSharp!FFI4Mmahuk2008-06-17 1:21
Christ, who let you people touch differentiation? You're all a bunch of retards. How hard is it to use the quotient rule? That's, like, the first fucking week of high school calculus. Jesus.
There is no discontinuity. The value of the function x^2 / x at 0 is 0. It was said in >>28. If one function equals another, how can one have a discontinuity and the other not at the same point?
If you were feeling rather pedantic you might want to try and argue there's a removable singularity there, but then it still doesn't matter because anybody who isn't a retard can see the analytic continuation.
That's the point, x^2/x is equal to x, except when x=0. If you take h(x)=x^2 and f(x)=x^2/x when x is not 0, and 0 elsewhere, then f and h are (Obviously) 2 diferent sets.
>>40
So you're saying x^2 / x != x?
Like >>37 said, the singularity is removable, so there is no discontinuity.
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Anonymous2008-06-17 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
>>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.
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Anonymous2008-06-17 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.
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Anonymous2008-06-18 1:51
>>45
a function is a set of what? I thought a function was a mapping
>>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.
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Anonymous2008-06-18 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?
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Anonymous2008-06-18 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.
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.
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Anonymous2008-06-18 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.
>>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.
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.
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Anonymous2008-06-19 8:15
>>58
What does being essential(sorry, not familiar with the term) give us?
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.
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Anonymous2008-06-19 12:02
I am extremly dumb in maths.
Should i kill myself?
I mean, i am so slow in understanding that it is useless.
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Anonymous2008-06-19 13:28
OK which one of you faggots brought complex analysis into this first
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Anonymous2008-06-19 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).
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.
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Anonymous2008-06-19 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?
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Anonymous2008-06-19 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.
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.
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4tran2008-06-20 4:57
>>68
ZOMG when was \TEX support introduced to 4chan?
\frac{\sqrt{169}}{3.1415926}