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/sci/ - Science & Math

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>> No.6700304 [View]

>>6700298
no they didn't, stable exotic elements are still hypothetical

>> No.6700296 [View]

>>6700294
there are a lot of really beautiful female scientists on CERN but no one is available.

>> No.6700271 [View]

>>6700266
yup they did, every one and then they replace all the parts of the laboratory.
>>6700269
Geneva definitely doesn't look sad, the city is more like a Mediterranean city in comparison of other swiss cities, but the people around there are very happy

>> No.6004281 [View]

>>6004212

Hello!

Haha, that was one of the most amusing scenes in that movie. It's just so...funny:

"Hmm, hmm, hmm...ah yes--I see you used Maclaurin there"

Thanks for making me laugh!

Best!

>> No.6004279 [View]

>>6003679

Hello!

As others have noted, this is a little muddled. You want to optimize functions from O^n (where O= Z[(1+sqrt(5))/2])) to--what? What kind of functions? Etc.

Moreover, what is the purpose of wanting to do this? That would greatly help us answer the question.

Thanks!

Best!

>> No.5996638 [View]

>>5996459

Hello!

I can see why one might think this, depending on their exposure to mathematics. Purely in terms non-tedious, thought driven, but also elementary mathematics, one needs look no further than the IMO competition.

To see some of the past exams, look here:

http://www.imo-official.org/problems.aspx


DISCLAIMER: I am not claiming that, in total generality, IMO problems don't have a bit of a formulaic nature. That said, they are definitely a good example of math problems which are simple to state, and hard to solve.


Best!

>> No.5991099 [View]

Hello!

I have no quantifiable evidence to back this up, but this would not be consistent with my experience. I know plenty of extremely mentally gifted people who have the attention span of a robin.

Best!

>> No.5990086 [View]

>>5990065

Hello!

Could you possibly give more context? Is it possible you are referring to phrases like "there is a NATURAL isomorphism" or "we can NATURALLY identify", etc?

Best!

>> No.5989420 [View]

>>5989416

Hello!

What if I phrased it to you this way. For every subset X of A, I am going to assign to you a person P_X which is made to perform a particular task. Namely, if you hand P_X an element of A he will examine it, and tell you whether or not that element of A actually belongs to X. Now, while P_X is not actually X (it's a person instead of a set!) for all things that sets care about (what are the sets members) P_X is just as good as X itself. Indeed, suppose that you wanted to tell if subsets X and Y were different. Well, in general you would show this by finding some element t of one of them, say X, and showing that t is NOT in Y. Well, this is the same thing as handing t to P_X who replies "t is in X" and handing t to P_Y who responds "t is not in Y". Thus, X is just as good as its assigned person P_X.

The indicator of a function, is just the set-theoretic way of encoding a sets associated person.

I hope that helps!

>> No.5985945 [View]

>>5985942

Hello!

I think you are somewhat confused by the aleph notation. Namely, the statement that #(R)=aleph_1 is the (unprovable in ZFC) continuum hypothesis. The alpehs are defined to be the successively smallest cardinals. So, aleph_1 is the smallest cardinal greater than aleph_0.

Does that clarify anything?

Best!

>> No.5985929 [View]

Hello!

What do you mean by "what would its cardinality be"? What more description than the cardinality of the power set of the reals were you looking for?

Best!

>> No.5983806 [View]

Hello!

The common notation for "apply f infinitely many times to x" would be lim f^{(n)}(x). Of course, if f is continuous and this converges, it must converge to a fixed point since

f(x)=f(lim f^{(n)}(x))=lim f(f^{(n)}(x))=lim f^{(n+1)}(x)=lim f^{(n)}(x)=x

For your second question, I am not sure what you mean by "solve". You shouldn't expect that the solution should have some closed form in terms of other standard mathematical notation. Your best bet is to approximate it. And, in fact, depending on the properties of your function f (e.g. if its a strong contraction) your best bet of approximating a fixed point might be by iterating the application of f to x (modulo some technical details).

Best!

>> No.5983787 [View]

>>5979166


Hello!

What exactly makes a specific Haar measure "abstruse"? The space on which it's defined is abstract? If so, then you might want to check out the Adele ring, common in algebraic number theory. That said, the Haar measure is just an invariant measure, so I don't know what makes it abstruse or not...


Best!

>> No.5983783 [View]

>>5982930

Hello!

It looks like you have found an answer to your question. For future reference, if you need to understand the details of something in alg top, and are stuck with a somewhat sloppy book (e.g. Hatcher) try the algebraic topology book by Tammo Tom Dieck--it's absolutely top-notch, and extremely homotopically minded.

Best!

>> No.5983775 [View]

>>5983679

Hello!

As another poster pointed out, the union you have written is the full ambient set (universe). If you change your union into logical statements, say A is "statement A" and B is "statement B" then you are asking what is the set of things for which

statement A is true OR statement A is not true OR statement B is true


Evidently everything satisfies the first two conditions, and so you get everything--regardless of what B is.

I hope that helps!

>> No.5982671 [View]

Hello!

I assume that x is an integer, and you mean the sum of all integers between 1 and x. Then, the classic formula is that this sum is precisely x(x+1)/2.

Best!

>> No.5982670 [View]

>>5982284

Hello!

I know this is most likely a troll, but regardless. I'd like it to be known that no serious person of mathematics thinks this way. Not only is this completely contravariant to the way a true academic perceives the world (some conceited hierarchy of 'subject difficulty'), but most deeply admire the workings of other subjects, including the inspiration they sometimes give to mathematics.

Best!

>> No.5982190 [View]

>>5981233

Hello!

Depending on your course, I think I may have the answer: discrete math *is* boring. When I think of discrete math I think of a hodgepodge of the extreme basics of set theory, graph theory, number theory, etc. There is no clear goal, no motivation, there are just the bare-bones facts about a variety of subjects. With this lack of cohesion it's difficult to latch onto the course.

So, I would not let this discourage you from taking more advanced math courses! They often times are working towards a well-defined goal, a concerted effort of theorems and lemmas to further your understanding of a specific thing.

Best!

>> No.5982099 [View]

>>5982051

Hello!

Do you really think that's the wisest course of action? Are you just saying words?


@OP It depends on what you want to do in math. If you have the desire to be a mathematician, I would say that the likelihood of success is somewhat low. It takes a long time to groom yourself, not only in terms of knowledge, but in terms of the "game" of academia (e.g. resume padding, etc.). Of course, if you are truly, truly passionate, it is not impossible. If it is something that really makes you happy pursue it, and if you decide later on that it isn't the right course of action, a math degree is extremely versatile.

Bes!

>> No.5981960 [View]

>>5980904

Hello!

Did you know that only recently did the mathematical world "field" mean what you think it means? It is historically common to call a division ring a "field" and a field a "commutative field".

Best!

>> No.5980233 [View]

>>5979292

Hello!

Not in the slightest. Don't ever think that. I know some people with absolutely stunning minds who had trouble with calculus 2.

I'm sure you're brilliant in your own way!

Best!

>> No.5980077 [View]

>>5980011

Hello!

Did you take any classes with Phong?

Best!

>> No.5980060 [View]

>>5980057

Hello!

You're welcome! Good luck with your linear algebra studies!

Best!

>> No.5980049 [View]

>>5980038

Hello!

It's not a lot easier, in fact, implicit in the other poster's statement is the fact that you already know your subset is a subspace. But, as I mentioned above, if you can actually see that your subset is the kernel of some linear map (any linear map) it is automatically a subspace, and this can be quicker (if you can see what the subset is the kernel of).

The other poster's comment was just justifying why all subspaces are kernels. To every space V and every subspace W there is an associated space U=V/W and a map t:V--->U such that ker t=W.

Best!

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