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If you want some help, there is a theorem which is quite useful for both Sylow theory and in general. I've scanned (briefly) through what I could get my hands on in Humphrey, but could not find this theorem.
Post #22 has the typo
But is fine otherwise.
From post #20
For 90, the Sylow 3-subgroup is unique.
You can not conclude this from direct application of the Sylow theorems. There could be 10 Sylow 3-subgroups. Even if you could, simply stating this is not sufficient.
Also, whenever you post numbers where prime factorization is important, you should also post the factorization. Instead of just writing 90, write
Post #19 can be simplified quite a bit. You know that the number of Sylow p-subgroups, call this n_p, must be congruent to 1 modulo p. Further, n_p must divide k, where our group is of order p*k with p > k. But if n_p is not 1, then it must be greater than p, hence greater than k, and therefore n_p can not divide k.
We conclude n_p must be 1.
Using | | for boxes, I got: |1| = |1| |0| = |10|
Under standard definitions, zero is an integer. There can be no debate about this because it is, after all, a definition.
this LaTex language is only for this forum right?
LaTeX is a typesetting language used for writing (math) documents. It is used universally to write papers for publication in almost any field of study, as well as entire books.
But it works much differently than Word. You type LaTeX in plain text, that is, no special formatting appears when you write. So if I wanted to do italic, my document would look like:
The quick \textit{brown} fox jumps over the lazy dog.
I then pass my document through an engine, and what pops out is that phrase with the word "brown" in italics. Compare this to Word: when you make something italic in Word, you see it in your original document as being italic. This is known as "rich text".
Because of licensing, Microsoft can't incorporate LaTeX into Word: they sell Word and LaTeX is free.
This way might be too informal, but here's how I see it:
I couldn't imagine of a more clear or rigorous way to do it.
Ok, that's what I figured. You probably won't see compact sets really used for a few months (with sequences of functions). But you can still get an idea about the usage.
Let X be a set, and let P be some property that holds locally. That is, for every point x there exists a delta such that
Well that's great, but we don't want to have our delta varying as our point x does. We want to be able to specify one delta for all x (this is like the concept of uniform continuity, if you're familiar). So we cover our set with these "delta balls". That is, for each x in X, let delta_x be the corresponding delta from above, and define
So for each point x, we have this delta ball centered at x where the property P holds throughout. As I said before we're trying to cover X, so the next logical thing to do is to take a union.
Now here is where compactness comes in. We have an open cover of X, it has to have a finite subcover. That is,
Where each U_i is really one of the delta balls U_x. Now for each i, we have the delta_i which is the radius of this ball. But there are only a finite number of i's. So we can take a minimum:
Now here is the big punch line. We went from:
To
Now once we pick a single delta, it works for the entire set X. This is much better than having to find a different delta for each x in X.
At least in the 2nd paper, they are talking about "compressed" data, an entirely different concept. I'd have to actually read the first to know about that one, but I have a feeling it is again a different use of the word compact.
ziggs, it would help us if you said what you got so far. Remember we have no idea what level you're at. I'm just going to list a couple of things, let me know if you want to know more about them.
For most geometry/topology/analysis applications, compactness lets you go from a local property to a global property.
It can also be seen as a finiteness condition.
While the definition is rather abstract, the Heine-Borel theorem tells you that compactness is very easy in Euclidean space: a set is compact and only if it is closed and bounded.
There are various ways to rephrase compactness in terms of sequences (every set of infinite terms has a limit point, every sequence has a convergent subsequence whose limit is contained in K)
Some related concepts are 2nd countable, paracompactness, locally compact spaces.
Pull out a square root of n from the radical and n. Approximate what's left.
But the company believes it can eventually be used in the consumer market place and home players.
For what?
there's maybe a 0.1% chance that you'd decide to write that down.
Why is it that people feel the need to quantify everything? It's rather unfortunate that when you don't know a quantity, humans tend to make things up instead of saying "a small amount". (Just a pet peeve, nothing more)
But you bring up a good point. The motives of the person writing it down are suspect to suspicion. And whenever you have psychology entering into a math problem, the typical answer is that the problem is not well-defined, at least from a mathematical point of view.
Of course, there are other reasons why this question is not well-defined. For example, as mathsyperson says each have an equal number of chances of occurring. However, a sequence with 20 1's and the rest 0's in it is much more likely to occur than a sequence with all zeros. The problem lies with statistics: statistics are mathematically generated, but they can not be mathematically interpreted.
Then you can get into the order of the coin flips and distributions. I think it would be hard for anyone to say a coin that flips 1000 heads and then 1000 tails is evenly weighted.
Because of all of this, my answer would be no, it is not possible to do better than random. That said, if I were a betting man, I'd have to go with the obvious choice.
coffeeking, as long as you are familiar with branch cuts, I'd say you don't need to show any more work than that.
But remember that log(z-1) has many analytic domains, depending on how you take your argument to be. It is true though that every single one of these domains does not include C \ {1}.
Post #20 is incorrect for groups of order 90. It is possible to have six Sylow 5-subgroups. I'll probably be checking the rest over once the semester is...
Sure, why not? We certainly make a good living, from what I remember +60,000 when coming out of grad school. Paper cuts and "white lung" (from chalk dust) are the only major health hazards. And perhaps most importantly, mathematicians are in great demand.
I taught that it should start with |x_n - x_m| and expand from there
This is what he needs to do, yes. But it is easiest to see how to do this by first figuring out the special case where n = m+2, and then generalizing. This is why I wrote, "Now generalize."
1. What is the difference between the nth and the mth term in that sequence?
2. Come up with an estimation for:
Using the triangle inequality. Now generalize.
(Hint: add 0 in a clever way)
By defining indented contour on a rectangle contour
Does this mean that your contour is defined so your singularities lie outside of it? If that's the case, then you have no residues.
A lot of times, you take f(z) to be different than f(x) (ignoring the abuse of notation). Make sure that isn't happening here. Otherwise, you seem to be correct.
That's just funny! Go Carl Sagan!
It is a rather clever and witty quote, and if that's what you meant by "funny" that's fine. But I hope you understand the concept that the quote is trying to convey. In other words, it isn't just funny; there is a point behind it, and a rather important one at that.
I am under the impression that there are methods to determine how accurate an approximation (such as a taylor series) is, and how many terms are needed fora given error bound. So if we have a calculator with n digits, take the error bound to be 10^(-n) and add as many as necessary. Do calculators seriously not do this? If not, that's depressing!
While we may approximate the error for Taylor series, you are ignoring the fact that what you are computing on can only use a finite number of bits (and typically 32 or 64). And before I go on, calculators use a variety of ways for computing functions, and I don't believe Taylor series is very often involved. The major disadvantage with Taylor series is that it depends upon the point you are calculating the function at, as error accumulates over distance.
The only error you're considering is truncation, which is not the only type of error that occurs. For example, take your function on your calculator to be e^x. Obviously for relatively small number sizes you get simply run out of bits and get overflow error. With the advanced definitions of floating numbers that are in use today this doesn't quite happen, but what does happen is that numerical results become less and less accurate the higher the number.
Seeing what exactly occurs is a bit complex, because we don't exactly do floating-point computations in a straightforward way. But it's just a question of information: you have 32 (or 64) bits, you obviously can't represent every number between -1.000.000.000.000 and 1.000.000.000.000 with up to 9 decimal digits. You would need to be able to represent 10^(12+9) numbers using only 64-bits which have 18.446.744.073.709.551.616 unique values. A little application of the pigeon-hole principle shows this is indeed impossible.
in your case, i believe there are 6 operations (square root, multiplication, addition, square root, division) and thus an accumulation of errors.
Actually, the only significant error that occurs in the above is when I add 1 to a really really big number. All the other operations (I believe, but would need to double check) are fairly safe... numerically.
Hasn't the four color theorem been proved with computers, and isn't the proof generally accepted? Certainly "proof by calculator" isn't unprecedented.
There is a major difference here. In the proof of the 4-color theorem, we are not talking about decimal representations which always have error. It is possible to make a program that doesn't contain any errors, it is not possible to exactly represent the decimal values 0.1 using a finite number of binary digits. Of course, as a CS professor of mine once said:
"Look engineers. All we're asking for is an infinite number of transistors on a finite-sized chip. You can't even do that?"
In fact, my calculator gives me 9 digits after the decimal place.. why would it bother to give me 9 if the first 2 were not reliable?
You seem to be under the impression that a calculator understands which decimals are correct and which are not.
Now i understand that a calculator only gives an approximation, but I at least trust the first n-1 digits if n are given. Is this unreasonable? what do YOU think?
Not only is it unreasonable, it is demonstratively false.
Compute:
Noting to follow my order of operations. This should give you 1 over root 2. My calculator gets the 5th digit wrong. Of course, much worse things can happen, and errors lead to more and more errors.
The calculations you are doing are not prone to these sorts of errors since log is a easy to compute and your numbers are not high (or low) enough to lead to them. But you should never trust a calculator to do a mathematicians job.