http://www.orgburo.com/pofoquiz/pofo.php

I took this a while ago, but with the few political quizzes running around I thought I'd choose a really big one.  The spectrum they have is also much bigger.

80 questions.  Get a snack.  I'll post mine sooner or later.

Overall, the PoliticsForum quiz considers you a socially-orientated, big-government, non-absolutist, controlled-market kind of person, who also seems quite Marxist.

These characteristics would put you in the overall category of Marxist. Your natural home at PoliticsForum would be the Communism area.

Individual vs Social

"The individualist believes that society works best through a focus on individual rights, freedoms, actions and responsibilities. The social thinker believes that the ideal state should focus more upon collective action and take a social approach to rights and responsibilities."

You scored 67 out of 100 on a scale of Individual vs Social. This means that politically you are more likely to value the need for group actions and group benefit over individual enterprise and benefit.

    * 76% of test takers were more individual than you.
    * 22% of test takers were more social than you.

Theist vs Materialist

"The materialist believes that all objective criteria to influence politics can be reasonably derived without recourse to the divine or the spiritual. The theist believes that spiritual beliefs are important and should influence government policy."

You scored 44 out of 100 on a scale of Theist vs Materialist. This means that politically you are neither more nor less likely to believe that religion and spirituality are superstitions that should not inform political debate.

    * 24% of test takers were more theist than you.
    * 75% of test takers were more materialist than you.

Big Government vs Small Government

"The big government advocate believes that governments should be responsible for regulating a wide array of social practices, even what might be considered personal decisions such as abortion, euthanasia, children's education and births. A small government advocate thinks that, wherever possible, these issues should be up to individuals or companies to direct."

You scored 35 out of 100 on a scale of Big Government vs Small Government. This means that politically you are less likely to believe that government should keep out of legislating social policies, leaving such decisions to individuals.

    * 8% of test takers were more big government than you.
    * 91% of test takers were more small government than you.

Nationalist vs Internationalist

"The nationalist believes in the sovereign rule of nation states, particularly his or her own. The internationalist believes that there should be more important international fora and perhaps, ultimately, international government."

You scored 58 out of 100 on a scale of Nationalist vs Internationalist. This means that politically you are neither more nor less likely to favour international bodies over national ones.

    * 44% of test takers were more nationalist than you.
    * 54% of test takers were more internationalist than you.

Protectionist vs Free Trader

"The protectionist believes in barriers against free trade most probably due to a belief that this is in his or her country's interests. The free trader rejects such notions, believing that the system ultimately suffers when tariffs, subsidies and other obstacles to free trade persist."

You scored 52 out of 100 on a scale of Protectionist vs Free Trader. This means that politically you are neither more nor less likely to favour free trade over protectionist policies.

    * 46% of test takers were more protectionist than you.
    * 53% of test takers were more pro free trade than you.

Absolutist vs Non-Absolutist

"The absolutist believes that either a divine presence or scientific laws provide absolute truths about the world, which can and should be applied in practise. The non-absolutist may be either a relativist, or simply someone who is more pragmatic."

You scored 67 out of 100 on a scale of Absolutist vs Non Absolutist. This means that politically you are less likely to believe that there is an absolute truth that may guide your ideological beliefs.

    * 83% of test takers were more absolutist than you.
    * 15% of test takers were more non-absolutist than you.

Controlled Market vs Liberal Market

"Both of these categories assume a capitalist system. Assuming this system, the controlled market believer holds that government should intervene in regulating a nation's economy: wage laws, environmental standards, privatised industries and workplace relations policy. A liberal market thinker believes that such regulation is unnecessary and often counter-productive."

You scored 25 out of 100 on a scale of Controlled Market vs Liberal Market. This means that politically you are more likely to believe that there is need for government regulation of industry.

    * 26% of test takers were more controlled market thinkers than you.
    * 72% of test takers were more liberal market thinkers than you.

Marxist vs Non-Marxist

"This scale purports to show to what extent you follow the thought and teachings of Marx. Marxists tend to be scientific, materialist and revolutionary, believe in class struggle and the laws of historical and dialectic materialism."

You scored 38 out of 100 on a scale of Marxist vs Non-Marxist. This means that politically you are more likely to follow the philosophies of Marx.

    * 26% of test takers were more Marxist than you.
    * 73% of test takers were more non-Marxist than you.

I would say Individual vs. Social was accurate as I'm more for benefitting society.
I would say Theist vs. Materialist is about accurate although I think I'd be more middlish.
Big Govt. vs Small Govt. was accurate, I believe in Big Govt. when necessary.
Nationalist vs. Internationals I thought was inaccurate, I'm more of a Nationalist yet this test rated me a slight Internationalist.
Protectionist vs Free Trader is pretty accurate, I'm down the middle and very picky on that.
Absolutist vs Non-Absolutist is pretty accurate, I believe you need to be flexible.
Controlled Market vs Liberal Market is accurate, I believe in control when necessary.
Marxist vs Non-Marxist I would say is about accurate although I wonder why the test would put me overall Marxist as according to it, I'm more extreme on Controlled Market vs. Liberal Market,  Absolutist vs Non-Absolutist, Big Government vs Small Government, and Individual vs Social. My Theist vs Materialist beliefs according to the test even contradict Marxism to an extent.

"I wish the UN had intervened in Iraq, rather than the coalition of the willing."

What if I would have preferred no militar intervention.. but if I had to choose, I would take the UN? :|

"We should welcome more state intervention, because the state sets standards, co-ordinates society and instils discipline and strong values."
Well, what does it mean when one clicks "Agree", when the statement is like "A because B"

What if I think B is correct, but A is not correct but state intervention is already enough in my country? :huh:

You scored 52 out of 100 on a scale of Individual vs Social. This means that politically you are neither more nor less likely to value the need for group actions and group benefit over individual enterprise and benefit.

You scored 73 out of 100 on a scale of Theist vs Materialist. This means that politically you are more likely to believe that religion and spirituality are superstitions that should not inform political debate.

You scored 53 out of 100 on a scale of Big Government vs Small Government. This means that politically you are neither more nor less likely to believe that government should keep out of legislating social policies, leaving such decisions to individuals.

You scored 50 out of 100 on a scale of Nationalist vs Internationalist. This means that politically you are neither more nor less likely to favour international bodies over national ones.

You scored 37 out of 100 on a scale of Protectionist vs Free Trader. This means that politically you are less likely to favour free trade over protectionist policies.

You scored 57 out of 100 on a scale of Absolutist vs Non Absolutist. This means that politically you are neither more nor less likely to believe that there is an absolute truth that may guide your ideological beliefs.

You scored 31 out of 100 on a scale of Controlled Market vs Liberal Market. This means that politically you are more likely to believe that there is need for government regulation of industry.

You scored 47 out of 100 on a scale of Marxist vs Non-Marxist. This means that politically you are neither more nor less likely to follow the philosophies of Marx.

Overall, the PoliticsForum quiz considers you a theist, small-government, liberal-market kind of person, who doesn't sound like a Marxist.

These characteristics would put you in the overall category of borderline social liberal. Your natural home at PoliticsForum would be the Liberalism area.

Below, on this page, is an analysis of the (up to) eight categories that you were sorted into above, and an explanation of how you scored on the eight scales used to determine these categorisations.

The pages following this one show the breakdown of results for all participants on each question. You can freely navigate through the pages by using the hyperlinks in the top right of each page.

Individual vs Social

"The individualist believes that society works best through a focus on individual rights, freedoms, actions and responsibilities. The social thinker believes that the ideal state should focus more upon collective action and take a social approach to rights and responsibilities."

Individual

Social

You scored 54 out of 100 on a scale of Individual vs Social. This means that politically you are neither more nor less likely to value the need for group actions and group benefit over individual enterprise and benefit.

    * 45% of test takers were more individual than you.
    * 53% of test takers were more social than you.

Theist vs Materialist

"The materialist believes that all objective criteria to influence politics can be reasonably derived without recourse to the divine or the spiritual. The theist believes that spiritual beliefs are important and should influence government policy."

Theist

Materialist

You scored 6 out of 100 on a scale of Theist vs Materialist. This means that politically you are less likely to believe that religion and spirituality are superstitions that should not inform political debate.

    * 1% of test takers were more theist than you.
    * 99% of test takers were more materialist than you.

Big Government vs Small Government

"The big government advocate believes that governments should be responsible for regulating a wide array of social practices, even what might be considered personal decisions such as abortion, euthanasia, children's education and births. A small government advocate thinks that, wherever possible, these issues should be up to individuals or companies to direct."

Big Gov

Small Gov

You scored 77 out of 100 on a scale of Big Government vs Small Government. This means that politically you are more likely to believe that government should keep out of legislating social policies, leaving such decisions to individuals.

    * 85% of test takers were more big government than you.
    * 13% of test takers were more small government than you.

Nationalist vs Internationalist

"The nationalist believes in the sovereign rule of nation states, particularly his or her own. The internationalist believes that there should be more important international fora and perhaps, ultimately, international government."

National

International

You scored 54 out of 100 on a scale of Nationalist vs Internationalist. This means that politically you are neither more nor less likely to favour international bodies over national ones.

    * 37% of test takers were more nationalist than you.
    * 61% of test takers were more internationalist than you.

Protectionist vs Free Trader

"The protectionist believes in barriers against free trade most probably due to a belief that this is in his or her country's interests. The free trader rejects such notions, believing that the system ultimately suffers when tariffs, subsidies and other obstacles to free trade persist."

Protection

Free-Trade

You scored 49 out of 100 on a scale of Protectionist vs Free Trader. This means that politically you are neither more nor less likely to favour free trade over protectionist policies.

    * 41% of test takers were more protectionist than you.
    * 58% of test takers were more pro free trade than you.

Absolutist vs Non-Absolutist

"The absolutist believes that either a divine presence or scientific laws provide absolute truths about the world, which can and should be applied in practise. The non-absolutist may be either a relativist, or simply someone who is more pragmatic."

Absolute

Non-Absolute

You scored 46 out of 100 on a scale of Absolutist vs Non Absolutist. This means that politically you are neither more nor less likely to believe that there is an absolute truth that may guide your ideological beliefs.

    * 16% of test takers were more absolutist than you.
    * 82% of test takers were more non-absolutist than you.

Controlled Market vs Liberal Market

"Both of these categories assume a capitalist system. Assuming this system, the controlled market believer holds that government should intervene in regulating a nation's economy: wage laws, environmental standards, privatised industries and workplace relations policy. A liberal market thinker believes that such regulation is unnecessary and often counter-productive."

Controlled

Liberal

You scored 74 out of 100 on a scale of Controlled Market vs Liberal Market. This means that politically you are less likely to believe that there is need for government regulation of industry.

    * 89% of test takers were more controlled market thinkers than you.
    * 10% of test takers were more liberal market thinkers than you.

Marxist vs Non-Marxist

"This scale purports to show to what extent you follow the thought and teachings of Marx. Marxists tend to be scientific, materialist and revolutionary, believe in class struggle and the laws of historical and dialectic materialism."

Marxist

Non-Marxist

You scored 63 out of 100 on a scale of Marxist vs Non-Marxist. This means that politically you are less likely to follow the philosophies of Marx.

    * 84% of test takers were more Marxist than you.
    * 14% of test takers were more non-Marxist than you.


trigonometry
function
algebra proposed
algebra

Let $B,C$ two angles of a triangle
Find  $k\in (0,\frac{\pi}{2})$ in function of B,C such that

$\frac{cos\ B}{cos\ C}=\frac{sin\left(B-k\right)}{sin\left(C+k\right)}$

$tg k =\frac{\sin (B+C)}{2\cos B\cos C}$.


Let $ a_i>0, i\equal{}{1,2,\dots n}$ distinct numbers such that $ a_1\cdot a_2\cdots a_n\equal{}1$ and the function $ f: [0,\infty) \rightarrow \mathbb R$ with $ f(x)\equal{}a_1^x\plus{}\cdots\plus{}a_n^x$. Is $ f$ increasing or decreasing?


Sorry, I'm new at this stuff. I know AM-GM, but what is GM-HM?

It is just that the geometric mean is greater than or equal to the harmonic mean

Namely

$\sqrt[n]{a_{1}a_{2}\cdots a_{n}}\ge \frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}$


calculus
integration
real analysis
function
derivative
topology
probability

Hi,

I don't see to get a grip on some unnatural concepts in abstract integration theory. In particular:

1)non-measurable sets,
2)regularity of measures, and why theorems fail when $ \mu$ is not regular.
3) the desire to get rid off non-complete measures
4) $ \mu*$-measurability
5) what if we drop $ \sigma$-finiteness. Why is the theorem doomed to fail.

I understand all the theorems concerning them and I know the classical counterexamples (Cantor examples). I also can consequently apply them in exercices. But my books (Rudin/Halmos) don't t give me a lot of intuition. I was wondering if there is (preferrably online, because I don't have too much money for new books), a list of classical examples when the above hold or fail to hold or how one can check them. (not really interested in theorems, i have them all here)

thanx in advance!

[quote="Jacobson"]
I don't see to get a grip on some unnatural concepts in abstract integration theory.
[/quote]
It is true that measure theory doesn't seem the most natural thing in the world :) In 1930's some famous Russian professor in Leningrad exclaimed "we'll not allow you to torture our students with Lebesgue integral!" when he heard of the proposal to introduce the measure theory into the regular university curriculum ;). Still, since then, many students have been through it and it was OK, though I agree with you that it takes a lot of time to grasp it in full. I do no know of any good online source but I'll try to give some brief answers to your questions. Perhaps, you know that much already but then you may ask for some particular examples.

1)non-measurable sets

For "simple measures" like Lebesgue measure on the line, this is the price you pay for accepting the Axiom of choice. If you reject it, there are none. Still, AC seems so attractive and reasonable that most mathematicians today are willing to pay this price. After all, the idea of the measure theory is not to ensure measurability of [b]everything[/b] but only of [b]everything useful[/b]. The notion of a $ \sigma$-algebra covers all the operations that you can do with your set having measure under control. If you request for the countable unions being measurable, this allows you to pass to limits, etc., and still you can say that the measure of the union does not exceed the sum of the measures and that is a useful information if the corresponding series converges, which is often the case. On the other hand, if you request that the unions of continuum many sets be measurable, it may please you aesthetically but is completely useless because no information about the measure of the resulting set can be obtained. So, countable unions/intersections is where you want to stop. Now, the Borel $ \sigma$-algebra becomes a very natural object: you want all continuous function measurable (because the class of continuous functions is, in a sense, one of the most natural and useful classes in analysis), so you want all open/closed sets measurable and the Borel $ \sigma$-algebra is the minimal one you can come up with that contains all open sets. Now, there is [b]absolutely no reason[/b] why the Borel sigma-algebra has to contain all sets and it does not. One can construct a non-Borel set without using axiom of choice explicitly or one can just show that the cardinality of all Borel sets on the line is merely continuum. So, if by "measurable" we mean "Borel", it is quite natural to assume that there exist non-measurable sets and not that hard to prove it.

2)regularity of measures, and why theorems fail when $ \mu$ is not regular.

As I said above, the most natural class of functions on metric spaces is continuous functions. What you want is that the measurable functions be not very far away from that class. That means that measurable functions should be well-approximable by continuous ones in some sense. Looking at characteristic functions of measurable sets, you can easily realize that their approximability is equivalent to regularity of measure. Surprisingly, the regularity often comes automatically: every finite Borel measure on a metric space is regular. By the way, abstract measure theory theorems like, say, Fubini theorem, do not require any regularity of measure, just ($ \sigma$)-finiteness. The regularity comes in when you want something like Lebesgue differentiation theorem, etc., when the main tool is the approximation of measurable functions by continuous ones.

3) the desire to get rid of non-complete measures

There is only one truly convincing reason for that and it is usually not presented in the standard university courses so I agree to you that this completion of measures procedure gets sort of suspended in the air and seems like something just invented to make students' lives more complicated. The reason is the following: the projection (or, more generally, a continuous image) of a Borel set is not necessarily Borel but it is still good enough to be measurable with respect to any finite [b]complete[/b] measure that lives at least on the Borel $ \sigma$-algebra. If you want to know more about this, look up Suslin sets and A-operation on the web.

4) $ \mu*$-measurability

I believe you mean here the sets $ E$ such that $ \mu^{*}(A\cap E)+\mu^{*}(A\setminus E)=\mu^{*}(A)$ for every set $ A$ where $ \mu^{*}$ is an outer measure. Let us start with the notion of outer measure. I claim that it is the most natural thing in the world to measure the size of a set. In almost all situations, the outer measure arises in the following way. You have some standard sets (like tiles you buy in Home Depot) and each set/tile has some price associated with it. Your task is to cover your set (your "room floor") using at most countably many such tiles. The outer measure just tells you the minimal price you'll have to pay for such cover. Now, as a rule, the outer measure is not additive at all. The question arises "on which sets do we have the additivity property?" and the answer is "at least on $ \mu^{*}$-measurable sets". By the way, I hate the name $ \mu^{*}$-measurable. I mention it in my courses but I always prefer the less official term "well-separating sets". This is a beatiful theorem due to Caratheodori (hope, I didn't misspell his name too badly) and that is the only place where the $ \mu^{*}$-measurability appears. It turns out that $ \mu^{*}$ measurable sets always form a $ \sigma$-algebra and $ \mu^{*}$ becomes countably additive and complete when restricted to that $ \sigma$-algebra.

5) what if we drop $ \sigma$-finiteness. Why is the theorem doomed to fail.

You start playing with infinities or more than countable sums of finite numbers then and in both cases you have problems. In general, you want your measure to be finite to do the analysis. The most important  theorem that fails without $ \sigma$-finiteness is the Fubini theorem. The Lebesgue differentiation theorem requires "local finiteness" (plus some covering lemma), etc.

[quote]
I understand all the theorems concerning them and I know the classical counterexamples (Cantor examples). I also can consequently apply them in exercices.
[/quote]
At this point it is good to ask "stupid questions" about the proofs like "Where did we use this assumption", "What if we drop it?", etc. but I suggest that at this stage of understanding you ask specific questions rather than general ones.

[quote]
But my books (Rudin/Halmos) don't t give me a lot of intuition.
[/quote]
Did you go through all the proofs in an inquisitive manner? If you did, I'm really surprised that you didn't develop your "intuition" yet (though what I mean by intuition may differ from what you mean by it). Both are really good books and I would have a hard time recommending something much better...

[quote]
I was wondering if there is (preferrably online, because I don't have too much money for new books), a list of classical examples when the above hold or fail to hold or how one can check them.
[/quote]
Unfortunately, I don't know of any such list and doubt that it exists but you are welcome to post more specific questions any time :)

[quote]
(not really interested in theorems, i have them all here)
[/quote]
If you have a theorem, you should always go through the proof, remove each particular assumption (one by one) and try to construct a counterexample to the conclusion yourself. Each theorem is a potential source of many examples, all you need is to devote some time and effort to them. Unfortunately, modern educational trends are to reduce the work with the theorems to memorizing the statements and "applying them in exercises", which is at most 1/3 of what one should really be doing if one wants to get a good grasp of the subject and 1/100 of what one should be doing if one wants to become a professional mathematician :(.

Here's another thought about #2, regularity:

It's a sort of meta-mathematical principle that when we say that an object has two mathematical structures, we almost always have some way in which these two structures should cooperate with each other. Some examples:

A ring is not merely a set with two algebraic operations; we must have cooperation between those operations, namely that multiplication distributes over addition.

An ordered field is not merely a field which is also a totally ordered set. (If it were so, we could always find some total order for $ \mathbb{C}.$) No, we must have some properties showing the order cooperating with the algebraic structure. (That is, the sum of two positives is positive and the product of two positives is positive.)

A topological group is not merely a group which has a topology; it wouldn't make sense unless the basic group operations made for continuous functions.

So what is the issue here? We have such familiar examples as $ \mathbb{R}$ or $ \mathbb{R}^{n},$ which are already known to us as topological spaces. We'd like to make them measure spaces. So what cooperation should we demand between a topological structure and a measure structure? This is where the notion of regularity comes from.

Note that is is possible to discuss measure spaces with no topology in sight. The most common setting for this would be probability theory. After all, what is a probability sample space? It's a measure space of total mass $ 1.$ A random variable is a (preferably measurable) function on that space, and an expected value is a Lebesgue integral. Only I've described, for a probability class, the idea that a probability sample space - the domain of some random variable - lies "behind the curtain." We talk about such concepts as the distribution of the random variable, but we don't really think about the points in the domain. And part of our reason for not looking (for keeping it behind the curtain) is that it has no natural topological structure.

Thanks fedja and Kent for clearing up those thing for me.

[quote="fedja"]If you have a theorem, you should always go through the proof, remove each particular assumption (one by one) and try to construct a counterexample to the conclusion yourself. Each theorem is a potential source of many examples, all you need is to devote some time and effort to them. Unfortunately, modern educational trends are to reduce the work with the theorems to memorizing the statements and "applying them in exercises", which is at most 1/3 of what one should really be doing if one wants to get a good grasp of the subject and 1/100 of what one should be doing if one wants to become a professional mathematician[/quote]
Don't worry. Since my school is a good example of your concerns, I rely on self-study to learn good mathematics (this is not even obligated :)). Rudin and Halmos are therefore the best; and the complete opposite of most modern education systems. And honestly... I would be crazy if I went through them only to solve some routine exercices.

[quote="fedja"]
Did you go through all the proofs in an inquisitive manner? If you did, I'm really surprised that you didn't develop your "intuition" yet (though what I mean by intuition may differ from what you mean by it). Both are really good books and I would have a hard time recommending something much better...
[/quote]
My intuition is not that bad as I let appaer. Since Halmos is rather old-school and builds everything from scratch I just have -from time to time- some trouble getting a grip on the abstract concepts (i.e. extensions/completions/...) And since Rudin is certainly not a stand-alone introductionary book, I don't have to expect much sympathy of it :) It contains some very heavy theorems from the start (Riesz/Urysohn for example) which are not obvious at all after a few reads...  And most of my questions arose when asking myself questions when reading proofs multiple times ;)

[quote="fedja"]
Unfortunately, I don't know of any such list and doubt that it exists but you are welcome to post more specific questions any time
[/quote]
I have some specific questions already (it is possible I'll add some more questions in the near future :)) Since Halmos is an old book, I refresh some notations for you. I suppose that will make life a bit easier.

[u][b]Some (slightly modified) Halmos notations and terminology:[/b][/u]
1) We call a class of sets $ E$ hereditary if $ F \in \bf E$ and $ G \subset F$ implies that also $ G \in \bf E$
2) If $ \bf R$ is a ring, then we denote $ H(\textbf{R})$ for the hereditary $ \sigma$-ring generated by $ \bf R$.
3) If $ \mu$ is a measure on a ring, then $ \mu$ extends to its generated hereditary $ \sigma$-ring $ H(\textbf{R})$ via the outer measure $ \mu^{*}$.
4) I will denote $ \bf S^{*}$ for the $ \sigma$-ring of all $ \mu^{*}$-measurable sets. When defining the measure $ \mu^{\circ}$ on $ S^{*}$ by $ \mu^{\circ}(E)=\mu^{*}({E})$, you have a complete measure.
5) $ S(\textbf{R})$ is the $ \sigma$-ring generated by $ \bf R$. It can be shown that $ S(\textbf{R}) \subset S^{*}$
6) An outer measure $ \mu^{*}$ regular if the outer measure induced by $ \mu^{\circ}$ (which is $ \mu^{\circ}^{*}$) coincides with $ \mu^{*}$. It appaers this is always the case for the outer measure $ \mu^{*}$ on a ring.

Now some questions  :blush:

[quote="[Halmos / p50 / Theorem C]"] If a set $ E \in H(\textbf{R})$ is of $ \sigma$-finite outer measure then there exists a set $ F$ in $ S(\textbf{R})$ such that $ \mu^{*}(E)=\mu^{\circ}(F)$ and that $ F$ is a measurable cover of $ E$.

[/quote]
I went through the proof. I noticed that the finiteness conditions was used somewhere in order to cancel both sides. In the exercices is outlined a *highly* non-trivial counterexample for when you drop $ \sigma$-finiteness. That doesn't really help me understand the true meaning of the condition...I would be better off with some intuitive answer, I guess.

[quote="Rudin / p41/ Theorem 2.14"]
In the construction of $ \mu$ in the representation theorem of Riesz (cf. p41) Rudin claims that:
i) For every $ E \in \mathfrak{M}$ that $ \mu(E)=\inf \{ \mu(V): E \subset V, V \ open \}$
ii) Also $ \mu(E)=\sup \{ \mu(K): K \subset E, K \ compact \}$ which holds for every open set $ E$, and for every $ E \in \mathfrak{M}$  with $ \mu(E)<\infty$
[/quote]
Later, he says that $ \mu$ is regular if the above equal each other. I don't see any link at all between the 'meaning of regular' in Halmos [check terminology 6 above]. What went wrong in Riesz' construction, or why does it fail? Didn't we form the outer measure on a ring too, and regularity implies $ \mu^{*}=\mu_{*}$?

 [quote="Rudin / p47,48/ Theorem 2.17"]
Suppose $ X$ is a locally compact, $ \sigma$-compact Hausdorff space. If $ \mathfrak{M}$ and $ \mu$ are as in theorem  2.14 (Riesz for positive linear functionals), then $ \mathfrak{M}$ and $ \mu$ have the following properties:
(a) If $ E \in \mathfrak{M}$ and $ \epsilon > 0$, then there is a closed set $ F$ and an open $ V$ such that $ F \subset E \subset V$, and $ \mu(V-F) < \epsilon$
(b) $ \mu$ is a regular Borel measure on $ X$.
[/quote]
I understand part $ (a)$. I have trouble seeing why it implies $ (b)$.

[quote="Rudin / p116 / Intro"]
A complex measure $ \mu$ on $ \mathfrak{M}$ is a complex function on $ \mathfrak{M}$ such that ${ \mu(E)=\sum_{i=1}^{\infty}\mu(E_{i}})$ for every partition of $ E$.
[/quote]
[quote="Halmos / p118 / Intro"]
We define a signed measure as an extende real valued , countably additive set function $ \mu$ on the class of all measurable sets of a measurable space $ (X,S)$. and such that $ \mu$ assumes at most one of the values $ +\infty$, $ -\infty$. Then exercice (2) says "A complex measure on the class of all measurable space is a set function $ \mu$ such $ \mu(E)=\mu_{1}(E)+i \mu_{2}(E)$ and where $ \mu_{i}$ are signed measures for $ i=1,2$."
[/quote]
I can assume they both have other terminology in mind?
[quote]
The construction of a finite measure $ |\mu|$ out of the complex measure $ \mu$ is a fact.
[/quote]
Later, I read things like "$ |\mu|(R^{k}) < \infty$", which I find rather strange. Are there any explicit examples why we should give $ \bf R^{k}$ (or something else very big for that matter) finite measure?

It's possible that the above contains some typos or non-sense. If something is unclear, let it know. Thanks in advance :)

A few quick answers (don't have time for more right now :()

The terminology in Rudin and Halmos is a bit different so the word "regular", indeed, seems to mean two different things completely unrelated to each other in these two books. I believe that nowdays most people use "regular" in the Rudin sense (i.e., in the sense that relates topology and measure).

As to complex measures, there is a difficulty with including infinities there. Indeed, you cannot consistently add infinity to the possible set of values of a complex measure if you want both additivity and possibility to multiply measures by complex numbers (the problem manifests itself even in the case of real valued signed measures in the way that you cannot allow $ -\infty$ and $ +\infty$ simultaneously). On the other hand, you are right when saying that the class of finite complex measures is too restrictive. It would be nice to have the Riesz representation theorem for any continuous linear functional $ \Phi$ on $ C_{0}(\mathbb R)$ (the class of continuous functions with compact support) and you, indeed, can say that once you fix the support, it is an integral with respect to some finite complex measure and when you extend the support, the new measure is just an extension of the old one, so, it seems like we have some object on the entire $ \mathbb R$ that represents our functional. Indeed, we have it, but it is not a complex measure in the usual sense of the world because it is defined not on a $ \sigma$-ring, etc. One way out is to say that the functional is represented by a positive locally finite measure $ \mu$ and a complex-valued locally integrable (or, if you prefer, unimodal) function $ f$ such that $ \Phi(u)=\int uf\,d\mu$.

More later :)

i wonder why the lebesgue integral is introduced via measure theoretical means, i mean i havent taken a formal course in measure theory but it seems alot of the fuss with it can be dropped if you adopt the riesz method (its how i first-and really only know so far learned the lebesgue integral) which seems highly intuitive and alot of classical results can be derived from it (i used a profs notes but they closely resemble a book called  Lebesgue Integration (or Integral im not sure) by Soo Bae Chang.

[quote="Jacobson"]
 [quote="Rudin / p47,48/ Theorem 2.17"]
Suppose $ X$ is a locally compact, $ \sigma$-compact Hausdorff space. If $ \mathfrak{M}$ and $ \mu$ are as in theorem  2.14 (Riesz for positive linear functionals), then $ \mathfrak{M}$ and $ \mu$ have the following properties:
(a) If $ E \in \mathfrak{M}$ and $ \epsilon > 0$, then there is a closed set $ F$ and an open $ V$ such that $ F \subset E \subset V$, and $ \mu(V \minus{} F) < \epsilon$
(b) $ \mu$ is a regular Borel measure on $ X$.
[/quote]
I understand part $ (a)$. I have trouble seeing why it implies $ (b)$.
[/quote]

I think that the following works:

- Let $ E \in \mathfrak{M}$. If $ \mu(E)< \plus{} \infty$, Rudin / p41/ Theorem 2.14/ (ii) gives the result, so that we can suppose $ \mu(E)\equal{} \plus{} \infty$. By (a), there is a closed set $ F \subset E$ such that $ \mu(F)\equal{}\plus{}\infty$.

- Since $ E$ is $ \sigma$-compact Hausdorff space, there is a sequence $ (K_n)$ of compact sets such that $ E\equal{} \cup_n K_n$. We can also suppose that this sequence is increasing (with respect to inclusion). Then $ \plus{} \infty\equal{}\mu(F)\equal{}\mu(\cup_n (K_n \cap F))\equal{}\lim_{n \rightarrow \plus{} \infty}\mu(K_n \cap F)$.

- Conclusion: $ (K_n \cap F)_n$ are compact subsets which lie in $ E$ such that  $ \lim_{n \rightarrow \plus{} \infty}\mu(K_n \cap F)\equal{}\mu(E)$. This shows (b).

We say that a probability measure $ \mathbb{P}$ on metric space $ S$ is regular if for every Borel set $ A$

$ \mathbb{P}(A)\equal{}\sup\{\mathbb{P}(C);C\subset A, C \textrm{ closed } (*)\}\equal{}\inf\{\mathbb{P}(U);A\subset U, U \textrm{ open } (**)\}$

($ \sigma$-algebra of Borel sets is defined as $ \sigma(\mathcal{U}_S)$ where $ \mathcal{U}_S$ is family of all open sets on $ S$).

I wanted to ask, why is it that, in $ (*)$, we require the sets to be closed, and respectively, in $ (**)$, sets to be open. I can't seem to figure out why is it so.

I hope no one minds I putted question in this certain topic, after all it is called "intuition" on measure theory. :wink:

Open and closed sets are the "nicest" sets in the metric spaces. You definitely have to include them into your $ \sigma$-algebra if you want all continuous functions to be measurable. So, the question arises how far other measurable sets are from them. It turns out that every set in the Borel $ \sigma$-algebra (the least $ \sigma$-algebra containing open and closed sets) can be approximated by a closed set from below and an open set from above with respect to any finite Borel measure (in the sense of the identities you posted). This, in turn, can be used to approximate measurable and integrable functions by continuous ones and the latter approximation plays quite an important role in the theory.

It is clear that you cannot hope to use the closed sets for the "upper" approximation because the closure of a small set can be quite large. For the same reason (the interior of a large set may be small or even empty), you cannot use open sets in the "lower" approximation. So, you have just one logical possibility left and the nice thing is that it works.


calculus
integration
geometry
trigonometry
calculus computations

For the curve $ C: y = \frac {1}{1 + x^2}$, Let $ A(\alpha ,\ f(\alpha)),\ B\left( - \frac {1}{\alpha},\ f\left( - \frac {1}{\alpha} \right)\right)\ (\alpha > 0).$ Find the minimum area bounded by the line segments $ OA,\ OB$ and $ C,$ where $ O$ is the origin.

Note that you are not allowed to use the integral formula of $ \frac {1}{1 + x^2}$ for the problem.

Let $ T\equal{}T(\alpha)$ be the area of the region in question.

Let $ D\equal{}(\alpha,0)$ and $ E\equal{}\left(\minus{}\frac1{\alpha},0\right).$

Area $ OAD\equal{}\frac12\cdot\frac{\alpha}{1\plus{}\alpha^2}.$

Area $ OBE\equal{}\frac12\cdot\frac{\frac1{\alpha}} {1\plus{}\frac1{\alpha^2}}\equal{} \frac12\cdot\frac{\alpha}{1\plus{}\alpha^2}$

Add these two triangles to get the area under the curve from $ \minus{}\frac1{\alpha}$ to $ \alpha.$

Hence, $ T\equal{}\int_{\minus{}\frac1{\alpha}}^{\alpha}\frac{dx}{1\plus{}x^2}\minus{} \frac{\alpha}{1\plus{}\alpha^2}$

By the Fundamental Theorem of Calculus:

$ \frac{dT}{d\alpha}\equal{}\frac1{1\plus{}\alpha^2}\minus{}\frac1{1\plus{}\frac1{\alpha^2}} \cdot\frac1{\alpha^2}\minus{}\frac{1\plus{}\alpha^2\minus{}2\alpha^2}{(1\plus{}\alpha^2)^2}$

$ \equal{}\frac{\alpha^2\minus{}1}{(1\plus{}\alpha^2)^2}$

Hence the critical point, which is a minimum, happens when $ \alpha\equal{}1.$

The minimum area is $ T(1).$ On the other hand, I don't see how to avoid knowing about the arctangent if we want a specific value for that. It does turn out that $ T(1)\equal{}\frac{\pi}2\minus{}\frac12.$

I have nothing to add your solution more than this. :)

Let me add one word, we set $ x\equal{}\tan \theta$, we are to evaluate $ \int_{\minus{}1}^1 \frac{1}{1\plus{}x^2}\ dx$

Because Japanese highschool students don't study arc tangent. I think that this situation is wrong, if we are restricted to such things, we shouldn't learn Calculus in highschool. My students are looking forward to see the forum and how people in abroad would solve Japanese university entrance exam, so i dare to post them with restrict to the solution, but those who hope to entrer Tokyo University or Kyoto University, they study university mathematics by themselves, so few students understand your solution.


Square A has side lengths each measuring $ x$ inches.  Square B
has side lengths each measuring $ 4x$ inches.  What is the ratio of
the area of the smaller square to the area of the larger square?
Express your answer as a common fraction.

I guess you just brute force? Or is there an intuitive way?

What's there to brute force?

$ \frac{\text{Area of Square A}}{\text{Area of Square B}}\equal{}\frac{x^2}{(4x)^2}\equal{}\frac{x^2}{16x^2}\equal{}\frac{1}{16}$

in general, if you take a polygon with side lengths $ a_1,a_2,a_3,...,a_n$, and multiply each of the side lengths by a factor of $ a$, then the ratio of areas is $ a^2$.  It should not be too hard to find out why this is true.


geometry
superior algebra
superior algebra unsolved

9. v 4etirexugolnike ABCD . O centr opisanoi okrujnosti E  . perese4eniya diagonalei. Vnutri 4etyrexugolnika vibrana . P takie 4to #PAB+#PCB=#PBC+#PDC=90
dokazat’, 4to .P lejit na pryamoi EO.  (# - eto ugol

At first glance I thought this is pure spam, but if one takes a closer look, it seems to be geometry...

What's wrong and why I lock it now:
- wrong language (English spoken here!)
- wrong forum (if it's geometry)
- wrong title (useless titles like this one say nothing)
- not given source (the number 9. at the beginning suggests a source)

Please repost it in an appropriate case (some place with geometry) and with correct language, title and source.


algebra
polynomial
superior algebra
superior algebra unsolved

Call a polynomial positive reducible if it can be written as a product of two nonconstant polynomials with positive real coefficients. Let $ f(x)$ be a polynomial with $ f(0)\not\equal{}0$ such that $ f(x^n)$ is positive reducible for some natural number $ n$. Prove that $ f(x)$ itself is positive reducible. [L. Redei]

[quote="mateivld"]It is simple. The term containing $X^k$ can vanish into the product. Look for example at $(X^2-X+1)(X^2+X+1)=X^4+X^2+1$. The RHS is a polynomial in $X^2$ whilst the terms in the LHS are not.[/quote]

huh ? :huh: The coefficients should be positive ! Have you read the problem ?

[quote="mahanmath"][hide="Solution"]Am I missing something here ? :huh:

Let $f(x^n) = h_1 (x) . h_2 (x) ... h_m (x) $ and suppose that in one of the polynomials in RHS (for example $h_1 (x)$) there exist a term like $ax^k$ such that $n \nmid k$ then $ax^k . h_2 (0) . h_3 (0)... h_m (0)$ appears in expansion of RHS and since all the coefficients are positive it never vanish and so remains .
Obviously it`s contradiction because exponent of terms in LHS are multiple of $n$ .

Hence $h_i (x) = g_i (x^n)$ and so $f(x) = g_1 (x) . g_2 (x) ... g_m (x) $ .[/hide][/quote]

Correct :)


Umbo has x money.  He gives 1/6 to his friend, Glorpus.  He then spends 1/5 of his remaining money.  After giving away 4/12 of his remaining money to charity, he has approx. $98.83 left over.  How much did he have to start with? (round to nearest cent)

EDIT: I wasn't thinking when I posted my answer the first time  :roll:

[hide]After takin 1/6 away, he has 5/6 left. 1/5 of that is 1/6. Now, he has a total of 1/3, or 2/3 remaning. 1/3 of that is 2/9. So 98.83 accounts to be 1-(1/6 + 1/6 + 2/9) or 4/9 of his remaining money, which is 98.83.

That means he started with 222.37.
[/hide]

[hide]
I would just do reverse operations.

Start: 98.83

$\frac {4}{12} = {1}{3}$

$98.83 \times 3 = 296.49$

Then, multiply that by 5:

$296.49 \times 5 = 1482.45$

1482.45 is the answer. Wait, though...that doesn't look right, either.
[/hide]

[hide]his friend gets [tex] \frac {x}{6} [/tex] dollars

he would then have [tex] \frac {5}{6} x[/tex] dollars left

he then spends [tex] \frac {1}{5} [/tex] of the remaining money

or [tex]\frac {1}{6} x[/tex] dollars. that would leave [tex] \frac {2}{3} x[/tex] dollars left over.

finally he gives [tex]\frac {1}{3}[/tex] of [tex] \frac {2}{3} x[/tex] dollars to charity
that means he gives [tex] \frac {2}{9} x[/tex] to charity

he has given away/spent a total of [tex]( \frac {x}{6} + \frac {x}{6} +  \frac {2}{9} x)[/tex]dollars or [tex]\frac {5}{9}x[/tex] dollars of his money
that means he has [tex] \frac {4}{9} x [/tex] dollars  or [tex] 98.83[/tex] dollars left

so [tex]x[/tex] would be [tex]\frac {9}{4} (98.83)[/tex] dollars or [tex]223.12[/tex] dollars[/hide]


Give the value for LHS:

$ (\sin x + 3x)(x\cos x + 4\sin x)(\sin x + \sin 2x) \geq 30x \sin \2x \sinh$

for $ x = \pi/6$

My calculator SUCKS;  :(  It gives me 6.940221725 and then gives me 6.141898374 after I got rid of the trig function to leave in terms of real numbers only.

I'm confused over occasions to put parenthesis or not; my calculator is driving me crazy as I can NEVER trust it...  :!:

I got $ 6.94022175$ using real numbers, you probably entered something wrong in the calculator.

$ \left(\sin{\frac{\pi}{6}}\plus{}3\left(\frac{\pi}{6}\right)\right)\left(\frac{\pi}{6}\cos{\frac{\pi}{6}}\plus{}4\sin{\frac{\pi}{6}}\right)\left(\sin{\frac{\pi}{6}}\plus{}\sin{2\left(\frac{\pi}{6}\right)}\right)$

$ \equal{}\left(\frac{1}{2}\plus{}\frac{\pi}{2}\right)\left(\frac{\pi}{6}\left(\frac{\sqrt{3}}{2}\right)\plus{}4\left(\frac{1}{2}\right)\right)\left(\frac{1}{2}\plus{}\sin{\frac{\pi}{3}}\right)$

$ \equal{}\left(\frac{1\plus{}\pi}{2}\right)\left(\frac{\pi\sqrt{3}}{12}\plus{}2\right)\left(\frac{1}{2}\plus{}\frac{\sqrt{3}}{2}\right)$

$ \equal{}\left(\frac{1\plus{}\pi}{2}\right)\left(\frac{24\plus{}\pi\sqrt{3}}{12}\right)\left(\frac{1\plus{}\sqrt{3}}{2}\right)$

At this point, I put this into my calculator and got...

6.940221725.

So, I agree with SnipedYou.

Here's what I put into my calc:

$ ((1\plus{}\pi)/2)((24\plus{}\pi\sqrt(3))/12)((1\plus{}\sqrt(3))/2)$

Try
$ (\sin x \plus{} 3x)(x\cos x \plus{} 4\sin x)(\sin x \plus{} \sin 2x)$

Set to Radians
$ /frac{\pi}{6}\minus{}>X$
Type this in and press enter
$ (\sin x \plus{} 3x)(x\cos x \plus{} 4\sin x)(\sin x \plus{} \sin 2x)$

I use a TI-84 (silver edition).

But as long as you have a good TI graphing calc, it shouldn't make a difference.

Enter the string:

[color=darkblue](sin(pi/6)+pi/2)*(pi/6*cos(pi/6)+4*sin(pi/6))*(sin(pi/6)+sin(pi/3))[/color]

into Google Search  :)

$ \blacktriangleright$ Here is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214488]link to the original thread[/url]
$ \blacktriangleright$ Attached is a graph of LHS-RHS for $ x \in \left[ 0, \frac {\pi}{2} \right]$


i can get the answer on my calculator but that doesnt help me. can someone explain how to do this step by step or get me on the right track. thx.

x/ (1-x-x^2)

do what?  write this as an infinite series?


invariant
function
real analysis
real analysis unsolved

Find all linear homogeneous differential equations with continuous coefficients (on the whole real line) such that for any solution $ f(t)$ and any real number $ c,f(t\plus{}c)$ is also a solution.

it has been discussed several times on this forum that the only shift-invariant closed subspace of continuous functions are exponential monomials. No complete proof has been posted nonetheless
Ps:  this particular problem should be much easier.

[quote="alekk"] the only shift-invariant closed subspace of continuous functions are exponential monomials. [/quote]

 :huh: ,Where can I find it ?

make a research for "shift invariant" in the forum. But as I said, no complete proof has been posted.


number theory
least common multiple
MATHCOUNTS
AwesomeMath
summer program

Alright. Sorry if I'm posting in the wrong section. I just want to know if I'm doing something wrong. I probably am.

A solar system contains four planets and a sun as shown. Planet A completes an orbit around the sun in 20 months; planet B completes an orbit in 36 months; planet C completes an orbit in 40 months; planet D completes an orbit in 48 months. If the four planets lie along a straight line in May of the year 2000, what is the next year they will again lie along the same line? Assume that the orbits are circular and that a year has exactly 12 months.

Sorry that I can't give the diagram. It pretty much looks like this:

(Sun)     (A)       (B)         (C)   (D)

My answer: 2060 (lcm of the months in 720, so my answer is (720/12)+2000)
Official MathCounts answer, mathcounts says there is no error: 2030

Where am I making a mistake?

well your answer would be the shortest time in which a,b,c,d are all on the same side of of the sun (in addition to in a straight line) as they were in 2000.

think about it: lets say a orbits in 1 yr and b orbits in 2. lcm(1,2)=2 however, in 1 year, they will be in straight line again (but on opposite sides of sun from which they started)

hurdler, thanks.

WHOA!!!! WRONG FORUM!! sorry. I was tired, mixed up AwesomeMath with MathCounts. Can a mod move this to the mathcounts forum please? Or delete if you can't? Sorry.


AMC
AIME
counting
derangement
Pythagorean Theorem
geometry
cyclic quadrilateral

I am looking for a few good AIME level contest problems with fantastic simple solutions that are at least mostly proofs without words.  All help would be appreciated, thanks.

I'll work on an animation proof for the proof of the Pythagorean Theorem, if we don't already have it.

8.  8 clay targets are arranged as shown. In how many ways can they be shot (one at a time) if no target can be shot until the target(s) below it have been shot.

[img]http://www.kalva.demon.co.uk/aime/soln/diag908.jpg[/img]

All you need is to find the derangements of LLLMMRRR

[quote="theone853"]don't you think that the pythagorean theorem would be the first thing they'd think of making into a proof without words

also,  i think there was a section in this site dedicated just for them

edit: i kinda found it

http://www.artofproblemsolving.com/Images/Side10.swf and you can change the numbers from 1 to 9.[/quote]

http://usamts.org/Gallery/G_Gallery.php includes a java proof without words.

Maybe some point, circle and line segment problems.  (these are fairly simple problems though, not sure if they are what you might be looking for)
Like $x^2 = (x-1)(x-2)$ Where x is the length of the line segment from a point to a tangent, x-2 is the length of a line segment from the point to where it intersects the circle and x-1 is the length of the secant made by that line segment.

You could show that in a cyclic quadrilateral $ABCD$ that $AB*CD + BC*AD = AC*BD$.  I'm not sure if this is advanced enough though.

This was on the 2004 BAMO T-shirt:

${n\choose2} + {m\choose2} \leq {n+m-1\choose2}$


Given an isosceles trapezoid with bases of 8 and 18 and an area of 156 square units, what is the number of units in the length of one of the non-parallel sides?

We first utilize the formula for the area of a trapezoid and plug in values:

$ A \equal{} \frac{1}{2} \cdot (b_1 \plus{} b_2) \cdot h$
$ 156 \equal{} \frac{1}{2} \cdot (26) \cdot h$

Given this, we can easily find the height of the trapezoid, which is $ 12$.

To find the difference between the bases, we subtract $ 8$ from $ 18$ to get $ 10$, and then divide that by $ 2$ to get $ 5$.  So the length of one of the non-parallel sides is $ \sqrt{5^2 \plus{} 12^2} \equal{} \boxed{13}$.


Iura is looking for a sequence $S = (a_1, a_2, \ldots, a_n)$ of $n$ natural numbers of the form:

\[a_1, \quad a_2 = a_1 + 1, \quad \ldots, \quad, a_n = a_{n-1} + 1\]

with the property that none of the numbers $a_1, a_2, \ldots, a_n$ has a sum of digits not being divisible by 5.

a.) What is the greatest length $n$ which such a sequence can have ?

b.) How many sequences $S$ of greatest length $n$ are there in $\mathbb{N}$ where the interval from 1 until 1000
is considered ?

if the last digit doesn't jump 9->0 it's 4. So we'll check if there is a longer one with jump. Obviously that means that sum(10k-1)= 4 mod 5 at best, if not possible 3,2,etc.

eg. 9,59,109... have requested property, they can attain length 8: 6-7-8-9-1-2-3-4 (numbers 6-13)

checking the same for higher numbers: 56-57-58-59-60-61-62-63 or in general for 50k+6 to 50k+13 shows that this 8 row is possible iff a_1=6 mod 50

So in 1-1000 there are 20 options. :)

Hi,

my attempt at the first part of the questions is as follows :

We can maximise the length of the sequence if we form the sequence in the following manner.

The sequence is like $a99...9b, a99...9(b+1),..., a99...99, (a+1)00...00, (a+1)00...01, ...., (a+1)00...0(5-a-1)$,

this is maximally long if the $(a+1) \equiv 1 \pmod{5}$, and if $b = 6$, and the first number $a_1$ is chosen keeping in mind these criteria.

So, the length of the sequence in this case is
 4 numbers before $(a+1)00...00$, and 4 numbers starting from $(a+1)00...00$ onwards.

Thus the maximum possible length of the sequence is 8.

*******************************************************************************************
Sambit

Hi Peter,

Sorry about that. I saw the problem and worked at solving it. While I was posting my solution, I saw yours only then. Sorry  :blush:

Sambit

really? you posted like half an hour after me :)

anyway no problem man. finally got my 1000th post and rid of that ugly yang-mill's theory :D

[quote="Peter VDD"] So in 1-1000 there are 20 options. :)[/quote]

Have one more try, please.  :)

[quote="orl"]with the property that none of the numbers $a_1, a_2, \ldots, a_n$ has a sum of digits not being divisible by 5.[/quote]
If we take the question as it is there, then the longest is 1 and the problem is trivial ;) so I guess we'll just leave away the "not".

Anyway I'm wrong indeed. Better try: not 50k+... but just according to the rest of the number :)

It must go from xy6 to x(y+1)3, unless we jump 100 then it's x96 to (x+1)03  (with y=9)
the number xy must now simply be 1 mod 5 to fulfill the requirements. And there are 18 options to do that :)

Greetz, Peter


I was wondering if anyone knows how to approximate or find the equation of a non-linear function of unknown degree given multiple points. Thanks.

To find the polynomial passing through those points, see [url=http://en.wikipedia.org/wiki/Lagrange_interpolation_formula]Lagrange's interpolation polynomial[/url]. There was a related [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=238484]recent thread here[/url].

In practice, it's easier to do interpolation in [url=http://en.wikipedia.org/wiki/Newton_polynomial]Newton form[/url].

Of course, the problem with exact polynomial interpolation is that a polynomial interpolation to a non-polynomial set of data often behaves very badly between the interpolation points.  Some alternatives:

- A collection of [url=http://en.wikipedia.org/wiki/B%C3%A9zier_curve]Bezier curves[/url].
- A [url=http://en.wikipedia.org/wiki/Fourier_series]Fourier series[/url] (of the piecewise linear function made by connecting the points, for example).
- The Bernstein polynomials I mentioned in the thread azjps linked to.


Solve each equation.  The letters a, b and c are constants.

(1) (a/x) + (b/x) = c, where c does not = 0.

(2) (b + c)/(x + a) = (b - c)/x - a), where c and a do not = 0.

For the first question:

(a/x) + (b/x) = c
(a+b)/x = c
x = (a+b)/c

For the second question:

(b + c)/(x + a) = (b - c)/(x - a)
(b + c)(x - a) = (b - c)(x + a)
bx -ab + cx -ca = bx + ab -cx - ca
2cx = 2ab
x = ab/c

For the second:

\begin{eqnarray*}\frac{{b+c}}{{a+x}}&=& \frac{{b-c}}{{x-a}}\hfill \\ \frac{{b+c}}{{b-c}}&=& \frac{{a+x}}{{x-a}}\hfill \\ \frac{b}{{b-c}}&=& \frac{x}{{x-a}}\hfill \\ \frac{{b-c}}{b}&=& \frac{{x-a}}{x}\hfill \\ 1-\frac{c}{b}&=& 1-\frac{a}{x}\hfill \\ \frac{a}{x}&=& \frac{c}{b}\hfill \\ x &=& \frac{{ab}}{c}\hfill \\ \end{eqnarray*}
 :rotfl:

Are we dealing with literal equations here?


How many cubic feet of wood are needed to build a solid door 7 feet high, 3 feet wide and 2 inches thick? Express your answer as a decimal to the nearest tenth.

$ 7 \times 3 \times \frac{2}{12} \equal{} \boxed{3.5 \mathrm{ft^{3}}}$

Hmm on FTW it says to the nearest hundredth for some reason :maybe:


trigonometry
logarithms
geometry
3D geometry
sphere
calculus
derivative

[quote="Hexaditidom"](Note: I = me, i = imaginary.)

1)
I tried to find a different way to express i (to solve problem #3 here) so I found that i = 1/i. Is this true?

2)
What is the square root of i? Is it defined?

3)
What is i^i? (My math book told me that this was a real number!)[/quote]

1)
1/i = -i --> in general, i^n  :equiv: i^(n mod 4)

So 1/i = i^-1  :equiv: i^3 = -i

2)
rep123max answered your question.

3) i^i

To begin, we must define i^x

Since e^i(pi/2) = i, it follows that i^x = e^(i*pi/2*x)

Thus, i^i = e^(i^2*pi/2) = e(-1*pi/2) = e^(-pi/2)

Hopefully this helps; I made use of Euler's relation numerous times here.  If you need any clarification, just ask.

#3:

Use [tex]e^{i\theta}=\mathrm{cis}~\theta[/tex] as follows:

[tex]\displaystyle i=e^{i(\pi/2+2\pi k)}[/tex] for any [tex]k\in\mathbb{Z}[/tex]

[tex]\displaystyle i^i = \left(e^{i(\pi/2+2\pi k)}\right)^i[/tex]

[tex]\displaystyle = e^{-\pi/2-2\pi k}[/tex]

So, [tex]i^i[/tex] takes infinitely many real values!

[quote]it follows that i^x = e^(i*pi/2*x)
[/quote]

I don't understand how "it follows."

[quote="ComplexZeta"][tex]\displaystyle i^x=\Big(e^{i\pi/2}\Big)^x=e^{i\pi x/2}[/tex][/quote]

It does look a lot better that way, doesn't it?

That's why learning to tex is the first thing on my summer to-do list.

#2.  This is a great example of the power of De Moivre's theorem, but for those students who haven't had trig yet, #2 can be solved algebraically:

Since we want the square root of i, start: z^2 = i and thus:  (a+bi)^2 = i
Expand and equate the real parts and the imaginary parts.

Not as elegant,  but no trig required.  Also, this way "looks like" square root.

Pertaining to number 2: One of the interesting things about i is that complex numbers are closed over the algebraic relations.  So you can add, multiply, add, subtract, exponent, take the logarithm of, take the nth root of, and all that other stuff you knew in middle school, of any two complex numbers.

I cannot off-hand think of any common relation that C is not closed under.

[tex]\mathbb{C}[/tex] is not closed under division unless you exclude 0.

But [tex]\hat{\mathbb{C}}[/tex] (the Riemann sphere) is. :)

Right....By C I definately meant the Riemann sphere....maybe.  :-)

The Riemann Sphere is the complex numbers plus a point "at infinity."  In other words, it makes infinity a number, sort of.

And the complex numbers aren't not closed under any common relation -- otherwise, you would have heard of some really useful extension of them, which there isn't.  (Quaternions don't count because you have to get rid of other nice properties that we like.  And they aren't [i]that[/i] useful, anyhow.)

The Riemann sphere is cool because you can divide by zero and turn the complex numbers into a compact surface. A lot of problems that are usually hard become easy when you look at them on the Riemann sphere.

[quote]A lot of problems that are usually hard become easy when you look at them on the Riemann sphere.[/quote]

Example?

I'd have to discuss a lot of background and convince people that the normal derivatives learned in calculus aren't necessarily the best way to think about things. But we might consider a derivative as a linear map. Then, for example, we could find out what the derivative of 1/z is at 0 just by doing a bit of geometry on the Riemann sphere.

This doesn't belong in Intermediate.

[quote="ComplexZeta"]This doesn't belong in Intermediate.[/quote]

No it doesn't.

It started that way though, and then all the advanced people came in. :)

Anyway.

And this is where the Mandelbrot and Julia sets came from?

The Riemann sphere is the same as the complex numbers with an extra point at infinity. Consider a sphere centered at the point (0,0,0), and let the complex plane be (x,y,0). Then we take each point (a,b,0) on the plane to a point on the sphere by drawing a line from (a,b,0) to (0,0,1) (the north pole of the sphere). The line will intersect the sphere at the north pole and one other point. We map (a,b,0) to the other point. Every complex number maps to exactly one point on the sphere, and every point on the sphere except the north pole maps to exactly one complex number. Therefore, we add a point at infinity and say that the north pole maps to the point at infinity.

[quote="mathnovice"]Out of curiosity, what other numbers are there? I've never heard of the Riemann sphere before.[/quote]

There are lots of them. There are the ones you already know, like the integers, the rational numbers, the real numbers, the complex numbers, numbers modulo other numbers, and matrices (you can add them and multiply them, so you can think of them as "numbers"); then there are the gaussian integers (complex numbers with integer real and imaginary parts), the quaternions, the cayley numbers, p-adic numbers, and Conway numbers - just to name a few. Most of these are defined by relaxing or tinkering with different properties of the familiar real numbers.

You can read about all of these things in a book called [u]The Book of Numbers[/u] by John Conway, or in a book called [u]Numbers[/u] by Ebbinghaus, et. al.


x,y,z are positive natural numbers such that :
$ x,(yz) \plus{}y,(zx)\equal{}z,(xy)$
 where : $ a,(b)\equal{} a,bbbbbb......$

(x,y,z)=(4.4.9)
 это правильно.  this is beatiful solution.
 i forget all English , because for long time i don't write english, do't say english...  only russian
сейчас, русский язык - это мой кровь.
очень приятно познакомиться с вами


modular arithmetic
number theory unsolved
number theory

Write $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\plus{}\frac{1}{p\minus{}1}$ where $ p$ is prime as a fraction reduced to lowest terms in the form $ \frac{A_p}{B_p}$

(i) Find the value of $ A_p\pmod{p}$ and prove the result

(ii) Make a conjecture for the value of $ A_p\pmod{p^2}$

Wolstenholme's theorem. See :
http://mathworld.wolfram.com/WolstenholmesTheorem.html

and many topics on mathlinks.

Pierre.

Lemma :
if $ p$ is prime then $ \frac {(p \minus{} 1)!}{k}\equiv \minus{}k^{p \minus{} 2}$ proof:
   Wilson's tehorem: $ (p \minus{} 1)!\equiv \minus{} 1\bmod{p}$      for $ p$ prime so
    $ (p \minus{} 1)!\equiv \minus{} 1\equiv \minus{}k^{p \minus{} 1}\bmod{p} \equal{} > \frac {(p \minus{} 1)!}{k}\equiv \minus{}k^{p \minus{} 2}\bmod p$

$ \sum_{k \equal{} 1}^{p \minus{} 1} \frac {1}{k} \equal{} \frac {\sum_{k \equal{} 1}^{p \minus{} 1} \frac {(p \minus{} 1)!}{k}}{(p \minus{} 1)!}$
$ \sum_{k \equal{} 1}^{p \minus{} 1} \frac {(p \minus{} 1)!}{k}\equiv \minus{}\sum_{k \equal{} 1}^{p \minus{} 1} k^{p \minus{} 2}\equiv \minus{}(\sum_{k \equal{} 1}^{p \minus{} 1} k)(\sum_{k \equal{} 1}^{p \minus{} 1} k^{p \minus{} 3} \minus{} \sum_{1\leq k,l \leq p \minus{} 1} (kl)^{\frac {p \minus{} 3}{2}})\equiv \minus{}p(\frac {p \minus{} 1}{2})(\sum_{k \equal{} 1}^{p \minus{} 1} k^{p \minus{} 3} \minus{} \sum_{1\leq k,l \leq p \minus{} 1} (kl)^{\frac {p \minus{} 3}{2}})\equiv 0\bmod p$
so one of divisor's is $ p$ , but $ p$ is not a divisor of $ (p \minus{} 1)!$ so after reducing $ A_{p}$ will be divisible by $ p$

[quote="Imp"]Lemma :

    $ (p \minus{} 1)!\equiv \minus{} 1\equiv k^{p \minus{} 1}\bmod{p} \equal{} > \frac {(p \minus{} 1)!}{k}\equiv k^{p \minus{} 2}\bmod p$

[/quote]

wouldn't it be $ (p \minus{} 1)!\equiv \minus{} 1\equiv \minus{}k^{p \minus{} 1}\bmod{p}$

You are right, sorry, but it don't changes anything :wink:


Se considera matricea $A = \left( {\begin{array}{*{20}c}
   0 & 1 & 0  \\
   0 & 0 & 1  \\
   1 & 0 & 0  \\
\end{array}} \right)$ si $C(A)=${$X \in M_3(C)|XA=AX$}.Sa se arate ca daca $Y \in C(A)$ si exista $p \in N,p>2$ astfel incat $Y^p=O_3$ , atunci $Y=O_3$.

A is a permutation matrix, let $Y\in C(A)$

$Y=\left[\begin{array}{ccc}a&b&c\\d&e&f\\g&h&w \end {array}\right]$

$AY=YA$ gives

$Y=a\left[\begin{array}{ccc}1&1&1\\1&1&1\\1&1&1 \end {array}\right]=aJ$

By induction $J^p=3^{p-1}J$

if there exist p>1 integer such that $Y^p=a^p3^{p-1}J=0$   then $a=0$

$Y=0$

Avem $Y = \left( {\begin{array}{*{20}c}
   a & b & c  \\
   d & e & f  \\
   g & h & w  \\
\end{array}} \right)
$

Si cum $AY=YA$ avem deci:$\left( {\begin{array}{*{20}c}
   a & b & c  \\
   d & e & f  \\
   g & h & w  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   0 & 1 & 0  \\
   0 & 0 & 1  \\
   1 & 0 & 0  \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
   0 & 1 & 0  \\
   0 & 0 & 1  \\
   1 & 0 & 0  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   a & b & c  \\
   d & e & f  \\
   g & h & w  \\
\end{array}} \right)
$
Care este tot una cu :$
\left( {\begin{array}{*{20}c}
   c & a & b  \\
   f & d & e  \\
   w & g & h  \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
   d & e & f  \\
   g & h & w  \\
   a & b & c  \\
\end{array}} \right) \Rightarrow Y = \left( {\begin{array}{*{20}c}
   a & b & c  \\
   c & a & b  \\
   b & c & a  \\
\end{array}} \right)
$
Cum demonstrez ca a=b=c.

Avand $A^p=0_3$ rezulta ca valorile propri ala matrici $Y$sunt $a_1=a_2=a_3=0$
deci  matricea $Y = \left( {\begin{array}{*{20}c}
a & b & c \\
c & a & b \\
b & c & a \\
\end{array}} \right)
$ are valorile propri $a_1=a_2=a_3=0(1)$,dar valorile matrici $Y$ sunt $a_1=a+b+c,a_2=b-a,a_3=c-a(2)$
deci din (1)(2) rezulta ca $a=b=c=0$ deci $Y=O_3$.

As dori sa mi se confirme daca este corecta sau nu solutia.

The entries of $Y$ you gave is correct

mine was not correct

[quote="silviu"]$Y = \left( {\begin{array}{*{20}c}
   a & b & c  \\
   c & a & b  \\
   b & c & a  \\
\end{array}} \right)$

[/quote]

Your expression is correct, from , now $Y$  is diaogonalisable it is similar to the diagonal matrix

$diag(a+b+c,a+jb+j^2c,a+j^2b+jc)$ where $j=exp(2i\pi/3)$

Suppose that there exist p>2 integer such that $Y^p=0$ then we get

(a+b+c)^p=0
(a+jb+j^2c)^p=0
(a+j^2b+jc)^p=0

a+b+c=0
a+jb+j^2c=0
a+j^2b+jc=0

sum this equation we have $a=0$ and solve the system $a=b=c=0$

$Y=0$

Remark: $A^p \neq 0$ since it is a permutation matrix but $A^3=I_3$

[quote="silviu"]$A^p=0_3$ [/quote]

This is not correct since $ A^3=I_3$  eigenvalues are $1,j,j^2$


Suppose that $a_1,a_2,a_3...a_{2n}$ are distinct such that the equation $(x-a_1)(x-a_2)...(x-a_{2n})-(-1)^n(n!)^2=0$ has an integer solution r.Show that $r=\frac{a_1+a_2+...+a_{2n}}{2n}$
I dont know if it was posted before.


I've noticed that most of the people on here seem to like math way more than regular people do. I mean, I don't really mind math, but I'm not totally OBSESSED with it, constantly doing it at every available opportunity.

So my questions are:

1. Why are you so obsessed with math?
2. If your answer has something to do with finding "ultimate truth" why do you do math instead of philosophy or some kind of science?
3. How come so many people think that mathy smart people are smarter than smart people who like other stuff?

I don't feel that this poll is very complete. Why is there no option along the lines of
"I like math because it's fun"
or
"I like math because it makes me feel good"
Sometimes it can be just as simple as that.

Ummm... I didn't vote for any of them... because I don't like math for any of those reasons... where's the "I like math because I think it's fun/interesting, and I sometimes like to do it in my freetime and for math competitions" option?

 lol and I'm not obsessed with math, I'm not even a mathy person

Yeah, all of these options are suprisingly negative.

1) I'm not [i]obsessed[/i] compared to most ppl on here, but compared to most ppl in my life I am. I do math because I enjoy it, period.
2) N/A
3) In my opinion, it's because a lot of math people show their smarts more than people who are smart in different areas.

Those are horrible pole options, I like math because its like solving a puzzle, it lets me learn things and its fun. Make a new pole with more options.

[quote="Kel_Varnsen"]How can you possibly not know how to spell "poll" when it's written all over the forum? I guess mathy people really ARE smarter than non-mathy people :rotfl::rotfl::rotfl:.[/quote]
Dang, people on AoPS are going to hate you now.  :P

Also, using spelling is the stupidest way to insult I know. Typos/Absentmindedness/etc. happen, Mr. God of Typing.

I think, as above posters have said, that most of us here do math b/c we think it's fun and we like it, not b/c of those reasons in the poll, which r very negative...

Yeap, this is a strange poll. Out of all the options only the first holds some water. The rest just show that OP has a very strange and vague idea of what mathematics really is. If you are really into it, it becomes as natural for you to think of abstract objects as to pay attention to your surroundings when walking along the street, so it is rather some kind of "awareness" than an "obsession". Also, doing mathematics has no more to do with memorizing formulae than writing poetry has to do with memorizing the backward spelling of the words.

As to the second question, I do not really believe in the ability of philosophy to discover the "ultimate truth" (it seems to me that it has already discovered a few too many) but my decision to go into mathematics rather than into physics or engineering was rather random. I do not think I would be any good in biology or chemistry though: it seems like there you need to pay a lot of attention to every small detail and that is not what I'm really good at.

As to the third question, math. people are "smarter" in one and just one respect: they are usually more capable of abstract thinking and drawing logical conclusions than an average person. In all other respects they may be as dumb (or bright) as everybody else. I do not believe that mathematicians tend to show off any more than other professionals not to mention that it is really hard for a mathematician to show off: if you try to tell a layman how exciting was what you just did and how proud you are of your discovery, most likely he won't even understand what you are talking about. So a lot of my non-mathematical acquaintances (like my wife's colleagues, etc.) tend to think of me as some kind of a rather  boring and stupid odd guy who may be somewhat good for teaching students but otherwise is totally useless. I gave up on trying to explain what I'm really doing to anybody who's not a scientist or engineer himself long ago and, when asked what my occupation is, just say "a mathematics teacher", which is usually good enough for them to place me into some niche in their social hierarchy. It may be an escapist position, but it spares my time and energy.

I once had a long conversation with a former mathematical teacher who later went into business management. He told me a few stories (sorry, JBL :P) and one of them was like this: his boss told him that he would like to get each part for several part orders shipped from the warehouse where that part is in the largest supply and the turnover on that part is the slowest. The teacher's response was immediate and the exact words were "Who's going to pay for the freight?". That is sort of smartness to expect from a "mathy" person and sometimes it may seem impressive but it is merely the ability to look one extra step ahead and it is by no means uncommon among people without mathematical training. As to the smartness you need for consoling a crying child or making somebody else do what you want, mathematical training is rather useless here and you should study some psychology instead.

[quote="mr.affable"]My views are most congenial with the "hyperbole" option.  Yet, isn't any phrase in rhetoric evaluated by the intention of the speaker?  If my intention in seeing math in such a grandiose manner is, at its heart, very candid, then I think the use of the term "hyperbolic" is a bit offensive, if not tinged with caustic manner.[/quote]

your writing manages to drudge up all of the repressed memories i have of reading horrible college admissions essays littered with inappropriate and obscure adjectives and painfully stilted sentence structures.  in fact, the perfect word to describe it (which you doubtless are very familiar with) is sophomoric.  if you could make your posts a little more clear and a lot less unreadable that would be peachy keen!

p.s: one might say that the above is tinged with caustic manner.

Excuse me?  Well very much for keeping our relations affable sir (or ma'am).

You quoted me as using "inappropriate and obscure" adjectives.  I will hereby list all of the "inappropriate" adjectives I used and define them, to show you that they represented my ideas quite well.

grandiose - characterized by affection of grandeur or splendor by an absurd exaggeration.

I believe that the word grandiose is the only word which effectively describes feelings redolent of "touching the face of God" as the OP suggested in one of the poll choices.

congenial - having same nature or disposition

I didn't compare myself as a whole to the said poll choice, I compared only my views on math.  Since my views on math had the same characteristic qualities, I felt that congenial would more fit the sentence than a more vague word such as "agreeing" or the like

candid - marked by honest sincere expression

I used this word to contrast against the seemingly "superficial" motives the OP seemed to ascribe to people who viewed Math in a grandiose manner.  This word fits the breadth of the idea I was trying to represent better than another word, such as honest, or sincere, since those words wouldn't represent my motives as well.

Of couse my post was "tinged with caustic manner", it was meant to berate the OP's choice of poll options as being unrepresentative of our actual views.

I used parentheses at all correct occurences (as I could see), and upon proofreading, the ideas I had concerning the subject were all represented effectively.

If you shrug me off as being "sophomoric" simply because my sentence structure and diction weren't precisely as you thought would be correct, then, perhaps, we can debate elsewhere.  I'm sorry if my ideas were not presented clearly and effectively, because that was the full intention of my post.

If you have any more choleric bile to spew at me, I suggest you respond.

so I'm now going to into a short explanation of why your post was whack. this is obviously way off the topic but the original topic is stupid as it was obviously posted by some troll, so I don't care

"any phrase in rhetoric" ? what the hell is this supposed to be? how about "anything you say" ?

"very candid" ? already too convoluted, independent of the rest of that horribly convoluted sentence structure

yeah okay I can't really being myself to go farther


Let $x_n$ be a sequence of positive reals that increases quickly (for instance, $x_{n+1}\ge x_n^2$ or , at least $x_{n+1}\ge e^{x_n}$ ). Prove that $\sum_{n=1}^\infty \frac{1}{x_n}$ is irrational.($\in \mathbb{R}\setminus \mathbb{Q}$ ).

I thought more of this problem. The correct form is:

[b]Problem [/b]

Let $(a_n)_n$ be a sequence of natural numbers s.t. $a_n\mid a_{n+1}$, and $a_{n+1}\ge 2 n(n+1)a_n$. Prove that :
1) $\sum_{i=1}^\infty \frac 1 {a_n}$ is convergent.
2) Denote by $\ell=\sum_{i=1}^\infty \frac 1 {a_n}$. Prove that $\ell\in\mathbb{R}\setminus \mathbb{Q}$.

The conditions can be made sharper. I mean that maybe $a_{n+1}\ge n a_n$ ,if not less, would suffice.


Let n be a positive integer and let T denote the collection of points
$ (x_{1}, x_{2}, , . . . , x_{n} ) \in {R^n}$ for which there exists a permutation $ \sigma$ of 1, 2, . . . , n
such that $ x_{\sigma(i)}\minus{} x_{\sigma(i\plus{}1)}\ge1$  for each i = 1, . . . , n − 1. Prove that
there is a real number d satisfying the following condition: For every
$ (a_{1}, a_{2}, , . . . , a_{n} ) \in {R^n}$  there exist points $ (b_{1}, b_{2}, , . . . , b_{n} )$ and $ (c_{1}, c_{2}, , . . . , c_{n} )$ in T such
that, for each i = 1, . . . , n,
ai =1/2$ (b_{i}\plus{}c_{i} )$, |ai-bi|$ \le$d and  |ai-ci|$ \le$d

we can also write $ T\equal{}\{(x_1,x_2,..,x_n)\in R^n|\ \forall i\neq j: \ |x_i\minus{}x_j|\ge 1\}$


vector
trigonometry
linear algebra
matrix
quadratics
analytic geometry
linear algebra theorems

Given two unit vectors $\vec a$ and $\vec b$ in $\mathbb R^{n}$, I can compute the angle between them: $\cos\theta=\vec a\cdot\vec b$. Suppose I have two orthonormal $k$-tuples of vectors: $(\vec a_{i})_{i=1}^{k}$ and $(\vec b_{i})_{i=1}^{k}$, and I want to compute the angle between the $k$-dimensional subspaces spanned by these vectors. Will the Grammian determinant $\begin{vmatrix}\vec a_{i}\cdot \vec b_{j}\end{vmatrix}_{i,j=1}^{n}$ do the trick?  :maybe:

We're trying to (maximize? minimize?)

$\left(\sum\alpha_{i}\vec{a}_{i}\right)\cdot \left(\sum\beta_{i}\vec{b}_{i}\right)$

subject to $\sum\alpha_{i}^{2}=1,\ \sum\beta_{j}^{2}=1$

That is, we're trying to find the (maximum? minimum? what?) value (for unit vectors) of the quadratic form generated by the grammian matrix:

$\sum_{i,j}\alpha_{i}\beta_{j}(\vec{a}_{i}\cdot\vec{b}_{j})$

Or in matrix form: we have $A$ and $B$ that are $n\times k$ matrices, with the condition that $A^{T}A=B^{T}B=I.$

We're trying to (maximize? minimize?) $\alpha^{T}A^{T}B\beta$ over unit column $k$-vectors $\alpha$ and $\beta.$

Hmm... I don't know where this is going.

Let $A=\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}$ and $B=\begin{bmatrix}1&0\\0&0\\0&1\end{bmatrix}$

That's two coordinate planes in $\mathbb{R}^{3}$; they have a nontrivial intersection. I guess you'd want to say they're orthogonal.

In that case, $A^{T}B=\begin{bmatrix}1&0\\0&0\end{bmatrix}.$

The quantity I wanted to maximize or minimize is just $\alpha_{1}\beta_{1}.$

OK, it looks like max or min is the wrong point of view. The determinant looks good so far. (But then so would eigenvalue closest to zero.)

I need another example.

[quote] (maximize? minimize?) [/quote]
Both. :) The angle between subspaces $S_{1}$ and $S_{2}$ can be defined as $\max_{x_{1}\in S_{1}\setminus\{0\}}\min_{x_{2}\in S_{2}\setminus\{0\}}\theta(x_{1},x_{2})$, where $\theta(x_{1},x_{2})$ is the angle between two vectors.

Edit: [url=http://www.emis.de/journals/BAG/vol.42/no.1/20.html]this paper[/url] seems to answer my question. But I really should read "Finite dimensional vector spaces" by Halmos.

OK, that makes sense to me. What I wanted to do was to find

$\min_{\alpha}\max_{\beta}\alpha^{T}ATB\beta,$ where we restrict $\alpha$ and $\beta$ to be column $k$-vectors of length 1.

For the example above

$\min_{\alpha}\max_{\beta}\alpha_{1}\beta_{1}= \min_{\alpha}|\alpha_{1}|=0$

And that's the cosine of the angle, so these two subspaces are orthogonal.

Now to relate that minimax of a quadratic form to properties of the matrix. But the determinant doesn't seem to be a natural. Why not? Because why should be be restricting to both subspaces being the same dimension? In other words, why restrict $A^{T}B$ to just the square matrices?

The paper that I linked above defines the angle between subspaces spanned by $(\vec a_{i})_{i=1}^{p}$ and $(\vec b_{j})_{j=1}^{q}$ by $\cos\theta=\frac{\sqrt{det(MM^{T})}}{\sqrt{\Gamma_{1}\Gamma_{2}}}$, where $M=(\vec a_{i}\cdot \vec b_{j})$, $\Gamma_{1}=det(\vec a_{i}\cdot \vec a_{j})$, and $\Gamma_{2}=det(\vec b_{i}\cdot \vec b_{j})$. A [url=http://www.emis.de/journals/BAG/vol.46/no.2/2.html]more recent paper[/url] offers a "correction" of this formula. :huh:

But I think I understand this now: given subspaces $P$ and $Q$ of dimension $p$ and $q$, $p\le q$, let $\pi\colon P\to Q$ be the orthogonal projection. The Jacobian of the linear map $\pi$ is the cosine of the angle between subspaces. If $M$ is a (rectangular) matrix representing $\pi$ in some orthonormal bases, then the Jacobian is $\sqrt{\det(M^{T}M)}$, if I got the transpose right. (I guess the above-mentioned correction was about the transpose.)


---------
Hi, I think it (subject) has no algebraic solution, am I right?

Help: [b]k[/b] and [b]r[/b] are coprime and [b]r[/b] is not a prime number.

Who don't know what does [i]mod[/i] means, than see [url]http://en.wikipedia.org/wiki/Modulo_operation[/url]

Any Ideas?

[quote="ZetaX"]You have an algorithm to find a corresponding integer. But what does "no algebraic solution" mean for you here¿[/quote]
This means that I'd like to express the variable only with arithmetics operators. An algorithm is typically a algorithmic solution. But, now (after some sessions with Maple  :P  ) I'm quite sure that it's impossible. Thanx anyway.

[quote="jiniusazn"]well, i got it down to r(nr+1)/k = g where n = positive integers.
idk if that's what you wanted.[/quote]

I agree, but it is not finished. There is a question what the [i]positive integers[/i] should be. Because it is not true for all positive integers. Only for the specific ones.


In a 3cm*3cm square box, there is a circular spot of diameter 1 cm located at the centre of the box.
If I throw a coin of diameter 1.8 cm into the box, how likely will the coin fully cover the spot???

 :?:
Seeking for help... :(

$\frac{\pi}{9}$ if I didn't make stupid mistakes (as I always)

Solution please? :) This is hard...where did you get it from?

I don't think $\frac{\pi}{9}$ is right, because the coin is bigger than just 1 cm in diameter, but I don't get the answer either. Do you use the center of the coin?

I tried that, but I couldn't figure out where the center would have to be.  :?

There is a possible area of 2.1^2 or 4.41 sq. cm. Now how much is desired? I have a feeling it's $\frac{9\pi}{49}$

i got the same answer as leepakhin.  i hope i'm not misunderstanding the problem...

regardless of where the coin is thrown, its center will be somewhere in a $1.2$ by $1.2$ box in the middle of the square.  the circle will only be entirely covered by the coin if the furthest point on the coin is at most $.8$ away from the circle.  The center of the coin must be at most .9 away from the other side of the circle, i.e. within a circle of radius $.4$.  so probability is $\frac{.16\pi}{1.44}=\boxed{\frac{\pi}{9}}$

Whoops, I messed up about 2949423 times, maybe next time i'll realize also that the area of a circle is pi times [b]radius[/b] squared  :roll: . Just one of 1000 mistakes.


\[\text{if }f(x) \text{ is an irreducible polynomial in }\mathbb{Q}[x] \text{ and }g(x) \text{ is a polynomial in }\mathbb{Q}[x], \text{then }f(g(x)) \text{ has at most deg }g \text{ irreducible}\] \[\text{factors in }\mathbb{Q}[x] \]

Suppose f(g(x)) has more than deg g factors. Then there is one factor a(x)
of degree less than deg p. If t is some root of a(x) than degree of t above
Q is <deg p, so degree of s=g(t) is <deg p. But p(s)=0 and p
is irreducible. Contradiction!


inequalities
symmetry
search
inequalities proposed

Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ a^4\plus{}b^4\plus{}c^4\plus{}d^4\plus{}2abcd \ge a^2b^2\plus{}a^2c^2\plus{}a^2d^2\plus{}b^2c^2\plus{}b^2d^2\plus{}c^2d^2.\]

lets assume $ a\plus{}b\plus{}c\equal{}1$ and by symmetry $ d\le \min\{a,b,c\}$
then we will prove the stronger $ 3(a^4\plus{}b^4\plus{}c^4\plus{}d^4)\plus{}4abcd\geq (1\plus{}d)(a^3\plus{}b^3\plus{}c^3\plus{}d^3)$
Since $ 3abcd\geq d(a^2\plus{}b^2\plus{}c^2\minus{}2(a^3\plus{}b^3\plus{}c^3))$ by Schur then all we need to prove is
$ \left(3(a^4\plus{}b^4\plus{}c^4)\minus{}(a^3\plus{}b^3\plus{}c^3)\right)\minus{}d\left(3(a^3\plus{}b^3\plus{}c^3)\minus{}(a^2\plus{}b^2\plus{}c^2)\right)\plus{}d\left(abc\plus{}2d^3\minus{}d^2\right)\geq 0$
everything is clear considering that
A) $ 3(a^4\plus{}b^4\plus{}c^4)\plus{}(a^2\plus{}b^2\plus{}c^2)\geq 4(a^3\plus{}b^3\plus{}c^3)$ by AM GM
B) $ abc\plus{}2d^3\minus{}d^2\geq d^3\plus{}d^2(a\plus{}b\plus{}c\minus{}3d)\plus{}2d^3\minus{}d^2\equal{}0$ by Bernoulli

also if $ a\geq b\geq c\geq d$ then Darij gave the expression
$ (c \minus{} d)(c (a\minus{}b)^2\plus{} d(a^2 \plus{} b^2 \minus{} cd \minus{} d^2)) \plus{} (b \plus{} c)(b \minus{} c)(a^2\plus{}(a\minus{}b)^2\minus{}c^2) \plus{} a(a \minus{} b)^2(a \plus{} 2b)\geq 0$ equivalent to Turkevici's Inequality

and if $ a,b$ are the two maximal among $ \{a,b,c,d\}$ then Sung-Yoon Kim gave
$ (ab \plus{} cd \minus{} c^2 \minus{} d^2)^2 \plus{} 2cd(c \minus{} d)^2 \plus{} (a \minus{} b)^2((a \plus{} b)^2 \minus{} c^2 \minus{} d^2)\geq 0$
equivalent to Turkevici's inequality.

My favorite proof is the one I gave in http://www.mathlinks.ro/Forum/viewtopic.php?t=73107 (the inequality proven there is stronger than Turkevici, notably). I don't remember where I have it from, but I definitely did not come up with it on my own. http://www.mathlinks.ro/viewtopic.php?t=204175 gives a similar proof.

  darij

I was talking about
http://www.mathlinks.ro/viewtopic.php?search_id=1090443480&t=6397
where there's also a little IMO strategy in the end.  :D

[quote="can_hang2007"]Dear friends,
I have just found a solution by AM-GM (with only 5 lines) for the following known inequality
Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 2abcd \ge a^2b^2 \plus{} a^2c^2 \plus{} a^2d^2 \plus{} b^2c^2 \plus{} b^2d^2 \plus{} c^2d^2.
\]
But before I post it here, I would like to see your solutions.
Regards
:)[/quote]
Thank you very much, guys. :) Here is mine (ah, I forgot that I also used Cauchy Schwarz  :blush: )
Assume that $ a \ge b \ge c \ge d$, then we can rewrite our inequality in the form
\[ (a^2\minus{}c^2)^2\plus{}(b^2\minus{}c^2)^2\plus{}(a^2\minus{}d^2)^2\plus{}(b^2\minus{}d^2)^2 \ge 2(ab\minus{}cd)^2\]
Now, the Turkevici' Inequality becomes trivial since
\[ (a^2\minus{}d^2)^2\plus{}(b^2\minus{}d^2)^2 \ge \frac{1}{2}(a^2\plus{}b^2\minus{}2d^2)^2 \ge \frac{1}{2}(2ab\minus{}2d^2)^2 \ge \frac{1}{2}(2ab\minus{}2cd)^2\equal{}2(ab\minus{}cd)^2\]
Done! :)
This also shows that the Turkevici is not tight at all.

Another one line solution is to write the inequality in the form
\[ \sum_{cyc} a^4 \plus{} 2abcd \minus{} \sum_{sym} a^2b^2 \equal{} (c \minus{} d)^2(a \plus{} c \plus{} d)(c \plus{} d \minus{} a) \plus{} (d^2 \minus{} b^2)(c^2 \minus{} b^2) \plus{} a(b \minus{} a)(2cd \minus{} ab \minus{} a^2)
\]
which is clearly nonnegative for $ 0 \le a \le b \le c \le d$. :)

[quote="BG Yoda"]how should  we pronounce [i]"Turkevici"[/i] ?  :maybe:[/quote]
I think "Turkevichi", but we must write it "Turkevici".
See here:
http://dxdy.ru/topic9750.html#90038

[quote="Albanian Eagle"]lets assume $ a \plus{} b \plus{} c \equal{} 1$ and by symmetry $ d\le \min\{a,b,c\}$
then we will prove the stronger $ 3(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4) \plus{} 4abcd\geq (1 \plus{} d)(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3)$
Since $ 3abcd\geq d(a^2 \plus{} b^2 \plus{} c^2 \minus{} 2(a^3 \plus{} b^3 \plus{} c^3))$ by Schur then all we need to prove is
$ \left(3(a^4 \plus{} b^4 \plus{} c^4) \minus{} (a^3 \plus{} b^3 \plus{} c^3)\right) \minus{} d\left(3(a^3 \plus{} b^3 \plus{} c^3) \minus{} (a^2 \plus{} b^2 \plus{} c^2)\right) \plus{} d\left(abc \plus{} 2d^3 \minus{} d^2\right)\geq 0$
everything is clear considering that
A) $ 3(a^4 \plus{} b^4 \plus{} c^4) \plus{} (a^2 \plus{} b^2 \plus{} c^2)\geq 4(a^3 \plus{} b^3 \plus{} c^3)$ by AM GM
B) $ abc \plus{} 2d^3 \minus{} d^2\geq d^3 \plus{} d^2(a \plus{} b \plus{} c \minus{} 3d) \plus{} 2d^3 \minus{} d^2 \equal{} 0$ by Bernoulli[/quote]

Sorry, but I think that

$ 3(a^4 \plus{} b^4 \plus{} c^4) \plus{} (a^2 \plus{} b^2 \plus{} c^2)\geq 4(a^3 \plus{} b^3 \plus{} c^3)$

is false.  :wink:

Yes Manlio thank you for pointing it out!
Indeed I thought that its enough to use $ d\le 1$ but indeed we can fix it by using $ d\le \frac{1}{3}$
and we need to use
A') $ 3(a^4\plus{}b^4\plus{}c^4)\minus{}(a^3\plus{}b^3\plus{}c^3)\geq (a^3\plus{}b^3\plus{}c^3)\minus{}\frac13(a^2\plus{}b^2\plus{}c^2)$
and now its true and trivial  :wink:

[quote="Albanian Eagle"]Yes Manlio thank you for pointing it out!
Indeed I thought that its enough to use $ d\le 1$ but indeed we can fix it by using $ d\le \frac {1}{3}$
and we need to use
A') $ 3(a^4 \plus{} b^4 \plus{} c^4) \minus{} (a^3 \plus{} b^3 \plus{} c^3)\geq (a^3 \plus{} b^3 \plus{} c^3) \minus{} \frac13(a^2 \plus{} b^2 \plus{} c^2)$
and now its true and trivial  :wink:[/quote]

Thank you very much. :)


Show that among any 9 points in a triangle of area 1, there are 3 points that form a triangle of area at most 1/4. (b) Show that given any 9 points in a triangle of area 1, there is a triangle of area at least 1/12 that does not contain any of those 9 points in its interior. (Can you improve 1/12?)

[hide="Solution"]
Given $ \triangle ABC$ choose $ D$ midpoint of $ AB$. Choose $ E$ midpoint of $ BC$. Choose $ F$ midpoint of $ AC$. Now there is a theorem that $ \triangle DBE = \triangle ADF = \triangle DEF = \triangle FEC$, so each one have area $ 1/4$. Given 9 point by strong pigeonhole $ [9/4]+1$ end up in the same "subtriangle". Now that subtriangle is enclosed by $ \triangle ADF$ (without lose of generality) so its area is [b]at most[/b] $ 1/4$.
[/hide]


whats the difference between general theory of relativity and special relativity?

Educate yourself. Dont stay in the dark!
http://en.wikipedia.org/wiki/Theory_of_relativity


is it possible to have
$a,b,c,d \in \mathbb{N}$ with $a \neq c$ and $b \neq d$
and also
$ab(a-b)=cd(c-d)$ ?

That should be "with $a\neq c$ and $b\neq d$". And also, are $a,b,c,d$ natural? integer? real? complex?

If $a,b,c,d$ are integer, then an example is $(-2,3,-5,1)$

how do you construct examples like that one farenhajt ?
yes $a \neq c$ and $b \neq d$

Simple. I assumed $a-b=c$, and after that I arbitrarily took $d=1$. This gives $b=\frac{a-1}{a+1}$

Generally, any quadruplet $(a, 2a^2+3a+1, -2a^2-2a-1, -a-1)$ or $(a, -2a^2+3a-1, 2a^2-2a+1, 1-a)$ is a solution. Of course, these are not all solutions.


[quote="GameBot"]An 8.5-by-11-inch piece of paper is folded in half repeatedly (never being unfolded), each time shortening the then longer side. What is the length of the longest side, in inches, immediately after the second fold? Express your answer as a decimal to the nearest tenth.[/quote]

0 folds: 8.5*11
1 fold: 8.5*5.5
2 folds: 4.25*5.5

So 5.5 is our answer.


5n^3 + 7n^5

where n is any natural number.

It's related to a problem in Apostol's Introduction to Analytical Number Theory; I can't seem to figure out why it's divisible by 12 for all n.


geometry
circumcircle
geometric transformation
reflection
power of a point
radical axis
geometry unsolved

Let be given a triangle $\Delta ABC$ in the plane. Find the locus of the points $M$ in the plane such that the power of $A$ with respect to the circumcircle of triangle $BMC$ equals to the power of $B$ with respect to the circumcircle of triangle $CMA$ and equals to the power of $C$ with respect to the circumcircle of triangle $AMB$.

Let D be the midpoint of BC. Let $(O_C), (O_B)$ be the circumcircles of the triangles $\triangle AMB, \triangle CMA$. The locus of points M, such that power $p(B, (O_B))$ of the vertex B to the circle $(O_B)$ is equal to the power $p(C, (O_C))$ of the vertex C to the circle $(O_C)$ is the median line AD. Indeed, let the circles $(O_C), (O_B)$ meet the line BC at K, L different from B, C. Since the midpoint D is on the radical axis of  the circles $(O_C), (O_B)$, the power of D to these 2 circles is the same and equal to

$DL \cdot DC = DK \cdot DB$

Since DC = DB, it follows that DK = DL and BK = CL. But the powers of the vertices B, C to the circles  $(O_B), (O_C)$ are $p(B, (O_B)) = BC \cdot BK$ and $p(C, (O_C)) = CB \cdot CL$, i.e., the same. By the same reasoning, if M is not on the median AD, but the points K, L are defined as before, then AM meets BC at a point D' different from D. Since $D'B \neq D'C$, it follows that $D'K \neq D'L$ and $BK \neq CL$ and consequently, $BC \cdot BK \neq CB \cdot CL$. Let $(O_A)$ be the circumcenter of the triangle $\triangle BMC$ and $p(A, (O_A))$ the power of the vertex A to this circle. From the transitivity of equivalence, it follows that the only point such that the powers $p(A, (O_A)) = p(B, (O_B) = p(C, (O_C)$ are equal is the triangle centroid G.
Q.E.D.

The power of one vertex, say A, to the circle $(O_A)$ is easy to calculate. Let the circle $(O_A)$ meet the median line $AG \equiv AD$ at a point P different from G. The power of the midpoint D to the circle $(O_A)$ is

$DB \cdot DC = DG \cdot DP$

$DP = \frac{DB \cdot DC}{DG} = \frac{3a^2}{4m_a}$

where a = BC is the side of the triangle $\triangle ABC$ opposite to the vertex A and $m_a = AD$ the A-median. The power of the vertex A to the circle $(O_A)$ is equal to

$p(A, (O_A)) = AG \cdot AP = AG \cdot (GD + DP) =$

$= \frac{2 m_a}{3} \left(\frac{m_a}{3} + \frac{3a^2}{4m_a}\right) = \frac{2 m_a^2}{3} + \frac{a^2}{2} = \frac{2b^2 + 2c^2 - a^2}{6} + \frac{a^2}{2} = \frac{a^2 + b^2 + c^2}{3}$

which is an expression symmetrical to the cyclic exchange of the sides a = BC, b = CA, c = AB, as expected.

Actually, there are 3 other points $M_1, M_2, M_3$ such that the power $p(A, (O_{A1})) = - p(B, (O_{B1})) = - p(C, (O_{C1}))$,  $p(B, (O_{B2})) = - p(C, (O_{C2})) = - p(A, (O_{A2}))$,  $p(C, (O_{C3})) = - p(A, (O_{A3})) = - p(B, (O_{B3}))$. The points $M_1, M_2, M_3$ are the reflections of the vertices A, B, C in the midpoints D, E, F of the sides a = BC, b = CA, c = AB, so that the triangle $\triangle ABC$ is the medial triangle of the triangle $\triangle M_1M_2M_3$. These do not satisfy the problem condition, because of the opposite sign of one of the powers, but they are interesting.

For example, the locus of points $M$, such that the powers $p(C, (O_{C})) = -p(B, (O_{B1}))$ of the vertices C, B to the circumcircles $(O_{C}), (O_{B})$ of the triangles $\triangle AMB, \triangle CMA$ have equal magnitudes but opposite signs is a line parallel to BC through the vertex A. Let M be a point on this parallel and let the circles $(O_{C}), (O_{B})$ meet the line BC at points K, L different from B, C. Since AMis the radical axis of the circles $(O_C), (O_B)$ and $BC \parallel AM$, the points B, K are symmetrical with respect to the center line $O_CO_B$ and the points C, L are also symmetrical with respect to this center line. Hence, for oriented segments, $BC = -KL,\ BL = BK + KL = BK - BC = CB + BK = CK$. Therefore, the power of the vertex C to the circle $(O_C)$ is equal to the power of the vertex B to the circle $(O_B)$ in magnitude. But since either the vertex C outside of the circle $(O_C)$ and the vertex B inside of the circle $(O_B)$ or the other way around (depending on which side of the pvertex A the point M lies on the parallel through A with BC), these powers always have opposite signs (unless the points $M \equiv A$ are identical, in which case the both the circles $(O_C), (O_B)$ coincide with the circumcircle of the triangle $\triangle ABC$, but then the circle $(O_A)$ is undefined):

$p(C, (O_C)) = CB \cdot CK = -BC \cdot BL = -p(B, (O_B))$

If the point M coincides with the reflection $M_3$ of the vertex C in the midpoint F of AB, which obviously lies on the parallel through A to BC, then

$p(C, (O_C)) = -p(B, (O_B)) = -p(A, (O_A))$

The power of the vertex C to the circle $(O_C)$ can be calculated similarly as before, when M was identical with the centroid G of the triangle $\triangle ABC$. Let Q be the other intersection of the circle $(O_C)$ with the median line $CG \equiv CF \equiv CM_3$. The power of the midpoint F to the circle $(O_C)$ is

$FA \cdot FB = FM_3 \cdot FQ$

$FQ = \frac{FA \cdot FB }{FM_3} = \frac{FA \cdot FB}{CF} = \frac{c^2}{4m_c}$

where $m_c = CF$ is the C-median of the triangle $\triangle ABC$. The power of the vertex C to this circle is then

$p(C, (O_C)) = CQ \cdot CM_3 = (CF - FQ) \cdot 2CF =$

$= 2m_c \left(m_c - \frac{c^2}{4m_c}\right) = 2m_c^2 - \frac{c^2}{2} = \frac{2a^2 + 2b^2 - c^2}{2} - \frac{c^2}{2} = a^2 + b^2 - c^2$