[quote="Interval"]Our alphabet contains 26 letters. Five of these letters are vowels (A, E, I, O, U). The remaining 21 letters are called consonants. Each letter of the alphabet is written on a card and you are asked to choose a card. What is the probability of choosing a consonant?[/quote]
[hide]There are 21 consonants and 26 letters total, so the probability is 21/26 which is in lowest terms.[/hide]

[quote="Interval"]Our alphabet contains 26 letters. Five of these letters are vowels (A, E, I, O, U). The remaining 21 letters are called consonants. Each letter of the alphabet is written on a card and you are asked to choose a card. What is the probability of choosing a consonant?[/quote]
that answer actually is [hide]
20/25 when y is considered a vowel which it is most of the time[/hide]
[quote="Interval"] cannot speak for other students. In MY case, the questions posted here are questions that I personally tried answering for over 30 minutes. There is a small quiz at the end of the study book that I own with 25 questions.

I simply post questions after getting most of the quiz questions right. The questions here are the ones I considered hard and thus, should be approached by people who are gifted in math or better prepared for upper level high school mathematics. [/quote]
umm and *cough cough* this is easy

[quote="funcia"]
that answer actually is
20/25 when y is considered a vowel which it is most of the time [/quote]
He specified what the vowels are so y is considered a consonant

[quote="ProtestanT"][quote="funcia"]that answer actually is [hide]
20/25 when y is considered a vowel which it is most of the time[/hide]
[/quote]
if y is considered a vowel, wouldnt it be $\frac{20}{26}$ instead of $\frac{20}{25}$?[/quote]
yeah typo
i am not using ten key
this is an annoying keyboard


inequalities
calculus
geometry
geometric transformation
algorithm
absolute value
superior algebra

Almost; what you did was show that if $ x \ge 0$ then $ x < y$, and that if $ x < 0$ then $ x > \minus{} y$. You did not show that $ x > \minus{} y$ when $ x \ge 0$, nor did you should that $ x < y$ when $ x < 0$; this is solved by remarking that $ y$ is necessarily positive given that $ |x| < y$.

ADDN: It is somewhat simpler to show that $ x \le |x|$, whence $ x \le |x| < y$, while $ \minus{} y < \minus{} |x| \equal{} \minus{} | \minus{} x| \ge \minus{} ( \minus{} x) \equal{} x$.
(Pf: $ |x| \ge 0$, so if $ x < 0$ then $ x < 0 \le |x|$. If $ x \ge 0$, then $ |x| \equal{} x$ so $ x \le |x|$.)

[quote="Nukular"]Almost; what you did was show that if $ x \ge 0$ then $ x < y$, and that if $ x < 0$ then $ x > \minus{} y$. You did not show that $ x > \minus{} y$ when $ x \ge 0$, nor did you should that $ x < y$ when $ x < 0$; this is solved by remarking that $ y$ is necessarily positive given that $ |x| < y$.

 [quote]

Why do you have to show that : x>-y ,when $ x\geq 0$,and

 x<y ,when x<0

Because you want to show that $ \minus{}y < x < y$, for every $ x$, which means you must show that $ \minus{}y < x$ and $ x < y$ for every $ x$. You appear to have shown $ x \ge 0 \implies x < y$ and $ x < 0 \implies \minus{}y < x$; this is not equivalent to $ \minus{}y < x < y$ for every $ x$. You only need some note as to $ x \ge 0 \implies \minus{}y < x$ being trivial (as is $ x < 0 \implies x < y$).

[quote="Nukular"]Because you want to show that $ \minus{} y < x < y$, for every $ x$, which means you must show that $ \minus{} y < x$ and $ x < y$ for every $ x$. You appear to have shown $ x \ge 0 \implies x < y$ and $ x < 0 \implies \minus{} y < x$; this is not equivalent to $ \minus{} y < x < y$ for every $ x$. You only need some note as to $ x \ge 0 \implies \minus{} y < x$ being trivial (as is $ x < 0 \implies x < y$).[/quote]

But [b]why[/b] must we show : -y<x<y for $ x\geq 0$,instead  of x<y ,and

again ,-y<x<y for x<0, instead of -y<x.

x<y and -y<x give us -y<x<y again.

I do not see any logical   or mathematical  mistake in my proof

Because, you showed that $ x < y$ when $ x \ge 0$, and you showed $ \minus{}y < x$ when $ x < 0$. For no value of $ x$ did you simultaneously show that $ x < y$ [i]and[/i] $ \minus{}y < x$.

The goal is to show that $ \minus{}y < x < y$. If your approach is to split into cases based on $ x$, you must show, [i] for every case [/i], this goal. You cannot just show half the goal for one case of $ x$ and the other half of the goal for some other case of $ x$.

[quote="Nukular"]Because, you showed that $ x < y$ when $ x \ge 0$, and you showed $ \minus{} y < x$ when $ x < 0$. For no value of $ x$ did you simultaneously show that $ x < y$ [i]and[/i] $ \minus{} y < x$.

The goal is to show that $ \minus{} y < x < y$. If your approach is to split into cases based on $ x$, you must show, [i] for every case [/i], this goal. You cannot just show half the goal for one case of $ x$ and the other half of the goal for some other case of $ x$.[/quote]

Surely , "half goal"  is not a logical or mathematical explanation

You showed $ x < y$ for $ x \ge 0$, and $ \minus{}y < x$ for $ x < 0$. You did not show $ \minus{}y < x$ for $ x \ge 0$, nor did you show $ x < y$ for $ x \ge 0$. Your claim was to show that $ \minus{}y < x$ and $ x < y$ for every $ x$. You did not do this. Therefore your proof is incomplete.

I really hope, for your case, you get past arguing about formalities, unless you want to argue about trivial absolute value inequalities for the rest of your mathematical career  :rotfl:.

[quote="Nukular"]

I really hope, for your case, you get past arguing about formalities, unless you want to argue about trivial absolute value inequalities for the rest of your mathematical career  :rotfl:.[/quote]

Again this is not a logical or mathematical explanation.

For someone to say that a mathematical proof is wrong ,he/she must pin point a certain law of logic or an axiom or a theorem or a definition in the proof which is wrong .

So i am asking you again which of the above is wrong in my proof?

[quote]
For someone to say that a mathematical proof is wrong ,he/she must pin point a certain law of logic or an axiom or a theorem or a definition in the proof which is wrong .
[/quote]

If you want me to judge your proof like it's a formal proof, then you should [i]provide[/i] a formal proof to be judged.

Also, [i]good luck[/i] trying to have a mathematically rigorous discussion of a proof; at the end of the day, there's no way to verify the correctness of the symbolic logic that this kind of discussion would reduce to, since people are fallible. We'd need to either feed it into a computer that we trust can do computations (or find some other near-infallible arbiter).

But, to humor you:

[quote]
Hence from (1) and (2) -y<x<y
[/quote]

For which values of $ x$ does this hold? (1) was derived in the case $ x \ge 0$, and (2) was derived from the case $ x < 0$.

[quote="Nukular"][quote]

But, to humor you:

[quote]
Hence from (1) and (2) -y<x<y
[/quote]

For which values of $ x$ does this hold? (1) was derived in the case $ x \ge 0$, and (2) was derived from the case $ x < 0$.[/quote][/quote]

Hence -y<x<y holds for all values of x

[quote="τρικλινο"][/quote][quote="Nukular"][quote]

But, to humor you:

[quote]
Hence from (1) and (2) -y<x<y
[/quote]

For which values of $ x$ does this hold? (1) was derived in the case $ x \ge 0$, and (2) was derived from the case $ x < 0$.[/quote]

Hence -y<x<y holds for all values of x[/quote][quote="τρικλινο"][/quote]

How, exactly, do the statements $ x \ge 0 \implies x < y$ and $ x < 0 \implies - y < x$ combine to form $ x \in \mathbb{R} \implies ( - y < x \; \mathrm{ AND } \; x < y)$?

EDIT: They do, by the way, but you just need to write it out to make the proof complete, otherwise you end up in hsiljak's situation.

[quote="Nukular"][/quote][quote="τρικλινο"][/quote][quote="Nukular"][quote]

But, to humor you:

[quote]
Hence from (1) and (2) -y<x<y
[/quote]

For which values of $ x$ does this hold? (1) was derived in the case $ x \ge 0$, and (2) was derived from the case $ x < 0$.[/quote]

Hence -y<x<y holds for all values of x[/quote][quote="τρικλινο"]

How, exactly, do the statements $ x \ge 0 \implies x < y$ and $ x < 0 \implies - y < x$ combine to form $ x \in \mathbb{R} \implies ( - y < x \; \mathrm{ AND } \; x < y)$?

EDIT: They do, by the way, but you just need to write it out to make the proof complete, otherwise you end up in hsiljak's situation.[/quote][quote="Nukular"][/quote]

Our difference is that you combine -y<x<y and [b] -y<x<y [/b] into -y<x<y,while i combine -y<x and x<y into -y<x<y

Both proofs are correct

B.T.W can you check both proofs thru the computer to see if they are correct??

[quote="τρικλινο"][quote="hsiljak"]Let A be a set of apples, O a set of oranges, and U the union of those two.

For each x, member of set A, x is an apple .................. (1)
For each x, member of set O, x is an orange ............... (2)

Hence from (1) and (2) every x from the set U is both an apple and an orange.[/quote]

Not a very good analogy.

Surely  each member   of U is an apple or an orange and not an orange and an apple[/quote]

Well, I could have made a better analogy, but this one serves the purpose, because it resembles your proof.

In your proof, you have two disjunctive sets $ \mathbb{R}^{\plus{}}\cup\{0\}$ and $ \mathbb{R}^{\minus{}}$ (just like my $ A$ and $ O$). You make a conclusion (1) about the members of the first, and conclusion (2) about the members of the second set, and then, in the end you say that conclusions (1) and (2) hold simultaneously for all members of the universe (their union).

[quote="τρικλινο"][/quote][quote="Nukular"][/quote][quote="τρικλινο"][/quote][quote="Nukular"][quote]

But, to humor you:

[quote]
Hence from (1) and (2) -y<x<y
[/quote]

For which values of $ x$ does this hold? (1) was derived in the case $ x \ge 0$, and (2) was derived from the case $ x < 0$.[/quote]

Hence -y<x<y holds for all values of x[/quote][quote="τρικλινο"]

How, exactly, do the statements $ x \ge 0 \implies x < y$ and $ x < 0 \implies - y < x$ combine to form $ x \in \mathbb{R} \implies ( - y < x \; \mathrm{ AND } \; x < y)$?

EDIT: They do, by the way, but you just need to write it out to make the proof complete, otherwise you end up in hsiljak's situation.[/quote][quote="Nukular"]

Our difference is that you combine -y<x<y and [b] -y<x<y [/b] into -y<x<y,while i combine -y<x and x<y into -y<x<y
[/quote]

You combine $ - y < x$ and $ x < y$ into $ - y < x < y$, when $ - y < x$ and $ x < y$ were arrived at [i] for different values of $ x$ [/i]. You're not allowed to do this.

[quote]
Not a very good analogy.

Surely each member of U is an apple or an orange and not an orange and an apple
[/quote]
I believe hsiljak's point was that your proof proceeds along the same lines; since his conclusion is clearly false, your proof is wrong.

[quote]
B.T.W can you check both proofs thru the computer to see if they are correct??
[/quote][quote="τρικλινο"][/quote]

If you provide me a formal proof that's computer checkable, I sure can, although I'm sure you know how to use a computer :). There are a variety of software packages that are capable of this, and if you provide it in a format that one will accept, I'll be perfectly happy to run the program with your proof as input.

ADDN:
Here, this should settle things:

From which statement, (1) or (2), does it follow that if $ |3| < y$ then $ 3 > - y$?

Certainly not (1), since (1) says $ 3 < y$. Certainly not (2), since $ 3 \not < 0$.

Or how about this?

For $ x \ge 0$, we know $ x > \minus{}1$
For $ x < 0$ we know $ x < 1$.

Therefore $ \minus{}1 < x < 1$ for every $ x$.

[quote="τρικλινο"]p=(x\geq 0) =======> r=( x<y)

q=(x<0) =======> s =( -y<x )

Again from logic we have the following law

(p or q)======> (r & s)[/quote]
Absolutely not. That "logic" is completely bogus. "Or" gives you "or", not "and" - and "r or s" is utterly worthless.

If you were paying attention, you'd notice that hsiljak and Nukular have already proven that, by deriving blatantly false results from that "logic" in posts 12 and 18, along with other comments.

Should I lock this thread now?

[quote="jmerry"]

Absolutely not. That "logic" is completely bogus. "Or" gives you "or", not "and" - and "r or s" is utterly worthless.[/quote]

Here is an example to show you the opposite:

Let

p= i go to the movies

q= i go to the theatre

r= i  buy a pepsi

s= i buy a chocolate

I say:

if i go to the movies i will buy a pepsi  ( p===>r)

if i go to the theatre i will buy a chocolate ( q===>s)

 Therefor if i go to the movies or to the theatre ( p or q) what would prevent me to buy a pepsi [b] and [/b] a chocolate (r & s)

So  (p or q) can imply ( r & s)

Because 'or' in our case does not exclude the possibility of "and" i.e going to the movies and to the theatre

So "or "does not [b] always [/b] give you "or"

you must check up your logic

First, I have to apologize to others reading this thread for feeding the troll, but I have to try explaining some things.

Triklino, your argument has no sense whatsoever.

OK: you went to the cinema, and you bought both pepsi and a chocolate. But that has nothing to do with the logical line you made there, the logic only implied that you have to buy a pepsi, it said nothing about the chocolate in that case whatsoever, i.e. the boolean variable "did I buy the chocolate" denoted by 's' doesn't play a role in the particular case. You didn't have to buy the chocolate, pepsi would have been enough.

If you have problems with this intuitive approach, take a truth table. It's exact, formal, and fast way of seeing whether a logical expression is a tautology or not.

[quote] Therefor if i go to the movies or to the theatre ( p or q) what would prevent me to buy a pepsi [b] and [/b] a chocolate (r & s)

So  (p or q) can imply ( r & s)
[/quote]
"Can imply" is not good enough :wink:

What you have proven is that your original claim [b]can[/b] be true. (See I'm using your own language). Now if you're satisfied with that why is your question in the original post asking for something more?
If I were you I would pause and take a break for some time from "formal" mathematics/logic or whatever it is you are claiming of achieving with your posts. You clearly have some logic gaps (and I mean middle school /high-school deductive logic not mathematical logic.) It is either that or you're just way too hardheaded, in which case it is probably time for you to learn that in mathematics or science in general, being able to conduct calm, logical, fruitful, sound discussions/correspondence is an important skill to have. I apologize if I sounded antagonistic but it's just that reading the above it seems that you make no effort at all into understanding other people's replies.

[quote="jmerry"]Should I lock this thread now?[/quote]  Yes: when someone insists (in the course of demanding that every statement be given a formal proof) that it's perfectly okay to replace $ a \vee b$ by $ a \wedge b$ because $ a \wedge b$ isn't necessarily in contradiction with $ a \vee b$, and then defends this position repeatedly, I think it's safe to say that productive conversation has ended :)

You all seem to be missing one important fact:

[b] It was not me who discovered ,formal proofs[/b].

All i did was to ask,thru my threads, if anyone of you can write a [b] single [/b] formal proof. To my astonishment none of you could .

Now instead  of learning or at least trying to learn how to write one ,which is the atitude of a well behaved mathematician you started mocking me and nastily criticising me ,take for example the recent comment of the Albanian Eagle:
"if i were you i would pause and take a break for some time from "formal" mathematics/logic,or [b] whatever[/b] ( the emphasis is mine) it is you are claiming of achieving with your posts".

B.T.W Albanian Eagle i challenge you in any topic in symbolic logic to find out who has logic gaps.

And other comments that one can find  in many of my threads.

[b]Do you not understand that all those comments and nasty remarks are against those brilliant men that discovered formal proofs and not against me ,making a ridicule of yourselfs [/b]?

What formal proofs were discovered for??

Now in this thread by presenting you with a proof ,which i found in a book, i just wanted to show that unless one writes a formal proof or at least a semi formal one can argue indefinitely.

And i decided to take the wrong side and argue successfully or unsuccessfully.

Remember i was the one who wrote the semi formal proofs ( skeleton proofs) for mine and Nukular's clearly showing the main difference between the two proofs .

No one detected that.

But leaded by passion and low incentives you decided it was time to attack me  and even lock the thread to show how unproductive i was, to use the expression of my friend JBL.

Yes only Nukular remained calm answering only to my arguing.

In mathematics and logic if one opposes you even with nonsenses you have to [b] prove that[/b],otherwise you running the risk of calling each other's arguing simple nonsenses

And proof in propositional calculus is not achieved by showing examples

Only hsiljak indicated a simentical proof by using truth tables

That being said, you propose the implication:

$ p \implies r (1)$
$ q \implies s (2)$

therefore

$ p \vee q \implies r \wedge s (3)$

I will let $ t$ denote the true statement, and $ f$ the false statement.

Then $ (f \implies f) \equal{} t$, and $ (t \implies t) \equal{} t$, therefore the assignment $ p \equal{} r \equal{} f$, $ q\equal{}s\equal{}t$, satisfies the hypotheses (1) and (2).

But then $ p \vee q \equal{} f \vee t \equal{} t$, and $ r \wedge s \equal{} f \wedge t \equal{} f$, so (3) becomes $ (t \implies f) \equal{} f$.

So there is an assignment of truth values to $ p,q,r,s$ that makes (1) and (2) true, but makes (3) false; this is a contradiction.

Clearly the point of "formal proofs" is to argue about false tautologies in propositional calculus  :roll:

[quote]
Now in this thread by presenting you with a proof ,which i found in a book, i just wanted to show that unless one writes a formal proof or at least a semi formal one can argue indefinitely.

And i decided to take the wrong side and argue successfully or unsuccessfully.

Remember i was the one who wrote the semi formal proofs ( skeleton proofs) for mine and Nukular's clearly showing the main difference between the two proofs .
[/quote]

That is where you and I disagree. I certainly recognize that without formalism, you can argue ad infinitum about the correctness of a proof. However, you seem to put forward that this means all proofs should be reduced to the formal proof level to begin with. I believe that proofs should be [i]rigorous[/i] -- I would define [i] rigorous [/i] as "being able to be reduced to further formalism." That is, everyone disagreed with your combinations of steps (1) and (2) into $ \minus{} y < x < y$; [i] at that point [/i], and no earlier, should you "open up" that part of the proof and provide the deeper formalism to convince us that you were correct, or to allow us to find which step in that opened up subproof is incorrect.

After that translation into a full formal proof, I didn't even have to look at all the formalism in the skeleton besides the one step I knew was wrong to begin with. It makes all that other formalism unnecessary -- the only [i] necessary [/i] piece of formalism was that which we asked you to elaborate on.

And that's what I (and many) consider to be a proof: an argument which [i] can be reduced [/i] to further formalism, if necessary, until reaching the level of axioms, logic, and established theorems and lemmas... because if it's easy for the reader to fill in the details, what's the point of doing all the extra writing, and forcing the reader to skim over what he already knows?

I spent many years making my proofs "too" rigorous, and they'd take up extra pages. It's just [i]not necessary[/i] for practical communication. Some level of formalism is necessary, and in fact our collective confidence in the truth-telling of such formalism is what allows mathematics to proceed forward, but it's a low-level concept, like binary operations done in the hardware of a CPU; programmers shouldn't have to worry about that when implementing algorithms, nor should mathematicians [b] have [/b] to worry about formalism when thinking about and manipulating high-level concepts. Yes, it needs to be there somewhere in the literature for everything to have a logical basis, but once it's established, move on and get proving!

(By the way, I [i] highly [/i] recommend reading Lakatos' [i] Proofs and Refutations [/i] for an thought-provoking discussion of the process of mathematical discovery and proof.)


The first figure in this pattern is a 2 by 2 square, with an area of 4 square units. In the second figure, a congruent square is placed behind the the first square such that the midpoints of the left and bottom sides intersect at the midpoints of the top and right sides, respectively, of the first square. In each successive figure, a congruent square is placed behind the preceding, intersecting in the same way. What will be the area of the complete region of the fifth figure in this pattern?

[asy]draw((0,0)--(0,10)--(10,10)--(10,0)--cycle,linewidth(1));
label("Figure 1",(5,0),S);

draw((20,0)--(30,0)--(30,5)--(35,5)--(35,15)--(25,15)--(25,10)--(20,10)--(20,0)--cycle,linewidth(1));
draw((25,5)--(30,5)--(30,10)--(25,10)--(25,5)--cycle,dashed);
label("Figure 2", (27.5,0),S);

draw((45,0)--(55,0)--(55,5)--(60,5)--(60,10)--(65,10)--(65,20)--(55,20)--(55,15)--(50,15)--(50,10)--(45,10)--(45,0)--cycle,linewidth(1));
draw((50,5)--(55,5)--(55,10)--(50,10)--(50,5)--cycle,dashed);
draw((55,10)--(60,10)--(60,15)--(55,15)--(55,10)--cycle,dashed);
label("Figure 2", (55,0),S);[/asy]

area of nth figure : 4+3(n-1)

n=5

4+3*4=16

answer:16

[quote="FantasyLover"]area of nth figure : 4+3(n-1)[/quote]

The unexplained reasoning being:  each time a square is added, you gain 3 units of area.

so for every square that is added, 3 more units is added so it would be 3n-3. You started out with 4 units, so the equation would be 4+3n-3 or 3n+1


Prove that for positive reals $ a,b,c$ with sum 2 ,
$ \frac {a}{b(a \plus{} b)} \plus{} \frac {b}{c(b \plus{} c)} \plus{} \frac {c}{a(a \plus{} c)} \ge 2$

$ LHS\ge \frac {4}{ \sum_{sym} a^2b}$.

So we only need to prove that:
$ 2\ge \sum_{sym}a^2b \Longleftrightarrow (a\plus{}b\plus{}c)^3 \ge 4(\sum_{sym}a^2b)$
which is true clearly true.

ok..
By AM-GM,
$ \sum_{cyclic}\frac {a}{b(a \plus{} b)} \plus{} (a \plus{} b)a \plus{} ab \ge 3a$
which is equicalent to
$ \sum_{cyclic}\frac {a}{b(a \plus{} b)} \ge 6 \minus{} 4 \equal{} 2$
the inequality also holds for the condition $ a\plus{}b\plus{}c\equal{}1$

[quote="Pain rinnegan"]Equality can't occur .[/quote]
right nice spotting pain.


A(-2,3,1) B(2,3,4) C(-8,11,1) find the outside bisector of angle A.

Hi anuonhu. I'm not sure that I understand the question, nor do I know a lot about geometry in three dimensions. However, drawing a sketch, it looks to me that the two angle bisectors, internal and external, are such that, for each of them, the perpendicular distance from a point $ (p, q, r)$ on it to each of the lines $ AB$ and $ AC$ are equal. It is easy to get vector equations of the lines $ AB$ and $ BC$. Then we could use the fact that the line joining $ (p, q, r)$ to $ AB$ is perpendicular to $ AB$ and so the dot product (scalar product) of their direction vectors is zero; and similarly for the line joining $ (p, q, r)$ to $ AC$. Hopefully (I haven't tried it yet) this will give simultaneous equations in $ p, q$ and $ r$ which should give two possibilities. Then the angle bisectors are the lines joining $ (p, q, r)$ to $ A$. I hope this is of some help.

If you consider that the external bissector of angle A will be in the plane of the triange then the answer is [x,(13-4x)/7,(3x+13)/7]


Prove that there are infinitely many natural numbers a with the following propoerty: The number n^4 + a is not prime for any natural number n.

I think this should go in the "Number Theory" section. Anyway, take $a=4k^4$. Then $n^4+4k^4=(n^2-2nk+2k^2)(n^2+2nk+2k^2)$. This can't be a prime unless $n^2-2nk+2k^2=1$, but this can be prevented from happening by choosing $k$ to be large enough ($n^2-2nk+2k^2=(n-k)^2+k^2\ge k^2$, so it's enough to have $k\ge 2$).


A palindrome is a number that reads the same forward and backward. The year 1991 was the last palindromic year of the twentieth century. How many years between 2000 and 3000 are palindrome?


linear algebra
matrix
linear algebra unsolved

I am only allowed to upload 3 images so i attached a pdf document.
there are 3 unsolved extra simple questions.
Can you please help me with these ones?
please  show me the way of the solution so I will actually learn something.
Thanks alot in advance.

Very strange.   :roll:   I just answered question number two last night on this same forum.  You can read my response here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161500[/url]

In general, if it is homework just admit to it up front.  I do not have an issue helping you, but I do not want to do all of your work for you either.

It is "homework".
I definitely dont want you to do all the black job,but i really need direction for question number 3.
thank you for your time.

Okay, so question $ 3$ basically asks one to conclude that the matrix:

$ \left[ \begin{array}{cc}A&B\\\noalign{\medskip}0&C\end{array}\right]$

is nonsingular if both $ A$ and $ C$ are.

Do you know how to take the determinant of a $ 2 \times 2$ block matrix?  If so, the answer is immediate.  If not, make a reasonable guess and try to prove it.


1. If set A ={ 26, 29, 34, ...., n^2+25,...}
      set B = {13501, 13504, ...., n^2+13500,....}
   how many common elements do the two sets have?

2. What is the smallest positive integer with 6 positive odd integer divisors and 12 positive even integer divisors?

3. In the equation, BIGGER+AND=BETTER, BIGGER and BETTER are 6 digit numbers and AND is a 3digit number. Each letter represents a particular digit (not necessarily distinct). If I =7 and E=R+2, find the value of D+A+R+E.

4. Compute the number of squares between 4^9 and 9^4.

5. Given any positive integer n, what are the possible last digits of n^5-n?[/hide]

[hide="1"]We are looking for $ n$ such that $ n^2\plus{}25\equal{}k^2\plus{}13500 \iff n^2\minus{}k^2\equal{}(n\minus{}k)(n\plus{}k)\equal{}13475$. $ 13475$ has a prime factorization of $ 5^2\cdot7^2\cdot11$, and has $ (2\plus{}1)(2\plus{}1)(1\plus{}1)\equal{}18$ factors. These form $ 9$ pairs of factors multiplying to $ 13475$, one of which corresponds to $ n\minus{}k$, and one of which corresponds to $ n\plus{}k$, and will give $ 9$ solutions for $ n$ total. (Note that if we had even factors involved, we would have to remove some of the total factors.)[/hide]
[hide="2"]First consider a number with 6 positive odd integer divisors only. It is either of the form $ p^2q$ or $ p^5$, where $ p,q$ are odd primes. Clearly, the smallest such number is $ 3^2\cdot5\equal{}45$. Multiplying this number by $ 2^2$ gives $ 2^2\cdot3^2\cdot5\equal{}180$. This number has 6 odd divisors (the divisors of 45), and 12 even divisors (2 times the divisors of 45, and 4 times the divisors of 45).[/hide]
[hide="4"]$ 4^9\equal{}(2^2)^9\equal{}(2^9)^2\equal{}512^2$, $ 9^4\equal{}(3^2)^4\equal{}(3^4)^2\equal{}81^2$. Every square from $ 82^2$ to $ 511^2$ is in between, making $ 511\minus{}82\plus{}1\equal{}430$ total.[/hide]

ans 5)
[hide]last digit is always 0.[/hide]
can u explain ans to q.4 a bit more

anyone for number 3?

How do you prove number 5?

The rest of the answers are correct. Salamat!

Number 5:

$ x^5 \equal{} x \bmod{10}$ for all positive integers $ x$.

#3
N=D=0
GG+A=TT
G=9
T=0
A=1
It is given that I=7, and since we carry a one over from GG+A,
E=8
R=6
D+A+R+E=0+1+6+8=15
im not so sure though...

[quote="Brut3Forc3"][hide="4"]$ 4^9 \equal{} (2^2)^9 \equal{} (2^9)^2 \equal{} 512^2$, $ 9^4 \equal{} (3^2)^4 \equal{} (3^4)^2 \equal{} 81^2$. Every square from $ 82^2$ to $ 511^2$ is in between, making $ 511 \minus{} 82 \plus{} 1 \equal{} 430$ total.[/hide][/quote]

We need to express $ 4^9$ in the form of $ k^2$, and $ 9^4$ in the form of $ r^2$, where $ k,r \in \mathbb{Z^{\plus{}}}$. Then we can find the number of squares between $ k^2$ and $ 9^4$, solving the problem.

Typically, we factorise the base of the exponent.


abstract algebra
superior algebra
superior algebra theorems

May I know if there is any classification of group of automorphisms? Sorry if this is a famous problem.

[quote="-oo-"]I think there's no general procedure ... even though the classification was successful in many cases. ;-)

Do you think of a certain group?[/quote]

finite group.... maybe it's still too general..

Yes, it is too general *g*. The classification of $Aut(G)$ for finite(ly generated), abelian $G$ is known, but for arbitrary finite $G$ ... I don't think so.


Consider an inversion of pole $A$. It turns circles into lines, so it turns the three circles into three lines concurrent at $B'$ (in general, let $T'$ be the image of $T$ under the inversion). For the lines $BA,BZ',BY',BX'$ we apply the cross-ratio and we get $\frac {AX'}{AZ'}:\frac {Y'X'}{Y'Z'}=$constant, so $\frac {AX'}{AZ'}:\frac {AX'-AY'}{AY'-AZ'}=$ constant. Now replace $AX',AY',AZ'$ by $\frac 1{AX},\frac 1{AY},\frac 1{AZ}$ respectively (remember the inversion). A brief computation gives what we want, namely $\frac {XY}{YZ}=$constant.

Grobber, your idea is nice, but the problem is much simpler. In fact, consider two points X and X' on the first circle. The line AX meets the other two circles at the points Y and Z (apart from A), and the line AX' meets the other two circles at the points Y' and Z' (apart from A again). Then we have to prove that XY / YZ = X'Y' / Y'Z'.

Indeed, we have < AXB = < AX'B (cyclic) and thus

  < BXY = 180° - < AXB = 180° - < AX'B = < BX'Y';

also we have < AYB = < AY'B (cyclic) and therefore

  < BYX = < AYB = < AY'B = < BY'X'.

Hence, from < BXY = < BX'Y' and < BYX = < BY'X', we conclude that the triangles BXY and BX'Y' are similar, so that XY / X'Y' = BY / BY'. Similarly, we find YZ / Y'Z' = BY / BY'. Therefore, XY / X'Y' = YZ / Y'Z', or, equivalently, XY / YZ = X'Y' / Y'Z'.

  Darij

Here is another proof

Let O1,O2,O3 be three centers of the three circles. O1I1,O2I2,O3I3 are perpendicular to AX (I1,I2,I3 lie on AX )

We can easily prove that

I1I2=XY/2
I2I3=YZ/2
I1I3=XZ/2

Therefore XY/YZ=I1I2/I2I3=O1O2/O2O3
               XY/XZ=I1I2/I1I3=O1O2/O1O3

[quote="Iris Aliaj"]Three fixed circles pass through the points $A$ and $B$. Let $X$ be a variable point on the first circle different from $A$ and $B$. The line $AX$ intersects the other two circles at $Y$ and $Z$ (with $Y$ between $X$ and $Z$). Show that the ratio $\frac{XY}{YZ}$ is independent of the position of $X$.[/quote]

Just to note that this problem coincides with Baltic Way 2004 Problem 20:

Three circular arcs $w_{1}$, $w_{2}$, $w_{3}$ with common endpoints A and B are on the same side of the line AB; hereby, $w_{2}$ lies between $w_{1}$ and $w_{3}$. Two rays emanating from B intersect these arcs at $M_{1}$, $M_{2}$, $M_{3}$ and $K_{1}$, $K_{2}$, $K_{3}$, respectively. Prove that $\frac{M_{1}M_{2}}{M_{2}M_{3}}=\frac{K_{1}K_{2}}{K_{2}K_{3}}$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=20174 for a secondary discussion of this problem.

  Darij


Suppose a, b, c, d >= 0 and ab + bc + cd + da = 1. Prove that

a/(b + c + d) + b/(c + d + a) + c/(d + a + b) + d/(a + b + c) :ge: 1/3.

Nice one :)

Always make sure you aren't working in reverse on inequalities.

Several solutions, again :

*A solution using Cauchy, see for example
http://www.kalva.demon.co.uk/short/soln/sh9023.html
*A solution using the lemma
a/A + b/B + c/C + d/D >= (a + b + c + d)/(4(A + B + C + D)).
See Maverick's solution.
*A solution using Jensen's inequality (my own solution).
If S > 0 then the function f(x) = x/(S - x) is convex.
Now put S = a + b + c  + d. Then the LHS equals
f(a) + f(b) + f(c) + f(d)
and using Jensen we obtain
f(a) + f(b) + f(c) + f(d) :ge: 4f(S/4).
So it suffices to show that 4f(S/4) >= 1/3.
This is quite straightforward.

using the solution by Jensen,

how do we know that 4f(s/4):ge:1/3

I know this would be true if f(s/4) is greater than 1/12, but how is that straightforward?

Well ...

4f(S/4) = 4 (S/4)/(S - S/4) = S/12

So we want to show that S/12 :ge: 1/3 or S :ge: 4. Now

S
= (a + b + c + d)
= a + b + c + d + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd
= (a + c) + (b + d) + 2(ab + bc + cd + da)
= (a + c) + (b + d) + 2
:ge: 2(a + c)(b + d) + 2
= 2(ab + ad + cb + cd) + 2
= 2(ab + bc + cd + da) + 2
= 2 + 2
= 4.

So we're done. Okay ?

Notice that (a + c) + (b + d) :ge: 2(a + c)(b + d) follows from AMGM.
Just take x = a + c, y = b + d in x + y :ge: 2xy.

ok i understand that, duh, i could have figured that out.. thank you for the explanation.

Similar problem
Find min of $ M\equal{}\frac{a^3}{b\plus{}c\plus{}d}\plus{}\frac{b^3}{a\plus{}c\plus{}d}\plus{}\frac{c^3}{b\plus{}a\plus{}d}\plus{}\frac{d^3}{a\plus{}b\plus{}c}$
where $ a;b;c;d$ be positive real numbers satisfying $ ab\plus{}bc\plus{}ca\plus{}da\equal{}1$

[quote="chien than"]Similar problem
Find min of $ M \equal{} \frac {a^3}{b \plus{} c \plus{} d} \plus{} \frac {b^3}{a \plus{} c \plus{} d} \plus{} \frac {c^3}{b \plus{} a \plus{} d} \plus{} \frac {d^3}{a \plus{} b \plus{} c}$
where $ a;b;c;d$ be positive real numbers satisfying $ ab \plus{} bc \plus{} ca \plus{} da \equal{} 1$[/quote]

I believe you mean $ ab \plus{} bc \plus{} cd \plus{} da \equal{} 1$...or $ (a \plus{} c)(b \plus{} d) \equal{} 1$

If so, the minimum can be found in the following way:

From Holder inequality we have that:

$ M\geq \frac {(a \plus{} b \plus{} c \plus{} d)^3}{4*3*(a \plus{} b \plus{} c \plus{} d)} \equal{} \frac {(a \plus{} b \plus{} c \plus{} d)^2}{12}\geq \frac {4(a \plus{} c)(b \plus{} d)}{12} \equal{} \frac {1}{3}$ which is the minimum of M  :wink:


geometry
trigonometry
Vieta
quadratics
trig identities
Law of Sines
similar triangles

I like what ragnarok is doing, and hope others will contribute geometry problems of their own. I wish the intermediate geometry marathon was like this. Here is my own problem:
In semicircle O with diameter AB and radius 4, OC of length 5 is drawn perpendicular to AB so C is opposite the semicircle. Point D is on OB such that OD=2, and E is on the perpendicular to AB at D on the same side as C such that DE=1. CE intersects the semicircle at G. Find CG.

No one can solve it? Perhaps I should have put it in intermediate (It was hard for me). Here are two hints:
[hide="1"]Similarity[/hide]
[hide="2"]Law of Sines[/hide]

[hide="Solution"]
Similar triangles give the diagram. By the Pythagorean Theorem, $ HF \equal{} \frac {5\sqrt5 \minus{} 2x}2$, so by Power of a Point on $ F$, $ \frac {39}4 \equal{} FG\left(\frac {5\sqrt5 \minus{} 2x}2\right)\implies FG \equal{} \frac {39}{2(5\sqrt5 \minus{} 2x)}$, and then by Power of a Point on $ C$, $ 9 \equal{} x\left(\frac {5\sqrt5}2 \plus{} \frac {39}{2(5\sqrt5 \minus{} 2x)}\right)\implies18(5\sqrt5 \minus{} 2x) \equal{} x(5\sqrt5(5\sqrt5 \minus{} 2x) \plus{} 39)$
$ \implies90\sqrt5 \minus{} 36x \equal{} 164x \minus{} 10x^2\sqrt5\implies x^2 \minus{} 4x\sqrt5 \plus{} 9 \equal{} 0$
$ \implies x \equal{} 2\sqrt5\pm\sqrt {11},\,x < 5\implies x \equal{} 2\sqrt5 \minus{} \sqrt {11}$
$ \implies CG \equal{} \frac9{2\sqrt5 \minus{} \sqrt {11}} \equal{} \boxed{2\sqrt5 \plus{} \sqrt {11}}$ (the last step can be done faster by Vieta's in the monic quadratic above).
[/hide]

[b]Edit:[/b] Thanks. :)

Nice solution! You mean $ 2\sqrt {5} \plus{} \sqrt {11}$ ,btw.
[hide="My solution"]
Since COF is right, we know $ CF \equal{} \frac {5\sqrt {5}}{2}$, so $ sin OCF \equal{} \frac {\sqrt {5}}{5}$. By law of sines on $ \triangle COG$, $ \frac {sin OCG}{4} \equal{} \frac {sinOCG}{5} \implies sin OCG \equal{} \frac {\sqrt {5}}{4}$. So by law of sines again, $ \frac {sin OCG}{4} \equal{} \frac {sin(sin^{ \minus{} 1}\frac {\sqrt {5}}{5} \plus{} sin^{ \minus{} 1}\frac {\sqrt {5}}{4})}{CG}$. Solving, $ CG \equal{} 2\sqrt {5} \plus{} \sqrt {11}$[/hide]
[hide="Second solution, very easy"]
Drop a perpendicular from O to CG. We easily find it's length, double Pythag finishes it. [/hide]
But I thought of it only after knowing the answer.


function
integration
inequalities
real analysis
real analysis unsolved

Let $S$ be the set of all $C^1$ functions $f$ on $[0,1]$ such that $f(0)=0$ and
\[ \int^1_0 f'(x)^2dx\leq 1  \]
Define
\[ J(f)=\int^1_0 f(x)dx  \]
Show that $J$ is bounded on $S$ and compute its supremum.  Is there a function $f_0$ such that $J$ achieves its supremum?  If so, find $f_0$.

I just "followed my nose" and did what seemed the most likely at each step, and it all works.

The bound is $J(f)\le\frac1{\sqrt{3}}$ and it is achieved.

$J(f)=\int_0^1f(x)\,dx$ (Now integrate by parts)

$=\left.(x-1)f(x)\right|_0^1-\int_0^1(x-1)f'(x)\,dx$

$=0+\int_0^1(1-x)f'(x)\,dx$ (Now use the Schwarz inequality)

$\le\left(\int_0^1(1-x)^2\,dx\right)^{\frac12}\left(\int_0^1(f'(x))^2\,dx\right)^{\frac12} \le\frac1{\sqrt{3}}\cdot 1=\frac1{\sqrt{3}}.$

The extremal function is $f_0(x)=\sqrt{3}\left(x-\frac{x^2}2\right).$

This is motivated simply by the equality conditions in the Schwarz inequality.

Neat.  I tried integrating by parts right away, but I should have been using an $x-1$ term instead of an $x$ term in view of the boundary conditions.


For  x,y,z>0  show that [tex] x^2+y^2+z^2+\sqrt
{3xyz(x+y+z)
}\geq2(xy+yz+zx) [/tex]

Thank you!

Nice one.

By cauchy
\[
\sqrt{3xyz(x+y+z)} = \sqrt{(xyz+xyz+xyz)(x+y+z)} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}
\]
By Schur, $\sum x(\sqrt{x} - \sqrt{y})(\sqrt{x} - \sqrt{z}) \geq 0$, so
\[
x^2 + y^2 + z^2  +x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy} \geq \sum_\mathrm{sym} x^{3/2} y^{1/2}
\]
and the rest follows by Muirhead.


Find all funtions $f: \mathbb R \to \mathbb R$ such that the following equality holds for all $ x,y \in \mathbb R$:
\[f(x^{3}-y)+2y(3f^{2}(x)+y^{2})=f(y+f(x)).\]

[hide="some useful moves!"]
by $ x=y=0$ we have: $ f(0)=f(f(0))$
by putting $ -y$ instead of $ y$ we have:
$ f(x^{3}+y)-2y\left(3f^{2}(x)+y^{2}\right)=f(f(x)-y)$ and by sum of this and the main one:
we have: $ f(x^{3}+y)+f(x^{3}-y)=f(f(x)-y)+f(f(x)+y)$
by putting $ y=f(x)$ we have:
$ f(x^{3}+f(x))-8f^{3}(x)=f(0)$
now this will help us to prve that $ f(0)=0$, by putting $ x=0$ we have: $ f(f(0))-8f^{3}(x)=f(0)\rightarrow-8f^{3}(x)=0\rightarrow f(0)=0$.
as a consequence we must have:
$ f(x^{3}+f(x))=8f^{3}(x)$
 :)
[/hide]

[quote="friendlist"]Find all funtions such that the following equality holds for all $ x;y \in R$
$ f(x^{3}-y)+2y(3f^{2}(x)+y^{2})=f(y+f(x))$[/quote]

Choose $ y$ such that $ x^{3}-y=y+f(x)$ which implies equality of the leftmost and rightmost summands in the equation :

$ y=\frac{x^{3}-f(x)}{2}$ $ \implies$ $ f(x^{3}-y)=f(y+f(x))$ and so $ y(3f^{2}(x)+y^{2})=0$ and so $ y=0$ (since $ y\in\mathbb{R}$ and $ f(x)\in\mathbb{R}$) and so $ f(x)=x^{3}$

And it is rather easy to check that this necessary condition is sufficient.

So the only solution is $ f(x)=x^{3}$


limit
integration
function
real analysis
real analysis unsolved

Give an equivalent for the sequence $a_n$ which verifies $\lim a_n\sum_{k=1}^na_k^2=1$.

consider a similar pb:
$\lim {f(x) \int_{0}^{x}f^2(x)dx}=1$. Consider the function $g(x)=(\int_{0}^{x}f^2(x)dx)^3$.
Then $\lim {g'(x)}=3$, $g(x)$ is equivalent to $3x$,$f(x)$ equivalent to $(3x)^{-1/3}$.
Do the same things with the sequence and you will get the result. (I thnik it has already been posted, perhaps by me ..)


Socialism and elitism have become seen as negatives in the view of the public.

Socialism is NOT communism.  Communism is EXTREME socialism.  Education systems and police forces are run by the government.  Why not health care, which has become "socialized" in most other first world countries?  And if that's not okay, then why is spending a trillion dollars to save the banks okay?

Elitism has become the equivalent of being well educated.  So now the general public is afraid of "book-learned-persuns"?  Would you entust an incompetent fool to run America, and a large part of your future?

Oh, and if you agree with me, or disagree, please take the effort to write (at minimum) what I have.  It only takes two minutes or so.

An incompetent fool.  You mean like George Bush?  We already did.  I took a course on socialism.  It is nothing to be afraid of.  In case people haven't noticed, we've already socialized some important things, such as public education and the postal service.  There's a difference between totalitarian dictatorship, and socialized medicine.

People generally prefer the status quo, and frankly, a black man who was president of the Harvard Law Review for the first time in history, who happens to be named Barack is not exactly status quo.  And honestly, Barack Obama does seem more, to use your language, "book-learned" than Sarah Palin.  Oh wait, he is!

A digression on the subject of "elite" from the point of view of a baseball fan.

David Eckstein has always been wildly popular with baseball fans and those who write about baseball. He gets called "gritty" and "scrappy" and a "blue-collar" player and various other things that praise him for personal qualities other than raw talent. Why so? Eckstein is short, unusually short for a major league baseball player. He's listed as being 5'6" tall and 170 pounds. His throwing arm isn't very strong, by the standards of major league shortstops, so he has to give a rather visible effort to get enough velocity on a throw across a baseball diamond. (But notice that qualification, "by the standards of major league shortstops" - I can assure you that he can throw a lot harder than I could ever throw - or anyone I knew in high school.) Because he's so small, he seems to encourage the fantasy in the minds of a lot of fans that he's some kind of "average Joe" who made it to the majors and if he can do it, than so could anyone. A lot of his popularity is that kind of identification.

Of course, this identification is romantic nonsense and wishful thinking. Eckstein [b]is[/b] an elite athlete. He possesses innate physical talents in such areas as hand-eye coordination, quickness, and body control that 99.99% of the population can only dream of. He made the major leagues at the age of 26 because he is a talented, elite athlete (who did work to improve his skills). And he's on he way out at the age of 33 because he's not as talented as the players who can keep playing at that level after their skills have started to erode with age.

Professional sports are a merciless meritocracy. The only people who ever make it to the top levels are "elite." Yes, some of them work harder than others, and some do a better job of utilizing the talents they have than others - but everyone that's there has some physical gifts that few others possess.

Few other areas of life - notably including politics - are anywhere close to being as meritocratic as sports. There's the old Ann Richards line she used to insult George H. W. Bush: "He was born on third base and thought he hit a triple." True, Sarah Palin wasn't "born on third base" - but neither was Barack Obama.

[quote="alexhhmun"]Socialism is NOT communism.  Communism is EXTREME socialism.[/quote]What makes you think many people don't know this?  [url=http://en.wiktionary.org/wiki/elitist]What is it called[/url] when one believes that 'the public' is too uneducated to know the difference between two words, and more generally which words are and aren't good for them?
[quote="alexhhmun"]Education systems and police forces are run by the governement.  Why not health care, which has become "socialized" in most other first world countries?[/quote]You misspelled government.  More to the point, what makes you think the portion of the public to which you previously refer is satisfied that the education system is being run by the government?  [url=http://en.wiktionary.org/wiki/arrogance]What is it called[/url] when one believes they understand the concerns and positions of the public, without bothering to inquire what said public actually thinks, and wishes to illuminate to it the correct solutions?[quote="alexhhmun"]And if that's not okay, then why is spending a trillion dollars to save the banks okay?[/quote]I don't know; do you think it's ok?  Do you think the general public thinks it's ok?  See above.[quote="alexhhmun"]Elitism has become the equivalent of being well educated.  So now the general public is afraid of "book-learned-persuns"?  Would you entust an incompetent fool to run America, and a large part of your future?[/quote]I would not, but [url=http://en.wiktionary.org/wiki/snob]what is it called[/url] when one believes that a person who is inadequately educated (by your standards?) is necessarily an 'incompetent fool'? (this is a great time to point out you misspelled entrust)

Though a proud member of the mindless and easily manipulated general public you've been implicitly deriding, I don't believe that being well educated is equivalent to being an elitist.  Unfortunately, I do believe that more often than not it's the most educated who fall into the delusion that they understand and see beyond the foolishness of the public, while in fact being out of touch with their reality.  Do I think it's a good idea for elitists to govern a Democracy?  No.  I also don't think either Barack Obama or John McCain have shown themselves to be elitists; but you've pulled it off in a few short sentences.

[quote="JRav"]I took a course on socialism....There's a difference between totalitarian dictatorship, and socialized medicine.[/quote]There's a difference between dictatorship and free health care?  This whole time I thought the simple folk down South are against universal health care because they feel it distorts the market, makes mandatory that which they believe should be a personal choice, reduces cost-effectiveness and possibly the quality of care, and acts as a redistributive tax.  But no, they're mostly just afraid that when health care comes the commies are not far behind.  Thanks for clearing that up.  :roll:

I didn't take a course on socialism, but feel I have a fairly good grasp on what it entails.  I don't like the idea of nationalizing private enterprise and means of production that don't need to be.  I very much don't like it, and I [u]do[/u] think there are plenty of things about it to be afraid of.  It may be shocking, but there are people who disagree with your views not because they are too stupid to understand the difference between the big words you use, but merely because they hold a different view.

[quote="alexhhmun"]Socialism is NOT communism.  Communism is EXTREME socialism.[/quote]

According to Karl Marx, socialism is the transitional stage between capitalism and communism. But it seems that no state ever achived the stage of "communism" (=each contributes according to his or her ability but consumes according to his or her needs). On the contrary, most socialist states of the 20th century were heading toward bankruptcy, not "communism".

So indeed, socialism is not communism, but i would say it is worse than communism: many of the horrors and terrible violence of the last century should be attributed to socialism (not to communism), even if they were done for the "communist ideal".

[quote="alexhhmun"]Two words the U.S. has taught the public to fear.
Socialism and elitism have become seen as negatives in the view of the public. [/quote]

[quote="JRav"]I took a course on socialism. It is nothing to be afraid of.[/quote]

I belive you are lucky to live in a country where you are taught to fear socialism. After 40 years of socialism and a violent revolution (many say it was actually a coup d’état) my country was literally a wreck. The consequences are still visible, even after 20 years, and some experts say it will last at least one generation for the society to recover. And it's an EU contry, I don't even want to imagine the real magnitude of destruction in countries like North Korea, Cuba or Venezuela.

Anyway, in my opinion intelligent people should not take courses on socialism, but rather read books by people like Karl Popper, Ayn Rand, Hazlitt, Walter Blocks etc.

Yep, socialism definitely has that unfortunate view in America. When my French teacher mentioned to our class that in the French presidential election of 2007, one of the candidates was from the Socialist Party, people were shocked.

And I don't think it's a good thing to fear socialism. I feel that whenever the government wants to do anything to help poor people, that becomes synonymous with socialism, and so it gets dispensed. If this is what socialism is, is there really anything wrong with socialism? Yes, some countries have gone violent in the name of socialism, but does that mean that socialism itself is bad?

As for elitism, I do have disdain for those who become too rich and out of touch with the average American, but unfortunately, as you point out, elitism has become a synonym for intellectualism today, and I of course have no problem with someone who is smart - as you say, they would probably be a better president!

JRav, I think alexhhmum agrees with you that being well-educated isn't the same thing as elitism. As well, alexhhmum isn't saying that people in the South are against "socialized medicine" because they think "the commies aren't far behind" - but I do agree that many people actually do oppose government intervention on the fear that it is socialist and therefore dangerous.

Again, where does this notion come from that a negative view of socialism is 'unfortunate'?  Don't you think it's a bit condescending to imply that a large group of people who disagree with you do so simply because they are unfortunately misinformed?  For instance, the poster from Romania on here doesn't seem thrilled by socialism;  does his view seem unfortunate to you?  Is it possible that it is you who is misinformed about what socialism is?  For instance, can you name a prospering country (or, more precisely, a country in which you believe the quality of life is better than in, say, the US, Germany, or England) that you believe has a socialist economy?

And the same question I posed to the OP: What makes you think people fear socialism because of the violence it has caused in the past, or because they fear it will somehow lead to dictatorship?  Is it inconceivable to you that people don't like the idea of nationalizing private industry simply because they think it will negatively impact their quality of life?

Edit: I deleted it, see the next post. Sorry for the confusion.  :oops:

[quote]Also, I realize your view of the subject is from a different vantage point, but the discussion here (I believe) was mostly about US politics. In the US, I don't believe the primary reason (or even a significant reason) for the opposition to socialism is the threat of violence or dictatorship.[/quote]

I must confess I don't really know how the americans understand socialism. But i"m sure some people realize that those two realities lie in the essence of socialism.

The questions were directed toward the previous poster, calc rulz.  My apologies for the ambiguity.

Also, I realize your view of the subject is from a different vantage point, but the discussion here (I believe) was mostly about US politics.  In the US, I don't believe the primary reason (or even a significant reason) for the opposition to socialism is the threat of violence or dictatorship.

Hmm, I would think that most people have a deluded perception of socialism in the United States b/c it's related to communism, which is related to dictators, totalitarian governments, purges of political enemies, etc.


algebra
polynomial
number theory unsolved
number theory

$a$ is a fixed positive integer and $ord_p(a)$ means the smallest $k$ such that $a^k\equiv 1$

now prove $\frac{p-1}{ord_p(a)}$ is not bounded

Please be more clear, and write a [b]rigorous[/b] statement :)
(what is $a$, what is $p$ what is $ord$, with respect to what variable is $\frac{p-1}{ord^a_p}$ unbounded, is $a$ fixed, and so on...)

$a$ is fixed, $p$ is prime and is what runs through different values.
$ord_p(a)$ is the smallest positive integer $k$ such that $a^k \equiv 1 \mod p$.
(I think)

Very nice :)

I hope the following is correct:
Take any positive integer $n$.
Then take prime $k$ such that $k>a$ and such that $km+1$ is not prime for $m=1,2,...,n$ (can be done by the Chinese Remainder Theorem and Dirichlet's theorem for primes in arithmetic progression).

Let $\Phi_k(z)$ by the $k$-th cyclotomic polynomial.
Now take a prime divisor $p \nmid k$ of $\Phi_k(a)$ (these divide either $k$ (excluded because of $k>a$ prime, so $a^k \equiv a \not\equiv 1 \mod k$) or have the property from the next sentence). So $ord_p(a) =k$ and then it follows $p \equiv 1 \mod k$ (since $p-1$ is the order of the group of residues $\mod p$), thus $p-1>nk+1-1=nk$ by the definition of $k$. Finally it follows $\frac{p-1}{ord_p(a)} > \frac{nk}{k} = n$.
As a result, the sequence is unbounded, since $n$ was arbitrary.

[quote="ZetaX"]
Now take a prime divisor $p \nmid k$ of $\Phi_k(a)$ (these divide either $k$ (excluded because of $k>a$ prime, so $a^k \equiv a \not\equiv 1 \mod k$) or have the property from the next sentence). So $ord_p(a) =k$....[/quote]

I don't get this one - you're probably using some property of cyclotomic polynomial I don't know - can you go more in detail in this one, please?

See my post in http://www.mathlinks.ro/Forum/viewtopic.php?t=62706
Note that this is independent from $k$ being prime, but the prove could be shortened a bit then.

Unless I'm entirely crazy...
We know that primes are unbounded, so $p-1$ is unbounded. We only need to prove now that $ord_p(a)$ is bounded. It is clearly bounded if $a=\pm1$. If we can prove that for some for all $a$, as $p$ increases, $a$ is not always a primitive root...

nevermind, harder than I expected.

While it is certainly motivated to use cyclotomic polynomials, I think [b]ZetaX[/b]'s proof can be done without them. The following is the same argument sans the CycPoly's. We will show that for given $k$ and infinitely many $n$, there is a $p>kn+1$ with $\text{ord}_p(a)=n$. Clearly if $\text{ord}_p(a)=n$ then $n\mid p-1$. Hence it suffices for $n+1,2n+1,\dots,kn+1$ to be composite. This is achievable by CRT with sufficiently many primes. Thus it follows that $\tfrac{p-1}{\text{ord}_p(a)}$ is unbounded.


Suppose we had an equation $ x^{2n}\equal{}k, \forall n \in \mathbb{N}$ and $ \forall x,k \in \mathbb{R}$. Suppose we take the $ 2n$-th root on both sides. We have $ \sqrt[2n]{x^{2n}}\equal{}\sqrt[2n]{k}$. Normally, we would add the $ \pm$ on the side with $ k$. Why is this so (other than the obvious fact that it's an even root and there are two distinct values for every even root)? Why does the $ \pm$ get added automatically? (I was pondering this when we were discussing quadratics today in class.)

[hide="My Thoughts"]
I'm not sure if these are valid or not, so please let me know if there are any invalid steps.

When considering the expression $ \sqrt[2n]{x^{2n}}$, we know that there are two values of $ x$ such that the outcome of the evaluation of the root is the same. This is because when there are an even number of $ x$ terms and when $ x < 0$, the outcome is a positive number. And if $ x > 0$, it will already be a positive number. So, in order to consider to take those cases into consideration, we cannot conclude that $ \sqrt[2n]{x^{2n}}\equal{}x$ because that doesn't work for values of $ x$ when $ x < 0$. Thus, we must have $ \sqrt[2n]{x^{2n}}\equal{}|x|$.

Now taking that into consideration and plugging that into the problem, we obtain:

$ |x|\equal{}\sqrt[2n]{k}$

$ x$ can either be a positive or negative number and satisfy the equation because the absolute value is positive. Thus we have $ x\equal{}\sqrt[2n]{k} \text{ or } x\equal{}\minus{}\sqrt[2n]{k}$.

So when $ x^{2n}\equal{}k$ (with the limitations of the values of $ x$,$ k$, and $ n$), $ x \equal{} \pm \sqrt[2n]{k}$.
[/hide]

A distinction has to be made between "the solutions to $ x^2 \equal{} k$" and "the value of $ \sqrt {k}$."  The expression $ \sqrt {k}$ is a [b]function[/b], and functions have to take [b]one value.[/b]  So $ \sqrt {x}$ is defined to be a function that takes non-negative real numbers to non-negative real numbers.  Hence when $ k$ is a non-negative real, one solution to $ x^2 \equal{} k$ is the non-negative real number $ \sqrt {k}$ and the other solution is the non-negative real number $ \minus{} \sqrt {k}$.

So you are absolutely correct that $ \sqrt{x^2} \equal{} |x|$ and that is also an absolutely valid way to think about this problem.  The important point is that $ x^2$ does not have an inverse because it fails the horizontal line test.


In $\Delta ABC$, the sum of the sides is $2s$ and the radius of the incircle is $r$. Three semicircles with diameters $AB$, $BC$ and $CA$ are drawn on the outside of $ABC$. A circle with radius $t$ touches all three semicircles. Prove that \[ \frac{s}{2} < t \leq \frac{s}{2} + \left(1 - \frac{\sqrt{3}}{2}\right)r. \]


I recently faced up to these three questions in my biology class:

[i]How is the frog's heart less efficient than our own?[/i]

[i]Why is the ventricle much more muscular than the atria?[/i]

[i]What are four orangs supplied with blood by the dorsal aorta?[/i]

I have ideas for first two but I'm not sure for the last one. I'm not sure whether my first and second are right so I'm asking here for the correct fact.

[quote][i]How is the frog's heart less efficient than our own?[/i][/quote]

[color=blue]The frog's heart is less efficient than our own because in frog, the two different blood from two atria comes into one ventricle. In human, two different blood from two atria goes into two different ventricles.  Our method of heart is more efficient than frog's even what frog does work for itself because two blood from two atria are different kind --- one is deoxygenated and one is oxygenated ---- and when those are mixed, the body won't get blood with one hundred percent oxygen. But with our human's method, we'll get one hundred percent oxygen blood. Frog was okay because it was cold-blooded animal.[/color]

[quote][i]Why is the ventricle much more muscular than the atria?[/i][/quote]

[color=blue]Similar as above. In ventricle, two different kind of blood (oxygenated and deoxygenated) are coming to itself. Because of this fact, the ventricle has to be more muscular than the atria to take in both kinds of blood. I know this is not best worded but idea is two atria and one ventricle means --> ventricle to be more muscular.[/color]

[quote][i]What are four orangs supplied with blood by the dorsal aorta?[/i][/quote]

[color=blue]I'm not sure.. I'll just guess.. Lung, heart, liver, and stomach? :roll: I don't know for this one so good help will be appreciated.[/color]

Thanks.

i think the dorsal aorta supplies blood to the alimentary tract, liver, kidneys and the hind limbs.

if i remember from doing some frog anatomy stuff, i think appasha is right

for the second question, i would put that the ventrical has to be more muscular as it has to pump blood longer distances (lungs or the body) whereas the atria only have to pump blood to a nearby ventricle.  thus a need to be more powerful is created and more muscle is the way to achieve the desired outcome.

Appasha ,Good job I juist remembered something I doid when Disceting the frog


Solve the system

$log_xe+log_ye=7/3$
$lnxy=7/2$

[quote="riddler"][quote="deej21"][quote="riddler"]
is there a value for $e$?
[/quote]

$e$ is euler's number.  $\ln x = \log_e x$[/quote]
euler? the "suppossed to be" greatest mathematician in the world?
can you guys post more log problems?[/quote]

Is $e$ euler ? I thought exponential ...

I think $ EULER $ was the greatest mathematician of all time. ;)

[quote="deej21"]Yes, Euler was one of the greatest mathematicians, and yes $e$ generally stands for Euler's number, or about $2.718281828459...$

There are good log problems in AoPS volume 2, the ones in volume 1 are fairly easy.  Here is a classic one, although posted here many times

If $p$ and $q$ are positive reals satisfying $\log_9 p = \log_{12} q = \log_{16} p + q$, find $\frac q p$[/quote]

[hide]
Let $\log_9 p = \log_{12} q = \log_{16} p + q$=$k$ where $k \in \text{R}$then

$p=9^k$ , $q=12^k$ , $p+q=16^k$  Then

$p+q=9^k + 12^k = 16^k$  .Divide by $9^k$ , $1+(\frac 43)^k = (\frac 43)^{2k}$ Letting $(\frac 43)^k$= $a>0$  we get $a^2-a-1=0$ and solve $a=\frac{1+\sqrt 5}{2}$ .Since we want to find $\frac qp=(\frac 43)^k=\frac{1+\sqrt{5}}{2}$ [/hide]

[quote="riddler"]shyong watch out pressing themouse button twice makes you double post. ;)  I got the same answer though and the same way, I hate logs though does it have to always take that much time? :mad: is there a see a sum and instantly get it way?[/quote]

Practise makes perfect I would say ... :P

[quote="mathmanman"][quote="shyong"]
Is $e$ euler ? I thought exponential ...[/quote]
Yeah, it is either exponential or euler.[/quote]

[deleted]

$e$ click my link for info [url=http://mathworld.wolfram.com/e.html]http://mathworld.wolfram.com/e.html[/url]

Euler constant $\gamma=\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{1}{k}-\ln n\right)$

Click link for more info [url=http://mathworld.wolfram.com/Euler-MascheroniConstant.html]http://mathworld.wolfram.com/Euler-MascheroniConstant.html[/url]

You can also take a look @ : http://en.wikipedia.org/wiki/E_(mathematical_constant)


What integer n has the property that the property that $ 5^{96}$ is greater than $ n^{72}$ and less than $ (n\plus{}1)^{72}$?

[hide="Hint."]
$ \sqrt [24]{}$

[hide="Solution."]
$ n^{72} < 5^{96} < (n \plus{} 1)^{72}$
Take the 24th root of both sides, which we may do since $ \sqrt [24]{x}$ is strictly increasing.
$ n^3 < 5^4 < (n \plus{} 1)^3$
$ n^3 < 625 < (n \plus{} 1)^3$
If you know your cubes well, you'll see that $ 625$ falls between $ 512$ and $ 729$, which are $ 8^3$ and $ 9^3$, respectively, so $ n \equal{} \boxed{8}$.
[/hide]

[/hide]

EDIT: Post after me deleted.


function
calculus
integration
geometry
analytic geometry
graphing lines
slope

Let f be a twice differentiable function function such that f(1)=2 and f(3) = 7.  Which of the following must be true for the function f on the interval 1 < x < 3?

1.  The average value of f is 9/2
2.  the average value of f' is 5/2

can you please show all work at how you arrived at the answer? thanks

for the interval, i fogot to add the equal signs..it's suppose to be less than or equal to.

1 doesn't have to be true, but 2 must be true.

[hide="Explanation"]
1. If you have already learned integrals, the average value of $f(x)$ is calculated by $\frac{\int_{a}^{b} f(x)dx}{b-a}$. Since we don't know the value of the definite integral, the average value is not uniquely determined.

In case you haven't learned integrals...
Basically, the idea is that the average value of $f(x)$ is $\frac{\text{(Area under the curve }y=f(x)\text{ from interval}x=a\text{ to }x=b)}{b-a}$.

Here's something similar. Say you have something with an uneven surface. The average "height" of the object is $\frac{\text{Volume}}{\text{Area of the base}}$.
This is a 3-dimentional example, but I hope you get the idea...

2. The average value of $f'(x)$ means the average rate of change of $f(x)$ from $x=a$ to $x=b$. (Recall that $f'(a)$ represents the instantanious rate of change at $x=a$.) What does this mean?- Well, it's the same thing as the slope of the line connecting the points $(a,f(a))$ and $(b,f(b))$, which is $\frac{f(b)-f(a)}{b-a}$.

Therefore, the average value of $f'(x)$ in the interval $[1,3]$ is $\frac{7-2}{3-1}=\frac{5}{2}$.
[/hide]


hi, i am a geometry student. I am currently in the most advanced class in my grade. I have exams soon, and im studying for it. here is a problem i couldn't solve. can someone help me? thx.

[img]http://akimatsu.mypicgallery.com/mathmath/q1_large.jpg[/img]

*image drawn by geometer's sketchpad*

The point is the ex-centre of the triangle ADE.

If you Ceva's Theorem, then you can prove it by concurrency.


trigonometry
geometry
similar triangles
geometry proposed

Two circles $K_{1}$ and $K_{2}$, centered at $O_{1}$ and $O_{2}$ with radii $1$ and $\sqrt{2}$ respectively, intersect at $A$ and $B$. Let $C$ be a point on $K_{2}$ such that the midpoint of $AC$ lies on $K_{1}$. Find the length of the segment $AC$ if $O_{1}O_{2}=2$

let a circle with diameter $ AO_{2}$ intersect with $ K_{1}$ at $ M$(it's clear that $ M$ is midpoint of $ AC$) let midpoint of $ AO_{2}$ be $ E$.hence we have $ O_{1}E\equal{}\sqrt{2}$.
$ AE^{2}\equal{}AO_{1}^{2}\plus{}EO_{1}^{2}\minus{}2EO_{1}.AO_{1}\cos\theta\Longrightarrow\cos\theta\equal{}\frac{5\sqrt{2}}{8}\Longrightarrow\sin\theta\equal{}\frac{\sqrt{14}}{8}$ hence $ AC\equal{}\frac{\sqrt{14}}{2}$

I solved this problem,but I don't know how post this, answer is a (x). and (x*x=3.5)

let P=midpoint of AO2, in a triangle AO1O2  PO1 is a median and PO1=x and x*x=2.  let R be a midpoint of AB,and M midpoint of AC, !=>  <BMO2=90,and B,R,M and O2 is on a one circle, and centre of this circle is P. let T be a midpoint of AM.  and in a triangle APO1  we know AO1,AP and PO1 !=> AT=y .and y*y=7/32.  and AT=AC/4. becouse AT=AM/2=AC/4   so AC=z and z*z=7/2

[quote="Amir.S"] hence $ AC \equal{} \frac {\sqrt {14}}{2}$[/quote]
I get everything but this

See that M (midpoint of AC) is on the circle $ \omega (P,AP)$, where P is the middle of $ AO_2$. Then see that $ \triangle AO_1P \sim \triangle AO_2O_1$, because  of $ O_1A^2 \equal{} AP \cdot AO_2$, hence $ O_1P \equal{} \sqrt 2$ . Because of M being on the circle $ \omega$, we have $ AC \perp O_1P$ and, calculating in 2 different ways the area of $ \triangle AO1P$ we get AN, where $ N \equal{} AM \cap O_1P$: from $ \triangle AO_1O_2$ we get $ cos A \equal{} \minus{} \frac {1} {2 \cdot \sqrt 2}$ , that is, $ sin A \equal{} \frac {\sqrt 7} {2 \cdot \sqrt 2}$ , hence $ AO_1 \cdot AP \cdot sin A \equal{} PO_1 \cdot AN$ , with $ AN \equal{} \frac {\sqrt 7} {2 \cdot \sqrt 2}$ and $ AC \equal{} 4 \cdot AN \equal{} \sqrt 14$.

But, it seems there is even a simpler proof:
1)	from $ \triangle AO_1P\sim \triangle AO_2O_1$ we get  $ PO_1$
2)	See that $ \triangle ABC\sim \triangle O_1PO_2$ and AB is twice the altitude from A in $ \triangle AO_1O_2$, hence from $ \frac {O_1O_2} {AC} \equal{} \frac {O_1P} {AB}$  ( from similar triangles) we easily get AC.

Best regards,
sunken rock


USAMTS
number theory
prime factorization

For each of n=84 and n=88 find the smallest interger multiple of n whose base 10 representation consists entirely of 6's and 7's. (USAMTS 1, aops book 1, pg.67)  Can anyone give me a [u]hint [/u]on how to start off with the problem?  you can post the answer too if you want, but please hide it.

Thanks

Look at the prime factorization of $n$. Let $N$ be an integer in the decimal system consisting only of the digits 6 and 7. If $N$ is divisible by $n = 88 = 2^3 \cdot 11$, then $8|N$ gives the last three digit of $N$ and $11|N$ is equivalent to the alternating sum of digits of $N$ is divisible by 11. Similary, if $n = 84 = 2^2 \cdot 3 \cdot 7$, consider the implications of $4|N,$ $3|N$ and $7|N$.

Consider which of the following last digits of the number you're looking for is divisible by 4.

66, 67, 77, 76

Hope that helps

[hide="solution and answer for 84"]Prime factorize $84=2^2\times3\times7$. Since it is divisible by 4, the last two digits must be 76. Since it is divisible by 3, and 6 is already divisible by 3, there should be 3k of 7's, otherwise the number is not divisible by 3. Assume k=1: there are 3 7's. Now test numbers to find the first number that is divisible by 7.

7776 no
67776 no
76776 works![/hide]

[hide="solution and answer for 88"]Prime factorize $88=2^3\times11$. Since the number is divisible by 8, the last three digits must be 776. From divisibility rule for 11, 6776 is the first number.[/hide]

wow lol  when i solved this problem, i thought that it meant divisible by 84 and 88...which is a good problem to do when you are done with this one

yay, I got the right answer.  Thanks for the help!

:wallbash_red: I have been trying to find a multiple of both 84 and 88 together consisiting only of 6s and 7s but no luck so far. Plz help.

They weren't looking for a multiple of both 84 and 88.  They were looking for one that is a multiple of 84 and another that is a multiple of 88.

I think he was just doing that as a follow up problem...

Btw, what is the divisibility rule for 11?

The difference of the sum of alternating digits is divisible by 11.

For example 9141 is divisible by 11 because 9+4 - (1+1) = 11, which is dvisible by 11.


geometry
3D geometry
calculus
integration
number theory
relatively prime

In a right triangle whose legs have integral side lengths $a$ and $b$ and whose hypotenuse has integral length $c$, if $a=3$ and $c=5$, then $a+c=8$, a perfect cube.  Find the least value of $c$ greater than $5$ for which $a$ and $c$ are relatively prime odd integers whose sum is the cube of an integer.

[quote="Altheman"]a^2+b^2=c^2
a+c=q^3

if we recall the pythagorean triple thing, a=(m^2-n^2), b=2mn, c=(m^2+n^2)...then we have

a+c=2m^2=q^3...so 2|m...(m/2)^2=(n/2)^3 =k...k is a perfect 6th power...so k=64, m=16, smallest c implies n=1

256+1=[b]257[/b]....the legs are 32 and 255[/quote][hide="uhh incorrect solution"]what if $b+c$? then $m^{2}+2mn+n^{2}=q^{3}$, which makes $(m+n)^{2}=64$ be the smallest perfect square/cube so $m+n=8$. we want to minimize $m^{2}+n^{2}$ where $m$ and $n$ are integers such that $m>n$, so we have $m=5$, $n=3$ so $m^{2}+n^{2}=34$ ($16,30,34$ triangle)[/hide]

EDIT oh oops :o  "relatively prime odd" hehe sorry  :D


inequalities
function
logarithms
inequalities unsolved

For b) Put $x=b+c,y=c+a,z=a+b$, then $x+y+z=2$ and $\frac{1}{a+b}+c=1+\frac{1-z^{2}}{z},$ etc. Now use Jensen's theorem for function $f(x)=\ln{(1+\frac{1-x^{2}}{x})}$.


algebra
polynomial
product of roots
sum of roots

1. A pit is 0,76 meter deep and diameter is Æ 0,66 m. A light pole with a diameter of Æ 0,34 m is
erected in the pit. After that the pit is filled with concrete. How many liters of concrete is needed
to fill up the pit? (1 l precision, the pit is as cylinder) (2 p)

2.Heat exchanger raises the temperature of air before blowing it into to the factory hall.
Transferred heat energy and the air temperature changes can be determined with the following
equation:
After inserting the values we get the following equation, where the unknown value is the air
temperature after the heat exchanger. Calculate the value of temperature t2. Give the answer in
one decimal precision. ( 2 p )
3. Calculate the sum and the product of the equation roots. Give the answer in fractions
(ex: x = 3 ¾). (1 + 1 p)
-3x2+2x+4=0

[hide="3"]
Suppose the equation has roots $ r_{1}$ and $ r_{2}$. Then $ (x\minus{}r_{1})(x\minus{}r_{2})\equal{}x^{2}\minus{}(r_{1}\plus{}r_{2})x\plus{}r_{1}r_{2}$. Dividing the given equation by $ 3$, we get $ x^{2}\minus{}\frac{2}{3}x\minus{}\frac{4}{3}\equal{}0 \Rightarrow$ product of roots $ \equal{}\minus{}\frac{4}{3}$ and sum of roots $ \equal{}\frac{2}{3}$.

[/hide]

[hide="1"]
Because you what to fill the hole up, the length of the pole will be the height of the hole which is .76 meters. The volume of concrete needed will be the volume of the hole minus the volume of the pole. This leads to
$ .76\pi(.33^2\minus{}.17^2)$
[/hide]


Calculati:

lim(x-(sqrt(x^2+x+1))*(ln(e^x+x)/x)) cand x->infinit

Limita ceruta este(LATEX):

$\lim_{x\to  \infty}(x-(\sqrt{x^{2}+x+1})\cdot\frac{ln(e^{x}+x)}{x})$

[quote="andyciup"]Limita ceruta este(LATEX):

$\lim_{x\to \infty}(x-(\sqrt{x^{2}+x+1})\cdot\frac{\ln(e^{x}+x)}{x})$[/quote]
  Pentru logaritm natural mai bine pui \ln, ca sa-ti apara $\ln x$.


Mexico should have won it last night, and Brazil.

Last year's Miss Universe was much better.

[quote="noneoftheabove"]what the heck is a miss universe?? :?:[/quote]

wow...i thought everyone knew what miss universe was

sorry... :blush:  :blush:  :blush:  :blush:

yeah, I definitely agree that Mexico should have won.  I think Canada must have had the best response to the question.

It's so silly to judge the final outcome based one who can make the most impractical but cliche response.  For example, "what can guys learn from girls?" --"the most important thing is to be yourself."
"what can countries do to make peace?"--"we need to remember how to help each other."

[quote="Chigr"]yeah, but she wasnt even hot enough to be miss Canada. Miss Mexico was definately the hottest, although all the Latinas taht advanced were beautiful.[/quote] I agree completely  :D  :D  :D

[quote="hello"]YOOOOOOOOOOO. ARE YOU LOSSERS:? Kidding

Canada's was the best because shne is Russian, she just lives in CAnada. She rocks and is better then all the other contestents. I think Russian women should win every year. They had one a couple of years ago but then she lost her crown.[/quote]
 :rotfl: Lol you have the hots for Russian ladies I see...

Vote for me!! I am the next year's Ms. Universe!!!  :w00t: lol jk.

lol too bad i didnt get to see it hehe
but really, theres no point in judging them based on their answers to questions like bubala said...

[quote="juicybooty911"]i agree, they should be judged only on how hoooottt  :w00t: they are, not how smart.[/quote]

Then latvia should have won because she was REALLY hot.

I don't agree with Hello.I don't think that Russian girls are the most beautiful girls in the world.Actually,my opinion
 is that Latin American are the best. In Europe, the girls from Poland and Hungary are the best.

wrong, the basic is this:
All hot girls are hot(doesn't matter where they're from)

[quote="cezar lupu"]Yes,you're right,but the question is: Where do you find these many beautiful girls ? ;)[/quote]

America ofcourse

that's bc hot girls are everywhere
the world well be a sad place if they are all in one place

If they're all in one place, the guys there won't think they're hot because there's no ugly girls to correlate with. Hotties should pay tribute to ugly girls cause then they can be actually beautiful.

[quote="hello"]Romanian guy, Russians are the best. HAHAHAH[/quote]

Russians female tennis plays are hot :wow:

A-hem.

[quote="The Beach Boys"]Well east coast girls are hip
I really dig those styles they wear
And the southern girls with the way they talk
They knock me out when I’m down there

The mid-west farmer’s daughters really make you feel alright
And the northern girls with the way they kiss
They keep their boyfriends warm at night

I wish they all could be california
I wish they all could be california
I wish they all could be california girls

<snip>

I been all around this great big world
And I seen all kinds of girls
Yeah, but I couldn’t wait to get back in the states
Back to the cutest girls in the world

I wish they all could be california
I wish they all could be california
I wish they all could be california girls

I wish they all could be california
(girls, girls, girls yeah I dig the)
I wish they all could be california
(girls, girls, girls yeah I dig the)
I wish they all could be california
(girls, girls, girls yeah I dig the)
I wish they all could be california
(girls, girls, girls yeah I dig the)[/quote]

Or, if you prefer,

[quote="The Beatles"]I'm back in the USSR
You don't know how lucky you are, boys
Back in the USSR

Well the Ukraine girls really knock me out
They leave the west behind
And Moscow girls make me sing and shout
They Georgia's always on my my my my my my my my my mind

Oh, show me round your snow peaked
mountain way down south
Take me to you daddy's farm
Let me hear you balalaika's ringing out
Come and keep your comrade warm
I'm back in the USSR
Hey, You don't know how lucky you are, boy
Back in the USSR[/quote]

Just for the record, though, anyone watching beauty pagents needs to find something better to do with his time.

[quote="noneoftheabove"]what the heck is a miss universe?? :?:[/quote]
miss universe is a beauty pageant, and they are discussing who should've won.

just for the record anyone with over 4000 posts on an internet forum needs something better to do with his time :)

[quote="Chigr"]just for the record anyone with over 4000 posts on an internet forum needs something better to do with his time [/quote]

well considering he got the account from 2003 its not that bad to get 4000

btw how do u make it so that u can say so and so wrote for the quotes??

If you want to quote just one post, instead of hitting the "Post Reply" button, hit the button at the post that says "quote" on it.  Alternatively, when you use the [ quote ] tags, you can write [ quote="quoted" ].

[quote="Quoted"]And it comes out like this.[/quote]

[quote="juicybooty911"]they are mostly in US, Europe, and South America(especially Brazil :D )[/quote]

Just a curiosity :
There  exists a   woman , mathematician, which won a Miss-(Local)  and also  a Medal at IMO ? :

[quote="noneoftheabove"]is that a question or a statement cuz i dont think there are that many girl or boy participants in the US who participate in the IMO let alone any miss whatever[/quote]

hehe UR right

puerto rican girls are the hottest ones in the world..

we have won a lot of miss universe.. i think the country with most miss univereses after US...

No they are not, Russians are the best! LOL They really are! Dont u know why US has the most winners, because they are the richest country that can afford to win, it is not a pure beauty contest.

[quote="Forsaken"][quote]All hot girls are hot(doesn't matter where they're from)[/quote]

Haha I agree!  :P

Did anyone happen to see some show where a Miss Universe tripped and her dress flew off?  :oops:  I think it was Miss Australia.[/quote]

I saw that somewhere....i didnt see the show but they showed somewhere later on one of those celebrity shows (i wasnt watching it, i just happened to see it when i was flipping through channels)

[quote="elvenchamp777"]u weird people... Chinese/Japanese/Korean girls are the hottest :D:D:D:D:D:D:D:D  Yes I'm asian.[/quote]

me too, but Asia girls are just not hot (except for 1 or 2 out of like severl billion) :(

wow, u guys need some new hobbies..................

OOOO, hold up a sec. here :!:
what do you mean :?:
hot girls are the few good things left in this world :D


If $ (A_n)_{n=1}^{\infty}\; ,\; (B_n)_{n=1}^{\infty}\; ,\; $ are  given sequences,  find a (non-trivial) particular solution of the  recursion
\[ X_{n+2}=A_nX_{n+1}+(B_{n+1}-A_n)B_nX_{n}\; \; ,\; \; \forall n\in {\mathbb
N}^*\; .\]

:blush: what do you mean by "particular" ?


Prove that the following function is equal to $ 1$ when $ j$ is prime and $ 0$ whenever $ j$ is composite:

$ F(j) = \left[\cos^{2}{\pi\frac{(j-1)!+1}{j}}\right]$

(The quantity in brackets is to be truncated to an integer)

[hide="Shortcut"]
Whenever $ j$ is prime then the expression $ \frac{(j-1)!+1}{j}$ is an integer and since $ n \in \mathbb{Z}\ , \ \cos \left( n \pi \right) \in-1 , 1 \implies \cos^{2}\left( n \pi \right) =1$
If $ j$ is composite then the expression becomes a fraction and using the floor funtion with $ x = \frac{p}{q}\not \in \mathbb{Z}\ , \ p,q \in \mathbb{Z}\ ,-1 < \cos \left( x \pi \right) < 1 \implies 0< \cos \left( x \pi \right) < 1$ yeilds $ 0$.
[/hide]


So I've been working on the same problem for this round for a few weeks now (won't say which one), and it was the last one I needed to solve. Just yesterday, I decided to wear the USAMTS shirt from last year and within an hour, I had it solved. Strange...  :|

Not nearly as good as I hope to do this year.  :oops:

[quote="mathemonster"]this is my first year participating as a 10th grader[/quote]
LOL!

Are there any other middle schoolers competing in USAMTS other than me???

[quote="FantasyLover"]me  :)

its my first time too[/quote]
well then good luck :D

[quote="Heero Kirashami"]I was a middle schooler last year...

 :ninja:[/quote]
Well then is this your first or second year participating in USAMTS?

I would've started this year as a 10th grader, but I'm in Japan. :(

lol wait dang.. I just noticed this thread essentially has no topic, thus everything is technically spam. But I don't have a USAMTS shirt either.

[quote="gauss1181"]I don't either. But this is gonna be my first year participating, as an 8th grader! :clap2:[/quote]
hi5, good luck to us.
[quote="gauss1181"]
LOL!

Are there any other middle schoolers competing in USAMTS other than me???[/quote]
i bet half of the ppl in USAMTS are elementary schoolers. but i was scared when i saw that last years "grade 8 and under" average was 23.something. And I think I failed

I'm in 7th grade and this is my first time also!!!!!!

I want a usamts shirt!!!!

EDIT: if i get an aops shirt, will i pwn on ftw?

i am in 9th grade and this is my second time doing USAMTS. last year i did not get a USAMTS t-shirt because i only did round 1.... i got a 19.

how do you get WOOT shirts, does everyone who participates in WOOT get a t-shirt

[quote="pythag011"]7th grader first time participating...

Note: WOOT shirts also work....[/quote]

you were in woot last year???

yeah, he and james4l made usamo as 6th graders....

*GASP*

and so did darryl wu.

Has he ever done usamts before?


function
calculus
derivative
calculus computations

Let the function $f$ be differentiable in the interval $[0,1]$ and $f(0)=0,f(1)=1$. Prove that there exists two different number $x,y$ such that: $f'(x)\cdot f'(y)=1$.

By the mean value theorem, $\exists x \in ]0,1[, f'(x)=1$. Whence if $f'(0)=1$, you are done. We then can assume $f'(0)>1$, without loss of generality. It is about trivial that we must have $f'(z)<1$ for some $z$ in $[0,1]$. Now using the fact that a derivative satisfies the intermediate value property, you can easily conclude.


[hide="hint"]
Use the trigonometric identity: $\sin x\sin y=\frac{\cos(x-y)-\cos(x+y)}{2}$
[/hide]

great very good solution thanks a lot :lol:


7. At Sandy's Skateboard Shop, eight of the employees are to be rewarded by a trip to the skateboard expo in St.Louis, Missouri. Sandy likes each employee equally, so she decides to determine who will go by having the 20 employees stand equally spaced on the circumference of a circle and successively count off 1,2,3,4,5,6,1,2,3,4,5,6.....repeatedly. Each person who says '6' steps out of the circle and picks up a travel ticket. What is the number of degrees in the measure of the central obtuse angle between the first and last person who receives the trip?

How would you guys do this? I actually drew, but still got it wrong..
[hide="Answer"]162 degrees[/hide]

well first by doing 360/20 you get 18 as the degrees between each person.

Then just draw a circle and cross out as you go. Label it like a clock, with each person at a spot, except with it having 20 numbers.

I'm not really sure what to say, just be sure to skip over the people who already got a trip.

By doing it out, you get 6 as the start, and 17 as the end. By counting, you get 9 spaces. so 9*18 is 162º

um if you want solutions many websites that have copies of the tests also have the solutions to the questions for example

[url]http://www.yashar.biz/mathcounts[/url]

is one of my favorites.


for you first problem:
[hide]$2^{2006} \times 5^{2007}$ $=$ $5 \times 10^{2006}$ $\rightarrow$ the sum of it digits is $5$. :D[/hide]

[quote="shadysaysurspammed"]1)Find the sum of the digits of $2^{2006}5^{2007}$

2)$a^2+2b=6$

$b^2+4c=-7$

$c^2+6a=-13$

$a,b,c$ are reals.

Find the value of $a^2+b^2+c^2$

 :)[/quote]#2 is flawed. According to hydro's solution, they don't satisfy the original equations. Try it.

[quote="10000th User"]According to hydro's solution, they don't satisfy the original equations. Try it.[/quote]
I was wondering why this happens. I don't understand where the proof is wrong...

Are there any extraneous roots somewhere?

hi aidan,haven't seen you since a long time :D

I will post what I made of the "flaw" as soon as every one has tried
EDIT:Yeah one more thing  [size=150]$WWWOHOOOOO$ my $1000th$ post[/size] (again)       :wow:  :agent:  :jump:

[quote="LordoftheMorons"]1. ${2^{2006}}{5^{2007}} = 5\cdot{10^{2006}}$
Clearly, the sum of the digits is $\boxed{5}$[/quote]
if you have 1000...000 and 5 wouldn't the sum be 6?
EDIT: i thought we were adding not mulitplying sorry...

[quote="smellyfart"][quote="kokosai"][quote="hydro"]why is my solution wrong?

[hide="2"]

sum the three equations obtaining $a^2+6a+b^2+2b+c^2+4c+1+9+4=0$
complete the squares $(a+3)^2+(b+1)^2+(c+2)^2=0$
the sum of three squares is $0$ iff all the squares are $0$
so $a=-3$ $b=-1$ $c=-2\rightarrow a^2+b^2+c^2=14$
[/hide][/quote]The implication only goes one way.  If you have $a$, $b$, and $c$ that satisfy the original three equations, they will satisfy the sum of the equations.  However, if you have $a$, $b$, and $c$ that satisfy the sum of the equations, they will not necessarily satisfy the original three equations.[/quote]

I think everyone figured that out but the question is $why$ does it happen?[/quote]No, I don't think everyone figured that out because the implication reveals a lot of things that can answer your $why$ question. See if you can find it.

[quote="smellyfart"]
I think everyone figured that out but the question is $why$ does it happen?[/quote]

Because $a, b, c$ don't satisfy the original separate equations, [hide="the implication is..."] that $a^2 + b^2 + c^2$ is not defined because the system has no solution. :) [/hide]

[quote="smellyfart"][quote="kokosai"][quote="hydro"]why is my solution wrong?

[hide="2"]

sum the three equations obtaining $a^2+6a+b^2+2b+c^2+4c+1+9+4=0$
complete the squares $(a+3)^2+(b+1)^2+(c+2)^2=0$
the sum of three squares is $0$ iff all the squares are $0$
so $a=-3$ $b=-1$ $c=-2\rightarrow a^2+b^2+c^2=14$
[/hide][/quote]The implication only goes one way.  If you have $a$, $b$, and $c$ that satisfy the original three equations, they will satisfy the sum of the equations.  However, if you have $a$, $b$, and $c$ that satisfy the sum of the equations, they will not necessarily satisfy the original three equations.[/quote]

I think everyone figured that out but the question is $why$ does it happen?[/quote]I'm talking about the nature of the logical implications here, smellyfart, not just the fact that the implications fail.  For example, here's a systems of equations from highschool Algebra:
$x+y=1$
$2x+y=2$.
There's only one solution: $(x,y)=(1,0)$.  However, say we add them up:
$3x+2y=3$.  Of course, $(x,y)=(1,0)$ is still a solution.  However, $(x,y)=(\frac{1}{3},1)$ works for the equation $3x+2y=3$, but not for the system.  The implications doesn't run both ways, because the original system is more specific; not only do they have to satisfy $3x+2y=3$, but $x$ and $y$ also have to meet certain other conditions.


trigonometry calculus function algebra domain

Hi everyone. I cannot be able to solve the question below. This is a slightly modified problem comes from the Stewart's Calculus Early Transcendentals, under the trig section. If anyone has an idea or solution to this question please show or explain. Thank you very much.

Question:
A pendulum swings back and forth. Its horizontal position x (in meters) at time t (in seconds) is given by:

x(t) = 1 + 2cos[pi/3(t+1)]

Q1: Find a formula for the inverse function t=t(x) when the domain of x(t) is restricted so that t is greater or equal to -1 and less or equal to 2. Express the answer in the form of A Arccos (B(x-a)) +b.

Q2: Find a formula for the inverse function t=t(x) when the domain of x(t) is restricted so that t is greater or equal to 2 and less or equal to 5.
Express the answer in the form of C Arcsin (D(x-c)) +d.

Hey you aren't supposed to post calculus problems here. This includes problems from calculus books.

Mods please move this to a forum that allows calculus problems. This thread doesn't belong here.

gauss1181, I don't think this is a calculus problem.

[quote="Ellen Anderson"]x(t) = 1 + 2cos[pi/3(t+1)]

Q1: Find a formula for the inverse function t=t(x) when the domain of x(t) is restricted so that t is greater or equal to -1 and less or equal to 2. Express the answer in the form of A Arccos (B(x-a)) +b.

Q2: Find a formula for the inverse function t=t(x) when the domain of x(t) is restricted so that t is greater or equal to 2 and less or equal to 5.
Express the answer in the form of C Arcsin (D(x-c)) +d.[/quote]

I won't solve it for you, but I'll give you a
[hide="hint"]Start with $ x\equal{}1\plus{}2\cos(\frac{\pi}{3}(t\plus{}1))$ and try to get $ t$ by itself.

Don't forget about the domain and range of the function. Between -1 and 2, x is decreasing; between 2 and 5, x is increasing.[/hide]


analytic geometry
geometry solved
geometry

This is a 5-minute quickie (disclaimer in case the next person who posts asks me whether I really meant to post this question because it is too easy):

A dead fly lies on a 1m by 1m square floor.  A spider, which is also on the same floor, wants to eat the fly.  The spider makes a move by selecting one of the corners of the floor, and moving to the midpoint of that corner and the spider's current position.  Prove that the spider can move in such a way that after exactly 8 moves, the spider is less than 1cm away from the fly.

The problem is equivalent to this:

We take one point (x,y) s.t. 0<=x,y<=1. After 1 step we replace (x,y) with the four points (x/2,y/2), (x/2,(y+1)/2), ((x+1)/2,y/2), ((x+1)/2,(y+1)/2). After 2 steps we do the same for each of the previous points to get 4^2 points. We continue this until we get 4^8 points. We have to show that given any point X inside the square there is a point from our set of 4^8 points at distance <=1cm from X.

I didn't exactly continue, but it's rather easy now (I think) because the 4^8 points have coordinates of the type (x/2^8+m/2^8, y/2^8+n/2^8), where 0,=m,n<=2^8-1, meaning that they are the vertices of a network of smaller squares of side 1/2^8.

That's it, you have the "standard" solution, grobber. :)

The quickie solution uses only 1 tiny calculation, and a brilliant imagination: (highlight to see)

[color=white]Pretend the fly is alive (that's where your imagination comes in ;) ) and that the fly can move in the same way as the spider.  If during every turn, the spider and the fly both move to the same corner, then the distance between the spider and the fly is halved each turn.

Now imagine that the fly's current position is actually its position after 8 moves.  Make the fly "backtrack" 8 moves, then make the fly "move ahead" to its final position, and each time the fly moves ahead the spider moves to the same corner as the fly.

Final distance between the spider and the fly < 100*sqrt2/256 < 1.[/color]

My god! This is too beutiful!  :D  :D

Where was this proposed?

It wasn't proposed anywhere.  When I went to China last year for an exchange programme, one of the maths olympiad trainers gave this problem during our lesson break.


A powderman set a fuse for a blast to take place in $ 30$ seconds. He ran away at a rate of $ 8$ yards per second. Sound travels at the rate of $ 1080$ feet per second. When the powderman heard the blast, he had run approximately:

$ \textbf{(A)}\ 200 \text{ yd.} \qquad\textbf{(B)}\ 352 \text{ yd.} \qquad\textbf{(C)}\ 300 \text{ yd.} \qquad\textbf{(D)}\ 245 \text{ yd.} \qquad\textbf{(E)}\ 512 \text{ yd.}$

[hide="Click for solution"]
Times are inversely proportional to rates, so $ \frac{t}{30\minus{}t}\equal{}\frac{\frac{1080}{3}}{8}\equal{}45 \implies t\equal{}30 \cdot \frac{45}{46}$ and $ d\equal{}8t\equal{}8 \cdot 30 \cdot \frac{45}{46} \approx 245$, or $ \boxed{\textbf{(D)}}$.
[/hide]


You *could* consider this as a calculus problem, but I think this is a good place for it.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous increasing function. Let $r \in \mathbb{R}$.

Show that the equation $x + f(x) = r$ has exactly one real solution $x$.

$x + f(x)$ is continuous and strictly increasing.

Since f is increasing, $lim\ f(x)$ when $x \rightarrow \pm \infty$ exists and is either finite or $\pm \infty$. In any case $lim\ x + f(x) = \pm \infty$ when $x \rightarrow \pm \infty$

Existence and unicity and $r$ is then a consequence of usual real-line calculus results.


Let $ x_i$ is reals. (i>=1, $ \sum^n x_i \equal{}n$) prove

$ \frac{\sum^n x_i^3}{\sum^n x_i^4} ( \sum^n (nx_i^4\minus{} x_i^3))\ge n^2\minus{}n$

[quote="4865550150"][quote="fjwxcsl"][quote="4865550150"]Let $ x_i$ is reals. (i>=1, $ \sum^n x_i = n$) prove

$ \frac {\sum^n x_i^3}{\sum^n x_i^4} ( \sum^n (nx_i^4 - x_i^3))\ge n^2 - n$[/quote]

The inequality does not hold for $ X = (x_1,x_2,\ldots,x_n)\in{R^n},$

But it holds for $ X\in{R^n_{ + }.}$[/quote]

if it is positive? The proof is?[/quote]

$ x_1^3+x_2^3+...+x_n^3=u$

$ x_1^4+x_2^4+...+x_n^4=v$

$ \textbf{1)} \quad x^3 \geq 3x-2$ for all $ x$ positive reals.

$ \Longrightarrow \; u \geq n$

$ \textbf{2)} \quad x^4 \geq x^3+x-1$ for all $ x$ positive reals.

$ \Longrightarrow \; v \geq u$

$ \Longrightarrow \; \frac{-u^2}{v} \geq -u$

$ un-\frac{u^2}{v} \geq un-u=u(n-1) \geq n(n-1)=n^2-n$

Solution is done. :wink:

[quote="crazyfehmy"][/quote][quote="4865550150"][quote="fjwxcsl"][quote="4865550150"]Let $ x_i$ is reals. (i>=1, $ \sum^n x_i = n$) prove

$ \frac {\sum^n x_i^3}{\sum^n x_i^4} ( \sum^n (nx_i^4 - x_i^3))\ge n^2 - n$[/quote]

The inequality does not hold for $ X = (x_1,x_2,\ldots,x_n)\in{R^n},$

But it holds for $ X\in{R^n_{ + }.}$[/quote]

if it is positive? The proof is?[/quote][quote="crazyfehmy"]

$ x_1^3 + x_2^3 + ... + x_n^3 = u$

$ x_1^4 + x_2^4 + ... + x_n^4 = v$

$ \textbf{1)} \quad x^3 \geq 3x - 2$ for all $ x$ positive reals.

$ \Longrightarrow \; u \geq n$

$ \textbf{2)} \quad x^4 \geq x^3 + x - 1$ for all $ x$ positive reals.

$ \Longrightarrow \; v \geq u$

$ \Longrightarrow \; \frac { - u^2}{v} \geq - u$

$ un - \frac {u^2}{v} \geq un - u = u(n - 1) \geq n(n - 1) = n^2 - n$

Solution is done. :wink:[/quote]

simple proof is the best. good job! congrats!


Consider the set $A=\{1,2,...,49\}$. We partitionate $A$ into three subsets. Prove that there exist a set from these subsets containing three distincts elements $a,b,c$ such that $a+b=c$

This problem was proposed by Romania at IMO1983 (you were not born Erdos at that time :P ).

Can somebody prove that if you color the edges a $K_{49}$ in three colors then there exists a monochromatic triangle?
which theorem on ramsey theory implies it?

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=acta+arithmetica&t=547

Pierre.


quadratics
number theory unsolved
number theory

a) Determine those odd primes $ p$ for which $ (\minus{}3 \mid p) \equal{} 1$ and those for which $ (\minus{}3 \mid p) \equal{} \minus{}1$.

b) Prove that 5 is a quadratic residue of an odd prime $ p$ if $ p \equiv \pm 1 (\text{mod } 10)$, and that 5 is a nonresidue if $ p \equiv \pm 3 (\text{mod } 10)$.

Use quadratic reciprocity. Nothing more to do.


linear algebra
matrix
ratio
function
probability and stats

Hi listers,
Does anybody has any references about how I would calculate the statistic of the likelihood ratio test in order to compare two covariance matrix from a multivariate normal sample!
I would like to obtain the likelihood ratio statistic from N(mean,sigma) p variables iid according to the hypothese H:Sigma=Sigma0
Thanks,
Marcio

We did this in Statistics 4 last spring.
I think the final ration was some monotonic function trace by determinant of the the covariance matrix.
If you can tell me how far you've done, I should be able to help.
Otherwise, I'll need some time to reconstruct the proof.
Have you done any of these tests for Normal before?
They follow a similar pattern.

My objective is to calculate the ratio of the likelihood function according to the two hypotheses.
ratio=H0 / H1
ratio=max L(sigma0) / max L(sigma)
Calculating the likelihood estimator of sigma I obtain trace(V)/N
My doubt is when I maximize the likelihood function over H0, I mean maximize sigma0
If I consider that the estimator of sigma0 is trace(V0)/N
I don't end up with a reasonable result in my opinion...
I think that there is something that I am not taking into account...
Thanks...


group theory
abstract algebra
algebra
function
domain
induction
linear algebra

Are all subgroups of $\mathbb{Z}^n$ isomomorphic to $\mathbb{Z}^k$.

Yes, this is a very well-known result. More generally, every submodule of a free module of rank $n$ over a principle ideal domain is free of rank at most $n$.

yes, of course it is well known ( for the case of $Z^n$ at least).
And as I know approximately notinig about module, could you post a proof of the more general statement you've just mentionned in your previous post ? I'm interested  :)
thanx

I don't know the proof by heart, and I don't have the time to think it through right now, sorry :). Too many other nice problems around here, so it's a busy time for me :).

The proof for the problem you posted is direct by induction, alekk.

alekk: Have you noticed that the administrators recently split algebra into two pieces? One is linear and matrix algebra; the other is groups/rings/fields. The linear and matrix algebra forum is busy enough already; it doesn't need problems that might be better placed in the other algebra forum.

Here's a proof, which is essentially Gaussian elimination and can be converted into a usable algorithm in a Euclidean domain.

Consider a submodule $M$ of $R^n$ for some p.i.d. $R$.
Step 0: Write a generating set for $M$, expressed with respect to the standard basis, as the rows of a matrix.

Step 1: The first coordinates of this set of generators generate an ideal in $R$, which is the set of possible first coordinates of elements of $M$. This ideal is generated by some element $d$. If $d$ is nonzero, find a linear combination of the rows with first coordinate $d$; this linear combination is our first basis element $v_1$. If $d$ is zero, skip step 2.

Step 2: Subtract an appropriate multiple of $v_1$ from every row of the matrix to zero the first coordinate.

Step 3: Ignore all zero columns, and repeat the process of steps 1 and 2 with the first nonzero column each time, adding on basis vectors and clearing the columns. The process terminates when all columns are zero.

Since at least one column is zeroed for every basis vector, the basis has at most $n$ elements. It is also clear that this basis generates $M$, and it is free because of its triangular structure.