quadratics
algebra
difference of squares
special factorizations

1. If $ 0<a<b<c<d\in\mathbb{Z}$ and $ 2b\equal{}a\plus{}c,c^2\equal{}bd,d\minus{}a\equal{}30$, find $ a\plus{}b\plus{}c\plus{}d$.

2. If $ n$ is an integer between 100 and 999 inclusive, and $ n'$ is the integer formed by reversing the digits of $ n$. How many possible values are for $ |n\minus{}n'|$?

[hide="#1"]
From the middle equation, we have $ (2b \minus{} a)^2 \equal{} b(a \plus{} 30)$. Let $ b \equal{} a \plus{} m$. We then have $ 4m^2 \plus{} (3a \minus{} 30)m \minus{} 30a \equal{} 0$. The discriminant of this quadratic must be a perfect square, so we have $ 9a^2 \plus{} 300a \plus{} 900 \equal{} k^2$. Complete the square to get $ (3a \plus{} 50)^2 \equal{} k^2 \plus{} 1600$. Use difference of squares to get $ (3a \plus{} k \plus{} 50)(3a \minus{} k \plus{} 50) \equal{} 1600$. So the sum of these two factors is $ 6a \plus{} 100$. We find (8 and 200) and (800 and 2) are the possible factor pairs whose sum is $ 6a \plus{} 100$. 800 and 2 gives $ a \equal{} 117$ which doesn't give an integer value for m. 8 and 200 gives $ a \equal{} 18$, $ m \equal{} 9$, $ b \equal{} a \plus{} m \equal{} 27$, $ d \equal{} 48$, $ c \equal{} 2b \minus{} a \equal{} 36$
So $ a \plus{} b \plus{} c \plus{} d \equal{} 129$[/hide]

[hide="#2"]
Let the numbers be $ 100a \plus{} 10b \plus{} c$ and $ 100c \plus{} 10b \plus{} a$. So the possible values are $ 99|a \minus{} c|$. $ |a \minus{} c|$ can range from 0 to 9, if we let $ n'$ start with a 0 digit. Thus, there are 9 possibilities. [/hide]
 :)

EDIT: O yeah, 0-9 is 10 numbers, not 9  :wallbash:

[hide]
2.Let $ n\equal{}100a\plus{}10b\plus{}c$, where $ a,b,c$ are the digits of $ n$. Then $ n'\equal{}a\plus{}10b\plus{}100c$.
$ |n\minus{}n'|\equal{}|99a\minus{}99c|\equal{}99|a\minus{}c|$.
There are 10 distinct values of $ |a\minus{}c| (0\minus{}9)$, so there are 10 distinct values of $ |n\minus{}n'|$.
The fact that $ a\not\equal{} 0$ isn't exactly pertinent to solving this problem.
[/hide]

For the second problem, I got 10 too (0 to 9), but the imomath test site says the answer is 9 (in fact only 32% people got this problem "correct")... or otherwise I wouldn't have posted such trivial problem.

EDIT: I might as well post what I got for #1

$ (2b\minus{}a)^2\equal{}b(a\plus{}30)$

$ 4b^2\plus{}a^2\minus{}4ab\equal{}ab\plus{}30b$

$ 4b^2\minus{}(5a\plus{}30)b\plus{}a^2\equal{}0$

$ discriminant\equal{}(5a\plus{}30)^2\minus{}16a^2\equal{}9a^2\plus{}300a\plus{}900\equal{}(3a\plus{}50)^2\plus{}650\equal{}k^2$

$ (k\plus{}3a\plus{}50)(k\minus{}3a\minus{}50)\equal{}2\cdot5^2\cdot13$

but the two factors must have the same parity, which is impossible since the power of 2 is 1 in the prime factorization...

(gosh I just wasted 30 minutes checking my work from the beginning to end but couldn't find any mistake... :wallbash_red:  :wallbash: )

If the answer is really $ 9$, then the only plausible reason I can think of is that both $ n$ and $ n'$ must be between $ 100$ and $ 999$ inclusive. Then $ |a\minus{}c|\not\equal{}9$, which leads to 8 values.


The wheel of a moving bicycle makes 20 revolutions and has a 27'' diameter. How many feet does the center of the wheel travel? Express your answer in terms of $ \pi$.

[quote="RunpengFAILS"]The center travels the same distance as the bike travels. The circumference of the wheel is $ 27\pi$. In 20 revolutions, the bike travels $ \boxed{540\pi}$[/quote]

if you are correct, then the real answer of $ \boxed{45\pi}$ is incorrect

but how do they get $ 45\pi$?

The discrepancy lies in the fact that Runpeng's answer is in inches, while the question asks for it in feet.


how do you caculate the how mnay digits for the decimal representation 1/2007 before it reapeats?

[quote="AstroPhys"]This was answered somewhere else before I think.[/quote]

uh i searched for it but it doesnt appear..

[hide=" possible :idea: "]
The maximum number of digits in the repeating block of $\frac{1}{n}$ is $n-1$.  Hope that helps...[/hide]

[quote="vishalarul"][url=http://mathworld.wolfram.com/MultiplicativeOrder.html]Try this link[/url][/quote]
ok using that we are trying to find what 10^x=1 mod 2007? wehre x is the first positive number satisfying that, so 10^1332 is 1 mod 2007, but how do I know if its not a multiple of 1332 (such as 222)?

[quote="AstroPhys"]This was answered somewhere else before I think.[/quote]
Thought so, thanks i_like_pie.


limit
real analysis
real analysis unsolved

if $ \lim_{n \to \infty}(a_n \plus{} a_{n\plus{}1})$ exist. of course $ \lim_{n \to \infty}|a_n \plus{} a_{n\plus{}1}|$

the problem would be more logicall if the question were

$ \lim_{n \to \infty}(|a_n |\plus{} |a_{n\plus{}1}|)$

[quote="EulerIntegral"]if $ \lim_{n \to \infty}(a_n \plus{} a_{n \plus{} 1})$ exist. of course $ \lim_{n \to \infty}|a_n \plus{} a_{n \plus{} 1}|$

the problem would be more logicall if the question were

$ \lim_{n \to \infty}(|a_n | \plus{} |a_{n \plus{} 1}|)$[/quote]

Typo, it was supposed to be a minus sign.

[quote="hadjimurad"]I have a problem from my analysis homework that I just can't solve. Let $ \{a_n\}$ be a sequence such that $ A \equal{} \lim_{n \to \infty}(a_n \plus{} a_{n \plus{} 1})$ and $ B \equal{} \lim_{n \to \infty}(a_n * a_{n \plus{} 1})$ both exist. Prove that:

a) $ A^2 \ge 4B$
b) $ \lim_{n \to \infty}|a_n \minus{} a_{n \plus{} 1}|$ exists.

My friends in the class solved a) by interchanging the limit operation with the square operation, but $ a_n$ is not guaranteed to converge, so I'm skeptical of whether this idea works. I'm pretty stumped on the second one. Any hints would be appreciated.[/quote]
The second part pretty much tells you what to do. Consider
\[ c_n \equal{}  (a_n \minus{} a_{n\plus{}1})^2 \equal{}  (a_n \plus{} a_{n\plus{}1})^2 \minus{} 4 a_n a_{n\plus{}1}  \, .\]
By assumption $ \lim_n c_n \equal{} A^2 \minus{} 4B$ exists and of course is non-negative. Hence also  $ \lim_{n \to \infty}|a_n \minus{} a_{n \plus{} 1}|$ exists.

Your mistrust of your friends' solution is justified. Take for example $ a_n \equal{} (\minus{}1)^n$. All assumptions and therefore all conclusions are correct, but the sequence does not converge.

[quote="hpe"][quote="hadjimurad"]I have a problem from my analysis homework that I just can't solve. Let $ \{a_n\}$ be a sequence such that $ A \equal{} \lim_{n \to \infty}(a_n \plus{} a_{n \plus{} 1})$ and $ B \equal{} \lim_{n \to \infty}(a_n * a_{n \plus{} 1})$ both exist. Prove that:

a) $ A^2 \ge 4B$
b) $ \lim_{n \to \infty}|a_n \minus{} a_{n \plus{} 1}|$ exists.

My friends in the class solved a) by interchanging the limit operation with the square operation, but $ a_n$ is not guaranteed to converge, so I'm skeptical of whether this idea works. I'm pretty stumped on the second one. Any hints would be appreciated.[/quote]
The second part pretty much tells you what to do. Consider
\[ c_n \equal{} (a_n \minus{} a_{n \plus{} 1})^2 \equal{} (a_n \plus{} a_{n \plus{} 1})^2 \minus{} 4 a_n a_{n \plus{} 1} \, .
\]
By assumption $ \lim_n c_n \equal{} A^2 \minus{} 4B$ exists and of course is non-negative. Hence also  $ \lim_{n \to \infty}|a_n \minus{} a_{n \plus{} 1}|$ exists.

Your mistrust of your friends' solution is justified. Take for example $ a_n \equal{} ( \minus{} 1)^n$. All assumptions and therefore all conclusions are correct, but the sequence does not converge.[/quote]

Thanks, that helped a lot. They put this question on McGill's Ph.D. entrance examination last year.


If $ a_{0}\equal{} x$ and for all positive integers $ n$, $ a_{n}\equal{} p\cdot a_{n\minus{}1}\plus{}q$. Find the explicit value of $ a_{n}$

What if $ a_{n}\equal{}p\cdot q^{a_{n\minus{}1}}\plus{}r$?

[hide="Part 1"]Let $ b_{n}: \equal{} a_{n}\minus{}k$ for some constant $ k$ that will be determined for convenience:
$ b_{n}\plus{}k \equal{} p b_{n\minus{}1}\plus{}pk\plus{}q$

Now, consider $ k :\; k \equal{} pk\plus{}q\implies k \equal{}\frac{q}{1\minus{}p}$
We now have: $ b_{n}\plus{}k \equal{} p b_{n\minus{}1}\plus{}k\implies b_{n}\equal{} p b_{n\minus{}1}\implies b_{n}\equal{} p^{n}b_{0}$

Now, recall that $ b_{n}\equal{} a_{n}\minus{}k \equal{} a_{n}\minus{}\frac{q}{1\minus{}p}\equal{} a_{n}\plus{}\frac{q}{p\minus{}1}$
Then, $ b_{n}\equal{} p^{n}b_{0}\implies a_{n}\plus{}\frac{q}{p\minus{}1}\equal{} p^{n}\left( a_{0}\plus{}\frac{q}{p\minus{}1}\right)$

Simplified, $ \boxed{a_{n}\equal{} p^{n}a_{0}\plus{}(p^{n}\minus{}1)\frac{q}{p\minus{}1}}$[/hide]

[hide="Part 2 Start"]
$ b_{n}: \equal{} a_{n}\minus{}r$, $ p q^{r}\equal{} k$
$ a_{n}\minus{}r \equal{} p q^{r}q^{a_{n\minus{}1}\minus{}r}\implies b_{n}\equal{} (p q^{r}) q^{b_{n\minus{}1}}\implies b_{n}\equal{} k q^{b_{n\minus{}1}}$

I don't know where to go from here?
[/hide]

[quote="The Zuton Force"]

$ b_{n}: \equal{} a_{n}\minus{}r$, $ p q^{r}\equal{} k$
$ a_{n}\minus{}r \equal{} p q^{r}q^{a_{n\minus{}1}\minus{}r}\implies b_{n}\equal{} (p q^{r}) q^{b_{n\minus{}1}}\implies b_{n}\equal{} k q^{b_{n\minus{}1}}$

I don't know where to go from here?
[/quote]
You can write $ \frac{b_{n}}{k}\equal{} (q^{k})^{\frac{b_{n\minus{}1}}{k}}$

And so $ c_{n}\equal{} u^{c_{n\minus{}1}}$

and now I think there are no more simplification available and no general explicit formula for $ c_{n}$.

Right, thanks, that's exactly what I was looking for  :)

But again, not possible to write compactly (unless $ c_{0}\equal{} u$, in which case we can use the notation for tetration).


I think you 12th std guys can help me here....
Solve for real a
$ 10^{a^3\plus{}5a^2\minus{}12a\minus{}2}\equal{}a$
Actual question was something else...i substituted $ log_{10}(x\minus{}2)\equal{}a$

[quote="pardesi"]can u post the question in the actual form[/quote]

Sure.
$ (x\minus{}2)^{log^3(x\minus{}2)\plus{}log(x\minus{}2)^5\minus{}12}\equal{}10^2log(x\minus{}2)$

all logs are in base 10.


abstract algebra
group theory
superior algebra
superior algebra solved

If $G$ is a group, and if $H_1,H_2,N$ are subgroups of $G$ such that $H_1 \lhd H_2$ and such that $H_2/H_1$ is Abelian, is it true that $(H_2 \cap N)/(H_1 \cap N)$ must be Abelian as well? :)

$(H_2\cap N)/(H_1\cap N)$ can be naturally identified to a subgroup of $H_2/H_1$. A subgroup of an abelian group is obviously abelian.

Let $Q'=(H_2\cap N)/(H_1\cap N)$ and $Q=H_2/H_1$. The map $f(a\cdot (H_1\cap N))=a\cdot H_1$ is a homomorphism from $Q'$ to $Q$, and it must be injective; if $a,b\in H_2\cap N$ and $a\cdot H_1=b\cdot H_1$, we have $b^{-1}\cdot a\in H_1$ and $b^{-1}\cdot a\in N$, so $b^{-1}\cdot a\in H_1\cap N$ and $a\cdot (H_1\cap N)=b\cdot (H_1\cap N)$.

This same argument works for quotients in essentially any algebraic structure; we just need to interchange some symbols.


Find all functions $f: \mathbb N \to \mathbb N$, for which the relation $ f (f (n)) + (f (n))^2 = n^2 + 3n + 3$ holds for all $n \in \mathbb N$.

[quote="tdl"]Find all functions $ f : N\rightarrow N$, for which the relation $ f (f (n)) \plus{} (f (n))^2 \equal{} n^2 \plus{} 3n \plus{} 3$ holds.[/quote]

If $ f(n)>n\plus{}1$, we have $ f(n)\geq n\plus{}2$ and so $ f(f(n))\leq n^2\plus{}3n\plus{}3\minus{}(n\plus{}2)^2\equal{}\minus{}n\minus{}1<0$, which is impossible. So $ f(n)\leq n\plus{}1$ $ \forall n$

If $ 0<f(n)<n\plus{}1$, we have $ f(f(n))>n^2\plus{}3n\plus{}3\minus{}(n\plus{}1)^2\equal{}n\plus{}2>f(n)\plus{}1$ and so $ f(m)>m\plus{}1$ with $ m\equal{}f(n)$ and this is impossible (see line above).

So we must have $ f(n)\equal{}n\plus{}1$ $ \forall n$ and it is easy to check that this necessary condition is sufficient.


There are 7 hat colors. 7 people get 1 hat each which can be any colors e.g. they can be all the same. Each person is able to see the colors of all the rest. When everyone is ready, they have to guess their color by saying out SIMULTANEOUSLY. They will win if at least 1 person guesses correctly. They can make a plan in advance.

How do they win?

What types of colors are the hats? I mean, what if one of them is like "my hat is orange" and the guy who checks if they said it right says "no, it's macaroni and cheese orange" :maybe:

In the original problem, it is rainbow color. It's easier to say color 1-7.

oh, oh, oh, i totally remember doing a problem like this in sixth grade, if only i could remember the answer.

I'll get back to you on this.

What are they allowed to do before everyone is ready? I'm assuming they aren't allowed to talk... I have an idea though.[hide]Suppose they are standing in a circle. Before hand, they number the colors 1-7. Before everyone is ready, they hold up the number of the color of the hat of the person to the right of them, and look to see what the person to the left of them held up for their own color. This way everybody gets it right.

I think something is wrong with my answer, but not sure what. :( [/hide]

[quote="laughinghead505"]Suppose they are standing in a circle. Before hand, they number the colors 1-7. Before everyone is ready, they hold up the number of the color of the hat of the person to the right of them, and look to see what the person to the left of them held up for their own color. This way everybody gets it right.

I think something is wrong with my answer, but not sure what. [/quote]

I don't think they are allowed to communicate to each other riight before the guess, but they can only plan before the hats are handed out.

OH! OH! I think I know!!! :lol: [hide]They all stand in a circle. Before hand, they chose one person to save. When they are all gathered, the other six will slouch or do something like that (decided before hand) if the color of the hat of the person who was chosen is the same as the color of the person next to them. That way the one person will know his color.

Is it right??? :) [/hide]

and if there isn't a person with the same color they will do nothing

but I do not think that this solution is allowed :D

They are not allowed to send any signal. Only thing they can do is saying out a color.

[hide="Hint"]Try 2 people with 2 colors first.[/hide]

[hide="2 people"]They decide before hand that one of them will say the color of the other person and the other person will say the color that is not the other person's color. For example: A and B are the two people and the colors of the hats are 1 and 2. They decided that A will say the color of B and B will say the color that is not of A. If A's hat is 1 and B's hat is 2, then A will say 2 and B will say 2.[/hide]

there's several strategies that give them a good probability of winning, but can they always win?
[hide="for example"]they could all say the color of the person to the right of them.  the probability this would work is the probability that two adjacent people have the same color.
[/hide]

There's a way to be sure that at least one person answers it correctly.

I think that it is impossible because no matter what color everyone says, they could just plain be wrong. Is this a trick question?

Well, with two people, no matter what they say they could just be plain wrong, but there's a way to do it... :maybe:

Oh...I see how to do it now.
[hide]
They together, assign each color a number, ie 1 2 3 4 5 6 0.  When they go out there, the first person says the color with number congruent to one plus the additive inverse of the sum of all the other hats (modulo 7).  The second person says 2 plus the sum of the additive inverse of the sum he sees.  And so on.  If the sum of all of the hats there is $k\mod{7}$, then the person with number congruent to $k\mod{7}$ will answer correctly (and only that person)

This is true because the first person is saying what color his hat will be if the sum is 1mod7, the second person is saying what color his hat will be if the sum is 2mod7, and so on.  One of them is right.
[/hide]

Finally got it.

Number the people 0 to 6. Number the colour 0 to 6. Let $a_{n}$ be the colour that #n person have.

Let N be the sum of products of the number of colour and the number of person :roll:
ie) $\Sigma n a_{n}$
Note that $a_{0}$ does not affect N.

This is how they guess:
person 1 will guess so that the N according to the guess and the others colour is $1+a_{0}$ mod 7. Likewise, person n will guess so that N is congruent to $n+a_{0}$ mod 7, where n>0. This is possible because (n,7)=1. person 0 will guess N mod 7. This is also possible, because this person knows N.

This is why it works:
There are 7 possible value of N mod 7, 0 to 6. Person 1 to 6 are covering every possible case but $a_{0}$ case. Yet, person 0 is guessing N mod 7 as his/her colour. Therefore, we have covered all cases.

warning:  This solution may not be clear, since I'm writing this near midnight :sleep2: (11:42 to be exact). Ask me to clear things up.

[hide="My solution"]Let the colors be 0,1,2,3,4,5,6. Each person finds the sum of the others' hat colors.
Person 1 says color that makes the sum congruent 1 mod 7.
Person 2 says color that makes the sum congruent 2 mod 7.
and so on.

This works because their hat colors sum has to congruent either 0,1,2,3,4,5 or 6.

For my solution, exactly one person answers correctly.[/hide]

[hide="stupid solution"]For the first person, one of the others gives him a secret and almost invisible sign, telling him what color his hat is. Then the others do that. Therefore, everyone is saved.[/hide]


function
geometry
geometric transformation
reflection
vector
MATHCOUNTS
number theory

How would you go about proving:

1) Additive and commutative properties of addition and multiplication?

2) Theorems related to the above (example, subtraction and division properties)?

3) $ a^{x}\times a^{y}= a^{x+y}$ ?

4) That the inverse of a function such as $ y = x^{2}+4$ can be found by interchanging the x and y variables?

I know it's quite possible that these are all accepted as definitions, but if there's the slightest chance that any of them can be derived from something, I want to know how. Hopefully no-one found that confusing  :oops:

Definition of an inverse function is that given $ y=f(x)$ and $ y=f^{-1}(x)$ is its inverse:

$ f(f^{-1}(x)) = f^{-1}(f(x)) = x$. So essentially $ y= f^{-1}(x)$ is the reflection of $ y=f(x)$ about the line $ y=x$. Which is why all you do is inter-change the variables.

For no.2)Subtraction and Division is defined as the inverse of addition and multiplication respectively, so theorems about subtraction and division comes from no.1)

[quote="Prime factorization"]How would you go about proving:

1) Additive and commutative properties of addition and multiplication?

2) Theorems related to the above (example, subtraction and division properties)?

3) $ a^{x}\times a^{y}= a^{x+y}$ ?

4) That the inverse of a function such as $ y = x^{2}+4$ can be found by interchanging the x and y variables?
[/quote]

The first 3 can be proved using a formal system with the Peano Axioms (check out the book Godel Escher Bach for a very lovely explanation of formals systems and the foundations of mathematics and lots of other cool math-philosophical subjects). The difficulty discovered in 1931 by Godel was that while you can prove arithmetic expressions are true using any suitable Formal System, there is no way to prove that any given formal system is both contradiction-free and complete.

The last does not require a proof. It is a definition. And as my theory of computation teach always told us, a definition is never wrong. It could be useless, but it's never wrong. The way you know it is a definition is that it simply gives a name to a statement. In the case of inverse functions, when you say:

$ y = f^{-1}(x)$, what is actually meant is the function which satisfies $ x = f^{-1}(y)$ (if one even exists). It is very important to be able to recognize the difference between definitions and theorems. Anyone can come up with definitions for any field. What matters is how useful those definitions are when it comes time to start proving theorems.

According to [url]http://en.wikipedia.org/wiki/Peano_axioms[/url], these axioms seem to only hold for natural numbers  :( I don't see why 4.7 + 3.8 wouldn't equal 3.8+4.7, which is the commutative property of addition if I'm not mistaken.

 :huh:

Also, from what I read, the concept seems a little beyond elementary as it involves number theory and set theory. I'm trying to avoid such topics until I'm older and prepared for such material  :blush:

You sure there isn't a simpler way to solve it? Perhaps you could show me a proof of just one axiom so I could pick up on the concept.

[quote="Prime factorization"]According to [url]http://en.wikipedia.org/wiki/Peano_axioms[/url], these axioms seem to only hold for natural numbers  :( I don't see why 4.7 + 3.8 wouldn't equal 3.8+4.7, which is the commutative property of addition if I'm not mistaken.[/quote]

Rational and Real numbers are just illusions of integers. A rational number is merely an ordered pair of Integers. A real number is just a function from Integers into the Rationals whose images tend to closer together as the argument of the function increases (Cauchy Sequences).

The math is no fun to work out. Proofs at this level are extremely mechanical and don't have the elegance of "higher level" proofs found in the rest of mathematics.

[quote]Also, from what I read, the concept seems a little beyond elementary as it involves number theory and set theory. I'm trying to avoid such topics until I'm older and prepared for such material  :blush:[/quote]

Learn as much as you can as soon as you have the time, and you'll surprise yourself someday with how much of it you actually understood.

[quote]You sure there isn't a simpler way to solve it? Perhaps you could show me a proof of just one axiom so I could pick up on the concept.[/quote]

There is a good example of the transitive property of integer addition in Godel Escher Bach, the book I spoke up. However, it takes two solid pages. These proofs aren't elegant. If you know anything about programming, Euclid's proof of the infinitude of primes would be analogous to a program written in Python, Java, or C, while elementary proofs like this would be like programming in assembler or raw machine code.

The proofs simply demonstrate the correctness, they do not paint a satisfactory picture of why those statements in arithmetic are true. The best proof for human consumption about the nature of arithmetic are probably given in terms of wooden blocks or small stones. Rigorous proofs are extremely useful when you can't visualize what you're talking about (for example, infinite dimension vector spaces or wave functions in Physics). They aren't really interesting when it comes to the fundamentals.

[quote="Prime factorization"]these axioms seem to only hold for natural numbers  :( I don't see why 4.7 + 3.8 wouldn't equal 3.8+4.7[/quote]

As I said in a previous MathCounts thread, definitions are only kept because they are useful.  The structure of the rationals and the reals is defined so that they continue to satisfy properties like associativity, commutativity, and distributivity.

This is distinct from the topic of the rigorous construction of the reals, which is fairly complicated and beyond the scope of this board.


algebra
polynomial
calculus
integration
linear algebra
matrix

Let R be an integral domain. and $ S \equal{} \{1,a,a^2,...\}, a\not \equal{} 0$
Is $ R_S$, the localization of $ R$ at $ S$ isomorphic to $ R[x] / ( \minus{} 1 \plus{} ax)$ ?

I am having trouble proving this. I got that $ R[X] /( \minus{} 1 \plus{} ax)$ is equivalent to $ F \equal{} \{rf \text{\big{|}} \minus{} 1 \plus{} af \equal{} 0, r\in R\}$ so $ f \equal{} a^{ \minus{} 1}$. So $ F \equal{} \{\frac ra\ | r\in R\}$. Is this $ R_S$?? I can't prove the isomophism part :(
  $ T[x] / (x^2 \plus{} 1)$ is equivalent to all elements of the form $ t_1 \plus{} t_2 \alpha$ where $ \alpha^2 \plus{} 1 \equal{} 0$, that is $ \alpha$ is a root of the polynomial we're modding with and is easily extended to problems like $ \mathbb{Z}_2[X] / x^2 \plus{} x \plus{} 1 \equal{} \{ a \plus{} b\beta \}$ where $ \beta$ is a root of $ x^2 \plus{} x \plus{} 1$ (Hence it has 4 elements =) ).

Thanks a bunch

For the picture to make sense, you have to include the higher-degree elements. I'd go with constructing the appropriate homomorphisms in both directions, and showing they're mutual inverses.

From $ R_S$, we set $ f\left(\frac{b}{a^n}\right)\equal{}[bx^n]$.
From $ R[x]$, we set $ g(x)\equal{}\frac1a$ and generate the rest from there; since $ g(ax\minus{}1)\equal{}a\cdot \frac1a\minus{}1\equal{}1\minus{}1\equal{}0$, $ g$ factors through the quotient $ R[x]/(ax\minus{}1)$; call the map from the quotient $ g'$

We have $ g'\circ f\left(\frac{b}{a^n}\right)\equal{}g'([bx^n])\equal{}b\cdot\left(\frac1a\right)^n\equal{}\frac{b}{a^n}$. That direction is the identity.
We have $ f\circ g'([b_0\plus{}b_1x\plus{}\cdots \plus{}b_nx^n])\equal{}f\circ g'(b_0)\plus{}f\circ g'(b_1x)\plus{}\cdots \plus{}f\circ g'(b_nx^n)$
$ \equal{}f(b_0)\plus{}f\left(\frac{b_1}{a}\right)\plus{}\cdots\plus{}f\left(\frac{b_n}{a^n}\right)\equal{}[b_0]\plus{}[b_1x]\plus{}\cdots\plus{}[b_nx^n]$
That's the identity, too.

You my want to check that these are homomorphisms as well. As a side bit, this shows that every class in the polynomial quotient has monomial representatives at high enough degree; those are the common denominator.

Here another more conceptual approach which does not boils down to perform calculations with elements over and over again. In the sequel every ring is meant to be unital and commutative.

Note that the localization $ R_S$ w.r.t. a multiplicative subset $ S$ fulfils a [b]universal property[/b] by which it is [u]uniquely[/u] (up to [u]unique[/u] isomorphism) determined:

Given any ring homomorphism $ \varphi: R\to T$ with $ \varphi(S)\subset T^*$ (the group of units of $ T$), then there is a [u]unique[/u] ring homomorphism $ \varphi_S: R_S\to T$ with $ \varphi_S\circ i_S\equal{}\varphi$ where $ i_S: R\to R_S$ denotes the (natural) ring homomorphism defined by $ i_S(r)\equal{}\frac{r}{1}$:
\[ \begin{matrix}
R& &\\
\downarrow_{i_S}& \searrow^\varphi &\\
R_S&\stackrel{\exists !\varphi_S}{\to}&T
\end{matrix}\]

After this general remark the claim boils down to show that $ R[X]/(aX\minus{}1)$ fulfils the universal property of $ R_S$. For this use the universal properties of (i.) the polynomial ring and (ii.) quotient rings. After some training this can be seen just by "ocular inspection". ;)

[color=blue]BTW: How can commutative diagrams be coded in this forum's postings? This above does not look too nice to me. [/color]:(

In fact, in general, it is simpler to construct [b]commutative[/b] localizations as quotients of polynomial rings since one can employ prior-known universal properties of polynomial rings and factor rings. For examples of this approach see e.g. Rotman, Advanced Modern Algebra or [1], though neither presentation is as simple as it could be. Note: The reason that the pair construction of fractions is ubiquitous is because it is needed in the noncommutative case -- not because the pair method is simpler or more natural. For more on the noncommutative case, see [2], esp. the survey by the renowned expert Paul M. Cohn: Localization in general rings, a historical survey, p.15. See also the many other interesting articles in that book, which is freely available online via the link [2] below.

[1] Voloch. Rings of fractions the hard way.
http://www.ma.utexas.edu/users/voloch/Homework/fractions.pdf

[2] Noncommutative localization in algebra and topology
ICMS Edinburgh 2002, Edited by Andrew Ranicki
http://www.maths.ed.ac.uk/~aar/books/nlat.pdf#page=15

Third Edition: There are packages for commutative diagrams, but they're not installed on the forum. Your best bet is to make a picture elsewhere and embed it.
See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=250471]this thread[/url].


If $0\le\theta\le2\pi$ find all the values of $\theta$ such that $\sin2\theta\le\sin\theta$
[i]Descartes 1982 3(b)[/i]

Oh, this is easy I thought to myself as I make a mistake I seem to do every time I'm confronted with a similar situation:

[hide="Wrong solution"]
$\sin2\theta\le\line\theta$
$2\sin\theta\cos\theta\le\sin\theta$
$2\cos\theta\le1$ <---- That's the kicker
$\cos\theta\le\frac{1}{2}$
Therefore $\frac{2}{3}\pi\le\theta\le\frac{4}{3}\pi$
[/hide]

This is the answer at the back of the book:
[hide]
$\sin2\theta=\sin\theta$
$2\sin\theta\cos\theta=\sin\theta$
$\sin\theta(2\cos\theta-1)=0$ <---- Should have done this
$\sin\theta=0$ or $\cos\theta=\frac{1}{2}$
The graphs intersect at $\theta=0,\frac{\pi}{3},\pi,\frac{5\pi}{3},2\pi$
Thus, $\sin2\theta\le\sin\theta$ if $\theta=0$ or $\frac{\pi}{3}\le\theta\le\pi$ or $\frac{5\pi}{3}\le\theta\le2\pi$
[/hide]

So what is conceptually wrong with the manipulation I made?

$2cos(\theta)$ is not $\le 1$.  Try $\theta = 0$.

Hmmm very true, I didn't actually look at the inequality while I was solving I just divided $\sin\theta$ throughout and kept on truckin'... So my question is, why doesn't that division work out to a correct solution? (Other than the fact that it does not in this specific case, or is it just this specific case?)

EDIT:
basically I'm asking why you can't divide by $\sin\theta$ throughout in $2\sin\theta\cos\theta=\sin\theta$ so that it results in $2\cos\theta=1$. Logically in my head there's no reason why this shouldn't work. Obviously I'm missing something.

UPDATE:
So I had an epiphany while eating dinner (if you can call something so simple epiphanic... eh):

As an example:

$\frac{x^{2}-4}{x+2}$ can be simplified to $x-2$, but the fact remains that $x\not=-2$ although it's not apparent in the simplification.

$2\cos x\sin x=\sin x$ can be simplified to $2\cos x=1$ but we are omitting the trivial case when $\sin x=0$ which is similarly unapparent after simplification like in the above example.

EDIT: I'm pretty sure this reasoning is still faulty... anyone care to help me over this ridiculous mental block  :P?

EDIT 2: After further consideration, I don't think this reasoning is faulty because if you plot $2\cos x\sin x-\sin x$ versus $2\cos x-1$ the intercepts are the same.

Could someone please just put me out of my misery and settle this? I hate being stumped on such trivial things.

I don't know what you are asking for, but I think you shouldn't have divided the inequality by $\sin x$ in the first place because $\sin x$ can be $0$ or negative. If it is negative, then the sign should face the opposite way.

$-3>-5$
divide by $-1$(which is also a possible sine value) and you get
$3>5$
first but you change signs to make it correct
$3<5$

The answer key solution you gave was very clear though. That's how my math teacher would do it, since she doesn't perform operations with inequalities at all :ninja:

[quote="#H34N1"][hide]What you did was that you should have brought the $\sin\theta$ over.

So, it'd be:

$2\sin\theta\cos\theta=\sin\theta$
$2\sin\theta\cos\theta-\sin\theta=0$
$\sin\theta(2\cos\theta-1)=0$ by factoring.

Also, you made a mistake since $2\cos\theta\not\leq 1$. It is only true if $\theta\neq 0$.[/hide]

edit: 2 responses by the time I post :roll:[/quote]
Uhhh, dead, consider $\theta=60$
$\sqrt{3}>1$


Find the number of real solutions of the equation $ (x\plus{}1)(3^x\minus{}2)\equal{}1$. This was asked in an objective test but I am not giving the options though. I am doubtful on this.

It just so happens that there are two real options. One is slightly less than 1, and the other is about -1.5, but neither is rational. Unfortunately, I can't tell you how to get those answers, as I used a calculator=)

You don't need to solve the equation to find the number of solutions; consider graphs. Rearranging the equation, we obtain:

$ 3^x \equal{} \frac {1}{x \plus{} 1} \plus{} 2.$

Clearly, $ 3^x$ will intersect once with the upper part of the hyperbola. The lower part of the hyperbola will also intersect the $ 3^x$ once because it is shifted up by $ 2$ and therefore, is above the $ x$-axis and must intersect with the exponent function. Thus, there are $ \boxed{2}$ real solutions to the equation.

Of course, this not very rigorous. I guess you could consider monotonicity.

Thanks for your replies and I too got 2 real solutions and using many applications of logarithms.Dont remember my solution and neither can I try now due to bad health.

Thanks again for your replies. Infact, I remember getting answers  near $ \plus{}$ or $ \minus{}$ $ \sqrt {2}$ goldenboy1.618 pointed it out too though he said one solution is slightly less than $ 1$ , but definitely my approach yielded two real solutions.


there are 15 subjects, 18 students.  Each student pick 4 subjects.
prove that there exist 2 student with 2 common subjects.

As easy as this problem may be, I'm not sure if proof questions belong in the High School Basics forum.

yes, it's a basic proof :)
but i dont know how to write it!
(don't think it belongs to the HSB but also the intermediate (too easy for that)

thought this would be easy but new tothe subject, I don't know how to write it out.

can someone show me how :(  :oops:

This is more of a conbinatorics problem, then the pigeonhole step is pretty trivial.

[hide]
There's 18 students, each have 4 subjects. There are $\binom{4}{2}= 6$ different pairs of subjects each person does.

If we could fill all the 18 * 6 = 108 pairs of subjects so that no two are equal, then there wouldn't exist 2 students with 2 common subjects.

There are $\binom{15}{2}= 105$ different pairs of subjects, which is less than 108 pairs required by the 18 students so that no 2 have 2 common subjects.
[/hide]


inequalities
logarithms
geometry
rectangle
limit
calculus
calculus computations

\[ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\cdots \plus{}\frac{1}{m} \leq 1 \plus{} \log m\]

thanks

$ lg(n\plus{}1)\minus{}lg(n) \geq \frac{1}{n}$

[quote="andreica"]$ lg(n \plus{} 1) \minus{} lg(n) \geq \frac {1}{n}$[/quote]

it should be $ lg(n \plus{} 1) \minus{} lg(n) \geq \frac {1}{n\plus{}1}$.

Draw a picture. This picture should include the following:

The curve $ y \equal{} \frac1x$ for $ 1\le x\le m.$

Boxes of width $ 1$ that lie under the curve. That is, draw the rectangles with sides parallel to the axes whose lower left and upper right vertices are, respectively:
$ (1,0)$ and $ \left(2,\frac12\right).$
$ (2,0)$ and $ \left(3,\frac12\right).$
$ \dots$
$ (m \minus{} 1,0)$ and $ \left(m,\frac1m\right).$

And, for good measure, the square with lower left and upper right vertices $ (0,0)$ and $ (1,1).$

Can you see the inequality in that picture?

A slightly different picture should get you
$ \ln m\le 1 \plus{} \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1{m \minus{} 1}.$

How to compute $ \lim_{m\to \infty} 1\plus{} \log m \minus{} 1\minus{}\frac{1}{2}\minus{}\cdots\minus{}\frac{1}{m}$ ?

It's $ 1\minus{}\gamma$, where $ \gamma$ is a famous constant defined by a very similar limit. Although it's not too hard to approximate $ \gamma$, very little is known about the properties of the exact number.

Don't look for an exact value here. Just estimate.


When a tag is added by clicking in the line of tag inputs above the message input area, and then manually deleted (meaning with backspace or delete), a closing tag is automatically inserted at the end of the post when submitted even though the original starting tag was already deleted. This causes some posts to unintentionally have a tag at the end of it dangling there in the middle of nowhere. The presence of these "dangling tags" is not that much of a problem, but is there a way to make the system check to make sure the start tag still exists before automatically implementing the end tag, so that users would not have to check and re-edit the extra tag out? Thanks for the thought input.

Or, you could spend an extra 15 seconds to preview your post, checking for such errors as well as other problems, which is something that everyone should probably do anyhow.  Not every minor inconvenience deserves a post in the support forum. :roll:

As JBL said, you can preview to make sure. Also, you can look at the top and if there is a * beside one of them, that means that it is going to show up at the end.

My solution: If you want to bold something, hide something, ... , you can actually type the code, or you could select the text then click the button. This adds both the start and end at the same time.


[hide="hint"]In order for $ 10|a$, the last digit of $ 33^{77} \plus{} 77^{33}$ must be a 0. So we only need to consider the last digits of $ 33$ and $ 77$, namely $ 3$ and $ 7$. Look for a pattern of the last digits of $ 3^k$ and $ 7^k$.[/hide]

hello, we get
$ 33^{77} \equiv 3 \mod 10$
and
$ 77^{33} \equiv 7 \mod 10$
so our sum $ 33^{77}\plus{}77^{33} \equiv 3\plus{}7 \equiv 0\mod 10$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we get
$ 33^{77} \equiv 3 \mod 10$
and
$ 77^{33} \equiv 7 \mod 10$
so our sum $ 33^{77} \plus{} 77^{33} \equiv 3 \plus{} 7 \equiv 0\mod 10$.
Sonnhard.[/quote]

Note that 33 and 77 are both congruent to 1 in mod 4, which is why they have their corresponding units digits.

powers of units digit 3 go in cycle: 3, 9, 7, 1 and repeats in cycles of 4
powers of units digit 7 go in cycle: 7, 9, 3, 1 and repeats in cycles of 4


In how many ways can two dice be rolled to yield a sum divisible by 3?


A packer sorts eggs by size. She finds $ 5\frac13$ dozen large eggs in the first 24 dozen eggs that she sorts. At this rate, how many large eggs can she expect to find in 60 dozen eggs?


Find all fundtionsf such that f(2008)=$ a^3$;f(2010)=$ (a\plus{}1)^3$   (with a is a integer)

I don't understand.  Without any restrictions on $ f$, the other values of $ f(n)$ are arbitrary.


linear algebra
matrix
invariant
geometry
geometric transformation
rotation
linear algebra unsolved

Let $ A\in\mathbb{R}^{n\times n}$ be an irreducible doubly stochastic matrix, how can I show that its cyclic index divides $ n$?

We use Perron-Frobenius theorem.
Let $ A$ be an irreducible and nonnegative $ n\times{n}$ matrix and let $ r$ its spectral radius. Then $ r$ is an eigenvalue of $ A$ and if $ \lambda$ is an other eigenvalue then $ |\lambda|\leq{r}$.
Moreover if $ A$ has exactly $ p$ eigenvalues of modulus $ r$ then the set of all its eigenvalues is invariant under rotations by $ 2\pi/p$ about the origin ($ p$ is the cyclic index). Thus the $ A$-eigenvalues of modulus $ r$ are vertex of a regular $ p$-polygon and are single eigenvalues.
    Moreover, on each circle of center the origin and of radius $ s<r$ there exist $ k_s\geq{0}$ eigenvalues where $ k_s$ is a multiple of $ p$. $ n$ is the sum of $ k_s$; then $ p$ divides $ n$.


A finite lattice L has n join-irreducible elements. What is the most number
f(n) of meet-irreducible elements L can have?

What do "join-irreducible" and "meet-irreducible" mean?

Thanks.. curse my ignorance >.<

[quote]oops...
i forgot to put this in advanced section.
can i move it?[/quote]

:lol: That explains my utter confusion when I read this. What subject do you learn about that in?


Is there a way I can convert a whole LaTeX document into multiple jpeg's, png's, etc.?

You can do this online at [url=http://sciencesoft.at/latex/?lang=en]ScienceSoft.at[/url] or do it yourself by the method that they use:
[quote="ScienceSoft.at"]LaTeX translates the input source text into the DVI-format (DeVice-Independent) - which is converted into the EPS-format Encapsulated PostScript) by means of the tool dvips (DVI-to-PostScript). Then, Ghostscript is used to convert the corresponding EPS-file into the PNG-format (Portable Network Graphics) which is fit for the world wide web.[/quote]


In how many ways can we $N$ [b]equal[/b] books put on $N+1$ different shelves?

Generaly, in how many ways can we $N$ [b]equal[/b] books put on $M$ different shelves?

The all you need is to avaluate the number of sequances $N$ letters $b$  and $M-1$ letters $|$, which separates books to stand them on shelves.
If there is no $b$ between two $|'s$ it means there is nothing to stand on corresponding shelve. As a result you need to choose $N$ places in the sequance for $b's$ or equivalently $M-1$ places for $|'s$.
If it is forbidden (?) to have empty shelves, just stand one book on the each shelve and reduce problem to above. Of course in this case $N$ must be no less than $M$.

In general, if you have $n$ books numbered $1$ to $n$ and you want to place them on $m$ shelves such that book $i$ comes at least $x$ shelves before book $(i+1)$ for all $1 \leq i < n$, then the number of ways is $\binom{m-(n-1)(x-1)}{n}$


Let $ a_1,a_2,...,a_n$ be $ n\geq 4$ positive integers whose sum is $ 2n \minus{} 1$. Prove that there are some consecutive terms $ a_i,a_{i \plus{} 1},...,a_j$ $ (j > i)$, whose average is an integer.

It is not true. For example $ n\equal{}2,a_1\equal{}1,a_2\equal{}2,a_1\plus{}a_2\equal{}3\equal{}2n\minus{}1$.


Using that $ (1\plus{}\frac{1}{t} )^t \rightarrow e$ as $ t\rightarrow \infty$, $ \alpha, \beta \in R$. Find the folowing limits

i) $ (\frac{n\plus{}\alpha}{n\plus{}\beta})^n$ as $ n\rightarrow \infty$
ii) $ (1\plus{}\frac{\alpha}{n})^{\beta n}$ $ n\rightarrow \infty$

hello, for i) write your term in the form
$ \left((1+\frac{\alpha-\beta}{n+\beta})^{\frac{n+\beta}{\alpha-\beta}\right)^{\frac{(\alpha-\beta)n}{n+\beta}}}$
the searched limit is $ e^{\alpha-\beta}$.
Sonnhard.


Is anybody here taking biochemistry classes? If so could i post some biochemistry topics?

I know Rushil is studying/taking organic, and I'm studying organic on my own, but I don't know if anybody is doing biochemistry.

Nothing/No one is saying you can't post them, though.  At least, I'm not. :lol:

As part of our general organic chem course, we cover Biomolecules - Carbohydrates and Amino acids. I am presently studying these ... I'd love to see questions on these. Which other topics are there in biochem??

Also, inteluser, I live in India where EVERYONE is supposed to take a standard chem course having a bulk of organic chem in Class XI and XII ... just for your info. Also, my favourite topics in Chem are (in order) m- Physical Chem.. (full advanced quantum etc. (not taught - just my super interest!) ), Organic and then (obviously) Inorganic!!!


algebra
function
domain
linear algebra
matrix
logarithms
calculus

Find the continous morphisms from the set of complex numbers to the set of nxn invertible complex matrices.

What operation or operations are you preserving with this morphism? Are you regarding the complex numbers as a group under addition?

Oh, sorry, in fact the domain of the morphism is the unit circle in the complex plane. Sorry again.

Find the continous morphisms from the set of REAL numbers to the set of nxn invertible REAL matrices

Again, here, by reals, you mean the multiplicative group $(\mathbb R^*,\cdot)$, right?

i mean $f(x+y)=f(x).f(y)$ for every real $x$ and $y$, $f(x)$ in $Gln(R)$, $f$ continuous.

Shame on me for not seeing what the answer was earlier :). All continuous $f$ group homomorphisms from $(\mathbb R,+)$ to $(Gl_n(\mathbb R),\cdot)$ are of the form $f(t)=\exp(tX),\ \forall t\in\mathbb R$ for a (unique) real $n\times n$ matrix $X$.

Let's prove the corresponding assertion for $f: \mathbb R\to Gl_n(\mathbb C)$ first, namely that all such continuous homomorphisms are of the form $f(t)=\exp(tX)$ for a unique complex $n\times n$ matrix $X$.

We have $f(0)=I$, so for small enough $t$, we'll have $\|f(t)-I\|<1$, where $\|\cdot\|$ is the [url=http://planetmath.org/encyclopedia/MatrixFNorm.html]Frobenius norm[/url]. This means that for small $t,t'$, the [url=http://planetmath.org/encyclopedia/MatrixLogarithm.html]principal logarithms[/url] $\ln f(t),\ln f(t'),\ln f(t+t')=\ln(f(t)\cdot f(t'))$ are defined, and they satisfy $\ln f(t+t')=\ln f(t)+\ln f(t')$ (in other words, $t\mapsto \ln f(t)$ is a continuous local homomorphism around $0$ from $(\mathbb R,+)$ to $(\mathcal M_n(\mathbb C),+)$). It's easy to conclude from here that there is a complex matrix $X$ such that $\ln f(t)=tX$ for all small enough $t$, which by exponentiation gives $f(t)=\exp(tX)$ (again, for small $t$). This relation is extended to the entire real line by continuity, and the existence of $X$ is proven. The uniqueness follows by observing that $t\mapsto \exp(tX)$ is differentiable, and $X$ is the derivative of this map at $t=0$, so $X$ is actually uniquely defined by $f$.

For our problem, $X$ has to be real, since $\exp(tX)$ is a real matrix for all $t$, so when we take the derivative of $t\mapsto \exp(tX)$ at $t=0$ we'll get a real matrix, which is $X$.

[quote="grobber"]What's your opinion, Harazi? I think they're of the form $f(z)=A^{-1}diag(z^{n_1},\ldots,z^{n_p})A$, where $A$ is a fixed invertible matrix, and $n_k$ are fixed integers.

I really don't have the necessary time right now, so I might be wrong because I'm in a hurry, but we can first look at those complex numbers which are roots of unity. Their images are diagonalizable matrices which commute, so they are simultaneously diagonalizable by some matrix $A$. After this, by continuity, we can show that all the matrices in the image are diagonalizable by $A$ and have complex numbers of modulus $1$ on the diagonal. After that, we break the homomorphism into $n$ distinct homomorphisms from the unit circle to itself, the $k$'th one taking $z$ to the $k$'th entry on the diagonal of $f(z)$. Since these are continuous, I believe they're of the form $f_k(z)=z^{n_k}$.

It looks Ok to me right now, but I bet it's not... :?[/quote]

I dont see why the continuous homomorphismos from the unit circle to itself have the form $f(z)=z^k$??

Let's identify the circle group with $\mathbb R/\mathbb Z$, in order to use additive notation. The statement then becomes: all continuous endomorphisms of $\mathbb R/\mathbb Z$ are of the form $x\mapsto kx$, for some integer $k$.

Choose a small neighborhood $U$ of $0$ (say, $U\subset\left(-\frac 12,\frac 12\right)$), and let $V\subset\left(-\frac 12,\frac 12\right)$ be a neighborhood of $0$ so small that whenever $x\in V,\ x$ is mapped into $U$. Now take $n$ so large that $\frac 1n\in V$. Since $n\cdot\frac 1n=0,\ \frac 1n$ must be mapped onto $\frac kn\in U$ for some integer $k$. Now, for positive integers $t,\ \frac 1^{2^tn}$ must be mapped onto $\frac k{2^tn}$ plus some integer, but since this image must also lie inside $\left(-\frac 12,\frac 12\right)$, that extra integer must be zero. Our homomorphism thus sends every element of the group of the form $\frac m{2^tn}\ (*)$ onto $\frac{km}{2^tn}$, and since the elements of the form $(*)$ form a dense set in our group, we're done: every element $x$ is sent to $kx$.


This is cute:

Find all primes $p, q$ such that $pq$ divides $2^p + 2^q$.

[quote="pbornsztein"]It is a vicious circle since I took it from here to propose it in our program... :D

Pierre.[/quote] :D :D :D


limit
logarithms
inequalities
Euler
number theory unsolved
number theory

[quote="pbornsztein"]It is known that for every $x \geq 2$, we have $\prod_{p \leq x} \frac p {p-1} \geq \log (x).$

It follows that the desired limit is $+ \infty .$

Pierre.[/quote]
Thx

pbornsztein, could you please give a proof of the inequality you used ?

Write $\prod_{p\le x}\frac p{p-1}$ as $\prod_{p\le x}\frac 1{1-\frac 1p}$, which is equal to $\prod_{p\le x}(1+\frac 1p+\frac 1{p^2}+\ldots)$, and this is the sum of the recirpocals of all the numbers having in their decomposition only primes less than $x$, which is larger than $\sum_{n\le x}\frac 1n$. Now, $\sum_{i=1}^n\frac 1i-\ln (n+1)$ tends to a positive limit (the Euler constant $\gamma$), so that suffices for the problem.

[quote="RobertuX"]Stimate the following product ($p_j$ represents the $j$-th prime number)
\[
\prod_{j=1}^m  \frac{p_j}{p_j-1},
\]
an use the stimation to calculate $\displaystyle \lim_{m\to \infty} \prod_{j=1}^m  \frac{p_j}{p_j-1}.$.

RX[/quote]
More easy saying that this product tends to $\zeta(1)=+\infty$. But thx a lot of ur comments.

[quote="grobber"]and this is the sum of the recirpocals of all the numbers having in their decomposition only primes less than $x$, which is smaller than $\sum_{n\le x}\frac 1n$. .[/quote]

Could you explain it for me ? I can't understand why   :?

I meant to say larger instead of smaller. I have edited my previous message.


function
algebra
polynomial
real analysis
real analysis unsolved

[quote="alexilic"]Prove that for every real continuous function $f$ on $[a, b]$, one can find polynomial with integer coeficients, such that for every $a > 0$, there exists $n \in N$ and $P_n$ with $\mbox{deg} P_n \leqslant n$, and
\[ |f (x) - P_n (x)| \leqslant a, \quad x \in [a, b].  \][/quote]
I think it is something wrong here with the double using of $a$, i think in [quote="alexilic"] such that for every $a > 0$, there exists [/quote] you meant some another $\epsilon$, i.e. you rpobvably meant that $f$ can be uniformly approximated by polynomials from $\mathbb{Z}[X]$.

For every $f \in C [a, b]$, prove that there exist array of polynomials $P_n (x) \in Z [X]$, such that
\[ P_n \rightarrow f, n \rightarrow \infty \] for every $x \in [a, b]$.

There are several ways of showing this. I'll use this:
WLOG, you may assume that $[a,b] \subset (0,1)$. Since the set $\{\frac{k}{2^n}, k,n \in \mathbb{N}_0 \}$ is dense in $(0,1)$ we only need to prove that $\frac{1}{2}$ can be approximated by polynomials with integers coefficients with arbitrarily small accuracy i.e. for any $\epsilon >0$ there exists $p \in \mathbb{Z}[X]$, such that $||\frac{1}{2}-p||_{[a,b]} < \epsilon$ (*). If we get this we can coclude that any real from $(0,1)$ can be approximated by polynomiasl from $\mathbb{Z}[X]$ and such all the polynomials from $\mathbb{R}_n[X]$ as well and know we can just use Weirstrass approximation theorem that states that our function $f$ can be approximated by some polynomial from $\mathbb{R}_n[X]$. So we have to prove (*):
$\frac{1}{2}=\frac{1-x}{1-(1-2(1-x))}= \sum_{k=0}^{\infty} (1-x)(1-2(1-x))^k$,  and as our approximation polynomials we can choose tha partial sum of this infinite series which as you may prove converges uniformly on $[a,b] \subset (0,1)$.

Yes, for the case, when the endpoints of our interval are integer as in this case of $[0,10]$ we also need the condition $f(a),f(b) \in \mathbb{Z}$.

[quote="alexilic"]So, can someone explain general solution?
Thx[/quote]
Well, as for me i don't know the answer in general. What is also that interests me, you didn't say anything about you $[a,b]$ and function $f$ except that it is continious. But not every function can be approximated with integer polynomials on every interval. I'll try to sum up the cases which i know about:

1). If $[a,b]$ doesn't contain integers then i showed a proof in post #4.

2). If $[a,b]$ does contain integer, then you must have at least these resctrictions on our fucntion $f$: $f(\mathbb{Z}\bigcap [a,b]) \in \mathbb{Z}$, i don't know whether these resctrictions are SUFFICIENT. For example if $[a,b]=[0,1]$ i know that they are sufficient(use Bernstein polynomials, to be accurate kinda integer part of them).
If $[a,b]=[-1,1]$ i know that they are also sufficient. But the general case...i really don't know.

Question to others: Who knows the result in general.

[quote="alekk"](beware : for $[a,b]=[0;10]$ it does not work)[/quote]

Ah, i can also prove that we the result of the problem is not true if $b-a \geq 4$. So, yopu must have at least add this condition to the problem statement. It is probably what Alekk was talking about.

This neccessary condition $b-a <4$ comes from the Chebysheff theoprem that $||p||_{[a,b]} \geq 2 (\frac{b-a}{4})^n$ and in the case of $b-a \geq 4$ we would have that $||p_n -p_m || \geq 2$ and this implies that  $p_n \to f$ uniformly is surely impossible due to Cauchy criterion for uniform convergence.


Assume $ a_{n} \geq a_{n+1}$, for $ \forall$ $ n \geq 1$

and $ \lim_{n \rightarrow \infty}a_{n}=0$

define: $ b_{n}=a_{n}-2a_{n+1}+a_{n+2}$, for $ \forall$ $ n \geq 1$

assume $ b_{n}\geq0$, for $ \forall$ $ n\geq1$

show that: $ \sum_{n=1}^{\infty}nb_{n}$ converge

[color=red]My Idea[/color]

I want to use Cauchy convergence principle....

so need to compute: $ |mb_{m}+(m+1)b_{m+1}+...+nb_{n}|$

the above equal:
$ |m[(a_{m}-a_{m+1})-(a_{m+1}-a_{m+2})]+(m+1)[(a_{m+1}-a_{m+2})-(a_{m+2}-a_{m+3})]+
...+n[(a_{n}-a_{n+1})- (a_{n+1}a_{n+2})]|$

$ =|m(a_{m}-a_{m+1})+a_{m+1}-a_{n+1}-n(a_{n+1}-a_{n+2})|$

But I can't go any further......

isn't $ 1/\sqrt n$ a counterexample?

[quote="Xingyuan"]Assume $ a_{n} \geq a_{n \plus{} 1}$, for $ \forall$ $ n \geq 1$

and $ \lim_{n \rightarrow \infty}a_{n} \equal{} 0$

define: $ b_{n} \equal{} a_{n} \minus{} 2a_{n \plus{} 1} \plus{} a_{n \plus{} 2}$, for $ \forall$ $ n \geq 1$

assume $ b_{n}\geq0$, for $ \forall$ $ n\geq1$

show that: $ \sum_{n \equal{} 1}^{\infty}nb_{n}$ converge

[color=red]My Idea[/color]

I want to use Cauchy convergence principle....

so need to compute: $ |mb_{m} \plus{} (m \plus{} 1)b_{m \plus{} 1} \plus{} ... \plus{} nb_{n}|$

the above equal:
$ |m[(a_{m} \minus{} a_{m \plus{} 1}) \minus{} (a_{m \plus{} 1} \minus{} a_{m \plus{} 2})] \plus{} (m \plus{} 1)[(a_{m \plus{} 1} \minus{} a_{m \plus{} 2}) \minus{} (a_{m \plus{} 2} \minus{} a_{m \plus{} 3})] \plus{} ... \plus{} n[(a_{n} \minus{} a_{n \plus{} 1}) \minus{} (a_{n \plus{} 1}a_{n \plus{} 2})]|$

$ \equal{} |m(a_{m} \minus{} a_{m \plus{} 1}) \plus{} a_{m \plus{} 1} \minus{} a_{n \plus{} 1} \minus{} n(a_{n \plus{} 1} \minus{} a_{n \plus{} 2})|$

But I can't go any further......[/quote]

[quote="Arrigo-Sacchi"]isn't $ 1/\sqrt n$ a counterexample?
[/quote]

I think $ \frac{1}{\sqrt{n}}$ is not a counterexample.......

because:

If $ a_{n}\equal{}\frac{1}{\sqrt{n}}$

then $ b_{n}\equal{}\frac{1}{\sqrt{n}}\minus{}\frac{2}{\sqrt{n\plus{}1}}\plus{}\frac{1}{\sqrt{n\plus{}2}}$

obviously, $ b_{n}\geq0$

and $ \sum_{n\equal{}1}^{\infty}nb_{n}\equal{}\sum_{n\equal{}1}^{\infty}n\left(\frac{1}{\sqrt{n}}\minus{}\frac{2}{\sqrt{n\plus{}1}}\plus{}\frac{1}{\sqrt{n\plus{}2}}\right)$

I guess the above series converges if and only if:

$ \sum_{n\equal{}1}^{\infty}\left(\sqrt{n}\minus{}2\sqrt{n\plus{}1}\plus{}\sqrt{n\plus{}2}\right)$ converges    $ (*)$

[color=red](I can't give a proof)[/color]

obviously, $ \sum_{n\equal{}1}^{\infty}\left(\sqrt{n}\minus{}2\sqrt{n\plus{}1}\plus{}\sqrt{n\plus{}2}\right)$ is converges

so if we work out $ (*)$, we can get the conclusion that its not a counterexample.....

[quote="Arrigo-Sacchi"][quote="Xingyuan"]

I think $ \frac {1}{\sqrt {n}}$ is not a counterexample.......
[/quote]

Yes, I made the error of considering $ \sum na_n$ instead of $ \sum nb_n.$[/quote]

But it is a good example :D

Hint: Let $ c_n \equal{} a_n \minus{} a_{n\plus{}1}$. Then $ b_n \equal{} c_n \minus{} c_{n\plus{}1}$. Use summation by parts on $ \sum_{n\equal{}1}^{N} n(c_n \minus{} c_{n\plus{}1})$.


$ (7\minus{}2z)^2\equal{}(2z\minus{}5)^3$

let $ w\equal{}2z\minus{}5$, then this looks like $ (2\minus{}w)^2\equal{}w^3$

$ 0\equal{}w^3\minus{}w^2\plus{}4w\minus{}4\equal{}w^2(w\minus{}1)\plus{}4(w\minus{}1)\equal{}(w\minus{}2i)(w\plus{}2i)(w\minus{}1)$

so.. $ z\equal{}\frac{w\plus{}5}{2}$, which are $ \frac{5}{2} \pm i, 3$


Which package is the bmatrix, pmatrix, vmatrix etc. commands in? I searched around and couldn't find it.

amsmath. See [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math Mode[/url].


Any1 got any tips on how to get hold of hacks or cheats for BattleOn AQ.

...Cheats? I think very few people cheat on this forum and I'm thinking very few even play that game/heard of it. I don't  think I've ever heard of this game but cheating and hacking is only for noobs.

[quote="math92"]i have never heard of this game either and what is noob?[/quote]

A noob is a player that either sucks, cheats/hacks, whines continuously, or any combination of the three.


In any triangle ABC, if $ BC\plus{}AC\equal{}2AB$, show that $ \cot(A/2)\plus{}\cot(B/2)\equal{}2\cot(C/2)$

$ a\plus{}b\equal{}2c \Longleftrightarrow 2s\minus{}(a\plus{}b)\equal{}2s\minus{}2c \Longleftrightarrow s\minus{}a\plus{}s\minus{}b\equal{}2(s\minus{}c)$

$ \Longleftrightarrow \frac{s(s\minus{}a)}{[ABC]}\plus{}\frac{s(s\minus{}b)}{[ABC]}\equal{}\frac{2s(s\minus{}c)}{[ABC]}$

Now using [u][b]Heron[/b]'s formula[/u] $ \implies \sqrt{\frac{s^2(s\minus{}a)^2}{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}}\plus{}\sqrt{\frac{s^2(s\minus{}b)^2}{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}}\equal{}$ $ 2\sqrt{\frac{s^2(s\minus{}c)^2}{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}}$

$ \Longleftrightarrow \sqrt{\frac{s(s\minus{}a)}{(s\minus{}b)(s\minus{}c)}}\plus{}\sqrt{\frac{s(s\minus{}b)}{(s\minus{}a)(s\minus{}c)}}\equal{}2\sqrt{\frac{s(s\minus{}c)}{(s\minus{}a)(s\minus{}b)}}$ $ \Longleftrightarrow \cot\frac A2\plus{}\cot\frac B2\equal{}2\cot\frac C2$ .

[color=darkblue]Other relations in triangles having [b]sides in aritmethic progression[/b]: $ \left\|\begin{array}{ccc}\frac{2}{h_{a}}  \equal{}  \frac{1}{h_{b}}\plus{}\frac{1}{h_{c}}\\\\
4\cos A\plus{}\cos\left(B\minus{}C\right)\equal{}3\\\\
\sin\frac{A}{2}\equal{}\frac{1}{2}\cos\frac{B\minus{}C}{2}\end{array}\right\|$ and $ \left\|\begin{array}{ccc}3\tan\frac{B}{2}\tan\frac{C}{2}\equal{}1\\\\
3\sin^{2}\frac{A}{2}\equal{}\sin B\sin C \\\\
2\cos A\plus{}\cos B\plus{}\cos C\equal{}2 \end{array}\right\|$ .[/color]

[quote="Mateescu Constantin"]$ a \plus{} b \equal{} 2c \Longleftrightarrow 2s \minus{} (a \plus{} b) \equal{} 2s \minus{} 2c \Longleftrightarrow s \minus{} a \plus{} s \minus{} b \equal{} 2(s \minus{} c)$

$ \Longleftrightarrow \frac {s(s \minus{} a)}{[ABC]} \plus{} \frac {s(s \minus{} b)}{[ABC]} \equal{} \frac {2s(s \minus{} c)}{[ABC]}$

Now using [u][b]Heron[/b]'s formula[/u] $ \implies \sqrt {\frac {s^2(s \minus{} a)^2}{s(s \minus{} a)(s \minus{} b)(s \minus{} c)}} \plus{} \sqrt {\frac {s^2(s \minus{} b)^2}{s(s \minus{} a)(s \minus{} b)(s \minus{} c)}} \equal{}$ $ 2\sqrt {\frac {s^2(s \minus{} c)^2}{s(s \minus{} a)(s \minus{} b)(s \minus{} c)}}$

$ \Longleftrightarrow \sqrt {\frac {s(s \minus{} a)}{(s \minus{} b)(s \minus{} c)}} \plus{} \sqrt {\frac {s(s \minus{} b)}{(s \minus{} a)(s \minus{} c)}} \equal{} 2\sqrt {\frac {s(s \minus{} c)}{(s \minus{} a)(s \minus{} b)}}$ $ \Longleftrightarrow \cot\frac A2 \plus{} \cot\frac B2 \equal{} 2\cot\frac C2$ .[/quote]

Can you show how do you get that
$ \sqrt{\frac{s(s\minus{}a)}{(s\minus{}b)(s\minus{}c)}}\equal{}\cot\frac{A}{2}$
P.S:Sorry for the ignorance

[quote="ridgers"][quote="Mateescu Constantin"]$ a \plus{} b \equal{} 2c \Longleftrightarrow 2s \minus{} (a \plus{} b) \equal{} 2s \minus{} 2c \Longleftrightarrow s \minus{} a \plus{} s \minus{} b \equal{} 2(s \minus{} c)$

$ \Longleftrightarrow \frac {s(s \minus{} a)}{[ABC]} \plus{} \frac {s(s \minus{} b)}{[ABC]} \equal{} \frac {2s(s \minus{} c)}{[ABC]}$

Now using [u][b]Heron[/b]'s formula[/u] $ \implies \sqrt {\frac {s^2(s \minus{} a)^2}{s(s \minus{} a)(s \minus{} b)(s \minus{} c)}} \plus{} \sqrt {\frac {s^2(s \minus{} b)^2}{s(s \minus{} a)(s \minus{} b)(s \minus{} c)}} \equal{}$ $ 2\sqrt {\frac {s^2(s \minus{} c)^2}{s(s \minus{} a)(s \minus{} b)(s \minus{} c)}}$

$ \Longleftrightarrow \sqrt {\frac {s(s \minus{} a)}{(s \minus{} b)(s \minus{} c)}} \plus{} \sqrt {\frac {s(s \minus{} b)}{(s \minus{} a)(s \minus{} c)}} \equal{} 2\sqrt {\frac {s(s \minus{} c)}{(s \minus{} a)(s \minus{} b)}}$ $ \Longleftrightarrow \cot\frac A2 \plus{} \cot\frac B2 \equal{} 2\cot\frac C2$ .[/quote]

Can you show how do you get that
$ \sqrt {\frac {s(s \minus{} a)}{(s \minus{} b)(s \minus{} c)}} \equal{} \cot\frac {A}{2}$
P.S:Sorry for the ignorance[/quote]

see here http://www.mathlinks.ro/viewtopic.php?t=322695


polynomial
algebra
Summation
equation
Diophantine equation
IMO Shortlist
IMO Longlist

Suppose that $p$ and $q$ are two different positive integers and $x$ is a real number. Form the product $(x+p)(x+q).$ Find the sum $S(x,n) = \sum (x+p)(x+q),$ where $p$ and $q$ take values from 1 to $n.$ Does there exist integer values of $x$ for which $S(x,n) = 0.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


Hi everyone...

Does anyone know where around or in MD (or which county/schools) offer the AMC B test. Please reply/PM me as soon as possible. I really want to take this test.

Thanks

http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml

I heard you have to register by the 14th or something.

[quote="E^(pi*i)=-1"]If you live in Maryland, you could probably take it with my math club, the Howard Area Homeschoolers.  PM CatherineAsaro for more information.[/quote]

thank you for your offer. I have emailed Catherine Asaro already, and I was wondering if I need to complete further registration or has everyone already been registered?

I'm not sure, but I would assume you wouldn't need to register.


Determine all solutions k, n1, n2, ..., nk to

n1|2n2 - 1
n2|2n3 - 1
...
nk|2n1 - 1.

I like this problem.

I think this lemma might help:

Lemma: [hide] if a is divisible by a smallest prime p and b is divisible by a smallest prime q, and p <= q, then a cannot divide 2^b - 1.

Proof:

first, clearly p!=2.

if a | 2^b - 1, then 2^b == 1 (mod a )  => 2^b == 1 (mod p).

Let k be the order of 2 (mod p).

Then k <= p-1, but k|b. But since the smallest prime factor of b is q > p, k cannot divide b unless k = 1. But k cannot equal 1 since otherwise 2==1(mod p) => p | 1.

This completes the proof.

[/hide]

From the lemma:

[hide]

The only solution is when all are equal to 1 (but k can be anything in this case).

(Assuming all are natural... otherwise we can also have all of them equal to 0)

Note we cannot have one value equal to one and another equal to something greater than one, since one equalling one implies all others are equal to one. (Just work backwards: 2^1 -1 = 1)

We assume that none are equal to one. Then each has a unique nonempty prime factorization.

suppsose all these conditions are true. Then

if p[i] is the smallest prime divisor of n[i]:

p[1] > p[2] > p[3] > ... > p[k] > p[1].

A contradiction.

[/hide]

Sorry for the sloppiness . Hope it works.

Is this the way you approached the problem? Or, did you have a different solution, because I would be interested to see it.

My solution is a little bit different, but it comes down to the same.

I defined
p' = min{p[b]|[/b]p is a prime such that p|ni for a certain 1 :le: i :le: k}.

Then we can show that one of the ni must have a prime divisor p" < p'.
(In fact, if that p'|nk, then p" < p' must divide nk + 1.
Actually it is the same method as in your lemma ...)

Hence contradiction, and the result follows if we assume k :ge: 2.

If k = 1 then the problem reduces to finding all n st n|2n - 1.
This is not so difficult.

actually u dont need the special case. if k = 1 then nk IS nk+1

Isn't that what  said? Or am I missing something ...

[quote="Arne"]Hence contradiction, and the result follows if we assume k :ge: 2.

If k = 1 then the problem reduces to finding all n st n|2n - 1.
This is not so difficult.[/quote]

Firstly its just me being picky.
My only "point" was that you dont have to assume k>=2.
</picky point>


A shoemaker died, leaving behind 17 pair's of men's shoes and boots. In his last will and testament he disposed of them as follows:

$ \frac{1}{2}$ of the collection to his youngest, hardworking, son; $ \frac{1}{3}$ to the middle son, and $ \frac{1}{9}$ to "Lazybones", the eldest son.

Understanding that it was not possible to divide 17 pairs into halves, thirds, and ninths, the exasperated executor of the will tossed in his own footwear, making it a total of eighteen pairs, whihc he divided as follows;

$ \frac{1}{2}$, or 9 pairs, to the youngest son; $ \frac{1}{3}$, or 6 pairs, to the middle son; and $ \frac{1}{9}$, to the eldest son.

Upon realising that adding up 9,6 and 2 made 17, the execytor elatedly retrieved his own pairs of shoes and closed the estate.

That’s because $ \frac {1}{2}$, $ \frac {1}{3}$ and $ \frac {1}{9}$ do not add up to 1. The agent was only splitting the 17 pairs of footwear items in the ratios $ \frac {9}{17}$, $ \frac {6}{17}$ and $ \frac {2}{17}$, so his own pair didn’t enter the calculation.

This property can be generalised by finding all values of n and a and b and c such that
(n+1)/a + (n+1)/b + (n+1)/c = 1  and 1/a + 1/b + 1/c not equal to 1


function
analytic geometry
trigonometry
quadratics
trig identities

Here are eight tough questions which I am unable to solve.
Please help me by explaining as well as showing the step\solutions.
Thank you in advance.

Question 1:
[hide]Given that $ Cosx = Cosy$ and $ Sinx = - Siny$, Show that $ Sin^2 \frac {x + y}{2} = 0$
[/hide]
Question 2:
[hide]The number of hours of day light in a certain town is given by $ y = 3sin\frac {2\pi}{365}(x - 79) + 12$,
where $ x$ is the number of days after January 1.
a) What is the amplitude of the function?
b) What is the period of the function?
c) How many hours of daylight are there on the longest day of the year?
d) How many hours of daylight are there on the shortest day of the year?
[/hide]
Question 3:
[hide]Let AC and BD are tow chords of a circle with centre O such that they intersect at right angles inside the circle at a point M. Suppose K and L are mid points of the arc AB and chords Cd respectively. Prove that OKML is a parallelogram.
[img]http://www.fileden.com/files/2006/11/16/384824/Question%203%20image.bmp[/img]
[/hide]
Question 4:
[hide]Determine the values of $ a$ and $ c$ for which
[img]http://www.fileden.com/files/2006/11/16/384824/question%204%20image.bmp[/img]
[/hide]
Question 5:
[hide]Two vertices of a triangle are (5,-1) and (-2,3). If the orthocentre of the triangle is the origin, find the coordinates of the third vertex.
[/hide]
Question 6:
[hide]If $ sec (\theta - \alpha), sec\theta$ and $ sec (\theta + \alpha)$ are in A.P., prove that $ cos\theta = \sqrt2 cos \frac {\alpha}{2}$
[/hide]
Question 7:
[hide]Show that $ 2tan^ - ^1 \frac {1}{5} + sec^ - ^1 \frac {5\sqrt2}{7} + 2tan^ - ^1 \frac {1}{8} = \frac {\pi}{4}$
[/hide]
Question 8:
[hide]Balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row of two balls, three in third row and so on. If 669 more balls are added then all the balls can be arranged in the shape of a square and each of sides then contains 8 balls less than the side of the triangle had. Determine the initial number of balls.[/hide]

[quote="CyanSnow"]Here are eight tough questions which I am unable to solve.
Please help me by explaining as well as showing the step\solutions.
Thank you in advance.

Question 1:
[hide]Given that $ Cosx \equal{} Cosy$ and $ Sinx \equal{} \minus{} Siny$, Show that $ Sin^2 \frac {x \plus{} y}{2} \equal{} 0$
[/hide]
[/quote]

[hide="Hint for 1"]
Consider the quadrants
What does that tell you about the bisector of x and y?
[/hide]

$ 1)$ We know that $ \cos x \equal{} \cos y$ and $ \sin x \equal{} \minus{} \sin y$.

[hide]So that must hold true for all $ x$ and $ y$.  Thus $ \cos \left( \frac {x}{2} \right) \equal{} \cos \left( \frac {y}{2} \right)$ and $ \sin \left( \frac {y}{2} \right) \equal{} \minus{} \sin \left( \frac {x}{2} \right)$.

Thus:

$ \cos \left( \frac {x}{2} \right) \cdot \sin \left( \frac {x}{2} \right) \equal{} \cos \left( \frac {y}{2} \right) \cdot \left[ \minus{} \sin \left( \frac {y}{2} \right)\right]$; substitute what we know is true
$ \iff \cos \left( \frac {x}{2} \right) \cdot \sin \left( \frac {x}{2} \right) \plus{} \cos \left( \frac {y}{2} \right) \cdot \sin \left( \frac {y}{2} \right) \equal{} 0$; bring over to other side
$ \iff \sin \left( \frac {x}{2} \plus{} \frac {y}{2} \right) \equal{} 0$; use trig identities
$ \iff \sin \left( \frac {x \plus{} y}{2} \right) \equal{} 0$
$ \iff \boxed{\sin^2 \left( \frac {x \plus{} y}{2} \right) \equal{} 0}$; just square it on this step[/hide]

$ 2)$ Given     $ y \equal{} 3 \sin \left[ \frac {2\pi}{365} \cdot (x \minus{} 79) \right] \plus{} 12$ where $ x$ is the number of days after January $ 1^{\text{st}}$ and $ y$ is the number of hours of daylight,

[hide]$ a)$ The amplitude of the function is $ 3$.  This is just the number that $ \sin$ is multiplied by.

$ b)$ When $ y\equal{}\sin (bx\plus{}c)$, the period is given by $ P\equal{}\frac{2\pi}{b}$.

Thus:

$ P \equal{} \frac{2\pi}{\frac{2\pi}{365}}$
$ \implies \boxed{P\equal{}365}$ which is obvious since there are 365 days in the year.

$ c)$ The maximum of the $ \sin$ function is $ 1$.  Thus the maximum of $ y$ is given by $ y \equal{} 3 \cdot 1 \plus{} 12 \equal{} \boxed{15}$

$ d)$ The mininum of the $ \sin$ function is $ \minus{}1$.  Thus the mininum of $ y$ is given by $ y\equal{}3 \cdot (\minus{}1) \plus{}12 \equal{} \boxed{9}$.[/hide]

Ok...Thank you for the help!

Now would you please guide me to question 5 and 6?

[quote="CyanSnow"]If $ \sec (\theta - \alpha), \sec\theta$ and $ \sec (\theta + \alpha)$ are in A.P., prove that $ \cos\theta = \sqrt2 \cos \frac {\alpha}{2}$[/quote]Some trig identities  :|
$ \frac{1}{\cos{\theta}\cos{\alpha}+\sin{\theta}\sin{\alpha}}$, $ \frac{1}{\cos{\theta}}$, $ \frac{1}{\cos{\theta}\cos{\alpha}-\sin{\theta}\sin{\alpha}}$ are in arithmetic progression.
\begin{eqnarray*}(\cos{\theta}\cos{\alpha})^2-(\sin{\theta}\sin{\alpha})^2&=&\frac{\cos{\theta}(\cos{\theta}\cos{\alpha}+\sin{\theta}\sin{\alpha})+\cos{\theta}(\cos{\theta}\cos{\alpha}-\sin{\theta}\sin{\alpha})}{2}\\
&=&\cos^2{\theta}\cos{\alpha}\end{eqnarray*}
$ (\cos{\theta}\cos{\alpha})^2-(1-\cos^2{\theta})\sin^2{\alpha}=\cos^2{\theta}\cos{\alpha}$
$ \cos^2{\theta}=\frac{\sin^2{\alpha}}{\cos^2{\alpha}+\sin^2{\alpha}-\cos{\alpha}}=\frac{\sin^2{\alpha}}{1-\cos{\alpha}}\implies \cos{\theta}=\frac{\sin{\alpha}}{\sqrt{1-(1-2\sin^2{\frac{\alpha}{2}})}}$
$ \implies\cos{\theta}=\frac{2\sin{\frac{\alpha}{2}}\cos{\frac{\alpha}{2}}}{\sqrt{1-(1-2\sin^2{\frac{\alpha}{2}})}}=\sqrt{2}\cos{\frac{\alpha}{2}}$

8. [hide]Let $ N\equal{}\frac{n(n\plus{}1)}{2}$, (where n is the no. of rows in the equilateral triangle) be the original no. of balls.so,we have
$ N\plus{}669\equal{}(n\minus{}8)^{2}$
solving the quadratic, we get $ n\equal{}55 \Rightarrow N\equal{}770$[/hide]

is the ans of ques 4 =
a=-3/2
c=1/2
is the ans of ques 5 =(-4,-7)


1.  64sin10sin30sin50sin70sin90 = ?
  2.  Let cos(A+B) = 3/root10  and cosA = 2/root5  determine cosB/cos(2A+B)

1.  [hide]  Apply the double angle identity many, many times to $ sin10$.  \[ sin10 \equal{} \frac {sin20}{2cos10} \equal{} \frac {sin40}{4(cos10)(cos20)} \equal{} \frac {sin80}{8(cos10)(cos20)(cos40)}.\]  Now use $ sin(x) \equal{} cos ( 90 \minus{} x)$.  We get \[ \frac {cos10}{8(cos10)(cos20)(cos40)} \equal{} \frac {1}{8(cos20)(cos40)}.\]

If we apply this previous identity to the $ sin50$ and $ sin70$.  You will see that all the bad angles cancel and we are left with $ 64/16 \equal{} 4$ .[/hide]

[hide="Hint for (1)"]
use $ \sin A. sin (60\minus{}A). \ sin(60\plus{}A) \equal{} \frac{1}{4} \sin 3A$

where the angles are in degrees.

[/hide]

[hide="hint (2)"]
using expansion of $ \cos (A\plus{}B)$ find $ \cos B$

now it should be simple!
[/hide]

[quote="Nora.91"][hide="Hint for (1)"]
use $ \sin A. sin (60 \minus{} A). \ sin(60 \plus{} A) \equal{} \frac {1}{4} \sin 3A$

where the angles are in degrees.

[/hide]

[hide="hint (2)"]
using expansion of $ \cos (A \plus{} B)$ find $ \cos B$

now it should be simple!
[/hide][/quote]

Just out of curiosity, did you know the formula $ \sin A. sin (60 \minus{} A). \ sin(60 \plus{} A) \equal{} \frac {1}{4} \sin 3A$ or did you just figure it out when you did the problem.

[quote="kurt.math"]
Just out of curiosity, did you know the formula $ \sin A. sin (60 \minus{} A). \ sin(60 \plus{} A) \equal{} \frac {1}{4} \sin 3A$ or did you just figure it out when you did the problem.[/quote]

i knew it.

$ \sin 3A \equal{} 3 \sin A \minus{} 4 \sin ^3 A \\ \equal{} 4 \sin A ( (\frac{\sqrt{3}}{2})^2 \minus{} \sin ^2 A ) \equal{} 4 \sin A (\sin (60 \minus{} A). \sin(60 \plus{} A) )$

64sin10sin30sin50sin70sin90  =   32cos80cos40cos20 = (16sin40cos40cos80)/sin20

 =  (8sin80cos80)/sin20   =  (4sin160)/sin20  =  4

  - -  5555     thank you everybody


[quote="vedran6"]If $ a$, $ b$, $ c$ are nonnegative reals satisfying $ a^2\plus{}b^2\plus{}c^2\equal{}1$, then prove that

$ \frac {1}{1 \minus{} a^2} \plus{} \frac {1}{1 \minus{} b^2} \plus{} \frac {1}{1 \minus{} c^2} \plus{} \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca} \geq 9$.[/quote]
My ugly proof:
$\sum_{cyc} \frac {1}{1 \minus{} a^2}+\sum_{cyc} \frac {1}{1 \minus{} ab}=\sum_{cyc} \frac {(3a+b+c)^2}{(3a+b+c)^2(1 \minus{} a^2)}+\sum_{cyc} \frac {(a+b)^2}{(a+b)^2(1 \minus{} ab)}\geq$
$\geq\frac{25(a+b+c)^2}{\sum\limits_{cyc}(3a+b+c)^2(1 \minus{} a^2)}+\frac{4(a+b+c)^2}{\sum\limits_{cyc}(a+b)^2(1 \minus{} ab)}\geq9$,
where the last inequality we can prove by $uvw$'s technique.


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

supposedly a very easy problem
<A,*> is a semi group.prove that there exists an element 'a' in the semigroup A s.t a*a=a.
best regards
vasu

Maybe it should be finite or something? Otherwise, it's clearly not true. Take the semigroup generated by $X$ in a polynomial ring $K[X]$, for instance.

Edit:

And if $A$ is indeed finite, then the problem is easy:

Take some $x\in A$ and two positive integers $m<n$ s.t. $x^m=x^n$, pick another positive integer $u$ s.t. $(u+1)n>(u+2)m$, and then take $a=x^{(u+1)(n-m)}$.


calculus
integration
trigonometry
logarithms
real analysis
real analysis solved

Evaluate the following definite integrals.

(1)$\int^{16}_1\frac{dx}{\sqrt{x}+\sqrt[4]{x}}$

(2)$\int^{\frac{\pi}{3}}_0 \frac{dx}{\sin x+\sqrt{3}\cos x} dx$

For the first one put $x=t^4$ so $t\in[1,2]$.
results our integral is equal to
$\displaystyle\int_1^2\frac{4t^3}{t+t^2}dt=4\int_1^2\frac{t^2}{t+1}dt=4\int_1^2\left(t-1+\frac1{t+1}\right)=8-1-(8-4)+4\ln\frac32=3+4\ln\frac32$

for the second one... rewrite $ \sin x+\sqrt3\cos x=2(\frac12\sin x+\frac{\sqrt3}2\cos x)=2\cos(x-\pi/6)$....

Thank you for your reply, Kuba.

For the first one, the answer is $2+4\ln \frac{3}{2}$.

kunny


Prove that the expression
\[ m^5 + 3m^4n - 5m^3n^2-15m^2n^3+4mn^4+12n^5 \] cannot have the value $33$, regardless of what integers are substituted for $m$ and $n$.

[hide]
Note that we have two sets of coefficients with $(1, -5, 4)$, thus we can factor to
$3n(m^4-5m^2n^2+n^4) + m(m^4-5m^2n^2 + n^4)$
$(m+3n)(m^2-n^2)(m^2-4n^2)$
For this expression to equal $33$, we either have to have 1 factor equal $33$, and the other two equal $1$, or two factors of $3$ and $11$, and another of $1$.  Regardless, one factor must be equal to $1$.
The 2nd and 3rd factors cannot equal 1, by contradiction.  The two possibilities are $m^2 = 1 + n^2$ and $m^2 = 1 + 4n^2 = 1 + (2n)^2$, both cases only have solutions $m = \pm 1, n = 0$, which never results in $33$.
Assume the first factor equals 1.  Then $m = 1-3n$, and plugging back into the expression, the remaining part is $(8n^2 - 6n + 1)(5n^2 - 6n + 1)$.  Setting both factors equal to 1 yields one solution, $n=0$, which does not work.  Thus the only possibility is that one factor is 3 and the other is 11.  Setting both equal to 3 yields one solution, $n=1$, for the first factor.  However plugging it into the second factor gives 0, which is not 11.
Therefore there are no integer solutions.
[/hide]


What sort of chemistry competitions are out there besides the USNCO and where can I find practice tests/previous tests for them and the USNCO?


If $a,b,c,d \in \mathbb{R}_{+}$ and $a+b +c +d =1$, show that \[ ab +bc +cd \leq \dfrac{1}{4}. \]


geometry
3D geometry
combinatorics proposed
combinatorics

A cube of edge $ n$ is made of $ n^{3}$ unit cubes, and there are $ 6n^{2}$ unit squares on the surface of the cube. At most how many of these squares can be chosen so that no two have a common edge?

Here is the idea for the case $ n = 2k$ is even.  The surface of the cube can be broken down into $ 4n$ disjoint "ribbons," each of which is an odd cycle encircling one of the vertices of the cube.  For example, eight of these ribbons consist of the three squares containing a given vertex of the cube.  The next eight (if $ k \geq 2$) consist of the six squares which border the first three, and the additional three squares needed to make a circuit out of them.  And so on.  (I'm sorry if this isn't clear.)  Thus, we have eight cycles of length 3, eight cycles of length 9, ..., and eight cycles of length $ 3(n-1)$.  The cycles of length 3 may have only one colored square; the cycles of length 9 may have only 4 colored squares; ...; and the cycles of length $ 3(n-1)$ may have only $ \frac{3(n-1)-1}{2}= 3k-2$ colored squares.  Thus, there are at most $ 8 \sum_{i = 1}^{k}3i-2 =12k^{2}-4k = 3n^{2}-2n$ colored squares.

It is still necessary to find a coloring for which this maximum is achieved (it's not too hard to do, but it's unpleasant to describe without a diagram) and to do the odd case.  I believe for $ n = 2k-1$ the situation is essentially the same; there is a little trick to be done at the end, since it is not possible to decompose the cube entirely into odd cycles in the same way, but the main idea is still good.


Evaluate the minimum value of $\frac{sen^{3}x}{cosx}+\frac{cos^{3}x}{senx}=-\frac{3}{4}$

it's easy to show that the result is periodic

So the minimun will be  ${-\infty }$

if you rewrite your question will be possible to solve


[quote="rcv"]Express $1+\frac{1}{2+\frac{1}{3}}$ as a common fraction.[/quote]

[hide]
Work your way up.

2 + 1/3 = 7/3

1 / (7/3) = 3/7

1 + 3/7 = 10/7
[/hide]

 :)

[quote="rcv"]Express $1+\frac{1}{2+\frac{1}{3}}$ as a common fraction.[/quote]

[hide]$1+\frac{1}{\frac{7}{3}}=1+\frac{3}{7}=\frac{10}{7}$[/hide]


Any prime number $ p \equiv 1\ mod  4$ can be expressed as a sum of two  squares.
That is the sequence $ 5 ,13,17,29,37.....................$ each term can be expressed as sum of two squares.
Is this true or not ?


Sheikh's Heritage
An Arab sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower will win. The brothers, after wandering aimlessly for days, ask a wise man for advise. After hearing the advice they jump on the camels and race as fast as they can to the city. What does the wise man say?

[hide="i knew this"]they switched camels[/hide]

I heard this too, even though I couldn't remember the answer :D .


inequalities
number theory proposed
number theory

Find all the nonegative integers satisfying m,n such that:

         1!3!5!.........(2n-1)!=m!

No,it isn't.It's very easy. :P It had been given at Russian Olympiad.I'll post a solution tomorrow. ;) Now I'm tired...

Suppose $n > 2$, whence $m > 3$
Let $p_n$ denote the $n_{%Error. "tmop" is a bad command.
{th}}$ prime. Let $p_k$ be the greatest prime divisor of $m$.
Then $p_k \leq m < p_{k + 1}$ (by definition) and $p_k \leq 2 n - 1 < p_{k + 1}$.
(If $2 n - 1 < p_k$, then $p_k \nmid \prod_{j = 1}^n \left( 2 j - 1 \right) !$ but $p_k |m!$. If $2 n - 1 \geq p_{k + 1}$, then $p_{k + 1} | \prod_{j = 1}^n \left( 2 j - 1 \right) !$ but $p_{k + 1} \nmid m!$. A contradiction in both cases). Moreover
\begin{eqnarray*} \prod_{j = 1}^{n - 1} \left( 2 j - 1 \right) ! & = & \frac{m!}{( 2 n - 1 ) !}\\ & = & \prod^m_{j = p_k + 1} j \end{eqnarray*}
is not divisble by $p_k$ (since $2 p_k > p_{k + 1} - 1 \geq m$). Therefore $2 ( n - 1 ) - 1 = 2 n - 3 < p_k$.
Since $2 n - 1 \geq p_k$ and $2 n - 3 < p_k$ we have $2 n - 1 = p_k$ or $2 n - 1 = p_k + 1$.
By asumption $n > 2$ hence $2 n - 2 = p_k$ is impossible. Therefore $2 n - 1 = p_k$.
Whence
\begin{eqnarray*} ( 2 n - 2 ) ! \prod^{n - 1}_{j = 1} \left( 2 j - 1 \right) ! & = & \frac{m!}{2 n - 1}\\ & = & \frac{m!}{p_k} \end{eqnarray*}

EDIT: Mistake, to be corrected here.

That is the proof of V.P.Beskarev, the author of the problem.I didn't read it very well, but I think it's correct, {x}.
Now, please solve this equation, composed by the same V.P.Beskarev:1!3!5!...(2m-1)!=(m(m+1)/2)!. ;)

Isn't this a particular case of the preceding problem ?
That is, it suffice to solve $( m, 1 + 2 + \ldots + m ) = ( 1, 1 ) \mbox{ or } (2, 3 )$. Hence $m = 1$ or $m = 2$.
Or is there some shorter, nicer solution we are supposed to find ?  :P (I guess yes, so I'll try to find one)
Also, did V.P Beskarev publish his results in some journal ? Do you have any reference ?

The solutions of this equation are m=1,2,3,4. ;) Do you want to know a different way to solve it?

Here it is.We'll use the Theoerem of Bertrand.Between a and 2a there are at least 1 prime number.We notice p the prime number which is between 2x-1 and 2(2x-1).If we had the inequality 2(2x-1)<x(x+1)/2, then we woull obtain that p is a prime factor of the right member of the equation, because it is smaller than x(x+1)/2 and it isn't a prime factor of the laft member.But this thing is impopssible.Therfore 2(2x-1)>x(x+1)/2=>x=<6.We only obtain the previous solutions.This is the solution of V.P.Beskarev in Kvant Magazine. ;)

Sorry, I've made a mistake, beginning with "If $m>p_k$".
I will correct it.


Given two points A and B, and some point C on segment AB which is not the midpoint of AB, construct some point C' on line AB such that

[tex]\displaystyle\frac{AC'}{C'B} = \frac{AC}{CB}[/tex]

using only a straightedge.

(NOTE: These are undirected edge lengths).

[hide]For this to be true, C' must be the same point as C.

Given a point C and another point C', if AC>AC', then CB<C'B.  This also applies to the inverse, if AC<AC', then CB>C'B.  Due to this inverse nature, the only way for the equation to hold true would be if C is the same point as C'.

[/hide]

The problem says C' lies on "line AB" not "segment AB."

Such a point C' does exist, as long as C is not the midpoint of segment AB.

So... construct it :).

EDIT: Och. I didn't see the "only with a straightedge" thing. This solution requires a compass. My apologies.

[hide]Suppose AC > CB. Let's have a = AC and b = CB. Construct a triangle with sides (a + b) and (a - b) (the angle between them doesn't matter - for convenience, let's suppose that these two sides are perpendicular). Construct a segment of length b that is parallel to the side (a - b), and touches both other sides of the triangle (or their extensions). Then, the segment between the vertice opposite length (a - b) and the intersection of the segments of length (a + b) (or its extension) and (b) (call it c) has the required length.

Proof: We need to construct length c such that

a : b = (a + b + c) : c

Rearrange to get

ac = ab + b2 + bc

ac - bc = ab + b2

(a - b)c = b(a + b)

c : b = (a + b) : (a - b)

This is the similar triangle construction in the diagram.

[/hide]

Coder wrote:Are we allowed to know any definite values during the construction?

[hide]Point C' can be defined as being AC/(CB-AC) away from point A if AC>CB (C' would be on the opposite side of point A from point B.  Point C' would have to be the same as point C if C' was in segment AB, and if it were on the other side of point B, the problem would become impossible), or it can be defined as CB/(AC-CB) away from point B in the same manner if AC<CB.

To explain why this works, I'm going to assume that AC>CB for simplicity's sake.  The same thing applies if AC<CB, but the referece point would be point B instead of point A.  The numerator of the difference between AC and CB can be defined as the number of "parts" that C'B needs to be divided into because C'A and C'B overlap.  The numerator of AC can then be defined as the number of the "parts" that point C lies from A in order to preserve the original ratio.
[/hide]

Here is how it goes... I give you the three points A, B, and C. You must construct C' using only a straightedge. No other aids are given.

Construction now, proof later.

Draw an arbitrary line through C with D and E on that line. AE and DB intersect at F, and AD and BE intersect at G. FG and AB intersect at C'.

Ok, proof time. See the diagram. (for clarification, h=m+n.) I'll prove that m/n=g/(g+h). The proof is basically Menalaus 20 times.

From menalaus, (g/h)(d/c)([a+b]/a) = 1, i.e. a/[a+b] = (g/h)(d/c).

Similarly, b/[a+b] = (g/h)(e/f), and adding these 2 equations and noting that a/[a+b]+b/[a+b]=1 gives

1=(g/h)(d/c)+(g/h)(e/f), i.e. h/g = e/f+d/c.

From symmetry, we also get i/j = b/a+c/d and k/L = a/b+f/e.

From menalaus again, (m/n)(d/c)([i+j]/j) = 1, which simplifies as follows:
(m/n)(d/c)(1+i/j) = 1
(m/n)(d/c)(1+b/a+c/d) = 1
[a+b]/a = (c/d)(n/m)-c/d
[a+b]/a = (c/d)(n/m-1)
(n/m-1)(a/[a+b]) = d/c.

From symmetry, we also get (n/m-1)(b/[a+b]) = (e/f). Adding these last two equations gives

n/m-1=d/c+e/f
n/m-1=h/g
n/m = h/g+1
n/m = [g+h]/g

as desired.

(There are 14 variables on the diagram and I used every single one of them in the solution. How scary! :))

[quote="zabelman"] (Proof) [/quote]

For those not in the know, Menelaus (spelled correctly) is a theorem similar to Ceva's.

http://planetmath.org/encyclopedia/MenelausTheorem.html

Knowing the Theorem,
Can't you just
Draw a random line CX and connect X to A and B
Draw a random line BP where P is on segment CX, the line intersect AX at point E,
draw AP, the line intersect BX at point F.
Connect EF, which must intersect intersect AB at point C'

Proof:
by Ceva's Theorem,
 (AC/BC)(BF/FX)(XE/AE)=1
by Melealus(or whatever it's called) Theorem,
(AC'/BC')(EX/AE)(FX/BF)=1
THus, AC/BC=AC'/BC'

Nice, beta! Our constructions are equivalent (by taking D and E on the same side of AB in my diagram), but yours makes the ceva solution much more visible. Very nice!

My intended solution was the one Beta gave... although this is apparently a well known construction for harmonic conjugates.


Not yet on forum, but here it is:
[url]http://www.youtube.com/watch?v=vXTO8S_WZV0[/url] :D

Posted AGES ago [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=300513]here[/url].

Locked.


trigonometry
inequalities
inequalities proposed

Let ABC be a triangle.Prove that:
$ \frac{1}{sinA}\plus{}\frac{1}{sinB}\plus{}\frac{1}{sinC} \ge \frac{6R}{\sqrt[3] {abc}}$, where R its circumradius.

Because of the sine law($ \frac{a}{sin{A}} \equal{} 2R$) the inequality can be written equally as:
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq\frac {3}{\sqrt [3]{abc}}$ which is true by AM-GM :wink:


I guess you mean RHS=$\sum^{n}_{k=0}\frac{(a_1+\ldots+a_n)^k}{k!}$.
Now $(a_1+\ldots+a_n)^k\ge k!\sum_{i_1<...<i_k}a_{i_1}\cdots a_{i_k}$. (We keep just that term when expanding it).
So $LHS=1+\sum a_i+\sum_{i<j}a_ia_j+...+ a_1\cdots a_n\le RHS$ . Done.

[quote="spanferkel"]I guess you mean RHS=$\sum^{n}_{k=0}\frac{(a_1+\ldots+a_n)^k}{k!}$.
Now $(a_1+\ldots+a_n)^k\ge k!a_1\cdots a_k$. (We keep just that term from expanding it).
So $LHS=1+\sum a_i+\sum_{i<j}a_ia_j+...+ a_1\cdots a_n\le RHS$ . Done.[/quote]
I dont get it  :?

yeah sorry, I edited. I forgot to write some non-neglectible details :blush:
I think now it is OK :roll:


The digits 1, 2, 3, 4 and 9 are each used once to form the smallest
possible $ {\bf even}$ five-digit number.  What is the digit in the tens place?

to make smallest 5 digit number we would start with 1, then 2, then 3, then 4 and finally 9 for 12349

but it has to be even. so we switch the 4 and 9 to get 12394 and 9 is in the tens


When will the WOOT syllabus be available?


There seems to be a single-minded focus in this forum on prestigious U.S. colleges, such as Harvard or MIT or Caltech or what have you. This focus entirely misses the fact that there are great universities in other countries that rival these for an undergraduate education. (There are universities in the U.S. that rival them as well, but that's another matter.)

Why should you care? Universities in other countries are typically public, so they are much less expensive than U.S. universities, even for international students. [See later posts.] And it can be much easier to get in to these places, though the education may be comparable.

I'm referring to English-language universities, such as the universities of Cambridge, Waterloo, Toronto, Oxford, British Columbia, etc.

The purpose of this thread is to get people to investigate these places as a possible alternative to U.S. schools. I can try to answer questions here and I'm sure there are others who are more knowledgeable.

(Disclosure: I went to school in the U.S. and am about to start my second year in math at the University of Waterloo.)

[quote="paradox"]Universities in other countries are typically public, so they are much less expensive than U.S. universities, even for international students.[/quote]Where did you get this information?

Some examples that I just looked up:

Waterloo is about 26,000 Canadian dollars / year for US students, which is about 23,000 US dollars.

Cambridge is about £12,000 / year, which is over 22,000 US dollars.

In my experience most "public" universities charge a much higher tuition rate for out-of-country students, and it is much more difficult to get financial aid.

Granted, these numbers are somewhat cheaper than elite US school (for example, MIT is 33,000/year), but factoring in the travel costs, costs associated with international currency transactions, getting the proper visa, and the extreme difficulty of getting financial aid as an international student, I wouldn't say that it's "much less expensive".

The other downside is that to a US employer, schools like Harvard, MIT, etc, are "known quantities", and non-US school may not necessarily be so.

To me, the main reason to consider non-US schools (if you are a US citizen) is for the experience of living abroad, and this is often more easily achieved by doing a year abroad during your undergraduate years.

I would also advise making the cost comparison very carefully. I certainly think Waterloo is a great school, and probably great value at its list price for foreign students, but as the first reply mentioned, the issue to check is what the net out-of-pocket cost is after financial aid awards are taken into account.

As to the value of having international experience, I am 100 percent in agreement that it is a good idea for Americans to spend significant time in other countries. There are some learning experiences you will have overseas that you just can't substitute book learning for. You have to see it to believe just how different other cultures, climates, cuisines, languages, and religions can be.

First, I should've said that the cost comparison was to elite private colleges, and that I'm indeed not factoring in financial aid. In my case, two terms (out of the three in a year) at Waterloo cost me about 25K U.S. in total expenses, including housing and food. As I would not have qualified for financial aid at U.S. schools, that's about half of what a U.S. college [url=http://web.mit.edu/finaid/tuition_fees/]would've cost[/url].

Waterloo's regular international student tuition [url=http://www.findoutmore.uwaterloo.ca/financing/expenses.htm]tops out[/url] at 18K Canadian (16K U.S.) for two terms. The co-op tuition is a bit higher, but that's offset by the money earned during the work terms. (More than half of the students in the Faculty of Math at Waterloo are in co-op.)

Since pretty much any Canadian city is decently close to the U.S. border, travel costs aren't significantly different. (Probably not the case for the U.K.) And in my experience, getting a study permit was not difficult.

The main point isn't that these places are cheap — anyway, it's hard to cost less than state universities in the U.S. It's that they can compete well on the academic front.

IIT is a very good college in India and for $6000, yea


LOL the last time someone posted on this forum was like last year, is this forum dead? I guess no one likes Mu Alpha Theta on AoPS LOL.

Its not about liking Mu alpha theta. people just have other things on their minds (math contests, etc). :)

[quote="ragnarok23"]The forum died because, well, MAO is not the most prominent competition, and there is not that many people from Florida on the forum. I think.[/quote]
Well, you got two or more people now. :P

[quote="computer_nuke"]Yeah, I'm from Florida, and here, EVERONE is obsessed with MAO.[/quote]

um....not from where i'm from....(davie/ft.lauderdale area)

yah...definitly not..but there's always me!

 :D

...and ONLY me....

in this forum at least

*tear*

 :(

but that's okay... cus now i don't have to hide from anyone..

 :ninja:

Hey, you're from Tampa! That's awesome, I live somewhere very close to there...

hmm... im from tallahassee.  were not OBSSESSED with mao but its definaty the bulk of our competitions.  did anyone here go to the rickards invitational? or vestavia?  and is anyone going to pensacola for the regional?

[quote="Elemennop"]Hey pratik you guys live in florida (the orange state) so all you guys want is MAO that is why you do not post on AOPS ok??

you and your go around the tree friend how about dig a hole[/quote]
wtf?

[quote="Elemennop"]Hey pratik you guys live in florida (the orange state) so all you guys want is MAO that is why you do not post on AOPS ok??

you and your go around the tree friend how about dig a hole[/quote]

yea, wtf.  go around the tree lives in your state anyways.  :rotfl:

so what is your problem?

can i pretend i live in Florida? what is MAO? :lol:

[quote="funcia"]MAT  :rotfl:
that is really funny
umm yeah can people in georgia do this?[/quote]

i dont know.  its like the "Florida Association of Mu Alpha Theta", but i think you can come to the invitiationals, just not regionals.  i know vhhs (vestavia) comes to a few invites.

Great! When I go on vacation I'll ask you guys if you will have a contest!

funcia, im glad you changed your avatar back

that avator with that woman was freaky with those red lips  :wink:

well i thought it look like...
yeah i like that avatar

[quote="ragnarok23"][quote="Walk Around The River"][quote="funcia"]i had lots of complaints but to tell the truth i would have kept that oneif people didn't tell me it was scary
 :lol:[/quote]

i thought it was pretty sexy.  :rotfl:[/quote]
I totally saw that coming[/quote]

ROFL  :rotfl:

[quote="Walk Around The River"][quote="now a ranger"][quote="Pakman2012"]whats that? a post in the florida section?

woo. go us. even if half of you arent from florida. thats alright[/quote]

Good job Florida Dancing Bears.[/quote]

yea saiem, whats up with the bear?

florida is an interesting state.  its the middle of december...and what? 85 degrees?  :ninja:[/quote]

I'm in Atlanta right now. It's 65 degrees GRRR... but it's better than in my hometown. It's like 40 degrees there

GO DANCING BEARS!!!!!

That bear is actually amusing.

It should be a Florida symbol slash mascot.  :D

[quote="Pakman2012"]whats that? a post in the florida section?

woo. go us. even if half of you arent from florida. thats alright[/quote]

 :ninja:

Perhaps if there was a private FL forum it would be a little more active...

No, it really wouldn't. There are like 7 of us on the forum. Pratik, Eli, bpms, me, david, yannik (no longer active), uhh.. i'm missing like one or two. Thats about it. Chen is on occasionally, but rarely for math.

if you think Florida is inactive, look at Alabama.

It's just Cicatriz(he only gets on once in a while), and Elemmnop, and me.  Talk about a lost cause.  :rotfl:

Florida also has Ckck, i think he got 9th or 12th at Vestavia, not sure.

But he's a fairview kid, so you didn't say him, so that's 8

and wasn't there a guy named Patrick or something? never gets on

how about not

florida people have intercourse with Teddy Bears.  :lol:

Wait, how does that happen?

The Florida God used magic to make this natural deed happen with Lovely Teddy Bears, like Saiem's Avatar

[quote="bpms"][quote="Pakman2012"]Well alabama people have intercourse with horses.

taste that.[/quote]
Don't forget their sisters...[/quote]

SAME WITH FLORIDA.

Except they have intercourses with Whiny Horses

Danny Kim, quit your spamming and do some work.

I'm from Florida.  See you at mao nationals guys.

i think mao rocks :lol:
other than mathcounts it's the only comp middle schoolers can do that i know of.....tell me if I'm wrong. :blush:
also anyone have any tips for a 8th grader going into geometry mao????
i don't want to place last like i did when...............  :D just kidding placed 20 something last year at blue key in the algebra 1 category.best i've done is 22nd....i think that stinks

and i even did previous tests on the mao website!!!!


Find the least integer n satisfys the following condition;
For any n integer(not necessarily distinct) $a_{1},a_{2},...,a_{n}$, there exist 9 integer $1\leqq_{1}<i_{2}<...<i_{9}\leqq$ and $b_{i}\in\{4,7\}(i=1,2,...,9)$ such that $\sum_{j=1}^{9}{b_{j}a{}_{i}{}_{j}}$ is a multiple of 9

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130463]Whole paper[/url]

I hope it is correct. Answer: 13. If $n \leq 12$, take two 1's, two 3's and the rest 0. Now suppose $n=13$. An observation we'll need is that among any 5 numbers, there are 3 whose sum is a multiple of 3. We will look at the $a_{i}$'s mod 3. We will find $i_{1},i_{2},...,i_{9}$ with that property which are different, not neccesarily increasing (it's perfectly equivalent).
If it is possible, take $i_{1},i_{2},i_{3}$ such that the sum of the corresponding $a_{i}$'s is a multiple of 3, but not all of them are multiples of 3. Among the 10 remaining numbers, there are 3 whose sum is a multiple of 3, and, after that, among the 7 remaining numbers, there are 3 whose sum is a multiple of 3. We have $3|4a_{i_{1}}+...+4a_{i_{9}}$. We can change some of the first 3 coefficients to 7 in order to get the desired divisibility by 9.
Now, if that choice of $i_{1},i_{2},i_{3}$ is not possible, at least 11 of the $a_{i}$'s are multiples of 3. An argument similar to the one in the previous paragraph shows that among any 11 integers, there are 9 whose sum is a multiple of 3 (indeed, we can extract 3 times a triple whose sum of elements is a multiple of 3). Apply this for the "thirds" of those 11 $a_{i}$'s and we are done.


vector
linear algebra
linear algebra unsolved

let $U\subset\mathbb R^{n}$ be a vector subspace. show that U is closed

It is finite-dimensional, whence complete, whence closed.
You have more general results in the context of topological vector spaces.

sorry but how does finite dimensional imply complete?

my approach was to split $\mathbb R^{n}$ as the direct sum of $U$ and $U^{p}$ and then use the continuity of the projection map onto $U^{p}$ .$U^{p}$ denotes the vectors which are orthogonal to every element of U.

[quote="kouboy"]sorry but how does finite dimensional imply complete?
[/quote]

If a sequence of elements of $\mathbb{R}^{n}$ is Cauchy, then the sequences of the components of the previous elements (over, say, the canonical basis of $\mathbb{R}^{n}$, are all Cauchy) in $\mathbb{R}$, whence convergent.