ratio
calculus
integration
geometric sequence

Let $ n$ be a composite natural number. Prove that there are integers $ a_{1}, a_{2}, ... , a_{k}$, all of them greater than 1, such that
[list]$ a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{k} \equal{} n\left(\frac {1}{a_{1}} \plus{} \frac {1}{a_{2}} \plus{} ... \plus{} \frac {1}{a_{k}}\right)$[/list]

[hide]If $ n$ is a square, then set $ k^2 \equal{} n$ and $ a_i \equal{} k$.

Otherwise, let $ k\plus{}2>0$ (since $ n$ is composite) be the number of divisors of $ n$ and identify each $ a_i$ one-to-one with a divisor of $ n$ which is neither 1 nor $ n$. Then two terms on both sides always cancel out.[/hide]

-- edited

@bertram: If n is a square, then by your definition, then $ k\equal{}nk$, thus $ n\equal{}1$.

[quote="bertram"][hide]If $ n$ is a square, then set $ k^2 \equal{} n$ and $ a_i \equal{} 1$.

Otherwise, let $ k$ be the number of divisors of $ n$ and identify each $ a_i$ one-to-one with a divisor of $ n$. Then two terms on both sides always cancel out.[/hide][/quote]$ a_i>1$ is a condition given in the problem. Notice that if $ n\equal{}p^2$ for some prime $ p$, then $ k\equal{}1$, making $ a_1\equal{}p$. This problem is equivalent to proving

\[ \sum_{1<d|n<n}d\equal{}\sum_{1<d|n<n}\frac{n}{d}\]

[hide]Because $ n$ is composite it may be written in the form $ n=r_1\cdot r_2$, where $ r_i\in\{\mathbb{Z}: 1<r_i<n\}$. So defined, the statement becomes \begin{eqnarray*}\displaystyle r_1+r_2
&=&n\left(\frac{1}{r_1}+\frac{1}{r_2}\right)\\
&=&n\left(\frac{r_1+r_2}{r_1\cdot r_2}\right)\\
&=&r_1+r_2\\
\end{eqnarray*} which holds for any composite number.[/hide]
Please critique my approach! Thanks.  :)

[hide="My approach"]In the exam, I wrote that, if the terms $ a_{1}, a_{2}, ... , a_{k}$ are equal, then $ n$ could equal $ a^{2}$, and the sum on both sides would equal $ (a)k$. Furthermore, if the terms were not equal, then they could form some sort of geometric progression. If the starting value is $ a_{1}$ and common ratio is $ a_{1}$, for example, the equation could be rewritten as

$ \sum_{i \equal{} 1}^{k} a_{1}^{i} \equal{} n \sum_{i \equal{} 1}^{k} \frac {1}{a_{1}^{i}}$,

from which we can derive one solution: $ n \equal{} a_{1}^{k \plus{} 1}$. The idea is that the terms form a geometric progression, and that $ r^{k \minus{} 1} \mid n$, where $ r$ is the common ratio. It's not as rigorous a proof as I'd like it to be, though; that's why I posted this problem.[/hide]

How does your approach work for 6, for example?

You realize you haven't proven the result for all $ n$.  bertram is asking you to prove the result for $ n \equal{} 6$.

The divisor-pair answer is well-known.  Do there exist other solutions?

The thing is, $ n$ is of the form $ a_{1}^{2}r^{k \minus{} 1}$ (in the cases I presented, at least), where $ a_{1}, a_{2}, \dots, a_{k} \in \mathbb {Z^{ \plus{} }}$ and $ 1 \leq a_{1} \leq a_{2} \leq \dots \leq a_{k}$, so $ n \equal{} 6$ iff $ a_{1} \equal{} 2$, $ k \equal{} 2$, and $ r \equal{} \frac {3}{2}$ (unless I'm missing something).

$ a_{1} \plus{} a_{2} \equal{} \sum_{i \equal{} 0}^{1} 2 \left(\frac {3}{2} \right)^{i} \equal{} 6 \sum_{i \equal{} 0}^{1} \frac {1}{2 \left(\frac {3}{2} \right)^{i}} \equal{} 6 \left( \frac {1}{2} \plus{} \frac {1}{3} \right) \equal{} 2 \plus{} 3 \equal{} 5$

[quote]Do there exist other solutions?[/quote]

Not really sure, that's why I'm asking for feedback. >,<

I don't see the rigor.  Can you show that a non-integer value of $ r$ always leads to integer values for $ a_k$, and can you show that all numbers $ n$ can be written in that form correctly?

If your argument is correct, it looks like you're just picking out a subset of the divisors anyway.

[quote="t0rajir0u"]The divisor-pair answer is well-known.  Do there exist other solutions?[/quote]

This might be trivial, but I think all proper divisors other than 1 should work too.

By the "divisor-pair" answer I mean [b]any[/b] collection of pairs of divisors $ d_1 d_2$ such that $ d_1 d_2 \equal{} n$.

A related question:  when $ n$ is not square, can you still find an answer where $ k$ is odd?

Yes, yes, I'm going to answer all that once I finish typing the proof — don't rush me. XP

[quote]If your argument is correct, it looks like you're just picking out a subset of the divisors anyway.[/quote]

Exactly. :P

[quote="t0rajir0u"]can you show that all numbers  can be written in that form correctly?[/quote][hide="Proof"]
[b]Proposition[/b]: Any composite number $ n$ can be expressed as $ a^{2}\left(\frac {m}{q}\right) ^{d} \equal{} a^{2}r^{d}$, where

1. $ a, d, m, q \in \mathbb {Z^{ \plus{} }}$;
2. $ a > 1$;
3. $ gcd(m,q) \equal{} 1$;
4. $ q^{d} \neq a^{2}$;

iff $ q^{d} \mid a^{2}$.

[b]Proof of necessity[/b]: Suppose $ q^{d} \nmid a^{2}$. Since $ q^{d} \nmid m^{d}$ for $ q > 1$ (if  $ q \equal{} 1$, then $ q^{d} \mid a^{2}$), $ \ q^{d} \nmid a^{2}m^{d}$, which means that $ \frac {a^{2}m^{d}}{q^{d}} \equal{} n$ is not an integer — but, by the definition of a composite number, $ n$ is an integer (contradiction). Thus, $ q^{d} \mid a^{2}$ is a necessary condition.

[b]Proof of sufficiency[/b]: Let $ n \equal{} a^{2}\left(\frac {m}{q}\right) ^{d} \in \mathbb {P}$, $ q^{d} \mid a^{2}$, and $ q > 1$ ($ q \equal{} 1 \Rightarrow n \equal{} a^{2}m^{d}$, and $ n$ is not prime, because $ a > 1$). Since $ q^{d} \mid a^{2}$, and $ a^{2} \mid n$, $ q^{d} \mid n$. However, $ q^{d} \neq a^{2}$, therefore $ q^{d}$ and $ a^{2}$ are two different divisors of $ n$ — which means that $ n$ is composite (widerspruch!). Thus, $ q^{d} \mid a^{2}$ is a sufficient condition.[/hide]

Nevertheless, as you suggested, my solution does not hold for all values of $ n$. I give you this to consider, instead:

[hide="A different solution?"]

We'll ignore the case where $ a_{1} \equal{} a_{2} \equal{} ... \equal{} a_{k}$, because we already know that this implies $ n \equal{} a_{k}^{2}\ \ \forall k \in \mathbb{Z^{ \plus{} }}$. Let us define $ S_{k}$, then, as the set of distinct integers greater than one that satisfy the equation presented in the first post.

Since $ a_{1}, a_{2}, ... , a_{k}$ are all positive integers, their sum must also be a positive integer, which means that either $ \forall a_{i} \in S_{k}, a_{i} \mid n$ or $ \sum_{i \equal{} 1}^{k}\frac {1}{a_{i}} \in \mathbb{Z}$. If the former is true, $ S_{k}$ must include all divisor-pairs of $ n$, so $ k$ must be even; otherwise, $ \exists a_{i} \in S_{k} : \frac {n}{a_{i}} \not\in S_{k}$, which means that $ \sum_{i \equal{} 1}^{k} a_{i} \neq n \sum_{i \equal{} 1}^{k} \frac {1}{a_{i}}$. Thus, we can say that $ \forall a_{i} \in S_{k},\ \exists a_{x} \in S_{k} : \frac {n}{a_{x}} \equal{} a_{i}$.  Note, also, that if at least two elements in $ S_{k}$ are equal, either $ n$ is the square of a composite number, or all the terms are equal (and $ n$ can be any perfect square, as mentioned before).

Now, if $ \sum_{i \equal{} 1}^{k}\frac {1}{a_{i}} \in \mathbb{Z}$, and $ \exists a_{m} \in S_{k} : a_{m} \nmid n$ (without loss of generality), then $ \forall a_{m} \in S_{k}, \exists S_{x} : (S_{x} \subseteq S_{k}, a_{m} \equal{} a_{x} \forall a_{x} \in S_{x}, \sum \frac {1}{a_{x}} \in \mathbb{Z^{ \plus{} }})$. Basically, for every integer $ a_{m} \equal{} y$ that does not divide $ n$, the set $ S_{x}$ contains $ M\cdot y$ equal elements, such that the sum of the reciprocals of the elements is a positive integer $ M$. Nevertheless, $ \sum a_{x} \equal{} M \cdot y^{2} \neq \sum \frac {1}{a_{x}} \equal{} nM$, so $ \sum_{i \equal{} 1}^{k} a_{i} \neq n \sum_{i \equal{} 1}^{k} \frac {1}{a_{i}}$ unless $ n \equal{} y^2$, which would mean that $ a_{m} \mid n$.

Finally, to prove that $ \forall n \not\in \mathbb{P}, \exists S_{k} : \sum_{i \equal{} 1}^{k} a_{i} \equal{} n \sum_{i \equal{} 1}^{k} \frac {1}{a_{i}}$, notice that $ S_{k}$ exists iff $ \forall a_{i} \in S_{k}, a_{i} \mid n$ (which we sort of proved earlier). $ \not\exists a_{i} \in S_{k} : a_{i} \mid n \rightarrow n \in \mathbb{P}$, since $ S_{k} \equal{} \{ a_{i} : a_{i} \in \mathbb{Z} \cap [2, \infty), a_{i} \mid n \}$. [/hide]

Comments would be appreciated. :\

Please pardon me if I dissected your post entirely with comments. But, before starting with that, I want to clarify the post I made earlier and that is the equivalence claimed. For such equivalence to hold, each $ a_i$ is distinct and that $ k$ is dependent on $ n$, ie $ k\leq d(n)\minus{}2$ where $ d(n)$ is the number of distinct divisors of $ n$. This is possible because $ d(n)>2$ if $ n$ is composite.

Now, let's begin with metafor's proof :D

[quote="metafor"][quote="t0rajir0u"]can you show that all numbers  can be written in that form correctly?[/quote]Proof
[b]Proposition[/b]: Any composite number $ n$ can be expressed as $ a^{2}\left(\frac {m}{q}\right) ^{d} \equal{} a^{2}r^{d}$, where

1. $ a, d, m, q \in \mathbb {Z^{ \plus{} }}$;
2. $ a > 1$;
3. $ gcd(m,q) \equal{} 1$;
4. $ q^{d} \neq a^{2}$;

iff $ q^{d} \mid a^{2}$.

[b]Proof of necessity[/b]: Suppose $ q^{d} \nmid a^{2}$. Since $ q^{d} \nmid m^{d}$ for $ q > 1$ (if  $ q \equal{} 1$, then $ q^{d} \mid a^{2}$), $ \ q^{d} \nmid a^{2}m^{d}$, which means that $ \frac {a^{2}m^{d}}{q^{d}} \equal{} n$ is not an integer — but, by the definition of a composite number, $ n$ is an integer (contradiction). Thus, $ q^{d} \mid a^{2}$ is a necessary condition.

[b]Proof of sufficiency[/b]: Let $ n \equal{} a^{2}\left(\frac {m}{q}\right) ^{d} \in \mathbb {P}$, $ q^{d} \mid a^{2}$, and $ q > 1$ ($ q \equal{} 1 \Rightarrow n \equal{} a^{2}m^{d}$, and $ n$ is not prime, because $ a > 1$). Since $ q^{d} \mid a^{2}$, and $ a^{2} \mid n$, $ q^{d} \mid n$. However, $ q^{d} \neq a^{2}$, therefore $ q^{d}$ and $ a^{2}$ are two different divisors of $ n$ — which means that $ n$ is composite (widerspruch!). Thus, $ q^{d} \mid a^{2}$ is a sufficient condition.[/quote]

First thing is the confusion in the way the proposition is stated. A clearer exposition is the following: Suppose $ n$ is a positive integer that can be expressed in the form $ a^2\left(\frac {m}{q}\right) ^{d}$ with the stated conditions. Then, $ n$ is composite iff $ q^{d} \mid a^{2}$. The reason for this small change is that both statements use the same conditions, which should be the 'global' properties for both statements.

Second, I don't know what the role of $ \mathbb{P}$ is (is that the set of primes?).

Third, there is a flaw in the proof: why does $ a^2\mid n$? Counterexample. Let $ n\equal{}6$. Then $ a$ is either 2 or 3, but $ a^2\nmid n$...

[quote="metafor"]Nevertheless, as you suggested, my solution does not hold for all values of $ n$. I give you this to consider, instead:

A different solution?

We'll ignore the case where $ a_{1} \equal{} a_{2} \equal{} ... \equal{} a_{k}$, because we already know that this implies $ n \equal{} a_{k}^{2}\ \ \forall k \in \mathbb{Z^{ \plus{} }}$. Let us define $ S_{k}$, then, as the set of distinct integers greater than one that satisfy the equation presented in the first post.[/quote]

$ S_k$ needs a small clarification about the role of that $ k$ in the subscript.

[quote="metafor"]Since $ a_{1}, a_{2}, ... , a_{k}$ are all positive integers, their sum must also be a positive integer, which means that either $ \forall a_{i} \in S_{k}, a_{i} \mid n$ or $ \sum_{i \equal{} 1}^{k}\frac {1}{a_{i}} \in \mathbb{Z}$.[/quote]

This part has no justification. How did you implicate that the sum being a positive integer implies $ a_i\mid n$? Or why would that sum of reciprocals being integral follows from the fact that the sum is integral?

[quote="metafor"]If the former is true, $ S_{k}$ must include all divisor-pairs of $ n$, so $ k$ must be even; otherwise, $ \exists a_{i} \in S_{k} : \frac {n}{a_{i}} \not\in S_{k}$, which means that $ \sum_{i \equal{} 1}^{k} a_{i} \neq n \sum_{i \equal{} 1}^{k} \frac {1}{a_{i}}$. Thus, we can say that $ \forall a_{i} \in S_{k},\ \exists a_{x} \in S_{k} : \frac {n}{a_{x}} \equal{} a_{i}$.  Note, also, that if at least two elements in $ S_{k}$ are equal, either $ n$ is the square of a composite number, or all the terms are equal (and $ n$ can be any perfect square, as mentioned before).[/quote]

The last statement needs a proof for such a claim, which I believe it's not true if $ a_i$'s are not distinct, ie take $ n\equal{}28$ and we can have multiple equal pairs of $ a_i$'s but 28 is not a square and not all $ a_i$'s are equal...

[quote="metafor"]Now, if $ \sum_{i \equal{} 1}^{k}\frac {1}{a_{i}} \in \mathbb{Z}$, and $ \exists a_{m} \in S_{k} : a_{m} \nmid n$ (without loss of generality), then $ \forall a_{m} \in S_{k}, \exists S_{x} : (S_{x} \subseteq S_{k}, a_{m} \equal{} a_{x} \forall a_{x} \in S_{x}, \sum \frac {1}{a_{x}} \in \mathbb{Z^{ \plus{} }})$. Basically, for every integer $ a_{m} \equal{} y$ that does not divide $ n$, the set $ S_{x}$ contains $ M\cdot y$ equal elements, such that the sum of the reciprocals of the elements is a positive integer $ M$. Nevertheless, $ \sum a_{x} \equal{} M \cdot y^{2} \neq \sum \frac {1}{a_{x}} \equal{} nM$, so $ \sum_{i \equal{} 1}^{k} a_{i} \neq n \sum_{i \equal{} 1}^{k} \frac {1}{a_{i}}$ unless $ n \equal{} y^2$, which would mean that $ a_{m} \mid n$.[/quote]

What's the motivation behind for using WLOG in the beginning of this part? The proof for this part ... to tell you the truth... I don't know what you are trying to prove  :huh:  So you define the subset $ S_x$. If you work carefully with your current proof for what you have defined, $ S_x\equal{}S_k$  :!:  Actually, what is $ m$ here? An index or a constant? It's really confusing.

[quote="metafor"]Finally, to prove that $ \forall n \not\in \mathbb{P}, \exists S_{k} : \sum_{i \equal{} 1}^{k} a_{i} \equal{} n \sum_{i \equal{} 1}^{k} \frac {1}{a_{i}}$, notice that $ S_{k}$ exists iff $ \forall a_{i} \in S_{k}, a_{i} \mid n$ (which we sort of proved earlier). $ \not\exists a_{i} \in S_{k} : a_{i} \mid n \rightarrow n \in \mathbb{P}$, since $ S_{k} \equal{} \{ a_{i} : a_{i} \in \mathbb{Z} \cap [2, \infty), a_{i} \mid n \}$.

Comments would be appreciated. :\[/quote]

This is quite redundant and that last statement is confusing and has no relevance for solving this problem (perhaps that could be something you can do over the summer as a research topic).

Having said all these comments, don't give up. It's just that the exposition needs to be clearer and simpler. Using simple ideas to solve a hard and complicated problem is the heart of 'the proof' that Erdos taught us. :)

PS: I'm waiting for t0rajir0u's comments  :lol:

It's no problem. I think I was actually more worried about using notation than about making any sense. :\

[quote="10000th User"]
Third, there is a flaw in the proof: why does $ a^2\mid n$? Counterexample. Let $ n \equal{} 6$. Then $ a$ is either 2 or 3, but $ a^2\nmid n$...[/quote]

Hahaha, you got me there.

[quote]$ S_k$ needs a small clarification about the role of that $ k$ in the subscript.[/quote]

$ k$ is the number of terms, as indicated by the last term in the ordered set, $ a_{k}$. :P

[quote]This part has no justification. How did you implicate that the sum being a positive integer implies $ a_i\mid n$? Or why would that sum of reciprocals being integral follows from the fact that the sum is integral?[/quote]

If all $ a_i$ in the set divide $ n$, then $ n\sum_{i\equal{}1}^{k} \frac {1}{a_{i}}$ is an integer; integer times integer equals integer.

[quote]The last statement needs a proof for such a claim, which I believe it's not true if $ a_i$'s are not distinct, ie take $ n \equal{} 28$ and we can have multiple equal pairs of $ a_i$'s but 28 is not a square and not all $ a_i$'s are equal...[/quote]

Ugh... lol. XDDD

[quote]What's the motivation behind for using WLOG in the beginning of this part? The proof for this part ... to tell you the truth... I don't know what you are trying to prove Huh? So you define the subset . If you work carefully with your current proof for what you have defined, Exclamation Actually, what is here? An index or a constant? It's really confusing.[/quote]

I was suggesting that $ a_{m}$ is any term in the set that doesn't divide $ n$ (therefore, $ 1  \leq m \leq k$).

[quote]
This is quite redundant and that last statement is confusing and has no relevance for solving this problem (perhaps that could be something you can do over the summer as a research topic).[/quote]

Explanation of the last statement: $ S_{k}$ is a subset of the set of integers greater than 2; thus, if you can't form a subset with elements that divide $ n$ and are not equal to $ n$, $ n$ is prime. I am a very redundant person, unfortunately. :P

Hmm... I should definitely stop trying to find proofs to math problems late at night. :S

(I think “efficient proof-writing” would be a more pertinent topic to research, haha.)


logarithms
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AoPSwiki
number theory
prime numbers
number theory unsolved

How can we prove that for sufficiently large values of $ n$, $ \pi(n)\equal{}\frac{n}{\ln n}$ , where $ \pi(n)$ is the number of prime numbers less than $ n$.

Search for [[prime number theorem]] on AoPSWiki (or on google)


Kudzu vine grows at a rate of one foot per day. When starting
with a length of three feet, to how many feet will a Kudzu vine
grow over a period of 28 days?

Being more formal, when planted, the vine was $3$ feet tall.   Then after each day, it grew $1$ foot tall.  Since it grew over a period of $28$ days, it grew $28\cdot1=28$ feet taller.  Therefore, the Kudzu vine grew to be $3+28=\boxed{31}$ feet after the period of $28$ days.

The point is, the question asks "how many feet will it [u]grow[/u] over a period of 28 days?" It [u]grows[/u] 28 feet. It's height is 31 feet, but it [u]grew[/u] 28 feet.

But the key word is [u]to[/u] how many feet the vine will grow.  So the answer is still 31.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $n>3$ be a natural number.
Does there exist a finite non-abelian group $G$ ($|G| \neq 2^{n}$)  such that $f(x)=x^{2^{n-1}}$ is a group-homomorphism ?

Let $G$ be the dihedral group of order $2^m$ for any $m \le n-1$: the map is the trivial homomorphism.

Sorry I meant $|G| \neq 2^{k}$ for each natural integer $k$.


[b]Consider the ODE X'''' = λ X with boundary conditions X(0)=X'(0)=X(1)=X'(1)=0. Find the eigenvalues and eigenfunctions.[/b]

General solution of the ODE is:
X = Aexp(kx) + Bexp(-kx) + C sin(kx) + D cos(kx)

[u]Case λ>0:[/u]
Let λ=k^4, with k>0
where k is solution cos(k) cosh(k) = 1. This gives all the positive eigenvalues.

To find the eigenfunctions, I got the following system of 4 equations in 4 unknowns by using the boundary condition:
0 = A+B+D
0 = A-B+C
0 = A exp(k) + B exp(-k) + C sin(k) + D cos(k)
0 = A exp(k) - B exp(-k) + C cos(k) - D sin(k)

Now how can I solve for A, B, C, D in the above system and find the [u]eigenfunctions[/u]?

[u]Attempt:[/u]
1st equation=> D = -A-B
2nd equation=> C = B-A
Put these into the 3rd and 4th equation, we get:
0 = A exp(k) + B exp(-k) + (B-A) sin(k) + (-A-B) cos(k)
0 = A exp(k) - B exp(-k) + (B-A) cos(k) - (-A-B) sin(k)

How should I continue??

Just wondering: In a system of 4 equations in 4 unknowns, is it POSSIBLE to have infinitely many solutions? or MUST the solution be unique?

(Since any nonzero multiple of an eigenfunction is again an eigenfunciton, I am expecting the solution of the system to have one arbitrary constant, i.e. infinitely many solutions, but is this possible in the above system?)

Any help is greatly appreciated!

What happens if you collect terms and continue to solve for A and B in the last two equations? I mean, the two last equations are
still two linear equations with two unknowns.


Diophantine equation
number theory unsolved
number theory

Let p be a prime number congruent 3 modulo 4 and consider this equation:
$(p+2)x^2-(p+1)y^2+px+(p+2)y=1$
Find all solutions in positive integers (x,y).

The problem gives as a clue to show that all solutions satisfy that x is dividable by p.

Using computer I found minimal nontrivial solutions:
$p=3$, $x=102$, $y=115$
$p=7$, $x=462$, $y=491$
What does it mean?

I got something but i need a book to finish it (some pell equation theory)

multyply the equation by 4(p+1)(p+2) and get:

(1)  (p+2)n 2 -(p+1)m 2 =p 2 .
we have:

n=2(p+1)y-p-2, and m=2(p+2)x+p.

I think from 1, we can show n and m are multiples of p (i hope) and it generates a sequence of solutions wich we must look wich of them works for m and n...

Somebody good in Pell equations could finish it up very easy from here i think.

What can we say about the equation $aX^{2}-bY^{2}=1$? If there is a solution, it odd powers generate infinitely many, but are they all?


let p be a prime and a and n be natural numbers such that (p^a -1 )/ (p-1) = 2 ^n
find the number of natural divisors of na. :)

I remeember it was the easiest problem an had least mark.
The answer is 3 or 4

the question is first a bit irritating, but it is real easy: show first that a is power of 2, then that a is not divisible by 4, thus a=1 or 2. in the first case n=0, so there is an infinite number of divisors  ;)  , in the second case show that n is prime, thus the number of divisors is 4.

Peter Scholze if n=a=2 and p=3 then n.a has 3 divisors.
Many students last year lost 1 mark because of this.

Can't we use Zsigmondy Theorem for this?
If we can,how?

[quote="oiviluhatef"]By Zsigmondy we show that a must be 2 and p should be a Mersenne prime number, so n must be prime. But an = 2n, has 4 natural divisors if n is different from 2 and 3 divisors if n=2 and we can control dhe n=2 gives p=3.[/quote]
Can you write the complete solution,please?


Ring Theory
calculus
integration
algebra
function
domain
abstract algebra

$R$ is a principle ideal domain. $p$ is a prime. Show that $\langle p\rangle$ is a prime ideal...

I just wonder, why do we need $R$ to be a PID? An integral domain is enough, I think. Do I miss something?

I don't think you're missing anything. However, the assumption that $R$ is a PID would be needed if we wanted to show that an ideal of $R$ is a prime ideal iff it's generated by a prime element.

Where is this problem from? What book are you studying algebra from, liyi?

From a book called "Modern Algebra". It is in Chinese.

And there is another similar problem.

"$R$ is an Euclidean ring. $p$ is a prime. Show that $R/\langle p\rangle$ is a field."

And as you pointed out, it holds when $R$ is a PID.

It does, and if $R$ is Euclidean, then it's a PID, so we're done :).

[quote="jmerry"]This got to "Solved Problems" quickly, didn't it?

A counterexample for why we need more than "integral domain".

Let $R=\mathbb{Z}[\sqrt{-5}]$
Then $2,3,1+\sqrt{-5},1-\sqrt{-5}$ are all distinct primes, and $2\cdot3=6=(1+\sqrt{-5})\cdot(1-\sqrt{-5})$.
Clearly, the ideal generated by $2$ is not a prime ideal.[/quote]

2 is not a prime. Since $2|(1+\sqrt{-5})\cdot(1-\sqrt{-5})$ implies neither $2|(1+\sqrt{-5})$ nor $2|(1-\sqrt{-5})$.

I think you mean that 2 is irreducible.

We're obviously not working from the same set of definitions, liyi.

The axioms of a ring, as I understand them:
The ring is an abelian group under addition.
Multiplication is associative and has a unit.
We have distributivity on both sides.
$1\neq0$.

You apparently leave out the part about the multiplicative identity, causing all sorts of unnecessary complications.

In my understanding, a prime element or an irreducible element is an element $p$ such that $p$ is not a unit, and if $ab=p$, either $a$ is a unit or $b$ is a unit.
A prime ideal is defined in the usual way; if a product of two elements is in the ideal, one of the elements is in the ideal.

Your objection to my example indicates that your definition of a prime element is "an element that generates a prime ideal". If so, why did you post this thread in the first place?

Actually, I'm pretty sure it's eomething like this:

a prime is a non-invertible element s.t. whenever $p|ab$, we have either $p|a$ or $p|b$. An irreducible element is a non-invertible element s.t. whenever $p=ab$, one of $a,b$ is invertible. You cannot equate primes and irreducibles (you say that an element is prime or irreducible if... whatever). These definitions are equivalent in special types of rings (Euclidean, PID's, factorial, for instance).

[quote="grobber"]a prime is a non-invertible element s.t. whenever $p|ab$, we have either $p|a$ or $p|b$[/quote]
This is immediately and obviously equivalent to "p generates a prime ideal". If that is the definition, why does this thread exist?

I don't like that definition anyway. Here's an example from liyi's definitions to drive everyone crazy: $R=2\mathbb{Z}$. The prime ideals are of the form $2p\mathbb{Z}$ for odd primes over $\mathbb{Z}$. There are no prime elements: $6$ is not prime because in the product $6\cdot6=36$, neither $6$ nor $6$ is divisible by $6$

Of course, that happens because we have no identity. I don't like that either about the problems that liyi has been posting lately :). While the definition of a ring does not technically include the multiplicative identity, that's usually assumed.

hmm. In my book,  :blush: we don't have multiplicative identity for a ring.
But when it comes to prime, the identity is assumed.


Gauss videos FTW ratio Mafia search geometry

eRepublik is a political/economic simulation world where you start your life as a citizen and can work your way up the corporate, political, media, or military ladders. it holds a remarkable amount of freedom for the player and responsibilities for the leaders, as well as potential for any citizen to succeed, since the game has plenty of space for player-powered improvement programs.

[url=http://www.erepublik.com/en/referrer/iLord+Solomon]eRepublik Here[/url]

It has a realistic economy with a balanced voting system, and has a remarkably complex foreign policy interactions, as well as a keenly developed player-powered media. Just reading over the articles lets you get a feel for the complexity of this society.

Here's the wiki for more information to see if this MMO is for you: http://wiki.erepublik.com/index.php/Main_Page

Do note that the game is pretty slow at the start - you will have to wait for about a week before you can fully become immersed in the world. Be patient and it's sure to pay off! This is both a pro and a con - it will take up at most 10 minutes of your day for a while, which is fine for more casual gamers, but leaves something to be desired by those that seek a more intensive game.

A discussion thread to see if anyone else out there is already playing, or is interested  :lol: !

i died... here's the text:
Login
Your citizen is dead

Your wellness level is equal to 0 and you do not have any food in your inventory - as a result, your citizen has died. Your eRepublik bank accounts, inventory and skills are the same as before.

III World War has begun... PEACE GC invading North America...

[quote="Bugi"][quote="EggyLv.999"]russia attacking alaska!!![/quote]

Alaska is defended, but battle for Nova Scotia was the greatest in the whole game (by numbers, damage etc), and France is getting into Canada...[/quote]

alaska attacked again!!!

Right now AOPS should focus on taking control of Singapore - it's a small country that doesn't have a lot of people, and if we can get a lot of people to join, we could easily get a lot of AOPS people elected into Congress. The voting population numbers about 100, and there's 10 members of Congress. I already have a few friends there, so we could really get a large percentage of the vote and control Singapore, like how 4Chan controlled Iran. We just need to get some more people.

Oh, and for those that don't know, if you are in the eUS, you can contact Chris Stanwick for free Moving Tickets, as long as you say that you are going to use them to move to Florida/Kansas.

It doesn't look good for Fortis + Eden in this war so far. Probably going to lose Montana today.

This is a good article series that describes the war fairly actively if you are interested.

[url]http://www.erepublik.com/en/article/a-non-paranoid-look-at-the-possible-war-856793/1/20[/url]

[quote="zapi2007"]okkkk, joined singapore now what?[/quote]

Seconded. Also, iLord, why aren't you in Singapore too?

What's your level on eR. guys? I am 14...

I'm temporarily in Hungary to communicate with Hungarian officials on their sympathies for allowing Singapore to join PEACE, which I feel should be our main platform. Sort of like an ambassador  :lol: .

If you follow what the little guy in the corner says, you should be able to get to Lv 3 on the second day. Just keep on working/training until you get to Lv 7, when you can join a party/vote. We'll set-up a party in Singapore in plenty of time for the next Congress elections.

Oh, and if anyone wants to temporarily move abroad to fight/communicate, after you get to Lv 6, you get 5 gold, which is about 250 SGD, which should be enough to buy moving tickets and such.

Place where you heal after losing wellness in war. You can go to only when you fight at least once, and you get wellness $ 10\times \text{hospital quality}$, so it's a maximum of 50 wellness gain.

Getting more gold: invite friends, when they are at Lv. 6, you and they get 5 gold each.

NEWS:

USA&Canada made a counter-offensive, bringing territories back.

Still some diversions... Like Ukraine vs. Romania, a lot of resistance wars...

Congress election day is in 5 days.

After you fight at least once, you can go to the hospital to raise your Wellness by 10*Quality of the Hospital. For Singapore, it's a Quality 5 Hospital, so you can raise your Wellness by 50. You can only use Hospitals once per day.

Singapore is currently not involved in the war, because the president has expressed a desire to stay neutral which is a mixed blessing - If Singapore helped the US, and Indonesia attacked Singapore in retaliation, Singapore would lose their Q5 Hospital, which would cost several thousand gold to replace, something the government can't afford right now. On the other hand, without the ability to fight, you won't be able to raise your Wellness up with Hospitals.

The best solution I see here is to join PEACE's side and fight against Fortis/EDEN, who looks like they're losing the war, and can't directly threaten Singapore in the event that they triumph.

EDIT: Yeah, Bugi got there before me. :P

The War's a mess right now - Bugi's right that there's a lot of Resistance Wars used as diversions. Can't tell who's going to invest resources fighting where.

EDIT2: I just published my first article in my newspaper:

[url]http://www.erepublik.com/en/newspaper/the-codex-193165/1[/url]

Vote and Subscribe it up so that more people will see it  :lol: !

btw, i'm flamingfingers
although i have no idea how ilord found me already

negative inflation is horrible: wage dropped from 12.5sgd to 6.5sgd and i was at 4.98 skill...

Well, deflation is typically coupled with decreased price of goods which, if the weapon industry is anything to show, has happened at a dramatic rate. On the whole, your buying power should be roughly the same.

Anyway, we're voting Neo organisitation now to take advantage of his 3 votes. We just need 2/3 more on him and then we just need a 10th person into Congress.

A great offer:
BUY IRON TODAY!!! QUALITIY FOUR IRON COSTS ONLY 2.3 SGD, AND I CAN SELL IN UNITS OF 20!!! IF INTERESTED, CONTACT ME EITHER ON AOPS OR EREP!!!

omg its a WAR!!! lol :rotfl:  :rotfl:  :rotfl:

I just had to emergency close down the AoPS guns company, seeing as demand for weapons in Singapore is going to DIE! We will likely just sell off all of our stuff, or possibly export to another country. Sorry if you lost your job due to this.

[quote="lotsofmath"]urg, I hate sites that don't let you join if you are 13 years old...  :mad:[/quote]

lots of people just fake their age...

Is it better to have your money as whatever the country uses or as gold?

I keep gold, because the value of gold isn't going to suddenly change, however, I have no idea what you are supposed to do.

Anybody playing or wishing to play eJahan? It's quite a similar game, and it's new, so we've got a chance to gain control as well.

For anybody that still cares, the AoPS Party will likely be merging with the Green Party, under their banner. I feel we gain a lot of good people this way.

not sure if anyone still cares, but it's cool.

the green party seems like the real deal.  I think it gives the parties some more power over stuff and stuff.

...  this topic has been dead for a while...

bllah the war games are not running today or yesterday

what do you need? its only dicussionless

blah there is an attempted PTO going on

I've came back about 2 months ago, then quit because of the restrictions that were going to be put on the AoPS.  I'm came back again yesterday, and am wondering how many people are still involved.


calculus
integration
function
calculus computations

By considering a reduction formula using
\[I_{n}=\int^{\infty}_{0}{x^{n}e^{-x}}\,dx \]
or otherwise, find $\int^{\infty}_{0}{x^{5}e^{-x}}\,dx$ and $\int^{\infty}_{0}{x^{100}e^{-x}}\,dx$


What is the distance from the origin to the graph of $y=\sqrt{6x-x^{2}-9}$?

How do you find the distance to a parabola?

It isnt a circle either, you forgot to restrict your domain

Infact, this is a point...

Oh yeah... :blush:

So then are you finding the distance to the center of the circle?

well its not really a circle, it's kind of like a circle with a "radius" of 0. It's just a point. So find the distance to that point.

just use the distance formula for distance to a point

Since the center of the circle is $(3, 0)$, the distance from the origin to the center is $3$.

What is the point going to be then?

Edit: nevermind, it was answered


Let $f(n)=n+\lfloor \sqrt{n}\rfloor$. Prove that, for every positive integer $m$, the sequence \[m, f(m), f(f(m)), f(f(f(m))), \cdots\] contains at least one square of an integer.

Induct on $ m$. The case $ m \equal{} 1$ is trivial. Now prove true for $ m$ given true for $ m\minus{}1$.

For all positive integers $ x > 1$, $ f(x)\minus{}f(x\minus{}1)$ equals 2 when $ x$ is a square and 1 otherwise. Suppose that the sequence $ f^{n}(m)$ does not contain a square. It then follows that $ f^{n}(m)\minus{}f^{n}(m\minus{}1) \equal{} 1$ for all $ n$. Now there exists $ N$ such that $ f^{N}(m\minus{}1) \equal{} a^{2}$ for some positive integer $ a$. We then get
\[ f^{N\plus{}1}(m\minus{}1)\equal{}a^{2}\plus{}a,\ f^{N\plus{}2}(m\minus{}1)\equal{}a^{2}\plus{}2a,\]

\[ f^{N\plus{}2}(m)\equal{}a^{2}\plus{}2a\plus{}1\equal{}(a\plus{}1)^{2}\]
a contradiction, as desired.

Largely empirically I found the following facts (see what you can prove and note that all letters denote positive integers, unless stated otherwise):

1) The [i]base units[/i] for the sequence $m, f(m), f(f(m)), ..$ are such that $m = n^2 + n - 1$. That is, if $m$ is not of that form, it occurs somewhere in the sequence $f(m'), f(f(m')), ..$ for some $m'$ smaller than $m$. So we can focus our attention on those base units; set $m = n^2 + n - 1$ for the remainder.
2) The smallest $x$ such that $x^2$ occurs in the sequence is equal to $2n - 1$ and occurs after $2n - 2$ steps; $f^{2n-2}(m) = (2n - 1)^2 = x^2$
3) $y^2$ occurs in the sequence if, and only if, $y = x2^l$ for some $l \ge 0$
4) If $f^k(m) = 4^lx^2$, then $f^{2k+2n+1-l}(m) = 4^{l+1}x^2$


Do you want to :$ f: N\to N$
[hide=hint]
$ f(n\plus{}2)\minus{}f(n)\equal{}f(n\plus{}3)(f(n\plus{}4)\minus{}f(n\plus{}2))$
So $ f(n\plus{}2)\equal{}f(n)$
then
$ f(2n\plus{}1)\equal{}f(1)\equal{}a$
$ f(2n)\equal{}f(2)\equal{}b$ then
$ a\plus{}b\equal{}ab\minus{}1996$
Easy to find out all $ (a,b)$

That solution has a mistake. The function $ f(2n\plus{}1) \equal{} 1$ and $ f(2n) \equal{} 1997(n\minus{}1) \plus{} c$ for any $ c \in \mathbb{N}$ is an answer also.

Result :

$ \left\{\begin{array}{ll}f(x) \equal{} 1\;\text{if x is odd} \\
f(x) \equal{} a \plus{} \left(\dfrac {x}{2} \minus{} 1\right)1997 \;\text{ other wise}\end{array}\right.$

$ \left\{\begin{array}{ll}f(x) \equal{} 2\;\text{if x is odd} \\
f(x) \equal{} 1998\;\text{ other wise}\end{array}\right.$

$ \left\{\begin{array}{ll}f(x) \equal{} 1998\;\text{if x is even} \\
f(x) \equal{} 2\;\text{ other wise}\end{array}\right.$

[quote="quangpbc"]Find all $ f: \mathbb{N}\to\mathbb{N}$ so that :

$ f(n) \plus{} f(n \plus{} 1) \equal{} f(n \plus{} 2)f(n \plus{} 3) \minus{} 1996$[/quote]
We have $ f(n\plus{}1)\plus{}f(n\plus{}2)\equal{}f(n\plus{}3).f(n\plus{}4)\minus{}1996$ with $ n \in N^*$
So $ f(n\plus{}2)\minus{}f(n)\equal{}f(n\plus{}3)[f(n\plus{}4)\minus{}f(n\plus{}2)]$
Then $ f(3)\minus{}f(1)\equal{}f(4).f(6)...f(2n).[f(2n\plus{}1)\minus{}f(2n\minus{}1)]$ with $ n \geq 2$
 $ f(4)\minus{}f(2)\equal{}f(5).f(7)...f(2n\plus{}1)[f(2n\plus{}2)\minus{}f(2n)]$ with $ n \geq 2$
Easy to prove that $ f(1) \leq f(3)$ and  $ f(2) \leq f(4)$
=>$ f(1)\equal{}f(3)$ or $ f(2)\equal{}f(4)$
[b]Case 1:[/b]$ f(1)\equal{}f(3);f(2)\equal{}f(4)$ then $ f(1)\equal{}f(2n\minus{}1)$ and $ f(2)\equal{}f(2n)$
Note that $ f(1)\plus{}f(2)\equal{}f(1).f(2)\minus{}1996$ =>$ [f(1)\minus{}1][f(2)\minus{}1]\equal{}1997$
=>$ f(1)\equal{}2;f(2)\equal{}1998$ or $ f(1)\equal{}1998;f(2)\equal{}2$
so $ f(n)\equal{}2$ if $ n$ is odd
and $ f(n)\equal{}1998$ if $ n$ is  even
or $ f(n)\equal{}2$ if $ n$ is even and $ f(n)\equal{}1998$ if $ n$ is odd
[b]Case 2:[/b]$ f(1)\equal{}f(3);f(2)<f(4)$
then $ f(1)\equal{}f(2n\minus{}1)$=>$ f(4)\minus{}f(2)\equal{}(f(1))^{n\minus{}1}[f(2n\plus{}2)\minus{}f(2n)];f(2n)<f(2n\plus{}2)$
So $ f(1)\equal{}1;f(2n\minus{}1)\equal{}f(2n\plus{}1)$=>$ f(4)\minus{}f(2)\equal{}1997$
Note that $ f(4)\minus{}f(2)\equal{}f(2n\plus{}2)\minus{}f(2n)$
So $ f(n)\equal{}1$ if $ n$ is odd and $ f(n)\equal{}a\plus{}1997(\frac{n}{2}\minus{}1)$ if $ n$ is even($ a \in N^*$)
[b]Case 3[/b]Similar,we have
$ f(n)\equal{}1$ if $ n$ is odd and $ f(n)\equal{}b\plus{}1997(\frac{n\minus{}1}{2})$ if $ n$ is even($ b \in N^*$)

[b]Approach by[/b] [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=29721]Erken:[/url]

Once again instead of considering some function we shall work with sequence,name it $ \{a_i\}_{i \equal{} 1}^{\infty}\in\mathbb{N}$
Let's try to change the given formula into something more explicit:
$ a_n \plus{} a_{n \plus{} 1} \equal{} a_{n \plus{} 2}\cdot a_{n \plus{} 3} \minus{} 1996$ and $ a_{n \plus{} 1} \plus{} a_{n \plus{} 2} \equal{} a_{n \plus{} 3}\cdot a_{n \plus{} 4} \minus{} 1996$.
Let's substract the first equation from the second one:
$ a_{n \plus{} 2} \minus{} a_{n} \equal{} a_{n \plus{} 3}(a_{n \plus{} 4} \minus{} a_{n \plus{} 2})$
Now let's see what happened if instead of $ n$ we will substitute $ 1$,after that $ 2$ and so on:
\[ a_3 \minus{} a_1 \equal{} a_4\cdot a_6\dots a_{2k}(a_{2k \plus{} 1} \minus{} a_{2k \minus{} 1})
\]

\[ a_4 \minus{} a_2 \equal{} a_5\cdot a_7\dots a_{2k \plus{} 1}(a_{2k \plus{} 2} \minus{} a_{2k})
\]
Suppose that neither $ a_3 \minus{} a_1$,nor $ a_4 \minus{} a_2$ are equal to zero.
Then obviously starting from some $ m$,we have that $ 1 \equal{} a_m \equal{} a_{m \plus{} 1} \equal{} a_{m \plus{} 2} \equal{} \dots$,because otherwise diffrence $ |a_{n \plus{} 2} \minus{} a_{n}|$ would infinitely decreased,as $ a_3 \minus{} a_1$ and $ a_4 \minus{} a_2$ are constants.
But if $ a_k \equal{} 1$ for all $ k\geq m$,then by our initial formula:
$ 2 \equal{} 1 \minus{} 1996$,contradiction.
Therefore,either $ a_3 \minus{} a_1 \equal{} 0$,or $ a_4 \minus{} a_2 \equal{} 0$.
It is not hard to check both cases and also the case when both of them are zero's.The last case gives us two answers,as there we need solve some easy equation,$ a \plus{} b \equal{} ab \minus{} 1996$.
So the finally we will obtain four answers:
[b]1.[/b]$ a_n \equal{} 1$,if $ n$ is odd and $ a_{n} \equal{} c \plus{} 1997\left(\frac {n}{2} \minus{} 1\right)$ if $ n$ is some even number.

[b]2.[/b]$ a_n \equal{} 1$,if $ n$ is even and $ a_{n} \equal{} c \plus{} 1997\left(\frac {n \plus{} 1}{2} \minus{} 1\right)$,where $ n$ is some odd number.

[b]3.[/b]$ a_n \equal{} 2$ if $ n$ is even and $ a_n \equal{} 1998$ if $ n$ is odd.

[b]4.[/b]$ a_n \equal{} 1998$ is $ n$ is even and $ a_n \equal{} 2$ if $ n$ is odd.


Ψάχνω εδώ και 4 μέρες στο ίντερνετ μία απόδειξη για το θεώρημα που λέει ότι μία τετραγωνική μορφή είναι θετικά ορισμένη αν και μόνο αν οι διαγώνιες υποορίζουσές του είναι θετικές και δεν βρίσκω κάτι..

Αν κάποιος μπορεί να βοηθήσει....

Καλή επιτυχία και από εμένα στα παιδιά που πάνε στο Βιετναμ!!!!!

Αν δε κάνω λάθος είναι γνωστό σαν Θεώρημα Sylvester.Λογικά όλο και κάτι θα υπάρχει στο google.Ελπίζω να βοήθησα...


trigonometry
geometry
parallelogram
analytic geometry
national olympiad

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=128303
A sequence $(a_{n})$ of positive integers is given by $a_{n}=[n+\sqrt{n}+\frac{1}{2}]$. Find all of positive integers which belong to the sequence.

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=128302
Find all $x,y,z\in\mathbb{N}_{0}$ such that $7^{x}+1=3^{y}+5^{z}$.

5. http://www.mathlinks.ro/Forum/viewtopic.php?p=727710#727710
Assume $A=\{a_{1},a_{2},...,a_{12}\}$ is a set of positive integers such that for each positive integer $n \leq 2500$  there is a subset $S$ of $A$ whose sum of elements is $n$. If $a_{1}<a_{2}<...<a_{12}$ , what is the smallest possible value of $a_{1}$?

6. http://www.mathlinks.ro/Forum/viewtopic.php?p=727720#727720
Let $m$ be equal to $1$ or $2$ and $n<10799$ be a positive integer. Determine all such $n$ for which $\sum_{k=1}^{n}\frac{1}{\sin{k}\sin{(k+1)}}=m\frac{\sin{n}}{\sin^{2}{1}}$.

be continous :D

2. http://www.mathlinks.ro/Forum/viewtopic.php?p=728301#728301
Let $E$ and $F$ are distinct points on the diagonal $AC$ of a parallelogram $ABCD$ . Prove that , if there exists a cricle through $E,F$ tangent to rays $BA,BC$ then there also exists a cricle through $E,F$ tangent to rays $DA,DC$.

4. http://www.mathlinks.ro/Forum/viewtopic.php?p=728362#728362
In the Cartesian plane, let $C$ be a unit circle with center at origin $O$. For any point $Q$ in the plane distinct from $O$, define $Q'$ to be the intersection of the ray $OQ$ and the circle $C$. Prove that for any $P\in C$ and any $k\in\mathbb{N}$ there exists a lattice point $Q(x,y)$ with $|x|=k$ or $|y|=k$ such that $PQ'<\frac{1}{2k}$.


let $ {f: Z ->Z}$ satisfying the conditions :
i/ f(0)=1,
ii/ f(f(x)) = x +4f(x) fr all ${ x  Z }$
 $ { find all natural numbers n \geq 1 such that fn(0) is divisible by 20^{11}^{2005}}

let $ {f: Z ->Z}$ satisfying the conditions :
i/ $f(0)=1$,
ii/ $f(f(x)) = x +4f(x)$ for all ${ x \in Z }$

Find all natural numbers $n \geq 1$ such that $f_n(0)$ is divisible by $20^{11^{2005}}$.

i want to  $n$
 i can't know
what  can  n take  value   ?????
                    do you help me?

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=298250&highlight=#298250]this[/url]. The main idea is there.


function
real analysis
Euler
complex analysis
complex analysis theorems

Does anyone know a proof of:

$ \sum_{n\equal{}1}^\infty \frac{\sigma(n)}{n^s} \equal{} \zeta(s) \cdot \zeta(s\minus{}1)$

?

Well, in the book An introduction to the theory of numbers by G. H. Hardy, E. M. Wright (a limited version of which is given [url=http://books.google.ba/books?id=3hTeH5VUheAC&dq=An+Introduction+to+the+Theory+of+Numbers+hardy&printsec=frontcover&source=bn&hl=bs&ei=QTcBSuriBpLMjAfZ-Oj3Bg&sa=X&oi=book_result&ct=result&resnum=4#PPA250,M1]here[/url]), theorem 290 (page 250) is what you're looking for. As you can see, the result is derived from other theorems and results in this book, so it is kind of complicated, and I'm not aware is there an easier way (btw, if some of the theorems the authors refer to aren't in the google preview, list them here and I can type them from my hardcopy).

Essentially, this is $ 1*\mathrm{Id} \equal{} \sigma$ in terms of Dirichlet convolutions.

Yes the Dirichlet convolution is nice that way. A popular identity;

$ \left( \sum_{n\equal{}1}^\infty \frac{f(n)}{n^s} \right) \left( \sum_{n\equal{}1}^\infty \frac{g(n)}{n^s} \right) \equal{} \sum_{n\equal{}1}^\infty \frac{[f*g](n)}{n^s}$

Do you know the definition of the dirichlet convolution?

Hi Wael!  :)

As ZetaX says, essentially the result is due to the fact that $ \zeta(s \minus{} 1) \equal{} \sum_{n \equal{} 1}^{\infty} \frac {n}{n^s}$, so multiplication by $ \zeta(s)$ produces the Dirichlet series of $ \sum_{d | n} d \equal{} \sigma(n)$.  This is due to the standard definition of the [url=http://en.wikipedia.org/wiki/Dirichlet_convolution]Dirichlet convolution[/url] and the identity in smile =)'s post.

Another way of thinking about this identity is to note that multiplicative functions are determined by their values at prime powers, which is another way of saying that their Dirichlet series are determined by their Euler products.  So we have the identities

$ \zeta(s) \equal{} \prod_{p \in \mathbb{P}} \frac {1}{1 \minus{} p^{ \minus{} s}}$
$ \zeta(s \minus{} 1) \equal{} \prod_{p \in \mathbb{P}} \frac {1}{1 \minus{} p^{ \minus{} s \plus{} 1}}$
$ \zeta(s) \zeta(s \minus{} 1) \equal{} \prod_{p \in \mathbb{P}} \frac {1}{(1 \minus{} p^{ \minus{} s})(1 \minus{} p^{ \minus{} s \plus{} 1})}.$

Letting $ x \equal{} p^{ \minus{} s}$, we have

$ \frac {1}{(1 \minus{} x)(1 \minus{} px)} \equal{} \left( \sum_{k \equal{} 0}^{\infty} x^k \right) \left( \sum_{k \equal{} 0}^{\infty} p^k x^k \right) \equal{} \sum_{k \equal{} 0}^{\infty} \left( 1 \plus{} p \plus{} ... \plus{} p^k \right) x^k \equal{} \sum_{k \equal{} 0}^{\infty} \sigma(p^k) p^{ \minus{} sk}$

from which it follows that $ \zeta(s) \zeta(s \minus{} 1) \equal{} \sum_{n \equal{} 1}^{\infty} \frac {\sigma(n)}{n^s}$ by another well-known property of Euler products.  We can also obtain the above identity by partial fraction decomposition.

I haven't looked at hsiljak's post, but I will note that the result is harder if you want to prove [b]convergence[/b]:  the proofs above are formal.


conics
ellipse
geometry
3D geometry
sphere
geometric transformation
rotation

Given is an ellipse $ e$ in the plane. Find the locus of all points $ P$ in space such that the cone of apex $ P$ and directrix $ e$ is a right circular cone.

E, F are the ellipse foci, EF meets $ \epsilon$ at A, B, AB is the ellipse major axis. The cone with directrix $ \epsilon$ is right $ \Longrightarrow \exists$ two Dandelin spheres $ \mathcal S_e,\ \mathcal S_f$ tangent to the cone, centered at $ I_e, I_f$ on its rotation axis ($ P,\ I_e,\ I_f$ are collinear), and touching the plane $ (\epsilon)$ at E, F. $ EI_e,\ FI_f \perp (\epsilon),$ $ PI_eI_f$ is in a plane $ \nu \perp (\epsilon),$ such that $ AB \equiv EF \equiv \nu \cap (\epsilon).$ The plane $ \nu$ cuts the spheres $ \mathcal S_e,\ \mathcal S_f$ in the incircle and P-excircle $ (I_e),\ (I_f)$ of the $ \triangle PAB.$ The incircle $ (I_e)$ touches PA, PB at U, V.

$ PB \minus{} PA \equal{} (PV \plus{} VB) \minus{} (PU \plus{} UA) \equal{} EB \minus{} EA \equal{}$
$ \equal{} EF \plus{} FB \minus{} EA \equal{} EF \equal{} \text{const}$

$ \Longrightarrow P$ is on a hyperbola $ h \in \nu$ with foci A, B and major axis EF.


How many different real roots does this polynomial have?
$x^{11} + x^{10} + x^9 + ...+ x + 1$

In tarquin's equation, $x^{12}-1$ has two real roots, $x=1$ and $x=-1$, but the first is extraneous, thrown in the mix after multiplying the expression by $(x-1)$. So we know that $x=-1$ is the only real root of $x^{11}+x^{10}+\ldots+x+1$.


vector
linear algebra
linear algebra unsolved

What is the quickest proof of the fact that $uv-vu=id$ has no solution, where $u,v$ are continuous endomorphisms of a normed vector space ?

Thank you Sam. Eugene made a typo in its solution: it should be $u^{n}v-vu^{n}=nu^{n-1}$.

Thank you, Julien, for noticing. I've edited it.


calculus
integration
logarithms
derivative
calculus computations

Find the minimum value of $\int_1^e \left|\ln x-\frac{a}{x}\right|dx\ (0\leq a\leq e)$

am i going to have to use partial derivatives to do this?

No, your answer is incorrect, adidasty. :(

$\int_1^e \left|\ln x-\frac{a}{x}\right|dx$

let the solution to xlnx = a be p

when x<p, a/x > lnx

therefore the integral can be rewritten as

$\int_1^p \frac{a}{x} - \ln x dx  + \int_p^e \ln x-\frac{a}{x} dx $

we get
2alnp - 2plnp + p - 1 - a

substitute a with plnp

2p(lnp)^2 - 3plnp + 2p - 1

take the derivative

2(lnp)^2 + lnp - 1 = 0
lnp=1/2 is when minimun value occur
p = sqrte

our desired minimun value is 2*sqrte*1/4 - 3sqrte*1/2 + 2sqrte - 1
=sqrte /2 - 3sqrte /2 + 2sqrte -1
= sqrte -1

$\sqrt{e} - 1$


function
integration
real analysis
Functional Analysis
real analysis unsolved

It is known that $ L^\infty \equal{} (L^1) '$, thus $ (L^\infty)' \equal{} (L^1)) ''$. Your question comes from the fact that $ L^1$ is not reflexible. Actually, $ L^1 \subset (L^1)''$

A common approach is to extend the evaluation functional $ f\mapsto f(a)$ from $ C(\mathbb R)$ to $ L^{\infty}(\mathbb R)$ (using the Hahn-Banach theorem), and then prove that any such extension does not correspond to an element of $ L^1(\mathbb R)$.

An other question for Doe: what is $ (L^\infty)'$?

@jmerry: Can you explain a bit further? I'm don't know about Hahn-Banach theorem.

@bookworm: $ (L^{\infty})'$ denotes a dual space of $ L^{\infty}$ http://en.wikipedia.org/wiki/Dual_space , although I'm pretty sure that's not what your asking me  :huh: .

I know these question's seem a bit silly, but I'm doing on my own intro to harmonic analysis, and this problem was left as an exercise, in my opinion a bit hard.

$ f\mapsto f(a)$ is a continuous linear functional on $ C(\mathbb R)$, which is a closed subspace of $ L^{\infty}$. The [url=http://en.wikipedia.org/wiki/Hahn-Banach_theorem]Hahn-Banach theorem[/url]  implies that there is a continuous functional $ \phi\in (L^{\infty}(\mathbb R))^*$ such that $ \phi(f)\equal{}f(a)$ for all $ f\in C(\mathbb R)$. It remains to prove that there is no function $ g\in L^1(\mathbb R)$ such that $ \phi(f)\equal{}\int_{\mathbb R}f\bar g\,dx$. For this, consider a nonzero element $ g\in L^1(\mathbb R)$, choose a Lebesgue point $ x_0\ne a$ such that $ g(x_0)\ne 0$, and let $ f_n(x)\equal{}\max(0,1\minus{}n|x\minus{}x_0|)$. Then $ \int_{\mathbb R}f_n\bar g\,dx\to \bar g(x_0)$, although $ f_n(a)\equal{}0$ for large $ n$.

[quote="Mr.Doe"]@bookworm: $ (L^{\infty})'$ denotes a dual space of $ L^{\infty}$ http://en.wikipedia.org/wiki/Dual_space , although I'm pretty sure that's not what your asking me  :huh: .quote]

No, I do not ask the definition of $ (L^{\infty})'$.


superior algebra
superior algebra solved

Prove that Z_n has an even number of generators if n > 2. What can we conclude about phi(n)

If $k$ is a generator, $-k$ is also one, and $k \neq-k$ since $2k \neq 0$ for $n > 2$. Whence $\phi(n)$ is even for $n>2$.

Please don't post new threads in "Solved Problems". The forum is for threads discussed elsewhere and then moved here.

[mod.: right, moved]


$ \mathop{\lim}\limits_{x\rightarrow4^{ - }}\displaystyle{\frac {{x^2 - 4}}{{x - 4}}} = \mathop{\lim}\limits_{x\rightarrow4^{ - }}\left({x^2 - 4}\right)\displaystyle\cdot\mathop{\lim}\limits_{x\rightarrow4^{ - }}{\frac {1}{{x - 4}}} = (16 - 4)\cdot( - \infty) = - \infty$.
$ \mathop{\lim}\limits_{x\rightarrow4^{ + }}\displaystyle{\frac {{x^2 - 4}}{{x - 4}}} = \mathop{\lim}\limits_{x\rightarrow4^{ + }}\left({x^2 - 4}\right)\displaystyle\cdot\mathop{\lim}\limits_{x\rightarrow4^{ + }}{\frac {1}{{x - 4}}} = (16 - 4)\cdot( + \infty) = + \infty$.
So the $ \mathop{\lim}\limits_{x\rightarrow4}\displaystyle{\frac {{x^2 - 4}}{{x - 4}}}$ does not exists.
But I saspect that maybe you mean ${ \mathop{\lim}\limits_{x\rightarrow4}}\displaystyle{\frac {{x^2 - 4^2}}{{x - 4}}}$ and that is another limit equal to $ 8$.

thank you dude, i appreciate it, no what i wrote is correct. thanks


[quote="mehdi cherif"]The equation  $ \iff$ :    $ (x \plus{} 1)^{2002} \equal{} (x \minus{} 1)^{2002}$
wish  $ S \equal{} \Phi$[/quote]

Wrong.
Assuming $ x\in \mathbb R$
$ (x \plus{} 1)^{2002} \equal{} (x \minus{} 2)^{2002}\implies (x \plus{} 1) \equal{} \minus{} (x \minus{} 2)\iff x \equal{} \frac {1}{2}$

yes it's not a wrong but a wrong of innatention  :P
[color=red]i've edited my post[/color]


Class Instructors:

Would it be possible to post the weekly problems as a single file (e.g. a PDF file) in addition to posting them on the class discussion board?  I, and probably many others in the class, end up copying and pasting all of them anyway to form a single weekly problem set.   Since I do this for transcripts and any other course material, I find I'm spending a lot of time copying and pasting, which leaves less time for problem solving!  This is complicated by software issues, such as TeX not appearing, etc., which only costs more time and frustration. These software issues can be overcome, of course, but I was hoping my suggestion might reduce some of these issues for students.

Thanks,

Thomas

In many of our classes there are students who are making the compilations you are requesting - post in your class and perhaps you can find a couple others to share the workload with.

Yeah I too spend a lot of time with copying the problems.  It doesn't bother me too much since I consider it as $\text{\TeX}$ practice.  If anyone who is in one of my classes would like to have a copy of my file of board problems I d be happy to send it to them.  I put them all in the same document and print that whole thing at the end.  I might just start doing it week by week though because that would seem like a better organizational method.


trigonometry FTW algebra polynomial inequalities quadratics

If $ x \plus{} \frac{1}{x} \equal{} 4$, then what is the value of $ x^3 \plus{} \frac{1}{x^3}$?

Spinning off what kunny just said, I will add that the solution lies in what he first pointed out:

a^3 + b^3 = (a + b)^3 -3ab(a+b)

x^3 + (1/x^3) = (x + 1/x)^3 - 3*x*1/x*(x + 1/x)

x^3 + (1/x^3) = (4)^3 - 3*1*(x + 1/x)

x^3 + (1/x^3) = 64 - 3*(4)

x^3 + (1/x^3) = 52

If you are familiar with trigonometry and De Moivre's Theorem, then you can set $ x\equal{} \cos(\theta) \plus{} i
\sin(\theta)$.  In the original equation, we can now see, by de Moivre's, that
 $ 4 \equal{} x^1 \plus{} x^{\minus{}1} \equal{} \cos(\theta)\plus{}i \sin(\theta) \plus{} \cos (\minus{}\theta) \plus{} i \sin(\minus{}\theta)$
which implies that
$ 2 \cos(\theta) \equal{} 4 \Rightarrow \cos(\theta) \equal{} 2$

On the other hand, $ x^3 \plus{} x^{\minus{}3}$ can be written as:
$ \cos(3\theta) \plus{} i \sin(3\theta) \plus{} \cos(\minus{}3\theta) \plus{} i \sin(\minus{}3 \theta) \equal{} 2 \cos(3 \theta)$

And by the triple angle formula for cosine
$ 2 \cos(3 \theta)\equal{} 2( 4 \cos^3(\theta) \minus{} 3 \cos( \theta )) \equal{} 2( 4(8) \minus{} 3(2) ) \equal{} \boxed{52}$.

In the time-intensive pressure of FTW, you would probably simply wish to memorize a few [url=http://en.wikipedia.org/wiki/Chebyshev_polynomial#Examples]Cheybshev Polynomials[/url]  Since the third Chebyshev polynomial is $ P_3 (a) \equal{} a^3 \minus{} 3a$, we set $ a\equal{}4$ to get $ P_3(4) \equal{} 4^3 \minus{} 3(4) \equal{} \boxed{52}$.

This one doesn't require memorized formulas except binomial cubing:

If x + 1/x = 4, then
(x + 1/x)^3
  = 64
  = x^3 + 3x^2/x + 3x/x^2 + 1/x^3
  = x^3 + 1/x^3 + 3(x + 1/x)
x + 1/x = 4, so we can substitute this in to get
64 = x^3 + 1/x^3 + 3*4
52 = x^3 + 1/x^3

Hmm, or we could do $(x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2} = 16$, so $x^2 + \frac{1}{x^2} = 14$.  Now we do $(x^2 + \frac{1}{x^2})(x + \frac{1}{x}) = x^3 + (x + \frac{1}{x}) + \frac{1}{x^3}$.  Substituting the right values in we have $4 \cdot 14 = x^3 + \frac{1}{x^3} + 4$ so our answer is $4 \cdot 14 - 4 = \boxed{52}$.

[quote="kunny"]$ a^3\plus{}b^3\equal{}(a\plus{}b)^3\minus{}3ab(a\plus{}b)$
$ a\plus{}b\plus{}c\equal{}0\Longrightarrow a^3\plus{}b^3\plus{}c^3\equal{}3abc$[/quote]

Using the second fact, $x+\frac{1}{x}+(-4)=0\Longrightarrow x^3+\frac{1}{x^3}-64=-12$, yielding the answer is $52$.

I think the easiest method is pretty much what catlover and professordad said. It's takes the least amount of time to do (in my opinion) and relatively easy to do, because cubing $x+\frac{1}{x}=4$ will give you $x^3+\frac{1}{x^3} + 3\left(x+\frac{1}{x}\right)=64$, so substituting yields $x^3+\frac{1}{x^3}+3(4)=64$. Subtract $12$ from both sides, resulting in our answer, $\boxed{52}$.

Hmmmmmm or you could solve for x.

x+1/x=4
x-4=-1/x
x^2-4x+1=0

Then quadratic formula  and plug in


calculus
integration
algebra
polynomial
derivative
calculus computations

Let $k$ be positive integers and $f(x)$ be the polynomial with integers coefficients. Find $k$ and $f(x)$ such that $2\int_{1}^{x}f(t)dt+x^{k}=xf(x),\ x\geq 1.$

I like this more than I thought I would:
[hide]Let $f(x) = a_{n}x^{n}+\ldots+a_{1}x+a_{0}$, $a_{n}\neq 0$.  Then $xf(x) = a_{n}x^{n+1}+\ldots+a_{1}x^{2}+a_{0}x$.

On the other hand,  $2\int_{1}^{x}f(t)dt+x^{k}=x^{k}+\frac{2a_{n}}{n+1}x^{n+1}+\ldots+\frac{2a_{1}}{2}x^{2}+2a_{0}x-c$ (where we know what $c$ is in terms of the $a_{i}$).

Comparing coefficients, we have that $c = 0$, $1 \leq k \leq n+1$, $\frac{2a_{k-1}}{k}+1 = a_{k-1}$ and for all $i \neq k$ $\frac{2a_{i}}{i+1}= a_{i}$.

From these, we have $a_{k-1}= \frac{k}{k-2}$ (and in particular $k \neq 2$) and otherwise $a_{i}= 0$ for $i \neq k-1, 1$.  Since $a_{n}\neq 0$ we have either $n+1 = k$ or $n =1$.

In the case $n = 1$ we have $2\left(\frac{a_{1}}{2}x^{2}+a_{0}x-\frac{a_{1}}{2}-a_{0}\right)+x^{k}= a_{1}x^{2}+a_{0}x$ and so $k = 1, a_{0}=-1, a_{1}= 2$.

In the case $n+1 = k$ we've done everything except set $c = 0$; it so happens that $c = \frac{1}{n-1}+\frac{a_{1}}{2}$ so $a_{1}=-\frac{2}{n-1}$ so $f(x) = \frac{n+1}{n-1}x^{n}-\frac{2}{n-1}x$ and $k = n+1$.[/hide]

Another possible approach - I haven't worked out the details though - would be note that when we differentiate both sides of the given equation w.r.t $x$, we obtain a first order linear differential equation whose solution can be obtained using the method of Integrating Factor.

After differentiation, we obtain $f'(x)-\frac1{x}f(x) = k x^{k-2}$ . I guess we could then proceed from here.

[quote="FieryHydra"]

After differentiation, we obtain $f'(x)-\frac1{x}f(x) = k x^{k-2}$ . I guess we could then proceed from here.[/quote]

That's right. :)


Well, there's a pattern so I assume the answer to be 1234?

It's not 1234.

Subtract the first equation from the second you get $ 3x_1 \plus{} 5x_2 \plus{} 7x_3 \plus{} ... \plus{} 13x_6 \plus{} 15x_7 \equal{} 11$

Now subtract the second from the third.  This yields $ 5x_1 \plus{} 7x_2 \plus{} 9x_3 \plus{} ... \plus{} 15x_6 \plus{} 17x_7 \equal{} 111$

Subtracting the equation that equals 11 from the one that equals 111 we get $ 2x_1 \plus{} 2x_2 \plus{} 2x_3 \plus{} ... \plus{} 2x_6 \plus{} 2x_7 \equal{} 100 \implies x_1 \plus{} x_2 \plus{} x_3 \plus{} ... \plus{} x_6 \plus{} x_7 \equal{} 50$

If $ 16x_{1} \plus{} 25x_{2} \plus{} 36x_{3} \plus{} 49x_{4} \plus{} 64x_{5} \plus{} 81x_{6} \plus{} 100x_{7}$ equaled 1234 as you said then we could subtract $ 9x_{1} \plus{} 16x_{2} \plus{} 25x_{3} \plus{} 36x_{4} \plus{} 49x_{5} \plus{} 64x_{6} \plus{} 81x_{7} \equal{} 123$ from it to get $ 7x_{1} \plus{} 9x_{2} \plus{} 11x_{3} \plus{} ... \plus{} 17x_6 \plus{} 19x_7 \equal{} 1111$.

 After that we would subtract $ 5x_1 \plus{} 7x_2 \plus{} 9x_3 \plus{} ... \plus{} 15x_6 \plus{} 17x_7 \equal{} 111$ from  $ 7x_{1} \plus{} 9x_{2} \plus{} 11x_{3} \plus{} ... \plus{} 17x_6 \plus{} 19x_7 \equal{} 1111$.  This would result in $ 2x_1 \plus{} 2x_2 \plus{} 2x_3 \plus{} ... \plus{} 2x_6 \plus{} 2x_7 \equal{} 1000 \implies x_1 \plus{} x_2 \plus{} x_3 \plus{} ... \plus{} x_6 \plus{} x_7 \equal{} 500$ but we already know $ x_1 \plus{} x_2 \plus{} x_3 \plus{} ... \plus{} x_6 \plus{} x_7 \equal{} 50$

My anwser is 334.
I am to lazy to post a solution so i will give you an outline
just let the expression equal an unknown say $ {a}$ then make it equation 4 and repeat the procedure of truffles.
I dare anybody to tell me what $ {x_1}$ to$ {x_7}$ is?

Gee thanks
okay what is x1 to x7?
so my mom was wrong

Woah you can't solve for all those values until you make seven equations in the way 21 7 15 made $ 16x_{1} \plus{} 25x_{2} \plus{} 36x_{3} \plus{} 49x_{4} \plus{} 64x_{5} \plus{} 81x_{6} \plus{} 100x_{7}$ and then you'd still have to solve for each individual variable.

See also the standard approach by finite differences, [url=http://www.artofproblemsolving.com/Wiki/index.php/1989_AIME_Problems/Problem_8]second solution here[/url]

There are no set values for $ x_1, \ldots, x_7$; in fact you can fix four of them to be arbitrary values (after that you get a 3 equation 3 unknown system).

Err is it just me or are those sums written wrong in that solution?  No I'm pretty sure it's not just me.

Why aren't there any set values for $ x_1,...,x_7$?

Because you need 7 equations for that. A general rule is that to determine the values of n variables, you need n equations to solve. (sometimes more when you get into squares/other powers/complex numbers/etc.)

Couldn't you just make seven equations?

Oh well is it that you wouldn't get anywhere making more equations like the three given in the problem because they're all similar?

Yeah. If you made new equations with the ones you have and tried to solve with them, you'd eventually end up with something like 2=2, x=x, or an equation you already had. It's kinda like if you only have x+y=5. You can double it to get 2x+2y=10, but minusing the first from it wont get you anywhere.

By set values, I mean that there is not only one specific set of solutions $ (x_1,x_2, \ldots, x_7)$, but there are actually infinitely many (which further, form a space of solutions with dimension $ 4$). You could pick any four of the $ x_i$s, say take $ \{x_4,x_5,x_6,x_7\}$, and assign to them [i]any[/i] values you like, and for each assignment there will exist a specific solution by solving for $ \{x_1,x_2,x_3\}$. [as an application to this problem, you can set $ x_4\equal{}x_5\equal{}x_6\equal{}x_7\equal{}0$, and immediately treat the problem as a standard 3 by 3 system - though given the coefficients it will be messy]

Indeed, these equations are "similar" - if we treat the problem as a system of four equations, due to finite differences, we can say that the four equations are "linearly dependent" [I'm abusing terminology a bit here]. This subject is a topic of linear algebra.

[hide="Edited solution to Truffles (bit unclear)"]
Subtracting the first from the second, you get $ 3x_{1} \plus{} 5x_{2} \plus{} 7x_{3} \plus{} ... \plus{} 13x_{6} \plus{} 15x_{7} \equal{} 11$.
Subtracting the second from the third, you get $ 5x_{1} \plus{} 7x_{2} \plus{} 9x_{3} \plus{} ... \plus{} 15x_{6} \plus{} 17x_{7} \equal{} 111$.

Noticing that both results have consecutive terms with constant terms in an arithmetic progression (common difference = 2), subtracting should give us something nice. Precisely, this: $ 2x_{1} \plus{} 2x_{2} \plus{} 2x_{3} \plus{} ... \plus{} 2x_{6} \plus{} 2x_{7} \equal{} 100$. (you can go ahead and divide by 2, but there is no need because of later)

Now subtracting the third equation from what we want, we get something similar to the first two results:
$ 7x_{1} \plus{} 9x_{2} \plus{} 11x_{3} \plus{} ... \plus{} 17x_{6} \plus{} 19x_{7} \equal{} n$
Retracing the steps to how we found the sum of all 7 terms, we subtract the second result from the third. This gives us an old equation! $ 2x_{1} \plus{} 2x_{2} \plus{} 2x_{3} \plus{} ... \plus{} 2x_{6} \plus{} 2x_{7} \equal{} 100 \equal{} n \minus{} 111$

Thus, $ n\equal{}211$, and what we want to find is the sum of $ n$ and the value of the third equation. This is $ 211 \plus{} 123 \equal{} \boxed{334}$.[/hide]

Thanks to Truffles for letting me copy some of his LaTeX. :P

Yeah thanks azjp and Ihatepie.

@dragon96: Lol I'm a girl and you're welcome?


Can we apply Ampere's Rule to find the magnetic field at the centre of a circular loop carrying a current. Why? or Why Not?

No for the magnitude (since Ameper's Rule says absolutely nothing about it :)) but yes for the direction. If you take a very short part of the circular loop, it can be approximated by a straight line segment carrying the same current as the whole loop. Apply the Ampere's Rule to this segment and find the direction of the magnetic field in the center of the loop due to the current flowing through it. If you divide the whole loop into small parts and approximate all of them by line segments, you may use the rule on every one of them and you'll get the same result for each segment, therefore the sum of the magnetic fields created by the small segments will be also in the same direction. (It's like integrating Ampere's Rule over the loop :).)


inequalities
induction
trigonometry
inequalities unsolved

Show for all $x\in(0,\pi)$ and $n\in N$ that $|sin(nx)| \le nsinx$.

It's not hard to prove (by induction) that :
$\forall n \in \mathbb{N}, \ \ \forall x \in \mathbb{R}, \\ |\sin(nx)| \leq n |\sin x|.$

So, it's clear for $n=0.$
Now, let $n \geq 0.$ Let's assume that $|\sin(nx)| \leq n |\sin x|.$
Then,
$|\sin(n+1)x| \leq |\sin nx|\cdot |\cos x|+|\cos nx| \cdot |\sin x|$
$\leq |\sin nx|+|\sin x|$
$\leq n |\sin x|+|\sin x|$
$= (n+1) |\sin x|.$

So we're done.

Actually, induction. It is a lemma in Complex Numbers from A to Z


Finjd all pairs of primes $ p$ and $ q$ such that $ p\minus{}q\equal{}3$.

First note that $ p$ and $ q$ must be of different parity to get a difference of an odd number. The only even prime number is 2. So if $ p \equal{} 2$ then $ q \equal{} \minus{} 1$, not valid. If $ q \equal{} 2$ then $ p \equal{} 5$, which is our only solution.


$ p^2\plus{}7pq\plus{}q^2$ is a perfect square.
Solve this for $ p$ and $ q$ in prime$ \mathbb{N}$

[hide="Casework"]Let the square be $ n^2$, without loss of generality assuming that $ n > 0$. $ (p\plus{}q)^2 \plus{} 5pq \equal{} n^2 \Longrightarrow 5pq \equal{} (n \minus{} p \minus{} q)(n \plus{} p \plus{} q)$.

So $ n \plus{} p \plus{} q \equal{} \{1,p,q,5,5p,5q,pq,5pq\}$, but from $ n > 0$ only $ 5p, 5q, pq, 5pq$ are possible.

If $ n \plus{} p \plus{} q \equal{} 5p, n \minus{} p \minus{} q \equal{} q$, then we find $ n \equal{} 4p \minus{} q \equal{} p \plus{} 2q \Longrightarrow 3p \equal{} 3q$, or $ p\equal{}q$ (same thing happens for $ 5q$).

If  $ n \plus{} p \plus{} q \equal{} pq, n \minus{} p \minus{} q \equal{} 5$, then the difference of those two equations yields $ pq \minus{} 2(p\plus{}q) \equal{} 5 \Longrightarrow (p\minus{}2)(q\minus{}2) \equal{} 9$, so either $ p \equal{} q \equal{} 5$ (which was covered) or $ \{p,q\} \equal{} \{3,11\}$.

If $ n \plus{} p \plus{} q \equal{} 5pq, n \minus{} p \minus{} q \equal{} 1$, then the difference of those two equations yields $ 5pq \minus{} 2(p\plus{}q) \equal{} 1$. WLOG $ p > q \ge 2$, then $ (5q \minus{} 2)p \minus{} 2q \ge 8p \minus{} 2q > 1$, contradiction.

So our solution set is $ \boxed{p\equal{}q, \{p,q\} \equal{} \{3,11\}}$.[/hide]


How many digits does $ 12^{60}$ have and what's the leftmost digit number of $ 12^{60}$?

You may use the values of $ \log_{10} 2 \equal{} 0.3010,\ \log_{10} 3 \equal{} 0.4771$.

[hide="Solution"]
$ 12^n$'s last digit repeats in cycle 2,4,8,6,...

Thus, $ 12^{60}$'s last digit is 6.

Also, $ 12^{60}$ has $ \lfloor\log_{10} 12^{60} \plus{} 1\rfloor \equal{} 65$ digits.

Answer: Last digit-6, 65 digits long[/hide]

EDIT: Oops, leftmost digit...  :oops:

I think writing it as $ (10 \plus{} 2)^{60}$ and using binomial expansion may help.
We are looking for the leftmost digit, not rightmost, and I don't think you can use a calculator...
If you can, that would work.

It asks for the leftmost digit, so the first digit, not the last.

The leftmost digit is.

Sorry I forgot to write the values of $ \log_{10} 2,\ \log_{10} 3$.

I have just edited.

Thank　you!


Here is the problem:
the sequence S(n) is defined as

S(n)= ∑([n^2+i^2]/n^3, i, 0, n)

find the limit of S(n) as n approaches infinity
...
then the solution follows like this:
∑([1+(i/n)^2]/n, i, 0, ∞)
set x = i/n
dx = 1/n
so

∑([1+(i/n)^2]/n, i, 0, ∞) = ∫((1 + x^2))dx, x, 0, 1)
= 4/3
...
Here is the question: what's the name of this method? (I've never seen this) and if anyone can provide some links that would be awesome (althought I could look it up - I don't have a problem with it - but i would appreciated), specially a proof.

Thanks in advance!!

It's from the definition of Riemann integral.

See Exercise 7.18 in Apostol's 'Mathematical Analysis'.

I know what the definition of the Riemann Integral is and its connection with Riemann sums, but I don't see the connection ...

This sequence is a sequence of Riemann sums for that integral. You know the function and interval; look for the partitions to match it.

The other posters have been pretty explicit.  For more on the Riemann Integral, see http://en.wikipedia.org/wiki/Riemann_integral


Find all functions $f$ defined on the set of positive reals which take positive real values and satisfy: $f(xf(y))=yf(x)$ for all $x,y$; and $f(x)\to0$ as $x\to\infty$.

Hmm i had a slightly different solution:

Consider $x=1$, which gives $$f(f(y))=yf(1)\implies f(f(f(y)))=f(yf(1))=f(y)f(1).$$  When we plug in $y=1$ into the last two equalities we get $f(f(1))=f(1)^2$ which instantly yields $f(1)=1$ due to the codomain restriction. Then due to the original expression that we derived we have $f(f(y))=y$, which we can very easily prove injectivity and surjectivity, and thus show that $f$ is bijective.
Now consider $x=y$, which yields $f(xf(x))=xf(x)$. However since we have shown bijectivity, we claim that $f$ is monotonically decreasing. (due to the second restriction) Then the only possible value for $xf(x)$ is identically $1$, so $xf(x)=1 \implies f(x)=\frac{1}{x}$. Checking gives that this is a valid solution.

can we use fixed points?i think that would provide the solution

[quote=e.lopes]If $f(k) = 1$, then $f(x) = f(xf(k)) = kf(x)$, so $k =1$.
Let $y = 1/f(x)$ and set $k = xf(y)$, then $f(k) = f(xf(y)) = yf(x) = 1$.
Hence $f(1) = 1$ and $f(1/f(x)) = 1/x$.
Also $f(f(y)) = f(1f(y)) = y$. Hence $f(1/x) = 1/f(x)$.
Finally, let $z = f(y)$, so that $f(z) = y$. Then $f(xy) = f(xf(z)) = zf(x) = f(x)f(y)$.

Now notice that $f(xf(x)) = xf(x)$. Let $k = xf(x)$.
We show that $k = 1$. $f(k^2) = f(k)f(k) = k^2$ and by a simple induction $f(kn) = kn$,
so we cannot have $k > 1$, or $f(x)$ would not tend to $0$ as $x$ tends to infinity.
But $f(1/k) = 1/k$ and the same argument shows that we cannot have $1/k > 1$.
Hence $k = 1$.

So the only such function $f$ is $f(x) = 1/x$.[/quote]

Good idea but wrong logic.
Do you consider the existence of $k$ such that $f(k)=1$?

- Taking $x=1$ gives $ff(y)=f(1)y$ hence $f$ is bijective.
- The number $xf(x)$ is a fixed point for all $x>0$. In particular $ff(1)=f(1)$. Then $f(1)=1$.
- If $p$ is a fixed point then so is $p^2$. Taking $x=1/p,y=p$ gives $f(1/p)=1/p$ hence $1/p$ is also fixed point.
- If $p>1$ then $p,p^2,p^4,\cdots$ are fixed points with limit $\infty$, contradiction. Similarly we can't have $0<p<1$.
- $p=1$ is the only fixed point and $f(x)=1/x$ for all $x$.

Let $P(x,y)$ be the given assertion.
$\lim_{y\to\infty}P(x,y)\Rightarrow\lim_{x\to0^+}f(x)=\infty$
Let $f$ have a fixed point $k$. If $k>1$, then $P(k,k)\Rightarrow f\left(k^2\right)=k^2$, so $f$ has arbitrarily large fixed points (simple induction). But then $\lim_{x\to\infty}f(x)=\infty$, contradiction. If $k<1$, then $f$ has arbitrarily small fixed points, so we reach a similar contradiction. Thus, any fixed point of $f$ must be equal to one, but $P(x,x)\Rightarrow f(xf(x))=xf(x)\Rightarrow xf(x)=1\Rightarrow\boxed{f(x)=\frac1x}$, which satisfies both conditions.

[hide=Solution]
Suppose that $f(a)=f(b)$ with $a\neq b$. Then, substituting $y=a$ and $y=b$ gives $af(x)=bf(x)$, which is a contradiction. Therefore, $f$ is injective. Now, substituting $y=1$ gives $f(xf(1))=f(x)$, so $f(1)=1$ by injectivity. Now, substituting $x=1$ gives $f(f(y))=y$, so substituting $y=f(y)$ gives $f(xy)=f(x)f(y)$. Consider the function $g(x)=\ln(f(e^x))$, which is possible since the domain and range of $f$ is positive reals. Then, $f(x)=e^{g(\ln x)}$, so the equation becomes $e^{g(\ln xy)}=e^{g(\ln x)+g(\ln y)}$, or $g(x+y)=g(x)+g(y)$ for all reals $x$ and $y$. Since $\lim_{x\to\infty}f(x)=0$, this means that $\lim_{x\to\infty}g(x)=-\infty$. This implies that $g$ is bounded below, so $g$ must be linear. Therefore, $g(x)=ax$, so $f(x)=e^{a\ln x}=x^a$. Since $f(f(x))=x$, we must have $a^2=1$, so $a=\pm1$. Since $a=1$ does not satisfy $\lim_{x\to\infty}f(x)=0$, the only solution is $\boxed{f(x)=\frac1x}$, which clearly works.

Consider the function $g(x)=xf(x)$. This has the same domain and codomain as $f(x)$. We claim that the solution is $f(x)=\frac{1}{x}$; that is, $g(x)=1$ for all positive reals $x$. This is easily verifiable, because $\frac{1}{\frac{x}{y}}=y\left(\frac{1}{x}\right)$. We can simplify the given equation into $$g(x)g(y)=g\left(g(y)\left(\frac{x}{y}\right)\right).$$ In particular, we have $$g(x)^2=g(g(x)).$$ Consider some number $a$ in the range of $g(x)$. Then $a^2=g(a)$ so that $f(a)=a$. Additionally, this means that $a^2$ is in the range of $g(x)$, so that $f(a^2)=a^2$, $f(a^4)=a^4$, and $f(a^{2^k})=a^{2^k}$ for all nonnegative integers $k$.

Suppose, for the sake of contradiction, that $a>1$. Then we know that $f(a^{2^k})=a^{2^k}$. If we let $k$ grow arbitrarily large, we get $f(a^{2^k})\to \infty$ as $a^{2^k}\to \infty$, which contradicts the problem statement.

Then we have that $a \leq 1$; that is, $g(x) \leq 1 \rightarrow f(x) \leq \frac{1}{x}$.

Again, suppose for the sake of contradiction that $a<1$. Then plugging $y=a$ into the original equation gives $$f(ax)=af(x).$$ Letting $x=1$ gives $f(1)=1$, and letting $x=\frac{1}{a}>1$ gives $\frac{1}{a}=f\left(\frac{1}{a}\right)$. But since $f(\frac{1}{a}) \leq a$, we have $\frac{1}{a} \leq a$, which is impossible as $a < 1$ and $\frac{1}{a} > 1$.

Then the only possible value for $a$ is $1$. This is equivalent to $g(x)=1$ for all positive reals $x$, or $f(x)=\boxed{\frac{1}{x}}$. $\blacksquare$

$y\to x \implies \exists f(z)=z \forall z=xf(x).$
Let $a$ be such that satisfies the condition on $z.$
$$\text{Then } f(a^2)= f(af(a))=af(a)=a^2.$$
By induction it follows $f(a^n)=a^n.$
Assume $a>1$. Thus $n\to \infty \implies a^n\to +\infty.$ A contradiction.
By similar argument, we can show that $f(a^{-n})=a^{-n}.$ And also get a contradiction with $a<1.$
So it must be that $a=1.$ Which follows that $$xf(x)=1\implies \boxed{f(x)=1/x},$$ which clearly fits.

I thought I didn't do it, so I resolved it.
Let $P(x,y)$ be the assertion.
$P(x,x)\implies f(xf(x))=xf(x),$ which is a fixed point.

Case 1: $\exists z>1: f(z)=z.$
$P(z,z)\implies f(z^2)=z^2,$ thus by induction, $f$ has infinite many fixed points. Then $\lim_{z\to \infty} z=\infty \implies \lim_{z\to \infty}f(z)=\infty,$ a contradiction.

Case 2: $\exists z<1: f(z)=z.$
$P(z,1/z)\implies f(1/z)=1/z>1,$ a contradiction.

Thus all fixed points $xf(x)=1\implies f(x)=1/x,$ which indeed fits.

Let $P(x,y)$ denote the given assertion.

The answer is $\boxed{f(x)=\frac{1}{x}}$, which clearly works.

Claim: $f$ is injective.
Proof: Suppose $f(a)=f(b)$.

$P(x,a): f(xf(a))=af(x)$.

$P(x,b): f(xf(a))=bf(x)$.

So $af(x)=bf(x)\implies a=b$. $\blacksquare$

$P(1,x): f(f(x))=xf(1)$.

$P(x,1): f(xf(1))=f(x)$.

So $f(f(f(x)))=f(xf(1))=f(x)$.

Now since $f$ is injective, we get $f(f(x))=x$.

$P(x,f(y)): f(xy)=f(x)f(y)$.

Set $g(x)=\ln f(e^x)$.

Note that $f(e^{x}e^y )=f(e^x)f(e^y)$, so $f(e^{x+y})=f(e^x)f(e^y)$. Take $\ln$ of both sides.

We get $g(x+y)=g(x)+g(y)$. Since $f(x)\to0$ as $x\to \infty$, $g(x)\to -\infty$ as $x\to \infty$. This implies that $g$ is bounded on some interval, so $g$ is linear.

Let $g(x)=cx$, we get $f(x)=x^c$.

So $f(x\cdot y^c)=x^c\cdot y\implies x^c\cdot y^{c^2}=x^c\cdot y\implies y^{c^2}=y\implies c^2=1$.

Clearly $f(x)=x$ doesn't satisfy $f(x)\to 0$ as $x\to \infty$, so $f(x)=\frac{1}{x}$.

We claim $f(x)=\frac{1}{x}$ is the only solution, which indeed satisfies both the functional equation and the limit condition.
-----
[b]Claim:[/b] $f(1)=1$.
[b]Proof:[/b] $P(1,1):f(f(1))=f(1)$
$P(\frac{1}{f(1)},1):f(1)=f(\frac{1}{f(1)})$
$P(\frac{1}{f(1)},\frac{1}{f(1)}):f(1)=(\frac{1}{f(1)})f(\frac{1}{f(1)})=1\Rightarrow f(1)=1$.
-----
[b]Claim:[/b] $x=1$ is the only fixed point.
[b]Proof:[/b] Assume $k\neq 1$ is fixed. Then from $P(k,k)$ we know $f(k^2)=k^2$ and from induction $f(k^n)=k^n$ for all positive integers $n$. But then $f$ has infinitely large (if $k>1$) or infinitely small (if $k<1$) fixed points, which contradicts the limit condition. Thus only $k=1$ is possible.
-----
We finish by noting that from $P(x,x)$ we get $f(xf(x))=xf(x)$, which means that $xf(x)$ is a fixed point for all $x$, but then $xf(x)=1$ and $f(x)=\frac{1}{x}$, as desired.

Let $P(x,y):=f(xf(y))=yf(x)$

[b][color=#FFD966]Claim: [/color][/b]$f$ is injective
[i]Proof:[/i]
Let $f(a)=f(b)$
$P(x,a)$ yields $f(xf(a))=af(x)$
$P(x,b)$ yields $f(xf(b))=bf(x)$
Thus $af(x)=bf(x)\Longrightarrow a=b$, therefore $f$ is injective $\square$.
[b][color=#93C47D]Claim:[/color][/b] $f$ is surjective
[i]Proof:[/i]
$P(x,\frac{x}{f(x)})$ yields $f\bigg(xf\left(\frac{x}{f(x)}\right)\bigg)=x$, therefore $f$ is surjective $\square$.
Since $f$ is injective and surjective, it is bijective.

$P(x,1)$ yields $f(xf(1))=f(x)\overset{\text{from injectivity}}{\Longrightarrow} xf(1)=x\Longrightarrow f(1)=1$
$P(1,f(x))$ yields $f(f(x))=x$
$P(x,f(y))$ yields $f(xy)=f(x)f(y)$ thus the function is multiplicative and is a Cauchy function.
Furthermore $\exists z\text{ such that } f(z)=z$
$P(z,z)$ yields $f(z^2)=z^2$
Furthermore by induction we obtain that $f(z^n)=z^n$

If $z>1$, we have that $\lim_{n\to\infty}f(z^n)=\lim_{n\to\infty}z^n=0$ which is clearly a contradiction, since $\lim_{n\to\infty}z^n$ diverges and explodes to infinity. Thus $z\le1\Longleftrightarrow xf(x)\le1\Longrightarrow f(x)\le\frac{1}{x}$
Therefore $f(x)\le1\text{ for }x\ge1$, thus $f$ is monotonically decreasing.
Furthermore, since $f$ is a Cauchy function and it's monotonically decreasing, $f(x)=x^c$ for some constant $c$.
By plugging into our original function we obtain $f(x\cdot y^c)=y\cdot x^c\Longrightarrow x^c\cdot y^{c^2}=y\cdot x^c\Longrightarrow y^{c^2}=y$
$\therefore c^2=1\Longrightarrow c=\pm1$
$c=1$ implies that $f(x)=x,\forall x\in \mathbb{R}^+$, however this is clearly a contradiction from our second condition since $\lim_{x\to\infty}x$ blows up to infinity, and does not converge to a finite value. Thus $c=-1$, $c=-1$ implies $f(x)=\frac{1}{x}$ which clearly satisfies both conditions.

So, to sum up  $\boxed{f(x)=\frac{1}{x}, \forall x \in \mathbb{R}^+}$ $\blacksquare$.


If  $ x_0\equal{}5$ and $ x_{ n\plus{}1}\equal{}x_n\plus{}\frac{1}{x_n}$ show that $ 45<x_{1000}<45.1$
This problem is taken from the list submitted for the 1975 Canadian Mathematics Olympiad


we define$\forall n\in N, f(n)\in N$ be the smallest number such that $n\mid \sum^{f(n)}_{k=1}k$.Show that $f(n)=2n-1$ if and only if $n$ be the power of $2$. :lol:

the "if part" is trivial, so here is the "only if" part:

Suppose n=p*2^q, where p > 1 is odd positive integer, q natural number. numbers {2^(q+1) - 1, 2*2^(q+1) - 1, 3*2^(q+1) - 1, ... , (p-1)*2^(q+1) - 1} assume the complete set of remainders modulo p, so f(n) < 2n - 1. Done. :D


Find the number of integral ordered triples (a,b,c) satisfying

$ |a\plus{}b|\plus{}c\equal{}19$
$ ab\plus{}|c|\equal{}97$.

My solution involved casework, and I'm wondering if there's a nicer one.

Absolute values tend to involve casework implicity.


a,b,c>0 and abc=1  prove that :
 a²+b²+c² +3 ≥ 2(ab+cb+ac);
it is not hard , but I didn't prove that by AM-GM , I used SCHUR inequality!!
who can prove that by AM-GM? :)

By schur ; $ x^3\plus{}y^3\plus{}z^3\plus{}3xyz \geq \sum_{sym}x^2y$
$ x\equal{}a^{\frac{2}{3}}$
$ \therefore x^2\plus{}y^2\plus{}z^2\plus{}3 \geq \sum_{sym}a^{\frac{4}{3}}b^{\frac{2}{3}} \geq 2(ab\plus{}bc\plus{}ca)$
By Am-Gm

(Hojoo Lee)Let $ t \in (0,3]$.for all $ a,b,c \geq 0$ prove that
\[ (3\minus{}t)\plus{}t(abc)^{\frac{2}{t}}\plus{}\sum_{cyclic}a^2 \geq 2\sum_{cyclic}ab\]


Let $(u_n)$ be the sequence difined $u_1 =\sqrt{2}, u_n = u_{n-1}.\sqrt[2^n]{n+1}, n>1$. Find $\large\lim_{n\to\infty}u_n$

In other words, we are asked to compute $\prod_{i=1}^\infty (i+1)^{2^{-i}}$.

[hide]Looking at the logarithm of this, it's clear that we have a convergent sum and thus a convergent product, in a strong sense.  Thus I should have no problem splitting this up into lots of little pieces; in particular, I want to count the total contribution of the prime $p$ to this product.  It shows up once in the $p - 1$st term, then again in the $2p - 1$st, and so on.  Thus, its initial contribution is with an exponent of $2^{-p + 1} + 2^{-2p + 1} + 2^{-3p + 1} + \ldots = 2\sum_{i = 1}^\infty (2^{-p})^i = 2\cdot \frac{2^{-p}}{1-2^{-p}} = \frac 2 {2^p - 1}$.  However, we've of course neglected those terms with higher powers of $p$.  The terms divisible by $p^2$ contribute an additional $2\sum_{i = 1}^\infty (2^{-p^2})^i = \frac 2 {2^{p^2} - 1}$, while the terms divisible by $p^3$ contribute an additional $\frac 2 {2^{p^3} - 1}$.  So the total contribution of the prime $p$ is $2\sum_{i = 1}^\infty \frac 1{2^{p^i} - 1}$.  Unfortunately, I don't know how to calculate this.  Calculation with Mathematica leads me to believe the answer should be about 2.76 (a bit larger than the "likely" answer $e$), although when I ask it to compute the infinite product it gives me the obviously wrong answer of 1.

[/hide]


An automobile went up a hill at a speed of $ 10$ miles an hour and down the same distance at a speed of $ 20$ miles an hour. The average speed for the round trip was:

$ \textbf{(A)}\ 12\frac {1}{2} \text{ mph} \qquad\textbf{(B)}\ 13\frac {1}{3} \text{ mph} \qquad\textbf{(C)}\ 14\frac {1}{2} \text{ mph} \qquad\textbf{(D)}\ 15 \text{ mph}$
$ \textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
Using the harmonic mean, we find $ A\equal{}\frac{2}{\frac{1}{10}\plus{}\frac{1}{20}}\equal{}\frac{40}{3}\equal{}13 \frac{1}{3}$, or $ \boxed{\textbf{(B)}}$.
[/hide]

[hide="In case someone wonders why use HM"]Let $ v_u$ be the uphill speed, $ v_d$ the downhill speed, $ v_a$ the average speed, and $ d$ the one-way distance.

Then, the total time traveled expressed in two different ways is

$ \frac{2d}{v_a}\equal{}\frac{d}{v_u}\plus{}\frac{d}{v_d}\implies\frac1{v_a}\equal{}\frac12\left(\frac1{v_u}\plus{}\frac1{v_d}\right)$

which is the HM of $ v_u$ and $ v_d$[/hide]


calculus
derivative
conics
parabola
calculus computations

Find the location of the indicated absolute extremum within the specified domain.
Maximum of f(x) = x^2 - 4 ; [-1, 2]

To find a an extremum (max or min), take a derivative and set equal to zero: $ f^{\prime}(x)\equal{}2x\equal{}0\Rightarrow x\equal{}0$ and since this is an upwards opening parabola, this point, namely $ \big( 0,f(0)\big)\equal{}(0,\minus{}4)$ is a minimum. For absolute exemum on an interval, check also the endpoints: $ f(\minus{}1)\equal{}\minus{}3$, and $ f(2) \equal{}8$, which is the absolute maximum.

That's clearly a mistake.

$ f(2) \equal{} 2^2 \minus{} 4 \equal{} 4 \minus{} 4 \equal{} 0$

$ f(2) \equal{} 0$


How do I place images I output to the GSview program into my latext documents?


Can someone please explain to me how the general solution of a Pell equation can be found???

It's all to do with continued fractions. If you write $\sqrt{n}$ as $\displaystyle{x_1 + \frac{1}{x_2 + \frac{1}{x_3 + \ldots}}}$ and truncate this fraction you can get two integers $p, q$ such that $\frac{p^2}{q^2} \approx n$ which for some values will give $p^2 - nq^2 = 1$.

Example, please? ( I don't know how to write a number in the form of a continued fraction)  :blush:

$\sqrt 5 = \displaystyle{2 + \frac{1}{4 + \frac{1}{4 + \ldots}}}$. Truncate at he first fraction tp get $\sqrt 5 \approx \frac{9}{4}$ which is indeed corresponds to a solution $(9,4)$ of the Pell equation $x^2 - 5y^2 = 1$.

Once you get the first solution ($x_{1},y_{1}$), you can easily get all other solutions ($x_{n},y_{n}$) by
\[ \left(x_{1}+y_{1}\sqrt{D}\right)^{n} = x_{n}+y_{n}\sqrt{D} \]
And then, you can get a recurrence relation and all.

However, there's no easy way to determine when the first solution will appear in the continued fraction (for sometimes you will end up with -1 instead of 1 on the RHS). Some numbers are terrible when it comes to find the first solution, notably 13 and 61.


Quadrilateral ABCD is cyclic. E and F are points on sides AD and AB such that DE = DC and BF = BC. G is the mid-point of EF.
Prove that <DGB is a right angle.
[img]http://www.idealmath.com/geomet4.gif[/img]
[url]http://www.idealmath.com[/url]

[color=darkblue]Let $ \omega_1, \omega_2$ be the circles with centers $ B,D$ and radii $ BC,CD$ respectively. Denote $ H \in \omega_1 \cap \omega_2$, $ M \in GD \cap EH, N \in BG \cap FH$.

Then $ \left\| \begin{array}{cc} \angle{EHC} \equal{} 180^\circ \minus{} \frac {1}{2}\angle{D} \\
 \\
\angle{FHC} \equal{} 180^\circ \minus{} \frac {1}{2}\angle{B} \end{array} \right\| \Rightarrow \angle{EHF} \equal{} 360^\circ \minus{} \big(360^\circ \minus{} \frac {1}{2}(\angle{B} \plus{} \angle{D})\big) \equal{} 90^\circ$

Because $ HG$ is median we have $ EG \equal{} GF \equal{} HG$. Quadrilaterals $ EDHG$ and $ HGFB$ are kite-shaped, so $ \angle{GNH} \equal{} \angle{GMH} \equal{} 90^\circ \Rightarrow MHNG$ is rectangle and $ \boxed{\angle{DGB} \equal{} 90^\circ}$.
[/color]
[hide="Image"][img]http://i055.radikal.ru/0806/7c/5b86688f0a3e.png[/img][/hide]


1.
A merchant has to take his sheeps to the neighbour country,to do that,he has to pass through thousands of guardroom...but all of the guard are very greedy,they want the merchant`s sheeps,so they set a rule:"each time the merchant pass the guardroom,he has to send in half of his sheeps and the guard will give him 1 sheep"
the merchant keep crossing those guardrooms with a very good mood,and when he get home,he still has some sheeps along with him..the question is how many sheeps does he have at the beginning and the end of the journey.

2.
a guest go to the bakery to order 9 cakes,and all of them must be placed in 4 boxes,each box must at least contains 3 cakes...Could you help the baker to do that?

3.
there are 2 light balls,have the same shape,each of them is hang by a very thin string,one of them is electricallly charged,the other is normal.
How can we know what is the normal ball and what is the electricallly charged ball?(without using tools)

4.
Two  balls have the same weight,same volume and same color,one of them is made from Copper(Cu),the other is made from aluminum(Al).Give the simpliest way to detect which ball is made from Cu and which ball is made from Al?

5.
Put the coin into the disk,poor the water into the disk to make sure the water is full fill the coin.can we take the coin out of the disk without getting wet?

6.
How to figure the temperature of :
a.the sun
b.the higher atmosphere
c.some insects
d.the moon
e.the continental shelf
f.liquid heli

7.
There are 13 balls,one of them have the different weight from the others,how to detect that ball with only 3 times measuring.

8.
Two cyclinder glasses have the same volume,store full of water inside.at first,the water`s temperature  in the 1st glass is higher than the 2nd glass.[b]Without[/b] giving energy from the outside,are there any way to make thet 2nd glass`s temperature higher than the 1st glass?

 -- sorry for my terrible english,hope you guys could read through it :blush:   --

[hide="problem 4"]
Put them into $\text{Hg}$
this is the first come to my mind...hope it's correct.[/hide]

woa,why doesnt anyone stop by this topic?i though these questions are interesting....i took it from a creative contest ,hope you guys could stop by and answer some of them  :D

[quote="TyphoonDC"]ok,lets discuss about it:
[hide="problem 1"]at first i think its 2,but then,i wonder if 1 is correct?and are there any more answers?[/hide]
[hide="problem 2"]in this question,my solution is to put each box into another box,so we have only 1 big box contain those 3 boxes(these contain each others too)[/hide]
[hide="problem 5"]i though of 2 ways:first is to use some clip to take the coin out,the second is to let the water to be dried by the sun light.but...is it too easy...i need a better solution.[/hide]
in problem 4,could you tell me why you put it in Hg?[/quote]
difference in density... ;)

hey thx,i solved all of them,but they are 'creative questions' so that i`m very eager to see new result...go go go  :lol:

#3.
you can use your hair!! or pieces of paper,
charge them by rubbing them to a wool, or some thing else, then see that to which light bulb they get attracted, that is the charged bulb.

#8,
just put some of the water from #1 to #2. ;)

#2.
put 3 cakes in 3 of the boxes. then put one of the box in the remaining box.

@7:2 disk scale.
@3:wrong pal,i told u NOT to use anything hehe (this mean not to interact with them) :D
@8:may be wrong,when u put like that,u also suply energy from 1 to 2.(my friend said so)
     i did a replacement,water from 1 to 3,2 to 1,and 3 to 2. :D

[hide="problem 4"]Since copper has a greater density than aluminum, the copper sphere must be hollow. Thus, when you roll them both down an incline, the aluminum sphere will reach the ground first, since the copper one has a greater moment of inertia.[/hide]

for the coin one, i haven't done it yet, but this will help
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=different+coins&t=34138


linear algebra
matrix
algebra
polynomial
vector
invariant
induction

Maybe Moubi should help us a little bit...  ;)

The ens student who sent me this though one, is very
busy with EGA Elements Géométries Algebriques written by
Grothendieck, so wait a little harazi for hints

Suppose we have that $ A_{1}, \ldots , A_{k} $ act on a linear space $ V $,
such that for every polynomial $ p(x_{1},x_{2}, \ldots , x_{k}) $
with non-commutative variables the matrix $ p(A_{1}, \ldots , A_{k})[A_{i},A_{j}] $
is nilpotent. It is enough to show that there exists a common eigenvector for all
$ A_{1}, \ldots , A_{k} $. If, first of all, $ [A_{i},A_{j}] = 0 $ for every $ i,j $,
then $ A_{1}, \ldots , A_{k} $ pairwise commute, so there exists a common eigenvector.
Now, if there exist $ i,j $, such that $ [A_{i},A_{j}] \neq 0 $, then there exists
a vector $ v $, such that $ w := [A_{i},A_{j}]v \neq 0 $, then we have that
for every $ p $, $ p(A_{1}, \ldots , A_{k})[A_{i},A_{j}] $ is a nilpotent, so
we cannot have $ p(A_{1}, \ldots , A_{k})[A_{i},A_{j}]v = v $. Now, if we denote by
$ W $ the set of all vectors of the form $ p(A_{1}, \ldots , A_{k})w $, where
$ p(x_{1},x_{2}, \ldots , x_{k}) $ runs over all polynomials, then it is clearly
a linear subspace, which is invariant under $ A_{1}, \ldots , A_{k} $, and which
is smaller than $ V $, because it does not contain $ v $. So by induction on the
dimension of $ V $, there exists a common eigenvector for all $ A_{1}, \ldots , A_{k} $.
This solution is short, so may be I made a stupid mistake on the way  :?  .


Karel can always solve the first 23 problems on the AMC correct, but never the last two. The story goes that one day, he decides to be bold and guess on the last two questions randomly to go for a perfect score. Soon after the test, he is told that he didn't get a perfect score. What is the probability that his score is above 140? (Assume each question has 5 choices, he answers the first 23 questions correctly, and that each question is worth 6 points).

I'm thinking this problem is about someone I know pretty well...

[hide]$ P \equal{} \frac{P_{\text{exactly one right}}}{P_{\text{not both right}}} \equal{} \frac{1\minus{}P_{\text{wrong}}^2 \minus{} P_{\text{right}}^2}{1\minus{}P_{\text{right}}^2} \equal{}\frac {1 \minus{} \left(\frac {1}{5} \right)^2 \minus{} \left(\frac {4}{5} \right)^2}{1 \minus{} \left(\frac {1}{5} \right)^2} \equal{} \frac {1}{3}$[/hide]

Ah but the AMC has a trend to have more answers be certain letters :) cough A test this year cough
So its still a good idea to guess, even if technically the probability of getting something right is 1/3 (w/ 2 guesses)

this is a good problem, since it is very relateable. i think someone went through this exact same situation, and didn't get over 140.  :P


I have to order 8 pizzas for a party.  The pizzas come in 4 different toppings (pepperoni, sausage, mushrooms, and veggies).  How many different combinations of the 4 types of pizzas can i order if i have to order atleast 1 of every kind?

[hide]p=a+1
s=b+1
m=c+1
v=d+1

so a+1+b+1+c+1+d+1=8 so a+b+c+d=4
and u have to do ((4+4-1)C(4-1)), which is (7C3) which is 35[/hide]

[quote="nebula42"][quote="Inspired By Nature"][quote="usaha"][hide]$p=a+1$
$s=b+1$
$m=c+1$
$v=d+1$
so $a+1+b+1+c+1+d+1=8$ so $a+b+c+d=4$
and u have to do $\left(\begin{array}{c}4+4-1 \\ 4-1 \end{array}\right)$, which is $\left(\begin{array}{c}7 \\ 3 \end{array}\right)$, which is $35$[/hide][/quote]
what are a, b, c, and d? please explain your solution.[/quote]
[hide]I think usaha means that $p$ is the number of pepperoni pizzas, $s$ is the number of sausage pizzas, etc. Since we have to have at least one of each, we can express them each as $1$ more than any nonnegative integer (which is what $a$, $b$, etc. are). Then the sum of $p$, $s$, etc. is $8$, so the sum of $a$, $b$, etc. is $4$, and we just calculate the number of solutions that solve that, which is $\boxed{35}$.[/hide]

I just did this:
[hide]Since there must be at least one of each, we can ignore $4$ of them such that one is a pepperoni, one is a sausage, etc. Then we have to choose the other $4$ from any of four different options. We can use $3$ _'s and $4$ |'s to form a sequence such that | means pick one pizza and _ means go to the next kind of pizza. For example, $1$ pepperoni and $3$ sausages is
| _ | | | _ _.
Thus our answer is $\binom{7}{3}=\boxed{35}$ ways total.[/hide][/quote]
oh, i see. Thanks nebula42 :)