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Let $ n$ be a positive integer. Given an integer coefficient polynomial $ f(x)$, define its [i]signature modulo $ n$[/i] to be the (ordered) sequence $ f(1), \ldots , f(n)$ modulo $ n$. Of the $ n^n$ such $ n$-term sequences of integers modulo $ n$, how many are the signature of some polynomial $ f(x)$ if
a) $ n$ is a positive integer not divisible by the square of a prime.
b) $ n$ is a positive integer not divisible by the cube of a prime.

Here is my idea, and I also get a solution. However my compution is poor so I possibly make it wrong.
I think there are 2 keys to queation a). We denote by $ d(n)$ the desired answer, and:
1. If $ m$ and $ n$ are coprime then $ d(mn)\equal{}d(m)d(n)$.
2. If for $ x\equal{}0,1, \ldots, p\minus{}1$, $ f(x) \equiv 0 \pmod p$ then there exists $ g(x)$ and $ r(x)$ with $ \deg r <p$ satisfying $ f(x)\equal{}(x^p\minus{}x)g(x)\plus{}pr(x)$.
And my answer to question a) is, if $ n\equal{}p_1p_2 \ldots p_t$ then there are $ p_1^{p_1} \ldots p_t^{p_t}$ $ n\minus{}$term sequences.
Similarly there is 2 keys to question b), saying
1. If $ m$ and $ n$ are coprime then $ d(mn)\equal{}d(m)d(n)$.
2. If for $ x\equal{}0,1, \ldots, p^2\minus{}1$, $ f(x) \equiv 0 \pmod {p^2}$ then there exists $ g(x)$ and $ r_1(x),r_2(x)$ with $ \deg r_1 <p, \deg r_2 <p$ satisfying $ f(x)\equal{}(x^p\minus{}x)^2g(x)\plus{}p(x^p\minus{}x)r_1(x)\plus{}p^2r_2(x)$.
My answer to question b) is, if $ n\equal{}p_1p_2 \ldots p_t q_1^2 \ldots q_l^2$ then there are $ p_1^{p_1} \ldots p_t^{p_t} q_1^{3q_1} \ldots q_l^{3q_l}$ $ n\minus{}$term sequences.

That is the right answer.  Here is a proof:
[hide="Solution"]
[b]Lemma 1: [/b] There is a polynomial for all signatures in $ \bmod p$ for any prime $ p$ (so there are $ p^p$ possible signatures in $ \bmod p$).

[i]Proof: [/i] Assume that there is not exactly one polynomial for every signature.  First, reduce all of the polynomials down to degree $ p-1$ (notice that $ p|\frac{(x+p-1)!}{(x-1)!}$ for all $ x$, so we can use this to reduce the degree of the polynomial) and all of the coefficients to be between $ 0$ and $ p-1$ inclusive.  Now, let us assume that $ f(x) \equiv g(x)\bmod p$ for $ x = 0, 1, 2, \cdots , p-1$.  This implies that for all $ x$, $ f(x) - g(x) \equiv 0 \bmod p$.  Let $ f(x) - g(x) \equiv h(x)\bmod p$.  Then, $ h(x)\equiv 0\bmod p$ where $ h(x)$ is a polynomial of degree $ p-1$, so $ h(x)$ has roots $ 1, 2, 3, \cdots , p$ in $ \mathbb {Z} p$.  Thus, $ h(x)$ has $ p$ roots in $ \mathbb {Z}p$, but only a degree of $ p-1$.  This is a contradiction by Lagrange's Theorem, so there is exactly one polynomial for each signature and there are $ p^p$ polynomials (in $ \bmod p$), so $ p^p$ signatures.

[b]Lemma 2:[/b] If we have a polynomial $ f(x)$ in $ \bmod m$ and a polynomial $ g(x)$ in $ \bmod n$, where $ m$ and $ n$ are coprime, then we can find an $ h(x)\bmod mn$ so that $ h(x)\equiv f(x)\bmod m$ and $ h(x)\equiv g(x)\bmod n$.

[i]Proof:[/i] Let $ z$ and $ y$ be integers so that $ mz\equiv 1\bmod n$ and $ ny\equiv 1\bmod m$ (by the coprime condition, this is possible).  Now, let $ h(x) \equiv mzg(x) + nyf(x)\bmod mn$, so $ h(x)\equiv f(x)\bmod m$ and $ h(x)\equiv g(x)\bmod n$.

[b]Lemma 3:[/b] $ f(x + p) - f(x) \equiv pf'(x)\bmod p^2$.

[i]Proof: [/i] Let $ f(x) \equiv \sum_{i = 0}^k {a_kx^k}\bmod p^2$.  This gives that \[ f(x+p) - f(x) \equiv \sum_{i=0}^k {a_k[(x+p)^k - x^p]} \equiv\] \[ \sum_{i=0}^k {a_k[pkx^{k-1} + p^2x^{k-2}\binom {k}{2} + \cdots ]} \equiv p\sum_{i=0}^k {ka_kx^{k-1}\bmod p\equiv pf'(x)\bmod p^2}\]

[b]Lemma 4:[/b] We can find a polynomial, $ f(x)$ which is congruent to $ 0\bmod p^2$ for $ p = 0, 1, 2, \cdots, p-1$, and having $ f(p+i)\equiv pa_i\bmod p^2$, where $ 0\le i\le p-1$,for any $ \{ a_0, a_1, \cdots,  a_{p-1} \}$.

[i]Proof: [/i] Let \[ f(x) \equiv g(x)[(x)(x-1)(x-2)\cdots (x-p+1)]\bmod p^2\] where $ g(x)$ is a polynomial of our choosing.  Clearly, $ f(x)\equiv 0\bmod p$ for $ x= 0, 1, 2, \cdots, p-1$.  Now, notice that \[ f(p+j)\equiv g(p+j)[(p+j)(p+j-1)\cdots (p+1)(p)(p-1)\cdots (j+1)] \bmod p^2\] Now, it suffices to have $ f(p+j)\equiv a_jp\bmod p^2$ for any $ j$ and $ a_j$.  Thus, we want for \[ [(p+j)(p+j-1)\cdots (p+2)(p+1)(p-1)(p-2)\cdots (j+1)]g(p+j)\equiv a_j\bmod p\]  Let \[ (p+j)(p+j-1)\cdots (p+2)(p+1)(p-1)(p-2)\cdots (j+1)\equiv k\bmod p\] Notice that since $ 0\le j\le p-1$, we have that $ k$ is not a multiple of $ p$.  This means that $ k$ and $ p$ are relatively prime, so $ k$ has an inverse $ \bmod p$.  Let this inverse be $ i_j$.  Thus, $ g(j)\equiv i_ja_j\bmod p$.  Let $ i_ja_j\equiv b_j\bmod p$.  It suffices to have $ g(j)\equiv b_j\bmod p$, where $ b_j$ is anything $ \bmod p$ and $ j$ ranges from $ 0$ to $ p-1$.  However, by Lemma 1, this is true, and so Lemma 4 is proven.

[b]Lemma 5:[/b] For any sequence of $ p$ numbers, $ a_0, a_1, \cdots a_{p-1}$, we can find a polynomial, $ f(x)$ that has its signature starting with $ a_0, a_1, \cdots , a_{p-1}\bmod p^2$.

[i]Proof: [/i] By Lemma 1, we can find a polynomial congruent to $ a_0, a_1, a_2, \cdots , a_{p-1}\bmod p$, say $ g(x)$.  Now, let $ g(i)\equiv b_i\bmod p^2$.  Then, notice that $ a_i\equiv b_i\bmod p$, so now consider a polynomial $ h(x)$ so that $ h(i)\equiv \frac{a_i-b_i}{p}\bmod p$ (this is possible by Lemma 1 again).  Now, let $ f(x) = ph(x) + g(x)$, which satisfies the conditions.

[b]Lemma 6: [/b] There are $ p^{3p}$ signatures in $ \bmod p^2$.

[i]Proof:[/i] By Lemma 3, we have that \[ f(x+ip) \equiv f(x+(i-1)p) + pf'(x+(i-1)p)\equiv f(x+(i-1)p) + pf'(x)\bmod p^2\]  Therefore, if we fix $ f(x)\bmod p^2$ for $ x=0,1,2,\cdots, 2p-1$, $ f(x)\bmod p^2$ is fixed for all of the other values of $ x$.  Since there are $ p^2$ possibilities for the first $ p$ values of $ p$ and then $ p$ values for the next $ p$ values of $ p$ (recall that $ f(p+i)\equiv f(i)+k_ip\bmod p^2$ for some $ k_i$), we have that there is a maximum of $ p^{2p+p}=p^{3p}$ solutions.  We now need to show that all of them are attainable.  Let us consider a sequence, $ s$, starting with $ a_0, a_1, \cdots , a_{2p - 1}$.  By Lemma 5, we can find a polynomial, $ g(x)$, starting with $ a_0, a_1, \cdots, a_{p - 1}$ and then having $ b_p, b_{p+1}, \cdots b_{2p-1}$.  By Lemma 4, we can find a polynomial, $ h(x)$, starting with $ 0, 0, 0, \cdots 0, a_p-b_p, a_{p+1}-b_{p+1}, \cdots , a_{2p-1}-b_{2p-1}$ with $ p$ zeroes (this is possible since $ a_{p+i}\equiv a_i\equiv b_i\equiv b_{i+p}\bmod p$).  Then, we simply consider $ f(x) = g(x) + h(x)$ as the desired polynomial.

[i]The Problem: [/i] For a), let $ n = p_1p_2\cdots p_k$.  Notice that $ f(x)\equiv f(p_i+x)\bmod p$.  Letting $ y_p$ be the residue of $ y\bmod p$, we have that $ f(y)\equiv f(y_{p_i})\bmod p_i$ for all $ i$.  Thus, when the first $ f(x)_{p_i}$ are fixed for $ x=0,1,2,\cdots , p-1$, $ f(x)_{p_i}$ is fixed for $ p\le x \le n-1$.  This means that fixing $ f(x)_{p_i}$ for all $ i$ fixes a system of congruencies for $ f(x)\bmod p_i$ for all $ i$, so $ f(x)\equiv f(x)_{p_i}\bmod p_i$ for all $ i$.  By Lemma 2 (applied repeatedly), there is a solution to this system of congruencies.  It suffices to find the number of signatures for $ f(x)$ in $ \bmod p_i$ for all $ i$.  However, this is just $ \prod_{i=1}^k {p_i^{p_i}}$, which is the answer to a).

For b), let $ n = p_1p_2\cdots p_kq_1^2q_2^2\cdots q_l^2$.  By similar reasoning as we used in a), the answer will be $ \prod_{i=1}^k {p_i^{p_i}}\prod_{i=1}^l {q_i^{3q_i}}$.  [/hide]

I think the general case can be solved in this way:
As mentioned in other proofs we should solve it for $n=p^{k}$.

lemma :let $k$ be a natural number,$p$ a prime and $f(x) \in \mathbb{Z}[x]$.
then for all $x\in \mathbb{Z}$ we have$ f
(x) \equiv 0 \bmod p^{k}$ $(*)$ iff $f(x)$ is in the form ${{{s_{p}(x)}^{k}g_{k}(x)+p^{1}s_{p}(x)}^{k-1}g_{k-1}(x)+...+p^{k-1}s_{p}(x)}^{1}g_{1}(x)+p^{k}g_{0}(x)$ with $\deg (g_{i})<p$ for $i<k$ and $s_{p}(x)=x(x-1)...(x-(p-1))$
Lemma can be proved by induction.

By lemma it is easy to see that  if we work in $\mathbb{Z} _{p^{k}}$ number of $f$ with property $(*)$ and degree less than $pk$  will be $(p^{k-1})^{p}.(p^{k-2})^{p}...(p^{0})^{p}=p^{\frac{k(k-1)}{2}p}$ $(**)$

Now the main question:we will work in $\mathbb{Z} _{p^{k}}$. By euclidean algorithm for any $f$ we have $f(x)=(s_{p}(x))^{k}g(x)+r(x),\deg (r)<kp$ and obviously for any $x$ we have $(s_{p}(x))^{k} \equiv 0 \bmod p^{k}$.so we only work on $f$ with $deg(f)<kp$.these $f$ form a group $|G|$ under addition.and the map $\phi :G \rightarrow {(\mathbb{Z} _{p^{k}})}^{p^{k}},\phi (f)=(f(1),f(2),...,f(p^{k}))$ is a homeomorphism.we have $|G|=p^{k^{2}p}$ and by $(**)$ $|ker( \phi )|=p^{\frac{k(k-1)}{2}p}$.so we have $|\phi (G)|=\frac{|G|}{|ker(\phi)|}=p^{\frac{k(k+1)}{2}p}$ and so the answer for $n=p^{k}$ is $p^{\frac{k(k+1)}{2}p}$

Another way to show that there are $p^{3p}$ signatures modulo $p^2$ is to write the (eventually terminating) Newton interpolation polynomial
\[f(x)=\sum_{k\ge0}\Delta^k[f](0)\binom{x}{k}=\sum_{k\ge0}a_k\binom{x}{k}.\]Clearly $(a_0,a_1,\ldots,a_{p^2-1})\pmod{p^2}\longleftrightarrow(f(0),f(1),\ldots,f(p^2-1))\pmod{p^2}$. Note that $f\in\mathbb{Z}[x]$ if and only if $k!\mid a_k$ for all $k\ge0$ (just consider the $[x^k]$ coefficients in [i]reverse[/i] order).

Analyzing $v_p(k!)$ for $k\ge0$, we see that for $k\ge 2p$, we have 1 choice for $a_k\pmod{p^2}$, for $p\le k\le 2p-1$, we have $p$ choices, and for $0\le k\le p-1$, we have $p^2$ choices. This gives us a total of $(p^2)^p p^p=p^{3p}$ working sequences, as desired. (It is easy to check such choices of $a_k\pmod{p^2}$ lead to a unique "class" of integer polynomials, as anything not divisible by $p$ is invertible modulo $p^2$).

[b]Edit (12/19/2013):[/b] Note that OldMath's lemma (and thus solution as well) fails for $k\ge p+1$:
[quote="OldMath"]I think the general case can be solved in this way:
As mentioned in other proofs we should solve it for $n=p^{k}$.

lemma :let $k$ be a natural number,$p$ a prime and $f(x) \in \mathbb{Z}[x]$.
then for all $x\in \mathbb{Z}$ we have$ f
(x) \equiv 0 \bmod p^{k}$ $(*)$ iff $f(x)$ is in the form ${{{s_{p}(x)}^{k}g_{k}(x)+p^{1}s_{p}(x)}^{k-1}g_{k-1}(x)+...+p^{k-1}s_{p}(x)}^{1}g_{1}(x)+p^{k}g_{0}(x)$ with $\deg (g_{i})<p$ for $i<k$ and $s_{p}(x)=x(x-1)...(x-(p-1))$
Lemma can be proved by induction.[/quote]
The correct version can be deduced from my Newton interpolation method above; we get a total count of $\prod_{t\ge0}p^{\max(0,r-v_p(t!))}$ for the $p^r$ case.

I am going to evaluate the number of such sequences for all positive integers $n$. Some part of my proof is similar to math154's post above.

I'll quantify the problem first. Define $\text{sgn}_n(f)= { (\overline{f(1)}, \overline{f(2)}, \cdots \overline{f(n)}) } $. Define $A_n= { \text{sgn}_n(f) | f \in \mathbb{Z}[x] }$. We need to find $|A_n|$

[i]Lemma:[/i] $|A_n|$ is a multiplicative function ([b]not[/b] completely) of $n$.

[i]Proof of Lemma:[/i] We shall equivalently establish a bijection between $ A_{mn}$ and $A_m \times A_n$, where $m$ and $n$ are coprime integers. Consider the map $X:A_{mn} \to A_m \times A_n$ satisfying \[ X(\text{sgn}_{mn}(f))= (\text{sgn}_m(f),\text{sgn}_n(f)) \]First, we show that the map is well-defined. Suppose that $\text{sgn}_{mn}(f)=\text{sgn}_{mn}(g)$ for some polynomials $f$ and $g$. But then, $mn|f(k)-g(k) \forall k \in [1,mn]$. Therefore, $m|f(k)-g(k) \forall k \in [1,m]$ and $n|f(k)-g(k) \forall k \in [1,n]$, implying $(\text{sgn}_m(f),\text{sgn}_n(f))=(\text{sgn}_m(g),\text{sgn}_n(g))$.

Next, we show that $f$ is injective. Suppose that, for polynomials with integer coefficients $f$ and $g$, $(\text{sgn}_m(f),\text{sgn}_n(f))=(\text{sgn}_m(g),\text{sgn}_n(g))$. Since $a-b|P(a)-P(b)$ for any polynomial $P$, therefore, $m|f(k)-g(k) \forall k \in \mathbb{N}$ and $n|f(k)-g(k) \forall k \in \mathbb{N}$. Since $(m,n)=1$, we also have $mn|f(k)-g(k) \forall k \in \mathbb{N}$. Consequently, $\text{sgn}_{mn}(f)=\text{sgn}_{mn}(f)$.

Finally, we are left to prove that $X$ is surjective. Since $(m,n)=1$, $\exists a,b \in \mathbb{Z}$ such that $am \equiv 1 \pmod{n}$ and $bn \equiv 1 \pmod{m}$. Define $P(x) = bnf(x)+amg(x)$. Then, $P(i) \equiv f(i) \pmod{m}$ and $P(i) \equiv g(i) \pmod{n} \forall i \in \mathbb{N}$. Therefore, $(\text{sgn}_m(f),\text{sgn}_n(g))=(\text{sgn}_m(P),\text{sgn}_n(P))=X(\text{sgn}_{mn}(P))$.

Consequently, $X$ is injective and surjective and therefore, bijective.

Now, it suffices to find out $|A_{p^k}|$, for primes $p$. Consider the Mahler expansion of a polynomial $f$. Clearly, the coefficients in the Mahler Expansion $(a_0,a_1, \cdots a_{p^k-1})$ uniquely define the values $(f(0),f(1), \cdots f(p^k-1))$ and vice-versa. Since $f \in \mathbb{Z}[x]$, we must have $m!|a_m \forall k \ge 0$.

Considering $v_p(m!)$, in $\mathbb{Z}/p^k\mathbb{Z}$, for $j \le k$, and $(j-1)p \le m < jp$, $a_m$ can take over exactly $p^{k-j+1}$ values. For $m \ge kp$, we have only one choice for $a_m$. By Multiplication Principle, the number of such sequences if $p^{0+p+\cdots (k-1)p+kp}=p^{pk(k+1)/2}$.

From here, we infer that if the canonical factorisation of $n$ is $\prod_{i=1}^rp_i^{a_i}$, we obtain \[ |A_n|= \prod_{i=1}^rp_i^{p_ia_i(a_i+1)/2} \]

Edit: the last half of the last paragraph is wrong but I'd imagine there being an easy fix.

Here's a relatively short elementary proof that there are $p^{\frac{a(a+1)}{2}p}$ signatures modulo $p^a$.

The crux of this solution is the following claim.

[b][color=#07f]Claim:[/color][/b] For any $k=0,1,\ldots,p-1$, if $(x_1,\ldots,x_{p^a})$ is the signature of $f(x)$, then $(y_1,\ldots,y_{p^a})$ is a signature, where $y_i=x_i$ if $i \equiv k \pmod{p}$, and $y_i=0$ otherwise.

[i]Proof:[/i] Let $S$ be the set of all $i$ in modulo $p^a$ such that $i \not\equiv k \pmod{p}$. Notice that if each element in $S$ was increased by a multiple of $p$, then $S$ would stay the same. Hence, if $r \equiv s \equiv k \pmod{p}$, then \[\prod_{i \in S} (r-i) \equiv \prod_{i \in S} (s-i) \pmod{p^a}.\] Set \[f_k(x)=\frac{\prod_{i \in S} (x-i)}{\prod_{i \in S} (k-i)} \cdot f(x).\] If $x \not\equiv k \pmod{p}$, then $\tfrac{\prod_{i \in S} (x-i)}{\prod_{i \in S} (k-i)} \equiv 0 \pmod{p^a}$, and if $x \equiv k \pmod{p}$, then we already proved that $\tfrac{\prod_{i \in S} (x-i)}{\prod_{i \in S} (k-i)} \equiv 1 \pmod{p^a}$. Therefore, $f_k(x)$ has signature $(y_1,\ldots,y_{p^a})$. $\square$

By this claim, the number of signatures modulo $p^a$ is the product of the number of possible values of \[(f(0),f(p),\ldots,f(p^a-p)),(f(1),f(p+1),\ldots,f(p^a-p+1)),\ldots,(f(p-1),f(2p-1),\ldots,f(p^a-1)).\] By shifting, it suffices to prove that the number of possible values of $(f(0),f(p),\ldots,f(p^a-p))$ is $p^{\frac{a(a+1)}{2}}$.

Notice that we only care about the coefficient of $x$ in modulo $p^{a-1}$, the coefficient of $x^2$ in modulo $p^{a-2}$, etc. We use this to define a [i]basic[/i] polynomial as a polynomial of degree at most $a-1$ such that the coefficient of $x^n$ is an element of $\{0,1,\ldots,p^{a-n}-1\}$ for all $n=0,1,\ldots,a-1$. By definition, any polynomial can be reduced to a basic polynomial with the same values of $f(0),f(p),\ldots,f(p^a-p)$. There are $p^a \cdot p^{a-1} \cdots p^1=p^{\frac{a(a+1)}{2}p}$ basic polynomials, so it suffices to prove that they all give distinct values of $(f(0),f(p),\ldots,f(p^a-p))$. Assume FTSOC that there exist two basic polynomials that give the same values. Then, we subtract them and shift the coefficient of $x^n$ of their difference by a multiple of $p^{a-n}$ to get a nonzero basic polynomial $f_0$ such that $f_0(0),f_0(p),\ldots,f_0(p^a-p)=0$. It's well known (proven by setting $f_0(x)=(x-k)q(x)+r$ for root $k$, quotient polynomial $q$, and remainder $r$) that $f_0$ must be divisible by $x(x-p) \cdots (x-(p^a-p))$. Suppose $f_0(x) \equiv q(x)x(x-p) \cdots (x-(p^a-p)) \pmod{p^a}$ for some nonzero polynomial $q(x)$ of degree $d$ with coefficients in $\{0,1,\ldots,p^a-1\}$. Then, the expansion of $f_0(x)$ must have a term with nonzero coefficient and degree $p^{a-1}+d$, contradicting the fact that $f_0$ is a basic polynomial (since basic polynomials have degree at most $a-1$ by definition). $\square$


Since the first kid gave 5 sandwichs, he should get 5 dollars and the second kid 3 dollars.  I am not sure...

first kid-7
second kid-1

First kid gives away 5-8/3 of his sandwich
Second kid gives away 3-8/3 of his sandwich

*8/3 is found by 8 sandwich total and shared by three people, since the first person has to keep some of his sandwich he shares 5-8/3

We have to find the value of a sandwich so that the shared ones equals 8 dollars. We have the equation

$ (5\minus{}\frac{8}{3}\plus{}3\minus{}\frac{8}{3})x\equal{}8$

When solving this you get x as 3.

When putting x back in we get:

$ (5\minus{}\frac{8}{3})3\equal{}7$
and
$ (3\minus{}\frac{8}{3})3\equal{}1$

And please, don't entitle everything "Need Help" next time, it gets annoying.

[quote="Ihatepie"]since they each get 2 2/3 sandwiches, the first gave away 2 1/3 while the second gave away 1/3. so the first gets 7, the next gets 1.[/quote]
And now the question is how do you explain that the first gave away 2 1/3 while the second gave away 1/3.
Shentang explained it a bit better, however, there still is the matter that the question did not specify how the sandwiches were distributed.
So instead of your way, say the third person got 2 sandwiches from the person with 5 sandwiches, and $ 1\frac{2}{3}$ sandwiches from the person with 3 sandwiches, and then the person with 5 sandwiches gave the person with 3 sandwiches 1 sandwich.
It evenly distributes the sandwiches out, but I guess under the circumstance that the problem did not specify how the sandwiches were distributed, your way is probably what the problem intended.

well with problems like this, where it doesn't say exactly how it should be carried out, your best bet is to assume that it is the same no matter how it is distributed. If it isn't, it is a badly worded problem and can be easily dismissed. But if it is, there is an easy way to get an answer.

@ shetang and all others here: he's posted other problems and asked for NNNNNNNEEEEEEEEEDDDDDDDDD HELP! He's using this as homework help. Stop posting full solutions to people like this.

well i think people who need help on problems should get the help they want, fishythefish. but i still dont get what the answer is...

Yes, but if the case is homework, then the person does not want help, per se; they just don't want to work out the solution for themselves. Ask around, most would agree.

[quote="dreambold22"]well i mean if they really NEED help, i think they should. otherwise, they're just being lazy.[/quote]

[hide]each person got 2 2/3 sandwiches. so saying, thats 8/3 sandwiches. assuming they did share equally, the kid who produced 5 sandwiches sacrificed more, he gets 5/8 of the money, the kid who produced 3 sandwiches gets 3/8 of the money. kind of obvious.
5 dollars to kid who gave shared 5 sandwiches
3 dollars to kid who shared 3 sandwiches.[/hide]

The boys have a total of 8 sandwiches to distribute equally among themselves. That means that each boy gets 8/3 sandwiches. Thus, the boy who brought 3 sandwiches gives the third boy 1/3 of a sandwich. Similarly, the boy who brought 5 sandwiches gives the third boy 7/3 sandwiches. Since the boy who brought 3 sandwiches gave the third boy 1/8 of his share, he deserves 1/8 of the 8 dollars, or 1 dollar. Similarly, the boy who brought 5 sandwiches gave the third boy 7/8 of his share, meaning that he deserves 7/8 of the 8 dollars, or 1 dollar.

Well the answer is clearly

[hide]

   Well...  Its actually quite simple. Since there were 8 sandwiches given away and there are three people each person got 8/3 sandwiches. So lets do person 1 first since person 1 gave away 5 you need to divide 8/3 by 5/8. And you would get: 64/15, which simplifies to 5 4/15. Which means he gave away 5 4/15 to each person. Giving person one 5 4/15 of the money paid by person 3. So leaving person 2 with 2 11/15

Person 1 will recieve 5 4/15  Dollars
Person 2 will recieve 2 11/15 Dollars

Of course this is just a complete guess could you guys tell me if you see anything wrong with this I would really like it if you guys told me what i screwed up so i can learn from my mistakes

Tyvm :lol: [/hide]

[quote="ZhangPeijin"][quote="Ihatepie"]since they each get 2 2/3 sandwiches, the first gave away 2 1/3 while the second gave away 1/3. so the first gets 7, the next gets 1.[/quote]
And now the question is how do you explain that the first gave away 2 1/3 while the second gave away 1/3.
Shentang explained it a bit better, however, there still is the matter that the question did not specify how the sandwiches were distributed.
So instead of your way, say the third person got 2 sandwiches from the person with 5 sandwiches, and $ 1\frac {2}{3}$ sandwiches from the person with 3 sandwiches, and then the person with 5 sandwiches gave the person with 3 sandwiches 1 sandwich.
It evenly distributes the sandwiches out, but I guess under the circumstance that the problem did not specify how the sandwiches were distributed, your way is probably what the problem intended.[/quote]

It doesn't matter how the sandwiches are given out. The proportion is still the same, of 7/1.

So say for example the guy with 5 is A and the guy with 3 is B. The guy with none is C. So say A gives 2 to C. B then gives the remaining 2/3 to C so he gets a 1/3rd of the 8 sandwiches. So now A has 3, B has 2 1/3, and C has 2 2/3. To even it all out, A gives B 1/3 of a sandwich. Adding it all up, you can see that A has given 2+1/3 or 7/3 sandwiches, while B has given 1/3. So therefore, since the ratio is 7/1, A should get 7 dollars and B should get 1. No matter how you arrange it, you get the same answer.

I don't know if this will help or not...

I have heard this problem before. I remember that there is a seemingly obvious way of solving it, but that the resulting proportion is wrong. There is a tricky way to get to the real solution, and it involves the number of sandwiches a lot more.

Might have been a little unclear, and I don't know if it'll help... :maybe:

[quote="mathmath1"]Remember to give a solution and how you got to the answer.

Three teenagers met after school to study.  Two of them bought sanwhiches, but the third was empty-handed.  He proposed that they should eat together and he would pay the value of his proportion.

This being agreed to, the first kid produced 5 sandwiches, the second kid 3 sandwiches, and the three of them shared the 8 sanwiches equally.  The third kid put down 8 dollars as the value of his share.  The others were satisfied with this but quarreled about how to distribute the money.  Thereupon the matter was referred to a teacher, who decided correctly how the money should be divided.  What was the decision?[/quote]

the first kid gave him 2 1/3 sandwiches and the second gave him 1/3 sandwiches because they all got 2 2/3 sandwiches.  so the first guy got 7 bucks and the other got 1

at fist i though the teacher would have decided, the first guy get 5 and the next get 3, but each gave away a different part so then i was confused :|

The 3 kids shared the 8 sandwiches equally so each kid ate 2 2/3 sandwiches. The first kid had 5 and ate 2 2/3 himself so he had 2 1/3 left to give to the third kid. The second kid had 3 and ate 2 2/3 himself so he had 1/3 left to give to the third kid. (1/3)/(2 1/3) = (1/3)/(7/3) = 1/7
The above ratio of the amount of sandwich(es) the first and second kid gave away show that to be fair, the first kid should get 7 dollars and the second kid should get 1 dollar.

first kid:5 dollars
second kid:3 dollars
a dollar each per sandwhich
im not that sure though


Prove that for every positive real numbers $a,b,c$, such that $abc=1$ the inequality holds:
$\frac{1}{1+2}+\frac{1}{b+2}+\frac{1}{c+2}\le 1$

I think you mean that:
$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\leq 1$

My solution is:
Since $(a+2)(b+2)(c+2)=abc+2ab+2ac+2bc+4a+4b+4c+8$ ,
$(a+2)(b+2)=2a+2b+ab+4$ and $(b+2)(c+2)=2b+2c+bc+4$ and $(a+2)(c+2)=2a+2c+ac+4$ we have to prove that:

$(a+2)(b+2)+(b+2)(c+2)+(a+2)(c+2) \leq (a+2)(b+2)(c+2)$ $\Rightarrow 2a+2b+ab+4+2b+2c+bc+4+2a+2c+ac+4 = 4a+4b+4c+ab+bc+ac+12 \leq abc+2ab+2ac+2bc+4a+4b+4c+8$
So we have to prove that $ab+bc+ac \geq 4-abc=3$.But that is very easy because $\left(\frac{ab+bc+ac}{3}\right)^{\frac{1}{2}}\geq (abc)^{\frac{1}{3}}$ $\Rightarrow \left( \frac{ab+bc+ac}{3}\right) \geq 1$ $\Rightarrow ab+bc+ac \geq 3$ Q.E.D. :D

...........................
Gabriel Ponce

[quote="Beat"]Prove that for every positive real numbers $a,b,c$, such that $abc=1$ the inequality holds:
$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\le 1$ (1)[/quote]
Let $a=\frac{x^{2}}{yz}$
(1) $\Leftrightarrow \sum\frac{yz}{x^{2}+2yz}\le 1\Leftrightarrow \sum\frac{x^{2}}{x^{2}+2yz}\ge 1$
And we have $\sum\frac{x^{2}}{x^{2}+2yz}\ge\frac{(x+y+z)^{2}}{(x+y+z)^{2}}=1$


geometry
geometric transformation
reflection
geometry proposed

Let $ A_1,B_1$ and $ C_1$ be the midpoints of the sides $ BC,AC$ and $ AB$ of the triangle $ ABC$. Take points $ E$ and $ F$ on $ B_1C_1$ and $ B_1A_1$ resp., such that the lines $ BE$ and $ BF$ are the bisectors of the angles $ \angle{AEB_1}$ and $ \angle{CFB_1}$ respectively. Prove that the bisectors of the angles $ \angle{ABC}$ and $ \angle{FBE}$ coincide.

Reflect $ A$ over $ BE$ is $ A'$, $ C$ over $ BF$ is $ B'$ we have: $ BA' \equal{} BA,BC' \equal{} BC$ and $ A'\in B_1C_1,C'\in A_1B_1$
Now:
$ \frac {sin\angle {BA'C_1}}{sin\angle {BC_1A'}} \equal{} \frac {BC_1}{BA'} \equal{} \frac {1}{2} \equal{} \frac {BA_1}{BC'} \equal{} \frac {sin\angle {BC'A_1}}{sin\angle {BA_1C'}}$
And $ \angle {BC_1A'} \equal{} \angle {BA_1C'}$ then $ \angle {BA'C_1} \equal{} \angle {BC'A_1}$
Then $ \angle {A'BC_1} \equal{} \angle {C'BA_1}\Leftrightarrow \angle {EBC_1} \equal{} \angle {FBA_1}$ (because $ \angle {A'BC_1} \equal{} 2\angle {EBC_1},\angle {C'BA_1} \equal{} 2\angle {FBA_1}$)
It is OK!
[img]http://i227.photobucket.com/albums/dd72/TPFC/1.png[/img]

It's correct. The idea to reflect $ A$ and $ C$ is nice!

Here's my solution:
As $ B$ lies on the bisectors of $ \angle{AEB_1}$ and $ \angle{CFB_1}$, the distances from $ B$ to the lines $ AE$ and $ B_1C_1$ are equal; and the distances from $ B$ to the lines $ CF$ and $ B_1A_1$ are equal. Since the distance from $ B$ to $ B_1C_1=\dfrac{h_a}2$ and from $ B$ to $ B_1A_1$ is $ \dfrac{h_c}2$, we get $ BM=\dfrac{h_a}2$ and $ BN=\dfrac{h_c}2$, where $ M$ and $ N$ are the projections of $ B$ on $ AE$ and $ CF$ respectively. Now, it's easy to see that the two right triangles $ \triangle{BMA}$ and $ \triangle{BNC}$ are similar, hence $ \angle{BAE}\equiv\angle{BCF}$. This easily implies $ \angle{ABE}\equiv\angle{CBF}$.

I know two more solutions for the problem. Try to find them!

[quote="freemind"][color=darkred]Let $ A_1,B_1$ and $ C_1$ be the midpoints of the sides $ [BC]$, $ [AC]$ and $ [AB]$ of the triangle $ ABC$. Take points $ E\in (B_1C_1)$ and $ F\in (A_1B_1)$ such that the lines $ BE$ and $ BF$ are the bisectors of the angles $ \widehat {AEB_1}$ and $ \widehat {CFB_1}$ respectively. Prove that the bisectors of the angles $ \widehat {ABC}$ and $ \widehat {FBE}$ coincide. [/color][/quote]
[color=darkblue][b]Proof.[/b] Denote $ \|\begin{array}{c}
X\in BE\cap AC\\\
Y\in BF\cap AC\end{array}$ Observe that $ \|\begin{array}{c}
\frac {XA}{XB_1}=\frac {EA}{EB_1}\\\\
\frac {YC}{YB_1}=\frac {FC}{FB_1}\end{array}$.
Apply the [b]Menelaus[/b]' theorem to the transversals

$ \|\begin{array}{ccccc}
\overline {BEX}\mathrm {\ in\ } \triangle\ AB_1C_1 & \Longrightarrow & \frac {BC_1}{BA}\cdot \frac {XA}{XB_1}\cdot\frac {EB_1}{EC_1}=1 & \Longrightarrow & EA=2\cdot EC_1\\\\
\overline {BFY}\mathrm {\ in\ } \triangle\ CB_1A_1 & \Longrightarrow & \frac {BA_1}{BC}\cdot \frac {YC}{YB_1}\cdot\frac {FB_1}{FA_1}=1 & \Longrightarrow & FC=2\cdot FA_1\end{array}$

$ \|\begin{array}{c}
\widehat {AC_1E}\equiv\widehat {CA_1F}\equiv \widehat {ABC}\\\\
\frac {EA}{EC_1}=\frac {FC}{FA_1}=2\end{array}\|$ $ \Longrightarrow$ $ \triangle AEC_1\sim\triangle CFA_1$ $ \Longrightarrow$ $ \|\begin{array}{c}
\widehat {BAE}\equiv\widehat {BCF}\\\\
\widehat {AEC_1}\equiv\widehat {CFA_1}\\\\
\widehat {AEX}\equiv\widehat {CFY}\end{array}\ .$

Therefore, $ \|\begin{array}{c}
\widehat {AEX}=\widehat {ABX}+\widehat {BAE}\\\\
\widehat {CFY}=\widehat {CBY}+\widehat {BCF}\end{array}\|$ $ \implies$ $ \widehat {ABX}\equiv\widehat {CBY}\ .$[/color]

Where can we find the Saint Petersburg problems?  :maybe:

[quote="Virgil Nicula"]
$ \left\|\begin{array}{c} \widehat {AC_1E}\equiv\widehat {CA_1F}\equiv \widehat {ABC} \\
 \\
\frac {EA}{EC_1} \equal{} \frac {FC}{FA_1} \equal{} 2\end{array}\right\|$ $ \Longrightarrow$ $ \triangle AEC_1\sim\triangle CFA_1$
[/quote]

Dear Virgil,

Are you sure we can conclude that?


topology
function
vector
Functional Analysis
advanced fields
advanced fields theorems

Hello,
I had a cursory introduction to topology many years ago in college as part of a control systems course, so my knowledge is a bit vague. I am coming back up to speed a bit.

If I remember my terminology correctly, I am wondering if anyone knows of any (practical) problems described and solved using a non-orthonormal basis (basis functions?)? I am also wondering how the span is described in this case?

How is a topological basis different than other bases (or is this the same thing)?

Thanks for your help. Enjoy the day.

Basis, span, and orthonormal are vector space words - this suggests that your interest is not so much in topology as in functional analysis (the study of topological vector spaces.) I won't try to answer your question (unless you make it much more specific), but I will give you some vocabulary to try to look up.

In a finite dimensional vector space, your concerns are primarily algebraic. After all, all reasonable topologies on a finite dimensional vector space are equivalent. You may, however, be concerned with infinite dimensional spaces. Here are four types of vector space. In each case, the one I list first also belongs to all of the types below:

Hilbert spaces (or, if not necessarily complete, inner product spaces)
Banach spaces (or, if not necessarily complete, normed vector spaces)
Locally convex topological vector spaces
Topological vector spaces

As for bases: there are number of different types. Here are three types to look up:

Hamel basis
Schauder basis
Orthonormal basis

You can only have an othonormal basis in an inner product space.

The very existence of Hamel bases depends on some equivalent of the Axiom of Choice. The use of a Hamel basis is thus not very constructive - perhaps not even imaginable - and often serves only the construction of strange examples and counterexamples. If you're looking for practicality or applicability, don't look at Hamel bases.

That leaves Schauder basis as a term for you to look up.

Also, a base for a topology has nothing to do with a basis for a vector space. A base is a set of open sets which generate the open sets by unions.
The standard example of a base for a topology is the set of balls in a metric space.


A ball is dropped from a height of 5 ft on a surface that causes it to bounce up 10% of the distance that it fell. If the probability that the ball is on its way up at any point in time is represented by the fraction $ (m/n)$ in lowest terms, where $ m$ and $ n$ are positive integers, what is $ m\plus{}n$?

[hide="Solution."]
Half the total distance traveled after the first drop can be found as the infinite geometric series with first term $ \frac {1}{2}$ and common ratio $ \frac {1}{10}$. Thus the total distance traveled by the ball is $ 5 \plus{} 2\cdot \frac {\frac {1}{2}}{1 \minus{} \frac {1}{10}} \equal{} \frac {55}{9}$.

The distance traveled by the ball on the way up is half the distance traveled by the ball after the first drop, which is $ \frac {\frac {1}{2}}{1 \minus{} \frac {1}{10}} \equal{} \frac {5}{9}$.

Thus the desired ratio is $ \frac {\frac {5}{9}}{\frac {55}{9}} \equal{} \frac {1}{11} \implies 1\plus{}11\equal{}\boxed{12}$.
[/hide]

[hide="Different Approach"]Each time the ball drops, it then bounces back up $ \frac {1}{10}$ of the distance that it just dropped. Thus, for every $ 11$ units of time, $ 1$ unit is spent traveling up. The probability is $ \frac {1}{11}$, so $ m \plus{} n \equal{} 1 \plus{} 11 \equal{} \boxed{12}$.[/hide]


Show that there exists a positive multiple of $1996$ whose sum of digits is $1996$.

[hide]
1996 factors into $4\cdot 499$. By Fermat's Little Theorem, $10^{498}\equiv 1\pmod{499}\Rightarrow \frac{10^{498}-1}{9}=\underbrace{11111\cdots }_{498}$ is divisible by 499.

Then the number
\[ N=\underbrace{11111\cdots}_{996}\underbrace{19961996\cdots}_{40} \]
will work. At the leftmost end are two strings of $\underbrace{11111\cdots}_{498}$ which have been multiplied by a large power of 10, so each is divisible by 1996. Adding the 1996's also retains divisibility by 1996, so $N$ is divisible by 1996. The sum of the digits of $N$ is $996+40(1+9+9+6)=1996$, so $N$ satisifies both conditions.
[/hide]

Actually, given [b]any[/b] $n >1$,  one can show that there exists a positive multiple of $1996$ whose sum of digits is $n$.  :)


If $ x$ is $ 40\%$ of $ y$, what percent of $ 2y$ is $ 3x$?

$ x$ is $ 40\%$ of $ y \Rightarrow 2x$ is $ 40\%$ of $ 2y \Rightarrow 3x$ is $ \boxed{60}\%$ of $ 2y$.


Nu stiu ce se intampla cu forumul...sau cu account-ul meu...In timpul zilei imi introduc user-ul si parola si nu imi intra in forum, ci revine la fereastra de introducere a user-ului si parolei, si tot asa. Pe de alta parte, seara revine la normal...Ce se intampla?

[color=darkred]Si mie mi se intampla acelasi lucru. Este infiorator. Trebuie comunicat moderatorului sau chiar administratorului. Eu am comunicat prin m.p., insa nu am primit nici un raspuns. Incearca si tu, poate ai mai mult succes.[/color]

Deci pana la urma cum ramane? A vazut totusi vreun admin/mod acest topic? Se poate remedia problema? Daca da, unde e problema, la account-ul meu sau la forum?

hmm, mie nu prea mi se intampla asa ceva, poate ca e de la browserul pe care il folosesti, incearca si tu mozilla.

nici eu nu am patit inca asa ceva . ar putea fi de la browser .. cum zice cosmin , sau de la firewall care iti  blocheaza cookie-urile .


You step into the elevator on the twenty-eighth floor of an office building and ride it down five floors to the print shop.  From there, you take the elevator up nine floors to the purchasing department.  On what floor number is the purchasing department, if floors are numbered with consecutive integers?

[hide]32.[/hide]Wasnt that a 1996 school competition question???

[hide="Solution and Answer"]28-5+9 = 28+4=[b]32nd flr.[/b][/hide]

thats easy  :lol:
[hide]28-5+9=32[/hide]

[quote="mbabbitt2003"]$28-5+9=32$[/quote]

Please hide ur answer next time.

I did the same thing that mbabbitt2003 did... ;)
[hide]
$28-5+9=32$ Thirty-two is my answer[/hide][/hide]

that was easy, just like the regular school math
[hide]28-5+9=32[/hide]


Find the maximum value of the expression:
         (c/a-b)(a/b-c)(b/c-a)(a 2 +b 2 +c 2 +ab+bc+ca) when a,b,c>0 add up to 1.

we will prove that

$(\frac{c}{a}-b)(\frac{a}{b}-c)(\frac{b}{c}-a)(a^2+b^2+c^2+ab+bc+ca) \leq \frac{16}{81}$

$\leftrightarrow \prod (a^2+ab+ac-bc)*(\sum a^2 +\sum ab) \leq 16\81 (a+b+c)^5 abc$

let $x=b+c, y=a+c, z=a+b, p=\displaystyle \frac 1{2(x+y+z)}$, we get

$\prod ((p-x)^2 p^2 - S^2)(x^2+y^2+z^2) \leq 32\81 p^6 S^4$

$\Leftrightarrow \prod ((p-x)^2 - r^2)(x^2+y^2+z^2) \leq 32\81 S^4$

$\Leftrightarrow \prod (1-tg^2 (A\2))(x^2+y^2+z^2) \leq 32\81 p^2$

$\Leftrightarrow \prod cosA (1+\prod cosA) \leq 64\81 \prod cos^4 (A\2)$

the triangle ABC must be acute, so let $x^2=n+p, y^2=m+p,z^2=m+n$
we will get

$81\sum_{sym} 8m^3 n^2 p+4m^2 n^2 p^2 \leq (\sqrt{m+n} + \sqrt{m+p} +\sqrt{n+p})^4 \sum_{sym} 2m^2np+m^2n^2$

it's enough to show that (\sqrt{m+n} + \sqrt{m+p} +\sqrt{n+p})^4 \geq 4*27(mn+np+pm)$
now we need to show that
$\sum_{sym} 6m^3 n^2 p+3m^2 n^2 p^2 \leq \sum_{sym} 6m^3 n^2 p+2m^2 n^2 p^2 +m^3 n^3$ which is right clearly.

 :D

It seems that I have another solution.
First,let see that if $ (c-ab)(a-bc)(b-ac) \leq 0 $ so $ f(a,b,c) \leq 0 $
If $ (c-ab)(a-bc)(b-ac) \geq 0 $ then three of them are positive.
[tex] (c/a-b)(a/b-c)(b/c-a)(a^2+b^2+c^2+ab+ca+bc) [/tex]
[tex] = \frac{(c-ab)(a-bc)(b-ac)(1-ab-ac-bc)}{abc} \ [/tex]
[tex] =(1-ab-ac-bc)(1- ab/c)(1-bc/a)(1-ac/b) [/tex]
[tex] \leq \frac 1{256}\ (4-ab-ac-bc-ab/c-bc/a-ac/b)^4 [/tex]
In fact, $ ab+ac+bc+(ab/c+bc/a+ac/b)(a+b+c) \geq \frac 43\(a+b+c)^2 $
It is equal to:
[tex] \sum a^2( \frac {b}{c}  \ + \frac {c}{b}\ ) \geq \frac43\ \sum a^2 + \frac 23\ \sum ab[/tex] which is right completely.
So [tex] 4-ab-ac-bc-ab/c-bc/a-ac/b \leq \frac83\ [/tex]
And notice that: $ 1-ab-ac-bc >0; 1-ab/c>0;1-bc/a>0;1-ac/b>0 $ so $ 4-ab-ac-bc-ab/c-bc/a-ac/b >0$
So [tex](c/a-b)(a/b-c)(b/c-a)(a^2+b^2+c^2+ab+ac+bc) \leq \frac{16}{81}[/tex] :)
The equality occurs if and only $ a=b=c= \frac13\ $  :lol:


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let G be any finite group and let T be a representation. Suppose there is a at least one element of g in G such that

det(T(g)) is not 1.

Prove that  |G/G'|>1

where G' denotes the derived group.

$ det\circ T$ is a nontrivial homomorphism $ G\to\mathbb{C}^x\equal{}GL_1(\mathbb{C})$ and hence a nontrivial representation of degree 1. The number of representations of degree 1 equals the index $ |G: G'|$ so this index is greater 1. (Or more elementary: Since there is a nontrivial homomorphism into an abelian group, the commutator subgroup must be nontrivial)


trigonometry calculus derivative function

Prove that if $a$ and $b$ are acute angles and a < b, then

$\frac{tan(a)}{a} < \frac{tan(b)}{b}$

[hide]
There's a solution by analyzing the derivative of $f(x) = \frac{\tan{x}}{x}$ but maybe there's a better algebraic solution.[/hide]

[hide="Solution"]
$\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \ge \tan x + \tan y$
$\tan (2x) \ge 2 \tan x$
$\frac{ \tan 2x}{2x} \ge \frac{\tan x}{x}$

From the continuity of $\frac{\tan x}{x}$ this implies that $\frac{\tan x}{x}$ is an increasing function in the range.  QED. [/hide]


Find all integers n such that n^2+n+2 is a perfect square.

[hide="solution"]

we can do one thing: cancel out the n term.

so we set $n+2 = 0$
$n = -2$ that works.

another thing: try to bracket into a square. Per say, $(n+1)^2$
only way to do that is to set $n+2 = 2n+1$, so that they are interchangeable, whcih would make the equation a square.

so $n=1$

those two are the only ones i have found so far.
[/hide]

errr i got $n = 1 or -2$

i ono the formal way to do it
to get the $n=1$ thing, i just thought of it as $n^2+n+2 = (n+1)^2$
because since you already have an $n^2$, you want to find a value for $n+2$ that will bring it to the next square
so
\[n^2+n+2 = n^2 + 2n +1\]
\[ n=1\]

then for the $n=-2$, I just kind of looked at it and realized that the $n$ and $2$ would cancel out if $n=-2$, and that $(-2)^2$ was equal to 4

so i dunno the .. formal way to do it... but thats how i did it

Here is how I did it:

[hide]Set n^2+n+2 equal to k^2 where k is an integer.  So we can write (n-k)(n+k)+n+2=0.   So in order for n and k to be integers this equation must be factorable.  So we can assume k is equal to plus or minus 2.  So solving the new quadratic n^2+n+2=4, we get solutions of -2, and +1.[/hide]

Is this method legit?  It kind of seems like a little trick to get you some solutions but it wont gurantee you all.

do you all know of a way where you can be sure that -2 and 1 are ALL the possible solutions?

Note that no perfect square exists between consecutive squares (duh!).  We also have

$n^2 + n + 2 = (n+1)^2 + 1 - n$

as well as

$n^2 + n + 2 = (n^2) + n + 2$

Therefore, if n > 1

$n^2 < n^2 + n + 2 < (n+1)^2$

Also, if n < -2, substitute n = -k for clarity of proof.  k is positive.

$n^2 + n + 2 = (k-1)^2 + 1 + k$

and

$n^2 + n + 2 = (k^2) - k + 2$

Therefore, $(k-1)^2 < n^2 + n + 2 < k^2$

We conclude that we must have $-2 \le n \le 1$, and the rest is simply trial and error.

That's really neat! How'd you think to do it that way?


Are there guidelines to predicting whether an oxide will exist as individual molecules?

Alot of them we just know: SiO2, P4O10,CO2. Do more nonmetal oxides exist as discrete molecules than metal oxides.

I'm not sure what you mean by "discrete molecule". Can you please give an example? :)

Thanks.

ah I actually found the answer to my question in a google search on an xkcd forum post.
Funny I didn't try that earlier.

http://echochamber.me/viewtopic.php?f=18&t=14398&view=next


\[ \begin{gathered} ZC \bot XY,AZ: ZB \equal{} 1: 2,\measuredangle BCY \equal{} \beta ,\measuredangle ACX \equal{} \alpha \hfill \\
Find\_\measuredangle AZC\_in\_terms\_of\_\alpha \_and\_\beta \hfill \\
\end{gathered}
\]

Apply the sin theorem:

call $ \widehat{AZC} \equal{} \gamma$

$ \frac {AZ}{\sin \widehat{ACZ}} \equal{} \frac {AC}{\sin \widehat{AZC}} \equal{} \frac {CZ}{\sin \widehat{BAC}} \Rightarrow \frac {AZ}{\cos \alpha} \equal{} \frac {AC}{\sin \gamma}$

$ \frac {BC}{\sin \gamma} \equal{} \frac {2AZ}{\cos \beta} \equal{} \frac {CZ}{\sin \widehat{ABC}}$

so

$ AZ \equal{} AC\frac {\cos \alpha}{\sin \gamma} \equal{} \frac {BC}{2}\frac {\cos \beta}{\sin \gamma} \Rightarrow \frac {AC}{BC} \equal{} \frac {\cos \beta}{2 \cos \alpha}$

$ \frac {CZ}{\sin \widehat{BAC}} \equal{} \frac {AC}{\sin \gamma}$ and $ \frac {CZ}{\sin \widehat{ABC}} \equal{} \frac {BC}{\sin \gamma}$ $ \Rightarrow \frac {AC}{BC} \equal{} \frac {\sin \widehat{ABC}}{\sin \widehat{BAC}}$

$ \frac {\sin \widehat{ABC}}{\sin \widehat{BAC}} \equal{} \frac {\cos \beta}{2 \cos \alpha}$

But $ \widehat{ABC} \plus{} (\pi \minus{} \gamma) \plus{} (\frac {\pi}{2} \minus{} \beta) \equal{} \pi$ and $ \widehat{BAC} \plus{} \gamma \plus{} (\frac {\pi}{2} \minus{} \alpha) \equal{} \pi$.

so, since supplementary angles have the same sin:

$ \frac {\sin \widehat{ABC}}{\sin \widehat{BAC}} \equal{}\frac{\sin \left [(\pi \minus{} \gamma) \plus{} (\frac {\pi}{2} \minus{} \beta) \right ]}{\sin \left [\gamma \plus{} (\frac {\pi}{2} \minus{} \alpha) \right ]}$ $ \equal{}\frac{\sin \gamma \sin \beta \minus{}\cos \gamma \cos \beta}{\sin \gamma \sin \alpha \plus{} \cos \gamma \cos \alpha}$ $ \equal{}\frac {\cos \beta}{2 \cos \alpha}$

$ 2\sin\gamma\sin\beta\cos\alpha \minus{} 2\cos\gamma\cos\beta\cos\alpha \equal{} \sin\gamma\sin\alpha\cos\beta \plus{} \cos\gamma\cos\alpha\cos\beta$

$ 3\cos\gamma\cos\alpha\cos\beta\equal{}\sin\gamma(\sin\alpha\cos\beta \minus{} 2\sin\beta\cos\alpha)$

thus $ \tan \gamma \equal{} \frac{3\cos\alpha\cos\beta}{\sin\alpha\cos\beta \minus{} 2\sin\beta\cos\alpha}$


search
complex analysis
complex numbers
puzzles

how do you solve for i^i
my friend brought this up in school, and when I put it in the calculator, I got 0.2078795764.
what I did was root(-1)^root(-1), then move the root over on the bottom, so that it's -1^(root(-1)/2), but that's just the same as -1..

this is for the complex analysis section... $ i \equal{} e^{i(\pi/2 \plus{} 2k\pi)}$ for all integers $ k$, so $ i^i \equal{} (e^{i(\pi/2 \plus{} 2k\pi)})^i \equal{} e^{ \minus{} (\pi/2 \plus{} 2k\pi)}$, so actually $ i^i$ has infinitely many possible values :)

... which is also true of any exponential $ w^z$ of two complex numbers if $ z$ is not both real and rational, due to the definition of the complex exponential and properties of the complex logarithm.  There have been several posts on the forum with more details -- if you search for posts by me containing the word "multivalued," you will find a few.  ($ i^i$ is kind of cool because all of those values are real, and (we can tell from pleurestique's expression) actually form a two-ended geometric sequence.  Your calculator just chose one of the many possible values to show you.)

Your calculator showed you this one:
$ i^i\equal{} (cis \dfrac {\pi}2)^i \equal{}(e^{i\dfrac {\pi}2})^i \equal{} e^{i^2 \dfrac {\pi}2} \equal{} e^{\minus{}\dfrac{\pi}2} \approx 0.2079$


Don't double posting, you have made double posting several times.
Be sure to that!

[quote="MJ GEO"]find all value of $ k$ that is a function like $ f$ from $ R$ TO $ R$ and $ f(k(x \plus{} y)) \equal{} f(x) \plus{} f(y)$,[/quote]

If $ k\equal{}1$, we get the classical Cauchy equation and its infinitely many solutions (with $ ax$ as continuous one).

If $ k\ne 1$, set $ x\equal{}\frac{k}{1\minus{}k}u$ and $ y\equal{}u$ and the equation implies $ f(\frac{k}{1\minus{}k}u) \equal{} f(\frac{k}{1\minus{}k}u) \plus{} f(u)$ and so $ f(u)\equal{}0$  $ \forall u$

exact and short solution of this easy problem.thanks


Sarah intened to multiply a two digit number and a three digit number, but she left out the multplication sign and simply placed the two digit number to the left of the three digit number, Thereby forming a five digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two digit number and the three digit number?

Does ACU have anything to do with computer programming. Maybe that is why no one answered this. Anyways you did not to write program and this was probably posted in the wrong section if you did not use program. Probably a Pre-Olympiad question because you need to use division formula.

So my solution is number theory style and I used division formula to solve for the numbers. Two digit number is 14 and three digit number is 112.

i don't know if this will work or not but give it a shot.
lets just say the two digit number was ab (this is not a times b).
and the three digit number was cde. then you can write ab as:
(10 a +b)and cde as:(100c+10d+e). then you can write ab*cde as: 1000ac+100ad+10ae+100bc+10bd+be then set this equal to: 90000a+9000b+900C+90d+9e.
then solve for 100c+10(a+d)+b+e


Let V and W be vector spaces over F. Let [tex](v_1, ..., v_n)[/tex] be a basis of V, and let [tex](w_1, ..., w_n)[/tex] be any list of vectors in W. Then we can define a function T from V to W such that [tex]T(a_1v_1+...+a_nv_n) = a_1w_1 + ... + a_nw_n[/tex], where each [tex]a_k[/tex] is any arbitrary element of F.

It can be easily checked that T is a linear map. However, I need to prove that [tex]Tv_k=w_k[/tex] for [tex]k=1, ..., n[/tex]. How can this be proved? In fact, I don't even think that this is true (even though my textbooks say that it is), since we can examine the case when [tex]n=2, Tv_1 = (a_2/a_1)w_2, Tv_2 = (a_1/a_2)w_1[/tex].

Any insight woud be appreciated, thanks.

Set $a_2=a_3=\ldots=a_n=0\in \mathbb{F}$

Then we get $T(a_1v_1)=a_1T(v_1)=a_1w_1$ and so on and so on.

i'm confused too now

if you have f(a1*v1+...an*vn)=a1*w1+...+an*wn for all a1,,an in the field

can't you just take a1=1 and all the rest zero
and so on for the others

or am I misunderstanding you

fred

Ah, silly me! Sorry, I was being stupid. I would explain why I thought that I couldn't do what you two did, but I don't quite understand why I thought so myself now. Anyways, thanks.


Just out of curiosity, what is the best way to solve this equation? Can I multiply by 2 to make .5  a 1, and if so why would arbitrary ideas like this be acceptable or usable?

10=.5+x(2+ 10/.5+x)
The 2 isn't part of the numerator.It's just a whole number.

My short answer is that seemingly arbitrary operations such as multiplying a whole equation by some number are precisely what algebra is about. Historically, algebra developed as a set of tools for solving equations when the answers to the equations weren't visible at a glance, as would be the case for an equation such as, for instance,

x=7+3

or the only slightly harder

x+3=13.

I sympathize with you because when I was a kid I thought it was bogus to be multiplying equations, adding or subtracting something from both sides, or, worse still, subtracting one equation from another to solve a system of equations. That all seemed so terribly arbitrary to me at the time. I didn't enjoy studying algebra until I learned about Cartesian coordinate graphs and solving systems of equations by graphing--then everything seemed to make sense to me. Maybe that is my learning style. My son doesn't seem to have had any problem learning algebra at a younger age than I did, probably because he had a much better prealgebra mathematics background.

What you'll have to do with the equation you started out with is getting x out of the denominator on the one side. Until then, the answer won't be easy (for most people).

I welcome other participants to jump in here with hints.

Yeah, I always had an issue with that too, though I just accepted it as truth.

I finally truly accepted it when I realized:
If a=b, then 2a=2b.  So, yes, you can.  Both sides are equal, so their doubles are equal.  And with adding systems, it helped to realize that I was adding equal quantities to both sides of the equation.


trigonometry
number theory theorems
number theory

Hi Does anyone have a solution for kroneckers's theorem ? :)

for The Theorem see:
http://mathworld.wolfram.com/KroneckersApproximationTheorem.html

PS.I will be vey thankfull to anyone who give a printable solution(I have to print it for my teacher)

This a corolary of the Weyl Equidistribution Theorem, that implies that $\{na\}$ is equidistributed mod 1, hence dense in the unit interval.
First look here: http://mathworld.wolfram.com/WeylsCriterion.html
Then look here: http://www.math.harvard.edu/~elkies/M259.02/weyl.pdf  for the proof of Weyl criterion, and the corrolary you are looking for.
I also found this: http://www.ias.ac.in/resonance/May2003/pdf/May2003p30-37.pdf which should provide a proof in one piece, but the document is unreadable. :(
Another solution, would be that I rewrite 4 pages from my book ...
(way too much for me ;))

Is there an elementary proof of Kronecker's theorem?

There is:
The sequence $\{n\alpha\}$ has a cluster point by the theorem of Bolzano-Weierstrass. Thus, for every $\epsilon >0$ we can find $m<n$ with $0< |\{m\alpha\}-\{n\alpha\}|< \epsilon$ (they can't be equal because $\alpha$ is not rational).
Then either $0<\{(n-m)\alpha\}< \epsilon$ or $1-\epsilon < \{(n-m)\alpha\}< 1$, depending on iff $\{m\alpha\} < \{n\alpha\}$ or not.
But in both cases, the numbers $\{k(n-m)\alpha\}$ ($1 \leq k \leq \frac{1}{\{(n-m)\alpha\}}$) divide $[0,1]$ into intervalls of length at most $\epsilon$, so every $x \in [0,1]$ has at most distance $\frac{\epsilon}{2}$ from some of them.
But since all this was independent of $\epsilon$, the numbers of the form $\{n\alpha\}$ are dense in $[0,1]$.

I'm forgetful today  :( I did in class this proof of the theorem (well, a special case of it: $\sin n \alpha$ is dense in $[0,1]$ for appropriate $\alpha$)

And I don't think you really need Bolzano-Weierstrass, you just need pigeon hole and $\alpha$ irrational


How much time did it take u to solve this??
I ordered a pizza. It arrived at my house the next year. Even though I told the pizza guy that I had ordered the pizza last year, I didn't get the pizza free by the 30 minutes or more rule and I had to pay the full price. There is no mistake by the pizza company or delivery guy. How did this happen?
:P

Yeah, this took me about 5 seconds also, maybe 6, because it's kind of obvious...

[quote="kstan013"]Hmmm...right after I got done reading it. That was easy.[/quote]

Same answer, almost immediately after I finished reading

[quote="RicOh3.14"][hide]My 5 seconds of thinking would tell me that you ordered it on the 31th of December after 23.30 h.
Is there something I am forgetting maybe?

RicOh[/hide][/quote]

what I got but please hide your answer.

[quote="mustafa"]A better time to ask a question like this would be June, or any time that isn't within a week of New Years  :wink: [/quote]
 :rotfl:
Yeah, I was thinking about making up a problem for the New Year and then I saw this problem and knew the answer.

um i got it a few seconds after i read the first answer lol

I got it immediately after I read the sms! lol....but one of my friends had to actually THINK for about AN HOUR!! so I wanted to see if she was the only person who found this problem "too tough"!!  :rotfl:  :rotfl:

[quote="Xantos C. Guin"][quote="ragnarok23"]I got it while I was reading the question[/quote]
Same here.

It took me longer to read the question than to figure it out. And It took me longer to make this post than to read the question.[/quote]I third that :D


On the trip home from the meeting where this AMC10 was constructed,
the Contest Chair noted that his airport parking receipt had digits
of the form $ bbcac$, where $ 0 \le a<b<c \le 9$, and $ b$ was the
average of $ a$ and $ c$. How many different five-digit numbers
satisfy all these properties?

We actually don't need to list anything out.  Note that either $a,c$ are both even or odd.  There are $5$ distinct even digits, namely $0,2,4,6,8$ and $\dbinom{5}{2} = 10$ ways to pick them.  The increasing order of those numbers is already uniquely determined.  Similarly there are $5$ odd digits and we see that we can iterate the same process and find that:
$$2 \cdot \dbinom{5}{2} = 2 \cdot 10 = \boxed{20}$$
is our final answer.

@ above: I actually love your solution! I tried listing them out, but I forgot to include two pairs of $a$ and $c$.

[hide=My Solution]Using a fairly straightforward reasoning, we can use 5 choose 2 is 10, and similarly, a since have an increasing by 2 iteration, as mattx pointed out, this is simply 10 * 2 = 20.[/hide]

[hide=And for those interested…][pywindow]count = 0
for i in range(10000, 99999):
    numlist = list(str(i))
    if (numlist[0] == numlist[1]) and (numlist[2] == numlist[4]):
        if (int(numlist[3]) >0 or int(numlist[3]) == 0) and (int(numlist[4]) < 9 or int(numlist[4]) == 9) and (int(numlist[0]) > int(numlist[3])) and (int(numlist[4]) > int(numlist[0])):
            if int(numlist[0]) == (int(numlist[3]) + int(numlist[4]))/2:
                count+=1

print(count)
[/pywindow] I made this AFTER solving the problem as writing all this code down for such a simple problem is quite stupid.....

and I am not stupid -.- ....

hence I did it after the problem….
[/hide]


[quote="hello"]HAHAAH. Another trick. Wanted to see how many people would check this after the day is almost done.[/quote]

I insists, people who wants to see how many people enters some specific topic does not think about people like me that just go from topic to topic pressing [b]Next Topic[/b] without being able to see the subject of the next topic!

I am not sure what you mean, djimenez? Well, will anyone else post?

[quote="hello"]Well, will anyone else post?[/quote]

No.  Hey, wait a minute.....
 :D

these kinda topics are sorta pointless cuz some people click them anyways to see what responses ppl put

I knew it was an April Fool's joke... if the USAMO was canceled, it would be on the front page of AoPS.

after first one , i understod it wasn't a joke
but i don't know anyone who will not believe the first one

it will take more people in if you post it on the Round Table!

If it were in the AMC forum and posted by the AMC Director then it really would have gotten ppl :D


show that the following equation has no solution
sinx + 2=4cosxcos2x

I think we can apply $ 2\cos{a}\cos{b} \equal{} \sin{(a \minus{} b)} \minus{} \sin{(a \plus{} b)}$ :maybe:

By that:
$ \sin{x} \plus{} 2 \equal{} 4\cos{x}\cos{2x}$
$ \Leftrightarrow \sin{x} \plus{} 2 \equal{} 2(\sin{x} \minus{} \sin{3x})$
$ \Leftrightarrow \sin{x} \plus{} 2 \equal{} 2(\sin{x} \minus{} (3\sin{x}\minus{}4\sin^3{x}))$
$ \Leftrightarrow \sin{x} \plus{} 2 \equal{} 2(4\sin^3{x}\minus{}2\sin{x})$
$ \Leftrightarrow 8\sin^3{x}\minus{}5\sin{x}\minus{}2\equal{}0$

This is followed by the result.

hello, i think we have
$ \sin(x\minus{}y)\minus{}\sin(x\plus{}y)\equal{}\minus{}2\cos(x)\sin(y)$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i think we have
$ \sin(x \minus{} y) \minus{} \sin(x \plus{} y) \equal{} \minus{} 2\cos(x)\sin(y)$.
Sonnhard.[/quote]

OK, that's right, Sonnhard.


A solution contains only water and alcohol.  The alcohol content of a 9 ounce solution is 50%.  How many ounces of water must be added to the 9 ounce solution so that the alcohol content is reduced to 30% in the enhanced solution?

Uh.. is it a whole number? I got 1.8....... Prety sure this is wrong. (I KNOW I SPELLED PRETY WRONG!)

If we set up an equation, it would look like this:

$ \frac{x\plus{}4.5}{9}\equal{}\frac{7}{10}$. So then we cross multiply and we get:

$ 10x\plus{}45\equal{}63$

Then after simple algebra, we get $ 10x\equal{}18$, and then $ x\equal{}9/5$....

I got 10.5 oz...
Since the 9 oz solution is 50% alcohol, there are 4.5 oz of alcohol. To find out how much water to add, just do a proportion thingy:
$ 4.5/x\equal{}3/10$ where x is the total mass of the solution.
so $ 45\equal{}3x$
so $ x\equal{}15$
then $ 15\minus{}4.5\equal{}10.5$ oz that must be added.

I got the same answer as jonathanchou711...

My equation was the exact same too.

[quote="modularmarc101"][quote="007math"]:wallbash_red:  :wallbash_red: thanks nick[/quote]

Tu parles le francais? Hablas espanol? Need help ?[/quote]
If you search his posts you will see that he can type English fluently.

How do you know I don't use Google Translate to translate from Spanish to English?


calculus
integration
trigonometry
logarithms
calculus computations

Calculate
$ I \equal{} \smallint \tan \left( {\frac{1}{{\arctan x \minus{} 2010}}} \right)dx$

It cannot be described by elementary functions.

See http://integrals.wolfram.com/index.jsp?expr=tan%5B1%2F%28Arctan%5Bx%5D-2010%29%5D&random=false   . Whose username 'milin' helped me finding this website.

Hhmm, thank you. How about this one
$ J \equal{} \int {arc\tan \left( {\frac{1}{{tanx \minus{} 2010}}} \right)dx}$

hello, Maple 13 gives the following answer
$ \minus{}1/2\,i\arctan \left(  \left( \tan \left( x \right) \minus{}2010 \right) ^{\minus{}1
} \right) \ln  \left( {\frac { \left( {\frac {1}{1010026}}\minus{}{\frac {
1005}{1010026}}\,i \right)  \left( \tan \left( x \right) \minus{}i \right) }{
\tan \left( x \right) \minus{}2010\minus{}i}} \right) \minus{}1/4\,{\it polylog} \left( 2,{
\frac { \left( {\frac {1010025}{1010026}}\plus{}{\frac {1005}{1010026}}\,i
 \right)  \left( \tan \left( x \right) \minus{}2010\plus{}i \right) ^{2}}{ \left(
\tan \left( x \right)  \right) ^{2}\minus{}4020\,\tan \left( x \right) \plus{}
4040101}} \right) \plus{}1/2\,i\arctan \left(  \left( \tan \left( x \right)
\minus{}2010 \right) ^{\minus{}1} \right) \ln  \left( {\frac {\minus{}{\frac {1}{1005}}\,i
 \left( \tan \left( x \right) \plus{}i \right) }{\tan \left( x \right) \minus{}2010
\minus{}i}} \right) \plus{}1/4\,{\it polylog} \left( 2,{\frac { \left( 1\plus{}{\frac {1}
{1005}}\,i \right)  \left( \tan \left( x \right) \minus{}2010\plus{}i \right) ^{2}}
{ \left( \tan \left( x \right)  \right) ^{2}\minus{}4020\,\tan \left( x
 \right) \plus{}4040101}} \right)$
Sonnhard.

Thx Dr Sonnhard Graubner. There is a "nice"  way to calculate J without using maple or mathematica


Eh how hard is it? My coach said I qualified even though I bombed the octm so I'm not really sure >.>

you can see how hard it is here

http://octmtournament.org/ohmio/previoustests.cfm

Mehh the problems are trivial.

Is it the time pressure that kills?

yeah of course

actually time pressure probably isn't that bad in individual; just in ciphering and team round

i made a ton of dumb mistakes last year anyway

argh

i made too many dumb mistakes on the octm and the amc lol

i fail under time pressure

Hey so are you both going this Saturday? Ive been stalking this forum for a few days, but Ive never really posted (I think). Im still fairly new to this site, and fairly new to the concept of math.

From what I hear, OHMIO seems like a lot of fun. Im glad I made it.


Can someone do this step by step. I would like to learn it.

Write a costrunction for the problem. Assume that the relationship is linear.

Fixed cost, $ 500 ; 5 items cost$3210 to produce?

You ran afoul of $ \text{\LaTeX}$; your dollar signs were interpreted as markup language.

A fixed version: [quote="ArmiAldi"]Fixed cost, $ \$$500 ; 5 items cost $ \$$3210 to produce?[/quote]


25 knights are seated at a round table and 3 are chosen at random. Find the probability that at least two of the chosen 3 are sitting next to each other.

solution from http://www.kalva.demon.co.uk/aime.html
[hide]
For two knights, there are 3 disjoint possibilities: (1) adjacent, prob 1/12, (2) one seat between them, prob 1/12, (3) more than one seat between them, prob 5/6. In case (2), prob third sits adjacent is 3/23, in case (3) it is 4/23. Hence 1/12 + 1/92 + 10/69 = 11/46.[/hide]

which is all good and dandy.. but i was wondering why what i did doesn't work.

# of ways at least 2 knights can sit next to each other = (25 ways 2 knights can sit next to each other)*(24 spots for the other knight) - (25 ways 3 knights can sit next to each other)... because each group of 3 was counted twice

P(at least 2 knights next to each other) = (# of ways at least 2 knights can sit next to each other)/(all possible groups of 3 knights) = (25*24 - 25)/(25C3) = 6/25 = wrong!!

[quote="saracampbellsoup"]
# of ways at least 2 knights can sit next to each other = (25 ways 2 knights can sit next to each other)*(24 spots for the other knight) - (25 ways 3 knights can sit next to each other)... because each group of 3 was counted twice
[/quote]

it would be *(23 spots for the other knight) because 2 are already taken. that gives you the right answer.


Let $z$ is a complex number such that $|z^{2}+1|=2|z+1|.$ Prove that $|z|\leq\sqrt{7}.$

ummm...if $z$ is complex, how can it be compared to a real?

Did you miss the absolute value signs, perhaps?

The absolute value signs denote the magnitude of the complex number, which is real.


What packages are there installed? (Although I realli don't think we need anything except the basic stuff to typeset most of the math)

Nothing fancy.  What are you looking for?

If possible, could the AMS packages (amsfonts, amsmath, amssymb, etc.)  and the xy package please be added? Thanks.

Sorry about that. I was checking which packages were included. I tried to do a Halmos square (\blacksquare), but it didn't recognize that because the AMS packages haven't been included.

I'm messing around with that now.  With any luck I'll have it done tonight or tomorrow.

[tex]\displaystyle{\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C} }[/tex]

Let [tex]\mathfrak{N}[/tex] be the nilradical of a ring [tex]R[/tex], and let [tex]\mathfrak{a}[/tex] and [tex]\mathfrak{b}[/tex] be ideals.

[tex]\blacksquare[/tex]

[quote="rrusczyk"][tex]\blacksquare[/tex][/quote]

What's the square used for?

Can you please install the Maple Style Sheet? (or let me email them to you)
My signature isn't working.

Can you send me a pointer to it and I'll check it out when I have time.

I know those posts were made days ago, but since nobody has responded, I was just wondering if any of the aforementioned packages were added to the forum?

Also, I have a question... where does one go about getting those packages for his/her home LaTeX setup. I currently use TeXnicCenter as my editor.

The AMS packages are installed.

Here's where I got the AMS packages:

[url]http://www.ams.org/tex/amsmath.html[/url]

Spoon: if you installed MiKTeX, the ams packages should've already been installed. Use

[code]\usepackage{amssymb}[/code]

to gain access to AMS symbols.

I start all my TeX files with this (after the documentclass, which varies):

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{epsf}
\usepackage{graphicx}
\usepackage{mathrsfs}
\usepackage[all]{xy}

is there a place that will explain what all the packages do?


Because this problem is given to manlio so I just put a hint on it!
[hide]
$ \frac {1}{2} \left( \sum_{cyc}\frac {1}{ab} \right) \ge \sum_{cyc} \frac {1}{c^2 \plus{} ab}$[/hide]

[quote="ductrung"]For all nonnegative real numbers $ a$, $ b$ and $ c$, no two of which are zero,
\[ \frac {a(b \plus{} c)(a^2 \minus{} bc)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)(b^2 \minus{} ca)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)(c^2 \minus{} ab)}{c^2 \plus{} ab}\ge 0
\]
[/quote]
Setting $ A \equal{} LHS,$ then we see that
$ A \equal{} \sum \frac {a^3(b \plus{} c)}{a^2 \plus{} bc} \minus{} abc\sum \frac {b \plus{} c}{a^2 \plus{} bc} \ge \sum \frac {a^3(b \plus{} c)}{a^2 \plus{} bc} \minus{} abc\sum \frac {1}{a}\equal{} \sum \frac {a^3(b \plus{} c)}{a^2 \plus{} bc} \minus{} \sum \frac {a(b \plus{} c)}{2} \equal{} \frac {1}{2} \sum \frac {a(b \plus{} c)(a^2 \minus{} bc)}{a^2 \plus{} bc} \equal{} \frac {1}{2}A,$
which shows that $ A \ge 0.$ :)

The following inequality holds too:
\[ \frac {a(b \plus{} c)(a^2 \minus{} bc)}{b^2 \plus{} c^2} \plus{} \frac {b(c \plus{} a)(b^2 \minus{} ca)}{c^2 \plus{} a^2} \plus{} \frac {c(a \plus{} b)(c^2 \minus{} ab)}{a^2 \plus{} b^2}\geq 0
\]
but it is easier than original.   :)

Thank you very much everyone. :)

Nice problem and solution. :)


MATHCOUNTS
geometry
geometric transformation
reflection
invariant
rotation
rectangle

First off, I apologize to the member known as the future because I was wrong on my last post about the shoelace method.  It turns out that the future's solution was correct; I simply didn't want to spam and waste a post saying "I'm sorry, you're right the future."

Now, for the real problem at hand.  Today I came across an easy problem that I knew the answer to, but apparently I didn't.  My MathCounts coach graded this problem and stated an answer that I very much disagree with.

Here's the problem (the ' signs indicate the beginning and end of the problem):
'How many different arrangements are possible for 5 keys on a circular key ring?'

Now, my initial answer to this I know for a fact is wrong; but my secondary answer I believe is correct.  At first I thought it was 5!, but considering the fact that there is a circular key ring, the true answer is 24.
However, my MathCounts coach's answer key says 12 is the answer.

I only got to form my secondary answer after MathCounts practice, so tomorrow I'll talk to my coach about it.  However, to convince her that my answer of 24 is correct and the answer key's answer of 12 is false, I'll need convincing proof.

Here's the best I can do to prove my answer:
Any such problem of this type always has the solution (n-1)!, where n is the number of rings/objects that you need to place on a circular object.  Why?  Well, n! won't work.  Consider the problem I posted a few paragraphs above.  Let's call the 5 respective keys 1, 2, 3, 4, and 5.  Now, the following 5 arrangements are actually the same due to the ring being circular:
1, 2, 3, 4, 5
2, 3, 4, 5, 1
3, 4, 5, 1, 2
4, 5, 1, 2, 3
5, 1, 2, 3, 4
If you're not convinced that the above arrangements are the same, draw them out, but always have 1 in the same spot.  It actually turns out that this method of having 1 in the same spot is what derives the formula (n-1)!.  Since 1 will always be in the same spot, that means you truly have only (n-1) or (5-1)=4 keys to move around.  That would be 4! (Fundamental Counting Principle states that) or (n-1)!=(5-1)!=4!.  Also, we can see that 4!=24.
Another way to arrive at 24 is considering the five arrangements that I listed above the above paragraph.  Let's pretend that there was no 'circular ring' in this problem.  Then the answer would be n!=5!=120.  However, back on the circular ring part of the problem, the 5 different arrangements I listed up above being equal to each other.  That means every 5 out of the 120 solutions are the same.  Therefore, performing 120/5, the answer is still 24.
A third (but mathematically 'incorrect') way to prove the answer is 24 is to be annoying.  I can achieve this by asking my coach to list all 12 possiblities (because she thinks the answer is 12 due to the faulty answer key).  Then, I will list 12 more possibilities, proving that 24 is the correct answer.

The above 3 ways requires some thought and is not easy to convey by typing, and therefore saying it orally will be much harder.  I want to have the most convincing argument that spends the least amount of time so I can prove that the answer is 12.

Thanks for the help.

You can flip the key chain to form the same arrangement, so you have to divide by 2.

That's what a lot of people said, alanchou.  But a lot of people also said you can't flip the chain.

I'm not really sure, but from a standard point-of-view, the question DOES say 'how many different arrangements,' implying that you can't do whatever you want to the thing.  You're saying that in a key ring of rings 1, 2, and 3, 123 as well as 132 are the same thing.  However, I disagree.  You have to manually replace the 2 and the 3 with each other, not flip it.

Think about basic geometry.  A term that you should be familiar with (if you're not go look it up) is 'invariant.'  One of the most basic facts about reflections is that the orientation of the resulting figure is NOT invariant to the original figure.  Therefore the figures are not the same.  If you perform a rotation or a translation, then the figures are invariant.  Think about vectors.  Draw a triangle with 3 arrows going from vertex to vertex either clockwise or counterclockwise.  If you reflect such a figure or 'flip' it, as you say, the direction of the arrows becomes reversed (counterclockwise rather than clockwise, and vice versa).  This is proof that the figure does NOT remain the same.

I definitely understand where you're going with the flipping thing, but flipping changes the figure's orientation and therefore distorts the entire thing.  Saying you can 'flip' the object is like saying you a diagonal of a rectangle is a line of symmetry for that rectangle; even though the resulting figures are congruent, they are definitely not the same in terms of order.

I propose one final reason for why you can't 'flip;' this reason is the best I can think of and I don't know how one can argue with it.  Let's say you paint the front of each key with a number; 1, 2, 3, 4, and 5.  Now, assuming that you have extremely good hands and timing, when you flip the key ring EXACTLY, the keys will not only be in different order, but you won't even see the numbers (because the key has been flipped).  How can you say that 1 2 3 4 5 is the same as (blank) (blank) (blank) (blank) (blank).
Have 5 people hold on to a tug-of-war rope and make a circle.  If you attempt to 'flip' the rope without having anyone let go, you'll find it's physically impossible unless you make the people stand on their heads!  How can the order of people standing on their heads be counted as the same order of the people standing right side up?

Actually, in your first example, just flip the key ring 1 2 3, and you should be able to get 1 3 2 :wink:

heres what i think:
first, if you have aops #1, then look in the index under counting, then circular counting.
                       A----------------------- A

            B                       C              is the same as C                             B.
This actually is fliiping but it still is on the same keychain.
Just try it yourself. If you have 3 keys on a keychain, there is only one way to do that. A,B,C. As above, just flip the keychain around. Even if you manually do it, this is still the same keychain. I believe that a different keychain is if you literally take keys off and put them in a different order. I think that is what the problem means. In this question, the question and questions like this (circular arrangments) just mean that you can manually flip it. You also could think about it this way. Put the keychain on a glass table and look on the bottom of it. I completley agree with the rest of the things you said about basic geometry, but in this question, i think you can manually flip it.

Ugh...
I suppose you guys are right in a way.

alanchou, your answer was completely irrelevant.  I typed those thousand words in my reply saying that you're not ALLOWED to flip it, so your answer makes absolutely no sense at all if you just took a few seconds to read the first sentence of my post.

pi guy, I think you're right.

All this flipping business is too troublesome.  The problem really should specify whether you can flip or not, because I've actually seen more people say you can't flip than the people who said you can flip.

My dad has a Korean math book that he used as a high school student, and it has this exact type of problem, and says that the answer is always (n-1)!.  Perhaps it's because Korea and the U.S. are two different countries, and so the way of looking at the problem can be different.

I really wish the problem specifically said something like 'flipping is allowed' or '123 and 132 are the same,' because all of the people in the world will look at it either one way or another; there isn't really a right answer except the one that you force to be the right answer.

[quote="MoreSubtleThan1+1"]
alanchou, your answer was completely irrelevant.  I typed those thousand words in my reply saying that you're not ALLOWED to flip it, so your answer makes absolutely no sense at all if you just took a few seconds to read the first sentence of my post.
[/quote]

If you take a look at one of your last sentences, where you said everyone looks at things differently, then you should know why I posted that.

Also, if the flipping business is troublesome, learn which problems include it and which problems don't.  It'll make yours, and ours, lives a whole lot simpler.

[quote="mathcrazed"]Please stop insulting other people... Saying their post was completely irrelevant is unnecessary.  In the other topic, you also said that the future was "flat-out wrong" and that he should check his math before posting it.  You don't put down other people for no reason.  If you don't agree with them, whatever.  You're asking for help, and everyone could have just as easily avoided giving you it.  Just be satisfied that he even bothered to reply.[/quote]

stop insulting MoreSubtleThan1+1, that was a little harsh.  :P


10.2 Prove for all $x>0$ and $n\in\mathbb{N}$ the following inequality \[1+x^{n+1}\geq \frac{(2x)^n}{(1+x)^{n-1}}.\] ([i]A. Khrabrov[/i])

This is an application of holder's inequality.
Since
$(1+x^{n+1})^{\frac{1}{n}}(1+x)^{\frac{1}{n}}...(1+x)^{\frac{1}{n}} \geq 1+x^2$
and
$ 1+x^2 \geq 2x$.

Indeed, it is only second problem in the first day. It should be quite easy ;)

Another thought is induction. For n=0 we have x+1=>x+1.

  We only have to prove that x^(n+3)  - x^(n+2) - x +1 =>0
   or that (x^(n+2)  -  1) * (x-1)=>0 which is obvious!!! (the equality holds iff x=1)

JJ-THE-KING


geometry
vector
analytic geometry
trapezoid
algebra
system of equations
geometry solved

Dear  all,

 In trying to solve geometry problems by vector method i found the follwing problem: (like G2-Imoshort list 1994 but it's not)

 Problem: Let ABCD  be a quadrilateral such that BC//AD . We call M is the midpoint of CD . P,Q are repectively the midpoints of BM and AM. Let N be the intersection of the lines CP and DQ. Prove that N are in the interior or sides of the triangle ABM if and only if
         $\frac13\leq\frac{BC}{AD}\leq 3$.

I bet i could show this by using vector method. You could? Or do you have another method? Tell me plz!

 Enjoin with the problem

http://www.kalva.demon.co.uk/short/soln/sh94g2.html

[color=red][Moderator edit: Of course, this doesn't solve the problem.][/color]

I have solved the problem using barycentric coordinates. Since it is most practical to have the triangle which one uses as base triangle for the barycentric coordinates denoted by ABC, I slightly change the notations of the problem:

[color=blue][b]Problem.[/b] Let ABXY be a convex trapezoid such that BX || AY. We call C the midpoint of its side XY, and we denote by P and Q the midpoints of the segments BC and CA, respectively. Let the lines XP and YQ intersect at a point N. Prove that the point N lies in the interior or on the boundary of triangle ABC if and only if $\frac13\leq\frac{BX}{AY}\leq 3$.[/color]

[i]Solution.[/i] We will use homogeneous barycentric coordinates (barycentrics) with respect to triangle ABC. First we look for a reasonable way to describe the barycentrics of the points X and Y.

Let R be the midpoint of the segment AB. Since the quadrilateral ABXY is a trapezoid with BX || AY, the line joining the midpoints C and R of its legs XY and AB is parallel to its bases BX and AY. In other words, CR || BX || AY. Hence, the lines CR, BX and AY share a common infinite point. Denote this infinite point by T; then, this point T is the point of intersection of the line CR with the line at infinity.

Let $\left(x_T:y_T:z_T\right)$ be the homogeneous barycentric coordinates of the point T with respect to triangle ABC.

Since the point R is the midpoint of the side AB of triangle ABC, it has the homogeneous barycentric coordinates $\left(x_R:y_R:z_R\right)=\left(1:1:0\right)$. Since the point T lies on the line CR, the points T and R lie on one line through the point C; thus, $x_T:y_T=x_R:y_R$. Consequently, $x_T:y_T=1:1$, so that $x_T=y_T$. On the other hand, since the point T lies on the line at infinity, the sum of its barycentric coordinates equals zero: $x_T+y_T+z_T=0$. Solving the system of equations $x_T=y_T$ and $x_T+y_T+z_T=0$, we get $x_T:y_T:z_T=1:1:\left(-2\right)$; hence, the homogeneous barycentric coordinates of the point T are $\left(x_T:y_T:z_T\right)=\left(1:1:-2\right)$.

Since the line BX passes through the point T, the points X and T lie on one line through the point B; hence, if $\left(x_X:y_X:z_X\right)$ are the homogeneous barycentrics of the point X, then $z_X:x_X=z_T:x_T$. Thus, $z_X:x_X=\left(-2\right):1$.

On the other hand, let the line XY meet the sideline AB of triangle ABC at a point U (in the case XY || AB, this point is infinite). Let p = BX and q = AY; then, by Thales, we have $\frac{UB}{UA}=\frac{BX}{AY}=\frac{p}{q}$. Thus, with directed segments, $\frac{BU}{UA}=-\frac{UB}{UA}=-\frac{p}{q}=\frac{-p}{q}$. Hence, the homogeneous barycentric coordinates of the point U with respect to triangle ABC are $\left(x_U:y_U:z_U\right)=\left(-p:q:0\right)$.

Since the points X and U lie on one line through the point C, we have $x_X:y_X=x_U:y_U$. Thus, $x_X:y_X=\left(-p\right):q$. Solving the system of equations $z_X:x_X=\left(-2\right):1$ and $x_X:y_X=\left(-p\right):q$, we get $x_X:y_X:z_X=\left(-p\right):q:\left(2p\right)$. Thus, the homogeneous barycentrics of the point X are

$\left(x_X:y_X:z_X\right)=\left(-p:q:2p\right)$.

On the other hand, since the point P is the midpoint of the side BC of triangle ABC, it has the homogeneous barycentrics

$\left(x_P:y_P:z_P\right)=\left(0:1:1\right)$.

Hence, the line XP, joining these two points X and P, must have the homogeneous line barycentrics

$\left(k_{XP}:l_{XP}:m_{XP}\right)=\left(y_Xz_P-z_Xy_P:z_Xx_P-x_Xz_P:x_Xy_P-y_Xx_P\right)$
$=\left(q\cdot 1-2p\cdot 1:2p\cdot 0-\left(-p\right)\cdot 1:\left(-p\right)\cdot 1-q\cdot 0\right)=\left(q-2p:p:-p\right)$.

Similarly, the line YQ has the homogeneous line barycentrics

$\left(k_{YQ}:l_{YQ}:m_{YQ}\right)=\left(q:p-2q:-q\right)$.

Thus, the point of intersection N of these lines XP and YQ has the homogeneous barycentrics

$\left(x_N:y_N:z_N\right)=\left(l_{XP}m_{YQ}-m_{XP}l_{YQ}:m_{XP}k_{YQ}-k_{XP}m_{YQ}:k_{XP}l_{YQ}-l_{XP}k_{YQ}\right)$
$=\left(p\cdot\left(-q\right)-\left(-p\right)\cdot\left(p-2q\right):\left(-p\right)\cdot q-\left(q-2p\right)\cdot\left(-q\right):\left(q-2p\right)\cdot\left(p-2q\right)-p\cdot q\right)$
$=\left(-p\left(3q-p\right):-q\left(3p-q\right):-2\left(p-q\right)^2\right)=\left(p\left(3q-p\right):q\left(3p-q\right):2\left(p-q\right)^2\right)$.

Now, we distinguish three cases:

[b]Case 1.[/b] We have $\frac13\leq\frac{BX}{AY}\leq 3$. Since p = BX and q = AY, this becomes $\frac13\leq\frac{p}{q}\leq 3$, so that $\frac13q\leq p\leq 3q$. This yields $3q\geq p$ and $p\geq \frac13q$, so that $3p\geq q$. Hence, $3q-p\geq 0$ and $3p-q\geq 0$. Thus, the homogeneous barycentrics $\left(p\left(3q-p\right):q\left(3p-q\right):2\left(p-q\right)^2\right)$ of the point N are all $\geq 0$. Hence, the point N lies in the interior or on the boundary of triangle ABC.

[b]Case 2.[/b] We have $\frac{BX}{AY}<\frac13$. Since p = BX and q = AY, this becomes $\frac{p}{q}<\frac13$. Hence, 3p < q, so that 3p - q < 0. On the other hand, 3p < q yields p < 3p < q < 3q, so that 3q - p > 0. Hence, among the homogeneous barycentrics $\left(p\left(3q-p\right):q\left(3p-q\right):2\left(p-q\right)^2\right)$ of the point N, the first one is > 0, the third one is $\geq 0$, but the second one is < 0. Hence, the point N lies outside of the triangle ABC; more precisely, the points N and B lie in different halfplanes with respect to the line CA.

[b]Case 3.[/b] We have $3<\frac{BX}{AY}$. Since p = BX and q = AY, this becomes $3<\frac{p}{q}$. Hence, 3q < p, so that 3q - p < 0. On the other hand, from 3q < p, it follows that q < 3q < p < 3p, so that 3p - q > 0. Therefore, among the homogeneous barycentrics $\left(p\left(3q-p\right):q\left(3p-q\right):2\left(p-q\right)^2\right)$ of the point N, the second one is > 0, the third one is $\geq 0$, but the first one is < 0. Hence, the point N lies outside of the triangle ABC; more precisely, the points N and A lie in different halfplanes with respect to the line BC.

Altogether, we see that the point N lies in the interior or on the boundary of triangle ABC if and only if $\frac13\leq\frac{BX}{AY}\leq 3$. This solves the problem.

  Darij

For a vector solution, we can use the following lemma:

Let $ABC$ be a triangle and $P$ a point in the plane such that $\overrightarrow{AP}=\alpha\overrightarrow{AB}+\beta\overrightarrow{AC}.$ Then $P$ lies in the interior or on the sides of triangle $ABC$ iff $\alpha ,\beta\geq0$ and $\alpha+\beta\leq1.$

Returning to the problem, let $\overrightarrow{AD}=a\overrightarrow{BC}.$ Note that $a>0.$ A short computation ;) shows that
\[
\overrightarrow{MN}=\frac{a\left(  3-a\right)  }{\left(  1+a\right)  ^{2}%
}\overrightarrow{MB}+\frac{3a-1}{\left(  1+a\right)  ^{2}}\overrightarrow{MA},
\]
therefore $N$ lies inside $ABM$ iff $a\left(  3-a\right)  \geq0,$ $3a-1\geq0,$
and $\frac{a\left(  3-a\right)  }{\left(  1+a\right)  ^{2}}+\frac
{3a-1}{\left(  1+a\right)  ^{2}}\leq1.$ The first two conditions give $\frac{1}{3}\leq a\leq3$ while the third one is equivalent to $\left(
a-1\right)  ^{2}\geq0,$ valid for all $a$ (this means that $N$ always lies in the halfplane delimited by $AB$ which contains the point $M$).


dates...

[quote]I just received my information on the AMC 10 today.  The February 16th date
is for makeups according to the instructions.  You most likely will need to
choose and it makes the most sense as a 9th and 10th grader to take the AMC
10; you'll score better and stand more chance of going on to the next level
and of gaining recognition.  Take the AMC 12 when you're a junior and
senior.  However, it's up to you and they're both available.  I need to know
in the next week or so, how many of you are going to take it so that I can
order the tests. [/quote]

This is what the teacher I e-mailed said.

However, being an autist myself, I have little skill in the art of persuasion. Can anyone please help me formulate a response? I need to avoid risking annoying the teacher. Should I even volunteer to pay for the second testing date?

Thanks

i had to go through the same thing last year and it ended up that my school still just took one test.  You can show your teacher

http://www.unl.edu/amc/b-registration/b1-archive/2003-2004/faq2003-04.html

You might simply point out that a lot of kids in other states take BOTH tests every year they are able to, taking one test (10 or 12) on the "A" date and the other test on the "B" date. That is quite routine and acceptable to the people at AMC headquarters.

I always pay for the AMC tests myself. So far I have been able to convince the friendly local public school to help with proctoring, but every year I have to ask again.

Good luck lining up the tests.

Thanks for the wonderful information! My reply...

Actually, it's perfectly acceptable for students to register for either AMC test on both dates, according to the AMC's website at http://www.unl.edu/amc/b-registration/b1-archive/2003-2004/faq2003-04.html. If you still would like to register for date "A," I'd be willing to pay for the costs incurred with registering for date "B".

I have actually only gathered three signatures so far (there doesn't seem to be so much enthusiasm for the AIME amongst my peers).

Thanks,

That is what the guy at my school said. Talk about depriving people of good old opportunities. So much for education at its best. Hehe.

And she makes an excuse not to register for both dates. Even if I volunteer to PAY THE FEE.

[quote]I just went over the rules and I have to give it to all students at the same
time.

On 09/23/04 6:56 PM, "Alex C" <hans_bethe@hotmail.com> wrote:

> I understand. Perhaps you could register for both dates (the latter date
> with the minimum amount of 10 tests) and I could pay the registration fee
> for the later date (there is a bus from Inglemoor to Northshore so I could
> just show up at NJH sometime soon).
>
> Thanks
>
> -Alex C
>
[/quote]

Can students simply register for the AMC 10 themselves? I'm tired of the pubic school bureacracy.

I've just come up with a serendipity, after taking the PSAT at another school. Could I try to take the AMCs at another school? Maybe I could e-mail the math teachers of every single nearby school (I've done that for the PSAT and was successful). That would be great!


function
number theory unsolved
number theory

Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a multiplicative function, i. e. $f(ab)=f(a)f(b)$ whenever $(a, b)=1$. And let $g(n)=\sum_{d|n}f(d)$. Prove or disprove that $g$ is multiplicative too.

$\displaystyle g(ab)=\sum_{d\mid ab}f(d)=\sum_{d_1\mid a, d_2\mid b}f(d_1d_2) = \sum_{d_1\mid a, d_2\mid b}f(d_1)f(d_2) = \sum_{d_1\mid a}f(d_1)\sum_{d_2\mid b}f(d_2)$

Generalized if f,t multiplicative (for all (a,b)=1,f(ab)=f(a)f(b),t(ab)=t(a)t(b)), then $g(x)=\sum_{d|x}f(d)t(\frac xd )$ is multiplicative. You consider $t(x)=1$ for all x.

I think $g(x)=\sum_{d\leq x}{f(d)t(\frac{x}{d})}$.  where $f(x)=0$ for all  $0<x<1$


Prove: If the sum of all positive divisors of $ n \in \mathbb{Z}^{\plus{}}$ is a power of two, then the number/amount of the divisors is a power of two.

lemma:if $ (a^n\minus{}1)\equal{}2^k(a\minus{}1)$ then $ n$ is a power of two.
suppose the inverse so exist odd prime $ p$ that$ p|n$ so $ (a^p\minus{}1)|(a^n\minus{}1)\equal{}2^k(a\minus{}1)$ so $ a^{p\minus{}1}\plus{}....\plus{}a \plus{}1$ is power of two .hence p is odd its impossible.so n is a power of two.
we now that if $ n\equal{}\prod p_{i}^{\alpha_{i}}$ then the sum of divisor of n is $ \prod (p_{i}^{\alpha_{i}}\plus{}...\plus{}p\plus{}1)$
because of lemma we have $ \alpha_{i}\plus{}1$ is power of two.so the number of divisor is power of two.

[hide="Solution"]
Decompose $n: \ \prod_i p_i^{e_i}$; then $\sigma(n) = \prod_i\left(\sum_{j=0}^{e_i} p_i^{j}\right)$
Immediately, we see that $p_i\not=2$, hence each $p_i$ is odd $(1)$.
Also, it is clear that \[\sigma(n) = 2^k \implies \sum_{j=0}^{e_i} p_i^{j}=2^{k_i}\]
Now $e_i+1$ is a power of two; [hide="indeed"]suppose for contradiction that it wasn't. Then pick an odd prime $q|e_i+1$:
\[2^{k_i}=\sum_{j=0}^{e_i} p_i^{j}=\left(\sum_{j=0}^{q-1} p_i^{j}\right)\left(\sum_{j=0}^{\frac{e_i+1}{q}-1} p_i^{qj}\right)\]
However, this yields a contradiction, since the quantity $\left(\sum_{j=0}^{q-1} p_i^{j}\right)$ is the sum of an odd number of odd terms (by $(1)$), and hence clearly odd.
[/hide]
Lastly, it is clear to see from above (and well-known) that the \[\text{number of divisors } = \prod_i (e_i+1)\]
which must be a power of two, since each $e_i+1$ is a power of two.
[/hide]


[color=red]ΑΝΙΣΟΤΗΤΑ ΑΡΙΘΜΗΤΙΚΟΥ-ΓΕΩΜΕΤΡΙΚΟΥ ΜΕΣΟΥ[/color]

Έστω $a_1,a_2,.....,a_n>0$  πραγματικοί αριθμοί .Τότε ισχύει

$\frac{a_1+a_2+....+a_n}{n}\geq\sqrt[n]{a_1a_2.....a_n}$.

[color=blue]Απόδειξη[/color]
Θα δώσω μια όχι και τόσο κλασσική απόδειξη .Η ανισότητα είναι ομογενής άρα μπορούμε να υποθέσουμε ότι $a_1a_2.....a_n=1$ άρα μένει νδο
$a_1+a_2+....+a_n\geq n$. Θα χρησιμοποιήσουμε επαγωγή .
Για $n=2$ είναι προφανής .Υποθέτουμε ότι ισχύει για $n=k$ θδο ισχύει και για $n=k+1$

Έχουμε $a_1....a_{n+1}=1$ υποθέτουμε ότι $a_1\geq 1\geq a_2$.
Άρα έχουμε ότι $(a_1-1)(a_2-1)\leq 0$   ή  $a_1a_2+1\leq a_1+a_2$  .

Τώρα από την υπόθεση της επαγωγής έχουμε ότι  $(a_1a_2)a_3.....a_n=1$ έχουμε ότι
$a_1a_2+a_3+....+a_n+a_{n+1}\geq n$  άρα  $a_1+a_2-1+a_3+....+a_n+a_{n+1}\geq n$  άρα η Π(n+1) αληθής άρα το ζητούμενο

[color=Green]Άμεση συνέπεια με μια παρατήρηση [/color]
Έστω $a,b>0$  και $m,n\in N$ . Θέτουμε $x_1=x_2=....=x_m=a$ και $x_{m+1}=....=x_{m+n}$
εφαρμόζοντας την ΑΜ-ΓΜ για τους $x_i>0$ παίρνουμε ότι
$\frac{ma+nb}{m+n}\geq (a^mb^n)^{\frac{1}{m+n}}$

Άρα για ρητούς  $w_1,w_2$ με $w_1+w_2=1$ έχουμε ότι
$w_1a+w_2b\geq a^{w_1}b^{w_2}$

[color=Green]Γενίκευση  άμεσης συνέπειας(ή Ανισότητα Βαρών)  [/color]
Έστω $w_1,....,w_n>0$  και $x_1+....+x_n=1$. Τότε  για κάθε $x_1,...,x_n>0$ έχουμε $w_1x_1+....+w_nx_n\geq w_1^{x_1}.....w_n^{x_n}$

[color=Cyan]Υπόδειξη   [/color]
Εφαρμόζουμε την weighted Jensen για τη συνάρτηση $ln x$

ΑΣΚΗΣΕΙΣ
1. Αν $x,y,z>0$  βρείτε το ελάχιστο της παράστασης

$\frac{x^5+y^5+z^5+3}{x+y+z}$

2.Αν $x,y,z>0$  δείξτε ότι
$3(x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)\geq xyz(x+y+z)^3$

3.Αν $x_1,.....,x_n>0$ τέτοιοι ώστε $\frac{1}{x_1}+....+\frac{1}{x_n}=n$
να βρείτε την ελάχιστη τιμή της παράστασης
$x_1+\frac{x_2^2}{2}+\frac{x_3^3}{3}+....+\frac{x_n^n}{n}$

4.Αν $x_1,.....,x_n>0$  με γινόμενο  1  να δείξετε ότι
$n^n\cdot \prod_{i=1}^n (x_i^n+1)\geq (\sum_{i=1}^n x_i+\sum_{i=1}^n \frac{1}{x_i})^n$

5. Αν $x_1,.....,x_n>0$  με γινόμενο  1  τότε δείξτε ότι
$\frac{1}{n-1+x_1}+\frac{1}{n-1+x_2}+.....+\frac{1}{n-1+x_n}\leq 1$

6. Αν $a,b,c>0$ τέτοιοι ώστε $21ab+2bc+8ca\leq 12$   να βρείτε το ελάχιστο της παράστασης  $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$

καλη παρουσιαση!

Καλο θα ηταν να στελνουμε και τις λυσεις

Ναι καλό θα ήταν .Θέλετε να τις βάλλω εγώ τις λύσεις ή θα το προσπαθήσετε .
Νομίζω ότι αυτή τη φορά το παράκανα λίγο με τις  ασκήσεις αφού είναι ως επί το πλείστον  <<tricky>>.

Ρε  Σιλουανέ  βάλε  την  λύση  μόνο  στην  3 και  στην  5  οι  άλλες  είναι  ποιο  έυκολες  :lol:  :lol:

Η [b]5. [/b]έχει ένα τυπογραφικό λαθάκι ή  όχι;

 [u]babis[/u]

Naι έχει ένα τυπογραφικό λαθάκη το διόρθωσα τώρα. Γιώργο δε νομίζω ότι οι ασκήσεις 4,6 είναι εύκολες. ;) Όσο για αυτές που μου είπες θα τις βάλλω το απόγευμα

Λύση για 3:

  Χρησιμοποιώντας ανισότητα Bernulli προκύπτει ότι:

   (x^k) / k =( (x-1 + 1)^k ) / k >= ( 1 + k(x-1) ) / k = 1 / k  +  x-1

  Άρα κάνοντας χρήση της πάνω έχω ότι η παράσταση είναι μεγαλύτερη ή ίση από

    χ1 + χ2 + .... + χn  -  n  +  1 +  1/2  + 1/3  +  1/4  + .... + 1/n.

   Όμως από AM-HM έχω ότι    χ1 + χ2 + ..... + χn >=  n   και άρα
   η παράσταση είναι μεγαλύτερη ή ίση από   1 + 1/2 + ..... + 1/n

   H ελάχιστη τιμή είναι αυτή καθώς για χ1=χ2=...=χn=1 παίρνει εκείνη την τιμή!

Η 5 όντως είναι δύσκολη. Ρε παιδιά προσπαθήστε και το πολυώνυμο που έγραψα στους χαιρετισμούς
και δείτε και την άσκηση που έβαλα με την σκακιέρα( Προέκυψε από πρόβλημα πληροφορικής!)

Sas grafo tin lisi mou stin (5) :

Paralipo tis tetrimenes periptoseis n = 1 i 2, kai ipotheto oti n > 2.
Theto $a_i = (x_i)^{\frac{n-1}{n-2}}$. Tote $\frac{1}{n-1+x_1} = \frac{(a_2 \cdot ...\cdot a_n)^{\frac{1}{n-1}}}{(n-1) \cdot (a_2 \cdot ...\cdot a_n)^{\frac{1}{n-1}}+a_1} \leq \frac{1}{n-1} \cdot \frac{a_2 +...+a_n}{a_2 +...+a_n +a_1}$. Epeidi i sinartisi $\frac{x}{ax+b}$ einai auksousa otan b >0 kai $a_1 \cdot ... \cdot a_n =1$ .

Stergios

Στέργιο υπέροχα .Θανάση όσο και αν φαίνεται παράξενο όλες λύνονται με ΑΜ-ΓΜ ;)
ή weighted AM-ΓΜ αφού εξάλλου αυτό είναι το θέμα του τόπικ

Λοιπόν να και η λυση που έχω βρει για την  5 η οποία μοιάζει σαν στυλ με αυτά που χρησιμοποίησε ο Στέργιος.

Γράφουμε την ανισότητα στην ισοδύναμη μορφή
$\frac{x_1}{n-1+x_1}+\frac{x_2}{n-1+x_2}+......+\frac{x_n}{n-1+x_n}\geq 1$

Παρατηρούμε ότι $\frac{x_1}{n-1+x_1}\geq\frac{x_1^{1-\frac{1}{n}}}{x_1^{1-\frac{1}{n}}+x_2^{1-\frac{1}{n}}+.....+x_n^{1-\frac{1}{n}}}$

kai όμοια για τα άλλα κλάσματα. ;)
Αυτές ισχύουν αφού για παράδειγμα η πρώτη ισοδυναμεί με
$x_1^{1-\frac{1}{n}}+x_2^{1-\frac{1}{n}}+.....+x_n^{1-\frac{1}{n}}\geq (n-1)x_1^{-\frac{1}{n}}$

η οποία ισχύει από την ανισότητα ΑΜ-ΓΜ  ;)

BTW,Για εμένα όταν τις έλυνα η πιο δύσκολη ήταν η  4 αλλά καιη 6 θέλει καλό τρικ

[quote="silouan"]
ΑΣΚΗΣΕΙΣ
1. Αν $x,y,z>0$  βρείτε το ελάχιστο της παράστασης

$A=\frac{x^5+y^5+z^5+3}{x+y+z}$

[/quote]

$minA=5 \sqrt[5]{\frac{1}{4^4}}$           $??$

 :)

Ακριβώς. Όλη η δουλειά είναι να σπάσεις το 3 σε 12 φορές το $\frac{1}{4}$  και μετά απλά ΑΜ-ΓΜ ;)

Έλυσε κάποιος από εδώ τα προβλήματα 4,6 .Αν ναι βάλτε τις λύσεις .
Ευχαριστώ.

βάζω τη λύση της 6 καθώς μου ζητήθηκε και για να μη μένουν άλυτες οι ασκήσεις.

Θέτουμε $\frac{1}{a}=x$  $\frac{2}{b}=y$ ,,... $\frac{3}{c}=z$

Τότε αρκεί να βρούμε το ελάχιστο της x+y+z όταν $2xyz\geq 2x+4y+7z$

Aπό την weighted AM-GM έχουμε  $x+y+z\geq (\frac{5x}{2})^{\frac{2}{5}}(3y)^{\frac{1}{3}}(\frac{15z}{4})^{\frac{4}{15}}$

και επίσης $2x+4y+7z\geq 10^{\frac{1}{5}}\cdot 12^{\frac{1}{3}}\cdot 5^{\frac{7}{15}}\cdot x^{\frac{1}{5}}\cdot y^{\frac{1}{3}}\cdot z^{\frac{7}{15}}$

το οποίο σημαίνει ότι $(x+y+z)^2(2x+4y+7z)\geq\frac{225xyz}{2}$

επειδή όμως $2xyz\geq 2x+4y+7z$  θα έχουμε

$(x+y+z)^2\geq\frac{225}{4}$  ή  $x+y+z\geq\frac{15}{2}$

 :lol:  ;)  :)


function
integration
geometry
geometric transformation
reflection
limit
trigonometry

Let $\gamma(x)=\int_{0}^{\infty}{e^{-t}t^{x-1}dt}$. Prove that $\gamma(x)\cdot\gamma(1-x)=\frac{\pi}{sin(\pi\cdot x)}$

This is really a famous formula.

Try to use series.

It is the famous "REFLECTION RULE". The $\Gamma$ function satisfies the reflection rule.
For the proof, use the following expression for $\Gamma(x)$ and $\Gamma(1-x)$:
\[ \Gamma(x)=\displaystyle\lim_{n\to +\infty}\dfrac{(n-1)!n^x}{x(x+1)\cdots(x+n-1)}\\\qquad\\\Gamma(1-x)=\displaystyle\lim_{n\to +\infty}\dfrac{(n-1)!n^{1-x}}{(1-x)(2-x)\cdots(n-x)}\]
Then:
\[\Gamma(x)\Gamma(1-x)=\left[x\prod_{n=1}^{ +\infty}\left(1-\dfrac{x^2}{n^2}\right)\right]^{-1}=\dfrac{\pi}{sin(\pi x)}\]
remembering the factorization of $\sin(\pi x)=x\prod_{n=1}^{ +\infty}\left(1-\dfrac{x^2}{(\pi n)^2\right)}$

Since the expression here only applies if the real part of $x$ is positive, I doubt this belongs here. On the other hand, if we use that definition for the right half plane and the recursion $x\Gamma(x)=\Gamma(x+1)$ on the left half-plane, we can extrapolate this equality from the strip of real parts between 0 and 1 to the whole plane except the integers, where it obviously has simple poles.

Or let us consider the Beta function $B(p,q)=\int_0^1 x^{p-1}(1-x)^{q-1}dx$.
If we substitute $x=\frac{y}{1+y}$, then we see that
$B(p,q)=\int_0^\infty \frac{y^{p-1}}{(1+y)^{p+q}}dy$
Thus
$B(p,1-p) = \int_0^\infty \frac{y^{p-1}}{1+y}dy$

It is well-known (and it is not difficult to prove) that $B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$
So $\Gamma(p)\Gamma(1-p) = \Gamma(1)B(p,1-p)=1\cdot\int_0^\infty \frac{y^{p-1}}{1+y}dy$

The only thing left is to calculate $\int_0^\infty \frac{y^{p-1}}{1+y}dy$. It is still not easy...(what about using residues?)

[quote="alekk"]how do you prove the last equality with $sin( \pi x)$ ? I'm interested  :)
[/quote]
It is too long to write out the complete proof.

You can see your mathematical analysis book.

[quote="alekk"]how do you prove the last equality with $sin( \pi x)$ ? I'm interested  :)
And what is the "reflection rule" ?[/quote]

Start with
\[I_n(x)=\int_0^{\pi/2}cos xt\,cos^nt\,dt\]
and integrate by parts to get
\[n(n-1)I_{n-2}(x)=(n^2-x^2)I_n(x).\]
Since $I_n(0)>0$ we obtain for $n\ge 2$
\[\dfrac{I_{n-2}(x)}{I_{n-2}(0)}=\left(1-\frac{x^2}{n^2}\right)\dfrac{I_n(x)}{I_n(0)}\]
Using the values $I_0(0)=\frac{\pi}2$ and $I_1(0)=1$ we have
\[\sin \frac{\pi x}2=\frac{\pi x}2\frac{I_0(x)}{I_0(0)}=\frac{\pi x}2\left(1-\frac{x^2}{2^2}\right)\frac{I_2(x)}{I_2(0)}=\frac{\pi x}2\left(1-\frac{x^2}{2^2}\right)\cdots\left(1-\frac{x^2}{n^2}\right)\frac{I_n(x)}{I_n(0)}\]
Now
\[\lim_{n\to+\infty}\frac{I_n(x)}{I_n(0)}=1\]
Infact
\[
\left|I_n(0)-I_n(x)\right|=\left|\int_0^{\pi/2}(1-cos xt)cos^nt\,dt\right|\]
\[            \qquad\;\;\;\;\;\le\frac12x^2\int_0^{\pi/2}t^2cos^nt\,dt\]

            \[ \qquad\le\frac12x^2\int_0^{\pi/2}tcos^{n-1}t\sin t\,dt\]

              \[ \qquad\;\;\;\;=\frac1nI_n(0)\]
Now replace $\pi x/2$ by t and all is done.
About the reflection rule go to the mathematics dictionary.

Wow, that's a very nice proof of the eulerian development of sinus. Very very nice.