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Let $ f: \mathbb{R}\to\mathbb{N}$ be a function which satisfies $ f\left(x \plus{} \dfrac{1}{f(y)}\right) \equal{} f\left(y \plus{} \dfrac{1}{f(x)}\right)$ for all $ x$, $ y\in\mathbb{R}$. Prove that there is a positive integer which is not a value of $ f$.

[i]Proposed by Žymantas Darbėnas (Zymantas Darbenas), Lithuania[/i]

I have only been able to do this one by brute force.  I'll post my solution in case it gives anybody an idea on how to finesse it...

Assume that for all $ n\in\mathbb{N}$, a real $ r$ exists such that $ f(r) \equal{} n$.  We may assume wlog that $ f(0) \equal{} 1$, since for any real constant $ r$, $ f'(x) \equal{} f(r \plus{} x)$ is a solution iff $ f(x)$ is a solution because if $ f$ is a solution, then $ f'\left(x \plus{} \frac {1}{f'(y)}\right) \equal{} f\left(x \plus{} r \plus{} \frac {1}{f(y \plus{} r)}\right) \equal{} f\left(y \plus{} r \plus{} \frac {1}{f(x \plus{} r)}\right) \equal{} f'\left(y \plus{} \frac {1}{f'(x)}\right)$, and taking $ r$ such that $ f(r) \equal{} 1$, then $ f'(0) \equal{} f(r) \equal{} 1$.

Now,
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For any real $ r$ and any positive integer $ n$, a positive integer $ m$ exists such that $ f\left(r \plus{} \frac {1}{m}\right) \equal{} n$.  It suffices to take $ y \equal{} r' \minus{} \frac {1}{f(r)}$ and $ m \equal{} f(y)$, where $ r'$ is the real such that $ f(r') \equal{} n$, hence $ n \equal{} f(r') \equal{} f\left(y \plus{} \frac {1}{f(r)}\right) \equal{} f\left(r \plus{} \frac {1}{f(y)}\right) \equal{} f\left(r \plus{} \frac {1}{m}\right)$.
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Moreover, if $ f(r) \equal{} f(r')$ for any two reals $ r\neq r'$,
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then for every positive rational $ q$ we have $ f(r \plus{} q) \equal{} f(r' \plus{} q)$; write $ q \equal{} \frac {m}{n}$ where $ m,n$ are positive integers, and clearly $ f\left(r \plus{} \frac {1}{n}\right) \equal{} f\left(r'' \plus{} \frac {1}{f(r)}\right) \equal{} f\left(r'' \plus{} \frac {1}{f(r')}\right) \equal{} f\left(r' \plus{} \frac {1}{n}\right)$, where $ r''$ is the real such that $ f(r'') \equal{} n$.  So the result is true when $ m \equal{} 1$, and by trivial induction it is true for all $ m$.  In particular, if $ f(q) \equal{} 1$ for any rational $ q$, then $ f(nq) \equal{} 1$ for all positive integers $ n$, and since a positive integer $ m$ exists such that $ f\left(0 \plus{} \frac {1}{m}\right) \equal{} 1$, then $ f\left(\frac {1}{m}\right) \equal{} 1$ for that integer $ m$, and $ f(1) \equal{} 1$, hence $ f(n) \equal{} 1$ for all positive integers $ n$.

Moreover, if $ f(r) \equal{} n$, then a positive integer $ m$ exists such that $ f\left(r \plus{} \frac {1}{m}\right) \equal{} n \equal{} f(r)$, or by trivial induction $ f(r) \equal{} f\left(r \plus{} \frac {m'}{m}\right)$ for any positive integer $ m'$, and in particular $ f(r) \equal{} f(r \plus{} n)$ for all real $ r$ and all positive integer $ n$.  Therefore, if $ r$ is the real such that $ f(r) \equal{} n$, then $ n \equal{} f(r) \equal{} f(r \plus{} 1) \equal{} f\left(r \plus{} \frac {1}{f(0)}\right) \equal{} f\left(0 \plus{} \frac {1}{f(r)}\right) \equal{} f\left(\frac {1}{n}\right)$.
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A contradiction is then found
[hide]by considering the positive integer $ n$ such that for a given odd prime $ p$, $ f\left(\frac {1}{p} \plus{} \frac {1}{n}\right) \equal{} 1$, which clearly exists.  We may then write $ \frac {1}{p} \plus{} \frac {1}{n} \equal{} \frac {m'}{m}$, where $ m,m'$ are coprime positive integers, and clearly $ m > 1$, since either $ 2 > \frac {m'}{m} > 1$ if $ n \equal{} 1$ or $ 1 > \frac {m'}{m} > 0$ if $ n > 1$.  It is well known that positive integers $ u,v$ exist such that $ um' \minus{} vm \equal{} 1$, or $ m \equal{} f\left(\frac {1}{m}\right) \equal{} f\left(\frac {um'}{m} \minus{} v\right) \equal{} f\left(u\frac {m'}{m}\right) \equal{} 1$, contradiction.
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After playing around for quite a while with the function, I have not been able to find any "smart shortcuts" to this solution, and finding the correct order of proving partial results was quite hard for me.  The idea of playing around with reals of the form $ r\plus{}\frac{1}{n}$ where $ r$ is any real and $ n$ is a positive integer, and from there to reals of the form $ r\plus{}q$ where $ q$ is a positive rational is more or less hinted by the form of the functional relation, but the way to a contradiction was still not obvious to me after this observation...  Hopefully someone smarter will come up with a more elegant solution.   :roll:

Hopefully this is right.
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First, let $ f(1)\equal{}w$. Next, note that if p is irrational then $ f(p)\equal{}f(p\minus{}\dfrac{1}{w}\plus{}\dfrac{1}{w})\equal{}f((p\minus{}\dfrac{1}{w})\plus{}\dfrac{1}{f(1)})\equal{}f(1\plus{}\dfrac{1}{f(p\minus{}\dfrac{1}{w})})$. Since $ f(p\minus{}\dfrac{1}{w})\in\mathbb{N}$ we have $ 1\plus{}\dfrac{1}{f(p\minus{}\dfrac{1}{w})}\in\mathbb{Q}$. Thus every image of an irrational input is also the image of a rational input. Therefore we now need to only consider rational inputs (that part I am sure of but this next part I'm not so sure of). Since $ \mathbb{N}$ and $ \mathbb{Q}$ have the same size, any function from $ \mathbb{Q}$ to $ \mathbb{N}$ that is onto must also be one to one. Therefore we need only to prove that $ f: \mathbb{Q}\to\mathbb{N}$ is not one to one. This is almost obvious; but I will still prove it. Take x=1 and y=10. Since $ f(1),f(10)\in\mathbb{N}$, we know that $ f(1),f(10)\geq1$ Therefore $ 0<\dfrac{1}{f(1)},\dfrac{1}{f(10)}\leq1$ and $ 10<10\plus{}\dfrac{1}{f(1)}\leq11$ and $ 1<1\plus{}\dfrac{1}{f(10)}\leq2$, so thus $ 10\plus{}\dfrac{1}{f(1)}\not\equal{}1\plus{}\dfrac{1}{f(10)}$. But $ f(10\plus{}\dfrac{1}{f(1)})\equal{}f(1\plus{}\dfrac{1}{f(10)})$ so $ f$ is not one to one and we are done.
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[quote="Bacteria"] Since $ \mathbb{N}$ and $ \mathbb{Q}$ have the same size, any function from $ \mathbb{Q}$ to $ \mathbb{N}$ that is onto must also be one to one.[/quote]

I think this is not necessarily true; consider the function such that, for any rational $ q \equal{} \frac {m}{n}$, where $ m,n$ are coprime integers, associates $ f(q) \equal{} |n|$, except for $ f(0) \equal{} 1$.  This function is suprajective, but not one-to-one.  The problem is that bijectivity of a suprajective function holds necessarily only when the cardinal of range and image are the same and finite, but is not necessarily true for infinite sets of the same cardinality...  Otherwise, my solution would also be complete as soon as I prove that
[hide] for any real $ r$ and any natural $ n$, a positive integer $ m$ exists such that $ f\left(r\plus{}\frac{1}{m}\right)\equal{}n$; take two different rational values of $ r$ with integer part differing in at least $ 2$, and you have found two distinct rational numbers with the same image...[/hide]

Very nice problem!

Here's the solution I got:

Let's assume that there are $ x_k$'s so that $ f(x_k)\equal{}k, \forall k\in \mathbb{N}$

We can assume that $ f(0)\equal{}f(1)\equal{}1$ (An affine modification and showing that $ f^{\minus{}1}(1)$ contains at least two elements does the trick)

This gives (*) $ f(x\plus{}1)\equal{}f(\frac{1}{f(x)})$

which implies $ f(k)\equal{}1 \, \forall k\in \mathbb{N}$

also $ f(x\plus{}1)\equal{}f(\frac{1}{f(x)})\equal{}f(\frac{1}{f(\frac{1}{f(x\minus{}1)})})\equal{}f(\frac{1}{f(x\minus{}1)}\plus{}1)\equal{}f(x)$ so $ f$ is $ 1$-periodic.

this implies together with (*) $ f(x)\equal{}f(\frac1{f(x)})$ and $ f(\frac1k)\equal{}k \forall k\in \mathbb{N}$

Now if $ a,b\in \mathbb{N}$ repeated use of our functional equation gives (**) $ f(x\plus{}\frac{a}{b})\equal{}f(\frac{1}{f(x)}\plus{}\frac{a}{b})$

So if $ f(\frac34)\equal{}m$ then $ f(\frac1m\plus{}\frac14)\equal{}1$ that means we can find $ x,y \in \mathbb{N} , (x,y)\equal{}1, y\geq 2$ so that $ f(\frac{x}{y})\equal{}1$ then let $ z\equiv x^{\minus{}1}\pmod{y}$ from repeated application of (**)

we know $ f(z \frac{x}{y})\equal{}1$ and by $ 1$-periodicity that $ f(\frac{1}{y})\equal{}1$ a contradiction as $ f(\frac1{y})\equal{}y$ :)

[quote="April"]Let $ f: \mathbb{R}\to\mathbb{N}$ be a function which satisfies $ f\left(x \plus{} \dfrac{1}{f(y)}\right) \equal{} f\left(y \plus{} \dfrac{1}{f(x)}\right)$ for all $ x$, $ y\in\mathbb{R}$. Prove that there is a positive integer which is not a value of $ f$.

[i]Proposed by Lithania[/i][/quote]
Does there exist function $ f$:$ R\rightarrow N \{0,a\}$( where $ a$ are positive integer)such that condition problem
I can't slove this problem.
Can anyone help me?

Suppose for the sake of contradiction that $\mathbb{Z}^+\subseteq\{f(x)\}_{x\in\mathbb{R}}$. By translation, WLOG $f(0)=1$.

[b]Lemma:[/b] If $f(a)=f(b)$, then $f(a+r)=f(b+r)$ for all $r\in\mathbb{Q}^+$.

[i]Proof:[/i] From $P(x,a),P(x,b)$, we have
\[f\left(a+\frac{1}{f(x)}\right) = f\left(x+\frac{1}{f(a)}\right) = f\left(x+\frac{1}{f(b)}\right) = f\left(b+\frac{1}{f(x)}\right).\]If $r=p/q$, then taking $x$ such that $f(x)=q$ and repeating this $p$ times gives the desired result.$\blacksquare$

[b]Corollary:[/b] If $f(r)=1=f(0)$ for some rational $r$, then $f(k|r|)=1$ for all $k\in\mathbb{Z}^+$.

[i]Proof:[/i] Strong induction on $k$ with the lemma.$\blacksquare$

Note that $P(-1,0)\implies f(0) = f\left(\frac{1}{f(-1)}\right)$, so taking $k=f(-1)$ in the corollary we have $f(1)=f(0)=1$.

Now by the lemma,
\[P(x,0)\implies f(x+1) = f\left(\frac{1}{f(x)}\right)\implies f\left(x-\frac{1}{f(x)}+2\right) = f(1) = 1.\]But in the same way,
\[P\left(x-\frac{1}{f(x)}+2,0\right)\implies f\left(x-\frac{1}{f(x)}+2-\frac{1}{f(1)}+2\right) = 1\]as well, so again by the lemma,
\[f(x+2) = f\left(x+4-\frac{1}{f(1)}\right) = f(x+3)\]for all $x\in\mathbb{R}$. Hence $f$ has a period of $1$, so
\[P(x,0)\implies f(x) = f(x+1) = f\left(\frac{1}{f(x)}\right)\implies n = f\left(\frac{1}{n}\right)\]for all positive integers $n$ (by assumption). Furthermore, the lemma is now obviously true for all $r\in\mathbb{Q}$ (not necessarily positive).

If $f(p/q)=1=f(0)$ where $p,q$ are coprime nonzero integers, then taking $kp\equiv1\pmod{q}$ in the corollary yields $q=f(1/q)=1$, i.e. $p/q$ must be an integer. Therefore
\[P(x,0)\implies f(x+1) = f\left(\frac{1}{f(x)}\right)\implies f\left(x-\frac{1}{f(x)}\right) = 1\implies x-\frac{1}{f(x)}\in\mathbb{Z}\]for all $x\in\mathbb{Q}$. If $x=p/q$ with $\gcd(p,q)=1$ and $q>0$, then it's easy to show that $q|f(x)$ and $f(x)|q$, whence $f(x)=q$. But then $f(1/3)=3=f(2/3)\implies 1=f(0)=f(1/3)=3$ by the lemma, a contradiction.

It is easy to see $f(x)=f(y) \Rightarrow f(x+r)=f(y+r)$ for any $r>0$ rational. Now, let's view $f$ as a surjective function $\mathbb{Q} \rightarrow \mathbb{R}$. It is surjective because if $f(a)=n$ and $a$ isn't rational, then take any $b=a-(1/f(y))$ for any rational $y$ and we get $n=f(a)=f(b+(1/f(y)))=f(y+(1/f(b)))$ and $y+(1/f(b))$ is rational. From now on I speak only about $\mathbb{Q}$.

Now assume $f(x)=f(x+r)$ for all sufficiently big $x$ (say $x>N$) and $r$ is fixed (call such an $r$ good). Then take any $q \le N$ and any $y>N$ and write $q=x+(1/f(y))$ and notice $z:=y+(1/f(x))>N$. Therefore, by our original observation, $f(q+r)=f(z+r)=f(z)=f(q)$ and so $f(x)=f(x+r)$ for all $x$, even if $x<N$. Now I prove that there exists some $n \in \mathbb{N}$ such that $r=1/n$ satisfies this.

Indeed, let $y=x-(1/f(x))$ for any $x$, and notice how $f(x+1/f(y))=f(x)$ and therefore for any $q>0$, $f(x+1/f(y)+q)=f(x+q)$ and if $r=1/f(y)$ then for any $x'>x$ we have $f(x')=f(x+q)=f(x+1/f(y)+q)=f(x'+r)$ and so for sufficiently big $x$ the equation holds and thus it holds for all $x$. Therefore, if $n=f(y)$, we have $f(x+1/n)=f(x)$ for all $x$. Notice $1/n$ is good.

Now let $k$ be any integer and let $f(x)=k$ for some $x$, then $f(y+1/k)=f(y+1/f(x))=f(x+1/f(y))=f(x+1/n)=f(x)=k$. Thus $f(y+1/k) \neq f(y+1/l)$ for any $y \neq l$, and thus $1/k-1/l$ is not good for any $k,l$. But $k=n-1, l=n(n-1)$ implies $1/k-1/l=1/n$ which is good. Thus $n=1$, and so the only good numbers are integers.

But by our first result, $x-(1/f(x))-(y-(1/f(y)))$ is good always, and thus $x-y=n+(1/f(x))-(1/f(y))$ for some $n$. So $x-y=n+1/m-1/k$ for some $n,m,k$. So any rational number is of the form $n+1/m-1/k$, which is clearly false. Take, for example $1/2-\epsilon$, for a really small positive $\epsilon$. Thus, we are done.

[hide=Solution]Let $P(x, y)$ denote assertion. Assume this function is surjective. It is easy to check that translating this function to the right by $c$ still keeps $P$ true, as $P(x-c, y-c)$ is true in the translated function if $P(x, y)$ is true in the original function. Thus, we can assume that $f(0)=0$.

Lemma 1: If $f(x) = f(y)$, then $f(x+m) = f(y+m) \forall m \in \mathbb{Q}^+$.
Proof: Let $m = \frac{p}{q}, p, q \in N$. Then, choose $a$ such that $f(a) = q$. Now,
$P(a, x) \implies f(a+\frac{1}{f(x)}) = f(x+\frac{1}{q})$
$P(a, y) \implies f(a+\frac{1}{f(y)}) = f(y+\frac{1}{q})$
Now, $f(x)=f(y) \implies f(a+\frac{1}{f(x)}) = f(a+\frac{1}{f(y)})$, so $f(x+\frac{1}{q}) = f(y+\frac{1}{q})$. Now, we can repeat this $p$ times to get $f(x+\frac{p}{q}) = f(y+\frac{p}{q})$.

Lemma 2: $f$ is periodic with period $1$
Proof: $P(x, x-\frac{1}{f(x)}) \implies f(x+\frac{1}{f(x-\frac{1}{f(x)})}) = f(x)$.
Now, let $f(x-\frac{1}{f(x)}) = q$. Then, $f(x) = f(x+\frac{1}{q}) = f(x+\frac{2}{q}) = \cdots = f(x+1)$ by Lemma 1.

Lemma 3: For all $x$, there exists $y \in \mathbb{Q}$ such that $f(y) = f(x)$
Proof: $P(0, x-1) \implies f(x) = f(\frac{1}{f(x-1)})$.

Thus, it suffices to prove this prove this problem for domain $\mathbb{Q}$. For the rest of the solution, we assume the domain is $\mathbb{Q}$.

Lemma 4: If $f(x) = f(y)$, then $f$ is periodic with period $y-x$
Proof: Choose $m \in \mathbb{Q}$. We show that $f(x+m) = f(y+m)$. Choose $k \in \mathbb{Z}$ such that $k < m$. Now, note that $f(x+k) = f(y+k)$, and by lemma $1$, since $m > k$, we must have $f(x+m) = f(y+m)$.

Lemma 5: If $f$ is not periodic with period $\frac{1}{p}$ for prime $p$, then $v_p(f(\frac{m}{n})) = v_p(n)$ where $\frac{m}{n}$ is in lowest terms.

Proof: Suppose otherwise. Then, we do
$P(0, x) \implies f(x+1) = f(\frac{1}{f(x)})$, and since $f(x+1) = f(x)$ by Lemma 2, $f(x) = f(\frac{1}{f(x})$. Thus, $f$ has period $x-\frac{1}{f(x)} \forall x$.

Now, we claim that in lowest terms, $x-\frac{1}{f(x)}$ has a denominator divisible by $p$ if $v_p(D(x)) \not = v_p(f(x))$, where $D(x)$ denotes the denominator of $x$ in lowest terms.

We show this by let $x = \frac{m}{n}$ in lowest terms and $f(x) = i$. Then $x-\frac{1}{f(x)} = \frac{m}{n} - \frac{1}{i} = \frac{mi-n}{ni}$. Now, note that $p \nmid m$, if $p|n$, so if $p|n$, then $v_p (D(\frac{mi-n}{ni})) = v_p(ni)-min(v_p(i), v_p(n)) = v_p(n) + v_p(i) - min(v_p(i), v_p(n)) = max(v_p(i), v_p(n)) \geq 1$(since $v_p(n+i) = min(v_p(i), v_p(n))$ if $v_p(n) \not = v_p(i)$. If $p \nmid n$ and $p \mid i $, then $p \nmid mi-n$, so $v_p(D(\frac{mi-n}{ni}) = v_p(i) \geq 1$. Thus, we must have a period that when written in lowest terms, has denominator divisible by $p$. Call the period $\frac{s}{t}$. Now, note that by bezout’s lemma, we can find $a, b \in \mathbb{Z}$ such that $as-bt = 1$. Now, note that $f(x) = f(x+\frac{as}{t})$ by repeatedly adding the period. Then, by Lemma 2, $f(x+\frac{as}{t}) = f(x+\frac{as-bt}{t}) = f(x+\frac{1}{t})$. Thus, $f$ has period $\frac{1}{t}$, and since $p|t$, $f$ also has period $\frac{1}{p}$.

Lemma 6: $f$ is not periodic with period $\frac{1}{p}$ for any prime $p$
Proof: Suppose otherwise. Then, choose $x$ such that $f(x) = p$.
Now, $P(0, x) \implies f(\frac{1}{p}) = f(x+1)$. SInce $f$ has period $\frac{1}{p}$, $f(\frac{1}{p}) = f(0) = 1$, and $f(x+1) = f(x) = p$, so we must have $1 = p$, contradiction.

Thus, by Lemmas $5$ and $6$, $f(x) = D(x) \forall x \in \mathbb{Q}$(where $D(x)$ denotes the denominator of $x$ in lowest terms). However, this function simply does not work, since
$P(\frac{2}{5}, \frac{1}{10}) \implies f(\frac{1}{2}) = f(\frac{3}{10})$, or $2 = 10$, a contradiction.
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Suppose for the sake of contraddiction that $f$ is suriective.
The invariance by traslation along $x$ axis, and suriectivity let us setting $WLOG$ $f\left(0\right)=1$.
Let $P\left(x,y\right)$ be the equality in the text.
Lemma 1: for all $k\in\mathbb{N}$ we have that $$f\left(j\right)=f\left(j+\frac{k}{f\left(j-\frac{1}{f\left(j\right)}\right)}\right)\ \ \forall\ j\in\mathbb{R}$$
We show this by induction on $k$. The base case $k=0$ is trivial. Now suppose the lemma 1 true for $k\le t$. So we have that $$f\left(j\right)=f\left(j-\frac{1}{f\left(j\right)}+\frac{1}{f\left(j\right)}\right)\stackrel{\text{inductive hp.}}{=}f\left(j-\frac{1}{f\left(j\right)}+\frac{1}{f\left(j+\frac{t}{f\left(j-\frac{1}{f\left(j\right)}\right)}\right)}\right)=f\left(j+\frac{t}{f\left(j-\frac{1}{f\left(j\right)}\right)}+\frac{1}{f\left(j-\frac{1}{f\left(j\right)}\right)}\right)=f\left(j+\frac{t+1}{f\left(j-\frac{1}{f\left(j\right)}\right)}\right)$$
Where the third equality is obtained by $P\left(j-\frac{1}{f\left(j\right)},\ j+\frac{t}{f\left(j-\frac{1}{f\left(j\right)}\right)}\right)$.

Now putting $k=f\left(j-\frac{1}{f\left(j\right)}\right)$ in the lemma 1 we have that $f\left(j\right)=f\left(j+1\right)$, so $f$ is periodic with period $1$.

By $P\left(x,\ 0\right)$ we have $f\left(x\right)=f\left(1+\frac{1}{f\left(x\right)}\right)\stackrel{\text{periodicity}}{=}f\left(\frac{1}{f\left(x\right)}\right)$, and by suriectivity we can put $f\left(x\right)=z$, so $f\left(\frac{1}{z}\right)=z\ \ \forall\ z\in\mathbb{Z}^+$.

Lemma 2: if there are $a,b\in\mathbb{Q}$ ($WLOG$ $a<b$) such that $f\left(a\right)=f\left(b\right)$, then $f$ is periodic with period $b-a$ in the domain $\mathbb{Q}$.
As math154 did, comparing $P\left(x,a\right)$ and $P\left(x,b\right)$ we have that $$f\left(a+\frac{1}{f(x)}\right)=f\left(b+\frac{1}{f(x)}\right)$$and so inductively $$f\left(a+\frac{n}{f(x)}\right)=f\left(b+\frac{n}{f(x)}\right)\ \ \forall\ n\in\mathbb{Z}^+,\ x\in\mathbb{R}$$hence by suriectivity of $f$ $$f\left(a+\frac{n}{m}\right)=f\left(b+\frac{n}{m}\right)\ \ \forall\ n,m\in\mathbb{Z}^+$$hence by periodicity $$f\left(a-i+\frac{n}{m}\right)=f\left(b-i+\frac{n}{m}\right)\ \ \forall\ n,m\in\mathbb{Z}^+,\ i\in\mathbb{Z}$$and finally, putting $u=a-i+\frac{n}{m}$ ($u$ is a generic rational)$$f\left(u\right)=f\left(u+\left(b-a\right)\right)\ \ \ \blacksquare$$
Now let's choose two positive integers $1<p<q$ such that $\gcd\left(p,q\right)=1$. Let's say that $f\left(\frac{p}{q}\right)=h=f\left(\frac{1}{h}\right)$. So $f$ on domain $\mathbb{Q}$ has period $\left|\frac{p}{q}-\frac{1}{h}\right|=\left|\frac{ph-q}{qh}\right|$, but we also know that it has period $1$, hence it has period $\frac{1}{l}$ where $1<l=\frac{qh}{\gcd\left(qh,ph-q\right)}$, so $f\left(0\right)=f\left(\frac{1}{l}\right)$, contraddiction because $f\left(0\right)=1$ and $f\left(\frac{1}{l}\right)=l>1$.

Hopefully I didn't mess anything up  :what?:. Suppose that $f$ is surjective.

[color=red][b]Claim 1:[/b][/color] For all $x\in\mathbb{R}$,
$$\left\{f(x+1), f\left(x+\frac 12\right), f\left(x+\frac 13\right),\hdots\right\}=\mathbb{N}.$$
[i]Proof:[/i] In the functional equation, if we fix $x$ and vary $y$, we see that the right hand side varies through all positive integers, due to surjectivity. Thus the left hand side must vary through all positive integers too, hence the claim. $\blacksquare$
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Thus the restriction of $f$ onto $\mathbb{Q}$ is also surjective. Hence we will consider the functional equation only in $\mathbb{Q}$. We prove a simple claim.

[color=red][b]Claim 2:[/b][/color] If $f(a)=f(b)$, then $f(a+t)=f(b+t)$ for any $t\in\mathbb{Q}^+$

[i]Proof:[/i] For each $x$, we have
$$f\left(a+\frac{1}{f(x)}\right) = f\left(x+\frac{1}{f(a)}\right) = f\left(x+\frac{1}{f(b)}\right) = f\left(b+\frac{1}{f(x)}\right).$$
By choosing $x$ so that $f(x)=n$, we deduce that $f\left(a+\tfrac 1n\right) = f\left(b+\tfrac 1n\right)$. Repeating this argument, we get $f\left(a+\tfrac mn\right) = f\left(b+\tfrac mn\right)$ for any $m,n>0$, hence the claim. $\blacksquare$
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[color=red][b]Claim 3:[/b][/color] [Main Claim] $f$ is periodic with period $1$.

[i]Proof:[/i] Fix a large positive integer $M$. Aiding with Claim 2, we will prove that $f\mid_{\geq -M}$ is periodic with period $1$. To that $n$, it suffices to show that for each prime $p$, $f$ has a period $t$ which $\nu_p(t)\leq 0$.

Note that by Claim 1, there exists a positive integer $n$ such that
$$f(-(M+2020)) = f\left(-\left(M+2020+p+\frac 1n\right)\right)$$
thus $f$ has a period $t = p+\tfrac 1n$. Clearly $\nu_p(t)\leq 0$ so we are done. $\blacksquare$
-----
Now, we note that, for any $a\in\mathbb{Q}$, the function $g(x) = f(x+a)$ satisfies the functional equation by plugging in $(x,y)$ with $(x+a,y+a)$. Thus, WLOG, assume that $f(0)=1$. Therefore $f(1)=f(2)=\hdots=1$. Moreover,
$$f\left(\frac{1}{f(x)}\right) = f\left(x+\frac{1}{f(0)}\right) =  f(x+1) = f(x)$$
, so as $f$ is surjective, $f\left(\tfrac 1n\right) = n$ for each positive integer $n$.

Now we claim that we are unable to determine $t=f\left(\tfrac 23\right)$. From the functional equation,
$$1 = f\left(\frac{2}{3}+\frac{1}{f\left(\tfrac 13\right)}\right) = f\left(\frac 1t + \frac 13\right)$$
, so $f$ is periodic with period $\tfrac 1t + \tfrac 13 = \tfrac{t+3}{3t}$. Since $\gcd(t+3,3t)\mid 9$, $f$ is periodic with period $\tfrac{\gcd(t+3,9)}{3t}$. This is a contradiction unless $3t=\gcd(t+3,9)$. This equation has no solutions.

Assume the contrary. By translating, WLOG $f(1)=f(0)=1.$

Lemma: $f(a)=f(b)\implies f(a+q)=f(b+q)~\forall a,b \in \mathbb{R}, q\in \mathbb{Q}.$
Proof. Note that $$f(a+\frac{1}{f(x)})=f(x+\frac{1}{f(a)})= f(x+\frac{1}{f(b)})=f(b+\frac{1}{f(x)}).$$ Let $f(x)=x \in \mathbb{N}$ then by induction we are done. As a corollary we also get $f(q)=f(q+1) ~\forall q\in \mathbb{Q}.$ $\square$

Setting $y=1$ and $f(x)=x: x\in \mathbb{N}$ yields $f(x)=f(\frac{1}{x}).$ Note also that $\forall y\in \mathbb{R}, x\in \mathbb{N},$ the set containing $f(y+\frac{1}{x})$ is $\mathbb{N}.$

Finish: Let $f(\frac{1}{x}+\frac{1}{3})=1$ where $x$ is natural. Also let $\frac{1}{x}+\frac{1}{3}=\frac{a}{b},$ where $a,b$ are co prime. It is clear that that $b>1.$ But $\exists c,d \in \mathbb{N}: ac-bd=1.$ It forces $b=1,$ a contradiction. Hence $f$ is not surjective. Q.E.D.

[quote=Mathboi489]Don't know if this is correct. We look at f(x + 1/f(y)). If we plug in x = y + f(y) so we get that f(x + 1/(x-y)) = f(y+f(y)+1/f(y).
Now if we plug in x = y + 1/f(y) we see that f(2x-y)=f(x+1/(x-y)).
Puting this together gives f(2x-y)=f(y+f(y)+1/f(y)) and so f(2(x+n)-y)=f(y+f(y)+1/f(y)) and what follows is that f(2(x+n)-y)=f(2x-y), so f(2x-y+2n)=f(2x-y). Since n can be any real number we get that f(x) must be a constant function. This satisfies our requirement so we are done because if f(x) is constant there is a number outside of it's range[/quote]

Your approach is wrong, plugging in $x=y+f(y)$ doesn't give that. Also there are other errors.

[b]S[/b]olved with [b]Siddharth03, AdhityaMV[/b]

[hide = solution]
Suppose for the sake of contradiction that $f(\mathbb{R}) = \mathbb{Z}_{ > 0}$. Replacing $f(x)$ with $f(x-c)$ for some $c \in \mathbb{R}$ if necessary, we assume $f(0) = 1$. Let $P(x, y)$ denote the given assertion.

Claim 1: If $f(u) = f(v)$ for some real numbers $u, v$ we have $f(u+r) = f(v+r)$ for all rationals $r > 0$.

Proof: Let $z$ be an arbitrary real number. We have the chain of equalities: \[f\left(u + \frac{1}{f(z)}) = f\left(z + \frac{1}{f(u)}\right)\right) = f\left(z + \frac{1}{f(v)}\right) = f\left(v+\frac{1}{f(z)}\right)\]
The claim follows after observing that $f(z)$ takes all naturals $q$ and for a given $r = p/q$ we can repeat the above $p$ times.

Claim 2: $f(0) = f(1)$

Proof: From $P(-1, 0)$ we have: \[f\left(0\right) = f\left(\dfrac{1}{f(-1)}\right)\]
Now using the above claim, we obtain: \[f(0) = f\left(\dfrac{1}{f(-1)}\right) = f\left(\dfrac{2}{f(-1)}\right) = \dots = f\left(\dfrac{f(-1)}{f(-1)}\right) = f(1)\]
This proves the claim.

Claim 3: For each positive integer $n$, there exists a positive rational $q$ with $f(q) = n$.
Proof: From our assumption, there exists some real $r$ with $f(r) =  n$. Now from $P(r-1, 0)$ we have: \[n = f(r) = f(1/f(r-1))\] and we're done since $1/f(r-1)$ is a rational.

Claim 4: $f(x) = f(1/f(x))$ for all $x \in \mathbb{Q}^{+}$.
Proof:
From $P(x, 0)$ we obtain that $f(x+1) = f(1/f(x))$. By Claim 1, we can get $f(x) = f(x+1)$ and so we're done. Observe that by varying $x$ we get $f(1/n) = n$ for all positive integers $n$.

Claim 5: $f$ is $1$-periodic
Proof: From $P(x, 1)$, we obtain: \[f(x+1) = f(1 + 1/f(x))\]
From Claim $1$ and Claim $3$, we obtain: \[f(x) = f(1/f(x)) = f(1+1/f(x)) = f(x+1)\] and thus $f$ is $1$-periodic. Observe that this also implies that we can subtract rationals in Claim $1$.

Claim 6. If $f(x) = 1$ for some rational $x$, we obtain $x \in \mathbb{Z}$.

Proof: Suppose $x = p/q$ with $q > 1$ and $p, q$ relatively prime. Observe that $f(x) = f(0)$ so from Claim $1$ any $f(nx) = f(0)$. Picking $k$ with $kp \equiv 1 \pmod q$, we obtain from Claim $4$ that $f(kp/q) = f(1/q) = 1$, contradicting $f(1/q) = q$.

Claim 7: $f$ is a mathematical impossibility.

Proof: We have from Claim $4$ for all rational $x$: \[f(x) = f(1/f(x)) \implies f(x-1/f(x)) = f(0) = 1\] From Claim $6$, $x-1/f(x) \in \mathbb{Z}$. Setting $x = 2/3$ gives a contradiction (remember that $f(2/3) \in \mathbb{Z}_{>0}$).
[/hide]


Watch it.

http://www.imdb.com/title/tt0293715/

GO WATCH IT NOW!

Hmm I think I remember this being released in Korea.
I used to think it was stupid.
Then again, I think most Korean movies are stupid.
Except "Tae Guk Ki".  that movie's even better than "saving private ryan" :lol: (seriously)

Hello Mozilla,

I think maybe this belongs in [url=http://www.mathlinks.ro/Forum/index.php?f=335]Music and Art[/url]
or perhaps

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34602]Movies you just saw[/url]

I understand the confusion though, officially movies should be discussed in Music and Art, but it has been taken over by Music and Dance people completely, and thus movie people come to Games and Fun, and the fact that our very own Valentin made "Movies you just saw" here only adds to that confusion  :o .

Korean movies can be weird, but I enjoyed
[url=http://www.imdb.com/title/tt0296658/]Take care of my cat[/url]


Eh, not too beautiful, but anyways...

If $a,b,c,d>0$, prove that

\[\sum_{\text{cyc}} \frac{a^9}{\sqrt{a^8(b^{10}+c^{10}+d^{10})+5b^6c^6d^6}} \geq \sqrt{2}.\]

[Edit] Actually, here's a nicer problem:

\[\sum_{\text{cyc}} \frac{a^3}{\sqrt{a^6+15b^2c^2d^2}} \geq 1.\]

Same as IMO 2001,it is solved by use $abcd=1$ and give $x=\frac{2}{a},y=\frac{2}{b},z=\frac{2}{c},w=\frac{2}{d}$.
Hmm.Let me think that $abcd$ can be $\frac{4}{\sqrt{15}}$ to solve this problem.


What is the easiest way to graph polar coordinates?

Samples:

Graph each point:

(1) (2, 5pi/6)

(2) (-2, -pi/6)

Try converting the radians to degrees if you are not comfortable.  Then just measure out the angle and the radius.  For example, $ \frac{5\pi}{6}=150^\circ$ .  Now we just measure out $ 150^\circ$ clockwise from the x-axis, and plot a point 2 units away from the origin.  If $ r$ is negative, then just go in the opposite direction.  If $ \theta$ is negative, the measure the angle clockwise, or just add $ 2\pi$ to the angle (Adding a full circle kmakes no difference) and then count as usual.

Might not be the quickest way, but the more precise way when you have to graph it is probably converting to Cartesian.
In polar coordinates $ \left( r, \theta \right)$ , then in Cartesian coordinates $ x = r \cos \theta \ , \ y = r \sin \theta \Rightarrow \left( r \cos \theta \ , \ r \sin \theta \right)$ .

How is converting to rectangular more precise?

Did not mean it generally really.  Just if you do not have a protractor.
Either way it is another way to graph it.


Let $ n$ be a positive integer such that $ n\geq 2$ and $ a_{1}, a_{2},...,a_{n}$ are $ n$ positive reals. Prove that:
$ \large\prod_{i \equal{} 1}^{n}(a_{i}^{3}\plus{}1)\geq\prod_{i \equal{} 1}^{n}(a_{i}^{2}a_{i\plus{}1}\plus{}1)$  (we treat $ a_{n\plus{}1}\equal{} a_{1}$)

My proof is first, we declare our inequality into sum of equal-degrees, then just apply Muirhead Theorem ^.^


At 1:20 PM, a clock’s hour hand is how many degrees from a vertical position?

A.30 degrees	B.36 degrees	C.40 degrees
D.42 degrees	E.45 degrees

[hide=" click to reveal hidden content"]

It is at the $ 5 \plus{} \frac{5}{3}$ minutes mark, that is $ \frac{20}{3}$ of the total $ 360^{\circ}$, that is $ 40^{\circ}$ from the vertical.

[/hide]

[hide]
*$ {cos^\minus{}1}$ is 'inverse of cosine' and $ {cos()}$ is 'cosine'
Using the equation:
$ {cos^\minus{}1(cos(5.5x))}$ to find the number of degrees between the minute and hand clock, whee x=minutes afte 12:00:

1:20 is 60+20 minutes after 12:00, so we have 80 minutes. 5.5*80=440, using $ {cos^\minus{}1(cos(440))}$ is 140 degrees. [calculated by degrees away from the next multiple OF 180]. Since a straight/vertical line is 180 degrees, 180-140 is $ \boxed{40 degrees}$[/hide]

[hide="Solution"]Just saying.. the PM wasnt really needed but anyways.. on to the solution
there  are 12 hours in a full clock so 12 parts in 360 degrees, so 30 degrees
for each hour and since it is only 1:20 you get 30 degrees, but then since 20 minutes
is 1/3 of 1 hour, you do 1/3 of 30 which is 10 so you add that to 30 to get C 40.[/hide]


i don't think cauchy gives you the minimum...it gives the maximum

:D lol thats what i meant... (heh) ok the problem has been changed.

wrong word... eh

here is a follow up, maximize

$\sqrt{4-x}+\sqrt{x-5}$

[hide="answer"]no solution[/hide]

[quote="Altheman"]here is a follow up, maximize

$\sqrt{4-x}+\sqrt{x-5}$

[/quote]

Err.. lol no $x$ makes the expression real. yay!

[quote="lotrgreengrapes7926"][quote="cincodemayo5590"]Find the maximum of the expression $\sqrt{x-4}+\sqrt{5-x}$ in as many ways as possible.

[hide=" One Way"]
Cauchy[/hide][/quote]
[hide="Cauchy"] $(\sqrt{x-4}+\sqrt{5-x})^{2}\le(x-4+5-x)(1+1)=1\cdot2=2$
$\sqrt{x-4}+\sqrt{5-x}\le\boxed{\sqrt2}$ [/hide]
EDIT: Never mind.

Can we find the minimum?[/quote]
[hide="I think the minimum is..."]Either $x=4$ or $x=5,$ which makes the expression 1. This is true: $4\le x\le5.$ Otherwise, it's imaginary.[/hide]


geometry
analytic geometry
trapezoid
similar triangles

I worked on this problem for a good bit and got nowhere. I worked on coordinates but that didn't help either. Well, now doing simple diagram shows that $ U$ and $ V$ definitely do not want to be at the same height. And obviously, $ U$ and $ V$ should not coincide with any of four vertices but that's as far as I got. Help please.  :|

I'm too lazy to check for divide by 0 errors, so let's just say that we discovered that when one of the corners was U or V, then max. occurs when the other one is at an adjacent corner. Now, forget about them being on corners for now.

$ ABV$ has constant area, so we need to minimize $ AUP+UBQ$

Now, $ AUD+UBC$ is constant, so we need to maximize $ APD+BQC$.

Since these triangles have a constant base, we only need to find the sum of the perpendicular from P onto AD + the perpendicular from Q onto BC.

Let the perp. from Q be R.

Let $ QR=k*VC=m*UB$. By similar triangles, we have $ k+m=1$

Now, $ (1-m)VC=m*UB$

$ m=VC/(VC+UB)$

Therefore, $ QR=UB*VC/(UB+VC)$

Similarly, let the perp. from P be S.

$ PS=AU*DV/(AU+DV)$

Therefore, we need to maximize $ UB*VC/(UB+VC)+AU*DV/(AU+DV)$

Which is equivalent to $ UB*VC/(UB + VC) + (1 - UB)*(1 - VC)/(2 - (UB + VC))$

So we need to maximize

$ xy/(x + y) + (1 - x)(1 - y)/(2 - (x + y)) = xy/(x + y) + (2 - (x + y) + xy - 1)/(2 - (x + y))$

So we need to maximize $ xy/(x + y) + (xy - 1)/(2 - (x + y))$ Now, for a fixed sum x+y, this equation is maximal when xy is as large as possible. By AM-GM this happens when x=y.

$ x/2 + (x^2 - 1)/(2 - 2x) = x/2 - (x - 1)/2 = 1/2$ So this occurs when x=y in general. Check that 2 adjacent corners work as well, and you're done. :D

I just did it the exact :!: same way, using the well-known fact that if you have two poles of height a and b and connect wires from the top of each to the bottom of the other, the intersection is a height of $ \frac {ab}{a \plus{} b}$ from the ground, no matter how far apart the poles are.

[quote="Silverfalcon"]In the diagram, $ ABCD$ is a square, with $ U$ and $ V$ interior points of the sides $ AB$ and $ CD$ respectively. Determine all the possible ways of selecting $ U$ and $ V$ so as to maximize the area of the quadrilateral $ PUQV$.

[img]http://i250.photobucket.com/albums/gg265/geometry101/CMO1992Number3.jpg[/img][/quote

$ S_{1}\equal{}[AUP];S_{2}\equal{}[PDV]$---> $ UADV$  trapezium----> $ [APD]\equal{}[UPV]\equal{}\sqrt{S_{1}S_{2}}$---->$ [UADV]\equal{}(\sqrt{S_{1}}\plus{}\sqrt{S_{2}})^{2}$

$ S_{3}\equal{}[UBQ];S_{4}\equal{}[VQC]$---> $ UVCB$  trapezium----> $ [UVQ]\equal{}[BQC]\equal{}\sqrt{S_{3}S_{4}}$---->$ [UVCB]\equal{}(\sqrt{S_{3}}\plus{}\sqrt{S_{4}})^{2}$

$ AB\equal{}a$--->
$ a^{2}\equal{}[ABCD]\equal{}[UADV]\plus{}[UVCB]\equal{}
(\sqrt{S_{1}}\plus{}\sqrt{S_{2}})^{2}\plus{}(\sqrt{S_{3}}\plus{}\sqrt{S_{4}})^{2}\geq 4\sqrt{S_{1}S_{2}}\plus{}4\sqrt{S_{3}S_{4}}$------>

$ \sqrt{S_{1}S_{2}}\plus{}\sqrt{S_{3}S_{4}}\leq\frac{a^{2}}{4}$

$ [UPVQ]\equal{}[UPV]\plus{}[UVQ]\equal{}\sqrt{S_{1}S_{2}}\plus{}\sqrt{S_{3}S_{4}}\leq\frac{a^{2}}{4}$----->

$ max[UPVQ]\equal{}\frac{a^{2}}{4}$     $ AU\equal{}UB;DV\equal{}VC$

[quote="rofler"]Now, $ AUD \plus{} UBC$ is constant, so we need to maximize $ APD \plus{} BQC$.[/quote]

I get why $ AUD \plus{} UBC$ is constant but I don't get why maximizing $ APD \plus{} BQC$ is necessary. Sorry, I'm not very good at geometry at all!  :blush:

Because $ ADU\plus{}BUC$ is constant value of 1/2. So Since $ AUP\plus{}APD\plus{}UBQ\plus{}BQC\equal{}AUD\plus{}BUC\equal{}1/2$

$ AUP\plus{}UBQ\equal{}1/2\minus{}(APD\plus{}BQC)$

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1992Problem2

Vo Duc Dien


geometry
analytic geometry
angle bisector

The locus of points equidistant from two intersecting lines, line 1 and line 2, is a pair of bisectors that bisect the angles formed by line 1 and line 2 .

What does the above mean in simple words?

In the plane of those two lines, all the points that are equidistant from the two lines are on the two angle bisectors of the angles formed at their intersection.

for example, if the two lines are the coordiante axes, then the lines x = y and x = -y, which bisect the right angles formed by the axes, are formed from points equidistant from the coordinate axes.

I'll try to simplify it further.

1.Imagine two intersecting Lines that form Angles of different measures; such as 150 and 30.
2.Imagine one line that bisects one angle; lets say it bisects one 150 into two 75 angles.
3.Imagine another line that bisects the other angle; bisecting to forming two 15 angles.

As you can see the locus is an angle bisector and a line perpendicular to it.


The temperature on the flat circular plate x^2 + y^2 <= 4 is given by T(x, y) = x^3 + 2y^2 - 4x. Determine
all local maxima and minima in the interior of the plate, and the absolute maxima and minima on the whole
plate.

Well first take the gradient to locate your critical points and then test them to make sure they're actually on your surface.  Then take the Hessian at those points to identify what type of extrema they are.

hello, i have found that $ T(x,y)\equal{}x^3\plus{}2y^2\minus{}4x$ has in $ x\equal{}\frac{2}{3}\sqrt{3},y\equal{}0$ a local minimum.
This lies in the interior of the plate.
Sonnhard.


Let $D$ be the point on side $AC$ of a triangle $ABC$ with $AB < BC$ such that  $AB = BD$. The incircle of $\Delta ABC$ touches $AB$ at $K$ and $AC$ at $L$, and $J$ is the incenter of triangle $BCD$. Prove that $KL$ bisects the segment $AJ$ .

It is an ISL 2006 problem, given also in the France TST, Day 2. See here: [url]http://www.mathlinks.ro/viewtopic.php?t=149173[/url]


[quote="Valentin Vornicu"]fleeting_guest, what browser do you use?! your post are very unusually wrapped ...  :?[/quote]

It is just a trick. It should just suggest that he writes much more than he actually does.  ;)  No offence meant we really appreciate what you have to say and deeply regret that your informative posts got lost recently. Maybe you could repost a subset of them...  :?

[quote="fleeting_guest"]orl and Valentin,

I use various machines with Internet Explorer.
I noticed the linebreak problem but don't know the
cause or how to fix it.    Also, any way to make the
browser load only a "minimalist" version of every
page (no movies with the proof-sans-words, etc)?
It is very slow for my connection.

I have no copies of the lost postings, sorry.
If the discussions become topical again I could
write similar (or new) comments.

Cheers.[/quote]Use the MathLinks skin instead. In some time there will be a xHTML 1.1 lo-fi version of the forum available (I'm working at it).

What resolution do you use with your computer?

[quote="rrusczyk"]To switch to the MathLinks skin, click on the Profile button after logging in - scroll down to Board Style and change that to MathLinks.[/quote]Either that or go to Options -> Board Layout -> Board Style :)

Tell us if it helps in some sort. If not, can you please specify your browser's details?


ba salam khedmat doostan  :)

morabayi be zele a darim. dar har ras an dayerei be shoaye a rasm mikonim. dar vasate morab fazaye khali vojood darad ke davayer ijad kardand. bishinie moraba kochaki hamrasata ba morabae bozorg dar fazaye khali ijad mikonim. 1 noghte daroone morabae bozorg gharar midahim. matloobast ehtemale noghte daroone morabae kochak :roll:  :!:

manzooret ine??? :?:

agar manzooret ine.
Ehtemalesh mishe $3-2\sqrt2$

salam arz mikonam khedmate said khan: na, yekkami bishtar roye masale fekr kon  ;)

shekli ke keshidi kamelan sahihe, vali javabet moshkel dare


MATHCOUNTS
calculus
articles
LaTeX
geometry
AoPS Books
\/closed

AoPS should hire a graphics person for their site and the brochures and other things they send out before mass-mailing season begins, since they're pretty darn ugly and I've heard of a few people who were just really turned off to the books/classes because of the design.

Just a suggestion...

[quote="ccy"]AoPS should hire a graphics person for their site and the brochures and other things they send out before mass-mailing season begins, since they're pretty darn ugly and I've heard of a few people who were just really turned off to the books/classes because of the design.
[/quote]

There would be anecdotal claims to this *no matter what design we used*.

Specific suggestions of what to include/exclude would be much more helpful.

I think that the cover of the books should not be in font Times New Roman, and the AoPS sign should not be in Papyrus. Using default Microsoft Office fonts= not good.

Now the inside of the books is designed quite nicely, especially the intro/intermediate. The thing about these kind of math books in general is that they all tend to have fairly bad design (the only really nice math book that has good design I've seen was the Art and Craft of Problem Solving, and that was by a major company like Wiley). Its true, they are still awesome math books, but many of those who simply use it without going to to the olympiads or competitions and use it just for classroom will probably just judge it by its design, especially cover design. It's a sad fact, but it's true in America that many who are unacquainted with real math (i.e. middle school teachers) would consider good cover design a sign of professionalism. The contest and problem solving sort of kids and mathcounts coaches will always use your books, but here are suggestions to spread your impact through schools:

-Create a hardcover edition for schools. Schools don't like buying textbooks every year (even though for this one it would cost the same as buying an expensive one). So that would mean glossy paper too. Add a bit more color, but don't become rediculous like the text book companies usually do. The "best looking" textbook that actually has real math that I've seen is Larson's Calculus books from Houghton Mifflin (8th edition). That has really good cover design and book design.

-Again, change your cover design in the next edition. The ball designs are really appealing to us nerds as they are fractalesque, but to appeal to school districts it needs more than that. Look at math and engineering science text books that have really nice design for inspiration.

Overall I think a hardback edition would be nice, but cover and book design for paperback is more pressing. The new ACoPS edition is paperback and has nice design.

Now web design is fairly decent. But at this point a major change would need a professional designer. And a major change in design can't be put off forever.

Brochures always use a professional designer because they are a good 3 days of work to be decent. 'Nuff said.

Book and web design aren't that pressing IMO. But when you do decide to change, change appropriately. For good fonts, go to Urban Fonts or DaFont. Both are really nice websites for free stuff.

P.S. A logo for the Wiki would be awesome ^.^

Also, a community contest would also be really cool for simple things like logos, typefaces, or even website intros.

If I may add to Anaxerxia's excellent comments, I believe the smaller size and less scary cover of the old AoPS books is far superior to the current format. Is it a handbook or a textbook? In my mind, the smaller size makes it a handbook, which is friendly as opposed to boring textbooks. And the dark color is all meh whereas the old cover was calming in my mind.

My copy of Vol. 2 is pretty beat up - yes, hardcover versions would be awesome.

I don't like the idea of a logo for the wiki, but I would like to see the spacing bugs in the ML wiki skin fixed.

I've never seen an AoPS brochure, so I can't comment on that...

[quote="Boy Soprano II"]Well, first, I would like to say that the rudeness of the first post shocked me.  Also, I think the logos, graphics, and whatnot look fine as they are.  And please don't model your books on any aspect of a mainstream calculus textbook.  I think some promising students already balk at the investment of 30 or 70 dollars.  (I waited until my birthday, myself.)[/quote]

It's true that in scientific community many of us do not pay attention to design. Content is very important, and AoPS's is the best, and that's why design is so important, it communicates to the world a way for your ideas. I'd like to say the brochures are one of AoPS's main ways of marketing, besides word of mouth ( :D  :D  :D ), so it's important that when making a first impression that you get people interested, and design is one of the best ways to do that. I mean, even the AMC people updated their design, and you do NOT want to see how crappy the scattered remnants of its previous form looked.

Even if you can't appreciate good design, you have to admit that design is crucial to getting the word out about AoPS and math in general. There are so many extremely intelligent students who would love math if they really saw what it was...

And your right, its really not a pressing issue right now at all, but one day an update will be needed and it should be done well.

As for the price, AoPS prices are a lot cheaper than most text books anyway, math or non math. And I don't see how slapping some good design on a paperback will make it more than $ \$$5 dollars cheaper, at the very extreme most$ \$$10. That is still pretty reasonable looking at content and comparing its price to the bloated crap on the market today.

[quote="Anaxerzia"]Even if you can't appreciate good design, you have to admit that design is crucial to getting the word out about AoPS and math in general. There are so many extremely intelligent students who would love math if they really saw what it was...[/quote]
Why do you think I do not appreciate design?  I think that design is quite important, and I also think that AoPS's designs are fine at the moment.  If you disagree, this does not mean that you do not appreciate design or that I do not appreciate design, but merely that we have different tastes.

[quote]It's true that in scientific community many of us do not pay attention to design.[/quote]
I do not think this is true.  Members of the scientific community must prepare well-organized research papers, which one cannot do without some concept of design.  Sorry if I seem quarrelsome, but I often find outsiders with stereotypes similar to this statement, and I find this frustrating in general because math is a creative subject in itself, and when it is applied to me in particular because I write often and strive to craft aesthetically appealing articles and elegant proofs in LaTeX, and because I am a musician as well.  In short, I dedicate a large portion of my life to aesthetics, so it rankles to be branded a philistine. :wink:

Basically the design is a matter of personal preference. You would be amazed as how people react to different designs (actually we have done some experiments in that area and they revealed interesting results).

As for the site design itself: considering the uniqueness of our site, we are striving to first make sure the right content is delivered to the already existent users, the speed of the site is satisfactory and well after that we will be worried with replacing some font on the main page.

@boy
[hide]1. You are being a bit quarrelsome
2. "you" $ \neq$ Boy Soprano II whenever it used
3. I am not a mathematical outsider
4. Employing good graphics design $ \neq$ being simply well-organized
5. If you say there is no terrible cover design in mathematical books, look at an Andreescu book
6. Using the word "many" $ \neq$ stereotype
[/hide]

I highly recommend: Barnes and Noble design section; sites like Before and After Magazine, Smashing Magazine, and looking for lots of design articles on the web for good ideas if you do decide to upgrade design in brochures and etc and want to do it yourself.. I'm sure your city has a good graphics design store too :).

Like I said, the issue isn't pressing at all at the moment, but I think its more than just fonts however, IMO. Eventually a new design should change a lot of things and make them more graphically appealing, as like I said the design can really change a lot. This would obviously be a hassle for both you guys and the designer since you guys need to have both  good design and functionality in a way that you can change anything you want without calling him. But I think, when you do do it, that it would be worth it and worth it to do it well.

And I think that design has come to a point that most people can agree that something is visually appealing and what is not, so I think the real preference is how a person wants to display their company through design.

Again, these are all just suggestions :)

[quote="#H34N1"]Replacing font? Because many do not have extra font sets installed, I think it would be safe to use the HTML safe fonts.[/quote]

I'm not talking about text font necessarily. I'm talking about logo font.

And its a lot more than font in design. Typography is just one aspect of a good design.

I think Mr. Vornicu is working on a new theme right now.


algebra
polynomial
search
algebra proposed

What is the condition for $a<b$ real numbers in order to exist a non-zero  polynomial $f$ with integer coefficients such that $ |f(x)|<1$ for all $x$ between $a$ and $b$.

Harazi, i think with every interval $[a, b]$, we can always find a polynomial $p(x)$ st $\mid p(x) \mid < 1$.
With every interval $[a, b]$, we choose $p(x)= d(x^n+x^{n-1}+...+x+1)$ and when $d$ is positive and sufficient small,  we have $\mid p(x) \mid < 1$
Is this right ? :?

But $d$ is supposed to be an integer so it can be as small as you want.

Pierre.

You are right. So i think the result is b-a around 3!!! :) :P

This is a very hard problem. Try to search near 4, not 3.  ;)


Find $ m \in Z^ \plus{}$ such that $ \phi(m) | m$

[hide="Not sure..."]Since $ \phi(m)\equal{}m\prod_{p|m}\left( \frac{p\minus{}1}{p} \right)$, where $ p$ is a prime number, we want the numerator of $ \prod_{p|m}\left( \frac{p\minus{}1}{p} \right)$ to be $ 1$. This happens if and only if $ \prod_{p|m}(p\minus{}1)$ is a factor of $ \prod_{p|m}p$. But since the only factors of a prime $ p$ are $ 1,p$, this can only happen for $ p\equal{}2$, so $ m$ must be a power of $ 2$. [/hide]

[hide="You missed some"]$ m \equal{} 6$?[/hide]

[hide="Oops!"]I missed that there might be consecutive primes, I think? So $ m\equal{}2^j3^k$ where $ j,k$ are integers. Since $ 2,3$ is the only case where $ p_1\minus{}1$ can result in another prime $ p_2$, I think that this accounts for all such $ m$. [/hide]

Your "there can't be any more" argument really fails to convince.  (Notably, I see no reason the second version should actually be more comprehensive than the first version: why does $ p_1 \minus{} 1$ need to be actually another prime, instead of just a product of other primes in the factorization?  For example, I don't see how your proof shows that $ \phi(42)$ fails to divide 42.)
[hide="Hint for a more convincing analysis"]How about looking at the highest power of 2 that divides $ m$ compared with the highest power of 2 that divides $ \phi(m)$?[/hide]

[hide="full solution"]Since $ \frac {\phi(p^k)}{p^k} \equal{} \frac {p \minus{} 1}{p}$ for all $ k\ge 1$, we assume that all primes are included at most once in $ m$.  Now we need to find when

$ p_1p_2p_3\ldots p_m \equal{} k(p_1 \minus{} 1)(p_2 \minus{} 1)(p_3 \minus{} 1)\ldots(p_m \minus{} 1)$

for some $ k$, where $ p_i > p_j$ if $ i > j$.  There can be at most one odd prime, because otherwise 2 would be in the right side more than once, and that is impossible on the left side.  So if $ p_2$ exists, then $ p_1$ must be 2 and

$ 2p_2 \equal{} k(p_2 \minus{} 1)$ or $ (k \minus{} 2)(p_2 \minus{} 1) \equal{} 2$.

Now this means that $ p_2 \equal{} 3$, and $ m \equal{} 6$.  If $ p_2$ does not exist, and $ p_1$ may or may not be 2, then

$ p_1 \equal{} k(p_1 \minus{} 1)$ or $ (k \minus{} 1)(p_1 \minus{} 1) \equal{} 1$.

This shows that $ p_1$ must be 2, and so must $ m$.  We must remember to include the values for $ m$ whose exponents exceed 1, and so the family of solutions that satisfies the given equation is only $ \boxed{2^{n \plus{} 1}3^{k}}$.

That is assuming that $ m$ has any prime factors at all.  If it doesn't, then $ m\equal{}1$ and it still works.[/hide]

[hide="Third try at a convincing argument"]Consider the exponent of $ 2$ in the prime factorization of $ m$. In $ \phi(m)$, any odd prime in $ m$ introduces an extra factor of $ 2$ in the numerator, since $ p \minus{} 1$ is always even for odd primes $ p$. Assume $ m$ has at least one odd prime factor. So $ m$ must have a factor of $ 2$ so that the $ \frac {1}{2}$ that results in $ \phi(m)$ cancels out the $ 2$ introduced by $ \frac {p \minus{} 1}{p}$. There will still be extra powers of two in the numerator if $ p$ is $ 1 \mod 4$, so $ p \equiv 3 \mod 4$. There can also be only one such $ p$ in the prime factorization of $ m$, since any more would result in a $ 4$ being in the numerator.

However, suppose that $ p \equiv 1 \mod 3$. Then $ p \minus{} 1$ is a multiple of $ 3$, so there will be more $ 3$'s in the prime factorization of $ \phi(m)$ than in $ m$. So $ p \equal{} 3$ or $ p\equiv 5 \mod 3$. (So, for example, $ 42$ has a factor of $ 7 \equiv 1 \mod 3$, so it doesn't work. )

But we can similarly show that $ p$ cannot be one more than any number ($ p \equal{} 13$ introduces a factor of $ 3$, $ p \equal{} 17$ introduces a factor of $ 2^4$, $ p \equal{} 19$ introduces a factor of $ 3$, $ p \equal{} 23$ introduces a factor of $ 11$, and so on. ) since $ p \minus{} 1$ either has a factors that are not equal to a power of two, or is equal to a power of two greater than two. In either case, $ \phi(m)$ is not a factor of $ m$.

So $ p \equal{} 3$.

Now, $ m \equal{} 2^k3^j$ where $ k,j$ are nonnegative.

EDIT: If $ j>0$, then $ k>0$. [/hide]

EDIT: Beaten, but is my argument rigorous? :maybe:


trigonometry
induction
absolute value
AMC

The sequence $ (a_n)$ satisfies $ a_0 \equal{} 0$ and $ \displaystyle a_{n \plus{} 1} \equal{} \frac85a_n \plus{} \frac65\sqrt {4^n \minus{} a_n^2}$ for $ n\ge0$. Find the greatest integer less than or equal to $ a_{10}$.

[quote="MellowMelon"][hide="Outline"]Divide the recurrence by $ 2^{n \plus{} 1}$, which suggests letting $ b_n \equal{} a_n/2^n$ and then $ \sin \theta_n \equal{} b_n$. Then the recurrence is just the sine addition formula.[/hide][/quote][hide]It's [b]almost[/b] the sine addition formula. The second term is in an absolute value, however, so when we are in $ [0,\frac{\pi}{2})$ we add but in $ [\frac{\pi}{2},\pi]$ we subtract. So once we cross the $ y$-axis the first time, $ b_n$ becomes periodic with a period of two. After some arithmetic, $ b_{10}\equal{}b_4$.[/hide]

Upon manipulation during the test, I got the recurrence:

\[ a_{n\plus{}2}\equal{}\frac{16a_{n\plus{}1}\minus{}20a_n}{5}\]

but I had no clue how to proceed from there.  I tried generating functions.  Anyone else get the same equation?

[hide="I believe this is what you did..."]
Multiply both sides by $ 5$ and rearrange to get $ 5a_{n\plus{}1}\minus{}8a_n\equal{}6\sqrt{4^n\minus{}a_n^2}\implies25a_{n\plus{}1}^2\minus{}80a_na_{n\plus{}1}\plus{}100a_n^2\equal{}36\cdot4^n$, so $ 25a_n^2\minus{}80a_{n\minus{}1}a_n\plus{}100a_{n\minus{}1}^2\equal{}36\cdot4^{n\minus{}1}\implies100a_n^2\minus{}320a_{n\minus{}1}a_n\plus{}400a_{n\minus{}1}^2$
$ \equal{}36\cdot4^n\implies25a_{n\plus{}1}^2\minus{}80a_na_{n\plus{}1}\plus{}100a_n^2\equal{}100a_n^2\minus{}320a_{n\minus{}1}a_n\plus{}400a_{n\minus{}1}^2$
$ \implies25(a_{n\plus{}1}^2\minus{}16a_{n\minus{}1}^2)\equal{}80a_n(a_{n\plus{}1}\minus{}4a_{n\minus{}1})$
$ \implies5(a_{n\plus{}1}\minus{}4a_{n\minus{}1})(a_{n\plus{}1}\plus{}4a_{n\minus{}1})$
$ \equal{}16a_n(a_{n\plus{}1}\minus{}4a_{n\minus{}1})\implies(a_{n\plus{}1}\minus{}4a_{n\minus{}1})(5a_{n\plus{}1}\minus{}16a_n\plus{}20a_{n\minus{}1})\equal{}0$. But for $ n\ge3$, $ a_{n\plus{}1}\equal{}4a_{n\minus{}1}$.
[/hide]

:wallbash_red:  :wallbash_red:

Can't believe I didn't notice that.  Time to organize my work better...

Better than what I did: looking back at my work, I got to $ 25a_{n \plus{} 1}^2 \minus{} 80a_na_{n \plus{} 1} \plus{} 100a_n^2 \equal{} 100a_n^2 \minus{} 320a_{n \minus{} 1}a_n \plus{} 400a_{n \minus{} 1}^2$
$ \implies25a_{n \plus{} 1}^2 \minus{} 80a_na_{n \plus{} 1} \equal{} \minus{} 320a_{n \minus{} 1}a_n \plus{} 400a_{n \minus{} 1}^2$...and then stopped.

To elaborate, divide by $ 2^n$ to get the recurrence $ b_{n \plus{} 1} \equal{} \sin (\theta \pm arctan3/4)$ We start at angle $ 0$ and keep adding $ \arctan3/4$ until we get to just before $ 90$ at which point we must subtract $ arctan3/4$. So $ b_{10} \equal{} \sin (2arctan 3/4) \equal{} \frac {24}{25}$.
So $ a_{10}$ = $ (1024)(24/25)$ which is $ 1025(24/25) \minus{} \frac {24}{25}$ which is a bit bigger than $ 983$.

Here's what I had in mind when I wrote this problem.

[hide]Rearranging the recursion formula gives
\[ a_{n+1} = \frac{8}{5}a_n + \frac{6}{5}\sqrt{4^n-a_n^2} =
2\left(\frac{4}{5}a_n + \frac{3}{5}\sqrt{4^n-a_n^2}\right).
\]

The part inside parentheses suggests a sum-of-angles trig identity, but the $ 4^n$ inside the square root means that instead of $ a_n = \sin \alpha$ or $ \cos \alpha$, we should have $ a_n = 2^n\sin\alpha$ or $ 2^n\cos\alpha$. Let $ \theta = \sin^{-1} (3/5)$. If we try to write $ a_n = 2^n\sin \alpha$, we get

\begin{align*}
a_{n+1} & =  2(2^n\cos\theta\sin\alpha + \sin\theta
\sqrt{4^n(1-\sin^2\alpha)} \\
& =  2^{n+1}(\cos\theta\sin\alpha + \sin\theta\cdot|\cos\alpha|)\\
& =  \left\{\begin{array}{ll} 2^{n+1}(\cos\theta\sin\alpha +
\sin\theta\cdot\cos\alpha) & \quad \textrm{if } \cos \alpha \geq 0\\
2^{n+1}(\cos\theta\sin\alpha - \sin\theta\cdot\cos\alpha) & \quad \textrm{if } \cos \alpha < 0
\end{array}\right. \\
& =  \left\{\begin{array}{ll} 2^{n+1}\sin(\alpha+\theta) & \textrm{if } \cos \alpha \geq 0 \\
2^{n+1}\sin(\alpha-\theta) &  \textrm{if } \cos \alpha < 0
\end{array}\right.
\end{align*}

Now, $ \cos \theta = \frac{4}{5}$. Since $ \cos 45^{\circ} = \frac{\sqrt{2}}{2} < \frac{4}{5} < \frac{\sqrt{3}}{2} = \cos 30^{\circ}$, $ 30^{\circ} < \theta < 45^{\circ}$. This means that the angle increases by $ \theta$ until it gets to $ 3\theta$, after which it bounces back and forth between $ 2\theta$ and $ 3\theta$.

So we get:
\[ a_n = \left\{\begin{array}{ll} 2^n\sin(n\theta) & n=
0, 1, 2 \\
2^n\sin(2\theta) &  n>2, \, \textrm{even}\\
2^n\sin(3\theta) &  n>2, \, \textrm{odd}
\end{array}\right.
\]

Thus $ a_{10} = 2^{10}\sin(2\theta) = 1024\cdot \frac{24}{25} =
\frac{24576}{25} = 983.04$.[/hide]

This took forever to format because AoPS apparently requires the \ [ and \ ] around an align* environment (which I had never seen before); and because having $ \$$-ed maths inside a \ textrm (like for writing English-language conditions with the occasional math symbol, inside an array) also causes issues.  :huh:

I didn't know about the stuff on sines and cosines forming some relationship at the time of writing the exam, so I calculated $ a_0$, $ a_1$, $ a_2$, $ a_3$, and $ a_4$, and noticed that $ a_2$ was 96/25 and $ a_4$ was 9600/625.

So I put down $ \frac{9600000000}{5^{10}}$ as the answer, and prayed for the best. Turns out, it was correct.

As was pointed out earlier (not just in my solution), the sequence $ b_n\equal{}\frac{a_n}{2^n}$ can be shown to be periodic with period two starting with $ b_4$. Good intuition  :lol:

Wow, this is an incredible problem! I tried some values and saw a bunch of Pythagorean triples, so I assumed it was a number-theoretic relation type problem. I tried prime factorizing $a_n,$ but that didn't get anywhere. But eventually I noticed the pattern that sdkudrgn88 pointed out and was able to get it by induction :P

[quote=shiningsunnyday]Very sorry for the revive. I got to the point where $b_n=\sin n \theta$ where $\sin \theta = \frac{3}{5}.$

[quote=generating]
Now, $ \cos \theta = \frac{4}{5}$. Since $ \cos 45^{\circ} = \frac{\sqrt{2}}{2} < \frac{4}{5} < \frac{\sqrt{3}}{2} = \cos 30^{\circ}$, $ 30^{\circ} < \theta < 45^{\circ}$. This means that the angle increases by $ \theta$ until it gets to $ 3\theta$, after which it bounces back and forth between $ 2\theta$ and $ 3\theta$.
[/quote]

Ok. So by the condition we know that $90 < 3\theta < 135.$ How do we know it starts bouncing back between $2\theta$ and $3\theta$ after that?[/quote]

It's because we are now in the second quadrant so $\cos{\theta}$ becomes negative thus you will start subtract out $\theta$ but we will end up in first quadrant again so time to add and in this way it becomes periodic

My progress so far was this: Firstly, we will compute some terms. We have $a_0 = 0$, so $a_1 = \frac{6}{5}, a_2 = \frac{48}{25} + \frac{6}{5} \cdot \frac{8}{5} = \frac{96}{25}, a_3 = \frac{936}{125}$.

How do I get more terms without extreme computation? Is there some sort of substitution that I need to make? I don't see any pattern in the numerators...

[hide=@above]I believe that if you do carry through the computation, you will find a pattern emerging at some point. However, if you want a slicker approach, try a trigonometric substitution.[/hide]


probability
Euler
blogs
logarithms
arithmetic sequence
puzzles

The first lateral thinking marathon died, so I guess I'll start a new one...

Explain how two twins can have birthdays four days apart, if they were born five minutes apart.

[hide="Big hint"]
Very special conditions.
[/hide]

One was smart and went to a math tournament and had to postpone his. :roll:

Well, it never says anything about the twins having to be related to each other, or even born from the same mother.  There was a baby born, with a twin on Jan. 1st.  Then another twin was born 4 days later from a different mother, who had twins, but the time was just 5 minutes apart, like 1 o'clock and 1:05.    Wait....

Does this have something to do with the wierd time change in the 1,000's where they skipped time because of their inaccuracy of calenders before then?

Here's an idea! Both were born 4 days before Chinese New Years in China. One continued to live in China while the other moved. All Chinese people celebrate their birthdays on New Years.

[hide="my answer"]
there are multiple possible solutions
[/hide]

The parents are counterculture and refuse to celebrate birthdays on the actual day of birth.

(If you can't tell, I hate lateral thinking puzzles.  Their solutions are too arbitrary.  Nothing like mathematics at all.)

JBL:  She was born on December 31st, 11:59 p.m., 1999, and she subtracts 2000 from the year?  (I used to do this because I was born in 1990.  :P )

[quote="t0rajir0u"] Their solutions are too arbitrary.  Nothing like mathematics at all.
[/quote]
What, that isn't true.... :)

[hide="The answer"]
Her birthday was at the turn of the century.
[/hide]

Neither of you is quite right, although you're both close.  After 1 minute, you have the date 1/1 ... I suppose she could always just subtract one month and one day and add one minute, but is that really what you want?

@tor: Some lateral thinking puzzles, certainly.  I think the two here actually encouraged a systematic line of thought -- what do I know that can be used to extend the time between two fixed calendar events, for example -- and stretch it as far as possible.

[quote="JBL"]The children were born on a cruise of the Arctic.  The first was born at 11:58 PM, February 28, 1999, off the western coast of Alaska.  The second was born five minutes later and a few miles to the west, off the eastern coast of Russia, at 12:03 AM, March 2, 1999.  The next year, the older son celebrated his first birthday a full 3 days before the younger son.  However, they again find themselves oddly placed on the globe, and so the actual celebrations take place even further apart.  I don't think I can get a spread larger than that.  :)

If this is the desired answer, I have another:  the daughter of a friend is an elementary school student.  She's quite a bright girl, a year ahead in mathematics, but not bizarrely talented or anything -- just smarter than most kids her age.  Certainly, not a savant or anything.  However, if you tell her the date and time, she can tell you almost instantaneously how old she is, down to the hour.  How can she do this?[/quote]

That's pretty much the same answer I was thinking.

She can calculate how old she is... she just doesn't get it right. :rotfl:

[hide="if this topic died also, click here to reveal hidden content. yep."]You flip a coin 1000 times and get heads each time. What is the probability of getting another heads on the next flip?[/hide]

it depends on how weighted the coin is. If its fair, then its still a 1/2 chance.

OK DO MY LATERAL THINKING PROBLEM:
a man is running down a dark hallway with lights hanging overhead. The lights flicker and he stop running and starts walking. WHY? you may ask yes or no questions.

[hide="Do not read this unless you want the answer spoiled for you"]
[hide="Seriously a spoiler"]
His wife was in intensive care and about to die, and he was running down the hall hoping to see her before she died. However, when the lights flickered, he knew that there had been a power outage and the artificial thingy failed. Now that there was no hurry, he began walking.
[/hide]
[/hide]

A plane crashes on the border of Nepal and Bangladesh. Where do they bury the DEAD?

[quote="laughinghead505"]Do you know if it is heavier or lighter?[/quote]

You don't know which.

[quote="SplashD"]first you weigh 3 and 3. this tells you which group of 6 the countefeit coin is in.

then switch one of the 3 with another 3. this will narrow the counterfeit coin to only one group and you will know if it's heavier or if it's lighter.

weigh 2 coins from that group of 3 to find the counterfeit coin[/quote]

What if you weigh 3 and 3 and they're the same. Then, you switch a group of 3, and they're still the same. You won't know if it's heavier or lighter.

assume the counterfeit coin isn't in the last group...

About the counterfeit coin:

 One way to think about this: Since you don't know if the  counterfit coin is heavy or lighter than a  true coin, is there are  $12*2 = 24$ choices, One weighing get us 3 choices ( = > <  )
so to get  theoretical  answer , we see  $2<\log_{3}24 < 3$
So, answer cannot be less than 3.

3 indeed is correct answer.
[hide]
For  up to $\frac{3^{k}-3}2$  coins can be done in $k$ times. (Eg  39 coins would  need 4 weighings.

If you have a set of good coins the number is  one more than above (Eg if you have 4 true coins (+ 13 coins out of which 1 is defective - you can still do in 3 weightings)

If you know the  counterfeit coin is heavy (or light) then , of course, $k$ weightings are sufficient for $3^{k}$ coins.

Any of these type of problem: principle  is always divide in 3 equal choices

Mehtod would work for any general case ..
 ... eg first weighing use 4 and 4

[/hide]

What eight letter word has kst in the middle, in the beginning, and at the end?

Answer:, but don't peek!
[hide][hide][hide]inkstand[/hide][/hide][/hide][/hide]

[quote="randomdragoon"]Every day, a cyclist crosses the border between Spain and France carrying a bag. No matter how much the customs officials investigate him, they do not know what he is smuggling. Do you?[/quote]

[hide]Bicycles? Or possibly bags...[/hide]

[quote="randomdragoon"]Every day, a cyclist crosses the border between Spain and France carrying a bag. No matter how much the customs officials investigate him, they do not know what he is smuggling. Do you?[/quote]

No, because no one told us.

[hide="answer?"]WINE!!! It's so popular in France![/hide]


For each positive integer $ n,$ let $ S(n)$ denote the sum of the digits of $ n.$ For how many values of $ n$ is $ n \plus{} S(n) \plus{} S(S(n)) \equal{} 2007?$

$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

[quote="DiscreetFourierTransform"][hide="Hint 1"] Whatever a number congrues to mod 9, that's the sum of it's digits.[/hide]

[hide="Hint 2"] The first one is greater than 1900.[/hide]

[hide="Hint 3"] The numbers are all multiples of 3.[/hide][/quote]

The first hint is wrong. $2007 \equiv 0 \mod 9$, but the sum of the digits of $2007$ is not 0 obviously.

Aha. Sorry. Shoulda made that more clear. The sum when you get 0 mod 9 IS 9.

why equivalent to mod 9 ? I got 4 numbers total (D)

just bound it...0<n<2007, so 0<s(n)<36, 0<s(s(n))<11, so we can strengthen our bound on n, 1960<n<2007, so we have
i)n=19ab, s(n)=10+a+b, s(s(n))=1+a+b, or a+b-8
checking gives a few solutions, make sure for carrys, 3 sol here
ii)200a, s(n)=2+a, s(s(a))=2+a, 1 sol here
so 4 solutions

That's correct.

Here's my solution:
[hide="Solution"]
First, I set some bounds on n.
For $n \leq 2007$, the value for $n$ that gives the greatest value for S(n) is clearly 1999.

S(1999) = 1999 mod 9 = 28.

Therefore, $n \leq 28$

S(28) = 10, so our lower bound is
2007 - 28 - 10, or
$n \geq 1969$

When taken mod 9,
n, S(n), and S(S(n)) all give the same result.

We're looking for N values that have the same value as $2007 mod 9$, or in other words N values that are multiples of 9.

This means that they also must be multiples of three.

Basically, I then brute forced all multiples of three between 1969 and 2007.

Took a while, but after the test I realized that there is also an upper bound between 1987 and 1999.
[/hide]

[hide="not brute force"]
Clearly, $n > 1900$. We can break this up into three cases:

Case 1: $n \geq 2000$

Inspection gives $n = 2001$.

Case 2: $n < 2000$, $n = 19xy$, $x+y < 10$

If you set up an equation, it reduces to

$4x+y=32$

which has as its only solution satisfying the constraints $x=8$, $y=0$.

Case 3: $n < 2000$, $n = 19xy$, $x+y \geq 10$

This reduces to

$4x+y=35$. The only two solutions satisfying the constraints for this equation are $x=7$, $y=7$ and $x=8$, $y=3$. The solutions are thus

$1977, 1980, 1983, 2001$

and the answer is $D) 4$.
[/hide]

@above

I'll ignore the bump, because I'm confused too.

These two equations are stated in the same solution:
$4x+y=32$

$4x+y=35$

That makes no sense.

[quote=OlympusHero]Those are for different cases.[/quote]

Ohhh i see. stupid me; didnt read the entire solution. :wallbash_red:

Repost of my C&P post, but I think this belongs better here.

ALTERNATE SOLUTION

This problem was originally a gigantic bash, but I found a rigorous solution that is quite bashy.

[quote]For each positive integer $n$, let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$ [/quote]

Claim. The only positive integers $n$ that satisfy the condition are perfect multiples of $3$.

Proof of claim:
We examine the positive integers mod $9$. Here are the cases.

Case 1. $n \equiv 1 \pmod 9$. Now, we examine $S(n)$ modulo $9$.
Case 1.1. The tens digit of $n$ is different from the tens digit of the largest multiple of $9$ under $n$. (In other words, this means we will carry when adding from the perfect multiple of $9$ under $n$.)
Observe that when we carry, i.e. Add $1$ onto $1989$ to obtain $1990$, the units digit decreases by $9$ while the tens digit increases by $1$. This means that the sum of the digits decreases by $8$ in total, and we have $-8 \equiv 1 \pmod 9$, so the "mod 9" of the sum increases by $1$. This means that, regardless of whether the sum carries or not, the modulo 9 of the sum of the digits always increases by $1$.

Case 1.2. The tens digits are the same, which is trivial since the units digit just increases by $1$ which means that the sum is also equivalent to $1 \pmod 9$.

This means that $S(n) \equiv 1 \pmod 9$ and similarly letting the next $n=S(n)$, $S(S(n)) \equiv 1 \pmod 9$. Summing these, we have $n+S(n)+S(S(n)) \equiv 3 \pmod 9$. Clearly, no integers of this form will satisfy the condition because $2007$ is a perfect multiple of $9$.

Case 2. $n \equiv 2 \pmod 9$.

In this case, we apply exactly the same argument. There is at most one carry, which means that the sum of the digits will always be congruent to $2$ mod $9$. Then we can apply similar arguments to get $S(n) \equiv 2 \pmod 9$ and $S(S(n)) \equiv 2 \pmod 9$, so adding gives $n+S(n)+S(S(n)) \equiv 6 \pmod 9$.

It is trivial to see that for $n \equiv k \pmod 9$, for $0 \leq k \leq 8$, we must have $n+S(n)+S(S(n)) \equiv 3k \pmod 9$. Only when $k=0, 3, 6$ is $3k$ a multiple of $9$, which means that $n$ must be a multiple of $3$.

Now, we find the integers. Again, consider two cases: Integers that are direct multiples of $9$ and integers that are multiples of $3$ but not $9$.

Case 1. $n$ is a multiple of $9$. An integer of the form $\overline{20ab}$ will not work since the least such integer is $2007$ which already exceeds our bounds. Thus, we need only consider the integers of the form $\overline{19ab}$. The valid sums of the digits of $n$ are $18$ and $27$ in this case.

Case 1.1. The sum of the digits is $18$. This means that $S(n)=18, S(S(n))=9$, so $n=2007-18-9=1980$. Clearly this number satisfies our constraints.

Case 1.2. The sum of the digits is $27$. This means that $S(n)=27, S(S(n))=9$, ,so $n=2007-27-9=1971$. Since the sum of the digits of $1971$ is not $27$, this does not work.

This means that there is $1$ integer in this case.

Case 2. $n$ is a multiple of $3$, not $9$.
.
Case 2.1. Integers of the form $\overline{20ab}$. Then $S(n)=3$ or $S(n)=6$; it is trivial to see that $S(n)=6$ exceeds our bounds, so $S(n)=3$ and $n=2007-6=2001$.

Case 2.2. Integers of the form $\overline{19ab}$. Then $S(n)=12, 15, 21, 24$ and we consider each case separately.

Case 2.2.1. Integers with $S(n)=12$. That means $n=2007-12-3=1992$ which clearly does not work.

Case 2.2.2. Integers with $S(n)=15$. That means $n=2007-15-6=1986$ which also does not work

Case 2.2.3. Integers with $S(n)=21$. That means $n=2007-21-3=1983$ which is valid.

Case 2.2.4. Integers with $S(n)=24$. That means $n=2007-24-6=1977$ which is also valid.

We have considered every case, so there are $\boxed{4}$ integers that satisfy the given condition.

Let $n=1000a+100b+10c+d$ for $a,b,c,d<10$. We get $1001a+101b+11c+2d+S(S(n))=2007$.

If $a=2$, we have $101b+11c+2d+S(S(n))=5 \implies 2d+S(S(n))=5$, where $b=c=0$. We can have $d=0,1,2$, and we see $d=1$ is the only solution for the case where $a=2$. $\boxed{n=2001}$ is a solution.

If $a=1$, we have $101b+11c+2d+S(S(n))=1006$. We have $b=9$. Therefore, $11c+2d+S(10+c+d)=97$. Two more cases:

[List]
[*] [b]Case $1$: If $c=7$[/b]

If $c=7$, we get $2d+S(17+d)$, and testing gives $\boxed{n=1977}$ as the only solution to this case.

[*] [b]Case $2$: If $c=8$[/b]

If $c=8$, we get $2d+S(18+d)=9$, and testing gives $\boxed{n=1980}$ and $\boxed{n=1983}$
[/List]

Therefore, we have $4$ solutions.

It is well-known that $n \equiv S(n)\equiv S(S(n)) \pmod{9}.$ Substituting, we have that $$n+n+n \equiv 2007 \pmod{9} \implies n \equiv 0 \pmod{3}.$$ Since $n \leq 2007,$ we must have that $\max S(n)=1+9+9+9=28.$ Now, we list out the possible vales for $S(n)$ in a table, noting that it is a multiple of $3$ because $n$ is a multiple of $3.$
[center]
\begin{tabular}{c|c c c c c c c c c c}
 S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\
\end{tabular}
[/center]
Then, we compute the corresponding values of $S(S(n)).$
[center]
\begin{tabular}{c|c c c c c c c c c c}
 S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\
\hline
S(S(n)) & 0 & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\
\end{tabular}
[/center]
Finally, we may compute the corresponding values of $n$ using the fact that $n=2007-S(n)-S(S(n)).$
[center]
\begin{tabular}{c|c c c c c c c c c c}
n & 2007 & 2001 & 1995 & 1989 & 1992 & 1986 & 1980 & 1983 & 1977 & 1971 \\
\hline
S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\
\hline
S(S(n)) & 0 & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\
\end{tabular}
[/center]
Notice how all conditions are designed to be satisfied except whether $S(n)$ is accurate with respect to $n.$ So, the only thing that remains is to check this. We may eliminate, for example, when $n=2007$ we have $S(n)=9$ while the table states that it is $0.$ Proceeding similarly, we obtain the following table.
[center]
\begin{tabular}{c|c c c c c c c c c c}
n & \cancel{2007} & \boxed{2001} & \cancel{1995} & \cancel{1989} & \cancel{1992} & \cancel{1986} & \boxed{1980} & \boxed{1983} & \boxed{1977} & \cancel{1971} \\
\hline
 S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\
\hline
S(S(n)) & 0 & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\
\end{tabular}
[/center]
It follows that there are $\boxed{4}\implies \textbf{(D)}$ possible values for $n.$

[hide=Solution]Clearly the minimum value of $n$ is $2007-28-10=1969$, from $1999$. Also, the maximum value is $2001$. We have $n \equiv S(n) \equiv S(S(n)) \pmod 9$, so $n \equiv 0 \pmod 3$. Our possibilities are $1969, 1972, 1975, \ldots 2001$. Testing all of these gives that only $1977, 1980, 1983, 2001 \implies 4 \implies \boxed{\text{(D)}}$.[/hide]


hello, we have $ 2xy(x\plus{}2y)\plus{}x^3\plus{}8y^3\equal{}2xy(x\plus{}2y)\plus{}(2y\plus{}x)(x^2\minus{}2xy\plus{}4y^2)\equal{}(x\plus{}2y)(4y^2\plus{}x^2)$.
Sonnhard.

Here's a simpler solution (in my opinion)

[hide="solution"]$ x^3\plus{}8y^3\plus{}2x^2y\plus{}4xy^2$
Rearrange to get
$ x^3\plus{}4xy^2\plus{}2x^2y\plus{}8y^3$
Factor out by grouping
$ x(x^2\plus{}4y^2)\plus{}2y(x^2\plus{}4y^2)$
$ (x\plus{}2y)(x^2\plus{}4y^2)$[/hide]


Phil rolls 6 dice. What is the probability that at least two dice show the same number?

The probability that no two dice show the same number is $ \frac{5!}{6^6}\equal{}\frac{120}{46656}\equal{}\frac{5}{1944}$.  Thus, the probability that at least two dice show the same number is $ \frac{1939}{1944}$.

[quote=sray070][quote="AIME15"]The probability that no two dice show the same number is $ \frac{5!}{6^{6}}$[/quote]It should be $ \frac{6!}{6^{6}}$

When six dice are rolled either at least two dice show the same number or no two dice show the same number (i.e all the dice show show different numbers).  The probability that all the dice show different numbers is $\frac{6}{6}\times\frac{5}{6}\times\frac{4}{6}\times\frac{3}{6}\times\frac{2}{6}\times\frac{1}{6}$ because the first dice can be any number, the second dice can be any of the five numbers that's not the first, and so on.  So the probability that at least two dice show the same number is $1-\frac{6}{6}\times\frac{5}{6}\times\frac{4}{6}\times\frac{3}{6}\times\frac{2}{6}\times\frac{1}{6}=\boxed{\frac{319}{324}}$.[/quote]

Please hide your solution

[hide=Complementary Probability]This is a fairly simple casework problem, but we could make it even more simple by finding the complement.

[b]Case I - No Matches:[/b] In order for this to happen, we'd have 6 possibilities for the first die, 5 for the second, and so on, giving us 6 factorial. This makes a total of 720 possible combinations.

[s][b]Case II - One Match:[/b][/s] It isn't possible to have only one matching die.

We have only 720 possible (complementary!) combinations. Without restrictions, there would be $6^6=46656$ possible combinations, so the probability of getting a valid combination is $$\dfrac{46656-720}{46656}=\dfrac{45936}{46656}=\boxed{\dfrac{319}{324}}.$$ $$\definecolor{color}{rgb}{0.2, 0.3, 0.1} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$

Why doesn't this solution work:

You could pick 2 dice out of 6 to be the same which is 15. The other 4 dice don't matter so you would have

15*6^4/6. Which would cancel out to 15/36 or 5/12?


calculus integration function trigonometry LaTeX

Anyway to put this on one line without removing the big brackets, parenthesis, and nicely formated $ |z-i|=2$ underneath the integrals?
\begin{align*}
    \mathop\oint\limits_{|z-i|=2}\bigg[\frac{1}{(e^{z+1}-1)z^2}-
    \text{Sin}\bigg(\frac{1}{z-1}\bigg)\bigg]dz &= \\
    &\mathop\oint\limits_{|z-i|=2}\frac{1}{(e^{z+1}-1)z^2}dz-\mathop\oint\limits_{|z-i|=2}\text{Sin}\bigg(\frac{1}{z-1}\bigg)dz
    \end{align*}

If I remove the \align environment, it's worst:

$ \mathop\oint\limits_{|z-i|=2}\bigg[\frac{1}{(e^{z+1}-1)z^2}-
    \text{Sin}\bigg(\frac{1}{z-1}\bigg)\bigg]dz=    \mathop\oint\limits_{|z-i|=2}\frac{1}{(e^{z+1}-1)z^2}dz-\mathop\oint\limits_{|z-i|=2}\text{Sin}\bigg(\frac{1}{z-1}\bigg)dz$

Thank,
Shaw

[color=green]Moderator edit: I shortened the width to make the thread more readable[/color]

This seems to work. I don't know why :D
[code]
\mbox{}{
\mathop\oint\limits_{|z-i|=2}\bigg[\frac{1}{(e^{z+1}-1)z^2}-
\text{Sin}\bigg(\frac{1}{z-1}\bigg)\bigg]dz =
\mathop\oint\limits_{|z-i|=2}\frac{1}{(e^{z+1}-1)z^2}dz-
\mathop\oint\limits_{|z-i|=2}\text{Sin}\bigg(\frac{1}{z-1}\bigg)dz
}
[/code]
$ \mbox{}{ \mathop\oint\limits_{|z \minus{} i| \equal{} 2}\bigg[\frac {1}{(e^{z \plus{} 1} \minus{} 1)z^2} \minus{} \text{Sin}\bigg(\frac {1}{z \minus{} 1}\bigg)\bigg]dz \equal{} \mathop\oint\limits_{|z \minus{} i| \equal{} 2}\frac {1}{(e^{z \plus{} 1} \minus{} 1)z^2}dz \minus{} \mathop\oint\limits_{|z \minus{} i| \equal{} 2}\text{Sin}\bigg(\frac {1}{z \minus{} 1}\bigg)dz }$

Alright, this is a tentative fix unless someone can suggest how to display the code on one line here on the forum (it displays nicely as one line in TeXnicCenter):  I just scrapped the $ |z\minus{}i|\equal{}2$ and used a $ C$:

\[ \mathop\oint\limits_{C}\bigg[\frac {1}{(e^{z \plus{} 1} \minus{} 1)z^2} \minus{} \text{Sin}\bigg(\frac {1}{z \minus{} 1}\bigg)\bigg]dz \equal{}\mathop\oint\limits_{C}\frac {1}{(e^{z \plus{} 1} \minus{} 1)z^2}dz \minus{}\mathop\oint\limits_{C}\text{Sin}\bigg(\frac {1}{z \minus{} 1}\bigg)dz\]
where $ C: z\equal{}i\plus{}2e^{it}$.

[quote="shawtoos"]Hi Yury.  That's too long though.  I suspect people are annoyed whenever someone makes a post that doesn't fit on the computer screen and you have to use the right arrows.
Thanks though. :)[/quote]

What do you mean? It is only 3% (20 pixels) wider than your formula.

He meant that until I edited your post, the code and the formula stretched well beyond the right edge of the screen :).

Why is \mathop necessary here - see also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1102282#1102282]Valentin's comment[/url]?

$ \oint\limits_{C}\left[\frac {1}{(e^{z \plus{} 1} \minus{} 1)z^2} \minus{} \text{Sin}\left(\frac {1}{z \minus{} 1}\right)\right]dz \equal{} \oint\limits_{C}\frac {1}{(e^{z \plus{} 1} \minus{} 1)z^2}\,dz \minus{} \oint\limits_{C}\text{Sin}\left(\frac {1}{z \minus{} 1}\right)dz$

I'm intrigued by the capital S in Sin; I presume it denotes a complex-valued function though I have only seen it used for the principal value of a multi-valued function like Log.

Alright Fedga, that code fits in one display.

Steve, is lower-case trig names the standard convention?  I've found often in reading posts, personally, that lower case sin, cos, etc., is slightly more tedious  to read; it blends into the surrounding text whereas Sin, Cos, etc., stand out and conveys the meaning more easily and quickly.  Also that's the standard convention in Mathematica, not that they have dibs on convention or nothing, but I use Mathematica a great deal and just carry it over into my writing style.  Hope that's not annoying anyone here.

sin, cos etc are standard mathematical conventions which is why \sin, \cos are used in LaTeX.
Mathematica however capitalises all its functions for programming reasons but I wouldn't want to follow those conventions nor the use of [] instead of () when writing mathematics. Even the Mathematica publishers don't use Sin[x] on its pages such as [url=http://demonstrations.wolfram.com/IntegratingOddPowersOfSineAndCosineBySubstitution/]Integrating Odd Powers of Sine and Cosine by Substitution[/url] and [url=http://mathworld.wolfram.com/Trigonometry.html]Trigonometry[/url].
Thus I think you should distinguish between [i]doing[/i] mathematics using Mathematica and [i]writin[/i]g mathematics.

Ok, thanks :)  I'll try to begin switching over:

\[ \mbox{}{ \mathop\oint\limits_{|z \minus{} i| \equal{} 2}\bigg[\frac {1}{(e^{z \plus{} 1} \minus{} 1)z^2} \minus{} \sin\bigg(\frac {1}{z \minus{} 1}\bigg)\bigg]dz \equal{} \mathop\oint\limits_{|z \minus{} i| \equal{} 2}\frac {1}{(e^{z \plus{} 1} \minus{} 1)z^2}dz \minus{} \mathop\oint\limits_{|z \minus{} i| \equal{} 2}\sin\bigg(\frac {1}{z \minus{} 1}\bigg)dz }\]


I do not have very much background in topology so this question might be very obvious. Why is the real extended number line homeomorphic to the unit interval? An explanation of this would also help me to understand why its compact (since its homeomorphic to a closed bounded subset). Any help would be greatly appreciated.

That being said, I believe the one-point compactification of $ \mathbb{R}^n$ is actually homeomorphic to $ S^n,$ so the extended real line (with a single point at $ \infty$) is homeomorphic to the circle. If you extend it in both directions, it's homeomorphic to the closed interval, presumably.

The real line is homeomorphic to an open interval; $ \tan$ and $ \arctan$ are a nice pair of homeomorphisms there. Then $ \infty$ and $ \minus{}\infty$ correspond to the endpoints of that interval.

When you're working in the purely topological realm, as in this problem, "bounded" has no meaning. Homeomorphisms don't preserve distance, only open and closed sets. We don't even have to have a distance function, and compactness is still meaningful without one.

Nukular, I see what your saying, however,I believe that the extended real number line is a two point compactification unlike the projectively extended number line (its my fault for not clarifying).

jmerry, in the definition I found on wikipedia, it stated that the extended real line was homeomorphic to the unit interval which is closed. This is what I'm having trouble understanding.

Thanks for the answers.

"Closed" is meaningless here. The real line is closed. The extended real line is closed. Every topological space is closed, and homeomorphisms operate on whole spaces.
The interesting point is that the extended real line is compact, unlike the non-extended version. How can we show that other than the homeomorphism? Any open cover must include a set $ U$ containing $ \infty$ and a set $ U'$ containing $ \minus{}\infty$ (These may be the same set). Then $ U$ contains some unbounded interval $ (M,\infty)$ and $ U'$ contains some unbounded interval $ (\minus{}\infty,M')$. If $ M<M'$, these two sets cover everything and we're done. Otherwise, $ [M',M]$ is a compact set, and we can find a finite subcover of it. Take this collection together with $ U$ and $ U'$ for a finite subcover of the extended line.

Thanks for the replies and your patience. I understand now.

Alternately, a sequence is either bounded (in which case it has a convergent subsequence by the compactness of bounded closed subsets of $ \mathbb{R}$) or unbounded (in which case it has a subsequence tending to $ \pm \infty$).


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

I have a question.  Is $tr(A)=tr(A^{*})$ where $A^{*}$ is the classical adjoint matrix?

$Tr(A)=\sum_{i=1}^n A_{i,i}$ ?? where is the problem ?

What is $A^*$ in terms of $A$, do you know... ? Maybe you don't know it, this is probably you problem.

Well, $\left( A^\star \right)_{ii}=$ the determinant of $B$, where $B$ is obtained from $A$ by eliminating the $i$th row and the $i$th column.

Thus, $\textrm{tr} \left( A^\star \right)$ is equal to the coefficient of $x^{n-1}$ in $p_A$, the characteristic polynomial of $A$.

Also, $\textrm{tr} \left( A \right)$ is equal to the coefficient of $x$ in $p_A$.

(well, a sign might differ here and there, but you get the idea)

Note that for $n \geq 3$, the property in question does not always hold.


[quote="Dr Sonnhard Graubner"]hello, we have
$ \frac {a^2(b^{ \minus{} 1} \minus{} c^{ \minus{} 1}) \plus{} b^2(c^{ \minus{} 1} \minus{} a^{ \minus{} 1}) \plus{} c^2(a^{ \minus{} 1} \minus{} b^{ \minus{} 1})}{a(b^{ \minus{} 1} \minus{} c^{ \minus{} 1}) \plus{} b(c^{ \minus{} 1} \minus{} a^{ \minus{} 1}) \plus{} c(a^{ \minus{} 1} \minus{} b^{ \minus{} 1})} \equal{} a \plus{} b \plus{} c$.
Sonnhard.[/quote]

In what way could you simplify the fraction?

[quote="vinskman"]$ \frac {\sum_{cyc}a^2(b^{ \minus{} 1} \minus{} c^{ \minus{} 1})} {\sum_{cyc} a(b^{ \minus{} 1} \minus{} c^{ \minus{} 1})} \equal{} ?$[/quote]
[hide="Solution"]
It is equivalent to simplifying\[ \frac {\frac {a^2}{b} \minus{} \frac {a^2}{c} \plus{} \frac {b^2}{c} \minus{} \frac {b^2}{a} \plus{} \frac {c^2}{a} \minus{} \frac {c^2}{b}}{\frac {a}{b} \minus{} \frac {a}{c} \plus{} \frac {b}{c} \minus{} \frac {b}{a} \plus{} \frac {c}{a} \minus{} \frac {c}{b}} \equal{} \frac {a^3c \minus{} a^3b \plus{} ab^3 \minus{} b^3c \plus{} c^3b \minus{} c^3a}{a^2c \minus{} a^2b \plus{} b^2a \minus{} b^2c \plus{} c^2b \minus{} c^2a} \equal{}\]\[ a \plus{} \frac {ab^3 \minus{} a^2b^2 \minus{} b^3c \plus{} ab^2c \plus{} c^3b \minus{} abc^2 \minus{} c^3a \plus{} c^2a^2}{a^2c \minus{} a^2b \plus{} b^2a \minus{} b^2c \plus{} c^2b \minus{} c^2a} \equal{}\]\[ a \plus{} b \plus{} \frac {ab^2c \minus{} a^2bc \plus{} c^3b \minus{} c^2b^2 \minus{} c^3a \plus{} c^2a^2}{a^2c \minus{} a^2b \plus{} b^2a \minus{} b^2c \plus{} c^2b \minus{} c^2a} \equal{}\]\[ a \plus{} b \plus{} c\] [/hide]


algebra
polynomial
inequalities
algebra proposed

Let $y$ be a positive real such that $y^2-y+4\neq \sqrt{3y^3+19}$.
Prove that the polynomial $P(x)=x^3-yx^2+3x-1$ has one and only one real root.

i think your problem is wrong...

lets say i has more than 1...(having 1 at least is obvious because the complex root come in pairs) so it must have 3 such that

$abc=1$, $ab+bc+ca=3$, $a+b+c=y$ but $ab+bc+ca=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$

and see that $a=2, b=\frac{4}{7-\sqrt{17}}, c=\frac{4}{7+\sqrt{17}}$ are roots of $x^3-\frac{15}{4}x^2+3x-1$ and for \frac{15}{4} we have:

$y^2-y+4=\frac{15^2-15*4+4*4^2}{64}=\frac{229}{64}$ and the other thing is far more different

maybe im wrogn in my counts but check the problem, i dont think is right since that only 4 y´s have that equality done

no, i don't think so... let me show you my proof...

[b]Lemma[/b]
If $x,y,z \in \mathbb{R}$ such that
$x+y+z>0 \\xy+xz+yz>0 \\xyz>0$
then $x,y,z>0$
[u]Proof[/u]
Suppose for the sake of contradiction the lemma is false, then WLOG suppose $y,z<0$, this because for $xyz>0$ we must have an even number of negative elements. Now, we have that
$x>-y-z>-y>0 \\\Rightarrow 0>\frac{1}{z}>\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$
but when multiplying the last inequality by $xyz$ we obtain
$0>xy+xz+yz$, which is a contradiction, hence the lemma is true.

now, returning to the problem... suppose $a,b,c$ are the roots of $P(x)$ and $a,b,c$ are all real (remember we cannot have just two real roots). by the Viete's formula we must have
$a+b+c=y>0 \\ab+ac+bc=3 \\abc=1$
then by the lemma we must have $a,b,c>0$, but an application of the am-gm gives
$3=ab+ac+bc\geq 3\sqrt[3]{(abc)^2}=3 \Rightarrow AM=GM \Rightarrow ab=ac=bc \Rightarrow a=b=c=1 \Rightarrow y=3 \Rightarrow 3^2-3+4\neq \sqrt{3\cdot 3^3+19}$, but we have that $3^2-3+4=10=\sqrt{3\cdot 3^3+19}$, which is a contradiction and we must have not all of them to be real.

nice, huh? ;)


calculus
integration
function
logarithms
trigonometry
quadratics
absolute value

Probably really simple for you guys, but the announcement said computations of integrals so I figured it was ok to post it here  :|

$\int \frac{2x}{x^{2}+6x+13}dx$

I tried substitution, completing square, and everything of those sorts and couldn't figure out what to do with the $x$ term in the numerator  :huh:

Step 1: complete the square in the denominator.

$x^{2}+6x+13=(x+3)^{2}+?$

Step 2: Make the substitution $u=x+3.$ Note that you have an $x$ in the numerator that must be dealt with. Take care of it by writing $x=u-3.$

Step 3: When you have a sum (or difference) in the numerator of a fraction, don't be afraid to pull it apart into two pieces and deal with the pieces separately, using different methods.

Reminder: three kinds of functions can occur as the antiderivatives of rational functions: rational functions, logarithms, and arctangents.

$\int \frac{2x}{x^{2}+6x+13}dx=\int \frac{2x}{(x+3)^{2}+4}dx$

You can use integal by parts.

$\int{\frac{2x}{x^{2}+6x+13}}\,dx =\int{\frac{2x}{(x+3)^{2}+4}}\,dx$

Letting $u=x+3$ yields

$\int{\frac{2u}{u^{2}+4}}\,du-6\int{\frac{1}{u^{2}+4}}\,du$

Make the substitution $u=2\tan{\theta}$ into the right hand integral:

$\int{\frac{2u}{u^{2}+4}}\,du-3\int{}\,d\theta$

Evaluating this and retracing the substitution we have:

$\int{\frac{2x}{x^{2}+6x+13}}\,dx=\ln{|x^{2}+6x+13|}-3\tan^{-1}{\left(\frac{x+3}{2}\right)}+c$
Where $c$ is the constant of integration.

[quote="coffeym"]$\dots\ \ln{|x^{2}+6x+13|}\ \dots$[/quote]
That's correct, but the absolute value signs aren't actually needed. The irreducible quadratic $x^{2}+6x+13$ is always positive anyway.

I tend to regard actually doing the trigonometric substitution as overkill and a habit to break, in the interests of time.

We know that $\int\frac{dx}{x^{2}+1}=\arctan x+C.$

Now, what do we make of $\int\frac{du}{u^{2}+a^{2}}\ ?$

We know it's going to be an arctangent. Think for a moment about units of measurement. Assume that $a$ carries some such unit - meters maybe, or some other unit of length. Then $u$ would also have to carry the same unit. But the argument of a transcendental function like arctangent should be pure unitless numbers. How are we going to get that out of $u$ and $a$? So we guess that we're looking for $\arctan\left(\frac{u}{a}\right).$ Differentiate that; compare it with what we want, make the necessary fixes, and arrive at the formula

$\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C.$

(And if we'd have pushed the "units of measurement" issue a little harder, we'd have understood why we need that $\frac1a$ out front.)

And then, we memorize this result - it happens often enough that it's worth memorizing.

Yeah I did realise that the mod signs were unnecessary, but out of force of habit i included them anyway.  Sorry  :oops:


Given that $m$ is an eigenvalue of the non-singular matrix $A$ .Show that $m^{-1}$ is an eigenvalue of $A^{-1}$ .
Also , deduce that $m+m^{-1}$ is an eigenvalue of $A+A^{-1}$

Im new in matrix ,so please explain abit detail .Thanks

no eigenvalue can be zero as $A$ is not singular

also note that if $A v =\lambda v$ then $A^{-1}v=\lambda^{-1}v$

you should see now that $v$ is also an eigenvector of $A+A^{-1}$

so you are saying that we add up both thing to get

$(A+A^{-1})v=(\lambda++\lambda^{-1})v$ and so we conclude $\lambda+\lambda^{-1}$ is eigenvalue of $A+A^{-1}$ ?

Yes, I think that would do it.
Of course make sure $v\neq 0$

How about multicipities ?
I think also if $\lambda$ is eigenvalue of $A$ with multicipity $k$ then $\lambda+\lambda^{-1}$ is eigenvalue of $A+A^{-1}$. Some proof similiar to grobber's proof  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88566]here[/url] could work.


Joann ate a total of 100 lollipops in five days. Each day after the first day she ate six more than she had eaten on the previous day. How many lollipops did she eat on the third day?

The number of lollipops eaten is an arithmetic progression, so on the middle day (3rd out of 5) it should be the average, or 100/5 lollipops/day = 20.