Prove inductively that $ 6 | 7^{n} \minus{} 1$ for all $ n \in \mathbb{N^{*}}$. I just need to see how induction is used for these types of problems. In case that one is too trivial, prove inductively that $ 5 | n^{5} \minus{} n$ (same conditions on n).

Nevermind, I think I just figured it out.

Let $ \frac {7^{n} \minus{} 1}{6} \equal{} I$ where $ I$ is an integer. The formula obviously holds for $ n \equal{} 1$ so we look at $ n \plus{} 1$.

$ \frac {7^{n \plus{} 1} \minus{} 1}{6} \equal{} 7 \left(\frac {7^{n}}{6}\right) \minus{} \frac {1}{6}$

$ \equal{} 7 \left(\frac {7^{n} \minus{} 1}{6}\right) \minus{} \frac {1}{6} \plus{} \frac {7}{6}$

$ \equal{} 7I \plus{} 1$

which is an integer, thus by inductive reasoning the formula holds true.

[hide="1"]base case is trivial.

Assume true for n=k.

$ 7^k\minus{}1\equal{}6x$

$ 7^{k\plus{}1}\minus{}1\equal{}7(7^k\minus{}1)\minus{}6\equal{}6(7x\minus{}1)$

Therefore, $ 6|7^n\minus{}1$.[/hide]

[hide="2"]base case is trivial.

Assume true for n=k.

$ 5|k^5\minus{}k$

$ k^5\minus{}k\equal{}5x$

$ (k\plus{}1)^5\minus{}k\minus{}1\equal{}k^5\plus{}5k^2\plus{}10k^3\plus{}10k^2\plus{}4k\equal{}k^5\plus{}5(k^2\plus{}2k^3\plus{}2k^2\plus{}k)\equal{}5(x\plus{}k^2\plus{}2k^3\plus{}2k^2\plus{}k)$

Therefore, $ 5|n^5\minus{}n$.[/hide]

Ok and what's the conventions for inequalities? Ex/ prove inductively that $ n \leq 2^{n\minus{}1}$ for all natural n.

[hide="inequality"]Base case is trivial.

Assume true for n=k.

$ k\leq 2^{k\minus{}1}$

$ k\plus{}1 ? 2^k$

$ 2^k\equal{}2*2^{k\minus{}1}\geq 2k \geq k\plus{}1$

Therefore, $ n\leq 2^{n\minus{}1}$[/hide]

It's basically the same thing.


Explain why in first case, deutirium exchange takes place while in second case, it does not

The answer is simple: the first is a simple ketone, and in the presence of water there is an equilibrium (shifted to the ketone) between the ketone and its enol. When the enol reverts to the ketone, the nucleophilic alpha carbon can attack a hydrogen (or deuterium). In the second case, the enolate ion is quite stable and does not appreciably revert to the (di)ketone.


integration
calculus
probability
function
real analysis
real analysis unsolved

Investigate:

$\int_a^{\infty} \frac{n^2xe^{-n^2x^2}}{1 + x^2} dx$

for a > 0 and a = 0.

I almost have it.
[hide="Partial Solution"]Your problem is equivalent to:
$n^2e^{n^2}\int^{\infty}_a\frac{x}{e^{n^2(x^2+1)}(1+x^2)}$ Let $u=x^2+1$
$\frac{n^2e^{n^2}}{2}\int^{\infty}_{a^2+1}\frac{du}{e^{n^2u}u}$ _____________Let $v=e^u$
$\frac{n^2e^{n^2}}{2}\int^{\infty}_{e^{a^2+1}}\frac{dv}{v^{n^2-1}ln(v)}$
I thought about this last step for a while, but couldn't come up with any ideas. Could you post a hint?[/hide]

[quote="Peter VDD"][quote="diophantus"][quote="Peter VDD"]As far as I know it has no primitive, it's known as Ei(1,n^2*(1+a^2)) ... Ei being the "exponential integral".[/quote]
Could you provide a brief definition of the exponential integral?  Perhaps with equations.[/quote]\[Ei(a,z) = \int^\infty_1 e^{-k*z}k^{-a}\ dk\]
sounds familiar right? And as far as I know, Ei (1, x ) has no simpler form... perhaps here it has, but it seems very unlikely to me.[/quote]
But the limit on the portion of my that agrees with the exponential integral is not $[1,\infty]$ but $[a^2+1,\infty]$.  Would it still be applicable?

[quote="diophantus"][/quote][quote="Peter VDD"][quote="diophantus"][quote="Peter VDD"]As far as I know it has no primitive, it's known as Ei(1,n^2*(1+a^2)) ... Ei being the "exponential integral".[/quote]
Could you provide a brief definition of the exponential integral?  Perhaps with equations.[/quote]\[Ei(a,z) = \int^\infty_1 e^{-k*z}k^{-a}\ dk\]
sounds familiar right? And as far as I know, Ei (1, x ) has no simpler form... perhaps here it has, but it seems very unlikely to me.[/quote][quote="diophantus"]
But the limit on the portion of my that agrees with the exponential integral is not $[1,\infty]$ but $[a^2+1,\infty]$.  Would it still be applicable?[/quote]
\[Ei(a,z) = \int^\infty_1 e^{-k*z}k^{-a}\ dk\]
In our case:
\[Ei(1,n^2(a^2+1)) = \int^\infty_1 \frac{e^{-k*n^2(a^2+1)}}{k}\ dk\]
Now do k * (1+a^2) = u
\[Ei(1,n^2(a^2+1)) = \int^\infty_{1+a^2} \frac{e^{-u*n^2} (1+a^2)}{u (1+a^2)}\ du = \int^\infty_{1+a^2} \frac{e^{-u*n^2}}{u}\ du \]

Well then that's the answer.
BTW what are some practical applications of Ei() besides just evaluating integrals.  Is it used in probability, etc?

I have to say I do not know. I know this function only from integration... but I'm pretty sure it does have "some" meaning in statistics...

[quote="sweetonect"]use $n^2x^2=y$
maybe works :([/quote]hmm... I don't see how that helps us here (and I think the previous shows there is no simple form)

could you explain how you would take it with this substitution? I only see it getting more complex...

The function you are talking about is a form of the normal distribution widely used in statistics..


(Also posted in the geometry section.)

Given circle with radius $r$ and midpoint $(x_1,y_1)$, let $A =$ the area of the circle in the first quadrant, $B =$ 2nd quadrant, $C$, 3rd, and $D$, 4th. Maximize $AC-DB$, or prove that it is invariant.

How would this be done? Any ideas?

Well, it can't be invariant - consider a circle entirely within one quadrant, then consider a circle tangent to the origin but entirely outside one quadrant.

I'm guessing that the maximum occurs in the second case with the line between the origin and the center on $y = x$ or $y = -x$.  As for a way to actually approach the problem, I'm inclined towards power of a point with the x- and y-axes, then deriving expressions for the area in each quadrant, but I'm far too tired to do it now  :)


$ \text{If } a\plus{}b\plus{}c\equal{}10 , \text{where}  a,b \text{  and } c \text{ are non\minus{}negative integers.}$

$ \text{ Find the max.value of }$

$ abc\plus{}ab\plus{}bc\plus{}ca$

[hide]abc \leq (\frac{a+b+c}{3})^3
ab+bc+ca \leq \frac{(a+b+c)^2}{3}
summing up ...[/hide]

$ abc \leq (\frac{a\plus{}b\plus{}c}{3})^3$
$ ab\plus{}bc\plus{}ca \leq \frac{(a\plus{}b\plus{}c)^2}{3}$
$ \text{summing up ... }$

$ \text{ I don't agree with yourwork ,HTA . Because}$

$ \text{If } abc \equal{} (\frac{a\plus{}b\plus{}c}{3})^3 , \text{then } \implies a\equal{}b\equal{}c  , \text{then }a\equal{}b\equal{}c\equal{}\frac{10}{3}$

$ \text{contradicts a,b,c are non\minus{}neagtive integers.}$

hello, i have found that the searched maximum is $ 69$, this will be reached for example if $ a\equal{}4,b\equal{}3,c\equal{}3$.
Sonnhard.

$ \text {Hello,Dr Sonnhard Graubner.}$

$ \text {That's right!!}$

$ \text {How do you find out the answer?}$

[quote="vinskman"]$ abc \leq (\frac {a \plus{} b \plus{} c}{3})^3$
$ ab \plus{} bc \plus{} ca \leq \frac {(a \plus{} b \plus{} c)^2}{3}$
$ \text{summing up ... }$

$ \text{ I don't agree with yourwork ,HTA . Because}$

$ \text{If } abc \equal{} (\frac {a \plus{} b \plus{} c}{3})^3 , \text{then } \implies a \equal{} b \equal{} c , \text{then }a \equal{} b \equal{} c \equal{} \frac {10}{3}$

$ \text{contradicts a,b,c are non \minus{} neagtive integers.}$[/quote]
oh , I'm sorry , i read the problem carelessly
here is my proof , hope that it is correct this time
note that if $ b\leq a\ leq c$ then $ f(a,b \plus{} 1,c \minus{} 1) \geq f(a,b,c) (since f(a,b,c) \minus{} f(a,b \plus{} 1,c \minus{} 1) \equal{} (b \minus{} c \plus{} 1)(a \plus{} 1) \leq 0$ <b can not be equal to c > )
so , if the maximum occurs when$ f(a,b,c) \equal{} f(a_0,b_0,c_0) b_0\leq a_0\ leq c_0$ then $ f(a_0,b_0,c_0) \equal{} f(a_0,b_0 \plus{} 1,c_0 \minus{} 1) \minus{} > b_0 \equal{} c_0 \plus{} 1 \equal{} a_0$
--> $ b_0 \equal{} a_0 \equal{} 3 ; c_0 \equal{} 4$
the maximum is 69


My teacher is confusing me in how to name Haloalkanes, I know this is simple but I seem to name them differently. Please name the following:

$ CCl_{3}CF_{2}Cl$
$ CCl_{2}BrCBrICBr_{3}$
$ CH_{3}CHFCHFCHICH_{3}$

1) 1,1,1,2-Tetrachloro-2,2-difluoroethane.

2) 1,1,1,2,3-Pentabromo-3,3-dichloro-2-iodopropane.

3) 2,3-Difluoro-4-iodopentane.

The purpose is to give the lowest numbers to the substituent that comes first in alphabetical order.

My answers exactly :!:  My teacher was saying something about electronegativity :?:

[quote="BanishedTraitor"]  My teacher was saying something about electronegativity :?:[/quote]
 :ninja:

Nomenclature of organic compounds doesn't have nothing to do with electronegativity. At most, atomic number arguments can be involved.


calculus
integration
limit
trigonometry
calculus computations

Let $k$ be a positive real number. Compute $\lim_{n \to \infty}\int_{0}^{k}\sin \left( x^{n}\right) \, dx$.

i dont know if it is uniformly convergent

but assuming it is

${\lim_{n\to \infty }\, \int_{0}^{k}\sin \left(x^{n}\right) \, dx=\int_{0}^{k}\lim_{n\to \infty }\, \sin \left(x^{n}\right) \, dx}$

Assuming

${0<x<1}$

${\lim_{n\to \infty }\, \sin \left(x^{n}\right)=0}$

${\int_{0}^{k}0 \, dx=0}$

it if is now uniformly convergent, need a another proof

pum pmu...

You have uniform convergence if $k<1$, and we can handle $k=1$ with a limit. For $k>1$, we need another technique.

$\int_{1}^{k}\sin(x^{n})\,dx=\int_{1}^{k}\frac1{nx^{n-1}}\cdot nx^{n-1}\sin(x^{n})\,dx= \left[-\frac1{nx^{n-1}}\cos(x^{n})\right]_{1}^{k}-\int_{1}^{k}\frac{n-1}{nx^{n}}\cos(x^{n})\,dx$
Both terms clearly go to zero.


trigonometry
geometry
circumcircle
trig identities
Law of Sines

[hide]
let $\angle A$ be opposite the side of length a and $\angle B$ be opposite the side of length b. Drop an altitude from C to AB.  Let the length of this altitude be h.

$\sin\angle A=\frac{h}{b}$
$\sin\angle B=\frac{h}{a}$
Divide one equation by the other
$\frac{\sin\angle A}{\sin\angle B}=\frac{\frac{h}{b}}{\frac{h}{a}}$
$\frac{\sin\angle A}{\sin\angle B}=\frac{a}{b}$
$\frac{\sin\angle A}{a}=\frac{\sin\angle B}{b}$
[/hide]

[quote="buzzer11"]Isn't there a third part to the equation? Of course, it easily follows from the proof given by tjhance, but...[/quote]
The so-called "Extended Law of Sines":

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$,

where $R$ is the circumradius of $\triangle{ABC}$.


inequalities
trigonometry
real analysis
triangle inequality
calculus
calculus computations

Show that if D is bounded and f is uniformly continuous on D, then f is bounded on D

[hide]Since $ f$ uniformly continuous, there exists a $ \delta$ such that if $ x,y$ are in $ D$ and $ |x\minus{}y|<\delta$, then $ |f(x)\minus{}f(y)|<1$.

Now pick $ x_0$ in $ D$. Since $ D$ is bounded, we can find $ M$ such that $ |x\minus{}x_0|<M$ for all $ x$ in $ D$. There exists a positive integer $ N$ such that $ N>\frac{M}{\delta}$. So for any particular $ x$, we can pick $ x_1,x_2,\cdots x_N$ (with $ x_N\equal{}x$) so that the $ x_k$ are evenly spaced. That means $ |x_k\minus{}x_{k\plus{}1}|\equal{}\frac{|x\minus{}x_0|}{N}<\delta$ so $ |f(x_k)\minus{}f(x_{k\plus{}1})|<1$.

Using the above for $ k\equal{}0,1,2,\cdots N\minus{}1$, adding, and using triangle inequality, $ |f(x)\minus{}f(x_0)|<N$ so $ f(x_0)\minus{}N<f(x)<f(x_0)\plus{}N$ for all $ x$ in $ D$. So $ f$ bounded (note that $ N$ is independent of $ x$).

(Note: the converse is not true, consider $ \sin \frac{1}{x}$ in $ (0,1)$.)[/hide]

The nature of D should be clarified. Is it a subset of some space with some structure? Is it connected or convex, or something else?


$S$ majmoe na motenahi az a'adade tabiei  ast.sabet konid do ozve $S$ vojood darand ke jameshan maghsoom elayhe avvali bozorgtar az $10^6$ darad. :)

yani hich kasi nemidone baraye halle in masale chi kar bayad bokonim? :huh:

bah,bah,yani in hame adam to in forum  hich kasi chizi be zehnesh nemirese? :)

benazaram yekam sakhte hala rosh dobare beshtar fekr mikonam

be nazare man kare dorosti mikonid chon in masale arzeshe fekr kardan dare. :)


inequalities
geometry
circumcircle
geometry proposed

Consider an $ A$-isosceles $ \triangle ABC$ with circumcircle $ w$ . Denote the midpoint $ D$ of $ [BC]$ and $ \phi =m(\angle BAC)$ .

Prove that for any point $ M\in w$ so that $ BC$ doesn't separate $ A$ , $ M$ there is the conditioned inequality :

$ \|\begin{array}{ccc}
DM\ \le\ DA & \mathrm{if} & \phi \le\ 90^{\circ}\\\\
DM\ \ge\ DA & \mathrm{if} & \phi\ \ge\ 90^{\circ}\end{array}$ . We have equality, i.e. $ DM=DA$ if and only if $ M: =A$ or $ \phi =90^{\circ}$ .

Let $ w'$ the circle with midpoint $ D$ and radius $ DA$. The statement follows from the fact that if $ \phi > 90^\circ$, $ w'$ is externally tangent to $ w$ at $ A$; if $ \phi \equal{} 90^\circ$, $ w' \equal{} w$ and if $ \phi < 90^\circ$, $ w'$ is internally tangent to $ w$ at $ a$.

edit: Have I misunderstood something? Just because I don't need the condition that $ BC$ doesn't separate $ A$, $ M$...


Simplify $2+\sqrt{2}+\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2}$.

Note : This was a team problem, but don't use a calculator.

[quote="bpms"][hide]I got positive 2...[/hide][/quote]
Me, too. :)

[hide]
So we start with $2+\sqrt{2}+\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2}$. From here, you rationalize the denominators and end up with $2+\sqrt{2}+\frac{2-\sqrt{2}}{2}+\frac{\sqrt{2}+2}{-2}$. Now, just put everything under the common denominator of 2, add it up, and you get $\frac{4}{2}= \fbox{2}$.[/hide]


[quote="arqady"][quote="Toan_VN_LC"]let a;b;c be non-negative real numbers,prove that:
$ \frac {{{a^2}}}{{{b^2} \minus{} bc \plus{} {c^2}}} \plus{} \frac {{{b^2}}}{{{a^2} \minus{} ac \plus{} {c^2}}} \plus{} \frac {{{c^2}}}{{{a^2} \minus{} ab \plus{} {b^2}}} \ge 2 \plus{} \frac {{abc({a^3} \plus{} {b^3} \plus{} {c^3})}}{{3({b^2} \minus{} bc \plus{} {c^2})({a^2} \minus{} ac \plus{} {c^2})({a^2} \minus{} ab \plus{} {b^2})}}$[/quote]
$ \sum_{cyc}\frac {{{a^2}}}{{{b^2} \minus{} bc \plus{} {c^2}}} \ge 2 \plus{} \frac {{abc({a^3} \plus{} {b^3} \plus{} {c^3})}}{{3({b^2} \minus{} bc \plus{} {c^2})({a^2} \minus{} ac \plus{} {c^2})({a^2} \minus{} ab \plus{} {b^2})}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(3a^6 \minus{} 3a^5b \minus{} 3a^5c \minus{} 3a^4b^2 \minus{} 3a^4c^2 \plus{} 6a^3b^3 \plus{}$
$ \plus{} 8a^4bc \minus{} 3a^3b^2c \minus{} 3a^3c^2b \plus{} a^2b^2c^2)\geq0\Leftrightarrow\left(\sum_{cyc}(a^3 \minus{} a^2b \minus{} a^2c \plus{} abc)\right)^2 \plus{}$
$ \plus{} \sum_{sym}(a^6 \minus{} a^5b \minus{} 2a^4b^2 \plus{} a^3b^3 \plus{} 3a^3b^2c \minus{} 2a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\left(\sum_{cyc}(a^3 \minus{} a^2b \minus{} a^2c \plus{} abc)\right)^2 \plus{} \sum_{cyc}(a \minus{} b)^2S_c\geq0,$
where $ S_c \equal{} a^4 \plus{} a^3b \minus{} a^2b^2 \plus{} ab^3 \plus{} b^4 \minus{} c^3(b \plus{} c) \plus{} 2c^2ab.$
Let $ a\geq b\geq c.$ Hence, $ S_c\geq0,$ $ S_b\geq0$ and we are done:
$ \sum_{cyc}(a \minus{} b)^2S_c\geq(a \minus{} c)^2S_b \plus{} (b \minus{} c)^2S_a\geq(b \minus{} c)^2(S_a \plus{} S_b)\geq0.$[/quote]
thank you,arqady ;)
here is my solution:

$ \Leftrightarrow 3\sum {{a^6}} \minus{} 3\sum {ab({a^4} \plus{} {b^4})} \minus{} 3\sum {{a^2}{b^2}({a^2} \plus{} {b^2})} \plus{} 6\sum {{a^3}{b^3}} \plus{} 8abc\sum {{a^3}} \minus{} 3abc\sum {ab(a \plus{} b)} \ge 0$
$ \Leftrightarrow 3\left( {\sum {{a^6} \plus{} abc\sum {{a^3} \minus{} \sum {ab({a^4} \plus{} {b^4})} } } } \right) \minus{} 3\left( {\sum {{a^2}{b^2}({a^2} \plus{} {b^2})} \minus{} 2\sum {{a^3}{b^3}} } \right) \plus{} 3abc\left( {2\sum {{a^3}} \minus{} \sum {ab(a \plus{} b)} } \right) \minus{} abc\left( {\sum {{a^3}} \minus{} 3abc} \right) \ge 0$
$ \Leftrightarrow 3\sum {{{(a \minus{} b)}^2}({a^4} \plus{} {b^4} \minus{} {c^4})} \minus{} 6\sum {{a^2}{b^2}{{(a \minus{} b)}^2}} \plus{} 6abc\sum {(a \plus{} b){{(a \minus{} b)}^2}} \minus{} abc(a \plus{} b \plus{} c)\sum {{{(a \minus{} b)}^2}} \ge 0$
$ \Leftrightarrow \sum {{{(a \minus{} b)}^2}(3{a^4} \plus{} 3{b^4} \minus{} 3{c^4} \minus{} 6{a^2}{b^2} \plus{} abc(5a \plus{} 5b \minus{} c)} ) \ge 0$
$ {S_a} \equal{} 3{b^4} \plus{} 3{c^4} \minus{} 3{a^4} \minus{} 6{b^2}{c^2} \plus{} abc(5b \plus{} 5c \minus{} a)$
$ {S_b} \equal{} 3{a^4} \plus{} 3{c^4} \minus{} 3{b^4} \minus{} 6{a^2}{c^2} \plus{} abc(5a \plus{} 5c \minus{} b)$
$ {S_c} \equal{} 3{a^4} \plus{} 3{b^4} \minus{} 3{c^4} \minus{} 6{a^2}{b^2} \plus{} abc(5a \plus{} 5b \minus{} c)$
we assume that: $ a \ge b \ge c$
then,
$ {S_c} \equal{} 3{({a^2} \minus{} {b^2})^2} \plus{} c(5{a^2}b \minus{} {c^3}) \plus{} abc(5b \minus{} c) > 0$
$ {S_b} \equal{} 3({a^4} \minus{} {b^4}) \plus{} 2a{c^2}(b \minus{} a) \plus{} 4{a^2}c(b \minus{} c) \plus{} abc(a \minus{} b \plus{} 3c)$
$ \equal{} (a \minus{} b)\left( {3(a \plus{} b)({a^2} \plus{} {b^2}) \minus{} 2a{c^2}} \right) \plus{} 4{a^2}c(b \minus{} c) \plus{} abc(a \minus{} b \plus{} 3c) > 0$
$ {a^2}{S_b} \plus{} {b^2}{S_a} \equal{} 3({a^6} \plus{} {b^6}) \plus{} 3{c^4}({a^2} \plus{} {b^2}) \minus{} 3{a^2}{b^2}({a^2} \plus{} {b^2}) \minus{} 6{c^2}({a^4} \plus{} {b^4}) \plus{} abc(5{a^3} \plus{} 5{b^3} \plus{} 5{a^2}c \plus{} 5{b^2}c \minus{} {a^2}b \minus{} a{b^2})$
$ \equal{} 3\left( {({a^6} \plus{} {b^6}) \minus{} {a^2}{b^2}({a^2} \plus{} {b^2})} \right) \plus{} 3{c^4}({a^2} \plus{} {b^2}) \plus{} 6({a^4}bc \plus{} a{b^4}c \minus{} {c^2}{a^4} \minus{} {b^4}{c^2}) \plus{} 5ab{c^2}({a^2} \plus{} {b^2}) \minus{} abc(a \plus{} b)({a^2} \plus{} {b^2})$
$ > 3{(a \minus{} b)^2}({a^2} \plus{} {b^2}){(a \plus{} b)^2} \plus{} ({a^2} \plus{} {b^2})(3{c^4} \plus{} 5ab{c^2} \minus{} abc(a \plus{} b)) \plus{} 6({a^4}c(b \minus{} c) \plus{} {b^4}c(a \minus{} c))$
$ > 3{(a \minus{} b)^2}({a^2} \plus{} {b^2}){(a \plus{} b)^2} \plus{} ({a^2} \plus{} {b^2})abc(c \minus{} a \plus{} c \minus{} b) \plus{} 2({a^4}c(b \minus{} c) \plus{} {b^4}c(a \minus{} c))$
$ \equal{} {(a \minus{} b)^2}(3{a^4} \plus{} 6{a^3}b \plus{} 6{a^2}{b^2} \plus{} {a^2}bc \minus{} 2{a^2}{c^2} \minus{} 2ab{c^2} \plus{} {b^2}ac \plus{} 6a{b^3} \plus{} 3{b^4} \minus{} 2{b^2}{c^2}) > 0$
So,we have done :D

Thank you for posting your own solution.

I think everyone who create an own problem ,at the end of the discussion the proposer should post own solution


Find the greatest real r such that some normal line to the graph of y=x^3+rx passes through the origin, where the point of normality is not the origin.

[hide]First, notice that if $ r>0$, there is trivially no solution, and if $ r=0$, we violate the conditions of the problem. Hence, $ r<0$.

We want $ f'(k)f(k)/k =-1 \implies f'(k)f(k)+k=0\implies (3k^{2}-r)(k^{3}+rk)+k=0$

Either $ k=0$ or $ (3k^{2}-r)(k^{2}+r)+1=0$ $ \implies 3(k^{2})^{2}+2r(k^{2})+(1-r^{2})=0$

Thus, $ \exists r : k^{2}= \frac{\sqrt{4r^{2}-3}-r}{3}\geq 0$ $ \implies \sqrt{4r^{2}-3}> r$.

If $ \sqrt{}\in \mathbb{R}\implies 4r^{2}-3 \geq 0$, then such a $ k$ exists. $ r^{2}\geq \frac{3}{4}\implies \boxed{r \leq-\frac{\sqrt{3}}{2}}$[/hide]


Hi all,

Can you post problems of Canadian Mathematical Olympiad (CMO) 2006?

Thanks  :wink:

Sorry! Because I finded it here  :blush:
http://www.cms.math.ca/Competitions/CMO/

I will post these problems in box National Olympiad later.

Are you agree?   :)

CMO 2006 with solutions can be found [url=http://www.math.utoronto.ca/barbeau/cmo2006sols.pdf]here[/url]

[quote="mathgeniuse^ln(x)"]Is there a link with problems only???[/quote]

Yes, here http://www.mathlinks.ro/Forum/viewtopic.php?t=130653

i know this is fairly specific, but for last year's CMO #3, does anyone know of another solution, not involving e.g. Yufei's ingenious double-counting? Thanks!

There is an induction proof provided here:
http://camel.math.ca/Competitions/CMO/solutions/sol_2006.pdf
(look at solution 1)

The problems are also posted here

[url]http://www.cms.math.ca/Competitions/CMO/reports/CMOReport2006.pdf[/url]


group theory
abstract algebra
superior algebra
superior algebra unsolved

If a group $ G $ is torsion-free, and $ H $ is a cyclic subgroup of $ G $ with finite index, show that $ G $ is cyclic.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43409]this thread[/url] from July.


Hello all. This is my 6th year of trainment to national math contests. I always used pencil (normal pencil), but now i'm willing to get a mech. pencil (i think).

Some friends h. suggested me a pentel sharp one. I already have one, but i'm having problems to use it, since it's, for me, a lot different of normal pencil.

So here comes my questions:

I) What do you think about what's the best? Should i continue using n-pencil? Do you know any famous person who uses (or used) pencil on big contests? (the last is just for curiosity).
II) Is pentel sharp a good option? Or it depends?

Well, i really would be grateful if someone help me. Thank you. (Sorry for my bad english).

I prefer a pen. There never is any use of erasing in contests because it slows you down. You can always use more scratch paper.

I've always used mechanical pencils for math contests, no sharpening needed, rarely have to change lead, though I did use pen for USAMO (it shows better on faxing) To me the only drawback is when tests like the SAT don't let you use them (citing that you might poke a hole in the answer sheet with the lead...)
As for a model, I'd go for something with a large eraser, and metal, that will probably be the most durable.

I have always used mechanical pencils for almost all my math contest so far. You don't have to sharpen them, and it always offers me a comfortable grip ;) I prefer to use the BiC (ones like this: http://content.etilize.com/Large/11960993.jpg) I mean you could still use it in USAMO or contests like that I suppose, they are pretty dark it used correctly.

I have always used PhD Ultra pencils - they are costly, roughly $ \$ 10$ per pencil (although I get them for half the price at a nearby Target), but they are highly durable, and I need only buy one per school year. They have nice grips, good erasers, and use 0.5 mm lead.

They're good for spinning too (significantly "fatter" and weighted than other pencils), which is what I frequently do under adrenaline and pressure. It's the spinning (and sometimes dropping) that probably wears my pencil down so much, but only having to buy one a year is worth it.

Using different pencils on examinations will not affect the outcome, so copying famous peoples' pencil uses will do you no good. You should use a pencil that you're comfortable with, not because someone scored a perfect on such and such contest.

[hide][img]http://www.kaboodle.com/hi/img/2/0/0/47/e/AAAAAlGU4C0AAAAAAEfqrA.jpg[/img][/hide]

I prefer pen, because with a pen, you don't have to worry about the tip breaking, and therefore, you can concentrate more on the problem. Also, you can always cross out the answer instead of having to go through all the erasing. However, there was one point where I ended up having to use a pink highlighter. Lol

I never use pencil (aside from filling in bubbles). I find pens much easier to read, and I like the way they feel against the paper.

Between "normal" pencils and mechanical pencils, I feel more comfortable with a normal pencil. I can't think of any benefit of using mechanical pencils except that:

(i) Some people are used to using them, and like the feel
(ii) They don't need to be sharpened

But (i) doesn't apply to you, and (ii) isn't a problem if you have multiple pencils, have a sharpener on hand, or are using pens.

[quote="TZF"](i) Some people are used to using them, and like the feel
(ii) They don't need to be sharpened[/quote]
I would also add they always have a nice fine point, which makes a more uniform and coherent "thickness" or "darkness" with the pencil marks and letters.
With regular pencils the fine tip can quickly smooth or break off and make lighter marks.

"Using different pencils on examinations will not affect the outcome, so copying famous peoples' pencil uses will do you no good. You should use a pencil that you're comfortable with, not because someone scored a perfect on such and such contest."

I know this. This is just a foolish that can make me feel better about my choices.

For example: some years ago i was playing some computer game and some friend told me i wouldn't go so far using the mouse i was using. I would never think about that, but, when i changed it, i really realized that the one i was using was limitated. That's why i'm asking about pencils.

For example, do the chinese guys often use pencils?

It's stupid, but some example would make me feel a little better.

And thanks to the answers.

Why are you so interested in what Chinese kids use?

Let's see why a certain mouse might be better than another mouse:
- Greater accuracy
- Appropriate/programmable sensitivity
- Faster response time
- More ergonomic to hold and use
- More features (e.g. programmable keys, greater scroll wheel functionality, etc)

We've pretty much covered any such qualities of pencils here. You're not going to find a pencil that'll give you "faster response time" or have programmable hotkeys. It depends on what feels best in your hand.

It's clear that some people value a pencil that doesn't need to be sharpened, or produces a fine line with consistent width, or has a certain grip, or whatever. A lot of these people settle for mechanical pencils (or fine pens). Many others settle for normal pencils or pens of their choice, according to what they like to use.

If you're seriously considering converting to mechanical pencils, try out different pencils with:
- Different graphite widths
- Different grip shapes
- Different mechanisms to release extra graphite (e.g. clicking the side, clicking the end, rotating the tip, rotating the end, etc)

and figure out which, if any, you'll want to use. Or, you might find out that you prefer normal pencils. (Or pens!)

Sorry for reviving an old thread, but I think that you might want to try a shaker. If you could go to East Asia, buy some The Shaker and then try them. They are expensive, but a perfect mix of precision and ergonomics. If you don't mind a really large grip and writing big, try the Dr.Grips. They are expensive, but I have written for hours on those- get the G-spec.

As for gaming peripherals, there is a marked change once you upgrade.
I upgraded from a regular Microsoft mouse to a Razer Lachesis. It was a very marked change, and I could feel it immediately. Not only was the OTF sensitivity very helpful, and the six macro buttons to be even better than I thought. (I want to upgrade to the Silverstone Raven-Mouse once I get Windows Vista simply because it exploits all the features and I'm too lazy to configure another mouse.)
As for the keyboard, it is up to you. I chewed through one keyboard in one year (A Creative Fata1ty, was not worth the 50 dollars I put up for it), which was really nice... except that you could see all the dust inside. Now, I have the Razer Lycosa. It's a perfect balance of ergonomics and performance, and I prefer to bang away on that keyboard for hours on end than using my stock keyboard. It's all about personal preference- is your eighty dollars worth more than your hands after a couple of hours on the computer?

[quote="mathfreak123"]Personally, I like mechanical pencils better because its lead is always uniform. I hate having to sharpen pencils or grab another one because it causes me to lose focus.[/quote]

and if you run out of lead?

Use a spare mechanical pencil, much like one would do after a regular pencil is used up.
Also, if there was enough time, one can also just refill the pencil with extra lead (and do the same with erasers if one has spare ones).
But if a mechanical pencil is jammed? When one has the time, it's easy to fix: just open the tip and push the broken piece of lead inside of the tip out, generally with the lead that's still part of the mechanical pencil in use.

I personally prefer mechanical pencils because they write more evenly than normal pencils.  I'm also ridiculously particular about having a sharp pencil. :) So a mechanical pencil can provide that constant sharpness, which is really hard to attain with a normal pencil.  My favorite mechanical pencil is probably the Papermate Expressions, although those are more girl-oriented.  Guys might not like them so much.....

Hey ppl, why don't you just assume there's a crazy guy like me that is wishing to know about pencil users?

I just want some good reference, like IMO GOLD medalist or something...

You have no idea about how it tranquilizes me.

I like mechanical because it is always sharp and you can just put a bunch of lead in it.

[quote="Mewto55555"]I for one prefer to just walk in to competitions with like 20 sharpened regular pencils. If one of them breaks or is too dull, I just grab another one, and I don't have to worry about running out.[/quote]

Exactly. Also, the lead of mechanical pencils break too fast, unlike regular pencils.


If ABC is an acute triangle with circumcenter O, orthocenter H and circumradius R, show that for any point P on the segment OH, PA+PB+PC <=  3R.

http://www.mathlinks.ro/Forum/viewtopic.php?t=19316

  darij


Let $ A,B\in \mathcal{M} _{3}( \mathbb R)$ satisfying $ det(A)\equal{}det(B)\equal{}det(A\plus{}B)\equal{}det(A\minus{}B)\equal{}0$
Prove that $ det(xA\plus{}yB)\equal{}0$ $ \forall x,y \in \mathbb R$

I tried by considering the polinomyal $ P(x,y)$ but I didn't manage to do it.

$ det(A\plus{}tB)\equal{}det(A)\plus{} a.t \plus{} bt^2 \plus{} (det B)^3.t^3$

use det(A+B)=det(A-B)=0

a+b=a-b=0 then a=b=0

det(A+tB)= 0 for any t real


logarithms
number theory unsolved
number theory

[i] For each positive integer n . Prove that there are only finite pair of positive integers $ (x;y)$ such that :

                 $ | x! \minus{} x^y| \ \leq \ n$ [/i]

Actually there are an infinite number given by $ x \equal{} 1$ and any $ y$.
However, with the extra condition that $ x > 1$,

For a given $ n$ there are $ n\minus{}1$ values of $ x$ such that $ 1 < x \le n$
For each of these values of $ x$ we must have $ y \le \frac{\log(x! \plus{} n)}{\log{x}}$

So there are only a finite number of solutions for which  $ x \le n$

If $ x > n$ then $ x$ divides $ |x! \minus{} x^y| \le n < x$
The only non-negative number less than $ x$ which is divisible by $ x$ is zero $ \implies x! \minus{} x^y \equal{} 0$

If $ x > 2$ then $ gcd(x \minus{} 1,x^y) \equal{} 1$ but $ gcd(x \minus{} 1, x!) \equal{} x\minus{}1$ so $ x! \ne x^y$
So the only solution to $ x! \equal{} x^y$ for $ x > 1$ is $ (x,y) \equal{} (2,1)$

Therefore the total number of solutions is finite.


[quote="Lovasz"]Let $x,y,z \geq 0$. Prove that:
\[x\sqrt{y+z \over (z+x)(x+y)}+y\sqrt{z+x \over (x+y)(y+z)}+z\sqrt{x+y \over (y+z)(z+x)}\geq 3\sqrt{3xyz \over 2(xy+yz+zx)}. \]
[/quote]

Got it

First, let us make the denominators all the same. We get

$\frac{x(y+z)+y(x+z)+z(x+y)}{\sqrt{(x+y)(y+z)(x+z)}}$ for the LHS

Also look at the $2xy+2yz+2zx$. We can write this as
$xy+yz+yz+zx+zx+xy$ = $y(x+z)+x(y+z)+z(x+y)$

So we have
$\frac{x(y+z)+y(x+z)+z(x+y)}{\sqrt{(x+y)(y+z)(x+z)}}\geq 3\sqrt{\frac{3xyz}{y(x+z)+x(y+z)+z(x+y)}}$

$\frac{x(y+z)+y(x+z)+z(x+y)}{3}\geq \sqrt{\frac{3x(y+z)y(x+z)z(x+y)}{y(x+z)+x(y+z)+z(x+y)}}$

Now let $a=x(y+z), b=y(x+z),$ and $c=z(x+y)$ and we get

$\frac{a+b+c}{3}\geq \sqrt{\frac{3abc}{a+b+c}}$

$\frac{(a+b+c)^{\frac32}}{3^{\frac32}}\geq \sqrt{abc}$

$\frac{(a+b+c)^{3}}{3^{3}}\geq abc$

$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$

Which is true by AM-GM

\[\begin{aligned}\sum_{\textrm{cyc}}x\sqrt{\frac{y+z}{(z+x)(x+y)}}& = \sum_{\textrm{cyc}}\frac{1}{2}\left(x\sqrt{\frac{y+z}{(z+x)(x+y)}}+y\sqrt{\frac{z+x}{(x+y)(y+z)}}\right) \\ & \geq \sum_{\textrm{cyc}}\sqrt{x\sqrt{\frac{y+z}{(z+x)(x+y)}}y\sqrt{\frac{z+x}{(x+y)(y+z)}}}= \sum_{\textrm{cyc}}\sqrt{\frac{xy}{x+y}}\\ & = \sqrt{xyz}\sum_{\textrm{cyc}}\frac{1}{\sqrt{z(x+y)}}\geq \sqrt{xyz}\cdot\frac{9}{\sqrt{zx+zy}+\sqrt{xy+xz}+\sqrt{yz+yx}}\end{aligned}\](First ineq.: AM-GM, Second ineq.: Cauchy.) So now we would like to prove that \[\begin{aligned}& \sqrt{xyz}\cdot\frac{9}{\sqrt{zx+zy}+\sqrt{xy+xz}+\sqrt{yz+yx}}\geq 3\sqrt{\frac{3xyz}{2(xy+yz+zx)}}\\ & \Longleftrightarrow\ \sqrt{3}\sqrt{2(xy+yz+zx)}\geq \sqrt{zx+zy}+\sqrt{xy+xz}+\sqrt{yz+yx}\\ & \Longleftrightarrow\ \sqrt{\frac{(zx+zy)+(xy+xz)+(yz+zx)}{3}}\geq \frac{\sqrt{zx+zy}+\sqrt{xy+xz}+\sqrt{yz+yx}}{3},\end{aligned}\]which is true by AM-QM.


algebra
polynomial
limit
induction
absolute value
algebra proposed

Does there exist a sequence $ a_0, a_1, a_2\dots$ such that for all $ n\in\mathbb{N}$,
\[ P_n(x) \equal{} a_0 \plus{} a_1x \plus{} a_2 x^2 \plus{} \dots \plus{} a_nx^n
\]
has precisely $ n$ real roots.

Yes.  The basic idea is that we can pick $ a_{n\plus{}1}$ small enough so that the existing roots are perturbed by a negligible amount but such that we can force the existence of an additional root.

$ a_0$ and $ a_1$ are arbitrary nonzero reals; $ p_1(x) \equal{} a_0 \plus{} a_1 x$ always has $ 1$ distinct real root, which is our base case.

Suppose that we have constructed a polynomial $ p_k(x) \equal{} a_0 \plus{} a_1 x \plus{} ... \plus{} a_k x^k$ with exactly $ k$ distinct roots $ r_1, r_2, ... r_k$.  Then between those roots $ p_k(x)$ alternates signs; that is, given reals $ q_0, q_1, ... q_k$ all not equal to zero such that $ q_0 < r_1 < q_1 < r_2 < ... < r_k < q_k$, the sequence $ p_k(q_i)$ alternates signs since we can write $ p_k(x) \equal{} a_k(x \minus{} r_1)(x \minus{} r_2)...(x \minus{} r_k)$.  Pick $ a_{k\plus{}1} \neq 0$ such that

$ |a_{k\plus{}1}| < \inf_{0 \le i \le k} \frac{p_k(q_i)}{q_i^{k\plus{}1}}$

and $ a_{k\plus{}1}$ has sign opposite $ a_k$.  We claim that

$ p_{k\plus{}1}(x) \equal{} p_k(x) \plus{} a_{k\plus{}1} x^{k\plus{}1}$

has $ k\plus{}1$ distinct roots.  Because of the bound on the absolute value of $ a_{k\plus{}1}$, we have guaranteed that $ a_{k\plus{}1} q_i^{k\plus{}1}$ is strictly less than $ p_k(q_i)$ for every $ i$, hence that $ p_{k\plus{}1}(q_i)$ has the same sign as $ p_k(q_i)$ for all $ i$, and since these signs alternate there exist $ k$ distinct real roots $ r'_1, r'_2, ... r'_k$ of $ p_{k\plus{}1}(x)$ satisfying $ q_0 < r_1' < q_1 < r_2' < ... < r'_k < q_k$ by the Intermediate Value Theorem.  It remains to find one additional root.

Note that since $ q_k > r_k$ we know that $ p_k(q_k)$ has the same sign as $ a_k$; without loss of generality, $ a_k$ is positive, so $ p_k(q_k) > 0$.  By construction, we still have $ p_{k\plus{}1}(q_k) > 0$.  On the other hand, since $ a_{k\plus{}1}$ is negative, we have $ \lim_{n \to \infty} p_{k\plus{}1}(x) \equal{} \minus{}\infty$, so for some $ q_{k\plus{}1} > q_k$ we have $ p_{k\plus{}1}(q_{k\plus{}1}) < 0$.  Again by the Intermediate Value Theorem, we conclude that there exists $ r'_{k\plus{}1}$ between $ q_k$ and $ q_{k\plus{}1}$ such that $ p_{k\plus{}1}(q_{k\plus{}1}) \equal{} 0$, so $ p_{k\plus{}1}(x)$ has $ k\plus{}1$ distinct roots $ r'_1, r'_2, ... r'_{k\plus{}1}$.

By induction, $ p_n(x)$ exists and has $ n$ distinct real roots for all $ n$ as desired.


I use WinEdt, and I want to insert a picture. I try to do this but I fail. I also activate [b]\usepackage{graphicx}[/b] but i don't know what i have to do from that point. Let's say that this is my picture. What I have to write in {caption}, {label}...and so on. I'll be very thankful if someone will explain me. :)

Put your picture file in the same directory as your tex file. Then it's as simple as saying
[code]\includegraphics{picture.png}[/code]See [[LaTeX:Pictures]] for more options.
To get a caption and label you need to use the figure environment. Again it's as simple as typing
[code]\begin{figure}
  \includegraphics{picture.png}
  \caption{This is my picture}
  \label{picture}
\end{figure}[/code]You can then reference it with \ref{picture}. For more about this see [url=http://www.andy-roberts.net/misc/latex/latextutorial6.html]Getting to Grips with Latex - Floats, Figures and Captions[/url].

Thanks a lot [b]stevem[/b]. I appreciate that. :lol:


function
limit
real analysis
real analysis unsolved

a<b reals

f,g in $C^{0}([a,b])$

$fog=gof$

$f_n=fo...of; g_n=go...og$

for all x in [a,b] suppose $f(x)>g(x)$

what can you said about $\lim_{n\rightarrow +\infty}(f_n-g_n)$ ?

I think there are some typos here. $f,g: [a,b]\to[a,b]$ continuous and $f\circ g=g\circ f$ is not compatible with $f(x)>g(x),\ \forall x$. It can be shown that $f(x)=g(x)$ for some $x$. So could you please correct it? Is it supposed to be $\ge$ instead of $>$?

Thanks.

Thank's for you post.
I don't know what is the correct version, a student sent it to me.

Let's try with $\geq$

Well, with $\ge$, the limit may easily fail to exist: just take $g(x)=a$ for all $x$ and let $f(x)$ be an arbitrary continuous mapping from $[a,b]$ to $[a,b]$ such that $f(a)=a$. Then, generally speaking, the iterations of $f$ can behave in almost any way (the orbits of generic points may be dense in $[a,b]$, etc.). I am quite inclined to believe that the original Moubinool's version is exactly what was asked but that it wasn't a problem by itself but rather a hint to some other problem and the aim was to show that the conditions $f\circ g=g\circ f$ and $f>g$ are incompatible. Indeed, if $f\ge g+\delta$ on $[a,b]$, then it is easy to show that $f_n\ge g_n+n\delta$ and, therefore, the corresponding limit is $+\infty$, which is impossible. Still, it is only my speculation...


LaTeX conics ellipse FTW geometry blogs inequalities

sup.

by reading this, you are now part of the You're Banned Game!!! Congratulations!!! oh, and by the way, if you don't participate in the You're Banned Game, i will send evil monkeys to harrass you.

so here's how you play the You're Banned Game!!!

you start by saying what the last person who posted on this topic did to get themselves banned. (by the way, we're not really banning people. it's just a game.  :wink: )

eg: So-and-So is banned because he/she wrote to much.

this is describing the post above your post!! get it?

ok. i'll start us off.

[color=green]winner is banned because his pure awesomness is wrecking the whole website.[/color]

Okay, this could get some insults going (BEWARE).

winner is banned because his pure awesomness is wrecking the whole website.

(how did you get it to be green????)

1=2 is banned since if 1=2, then 1=0, then 42=0, then the universe=0, and the universe doesn't exist!

HE IS TRYING TO DESTROY THE UNIVERSE.

abcak is banned because s/he made me lose THE GAME!

myyellowducky is banned for attempting to reveal abcaks gender(s/he :o ?)

algebra2000 is banned because i have to write an ABC story.

Here's an example of a partial (really stupid) ABC story:

[b]A[/b]lgebra2000 was walking along the hallway. [b]B[/b]ut suddenly, a vampire with a huge ax jumped out and tried to cut of his head. [b]C[/b]razily enough, algebra manged to duck in time to escape the blow. [b]D[/b]ropping his books, he ran away. [b]E[/b]mitting a grunt the ax murderer ran off in pursuit of algebra.

ytrewq is banned for having an (is that an exploding volcano?) for an avatar. That user is trying to blow up the world!

(colors can be reached through the Font Colour option right below Bold, Italic, and Underline.)

pianogirl is banned for thinking that [url=http://blog.lib.umn.edu/rein0328/architecture/puzzle.bmp]this[/url] is an exploding volcano

ytrewq is banned for thinking that I thought that that was an exploding volcano. I asked if it was an exploding volcano. I never said it. :P

ytrewq is banned for posting external links to a giant version of the avatar.
Never mind
Pianogirl is banned for posting right before I did and messing me up. :mad:
Never mind
ytrewq is banned for the same reason as pianogirl :mad:

FlyAgaric is banned for saying that I'm banned not pianogirl

flyagaric is banned for having a username i cant understand. he's obviously trying to make our brains explode...messily...very messily...

ytrewq is banned for having an un-pronouncable username.

ytrewq is banned for making it look like [b]I banned flyagaric after him with evil dark magic...[/b]

read it again

the following person is banned for having the 1236 post =)

mathwiz85 is banned for reversing the rules.

mathelete is banned for using  :ddr: in a signature.

pii is banned for having 42 posts ITS 2 AWSOME!!!

mathwiz85 is banned for posting that AS I POSTED MY 43RD.  :P

pii is banned for misspelling pi for a username

mathwiz85 is banned for not seeing that it my username is pi*i, as clearly indicated in my signature.

mathwiz85 is banned again because I took pi*i's side.

What! mathlete is banned for taking misspelled pi's side!!!

mathwiz85 is banned for misspelling pi*i.

mathelete is banned for having DDR in his sig cus i suck at it X|

walkingtaco is banned for sucking at DDR

pi*i is banned for joining over 3 years ago and having only 44 posts.

wishfulthinking360 is banned for having a signature with the same number and type of emoticons as mine

wishfulthinking360 is banned again for the same reasons as mathwiz85 was double banned.


$ a_n$ is the last digit of $ [10^{\frac {n}{2}}]$ and $ b_n$ is the last digit of $ [2^{\frac {n}{2}}]$. Are the sequence $ a_n,b_n$ periodic?

[hide="for $ a_n$:"]
If $ n \equal{} 2m$ is even, then $ [10^{\frac {2m}{2}}] \equal{} 10^m$. Thus $ a_{2m} \equal{} 0$.
If $ n \equal{} 2m \plus{} 1$ is odd, then  $ [10^{\frac {2m \plus{} 1}{2}}] \equal{} [10^m \cdot \sqrt {10}]$.
Let $ \sqrt {10} \equal{} d_{0}.d_{1}d_{2} \ldots$ where the $ d$'s are digits in base $ 10$.
$ a_{2m \plus{} 1}$ is the last digit of $ [10^m \cdot \sqrt {10}] \equal{} [d_{0}d_{1}\ldots d_{m}.d_{m \plus{} 1} \ldots] \equal{} d_{0}d_{1}\ldots d_{m}$. Thus $ a_{2m \plus{} 1} \equal{} d_{m}$.
Thus, the sequence $ a_n$ is $ d_0,0,d_1,0,d_2,0, \ldots$
Since $ \sqrt {10}$ is irrational, the sequence $ d_0,d_1,d_2, \ldots$ is not periodic.
Therefore, the sequence $ d_0,0,d_1,0,d_2,0, \ldots$ is also not periodic. $ \frac {1}{2}(Q.E.D.)$
[/hide]


Hey everyone,
I've searched this up in the forums, but I haven't found a definitive answer:
How exactly do you participate in the screening round? My school has horrible physics overall and doesn't know about the USAPHO, so I'm not really sure how to take the test. It's passed this year, but how do I get it for our school next year?
I've heard some people say there's a form, and some say to just bother the teacher until she registers for you, is there a specific place/form/application you have to fill out?

Thanks guys  :)

I posted on this before; consult this website:

http://www.aapt.org/contests/olympiad.cfm

And this one is for this year's team:

http://www.aapt.org/olympiad2008/

Also show these to your physics teacher.  Your physics teacher has to contact AAPT and register you.  In my case, my teacher took care of all the registration and I didn't have to fill out any forms.

Thanks genericme;
do you know when this should take place? The tests take place in spring, so do the teacher apps have to be sent earlier?

Yeah, the deadline was sometime in December or January this year.  The 2008 website used to have this information, but now it's being redesigned.  If you check for the 2009 website sometime later this year, though, it should have that information (i.e., just switch the "8" for a "9" in the URL).

I can't find where to find information for the 2009 contest.

The website http://www.aapt.org/olympiad2009/ will presumably be up sometime later this year.


You are given an oriented graph. Each edge of the graph is colored in one of the three colors. Your task is to find the length of the shortest path from the first vertex to the N-th. Note that any two successive edges in the path can't have the same color. 1<=N<=200

Hint: try making 3 nodes for every node in the graph.. if you connect the nodes in logical way, then you can solve it easily.

Do you mean how do you decide what edges to use? I mean, just define 3*the number of nodes..
(as for the edges, well just think about it for a little bit, theres not too many possibilities, and I don't want to give the whole thing away ;))

Thanks a lot. I figurate how to solve the problem.

i thinbk i havenot understood the problem and soln.

the problem said, a graph having edges with only colored in 3 colors. we have to fnd out the shortest path from a node to another right?

if so then why do we need three nodes for every node?

[does oriented graph diffrent from normal graph?]

if i have understand the problem correctly then can't it be solved by BFS?

You need three nodes because two succesive edges in the graph can't be the same color.

[quote="mahbub"]well i think u r talking about three index in a same node. three index for min d coming from three color edges.[/quote]

I'm sure theres several ways of doing it, but creating 3 times the number of vertices takes a couple of lines of code and means you can use the very basic form of BFS to actually solve it, with no changes necessary. Was the fastest way to do it, codewise, that I could think of.

actually which seems ez for whom, he will use that ;)


usually non metals excpt noble gasses form diatomic molecules

esp of Group 15,16,17 :)

[quote="ThetaPi"]Which elements form diatomic molecules? What group's elements are diatomic?[/quote]

Group 7A elements (F, Cl, Br, and I), nitrogen, oxygen, and hydrogen all form diatomic molecules.

@rituraj007: Carbon and silicon are non-metals but they would require a quadruple bond to be diatomic, which isn't possible. :wink:

I think $ \ce{C2}$ does exist actually. Drawing the MO diagram gives a bond order of $ 2$ so I guess it probably can exist.

Much more stable is its dianion $ \ce{C2^{2 - }}$. Even if it exists, it much have only a fleating existence: note that for such a species one contributing structure would be $ : C=C:$, a dicarbene structure, which gives a taste of how reactive is must be.

well as long as i kno,if u vaporise C in absence of oxygen ,it wud give C2 ...isn't this correct??? :maybe:

and yes ofc extent of overlap wud b 2 poor in Si2 so it just can't exist... :)


calculus
integration
LaTeX
trigonometry
calculus computations

Hi. This is the last problem on my Calculus Final review sheet.

(integrate) (tan^2x sec^2x)dx

I set
u = (tan^2x)
du = (sec^2x)

and got

(1/3) * (tanx)^3  as my answer. Is this correct?

Btw, the site to download the miktex typesetter (for using latex) was down atm. I'll try and get it tommorow before posting any further.

Yes, but your $u$ is wrong...

$\int \tan^2(x)\sec^2(x) \ dx$

$u=\tan(x)$

$du=\sec^2(x) \ dx$

$\int u^2 \ du = \frac{u^3}{3}$

$\int \tan^2(x)\sec^2(x) \ dx = \frac{\tan^3(x)}{3}$

1. This forum - Calculus Computations and Tutorials - is the right place for this kind of material.

2. You don't need ANY TeX implemetation or typesetting program to make use of LaTeX as implemented on this site. You can just type in the code directly (surrounded by dollar signs); the site software handles is. To find out if you're doing it right, use the preview button early and often.

3. Why ask us whether a proposed antiderivative is correct? Check it yourself by differentiating it!


You should finish this off with infinite descent (somehow). There is a unique solution :D


inequalities USAMTS algebra polynomial trigonometry induction geometry

How do you guys think the difficulty of Round 3 was? I thought it was fairly easy. I was able to do #1,3,4, and hopefully I will get partial credit on #2 and #5, since part of my solutions on those questions were kind of the same as the official one.

My #1 solution involved setting some bounds and some graphing
My #2 solution was pretty bad, but it had some parts that were the same as the official solution
My #3 solution was basically like the official one
My #4 solution was kind of like the official one, but I didnt prove the Frobenius Coin Problem. Its an extremely well known theorem, so I didnt think we would need to.
My #5 solution was OK, I guess. I proved that $ a_n \plus{} b_n \plus{} c_n \equal{} n \plus{} 1$ for all $ n \ge 0$, and then I made up some junk about limits

I thought it was pretty hard. I only got #1 and 2, and I probably lost a point on #2 for style. I BS'ed #3, so I probably got a 0 on that one too, for a total score of 9.

I disliked this round... I solved all the problems but neither of the official #5's seemed too motivated. [hide="Mine"]We are given that $ a_0 = 1,b_0 = 0,c_0 = 0$, and
\[ a_n = a_{n - 1} + \frac {c_{n - 1}}{n},\quad b_n = b_{n - 1} + \frac {a_{n - 1}}{n},\quad c_n = c_{n - 1} + \frac {b_{n - 1}}{n}.\]
First we plug in $ a_0,b_0,c_0$ into the recursions for $ n = 1$ and find that $ a_1 = 1,b_1 = 1,c_1 = 0$. Letting
\[ x_n = \sqrt {3n}\left(a_n - \frac {n + 1}{3}\right)\]
and so on, we have $ x_1 = y_1 = 1/\sqrt {3}$ and $ z_1 = - 2/\sqrt {3}$. Then we now want to prove that, for $ n\ge1$,
\[ \left|a_n - \frac {n + 1}{3}\right| < \frac {2}{\sqrt {3n}}\Longleftrightarrow\left|\sqrt {3n}\left(a_n - \frac {n + 1}{3}\right)\right| < 2\Longleftrightarrow|x_n| < 2.\]
First note that
\[ x_n = \sqrt {3n}\left(a_n - \frac {n + 1}{3}\right)\Longleftrightarrow a_n = \frac {x_n}{\sqrt {3n}} + \frac {n + 1}{3}.\]
(Cyclic relations hold as well.) Hence the original recursions can be rewritten as
\begin{align*}\frac {x_n}{\sqrt {3n}} + \frac {n + 1}{3} = a_n & = a_{n - 1} + \frac {c_{n - 1}}{n} \\
& = \frac {x_{n - 1}}{\sqrt {3(n - 1)}} + \frac {n}{3} + \frac {z_{n - 1}}{n\sqrt {3(n - 1)}} + \frac {1}{3} \\
x_n & = \frac {x_{n - 1}\sqrt {n}}{\sqrt {n - 1}} + \frac {z_{n - 1}}{\sqrt {n(n - 1)}} \\
y_n & = \frac {y_{n - 1}\sqrt {n}}{\sqrt {n - 1}} + \frac {x_{n - 1}}{\sqrt {n(n - 1)}} \\
\quad z_n & = \frac {z_{n - 1}\sqrt {n}}{\sqrt {n - 1}} + \frac {y_{n - 1}}{\sqrt {n(n - 1)}}.\end{align*}
Summing the three new recurrence relations up, we have
\[ x_n + y_n + z_n = \left(\frac {\sqrt {n}}{\sqrt {n - 1}} + \frac {1}{\sqrt {n(n - 1)}}\right)(x_{n - 1} + y_{n - 1} + z_{n - 1}).\]
But $ x_1 + y_1 + z_1 = 1/\sqrt {3} + 1/\sqrt {3} - 2/\sqrt {3} = 0$, so we can easily see that in general $ x_n + y_n + z_n = 0$ as well.

We now claim that $ x_n^2 + y_n^2 + z_n^2 < 16/3$, which would imply by the Cauchy-Schwarz Inequality that
\[ 16 > \left(1^2 + 1^2 + 1^2\right)\left(|x_n|^2 + |y_n|^2 + |z_n|^2\right)\ge(|x_n| + |y_n| + |z_n|)^2,\]
or taking the square root (note that the right hand side parentheses contains a nonnegative expression) and using the triangle inequality,
\begin{align*}4 > |x_n| + | - y_n| + | - z_n| & \ge|x_n - y_n - z_n| \\
& = |x_n + (x_n + y_n + z_n) - y_n - z_n| = |2x_n| \\
2 & > |x_n|,\end{align*}
as we need to prove for the original problem.

To prove this claim, notice that squaring our recursions for $ x_n,y_n,z_n$ we have
\begin{align*}x_n^2 & = \frac {n}{n - 1}x_{n - 1}^2 + \frac {1}{n(n - 1)}z_{n - 1}^2 + \frac {1}{n - 1}2x_{n - 1}z_{n - 1} \\
y_n^2 & = \frac {n}{n - 1}y_{n - 1}^2 + \frac {1}{n(n - 1)}x_{n - 1}^2 + \frac {1}{n - 1}2y_{n - 1}x_{n - 1} \\
z_n^2 & = \frac {n}{n - 1}z_{n - 1}^2 + \frac {1}{n(n - 1)}y_{n - 1}^2 + \frac {1}{n - 1}2z_{n - 1}y_{n - 1}.\end{align*}
Let $ s_n = x_n^2 + y_n^2 + z_n^2$, so $ s_1 = x_1^2 + y_1^2 + z_1^2 = 1/3 + 1/3 + 4/3 = 2$. Adding these three squared quantities together and using
\begin{align*} & 2x_{n - 1}y_{n - 1} + 2y_{n - 1}z_{n - 1} + 2z_{n - 1}x_{n - 1} \\
= & (x_{n - 1} + y_{n - 1} + z_{n - 1})^2 - (x_{n - 1}^2 + y_{n - 1}^2 + z_{n - 1}^2) \\
= & (0)^2 - s_{n - 1} = - s_{n - 1}\end{align*}
we have
\begin{align*}s_n & = x_n^2 + y_n^2 + z_n^2 = \left(\frac {n}{n - 1} + \frac {1}{n(n - 1)}\right)s_{n - 1} + \frac1{n - 1}( - s_{n - 1}) \\
& = \left(\frac {n}{n - 1} + \frac {1}{n(n - 1)} - \frac {1}{n - 1}\right)s_{n - 1} = \left(1 + \frac {1}{n(n - 1)}\right)s_{n - 1} \\
& = \left(1 + \frac {1}{n(n - 1)}\right)\cdots\left(1 + \frac {1}{2\cdot1}\right)s_{1} \\
& = 2\left(1 + \frac {1}{n(n - 1)}\right)\cdots\left(1 + \frac {1}{2\cdot1}\right).\end{align*}
Therefore it is sufficient to prove that for all positive $ n$ (when $ n = 0$, $ s_{0 + 1} = s_1 = 2 < 16/3$),
\begin{align*}\frac {16}{3} > s_{n + 1} & = 2\left(1 + \frac1{1\cdot2}\right)\cdots\left(1 + \frac {1}{n(n + 1)}\right) \\
& = 2\prod_{k = 1}^{n}\left(1 + \frac {1}{k(k + 1)}\right) \\
& = 2\prod_{k = 1}^{n}\frac {k^2 + k + 1}{k^2 + k},\end{align*}
or dividing both sides by $ 2$ and taking the reciprocal (reversing the direction of the inequality),
\[ \frac {3}{8} < \prod_{k = 1}^{n}\frac {k^2 + k}{k^2 + k + 1} = \prod_{k = 1}^{n}\left(1 - \frac {1}{k^2 + k + 1}\right).\]
Since $ s_1 = 2$, we can calculate for $ n = 1,2$ from the recursion $ s_{n + 1} = (1 + 1/(n(n + 1)))s_{n}$ that $ s_2 = (1 + 1/(1\cdot2))s_1 = 3/2\cdot2 = 3$ and $ s_3 = (1 + 1/(2\cdot3))s_2 = 7/6\cdot3 = 7/2$, both of which are less than $ 16/3$. Otherwise, $ n\ge3$, so because $ 1 - 1/(k^2 + k + 1) > 1 - 1/k^2\Longleftrightarrow k^2 + k + 1 > k^2$ is true for all positive $ k$ we have
\begin{align*} & \prod_{k = 1}^{n}\left(1 - \frac {1}{k^2 + k + 1}\right) \\
> & \left(1 - \frac {1}{1^2 + 1 + 1}\right)\left(1 - \frac {1}{2^2 + 2 + 1}\right)\prod_{k = 3}^{n}\left(1 - \frac {1}{k^2}\right) \\
= & \left(\frac {2}{3}\right)\left(\frac {6}{7}\right)\prod_{k = 3}^{n}\left(\left(1 - \frac {1}{k}\right)\left(1 + \frac {1}{k}\right)\right) = \frac {4}{7}\prod_{k = 3}^{n}\frac {k - 1}{k}\prod_{k = 3}^{n}\frac {k + 1}{k} \\
= & \frac {4}{7}\left(\frac {\prod_{k = 3}^{n}(k - 1)\prod_{k = 3}^{n}(k + 1)}{\prod_{k = 3}^{n}k\prod_{k = 3}^{n}k}\right) = \frac {4}{7}\left(\frac {\left(\frac {2}{n}\prod_{k = 3}^{n}k\right)\left(\frac {n + 1}{3}\prod_{k = 3}^{n}k\right)}{\prod_{k = 3}^{n}k\prod_{k = 3}^{n}k}\right) \\
= & \frac {4}{7}\left(\frac {2}{3}\right)\left(\frac {n + 1}{n}\right) = \frac {8}{21}\left(\frac {n + 1}{n}\right) = \left(\frac {3}{8} + \frac {1}{168}\right)\left(1 + \frac {1}{n}\right) > \frac {3}{8}.\blacksquare\end{align*}
[/hide] was fairly contrived too. :(

But yeah at first I didn't know to take the reciprocal of the $ 1+1/(k(k+1))$ thing and wrote a ten page solution which can be downloaded [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=324691]here[/url].

the official 1st solution for #5 makes me lol xD

I solved all five problems, but I'm wondering if we need to prove the Chicken McNugget threorem (#4)....

also, I'm wondering if they will take of points for me saying that pi<something (decimal) after what happened last round...  :maybe:

mieh, here's my solution for #5:

[url=http://usamts.org/Papers/Year21/Round3/9877-1262017234-solutions.pdf]Here's mine.[/url]

I know I screwed up the square problem, don't remind me.  It reminded me of the standard problem AOPS uses about the ring with the inner radius of 2 and outer radius of 3 where you have to show that there is some spot in which the ring can cover 10 dots.  I kinda went down those same lines, and so did the official solutions, but not well enough.  Maybe a 4, hopefully.

Then the others were all fairly easy except #1 and #5.  I used inversion in my #1, and the cool part is that inversion is actually applicable to the problem!

For #2, that was the shortest proof for a USAMTS problem I've ever written.  Hope it is accepted, and that some of the stuff that the official solution showed doesn't need showing.

#3, I told you about already.  I forgot to include the condition that the circle had to be completely inside.

#4, once you knew to use the chicken mcnugget theorem, was trivial (To take a phrase from someone, I don't know who, but probably pythag011).

#5.  This problem gave me so many problems.  I wasn't sure how many people would actually answer this.  My solution is remarkably like the official solution's, which is crazy to me.

I hope for a 5/5/4/5/5 or a 24.

Edit: Yeah I messed up the problem numbers.  I was testing you, and v_Enhance passed the test.  Now let's see if he reads my post again: v_Enhance, PM me to claim your prize.

Hopefully the fact that I'm not yet in high school will be enough of an excuse for the amount of success I had, because overall this round was far more difficult than the first two rounds.

#1- This was okay.. My solution was the same as the official one up to the point where they take the reciprocal of the equation.  I failed to notice that this would have saved a lot of work and proceeded to launch an algebraic nightmare.  In the end I still got 231 though.

#2- Pretty easy.  I expanded the numerator to get a massive polynomial then showed that $ n\equal{}500^2$ would cause the fraction to leave a remainder of 1 when divided by any prime less than 500.

#3- Junk proof.  It looks valid at first glance.  It probably isn't.  I failed this one.

#4- Okay.  I got a totally different solution using floor summations, but I still ended up with the formula $ \frac{(a\minus{}1)(b\minus{}1)}{2}$.  So my solution is probably right considering I got the same set $ (a,b)$ and a final answer of 133.  I have never heard of the Chicken McNugget Theorem.  Sad.

#5- This problem was insane.   :dry: I only found $ a_n\plus{}b_n\plus{}c_n \equal{} n\plus{}1$ one week before the answers were due and didn't get much further (well okay, nothing at all) than that.

Overall, my final score would be something like:
5/5/2/5/1 = ~18

This is my first time submitting junk proofs-- on Rd 1 I left #4 blank and on Rd 2 this was not necessary.  Therefore I am not sure how generous the graders are with nontrivial progress.

I hope I got a $ 25$. I did #3 using a decimal approximation for absolutely no reason whatsoever, so maybe not.

I also wonder if they'll take off points for giving a number not in closed form for #2.

My solutions are at http://www.usamts.org/Papers/Year21/Round3/12374-1263103840-solutions.pdf.

[quote="tenniskidperson3"][url=http://usamts.org/Papers/Year21/Round3/9877-1262017234-solutions.pdf]Here's mine.[/url]
#3, once you knew to use the chicken mcnugget theorem, was trivial (To take a phrase from someone, I don't know who, but probably pythag011).

#4, I told you about already.  I forgot to include the condition that the circle had to be completely inside.
[/quote]
Is it just me or did you mix up the question numbers?

I thought it was kind of hard, but I didn't put as much time into it as I should have.

1. It looked like a trig bash at first, but then I saw similar triangles/area and Pythagorean Theorem. My solution involved letting $ AP \equal{} a$ and solving for itself (going around the quadrilateral) in terms of $ x$. The $ a$s canceled, leaving $ x \equal{} 231$. Though it wasn't as elegant as the official solution, it was the same equations in a different form.

2. All I did was say that it could be changed into $ \binom{n \plus{} 500}{500}$. I don't expect anything for this.

3. Got this after about 10 minutes; the easiest problem in my opinion. I showed that in the worst case scenario (pretty much the same as official), there exists room for the center of the circle to be. Thus, a circle can be drawn in any situation.

4. I stated that $ a$ and $ b$ are relatively prime and the largest inexpressible is $ ab \minus{} a \minus{} b$. If I had known that $ S$ contains exactly half of the integers from $ 0$ to $ N$, this would have been easy.

5. I tried induction, but my solution was kind of weak and included graphing. I did not understand much of the official solution.

Overall, my highest hope is 5/1/5/1/3 = 15, but I expect 5/0/5/0/2 = 12. I have 34 right now and my goal is bronze, so a 12+34=46 would need 14 more points.

1,2,3,4 for me were fairly straightforward, though I'm horrendous at geometry and bashed number 1 with two pages of algebra.  The other three solutions of mine were pretty much the same as the official ones.

As for number 5, I found a recursion for $ a_n$ based on its two preceding terms and then used induction, which eventually culminated in solving a massive polynomial using Mathematica to demonstrate that the bounds I got for $ a_{n\plus{}1}$ based on $ a_n$ and $ a_{n\minus{}1}$ were sharper than $ \frac{2}{\sqrt{3n\plus{}3}}$.  It was four pages of algebraic ugliness, but I think I proved it legitimately...  Emphasis on "think"....... :( I'm so glad I had Mathematica while working this problem out.

[quote="dysfunctionalequations"]I also wonder if they'll take off points for giving a number not in closed form for #2.[/quote]

I unfortunately did the same thing except I wrote out the prime factorization and the actual number (434 digits) explicitly. :wink:  :(

[url=http://www.usamts.org/Papers/Year21/Round3/12437-1263239952-Solutions09-Round3.pdf]My Solutions[/url]

Probably 5/5/5/5/1 = 21

#1: Harder than I expected for a #1.
#2: Easiest problem. My solution's about half a page.
#3: I thought I solved it, but then the day before it was due (before I had typed up most of my solutions), I realized it didn't work, and spent a whole bunch of time on the wrong track. I finally solved it thinking about it in bed just before I went to sleep.
#4: Fairly easy to find the answer, but I had more trouble trying to prove it.
#5: I thought problem 5 was by far the hardest problem so far. I solved all the problems in the first two rounds fairly easily, but I spent hours trying everything I could possibly think of on #5 and still didn't get a solution. I also found a recursion (and a useless closed form) for $ a_n$, but I think I spent too much time with it on Mathematica, trying to get something to work (somehow nothing I tried with Mathematica seemed to be very successful). I tried induction, but $ \left|a_n\minus{}\frac{n\plus{}1}{3}\right|<\frac{2}{\sqrt{3n}}$ does not imply $ \left|a_{n\plus{}1}\minus{}\frac{n\plus{}2}{3}\right|<\frac{2}{\sqrt{3(n\plus{}1)}}$, and I couldn't find a way to sharpen the bounds to make it work.

@dysfunctionalequations: How'd you get \[ x_n\equal{}a_n\plus{}\frac{n\plus{}1}{3}\] if you claimed originally that $ x_n\equal{}a_n\minus{}\frac{n\plus{}1}{3}$?  Maybe I'm just misreading this... :maybe:

Those should all be "-" signs. He has the right recursion though, so his solution still works.

Hm so it seems that substitutions like $ x_n\equal{}a_n\minus{}(n\plus{}1)/3$, etc. are pretty natural.

Yes, that was a typo. It doesn't affect the rest of the solution (I hope).

[quote="pythag011"]On 5:

Let x_n = a_n + z*b_n + z^2*c_n, where z is a 3rd root of unity. Then, you can prove that x_n+1 = (1+z/n), and apparently some inequalities on the magnitude solve it somehow, but I don't know the details.
[/quote]

This is basically what I did--Basically I did a root of unity filter on $ a_n\plus{}b_nx\plus{}c_nx^2$ and used the recursion, then bounded the magnitude.  (Also I think you mean $ (1\plus{}\frac{z}{n})x_n$.)

Roots of unity filter is what I used, and even though it seems unmotivated as someone on this thread has already said, once you decide to find the first values of $ a_n, b_n, c_n$ as not collapsed but expanded as products of numbers from 1 to $ n$, the root of unity filter comes naturally.  As an aside, I didn't solve #3 in bed but I did solve #5 in bed!

Pythag011: No, I think it was me you are thinking of; dysfunctionalequations's solution is the one about $ a_n^2\plus{}b_n^2\plus{}c_n^2$.  Unless (s)he came up with another solution?

123456789: [url=http://www.artofproblemsolving.com/Books/Excerpts.pdf]Look on page 43.[/url]

[quote="tenniskidperson3"]123456789: [url=http://www.artofproblemsolving.com/Books/Excerpts.pdf]Look on page 43.[/url][/quote]
Darn... that would have made my life so much easier....  I guess I just haven't encountered enough challenging problems yet.    :|

[quote="CatalystOfNostalgia"]Yeah guys #5 is probably the hardest USAMTS problem ever...at least two black MOPpers were not able to solve it.[/quote][quote="pythag011"]So 5 is really hard in my opinion.[/quote]Wow, that's reassuring. I spent a ton of time on #5, and was disappointed not to have been able to solve it.

[quote="timwu"]Note $ a_n \plus{} b_n \plus{} c_n \equal{} n \plus{} 1$. Redefine each sequence by subtracting $ (n \plus{} 1)/3$ which makes $ a_n \plus{} b_n \plus{} c_n \equal{} 0$. Note the recurrences are still satisfied. Now we need to show $ |a_n| < \frac {2}{\sqrt {3n}}$. Let $ s_n \equal{} a_n^2 \plus{} b_n^2 \plus{} c_n^2$. Squaring the recurrences and simplifying (using $ (a_n \plus{} b_n \plus{} c_n)^2 \equal{} 0$) we obtain $ s_n \equal{} \frac {n^2 \minus{} n \plus{} 1}{n^2}\cdot s_{n \minus{} 1}$. Note $ b_n^2 \plus{} c_n^2 \geq \frac {(b_n \plus{} c_n)^2}{2} \equal{} \frac {a_n^2}{2}$ so $ a_n^2 \leq \frac {2}{3}s_n$. It suffcies to show $ s_n < \frac {2}{n}$ for all $ n\geq 2$ (manually check for $ n \equal{} 1$). Strengthen the inequality to $ s_n \leq \frac {2}{n} \cdot \frac {n \minus{} 1}{n}$ and use induction. Then $ a_n^2 \leq \frac {3}{2} s_n < \frac {4}{3n}$ and we are done.[/quote]
I did all of this up to the part about proving $ s_n<\frac{2}{n}$! (Somehow I thought I needed to prove $ s_n<\frac{6}{n}$, but I couldn't even do this.) However, it took me a long time to get this far; I wasted a lot of time trying to get a closed form or bash it any other way with Mathematica.

I also got a not closed form number for number 2, though I knew there had to be a nice closed form number as well, but that was just not the first thing that occurred to me. I put $ (\prod_{p < 500} p)^9$ even though that was unnecessarily big.

I think I got the rest except 5, although my number 4 was written poorly, so I'm hoping for 18-19.

[quote="tenniskidperson3"]Roots of unity filter is what I used, and even though it seems unmotivated as someone on this thread has already said, once you decide to find the first values of $ a_n, b_n, c_n$ as not collapsed but expanded as products of numbers from 1 to $ n$, the root of unity filter comes naturally.  As an aside, I didn't solve #3 in bed but I did solve #5 in bed!

Pythag011: No, I think it was me you are thinking of; dysfunctionalequations's solution is the one about $ a_n^2 \plus{} b_n^2 \plus{} c_n^2$.  Unless (s)he came up with another solution?

123456789: [url=http://www.artofproblemsolving.com/Books/Excerpts.pdf]Look on page 43.[/url][/quote]

Yes he came up with another solution:

Basically, time line:

Some day near the end: BACKHOOOOOO tells me about problem #5

That day: Hamster and me find the complex number trick, don't really continue with the solutions

Half a Week Later: We show the problem to dysfunctionalequations

That day: He says he has a solution with magnitude, I didn't hear what he said after that.

@The motivation:

Well, its seminatural I think to express it in terms of vectors, and then you see its just adding a cyclic permutation of the vector to itself, which seems pretty complex numbery.

Hmm well it turns out to be a really similar solution to the $ a_n^2 \plus{} b_n^2 \plus{} c_n^2$ one, unless I did something wrong...

I had a hilarious time solving #5 at 4:00 AM in the morning because I started it the morning of the due date. In the end I came up with a solution essentially identical to dnkywin's and dysfunctionalequation's.

#5 can be solved using arbitrary even powers, so any kind of even power mean is satisfactory. However, it is obvious that quadratic is easiest.

Also, am I the only one who thinks that the "USA Mathematical Talent Search" title is too far to the right on the official problem sheet? And that it should be moved left a bit?

Darn so I kinda suck at induction, I had gotten all the non-induction parts of the $ x_n\equal{}a_n\minus{}(n\plus{}1)/3$ solutions. My solution is basically equivalent (I think...) to them except it just avoids the induction. :P But I came up with the crux of my solution (realizing it's equivalent to proving $ 8/3 > \prod(1 \plus{} 1/(k(k \plus{} 1))$) while sleeping in front of my computer right after coming home for winter break...

[quote="math154"]Darn so I kinda suck at induction, I had gotten all the non-induction parts of the $ x_n \equal{} a_n \minus{} (n \plus{} 1)/3$ solutions. My solution is basically equivalent (I think...) to them except it just avoids the induction. :P But I came up with the crux of my solution (realizing it's equivalent to proving $ 8/3 > \prod(1 \plus{} 1/(k(k \plus{} 1))$) while sleeping in front of my computer right after coming home for winter break...[/quote]

I think everyone did the same thing; how do we get rid of the $ \frac{k}{k\plus{}1}$ trick? Theoretically the manipulation does not change the expression.

[quote="math154"]OK darn yeah sorry.[/quote]

Is that for me? I'm not complaining about anything you did; I'm asking a question about all of us induction people on how we could have improved our proofs. We all multiplied (divided) that expression by $ \prod\frac {k}{k \plus{} 1}$ which seems like an unnecessary step.

Here is my set of solutions: http://usamts.org/Papers/Year21/Round3/9665-1263209655-solutions.pdf

arg, my solutiosn suck. I didnt even give a diagram for #3. Also, i used 25*20 instead of 24*19 so i will be deducted D:.

I thought this was the hardest round so far.... I did all of them but #5, but I don't think my solutions were particularly great or anything so I probably lost points here and there. For #5 I just proved that an+bn+cn = n+1 like some other people did... i hope that's a point for nontrivial progress.

wow for #5 that's exactly what i did too...couldn't get anywhere else with it  :(


quadratics
modular arithmetic
number theory unsolved
number theory

$ p$ is a prime number and $ p\equal{}a^2\plus{}5b^2$, and $ a$ is  odd number. Prove
 $ a$ is quadratic residue (mod p) $ \iff \ p \equiv\ 1(mod5)$

Because $ a$ is odd, by properties of Jacobi Symbol (generalization of Legendre's symbol), we get:
$ \left( \frac{a}{p} \right) \left( \frac{p}{a} \right)\equal{}(\minus{}1)^{\frac{(a\minus{}1)(p\minus{}1)}{4}}$
But $ a$ is odd and $ p \equiv a^2 \equiv 1 \pmod 4$ (because $ b$ must be even), so $ (\minus{}1)^{\frac{(a\minus{}1)(p\minus{}1)}{4}}\equal{}1$.
Now we know that $ \left( \frac{a}{p} \right)\equal{}\left( \frac{p}{a} \right)\equal{}\left( \frac{5b^2}{a} \right)\equal{}\left( \frac{5}{a} \right)\equal{}\left( \frac{a}{5} \right)$.
But $ \left( \frac{a}{5} \right)\equal{}1 \leftrightarrow a^{\frac{5\minus{}1}{2}} \equiv 1 \pmod 5  \leftrightarrow a^2 \equiv 1 \pmod 5  \leftrightarrow p \equiv 1 \pmod 5$, and the result follows.


Since nobody else has posted this here, I guess I will:

(A) A press release from the NSF about the US team:
[url]http://nsf.gov/news/news_summ.jsp?cntn_id=112073&org=NSF&from=news[/url]

(B) More generally,  the full results from the ILO (or, as it seems to be called now, the IOL)
web site:  [url]http://iol6.linguistics-bg.com/results.php[/url]

The [url=http://iol6.linguistics-bg.com/problems.php]questions and solutions[/url] are pretty interesting, too!

the qustions are really interesting. thanks for posting.and i didnt even know ilo existed.!!  :maybe:

i didnt know this forum existed....they have an international competition for everything now?!?!?!


Βρέθηκε περιττός τέλειος αριθμός:
$3^{2}\cdot 7^{2}\cdot 11^{2}\cdot 13^{2}\cdot 22021$
από ένα μαθητή λυκείου.
Είναι αξιοσημείωτο το γεγονός ότι ο συγκεκριμένος αριθμός δεν είχε ανακαλυφθεί μέχρι σήμερα παρότι δεν είναι ιδιαίτερα μεγάλος, αλλά έτσι δεν είχε γίνει και με το δεύτερο μικρότερο ζευγάρι φίλιων αριθμών?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141540

Ola simera eginan etsi? Mipws lithike kai i eikasia tou Goldbach? :) :)

Wraio psemataki pantw... tsimpaei!!!  Hdi iparxei ena thima sto opoio to eipa!!

Alexandros

Ένα αθώο ψεματάκι ήταν ναι...αλλά αν δεν έβλεπα το ποστ του cretanman ήμουν έτοιμος να υπολογίσω τον αριθμό... :wink: Ωραίος thatin. :lol:

:lol:  elega na pw pws tha kleisoun to elliniko subforum alla auto itan kalitero an kai oxi prwtotipo...!


Hi,

I am having trouble fully understanding the basics of MA and CRT. I am a high school student looking to learn these things to solve math problems to one day make the IMO :)

I think one of my main problems is that i dont know where to find material to learn it. I can find anything online good, any suggestions?

Some of the stuff that i dont understand is like how, for example, $ 2 mod(3)$ is equivelent to $ \minus{}1 mod(3)$???

Anyway, please help???

[quote="TZF"]For integers $ a$, $ b$, and $ c$, the statement $ a \equiv b \pmod{c}$ is true if and only if there exists an integer $ k$ such that $ a \equal{} b \plus{} kc \iff a \minus{} b \equal{} kc$.

In your example, we have that $ 2 \equiv \minus{} 1 \pmod{3}$ if and only if there exists an integer $ k$ such that $ 2 \equal{} \minus{} 1 \plus{} 3k$. There is; namely, $ k \equal{} 1$. Therefore the statement is true.

I think you should get very comfortable with using modular arithmetic in simple cases (ie, some basic proofs) before even touching the CRT.[/quote]

Thanks alot for that. I will take your recommendation about learning modular arithmetic first. Could you tell me where you learnt modular arithmetic?

Try reading aops wiki on modular arithmetic: [url]http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic[/url]

Aops 1 also has good material on modular arithmetic.

order intro to NT and aops v2. aops v1 is unnecessary


[quote="trm620"]Hello all, I am a first year student in Mt. Sierra College in the field of Information Technology. Below is a beginners level Venn Diagram that must be coverted to a sentence. I have been struggling with the equations for about a week now, and was wondering if anyone on this forum might be able to help me out.

[b][u]PROBLEM[/u][/b]

Convert the equations into sentences:

1.  A is subset of (B n C')

2.  (A n B)' is not a subset of C

3.  (A u C) is a subset of (B n C)'

Where A = {People who like apples}, B = {People who like bananas}, C = {People who like cherries}[/u][/b][/quote]

Apparently either no one knows or no one is willing to offer any help. Maybe I will try to find a better forum to assist me with my studies.

1.  Every person who likes apples also likes bananas, but doesn't like cherries.

2.  There exists a person who dislikes cherries and dislikes either apples or bananas.

3.  Every person who likes apples or cherries dislikes either bananas or cherries.

this question should probably be in another section.  i don't think it falls under "advanced fields"


Find all nonnegative real numbers $ a_1 \le a_2 \le ... \le a_n$ satisfying:

$ \displaystyle\sum_{i\equal{}1}^{n}a_i\equal{}96, \displaystyle\sum_{i\equal{}1}^{n}a_i^2\equal{}144, \displaystyle\sum_{i\equal{}1}^{n}a_i^3\equal{}216.$

Cauchy-Schwarz inequality gives

$ ( \sum_{i\equal{}1}^n \ a_i^{1/2}.a_i^{3/2} )^2 \leq ( \sum_{i\equal{}1}^n \ a_i )( \sum_{i\equal{}1}^n \ a_i^3 )$

or

$ ( \sum_{i\equal{}1}^n a_i^2 )^2 \leq ( \sum_{i\equal{}1}^n a_i )( \sum_{i\equal{}1}^n a_i^3 )$

from given values,

LHS = $ 144^2 \equal{} 20736$

RHS = $ 96 \times 216 \equal{} 20736$

This means there is the equality sign in Cauchy schwarz

so all terms are proportional.

$ \Rightarrow \ \frac{a_i^{1/2}}{a_i^{3/2}} \equal{} constant \ \forall \ i$

$ a_1 \equal{} a_2 \equal{} a_3 \equal{} ... \equal{} a_n$

[b]
This gives n = 64 and $ a_i$ s = 3/2[/b]

[color=red]Hopefully no calc or logical mistake [/color] :)

The results are [hide="these"]$ n \equal{} 64, a_1 \equal{} a_2 \equal{} ... \equal{} a_n \equal{} a \equal{} \frac {3}{2}$[/hide]

@ moldovan : Clearly n can't be 32...

that wouldn't satisfy

$ \sum_{i\equal{}1}^n a_i \equal{} 96 \ as  \\  32 \times \frac{3}{2} \neq 96$

[color=blue]Correct answer :[/color] [b]n = 64, a = 3/2 [/b]

These are the results from the book. It seems to be correct what you say, so maybe the results from the book contain an error. However, the solution is the most important.


donbaleie zir az adade tabiee ra dar nazar begirid.aia adade 7 dar in donbale mi aiad :?:

1,2,4,8,1,3,6,1,2,5,1,2,4,8,...

 :|  :blush:  :huh:

salam
baba man ie tazevaredam amir agha.mizashtin ie chand nafar ie negahi bokonan bad...mishe beporsam koja dide boodinesh?


[quote="CatalystOfNostalgia"]We are given $ a_{1},a_{2},\ldots,a_{n}\in S$ where $ S$ is a set in which communitivity and associativity over addition both hold. Let $ \{b_{1},b_{2},\ldots,b_{n}\}$ be a permutation of the set $ \{a_{1},a_{2},\ldots,a_{n}\}$. Show that $ \displaystyle\sum_{i \equal{} 1}^{n}a_{i} \equal{} \displaystyle\sum_{i \equal{} 1}^{n}b_{1}$.[/quote]
isn't this kind of trivial if we're given that commutativity holds?

[hide="my wacky proof"]From computer science, "[url=http://en.wikipedia.org/wiki/Bubble_sort]bubble sort[/url]" each of the permutations. Bubble sort only involves swapping adjacent elements, so because commutivity holds the sum of each permutation remains the same. However, now you have two sorted lists, which are the same, so they have the same sum. Therefore the original permutations have the same sum.[/hide]


E=mc 2 Energy=Mass x The speed of light squared
This problem was easy and I can't believe that someone actually posted this. I am taking English class to help with my English.

?  ARe you inquiring about something or just chatting out of the sake of it?


linear algebra
matrix
function
ratio
calculus
trigonometry
Euler

hello, here will you find the answer
http://en.wikipedia.org/wiki/Euler's_formula
Sonnhard.

@ Dr. Graubner
There exists  no such page in wikipedia.
What i want to know is the different no. of ways in which the Euler's formula can be proved.
For eg, it can be proved by equating the respective expansions and any other method...

I think that this copy of the best-seller of Hairer and Wanner (Analysis by Its History) pages 57 and 58 might be usefull ! First A and then B !

It's my opinion that this formula is best understood via the [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=216343]matrix exponential[/url] on the Lie algebra of the circle group, but some might disagree.

Something I think you need in order to justify what is in the attachment in post #4: what good reason do we have to accept the the function $ f(z)\equal{}\sum_{n\equal{}0}^{\infty}\frac{z^n}{n!}$ as an exponential?

We start by going through the theory of the convergence of series for complex numbers, including that absolutely convergent series converge. Then we note (the ratio test helps) that the series we want converges absolutely for all $ z\in \mathbb{C}.$

Then we prove some things about double series and products of series, all with the hypothesis of absolute convergence.

All of this sets up the key point: if $ f(z)\equal{}\sum_{n\equal{}0}^{\infty}\frac{z^n}{n!},$ then prove that $ f(z\plus{}w)\equal{}f(z)f(w)$ for all $ z,w\in\mathbb{C}.$

That's the key to the laws of exponents. With that in place, we start writing $ e^z$ for $ f(z),$ and we have other things, like $ e^{nz}\equal{}(e^z)^n$ and $ e^{\minus{}z}\equal{}\frac1{e^z}$ (with the corollary that $ e^z\ne 0.$)

And then we say that if $ z\equal{}x\plus{}iy,$ then $ e^z\equal{}e^{x\plus{}iy}\equal{}e^xe^{iy}.$ We recognize the first factor as the already reasonably well understood real exponential. If we've already developed the calculus of the trigonometric functions, then we do what is done in post #4 to understand $ e^{iy}.$ Alternatively, if we [i]don't[/i] already have the calculus of the trigonometric functions, we define those functions in terms of our newly established complex exponential.

[quote="Kent Merryfield"]what good reason do we have to accept the the function $ f(z)\equal{}\sum_{n\equal{}0}^{\infty}\frac{z^n}{n!}$ as an exponential?[/quote]
Can we use the fact that $ f(z)\equal{}e^{z}$ is an entire function so it has a Taylor series expansion?  Or must we actually  prove that $ e^{z}$ is analytic first?

[quote="JRav"]
Can we use the fact that $ f(z) \equal{} e^{z}$ is an entire function so it has a Taylor series expansion?  Or must we actually  prove that $ e^{z}$ is analytic first?[/quote]

in what kent is talking about, we make the [i]definition[/i] $ e^z \triangleq \sum^\infty_{j\equal{}0} \frac{z^j}{j!}$.

$ f(x)\equal{}\frac{\cos(x) \plus{} i \sin(x)}{e^{ix}}$
$ f'(x)\equal{}\frac{\minus{}\sin x e^{ix}\plus{}i\cos x e^{ix}\minus{}i e^{ix} \cos x\plus{}e^{ix}\sin x}{e^{2ix}}\equal{}0$
$ f(x)\equal{}f(0)\equal{}1$.

I believe the proof of Euler's formula has been posted several times in MathLinks, but I can't find my post.

Anyway, this is the proof that I have posted.

Let $ f(x)\equal{}\cos x\plus{}i\sin x$, we have $ f'(x)\equal{}if(x)\longrightarrow f(x)\equal{}f(0)e^{ix}$, yielding desired reault.

[quote="JRav"][quote="Kent Merryfield"]what good reason do we have to accept the the function $ f(z) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {z^n}{n!}$ as an exponential?[/quote]
Can we use the fact that $ f(z) \equal{} e^{z}$ is an entire function so it has a Taylor series expansion?  Or must we actually  prove that $ e^{z}$ is analytic first?[/quote]

It seems to me that proving $ e^z$ is analytic begs the question.

In order to show that $ e^{x \plus{}i y}$ satisfies the Cauchy-Riemann conditions doesn't one have to either assume the Euler formula, $ e^{i y} \equal{}\cos y \plus{} i\sin y$ -- what we wish to prove in the first place -- or at least assume that $ e^{i y} \equal{}\sum_{n \equal{} 0}^{\infty}\frac{(iy)^n}{n!}$, which of course is tantamount to the same thing?

Dr. No is responding to JRav rather than to me, but to clarify my position:

Proving the analyticity of $ e^z$ was never part of my plan at all. I don't need it. The centerpiece of the whole argument is the proof, based solely on the power series, that $ e^{z \plus{} w} \equal{} e^ze^w.$ Once we have that, many other things flow from that one result (including the world of trig identities).

But as a side issue: we [i]defined[/i] $ e^z$ as the sum of a power series. It's pretty straightforward to show that a power series in $ z$ gives an analytic function, provided you stay strictly inside the radius of convergence (and this particular series has an infinite radius of convergence). Hence, we have analyticity if we want it - and it's not begging the question.

JRav wanted $ e^z$ to be analytic so he could justify it having a power series. That's [i]not[/i] the direction of my argument. In my version, the power series defines a function just because it's a convergent power series. What we then need to do is to prove the algebraic laws that would leave us comfortable with using the exponential name and notation for this function.

[quote="Kent Merryfield"]It's pretty straightforward to show that a power series in $ z$ gives an analytic function, provided you stay strictly inside the radius of convergence (and this particular series has an infinite radius of convergence). Hence, we have analyticity if we want it - and it's not begging the question.[/quote]

Kent,

If analyticity is a given for the series, in as much as we don't have to show satisfaction of the C-R conditions but can infer it from the ROC as you say, then doesn't that mean that all the properties of the exponential function come along for the ride? and if so, what is there left to prove?

[quote="Dr. No"]If analyticity is a given for the series, in as much as we don't have to show satisfaction of the C-R conditions but can infer it from the ROC as you say, then doesn't that mean that all the properties of the exponential function come along for the ride? and if so, what is there left to prove?[/quote]

The only real thing that you have to prove is that $ \exp(x\plus{}y)\equal{}\exp(x)\exp(y)$.  This is not quite trivial, but can be accomplished through use of standard facts on convergent power series.  That, combined with the fact that $ \exp(1)\equal{}e$ and $ x\to\exp(x)$ is continuous, completely characterizes the exponential function, so it's a perfectly good definition.

[quote="Kent Merryfield"]JRav wanted $ e^{z}$ to be analytic so he could justify it having a power series.[/quote]
I understand this is not your argument, but would this be a suitable way to justify that power series expansion?

[quote="blahblahblah"]The only real thing that you have to prove is that $ \exp(x + y) = \exp(x)\exp(y)$.  This is not quite trivial, but can be accomplished through use of standard facts on convergent power series.  That, combined with the fact that $ \exp(1) = e$ and $ x\to\exp(x)$ is continuous, completely characterizes the exponential function, so it's a perfectly good definition.[/quote]

Fair enough, let's get on with it then.

\[ \exp(z + w) =\sum_{m=0}^{\infty}\frac{(z + w)^m}{m!}\\ \\ \\ =\sum_{m=0}^{\infty}\frac{\sum_{k =0}^{m} \binom{m}{k}z^{m - k}w^k}{m!} \\ \\ \\ =\sum_{m=0}^{\infty}\frac{\sum_{k =0}^{m} z^{m - k}w^k}{(m-k)! k!} \\ \\ \\ = \exp(z)\exp(w)\]

That done, let's consider $ \log(\cos x + i\sin x)$.

\[ \frac{d}{dx}\{\log(\cos x + i\sin x)\} =\frac{-\sin x + i\cos x}{\cos x + i\sin x}\\ \\=\frac{-\cos x\sin x +\cos x\sin x + i(\cos^2 x +\sin^2 x)}{\cos^2 x +\sin^2 x}\\ = i\]

\[ \implies \int \mathrm{d}\log(\cos x + i\sin x) = i\int \mathrm{d}x \\ \\ \implies \log(\cos x + i\sin x) = i x +\log C\]

\[  \log(\cos x + i\sin x)|_{x = 0} =  i x +\log C|_{x = 0} \\ \\ \implies \log(\cos 0 + i\sin 0) = i\cdot 0 +\log C \\ \\ \implies \log 1 = \log C = 0\]

\[ \log(\cos x + i\sin x) = i x \\ \\ e^{\log(\cos x + i\sin x)} = e^{i x} \\ \\ \boxed{\cos x + i\sin x = e^{i x}}\]

[quote="praveen97uma"]I want to know all the possible ways of proving e^ix = cos(x) + i sin(x).[/quote]

There are several proofs, but they're all equivalent; furthermore, it's a proof so you only need one.

[quote="t0rajir0u"]That proves nothing except formally; among many other things, until you do something like what Kent did you have no guarantee that $ e^{ix}$ even makes sense as a complex function, and you haven't made it at all clear what values $ x$ takes and therefore what kind of differentiability you're looking for (real or complex).[/quote]

For reals of course. Every complex number can be written as $ \cos x \plus{} i \sin x$ where $ x\in \mathbb{R}$.Therefore, there was no sense to write a proof of the formula for arbitrary numbers (including complex). :)

[quote="BG Yoda"]...Every complex number can be written as $ \cos x \plus{} i \sin x$ where $ x\in \mathbb{R}$.Therefore, there was no sense to write a proof of the formula for arbitrary numbers (including complex). :)[/quote]

This bit is only strictly true for complex numbers having modulus $ 1$.

Other than that, yes there is a reason to write a proof of "the formula" for arbitrary numbers.

Complex numbers do not behave the same way real numbers behave. For example, a function can be differentiated with respect to a complex variable only if said function is analytic. If the function is not analytic in the region of the Argand plane where one is attempting to find the derivative then the idea of its having a derivative in that region is nonsense.

[quote="t0rajir0u"][quote="Dr. No"]There are several proofs, but they're all equivalent; furthermore, it's a proof so you only need one.[/quote]
This is misleading.  It would be more accurate to say that there are several ways to define the functions $ e^z, \sin z, \cos z$ and that the work consists in proving that each of these definitions are equivalent.  Once you have proven these equivalences, Euler's formula falls out.[/quote]

I agree that it would be more exhaustive to prove the various possible definitions of $ e^z, \sin z,\mbox{and } \cos z$ are equivalent. However, that is not the proof that the author asked for; he or she asked for all the ways one could prove that $ \cos x \plus{} i\sin x \equal{} e^{i x}$ (with the unstated assumption that $ x \in \mathbb{R}$.) That is what I proved, there are many proofs of it and they are all equivalent; therefore, you do only need one!

[quote="t0rajir0u"]Yes, but where you start the proof depends on how you're planning to define the function $ e^{ix} : \mathbb{R} \to \mathbb{C}$.  The simplest way is via its power series expansion, and there it requires no extra effort to extend your definition to $ x \in \mathbb{C}$.  But you can't just claim that this function is well-defined because $ e^x : \mathbb{R} \to \mathbb{R}$ is well-defined.[/quote]

Yes, well I started with Kent Merryfield's power series definition of $ e^z$, which is perfectly well defined, to guarantee that is why I went to the trouble to show that $ e^{z \plus{} w} \equal{} e^z e^w$, and a function $ \log z$, which is itself demonstrably analytic everywhere.

As a consequence, everything I did in generating the theorem was legitimate. Hence, we have proof of Euler's formula.  Most importantly, nothing I've said is in any way misleading.


A $ \textit{composite number}$ is a number that has two or more prime factors. The number 87 can be expressed as the sum of two
composite numbers in many ways. What is the minimum
positive difference between two such numbers?

The wording here is ambiguous; it should read "two (not necessarily distinct) prime factors."

It still shouldn't affect the answer: 45-42=3. (Both have two or more indistinct prime factors.)

[quote=Tuktuk24]9 can't be expressed as the some of 2 composite numbers.[/quote]

It can, if 1 is composite (is it?)

No, 1 is defined as neither composite or prime.
[quote=Aops Wiki]Note that the number one is neither prime nor composite.[/quote]


[quote="PluTo"]Let $ G$ be a finite group of odd order, and let $ p$ be the smallest prime dividing $ |G|$. Suppose that $ G$ has a normal $ p$-Sylow subgroup $ P$ of order $ \leq p^2$. Prove that $ P$ is contained in the center of $ G$.[/quote]

Compute the order of the automorphism group of every group of order p or p^2, and notice that the order is not divisible by any primes greater than p.  Here it is important that p is odd, so that p+1 is divisible by 2.

This is related to the earlier question on "conjugation".  G/C_G(P) is isomorphic to a subgroup of Aut(P) when P is normal.


geometry
3D geometry
tetrahedron
perpendicular bisector
IMO Shortlist
IMO Longlist

Let $ABCD$ be a regular tetrahedron. To an arbitrary point $M$ on one edge, say $CD$, corresponds the point $P = P(M)$ which is the intersection of two lines $AH$ and $BK$, drawn from $A$ orthogonally to $BM$ and from $B$ orthogonally to $AM$. What is the locus of $P$ when $M$ varies ?

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

This is the first serious stereometry problem I am solving, so I would be very glad to hear whether it is correct or not.

I will use a lemma from plane geometry (a theory where I feel rather at home :) ):

[b]Lemma 1.[/b] If H is the orthocenter of a triangle ABC, and N is the foot of the altitude from the vertex A, then we have $NH\cdot NA = BN\cdot NC$, where the segments are directed.

[i]Proof.[/i] I won't use directed segments in the proof, so I will only show the equality $NH\cdot NA = BN\cdot NC$ for non-directed segments. However, showing that it also holds for directed segments will be then just a matter of looking at the sketch.

If O is the foot of the altitude of triangle ABC from the vertex B, then the two lines AN and BO are altitudes of the triangle ABC and thus meet at its orthocenter H. Consequently, < HBN = < OBC = 90 - < OCB = 90 - < ACN = < CAN. On the other hand, clearly < HNB = 90 = < CNA. Thus, the triangles HBN and CAN are similar, and we can conclude that HN : BN = CN : AN. In other words, NH : BN = NC : NA, or, equivalently, $NH\cdot NA = BN\cdot NC$. This proves Lemma 1.

Lemma 1 has a converse:

[b]Lemma 2.[/b] Let ABC be a triangle, let N be the foot of the altitude from the vertex A, and let H' be a point on the line AN such that the equation $NH^{\prime}\cdot NA = BN\cdot NC$ holds with directed segments. Then, this point H' is the orthocenter of triangle ABC.

[i]Proof.[/i] If H is the orthocenter of triangle ABC, then Lemma 1 yields $NH\cdot NA = BN\cdot NC$. Comparing this with $NH^{\prime}\cdot NA = BN\cdot NC$, we see that NH = NH'; since we are using directed segments and the points H and H' both lie on the line AN, we can conclude that H = H', so that the point H' is the orthocenter of triangle ABC. Lemma 2 is proven.

Now let's come to the solution of the problem. For any two points U and V in space, I will denote the plane passing through the midpoint of the segment UV and being perpendicular to the line UV as the [b]perbisplane[/b] of the segment UV. This is, of course, an abbreviation for "perpendicular bisector plane"; we will often use the fact that the perbisplane of a segment UV is the locus of all points W such that UW = VW.

Since the tetrahedron ABCD is regular, the triangles ABC and ABD are equilateral. Let X and Y be the centers of these equilateral triangles. Then, of course, AX = BX and AY = BY; thus, the points X and Y lie on the perbisplane of the segment AB. Also, since the triangles ABC and ABD are equilateral, we have AC = BC and AD = BD, so that the points C and D lie on the perbisplane of the segment AB. The point M, lying on the line CD, also lies on the perbisplane of the segment AB. Hence, AM = BM.

Trivially, another point lying on the perbisplane of the segment AB is the midpoint S of the segment AB.

The definition of the point P that was given in the problem is just a complicated way to say that the point P is the orthocenter of the triangle ABM. Of course, since the triangle ABM is isosceles (we have AM = BM), we thus must have AP = BP, and consequently, the point P lies on the perbisplane of the segment AB. Altogether, the points X, Y, C, D, M, S and P all lie on the perbisplane of the segment AB. This will be very useful for us.

Now, I am not really sure what the problem means when it says that the point M must lie on the edge CD: should it mean that the point M lies on the [i]line[/i] CD, or should it mean that the point M lies on the [i]segment[/i] CD ? In the following, I will care for both of these cases. In fact, I will show:

[b](a)[/b] If the point M traces the whole line CD, the locus of the point P is the circumcircle of triangle SXY (except for the point S, which corresponds to an infinite position of the point M).

[b](b)[/b] If the point M traces just the segment CD, the locus of the point P is the arc XY of the circumcircle of triangle SXY.

At first, I will prove part [b](a)[/b]:

Since the point S is the midpoint of the segment AB, the line MS is a median of triangle ABM. But since this triangle ABM is isosceles (AM = BM), this median is also an altitude, so that the point S can be considered as the foot of the altitude of triangle ABM from the vertex M. Since the point P is the orthocenter of triangle ABM, it follows that the point P lies on the line MS, and by Lemma 1 we have $SP\cdot SM = AS\cdot SB$. Since the point S is the midpoint of the segment AB, we have $AS=SB=\frac{AB}{2}$; denoting $\frac{AB}{2} = s$, we get AS = SB = s, and thus $SP\cdot SM = s^2$.

Now, if we consider the circle k which has center S and radius s and which lies in the perbisplane of the segment AB, then the fact that the point P lies on the line MS and satisfies $SP\cdot SM = s^2$ yields that the point P is the inverse of the point M in the circle k.

Similarly, taking into account the points X and Y, which are the centers - in particular, the orthocenters - of the equilateral triangles ABC and ABD, we see that the points X and Y are the inverses of the points C and D in the circle k. Hence, the inverse of the line CD in the circle k is the circumcircle of triangle SXY (in fact, this inverse must be, like the inverse of any line which doesn't pass through the center of the inversion, a circle through the center of the inversion, and it must pass through the points X and Y since they are the inverses of the points C and D in the circle k).

Now, since the point M lies on the line CD, the inverse P of the point M in the circle k must lie on the inverse of the line CD, i. e. on the circumcircle of triangle SXY. Conversely, if we take any point on the circumcircle of triangle SXY (except of the point S) and denote it by P', then by redoing our argumentation above backwards, we see that there is a point M on the line CD such that the point P' is the orthocenter of the triangle ABM (in fact, just call M the inverse of the point P' in the circle k; it is then easy to show that $SP^{\prime}\cdot SM = AS\cdot SB$, and applying Lemma 2 instead of Lemma 1 we get what we need). Hence, the locus of the point P while the point M traces the line CD is the circumcircle of triangle SXY. This proves assertion [b](a)[/b].

For the proof of assertion [b](b)[/b], just look at the picture (you needn't draw the points A and B, just consider the points X, Y, C, D, M, S and P which all lie on the perbisplane of the segment AB). The point P lies on the arc XY of the circumcircle of triangle SXY if and only if the line SP divides the angle XSY internally, but this is clearly equivalent to saying that the point M lies inside the segment CD.

So the proof is complete.

Thanks for reading this (if you have :D ). Also thank you in advance if you find mistakes in the above solution (as I've said, I'm new to stereometry).

  Darij


Can anybody help me with this question??

Class A, with 15 students in the class, scored an average of 94 marks in a mathematics test. The maximum possible score of the test is 100 marks. What is the lowest possible score that any of the 15 students could have scored??

Thanks..... :blush:

Well, you want one studen't to be the lowest, so set the rest as the highest possible, or 100.  The total sum is 94*15, or 1410.  The highest scoring kids get 100*14, or 1400 total.  Subtract to get: 10.

Wow that kid sucks (the one who got 10)

This seems to be a problem from the Singapore Mathematical Olympiad.

$ 15 \times 94\minus{}14 \times 100 \equal{} 10$.

For an average of 94 for the 15 kids, there must be 1410 points total.
Now for the lowest score possible, 14 kids must get 100.
1410-1400=$ \boxed{10}$

Thanks very much for all your replies.....

THANKS