Let $ a$ be a positive number. Find all function $ f: R^ \plus{} \to R^ \plus{}$ such that: $ f(f(x)) \equal{} ax \minus{} f(x)$ for all $ x > 0$ :)

Hm, :( Noone can solve it? If that;s right, i'll Suggest! :)

We'll write $ a \equal{} c^2 \plus{} c$, with $ c$ is a positive real number, and now, we'll show that $ f(x) \equal{} cx$ for all $ x$! :)

[quote="Rust"]There are infinetely many solutions.[/quote]
Rust, I donn't think like that, we can conclude that "There are infinetely many solutions" only in case the function $ f: R \to R$, but here, $ f: R^ \plus{} \to R^ \plus{}$.
If in the case $ f: R^ \plus{} \to R^ \plus{}$, you say too "There are infinetely many solutions", can you denote the way to construct a class of function $ f: R^ \plus{} \to R^ \plus{}$ what satisfy this problem! :)

Let $ 0 < a\le 1$, we can find $ f(x) \equal{} xc(x)$, then $ c(xc(x)) \equal{} \frac {a}{c(x)} \minus{} 1$.
We can consider $ c(x)\equiv c_i , \ x_i\le x < x_{i \plus{} 1}$, then $ c_{i \minus{} 1} \equal{} \frac {a}{c_i} \minus{} 1, x_i \equal{} c_ix_{i \plus{} 1}$.
It give infinetely many solutions.
 I am sorry. It work only for $ 0 < a < 2$ and $ f: (0,A)\to (0,A$ or
$ a > 2$ and $ f: (A,\infty)\to (A,\infty )$.
In these case $ f: R_\plus{}\to R_\plus{}$ unique solution for $ a>0$.

Ohlala, and with case $ a$ is a enough positive number, example $ a\equal{}10^9$ :) How do we solve it? :P

Let $ a>0$ and $ c$ is root of $ c^2\plus{}c\equal{}a$. Let exist $ x_0$, suth that $ c(x_0)\equal{}c_0\not \equal{}c$. Let $ g(c)\equal{}\frac ac \minus{}1.$
Then $ sign(g(x)\minus{}c)\equal{}sign(c\minus{}x)\to c_0>c>c_1\equal{}g(c_1) \ or \ c_0<c<c_1$.
Let $ c_2\equal{}g(c_1)\equal{}\frac{ac_0\plus{}c_0\minus{}a}{a\minus{}c_0}\equal{}c_0\plus{}\frac{c_0\plus{}c_0^2\minus{}a}{a\minus{}c_0}$.
If $ c_0<c$ then $ c_2<c_0$. If $ c<c_0<a$ then $ c_0<c_2\equal{}g_1(c_0)$.
Let $ x_{i\plus{}1}\equal{}x_ic_i,c_{i\plus{}1}\equal{}g(c_i).$ If $ c_i\ge a$, then $ f(f(x_i))\le 0$ - contradition.
Tlse we get $ c>c_0>c_2>...\ge 0$ or $ c<c_0<c_2<....\le a$. It give $ lim_i c_{2i}\equal{}c_*\not \equal{}c$ and $ g_1(c_*)\equal{}c_*$.
We prove that $ g_1(c_*)\equal{}c_*\to c_*\equal{}c$. Therefore $ c_0\equal{}c \ \forall x_0>0$ or $ f(x)\equiv cx$.


inequalities
induction
inequalities solved

Given a1,a2,...,an real numbers define bk=(a1+...+ak)/k for k=1,2,..,n.

If C=(a1-b1)^2 + (a2-b2)^2 +...+ (an-bn)^2

and D=(a1-bn)^2 + (a2-bn)^2 +...+ (an-bn)^2

prove that  C \leq D \leq 2C

Let f(x)=(x-a_1)^2+(x-a_2)^2+\dots+(x-a_n)^2 then

f(x)=n(x-b_n)^2+f(b_n). (*)

If n=1 then inequalitys are trivial (C=D). To prove inequalities by induction it is enough to prove that

0\leq f(b_{n+1})-f(b_{n})\leq (a_{n+1}-b_{n+1})^2.

First ineq follows imediatly from (*) (x=b_{n+1}). Second follows from equalities

(n+1)b_{n+1}=nb_n+a_{n+1}, n(b_{n+1}-b_n)=a_{n+1}-b_{n+1},

f(b_{n+1})-f(b_{n})=n(b_{n+1}-b_n)^2=(a_{n+1}-b_{n+1})^2/n


analytic geometry
graphing lines
slope
algebra
system of equations

I appericate all the help and want to thank you all in advance..
Im just a little stumped on how to figure these out and show the work for it..Espsically questions #3-6

1.  Find the slope of the line that passes through the points (2, 3) and (5, 8).

2.  Find the equation of the line that passes through the points (3, -2) and (4, -2).

3.  Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5).

4.  Solve the system of equations using the substitution method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
-3x + y = 1
5x + 2y = -4

5.  Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
-7x + y = 8
2x – y = 2

6.  Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
3x – 2y = -7
-9x + 6y = 21

Are you sure this isn't a homework?

3. The slope of perpendicular lines are negative reciprocals.

4-6. Multiply both sides by numbers which will cancel out one variable. For example, $ 2x+a$ and $ -3x+2a$, multiply the expression 1 by 3 and expression 2 by 2. That will eliminate $ x$.

[quote="mountain.dew"]These should be in HSB...
[hide]1. Slope is $ \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$. So, the answer is $ \frac{5}{3}$.
2. Using the previous method, the answer is $ 0$.
3. The line has slope $ -\frac{1}{3}$, so its perpendicular would have slope $ 3$. Using point-slope form, we have
$ 3(x+3)=y-5$
$ 3x-y=-14$.
4. From the first equation, $ y=3x+1$. Putting this into the second equation, we have
$ 5x+2(3x+1)=-4$
$ x=-\frac{6}{11}$
$ y=-\frac{7}{11}$
$ (-\frac{6}{11},-\frac{7}{11})$
5. Adding the equations, we have
$ -5x=10$
$ x=-2$
$ y=-6$
$ (-2,-6)$.
6. Multiplying the first equation by 3, we have $ 9x-6y=-21$. Adding the equations, we have $ 0=0$, which is true for all $ x$'s and $ y$'s. The system has infinitely many solutions.[/hide][/quote]

Please avoid answering questions directly. Hints are better, it will make them think and work on their own. Anyway, these are very easy questions.  :)


All face of a convex polyhedron are parallelograms.Can the polyhedron have exactly 1992 faces?

[hide]
Consider a set of parallelograms faces that goes all the way "around" such a polyhedron. Obviously, two adjacent faces in such a set share an edge. Since they're parallelograms, the sides are all parallel. We can see that any two of these sets intersect at two faces, and conversely, that any face belongs to two rings. So if there's $ n$ distinct sets of these, then there's $ n(n\minus{}1)$ faces. However, there's no integer solutions to $ n^2\plus{}n\minus{}1992\equal{}0$, so it can't have 1992 pages.

Wow, that's a tricky problem.[/hide]


function
LaTeX
calculus
calculus computations

[color=darkblue]prove that this equation: $ y' \equal{} x \plus{} y^2$ have no root.[/color]

What do you mean "roof"?  Are you asking about roots/zeroes?

this is a differential equation; do you mean 0 = -y' + x + y² has no solution in x where y is a function of x, or do you mean the differential equation has no general solution?

You can always at least make an approximation using power series.  Even if it isn't in closed form, you get something that is correct.

[quote="JRav"][b]What do you mean "roof"? [/b] Are you asking about roots/zeroes?[/quote]

 :rotfl:    :D

[b]thanhnam2902[/b]
Do you mean y=f(x) ?

When y=x  , we have
y'=x+y²
1=x+x²
Which has 2 real roots.
Or when y=x²
2x=x+x^4 , which also has a real root x=1 and x=0.

Maybe I'm just missing something, but I don't see why $ y\equal{}x$ is a solution.
$ y'\equal{}1\neq x\plus{}x^{1}$

By the existence-uniqueness theorem, this differential equation has local solutions. What you want is to show that these solutions cannot be extended indefinitely; all such solutions blow up to $ \infty$ in finite time.

As a first step, note that all positive solutions $ \frac1{c \minus{} x}$ to $ y'' \equal{} y^2$ blow up in finite time. You want to show that each solution to your differential equation is eventually bigger than one of these, and hence blows up faster.

I absolutely agree with jmerry.
to ლევანი
you can use latex ,
see here
http://www.mathlinks.ro/LaTeX/AoPS_L_About.php


algebra
polynomial
inequalities open
inequalities

This one is still a special case of the one I posted in http://www.mathlinks.ro/Forum/viewtopic.php?p=580614#580614 middle of first page (Tue Dec 06, 2005) :lol: :
[quote="spanferkel"]
[b] For partitions $\alpha,\beta\vdash n$ and $0\leq t\leq1$ we have $[\alpha]+[\beta]\geq [(1-t)\alpha\oplus t\beta]+[t\alpha\oplus(1-t)\beta]$. (*)[/b][/quote]  :)  :)


Someone on this forum gave me an invitation on Yahoo Messenger, can yo do that again? I have pressed Cancel by mistake! Sorry! :(

cheers! :D :D

...Lol? I feel like this is the wrong forum...I may be wrong though.


A most unusual problem with one of my computers:

When on AOPS (ONLY FOR AOPS) I can only select a textbox once, or else the browser forcequits/freezes/kills the computer. This happens for every browser, e.g. Firefox, I.E., AOL, Netscape, Chrome, Opera, Safari, etc. I try, and currently I am using a custom browser (Avant) which freezes as well at certain random times.

I don't understand what you mean by "I can only select a textbox once" ...

As in, when I select a textbox, there is a flashing line (|) showing up. If I click on the textbox again, then it freezes.


Functional Analysis
real analysis
real analysis unsolved

i have a hard problem need to help.
 Given 2  bounded linear operators A, B satisfied AB is compact. Does A or B must be compact?

Let's try this one, using operators on the Hilbert space $\ell^{2}:$

Define $A$ by $A: (x_{1},x_{2},\dots)\mapsto(x_{1},0,x_{3},0,x_{5},0,\dots).$

Define $B$ by $B: (y_{1},y_{2},\dots)\mapsto(0,y_{1},0,y_{2},0,y_{3},\dots).$

Neither $A$ nor $B$ is compact. $A$ is an orthogonal projection onto a one-dimensional subspace, and $B$ is an isometry.

But $AB=0,$ and the zero operator is compact.


function
calculus
derivative
calculus computations

Hi, can anybody help me with this?

I have a function y=arctg (1/x)
In which point has a tangent to this function "k = -1/2"

Thanx,
Tyrrel

Solve of a tangent = derivative at the point of tangency. Solve $\frac{-1}{1+x^{2}}=-\frac{1}{2}$ for $x$.

Not so advanced after all.


[quote="Hong Quy"]Let $ a,b,c,d$  be nonnegative real numbers. Prove that:
$ \sum\limits_{cyc}\frac {a^2}{b \plus{} c \plus{} d} \leq 4\frac {(a \minus{} b)^2}{a \plus{} b \plus{} c} \plus{} \frac {4(a \plus{} b \plus{} c \plus{} d)}{3}$[/quote]
Let $ a\equal{}b\equal{}0$, the above inequality becomes
\[ \frac{c^2}{d}\plus{}\frac{d^2}{c} \le \frac{4}{3}(c\plus{}d)\]
which does not always hold, for example with $ d\equal{}1,c\equal{}100$.

Let $ a,b,c,d$  be nonnegative real numbers. Prove that:
$ \sum\limits_{cyc}\frac {a^2}{b \plus{} c \plus{} d} \leq 4max \left( \frac {(a \minus{} b)^2}{a \plus{} b \plus{} c},\frac {(c \minus{} b)^2}{d \plus{} b \plus{} c},\frac {(c \minus{} d)^2}{d \plus{} a \plus{} c},\frac {(d \minus{} a)^2}{a \plus{} b \plus{} d}\right) \plus{} \frac {4(a \plus{} b \plus{} c \plus{} d)}{3}$


function
integration
algebra
linear equation
real analysis
real analysis unsolved

[b]I'm aware this is a long problem but I would really appreciate some help/hints especially for getting the inhomogeneous ODE of the first part:[/b]

The function $ y(x)$ satisfies the linear equation
\[ y'' \plus{} p(x)y' \plus{} q(x)y \equal{} 0.\]
The Wronskian of two independent solutions, denoted $ y_1(x)$ and $ y_2(x)$, is defined to be
\[ W(x) \equal{} \left| \begin{array}{cc} y_1 & y_2 \\
y_1' & y_2' \end{array} \right|.\]
Let $ y_1(x)$ be given. Use the Wronskian to determine a first-order inhomogeneous differential equation for $ y_2(x)$. Hence, show that
\[ y_2(x) \equal{} y_1(x) \int_{x_0}^{x} \frac {W(t)}{y_1(t)^2} \, dt. \qquad \qquad (*)\]
Show that $ W(x)$ satisfies
\[ \frac {dW}{dx} \plus{} p(x)W \equal{} 0.\]
Verify that $ y_1(x) \equal{} 1 \minus{} x$ is a solution of
\[ xy'' \minus{} (1 \minus{} x^2)y' \minus{} (1 \plus{} x)y \equal{} 0. \qquad \qquad (\dagger)\]
Hence, using $ (*)$ with $ x_0 \equal{} 0$ and expanding the integrand in powers of $ t$ to order $ t^3$, find the first three non-zero terms in the power series expansion for a solution, $ y_2$, of $ (\dagger)$ that is independent of $ y_1$ and satisfies $ y_2(0) \equal{} 0, y_2''(0) \equal{} 1$.

hello, the solution of your equation is
$ y(x)\equal{}C_1(x\minus{}1)\plus{}C_2(x\minus{}1)\cdot\int\frac{xe^{\minus{}\frac{1}{2}x^2}}{(x\minus{}1)^2}\,dx$
Sonnhard.

To determine the first-order differential equation for y_2, just evaluate the Wronskian.

I got $ y_1(x)y_2'(x) \minus{} y_1'(x)y_2(x) \equal{} W(x)$, so that $ \frac{y_1(x)y_2'(x) \minus{} y_1'(x)y_2(x)}{y_1(x)^2} \equal{} \frac{W(x)}{y_1(x)^2}$.

Thus $ \left( \frac{y_2(x)}{y_1(x)} \right)' \equal{} \frac{W(x)}{y_1(x)^2}$. From here I think you integrate wrt $ x$, but how does the $ t$ variable and the lower limit $ x_0$ come in?

[quote="aidan"]I got $ y_1(x)y_2'(x) \minus{} y_1'(x)y_2(x) \equal{} W(x)$, so that $ \frac {y_1(x)y_2'(x) \minus{} y_1'(x)y_2(x)}{y_1(x)^2} \equal{} \frac {W(x)}{y_1(x)^2}$.

Thus $ \left( \frac {y_2(x)}{y_1(x)} \right)' \equal{} \frac {W(x)}{y_1(x)^2}$. From here I think you integrate wrt $ x$, but how does the $ t$ variable and the lower limit $ x_0$ come in?[/quote]

You've gone a step too far. The question says that y_1 is given - so treat it as a given polynomial. You have a first order differential equation in y_2 - solve it as you normally would (e.g. integrating factor), in terms of y_1.

[quote="aidan"]I forgot about that.. so the integrating factor is $ e^{y_1(x)}$, I presume?[/quote]

If you have the equation $ y' \plus{} p(x) y \equal{} f(x)$, the integrating factor is $ e^{\int p(x) \ dx}$

In this example, $ p(x) \equal{} \minus{} \frac {y_1'(x)}{y_1(x)}$ (the coefficient of y_2' must be 1).

So the integrating factor is $ e^{\int \minus{} \frac {y_1'(x)}{y_1(x)} \ dx} \equal{} \frac {1}{y_1(x)}$

Made a mistake there because of quick calculation  :blush:  Thanks :)


On almost winning Mathcounts.

(I'm serious though, congrats, you guys (EDIT: and girl) did well.)

Man I am still wondering how this happened...oh well I suppose Lewis Chen didn't try hard enough.

Gj to max for getting a 44.  Also CA failed...only 1 person in cd.  Hmph.

Blah another question: is it because you're a mod that you can post all caps in the title?

so this is like the one of the most nonspammy topics on mo forum ever...

Dude what? Is he your hero or something now? MO people must be so proud of him :O

[quote="jonathanchou711"][quote="LilyBreeze"]Dude Max looked soo hot at the countdown round![/quote]

No comment.

Max is like my height! He's kind of short too. No offense. I'm short too.

Gj. Max. You're awesome.[/quote]

Er, Max isn't that short.

So of course, no one has denied that max is pretty hot. He's pretty hot when he's doing math/meth whatever.

OH MAN MAX LIKE PWNED MATHCOUNTS :!:  :!:  :!:

We totally should've been first...

[quote="pythag011"][quote="suma_milli"][quote="pythag011"]David lost in the first round at state. Proves how much harder CA is too win then nats (jk.)[/quote]

Didn't David skip cd to go to ARML practice?[/quote]

That was chapter.[/quote]

:gasp   chapter < ARML practice !!!!!!!!!!!!!!!

[quote="pythag011"][quote="suma_milli"][quote="pythag011"]David lost in the first round at state. Proves how much harder CA is too win then nats (jk.)[/quote]

Didn't David skip cd to go to ARML practice?[/quote]

That was chapter.[/quote]

Ohh right. D:

[quote="pythag011"]David lost in the first round at state. Proves how much harder CA is too win then nats (jk.)[/quote]

Dude well nats cd was way too slow.  took like 30 seconds to realize that a equilateral triangle has 60 degree angles and regular hexagon has 120? :P then answered the wrong one

so... the mo arml t-shirt design is gonna be pretty awesome, depending on what color shirt they choose.

basically, just fill out an application.

wait so basically what you're saying is if someone from texas fills in an application it should be okay right

er how... victor?

let me edit that statement. if you can fill out an application, pay the fee, do it on time (or near enough, they're somewhat lenient), and you actually live in this region (basically, KS, MO, and southern IL, ie. non-Chicago), you're in. there's no requirement to attend practices, and there is no tryout of any sort. you're asked for test scores, but they're largely irrelevant, since the applicant pool is usually around 35ish students, and MO sends two teams + alternates.

i forgot that you have to fill out a form to make arml

clearly this is a requirement

and i failed it

well that was dumb...

i even made a thread about that.

well actually i just realized i filled out the form but never turned it in for reasons

meh, you shoulda emailed, they'd probably would've let you in a week or two ago.


function
abstract algebra
algebra
polynomial
integration
calculus
group theory

prove that: (in this case $ G$ is  discrete LCA group )
for every measurable function $ f$ on $ G$  and every positive number $ \epsilon >0$ there exist a function $ g$ with uniformly convergent
Fourier series satisfying the condition :
$ m\{f \neq g\}<\epsilon$
also what can you say about spectrum of $ g$?

Discrete locally compact abelian group? You mean something like $ \mathbb Z$? Then what is "the Fourier series" of a function defined on $ G$? :?

There is a nice theorem like that on the circle but the known proof is by no means simple (at least, too long and involved to post here). Do you have a better approach?

Yes, Discrete locally compact abelian group. One can say that, every discrete group is locally compact  .
Yes, I mean something like $ \mathbb{Z}$, but not only .
The fourier transform for $ \mathbb{Z}$ will be :
$ \widehat{f}(z) \equal{} \sum_{n\in\mathbb{Z}}f(n)z^{\minus{}n}$ where $ z\in \mathbb{T}$
and to define fourier series we need to define some "basis" for the dual group of $ \mathbb{Z}$ which is isomorphic  to $ \mathbb{T}$
I try to write it by this way (but it's not correct ! ):
$ \sum_{\gamma \in T}\widehat{f}(\gamma)\gamma$
where $ \gamma \in T$  we  understand in some reasonable sense to understand uniformly convergent of fourier series.
I have some idea how to define this "basis" and  also I can to say that it will be include uniformly convergent in any reasonable sense.
Theorem, what are you talking is in exact Men'shovs theorem which is true only for $ \mathbb{G}\equal{}\mathbb{T}$ and the proof is too long.
There is more better approach (also is too long) of S.V Kislyakov for any LCA group where  they are not discrete  :(

OK. Let $ G\equal{}\mathbb Z$ and $ f(z)\equal{}z$. Show me your $ g$ for this case. :P

By the way, what measure $ m$ are you considering? (if the Haar measure on $ G$, i.e., the counting measure, your claim is certainly false unless [b]all[/b] functions have "uniformly convergent Fourier series" in your sense, when it is trivial; if some finite measure on a countable group, then the claim reduces to the statement that you can interpolate any function at finitely many points by a trigonometric polynomial) :?

So, put it this way or that, it still seems to me that you are talking nonsense :(. As long as you have just some vague ideas and do not bother to tell any details, nobody will understand you even if what you say makes some sense.

This is not to discourage you but merely to remind you (once again :roll:) of the accepted standards of mathematical communication :).

Yes, Haar measure .
Ok. This is what I was thinking :
Let's $ m$ will be Haar measure  on LCA group $ G$ where $ G$ is discrete (for example $ G \equal{} \mathbb{Z}$) and $ m_{1}$ on $ \mathbb{T}$
Haar measure on $ \mathbb{T}$is normalized so that Parseval's equality holds :

$ \parallel{}\widehat{f}\parallel{}_{L^{2}(\mathbb{T})} \equal{} \parallel{}f\parallel{}_{L^{2}(\mathbb{Z})}$
where by $ \widehat{f}$ I denote :
\[ \widehat{f}(\gamma) \equal{} \int_{\mathbb{Z}} f(x)(\gamma^{ \minus{} 1},x)d m(x)
\]
consider partial integrals  ("partial sums" ) :
\[ f(x) \equal{} \int_{\mathbb{T}} \widehat{f}(\gamma) (\gamma,x)1_{E}d m_{1}(\gamma)
\]
where $ \gamma \in \mathbb{T}$, $ E$ is measurable subset of $ \mathbb{T}$ and $ 1_{E}$ is characteristic function.
in my first post by $ \sum_{\gamma \in \mathbb{T}}$ I was meaning $ \int_{\mathbb{T}}$.
Now, I think we already understand uniformly convergent of fuorier series. (it's same as uniformly convergent of partial integrals.), if we can say about choosing of $ E$ something good (in my first post I was talking about "basis").

And, if not a secret, what is "uniform convergence of partial integrals"? Note that if $ E$ is sufficiently close to $ \mathbb T$, then $ f$ and $ \mathcal F^{\minus{}1}[\widehat f\chi_{{}_E}]$ are close in $ \ell^2$ and, thereby, uniformly, so, unless you mean something exotic, all $ f\in \ell^2$ have "uniformly convergent Fourier series" in this sense...


probability
geometry
3D geometry
floor function
AMC
AIME
logarithms

1. If a fair coin is flipped an infinite number of times, what is the probability three heads will appear before two tails?

2. How many numbers in the 100th row of the Pascal triangle (the one starting with 1, 100...) are not divisible by 3?

3. Among the first billion positive intagers, consider the sets of:

a. palindromic numbers, e.g. 22, 121, 1145411
b. prime numbers
c. perfect cubes

Arrange these in order of decreasing size.

These problems were taken from the 2003 UGA math contest.

2) ${100 \choose n}=\frac{100!}{n!(100-n)!}$.
The largest power of 3 that divides 100! is
$\lfloor \frac{100}{3}\rfloor+\lfloor \frac{100}{9}\rfloor+\lfloor \frac{100}{27}\rfloor+\lfloor \frac{100}{81}\rfloor$
So 3 does not divide ${100 \choose n}$ iff
$\sum_{n=1}^{4} \lfloor \frac{n}{3^n}\rfloor+\sum_{n=1}^{4} \lfloor \frac{100-n}{3^n}\rfloor=\sum_{n=1}^{4} \lfloor \frac{100}{3^n}\rfloor$
So
$\lfloor \frac{100}{81}\rfloor=\lfloor \frac{n}{81}\rfloor+\lfloor \frac{100-n}{81}\rfloor$
and that holds iff $n(\mod 81) \leq 19$, so there are 40 n's.
Also$\lfloor \frac{100}{9}\rfloor=\lfloor \frac{n}{9}\rfloor+\lfloor \frac{100-n}{9}\rfloor$, and that holds iff $n(\mod 9) \leq 1$. That eliminate 28 n's of the 40. So 40-28=12.

#1.  there was an aime question that was very similar to this.  i'm not sure which

[hide]

let's have H=the strings of heads and tails that fits our discription which begins with a H.

let's have T=the strings of heads and tails that fits our requirements that begins with a T

we know that (probability of getting a H)=2(probability of getting a T) because when a string begins with tails, the next flip must be heads.

we also know that the number of flip sequences that fits our description must fit either start with tails followed by a H, {start with 1-2 heads followed by a T, or the a sequence of 3 heads right at the beginning}.

notice that the possibilities enclosed b the {} are all parts of H.  we also realize that these 2 statements include all the possibiities which are a part of H.

we let Pt=probability of getting a T string, and Ph=probability of getting an H string

so this gives us another equation: (Ph)=(Pt/2+Pt/4)+1/8

solving the two equations gives us Pt=1/10, Ph=1/5.  so add them together, and the answer is 3/10[/hide]

[hide]

Q3: this question isn't a very good question but anyway:

Consider the set of palindromic numbers less than 10^9.

We can choose the first 5 digits, each corresponds to a unique such number. So there are 10^5-1 palindromic numbers.

Consider the set of cubes. We have 1^3, 2^3 ... (10^3)^3.

This gives us 10^3 cubes.

Consider the set of primes. There are approximately 10^9 / In (10^9) primes

which is approximately 48,254,942 = 1/2 * 10^8.

So the sets in decreasing order are: the set of primes, the set of palindromes then the set of the cubes.

[/hide]

i have no idea how this got double posted.

[quote="vinoth_90_2004"]
Q3: this question isn't a very good question but anyway:
[/quote]

Out of curiosity, why do you believe this to be a poor question?

I agree with Vinoth's conclusion.

However, I see his method of counting palindromes only counting those with an odd number of digits.  Eg, selecting the five digits 00014 would generate the palindrome 141; but I don't see how would 1441 be generated by this method.

Extending Vinoth's method, I get [hide](10^5 - 1) + (10^4 - 1) = 109998[/hide]

[quote="boris"]The solution provided with the test runs as follows:

Let us estimate how many of these types of numbers we have which are less than or equal to $n$ for some large $n$. For palindromic numbers, it is about $\sqrt{n}$ since only the last half of the digits matter. For perfect cubes, the answer is certainly $\sqrt[3]{n}$. For prime numbers, the answer is known to be $n/\ln n$, from number theory. Since $\ln n \ll n^c$ for any $c>0$ (for large $n$), we have
\[
\frac{n}{\ln n} \gg \sqrt{n} \gg \sqrt[3]{n}
\][/quote]

Of course, using the PNT on a high school math competition is a little weird, but whatever...

BTW, were any of those questions yours?

The provided solution to the Pascal's Triangle question runs as follows:

We need to find the number of coefficients in the polynomial
\[ (1+x)^{100}= 1 + {100\choose 1}x + {100\choose 2}x^2 + \dots + x^{100} \]
which are not equal to 0 modulo 3. Note that modulo 3 one has
\[(1+x)^3= 1+ 3x+3x^2+ x^3 \equiv 1+x^3\]
(this is called Freshman's Dream sometimes, or the high school student's binomial theorem), and so also
\[(1+x)^9 \equiv (1+x^3)^3 \equiv 1+x^9,\]
etc, for any power of 3. Now, $100= 81+2\cdot 9 + 1$. Therefore, modulo 3 one has
\[ (1+x)^{100}= (1+x)^{81} \left((1+x)^9\right)^2 (1+x) =
(1+x^{81}) (1+ 2x^{9} + x^{18}) (1+x) \]
In this product all $2\cdot 3 \cdot 2=12 $ powers of $x$ are different (because every integer can be written in base 3 in a unique way), and the coefficients are all nonzero modulo 3. So, the answer is 12.


Prove that

$1+10+10^{2}+10^{3}+\cdots+10^{n}$

can never be a perfect square.

[hide]If you put the expression in (mod 4) you get
$1+2+0+0+0+...0 \equiv 3 (mod 4)$

Which cannot be a perfect square.

(When n=0, the expression is a perfect square, but that's not really relevant) [/hide]

[hide]Every term $10^{k}$, for k>1, is $\equiv 0 \pmod{4}$.
Thus, expression is $\equiv 1+10\equiv 3 \pmod{4}$, but quadratic residues mod 4 are 0 and 1.[/hide]


modular arithmetic
induction
number theory unsolved
number theory

Let $ p$ be an odd prime. Prove that:
(1) $ (1 \plus{} p)^{p^{n \minus{} 1}} \equiv 1 \pmod{p^n}$
(2) $ (1 \plus{} p)^{p^{n \minus{} 2}} \neq 1 \pmod {p^n}$

to prove (1), it's sufficient to show that $ p^{n \minus{} i}$ divides $ \binom{p^{n \minus{} 1}}{i}$ for 0<i<n, but I don't see the light!

OK, induction did help (I tried it before but gave up to easily), thanks. I proved (1) using Legendre's formula and some identities of the floor function. But it's ugly... did anybody find a nicer solution?

Now I'm working on (2). (1) is true even if p=2, so we probably need to use the fact that p is odd now. I am kind of stuck...

$ (Q_n)\equal{}[\exists k_n\in\mathbb{N}: \ (1\plus{}p)^{p^{n\minus{}1}}\equal{}1\plus{}k_np^n;\ p\not|k]$

$ (Q_2)$:[ $ (1\plus{}p)^{p^{2\minus{}1}}\equal{}1\plus{}p^2(1\plus{}\sum_{j\equal{}2}^{p}C_{p}^{j}p^{j\minus{}2})$]
for $ n\ge 2$
$ [(Q_{n})\to (Q_{n\plus{}1})]$ :
$ \exists k_n\in\mathbb{N}: \ (1\plus{}p)^{p^{n\minus{}1}}\equal{}1\plus{}k_np^n;\ p\not|k$
$ \Rightarrow (1\plus{}p)^{p^{n}}\equal{}(1\plus{}k_np^{n})^p\equal{}1\plus{}p^{n\plus{}1}(k_n\plus{}p\sum_{j\equal{}2}^{p}C_{p}^{j}k_n^jp^{n(j\minus{}1)\minus{}2})$
so $ k_{n\plus{}1}\equal{}k_n\plus{}p\sum_{j\equal{}2}^{p}C_{p}^{j}k_n^jp^{n(j\minus{}1)\minus{}2}$
gives $ p\not|k_{n\plus{}1} and (1\plus{}p)^{p^{n}}\equal{}1\plus{}k_{n\plus{}1}p^{n\plus{}1}$
gives $ (Q_{n\plus{}1})$

so $ \forall n\ge 2: \ \exists k_n\in\mathbb{N}: \ (1\plus{}p)^{p^{n\minus{}1}}\equal{}1\plus{}k_np^n;\ p\not|k$
and we have $ (1\plus{}p)^{p^{n\minus{}1}}1\plus{}k_np^{n}\equiv 1[p^n]$
and $ (1\plus{}p)^{p^{n\minus{}2}}\equal{}1\plus{}k_{n\minus{}1}p^{n\minus{}1}\not\equiv 1[p^n]$


modular arithmetic
pigeonhole principle
number theory
prime numbers

1. Show that for any five primes greater than $ 2$ there exists two numbers, $ p$ and $ q$, such that $ 8$ divides $ p\minus{}q$.

2. Let $ a, b, c$ be nonzero integers such that $ a$ doesnt equal $ c$ and

$ \frac{a}{c}  \equal{}  \frac{a^2\plus{}b^2}{c^2\plus{}b^2}$

Show that $ a^2\plus{}b^2\plus{}c^2$ is not a prime number.

3. Let $ p$ be a prime number. Find all numbers $ k$ such that $ \sqrt{k^2\minus{}pk}$ is an integer.

4. Show that if $ p$ and $ q$ are two consecutive odd prime numbers, then $ p\plus{}q$ is the product of at least three prime numbers (not necessarily distinct).

5. Show that if $ p$ an $ 8p\minus{}1$ are prime numbers, then $ 8p\plus{}1$ is composite.

1- Taking congruences $ (mod 8)$  we have that the possible values por the odd primes are $ 1$, $ 3$, $ 5$, $ 7$ $ (mod 8)$ Since we choose five numbers there will be two who have the same rest $ mod 8$ so $ p\minus{}q\equal{}0 ( mod 8)$
4-If you mean that consecutive is $ p$ and $ p\plus{}2$ then $ p\plus{}q\equal{}4k$ and since $ p\plus{}q>4$ then it has a third prime factor
5-If $ p\equal{}3$ $ 8p\minus{}1\equal{}23$  $ 8p\plus{}1\equal{}25$ not prime...
If $ p\equal{}1(mod 3)$  $ 8p\plus{}1\equal{}0 (mod 3)$ and because $ 8p\plus{}1>3$ it can't be prime
If $ p\equal{}2 ( mod 3)$ $ 8p\minus{}1\equal{} 0( mod 3)$ and since $ 8p\plus{}1>3$ it is not prime so there's nothing to prove

When I have more time I will post 2 and 3

Daniel

Well, for number 4), consecutive primes such as , $ 7\plus{}11, 13\plus{}17, 31\plus{}37$ work also so I don't think it is necessarily $ p \plus{} (p\plus{}2)$.

I think its more $ p \plus{} (p\plus{}2k)$
so $ 2(p\plus{}k)$... Guess it remains to be proven for 2 cases, if $ k\equiv 0 \pmod{2}$ and $ k \equiv 1 \pmod{2}$

[hide="1"] All primes must be either 1,3,5 or 7 (mod 8). By Pidgeonhole, at least two of the primes are the same when taken (mod 8). Thus, their difference is 0 (mod 8), and we are done.[/hide]
[hide="4"] WLOG, let p<q. Then q=p+2, and p+q=2p+2=2(p+1). Since p is odd, p+1 is even and must have at least 2 prime factors. Thus, we have the three prime numbers needed.[/hide]

[hide="2"] $ \frac{a}{c} \equal{} \frac{a^2\plus{}b^2}{c^2\plus{}b^2} \\
\Rightarrow \frac{a\minus{}c}{a\plus{}c} \equal{} \frac{a^2\minus{}c^2}{a^2\plus{}c^2\plus{}2b^2} \\
\Rightarrow a^2\plus{}c^2\plus{}2b^2\equal{}(a\plus{}c)^2 \\
\Rightarrow a^2\plus{}c^2\plus{}b^2 \equal{}(a\plus{}c\plus{}b)(a\plus{}c\minus{}b) $

hence not prime [/hide]

[hide="4"] Clearly $ p \plus{} q$ is even, so it has $ 2$ as a factor. Then $ \frac {p \plus{} q}{2}$ is an integer. Since it is between $ p$ and $ q$, then it must be composite, since $ p$ and $ q$ are consecutive primes. Hence it can be broken into at least two prime factors, which along with our original factor of $ 2$, make at least three prime factors. [/hide]

[hide="5"] If $ p \equal{} 0 (mod 3)$ then $ p \equal{} 3$ and $ 8p\plus{}1\equal{}25$, which is composite.

If $ p \equal{} 1 (mod 3)$ then $ 8p\plus{}1\equal{}0 (mod 3)$, so since $ 8p\plus{}1>3$, $ 8p\plus{}1$ will be composite.

If $ p \equal{} 2 (mod 3)$ then $ 8p\minus{}1\equal{}0 (mod 3)$, so since $ 8p\minus{}1>3$, $ 8p\minus{}1$ will not be prime, so it doesn't matter what $ 8p\plus{}1$ is in this case. [/hide]

[quote="nayrb"]The sum of all pairs of twin primes will have a factor of twelve. Every prime can be represented as 6k-1 or 6k+1. A twin prime pair must be represented as p=6k-1 and q=6k+1, otherwise the difference between them wouldn't be 2. Thus, the sum is p+q=12k.

Do you think the fact that p=6k-1 or 6k+1, or, p=4k-1 or 4k+1, can be used to prove the case that the sum of any two consecutive prime has three prime factors?[/quote]

I think after that you must mechanically show the sum of the twin primes $ 3 and 5$ have 3 prime factors, namely $ 2^3$. :)

[hide="3"]
$ \sqrt {k^2 \minus{} pk} \equal{} n \implies k(k \minus{} p) \equal{} n^2$.  Hence we need both $ k$ and $ k \minus{} p$ to be perfect squares.  Let $ k \equal{} r^2$ and $ k \minus{} p \equal{} s^2$.  Then we have $ r^2 \minus{} p \equal{} s^2 \implies r^2 \minus{} s^2 \equal{} p \implies (r \minus{} s)(r \plus{} s) \equal{} p$ so equating factors:

Case 1: $ r \minus{} s \equal{} 1$ and $ r \plus{} s \equal{} p$
so $ r \equal{} \frac {p \plus{} 1}{2}$ and $ s \equal{} \frac {p \minus{} 1}{2}$.  $ k \equal{} r^2 \implies k \equal{} \frac {(p \plus{} 1)^2}{4}$

Since $ n \equal{} rs$, $ n \equal{} \frac {(p \plus{} 1)(p \minus{} 1)}{4}$, which is an integer as long as $ p \neq 2$

Case 2: $ r \minus{} s \equal{} p$ and $ r \plus{} s \equal{} 1$
Then $ r \equal{} \frac {p \plus{} 1}{2}$ and $ s \equal{} \frac {1 \minus{} p}{2}$.  $ k$ is the same as in case 1.  $ rs < 0$ for all $ p$ but $ n \ge 0$, so this case gives no solutions.

Case 3: $ r \minus{} s \equal{} \minus{} 1$ and $ r \plus{} s \equal{} \minus{} p$
$ r \equal{} \minus{} \frac {p \plus{} 1}{2}$ and $ s \equal{} \minus{} \frac {p \minus{} 1}{2}$ which gives the same solutions as case 1.

Case 4: $ r \minus{} s \equal{} \minus{} p$ and $ r \plus{} s \equal{} \minus{} 1$
Then $ r \equal{} \minus{} \frac {p \plus{} 1}{2}$ and $ s \equal{} \frac {p \minus{} 1}{2}$ so $ rs < 0$ for all $ p$ but $ n \ge 0$, so this case gives no solutions.

So our only solution so far is $ \boxed{k \equal{} \frac {(p \plus{} 1)^2}{4}}$ as long as $ p \neq 2$.  But we only considered the case where both factors are perfect squares, which is only necessary when neither are $ 0$.  Thus we have to add in the solutions $ \boxed{k \equal{} p}$ and $ \boxed{k \equal{} 0}$.

[/hide]

[hide="oops.. some corrections for 3"]
As I noted, we have $ k(k\minus{}p)\equal{}n^2$.  But this does not necessarily imply that $ k$ and $ k\minus{}p$ are perfect squares.  Instead, note that every whole number can be written as the product of a perfect square and a square-free number.  Thus, $ k\equal{}r^2t$ and $ k\minus{}p\equal{}s^2t$ - the squarefrees have to be the same to give a perfect square product.  Hence we have $ r^2t \minus{} s^2t \equal{} p$, so $ t|p$ and thus $ t\equal{}\pm 1$ or $ t\equal{}\pm p$.

In the case $ t\equal{}1$, we proceed as in my last post.

In the case $ t\equal{}\minus{}1$, we have $ s^2 \minus{} r^2 \equal{} p$ so equating factors we have
(1) $ s\minus{}r\equal{}1$, $ s\plus{}r\equal{}p$ so $ r\equal{}\frac{p\minus{}1}{2}$ and $ k\equal{}\minus{}\frac{(p\minus{}1)^2}{4}$
(2) $ s\minus{}r\equal{}\minus{}1$ $ s\plus{}r\equal{}\minus{}p$ so $ r\equal{}\minus{}\frac{p\minus{}1}{2}$ and thus $ k\equal{}\minus{}\frac{(p\minus{}1)^2}{4}$
(3) $ s\minus{}r\equal{}p$, $ s\plus{}r\equal{}1$ so $ r\equal{}\minus{}\frac{p\minus{}1}{2}$ and thus $ k\equal{}\minus{}\frac{(p\minus{}1)^2}{4}$
(4) $ s\minus{}r\equal{}\minus{}p$, $ s\plus{}r\equal{}\minus{}1$ so $ r\equal{}\frac{p\minus{}1}{2}$ and thus $ k\equal{}\minus{}\frac{(p\minus{}1)^2}{4}$

In the case $ t\equal{}p$, we have $ r^2 \minus{} s^2 \equal{} 1$ and equating factors (1) $ r\plus{}s\equal{}1$ and $ r\minus{}s\equal{}1$ or (2) $ r\plus{}s\equal{}\minus{}1$ and $ r\minus{}s\equal{}\minus{}1$.  In case (1), $ r\equal{}1$ and $ s\equal{}0$, giving $ k\equal{}p$ (which I noted was a solution earlier).  In case (2), $ r\equal{}\minus{}1$ and $ s\equal{}0$, which also gives $ k\equal{}p$

In the case $ t\equal{}\minus{}p$, we have $ s^2\minus{}r^2 \equal{} 1$ and equating factors
(1) $ s\minus{}r\equal{}1$ and $ s\plus{}r\equal{}1$ in which case $ r\equal{}0$ and thus $ k\equal{}0$
(2) $ s\minus{}r\equal{}\minus{}1$ and $ s\plus{}r\equal{}\minus{}1$ in which case $ r\equal{}0$ and thus $ k\equal{}0$

In summary: $ \boxed{k\equal{}\minus{}\frac{(p\minus{}1)^2}{4}, 0, p, \frac{(p\plus{}1)^2}{4}}$
[/hide]


$ \frac{\minus{}x\minus{}9}{(x\plus{}1)(x\minus{}3)}\geq 0$

I tried to graph these, the solution to where both in the denominator is greater than $ 0$ is $ (\minus{}\infty, \minus{}1) \cup (3, \plus{}\infty)$. Both in the denominator to be less than zero is $ (\minus{}1,3)$ and the one in the numerator to be always positive $ (\minus{}\infty, \minus{}9]$. I don't know how to tie all these in together for the final answer.

[hide="sign chart ftw"]
You forgot about if the top is negative.

<~~~-9~~~~~~~~~-1~~~~3~~~>
-1 and 3 are asymptotes, -9 is a zero.
Try a random value from each segment.
-10 returns positive.
-2 returns negative.
0 returns positive.
4 returns negative.

So the segments that work are $ (\minus{}\infty,9] \cup (\minus{}1,3)$
The second part has round brackets because -1 and 3 are asymptotes.

[/hide]

I was so lost with 7 or so of my graphs for this problem.

your method is really simple and clear. Would it work for any types of inequality equations?

Yep, just graph all the x-intercepts and asymptotes, take any value from each section, then put it together - asymptotes with round brackets and x-intercepts with square brackets if the equation is $ \ge$ or $ \le$

What if the equation is something like:

$ \frac { \minus{} x \minus{} 9}{|x \plus{} 1|(x \minus{} 3)}\geq 0$

The sign chart will change so that the sections to the right of -1 and the left of -1 are merged, but -1 remains out of the domain.   ... I think.


Has anyone here ever seriously tried using it to compose and create high-quality music? While the software itself is a little hard to use, I find that it has a great deal of potential, as it not only removes the human element from song production but also allows extremely wide vocal ranges to be achieved without overzealous autotuning. Many very fine adjustments can be made within the UI itself if needed, and many excellent creations such as the below can be found on youtube amongst other video sharing sites. (I'm not going to bother with the 3DPVs here since a lot of them have audio that were further enhanced.) The English vocaloids are a little bit "noisier", but still quite excellent nevertheless.

[youtube]qeWlez-s-2E[/youtube]

[youtube]HGMd1w-T6rg[/youtube]

Heh; I was just listening to vocaloids. http://www.youtube.com/watch?v=teA-BD5XSQk
I would use vocaloids, but I don't have too much time to look for a download, and I'm paranoid of the virus factor.
EDIT: Another offsetting topic is that generally, only the Japanese versions of vocaloids are good, meaning that if you wanted to make a good song with a vocaloid, then it'd either have to have an accent, or be in Japanese.

I've never really tried but I do love Vocaloid
Hatsune Miku, Kaito...
:) Gotta love Japan!

i actually am a vocaloid producer! i've never published songs on nico nico or youtube but i do it as more of a personal hobby.

[quote=UnknownUser123]I like vocaloids, tried downloading it to use in my songs but didn’t work[/quote]

did you pay? if it doesnt work maybe you can try utau, the free version of vocaloid. it works pretty much the same, but it's free. there are some pretty good utauloids out there too. (sorry if i'm being unhelpful)


analytic geometry
graphing lines
slope
vector
calculus
integration

[quote="darkmatter47"]If Bob makes a round trip from point A to B in 12 hours and travels at 3 mph while going on a upwards slope, 9 mph on a downwards slope, and  9/2 mph on a horizontal surface, what is the distance between A and B?

What do you think? (Before I hand it in)

[hide="explanation"] Denote the upwards distance when going towards point B as a, the horizontal distance as b, and downwards distance as c, and D as the distance between A and B. Then:

D = a + b + c.

By the d=rt,

a+b=3t [size=59]1[/size]  , since while going back to A, Bob will go up the distance of b.

a+b=9t [size=59]2[/size]

2c=4.5t [size=59]3[/size]
Dividing into both sides on all three equations yields

(a+b)/3 = t [size=59]1[/size]
(a+b)/9 = t [size=59]2[/size]
2c/4.5 = t [size=59]3[/size]

If we add t [size=59]1[/size], t [size=59]2[/size], and t [size=59]3[/size] together, we know it equals 12, since we were given in the problem that it takes Bob 12 hours.
So,

(a+b)/3 + (a+b)/9 + 2c/4.5 = 12
                                                Multiplying by 9...
3a+3b+a+b+4c=108
4(a+b+c)=108
a+b+c=27
D=a+b+c=27 miles. [/hide][/quote]
If you look at the rates, you will notice that no matter what terrain in his way, the distance would always be 27 miles.

I realize this might be beyond the scope of middle school math, but this is quite similar to conservative vector fields and path integrals in calculus and physics (it doesn't matter how you get there, if the two endpoints are the same, the distance to get there is the same).


Show that for any $ r\in \mathbb{Q}$ there exist a triangle for which it's sidelengths are rational numbers and it's area is equal to $ r$.

It is interesting problem, bu i have not solved yet. Who can help me....


If n is an integer, what is the greatest even number of the form 2^n+3^n?

yes, explain your reasoning. should only take a few secs

[quote="Syntax Error"]so whats your answer?[/quote]
2, because 2^n is always even (even*even=even) and 3^n always equal odd (odd*odd=odd) and even*odd=odd, so 0 is the only one in which both components equal "an odd number!" :) (from SBSP) giving us an even answer.

(also negative n would give you something <2)

read the question again. 2^n+3^n, where n is 0, would bring 2

Right, I mean n=0, I mean, I'm not [b]that[/b] stupid :)

...on second thought, you might want to purchase the "I'M WITH THE DUMMY :arrow: " shirt from Patrick :)

Aaaaaaarg! (I bet you won't even understand that one :) )

Patrick Star? from Spongebob Squarepants? Does no one watch that show!? I mean, at least half of us at my school sees it, even some of the teachers :)

i did for a very very short while. but i dont like to talk about that

Why not? I mean, he's like the symbol of goodness, happiness, and friendliness. If everyone was like Spongebob, then there wouldn't need a need for Heaven.

has to be 2 else 3^n is odd and 2^n is even, producing an odd sum


2^6+2^6= 4 to the power of x.

solve for x.

how would you do this?

[quote="bpms"]Are logs beyond HSB?[/quote]

You don't need logs. :D


function
calculus
logarithms
factorial
integration
algebra
functional equation

I was playing around with functions and saw this: let *f*(x)= f(x+1/f(x))-f(x). Then when I plot *x^x* it looks like log(x)+1 for the values my computer can plot (I did not have enough memory for x>9). But when I plot *x!* the graph looks like log(x).  How can this be? Is this related to the stirling formula?

It's straightforward calculus, at least in the first case; your $f^{*}(x)$ is approximately $\frac{f'(x)}{f(x)}=\frac{d}{dx}\ln f(x)$.
This is indeed related to the basic form of the Stirling formula.

For the factorial, you need to interpolate with the $\Gamma$ function; $\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}\,dt$, and $\Gamma(n+1)=n!$ for integers $n$. We have the functional equation $\Gamma(x+1)=x\Gamma(x)$, so $\int_{x}^{x+1}\frac{d}{dt}\ln\Gamma(t)\,dt=\ln x$.

This definition is very smooth, and we have that $\ln\Gamma$ is actually convex. That gives $\ln(x-1)<\Gamma^{*}(x)<\ln x$.

For that convexity: $\frac{d^{2}}{dx^{2}}\Gamma(x)=\frac{d}{dx}\frac{\int_{0}^{\infty}\ln t\cdot t^{x-1}e^{-t}\,dt}{\int_{0}^{\infty}t^{x-1}e^{-t}\,dt}= \frac1{\Gamma(x)^{2}}\left(\int_{0}^{\infty}(\ln t)^{2}t^{x-1}e^{-t}\,dt\cdot\int_{0}^{\infty}t^{x-1}e^{-t}\,dt-\left(\int_{0}^{\infty}\ln t \cdot t^{x-1}e^{-t}\,dt\right)^{2}\right)$

This is positive because it is a variance; if $\gamma$ is a random variable with density function $\frac1{\Gamma(x)}t^{x-1}e^{-t}$, this is the variance of $\ln\gamma$. We could also use Cauchy-Schwarz to prove it.
Incidentally, $\frac{d}{dx}\Gamma(x)$ is the expected value of $\ln\gamma$, and the expected value of $\gamma$ is $x$.


Andre is a butcher and president of a committee, which also includes a grocer, a baker, and a candlestick maker. All are sitting around a square table, one on each side of the table. Andre sits on Charmin’s left, Barton sits at the grocer’s right, and Duclos, who faces Charmin, is not the baker. Who sits across the table from the butcher?

A. the baker B. Duclos C. the grocer D. the candlestick maker E. Charmin

[hide="Solution"]By symmetry, assume WLOG Andre is at the top of the square (if we look down upon it). Charmin is on the left of the square. Duclos is on the right of the square. Barton faces Andre. We know Charmin must therefore be the grocer, so Barton is the baker. Answer: $ \boxed{A}$.

[/hide]


treat benzene with nitrating mixture H2SO4 HNO3 to get nitrobenzene.
React it with CH3COCl in presence of AlCl3 to get a meta substituted product.
SnCl2/HCl to reduce nitro group to amino (not sure if it would reduce carbonyl group)
Diazotization followed by addition of water to convert the amino group to hydroxy group

[quote="Zimrock"]React it with CH3COCl in presence of AlCl3 to get a meta substituted product. SnCl2/HCl to reduce nitro group to amino (not sure if it would reduce carbonyl group)
Diazotization followed by addition of water to convert the amino group to hydroxy group
[/quote]

1) Fridel-Crafts reactions don't work when a meta directing group is on the ring.
2) The carbonyl would also probably be reduced, and what would happen between an amine and a ketone in the reaction mixture?

Outline a synthesis of 3-hydroxyacetophenone from:
1) Benzene.
2) Open chain precursors.

Is this right?
1. convert benzene into toluene by friedel and crafts.
2. convert toluene into benzoic acid by oxidation.
3. sulphonate using fuming H2SO4 in the meta posn.
4. Use Ruzicka rn ( CH3COOH ) to convert COOH into COCH3
5. Use i) aq. KOH ii) H+ to convert into the reqd. compound.

Alternately
Can we just acylate benzene into Acetophenone and then Sulphonate and complete the rn?

Madness..........
In wikipedia it is given that Ruzcika reaction is the conversion of a dicarboxylic acid into a cyclic ketone...
How is it applicable here???
can you please explain????????????????? :(

[quote="Madness"]Alternately Can we just acylate benzene into Acetophenone and then Sulphonate and complete the rn?[/quote]

You could, but what happen with acetophenone in the presence of base?

Would it undergo aldol?  :wink:
So can i just protect the C=O grp using ethylene glycol and then carry out the process?

hey madness......................
explain how Ruzicka reaction is applicable here........

You dont even have a jug in your house then how will you know?  :wink:   :D

[quote="valeriummaximum"]I am asking because in wikipedia it is given that it is conversion of a dicarboxylic acid into a cyclic ketone...[/quote]
hey but Ruzika reaction is given in pradeeps itself no???dunno abt schoo textbook!!! but it is the preparation of aldehyde or ketone from  dicarboxylic acid using MnO2 no???when you use formic acid you get an aldehyde??? :D

[quote="valeriummaximum"]see in wikipedia srinitrs......
the catalyst is ThO2[/quote]
but MnO2 is also correct no??

cut out the nos, will you. as far as i no( :P ),tho2 is used as the catalyst


geometry
summer program
Mathcamp
PROMYS
AMC
AMC 10
number theory

I heard that last year AMSP let you do 4 courses of each discipline (number theory, geometry, algebra, counting). However this year they are only letting you pick 2 courses, to go more indepth. But are you allowed to choose 4 classes if you want to, or is that not possible?

ummmmm...
last year we could only take two courses as well... we could choose 2 out of the four categories though.

By the way, is there some way I can select these courses beforehand?

well,
you select your courses before camp. packet should come..
you get to camp. you take the classes and you can change if you want.

btw does anyone know the acceptance rating of AMSP? And also for other camps like Mathcamp PROMYS and such?

I am going to attend AMSP but im just wondering how the acceptance rating is related to other camps. (also with the number of people applying)

First of all, about the courses, I e-mailed Dr. Andreescu few days earlier...

He said they are still finalizing academic plans for the 2009 AMSP, but only 2 courses is right.

Also, about the acceptance rate.

All camps are pretty loose on their acceptance rates.. I mean it is a CAMP, not like a private school system or something. Most, if not all, people apply, get in. HOWEVER, you might have a tough time in camp, but you're there to learn, and it's still good for you, unless you don't even know what the prof is talking about.. Like you got a 90 on a AMC10 and taking the advanced courses.. That would be silly.

[quote="sjyoo93"]
All camps are pretty loose on their acceptance rates.. [/quote]

not necessary. i mean like camps like RSI are extremely hard to get into.
and mathcamp's acceptance rate is about ~30-50% according to their website...

If you think about it 50% is already very high. You don't need to do that well on the Mathcamp test to get in as well, so I heard. It's just to make sure there is SOME kind of minimum level at the camp.
The only truly selective camps out there are research camps like RSI, which there are for other subjects I believe, and the summer training camps for various olympiads.

Hmmm...so if the classes are what it is on the website, I think I will do olympiad geo and nt?


analytic geometry
geometry
circumcircle
geometry unsolved

[color=darkblue]Let $ (I; r)$ is incribed circle of $ ABC$ triangle. Let $ BC \equal{} a, CA \equal{} b, AB \equal{} c$. Prove that: $ \boxed{\frac{IA^2}{bc}\plus{}\frac{IB^2}{ca}\plus{}\frac{IC^2}{ab}\equal{}1}$[/color].

[color=darkblue][b]The easy proof.[/b] Prove easily that $ \boxed {\ IA^2 \equal{} \frac {bc(p \minus{} a)}{p}\ }$ a.s.o.

[b]The high proof.[/b] The relation $ a\cdot IA^2 \plus{} b\cdot IB^2 \plus{} c\cdot IC^2 \equal{} abc$ is a particular case of the [b]Lagrange's identity[/b].[/color]

[color=darkred]If $ M$ has the barycentrical coordinates $ (x,y,z)$ w.r.t. the triangle $ ABC$ , $ x \plus{} y \plus{} z \equal{} 1$ , then for any $ X$ from the same plane, there is

identity $ \boxed {\ x\cdot XA^2 \plus{} y\cdot XB^2 \plus{} z\cdot XC^2 \equal{} XM^2 \minus{} p_w(M)\ }$ , where $ p_w(M)$ is the power of  $ M$ w.r.t. the circumcircle of $ \triangle ABC$ .

Particularly, for $ X: \equal{} M(x,y,z)$ obtain $ \boxed {\ x\cdot MA^2 \plus{} y\cdot MB^2 \plus{} z\cdot MC^2 \equal{} \minus{} p_w(M)\ }$ .[/color]

[color=darkblue]If $ M: \equal{} I\left(\ \frac {a}{2p}\ ,\ \frac {b}{2p}\ ,\ \frac {c}{2p}\ \right)$ , then  $ p_w(I) \equal{} \minus{} 2Rr$ $ \implies$ $ \sum a\cdot IA^2 \equal{} \minus{} 2p\cdot ( \minus{} 2Rr) \equal{} 4Rpr \equal{} abc$ $ \implies$ $ \sum\frac {IA^2}{bc} \equal{} 1$ .

Now you can find many another "old" identities for particular cases of the points $ M$ and $ X \ !$[/color]

[color=darkblue]Thank Virgil Nicula very muck.[/color]


is the gravitational potential energy of two particles:

A) $FG= m1,m2/r^2$ B) $w = %Error. "intergral" is a bad command.
r f(r) * dr$ C) $U = - GMm/r$ D) $T^2 = (4\pi^2)/(GM)$

[color=blue] I think the answer is [b]C[/b], $U=-\frac{GMm}{r}$ and $U=0$ at the $\text{infinity}$[/color]


Let $p_{n}$ denote the $n$th prime number. For all $n \ge 6$, prove that \[\pi \left( \sqrt{p_{1}p_{2}\cdots p_{n}}\right) > 2n.\]

What is $ \pi$ here? It's not $ 3.14...$ I guess, since then the problem is trivial (even for $ n<6$) so what is it then?

[quote="Peter"]What is $ \pi$ here? It's not $ 3.14...$ I guess, since then the problem is trivial (even for $ n < 6$) so what is it then?[/quote]

Yeah, it is not. Here, $ \pi(x)$ denotes the number of primes $ \leq x$.

For $ n \equal{} 7,8,9$ this can easily be checked. So assume $ n\ge 10$. Then we have that $ \sqrt{p_{1}p_{2}\cdots p_{n}}> 3^{n}$.

Now, we know that for all positive integers $ m$, there is at least $ 2$ primes in $ ]m,3m]$ (strictly more for $ m>3)$. (I have to use this fact, I see no solution without?)

So, there are more than $ 2n$ primes in $ ]1,3]\cup]3,9]\cup\cdots\cup]3^{n\minus{}1},3^{n}] \equal{} ]1,3^{n}]$, thus $ \pi\left(\sqrt{p_{1}p_{2}\cdots p_{n}}\right)\ge\pi (3^{n})> 2n$.

We can get by using the fact that $ p_{n\plus{}1}<2p_n$. It follows that $ p_{2n\plus{}2}<4p_{2n}$. For $ n\ge 6$, since $ p_{n\plus{}1}\ge 17$, it follows that $ \sqrt{p_{n\plus{}1}}>4>\frac{p_{2n\plus{}2}}{p_{2n}}$. Thus, if it is true that $ \sqrt{p_1p_2\ldots p_n}>p_{2n}$, it is also true that $ \sqrt{p_1p_2\ldots p_np_{n\plus{}1}}>p_{2n\plus{}2}$. Since this is indeed true for $ n\equal{}6$ ($ \sqrt{30030}>37$), it is true for all $ n\ge 6$ by induction.

Luke
See my puzzle blog at [url]http://bozzball.blogspot.com[/url]


abstract algebra
superior algebra
superior algebra unsolved

let I a finitely generated ideal. Are there any sequence in I like this: i_1, i_2, i_3, ...    witch i_n is not in <i_1, i_2, ..., i_n-1> and i_n in I ?

<i_1, i_2, ..., i_n> =: the generated ideal by i_1, i_2, ..., i_n.[hide]i[/hide]

yes, in the unit ideal of a non-noetherian ring.


logarithms trigonometry conics ellipse geometry probability trapezoid

1. A circle centered at the origin is tangent to the graph $ 25(x\minus{}1)^2 \minus{} 36y^2 \equal{} 900$. Find the minimum radius of this circle.

2. Find the period of the graph of $ y \equal{} 12 \sin(4\pi x/7) \plus{} 13 \cos(2\pi x/5\plus{}\pi/2)$.

3.  Evaluate $ \sum_{k \equal{} 1}^{9999} \log(\sin(\pi k/10000))$ to the nearest integer given that log(2) ≈ 0.30103.

4. An ellipse has major axis of length 122 and minor axis of length 120. A circle centered at a focus of the ellipse is tangent to it. What is the sum of all possible diameters of the circle?

5. 17 points are spaced at equal intervals around a unit circle. From one point, segments are drawn to every other point. Find the sum of the squares of the lengths of these segments.

6. In a two-person game, one person says an integer, and the other person must say an integer greater than or equal to 1/3 of the previous person's integer, but less than the previous person�s integer. They go back and forth following this rule until one person says 1, and that person loses. If Johnny says 2004, what number should Bernie counter him with to guarantee a win for himself?

7. ABC has AB = 75, BC = 78, CA = 51. DEF fully contains ABC and has DE parallel to AB, EF parallel to BC, and FD parallel to CA with a distance of 3 between corresponding sides. What is the area of DEF?

[hide="6."]
Let's start off by trying to find a general pattern for wins and losses. Let's say you start off the game with any number you wish. If you say 1, obviously, you lose. However, if you say 2, your opponent is forced to say 1 and you win. If you say any number from 3-6, you lose because then your opponent says 2 and wins. If you say 7, your opponent is forced to say any number from 3-6 and loses. If you say any number from 8-21, you lose because then your opponent says 7 and wins, etc. Note that the pattern of winning numbers starts with 2, then you multiply by 3 and add 1 to obtain the next winning number and so on. Therefore, the sequence of winning numbers is 2, 7, 22, 67, 202, 607, 1822. 1822 is greater than one-third of 2004 and is less than 2004, and therefore, is Bernie's ideal choice.
[/hide]

Here are 3 more:
8. Let distinct x and y be drawn from $ \{ 1,2,3,\ldots,99\}$ such that each ordered pair (x,y) is equally likely.  If the probability that the sum of the units digits of x and y is less than 10 is $ \frac {a}{b}$,where a and b are relatively prime, what is a+b?

9. What is the smallest number d such that fewer than half of the positive integers with d-digits have all distinct digits?

10. In trapezoid ABCD, $ AB\parallel DC$ and $ AB < DC$, the lines AC and DB intersect at P.  The sum of the areas of $ \triangle APD$ and $ \triangle BPC$ is $ 1996$.  If the lengths of the bases $ \overline{AB}$ and $ \overline{CD}$ are integers and the distance between them is an integer, what is the minimum area of ABCD?

For #8 I got $ 109/198\implies a \plus{} b \equal{} 307$ but that doesn't agree with the official answer...

[hide="9."]
We're just looking for the smallest number $ d$ which satisfies
$ \frac{9P^9_{d\minus{}1}}{9*10^{d\minus{}1}} < 0.5$
$ P^9_3$ is just barely greater than $ \frac{10^3}{2}$, so $ d\minus{}1 \equal{} 4$, $ d \equal{} 5$.
[/hide]

Wow, I was being really careless (and probably stupid) - the denominator in my original solution was $ \binom{100}2$ when it should have been $ \binom{99}2$ and that's why I got $ 1 \minus{} \frac {2000 \plus{} 450/2}{50\cdot99} \equal{} 109/198$...
Here's my reasoning anyway: find all possible pairs mod 10 where the sum is at least 10 - if the two pairs are distinct, we have $ (9,1),...(9,8),(8,2),...(8,7),... ...(6,4),(6,5)\implies 20$ pairs. So the overall total pairs with distinct unit digits is $ 20\cdot10\cdot10$. If the unit digits are the same we have $ (9,9),(8,8),...,(6,6),(5,5)\implies 5$ pairs. So in this case we have $ 5\cdot 10\cdot9/2$ - that's how I got the numerator.

11. The sides of triangle $ ABC$ have length $ AB \equal{} 7$, $ BC \equal{} 8$, and $ AC \equal{} 9$.  What is the distance between the incenter and the orthocenter of $ \triangle ABC$?

12. How many triples of natural numbers, $ (a,b,c)$ are there such that $ a \plus{} b \plus{} c \equal{} 111$ and $ a$, $ b$, and $ c$ are in geometric progression?

13. Solve the system of equations

$ 4y\plus{}gx\equal{}34$
$ \frac{g^{4}x^{2}\minus{}16y^{2}}{g^{2}x\plus{}4y}\equal{}\minus{}10$

in integers $ (g,x,y)$

For #11 I got $ \sqrt {105}/10$ but that does not agree with the official answer... For #12 I think there are 4 but I don't know of a good way to find them all...
BTW @ rpond: I did exactly the same way but I find it difficult to find $ d \equal{} 5$ without a calculator... (or the fact that $ P_3^9$ is just barely $ > 10^3/2$) how did you find it anyway?

$ P^9_3$ is pretty easy to calculate. You can think of it like this:
$ P^9_3 \equal{} 9*8*7 \equal{} 8*(8 \plus{} 1)*(8 \minus{} 1) \equal{} 8*(8^2 \minus{} 1) \equal{} 512 \minus{} 8$ or $ 8*63 \equal{} 504$
And of course $ \frac {10^3}{2} \equal{} \frac {1000}{2} \equal{} 500$.
From there, $ P^9_4 \equal{} 6*P^9_3$ and $ \frac {10^4}{2} \equal{} 10*\frac {10^3}{2}$, so it's safe to assume that the latter will be greater.

EDIT:
[hide="13."]
$ 4y\plus{}gx\equal{}34$
$ \frac{g^{4}x^{2}\minus{}16y^{2}}{g^{2}x\plus{}4y}\equal{}\minus{}10$
$ g^4x^2\minus{}16y^2 \equal{} (g^2x\minus{}4y)(g^2x\plus{}4y) \equal{} \minus{}10(g^2x\plus{}4y)$
$ g^2x\minus{}4y \equal{} \minus{}10$
Adding to the first equation yields
$ g^2x\plus{}gx \equal{} gx(g\plus{}1) \equal{} 24$.
$ g(g\plus{}1)$ has to be a factor of $ 24$. $ 12 \equal{} 3*4 \equal{} \minus{}4*\minus{}3$, $ 6 \equal{} 2*3 \equal{} \minus{}3*\minus{}2$, and $ 2 \equal{} 1*2 \equal{} \minus{}2*\minus{}1$ are all factors of $ 24$, so we have a grand total of $ 6$ $ g$ values:
$ \minus{}4, \minus{}3, \minus{}2, 1, 2, 3$.
Additionally, $ (g, x) \equal{} (\minus{}4, 2), (\minus{}3, 4), (\minus{}2, 12), (1, 12), (2, 4), (3, 2)$.
Now, looking at the first equation in $ mod 4$, $ 0 \plus{} ? \equal{} 2$. Clearly $ gx$ cannot be divisible by $ 4$. Scanning through our pairs $ (g, x)$, we find that only one of them, $ (3, 2)$, will result in a value of $ gx$ that isn't divisible by $ 4$. Plugging these values in and solving for $ y$, we get our answer:
$ (g, x, y) \equal{} (3, 2, 7)$
[/hide]

[hide="#1"]the graph of $ 25(x \minus{} 1)^2 \minus{} 36y^2 \equal{} 900$ is a hyperbola
the center is located at (1,0) and the tips of the hyperbola are 6 units horizontally from the center. Now we know the tips are at (-5,0) and (7,0). The nearest tip is (-5,0) so the radius of the circle must be 5[/hide]
Hope I got it right, I'm new to this.

14. In isosceles triangle $ ABC$ (AB=AC), the angle bisector of $ B$ intersects with $ AC$ at $ D$; we have $ BC \equal{} BD \plus{} AD$ , what is $ \angle A$?

15. From point $ O$ inside $ \triangle ABC$ we draw perpendicular  lines $ OM,ON, OP$ on to $ AB,AC,BC$ respectively. If $ \angle AOB \equal{} 60^{\circ}, MN \equal{} \frac {\sqrt {5}}{3}, OP \equal{} 3, NP \equal{} \sqrt {3}, OM \equal{} \sqrt {5}$ what is $ \angle BOC$?

16. In $ \prod_{1\le i < j\le 9}{x_i \minus{} x_j}$ what is the coeffiecient of $ (x_1^4x_2^4\cdots x_9^4)$?

17. Find all n such that $ d(d(d(n)) \equal{} d(d(n))$, where $ d(n)$ is the number of divisors of n.

18. 7 people numbered 1 through 7 are standing in a line. We know the following:
between the numbers 1 and 5, there is one person between the numbers 1 and 7, there are 3 people
between the numbers 1 and 3 there is one person between the numbers 2 and 4 there is one person
and between the numbers 6 and 4 there is one person. In how many ways could these 7 people be standing in the line?

19. What is the maximum number of terms that can be picked from the following set of numbers such that the mean of the picked numbers would be $ 15$?
$ \{21,14,13,17,15,16,23,12,8,11,12,9,5,4,16,2,14,15,18,8,3,16\}$

20. how many natural numbers in a form of $ \overline{abcd}$ can be found such that: $ \overline{abcd} \equal{} \overline{ad}\cdot\overline{ada}$.

21. in the isoceles triangle $ ABC$, $ AB \equal{} AC$ and $ \angle A \equal{} 100$.  Along the side $ AB$ (from $ B$), we pick the point $ M$ such that $ AM \equal{} BC$. $ \angle BCM$ equals?

22. how many numbers, $ n$, between $ 1$ and $ 100$ are there such that $ n$ is sum of some positive integers, where every digit appears exactly once.(for example: $ 90 \equal{} 0 \plus{} 1 \plus{} 52 \plus{} 3 \plus{} 4 \plus{} 6 \plus{} 7 \plus{} 8 \plus{} 9$)

Here is one more problem (though it seems simple):
23. Trapezoid ABCD has $ \angle D \equal{} 90, \angle B \equal{} 60$, and AB = BC. If DC = 9, find the sum of the two possible lengths for AB.
I must be overlooking something, but it seems to me that there is only one unique answer - 18. I don't see how the official answer got $ 18\plus{}6\sqrt3$...

According to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1713931#1713931[/url], the distance is $ d\equal{}\sqrt{R(R\minus{}2r)}$ with $ R\equal{}(21/10)\sqrt5,r\equal{}\sqrt5\implies d\equal{}\sqrt{105}/10$. What's wrong?

$ d \equal{} \sqrt {R(R \minus{} 2r)}$ is for the distance from the incenter to the circumcenter. But the question asks for the distance from the incenter to the [b]orthocenter[/b].


[quote="greatestmaths"][img]C:\Documents%20and%20Settings\Administrateur.XPPRO\Mes%20documents\Mes%20images[/img][/quote]

What is it, greatestmaths??  :rotfl:


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let $C_1$ be a circle inside the circle $C_2$ and let $P$ in the interior of $C_1$, $Q$ in the exterior of $C_2$. One draws variable lines $l_i$ through $P$, not passing through $Q$. Let $l_i$ intersect $C_1$ in $A_i,B_i$, and let the circumcircle of $QA_iB_i$ intersect $C_2$ in $M_i,N_i$. Show that all lines $M_i,N_i$ are concurrent.

Let $\Gamma_i$ be the circumcircle of $\triangle QA_iB_i$.  Then the line $M_iN_i$ is the radical axis of $C_2$ and $\Gamma_i$.  It suffices to observe that $PQ$ is the radical axis of any two $\Gamma_i$ and $\Gamma_j$.  Hence $M_iN_i$, $M_jN_j$ and $PQ$ are concurrent at the radical center of $C_2$, $\Gamma_i$ and $\Gamma_i$.  Now fix $\Gamma_i$ and let $\Gamma_j$ vary, and vice versa.

To finish this solution: Denote by $R_i$ the point where $(Q,M_i,N_i)$ intersect $PQ$ . We have $PQ\cdot PR_i=PA_i\cdot PB_i$. the last quantity is constant ($=R^2-OP^2$ where $O$ is the center of $C_1$), hence $R_i$ is fixed, hence $PQ$ belongs to all circles $\Gamma_i$, hence it is their radical axis.


Prove that if the integers $a_{1}$, $a_{2}$, $\cdots$, $a_{n}$ are all distinct, then the polynomial \[(x-a_{1})^{2}(x-a_{2})^{2}\cdots (x-a_{n})^{2}+1\] cannot be expressed as the product of two nonconstant polynomials with integer coefficients.

Let's assume that $P$ is reducible in $\mathbb{Z}[X].$ There exist $F, G \in \mathbb{Z}[X], \text{deg}(F) \geq 1$ and $\text{deg}(G) \geq 1,$ such that
\[P = (X-a_{1})^{2}(X-a_{2})^{2}\cdots (X-a_{n})^{2}+1 = F(X)G(X).\]
We may take $F$ and $G$ to be monic. Let $k = \text{deg}(F), l = \text{deg}(G),$ so that $k+l=2n.$
Clearly, for any real $x, P(x) \geq 1.$ Hence $F$ and $G$ take positive values on $\mathbb{R}.$
For any $i, 1 \leq i \leq n, F(a_{i})G(a_{i}) = 1 \ \ (*).$ Therefore, for any $i, F(a_{i}) = G(a_{i}) = 1.$
If $k<l,$ then $k<n.$ But the polynomial $F-1$ is the zero polynomial. Hence $F=1,$ contradiction. Similarly $l<k$ is impossible. Thus $k=l=n.$
Thanks to $(*),$ for any $i, 1 \leq i \leq n, F(a_{i}) = G(a_{i}).$ So $F-G = 0,$ that is to say $F=G,$ so that $P(X) = F(X)^{2},$ which means that
\[[F-(X-a_{1}) \cdots (X-a_{n})][F+(X-a_{1}) \cdots (X-a_{n})] = 1,\]
which is impossible.
Therefore $P$ is irreducible in $\mathbb{Z}[X].$

[b]Remark :[/b] Actually, this polynomial is irreducible in $\mathbb{Q}[X],$ thanks to Gauss' lemma.

[quote="mathmanman"] Hence $F$ and $G$ take positive values on $\mathbb{R}.$
[/quote]

Why is this true?

Isn't it possible that $F(X)$ and $G(X)$ are both negative for some $X\in{R}$?

They are both monic polynomials without real roots, thus they are everywhere positive (otherwise the intermediate value theorem would give some root).


calculus
integration
inequalities
trigonometry
function
real analysis
real analysis unsolved

problem27: prove that for every a>0 integral(0 to 1)[(sin1/x)/(1+x^a)]dx >0.

Here is an idea of how, I think, this problem can be attached.

We have

$I=\int_{1}^{\infty}\sin x\frac{y^{a-2}}{1+y^{a}}dy$.

It is not hard to see that the function $f(x)=\frac{x^{a-2}}{1+x^{a}}$ is a decreasing function for all $x\geq b$ where $b$ is some positive quantity, however if $0<a\leq 4$ we get that $f$ is decreasing on the interval $(1,\infty)$, hence I will treat only this case.

Then

$I=\int_{1}^{\pi}\sin x\frac{x^{a-2}}{1+x^{a}}dx+\sum_{k=1}^{\infty}\int_{2k\pi}^{(2k+1)\pi }\sin x f(x)dx+\sum_{k=0}^{\infty}\int_{(2k+1)\pi}^{(2k+2)\pi}\sin x f(x)dx\geq \int_{1}^{\pi}f(x)\sin x dx+2\sum_{k=1}^{\infty}f((2k+1)\pi)-2\sum_{k=0}^{\infty}f((2k+1)\pi)$

$=\int_{1}^{\pi}f(x)\sin x dx-2f(\pi)>0$.

$\int_{1}^{\pi}\sin x\frac{x^{a-2}}{x^{a}+1}dx>2\frac{\pi^{a-2}}{\pi^{a}+1}$ for all $0<a\leq 4$.


function
integration
analytic geometry
graphing lines
slope
inequalities
real analysis

$ 0<f'(t)<1$ and $ f'(t)$ is continuous function for $ t\in[0,1]$
And $ f(0)\equal{}0$

Prove that

$ \left(\int^1_0f(t).dt\right)^2\ge\int^1_0\left(f(t)\right)^3.dt$

[quote="pqrs"]$ 0 < f'(t) < 1$ and $ f'(t)$ is continuous function for $ t\in[0,1]$
And $ f(0) \equal{} 0$

Prove that

$ \left(\int^1_0f(t).dt\right)^2\ge\int^1_0\left(f(t)\right)^3.dt$[/quote]
I tried to crack this. One obvious conclusion I've drawn, and everyone who's read this has, is that the function increases on $ [0, 1]$, and that its slope is [b]not[/b] $ >1$ or $ <0$ throughout this interval. So, as a last resort, I tried the hit and try method and found out the functions satisfying the condition.
One good one, for example is $ \frac {x^2}{2}$ but that hardly solves the problem concretely.

Everything is $ \ge 0;$ I won't state that at every step.

Start with $ f'(x)\le 1.$

$ f(x)f'(x)\le f(x)$

$ \int_0^xf(t)f'(t)\,dt\le \int_0^xf(t)\,dt$

$ \frac12f(x)^2\le\int_0^xf(t)\,dt$

$ f(x)^3\le 2f(x)\int_0^xf(t)\,dt.$

Now note that $ \frac{d}{dx}\left(\int_0^xf(t)\,dt\right)^2\equal{}2f(x)\int_0^xf(t)\,dt.$

So $ f(x)^3\le \frac{d}{dx}\left(\int_0^xf(t)\,dt\right)^2$

With both sides of this being zero at zero, we have

$ \int_0^xf(t)^3\,dt \le \left(\int_0^xf(t)\,dt\right)^2$

Now let $ x\equal{}1.$

In fact, we've proved a stronger inequality than asked for, since this is valid for all $ x>0,$ not just $ x\equal{}1.$

=======================

Let's take an instructive example: if we let $ f(t)\equal{}at$ and compute with that, we get

$ \frac{a^3x^4}4\le\frac{a^2x^4}4$ for all $ x>0.$ Since $ a\le1,$ this is always true. But we see that if $ a\equal{}1$ - that is, if $ f(t)\equal{}t$ - then we have a sharp extremal function for this inequality.

[quote="ith_power"]By Lagranges mean value theorem we get $ f(t)\le 1$ and $ f'(x)\ge 0$ for $ 0\le x\le 1$.
so, $ f(t)^2 \ge f(t)^3$. so,
$ \left(\int^1_0f(t).dt\right)^2\ge\int^1_0\left(f(t)\right)^2.dt\ge\int^1_0\left(f(t)\right)^3.dt$[/quote]

ith_power  is wrong  :D . The first inequality is  manifestly false because of Cauchy-Schwartz inequality : $ \left(\int^1_0f(t).dt\right)^2\leq\int^1_0\left(f(t)\right)^2.dt$
...
Nice solution from Kent Merryfield . :first:  .
In the same way , if $ \ f'(x) \leq g(x),\  0 \leq g(x),\  0\equal{}f(0) \leq f(x),\  0 < b, \ 0 \leq x$,
you can prove  $ \ \frac 2{b\plus{}1}\int_0^x(f_{(t)})^{2b\plus{}1}g_{(t)}dt \le \left(\int_0^x(f_{(t)})^bg_{(t)}dt\right)^2$ .
 :cool:

sorry  :P didn't check properly :blush:
but still explain the following:

$ (lim_{m\minus{}>inf}\sum_{r\equal{}0}^n \frac{r}{n})^{2} \ge lim_{m\minus{}>inf}\sum_{r\equal{}0}^n (\frac{r}{n})^{2}$ because the 2.(i/n)(j/n) terms are extra on l.h.s.
isn't that true? :maybe:


Compute: $ 27 \times 3 \plus{} 27 \times (\minus{}4)$.

$ 27 \times 3 \plus{} 27 \times (\minus{}4) \equal{} 27 \times (3\plus{}(\minus{}4)) \equal{} 27 \times (\minus{}1) \equal{} \boxed{\minus{}27}$.


Please guy I want to know some important identity and factoring. Some useful .

:huh:
Just in general?

Well, here's a quick (redundant) list. Far from comprehensive though.

$ a^2 - b^2 = (a+b)(a-b)$
$ a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \ldots + b^{n-1})$
$ a^{2n+1} + b^{2n+1} = (a+b)(a^{2n} - a^{2n-1}b + a^{2n-2}b^2 - \ldots + b^{2n})$
$ ab + xa + yb = (a+y)(b+x) - xy$ (Simon's Favorite Factoring Trick)
\begin{eqnarray*}a^3 + b^3 + c^3 - 3abc &=& (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \\ &=& (a+b+c)\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} \\ \end{eqnarray*}
$ (a^n + a^{n-1} + \ldots + 1)^2 = a^{2n} + 2a^{2n-1} + 3a^{2n-2} + \ldots  + na^n + (n-1)a^{n-1} + \ldots + 1$

See [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2007-ua/2007USAMOsoln.pdf]problem 5[/url] in here for a totally wacky one that will probably never appear on a contest again.
http://mathworld.wolfram.com/CyclotomicPolynomial.html for information on factorizations of $ x^n - 1$ for varying $ n$, as well as some theory. More useful in later algebra though.

The best way to develop a factoring sense is to solve a lot of related problems. Eventually you get to the point where you can draw up a solution that makes people accuse you of pulling stuff out of your ***.

Hope that's what you were looking for. This is definitely something you probably could've [url=http://www.artofproblemsolving.com/Forum/search.php]search[/url]ed the forum for though.

A recent post by darij was extremely good on this topic (although I couldn't find it using the search)

[quote="darij grinberg"][b]Some nice formulae for three numbers.[/b]

If $ a$, $ b$, $ c$ are three elements of a field $ K$, then
$ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} \left(a \plus{} b \plus{} c\right)\left(a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} ca \minus{} ab\right)$;
$ \left(b \plus{} c\right)\left(c \plus{} a\right)\left(a \plus{} b\right) \plus{} abc \equal{} \left(a \plus{} b \plus{} c\right)\left(bc \plus{} ca \plus{} ab\right)$;
$ \left(a \plus{} b \plus{} c\right)^5 \minus{} \left(a^5 \plus{} b^5 \plus{} c^5\right) \equal{} 5\left(b \plus{} c\right)\left(c \plus{} a\right)\left(a \plus{} b\right)\left(a^2 \plus{} b^2 \plus{} c^2 \plus{} bc \plus{} ca \plus{} ab\right)$;
$ \left(b \minus{} c\right)^2 \plus{} \left(c \minus{} a\right)^2 \plus{} \left(a \minus{} b\right)^2 \equal{} 2\left(a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} ca \minus{} ab\right)$;
$ \left(b \minus{} c\right)^4 \plus{} \left(c \minus{} a\right)^4 \plus{} \left(a \minus{} b\right)^4 \equal{} 2\left(a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} ca \minus{} ab\right)^2$. [/quote]

[url]http://www.mathlinks.ro/viewtopic.php?t=192701[/url]

I'm not sure what you mean.

In any case, there's always the lovely Sophie Germaine identity $ x^4 \plus{} 4y^4 \equal{} (x^2 \plus{} 2xy \plus{} y^2)(x^2 \minus{} 2xy \plus{} y^2)$, which is in turn a special case of the "incomplete" factorization $ a^2 \plus{} b^2 \equal{} (a \plus{} b)^2 \minus{} 2ab$.  It becomes "complete" if $ a, b$ are chosen so that $ 2ab$ is a square.


inequalities
search
inequalities proposed

Let $ a,b,c>0$. Prove that:
$ \sqrt{\frac{2a}{a\plus{}b}}\plus{}\sqrt{\frac{2b}{b\plus{}c}}\plus{}\sqrt{\frac{2c}{c\plus{}a}}\leq 3$.
[hide="own solution(sorry if not new)"]After bringing to a common denominator, and squaring both sides the ineq. becomes
$ (\sum{\sqrt{a(b\plus{}c)(c\plus{}a)}})^2\leq \frac{9}{2}(a\plus{}b)(b\plus{}c)(c\plus{}a)$
From Cauchy-Schwartz we infer that:
$ (\sum{\sqrt{a(b\plus{}c)(c\plus{}a)}})^2\leq (\sum{a(b\plus{}c)})(\sum{(c\plus{}a)})\equal{}4(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)$
We are left to show that $ 8(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\leq 9(a\plus{}b)(b\plus{}c)(c\plus{}a)$, which is trivial with AM-GM.[/hide]

I saw that the oficial solution is too long, so I decided to post a simple 2-line solution.
If this solution was posted, then I'm sorry for creating an useless topic.

This has been posted many many times in the past. For example take a look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=158729]here[/url]. In that link there are two more links which contain more solutions. :wink:

Sorry, nayel, for a double post.
When I solved this inequality (sincerly, I didn't copy your solution),
I saw in the book that this problem it has a much harder and long solution,
so I decided to post the easy solution here, so people could see.
Although I was convinsed that someone found this solution earlier than me.
Congratulations, nayel, for being the first one! :thumbup:

No, it's okay. We often don't know which problems have been posted and which have not. Already posted inequalities are hard to find using the search function. And I know you didn't copy my solution. When I found my solution I also didn't know that this solution was already known. It is not unusual that we sometimes discover solutions/problems already found by others. So the credit for your solution does go to you. :)


induction
combinatorics unsolved
combinatorics

I hope this didn't appear on the forum before!
We are given 100 sets, so that any of them have intersection an even number, but all of them have intersection an odd number.
Find the minimal cardinality of their reunion!

[Edit: As grobber asked, we are allowed to select any k of them ($100> k\geq 2 $) and we know that their intersection will be even]
[Edit 2: Minimal cardinality! Sorry for my mistake ] :blush:

What do you mean by "any of them have intersection an even number"? Any two of them? Or the intersection of any proper subset of the set of our sets? (I know that's confusing :)). Anyway, I don't think it's clear :?.

yah it is quite unclear as to what the problem wants;

suppose you have a family $\mathcal{A}$.  let $X=\bigcup_{A \in \mathcal{A}}A$.let $B \cap X = \emptyset,B$ an even set.then taking the sets $A \cup B,A \in \mathcal{A}$ the same property is true,so now this new family has a larger union.so there cannot be  a maximum.
if the question asks for minimum size of the union then the problem becomes more nontrivial.

Let me try:
For a given set of indices ${I=\{i_1,i_2,\cdots,i_k}$ denote $S_I$ be the set of elements belonging [b]only[/b] to $A_{i_1},\cdots, A_{i_k}$. Let's prove $S_I$ is odd for any $I \neq \{\}$.
The proof is by induction on $N-|I|$ ($N$-the number of elements in the union)
Basis: The whole intersection is odd:
Induction step:
WLOG suppose $I=\{1,2,\cdots,k\}$.
Then working in $Z_2$ we get $0=|A_1 \bigcap A_2 \cdots \bigcap A_k|=\sum_{I \subset J} S_J=
2^{n-|I|}-1+S_I$ by induction hypothesis so $S_I=1$ in $Z_2$ so $|S_I|$ is odd. The claim is proved.
The the sets $S_I$ all are disjoint and non-empty, giving at least $2^n-1$ elements (here we have $n=100$). The example with $2^n-1$ elements is obvious from the solution.

By the way, this was problem $10.5$ at the eliminatory round, proposed by S. Ivanov, if someone's interested


And?  :maybe:

What's the question :wink:

Sorry
Let $ (x_n) (n \equal{} 1,2,...)$ be a bounded-above sequence such that
$ x_{n \plus{} 2}\geq\frac {1}{4}x_{n \plus{} 1} \plus{} \frac {3}{4}x_n$, for all $ n\equal{}1,2,...$
Prove that this sequence has limit

[hide="Hint"]Since it is bounded above, it is enough to show that the sequence is increasing.[/hide]

[quote="SimonM"]it is enough to show that the sequence is increasing.[/quote]
Yes, but the sequence needn't be: if $ x_2 < x_1$ and all inequalities are equalities then the sequence oscillates.


Sean $x_0, x_1,x_2 \cdots x_n$ reales positivos tales que $x_ix_{i+2} \leq x_{i+1}^2$ para todo $0 \leq i \leq n-2$ demostrar que
$\frac{\sum_{i=0}^n x_i}{n+1} \frac{\sum_{i=1}^{n-1} x_i}{n-1} \geq \frac{\sum_{i=0}^{n-1}x_i}{n} \frac{\sum_{i=1}^n x_i}{n}$

[quote="Claudio Espinoza"]Disculpa,solo dos preguntas:
La condicion del problema, para que valores de $i$ se cumple?
En la sumatoria aparece $x_0$, debo entender que es $x_n$?, o es un nuevo valor?[/quote]

Hay que demostrar que si $x_{i}^2 \ge x_{i-1} \cdot x_{i+1}$ para todo $1\le i \le n-1$ entonces la desigualdad se cumple...  el problema deberia decir: sean $x_0, x_1, x_2....,x_n$ reales positivos.

Sea $S=x_0+x_1+...+x_n$. La desigualdad a demostrar es equivalente a $(\frac{S}{n+1})(\frac{S-x_0-x_1}{n-1})\ge (\frac{S-x_n}{n})(\frac{S-x_0}{n})$
$\equiv S(S-x_0-x_1)\ge (n^2-1)x_0x_n$
$\equiv (S-x_0)(S-x_n)\ge n^2x_0x_n$
$\equiv (x_1+x_2+...+x_n)(x_0+x_1+...+x_{n-1})\ge n^2x_0x_n$
$\equiv (\frac{x_1+x_2+...+x_n}{n})(\frac{x_0+x_1+...+x_{n-1}}{n})\ge x_0x_n$
$\equiv (\frac{\frac{x_1}{x_0}+\frac{x_2}{x_0}+...+\frac{x_n}{x_0}}{n})(\frac{\frac{x_0}{x_n}+\frac{x_1}{x_n}+...+\frac{x_{n-1}}{x_n}}{n})\ge 1$
Esta última desigualdad es más fácil de demostrar. Veamos:

Por un lado aplicando la desigualdad de la MA-MG a los $n$ números $\frac{x_1}{x_0},\frac{x_2}{x_0},...,\frac{x_n}{x_0}$. Y también a los $n$ números $\frac{x_0}{x_n},\frac{x_1}{x_n},...,\frac{x_{n-1}}{x_n}$. Y multiplicando ambas desigualdades obtenemos:

$(\frac{\frac{x_1}{x_0}+\frac{x_2}{x_0}+...+\frac{x_n}{x_0}}{n})(\frac{\frac{x_0}{x_n}+\frac{x_1}{x_n}+...+\frac{x_{n-1}}{x_n}}{n})\ge \sqrt[n]{\frac{x_1}{x_0}\frac{x_2}{x_0}...\frac{x_n}{x_0}\frac{x_0}{x_n}\frac{x_1}{x_n}...\frac{x_{n-1}}{x_n}}=\sqrt[n]{(\frac{x_1x_{n-1}}{x_0x_n})(\frac{x_2x_{n-2}}{x_0x_n})...(\frac{x_{n-1}x_1}{x_0x_n})}$

Bastaría con demostrar que la última expresión es mayor o igual que 1. Con las $n-1$ desigualdades dadas realizamos las siguientes operaciones y obtenemos los siguientes resultados:

Multiplicamos las $n-1$ desigualdades: $x_1x_{n-1}\ge x_0x_n$
Multiplicamos las $n-1$ desigualdades excepto la primera y la última:  $x_2x_{n-2}\ge x_1x_{n-1}$
Multiplicamos las $n-1$ desigualdades excepto las 2 primeras y las dos últimas:  $x_3x_{n-3}\ge x_2x_{n-2}$
Repitiendo este procedimiento obtenemos la siguiente secuencia:
$x_0x_n\le x_1x_{n-1}\le x_2x_{n-2}...$, donde la forma del último termino de la secuencia depende de la paridad de n.
De esto podemos concluir por transitividad que: $x_kx_{n-k}\ge x_0x_n; k=0,1,2,...n$.
O su equivalente: $\frac{x_kx_{n-k}}{x_0x_n}\ge 1; k=0,1,2,...,n$
Con lo cual cada factor de la expresión que buscabamos es mayor o igual que 1 y por tanto dicha expresión es mayor o igual que 1. Lo cual termina la demostración.


geometry
circumcircle
perpendicular bisector
cyclic quadrilateral
geometry proposed

Let $ AA_1$ is the bisector of the acute-angled triangle $ ABC$, point $ M$ is the midpoint of $ AA_1$. Such point $ P$ is taken on the $ BM$ that $ \angle APC \equal{} 90^{\circ}$ and such point $ Q$ is taken on the $ CM$ that $ \angle AQB \equal{} 90^{\circ}$. Prove that points $ A_1$, $ P$, $ M$ and $ Q$ are cyclic.

I know three different solutions of this problem and is interesting - which of them you could find?

Let $ D$ be the foot of the altitude from $ A$, then $ MA = MD = MA'$.
Consider the inversion $ I$ of pole $ M$ and power $ MA^2$, and let $ I(C) = Q'$ (obviously $ I(C)\in{MC}$).
Then $ \angle{AQ'M} = \angle{MAC} = \frac {\angle{A}}{2}$ and ${ \angle MQ'D} = \angle{MDA'} = \angle{C} + \frac {\angle{A}}{2}$.
Thus $ \angle{ABD} + \angle{AQ'D} = 180$ and hence $ Q' = Q$ (since $ ABDQ$ is cyclic and $ \angle{AQ'B}=\angle{ADB}=90$).
Finally, since $ I(BC) = C_{\triangle{MDA'}}$ we get that $ MDA'Q$ is cyclic.
Analogously, we prove that $ MPDA'$ is cyclic and the conclusion follows.

What a nice problem! I want to use Inversion here, :wink:
Consider an inversion by center $ A$,and an arbitrary radius $ k$.let me  analyzes the changes,
$ AB,AC,AP,AQ,AM,AA_1 \longrightarrow AB',AC',AP',AQ',AM',AA_1'$(repectively)
$ BM\longrightarrow$arc $ B'M'$of the circumcircle of $ \triangle AB'M'$
$ CM\longrightarrow$arc $ C'M'$of the circumcircle of $ \triangle AC'M'$
$ BQ\longrightarrow$arc $ B'Q'$of the circumcircle of $ \triangle AB'Q'$
$ CP\longrightarrow$arc $ C'P'$of the circumcircle of $ \triangle AC'P'$
in this case we should prove the quadrilateral$ A_1'Q'P'M'$ is cyclic.
in the inversive shape,$ B'Q'\cap C'P'\equal{}H'$,which$ H'$ is the inverse of $ H$,the foot of the altitude from $ A$to $ BC$.and$ \angle AA_1'H'\equal{}\frac{\pi}{2}$
and $ A_1'$ is the midpoint of $ AM'$,then $ A_1'H'$ is the perpendicular bisector.let $ A_1'H' \cap$arc$ B'M'\equal{}R$. and$ A_1'H' \cap$arc$ C'M'\equal{}S$
$ \angle B'AA_1'\equal{}\angle B'H'R\equal{}\angle C'AA_1'\equal{}\angle C'H'S$,thus $ RS$ is the bisector of $ \angle C'A_1'Q'$ and it means:arc$ C'S\equal{}$arc$ SQ'$
and arc$ B'R$=arc$ RP'$,and because of $ SR$ is the perpendicular bisector,$ C'$ and $ Q'$ turn into each other with a respect to$ RS$,
$ B'$ and $ P'$,also turn into each other with a respect to $ RS$,we also know that $ A$ and $ M'$,turn into each other with a respect to $ RS$,
the conclusion is the quadrilateral $ AC'A_1'B'$turnes into quadrilateral  $ A_1'Q'P'M'$ with a respect to$ RS$,and also
these two quadrilaterals are equal to each other,and because $ AC'A_1'B'$ is cyclic,($ B,A_1,C$ are collinear),the quadrilateral $ A_1'Q'P'M'$
will also be cyclic,and it is proved!

We can also mix these two solutions. Apply an inversion with center $ A$ and radius $ r$ and assume it sends point $ X$ to $ X'$. As [b]Ashegh[/b] wrote, we have $ AA_1'\equal{}A_1'M'$ and $ A_1'H'$ is perpendicular bisector of $ AM'$. Since points $ M,Q,C$ are collinear w see that $ AC'Q'M'$ is cyclic quadrilateral hence $ A_1'C'\equal{}A_1'Q'$ and using similar arguments $ A_1'C'\equal{}A_1'B'\equal{}A_1'P'$ which means that $ A_1'$ is circumcenter of $ P'Q'C'B'$ which means that points $ PQCB$ are concyclic (1). [b]Terryjohn[/b] proved that $ MA'^2\equal{}MQ\cdot MC$ hence $ \angle MCB\equal{}\angle MA_1Q$ and using this fact and (1) we obtain $ \angle MPQ\equal{}MA_1Q$ qed

You can find another very nice solution without using inversion by [b]Huyen Vu[/b] here: http://www.mathlinks.ro/viewtopic.php?t=141416 .

BTW, could you please post all problems from Ukraine MO on MathLinks? I found some geometry problems from Ukraine are very interesting  :thumbup:

Strangely, but this year the same configuration of points again appeared on Ukrainian Mathematical Olympiad. With the only difference that the final question was to prove that the points B, C, P, Q are concyclic (see the post of [b]limes123[/b] with the proof of this statement).

Link: http://matholymp.org.ua/_files/07cca67b94/2013-ukr-sols-2.pdf , problem 10-8.


Factorize in $\mathbb Z[x]$ the following polynom:
$x^{8}+98x^{4}+1$

[hide]expand $(x^{4}+ax^{3}+bx^{2}+cx+1)(x^{4}+ax^{3}+bx^{2}+cx+1)$ and equate coefficients (think if t is a root, then -t is a root)

then $2b-c^{2}=2b-a^{2}=0$ and $-2ac+b^{2}+2=98$, solve a quartic for c, you get rational solutions for $c=\pm 4$, then put it back and get $a=\mp 4$ and $b=8$

$(x^{4}+4x^{3}+8x^{2}-4x+1)(x^{4}-4x^{3}+8x^{2}+4x+1)$[/hide]


$A$ and $B$ are two point .
Find the geometric locus of $M$ such that $(MA,MB)$=$45$°

[hide]
Construct a square $AOBO'$ whose diagonals are $AB$ and $OO'$.

Draw two circles that go through $A$ and $B$ whose centers are $O$ and $O'$ respectively.
(The two circles intersect at $A$ and $B$.)
The locus of $M$ such that $\angle AMB=45^\circ$ is made of two major arcs of the circles (the parts of the circumferences that are outside of square $AOBO'$).

I hope I described the locus clearly...
[/hide]

sorry, i was a little rushed, i was posting this at school, it should be 2 circles that intersect at A and B such that arc AB = 90, and M is not on either minor arc, but is on the major arc AB on both circles,

this works because the angle is inscribed in the circle and intercepts 90 degrees, so 90/2=45, and if it were not on these circles, the secant-secant and chord-chord angle theorems would show that it would be less than 45 if M is ouside, and greater if M is in the insides of the circles


Re(iz)=1, |z+i|=2. Find all complex numbers z

Let $ z \equal{} a\plus{}bi$ for reals $ a,b$. The first given becomes $ \text{Re}(\minus{}b\plus{}ai) \equal{} 1$, and we get $ b \equal{}\minus{}1$. Then the second given becomes $ |a\minus{}i\plus{}i| \equal{} 2$, and we get $ a \equal{} 2$. So $ z \equal{} 2\minus{}i$.

EDIT: Derr! $ a \equal{}\minus{}2$ also, so $ z \equal{}\minus{}2\minus{}i$ is the other solution.


Let $n$ be a natural number greater than 6.  $X$ is a set such that $|X| = n$.  $A_1, A_2, \ldots, A_m$ are distinct 5-element subsets of $X$.  If $m > \frac{n(n - 1)(n - 2)(n - 3)(4n - 15)}{600}$, prove that there exists $A_{i_1}, A_{i_2}, \ldots, A_{i_6}$ $(1 \leq i_1 < i_2 < \cdots, i_6 \leq m)$, such that $\bigcup_{k = 1}^6 A_{i_k} = 6$.

Ok here we go, Use the following double counting to stablish some lemmas.

Call $m_k$ partial bound for the max number of subsets such that we cant find six $A_i$ with the desired property.

First, count the number of sets of $6$ elements such that $5$ of them belong to one $A_i$, using this for $n=7$, we get that for $n=7$, $5{\binom{7}{6}}> 2m$, so we get  $m_7=17$.

Now counting the number of $7$ elements set, such that $5$ of them belong to an $A_i$, for $n=8$, we get $m_8=45$,

and Finally for the other $n$, counting the number of $8$ element sets, $5$ of them in an $A_i$, we get a better bound that the desired one.

I really dont understand the solution  :maybe:
Could someone explain more please.


ok.. this was an olymon problem from 3 months ago. I think I can post them here now. I'm dying to see the solution; please, someone help!

Prove there doesn't exist consecutive perfect numbers.

[quote="N.T.TUAN"]In the http://www.mathlinks.ro/Forum/viewtopic.php?p=932109&sid=d760931f0812469cb550fdbed14ce957#932109 has a hint.  :wink:[/quote]

Ya, I got that far, but still couldn't do it. Sorry  :(

Could you help me some more? thanks!

I wrote out a solution that used several properties of perfect numbers a few months ago, but never found time to write up the others so I could send them in.  I'll try to find (or recall) the solution.  The problem was fun.

So I think you might find it useful to know for any odd perfect number N, there must exist a prime P such that NP ( :rotfl: ) is a perfect square.  Then wonder what N can be mod 4 and 3.


calculus integration conics parabola inequalities symmetry geometry

how many integral points are in the parabola, in the region formed by the parabola $ y^{2}\equal{}4ax$ and circle  $ x^{2}\plus{}y^{2}\equal{}16$
ans:17
please solve this

are u sure that is $ a$?
well the basic idea is to solve the following inequalities
$ x^{2}\plus{}y^{2}\leq 16$ -------------#
$ y^{2}\leq 4ax$
$ x\geq 0$
# gives $ 0\leq x\leq 4$
by symmetry consider $ y\geq 0$ then u can reflect the points obtained about x-axis.

yes that is a ,hey but where's the final answer

[quote="veryfriendly"]ITs good to see some problems on the Indian forum. You guys need them.[/quote]
what do u mean by "you guys". :P

[quote="SOUMYASHANT NAYAK"][quote="veryfriendly"]ITs good to see some problems on the Indian forum. You guys need them.[/quote]

It sounds somewhat rude as if Indians were a bunch of retards. :P  :lol:[/quote]

Ok, I never meant to say that. Um if you notice this is one of the only problems that has been posted in the Indian forum.

yes that sounda logical because posted elesewhere u r assured of getting it solved


function
linear algebra
matrix
calculus
derivative
advanced fields
advanced fields unsolved

Let $ u \in C^2(B_1^2(0)) \cap C( \overline {B_1^2(0)}); u_{xx}\plus{}u_{xy}\plus{}u_{yy}\equal{}1, t\equal{}(x,y) \in B_1^2(0).$ The question is if $ u(t)$ can have 1)maximum; 2)minimum inside $ B_1^2(0)$?

As for minimum we have that function $ u(x,y)\equal{}\frac 1 4(x^2\plus{}y^2)$ satisfies the equation and $ u$ has minimum at $ (0,0)$. So I think that answer for b) is positive. And what about a)?

Just look at the Hessian at the possible point of maximum. What can you say about its entries? Now, what can you say about its determinant? (or just see whether the word "elliptic" brings up any associations ;) )

I guess $ u_{xx}<0, u_{yy}<0$ at the point of maximum, what about $ u_{xy}$ I don't know(is it equal to zero at the point of maximum? - if so, hessian is positive at the point of max and it is wrong, because it should be negative, so the function can't have maximum).

$ u_{xy}\equal{}1\minus{}u_{xx}\minus{}u_{yy}$, remember? So your matrix at the point of maximum should be of the form $ \begin{bmatrix}\minus{}a&1\plus{}a\plus{}b\\1\plus{}a\plus{}b&\minus{}b\end{bmatrix}$ for some $ a,b\ge 0$. Can it be non-positive definite?

Why at the point of maximum $ u_{xx}=-a$, ${ u_{yy}}=-b$, where $ a \ge 0, b \ge 0$.?

Second derivative test: if you have a maximum in two variables, you certainly have a maximum in each one separately (but not vice versa!).


[quote="jli"][quote="Quevvy"]A student ID number at a certain university consists of a 6-digit number, such as 023457. How many student numbers can be chosen such that no two of its digits are the same?

I thought it was just 10! but it was wrong.[/quote]

[hide=" :maybe: "]ummm... wouldnt it just be $10\times 9\times 8\times 7\times 6\times 5=151200$?[/hide][/quote]
Oh yeah..... I'm stupid!
I forgot that there were only 6 digits. :wallbash:

wait... so what is it?

10! x 9! x 8! x 7! x 6! x 5! ? or something?

[quote]10! x 9! x 8! x 7! x 6! x 5! ? or something?[/quote]
No because its not 10!*9!... its just simply 10*9*8*7*6*5, because you have 10 choices for the first spot, 9 fro the second...


function
complex analysis
complex analysis unsolved

$D=\{z\in\mathbb C: |z|<1$ and $f: D\rightarrow\mathbb C$ is an analytic function which for $n\geq 2$ we know that $f(\frac 1n)\in\mathbb R$. Prove that for each $n$, $f^{(n)}(0)\in\mathbb R$

$f(z)=a_0+a_1z+a_2z^2+...$ with $a_i \in \mathbb{C}$, and one wants to prove that those $a_i$ are in fact real numbers.
Suppose that it is not the case and that $N \in \mathbb{N}$ is the smallest integer such that $a_N$ is not real. Then $g(z)=a_Nz^N+a_{N+1}z^{N+1}+..=z^{N}(a_N+a_{n+1}z+..)$ also satisfies $g(\frac{1}{n}) \in \mathbb{R}$, and also $h(z)=a_N+a_{n+1}z+..$. This is clearly a contradiction because $h(0)$ is not real and because of the continuity of $h$, $h(z)$ is not real on a neighborhood of $0$.


function
integration
calculus
calculus computations

If $ y\equal{} 5 \plus{} \int_2^{2x} e^{\minus{}t^2} dt$, which of the following is true?

A) $ \frac{dy}{dx} \equal{} e^{\minus{}x^2}$ and $ y(0)\equal{}5$

B) $ \frac{dy}{dx} \equal{} e^{\minus{}x^2}$ and $ y(1)\equal{}5$

C) $ \frac{dy}{dx} \equal{} e^{\minus{}4x^2}$ and $ y(1)\equal{}5$

D) $ \frac{dy}{dx} \equal{} 2e^{\minus{}4x^2}$ and $ y(0)\equal{}5$

E) $ \frac{dy}{dx} \equal{} 2e^{\minus{}4x^2}$ and $ y(1)\equal{}5$

Hi :)

I thinks that the last choose is correct (E): $ \frac {dy}{dx} \equal{} 2e^{ \minus{} 4x^2} \ and \ y(1) \equal{} 5$

if you want a prof say to me i'm present
thanks

Yeah the little integral part confuses me

let $ F(x)$ be an antiderivative of $ e^{\minus{}x^2}$.  Then we have $ y\equal{}5\plus{}F(2x)\minus{}F(2)$.  Observe when $ x\equal{}1$, $ y\equal{}5$.  Now differentiate: $ y' \equal{} 2F'(2x)$ by chain rule, so $ y'\equal{}2e^{\minus{}(2x)^2}$.


calculus
derivative
inequalities unsolved
inequalities

slove the image of  $ y \equal{} \sqrt {2x^2 \minus{} 6x \plus{} 9} \plus{} \sqrt {2x^2 \minus{} 10x \plus{} 17}\;$

What do you mean? Should $ y$ be an integer?

If $ x\in \mathbb{R}$ then
$ A\equal{}\{ y \in \mathbb{R} | y\ge 2\sqrt{5}\}$

If you don't want to use derivatives to find the minimum of $ \sqrt{2x^{2}\minus{}6x\plus{}9}\plus{}\sqrt{2x^{2}\minus{}10x\plus{}17}$ then think of a geometric interpretation

[quote="Rafikafi"]If $ x\in \mathbb{R}$ then
$ A \equal{} \{ y \in \mathbb{R} | y\ge 2\sqrt {5}\}$

If you don't want to use derivatives to find the minimum of $ \sqrt {2x^{2} \minus{} 6x \plus{} 9} \plus{} \sqrt {2x^{2} \minus{} 10x \plus{} 17}$ then think of a geometric interpretation[/quote]

How to use derivatives to find the minimum ?
can you tell me ?
thanks

Let $ f(x)\equal{}\sqrt{2x^2\minus{}6x\plus{}9}\plus{} \sqrt{2x^2\minus{}10x\plus{}17}$ and find the unique $ x$ such that $ f'(x)\equal{}0$.

[quote="FelixD"]Let $ f(x) \equal{} \sqrt {2x^2 \minus{} 6x \plus{} 9} \plus{} \sqrt {2x^2 \minus{} 10x \plus{} 17}$ and find the unique $ x$ such that $ f'(x) \equal{} 0$.[/quote]
How to use derivatives to find the minimum ?
can you tell me ?
thanks

$ \sqrt {2x^2 \minus{} 6x \plus{} 9} \plus{} \sqrt {2x^2 \minus{} 10x \plus{} 17} \equal{} \sqrt {2}\left ( \sqrt {(x \minus{} \frac {3} {2})^2 \plus{} (\frac {3} {2})^2} \plus{} \sqrt {(x \minus{} \frac {5} {2})^2 \plus{} (\frac {3} {2})^2} \right )$.

Take $ A(\frac {3} {2}, \frac {3} {2})$, $ B(\frac {5} {2}, \frac {3} {2})$ and $ X(x, 0)$, then the expression is $ \sqrt {2}(XA \plus{} XB)$. Its minimum occurs for $ x \equal{} 2$, so the range of values is $ [2\sqrt {5}, \plus{} \infty)$.