[quote="mathgeniuse^ln(x)"]In terms of percent, which fits better-a round peg in a square hole or a square peg in a round hole? (Assume a snug fit in both cases.)[/quote]

[hide]OK, first, a round peg in a square hole:

Say the hole is 2 units by 2 units in area, then the radius of the peg would be one unit. So the fit here is $\frac{\pi}{4}$

Now square peg in round hole. Say the radius of the hole is 1, then the area of the hole is $\pi$.
Then the diagonal of the square is 2, so the side of the square is $\sqrt2$, so the area of the square is 2, so the fit here is $\frac{2}{\pi}$

Comparing, we know $\pi^{2}$ is greater than 8, so the better fit is round peg in a square hole.[/hide]

[hide]first, square in circle.  If a side of the square is $1$, then the area of the circle is $(\frac{1}{2})^{2}\pi$, for a percent of $25\pi \approx 79$.  For the circle in the square, assume the radius of the circle is $1$.  Therefore, the diagonal of the square is $2\times1=2$ and the sides are equal to $\sqrt{2}$.  So, the area of the square is $\sqrt{2}^{2}=2$ and the percentage is $\frac{200}{\pi}\approx 64$  Therefore, the answer is a circle in a square.[/hide]


Is there any way to customize your account such that your signature is automatically posted on every post( i keep forgetting to check off the box)?

Click on the blue button "Profile," then "Preferences," then "Posting a message," then you should see a "Always attach my signature" option.


Putnam
calculus
integration
function
trigonometry
real analysis
real analysis solved

I saw on a previous putnam exam that int(e^[-x^2])dx evaluated from -infinity to infinity = sqrt[pi].  Can someone show me a proof for this?
Also, my teacher tells me that this integral can be evaluated using a polar transformation.  If this is true, please tell me how, because I thought that this function did not have an elementary primitive.

Thanks.

No, it does not have an elementary primitive.

You need to know knowledge about double integral.

Let $I = \int_0^\infty e^{-x^2}dx$.

$I^2 = \left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2}dy\right) =
\int_{[0,\infty)\times[0,\infty)} e^{-(x^2+y^2)} dxdy$

Now take the standard polar transformation $x=r\sin\theta$, $y=r\cos\theta$.
$x^2+y^2=r^2$ and $dxdy=r\drd\theta$.

And the integral above becomes

$\int_{[0,\infty)\times[0,\infty)} e^{-(x^2+y^2)} dxdy =
\int_{[0,\infty)\times[0,\pi/2]} re^{-r^2} drd\theta
= \pi/2\int_{[0,\infty)} re^{-r^2} dr = \pi/4
$

Hence $I=\sqrt{\pi}/2$.

Taking the (rather famous) idea that liyi just posted and trying to use it in 3 or more dimensions is one possible path to the "hypervolume" of the "hypersphere", as in ASquare's topic on this forum. (Note: to follow up on this suggestion, you must know the Gamma function.)

Thanks, your explanation really helped.

On the note of hypervolume (in 4 or more dimensions), it has an interesting relationship to pi^2/6(which is the sum of the recipricols of the squares of the counting numbers).


[hide="Part a"]
$\sin{2\theta}+\cos{\theta}<0$
$\Leftrightarrow 2\sin{\theta}\cos{\theta} +\cos{\theta}<0$
$\Leftrightarrow \cos{\theta} (2\sin{\theta}+1)<0$

Case 1: $\cos{\theta}>0$ and $2\sin{\theta}+1<0$ ($\therefore \sin{\theta} <-\frac12$)
$\longrightarrow \frac{3\pi}{2} <\theta <\frac{11\pi}{6}$

Case 2: $\cos{\theta}<0$ and $2\sin{\theta}+1>0$ ($\therefore \sin{\theta} >-\frac12$)
$\longrightarrow \frac{\pi}{2} <\theta <\frac{7\pi}{6}$[/hide]

[hide="part b"]
$\sin \theta \sec \theta = \tan \theta = \frac 1 2$
$\implies \sin \theta = \frac{ \sqrt 5 }{5}, \cos \theta = \frac{ 2\sqrt 5 }{5}$
$\frac{-k}{2} = \sin \theta + \sec \theta = \frac{ \sqrt 5 }{5} + \frac{ \sqrt 5}{2}$
$\implies k = \frac{-7\sqrt 5 }{5}$
[/hide]


Suppose $ A$ is the set of all odd numbers between 10 and 16, and $ B$ is the set of all prime numbers between 10 and 16. How many distinct results are possible if a number from set $ B$ is subtracted from a number in set $ A$?

Set $ A$ consists of {$ 11$, $ 13$, $ 15$}. Set $ B$ consists of {$ 11$, $ 13$}. Therefore, we have $ \boxed{4}$ distinct results. ($ \minus{}2$, $ 0$, $ 2$, $ 4$).


inequalities
combinatorics proposed
combinatorics

Let P be a set of n points in the plane, no three collinear. Let d_1,...,d_k  denote distinct distances determined by P, and let s_i be the number of point pairs whose distances is d_i.
Prove that

s 2 _1+s 2 _2+...+s 2 _k <= 3/4 n 2 (n-1)

Pff... It took me a long time to understand how to make appear $s_i^2$... ;)

Let $P_1, \cdots, P_n$ be the points.
For each $i,j$ let $n_i$ be the number of points $M \in P$ such that $MP_j = d_i$.

Let's count the number $T$ of triples $(P_a,P_b,P_c)$ such that $a,b,c$ are pairwise distinct and $P_aP_b=P_aP_c.$ It is the number of isoceles triangles but counted with multiplicities (an equilateral is counted three times).

Let $M$ be fixed in $P$. The number $T(M)$ of triples $(M,X,Y)$ such that $MX=MY$ is $T(M) = \sum_{i=1}^k C_{n_i(M)}^2$, where $n_i(M)$ denotes the number of points $Q$ such that $MQ=d_i.$
Clearly $\sum_{i=1}^k n_i = n-1$ and $2s_i = \sum_{M \in P} n_i.$
Thus :
$T(M) = \frac 1 2 ( \sum_{i=1}^k n_i^2 )- \frac {n-1} 2.$

Therefore :
$T = \sum_{M \in P} T(M) = - \frac {n(n-1)}2 +  \frac 1 2 \sum_{M \in P} \sum_{i=1}^k n_i^2 .$ (1)

In another hand, since no three points are collinear, it follows that for each  pair of points $X,Y$ from $P$ there are at most two points $M$ such that $MX=MY$ (otherwise, there would be at least three points on the perpendicular bissector of $XY$). Thus, $T \leq 2C_n^2 = n(n-1).$ (2)

From (1) and (2), we have :
$\frac {3n(n-1)} 2 \geq \frac 1 2  \sum_{i=1}^k \sum_{M \in P} n_i^2 \geq  \frac 1 2 \sum_{i=1}^k n(\frac {2s_i} n )^2$,
the last inequality following from the convexity of $f(x) = x^2.$
This immediately gives :
$\sum_{i=1}^k s_i^2 \leq \frac {3n^2(n-1)} 4$, as desired.

Pierre.


algebra
polynomial
superior algebra
superior algebra solved

Three matrices A, B, C are commutative, i.e AB = BA, AC = CA, BC = CB. Show that there exist a, b, c not all equal to zero scuh that det(aA+bB+cC) = 0

<i>Source: From the Book "Olympiad problems for Undergraduate students" (in Vietnamese)</i>

I am really sure that I did not understand corectly the problem:
  If A is not inversable then take b=c=0 and a=1. The same if B or C is not inversable. So, let A, B and C inversable. Let T=BA^-1 and K=CA^-1. We have to prove  that we can find a,b,c not all 0 such that det(aI+bT+cK)=0.Fix b and c, at least one different from 0 and regard LHS as a polynomial in a.It has a complex root a. Then (a,b,c) is what we wanted.

A, B, C \in Mn(R) commute 2 by 2, we can see these matrices in Mn(C).
In appropriate basis
A, B, C are simultanously triangular superior

det(aA+bB+cC)=
product(a \lambda _A + b \lambda _B + c \lambda _C)

where  \lambda _A eigenvalue of A,...
There exist (a,b,c)=/=(0,0,0) s.t
a \lambda _A + b \lambda _B + c \lambda _C = 0

Remark: If we work in Mn(K) where K is a commutative field,
consider Mn(ac(K)) where ac(K) is the algebraic closure of K


Hi. I have a problem. My scores are not showing up for round 4, and My USAMTS says that I am a silver winner. Is this just a fluke? Anyone else having this problem? Sorry if this is the wrong place to post, but I didn't see this topic fitting under Grading Disputes.
Thanks for the help.

I can't do anything until you let me know what your USAMTS ID # is.

You can send it to me via a private message.


graph theory
combinatorics unsolved
combinatorics

This is a strange graph problem. Maybe it is famous, but I don't solve it...

Consider a complete graph with n vertex.
Its edges are coloured with two colors (black and white) such that:

--If A,B,C are vertices such that AB and AC are black then BC is white;

If A,B are vertices such that AB is white then there exists two (and only two) vertices X,Y such that AX,AY,BX,BY are black.

Show that n-1 is a triangular number, and this graph is regular.

Let's change the language a bit: we say that there is an edge between two points if the respective edge is black, and we say that there is no edge between them if the edge is white. The conditions are translated as follows:

If $x,y$ are connected, then they have no common neighbour; If $x,y$ are not connected, then they have precisely two common neighbours.

It looks very familiar. I'm almost sure I've solved it before (not necessarily on the forum though :)).

Take any vertex $x$ of our graph, and let $d$ be its degree. Each two of its neighbours have another common neighbour among the vertices which are not its neighbours, and, conversely, each non-neighbour of $x$ has two neighbours in common with $x$. This means that there are precisely as many non-neighbours of $x$ as there are pairs of neighbours of $x$. This translates to $n-1-d=\frac{d(d-1)}2$, and we get $n-1=\frac{d(d-1)}2\ (*)$. This shows that $n$ is indeed triangular, and, since $(*)$ holds for the degree of an arbitrary vertex of our graph, it means that all the vertices have the same degree.


group theory
abstract algebra
induction
function
superior algebra
superior algebra solved

a) Show that if G is isomorphic to a subgroup of G (different from G), then it's isomorphic to infinitely many such subgroups.
b) Let C(R)={f:R->R|f is continuous}. Show that the group (C(R),+) has the property syayed at point (a).

For the moment I have only part a)

Let f be the isomorphism between G and the subgroup
Denote f[n]=f o f o ... o f (n times). and Sn=Im f[n]
Also take S0=G

If f is a morphism then f[n] is also a morphism and its image is a group.

Then by the method of induction it can be proved that for every n we have Sn strictly included in Sn+1 and isomorphic to it, the isomorphism between them is f

For n=0 it is clearly true.
n->n+1
f is the isomorphism from Sn to Sn+1. Applied again from Sn+1, a proper subset of Sn,  to Sn+1 the function is injective, but cannot be surjective, so its image, Sn+2, is strictly included in Sn+1, and between the 2 sets the function is bijective so they are isomorphic. :)

Got an idea for b)  :D

Take F(f) to be the primitive of the function f that becomes 0 in 0. This function is an isomorphism from the set of continuous functions to the set of differentiable functions that are 0 in 0. :)

Do you know how long it took me to get that idea for (b)?:D


In my Algebra class we proved the following theorem:

"Theorem. Let $K$ be a field and let $f$ be an irreducible polynomial in $F[X]$.
(a) If $\textrm{char}\ K = 0$, then $f$ doesn't have any multiple roots.
(b) If $\textrm{char}\ K = p$, then $f$ has a multiple root iff $f\left(X\right) = g\left(X^p\right)$ where $g \in K[X]$."

After the proof, there is a remark:

"Remark:
One can prove that an irreducible polynomial over a finite field of characteristic $p \neq 0$ doesn't have multiple roots."

Why isn't this remark contradicting (b) in the theorem above?

[quote="Arne"]
(b) If $\textrm{char}\ K = p$, then $f$ has a multiple root iff $f\left(X\right) = g\left(X^p\right)$ where $g \in K[X]$."

After the proof, there is a remark:

"Remark:
One can prove that an irreducible polynomial over a finite field of characteristic $p \neq 0$ doesn't have multiple roots."

Why isn't this remark contradicting (b) in the theorem above?[/quote]

Because in a finite field of characteristic $p>0$, a polynomial of the form $g(x^p)$ is actually a $p$'th power of a polynomial, so it cannot be irreducible: $x\mapsto x^p$ is an injection, and in a finite field this also means that it's a surjection. We thus have $a_0+a_1x^p+\ldots+a_nx^{np}=b_0^p+b_1^px^p+\ldots+b_n^px^{np}=(b_0+b_1x+\ldots+b_nx_n)^p$.

If $\mbox{char}\ K=0$ then if $\alpha$ is a multiple root of $p(x)$ then $\alpha$ is also root of $p'(x)$. You could see easily that $deg\ p'<deg\ p$ and $p'(x)\neq0$. This is a contradiction but in the field that $\mbox{Char}\ K=p$, then if $p'(x)\neq0$ there is a similiar contradiction, So $p'(x)=0$. Easily you could see that all coffeicents of $x^i$ that $pk+r,r\neq0$ are zero. So $p(x)=g(x^p)$

I did get that part, you know :) I was just confused by the remark.


complex analysis
algebra
function
domain
topology
real analysis

hey,

I am a beginner in complex analysis and I hope that someone of you help me with this  question (may be too easy for you)

 $ f$ is holomorphic in a connected open set $ \Gamma$ prove that:
$ f$ is constant in $ \Gamma$ is equivalent to $ \forall z\in\Gamma,f'(z)\equal{}0.$

This equivalence is it true if $ \Gamma$  is an open non-connected?

This statement is of course false for non-connected region. Just take two open circles wich do not interesect each other. In the first circle let $ f(z) \equal{} a$ and in another one $ f(z) \equal{} b$ for $ a\ne b$. Now to prove your statement for connected region just...integrate :wink:

could you please just use elementary proof for the first part of my question!

What do you mean when you say "elementary"?What arguments(theorems, definitions) can I use?

oppps,I mean by this,just elementary definitions (cauchy's conditions,cauchy's theorem)! if it is possible of course!

thanks for your response!

This isn't really a complex analysis problem.
Show that $ S\equal{}\{z: f(z)\equal{}f(z_0)\}$ is both open and closed, relative to the domain of $ f$. The tools are entirely those of real analysis and elementary topology.

jmerry is right but I want to add that the proof is very clever, you can find a proof in Conway's book !


induction
number theory unsolved
number theory

Can a total of two or more powers of ten with whole number exponents which don't recur be an even power of a whole number?

An even power of a whole number is always a square.

The number in base ten consists of digits 1 and 0 only with at least two times the digit 1.
The end of the number will be 1...1... where the dots indicate any number of zeros
In fact it must end in an even number of zeros since it is square and if we drop the end zeros we still get a square
So we need a square that ends 1...1

It can be checked by induction that a square that ends in a string of n zeros followed by a one is either the square of
another number that ends in n zeros followed by a one, or a number that ends in n+1 nines.
It is also then easy to show that the number before the zeros in the square is even, so it can't be a one.

Therefore no such number exists.


[quote="arqady"]Prove that for all $a,$ $b,$ $c$ and $d$ the following inequality holds:
$\sum_{sym}(a^{2}b^{2}-2a^{2}bc+abcd)\geq0.$[/quote]

Nice ineq. This is equivalent to $(\sum_{cyc}(b-c)^{2})a^{2}-2(\sum_{cyc}d(b-c)^{2})a+\sum_{cyc}d^{2}(b-c)^{2}\geq 0$, which is true as $\frac{D}{4}=(\sum_{cyc}d(b-c)^{2})^{2}-(\sum_{cyc}(b-c)^{2})(\sum_{cyc}d^{2}(b-c)^{2})\geq0$ by Cauchy-Schwart ineq. :)

Was it your proof to rewrite the inequality as $\left(a-d\right)^{2}\left(b-c\right)^{2}+\left(b-d\right)^{2}\left(c-a\right)^{2}+\left(c-d\right)^{2}\left(a-b\right)^{2}\geq 0$ ?

  darij

Yes, darij!  But not in it interest. This inequality is very strong.


trigonometry
calculus
derivative
inequalities proposed
inequalities

Have you got a proof without any trigonometry, derivative for the following?

Let $x,y,z$ be real positive numbers such that $xy+yz+zx=1$, prove that
\[ \frac 1{1+x^2}+\frac1{1+y^2}+\frac1{1+z^2}\leq\frac94. \]
and
\[ \frac x{1+x^2}+\frac y{1+y^2}+\frac z{1+z^2}\leq\frac{3\sqrt3}4. \]

Thank you Trung. We can use the condition $xy+yz+zx=1$ this way
\[ 1+x^2=xy+yz+zx+x^2=(x+z)(x+y). \]

trigonometric substitution here is natural.
x=ctgA\2 etc. give a very simple solution to both.


Find all positive integers n for which for which both $ 2n\plus{}1$ and $ 3n\plus{}1$ are perfect squares. Prove that all such integer are divisible by 40.

$ n \equal{} \frac {x^2 \minus{} 1}{2}$ where $ (x,y)$ solution of the pell equation $ 3x^2 \minus{} 2y^2 \equal{} 1$

here's another problem

Let $ {u_n}_n$ and $ {v_n}_n$ be two sequences that satisfying the relations $ u_{n\plus{}1}\equal{}3u_n\plus{}4v_n$ and $ v_{n\plus{}1}\equal{}2u_n\plus{}3v_n, u_1\equal{}3, v_1\equal{}2.$ Define $ {x_n}_n$ and $ {y_n}_n$ by $ x_n\equal{}u_n\plus{}v_n, y_n\equal{}u_n\plus{}2v_n.$ Prove that for all $ n$, $ y_n\equal{}\lfloor{x_n \sqrt{2}}\rfloor$.

[hide="Solution"]
$ u_n > v_n,\ \forall n \in \mathbb{N}$

We must show that,

$ x_n\sqrt{2} \minus{} 1 < \lfloor x_n\sqrt{2} \rfloor \le x_n \sqrt{2}$

$ \Leftrightarrow x_n\sqrt{2} \minus{} 1 < y_n \le x_n \sqrt{2}$

[b]For the right hand side:[/b]

Furthermore it's easy to show by induction that,

$ 2v_n^2 \le u_n^2 \implies \frac{2v_n}{u_n\plus{}v_n} \plus{} \frac{v_n^2}{(u_n\plus{}v_n)^2} \le 1$

$ \implies \left(\frac{y_n}{x_n}\right)^2 \le 2 \implies y_n \le \sqrt{2}x_n$

[b]For the left hand side:[/b]

We must show, $ (y_n\plus{}1)^2 > 2x_n^2$

$ \Leftrightarrow 2v_n^2 \plus{} 2u_n \plus{} 4v_n \plus{} 1 > u_n^2$ which can be established by induction.
[/hide]


I will be a Junior as of Fall 2005. I will be taking Algebra 2 and I was wondering which books will best prepare me for both the 1c and 2c exam I will be taking in the spring. I heard that the 2c contains materials from pre-calc which I have NO background knowledge of. Will the barrons book best prepare me for an 800 on both test?

princeton review

that's what i used, 790 on IC
taking IIC around dec.
but i only used PR as a review guide to refreash my memories

Thanks for the first reply. I'd like to hear other replies in this thread. I thought that there was a consensus about what is the best brand of test prep material for the SAT II Level 2 math test, but I'd like to hear current opinions to refresh my memory. There seem to be especially many brands of test prep books for that test.

I've heard Barrons is good, because their tests are usually harder than the real ones, so it's good preparation.

I use the SparkNotes books with my students.  They have two books for Math Level 1 and two books for Math Level 2.  I feel their practice tests are pretty close in difficulty to the real thing.  (I've taken both tests several times as part of my job.)

You should also get, of course, the Real SAT II book published by the College Board.

Will the PR books teach me the materials on the math 2c test?...because I've never learned the things tested in there and I need a book that would do that.

The sparknotes SAT IIC test are pretty on par with the real test. I took one to see and it was quite similar to the test i took on test day.


trigonometry
calculus
integration
function
real analysis
real analysis unsolved

Hi!
1) Calculate
\[ \int_0^{2\pi} \sin^{2s}\theta \ d \theta\]

where $ s \in \mathbb{N}$
2) Find the asymptotic expansion of this integral for $ s \gg 1$.

3) Show that (1) and (2) agree.

Cheers

here is a solution to the first part [hide] There are many ways to calculate this integral, namely the Beta function, a recurrence relation and use of the calculus of residues immediately come to mind. I'll use the calculus of residues. We let $ z\equal{}e^{i\theta}$ and find that
\[ I \equal{} \oint \minus{}\frac{i}{(2i)^{2s}}\frac{(z^2\minus{}1)^{2s}}{z^{2s\plus{}1}} \ dz\]

The residue of this integral will be $ 2 \pi i \  (\minus{}1)^s \ \binom{2s}{s} \frac{\minus{}i}{(2i)^{2s}}$
$ \therefore \boxed{ I \equal{}  \frac{2 \pi}{2^{2s}}\binom{2s}{s}}$

[/hide]


Let $ a,b,c$ are positive reals such that $ a\plus{}b\plus{}c\equal{}a^{2}\plus{}b^{2}\plus{}c^{2}$.
Find the biggest real k such that $ a^{k}\plus{}b^{k}\plus{}c^{k}\geq 3$ holds.

[quote="stephen38"]Let $ a,b,c$ are positive reals such that $ a \plus{} b \plus{} c \equal{} a^{2} \plus{} b^{2} \plus{} c^{2}$.
Find the biggest real k such that $ a^{k} \plus{} b^{k} \plus{} c^{k}\geq 3$ holds.[/quote]
$ a^{k} \plus{} b^{k} \plus{} c^{k}\geq 3\Leftrightarrow(a^{k} \plus{} b^{k} \plus{} c^{k})(a \plus{} b \plus{} c)^k\geq 3(a^2 \plus{} b^2 \plus{} c^2)^k.$
If $ k > 0$ it's wrong since, if $ a \equal{} 1$ and $ b \equal{} c\rightarrow0^ \plus{}$ then
$ (a^{k} \plus{} b^{k} \plus{} c^{k})(a \plus{} b \plus{} c)^k\rightarrow1$ and $ 3(a^2 \plus{} b^2 \plus{} c^2)^k\rightarrow3.$
For $ k \equal{} 0$ it's true. Thus, $ \max k \equal{} 0.$

Thanks,arqady. :) You make the problem looked easy. :D

Dear arqady,sorry for my foolishness.:wallbash_red:
When i write your solution in my note,i found a mistake.
If we put $ a\equal{}1,b\equal{}c$,then we must have $ b\equal{}c\equal{}1$,so it can't be the case $ a\equal{}1,b\equal{}c\rightarrow 0^{\plus{}}$.Moreover,if we only put $ b\rightarrow 0^{\plus{}}$,we must have $ c\rightarrow 1^{\plus{}}$,isn't it?? :(

[quote="stephen38"]
If we put $ a \equal{} 1,b \equal{} c$,then we must have $ b \equal{} c \equal{} 1$,so it can't be the case $ a \equal{} 1,b \equal{} c\rightarrow 0^{ \plus{} }$[/quote]
OK! $ a\rightarrow1$ and $ b\equal{}c\rightarrow0,$ where $ a>0$ $ b>0$ and $ c>0$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}a\plus{}b\plus{}c.$ :wink:

Thanks.How about this one :
Find the biggest real k such that $ a^{k} \plus{} b^{k} \plus{} c^{k} \leq 3$ holds. :wink:


Let $f(x)$ be a polynomial with degree $5$. Suppose $f(k)=2^k$ for $k=1,2,3,4,5$.
Find the value of $f(6)$

Are you sure you didn't miss some condition?

[quote="Singular"]Impossible: 6 points uniquely defines a polynomial of degree 5. So there are infinitely many values of f(6).[/quote]
then what about this thread?
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32417
which needs only 2 points to determine a polynomial under the condition mentioned

That thread assumed nonnegative coefficients...

i get 62 by finite differences (sometimes called polynomial interpolation i think)

No, the answer is only actually $63$ if it has degree $4$ or you set $f(6)=63$.
Like Magnara, I would say:
$f(6) = \frac{\zeta (3+i)}{\lim_{n \to \infty} \zeta \left( 1+\frac{1}{n} \right) - n} + e^i \cdot \vartheta (1+\Gamma (\sqrt i )) $  ;)  :lol:

What about this solution I found? Is there something wrong with it?

Consider the polynomial\[g(x)=1+{x\choose 1}+{x\choose 2}+{x\choose 3}+{x\choose 4}+{x\choose 5}\]
Using the expression $2^n=\sum^n_{r=0}{n\choose r}$, it is easy to check that $g(k)=2^k$ for $k=1,2,3,4,5$
Since both $f(x)$ and $g(x)$ are polynomials of degree 5 and they agree at 6 distinct points, we must have $f(x)\equiv g(x)$
Thus $f(6)=g(6)=2^6-1=63$

[quote="digitalboy1"]What about this solution I found? Is there something wrong with it?

Consider the polynomial\[g(x)=1+{x\choose 1}+{x\choose 2}+{x\choose 3}+{x\choose 4}+{x\choose 5}\]
Using the expression $2^n=\sum^n_{r=0}{n\choose r}$, it is easy to check that $g(k)=2^k$ for $k=1,2,3,4,5$
Since both $f(x)$ and $g(x)$ are polynomials of degree 5 and [b]they agree at 6 distinct points[/b], we must have $f(x)\equiv g(x)$
Thus $f(6)=g(6)=2^6-1=63$[/quote]

what is the 6. point on that they are equal¿


Let $ a\plus{}ar_1\plus{}ar_1^2\plus{}ar_1^3\plus{}\cdots$ and $ a\plus{}ar_2\plus{}ar_2^2\plus{}ar_2^3\plus{}\cdots$ be two different infinite geometric series of positive numbers with the same first term.  The sum of the first series is $ r_1$, and the sum of the second series is $ r_2$.  What is $ r_1\plus{}r_2$?

$ \textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ \frac{1\plus{}\sqrt{5}}{2}\qquad \textbf{(E)}\ 2$

[hide="Click to reveal solution"]The sum of each series is $ \displaystyle \frac {a}{1 \minus{} r_i}$, where $ i \equal{} 1,2$.  Thus,
\[ \frac {a}{1 \minus{} r_i} \equal{} r_i \Rightarrow r_i^2 \minus{} r_i \plus{} a \equal{} 0 \textrm{.}
\]The roots of this quadratic sum to $ 1$, so $ \fbox{C}$.
[/hide]


co ai vao trang http://www.maths.ru chua? trong do toan tieng Nga ko ha. Co ai giup em doi qua tieng Anh duoc ko? hay lam cach nao de doc duoc. Cam on nhieu! :roll:  :(

Tat nhien may trang web Toan cua Russia thi toan la Russian . Ve chuyen dich gium ban thi co le chi con cach nho may thay truoc kia hoc o Nga (thay Nam Dung chang han). O ML , Darij cung co post nhieu bai tren Kvant lam , search with keyword " Kvant " cung duoc khoi. Minh nghi ban khong can xem may trang web do dau , vao DDTH xem tap chi AMM , Cruz cung hay lam do !


Define sequences of integers by
$ p_1 \equal{} 2, q_1 \equal{} 1, r_1 \equal{} 5, s_1\equal{} 3, p_{n\plus{}1} \equal{} p_n^2 \plus{} 3q_n^2, q_{n\plus{}1}\equal{} 2 p_nq_n, r_n \equal{} p_n \plus{} 3 q_n,$
$ s_n \equal{} p_n \plus{} q_n$.
Show that $ \frac{p_n}{q_n} > 3 > \frac{r_n}{s_n}$ and that $ \frac{p_n}{q_n}$ differs from $ \sqrt{3}$ by less than $ \frac{s_n}{2 r_nq_n^2}$.

Sorry, my problem is show that $ \frac{p_n}{q_n}>\sqrt{3}>\frac{r_n}{s_n}$! :oops:


number theory
relatively prime
equation
polynomial
Diophantine equation

Solve the following equation in integers (x, y, z):

$2x^4 + 2x^2y^2 + y^4 = z^2$

Does anyone have any ideas on this problem? It appears that the only possible solutions are when x = 0, but I am not sure about this. I have tried writing 2x^4 + 2x^2y^2 + y^4 = z^2 as

(x^2)^2 + (x^2 + y^2)^2 = z^2

and assuming that GCD(x, y) = 1. Assuming x,y are positive integers, y must be odd, looking at the equation (mod 4). Thus, 2mn = x^2,  m^2 - n^2 = x^2 + y^2,  and m^2 + n^2 = z for some relatively prime integers m and n. However, I have not been able to manipulate this any further. Can anyone see a solution based on this? (Maybe Pell's equation could be of help????)

[quote="ignorance"]I think that it's sufficient to show that left side lies between two successive square: (x^2+y^2+1)^2 and (x^2+y^2+2)^2 -> so the equation is impossible. Is this correct?[/quote]

No way.

well keep aplying the pitagorean triples thing and you ill arrive to another equation of the form (a 2 ) 2 +(a 2+b 2 ) 2 =c 2   with a,b smaller than x,y unless x=0

BTW it is honk kong 2004 problem 3 i think

pascual2004, i cant see how your method works - because we get there exists a,b,c s.t.  a^2 - b^2 - 2ab= y^2 and 2ab=x^2. how do we get another pythagorean triple out of this? (BTW cauchyguy you seem to have ignored the case in which x, y and z are all odd, but this is trivial mod 16).

I hope this will be a solution. first we cancel any common divisor of all of the x,y,z and if x,y have a common divisor, then z is divisible by that one too, so x,y have no common divisor. x,y cannot both be odd, as vinoth recognized(it even suffices to look at it mod 8 I think :) ), but y must be odd(if it was even, then x had to be odd, but then the left side is 2 mod 8, so it is no square). thus x is even. let x,y,z wlog all be positive.
Thus, we have a pythagorean triple $(x^2+y^2, x^2, z)$, with $x^2+y^2$ odd and $x^2$ even, so $x^2+y^2= m^2-n^2$, $x^2=2mn$. now, m and n have no common divisor since x and y have none. thus, m and n must be squares($x^2=2mn$) up to one 2 which is in one of them. since $x^2+y^2$ is 1 mod 4, we must have m odd, n even:
$m=p^2$, $n=2q^2$. then $y^2=m^2-n^2-2mn=p^4-4q^4-4p^2q^2$. this is equivalent to $(p^2+2q^2)^2+y^2=2(p^2)^2$. but all triples $(a,b,c)$ satisfying $a^2+b^2=2c^2$ are of the form $a=s^2+2st-t^2$,$b=-s^2+2st+t^2$,$c=s^2+t^2$ for some integers s,t. this is quite easy to prove(I leave it as an exercise :) ). thus $p^2=s^2+t^2$, $p^2+2q^2=s^2+2st-t^2$, that means $2q^2=s^2+2st-t^2-(s^2+t^2)=2t(s-t)$ or $q^2=t(s-t)$. since $p^2+2q^2$ and $p^2$ have no common divisor(since m and n have none), s and t have no common divisor, thus t and s-t have no common divisors, thus are squares: $t=a^2$, $s-t=b^2$. but $p^2=s^2+t^2=(s-t)^2+2st=(s-t)^2+2(s-t)t+2t^2=b^4+2a^2b^2+2b^4$, which is another solution to our original equation, but with smaller integers(some trivial estimates should suffice to show that I hope :) ). thus there is only a solution if $x=0$.

Peter

here are the trivial estimates: suppose $x>0$. take a solution with smallest z. then $z>x^2$. but $m<x^2$ and $0<=p<=m$, so $0<=p<=m<x^2<z$, so we get a solution with smaller(and nonnegative) z. Thus we must have $x=0$.

2 things. first it is korea 2004 problem 3.

now, that was wath i was saying one should do. I thing to prove the exercise you left you use the following: 2a 2 =b 2 +c 2  then 4a 2 =(b+c) 2 +(b-c) 2  and shows (locking mod 4 that b+c and b-c are both even) from were results follows aplying the pitagorean triples thing.

i would like to say that my proof is correct but the thing with $a^2+b^2=2c^2$ doesn't actually work(consider, e.g. the solution (1,1,1): we have $s^2+t^2=1$, so $s=0,t=1$ or the other way round, so $a=-1,b=1$ or the other way round). actually, it is not even true that ALL pythagorean triples are of the form $(m^2-n^2,2mn,m^2+n^2)$, only some of them are, including all relative prime triples. but one can help by saying that all relative prime solutions to $a^2+b^2=2c^2$ are either of the form $(s^2+2st-t^2,-s^2+2st+t^2,s^2+t^2)$ with one of s and t even and the other one odd, or of the same form with a 2 in the denominator and s, t odd. using that in our proof, we can finish it with a slight change:
we must consider the three equations $(x^2+y^2)^2+(x^2)^2=z^2$, $(2x^2+y^2)^2+(2x^2)^2=z^2$ and $(x^2+2y^2)^2+(x^2)^2=z^2$ at one time, choose the solution with smallest z to one of them and then do the infinite descend. i think you should be able to do it having read my argument above.


Draw a diagram of all of the forces acting on each object.

The vector sum of the all of the forces will give you the direction and magnitude of the force. From there, it is easy to see the solution.

Don't subtract acccelerations. Subtract forces. It is more systematic.

For any physics force problem:

1) Draw a diagram with all angles and forces.
2) State your givens.
3) Add all of the force vectors.

[quote="Phelpedo"]Body A weighs 102 N and is on an incline plane of 40 degrees. Body A is connected by a frictionless, massless pulley to body B, which is hanging vertically and has weight 32N. The coefficients of friction between A and the incline plane are $u_{s}=.56$ and $u_{k}= .25$. Find the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline.

In general, for problems like these, do I calculate the friction first and then calculate the acceleration of the system as if it were a frictionless plane, or do I calculate the acceleration of the system as if it were a frictionless and then apply friction?[/quote]
1. Draw a diagram of the system showing [u]all[/u] the forces acting on the system (including friction).

2. Using your diagram for reference write the equations of motion for the system.

3. Solve the equations of motion for the required acceleration.

Hint: Friction always acts parallel to surfaces in contact with each other, and is directed opposite the direction of motion.


If $ p$ is an odd prime $ >3$ prove that the product of the primitive roots mod $ p$ is congruent to $ 1$ mod $ p$

Let $ g$ be a primitive root.  ${ \prod_{(k, p-1) = 1} g^k = g^{ \sum(k, p-1) = 1} k}$, and we know that $ (k, p-1) = 1 \Leftrightarrow (p-1-k, p-1) = 1$, so the exponent is equal to $ \frac{(p-1) \varphi(p-1)}{2}$, and $ \varphi(p-1)$ is even for $ p > 3$.  The conclusion follows.

Or note that this is the constant coefficient of the $ (p\minus{}1)$-th cyclotomic polynomial.


I found this on the forum, but I didn't find a proof of it:
Let $A$ be a real matrix with $tr (A)=0$, prove that $A$ is similar to a matrix which has zeros on the main diagonal.

This problem is driving me crazy... I think it beats Iran '96 and "BMO 2006" by a long shot.

* slams head against the keyboard * (see spix's avatar)

The proofs:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=505708#505708]Proof 1.[/url] (pretty technical)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=219655&highlight=basis#219655]Proof 2.[/url] (nice, but doesn't deal with the complex case)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=443239&highlight=convex#443239]Proof 3.[/url] (for advanced users only, i.e. not for me)


let a, b, c and d be relatively prime to m = ad - bc.
prove that the pairs of integers (x, y) for which ax+by is divisible by m are identical with those for which cx+dy is divisible by m

consider some x,y, satisfying m divides ax+by. that is, (ad-bc) divides (ax+by), and therefore, (ad-bc) divides d(ax+by)-x(ad-bc), so (ad-bc) divides b(cx+dy), but since b is relatively prime to (ad-bc), (ad-bc) divides (cx+dy). a similar argument holds in reverse, and this shows that (m divides ax+by) iff (m divides cx+dy).

nice solution mathking.

here is another solution:

first we note that a and b are relatively prime. if they had a common divisor greater than 1, this divisor would divide m = ad - bc and a and m would not be relatively prime which is contrary to the hypothesis.

let for some (x, y)

ad-bc divides (ax + by)

=> ax + by = n (ad - bc)

=> a(x - nd) + b(y + nc) = 0

=> b(y+nc) = a(nd - x)

since a and b are relatively prime

y + nc = kb..............(1)
-x + nd = ka................(2)

we multiply (1) with x and (2) with y and then add the equations getting:

n(cx + dy) = k(ax + by)

                = k (ad - bc)n

therefore,

(cx + dy) = k(ad - bc).

by the same procedure, we can prove the converse.

hence proved  :)


geometry
3D geometry
linear algebra
matrix
linear algebra unsolved

Let $ A, B \in M_{n\times n}(\mathbb R)$ be matrices satisfying $ A^2\plus{}B^2\equal{}AB.$ Prove that if $ AB\minus{}BA$ is invertible then $ n$ is divided by 3.

Hint $ (A\plus{}jB)(A\plus{}j^2B)$

with $ j$ cube root of 1 in C

Amer. Math. Monthly problem 11196 is a similar one.  :)

1) following Moubi:
$ C\equal{}(A\plus{}jB)(A\plus{}j^2B)\equal{}j(BA\minus{}AB)$; $ det(C)\equal{}j^ndet(BA\minus{}AB)\geq{0}$. If $ det(AB\minus{}BA)\not\equal{}{0}$ then $ j^n$ is a real and $ 3$ is a divisor of $ n$.
2) Show that if $ A^2\plus{}B^2\equal{}AB$ then $ trace((AB\minus{}BA)^2)\equal{}0$; if moreover $ n\equal{}3$ then $ AB\minus{}BA$ is similar to $ a\begin{pmatrix}1&0&0\\0&0&\minus{}1\\0&1&\minus{}1\end{pmatrix}$ with $ a<0$.
3) If $ n\equal{}3$ then find explicitly $ A,B$ s.t. $ A^2\plus{}B^2\equal{}AB$ and $ AB\minus{}BA$ invertible.

For the point 3) (cf my post #4), there are solutions $ A,B\in\mathcal{M}_3(\mathbb{Z})$.


function
real analysis
real analysis unsolved

For differentiable function.

We consider one function
$ f: \left[ {a,b} \right] \to \mathbb{R},$
continuous on the space [a, b] and differentiable on the space (a, b) .Prove that for every function
$ g: \left[ {a,b} \right] \to \mathbb{R},$ differentiable on the space [a, b] there are
$ x_1 ,x_2 \in \left( {a,b} \right): f'\left( {x_1 } \right)f'\left( {x_2 } \right) \plus{} g'\left( {x_1 } \right)g'\left( {x_2 } \right)f\left( {x_1 } \right)f\left( {x_2 } \right) \equal{} 0,$
(Consider  $ f_1 \left( x \right) \equal{} e^{g\left( x \right)} f\left( x \right): \left[ {a,b} \right]..... \wedge f_2\left(x \right) \equal{} .....$)

S.E.Louridas

Are you sure that's true? Let's consider these functions:
$ f,g: [0,1]\rightarrow \mathbb{R}$, $ f(x) \equal{} g(x) \equal{} x$ for all $ x\in[0,1]$. Obviously, both $ f$ and $ g$ are differentiable and therefore continuous.

Now, $ f'(x) \equal{} g'(x) \equal{} 1$ for all $ x\in[0,1]$, so we have to find $ x_1$ and $ x_2$ such that $ 1\cdot 1 \plus{} 1\cdot 1\cdot x_1\cdot x_2 \equal{} 1 \plus{} x_1x_2 \equal{} 0$. However, since $ x_1,x_2\geq 0$, $ 1 \plus{} x_1x_2\geq 1 > 0$ and there are no such $ x_1$ and $ x_2$.

I may be doing something wrong here, but if I'm not, then we've got a problem :maybe:

YES ,there is and the doc. f(a)=f(b)=0.

S.E.Louridas

Oh well, using this extra condition, it's easy then, given your hint.

We will make use of the function $ f_1(x) \equal{} e^{g(x)}f(x)$. This function is differentiable on $ \langle a,b\rangle$ and continuous on $ [a,b]$ and f_1(a)=f_1(b)=0. Therefore, it satisfies the condition of Rolle's theorem and we conclude that there exists $ x_1\in\langle a,b\rangle$ such that $ f_1'(x_1) \equal{} 0$, which means $ e^{g(x_1)}f'(x_1) \plus{} f(x_1)g'(x_1)e^{g(x_1)} \equal{} 0$. Since $ e^t > 0$ for all $ t\in\mathbb{R}$, we therefore have $ f'(x_1) \equal{} \minus{} f(x_1)g'(x_1)$.

Similarly, we use the function $ f_2(x) \equal{} e^{ \minus{} g(x)}f(x)$. All the conditions for Rolle's theorem obviously hold for $ f_2$ too. So, there exists $ x_2\in\langle a,b\rangle$ such that $ f_2'(x_2) \equal{} 0$, that is $ e^{ \minus{} g(x_2)}f'(x_2) \minus{} e^{ \minus{} g(x_2)}g'(x_2)f(x_2) \equal{} 0$ or in other words $ f'(x_2) \equal{} g'(x_2)f(x_2)$.

Multiplying the two expressions we got, we have $ f'(x_1)f'(x_2) \equal{} \minus{} g'(x_1)g'(x_2)f(x_1)f(x_2)$, which is precisely what we were looking for.


superior algebra
superior algebra theorems

i have started a course on abstract algebra. We will be covering
groups,rings and fields.
could someone please suggest a rigorous book for the course?
thanks in advance :)

For my class, we use "Introduction to Abstract Algebra" by Nicholson.  It's a pretty good book, and it's meant for self-study as well as classroom use.  I prefer Dummit and Foote's book.  It's more on the level of a graduate level book but it's very, very good.  Lang's book is also pretty popular, but I know many people who absolutely hate it.

if you are starting abstract algebra in undergraduate level then artin's book might help you and herstien's book is also like a good hand book of abstract algebra


Prove that if $ a,b,c$ are nonnegative real numbers, we have
\[ \sqrt [3]{\frac {a(a \plus{} b)}{a^2 \plus{} 2b^2}} \plus{} \sqrt [3]{\frac {b(b \plus{} c)}{b^2 \plus{} 2c^2}} \plus{} \sqrt [3]{\frac {c(c \plus{} a)}{c^2 \plus{} 2a^2}}\ge 3\sqrt [3]{\frac {2abc}{(a^3 \plus{} b^3 \plus{} c^3)}}
\]

[quote="manlio"]Prove that if $ a,b,c$ are nonnegative real numbers, we have
\[ \sqrt [3]{\frac {a(a \plus{} b)}{a^2 \plus{} 2b^2}} \plus{} \sqrt [3]{\frac {b(b \plus{} c)}{b^2 \plus{} 2c^2}} \plus{} \sqrt [3]{\frac {c(c \plus{} a)}{c^2 \plus{} 2a^2}}\ge 3\sqrt [3]{\frac {2abc}{(a \plus{} b \plus{} c)}}
\]
[/quote]
The problem must be
\[ \sqrt [3]{\frac {a(a \plus{} b)}{a^2 \plus{} 2b^2}} \plus{} \sqrt [3]{\frac {b(b \plus{} c)}{b^2 \plus{} 2c^2}} \plus{} \sqrt [3]{\frac {c(c \plus{} a)}{c^2 \plus{} 2a^2}}\ge 3\sqrt [3]{\frac {2abc}{a^3 \plus{} b^3 \plus{} c^3}}
\]
Manlio :)

Thank you very much. I have corrected it  :)

By Am-Gm ,we need to prove :
$ (a^3\plus{}b^3\plus{}c^3)^3 \ge abc( a^2\plus{}b^2\plus{}c^2)^3$
It can prove by Am-Gm :D .

[quote="mitdac123"]By Am-Gm ,we need to prove :
$ (a^3 \plus{} b^3 \plus{} c^3)^3 \ge abc( a^2 \plus{} b^2 \plus{} c^2)^3$
It can prove by Am-Gm :D .[/quote]

Can you please post your solution?

Thank you very much.


find $ f$ on $ R$ such that:
$ (\forall(x,y)\in R^2)yf(x)\minus{}xf(2y)\equal{}8xy(x^2\minus{}y^2)$

for y=1 ,$ f(x)\equal{}8x^3\plus{}bx$ ,try to search the constant $ b$

[quote="drapt@"]for y=1 ,$ f(x) \equal{} 8x^3 \plus{} bx$ ,try to search the constant $ b$[/quote]

Which leads to no solution possible.

The original problem was certainly $ yf(x)\minus{}xf(y)\equal{}8xy(x^2\minus{}y^2)$ which gives (with drapt@ method) $ f(x)\equal{}8x^3\plus{}ax$  and which may be solved also by :
$ yf(x)\minus{}xf(y)\equal{}8xy(x^2\minus{}y^2)$ $ \Leftrightarrow$ $ \frac{f(x)\minus{}8x^3}{x}\equal{}\frac{f(y)\minus{}8y^3}{y}\equal{}a$ $ \forall x\neq0,y\neq 0$

1) for y=1 then f(x) = 8x^3 +bx (1)

2) For y=x then f(x) = f(2x) (2)

So , 8x^3 + bx = 64x^3 + 2bx

      56x^3 + bx = 0 => 56 = 0 . So we have no any f(x) for this

According to my solution there is no $ f(x)$ exist for which the given condition is satisfied.
My solution is following
given equation is $ yf(x) \minus{} xf(2y) \equal{} 8xy(x^2\minus{}y^2)$
by substituting x= y
we get the following
$ x(f(x) \minus{} f(2x))\equal{}0$
which says that $ f(x) \equal{} f(2x)$
and hence using this result we substitute this equation in the main equation
$ yf(x) \minus{} xf(y) \equal{} 8xy(x^2\minus{}y^2)$
and then substitue y=2x
$ 2xf(x) \minus{} xf(2x) \equal{} 8x(2x)(x^2\minus{}(2x)^2)$
using $ f(2x) \equal{} f(x)$
we can say that
$ f(x) \equal{} \minus{}48x^3$
but the result is contradicting $ f(2x) \equal{} f(x)$
hence we can say that no $ f(x)$ exist...


A 15-g bullet is fired from a rifle.  It takes 2.50 x $ 10^{\minus{}3}$ s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m/s.  Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

--

My free body diagram:

 [geogebra]c1b40b6e195f451668b59a51ff0cf38758d460cd[/geogebra]

Is the above correct?  If so, how do I use it to solve the problem?  Thanks!

You don't really need a free-body diagram to solve this problem.  All you need is a statement of the second law, and the definition of acceleration.  It's fair to assume that the bullet started from rest, and you're given that it has a final speed of 715 m/s.  Hence
$ a\equal{}\frac{715}{2.5\times 10^{\minus{}3}}\equal{}286000$ (in meters per second squared).
You know the mass, which is .015 kg, so the net force is just $ m\vec{a}\equal{}4290$ Newtons.


MATHCOUNTS AMC USA(J)MO USAMO AIME geometry

If you're intested in easier math chat, please sign-up here.

Why do people have to "sign up" for an AIM-based math chatroom?  Can't they just send you an instant message at the appropriate time?

It makes easier to organize on who's on and who's not

Like some basic algebras plus further geometry...

I'll try it but I dunno if I'll be able to follow it.

The difficulty of the math chats has varied the times that I have went.  At first it was mathcounts-amc.  Then that chat elevated to aime.  I'm ok with aime(the first few at least).  Then this last one was pretty much crazy.  I know that we had a lot more experienced people but it still was too hard for what it was intended to be.  I mean there were USAMO and aime #15's etc.  So next time we will have to know who are going and what levels they are at ahead of time.

Most of the chats ive been to had a few at mathcounts to amc level, a few at amc to aime level, and some at above that.  So i suggest staying at the level of first few aime and below.

I think that we should also do a better job of putting together problems ahead of time.  I'll help with that if you need me to.

Yes, it was quite amazing than fun..

For the easy math chat, I'll take charge on that. I'll be instructor and we'll began to actually have classlike math chat than competition prepare.  Like from my experience, some people are seem to good at competition but actually isn't good competition person.  The reason, I found, was that they seemed to know the cover of math, not the inside.

I'll be really happy if joml88 can attend in easy chat and be co-coach.  I'm pretty good at Algebra II because it's the class I took this year but further geometry is out of my hand so.. joml88, how about you pm or aim me with kind of questions in basic mathematical areas? Then we can sort them and like teach other people about it...


geometry
parallelogram
geometry unsolved

A chord $XY$ of a circle is perpendicular to its diameter $BC$. Points $P,M$ are taken on $XY,CY$ so that $CY\parallel PB$ and $CX\parallel MP$. The lines $CX,PB$ intersect at $K$. Show that $PB$ is perpendicular to $MK$.

$CYX$ is isosceles and $MP \parallel CX$, then $MY = MP$.

$CMPK$ is a parallelogram, then $MP = CK$.

$KB \parallel XY$, then $\angle KBC = \angle YCB = \angle BCK$, then CKB is isosceles, then $CK = BK$.

Finally, $MY = BK$, but $MY \parallel BK$, then $MYBK$ is a parallelogram. But $\angle MYB = 90^{\circ}$, then $MYBK$ is a reactangle, and the thesis follows.

Nice  :wink:


Prove that
$ (a\plus{}b)(a^2\plus{}b^2)(a^3\plus{}b^3) \le 4(a^6\plus{}b^6)$
 :lol:

By Muirhead or rearrangement it is obvious:
The inequality is equivalent to $ [6,0] \plus{} [3,3] \plus{} [4,2] \plus{} [5,1]\leq 4[6,0]$ which is trivial... :wink:

[quote="Attila the Hun"][quote="TRAN THAI HUNG"][quote="SHIRO_DHKT"]Prove that
$ (a \plus{} b)(a^2 \plus{} b^2)(a^3 \plus{} b^3) \le 4(a^6 \plus{} b^6)$ :)
 :lol:[/quote]
$ (a \plus{} b)(a^2 \plus{} b^2) \leq 2(a^3 \plus{} b^3)$
And $ 2(a^3 \plus{} b^3)^2\leq 4(a^6 \plus{} b^6)$
Thus, $ (a \plus{} b)(a^2 \plus{} b^2)(a^3 \plus{} b^3) \le 4(a^6 \plus{} b^6)$[/quote]

I don't understand how you got the second inequality. can you please elaborate?[/quote]
Cauchy:
$ (1\plus{}1)(a^6\plus{}b^6)\ge (a^3\plus{}b^3)^2$

[quote="Kyoshiro"][/quote][quote="Attila the Hun"][quote="TRAN THAI HUNG"][quote="SHIRO_DHKT"]Prove that
$ (a + b)(a^2 + b^2)(a^3 + b^3) \le 4(a^6 + b^6)$ :)
 :lol:[/quote]
$ (a + b)(a^2 + b^2) \leq 2(a^3 + b^3)$
And $ 2(a^3 + b^3)^2\leq 4(a^6 + b^6)$
Thus, $ (a + b)(a^2 + b^2)(a^3 + b^3) \le 4(a^6 + b^6)$[/quote]

I don't understand how you got the second inequality. can you please elaborate?[/quote][quote="Kyoshiro"]
Cauchy:
$ (1 + 1)(a^6 + b^6)\ge (a^3 + b^3)^2$[/quote]

Or dividing by 8 we have the power mean inequality $ \sqrt[6]{\frac{a^{6}+b^{6}}{2}}\geq \sqrt[3]{\frac{a^{3}+b^{3}}{2}}$ :wink:

Power Mean throughout:

[hide]
By Power Mean Inequality, the following are true:

$ \left(\frac{a^6\plus{}b^6}{2} \right)^{\frac{1}{6}} \ge \left(\frac{a\plus{}b}{2} \right)$

$ \left(\frac{a^6\plus{}b^6}{2} \right)^{\frac{2}{6}} \ge \left(\frac{a^2\plus{}b^2}{2} \right)$

$ \left(\frac{a^6\plus{}b^6}{2} \right)^{\frac{3}{6}} \ge \left(\frac{a^3\plus{}b^3}{2} \right)$

Multiplying them gives the desired inequality[/hide]


abstract algebra
algebra
function
domain
group theory
superior algebra
superior algebra unsolved

I need  :help:
I must prove that group homomorpism from Integers to Unit circle of Komplex numbers is 1)one-to-one and 2) not onto.
I understand that I must use Ker(f)=<2 Pi integers> to prove one-to-one.
But I cannot write all this in beatiful mathematic language. :blush:

I will be very grateful for complete answer. :oops:

Best regards
Tanja

[quote="fredbel6"]I do not understand.  What is the homomorphism you speak about?[/quote]

group homomorphism
f : Z--> S^1 , where
Z- integers
S^1 - unit circle group
such as f(k)=e^(ik), where k  is integers, i =sqrt(-1)
 :blush:

Since $2\pi$ is irrational, the kernel is trivial. There are uncountably many elements in the codomain and countably many in the domain, so the function cannot be surjective.
Also, we can find specific values such as $e^{\frac i2}$ that are not achieved.

I read  :read: your answer.
Thank you very much for help. : :lol:

Best regards
Tanja


If $ a,b,c$ are nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$, then

$ \frac 1{4a^2\plus{}b^2\plus{}c^2}\plus{}\frac 1{4b^2\plus{}c^2\plus{}a^2}\plus{}\frac 1{4c^2\plus{}a^2\plus{}b^2}\le \frac 1{2}$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac 1{4a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{4b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{4c^2 \plus{} a^2 \plus{} b^2}\le \frac 1{2}$.[/quote]
Turned out easy:
$ \frac 1{4a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{4b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{4c^2 \plus{} a^2 \plus{} b^2}\le \frac 1{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(a^6\minus{}4a^5b\plus{}13a^4b^2\minus{}2a^4bc\minus{}6a^3b^3\minus{}12a^3b^2c\plus{}10a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2(2c^4\plus{}2(a^2\minus{}4ab\plus{}b^2)c^2\plus{}a^4\minus{}2a^3b\plus{}4a^2b^2\minus{}2ab^3\plus{}b^4)\geq0,$
which true because $ (a^2\minus{}4ab\plus{}b^2)^2\minus{}2(a^4\minus{}2a^3b\plus{}4a^2b^2\minus{}2ab^3\plus{}b^4)\equal{}$
$ \equal{}\minus{}(a\minus{}b)^2(a^2\plus{}6ab\plus{}b^2)\leq0.$

Nice, Arqady.  :)
 But I think there is a nice solution without expansion, too. :o  :!:

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac 1{4a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{4b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{4c^2 \plus{} a^2 \plus{} b^2}\le \frac 1{2}$.[/quote]
My solution:
Set $ s\equal{}ab\plus{}bc\plus{}ca,A\equal{}abc \Rightarrow s \le 3, A\le 1$
We have:
$ \sum \frac 1{4a^2 \plus{} b^2 \plus{} c^2}\equal{} \sum \frac{1}{9\plus{}3a^2\minus{}2s}\equal{}\frac{9(9\minus{}2s)^3\plus{}9(s^2\minus{}6A)}{7(9\minus{}2s)^3\plus{}9(9\minus{}2s)(s^2\minus{}6A)\plus{}27A^2}$
So we need to prove:
$ 7(9\minus{}2s)^3\plus{}9(9\minus{}2s)(s^2\minus{}6A)\plus{}27A^2 \ge 2(9(9\minus{}2s)^3\plus{}9(s^2\minus{}6A))$
$ \Leftrightarrow F(A)\equal{}27A^2\minus{}54(7\minus{}2s)A\plus{}(\minus{}50s^3\plus{}423s^2\minus{}126s\plus{}1458) \ge 0$
$ \Delta'_{A}\equal{}50s^3\minus{}419s^2\plus{}1266s\minus{}1409<0$ (because $ 0<s \le 3$)
So $ F(A) \ge 0$ :oops:

[quote="NguyenDungTN"] $ \Leftrightarrow F(A) \equal{} 27A^2 \minus{} 54(7 \minus{} 2s)A \plus{} ( \minus{} 50s^3 \plus{} 423s^2 \minus{} 126s \plus{} 1458) \ge 0$
$ \Delta'_{A} \equal{} 50s^3 \minus{} 419s^2 \plus{} 1266s \minus{} 1409 < 0$ (because $ 0 < s \le 3$)
So $ F(A) \ge 0$ :oops:[/quote]
If $ \Delta'_{A}<0$, then $ F(A)>0$, which is is wrong because $ F(A)\equal{}0$ for $ a\equal{}b\equal{}c\equal{}1$.  :?:

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac 1{4a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac 1{4b^2 \plus{} c^2 \plus{} a^2} \plus{} \frac 1{4c^2 \plus{} a^2 \plus{} b^2}\le \frac 1{2}$.[/quote]
$ \sum_{cycl}\frac {1}{4a^2 \plus{} b^2 \plus{} c^2}
\equal{}\frac{1}{9}\sum_{cycl}\frac {(a\plus{}b\plus{}c)^2}{4a^2 \plus{} b^2 \plus{} c^2}$
$ \le \frac{1}{9}\sum_{cycl}\left( \frac{b^2}{a^2\plus{}b^2}\plus{}\frac{c^2}{a^2\plus{}c^2}\plus{}\frac{a^2}{2a^2} \right)\equal{}\frac{1}{2}$

Indeed, Ductrung, this is the nice solution. :lol:

The following generalization is true.

If $ a_1,a_2,...,a_n$ are nonnegative numbers such that $ a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2 \equal{} n$, then

$ \sum \frac 1{a_1^2 \plus{} 1} \le \frac {n^3}{2(\sum a_1)^2}$.

[quote="Vasc"]The following generalization is true.

If $ a_1,a_2,...,a_n$ are nonnegative numbers such that $ a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2 \equal{} n$, then

$ \sum \frac 1{a_1^2 \plus{} 1} \le \frac {n^3}{2(\sum a_1)^2}$.[/quote]
Indeed, by the same idea,
$ \frac {a_1^2}{2a_1^2} \plus{} \frac {a_2^2}{a_1^2 \plus{} a_2^2} \plus{} \cdots \plus{} \frac {a_n^2}{a_1^2 \plus{} a_n^2}\ge\frac {(a_1 \plus{} \cdots \plus{} a_n)^2}{na_1^2 \plus{} (a_1^2 \plus{} \cdots \plus{} a_n^2)}$
Adding this and similar inequalities, the result follows.

Variation on the same theme: If $ a_1,a_2,...,a_n$ are nonnegative numbers such that $ a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2 \equal{} n$, then
\[ \sum_{k\equal{}1}^n\frac{1}{a_k^2\plus{}n\minus{}1}\le\frac{n^2}{(a_1\plus{}\cdots\plus{}a_n)^2} \]


If real numbers $x,y,z$ satisfy $x^2+y^2+z^2=4x+6y+12z-49$, find the value of $xyz$.

[hide]
You can complete the square a bunch of times to get $(x-2)^2+(y-3)^2+(z-6)^2=-49+4+9+36$ where the RHS just happens to be $0$.
So real numbers that satisfy this are $(x,y,z)=(2,3,6)$ and the product is $36$[/hide]


The Newport Crab Company marks the location of its crab pots
with colored balloons. They mark every fourth pot with a red
balloon, every sixth pot with a blue balloon, and every tenth pot
with a yellow balloon. After placing 600 crab pots in the bay,
how many pots have three different colored balloons attached?

The Least Common Multiple of 4, 6, and 10 is 60, so our answer is $ 600 \div 60 \equal{} 10$.

[quote="markoboy"]ill answer this in a series of steps

[hide="this is my explanation"]1.we know the follwing facts

 I.They mark every fourth pot with a red
balloon

 II.They mark every sixth pot with a blue
balloon

 III.they mark every tenth pot with a yellow
balloon

2.this is a LCM question so we can just find the LCM for $ 4,6,10 $

3.we get 10

4.10[/hide]

[hide="."][size=200][color=#FF80FF]PLEASE RATE MY POST THANK YOU SO MUCH

 [size=50][color=#BF8000][b][i][u]please be fair when you're rating my post and please PM me if there is a better way to explain this thank you so much !!!\[/u][/i][/b][/color][/size][/color][/size][/hide][/quote]
You realize that you basically just repeated exactly what Izzy said?
There's a tip that sometimes pops up before you post something:
Tip: Check if someone else has already posted something similar


I use Windows on our laptop because my parents can't use linux/FreeBSD. But, on my desktop which no one else uses, FreeBSD FTW!

Also, if you don't know what FreeBSD is, go to: [url=freebsd.org]here[/url]
I highly highly recommend it to any experienced linux user, if you're not so experienced, go [url=ubuntu.com]here[/url]

[quote="Poincare"]I use Windows on our laptop because my parents can't use linux/FreeBSD. But, on my desktop which no one else uses, FreeBSD FTW!

Also, if you don't know what FreeBSD is, go to: [url=http://freebsd.org]here[/url]
I highly highly recommend it to any experienced linux user, if you're not so experienced, go [url=http://ubuntu.com]here[/url][/quote]

Windows Vista.  :|

What if I use more than one? I dual-boot Windows and Kubuntu (well, I'm trying out Fedora, but...) on my computer.

[quote="Elixir10"]I use Vista because it came with my laptop. It isn't bad.

Re double-booting: I keep reading stuff about why it isn't recommended, though my dad does it all the time anyway.[/quote]

The only reason I wouldn't recommend double booting is because it takes up space. If it's for OEM concerns, by all means. As long as you have an OS reinstall disk if the whole thing goes wonky.

XP.

I know it sounds crazy, but I'd rather have Vista.  The graphics seem cool, it has a lot of good features, and I'd simply install a good anti virus software to get rid of the bugs.

I'm the only one who uses a Mac OS X so far?  :o

Windows XP, because my uncle works for Microsoft, so he just let us copy the OS.

Speaking of operating systems, last Friday my computer programming teacher asked us whether we thought PC's or Macs were better. About half the class instantly responded "Mac."

I go to school in Redmond, Washington. :wink:

Mac is based on FreeBSD so it has a REALLY good core and its pretty too! So, if it was free, i'd definitely have it. But its dumbed down too, no GCC, no perl and you can't customize the GUI and no open source so :(

Windows XP,
my parents keep on buying different monitors;
yet we keep the same Hard Drive.

[quote="evilhamster"]I go to school in Redmond, Washington. :wink:[/quote]

Wasn't it once said, "a software developer is without honour in his own city"?

Linux has gotten a lot easier, sometimes maybe easier than windows. An example for this is software installation. On Windows, you have to hunt down a common software you want, but on a polished distribution of Linux, you just have to open the package manager GUI and search for it. It is usually a one-click install after that.


inequalities
trigonometry
inequalities unsolved

[color=blue][b]May this problem not have any solution?????????????????????????[/b][/color]

[quote="quocbao153"][color=blue][b]With $ \forall \Delta ABC$. Prove that:

$ \sqrt {\frac {{\cos A\cos B}}{{\cos \frac {A}{2}\cos \frac {B}{2}}}} \plus{} \sqrt {\frac {{\cos A\cos C}}{{\cos \frac {A}{2}\cos \frac {C}{2}}}} \plus{} \sqrt {\frac {{\cos C\cos B}}{{\cos \frac {C}{2}\cos \frac {B}{2}}}} \le \frac {2}{{\sqrt 3 }}\left( {\sin \frac {A}{2}\sin \frac {B}{2} \plus{} \sin \frac {A}{2}\sin \frac {C}{2} \plus{} \sin \frac {C}{2}\sin \frac {B}{2}} \right) \plus{} \frac {{\sqrt 3 }}{2}$[/b][/color][/quote]
The triangle must be acuted.
By AM-GM Inequality, we have that
\[ 2\sqrt {\frac {\cos A\cos B}{\cos \frac {A}{2} \cos \frac {B}{2}}} \equal{} 2\sqrt {4\sin \frac {A}{2}\sin \frac {B}{2}\cdot \cot A\cot B} \le \frac {4}{\sqrt {3}} \sin \frac {A}{2}\sin \frac {B}{2} \plus{} \sqrt {3} \cot A\cot B
\]
Applying this to the similar terms and adding up with notice that $ \cot A\cot B \plus{} \cot B\cot C \plus{} \cot C \cot A \equal{} 1$, we get the result.


symmetry
geometry
parallelogram
geometry unsolved

Two circles C and C' intersect at points A and B. A straight line passing through B intersects the circles C and C' respectively at M and N, with B between M and N. Let I be the midpoint of the arc AM of circle C not containing B, and J the midpoint of the arc AN of circle C' not containing B. If K is the middle of MN, prove that KI is perpendicular to KJ.

Denote I’ is the symmetry point w.r.t K
We have I’NIM is a parallelogram
∠IAJ=360-∠IAB-∠JAB=180-∠IAB+180-∠JAB=∠MNI’+∠MNJ=∠I’NJ
And we have MI=IA=I’N   NJ=JA
△IAJ≌△I’NJ   =>  JI=JI’ and K is the midpoint of II’ =>IK⊥JK

I found a problem to which this one is a particuliar case, in a short list of IMO 2002 :

http://www.mathlinks.ro/resources.php?c=1&cid=17&year=2002

 problem nr 8 of geometry, aldo available in the book "IMO compendium".


How do you convert fractions in different bases?

For example, what is $ \frac{3}{4}$ in base $ 8$?
and what is $ \frac{3}{4}_{8}\minus{}3/4 in base 8\minus{}$in base $ 10$?

Divide in the appropriate base.  It's exactly the same algorithm as dividing in base $ 10$, but... in base $ 8$.

In this case, $ \frac{3}{4}$ has a very simple terminating base expansion:  it is $ 0.6_{8} \equal{} \frac{6}{8}$.

just like in base 10, the place values are:

10, 1, 1/10, 1/100, so

8, 1, 1/8, 1/64

because 4 and 8 are so closely related, 3/4 to base 8 is easy, but if you wanted to do something slightly more complicated, like 2/5...

2/5=3/8+1/64+...

I'll try to come up with a more neat way to convert.

I'm sorry for not knowing any of this, but how do you convert bases in the first place?

its fine.  so, take a decimal (that means base 10 number) say, 23.  There's a 1's place and a 10's place.  2*10+3*1.  It's divided into parts that are divisible by 10 and the part it's not.  Its the same with other bases.  45[size=75]8[/size]=4*8+5*1 In fact, that's how you convert from one base into base 10. Just write it as the sum of its parts.

Converting from base 10 to another base I'll show the long way first.  23 base 10 into base 8.  23=2*8+7=27 base 8.

The short way[hide] 23/8=2 R7.  The 7 is the part that belongs to the column with no powers of 8 divisors. The 2 belongs to the column with one power of 8 as a divisor, so 23=27 base 8.  That's hard to see with a small example, so i'll do 105.

105=13*8 R 1
13=1*8+5

So 1*64+5*8+1*1
105= 151 base 8[/hide]

[quote="allyrb"]Converting from base 10 to another base I'll show the long way first.  23 base 10 into base 8.  23=2*8+7=27 base 8.
[/quote]

Thanks a lot for helping, but I don't get one part. Wouldn't 23 in base 8 be

23=2*8+3*1=19?

It depends on what you mean by "$ 23$ in base $ 8$."  allyrb means that

$ 23_{10} \equal{} 2 \cdot 10 \plus{} 3 \cdot 1 \equal{} 2 \cdot 8 \plus{} 7 \cdot 1 \equal{} 27_{8}$

whereas you mean that

$ 23_{8} \equal{} 2 \cdot 8 \plus{} 3 \cdot 1 \equal{} 19_{10}$.

Both calculations are correct, but without the subscripts it's not clear what "$ 23$" means.


These are 3 problems in which the inclusion-exclusion principle can be applied. I don't know the correct answers for them, but I tried to do them and would like to see whether my answers are correct.

1.  Each of a class of 60 students studies at least one of mathematics and English. If 40 study mathematics and 37 study both, how many study English?

2.  How many integers from 1 to 1000 are divisible by none of 7, 11 or 13?

3.  A survey carried out over a large number of pop music fans reveals that 95% of them like at least one of Madonna, Sting and Elvis. A total of 53% like Madonna, 42% like Sting and 43% like Elvis. If 18% like only Elvis, and 7% like all three, what percentage like Madonna and Sting but not Elvis?

you could draw a venn diagram for 1 and 3

for number 2, you could find all divisible by 7,11 and/or 13, and subtract that from 1000

[quote="Altheman"]for number 2, you could find all divisible by 7,11 and/or 13, and subtract that from 1000[/quote]
I don't think that would work, because some are divisible by two or more of 7, 11 and 13, and then you would be subtracting them twice or three times ;)

[quote="Melissa"]2.  How many integers from 1 to 1000 are divisible by none of 7, 11 or 13?[/quote]

[hide]$\lfloor 1000/7 \rfloor = 142$ multiples of 7 less than or equal to 1000.
$\lfloor 1000/11 \rfloor = 90$ multiples of 11 less than or equal to 1000.
$\lfloor 1000/13 \rfloor = 76$ multiples of 13 less than or equal to 1000.

$\lfloor 1000/77 \rfloor = 12$ multiples of 77 less than or equal to 1000.
$\lfloor 1000/91 \rfloor = 10$ multiples of 91 less than or equal to 1000.
$\lfloor 1000/143 \rfloor = 6$ multiples of 143 less than or equal to 1000.

$\lfloor 1000/1001 \rfloor = 0$ multiples of 1001 less than or equal to 1000.

Our answer is $x$ where $1000=(142+90+76)-(12+10+6)+(0)+x$, so $x=664$.[/hide]

[quote="chess64"]Our answer is $x$ where $1000=(142+90+76)-(12+10+6)+(0)+x$, so $x=664$.[/quote]
I also got to this, except that I got $x=720$. I think you just made a calculation error ;)

Woops, you're right; don't know what happened there :blush:


function
geometry
geometric transformation
algebra
polynomial
Support
Ring Theory

Let $ V$ be an afine algebraic irreducible variety of dimension at least $ 2$ over an algebraically closed field and let $ P$ be a normal point on $ V$. Assume that $ f$ is a regular function on $ V\setminus\{P\}$. Prove that $ f$ extends uniquely to a regular function on the whole of $ V$.

Algebraic translation:
Let $ k$ be an algebraically closed field. Let $ A \equal{} k[X_{1},\ldots,X_{n}]/I \equal{} k[x_{1},\ldots,x_{n}]$ where $ I$ is a prime ideal of $ k[X_{1},\ldots,X_{n}]$. Assume that the Krull dimension of $ A$ is at least $ 2$. Let $ K$ be the quotient field of $ A$.
Assume that all the polynomials in $ I$ vanish in $ (0,\ldots,0)$ and that $ A_{(}x_{1},\ldots,x_{n})$ is integrally closed.
Then $ A \equal{} \cap_{m\neq (x_{1},\ldots,x_{n})}A_{m}$, where the intersection is taken over all the maximal ideals of $ A$ and $ A_{m}$ denotes the localization of $ A$ with respect to the multiplicatively closed system $ A\setminus m$. The intersection is taken in $ K$.

This follows from normal => $ (S_2)$ and the local cohomology/cohomology with supports exact sequence.


I didn't do state, but I got 40 unofficially at home. Anyone know the cutoff?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1462150#1462150[/url].

Tied for fourth.

yeah, you probably would have made nats even if you didnt get the tie breaker because some of the ppl were rather unfast at cd but maybe they were nervous.


Find all $ x,y,z\in\mathbb{N}_{0}$ such that $ 7^{x} \plus{} 1 \equal{} 3^{y} \plus{} 5^{z}$.

[i]Alternative formulation:[/i] Solve the equation $ 1\plus{}7^{x}\equal{}3^{y}\plus{}5^{z}$ in nonnegative integers $ x$, $ y$, $ z$.

Consider by mod (8) give $x=y=z(mod \ 2)$.
Obviosly solutions $x=y=z=0$ and $x=y=z=1$.
I think there are not another solutions.

I don't understand you, @Rust.
Who can help me?

Let  $y=0.$ Then $7^x=5^z\to x=z=0$.
Let $y=1$. Then $7^x=5^z+2.$ Therefore $z\ge 1, x=1\mod 4$ (consider by mod 5).
If $z=1$, then $x=1$. If $z>1$, then $7^4=1\mod 25$, therefore $7^x=7\mod 25$ and $ 7\not =2\mod 25.$
There are not solutions, suth that $z>1$.
Now consider $y>1$. Must be $7^x+1=3^y+5^z\mod 9, x=1\mod 4$.
Because $7^x+1=2\mod 3\to z -odd \to z=3\mod 6$.
Consider by mod 7 $1=3^y-1\mod 7\to y=2\mod 6$.
If $y=2\mod 6, z=3\mod 6$ $3^y+5^z=6\mod 8$. But never $7^x+1=(-1)^x+1\not =6\mod 8$.
Therefore only $x=y=z=0$ and $x=y=z=1$ are solutions.

We wil show that the only solutions are $(0,0,0)$ and $(1,1,1)$. I can follow the proof of Rust up to the point
where he concludes that z=3 mod 6, which I don't follow. I think he wants to use $x\equiv 1(4)\Rightarrow
7^x\equiv 7(9)$ but that is not true, because the order of 7 modulo 9 equals 3 and not 4.
Hence we will follow a slightly different method.
As in his proof another solution has $x,y,z>0$ and $y>1$, which we assume.
Modulo 3 one has $2\equiv 5^z$, hence $z$ is odd.
Modulo 4 one has $(-1)^x+1\equiv (-1)^y+1$, hence $x\equiv y(2)$.
Modulo 5 one has $2^x+1\equiv(-2)^y$. If $x,y$ are even with $x=2a,y=2b$, we have $(-1)^a+1\equiv(-1)^b$.
Since $\{0,2\}\cap \{-1,1\}=\emptyset$ this is a contradiction, hence $x,y$ are odd.
Reading again $2^x+1=3^y$ and noting that $(\{2,-2\}+1)\cap \{3,-3\}=\{3\}$ , we find $x\equiv y\equiv 1(4)$.
Modulo 16 we have $7^2\equiv 1,3^4\equiv 1$, hence $7^x\equiv 7,3^y\equiv 3\Rightarrow 5^z\equiv 5\Rightarrow
z\equiv 1(4)$. Hence $x,y,z\equiv 1(4)$.

We now combine $3^y+5^z\equiv 1(7)$ with $7^x+1\equiv 5^z(9)$ (here we use $y>1$) and consider the
possible residues of $x\mod 3$.
If $x\equiv 0(3)$ then $7^x\equiv 1(9)\Rightarrow 5^z\equiv 2(9)\iff z\equiv 5(6)\Rightarrow
5^z\equiv 3(7)\Rightarrow 3^y\equiv 5(7)\iff y\equiv 5(6)$, hence $(x,y,z)\equiv (9,5,5)\mod 12$.
But then we can read $7^x+1=3^y+5^z$ modulo 13 and obtain $7^9+1\equiv 3^5+5^5$, not true.
The other cases, $x\equiv 1,2(3)$ are similar and left to the reader.


Let $ a,b \in \mathbb{Z}$ and
\[ \frac {(2n)!}{n! \cdot (n \plus{} 1)! } \equal{} a \cdot \binom{2n \minus{} 1}{n} \plus{} b \cdot \binom{2n \plus{} 1}{n} .
\]
Find $ a \text{ and } b$ .

[hide]Write the binomial coefficients on the RHS using the $ \frac{n!}{(n\minus{}k)!k!}$ formula.  Get the one with coefficient $ a$ to the least common denominator $ n!(n\plus{}1)!$ and then multiply through by that since it's on all terms. Finally, cancel out a $ (2n\minus{}1)!$. Doing all of this gives
\[ 2n \equal{} a \cdot n \cdot (n\plus{}1) \plus{} b \cdot (2n) \cdot (2n\plus{}1)\]
\[ 2 \equal{} a \cdot (n\plus{}1) \plus{} b \cdot (4n\plus{}2)\]
We need to cancel $ n$s out, so $ a \equal{} \minus{}4b$.
\[ 2 \equal{} \minus{}4bn \minus{} 4b \plus{} 4bn \plus{} 2b\]
\[ 2 \equal{} \minus{}2b\]
So $ b \equal{} \minus{}1$ and thus $ a \equal{} 4$.[/hide]


calculus
integration
function
real analysis
algebra
partial fractions
calculus computations

I was able to get as far as x'(t)=2-x(t) and y'(t)=2x(t)+y(t) before my total lack of differential equations knowledge betrayed me. So I sought Mathematica's help:

$\begin{array}{l}x(t) = 2+Me^{-t}\\ y(t) = 2e^{t}(-2e^{-t}-\frac{M}{2}e^{-2t})+ce^{t}\\ \end{array}$

Didn't try plugging it back in though.
:)

With those convolutions, I'll go with Laplace transforms again:

$X(s)=\frac1s+\frac1{s+2}X(s)$

$Y(s)=\frac1{s+2}(2X(s)+3Y(s))$

This is two linear equations (already in triangular form) for the two unknowns $X$ and $Y.$ Solving and then using partial fractions we get:

$X=\frac{s+2}{s(s+1)}=\frac2s-\frac1{s+1}$

$Y=\frac{2s+4}{s(s+1)(s-1)}= \frac{-4}s+\frac1{s+1}+\frac3{s-1}$

These in turn can be read back through the transform to get

$x(t)=2-e^{-t}$

$y(t)=-4+e^{-t}+3e^{t}$

These do have $x(0)=1,y(0)=0,$ as they should.

The solution §outh§tar wrote needed to have initial conditions applied so as to evaluate $M$ and $c.$ Given that, it would have been correct.


The sum of the digits of a two-digit number is $ 12$. When the digits are reversed, the original number is increased by $ 54$. What is the product of the digits?

We have that $ x\plus{}y\equal{}12$ and $ 10x\plus{}y\plus{}54\equal{}10y\plus{}x$.  Therefore, $ 10x\plus{}y\plus{}54\plus{}10y\plus{}x\equal{}11(x\plus{}y)\plus{}54\equal{}11(12)\plus{}54\equal{}186\equal{}20y\plus{}2x\Rightarrow 93\equal{}10y\plus{}x$.  So $ y\equal{}9$ and $ x\equal{}3$ so the product is 27 and in the original number was 39.

Guess and check might be a faster way, just saying.


这个问题也许比较傻，但我确实?会 :( 。我已下载了DVI Viewer 的setup，但?会安装。说明书上的步骤为：
1.       Open a “Command Prompt? or “MS-DOS Prompt?.

2.       Create a temporary directory. For example,

   c:
   cd \
   md \instdvi
   cd \instdvi

3.       Click here and save dvi_view_setup.exe into the directory you created. It is about 2 MB in size.

4.       Go to the Command Prompt. Unpack the setup program by running it:

   dvi_view_setup

5.       Install it by running the setup file:

   setup

我?是很明白，尤其是第一步里“Command Prompt? or “MS-DOS Prompt?，我?试了，但跳出了一个对?框，说C:\winnt\system32\autoexec.nt 系统文件?适用于?行MS-DOS 与 Microsoft Windows 应用程?，怎么回事？
还有第四步怎么?作？

在哪里?以下载DVI Viewer 的setup？能告诉我?？网上此类文件还?少，就是没法看。

我奋斗了2?时, 总算?出了点??堂. 但似乎?能读出简体汉字, ?知?那里有相关的组件下载?

[quote="liyi"]你用cct还是CJK?

Visit http://www.ctex.org[/quote]

太感谢您为我??供了一?这么好的网?了? :D  :D  :D  :)  :)  :)

?过我目?还?懂cct或CJK。现正在研究。

用CJK之??调个包
\usepackage{CJK}

之?正文中在
\begin{CJK*}{GBK}{song}
\end{CJK*}
之间就能用中文了 :)


An observer is standing next to a lightning rod that's in an inertial reference frame at rest with respect to a barn 9.3 * 10^8 meters away. The observer sees lighting strike the vane 4.00 seconds before it strikes the rod. Find the proper time between the events.

I don't quite understand your question. Where does the vane enter the problem? Is it on the lightning rod or is it at the barn?

It's on the barn. I said the light takes $\frac{9.3*10^8 m}{3.00*10^8 m/s}$ to reach you, so just subtract that from 4.00 seconds, but that's not right. :(

How's this:

[code]Barn---------------Clock---------------Rod
     4.65 * 10^8 m       4.65 * 10^8 m[/code]

Let's say lightning strikes the barn at $t=0s$.  It takes the
light $\frac{4.65\cdot10^8m}{3.00\cdot10^8m/s}=1.55s$ to travel to the clock, so
the clock records the strike at $t=1.55s$.  Meanwhile, the
light takes $\frac{9.30\cdot10^8m}{3.00\cdot10^8m/s}=3.10s$ to reach the observer,
so the observer records the event at $t=3.10s$.  Four seconds
later, at $t=7.10s$, the lightning strikes the rod.  This light also
takes $\frac{4.65\cdot10^8m}{3.00\cdot10^8m/s}=1.55s$ to travel to the clock, so
the clock records this strike at $t=8.65s$.  Subtracting the time
between the events, we have an unbiased time interval of $8.65s-1.55s=7.10s$

Note that this is the sum of the rod's time interval and the traveling time
from the barn to the rod.

(Not that I'm sure that's correct, but it sounds good.)

yea... i think you just add 4 from what townsfreak had... except he subtracted it...should be somethign like $7.15$?... i dont have my calculator, so i am estimating as i go...

EDIT... $7.10$. I didnt see inteluser's answer...do'h ;)


Let $ a, b, c$ be distinct real numbers. Prove or disprove that

$ \sum[\frac{a(3b \plus{} 3c \minus{} 2a)}{(b \minus{} c)^2}]\ge 4$ .

__________
 :roll:

[quote="Ligouras"]Let $ a, b, c$ be distinct real numbers. Prove or disprove that

$ \sum[\frac {a(3b \plus{} 3c \minus{} 2a)}{(b \minus{} c)^2}]\ge 4$ .

__________
 :roll:[/quote]
Try $ a\equal{}2,$ $ b\equal{}1$ and $ c\rightarrow\plus{}\infty.$

Perfect [b]arqady[/b], thank you.  :wink:


calculus
function
integration
geometry
calculus computations

What is the average value of the function

$F(x)=\sqrt{9-x^{2}}$

That is, consider only the top half of the circle.

All help is appreciated!

The average value of a function $F(x)$ on an interval $[a, b]$ is given by

$\frac{1}{b-a}\int_{a}^{b}F(x) \, dx$

(Can you see why?)  You shoud be able to evaluate the integral.  (Hint: don't evaluate the integral.)

Edit:  Oh, intro to calc.  In that case, you should know that

$\int_{a}^{b}F(x) \, dx$

Means "the area between $F(x)$ and the $x$-axis from $a$ to $b$."

(base) x (average height) = (area)

Base is $2r = 6$
Area is $\frac12\pi r^{2}=\frac92\pi$


I had  a long  debate towards this problem, so I would like a third ( or n-th) opinion on it:

if log(108,N)=a
log(72,N)=b

compute log(2,N)

I am quite curious what u can tell me...the solution interests me especially.

All you do is use basic properties of logs:

\[\log_N32=\log_N\left(\frac{72^3}{108^2}\right)=3\log_N72-2\log_N108=\frac3a-\frac2b\]
Therefore,
\[
\log_2N=\frac{5}{\log_N32}=\frac{5}{\frac3a-\frac2b}
\]


Let C be a circle of radius 3 with fixed diameter AB. What is the probability that a random point P inside of C is such that :ang: APB :le: 3:pi:/4? (Where :ang: APB is measured in radians between 0 and :pi: inclusive).

Interesting problem. My solution: [hide]an arc could be drawn in circle C that has endpoints AB that separate the area that P lies in which will make an 3 :pi: /4. 3 :pi: /4 = 135 degrees. Let's call the circle which minor arcAB lies on circle with center D. Since we have an inscribed angle 135 degrees, the intercepted arc must be 135*2=270, there therefore the central angle ABD must be 360-270=90. The radius(midpt. of AB) of circle C =3, then the radius of circle D=3 :rt2: . The area of the desired segment of circle = Area of circle D/4-(3 :rt2:)^2/2 = (9 :pi: +18)/2. Subtract it from semi-circle AB we have 9, which is the area P lies in so angle APB<135, 9/area of circle C which is 1/ :pi: is the probability[/hide]

Wait, I think I forgot to multiply by 2, the answer is[hide] 2/ :pi: [/hide]

I believe that the probability is simply 2(135)/360 = 270/360 or 3/4. But I seems that this question would probably be more complicated than this.

Where did you get that calculation, Jobro?

I'm going to wait until some more people work on it before I announce what the answer is. This is a good problem for people to work on if they haven't tried it already.


Let $ (a_n)_{n>\equal{}1}$ be a sequence of real numbers so that $ |a_{n\plus{}1}\minus{}a_n|<\equal{}1$ for $ n\in\mathbb{N}^*$ and $ (b_n)_{n>\equal{}1}$ the sequence defined as $ b_n\equal{}\frac{a_1\plus{}a_2\plus{}\cdots \plus{}a_n}{n}$.

Prove that $ |b_{n\plus{}1}\minus{}b_n|<\equal{}\frac12$

$ |b_{n\plus{}1}\minus{}b_n| \equal{}\frac {|na_{n\plus{}1}\minus{}na_n \plus{} (n\minus{}1)a_n\minus{}(n\minus{}2)a_{n\minus{}1}\plus{} ...\plus{}2a_3\minus{}2a_2 \plus{}a_2\minus{}a_1|}{n(n\plus{}1)}$$ \leq \frac {n\plus{}(n\minus{}1)\plus{}...\plus{}2\plus{}1}{n(n\plus{}1)}\leq \frac 12$
 :cool:


In the Tour de France, what is the position of a rider, after he passes the second placed rider?

You have five pieces of chain, each consisting of three links. You want to make one long chain of these five pieces. Breaking open a link costs 1 dollar , and welding an open link costs 3 dollars.
Is it possible to make one long chain of the five pieces, if you have just 15 dollars?

[hide="first one"]you're in the second place, until you pass by the first placed rider[/hide]
[hide="second one"]Yes, it is possible[/hide]

Second.

Well, to break open the last link in each chain and then weld it back together would cost a total of $16, so it's not possible that way... I don't know if there's any other way to do it, but I would think not.

Break one link, attach two pieces, weld link again.  Break second link, attach other two pieces, weld link again.  Break last lone link, attach two big pieces, weld link again.  Weld it again with the extra 3 dollars for fun.


A block of mass m is moving with a speed v(0) on a straight surface. Then the block starts moving along a circular guide of radius R on a trolley (truck is better perhaps). The block m reaches the point Q where it's at rest relative to the truck. The block is to be considered as a point, there's no friction.
Find:

a) Find the speed v(0) using m and R;

b) Find the work of the reaction force F applied by the block on the truck from the starting point to Q;

c) Find the value of the F in Q.

My doubt is on the reaction force.
Enjoy!

a look into ur figure gives an idea that when the block is at $ Q$ it has velocity only in the upward direction.then how can it be at reslative rest with the trolley

edit:now its seems assuming the problem ststement is right the horizontal componenents of the trolley and block are the same

my english is bad...but the problem says exactly what I've written...maybe we must consider the acceleration
otherwise, even if the speeds have different directions, that means their values are equal...let's try a solution

My results:
[hide]
a) $ V_{0}\equal{}\sqrt{2gR(1\plus{}\frac{m}{M})}$
b) $ N \equal{}\frac{mg}{1\plus{}\frac{m}{M}\sin\alpha}\Bigg(\frac{2(1\minus{}\sin\alpha)}{1\minus{}\frac{m\cos^{2}\alpha}{m\plus{}M}}\plus{}\cos\alpha\Bigg)$
alpha is the angle between the vertical direction.
c) $ N(\pi /2) \equal{} 0$
[/hide]

Isn't answer to b) zero as F is always perpendicular to motion?

[b]@golduck[/b] what have u been taking the definition of relative rest at the highets point

well can you show which is the reaction force?
@ pardesi: I have considered v(trolley)=v(block) in Q, then i've the same result for part a)..using the conservation of momentum and energy...see u

sorry i misundrstood the statement... :rotfl:

[b]@pardesi[/b] the definition of relative rest at the highest point means that the mass is at rest in the frame of reference connected with the trolley - they have got the same (horizontal) velocity.
[b]@shreyas_patankar[/b] My answer to b) is incorrect, forget about it - I thought I have to calculate $ F$ not the work done by $ F$ on the trolley  :oops:  But anyway the work done by the reaction force F applied by the block [u]on the truck[/u] shouldn't be zero, I'll try to calculate it...

[hide="My solution"]
a) Let $ u$ be the velocity of the trolley and the mass, when the mass is an point Q. Using conservation of energy and conservation of momentum in horizontal direction we get:
\[ (M+m)u=mv_{0}\]
\[{ \frac{1}{2}mv_{0}^{2}=\frac{1}{2}mu}^{2}+\frac{1}{2}Mu^{2}+mgR\]
Solving these equations we get:
\[ v_{0}=\sqrt{2gR(1+\frac{m}{M})}\]
b) The work done by the reaction force F applied by the block on the trolley from the starting point to Q equals the change of energy of the trolley:
\[ W=\frac{1}{2}Mu^{2}\]
and $ u=\frac{m}{m+M}v_{0}$ so
\[ W=\frac{m}{m+M}mgR\]
c) In the non inertial frame of reference connected with the trolley the mass is moving on a circular track. In each point of the track we have got:
\[ \frac{m(v')^{2}}{R}=ma\sin\alpha+F-mg\cos\alpha\]
Where $ v'$ is the velocity of the mass in the trolley's frame, $ a$ is the acceleration of this frame ($ a=\frac{F}{M}\sin\alpha$) and $ \alpha$ is the angle between the vertical direction. For $ \alpha=\pi / 2$ we get:
\[ \frac{m(v')^{2}}{R}=ma+F=F(\frac{m}{M}+1)\]
But as the mass is at rest in the troley's frame, so $ v'=0$ and also $ F=0$.
[/hide]


(C) is a circle and AB is a chord of it.(C') is a variable circle wich is tangent to (C) and AB,in points Pand

Q,respectivly.

prove that:  PQ goes through a fixed point.

the fixed point,will be the midpoint of arc AB :?


Why those smilies? And why posting it in the geometry section? And didn't you post this before? I'm not a mod here, so can some other mod move or remove this?

[quote="Zao1989"]For given $u_0$,find the general form of $u_n$ if $u_n=a u_{n - 1}^2 + b u_{n - 1} + c$.[/quote]

At first, this (geometry proposed&own problems - now I moved it out) is not the right place to post it, and you already posted it at http://www.mathlinks.ro/Forum/viewtopic.php?t=47857 ; but then, do you really think your sequence has a reasonable general form?

  darij


f(x) is a cubic equation of the from $f(x)=ax^3+bx^2+cx+d$
f(1)=1
f(4)=2
f(9)=3
f(16)=4

find 15b+3c

[hide]Plug in one.  a+b+c+d=1
Plug in 4.  64a+16b+4c+d=2
Plug in 9.  729a+81b+9c+d=3
Plug in 16.  4096a+256b+16c+d=4
Subtracting equation 1 from 2, 63a+15b+3c=1.  Since we were told to find 15b+3c, 63a must be found and subtracted.
Subtracting equation 2 from 3, 665a+65b+5c=1.
Subtracting equation 3 from 4, 3367a+175b+7c=1.
Now, to cancel c.  Doing so with the first and second members of the new set of equations yields -1680a-120b=2.  Doing so with the second and third members of the new set of equations yields -12180a-420b=2.
Now, cancelling b yields 12600a=10 or $a=\frac{1}{1260}$.  Plug that in for a and subtract.  I get 15b+3c=19/20.[/hide]
In fact, the cubic polynomial in this problem is $\frac{1}{1260}x^3-\frac{1}{36}x^2+\frac{41}{90}x+4/7$

often you want to look for more efficient ways of doing these problems than a huge system

[hide="some steps"]we have $f(n^2)=n$ for $n=1,2,3,4$

$g(x): =f(x^2)-x=k(x-1)(x-2)(x-3)(x-4)$ since those are the roots of $g(x)$

$f(x^2)=x+k(x-1)(x-2)(x-3)(x-4)$[/hide]


ratio
number theory
prime numbers
real analysis
real analysis unsolved

im trying to read the proof [url=http://secamlocal.ex.ac.uk/~rjc/etc/dirichlet.pdf]here[/url] about dirichlet's theorem on prime numbers in
arithmetic progressions.
theorem 4(at page 5 up) is unclear to me in the sense that i do not understand
its meaning.
in particular i do not understand the statement
"Each character csi of H is the restriction of |G:H| characters of G"(1).

What i probably think it means is the following:
if we have

$|G|=nm$ and $|H|=n$ and $|G: H|=m$ for each
$\psi \in \hat{H}$ there are at least $m$ other characters in $\hat{G}$
wich we denote by $\chi_{_{i}}$ with $1\leq i \leq m$ so that
$(\chi_{_{i}})|_{_{H}}= \psi$

Is this the meaning of (1) ?
I need this so i can go on reading the proof.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=135418]this topic[/url].
[quote="spx2"]
sorry for reposting in Number Theory section,if a moderator finds
appropriate he/she can delete the one from Real Analysis ive previously
posted.
[/quote]

there's no "exactly" in the statement of theorem 4 as it is written in
the link above.
is it by default , "exactly" ?
to me it sounds like "at least"

perfect_radio are you beeing fun in a subtle way and making fun of me ?

I think he was just trying to do what I would have done if I noticed a double post: direct all discussion to the thread that it is in the more appropriate place, which, for this particular thread is Number Theory. If you really wanted to move it to Number Theory yourself, you could do one of three things:

1) Delete your post (you can do it as long as there are no replies) and repost it there.

2) Ask a moderator of the Real Analysis forum to move it

3) Add a remark to the [b]Real Analysis[/b] post.

The way you did it (by apologizing in the Number Theory post) would by no means allow the people who frequent the Real Analysis subforum to realize that the topic was reposted. Also, the moderators of Number Theory can not delete posts in Real Analysis and vice versa (except Myth, who moderates both, but he is not a frequent visitor nowdays :(). So, perfect_ratio just corrected that little logical mistake of yours, no offence being intended. :)


(forma putin simplificata)
Fie $f\in Z[X]$ un polinom. Se se afle f stiind ca functiile $\hat{f(x)}$, sunt surjective, pentru orice $Z_n,n\geq 2$

Daca $\deg{f}>1$, fie $g(x)=f(x+1)-f(x)$. Atunci $\deg{g}\ge 1$ si, deci, exista $a$ aî $|g(a)|>1$. Fie $n=|g(a)|=|f(a+1)-f(a)|$; rezulta ca in $\mathbb{Z}_n$ avem $\hat{f}(\widehat{a+1})=\hat{f}(\hat{a})$, adica $\hat{f}$ nu e injectiva, deci nici surjectiva.
Daca $f(x)=px+q$ si $|p|\ge 2$, in $\mathbb{Z}_{|p|}$ functia $\hat{f}$ e constanta, deci nesurjectiva. Solutia e $f(x)=\pm x+q$, cu $q\in \mathbb{Z}$.


topology
calculus
integration
function
modular arithmetic
advanced fields
advanced fields unsolved

The subject doesn't quite cover what I'm asking about actually.  This is from one of my psets and I want to make sure that the problem makes sense:

$\mathbb{Q}_{p}$ is a field and has a topology.  Determine all continuous characters (ie homomorphisms) $\psi: (\mathbb{Q}_{p},+) \mapsto \{z: |z| =1, z \in \mathbb{C}^{*}\}$ and show that as a topological group $\widehat{\mathbb{Q}_{p}}\cong \mathbb{Q}_{p}$.
where $\widehat{\mathbb{Q}_{p}}$ is the group of these continuous characters under pointwise multiplication.

I don't think the problem is completely bogus because I have proven that the two are actually isomorphic just as groups.  I am just a bit confused about this "topological group" thing.  What sort of topology does $\widehat{\mathbb{Q}_{p}}$ have?

The fairly natural isomorphism I have between $\widehat{\mathbb{Q}_{p}}$ and $\mathbb{Q}_{p}$, doesn't seem to be continuous if we define the distance between two characters to be just the maximum distance between points in the function.  I have no idea how integration of the difference might work.

By analogy, consider the continuous characters on $\mathbb{R}$: they're the maps $f_{c}(t)=e^{ict}$ for real $c$. The obvious map $c\mapsto f_{c}$ should be a homeomorphism, but the uniform topology doesn't work; the $f_{c}$ are discrete in that norm.

What's the fix? The topology of pointwise convergence works, but it isn't necessarily a metric topology: it's the product topology on the space of functions, restricted to the subset we're looking at. An intermediate version is the topology of uniform convergence on compact sets: $f_{k}$ converges to $f$ if the $f_{k}$ converge uniformly to $f$ on each compact set $K$. On a $\sigma$-compact space such as $\mathbb{R}$ or $\mathbb{Q}_{p}$, this is a metric topology (but it's not worthwile to use an explicit metric).

It should have been specified in the statement, of course.

Wow, thank you for the quick response.

I actually unfortunately missed quite a lot of lecture, so this problem is taken from somebody's notes, it's entirely possible that the professor explained the problem more completely than what the notes suggest.  The problem was just strange since we hadn't talked much about topology at all in the class and I definitely haven't taken any topology before, so that's why my question might sound naive.  The only bit of intuition about topology that I have been able to pick up is that metrics are useful to determine an idea of "closeness", but I guess there are other ways.  Your response makes it clear to me that I need to know a lot more topology, haha.  Can you suggest a good book to read on introductory topology?

Also, is it possible that you could explain how my isomorphism might work out for the possible topologies you have suggested?
Basically, the gist of the isomorphism is that if you consider these characters, they are determined by their values at $p^{k}$ for all $k \in \mathbb{Z}^{+}$.  I spend some time proving that for this to be continuous, there must a $k$ such that $\psi(p^{k}) = 1$.  Clearly, then for $l \ge k$, we have that $\psi(p^{l}) =1$.  Now we find the minimal such $k$ for our $\psi$.  If it doesn't exist, we clearly have a trivial character.  Otherwise, we find the minimum $k$.  We then consider $\psi(p^{k-t})$ for $t \in \mathbb{Z}^{+}$.  We can see that $\psi(p^{k-t}= exp(a_{t}/p^{t})$ where $exp(\alpha) = e^{2\pi i \alpha}$ where you then produce a chain of $a_{t}$ such that the satisfy the sort of thing for the definition of $\mathbb{Z}_{p}$ ie $a_{t+1}\equiv a_{t}\pmod{p^{t}}$.  Thus this gives us the very natural isomorphism of basically "flipping"  our character in some sense and associating the p-adic number $p^{-k}(a_{1},a_{2}, \dots)$ and then associating the trivial character with 0.  $(a_{1},a_{2},\dots) \in \mathbb{Z}_{p}$ with $|a|_{p}= 1$.

To prove the continuity, I was going to just prove continuity at 0.  Notice that as $x \rightarrow 0$, we have that the corresponding characters have $\psi(p^{k})$ for smaller and smaller k since higher and higher powers of p are dividing x.
So for any fixed m, $\psi(p^{m})$ will eventually be equal to 1 for all characters assocated with the x that is converging to 0.  Is that enough to get any of those topologies you mentioned?  It seems that maybe the compact set one sounds attractive, but I don't really understand it.

Thank you so much for any help you can give.

I learned topology in an analysis class; there's enough topology in standard analysis texts for most practical needs.

The characters are determined by their values at $p^{-k}$, not $p^{k}$. If we know the value at 1, we already know it at $p^{k}$ for all positive $k$.

Working out the topologies:
(Note: I'm looking at characters with values in $\mathbb{R}/\mathbb{Z}$; this is naturally isomorphic to your version, but the operation is addition rather than multiplication)
Pointwise convergence: This is the product topology on $\left(\mathbb{R}/\mathbb{Z}\right)^{\mathbb{Q}_{p}}$. The basic open sets in this topology look like $\{f: f(x_{i})\in U_{i}\}$ for some finite list of points $x_{i}\in \mathbb{Q}_{p}$ and open intervals $U_{i}\in \mathbb{R}/\mathbb{Z}$.
What characters are in a neighborhood of the identity character $0$? There are all of the characters that should be there (the ones which are equal to zero at each $x_{i}$, but we can't eliminate all of the characters that should be far away: for sufficiently large $n$, the character with values $\frac{p^{k}}{p^{n}}\pmod 1$ at each integer $k$ is in the neighborhood. This character corresponds to the $p$-adic number $\frac1{p^{n}}$.

That didn't work; it's too coarse.
Uniform convergence on compact sets does work. Since $\mathbb{Q}_{p}$ is the union of the compact balls $B_{j}$ of radius $p^{j}$ ($j\ge 0$) around the origin, we only need to check for uniform convergence on these balls.
To understand $f$ on $B_{j}$, we need only know its value at $p^{-j}$; every element of $B_{j}$ is a limit of integer multiples of $B_{j}$. Consider the neighborhood $U=\{f: |f(x)|\le\frac1{2p}\text{ for all }p\in B_{j}\}$. This is a neighborhood of the identity, and we can verify that all of its elements must be zero on $B_{j}$. The set of characters which are zero on $B_{j}$ corresponds to a $p$-adic ball around the origin, and the diameters of these balls go to zero as $j\to\infty$. These neighborhoods translate; the set $V=\{f: |f(x)-g(x)|\le\frac1{2p}\text{ for all }p\in B_{j}\}$ is a ball around $g$ by the same argument applied to $f-g$. Every open set in the space of functions contains one of these, so these neighborhoods form a base for the topology, and we have a homeomorphism.

Since we're mapping to a bounded metric space, we can write an explicit metric for the preceding topology: choose your favorite convergent series $\sum a_{j}$, and let $d(f,g)=\sum_{j}a_{j}\cdot\sup_{x\in B_{j}}|f(x)-g(x)|$. If $\sum a_{j}$ converges rapidly enough, the distances will sort into bands just like $p$-adic distances.


geometry
trigonometry
trig identities
Law of Cosines
geometry unsolved

I'm new to these forums, and I hope I'm posting this in the right forum area.

My sister gave me a coffee cup for Christmas with the problem shown in the attachment.  Please help if you can.  It's been driving me nuts.   :(

Welcome!  In general, geometry problems where you must find some measurement as opposed to proving something go in the Pre-Olympiad and Intermediate forums.  But maybe that's just my feeling.

[hide="hint"]Can you find $ m\angle{AGC}$?[/hide]
[hide="Solution"] $ m\angle{AGC}\equal{}m\angle{B}\plus{}\frac{1}{2}m\angle{A}\plus{}\frac{1}{2}m\angle{C}\equal{}135^{\circ}$.  So, using the law of cosines, we get
$ AC^2\equal{}9^2\plus{}2 \cdot 8^2\minus{}2 \cdot 9 \cdot 8\sqrt{2}\cos{135^{\circ}}\equal{}81\plus{}128\plus{}144\equal{}353$, if I'm not mistaken.  So we then have $ AC\equal{}\sqrt{353}$...[/hide]

[b]K81o7[/b], we are not given the length of $ AG$ and $ CG$. The numbers on the picture refer to $ AE$ and $ CD$.

That's correct, mlok.  I'm sorry to have posted a misleading picture, and also sorry to have posted this problem in the wrong forum.


Prove that for any integer $ n\geq2$ there is a set $ A_n$ of $ n$ distinct positive integers such that for any two distinct elements $ i,j\in A_n$,  $ |i\minus{}j|$ devides $ i^2\plus{}j^2$

$ A_n \equal{} \{1\cdot (n \minus{} 1)!,2 \cdot (n \minus{} 1)!,3\cdot (n \minus{} 1)!,\cdots,n \cdot (n \minus{} 1)! \}$ works because:

$ |i \minus{} j| \in \{1\cdot (n \minus{} 1)!,2 \cdot (n \minus{} 1)!,3\cdot (n \minus{} 1)!,\cdots,(n \minus{} 1) \cdot (n \minus{} 1)! \}$

and $ (n \minus{} 1)! ^2 | i^2 \plus{} j^2$

How about finding $ \min{(\max{\{a_i | a_i \in A_n}\})}$?

Given like this the problem is trivial. Take arbitrary numbers and if they don't work multiply them by some "big" number and they will work. I think there is something missed. Maybe the numbers in $ A_{n}$ must be coprime.


calculus integration derivative trigonometry function algebra polynomial

Hints:

1. $ \frac1{\sin^2x}\equal{}\csc^2x.$ I hope you remember the antiderivative of $ \sec^2x.$ The cosecant should be like the secant, on with the appropriate co-function and a minus sign somewhere. Make a guess and check your guess by differentiation.

2. Divide out the polynomials.

2a. You can make the substitution $ u\equal{}x\minus{}2,\ x\equal{}u\plus{}2,\ du\equal{}dx$ first, then do the division. That's probably what I'd do.

3. Integrate by parts. Don't bother splitting the integral up first - just do the integration by parts in the "usual" direction (differentiating the polynomial and integrating the exponential).

3a. Note that the function is negative on the interval of integration. Your final numerical answer should be negative.

hello, for question two, show, that $ \frac{x^2\plus{}3}{x\minus{}2}\equal{}x\plus{}2\plus{}\frac{7}{x\minus{}2}$ for all $ x$ with $ x \ne 2$.
Sonnhard.

hello, for 3) setting $ u\equal{}x\minus{}3$ and $ v^{'}\equal{}e^{3x}$ and using integration by parts you will get
$ \int(x\minus{}3)e^{3x}\,dx\equal{}(x\minus{}3)\frac{e^{3x}}{3}\minus{}\frac{1}{3}\int e^{3x}\,dx$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, for question one, what is the first derivative of $ \cot(x)$?
Sonnhard.[/quote]

That should be $ \minus{}\cot x.$

That's actually the point of my statement "Make a guess and check your guess by differentiation." Do that, and you'll see where you need the minus sign.

hello
[hide="Solutions for 3rd ques"]
[hide="I method "]
$ \int e^{\alpha x} p_{n} (x) dx$
$ \equal{} e^{\alpha x}[ \frac{p_{n} (x)}{\alpha}\minus{}\frac{p_{n}'(x)}{{\alpha}^2}\plus{}...........\plus{}(\minus{}1)^{n\plus{}1}\frac{p_{n}^{n}(x)}{{\alpha}^{n\plus{}1}}]\plus{}C$
where $ p_{n} (x)$ is a polynomial of degree $ n$.
So
$ \int e^{3x} (x\minus{}3) dx$
=$ e^{3x} [ \frac{x\minus{}3}{3} \minus{} \frac{1}{9}] \plus{} C$ [/hide]
[hide=" II method"]
integrals of the form $ \int e^{\alpha x} p_{n} (x) dx$ , we always obtain a function of the form $ q(x)e^{\alpha x}$ where $ q(x)$ is a polynomial of the same degree as the polynomial $ p_{x}$.
This allow us to calculate the integrals using the method of indefinite coefficients.
$ \int e^{3x} (x\minus{}3) dx \equal{} e^{3x} (Ax\plus{}B) \plus{}C$
Differentiating the above eq. and canceling $ e^{3x}$
$ (x\minus{}3)\equal{} ( A \plus{} 3 (Ax\plus{}B) )$
Now just by comparison u can find out $ A$ and $ B$.
[/hide]
[hide="III method "]
The common method - Integration by parts which is already explained by Dr. So I probably no need to explain it again. [/hide]
[/hide]
thank u


Prove that all positive integers can be written as the difference of two positive integers which have the same number of prime divisors.

We can have 3 cases.
1) a positive integer=n divisible by both 2 and 3 or none of them:
then n=3n-2n(both of 3n and 2n  have the same number of prime divisors.

2)n is divisible by 2 but not for 3:
then n=2n-n

3)n is divisible by 3 but not for 2:
we prove it indirectly

(*)suppose that there exist a number which can't be represented as the difference of two positive integers which have the same number of prime divisors,then
however n=5n/3 -2n/3 so n is divis. by 5(if not then it contradicts with (*) )

We claim that n is divisible by 3=p_1<p_2<...<p_k for all k,where p_k is the  (k+1)st prime.It's true for k=1,2.Suppose that n is divisible by all primes <p_k
then n=p_k*n-2m*n for some m not divisible by primes >=p_k ,hence if n is not divisible by p_k then it contradicts with (*).
So n is divisible by all primes,contradiction.
Hence n can be written as the difference of two positive integers which have the same number of prime divisors.