Can anyone help me with the following?

If $ P$ is a projection matrix, show from the infinite series that $ e^P \sim  I \plus{} 1.718P$?

For this problem, I wrote out the infinite series expansion $ e^P \equal{} I \plus{} P \plus{} \frac{P^2}{2!} \plus{} \frac{P^3}{3!} \plus{} ... \plus{} \frac{P^n}{n!}$, but the $ P$s are matrices and I need 1.718, a number. Also, how can I show that what I get is independent of any projection matrix?

Remember that $ P^n \equal{} P$, so it's $ P\left(\sum_1^\infty \frac{1}{n!}\right)$


Yeah, exchanging the variables would work. Also remember that if g is the inverse of f, then
f(g(x)) = x and g(f(x)) = x.

Fierytycoon

Ex: find the inverse of f(x) = 3x + 2

Here's the sequence:
f(x) = 3x + 2
y = 3x + 2
x = 3y + 2
x - 2 = 3y
y = (x-2)/3
f^-1(x) = (x-2)/3

So f(f^-1(x)) = x

[quote="mathfanatic"]So f(f^-1(x)) = 1[/quote]

f(f^-1(x))=x actually.

Ug.  I am having a bad day...

Hmm...maybe I can make a list of short phrases that I use to express disgust: "ug, oops, erp, eek, etc."

Soon you realise that for most functions it is not easy to find an inverse function. Sometimes specific inverse functions have been created, i.e. arcsin, arccos, arctan. :)

Well, we were referring to polynomial functions.  Exponential functions (and their inverses logarithmic functions), trigonometric functions, etc. naturally don't use this rule.

Find the inverse of a generic quintic equation that has df/dx>=0 for all x. That's still really hard  (read: impossible with ``elementary'' functions).


[quote="yeppyyep"]coeff of $x^{5}y^{4}$ in $(2x-3y^{2})^{7}$?[/quote]

[hide="solution"]$\text{Using the binomial theorem, the coefficient of}\ x^{5}y^{4}\ \text{is}$ ${7\choose 2}2^{5}(-3)^{2}=21\cdot 32\cdot 9=\boxed{6048}$[/hide]

[quote="yeppyyep"]coeff of $x^{5}y^{4}$ in $(2x-3y^{2})^{7}$?[/quote]
Obviously his way works but another way of looking at is (just taking his way to the next step) is:

[hide]$\frac{7! \cdot (2x)^{5}\cdot (-3y^{2})^{2}}{5! \cdot 2!}$; Just take the exponents which each of your variables are raised to and put them as factorials on the bottom while their sum as a factorial on the top.
$=6048 x^{5}y^{4}$

Thus it is $\boxed{6048}$[/hide]


1.$ ABCDEF$ is a convex hexagon such that all of its vertices are on a circle. Prove that $ AD$, $ BE$ and $ CF$ are concurrent if and only if  $ \frac {AB}{BC}\cdot\frac {CD}{DE}\cdot\frac {EF}{FA} = 1$.

2. Determine whether there are two infinite point sequences $ A_1,A_2,\ldots$ and $ B_1,B_2,\ldots$ or not, such that for all $ i,j,k$ that satisfies $ 1\le i < j < k$,
    i.$ B_k$ is on the line that passes through $ A_i$ and $ A_j$ if and only if $ k = i + j$.
    ii.$ A_k$ is on the line that passes through $ B_i$ and $ B_j$ if and only if $ k = i + j$.

3. $ n$ is a positive integer. Calculate the sum $ \sum_{k = 1}^{n}{\sum_{1\le i_1 < \ldots < i_k\le n}^{}{\frac {2^k}{(i_1 + 1)(i_2 + 1)\ldots (i_k + 1)}}}$

4. The sequence of polynomials $ (a_n)$ is defined as, $ a_0 = 0$, $ a_1 = x + 2$ and for all $ n > 1$ integers, $ a_n = a_{n - 1} + 3a_{n - 1}a_{n - 2} + a_{n - 2}$.
    a.Show that for $ k,m$ positive integers, if $ k$ divides $ m$; $ a_k$ divides $ a_m$.
    b.Find all $ n$ positive integers such that sum of the roots of polynomial $ a_n$ is an integer.

number $ 1$ is VERY old,it has been discussed here to. :huh:
here for example:
[url]http://www.mathlinks.ro/viewtopic.php?t=182839[/url]

1. It is very well-known :D
2.[url]http://www.mathlinks.ro/viewtopic.php?t=197870[/url]
3.[url]http://www.mathlinks.ro/viewtopic.php?p=1088260[/url]
4.[url]http://www.mathlinks.ro/viewtopic.php?p=1088257[/url]


Find the number with $97$ digits, all different from $0$, such that it is multiple of the sum of all its $97$ digits.


let $\left\{ x_n \right\}$ be a sequence given by : $x_1 = 1$ and $x_{n+1} = \frac{x_n}{n} + \frac{n}{x_n}$
prove that :$[x_n^2] = n$

$x_{n+1} = \frac{x_n}{n} + \frac{n}{x_n} \Longleftrightarrow x_2=2$$???$

Actually, this statement is true for all $n \ge 4$.


function
real analysis
real analysis unsolved

Is every continuous (resp. smooth) function $ f(x, y, z) : \mathbb{R}^3 \to \mathbb{R}$ a finite composition of continuous (resp. smooth) two-variable functions $ \mathbb{R}^2 \to \mathbb{R}$?  (These include, for example, the functions $ x \plus{} y$ and $ xy$.)  If so, what's the weakest condition we can replace "continuous" (resp. smooth) with so that this still holds?

For continuous functions, the answer is "Yes". It was proved by Kolmogorov (I would like to give you a free online reference to some readable proof but the best I could find was [url=http://sfb611.iam.uni-bonn.de/uploads/342-komplett.pdf]this[/url]). I don't remember what happens in the smooth case (probably the answer is "No").


Congratulations to all winners on this year's CMO.

Details can be found here:

[url]http://www.newswire.ca/en/releases/archive/May2008/26/c3858.html[/url]

Congratulations winners!!!  :D
On another note, when will the full results be released?

2008 CMO WINNERS

FIRST PRIZE - [b]Chen Sun,[/b] A.B. Lucas Secondary School, London, ON;
SECOND PRIZE - [b]Jonathan Schneider,[/b] University of Toronto Schools, Toronto, ON;
THIRD PRIZE - [b]Yan Li[/b], Dr. Norman Bethune Collegiate Institute, Toronto, ON;

HONOURABLE MENTIONS were awarded to:
[b]Dimitri Dziabenko[/b], Don Mills Collegiate Institute, Toronto, ON;
[b]Neil Gurram[/b], Indus Center for Academic Excellence, Troy, MI, USA;
[b]XiaoLin Shi[/b], Sir Winston Churchill High School, Calgary, AB;
[b]Chengyue (Jarno) Sun[/b], Western Canada High School, Calgary, AB;
[b]Tianyao Zhang[/b], Sir John A. Macdonald Collegiate Institute, Toronto, ON.

WOW .. Good job to u all ^^~ CONGRADULATIONS!

finally, the results are out

http://www.math.ca/Competitions/CMO/reports/CMOReport2008.pdf

Hmm, I see some familiar names.  Very cool.


trigonometry
integration
complex analysis
complex analysis unsolved

I can't find any information on the behaviour of $ w(z)\equal{}\sum_{n\equal{}0}^{\infty}\sqrt{z}e^{\minus{}zn^2}$ at $ z\equal{}0$, besides that as $ z\rightarrow 0^{\plus{}}$ over the reals one has $ w(z)\rightarrow \frac{\sqrt{\pi}}{2}$.

What do you mean by $ \sqrt{z}$? That has a branch point, and there's no particular reason to choose one sign or the other.

For every complex number $ z\neq 0$ there exists

a unique polar representation $ z\equal{}r\, e^{i\varphi}$ with $ \minus{}\pi<\varphi\le \pi$ .

I assume by $ \sqrt{z}$ you mean the principle value $ \sqrt{r}\, e^{i \frac{\varphi}{2}}$ .

$ w(z)\equal{}w(r e^{i\varphi})\equal{}\sum_{n\equal{}0}^\infty \sqrt{r}\, e^{i\frac{\varphi}{2}} \, e^{\minus{}r e^{i\varphi} n^2}\equal{}e^{i\frac{\varphi}{2}} \sum_{n\equal{}0}^\infty e^{\minus{}e^{i\varphi}\, (\sqrt{r}\, n)^2}  \sqrt{r}$

If $ \varphi\notin \left]\minus{}\frac{\pi}{2},\frac{\pi}{2}\right[$ and $ r\neq 0$ then $ \left|e^{\minus{}e^{i\varphi}\,(\sqrt{r} \, n)^2 }\right|\equal{}\exp\left(\underbrace{\minus{}\cos\varphi}_{\ge 0} \, (\sqrt{r} \, n)^2\right)$

doesn't tend to $ 0$ for $ n\to\infty$ . Therefore $ w(z)$ cannot exist.

If $ \varphi\in \left]\minus{}\frac{\pi}{2},\frac{\pi}{2}\right[$ and $ r\neq 0$ then $ w(z)\to e^{i\frac{\varphi}{2}} \int_0^\infty e^{\minus{}e^{i\varphi}\, x^2} dx$

$ \equal{}e^{i\frac{\varphi}{2}} \frac12 \frac{\sqrt{\pi}}{\sqrt{e^{i\varphi}}}\equal{}\frac{\sqrt{\pi}}{2}$ for $ r\to 0^\plus{}$ , $ r\neq 0$ .

Let $ D$ be $ \left\{z\in\Bbb{C} \, : \, \text{Re}(z)>0 \right\}$ .

For every sequence $ (z_n)$ with $ z_n\in D$ and $ z_n\to 0$

$ w(z_n)\to \frac{\sqrt{\pi}}{2}$ and for all other $ z\notin D \quad w(z)$ is either undefined or $ 0$ iff $ z\equal{}0$ .


Let $ f: R\minus{}>R$ be a nonconstant function, periodical and with primitives. Prove that $ f$ has a minimal positive period.

since f has primitives, f is continuous. since f is periodical and nonconstant it has a positive period $ b$. Suppose f does not have a minimal positive period then $ b/2 ,...,b/n$ is a period of f for all integer n. $ \forall x f(x\plus{}b/n)\equal{}f(x)$ then $ f(\frac{kb}{n})\equal{}f(0)$ | $ \forall (n,k) \in N^{*}.N$
We can suppose $ b \in Q^{\plus{}}$ , the case $ b \notin Q$ is similar. So for all rational number x f(x)=f(0) and we conclude by the density of Q in R and the continuity of f, which extends the result for every real number x , f(x)=f(0). That is in contradiction with the nonconstantness of f.
Hence f has a minimal positive period

[quote="jedusor"]since f has primitives, f is continuous[/quote]
This is wrong. There are discontinuous functions that have antiderivatives.

oh , the stupid mistake, the problem is not as trivial as i thought.  :oops_sign:

[quote="jedusor"] Suppose f does not have a minimal positive period then $ b/2 ,...,b/n$ is a period of f for all integer n. [/quote] I may put a stupid question too but why is $ b/n$ also period? :roll:

because when supposing that f doesn't have a minimal positive period every positive real number less than b is a period of f , it's the case for b/n. For example $ ]0;1]$ doesn't have a minimal positive element $ \implies \forall b \in ]0;1] ,b/n \in ]0;1]$


superior algebra
superior algebra unsolved

if $ f: G\rightarrow H$ is a homormorphism with kernal $ N$ and $ K\leq G$ , then prove that $ f^{\minus{}1}(f(K)\equal{}KN$.

 :)

What progress do you have? Where are you stuck?
[hide="hint"]Just show that $ f^{\minus{}1}(f(K) \subseteq KN$ and $ KN \subseteq f^{\minus{}1}(f(K)$[/hide]

Finally , I got the usual way .

Is there another solution with subsets argument ? :wink:

What algebraic structures are those $ K,G,N,H$? Monoids or groups or something else?


linear algebra
matrix
algebra
polynomial
calculus
derivative

its easy to see that $c=0$ is a solution

since  $f(x)$ is polynomial, its derivative is too.

So it need to have a root, real or complex.

Since the determinat dont have a diagonal with just constants. so.   $f(0) = 0$

and the same thing to his derivative, because of the product of xs.... etc.. etc...

post your solution, please!  :D I not understand.

This is my solution

We have $f(0)=f(1)=0$ and $f(x)$ is the polynomial therefore by Mean value theorem we have done :D

i did the same thingh and u dont understand

oh......

this is polynomial

i said it!!!!

afff......

dont joke with me

[quote="Thales418"]

dont joke with me[/quote]
I not joke with you, I not understand what you want speaked. That is truth :)


vector
linear algebra
linear algebra unsolved

Let $ V$ be a finite dimensional Euclidean or unitary space and $ W_1$
and $ W_2$ two vector subspace of $ V$ . Prove that:

a) $ (W_1 \plus{} W_2)^\perp \equal{} W_1^ \perp \cap W_2^\perp$
b) $ (W_1 \cap W_2)^\perp \equal{} W_1^ \perp \plus{} W_2^\perp$

Fitim

Don't you mean the ortho-complements of the first vector spaces mentioned in (a) and (b) to have the claim be true? ;)

I forgot to write "T" to. Look again please :)

Thank you
Fitim

To prove two sets are equal, show that each is a subset of the other.

Use transseries' hint to show (a) - here you do not need the assumption that $ V$ is finite dimensional.

For (b) show $ LHS\subseteq RHS$ first, then that $ \dim LHS \equal{} \dim RHS < \infty$ using
\[ \dim W \plus{} \dim W^\perp \equal{} \dim V < \infty\]
for every sub-vector space $ W$ of $ V$ having [b]finite[/b] dimension.


Dividing through by $a_{n-1}$, we get

$b_{n-1}=1+\frac {1}{b^2_{n-2}}$

Taking limits, we get

$L=1+\frac {1}{L^2}$ or $L^3-L-1=0$.  This is a cubic, and hence it has a real root.  We want to show that there's only one, and that it lies in the desired interval.

Now, $x^3-x=0$ has a maximum for $x\leq 0$ of $\frac {2\sqrt{3}}{9}<1$, so we must have $L>0$

$\frac {d}{dL} L^3-L-1=3L^2-1$, so that it is increasing for $|L|^2>\frac {1}{3}$.

Furthermore, $L^3-L<0$ for $0<L<1$.  So $L>1$.  Setting $L=\frac {3}{2}$, we see that $L^3-L-1=\frac {27}{8}-\frac {3}{2}-1>0$, so

$1<L<\frac {3}{2}$

as desired.  The way to proceed is obvious here, there was just a bunch of boring grunt work to be done.


[quote="zptlimyw"]Just a quick question about the due date/time.

I realize that, first of all, USAMTS Round 4 is due tomorrow; that is, Monday, March 12, 2007.

What I am wondering right now is this:

Will it due by PM 12:00 in PST or EST?[/quote]
PST

That's 3:00 EST for east coast procrastinators such as myself  :blush:

Probably PDT, actually. That's one hour less for you procrastinators.


[quote="kevin gong"]If we submit solutions through email and we typed them on a TXT file, does it have to be in latex or can we write like we were writing on paper (as in typing on the file as if we were writing our solutions by hand)

basically, if we send something by email does it have to be latex or something scanned?[/quote]
I'm not sure if .txt is a USAMTS-approved format; it says on the website it is, but it also says .rtf is, and I know they no longer accept that.  If possible, convert your .txt to a pdf and then submit that (for example, by copy-pasting into Word and then using Go2PDF).  Also, why did you type up your solutions in a text file?  There's no way to display equations well in a text file.

To answer the second part of your question, no, it doesn't have to be LaTeX for them to accept an e-mailed file.

Uhh I kinda got lost on the Go2PDF part so I sent it as a TXT file.

Since I'm not exactly "fluent" in latex it would take me a very long time to type it out in latex, so the equations I typed are in the long (and confusing) way (ie: x^2). Hopefully the graders will be able to make sense of it...hopefully  :maybe: .

Message sent Monday, November 20, 2006 11:57:22 PM

Whew... :oops:

txt should be ok, but we strongly recommend learning LaTeX.  Reading plain text is going to be very painful for the graders!


[quote="math92"]Game birds such as ducks and pheasants that are raised on a farm need a brooder house temperature of 95F to start. The temperarture is then decreased five degrees per week until the outside temperature is reached.  How many weeks will it take for the brooder house temperature to reach 60F?[/quote]

[hide="my answer"]95=start
60=need
5 per week

so (95-60)/5= 35/5= 7 weeks!!![/hide]


compute the largest integer n such that
(2005^2^100)-(2003^2^100) is divisible by 2^n.

[hide="hint"] Factor out $2005-2003$. [/hide]

[hide="solution"]Like paladin8 said, we factor out 2005-2003 a bunch of times and get:

$(2005^{2^{99}}+2003^{2^{99}})((2005^{2^{98}}+2003^{2^{98}}) \cdots (2005^{4}+2003^{4})(2005^{2}+2003^{2})(2005+2003)(2005-2003)$

If you expirement with the first few terms that have exponents, you can see that they only have 1 2 in their prime factorization.

So 99+(x)=n. We are trying to find x, which is how many 2's there are in 2*4008. We multiply it out, and see that there are only 4. 99+4=[b]103[/b][/hide]

I hope that solution is correct... :oops:


For $ 2 \leq k < p$ with $ kp$ even, let $ C(p, k)$ be the maximum number of connected components of any $ k$-regular graph with $ p$ vertices. Determine
the value of $ C(p, k)$.
=)

(Hint: obtain an upper bound for C(p, k), then show by construction that
this bound is attained by some graph.)


Hello! :D

Find
$ \lim_{n \to \infty}\sum_{i\equal{}0}^{n}\frac{1}{n\plus{}\sqrt{i}}$.


The diameter of the base of a right circular cone is 16 inches and the height is 9 inches. What is the volume of the cone in cubic inches? Express your answer in terms of $ \pi$.

[quote="nikeballa96"]$ \frac {16^2\pi\cdot9}{3} \equal{} 256\pi\cdot3 \equal{} \boxed {768\pi}$[/quote]

whoops, diameter = 8, radius=4.

$ \frac{16\pi\cdot3}{1}\equal{}\boxed{48\pi}$

sorry bout that...

Sorry for reviving (is it reviving if it's a FTW prob?)...but I think the answer would be $192\pi$.

What I did was I applied the volume of a cone formula to get:

$\frac{8^2\pi \cdot 9}{3}=192\pi$

Please correct me if I'm wrong...

[quote="nikeballa96"]

whoops, diameter = 8, radius=4.

$ \frac{16\pi\cdot3}{1}\equal{}\boxed{48\pi}$

sorry bout that...[/quote]

The diameter is actually $16$ and the radius $8$ as stated in the problem. Canton is right.


can someone give me the solutions to 2002-03 warmup 18???
even better can someone pm the handbook?
am willing to trade..

I will solve the questions if you pm them to me!

If you look in the stickies, you'll find a whole topic dedicated to trading old tests.
I also have the solutions; maybe I can e-mail them to you.  :wink:


If $ \{A_{\lambda}: \lambda \in J \}$ is a partition of the set A into disjoint subsets whose union is A, and $ X_{\alpha}$ is a topological space for each $ \alpha \in A$, then $ \prod_{\lambda \in J}\left(\prod_{\alpha \in A_{\lambda}}X_{\alpha} \right)$  is homeomorphic to $ \prod_{\alpha \in A}X_{\alpha}$

[quote="Third Edition"]$ \prod_{\lambda\in J}\left(\prod_{\alpha\in A_{\lambda}}X_{\alpha}\right)$ fulfils the universal property of $ \prod_{\alpha\in A}X_{\alpha}$. And this simply boils down to checking that the two products are naturally bijective forgetting the topology.[/quote]

Personal interjection: this is a really awesome way of checking isomorphisms involving representative objects, since it reduces things to the category of sets, which is often much easier to prove things about.

Got it!!  :wink: This is a really nice way to check [b]natural[/b] isomorphisms, since all the properties are already incorporated in the (most of the time nasty proof of the) existence of the representing object. This generalizes of course to situations of categories where you have considerable universal constructions under control. What is behind that is Yoneda's lemma (of category theory) and - what I know as - categorical points of functors.

Thank you for the solutions. Although the category language generalizes and simplifies certain problems, the book [Willard] from which I get this problem does not have a mention of a Universal Property.

Let's define $ \phi: \prod_{\alpha \in A}X_{\alpha} \to \prod_{\lambda \in J}\left(\prod_{\alpha \in A_{\lambda}}X_{\alpha}\right)$ by letting $ \phi(\lambda) \in \prod_{\lambda \in J}\left(\prod_{\alpha \in A_{\lambda}}X_{\alpha}\right)$ such that $ \pi_{\alpha}\left(\pi_{\lambda} \phi(f)\right) \equal{} f_{\alpha}$ for each $ \alpha \in A$.

OK, then $ \phi$ is certainly a bijection. But you need to show that it is a homeomorphism, i.e. $ \phi$ and $ \phi^{\minus{}1}$ are continuous.

1. Let U be open in $ \prod_{\alpha\in A}X_{\alpha}$. Then we have: $ \phi(U) \equal{} \pi_{\lambda}^{\minus{}1}\left(\pi_{\alpha}^{\minus{}1}(U_{\alpha})\right)$, an open set in the range.

I am not even sure if I am doing this right and so I will rather see someone else do the second part.

I would prefer I you could do that ;) - perhaps with the following hints:

1 - Can you provide generating systems of the topology of the two products, resp.,  and show that the preimages of them are open?

2 - Why is this sufficient to show that your set-theoretic bijection is a homeomorphism?

(1). Can you provide generating systems of the topology of the two products, resp?

Set $ X_{\lambda} \equal{} \prod_{\alpha\in A_{\lambda}}X_{\alpha}$ Then we have:

i) The topology on $ X_{\lambda}$ is generated by the elements of the form: $ \pi_{\alpha}^{\minus{}1}\left(U_{\alpha}\right)$, where $ U_{\alpha}$ is open in $ X_{\alpha}$ for each $ \alpha \in A_{\lambda}$

ii) The topology on $ \prod_{\lambda \in J}X_{\lambda}$ is generated by the elements of the form: $ \pi_{\lambda}^{\minus{}1}\left(V_{\lambda}\right)$, where $ V_{\lambda}$ is open in $ X_{\lambda}$ for each $ \lambda \in J$

Now we take the pre-images of the sub-base elements and we get images of the projections, which are open. Since pre-images preserve intersection we are going to have it. This is the answer to your question number (2)


logarithms
geometry
conics
analytic geometry
probability
trigonometry
algebra

Not now, but soon in summer, I'll start to instruct some mathematics through AIM.  Before we start, I need people who want to join plus the things that we should cover on.  I personally think everyone should be able to do basic algebra, logarithrm, series and sequences, geometry, and if advanced, binomial theorem and Pascal's triangle.  But mostly I need idea for topics to cover so if anyone has idea, please post here.

Sincerely,

Silverfalcon

I personally need a good lesson on modular arithmetic.  I don't know if you know enough to teach it, silverfalcon, but maybe someone else could help.  Meanwhile I 'll try to learn it on my own.  I think number bases should be covered and also basic problem solving tactics.

Btw, I'll help you with the instructing if you want me to.  Finally who would be the ones learning?  That's more of a question directed to the ppl in this forum than silverfalcon.

I am also someone who for some reason does not get modular arithmetic, even though I read about it in AoPS books.

For me I would say that I am really good at algebra and pretty good at geometry, because I have had formal instruction of these topics in HS.  So as a result I know a lot of the topics in AoPS2 already like matrices, conics, polar coordinates, etc. but don't know some of even the easy number theory stuff.  *sigh*  I am working on teaching myself this though.  Thus, I know all of the topics that you have listed very well.  Number theory, probability, counting problems, etc. get more shakey as they get harder.  Thanks Rep for the offer btw.

I guess I can join.  I think you should cover some proof-writing techniques.

Proof would be good. I'm trying to cover geometry, trig, and precalc during summer by software I have. If I get pretty well, I think I can do some proofs...

number theory definately, polynomials, modular arithmetic, functions, AIME stuff


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Show counterexamples for the following:

[b](i)[/b] A subring of a noetherian ring is not necessarily a noetherian ring.
[b](ii)[/b] If $ R$ is a commutative ring with $ 1$ and $ J$ is an ideal strictly contained in $ R$ which is finitely generated, then an ideal $ I$ of $ R$ which is strictly contained in $ J$ is not necessarily finitely generated.

Danken!

[hide="(i)"]Subring $ S \equal{} \mathbb{Z} \plus{} pX\mathbb{Z}[X]$ (where $ p$ is prime number) of ring $ \mathbb{Z}[X]$. Since $ \mathbb{Z}$ is noetherian then from Hilbert Basis Theorem $ \mathbb{Z}[X]$ is noetherian, but  $ (pX)\subsetneq (pX, pX^{2})\subsetneq \ldots \subsetneq (pX, \ldots, pX^{n})\subsetneq \ldots$ is infinite ascending chain of ideals in $ S$[/hide]
[hide="(ii)"]Take $ R \equal{} \mathbb{Z} \plus{} pX\mathbb{Z}[X]$, where $ p$ is prime number, $ J \equal{} (q)$, where $ q\in \mathbb{Z}\setminus \{ \minus{} 1,1\}$ and $ q$ is not divisible by $ p$, $ I \equal{} (pqX^{i} \ : \ i \in \mathbb{N}_{1} \equal{} \{1,2,3,\ldots\})$[/hide]


integration
function
trigonometry
real analysis
real analysis solved

how do you show that $f(x)=\sum_{n=1}^{\infty}\frac{\sin(2\pi nx)}{\sqrt n}$ is not continuous?

There are several ways this could be done. Here's one: $f\not\in L^2,$ so it can't possibly be continuous.

More specifically, by Parseval's identity,

$\int_0^1|f(x)|^2\,dx=\frac12\sum_{n=1}^{\infty}\left(\frac1{\sqrt{n}}\right)^2 =\frac12\sum_{n=1}^{\infty}\frac1n=\infty.$

A continuous function on $[0,1]$ would be bounded and would have a finite integral.

That suffices. With some more sophisticated work, it should be possible to show that as $x\to0,\,|f(x)|$ looks like $\frac C{\sqrt{|x|}}.$

that seems the quick answer, but is not good enough :lol:

Parseval identity presuposes that the series is *the* Fourier series of $f(x)$, which is not obvious. ($f$ is the sum of the series, which converges by abel criterion..)

so could you tell me about the more sophisticated work please..

If anyone cares: as k. suggested, one should prove that $|f(x)|\gg\frac{1}{\sqrt x}$ as $x\to0^+$, hence $f$ blows up at $x=0$.
To see this:

Summing by parts and using identity $\sum_{k=1}^n\sin(kx)=\frac{\sin(\frac{n+1}{2}x)\sin(\frac{nx}{2})}{\sin\frac{x}{2}}$ gives:

$f(x)=\frac{1}{\sin\frac{x}{2}}\sum_{n=1}^{\infty}\sin(\frac{n+1}{2}x)\sin(\frac{nx}{2})\big(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\big)$.

Formula :

$\sin(\frac{n+1}{2}x)\sin(\frac{nx}{2})=\frac{1}{2}(\cos(\frac{x}{2})-\cos(nx+\frac{x}{2}))$
$=\frac{1}{2}(\cos(\frac{x}{2})-\cos(nx)\cos(\frac{x}{2})+\sin(nx)\sin(\frac{x}{2}))$
$=\frac{1}{2}\sin(\frac{x}{2})\sin(nx)+\cos(\frac{x}{2})\sin^2(\frac{nx}{2})$

yields, with the notation $a_n=\frac{1}{\sqrt n}-\frac{1}{\sqrt{n+1}}$, the following formula

$f(x)=\frac{1}{2}\sum_{n=1}a_n\sin(nx)+\cot(\frac{x}{2})\sum_{n=1}^{\infty}a_n\sin^2(\frac{nx}{2})$
$=g(x)+\cot(\frac{x}{2})h(x)$

On the one hand, $a_n=O(n^{-3/2})$, so the series defining $g(x)$ converges uniformly, thus $g(x)$ is continuous.  In particular $g(x)=O(1)$ as $x\to 0$.

On the other, $h(x)$ is defined by a positive series so

\[ h(x)\geq \sum_{\frac{\pi}{2x}\leq n\leq\frac{3\pi}{2x}}a_n\sin^2(\frac{nx}{2})  \]

But in this range $\sin^2(nx)\geq \frac{1}{2}$. Also,  $a_n\gg n^{-3/2}$
[with notation: $x_n\gg y_n$ if there exists $C>0$ such that $x_n\geq C y_n$ for all $n$]

Since the length of the range is greater than a constant times $x^{-1}$, we have

\[ h(x)\gg x^{-1}\cdot x^{3/2}=x^{1/2}  \]

But $|\cot(\frac{x}{2})|=\left|\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}\right|\gg\frac{1}{|x|}$ as $x\to 0$.  Therefore $\cot(\frac{x}{2})h(x)\gg |x|^{-1/2}$ is the bad guy.

Conclusion:

$|f(x)|=|g(x)+\cot(\frac{x}{2})h(x)|\gg |x|^{-1/2}$ as $x\to 0$, which is what was needed.

More summation by parts probably gives $h(x)=O(|x|^{-1/2})$ as well,   so we see that $|x|^{-1/2}$ is indeed the true order of magnitude of $f(x)$ as $x\to 0$ (and not just a lower bound).


[hide]there is no one except 0,0,0...
the proof is that we want $\frac{x+d}{x}=\frac{x}{x-d}$ thus $d=0$ so k,k,k..satisfy this[/hide]

[quote="king_23"]I'm not sure how to create a sequence that is both arithmetic and geometric.[/quote]

Well, I can only consider the trivial cases.

{a}
{a,b}
and
{a,a,a,...}

Where a and b are real numbers.  But I don't know if they could be imaginary.

suppose the first 3 terms are:

$a-b$, $a$, $a+b$

then it is aritmetic, now for geometric

$(a-b)(a+b)=a^2$
$b=0$

therefore the sequence is a constant

i think there is such sequence that is both arithmetic and geometric since randomness and chaos exists.in my opinion everything has a sequence and that sequence must be arithmetical,geometrical,.....etc, loads of them in one sequence,
that makes it difficult for us to predict future.

I admit to being somewhat baffled as to the relevance of "chaos," "randomness," or "predict[ing the] future" to this question.  "Arithmetic" and "geometric" are not some mystical phrases, the extent of which are unknown and unknowable -- they mean something very specific and well-defined.  The post seems particularly odd because it follows directly a proof that the only arithmetic-geometric sequences with 3 or more terms are constant.

How is something like $\frac{1}{2}, \frac{2}{4}, \frac{3}{8}, ..., \frac{n}{2^n}$ categorized?


[quote="Sunkern_sunflora"]If there be three real nos. a,b,c lying in the interval [0,1], prove that

               [b] [a/(b+c+1)]+[b/(c+a+1)]+[c/(a+b+1)]+(1-a)(1-b)(1-c) is less than equal to 1.[/b][/quote]
Suppose that $ a \leq b \leq c$
then $ \frac{a}{b\plus{}c\plus{}1}\plus{}\frac{b}{c\plus{}a\plus{}1}\plus{}\frac{c}{a\plus{}b\plus{}1} \leq \frac{a\plus{}b\plus{}c}{a\plus{}b\plus{}1}$
$ (1\minus{}a)(1\minus{}b)(1\plus{}a\plus{}b) \leq 1$
so $ (1\minus{}a)(1\minus{}b) \leq \frac{1}{1\plus{}a\plus{}b}$
=>$ (1\minus{}a)(1\minus{}b)(1\minus{}c) \leq \frac{1\minus{}c}{1\plus{}a\plus{}b}$
=>$ \sum_{cyc} \frac{a}{b\plus{}c\plus{}1} \plus{}(1\minus{}a)(1\minus{}b)(1\minus{}c) \leq \frac{a\plus{}b\plus{}c}{a\plus{}b\plus{}1}\plus{}\frac{1\minus{}c}{a\plus{}b\plus{}1}\equal{}1$
Done!

Oh! That was a beauty chien than! Just I didn't understand the very first step, i.e.
         [a/(b+c+1)]+[b/(c+a+1)]+[c/(a+b+1)] < [(a+b+c)/(a+b+1)]
Could you explain it?
Here's another inequality for you, chien than!

[i]Prove, for positive reals, [/i] [b](ab)^1/2+(cd)^1/2 < [(a+d)(b+c)]^1/2[/b] :wink:

[quote="Sunkern_sunflora"]Oh! That was a beauty chien than! Just I didn't understand the very first step, i.e.
         [a/(b+c+1)]+[b/(c+a+1)]+[c/(a+b+1)] < [(a+b+c)/(a+b+1)]
Could you explain it?
[/quote]

note that $ a\leq b\leq c$ so:

$ b\plus{}c\plus{}1\geq a\plus{}b\plus{}1\Rightarrow\frac{a}{b\plus{}c\plus{}1}\leq\frac{a}{a\plus{}b\plus{}1}$

$ c\plus{}a\plus{}1\geq a\plus{}b\plus{}1\Rightarrow\frac{b}{c\plus{}a\plus{}1}\leq\frac{b}{a\plus{}b\plus{}1}$

$ \Rightarrow\frac{a}{b\plus{}c\plus{}1}\plus{}\frac{b}{c\plus{}a\plus{}1}\plus{}\frac{c}{a\plus{}b\plus{}1}\leq\frac{a\plus{}b\plus{}c}{a\plus{}b\plus{}1}$

[quote="Sunkern_sunflora"][i]Prove, for positive reals, [/i] [b](ab)^1/2+(cd)^1/2 < [(a+d)(b+c)]^1/2[/b] :wink:[/quote]

I think it should be:

$ \sqrt{ab}\plus{}\sqrt{cd}\leq\sqrt{(a\plus{}d)(b\plus{}c)}$

anyway,by squaring both sides we have to prove that:

$ ab\plus{}cd\plus{}2\sqrt{abcd}\leq (a\plus{}d)(b\plus{}c)$

$ \iff 2\sqrt{abcd}\leq ac\plus{}bd$

which is a directly consequence of AM-GM.


Can you find a formula for the sum from i=1 to i=n of

$\Sigma\frac{1}{(i)(i+1)(i+2)}$

Try decomposing it into partial fractions and then telescoping the terms to simplify things.

Yes, telescoping is one of my favorite algebraic tools.

In case you don't know what that is, its when you write out a sum, and lots of terms cancel, leaving like some amount of terms to add that you can deal with, like

$(a-b)+(b-c)+(c-d)$

or something.

And in general we know what

$\sum_{i=1}^{\n}\frac{1}{(i)(i+1)(i+2)\cdots(i+m)}$

is. Its one of the more well know formulas.


Let $ a,b,c$ be positive numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove that $ (1 \plus{} a)\sqrt {\frac {1 \minus{} a}{a}} \plus{} (1 \plus{} b)\sqrt {\frac {1 \minus{} b}{b}} \plus{} (1 \plus{} c)\sqrt {\frac {1 \minus{} c}{c}}\geq\frac {3\sqrt {3}}{4} \frac {(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}{\sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}}$       :?:

Anyone has a solution...it looks rather difficult...

[quote="thepunisher123"]It's very easy  :rotfl:[/quote]

I hope you can post your solution....Thanks in advance.


For every positive numbers $ a,b,c,x,y,z$, prove that

$ \sqrt[3]{a(b\plus{}1)yz}\plus{}\sqrt[3]{b(c\plus{}1)zx}\plus{}\sqrt[3]{c(a\plus{}1)xy}\leq \sqrt[3]{(a\plus{}1)(b\plus{}1)(c\plus{}1)(x\plus{}1)(y\plus{}1)(z\plus{}1)}$

[quote="radio"]For every positive numbers $ a,b,c,x,y,z$, prove that

$ \sqrt [3]{a(b \plus{} 1)yz} \plus{} \sqrt [3]{b(c \plus{} 1)zx} \plus{} \sqrt [3]{c(a \plus{} 1)xy}\leq$
$ \leq\sqrt [3]{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)(x \plus{} 1)(y \plus{} 1)(z \plus{} 1)}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89595


inequalities
function
logarithms
inequalities unsolved

$f(x,y)={1 \over \sqrt{1+xy}}$ in $[0,1]^2$
${d^2f \over dxdy}={1-xy \over 4(xy+1)^2\sqrt{xy+1}}\geq 0 $

so we can use Tschebyschev inequality:
$f(x,y)+f(x,y)\geq f(x,x)+f(y,y)$

Or i have mistake in my calculation

[quote="erdos"]Prove the inequality :$\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}\leq \frac{2}{\sqrt{1+xy}}$
for $0\leq x,y\leq 1$[/quote]
Actually, if I correctly remember we proved somewhere on the forum that it holds for $xy\leq 1$. Try it! :) It is more hard, in my opinion, but who knows ;)

A trigonometric substitution $x=tg\alpha,y=tg\beta$ would be very helpful.Between can we use Weiershtrass theorem that the extremum must be at the  endpoints and check putting just the values $(x,y)=(1,1);(1,0);(0,0).$

An other way :  :lol:
for $x=0$ or $y=0$ the inequality is obvious
suppose $0<x\leq 1$ and $0<y\leq 1$ then there exist $u$ and $v$ such as :
$x=e^{-u}$ and $y=e^{-v}$ with $u,v\geq 0$
the inequality becomes $\frac{1}{\sqrt{1+e^{-2u}}}+\frac{1}{\sqrt{1+e^{-2v}}}\leq \frac{2}{\sqrt{1+e^{-\left( u+v\right) }}}$
considering the function $f\left( x\right) =\frac{1}{\sqrt{1+e^{-2x}}}$ for $0<x\leq 1$
we have $f^{\prime \prime }\left( x\right) =\allowbreak e^{-2x}\frac{e^{-2x}-2}{\left( \sqrt{\left( 1+e^{-2x}\right) }\right) ^{5}}\geq 0$
so $f$ is convex.
Applying Jenson inequality we get the result

Indeed, it is Jensen! But it fails for the generalization :D ;)

[quote="erdos"]Prove the inequality :$\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}\leq \frac{2}{\sqrt{1+xy}}$
for $0\leq x,y\leq 1$[/quote]
it remains to prove that
$\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}\leq \frac{2}{1+xy}.$

[quote="Myth"][quote="erdos"]Prove the inequality :$\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}\leq \frac{2}{\sqrt{1+xy}}$
for $0\leq x,y\leq 1$[/quote]
Actually, if I correctly remember we proved somewhere on the forum that it holds for $xy\leq 1$. [/quote]
Are you mean for $-1<xy\leq1$?

[quote="erdos"]Prove the inequality :$\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}\leq \frac{2}{\sqrt{1+xy}}$
for $0\leq x,y\leq 1$[/quote]
This inequality is true for all reals $x$ and $y$ such that $-1<xy\leq2$.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=576617
Let $a, b $  be positive real numbers such that $ab\ge 3$ , prove that\[\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\ge \frac{2}{\sqrt{1+ab}}.\]

I think Jensen kills the inequality. It is not difficult to see:

$\frac{1}{\sqrt{1+ xy}} \geq \frac{1}{\sqrt{1+ (\frac{x+y}{2})^2}}$

Because $(x + y)^2 \geq 4xy$ and

 $ \frac{1}{2.\sqrt{1+ x^2}}  + \frac{1}{2.\sqrt{1+ y^2}}   \leq \frac{1}{\sqrt{1+ (\frac{x+y}{2})^2}} $

By Jensen taking $f(x) = \frac{1}{\sqrt{1+ x^2}}$

thus $f'(x) = \frac{- x}{\sqrt{(1+ x^2)^3}}$

And finaly

 $f''(x) = - \frac{x^6 + 3x^4 + 1}{\sqrt{(1+ x^2)^9}} < 0 $  obviuosly.


vector
trigonometry
Pythagorean Theorem
geometry

can you help me to trancelate the numbers to polar form?

1)  1+j
2)  -4+3j
3)  (1+j)(-4+3j)
4)  e^(j*π/4) + 2e^(-j*π/4)
5)  e^j + 1

      thank you!!

I think it is the unit vector $ \hat{\mathbf{j}}$.

It may as well be $ i$.

[hide="Hints"]1. The argument (angle) should be pretty obvious for this one. Pythagorean theorem for the radius.
2. Same thing, though you'll need either the answer in terms of $ \arctan$ or a calculator.
3. De Moivre's Theorem. The answer to this is obtained if you multiply the radii of the answers to 1 and 2 and add their arguments.
4. Simplify it into $ a \plus{} bi$ form.
5. $ \cos 1$ and $ \sin 1$ ($ 1$ radian) are pretty much meaningless numbers, so I'd put the answer in terms of them.[/hide]

In electrical engineering and some other applied fields, the imaginary unit is often called $ j$ instead of $ i$ to avoid confusion with current.

In specialized cases, $ \iota$ is sometimes seen to denote the imaginary unit.

thank you a lot!  Can you help me to transform please!!

A complex number in cartesian form is given by $ a \plus{} ib$. Now, looking at this number in the complex plane as an oriented vector from the origin to the point $ a \plus{} ib$, then its polar form is given $ \sqrt{a^2\plus{}b^2} \cdot cis(\theta)$, where $ cis$ means $ \cos \theta \plus{} i \sin \theta$, and the angle $ \theta$ is defined as $ \theta \equal{} \arctan \frac{b}{a}$, for a > 0, and as $ \theta \equal{} \pi \plus{} \arctan \frac{b}{a}$, for a < 0.

1) 1+j
2) -4+3j
3) (1+j)(-4+3j)
4) e^(j*π/4) + 2e^(-j*π/4)
5) e^j + 1

I would really appreciate your help on exersise above.
Any suggestions-entries are most welcome.

???

We've given you plenty of help. We're not going to do the problems for you if that's what you're expecting.


algebra
polynomial
induction
function
factorial
abstract algebra
number theory

[quote="Elio (n)"]If the polynom $ ax^{3} \plus{} bx^{2} \plus{} cx \plus{} d$ is divisble by 5 for all x prove that $ 5|a,b,c,d$  :)[/quote]
hi $ p(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$
$ \forall x \in R      p(x)\equiv 0(5)$ so
$ p(0) \equal{} 0(5)$ ==> $ d \equal{} 0(5)$
$ p(2) \equal{} 0(5)$ ==> $ 4a \plus{} 2b \plus{} c \equal{} 0(5)$  (1)
$ p(1/2) \equal{} 0(5)$ ==> $ a \plus{} 2b \plus{} 4c \equal{} 0(5)$  (2)
from (1) and (2) $ b \equal{} 0(5)$  and $ a \minus{} c \equal{} 0(5)$
$ p(1) \equal{} 0(5)$==> $ a \plus{} b \plus{} c \equal{} 0(5)$ ==> $ (a \plus{} c) \equal{} 0(5)$
finally 5/(a.b.c.d)

What if we try to generalize?

Given a positive integer $ k$ and a polynomial $ P_n(x) \equal{} a_0x^n \plus{} a_1x^{n \minus{} 1} \plus{} \dots \plus{} a_{n \minus{} 1}x \plus{} a_n$ with integer coefficients such that $ k\mid P_n(x)$ for all $ x$, prove that $ k\mid a_i\,\forall i$.

Lemma: If $ k\mid mn$ for all $ m$ then $ k\mid n$.

Proof: Choose an $ m$ such that $ (k,m) \equal{} 1$ and the conclusion follows.

Let us proceed by induction on the degree of the polynomial. This is clearly true for a linear polynomial, since $ k\mid P_1(0) \equal{} a_1$ and so $ k\mid a_0x\,\forall x\implies k\mid a_0$, by the lemma. Now assume that it is true for any polynomial $ P_{n \minus{} 1}(x)$ of degree $ n \minus{} 1$. Then $ k\mid P_n(0) \equal{} a_n$ implies $ k\mid a_0x^n \plus{} a_1x^{n \minus{} 1} \plus{} \dots \plus{} a_{n \minus{} 1}x\,\forall x$. But this implies $ k\mid x(a_0x^{n \minus{} 1} \plus{} a_1x^{n \minus{} 2} \plus{} \dots \plus{} a_{n \minus{} 2}x \plus{} a_{n \minus{} 1}) \equal{} xP_{n \minus{} 1}(x)$. Hence by the lemma we conclude that $ k\mid P_{n \minus{} 1}(x)$ for all $ x$ and thus by the inductive hypothesis $ k\mid a_i$ for $ i \equal{} \overline{0,n \minus{} 1}$. Therefore $ k\mid a_i$ for all $ i \equal{} \overline{0,n}$.

EDIT: I think we can generalize even further.

If $ k\mid P_n(x)$ for $ n \plus{} 1$ consecutive integer values of $ x$ then $ k\mid a_i\,\forall i$.

The proof should be similar I guess.

[quote="nayel"]Given a positive integer $ k$ and a polynomial $ P_n(x) \equal{} a_0x^n \plus{} a_1x^{n \minus{} 1} \plus{} \dots \plus{} a_{n \minus{} 1}x \plus{} a_n$ with integer coefficients such that $ k\mid P_n(x)$ for all $ x$, prove that $ k\mid a_i\,\forall i$.[/quote]
This is far too strong a result.  Let $ k \equal{} r!$ and let $ P(x) \equal{} x(x \minus{} 1)...(x \minus{} (r \minus{} 1))$.  Similarly, let $ k \equal{} p$ (a prime) and let $ P(x) \equal{} x^p \minus{} x$.  (Your proof is invalid because $ P_{n \minus{} 1}(x)$ is not a constant, so your lemma is not applicable.)

In generalizing this result, it is essential to compare $ k$ to $ n$.  Restricting $ k$ to prime values, the problem is this:  we are given a polynomial with $ p$ roots in $ \mathbb{Z}_p$ and we wish to prove that it is the zero polynomial.  Since $ P(x) \equal{} x^p \minus{} x$ is a polynomial with $ p$ roots that is [b]not[/b] the zero polynomial, we know that $ n < p$ is a necessary condition.  Now, can you prove that it is sufficient?  How about generalizing to non-prime values of $ k$?

Edit:  The necessary lemma for prime values of $ k$ is [url=http://en.wikipedia.org/wiki/Lagrange%27s_theorem_%28number_theory%29]Lagrange's theorem[/url].

I see. Thank you very much for pointing it out. :)

The problem for non-prime values of $ k$ is quite interesting.  The fact that Lagrange's theorem fails makes it harder to get a grip on $ n$, although we know that for $ k \equal{} r!$ the smallest $ n$ that does not satisfy the problem criterion (which is $ r$) can be quite small.

[hide="Further discussion"] First of all, an explicit counterexample for some $ n$ for any value of $ k$ is as follows.  Let $ e$ be the largest exponent of a prime in the prime factorization of $ k$.  Then

$ P(x) \equal{} \left( x^{\lambda(k)} \minus{} 1 \right) x^e$

is divisible by $ k$ for all values of $ x$ (where $ \lambda(k)$ is the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url]), so we know that $ n < \lambda(k) \plus{} e$.  The behavior of $ \lambda(k)$ depends strongly on the prime factorization of $ k$, so while this bound is sharp when $ k$ is prime it does not guarantee much when $ k$ is composite.  For example, when $ k \equal{} 5! \equal{} 2^3 \cdot 3 \cdot 5$ we get $ \lambda(k) \plus{} e \equal{} 4 \plus{} 3 \equal{} 7$ but we know that $ 5$ is a stronger upper bound.

Having the factorial example at hand presents a second idea.  Let $ M$ be the smallest positive integer such that $ k | M!$.  Then

$ P(x) \equal{} x(x \minus{} 1)...(x \minus{} (M \minus{} 1))$

is divisible by $ k$ for all integer $ x$.  Then $ n < M$.  I believe this bound is tight when $ k \equal{} M!$ (the case we considered earlier), and it is also tight when $ k$ is prime (because $ M \equal{} p$), but the growth rate of $ M$ when $ k$ is not of either form again strongly depends on the prime factorization of $ k$ and I am again unsure whether it is tight in general.  [/hide]
The next step after the case $ k \equal{} p$ is the case $ k \equal{} p^m$, which I think might be enough to solve the composite case.


modular arithmetic
number theory unsolved
number theory

Determine all quadruples $ (a,b,c,d)$ of nonnegative integers satisfying:

$ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}a^2 b^2 c^2$.

[quote="moldovan"]Determine all quadruples $ (a,b,c,d)$ of nonnegative integers satisfying:

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} a^2 b^2 c^2$.[/quote]

Clearly $ (a,b,c,d)\equal{}(0,0,0,0)$ is a solution.

Assume that other solution $ (a_1,b_1,c_1,d_1)$ exists with $ \gcd(a_1,b_1,c_1,d_1)\equal{}1$.

Then any one of $ a_1,b_1,c_1$ even $ \Rightarrow$ all of $ a_1,b_1,c_1,d_1$ even after mod 4 both sides of the equation, contradiction as now $ 2|\gcd(a_1,b_1,c_1,d_1)$.

all $ a_1,b_1,c_1$ odd $ \Rightarrow a_1^2\plus{}b_1^2\plus{}c_1^2\plus{}d_1^2 \equiv 3\plus{}d_1^2 \equiv 3,0 \pmod{4}$, contradiction as $ a_1^2b_1^2c_1^2 \equiv 1 \pmod{4}$

So $ (a,b,c,d)\equal{}(0,0,0,0)$ is the only solution.


Two dice are thrown.  What is the probability that the product
of the two numbers is a multiple of 5?  Express your answer as a common fraction.

That means one of the numbers is 5.
The other number we can choose in 6 ways, and we divide by 2 because there's no matter of order.
15

The probability of no 5s is 5/6 ^2. This its 1-25/36=11/36

don't get something- if you just think about it, as long as there is a 5 for one dice, the product will automatically be a multiple of 5. So the prob is 1/6 then accounting for both its [u][i][b]1/3[/b][/i][/u]

Because you are overcounting when we get 5 on both dice.

[quote=Tuatramxo]Let's use...
[b][size=150][center]BRUTE FORCE[/size][/center][/b]
[hide]Dice combinations for $5$: 1 & 5 and 5 & 1 (2)
           Dice combinations for $10$: 2 & 5 and 5 &1 (2)
           Dice combinations for $15$: 3 & 5 and its inverse (2)
           Dice combinations for $20$: 5 & 4 & inverse (2)
           Dice combinations for $25$: 5 & 5 (1)
           Dice combinations for $30$: 5 & 6 & inverse (2)
           Add them all, and you get 11 in total.
           We have $6*6$ total cases, which gives us $[box]11/36[/box]$ cases[/hide][/quote]

Agreed.


calculus
integration
function
calculus computations

Find the indefinite integral $ \int \frac{x^3}{(x\minus{}1)^3(x\minus{}2)}\ dx$.

i think now it is right ... if not plz say my fault and correct answer

Partial fraction decomposition gives a function of the form:

$ \frac{A}{(x-1)^3}+\frac{B}{(x-1)^2}+\frac{C}{x-1}+\frac{D}{x-2} = \frac{x^3}{(x-1)^3(x-2)}$

Putting over a common denominator and solving the resulting linear system gives:

$ A=-1$
$ B=-4$
$ C=-7$
$ D=8$

The integral is then:

$ \int{\frac{x^3}{(x-1)^3(x-2)}=\frac{1}{2(x-1)^2}+\frac{4}{x-1}-7\ln|x-1|+8\ln|x-2|}$


Prove there exists infinitely many squares not of the form 4xy+y+x. I think this is the same as proving that there exists infinitely many primes of the form kk+1 so a simple proof may not be possible.

Closely related to http://www.mathlinks.ro/Forum/viewtopic.php?t=90831 ;)

Eqivalent this time: $(2z)^2+1 = (4x+1)(4y+1)$.
From that point, it's easy to see that is equivalent to $\infty$ many primes of type $(2z)^2+1$.


How many ordered triples of primes $ (a,b,c)$ exist such that $ a\le b\le c$ and $ a\plus{}b\plus{}c\equal{}26$?

[quote="GameBot"]How many ordered triples of primes $ (a,b,c)$ exist such that $ a\le b\le c$ and $ a \plus{} b \plus{} c \equal{} 26$?[/quote]

One of the primes[b]must[/b] be 2 or else you couldn't have an even sum.  Now we need two primes summing to 24.  They are:

(5,19)
(7,17)
(11,13)

So we have:

$ 2 \le 5 \le 19\\
2 \le 7 \le 17\\
2 \le 11 \le 13$

Thus our answer is $ \boxed{3}$


(US) I just heard that the NSA is coupling up with companies like Horizon and spying on Americans' phone calls...We've got do have some sort of privacy around us, don't you think?

You can't. For all that bloody US Govt knows you could well be plotting against it. (sorry for the expletive)


it's an  easy problem :
if we have :

xy / (x,y)^2  =  z
and xz / (x,z)^2 = y
then prove that : (y / (x,y) , (x,y) ) = 1

Denote $(x,y)=d, x=da, y=db, (a,b)=1$

Then $xy=z(x,y)^{2}\Rightarrow adbd=zd^{2}\Rightarrow z=ab$

$xz=y(x,z)^{2}\Rightarrow adab=bd(ad,ab)^{2}\Rightarrow (b,d)=1\Rightarrow (\frac{y}{(x,y)}, (x,y) ) = 1$


What are the possible lenths of horizontal line segments that start and finish on the graph of $ x^3 \minus{} x$

It is $ 2$. Let $ f(x) \equal{} x^3 \minus{} x$. The line that passes through the points $ (x, f(x))$ and $ (y, f(y))$ is horizontal iff $ f(x) \equal{} f(y)$. Then its lenght is $ |x \minus{} y|$. Suppose by absurd $ |x \minus{} y| > 2$ and $ x^3 \minus{} x \equal{} y^3 \minus{} y$. We have

$ x^3 \minus{} y^3 \equal{} x \minus{} y \Longleftrightarrow x^2 \plus{} xy \plus{} y^2 \equal{} 1 \Longleftrightarrow (x \minus{} y)^2 \plus{} 3xy \equal{} 1$

We have $ (x \minus{} y)^2 > 4 \Longrightarrow 3xy < \minus{} 3$. Then $ \minus{} 1 < \frac {1}{xy} \equal{} 1 \plus{} \frac {x}{y} \plus{} \frac {y}{x} \Longrightarrow \frac {x}{y} \plus{} \frac {y}{x} > \minus{} 2$. So $ x$ and $ y$ have the same sign (otherwise we would have $ \frac {x}{y} \plus{} \frac {y}{x} \le \minus{} 2$). But this contradicts $ xy < 0$.

We have $ |x \minus{} y| \equal{} 2$ for $ x \equal{} 1$ and $ y \equal{} \minus{} 1$.

If you look at the graph of $ x^3\minus{}x$, you will notice it is increasing on $ (\minus{}\infty,\minus{}\frac1{\sqrt{3}})$ and $ (\frac1{\sqrt{3}},\infty)$ and decreasing otherwise. So $ x^3\minus{}x\equal{}a$ has three real solutions when $ a\in(\minus{}\frac{2\sqrt{3}}9,\frac{2\sqrt{3}}9)$ and has two solutions when $ a\equal{}\pm\frac{2\sqrt{3}}9$. My calculation is this line segments may have any length in $ (0,2]$.


Has anyone from here gone to the arml tryouts of East MA 2006? If so post here!
I did pretty well but I made stupid mistakes on two of the questions. like for example there was one qeustion that asked for the area of a circle, but instead I put down the radius...

(This forum looks empty)

yea i took the tryout

i got a 15  :( , which would have a been a decent score if the problems were actually hard

i hate those problems were you have to use a lot of ingenuity to get to the last step but then out a nowhere there is an extraneous solution that wastes ur effort

i hope the problems at the practices are harder

hmm.. i hear lexington might not coem with us

if so...[hide]omg we're screwed!!!!!!!!![/hide]

we're all from Ma in this forum anyways, so no fear in using real names, my name is Norman Yu, Boston Latin School, Grade 11

Heh, yeah, I'm from Lex, and we would have gladly come had someone told us about the tryout. hmm . . .

But anyways, we might have our coach just try us out for the team at Lex, so try to not discuss problems on the forum.

We also have an extremely small chance of forming our own ARML team, but that probably won't happen.

Oh, yeah, if we make the team are practices required, and when are practices?

Hmm.. steve xu and sway chen did ARML last year... they should;ve known

mr curry sent the tryouts to mr rahman, so he ahs the problems and should be able to admiinister them

ask sway about the practice schedule, to lazy to type

Yeah, none of us unfortunately can go to the practices, so we can't try out.

o man, so no one from lex is going?!?!??

Nope. No one at all. Unless someone has a change of heart, which they probably won't.

hello, Norman :D
It's me, James Sun from Brookline high.

My school also didn't receive an email for the tryouts (on time), so I would have missed it if I did not hear about it. that's a shame too, because I know a few people at my school who would have performed just as well..

Anyways, norman, how do you find out that z = -15 is extraneous?
Also, I can't figure out the dollar chang problem, tell me why the answer is 9 (or tell me what the problem asked for)

well... we shouldn't discuss the solutions to the problems yet, but i dont really care

you figure out z=15 is extraneous because on of the equations was x^2+y^2+z^2 = 49 or something, if z=15, then x^2+y^2 will be negative which cant happen since its states x,y,z are real

the dollar change one is just counting... hard to explain thr\ourougjly

Noah, the EMass ARML tryouts were announced at the end of the 2/2 MML meet at Westford Academy.  I heard the announcement.  I know the Worcester Academy kids heard it, because those of them trying out for WMass ARML asked what it was all about.

Our school heard about it, but I thought it would be too hard to qualify. Turns out everyone from my school who went qualified, and I should have gone  :?


I was wondering about the proof that the sum of an infinite geometric series being $\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio.  What is it?

[quote="varun"]while considering infinite geo. sum you should always mention that the mod of common ratio is less than 1
i.e. $\mid r\mid \leq 1$[/quote]

or we can just stick with the general formula, $\frac{a(1-r^{n})}{1-r}$.

Personally, I like to prove geometric with the identity $x^{n}-1 = (x-1)(x^{n-1}+x^{n-2}+\dots+1)$

[quote="varun"]while considering infinite geo. sum you should always mention that the mod of common ratio is less than 1
i.e. $\mid r\mid \leq 1$[/quote]
Well, duh.  Otherwise the terms would grow and their sum would be plus or minus infinity (depepnding on the value of $r$).  And if $r=-1$, you can't really determine it.

[quote="eryaman"]Basically you take the formula for regular geometric series $\frac{a(1-r^{n})}{1-r}$, take the limit as $n\to \infty$ where $|r|<1$.  What happens is $r^{n}$ "becomes" $0$, and we are left with $\frac{a}{1-r}$.[/quote]
Or, using mathematical syntax: When $|r|<1$, $\lim_{n\to\infty}{r^{n}}=0$. Therefore, $\lim_{n\to\infty}{\frac{a(1-r^{n})}{1-r}}=\frac{a}{1-r}$.

Okay, thanks, everyone.  I was wondering about it.

A proof without really using limits:

[quote="manuel"]Consider $S-Sr$[/quote]

Let $S=a+ar+ar^{2}+ar^{3}+...$
Then $Sr=ar+ar^{2}+ar^{3}+...$
Thus $S-Sr=a$
$S(1-r)=a$
$S=\frac{a}{1-r}$
Q.E.D.

the one I have learned goes:

Let S be the sum of the series.
Let r be the rate.
Let a be the first term.

$S = a+ar+ar^{2}+ar^{3}+...$
Factor out the r from $ar+ar^{2}+...$

$S = a+r(a+ar+ar^{2}+ar^{3}+...)$

Since $a+ar+ar^{2}...$ has already been accepted as S, substitute S in for the quantity in parenthesis.

$S = a+rS$
So $S-rS = a \Rightarrow S(1-r) = a \Rightarrow S = \frac{a}{1-r}$

I hope it makes sense :)

yes the above 2 posts show the most common technique for this problem, however, you have to be careful of the way that you manipulate $S$, by subtracting, etc, you implicitly assume that S has a finite value, which is only true iff $|r|<1$

Oops sorry I didn't include it in my proof!  I had figured that all of the proofs above mine had made it clear that |r| must be less than one.

[quote="mathnerd314"]A proof without really using limits:[/quote]
I know all I do in the intermediate forum is whine about lack of rigour, but anyway: The very definition of an infinite sum depends on limits. You can not just add up an infinite amount of numbers with the traditional addition.

Yeah, you can't really get this proof without limits.

Here's how I'd prove it:
$S=a+ar+ar^{2}+ar^{3}+\cdots$
$Sr=ar+ar^{2}+ar^{3}+ar^{4}+\cdots$

So $S-Sr=a-ar^{\infty+1}$.  Since $\lim_{k\to\infty+1}ar^{k}=0$ (since $|r|<1$), we have $S-Sr=a$.  The rest is pretty easy.


Find all $ m,n$ such that $ 1\plus{}5.2^{m}\equal{}n^{2}$

$ 5 \cdot 2^m\equal{}(n\plus{}1)(n\minus{}1), 2\nmid n, (n\plus{}1,n\minus{}1)\equal{}2$, so $ n\plus{}1$ is the form $ \{5\cdot 2,5\cdot 2^{m\minus{}1},2,2^{m\minus{}1}\}$ that leads to $ n \in \{9,3,1,11\}$, but  $ (m,n)\equal{}(4,9)$ is the unique solution.


trigonometry
algebra
polynomial
Pythagorean Theorem
geometry
trig identities
Law of Cosines

A machine-shop cutting tool has the shape of a notched circle, as shown.  The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is 6 cm, and that of $BC$ is 2 cm.  The angle $ABC$ is a right angle.  Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy]
size(150); defaultpen(linewidth(0.65)+fontsize(11));
real r=10;
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;
path P=circle(O,r);
C=intersectionpoint(B--(B.x+r,B.y),P);
draw(Arc(O, r, 45, 360-17.0312));
draw(A--B--C);dot(A); dot(B); dot(C);
label("$A$",A,NE);
label("$B$",B,SW);
label("$C$",C,SE);
[/asy]

[hide]Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Extend a perpendicular from $O$ to $AB$ and label it $D$. Additionally, extend a perpendicular from $O$ to the line $BC$, and label it $E$. Let $OE=x$ and $OD=y$. We're trying to find $x^2+y^2$.

Applying the Pythagorean Theorem, $OA^2 = OD^2 + AD^2$, and $OC^2 = EC^2 + EO^2$.

Thus, $(\sqrt{50})^2 = y^2 + (6-x)^2$, and $(\sqrt{50})^2 = x^2 + (y+2)^2$. We solve this system to get $x = 1$ and $y = 5$, resulting in an answer of $1^2 + 5^2 = 26$.[/hide]

interesting solution...I actually destroyed this problem w/ trig, creating a right triangle with the hypotenuse as the diameter etc.

[quote="NeverOddOrEven"]Wouldn't it be$\sqrt{26}$? That's also how it is on Kalva.[/quote]

Find the square of the distance.

Kalva changes the questions a little.

With Pythagorean theorem we get $ |AC|\equal{}\sqrt{40}$
Lets say $ \angle BAC\equal{}\beta$
             $ \angle OAC\equal{}\alpha$
             $ \angle OAB\equal{}\phi$

Then by Law of Cosines we get:
$ \cos \beta\equal{}\frac{4\minus{}36\minus{}40}{\minus{}2*6*\sqrt{40}}\equal{}\frac{3}{\sqrt{10}} \Rightarrow \sin \beta\equal{}\frac{1}{\sqrt{10}}$

$ \cos \alpha\equal{}\frac{50\minus{}50\minus{}40}{\minus{}2*\sqrt{50}*\sqrt{40}}\equal{}\frac{1}{\sqrt{5}} \Rightarrow \sin \beta\equal{}\frac{2}{\sqrt{5}}$

Then $ \cos \phi\equal{}\cos (\alpha\minus{}\beta)\equal{}cos\alpha \cos\beta\plus{}sin\alpha \sin\beta\equal{}\frac{5}{\sqrt{50}}$

Finally $ |OB|^2\equal{}|OA|^2|AB|^2\minus{}2|OA\parallel{}AB|\cos\phi\equal{}50\plus{}36\minus{}2*6*\sqrt{50}*\frac{5}{\sqrt{50}}\equal{}
26$
Therefore answer is 26.

Draw $ \overline{AC}$ and $ \overline{OC}$. Now it's Law of Cosines, and then $ \sin x \equal{} \sqrt {1 \minus{} \cos^2x}$ in triangles (since $ 0^\circ<x<180^\circ$, and $ \sin x$ is positive for these values).

[quote="brightzhu"]Sorry for reviving the old thread, but how did you find the values for the sines of the angles?[/quote]
Don't apologize. If you have a question about a problem posted in an old topic, go ahead and revive it. It's better to keep all of the discussion consolidated, rather than having $ n$ copies of the problem posted all over the forum.

[quote="4everwise"]A machine-shop cutting tool has the shape of a notched circle, as shown.  The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is 6 cm, and that of $BC$ is 2 cm.  The angle $ABC$ is a right angle.  Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy]
size(150); defaultpen(linewidth(0.65)+fontsize(11));
real r=10;
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C;
path P=circle(O,r);
C=intersectionpoint(B--(B.x+r,B.y),P);
draw(Arc(O, r, 45, 360-17.0312));
draw(A--B--C);dot(A); dot(B); dot(C);
label("$A$",A,NE);
label("$B$",B,S);
label("$C$",C,SE);
[/asy][/quote]
[hide="Solution"]
Let $O$ be the center of the circle. Let the foot of the perpendicular from $O$ to $AB$ be $X$, and let $OX = x$. Note that by the Pythagorean Theorem, $AX = \sqrt{50-x^2}$, so $BX = 6 - \sqrt{50-x^2}$. Also, the horizontal distance between $O$ and $C$ is $x+2$. Therefore, by the Pythagorean Theorem, $OC = \sqrt{50} = \sqrt{(x+2)^2 + (6 - \sqrt{50 - x^2})^2}$.

We get $50 = x^2 + 4x + 4 + 36 + 50 - x^2 - 12\sqrt{50-x^2}$, or $0 = 4x + 40 - 12\sqrt{50-x^2}$, so $x+10 = 3\sqrt{50-x^2}$, meaning $x^2 + 20x+100 = 450 - 9x^2$, or $10x^2 + 20x - 350 = 0$, so $x^2 + 2x - 35 = 0 \implies (x-5)(x+7) = 0$, from which we get $x = 5$. Therefore, $x^2 = 25$, and $(6 - \sqrt{50-x^2})^2 = 1$, so our desired answer is $25 + 1 = \boxed{026}$.
[/hide]

[hide="Long Solution"]Let $O$ be the center of the circle. Let the foot of the perpendicular from $O$ to $AB$ be $E$. Let us extend $AB$ to intersect circle $O$ at $D$. Let $BD=x$.
[asy]
size(150); defaultpen(linewidth(0.65)+fontsize(11));
real r=10;
pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C,D=r*dir(315),E=foot(O,A,D);
path P=circle(O,r);
C=intersectionpoint(B--(B.x+r,B.y),P);
draw(Arc(O, r, 0, 360));
draw(A--B--C); dot(O); dot(A); dot(B); dot(C); dot(D); dot(E);
draw(B--D); draw(O--A); draw(O--D); draw(O--E);
label("$O$",O,W);
label("$A$",A,NE);
label("$B$",B,SE);
label("$C$",C,SE);
label("$D$",D,SE);
label("$E$",E,NE);[/asy]
First, we see that $OC=\sqrt{\left(OE+BC\right)^2+EB^2}$. Next, we notice that $OB=\sqrt{OE^2+EB^2}$ and finally, $OE=\sqrt{OA^2-AE^2}$. We also know that $AE=\dfrac{x+6}{2}$ and $EB=\dfrac{-x+6}{2}$. We can substitute $OE$ into $OB$, \begin{align*}OB&=\sqrt{EB^2+OE^2}\\ &=\sqrt{EB^2+OA^2-AE^2}\\&=\sqrt{\left(\dfrac{-x+6}{2}\right)^2+50+\left(\dfrac{x+6}{2}\right)^2}\\&=\sqrt{\left(\dfrac{x^2-12x+36}{4}\right)+\left(\dfrac{x^2+12x+36}{4}\right)+50}\\&=\sqrt{50-6x}\end{align*}
Now, we need to find $x$. We substitute $OE$ into $OC$, \begin{align*}OC&=\sqrt{\left(OE+BC\right)^2+EB^2}\\&=\sqrt{\left(\sqrt{OA^2-AE^2}+BC\right)^2+EB^2}\\&=\sqrt{\left(\sqrt{50-\left(\dfrac{x+6}{2}\right)^2}+2\right)^2+\left(\dfrac{-x+6}{2}\right)^2}\end{align*}
To make calculations easier, let $y=\dfrac{x}{2}$, \begin{align*}&=\sqrt{\left(\sqrt{50-\left(y+3\right)^2}+2\right)^2+\left(-y+3\right)^2}\\&=\sqrt{\left(\sqrt{50-y^2-6y-9}+2\right)^2+\left(y^2-6y+9\right)}\\&=\sqrt{\left(\sqrt{-y^2-6y+41}+2\right)^2+\left(y^2-6y+9\right)}\\&=\sqrt{\left(-y^2-6y+41+4\sqrt{-y^2-6y+41}+4\right)+\left(y^2-6y+9\right)}\\&=\sqrt{-y^2-7y+48+4\sqrt{-y^2-6y+41}}\end{align*}
We know $OC=\sqrt{50}$, \begin{align*}-y^2-7y+48+4\sqrt{-y^2-6y+41}&=50\\y^2+7y+2&=4\sqrt{-y^2-6y+41}\\(y^2+7y+2)^2&=16(-y^2-6y+41)^2\\y^4+14y^3+53y^2+28y+4&=-16y^2-96y+656\\y^4+14y^3+69y^2+124y-652&=0\end{align*}
We know from the Factor Theorem that $y|-652$, but $y>0$, so $y|652$. We try testing small values of $y$ first, and we see that $y=2$. When we divide the LHS by $(y-2)$, \[ (y-2)(y^3+16y^2+101y+326)=0 \]
The polynomial to the right of $(y-2)$ must have the bound $y<0$ because it has positive coefficients. Therefore, there are no more positive solutions for $y$. And if $y=2$, then $x=4$. Substituting $x$ back into $OB=\sqrt{50-6x}$ gives $OB=\sqrt{26}$. The answer is $OB^2=\boxed{26}$[/hide]

Refer to attached diagram.

Drop perpendiculars from $O$ to $AC$ and $AB$ and call their feet $D$ and $E$, respectively. Also connect $C$ to $A$ and $O$, and $O$ to $A$ and $B$. We will find $AE$ in two different ways.

Observe that: \[\triangle{ADP}\sim \triangle{ABC}\sim \triangle{OEP}\quad (*)\] By the Pythagorean Theorem, $AC$ $=\sqrt{40}$. Further, since $OA=OC=\sqrt{50}$, it follows that $\triangle{AOC}$ is isosceles, therefore $AD=AC/2=\tfrac{\sqrt{40}}{2}$. Thus, going back to (*) we have $\tfrac{AD}{AB}$ $=\tfrac{AP}{AC}$ $\Rightarrow AP$ $=\tfrac{10}{3}$. Therefore, applying the Pythagorean Theorem to $\triangle{ADP}$ yields $PD=\tfrac{\sqrt{40}}{6}$; in the same way, $\triangle{ADO}$ gives $OD=\sqrt{40}$. As a result, $OP=\tfrac{5\sqrt{50}}{6}$, and consequently (*) implies $\tfrac{PE}{PD}=\tfrac{OP}{AP}$, that is, $PE=\tfrac{5}{3}\Rightarrow AE=5$.

Now, consider the right triangles $\triangle{OAE}$ and $\triangle{OBE}$. By the Pythagorean Theorem, \[OA^2-AE^2=OB^2-(AB-AE)^2\] \[\Leftrightarrow 50-AE^2=OB^2-(6-AE)^2\] \[\Leftrightarrow AE=\frac{86-OB^2}{12}.\] Lastly, equating our two values for $AE$ we obtain $5=\tfrac{86-OB^2}{12}$ or $OB^2=\boxed{26}$.

Here's a solution with coordinates.

[hide="Solution"]

Let $B$ be at $(0,0)$ and consequently $A(0,6)$ and $C(2,0)$ are on a circle defined by $(x-h)^2+(y-k)^2=50$.
\[h^2+(6-k)^2=h^2+k^2-12k+36=50 \Rightarrow h^2+k^2-12k=14.\]
\[(2-h)^2+k^2=h^2+k^2-4h+4=50 \Rightarrow h^2+k^2-4h=46.\]
\[\Rightarrow 3k-h=8 \Rightarrow h=3k-8.\]
\[\Rightarrow (3k-8)^2+k^2-12k=14 \Rightarrow 10k^2-60k+50=0 \Rightarrow (k-1)(k-5)=0 \Rightarrow k=1, h=5\] and $k$ is not $5$, which is obvious by referring to a diagram (I won't prove it).

So by the distant formula, the square of the distance is $5^2+1^2=26$.

This is AIME (1983).
https://www.artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems#Problem_4

[hide=power of a point]
Extend $AB$ through $B$ to intersect the circle at point $D$. and Extend $CO$ through $O$ to intersect the circle at $E$. Now, $\angle CED = \angle CAB$ because they intercept the same arc, and both $\triangle CED$ and $\triangle CAB$ are right, so $\triangle CED \sim \triangle CAB$.

Then, $\dfrac{CE}{CA} = \dfrac{ CD}{CB} \implies \dfrac{2\sqrt{50}}{\sqrt{40}} = \dfrac{CD}{2} \implies CD = \sqrt{20}$. So, $BD = \sqrt{20 -  4} = 4$.

Let $x$ be the desired length $OB$. Extend segment $OB$ to intersect the circle twice at $F$ and $G$ ($F$ further from $B$ than $O$), and applying power of a point to segments $FG$ and $AD$ gives us $(FB)(BG) = (AB)(BD) \implies (\sqrt{50} + x)(\sqrt{50} -x) = (6)(4) = 24 \implies 50 - x^2 = 24 \implies x^2 = \boxed{26}$.
[/hide]

[hide]Refer to scuffed diagram below. We start off by letting $O$ be the center of the circle, drawing $\overline{OC}$ and $\overline{OA}$, and dropping the perpendicular from $O$ to $AB$, with their intersection being $H$, and doing the same for $O$ and $BC$, with their intersection being $J$. The reason we do this is that we can now get $\overline{AH}$ and $\overline{BH}$ in terms of $\overline{OH}$, so we can get an expression for $\overline{AB}$ in terms of $\overline{OH}$ which we know is equal to $6$. Therefore, let $\overline{OH}=x$. Then, $\overline{AH}=\sqrt{50-x^2}$ and $\overline{JB}=x$, so $\overline{OJ}=\overline{HB}=\sqrt{50-(x+2)^2}$. So we have the equation $\sqrt{50-x^2}+\sqrt{50-x^2-4x-4}=6$. We can move $\sqrt{50-x^2}$ to the other side and square to get $50-x^2-4x-4=50-x^2+36-12\sqrt{50-x^2}$. Cancelling, isolating the square root, and squaring again, we get $x^2+20x+100=450-9x^2$, which we can solve to yield the solutions of $-7$ and $5$; $5$ clearly being the one we want to use in this situation. Therefore, $OB=\sqrt{26}$ so our answer is $26$.  [/hide]


geometry
inradius
circumcircle
geometry proposed

This problem based on that identity(i will try to post proof later):
$ \sum_{cyclic}{XA^2\cdot a}\equal{}XI^2\sum_{cyclic}{a}\plus{}abc$

so if $ P$ is a point in the inscribed cirumference of a $ \triangle ABC$ then we have:
$ aPA^2 \plus{} bPB^2 \plus{} cPC^2 \equal{} r^2(a\plus{}b\plus{}c)\plus{}abc\equal{}2r[ABC]\plus{}4[ABC]R\equal{}(4R\plus{}2r)[ABC]$

if we have:
$ aPA^2 \plus{} bPB^2 \plus{} cPC^2 \equal{} (4R \plus{} 2r)[ABC]$
$ \leftrightarrow$
$ PI^2(a\plus{}b\plus{}c)\plus{}abc\equal{}4R[ABC]\plus{}2r[ABC]$
$ \leftrightarrow$
$ PI^2(a\plus{}b\plus{}c)\plus{}abc\equal{}abc\plus{}r^2(a\plus{}b\plus{}c)$
$ \leftrightarrow$
$ PI\equal{}r$
So point $ P$ is in the inscribed circumference of $ \triangle ABC$

[quote=SaYaT]This problem based on that identity(i will try to post proof later):
$ \sum_{cyclic}{XA^2\cdot a}\equal{}XI^2\sum_{cyclic}{a}\plus{}abc$
[/quote]

Is there any proof of this identity without using the barycentic coordinates?

[quote=Luis González]Show that a point $P$ lies on the incircle $(I)$ of $\triangle ABC$  if and only if

\[a \cdot PA^2 \plus{} b \cdot PB^2 \plus{} c \cdot PC^2 \equal{} (4R \plus{} 2r)[\triangle ABC] \]
$R,r$ denote the inradius and circumradius of $\triangle ABC$ and $[\triangle ABC]$ stands for its area.[/quote]

This condition can be rewritten in the following form: $$|a\cdot \overline{PA}+b\cdot \overline{PB}+c\cdot \overline{PC}|=2S(ABC).$$ This condition holds iff $P$ lies on the incircle... Very interesting...


If the price of a stamp is 33 cents, what is the maximum number
of stamps that could be purchased with $ \$32$?

32 dollars is 3200 cents. Since each stamp is 33, we divide $ 3200 \div 33 \approx \boxed{96}$.


For some odd reason, we don't have chapter until tomorrow.  But the problem is that we only have 2 weeks to prepare for state.

[quote="NathanSkinner"]I am very nervous about state....grrrr....[/quote]

what grade are you in?

[quote="SoccerBrainy40"]My States are next Saturday. I've found that nervousness does make me do worse, so I try not to be nervous. But most of the time, it doesn't work.  :D[/quote]

Wow, my state isn't for a month

Lol, I guess MA States are really early. Lucky, you get more time to study! It's not like I'm even going to make it to the top 10, anyway.  :blush:

[quote="hockeyadi23"]Our states are on March 12, [size=200]one day before my birthday[/size], and I can't ask for a better birthday present than making the nationals!!!

I think two weeks are pretty much enough for anyone to study, because its in those last two weeks when you're most efficient in your study habits..[/quote]

SAY WHAAAAAAAT? States are on the 12th??? Damn, it used to be late March.

Just like Chapter, the dates for States vary from state to state.  Ours is the 12th, and it is the first year it hasn't been the first Saturday in March.

My dilemma is that I have a very competitive, hard worked for history Day project, and it's on the SAME DAY SAME TIME as mathcounts STATE (Mar. 5 here)! The only stroke of luck is my partner can present without me, but he's almost clueless on our project and was pretty much deadweight during the task =/

we just had chapter today
=P state in like two weeks

[quote="theaznkungfu"]I'VE BEEN DENIED MY CHANCE TO GO TO STATES!!!!!! grrr :( our school is going on the east coast trip (yea us californains get to "explore our rich history") so i cant go to states... boohoo AND i was on the team... so our 6th graders are going on the team to replace two of us going on the trip[/quote]
Wait... I thought that the only substitutions allowed are for medical reasons... at least that's how it is here...

I was first in my school competition, and then our team made 2nd at our chapter competition (thanks to a person whose score of 14 out of 46 killed our team average). Someone now replaced me for the state team because I have travel to China, starting right on the day the competition is held! Boo hoo!  :(


Can anybody prove or disprove the following inequality
If a, b and c are positive real numbers then
a*2b + b*2c +c*2a<= 1/9(a+b+c)*3
(Here x*2 means x square)

[quote="delegat"]If a, b and c are positive real numbers then
a*2b + b*2c +c*2a<= 1/9(a+b+c)*3
(Here x*2 means x square)[/quote]

Wrong for a = 2, b = 2, c = 1.

  darij


Suppose $ k\in \mathbb N$ be an arbitrary number and $ N$ be the set of numbers that $ k$ does not occur in their decimal representation. Prove that:
\[ \sum_{n\in N}\frac1n\]
is finite.

This proves that there are infinitely many primes such as $ p$ that $ k$ occurs in their decimal representation as $ \sum\frac 1p$ is infinite.

Hi Omid! I posted my proof here:

[url=http://bboyjordan.wordpress.com/2009/10/31/every-number-m2m-nm-has-all-digits/]Problem 11.Only a few element in harmonic sum determine its divergence[/url]

However I am not sure that I have understood the quoted corollary:
[quote="Omid Hatami"]This proves that there are infinitely many primes such as $ p$ that $ k$ occurs in their decimal representation as $ \sum\frac 1p$ is infinite.[/quote]
By Dirichlet Theorem if $ 10^{m\minus{}1}>k$ then $ \displaystyle \sum_{p \in \mathbb{P}\text{ s.t. }10^m \mid p\minus{}10k\minus{}1}{p^{\minus{}1}}$ diverges, but I don't see the relation with the divergence of $ \displaystyle \sum_{p \in \mathbb{P}}{p^{\minus{}1}}$ and with the above problem.. :huh:

Yes, Obviously Dirichlet proves the fact.

But this is an elementary proof. If except finitely many primes all other primes don't contain $ k$ in their decimal representation, then $ \sum_{n\in N}\frac1n$ will be infinite as $ \sum\frac1p$ is infinite...


algebra
polynomial
factorial
calculus
derivative
algebra unsolved

Prove that the polynomial $p(x) = \sum_{k=1}^{n} \frac{x^k}{k!}$ has $n$ different complex roots for every $n\in \mathbb{N}$

Clearly $p(x)-p'(x)=\frac {x^k}{k!}$

If a polynomial has a double root $x=k$, then $p(k)-p'(k)=0$.  But that is only possible at $x=0$, and $p$ does not vanish there.

What do you mean with $p'(x)$

can you explain your solution a little more

$p'(x)$ is the derivative of $p(x)$.  If $p$ is a polynomial and it has a double root at $x=k$, then $p'(k)=0$.

This proves that there are no double roots.  The fundamental theorem of algebra states that $p$ has $n$ roots in the complex plane.  Therefore they are $n$ distinct roots, which is the desired conclusion.

Yes, i had tried that.. but just notice that $p(x)-p'(x) = \frac{x^n}{n!}-1$ and didn't know what to do with that $-1$... Also, $c=0$ is a root of $p$.

Sorry, I misread the bounds on the summation.  I have no idea how to proceed, save for showing that the $n$-th roots of $n!$ are not roots of $p$.  I tried using Rouche's theorem, but had no success.


Hi, I am looking for a book that will help me with my Problem Solving so I am considering buying AoPs Vol. 1&2 + Solutions for $ \$$73.

Firstly, what currency is this? AUD or US? (I live in Australia)

Also, do you think these books would be suitable for me? I am in year 10 at the moment and am mostly competent with most of the syllabus, although I am weak in number theory (no lessons on it) and geometry as far as I know. I usually get distinctions in the AMC(Australian), even getting top 1% in year 7, but since then I've dropped to about top ~10% (either im getting dumber or examiners are separating the wheat from the chaff).
Thanks.

I think you have the proper level for those books. We also offer a series of subject-orientated books, both Intro and Intermediate levels.

I'm pretty sure the prices are in US currency.

I think AOPS Volume 1 would be great for your level.  I remember first working out of Volume 1 when I was about 12 and I loved it.  The introduction books would also be good for you if you really feel weak in a subject, but I recommend just working through Volume 1.  Volume 2 is probably too much now but you'll want it later.

So a few days ago I ordered them, they should hopefully come in a few weeks... which is pretty awesome.  :D Still a bother ordering overseas though, but whatever

One nice thing for our non-US customers right now is that the US Dollar is historically weak relative to most other currencies.  So goods priced in US Dollars (like our books) are relatively cheap.

For example, our new [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]Intermediate Counting & Probability[/url] books are 47 USD, which is only 23.12 UK Pounds or 33.45 Euro!  What a deal! :)

And to our Canadian customers -- the Canadian dollar is now actually worth [b]more[/b] than the US dollar.  (I never thought I'd see this in my lifetime -- I grew up in Buffalo, NY, and I remember when 1 USD = 1.50 CAD.)

On the down side, yes, overseas shipping is expensive for us.  But we've recently switched our international shipping to the US Postal Service's new Priority Mail International service, so our overseas shipments should be both faster and more reliable than they were in the past.

Cool, they arrived, America to Australia in a week! :lol:


Let $P$ be a point inside an equilateral triangle $ABC$, and let $d_{a}, d_{b}, d_{c}$ denote the distances of P from edges, respectively, of triangle. Prove or disprove
$(1+\frac{d_{a}}{PA})(1+\frac{d_{b}}{PB})(1+\frac{d_{c}}{PC})\ge\frac{27}{8}$

$d_{a}+d_{b}+d_{c}$ is the hight of the equilateral triangle.


analytic geometry
graphing lines
slope
Pythagorean Theorem
geometry
absolute value

What is the x-coordinate of the point on the x axis which is equidistant from (8,5) and (-2,3)?

Source: Math Team team round quesiton

feel free to correct

[hide]Ok the way i would do this is first find the midpoint of the points (8,5) and (-2,3) which is $ ({\frac {8 \minus{} 2}{2} , \frac {5 \plus{} 3}{2}})$ which is $ (3 , 4)$ then we figure out the line that passes through that point and is perpendicular to the line that contains (8,5) and (-2,3).  The formula of this line is $ y \equal{} 5x \minus{} 11$ then we find the zeroes of this equation so we get $ 0 \equal{} 5x \minus{} 11$ therefore $ \boxed{x \equal{} \frac {11}{5}}$[/hide]

EDIT: didnt actually edit accidentally clicked edit box

the answer the teacher gave us was[hide]19/5[/hide] but a lot of times, she makes silly mistakes, so you might be right.

I have no idea how to do this problem, so I have no idea myself.  :(

[quote="zolojetto"]feel free to correct

[hide]Ok the way i would do this is first find the midpoint of the points (8,5) and (-2,3) which is $ ({\frac {8 \minus{} 2}{2} , \frac {5 \plus{} 3}{2}})$ which is $ (3 , 4)$ then we figure out the line that passes through that point and is perpendicular to the line that contains (8,5) and (-2,3).  The formula of this line is $ y \equal{} 5x \minus{} 11$ then we find the zeroes of this equation so we get $ 0 \equal{} 5x \minus{} 11$ therefore $ \boxed{x \equal{} \frac {11}{5}}$[/hide][/quote]

its the negative recipricol, not the recipricol that is the slope of a perpendicular line.

Your teacher is correct.  Let x = the x-coordinate of the desired point.  Then (x, 0) is the desired point.  Let d1 = the distance from (x, 0) to (8, 5).  Let d2 = the distance from (x, 0) to (-2, 3).  From the Pythagorean theorem or, alternatively, the definition of distance:

d1 = ((x - 8)^2 + (0 - 5)^2)^(1/2)
d2 = ((x - -2)^2 + (0 - 3)^2)^(1/2)

Since d1 = d2 by hypothesis, set the two right-hand sides above equal to get
((x - 8)^2 + (0 - 5)^2)^(1/2) = ((x - -2)^2 + (0 - 3)^2)^(1/2).

Square both sides to get
(((x - 8)^2 + (0 - 5)^2)^(1/2))^2 = (((x - -2)^2 + (0 - 3)^2)^(1/2))^2.

From the definition of absolute value, (x^(1/2))^2 = |x|.  Therefore:
|(x - 8)^2 + (0 - 5)^2)| = |(x - -2)^2 + (0 - 3)^2)|

From the definition of distance, d1 >= 0 and d2 >= 0.  Therefore, |x| = x, and
(x - 8)^2 + (0 - 5)^2 = (x - -2)^2 + (0 - 3)^2.

Simplify to get:
(x - 8)^2 + 25         = (x + 2)^2 + 9
x^2 - 16x + 64 + 25 = x^2 + 4x + 4 + 9
      - 16x + 89         =          4x + 13
                 76         =         20x

Therefore, x = 76⁄20 = 19/5.


[quote="Totally Zealous"]The positive integers A, B, A-B, and A+B are all prime numbers. The sum of these four primes is
(A) even
(B) divisible by 3
(C) divisible by 5
(D) divisible by 7
(E) prime[/quote]

[hide]The sum is 17 which is E[/hide]

[hide]Let A=5, B=2 A+B=7, A-B=3 The sum is 17, which is a prime. THE ANSWER IS E.[/hide]

[hide]SUM IS 17 so the answer is E

$E$[/hide]


Find continous functions $f: \mathbb{R}\to\mathbb{R}$ satisfying:
$f(f(x))=f(x)+2x$

[color=red]Edit by Megus: splitted from another topic (don't post two problems in one topic!) and LaTeXed[/color]


A homogenous ladder of length L and mass M rests against a smooth, vertical wall. If the coefficient of static friction between the ladder and the ground is $\mu_{s}$, find he minimum angle $\theta_{min}$, such that the ladder does not slip. Express your answer in terms of L, M, $\mu_{s}$ and g.

Hi,

The force from gravity would then be M(g)
but the frictional force on the ladder is :[mu_{s}]M * g

The weight of the ladder is the y-component of the resultant force of the ladder on the ground (Which is applied at a direction of {theta} below the horisontal)

thus weight / sin({theta}) = (x-component) /cos({theta})

(x -component)= (weight*cos({theta}))/sin({theta})
(x component) = (M*g*cos(theta))/sin(theta)

but x-component must have a greater magnetude than [mu_{s}]*M*g

thus (M*g)*[(cos(theta)/sin(theta))-[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f14c87a6a6fe31a1391c59a761bbc511.gif[/img] ] > 0

hence cos(theta)/sin(theta) - [mu_{s}] > 0

angle before slipping gives cos(theta)/sin(theta) = [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f14c87a6a6fe31a1391c59a761bbc511.gif[/img]
theta = arccot([img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f14c87a6a6fe31a1391c59a761bbc511.gif[/img])


Anything which makes the exponent 0 makes the number 1.  Hence, $x=4,5$ work.
Also, you get 1 if the base is 1.  Thus, $x=1, 4$ fulfill this property, and so the answers are $x=1,4,5$, repectively yielding $1^{12}$, $1^0$, and $5^0$.

So you say roots are ( $1,4,5$ ) only .. But why not $x=2$ ? and for that matter $x=3$ They work too!  :)

(Hope that helps - )

(Actually if you don't restrict $x$ to real values (eg if complex values are allowed for x there are infinite solutions ...:)

[quote="Ijo07"]Can anyone help me with this problem?

(x2-5x+5)x2-9x+20=1

I think you're supposed to start with logarithms, but I'm not sure. I'd appreciate any assistance.[/quote]

Alright so when does something equal 1?  $y^0$ when $y \not= 0$, $1^y$ for all $y$, and $-1^{2y}$ for all $y$.
This means $x^2-9x+20 = 0 \rightarrow x=5,4$ and at $x=5,4$, $x^2-5x+5 \not= 0$.  $x^2-5x+5 = 1 \rightarrow x = 4,1$.  Finally, $x^2-5x+5 = -1 \rightarrow x=3,2$, but we need to check that $x^2-9x+20$ is even for $x=2,3$, which is true so the roots are $x=1,2,3,4,5$.

Thanks, FMako. I didn't even conceive that you could take this approach to the problem. When I plugged this into [url=http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic]this algebra equation solver[/url], I only got 1, 4, and 5, and Gyan said that it could also be 2 and 3, but you proved why it could be 2 and 3. I guess the calculator didn't give me 2 and 3 because it can't calculate -12yas being positive.

For those of you who are interested, this question was a NYSML tiebreaker from two years ago. I've also been told it was on an AMC.


[color=brown]For any $ x,y,z> 0$ prove that

$ \sum \dfrac{1}{(x\plus{}y)^2}\ge \dfrac{9}{4}\cdot \dfrac{1}{\sum xy}$[/color]

Er....1996 Iran with have posted many times ....

[quote="sophie germain"]i am nnot sure if my solution is right but using titu's lemma ,i think the problem can be done. :maybe:[/quote]
I wish to see it. Thank you!  :)

[quote="sophie germain"]i am nnot sure if my solution is right but using titu's lemma ,i think the problem can be done. :maybe:[/quote]
I don't think its right :wink:
By titus lemma we have :
$ LHS \geq \frac{9}{2(x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx)}$,which is not bigger from $ RHS$...
If you mean something else i would like to see it too :)

yes ,i did get that. but after that  i got
$ x^2\plus{}y^2\plus{}z^2$ $ \ge$ $ xy\plus{}yz\plus{}zx$
then arent u done? :maybe:

[quote="sophie germain"]yes ,i did get that. but after that  i got
$ x^2 \plus{} y^2 \plus{} z^2$ $ \ge$ $ xy \plus{} yz \plus{} zx$
then arent u done? :maybe:[/quote]
Look it better my friend :wink:
It is the opposite:
$ \frac{9}{2(x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx)} \leq \frac{9}{4(xy\plus{}yz\plus{}zx)}\Leftrightarrow
x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx \geq  2(xy\plus{}yz\plus{}zx)$

oh yeah, i realise the mistake,.thank you :)


Hey i just want to know how can i get all my scores from past years if our school didn't keep everything?

is there an email address i am email to ask for my scores?

For what it's worth, I have records of the last four years of AIME qualifiers in California - but only California, and only AIME qualifiers.

Can't you buy entire packets of info on AIME qualifiers from the AMC people?

Write a letter or an email to the AMC office with your request.  Our postal address and our email address is on all of our literature.

Be sure to specify your name, exactly as it was used on the contests you took, the years that you took the contests, the name of the school where you took the contest (the CEEB number will help us look it up faster, if you know the CEEB number) and the city and the state.

Requests for scores are answered only for the individual requesting the scores.

Plase allow up to two weeks for an answer.

Steve Dunbar
MAA Director, American Mathematics Competitions

thanks ! i just sent an email to amcinfo @ unl . edu

( i hope that is the right email XP)

Yes, we are still sending the plaques.

Most of them were sent last week.  If you do not receive the correct plaque, first contact your teacher to determine if your teacher or contest manager received the plaque, then contact our office.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="xxazurewrathxx"]When are amc 12 perfect score plaques being delivered? Or are they still doing that this year?[/quote]

Must it be perfect score?  I thought it was just winner of the region.

Also, are there plaques for state winners on 10A/10B as wel

I think it's just for AMC 12, but I could be wrong. Also, I think winners of regions get awards too, but aren't they different from perfect score plaques? I seem to recall that the top score in the region where there are no perfect scores receive some kind of special award, and everyone else get subscriptions or books and stuff. Are awards delivered to the home or the school?

EDIT: Oh by the way, it looks like I posted my question on the wrong thread by accident. It was supposed to go on the AMC Awards thread. Sorry about that.

[quote="rainynightstarz"]i sent an email to amcinfo @ unl . edu  two weeks ago about getting all my scores
but i haven't heard any replies yet[/quote]

The two regular answerers to that email address are 1) extremely busy with MOSP and 2) on vacation.  MOSP ends Saturday, so give us another week to catch up on the backlog, and if you haven't heard anything, email again.  I'd look into it for you, but I'm not involved in the registration/score processing end of things.

Elizabeth Claassen
AMC Senior Accounting Clerk


geometry
3D geometry
sphere
calculus
integration
trigonometry

A uniform sphere (Mass M, radius R) is rolling w/o slipping. Its centre of mass translates with a speed $ v_0$ The sphere encounters a "stair"(like you climb steps/stairs) of height $ h(<R)$. Find bounds on $ v_0$ for which the sphere climbs the stair. Assume friction is sufficient to prevent slipping.

[hide="my attempt"]

hi hell  :D ,nice to see you

let $ \omega_0,v ,\omega$ be initial ang velocity,velocity and angular vel. after collision

Using COAM about horizontal axis passing to pt of contact at the time of collision,we have:
$ \frac{2}{5}MR^2{\omega_0} +mv_0(R-h)=(\frac{2}{5}MR^2+MR^2 ){\omega}$
now,simplifying and using $ v={\omega} R$ as there is no slipping,we get,
$ \frac{2}{5}v_0 +\frac{v_0(R-h)}{R}=\frac{7v}{5}$
again simplifying,we get,
$ v=v_0(\frac{7R-5h}{7R})$  .........(i)

[b]bound 1[/b]
Now,to clear stair,total K.E just after collision must be greater than $ mgh$(using Conservation of energy),so

$ \frac{7}{10}Mv^2 > mgh$     (K.E of solid sphere rolling without slipping is $ \frac{7}{10}Mv^2$)

$ \Longrightarrow v^2>\frac{7}{10}gh$

now,using (i),
$ \Longrightarrow v_0(\frac{7R-5h}{7R}) >\sqrt{\frac{10}{7}gh}$
$ \Longrightarrow v_0>\frac{R\sqrt{70gh}}{7R-5h}$ .......(ii)

[b]bound 2[/b]
to roll without slipping,total inward force must be greater than total outward force,
so,
$ Mgcos{\theta}>\frac{Mv^2}{R}$
$ \Longrightarrow v<\sqrt{gRcos{\theta}}$
$ \Longrightarrow v<\sqrt{gR(R-h)}$    as $ (cos{\theta}=\frac{R-h}{R})$

now using (i),
$ \Longrightarrow v_0(\frac{7R-5h}{7R}) <\sqrt{gR(R-h)}$
$ \Longrightarrow v_0 < \frac{7R}{7R-5h}\sqrt{gR(R-h)}$ .........(iii)

using (ii) and (iii),we can get bounds  :)

but second bound may be wrong(its just logical thinking)
$ {\theta}$ is angle made by pt of contact with vertical

srry for latexing
[/hide]

Question about part 1 of question: (to Skand and anyone who read his solution).

- I guess you used the system being the sphere.  If so , how did you use conservation of angular momentum in part 1?  There is a net torque due to gravity on the sphere between the initial rolling and final rolling, which causes a change in angular momentum?

Um... that solution looks very sketchy to me, for several reasons:

1. COAM is applied incorrectly, as integration pointed out.

2. In the limiting case $ h \rightarrow R$, we clearly should have that the minimum velocity approaches infinity.  Instead, skand's upper bound gives $ v < 0$ and $ v > \frac {1}{2}\sqrt {70gR}$, a finite and positive quantity.  Huh?

3. It doesn't take into account the fact that when the sphere impacts the stair, the stair will "absorb" most of its velocity, leaving only a component normal to the line connecting the contact point and the center of the sphere.

This is an interesting problem, but I believe it can be interpreted many different ways.

At the moment of impact, the step exerts a force on the ball. But the ball's angular velocity about the axis through the point of impact is not affected by the impact. Thus the ball's initial angular velocity (at and just after impact) about this axis is $ \omega_0 \equal{} \frac{v_0 \sin{\theta}}{R} \equal{} \frac{R\minus{}h}{R^2} v_0$ where $ \theta$ is the angle the line between the center of the ball and the point of impact makes with the horizontal. After impact, the only torque affecting the system is due to gravity, so we just need to make sure that the initial rotational energy is sufficient for the ball to get over the step: $ \frac{1}{2} I \omega_0^2 \geq M g h$. Solving we get  $ v \geq \frac{R \sqrt{5 g h}}{R \minus{} h}$.

Yup, I totally agree with that solution.

Haha, when you said integration generic, I thought What?  There's not even an integral in the solution?? :D

Anyway, what's COAM?

Nice solution durt... I was stuck for a while on this one.

some explanations  :)
[hide="1"]$ mg$is a constant(non impulsive) force,so,torque due to gravity can be neglected,here the axis is fixed to ground and not to the ball,that also explains the fact why torque due to pseudo forces are not considerd(there is no pseudo force)[/hide]

[hide="3"]it does take,the Conservation of energy is applied after change in velocity and angular velocity,and the values are after the impact(that are less than initial values)
the total Kinetic energy lost in collisionis $ \frac{7}{10}M(v_0^2\minus{}v^2)$[/hide]

[hide="2"]i couldn't understand what you are saying  :( ,please explain,and i couldn't understand durt's soln(i am just a beginner)[/hide]

COAM = conservation of angular momentum.

In no. 2, skand, I'm just saying that it might be helpful to see whether your answer makes sense in special cases that are easy to analyze.

[quote="befuddlers"]@ skand plz explain ur answer with some more detail, and also plz add a picture so that i can understand easily :D[/quote]
[quote="durt"]so we just need to make sure that the initial rotational energy is sufficient for the ball to get over the step: $ \frac {1}{2} I \omega_0^2 \geq M g h$.[/quote]
Your method is alright. You also need to consider the translational Kinetic energy, besides the rotational energy here. One mistake that you have made, I believe, is considered the translational velocity of the ball to become $ 0$ on hitting the step. Note that the collision causes a change in only that component of velocity which is perpendicular to it. Hence $ v_0 \sin \theta$ remains the velocity of the ball just after collision and this velocity is perpendicular to Normal Reaction $ N$.

My solution goes as follows:

(The ball should collide inelastically with the step, else it will bounce.)
Now, as it hits the step, the velocity in the direction of the Normal at the point of contact will disappear $ (%Error. "since" is a bad command.
e = 0)$.
The velocity that will remain is $ v_0 \sin \theta$ (see figure) perpendicular to the Normal reaction $ N$.

Now, the ball climbs the step to gain height $ R(1 - \sin \theta)$.

$ \begin{align*} & \therefore mgR(1 - \sin \theta) = \frac {1}{2}mv_0^2 \sin^2 \theta + \frac {1}{2} I \omega ^2 \\
& \therefore gR(1 - \sin \theta) = \frac {7}{10} v_0^2 \sin^2 \theta \\
\left( I = \frac {2}{5} mR^2, \; \omega = \frac {v_0^2 \sin^2 \theta}{R}\right)$

$ \therefore \boxed {\color{red}{v_0 \ge \frac {R}{R - H} \sqrt {\frac {10Hg}{7}}}}$

Hey! No comments/suggestions about this one?


quadratics
search
number theory
number theory open

Let $ C_N(p)$ be the number of $ N$ consecutive quadratic residues modulo some prime $ p$.

[hide="For Example"]
$ C_1(p) \equal{} (p \minus{} 1)/2$
$ C_2(p) \equal{} \frac {1}{4}(p \minus{} 4 \minus{} ( \minus{} 1)^{(p \minus{} 1)/2})$ ([i]Number Theory[/i], Andrews)
[/hide]

Prove (or disprove) that $ C_N(p) \equal{} O \left (\frac {p}{2^N} \right )$.

Hopefully I am making the problem statement clear. It is not unrelated to [url=http://www.mathlinks.ro/viewtopic.php?search_id=1670343724&t=253789]This Problem[/url] and, if true, provides an alternate proof to it. I lack the ability to get near a solution, but would be interested in seeing one.

Hmm, [url=http://citeseer.ist.psu.edu/viewdoc/summary?doi=10.1.1.29.5334]yes, apparently quadratic residues really do behave pretty much like coin flips...[/url]

Do you really want $O$-notation, as this gives only upper bounds¿
I think you intended $C_N(p) \sim \frac{p}{2^N}$.


Ring Theory
algebra
polynomial
abstract algebra
search
calculus
integration

$ A$ is subring of $ \mathbb{C}$ generated by mth roots of unity, $ p$ is a prime number. Prove that there are finitely many prime ideals $ \mathfrak{p}_1,....\mathfrak{p}_h$ of $ A$ containing $ p$. The quotents $ A/\mathfrak{p}_i$ are finite fields of characteristic $ p$, and there exists an integer $ N$ such that $ pA\supset (\mathfrak{p}_1\cap...\cap \mathfrak{p}_h)^N$. Help me. Thank!

the $ \mathfrak{p_i}$ correspond to prime ideals of $ A$/ $ pA$, which is a finite ring of characteristic $ p$. The first two assertions folow from this. This third follows from the fact that $ (\mathfrak{p_1}\cap{\mathfrak{p_2}} \cdots \cap{\mathfrak{p_n}})/pA$ is the radical of the artinian ring $ A/pA$, thus is nilpotent.

For the first part let $ A\equal{}\mathbb{Z}[u]$, where $ u$ is an mth primitive root of the unit and let $ \phi_m$ be the mth cyclotomic polynomial. We know that $ \phi_m$ is the irreducible monic polynomial of low degree in $ \mathbb{Z}[x]$ that annihilates $ u$. If $ f\in\mathbb{Z}[x]$ satifaces $ f(u)\equal{}0$ theh $ \phi_m$ divides $ f$. This shows that the map $ \mathbb{Z}[x]\rightarrow \mathbb{Z}[u]$ given by $ f\rightarrow f(u)$ is an epimorphism whose kernel is $ (\phi_m)$. Then the prime ideals in A that contian $ p$ are in correspondence with the prime ideals in $ Z[x]$ that contains $ p$ and $ (\phi_m)$. Now it is convenient to read http://www.mathlinks.ro/viewtopic.php?search_id=1643396135&t=72993).

Let $ \phi_m\equal{}f_1f_2...f_n$ the factorization in irreducbles in $ \mathbb{Z}_p[x]$. Then $ f_i$ is irreducible in $ \mathbb{Z}[x]$ since $ f$ is monic and is irreducible as a polynomial in $ \mathbb{Z}_p[x]$. We claim that $ (p,f_1),....,(p,f_n)$ are all the prime ideals that contain $ (p,\phi_m)$. In fact, any other prime ideal that contains $ (p,\phi_m)$ is of the form $ (p,f)$ with $ f$ irreducible as a polynomial in $ \mathbb{Z}_p[x]$ (see the link). Then $ f$ divides $ \phi_m$ in $ \mathbb{Z}_p[x]$. This implies $ (f)\equal{}(f_i)$ for some $ f_i$ as ideals in $ \mathbb{Z}_p[x]$ and then $ (p,f)\equal{}(p,f_i)$ as ideals in $ \mathbb{Z}[x]$

Thank you,
But I don't understand why  $ A/pA$ is an Artin ring?

$ \mathbb{Z}/p \mathbb{Z} \to A/pA$ is integral and of finite type, thus finite.


vector
abstract algebra
group theory
linear algebra
linear algebra unsolved

Find an abelian group V and a field F for which V is a vector space over F in two
different ways, that is, there are two different definitions of scalar multiplication
making V a vector space over F.

you can take any vector space and define scalar operation to take everything into the zero vector.

I'll assume that we have an axiom that $ 1\cdot x\equal{}x$- that rules out zero multiplication.

$ \mathbb{R}^2$ as a vector space over $ \mathbb{C}$:
(1) $ (a\plus{}ib)\cdot (x,y)\equal{}(ax\minus{}by,ay\plus{}bx)$
(2) $ (a\plus{}ib)\cdot (x,y)\equal{}(ax\plus{}by,ay\minus{}bx)$

This is impossible if the field is $ \mathbb{Q}$ or the field with $ p$ elements for some prime $ p$. For other fields, the two structures will always be related by a field automorphism- in the above case, the conjugate.

For a related problem, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=166604]this topic[/url].

my suggestion is wrong but i cannot edit. sorry


limit
function
calculus
calculus computations

Let be sequence $ a_0 \equal{} 1,\,\,a_{n \plus{} 1} \equal{} \sum\limits_{i \equal{} 0}^n {a_i a_{n \minus{} i} }$, ($ n\ge 0$ )
Find limit of sequence $ a_n$:
$ \mathop {\lim }\limits_{n \to \infty } \frac {{a_n }}{{a_{n \minus{} 1} }}$

[hide="Hint 1"]Consider the generating function $ f(x) \equal{} \sum_{k\equal{}0}^\infty a_k x^k$. [/hide]
[hide="Hint 2"]Show that $ f(x) \equal{} x f(x)^2 \plus{} 1$. Conclude that
\[ f(x) \equal{} \frac{1 \minus{} \sqrt{1\minus{}4x}}{2x}.\]
[/hide]
[hide="Hint 3"]Therefore, for $ k>0$
\[ a_k \equal{} 2^k \frac{(2k\minus{}1)!!}{(k\plus{}1)!},\]
and $ \lim \frac{a_{k\plus{}1}}{a_k} \equal{} 4$.
[/hide]


Does anyone know the website with all the pumac problems of previous years?


Galois Theory
superior algebra
superior algebra unsolved

An exam typo suggested this cute (though not particularly deep) problem.  I enjoyed solving it, and hope others will as well. :)

Let $ K \equal{} \cup_{m \equal{} 0}^\infty \mathbb{F}_{2^{2^m}}$, and let $ \sigma: K\to K$ be the Frobenius automorphism $ x\mapsto x^2$.  Prove that $ \{\sigma, \sigma^{1 \plus{} 2}, \sigma^{1 \plus{} 2 \plus{} 3},\ldots \}$ is dense in the topological group $ \rm{Gal}(K/\mathbb{F}_2)$.

For a bonus, show that this is false over $ \mathbb{F}_p$ for odd $ p$!

Hints:

(1) If $ \tau: K\to K$ is an arbitrary automorphism, an elementary open neighborhood of $ \tau$ consists in those automorphisms of $ K$ that agree with $ \tau$, for some fixed $ \mathbb{F}_{2^{2^m}}$.

(2) In particular, every open neighborhood of $ \tau$ contains some power of $ \sigma$.  Hence we may assume without loss of generality that $ \tau \equal{} \sigma^k$.

(3) The conclusion follows if we can find some $ j$ such that $ \sigma^{\frac {j(j \plus{} 1)}{2}} \equal{} \sigma^k$ on $ \mathbb{F}_{2^{2^m}}$.  This is entirely a number-theoretic question.

The Galois group is $ \mathbb Z_2$ ($ 2$-adics) and $ \sigma$ corresponds to $ 1$. Thus it's equivalent to showing that $ n^2\plus{}n$ is dense in $ 2\mathbb Z_2$, which follows from Hensels Lemma.

But I never needed $ p\equal{}2$ here...  :maybe:

Well, you need $ p\equal{}2$ to guarantee that $ (n^2\plus{}n)'\equal{}2n\plus{}1$ is nonzero... but also you need to know first that $ n(n\plus{}1)/2$ is dense mod p, which is not true for $ p>2$.

For those not familiar with the p-adics: you can alternatively just show that $ n(n\plus{}1)/2 \equal{} ((2n\plus{}1)^2 \minus{}1)/8$ is dense mod $ 2^k$ for all $ k$, which is equivalent to showing that $ 8l\plus{}1$ is always a square mod $ 2^{k\plus{}3}$.  This can be done with, e.g. a counting argument.

No it's $ \mathbb Z_2$ independent of $ p$ (if you work with $ K \equal{} \cup_{m \equal{} 0}^\infty \mathbb{F}_{p^{2^m}}$). So the value of $ p$ is not needed.

I suppose I was thinking of the fields $ \mathbb{F}_{p^{p^m}}$.  You're right.


P(x) is of degree 7. P(x) is odd and monic. P(x) has a factor of $ (x\minus{}5)^{2}$. The curve y=P(x) passes through $ (1,1152)$. P(x) has 5 real roots.

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[hide="part of the soln. i dont understand"] $ (x\minus{}5)^{2}$ is a factor and since its odd then $ (x\plus{}5)^{2}$ is a factor.
Odd functions pass through the origin therefore x is also a factor. $ P(x) \equal{} x(x\minus{}5)^{2}(x\plus{}5)^{2}$.
Then the soln. says that P(x) cannot have any real fns hence let the final factor be $ (x^{2}\plus{}k^{2})$. Why can't it have any more factors?[/hide]
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alternative soln. are wlecome

Since $ P(x)$ is an odd function so we have that $ P(\minus{}5) \equal{}\minus{}P(5) \equal{} 0$ hence $ (x\plus{}5)^{2}$ is a factor as well. Because it is an odd function so $ P(0) \equal{}\minus{}P(0)$ hence $ x$ is a factor. The question says that it has $ 5$ real roots and since we have already got $ 5$ factors, so no more real roots exist. So let us represent the last factor as $ x^{2}\plus{}k^{2}$ (which we are assuming the roots are purely imaginary). $ P(x) \equal{} x(x\plus{}5)^{2}(x\minus{}5)^{2}(x^{2}\plus{}k^{2})$ an substitue $ x \equal{} 1$ in and we see that $ k^{2}\equal{} 1$.

$ P(x) \equal{} x(x\plus{}5)^{2}(x\minus{}5)^{2}(x^{2}\plus{}1)$

[quote="BanishedTraitor"]Since $ P(x)$ is an odd function so we have that $ P(\minus{}5) \equal{}\minus{}P(5) \equal{} 0$ hence $ (x\plus{}5)^{2}$ is a factor as well. [/quote]

This argument only proves that $ (x\plus{}5)$ is a factor.  To prove that $ (x\plus{}5)^{2}$ is a factor, we must do more work:

$ P(x) \equal{} x(x\plus{}5)^{2}(x\minus{}a)(x\minus{}b)(x\minus{}c)(x\minus{}d) \equal{}\minus{}P(\minus{}x) \equal{} x(x\minus{}5)^{2}(x\plus{}a)(x\plus{}b)(x\plus{}c)(x\plus{}d)$

Alternatively, if $ P(x)$ is odd then $ P'(x)$ is even, and $ P'(\minus{}5) \equal{} P(5) \equal{} 0$.

since there is a double root at 5, there is a double root at -5 since it is an odd function... [we have reflective symmetry wrt. the origin...]