2. Set A consists of m consecutive integers whose sum is 2m, and set B consists of 2m consecutive integers whose sum is m. The absolute value of    the difference between the greatest element of A and the greatest element of B is 99. Find m.

In the solution here [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=26]www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=26[/url] they say [hide]that the medians are 2 and 1/2.  It's not clear how they find that out.  Anyone?[/hide]

Well in A, the mean is 2m/m = 2. Since they're consecutive, that's also the median. Similarly, the mean/median of B is m/2m = 1/2.

[hide=solution]
well.. in element A, the median is 2m/m =2
in element B, the median is m/2m = 1/2

so when finding the biggest number of A and B, we have to find out how many numbers are bigger than the median.
for A, it is (m-1)/2, and for B, it is m-1/2
so the biggest term of A is 2+(m-1)/2 or (m+3)/2, and the biggest term of B is m.

subtract the biggest term of B from A, and we get
$\abs{\frac{3}{2} - \frac{m}{2}} = 99$
$\abs{3 - m} = 198$
$m = 201 or -195$, but it cannot be -195 because the answer has to be in a positive integer form. so m=201.

[hide="solution"]
well.. in element A, the median is 2m/m =2
in element B, the median is m/2m = 1/2

so when finding the biggest number of A and B, we have to find out how many numbers are bigger than the median.
for A, it is (m-1)/2, and for B, it is m-1/2
so the biggest term of A is 2+(m-1)/2 or (m+3)/2, and the biggest term of B is m.

subtract the biggest term of B from A, and we get
${\ |\frac{3}{2} - \frac{m}{2}}| = 99$
$\ |{3 - m}| = 198$
$m = 201 or -195$, but it cannot be -195 because the answer has to be in a positive integer form. so m=201.[/hide]

You guys beat me to it. It's really nice how AIME problems can be reduced to simple algebra by basic brainwork, isn't it?


assuming that the inverse exists, is the inverse of the adjoint of a linear map the adjoint of the inverse? And if yes, could someone go through a simple proof of that fact? It seems intuitive when working with the matrices, but I am havoung trouble expressing it properly.

I think I've got it actually:

Let $ T$ be an invertible linear map, $ T^*$ its adjoint, $ T^{ - 1}$ its inverse.

Then:

$ (TT^{ - 1})^* = T^{ - 1}^*T^*$

But $ (TT^{ - 1})^* = I^* = I$

So we have:
$ T^{ - 1}^*T^* = I$, which shows that $ T^{ - 1}^*$ is a left-inverse of $ T^*$

Right inverse can be proved the same way, just reversing the order of $ T$ and $ T^{ - 1}$ initially

Is that correct?


logarithms
algebra
polynomial
quadratics
quadratic formula

48^(x/2) = 2* 3^(x/2)   X = ?

6x^2 -7x-10

7x+5
3x-10
2n-5

Which 1 of these is the factors?  what is the easiest way to find it?

[hide="1"]As archimedes1 said, $ \left(\frac{48}{3}\right)^\frac{x}{2}=2$. Taking $ \log$ of both sides and using a $ \log$ property, we have
$ \frac{x}{2}\cdot\log\frac{48}{3}=\log2$
$ x=\frac{2\log2}{\log\frac{48}{3}}$[/hide]
[hide="2"]I think you might have made a mistake. $ 6x^{2}-7x-10$ factors into $ (6x+5)(x-2)$.[/hide]

[quote="kdog3682"]HOw did u factor the polynomial so quickly?  Do you just do guess+check or is there a method?[/quote]

Quicker method:

$ 6x^{2}-7x-10$

Factors of 6: 1, 2, 3, 6
Factors of 10: 1, 2, 5, 10

Let's do:

[color=green]6[/color]        [color=cyan]5[/color]
[u]1     [color=red]-2[/color][/u]

(Put two numbers that multiply to 6 in the first column and two nubmers that multiply to -10 in the second)

Cross multiply and you get 5 and -12. If these add up to -7 (the "b" in $ ax^{2}+bx+c$) then your factors are

([color=green]6[/color]x+[color=cyan]5[/color])(1x-[color=red]2[/color]).

So you don't have to do the quadratic formula  :wink: .

[quote="Quevvy"][quote="nayel"][quote="mountain.dew"][hide="1"]As archimedes1 said, $ \left(\frac{48}{3}\right)^\frac{x}{2}=2$. Taking $ \log$ of both sides and using a $ \log$ property, we have
$ \frac{x}{2}\cdot\log\frac{48}{3}=\log2$
$ x=\frac{2\log2}{\log\frac{48}{3}}$[/hide][/quote]

We need not do all that. As $ 2=\left(\frac{48}{3}\right)^{\frac{x}{2}}=(16)^{\frac{x}{2}}=4^{x}$, we easily get $ x=\frac{1}{2}$.[/quote]
Shouldn't it be x=1?[/quote]
No, check $ 1/2$ in the original equation if you're unsure.

Yeah..  :blush: I forgot $ \frac{48}{3}=16$.

[quote="Quevvy"][quote="nayel"][quote="mountain.dew"][hide="1"]As archimedes1 said, $ \left(\frac{48}{3}\right)^\frac{x}{2}=2$. Taking $ \log$ of both sides and using a $ \log$ property, we have
$ \frac{x}{2}\cdot\log\frac{48}{3}=\log2$
$ x=\frac{2\log2}{\log\frac{48}{3}}$[/hide][/quote]

We need not do all that. As $ 2=\left(\frac{48}{3}\right)^{\frac{x}{2}}=(16)^{\frac{x}{2}}=4^{x}$, we easily get $ x=\frac{1}{2}$.[/quote]
Shouldn't it be x=1?[/quote]
you guys are making this too complicated

divide both sides by $ 3^{(x/2)}$ and get $ 16^{(x/2)}=2$ so $ x=2(\log_{2}{16})$ therefore $ x=8$
right?

[quote="abacadaea"][/quote][quote="Quevvy"][quote="nayel"][quote="mountain.dew"][hide="1"]As archimedes1 said, $ \left(\frac{48}{3}\right)^\frac{x}{2}=2$. Taking $ \log$ of both sides and using a $ \log$ property, we have
$ \frac{x}{2}\cdot\log\frac{48}{3}=\log2$
$ x=\frac{2\log2}{\log\frac{48}{3}}$[/hide][/quote]

We need not do all that. As $ 2=\left(\frac{48}{3}\right)^{\frac{x}{2}}=(16)^{\frac{x}{2}}=4^{x}$, we easily get $ x=\frac{1}{2}$.[/quote]
Shouldn't it be x=1?[/quote][quote="abacadaea"]
you guys are making this too complicated

divide both sides by $ 3^{(x/2)}$ and get $ 16^{(x/2)}=2$ so $ x=2(\log_{2}{16})$ therefore $ x=8$
right?[/quote]
$ 16^{(x/2)}=2$
$ \frac{x}{2}=\frac{1}{4}$
$ x=\frac{1}{2}$


In $\triangle ABC$ with right angle at $C$, altitude $CH$ and median $CM$ trisect the right angle.  If the area of $\triangle CHM$ is $K$, then the area of $\triangle ABC$ is
\[ \text{(A)}\ 6K \qquad \text{(B)}\ 4\sqrt{3} K \qquad \text{(C)}\ 3 \sqrt{3} K \qquad \text{(D)}\ 3K \qquad \text{(E)}\ 4K \]

[quote="SnowStorm"]How do you get E?

If it trisects the angle, doesn't it also trisect the opposite side?[/quote]

[hide="SnowStorm"]No, not necessarily. In fact, the median of a right triangle extending from the right angle to the hypothenuse bisects the hypothenuse.[/hide]

[hide="Hint"]
Let $HM=x$. Then using your knowledge of 30-60-90 triangles, solve for the other sides of $\triangle CHM$. After that, substitute your area in terms of $x$ so that it is expressed in terms of $k$.

And yes, the answer is (E).
[/hide]

[hide]Since $\angle ACH = \angle HCM$ and since $M$ is the midpoint of $AB$, $AH = HM = \frac 14 AB$. We also know that $ACB$ must be a 30-60-90 triangle. Because $\triangle ACH$ is congruent to $\triangle MCH$, and the ratio of $AH$ to $AC$ is $1:2$ because $\triangle AHC$ is 30-60-90, the ratio of the dimensions of $\triangle MCH$ to $\triangle ABC$ is $1:2$ and their areas $1:4$ making the area of $\triangle ABC$ equal to $4K$.[/hide]

it is obvious that HM is one forth of AB, and the area of  :tri: CHM is CH.HM/2 and the area of  :tri: ABC is CH.AB/2 so the area of  :tri: ABC is four times the area of  :tri: CHM which equals 4K

Consider a semi circle with diameter $AB$, then $\angle ACB=\angle R$ and $H$ is the midpoint of $MB$.
Note that $\triangle CMB$ is a regular triangle.


The arithmetic mean of eight numbers is 23. What is their sum?

Because there are a total of $ 8$ numbers with an average value of $ 23$, their sum is thus $ 8 \cdot 23 \equal{} \boxed{184}$.


calculus
integration
inequalities
search
linear algebra
real analysis
real analysis unsolved

Let $ f: [0,1]\rightarrow\mathbb{R}$ be such that

$ \int_{0}^{1}f(x)dx\equal{}\int_{0}^{1}xf(x)dx\equal{}1$.

Prove that

$ \int_{0}^{1}f^2(x)dx\geq 4$.

Reformulation as linear algebra:

Suppose $ u,v,w$ belong to a real inner product space. Suppose $ \|u\|^2\equal{}1,\|v\|^2\equal{}\frac13,$ and $ \langle u,v\rangle\equal{}\frac12.$ If $ \langle u,w\rangle\equal{}\langle v,w\rangle\equal{}1,$ then prove that $ \|w\|^2\ge 4.$

(Note the appearance of a $ 2\times 2$ Hilbert matrix.)

Take a look here: http://www.mathlinks.ro/viewtopic.php?search_id=2021356052&t=77089


I'll be getting my BS in math this May from a fairly small private university in Ft. Worth and have applied to a couple PhD programs in mathematics.  So far I have received a fellowship to Southern Methodist University and am currently ranked #1 on Rice University's waitlist.  Thus, I will probably be admitted to Rice with full funding.  Here is the problem:  I have little ambition to pursue academics beyond graduate school.  That is, I would rather work in industry than in education.  I didn't come to this realization until after I applied to these schools.  SMU's program is in applied math and will probably help me seek the kind of career I desire.  Conversely, Rice's program is in pure math, which is something I am [i]much[/i] more interested in.  Also, Rice is a more powerful school than SMU and will perhaps look a bit better on the resume.  However, if I were to pursue a graduate degree in pure mathematics, I would be worried that there would be no place for me outside academics.  If I were more certain that I could obtain a non-academic job with a PhD in pure math, I'd certainly go to Rice.  I'm having a very difficult time deciding which path would be better for me.

    I have read posts on this forum before, and I know that there are very experienced and wise people who give advice regularly.  I am not asking for anyone to tell me what to do.  I'm just curious as to what some of your thoughts are.  Any help would be appreciated.

Thanks

Many, many people get "pure" math Ph.D. and then go on to non-academic careers.

Plus, you'll be spending 4+ years getting your Ph.D. and you might change you mind about what you want to do after graduation.

If at this moment in time, you seem inclined to a pure math program, then my advice would be to do that.  The fact that you get to go to Rice -- an excellent school and one that will, admittedly, look a lot better on a resume than SMU -- is the icing on the cake.

I agree with DPatrick.  Plus, getting a Ph.D. in pure mathematics is not exclusive of getting training in applied mathematics.  Rice has a department of computational and applied mathematics; as a pure-math grad student, you'll probably be welcomed in any c.a.a.m. courses you want to take.

Thanks for the advice guys.  You are correct in saying that I can participate in the c.a.a.m. dept. even as a pure math student.  I have just done some research in this.  I think this  seems to be the best route.


The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:

$\textbf{(A)}\ 6\pi \qquad
\textbf{(B)}\ 6 \qquad
\textbf{(C)}\ 12 \qquad
\textbf{(D)}\ 36 \qquad
\textbf{(E)}\ 36\pi$

n=36.
1/4n=3by3n=4by9,n=36[quote]i hate maths[/quote][/i]


find the volume of the set:

$ \{(\frac{x}{1\minus{}z})^2\plus{}(\frac{y}{1\plus{}z})^2<1,\minus{}1<z<1\}$


Let $p$ be a prime number such that $2p-1$ is also prime. Find all pairs of natural numbers $(x,y)$ such that \[\left( xy-p \right)^{2}= x^{2}+y^{2}.\] [i]Dumitru Acu[/i]

Simplifying the above equation we get
$(x+y-xy-1+p)(x+y+xy+1-p)=2p-1$
So now one of then will be $2p-1$ and the other will be $1$ as it is given that $2p-1$ is a prime.
Taking $x+y-xy-1+p=2p-1$ and $x+y+xy+1-p=1$ we get $x=0$ and $y=p$
Now taking $x+y-xy-1+p=1$ and $x+y+xy+1-p=2p-1$
Eliminating $p$ we get $(x-2)(y-2)=2$. So $x=4$,$y=3$ and
$x=3$,$y=4$ are the solutions.


calculus
integration
real analysis
real analysis unsolved

Let's divide the problem into 3 parts....

$ \int_{ \minus{} 100}^{ \minus{} 10}\left( \frac {x^{2} \minus{} x}{x^{3} \minus{} 3x \plus{} 1}\right) ^{2}dx \equal{} \allowbreak \int_{10}^{100}\frac {x^{2}\left( 1 \plus{} x\right) ^{2}}{\left( x^{3} \minus{} 3x \minus{} 1\right) ^{2}}\,dx$

$ \int_{1/101}^{1/11}\left( \frac {x^{2} \minus{} x}{x^{3} \minus{} 3x \plus{} 1}\right) ^{2}dx$

$ \equal{} \allowbreak \int_{11}^{101}\frac {\left( x \minus{} 1\right) ^{2}}{\left( 1 \minus{} 3x^{2} \plus{} x^{3}\right) ^{2}}dy\allowbreak \equal{} \allowbreak \int_{10}^{100}\frac {x^{2}}{\left( \allowbreak x^{3} \minus{} 3x \minus{} 1\right) ^{2}}\,du$

$ \int_{101/100}^{11/10}\left( \frac {x^{2} \minus{} x}{x^{3} \minus{} 3x \plus{} 1}\right) ^{2}dx$

$ \equal{} \int_{1/100}^{1/10}\frac {x^{2}\left( x \plus{} 1\right) ^{2}}{\left( x^{3} \plus{} 3x^{2} \minus{} 1\right) ^{2}}dy \equal{} \allowbreak \int_{10}^{100}\frac {\left( 1 \plus{} x\right) ^{2}}{\left( x^{3} \minus{} 3x \minus{} 1\right) ^{2}}\,dx$

Summing the 3 transformed integrals, the given problem simplifies to

$ \int_{10}^{100}\frac{x^{2}\left( 1\plus{}x\right) ^{2}\plus{}x^{2}\plus{}\left( 1\plus{}x\right) ^{2}}{\left( x^{3}\minus{}3x\minus{}1\right) ^{2}}\,dx\equal{}\allowbreak \frac{11\!131\!110}{107\!634\!259}$


calculus
integration
function
limit
trigonometry
real analysis
abstract algebra

Let $f: [0,\pi] \rightarrow \mathbb{R}$ be a function twice differentiable with $f''$ continuous. Find
\[ \lim_{n\to \infty} \int_0^{\pi} \frac{\sin (nt+\frac t 2 )}{\sin \frac t 2}f(t)dt \]

i think we need only that f is one time differentiable ]0,pi[ and the limite could be f'(0)*pi
the limite of this integral is the same if we replace sin(t/2) by t/2 (because the difference tend to zero by Riemann Lemma ) so using the fact that (f(x)-f(0))/x is extented to a continuos function on [0,pi] and that int(sin(u)/u,0,+00)=pi/2 we get the result
 i hop i didn't make stupid mistakes  :D

i want to say that the limite is f(o)*pi and we need only that f is defferentiable in 0

The limit is actually $\pi f(0),$ or if you want to weaken the hypotheses on $f$ some, $\lim_{t\to0^+}f(t)$ (usefully abbreviated as $f(0^+).$) (OK, I just saw yassinus's correction.)

yassinus is right in that there's no need to assume anything about a second derivative. The stated hypotheses are overkill.

Here's how I would organize things:

Let $D_n(t)=\frac{\sin\left(\left(n+\frac12\right)t\right)}{\sin\left(\frac t2\right)} = \sum_{k=-n}^ne^{ikt}=1+2\sum_{k=1}^n\cos(kt).$

Then we have $\int_0^{\pi}D_n(t)\,dt = \int_0^{\pi}\left(1+2\sum_{k=1}^n\cos(kt)\right) \,dt=\pi.$

Now define $V_n(t)=\int_0^tD_n(s)\,ds$

I claim three things about the sequence of functions $V_n:$

(1) For each $t>0,\,\lim_{n\to\infty}V_n(t)=\pi.$

(2) For each $\delta>0,\,V_n(t)\to\pi$ uniformly on $[\delta,\pi].$

(3) There is an absolute constant $B$ such that $|V_n(t)|\le B$ for all $n$ and all $t\in[0,\pi].$

I would get (1) and (2) by writing $\pi-V_n(t)=\int_t^{\pi}D_n(s)\,ds$ and then integrating by parts, differentiating the $\frac1{\sin(s/2)}$ factor and integrating $\sin\left(\left(n+\frac12\right)s\right)$

(3) is trickier. In this case, consider $\int_0^tD_n(s)\,ds$ do what yassinus did: compare it to the integral with $\frac s2$ in the denominator instead of $\sin\left(\frac s2\right).$ Show that the difference is bounded and by making a dilating change of variables, show that the integral with $\frac s2$ in the denominator is uniformly bounded.

Now for the main part - another integration by parts:

$\int_0^{\pi}D_n(t)f(t)\,dt=\left.(V_n(t)-\pi)f(t)\right|_0^{\pi}- \int_0^{\pi}(V_n(t)-\pi)f'(t)\,dt$

Since $V_n(\pi)=\pi$ and $V_n(0)=0,$ the boundary terms evaluate to $\pi f(0^+).$

Conditions (2) and (3) above suffice for a (Reimann-integral style) argument that the remaining integral tends to zero as $n\to\infty,$ as long as $f'$ is bounded.

but kent , we no need the condition (2) , only (1) and (3) can make the deal (using theorem of bounded convergent of Lebesgue ) , are you ok that we need only differentiability in 0+ , and continuity by part on [0,pi], and we will get pi*f(0+)?

Here is my proof.

$\lim_{n\to \infty} \int_0^{\pi} \frac{\sin (nt+\frac t 2 )}{\sin \frac t 2}f(t)dt - \lim_{n\to \infty} \int_0^{\pi} \frac{\sin (nt+\frac t 2 )}{\frac t 2}f(t)dt =$
$\lim_{n\to \infty} \int_0^{\pi} \sin (nt+\frac t 2 )f(t) \left( \frac{1}{\sin \frac t 2 }-\frac{1}{\frac t 2} \right)dt=$
$\lim_{n\to \infty} \int_0^{\pi} \sin (nt+\frac t 2 )f(t) \frac{\frac t 2-sin (\frac t 2)}{sin (\frac t 2)\frac t 2} dt=$
$\lim_{n\to \infty} \int_0^{\pi} \sin (nt+\frac t 2 )f(t) \frac{O(t^3)}{O(t^2)}dt=$
$\lim_{n\to \infty} \int_0^{\pi} \sin (nt+\frac t 2 )f(t) O(t)dt=$
$\lim_{n\to \infty} t_n \int_0^{\pi} \sin (nt+\frac t 2 )f(t)dt \to 0$ because $t_n$ is bounded and $\int_0^{\pi} \sin (nt+\frac t 2 )f(t)\ to 0$ from the following lemma.
 (yassinus, what is the proof with "Riemann Lemma")
 Lemma: $\lim_{n\to\infty}\int_0^{\pi} \sin (n t) f(t)=0$. Where $f$ is an integrable function.

We define $g(t)=\begin {cases} \frac{f(t)-f(0)}{t} t\neq 0\\ f'(0) t=0\end{cases}$
We have that $g$ is continuous.

Now I caculate:
$I=\lim_{n\to \infty} \int_0^{\pi} \frac{\sin (nt+\frac t 2 )}{\frac t 2}(f(t)-f(0))dt=$
$\lim_{n\to \infty} \int_0^{\pi} \sin (nt+\frac t 2 )\frac{f(t)-f(0)}{\frac t 2}dt =$
$\lim_{n\to \infty} 2 \int_0^{\pi} \sin (nt+\frac t 2 )g(t)dt =$
$\lim_{n\to \infty} 2 g(x_n) \int_0^{\pi} \sin (nt+\frac t 2 )dt \to 0$
Because $g(x_n)$ is bounded and $\int_0^{\pi} \sin (nt+\frac t 2 )dt \to 0$ from the lemma.

Now I can say that $\lim_{n\to \infty} \int_0^{\pi} \frac{\sin (nt+\frac t 2 )}{\sin \frac t 2}f(t)dt= \lim_{n\to \infty} \int_0^{\pi} \frac{\sin (nt+\frac t 2 )}{\sin \frac t 2}f(0)dt= f(0) \lim_{n\to \infty} \int_0^{\pi} \frac{\sin (nt+\frac t 2 )}{\sin \frac t 2}dt$

I know that $\frac{\sin (nt+\frac t 2 )}{\sin \frac t 2}=1 +\sum_{k=1}^n \cos(ky)$  Integrating we get that  $\int_0^{\pi} \frac{\sin (nt+\frac t 2 )}{\sin \frac t 2}dt=\pi$.
So the limit is $\pi f(0)$. I used only that $f$ is differentiable in 0 and $f'$ is bounded near 0.

Riemann lemma is : int(ei(tx)*f(t)dt,t=a,b) -->0 when |t|-->+00 where f is  continuos by part on [a,b]

Questions:

1) yassinus: What is the Riemann Lemma?
2)yassinus: What means: "int(sin(u)/u,0,+00)=pi/2"?
3) Kent: Are you sure about your forumla : $\frac{\sin\left(\left(n+\frac12\right)t\right)}{\sin\left(\frac s2\right)} = \sum_{k=-n}^ne^{ikt}=1+2\sum_{k=1}^n\cos(kt).$ In a book it writes something else.
4) yassinus: What is the "theorem of bounded convergent of Lebesgue "?
5) yassinus: What do you mean by "continuity by part "?
(6) yassinus: How would you demonstrate that convergence without condition (2) )

1) for Riemann lemma you can see 7 in this topic .
2) i mean integral( sin(t)/t dt,t=0...+oo) is equal to pi/2 it's  easy to show
3)theorem of bounded convergent of Lebesgue : if you have a sequence of functions  (V_n )continuos by part ( in french continue par morceau ) ojn a [a,b] and satisfy
   (*) simple convergence on [a,b] to a function V
   (*) condition 3 in Kent's proof
 then we can say that integral(V_n,a,b) tends to integral(V,a,b)
4) Kent's formula is true ( just multiply by sin(t/2) his sum and after transforming product into sum you will get Kent's formula

Did you mean that:

\[ \lim_{x \to \infty }\int_a^b e^{ixt}\cdot f(t)dt =0 \text{ ?, if f is continuous almost everywhere and bounded} \]

I think that I used a particular case of this without knowing :D.(Lemma) But I see that that particular case also gives a proof for Riemann Lema.

\[ \lim_{x \to \infty }\int_a^b e^{ixt}\cdot f(t)dt -\lim_{x \to \infty }\int_a^b e^{i[x]t}\cdot f(t)dt = \]
\[ \lim_{x \to \infty }\int_a^b e^{i[x]t}\cdot f(t) \cdot (e^{i\{x\}t}-1)dt= \]
\[ \lim_{x \to \infty } (e^{i\{x\}t_{n,x}}-1) \int_a^b e^{i[x]t}\cdot f(t) \]
Since $(e^{i\{x\}t_{n,x}-1)}$ is bounded and $\int_a^b e^{i[x]t}\cdot f(t)\to 0$ we have that
\[ \lim_{x \to \infty }\int_a^b e^{ixt}\cdot f(t)dt -\lim_{x \to \infty }\int_a^b e^{i[x]t}\cdot f(t)dt \to 0\Rightarrow \lim_{x \to \infty }\int_a^b e^{ixt}\cdot f(t)dt \to 0 \]

I do not have a nice demonstration for my lemma? do you have one for Riemann lema?

you can prove it by density theorems ,it will be very easy  :)

xirti: once you fix my typo - that should be $\sin\left(\frac t2\right)$ in the denominator, not $\sin\left(\frac s2\right)$ - the way I would prove the validity of that formula is to start with the right hand side and treat it as the geometric series that it is:

$\sum_{k=-n}^ne^{ikt}=\frac{e^{-int}-e^{i(n+1)t}}{1-e^{int}}= \frac{e^{i(n+1)t} -e^{-int}}{e^{int}-1}$

Now multiply by $\frac{e^{-i\frac t2}}{e^{-i\frac t2}}$

$=\frac{e^{i\left(n+\frac12\right)t}-e^{-i\left(n+\frac12\right)t}}{e^{i\frac t2}-e^{-i\frac t2}} =\frac{\sin\left(\left(n+\frac12\right)t\right)}{\sin\left(\frac t2\right)}$

Because this formula - the Dirichlet kernel - is such an important object in the theory of Fourier series, I have this computation committed to memory.

yassinus: I did say, "Conditions (2) and (3) above suffice for a (Reimann-integral style) argument that the remaining integral tends to zero ..."

I was intentionally trying to pitch my answer at the level of a course in which the Riemann integral is developed (along with uniform convergence) but not the Lebesgue integral. Yes, if you're willing to use the tools of that later course, such as the Dominated Convergence  Theorem, then we can relax the hypothesis yet further. After all, this is basically the a claim that the partial sums of a Fourier series converge at a particular point.

In fact, a sufficient condition on $f$ for this result would be that $f$ is of bounded variation.

The theorem you're calling the "Riemann lemma" I know as the "Riemann-Lebesgue lemma." I didn't quote that, again, because of the level of course I was trying to stay within.

By bouded varioation in x do you understand that $\frac{f(t)-f(x)} {t-x}$ is bounded on a neibourghood of x or $f(t)-f(x)$ is bounded on a neibourghood of x?
Try to post a solution for this hipotesys. Maybe we will understand it. Please. :D

No. "Bounded variation" is a specific technical term, and not any of the things you are talking about. Try looking it up.

many thinks to kent for all the clarifications  :)

I looked it up and found that a ufnction with bounded variation on an interval $[a,b]$ means that there exists $M>0$ such that $\sum_{i=1}^{n}|f(x_{i+1})-f(x_i)|<M$ for all $a=x_0<x_1<...<x_n=b$ and for all $n$. Jordan sais that a function with bounded variation can be written as a difference af two monotone functions. And since any monotone function is differentiable almost everywhere results  that [b]a function with bonuded variation is differentiable almost everywhere[/b].

   In fact, we only need that $g(t)=\frac{f(t)-f(0)}{t}$ is integrable. (I set g(0)=0).
Indeed, the only place i used the differentiability of f is here $\lim_{n\to \infty} \int_0^{\pi} \sin (nt+\frac t 2 )\frac{f(t)-f(0)}{\frac t 2}dt =0$. But from the lemma it suffices that $g$ is integrable.
For $g$ to be integrable it suffices that $f$ to be integrable $-\infty<\liminf \frac{f(t)-f(0)}{t} <\limsup \frac{f(t)-f(0)}{t} <\infty$.

By the bounded variation condition we restrict f to be differentiable almost everywhere and we only need f to be continuous almost everywhere and to be $O(t)$.
So I weakened the condition without using any superior calculus. Even the demonstration of the lemma do not take more than five rows.
Correct me if i'm wrong. :D

xirti: as you noted, a function of bounded variation is continuous [i]almost[/i] everywehere and differentiable [i]almost[/i] everywhere. But almost everywhere isn't everywhere. If you name one particular point, such as zero, you are not entitled to assume that the function is differentiable at that point.

Take, for instance, $f(t)=\sqrt{t}.$ This function is certainly of bounded variation on $[0,\pi];$ in fact, it is monotone.

But $\lim_{t\to0^+}\frac{f(t)-f(0)}t=\infty.$

My plan for proving it using bounded variation as a hypothesis: Do the integration by parts that I did in #4, only with the integrals interpreted as Riemann-Stieltjes integrals. We now have to show that

$\lim_{n\to\infty}\int_0^{\pi}(V_n(t)-\pi)\,df(t)=0.$

We do that by interpreting it as a Lebesgue-Steltjes integral and invoking the Dominated Convergence Theorem.

The tools used make that less elementary than what I claimed in #4.

please , do you mean by "continuos almost everywhere"what we call it in frensh " continue par morceaux " ( that's what i translated to  continuos by part) ? if it is , how can we show that it's must be differentiable almost everywhere ? or juste is it simple to prove ? and what is integrability of Riemann-Stieltjes ?

I said that f needs also to be $O(t)$ so your counterexample doesnt work, kent. My function g is continuous almost everywhere and bounded and so, integrable.

I don't know how to spell it, but the French abreviation for "almost everywhere" is "p.p." Something like "presque partout."

This other concept you're talking about may be "piecewise continuous," which is a different condition: you can partition the interval into finitely many subintervals, and the function in question is continuous in the interior of each subinterval and has one-sided limits at each end.

But a piecewise continuous function may not be a function of bounded variation. A good example is $x\sin\left(\frac1x\right)$ on some interval containing zero. Continuous (hence piecewise continous) but not of bounded variation.

In the other direction, we could have a function of bounded variation that is not piecewise continuous. For an example, let $(r_n)$ be an enumeration of the rational numbers in $(0,1).$

Let $f(x)=\sum_{\{n: r_n<x\}}2^{-n}.$

Then $f$ is not piecewise continuous because it is discontinuous at every rational, but it is of bounded variation because it is monotone and bounded.

This entire topic is really about the convergence of Fourier series. In that context, bounded variation is a more useful hypothesis than continuity (or piecewise continuity).

There exists a continuous function whose Fourier series diverges at infinitely many points.

But for a function of bounded variation, the Fourier series converges at every point to the average of the right and left hand limits.

think you kent , from the conditiond it's what we call it "continue par morceaux"


I am using TeXnicCenter 1.0, MikTeX2.8, and Asymptote 1.91.
When I try to include an Asymptote EPS file with the build profile LaTex => PDF, I get the error message 'LaTeX error: Unknown graphics extension: .eps'

I read somewhere that replacing pdflatex with xelatex in the output profile would help, but no such luck. I get the following messages:
 Warning ** No image converter available for converting file
 Warning ** Please check if you have 'D' option in config file
 Warning ** pdf: image inclusion failed

I have no idea how to deal with this (neither for the image converter, nor do I know which config file the warning refers to.

As you have MikTeX 2.8 the answer is easy. Background is that earlier versions of LaTeX could only use eps files when compiling to DVI, then along came pdflatex for PDF which could only use formats like png, jpg, pdf but not eps. So the only solution was to go to pdf via dvi or use a package like eps2pdf.

But as I say MikTeX 2.7 and later makes this all history because you can use XeTeX instead:
[list=a][*]In TexnicCenter go to Build, Define Output Profiles
[*]Click on LaTeX=>PDF
[*]Click on Copy and change the name to XeLaTeX=>PDF
[*]Click on XeLaTeX=>PDF and look at the box labelled "Path to (La)TeX Compiler"
[*]Don't disturb the path but change pdflatex.exe to xelatex.exe
[*]Click OK to finish[/list]Now you will have XeLaTeX=>PDF in the dropdown box. Choose that and use that for compiling.

I feel that I followed the instructions, but it still does not work, and have attached a screenshot with the messages.
There are three messages at the bottom (about \citation, \bibdata, \bibstyle), which I have ignored so far, since they did not create problems with 'normal' TeX files. But maybe...

Thanks in advance for your help.

[quote="ryanr"]
There are three messages at the bottom (about \citation, \bibdata, \bibstyle), which I have ignored so far, since they did not create problems with 'normal' TeX files. But maybe...[/quote]They aren't really errors as they are telling you that you haven't used citations or a bibliography which is correct and you will always get these messages if you don't need references.

If you haven't already done so, put the eps file in the same directory as the tex file. To eliminate any other possiblities, try a simple document such as
[code]\documentclass{article}
\usepackage{graphicx}
\begin{document}
\includegraphics{filename.eps}
\end{document}[/code]Try a different eps file - ones exported from OpenOffice Draw usually work well. There could be a problem with your file.  You may also need Ghostscript to do the transformation of the eps file in the background so you should try installing that as well.

My TeX code looks exactly like your proposal (the file name is different, of course). The EPS file is most definitely in the same directory.
I will try to create to use a EPS file from a different source. However, the warning message seems to indicate that this root cause is different - that EPS is not supported by XeTex.
On a XeTex website (http://scripts.sil.org/cms/scripts/page.php?site_id=nrsi&id=xetex) I found the following answer to a problem:

(D. Venu Gopal", Tue, Jan 19, 2010 22:45 (CST)
Re: [XeLaTeX] includegraphics problem

XeLaTeX output is always pdf. So like pdflatex, in XeLaTeX you cannot use eps files with XeLaTeX. You can use jpg, pdf and png formats. Try them.

Are you absolutely sure that XeTex allows to include EPS files?

It is [b]XeLaTeX[/b] (not xetex) that supports eps and I have just successfully compiled such a file. You did put xelatex.exe (not xetex.exe) in the profile in TexnicCenter? Have you got Ghostscript installed?

Here is the first line of my log file - notice that it mentions [i]format=xelatex[/i] which yours is missing
[code]This is XeTeX, Version 3.1415926-2.2-0.9995.1 (MiKTeX 2.8) (preloaded format=xelatex 2009.9.16)  26 JAN 2010 20:58
entering extended mode[/code]

Here is the line from my output profile:
C:\Program Files (x86)\MiKTeX 2.8\miktex\bin\xelatex.exe
The next line (command line argument) is:
-interaction=nonstopmode "%pm"

What do you use as the command line argument? (Someone recommended on a different forum to add '-output-driver=xdv2pdf' as a command line option, but that was not successful.)

I tried a EPS file that I produced with Adobe Photoshop Elements, and that resulted in the same error message.

I have GSview and Ghostscript installed, and they work (I get the display of the file from Asymptote).

I have no idea where else to look. I still think that there is a configuration file somewhere that results in not finding Ghostscript, but I just do not know where that could be.  (I had to point GSview to the directory of Ghostscript.)

I have -interaction=nonstopmode -max-print-line=120 "%pm"
1) Try the LaTeX=>PS=>PDF profile
2) Try LaTeX=>PDF with \usepackage{eps2pdf}
3) Try eps2pdf from the command line to convert the eps
All methods use Ghostscript somewhere. In fact
4) load the eps into GSView and see if it will convert it to pdf
Can you post a small sample eps so I can see if it's a problem with the type of eps?

The good news: I can (finally) include EPS files, and the output looks a lot better than it did with the graphs converted to PNG.

Here is what happened: when I installed Asymptote and integrated it with MiKTeX, I had to maintain the MiKTeX settings (with mo.exe). I remember that first it did not seem to work, and that I ran both mo.exe and mo_admin.exe.
It turn out that the file dvipdfmx.cfg was created in C:\Users\Ryan\AppData\Roaming\MiKTeX\2.8\dvipdfm\config\. The file contained only one line (p 612bp,792bp).

Once I removed this directory, and maintained the output profile according to your original instructions,  everything was fine.

I really appreciate your help!


calculus
integration
limit
trigonometry
calculus computations

Find $ \lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \sum_{k\equal{}1}^n \sin kx \sin x\ dx$.


vector
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ R$ be a commutative local ring with identity and $ M,N$ are two finitely generated $ R$-modules. Prove $ M\otimes_R N\equal{}0$ implies $ M\equal{}0$ or $ N\equal{}0$.

Thanks

[quote="myan"]Tensor with the residue field, note that it is true for $ R$ a field, and use Nakayama's lemma.[/quote]

Please explain for me exactly!!

Thanks

Let $ m\subset R$ be the maximal ideal, and let $ k = R/m$ be the residue field.

Define $ M_k = k\otimes_R M$ to be as in the hint, tensoring with the residue field. Now we have an iso $ M_k\cong M/(mM)$ (try to prove this, it isn't immediate).

Note that $ m = J(R)$ and $ M$ finitely generated gives us the hypothesis of Nakayama, so if $ M_k\cong M/(mM) = 0$, then $ M = mM$ so $ M = 0$.

So we will work towards that.

Localization will preserve "being 0", so:

\begin{eqnarray*} 0 = (M\otimes_R N)_k & = & (k\otimes_R M)\otimes_R N \\
& = & M_k \otimes_R N \\
& = & (M_k\otimes_R N) \otimes_k k \\
& = & M_k \otimes_k (N\otimes_R k) \ (associates \ by \ bimodule, \ and \ commutes) \\
& = & M_k\otimes N_k \end{eqnarray*}

But $ M_k$ and $ N_k$ are vector spaces over the field $ k$, so either $ M_k = 0$ or $ N_k = 0$. But by the above Nakayama argument, we have what we wanted to show.

Edited: Sorry, I used $ A$ not $ R$ for my ring, it is fixed now.

Let $ (R,\mathfrak{m},k)$ be our local ring.

If $ P$ is an $ R$-module let $ P_k = k\otimes_{R}P$. This is isomorphic to $ P/\mathfrak{m}P$.

To prove this, tensor the exact sequence of $ R$-modules
\[ 0\longrightarrow \mathfrak{m}\stackrel{j}{\longrightarrow}R\stackrel{\pi}{\longrightarrow}R/\mathfrak{m}\longrightarrow 0\]
with $ P$ to get the exact sequence
\[ \mathfrak{m}\otimes_R P\stackrel{j\otimes 1}{\longrightarrow}R\otimes_R P\stackrel{\pi\otimes1}{\longrightarrow}R/\mathfrak{m}\otimes_R P\longrightarrow 0\]
Since the map $ f: R\otimes_R P\to P$ defined by $ r\otimes p\to rp$ is an isomorphism of $ R$-modules, define
\[ g = (\pi\otimes 1)\circ f^{ - 1}: P\to k \otimes_R P\]
Then $ \text{Im}(g) = \text{Im}(\pi\otimes 1) = P_k$ and $ \text{Ker}(g) = f(\text{Ker}(\pi\otimes 1)) = f(Im(j\otimes 1)) = \mathfrak{m}P$. So we have the isomorphism we wanted.

Now suppose we have $ M\otimes_R N = 0$ where $ M,N$ are finitely generated $ R$-modules.
Then
\[ M_k\otimes_k N_k = (M\otimes_R k)\otimes_k(N\otimes_R k)\cong M\otimes_R(k\otimes_k(N\otimes_R k))\]

\[ \cong M\otimes_R(k\otimes_k(k\otimes_R N))\cong M\otimes_R((k\otimes_k k)\otimes_R N)\cong (M\otimes_R N)_k = 0\]
Therefore $ M_k\otimes_k N_k = 0$ but because $ M_k,N_k$ are vector spaces we will either have $ M_k = 0$ or $ N_k = 0$.
In other words either $ \mathfrak{m}M = M$ or $ \mathfrak{m}N = N$. The result follows from Nakayama's lemma.

EDIT: HilbertThm90 beat me :P. I wrote basically the same

Excellent. You even proved the iso I just claimed!  :lol:

I had seen this solution in Atiyah-Macdonald book's (p. 31), but in my problem $ M$ is a right $ R$-module and $ N$ is a left, No bi-modules.


search
function
algebra unsolved
algebra

Some users don't want to learn, others simply ignore advises.
But please read and respect the following guideline:

To make it short: [b]ALWAYS USE YOUR COMMON SENSE WHEN POSTING![/b]
If you don't have common sense, don't post.

More specifically :

[u]For new threads :[/u]

a) [b]Good, meaningful title[/b]
b) [b]Use search function[/b]
c) [b]Good problem statement[/b]

[u]For answers to already existing threads :[/u]

d) [b]Of any interest and with content[/b]
e) [b]Well written and checked answers[/b]

To repeat it: [b]ALWAYS USE YOUR COMMON SENSE IF POSTING![/b]

Everything definitely out of range of common sense will be [u]locked or deleted[/u] (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from tomorrow on (April the 19th, 2007).
Feel free to discuss this in here.

For more explanations, see : http://www.mathlinks.ro/Forum/viewtopic.php?t=135914

This is spam! We don't need to look the same thing three times at a time! :mad:

Japanese Communities Moderator:

kunny

If by "three times" you meant the three same threads I created :

Firstly, they are here just because some people do not look at each and every single section, and I want to make sure everyone sees that.
Secondly, two of them are locked since I don't want them to be spammed, and this one is the main one (you surely have noticed a link to it in the two other threads), in order to discuss those guidelines (but I'm repeating myself as this is said in my previous post).
Thirdly and lastly, you had better look at the posts in this section, see how spammy they are, and come back to discuss about [i]it[/i].

Regards,
-mathmanman.

[quote="derezins"][quote="mathmanman"]
Everything definitely out of range of common sense will be [u]locked or deleted[/u] (exept for new users having less than about [b]42[/b] posts, they are newbies and need/get some time to learn).

The above rules will be applied from tomorrow on (April the 19th, 2007).
Feel free to discuss this in here.

For more explanations, see : http://www.mathlinks.ro/Forum/viewtopic.php?t=135914[/quote]

[b]What is so magic in the number 42????????????????????????????????[/b][/quote]

42 is the higher score at IMO  :rotfl:

[quote="Meatman"]42 is the [color=red]higher[/color] score at IMO  :rotfl:[/quote]

You meant [i]highest[/i]  :D

No! 42 is the number of the UNIVERSE (Hitchhicker's guide to the galaxy), then IMO has 42 as the higher score! :rotfl:

All posts here, except mathmanmans, give this thread a very ironic and selflegitimating touch... :roll:


I am a first year student doing analysis and was wondering if someone could help me prove

given 2 real number a and b, and that there exsist a number x such that ax = b
if b /=(cannot equal) 0, explain why b/0 is not defined.


two regular ploygons of the same number of side has sides 48 meters and 55 meters in length respectively. What is the length of the side of another polygon of the same number of sides, of its area is equal to the sum of the two?

The area of a regular polygon is something times (side length)^2
a*48^2+a*55^2=a*x^2
x=$ \sqrt{5329}\equal{}\boxed{73}$


I wonder if there is an easy answer for that question:
What is the minimum value of b, such that for all a
$ i(a) \cdot b>a!$, where $ i(a)\equal{}p_1 \cdot p_2 \cdot ... \cdot p_{\pi(a)}$, and $ a\equal{}p_1^{x_1} \cdot p_2^{x_2} \cdot ... \cdot p_{\pi(a)}^{x_{\pi(a)}}$, and $ p_i$ is prime.

[quote="tkrass"]... and $ a \equal{} p_1^{x_1} \cdot p_2^{x_2} \cdot ... \cdot p_{\pi(a)}^{x_{\pi(a)}}$, and $ p_i$ is prime.[/quote]

I am unclear on what each $ x_i$ can be.  Can $ a$ be prime or a
composite, based on its exponents?
Can you give a few begiining examples to illustrate your problem?

It seems to me that $ i(a)$ is the product of the (first powers of the) distinct primes which ocur in the prime factorisation of $ a$.

Basically, he wants a minimum b such that $ b$ times the prime factors of $ a$ is greater than $ a$.

The answer: There is no such b.

Note that $ i(a)\equal{}i(2a)$ if a is even.

Assume that such a $ b$ exists. Then, $ bi(a)>a>(b\minus{}1)i(a)$

Note that this would imply that $ bi(2^n a)\equal{}bi(a)>2^n a$ ifor all even a.

But, $ a>(b\minus{}1)i(a) \implies bi(a)>2^n (b\minus{}1)i(a) \implies \frac{b}{b\minus{}1} > 2^n$. However, we can always pick a value of $ n$ such that this is false.

[quote="tkrass"]I wonder if there is an easy answer for that question:
What is the minimum value of b, such that for all a
$ i(a) \cdot b > a!$, where $ i(a) \equal{} p_1 \cdot p_2 \cdot ... \cdot p_{\pi(a)}$, and $ a \equal{} p_1^{x_1} \cdot p_2^{x_2} \cdot ... \cdot p_{\pi(a)}^{x_{\pi(a)}}$, and $ p_i$ is prime.[/quote]

No b. let $ a\equal{}p^x$, p prime.Suppose there is such b.if we increase $ x$ enough, $ a!$ will eventually be bigger than $ i(a)b$.


Well, first of all, can you guys see my signature??
I can't seem to see my own signature after I post..
I didn't put anything offensive like Valentin Vornicu said and I really don't think he deleted mine.. or Did you??
Well I have $ \infty^{\infty}$ for now, but again, I can't see mine.
Did I do something wrong or is it my computer that is causing problems..??
Please Help me Out! :help:

Make sure you activated the CheckBox under the post (third CheckBox from the top, that one with "Attach signature (signatures can be changed in profile)").

Oh! Oops.. :oops:
Now can you see it?(Well, that is pretty stupid question..cuz I probably can see it myself.. we'll see..)
Okay
Thank you ZetaX!
OH! One more thing.
Does the check on the box stay?

[quote="jmerry"]Look at the Preferances-Posting a Message tab in your profile. There's an option "Always attach my signature". If that's checked, you'll include it by default; if it's not, you won't. Either way, you can override it for any particular post.[/quote]

Okay. Thank you!

i don't see a checkbox, I just see the   Disable BBCode in this post
  Disable Smilies in this post
  Notify me when a reply is posted
ones.

... in your profile.

You're looking in the wrong place.

when i do that though, it still doesn't show...

You don't have a signature. There's nothing to show.

To make a signature, go [url=http://www.artofproblemsolving.com/Forum/usercp.php?mode=editprofile&sub=sig]here[/url].

It probably has nothing to do with your platform. Look in the FAQ section in this forum.

[quote="yunhua98"]I tried that, but it deletes it everytime I look!
this might have nothing to do with this, but I use Windows Vista, and it's supposed to be missing several features that XP has.[/quote]
You need to type in a signature and press save and check that little checkbox.


By melting a certain amount of gold I can make 12 coins, but no more (some gold may be left over), each weighing 15 grams; or, I can make 16 coins, but no more, each weighing 12 grams. If I decide to make coins each weighing 14 grams, what is the maximum number of coins I can make?

By making 12 coins with a weight of 15 grams each, you have at least $ 12\times15\equal{}180$ grams of gold. By making 16 coins with a weight of 12 grams each, you have at least $ 16\times12\equal{}192$ grams of gold. Thus, you can make a maximum of $ \frac{192}{14}\approx13$ gold coins each weighing 14 grams.

More rigorous:

Making 12 coins each 15 g gives the following bounds: 180-195
Making 16 coins each 12 g gives the following bounds: 192-204
The intersection is 192-195 g.

Divide by 14: 13.71-13.93 coins, so [b]13[/b].


inequalities
trigonometry
inequalities unsolved

Find the maximum value of:
$P=\cos^2\alpha-\cos^2(\alpha+\beta)+\frac{3}{2}\cos^2\beta$ when $\alpha,\beta\in(0,\frac{\pi}{2})$


$8$ singers take part in a festival. The organiser wants to plan $m$ concerts. For every concert there are $4$ singers who go on stage, with the restriction that the times of which every two singers go on stage in a concert are all equal. Find a schedule that minimises $m$.

Assume the singers are $A,B,C,D,E,F,G,H$

Let $S_{A,B}$ be the numbers of times $A$ and $B$ go on stage together.
Let $S = \sum S_{A,B}$ for all possible pairs of singers. (There are C(8,2) = 28 pairs)

Since the times of which every two singers go on stage in a concert are all same, we have that $S = 28S_{A,B}$

On the other hand, in each concert six pairs of singers go on stage together.
So, $S = 6m$
So $28S_{A,B} = 6m$
$14S_{A,B} = 3m$
$m = \frac{14S_{A,B}}{3}$
$m\geq 14$

14 concerts is possible:
$(A,B,C,D),(A,B,E,F),(A,B,G,H),(C,D,E,F),$ $(C,D,G,H),(E,F,G,H),(A,C,E,G),(B,D,F,H)$ $,(A,C,F,H),(B,D,E,G),(A,D,E,H),(B,C,F,G),(A,D,F,G),(B,C,E,H)$

The minimum is 14. Let the equal quantity be $k$, such that $${8 \choose 2} \cdot k = {6 \choose 2} \cdot m \iff 14k = 3m,$$ which implies $14 \mid m$ and $m \geq 14$.

Now we give a construction for 14. Label the singers $a_1, a_2, \cdots, a_8$ in some order. Consider the set of constructions $(a_1, a_{2+i}, a_{3+i}, a_{5+i})$ and $(a_{2+i}, a_{3+i}, a_{4+i}, a_{6+i})$ for $0 \leq i \leq 6$. Indeed, in this construction, each pair of singers appears on-stage simultaneously precisely three times, so we are done.


Suppose that there is a point X in the plane of a given quadrilateral ABCD such that the perimeters of ABX, BCX, CDX, ADX are equal. Prove that the quadrilateral is cyclic.

It is not cyclic, but circumscribed. The proof is trivial: Since the perimeters of the triangles ABX, BCX, CDX, DAX are equal, we have AB + AX + BX = BC + BX + CX = CD + CX + DX = DA + DX + AX, so that

(AB + CD) + (AX + BX + CX + DX) = (AB + AX + BX) + (CD + CX + DX) = (BC + BX + CX) + (DA + DX + AX)
= (BC + DA) + (AX + BX + CX + DX),

so that AB + CD = BC + DA, and thus the quadrilateral ABCD is circumscribed.

  Darij

Well I obtained that it is circumscribed,but that is all you can get out of it?The statement is that it is also cyclic.

[quote="mateivld"]The statement is that it is also cyclic.[/quote]

Then it is wrong...

  Darij

Can tou tell me why?I don't think that it is wrong.

AB=BC=CD=DA. Then ABCD is not always cyclic.

(a) Suppose that there is a point $X$ in the plane of a given convex quadrilateral $ABCD$ such that the perimeters of triangles $ABX,BCX,CDX, DAX$ are equal. Show that $ABCD$ is a tangential quadrilateral.

(b) If a convex $ABCD$ is tangential, does there necessarily exist a point $X$ such that the perimeters of triangles $ABX,BCX,CDX,DAX$ are equal?

[hide]problem originates from Belarus NMO 2004 B. 10.6, the wording comes from [url=https://imomath.com/othercomp/Blr/BlrMO04.pdf]Imomath [/url],
their translation is incorrect, tangential is the word, they miss the convexity also,
here is the correct wording with the second part also

problem was also 2005 Brazil Cono Sur TST 2.3[/hide]


There are three square boards, each having an integral number of centimetres in its side. The second board is $ 5$ $ cm^{2}$ larger than the first and as much smaller than the third. What are the dimensions of the three boards?

[hide="Meh"]Let the second have an area of $ x^2$, so that the first has area $ x^2\minus{}5$ and the third has $ x^2\plus{}5$.

Let the side length of the first be $ (x\minus{}k)^2\equal{}x^2\minus{}2xk\plus{}k^2\equal{}x^2\minus{}5$.

Then $ k(2x\minus{}k)\equal{}5$ so $ k\in\{1,5\}$. Trying $ k\equal{}1$, we get $ 2x\minus{}1\equal{}5$ or $ x\equal{}3$. Trying $ k\equal{}5$, we get $ 2x\minus{}5\equal{}1$ so that $ x\equal{}3$ once again, but in this case the side length, $ x\minus{}k$, would be negative. So we choose $ x\equal{}3$ and $ k\equal{}1$, making the side length of the first one $ 2$ and that of the second one $ 3$.

But then the last board must have area $ 9\plus{}5\equal{}14$ which is a nonsquare, contradiction. So there are no solutions?[/hide]

Did I go wrong somewhere?

Here's my thinking. I don't thnk it can be done with integral lengths.
If you look at areas of the first few integers, $ 1^2\equal{}1, 2^2\equal{}4, 3^2\equal{}9, 4^2\equal{}16, 5^2\equal{}25$, then each square increases by an arithmetic sequence. $ 4\minus{}1\equal{}3$, then $ 9\minus{}4\equal{}5$, then $ 7$, $ 9$, and so on.
Of course, we can disregard $ 4^2$ and up since the differences of the squares, starting with $ 7$, only get larger. So a possibility for two lengths are $ 2$ and $ 3$, However, since you can't have negative lengths, it can't go lower than $ 1$, but $ 2^2\minus{}1^1$ is clearly not $ 5$, so there can't be any integral lengths.

Sorry. This should have said:

There are three square boards, each having an integral number of inches in its side. The second board is five square feet larger than the first and as much smaller than the third. What are the dimensions of the three boards? [$ 12$ inches $ \equal{} 1$ foot.]


Let F(x,y)=4e^ [(-x/4)^2 - (-y/4)^2]
be the function that assigns to the point (x,y) the elevation of the ground at that point.

-The temperature at the point (0,0,0) is 15 degrees C and it falls by 1 degree C for every unit of elevation.
- The temperature decreases by 5 degrees C for each unit in the positive x direction and by 1 degree C for each unit in the positive y direction.
- The temperature is determined by the sum of the effects of elevation and position.

Questions:
1) Let T(x,y) denote the function that assigns to the point (x,y) the temperature at that point.
Write down the equation that describes this function.
2) DEtermine the direction in which Bill and Jane should begin to walk, away from the point (5,-1), in order for the temperature to rise most quickly.
------------------------------------------------------
My answer for 1) would be:
(Given dT/dx=-1, dT/dy=-5, dT/dF=-1)
so, T(x,y)=(dT/dF)(F(x,y)) + (dT/dx)x + (dT/dy)y
 T(x,y)=-F(x,y)-5x-y

Am I correct so far??
But then I have no idea on how to do #2)...
Can someone help??

close, but you want T(0,0)=T(0,0,4)=15-4=11 so

$ T(x,y)\equal{}15\minus{}y\minus{}5x\minus{}4e^{\frac{x^2\minus{}y^2}{16}}$

then $ T_x\equal{}\minus{}5\minus{}\frac{x}{2}e^{\frac{x^2\minus{}y^2}{16}},T_y\equal{}\minus{}1\plus{}\frac{y}{2}e^{\frac{x^2\minus{}y^2}{16}}$

so they should head in the direction of the gradient

$ (T_x(5,\minus{}1),T_y(5,\minus{}1))\equal{}(\minus{}\frac{10\plus{}5e^{\frac{3}{2}}}{2},\minus{}\frac{2\plus{}e^{\frac{3}{2}}}{2})$


In my school, sixth graders are not allowed to count their scores in mathcounts

they say sixth graders are not allowed, only seventh and eighth graders

is this wrong

and also:

can the teacher of the club change the ranks of students depending on grade

like favor an eighth grader for #1, but the seventh grader made #1

thanks!!

Are you serious? That is definitely not fair: the coach should use the official tiebreaker rules in MATHCOUNTS to determine places, etc., and should base everything on math skill by itself. Sadly, this is not enforceable.

Not only math skill, but how well each member of the team works with the others

Sixth graders are eligible to participate in MathCounts, and they are the ones that keep the team going after the probably stronger 7th and 8th graders leave, so I really don't see the logic between the teacher's decision

However, in terms of team selection, the coach may use any method that doesn't directly violate any of the rules set by MathCounts, whatever they may be.  They could go like "You four kids are on the team" if they wanted to do that.

There were many smart 6th graders at Miller Middle School last year, but their school was like yours; no 6th graders on the team...a few 6th graders that barely missed USAMO and have some speed couldn't compete.

This is really unfair. Teachers should stick to the in-school competition scores, regardless of age. I know that if two students have the same score on the school test, and one is older than the other, the older one places higher than the younger one, which is totally unfair. i was in 6th grade last year, and I wasn't the best on the team, but I did pretty well @ chapter and state, and the team was better than it would have been if they had only let in 7th and 8th graders.

Teachers should not show partiality; they must decide for the good of the team. Personally, I would put many 7th graders. "Freshman" 8th graders are gone the next year and unless you're not an average person, 7th graders are probably better. For next year, they will have better coordiation that comes with experience.

8th graders with previous are experience are good though.

[quote="akhil0422"]Teachers should not show partiality; they must decide for the good of the team. Personally, I would put many 7th graders. "Freshman" 8th graders are gone the next year and unless you're not an average person, 7th graders are probably better. For next year, they will have better coordiation that comes with experience.[/quote]

That's pretty much at my school. Last year, with the exception of one 8th grader, the top 5 or 6 were all 7th graders, including me.

This year, with that 8th grader gone, all of the top 4 are 8th graders. Its just like this one generation.

[quote="Sly Si"]Teachers can pick their teams pretty much however they want.  They can give the school competition (or part of it--when I was in MC we only ever took the sprint round) and base it entirely on that, or they can go on subjective assessments of who's going to work well together, or who will perform better under the pressure of competition day.  They are also allowed to be biased towards older students (perhaps to "reward" them for several years of dedication to the team), or younger students (to rebuild for next year).[/quote]

It is not fair. Mathcounts is a competition. Best of best in math (not in age, grade, or teacher"s favor) goes up to Chapter, States, and Nationals.

I'm sure you can get another school to let you take it as an individual.

This is terrible. Sixth graders have every right to participate. Really, try hard to get your teachers and coaches to let you on the team.

In many schools, 8th grade takes priority. Remember, you dont have to be on the official team to make state or nats. Being among the four individuals (top 8 from your school overall) is good enough. I made nats last year by being on my school's "B" team.

If you can't take it, make sure you register as an individual. Competitions are soon, so take action quick!!

usually 8th graders are chosen because it is their last year in mc...

Eh, here we have a "junior high" format and not a "middle school" format so we start at 7th grade...

[quote="ghertfort"]It is not fair. Mathcounts is a competition. Best of best in math (not in age, grade, or teacher"s favor) goes up to Chapter, States, and Nationals.[/quote]

Yes, it's a competition.  But it's also a competition between schools, so teachers have every right to do what they believe is best for their team's chances.  If that means taking younger students to try to put together an excellent team next year, or giving a spot to someone who clearly just had a bad day on the school round, then IMHO, so be it.


Note that on all other forums, I go by the name drspyder.

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[url=http://www.putfile.com/media.php?n=MapleStory%202005-02-17%2010-57-28-59]The Linky[/url]
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[size=150]Enjoy![/size]

[i]Edited by admin to: (a) delete the objectionable ones (because we do have have younger kids on the forum) and (b) to move to Fun & Games[/i]

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Aren't these two the same?

this is mine :blush: ...i know, i'm pathetic


algebra
polynomial
induction
binomial coefficients
number theory proposed
number theory

We know that :
 $ x \plus{} 1 \minus{} x \equal{} 1$
 $ (x \plus{} 1)^2 \minus{} 2x^2 \plus{} (x \minus{} 1)^2 \equal{} 2$
 $ (x \plus{} 1)^3 \minus{} 3x^3 \plus{} 3(x \minus{} 1)^3 \minus{} (x \minus{} 2)^3 \equal{} 6$
 Formulize and prove that formula  ?

Formula:
For a polynomial $ p$ define the polynomial $ \Delta (p)$ as $ \Delta (p) (x) : \equal{} p(x \plus{} 1) \minus{} p(x)$ and then recursivly $ \Delta^0 (p) : \equal{} p ; \Delta^n (p) : \equal{} \Delta (\Delta^{n \minus{} 1} (p))$.
Then by induction you get $ \Delta ^n p(x) \equal{} \sum_{k \equal{} 0}^n ( \minus{} 1)^{n \minus{} k} \binom{n}{k} p(x \plus{} k)$.
Now $ \deg ( \Delta^n (p) ) \equal{} \deg (p) \minus{} n$ as long as $ \deg (p) \geq n$.
Now you can state your formula (after substitution $ x \to x \plus{} n \minus{} 1$) as $ \sum_{k \equal{} 0}^n ( \minus{} 1)^{n \minus{} k} \binom{n}{k} (x \plus{} k)^n \equal{} \Delta^n (x^n) \equiv n!$, which follows directly from the previous (it has degree $ 0$, so is constant, and the constant can be found trivially).

My English is not good and indeed i can't understand your solution , Zetax .
 Can you write explicit formula of $\Delta^n(p)$ ? ($\Delta^0=p$ , $\Delta^1(p)=p(x+1)-p(x)$ , $\Delta^2(p)=p(x+2)-2p(x+1)+p(x)$ , am I right ? )
  What is polymomial $p$ in your solution  ? Can you explain it for me ?

My english isn't good, too...
I define it for every polynomial $p$, and yes, you are right:
The written out version is the formula $\Delta ^n p(x) = \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} p(x+k)$ (here stood an $f$ instead of $p$ on the original RHS, sorry, I corrected it), since on the RHS only binomial coefficients and $p(x+k)$ occurs, so e.g. $\Delta ^4 (p)(x) = p(x+4)-4 p(x+3)+ 6 p(x+2) - 4 p(x+1) + p(x)$.
And at the end, I set $p(x)=x^n$ (you could also use every polynomial of degree $n$ having leading coefficient $1$) into this formula to get $\sum_{k=0}^n (-1)^{n-k} \binom{n}{k} (x+k)^n \equiv n!$.


geometry
function
calculus
derivative
calculus computations

The hypotenuse of a right triangle has one end at the origin and one end on the curve $y = x^{2}e^{-3x}$, with x [u]>[/u] 0. One of the other two sides is on the x-axis, the other side is parallel to the y-axis. Find the maximum area of such a triangle. At what x-value does it occur?

All right, I know that I'll probably need $A = \frac{1}{2}bh$. I know I have to in some way find out where the curve intersects either the vertical side of the triangle or the hypotenuse to maximize the area? I think? But I don't know if i'm thinking the right way, and I don't know how to go about setting this up.

Let $P(t,\ t^{2}e^{-3t})\ (t>0)$ on the curve $y=x^{2}e^{-3x},$ and the point $Q$ such that $PQ$ is perpendicular to $x$ axis. The area of the traingle $OPQ\ : =S(t)$ is $S(t)=...$ then $S'(t)=...$

all right.. let's see if i understand what you just said.

$A = \frac{1}{2}bh$
The base is just x, and the height is the function $x^{2}e^{-3x}$ (because the vertical side has to end on a point on the curve).

So, the area can be written as
$A = \frac{1}{2}x(x^{2}e^{-3x}) = \frac{1}{2}x^{3}e^{-3x}$

To maximize the area, we will take the derivative
$\frac{dA}{dx}= \frac{3}{2}x^{2}e^{-3x}-\frac{3}{2}x^{3}e^{-3x}$
$\frac{dA}{dx}= \frac{3}{2}x^{2}e^{-3x}(1-x) = 0$
$x = 0, x = 1$

x = 0 is useless because if we have a base of 0, then there will be no area, and we won't even have a triangle anymore.

So, to test if x = 1 is a maximum i'll use the first derivative test and test x = 1/2 and x = 2.

$A'(\frac{1}{2}) = \frac{3}{2}(\frac{1}{2})^{2}e^{-3* \frac{1}{2}}(1-\frac{1}{2}) ....$  increasing.
$A'(2) = \frac{3}{2}2^{2}e^{-3* 2}(1-2) ....$  decreasing

So, at x = 1 there is a maximum.

Now, let's find the value of the maximum area
$A = \frac{1}{2}1^{3}e^{-3*1}= \frac{1}{2}e^{-3}$

So, the maximum area of the triangle is $\frac{1}{2}e^{-3}$ square units, and the maximum occurs at x = 1.


vector
analytic geometry
linear algebra
matrix

Let $k$ be a real variable and let $\beta = \{e_1,e_2,e_3\}$ be the standard base for $\mathbb{R}^3$. Consider the vector $v$ having $(1,1,-1)$ as coordinates with respect to $\beta$. Define $f_k$ as the linear transformation of $\mathbb{R}^3$ satisfying \[ \left\{ \begin{aligned} f_k\left(e_1\right) & = ke_1 - ke_2 + ke_3 \\ f_k\left(e_2\right) & = -e_1 + (2k - 1)e_2 + e_3 \\ f_k\left(e_3\right) & = 2e_1 - 2ke_2 - 2e_3 \end{aligned} \right. \] For which values of $k$ does $v$ belong to the image of $f_k$?

It's a system of equations 3 x 3 you need to solve  ...  :?

[quote="Arne"]I know, I know, I just posted this here for everyone who wants to have a simple practice problem on matrices and linear transformations.[/quote]Yes, but isn't this site supposed for problem solving? This looks merely calculation... :(
[unless I'm wrong on this of course]

Peter, this is the Matrix Algebra - COMPUTATIONS & TUTORIALS section. :mad:


Let $a,b,c$ be three sides of a triangle.. Suppose $a,b$ are the roots of the equation $x^2+4(c+2)=(c+4)x$ and the largest angle of the triangle is $x$ degrees. Find $x$

even though pontios solved this problem correctly, the original problem has very confusing mix up with the variables, $x$ is a variable in that equation and then $x$ is a constant in which it is the $x$ degrees of the largest angle of the triangle  :?


Let nonnegative real numbers $ a_{1},a_{2},a_{3},a_{4}$ satisfy $ a_{1} \plus{} a_{2} \plus{} a_{3} \plus{} a_{4} \equal{} 1.$ Prove that
$ max\{\sum_{1}^4{\sqrt {a_{i}^2 \plus{} a_{i}a_{i \minus{} 1} \plus{} a_{i \minus{} 1}^2 \plus{} a_{i \minus{} 1}a_{i \minus{} 2}}},\sum_{1}^4{\sqrt {a_{i}^2 \plus{} a_{i}a_{i \plus{} 1} \plus{} a_{i \plus{} 1}^2 \plus{} a_{i \plus{} 1}a_{i \plus{} 2}}}\}\ge 2.$
Where for all integers $ i, a_{i \plus{} 4} \equal{} a_{i}$ holds.

Do any people want to have a try?This is the unique small-inequality in China TST.It looks terrible,right?But it indeed have a two line proof,try and solve it mathlinkers :)

p.s. I failed to solve it in the exam,but when I see the two line proof,I was wondering how stupid I am.

This is the solution I came up with during the test.maybe it is the "two lines solution"

Lemma:for non-negative real numbers a,b,c, $ \sqrt {b^2\plus{}a^2\plus{}ba\plus{}bc}\plus{}\sqrt {b^2\plus{}c^2\plus{}ba\plus{}bc} \ge 2b\plus{}a\plus{}c$
Proof: square both sides, it is equivalent to $ {(b^2\plus{}a^2\plus{}ba\plus{}bc)}*{(b^2\plus{}c^2\plus{}ba\plus{}bc)} \ge {(b^2\plus{}ac\plus{}ba\plus{}bc)}^2$, which is just an application of the famous Cauchy Inequality.

Back to the problem, take the sum of the two terms in the blanket.It suffices to prove the sum is not less than 4.Using the lemma four times we get the sum is not less than $ 4(a_{1}\plus{}a_{2}\plus{}a_{3}\plus{}a_{4})\equal{}4$.
Q.E.D

You're right !

p.s. congratulate you on being selected to be the member of our national team,wish you a perfect performance on 2009 IMO :lol:

[quote="linboll"]thanks
PS: what's your name?could you tell me via short message?[/quote]
Thank you for attention,and I have sent you a message,please check your inbox :)


we have $ a,b,c,d \leq 2$ we have the conclusion

abc <= (a^3 + d^3 + c^3) / 3

So ... using this and a, d, c, d <= 2 we have the answer

[quote="hanschinaman"]$ a,b,c,d\geq 0$ such that $ a^{2}+b^{2}+c^{2}+d^{2}=4$, prove
$ a^{3}+b^{3}+c^{3}+d^{3}+abc+bcd+cda+dab\leq 8$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99927


[hide="Hint"]You're picking 51 out of 100 numbers. Consider all the pairwise differences. What has to be the smallest one?[/hide]

For $ n>1$ consider the set $ \{ 1,2, ... , 2n\}$. Then any $ n\plus{}1$ element subset [b]must[/b] contain two numbers which are relatively prime.

[b]Proof:[/b] Let the pigeons be $ \{1 , 2\},\{3,4\}, ... ,\{2n\minus{}1,2n\}$ and use the fact that $ \gcd (k,k\plus{}1)\equal{}1$.  :wink:

[quote="mdk"]51 numbers are to be chosen from 1, 2, 3, . . ., 99, 100. Show that no matter how these
numbers are chosen, we can find two that have no common prime divisor.[/quote]
if we choose $ a_{1}<a_{2}<...<a_{51}$ and take for $ i\equal{}1...50:\ v_{i}\equal{}a_{i\plus{}1}\minus{}a_{i}$
we have v_1+v_2+...+v_{50}=a_{51}-a_1\le 99
so $ \exists i:\ v_{i}\le\frac{99}{51}\ gives\ v_{i}\equal{}1$ then $ gcd(a_{i},a_{i\plus{}1})\equal{}gcd(a_{i},v_{i})\equal{}gcd(a_{i},1)\equal{}1$.


topology
vector
real analysis
real analysis unsolved

we say that $M$ ,m-dimensional manifold ,is paralleled manifold if $TM$ is homemorphic to $M\times \mathbb{R}^{m}$.
1)prove that $M$ is paralleled manifold iff it would be possible to find $C^{0}$vector field $X_{1},\dots,X_{m}$ on $M$ such that all $x\in M$ ,all vectors $X_{1}(x),\dots,X_{m}(x)$ in $TM_{x}$ are linear independent .
2) prove that $S_{3}$ is paralleled .
3)can you prove $S_{7}$ is paralleled .
4) if $M$ is paralleled ,prove that $TM$ is also paralleled.

1) See the general result for nowhere dependent sections, for example, at the "Characteristic classes" by Milnor- Stasheff, p. 18, theorem 2.2.

2) $S^3 \subset \mathbb{R}^4$ admits three nowhere dependent vector fields $s_i(x)=(x, \overline{s}_i(x))$ where

$\overline{s}_1(x)=(-x_2, x_1, -x_4, x_3), \; \overline{s}_2(x)=(-x_3, x_4, x_1, -x_2), \;  \overline{s}_3(x)=(-x_4, -x_3, x_2, x_1).$

These were come from the the quaternions multiplication in $\mathbb{R}^4.$

3) The same approach as for $S^3$ with a point that we should consider the [url=http://mathworld.wolfram.com/Octonion.html]Octonions[/url] multiplication in $\mathbb{R}^8.$

4) Suppose $M$ is of dimension $m$, then we have $T(TM) \simeq T(M \times \mathbb{R}^m) \simeq TM \times T(\mathbb{R}^m) \simeq TM  \times \mathbb{R}^{2m}. $


Find all functions f: N to N such that:
1) f(m)=1 if and only if m=1
2) if d=gcd(m;n) then f(mn)=f(m).f(n)/f(d) and
3) for every $\large m \in N$ we have $\large f^{2000}(m)=m$
...help me...

Let $x,y$ be positive integers such that $y\geq 2$. Then
$f(x^y)=\frac{f(x^{y-1})f(x)}{f(x)}=f(x^{y-1})=\cdots =f(x)$
Hence $x^y=f^{2000}(x^y)=f^{1999}(f(x^y))=f^{1999}f(x)=f^{2000}(x)=x$ for all $y\geq 2$ which is obviously wrong.

Quite a strange problem since obviously false...
http://www.mathlinks.ro/Forum/viewtopic.php?t=65863,

i did a proof similar to leepakhin's, but working on geometric series and expecially on f(2)...are you sure the text is right?


quadratics
modular arithmetic
algebra
algebra proposed

Consider the quadratic equation $ x^2\minus{}ax\minus{}1\equal{}0$ where $ a$ is a positive integer.Let $ \alpha$ be a positive root of this equation and construct a sequence $ (x_n)$ by the following recursive rule
$ x_0\equal{}a,x_{n\plus{}1}\equal{}[\alpha.x_n],n\equal{}0,1,2,...$
Prove that there exists infinitely many integer $ n$ such that $ x_n$ is a multiple of $ a$

can you find $ b,c$ so that $ x_{n \plus{} 1} \equal{} bx_n \plus{} cx_{n \minus{} 1}$  ? :wink:

prove that $ x_{n\plus{}1}\equal{}ax_n\plus{}x_{n\minus{}1}$ with $ x_0\equal{}1$ and $ x_1\equal{}a$
by induction.
then look at the sequence $ (x_k \pmod{n})$, does it have to be periodic? why? how can we conclude?

I hope I'm wrong, but this problem looks like something from a Vietnamese magazine (probably ongoing competition)...

(it was posted here in two separate threads, by people who had already been "caught" posting such active problems. Can anyone from VN check this, please? Thank you.)


induction
vector
geometry
geometric transformation
reflection
perpendicular bisector
advanced fields

Show that any isometry of Euclidean space is a product of 1, 2 or 3 reflections. Furthermore, every orientation preserving isometry of S (n-phere) is a rotation.

Thanks,

Rosa

An isometry $T$ of the Euclidean $n$-space is a product of at most $n$ reflections. You can prove this by induction on $n$.

The case $n=1$ is trivial. Now suppose it's proven for dimensions $\le n-1$ and you want to show it for $n$. If the vector $v$ is taken to $Tv$, then let $U$ be the identity if $v=Tv$ and the reflection in the perpendicular bisector hyperplane of $v,Tv$ if $v\ne Tv$. Now $UT$ is an isometry which fixes $v$, so it fixes the orthogonal complement $v^\perp$ of $v$ as well. By the induction hypothesis, when regarded as a self-map of the hyperplane $v^\perp,\ UT$ is a product of at most $n-1$ reflections in hyperplanes of $v^\perp$. In $\mathbb R^n$ we regard each such reflection as a reflection in the $n-1$-dimensional hyperplane generated by the respective $n-2$-dimensional hyperplane of $v^\perp$ and $v$, and that's that.


I want to ask if I should get volume 2: and beyond or the Introduction to Number Theory book.

Actually, I should rephrase that: I already got the volume 2, but a lot of people suggested the intro to nt to me, so I was wondering what each book was good for.

I've heard that the nt book is VERY interesting, but so far, the volume 2 was pretty interesting also.

Um...well, it depends on what you want to work on. The AOPS Vol. 2 talks about a wide range of topics, but if you want to focus on a specific topic, and want to learn it more in depth, an intro book would be better.

Take a look at the Table of Contents in the Intro to NT book and see how much of it you actually know. If you find that a significant portion of the book is in your knowledge you can consider passing over (when there's vol 2 to consider).

I took one look at vol 2 and didn't know half of what was going on so I ordered that instead. I felt that vol 2 expanded my math knowledge, even though it didn't go as deep as the Intro/Inter books.

All of our Introduction-series books have diagnostic tests [url=http://www.artofproblemsolving.com/Books/AoPS_B_Recs.php]on this page[/url].  In particular, take a look at [url=http://www.artofproblemsolving.com/Books/IntroNumb/posttest.pdf]the post-test for the Intro Number Theory book[/url].  If you can solve most of the post-test without too much difficulty, then you don't really need the Number Theory book.

thanks to all of you guys! I realized that I knew a lot of stuff in the intro to NT book, so I think i made a good decision of getting a vol. 2 book. Thanks for your reviews!


When report cards were distributed in a certain homeroom class, it was found that 13 students recieved at least one grade of A, 15 students received at least one grade of B and 6 students received BOTH A's and B's.  If 1/3 of the students in the homeroom receive NO A's and NO B's, how many students are in the homeroom class?

[hide]
By PIE, the number of students who received A's or B's is $13+15-6=22$.  That's $\frac{2}{3}$ of the homeroom, so there are $33$ people in the homeroom.[/hide]

[hide="VENN DIAGRAM!!! (CLICK ME)"]Look at the Venn diagram. It is clear that $7+6+9 = 22$ students got A's or B's.

Let $x$ be the number of students in the homeroom.

Then $\frac{x}{3}$ students receive no A's nor B's. This implies that $x-\frac{x}{3}= \frac{2x}{3}$ students receive A's or B's. As mentioned earlier, this is 22 students. These quantities must be the same, so:

\[\frac{2x}{3}=22\]

\[x = \frac{3\cdot 22}{2}= 33\]

$\fbox{\textrm{33 students}}$[/hide]


Triangle ABC is inscribed in the unit circle. Let P be arbitrary point inside triangle. Show that |PA|.|PB|.|PC|<32/27.

I find out that this problem was proposed in
“Crux Mathematicorum”  ( somewhere from vol. 1, no. 1 to vol. 4, no. 2 known as “Eureka”),
the problem was numbered as [b]1895[/b] but unfortunately i dont have appropriate issue and i can' t find it on web (there is no old issues). So, if anybody have that issue, please be kind and help me   :roll:


He probably means interested in taking it.

this thread is totally spam, a mod should delete this

umm ur gonna have to give us a bit more information than that.
like:
grade
nation of residence

also, a LOT of ur questions can be answered at [url]http://www.unl.edu/amc/[/url]
im pretty sure that you can also find contact info (phone, adress) on the website. im sure they can answer your questions much better than we can..


inequalities
Euler
arithmetic sequence
number theory solved
number theory

1.
Prove that there exist arbitrarily long arithmetic progressions with all the terms being powers of integers.

2
Does there exist an infinite arithmetic progression with all the terms being powers of integers ?

3.
Do there exist arbitrarily long arithmetic progressions with all terms being equal powers of integers ?
If so, for which positive integers n do there exist arbitrarily long arithmetic progressions with all terms being n-th powers of integers ?

Good problem, Arne!

For 1.
By rk denote member of the finite arithmetic progression n!+1+k(n-1)!, k=1..n.
It is obvious gcd(ri,rj)=1.
Let ai=si*r1*r2*...*rn/ri, si is positive integer and satisfies the condition
ri|(ai+1) (gcd(r1*r2*...*rn/ri,ri)=1 ==> si exists).

Denote m=r1a1*...*rnan.
The sequence r1*m, r2*m,..., rn*m is required since rk*m=b^rk for some positive integer b.

For 2.
Suppose such arithmetic progression a+nd exists.
It that case the following inequality holds for all n=1,2,...
a+nd \leq [(a+nd)1/2]+[(a+nd)1/3-1]+[(a+nd)1/4-1]+...=[(a+nd)1/2]+[(a+nd)1/3-1]+[(a+nd)1/4-1]+...+[(a+nd)1/s-1]=A, where s=[ln(a+nd)/ln(2)] ==> a+nd \leq A\leq (a+nd)1/2*ln(a+nd)/ln(2) for all n=1,2,... It is absurd.

For 3.
I don't know.

Wow! That's a nice soln! The first 2 you can also find in some MOP handouts you can download here (they are combined in 1 prb). For the 3'rd one.. :D

What is MOP? And from where can it be downloaded?

Do you know how to solve 3, grobber?

MOP is the preparation for the IMO (I think  :D ). You can download those handouts (they contain some problems used for preparation in Romania) in the downloads section of this site-just look up and to the right. No, I don't know how to solve 3. I didn't even know how to solve 1 or 2, for that matter. Anyway, cool proof ! (I know I said it before but..).

See ya! :D

[b]MOP [/b]is the acronym for Math Olympiad Program - it's the name of the US training program for the International Mathematical Olympiad. I've named the similar Romanian Training program [b]RoMOP [/b]:D.

the first problem was proposed at the Balkan Olympiad in 2000. The third is super cool, and I was thinking about it. I've made it through for n=2 (that is squares :D). Further on, I don't see any breakthroughs soon :D

Valentin! Did you mean you found arbitrary long arithmetic progressions with all terms being squares? If you did, please, write proof.

So Valentin, what about the case n = 2 ? :D

for n=2

In fact, it is easy to prove that the only arithmetical sequences of squares of length 3 are :
(2n^2-1)^2  ;  (2n^2+2n+1)^2 ;  (2n^2+4n+1)^2    .

Then, it is possible to check that the next term of this arithmetical sequence is never a square, which proves that there doesnt exist arithmetical sequences of squares of length 4.

ool

I think you are wrong.

All arithmetical sequences of squares of length 3 with coprime items are
(m2+2mn-n2)2
(m2+n2)2
(n2+2mn-m2)2

It is known from Euler (1780) that there is no set of 4 distinct squares in arithmetic progression.

Pierre.

And where is the proof, pbornsztein?
I think it is incorrect to make such links about non well known facts.

I did not say that the proof is easy....I've got it and it is long.
If you really want I can put it here, but not now because as usual I'm in a hurry...

Pierre.


Prove that equation  $ x \equal{} \cos(x)$   has unique solution $ x_0$,  and that sequences  $ \{x_n\}_{n \equal{} 1}^{\infty}$,  such that   $ x_1 \equal{} 20,\,x_n \equal{} \cos(x_{n \minus{} 1})\ (n\geq 2)$  converges to  $ x_0$.

[url=http://en.wikipedia.org/wiki/Banach_fixed_point_theorem]Banach fixed point theorem[/url].


Find all integer solutions for $ n$, $ x$ and $ y$ of $ 2^{x}+1=y^{n}$.

It is number theory problem. You posted it in a wrong place.

Anyway, I give a solution.
If $ n$ is negative there's no solution.
If $ x=0$ then $ y=1$ and $ n$ can be anything, or $ y=-1$ and $ n$ is even. If $ n=1$ then $ y=2^{x}+1$ for every $ x$. Now suppose $ n\neq 1$ and $ x\neq 0$.
$ 2^{x}=y^{n}-1=(y-1)(1+y+...+y^{n-1})$. $ y$ is odd, but $ (1+y+...+y^{n-1})$ is even so $ n$ must be even, $ n=2m$. Now $ y^{2m}-1=(y^{m}-1)(y^{m}+1)$ is a power of two. But, the factors are not both divisible by 4. So $ y^{m}-1=2$, implying $ y=3$, $ n=2$, $ x=3$.


Suppose i had a life-size world map and its thickness was .1 millimetres.Then i folded it in half and in half angain,etc...In total i folded it in half 40 times.What would the thickness be in metres?

We coudl use geometric series: $T=\frac{0.1}{1000}\cdot 2^{40}$, where $T$ is thickness.  The common ratio is 2 because when you fold something the thickness doubles.

Masoud Zargar


geometry
trigonometry
parallelogram
cyclic quadrilateral
geometry proposed

Let $ ABC$ be a triangle. Its incircle touches the sides $ BC,CA,$ at $ D,E$ respectively. Denote by $ M,N$ the tangency points of $ A$-excircle of triangle $ ABC$ with its sides $ BC,AB$ respectively. Line $ DE$ meets $ MN$ at $ P.$ Show $ |PA|$ is equal to the radius of $ B$-excircle of triangle $ ABC.$

dear Fang:
according to my figure,based on the porperty:LA=IaT  L,A,S are collinear
AS/AL=AG/AE=BT/BP=IcP/IaT
we get the result

ps:LA/MIb=AY/YIb=AQ/QB=TC/CM=IaT/IbM   =>  LA=IaT

[color=darkblue]Firstly, we will prove $ AP\perp BC$.

Suppose $ b\ge c$.
We have

$ BD\equal{}MC\equal{}p\minus{}b \Longrightarrow DM\equal{}a\minus{}2(p\minus{}b)\equal{}b\minus{}c$

$ \angle BMN\equal{}\frac{\angle B}{2}$

$ \angle DPM\equal{}180^\circ\minus{}\bigg(90\plus{}\dfrac{\angle C}{2}\bigg)\minus{}\frac{\angle B}{2}\equal{}\frac{\angle A}{2}$

By sine law in $ \triangle PDM$ we have

$ PD\equal{}(b\minus{}c) \cdot \dfrac{\sin \frac{B}{2}}{\sin \frac{A}{2}}$

By cosine law in $ \triangle ABD$ and $ \triangle BDP$ we have

$ AD^2\equal{}c^2\plus{}(p\minus{}b)^2\minus{}2c(p\minus{}b)\cos B$

$ BP^2\equal{}(p\minus{}b)^2\plus{}PD^2\minus{}2PD(p\minus{}b)\sin \frac{C}{2}$

Our perpendicularity is equivalent to

$ PD^2\minus{}PB^2\equal{}AD^2\minus{}c^2$

or

$ PD^2\minus{}\bigg( (p\minus{}b)^2\plus{}PD^2\minus{}2PD(p\minus{}b)\sin \frac{C}{2} \bigg)\equal{}\bigg( c^2\plus{}(p\minus{}b)^2\minus{}2c(p\minus{}b)\cos B \bigg)\minus{}c^2$

$ \minus{}(p\minus{}b)^2\plus{}2PD(p\minus{}b)\sin \frac{C}{2}\equal{}(p\minus{}b)^2\minus{}2c(p\minus{}b)\cos B$

$ (b\minus{}c) \cdot \dfrac{\sin \frac{B}{2}}{\sin \frac{A}{2}}\cdot  \sin \frac{C}{2}\equal{}(p\minus{}b)\minus{}c\cos B$

$ (b\minus{}c) \cdot \dfrac{\sqrt{\frac{(p\minus{}a)(p\minus{}c)}{ac}}}{\sqrt{\frac{(p\minus{}b)(p\minus{}c)}{bc}}}\cdot  \sqrt{\frac{(p\minus{}b)(p\minus{}a)}{ab}}\equal{}(p\minus{}b)\minus{}c\cos B$

$ (b\minus{}c) \cdot \dfrac{p\minus{}a}{a}\equal{}(p\minus{}b)\minus{}c\cdot \dfrac{a^2\plus{}c^2\minus{}b^2}{2ac}$

$ 2(b\minus{}c) (p\minus{}a)\equal{}2(p\minus{}b)a \minus{} (a^2\plus{}c^2\minus{}b^2)$

Can be checked by opening brackets.....

Let $ H$ be foot of the altitude from $ A$.
By sine law in $ \triangle PDM$ we have

$ PM\equal{}(b\minus{}c)\dfrac{\cos \frac{C}{2}}{\sin{\frac{A}{2}}}$

$ HP\equal{}PM \sin \frac{B}{2}$

We need to prove that

$ AH\plus{}HP\equal{}r_a$

or

$ \dfrac{2S}{a}\plus{}(b\minus{}c)\dfrac{\cos \frac{C}{2}\sin{\frac{B}{2}}}{\sin{\frac{A}{2}}}\equal{}\dfrac{S}{p\minus{}b}$

$ (b\minus{}c)\dfrac{\cos \frac{C}{2}\sin{\frac{B}{2}}}{\sin{\frac{A}{2}}}\equal{}S\bigg(\dfrac{1}{p\minus{}b}\minus{}\dfrac{2}{a}\bigg)$

$ (b\minus{}c)\dfrac{\sqrt{\frac{(p\minus{}a)(p\minus{}c)}{ac}}\cdot  \sqrt{\frac{p(p\minus{}c)}{ab}}}{\sqrt{\frac{(p\minus{}b)(p\minus{}c)}{bc}}}\equal{}S\cdot \dfrac{a\minus{}2p\plus{}2b}{a(p\minus{}a)}$

$ (b\minus{}c)\sqrt{\frac{p(p\minus{}a)(p\minus{}c)}{a^2(p\minus{}b)}}\equal{}S\cdot \dfrac{b\minus{}c}{a(p\minus{}b)}$

$ (b\minus{}c)\frac{S}{a(p\minus{}b)}\equal{}S\cdot \dfrac{b\minus{}c}{a(p\minus{}b)}$

Problem is solved....[/color]

[url=http://www.mathlinks.ro/viewtopic.php?t=278439]From here[/url] we have AP, DE, MN are concurrent.
And see my solution at there, we have $ \angle HLK\equal{}\angle HPK$ then HPLK is a cyclic quadrilateral.
But $ PSCL$ is a cyclic quadrilateral then $ SC//HK$
Moreover, $ \angle ASC\equal{}90^o\Rightarrow SC//AI_b$
Then $ HK//AI_b$
And $ AH//I_bK$
So $ AHKI_b$ is a parallelogram therefore $ AH\equal{}I_bK\equal{}r_b$


Which are the methods to identify ammonia in products?

Thank you.  :lol:

I ran to Wikipedia and found that [url=http://en.wikipedia.org/wiki/Nessler%27s_solution]Nessler's Reagent[/url] is commonly used for that purpose... Also, you should probably be more specific as to what you are doing.  For example, if you have access to a GC that would obviously work as well.

Well, Nessler's reagent is the most common thing for checking the presence of the $NH_4^+$ ion..Actually speaking, even introducing a rod dipped in HCl into a few ammonium compounds might give you dense white fumes of ammonium chloride, though Nessler's reagent is a more conclusive test...

Additionally,

If you are looking in a water solution, or can convert your sample to a water solution, try using the dip strips sold for aquariums.  They detect ammonia at given concentrations.  With a dilution/concentration system, you can easily increase the senstivity -- all at a reasonable cost (particularly if your brother has an aquarium).


Let $F_1,F_2,G,O,R$ be the Fermat-Torricelli points, the centroid, the circumcenter and the circumradius of a triangle ABC. Prove that

\[
\left( {GO} \right)^2  + \left( {GF_1 } \right)^2  + \left( {GF_2 } \right)^2  = R^2
\]

What are Fermat-Torichelli points?
I know a Fermat-Torichelli point, but there is only one such point.

[quote="grobber"][url=http://mathworld.wolfram.com/FermatPoints.html]Here they are[/url]. :)[/quote]

Unfortunately, this link doesn't help against the traditional confusion about the definition of the Fermat points. The first Fermat point of triangle ABC is [b]not[/b] defined as the point P minimizing the sum PA + PB + PC.

Actually, the definitions are as follows:

The [b]first Fermat point[/b] of triangle ABC is the point of concurrence of the lines AA', BB', CC', where A', B', C' are the apices of the equilateral triangles erected [i]outwardly[/i] on the sides BC, CA, AB of triangle ABC.

The [b]second Fermat point[/b] is defined the same way, with the equilateral triangles being erected [i]inwardly[/i] rather than outwardly on the sides BC, CA, AB of triangle ABC.

Instead of "first Fermat point", one often just says "Fermat point". Also, "Fermat point" and "Fermat-Toricelli point" can be considered as synonyms.

Now, if all angles of triangle ABC are $\leq 120^{\circ}$, then the first Fermat point of triangle ABC is really the point P minimizing the sum PA + PB + PC. But if triangle ABC has an angle > 120°, then the first Fermat point lies outside triangle ABC, whereas the point P minimizing the sum PA + PB + PC is the vertex with this angle. Thus, there is only a limited connection between two Fermat points of triangle ABC and the problem of finding the minimum of the sum PA + PB + PC, although this problem is, historically seen, the first occurence of the (first) Fermat point of the triangle.

  Darij

[quote="darij grinberg"]

The [b]first Fermat point[/b] of triangle ABC is the point of concurrence of the lines AA', BB', CC', where A', B', C' are the apices of the equilateral triangles erected [i]outwardly[/i] on the sides BC, CA, AB of triangle ABC.

The [b]second Fermat point[/b] is defined the same way, with the equilateral triangles being erected [i]inwardly[/i] rather than outwardly on the sides BC, CA, AB of triangle ABC.

  Darij[/quote]

They are also called isogonic points. They are the points X(13) and X(14) in ETC.

Leon

[quote="Leon"]Let $F_1,F_2,G,O,R$ be the Fermat-Torricelli points, the centroid, the circumcenter and the circumradius of a triangle $\triangle ABC$. Prove that

\[ \left( {GO} \right)^2  + \left( {GF_1 } \right)^2  + \left( {GF_2 } \right)^2  = R^2  \][/quote]

For the solution, I will use two lemmas

[b]Lemma 1.[/b] Let $O, G$ be the circumcenter and the centroid of triangle $\triangle ABC$. If $P$ is any point in the plane of $\triangle ABC$ then

$PA^2  + PB^2  + PC^2  = GA^2  + GB^2  + GC^2  + 3 \cdot GP^2  \ \ \ \ $  (Leibnitz theorem)

$OG^2  = R^2  - \frac{1}{9}\left( {a^2  + b^2  + c^2 } \right)$

$AG^2  + BG^2  + CG^2  = \frac{1}{3}\left( {a^2  + b^2  + c^2 } \right)$

[b]Proof.[/b] See  [i]N. Altshiller-Court, College geometry, second edition, pag 70,71.[/i]

[b]Lemma 2.[/b] If $F_1$ and $F_2$ are the first and second Fermat points of a triangle $\triangle ABC$ then

$\left( {AF_1 } \right)^2  + \left( {BF_1 } \right)^2  + \left( {CF_1 } \right)^2  = \frac{1}{2}\left( {a^2  + b^2  + c^2 } \right) - \frac{{2\Delta }}{{\sqrt 3 }}$

$\left( {AF_2 } \right)^2  + \left( {BF_2 } \right)^2  + \left( {CF_2 } \right)^2  = \frac{1}{2}\left( {a^2  + b^2  + c^2 } \right) + \frac{{2\Delta }}{{\sqrt 3 }}$

[b]Proof.[/b] See  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=26139[/url]

[b]Theorem.[/b] If $F_1$ and $F_2$ are the first and second Fermat points of a triangle $\triangle ABC$ then

$\left( {GO} \right)^2  + \left( {GF_1 } \right)^2  + \left( {GF_2 } \right)^2  = R^2 $

[b]Proof.[/b]  From Lemma 1 and 2  we have

$\ \ \  \begin{array}{l}
 3\left( {GF_1 } \right)^2  = \left( {AF_1 } \right)^2  + \left( {BF_1 } \right)^2  + \left( {CF_1 } \right)^2  - AG^2  - BG^2  - CG^2  =  \\  \\
  = \frac{1}{2}\left( {a^2  + b^2  + c^2 } \right) - \frac{{2\Delta }}{{\sqrt 3 }} - \frac{1}{3}\left( {a^2  + b^2  + c^2 } \right) = \frac{1}{6}\left( {a^2  + b^2  + c^2 } \right) - \frac{{2\Delta }}{{\sqrt 3 }} \\  \\
 \end{array}$

and

$ \ \ \  \begin{array}{l}
 3\left( {GF_2 } \right)^2  = \left( {AF_2 } \right)^2  + \left( {BF_2 } \right)^2  + \left( {CF_2 } \right)^2  - AG^2  - BG^2  - CG^2  =  \\  \\
  = \frac{1}{2}\left( {a^2  + b^2  + c^2 } \right) + \frac{{2\Delta }}{{\sqrt 3 }} - \frac{1}{3}\left( {a^2  + b^2  + c^2 } \right) = \frac{1}{6}\left( {a^2  + b^2  + c^2 } \right) + \frac{{2\Delta }}{{\sqrt 3 }} \\  \\
 \end{array}$

Therefore

$\left( {GO} \right)^2 + \left( {GF_1 } \right)^2  + \left( {GF_2 } \right)^2  = \\ \\
= R^2  - \frac{1}{9}\left( {a^2  + b^2  + c^2 } \right) + \frac{1}{{18}}\left( {a^2  + b^2  + c^2 } \right) - \frac{{2\Delta }}{{\sqrt 3 }} + \frac{1}{{18}}\left( {a^2  + b^2  + c^2 } \right) + \frac{{2\Delta }}{{\sqrt 3 }} = R^2$

$\blacksquare$


linear algebra
matrix
inequalities
linear algebra unsolved

So i have to prove that every linear map from $ \mathbb{R}^n$ to $ \mathbb{R}^m$ is continuous. The professor said it would be easier to prove that it is Lipschitz and this boils down to proving that there exists $ M$ so that $ \forall v \in \mathbb{R}^n$, $ \|Av\|\le M\|v\|$ where $ A$ is the matrix representing the linear transformation. I have managed to prove this by proving it first for the standard basis on $ \mathbb{R}^n$ and then generalzing to any $ V$ but the proof is pretty long and tedious. Is there another more straightforward way of proving the existence of $ M$?

With $ M\equal{}\|Ae_1\|\plus{}\dots \plus{}\|Ae_n\|$, you have
$ \|Av\| \equal{} \|A(v_1e_1\plus{}\dots\plus{}v_ne_n)\|\le |v_1|\|Ae_1\|\plus{}\dots \plus{} |v_n|\|Ae_n\|\le M \max(|v_1|,\dots,|v_n|) \le M\|v\|$.

Let's be crude. We have the following norm inequalities in $ \mathbb{R}^n$: $ \|v\|_\infty \le \|v\|_2 \le \|v\|_1$; the Euclidean norm is bigger than the largest entry in $ x$, and smaller than the sum of the entries.

Now, $ \left|(Av)_i\right| \equal{}\left|\sum_j A_{ij}v_j\right|\le \sum_j |A_{ij}|\|v\|_\infty$. Sum over $ i$, and $ \|Av\|_1\le \left(\sum_{i,j}|A_{ij}|\right)\|v\|_\infty$. Mix that with the norm inequalities above, and
$ \|Av\|_2\le \|Av\|_1\le \left(\sum_{i,j}|A_{ij}|\right)\|v\|_\infty\le \left(\sum_{i,j}|A_{ij}|\right)\|v\|_2.$
We can take $ M$ to be the sum of the absolute values of the entries of $ A$. The actual operator norm is of course much smaller most of the time.

Note that this estimate applies for any of the standard power norms, as they all fit between the $ 1$ and $ \infty$ norms.


My ineq is true. And it is very easy. $LHS \leq 7$ and $RHS \geq 7$  :)


I've found this result, which is a generalisation of Schur:

$ x(a \minus{} b)(a \minus{} c) \plus{} y(b \minus{} a)(b \minus{} c) \plus{} z(c \minus{} a)(c \minus{} b) \ge 0$ for all $ a,b,c \in \mathbb{R}$ such that $ a \ge b \ge c$ or $ a \le b \le c$.
[u]If and only if:[/u] $ x,z \ge 0$ and $ (\sqrt {x} \plus{} \sqrt {z})^2 \ge y$.

Is this well-known or was I the first to discover it ? :) (I have searched and found a proof if $ x \plus{} z \ge y$, but $ (\sqrt {x} \plus{} \sqrt {z})^2 \ge y$ is the strongest possible!)

I think it's here:[url]http://www.mathlinks.ro/viewtopic.php?t=49427[/url]


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_{\frac{\pi}{8}}^{\frac{3}{8}\pi} \frac{11\plus{}4\cos 2x \plus{}\cos 4x}{1\minus{}\cos 4x}\ dx.$

[hide]
Let $ I$ be the given integral, then under the substitution $ u\equal{}\frac{\pi}{2}\minus{}x$ (so that $ du \equal{}\minus{}dx$ and the limits of integration switch, which undo each other),
\[ I\equal{}\int_\frac{\pi}{8}^\frac{3\pi}{8} \frac{11\minus{}4\cos 2u \plus{}\cos 4u}{1\minus{}\cos 4u}\ du\]
Adding to the original and dividing by 2 gives
\[ I\equal{}\int_\frac{\pi}{8}^\frac{3\pi}{8} \frac{11\plus{}\cos 4x}{1\minus{}\cos 4x}\ dx\]
\[ \equal{}\minus{}\frac{\pi}{4}\plus{}12\int_\frac{\pi}{8}^\frac{3\pi}{8} \frac{dx}{1\minus{}\cos 4x}\]
To evaluate the remaining integral, we have $ 1\minus{}\cos 4x\equal{}2\sin^2 2x$, so that $ \frac{1}{1\minus{}\cos 4x}\equal{}\minus{}\frac{1}{4}(\minus{}2\csc^2 2x)$. As a result,
\[ \int_\frac{\pi}{8}^\frac{3\pi}{8} \frac{dx}{1\minus{}\cos 4x}\equal{}\minus{}\frac{1}{4}\left[ \cot 2x\right]_\frac{\pi}{8}^\frac{3\pi}{8}\equal{}\frac{1}{2}\]
\[ \Rightarrow I\equal{}\boxed{6\minus{}\frac{\pi}{4}}\]
[/hide]


In how many ways can a positive integer $n$ be expressed as a sum of positive integers, if we take order into account?  Of course, try to give a reasonable explanation for your answer.

(for example, $3$ can be expressed as $3$, $1+1+1$, $1+2$, $2+1$)

[hide]
Consider $n$ marbles, to be divided among up to $n$ groups.
There are $\binom{n-1}{x-1}$ ways to divide $n$ marbles into $x$ groups.
So then the desired result is equivalent to $\sum_{i=1}^{n}\binom{n-1}{i-1}$ since there cannot be less than one group or more than $n$ groups.
Or simplified, $2^{n-1}$.
[/hide]

hmm.. not sure, but this what i would do.
assume you want to write the number $n$ as sum of $k$ numbers. then you can write $n$ ones. then put $k-1$ walls in between the ones, (assume that the walls can move).
then the sum would be
$\sum_{k=1}^{n} \binom{n+k-1}{n} $ or $\sum_{k=1}^{n} \binom{n+k-1}{k-1} $  ;)

[hide]Break $n$ up into $n$ unit pieces that are placed in a row.  Now,  there are $n-1$ spaces between the pieces.  You can either put a divider in the space or not.  Thus there are $2^{n-1}$ to express $n$ as a sum of positive integers.[/hide]

now, what if we wanted to eliminate the repeats?

Now you use generating functions.  This is a famous problem.  By the way, your initial solution is wrong.


Let $ p$ be a prime number.

Find the sum:

$ \sum_{n=1}^{\infty}t^{np}\left(\zeta(np+1)-1\right)=?$.

Of course, $ t$ is a real number for which the sum converges.

Just to do something obvious for the sake of doing it:

$ \sum_{n=1}^{\infty}t^{np}\sum_{k=2}^{\infty}\frac{1}{k^{np+1}}=\sum_{k=2}^{\infty}\frac1k\sum_{n=1}^{\infty}\left(\frac{t}{k}\right)^{np}$

$ =\sum_{k=2}^{\infty}\frac1k\cdot \frac{\left(\frac{t}{k}\right)^{p}}{1-\left(\frac{t}{k}\right)^{p}}=\sum_{k=2}^{\infty}\frac{t^{p}}{k(k^{p}-t^{p})}$

All right, I admit that it's not that obvious what to do next, or whether this was useful, or what $ p$ being prime has to do with anything.


[quote="xiaohui"]$ \frac {3}{3 \plus{} 2} \plus{} \frac {3^2}{3^2 \plus{} 2} \plus{} \cdots \plus{} \frac {3^n}{3^n \plus{} 2} > \frac {n^2}{n \plus{} 1}.$[/quote]
$ \Longleftrightarrow\frac {1}{3 \plus{} 2} \plus{} \frac {1}{3^2 \plus{} 2} \plus{} \cdots \plus{} \frac {1}{3^n \plus{} 2} < \frac {n}{2(n \plus{} 1)},$

$ \Longleftarrow\frac {1}{3 \plus{} 2} \plus{} \frac {1}{3^2 \plus{} 2} \plus{} \cdots \plus{} \frac {1}{3^n \plus{} 2} < \frac {11}{32}.$

We could easily check for n=1 or 2.
For $ n \geq 3$,
$ \sum_{i\equal{}1}^{n} \frac{3^i}{3^i\plus{}2}\equal{}\sum_{i\equal{}1}^{n} 1\minus{}\frac{2}{3^i\plus{}2}\equal{}
n\minus{}\frac{2}{5}\minus{}\sum_{i\equal{}2}^{n} \frac{2}{3^i\plus{}2} > n\minus{}\frac{2}{5} \minus{} \sum_{i\equal{}2}^{n} \frac{2}{3^i}
>
n\minus{}\frac{2}{5} \minus{} \sum_{i\equal{}2}^{\infty} \frac{2}{3^i}
\equal{}n\minus{}\frac{11}{15}>n\minus{}\frac{n}{n\plus{}1}\equal{}\frac{n^2}{n\plus{}1}$
since $ \frac{n}{n\plus{}1} > \frac{11}{15}$ for $ n \geq 3$.

Alternatively, we can proceed by induction.
We can readily verify the statement for n=1 and n=2.
Suppose it is true for n. Then,
$ \sum_{i\equal{}1}^{n\plus{}1} \frac{3^i}{3^i\plus{}2}>\frac{n^2}{n\plus{}1}\plus{}1\minus{}\frac{2}{3^{n\plus{}1}\plus{}2}$.
Now,
$ \frac{n^2}{n\plus{}1}\plus{}1\minus{}\frac{2}{3^{n\plus{}1}\plus{}2} >
\frac{(n\plus{}1)^2}{n\plus{}2}
\Longleftrightarrow
n^3\plus{}3n^2\plus{}3n\plus{}2\minus{}(n\plus{}2)(n\plus{}1)\frac{2}{3^{n\plus{}1}\plus{}2} >
n^3\plus{}3n^2\plus{}3n\plus{}1
\Longleftrightarrow
1>\frac{2(n\plus{}1)(n\plus{}2)}{3^{n\plus{}1}\plus{}2}$,
which is true for $ n \geq 2$. Thus, we may use induction to prove the statement for $ n \geq 3$.

it's trivial ,we can use cauchy schwartz's inequality  :)


first, you throw away the outside, then keep the inside.
next, you eat the outside, then throw away the inside.

what is it?

[hide]A peach? Though some people eat the skin...[/hide]

[hide="to keep up the fruit pattern"]MANGO[/hide]

[hide]chicken??
sorry I broke the fruit pattern. :lol: [/hide]

yes, it was corn. not mango. or chicken.

that was pretty good. :)

Why isn't it chicken? You throw away the feathers, and eat the meat, and throw away the bones.

[quote="Bill The Maniac"]Another maple story![/quote]

giving away level 70 chief bandit on scania! it has no moneys though

[quote="randomdragoon"]wtf how did this just go to maple story

And I think this riddle applies to any food with an inedible peel/shell/skin and an inedible core/pit. Like chicken.[/quote]

answer given already
it was corn


How many distinct triangles can be formed by selecting three vertices of a regular octagon?

you can choose 3 vertices from 8.

$ \binom{8}{3}\equal{}56$

answer : 56


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number theory unsolved
number theory

Find all odd natural number $ n$ such that for any  divisors $ a$ and $ b$ of $ n$ with $ \gcd(a,b)\equal{}1$, The number $ a \plus{} b \minus{} 1$ is also a divisor of $ n$

[quote="Evariste-Galois"]Find all odd natural number $ n$ such that for any  divisors $ a$ and $ b$ of $ n$ with $ \gcd(a,b) \equal{} 1$, The number $ a \plus{} b \minus{} 1$ is also a divisor of $ n$[/quote]

if n is 1 >> definitely ok.

if $ n\equal{}p^{a}$ (p is prime num, a is natural) >> divisors of n is $ 1,p,p^2 , \cdots, p^a$. Therefore, because any (a,b) must be $ (1, p^i )$, ok.

otherwise >> there are two prime factors of n -> let p,q.
Then p,q are divisors of n and $ \gcd (p,q)\equal{}1$. Also $ pq|n$. If we assume $ p\plus{}q\minus{}1|n$, then $ pq\minus{}p\minus{}q\plus{}1 \equal{} (p\minus{}1)(q\minus{}1) |n$. However, because p,q>1 and odd, $ 4|(p\minus{}1)(q\minus{}1)$. therefore $ 4|n$, it makes contradiction.

So answer is 1 and  all powers of all odd prime numbers.

[hide="What I did"]Suppose there were two distinct prime factors $ p$, $ q$, of $ n$. WLOG $ p > q$. Observe now that $ p \not | q \minus{} 1$, and thus, $ (p, q \minus{} 1) \equal{} 1$.

Now, clearly $ (p, q) \equal{} 1$. Thus, $ p \plus{} q \minus{} 1 | n$. Since $ (p \plus{} q \minus{} 1, p) \equal{} (q \minus{} 1, p) \equal{} 1$, we also have that $ 2p \plus{} q \minus{} 1 | n$. We can thus inductively prove that $ kp \plus{} q \minus{} 1 | n$ for any $ k \geq 1$. But $ kp \plus{} q \minus{} 1$ is unbounded, so $ n$ must be arbitrarily large, a contradiction. Thus, $ n$ must be a power of a prime.

It is easy to verify that all powers of odd primes satisfy the given properties.

Hmm I don't think I even used the property that $ n$ is odd. Did I do something wrong? [/hide]

@cadiTM: If $ p\plus{}q\minus{}1 \mid n$ and $ pq \mid n$, then it does not follow that $ pq\minus{}p\minus{}q\plus{}1 \mid n$. For example, take $ p\equal{}3$, $ q\equal{}5$ and $ n\equal{}105$.

@Zhero: You have $ 2p\plus{}q\minus{}2\mid n$.

See here http://www.mathlinks.ro/viewtopic.php?search_id=125209069&t=205140 for a sketch of a solution

My solution is the same of Zhero's one  But  $ kp\plus{}q\minus{}k \mid n$ not $ kp\plus{}q\minus{}1 \mid n$  :maybe:   Indeed $ (k\plus{}1)p\plus{}q\minus{}(k\plus{}1)>kp\plus{}q\minus{}k$ so it's enough to choose a $ k$  to have a large number and we'll have a contradiction
 :)

Hmm, my hole was slightly big. I considered the fix you mentioned, Evariste-Galois, since we can generalize to $ (p, q \minus{} k) \equal{} 1$. This is always true. Unless, of course, $ k \geq q$...

When we hit $ pq$, i.e., $ k \equal{} q$, we cannot repeat this argument. $ pq$ and $ p$ are no longer relatively prime.

Correct me if I'm wrong, because I have a tendency to be wrong a lot nowadays (just look at my other solution :P), but I'm pretty sure we've hit a dead end here.


[quote="Thaakisfox"]Given $a_{1},a_{2},......,a_{n}$ non-negative real numbers.
Prove that:
$(a_{1}+a_{2}+a_{3})^{2}(a_{2}+a_{3}+a_{4})^{2}\cdot ... \cdot (a_{n-1}+a_{n}+a_{1})^{2}(a_{n}+a_{1}+a_{2})^{2}\geq 2^{n}(a_{1}+a_{2})^{2}(a_{2}+a_{3})^{2}\cdot ... \cdot (a_{n-1}+a_{n})^{2}(a_{n}+a_{1})^{2}$
if $n\geq 4$[/quote]
It is an inequality of Olympiad 90-s' years of 239 school by Sankt-Petersburg .
The Harazi's hint: for $x,$ $y$ and $z$ are non-negative numbers the following inequalities are true:
$(x+y+z)^{2}\geq(2x+y)(y+2z)$ and $(x+2y)(2x+y)\geq2(x+y)^{2}.$ :wink:


What is the best pre-calculus textbook for self-study (along with a probably poor class)?

I like David Cohen's Precalculus with Unit-Circle Trigonometry, but I didn't use it for self study, personally. It has good coverage of a lot of interesting topics. Any of the books about high school mathematics by I.M. Gelfand or S.I. Gelfand would also be helpful.

The Art of Problem Solving, Volumes 1 and 2. Order them here:

http://www.artofproblemsolving.com/Books/AoPS_B_About.php

:D A very thorough coverage of pre-calculus and more...

The one we used was [u]Precalculus Mathematics[/u] By. Demana Waits Clemens   i liked it well enough