limit
algebra
polynomial
inequalities
real analysis
real analysis solved

Let $\{U_n\}_{n=1}^\infty$ be sequence of real numbers, $I$ is an arbitrary interval. We define $P_n(I,U)=\frac{1}{n}\left|\left\{k\,:\,k\leq n,\ U_k\in I\right\}\right|$.
We say that a sequence $\{U_n\}_{n=1}^{\infty}\subset [0,1]$ is well distributed if it satisfies $\forall\,I\subset[0,1]\ \exists\lim_{n\to\infty}P_n(I,U)$.

Let $f(x)=a_sx^s+a_{s-1}x^{s-1}+\dots+a_1x+a_0$, $a_i\in\mathbb{R}$, $i=0,1,\dots,s-1$, $a_s\in\mathbb{R\setminus Q}$.
Prove that sequence $U_n=\{f(n)\}$, $n\in\mathbb{N}$, is well distributed.

Please, use more readable notations next time.

If $s$ is anything you want, and all coefficients are $0$ except for $a_s$, which is $1$, then you get $\{P(n)\}=0,\ \forall n$. I'm sure you meant $a_s\in\mathbb{R\setminus Q}$. This is a weaker version (because the strong version should be really tough to prove) of Weyl's theorem: if $P$ is a polynomial with at least an irrational coefficient, then $\{P(n)\}$ is [b]uniformly[/b] distributed in [$0,1$], i.e. the limit of that sequence is actually the length of the interval ($\lim_{n\to\infty}P_n(I,U)=|I|$, where if the endpoints of $I$ are $a<b$, then $|I|=b-a$).

I've never seen a proof of Weyl's theorem, but I hear it's hard to prove. Anyway, this doesn't mean the problem posted here is as tough as the actual theorem.

You're right. $a_s\in\mathbb{R\setminus Q}$.

I think that Weyl's theorem is not that hard if you use weierstrass approximation theorem .. I will post a proof if you want.

Hi there!
I am new one here.

I saw a discussion on distribution of polynomial values.
A long ago I have found an elementary proof for denseness of values
of a*n^2 (mod 1) whare a is irrational.
I attach the file with a proof ( I am soryy, it contains errors in English,
but the proof is ok).

Thw Weyl theorem is an easy corollary of van der Corput's theorem: if, for every $k \in \mathbb{N}$, the sequence $\{\nu_{n+k} - \nu_n\}_{n \geq 1}$ is equidistributed $\mod{1}$, then $\{\nu_n \}_{n \geq 1}$ is also equidistributed $\mod{1}$.

Proof of the van der Corput theorem or of the implication [i]van der Corput[/i] $\Rightarrow$ [i]Weyl[/i]?  :)

I would like to see a solution that does not use his monstruous inequality and Weyl's criterion. Vess, do you know such solution?

No, I don't know a solution that does not essentially depend on Weyl's criterion ... .

In this topic I already attached a file with elementary solution of a
very particular case of the Weyl theorem. Despite the fact that this is a very
particular case, this is the first non-trivial case and my solution is elementary.
If somebody is interested in it, please read it.


algebra
polynomial
trigonometry
complex numbers
absolute value

[hide]
Factors are $x-e^{2\pi ik/5}$, $k\in \{0,1,2,3,4\}$. What do you mean by special? I probably didn't do what you asked... There are a lot of special things about roots of unity.
[/hide]

[quote="Altheman"]Factor over complex numbers $P(x)=x^5-1$. Furthermore, what is "special" about these factors?[/quote]

what do you mean by "Factor over complex numbers"???, If you just mean factoring including complex numbers in expression:

[hide]
well, we definitely get $(x-1)(x^4 + x^3 + x^2 + x + 1)$, to factor the right term further, we note that these are the 5 5th roots of unity by setting $P(x) = 0$.  Therefore, the factor it as $P(x) = \left(x-1\right)\left(x - cis(\frac{2\pi}{5})\right)\left(x - cis(\frac{4\pi}{5})\right)\left(x - cis(\frac{6\pi}{5})\right)\left(x - cis(\frac{8\pi}{5})\right)$. The special thing is just they are all 5th roots of unity.
[/hide]

[quote="Altheman"][quote="pkerichang"][quote="Altheman"]i suppose i was unclear, i want if factored where you have an exact value, the point of this is to derive the exact value for sin72, cos72, etc[/quote]

so we have to find $\sin{72}$ and $\cos{72}$ in radicals, fractions, etc.???[/quote]

yes, but again, the idea is to [b]derive[/b] those values by factoring[/quote]

in that case...

[hide]If $r$ is a root of $x^4+x^3+x^2+x+1$, then so is $r^{-1}$. This can be taken advantage of through a substitution $y=x+x^{-1}$. After dividing both sides of $x^4+x^3+x^2+x+1=0$ by $x^2$ and using $y^2=x^2+x^{-2}+2$,

\[ y^2+y-1=0\Rightarrow y=\frac{-1\pm \sqrt{5}}{2} \]

Then solving for $x$ in terms of $y$, because $x^2-xy+1=0$, $x=\frac{y\pm \sqrt{y^2-4}}{2}$. So after substituting, the four roots are given by

\[ \boxed{\frac{-1+\sqrt{5}}{4}\pm \frac{\sqrt{10+2\sqrt{5}}}{4}i,\ \frac{-1-\sqrt{5}}{4}\pm \frac{\sqrt{10+2\sqrt{5}}}{4}i} \][/hide]

good  :)
[hide="i was thinking of a similar way"]but it is essentially the same as yours, the problem lies on the fact that the roots come in conjugate pairs (we see this using trig solution)
since there are 2 pairs of conjugate roots, and the have an absolute value of 1, we can say that the polynomial factors into
$x^5-1=(x-1)(x^4+x^3+x^2+x+1)=(x-1)(x^2+ax+1)(x^2+bx+1)$, then expand and compare coefficients to solve for $a$ and $b$[/hide]


abstract algebra
superior algebra
superior algebra unsolved

Let F be finite field with n elements.
Prove that x^(n-1) = 1 for all non-zero x in F.

[quote="ZetaX"]Just use Lagranges theorem.[/quote]

That's what I have been trying to use and I am doing it this way:
x^(n-1) = x^n * x^(-1) = 1*x^(-1). Now what do I do with the remaining x^(-1)?

Lagrange's theorem says Let G be a finite group, and a be in G. Then a^|G| = e. In this case e is 1, right?

Yes, $e=1$, but $|G|=n-1$ here, since we are looking at the multiplicative group (so $0$ is excluded).

Exactly

it is the other way around

first you prove $x^{n-1}=1$ holds, and then $x^n=x$ is a logic consequence


Prove that for all natural numbers $ n \ge 3$ there exist odd natural numbers $ x_n, y_n$ such that $ 7x_n^2\plus{}y_n^2\equal{}2^n$.

This is problem 1, Bulgaria National Contest 1996
[url]http://www.mathlinks.ro/viewtopic.php?t=150647[/url]


prove that
$ \sin(y)\equal{}\lambda\sin(2x\plus{}y) \Leftrightarrow \tan(x\plus{}y)\equal{}\frac{1\plus{}\lambda}{1\minus{}\lambda} \tan(x)$

[hide="Hint"]$ \Leftrightarrow sin[(x \plus{} y) \minus{} x] \equal{} \lambda sin[(x \plus{} y) \plus{} x]$
$ \Leftrightarrow sin(x \plus{} y)cosx \minus{} sinxcos(x \plus{} y) \equal{} \lambda[sin(x \plus{} y)cosx \plus{} sinxcos(x \plus{} y)]$
$ \Leftrightarrow (1 \plus{} \lambda)sinxcos(x \plus{} y) \equal{} (1 \minus{} \lambda)cosxsin(x \plus{} y)$[/hide]


[quote="chien than"]Prove that:
$a^{2}+b^{2}+c^{2}+3abc \geq 6$
where $a;b;c$ are positive real numbers satisfying $ab+bc+ca+abc=4$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117363


topology
group theory
abstract algebra
advanced fields
advanced fields unsolved

Let $ O(n)$ be the (real) orthogonal group. Suppose we have already proved that $ O(n)$ acts transitively on $ S^{n\minus{}1}$ and the stabilizer of $ N\equal{}(0,...0,1)$ is $ O(n\minus{}1)$.
Does it follow that $ \dim O(n)\minus{}\dim O(n\minus{}1)\equal{}\dim S^{n\minus{}1}$?

You get from what you wrote $ O(n)/O(n\minus{}1)\cong S^{n\minus{}1}$. So if you know that $ dim(O(n)/O(n\minus{}1))\equal{}dim(O(n))\minus{}dim(O(n\minus{}1))$. Then you have it. (Note that I didn't say that is true, just that if you know it to be true. I'm not certain that dim of a quotient is subtracting the dimensions.)

[quote="HilbertThm90"]I'm not certain that dim of a quotient is subtracting the dimensions.[/quote]
It is. In the Lee text, it's the "Quotient Manifold Theorem", in the chapter on Lie group actions. Full statement:
Suppose a Lie group $ G$ acts smoothly, freely, and properly on a smooth manifold $ M$. Then the orbit space $ M/G$ is a topological manifold of dimension $ \text{dim }M\minus{}\text{dim }G$, and has a unique smooth structure with the property that the quotient map $ \pi: M\to M/G$ is a smooth submersion.

Here, we have a copy of $ O_{n\minus{}1}$ acting on $ O_n$ by left-multiplication, and we identify the orbit space with $ S^{n\minus{}1}$ using the preimage of the "north pole" $ (0,\cdots,0,1)$ under any element of the orbit. The action is smooth and free since it's multiplication in a Lie group, and it's automatically proper because $ O_{n\minus{}1}$ is compact.
We don't, however, get any kind of group structure on this quotient. That requires a normal subgroup.

Right. Well, since only $ S^0$, $ S^1$ and $ S^3$ admit Lie structures (so the group aspect is all that is missing), I wouldn't expect a group structure in general.

After thinking about it today, I figured the dimension thing was probably true. I couldn't think of a reference, though.

Well, that's why I referenced a textbook we both like. Theorem 9.16, page 218-223 (first edition).


Has anyone ever been here:
http://www.mathmuseum.org/
(I mean the place not its website)

no, but it looks like interesting place.

have you? can you tell us more about it?

I never have visited the place. Was hoping someone did and would tell me about it. :)


Suppose that    ai(i=1,2,3```2009)   are positive real number.
Prove that exist real number  bi(i=1,2,3```2009)
such that(1)0.5<bi/bi+1<2
(2)b1+b2+b3+```+b2009<3(a1+a2+a3+```+a2009)[/code]

Please state the problem more clearly. Use $ \text{\LaTeX}$, i.e. put dollar signs around the formula. For example,
you'll get $ \frac {1}{2} < \frac {b_i}{b_{i + 1}} < 2$ by writing
[code]$\frac{1}{2} < \frac{b_i}{b_{i+1}} <2$[/code].  :wink:
Moreover, if you want to see a $ \text{\LaTeX}$-code, just click on the formular and it will apear.

But I guess that there is something wrong with your problem in general. Take $ b_1 = \dots = b_{2009} = \frac{\sum a_i}{2009}$.

[quote="FelixD"]Please state the problem more clearly. Use $ \text{\LaTeX}$, i.e. put dollar signs around the formula. For example,
you'll get $ \frac {1}{2} < \frac {b_i}{b_{i + 1}} < 2$ by writing
[code]$\frac{1}{2} < \frac{b_i}{b_{i+1}} <2$[/code].  :wink:
Moreover, if you want to see a $ \text{\LaTeX}$-code, just click on the formular and it will apear.

But I guess that there is something wrong with your problem in general. Take $ b_1 = \dots = b_{2009} = \frac {\sum a_i}{2009}$.[/quote]

sorry   ai<bi

Come on, why do I tell you how to use $ \text{\LaTeX}$?  :wink:


A Heronian triangle is a triangle whose side lengths and area are integers. Prove: the area of every heronian triangle is divisible by 6.

Let $\triangle ABC$ be a Heronian triangle with sides $a$, $b$, $c$ ($a$ opposite to $A$). Let $h_{A}$, $h_{B}$, $h_{C}$ be the heights from $A$, $B$, $C$, respecively, and $x$, $y$ - lengths of the segments into which the foot of $h_{A}$ divides $BC$.
Obviously, $S=|ABC|=\frac{ah_{A}}{2}=\frac{bh_{B}}{2}=\frac{ch_{C}}{2}$. Hence,  $h_{A},h_{B},h_{C}\in N$.
 Suppose, $x \not\in N$. As $a=x+y$, we have $x^{2}=a^{2}-2ax+y^{2}$, so $x^{2}\not\in N$, contradiction, because $x^{2}=c^{2}-h_{A}^{2}$. Therefore, $x,y \in N$ .
Suppose, $3 \nmid S$. We have $x^{2}+h_{A}^{2}=c^{2}$, so one of $x$, $h_{A}$, $c$ must be divisible by $3$. As $3 \nmid S$ it follows that $3|x$. Similarly, $3|y$. So $3|a$, contradiction. Therefore, $3|S$.
Suppose, $2 \nmid S$. Using similar argumentation ($4$ instead of $3$) we reach contradiction. Therefore, $6|S$.

It is not true that the heights of a heronian triangle must be integers.

Of course, you are right. I don't know yet how to correct the proof.

$16S^{2}=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)$

If $a+b+c$ is odd, then $a+b-c, a-b+c,-a+b+c$ is odd, too. It will lead to a contradiction. Then $a+b+c, a+b-c, a-b+c,-a+b+c$ is even.

Let $x=\frac{a+b-c}{2}, y=\frac{a-b+c}{2}, z=\frac{-a+b+c}{2}$

Then we have $S^{2}=(x+y+z)xyz$.

We assume that $S$ is odd, then $(x+y+z)xyz\equiv 1\mod{4}$

We have four cases $(x,y,z)=(1,1,1),(1,1,-1),(1,-1,-1),(-1,-1,-1)\mod{4}$, for each case we finally have the same conclusion $(x+y+z)xyz\equiv 3\mod{4}$, it is a contradiction!

Then we assume that $S$ is not the divisible by $3$, then $(x+y+z)xyz\equiv 1\mod{3}$

We have only two cases $(x,y,z)=(1,1,-1),(1,-1,-1)\mod{3}$, for each case we finally have the conclusion $(x+y+z)xyz\equiv 2\mod{3}$, it is also a contradiction!


When Erica was two years old, Leroy was four times as old as Miriam. When Miriam was twice as old as Erica, Leroy was three times as old as Miriam. How old was Erica when Leroy was twice as old as Miriam?

Hi calc990.

When Erica was $ 2$, looking at the ratios, I tried $ 6, 7$ and $ 8$ for Miriam. Thus, if Miriam was $ 8$ when Erica was $ 2$, Leroy must have been $ 32$. Then, in another $ 4$ years, when Miriam was twice as old as Erica, Miriam was $ 12$ and Erica was $ 6$, so Leroy was $ 32$. Then, in another $ 12$ years, when Miriam was $ 24$, Leroy was twice as old as her, being $ 48$ and so Erica was $ 18$.


integration
function
inequalities
calculus
complex numbers
calculus computations

Take $ f(x) : \equal{} \begin{cases} \minus{}1 \,\qquad x \in [a,\frac{a\plus{}b}2]\\
1 \quad\qquad x \in (\frac{a\plus{}b}2, b]\end{cases}$.

[quote="Ingemat"]let $ f$ be a continuous function, then the following inequality holds: $ \left| \int_{a}^{b}{f} \right|\le \int_{a}^{b}{\left| f \right|}.$[/quote]That is what I want. The proof of that inequatily.

My post was to show that at least

$ |\int_{a}^{b}f | >  \int_{a}^{b}|f|$ is false.

Obviously equality can hold (take $ f(x) = 0$).

So maybe it is true that:

$ |\int_{a}^{b}f | \leq  \int_{a}^{b}|f|$.

Look at the definition of the Riemann integral.. this should be a doable exercise.

The proof in the case that $ f$ is complex-valued. (Real-valued would be a little easier; I'll go for the whole thing.)

Suppose $ f$ is a complex-valued integrable function on a set $ E.$

Let $ z \equal{} \int_Ef.$

If $ z \equal{} 0,$ then trivially, $ \left|\int_Ef\right| \equal{} |z| \equal{} 0\le\int_E|f|.$ So assume $ z\ne 0.$

Let $ w \equal{} \frac {\overline{z}}{|z|}.$ Note that $ |w| \equal{} 1.$

$ wz \equal{} \frac {\overline{z}z}{|z|} \equal{} \frac {|z|^2}{|z|} \equal{} |z|,$ so

$ \left|\int_Ef\right| \equal{} w\int_Ef \equal{} \int_Ewf \equal{} \int_E\text{Re}\,(wf).$

The last equality comes from noting that $ \int_Ewf$ is real, and the real part of an integral is the integral of the real part.

For any complex number $ a,$ we have $ \text{Re}\,(a)\le|a|,$ so on a pointwise basis, $ \text{Re}\,(wf)\le|wf| \equal{} |f|.$

Hence $ \int_E\text{Re}\,(wf)\le\int_E|f|$ and we are done.

Equality holds exactly when $ \text{Re}\,(wf) \equal{} |f|$ almost everywhere, which is the same thing as $ \text{Re}\,(wf)\ge 0$ almost everywhere. That is, we have equality in this inequality whenever (almost) all of the nonzero values taken on by $ f$ have the same complex argument (lie on the same ray from the origin). (We also have equality for the function that is zero almost everywhere.)

[quote="enb081"]Which one is greater:

$ \int_{a}^{b}\left|{f{(x)} \right|}dx$      or    $ \left|\int_{a}^{b}f{(x)}dx \right|$[/quote]

$ -|f(x)|\leq f(x)\leq |f(x)|\Longrightarrow -\int_a^b |f(x)|dx\leq \int_a^bf(x)dx\leq \int_a^b|f(x)|dx$

$ \Longleftrightarrow \left|\int_a^b f(x)dx\right|\leq \int_a^b |f(x)|dx.$

[quote="kunny"][quote="enb081"]Which one is greater:

$ \int_{a}^{b}\left|{f{(x)} \right|}dx$      or    $ \left|\int_{a}^{b}f{(x)}dx \right|$[/quote]

$ - |f(x)|\leq f(x)\leq |f(x)|\Longrightarrow - \int_a^b |f(x)|dx\leq \int_a^bf(x)dx\leq \int_a^b|f(x)|dx$

$ \Longleftrightarrow \left|\int_a^b f(x)dx\right|\leq \int_a^b |f(x)|dx.$[/quote] Very concise solution. Thank you.


Let $ (a_n)_{n\geq1}$ be an ascending sequence of positive numbers.Prove that $ \lim_{n\to\infty}\sqrt[n]{a_1^n\plus{}a_2^n\plus{}...\plus{}a_n^n}\equal{}\lim_{n\to\infty}a_n$

$ a_n\equal{}\sqrt[n]{a_n^n}\leq\sqrt[n] {a_1^n\plus{}a_2^n\plus{}\dots \plus{}a_n^n}\leq\sqrt[n]{na_n^n}$

     $ \Rightarrow$  $ \lim_{n\to\infty}\sqrt[n] {a_1^n\plus{}a_2^n\plus{}\dots \plus{}a_n^n}\equal{}\lim_{n\to\infty} a_n$

Imam Jafar Sadegh (A. S.)            ((Vasael, vol. 6, p. 321))

The believers have four signs: good humor, tactfulness, kind heartedness and openhandedness


algebra
function
domain
Ring Theory
superior algebra
superior algebra unsolved

Let $ R$ be a unital ring which has no non-zero nilpotent element and $ P$ is a minimal prime ideal(There is no prime ideal contained $ P$).Prove that $ \frac{R}{P}$ is domain (It has no non-zero zero divisor).

you don't assume $ R$ to be commutative, right?

Askar A. Tuganbaev, Rings Close to Regular, p. 39, available on google books


function
trigonometry
complex numbers
real analysis
real analysis unsolved

Hi!

I'm searching for a proof that all contious functions from R to R, wich satisfy f(x+y)+f(x-y)=2*f(x)*f(y) are cos(cx), cosh(kx) or f==1 or f==0.

The recursion you get for x=n ( n natural number) , y=1 seem similar to the definition of tschebycheff polynoms. But I don't know if this is useful.

I am glad for any ideas thx.

Nod.

It's easier to assume complex values here and specialize to real values at the end.

First, either $ f(0)\equal{}1$ or $ f$ is identically zero. Assume the former.
If $ f(c)\equal{}\cos \alpha$ for some complex $ \alpha$, we can show that $ f(nc)\equal{}\cos n\alpha$ for all integers $ n$ by induction. Since $ \cos$ is surjective on $ \mathbb{C}$, this is universal.

For each $ k$, let $ \alpha_k\equal{}\cos^{\minus{}1}(f(2^{\minus{}k}))$., taking the branch with real part between zero and $ \pi$, along with the positive imaginary axis. We have $ \alpha_k\equiv 2\alpha_{k\plus{}1}\mod \pi$. By continuity at zero, $ \alpha_k\to 0$, and thus $ \alpha_k\equal{}2\alpha_{k\plus{}1}$ for large enough $ k$. Let $ M$ be this threshold, and define $ a\equal{}2^M\alpha_M$. We get $ f(x)\equal{}\cos(ax)$ for all dyadic rationals $ x$. This is a dense set, so $ f(x)\equal{}\cos(ax)$ for all real $ x$.

If $ f$ is actually real-valued, $ a$ is either pure real or pure imaginary. In the first case, we get $ \cos$, and in the second we get $ \cosh$.

Thank you a lot.
I know I'm a beginner in the art of problem solving, but I'm quite impressed.

One detail to correct: We don't have $ \alpha_k\equiv 2\alpha_{k\plus{}1}\mod \pi$; instead either $ \alpha_k\equiv 2\alpha_{k\plus{}1}\mod 2\pi$ or $ \alpha_k\equiv 2\pi\minus{}2\alpha_{k\plus{}1}\mod 2\pi$. The conclusions are the same.


Sorry, I don't understand your question...  :maybe:

[quote="linzk3"]when you use a radical sign is it still a rational number or an integer?[/quote]
$\sqrt{k}$ is an integer iff $k$ is a perfect square (k is the square of an integer).

If you're talking about a rational number under the radical, say $\frac{p}{q}$, $\sqrt{\frac{p}{q}}$ is a rational number iff $p$ and $q$ are both perfect squares.

Assumed that $p,q$ are coprime.

This indeed extends to $n$-th powers:

If integers $p,q$ are coprime, $\sqrt[n]{\frac{p}{q}}$ is rational iff $p,q$ are each $n$-th powers of integers.


function
calculus
derivative
integration
calculus computations

Hello everybody

Find all dérivable function f such as   $ (f'(x))^2 \equal{}f(x)$   for all $ x \in R$

hello, it is clear that $ f(x)\equal{}0$ is one solution, in the other case we have
$ \pm\frac{dy}{\sqrt{y}}\equal{}dx$
$ \pm2\sqrt{y}\equal{}x\plus{}C$
squaring both sides we get
$ y\equal{}\frac{1}{4}x^2\plus{}\frac{C^2}{4}\plus{}\frac{1}{2}xC$
Sonnhard.

Write that as $ (y')^2=y$; away from zero, we can solve and get $ y'=\pm\sqrt{y}$. Since the derivative of a function which is differentiable everywhere can't have a jump discontinuity, we choose one sign and stick to it, except possibly at points where $ y=0$.

Solving the separable equation, the real solutions of $ y'=\sqrt{y}$ are $ \int \frac{dy}{\sqrt{y}}=\int dx$, or $ x=2\sqrt{y}+c$. Solving for $ y$, that's $ y=\frac14(x-c)^2$ for real constants $ c$ and $ x\ge c$. Similarly, the solutions of $ y'=-\sqrt{y}$ are $ y=\frac14(c-x)^2$ for real $ c$ and $ x<c$.

The first group of solutions run to $ x=\infty$, while the second run to $ x=-\infty$. The constant zero function is also a solution; to get a solution on all of $ \mathbb{R}$, we splice them as follows:
$ f(x)=\begin{cases}\frac14(a-x)^2& x\le a\\ 0& a\le x\le b\\ \frac14(x-b)^2& b\le x\end{cases}$, where $ a,b\in [-\infty,\infty]$ (the extended reals) with $ a\le b$.

Thank you [b]Dr Sonnhard Graubner [/b] & [b]jmerry [/b] for your precious help


calculus
derivative
LaTeX
advanced fields
advanced fields unsolved

I have been studying for the upcoming midterm by solving the questions at the end of the chapters and i could not solve some of them. Can you help me with these :
(Note: x't,y't and u't denote the partial derivative of x,y and u with respect to t; u'x and u'y denote the partial derivative of u with respect to x and y.)

1. Solve x(u'y) - y(u'x) + u = 0 subject to the initial condition u(x,0) = 1,x>0.

Here we have x't = -y, y't = x, and u't = -u with x(0,s) = s, y(0,s) = 0, and u(0,s) = 1. Now what can we do in order to find x,y and u in terms of s and t ?

2. Solve (x+y^2)u'x + (y)u'y + [(x/y) - y]u = 1, u(x,1) = 0.

Now we get x't = x+(y^2), y't = y, and u't = 1 + u[y - (x/y)]. Then i have obtained y(t,s) = e^t and x(t,s) = e^2t  + (s-1)*(e^t). Also i have found
u't + (s-1)u = 1, but could not combine it with the related initial condition.

3. Solve (u'x)^2 + u'y = 0, u(x,0) = x.

I solved it through 'prediction' and obtained u(x,y) = x, but is there a better way to solve this by applying the methods above?

[quote="cooley"]3. Solve (u'x)^2 + u'y = 0, u(x,0) = x.

I solved it through 'prediction' and obtained u(x,y) = x, but is there a better way to solve this by applying the methods above?[/quote]

Well, I think you need to work on your predicting, or perhaps your typing.  Also, for the sake of readability, [i]please[/i] learn and use Latex.  Even a little bit is much kinder on my brain.

[quote]$x\frac{\delta u}{\delta y}-y\frac{\delta u}{\delta x}+u = 0$ subject to the initial condition $u(x,0) = 1$, $x>0$.

$(x+y^{2})\frac{\delta u}{\delta x}+y\frac{\delta u}{\delta y}+\left(\frac{x}{y}-y\right)u = 1$, $u(x,1) = 0$

$\frac{\delta u}{\delta x}^{2}+\frac{\delta u}{\delta y}= 0, u(x,0) = x$.
[/quote]

Personally I like partial derivative symbols more than deltas...
[quote]$x\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}+u = 0$ subject to the initial condition $u(x,0) = 1$, $x>0$.

$(x+y^{2})\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}+\left(\frac{x}{y}-y\right)u = 1$, $u(x,1) = 0$

$\left(\frac{\partial u}{\partial x}\right)^{2}+\frac{\partial u}{\partial y}= 0, u(x,0) = x$.
[/quote]

And for the last one, how about $u(x, y) = x-y$? But of course I have no evidence that this is the only solution...

Sorry, wrongfully i gave my prediction about a similar question,not this one, which is correct but of course for this my answer seems quite stupid. In the third one, thanks, u(x,y) = x-y is a solution but how to find it by using characteristic method?

Lastly, what is the best site to start learning Latex?

[quote="Xevarion"]Personally I like partial derivative symbols more than deltas...[/quote]

Yah, me too, but there are so many extra letters to type ... ;)

Also, for using LaTeX on this site, you don't need to do any of the downloading stuff or document-level formatting; the only thing you need to learn is the basic codes and then slap them between dollar signs (e.g., \partial gives $\partial$ and \frac{a}{b} gives $\frac{a}{b}$ so \frac{\partial u}{\partial x} gives $\frac{\partial u}{\partial x}$).

Unfortunately, I don't have any more help to offer you for the mathematics at hand  :oops:


I've been having trouble compltely understanding significant digits and the rules that apply to them. Could anyone give me some websites or information to help?

It is just a way to insure that precision is not perserved.
When using significant digits the decimal point and every non-zer digits, except past the decimal, represent a degree of "sure-ness".

Examples
$ 1000$ Has only one [url=http://en.wikipedia.org/wiki/Significant_figures]significant digit[/url] since the zeros are not taken as being "definite".
$ 1000.$ Though when the decimal point is added the number of significant digits is increased to four.
Similairly $ 1230$ has only three significant digits and $ 1230.$ has four.
When past the decimal place the zeros automatically significant so $ 1230.0$ and $ 1230.1$ have the same number of significant digits.

When [url=http://en.wikipedia.org/wiki/Rounding]rounding[/url] in using significant digits, the rules for "regular" rounding exist EXCEPT when the when there is a five in the place to the right of the digit being rounded to.
Examples
To two significant digits
$ 0.125000$ is $ 0.12$ since there are not non-zero digits to its right, round to the even digit and $ 0.175000$ is $ 0.18$.
$ 0.035001$ is $ 0.04$ since there is a non-zero digit.

When adding (or subtracting) and multiplying (or dividing) there are also rules with significant digits.
The least number of significant digits after the decimal in any single one of the numbers being added or subtracted, must be the number of digits past the decimal after adding or subtracting the numbers.
Examples
$ 0.5+0.43=0.93$ The least number of significant digits past the decimal of any of the numbers being added is one, so the answer must have one. Making the answer $ 0.9$, so it is not more precise than any of the other added numbers.
$ 10.5+0.43=10.9$ After the significant digits were considered.
(Notice that in each of those cases the zero before the decimal like in $ 0.43$ is not significant since there are not any digits to its left.)

The least total number of significant digits in any of the numbers being multiplied or divided must the number in the final answer.
Examples
$ 0.5 \cdot 10.1=5.05$ Which is $ 5.$ after adjusting to the limit of one significant digit.
$ 1230. \cdot 1000 = 1230000$ Which is $ 1000000$.


Can you do a solution that is entirely solved by a computer?

For example, problem 2b of round 1

If somebody made a program to find every permutation of 1,2,3,4,5,6,7,8 with 1 always first and tested if the second was greater than the first and third the fourth was greater than the third and fifth...So that the computer does the brute force work and you just write something that tests every single possibility.

My understanding is that you [i]can[/i]. But I personally don't consider this math (although I did use a computer to verify my answer)

But when submitting your program as a solution, you must save the results and prove [i]why[/i] your program solves the problem accurately (avoiding double-counting, etc.)

Also, I don't know if you actually submit all of your code, just an outline of what the code does, or both. There is a brief section about computer solutions in the "submitting answers" section of the USAMTS website, although I don't have a link for that.

Bottom line, yes you can, but you must prove why it works.

I think.

I really don't think that a code for 2b would be very long, so I would just copy and paste it...And I don't think it would be hard to prove it does not double count if I try every single permutation once and test it.  But the good thing is that I have a reason to not have done that (I missed something in my casework):  I just started computer programming this year and would not have been able to.

here is a link for your edification: [url=http://www.usamts.org/TipsFAQ/U_FAQ.php#computer]clicky[/url]


[b]Prolem [ZaiZai-Hoang][/b]
Let $ a,b,c,d,e\ge 0$ and such that $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 5$. Prove that
\[ A\ge B\]
with:
\[ A\equal{}2(abcd \plus{} abce \plus{} abde \plus{} acde \plus{} bdce) \plus{} 80\]
\[ B\equal{} 9(ab \plus{} ac \plus{} ad \plus{} ae \plus{} bc \plus{} bd \plus{} be \plus{} cd \plus{} ce \plus{} de)\]

In my opinion, it is really nice  :)

$ a,\ b,\ c,\ d,\ e$ are the roots of

$ x^5\minus{}5x^4\plus{}\frac{B}{9}x^3\minus{}(abc\plus{}abd\plus{}abe\plus{}acd\plus{}ace\plus{}ade\plus{}bcd\plus{}bce\plus{}bde\plus{}cde)x^2$
$ \plus{}\frac{A\minus{}80}{2}x\minus{}abcde\equal{}0.$


calculus
integration
limit
calculus computations

Let $ \alpha ,\ r$ be real numbers such that  $ r > 1,\ r\neq 3,\ r\neq 4$.

Find the values of $ \alpha ,\ r$ such that $ \lim_{n\to\infty} \sum_{k \equal{} 1}^n \frac {n^{\alpha}}{(n \plus{} k)^{r}} \equal{} \frac {r \minus{} 3}{(r \minus{} 1)(r \minus{} 4)}$.

$ r\equal{}2$ and $ \alpha \equal{}1$

[hide="answer"]$ r\minus{}\alpha \equal{}1$

and $ LHS\equal{}\int_0^1 (1\plus{}x)^{\minus{}r}dx\equal{}\frac {1\minus{}2^{1\minus{}r}}{r\minus{}1}$[/hide]

We have $ 1 \minus{} 2^{1 \minus{} r} \equal{} \frac {r \minus{} 3}{r \minus{} 4}\Rightarrow 2^{1 \minus{} r} \equal{} \frac {1}{4 \minus{} r}$

$ y \equal{} 2^{1 \minus{} x}$ and $ y \equal{} \frac {1}{4 \minus{} x}$ intersect only at one point and it is easy to see that they do so at x=2


10 cm
10 cm  are dimensions
10 cm

_______________ Find the length of each edge of a cube that has a volume
twice that of the cube shown. Give your answer to the nearest tenth of a
centimeter. You may use a calculator.

Next time, try not to post the same subject twice.
This happens when your computer submits your post slow, and you click submit more than once, making the computer think you want more than one post.

Actually, this problem is different from the other one.  It just has the same title(3 cube for you cgayo).

Here's the solution:

10x10x10 is 1,000 so the volume of that cube is 1,000. 2x1,000 is 2,000 so our answer is $ \sqrt[3]{2,000}$, which, according to my calculator:P, is approximately 12.6.

Oh sorry, with the same title, i made a bad assumption.

[quote="maybach"]so our answer is $ \sqrt [3]{2,000}$, which, according to my calculator:P, is approximately 12.6.[/quote]
Or can be simplified to $ 10\sqrt[3]{2}$.


Microsoft Windoze has a Task Scheduler built into the operating environment, which you can set to run a program, and the "program" could be displaying a reminder. I usually use the do-list function of whatever email program I use to do what you're asking about, but that would be the more system-wide way to do it.

Don't forget to . . .

[quote="Rep123max"]Also, what are the folders "My EBooks" and "Filelib"? I keep deleting them and they keep coming back but they are always empty?[/quote]
Give up. They're windows folders.

[quote="Dark_Shadow"]You can stick things on your desktop...[/quote]

you can stick things on your screen as well :roll: that way they're best visible :D

[quote="nr1337"][quote="Rep123max"]I know this proves my compouter illiteracy, but is there some way to set a reminder so that like a reminder pops up? I know I can do it with email, but if I am not checking me email at that moment, it doesn't really help.

THANKS[/quote]

Macintosh:  Stickies is a great program!

Windows:  Put shortcuts that do nothing all over your desktop and put words in them.[/quote]

NotesHolder Lite is a free program with limited capability that has almost as many features as random shortcuts. For example, the Lite version includes a banner and doesn't feature catagories, something that can be easily done with shortcuts

[quote="h_s_potter2002"][quote="churchilljrhigh"]Get Microsoft Outlook. Pratically everybody has it. U can get reminders, check email, etc.
I got My Ebooks either when I installed Adobe Acrobat Profesional or Microsoft Ebook reader. There's so many good free ebooks to download on the net like My Life, Harry Potter, etc.[/quote]

How do you use Outlook with a yahoo account?[/quote]

You have to pay money (Since free mail has to have ads). However, yahoo used to have free pop3 email. However, your ISP often provides you with a Pop3 acces or HTTP. Like MSN, or Yahoo!


search
number theory unsolved
number theory

show product of consecutive integers cannot be perfect power.

Nobody said [b]two[/b] consecutive numbers. The result is this:

If $n\ge 1,k\ge 2$ are positive integers, then $(n+1)\cdot\ldots\cdot(n+k)$ is not a perfect power.

There is http://www.mathlinks.ro/Forum/viewtopic.php?t=2780.
What grobber didn't say: not every perfect power is a perfect square ;)

is there any elementary sol.? :?:
 t. n.shorey has worked and is working on analogous results where we take pro. of con. terms in a.p. google search will give alot. :|


A number is called ascending if each digit is greater the digit before it (ex. 12345). How many 5 digit numbers are ascending?

[hide]first numbers are "1234" : 1 + 1 + 1 + 1 + 1 = 5 cases
first numbers are "1235" : 1 + 1 + 1 + 1 = 4 cases
first numbers are "1236" : 1 + 1 + 1 = 3 cases
...
we add them all together to generalize
first numbers are "123" : 5 + 4 + 3 + 2 + 1 = 15 cases
first numbers are "124" : 4 + 3 + 2 + 1 = 10 cases
...
generalize again
first numbers are "12" : 15 + 10 + 6 + 3 + 1 = 35 cases
first numbers are "13" : 10 + 6 + 3 + 1 = 20 cases
...
generalize final time
first number is "1" : 35 + 20 + 10 + 4 + 1 = 70 cases
first number is "2" : 20 + 10 + 4 + 1 = 35 cases

add all together
70 + 35 + 15 + 5 + 1 = 126 cases

i'm not sure if its the right solution... and i can't find a more elegant way using probability formulas[/hide]

There is in fact a easy way to do this problem.

It was posted before, but don't worry about it.

[hide="EasyWay"]
Notice that whenever you choose five distinct digits, there is exactly one way to arrange them such that the number is ascending.

Then, you subtract the numbers which begin with 0. There are 9 other digits from which you must choose 4 to fill in the digits after 0.

So, it's C(10,5) - C(9,4) = 126.
[/hide]


Has anyone made a batch file yet that compiles your LaTeX document with the asymptote code, compiles the asymptote code, and compiles the LaTeX document again? I've been attempting to do this, but I don't know what arguments I need to send to TeXnicCenter in order for it to build the LaTex document.

You don't say how you are running the batch file.
1. You can run it as an external file in the Tools comand - see [[Asymptote:_Advanced_Configuration#Showing_Asymptote_error_messages_in_TeXnicCenter]] for an example of such a batch program
or
2. You can run it as a postprocessor in a profile, and the first part (compiling with LaTeX) has been done for you. For more help wth this see the LaTeX => PS => PDF profile and [i]Placeholders for single files[/i] in TexnicCenter's help.
Remember you should build in error checking into your batch file since if there's an error at any stage you don't want it to carry on with the next stage, but should say what is wrong.

Ok I had a postprocessor that called asy.exe with the asy file associated with the tex file using asymptote. Thanks a lot! It worked! All I have to do now when running a new asy environment is building once (I edited it so that it calls the bibtex, pdflatex, and makeindex after compiling asymptote). I don't think postprocessors work with batch files because I tried to and it didn't appear to execute it.

EDIT: How can I have asy.exe pause without a batch file? Is it possible to run multiple exe files to make that part work?


trigonometry
vector
geometric series
trig identities
Law of Cosines

Let $ n \geq 3$ an integer finit number and $ \omega \equal{} \cos \frac {2\pi}{n} \plus{} i\sin \frac {2\pi}{n}$.Prove that for any $ a \in C$ and $ r \in R$,$ r > 0$ the number $ \sum_{k \equal{} 1}^n \frac {|a \minus{} r\omega^k|^2}{|a|^2 \plus{} r^2}$ is an integer number.

$ z\cdot\overline{z}\equal{}|z|^2$. $ \overline{\omega}\equal{}\omega^{\minus{}1}$. Also $ \omega\plus{}\omega^2\plus{}...\plus{}\omega^n\equal{}0$. Now we can simplify the sum \[ \sum_{k\equal{}1}^n |a\minus{}r\omega^k|^2\equal{}\sum_{k\equal{}1}^n a\cdot\overline{a}\minus{}r(a\omega^{\minus{}k}\plus{}\overline{a}\omega^{k})\plus{}r\omega^k r\omega^{\minus{}k}\equal{}n\cdot (|a|^2\plus{}r^2)\]
Dividing by $ |a|^2\plus{}r^2$ finishes the problem. Also note that $ r$ could be complex.

Wow your solution kinda pwns mine... very nice  :)

ah I kinda skipped over that $ a$ is a complex number so otherwise...I got what Altheman got but taking a bit more lines.

@Anaxerzia, your geometrical approach is neat, though it may much more work, I like the interpretation.

[quote="Anaxerzia"]Very nice! Fairly simple, but very nice! :)

[hide] Let $ a$ have a magnitude $ m$ and a argument $ \theta$. Now consider the following complex mapping:

[img]http://i193.photobucket.com/albums/z278/Anaxerzia/hi4-1.png[/img]

The green lines are equivalent to $ rw^k$ for all $ 1\le k\le n$ and the red line is equivalent to $ a$. The black lines must obviously represent each vector $ a - rw^k$. Now consider the case for $ a$ with $ 0\le\theta < \frac {2\pi}{n}$ (we only need to consider this case as we can keep rotating the figure by $ \frac {2\pi}{n}$ to get any desired value of $ \theta$ while the magnitudes shall remain the same).

We shall express the magnitudes of each $ |a - rw^k|^2$ with the law of cosines from the triangle of the vector, $ a$, and $ rw^k$. For even $ n$ (which we shall express as $ 2t$), one half of the $ rw^k$'s will have their argument increased by $ \theta$ and the other decreased. Thus the magnitudes sum to $ \sum_{k = 1}^t m^2 + r^2 - 2mr \cos{(\frac {\pi k}{t} - \theta) + \sum_{k = 1}^t m^2 + r^2 - 2mr \cos{(\frac {\pi k}{t} + \theta)}}$ for both halves. It becomes:
$ \sum_{k = 1}^t 2(m^2 + r^2) - 2mr \cos{(\frac {\pi k}{t} - \theta) + \cos{(\frac {\pi k}{t} + \theta)}}$
$ \sum_{k = 1}^t 2(m^2 + r^2) - 4mr (\cos{\theta}\cos{k\pi/t})$[/hide]

So yeah after trudging through all (even a cosine geometric series) of that we get the sum is $ 2t(m^2 + r^2) = n(m^2 + r^2)$, which is obviously divisible by $ m^2 + r^2$. Similar for odd $ n$ I suppose :)

EDIT: forgot diagram  :oops:[/quote]
you have the same solution as mine... :P

Here is my proof similar with Anaxerzia[hide]$ a = |a|(cos\theta + isin\theta)$ we obtain that $ a - r \omega^k = (|a|cos \theta - rcos\frac {2k\pi}{n}) + i(|a|sin\theta - rsin\frac {2k\pi}{n})$ and then $ |a - r\omega^k|^2 = |a|^2 + r^2 - 2r|a|cos(\frac {2k\pi}{n} - \theta)$
therefore
\[{ \sum_{k = 1}^n \displaystyle\frac {|a - r\omega^k|^2}{|a|^2 = r^2} = n - \displaystyle\frac {2r|a|}{|a|^2 + r^2} \cdot \sum_{k = 1}^n cos(\frac {2k\pi}{n} - \theta})
\]
In the well known identity
\[ \sum_{k = 1}^n cos(x + k\varphi) = \displaystyle\frac {sin\frac {n\varphi}{2}}{sin\frac {\varphi}{2}} \cdot cos(x + \frac {(n + 1)\varphi}{2})
\]
if we take $ x = - \varphi$ and $ \varphi = \frac {2\pi}{n}$ we obtain that
\[ \sum_{k = 1}^n cos(\frac {2k\pi}{n} - \theta) = 0
\]
from where the sum from our problem is $ n$.
If $ A_k$ is the geometrycal image for the compelx number $ r\omega^k$ $ (1 < = k < = n)$,$ A$ is the geometrycal image for the complex number $ a$,and $ B$ is a point on the circle with the center in origin and the diameter $ 2r$,suche as $ AO\bot OB$ and then the relation from the problem becomes:
\[ \sum_{k = 1}^n AA_k^2 = n AB^2
\]
[/hide]

Nice proof Altheman!  :)


geometry
geometric transformation
invariant
topology
Functional Analysis
real analysis
real analysis unsolved

what is the measure assigned to each non-empty open set by any non-trivial
translation invariant Borel measure on a infinite dimensional separable Hilbert
space? prove your answer

Let $\{e_n\}$ an orthonormal sequence, and let $B_n:=B\left(e_n,\frac{\sqrt 2}2\right)$. If $x\in B_n\cap B_m$ with $m\neq n$ then $\sqrt 2=||e_m -e_n||\leq ||e_m-x||+||x-e_n||<2\frac{\sqrt 2}2=\sqrt 2$, hence these ball have to be pairwise disjoint. Hence we should have for such a measure $m$: $m\left(B\left(0,1+\frac{\sqrt 2}2\right)\right)\geq \sum_{n\in\mathbb N}m(B_n)$. But since $m$ is invariant by translations, $m(B_n)=m(B_0)$, therefore either each ball has a $0$ measure or each ball has an infinite measure. Since we want a non-trivial measure, each open set has an infinite measure.


in a set of $2n+1$ persons,for any $n$ persons there exist a person defferent from them who knows each of them.Prove that there is a person in this set who knows all the others. :D

[quote="DK"]I have a solution with contradiction that is more "algorithmic". My friend though presented a much simpler solution:

There are two persons that know each other (if not we get a contradiction). So let those persons be in the set A. Now there is a third person outside A that knows both of those persons in A. Include that third person in A. Again there is a fourth person outside A that knows everyone in A, so include him in A too. You can continue in this way untill you get n+1 persons in A where everyone knows everyone. But now look at the n persons that are outside A. According to the problem, there is a person in A that knows all of them, and because that person knows everyone in A he knows every person in the original set.

My proof was by considering non-friendships and a graph with those persons as vertices. By way of contradiction, every person has at least one edge eminating from him. Now I consider an arbitrary set of n persons and the rest of n+1 persons. Now I just construct a set with n+1 persons in which every person in that n+1-set has en edge to someone outside that set. That's quite easy to do.[/quote]
really nice and easy solution.excellent. :D


[quote="Carcul"]How can we convert aniline into 3-chloroaniline?[/quote]
convert aniline to anilinium chloride by adding HCL and then do normal friedel craft chlorination to get it in the 3rd position!!!

no anilinium chloride is unreactive towards chlorination

What is a " normal friedel craft chlorination"? And why is "anilinium chloride is unreactive towards chlorination"?

well isnt anilinium chloride a powerfully deactivating grp?  :maybe:

And what does that have to do with chlorination? Aren't you making a confusion with the lack of reactivity of benzene rings bearing deactivating substituents in Friedel-Crafts reactions?

Well doest the mechanism of the reaction concern Cl+ and a sigma complex ?
And if it does how will it be stabilised. wont it be deactivated by the anilinium grp?

1) What is a sigma complex?
2) How is a sigma complex formed in an aromatic chlorination?

Well wat i mean is if Chloronium ion is going to attack the anilinium ring (if i am right if i am not please tell), then wont the intermediate carbocation be destabilised by the anilinium grp?

The Wheland intermediate is always destabilized when a deactivating group is on the ring: however, the meta position is the least destabilized, and so deactivating groups are meta directors. On the other hand, a meta directing group does not inhibit halogenation, just the Fridel-Crafts reactions.

sigma complex is a jargon for the Wheland intermediate

Hey carcul.......
will the yield of the method mentioned by srinitrs be high??????
I just want to know?

[b][i]
1.Protect with CH3COCl
2.Conc.HNO3/Conc.H2SO4
3.H2O
4.CH3Cl/AlCl3(Here one of the groups is -NH2 which is activating)
5.NaNO2/HCl
6.H3PO2
7.Sn/HCl[/i][/b]

See this method

1.Oxidise using CF3CO3H.
2.Nitrate using Nitration mixture.
3.use ammonium sulphide(partial reduction)
4.Diazotise
5.treat with CuCl/HCl
6.treat with Fe/HCl

[quote="Chemrock"]4.CH3Cl/AlCl3(Here one of the groups is -NH2 which is activating)[/quote]

One more time, you cannot make Fridel-Crafts reactions with -NRR' substituents on the ring.

[quote="Valeriummaximum"]1.Oxidise using CF3CO3H.[/quote]

Not with that oxidizing agent. Also, if you have already a nitro group on the ring, then why do you use that unnecessarily long method for introduction of a meta chloro, if Fe/Cl2 is enough?


calculus
integration
calculus computations

Find a general solution to the differential equation $ yy' \equal{} xe^{\frac{x}{y}}$.

rewrite it as $ y'\equal{}\frac xy e^{\frac xy}$ and put $ y\equal{}xz.$

[quote="Integrability"]rewrite it as $ y' \equal{} \frac xy e^{\frac xy}$ and put $ y \equal{} xz.$[/quote]

Yes I've tried that, but the result will be $ \frac{dz}{z(1\minus{}z^2e^z)} \equal{} \frac{dy}{y}$ and I can't simplify the integration on the left part, or is there something wrong with my calculation ?

hello, i think it doesn't exist an explicit solution of your equation.
Sonnhard.

Hmm.... that's interesting, RDeepMath91. You could not solve the problem but somehow your title for this posting is "Simple Differential Equation"? :rotfl:


Find all prime numbers $ p$ with $ 5^p\plus{}4p^4$ is the square of an integer.

this is really easy... $ p \equal{} 5$ only.
In fact, if $ n^2 \equal{} 5^p \plus{} 4p^4$, $ n \minus{} 2p^2 \equal{} 5^a$, $ n \plus{} 2p^2 \equal{} 5^b$, $ a < b$ and $ a \plus{} b \equal{} p$. If $ a \equal{} 0$, $ 5^p \minus{} 1 \equal{} 4p^2$. This has no solutions since LHS grows a lot more faster than RHS and we can check by hand cases where $ p\equal{}2,3$. So, we may assume $ a\geq 1$, so that $ 5$ divides $ 4p^2$ and thus $ p \equal{} 5$. Conversely, $ 5^5 \plus{} 4\times 5^4 \equal{} 75^2$.


"Mrs. Jewls found four answer sheets from a spelling test crammed in the back of a drawer. Jenny:rode,toad, towed, road, rowed. Dana:road, rode, towed, toad, rowed. Terrence: rowed, toad, towed, rode, road. Stephen:rode, towed, toad, rowed, road. She remembered that one of them had all five correct. Another student got all five wrong. And she knew Jenny got a better grade than Terrence. What were the correct answers?"

You know that either Dana or Stephen got all of the answers correct (and the other missed them all) because Jenny and Terrence agreed on one, making it impossible for one to get that question correct, and the other person missing it.  Assume Dana has all the correct answers.  Then, Jenny would have 2 correct answers and Terrence would have 1 correct answers, giving Jenny a higher score.  Therefore, the correct answers are:
[hide="Correct Answers"]
Dana's answers
road, rode, towed, toad, rowed
[/hide]

[hide]First we identify two lists that have nothing in common, since one got all right and another all wrong. Dana and stephen have completely different lists. Let's say that Stephen got a perfect score. Then Jenny and Terrence will both have one right, which is against the rules. If Dana was perfect, Jenny and Terrence would have 2 and 1 right, respectively, which follows the given rules. Thus the answers are :

1. road
2. rode
3. towed
4. toad
5. rowed[/hide]

[quote="easyas3.14159..."][quote="dogseatcheese"][hide]The only thing that everybody got correct except 1 person was Stephen and it was #3. Thus, the only person that has nothing in common with Stephen and got it all correct was Dana. If Dana got it all correct, then Jenny DOES have a better score than Terrance.

THus the correct answers are Road, rode, towed, towed, rowed.[/hide][/quote]

dogseatcheese, dont you mean [hide]towed, toad?[/hide][/quote]

 :blush: oops. Yea.....what she said......

[hide]Supposing Jenny got them all right;
Jenny:5/5
Dana:2/5
Terrence:2/5
Stephen:1/5
No one got them all wrong so Jenny didn't get perfect.

Supposing Dana got them all right;
.......
[b]So Dana got perfect on her test.[/b][/hide]

[hide]1. road 2. rode 3. towed 4. toad 5.rowed [/hide]
I love those problems![/hide]


A rhombus is inscribed in a circle. If the length of one diagonal of the rhombus is 8x, find the length of the other diagonal.

is there enough info to solve your problem? the rhombus could be a square, or a diamond shape......

I agree with h78reg, this problem isn't very clear, or am I missing something?  :fool:

There's enough information.

Hint: [hide] It's inscribed in a circle. Therefore, the rhombus must be a _______[/hide]

My answer
[hide] 8x? If it's inscribed in a circle and a rhombus than it has to be a square. Therefore both diagonals are the same [/hide]

[hide]
if the rhombus is inscribed in a circle then it must be a square, so 8x is one diagonal and the other must be 8x.[/hide]


A regular octagon is to be formed by cutting equal isosceles right triangles from the corners of a square. If the square has sides of one unit, the leg of each of the triangles has length:

$ \textbf{(A)}\ \frac{2 \plus{} \sqrt{2}}{3} \qquad
\textbf{(B)}\ \frac{2 \minus{} \sqrt{2}}{2}\qquad
\textbf{(C)}\ \frac{1 \plus{} \sqrt{2}}{2}\qquad
\textbf{(D)}\ \frac{1 \plus{} \sqrt{2}}{3}\qquad
\textbf{(E)}\ \frac{2 \minus{} \sqrt{2}}{3}$

[hide="My Solution"]
Let $ x$ be the side length of the regular octagon. We know that the line segments between each of the four triangles must have a length of $ x$ and the hypotenuses of the isosceles right triangles must also have a length of $ x$ in order for the octagon made to be a regular octagon. Each leg of the isosceles right triangle must then be $ \frac{x}{\sqrt{2}} \equal{} \frac{x\sqrt{2}}{2}$. Adding up all the lengths on one side of the square, we obtain

$ \frac{x\sqrt{2}}{2} \plus{} \frac{x\sqrt{2}}{2} \plus{} x \equal{} (\sqrt{2} \plus{} 1)x \equal{} 1 \implies x \equal{} \sqrt{2} \minus{} 1\implies \frac{x\sqrt{2}}{2} \equal{} 1 \minus{} \frac{\sqrt{2}}{2} \equal{} \frac{2 \minus{} \sqrt{2}}{2}.$

Therefore, the answer is $ \fbox{(B)}$.
[/hide]

[hide]
We can make $ x$ = the leg of each triangle. Since the triangles are isosceles right triangles, each triangle is a 45-45-90 triangle. So the hypotenuse is $ x\sqrt{2}$. This is one of the sides of the regular octagon.
Since the square has sides of one unit, the side of the octagon that is along the side of the square has length $ 1-2x$. (The isosceles triangles that were cut out have leg length $ x$, and there are two of them on each side of the square.)
Because a regular octagon's sides have the same length, $ x\sqrt{2}=1-2x$. Adding $ 2x$ to both sides and then factoring out $ x$,
$ x(2+\sqrt{2})=1$
Dividing by $ 2+\sqrt{2}$,
$ x=\frac{1}{2+\sqrt{2}}$
Multiplying by $ \frac{2-\sqrt{2}}{2-\sqrt{2}}$,
$ x=\frac{1}{2+\sqrt{2}}\cdot{\frac{2-\sqrt{2}}{2-\sqrt{2}}}$
$ x=\frac{2-\sqrt{2}}{4-2}$
$ x=\frac{2-\sqrt{2}}{2}$
So the answer is $ \fbox{B}$.
[/hide]

AHSME 1957 Problem 31

Wasn't the AHSME a 30-problem test?  :maybe:

EDIT: Just saw that it was 50 problems.  :blush:  :wallbash:


calculus
integration
limit
logarithms
real analysis
real analysis unsolved

Please help me with this:

How to compute that integrals:
$ 1.)\int_{0}^{1} \frac {2^{x}}{x \plus{} 1}dx$
$ 2.)\int \left( \frac {x}{\arctan{x}} \minus{} 1\right)^{ \minus{} 2}dx$

I believe the first integral would be somewhat nicer if instead of 2 there was an $ e$.

$ \lim_{n \to \plus{} \infty} \sum_{i\equal{}0}^n \dbinom{n}{i} \frac{1}{n^i(i\plus{}k\plus{}1)} \equal{} ...$

Hi :) !!

for the first integral we can prove that:

$ \int_0^1 \frac{2^x}{1\plus{}x}dx \equal{} \left( \sum_{n \geq 0} \frac{ln \left(\frac{1}{2}\right)^n}{n!}\sum_{k\equal{}1}^{n\minus{}1}\frac{(\minus{}1)^{k\minus{}1}}{k} \right) \minus{} ln(2) \equal{} L$

with $ 0,96 \leq L \leq 0,961$

hello, for the first integral Maple 12 gives the following answer
$ \int_{0}^{1}\frac{2^x}{x\plus{}1}\,dx\equal{}1/2\,{\it Ei} \left( 1,\minus{}\ln  \left( 2 \right)  \right) \minus{}1/2\,{\it Ei}
 \left( 1,\minus{}2\,\ln  \left( 2 \right)  \right)$.
Sonnhard.


Let $p_n=\sqrt{a_nb_n}$; then in general
\begin{eqnarray*} p_{n+1}&=&\sqrt{(a_n+\frac{1}{b_n})(b_n+\frac{1}{a_n})}\\ &=&\sqrt{p_n^2+\frac{1}{p_n^2}+2}\\ &=&p_n+\frac{1}{p_n} \end{eqnarray*}
Also, if $s_n=a_n+b_n$ then $s_n=s_{n-1}(1+\frac{1}{p_{n-1}^2})=s_{n-1}\frac{p_n}{p_{n-1}}$.  So $s_{25}=s_1\frac{p_{25}}{p_1}\geq 2p_{25}$ by AM-GM.

$p_{25}$ is uniquely determined as $f^{24}(p_1)$ by our recurrence relation, where $f(x)=x+\frac{1}{x}$.  Note $f(x)\geq 2$ always, and $f(x)$ is increasing and greater than 2 for $x$ greater than or equal to 2.  So $p_{25}=f^{24}(p_1)\geq f^{23}(2)$.  By brute force calculation (though I'm sure there's an easier way) $f^{23}(2)\approx 7.17>7.07\approx 5\sqrt{2}$.  So $s_{25}>10\sqrt{2}$ as required.


We roll a fair 6-sided die 5 times.  What is the probability that we get a 6 in at most 2 of the rolls?

I thought the answer should be the probability of rolling two 6s + the probability of rolling one 6, but the solution also added the probability of rolling no 6s, and I don't get why.

Of course, you meant 2 sixes, 1 six, and 0 sixes :wink:

Oh, it said six? I thought two. Well, same thing.

I don't understand why you do case 3. The question asks "What is the probability that we get a 6...?", and case 3 counts how many ways we can't get a 6.

It said "at most two times", and zero times is less than two.

I understand that zero times is less than two, but in that case you aren't getting a 6. Why would you count this case?

Do you agree, that if you roll 0 6's, there's less than 2 6's?

Well then we must count it, because 0 6's is less than 2 6's...

but it asked how many times we could roll a 6 given some condition, and getting a 6 zero times doesn't count as rolling a 6

It never said that. It said how many ways can you roll [b]at most[/b] two 6's.

I got this wrong....it brought down my rating by 12...:(

Because 0,1, and 2 are all below or 2. [quote=AceOfDiamonds]don't get why you count no sixes :( :([/quote]

[quote=isabella2296]That is a correct method, so perhaps you calculated the probabilities incorrectly.

Case 1: 2 two's

$ \left ( \frac{1}{6} \right ) ^2 \cdot \left ( \frac{5}{6} \right )^3 \cdot \binom{5}{2} \equal{} \frac{625}{3888}$

Case 2: 1 two

$ \left ( \frac{1}{6} \right ) \cdot \left ( \frac{5}{6} \right)^4 \cdot \binom{5}{1} \equal{} \frac{3125}{7776}$

Case 3: 0 two's

$ \left ( \frac{5}{6} \right )^5 \equal{} \frac{3125}{7776}$

Adding these all together, we get a result of $ \frac{625}{648}$.[/quote]

Why would you multiply each by like $C(5,2)$ or something?

[quote=AceOfDiamonds]don't get why you count no sixes :( :([/quote]

We count the no sixes case because it said that we should get a 6 in [b]at most 2[/b], which means no more than two. Therefore, we count 0,1 or 2 sixes.

[quote=Element15-Phosphorus][quote=isabella2296]Case 1: 2 two's

$ \left ( \frac{1}{6} \right ) ^2 \cdot \left ( \frac{5}{6} \right )^3 \cdot \binom{5}{2} \equal{} \frac{625}{3888}$

Case 2: 1 two

$ \left ( \frac{1}{6} \right ) \cdot \left ( \frac{5}{6} \right)^4 \cdot \binom{5}{1} \equal{} \frac{3125}{7776}$

Case 3: 0 two's

$ \left ( \frac{5}{6} \right )^5 \equal{} \frac{3125}{7776}$

Adding these all together, we get a result of $ \frac{625}{648}$.[/quote]

I am confused on how you are multiplying each by $\dbinom{5}{1}$ or $\dbinom{5}{2}$ and not other. I get how you are multiplying those because you are choosing 2 slots in the 5 available to become 2, but are you counting other repetitions within the 5 spins?[/quote]

For the one sixes case, we are choosing 1 out of the 5 total rolls, so we must multiply the case by $\binom{5}{1}$. We apply the same reasoning for the 2 sixes case.


$a$,$b$,$c$ are positif real numbers
Prove that : $\left( \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) ^{2}\geq \left( a+b+c\right) \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) $
It's not very hard I guess  :P

Multiply it out and rearrange some terms:
$\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)+2\left(\frac ac+\frac cb+\frac ba\right)-\left(\frac ac+\frac cb+\frac ba\right)-\left(\frac ab+\frac bc+\frac ca\right)-3$
$=\left(\frac{a^2}{b^2}-\frac ab-1+\frac ba\right)+\left(\frac{b^2}{c^2}-\frac bc-1+\frac cb\right)+\left(\frac{c^2}{a^2}-\frac ca-1+\frac ac\right).$
Each term is of the form $u^2-u-1+\frac1u=(u-1)^2\left(1+\frac1u\right),$ which is clearly greater than or equal to zero.

It's even a different solution than the ones in the linked thread.

this is my solution :
we have $2\frac{a}{b}+\frac{b}{c}=\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\frac{a}{\root{3}\of{abc}}$
the same $2\frac{b}{c}+\frac{c}{a}\geq 3\frac{b}{\root{3}\of{abc}}$
and $2\frac{c}{a}+\frac{a}{b}\geq 3\frac{c}{\root{3}\of{abc}}$
by suming the three we get $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{\left( a+b+c\right) }{\root{3}\of{abc}}$ (1)
on the other hand : $\frac{a}{b}+2\frac{b}{c}=\frac{a}{b}+\frac{b}{c}+\frac{b}{c}\geq 3\frac{\root{3}\of{abc}}{c}$
the same : $\frac{b}{c}+2\frac{c}{a}\geq 3\frac{\root{3}\of{abc}}{a}$
and $\frac{c}{a}+2\frac{a}{b}\geq 3\frac{\root{3}\of{abc}}{b}$
by suming the three we get $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \root{3}\of{abc}\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) $ (2)
finally by multiplying (1) and (2) we get the inequality desired.


group theory
abstract algebra
number theory
prime numbers
superior algebra
superior algebra unsolved

if a group G of order n has for each divisor d of n  a unique subgroup of order d. Then G must be cyclic. How to prove this?
Do I to use prime numbers or something such that?
thankss

In such a G, all subgroups are characteristic and in particular normal. Hence all sylowgroups are normal, hence G is the direct product of its sylowgroups. So we can wlog assume that n is a prime power $ p^r$.

Because there is a unique subgroup of order $ p^{r\minus{}1}$, there is exactly one maximal subgroup. This implies, that G is cyclic: Choose a $ g\in G$ that is not in this maximal subgroup and consider the subgroup $ \langle g\rangle$. If it would be a proper subgroup, it is contained in a (=the) maximal subgroup, which is a contradiction to the choice of g. Hence $ G\equal{}\langle g\rangle$.


algorithm
combinatorics unsolved
combinatorics

100 people from 25 countries, four from each countries, stay on a circle. Prove that one may partition them onto 4 groups in such way that neither no two countrymans, nor two neighbours will be  in the same group.

we use an idea from one of the other ARO problems as a first step: if there are $100$ people from $50$ countries in a circle, we can partition them into $2$ groups of $50$ each so that there are no $3$ consecutive people(in the circle) within the same group, nor are there two representatives of the same country.
each country $C$ has $4$ members,$C_1,C_2,C_3,C_4$, say.suppose we regard $C_3,C_4$ as members of another country, then we have $2$ representatives from $50$ different countries and so by the above result(which is also there in this forum) we can get $2$ groups of $50$ people each such that the above holds. now consider the first of these groups and consider the following graph: the vertices are the members of the first group and for 2 members $v,w$ we have $v\leftrightarrow_{1} w$ if they are adjacent on the circle;$v\leftrightarrow_{2} w$ if they are both of the same country originally(we'll call these as $1$-edges and $2$-edges respectively). then the previous problem implies that for any $v$ its $1$-degree and $2$-degree are both $\leq 1$. consequently this graph cannot have odd cycles(else there are two consecutive edges of the same color $1$ or $2$) and so is bipartite.
hence this graph can be properly vertex-2-colored and each of these color classes will again contain precisely one member from each country.
doing the same from the other group, we get $4$ groups as desired.

Congratulations. I am deeply impressed by Vasya Astakhov, who solved it without knowing 9.8. I tried a lot (somemonthes ago), and had no thoughts.

well, i first tried this as a consequence of the Gallai-Roy theorem(if a graph has an orientation with longest oriented path of length $l$, then the chromatic number of the graph is atmost $l+1$), because this must be the most general way of setting this problem up. since the Gallai-Roy theorem also assures that if the chromatic number is $l$ then there also exists an orientation that achieves the bound specified.
unfortunately for me it did not come through simply enough, though i believe that one must be able to describe an orientation appropriately. then i fiddled with the idea of atleast splitting into 2 groups and then remembered having seen this other problem too,so it came out quickly after that....
but i do admit that it seems quite difficult to try it without that idea at all...

By the way, what is Astakhov's solution??

Ok, so this is my attempt at a solution  :maybe: Hope it's correct/rigorous.

It's an algorithm.

First we partition the 100 people into the four groups so that each partition has one member of each nationality.

Now, consider the circle of people and label each $ P_1, P_2, ...P_{100}$ going clockwise. Now, start from an arbitrary person, $ P_n$, and 'locate' $ P_{n \plus{} 1}$ (that is, find out which partition he is in).

We have two possible cases:

$ P_{n \plus{} 1}$ is not in the same initial partition as $ P_n$: leave the two alone and move on to find $ P_{n \plus{} 1}$ and check to see if $ P_{n \plus{} 2}$ is in the same initial partition and repeat the algorithm.

$ P_{n \plus{} 1}$ is in the same initial partition as $ P_n$: in this case, we must switch $ P_{n \plus{} 1}$ with a countryman in another partition. The only restriction we have is that this countryman he switches with cannot be the other neighbor of $ P_n$ $ (P_{n \minus{} 1})$ or the neighbor of $ P_{n\plus{}1}$, which would be $ P_{n\plus{}2}$. However this will always leave at least one option to switch.

We make our way around the circle from $ P_1$ to $ P_{100}$ making the switches when needed. Because this method of partition separates the countrymen so that no two are in the same partition, the first condition is met. And because the algorithm also ensures that for every $ P_n$, $ P_{n \plus{} 1}$ is in a different partition, we are done.

Is this correct to any degree, or am I just going around in circles?

I see no mistake here. You tackle such a hard pro with a both rigorous and simple-minded approach. :police:

Congrats :omighty:

[quote="holmes.roams"]
$ P_{n \plus{} 1}$ is in the same initial partition as $ P_n$: in this case, we must switch $ P_{n \plus{} 1}$ with a countryman in another partition. The only restriction we have is that this countryman he switches with cannot be the other neighbor of $ P_n$ $ (P_{n \minus{} 1})$ or the neighbor of $ P_{n \plus{} 1}$, which would be $ P_{n \plus{} 2}$. However this will always leave at least one option to switch.[/quote]

Suppose you switch Pn+1 with, says, Pm, where m is not n or n+2. There is possibility that m<n, right? If so, I think it might destroy the structure you maintain previously, because now the new Pm may be in a same partition with Pm+1 or Pm-1.

[quote="filosofimenos"]Hmm, what your saying is right, i didn't notice. However, in this case you have again the same number of pairs to "decompose"(know what i am telling). So what should now be proved is that after finite rearrengements of the type holmes.roams mentions the situation that you mention doesn't occur. Is this obvious?[/quote]

Maybe you will get even more pairs to decompose, because Pm can be in the same partition with BOTH Pm-1 and Pm+1.

And there doesn't seem to be anything immediately visible that could lead to a fixing of the proof. So how would you go about an elementary solution to this? IS there one?

[quote="filosofimenos"]Your right greentreeroad. Yea, i would like to see an elementary proof too, without considering the knowledge of the other problem. Do we have a link of the official solution?

Edit: I have an idea but it seems impractical. We could give a counting approach. It's like we have a table of 25 rows(countries) and 25 rows(groups). What we want to do is to fill each row of the table with the four representatives of each country, in a way that in no column there exist 2 neighbours. Its easy to show that the number of ways to feel the table is (4*3*2)^25. What we shall show now is that the maximum number of ways to fill the table such that there exists at least one pair of neighbours to at least one column is less than (4*3*2)^25. Can you count this number???[/quote]

Here is the original problem: http://www.mathlinks.ro/Forum/viewtopic.php?p=220234#220234

In Russia MO they often gives these kind of paired problems(maybe one in lower grade and the other in higher grade). And often to solve the harder one you need the knowledge of the easier one.(this is why the harder ones are always very hard because you need to "guess" out the easier problem).

I think the combination of the solution in the link and the solution in this post is elementary enough ,AND BEAUTIFUL!


[quote="mhchow"]$ \frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+....+\frac{1}{19\cdot 21}=$ What is the answer?? [/quote]

Anyway, I don't think that one should be intelligent to solve this. It's enough to apply a trick

[hide]$ \boxed{\frac{1}{n(n+2)}= \frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)}$

Applying the above formula we have:

$ \frac{1}{3\cdot 5}+....+\frac{1}{19\cdot 21}= \frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right) =$

$ = \frac{1}{2}\left(\frac{1}{3}-\frac{1}{21}\right) = \frac{1}{2}\cdot \frac{1}{3}\left(1-\frac{1}{7}\right) = \frac{1}{2}\cdot \frac{1}{3}\cdot\frac{6}{7}= \frac{1}{7}$[/hide]

Yeah... that is also how i worked it out... Is there another way though?


I am just starting to use Latexit (mac version) and I want to use symbols for math such as union, intersection, element of and so forth...

is there any resourse where i may find how to typeset most math symbols?

Just click on "Latex Help" in front of the page ! :roll:


In a table $ 100X100$ some fields are coloured.It is known that every coloured field is the only one coloured or in its row ,or in its column.Find the maximum number of coloured fields.

Actually,I think the answer is $ 198$,but how to prove?

Observe maximal coloring. In that coloring, it isn't posible that every column has exactly one colored cell (because, if so, then there are 100 colored cells, but as you saw, you can color 198 of them). The same reason, you can't have exactly one colored cell in every row. Now, you make these moves:

You pick one colored cell, and color it's row if it is only only one colored in its row, or column for same reason. If you color one row or column, you are sure that it wasn't more then one already colored cell in it. So you can color at most 99 rows and 99  columns, hence you have at most 198 colored cells.
It;s easy to find an example for 198

bye


inequalities
real analysis
real analysis unsolved

Prove that if $(a_n)_n$ is a decreasing sequence of positive reals s.t. $\sum a_n<\infty$, then $\sum\left|(n+1)a_{n+1}-na_n\right|<\infty$.

This seems too simple. By modulus inequality and the fact that $\sum{a_n}$ converges, it suffices to prove that $\sum{|na_{n+1}-na_n}|$ converges. But the last sum is $\sum{n(a_n-a_{n+1})}$ and the convergence is clear (just write the general term).


trigonometry
function
algebra
domain
inequalities
inequalities proposed

Let $x, y \in R$ such that : $x^2+y^2 =1$ . Find the minimum value of the expression :
  $x\sqrt{1+y}+y\sqrt{1+x}$

  I guess it's $-1$ ?

Let $x=\cos 2a=\sin (90-2a)$ and $y=\sin 2a=\cos(90-2a)$, where a is any real number.

We have,
$x\sqrt{1+y}+y\sqrt{1+x}=\sin (90-2a)\sqrt{1+\cos(90-2a)}+\sin 2a \sqrt{1+\cos2a}$
$\geq -\sin (90-2a)\sqrt{2}\cos(45-a)-\sin 2a \sqrt{2}\cos a$ (since $\pm \sqrt{\dfrac{\cos 2a +1}{2}}=\cos a$)
$=-\sqrt{2}(\sin (90-2a) \ cos(45-a)+\sin 2a \cos a)$
$=-\dfrac{\sqrt{2}}{2}(\sin (135-3a)+\sin (45-a)+\sin 3a +\sin a)$
$=-\dfrac{\sqrt{2}}{2}(\sin (135-3a)+\sin 3a + \sin (45-a)+\sin a)$
$=-\sqrt{2}(\sin (135/2)\cos (135/2-3a) + \sin (45/2)\cos (45/2-a))$

Now $135/2-3a$ is 3 times of $45/2-a$, which means $\cos (135/2-3a)$ can be written in terms of $\cos (45/2-a)$.
Let x be $\cos (45/2-a)$, then we get the whole thing in terms of x. All we need to do is to find the minimum of the third-degree function in domain of $-1 \geq x \geq 1$.

This is too complicated. There must be a better solution. :(

Your "solution" looks very terrible . Try other ways !

Because of $x^2 + y^2=1$ we know that there exist such $\alpha$ such that $x=sin \alpha$ and $y=cos \alpha$. Then:
$x\sqrt{y+1} + y\sqrt{x+1}=\sin \alpha \sqrt{cos \alpha + 1} + \cos \alpha \sqrt{sin \alpha + 1}$
We will show that $\sqrt{cos \alpha + 1} \geq cos \alpha$
If $cos \alpha \geq 0$ then the LHS is $\geq 1$ and the RHS is $\leq 1$. If $cos \alpha < 0$ then the LHS is positive, and the RHS is not.
Similary, we prove that $\sqrt{sin \alpha + 1} \geq sin \alpha$
Using these inequalities we get:
$\sin \alpha \sqrt{cos \alpha + 1} + \cos \alpha \sqrt{sin \alpha + 1} \geq \sin \alpha \cos \alpha + \cos \alpha \sin \alpha = 2sin \alpha cos \alpha= sin 2\alpha \geq -1$
Equality holds for $\alpha=135^\circ$.

If $\sin \alpha$ or $\cos \alpha <0$ , you can't get :
  $\sin \alpha \sqrt{cos \alpha + 1} + \cos \alpha \sqrt{sin \alpha + 1} \geq \sin \alpha \cos \alpha + \cos \alpha \sin \alpha = 2sin \alpha cos \alpha= sin 2\alpha \geq -1$


Let c1={(x,y|x^2+y^2=a^2}and c2=(x,y|x^2+y^2=b^2}find abijection between  c1 and c2   ????????

I don't really see how this would belong to Number Theory :S

Anyways just draw the two circles on a plane, and draw any straight line through the origin. This will intersect both circles in a bijective manner ;)


Let $a,b,c$ be positive numbers. Prove that
(1)     $\sum \frac{a^2-bc}{\sqrt{a^2+bc}} \geq 0$;
(2)     $\sum (a^2-bc)\sqrt{a^2+4bc} \geq 0$.

Vacs,I have solved the problem a) by use SOS theorem.
Here are more nice inequalities I found from your inequality:
a)Give $k \geq \frac{2}{3}$ and $a,b,c >0$.Prove that:
           $\sum \frac{a^2-bc}{\sqrt{a^2+kbc}} \geq 0$
b)$n \in N^{*} \geq 2$ and $a,b,c >0$.Prove that:
            $\sum \frac{a^2-bc}{\sqrt[n]{a^2+bc}} \geq 0$
I will post the proof to these inequalities later because I don't have much time now. :)  :)

[quote="nhat"]
a)Give $k \geq \frac{2}{3}$ and $a,b,c >0$.Prove that:
           $\sum \frac{a^2-bc}{\sqrt{a^2+kbc}} \geq 0$
[/quote]I have a transformation:
Let$D_a=\sqrt{a^2+kbc}$
$2\sum\frac{a^2-bc}{\sqrt{a^2+kbc}}=\sum(a-b)(\frac{a+c}{D_a}-\frac{b+c}{D_b})\newline=\frac{c(a-b)^2.M_c}{D_aD_b((a+c)D_b+(b+c)D_a)}$
$M_c=k(a^2+b^2+c^2)+(k-2)ab+(2k-1)c(a+b)\ge(3k-2)ab+kc^2+(2k-1)c(a+b)\ge 0$
 :)  :)

[quote="nthd"][quote="nhat"]
a)Give $k \geq \frac{2}{3}$ and $a,b,c >0$.Prove that:
           $2\sum \frac{a^2-bc}{\sqrt{a^2+kbc}} \geq 0$
[/quote]I have a transformation:
Let$D_a=\sqrt{a^2+kbc}$
$\sum\frac{a^2-bc}{\sqrt{a^2+kbc}}=\sum(a-b)(\frac{a+c}{D_a}-\frac{b+c}{D_b})\newline=\sum\frac{c(a-b)^2.M_c}{D_aD_b((a+c)D_b+(b+c)D_a)}$
$M_c=k(a^2+b^2+c^2)+(k-2)ab+(2k-1)c(a+b)\ge(3k-2)ab+kc^2+(2k-1)c(a+b)\ge 0$
 :)  :)[/quote]

Your solution is very nice.Here is more general :
  Give $k \geq \frac{2}{3(n-1)}$;$n \in N^{*} \geq 2$;$a,b,c >0$.Prove that:
                      $\sum \frac{a^2-bc}{\sqrt[n]{a^2+kbc}} \geq 0$ :)  :)

[quote="Vasc"]Let $x,y,z$ be positive numbers. Prove that $\sum \frac{x^{2}-yz}{\sqrt{x^{2}+yz}}\geq 0$.[/quote]
\[\sum\frac{x^{2}-yz}{\sqrt{x^{2}+yz}}\geq\sum\frac{3(2x^{2}-y^{2}-z^{2}-6yz+3zx+3xy)}{5\sqrt{2}(x+y+z)}=0. \]
[hide="Remark."][quote="nhat"]Give $k \geq \frac{2}{3}$ and $x,y,z>0$. Prove that $\sum\frac{x^{2}-yz}{\sqrt{x^{2}+kyz}}\geq 0$.[/quote]
For positive numbers $x,y,z$, the following inequality
\[\sum\frac{x^{2}-yz}{\sqrt{x^{2}+kyz}}\geq 0 \]
holds if and only if $k\geq\frac{4}{\sqrt{7}}-1=0.5118578920369089088580\cdots$; with equality if $x=\frac{2\sqrt{7}-1}{3}=1.4305008740430603936677\cdots,y=z=1,k=\frac{4}{\sqrt{7}}-1$.[/hide]

[quote="Vasc"]Let $x,y,z$ be positive numbers. Prove that $\sum(x^{2}-yz)\sqrt{x^{2}+4yz}\geq0$.[/quote]
$\sum(x^{2}-yz)\sqrt{x^{2}+4yz}\geq\sum\frac{2x^{4}-y^{4}-z^{4}-6y^{2}z^{2}+3z^{2}x^{2}+3x^{2}y^{2}-2(5-3\sqrt{5})xyz(2x-y-z)}{2(x+y+z)}$

$=0$.

[hide="Remark."]For nonnegative numbers $x,y,z$, the following inequality
\[\sum(x^{2}-yz)\sqrt{x^{2}+kyz}\geq 0 \]
holds if and only if $0\leq k\leq4$; with equality if $x=0,y=z=1,k=4$.[/hide]


geometry
circumcircle
Euler
inequalities
triangle inequality
geometry unsolved

PROBLEM 1: $ \bigtriangleup ABC$. AD, BE, CF are three altitudes and H is the orthocenter. $ X\in XD$, $ Y\in BE$, $ Z\in CF$ such that $ \frac {XA}{AD} \plus{} \frac {YB}{BE} \plus{} \frac {ZC}{CF} \equal{} 1$. Prove that: HXYZ is cyclic
PROBLEM 2: Let a, b, c be three sides of a triangle. R and r are the radius of circumcircle and incircle. prove that
$ a \plus{} b \plus{} c < 4(r \plus{} R)$

i think that the sum of fractions in the first problem must be $ 2$,(sorry if i'm wrong)

at first note that the equation can be transformed to:
$ \frac{XA}{\frac{2S_{\Delta ABC}}{BC}}\plus{}\frac{YB}{\frac{2S_{\Delta ABC}}{AC}}\plus{}\frac{ZC}{\frac{2S_{\Delta ABC}}{AB}}\equal{} 2$

or in other words :

$ XA . BC \plus{} YB .AC \plus{} ZC .AB \equal{}  4S_{\Delta ABC}$

replace $ XA$ by $ AD\minus{}XD$ and similarly for $ YB,ZC$ to get

$ 2S_{\Delta ABC}\equal{}XD.BC\plus{}YE.AC\plus{}ZF.AB \equal{} 2S_{\Delta BXC}\plus{}2S_{\Delta ACY}\plus{}2S_{\Delta ABZ}$

$ S_{\Delta ABC}\equal{}S_{\Delta BXC}\plus{}S_{\Delta ACY}\plus{}S_{\Delta ABZ}$
now see [url=http://www.mathlinks.ro/viewtopic.php?p=1636376#1636376]here[/url]
the second problem must be well-known...

I have just edited the problem.
What about the problem 2. Can anyone give me the solution?

[quote] $ \bigtriangleup ABC.$ $ AD, BE, CF$ are three altitudes and $ H$ is the orthocenter. $ X\in XD,$ $ Y\in BE,$ $ Z\in CF$ such that $ \frac {XA}{AD} + \frac {YB}{BE} + \frac {ZC}{CF} = 2.$  Prove that $ HXYZ$ is cyclic[/quote]
Assume WLOG that $ Y$ lies on the ray $ HE.$ By Gergonne-Euler theorem for $ H,$ we have

$ \frac {AH}{AD} + \frac {BH}{BE} + \frac {CH}{CF} = 2 \Longrightarrow  \frac {AX + XH}{AD} + \frac {YB - YH}{BE} + \frac {CZ + ZH}{CF} = 2$

$ \frac {XH}{AD} + \frac {ZH}{CF} - \frac {YH}{BE} = 0 \Longrightarrow BC \cdot XH + AB \cdot ZH = AC \cdot YH \ (\star)$

Let the circle passing through $ H,X,Z$ cut the ray $ HE$ at $ Y'.$  We'll prove that $ Y \equiv Y'.$

Since $ \angle Y'XZ = \angle EHC = \angle BAC$  and $ \angle Y'ZX = \angle EHA = \angle ACB,$ we have that $ \triangle ABC \sim \triangle XY'Z$ and let $ k$ be their similarity coefficient. Thus

$ XY' \cdot HZ = k \cdot AB \cdot HZ$

$Y'Z \cdot HX = k \cdot BC \cdot HX$

$ Y'H \cdot XZ = k \cdot AC \cdot Y'H$

By Ptolemy's theorem for $ HXY'Z$  $\Longrightarrow$ $ k(AB \cdot HZ + BC \cdot HX) = k \cdot AC \cdot YH' $ $ \Longrightarrow$ $ AB \cdot HZ + BC \cdot HX = AC \cdot YH'.$

Due to $ (\star),$ we deduce that $ Y \equiv Y'$ and the proof is completed.

[quote="Yafmuzz"]Let $ a, b, c$ be three sides of a triangle. $ R$ and $ r$ are the radii of circumcircle and incircle, prove that $ a + b + c < 4(r + R)$[/quote]
Assume that $ \triangle ABC$ is acute and let $ H$ be the orthocenter of $ \triangle ABC.$ We use that the sum of the distances from $ H$ to the vertices of  the acute triangle $ \triangle ABC$ equals the sum of the diameters of its incircle and circumcircle

By triangle inequality for  $ \triangle HAB , \triangle HBC$ and  $ \triangle HCA,$ we obtain

$ HA + HB > AB \ , \ HB + HC > BC \ , \ HC + HA > AC$

$ \Longrightarrow 2(HA + HB + HC) = 4(R + r) > a + b + c.$

If $ \triangle ABC$ is right the inequality is trivial, but if $\triangle ABC$ is obtuse, then the argument fails.

[quote="Luis González"]
We'll use that the sum of the distances from $ H$ to the vertices of  the acute triangle $ \triangle ABC$ equals the sum of the diameters of its incircle and circumcircle[/quote]
Can you prove this?
The sum of the fractions in problem 1 is 1

The sum is 1.
My teacher has just given me the solution of problem 1.
Anyway, thanks for your solution.

sorry,it seems that i wasn't thinking right but...
but there is sth wrong again(correct me if i'm wrong!),
all the equations in my proof [url=http://www.mathlinks.ro/viewtopic.php?p=1636376#1636376]here[/url] are $ \Longleftrightarrow$ !!plz take a look at it!
which implies that if $ HXYZ$ is cyclic then the sum of the fractions must be $ 2$!

Here is my solution if the sum of the fractions is 1:
Let N be the point in the triangle such that $ NZ\perp CF$, $ NY\perp BE$, $ AN\cap BC\equiv L$, $ CN\cap AB\equiv M$, $ BN\cap CA\equiv K$.

We have: $ \frac {CZ}{CF} \equal{} \frac {CN}{CM}, \frac {BN}{BK} \equal{} \frac {BY}{BE}$ and notice that $ \frac {AN}{AL} \plus{} \frac {BN}{BK} \plus{} \frac {CN}{CM} \equal{} 1 \equal{} \frac {XA}{AD} \plus{} \frac {YB}{BE} \plus{} \frac {ZC}{CF}$.

Hence $ \frac {AN}{AL} \equal{} \frac {AX}{AD}\Longrightarrow\frac {AN}{AL} \equal{} \frac {AX}{AD}\Longrightarrow NX\parallel BC\ \Longrightarrow NX\perp AD$.

 Thus $ \angle NYH \equal{} \angle NZH \equal{} \angle NXH \equal{} 90^{o}$ and $ HXYZ$ is cyclic.
If N is not in the triangle, may be my solution is wrong.
Can anyone show me how to post image in the message. :blush: I'm a new here.

[quote="Nam Luu"]$ \frac {AN}{AL} \plus{} \frac {BN}{BK} \plus{} \frac {CN}{CM} \equal{} 1 \equal{} \frac {XA}{AD} \plus{} \frac {YB}{BE} \plus{} \frac {ZC}{CF}$ [/quote]
This relation is not true, it actually should be

$ \frac {AN}{AL} \plus{} \frac {BN}{BK} \plus{} \frac {CN}{CM} \equal{} \frac {XA}{AD} \plus{} \frac {YB}{BE} \plus{} \frac {ZC}{CF} \equal{} 2.$


let the length of the rectangle be l and the width w. l*w=12, and the area of the triangle is 25. express as a sum of parts:
25=(5-w)*l/2+(10-l)*w/2+12. simplifying, 5l-wl+1ow-lw=26. substituting and simpliplying, 5l+10w=50, l+2w=10. l=10-2w, and (10-2w)*w=12, giving w=2 and l=6.

[quote="xcenario"]A rectangle is inscribed in a triangle with base $ b \equal{} 10$ and height $ h \equal{} 5$, two vertex's of the rectangle touch the base of the triangle, and the two others touch the other sides. If the area of the rectangle is $ 12$, what is the measure of its sides?[/quote]

Let the base of the triangle $ \triangle ABC$  be $ BC$ , vertex of the rectangle touching $ AB$ be $ P$, and $ P$ divides $ AB$ in ration $ x: (1\minus{}x)$, then $ \frac{13}{25}\equal{}\frac{[\triangle ABC]\minus{}Area of the rectangle}{[\triangle ABC]}\equal{}x^2\plus{}(1\minus{}x)^2$

Solve it we get $ x\equal{}\frac{3}{5}$ or $ \frac{2}{5}$

So height of the rectangle (let the base of the rectangle be on $ BC$) =$ 5 \times (\frac{3}{5}\equal{}3$ or $ \equal{}5 \times (\frac{2}{5}\equal{}2$

The two possible dimensions are $ 3 \times 4$ or $ 2 \times 6$.

Let the base of $ \triangle ABC$ is $ BC$, and the sides of the rectangle which lie on $ BC$ is $ QR$ then $ P$ is on $ AB$ and $ S$ is on $ AC$

Since $ \triangle APS\sim\triangle ABC$, we have
$ \frac{5\minus{}PQ}{5}\equal{}\frac{QR}{10}$
$ 10\minus{}2PQ\equal{}QR$
From the problem, we also have $ PQ.QR\equal{}12$
Solving both equations, we get $ PQ\equal{}2,QR\equal{}6$ and $ PQ\equal{}3,QR\equal{}4$

So the dimentions of the rectangle are $ 2\times 6$ and $ 3\times 4$

[quote="lowtopology"][quote="stephencheng"]Let the base of the triangle  be  , vertex of the rectangle touching  be , and  divides  in ration , then $ \frac {13}{25} \equal{} \frac {[\triangle ABC] \minus{} Area of the rectangle}{[\triangle ABC]} \equal{} x^{2} \plus{} (1 \minus{} x)^{2}$[\quote]

How did you get that?  I just need a little bit of clarification... :blush:[/quote][/quote]

[geogebra]f339cc5df5a7df101586e3feb279a6e5a09e95eb[/geogebra]

We stick $ \triangle EFC$ to $ \triangle DGB$ so that $ D\equal{}E$, $ G\equal{}F$,

then $ \triangle B(D,E)C \sim \triangle DAE \sim \triangle BAC$ and the rest you should know.


If all the edges of a cube are replaced with $ 10 \Omega$ resistors and points X and Y are opposite vertices (i.e. those on the long diagonal of a cube), what is the equivalent resistance between X and Y?

How do you solve the problem in general if the resistances are all different?

You have $ 3$ resistors in parallel (call it $ A$), $ 6$ resistors in parallel (call it $ B$) and $ 3$ resistors in parallel (call it $ C$). The equivalent circuit is to have $ A,B,C$ in series (sorry I can't post a diagram).

[hide="Answer"] $ \frac {25}{3} \Omega$[/hide]

I do not think you can do this without using symmetry in current distribution. At every junction currents split symmetrically Then use the fact that Potential difference across any path from $ X$ to $ Y$ is constant and Ohm's law.
[hide="Answer"]$ R_{eq}\equal{}\frac{5}{6}\Omega$[/hide]

[quote="Carcul"]I don't think the answer will be so straighforward, BanishedTraitor. Can you please tell me where do you have 3 resistors in parallel, 6 resistors in parallel, and 3 resistors in series in the cube?[/quote]

Really? I am quite sure my reasoning is correct (it is a bit hard to explain without a diagram). But here it goes:

First draw a cube and label the top square from its top left hand corner in a clockwise fashion $ E,C,F,D$ and label the bottom square $ G,H,B,A$. Say each resistor has a resistance of $ r$ and we seek the equivalent resistance when the current flows from $ A$ to $ C$.

We see that the points $ B,G,D$ have the same electric potential and the points $ E,F,G$ have the same electric potential. That how I got my answer and calculated the $ R_{eq} \equal{} \frac {5}{6} r$.

Edit: I made a mistake in my original post it should read "and $ 3$ resistors in 'parallel' (call it $ C$)."

Yes the diagram would really help =P BanishedTraitor, don't you mean that $ E$, $ F$ and $ H$ have the same electric potential? And do you apply Kirchoff's law? I find it pretty hard to redraw it as a 2D circuit.

BanishedTraitor, you do not have in this circuit resistors in parallel or in series.

[quote="Aidan"] I find it pretty hard to redraw it as a 2D circuit.[/quote]

That's natural, since this is a non-planar circuit. In my humble opinion, the most suitable technique to be used in circuits like this is nodal analysis.

BanishedTraitor means to say that:
since B,G,D have same potential, the 3 resistors in the legs AB,AG,AD can be taken as if they were in parallel.
Similarly, since E,F,H are again having same potential, the legs DE,GE,DF,BF,BH,GH can be taken as if they were in parallel.
Finally, the three legs that remain which are also in parallel are: EC,FC,HC.
You can draw it all now..
Finally ofcourse, $ R_{A,C}\equal{} \frac{R}{3} \plus{} \frac{R}{6} \plus{} \frac{R}{3} \equal{} \frac{5R}{6}$

[quote="aidan"]How do you solve the problem in general if the resistances are all different?[/quote]

May be using equivalence of delta and wye connected resistances, or, least Kirchoffs laws, but that could become quite tedious...

Why not starting with that one:
http://www.mathlinks.ro/Forum/viewtopic.php?t=168768

There are other interesting problems [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=166292]here[/url].


Given 3-digits number by $ \overline {abc}$. Give $ \left( {\overline {abc} } \right)^2  \equal{} 1000x \plus{} y$ $ x,y\in Z^\plus{}$
Find all possible $ \overline {abc}$ such that $ x\plus{}y$ is a perfect square number.
For example:
$ 956^2\equal{}913936$ ,$ 913\plus{}936\equal{}1849\equal{}43^2$


Is the application fee also due tomorrow (May 1st)?

There is only an application fee if you are not applying online.  If you are applying online, then there is no fee.

If you are applying by mail, then please enclose a check for the fee with your application.


symmetry
inequalities proposed
inequalities

$ x+y+z=1$. Find max $ \sum \frac{x}{1+x^2}$

See here: http://www.mathlinks.ro/Forum/viewtopic.php?t=27450 and http://www.mathlinks.ro/Forum/viewtopic.php?t=1831 . By the way, do you have non-casebash a solution for real x, y, z ? Or do you understand Hatucdao's solution?

  Darij

How about the following method?  By symmetry, we may assume that $x \le y \le z$.  We split into three cases, depending on the smallest number.

[b]Case 1: $x \le -3$:[/b]  In this case, we have
\[
  z \ge \frac{y + z}{2} = \frac{1 - x}{2} \ge 2.
\]
Thus we have
\begin{eqnarray*}
\frac{x}{x^2 + 1} & < & 0 \\
\frac{y}{y^2 + 1} & \le & \frac{1}{2} \\
\frac{z}{z^2 + 1} & \le & \frac{2}{5}.
\end{eqnarray*}
Adding gives the desired inequality.

[b]Case 2: $-3 \le x \le -\frac{1}{3}$:[/b]  In this case, we have
\begin{eqnarray*}
\frac{x}{x^2 + 1} & \le & -\frac{3}{10} \\
\frac{y}{y^2 + 1} & \le & \frac{1}{2} \\
\frac{z}{z^2 + 1} & \le & \frac{1}{2}.
\end{eqnarray*}
Adding gives the desired inequality.

[b]Case 3: $x \ge -\frac{1}{3}$:[/b]  This case was already solved by Arne and others, by bounding $\frac{t}{t^2 + 1}$ by a linear function.

[quote="Ravi B"]We split into three cases, depending on the smallest number.
[/quote]

Nice. I don't think it's possible to solve the problem without having to distinguish some cases... *sigh*


1. You have a square piece of cardboard that has a perimeter of 12 centimeters. What is the area of the cardboard?

john lai, such problems are much more suited for Middle School forum, and that's where you should post them.

This should be in Classroom Math.

[hide]Side length is 3, so area is 9.[/hide]


I just need help on this because way my book has doesn't make sense to me. :?

Problem was:

Find value of $x$ in $6 \cos^2 x - 13 \cos x + 6 = 0$.

My answer:

$(2 \cos x - 3)(3 \cos x -2) = 0$

And $x = \arccos \frac {3}{2}$ and $x = \arccos \frac {2}{3}$ except on adding $+2n\pi$ to both answer.

But when I put $\arccos \frac {3}{2}$ in the calculator, it's DOMAIN. But the answer in the book has 5.4421.

How?

Same thing for the equation where I have $y = \arcsin -5$ and $y = \arcsin 4$

Thank you.

The range of a sin or cosine function is from 0 to 1, thus the domain of a arcsin or arccos fuction is from 0 to 1. That is why for arccos 3/2, you get a domain error. Same for arcsin-5 and arsicn 4. They are simply not defined out of the domain. I'm guessing the book's answer referred to arccos 2/3.

Edited

What is the difference between $\text {arcos}$ and $\text {arccos}$ then?

You said:

[quote]I'm guessing the book's answer referred to [b]arcos 3/2[/b][/quote]

There is no such thing as arcos...he meant arccos.  And I think he meant arccos 2/3 on the last line but I'm not sure.

[quote="deej21"]This is weird... the 5.4421 you said the book had, it works if that is in radians.  However since that is greater than $2\pi$ it is the same as -.84108 radians.  When I graphed the equation and set it equal to zero, I got solutions +- .84108, and plugging that back in, they both work.
For one, I don't know why they'd have a solution in radians that is greater than $2\pi$.  But your solution works still.
Of course, you can't do $arccos \frac 3 2$ so that gives us no solution, but the other one, $arccos \frac 2 3$ does work.  In radians, the solution is, you guessed it, .84108.  But since cos functions are positive in the 1st and 4th quadrant, $\cos x = \cos -x$, and therefore -.84108 is a solution also.[/quote]

$5.4421 < 2\pi$ btw...

If you get something like $2\cos{x}-3=0$ and you want to solve for $x$ you can't because the range of $\cos{x}$ is $-1 \le \cos{x} \le 1 \Rightarrow 2\cos{x}-3 \le -1$. It's like solving $x^2+1=0$ in the reals - there aren't any solutions.

Oh, another thing. Plugging $\arccos{\frac{2}{3}}$ into your calculator will give you only one solution. You can find the other one like deej21 said. So I'm assuming you plugged that in and got one of the solutions and wondered where the $5.4421$ came from - it's merely the other point within $0 \le x \le 2\pi$ that will give you $\cos{x}=\frac{2}{3}$.

The strangest thing just happened. I solved for arccos(2/3) in my calculator, took the negative, then added 2pi and got to the answer you reported, 5.4421.


trigonometry
function
rotation
geometry
parallelogram
ratio
similar triangles

How do you prove the angle angle similarity theorem for triangles?

AA implies AAA (hopefully you see why this is true). And if all of the angles are equal, then this is the definition of similar triangles.

I thought that the definition of similar triangles was that all three angles are congruent [b]and [/b]all three sides are proportional.  So prove that if all three angles of two triangles are congruent, then the sides are proportional.

I think you can use the Law of Signs to show this, although formally I'm not sure what the proof would be. Since the sin(A), sin(B), sin(C) will be constants in the formula when we consider it for AAA triangles, then the only way to maintain the equality of $ \frac{a}{\sin{A}} \equal{} \frac{b}{\sin{B}} \equal{} \frac{c}{\sin{C}}$ when one of a, b, or c is changed is to multiply each of the others by the factor that the side was changed by. In other words, all triangles with the same angles will be in the same proportion.

I don't no very much trigonometry, but I'm pretty sure that using sines in a proof is circular reasoning, because the trigonometric functions are based off of the AA similarity theorem.

But since the Law of Sines is true, the reasoning leading to it (i.e., AA similarity) must also be true. If AA similarity didn't exist, then the trigonometric laws wouldn't work.

Also, AA similarity is called a "postulate" or "axiom," meaning that its mathematical truth is considered to be self-evident.

[quote="KrazyNerd"]But since the Law of Sines is true, the reasoning leading to it (i.e., AA similarity) must also be true.[/quote]  This argument can be used to prove a really impressive collection of facts:
Ex:
Premise 1: Pink unicorns exist.
Premise 2: Pink unicorns, if they exist, cause the Earth to rotate around the Sun.
Conclusion 1: The Earth rotates around the Sun.

Since the conclusion is obviously correct, it follows that the reasoning leading to it must be valid as well ....

AA similarity is only called a postulate because some texts are too lazy to include its proof or too sloppy about naming statements.

Yeah but if 'AA ----> Law of Sines' is true, and 'Law of Sines' is true... isn't it inconclusive about the truth of AA? F -> T is a true statement though (in the mathematical logic sense)  :maybe: ... I would like to see a full geometric exposition ;)

Here's my best argument right now, but it is only valid when $ r$, the scale factor between the two similar triangles, is rational. (I suppose we could view the irrational case as the limiting case we we use a sequence of rational approximations)

Given: two triangles with corresponding angles congruent

First, let the larger triangle be $ ABC$.  Now lets superimpose the smaller triangle.  Let the angle corresponding to $ \angle A$ lie on $ A$ such that the opposing side is parallel to $ \overline{BC}$ (which is possible since $ \angle B$ is congruent to one of the angles of the smaller triangle).  Lets call the other two vertices of the smaller triangle $ D$ and $ E$, with $ D$ on $ \overline{AB}$ and $ E$ on $ \overline{AC}$ (which makes $ \angle ADE$ a right angle).

Now lets add in another triangle, $ \triangle AFG$, with $ \angle F \cong \angle B \cong \angle D$ and $ \angle G \cong \angle C \cong \angle E$ with $ F$ on $ \overline{AB}$ and $ G$ on $ \overline{AC}$ such that the length of $ \overline{AG}$ is the largest real number that there Iexist positive integers $ x$ and $ y$ such that $ x \cdot AG \equal{} AE$ and $ y \cdot AG \equal{} AC$ (these numbers only exist when the scale factor between $ \triangle ABC$ and $ \triangle ADE$ is rational)

Basically what we will do is superimpose $ \triangle AFG$ multiple times along the length of $ AC$.  Each time, move the triangle so that $ A$ now lies where $ G$ just was, until we get that $ G$ coincides with $ C$.  All the lines created by extending the various $ \overline{AF}$'s and $ \overline{FG}$'s are parallel, creating a huge grid of parallelolgrams - and opposite sides of parallelograms are congruent.  Since we know that some integer multiples of $ AG$ give $ AC$ and $ AE$, we now know that $ x \cdot AF \equal{} AD$, $ x \cdot FG \equal{} DE$, $ y \cdot AF \equal{} AB$, and $ y \cdot FG \equal{} BC$.  Now we have that $ \triangle ABC \sim \triangle AFG$ and $ \triangle AFG \sim \triangle ADE$; by transitivity, $ \triangle ABC \sim \triangle ADE$ as desired.

Picture is on the way :)

[quote="KrazyNerd"]
Also, AA similarity is called a "postulate" or "axiom," meaning that its mathematical truth is considered to be self-evident.[/quote]
The AA similarity can be proved as shown below.  You will, however, have to assume the truth of SAS congruence.
[hide="Solution"]
Let the triangles be $ ABC$ and $ DEF$ with $ \angle A \equal{} \angle D$, $ \angle B \equal{} \angle E$, and $ \angle F \equal{} \angle C$.  Let $ P$ and $ Q$ be on $ AB$ and $ AC$ so that $ AP \equal{} DE$ and $ AQ \equal{} DF$.  Then, since $ \angle PAQ \equal{} \angle DEF$, we have that $ \triangle PAQ\cong \triangle EDF$.  Thus, $ \angle APQ \equal{} \angle DEF \equal{} \angle ABC$, and so $ PQ\parallel AB$.  This entails that $ [AQB] \equal{} [APC]$.  Comparing side and area ratios, we see that $ \frac {AP}{AB} \equal{} \frac {[APC]}{[ABC]} \equal{} \frac {[AQB]}{[ABC]} \equal{} \frac {AQ}{AC}$.  Thus, $ \frac {DE}{AB} \equal{} \frac {DF}{AC}$.  Similarly, we have that $ \frac {DE}{AB} \equal{} \frac {DF}{AC} \equal{} \frac {EF}{BC}$, which concludes the proof.  [/hide]

EDIT: Sorry for the typo, [b]facis[/b]  :oops:

Quatto, how do you get $ \triangle PAQ \sim \triangle BAC$?

Nice proof.  I like your methodology, but I have 1 question: how do you get to $ EF/BC$ at the end?

Also it looks like you could use almost the exact same proof to prove SAS similarity.

[quote="facis"] how do you get to $ EF/BC$ at the end? [/quote]
To prove $ \frac {EF}{BC}$ at the end, you can construct other points, say $ X$ and $ Y$ on $ BA$ and $ BC$ respectively so that $ \triangle XBY\cong \triangle DEF$.  In the same fashion as before, and noting our previous result that $ \frac{DF}{AC}\equal{}\frac{DE}{AB}$, we can get that\[ \frac {DF}{AC} \equal{} \frac {DE}{AB} \equal{} \frac {XB}{AB} \equal{} \frac {BY}{BC} \equal{} \frac {EF}{BC}\]

Right.  I guess I just wanted to keep using P and Q instead of defining new points.

Anyway, along the same line of thinking, some more problems
-Prove SAS congruence
-Prove ASA congruence
-Anything else that is introduced as an axiom and shouldn't be


search
logarithms
calculus
calculus computations

Prove that the sequence $ (x_n)_{n \ge 1}$ given by: $ \left (1\plus{}\frac{1}{n} \right )^{n\plus{}x_n}\equal{}1\plus{}\frac{1}{1!}\plus{}\frac{1}{2!}\plus{}...\plus{}\frac{1}{n!}$ is convergent and find its limit.

As $ n\rightarrow\infty$, that sequence tends to $ e$, by definition.

It is true that:

$ \left (1\plus{}\frac{1}{n} \right )^{n} \rightarrow e$

$ 1\plus{}\frac{1}{1!}\plus{}\frac{1}{2!}\plus{}...\plus{}\frac{1}{n!} \rightarrow e$

But that's not enough to prove that $ x_n \rightarrow e$. Infact, in the solution from the book, the limit is $ \frac {1}{2}$. There, in my book, is given only the result.

Hi it's been posted before (with solution)
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=983987399&t=136676

Daniel

Thanks for the adress, but it's not the same problem, because the right member is a variable. There exist a solution with Stolz-Cesaro, because this theorem is mentioned in the "very short" solution from the book. I will read the replies from that topic because some of them are helpful.

Let $ \textstyle r_n \equal{} e \minus{} \sum_{k \equal{} 1}^{n} 1/k!$. From its rapid decaying behavior, for any $ \varepsilon > 0$, if $ n$ is large, we have $ 0 < r_n < e \varepsilon / n$.

Since

$ x_n \equal{} \frac {\log(e \minus{} r_n) \minus{} n \log \left( 1 \plus{} \frac {1}{n} \right) }{\log \left( 1 \plus{} \frac {1}{n} \right) },$

we have

$ \frac { 1 \plus{} \log \left( 1 \minus{} \frac {\varepsilon}{n} \right) \minus{} n \log \left( 1 \plus{} \frac {1}{n} \right) }{\log \left( 1 \plus{} \frac {1}{n} \right) } \leq x_n \leq \frac { 1 \minus{} n \log \left( 1 \plus{} \frac {1}{n} \right) }{\log \left( 1 \plus{} \frac {1}{n} \right) }.$

Taking $ n \to \infty$, we have $ \tfrac{1}{2} \minus{} \varepsilon \leq \liminf_{n\to\infty} x_n \leq \limsup_{n\to\infty} x_n \leq \tfrac{1}{2}$.

Since this is true for any $ \varepsilon > 0$, the limit of $ x_n$ must be equal to $ \textstyle \frac {1}{2}$.

I have to make a correction. I think that: $ \textstyle r_n \equal{} e \minus{} (1\plus{}\sum_{k \equal{} 1}^{n} 1/k!)$. Now, I have two questions:

$ 1.$ How did you deduce that for any $ \varepsilon >0$, if $ n$ is large, we have $ 0<r_n<e \varepsilon /n$ ?

$ 2.$ How did you obtain this relation: $ \tfrac{1}{2} \minus{} \varepsilon \leq \liminf_{n\to\infty} x_n \leq \limsup_{n\to\infty} x_n \leq \tfrac{1}{2}$?

The rest of the solution seems to be correct. Give more details about these two parts if you have time  :D

[quote="moldovan"]I have to make a correction. I think that: $ \textstyle r_n \equal{} e \minus{} (1 \plus{} \sum_{k \equal{} 1}^{n} 1/k!)$.[/quote]

Oh, you're right  :blush:

A1. It follows from a simple observation: $ 0 < r_n \equal{} \sum_{k \equal{} n \plus{} 1}^{\infty} \frac {1}{k!} < \sum_{k \equal{} 1}^{\infty} \frac {1}{(n \plus{} 1)^{k} \cdot n!} \equal{} \frac {1}{n \cdot n!}$

A2. One can do like this: substitute $ x \equal{} 1/n$ and consider the corresponding functional limit. Then apply L'hospital's rule or Taylor's theorem or anything you want. Personally, I prefer using Taylor's theorem combined with Little-Oh notation.

Here is what I deduced from the topic indicated by [b]lambruscokid[/b] (Daniel). I will use Taylor expansions.

[b][u]OBS[/u][/b]:$ ln x= log_e x$

$ x_n=\frac{ln \left side ( 1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!} \right side ) -n ln(1+\frac{1}{n})}{ln(1+\frac{1}{n})}$

$ \Rightarrow x_n=\frac{ln \left side ( 1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!} \right side ) -n (\frac{1}{n}-\frac{1}{2n^2}+o(\frac{1}{n^3}))}{\frac{1}{n}+o(\frac{1}{n^2})}$

Now, it is easy to deduce that the limit of this sequence is $ \frac{1}{2}$.


[quote="MUN"]Q1 Find probability of getting 3 consecutive sixes in 10 throws of a die.

Q2 A die is thrown again and again until we obtain 2 consecutive sixes are obtained .find the probabiltiy of getting 2 consecutive sixes[/quote]

[hide="1"]

THere are $ 8$ ways to get 3 consecutive $ 6$s, and the probability is:
\[ \dfrac{1^35^74}{6^{10}}
\]
...which I'm too lazy to calculate.[/hide]

Is number 2 like the coin paradox?

BTW: is my #1 right?

[b]EDIT:[/b]  Never mind, my #1 is wrong.  You could have 6, 6, 2, 3, 3,4 6, 6, 6....so you can have more than 3 sixes.

[quote="MUN"]Q1 Find probability of getting 3 consecutive sixes in 10 throws of a die.

Q2 A die is thrown again and again until we obtain 2 consecutive sixes are obtained .find the probabiltiy of getting 2 consecutive sixes[/quote]

I: err, i think it's 1/6^3 = 1/216.

II: 1/36