Find the unit vector perpendicular to both $ u\equal{}2i\plus{}6j\plus{}k$ and $ v\equal{}3i\plus{}k$.

Aren't there multiple vectors that satisfy? Or am I just missing something...?

[quote="buzzer11"]Find the unit vector perpendicular to both $ u \equal{} 2i \plus{} 6j \plus{} k$ and $ v \equal{} 3i \plus{} k$.

Aren't there multiple vectors that satisfy? Or am I just missing something...?[/quote]
There are two such vectors. One is the negative of the other. The question asks for the unit vector, i.e. the vector whose length is $ 1$. Further since it is perpendicular to both $ \mathbf{u}$ and $ \mathbf{v}$ it does not lie in the plane defined by those vectors.

[hide]
\[ \mathbf{u}\times\mathbf{v} \\ \equal{} \minus{}18k \plus{}3j \minus{}2j \plus{}6i \\ \equal{} 6i \plus{} j \minus{}18k\]

\[ \mathbf{\hat{n}} \equal{}\frac{\mathbf{u}\times\mathbf{v}}{\parallel{}\mathbf{u}\times\mathbf{v}\parallel{}} \equal{} \frac{6}{19}i \plus{}\frac{1}{19}j \minus{}\frac{18}{19}k\]
[/hide]

[quote="buzzer11"]Thank you. So when it asks for the normal vector, does it matter which one we put down? They are not equivalent vectors, but in the context of the problem, is it okay?[/quote]

One is the opposite of the other, i.e.

\[ \mathbf{u}\times\mathbf{v} \equal{} \minus{}\mathbf{v}\times\mathbf{u}\]

So, no, it doesn't really matter which one you write unless the problem explicitly asks you to compute the cross product in a particular order.

Okay, they didn't ask for it in any particular order, so I'll use $ \pm\mathbf{\hat n}$. Thank you!


integration
function
Support
real analysis
real analysis unsolved

Let $ \mu$ be $ \sigma$-finite and $ 1\leq p\leq\infty$ with $ q$ conjugate to $ p$. If $ fg\in L^1(\mu)$ for all $ f\in L^p$ prove that $ g\in L^q$.

Note: This is sort of a converse to the more common theorem (which I can prove).

So anything I try seems to land me into saying, well, that depends on whether or not $ g\in L^q$, which as I am trying to prove that it is, I can't say anything about it.

I tried letting $ \phi (f) \equal{} \int fgd\mu$. So now we have a linear functional on $ L^p$. If we had that it was bounded, then by theorem $ g\in L^q$, but if you try it directly or even try to get the functional to be continuous, you always end up with something bounded times $ \|g\|_q$ which we don't know about.

Edit: Woops.

[hide="Ignore this - settled above."]There are some quantifiers missing in your statement - there must be a "for all" somewhere that you didn't state. If it's just one $ f$ and one $ g,$ then the statement is false in silly ways.

For instance: on $ \mathbb{R},$ let $ f(x) \equal{} 1$ on $ [ \minus{} 1,0]$ and zero otherwise, and let $ g(x) \equal{} x^{ \minus{} 1/q}$ on $ (0,1]$ and zero otherwise.[/hide]

It seems easier to prove the contrapositive: Let $ \mu$ be $ \sigma$-finite and $ 1 \leq p \leq \infty$ with $ q$ conjugate to $ p$. If $ g \notin L^q$, then there is $ f \in L^p$ such that $ fg \notin L^1 (\mu)$.

[hide="My solution"]
If $ \mu \{ |g| \equal{} \infty \} > 0$, this is clear by choosing appropriate characteristic function $ f \equal{} \chi_E$. So we may assume $ g$ is finite a.e..

By assumption, there is a countable partition $ \{ X_i \}$ of $ X$ with finite measure.

[b]Case 1)[/b] Suppose $ p \equal{} 1$ and put $ E_n \equal{} \{ n\minus{}1 < |g| \leq n \}$. If $ g \notin L^{\infty}$ (that is, if $ \| g \|_{\infty} \equal{} \infty$), then $ \mu(E_n) > 0$ for infinitely many $ n$. Also, $ \mu(E_n) \equal{} \sum_{i\equal{}1}^{\infty} \mu(E_n \cap X_i)$ with each summand finite. These facts together shows that we can choose a disjoint countable subsets $ \{ A_i \}$ of $ X$ such that (1) $ \infty > \mu(A_i) > 0$ and (2) $ |g| > i$ on each $ A_i$.

Now put $ f \equal{} \sum_{i\equal{}1}^{\infty} \frac{1}{\mu(A_i) i^2} \chi_{A_i}.$ Since $ \int_X |f| \equal{} \sum_{i\equal{}1}^{\infty} \frac{1}{i^2} < \infty$, $ f \in L^1$.

But $ \int_X |fg| \geq \sum_{i\equal{}1}^{\infty} \frac{1}{i} \equal{} \infty$, so $ fg \notin L^1$.

[b]Case 2)[/b] Suppose $ 1 < p < \infty$ and $ g \notin L^q$. If we let $ E_{n,m} \equal{} \{ x \in X_m \, | \, n\minus{}1 < |g| \leq n \}$, then $ \textstyle \int_{E_{n,m}} |g|^q < \infty$ and $ \textstyle \sum_{n,m \equal{} 1}^{\infty} \int_{E_{n,m}} |g|^q \equal{} \int_{X} |g|^q \equal{} \infty$. Thus we may choose a disjoint countable subsets $ \{A_i \}$ of $ X$ such that $ \textstyle 1 < \int_{A_i} |g|^q < \infty$.

Put $ \textstyle a_i \equal{} \int_{A_i} |g|^q$ and $ \textstyle f \equal{}  |g|^{1/(p\minus{}1)} \sum_{i\equal{}1}^{\infty} (1/i) a_i^{\minus{}1/p} \chi_{A_i}$. Since $ \textstyle \int_{X} |f|^p \equal{} \sum_{i\equal{}1}^{\infty} 1/i^p < \infty$, $ f \in L^p$. (Here, we used the fact that $ q \equal{} \tfrac{p}{p\minus{}1} \equal{} \tfrac{1}{p\minus{}1} \plus{} 1$.)

But $ \textstyle \int_{X} |fg| \equal{} \sum_{i\equal{}1}^{\infty} (1/i) \, a_i^{1/q} \geq \sum_{i\equal{}1}^{\infty} (1/i) \equal{} \infty$

[b]Case 3)[/b] Suppose $ p \equal{} \infty$ and $ g \notin L^1$. Then as in Case 2, we may choose a disjoint countable subsets $ \{A_i \}$ of $ X$ such that $ \textstyle 1 < \int_{A_i} |g| < \infty$. Then put $ f$ the characteristic function of the union of $ A_i$. Clearly, $ \| f \|_{\infty} \equal{} 1$ but $ \| fg \|_{1} \equal{} \infty$.
[/hide]

In Case 1 I don't really see how the $ A_i$ were chosen. Let me try to flesh that out in my own way and see if it works.

Instead of defining $ E_n$ the way you did, let's just shift the indices: $ E_n\equal{}\{x: n<|g(x)|\leq n\plus{}1\}$. Then this doesn't change anything, so we get $ \mu(E_n)\equal{}\sum_{i\equal{}1}^\infty \mu(E_n\cap X_i)$.

Now we can construct $ A_k$ by saying that there exists some $ X_i$ such that $ \mu(E_k \cap X_i)>0$ since $ \mu(E_k)>0$. So explicitly pick $ A_k\equal{}E_k\cap X_i$. These are all clearly disjoint and by definition of $ E_k$ we have $ |g|>k$ on $ A_k$.

I guess my confusion came from the fact that if you do this, then the $ A_k$ can be sparse, i.e. $ \cup A_k \subsetneq X$. So when defining $ f$ do should we state that we want $ f$ on $ X\setminus \cup A_k$ to be $ 0$? Or maybe your construction of $ A_k$ was different?

My construction of $ A_k$ is based on the idea similar to yours. Here is an explicit construction of $ A_i$ which I thought:

Choose a subsequence $ n_k$ of $ ( 1, 2, 3, \cdots )$ so that $ n_k > k$ and $ \mu(E_{n_k}) > 0$ for all $ k$. (Note that all the $ \mu(E_n)$ need not be positive.) Then for each $ k$, there is at least one $ i_k$ such that $ \mu(E_{n_k} \cap X_{i_k}) > 0$. Put $ A_k \equal{} E_{n_k} \cap X_{i_k}$. Since $ n_k \geq k \plus{} 1$, $ |g| > k$ on $ A_k$.

As you can see, in general $ A_k$ does not exhuast all the space $ X$. But this does not matter. And... you don't have to worry about this fact. A characteristic function $ \chi_{E}$ of a measurable set $ E \subset X$ is defined on $ X$. That is, $ f$ is automatically defined to be identically zero outside $ \textstyle \bigcup_k A_k$. You don't have to mention that.

What will happen if the $ \sigma$-finiteness assumption is dropped?

That makes no difference. Note in particular sos440's comment that for his argument, he doesn't care whether his sets $ A_k$ exhaust the space or not.

There is one thing that $ \sigma$-finiteness matters for: we can't allow "atoms" with infinite measure. If $ S$ has infinite measure and every measurable subset of $ S$ has either measure zero or measure $ \infty$, we have the counterexample $ g\equal{}\chi_S$ (for $ 1\le p<\infty$), since for every $ L^p$ $ f$, the intersection of the support of $ f$ and $ S$ has measure zero.

In the following, suppose we avoid that and the space is a union of (possibly uncountably many) disjoint finite-measure sets.
The set on which any particular $ L^p$ function $ f$ is nonzero (for $ p\neq \infty$) is $ \sigma$-finite. Given any $ g$ with larger support, it's greater than $ \epsilon$ on some infinite measure subset $ S$, and this lets us construct $ f\in L^p\setminus L^1$ (unless $ p\equal{}1$) supported on $ S$ for which $ \int fg$ diverges.

So, in the $ 1<p\le\infty$ case, $ g$ is supported on a $ \sigma$-finite subset of $ \mu$  anyway. The $ p\equal{}1$ special case allows very large support for $ g;$ let's see how that case works.
Following the contrapositive path, we choose some $ g$ with $ |g|_\infty\equal{}\infty$. Define $ E_n\equal{}\{x: n\le |g(x)|< n\plus{}1\}$. The union of the $ E_n$ is the whole space, and the infinite $ \infty$-norm implies that $ \mu(E_{n(k)})>0$ for some increasing sequence $ n(k)$ (Start at $ k\equal{}1,n(1)\ge 1$). Define $ f\equal{}\sum_k \frac1{k\cdot n(k)\cdot \mu(E_{n(k)})}\chi_{E_{n(k)}}$. We get $ \int |f|\equal{}\sum \frac1{k\cdot n(k)}\le \sum\frac1{k^2}<2$ and $ \int |fg|\ge\sum \frac1k\equal{}\infty$.


[quote="BlAcK_CaT"]Let $ a$ and $ b$ be integers such that $ \frac {a}{b} \equal{} 1 \plus{} \frac12 \plus{} ... \plus{} \frac {1}{p \minus{} 1}$ and $ \left(a,b\right) \equal{} 1$, where $ p$ is a prime greater than $ 2$. Prove that $ p$ divides $ a$.[/quote]
$ 1 \plus{} \frac {1}{2} \plus{} ... \plus{} \frac {1}{p \minus{} 1} \equiv 1^{ \minus{} 1} \plus{} 2^{ \minus{} 1} \plus{} ... \plus{} (p \minus{} 1)^{ \minus{} 1}$ $ \equal{} 1 \plus{} 2 \plus{} ... \plus{} p \minus{} 1 \equal{} p \cdot \frac {p \minus{} 1}{2} \equiv 0 \bmod p$ (the last part only works when $ p \neq 2$) The proof builds on the fact that for every $ n$ there is a unique inverse $ n^{ \minus{} 1}$. So $ (1^{ \minus{} 1},2^{ \minus{} 1},...,(p \minus{} 1)^{ \minus{} 1})$ is a permutation of $ (1,2,...,p \minus{} 1)$

[quote="peine"]we have:
$ \frac {2a}{b} \equal{} \left(1 \plus{} \frac {1}{p \minus{} 1}\right) \plus{} \left(2 \plus{} \frac {1}{p \minus{} 2}\right) \plus{} ....\left(1 \plus{} \frac {1}{p \minus{} 1}\right)$
therefor $ \frac {2a}{b} \equal{} \frac {p}{p \minus{} 1} \plus{} \frac {2p}{p \minus{} 2} \plus{} .... \plus{} \frac {p}{p \minus{} 1} \equal{} p\left(\frac {1}{p \minus{} 1} \plus{} \frac {2}{p \minus{} 2} \plus{} .... \plus{} \frac {1}{p \minus{} 1}\right)$ then $ p|a$[/quote]

Sorry, i don't understand this steps :(
Can someone explain to me how to do it?

Peine's proof modulo typos:

[quote="peine"]we have:
$ \frac {2a}{b} \equal{} \left(\frac {1}{1} \plus{} \frac {1}{p \minus{} 1}\right) \plus{} \left(\frac {1}{2} \plus{} \frac {1}{p \minus{} 2}\right) \plus{} ....\left(\frac {1}{p \minus{} 1} \plus{} \frac {1}{1}\right)$
$ \equal{} \frac {p}{1\left(p \minus{} 1\right)} \plus{} \frac {p}{2\left(p \minus{} 2\right)} \plus{} .... \plus{} \frac {p}{\left(p \minus{} 1\right)1}$
$ \equal{} p\underbrace{\left(\frac {1}{1\left(p \minus{} 1\right)} \plus{} \frac {1}{2\left(p \minus{} 2\right)} \plus{} .... \plus{} \frac {1}{\left(p \minus{} 1\right)1}\right)}_{\text{this sum can be written as a fraction }x/y\text{ with }p\nmid y}$, thus $ p\mid 2a$, thus $ p\mid a$[/quote]

  darij


ABC triangle. Prove that:

 \sum (a 2 +b 2 )/mc \leq 12Rr

What's mc? If it's the median corresponding to side c, then we have a degree problem: the left term has deg 1 and the right one has deg 2. Maybe I'm wrong..

Yes mc is the median. Maybe I copied it wrong but I don't think so. I'll look again. Of course I'll have to find my materials first(which is a challenge :)  they're spread all over the house). Now seriosly I'll look again.

The problem is  \sum (a^2+b^2)/mc \leq 12R

We have the inequality (a 2 +b 2 )/(4mc) \leq R, by problem 11, Baltic Way contest, '98. By adding this inequality with the analogous ones we obtain the desired inequality.


In acute triangle $ ABC$, $ H$ is the orthocenter. The tangents from $ A$ to the circle with diameter $ BC$ touch the circle at $ M$ and $ N$. Prove that $ M$, $ H$, $ N$ are collinear.

Is this a well-known problem?  :(
Thanks  :)

Yes, it is.

http://www.mathlinks.ro/Forum/viewtopic.php?t=68425 (with some more links).

  darij

Yes, this problem is rather familiar. You should extend it.
Let a circle (O). From a point A outside circle, we draw two tangents AM,AN to (O) and two transversals AEC, AFB to (O) (B,C,E,F lie on (O)). BE meets CF at H. Then M,H,N are collinear.
Proof: It's easy to show that A,H are inverse wrt (O), so H lies on the polar MN of A wrt (O).

[hide="Another simple proof for your problem"]$ AM^2\equal{}AF.AB\equal{}AH.AD\Rightarrow\triangle AMH\sim\triangle ADM\Rightarrow\angle AHM\equal{}\angle AMD$
Analogously, $ \angle AHN\equal{}\angle AND, so \angle AHM\plus{}\angle AHN\equal{}\angle AMD\plus{}\angle AND\equal{}180^o$ (because A,M,D,N all lie on a circle whose diameter AO where O is the midpoint of BC)[/hide]


$a_1,a_2,...,a_n$are all positive real numbers.$a_1^{b_1}a_2^{b_2}...a_n^{b_n}$is a term of $(a_1+a_2+...+a_n)^m$.Find the sufficient and necessary condition of $b_1,b_2,b_3...b_n$,lead to the existance of an $k$,which satisfies the inequality $(a_1+a_2+...+a_n)^m \geq k\sum_{cyclic}{a_1^{b_1}a_2^{b_2}...a_n^{b_n}}$,which the equality satisfies when and only when $a_1=a_2=...=a_n$.

Still no one?
Is my traslation too bad?I make it by myself.

it is hard. i think. maybe we can do n=3 at first,

I don't think I can solve it by myself.I just want to have a reserch with everyboby here.
I think if we can solve it,we can save much time when we face symmetric inequalities.
And what about n=4?


geometry
circumcircle
trigonometry
function
inequalities
perimeter
cyclic quadrilateral

Prove or disprove that betwin all the inscribed quardiletarls the square has the maximum perimeter.

Please your reply, thanks.

If I correctly understand you, you want to see a proof of the following fact:

[color=blue][b]Theorem.[/b] Let R be a positive number. Among all cyclic quadrilaterals with circumradius R, the square has the greatest perimeter.[/color]

In this case, the [i]proof[/i] is easy: Let ABCD be a cyclic quadrilateral with circumradius R. Let O be the circumcenter of the quadrilateral ABCD, and let $\alpha=\frac12\measuredangle AOB$, $\beta=\frac12\measuredangle BOC$, $\gamma=\frac12\measuredangle COD$ and $\delta=\frac12\measuredangle DOA$ be the chordal angles of the sides AB, BC, CD and DA, respectively, in the circumcircle of the quadrilateral ABCD. Then, since the length of a chord in a circle equals $2\cdot\text{ radius of the circle }\cdot\text{ sine of the chordal angle of the chord}$, we have $AB=2R\sin\alpha$, $BC=2R\sin\beta$, $CD=2R\sin\gamma$ and $DA=2R\sin\delta$. On the other hand,

$\alpha+\beta+\gamma+\delta=\frac12\measuredangle AOB+\frac12\measuredangle BOC+\frac12\measuredangle COD+\frac12\measuredangle DOA$
$=\frac12\cdot\left(\measuredangle AOB+\measuredangle BOC+\measuredangle COD+\measuredangle DOA\right)=\frac12\cdot 360^{\circ}=180^{\circ}$.

As the angles $\alpha$, $\beta$, $\gamma$ and $\delta$ all lie in the interval [0°; 180°], and the function $f\left(x\right)=\sin x$ is strictly concave in this interval, the Jensen inequality yields

$\sin\alpha+\sin\beta+\sin\gamma+\sin\delta\leq 4\sin\frac{\alpha+\beta+\gamma+\delta}{4}=4\sin\frac{180^{\circ}}{4}=4\sin 45^{\circ}$,

with equality if and only if $\alpha=\beta=\gamma=\delta$, what is equivalent to saying that the quadrilateral ABCD is a square (in fact, $\alpha$, $\beta$, $\gamma$ and $\delta$ are the chordal angles of the sides AB, BC, CD and DA of the cyclic quadrilateral ABCD, and if these chordal angles are equal, the sides are also equal, but a cyclic quadrilateral with equal sides is a square). Hence, we get the following inequality for the perimeter AB + BC + CD + DA of the quadrilateral ABCD:

$AB+BC+CD+DA=2R\sin\alpha+2R\sin\beta+2R\sin\gamma+2R\sin\delta$
$=2R\cdot\left(\sin\alpha+\sin\beta+\sin\gamma+\sin\delta\right)\leq 2R\cdot4\sin 45^{\circ}$,

with equality if and only if the quadrilateral ABCD is a square. Thus, the perimeter of the quadrilateral ABCD is maximal if the quadrilateral ABCD is a square. This proves Theorem 1.

  Darij

[b]Lemma.[/b] Let $ABC$ be a isosceles triangle $(CA=CB)$ inscribed in the circle
$w=C(O,R)$, the point $D\in w\cap (CO$ and a mobile point $M\in w,$
$\ AB\cap MC\ne \emptyset$. Then the sum $MA+MB$ is at the most if only if $M: =D$.

[u]A old solution.[/u] I apply the Ptolemeu's theorem to the cyclic quadrilateral $MACB\ :$
$MA\cdot BC+MB\cdot AC=AB\cdot MC\Longrightarrow MA+MB=\frac{BA}{BC}\cdot MC$.
$MA+MB\ -\ max.\Longleftrightarrow MC\ -\ max.\Longleftrightarrow M: =D$.

[u]The solution of the proposed problem.[/u]
Let $ABCD$ be a convexe quadrilateral inscribed in the circle $w=C(O,R)$.
I note $D_1\in w,\ B_1\in w,\ B_1D_1\perp AC,\ O\in (B_1D_1),\ BD_1\cap AC\ne \emptyset$.
From the previous lemma result:
$AD+DC\le AD_1+D_1C,\ AB+BC\le AB_1+B_1C\Longrightarrow$
$AB+BC+CD+DA\le AB_1+B_1C+CD_1+D_1A$.
Thus, $D: =D_1,\ B: =B_1$ and for the quadrilateral $AB_1CD_1$
I will take again the process with the diagonal $B_1D_1$ and will obtain the points
$A_1\in w,\ C_1\in w,\ A_1C_1\perp B_1D_1,\ O\in (A_1C_1), AC_1\cap B_1D_1\ne\emptyset$.
In conclusion, the square $A_1B_1C_1D_1$ has the maximum perimeter.

[u]Remark[/u] (A conection with the Darij's solution).
$x=m(\widehat {ABM}),\ y=m(\widehat {BAM})\Longrightarrow m(\widehat {ADB})=m(\widehat {BDA})=\frac{x+y}{2}$.
$MA+MB\le 2\cdot DA\Longleftrightarrow \sin x+\sin y\le 2\sin \frac{x+y}{2}\Longleftrightarrow$
$2\sin \frac{x+y}{2}\cos \frac{x-y}{2}\le 2\sin \frac{x+y}{2}\Longleftrightarrow \cos \frac{x-y}{2}\le 1$; it is true.


Solve the equation
\[ \sqrt {x \plus{} \sqrt {4x \plus{} \sqrt {16x \plus{} \sqrt {\dotsc \plus{} \sqrt {4^{2008}x \plus{} 3}}}}} \minus{} \sqrt {x} \equal{} 1.
\]
Express your answer as a reduced fraction with the numerator and denominator written in their prime factorization.

[hide="Solution"]Add $ \sqrt {x}$ to both sides. Repeatedly square and simplify to see that the RHS follows the pattern $ \sqrt {x} \plus{} 1, \; 2 \sqrt {x} \plus{} 1 , \; 4 \sqrt {x} \plus{} 1, \ldots$ and realize that inductively, we will eventually get $ \sqrt {4^{2008} \plus{} 3} \equal{} 2^{2008} \sqrt {x} \plus{} 1$

Letting $ y \equal{} 2^{2008} \sqrt {x}$, we have $ \sqrt {y^2 \plus{} 3} \equal{} y \plus{} 1 \implies y^2 \plus{} 3 \equal{} y^2 \plus{} 2y \plus{} 1 \implies y \equal{} 1 \implies x \equal{} (2^{ \minus{} 2008})^2$

Thus, $ \boxed{x \equal{} \frac {1}{2^{4016}}}$

[b]Note:[/b] [i]This question was quite guessable if you realized that you were taking a couple thousand roots with rational numbers, which doesn't turn out pretty unless you start off clean.[/i][/hide]

On a side note, a recurrence related to this problem is: $ a_{n}^2 \minus{} a_{n \minus{} 1} \equal{} \frac {C}{4^n}$, $ a_0 \equal{} 3$. Does anyone know of a way to solve such a recurrence?


Hint:
[hide]Cauchy[/hide]

Yeah, it's a nice problem.  Here's another cool (and kind of similar) one:

Let $a,b,c$ be positive reals with product 1.  If $a^k+b^k+c^k \geq \frac{1}{a^k}+\frac{1}{b^k}+\frac{1}{c^k}$ holds for $k=1$, then prove that it holds for all $k>0$.

[quote="ThAzN1"]Let $a,b,c$ be positive reals with product 1.  If $a^k+b^k+c^k \geq \frac{1}{a^k}+\frac{1}{b^k}+\frac{1}{c^k}$ holds for $k=1$, then prove that it holds for all $k>0$.[/quote]
[i]Source:All-Russia Mathematical Olympiad 1999, grade 9, pr.3[/i]

[quote="blahblahblah"]Let $P(x)$ be a polynomial with positive coefficients.

Prove that if $P\left(\frac{1}{x}\right)\geq \frac {1}{P(x)}$ for $x=1$, then it holds for all $x>0$[/quote]

We had this one on http://www.mathlinks.ro/Forum/viewtopic.php?t=5443 ...

[quote="ThAzN1"]Let $a,b,c$ be positive reals with product 1.  If $a^k+b^k+c^k \geq \frac{1}{a^k}+\frac{1}{b^k}+\frac{1}{c^k}$ holds for $k=1$, then prove that it holds for all $k>0$.[/quote]

... and this on http://www.mathlinks.ro/Forum/viewtopic.php?t=18603 .

  darij


Let $ ABCD$ be a trapezium with $ AB\| CD$.Consider two circles with diameters $ AD$ and $ BC$.Let $ M,N$ be their intersection point and $ O\in AC\cap BD$.Prove that $ M,O,N$ are collinear.

It's a problem from Romanian JBMO TST . See here http://www.mathlinks.ro/viewtopic.php?p=818557#p818557


quadratics
linear algebra
matrix
geometry
geometric transformation
linear algebra unsolved

Prove tha tquadratic form is positive defined iff $\exists C: \: A=C^{\text{tra}}C$ where C is the matrix with non-zero determinant and A is the matrix of this quadratic form

[hide]I'm not sure about my translation of the terminology so here smth what may be u haven' understood[/hide]:
[hide]The quadratic form with the n unnouns is positive defined if its rank and signature is equals to n.
The quadratic form $f(x_1,...,x_n)$ with the n unnouns is positive defined iff $\forall \alpha_1,\alpha_2,...,\alpha_n: \: \alpha_1^2+...\alpha_n^2 \ne 0 s.t. f(\alpha_1,...,\alpha_n)>0$.[/hide]

unnoun?

we are working over the reals right?

Anyway : if there is such C, then

for all $x\neq 0 : C x\neq 0$ and thus $x^t A x=x^t C^t C x=(C x)^t(Cx)>0$

for the converse : there is a theorem stating that for every symmetric matrix A there are matrices such that $B^t B= I$ and $B^t A B= B^{-1} A B=D$ is diagonal matrix

realise now for yourself that $D$ has a strictly positive diagonal, and take a square root of that matrix E

$B E B^{-1}$ is the matrix C you seek

[quote="fredbel6"]Anyway : if there is such C, then

for all $x\neq 0 : C x\neq 0$ and thus $x^t A x=x^t C^t C x=(C x)^t(Cx)>0$[/quote]

what u mean by saying $Cx\ne 0$? the determinant?
the quadratic form is positive iff all the main minors are positive, but u have $(C x)^t(Cx)>0$
please write more details, as u know I'm  not familiar with many terms

I think I got the second part,Thanx a lot. :jump:

the first part is the easy part!
a quadratic form defined by A is positive definite if for all nonzero x : $x^t A x>0$

because $A=C^t C$ with C assumed invertible, we get that IF $x\neq 0\Longrightarrow : Cx\neq 0$ and thus

$x^t A x = x^t C^t C x = (C x)^t (C x) >0$

Thanx a lot fredbel6.
I understood it when I was in the english class doing math( :D  To be honest I should learn english more and not to solve the math problem during this lesson!)


$ \phi^{11} \equal{} \phi^{10} \plus{} \phi^9 \equal{} 2 \phi^9 \plus{} \phi^8 \equal{} 3 \phi^8 \plus{} 2 \phi^7 \equal{} 5 \phi^7 \plus{} 3 \phi^6 \equal{} 8 \phi^6 \plus{} 5 \phi^5$

$ \equal{} 13 \phi^5 \plus{} 8 \phi^4 \equal{} 21 \phi^4 \plus{} 13 \phi^3 \equal{} 34 \phi^3 \plus{} 21 \phi^2 \equal{} 55 \phi^2 \plus{} 34 \phi \equal{} 89 \phi \plus{} 55$,

or, much more simply by using the relation with Fibonacci numbers, $ \phi^n \equal{} F(n) \phi \plus{} F(n \minus{} 1)$, and looking for a list of the first few Fibonacci numbers,

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377

then we get immediatly $ \phi^{11} \equal{} 89 \phi \plus{} 55$.

Therefore, $ 2 \phi^{11} \equal{} 178 \phi \plus{} 110$.


[quote="Walk Around The River"][quote="chien than"]Solve the equation with integer roof:
$x^{2}+y^{3}+27-9y^{2}-36=0$[/quote]

for some reason, this is screaming "not classroom math"

or do i just not know what interger roof means...[/quote]
why? Because you can't solve it? :P
Just Joking. I agree with eli. This doesn't belong in classroom math. At the very least this should be in HSB. Can a mod move this?

could you please explain your "logic".
 so far i have gotten to $x^{2}+(y+3)(y-6)^{2}=14^{2}-7^{2}=147$

[quote="machack"]i believe he was refering to root when he said roof
the answer would be x=3 and y=9

just use logic to get this :P (guessing and check is for noooobs!!)[/quote]
Or y=0 for that matter..
Not to mention x=-3 or x=+3

[hide="The beginning"] We can rewrite as

$(x-3)(x+3) = y^{2}(9-y)$

There are only a few ways to match the factors.  Casework on $x \bmod 3$ should work. [/hide]


Hi,

I simply want to rename my bibligraphy, how can I do that ? I tried ;

\renewcommand{\bibliographyname}{x}

It didn't worked, don't know why.

Also, when I'm making a bibliography (automated) with an author, in the "author" section of my bib, I write "Gillespie, J.H.", in my bibliography he doesn't care about the "H", it'll always be "Gillespie, J. or J. Gillespie." Why ?

Thx

Well, I solved the first problem by modyfing the .sty of my bibliography. However, I still don't know why an author named "West, G.B." in my bib would always appear "West, G.".


function
limit
trigonometry
calculus
calculus computations

Discuss the continuity of $f$
$f(x)=\lim_{n \to \infty}\left\{ \begin{array}{ll}\cos^{2n}{x}&{ \text{if}\ \ x < 0}\\ \sqrt[n]{1+x^{n}}&{ \text{if}\ \ 0 \leq x \leq 1}\\ \frac{1}{1+x^{n}}&{ \text{if}\ \ x > 1}\end{array}\right.$

The condition $\sqrt[n]{1+x^{n}}$ produces a limit of 1 for all numbers less than 1, and the condition $\frac{1}{1+x^{n}}$ produces a value of 0 for all inputs greater than 1. The function cannot be continuous because if it were, the Intermediate Value Theorem would imply that the function has multiple outputs for the input 1, contradicting the definition of a function.

what about the continuity at point $x=0$? and for $x<0$?

For $x>1,\,\lim_{n\to\infty}\frac1{1+x^{n}}=0.$

For $0\le x\le 1,\ 1\le\sqrt[n]{1+x^{n}}\le\sqrt[n]{2}.$ Since $\lim_{n\to\infty}\sqrt[n]{2}=1,$ we then have that $\lim_{n\to\infty}\sqrt[n]{1+x^{n}}=1.$

For $x<0,\ f(x)=\lim_{n\to\infty}\cos^{2n}x$ equals $1$ whenver $x=k\pi$ for some negative integer $k$ and equals zero otherwise.

So in the end, $f$ is discontinuous at nonpositive integer multiples of $\pi$ and at $1$ and continuous everywhere else. $f$ takes on only the values $1$ and $0.$

Thanks, very nice! Now it's clear. I was wondering in indeterminate forms like $0^{\infty}$ and $\pm 1^{\infty}$ but now i understand it, the $2n$ in $\cos^{2n}x$ is not there for nothing... By the way, there are any other way to compute $\lim_{n \to \infty}\sqrt[n]{1+x^{n}}\ \ \text{for}\ 0\leq x \leq 1$ without squeeze theorem?

[quote="math_music"]By the way, there are any other way to compute $\lim_{n \to \infty}\sqrt[n]{1+x^{n}}\ \ \text{for}\ 0\leq x \leq 1$ without squeeze theorem?[/quote]

For $0\le x< 1,\ \lim_{n\to\infty}x^{n}=0$ so $1+x^{n}\to 1.$ Take the $n$th root and we have a "$1^{0}$" case. Think about that for a moment and you'll realize that it's not indeterminate. Then handle $x=1$ separately.


superior algebra
superior algebra unsolved

Show that if $ a \in \mathbb{Z}$ and $ x^p \minus{} a$ has a solution modulo $ q$ for all primes but a finite set of primes $ q$, then $ x^p \minus{} a$ has a solution in $ \mathbb{Z}$.


Say $a_1,a_2 ... a_n$ are integers, with at least some two of them being unequal. Prove we can find infinitely many primes $p$, s.t. there exists an integer $k$ with the number $a_1 ^ k + a_2 ^ k ... + a_n ^ k$ being divisible by $p$.

EDIT: Stupid post :) See the remark by ZetaX below.

Note that the case $n=2$ is a simple corrollary of a corollary of Zsigmondy theorem :)
Let $n = 2$. We let $f ( k ) = a_1^k + a_2^k$.
Suppose to the contrary that there is some $\alpha_i : \mathbb{N} \longrightarrow \mathbb{N}$ such that for all positive integer $k$
\begin{eqnarray*} a_1^k + a_2^k & = & \prod_{j = 1}^m p_j^{a_j \left( k \right)}\\ & & \end{eqnarray*}
With $m$ a positive integer and $p_1, \ldots, p_m$ primes. Then
\begin{eqnarray} \varphi ( f ( k ) ) & = & \prod_{a_j ( k ) \geq 1} \left( p_j - 1 \right) \cdot \prod_{\alpha_j ( k ) \geq 1} p_j^{\alpha_j ( k ) - 1} \nonumber\\ \varphi ( f ( k ) ) & | & \prod_{j = 1}^m ( p_j - 1 ) \cdot \prod_{\alpha_j ( k ) \geq 1} p_j^{\alpha_j ( k ) - 1} \end{eqnarray}
Homever by a corollary of Zsigmondy (see http://mathworld.wolfram.com/TotientFunction.html equation 18) theorem, we have for all sufficently large $k$
\begin{eqnarray*} k & | & \varphi ( a_1^k + a_2^k ) = \varphi ( f ( k ) ) \end{eqnarray*}
A contradiction with (1).

Why using corollaries of it when $n=2$ is just a weaker statement than Zsigmondy¿  :?

Yes, yes you are correct :) When writting the post I was aware only of the corrollaries of Zsigmondy theorem not of the theorem itself. My ignorace of Zsigmondy theorem itself, led me to write this (as follows) useless post...


calculus
derivative
LaTeX
function
logarithms
calculus computations

Hello , i had some problems which i thought you all should take a look at. They are nice , though ive not got 2 of them.
All based on calculus , could use a hand. Please bear the pain of downloading the simple .doc file as to view them , I am not much of a latex user , too complex :D.

In $ LaTeX$

[b]Q1: [/b] If $ 2x\equal{}y^{\frac{1}{5}}\plus{}y^{\minus{}\frac{1}{5}}$ then express $ y$ as an explicit function of $ x$ and prove:

$ (x^2\minus{}1)\frac{d^2 y}{dx^2}\plus{}x \frac{dy}{dx}\equal{}25y$

[b]Q2:[/b] Given $ y\equal{}1\plus{}\frac{a_1}{x\minus{}a_1}\plus{}\frac{a_2x}{(x\minus{}a_1)(x\minus{}a_2)}\plus{}\frac{a_3x}{(x\minus{}a_1)(x\minus{}a_2)(x\minus{}a_3)}\plus{}...$ to $ (n\plus{}1)$ terms.

Show that , $ \frac{dy}{dx}\equal{}\frac{y}{x}[n\minus{}\frac{x}{x\minus{}a_1}\minus{}\frac{x}{x\minus{}a_2}\minus{}...\minus{}\frac{x}{x\minus{}a_n}]$

[b]Q3:[/b] If $ x<1$, prove that :

$ \frac{1\minus{}2x}{1\minus{}x\plus{}x^2}\plus{}\frac{2x\minus{}4x^3}{1\minus{}x^2\plus{}x^4}\plus{}\frac{4x^3\minus{}8x^7}{1\minus{}x^4\plus{}x^8}\plus{}.......\equal{}\frac{1\plus{}2x}{1\plus{}x\plus{}x^2}$

[quote="stephencheng"]In $ LaTeX$

[b]Q1: [/b] If $ 2x \equal{} y^{\frac {1}{5}} \plus{} y^{ \minus{} \frac {1}{5}}$ then express $ y$ as an explicit function of $ x$ and prove:

$ (x^2 \minus{} 1)\frac {d^2 y}{dx^2} \plus{} x \frac {dy}{dx} \equal{} 25y$
[/quote]

[b]Q1: [/b] $ y\equal{}(x \pm \sqrt{x^2\minus{}1})^5$,  $ \frac{dy}{dx}\equal{}\pm \frac{5y}{\sqrt{x^2\minus{}1}}$,

$ (x^2\minus{}1) \frac{d^2y}{dx^2}\equal{}\pm \sqrt{x^2\minus{}1}(5)\frac{dy}{dx} \mp \frac{5xy}{\sqrt{x^2\minus{}1}}$

Now it's clear that $ L.H.S.\equal{}25y$

Your $ y$ in question (2) is stated wrongly. Instead, it should be

$ y\equal{}1\plus{}\frac{a_{1}}{x\minus{}a_{1}}\plus{}\frac{a_{2}x^{2}}{(x\minus{}a_{1})(x\minus{}a_{2})}\plus{}\frac{a_{3}x^{3}}{(x\minus{}a_{1})(x\minus{}a_{2})(x\minus{}a_{3})}\plus{}...\plus{}\frac{a_{n}x^{n}}{(x\minus{}a_{1})(x\minus{}a_{2})(x\minus{}a_{3})...(x\minus{}a_{n})}$

Then observe that

$ (x\minus{}a_{1})(x\minus{}a_{2})(x\minus{}a_{3})...(x\minus{}a_{n})y\equal{}x^{n}$

$ \ln (x\minus{}a_{1})(x\minus{}a_{2})(x\minus{}a_{3})...(x\minus{}a_{n})y\equal{}n\ln x$

Hence

$ \frac{dy}{dx}\equal{}\frac{y}{x}\left( n\minus{}\frac{x}{x\minus{}a_{1}}\plus{}\frac{x}{x\minus{}a_{2}}\plus{}...\plus{}\frac{x}{x\minus{}a_{n}}\right)$

Q3

Briefly, you can verify with proper limiting process that for $ |x| < 1,$

$ \left( x^{2} \plus{} x \plus{} 1\right) \left( x^{2} \minus{} x \plus{} 1\right) \left( x^{4} \minus{} x^{2} \plus{} 1\right) \left( x^{8} \minus{} x^{4} \plus{} 1\right) .... \equal{} 1$

Hence,

$ \ln \left( x^{2}\plus{}x\plus{}1\right) \equal{}\minus{}\ln \left( x^{2}\minus{}x\plus{}1\right) \minus{}\ln \left( x^{4}\minus{}x^{2}\plus{}1\right) \minus{}\ln \left( x^{8}\minus{}x^{4}\plus{}1\right) \minus{}.....$

Differentiating wrt $ x$ yields the result desired.


linear algebra
matrix
inequalities
induction

Let $ P$ be the determinant of the following real matrix:

$ \left[\begin{array}{cccc}1 & 1 & 1 & 1 \\
\sqrt {2} & \sqrt {3} & 2 & x \\
2 & 3 & 4 & x^2 \\
2\sqrt{2} & 3\sqrt{3} & 8 & x^3\end{array}\right]$

Solve the inequality $ P < 0$ in $ \mathbb{R}$

Correction: the exercise was typed incorrectly in the book, 4 and 9 are supposed to be $ 2\sqrt {2}$ and $ 3\sqrt {3}$ , and now I got it.

Sorry if this wasted anybody's time.

[color=green][Original post has now been edited. It is Vandermonde.][/color]

well i'd take a look at column 3 and start by making zeros in positions $ a_{13},a_{23},a_{33}$ and then expand by cofactors.

the same procedure can be applied so that you can reduce the matrix into a square one of order 2 which determinant is easy to compute; finally you need to solve an inequality.

[quote="Christiano89"]It is possible to somebody to explain how Vandermonde Determinant works?[/quote]

[url]http://mathworld.wolfram.com/VandermondeDeterminant.html[/url]

Oh I had forgotten mathworld. Many thanks b555

For those still wondering, the key to the Vandermonde determinant concept is that the determinant of a matrix with two equal rows or columns is 0. Keeping that in mind, a quick induction or perhaps even direct computation gives you an explicit formula.


In 1978 Peter Dowdeswell ate 13 raw eggs without shells in 2.2 seconds. How fast did he eat 1 egg?

My work:

13 eggs/1 egg = 2.2/x

I crossed multiplied and got:

13x = 2.2

x = 2.2/13

x = 0.2.

Am I right or wrong?

If wrong, why?

that's right.
it might be easier to think of it as

13 eggs per 2.2 seconds = 1 egg per 2.2/13 seconds.

Your method is correct because it states the number of eggs he eats is directly proportional to the time taken.

Here's another way to think of it:
$\frac{13\ \text{eggs}}{2.2\ \text{s}}= \frac{1\ \text{eggs}}{x\ \text{s}}$
This equation basically says that the rate of eggs/second is constant.

you are completely correct, except...

$\frac{2.2}{13}\ne{.2}$  im assuming you knew that and were just rounding but in case you didnt...

but good approach.  but once again, nothing against you, but these problems are better suited for classroom math.


1.Fie $A,B\in M_2(Z)$ cu $A^{-1}\in M_2(Z)$ si multimea $M=\{k\in Z| (A+kB)^{-1} \in M_2(Z)\}$.
Sa se arate ca $card M\neq 2006$.
2.Fie $A\in M_3(R)$ inversabila astfel incat $tr(A)=tr(A^2)=0$. Sa se arate ca $\det(A+A^{-1})=\det(A)+\det(A^{-1})$.
3.Fie sirurile $a_n,b_n,c_n$ definite prin:
$\sqrt{a_{n+1}+a_{n+1}b_n}+\sqrt{a_{n+1}+a_{n+1}c_n}=\sqrt{b_n+b_nc_n}+\sqrt{c_n+c_nb_n}$
$\sqrt{b_{n+1}+b_{n+1}a_n}+\sqrt{b_{n+1}+b_{n+1}c_n}=\sqrt{a_n+a_nc_n}+\sqrt{c_n+c_na_n}$
$\sqrt{c_{n+1}+c_{n+1}b_n}+\sqrt{c_{n+1}+c_{n+1}a_n}=\sqrt{b_n+b_na_n}+\sqrt{a_n+a_nb_n}$
oricare ar fi $n\in N$, where $a_0,b_0,c_0\in (0,\infty)$. Sa se arate ca:
  $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\lim_{n\to\infty}c_n=tg^2(\frac{arctg\sqrt{a_0}+arctg\sqrt{b_0}+arctg\sqrt{c_0}}{3}$.

prima si a doua sunt simple :P iar a treia e bestiala. sa imi spuna si mie cineva de unde pot sa iau LATEX ca sa nu ma chinui p-aici . 10x anticipat

:welcomeani: La a 3-a problema daca nu ar fi fost indicatia din concluzia ce trebuie demonstrata problema ar fi fost grea! ;)
$\LaTeX$ e un cod .... nu trebuie sa ti-l iei.... trebuie  sa il inveti!

aha ... am inteles ... deci nu e un editor sau ceva de genu nu ? e incorporat ... cum ar fi JavaScript or smthn ... pai a treia si cu indicatie tot mi se pare grea :) bine, eu sunt mai noob in alte matematicii :P am inceput si eu de anu asta sa mai misc cate ceva ... si nici aia bine . a doua se face cu polinom caracteristic ... iar la prima dupa mine e cam ambigua

La prima problema 2006 e pus asa.... ca suntem in 2006... eu am aratat ca maxim cardinalul poate fi 4. Cateva notiuni simple de valori proprii rezolva problema 2, iar la a treia datele problemei sugereaza notatia $\sqrt{a_0}=tg(x_0)$

aha, de ce sunt doar 4 valori? sa fi mai explicit la partea asta . in rest, am inteles ...  vad ca esti sau cel putin te dai meserias , de cand te duci la mate? pt tine toate sunt simple ?

Nu ma dau meserias... si nu prea sunt meserias. Nu sunt toate simple.
Pentru a usura munca voi nota $C_k=A+kB$ . E clar ca daca $k\in M$ atunci $\det(C_k)\in\{+1,-1\}$.
Mai departe am procedat prin reducere la absurd ... adica am presupus ca cardinalul ar fi 2006. Atunci pot sa impart $M$ in doua submultimi disjuncte $X,Y$ astfel $X$ contine doar $C_k$ cu proprietatea ca $\det(C_k)=1$ si $Y$ invers. Atunci cel putin una din multimi va avea cel putin $1003$ elemente, adica $\det(C_k)=1$ pentru cel putin 1003 valori distincte ale lui $k$ . Dar relatia este una polinomiala de grad 2 cu cel mult $2$ radacini distincte care in cazul nostru are cel putin 1003 radacini distincte deci polinomul $\det(C_x)-1$ este nul petru orice $x\in R$ in particular pentru toate numerele intregi deci cardinalul acelei multimi este infinit contradicite! .... de aici si ideea ca 5 ar fi cel mai mic pentru care se poate demonstra.

da .. f bine aplicata proprietatea asta la polinom ... pe cand se predau polinoamele stateam si invatam chimie. acuma imi pare rau, dar asta e viata domne ce sa-i faci . am mai vazut ceva asemanator la unele exercitii ... ideea de obicei e sa obtii un polinom constant ... si apoi particularizezi ca sa obtii concluzia dorita . oricum, well done  :idea: scoala buna la Suceava acolo, stiu si eu un baiat d-asta a fost la chimie... Mancas Alexandru, il stii ?


MATHCOUNTS
ratio
algebra
polynomial
geometry
analytic geometry

the answer is always "C" if Multiple Choice.
$ 42$ if the answer is numerical.

the probabliity of guessing a right (mathcounts) question is almost 1/infinity. How ever, there are some hints that can help you get a good guess (not necessarialy right)....
if its a ratio, try putting part:whole.. I did that 2day on a target and got it right.
remember not to waste too much time into guessing; if you dont know it, go to an easier problem.
Also, I've heard some problems have 0,1-,1 as answers

[quote="unimpossible"]How do you guys determine what answer to guess on a problem?[/quote]
I denote sides of my pencil as the answer choices and roll them.. :lol:

Well, first use what clues you have and then make the most reasonable guess. Very important for problems that require brute force guess and check.

For problems with complicated polynomials or fractions or rational polynomials, guess $ x$ or $ 1$ or $ 3$ or $ 3x$. I think they like the number $ 3$. Sometimes there's a chance that it's supposed to be $ 7$ if its some sort of counting problem.

Sometimes it uses the year as part of the answer, like maybe $ \frac{2007}{2008}$

on the lame ones, the answer is sometimes the number of the problem

[quote="andrewliu"][quote="Nerd_of_the_Ages"][quote="ProtestanT"]
Sometimes it uses the year as part of the answer, like maybe $ \frac {2007}{2008}$[/quote]
Lame as that sounds, it actually is sometimes true. I guess that some people are just really bored when they make questions and make it involved in a year.[/quote]
Yeah, like the 2007 chapter rounds.
One of htem was like
area of a figure on a coordinate plane... I think
yeah well the answer was 3008.... how lame....
the year+1000+1......
I think this was a target #8[/quote]
 I think it was #6, and further, that isn't too lame. I mean, if the answer is 5, you could say it's the year minus 2002.

Plenty of responses have been posted, and I don't think 'guessing skillz' will help anyone learn how to solve problems  :wink: , so I'm going to lock this topic.


How many quadrilaterals of any size are in the diagram?

[asy]unitsize(1cm);
for (int i=0; i<5; ++i) {
draw((i,0)--(i+1,0)--(i+0.6,0.8)--cycle,linewidth(0.7));
}
draw((0.6,0.8)--(4.6,0.8),linewidth(0.7));[/asy]

There are $ 8\plus{}7\plus{}6\plus{}5\plus{}4\plus{}3\plus{}2\plus{}1\equal{}\frac{8(8\plus{}1)}{2}\equal{}\fbox{36}$. Note that with each increasing degree of size, there is one less quadrilateral, thus the sequence done above.

Another way to do this problem is to find out how many ways we can create a quadrilateral.

Notice that a quadrilateral can be formed with two triangles, the smallest being two triangles  next to each other and the largest being the two triangles at the end of the trapezoid.  There are $9$  triangles, and we need to choose $2$ of them to create a quadrilateral; that makes a total of $\binom{9}{2}=\boxed{36}$ quadrilaterals of any size are in the diagram.


calculus
integration
function
derivative
complex analysis

I am not sure why, but I am having a hard time fully seeing what is going on with what follows, though I am sure it is analogous to something very simple that I am probably quite familiar with. This is from pages 481-483 of Whittaker and Watson's "A Course of Modern Analysis." They are integrating functions $ f$ and $ g$, which among other properties, satisfy $ f(\tau')\equal{}g(\tau)$, where $ \tau'\equal{}\minus{}1/\tau$ (and thus also $ g(\tau')\equal{}f(\tau)$). They then integrate over a path starting a point $ B$ and ending at a point $ C$, which transforms to a path starting at $ E'$ and ending at $ D'$ under the above change of variables (this much I understand). They then write...

As $ \tau$ describes $ BC$, $ \tau'$ describes $ E'D'$, so $ \int_{BC}\frac{1}{f(\tau)\minus{}c} \frac{df(\tau)}{d\tau}d\tau\equal{}\int_{BC}\frac{1}{g(\tau')\minus{}c} \frac{dg(\tau')}{d\tau}d\tau\equal{}\int_{E'D'}\frac{1}{g(\tau')\minus{}c} \frac{dg(\tau')}{d\tau'}d\tau'$.

It seems as though they simply applied the transformation law in the first step, but I am not clear on the second step, why it is valid to switch $ \tau$ to $ \tau'$ and only change the path, without worrying about $ g$.

Later on, they are considering a similar looking integral $ \int_{B'AB}\frac{1}{h(\tau)\minus{}c} \frac{dh(\tau)}{d\tau}d\tau$, where the function $ h(\tau)\equal{}\minus{}f(\tau)/g(\tau)$, which implies that $ h(\tau')\equal{}1/h(\tau)$. They then write...

As $ \tau'$ describes $ B'AB$, $ \tau$ describes $ EE'$, so the above becomes
$ \int_{EE'}\frac{1}{h(\tau')\minus{}c} \frac{dh(\tau')}{d\tau}d\tau\equal{}\minus{}\int_{EE'}\frac{1}{h(\tau)(1\minus{}ch(\tau))} \frac{dh(\tau)}{d\tau}d\tau$.

Here this does not look like the first step... the $ d\tau$ never changes. Unfortunately, I can even produce a calculation that results in the second integrand! I just have NO clue what any of it really means.

I think I am confused with the mixing of the differentials in the notation of derivatives and integrals, and how they work with change of variables. I would be most grateful for an explanation as to what is going on here, or a good reference regarding change of variables and working with differentials and differential forms. Again, I am sure it is simple and I am being thickheaded. Any reply is certainly most appreciated... thanks for reading!

Basically, that step you don't understand is hiding some calculus manipulations that are actually pretty specific to this integral, with the derivative in it. The whole thing, in different notation:

Let the curves $ \gamma$ and $ \gamma'$ be related by the function $ T$; as $ \tau'$ traverses $ \gamma'$, $ \tau\equal{}T(\tau')$ traverses $ \gamma$. We consider the integral
$ I\equal{}\int_{\gamma}\frac{f'(\tau)}{f(\tau)\minus{}c}\,d\tau$.
Perform the substitution $ \tau\equal{}T(\tau'),d\tau\equal{}T'(\tau')\,d\tau'$. The integral becomes
$ I\equal{}\int_{\gamma'}\frac{f'(T(\tau'))}{f(T(\tau'))\minus{}c}\cdot T'(\tau')\,d\tau'$
$ I\equal{}\int_{\gamma'}\frac{f'(T(\tau'))\cdot T'(\tau')}{f(T(\tau'))\minus{}c}\,d\tau'$.
Now, define $ g\equal{}f\circ T$; we have $ g'(\tau')\equal{}f'(T(\tau'))\cdot T'(\tau')$ and we get
$ I\equal{}\int_{\gamma'}\frac{g'(\tau')}{g(\tau')\minus{}c}\,d\tau'$.

I'm really not comfortable ever writing down half-transformed integrals. The change-of-variables process is all or nothing.


salam
aisa kiun hai k yeh humare pakistan wali foroum itni unactive hai , or just 4 topics hain? y  :o common guys make it active

I agree. Also, could we talk in english a little more often, it's easier to understand.

i think u should post some imo or problems for different national or regional olympiads that might help in regaining the activeness of  :)  :)  :P  the forum

hey post some problems
i'll give an example

1. Two boxes contain between them 65 balls of several different sizes. Each ball is white, black,
red or yellow. If you take any 5 balls of the same colour at least two of them will always be of the same size (radius). Prove that there are at least 3 balls which lie in the same box have the same colour and have the same size (radius).

5. Let A be a set of 16 positive integers with the property that the product of any two distinct numbers of A will not exceed 1994. Show that there are two numbers a and b in A which are not relatively prime.

6. Let AC and BD be two chords of a circle with center O such that they intersect at right angles inside the circle at the point M. Suppose K and L are the mid-points of the chord AB and CD respectively. Prove that OKML is a parallelogram.

he asked wat do u talk!!
not the 1st timers

bhaiyon salaam!

lagta hain aap is community ko refresh nahin kar rahein hain

mere kehne ka matlab hain ki humme wisconsin aur virginia ki tarah nahin hona hain

plz guys and galz be active

and post problems

....................................

merei urdu bohot bura hei

leken ye kya hei? last post [hide=ath]at-h lke as in the taa sound[/hide] saal taa is thread mei?

[hide=jee] for some reason this sounds cringy ;)[/hide]


It follows from my inequality:

\[ \ \frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{(1\plus{}b)^2}\plus{}\frac{1}{(1\plus{}c)^2}\plus{}\frac{3}{4(a\plus{}b\plus{}c)} \ge 1\]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185049
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184726

Thank you so much.
I'm new here,so I don't know the problem have been discuss here.
I think I have to scan the forum about equation.


function
limit
induction
algebra unsolved
algebra

[img]http://img011.photo.21cn.com/photos/user/24454028/20070331224152/o/8338508.jpg[/img]

I can prove only the first two equations,as for the last three ones,I have not any idea.So is there anyone could help me to prove them?

Thanks

[hide="#5"]For number 5, we use e=1/0!+1/1!+1/2!...

so we have e-2=1/2!+1/3!...   (1)
(e-2)lnx=(1/2!)lnx+(1/3!)lnx...   (2)
(e-2)lnx=1/2ln(x(cubert(x(fourthroot(...))))) (3)

(e-2)lnx= ln(sqrt(x(cubert(x(fourth root(x...)))) (4)

L.H.S.= ln(x^(e-2))

take e^ of both sides gives desired results. The step from  to three is more obvious going backwards from 3 to 2.[/hide]

[hide="#3"] Multiplying both sides by sqrt(2), and whenever you have a part of the right hand side that looks like k*sqrt(...) you replace by sqrt(k^2*...), you get

2=sqrt(2+sqrt(2+sqrt(2+...)))

let R.H.S.=x

x^2-2=x

x^2-x-2=0

(x-2)(x+1)=0
regecting the negative root, R.H.S.=2 L.H.S=2,

R.H.S.=L.H.S.[/hide]

Let me ask you, how to define those infinite iterations. :huh:
If one has a sequence of functions $f_{n},n\in\mathbb{N}$, then what is $f_{1}(f_{2}(f_{3}(\cdots)))$?

it is the limit of taking the functions nested within one another to infinity.

That's obvious, but every function needs to be taken on some real number.
Think about this. There is no natural way to define this generally.

I don't understand how that contradicts what I said, we have a set of function, f1,f2,f3,f4,f5,f6,... and we produce the sequence f1(x),f2(f1(x))...Now, that limit either exists or it doesn't, if it does, the answer is g(x), if it doesn't, the answer is infinity.

Yes, but $g(x)=\lim_{n\to \infty}f_{1}(f_{2}(\cdots f_{n}(x)\cdots))$ is a fucntion (if it exists) and not a real number,
and those $5$ terms above are not defined naturally and uniquely.
Before we can even start calculating around, we need to define them first explicitly as limits of well-defined sequences,
and then use all the same definition.

So much from my part. Just want you to take care with this. :)

I see what you mean, are my proofs still valid though?

P.S. I could not get number 4 for general x, induction is obvious, but does not work for x E R

[quote="rofler"]are my proofs still valid though?[/quote]
Almost, if we define
${\sqrt{{2\over 2^{2^{0}}}+\sqrt{{2\over 2^{2^{1}}}+\sqrt{{2\over 2^{2^{2}}}+\cdots}}}=\lim_{n\to\infty}\sqrt{{2\over 2^{2^{0}}}+\sqrt{{2\over 2^{2^{1}}}+\cdots \sqrt{{2\over 2^{2^{n}}}}}}}\\ \hspace*{1.95in}={1\over \sqrt{2}}\lim_{n\to\infty}\underbrace{\sqrt{2+\sqrt{2+\cdots\sqrt{2}}}}_{n+1\text{ roots}}$
and
$\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\cdots}}}}=\lim_{n\to\infty}\sqrt{x\sqrt[3]{x\cdots \sqrt[n]{x}}}=\lim_{n\to\infty}x^{{1\over 2!}+{1\over 3!}\cdots{1\over n!}}$.
For $\lim_{n\to\infty}\underbrace{\sqrt{2+\sqrt{2+\cdots\sqrt{2}}}}_{n+1\text{ roots}}=2$ you need to show,
that the sequence $a_{n}=\underbrace{\sqrt{2+\sqrt{2+\cdots\sqrt{2}}}}_{n+1\text{ roots}}$ is increasing with upper bound $2$,
and so has a limit $t\geq \sqrt{2}$ with $t=\sqrt{2+t}$. So $t=2$.

#4 is a little technical.
Assume $\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}=\lim_{n\to \infty}f_{n}(x)$,

where $f_{n}(x): =\sqrt{1+x\sqrt{1+(x+1)\cdots \sqrt{x+n}}}$ for $x\geq 0$ and $n\in\mathbb{N}$.
Then $f_{n+1}(x)=\sqrt{1+xf_{n}(x+1)}$ for all $x\geq 0$ and $n\in\mathbb{N}$, which gives easily by induction

$x+1-f_{n}(x)={x\over (1+{f_{n}(x)\over x+1})(1+{f_{n-1}(x+1)\over x+2})\cdots(1+{f_{1}(x+n-1)\over x+n})}$ for all $x\geq 0$ and $n\in\mathbb{N}$.

Now clearly $f_{n}(x)\geq \sqrt{x+1}$ for all $x\geq 0$ and $n\in\mathbb{N}$.
So
$0\leq x+1-f_{n}(x)\leq{x\over (1+{1\over\sqrt{x+1}})(1+{1\over\sqrt{x+2}})\cdots(1+{1\over\sqrt{x+n}})}$
and there exists $N\in\mathbb{N}$ with $\sqrt{x+n}\leq n$ for all $n\geq N$.
Then for fixed $x\geq 0$ set
$C: ={x\over (1+{1\over\sqrt{x+1}})(1+{1\over\sqrt{x+2}})\cdots(1+{1\over\sqrt{x+N-1}})}$
and then for all $n\geq N$ we get
$0\leq x+1-f_{n}(x)\leq{C\over (1+{1\over\sqrt{x+N}})(1+{1\over\sqrt{x+N+1}})\cdots(1+{1\over\sqrt{x+n}})}\\ \hspace*{1.26in}\leq{C\over (1+{1\over N})(1+{1\over N+1})\cdots(1+{1\over n})}\\ \hspace*{1.26in}={C\over ({N+1\over N})({N+2\over N+1})\cdots({n+1\over n})}\\ \hspace*{1.26in}={CN\over n+1}\mathop{\longrightarrow}_{n\to\infty}0$
which gives $f_{n}(x)\mathop{\longrightarrow}_{n\to\infty}x+1$ for all $x\geq 0$.

olorin , good job , well prove . Thanks :roll:


induction
algebra
binomial theorem
algebra solved

Let $L_n$ be the sequence of Lucas Numbers, satisfying $L_0 = 2, L_1 = 1, L_{i+1} = L_i+L_{i-1}$ for $i \ge 1$. It's not hard to see that $a_{i+1} = 10L_{2i}$ for $i \ge 0$ and $L_n = \phi^n+(-\phi)^{-n}$ for $\phi = \frac{1+\sqrt{5}}{2}$.
So $5a_{n+1}a_{n+2}+1=500\left(L_{2n}L_{2n+2}\right)+1$ $\\=500\left(\phi^{2n}+(-\phi)^{-2n}\right)\left(\phi^{2n+2}+(-\phi)^{-2n-2}\right)+1$ $\\= 500\left(\phi^{4n+2}+(-\phi)^{-4n-2}+3\right) = 100\left(5\left(\phi^{2n+1}+\phi^{-2n-1}\right)^2+5\right)$

Now we show that $\left(\phi^{2n+1}+\phi^{-2n-1}\right)$ (let's call it $S$) is of the form $k\sqrt{5}$ where $k$ is an integer. Firstly, if you expand out the binomial theorem, you can easily that $S$ is of the for $q\sqrt{5}$ for a rational $q$. But also note that $\phi$ and $\phi^{-1}$ are both algebraic integers, so $S$ is an algebraic integer, so $q$ is an integer.

Returning to the problem, write $S = k\sqrt{5}$ are substitute is in the get $100(5(k\sqrt{5})^2+5)+1=x^2$ for some integer $x$ by the condition. So we get that $x^2-2500k^2=501 \implies (x-50k)(x+50k)=501$. Now some easy casework gives $x=251, k=5$. Going back, $\left(\phi^{2n+1}+\phi^{-2n-1}\right) = 5\sqrt{5}$ so its easy to find that $n = 2$. For the problem they were asking for $1+a_na_{n+1}$, so I guess the answer is $\boxed{n=3}$.

[quote="SuperMath"]Let the sequence $ \{a_n\}_{n\geq 1}$ be defined by $ a_1 \equal{} 20$, $ a_2 \equal{} 30$ and $ a_{n \plus{} 2} \equal{} 3a_{n \plus{} 1} \minus{} a_n$ for all $ n\geq 1$. Find all positive integers $ n$ such that $ 1 \plus{} 5a_n a_{n \plus{} 1}$ is a perfect square.[/quote]

I have a pretty nice solution.

It is obvious to see that $a_n<a_{n+1}$ (Condition 1)

Suppose that $a_k$ and $a_{k+1}$ have a common factor other than $2$ or $5$. Then $a_k$ and $a_{k-1}$ have a common factor other than $2$ or $5$. And by repeating this, $a_1=20$ and $a_2=30$ have a common factor other than $10$.

The above means than $a_n$ is coprime with $a_{n+1}$ (Condition 2)

We let $a_n=10b_n$

It is easy to verify that $\frac{b_{n+1}}{b_n}<3$ (Condition 3)

Now suppose that $1+5a_ka_{k+1}=m^2$.

Then $500b_{k}b_{k+1}=(m-1)(m+1)$. Since the LHS is even, $m$ is odd.

Let $m=2p+1$. Then $125b_{k}b_{k+1}=p(p+1)$.

Using condition 2, we see that we can only these 4 cases

Case 1:
$125=p,b_{k}b_{k+1}=p+1$. This gives $n=3$ as solution. (By condition 1, when $k$ gets big, there is no solution)

Case 2:
$b_{k}b_{k+1}=p,125=p+1$. This gives no solution . (By condition 1, when $k$ gets big, there is no solution)

Case 3:
$125b_{k}=p,b_{k+1}=p+1$. (By condition 3, this is not possible)

Case 4:
$125b_{k}=p+1,b_{k}=p$ (By condition 1, this is not possible).

So $n=3$ is the only solution.

according to method of induction mathematically;we have $(u_n+u_{n+1})^{2}+501=1+5.u_n.u_{n+1}=m^{2} \Rightarrow (m-u_n-u_{n+1})(m+u_n+u_{n+1})=501=1.501=3.167$
case 1: $m-u_n-u_{n+1}=1$ and $m+u_n+u_{n+1}=501$ so $m=251;u_n+u_{n+1}=250 \Rightarrow n=3$ (because of a strictly increasing sequence)
case 2:  $m-u_n-u_{n+1}=3$ and $m+u_n+u_{n+1}=167$ so $u_n+u_{n+1}=82 \Rightarrow u_1+u_2<u_n+u_{n+1}<u_2+u_3$ (no solution)


any 4 points on the plane.
If the length of the line segments joining any 2 points are POSITIVE INTEGER, then
prove:
One of the length must be multiple of 3. :D

Well, since all the lengths are integer, I think you can prove that all the angles are multiples of $90$ or $60$.

well...cosinator...i wanna know how to prove it at least...
I heard the hint like what u told before....
nevertheless....How?How?How?How?How?How?(echoing...) :(
I really need someone can help me solve this question plz.... :(

[u]SOLUTION[/u]

[hide]HERE I AM GOING TO USE PIGEON HOLE PRINCIPLE
ITS QUITE OBVIOUS THAT THE 4 PTS CANNOT LIE ON A SQUARE (AS IF THE LENGTHS OF THE SIDES ARE INTEGERS THEN THE LENGTHS OF THE DIAGONALS AREN'T INTEGERS)
WHICH NOW IMPLIES THAT THERE IS ATLEAST ONE LENGTH WHICH IS UNEQUAL TO THE OTHER LENGTHS.
NOW CONSIDER AN EQUILATERAL TRIANGLE
HERE THERE ARE ALREADY 3 EQUAL LENGTHS. IF WE PLACE THE 4th POINT ON THE PERPENDICULAR BISECTOR OF ANY SIDE OF THE TRIANGLE THERE ARE 5 EQUAL LENGTHS. BUT IN THIS CASE THE LENGTH OF THE VERTEX FROM WHERE THE PERPENDICULAR BISECTOR WAS DRAWN WON'T BE AN INTEGER (AS IT WILL BE EQUAL TO (sqrt3)*(length of side of equi triangle)/2 + cos60*integer side WHICH IS IRRATIONAL + RATIONAL = IRRATIONAL.
HENCE THERE CAN'T BE 5 EQUAL LENGTHS.
WHICH IMPLIES THERE CAN BE A MAX OF 4 EQUAL LENGTHS.
WHICH IMPLIES THERE ARE 3 DIFFERENT LENGTHS OR 3 INTEGERS OF DIFFERENT MODULO 3
BY PHP THERE IS ATLEAST ONE LENGTH WHICH IS A MULTIPLE OF 3[/hide][/u]

[quote="rohit"]THERE ARE 3 DIFFERENT LENGTHS OR 3 INTEGERS OF DIFFERENT MODULO 3[/quote]

The first conclusion certainly doesn't imply the second.  Why would they need to be different $\bmod 3$ to be different lengths?

[b]
SORRY IT WAS MY MISTAKE.
BY ANY CHANCE DO YOU KNOW WHAT THE COSINATOR MEANT?????[/b]


calculus
logarithms
inequalities
limit
geometric sequence
calculus computations

This was my warmup exercise for today's discussion section of my calculus class - I had it on the board as people walked in.
\[ \sum_{n\equal{}3}^{\infty}\frac{n^{\ln n}}{(\ln n)^n}\]

The followup question: find the radius of convergence of $ \sum_{n\equal{}3}^{\infty}\frac{n^{\ln n}}{(\ln n)^n}\,x^n.$

We have:

$ 0 < \frac{n^{\ln n}}{(\ln n)^{n}}  \leq \frac{e^{\alpha^2}}{\alpha^n} \qquad \forall n\geq 3 \text{ and } \forall \alpha > 0$

and since the sum of the RHS of the inequality converges, so does the middle.

Huh?  For $ n \equal{} e^2$, we have $ \frac {n^{\ln n}}{(\ln n)^n} \equal{} \frac {e^4}{2^{e^2}} \approx 0.3257\ldots$ while for $ \alpha \equal{} 1.9$ we have $ \frac {e^{1.9^2}}{1.9^{e^2}} \approx 0.32216\ldots$.

Rewrite $ \frac {n^{\ln n}}{(\ln n)^n} \equal{} e^{(\ln n)^2 \minus{} n \ln \ln n}$.  For large enough $ n$, this is bounded above by $ e^{\sqrt {n} \minus{} n}$, which in turn is bounded above by $ e^{ \minus{} n/2}$, so we are comparable to a convergent geometric series.

For the latter question, I'm thinking root test is the way to go?  (My previous answer gave $ R \geq \sqrt {e}$, but of course I gave away a ridiculous amount -- I could have gotten $ R \geq e$ without too much trouble.  And that $ \ln \ln n$ is small enough that it isn't going to make any difference to the radius of convergence.)

Root test? Fire it up and see what happens.

Oof.
\begin{align*}
\lim_{n \to \infty} \sqrt[n] {\frac{n^{\ln n}}{(\ln n)^n}} & = \lim_{n \to \infty} \frac{n^{(\ln n)/n}}{\ln n} \\
& = \lim_{n \to \infty} e^{(\ln n)^2 / n - \ln \ln n} \\
& = e^{0 - \infty} \\
& = 0,
\end{align*}
which means actually I was badly wrong in saying the $ \ln \ln n$ didn't matter: it actually makes all the difference!

JBL, you seem to be right.. I missed that.. But can we say this:

$ \forall\alpha > 0 \, \exists n_0  \geq 3  \, \forall n \geq n_0 \left(0 <\frac{n^{\ln n}}{(\ln n)^{n}}\leq\frac{e^{\alpha^{2}}}{\alpha^{n}}\right)$ ?

Well, it ultimately decays faster than any fixed geometric sequence, so we can say that given $ \alpha,$ there exists $ n_0$ such that for $ n>n_0,\ 0<\frac{n^{\ln n}}{(\ln n)^n}<\frac1{\alpha^n}.$

But I don't get the point of writing $ e^{\alpha^2}.$

No point.. It just remained there from the approach to get the fact that it does decays faster..


probability
number theory
prime factorization

Four dice are thrown. What is the probability that the product of the number equals $ 36?$

[hide="Solution"]
It can be found that the only combinations of 4 numbers from 1 to 6 whose product is 36 are of the form:
$ 1\cdot1\cdot6\cdot6$
$ 1\cdot2\cdot3\cdot6$
$ 1\cdot3\cdot3\cdot4$
$ 2\cdot2\cdot3\cdot3$

The are $ \frac{4!}{2!2!}\equal{}6$ ways to arrange the factors in the first basic combination. Similarly, there are $ 24$, $ 12$, and $ 6$ ways to arrange the second, third, and fourth basic combinations, respectively. Thus, there are $ 6\plus{}24\plus{}12\plus{}6\equal{}48$ possible combinations of 4 numbers from 1 to 6 whose product is 36. There are $ 6^4$ combinations of any 4 numbers from 1 to 6, so the probability is $ \frac{48}{6^4}\equal{}\boxed{\frac{1}{27}}$.
[/hide]

[hide="proof that they are the only combinations"]The prime factorization of 36 is $ 2*2*3*3$. Hence we want two twos and two threes. We can put two twos together and a 2 and a 3 together, but we can't put 2 3's together. Hence if the two twos are together then we have the other three spaces as 3-3-1, so we have 4-3-3-1. This is the only combination that has a 4 in it.

We now count the groups of two-threes (sixes). If there are two 6's the other two numbers are ones (6-6-1-1). If there is just 1 then we have 6-3-2-1. If we don't have any 6's then we completely split up the primes into 2-2-3-3. These are all the possibilities.[/hide]


Please Help, Thx!!

Suppose $ f(x)$ is Lebesgue measurable on $ [0,1].$ Show that $ g(x,y)\equal{}f(x)\minus{}f(y)$ is measurable on $ [0,1]\times[0,1]$ with respect to the two dimensional Lebesgue measure.

How about this.  Let $ g_1(x,y) \equal{} f(x)$.  Let $ g_2(x,y) \equal{} f(y)$.  Prove that $ g_1$ and $ g_2$ are measurable, then conclude that their difference is measurable.


algebra
function
domain
polynomial
quadratics
LaTeX
complex analysis

Find domain of q(x) = sqrt{x^2 - x - 2}

Note:

What are the steps to find the domain of functions within radicands?

Can the range be found in such functions?

If so, how do I find the range of the above function?

For this problem, we see that the polynomial has a square root above it. So that means the polynomial must be positive in order for the resulting number to be a real number. So, just have the entire function equal to 0, find the x values (x = -1,2), and there you have it, the domain of the function.
The range has to be all positive numbers, including 0, since any number substituted for x under this situation will result in a positive number. Try it yourself.

Are you saying that I need to equate the square root to = 0 or what lies in the radicand?

Choice A: x^2 - x - 2 = 0

OR

Choirce B: sqrt{x^2 - x - 2} = 0

If it is choice B above, then should I treat it as a radical equation?

[hide="hint"]choice A = choice B[/hide][/hide]

We have:

x^2 - x - 2 = 0

Factoring:

(x - 2) (x + 1) = 0

Then:

x = 2 and x = -1

I got it now.

Well, finding out the numbers is just one step.

[hide]
Next, you have to find out for which set of numbers the polynomial will be negative. An easy way for one to get this is to make a little table and substitute.

(-∞,-1) +
(-1, 2) -
(2, ∞) +

So the domain is (-∞, -1) U (2, ∞) and the range is [0, ∞).
[/hide]

I know that this was already pointed out. I merely pointed it out because in some problems the set between the numbers can be the valid set, instead of the other two sets.

[hide="homework problem??  HINT"]
hint: use the property that $x^{2}-x-2>0$
or else sqrt cannot exist.[/hide]

[hide="."]Here's one way to look at it: factor $x^{2}-x-2>0$, and draw a graph. It's a quadratic, and crosses the x-axis at (2,0) and (-1,0). So you have to find where the graph is above the x axis. This is on the left and right sides, so the domain is $(-\infty,-1)\text{U}(2, \infty)$[/hide]

I don't think I'm crazy. That is at least, I [hide="hope."]
$\sqrt{x^{2}-x-2}$
$x = 0$
$\sqrt{(0)^{2}-(0)-2}$
$\sqrt{-2}= i\sqrt{2}$
[/hide][b][i]NICE[/i][/b] positive return value.
A little input on domain.[hide="YOU JUST WON A GREAT ONLINE PRIZE CLICK HERE TO CLAIM"]
let $n$ be any positive real number.
$(\pm\sqrt{n})^{2}= n$
Following the laws of logic...
let $f(n)$ be any [i]positive real[/i] function.
$ \left(\pm\sqrt{f(n)}\right)^{2}= f(n)$
[/hide]Actually, that seems pretty solid.
Is it blasphemous to say that 0 is not included here....Cuz i just don't see it doing down.

EDIT: ok whew im sane. didn't see jecruz.  :blush: I also made it pretty.

[quote="qwang"]Here's one way to look at it: factor $x^{2}-x-2>0$, and draw a graph. It's a quadratic, and crosses the x-axis at (2,0) and (-1,0). So you have to find where the graph is above the x axis. This is on the left and right sides, so the domain is $(-\infty,-1)\text{U}(2, \infty)$[/quote]

$\LaTeX$ note: It looks better if you use $\cup$ instead of $\text{U}$.

$\sqrt[even]{\text{more than or equal to 0}}$

${\sqrt[odd]{\text{arbiterary real numbers}}}$


algebra
polynomial
vector
linear algebra
matrix

Let $ T$ be the linear operator on $ M_{nxn}(\mathbb R)$ defined by $ T(A)=A^{t}$

a) Show that $ \pm 1$ are the only eigenvalues of $ T$
b) describe the eigenvectors corresponding to each eigenvalue of $ T$
c) Find an ordered basis $ \beta$ for $ M_{2x2}(\mathbb R)$ such that $ [T]_\beta$ is a diagonal matrix.

I have already done part a but could someone please give me a few pointers on how to do parts b and c.
Thanks.

$ T$ is an example of an [b]involution[/b]; a map that is its own inverse. Since $ T^{2}=I$, each eigenvalue $ \lambda$ must satisfy $ \lambda^{2}=1$.
b) $ T(A)=A$ means that $ A$ is symmetric; $ T(A)=-A$ means that $ A$ is antisymmetric.
c) find a basis for the space of all symmetric matrices (it has dimension $ n(n+1)/2$) and a basis for the space of all antisymmetric matrices (dimension $ n(n-1)/2$).

$ T^{2}(A)=A.$ Therefore, the operator $ T$ satisfies the polynomial $ x^{2}=1.$ If a linear operator satisfies a polynomial, then the only possible eigenvalues are roots of that polynomial. Hence, the only possible eigenvalues of $ T$ are $ 1$ and $ -1.$ We haven't yet shown that those are eigenvalues, but that's coming, below.

In what follows, I will assume we are working over a field of characteristic other than $ 2.$ Anyone who's interested may pursue the case of characteristic $ 2$ for himself.

The eigenspace corresponding to the eigenvalue $ 1$ consists of all symmetric matrices - that is, matrices that satisfy $ A^{T}=A.$

The eigenspace corresponding to the eigenvalue $ -1$ consists of all skew-symmetric matrices - that is, matrices that satisfy $ A^{T}=-A.$

If $ n\ge2,$ we can construct nontrivial examples of both symmetric and skew symmetric matrices. In fact, the dimension of the subspace of symmetric matrices is $ \frac{n(n+1)}2$ while the dimension of the subspace of skew-symmetric matrices is $ \frac{n(n-1)}2.$ Since these dimensions add up to $ n^{2},$ which is the dimension of the vector space $ M_{n\times n},$ the operator $ T$ is diagonalizable.

In $ M_{2\times 2},$ take the following basis:

$ \left\{\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}, \begin{bmatrix}0&1\\1&0\end{bmatrix}, \begin{bmatrix}0&-1\\1&0\end{bmatrix}\right\}$

In that ordered basis, the matrix for $ T$ would be

$ \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-1\end{bmatrix}.$

The basis is, as with any nontrivial basis, not uniquely given. If you like a different presentation, you're welcome to it.


geometry
circumcircle
IMO
tangent
reflection
IMO 1969

$C$ is a point on the semicircle diameter $AB$, between $A$ and $B$. $D$ is the foot of the perpendicular from $C$ to $AB$. The circle $K_1$ is the incircle of $ABC$, the circle $K_2$ touches $CD,DA$ and the semicircle, the circle $K_3$ touches $CD,DB$ and the semicircle. Prove that $K_1,K_2$ and $K_3$ have another common tangent apart from $AB$.

Denote the triangle sides a = BC, b = CA, c = AB. Let (O) be the circumcircle of the right angle triangle $\triangle ABC$ centered at the midpoint O of its hypotenuse c = AB. Let R, S, T be the tangency points of the circles $K_1, K_2, K_3$ with the line AB. In an inversion with the center A and positive power $r_A^2 = AC^2 = b^2$ ($r_A$ being the inversion circle radius), the line AB is carried into itself, the circle (O) is carried into the altitude line CD and the altitude line CD into the circle (O). This implies that the circle $K_3$ intersecting the inversion circle (A) is carried into itself, but this is possible only if the circle $K_3$ is perpendicular to the inversion circle (A). It follows that the tangency point T of the circle $K_3$ is the intersection of the inversion circle $(A, r_A = b)$ with the line AB. Similarly, in an inversion with the center B and positive power $r_B^2 = BC^2 = a^2$ ($r_B$ being the inversion circle radius), the line AB is carried into itself, the circle (O) is carried into the altitude line CD and the altitude line CD into the circle (O). This implies that the circle $K_2$ intersecting the inversion circle (B) is carried into itself, but this is possible only if the circle $K_2$ is perpendicular to the inversion circle (B). It follows that the tangency point S of the circle $K_2$ is the intersection of the inversion circle $(B, r_B = a)$ with the line AB. :cool:

The distance between the tangency points S, T is the equal to ST =  AT - AS = AT - (AB - BS) = AC - (AB - BC) = a + b - c. The radius r of the incircle $K_1$ of the right angle triangle $\triangle ABC$ is equal to

$r = \frac{|\triangle ABC|}{s} = \frac{ab}{a + b + c} = \frac{a + b - c}{2} = s - c$

where $|\triangle ABC|$ and s are the area and semiperimeter of the triangle $\triangle ABC$, for example, because of an obvious identity

$(a + b + c) (a + b - c) = a^2 + 2ab + b^2 - c^2 = 2ab$

or just because the angle $\angle C = 90^\circ$ is right. Therefore, ST = 2r. Let R' be the midpoint of ST. Then

$AR' = AT - \frac{ST}{2} = b - r = b - \frac{a + b - c}{2} = \frac{c + b - a}{2} = s - a = AR$

Therefore, the points $R' \equiv R$ are identical and the midpoint of the segment ST is the tangency point R of the incircle $K_1$ with the triangle side c = AB. It follows that the normals to the hypotenuse AB at the tangency points S, T of the circles $K_2, K_3$ are tangent to the incircle $K_1$. Radii $r_2, r_3$ of the circles $K_2, K_3$ are now easily calculated:

$r_2 = SD = BS - BD = a - \frac{a^2}{c}$

$r_3 = TD = AT - AD = b - \frac{b^2}{c}$

Denote $I, I_2, I_3$ the centers of the circles $K_1, K_2, K_3$. The line $I_2I_3$ cuts the midline RI of the trapezoid $STI_3I_2$ at the distance from the point R equal to

$\frac{SI_2 + TI_3}{2} = \frac{r_2 + r_3}{2} = \frac{a + b}{2} - \frac{a^2 + b^2}{2c} = \frac{a + b - c}{2} = r = RI$

As a result, the centers $I_2, I, I_3$ are collinear (in fact, I is the midpoint of the segment $I_2I_3$). The common center line $I_2I_3$ and the common external tangent AB of the circles $K_1, K_2, K_3$ meet at their common external homothety center $H \equiv I_2I_3 \cap AB$ and the other common external tangent of the circles $K_2, K_3$ from the common homothety center H is a tangent to the circle $K_1$ as well.

It directly follows from Sawayama-Thebault's Theorem([url]http://en.wikipedia.org/wiki/Th%C3%A9bault%27s_theorem[/url])

An outline of my solution (I wrote it twice, but the system restarted unexpectedly before sending it):

Let $M, N, P$ be contact points of the 3 circles with $AB$, and $I, Q, R$ their centers. As shown before and well known, $BN=BC, AP=AC$; also $\Delta CNP$ has the circumcenter $I$ and $\angle NCP=45^\circ$, hence $IM=MN=MP$.
With $QN=ND, PR=PD$, we get $QN+PR=ND+PD=NP=2MI$, so $MI$ is the midline of the trapezoid $QNPR$ and, $Q-I-R$ are collinear, hence the reflection of $AB$ about $QR$ is the second common tangent of the three circles.

Best regards,
sunken rock


Prove that for any integer $a$ there exist infinitely many positive integers $n$ such that $a^{2^n}+2^n$ is not a prime.

[b]proposed by S. Berlov[/b]

um it's not new,
For a=0,1 okey!
Now $|a|\ge2$
Chose n=2m

$a^{2^n}+2^n=(a^{2^{n-1}}+2^m)^2-2^{m+1}a^{2^{n-1}}$
Take m=odd then this it's easy to get the result!


integration
logarithms
function
calculus
calculus computations

For 2., write $ x^{-x}= e^{-x \ln x}$ and then use the power series representation of the function $ e^{y}$, with $ y =-x \ln x$.

[quote="mdk"]Prove that

1. $ \int_{0}^{1}(\frac{1}{1-x}+\frac{1}{lnx})dx= \gamma$ where $ \gamma= lim_{k\to\infty}(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k}-ln(k)))$

[/quote]

$ \int_{0}^{1}\frac{1}{1-x}+\frac{1}{\ln x}dx = \int_{0}^{1}1+x+x^{2}+...+\frac{1}{\ln x}dx$

Now, $ \int_{0}^{1}\frac{1}{\ln x}dx = \int_{-\infty}^{0}\frac{e^{t}}{t}= \int_{-\infty}^{0}\frac{1+t+t^{2}/2!+...}{t}dt$
$ =\ln t+t+t/2!+t^{3}/3!+... \big|_{-\infty}^{0}$ which is (abusing notation and formallness) $ \ln 0 =-\ln \infty$.

Now return to first line and integrate term-by-term,
$ 1+\frac{1}{2}+\frac{1}{3}+...-\ln \infty = \gamma$

By the way: Isnt this the ugliest looking proof?  :rotfl:


Λοιπον,αρχικα γεια σε ολους!
Προχθες συζητουσαμε με τον στελιο (stedes)για ενα παλιο προβλημα που ειχε βαλει ο θανος(thatin) kai ελεγε
0 αριθμος 6 ειναι τελειος ο αριθμος 4 ειναι τετραγωνο και ο 64 ειναι τελειο τετραγωνο!
Αργοτερα ο στελιος βρηκε και το 28 τελειος ,9 τετραγωνο,289τελειο τετραγωνο!
Και δημιουργηται το ερωτημα <να βρεθουν ολοι οι αριθμοι με αυτη την ιδιοτητα>

Καθως προσπαθουσαμε να βρουμε ολους αυτους τους αριθμους καθισμενοι σε ενα παγκακι ,χωρις χαρτι,και μονο στα λογια,αποδειξαμε κατι αλλο.
<να εξετασετε αν υπαρχει τελειο τετραγωνο το οποιο να ειναι τελειος αριθμος>

Θετω αυτα τα δυο προβληματα λοιπον γιατι ειναι πραγματικα ενδιαφεροντα!

Σημ.Ηαποδειξη στο δευτερο ειναι πολυ ευκολη γι αυτο προσπαθηστε.....χωρις χαρτι!!!

ΡΑΛΛΗΣ :D

Ραλλη μηπως ο τιτλος θα επρεπε να ειναι στο παγκακι??? :P  πολυ καλα ερωτηματα...θα τα προσπαθησω...

[quote="RAKAR"]
0 αριθμος 6 ειναι τελειος ο αριθμος 4 ειναι τετραγωνο και ο 64 ειναι τελειο τετραγωνο!
Αργοτερα ο στελιος βρηκε και το 28 τελειος ,9 τετραγωνο,289τελειο τετραγωνο!
Και δημιουργηται το ερωτημα <να βρεθουν ολοι οι αριθμοι με αυτη την ιδιοτητα>
[/quote]

Δεν προσδιόρισα ακόμη όλους τους αριθμούς με αυτήν την ιδιότητα, απέδειξα όμως ότι είναι άπειροι (... κι αυτό χωρίς χαρτί...).  :cool:
 :clap:  :clap2:

P.S. Και ένας ακόμη που βρήκε ο Θάνος είναι ο 625.  :wink:

παιδια μηπως για καθε τελειο αριθμο υπαρχουν απειροι αριθμοι ωστε οταν τοποθετηθουν δεξια του να σχηματιζεται τελειο τετραγωνο? δεν εχω και χρονο να το ελεγξω

Λοιπον και κατι αλλο που μου ηρθε μετα....
Απο το 6=τελειος,4=τετραγωνο,64=τετραγωνο
αν βαλουμε οπου τελειος=τετραγωνο και οπου τετραγωνο=τελειος
εχουμε 49=τετραγωνο,6=τελειος,496=τελειος  !!!
και 81=τετραγωνο,28=τελειος,8128=τελειος!!!
<Να βρεθουν ολοι οι αριθμοι μ αυτη την ιδιοτητα>

Τελικα οι τελειοι αριθμοι,και τα τετραγωνα εχουν πολυ τελειες ιδιοτητες!!!! :P

ΡΑΛΛΗΣ :D

Μιας και αρχίσαμε τις λογοτεχνικές περιγραφές... μόλις έλαβε το μήνυμα του stedes o thatin έφυγε από Αθήνα για να δει αν εννοούσε αυτά που έλεγε και έτσι έχασε τον τελικό :mad:  Ευτυχώς το πρώτο μήνυμα του RAKAR καθησύχασε όλους τους φόβους του και έτσι άφησε το topic αυτό και αφού είδε όλα τα άλλα topics που τον ενδιέφεραν επέστρεψε σε αυτό... Λοιπόν:
1)Οι εν λόγω αριθμοί ονομάζονται S-ps (S-perfect squares) ελληνιστί S-ττ (S-τέλειο τετράγωνο). (δική μου ονομασία όπως και ιδέα :rotfl: )
2)Μία επιπλεόν ιδιότητα που πρέπει να έχουν είναι τα 2 τελευταία ψηφία τους να μην είναι όλα μηδέν
3)137438691328, 2305843008139952128 υπάρχουν άλλη αριθμοί με αυτή την ιδιότητα? (και ποια είναι αυτή?)

Λοιπόν δεν ξέρω αν είναι τυχαίο αλλά από τους δέκα πρώτους τέλειους αριθμούς,6 έχουν σαν τελικό ψηφίο το 6,δηλαδή τον πρώτο τέλειο αριθμό και οι υπόλοιποι 4 τελειώνουν με τον αριθμό 28,δηλαδή τον δεύτερο τέλειο αριθμό! :o

[quote="thatin"]
2)Μία επιπλεόν ιδιότητα που πρέπει να έχουν είναι τα τρία τελευταία ψηφία τους να μην είναι όλα μηδέν
[/quote]
Διαφωνώ.Επειδή δηλαδή είναι δικής σου εμπνεύσεως θα τους κάνεις ότι θες; :P
Επίσης συμφωνώ με την ονομασία τους ως S-ps και αναρωτιέμαι από που εμπνεύσθηκες αυτήν την ονομασία. :rotfl:

[quote="thatin"]
3)137438691328, 2305843008139952128 υπάρχουν άλλη αριθμοί με αυτή την ιδιότητα? (και ποια είναι αυτή?)
[/quote]
Είναι προφανές νομίζω. :P

[quote="stedes"]

[quote="thatin"]
2)Μία επιπλεόν ιδιότητα που πρέπει να έχουν είναι τα τρία τελευταία ψηφία τους να μην είναι όλα μηδέν
[/quote]
Διαφωνώ.Επειδή δηλαδή είναι δικής σου εμπνεύσεως θα τους κάνεις ότι θες; :P
[/quote]

ακριβώς :rotfl:
αναρωτιέσαι από που εμπνέυστηκα αυτή την ονομασία γιατί κατάλαβες από που ή γιατι δεν κατάλαβες?

μπορει κανεις να γραψει την αποδειξη οτι οι S-pt ειναι απειροι.μηπως ο Στελιος ? :roll:

S-pt??
αν εξαιρέσουμε το 2 όρο τότε 64.00.00.00.00...
και το πρόβλημα γίνεται για  :rotfl: το ίδιο και οι αριθμοί...

Κώστα,αρκεί να κάνεις την παρατήρηση ότι οι αριθμοί 64,6400,640000,.... είναι όλοι S-ps.   :wink:

Θάνο το πρόβλημα ήταν από την αρχή για  :rotfl: το ίδιο και οι αριθμοί... .  :P

αααα, αγαπητε οι 64000000..., τελειωνουν σε 0. αλλη ακολουθια χωρις τοσα πολλα 0 υπαρχει?  :)

Και πού το πρόβλημα;

Δεν ξέρω αν υπάρχει άλλη.  :maybe:

[quote="thatin"]
το πρόβλημα είναι προφανές :wink:  καποιοι όμως :sleep2:
θα δούμε... στα λόγια!!!
[/quote]

 :huh:
Κάποιοι  :sleep2: ,αλλά κάποιοι άλλοι  :sleeping: !

Tί να δούμε στα λόγια; Αφού όλοι "όχι" θα ψηφίσουν!
Είναι προφανές ότι η απόδειξή μου είναι σωστότατη και εσύ για να μου κάνεις σαμποτάζ επινόησες έναν όρο ώστε να χαλάσεις την απόδειξή μου.
Λες και τα μαθηματικά είναι κι εγώ δεν ξέρω τι και βγάζουμε κανόνες όπως μας συμφέρει.  :mad:

P.S. :rotfl:

wie du willst, wir mussen wissen wir werden wissen
ως τότε ας τελειώνουμε με το όνομά σου, ok S6?

Άρχισες τώρα και τα γερμανικά;

Ψηφίστε όλοι "όχι". Είναι φανερό το τι θέλει να καταφέρει με το να επιβάλει τους κανόνες του. Δεν είναι τυχαίο το ότι μιλάει στα γερμανικά. :agent:

Ο όρος σου δεν θα περάσει που να :wallbash_red: .

λέγε ότι θέλεις... δε θα μου χαλάσεις τη βραδιά...
Riemann Hypothesis!!!!!!!!!!!!!!!!!!
(α, πρόσθεσα ένα όρο και σε αυτή, επανεξέτασε την απόδειξή σου, νομίζω υπάρχει πρόβλημα)
Α, και δεν απέδειξες οτι υπάρχουν άπειρα S-ps, έγω το έκανα (εσύ είπες απλώς ότι το απέδειξες :rotfl: )
Τελικά αναρωτιέσαι από που εμπνέυστηκα αυτή την ονομασία γιατί κατάλαβες από που ή γιατι δεν κατάλαβες?

:spam:
[color=#f8f8f8]...μας ντροπιάζεις όλους... τι θα λέει ο κόσμος για το επίπεδο της ολυμπιάδας μας όταν η κορυφή μας κάνει τέτοια? [/color]

[color=white]Δε βαριέσαι...; Σιγά τώρα,ας πουν ότι θέλουν.[/color]:roll:

σου άρεσε το δώρο μου?
[color=#f8f8f8]δεν πέτυχες το χρώμα... hey για να μην το καταλάβουν γράψε κάτι άσχετο με κανονικό και τα υπόλοιπα με f8f8f8 [/color]

Πολύ ωραίο! Από πού το αγόρασες;  :P
[color=#f0f0f0]Xα,χα καλό κόλπο!!!![/color]

athens... ΕΜΠ... τι να γίνει το ΑΠΘσου δεν έχει τέτοια...
[color=#f8f8f8]πάλι φαίνονται τα γράμματά σου λίγο... και τα δικά μου φαίνονται? τώρα δε φαίνονται... τι γίνεται? άλλαξε και το πρώτο σου μήνυμα[/color]

Έχει πολύ καλύτερα!  :P
[color=#f0f0f0]Πώς μπορεί κάποιος να βρει τον κωδικό κάθε χρώματος; Το Photoshop βοηθάει καθόλου;[/color]

ναι για συναχωμένους...

[color=#f8f8f8][cοlor=white]Δε βαριέσαι...; Σιγά τώρα,ας πουν ότι θέλουν.[/cοlor]:roll:
με αυτό τι θα γίνει?ναι βοηθάει...[/color]

για συναχωμένους;;;;;;;;;;;;;; :huh:
[color=#f0f0f0]Αλλάζουμε χρώματα; Με βολεύει περισσότερο το δικό σου.[/color] :roll:

αψου...
[color=#f8f8f8]τι εννοείς? αφού στο έκανα δώρο ... εσύ τι χρώμα έχεις? [/color]

γεια σου...
[color=#f0f0f0]Ρε συ,εννοώ το χρώμα του φόντου σου.Τί μου λες ότι μου το έκανες δώρο;;;;;[/color]

KN!
[color=#f8f8f8] το χρώμα!! για να μην το ψάχνεις!εσύ τι χρώμα βάζεις? &sd [/color]

[size=200][color=#f1f1f1] Το κάνατε θέμα [/color][/size]  :D

[color=#f1f1f1]Υπάρχουν και pm's... [/color]:police:

[color=#f0f0f0]Σωστά, &sd![/color]  :lol:

Όχι,όπου 8 βάλε 0.   :P
Ένας Ε μόνο Τ και δεν είσαι εσύ. :rotfl:

Επιτέλους τί;

επιτέλους ΕΤ!!!!!!!!!!!!!! [color=#f8f8f8] τώρα λες σαχλαμάρες και θα στο αποδείξω...[/color]
Aopsgrατες πόσοι τέλειοι υπάρχουν? 1 ή περισσότεροι?

Aopsgrατες:Περισσότεροι!

Ε και; Άλλο τέλειοι, άλλο Τ !  :wink:


We are given two triangles. Prove, that if $\angle{C}=\angle{C'}$ and $\frac{R}{r}=\frac{R'}{r'}$, then they are similar.

By $r=4R\prod\sin{\frac{A}{2}}$ and hypothesis we have $\sin{\frac{A}{2}}\sin{\frac{B}{2}}=\sin{\frac{A'}{2}}\sin{\frac{B'}{2}}\Rightarrow \cos{\frac{A-B}{2}}=\cos{\frac{A'-B'}{2}}$
$\Rightarrow |\frac{A-B}{2}|=|\frac{A'-B'}{2}|$, from this and $\frac{A+B}{2}=\frac{A'+B'}{2}$ we have done !


function
algebra
polynomial
Gauss
Support
LaTeX
rational function

Suppose $ F$ is a field and $ t$ is an indeterminate. In the rational function field $ F(t)$, let $ s \equal{} \frac{t^3}{t\plus{}1}$.

$ a)$ Prove that $ s$ is transcendental over $ F$.
$ b)$ Prove that $ F(t)$ is a simple algebraic extension of $ F(s)$
$ c)$ Compute $ [F(t): F(s)]$

[hide="how do I fix this for this forum?"]
\begin{enumerate}
\item Prove that $ s$ is transcendental over $ F$.
\item Prove that $ F(t)$ is a simple algebraic extension of $ F(s)$
\item Compute $ \left[ F(t): F(s) \right ]$
\end{enumerate}  [/hide]

The key to this whole problem is that $ t$ is a root of the polynomial $ P(X) \equal{} X^3 \minus{} sX\minus{}s$ in $ F(s)$.  b) follows immediately, c) requires a proof that $ P$ is irreducible, (hint: switch the variables and use Gauss' lemma) and a) follows from b).

And: The forum doesn't support full LaTeX as far as I know, so \begin{enumerate} is probably too ambitious.


limit
trigonometry
real analysis
real analysis unsolved

Find the limit of the following expression:
\[ \lim_{n \to \infty}{\sin{\sqrt[3]{8n^3\minus{}2n^2\plus{}7}} \cdot \pi }.\]

hello, do you meant $ \lim_{n \to \infty}\left(\sin\sqrt[3]{8n^3\minus{}2n^2\plus{}7}\cdot\pi\right)$?
Sonnhard.

[hide]$ \lim \,\, \sin(\pi \sqrt [3]{8n^3 \minus{} 2n^2 \plus{} 7}) \equal{} \lim \,\, \sin (\pi \sqrt [3]{8n^3 \minus{} 2n^2 \plus{} 7} \minus{} 2n \pi)$

$ \equal{} \lim \,\, \sin \left[\pi \left(\sqrt [3]{8n^3\left(1 \minus{} \frac {1}{4n} \plus{} \frac {7}{8n^3}\right)} \minus{} 2n\right) \right]$

$ \equal{} \lim \,\, \sin \left[\pi \left(2n \sqrt [3]{1 \minus{} \frac {1}{4n} \plus{} \frac {7}{8n^3}} \minus{} 2n\right) \right]$

$ \equal{} \lim \,\, \sin \left[\pi \frac {\sqrt [3]{1 \minus{} \frac {1}{4n} \plus{} \frac {7}{8n^3}} \minus{} 1}{\frac {1}{2n}} \right]$

$ \equal{} \lim \,\, \sin \left[\pi \frac{1 \minus{} \frac {1}{4n} \plus{} \frac {7}{8n^3} \minus{} 1}{\frac{1}{2n}\left[1 \plus{} \sqrt [3]{1 \minus{} \frac {1}{4n} \plus{} \frac {7}{8n^3}} \plus{} \left(1 \minus{} \frac {1}{4n} \plus{} \frac {7}{8n^3}\right)^{2/3} \right]} \right]$

$ \equal{} \lim \,\, \sin \left[\pi \frac{\frac {7}{4n^2} \minus{} \frac12}{1 \plus{} \sqrt [3]{1 \minus{} \frac {1}{4n} \plus{} \frac {7}{8n^3}} \plus{} \left(1 \minus{} \frac {1}{4n} \plus{} \frac {7}{8n^3}\right)^{2/3}} \right]$

$ \equal{} \sin \left(\minus{} \frac{\pi}{6}\right)$

$ \equal{} \minus{} \frac12$
[/hide]


Prove that the group defined by the following presentation
$ \left<x,y,xy^2\minus{}y^3x,yx^2\minus{}x^3y\right>$ is trivial.

A small typo, it should be '=' instead of '-' :
$ \left < x,y;xy^2 \equal{} y^3x,yx^2 \equal{} x^3y\right >$


inequalities
search
symmetry
inequalities proposed

I went to Turkey Olympiad Studying Winter Camp in February. This question was asked to us in camp finish exam.

i can't believe this is a TST problem.

from chebyshev we have \[4(a^{3}+b^{3}+c^{3}+d^{3})\ge (a^{2}+b^{2}+c^{2}+d^{2})(a+b+c+d)\]and from power mean \[\frac{a^{3}+b^{3}+c^{3}+d^{3}}{4}\ge \left(\frac{a+b+c+d}{4}\right)^{3}\]that is \[2(a^{3}+b^{3}+c^{3}+d^{3})\ge \frac18\]adding we get the result.

[quote="Aravind Srinivas L"]A doubt please explain
Tchebychev's inequality can be applied only if $ a_1\ge a_2\ge .......\ge a_n$ and similarly for $ b_n$.
Then how come you have applied directly?[/quote]
Now I'm not sure about the Tchebychev inequality. But i think it is something like
$ (a_1 \plus{} a_2 \plus{} ... \plus{} a_n)(b_1 \plus{} b_2 \plus{} ... \plus{} b_n) \le n(a_1b_1 \plus{} a_2b_2 \plus{} ... \plus{} a_nb_n)$
If i'm correct it doesn't matter if $ a_1 \ge a_2 \ge .. \ge a_n$. We just have to have $ a$ similar sorted as $ b$.

[quote="Aravind Srinivas L"]A doubt please explain
Tchebychev's inequality can be applied only if $ a_1\ge a_2\ge .......\ge a_n$ and similarly for $ b_n$.
Then how come you have applied directly?[/quote]

Chebyshev's inequality works here because $ (a^2,b^2,c^2,d^2)$ and $ (a,b,c,d)$ are always sorted similarly. In fact you can assume $ a\ge b\ge c\ge d$ due to symmetry and verify the fact.

this problem can be solved easily with ilham's cauchy.

$ 48(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \ge 8(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)(a \plus{} b \plus{} c \plus{} d) \plus{} (a \plus{} b \plus{} c \plus{} d)^3$ so we are done by muirhead

mm.. maybe i have another solution

by the cauchy-schwarz inequality $(a^3+b^3+c^3+d^3)(a+b+c+d) \geq (a^2+b^2+c^2+d^2)^2$
hence $(a^3+b^3+c^3+d^3) \geq (a^2+b^2+c^2+d^2)^2$
now we set the $LHS=x$ and $RHS=y$ and we have a new system of inequalities

$x\geq y^2$

$6x\geq y+  1/8$

$\Rightarrow 6x\geq 6y^2 \geq y+1/8$

thus we have $6y^2-y-1/8 \geq 0 $ from wich follows $y\leq -1/12 , y \geq 1/4$ since $y$ is always positive

we just have to show that $y \geq 1/4$ wich follows from cauchy schwarz :

$ 4(a^2+b^2+c^2+d^2) \geq (a+b+c+d)=1$ , can someone check if it's correct ? it's a new topic for me, thanks!!

Using $a+b+c+d=1$, we get

$156S(3,0,0) \geq 120S(2,1,0)+24S(1,1,1)$

Obviously this holds by Muirhead inequality :)

[quote=Igor]Let $a,b,c,d$ be positive reals such taht $a+b+c+d=1$.

Prove that: \[6(a^{3}+b^{3}+c^{3}+d^{3})\geq a^{2}+b^{2}+c^{2}+d^{2}+\frac{1}{8}.\][/quote]
https://artofproblemsolving.com/community/c6h1531976p9218816

Let $f(x)=6x^3-x^2-\frac1{32}$. The inequality is equivalent to $f(a)+f(b)+f(c)+f(d)\ge0$. Now by tangent-line trick, we have:
$$f(x)-\frac58x+\frac5{32}=(4x-1)^2(3x+1)\ge0\Rightarrow f(x)\ge\frac58x-\frac5{32}$$
and summing this over $a,b,c,d$ yields:
$$f(a)+f(b)+f(c)+f(d)\ge\frac58(a+b+c+d)-\frac58=0$$
as desired.

[quote=dudeldai]
$x\geq y^2$

$6x\geq y+  1/8$

$\Rightarrow 6x\geq 6y^2 \geq y+1/8$

thus we have $6y^2-y-1/8 \geq 0 $[/quote]

If $a \geq b$ and $a \geq c$, this does not say anything about whether $b \geq c$...

This can be solved just by using Chebyshev's inequality 2x lol

[quote=eeshift]Symmetric inequality of Degree 3 theorem - SID theorem

see http://www.mathlinks.ro/viewtopic.php?search_id=597762145&t=148928[/quote]

^^

Let $f(a,b,c,d)=6a^3+6b^3+6c^3+6d^3-a^2-b^2-c^2-d^2\text{ and }g(a,b,c,d)=a+b+c+d-2$, furthermore, define $L=f(a,b,c,d)-\lambda g(a,b,c,d)$

$$\frac{\partial L}{\partial a}=18a^2-2a-\lambda=0$$
$$\frac{\partial L}{\partial b}=18b^2-2a-\lambda=0$$
$$\frac{\partial L}{\partial c}=18c^2-2c-\lambda=0$$
$$\frac{\partial L}{\partial d}=18d^2-2d-\lambda=0$$
Thus $18a^2-2a=18b^2-2b=18c^2-2c=18d^2-2d\Longrightarrow 9a^2-a=9b^2-b=9c^2-c=9d^2-d$ and let [color=#6AA84F](1)[/color],[color=#45818E](2)[/color],[color=#674EA7](3)[/color] and [color=#c00](4)[/color] denote the equations respectively.
Furthermore from [color=#6AA84F](1)[/color] and [color=#45818E](2)[/color] we obtain $9a^2-a=9b^2-b\Longrightarrow 9(a^2-b^2)=a-b\Longrightarrow 9(a+b)(a-b)=a-b$ thus $9(a+b)=1\Longleftrightarrow a+b=\frac{1}{9}$ if and only if $a\neq b$
Also, from [color=#674EA7](3)[/color] and [color=#c00](4)[/color] we obtain $9c^2-c=9d^2-d\Longrightarrow 9(c+d)(c-d)=c+d$ thus $9(c+d)=1\Longleftrightarrow c+d=\frac{1}{9}$ if and only if $c\neq d$
Therefore, if $a\neq b\text{ and }c\neq d$, we get that $a+b+c+d=\frac{1}{9}+\frac{1}{9}=\frac{2}{9}<1$ which is clearly a contradiction, thus $a=b\text{ and }c=d$ (we obtain the same contradiction from the other equations)
Since we can obtain the same contradiction from any other permutation, we can conclude that $a=b=c=d$, thus $4a=1\Longrightarrow a=b=c=d=\frac{1}{4}$
Therefore the minimum of the function is achieved at $f\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)=\frac{1}{8}$ $\blacksquare$.


Firefox 2 (most recent version) here.

http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=20845 (SCHNOGGINVOGGINSCHWITZERW)

Any forum with this user's topics or any topic with his posts causes the table column containing the username (in topics the leftmost one with profile info, and in forums the topic creator column) to be incredibly wide, thus squishing all of the more relevant subject matter into a far smaller size.

I mean no ill will towards the user, but I am hoping to see a change.

Then some usernames (and maybe long topic names) will not be visible...

I suggest a character limit in usernames. :)

I wouldn't like to impose restrictions on the size of the usernames (the ones with spaces in them are fine) but rather to find a solution for this. I'm looking into some CSS options, but nothing is looking good so far.

Could you make them wrap within the column? This will make the username hard to read, however.

[quote="i_like_pie"]This will make the username hard to read, however.[/quote]

But it won't be indistinguishable. Username wrapping sounds like a nice idea.

[quote="Temperal"]How about auto overflow for individual cells?[/quote]

Ugly.

There are very few usernames that would require overflow, though. Also, hide-overflow would probably confuse some users.

[quote="MustItAlwaysBeMe"]Wouldn't it be best to hide the overflow?  That's the only non-ugly solution.  Auto-overflow for every post of a person would be terrible.  It's not like there's a way you can wrap the text without spaces.

(or make a server script that replaces abcdefghijkl with abcdef- ghijkl or w/e)[/quote]

Your signature also stretches the page

[quote="moogra"]Your signature also stretches the page[/quote]

In common with long usernames, obnoxious and large LaTeX signatures also annoy me. I shake my fist.

Actually the page stretching doesn't affect me on my 22" widescreen but it does on my regular 19.

[quote="Temperal"]Signatures actually annoy me in general nowadays. What's the point?

There should be an area-limit (like x by y pixels) instead of a character limit. Of course, this can't be enforced automatically, though.[/quote]The character limit is due to the nature of the database field the signature is kept on (255 varchar :D).

Oh, okay. I think it would be nice if you put a suggested pixel limit, though...

Oh, it's perfectly enough that people don't reply to posters who like to annoy them with their signatures... ;)

  darij

Well you could try inserting this in between some letters in the username to make them wrap:
[code]<span style="font-size:0;"> </span>[/code]
If the line's going to wrap, then it does, which is good, but if it doesn't, it just doesn't show up, so it wouldn't matter.  Maybe there's a better solution, but I dunno...

That's sweet. Adds extra markup but hey untill CSS3 standards are approved, we have to :)

Please see this guys: http://www.mathlinks.ro/viewtopic.php?p=1071221#1071221

This is related, I guess...

Maybe the Location field should be wrapped, also? Somebody has http://www.artofproblemsolving.com as their location, and it stretches the column on the left like crazy. (Still looking for that user...)

Hmm perhaps a more elegant solution is the soft hyphen character I just stumbled upon ()...

For example, you can test:
[code]javascript:document.body.innerHTML="fdasfdsafdasfdsafdsafdsafdsa";[/code]and change your window width.


Can someone please prove the following statement:

Let k be a variable and n a constant.

Then,

Summation (nCk * 2^k) = 3^n , from k=0 to k=n

I apologise for not using LaTeX

Thanks a lot (although you make me look a bit stupid with the simplicity of your proof!)

You can also prove this using a combinatorial argument.

Suppose that you have $n$ people and three colors red, green, and blue.  You want to color each person with these colors.  Clearly, the number of ways to do so is $3^{n}$.  This is also equivalent to choosing $k=1,\hdots, n$ people from $n$ and coloring them using $2$ colors (the other $n-k$ people will be of the third color).  Summing the values for all $k$, we get the desired result.


group theory
abstract algebra
search
number theory unsolved
number theory

Given $ p_1 ,..,p_k$ are primes and $ \alpha _1 ,..,\alpha _k$ are rationals. Prove that if $ \alpha _1 \sqrt {p_1 }  \plus{} .. \plus{} \alpha _k \sqrt {p_k }  \equal{} 0$ then $ \alpha _1  \equal{} .. \equal{} \alpha _k  \equal{} 0$.

I believe you also meant to give the condition that $ p_i$ are distinct.

You don't really need primes. The most meaningfull condition is: $ p_i$ generate a subgroup of $ \mathbb Q^\times / (\mathbb Q^\times)^2$ of maximal order (that is: $ 2^k$). By this formulation, this gets a consequence of Kummer theory.
For more direct approaches, you can search the forum, this problem occured a couple of times.


Please post the ones you have found best.
Here are the ones I know-

Physics-
1 Fundamentals of Physics
2 Concepts of Physics

Maths-
1 S.L. Loney(Trig/Co-ord)
2 M.L. Khanna

Chem-

Org-
1. Morrison & Boyd
2. Solomons
Phy-
1. P.W. Atkins
2. R.C. Mukherjee
Inorg-
J.D. Lee

Phy--
Resnik Halliday is a really good one

maths--
R.D Sharma obj (is full of concepts) :roll:

I really appreciate creating this thread :) . I can't participate here now of course ...
However it will come very handy to me a few months later perhaps. So keep pouring in suggestions.  :roll:  :D

BTW isn't it Resnick Halliday Walker together wrote Fundamentals of Physics?

[quote="rituraj007"][quote="nuclear_alchemist"]1 Fundamentals of Physics
[/quote]i think this is Resnick only!! :P

and u forgot the good old Irodov
And maths ..TMH[/quote]
Irodov? Isn't it above JEE level?  :huh:
And Akshanil, yes they wrote it together.
P.S. I think the newer versions aren't available easily in India. I'm studying from the ebook. Got the hardcopy of 6th Edition.

hello
books that i have read in xi class for JEE for Mathematics::
1.SL Loney
2.I A Maron
3.A Das
4.PCM

[quote="rituraj007"]Irodov is really good..not too above JEE (ofc dont think i try it)

Resnicks old editions r much better[/quote]
I never try Irodov myself. But our tutorial sir, Mr. Atul gets them done in the class. He teaches at AAKASH Delhi too so you might have heard of him :)

I like Irodov really coz most of the problems test  mathematics skill.
(mainly calculas)

I suck at using calculus in physics.  :(  I dunno what to do!

you may try a book of DYNAMICS via calculas(Intermediate and engineering entrance examination)
by K.C Sinha and U.C Sinha.
However I doubt u will get it .
Its a very old book(my book has an edtion of 1984)

Did you mean http://books4u.in/book_detail.php?book_id=8105
I guess not :( There were no other search results related to it.
Anyways, thanks :)

I do not know personally.. as I dont put my head a lot into these....
by what i have heard from others... this is my opinion

H.C.Verma has a good collection of Problems.. but the theory is really pathetic...
Irodov is a really magnificent book.

S.L.Loney is by far the best for co-ordinate geometry...

No idea about chemistry

[quote="Rijul saini"]@kabi-- i dont think pre college maths will be of much use in jee preparation......[/quote]
well i have done P & C from that book only.It was very helpful.

Physics:

HC Verma
Irodov
DJ griffith(electro)
Kleppner and Kolenkow(mechanics)

Chem:

Morrison Boyd
Chemical Principles (Atkins and jones)

Maths:

TMH
Everything else is waste, you wont find conceptual IIT type problems anywhere in bulk.

my views(though i am not in situation to advise)

Maths:
TMH(i like it very vey much and don't think i could manage anything else)

Phy:
Resnick,halliday Krane(theory)
D.C.Pandey (irodov don't need to ne named)

Chem:
phy:
a big ques mark(i have been doing RCM but now being advised by many to leave it,donno what to do)

Inor:
J.D lee
NCERT

org:
Morrison boyd
solomons

so any advise on Phy chem?????????

[quote="skand"]
so any advise on Phy chem?????????[/quote]
I can most certainly say that Paula Atkin's Physical Chemistry is an excellent book :D And rest O.P. sir's notes WOW! But sadly not everyone can have those..
P.S. Thanks to everyone for their book suggestions!

Atkins Physical Chem is not a good buy as 2/3rd of it is quantum mechanics and other unneeded stuff even by olympiad standards. See the book I mentioned.

The likes of Griffith and Atkins are not really very ideal as tools for JEE preparation, especially as JEE's scope is highly limited. Learning too much extra doesnt really help. They are however, great books if you want to learn the subject.

Learning too much extra doesnt really help.

IS IT? :P
Knowledge always pays off.......
I personally doesnat believe in exam oriented preparation of the subjects I like.
However its my personal thought. :roll:

In my opinion books like atkins and griffiths are anywhere there for your first year portions inside IIT. so just study whatever is required bcs the time will be very short and u have to give equal weightage to all the 3 subjects.

For organic i think books like morrison and boyd or solomons is too high a level to start with and they are apt books for the olympiads and not JEE. But the book by "Carey" is good enough for JEE level, it covers all the elementary concepts and some high stuff too.

@ murgi
Chronoz or Rak or Conjurere never said that u shud learn for an exam.
IIT JEE has hell lot of syllabus to cover and that too 3 subjects. Interesting books like atkins, etc which have quantum mechanics can be read even after JEE . there is no problem in doing that. If u wanna get a good rank in JEE, then, doing what s required properly suffices. If u want to do it for the interest of the subject
then no problem. Any one can go ahead. The topic is  named as Books for JEE. so I think the advice of the seniors very much suffices.

Morrison is  a must to start with. ITs the book that got me interested in organic in the first place(in mid 12th infact).

I guess the senior members here are right. I don't even know the full scope of JEE syllabus at the moment. That's probably why I created this thread in the first place. But I'd still say Atkins is an irresistible read for people like me who can never manage to get too serious about marks. But as far as the purpose of this thread is concerned, I guess its a thumbs down to it.  :)

[hide="The best bOoks are"][hide="open"][hide="go on"]vidyamandir classes modules  :P  :rotfl:  :rotfl: [/hide][/hide][/hide]


If it takes Jeremy 18 hours to dig a 2 meter by 2 meter by 2 meter hole, how many hours would it take three men (each working at the same rate as Jeremy) to dig a 4 meter by 4 meter by 4 meter hole?

A.	12	B.	18	C.	24	D.	36	E.	48

[hide="click to reveal hidden content"]

One man takes $ 18$ hours to dig a volume of $ 8$, so three men would take $ 6$ hours to dig a volume of $ 8$ and $ 6 \times 8 \equal{}48$ hours to dig a volume $ 8$ times as large.

[/hide]


calculus
integration
limit
real analysis
real analysis unsolved

Prove that $ \lim_{n\to\infty}\frac{1}{n}\sum\limits_{k\,\equal{}\,0}^{n}{\left\{ (n\minus{}k)\int_{\frac{k}{n}}^{\frac{k\plus{}1}{n}}{f(x)\,dx} \right\}}\equal{}\int_{0}^{1}{(1\minus{}x)f(x)\,dx}.$

[quote="Black"]Prove that $ \lim_{n\to\infty}\frac {1}{n}\sum\limits_{k\, \equal{} \,0}^{n}{\left\{ (n \minus{} k)\int_{\frac {k}{n}}^{\frac {k \plus{} 1}{n}}{f(x)\,dx} \right\}} \equal{} \int_{0}^{1}{(1 \minus{} x)f(x)\,dx}.$[/quote]

Suppose $ f$ is continous on $ R$

after some simplification

$ u(n)\equal{}\frac {1}{n}\sum\limits_{k\, \equal{} \,0}^{n}{\left\{ (n \minus{} k)\int_{\frac {k}{n}}^{\frac {k \plus{} 1}{n}}{f(x)\,dx} \right\}} \equal{}\frac{1}{n}F(k/n)$ Riemann sum

with $ F(x)\equal{}\int_{0}^{x}f(t)dt$

$ u(n)$ tends to $ \int_{0}^{1}F(t)dt\equal{}\int_{0}^{1}(1\minus{}x)f(x)dx$ with an integration by parts


integration
trigonometry
function
real analysis
real analysis unsolved

Let $f,g,h: \mathbb{R}\to\mathbb{R},\ g(x)=f(x)\sin(x), \ h(x)=f(x)\cos(x)$ for any real $x$. If $g,h$ have primitives on $\mathbb{R}$ show that $f$ has too.

you say that if $f^2$ has primitives so does $f$? Your answer (2) isn't clear enough to since my knowledge about primitives are small..

AYMANE's post doesn't make much sense to me either: if $g$ has a primitive, it doesn't follow that $g^2$ has a primitive and if $f^2$ has a primitive, then it doesn't follow in general that $f$ has a primitive. The solution I know runs as follows. Let $G$ and $H$ be the primitives of $g$ and $h$ respectively. Let $F(x)=G(x)\sin x+H(x)\cos x$. Then $F$ is differentiable and $F'(x)=f(x)+[G(x)\cos x-H(x)\sin x]$. But the function in brackets is continuous and, therefore, has an antiderivative. It remains to subtract this antiderivative from $F$ to get an antiderivative of $f$.

That's better. Taking a function $F(x)$ like you did isn't so obiouvsly for me. Maybe the problem is classical so there is an explination :)

maybe i'm wrong, but:

we know (edit: i was wrong)

:)

You forgot $f'(x)$ coming from the chain rule (which may fail to exist, by the way) ;)