Fie $A$ o multime inclusa in $[0,1].$ Atunci multimea este Lebesgue masurabila daca si numai daca
  \[ \lambda^{*} (A) +\lambda^{*} ([0,1]-A) = 1. \]

The direct implication is obvious.
For the indirect one, we have to show that: $\lambda^{*}(E) \geq \lambda^{*}(E \cap A) + \lambda^{*}(E-A)$ for every set $E$ of reals (the subadditivity of $\lambda^{*}$ giving the reverse inequality). We can assume $E \subseteq [0,1]$ since the $\sigma$-algebra on $[0,1]$ given by the trace of the Lebesgue $\sigma$-algebra on $\mathbb{R}$ is exactly the Lebesgue $\sigma$-algebra on $[0,1]$.
Suppose the result to be false. Then $\exists E \subseteq [0,1], \lambda^{*}(E) < \lambda^{*}(E \cap A) + \lambda^{*}(E-A)$. Hence by definition of $\lambda^{*}$ there exists an open (in $[0,1]$) set $O$ (containing $E$) satisfying $\lambda^{*}(O) < \lambda^{*}(O \cap A) + \lambda^{*}(O-A)$. But then:
$\lambda^{*}(O^{c} \cap [0,1])+\lambda^{*}(O) < \lambda^{*}(O^{c} \cap A) + \lambda^{*}(O^{c} \cap [0,1]-A) + \lambda^{*}(O \cap A) + \lambda^{*}(O-A)$,
$1 < [\lambda^{*}(O^{c} \cap A)+\lambda^{*}(O \cap A)]+[\lambda^{*}(O^{c} \cap [0,1]-A)+\lambda^{*}(O-A)]$,
$1 < \lambda^{*}(A)+\lambda^{*}(A^{c}\cap [0,1]) = 1$, which is absurd.
For the last inequality we used the measurability of $O$ and $O^{c}\cap [0,1]$.

A similar one: define the inner measure $m_{*}(E)=sup(\{m(U); \ U \subseteq E, \ U \ measurable\})$ where $m$ is the Lebesgue measure on $\mathbb{R}$. Show that $m_{*}(E)=1-m^{*}([0,1]-E)$ for all $E \subseteq [0,1]$, where $m^{*}$ denotes the Lebesgue outer measure.


vector
probability
linear algebra
matrix
function
integration
probability and stats

how would you do to generate a random vector $ (X_1, X_2, \ldots, X_n)$ such that
1/ $ P(X_i\equal{}1) \equal{} 1\minus{}P(X_i\equal{}0)\equal{}p_i$ is a given
2/ $ Cov(X_i X_j)\equal{}\sigma_{i,j}$ is given
as efficiently as possible ?

I will assume for simplicity that the probability of success for all $ X_i$ is $ p$. One possible approach to simulate vector $ X$ is this: let $ \xi_1,\ldots,\xi_n$ be iid Bernoulli with probablity of success $ p$. Let $ X_1\equal{}\xi_1$. Let $ \eta_2$ be independent of $ \xi_i$ s.t. $ P(\eta_2\equal{}1)\equal{}q_1, \ P(\eta_2\equal{}2)\equal{}1\minus{}q_1$, and define $ X_2\equal{}\xi_{\eta}$. It is easy to check that if $ q_1\equal{}\frac{\sigma_{12}}{p(1\minus{}p)}$, then we get the desired covariation. Now do the same for $ X_3$: this time $ \eta_3$ can take 3 values, and the probabilities assigned to them are the solution of the system of linear equations of order 3.
Anyway, this is pretty straightforward and probably not the most effective procedure.

Yustas, I must admit that I don't understand what have you written  :(. Can you ( or someone else ) please explain in detail how to solve alekk's problem?

I would also like to add another question; suppose that we want to generate a random vector $ (X_1,\dots,X_n)$, but let's assume that $ X_i$  (*) are continuous variables and we are given the covariance matrix $ \Sigma$. How would you do it?
We could express the distribution function $ \mathbb{F}_{(X_1,\dots,X_n)}(x_1,\dots,x_n) \equal{} \int_{t_i\leq x_i,\i \equal{} 1,\dots n}f(t_1,\dots,t_n,\Sigma)dt_1\dots dt_n$, then simulate random variable $ U(\omega)$ from uniform distribution and invert $ \mathbb{F}_{(X_1,\dots,X_n)}^{ \minus{} 1}(U(\omega))$.
But that seems to complicated, and it would be challenging to write a program that would do that in general situations...¸

Edit: correction: (*) we assume that $ (X_1,\dots,X_n)$ is continuous, that is, it has a density function.

Basically, the idea is to generate our sequence by transforming the array of independent random variables $ \xi_i$ by assigning the random numbers $ \eta_i$, that is to look at $ \xi_{\eta_i}$. I gave some details above.

1) The problem is NP-hard. Moreover, even determining whether there exists a distribution of $ (X_1,\dots, X_n)$ with the given covariance matrix is NP hard (each $ X_i$ is a 0-1 random variable).

2) Also the covariance matrix doesn't uniquely define the distribution of $ (X_1,\dots, X_n)$.

3) The problem becomes easy if we don't require that each $ X_i$ is a 0-1 random variable. Then we can sample vector $ X$ from a multivariate Gaussian distribution ( http://en.wikipedia.org/wiki/Multivariate_normal_distribution ).

do you have any reference for the NP-hard thing ?[/youtube]

I will sketch the proof. Let us consider a special case when all $ p_i \equal{} 1/2$. For simplicity (of notation), we assume that each $ X_i$ takes two values -1 and 1 --- each w.p. 1/2.

Consider the set $ C$ of covariance matrices corresponding to different joint distributions of $ (X_1,\dots,X_n)$. Note that the set $ C$ is convex: if $ A\in C$ is the covariance matrix of vector $ X$ and $ B\in C$ is the covariance matrix of vector $ Y$, then $ (A \plus{} B)/2$ is the covariance of the random vector
\[ Z \equal{} \begin{cases} X, & \text{w.p. } 1/2; \\
Y, & \text{w.p. } 1/2. \end{cases}
\]
Assume that we are given a graph $ G \equal{} (V,E)$ on the set $ V \equal{} \{1,\dots,n\}$. Note that each vector $ X$ defines a cut on $ G$ (i.e. partitioning of its vertices into two parts): $ S \equal{} \{i: X_i \equal{} 1\}$ and $ T \equal{} \{i: X_i \equal{} \minus{} 1\}$.

Moreover, for every cut $ V \equal{} S\cup T$ in $ G$, there is a corresponding random vector $ X_{S,T}$:
\[ X_{S,T} \equal{} \begin{cases} \chi, & \text{w.p. } 1/2; \\
\minus{} \chi, & \text{w.p. } 1/2. \end{cases}
\]
where
\[ \chi_i \equal{} \begin{cases} 1, & \text{if } i\in S; \\
\minus{} 1, & \text{if } i\in T. \end{cases}
\]
We compute the covariance matrix $ A^{S,T}$ of $ X_{S,T}$:
\[ a_{ij} \equal{} \begin{cases} 1, & \text{if } i\in S, j\in S \text{ or } i\in T, j\in T \\
\minus{} 1, & \text{if } i\in S, j\in T \text { or } i\in T, j\in S. \end{cases}
\]
It is easy to see that matrices $ A^{S,T}$ form the set of vertices of $ C$. Thus the set $ C$ is called [b]the cut polytope[/b] in the literature.

Finally, we compute the number of cut edges (i.e. edges going from $ S$ to $ T$):
\[ f(A) \equal{} \sum_{(i,j)\in E} \frac {1 \minus{} a_{ij}}{2}
\]
Note that the number of cut edges is given by a linear function.

If we could check whether a point belongs to $ C$ we would be able find the maximum value of a linear function over $ C$ (see e.g. http://www-math.mit.edu/~vempala/papers/adamanneal.pdf ). In particular, we would be able to find the vertex of $ C$ that maximizes $ f$ --- or in other words, we would be able to find the maximum cut in $ G$. But the Maximum Cut Problem is NP-hard [Richard M. Karp (1972). "Reducibility Among Combinatorial Problems"].

By the way, there is a lot of research on the cut polytope and its "relaxations" in computer science.

thank you very much! that was very interesting !


There are $ n$ persons and $ n$ seats available for them.They are all assigned seat numbers written on a ticket given to them.Now there is one person(say $ Mr.X$) on whose ticket the number is not mentioned.

The people are told to go and sit one by one,$ Mr.X$ being the first person.

The rule is that if $ Mr.X$ doesn't know the number on the ticket then he can sit on any seat chosen at random by him.

And if the person who comes after $ Mr.X$(say $ Mr.Y$) sees that $ Mr.X$ is sitting on the seat meant for him then $ Mr.Y$ can sit on any seat he likes chosen at random, otherwise $ Mr. Y$ will sit on the seat assigned to him on the ticket.Then $ Mr.Z$ comes and if he sees that his seat is occupied then he can sit on any seat he likes otherwise he will sit on the seat assigned to him.This process continues till the last person has been seated.

Find the probability that the last person is sitting on the seat assigned for him.

I think the above problem is akin to the one posted in the thread below:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=278908[/url]

P.S.: A solution had been posted in the 'link' that I had provided.


geometry
incenter
circumcircle
angle bisector
geometry proposed

Given quadrilateral $ABCD$ inscribed $(O)$ $AC\cap BD=I$ the circle $(C)$ tangent the rays (not segment )$ID,IC$ at $M,N$ and tangent external $(O)$, Let $I_{a},I_{b}$ be excenters of triangle $ADC$ and $BDC$ corespond angle $A,B$ resp,Prove that $M,N,I_{a},I_{b}$ are colinear. :D

Applying Thebault theorem for triangle $ADC$ and $BDC$we have the above result. This problem is still right when the circle $(C)$ tangent internal $(O)$.
Then $M,N$ and the incenters of triangle $ADC$ and $BDC$ are collinear.

Do you understand my problem,Huyền Vũ ?
[quote="Wolfram MathWorld"]Let $ABC$ be a triangle and $D$ a point on the side $BC$. Let $I$ be the incenter, $P$ the center of the circle tangent to the circumcircle and segments $AD$ and $BD$, $Q$ the center of the circle tangent to the circumcircle and segments $AD$ and $CD$. Then Thébault's theorem states that $P, Q$ and $I$ are collinear.[/quote]
Actually "Thébault's theorem" is only corollary of mine, It is three colinear centers while my problem need to prove tangents points and centers, more the relations theorem is for internal tangent circles while mine is for external!

Oh sorry .The problem that I wrote is the lemma to prove Theault theorem.
You can see the lemma here. This lemma is right when (C') tangent external or internal the circle (O). http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf
When (C') tangent external (O) the tangents point and excenter are collinear (*)
When (C') tangent internal (O) the tangents point and incenter are collinear (**)
You can solve (*) similarly the solution for lemma in the file or use (**) to solve (*)
This is my solution for (*)
[b]Lemma:[/b]
Let $I$ be the point on $AC$ of triangle $ADC$. The circle $(C')$ tangents the ray $ID,IC$ at $M,N$ and tangents external the circumcircle $(O)$ of triangle $ADC$ at $K$.Let $Ia$ be excenter of triangle $ADC$ corespond angle $A$. Prove that $M,N,Ia$ are collinear

[b]Solution[/b]
$KN,KM$ cuts $(O)$ at $P,Q.$
$K$ is the similar point of $(O)$ and $(C')$ $(1)$ so $C'N // OP$. but $C'N \bot AC$
 so $OP \bot AC$. So $P$ is the midpoint of arc $AC(2)$
$MN$ cuts $DP$ at $Ia$
$\angle PCN=180-\angle PCA=180-\angle PAC$ (because of  (2)) $=\angle PKC$
So Triangle $PKC$ and $PCN$ are similar
$PC^{2}=PK*PN (4)$
we have $MN//PQ$ (because of (1))
So $\angle IaMK=\angle PQK=\angle PDK=\angle IaDK$
Thus $M,D,K,Ia$ are cyclic. So $\angle DIaK=\angle DMK=\angle KNM$
So that triangle $PIaK$ and $PNIa$ are similar
so $PIa^{2}=PK*PN (5)$
From (4) and (5) $PIa=PC=PA.$
So $A,C,Ia$ are on the circle center $P$ and the radius $PA.$
We have $\angle CAIa=1/2 \angle CPIa=1/2 \angle CAD$
So $AIa$ is the angle bisector $CDA$ or $Ia$ is the excenter of triangle $ACD$  $q.e.d$
Applying this lemma for the main problem we have solution

See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41667]Concyclic points with triangle incenter[/url] (external tangency and excenter) and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=118385]Circles and circles[/url] (internal tangency and incenter).


In how many ways can 36 be written as the product
a × b × c × d, where a, b, c and d are positive integers such
that a ≤ b ≤ c ≤ d ?

[quote="HiDN428"]We are asked to find in how many
ways 36 can be written as the
product a × b × c × d, where a, b, c
and d are positive integers such that
a ≤ b ≤ c ≤ d.
First factor 36:
36 = 2 × 2 × 3 × 3
The factors of 36 are:
1, 2, 3, 4, 6, 9, 12, 18 and 36
Suppose the highest value we use is
36.
1 × 1 × 1 × 36
Can’t do anything more with 36.
Now 18.
1 × 1 × 2 × 18
Now 12
1 × 1 × 3 × 12
Now 9:
1 × 1 × 4 × 9
1 × 2 × 2 × 9
Now 6:
1 × 1 × 6 × 6
1 × 2 × 3 × 6
Now 4:
1 × 3 × 3 × 4
Now 3:
2 × 2 × 3 × 3
We’re done. Let’s count them up: There are 9. so the ans is 9, i think.[/i][/quote]

Meh..I kinda feel stupid replying to my own post but.. This is how I did it.

First factor 36:
36 = 2 × 2 × 3 × 3
The factors of 36 are:
1, 2, 3, 4, 6, 9, 12, 18 and 36
Suppose the highest value we use is
36.
1 × 1 × 1 × 36
Can’t do anything more with 36.
Now 18.
1 × 1 × 2 × 18
Now 12
1 × 1 × 3 × 12
Now 9:
1 × 1 × 4 × 9
1 × 2 × 2 × 9
Now 6:
1 × 1 × 6 × 6
1 × 2 × 3 × 6
Now 4:
1 × 3 × 3 × 4
Now 3:
2 × 2 × 3 × 3
So there are a total of 9 ways.
Ok, is there an easier way..lol?

haha, idk, i think the best way is just to list them out because there really isn't that much...


A COMPUTER SOFTWARE PACKAGE GENERATES RANDOM NUMBERS THAT ARE NORMALLY DISTRIBUTED WITH A MEAN AND STANDARD DEVIATION THAT ARE ENTERED BY THE USER. SUPPOSE THAT THE USER ENTERS A MEAN OF 400 AND A STANDARD DEVIATION OF 100. FIND THE PROBABILITY THAT A VALUE WILL BE

1.) BETWEEN 525 AND 650
2.) ABOVE 490
3.) BELOW 535
4.) BELOW 357
5.) ABOVE 619
6.) BETWEEN 340 AND 714

Please help me how can I answer with this one. Even steps will help. Thanks in advance!

Ew, statistics. Brings back memories of statistics videos of MathPath. If only I had payed attention once in a while during those videos, I would most likely remember the formula for standard deviations.

I guess my post is spam then.

Normalize the scores by $\frac{X-\mu}{\sigma}$ and then look up the p-value in a z table.

Hokkage, can you show me one example please? I have a z table here..

Hokkage means to find the z-scores, by using $z=/frac{x-\mu}{\sigma}$, where $x$ is the given values. For example, for $x=525$, $z=\frac{525-400}{100}=1.25$. So part 1) becomes: Find $P(1.25<z<2.5).$
I don't mean to be condescending but this is a very basic concept. You should show us your attempts so we can tell where the problem is. We can tell you the steps, but so can your text.


$x,y$ satisfy $x>y>0$ and $2(x+y)\geq 5\sqrt{xy}$. Prove that $x\geq 4y$.

yes,it is small.
just $\sqrt{\frac x y}=a$,$a>1$
we have $a+\frac 1 a \ge \frac 5 2$
so $a \ge 2$
...

You're right zhaobin
$\ 2(x+y) \geq 5 \sqrt{xy} \Rightarrow x-\frac{5}{2} \sqrt{xy}+y \geq 0$
$\ \rightarrow (\sqrt{x}-2\sqrt{y})(\sqrt{y}-2\sqrt{x}) \geq 0$
which shows your idea: $\ \sqrt{\frac{x}{y}} \geq 2$


I have been making suggestion for this Section for a while  :?
Nobody seems to notice :)
Anyway, who would like to see something like this, please vote.

Yeah, could be a good idea. At least, considering it would be mainly about algorythms and not really about coding language, as the algorythm search comes pretty close to the thinker-math questions :D

Especially graph theory and other counting stuff :)

You guys must convince more than 9 people to vote :) (and especially to vote yes :P) ...

I don't get people that vote NO. Does it trouble them in any way? I know that I am wrong , but when I put the option No, I didn't think there would be such votes. Otherwise, the only answer I would have allowed would be YES :) :D

well I have no idea, but not even half the people said yes, so :( sorry ..

Though programming is a part of my profession (I lecture some programming cources) I think MathLinks is not right place for subj.

[quote="Valentin Vornicu"][quote="PiDeltaPhi"]Why not Myth?
Shouldnt Mathlinks expand?

I dont understand the people who voted no...i'm sorry...[/quote] Well for the moment it's hard maitaining the site with only math people :) expanding to CS would mean a major increase in web traffic, and I am not (yet) prepared for this :)

However, people can discuss CS subjects in the Bable Tower :) :D[/quote]

I understand that  :)

A forum for computer sciences and informatics already exists. And once again, please do not revive old threads.

Locked.


calculus
function
limit
logarithms
floor function
calculus computations

This is a small arithmetic trifle, not a serious problem.

I was writing up a classroom handout for a calculus class on the "Hierarchy of Size," which is something I think quite important to the understanding of calculus. As an example, I talked about the function $ f(x) \equal{} x^{1000}e^{ \minus{} .001x}$ on $ [0,\infty).$ I wrote that although the function increases at first and reaches an impressively high maximum value, it still eventually decreases and $ \lim_{x\to\infty}x^{1000}e^{ \minus{} .001x} \equal{} 0.$

But then, to get the effect I intended, I needed to name that maximum value.

So: Let $ M \equal{} \max_{x\in[0,\infty)}x^{1000}e^{ \minus{} .001x}.$

1. Using only pencil-and-paper arithmetic - no calculator or computer - estimate the order of magnitude of $ M.$

2. Using only an ordinary floating-point device (calculator or spreadsheet), give a reasonable floating-point approximation of $ M.$

Do not use anything (Mathematica, Maple, TI-89, etc.) that uses arbitrary-precision arithmetic; the restriction is to ordinary floating-point arithmetic.

[hide="Hint for 1."]Two of the many approximations I have memorized over the years: $ \ln 10\approxeq2.303$ and $ \log_{10}e\approxeq.4343.$[/hide]

Ah, but the point was a base-10 approximate representation of that number.

You are sligtly better off with $ e^7\approx 1.1\cdot 10^3$, $ 1.1^{10} \approx e$, so
\[ e^{1000}\approx 1.1^{143}\cdot 10^{429} /e\approx 1.1^3 e^{13}\cdot 10^{429}\approx 1.1^5\cdot 10^{435}/e\]
so
\[ \frac{10^{6000}}{e^{1000}}\approx 10^{5565}\sqrt e\approx 1.6\cdot 10^{5565}.\]
The (relative) error is about $ \frac{1}{250}\cdot 143$ coming mainly from the estimate of $ e^7$, which is really $ 1096.\text{something}$ (the thing I memorized long ago when trying to find a power of $ e$ close to a power of $ 10$), so one may be pretty sure that $ 10^{5565}$ is a fairly close guess (unless I made some stupid mistake in my arithmetic).

Both of you have been going after question 1. Here's my version of that, depending on my long-ago-memorized $ \log e\approxeq.4343.$ (I'm using $ \log$ here to mean $ \log_{10}.$)

$ M\equal{}(10^6)^{1000}e^{\minus{}1000}\equal{}10^{6000}e^{\minus{}1000}.$

$ \log M\equal{}6000\minus{}1000\log e\approx 6000\minus{}434.3\equal{}5565.7.$

Since $ \log 2\equal{}.3$ (which is the same fact as $ 2^{10}\equal{}10^3$) implies that $ \log 5\equal{}.7,$ that gives us an estimate for $ M$ of $ 5\times 10^{5565}.$

So fedja definitely has the correct order of magnitude. To settle the issue of whether $ 1.6$ or $ 5$ is the better number up front requires us to move on to question 2, in which we do get to use an ordinary calculator or spreadsheet. There, the computation I want to do this: let $ a\equal{}(6000\minus{}1000\log e)\minus{}\lfloor 6000\minus{}1000\log e\rfloor,$ and then we let $ b\equal{}10^a.$ We report the answer as $ b\times 10^{5565}.$ This gives us $ 5.07596\times 10^{5565}.$


Let $ a,b,c$ be nonnegative real numbers.

(a) If $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac {7a^2 \plus{} 9}{b^2 \plus{} c^2} \plus{} \frac {7b^2 \plus{} 9}{c^2 \plus{} a^2} \plus{} \frac {7c^2 \plus{} 9}{a^2 \plus{} b^2}\ge 24$;

(b) If $ ab \plus{} bc \plus{} ca \equal{} 3$, then

$ \frac {4a^2 \plus{} 3}{b^2 \plus{} c^2} \plus{} \frac {4b^2 \plus{} 3}{c^2 \plus{} a^2} \plus{} \frac {4c^2 \plus{} 3}{a^2 \plus{} b^2}\ge \frac {21}{3}$;

(c) If $ abc \equal{} 1$, then

$ \frac {3a^2 \plus{} 1}{b^2 \plus{} c^2} \plus{} \frac {3b^2 \plus{} 1}{c^2 \plus{} a^2} \plus{} \frac {3c^2 \plus{} 1}{a^2 \plus{} b^2}\ge 6$.

These are very nice (see point (a) and (b) above)

Let $ a,b,c$ be nonnegative real numbers.

(a) If $ a\plus{}b\plus{}c\equal{}3$, then

$ \frac {4a^2 \minus{} 27}{b^2 \plus{} c^2} \plus{} \frac {4b^2 \minus{} 27}{c^2 \plus{} a^2} \plus{} \frac {4c^2 \minus{} 27}{a^2 \plus{} b^2} \plus{} \frac {69}{2}\ge 0$;

(b) If $ ab \plus{} bc \plus{} ca \equal{} 3$, then

$ \frac {a^2 \minus{} 18}{b^2 \plus{} c^2} \plus{} \frac {b^2 \minus{} 18}{c^2 \plus{} a^2} \plus{} \frac {c^2 \minus{} 18}{a^2 \plus{} b^2} \plus{} \frac {51}{2}\ge 0$.

Equality in each inequality occurs for $ a \equal{} b \equal{} c$, and also for  $ \frac a{5} \equal{} b \equal{} c$ or any cyclic permutation.

[quote="Vasc"]These are very nice (see point (a) and (b) above)

Let $ a,b,c$ be nonnegative real numbers.

(a) If $ a \plus{} b \plus{} c \equal{} 3$, then

$ \frac {4a^2 \minus{} 27}{b^2 \plus{} c^2} \plus{} \frac {4b^2 \minus{} 27}{c^2 \plus{} a^2} \plus{} \frac {4c^2 \minus{} 27}{a^2 \plus{} b^2} \plus{} \frac {69}{2}\ge 0$;

(b) If $ ab \plus{} bc \plus{} ca \equal{} 3$, then

$ \frac {a^2 \minus{} 18}{b^2 \plus{} c^2} \plus{} \frac {b^2 \minus{} 18}{c^2 \plus{} a^2} \plus{} \frac {c^2 \minus{} 18}{a^2 \plus{} b^2} \plus{} \frac {51}{2}\ge 0$.

Equality in each inequality occurs for $ a \equal{} b \equal{} c$, and also for  $ \frac a{5} \equal{} b \equal{} c$ or any cyclic permutation.[/quote]
(a) and (b) in this problem are equivalence. Similar, (a) and (b) in the first problem are equivalence. :) I proved them by mixing variable methods. :)

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 3$. Then

$ \frac {a^2 \minus{} 18}{b^2 \plus{} c^2} \plus{} \frac {b^2 \minus{} 18}{c^2 \plus{} a^2} \plus{} \frac {c^2 \minus{} 18}{a^2 \plus{} b^2} \plus{} \frac {51}{2}\ge 0$,

with equality for $ a \equal{} b \equal{} c$, and also for  $ \frac a{5} \equal{} b \equal{} c$ or any cyclic permutation.[/quote]
The general problem is to find $ m$ for given nonnegative $ k$ such that

$ \sum \frac {ma^2 \plus{} ab \plus{} bc \plus{} ca}{ka^2 \plus{} b^2 \plus{} c^2} \le \frac {3(m \plus{} 3)}{k \plus{} 2}$.

Other two interesting particular cases (for $ m\equal{}1$) are the following:

$ (1) \ \ \ \ \ \sum\frac {(a \plus{} b)(a \plus{} c)}{7a^2 \plus{} b^2 \plus{} c^2} \le \frac 4{3}$,

with equality for $ a \equal{} b \equal{} c$, and also for  $ \frac {5a}{4} \equal{} b \equal{} c$ or any cyclic permutation;

$ (2) \ \ \ \ \ \sum\frac {(a \plus{} b)(a \plus{} c)}{a^2 \plus{} 4b^2 \plus{} 4c^2} \le \frac 4{3}$,

with equality for $ a \equal{} b \equal{} c$, and also for  $ \frac {2a}{7} \equal{} b \equal{} c$ or any cyclic permutation.

The case $ m \equal{} 0$ is also interesting.


vector
absolute value
linear algebra
linear algebra unsolved

Suppose that $u_1,...,u_k$ are some unitary vectors in an euclidian space of dimension $n$ such that the absolute value of $<u_i,u_j>$ is the same for all $i<>j$ and it is different from 0 and 1. Prove that $k\leq n(n+1)/2$. Can you find an optimal bound ( I don't have answer for the last question and I believe it is very difficult).

Agregation Composition Mathématiques Générales 1979

Harazi there were severals questions before asked to prove this inequality.

I know, I know and I give you the same answer as in the post with hyperbolic matrices. You know, solving problems is not just giving the reference.  ;)


(1)
shazde koochooloo rooye sayyareye koochaki zendegi mikonad ke shoaaash $60$ metr ast.roozi ba shoroo az rooye khatte esteva,$30\pi$ metr be shargh,$20\pi$ metr be shomal,$30\pi$ metr be gharb va saranjam $20\pi$ metr be jonoob miravad.shazde koochoolo ta makaane avvalash chand metr faasele darad?
1)sefr
2)$30\pi$
3)$20\pi$
4)$40\pi$
5)$15\pi$

(2)
zaraayebe chand jomleie $P$ sahih ast.tahte kodaam sharaayete zir $P$ nemitavaanad risheye sahih dashte bashad?
1)$P(5)=5,P(6)=6$
2)$P(6)=5,P(5)=5$
3)$P(5)=6,P(6)=6$
4)$P(6)=5,P(5)=6$
5)tahte harkodam az in sharayet ,$P$ mitavanad risheye sahih dashte bashad.

salam  narahat nabash inshalah ghabooli man va  doostam sare olampiad basij intori shodim
albate soalate emrooz benazaram sakhttar as ghabl bood
7)mostatili dar safhe ba roose (0,0),(0,100),(150,0),(150,100) hast. chand khate movazi ba ghotr gozara az (0,0)azlae motatil ra dar do noghtey ba moghtasate sahih ghat mikonad? khode ghotr ham hast?
1)99   2)100    3)199    4)200     5)300
 man khodam migam 1)99      chera ke :     x/100 =y/150     x= 2,4,6,.....100  dar natige tedadeghtrha mishe
                                                                                50 +50 -1(ghotr dobar) =99

Ber nazare manam kheili fekre khoobie.inke javaabaaro injaa benevisim.Faghat yeki zahmat bekeshe hamashoono injaa benevise..Dar zemn be nazare manam soaalaaye emsaal ye meghdaar gheire moteaaref bood.fek mikonam nomreye ghabooli 7-8taa soaale doros baashe.

2 ta az soalha ra ke midoonam too emtehan oomade bood male man bood yeki jame adade jadvale $10\times10$ va digari ham tamame $n$ hayi ra biyabid ke $\sqrt{n-1}+\sqrt{n+15}$ sahih bashad. Har do soalhaye asooni boodand shayad soalate digari az manham too emtehan oomade bashe. :)  :)

dar shekle roo be roo mikhahim az noghte A be B beravim.masir ha dar jahate felesh yek tarafe hastand.be chand tarigh mitavanim in kar ra anjam dahim?
man khodam 58 be dast avordam.idamam ine ke az noghte B be aghab bargardim yani hesab konim az har kodam az noghat be chand tarigh mitavan be B raft. :)

kodam bozorgtar ast?
1)2 be tavane 431
2)3 be tavane 421
3)4 be tavane 321
4)21 be tavane 43
5)31 be tavane 42
agar eshtebah nakarde basham gozine 2 dorost hast. :)

The answer is $\left(\begin{array}{c}6\\2,2,2\end{array}\right)=90$

For the number of paths from $A$ to $B$.

Etefghan jove gozineha hast gozineha ebaratand az :

$a)60\\ b)90\\ c)36\\ d)54\\ e)45$

(3)algorithme zir ra rooye chand jomleie $P(x)$ ejraa kardeim.
1.$d$ ra baraabare darajeye $P$ gharaar bede va agar $d<1$ be satre 4 boro.
2.$a$ ra baraabare zaribe $x^d$ dar $P$ gharaar bede va $P(x)-a{x^{d-1}}(x+2)$ ra be jaye $P(x)$ bogzaar.
3.be satre 1 boro.
4.$P$ ra chap kon.
pas az ejraaye algorithm adade $1384$ chap shode ast.$P$ dar aghaaz kodaam boode ast?
1)$x^{10}-45x^3$
2)$13{x^7}-2$
3)${x^11}-83{x^3}$
4)$16{x^8}+x$
5)$-x^{11}+84$

bebakhshid ke man halle khodamo nemizaram chon man aslan  khodamo ghabool nadaram.man dochare yek jange darooni hastam....

kermi shekamoo mikhahadhameye mivehaye derakhte rooberoo ra bekhorad( :blush: ) va dar eyne hal kamtarin masaafat ra tey konad kodam makan behtar ast ?
1)a   3)b  3)c   4)d    5)e
  man migam b  aya doroste?!

masaleye aakhar:
shekle rooberoo majmooe javabhaye kodam yek az moaadelaate zir ast :
 man zadam 4  :
1) y=0 , x yek javaab dare
2)ba taghire x be -x moaadele sabet mimoone vali dar shekl ekhtelafe noghate barkhorde nemoodar ba mehvar moghtasaat ta markaz mashhoode
3)y sefr nemishavad valiiiiiiii
4)+
5)x va y motagharenand vali be nemoodar nemikhore !!

bacheha in soal chi mishe?
tasaode hesabi az addade avvale ba ghadre nesbate n^2+1   ke n addadi tabii ast hadeaksar chand ozv darad???
1)3   2)4   3)5  4)6  5)tasaode ba har teddade ozv vojood darad

az shahre a jadeii mostaghim kharej shode ke  2 shahre b va c dotarafash gharar darand majmooe faseleye b va c az jadee hadeaksar chand kilometr ast midanim ke faseleye shahre a az 2 shahre digar 60 va 50 kilometr ast va faseleye do shahre b va c 40 kilometr ast?
1)30 2)40     3)50  4)60  5)70

dorbini zire havapeyma nasb shode havapeyma rooye masir khatti dar hal oj giri ast zamin ra mosatah farz konid masahate filmbardari shode tabee daraje chandi az jabejayi makani havapeyma ast?
1)1  2)2  3)3  4)4  5)chand jomleii nist

nimsaz haye dakhelli mosalas ABC dayere mohiti an ra dar noghat A' ,B',C' ghat mikonad agar I' markaze dayere mohati dakhelli mosalase A'B'C'  bashad andaze zaviye B'I'C' barabare kodam gozine hast?
1)90+ B+C/4     2)180-B+C/4  3)2A+B-C   4)180-A    5)2A

javaabhaaro injaa neveshtam :)

[url=http://math-olympiad.persianblog.com/]math-olympiad.persianblog.com/[/url]


hey
  I found a website where you could buy the fundamentals of physics full solution manual for just 20 dollars. It is really cool. All problems are solved and no one is missing. I was able to do all my assignments and prepare for my exams. I recommend this to everyone. Just click the link   [u]http://windsor.kijiji.ca/c-buy-and-sell-books-Fundamentals-of-physics-full-solution-manual-8TH-EDITION-W0QQAdIdZ76404065[/u]

That's pretty cool.  Can you by any chance find the solutions manual for the Halliday/Resnick/Krane book?

I downloaded the full solution manual (maybe not) from eSnips for free as well as the main text. I'm not sure if it's complete though.

ofc
soln ,book, and other good books
all r available at http://www.4shared.com


Up to my knowledge, this problem is open (see http://mathworld.wolfram.com/CarmichaelsTotientFunctionConjecture.html ).

That is strange... Hojoo, do you have the book in which this appears, or could you check it with the person that contributed this?

[quote="Peter"]Show that if the equation $\phi(x)=n$ has one solution, it always has a second solution, $n$ being given and $x$ being the unknown.[/quote]

I'm not sure... but it seems that I also included some open problems :)

I have asked this problem on ML.Actually,I have tried to solve this one for a day.so disppointed.I think I have no ability to solve an open problem...


email
group theory
abstract algebra
superior algebra

Find all groups that have exactly 6 subgroups. Our algebra teacher from Bucharest said last year that this is a monstruous problem. And I think he's right.

You can send your answer in french or english

Alain Tissier (Professeur MP*)
17, avenue des Ormes, 93370 Montfermeil
France

or by email

al.tissier@laposte.net

You must mention this is the Question 488 in
RMS  n2; Mars 2004; page 173

If your answer is nice RMS will publish it.

Assuming the $6$ subgroups do include the trivial subgroup and the group itself, I believe the only solutions are $\mathbb Z_3\times\mathbb Z_3,S_3$, and all groups of the form $\mathbb Z_{p^5}$, where $p$ is prime. I was on my way to post what looked like a correct solution, when my lights went out and I lost my message, so I won't be posting it again any time soon :). There's nothing fancy about it anyway: we just need to check a few cases.

[quote="grobber"]Do the $6$ subgroups include the improper ones? (i.e. the group itself and the group formed only by the unit)[/quote]
Wrong terminology! Only the group itself is called an improper subgroup. The subgroup formed only by the unit is the trivial subgroup and is a proper subgroup, unless $G=1$.

[quote="jmerry"]One more set of obvious solutions: $\mathbb{Z}_{pq^2}$[/quote]

Oh, yes, I forgot about those. I hope there are no more. Actually, I kind of stormed through the cases that needed to be checked, and there might have been some slip-ups such as this one.

You did miss one: the quaternion group.

Here's an argument:
First, we consider cyclic groups. It is clear that a cyclic group has six subgroups when it has order $p^5$ or $p^2q$ for distinct primes $p,q$.

Now, suppose $|G|=n$ and the largest order of an element is $m$, with $m<n$. The number of subgroups is at least $2+\frac{n-1}{m-1}$, so $n<4m$.
Case 1: $n=3m$. In this case, we have a subgroup of index 3, and the elements outside this subgroup generate at least 3 cyclic subgroups. Including the trivial subgroups and the subgroup of order $m$, $m$ is prime. For $m=2$, we get $S_3$. For $m=3$, we get $\mathbb{Z}_3^2$ (which technically doesn't belong here). For $m>3$, the subgroup of order $m$ is normal and we have $3+3\frac{m-1}2$ subgroups, which is too many.
Case 2: $n=2m$. Here, the elements outside the subgroup generate at least two cyclic subgroups, so either $m$ is prime or $m$ is the square of a prime. Considering $n=2p$ with $p>2$, we have the dihedral group, with $p+3$ subgroups. This works for $p=3$, giving us $S_3$.
In the $n=2p^2$ case, with $p>2$, the group is again dihedral, with more than $p^2+3$ subgroups; this doesn't work.
This leaves $n=4$ or $n=8$ to check separately. This check gives us the quaternion group, and no others.

Recapping the full list:
$\mathbb{Z}_{p^5},\mathbb{Z}_{p^2q},\mathbb{Z}_3^2,S_3,Q$.


Hi,
I have a question which is about the positive definite. Wonder if anyone of you could give a little help, thx.

If $A$ and $X$ are square matrice, $A$ is posotive definite , and $X$ is nonsingular then the matrix $B=X^T A X$ is also positive definite.

I guess I might need to use the Cholesky Theorem to prove it, right? Any hint to help?
Thx

No need. For any $y\neq0$, $y^T(X^TAX)y=(Xy)^TA(Xy)>0$ and $A$ is positive definite.

Never overlook the power of the definition.


let (C) be a circle inwich its center is inaccessible .only by using compass find the center of (C).

Come one, this is the third time you're posting this today!!

And you keep confusing "campos" and "compass".

A little search will do it :

http://www.mathlinks.ro/Forum/viewtopic.php?t=24888 .

oh.. sorry Arne :blush: .

u know.thats because of that i posted all them in one minute.and by that time i didnt know the different betwin

(campos)and (compass) :rotfl:

to see the solution here is alink: http://www.mathlinks.ro/Forum/viewtopic.php?t=57006


induction
function
vector
calculus
derivative
real analysis
real analysis theorems

show:
$ \partial^{\alpha}(fg)\equal{}\sum_{\beta\plus{}\gamma\equal{}\alpha}\frac{\alpha !}{\beta ! \gamma !} \partial ^{\beta}f\cdot\partial ^{\gamma} g$

information on Multi index notation:  http://en.wikipedia.org/wiki/Multi-index_notation

[quote="mlok"]Information on mathematical induction:
http://en.wikipedia.org/wiki/Mathematical_induction[/quote]

I never said that f and g were functions of one variable.
If you want to use induction, there's two things you have to show, first it holds for all n, where n is the number variables. but you also have to show that this holds for all vectors $ \alpha$ no matter its components.

please correct me if im wrong, but this doesnt seem trivial to me.

It's still induction- we're just using the partial ordering on the index set rather than the strict ordering on $ \mathbb{N}$. Assume it holds for all smaller indices, differentiate, and get the recursion using the ordinary product rule. The base case here is no derivatives at all.

Alternately, you could reduce to several copies the single-variable version; partial derivative operators commute, so we can decompose each $ \frac{\partial^{n_i}}{\partial x_i^{n_i}}$ and combine the results.

This is Exercise 1.5.3 from Evans, PDE book.

Let me elaborate on the inductive step. Assume, this formula

$ \partial^{\alpha}(fg)\equal{}\sum_{\beta\plus{}\gamma\equal{}\alpha}\frac{\alpha !}{\beta !\gamma !}\partial^{\beta}f\cdot\partial^{\gamma}g \equal{}
\sum_{\beta \le \alpha}{\alpha \choose \beta}\partial^{\beta}f\cdot\partial^{\alpha \minus{} \beta}g$

holds. Remark: there are $ (\alpha_1 \plus{} 1) \cdot ... \cdot (\alpha_n \plus{} 1)$ summands in this sum. Now we prove that

$ \partial^{\alpha \plus{} e_i}(fg) \equal{}
\partial_{x_i}\sum_{\beta \le \alpha}{\alpha \choose \beta}\partial^{\beta}f\cdot\partial^{\alpha \minus{} \beta}g \equal{}
\sum_{\beta \le \alpha} \left[ {\alpha \choose \beta} \partial^{\beta\plus{}e_i}f\cdot\partial^{\alpha \minus{} \beta}g \plus{} {\alpha \choose \beta}\partial^{\beta}f\cdot\partial^{\alpha \minus{} \beta \plus{} e_i}g \right] \equal{}
\left. \sum_{\beta \le \alpha \plus{} e_i}{\alpha \plus{} e_i \choose \beta}\partial^{\beta}f\cdot\partial^{\alpha \plus{} e_i \minus{} \beta}g \right.$.

Indeed, the equality is true because each term of the form $ \partial^{\beta}f\cdot\partial^{\alpha \plus{} e_i \minus{} \beta}g$ comes from either a term $ \partial^{\beta \minus{} e_i}f\cdot\partial^{\alpha \minus{} (\beta \minus{} e_i)}g$ or a term $ \partial^{\beta}f\cdot\partial^{\alpha \minus{} \beta}g$ in the large sum expression (except two terms of a view $ f\partial^{\alpha \plus{} e_i}g$ and $ g\partial^{\alpha \plus{} e_i}f$). And multipliers near those terms are combined as

$ {\alpha \choose \beta \minus{} e_i} \plus{} {\alpha \choose \beta} \equal{} {\alpha \plus{} e_i \choose \beta}$.

The last is verified in a straightforward fashion, remembering the definition

$ {\alpha \choose \beta} \equal{} \frac {\alpha!} {\beta! (\alpha \minus{} \beta)!}$, where $ \alpha! \equal{} \alpha_1! \cdot ... \cdot \alpha_n!$


Let triangle ABC.Prove that $sin^{3}A+sin^{3}B+sin^{3}C\leq\frac{9\sqrt{3}}{8}$

Wrong: take $A\to 90^{\circ}$, $B\to 90^{\circ}$, $C\to 0^{\circ}$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=33518 or http://www.mathlinks.ro/Forum/viewtopic.php?t=48222 for the best bound.

  darij


[quote="da ban man"]If $x>y$, then how many values of $(x,y)$ can there be if

$\frac{1}{x}+\frac{1}{y}=\frac{1}{24}$[/quote]

[hide]Getting rid of the denominators,
$24x+24y = xy$
$xy-24x-24y+484 = 576$
$(x-24)(y-24) = 24^{2}$,
from SFFT.

Notice that some pair of factors produce a solution; if $f_{1}f_{2}= 24^{2}$, $(x,y) = (24+f_{1}, 24+f_{2})$, where $f_{1}>f_{2}$.

Notice that the right has $21$ positive factors, and $21$ negative factors. So the number of factors where $f_{1}>f_{2}$ is $20$. Thus, there are $\boxed{20\textrm{ solutions}}$[/hide]

EDIT: thanks, 13375P34K43V312.

[quote="vishalarul"][quote="da ban man"]Alright, yea thats right,

Now what if (x,y) are positive integers also satisfying x>y.[/quote]

It would by $10$ solutions, as I mentioned earlier.[/quote]
positive and [b]integers[/b]. (and x>y)


geometry
rectangle
calculus
integration
derivative
quadratics
modular arithmetic

A rectangular piece of of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called [i]basic [/i]if

(i) all four sides of the rectangle are segments of drawn line segments, and
(ii) no segments of drawn lines lie inside the rectangle.

Given that the total length of all lines drawn is exactly 2007 units, let $N$ be the maximum possible number of basic rectangles determined. Find the remainder when $N$ is divided by 1000.

[hide]So we realize that drawing x vertical lines and y horizontal lines, the number of basic rectangles we have is (x-1)(y-1). The easiest possible case to see is 223 vertical and 223 horizontal lines  as (4+5)223 = 2007. Now, for every 4 vertical lines you take away, you can add 5 horizontal lines, so you basically have the equation $(222-4x)(222+5x)$ which you want to maximize. Expanded, this gives $-20x^{2}+222x+222^{2}$. From $-\frac{b}{2a}$ you get that the vertex is at $x=\frac{111}{20}$. This is not an integer though, so you see that when x=5, you have $-20*25+222*5+222^{2}$ and that when x=6, you have $-20*36+222*6+222^{2}$. 222 is greater than -20*11, so the maximum integral value for x occurs when $x=6$. Now you just evaluate $-20*36+222*6+222^{2}\mod 1000$ which is $\boxed{896}$.[/hide]

Edit: Fixed missing backslash

counting the edges of the piece of paper as lines that have been drawn seems unecessary but since i counted only the lines inside i lost the point.

actually, since you draw x horizontal lines, it partitions the paper into x+1 partitions. 2 don't count, so you get x-1. same goes for y.

Your solutions make sense, except I have one vital question: Where in the problem does it say that the lines you draw have to be from one side of the piece of paper to the other? I read this problem several times and couldn't understand it, because it only gives that the lines are "parallel" to the edges of the piece of paper, and as far as I could tell, you could have a bunch of infinitely small lines intersecting in a tiny grid, and there is no upward bound for the number of basic rectangles formed. I'm very mad at what I percieve to be an inaccurately written question, though I'd appreciate it if someone would explain what I missed. Good math after that question, though.

if your'e gonna distinguish between line and line segment, length wouldn't really make sense for an infinite line...

I thought this was a horribly written problem.... I just could not understand it, and it's not like I'm *that* much of a noob. Usually I understand enough of the problem to know what part of it makes it too hard for me to do, but with this one it was vaguely worded enough that I couldn't be sure.

I got $a$ in terms of $b$ and took a derivative to minimize :ninja: but it worked, so I'm not complaining.

[hide="My solution"]

$4a+5b=2007\Rightarrow a=\frac{2007-5b}{4}$ from the condition

so the # of squares is $\left(\frac{2007-5b-4}{4}\right)(b-1) = \left(\frac{2003-5b}{4}\right)(b-1)$

now $\frac{d}{db}$(expression for # of squares)$=\frac{-5b+1004}{2}\Rightarrow b=\frac{1004}{5}, a=\frac{1003}{4}$

Now we check the two closest things that are 3 mod 4 to 201 (since b must be 3 mod 4 because 2007=3 mod 4) which are 199 and 203, and $b=199$ gives the maximum # of squares which is

\[252\times198=\fbox{49896}\][/hide]

Yeah, I originally thought about the difference between lines and line segments, but then it started talking about the total "length" of the lines, so clearly it meant line segments (as krutyteklown said). So I am still baffled.
And mrosett and I agree again! This is kind of cool.

[quote="not_trig"]I got $a$ in terms of $b$ and took a derivative to minimize :ninja: but it worked, so I'm not complaining.

[hide="My solution"]

$4a+5b=2007\Rightarrow a=\frac{2007-5b}{4}$ from the condition

so the # of squares is $\left(\frac{2007-5b-4}{4}\right)(b-1) = \left(\frac{2003-5b}{4}\right)(b-1)$

now $\frac{d}{db}$(expression for # of squares)$=\frac{-5b+1004}{2}\Rightarrow b=\frac{1004}{5}, a=\frac{1003}{4}$

Now we check the two closest things that are 3 mod 4 to 201 (since b must be 3 mod 4 because 2007=3 mod 4) which are 199 and 203, and $b=199$ gives the maximum # of squares which is
\[252\times198=\fbox{49896}\]
[/hide][/quote]

Rather than take the derivative, a pre-calc method is to expand it and use $\frac{-b}{2a}$.

[hide="Solution"]
FOIL this to get a quadratic, $-\frac{5}{4}x^{2}+502x-\frac{2003}4$ (changed bs to xs b/c thats how i typed it up on the wiki). Use $\frac{-b}{2a}$ to find the maximum possible value of the quadratic: $x = \frac{-502}{-2 \cdot \frac{5}{4}}= \frac{1004}5 \approx 201$. However, this gives a non-integral answer for $y$. The closest two values that work are $(199,253)$ and $(203,248)$.

We see that $252 \cdot 198 = 49896 > 202 \cdot 247 = 49894$. The solution is $896$.
[/hide]

[hide="another solution"]Orient the rectangle so its width is 5. Suppose we have $a$ vertical lines and $b$ horizontal lines. We are given that $4a+5b=2007$, and that we want to maximize $(a-1)(b-1)$. Consider this equation modulo 4. This makes $b\equiv3\pmod{4}$, so $b=4y+3$ for $y\in\mathbb{Z}$. Likewise, modulo 5 gives $a\equiv3\pmod5$, so set $a=5x+3$ with $x\in\mathbb{Z}$.

Substituting, our equation becomes $(20x+12)+(20y+15)=2007$, so $x+y=99$. We want to maximize $(5x+2)(4y+2)$. We can rewrite this as $20(x+.4)(y+.5)$. These last two numbers sum to a constant, so their product will be maximized when their difference is minimized. This occurs for $x=50$ and $y=49$, which makes $(5x+2)(4y+2)=(252)(198)=50400-504=49896$, which ends in $\fbox{896}$.[/hide]

wow that's a good one :D

very nice... much better than mine, lol

Dangit... I missed that the rectangles touching the sides of the rectangle aren't basic rectangles, so I got $800$ as the answer.

i still don't get it. is there a list of solutions on the AMC website? or even better, could someone come up with a down to earth, dumbed-down explanation for me? thanks

This is completely after the fact, and I didn't even take the AIME II (took I), but I think that they wanted the fact that you draw "lines" and not "line segments" to mean they extend infinitely in each direction, but you only draw on the paper, so you'd be drawing from one side of the paper to the other.

ya, that's what i did. then i got the equation $4x+5y = 2007$ where $x$ and $y$ are the # of lines drawn in each direction. is that even right?

Yeah, that's how they got the 223 lines in each direction that they used to derive their quadratic...

Sorry to revive this thread but I think my solution does not use any messy quadratic formulas and can be done very quickly.
[hide="Solution"]
Call the number of horizontal lines $x$ and the number of vertical lines $y$. Then we have that $5x+4y=2007$ and we are trying to maximize $(x-1)(y-1)$. Now, let $a=x-1$ and $b=y-1$, then we have that $5a+4b=1998$ and we are trying to maximize $ab$. By $AM-GM$, we have that $\frac{5a+4b}{2}\ge\sqrt{(20)(xy)}$ with equality when $4a=5b$. Therefore, we wish to maximize the geometric mean so we want to be as close to equality as possible. Since $a$ and $b$ are integers, $a=198, b=252$. Therefore, the answer is $896$.
[/hide]


inequalities
algebra
polynomial
linear algebra
linear algebra unsolved

[quote="Moubinool"]$A,B,C \in M_n(C)$ such that $\forall k \in \{1;2;3\}, \; A^k=B+kC$
Prove that $rank(A)+rank(A^2-2A+I_n)= n$[/quote]

Hey, beautiful... I liked the problem, very smart (but somehow easy)

First than all, note that the statement is equivalent to proof that $null(A)+null(A^2-2A+I_n)= n$. Now, note that if $F,G\in M_n$ such that $FG=0$ then $null(F)+null(G)\geq n$ because, suppose that $null(B)=p$, then, $ker(B)^\perp\subseteq ker(A)$ (it follows also $rank(B)\leq null(A)$), and the inequality follows.

Now, note that $A(A^2-2A+I_n)=(A^3-A^2)-(A^2-A)=C-C=0$, hence $null(A)+null(A^2-2A+I_n) \geq n$.

Now, note that if $x\neq0$ ($x\in\mathbb{C}^n$), then, if $Ax=0$, then $(A^2-2A+I_n)x=x\neq0$, hence, $null(A) \leq rank(A^2-2A+I_n)$. Also, if $(A^2-2A+I_n)x=0\Rightarrow (A-2I_n)Ax=x\neq0\Rightarrow Ax\neq0\Rightarrow  null(A^2-2A+I_n) \leq rank(A)$, hence

\[null(A)+null(A^2-2A+I_n)\leq rank(A)+rank(A^2-2A+I_n)\]

\[\Rightarrow null(A)+null(A^2-2A+I_n) \leq n\]

and therefore

$null(A)+null(A^2-2A+I_n)=rank(A)+rank(A^2-2A+I_n)=n$.

There are some nice properties about this matrices, like the fact that $AB=A$ and $AC=C$, and some others.

Best regards,

$A,B,C \in M_n(C)$ such that $\forall k \in \{1;2;3\}, \; A^k=B+kC$

Prove that

$(P_{A}(1)-det(A))B+P'_{A}(1)C+det(A)I_n=0$ where $P_{A}$ is characteristic

polynomial of $A$


geometry calculus trigonometry MATHCOUNTS AMC USA(J)MO USAMO

I'm learning geometry right now.
Next year I get to take Algebra 2, but I have a chance to take a class over the summer or take a test to pass it and move onto Pre-calc(I think..)
I need you guys' help who has faced or know what to do.
Is Algebra 2 important? or Would you say that it's not as necessary or important.

Algebra 2 is extremely important, particularly in calculus but as well in pre-calc/trig. I would skip Geometry though  :D .

If you can (and actually will) take Algebra II over the summer I say go ahead and do it, and move on to pre-Calc next year.  Although I agree that Geometry pretty much sucked, if you're in the US and plan on going to college, there are many random geometry parts to the ACT and SAT that you should know, so don't skip it...  Besides, it keeps popping up in both my Physics and Calc classes this year.  Argh...

Algebra II is important for pre-Calc and Calc.  You need it.  If you think that you can learn it in it's entirety over the summer, go for it.  Just testing out of it only knowing the material on the test isn't such a good idea though - there are likely things not on the test that you need to know...  If you plan to teach yourself (which works, I've done it before) make sure that you really do it, don't just kinda read the book and answer a couple of the easier questions...  I'm sure that not everyone is a slacker like me, but be warned.  Without deadlines, sometimes things don't get done for some people (like me!).

If there's a class your school is offering over the summer, and you want to do it, by all means go for it!  Good luck!

Uh. I never actually learned Algebra II, in school or otherwise, and I'm doing quite fine in BC Calc, thank you very much. I picked up what I didn't know from Algebra II after I got into Precalc by doing AoPS II, haha. Though you should probably go for the summer thing anyway, just in case you don't want to/can't BS through the Precalc entrance exam.

I wouldn't reccomend spending the time during the summer learning Algebra II, you could be doing much better things with your time.

I'd say if you really are like doing your geometry homework in like 10 minutes every night, then you should spend maybe an hour a week studying Algebra II material in place of the excess time you have after finishing geometry homework.

Don't have much that much experience on this, but I'd say go ahead, if you want to spend the time over the summer preparing. My brother skipped Alg II by taking a summer course.

Just saying, Pre-Cal is just a big review of Algebra II and Geometry. So you really shouldn't take it unless you have too.

It's not a review. You do trig and a little bit of calc.

In my Algebra II class we did trig and Simple Calc like summation. Maybe it was different from the other Algebra II classes.

Hmmm... So what you guys all agree upon is that Algebra 2 is essential for students like me to get to the higher level of Math, and shouldn't think that it's really not important to skip it, right?

Well, in my opinion, I think Algebra II is important for you to graduate from high school and get into a good college. It's one of the things that determine the foundation of your future college admission. (and for a good score on the SAT/ACT)

At my school, Algebra II is pretty ignorant.

[quote="jmerry"]Algebra 2 is essential material for anyone. For some exceptional students (like many of us), it's possible to learn that material without actually taking the class. That's not the same as skipping it.
I learned Alg. 2 and trig while I was taking geometry in 9th grade. The teacher let me borrow the textbook for algebra, and I derived most of the trigonometry myself.

After that, I studied some precalculus informally in the summer, and took calculus in 10th grade.[/quote]

Wow... You must have spent bunch of your time doing those 3 subjects at once..

I would have thought jmerry had mostly learned the material of Algebra II while in Algebra I in the 8th grade. Also: he was at National Mathcounts as an 8th grader and a USAMO qualifier as a 9th grader, so this isn't necessarily something that just anyone could do.

[quote="D3m0n Shad0w"]You're in 8th grade, right?

You shouldn't skip Algebra II because it makes the colleges look like your wayy too serious :P Like other people said, you can always spend time doing fun, cool things during summer. Or, if you take Algebra II you can study Geometry (if you really wanted to), Pre-Calc, Trig, anything higher than your level over the summer.

It's your choice. I'm just recommending you don't skip it. You'll regret it. Geometry wise :D[/quote]

Yes, I am an 8th grader..and I'll take your word along with others.
Thanks!

[quote="D3m0n Shad0w"]You're in 8th grade, right?

You shouldn't skip Algebra II because it makes the colleges look like your wayy too serious :P Like other people said, you can always spend time doing fun, cool things during summer. Or, if you take Algebra II you can study Geometry (if you really wanted to), Pre-Calc, Trig, anything higher than your level over the summer.

It's your choice. I'm just recommending you don't skip it. You'll regret it. Geometry wise :D[/quote]

*Cough* says the fifth grader.

As a general rule of thumb, don't skip a math subject unless your schedule demands it. You'll have an easy year for math so you can concentrate on the subjects you need work on.

I plan to take AP physics in junior year but to do that I need to be finished calc AB. So in a sense, I [i]have[/i] to skip a math course. I prefer skipping calc than pre-calc because personally, I suck at trig.

[quote="ProtestanT"][quote="DonkeyKong"][quote="ProtestanT"]Algebra 2 was a really fun year. (everything you learn is fun) I don't suggest you skip it.
Maybe skipping Precalculus would be a better idea, if you need to skip a year so that you could put credits in certain electives.[/quote]
Can you briefly tell me what you learn in there?
That would be really helpful.[/quote]
briefly, we did alot of graphing, polynomial functions, finding roots, minimums and maximums, factoring, synthetic division, rational functions, conic sections (starting with parabolas), transcendamo or hwoever u spell it (exponential + logarithmic) functions, systems of linear equations, simple matrices (finding determinants, operations on matrices, gaussian elimination, inverses, etc), probability, and enumerative combinatorics (counting).
Our teacher did not teach us trignometry.[/quote]

Those sound important..

I think you take a peek at Trig. in the Geometry...

From my experience, the first semester of Alg. 2 is review of Alg. 1, and the first semester of Pre-Calc (or Analysis, whatever they call it where you live) is review of Alg. 2

If you really learn the material for Algebra 2 and feel like you could be spending more time on math HW, go ahead. I tested out of Geometry and Analysis, and i've been getting along fine (but i did quite a bit of extra-curricular Geometry)

In fact, i got a B- on the Analysis test-out, but still got an A in CalcBC

I say- Dont skip it, unless you have taken algebra 1; 2 times. Because its good practice(I bet) and you learn some new stuff. :ninja:

In general, I think you should never skip algebra 2. Its important for your college future.

my precalc class was basically my alg 2 class+trig, just a bit faster and less explicated. explain exactly what ur schools alg 2 class and precalc class cover 4 accurate suggestions, b/c w/o that info theres no way 2 know if you'd be hurt by skipping it.


Prove or disprove that there exists a positive real number $u$ such that $\lfloor u^n \rfloor -n$ is an even integer for all positive integer $n$.

Let $ u\equal{}1\minus{}\sqrt{2}$
$ [u^{2n}]\equal{}0$
$ [u^{2n\plus{}1}]\equal{}\minus{}1$
Imply that $ [u^n]\minus{}n\equiv 0(\mod 2)$

Why $ [u^{2n}]\equal{}0$ and $ [u^{2n\plus{}1}]\equal{}\minus{}1$?  :maybe:

It's since $ \minus{}1<u<0$, so then $ 0<u^{2n}<1$ and $ \minus{}1 < u^{2n\plus{}1} < 0$ which means $ [u^{2n}]\equal{}0$ and $ [u^{2n\plus{}1}]\equal{}\minus{}1$.

Yes, in the litteral form of course, but $ 1\minus{}\sqrt{2}<0$. I thought he at least meant $ \sqrt{2}\minus{}1$ instead. Because the problem really isn't solved with this.

[quote="Peter"]Prove or disprove that there exists a positive real number $ u$ such that $ \lfloor u^n \rfloor \minus{} n$ is an even integer for all positive integer $ n$.[/quote]

Let $ x_0 \equal{} 2,\ x_1 \equal{} 3,\ x_n \equal{} 3x_{n \minus{} 1} \plus{} 2x_{n \minus{} 2}$ for $ n\geq 2$. It is easy to see that $ x_n$ is an odd positive integer for all $ n\geq 1$. It is also easy to obtain an explicit formula for $ x_n$:
\[ x_n \equal{} \left(\frac {3 \plus{} \sqrt {17}}{2}\right)^n \plus{} \left(\frac {3 \minus{} \sqrt {17}}{2}\right)^n.
\]
Since the second summand in this formula is negative, for $ u \equal{} \frac {3 \plus{} \sqrt {17}}{2}$ and all $ k \equal{} 1, 2, \dots$ we have:
$ \lfloor u^{2k}\rfloor \equal{} x_{2k} \minus{} 1$ is even;
$ \lfloor u^{2k \minus{} 1}\rfloor \equal{} x_{2k \minus{} 1}$ is odd.
Therefore, $ u \equal{} \frac {3 \plus{} \sqrt {17}}{2}$ gives affirmative answer to the original problem.

P.S. The same solution I gave back in 2006 in Russian forum: http://dxdy.ru/post25766.html#p25766

$\frac{2a+1+\sqrt{(2a+1)^2+b}}{2}$ is a solution
for all naturals $(a,b)$ with the condition $2a\geq b$. And also all quadratic irrationalities that satisfy the conditions of the problem are in this form.


inequalities solved
inequalities
n-variable inequality
Symmetric inequality
Korea
2001
CS inequality

Let $x_i ,y_i$  be  real numbers satisfying $\sum_{i=1}^n x_i^2=\sum_{i=1}^n y_i^2=1$ for $i=1,2,\cdots,n,$ where $n$ is a natural number. Show  that
\[(x_1y_2-x_2y_1)^2\leq  2\cdot\left|1-\sum_{i=1}^n x_iy_i\right|\]

We obviously have \[\begin{aligned}(x_1y_2-x_2y_1)^2 & \le\sum_{i<j} \left(x_iy_j-y_jx_i\right)^2 \\ &=\left(\sum_{i=1}^n x_i^2\right)\left(\sum_{i=1}^n y_i^2\right)-\left(\sum_{i=1}^n x_iy_i\right)^2\\&=\left(1-\sum_{i=1}^n x_iy_i\right)\left(1+\sum_{i=1}^n x_iy_i\right)\leq2\cdot\left(1-\sum_{i=1}^n x_iy_i\right),\end{aligned}\]
Since $\sum_{i=1}^n x_iy_i\leq1$.


Please help me clearly define:

-Thermodynamics
-Thermochemistry
-Thermodyanmic/Kinetic Control
-Thermodynamic/Kinetic Stability

I have an intuitive understanding but not an objective understanding of these terms.

You could check the Wikipedia pages. Anyway, here's what I think:

- Thermodynamics is the branch of Physical Chemistry that studies the macroscopic properties of systems, their interrelationships, and how they affect the state of a system. Sometimes Thermodynamics is also defined as the branch of science the studies the relations between temperature and the other properties of a system.

- Thermochemistry is the application of the thermodynamic formalism to chemical reactions.

- A reaction is said to be in thermodynamic control is the more stable product is formed. In kinetic control, the product with the lowest activation energy is preferantly formed.

- By "thermodynamic stability" you may want to mean thermodynamic equilibrium, which is a situation in which there is simultaneously mechanical equilibrium, material equilibrium, and thermal equilibrium.


probability
real analysis
real analysis unsolved

Give an example to show that if a measure takes the value $ \plus{} \infty$, it does not follow in general that countable additivity implies continuity at $ \emptyset$.

Moderator's ruling: if a measure can take the value $ \plus{}\infty,$ then the topic does not belong in the probability forum.


find x,y,z is positive integer so that  (x 2 +1)(x 2 +2 2 ).....(x 2 +y 2 )=z 2


projective geometry
conics
ellipse
geometry
angle bisector
geometry solved

For the record (it is a gem):

[quote="pestich"]In quadrilateral ABCD BCxAD at P, ABxCD at Q.

   Find locus of points X so that  <DXP = <BXQ.
   Also, show orthogonality of angle bisectors of
   <AXC  and  <DXB.

    Lt. Pestich[/quote]

Let's look at the big picture:

There is 1 complete quad. It looks as if it has 4 sides. Using Desargues show that opposite sides intersect. Or something else similar to this.

It is known that 2 intersecting circles intersect at their intersection. The intersection is pulsating about ellipse centered at its center. The whole configuration is projective, so projecting line BQ and its parallel thru D to infinity, which is more than megaparsec, 2 circles become concentric and their intersection stops pulsating, because it isn't there anymore. So the locus must be hedgehog!

For AXP = BXQ, P is on AD, Q is on CD, so that angle bisector of BXD is perpendicular to its perpendicular and angle bisector of AXC is perpendicular to its perpendicular. According to Offenbach, when altitudes are perpendicular to their perpendiculars, angle bisectors are also perpendicular to their perpendiculars. Isn't it obvious?

What happens if we add X7 to the picture?

I claim that these ideas are 95% crackpot and the remaining say < 5% are also crackpot and it is up to you guys to separate crackpot from crackpot.

Thanks, Pestich, geometry has never been so much fun to me as now, reading your non-stop spams and Yetti's replies here...

:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

Now, let's talk seriously about the problem, which I rewrite in a more general form:

[color=blue][b]Problem.[/b] We will use directed angles modulo 180°.

Given a quadrilateral ABCD. Let the lines BC and DA intersect at P, and let the lines AB and CD intersect at Q.

Let X be a point in the plane.

[b](a)[/b] Prove that < DXP = < QXB if and only if the angle bisectors of the angles AXC and DXB are parallel or perpendicular to each other. (If the point X lies inside the quadrilateral ABCD, then "parallel or perpendicular" can be replaced by just "perpendicular", but for points X lying elsewhere in the plane, the angle bisectors can also be parallel.)

[b](b)[/b] Find the locus of all points X in the plane such that < DXP = < QXB.[/color]

[i]Solution.[/i] The condition < DXP = < QXB rewrites as < DXP = - < BXQ, what is equivalent to < DXP + < BXQ = 0°. Now, consider the complete quadrilateral formed by the lines AB, BC, CD, DA. The points D and B are opposite vertices of this complete quadrilateral; so are the vertices P and Q, and so are the vertices A and C. Thus, by http://www.mathlinks.ro/Forum/viewtopic.php?t=15103 problem [b]b)[/b] (see post #3 for a solution), we have < DXP + < BXQ = 0° if and only if < AXD + < CXB = 0°. Hence, we find that < DXP = < QXB if and only if < AXD + < CXB = 0°.

Now, let v and w be the angle bisectors of the angles AXC and DXB. Then, < (AX; v) = < (v; CX) and < (DX; w) = < (w; BX). Hence,

  2 < (v; w) = < (v; w) + < (v; w)
      = (< (v; AX) + < (AX; DX) + < (DX; w)) + (< (v; CX) + < (CX; BX) + < (BX; w))
      = (- < (AX; v) + < AXD + < (DX; w)) + (< (v; CX) + < CXB - < (w; BX))
      = (- < (v; CX) + < AXD + < (w; BX)) + (< (v; CX) + < CXB - < (w; BX)) = < AXD + < CXB.

Hence, < AXD + < CXB = 0° if and only if 2 < (v; w) = 0°. But we have 2 < (v; w) = 0° if and only if the angle < (v; w) is either 0° or 90°; in other words, we have 2 < (v; w) = 0° if and only if the angle bisectors v and w of the angles AXC and DXB are either parallel or perpendicular to each other. We also know that we have < DXP = < QXB if and only if < AXD + < CXB = 0°. Thus, we have < DXP = < QXB if and only if the angle bisectors v and w of the angles AXC and DXB are either parallel or perpendicular to each other. This solves part [b](a)[/b] of the problem.

For part [b](b)[/b], note that, as a consequence of the remarks at the very end of post #3 at http://www.mathlinks.ro/Forum/viewtopic.php?t=15103 , the locus of all points X such that < DXP + < BXQ = 0° is a cubic curve, namely the so-called [b]isoptic cubic[/b] of the complete quadrilateral formed by the lines AB, BC, CD, DA. Since we saw above that < DXP = < QXB is equivalent to < DXP + < BXQ = 0°, it follows that the locus of all points X such that < DXP = < QXB is the isoptic cubic of the complete quadrilateral formed by the lines AB, BC, CD, DA. This solves part [b](b)[/b] of the problem.

[b]PS.[/b] @Pestich: Look, elliptic curves out here! Don't you see the relation to FLT?

  Darij


What coordinate on the number line is equidistant from the
coordinates -5.65 and 8.75? Express your answer as a decimal
to the nearest hundredth.

We want the midpoint, which is the mean of the coordinates.

(-5.65+8.75)/2

$ \boxed{1.55}$


[code]\[
\left[ \int_a^b \phi_1(x) \phi_2(x) \, dx \right]^2 \leq \left( \int_a^b [ \phi_1(x) ]^2 \, dx \right) \left( \int_a^b [ \phi_2(x) ]^2 \, dx \right)
\][/code]

gives
\[ \left[ \int_a^b \phi_1(x) \phi_2(x) \, dx \right]^2 \leq \left( \int_a^b [ \phi_1(x) ]^2 \, dx \right) \left( \int_a^b [ \phi_2(x) ]^2 \, dx \right)
\]
Can we get this in another way ? I mean this doesn't seem to work everywhere..

[quote="Euclid_great"]I mean this doesn't seem to work everywhere..[/quote]
How do you mean?


function
calculus
integration
derivative
real analysis
real analysis solved

[quote="galois"]here goes the problem
  Find all real-valued continuous functions defined on the interval (-1, 1) such that (1 - x^2) f(2x/(1 + x^2) ) = (1 + x^2)^2 f(x) for all x.

well here is wot i did.let F be the primitive of f(x).and the computed the definite integral of f(x) from 0 to "a" in two ways.note that
2(1-x^2)/(1+x^2)^2 is the derivative of 2x/(1 + x^2) so we take it as some other variable t and integrate.then i performed the substitution x=tanh u which duz help.but i'm having  problems proving F(0)=0 which i think must hold.and i guess the function is m/(1-x^2) for arbitrary real m.
please help me
cheers!!![/quote]
==========
Put g(x) = 2x/(1+x^2), h(x) = (1-x^2)f(x) then f is continuos on (-1,1) and satisfy the equation iff h(x) is continuos on (-1,1) and satisfy
        h(g(x)) = h(x)  (1)
Now, we note that u(x) = (1-x)/(1+x) is bijective map from R+ to (-1,1). So we can rewrite (1) to:
       h(g((1-x)/(1+x)) = h((1-x)/(1+x)) x from R+
or
      h((1-x^2)/(1+x^2)) = h((1-x)/(1+x))  x from R+
Put k(x) = h((1-x)/(1+x)) x from R+, then h is contionuos on (-1,1) and satisfy (1) iff k(x) is continuos on R+ and satisfy k(x^2)=k(x). Now, we can easily show that k (and h also) is constant function from that we get f(x) = m/(1-x^2).

Have joy.

Namdung


my question reads.  " if two planes only shared one coplanar point, what ight these two planes look like?"

I think that it is impossibile for two planes to intersect and only share one coplanr point because points are so small and aren't a unit of measure, thus making it impossible for two planes to only share ONE coplanar point.

Anyone have any corrections or ideas towards this?

Two planes are either parallel or intersect in a line. It is impossible for two planes to intersect in a point.

ok so my original statement would be correct then?

[quote="iownatmath"]points are so small and aren't a unit of measure, thus making it impossible for two planes to only share ONE coplanar point.    [/quote]

This is a rather shady justification.  It's not clear at all what you mean by "points are small."  The important part, I'd say, is that [i]planes are big.[/i]  They are, in fact, infinite, and that's why two planes can only intersect in a line or not at all.  (Actually, I'd prefer justifying this algebraically, and the algebra works out that way, so... yeah.)

thanks, I'm just trying to grasp the concept we're learning in freshman geometry... it's so annoying because we get tought text book definitons leaving me thinking after every word that is said.  Thanks for the information thouhg. You'll probably get some unclear questions for a while because I just don't know any better.  Thanks for the help though.

[quote="graveyard destroyer"]Three planes could intersect at one line, but two cannot. Its like trying to Cut an Apple with 2 intersecting cuts and not end up with 4 pieces.[/quote]

Consider the trivial case of $ 3$-dimensional Euclidean space.  The $ xy$-plane and the $ yz$-plane intersect exactly at the $ y$-axis, by definition (and similarly for the $ x$- and $ z$-axes).  Generally, two planes are defined by two equations

$ ax\plus{}by\plus{}cz \equal{} d$
$ ex\plus{}fy\plus{}gz \equal{} h$

This is a system of two equations in three variables, and the best you can do is eliminate one variable.  This leaves the equation of a line unless the planes are parallel, in which case the intersection is either null or the planes themselves (if they are the same).

It's interesting that you bring up three planes.  Three planes [b]can[/b] intersect at a point.  This is a system of three equations in three variables, which generally has one solution.

And I'm not sure why you're talking about pieces.  It doesn't really relate to the shape of the intersection.

In four dimensions they can meet at one point. For example, the planes

x=y=0 and z=w=0

Their only intersection is (0,0,0,0).

However, in three dimensions (which is probably what you want), there's no example like this.

[quote="Hamster1800"]In four dimensions they can meet at one point. For example, the planes

x=y=0 and z=w=0

Their only intersection is (0,0,0,0).[/quote]

True.  The line analogue to this is that two lines in the plane can only either intersect at a point, be parallel, or be concurrent.  Two lines in space, however, can be [i]skew[/i], that is, none of the above.  Among other things, this is because a plane in three dimensions is defined by a single equation in three variables, whereas a plane in four dimensions is defined by two equations in four variables.

This is actually the simplest example of what's known as an [i]algebraic variety.[/i]  Algebraic geometry is a fascinating field of study; I highly recommend it for those of you who are bored with high school mathematics ;)


topology
function
advanced fields
advanced fields unsolved

Consider the sequence of continuous functions $f_{n}: (-1,1)\rightarrow\mathbb{R}$ defined by $f_{n}(x) = \sum_{k=1}^{n} kx^{k}$.
Show that $(f_{n})$ converges in the topology of compact convergence; conclude that the limit function is continuous.

The radius of convergence is $1$ whence the convergence is uniform on any closed interval contained in the interval of convergence; also the uniform limit of continuous functions is continuous.


Let the function $ f(x)$ twice differentiable function for which: $ f''(x) \plus{}f(x)\equal{}\minus{}x g(x)f'(x)$  where $ g(x) \geq 0$  forall $ x$.
Prove that, $ f(x)$ the function is bounded

If $ f'(x)\equal{}0$ for all $ x$, we are done. So let us assume that $ f'(x)$ is not zero for all $ x$. Multiplying by $ 2f'(x)$ throughout, we get
$ 2f'(x)f''(x)\plus{}2f(x)f'(x)\equal{}\minus{}2xg(x)(f'(x))^2$
that is
$ ({f'}^2(x)\plus{}f^2(x))^\prime \equal{} \minus{}2xg(x)(f'(x))^2$
Let $ h(x)\equal{}{f'}^2(x)\plus{}f^2(x)$. Then for $ x\geq0$, $ h'(x)\le0$. So $ h(x)$ is non-increasing for $ x\geq 0$. So, $ h(x)\leq h(0)$ for all $ x\geq 0$.
For $ x<0$, $ h'(x)\geq 0$. So $ h(x)\leq h(0)$ for all $ x<0$.
Hence, for all $ \in\mathbb{R}$, the function $ h(x)\leq h(0)\equal{}C$ (say).
That is $ {f'}^2(x)\plus{}f^2(x)\le C$. But that means
$ f^2(x)\le C$ which finally gives $ |f(x)|\leq \sqrt{C}$. So $ f$ is bounded.

[quote="kaymant"]If $ f'(x) \equal{} 0$ for all $ x$, we are done. So let us assume that $ f'(x)$ is not zero for all $ x$. Multiplying by $ 2f'(x)$ throughout, we get
$ 2f'(x)f''(x) \plus{} 2f(x)f'(x) \equal{} \minus{} 2xg(x)(f'(x))^2$
that is
$ ({f'}^2(x) \plus{} f^2(x))^\prime \equal{} \minus{} 2xg(x)(f'(x))^2$
Let $ h(x) \equal{} {f'}^2(x) \plus{} f^2(x)$. Then for $ x\geq0$, $ h'(x)\le0$. So $ h(x)$ is non-increasing for $ x\geq 0$. So, $ h(x)\leq h(0)$ for all $ x\geq 0$.
For $ x < 0$, $ h'(x)\geq 0$. So $ h(x)\leq h(0)$ for all $ x < 0$.
Hence, for all $ \in\mathbb{R}$, the function $ h(x)\leq h(0) \equal{} C$ (say).
That is $ {f'}^2(x) \plus{} f^2(x)\le C$. But that means
$ f^2(x)\le C$ which finally gives $ |f(x)|\leq \sqrt {C}$. So $ f$ is bounded.[/quote]
Thanks, kaymant!!!


inequalities
calculus
derivative
inequalities proposed

As Vasc said there are solutions of these inequality witout using SOS, derivatives or full expansions.

\[\frac{a^{2}+b^{2}}{a+b}+\frac{b^{2}+c^{2}}{b+c}+\frac{c^{2}+a^{2}}{c+a}\le \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}. \]

I am trying this solution for three weeks but I cannot, so I ask anyone to help me.  :blush:

P.S. If Vasc want to help us.........we thank for him. :)

[quote="manlio"]witout using SOS, derivatives or full expansions.
\[\frac{a^{2}+b^{2}}{a+b}+\frac{b^{2}+c^{2}}{b+c}+\frac{c^{2}+a^{2}}{c+a}\le \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}. \]
[/quote]

Hi manlio,

Multiply both sides by $a+b+c$:
$\sum\frac{(a^{2}+b^{2})(a+b+c)}{a+b}\le 3(a^{2}+b^{2}+c^{2})$
which is reduced to
$\sum\frac{c(a^{2}+b^{2})}{a+b}\le a^{2}+b^{2}+c^{2}$
Now add $\sum\frac{2abc}{a+b}$ to both sides:
$\sum\frac{c(a^{2}+b^{2})}{a+b}+\sum\frac{2abc}{a+b}\le a^{2}+b^{2}+c^{2}+\sum\frac{2abc}{a+b}$
which is equivalent to
$\sum\frac{c(a+b)^{2}}{a+b}\le a^{2}+b^{2}+c^{2}+\sum\frac{2abc}{a+b}$
or to
$2(ab+bc+ca) \le a^{2}+b^{2}+c^{2}+\succ\frac{2abc}{a+b}$
This is true since, by Cauchy and Schur inequalities
$a^{2}+b^{2}+c^{2}+\sum\frac{2abc}{b+c}\ge a^{2}+b^{2}+c^{2}+\frac{9abc}{a+b+c}\ge 2(ab+bc+ca)$

Regards,

Thank you very nice, Ductrung. Your proof is very NICE.  :)

See my post at http://www.mathlinks.ro/Forum/viewtopic.php?t=32967 - would you call this SOS?

  darij

Very Nice proof, Darij. Thank you very much. :)


limit
inequalities
ratio
triangle inequality
calculus
calculus computations

Prove that $ \lim_{x\to{3}}{\frac{2}{x\plus{}3}}\equal{}\frac{1}{3}$.

I can get up to the fact that $ \frac{1}{3}|x\minus{}3|\frac{1}{|x\plus{}3|}<\varepsilon$ but then I get really lost. Why is it valid to assume that $ \delta\le{1}$? The most confusing part for me right now is how I can find an inequality to "replace" $ \frac{1}{|x\plus{}3|}$. I have the detailed solution but I don't really get how you can conclude that:

\[ \frac{1}{3}|x\minus{}3|\frac{1}{|x\plus{}3|}<\frac{1}{3}|x\minus{}3|\frac{1}{5}<\varepsilon\] (basically, the addition of the second expression into the inequality and how to find it). Thanks.

Let $ \delta \equal{} min({1, 15\epsilon})$. Then for $ \epsilon >0$, we have
$ |f(x) \minus{} f(3)|$
$ \equal{}\frac{|x\minus{}3|}{|3x \plus{}9|}$
$ \leq \frac{|x\minus{}3|}{3|x| \plus{}9}$
$ <\frac{|x\minus{}3|}{15}$ if $ |x\minus{}3|<1$
$ <\frac{15\epsilon}{15}$ if   $ |x\minus{}3|<15\epsilon$
$ <\epsilon$.

Since we wish to consider values of x close to 3, putting $ \delta< 1$ is reasonable since that puts x within 1 of 3. Also,  $ 0<|x\minus{}3|<\delta< 1$ implies that $ \frac{1}{|x\plus{}3|}<\frac{1}{5}$.

Okay, thanks I see the whole $ \delta < 1$ thing now. And I see how to get that $ \frac {1}{7} < \frac {1}{|x - 3|} < \frac {1}{5}$ but I'm [i]really[/i] confused as to how you can just add that to the previous inequality to get:
\[ \frac {1}{3}|x - 3|\frac {1}{|x + 3|} < \frac {1}{3}|x - 3|\frac {1}{5} < \varepsilon
\]
It seems to me like in the "middle expression" ($ \frac {1}{3}|x - 3|\frac {1}{5}$), $ \frac {1}{|x + 3|}$ was simply replaced by $ \frac {1}{5}$. I simply don't get the logic behind that...

Edit: I think I've got the hang of it now and a few practice problems are helping alot. I know $ \delta=1$ and $ \delta=15\varepsilon$ but I don't get why $ \delta=\min{\left\{1,15\varepsilon\right\}}$, mainly why we want to take the minimum value of $ \delta$ in these types of cases.

[quote="akech"]
$ \frac {|x \minus{} 3|}{|3x \plus{} 9|} \leq \frac {|x \minus{} 3|}{3|x| \plus{} 9}$
[/quote]

So you are claiming $ 3|x| \plus{} 9 \leq |3x \plus{} 9|$? Doesnt that go against triangle inequality?

[quote="Blkmage"]Okay, thanks I see the whole $ \delta < 1$ thing now. And I see how to get that $ \frac {1}{7} < \frac {1}{|x - 3|} < \frac {1}{5}$ but I'm [i]really[/i] confused as to how you can just add that to the previous inequality to get:
\[ \frac {1}{3}|x - 3|\frac {1}{|x + 3|} < \frac {1}{3}|x - 3|\frac {1}{5} < \varepsilon
\]
It seems to me like in the "middle expression" ($ \frac {1}{3}|x - 3|\frac {1}{5}$), $ \frac {1}{|x + 3|}$ was simply replaced by $ \frac {1}{5}$. I simply don't get the logic behind that...

Edit: I think I've got the hang of it now and a few practice problems are helping alot. I know $ \delta = 1$ and $ \delta = 15\varepsilon$ but I don't get why $ \delta = \min{\left\{1,15\varepsilon\right\}}$, mainly why we want to take the minimum value of $ \delta$ in these types of cases.[/quote]From $ \frac{1}{|x+3|}<\frac15$, multiply both sides by $ \frac13|x-3|$.

[quote="isomorphism"]So you are claiming $ 3|x| \plus{} 9 \leq |3x \plus{} 9|$? Doesnt that go against triangle inequality?[/quote]

Since everything is positive, $ 3|x| \plus{} 9 \geq |3x \plus{} 9|$ implies $ \frac{1}{3|x| \plus{} 9} \leq \frac{1}{|3x \plus{} 9|}$.  When you have a ratio of two positive quantities, making the denominator larger makes the fraction smaller.

[quote="JBL"][quote="isomorphism"]So you are claiming $ 3|x| \plus{} 9 \leq |3x \plus{} 9|$? Doesnt that go against triangle inequality?[/quote]

Since everything is positive, $ 3|x| \plus{} 9 \geq |3x \plus{} 9|$ implies $ \frac {1}{3|x| \plus{} 9} \leq \frac {1}{|3x \plus{} 9|}$.  When you have a ratio of two positive quantities, making the denominator larger makes the fraction smaller.[/quote]

Yes but see #2, he claims otherwise  :(

You are right, isomorphism. The mistake there is acknowledged and that step was not needed.


If a three-digit number of the form $ 1N1$ is divided by $ N$, the quotient is of the form $ 2N$ remainder 5. What is the value of $ N$?

We see that $ 1N1$=$ N(2N)\plus{}5$, so $ 100\plus{}10N\plus{}1\equal{}2N^2\plus{}5$, and our answer is $ \boxed{N\equal{}6}$


we have circumscribed quadrilateral $ ABCD$ that the center of  its circle is $ I$.$ M$ is on $ DI$ such that $ AM$ is perpendicular to $ AB$ and $ N$ is on $ BI$ such that$ AN$ is perpendicular to $ AD$.prove that $ AC$ is perpendicular to $ MN$

i find out too solution for my problem.does  any body have nice solution

Hi,

Can you post your solution? Thanks.

Maybe i missed something , the result seems false,  changing positon of C on the circle doesn't affect the rest of the configuration so it can't be $ MN$ perpendicular on $ AC$

solution 1. lemma. if $ ABCD$ be a circumscribed quadrilateral and the circle  tangent to $ AB,BC,CD,AD$ at $ X,Y,Z,W$ then $ AC,BD,XZ,WY$ are concur at one point and if the intersection named $ P$ then $ (AP)(CW)\equal{}(CP)(AX)$ and $ (BP)(DX)\equal{}(DP)(BY)$
from this lemma we can conclude that if $ L,Q$ be the projections of $ B,D$ on $ AC$ then (with Thales) $ (BL)(DX)\equal{}(DQ)(BY)$.so if in our problem the projection of $ M$ on $ AC$ named $ R$ then $ (AR)(BY)\equal{}r(BL)$ (by Thales) and if  $ R_1$ be similiary like $ R$ for $ N$ then $ (AR_1)(DX)\equal{}r(DQ)$.awe proved that $ AR\equal{}AR_1$ so the problem is proved
sol 2.we must prove that $ (AM)^2\minus{}(CM)^2\equal{}(AN)^2\minus{}(CN)^2$.but its not easy to prove it without anything. we must draw perpendicular from $ A$ to $ CD$ and $ CB$


It is reqd to prove on simplification that $ a^2 \plus{} b^2 \plus{} c^2 \geq a \cdot b \plus{} b \cdot c \plus{} c \cdot a$
This is proved by the AM GM inequality  :)

It's obviously true by Cauchy-Schwarz :wink:

hmm................... I didn't know that .

What are this AM and GM inequality? Can you tell me please?

The AM-GM (arithmetic mean-geometric mean) inequality states that for non-negative reals $ r_1, r_2, ... r_n$ we have

$ \frac{r_1 \plus{} r_2 \plus{} ... \plus{} r_n}{n} \ge \sqrt[n]{r_1 r_2 ... r_n}$.

madness is talking about the sum of the inequalities

$ a^2 \plus{} b^2 \ge 2ab$
$ b^2 \plus{} c^2 \ge 2bc$
$ c^2 \plus{} a^2 \ge 2ca$

which proves the desired inequality.  The inequality he gives is well-known and can also be proven by Rearrangement.  apollo gives the direct solution

$ (1^2 \plus{} 1^2 \plus{} 1^2)(a^2 \plus{} b^2 \plus{} c^2) \ge (1 \cdot a \plus{} 1 \cdot b \plus{} 1 \cdot c)^2$

by [url=http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality]Cauchy-Schwarz[/url].  These are the first techniques introduced in dealing with inequalities.

See this for the inequality http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

For this case, it will be like

$ \sum (a^2 \plus{} b^2) \geq 2 \sum \sqrt {a^2 \cdot b^2}$

$ \implies \sum (a^2 \plus{} b^2) \geq 2 \sum a \cdot b$

Hence $ \sum a^2 \geq \sum a \cdot b$

EDIT: Sorry t0rajir0u didnt see that you had posted it  :)

By Cauchy Schwarz Inequality

$ (a^2 \plus{} b^2 \plus{} c^2) (1^2 \plus{} 1^2 \plus{} 1^2) \geq (a \plus{} b \plus{} c)^2$

By Chebyshev's Inequality, assuming WLOG a>b>c,

$ 3(a*a \plus{} b*b \plus{} c*c) \geq (a \plus{} b \plus{} c) (a \plus{} b \plus{} c) \equal{} (a \plus{} b \plus{} c)^2$

Or by simple transformations:
$ 3(a^{2} \plus{} b^{2} \plus{} c^{2})\geq (a \plus{} b \plus{} c)^{2}$
$ 2a^{2} \plus{} 2b^{2} \plus{} 2c^{2}\geq 2ab \plus{} 2bc \plus{} 2ca$
$ a^{2} \minus{} 2ab \plus{} b^{2} \plus{} b^{2} \minus{} 2bc \plus{} c^{2} \plus{} c^{2} \minus{} 2ca \plus{} a^{2}\geq 0$
$ (a \minus{} b)^{2} \plus{} (b \minus{} c)^{2} \plus{} (c \minus{} a)^{2}\geq 0$ :wink: with equality if and only if $ a \equal{} b \equal{} c$

Yup, that's what 2-var AM-GM is about  :)

By the way, the last $ c^2$ in your 3rd line of math should be an $ a^2$

Similar one:
Prove that for all reals $ a,$ $ b$ and $ c$ the following inequality holds:
\[ 9(a^4\plus{}b^4\plus{}c^4)^2\geq(a^5\plus{}b^5\plus{}c^5)(a\plus{}b\plus{}c)^3\]

I think only Chebyshev and power mean is required:
We have to prove:
$ 9(a^4 \plus{} b^4 \plus{} c^4)^2\geq(a^5 \plus{} b^5 \plus{} c^5)(a \plus{} b \plus{} c)^3$
By Chebyshevs inequality we have:
$ \frac {a^{4} \plus{} b^{4} \plus{} c^{4}}{3} \equal{} \frac {a^{5}*\frac {1}{a} \plus{} b^{5}*\frac {1}{b} \plus{} c^{5}*\frac {1}{c}}{3}\geq \frac {a^{5} \plus{} b^{5} \plus{} c^{5}}{3}*\frac {\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}{3}$
thus it remains to prove:
$ (a^{5} \plus{} b^{5} \plus{} c^{5})(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^{2}\geq (a \plus{} b \plus{} c)^{3}$
by AM-HM we get:
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq \frac {9}{a \plus{} b \plus{} c}$ thus it suffices to prove that:
$ (a^{5} \plus{} b^{5} \plus{} c^{5})*9^{2}\geq (a \plus{} b \plus{} c)^{5}$ which is true, because the last inequality is equivalent to:
$ \sqrt [5]{\frac {a^{5} \plus{} b^{5} \plus{} c^{5}}{3}}\geq \frac {a \plus{} b \plus{} c}{3}$ which is a direct conclusion from the generalizec power mean Q.E.D. :wink:

EDIT: Nice solution from Cachy-Schwarz   :)

[quote="SimonM"]

$ (s_5s_1^3)^4 \& \leq s_1^4s_2^{10}s_3^29^2$
[/quote]
I think, it's wrong. :wink:
[quote="polskimisiek"]
By Chebyshevs inequality we have:
$ \frac {a^{5}*\frac {1}{a} + b^{5}*\frac {1}{b} + c^{5}*\frac {1}{c}}{3}\geq \frac {a^{5} + b^{5} + c^{5}}{3}*\frac {\frac {1}{a} + \frac {1}{b} + \frac {1}{c}}{3}$
[/quote]
I think, it's wrong. :wink:

Aaah, I see. I totally forgot aout the necessary condition $ a\geq b\geq c$ and $ \frac {1}{a}\geq \frac {1}{b}\geq \frac {1}{c}$ which unfortunately cannot be true  :oops:  :ninja:


topology
advanced fields
advanced fields theorems

In R[sup]n[/sup] and S[sup]n[/sup], when we have a homeomorphism on a space X, we can create a homotopy whose endpoints match the homeomorphism (image of Xx{0} is the identity; image of Xx{1} matches image of homeomorphism).

In T^2 (the 2-dimensional torus), many homeomorphisms cannot be "traced" by a homotopy (e.g., any combination of meridional and longitudinal twists).

1. In which spaces can any homeomorphism be "traced" by a homotopy, and why?

2. Can we identify a certain group of homeomorphisms for each space outside of which any homeomorphism can be "traced" by a homotopy?

I have a suspicion that the first question is something relatively basic and that I've seen the reason before, and a suspicion that the second question is not well-defined.  Anyone who can point me in a helpful direction gets a cookie.  : D

Edit:[i]I suppose this is technically inherently topological, and solved, so doesn't belong in the unsolved algebra questions folder.  Oops.[/i]  >.>

Further note: [i]This question came up when thinking about knots on a torus, while working through Rolfsen's [u]Knots and Links[/u][/i].

I think you are interested in the [url=http://en.wikipedia.org/wiki/Mapping_class_group]mapping class group[/url].

Ah, thank you, exactly what I was looking for.


Find n such that
$\lfloor\log_{2}1\rfloor+\lfloor\log_{2}2+\rfloor+\lfloor\log_{2}3\rfloor+...+\lfloor\log_{2}n\rfloor=1994$


geometry
3D geometry
topology
real analysis
real analysis unsolved

Assume that M is compact, non-empty, perfect, and homeomorphic to its Cartesian square, M =~ M x M. Must M be homeomorphic to the Cantor set, the Hilbert cube, or some combination of them?

I'm just wondering. Do you consider these three problems in topology boring or should I post some more?

Please post as many topology problems as you want. Topology is not boring.
What is the Hilbert cube?
What do you understand by "some combination of them"?

I found out what Hilbert cube is. It is [0,1] <sup>|N|</sup>.

Hmm, I read your post before, in school, I was going to write a reply at home, sorry that I didn't! I still don't have any answer to this question, seems hard.

Which topological set do you call to be perfect?


4n points are arranged around a circle. The points are colored alternately yellow and blue. The yellow points are divided into pairs and each pair is joined by a yellow line segment. Similarly for the blue points. At most two segments meet at any point inside the circle. Show that there are at least n points of intersection between a yellow segment and a blue segment.

[i]This is from the Swedish Math Olympiad in 1989.[/i]

Consider two intersecting yellow segments AC and BD. Then the number of points of intersection of any line with AB abd CD does not exceed the number of intersection points of the line with AC and BD.
So if we substitute of yellow segments AC and BD with AB и CD the number of intersection points of yellow and blue will no increase and the number of intersection points of yellow lines will decrease.
If we make certain number of such operations all yellow segments will not meet each other. It is sufficient to prove the statement only in this case.
Look at any yellow segment L. Other yellow segments do not meet it, so both semiplanes created by L contain an even number of yellow points and as consequence an odd number of blue points. Therefore there is a blue segment that intersects L. So the number of yellow-blue intersections is at least the number of yellow segments that is n.


inequalities
geometry
geometric transformation
reflection
inequalities proposed

Let $a,b,c$ be positive real numbers. Prove that

$\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}-3 \ge 18(\frac a{b}+\frac b{c}+\frac c{a}-\frac b{a}-\frac c{b}-\frac a{c})$,

with equality for $a=b=c$, and also for $a=\frac b{2}=\frac c{4}$ or any cyclic permutation thereof.

I think this one is just another way of formulating the well known.

Let $a,b,c$ be positve real numbers such that $abc=1$. Prove that
\[a^{2}+b^{2}+c^{2}\geq 3+18(a+b+c-ab-bc-ca).\]

yes but why is it well known?
has it been posted before? :maybe:

See the article "Four Applications of RCF and LCF Theorems" in Mathematical Reflections, 1, 2007:
http://reflections.awesomemath.org/2007_1.html
I am still waiting a nice solution.

OK, VASC. It is a typical problem of mixing variable method. I solved it here

Dear hungkhtn: How can i buy your book?  :?:

Only the Vietnamese are available now. It was published in 2006, but the English version is totally different. I think that Gil Publishing house will announce this book before IMO 2007 in Vietnam, so please waiting. Thank you for concern.  :lol:

[quote="The soul of rock"]So how can Vietnamese orders your book? It's very hard to buy book from Gil in our country, hungkhtn![/quote]

I think in this case, we need help of Gil. Anyway, it is not a right place to discuss about my book, so let me say that we should stop now. Please waiting when Mircea Lascu announces it, or we may discuss through PM Message.


Find a possible formula for a function $ m(x)$ such that $ m'(x) \equal{} x^5 \cdot e^{x^6}$.

[hide]I came up with m'(x) = e^{x^5}^{6} but I don't think it works[/hide].

[quote="jeez123"]Find a possible formula for a function $ m(x)$ such that $ m'(x) \equal{} x^5 \cdot e^{x^6}$.
[/quote]

$ m'(x) \equal{} x^5 \cdot e^{x^6}$ becomes: $ m'(u) \equal{} \frac {1}{6}e^{u}du$, with $ u \equal{} x^{6}$
You now integrate $ m'(u)$ to get: $ m(u) \equal{} \frac {1}{6}e^{u} \plus{} c$, where $ c$ is a constant. Now replace $ u$ with $ x^{6}$ to get the desired result.


Rationalize the denominator: $ \frac1{2\sqrt7}$.


calculus
integration
function
algebra
polynomial
real analysis
real analysis unsolved

1a. Let S be the solid 0 <= x <= 2, −x <= y <= x, 0 <= z <= x^2 − y^2. Find the volume of S
 b. A scoop is in the shape of the parabolic surface −1 <= x <= 1, y = x2, 0 <= z <= 2, with
one end covered by the region −1 <= x <= 1, x^2 <= y <= 1, z = 0. How much water can the
scoop hold?

2. Let f be differentiable on [a,b] with |f '(x)| <= M for all x in [a,b]. Let I = int (a to b)f and let L[size=75]n[/size] be the Left Endpoint Approximation for I on n equal-sized subintervals. Prove that
|I - L[size=75]n[/size]| <= (M(b-a)^2)/2n

The general method for proving accuracy of numerical integration methods:
Step 1: Do it for one interval, not subdivided. Use Taylor's theorem or something similar to estimate the function with a polynomial that the numerical method gets right.
Step 2: Use the result on each subinterval in your new division. Add up the estimated errors for the total.


calculus
derivative
function
trigonometry
calculus computations

Obtain the derivative of the following function:

f(x) = exp(sin2x)

This thread is in the wrong location. The [b]High School Basics[/b] forum is for algebra/geometry/precalculus topics mainly. This should probably be placed in the [b]Calculus Computations and Tutorials[/b] subforum.  :)

all right I didn't know...I will post it there instead....thanks for telling me...


geometry
3D geometry
function
trigonometry

I don't know how to solve this one \[x^{6}=20000\]

Any help or hint?

In wich set you want to solve that equation ? $R$ , or $C$, or $Z$ ? :roll:

are you just trying to solve for x? Cause that's all I see, just use logs.

For $R$, I would just take $\pm\sqrt[6]{20000}$, for $C$ I would use De Moive's theorem.

I suppose it is impossible for $Z$ so i would like a $R$ solution and then a $C$ one.

@besttate I'm afraid $\pm \sqrt[6]{20000}$ wouldn't help. And i am not prepared enough to use such theorems like De Moivre's.

Can you help me?

In reals, note that the cube root function is one-to-one, so we want to solve

$x^{2}= \sqrt[3]{20000}= 10 \sqrt[3]{20}= 27.144...$

Which has roots

$x = \pm \sqrt[6]{20000}= \pm 5.2100...$

The other four complex cube roots are complex multiples of this one, but their form is not easy to derive without the use of de Moivre's theorem.

I do not know how to do the problem in any other way, below is using the methods described above. Hopefully these can help you in some way.

In $R$, simplifying $\pm\sqrt[6]{20000}$, we have $\pm2\sqrt[6]{3125}$

In $C$, using De Moivre's theorem, we write $20000$ in polar notation as $20000\cos0+20000i\sin0$, or "$20000$cis$0$" ($A$cis$\theta$ simply means $A\cos\theta+Ai\sin\theta$).

First, we calculate $20000^{1/6}$cis$0\pi/6$, which becomes $2\sqrt[6]{3125}$.

Next, we calculate $20000^{1/6}$cis$2\pi/6$, which becomes $\sqrt[6]{3125}+i\sqrt{3}\sqrt[6]{3125}$

Next, we calculate $20000^{1/6}$cis$4\pi/6$, which becomes $-\sqrt[6]{3125}+i\sqrt{3}\sqrt[6]{3125}$

Next, we calculate $20000^{1/6}$cis$6\pi/6$, which becomes $-2\sqrt[6]{3125}$

Next, we calculate $20000^{1/6}$cis$8\pi/6$, which becomes $-\sqrt[6]{3125}-i\sqrt{3}\sqrt[6]{3125}$

Finally, we calculate $20000^{1/6}$cis$10\pi/6$, which becomes $\sqrt[6]{3125}-i\sqrt{3}\sqrt[6]{3125}$

In summary, the 2 real solutions are $\pm2\sqrt[6]{3125}$ and the 4 complex solutions are the 4 combinations of $\pm\sqrt[6]{3125}\pm i\sqrt{3}\sqrt[6]{3125}$.

Hope this helps a little  :D .


I need to display a union over a set; however the \bigcup{}{} command won't let me put i \in I in the lower argument. Can someone tell me how to get around this, please?

Use the same syntax as for \int and \sum with _ for lower limit and ^ for upper limit.
\bigcup_{i \in I} A_i gives $ \bigcup_{i \in I} A_i$ and \bigcup_{i =1}^n A_i gives $ \bigcup_{i \equal{}1}^n A_i$

Thanks, stevem.

I was putting the second argument on, but left it empty; and I didn't use the subscript!---I got something ugly.


vector
linear algebra
matrix
Support
algebra
polynomial

Hey, I got some questions, please help

1. I remember in first year LA we preformed ERO's (elementary row operations) on matrices without linear transforms in mind. if we have $ Tx\equal{}s$ with $ x$ our unknown vector ERO's change our $ T$ and $ s$ such that $ x$ stays the same, if I remember correctly.
What is the relevance of this to the study linear transforms?
How would you prove an ERO leaves the solution to the equation $ Tx\equal{}s$ unchanged in the language of linear transforms?

2. Given a matrix T: say  $ \begin{pmatrix}
1 & 1 \\
\minus{}1 & 2
\end{pmatrix}$ how do we find its upper-triangular form, if it exists.

3. How would we use the exponential of a matrix to solve the DE:

$ u'\equal{}5u\minus{}2v$
$ v'\equal{}7u\minus{}3v$

1: A row operation is equivalent to multiplication on the left by some specific matrix $ R$. If $ Tx \equal{} s$, then $ RTx \equal{} Rs$. Conversely, if $ R$ is invertible and $ RTx \equal{} Rs$, $ Tx \equal{} s$. Row interchanges, shears in which we add a multiple of one row to another, and multiplying a row by a nonzero constant are all invertible.

2: What do you mean, "its upper-triangular form"? There are many ways for matrices to be related to each other, which will give wildly different answers.
You're probably talking about row/echelon form, which is intrinsically linked to row operations. Even then- that's not really unique, unless you have very strict rules about when to stop. Flesh that out, and it leads to LU-factorization; how to write a matrix as a product $ LU$ with $ L$ lower triangular and $ U$ upper triangular, with one of them having ones on the diagonal. That form isn't quite universal, either; if you needed row interchanges, there's also a multiplication by a permutation matrix $ P$.

3. The solution to $ y'(t) \equal{} Ay$ ($ A$ constant) is $ y \equal{} e^{At}y_0$. Now, how do you actually calculate that matrix exponential? Diagonalize, or at least put it in Jordan form. Similarity works well with everything.

1. A row operation (elementary or not) corresponds to left-multiplying by an invertible transformation. So row operations turn $ Tx \equal{} s$ into $ PTx \equal{} Ps.$ To support the claim that this leaves the solutions set unchanged, we should explicitly invoke the fact that $ P$ is invertible.

2. When you say "its upper triangular form" are you talking about similarity? If you are, then we should find the characteristic polynomial and hence the eigenvalues. For a $ 2\times 2$ matrix $ A$ the characteristic polynomial is $ \lambda^2 \minus{} (\text{trace}\,A)\lambda \plus{} \det A.$ In this case, that's $ \lambda^2 \minus{} 3\lambda \plus{} 3.$ This has only complex roots, so this matrix cannot be diagonalized or triangularized over the reals, but it can be diagonalized over the complex numbers.

3. If the differential equation is $ y' \equal{} Ay$ with the initial condition $ y_0$ (where $ y$ and $ y_0$ are vectors and $ A$ is a square matrix), then the solution is $ y(t) \equal{} e^{tA}y_0.$

For a formula for finding $ e^{tA}$ for any $ 2\times 2$ matrix $ A$, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310425]here.[/url]

Edit: partly redundant with jmerry's post; composed simultaneously. Except that we came to different interpretations of "its upper triangular form", which just serves to emphasize jmerry's point about that being ambiguous.


show that the co-efficient of $ z^{2}$ in the series expansion about $ z\equal{}0$ of

$ \frac{1}{z\cdot (z\plus{}1)\cdot\Gamma{(z)}}$

is $ \boxed{1\minus{}\frac{1}{2}\,\gamma\,(2\minus{}\gamma)\minus{}\frac{1}{2}\,\zeta{(2)}}$  :D

read Ga/7 .( first relation)
http://dlmf.nist.gov/Contents/GA/7/
I think after you can do your problem yourself


linear algebra
matrix
geometry
rectangle
vector
logarithms

[quote="now a ranger"]Can someone explain exactly what a matrice is and how to do  magic with it?  :D[/quote]
A [b]matrix[/b] is a rectangle array of numbers. (e.g. 1 3 7
                                                                        6 8 3)

dude
by magic, do you mean magic squares or something?
i haven't encountered real matrix magic.
funcia

can someone elaborate further? Not getting anything!

:lol:
how about this:
http://en.wikipedia.org/wiki/Matrix_%28mathematics%29
i would google matrix and magic if i were you

yeah well i have no idea if you have access
to a library but if one nearby had
this book i liked :
Introduction to matrices and vectors
 by Jacob Schwartz

A matrix is like a data table, except you can preform operations with it:

$\begin{bmatrix}a & b \\ c & d \end{bmatrix}\times 2$

$\begin{bmatrix}a & b \\ c & d \end{bmatrix}\times \begin{bmatrix}e & f \\ g & h \end{bmatrix}$

We were told that matrices and logarithms can greatly simplify calculations. I know that logs can come in handy but even after learning matrices, I'm not sure of how they can help u in big calculations. How do they??

Matricies help with solving systems of equations. Like with 4 variables  :wink: .

But how do they do that?? :huh:
If you don't mind, can you show me an example?
Thanx!!

Say for example:

$2x+3y=7$

$x-9y=13$

I'm only going to use 2 variables because i don't have a lot of time =)

set up the matrix as follows

$\begin{bmatrix}2 & 3 \\ 1 &-9 \end{bmatrix}$

now the equation is Ax=B, where A is the above matrix. and B is the solutions:

$A = \begin{bmatrix}2 & 3 \\ 1 &-9 \end{bmatrix}$

and

$B= \begin{bmatrix}7 \\ 13 \end{bmatrix}$

To find x, which is your variables-

$X=\begin{bmatrix}x \\ y \end{bmatrix}$.

Solve for x using $A^{-1}X=B \end{bmatrix}$.

sorry for not explaining this very well. i have to go now I will make it better when i come back


calculus
integration
trigonometry
calculus computations

how do you integrate y=$ \sqrt{4\minus{}x^2}$?

thanks  :)

hello, substitute $ x\equal{}2\sin(t)$
Sonnhard.

[hide="Solution?"]\begin{align*} \int\sqrt {4 - x^2}\;dx & = \int\sqrt {4 - 4\sin^2t}\;dt \\
& = 2\int\sqrt {\cos^2t}\;dt \\
& = 2\int\cos{t}\;dt \\
& = 2\sin{t} + C \\
& = \boxed{x + C} \end{align*}[/hide]

$ dx \neq dt$.  You forgot to change that part of the integrand.

EDIT: oops...

$ x \equal{} 2\sin t \implies dx \equal{} 2\cos t\; dt$

$ \int \sqrt {4 \minus{} x^2} \; dx \equal{} \int 2\sqrt {1 \minus{} \sin^2 t} \; 2\cos t \; dt \equal{} \int 2 \cos t \; 2\cos t\; dt$

which is $ \int 4\cos^2 t \; dt$

how do you go from there?

You substituted $ \cos$ instead of $ \sin$ under the square root sign. You'll need to use the double angle formulae to evaluate the $ \cos^2$

ugh im lost...

could somebody please post a solution??

Wait! Is $ \sqrt{A^2}\equal{}A$ for real number $ A$ always hold?


linear algebra
matrix
linear algebra unsolved

Let $ A$ be a $ nxn$ matrix. Show that $ \det\left[\begin{array}{cc} A & A^2\\
A^2 & A^3
\end{array} \right]$ is zero.

This is a simple application of the following lemma.

$ \textbf{Lemma}$ Let $ C$, $ D$, $ X$, $ Y$, be $ n\times n$ complex matrices. If $ C$ commutes with $ X$ then

$ \det\left[\begin{array}{cc} C & Y\\
X & D
\end{array} \right]\equal{}\det(CD\minus{}XY).$

Let $ B$ be the $ 2n\times 2n$ matrix $ \begin{bmatrix}A&A^2\\A^2&A^3\end{bmatrix}$.

Then not only is $ B$ singular, but $ \text{rank}(B)\equal{}\text{rank}(A)\le n.$

Proof by row and column reduction:

$ \begin{bmatrix}I&0\\\minus{}A&I\end{bmatrix} \begin{bmatrix}A&A^2\\A^2&A^3\end{bmatrix} \begin{bmatrix}I&\minus{}A\\0&I\end{bmatrix}\equal{} \begin{bmatrix}A&0\\0&0\end{bmatrix}$