[hide="AM-GM"]
Expand out, giving
$8abcd+8 > abcd + abc+abd+...+ab+bc+...+a+b+c+d+1$ (RHS is symmetric so all combinations of 1,2,3,4 variables are summed)

$7abcd+7> abc+abd+...+ab+bc...+a+b+c+d$. Let $S=RHS$.

By AM-GM $S \geq 14 \sqrt[14]{a^7 b^7 c^7 d^7}=14\sqrt{abcd}$.
$7abcd+7>S \geq 14\sqrt{abcd} \Rightarrow abcd+1 > 2\sqrt{abcd}$. This is true by AM-GM again, with equality at $abcd=1$, but $a,b,c,d>1$ so we are done. [/hide]

(edit: changed $\iff$ to $\Rightarrow$).. no wait, its not valid. darn.

[hide="Spoiler"]
Use the substitution $a = w + 1$, $b = x + 1$, $c = y + 1$, $d = z + 1$, so $w$, $x$, $y$, $z > 0$.

Then the inequality becomes
\[ 7wxyz + 6(wxy + wxz + wyz + xyz) + 4(wx + wy + wz + xy + xz + yz) > 0, \]
which is trivial.
[/hide]

[quote="JBL"][quote="cincodemayo5590"]$7abcd+7>S \geq 14\sqrt{abcd} \iff abcd+1 > 2\sqrt{abcd}$[/quote]

This is false.  The inequality on the left implies the inequality on the right, but not the other way around.[/quote]

oops, yes you are right.. is my solution still valid if i change that to just an arrow? is the method ok?

[quote="JBL"]If you change that to just an arrow, what you've done is still correct, but it no longer proves what you want it to prove -- it becomes a proof of something true given the problem as an assumption, rather than a proof of the problem given true things as the assumption.

One problem with inequalities is that people often solve them "backwards": your solution begins from the thing you're trying to prove and reduces it to something true.  I think as a general rule it's good to write solutions to inequalities the other way -- start with the true things and then combine them to get the result you want.  That helps avoid the kind of mistake you've made here, which for some reason is a very easy sort of mistake to make when solving inequalities.[/quote]

Ok, i understand what you're saying.. that makes sense that its not the best to work from the given inequality to something true but rather to go the other way. Sweet.. thanks!


Let $ a,b,c\geq0$.Prove that:
$ \dfrac{1}{5(a^2\plus{}b^2)\minus{}ab}\plus{}\dfrac{1}{5(b^2\plus{}c^2)\minus{}bc}\plus{}\dfrac{1}{5(c^2\plus{}a^2)\minus{}ca}\geq\dfrac{1}{a^2\plus{}b^2\plus{}c^2}$

$ \dfrac{1}{5(a^2 \plus{} b^2) \minus{} ab} \plus{} \dfrac{1}{5(b^2 \plus{} c^2) \minus{} bc} \plus{} \dfrac{1}{5(c^2 \plus{} a^2) \minus{} ca}\geq\dfrac{1}{a^2 \plus{} b^2 \plus{} c^2}$
It is clear that numbers $ 5(a^2 \plus{} b^2) \minus{} ab$, $ 5(b^2 \plus{} c^2) \minus{} bc$, $ 5(c^2 \plus{} a^2) \minus{} ca$ are pozitive.
Then apllying Cauchy's Lemma we obtain:
$ \dfrac{1}{5(a^2 \plus{} b^2) \minus{} ab} \plus{} \dfrac{1}{5(b^2 \plus{} c^2) \minus{} bc} \plus{} \dfrac{1}{5(c^2 \plus{} a^2) \minus{} ca}\geq\dfrac{9}{10(a^2\plus{}b^2\plus{}c^2)\minus{}(ab\plus{}bc\plus{}ca)}$
So, we should prove that:
$ \dfrac{9}{10(a^2\plus{}b^2\plus{}c^2)\minus{}(ab\plus{}bc\plus{}ca)}\geq\dfrac{1}{a^2 \plus{} b^2 \plus{} c^2}$ or
$ 9(a^2\plus{}b^2\plus{}c^2)\geq10(a^2 \plus{} b^2 \plus{} c^2)\minus{}(ab\plus{}bc\plus{}ca)$ or
$ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$, which true for $ a,b,c\geq 0$.

You are wrong . This ineq isn't easy like that ^^
[b]Note:[/b] Your solution was wrong ^^

[quote="Inequalities Master"]Let $ a,b,c\geq0$.Prove that:
$ \dfrac{1}{5(a^2 \plus{} b^2) \minus{} ab} \plus{} \dfrac{1}{5(b^2 \plus{} c^2) \minus{} bc} \plus{} \dfrac{1}{5(c^2 \plus{} a^2) \minus{} ca}\geq\dfrac{1}{a^2 \plus{} b^2 \plus{} c^2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158697
But the Vask's proof  nicer.

[quote="Inequalities Master"]Let $ a,b,c\geq0$.Prove that:
$ \dfrac{1}{5(a^2 \plus{} b^2) \minus{} ab} \plus{} \dfrac{1}{5(b^2 \plus{} c^2) \minus{} bc} \plus{} \dfrac{1}{5(c^2 \plus{} a^2) \minus{} ca}\geq\dfrac{1}{a^2 \plus{} b^2 \plus{} c^2}$[/quote]
It's very easy.
We need prove
$ \sum {\frac{{5c^2  \plus{} ab}}{{5(a^2  \plus{} b^2 ) \minus{} ab}}}  \ge 2$
Using Cauchy Schwarz,we have:
$ \sum {\frac{{5c^2  \plus{} ab}}{{5(a^2  \plus{} b^2 ) \minus{} ab}}}  \ge \frac{{(5\sum {a^2 }  \plus{} \sum {ab} )^2 }}{{\sum {(5(a^2  \plus{} b^2 ) \minus{} ab)(5c^2  \plus{} ab)} }}$(1)
Setting q=ab+bc+ca,r=abc,p=a+b+c=1.We have
(1)$ \Leftrightarrow (5\sum {a^2 }  \plus{} \sum {ab} )^2  \ge \sum {(5(a^2  \plus{} b^2 ) \minus{} ab)(5c^2  \plus{} ab)}$
$ (5 \minus{} 9q)^2  \ge 2( \minus{} 108r \plus{} 39q^2  \plus{} 5q)$
 $ \Leftrightarrow (q \minus{} 1)(3q \minus{} 1) \ge 0$ :)

tronghieu, it's the Vasc's proof too. :wink:
By the way, expanding gives $ \sum_{sym}(25a^6\minus{}10a^5b\minus{}50a^4b^2\plus{}15a^3b^3\plus{}16a^4bc\plus{}12a^3b^2c\minus{}8a^2b^2c^2)\geq0,$
which does not triviality.


Let $a,b,c,d$ be four side of a quadrilateral. Prove that
\[\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{c+d+a}}+\sqrt{\frac{c}{a+b+d}}+\sqrt{\frac{d}{a+b+c}}\le 1+\frac{3}{\sqrt{5}}.\]

This is a very hard inequality and chosen as a problem of week of Vietnam Ineq Forum. Detailed here http://www.ch06k.com/forum/showthread.php?t=134.

By LCF-Theorem, it suffices to prove the inequality for $a=b=c=1$ and $0\le d \le 3$.

But I want to see another solution ... Non-LCF and Non-any theorem  :blush:
Can you help me, Vasc or Hungkhtn?  :roll:

[quote="hungkhtn"]Let $ a,b,c,d$ be four side of a quadrilateral. Prove that

$ \sqrt {\frac {a}{b \plus{} c \plus{} d}} \plus{} \sqrt {\frac {b}{c \plus{} d \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b \plus{} d}} \plus{} \sqrt {\frac {d}{a \plus{} b \plus{} c}} < 1 \plus{} \frac {3}{\sqrt {5}}.$[/quote]Without loss of generality, we can assume $ a \plus{} b \plus{} c \plus{} d \equal{} 6$ by the homogeneity.

$ \left(\frac {25 \minus{} 11\sqrt {5}}{100}a^3 \minus{} \frac {25 \minus{} 11\sqrt {5}}{25}a^2 \plus{} \frac {125 \minus{} 43\sqrt {5}}{100}a \plus{} \frac {3 \sqrt {5} \minus{} 5}{10}\right)^2 \minus{} \frac {a}{6 \minus{} a}$

$ \equal{} \frac {(3 \minus{} a)(a \minus{} 1)^2[a^4 \minus{} 9a^3 \plus{} 2(44 \plus{} 15\sqrt {5})a^2 \minus{} 10(31 \plus{} 13\sqrt {5})a \plus{} 200(9 \plus{} 4\sqrt {5})]}{250(123 \plus{} 55\sqrt {5})(6 \minus{} a)}$

$ \geq0\Longrightarrow$

$ \sum{\sqrt {\frac {a}{b \plus{} c \plus{} d}}} \equal{} \sum{\sqrt {\frac {a}{6 \minus{} a}}} < \sum{\left(\frac {25 \minus{} 11\sqrt {5}}{100}a^3 \minus{} \frac {25 \minus{} 11\sqrt {5}}{25}a^2 \plus{} \frac {125 \minus{} 43\sqrt {5}}{100}a \plus{} \frac {3\sqrt {5} \minus{} 5}{10}\right)}$

$ \equal{} \sum{\left[\frac {25 \minus{} 11\sqrt {5}}{100}a^3 \minus{} \frac {25 \minus{} 11\sqrt {5}}{25}a^2\left(\frac {a \plus{} b \plus{} c \plus{} d}{6}\right) \plus{} \frac {125 \minus{} 43\sqrt {5}}{100}a\left(\frac {a \plus{} b \plus{} c \plus{} d}{6}\right)^2 \plus{} \frac {3\sqrt {5} \minus{} 5}{10}\left(\frac {a \plus{} b \plus{} c \plus{} d}{6}\right)^3\right]}$

$ \equal{} \left(1 \plus{} \frac {3}{\sqrt {5}}\right)\left(\frac {a \plus{} b \plus{} c \plus{} d}{6}\right)^3$

$ \minus{} \frac {(a \plus{} b \plus{} c \minus{} d)(5d^2 \minus{} a^2 \minus{} b^2 \minus{} c^2 \plus{} 2bc \plus{} 2ca \plus{} 2ab) \plus{} 4(a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} ca \minus{} ab)(d \minus{} c) \plus{} 4(2c \minus{} a \minus{} b)(a \minus{} b)^2}{60(25 \plus{} 11\sqrt {5})}$

$ < 1 \plus{} \frac {3}{\sqrt {5}},$ which is clearly true for $ a\leq b\leq c\leq d < a \plus{} b \plus{} c.$


Αν $ a,b,c >0$ και $ a\plus{}b\plus{}c \geq abc$ , να αποδειχθεί ότι :

    $ \frac {bc}{a^3(b\plus{}c)} \plus{}  \frac {ca}{b^3(c\plus{}a)} \plus{}\frac {ab}{c^3(a\plus{}b)} \geq \frac{1}{2}$

[quote="stergiu"]Αν $ a,b,c > 0$ και $ a \plus{} b \plus{} c \geq abc$ , να αποδειχθεί ότι :

    $ \frac {bc}{a^3(b \plus{} c)} \plus{} \frac {ca}{b^3(c \plus{} a)} \plus{} \frac {ab}{c^3(a \plus{} b)} \geq \frac {1}{2}$[/quote]

Θέτουμε $ a\equal{}\frac{1}{x},b\equal{}\frac{1}{y},c\equal{}\frac{1}{z}$ οπότε έχουμε $ \frac{bc}{a^{3}(b\plus{}c)}\plus{}\frac{ac}{b^{3}(a\plus{}c)}\plus{}\frac{ab}{c^{3}(a\plus{}b)}\equal{}\frac{x^{3}}{y\plus{}z}\plus{}\frac{y^{3}}{z\plus{}x}\plus{}\frac{z^{3}}{x\plus{}y}$.

Από την ανισότητα Andreescu έχουμε όμως ότι $ \frac{x^{3}}{y\plus{}z}\plus{}\frac{y^{3}}{z\plus{}x}\plus{}\frac{z^{3}}{x\plus{}y}\geq \frac{(x\plus{}y\plus{}z)^{3}}{6(x\plus{}y\plus{}z)}\equal{}\frac{(x\plus{}y\plus{}z)^{2}}{6}\geq \frac{3(xy\plus{}yz\plus{}zx)}{6}$  (1)

Όμως $ a\plus{}b\plus{}c\geq abc \Leftrightarrow xy\plus{}yz\plus{}zx\geq 1$   (2)

Άρα αό τις σχέσεις (1) και (2) έχουμε το επιθυμητό αποτέλεσμα. :)

Η ανισοτητα που χρησιμοποιεις ειναι η γενικευμενη ανισοτητα ADRESSCU ΓΝΩΣΤΗ Ως ΑΝΙΣΟΤΗΤΑ hoLDER


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algebra
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Here's a thought I had recently while looking over a book: in many problems we are asked to evaluate expressions such as $ \sqrt{a\plus{}\sqrt{b}}$. Usually, we go about solving this by squaring and finding the values of a and b by solving a system of algebraic expressions. However, why not use the general binomial theorem. For example, in the above problem, why don't we just do $ (a\plus{}\sqrt{b})^{1/2}$ and proceed with the general binomial theorem. i understand that one main problem arises from the start: the method of using the general binomial theorem consists of a series of calculations that do not terminate! But in theory, could this way not work?

Any thoughts would be appreciated!

The general binomial theorem gives a series with ugly terms.  We don't want a series or ugly terms.  What we are really interested in is a factorization of $ a \plus{} \sqrt{b}$ over the ring of integers in an appropriate [url=http://en.wikipedia.org/wiki/Algebraic_number_field]algebraic number field[/url].

thanks for the help, torajiru. that article was a little over my head. what exactly was it saying?
also, in theory, if you were to evaluate that long string of infinite terms in the expansion of the general binomial theorem as let's say, some sort of sequence or series, than would they tend to the actual factor of $ \sqrt{a\plus{}\sqrt{b}}$?
in other words, summing up the terms we get from the binom. theorem, could we get the actual factor? i know it's impractical, but I want to see if there is any relation.

Summing up the terms from the binomial theorem, we get ...

$ \sqrt{a \plus{} \sqrt{b}}$.

That's an analytical approach to what is essentially a number-theoretic problem, and while it's true that analysis and number theory have far more in common than you might think, those are two very different kinds of square roots.

When we use the binomial theorem to expand a square root, we are looking at $ f(x) \equal{} \sqrt{x}$ analytically, as a function.  That's one way of looking at the square root.

When we are trying to express $ \sqrt{a \plus{} \sqrt{b}}$ using radicals, we are factoring.  That's another way of looking at a square root.

It's the difference between calculating $ \sqrt{64}$ and estimating $ \sqrt{64.1}$.

As far as the "bigger picture" (which is what I was trying to get at with that link) of the second way of looking at a square root, here is the sequence of ideas:

- When we take the square root of an integer to find another integer, we are essentially factoring.  That is, $ \sqrt{64} \equal{} 8$ because $ 64 \equal{} 2^6 \equal{} (2^3)^2$.  This is a factorization of $ 64$ over the integers, which are denoted $ \mathbb{Z}$.
- When we take the square root of something like $ 3 \plus{} 4 \sqrt{2}$ to find a number of the form $ a \plus{} b \sqrt{2}$, we are still factoring, but [i]not over the integers![/i]
- It turns out the right way to express this idea is in terms of algebraic number fields.  We are factoring over the set of numbers of the form $ a \plus{} b \sqrt{2}$, where $ b$ is an integer.  This is denoted $ \mathbb{Z}[ \sqrt{2} ]$, and it is the ring of integers in the algebraic number field $ \mathbb{Q}[\sqrt{2}]$.  (I was being unnecessarily cryptic in the previous post; don't worry about what a ring or a field is for the time being.)
- It also turns out that $ (1 \plus{} \sqrt{2})^2 \equal{} 3 \plus{} 4 \sqrt{2}$, so we have our factorization and hence our square root.  The general problem of factoring some number of the form $ x \plus{} y \sqrt{2}$ is similar to the problem of integer factorization, but that's more detail than we need to get into right now.


inequalities quadratics function conics parabola algebra domain

Let $a$ be an arbitrary constant number.
Solve the following inequality.

\[a(x^2+1)<x(a^2+1)\]

If a = 0, then ineq is x >0
If x = 0, then ineq is a < 0

else, suppose a and x are not zero.

We can write as (a-x)(1 - 1/ax) > 0.

Then a > x and 1 > 1/ax or x>a and 1/ax > 1.

These are conditions for x in terms of a.

[hide="Hint"]For a quadratic function in $x$, $f(x) = ax^2+bx+c$, with $a>0$, the condition for $f(x)>0$, for all $x$, is $b^2-4ac<0$[/hide]

if a=0, S=R+*
else,
let solve ax 2 -x(a 2 +1)+a=0
the solutions are {((a 2 +1)+/-  |a 2 -1 | )/(2a)}
then look at the intervalles.....

it s too easy!!!!

[quote="no_one_2000"]Do those inequalities reduce?  I got a different answer, but this is the first time I've ever worked an inequality problem, so I could be wrong.

I said $a(x^2+1)<x(a^2+1)$ factors to $(ax-1)(x-a)<0$.

That gives us the inequalities $x<\frac{1}{a}$ and $x<a$. (Right?)

Now here's the part I'm not sure on... I said

a >= 1:   x<a
0 < a < 1:   x<1/a
a = 0:   x>0
-1 < a < 0:   x<a
x <= -1:   x<1/a

just from plugging in a few values for a.

... But that doesn't look anything like what you guys got.  Sorry if this is a stupid post, but I'm just trying to make sense of it all. :)[/quote]

At this point your solution is the nearest to the correct answer. :)

You need to examine which is greater $a$ and $\frac{1}{a}$, everyone ;)

yes, that is ture. so you will have to define three dituations:
$-1\le a$
$-1<a<0$
$a=0$
$0<a<1$
$1\le a$

a(x^2+1)<x(a^2+1) is as :  a/(a^2+1)<x(x^2+1)

study the function f/ f(x)= x(x^2+1)

and u'll have the solution.

Wait, wait, I had my inequalities backwards.  *punches self*

Move back to the part where we have x<a and x<1/a.

Okay, it goes like THIS.

1<=a : x<1/a
0<a<1 : x<a
a=0 : x>0
-1<a<0 : x<1/a
a<=1 : x<a

Sorry guys, I can be an idiot sometimes :P

Is this the right answer?

Ouch, wrong twice in a row. :oops:  That's embarrassing. lol

I don't get the hint either, but I just broke out the graphing calculator and graphed a few functions.  Hopefully this time I'm right, but with me, you can never be sure.

Okay, I just took the equation $x(a^2+1)-a(x^2+1)>0$.  I decided to give a random values to $a$ and look at the graph it produced.  My [i]new[/i] values for the intervals of $a$ are...

a >= 1 : 1/a<x<a
0 < a < 1 : a<x<1/a
a = 0 : x>0
-1 < a < 0 : x<1/a, x>a
a <= 1 : x<a, x>1/a

So my questions are:

- Is [b]this[/b] correct? :\
- If so, is there a beter way to solve it?  My answer was based off viewing a bunch of parabolas, which hardly seems efficient.
- If it isn't right, how do you do it?
- Eh, what's the hint mean?

Hello no_one_2000.[quote="no_one_2000"]

a >= 1  :(  : 1/a<x<a
0 < a < 1 : a<x<1/a  :)
a = 0 : x>0  :)
-1 < a < 0 : x<1/a, x>a  :)
a <= 1 : x<a, x>1/a  :(

So my questions are:

- Is [b]this[/b] correct? :\

- If so, is there a beter way to solve it?  My answer was based off viewing a bunch of parabolas, which hardly seems efficient.
- If it isn't right, how do you do it?
- Eh, what's the hint mean?[/quote]

Sorry for confusing you, How do you pronounce $a$, the answer is $x$ (it's joking!).That is to say, we regard $a$ as

variable $x$, which is I wanted to say. :)

kunny

Heh, EUCLA's solution looks a lot like mine.  Anyway, I give up, how do you solve the problem?  Can you post the solution, kunny?

[quote="kunny"]Your solution is wrong, EUCLA :?[/quote]
could you explain me why? :(

[quote="kunny"]First for $a\neq 0$, solve inequality of second degree for $x$, $(x-a)(ax-1)<0$.

Solving the quadratic equation for $x$, $ax^2-(a^2+1)x+a=0\Longleftrightarrow (x-a)(ax-1)=0$,

we have the roots $x=a,\frac{1}{a}$. Then we consider which is greater or equal, $a$ and $\frac{1}{a}$.

Using $a-b$ plane, drawing the two graphs of $b=a$ and $b=\frac{1}{a}$ on the same plane,....

Let's try again, EUCLA and no_one_2000.

kunny[/quote]
but i didn't write anything wrong, did I? if no i only have to change it a little bit... :huh:

EUCLA,Sorry for exaggerating of impossibility for your souition. I should have said your answer was incorrect.

kunny

Can anyone solve this problem completely?

i still find my solution you've said is wrong :?

we are to classify 7 cases of the domain for a.

kunny

No one seem to be posted any more.

Here is the answer.

For $a<-1$, we have $x<a,\frac{1}{a}<x$

For $a=-1$, we have $x\neq -1$

For $-1<a<0$, we have $x<\frac{1}{a}, a<x$

For $a=0$, we have $x>0$

For $0<a<1$, we have $a<x<\frac{1}{a}$

For $a=1$,  there don't exist the set of roots.

For $a>1$,  we have $\frac{1}{a}<x<a$

kunny


conics
ellipse
symmetry
geometry
geometric transformation
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Find the foci of the ellipse $ 25x^2\minus{}14xy\plus{}25y^2\minus{}288\equal{}0$.

[hide="Symmetry"]Note that interchanging $ x,y$ does not change the equation, so the ellipse is symmetric over the line $ x \equal{} y$. Also, negating both $ x,y$ does not change the equation, so the ellipse is point-symmetric about the origin, and is thus origin-centered.

We can thus infer that the endpoints of the two axes will take the form $ \pm (a,a)$ and $ \pm (b, \minus{} b)$. Let's find $ a,b$ to determine which is the major axis.

Plugging in $ (x,y) \equal{} (a,a)$ gives $ 36a^2 \equal{} 288 \implies a \equal{} 2 \sqrt {2}$
Plugging in $ (x,y) \equal{} (b, \minus{} b)$ gives $ 64 b^2 \equal{} 288 \implies b \equal{} (3/2) \sqrt {2}$

Since $ a$ is larger, the major axis is along the line $ x \equal{} y$, and by symmetry, the foci will be $ \pm (c,c)$.

The point $ (b, \minus{} b)$ is equidistant from both foci, so the sum of its distances from the two foci is:
$ 2 \sqrt {(c \plus{} 3/ \sqrt {2} )^2 \plus{} (c \minus{} 3 / \sqrt {2} )^2}$

The point $ (a,a)$ is co-linear with the two foci. Verify that the sum of its distances from the two foci is equal to the length of the major axis of the ellipse (in general). In this case, this gives:
$ 2 \sqrt {2}a \equal{} 8$

By the definition of an ellipse, the sum of the distances from any point on the ellipse to the two foci is constant. Thus:
$ 2 \sqrt {(c \plus{} 3/ \sqrt {2} )^2 \plus{} (c \minus{} 3 / \sqrt {2} )^2} \equal{} 8$
$ 2c^2 \plus{} 9 \equal{} 16$
$ c \equal{} \pm \sqrt {7/2}$

Thus, the two foci are $ \boxed{F_{1,2}\equal{}\pm \left(\sqrt {\frac {7}{2}}, \; \sqrt {\frac {7}{2}} \right)}$[/hide]

There are many other solutions! For example, [hide="you could use"] rotations[/hide].

TZF, thank you so much. Symmety! How neat a method it is. If we do this in another way (rotation), will it be a lot complicated?

The rotation solution would be to replace $ x\equal{}\frac{1}{\sqrt{2}} x'\minus{} \frac{1}{\sqrt{2}} y'$ and $ y\equal{}\frac{1}{\sqrt{2}}  x' \plus{} \frac{1}{\sqrt{2}}  y'$, or in any other way, rotate the entire ellipse about the origin by 45 degrees, so that the axes of the ellipse become parallel to the coordinate axes. Then you can get the ellipse in standard form, and use the usual $ c^2\equal{}a^2\minus{}b^2$ formula (where $ a$, $ b$, and $ c$ are the lengths of the semimajor axis, semiminor axis, and center-to-focus, respectively, UNLIKE in my previous solution).

Then you would get the focus in $ (x',y')$ coordinates, and you could convert those back to $ (x,y)$ coordinates. Of course, if you realized that the ellipse was rotated 45 degrees in the first place, you may as well have used the solution I posted.


Given a triangle ABC, construct (using ruler and compass)  D on AB and E on AC so that AD=DE=EC.

PS
inspired by http://www.mathlinks.ro/Forum/viewtopic.php?t=71581

assume that $D,E$ are on the [u]sides[/u] $AB,AC$

$AD=AE\Rightarrow \triangle DAE$ isosceles $\Rightarrow \angle DEA=\angle A$

a first condition is that $A < 90^o$

$ED=EC\Rightarrow \triangle EDC$ isosceles

The angles of base are equal to $\frac{\widehat A}{2}$ (because $\angle DEA=\angle A$)

Another condition is $\widehat C > \frac{\widehat A}{2}$ (because we want $D$ on the side $AB$)

[size=134][color=blue]Construction[/color][/size]

So, we choose $D$ on $AB$ such that $\widehat{ACD}=\frac{\widehat A}{2}$

Then we find the point $E$ on $AC$ such that $DA=DE$.

Let's check if it is correct

$\triangle ADE$ is isosceles and for the triangle $EDC$ we have

$\widehat{AED} =\widehat{EDC}+\widehat{ECD}\Rightarrow$

$\widehat{A}=\widehat{EDC}+\frac{\widehat A}{2}\Rightarrow$

$\widehat{EDC}=\frac{\widehat A}{2}\Rightarrow$

$\triangle EDC$ isosceles $\Rightarrow ED=EC$

if we allow the point to be on the [u]lines[/u] $AB,AC$, then we have always a solution (no conditions).

Even if $A$ is a right angle, then $E \equiv A$

...

Sorry I misstyped the intention was this:

Given a triangle ABC, construct (using ruler and compass) D on AB and E on AC so that BD=DE=EC.


Ok, I have been trying to figure this out for hours, can't do it.

There are four marking periods.  Each counts as 22%.  There is also a final grade, worth 12% to make the full 100%.  Going into the final, the average of the four 22% marking periods is 59.  What would I need to get on the 12% final to get the 59 up to a 65?  Thanks.

Ok, you have an average of $59$ for 4 marking periods, meaning that $88$% of your grade is a $59$, letting $x$ represent your grade on the final, we have

$.88\cdot 59+.12x=65$

$x=109$, which (without extra credit) is impossible.

so you have 4 marking periods of 22%with an average of 59. Basically, 88% of the grade is 59. Take 59* 0.88. 65 of 100% is obviously 65. 59*0.88 = 51.92. Now, subtract 51.92 from 65. Then you get 13.08. However, you need to divide this by 0.12 since this is weighted. Therefore, you'd need a grade of 109.

[quote="amirhtlusa"]EDIT.
it's impossible. :)[/quote]

I was wondering how you ended up with 50% :D


[hide="2"]i'm getting $>$ here.

$\frac{1}{\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2\left( \sqrt{k+1}-\sqrt{k} \right)$

So if $\sum_{k=1}^n \left( \sqrt{k+1}-\sqrt{k} \right)=\sqrt{n+1} -1$ then the inequality is true.

This is easy with induction, for the base case clearly works (only 1 term) and $\sum_{k=1}^{n+1} \left( \sqrt{k+1}-\sqrt{k} \right)=\sqrt{n+1}-1+\sqrt{n+2}-\sqrt{n+1}=\sqrt{n+2}-1$ so the inductive step works.[/hide]


Can you say where are $A,H,J,K$ ?

I tried to guess what you mean So

We have $EBFG$(Not EBGF) as a quadrilateral with $\angle GEB=\angle BFE=90$ and $C$ belongs to $EG$ that $EC=EB$.

PS. Hmmm darij post was not there when i Wanted to post this ,
Darij, you are fast man .

Download the document to see the picture of the problem

First, please write problems in words (a picture is good, but the text should suffice for understanding the problem); then, I don't understand your picture (e. g. the point E doesn't occur there at all).

  darij

$E$ is the foot of the perpendicular from $A$ to $EB$.


one pair of zeroes, one pair of decimal points, one pair of ones,
one pair of parentheses, one "^" for exponentiation, one "-" for
subtraction or for negation (depending on the particular expression), and one exclamation point for a factorial sign?
(There are eleven characters in total.)

1) You must use all of these to form your expressions each time,
    and not ones from Latex.  You may not use any other
    characters besides these.

2) The following list shows number forms with leading zeroes that  are not allowed for my purposes*:
    01, 001, 010, 011, 0011, 0101, 0110, 00.1, 00.11, 01.0, 01.1,
    01.01, 01.10, 001., 001.1, 010., 011., 010.1, 011.0, 0011.,
    0101., 0110., 01., 00, .00, 00., 0.0, .0, 0.

3) In addition to these imposed restrictions, the expressions must
be valid under no higher than a TI-83 Plus Graphics for their order of operations.  Regardless of this, the indeterminate form, 0^0, is not allowed here.

4) You must work with a factorial of only a nonnegative integer
in an expression.

* Although the calculator would work with them.

Your expression given is a valid one.

I ultimately want the total number of ways.

Open hint for all:  There are more than 10 ways of expressing 1 using these

rules, including the expression lingomaniac88 donated.

Update:

To reduce the number of ways and because at least the TI-83 model recognizes these, please disregard any expressions that make use of
(if they would be possible) the following list:

0!0, 0!1., 0!1.0, 1!0, 1.!1., 1.!1.0, 1.!1.00, 1.0!0, 1.0!1., 1.0!1.0, 1.00!1.,
10.!0, 10.!1., 10.!1.0, 10.!10., or 11.!0

Also disregard expressions that might have made use of
(if they would be possible) of this shorter list:

100.!1., 101.!0, or 110.!0

There IS an answer (the number of ways), but it is much greater than I wanted it to be.  I want this thread to end.


The equations are equivalent to $a^2+ab+b^2 = c^2+ cd+d^2$ and $a^4+2a^3b+3a^2b^2+2ab^3+b^4 = c^4+2c^3d+3c^2d^2+2cd^3+d^4$.
And the right one is clearly the square of the left one.

I have seen this problem on this site. This problem is posted by Arne, I think.
I posted the solution, but I can't find it. :(

I could find it, Beat. Here is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=51048]www.mathlinks.ro/Forum/viewtopic.php?t=51048[/url]


This seemed the best board to post this (please move if wrong) ;

I am new to problem solving in the sense of the goal of the site, and I was wondering which, if either, of the AoPS would be suitable for me to hopefully get a good grasp on the great hobby  :)

I am 17 and have found school maths too easy here in England. I got a high A* in GCSE, and am currently doing the Further Maths A Level, just about to take my AS Maths next week (expected to get 100% on the tests)

In terms of competitions, I have done the Junior, Intermediate and Senior UK Maths Challenges, typically scoring Gold but never prepared in any way for this, and not scoring high enough for BMO, just turned up and did the test.

Are either of the books any use to me? I have read Polya's book and have Mathematical Olympiad Challenges (which I am finding too difficult)

Any suggestions are welcome.

Hmm one idea might be to take a practice AMC 12 or AHSME (you can find some here: http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm ) and if you get considerably over 100 (like 110+) then you should get AoPS vol 2 and if you get less, you might want to get AoPS vol 1.

Oh, 6 pts (5 for AHSME) for each right answer, 2.5 for each skipped problem, and 0 for incorrect problems.

Or you can just wait for somebody else's more wise advice. :dry:

It sounds like AoPS 2 is probably at your level.  It is a good book for working your way up to being able to read a book like Math Olympiad Challenges and other Olympiad level material (as well as undergraduate math preparation in general).

Ah!

If you scored fairly well to get Gold in UK Senior Math Challenge, AoPS V.2 will be good for you. But I also suggest buying first one too because even though many concepts will be review, geometry section in Volume 1 is very excellent. Remember that most of the geometry problems in Olympiad level can be solved with pure geometry (Trig is sometimes used but there are some that can be solved by pure geometry).

Only reason I can answer your question is because I took look at couple of UK Senior and Junior Math Challenges.  :)

Hope this helps!


hmm, looks difficult - maybe yetti or someone else can respond?

[hide]not really... just remember that since it is a right triangle, the median is half the hypotenuse, or 10, let the median touch the hypotenuse at M, then find EM and FM with the angle biscector theorem, and subtract them[/hide]

Can you write the solution because I dont exactly see what you are saying. I have a hard time visualizing.

Since this is a right triangle, then the median, let's call it $XM$ where $M$ is on the hypotenuse$=10$, from there, using angle bisector theorem, we can see that $5x+6x=10 \Longrightarrow x=\frac{10}{11}$, thus $EM=\frac{50}{11}$, now we can find out $EM$ and using angle bisector again, we have $5x+8x=10 \Longrightarrow x=\frac{5}{7}$, thus $FM=\frac{25}{7}$, since $EF+FM=EM$, we see that $\boxed{EF=\frac{75}{77}}$.

MarioKart, angle bisector says that the rato of the sides of the segment where the angle bisector touches is the same ratio of the sides adjacent to the angle bisector  :)

[hide]The median of $\angle {B}$ cuts $AC$ at $M$.

Triangle $ABC$ has angle $B=90degrees$, so $AM=BM=CM=10$

In triangle $ABM$, by angle bisector theorem, we have:

$\frac{BE}{EM}=\frac{AB}{AM}=\frac{6}{5}$

$\implies EM=\frac{5BE}{6}$

Similarly in triangle $BCM$ we will have: $FM=\frac{5BF}{8}$

Now: $BE+EM=BM=10$

$\implies BE+\frac{5BE}{6}=10$

$\implies BE=\frac{60}{11}$

Also: $BF+FM=BM=10$

$\implies BF+\frac{5BF}{8}=10$

$\implies BF=\frac{80}{13}$

$\boxed {EF=BF-BE=\frac{80}{13}-\frac{60}{11}=\frac{100}{143}}$

[/hide]

[hide]I won't bother to prove a few things and will skip a few basic steps:
Let $AB=12, AC=16, BC=20; A=(0,0), B= (12,0), C=(0,16)$
The median out of $A$ intersects $BC$ at $D= (6,8)$
The angle bisector out of $B$ intersects $AC$ at $X=(0,6)$.
The angle bisector out of $C$ intersects $AB$ at $Y=(\frac{16} {3},0)$
[u]Equations for certain lines:[/u]
$AD$: $y=\frac{3}{4}x$
$BX$: $y=-\frac{1}{2}x+6$
$CY$: $y=-3x+16$
$AD$ and $BE$ intersect at $E=(\frac{24}{5},\frac{18}{5})$
$AD$ and CF intersect at $F=(\frac{64}{15},\frac{16}{5})$
So:
$EF=\frac{2}{3}$
Done! [/hide]
That took a while to do! :)

[quote="Mario_KaRt"]so which answer is correct? 75/77 or 100/143?[/quote]

What we have done basically tells you how to do this problem, so maybe one of us made stupid mistake in doing some arithmetics. You can try again and figure it out what we have done wrong  :D .

[quote="Mario_KaRt"]so which answer is correct? 75/77 or 100/143?[/quote]
100/143 is the correct answer.  I solved it using the brain-dead technique, so I'm sure it's right.


Let be $ AM$ a bisector in the triangle $ ABC\ ,\ M\in (BC)$ . Show that :

$ AM^2\equal{}AB\cdot AC\minus{}MB\cdot MC$

hello, we have $ \overline{MC}\equal{}\frac{ab}{b\plus{}c}$ and $ \overline{MB}\equal{}\frac{ac}{b\plus{}c}$, inserting this in your equation we get
$ \overline{AM}^2\equal{}bc\left(1\minus{}\frac{a^2}{(b\plus{}c)^2}\right)$ this is the equation for the bisector.
Sonnhard.

hello, for the equation of an angle bisector see here
http://mathworld.wolfram.com/AngleBisector.html
Sonnhard.

Ray $ AM$ cuts the circumcircle of $\triangle ABC$ again  at $ P.$ Since $AP$ is self-isogonal with respect to $\angle BAC,$ it follows that $ AP \cdot AM \equal{} AB \cdot AC \ (\star),$  but from power of $ M$ with respect to the circumcircle $\odot(ABC)$ we get

$BM \cdot MC \equal{} AM \cdot MP \equal{} AM(AP \minus{} AM) \equal{} AM \cdot AP \minus{} AM^2$

Combining with $(\star)$ yields  $AM^2 \equal{} AB \cdot AC \minus{} MB \cdot MC.$


Let $a,b,c$ be positive real numbers such that $abc=1$, find the least positive constant $r$ such that the following inequaltiy holds
\[ a+b+c \leqslant a^r+b^r+c^r. \]

[quote="Davron"]I found the solution !!!

 r=4k for k any Natural number.

 And then use the Muirheads Inequality. Hope that you will understand![/quote]
I think, also solution is $r=3k+1$, since $a^{3k+1}+b^{3k+1}+c^{3k+1}\geq a^kb^kc^k(a+b+c)$  and then use the Muirheads Inequality.

I think you mean r = 1. It is the best constant by muirhead, since the ineq is true for any $r \ge 1$

And I think the right answer is $r=1$.
If $r<1$, then using the Muirheads Inequality we get $a+b+c\geq a^{\frac{1-r}{3}}b^{\frac{1-r}{3}}c^{\frac{1-r}{3}}(a^r+b^r+c^r)=a^r+b^r+c^r$ for all $a,b,c\geq 0$.
If $r=1$ we have equality.

Yes, pavel is right but pvthuan, what's the condition of $r$ (in $R$, $N$...)

I don't think this is a hard inequality If you use power-mean Ineuqlity :)

Indeed zhaobin, and... :
http://www.mathlinks.ro/Forum/viewtopic.php?t=68147 ;)

Oh i did not read the problem carefully and sure the answer is 1 about the equality.
But the general answer is 4k=r for kE Z+


$ \text{1. Let}$ $ f(x) \equal{} x^{x\plus{}1} \minus{} (x\plus{}1)^x.$ $ \text{Find all zeros of}$ $ f(x).$ $ \text{Prove that the list is complete.}$ $ \text{Let}$ $ \alpha$ $ \text{be the largest zero of}$ $ f(x).$ $ \text{Show that}$ $ f(x) > 0$ $ \text{for}$ $ x > \alpha$.
$ \text{2. Now, for all natural number}$ $ n,$ $ \text{let}$ $ f_n(x) \equal{} x^{x\plus{}n} \minus{} (x\plus{}n)^x.$ $ \text{Find all zeros of}$ $ f_n(x).$ $ \text{Let}$ $ \alpha_n$ $ \text{be the largest zero of}$ $ f_n (x).$ $ \text{Is it necessarily true that}$ $ f_n(x) > 0$ $ \text{for}$ $ x > \alpha_n$ $ \text{?}$


im considering getting one of these books.

how do these two books compare to one another?

and what are the strengths/weaknesses of each book in terms of the topics covered.

also, how good are the problems?

ACoPS is crazy good. I have not read the other book, but i certainly recommend ACoPS. It will change your way of approaching math problems, plus it has a hoard problems.(as in math)

all in all, ACoPS is a must. it's only flaw is that it doesn't have solutions.

You can email Mr. Zeitz for a solutions manual and he'll send one (I think). As the previous poster, I have not read Mathematical Olympiad Challenges so I cannot comment on it. ACoPS is really good though (you might have figured that out by reading other threads :)).

Ditto. A unique perspective is given by ACoPS. Hmm, can you actually get a solutions manual? If so, why does he not publish/sell it?

Mathematical Olympiad Challenges is great too. I think it's also a little more difficult.

[quote="pythag011"][quote="pinkmuskrat"]Ditto. A unique perspective is given by ACoPS. Hmm, can you actually get a solutions manual? If so, why does he not publish/sell it?[/quote]

1. Because IMO, not having a solution manual is a benefit.

2. Because some teachers use it as textbooks.

Both are good books, really depends on what you are looking for: If you are someone who is starting usamo and has no idea how to approach problems, ACoPS is better. If you already have a good way of starting on problems, or you prefer just doing whatever you think of (which is actually a very good way, i'm serious.) By approach, i mean you don't know how to find a way to start it. Mathematical Olympiad Challenges is better if you already know how to approach most problems and want to expand your knowledge in subjects.[/quote]

I agree with pythag. But, one more thing, mathematical olympiad treasures is a GREAT book. Even though its is not extremely thorough, it teaches "tricks" consciously and you also learn (subconsciously) how to go about coming up with lemma and corollaries.

Out of curiosity, how does Problem Solving Strategies compare to the two books?

[quote="NuncChaos"]And as a follow up, what is the difficulty level of ACoPS?

It's geared more towards USAMO preparation than AIME preparation, rite?[/quote]
Well, if you read the "tools" sections first then it'll help you for AIME, but if you read the strategies section first you may have a tough time going along.


algebra
polynomial
modular arithmetic
absolute value
algebra proposed

Prove that for all $ n\geq 2,$ there exists $ n$-degree polynomial $ f(x) \equal{} x^n \plus{} a_{1}x^{n \minus{} 1} \plus{} \cdots \plus{} a_{n}$ such that
(1) $ a_{1},a_{2},\cdots, a_{n}$ all are unequal to $ 0$;
(2) $ f(x)$ can't be factorized into the product of two polynomials having integer coefficients and positive degrees;
(3) for any integers $ x, |f(x)|$ isn't prime numbers.

For $ n>3$, $ f(x)\equal{}x^n\plus{}12x^{n\minus{}1}\plus{}3x^{n\minus{}2}\plus{}12x^{n\minus{}3}\plus{}\ldots \plus{}12$. It is irreducible due to Eisenstein Criterion and $ 4$ divides $ f(x)$ for all integer $ x$ so it can't be prime.
For $ n\equal{}3$, $ f(x)\equal{}x(x\plus{}1)(x\plus{}2)\plus{}12$. It is irreducible because if it is not it must have an integer root. $ 6$ divides $ f(x)$ for all integer $ x$ so it can't be prime.
For $ n\equal{}2$, $ f(x)\equal{}x^2\plus{}x\plus{}100$

I am not sure I understand the example for $ n > 3$. How is it defined for even $ n$? with 3 as the coefficient of $ x^0$, or 12?
Also, why is it divisible by $ 4$? For $ n \equal{} 5$, $ f(x) \equal{} x^5 \plus{} 12x^4 \plus{} 3x^3 \plus{} 12x^2 \plus{} 3x \plus{} 12 \equiv x^5 \minus{} x^3 \minus{} x \equal{} x(x^4 \minus{} x^2 \minus{} 1) \pmod 4$.
If we take x=1, we get $ \minus{} 1 \equiv 0 \pmod 4$.

If $ n\equal{}2$, consider $ f(x)\equal{}x^2\plus{}x\plus{}100$.

If $ n$ is odd, consider $ f(x)\equal{}x^n\plus{}21x^{n\minus{}1}\plus{}21x^{n\minus{}2}\plus{}\ldots\plus{}21x^2\plus{}14x\plus{}42$.

If $ n$ is even and greater than $ 2$, consider $ f(x)\equal{}x^n\plus{}21x^{n\minus{}1}\plus{}21x^{n\minus{}2}\plus{}\ldots\plus{}21x^3\plus{}14x^2\plus{}42x\plus{}42$.

bambaman, the rest of the coefficients are $ 12$. sorry for this unclear notation.

$ f(x) \equal{} x^n \plus{} 4n x^{n \minus{} 1} \plus{} 4 x^{n \minus{} 2} \plus{} ...$ is irreducible by Perron's criterion for $ n \ge 2$.

The official solution:
Let $ f(x)\equal{}x^n\plus{}210(x^{n\minus{}1}\plus{}x^{n\minus{}2}\plus{}\cdots \plus{}x^2)\plus{}105x\plus{}12$

First $ f(x)$ is irreducible by Eisenstein Criterion for $ P\equal{}3$
Second $ f(x)$ is even for all integer $ x$

If $ f(x)\equal{}2$ has integer root $ x$, then $ g(x)\equal{}x^n\plus{}210(x^{n\minus{}1}\plus{}x^{n\minus{}2}\plus{}\cdots \plus{}x^2)\plus{}105x\plus{}10$ has integer root, but $ g(x)$ is irreducible by Eisenstein Criterion for $ P\equal{}5$, a contradiction.

If $ f(x)\equal{}\minus{}2$ has integer root $ x$, then $ g(x)\equal{}x^n\plus{}210(x^{n\minus{}1}\plus{}x^{n\minus{}2}\plus{}\cdots \plus{}x^2)\plus{}105x\plus{}14$ has integer root, but $ g(x)$ is irreducible by Eisenstein Criterion for $ P\equal{}7$, a contradiction, too.

So for any integer $ x$, $ f(x)$ is even and its absolute value is not $ 2$, which means $ f(x)$ is a composite number.
Q.E.D

For $n = 2$ use $f(x) = x^2 + x + 4$. For $n \ge 3$, choose a large prime $q$ and $k$ such that $q| 3k-1, 3k-1 > q^{1778}$.
Take $f(x) = x^{n-2}(x^2 -1 + 3k) + 3q$ which is always divisible by $3$. Moreover, by Eisenstein's Criterion on prime $q$, $f$ is irreducible. It suffices to check that $|f(x)|$ is not prime for any integer $x$. Suppose otherwise, then $3|f(x) \implies f(x) = \pm 3$, i.e. $x^{n-2}(x^2 - 1 + 3k) = -3q \pm 3$. This is false for $x = 0$ as we've chosen $3k-1 > q^{1778} > 3q \cdot 1778$. Thus $x \ne 0$. But then the absolute value of LHS is at least $x^2 - 1 + 3k > -1 + 3k > q^{1778} > 3q\cdot 1778$ again, contradiction.

Does this work?

Case 1: $n$ is even.
Consider $a_1 = 239$, $a_2 = a_3 = \cdots = a_{n-1} = 1434$, and $a_n$ is a sufficiently large even multiple of $239$ but not $239^2$ (clearly infinitely many such positive numbers exist). Such a polynomial is irreducible over the integers by Eisenstein with $p=239$. We see that $f(x)$ is always even because $f(0) = a_n\equiv 0\pmod 2$ and $f(1)  = 1 + \sum a_i \equiv 0\pmod 2$. It remains to show that $f(x)\ne \{-2,2\}$. Since $f(x) - a_n$ has a lower bound, we may choose $a_n$ large enough such that $f(x) >2$ for any integer $x$.

Case 2: $n$ is odd.
Consider $a_1 = 69$, $a_2 = 92$, $a_3 = \cdots = a_n = 276$. Such a polynomial is irreducible over the integers by Eisenstein with $p=23$. Now we show that $|f(x)|$ is never prime for any integer $x$.

Claim: $f(x)$ is a multiple of $6$ for any integer $x$
Proof: It suffices to show that $2$ divides $f(0)$ and $f(1)$, and also that $3$ divides $f(0), f(1), $ and $f(2)$.

We have $f(0) = a_n = 276\equiv 0\pmod 6$.

We have $f(1) = 1 + \sum a_i$, which is even because it's equivalent to $1 + a_1 $ modulo $2$, and a multiple of $3$ because it's equivalent to $a_2 + 1 \pmod 3$.

We have $f(2) \equiv 2^n + a_2 2^{n-2} \equiv 2 + 2a_2 \equiv 0\pmod 3$, as desired. $\square$

Therefore, $|f(x)|$ is never prime since there are no prime multiples of $6$.

We give the following constructions.

For $n \ge 4$, we can consider $x^{n} - 21x^{n-2} - 12$ which is irreducible by Eisenstein's criterion and always divisible by $4$.

For $n = 2, 3$, we have $x^n - 5x - 30$ which is always divisible by $2$ and never $2$ or $-2$ at an integer. Once again, it is irreducible.

There are lots of possible approaches to this; the main idea is to try and force $f$ to be even. In particular, set $a_n = 3$ and $a_1 = 15$, with $a_i = 0$ for all other $i$. Notice that $f(x)$ is always even and never equal to two, but by Eisenstein with $p=3$, $f$ is irreducible.


modular arithmetic
algebra
binomial theorem

What is the one's digit of $ (2 \plus{} \sqrt 5)^{2008}?$

[hide]By the binomial theorem, $ (\sqrt 5 \plus{} 2)^{2008}\plus{}(\sqrt{5} \minus{} 2)^{2008} \equiv 2(5^{1004} \plus{} 2^{2008} \plus{} 10[\cdots]) \equiv 2^{2009} \pmod{10}$. We have $ 2^{4} \equiv 1 \pmod{5}$, $ 2^{2009} \equiv 2 \pmod{5}$, and by CRT, $ 2^{2009} \equiv 2 \pmod{10}$. Since $ 0 < (\sqrt{5} \minus{} 2)^{2008} < 1$, we know that $ (\sqrt 5 \plus{} 2)^{2008}$ is very slightly less than a number with last digit $ 2$, eg it ends in a $ 1$.[/hide]


linear algebra
matrix
algebra
polynomial

Let $ A$ be a $ 2\times 2$ matrix and $ k$ be a positive integer number, $ k > 2$. Prove that if $ A^k \equal{} 0$ then $ A^2 \equal{} 0$.

The minimal polynomial of a $ 2 \times 2$ matrix has degree either one or two.

Note that since $ A$ is a $ 2 \times 2$ matrix, it must be the case that:

$ A^2 \minus{} tr(A) \plus{} \det(A) \equal{} 0$.

We realize that:

$ 0 \equal{} \det(A^k) \equal{} \det(A) ^k \Rightarrow \det(A) \equal{} 0$

and $ 0 \equal{} tr(A^k) \equal{} tr(A) ^k \Rightarrow tr(A) \equal{} 0$.

Hence, $ A^2 \equal{} 0$.

[quote="mathangel"]Note that since $ A$ is a $ 2 \times 2$ matrix, it must be the case that:

$ A^2 \minus{} tr(A) \plus{} \det(A) \equal{} 0$.[/quote]
This is a special case (easier to prove than the general case) of the [url=http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem]Cayley-Hamilton Theorem.[/url]

[quote="mathangel"]

$ tr(A^k) \equal{} tr(A) ^k$

[/quote]
This is not always true.

In order to prove that $ tr(A)\equal{}0$ use  the fact that $ tr(A^k)\equal{}\lambda_1^k\plus{}\lambda_2^k$ and $ tr(A^{k\plus{}1})\equal{}\lambda_1^{k\plus{}1}\plus{}\lambda_2^{k\plus{}1}$, where $ \lambda_1,\lambda_2$ are the eigenvalues of $ A$.

$ A^k \equal{} 0$ means that A is nilpotent. Therefore, all its eigenvalues are zero. Since the trace is the sum of the eigenvalues and the determinant is their product, both these quantities are zero. Therefore, from $ A^2 \minus{} tr(A) \cdot A \plus{} \det(A) \equal{} 0$ we conclude that $ A^2 \equal{} 0$. In fact, this must be the case with any nilpotent matrix: if A is a nilpotent matrix of order n then $ A^n \equal{} 0$.


Let $ x_0$ be a root of the equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$, whth $ ab \ne0$. Prove that:
\[ \left| {{x_0}} \right| \le \frac {{{b^2} \plus{} ac}}{{\left| {ab} \right|}}\]
:idea:

hello, i think your inequation is wrong, consider the equation $ x^2\plus{}2x\minus{}3\equal{}0$.
Sonnhard.

Sory, my inequality is :oops:

Let $ x_0$ be a root of the equation $ ax^2 \plus{} bx \plus{} c \equal{} 0$, whth $ ab \ne0$. Prove that:
\[ \left| {{x_0}} \right| \le \frac {{{b^2} \plus{} \left| {ac} \right|}}{{\left| {ab} \right|}}\]

[quote="drst123"]WLOG a>0

First, just check that $ \max(\left|x_0\right|) \equal{} \frac {\left|b\right| \plus{} \sqrt {b^2 \minus{} 4ac}}{2a}$

So prove a inequality $ \frac {\left|b\right| \plus{} \sqrt {b^2 \minus{} 4ac}}{2a} \le \frac {b^2 \plus{} a\left|c\right|}{a\left|b\right|}$

if and only if $ b^2 \plus{} \left|b\right|\sqrt {b^2 \minus{} 4ac} \le 2b^2 \plus{} 2a\left|c\right|$

if and only if $ \sqrt {b^2 \minus{} 4ac} \le \left|b\right| \plus{} 2a\left|\frac {c}{b}\right|$

if and only if $ b^2 \minus{} 4ac \le b^2 \plus{} 4a\left|c\right| \plus{} 4a^2\frac {c^2}{b^2}$

So proof is end (equality holds when c=0 and $ x_0$ isn't 0)[/quote]

Yes Nice Proof ,drst123. My Way for this Problem is the same as your and i think there is only it :)


There are $ a,b,c \in \mathbb{N}$ such that
$ \begin{align*} a & = b + c \\
b & = \dfrac{10c}{a}$
Find all possible values of $ c$. (Source: Myself)

3 variables, 2 equations... blah it becomes indefinite equation

[hide="Big Hint"]
Put $ c$ on one side of the equation for both equations. See anything interesting?
[hide="Bigger"]
Divide both equations after doing the previous step and then cross multiply.
[hide="Biggest"]
Desperate are we?

Factor. That's all I'm going to say.  :P

[/hide]
[/hide]
[/hide]

[quote="corps0507"]I'll just post my answer. :lol:

$ c \equal{} 5, 9, 32$ and $ 81$[/quote]
Can you post your full solution please ?
I don't get it.

[b]@FantasyLover:[/b] You actually have at least 3 conditions. Suppose you rewrite the condition $ a,b,c \in \mathbb{N}$ as the equation: $ \{ a \} \equal{} \{ b \} \equal{} \{ c \} \equal{} 0$, where braces determine the fractional part of a real.

[b]Approach:[/b] [hide]We know by the first equation that if $ a$ and $ b$ are integers, $ c$ is an integer. So for simplicity, suppose there are no restrictions on $ c$, apart from the two equations.

Multiply the second equation by $ a$ and substitute for $ c$ to get $ ab \equal{} 10a \minus{} 10b \implies ab \minus{} 10a \plus{} 10b \equal{} 0 \implies (a \plus{} 10)(b \minus{} 10) \equal{} \minus{} 100$[/hide]

[quote]Multiply the second equation by  and substitute for  to get ab = 10a - 10b \implies ab - 10a + 10b = 0 \implies (a + 10)(b - 10) = - 100
[/quote]

I think you gave another equation to be satisfied. It is correct, but this equation looks for the solutions for $ a$ and $ b$, and the author wants to find $ c$. Remember that we have to obtain something (a solution in this case) in the least possible time needed or required.  :wink:

Here's my solution.

Multiply the first equation by $ b$ and substitute $ ab \equal{} 10c$
$ ab \equal{} b^2 \plus{} bc$
$ 10c \equal{} b^2 \plus{} bc$

Solve for c gives you

$ c \equal{} \dfrac{b^2}{10 \minus{} b}$

Note that $ b$ is restricted on the interval $ \left[1, 10\right).$ Also, we want $ c$ to be a natural number.

Try $ b \equal{} 5, 6, 8, 9$. I'm sure $ c$ is a natural number and is equal to $ 5, 9, 32, 81$

You can also do it in terms of $ a$ and $ c$, but $ a$ is not restricted and is infinitely many. It consumes more time when it comes to finding a natural number $ c$. Checking the solution gives you the set of triples $ \left(a,b,c\right)$, particularly: $ \left(10,5,5\right)$ $ \left(15,6,9\right)$ $ \left(40,8,32\right)$ and $ \left(90,9,81\right)$

b=10c/a => a=10c/a + c   or   a^2=10c + ac
=> a^2 - ac - 10c =0
For a is a integer so c^2 + 40c = k^2
=> (c+20)^2 - 400 = k^2
We can easily find the value of c are 5,9,32,81.

Okay, the equations are

a = b + c
and
b = 10c/a

Using sunehra's BIG HINT we get

c = a - b
c = ab/10

Using sunehra's BIGGER HINT

well, a - b = ab/10, and we can divide both sides by 1 to get
(a-b)/1 = ab/10, and if we cross multiply we get

10a - 10b = ab

And then (I honestly didn't have to use sunehra's last hint) we can use Simon's Favorite Factoring Trick, which I just so happened to look at earlier today. (What a coincidence).

We want to get the variables on one side like below;

ab + 10b - 10a = 0

Now we can try to factor;

(a _) x (b _)
We know that we need a positive ten after the a to get us the 10b, so we have

(a + 10) x (b _)

We also need a negative ten after the b to get us the -10a, so we have

(a + 10) x (b - 10)

But that also gives us a negative 100, so we have to subtract 100 from the other side to get;

(a + 10) x ( b - 10) = -100

Now we can have a+ 10 = c, and b - 10 = d, and then -100=cd, and all we have to do from there is find all the possibilities for c and d, and then when we find all the values of c, we subtract ten from all of them to get a, and add 10 to all the values of d to get b. And then we use the values of a and b to find the values of c.


induction
function
algebra unsolved
algebra

Let $ f: R \to R$ such that
$ i)f(0)\equal{}1$
$ ii)2f(u)f(v)\equal{}f(u\plus{}v)\plus{}f(u\minus{}v)$ for all $ u,v,\in R$
$ iii)$ Exist $ c\in R$ such that $ f(x)\le c$ for all $ x\in R$
Prove that $ |f(x)|\le 1$

Solution...
[hide]
Suppose $ |f(t)| \equal{} A > 1$. Putting $ u \equal{} v \equal{} t$ in $ ii$ we will have:
$ 2A^2 \equal{} f(2t) \plus{} 1$
$ f(2t) \equal{} A^2 \plus{} (A^2 \minus{} 1)$;

But $ \frac {|f(2t)|}{|f(t)|} \equal{} \frac {A^2 \plus{} (A^2 \minus{} 1)}{A} \equal{} A \plus{} \frac {A^2 \minus{} 1}{A} > A$.

So, it easy to see (you can prove it by induction) that $ \frac {|f(2^{n}t)|}{|f(t)|} > A^{n}$.
Now, $ |f(2^{n}t)| > A^{n\plus{}1}$ and function cannot be bounded. Contradiction.[/hide]


Does anyone here do Science Bowl? And if so, have you gone to nationals before? Just curious...I was involved with my school's team as a freshman and might be on the team next year. We made it to nats and I hope we do so again next year...

I was on my school's science bowl team three times, but we never made it to nationals. The closest we came was in 2002 when we lost to one school twice, the first time by two points, and then second time by four points (because they guessed correctly on a multiple-choice tossup without hearing the question when four seconds were left and we were up 10 points. Then they proceeded to get the bonus question correct. AAAAAACK), Bay Area regionals are tough.

Bay Area regionals are really tough...Lynbrook/Monta Vista/Mission San Jose <-especially that one...and the strange thing is that CA has the most regionals. We have 8 or 9 and most other states have 1. Or 2. They got the bonus right? Ouch. Painful. Heh. 2003 was no contest...Lynbrook swept all the rounds...=) MSJ is going to be after us with a vengeance...

Don't worry too much about MSJ. They were incredible in 2001 and 2002 (and maybe earlier) because of their captain, who was amazing. In 2003 they were a bunch of new people. I watched some of their games because I scouted out in the Las Positas regional to remind myself of some science bowl tactics a week before our regional at Oakland Operations Office. Then again, Lynbrook won't exactly be a powerhouse next year without Jimmy. Come to think of it, I can't recall any teams that will be really strong next year.

I know. Jimmy steamrollered the other teams. Although Lisa was getting the hang of it by Nats. Next year...next year is very worrying. You were at Las Positas? I didn't see you. Hm...I ought to PM this...


algebra
polynomial
number theory proposed
number theory

Determine $ x$ irrational so that $ x^2\plus{}2x$ and $ x^3\minus{}6x$ are both rational.

$ x$ is irrational but a common root of the rational polynomials $ x^2 \plus{} 2x \minus{} p, x^3 \minus{} 6x \minus{} q$.  The former must be the minimal polynomial of $ x$ (over $ \mathbb{Q}$), hence it divides the latter.  Expanding,

$ (x^2 \plus{} 2x \minus{} p)(x \minus{} r) \equal{} x^3 \plus{} (2 \minus{} r)x^2 \plus{} (\minus{}2r\minus{}p)x \plus{} pr \equal{} x^3 \minus{} 6x \minus{} q$.

We conclude that $ r \equal{} 2, p \equal{} 2, q \equal{} \minus{}4$.  This gives

$ x \equal{} \frac{\minus{}2 \pm \sqrt{4 \plus{} 4}}{2} \equal{} \boxed{\minus{}1 \pm \sqrt{2} }$.


induction
combinatorics proposed
combinatorics

The squares of an $n\times n$ board are colored with $3$ colors, and we know that each square has neighbours colored in both the other two colors (neighbours are squares sharing a side). Show that there are two neighbours colored the same.

Assume the contrary. Let's denote the colors $1,2,3$.
Let $a_{i,j}$ be the number in the common square of the $i^{th}$ row and $j^{th}$ column.
Wlog we may assume that the upper left square is $a_{1,1} = 1$. Since its two neighbors must use the two other colors, we may assume that $a_{1,2} = 2$ and $a_{2,1} = 3$.
Then, from our hypothesis, $a_{2,2}$ cannot be $2$ or $3$, so that $a_{2,2} = 1$.
It follows that $a_{1,3} = 3$ otherwise the neighbors of $a_{1,2}$ would not use the two colors $1$ and $3$.
Similary, we deduce that $a_{2,3} = 2$ and so on. By an easy induction, we dedduce that $a{1,k} = k$ mod[3] and $a_{2,k} = a_{1,k}-1$ mod[3].

It follows that the two neighbors of $a_{1,n}$ have the same color contradicting the statement of the problem.

Pierre.

I did the same thing, and I have a solution here (in a book) which is exactly the same :). Apparently, it's the only way.. Thanks!

Well, i don't know if it is the only way, but it is the obvious way when you draw a figure and start to fill the $n \times n$ square...

Pierre.


A number $ x$ is 5 less than number $ y$. Three times $ x$ is 10 less
than twice $ y$. What is the value of $ x \minus{} y$?

$ x \equal{} y \minus{} 5\implies2x\equal{}2y\minus{}10$ (1)
$ 3x \equal{} 2y \minus{} 10$ (2)
$ x\equal{}0$ (2)-(1)
$ y\equal{}5$
$ 0\minus{}5\equal{}\boxed{\minus{}5}$.

it's a trick question: it tells you
$x = y-5$

then it asks what $x-y$ is

$\begin{align*}x&= y - 5\\ x -y& = y- 5 -y \\ x- y& =-5$

you don't actually have to solve for y or x


vector
geometry
geometric transformation
real analysis
real analysis unsolved

$ X$ and $ U$ are banach spaces,$ M: X \to U$ a bounded linear map whose range $ R_M$ is a subset of $ U$ of second category.Then the range of $ M$ is all of $ U$.

No its not true.
Let   $ X: \equal{} C[0,1] \cap \{f: R \rightarrow R: f(0) \equal{} 0 \}$ and $ U: \equal{} C[0,1]$ and let $ M$ is an inclusion from $ X$ to $ U$.
$ M$ is of course bounded and $ range M \equal{} X$ thus $ range M$ is Banach and infinity dimensional hence is a second category set.

Thank you.But in your example,$ X$ is closed in $ U$.So you are easily check that $ X$ is nowhere dense set of $ U$.Thus $ X$ is a first category set of $ U$. :wink:

i'm sorry, i made mistake.
It's always true yet.
range M is closed in U hence  it has nonempty interiror (because $ range M$  mustn't be nowhere dense). Thus $ range M$  contains a ball $ B$.
Assume that $ B$ has the origin $ a$ and radius $ r$.
range M is a vector space hence $ \minus{} a \in range M$. Hence,  translation by vector $ \minus{} a$ moves origin of $ B$ to  point $ 0$.
Thus $ \ A : \equal{} \{x \in U: \parallel x \parallel \leq r \} \subset range M$
Hence $ range M \equal{} U$ since $ \frac {x}{\parallel x \parallel} r \in A$ for every $ x \in U$.

because it's Banach space (as image of Banach by bounded linear map) thus complete

[quote]because it's Banach space (as image of Banach by bounded linear map)[/quote]

Let $ A: \ell_1\to \ell_1$, if $ x\equal{}(x_1,x_2,...)$ then $ Ax\equal{}(\frac{x_1}{1},\frac{x_2}{2},...)$.
Then $ \|x\|\equal{}\sum |x_i| \ge \sum|\frac{x_i}{i} | \equal{} \|Ax\|.$ thus $ \|A\|\le 1$, so $ A$ is bounded.
Now let $ x_n\equal{}(1,\frac12,\frac13,...,\frac{1}{n},0,0,...)$. $ Ax_n\equal{}(1,\frac{1}{2^2},\frac{1}{3^2},...,\frac{1}{n^2},0,0,...)$.
and we have $ Ax_n \to y\equal{}(1,\frac{1}{2^2},\frac{1}{3^2},...)$, but $ y\notin R(A).$
So $ R(A)$ is not closed..

Copy the proof of the open mapping theorem


Prove that

$ 8(a^2\plus{}b^2\plus{}c^2)^3\geq27(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)$

This inequality is true for all $ a,b,c>0$ .

It is well-known that $ a^2\plus{}b^2\plus{}c^2\ge ab\plus{}bc\plus{}ca$, so $ 2(a^2\plus{}b^2\plus{}c^2)\ge\sum(a^2\plus{}bc)$ .

$ \Longrightarrow\ 8(a^2\plus{}b^2\plus{}c^2)^3\ge \left(\sum(a^2\plus{}bc)\right)^3\stackrel{AM\minus{}GM}{\ge}$ $ \left(3\sqrt[3]{(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)}\right)^3\equal{}27\prod(a^2\plus{}bc)$ .


algebra
polynomial
trigonometry
inequalities
search
algebra unsolved

[hide="hint"]square both leg and put $|x-1|=y$.[/hide]

If you substitute y for sqrt(2x - x^2), and note that
(x - 1)^4 = (1 - (2x - x^2))^2
You'll get a simpler equation.  Then, in that, substitute z = 1 - y^2, a further simplification.

But you still wind up with an 8th degree polynomial as far as I can see.  I'm still recovering from a major operation and my not have that quite right, but see what you can do from there.

[hide]

take $y = \sqrt{2x-x^{2}}$
first note that $-1+2x+x^{2}\le 0$, hence $1 \ge y^{2}$ and $1 \ge y$.

then let $y = \cos{\theta}$.

$\sqrt{2}\left(\sin{\frac{\theta}{2}}+\cos{\frac{\theta}{2}}\right) = 2sin^{4}{\theta}\left(1-2\cos^{2}{\theta}\right)$

The LHS is equal to $\sin{\frac{\theta}{2}+\frac{\pi}{4}}$.

The RHS = $-\sin^{4}{\theta}\cos{2\theta}$

Now analyze by intervals to get reasonable bounds and work from here - maybe use double-angle formulas to get all in terms of $\frac{\theta}{2}$. the current form is kind of ugly, but less so than the original.

yay for trig substitution.[/hide]

[b]lemma[/b]: For all $t\in[-1;1]$ then $\sqrt{1+t}+\sqrt{1-t}\ge 2-t^{2}$
[b]Solve problem[/b]
The condition $0\le x\2$
Put $t=\sqrt{2x-x^{2}}$ then $0\le t\le 1$
The equation becomes:
$\sqrt{1+t}+\sqrt{1-t}=2(1-t^{2})^{2}(1-2t^{2}$
Applying above lemma for left hand of equation we get:
$2(1-t^{2})^{2}(1-2t^{2}\ge 2-t^{2}\Leftrightarrow t^{2}(-4t^{4}+10t^{2}-9)\ge 0$
Alternatively $-4t^{4}+10t^{2}+9 <0$ for all $t\in[0;1]$
Therefore from above inequality we conclude $t=0\Rightarrow x=0$ or $x=2$

Done :o

Oh dear, while i search some equations in this forum, i have detected it's has been posted before.
It's here: http://www.mathlinks.ro/Forum/viewtopic.php?t=130671
but there is still remain problem in that post not solved up to now


inequalities quadratics function calculus derivative

I posted this in the calculus forums, because the original statement of the problem involved derivatives, but so far no one has answered it yet. This forum has a different (younger) audience, but I think someone might be able to solve it here :) (I have the solution, no college knowledge is required!)

Anyway, here is the problem:
Let $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$ for some $ a,b,c \in \mathbb{R}$ such that $ |f(x)| \leq 1$ for $ |x| \leq 1.$
Prove that $ |2ax \plus{} b| \leq 4$ for $ |x| \leq 1.$

[hide="Note:"]
For those who know a thing or two about derivatives, you may recognize $ 2ax \plus{} b$ as $ f'(x)$. Indeed, that is the notation I used. I assure you, though, that there is no need to know that, or any calculus for the matter, in order to solve the problem.[/hide]

WLOG let $ a,b \ge 0$. (Consider one of $ f(x),f(\minus{}x),\minus{}f(x),\minus{}f(\minus{}x)$.)
Note that $ 2ax\plus{}b$ is monotonic in $ x$, so we need only check $ x\equal{}\minus{}1,1$. (And we do not have to check $ x\equal{}\minus{}1$ since we assumed $ a,b\ge 0$.)
We have $ |2a\plus{}b|\equal{}2a\plus{}b\equal{}\frac{3}{2}f(1)\plus{}\frac{1}{2}f(\minus{}1)\minus{}2f(0)\le 4$ by the triangle inequality.


Give
$ u_{1}=1 , u_{2}=0$

Find the general term of this sequence

[hide]Hint : Setting $ a_{n}=u_{n+1}-u_{n}$ and use telescopic sum!  :wink: [/hide]


A point on the extension of a trapezoid's diagonal is joined to the midpoints of the bases. Prove that the line through the intersection points of these lines with the other two sides is parellel to the bases.

[img]http://www.mathcenter.net/webboard/upload_files/2-000145-000000.jpg[/img]
From this picture. Prove that $HI//AB//CD$

By Menalaus, we get:

AH/HD=GA/GD=HI/HC

, from which we get HI is paralllel to the lines we want.

Bomb

To find Menalaus, serach it in Geomtry throrem section.  :)
Its basically an iff condition for collinearity of 3 points.

[hide="Here is a easy extension."]
[color=darkblue][b]Generalization.[/b] Let $ABCD$ be a trapezoid ($AB\parallel CD$, $E\in AD\cap BC$) and let $L\in AC$, $M\in AB$, $N\in CD$ be three mobile points. Define the intersections $X\in LM\cap BC$, $Y\in LN\cap AD$. Prove that $\boxed {\ XY\parallel AB\Longleftrightarrow E\in MN\ }$.

[b]Proof.[/b] I shall use the orientated segments. Apply the Menelaus' theorem to the transversals $\overline {LMX}$, $\overline {LNY}$ for the triangles $ABC$, $ADC$ respectively: $\frac{LA}{LC}\cdot \frac{XC}{XB}\cdot \frac{MB}{MA}=1$; $\frac{LA}{LC}\cdot \frac{NC}{ND}\cdot \frac{YD}{YA}=1\ .$ Therefore, $XY\parallel AB$ $\Longleftrightarrow$ $\frac{XC}{XB}=\frac{YD}{YA}$ $\Longleftrightarrow$ $\frac{MB}{MA}=\frac{NC}{ND}$ $\Longleftrightarrow$ $E\in MN\ .$[/color][/hide]


function
integration
limit
real analysis
real analysis unsolved

Got this question, but dont know how to prove it right. Look forward to seeing if anyone knows how to show it. Thx.

Let $0<a<b$

1. Prove that $\int_{0}^{\infty} \frac{f(bt)-f(at)}{t} dt$=$\lim_{k\to\infty} \int_{Ka}^{Kb} \frac{f(t)}{t} dt - \lim_{\epsilon\to 0^+} \int_{\epsilon a}^{\epsilon b} \frac{f(t)}{t} dt$

2. Show that $\lim_{k\to\infty} \int_{Ka}^{Kb} \frac{f(t)}{t} dt$ =$M_\infty(f) log(b/a)$

3. Show that $\lim_{\epsilon\to 0^+} \int_{\epsilon a}^{\epsilon b} \frac{f(t)}{t} dt$=$M_0(f)log(b/a)$

What hypotheses does the function $f$ satisfy and what do you mean by $M_0(f)$ and $M_{\infty}(f)$ ?

Sorry my mistake that I forgot to state the hypothesis,
Here is:

$F(x)= \frac{1}{x} \int_{0}^{x} f(t) dt$
$G(x)= x \int_{x}^{\infty} \frac{f(t)}{t^2} dt$

$M_0(f) =\lim_{x\to0^+}F(x)$
$M_\infty(f)= \lim_{x\to\infty} G(x)$

Does anyone have a chance to take a look, really need some help with that. Thx alot.


calculus
integration
trigonometry
calculus computations

Calculate the following indefinite integrals.

(1) $ \int \frac {3x \plus{} 4}{x^2 \plus{} 3x \plus{} 2}dx$

(2) $ \int \sin 2x\cos 2x\cos 4x\ dx$

(3) $ \int xe^{x}dx$

(4) $ \int 5^{x}dx$

[hide="1)"]
$ \int\frac{3x\plus{}4}{x^{2}\plus{}3x\plus{}2}dx \\ \equal{}\int\frac{3x\plus{}3\plus{}1}{(x\plus{}1)(x\plus{}2)}dx \\ \equal{}\int\frac{3(x\plus{}1)}{(x\plus{}1)(x\plus{}2)}dx\plus{}\int\frac{4}{(x\plus{}1)(x\plus{}2)}dx \\ \equal{}\int\frac{3}{x\plus{}2}dx\plus{}\int\frac{4}{x\plus{}1}dx\minus{}\int\frac{4}{x\plus{}2}dx \\ \equal{}3ln|x\plus{}2|\plus{}C_1\plus{}4ln|x\plus{}1|\plus{}C_2\minus{}4ln|x\plus{}2|\plus{}C_3 \\ \equal{} 4ln|x\plus{}1|\minus{}ln|x\plus{}2|\plus{}K$[/hide]

[hide="2)"]$ \int\sin 2x\cos 2x\cos 4x\ dx \\ \equal{}\frac{1}{2}\int 2\sin 2x \cos 2x \cos 4x dx  \\ \equal{} \frac{1}{4}\int 2 \sin 4x \cos 4x dx \\ \equal{} \frac{1}{4} \int \sin 8x dx \\ \text{Let 8x}\equal{}t \\ \Longrightarrow dx\equal{}\frac{dt}{8} \\ \Longrightarrow \frac{1}{32} \int sin(t) dt \\ \equal{} \minus{}\frac{1}{32}\cos 8x \plus{}C$[/hide]

[hide="3)"]$ \int xe^{x}dx \\ \text{Using integration by parts,we get} \\\equal{} xe^x\minus{}\int e^x dx \plus{}C \\ \equal{} e^x(x\minus{}1)\plus{}C$[/hide]

Sorry, the answer to (1) was incorrect. The third line can see the typo.

[hide="(1)"]
 $ \int \frac{3x\plus{}4}{x^2\plus{}3x\plus{}2}dx
\equal{}\int (\frac{1}{x\plus{}1}\plus{}\frac{2}{x\plus{}2})dx
\equal{}\ln|x\plus{}1|\plus{}2\ln|x\plus{}2|\plus{}C$[/hide]

[hide="(4)"]
$ \int 5^{x}dx \equal{}5^x\frac{1}{\ln5}\plus{}C$[/hide]

Now the all problem has been solved. :lol:


pigeonhole principle
combinatorics proposed
combinatorics

Yesterday we had an exam at school(4 problem in 100 minutes ) and the combinatoric`s problem was this(indeed its easy,but its a little fuuny problem):

[color=blue]A Social math Problem :D [/color]we have $60$ boys in a room.between each $10$ of them ,$3$ of them have a same girl friend.
Prove that $15$ of them have a same girl friend(What a disaster  :) )

Let m be the total number of girlfriends. If $m \geq 9$ then choose a boyfirend of each of them and one another boy chose at random. Among the ten boys there are 3 with the same girlfriend. But from the way we choosed them we cannot have three with the same girlfriend. If $m \leq 4$ from the pigeonhole theorem we have a girl who has at least  15 boyfriends. So,$5 \leq m \leq 8$.

1)   If there are only four girls with more than one boyfriend, those three girls have a total number of $60-(m-4)$ boyfriends, because the rest of the girls have only one boyfriend each.
If $m \leq 7$  from pigeonhole theorem there is a girl with at least $[\frac{57}{4}+1]=15$ boyfriends.
If $m=8$ , assume that there isn't any girl with 15 boyfriends.So, every girl have at most 14 boyfriends.
But since the total number of their boyfriends is 56, each of the four girls have 14 boyfriends. And the rest of girls have 1 boyfriend each. Now choose 2 boyfriends of each of the first 4 girls(those who have more than one boyfriend) and the boyfriends of two of the last 4 girls(those who have only one boyfriend). So, there are 10 boys among who there aren't three with the same girlfriend.
2)  If there are at least 5 girls with more than one boyfriend, choose two boyfriends of each of th 5 girls and among those 10 boy there aren't three with the same girlfriend.

We start with 10 boys $b_1$ to $b_10$ .

Lets say $b_1,b_2,b_3$ have the same girlfriend (we call them group A) . $b_4,b_5,b_6$ another group B . $b_7,b_8$ is group C . $b_9,b_{10}$ is group D .

Now lets take away $b_1$ and substitute with $b_{11}$ . We wish to minimise the number of person in each group so we would put $b_{11}$ in group C . Then take away $b_4$ from B and substitute with $b_{12}$ into group D . So now , each group have 3 boys (girl in each group know 3 boys) . Then for the rest 48 boys , we just divide them equally into the four groups using the same way ( take out one boy and substitute another) until you have each group have 15 boys . :)


If z is a complex number such that |z| = 1, compute |1 + z|2 + |1 - z|2.

It may be also useful to think geometrically :

M(z) is on circle (center=0,radius=1).

|1-z| is the distance to point (1,0)
|1+z| is the distance to point (-1,0)

By Pythagoras, |1-z| + |1+z| is the square of twice the radius = 4.


I think I'm making this problem too complicated, but I keep getting bogged down in tons of algebra, so if anyone has a nice solution let me know.

Write the function $f(z) = z+\frac{1}{z}$ in the form of $f(z) = u(r,\theta)+iv(r,\theta)$.

Thanks.

EDIT:  Nevermind, I figured it out (through a lot of tedious algebra), and I can't seem to delete this post.

$f(z) = \left(r+\frac{1}{r}\right)\cos \theta+i\left(r-\frac{1}{r}\right) \sin\theta$.

Just find each piece, then add them up. You're only half polar here.

For the $\frac1z$, I'd use the fact that $\frac1z=\frac{\overline{z}}{|z|^{2}}$.


algebra
polynomial
geometry
3D geometry
quadratics
Rational Root Theorem

Find the real roots of these polynomials:

(a) $x^4-15(x^2)+25$

(b) $((x-2)(x-3)-2)((x-2)(x-3)-3)-x$

(c) $x^3-3x^2-3x-1$

[hide="hint on (c)"]Try writing $x^3-3x^2-3x-1=0$ as $\text{(something)}^3=\text{(something else)}^3$.
There will be cube roots involved but as long as there's no $\sqrt[3]{x}$ or $x^\frac{2}{3}$ term it will work.[/hide]

[hide="a"]let $a=x^2$, then $a^2-15a+25 = (a-\frac{15+5\sqrt{5}}{2})(a-\frac{15-5\sqrt{5}}{2})$
Therefore $x=\pm \sqrt{a}$, I won't bother simplifying such a thing.[/hide]
and the others are my homework for the weekend...[/hide]

Both a and b factor into irreducible quadratics.  c does not factor.  You probably want to give its linear term a positive coefficent.

c is

$(x-1)^3-6x$ if that helps

or if the 3rd term is positive, then it is just $(x-1)^3$

no, the coefficients on (c) are all negative (except for the leading one), which means $x^3-3x^2-3x-1$ [b]is[/b] irreducible.
It can be solved without messing with the dreadful general cubic method though...

Use rational root theorem/synthetic substitution?

For 3, you can use Eisenstein.  Substitute $x=y+1$ to get that it's equivalent to $y^3-6y-6$, which satisfies the conditions of Eisenstein for $p=3$ and has trailing coefficient not divisible by $3^2=9$.

Of course, since any cubic factors as a linear factor times a quadratic factor, it's enough to use the rational root theorem.  But Eisenstein has plenty of applications to which the rational root theorem is blind.

I have never heard of the Eisenstein method. How does it work?


[quote="Nam Luu"]Find the minimum value of $ A \equal{} x^{2} \plus{} 2y^{2} \plus{} xy \plus{} 5x \plus{} 9y \plus{} 18$ when $ x, y \in Z$.
The hardest part of this problem is that we must find the maximum value when $ x, y \in Z$.[/quote]

$ A\equal{}(x\plus{}\frac 12y\plus{}\frac 52)^2\plus{}\frac 74(y\plus{}\frac{13}7)^2\plus{}\frac{40}7$ and so $ A\ge \frac{40}7$ and, since $ A\in\mathbb Z$, we get  $ A\ge 6$

And, since $ (x,y)\equal{}(\minus{}1,\minus{}2)$ $ \implies$ $ A\equal{}6$ : the minimum value of $ A$ is $ 6$.


Hi! Can someone compare PhD programs in Rutgers and Wisconsin? Of which of these universities do you consider mathematics department to be more "powerful". I am interested in partial differential equations and nonlinear functional analysis. [Rutgers has H. Brezis and Wisconsin has P. Rabinowitz]

Qualifying exams of Wisconsin seem harder to me than those of Rutgers. From which university it is harder to receive an admission offer? And how many applicants have publications? What are the "key reasons" for commite to decide whom to send an admission offer and whom not?

I'm not familiar with Rutgers program but I know that Wisconsin has a lot of strong PDE people, both "pure" and "applied". On the other hand, if I were a PDE person choosing a graduate school in the USA, I would seriously think of Minnesota.

Thank you for information, Fedja. Do you know percentage (approximately, of course) of applicants to "good" graduate schools who have publications?


show that the number N=(3^n+2)*(2^n+3)+(3^n+3)*(4^n+2) is divisible by 63.


If your proctor didnt know one of the rules which, had the proctor known it, would have made a very large difference of your score on AIME, what happens to you?

hmmm; used a calculator perhaps?

i mean scores on AIME dont' really drop; unless he didn't let you write or something....

I believe that scrambled is referring to the fact that the proctor included the bubbling of personal information in the three hour time limit.

But I could be wrong. I am a lot.

I expect if knowing the rule would've dropped your score your school would be disqualified, and if knowing the rule would've raised your score nothing happens.  I don't think they do appeals  :(

[quote="The Zuton Force"]That you can't talk

From our school, 3 sophmores (including me) and 3 seniors got selected for AIME. There was only one proctor and 2 small rooms, so the proctor monitored us 3 sophmores and the 3 seniors cheated the whole time (and still failed)[/quote]

THAT IS SAD....

But even if our school cheated together, we will still get at most 9...(If only I didn't mess up the square root of 27...)


Kevin Chen, the 2007 MATHCOUNTS National Champion, is scheduled to appear on "Live with Regis & Kelly" tomorrow, Wednesday, May 30.

The show generally airs at 9 a.m. Eastern/Pacific (live in the east), but, as they say, check your local listings.

setting the timer right now

thanks mr. patrick!!!

Hmm, so it airs at 9 am throughout the US.
Well, looks like I woke up an hour too early.

Well, that was underwhelming. I thought they would at least show a few clips of countdown or something.

[quote="Vaele"]Wow...this is making Kevin look stupid to anyone who knows even a slight bit about math...they should put more challenging problems...

And Regis and Kelly fail at life if they can't answer some of those questions.  :P[/quote]

The show looked just right! Remember, this show was designed for average viewers, so in order for most of the viewers to follow, they could not have more difficult questions.

Kevin looked confident and great ! You go, Kevin.

Is there a video anywhere? Just wondering

i heard a friend of mine say
that kevin said that there would be a video
on regis n kelly's webpage, in case anyone missed it!
unfortunately, i have not verified as such, so if it isn't
there, i'm extremely sry for sending on wild-goose chase

P.S. what happened on the show? i didnt see it...
 :(
kevin says he didnt really like the show

[quote="zserf"][quote="3.14159man"]
kevin says he didnt really like the show[/quote]

He didn't look like he was enjoying it. You could tell he was either thinking "Wow, they're really stupid" and/or "Get me out of here"[/quote]

lol

yeah i was probably thinking the second

although the show was somewhat interesting

and i did get another cool trophy

were regis and kelly nice to you (i didn't watch it) like out of air-time (maybe you didn't even see them then lol)?

nice job kevin!


vector
linear algebra
linear algebra unsolved

let $V$ be a vector space of finite dimension. Let $f$ be a bilinear application
from $V*V$ to $V$. Suppose that $f(x,y)$ is proportional to $f(y,x)$ for any $x$,$y$ in $V$. Prove that
$f$ is either symetric or antisymetric.

this strongly reminds me of the theorem that reflexivity implies symmetric or symplectic

however, what is your definition of symmetric
it seems that your problem already states that it is symmetric or am i wrong????

of course I made a mistake, I will edit my post  :D

:D so f(x,y)=k*f(y,x) for all x and y,with k fixed

if k=0 then it is always zero and thus symmetric

if it isn't, then f(x,y)=0<->f(y,x)=0 making it reflexive and thus implying symmetric or symplectic again

i don't think you need finite dimension

fred

I don't think $k$ is fixed. You only know that it depends on $x,y$.

of course $k$ is not fixed otherwise it's not that interesting  :D

:blush:

pardon me i need to read questions more thoroughly : it's an application to V not the field

sorry! :blush:

Some steps to be proved:
1)Show that $\" f(x,y)=0$ then $f(y,x)=0\"$ implies $f$ is symetric or $f$ is antisymetric.
2)Let $g(x)=f(x,x)$ if $g=0$ then we get again the conclusion
3)Show for a particulier case:$dim(V)=2$.Use $1)$

Now conclude using all of this.


[quote="jimli"]3 min.
how many sets of 2 primes add up to 36?[/quote]
$3+33$
$5+31$ yes
$7+29$ yes
$11+25$
$13+23$ yes
$17+19$ yes

that's all - 4 is the answer.

[quote="236factorial"]How long did that take you? Nicely done.[/quote]

About a minute.

Well, I had to type it  :P Without that, about 30 seconds would be right.

i didn't really time myself, but i'm sure i got under 3 mins. and yes, i got 4.

10 min?  that's really slow.
[hide="da answer"]4[/hide]

I can finish it in 10 minutes writing all the work with my left hand (not dominant), not to hurt you feelings or anything...

while some mental math abilities are OK especially in contests, you guys probably dont want to concentrate on doing things like this... logic is more imporatnt  :D.  you'd never see thsi kind of question on mathcounts!

easy, you find all the primes 1-17, then try out a number on each one.
2-34
3-33
5-31 :idea:
7-29 :idea:
11-25
13-23 :idea:
17-19 :idea:
so there are 4 solutions :idea:


Let $ A \in M_{3x4}(\mathbb{C}),B \in M_{4x2}(\mathbb{C}),C \in M_{2x3}(\mathbb{C}),M \in M_{3x3}(\mathbb{C}),$
Such that :
        $ M \equal{} \begin{pmatrix} 0 & \minus{} 1 & \minus{} 1 \\
\minus{} 1 & 0 & \minus{} 1 \\
1 & 1 & 2 \end{pmatrix}$
         $ (A\otimes{A})(B\otimes{B})(C\otimes{C}) \equal{} M\otimes{M}$
Calculate $ CAB \equal{} ?$
[hide="Answer of me"]$ CAB \equal{} I_2$ or $ CAB \equal{} \minus{} I_2$    [/hide]

$ ABC\otimes{A}BC \equal{} M\otimes{M}$ then $ ABC\equal{}\pm{M}$. For example $ ABC\equal{}M$.
$ spectrum(M)\equal{}\{0,1,1\}$ then $ spectrum(CAB)\equal{}\{1,1\}$ because $ spectrum(CAB)\equal{}spectrum(ABC)$ up to multiplicity of the eigenvalue $ 0$.
Moreover the eigenvalues $ \not\equal{}0$ of $ UV$ and $ VU$ have the same Jordan structure in both cases. $ M$ is diagonalizable then $ CAB$ is diagonalizable and $ CAB\equal{}I_2$.

[quote="loup blanc"]$ ABC\otimes{A}BC \equal{} M\otimes{M}$ then $ ABC \equal{} \pm{M}$. .[/quote]
 Can You  make clear it ?

If $ A\otimes{A} \equal{} B\otimes{B}$ then $ A\equal{}\pm{B}$.
Proof:  for all $ i,j,k,l$:  $ a_{ij}a_{kl}\equal{}b_{ij}b_{kl}$.
case 1: $ A\equal{}0$. Then for all $ i,j$: $ b_{ij}^2\equal{}0$ and $ B\equal{}0$.
Case 2: there exists $ i,j$ s.t. $ a_{ij}\not\equal{}0$. Then $ a_{ij}^2\equal{}b_{ij}^2$ and $ a_{ij}\equal{}\pm{b_{ij}}$. For example $ a_{ij}\equal{}b_{ij}$. Thus for all $ k,l$: $ a_{kl}\equal{}b_{kl}$ and $ A\equal{}B$.   QED.

[quote="loup blanc"]If $ A\otimes{A} \equal{} B\otimes{B}$ then $ A \equal{} \pm{B}$.
Proof:  for all $ i,j,k,l$:  $ a_{ij}a_{kl} \equal{} b_{ij}b_{kl}$.
Case 2: there exists $ i,j$ s.t. $ a_{ij}\not \equal{} 0$. Then $ a_{ij}^2 \equal{} b_{ij}^2$ and $ a_{ij} \equal{} \pm{b_{ij}}$. For example $ a_{ij} \equal{} b_{ij}$. Thus for all $ k,l$: $ a_{kl} \equal{} b_{kl}$ and $ A \equal{} B$.   QED.[/quote]
  Note that here the exist $ a_{ij} \equal{} {b_{ij}}$ and $ a_{kl} \equal{} \minus{} b_{kl}$ then  $ A \neq B$  :?:
Example for $ A \equal{} \begin{bmatrix}1 & 0 \\
0 & 0\end{bmatrix},B \equal{} \begin{bmatrix} \minus{} 1 & 0 \\
0 & 0\end{bmatrix}$
 Then $ A\otimes{A} \equal{} B\otimes{B} \equal{} \begin{bmatrix}1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\end{bmatrix}$ with $ A,B \in M_2{\mathbb{(R)}}$ then $ A \neq B$  :(
 or  $ A \equal{} \begin{bmatrix}i & 0 \\
0 & 0\end{bmatrix},B \equal{} \begin{bmatrix} \minus{} i & 0 \\
0 & 0\end{bmatrix}$
 Then $ A\otimes{A} \equal{} B\otimes{B} \equal{} \begin{bmatrix}\minus{}1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\end{bmatrix}$ then $ A \neq B$

Let $ A \in M_{4x2}{\mathbb{(C)}},B \in M_{2x4}{\mathbb{(C)}}$ such that :
  $ AB \otimes {AB} \equal{} \begin{bmatrix}1 & 0 & \minus{}1 & 0 \\
0 & 1 & 0 & \minus{}1 \\
\minus{}1 & 0 & 1 & 0 \\
0 & \minus{}1 & 0 & 1\end{bmatrix}\otimes {\begin{bmatrix}1 & 0 & \minus{}1 & 0 \\
0 & 1 & 0 & \minus{}1 \\
\minus{}1 & 0 & 1 & 0 \\
0 & \minus{}1 & 0 & 1\end{bmatrix}}$
Calculate $ BA \equal{} ?$
[hide="I hope that you have answer same as me"]$ BA \equal{} 2I_2$ or $ BA \equal{} \minus{} 2I_2$[/hide]

Hi Kybac,
you have not understood my proof; if there exist  $ i,j$ s.t. $ a_{ij}\not\equal{}0$ then:
Case 1: $ a_{ij}\equal{}b_{ij}$ and $ A\equal{}B$.
Else:
case 2: $ a_{ij}\equal{}\minus{}b_{ij}$ and $ A\equal{}\minus{}B$.

Thus my conclusion  (for your exercise) is, by the same way, $ CAB\equal{}\pm{I_2}$.
When I write "for example" I want to say that I work about the first case but obviously there exist 2 solutions.

Hy Quybac,
for your second exercise:
$ AB\equal{}\pm{\begin{bmatrix}1 & 0 &\minus{}1 & 0\\ 0 & 1 & 0 &\minus{}1\\ \minus{}1 & 0 & 1 & 0\\ 0 &\minus{}1 & 0 & 1\end{bmatrix}}\equal{}\pm{M}$. $ spectrum(M)\equal{}\{2,2,0,0\}$ and $ M$ is  diagonalizable (it is symmetric) . Thus (see my first post) $ BA$ is diagonalizable and $ BA\equal{}\pm{2}I_2$.

A ha , now I understand you want talk here (Sorry english of me so sad  :blush: )
 I will make clear a problem (IMC2004) with above exercise are same .
 Let $ M = \begin{bmatrix}1 & 0 & - 1 & 0 \\
0 & 1 & 0 & - 1 \\
- 1 & 0 & 1 & 0 \\
0 & - 1 & 0 & 1\end{bmatrix} = \begin{bmatrix}I_2 & - I_2 \\
- I_2 & I_2\end{bmatrix}$ then easy see that $ M^2 = 2M$ and $ rank(M) = 2$ , use $ rank(U\otimes{V}) = rank(U)rank(V)$
 We have $ (AB)^2\otimes{(AB)^2} = M^2\otimes{M^2} = 4M\otimes{M} = 4AB\otimes{AB}$
or $ 4M\otimes{M} = (A(BA)B)\otimes{A(BA)B} = (A\otimes{A})(BA\otimes{BA})(B\otimes{B})$
frome here easy see $ rank(4M\otimes{M}) \leq rank(BA\otimes{BA})$ or $ rank^{2}(M) = 4 \leq rank^2(BA)$
Hence $ rank(BA) \geq 2$  by $ BA \in GL_2{\mathbb{(C)}}$ then $ rank(BA) = 2$ or $ BA \in GL_2(\mathbb{C})$ and $ BA \otimes{BA} \in GL_4(\mathbb{C})$
 We have $ (BA)^3\otimes{(BA)^3} = B(AB)^2A\otimes{B(AB)^2A}$ = $ (B\otimes{B})((AB)^2\otimes{(AB)^2})(A\otimes{A}) = 4(B\otimes{B})(AB\otimes{AB})(A\otimes{A})$ = $ 4(BA)^2\otimes{(BA)^2}$  or $ (AB\otimes{AB})^3 = 4(AB\otimes{AB})^2$ don't dificult by $ BA \otimes{BA} \in GL_4(\mathbb{C})$ to have $ BA\otimes{BA} = 4I_2\otimes{I_2}$ (1)
 Put $ BA = \begin{bmatrix}a & b \\
c & d\end{bmatrix}$ then from (1) we have :
 $ \begin{bmatrix}a^2 & ab & ba & b^2 \\
ac & ad & bc & bd \\
ca & cb & da & db \\
c^2 & cd & dc & d^2\end{bmatrix} = 4I_4$
 easy see that $ b = c = 0$ ,$ a^2 = d^2 = 4,ad = 4$ Implies $ a = d = - 2$ or $ a = d = 2$
Or $ BA = 2I_2,BA = - 2I_2$  :rotfl:


In order to find rank(A), can one do elementary transformations on rows and columns at the same time?

Try to do as much operations you can. Then count the number of non-zero columns/rows.

example :

1 5 0 1
0 1 5 0
0 0 0 0     --->> it's can't be a rank
0 0 3 1

that matriks just have 3 rank, because there is one row which just have zero as a part..

example again :

1 4 6 0 8
0 1 5 7 8
0 0 1 4 4
0 0 0 1 1
0 0 0 0 1

that matriks have 5 ranks... :D:D:D

It appears as if the question was actually whether you can do BOTH elementary column operations AND elementary row operations in order to find the rank of a matrix. The answer is yes, since neither change the underlying system of equations. The catch is that it's faster to stick to Gaussian elimination using just one type of operation (ERO or ECO).

hai

Rank of a matrix is just the maximum number of linearly independent rows or cloumns

hence we can operate with the rows and columns because the determinant of the matrix is not going to change unless the entries of the actual matrix is changed.....


tz now i will steal your avatar since you didn't give it to me of your free will


Triangles ABC and DEF are similar but not congruent.
[AB] // [DE], [AC]// [DF], [BC]//EF].
Prove that (AD), (BE), (CF) intersect at one point.

its easy by homotecy...

i will post the solution as soon as possible ;)

;) I NEED IT BEFORE TOMORROW...

PS: If possible find another proof ;)

Regards

[b]A Hint[/b]
If AD and BE meet at O , then  OD/OA =DE/AB  = DF / AC  , so triangles ODF and OAC are similar . Hence OD/OA = DF/AC = OF / OC. But then points O , E , F are on the same line ( that is on OC)  and we are done ( please see also  that OE/OB =DE/AB = EF / BC ect) . This is the main idea . Write down the detales.

[u]babis [/u]


superior algebra
superior algebra unsolved

Let $ M$ be a finitely presented module. Is it true that every exact sequence
\[ 0\to K\to N\to M\to 0\]
splits? Give proofs, please.

$ 0\to\mathbb Z\to\mathbb Z\to\mathbb Z/2\mathbb Z\to0$. owk


A particle with mass $ m$ is in a unidimensional potential $ V(x)$:
$ V(x)\equal{}\plus{}\infty$ if $ x<0$
$ V(x)\equal{}\frac{1}{2}m\omega ^2 x^2$ if $ x\geq0$

What are the eigenvalues of the Hamiltonian?

In $ t\equal{}0$ the potential becomes $ V(x)\equal{}\frac{1}{2}m\omega ^2 x^2$ for $ \minus{}\infty<x<\infty$

What possible energy values can we find for $ t>0$? What are their probabilities?
What's the expected value of x for $ t>0$?

I've got a 20+ year old write-up I did for a differential equations class. It's a Word document - let me see how much I can rescue for posting here.

Hmm - porting it over to $ \text{\TeX}$ may take a while. Maybe this weekend? Maybe for now I'll just upload the Word document. There were some pictures to go with this - graphs of the solutions for various values of $ \lambda,$ posted on top of the pseudo-classical probability distribution on the first page of this.

Thanks, but unfortunately i can't see the formulae...

Is this a homework you need help to do, or you're studying the matter by your self?
I'm asking because I don't know how much you know about and how could we help you.
Btw, dead you solve the other question: "Potential Well" ?


The function f is defined on the set of positive integers and its values are positive integers. Given that f(n+1) > f(f(n)) for all n, prove that f(n) = n for all n.

Let $d$ be the least element of the range of $f$ .
  We can conclude that $d=f(1)$
  Consider the set $f(n): n\geq 2$ . We have $f(1) < f(2)$
  So $f(1) <f(2)<...<f(n)<...$
  Because $f(1) \geq 1$ , $f(k) \geq k$ for all k
 Suppose $f(k) \geq k+1$ . So $f(f(k)) \geq f(k+1)$ : contradiction !
  This infers $f(k)=k$


geometry
geometric transformation
reflection
analytic geometry

The equation of a line on an xy-coordinate graph is $y=2x+5$
(a) Find the equation of the line perpendicular going through the x intersection point.
(b) Find the equation of the line that reflects over the x axis on the x intersect point.
(c) Find the equation of the line that reflects over the y axis on the y intersect point.

Aren't there infinite amounts of lines for a?

Yeah, sorry, going through the x intersection point. I fixed it.  :blush:


My goal this coming year is to qualify for USAMO. About how high of an index score should I aim for?

220 to be safe I would say. If you take AMC 10, then probably an 8 on aime.

I would go for 224+.

Also, I believe if you are a 10th grader or lower, you should go for 8+ on AIME.

[quote="calc rulz"]It often depends greatly upon the difficulty of the contest. I have in fact noticed a trend of easier AIME's, so the index would probably be similar to those in recent years, so 220 would make sense. As well, though, the new AMC scoring system might lower people's scores, making the index a bit lower.[/quote]

I heard that to make up for the lower scores, the questions will be a little easier.

If that's true though, will that have any effect on the difficulty of the AIME and the USAMO?

And does anyone know if the number of USAMO participants will be the same as last year?


as a produce manager ,you have purchased 50 pounds of grapes at a whole sale price of 75 dollars. if you want to make a profit of 40 cents  per pound , at what retail price should you sell the grapes ?

You bought a pound of grapes for $ \$ \frac{75}{50}$ or $ \$1.50$.  If you want to make a $ \$0.40$ profit, you must sell the grapes for $ \$1.50 \plus{} \$.40 \implies \boxed{\$1.90}$.


Given $ a,b,c \ge 0$
Prove that
$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 4\sqrt[3]{abc(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)}$
Thank you,arqady,you jion almost my problem

We have: $ (a\plus{}b)(b\plus{}c)\equal{}b^2\plus{}ca\plus{}b(c\plus{}a) \geq\ 2. \sqrt{b(c\plus{}a)(b^2\plus{}ca)}$
Indeed, it's my old problem, babyloverain  :P

Nice inequality! :lol:
After expanding we obtain
$ \sum_{sym}(a^6b^3\plus{}3a^5b^4\plus{}3a^6b^2c\plus{}12a^5b^3c\minus{}23a^4b^4c\minus{}23a^5b^2c^2\plus{}39a^4b^3c^2\minus{}12a^3b^3c^3)\geq0,$
which is not so obvious.

[quote="nguoivn"]We have: $ (a \plus{} b)(b \plus{} c) \equal{} b^2 \plus{} ca \plus{} b(c \plus{} a) \geq\ 2. \sqrt {b(c \plus{} a)(b^2 \plus{} ca)}$
Indeed, it's my old problem, babyloverain  :P[/quote]
Sr to nguoivn,almost my problem is of you :((

[quote="babyloverain"]Sr to nguoivn,almost my problem is of you :(([/quote]
If you want to creat a new ineq, you should start with a new hint, and try to transform the hint to many another form.
Example, when I had a simple hint: $ (a \plus{} b)(b \plus{} c) \equal{} b^2 \plus{} ca \plus{} b(c \plus{} a) \geq\ 2. \sqrt {b(c \plus{} a)(b^2 \plus{} ca)}$
I thought immediataly about the another equality like: $ (a^2 \plus{} bc)(b^2 \plus{} ca) \equal{} ...; (a^2 \plus{} bc)(b \plus{} c) \equal{} ...$
And I had 2 ineq like this:
$ (a \plus{} \frac {bc}{a})(b \plus{} \frac {ca}{b})(c \plus{} \frac {ab}{c}) \geq\ \sqrt [3]{(a^3 \plus{} b^3)(b^3 \plus{} c^3)(c^3 \plus{} a^3)}$

$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc} \geq\ \frac {(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}$

Besides, I also improve it to many another ways and I obtained many interesting.
This is a small experience I want to share with you. Hope that you'll like it  :)

thank you,and I try to creat new ineq.this problem is delicate to you and arqady.
Give $ a,b,c > 0$ and abc=1.Find the minimum value:
$ A\equal{} \frac{a^5}{b}\plus{} \frac{b^6}{c}\plus{} \frac{c^5}{a}$
I think this problem is not easy.

[quote="babyloverain"]
Give $ a,b,c > 0$ and abc=1.Find the minimum value:
$ A \equal{} \frac {a^5}{b} \plus{} \frac {b^6}{c} \plus{} \frac {c^5}{a}$
I think this problem is not easy.[/quote]
The minimum is equal to $ \frac{104}{\sqrt[104]{37^{37}31^{31}36^{36}}}.$

Substiute
$ a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$
$ %Error. "hence" is a bad command.
A=\sum\frac{a^5}{b}=\sum\frac{x^5z}{y^6}\ge 3\cdot\sqrt[3]{\frac{x^6y^6z^6}{x^6y^6z^6}}=3$
thus minimum value of A=3
 :huh:  :huh:

We have also a ineq( but easy): with same conditional.Find Min:
  $ A\equal{}\frac {25a^5}{b} \plus{} \frac {16b^6}{c} \plus{} \frac {25c^5}{a}$


I must be tired...

well, i have to points A(xA,yA,zA) and B(xA,yB,zB)
i want to find the coordinates of the point O so that AO=AB/3
ps: vector AO is colinear to vector AB..... O is on AB...

would you help me ?

Thank you

Its a weighted average.  If you want the point towards A, you give A a bigger weight.  Here you are.
$(O_x,O_y,O_z)=(\frac{2A_x+B_x}{3},\frac{2A_y+B_y}{3},\frac{2A_z+B_z}{3})$


I think I can safely post this now that I have turned it in...

$M\subseteq N\subseteq V$ (all vector spaces), show that $(V/M)/(N/M)$ is isomorphic to $V/N$.

Here's the way I'm interpreting the problem:

$M\subseteq N$ means $M$ is a subspace of $N$.
$V / M = \{ M + v | v \in V \}$ where $M$ is a subspace of $V$.

Let $f:V/M\to V/N$ defined by $f(v+M)=v+N$.
$v+M=v'+M\Leftrightarrow v-v'\in M\Rightarrow v-v'\in N\Rightarrow v+N=v'+N$, so $f$ is well defined.
$f$ is obviously surjective.
$\ker f=\{(v+M)\in V/M| v\in N\}=N/M$

Now just apply the fundamental theorem of isomorphism.


calculus
integration
geometry
3D geometry
sphere
function
derivative

How do I find the surface area on a unit sphere inside the region $ y^2\minus{} 4xz <0$?

A sphere can ge expressed as a function $ f(x,y)\equal{}z$ so we use the following formula:

$ \iint_{D}\sqrt{1\plus{}f_{x}^{2}\plus{}f_{y}^{2}}\ dx\ dy$

where $ f_{x}$ and $ f_{y}$ represent the partial derivatives of $ f$ with respect to the subscripted variable.

[quote="JRav"]A sphere can ge expressed as a function $ f(x,y) \equal{} z$ so we use the following formula:

$ \iint_{D}\sqrt {1 \plus{} f_{x}^{2} \plus{} f_{y}^{2}}\ dx\ dy$

where $ f_{x}$ and $ f_{y}$ represent the partial derivatives of $ f$ with respect to the subscripted variable.[/quote]

how do you apply the restriction?
what is $ D$?

Start by sketching the surface $ z \equal{} \frac{y^2}{4x}$. D is the common projection of both surfaces on the xy plane.


algebra polynomial calculus derivative AMC

In the Wiki, the solution says "there are no more zeros of this polynomial, as there already have been three turns in the curve." Why?

http://www.artofproblemsolving.com/Wiki/index.php/2000_AMC_12_Problems/Problem_22

Each "turn in the curve" is a local maximum or minimum of P.  As such it is a zero of the derivative of P.  The derivative of P is a cubic, so it cannot have more than 3 roots.


8 chocolate chips are randomly distributed among 4 cookies. What is the probability that every cookie has at least one chocolate chip in it?

BALLS
AND WALLS

SEVEN CHOOSE FOUR OVER EIGHT TO THE FOURTH
THIRTY-FIVE OVER FOUR-OH-NINE-SIX

DARN WHY NOT FOUR-OH-SIX-NINE

MIRITE?

[quote="DiscreetFourierTransform"]Can you explain this?
I'm interested to see an elegant solution.
Does it use balls and URNS, perhaps?
I'm only passingly familiar with that one.[/quote]
[hide]
yes.  ive also heard it called coins and purses but whatever

There are 8 chips (balls) and 4 cookies (urns).  If we have 3 dividers, that splits the region int 4 spaces which makes th urns.  for example, OO|O|OOOO|O or ||O|OOOOOOO are possibilities.  The number or ways to arrange the balls and dividers is ${11\choose 3}=165$.

To find the nubmer of ways each cookie has a chip in it, assume that there already is.  So we need to place 4 more chips in 4 cookies.  Using the same method, we get  ${{7\choose 3}=35}$

$\frac{35}{165}=\frac{7}{33}$[/hide]

i have no idea what treething did tho

oh right

brain slipped, yeah, demoninator shouldn't be 4096, it should be 165.

and no no no no no no no
it's not balls and urns
it's balls and walls
it rhymes
poetry

balls and walls
balls and urns
sticks and dots
coins and purses
ls and os
Is an xs
Call it what you want!


How many square units are in the shaded area of the rectangle shown?
[asy]import olympiad; import geometry_dev; import graph; size(150); defaultpen(linewidth(0.9));
draw(origin--(11,0)--(11,6)--(0,6)--cycle);
fill((3,0)--(11,0)--(11,6)--(8,6)--cycle);
label("$6$",origin--(0,6),W);label("$3$",origin--(3,0),N); label("$8$",(0,6)--(8,6),N); label("$3$",(8,6)--(11,6),N);[/asy]

Notice that the unshaded and shaded regions are congruent, and thus equal in area, so our answer is $ \frac12\cdot6\cdot11\equal{}\boxed{33}$.


function
topology
Support
integration
real analysis
real analysis unsolved

[quote="Amazigh"]If a and b are real numbers and a<b. Does there exist an increasing function g:R->R which is infinitely times differentiable and g(x)=0 for x=<a, g(x)=1 for x>=b.

I started to use lagrange, we then find  c between a and b so that g"(c)=0. But i have no ideas anymore :(

Can you help me ?

thanks[/quote]

Dear Amazigh,

what about the following function:

$ f(x) = e^{ - e^\cot(x)}$ for $ 0 < x < \pi$

Also $ f(x) = 0$ for $ x < 0$ and $ f(x) = 1$ for $ x > \pi$.

[img]http://i13.photobucket.com/albums/a287/Gaz_v_pol/sample.jpg[/img]

If you like this problem, here is another one. Does there exist an increasing function $ g: R \to R$ which is infinitely times differentiable and $ g(x) = - 1$ for $ x \leq - 1$, $ g(x) = \sin(x)$ for $ 0 < x < 1$ and $ g(x) = e^x$ for $ x\geq2$?

Sergey Markelov
markelov@mccme.ru

Using a partition of unity seems a bit circular.  After all, the OP is asking how to construct a bump function, which are themselves used to construct partitions of unity to begin with.

As far as the original question, see: http://en.wikipedia.org/wiki/Bump_function

[quote="Xixas"]Consider $ \mathbb R$ as a smooth manifold. Consider an open cover of this manifold consisting just of two open sets $ A \equal{} \{ x | x > a\}$ and $ B \equal{} \{x | x < b\}$. Let $ \{\phi_a, \phi_b\}$ be a $ C^\infty$ partition of unity subordinate to this open cover. Then since $ \phi_a \plus{} \phi_b \equiv 1$ and $ \text{supp } \phi_a \subset A$, $ \text{supp } \phi_b \subset B$ we can take $ g \equal{} \phi_a$.[/quote]

That language is too advanced for me. Thanks anyway

Markelov, your example is hard to come up with, I think there is a much simpler one.

how explicitly do you want it?

the function which agrees with $ \exp\left(\frac { \minus{} 1}{1 \minus{} x^2}\right)$ on $ [\minus{}1,1]$ and is $ 0$ everywhere else is $ C^\infty$ and has support equal to $ [\minus{}1,1]$.

next we translate and scale it so that it has support in $ [a,b]$ instead of $ [\minus{}1,1]$.  call the resulting function $ \varphi$.

next we multiply it by a constant $ C$ so that
\[ \int_\mathbb{R} C\varphi(t)dt \equal{} 1
\]
now we define a function $ \psi$ by
\[ \psi(x) : \equal{} \int^x_{ \minus{} \infty} C\varphi(t)dt
\]
$ \psi$ has all the properties you're looking for.

Thanks, if you take $ a\equal{}\minus{}1,\ b\equal{}1$, it is already good. Thanks all!


function
real analysis
real analysis unsolved

a).Let $ g(t)$ be a continuous function in $ L^{2}(\mathbb{R})$. Assume that $ g(t) \neq 0$ for almost every $ t \in \mathbb{R}$. Show that the linear span of the functions $ \{ g(t)\exp(itx) x \in \mathbb{R}\}$ is dense in $ L^{2}(\mathbb{R})$.
b). Suppose $ f \in L^{2}(\mathbb{R})$ and let $ Ff$ denote its $ L^{2}$ Fourier transform. Assume that $ Ff \neq 0$ for almost every $ t \in \mathbb{R}$. Show that the linear span of the set of functions $ \{ f_{y}(x)=f(x+y : y \in \mathbb{R}) \}$ is dense on $ L^{2}(\mathbb{R})$.

a) Suppose not. Then there is a nonzero function $ h\in L^{2}$ that is orthogonal to all elements of that linear span. The product $ \bar gh$ is a nonzero $ L^{1}$ function whose Fourier transform is identically zero, a contradiction.

b) is just a reformulation of a).


function
trigonometry
modular arithmetic

i was wondering if there was a solution to a problem like $-1 = \sqrt{x}$

i see that when you square, you cancel out the negative but there should be a way to solve this (even if there is another variable that would need to be made like $i$

The square root of a number can be either negative or positive, for example: $\sqrt{x^2}=x,-x$, In your specific problem, $x=1$, just square both sides!!

I think the answer is "no solution".
LHS ($-1$) is negative, while RHS ($\sqrt{x}$) is non-negative...

[quote="Iversonfan2005"] The square root of a number can be either negative or positive, for example: $\sqrt{x^2}=x,-x$ In your specific problem, $x=1$, just square both sides!![/quote]
Strictly speaking, it's $\sqrt{x^2}=x$ for $x\ge 0$ and $\sqrt{x^2}=-x$ for $x<0$ ;)
And $x=1$ doesn't work...

In order for the square root to be a function, it can only return one value, which by convention is the positive value, and thus the equation $\sqrt{x} = -1$ has no solutions.

[quote="Iversonfan2005"]But why? $\sqrt{1}$ can also be $-1$ :?: Why does it HAVE to be positive?[/quote]

Well every complex number has 2 square roots, but when we use the square root function, we only want the principle value.  Think about the arcsin function, there are infinetly many solution to $\sin x = \frac{1}{2}$, but $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$, which is the principle value.

i understand that there are no solutions but why is there no variable assigned to such a case. for a long time, people thought that there was no solution to $x^2 + 1 = 0$ but they made a new variable to fit the situation. so why could there not be a variable $v$ such that it satisfies $\sqrt{v} = -1$

Keep in mind that the concept of $\sqrt{-1}$ is vastly important as it makes our number system complete.  Thus, it makes sense to assign a letter to such an important number to make calculations using that number easier to understand.

Also, $i$ was different from all real numbers.  It was a new number.  The solution to $\sqrt{x}=-1$ is either $x=1$ or there are no solutions (depending on whether or not you regard the principle value as the only value of a square root).  Therefore, what really would you be assigning to a variable?

[quote="yunorman"]If you're looking into more advanced math, like rings of $\mathbb{Z}$, $\sqrt{-1}$ is important and does exist.
And more importantly you need to figure when it exists,
$\sqrt{-1}\equiv 3\text {     in      }\mathbb{Z}_5$
Why?
[hide]$\Let x=\sqrt{-1}$
$x^2=-1$

So we're searching for a number $x\equiv -1 \bmod 5$

$3^2 \equiv -1\bmod5$[/hide][/quote]

It is also easy to show that -1 is a square mod p whenever $p \equiv 1 \pmod 4$

thanks for the helpful responses, i understand now, i am not necessarily going to get into more advanced math just yet, i just thought of this based on a problem in my algebra 2 textbook

There's an important distinction to make here.  $\sqrt{1}$ does not equal both 1 and -1.  It only equals 1.  Only only only.  Because $\sqrt{x}$ is a function and functions only return one value.  But.  The equation $x^2=1$ has two solutions, 1 and -1.  It's a very subtle distinction that people confuse frequently.