If we have dooble quadratic roots we have the following method.

If  $A^2-B=C^2$  then  $\sqrt{A\pm \sqrt{B}}=\sqrt{\frac{A+C}{2}}\pm\sqrt{\frac{A-C}{2}}$

But if we have  $\sqrt[3] {A\pm\sqrt{B}}$   do you know a method to cope with it imediately? Thanks

There would probably only be a chance of simplyfying if $A^2-B=C^3$.

In this case, if $\alpha=\sqrt[3]{A+\sqrt{B}}$ and $\beta=\sqrt[3]{A-\sqrt{B}}$, then $\alpha \beta=C$. Then a way to find $\alpha$ and $\beta$ would be to evaluate $\alpha +\beta$, because then the sum and product of the roots of a quadratic can be used.

If $x=\alpha +\beta$, then expanding and cancelling the radicals, $x^3=2A+3Cx$ which then gives a cubic polynomial. If the polynomial has rational roots then $\alpha$ and $\beta$ can be simplified. This isn't always the case: for example $\sqrt[3]{2+\sqrt{3}}+\sqrt[3]{2-\sqrt{3}}$ doesn't simplify nicely.


[hide]
2x + y = 8
4x + 2y = a

distribute a 2 across and get them to look the same

2[2x + y] = 2[8]
4x + 2y = 16
4x + 2y = a

so a = 16
[/hide]

Correct. You didn't have to be so detailed

[hide]
Just multiply the first equation by 2. The second equation is therefore $16$[/hide]

[hide="solution"]We multiply both sides by two to get 16[/hide]

[hide="solution"]x=3 y=2, 12+4=[color=red]16[/color]
 :10: [/hide]

[hide]If 2x + y = 8, what is the value of 4x + 2y?

its doubled so 16[/hide]


A five-digit number is called a  mountain number  if  the first three digits increase and the last three digits decrease. For example, 35,763 is a mountain number but 35,663 is not. How many five-digit numbers
greater than 70,000 are mountain numbers?

[hide] Clearly, in order for the number to be a mountain number, the number must be of the form $ 789xy$, where $ x$ and $ y$ are digits. Now, if $ x$ is $ 8$, there are $ 8$ possibilities for $ y$ (the digits from $ 0$ to $ 7$). In the same way, there are $ 7$ possibilities if $ x\equal{}7$, $ 6$ possibilities is $ x\equal{}6$ , etc. So, $ 8\plus{}7\plus{}6\plus{}5\plus{}4\plus{}3\plus{}2\plus{}1\equal{}\frac{8\times{9}}{2}\equal{}\boxed{36}$.[/hide]


View the stickie located in this exact same forum.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146659[/url]


My friend came to my house,
"Your Sons must be growing up," he remarked.
"Yes," I said, "the product of their ages is 72."
"Thats not very informative; tell me more."
"Well, the sum of their ages is my house number."
"That is still not enough."
I replied thoughtfully, "You are right. The oldest one looks
a lot like his mother."
Whereupon, he was able to deduce what the three ages (all whole numbers) were.  List the ages of the three children.

OK the solution!!
[hide]The clue that "my OLDEST daughter..." so, you see that he has only one oldest daughter... now try and find all possible sums... if you list them out, you get two 14's, which is the only repeated one, but one of them is 6,6,2. so the answer is 8,3,3[/hide]
[size=75][color=red][Edit:  Please hide your solutions  -- Moderator RCV.][/color][/size]

mathcrazed, that's right

[quote] kyyuanmathcount - Ah, I remember a very similar one about a census taker in Paul Zeitz's Art and Craft of Problem Solving. (like, chapter 1?) [/quote]

Could be.  I got that book for Xmas but haven't cracked it open yet :lol: I've heard the question before but saw it online and thought I'd put it into a mathcounts test.

[quote="philB"]Why can't the answer be 3,4,6?[/quote]
[hide]
If that was the answer, then the friend would know the answer when told the sum of their ages was the same as the house number since 3+4+6 = 13 and no other possibilities add up to 13.  The friend would not have needed more info.  Remember the friend is at my house and so they know the house number.

2+6+6 = 14 and 3+3+8 = 14.  14 is the only sum that comes up twice, so it is the only sum(house number) that require the friend to ask for more help.  3,3,8 is the only one of those 2 with an oldest kid.[/hide]


inequalities
geometry
circumcircle
trigonometry
trig identities
Law of Sines
inequalities unsolved

http://www.mathlinks.ro/Forum/viewtopic.php?t=6073

  darij

[quote="scpajmb"]Let ABC be a triangle inscribed in a circle of radius R, and let P be a point in the interior of ABC. Prove that
$\frac{{PA}}{{BC^2 }} + \frac{{PB}}{{CA^2 }} + \frac{{PC}}{{AB^2 }} \ge \frac{1}{R}$

CHINA: High-School Mathematics ，1993(2) .[/quote]

generalize:
Let ABC 、$A_1 B_1 C_1$  are two triangle,  riangle $A_1 B_1 C_1$ inscribed in a circle of radius $R_1$, and let P be a point in space. Prove that
$\frac{{PA}}{{BC*B_1 C_1 }} + \frac{{PB}}{{CA*C_1 A_1 }} + \frac{{PC}}{{AB*A_1 B_1 }} \ge \frac{1}{R_1}$

A really beautiful inequality!

[color=blue][b]Theorem 1.[/b] Let ABC and $A_{1}B_{1}C_{1}$ be two triangles, and let $R_{1}$ be the circumradius of triangle $A_{1}B_{1}C_{1}$. Let P be a point in space. Then,

$\frac{PA}{BC\cdot B_{1}C_{1}}+\frac{PB}{CA\cdot C_{1}A_{1}}+\frac{PC}{AB\cdot A_{1}B_{1}}\geq\frac{1}{R_{1}}$.[/color]

[i]Proof of Theorem 1.[/i] We will use two lemmas:

[color=blue][b]Lemma 2.[/b] If ABC is a triangle and P is a point in space, then

$\frac{PB\cdot PC}{CA\cdot AB}+\frac{PC\cdot PA}{AB\cdot BC}+\frac{PA\cdot PB}{BC\cdot CA}\geq 1$.[/color]

[i]Proof of Lemma 2.[/i] Let Q be the orthogonal projection of the point P on the plane ABC. Then, the triangles PQA, PQB, PQC are right-angled at Q, and since the hypotenuse of a right-angled triangle is $\geq$ to any of its catets, we thus have $PA\geq QA$, $PB\geq QB$, $PC\geq QC$. Thus, the inequality

$\frac{PB\cdot PC}{CA\cdot AB}+\frac{PC\cdot PA}{AB\cdot BC}+\frac{PA\cdot PB}{BC\cdot CA}\geq 1$

would immediately follow from the inequality

$\frac{QB\cdot QC}{CA\cdot AB}+\frac{QC\cdot QA}{AB\cdot BC}+\frac{QA\cdot QB}{BC\cdot CA}\geq 1$.

But since Q is a point on the plane ABC, this is a well-known inequality (Lemma 2 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=136745#136745]http://www.mathlinks.ro/Forum/viewtopic.php?t=21016 post #14[/url]). Hence, Lemma 2 is proven.

[color=blue][b]Lemma 3.[/b] If x, y, z are nonnegative reals, and A, B, C are the angles of a triangle ABC, then

$\left(x+y+z\right)^{2}\geq 4\left(yz\sin B\sin C+zx\sin C\sin A+xy\sin A\sin B\right)$.[/color]

(Notice the similarity to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=590661#590661]http://www.mathlinks.ro/Forum/viewtopic.php?t=69200 post #11[/url] Theorem 2.)

[i]Proof of Lemma 3.[/i] We have

$\sin B\sin C=\frac12\left(\cos\left(B-C\right)-\cos\left(B+C\right)\right)$,

but B + C = 180° - A (by the sum of angles in triangle ABC) yields cos (B + C) = - cos A, so this becomes

$\sin B\sin C=\frac12\left(\cos\left(B-C\right)-\left(-\cos A\right)\right)=\frac12\left(\cos A+\cos\left(B-C\right)\right)$.

Now, using the notation $\sum$ for cyclic sums, the inequality of Lemma 3 rewrites as $\left(x+y+z\right)^{2}\geq 4\sum yz\sin B\sin C$. Since $\left(x+y+z\right)^{2}=\left(x^{2}+y^{2}+z^{2}\right)+2\sum yz$ and $\sin B\sin C=\frac12\left(\cos A+\cos\left(B-C\right)\right)$, this inequality is equivalent to

$\left(x^{2}+y^{2}+z^{2}\right)+2\sum yz\geq 4\sum yz\cdot\frac12\left(\cos A+\cos\left(B-C\right)\right)$.

This rewrites as

$\left(x^{2}+y^{2}+z^{2}\right)+2\sum yz\geq \sum 2yz\cos A+2\sum yz\cos\left(B-C\right)$.

And this is obviously the sum of the two inequalities

$x^{2}+y^{2}+z^{2}\geq\sum 2yz\cos A$ and
$2\sum yz\geq 2\sum yz\cos\left(B-C\right)$.

The first of these inequalities follows from Theorem 1 from http://www.mathlinks.ro/Forum/viewtopic.php?t=69200 because A + B + C = 180° (by the sum of angles in triangle ABC); the second is trivial (we have $yz\geq yz\cos\left(B-C\right)$, since $1\geq\cos\left(B-C\right)$, since every cosine is $\leq 1$). Thus, Lemma 3 is proven.

In order to prove Theorem 1, we denote by $A_{1}$, $B_{1}$, $C_{1}$ the angles of triangle $A_{1}B_{1}C_{1}$; then, the extended law of sines yields $B_{1}C_{1}=2R_{1}\sin A_{1}$, $C_{1}A_{1}=2R_{1}\sin B_{1}$, $A_{1}B_{1}=2R_{1}\sin C_{1}$ (since $R_{1}$ is the circumradius of triangle $A_{1}B_{1}C_{1}$), so that the inequality in question,

$\frac{PA}{BC\cdot B_{1}C_{1}}+\frac{PB}{CA\cdot C_{1}A_{1}}+\frac{PC}{AB\cdot A_{1}B_{1}}\geq\frac{1}{R_{1}}$,

becomes

$\frac{PA}{BC\cdot 2R_{1}\sin A_{1}}+\frac{PB}{CA\cdot 2R_{1}\sin B_{1}}+\frac{PC}{AB\cdot 2R_{1}\sin C_{1}}\geq\frac{1}{R_{1}}$,

what is equivalent to

$\frac{PA}{BC\sin A_{1}}+\frac{PB}{CA\sin B_{1}}+\frac{PC}{AB\sin C_{1}}\geq 2$.

We can square this:

$\left(\frac{PA}{BC\sin A_{1}}+\frac{PB}{CA\sin B_{1}}+\frac{PC}{AB\sin C_{1}}\right)^{2}\geq 4$.

But, applying Lemma 3 to $x=\frac{PA}{BC\sin A_{1}}$, $y=\frac{PB}{CA\sin B_{1}}$, $z=\frac{PC}{AB\sin C_{1}}$ and to triangle $A_{1}B_{1}C_{1}$ instead of triangle ABC, we get

$\left(\frac{PA}{BC\sin A_{1}}+\frac{PB}{CA\sin B_{1}}+\frac{PC}{AB\sin C_{1}}\right)^{2}$
$\geq 4\left(\frac{PB}{CA\sin B_{1}}\cdot\frac{PC}{AB\sin C_{1}}\cdot\sin B_{1}\sin C_{1}+\frac{PC}{AB\sin C_{1}}\cdot\frac{PA}{BC\sin A_{1}}\cdot\sin C_{1}\sin A_{1}\right.$
$\left.+\frac{PA}{BC\sin A_{1}}\cdot\frac{PB}{CA\sin B_{1}}\cdot\sin A_{1}\sin B_{1}\right)$
$=4\left(\frac{PB\cdot PC}{CA\cdot AB}+\frac{PC\cdot PA}{AB\cdot BC}+\frac{PA\cdot PB}{BC\cdot CA}\right)$.

By Lemma 2, we have $\frac{PB\cdot PC}{CA\cdot AB}+\frac{PC\cdot PA}{AB\cdot BC}+\frac{PA\cdot PB}{BC\cdot CA}\geq 1$, so this becomes

$\left(\frac{PA}{BC\sin A_{1}}+\frac{PB}{CA\sin B_{1}}+\frac{PC}{AB\sin C_{1}}\right)^{2}\geq 4$,

and Theorem 1 is proven.

Note that a corollary of the above is:

[color=blue][b]Theorem 4.[/b] Let ABC be a triangle, and P a point in space. Then,

$\frac{PA}{BC\cdot\cos\frac{A}{2}}+\frac{PB}{CA\cdot\cos\frac{B}{2}}+\frac{PC}{AB\cdot\cos\frac{C}{2}}\geq 2$.[/color]

[i]Proof of Theorem 4.[/i] During the proof of Theorem 1, it was shown that

$\frac{PA}{BC\sin A_{1}}+\frac{PB}{CA\sin B_{1}}+\frac{PC}{AB\sin C_{1}}\geq 2$,

where $A_{1}$, $B_{1}$, $C_{1}$ are the angles of an arbitrary triangle $A_{1}B_{1}C_{1}$.

Now, the angles $90^{\circ}-\frac{A}{2}$, $90^{\circ}-\frac{B}{2}$, $90^{\circ}-\frac{C}{2}$ are the angles of a triangle (in fact, these angles are positive, since A < 180° (because A is an angle of triangle ABC) yields $\frac{A}{2}<90^{\circ}$ and thus $90^{\circ}-\frac{A}{2}>0$, and similarly $90^{\circ}-\frac{B}{2}>0$ and $90^{\circ}-\frac{C}{2}>0$, and the sum of these angles is

$\left(90^{\circ}-\frac{A}{2}\right)+\left(90^{\circ}-\frac{B}{2}\right)+\left(90^{\circ}-\frac{C}{2}\right)$
$=270^{\circ}-\frac{A+B+C}{2}=270^{\circ}-\frac{180^{\circ}}{2}$        (since A + B + C = 180° by the sum of angles in triangle ABC)
$=180^{\circ}$).

Hence, we can apply the inequality

$\frac{PA}{BC\sin A_{1}}+\frac{PB}{CA\sin B_{1}}+\frac{PC}{AB\sin C_{1}}\geq 2$

for $A_{1}=90^{\circ}-\frac{A}{2}$, $B_{1}=90^{\circ}-\frac{B}{2}$, $C_{1}=90^{\circ}-\frac{C}{2}$, and obtain

$\frac{PA}{BC\sin\left(90^{\circ}-\frac{A}{2}\right)}+\frac{PB}{CA\sin \left(90^{\circ}-\frac{B}{2}\right)}+\frac{PC}{AB\sin\left(90^{\circ}-\frac{C}{2}\right)}\geq 2$,

what rewrites as

$\frac{PA}{BC\cos\frac{A}{2}}+\frac{PB}{CA\cos\frac{B}{2}}+\frac{PC}{AB\cos\frac{C}{2}}\geq 2$,

and Theorem 4 is proven.

  Darij

see
Jiang Ming-bin.     Generalization  of   some Triangle Inequality.

[i]Hunan Annals of Mathematics [/i]Vol.15 No.4 Dec. 1995

(Chinese)


This forum is for the discussion of [b]Alcumus[/b], a new adaptive learning program from Art of Problem Solving.  Alcumus consists of a database of problems with and video lessons backed by an algorithm that chooses problems for students based on their past performance.  Alcumus currently complements our [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202]Introduction to Counting & Probability[/url] textbook.  We plan to add material in support of our other texts in coming years.

Please use this forum for questions about how to use Alcumus, for bug reports, and for feature requests.  We look forward to hearing your ideas about how we might develop Alcumus in the future.

[url=http://www.artofproblemsolving.com/Alcumus/Introduction.php]Click here to get started with Alcumus[/url]!

Is there any specific reason it's called "Alcumus?"
i.e., does it stand for anything or mean anything?
(just curious)

I don't think I understand the question.  We don't intend to take it down, if that's what you're asking.

[quote="ReseesPieces"]huh is this like the OPLET thing that mathcounts made up[/quote]

In OPLET, you can choose your parameters, and it has all kinds of topics.  Alcumus, right now, is just counting and probability and the questions you get are randomized.

Will there ever be any organization of grade levels in Alcumus? For example, is Alcumus going to have the feature of clicking what grade level difficulty you want and then have you solve questions?

Alcumus focuses more on subjects and where you are at in that subject rather than arbitrary grade levels which can differ greatly from region to region.

Is there only counting/probability?
Just wondering

Hello,

Thank you for using Alcumus. At present there are only Introduction to Counting & Probability level problems. We expect some of our staff to be adding Introduction to Algebra type problems into Alcumus this summer, however there are a lot of issues that remain before launching Algebra, mostly due to the wide variety of answer types and different forms of inputting answers.

For more information on Alcumus, please view the [url=http://www.artofproblemsolving.com/Alcumus/Introduction.php]Alcumus Introduction[/url], [url=http://www.artofproblemsolving.com/Alcumus/Instructions.php]Alcumus Instructions[/url], and [url=http://www.artofproblemsolving.com/Alcumus/FAQ.php]Alcumus FAQ[/url].

Is there any chance you will add problems leveled at the Intermediate Algebra book?  :blush:

It's (obviously) on the waiting list, at most.

[img]http://aceksemedia.com/wp-content/uploads/web_tasarim_seo.png[/img]
Thank you :)

Thank you :)
[img]http://aceksemedia.com/wp-content/uploads/web_tasarim_seo.png[/img]

You're welcome.

Now what exactly does that diagram mean, though?

Hi everyone,

Please try to keep your posts relevant to support/requests of that nature. :)

Does anyone know how many focuses there are in alcumus

I want to know so I can figure out how many I have to do a day to get ready for my mathcounts competition.

It depends on how hard the questions are.

[quote=YEGSwim]Alcumus should make it harder to master focuses because I mastered one in 4 questions. My little brother does other math and it takes him 28+ questions (28 if they're all right, more if he gets them wrong)[/quote]

Yes, your little brother is probably doing it in easy, and you in Insanely Hard

how do you get the achievements like 13:37

They are called "secret achievements"
They relate to putting in answers that are weird, or answering a quesion at a certain time. You have to figure them out ;)

does anybody know how to get the achievement 007?

there's a topic in Fun Factory that has a list of all the secret achievements and how to get it, search up secret achievements in Fun Factory and it'll show up

007 is do it when the clock is at x:y0:07

I think

x and y are variables

It's called James Bond or something


for $ x \equal{} 3\minus{}\sqrt5,y\equal{}3\plus{}\sqrt5$
x has at least ask how many digits from 0 at the end?
 it follows that y has at least as many digits in the last 0

Frankly speaking, I completely fail to understand what is being asked and I have a feeling that I'm not alone with this problem :(.


S is any set of n points in the plane, n \geq 2. Let D and d, respectively, denote the greatest and the least distances determined by two points of S. Prove that D > 3^0.5/2 * (n^0.5-1)d.

Ok...a classical problem for the area method...

Let M_1,....,M_n be the points.
Let C be the disk with least radius which encloses the n points, and let O be its center and R its radius.

For each of the M_i, consider the disk C_i with center M_i and radius d/2. No two of these disks have interior point in common, and each of them is contained in the disk C' with center O and radius R + d/2.
It follows that the sum of the areas of the C_i's is less than the area of C',  that is :
\pi (R + d/2) 2   >   n\pi (d/2) 2
which leads to R > ( \sqrt n - 1)d/2.

It is a classical result to prove that two ot the points, say M_1,M_2 form a diameter of C or three of the points , say M_1,M_2,M_3 are on the boundary of C and form an acutangle triangle.
In the first case, we have D = M_1M_2 = 2R > 2( \sqrt n - 1)d/2 >  \sqrt 3( \sqrt n - 1)d/2
In the second case, wlog we may assume that <M_1OM_2  \geq 2 \pi /3. Thus (M_1M_2) 2   \geq 3R 2 . From which it follows that D  \geq  \sqrt 3R >  \sqrt 3( \sqrt n - 1)d/2.

Pierre.


I need help proving/disproving a statement.
Given:
$ N\implies \sim J\\
J\implies D\\
D\implies \sim A$.

Prove: $ \sim A \vee \sim N$.

I know the conclusion would be more obvious if the first given were $ N\implies J$, but I'm not sure if the conclusion actually follows from the givens.
Please show me the steps taken and why they work (Laws, Definitions, etc.), or why the conclusion does not work. Thanks in advance!

Take $ D \equiv \sim A$, $ J \equiv D$, $ N \equiv \sim J$, hence $ \sim A \equiv \sim N$. The conclusion now writes $ \sim N \vee \sim N \equiv \sim N$, clearly not a tautology.


If the tens digit of $(4A1)^{1A4}$ is $2$, compute all possible values for the digit $A$.

[hide]
Looks like:
$A^{1A4} \equiv 2 \pmod{10}$

A must be even. 2 works, and 4, 6, 8 don't work. So the only answer is 2.
[/hide]


hello,

i'm writing my thesis at this moment and i noticed that the bottom of the + sign appears a litle bit below the rest of a sentence...

how can i fix this???

If you don't know what i mean, just try to type something like blabla+blalab, then zoom a bit and you'll see what i mean

im using book style 11pt, and the sentence is outside a math environment or so, just plain text

[quote="aktano"]try to type something like blabla+blalab[/quote]

$blabla+blalab$  maybe because of this "new" and provisory latex


The formula expressing the relationship between $ x$ and $ y$ in the table is:
\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & 2 & 3 & 4 & 5 & 6 \\
\hline y & 0 & 2 & 6 & 12 & 20 \\
\hline \end{tabular}
\]
$ \textbf{(A)}\ y \equal{} 2x \minus{} 4 \qquad\textbf{(B)}\ y \equal{} x^2 \minus{} 3x \plus{} 2 \qquad\textbf{(C)}\ y \equal{} x^3 \minus{} 3x^2 \plus{} 2x$
$ \textbf{(D)}\ y \equal{} x^2 \minus{} 4x \qquad\textbf{(E)}\ y \equal{} x^2 \minus{} 4$

[quote="Brut3Forc3"]The formula expressing the relationship between $ x$ and $ y$ in the table is:
$ \begin{tabular}{|c|c|c|c|c|c|} \hline x & 2 & 3 & 4 & 5 & 6 \\
\hline y & 0 & 2 & 6 & 12 & 20 \\
\hline \end{tabular}$

$ \text{(A)}\, y \equal{} 2x \minus{} 4 \qquad\text{(B)}\, y \equal{} x^2 \minus{} 3x \plus{} 2 \qquad\text{(C)}\, y \equal{} x^3 \minus{} 3x^2 \plus{} 2x \\
\text{(D)}\, y \equal{} x^2 \minus{} 4x \qquad\text{(E)}\, y \equal{} x^2 \minus{} 4$[/quote]
It's easy.That's B


We have a new book from the AMC available on our site: [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=52]First Steps for Math Olympians[/url] by J. Douglas Faires.

Professor Faires has been the chairperson of the AMC 10 since it began.  In his new book he gives lessons, examples, and exercises on a comprehensive variety of topics that appear on the AMC 10 and AMC 12.

My school ordered a copy of this book and it is EXCELLENT.
The honors precalc teacher at my school also loves it because it has interesting questions arranged by topic - perfect for bonus problems on tests!

If a student wants a copy and doesn't want to spend the money, you could ask your school to order it.  It's a great book for a school library and your teacher might like it for bonus questions.


logarithms
inequalities
limit
function
algebra
polynomial
inequalities proposed

Prove that:

$ \ln \left( {e^3  \plus{} 1} \right) \plus{} 1 > e \plus{} \ln \left( {e \plus{} 1} \right).$

S.E.Louridas

We have that $ \ln \left( {e^3 \plus{} 1} \right) \plus{} 1 > 4.04$ and $ e \plus{} \ln \left( {e \plus{} 1} \right) < 4.04$. I used Sage but one can form Taylor series of $ \ln x$ and prove that the error term is small enough.

Actually I'm not sure if the method I explained works at all. The series
$ \ln (1 \plus{} x) \equal{} \sum_{n \equal{} 1}^{\infty}( \minus{} 1)^{n \plus{} 1}\frac {x^n}n$ converges if $ |x| < 1$ but $ e^3 \plus{} 1 > 1$

You could write the inequality as $ \ln((e \minus{} 1)^2 \minus{} (e \minus{} 1) \plus{} 1) > e \minus{} 1$ or $ (e \minus{} 1)^2 \minus{} (e \minus{} 1) \plus{} 1 > e^{e \minus{} 1}$. $ RHS \equal{} 1 \plus{} (e \minus{} 1)/1! \plus{} (e \minus{} 1)^2/2! \plus{} (e \minus{} 1)^3/3! \plus{} \ldots$. Therefore one has to prove that $ 1 > \frac{1}{2!} \plus{} \frac {e \minus{} 1}{3!} \plus{} \frac {(e \minus{} 1)^2}{4!} \plus{} \frac {(e \minus{} 1)^3}{5!} \plus{} \ldots$. I'm not sure if this helps.

One theoretical solution :

The problem is equivalent to this one:
Prove that :
$ e\left( {e \minus{} 1} \right) > e^{e \minus{} 1}  \minus{} 1$ and then we have that,
$ \left( {e \minus{} 1} \right)e > e^{e \minus{} 1}  \minus{} 1 \Leftrightarrow e^2  \minus{} e > e^{e \minus{} 1}  \minus{} 1 \Leftrightarrow e^2  \minus{} e^{e \minus{} 1}  \minus{} e \plus{} 1 > 0.$
We know that:
$ \left( {1 \plus{} \frac{a}
{n}} \right)^n  \equal{} 1 \plus{} a \plus{} \frac{{n \minus{} 1}}
{{2! \cdot n}}a^2  \plus{} \frac{{\left( {n \minus{} 1} \right)\left( {n \minus{} 2} \right)}}
{{3! \cdot n^2 }}a^3  \plus{} ... \plus{} \frac{{a^n }}
{{n^n }}.....\left( \sigma  \right).$  Therefore we can to consider :
$ u_n  \equal{} \left( {1 \plus{} \frac{2}
{n}} \right)^n  \minus{} \left( {1 \plus{} \frac{{e \minus{} 1}}
{n}} \right)^n  \minus{} \left( {1 \plus{} \frac{1}
{n}} \right)^n  \plus{} 1....(\alpha )$  for this we want, finally, to prove
$ \mathop {\lim }\limits_{n \to  \plus{} \infty } u_n  > 0.$ we remand that:
 $ \mathop {\lim }\limits_{n \to  \plus{} \infty } \left( {1 \plus{} \frac{x}
{n}} \right)^n  \equal{} e^x ,x \in \mathbb{R}.$ From the previews ,from the second part of the  (α) its enough to prove:
$ 2 \minus{} e \plus{} \mathop {\lim }\limits_{n \to  \plus{} \infty } \left[ {\frac{{n \minus{} 1}}
{{2! \cdot n}}\left( {2^2  \minus{} \left( {e \minus{} 1} \right)^2  \minus{} 1} \right) \plus{} ... \plus{} \frac{{2^n  \minus{} \left( {e \minus{} 1} \right)^n  \minus{} 1}}
{{n^n }}} \right] > 0,$ with
$ 2 \minus{} e \equal{} \mathop {\lim }\limits_{n \to  \plus{} \infty } \left[ {2 \minus{} \left( {1 \plus{} \frac{1}
{n}} \right)^n } \right] \equal{} \mathop {\lim }\limits_{n \to  \plus{} \infty } \left( {\frac{{n \minus{} 1}}
{{2!n}} \plus{} \frac{{\left( {n \minus{} 1} \right)\left( {n \minus{} 2} \right)}}
{{3!n^2 }} \plus{} ... \plus{} \frac{1}
{{n^n }}} \right),$ that is its enough
$ \mathop {\lim }\limits_{n \to  \plus{} \infty } u'_n\  > 0,when$
$ u'_n\  \equal{} \frac{{n \minus{} 1}}
{{2! \cdot n}}\left[ {2^2  \minus{} \left( {e \minus{} 1} \right)^2  \minus{} 2} \right] \plus{} \frac{{\left( {n \minus{} 1} \right)\left( {n \minus{} 2} \right)}}
{{3! \cdot n^2 }}\left[ {2^3  \minus{} \left( {e \minus{} 1} \right)^3  \minus{} 2} \right] \plus{} ... \plus{} \frac{1}
{{n^{n \minus{} 2} }}\left[ {2^n  \minus{} \left( {e \minus{} 1} \right)^n  \minus{} 2} \right] \plus{} \frac{{a^n }}
{{n^n }}.$ We consider the function
$ f: \left[ {2, \plus{} \infty } \right) \to \mathbb{R} \wedge f\left( x \right) \equal{} 2^x  \minus{} \left( {e \minus{} 1} \right)^x  \minus{} 2 \Rightarrow f^{\prime} \left( x \right) > 0 \wedge f\left( 3 \right) > 0,$ because
$ 2 > e \minus{} 1 > 0.$ This step ,for the monotony of f given the certainty of the positive part,
$ \forall n \in \mathbb{N} \wedge n > 2$ and this step is significant because all the brackets except of the first ,are positives . We remark that:
$ \mathop {\lim }\limits_{n \to  \plus{} \infty } \frac{{n \minus{} 1}}
{{2! \cdot n}}\left[ {2^2  \minus{} \left( {e \minus{} 1} \right)^2  \minus{} 2} \right] \approx  \minus{} 0,4762... >  \minus{} 0,48,$
$ \mathop {\lim }\limits_{n \to  \plus{} \infty } \frac{{\left( {n \minus{} 1} \right)\left( {n \minus{} 2} \right)}}
{{3! \cdot n^2 }}\left[ {2^3  \minus{} \left( {e \minus{} 1} \right)^3  \minus{} 2} \right] > 0,15,$
$ \mathop {\lim }\limits_{n \to  \plus{} \infty } \frac{{\left( {n \minus{} 1} \right)\left( {n \minus{} 2} \right)\left( {n \minus{} 3} \right)}}
{{4! \cdot n^3 }}\left[ {2^4  \minus{} \left( {e \minus{} 1} \right)^4  \minus{} 2} \right] > 0,22,$
$ \mathop {\lim }\limits_{n \to  \plus{} \infty } \frac{{\left( {n \minus{} 1} \right)\left( {n \minus{} 2} \right)\left( {n \minus{} 3} \right)\left( {n \minus{} 4} \right)}}
{{5! \cdot n^4 }}\left[ {2^5  \minus{} \left( {e \minus{} 1} \right)^5  \minus{} 2} \right] > 0,12,$ with the observation:
$ \minus{} 0,48 \plus{} 0,15 \plus{} 0,22 \plus{} 0,12 > 0.$ Now we can to speak what its true the following inequality    $ \mathop {\lim }\limits_{n \to  \plus{} \infty } u_n^{\prime}  > 0$.

S.E.Louridas

Okay. I tried also find a proof without a calculator or a computer. My approach:

It is sufficient to show that $ \ln(e^3 \plus{} 1) \plus{} 1 > 4.04$ and $ e \plus{} \ln(e \plus{} 1) < 4.04$.
It is sufficient to show that $ e^3 \plus{} 1 > 21.08$ and $ \ln(21.08) > 3.04$ and $ e < 2.72$ and $ \ln(e \plus{} 1) < 1.32$.
It is sufficient to show that $ e > 2.7181$ and $ \ln(21.08) > 3.04$ and $ e < 2.72$ and $ \ln(3.72) < 1.32$.
It is sufficient to show that $ e > 2.7181$ and $ 21.08 > e^{3.04}$ and $ e < 2.72$ and $ 3.72 < e^{1.32}$.

Now, as a consequence of Taylor’s theorem, if $ P_n$ is the $ n$th order Taylor polynomial for $ f \equal{} 1 \plus{} x \plus{} x^2/2! \plus{} \ldots$ at $ 0$, then
\[ |f(x) \minus{} P_n(x)|\leq \frac {M}{(n \plus{} 1)!}|x|^{n \plus{} 1},\]
where $ M$ is the maximum value of $ f^{(n \plus{} 1)}$ on the closed interval from $ 0$ to $ x$. Therefore
\[ |f(x) \minus{} P_n(x)|\leq \frac {M}{(n \plus{} 1)!}|x|^{n \plus{} 1},\]
From this we see that to get an approximation of $ e$ by four digits we have to find a positive integer $ n$ such that $ \frac {3}{(n \plus{} 1)!} < 0.00005$. This inequality holds if $ n > 8$. Therefore $ e\approx 1 \plus{} 1 \plus{} 1/2! \plus{} 1/3! \plus{} 1/4! \plus{} 1/5! \plus{} 1/6! \plus{} 1/7! \plus{} 1/8! \equal{} 109601/40320$. But as $ 2.71827 < 109601/40320 < 2.71828$, we have that $ e\approx 2.7182$ has at least four correct decimals. In particularly $ 2.7181 < e < 2.72$. Therefore $ e^3 \plus{} 1 > 2.7181^3 \plus{} 1 > 21.08$.

Let us now show that $ e^{3.04} < 21.08$. To find a degree of a suitable Taylor's polynomial, we have to choose a positive integer $ n$ such that $ \frac {e^{3.04}}{(n \plus{} 1)!}|e^{3.04}|^{n \plus{} 1} < 0.1$. But $ \frac {e^{3.04}}{(n \plus{} 1)!}|e^{3.04}|^{n \plus{} 1} < \frac {3^4}{(n \plus{} 1)!}|3^4|^{n \plus{} 1} \equal{} \frac {81}{(n \plus{} 1)!}|81|^{n \plus{} 1}$. But http://www.mathlinks.ro/Forum/viewtopic.php?t=299439 shows that this is true for $ n > 82$ and evaluating a polynomial of degree $ 83$ is quite boring by pencil and paper. Or if you want to use transcendental functions, you see that $ e^{3.04}\approx 20.9$ so you have to evaluate a polynomial of degree $ 21$.

Actually I can complete my idea using only huge precision rational arithmetic such that I don't need any Neper constant or ln's. To show that $ 21.08 > e^{3.04}$ one can show that $ e^{76/25} < 21.08$ or $ e^{76} < 21.08^{25} \approx 1.249\cdot 10^{33}$. But $ e^{76} < 2.72^{76} < 1.1\cdot 10^{33}$. Also $ e^{1.32} \equal{} 1 \plus{} 1.32 \plus{} 1.32^2/2 \plus{} 1.32^3/3! \plus{} \ldots$. By taking the first five terms gives you a number bigger than $ 3.72$.


Why is D a better answer than A? I get the same rate law for A.
See pic.

A doesn't satisfy the balanced equation (the mechanism gives 2B+A->C). D does, and gives the correct rate law.


please answer me this question:

are there some people who found notrio particles??

please i"m very interested in it ;)

uh...i think you mean "neutrino" particles


ratio
probability
geometry
3D geometry
prism
pyramid
trigonometry

For number 1, it's simple algebra. Well you know that the angle inscribed is proportional to the arc formed from it...so, you know that 3x+4x+5x=360, because all the inscribed angles in the circle totals 360 degrees. So solve the equation angle get 12x=360, thus x=30, since it's asking for the largest angle, it has to be 5x or 5(30) because x=30. Therefore the largest angle is 150 degrees, pretty sure it's not 75?

I'll answer more in a sec.

[quote="returnofkiladi"]1) A circle is divided into 3 arcs in the ration of 3:4:5. If an inscribed angle of the circle intercepts the largest of he 3 arcs, what is the measure of the inscribed angle.
Answer: [hide]
75 degrees
[/hide]

2) Steve did not study for his social studies quiz, so he had to guess the answers to all ten questions. If the questions were all true/false questions, what is he probability that Steve correctly guessed at least 5 answers?
Answer: [hide]
319/512
[/hide]

3) An empty reservoir in the shape of a hemisphere begins to fill with water at a rate of 3000 gallons/minute. If the radius of the reservoir is 15 meter and there are .0038 meters cubed in a gallon, how long will it take for the reservoir to fill completely (to the nearest minute)?
Answer: [hide]
620
[/hide]

4) What is the mean deviation of the set  {12, 14, 14, 17, 18, 20}?
Answer: [hide]
2.5
[/hide]

5) Pete rolls three 6-sided dice. What is the probability that at leas one of the dice shows a 4.
Answer: [hide]
91/216
[/hide]

6) A prism has an equilateral triangle with sides of length 10 inches at its base. If the area of one of the lateral sides of the prism is 70 sq inches, what is the volume of he prism in cubic inches?
Answer: [hide]
175 sqrt 3
[/hide]

7) A small triangular pyramid is cut off at each vertex of a cube and discarded. The volume of each triangular pyramid is approximately one thousandth the volume of the cube. How many vertices does the remaining polyhedron have?
Answer: [hide]
24
[/hide]

8) In the land of Alpha, (1/4)X32 = 5. What is (1/3)X10 in Alpha?
Answer: [hide]
2
[/hide][/quote]

#8: Think base six.

For 2:

There are 2^10 total possibilities. So first find out the probability you can get 5 right. So 10C5=126/1024. Sine 10c4=10c6, 10c7=10c3, add half to it to get 638/1024 or 319/512

Also: mean deviation isnt on mathcounts.

[quote="returnofkiladi"]1) A circle is divided into 3 arcs in the ration of 3:4:5. If an inscribed angle of the circle intercepts the largest of he 3 arcs, what is the measure of the inscribed angle.
Answer: [hide]
75 degrees
[/hide]

2) Steve did not study for his social studies quiz, so he had to guess the answers to all ten questions. If the questions were all true/false questions, what is he probability that Steve correctly guessed at least 5 answers?
Answer: [hide]
319/512
[/hide]

3) An empty reservoir in the shape of a hemisphere begins to fill with water at a rate of 3000 gallons/minute. If the radius of the reservoir is 15 meter and there are .0038 meters cubed in a gallon, how long will it take for the reservoir to fill completely (to the nearest minute)?
Answer: [hide]
620
[/hide]

4) What is the mean deviation of the set  {12, 14, 14, 17, 18, 20}?
Answer: [hide]
2.5
[/hide]

5) Pete rolls three 6-sided dice. What is the probability that at leas one of the dice shows a 4.
Answer: [hide]
91/216
[/hide]

6) A prism has an equilateral triangle with sides of length 10 inches at its base. If the area of one of the lateral sides of the prism is 70 sq inches, what is the volume of he prism in cubic inches?
Answer: [hide]
175 sqrt 3
[/hide]

7) A small triangular pyramid is cut off at each vertex of a cube and discarded. The volume of each triangular pyramid is approximately one thousandth the volume of the cube. How many vertices does the remaining polyhedron have?
Answer: [hide]
24
[/hide]

8) In the land of Alpha, (1/4)X32 = 5. What is (1/3)X10 in Alpha?
Answer: [hide]
2
[/hide][/quote]

[hide="2"]
Ew, casework.

Case 1: He gets exactly 5 right.

$ \frac{1}{2}^{10} \cdot \binom{10}{5}$

Case 2: 6 right

$ \frac{1}{2}^{10} \cdot \binom{10}{6}$

etc.

Add them all up and you get 319/512.
[/hide]

[hide="5"]
The probability that he gets ZERO 4's is $ \frac{5}{6}^3 \equal{} \frac{125}{216}$. Thus our answer is $ 1 \minus{} \frac{125}{216} \equal{} \frac{91}{216}$.
[/hide]

For number 2, i know that the possible outcome is 1024, because 2^10, and u say the answer is 319/512, so the successive outcome must be 638, i don't know how to arrive at 638 though..

For number 3, well, first we need to find the volume of the hemisphere, which is (2/3)*r^3*pi, plugging the radius in, you get that the volume=2250*pi, which is about 7068.58 meters cubed. Using that number, and the fact that there is .0038 meters cubed in a gallon, you divided 7068.58 by .0038, and get about 1860153.55. Finally take that number and divide by 3000 because the problem says that the hemisphere is getting filled at the rate of 3000 gallons per minute. By doing that, you get 620.1 minutes, which is about 620 minutes.

For number 4, well, i bet you dont know what mean deviation means do you, let me explain it shortly. To find the mean deviation, you take the average of the numbers, then take that average and find the absolute value difference of it from each number in the list and add it all up and divide by the number of numbers. confusing? well, lets work with this set {12,14,14,17,18,20}. The average of it is(12+14+14+17+18+20)/6= 15.8333..Now, take this number and find the absolute value difference of it and every number in the set.
|12-15.83|=3.83
|14-15.83|=1.83
|14-15.83|=1.83
|17-15.83|=1.07
|18-15.83|=2.17
|20-15.83|=4.17
Now take all the numbers from this new set {3.83,1.83,1.83,1.07,2.17,4.17} and find the average. The average is about 2.5

[quote="returnofkiladi"]1) A circle is divided into 3 arcs in the ration of 3:4:5. If an inscribed angle of the circle intercepts the largest of he 3 arcs, what is the measure of the inscribed angle.
Answer: [hide]
75 degrees
[/hide]

2) Steve did not study for his social studies quiz, so he had to guess the answers to all ten questions. If the questions were all true/false questions, what is he probability that Steve correctly guessed at least 5 answers?
Answer: [hide]
319/512
[/hide]

3) An empty reservoir in the shape of a hemisphere begins to fill with water at a rate of 3000 gallons/minute. If the radius of the reservoir is 15 meter and there are .0038 meters cubed in a gallon, how long will it take for the reservoir to fill completely (to the nearest minute)?
Answer: [hide]
620
[/hide]

4) What is the mean deviation of the set  {12, 14, 14, 17, 18, 20}?
Answer: [hide]
2.5
[/hide]

5) Pete rolls three 6-sided dice. What is the probability that at leas one of the dice shows a 4.
Answer: [hide]
91/216
[/hide]

6) A prism has an equilateral triangle with sides of length 10 inches at its base. If the area of one of the lateral sides of the prism is 70 sq inches, what is the volume of he prism in cubic inches?
Answer: [hide]
175 sqrt 3
[/hide]

7) A small triangular pyramid is cut off at each vertex of a cube and discarded. The volume of each triangular pyramid is approximately one thousandth the volume of the cube. How many vertices does the remaining polyhedron have?
Answer: [hide]
24
[/hide]

8) In the land of Alpha, (1/4)X32 = 5. What is (1/3)X10 in Alpha?
Answer: [hide]
2
[/hide][/quote]

Thanks for your help so far, guys. I'm kind of confused on #2, I don't understand casework, and what does 10c5 mean?
In addition, I got 150 for number 1, too, but the answer was 75. I wasn't quite sure, I doubt the book is wrong, but then again it's possible. My only guess is to set everything equal to 180 instead of 360.
I also don't at all get #8, it confuses me what the x even means. Can someone explain that to me?
#2, #3, #4, #6, #7, #8 are all the problems I don't understand thus far and #1 is still questionable.

For problem #1 the central angle is 150 degrees but the question asks for the inscribed angle. The inscribed angle is half of the central angle, so the answer is 75 degrees.


in a isosceles triangle median are perpendicular to each other. find the measure of the vertex angle and base angles.

As I solved it is not rational, it $ \cos$ is equal to $ \frac{4}{5}$


Give an example of a compact Hausdorff topology on $ \mathbb{R}$.

Not difficult enough to link away  :D

Just take a bijection (as sets) $ \mathbb R \to [0,1]$; the later has a obvious compact hausdorff structure, so we get one for the reals, too.


From a deck of fifty-two cards, twelve cards are dealt. six are dealt
face down and six face up. the visible cards are a four, a five, a
seven, a jack, and two kings. what is the probability that the next card
dealt is a four or less? (consider aces to be high, not low).

[quote="anyone"][hide]First off, you need to find the number of possible outcomes. You have already turned over 6 cards, so there are $52-6=46$ total possible outcomes. Now you need to know the number of possible outcomes that satisfy the conditions. In a deck of cards there are 16 cards that are $\le{4}$. However, you've already turned over a four so there are only 15 possible cards that satisfy your conditions. So $P(\le4)=\boxed{\frac{15}{46}}$[/hide][/quote]

Aces are high, not low.  So you counted four cards too many in your total.

Wasn't this question deleted already because it is part of the application for SIMUW?

Yeah, I overcounted with the aces thing. I won't fix my solution since the problem is part of an application.

[quote="sharkman"]I was wondering what happened to my question.[/quote]

Yeah, it probably got deleted because of that.  I'm not sure whether or not you are applying to the program in question, but it might be better to wait on this one unless it's homework or someone is threatening you at gunpoint or something.


integration
function
complex numbers
real analysis
real analysis unsolved

Let $f,f_{k}$ be in $L^{p}$ and $1 \leq p < \infty$. Assuming that $f_{k}\to f$ pointwise a.e. and that $\| f_{k}\|_{p}\to \| f\|_{p}$, prove that $\| f_{k}-f\|_{p}\to 0$

He wasn't being particular about the underlying measure space. This is true in both finite and infinite measure spaces. It is false for $p=\infty$ in an infinite measure space, but the claim is only for $1\le p<\infty.$

Here's an argument for the $p=1$ case.

First, assume WLOG that $f_{k}$ and $f$ are nonnegative functions.  Once we prove it under those conditions, we split the more general $f_{k}$ and $f$ into positive part of real part, negative part of real part, positive part of imaginary part, and negative part of imaginary part.

$\int|f_{k}-f|=\int(f_{k}-f)_{+}+\int(f_{k}-f)_{-}$

Since $(f_{k}-f)_{-}\le f$ and $f$ is integrable, the Dominated Convergence Theorem tells us that

$\int(f_{k}-f)_{-}\to 0.$

Then $\int(f_{k}-f)_{+}=\int(f_{k}-f)-\int(f_{k}-f)_{-}$

But $\int(f_{k}-f)=\int f_{k}-\int f\to 0$ by the hypothesis that the norms converge. That together with what we had already proved about the negative part implies that

$\int(f_{k}-f)_{+}\to 0.$

With that we are done - at least for the $p=1$ case.

I'm still not remembering exactly how to get the case of general $p$ out of the same argument.

Assume we are working on a $\sigma$-finite measure space $(X,m)$. We can then write $X=\cup{X_{n}}$, where $(X_{n})$ is increasing and $\forall n, m(X_{n})>0$. Since $E \mapsto \int_{E}|f|^{p}dm$ is a measure, and since $X_{n}\rightarrow X$ monotonously, $\int_{X_{n}}|f|^{p}dm \rightarrow \int_{X}|f|^{p}dm$. In particular, we can find $N \in \mathbb{N}$ such that $\parallel f\parallel_{p}< \epsilon$. Since $\parallel f_{k}\parallel_{p}\rightarrow \parallel f\parallel_{p}$, we also have $\parallel f_{k}\parallel_{p}<\frac{3}{2}\epsilon$ when $k$ is large enough. This means that on $X-X_{n}$ we have $\parallel f_{l}-f_{k}\parallel_{p}\leq 3 \epsilon$ when $k,l$ are large enough.

Now on $X_{n}$ what happens is clear: by Egoroff's theorem you can find a measurable subset $Y$ of $X_{n}$ such that $(f_{k})$ is almost uniformly convergent to $f$ on $Y$ (i.e, in particular, $\parallel f_{k}-f_{l}\parallel_{p,Y}<\epsilon$ when $k,l$ are large enough), and $\int_{X_{n}-Y}|f|dm<\epsilon$ ( so that like above we can write $\int_{X_{n}-Y}|f_{k}-f_{m}|dm<3 \epsilon$ when $k,l$ are large enough).

In conclusion, the sequence of integrable functions $(f_{k})$ is Cauchy in measure for the $L_{p}$ norm, and $(f_{k}) \rightarrow f$ a.e, whence $|f|^{p}$ is integrable and $\parallel f_{k}-f\parallel_{p}\rightarrow 0$.

[quote="Kent Merryfield"]First, assume WLOG that $f_{k}$ and $f$ are nonnegative functions.  Once we prove it under those conditions, we split the more general $f_{k}$ and $f$ into positive part of real part, negative part of real part, positive part of imaginary part, and negative part of imaginary part.[/quote]

How do you know that if $\parallel f_{k}\parallel _{p}\rightarrow \parallel f|_{p}$ then the four functions you defined above will still satisfy this condition (and hence that we can assume $f$ and the $f_{k}$ to be positive) ? For the two other statements of the problem I have no objection, but I'm stuck here.

[quote="Kent Merryfield"] Since $(f_{k}-f)_{-}\le f$.[/quote]

How do we know this, even assuming $k$ is large ?


Wow, that's freaky!  How'd you guys figure that out?  Lots of free time on your hands? lol just kidding

[quote="miraculouspostmaster"]I think that representing such a series (which would continue into infinity) with an algebraic variable would not be permissible (x, infinity, minus two would still be infinity).  Not quite sure though.[/quote]

Is it possible to have a number infinity?  If infinity were to be represented by an actual number (probably can't be), what would it be? Like 999,999,999,999,999,999,999,999.....? If it were, and you subtracted 2 from it, it would no longer be infinity, but 2 less than infinity.  Same for if you added something to infinity.  Or maybe even like infinity times two.  Ugghhh this is getting confusing.

infinity minus two = infinity
[url]http://www.answers.com/topic/hilbert-s-paradox-of-the-grand-hotel[/url]

[quote="pkothari13"]
(x-2)/2 = 2 + 4 + 8 + 16 + 32 + 64 + ...
(x-2)/2 = x[/quote]
this step, i believe is incorrect
what you did is like this
you have the last term, 2^infinity, removed, which is infinitely large

[quote="pkothari13"]yeah basically, in both cases, neither of the series converges, so you can't set a definite value to x or a.

Check this one out too:

$x = 1+2+3+4+...$
$n = 1+3+5+7+...$
$n = 1+2+3+4+... - (2+4+6+8...)$
$n = x - 2x$
$n = -x$[/quote]

Lol. That's a pretty good one. It's semi-convincing, too.


inequalities calculus derivative function ratio

$\forall a,b,c \in \mathbb{R}$ and $a,b,c \geq \frac{1}{4}$ and $a+b+c=1$, prove that \[\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\leq 5\]

[quote="Peter VDD"]Aren't the absurd conditions marvellous?  :whistling:[/quote]

what do you mean by that?

[quote]Aren't the absurd conditions marvellous?  :whistling: [/quote]

 :lol:  That's exactly why I posted it here. I found a solution, but the condition and the strange boundary made me hesitate, so I posted it here to check

[quote="maokid7"][quote="Peter VDD"]Aren't the absurd conditions marvellous?  :whistling:[/quote]what do you mean by that?[/quote]This is rather a (very mathematical) joke than a serious inequality. :)

I have a more interesting exercise :
$a+b+c=1$.max of $\sqrt{2x+1}+\sqrt{3y+1}+\sqrt{4z+1}$

it was with cauchy also.
(2+3+4)((x+1/2)+(y+1/3)+(z+1/4)) >= (sqrt(2x+1)+sqrt(3y+1)+sqrt(4z+1))^2
i dont wanna make the calculations.

I was going to ask how you found those strange values but then I realized that the C-S equality condition is well known...

Actually, at first I didn't think about the C-S equality condition, but instead sort of cheated and found the maximum by solving a system of partial derivatives of the function
$\sqrt{2x+1}+\sqrt{3y+1}+\sqrt{5-4x-4y}$

pssch! cheating with calculus when there are easier ways... *frowns upon*
 the equality condition for C.S. is when the ratio of the a's to the b's is equal

If you only want a solution, not caring wether it is nice or not: $(a+b+c)^{4}(ab+bc+ca)\ge 8(a^{2}+b^{2}+c^{2})(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$ by muirhead.


Let $M$ be the set of all polynomials $P(x)$ which satisfy the following conditions:
[list]
[*] all roots of $P(x)$ are pairwise different integers
[*] the absolute values of all coefficients of $P(x)$ are nonnegative integers less than $2007$.[/list]
Find the maximum degree of the polynomials in $M$.


topology
Functional Analysis
advanced fields
advanced fields unsolved

Let $ E, F$ be the Banach spaces, let $ \mathcal{L}(E,F)$ be the set of all continuous linear operators $ E\to F,$ and let $ \mathcal{S}$ denote the set of all surjections in $ \mathcal{L}(E,F).$

Is it true that $ \mathcal{S}$ is open set in $ \mathcal{L}(E,F)$ (with respect to the operator norm)?

[hide="Some thoughts"]I know that the set of right-invertible operators in $ \mathcal{L}(E,F)$ is open, so the answer is positive when $ E,F$ are Hilbert Spaces, or $ F$ has finite dimension. However not every continuous epimorphism between Banach spaces is right-invertible and I wonder if it is still true in general.[/hide]

Well, a bounded linear operator $ T: E \to F$ is surjective if and only if the ajoint operator $ T^*: F^* \to  E^*$ is bounded from below, i.e. if there is a $ c > 0$ such that $ \|T^* f\| \ge c \|f\|$ for all $ f \in F^*$. But if one adds a small norm operator $ U: E \to F$ (with, say, $ \|U\| \le \frac{c}{2}$), $ T^* \plus{} U^*$ remains  bounded from below (with a smaller constant), so $ T\plus{}U$ remains surjective.

It seems that there are a lot of nice things in Functional Analysis which I should learn before I start asking stupid questions ;)

Thank You volodya:)


Suppose not all four integers, $a,b,c,d$, are equal. Start with $(a,b,c,d)$ and repeatedly replace $(a,b,c,d)$ by $(a-b, b-c, c-d, d-a)$. Prove that at least one number of the quadruple will eventually become arbitrarily large.

[hide]
If anybody has Engel, how do they get from the second to last inequality to the last inequality?
[/hide]

Well, they do it recursively.

You can make one term arbitratily large because the distance from the origin is at least two times greater than the distance before the iteration.

So after n-1 iterations,

$a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\geq (a_{1}^{2}+b_{1}^{2}+c_{1}^{2}+d_{1}^{2})$


geometry
incenter
circumcircle
trigonometry
angle bisector
geometry proposed

$ A,B$ and $ C$ are three points on a circle.The chord joining the midpoints of arcs $ AB$ and $ AC$ intersects the Chord $ AB$ at $ D$.The Chord joining the midpoints of arcs $ AC$ and $ BC$ intersects the Chord $ BC$ at $ E$.
(a)Prove that $ DE$ passes through the incenter of triangle $ ABC$.
(b)Prove that $ DE$ is parallel to $ AC$.

[hide="a trigonometric proof"]call $ \Gamma$ the circumcircle, call M the midpoint of the anch AC and $ F: DM \cap AC$ and $ G: EM \cap AC$, call $ K: DM \cap \Gamma$ and $ H: EM \cap \Gamma$ . Call I the incenter of ABC, so we have:
$ \angle EGC \equal{} 90 \minus{} \frac{C}{2}$ and $ \angle DFA \equal{} 90 \minus{} \frac{A}{2}$
$ \frac{KM}{HM}\equal{}\frac{BM \sin{\frac{B}{2}} \cos{\frac{C}{2}}}{BM \cos{\frac{A}{2}} \sin{\frac{B}{2}}} \equal{} \frac{\cos{\frac{C}{2}}}{\cos{\frac{A}{2}}} \equal{} \frac{FM}{GM}$ Then for thales $ DE \parallel AC$.

Call J the projection of D on AC, so we have:

$ DJ \equal{} AM \left [ \frac{\sin{\left ( C \plus{} \frac{B}{2} \right )}\sin{\frac{B}{2}}\cos{\frac{A}{2}}}{\sin{\frac{B}{2}} \cos{\frac{A}{2}}} \minus{} \sin{\frac{B}{2}} \right ] \equal{}$
$ \equal{} AM \left [ \sin{\left ( C \plus{} \frac{B}{2} \right )} \minus{} \sin{\frac{B}{2}} \right ] \equal{} 2AM \sin{\frac{C}{2}}\sin{\frac{A}{2}} \equal{}$
$ \equal{} \frac{b  \sin{\frac{C}{2}}\sin{\frac{A}{2}}}{\cos{\frac{B}{2}}} \equal{} 4R  \sin{\frac{A}{2}}\sin{\frac{B}{2}} \sin{\frac{C}{2}} \equal{} r$[/hide]

[quote="plane geometry"]Notice IF^2=FA^2=FD*FH  GI^2=GB^2=GE*GH  So ∠FID=∠IHF   ∠EIG=∠GHI
∠FID+∠EIG+∠FIG=180[/quote]
∠DIA=∠GIE=∠IHG=∠IAC So DE ‖AC

It is easy to show the collinearity of the points $ D\equiv AB\cap FH, \ I\equiv AG\cap CF,\ E\equiv BC\cap HG,$ by [b][size=100]Pascal theorem[/size][/b], in the configuration of the cyclic non-convex hexagon $ ABCFHG.$

Because of now, $ HD$ is the angle bisector of the angle $ \angle AHB$ of the triangle $ \bigtriangleup HAB,$ we have that $ \frac {DA}{DB} \equal{} \frac {HA}{HB}$ $ ,(1)$

Similarly, from the triangle $ \bigtriangleup HBC,$ we have that $ \frac {EC}{EB} \equal{} \frac {HC}{HB}$ $ ,(2)$

From $ (10,$ $ (2)$ $ \Longrightarrow$ $ \frac {DA}{DB} \equal{} \frac {EC}{EB}$ $ ,(3)$ ( because of $ HA \equal{} HC$ ).

From $ (3),$ we conclude that $ DE\parallel AC$ and the proof is completed.

Kostas Vlittas.

My solution is follow:
Call M is intersection of GH and AC. Since IC=GC, and GH is perpendicular with IC then IE=EC, IM=MC. So <ICM=<ICE=<EIC=<MIC. Then IE// MC.
Analogously, we have ID//AN (N is intersection of AC  and FH).
So I, D, E are in a line, which is parallel with AC.

Call $ C_1$ and $ A_1$ the midpoints of arcs $ AB$ and $ BC$ respectively. $ \frac {BD}{DA} \equal{} \frac {BC_1}{C_1A} \cdot \frac {\sin{\angle{DC_1B}}}{\sin{\angle{DC_1A}}}$ analogously $ \frac {BE}{EC} \equal{} \frac {BA_1}{A_1C}\cdot\frac {\sin{\angle{EA_1B}}}{\sin{\angle{EA_1C}}}$ but $ \pi \minus{} \angle{DC_1B} \equal{} \angle{EA_1B}$ and $ \angle{AC_1D} \equal{} \angle{EA_1C}$ then $ \frac {BD}{DA} \equal{} \frac {BE}{EC}$ and $ DE\parallel{AC}$


LaTeX
number theory unsolved
number theory

Let $n$ be a natural number, $r(n)$ is the sum of the $n$ remainds obtained by dividing $n$ with $1,2,..,n$. Find a number $n$ greather than $1 000 000$ such that $r(n)=r(n+1)$.

(n+1)%k=n%k+1 if k is not divisor (n+1), else (n+1)%k=n%k-(k-1). Therefore
$r(n+1)=r(n)+2n+1-\sigma(n+1)$, were $\sigma(x)=\sum_{k|x} k$.
If $n+1=2^k$, then $\sigma(n+1)=\sum_{i=0}^k 2^i=2n+1$. Therefore $n=2^k-1$ works.

@Rust: something with your $\LaTeX$-code is not correct, please don't use % in it.


geometry
circumcircle
trigonometry
geometry proposed

let ABC be a triangle and A' the midpoint of BC, let Ta be the tangency point of the incircle with BC and let Sa be the circle with center A' and wich passes trought Ta.
Define analogously Sb and Sc. prove that if Sa is tangent to the circuncircle of ABC then at least onbe of Sb and Sc also satisfies this property

Is $ABC$ supposed to be non-obtuse? I've managed to construct a counterexample with the aid of dynamic software. However, I can prove the assertion given the extra hypothesis that all the angles of $ABC$ are $\le \frac {\pi}2$ (i.e. non-obtuse). The triangle I built is obtuse and $S_a$ touches the circumcircle in the semiplane determined by the line $BC$ which contains $A$.

again my fault......Sa is tangent to the circuncircle of ABC in the arc BC wich doesn't contains A....well i forgot this info...sorry, i dont have the exam papers so I forgot things like that but thanks anyway :blush:

Thanks. In that case, that condition is equivalent to one of the angles $\angle B,\ \angle C$ being $\frac {\pi}2$, which implies, as I've said, the conclusion.

To show that it holds for right-angled triangles (for $\angle A=\frac {\pi}2$): we must show that $\b\cdot \tan{\frac {\angle B}2}=|a-c|=a-C\iff br=(a-c)(p-b)\iff br=\frac{a^2-c^2-(a-c)b}2\iff br=b(p-a)$, which is obviously true. Next we use the argument I've told you about, moving the triangle a little and showing that the condition can't be true anymore.


Solve

$ x\plus{}y\plus{}z\equal{}9$

$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}1$

$ xy\plus{}yz\plus{}zx\equal{}27$.

Because $ (x\plus{}y\plus{}z)^2 \equal{} x^2\plus{}y^2\plus{}z^2\plus{}2(xy\plus{}yz\plus{}zx)\implies x^2\plus{}y^2\plus{}z^2\equal{}81\minus{}2\cdot 27\equal{}27$ .

But $ xy\plus{}yz\plus{}zx\equal{}27$ so from the inequality $ x^2\plus{}y^2\plus{}z^2\ge xy\plus{}yz\plus{}zx$ it follows that $ x\equal{}y\equal{}z\equal{}3$ .

PS : We don't need the second relation .

hello, it is clear that $ xyz\ne 0$, so we can divide the third equation by $ xyz$ and we get
$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}\frac{27}{xyz}$
now we use the second equation, we get
$ xyz\equal{}27$
from here we have
$ xy\equal{}\frac{27}{z}$
plugging this in the third equation we get
$ \frac{27}{z}\plus{}z(9\minus{}z)\equal{}27$
and this equation can we solve for $ z$, the solution is $ (3,3,3)$ of our system.
Sonnhard.

let $ p\equal{}x\plus{}y\plus{}z , q\equal{}xy\plus{}yz\plus{}zx , r \equal{} xyz$

x , y & z are the solution of  $ X^3\minus{}PX^2\plus{}QX\minus{}R\equal{}0$which admet  only one solution $ (3,3,3)$


I am not a math wiz but know I am missing something very simple... my co-worker gave me this puzzle can anyone help me???

	                    a = b
	                  a2 = ab
	           a2 – b2 = ab – b2
	(a + b) (a – b) = b (a – b)
	           (a + b) = b
	              a + a = a
	                  2a = a
	                    2 = 1

Since 2 does not equal 1 in the real world, there is obviously something wrong with the above problem.  Do you know what is wrong?

Here is a clue as to how to solve this: Every step, taken one step at a time, is mathematically correct; the problem is found when you look at it as a whole rather than one step at a time.

[quote="bluespider"]Here is a clue as to how to solve this: Every step, taken one step at a time, is mathematically correct.[/quote]

Not really

[quote="plokoon51"][quote="captcha000"]perhaps it is because you divide by a-b, but this is zero since a=b?[/quote]

thats exactly right. i love giving this problem to my math class and watch everyone struggle to find what is wrong :P[/quote]

Thanks, I really appreciate the insight.. I myself was spinning my wheels with the a=0 but think I see the "big picture" :-)

[quote="bluespider"]

	(a + b) (a – b) = b (a – b)
	           (a + b) = b

You see the step where you are dividing by (a-b)? Well, since a=b, (a-b) is the same as (a-a) which is 0. And you cannot divide by zero because it is undefined. That is why the proof does not work.

This made me remember something that happened a few months ago. A girl that I met wrote me from Bogota, because her ninth grader sister had a problem. Her math professor told them that [i]"Something that shows that math was always non applicable to reality is that there were a bunch of proofs that gives you 2 = 1"[/i] and that we would give 5 (maximum grade on her school, say, an A+) in the semester to anyone who could find a proof about it. Well, I explain my friend that any proof you can find about this is false, it has some mistake (at least under regular cardinality theory), but I showed her the one above, and also tis one,

$\frac{9}{4}-6=\frac{25}{4}-10$
$\Rightarrow\frac{9}{4}-6+4=\frac{25}{4}-10+4$
$\Rightarrow\left(\frac32-2\right)^2=\left(\frac52-2\right)^2$
$\Rightarrow\frac32-2=\frac52-2$
$\Rightarrow0=1$
$\Rightarrow1=2$

Of course, it is not hard to see the mistake. They guy didn't tell her why her two solutions were wrong, but didn't give her the 5, but he gave them this

$x^2=x+x+\cdots+x$ where the sum has $x$ terms. Then, you take derivative to both sides and get $2x=1+1+\cdots+1=x$, then cancelating the $x$'s you have $2=1$. Of course, ninth greaders, nobody understood a sh..., but the guy said that the demonstration was completely correct! Can you believe it?

[quote="djimenez"]but the guy said that the demonstration was completely correct! Can you believe it?[/quote]

How can a math teacher say that that is completely correct? That is blasphemy!! :D

You can find a discussion [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=proof+1%3D2+derivative&t=8790]here[/url] about why the proof using the derivative is wrong.

[quote="djimenez"]
$\frac{9}{4}-6=\frac{25}{4}-10$
$\Rightarrow\frac{9}{4}-6+4=\frac{25}{4}-10+4$
$\Rightarrow\left(\frac32-2\right)^2=\left(\frac52-2\right)^2$
$\Rightarrow\frac32-2=\frac52-2$
$\Rightarrow0=1$
$\Rightarrow1=2$[/quote]

[hide]Note that every number has [i]two[/i] square roots, one positive and one negative[/hide]

[quote="adidasty"]also a + a doesnt equal a. it equals 2a.[/quote]

But that's not the false step of the proof.  Assuming all the previous steps in the proof are correct (which of course they are not) a+a=a is a correct statement.

Also, saying that the a+a=a part is wrong is equivalent to saying that 2=1 is wrong...

[quote="joml88"]Also, saying that the a+a=a part is wrong is equivalent to saying that 2=1 is wrong...[/quote]
well saying 2=1 is wrong, it doesn't matter how you got that.
and for the record, 2 doesn't equal 3.

Well, actually it does matter how you got it.  You could define your own number system in such a way that 2=1 is a correct statement.

And saying that $2\not=1$ does not explain [b]why[/b] that proof is incorrect neither does saying that $a+a\not=a$ for any $a$.

[quote="paladin8"][quote="djimenez"]
$x^2=x+x+\cdots+x$ where the sum has $x$ terms. Then, you take derivative to both sides and get $2x=1+1+\cdots+1=x$, then cancelating the $x$'s you have $2=1$. Of course, ninth greaders, nobody understood a sh..., but the guy said that the demonstration was completely correct! Can you believe it?[/quote]

I remember discussing this one on AoPS a while ago. Every time I show it to my friends they get confused (the ones that know calculus, that is).[/quote]

The ones that don't know calculus aren't confused?   :P


superior algebra
superior algebra unsolved

Let $ \sigma, \tau \in S_n$ such that $ \sigma \tau \equal{} \tau \sigma$. Show that if $ \sigma$ is a $ n$-cycle then $ \tau$ must be a power of $ \sigma$.

Without loss of generality, $ \sigma\equal{}(1\ 2\ldots n)$.

What do powers of $ \sigma$ look like?  Show that if $ \tau$ sends $ 1$ to $ k$, it must send $ 2$ to $ k\plus{}1$, etc.


USAMTS ARML AMC AIME USA(J)MO USAMO search

What is the total number of math contests you have, or will be, competing in this school year (Aug. 08-July 09)? Include out-of-school and in-school contests. If a contest has multiple rounds, like the USAMTS or ARML Power Contest, count them as one contest. But count the AMC A, B, AIME, and USAMO each as separate contests.

Hmm. I've never counted, this should be interesting:
State Talent Search
Mandelbrot Regional
AMC 10A
AMC 12B
AIME
USAMO
ARML
UVM High School Math Exam
Area Math League
I'm probably missing some...

iTest
KU
GPMLx2
USAMTS
AMC 12A
AMC 12B (fail)
AIME
USAMO
ARML

(Didn't do KSU or KCATM this year).

[quote="Ihatepie"]Does it have to be purely math or can it be a trivia show that has math as a quarter of it?[/quote]
Purely math.

Then I only have a few:
Math League
AMC12A
AMC10B
AIME
ARML

Hmmm....
Mandelbrot Nationals
Mandelbrot Teamplay
ARML Team Power (or whatever it's called)
Santa Clara University High School Math Contest
Stanford Math Tournament
Bay Area Math Olympiad
CAML (most of it, gave up because it was silly)
AMC A
AMC B
AIME
Will be USAMO

Last year I had Purple Comet, Santa Clara Valley Math Field Day, ARML team selection (didn't end up going) as well, and probably something else I am forgetting.

Here's what I have:

USAMTS
Math League
Some middle school math contest sponsored by my local high school, which I'm not gonna say the name of since that'd give away the name of the exact city that I live in
AMC 10A
AIME
MATHCOUNTS Chapter
MATHCOUNTS State
MATHCOUNTS Nationals

...Yeah...

I have

Mathcounts
AMC 8
AMC 10B
Math League
Purple Comet (later this year)
Gauss 8 (later this year if my math teacher hopefully didn't forget...)

I did qualify for AIME, but I didn't take it b/c of MC States (which I failed).

MC
AMC8
AMC10 (which I failed)
UNM-PNM
Purple Comet (later)

YAY

Fall Startup Event
CAML
ANother test from nats assessment
Another one from nats assessment
MC
AMC 8
AMC 10A
AMC 10B
AIME
ARML
NCCQM (if you could call that a contest...)
Purple Comet
BAMO
SMT
I think I'm missing some, but this puts me at 14+.

AMC
MC
USAMTS
ARML
final exam in math class (jk)

Im pretty young so i have.....
Math Kangaroo, (anyone heard of it?)
AMC 8
and KSEA (for Korean Americans) :blush:

In Vietnam,maybe because of the problem of money,there is very few math contest,I remmemeber when in highshool, each year I only could take one maths contest :the math contest of my province, I was rather sad that almost fields of maths in this contest are not concerned my interest,so  I got very few academic achivements .

iTest
Mandelbrot
Mandelbrot Team Play
NYML
NYCIML
NYSML
ARML
ARML Power Contest
Purple Comet
AMC 12 (A&B)
AIME
USAMO
PUMaC
HMMT

Probably forgetting some, but that brings me up to "14+"

Oh, I guess I had more than I thought...
Unless this is upcoming events too

County Math Field Day
Regional Math Field Day
Math League
Regional Math and Science Day (lol, more math than science)

And...
State Math Field Day coming up April 26
and if I get top 15 at State MFD,
ARML @ PSU

So, possibly 6?

Not really sure. Atleast 11 I can think of off the top of my head, so probably 14+. It's not about quantity, but quality, which I do not have. :(

It must be nice to be able to participate in so many contests that you forget what they were....

I only competed in one, but I picked 11 in the poll based on how many I've "participated" in.  That's one as a competitor, eight as a writer, one as a coordinator, and one as a coach.

uhh...
Mathcounts
Math Is Cool
Washington state Math Championship
Washington State Math Olympiad or something like that
Uhh...I guess that's all :)

Hmmm...

Mathcounts
Math is Cool
AMC 8
State championships
RCML

A few years back I would have had like 10 or something. Where did all that go?

iTest
HMMT
PUMaC
NYSML (2 days ago)
ARML
AMC 12A
AMC 10B
AIME
USAMO (tomorrow)

Next year I'm adding USAMTS and hopefully Mandelbrot to my mathematical competition itinerary.

same as fooblah, plus I think he/she (not giving it away) forgot Math League.

Oops...whatever...We were going to do Rocket City Math League except that the person who organized the math team didn't start the math club until soccer season was over, and by then, the beginning of RCML was over too...:( whatever. next year i'm also gonna do AMC 8.


calculus
integration
real analysis
real analysis unsolved

If $ f(0)\equal{}1$ find $ f$ which satisfays $ (f''(x))^2\equal{}f(x)f'(x)$

Let $ f'(x)\equal{}p$. Then: $ f''(x)\equal{}p\frac{dp}{df}$. We will have:
$ \left(p\frac{dp}{df}\right)^2\equal{}fp$.

$ p\frac{dp}{df}\equal{}\sqrt{fp}$ or $ p\frac{dp}{df}\equal{}\minus{}\sqrt{fp}$.

If $ p\equal{}0$ then we have solution $ f(x)\equal{}1$ else we have:

$ \sqrt{p}dp\equal{}\sqrt{f}df$ or $ \sqrt{p}dp\equal{}\minus{}\sqrt{f}df$. After integration, we will have:

$ \frac{2}{3}p^{\frac{3}{2}}\equal{}\frac{2}{3}f^{\frac{3}{2}}\plus{}\frac{2}{3}C$ or $ \frac{2}{3}p^{\frac{3}{2}}\equal{}\minus{}\frac{2}{3}f^{\frac{3}{2}}\plus{}\frac{2}{3}C$, where $ C$ is a constant.

$ C\equal{}f'(0)^{\frac{3}{2}}\minus{}1$.

$ p^{\frac{3}{2}}\equal{}f^{\frac{3}{2}}\plus{}C$ or $ p^{\frac{3}{2}}\equal{}\minus{}f^{\frac{3}{2}}\plus{}C$

$ p\equal{}\left(f^{\frac{3}{2}}\plus{}C\right)^{\frac{2}{3}}$ or $ p\equal{}\left(\minus{}f^{\frac{3}{2}}\plus{}C\right)^{\frac{2}{3}}$

Remember that $ p\equal{}\frac{df}{dx}$. We will have:

$ \frac{df}{\left(f^{\frac{3}{2}}\plus{}C\right)^{\frac{2}{3}}}\equal{}dx$ or $ \frac{df}{\left(\minus{}f^{\frac{3}{2}}\plus{}C\right)^{\frac{2}{3}}}\equal{}dx$

$ \int{\frac{df}{\left(f^{\frac{3}{2}}\plus{}C\right)^{\frac{2}{3}}}}\equal{}x \plus{} D$ or $ \int{\frac{df}{\left(\minus{}f^{\frac{3}{2}}\plus{}C\right)^{\frac{2}{3}}}}\equal{}x\plus{}D$

Last integrals we can't calculate in general. We should know the value of constant $ C$. If $ C\equal{}0$ we will have the solution in elementary functions. It will be $ p\equal{}f$ or $ f'(x)\equal{}f(x)$ or $ f(x)\equal{}e^x$.


geometry
circumcircle
geometric transformation
reflection
trigonometry
perpendicular bisector
geometry unsolved

Let N be a point on the longest side AC of a triangle ABC. The perpendicular bisector of AN and NC intersect AB and BC respectively in K and M. Prove that the circumcenter O of ABC lies on the circumcircle of triangle KBM.

[hide="Solution"]
Denote $ P$ as the reflection of $ N$ over $ KM$.  Notice that $ PM \equal{} MN \equal{} MC$ so $ M$ is the circumcenter of $ \triangle PNC$ and similarly, $ K$ is the circumcenter of $ \triangle APN$.  It follows that $ \angle APC \equal{} \angle APN \plus{} \angle NPC \equal{} \frac {\angle NMC \plus{} \angle AKN}{2}$.  Also, $ \angle AKN \equal{} 180 \minus{} \angle KAN \minus{} \angle KNA \equal{} 180 \minus{} 2\angle BAC$ since $ KN \equal{} AK$.  Similarly, $ \angle NMC \equal{} 180 \minus{} 2\angle BCA$, which means that $ \angle APC \equal{} 180 \minus{} \angle BAC \minus{} \angle BCA \equal{} \angle ABC$, so $ BPCA$ is cyclic.  This means that $ OP \equal{} OC$.  This, coupled with the fact that $ OM \equal{} OM$ and $ PM \equal{} MC$ means that $ \triangle POM\cong \triangle COM$, so $ \angle OBM \equal{} \angle OCM \equal{} \angle OPM$, so $ PBOM$ is cyclic.  Similarly, $ PBKO$ is cyclic, so $ BPMOK$ is cyclic and we conclude that $ BMOK$ is cyclic, as desired. [/hide]

[hide="My solution"]Call D the midpoint of BC and F the midpoint of AB and let $ AN=x$ and call $ \Gamma$ the circumcircle of KBM. Then

$ O \in \Gamma \ \Longleftrightarrow \ \angle KOM = 180 - \angle ABC = \angle DOF \ \Longleftrightarrow \ \angle FOK = \angle DOM \ \Longleftrightarrow$

$ \Longleftrightarrow \ \frac{KF}{OF}=\frac{MD}{OD} = \frac{\frac{c}{2}-\frac{x}{2\cos{\alpha}}}{R\cos{\gamma}} = \frac{\frac{b-x}{2\cos{\gamma}}-\frac{a}{2}}{R\cos{\alpha}} \ \Longleftrightarrow \ c\cos{\alpha}+a\cos{\gamma} = b$[/hide]


Hello everybody!

Again I have a little problem with some exercises. Ok, here it goes.

Cone's axiscut ( I really don't know how to write this word in english, sorry if I am wrong ) is [b]12 cm[/b][size=59]2[/size] and height is [b]3cm[/b]. I need to calculate cone's volume.

I tried to do it but i stopped at Pytagor formul where my radius is square root from 55.

V=πR x R x H ( height )
    ------------------------
                  3

And I attached sketch becouse my english in math isn't that good.

Here is the sketch: http://i28.photobucket.com/albums/c201/Cipruss/123.jpg

CSX area is 12 and SO ( heght ) is 3 cm. And i need to calculate the volume. Like I said i stopped at Pytagor formula where my radius is square root from 55. And i don't think that it could be like that. Somewhere but i don't know where i have a mistake.

So, please help me!

Nevermind. Just close this topic please. I got the answer.


geometry
trapezoid
rectangle
LaTeX
trigonometry
ratio
Pythagorean Theorem

Here are some geometry problems to celebrate the new year :)

1. In quadrilateral $ABCD$, $\angle BAD=\angle BCD=90^\circ$. $E$ is a point on $AB$ such that $AE=5$ and $BE=3$. $F$ is a point on $CD$ such that $FD=2$. If $AD=8$ and $BC=10$, find the area of quadrilateral $BFDE$.

2. $\triangle ABC$ is inscribed in a circle and $AB$ is its diameter. (i.e., $\angle ACB=90^\circ$) Circle $D$ is inscribed in $\triangle ABC$. Find $\angle ADB$.

3. In quadrilateral $ABCD$, $\angle ABC=\angle ACD=90^\circ$. Segment $AC$ bisects $\angle BAD$. $E$ is a point on $AB$ such that $DE\perp AB$. If $AB=4$ and $BC=3$, find $BE$.

4. $ABCD$ is an isosceles trapezoid with $AB=10$ and $CD=8$. $E$ is the midpoint of $AB$ and $F$ is the midpoint of $CD$, with $EF=12$. Point $R$ is on segment $EF$ such that $\angle CRB=90^\circ$. Find all possible lengths for $RE$.

5. $ABCD$ is a square with $AB=20$. $P$ is on $CD$ and $Q$ is on $AD$ such that $AP\perp BQ$. Let $R$ be the intersection of $AP$ and $BQ$. If $AR=12$, find the area of quadrilateral $BCPR$.

does question 1 work?

because like seriously..

is it just me

or does question 1 just not work?

because firstly, <BAD = < BCD = 90.. opposite angles in quad are right ==> rectangle..

and then we have

AD = 10 BC = 8...

arents opposite sides in a rectangle suppose to be equal?

or am i just lost?

opposite angles being equal to 90 doesnt mean the quad is a rectangle. ;)

[hide="1"]
In a hurry sorry,
Is the answer $22$?
[/hide]

sorry cant do 1.. im trying thoo.. still cant get the diagram =S

2 [hide] 135 degrees [/hide]

3 [hide] 2 1\4 [/hide]

4 [hide] 2 or 10 [/hide]

5 [hide]154 [/hide]

sorry i havnt posted up how i got solutions, i needa get latex so it wont be confusing..

oh btw.. lol.. opposite angles ==> rectangle?!? sorry sorry sorryy... dnoo.. its js relli late.. and sorta tired.. >< =S

okaiis now..

i got question 1..

frkn helLl.. arghh at last.. im going to sleep now.. its 133am.. ><

1 [hide] 22 .. yeh ur right ^^

firstly find the length of diagonal BD.. it is root 128.. and using this u can solve for the length of CF, knowing that BC = 10 and DF = 2 using simple algebra

CF = V28 - 2

find area of entire figure ABCD, just add the areas of triangle ABD and BCD

total area of abcd = 5V8 + 32

next find the area of triangle CBF and AED, 5V8 + 10 and 20 respectively

minus from the total area of abcd

and voilaa

u get 22..

[/hide]

1.
[hide]$ABCD$ is cyclic, so $BD$ is the diameter of the circumscribed circle.
$128 = 100 + (FC + 2)^2$ $\Longrightarrow$ $FC = \sqrt{28} - 2$.
$\left[ABCD\right] = 32 + 5\sqrt{28}$.
So $\left[BFDE\right] = 32 + 5\sqrt{28} - (5(\sqrt{28} - 2)) - 20 = 22$.[/hide]

[hide="3"]
I get a really wierd answer :?  :(
$\frac{\sqrt{161}}{4}$
[/hide]
[hide="4"]
$6-\sqrt{15}$
and
$6+\sqrt{15}$
[/hide]

For the second I don't understand the text. What is $D$?

I would say it was the centre of the other circle

[hide]
#2

D is the center of the incircle of triangle ABC, so D is where the angle bisectors of triangle ABC meet.

Since angle ACB = 90,  angle CAB and CBA must add up to 90.

When the angle bisectors are drawn to D from each angle, angle DAB plus angle DBA is 45, half of their original total.  The third angle in the triangle formed by A B and D, angle ADB, must be the rest of 180 degrees: 180 - 45 = 135.

[/hide]

[hide="1"]
Inscribe $\triangle{ABD}$ and $\triangle{BCD}$ in circles.  $BD$ is the diameter for both circles, because $\angle{C}=\angle{A}=90^{\circ}$, so $ABCD$ is cyclic.  Because angle A is right, we use the pythagorean theorem to find that $BD=8\sqrt{2}$.  Now, again using the Pythagorean Theorem, we get $CD=2\sqrt{7}$.  Thus, \[ \triangle{BDE}=\frac{3}{8}\triangle{ABD}=\frac{3}{8}32=12  \] because $\frac{BE}{AB}=\frac{3}{8}$.  Similarly, \[ \triangle{BDF}=\frac{1}{\sqrt{7}}\triangle{BCD}=\frac{1}{\sqrt{7}}10\sqrt{7}=10  \]  Thus, the area in question is $\triangle{BDE}+\triangle{BDF}=10+12=\boxed{22}$[/hide]

[hide="2"]
We see that \[ \measuredangle{ADB}=180^{\circ}-\measuredangle{DBA}-\measuredangle{DAB}  \]  Because $DB$ and $DA$ are the angle bisectors of A and B, we now have \[ 180^{\circ}-\frac{\measuredangle{A}+\measuredangle{B}}{2}  \]  Because C is right, A and B are complementary, so now we have \[ 180^{\circ}-\frac{90^{\circ}}{2}=\boxed{135^{\circ}}  \][/hide]

[hide="3"]
By the Pythagorean Theorem, $AC=5$.  Let $\angle{BAC}=\angle{CAD}=\theta$.  Then, $AD=\frac{5}{\cos{\theta}}=\frac{25}{4}$.  Thus, \[ \cos{\theta}=\frac{4}{5}, \sin{\theta}=\frac{3}{5} \implies \cos{2\theta}=\frac{16-9}{25}=\frac{7}{25}  \]  Then \[ AE=AD\cos{2\theta}=\frac{7}{4} \implies BE = 4-\frac{7}{4}=\boxed{\frac{9}{4}}  \][/hide]

[hide="4"]
Let $O$ be the midpoint of $BC$.  Draw a circle, with $O$ as center, which passes through $B$ and $C$.  $\angle{BRC}$ is right iff it lies on this circle.  We see that $BC=\sqrt{145}$, so \[ RO=\frac{\sqrt{145}}{2}  \]  The distance from $O$ to $EF$ is $\frac{9}{2}$.  Let the circle intersect $EF$ at $X$ and $Y$, and let the midpoint of $EF$ be $N$.  Then $ON \perp EF$, and $OX=OY=\frac{\sqrt{145}}{2}$.  So, by the Pythagorean Theorem, \[ XN=\sqrt{OX^2-ON^2}=\sqrt{\frac{145-81}{4}}=4 \implies XE=6-4=2, YE=6+4=10  \] Thus, $RE \in \boxed{(2, 10)}$[/hide]

[hide="5"]
If $AR=12$, then, by the Pythagorean Theorem, $BR=16$.  Then \[ \angle{BAR} \cong \angle{RAQ} \implies \triangle{BAR} ~ \triangle{AQR} \implies QR=9, AQ=15  \] Thus, $AD=20-AQ=5$.  Because they have $\angle{DAP}$ in common, \[ \triangle{RAQ} ~ \triangle{DAP} \implies PD=15, AP=25 \implies PR=13  \]  We know that the area in question is $ABCD-\triangle{ABR}-\triangle{APD}=400-96-150=\boxed{154}$[/hide]

Oh wow... thanks for lots of participations :)

Here are the correct answers.
[hide]
1. $22$
2. $135^\circ$
3. $\frac{9}{4}$
4. $2$, $10$
5. $154$
[/hide]
sen and K81o7 got all of them :10:

If you can figure out, #1 can be solved in 10 seconds. You don't even need to use Pythagorean theorem or cyclic quadrilateral stuffs ;)  Can you find the "easier" way?

K81o7, try to find another way to solve #3. I like your way to use trig, but this problem can be solved without trig.

Also, regarding #2, yes, I meant $D$ is the center of the incircle.

Hmm...come to think of it, I didn't even need to put the circle in #1.  But is there a way without the pythagorean theorem?  I used that to find areas, and then did a ratio...

For #3...hm i think i see it...
[hide]$\angle{CDE} \cong \angle {BAC} \implies \frac{BE}{CD}=\frac{3}{5}$.  Then because of similar triangles, and ratios, we find that $CD=\frac{15}{4}$, and thus, $BE=\frac{9}{4}$[/hide]
Is that it?

Yeah, that's probably the fastest way for #3.

If you can't figure out the easy way for #1...
[hide="Hint"]
Divide $BFDE$ into two triangles and find the area of each triangle ;) [/hide]

[quote="K81o7"][quote="frt"]
Divide $BFDE$ into two triangles and find the area of each triangle ;) [/quote]

But that's what I did...or do you mean the other way?[/quote]
You don't even need to use area ratios.

[hide]
$BFDE=\triangle BDE+\triangle BDF=\frac{1}{2}\cdot 3\cdot 8+\frac{1}{2}\cdot 2\cdot 10=\boxed{22}$ ;) [/hide]

Oh yeah...i completely forgot that you could calculate area that way... :D


Compare each member of D to the corresponding member of N. For example, compare 1 to 2, 3 to 4, etc. There are 50 numbers in each sequence. Each of the members of D is 1 smaller then the corresponding member of N, so there is a difference of 50*1=50 between D and N.

you can group it into 49 groups that have D+1=N so after those we have 49 then we subtract 99 because D has one more number and we get 50 as our answer :idea: solidifarms, you did this problem like it was 1-99 and 2-100 instead of 1-99 and 2-98, yet you got lucky and got the right answer anyway :rotfl: i guess in mathcounts the wouldnt care, but watch out for tricks like that in the future :)

:blush: ooops....I think I remember doing this one the same way last year...  :blush:

[quote="bryantacademy"]D is the sum of all the odd numbers from 1 to 99 inclusive, and N is the sum of all the even numbers from 2 to 98 inclusive:
D=1+3+5+7+9...+99 asnd N=2+4+6+8...+98
Which is greater, D or N and by how much?

Thanks
 :huh:[/quote]First, note that N = 2*(1+2+3+...+49).
Gauss discovered that the sum of the first n natural numbers $=\frac{n(n+1)}{2}$.

With that in mind $D = \frac{99(100)}{2}$ and

$N = 2\frac{49(50)}{2}$

$D = 99(50)$

$N = 49(50)$

$D-N = 50(50) = 2500$


$\frac{n^2}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}>\frac{2n-1}{a_1+\dots+a_n}$

[edited] by myself

it's $(2n-1)^2$ instead of $2n-1$ but this is hardly an olympiad problem


Let's say we have a hospital room holding 4 newly born babies (A,B,C,D), each with name tags.
Some stranger comes in and takes off all the name tags and runs away.
Now the nurses come in, and they can't remember which baby belongs to which couple
of parents (a,b,c,d). what are the chances that they, when giving the babies to the couples, botch up the entire situation and none of the babies make it to the correct set of parents.

this was nice since there were only 24 cases to look at, but how would one actually calculate this? Also, how
would one change the calculations to consider matching 1 baby-parents, 2 baby-parents?

i realized that the denominator was 4P4, but i couldn't figure out the numerator-- it also hit me that you couldn't
match 3 without matching all 4.

i felt as though this'd be really simple, but this counting has me stumped.

You're right it does get a bit tricky. How about using complementary counting? There are 24 cases, 1 4-correct case, 0 3-correct cases, $ \binom{4}{2} \equal{} 6$ 2-correct cases, $ 2\binom{4}{1} \equal{} 8$ 1-correct cases, thus $ 24\minus{}1\minus{}0\minus{}6\minus{}8 \equal{} 9$ incorrect cases. Or you could just count the positive cases. Assuming each baby $ (w,x,y,z)$ goes to the corresponding parent in $ (a,b,c,d)$ then the completely incorrect ones are
$ (B,C,D,A), (B,A,D,C), (B,D,A,C)$ along with the corresponding permutations when C and D are in the first spot. Thus we have $ (3)(3) \equal{} 9$ and our probability is $ \frac{9}{24} \equal{} \frac{3}{8}$. What do you mean by 2-baby parents?


calculus
derivative
calculus computations

Are there any formulas that could be used to find the nth derivative of a function... even if it only works for a certain function.  I know that patterns often exist but i was wondering if there was a quicker way to the solution.


geometry
circumcircle
incenter
geometry proposed

Let $ (O) $ be a circumcircle of triangle $ ABC $ . A circle $ (J) $ tangents to $ AB $ at $ D $ and internally tangents to $ (O) $ at $ F $ . A line through $ C $ and tangents to  $ (J) $ at $ E $ . Prove that : $ DE $ goes through $ I $ , which is the incenter of triangle $ ABC $

Which tangent from C to (J) do you choose?

The enunciation of problem is wrote negligently. I suppose that you will use this problem:
" Let $C_1=C(O_1,r_1),\ C_2=C(O_2,r_2)$ be two circles such that the circle $C_2$ is interior tangent in the point $A$ to the circle $C_1$. Then, for a tangent $BC$ in the point $D$ to the circle $C_2$, where $B,C\in C_1$,  we have $\widehat {BAD}\equiv \widehat {CAD}$. "

To Levi , The lemma you showed is used to solve a similar problem to this problem . I think it can work here (I'm not sure) .
   To EvaristeG : $ E $ satisfy : the segment $ EF $ intersects $ BC $

I have an idea , but I can't prove it :
 $ DE $ intersects $ AC $ at $ G $ . Then , $ FG $ is the bisector of $ \angle AFC $

Help me to complete this problem , Mlikers !  ;)

I found a topic mentioning a similar problem :
   [url]http://www.mathlinks.ro/Forum/topic-22925.html[/url]

I think that the problem should change "I -> I_b"....

I have a solution for this problem (just by angle chasing) .


Well,

        After all this hard work and what not its finally here people! I just wanted to wish everyone whos taking the USNCO (or has taken it already) good luck. I hope to make camp and see you people there too!

OH NOES...SECOND TIME TAKING THIS TEST!!!

i hope i'm prepared for this...

Good luck everyone!  I took it today and I think I have quite a shot at making camp...hope to maybe see you guys there ;)

Hopefully everyone who has posted here so far makes camp.  :D

Looks like I'm not going to make the camp...  :(


I didn't see this posted before, so yea...

Yep! :D
How many times have y'all qualified?

I don't know what to put! I was really scared of not making AIME this year since I failed last year, and now... not only did I make AIME for the first time, I got a perfect too! I'm befuddled.

To 1=2: my second time.  However, I've only taken teh AMC's twice...

Unless I epicly misbubbled.

lets just say, every time i have taken any AMC, i have qualified for AIME. so this is my third time.

hmm I suspect the large majority of (active) AoPSers make AIME >.> [unless you're spending too much time socializing on AoPS and NOT ENOUGH MATH]

This is actually the first time I've made AIME, and its the second time I took the AMC.

Well, if we count by different tests, I didn't make it from either time in 6th grade, from both tests in 7th, and one so far in 8th. So qualified from the 3/5 tests I've taken.

If you're talking about yearwise, I made it in 7th and 8th but not 6th. So two out of three.

This is the first year our middle actually did AMC10 and I qualified with a failing score of 144...

I made it, but only with 130.5. But this is my first AMC 10 so I am happy.  :)

I made it with a solid 100.5! (Assuming no scantron errors)

Even if I hadn't made two dumb mistakes, I would still need to way improve that, but I have two more years to, so that's good.

ONTO FAILING AT THE AIME! WOOOOO!

I definitely made AIME but only with a 129 on AMC 10...I keep making stupid mistakes. :(

Made it but will be taking B to get a higher index. This is my first time.

2nd time making AIME, 4th time taking AMC's. Coincidentally, the first time I mad AIME was also the first year I discovered AoPS.

Interesting choices for the poll. This is the first time I have made AIME. I hope to make it off the B test also.

I got a 97.5 last year, which tied for first at my school. Made a couple stupid mistakes, I hope to make it this year.

First time taking the AMC10 and got a 114. Hoping I can do better at the 10B.

4th time qualifying. Got a 144 on the 10.

2nd time AMC-10, second time AIME, went from 120 to 141

second time taking AMC-10, first time to AIME

failed in 7th grade.....91 =(
this year.....132....better =)


linear algebra
matrix
algebra proposed
algebra

[quote="spanferkel"]it is easy to show that $\sum_{i=1}^{1987}\sum_{j=1}^{1987}a_{ij} =\sum_{i=1}^{1987}(-1)^{i+1}a{}_i{}_i\le \sum_{i=1}^{1987}|a{}_i{}_i|\le1987$: To see this, observe that for odd $n$, $\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}= \sum_{i=1}^{n-2}\sum_{j=1}^{n-2}a_{ij}+a{}_n{}_n-a_{n-1,n-1}$ by 'taking away' two rectangular 2x(n-1) strips of sum 0 that overlap at $a_{n-1,n-1}$ :)[/quote]
hi spanferkel,your solution is good.
but the condition is $a_{ij}\leq 1$,not $|a_{ij}|\leq 1$
this is not your problem.I think based on the condition $a_{ij}\leq 1$ we cannot gain the required result.conversely,we only can proove$\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}\leq 2n-1$
and if let$a_{ij}=-3$for all$i,j$both odd,and $a_{ij}=1$for $ij$is even.then $\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}=2n-1$ so the constant $2n-1$is the best.
so I think we shoud alter the condition or the result of this problem to make it true.

[quote="mathematica"][quote="Hawk Tiger"]
I think based on the condition $a_{ij}\leq 1$ we cannot gain the required result.[/quote]
We can gain the result. I've seen the solution.[/quote]
Why?
Do you read my post?I give a contradictory example,and I pointed out where the mistake is in the solution.
Maybe I was wrong.But I believe myself.

I think you have made a mistake. Because in your example the value of $\sum_{i=1}^{n}\sum_{j=1}^{n}{a_{ij}}$ is $-2n-1$ and not $2n-1$. For example take $\left( \begin{array}{ccc}-3 & 1 &-3 \\ 1 & 1 & 1 \\-3 & 1 &-3 \end{array}\right)$ where the sum is $-7$.

OK Hawk Tiger is essentially right, he should just have said "let $a_{ij}=-3$ for $i,j$ both [b]even[/b],and $a_{ij}=1$ elsewhere". (That must be what he meant and yields $2n-1$ for the sum. Maybe he just started his numbering with $0$ instead of $1$ :) )
And it is easy to see that this is best possible given $a{}_{i}{}_{j}\le 1$.

[quote="mathematica"]Yes you are right. Now that I have read the proof again, it seems that even though that in the question (which I saw in a book and not the original Quant magazine) it said that $a_{ij}\leq 1$, in the solution it used the fact that $|a_{ij}| \leq 1$. Now I'll fix the first post so other people won't be mislead. :wink:[/quote]Nobody knows what the Quant magazine originally stated? anyway let's just state, there are two different (but of course very closely related) problems for $n\ge3$,
one with condition $a_{ij}\leq 1$ claiming RHS $\le 2n-1$, and
one with condition $|a_{ij}| \leq 1$ claiming RHS $\le n$. :) :roll:

[quote="spanferkel"]
one with condition $a_{ij}\leq 1$ claiming RHS $\le 2n-1$[/quote]
this wasn't proved...was it? :?:

well it is very easy. Consider a 2*2 submatrix along the diagonal (or wherever). It's least element must be $\ge-3$ and thus the alternating sum of it's two main diagonal elements between -4 and 4. It follows easily that $-2n-1\le\sum\le 2n-1$. :)


function
topology
real analysis
real analysis unsolved

Prove or find a counterexample:

Let $ S \subseteq \Re^2$ be connected and compact. Then there is a continuous function $ f$ from the interval $ [0,1]$ into $ \Re^2$ such that the image of $ [0,1]$ under $ f$ is the boundary of $ S$.

An annulus is an obvious counterexample. It is much more interesting to ask if $ S$ itself is always such an image ;)

$ S$ can always be represented as a continuous image of $ [0,1]$

Let $ \varepsilon > 0$ be given
$ \forall c\in S$ consider open set $ \alpha$ of the points in $ \varepsilon$-neighbourhood of $ c$.
The collection $ G_{\alpha} = \bigcup\alpha$ is an open cover of $ S$, and since $ S$ is compact, it has a finite subcover, given by $ \bigcup_{i = 1}^n V_{\varepsilon}(c_i)$ (say)

As $ [0,1]$ is itself compact, $ \exists \{t_1,t_2,\ldots,t_n\}$ such that $ [0,1]\subset\bigcup_{i = 1}^n V_{\delta}(t_i)$ where $ \delta > 0$

As $ S$ is connected, we can arrange $ c_i$ such that $ d(c_k,c_{k + 1}) < \varepsilon$, thus if we take a function $ f$ such that $ f\left( V_{\delta}(t_i)\bigcap [0,1]\right) = V_{\varepsilon}(c_i)\bigcap S$

(If $ I$ is any interval and $ S\subset\mathbb{R}^2$ is bounded, then there exists surjective $ f: I\rightarrow S$)

Honestly, I don't quite understand what you wrote. :( I do not see why $ f\left( V_{\delta}(t_i)\bigcap [0,1]\right) \equal{}V_{\varepsilon}(c_i)\bigcap S$ is possible to achieve: the left hand side is always connected and the right hand side can easily be not connected. Actually, the continuous image of an interval is always path connected and there are plenty of connected but not path connected sets, so the answer to the question as I posed it is "No". The next question is whether every compact path connected set is a continuous image of $ [0,1]$.

Sorry, my explanation was not clear

$ V_{\varepsilon}(c_i)\bigcap S$ may not be connected, but that does not matter as the criterion for continuity is still satisfied.

The problem is with [quote]As $ S$ is connected, we can arrange $ c_i$ such that $ d(c_k,c_{k \plus{} 1}) < \varepsilon$[/quote]
which is actually true only for path-connected sets.

Hmm, I probably should have included the criterion that S be simply-connected.

It wouldn't help. You still can create a region whose boundary is connected but not path connected:
[asy]size(100);
int n=300;
pathpen=blue+linewidth(1);
path P=nullpath;
real r=0.27;
for(int k=n;k>10;k=k-1) P=P--(k/n,r*sin(3.1415*sqrt(n/k)));
D(P);D(shift(0,-2)*P);
D((0,-r)--(0,r)); D((0,-2-r)--(0,-2+r));
D((1,0)--(1,-2));
D((0,0)--(-1,0)--(-1,-2)--(0,-2));[/asy]


calculus function integration derivative trigonometry algebra polynomial

This is a straightforward computational exercise. I'm putting it here instead of "Calculus computations and tuturials" becuase one does need to know something not in a typical first calculus course.

Compute, as a function of $s\in\mathbb{R},$

$\int_{-\infty}^{\infty}\frac{e^{-2\pi isx}dx}{x^4+4}.$

Some things to expect:

Let $f(x)=\frac1{x^4+4}$ and denote the object we're calculating by $\hat{f}(s).$

$f$ is real and even; hence $\hat{f}$ will be real and even.

$x^2f(x)\in L^1$ but $x^3f(x)\not\in L^1.$ That, with the exact rate of decay, leads us to expect that $\hat{f}\in C^2\setminus C^3$, and that the third derivative of $\hat{f}$ should have a jump discontinuity.

$f\in C^{\infty}$ and every derivative is integrable. This leads us to expect that $\hat{f}(s)$ goes to zero as $|s|\to\infty$ faster than any rational function of $s.$

We can compute this using contour integration, by considering a semi circle on the real axis and in the lower half plane, with orientation counterclockwise along the semicircle, and towards negative infinity on the real axis. So with this contour, for $\xi > 0$, consider the integral

$\int_{C^{-}[-R,R]} \frac{1}{z^{4} + 4} e^{-2 \pi i z \xi}\,dz =$

$ -\int_{-R}^{R} \frac{1}{x^4+4} e^{-2 \pi i x \xi} \,dx + \int_{0}^{-\pi} \frac{1}{R^4e^{4 i \theta} + 4} e^{-2 \pi i R (\cos{\theta} + i \sin{\theta}) \xi} Re^{i \theta} \,d\theta$.

Now, we have that

 $\left|\int_{0}^{-\pi} \frac{1}{R^4e^{4 i \theta} + 4} e^{-2 \pi i R (\cos{\theta} + i \sin{\theta}) \xi} Re^{i \theta} \,d\theta \right| \leq \int_{0}^{-\pi} \frac{1}{R^{4} - 4} e^{-2 \pi R i \sin{\theta} \xi} R \,d\theta$

$\leq \frac{R}{R^{4}-4} \pi \rightarrow 0$ as $R \rightarrow \infty$.

For $f(x) = \frac{1}{x^4+4}$, we have 4 poles, 2 of which are inside our region of integration. So $2 \pi i \cdot \sum_{\text{Residues}}\left(\frac{1}{x^4+4}e^{-2 \pi i x \xi},-1-i, 1-i \right) = -\int_{-\infty}^{\infty} \frac{1}{x^4+4} e^{-2 \pi i x \xi} \,dx$.

So we finally have

$- \int_{-\infty}^{\infty} \frac{1}{x^4+4} e^{-2 \pi i x \xi} \,dx = 2 \pi i \cdot \frac{1}{8} i e^{-2 \pi \xi} (\cos{2 \pi \xi} + \sin{2 \pi \xi})$

So $\int_{-\infty}^{\infty} \frac{e^{-2 \pi i x \xi}}{x^4+4} \,dx = \frac{1}{4} e^{-2 \pi \xi} \pi (\cos{2 \pi \xi} + \sin{2 \pi \xi})$.

Use the same argument for $\xi < 0$. and so we have

$\int_{-\infty}^{\infty} \frac{e^{-2 \pi i x \xi}}{x^4+4} \,dx  = \frac{\pi}{4} e^{-2 \pi |\xi|} (\cos{2 \pi |\xi|} + \sin{2 \pi |\xi|})$.

This is the right way to do it (or, at least, one of the best possible ways), but the answer is completely wrong (actually it is even meaningless because $x$ appears in the answer that should depend on $\xi$ only). Everything before the words "So we finally have" seems OK though. The mistake is in the computation of the residues.

If you know the Jordan lemma, it's really easy to see that the integral over the lower curve goes to $0$

[quote="neoscient"]$\int_{-\infty}^{\infty} \frac{e^{-2 \pi i x \xi}}{x^4+4} \,dx = \frac{\pi}{4} e^{-2 \pi |\xi|} (\cos{2 \pi |\xi|} + \sin{2 \pi |\xi|})$[/quote]
Let's check this against the set of expectations I put in post #2:

Real and even? Yes.

Goes to zero as $\xi\to\infty$ faster than any power of $\xi$? Yes, we have that exponential $e^{-2\pi|\xi|}$ to take care of the decay.

$C^2$ but with a jump discontinuity in the third derivative? Time for some Taylor polynomial work:

$e^{-u}(\cos u+\sin u)=$

$=\left(1-u+\frac12u^2-\frac16u^3+O(u^4)\right)\left(1+u-\frac12u^2-\frac16u^3+ O(u^4)\right)$

$=1-u^2+\frac23u^3+O(u^4)$

So, $\frac{\pi}4 e^{-2 \pi |\xi|} (\cos{2 \pi |\xi|} + \sin{2 \pi |\xi|})= \frac{\pi}4\left(1- 4\pi^2\xi^2+\frac{16}3\pi^3|\xi|^3+O(\xi^4)\right)$

That checks as well. The jump discontinuity of the third derivative happens at zero.

Although these checks don't, by themselves, prove the answer is correct, they increase our confidence. And the answer is correct.

I like this as a homework exercise in learning how to really use the residue theorem.


linear algebra
matrix
superior algebra
superior algebra theorems

Theorem: Let A \in M_n(C) then A is similar to a matrix
triangular superior, this means there exist P matrix invertible such that
P^(-1)AP=T where
T=
(t_1 *............*)
(0   t_2*.........*)
........................
(00................t_n)

fairly easy.....

there exist P invertible such that T=P^{-1}AP is Jordan canonical form...
and T is lower triangle here...
Reverse the basis, and T becomes upper triangle...


I need help with some logarithms please.

1.$log_{15}1$

2.$log_{4}256$

3.$log_{4}4^{2x-1}$

4.$log_{5}15$

5.$5^{log_{5}17}$

6.$log_{6}(1/6^{4})^{3}$

7.$log_{9}27^{4}$

8.$log_{4}16^{x-1}$

9.$log_{4}(1/16)$

10.$ln e^{5x-3}$

11.$e^{4 ln(x-2)}$

12.$y^{\sqrt{2}}* y^{\sqrt{18}}$

13.$125^{\sqrt{3}}/ 5^{\sqrt{3}}$

14.$log_{2}256-\log_{2}64$

15.$9^{2x}=27^{x+4}$

16.$log_{5}(4x-5)^{2}=6$

17.$log (x^{2}-9)=\log (5x+5)$

18.$log x-log 8=3$

19.$log_{4096}x =-3/4$

20.$3^{2x}\ge 27^{x+2}$

21.$5^{2x}=20$

22.$A=P(1+r)^{n}$ gives the ammount,$A$, in an account after $n$ years for an initial investment, $P$, that earns interest at an annual rate, $r$. How long will it take for  $\$250.00$ to increase to  $\$500$ at 4% annual interest? (Show work please)

23. Use this formula $A=Pe^{rt}$ to compute the total ammount for an investment of  $\$4500$ at 5% interest compounded continuously for 6 years. $A$ is the ammount after $t$ years, $P$ is the initial investment, $r$ is the rate of interest.

Thank you guys so so much!

If you use the code [code]\log[/code] instead of the code [code]log[/code] it comes out looking much better.

Here are the solutions to the first few, see if you can solve the rest yourself:
[hide] By the definition of logarithm, $log_{x}y=z$ means that $x^{z}=y$.
1.  Since $15^{x}$=1, we can say that x=0, so $log_{15}1=0$
2. Since $4^{x}=256$, we can say that x=4.  Thus, $log_{4}256=4$
3.  Since $4^{y}=4^{2x-1}$, we can say that y=2x-1.  Hence, $log_{4}(4^{2x-1})=2x-1$.
See if you can solve the rest.   [/hide]

[hide]
and another tip is that ln stands for natural logarithm and is the inverse function of e
thus for #10 the answer would simply be 5x-3, because the ln and e cancel each other.
  [/hide]

[hide="20"]
Take the log of both sides.

$\log 3^{2x}\geq \log 27^{x+2}$ Use the Rule of Logarithms that you can bring the power outside of the log:$2x \log 3 \geq x+2 \log 27$

Distribute:

$2x \log 3 = x \log 27+2 \log 27$
Get the X's on one side:
$2x \log 3-x \log 27 = 2 \log 27$
Factor out x from the LHS:

$x (2 \log 3-log 27) = 2 \log 27$
Divide:
$x = \frac{2 \log 27}{2 \log 3-log 27}$
$x = \frac{\log 729}{\log \frac{3^{2}}{27}}$
$x = \boxed{\frac{\log 729}{\log \frac{1}{3}}}$
[/hide]

[quote="helpmeplease"]20. $3^{2x}\ge 27^{x+2}$[/quote]

\begin{eqnarray*}3^{2x}&\geqslant& 27^{x+2}\hfill \\ &\geqslant& (3^{3})^{x+2}\hfill \\ &\geqslant& 3^{3x+6}\hfill \\ 2x &\geqslant& 3x+6 \hfill \\ x &\leqslant&-6 \hfill \\ \end{eqnarray*}


find the minimum of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real $x$.

This is Putnam 2003 question A3, and it surely exists already on the forum.


I wanted to know how to install LaTeX in a wiki like the one of AoPs. I have been trying to do it, but havn't got much further. Any help from anyone would be welcome.

[quote="stevem"]It depends on which Wiki, which renderer, if you have installed LaTeX on the server, your access to the server etc etc. You would need to much more specific about what you are doing and what you have tried.[/quote]

I am using MediaWiki and no I havn't installed LaTeX in the server... :(

[quote="stevem"][url=http://www.mediawiki.org/wiki/Mediawiki_and_LaTeX_on_a_host_with_shell_access]Mediawiki and LaTeX on a host with shell access[/url][/quote]

Thanks very much.....