$n$ nafar dore yek miz neshaste and.be chand tarigh mitavanim $r$ nafar ro entekhab konim ta beistand be sharti ke hich do nafarre mojaveri ba ham boland nashode bashand?

agha ma da moghabel ostad harfi baraye goftan nadarim :wacko:
syboy  :ninja:

[quote="BaBaK Ghalebi"]vala in masale motanazer ba masalei ke mige az adade 1 ta n mikhahim yek zir majmooe r ozvi tashkil bedeim ke hich do adade mojaveri entekhab nashode bashad ba in khasiat ke 1 va n ra niz motavali migirim
dar haghighat adad ra be peimane n misanjim
hale in masale sade ro ham be aghaye yazdanbod mokool mikonim
merC[/quote]
very nice. :D


Three ways:
1. Use a TI-83 calculator's list
2. There is a formula that I can't remember and can't be bothered to derive using sums of squares and some other things.  Some people think it is easier than the formula in the original form.
3. Use excel.

4. Do it by hand???

There is no "easiest way" -- you just have to calculate it.  It's appreciably easier to use a computing device than to do it by hand, I suppose, but the calculation will be the same.  Obviously, if you have either the std. dev. or the variance, you know how to get the other one.

$\sigma_x=\sqrt \frac{\sum^n_{i=1}\left(\bar{x}-x_i\right)^2}{n}$
$S_x=\sqrt \frac{\sum^n_{i=1}\left(\bar{x}-x_i\right)^2}{n-1}$

Use the top for whole populations and the bottom for samples. I think you should use the top if you don't konw the difference.
If you already know the formula, there is nothing to do but work it out.
If you don't have to show [i]any[/i] work at all, stick the "word length x frequency" into one of the lists, and use 1-Var Stats. The screen shows $S_x$ and $\sigma_x$. Again, pick the one you need.

[quote]2. There is a formula that I can't remember and can't be bothered to derive using sums of squares and some other things. Some people think it is easier than the formula in the original form.[/quote]
We have a random variable $X$ which takes on a finite list of values. The probabilty that this takes on the value $x_j$ is $p_j$. Of course, $\sum_jp_j=1.$

The expected value of $X$ is $E(X)=\sum_jp_jx_j.$ The table in the first post is set up to compute this. Note that the sum of the frequencies is 500, so if you just divide by 500, you'll have probabilities, and $E(X)=\frac{2383}{500}=4.766.$

The variance is $\text{Var}(X)=E((X-E(X))^2)=E(X^2-2E(X)X+E(X)^2)$

$=E(X^2)-2E(X)^2+E(X)^2=E(X^2)-E(X)^2.$

You already have $E(X)$ and thus $E(X)^2$ but you need $E(X^2).$

$E(X^2)=\sum_jp_jx_j^2.$

So you need one more column on your table: "(word length)^2 x frequency."

Add that up (and divide by 500): $E(X^2)$. Subtract $E(X)^2$ to get the variance. The standard deviation is the square root of the variance.

You must be sitting at a computer if you're typing this message. Use your spreadsheet program to do this calculation.

[Note: I have glossed over the statistical difference between a population mean and variance and a sample mean and variance; Bictor717 is trying to preserve that distinction.]


If x,y,z>0 then prove that the following inequality holds:
sqrt((y+z)/x)+sqrt((x+z)/y)+sqrt((y+x)/z)>=2[sqrt(x/(y+z))+sqrt(y/(z+x))+sqrt(z/(x+y))].

This ineq had been posted in forum yet (and had been solved). But I have another sln, so I will post it here:

The initial ineq is equivalent to
Sum (y+z-2x)/sqrt(x(y+z)) >=0
WLOG, suppose that  x <= y <= z. Then y+z-2x >=0, x+y-2z<=0 and x(y+z)<=y(x+z)<=z(x+y)
So
Sum (y+z-2x)/sqrt(x(y+z)) >= (y+z-2x)/sqrt(y(x+z)) + (x+z-2y)/sqrt(y(x+z)) + (x+y-2z)/sqrt(y(x+z)) = 0.

Namdung


Points L, M, and N lie in a plane. If the distance between L and M is 4, and the distance between M and N is 9, which of the following could be the distance between L and N?

The choices are 5, 8, and 3

I set up the points to form a triangle (LMN). It's a special right triangle.

The base = 4

One side = 9

(2x)^2 = 81 + 16

4x^2=97

x^2 = 24.25

x = 4.92

The answer is 5. If that is right, is 5 the only possible answer?
Can it be set up another way?

[quote="Bictor717"]I'm not sure why you assume the points form a right triangle, or why you use 2x instead of x.
This looks like a simple application of the triangle inequality.[/quote]

I'm confused and need help. It's not so simple for me.
Sorry.

[quote="Zhero"]Just note that they always form a triangle of some sort (including degenerate triangles). Hence, they satisfy the triangle equality, i.e., $ ML \plus{} LN \geq MN$, i.e., $ 4 \plus{} LN \geq 9$, i.e., $ LN \geq 5$. Hence, 5 and 8 are solutions, while 3 is not.

In the case that $ LN \equal{} 5$, the points are colinnear (on the same line). When $ LN \equal{} 8$, we simply have a triangle with side lengths 8, 9, and 4.[/quote]
Thank you for your kindness.


Prove that :                                                                                $\displaystyle\frac {a-b}{lna-lnb} < \frac{1}{3}(2\sqrt{ab} + \frac{a+b}{2})\forall a,b>0$

hello, it is equivalent to
$\frac{x-1}{\ln(x)}<\frac{1}{3}\left(2\sqrt{x}+\frac{x+1}{2}\right)$ with $x=\frac{a}{b}$
Sonnhard.


I'm interested in seeing how people submit their solutions.

Handwritten...I write huge so I used 24 pages for round 1, and I didn't even come close to complete solutions for #4 and #5...Imagine if I got every problem, maybe like 40 pages.

I hand written last year mostly because I didn't know how to write in $\LaTeX$, but I learned, so this year I'm typing and emailing my solutions.

I did neither ... I didn't handwrite b/c my handwriting is awful, but i was having trouble getting TeX to work.  So I typed them in word and used te TeXeR to typeset my math.

I know they dont want Word emailed to them, so i printed it and snail-mailed it.

JB

I latexed my solutions. I had like 50 million problems at first, but eventually I figured out how to fix them and now I am good at latex...

I latexed my solutions, but I had some problems at first. Now, however, I know how to compile the solutions file correctly, so it will take a lot less time to type now.

I really suggest you learn LaTeX and how to use LaTeX commands in TexnicCenter.  It is useful for USAMTS as well as general AoPSing.  Honestly, the hardest part is installing TexnicCenter, and I can help you with that.

TexnicCenter is great. Before, I had been doing it on a Unix server, but it took a long time with SSH and WinSCP; now, it's all on my computer. Yay!

BTW, MikTex didn't work at first, and then it suddenly did... ???

[quote="matt276eagles"]This poll doesn't accurately represent the average USAMTS participant, since most AoPS users already know how to use LaTeX from posting on the forums.  I would guess that well over half of the people who participate handwrite their solutions.[/quote]
Well, if he wants to know for the average USAMTS participant, I think Mr. Patrick or Rusczyk could answer that. Or maybe not...

[quote="matt276eagles"][quote="Boy Soprano II"]Eventually, statistics from each round will be posted on the "Leaders" page of the USAMTS website.[/quote]

Yes, but I doubt the statistics on the leaders page will include the percentage of people who used LaTeX.  :lol:[/quote]
Oh, right, whoops.  (Shows what happens when you don't consider what thread you're posting in, doesn't it?)


Consider a sequence $ x_1, x_2, x_3, ...$, defined by
\begin{align*}x_1 &= \sqrt [3]{3}\\
x_2 &= \sqrt [3]{3} ^ {\sqrt [3]{3}},\end{align*}and in general
\[ x_n \equal{} (x_{n \minus{} 1}) ^ {\sqrt [3]{3}}\,\,\text{ for }\,\,n > 1.
\]What is the smallest value of $ n$ for which $ x_n$ is an integer?

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 27$

[hide="A Little More Explanation?"]$ a^{bcd} \equal{} a^{{{b}^c}^d}$.  Obviously, $ b \equal{} c \equal{} d \equal{} \sqrt[3]{3}$, so we have $ bcd \equal{} (\sqrt[3]{3})^3$, or just $ 3$.  So we end up with $ (\sqrt[3]{3})^3$, which is indeed the smallest possible integer in the sequence.[/hide]

Sorry for the bump. Can anyone give a full solution? I'm struggling with this problem.


number theory
relatively prime
number theory unsolved

If $ g$ and $ g'$ are primitive roots $ (\bmod p)$ for some odd prime $ p$, can $ gg'$ be a primitive root $ (\bmod p)$?

If $ g$ and $ g'$ are primitive roots, then $ g^{\frac{p\minus{}1}{2}}\equal{}\minus{}1$ and $ g'^{\frac{p\minus{}1}{2}}\equal{}\minus{}1$, so $ (gg')^{\frac{p\minus{}1}{2}}\equal{}(\minus{}1)^2\equal{}1$, so $ gg'$ cannot be a primitive root.

If $ g$ is a primitive root, then $ g' \equal{} g^k$ for some $ k$ relatively prime to $ p\minus{}1$, hence $ k$ is odd.  But then $ k\plus{}1$ is even.  (This is essentially CaptainCanada's solution.)


[color=red]Let $ \triangle ABC$ be a triangle and $ \omega_{B},\omega_{C}$ be the $ B,C$-excircles respectively.
Let $ A_{1},B_{1},C_{1}$ be the points of tangency $ \omega_{C}$ with $ CB,CA,AB$ respectively and $ A_{2},B_{2},C_{2}$ be the points of tangency $ \omega_{B}$ with $ CB,CA,AB$ respectively.
Denote $ P\equal{}A_{1}B_{1}\cap A_{2}B_{2}$,$ Q\equal{}A_{1}C_{1}\cap A_{2}C_{2}$.Prove that $ A,P,Q$ are collinear.[/color]

I think there is a typo.

The points $ A_{2},$ $ B_{2},$ $ C_{2},$ you denoted, must be on $ CB,$ $ AB,$ $ CA,$ respectively, instead of $ CB,$ $ CA,$ $ AB.$

So, this problem has already been posted lately as [b][size=100]Lemma[/size][/b], at http://www.mathlinks.ro/Forum/viewtopic.php?t=164092 , and the points $ P,$ $ Q$ ( corresponding to the points $ P,$ $ R,$ there ), lie on the $ A$- altitude of $ \bigtriangleup ABC.$

However, if there is not any typo, then your points $ P,$ $ Q$ ( corresponding to the points $ K,$ $ L,$ there ), lie on the line connecting the centers of the excircles $ (I_{b}),$ $ (I_{c}).$

Kostas Vittas.


How many four-digit numbers can be written with the digits 1, 2, 3 so that no two of them compared coincide at more than one position? For example, 2131 and 2233 are not included because they coincide at both the thousands and tens positions.


Pascal\u0027s Triangle
binomial coefficients

How do I find out the summation equation for the sum of the terms in the 80th row of pascal's triangle?

One approach is to experiment.

$ 1 \equal{} 1$
$ 1 \plus{} 1 \equal{} 2$
$ 1 \plus{} 2 \plus{} 1 \equal{} 4$
$ 1 \plus{} 3 \plus{} 3 \plus{} 1 \equal{} 8$
$ 1 \plus{} 4 \plus{} 6 \plus{} 4 \plus{} 1 \equal{} 16$

Do you see a pattern?  Evidently, if we start counting from $ 0$, the sum of the terms in the $ n^{th}$ row of Pascal's triangle is $ \boxed{ 2^n }$.

This can be proven in several equivalent ways.

One way is to consider how Pascal's triangle is generated:  each subsequent row is formed by adding pairs of terms from the above row, so [i]every term in the $ k^{th}$ row appears exactly twice in the sum of the $ (k \plus{} 1)^{th}$ row.[/i]  Conclusion: the sum doubles every time we move to a higher row.

Another is directly:  knowing that the terms in Pascal's triangle are binomial coefficients,

$ (1 \plus{} 1)^n \equal{} {n \choose 0} \plus{} {n \choose 1} \plus{} {n \choose 2} \plus{} ... \plus{} {n \choose n} \equal{} 2^n$

by the Binomial Theorem.

Yet a third way is by a counting argument.  Since binomial coefficients give the number of ways to choose a subset of size $ k$ from a set of size $ n$, given a set of $ n$ distinct elements, we can calculate the total number of subsets of that set in two different ways:

- Any element of the set is either in a given subset, or it is not in a given subset.  Since there are $ 2$ choices for each of $ n$ elements, there are a total of $ 2^n$ subsets.
- There are $ {n \choose 0}$ subsets of size $ 0$, $ {n \choose 1}$ subsets of size $ 1$, $ {n \choose 2}$ subsets of size $ 2$, and so forth.

This produces the above identity.


Gauss
geometry
circumcircle
geometry proposed

Given a cyclic quadrangle $ABCD$ . Let $AD\cap BC\equiv F$ (outside the quadrangle) and $AC\cap BD\equiv E$ . Let $I,M,N$ be the midpoint of $EF,AB,CD$ respectively. Prove that : $\triangle IME$ is similar to $\triangle IEN$

In other words, we'll have to show that $IM\cdot IN=IE^2$ (this is easily seen to be equivalent to the conclusion if we keep in mind that $I,M,N$ are colinear; since this is the Gauss line of the quadrilateral $ACBD$), or prove that $M,N$ are the inverse of one another in the circle with diameter $EF$.

Let $T=AB\cap CD$, and let $O$ be the circumcircle of $ABCD$. We must show that the circle with diameter $OT$ is invariated by the inversion through the circle with diameter $EF$. This can be achieved by showing that if $T'$ is the image of $T$ through this inversion, then $OT'\perp IT$.

I'll use some facts which I won't prove, either because they are not hard to prove or because they have been proven before on the forum. First of all, the circle with diameter $EF$ is orthogonal to the circumcircle of $ABCD$ (as are the circles with diameters $FT,TE$). Second of all, given a point $X$ not on a circle $\mathcal C$ with center $O$, then the inversion of pole $X$ which invariates $\mathcal C$ turns the polar of $X$ wrt $\mathcal C$ into a circle with diameter $OX$. We apply this with $X=I$ and use the fact that $T$ lies on the polar of $I$ wrt $(ABCD)$ because the polar of $T$ wrt this circle is $EF$, which contains $I$.

Sorry for the lack of details.

Here is the diagram of my solution. I will write up the details later, because now I'm busily revising for my biology test on friday... :(
Edit:
First, note that I,M,N are collinear, EM/EN=AB/CD=FM/FN and $\angle ENF=\angle EMJ$. Thus by similar triangle, $\angle IFN=\angle EJM=\angle IMF$. The conclusion is trivial then.


Find all those 2-digit numbers that are divisible by both the sum and the product of their digits.

  Heres a question that is unrelated to the problem:  Have any of you read "Mathematics of Choice; how to count without counting" by Ivan Niven?

[hide]Let's see.... 12, 24, and 36.... ummm.... my method.... I just went through all of the numbers in my head. [/hide]

By the way, those two are same book. ;)

It's "Mathematics of Choice or How to Count Without Counting" by Ivan Niven.

As for answering the question, I'm getting that book to self-study for the counting. I heard it's good but unless you had some basic counting, I suggest taking Intro AoPS class instead.

Since you're probably (correct me if I'm wrong) in level of middle school, just take AoPS class and look over AoPS book.

If you want to know more about the book, search the forum. I remember posting about this book sometime ago.

I have a rather clumsy method of arriving at the same solution.  My reasoning runs as follows :
[hide]All the numbers must be even, because if it is an odd number and the tens' digit is odd, the sum is even.  On the other hand, if it has an even tens' digit, the product will be even.
No multiples of eleven can fit, because if [i]x[/i] is the digit, then 11[i]x[/i] must be divisible by 2[i]x[/i].
No multiples of five, because no number is divisible by zero.
If the tens' digit is odd, the ones' digit must be 2 or 6, since 4 or 8 would make it divisible by four, and that cannot occur in that form.  If the tens' digit is even, the ones' digit must be 4 or 8, since it must be divisible by four.  This implies that the number must be divisible by eight, which means that if the tens' digit is a multiple of four, the ones' digit must be 8, but if it is an even non-multiple of four, the ones' digit must be 4.
Taking the first, third, and second arguments, we can eliminate all but 14 of the 90 original numbers, one of which (88) can quickly be eliminated because it is a multiple of 11.  Testing these thirteen numbers, we arrive with the conclusion that 12, 24, and 36 are the only numbers that fulfil these conditions.[/hide]
Does anybody have a better solution?
No, I have not read that book.


geometry
inequalities proposed
inequalities

[quote="gemath"]Given triangle $ABC$ area $S$ prove that
$a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\ge 4\sqrt{3}S$
it is stronger than
$ab+bc+ca\ge 4\sqrt{3}S$  :)[/quote]
$a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\geq2ab+2ac+2bc-a^{2}-b^{2}-c^{2}\geq4\sqrt{3}S.$ :wink:

ok that's right it is too simple :D, but it will have more meanings if we know it is a special case of the following:
$xa\sqrt{bc}+yb\sqrt{ca}+zc\sqrt{ab}\ge 4\sqrt{xy+yz+zx}S,\forall x,y,z>0$,


I have put together a contest out of different questions that are from various gauss contests, which I will post later. Please sign up for the contest here. When we get about 10-20 people, I will post the questions. Pm the answers to me. You have 1week after I post them to finish the contest and pm me the answers. Any type of calculator may be used, But you may not look for question answers on the internet. You also may not work wih other people. And finnaly, you must be in grade 8 or lower. Sign Up!

[u]Users Signed Up[/u]
1.shentang
2.tinytim
3.Yongyi781
4.miller4math
5.leoxnlin
6.hm29168
7.dynamo729

It is a canadian math contest, multiple choice, for grade 7 and 8 students.
[url]http://cemc.math.uwaterloo.ca/english/contests/gauss.shtml[/url]

/me signs up.

Oh wait... let me try again.

Yongyi781 signs up.

 :D

I remember this... I got a perfect on Gauss 7 last year.
Hmm, I don't really want to sign up since I've already done most of the old questions.

meh. doesn't seem too hard. (AMC 8/ 10?)
anyways, i'll do it

@xpmath
This will be a lot harder then the normal gauss, as I am doing nothing but part c questions, and I am including some grade 8 ones.

SIGN ME UP PLEASE!!!! (I am in 7th grade)

I don't think he's gotten on since the due date. My test still is in my outbox.

I've been on since this thing started, every day.

don't worry, I'm marking right now.
The answers were at my school, and I didn't get a chance to mark untill know.

Alright, the scores are in!
IN order from last to first:
In 9th place: neurosurgeon1118
In 8th place: Seventh Secret
In 7th place: Shentang
In 6th place: 12Markkram34
In a tie for second place, with a total of 23 out of 25, Ajain, Lotsofmath, Ac-King, and yongi741

And finnaly, in first place, with a total of 24
[hide="And the winner is..."]Moogra![/hide]

Congratulations to anyone who took the time to write the contest. If anyone wants to know there score, then just ask!

Also, Bogtro, I was unable to get the contest to format correctly, whenever I posted the whole thing, it took out the first half of the questions. sorry.

The answers are still at school, on the computer, but I will tell you when I see them again.

What was my score? and can i get my test back with the ones i got wrong?

I will send out everyone score and paper tommorow

Heh I already know that I got 22 correct. I solved those 22 then guessed randomly on the other 3 because I didn't want to take the time to solve them.

Wow I won? I didn't see that coming, I thought everyone beat me once you announced that the answers were on the web

[quote="neurosurgeon1118"]I noticed that it said contest number one in the title, so I was wondering if their are going to be more, somehow considering the amount of time you have already put into this, this is the only one.[/quote]
I am in the prcess of making a knew one right now.

So are the tests going to be out today also, do they have memorial day in Canada?

You guys, all of the answers/questions have been erased from my school (stupid deep freeze) so I can't give any more answers out. Sorry.


John A. Wheeler, Physicist Who Coined the Term ‘Black Hole,’ Is Dead at 96

John A. Wheeler, a visionary physicist and teacher who helped invent the theory of nuclear fission, gave black holes their name and argued about the nature of reality with Albert Einstein and Niels Bohr, died Sunday morning at his home in Hightstown, N.J. He was 96.

Full story here:  http://www.nytimes.com/2008/04/14/science/14wheeler.html?_r=1&oref=slogin

I heard about it from a friend.  Very sad.


geometry
area of a triangle
geometry solved

Hi all,

I'm working on the review problems in Chapter 11 of the AoPS, Vol: 1 book. I've solved all of them but one, and am having much difficulty in solving this last one. It basically asks that given an interior point (let's call it P) in an equilateral triangle, prove that the sum of the lengths of the perpendiculars from P to the three sides is equal to the altitude of the triangle.

I was able to prove it easily using the centroid(/orthocenter/circumcenter/incenter, since it's equilateral), but I doubt that this is strong enough to hold as a general theorem for any interior point, since these points have so many special properties. I was hoping someone could give me some guidance in starting this problem? Any help is greatly appreciated. Thank you! :)

It is so easy :) . You just consider the area of the triangle: it is s/2 where s is the sum of perpendiculars and the side length is 1. On the other hand the area is 1*h/2. Now you have h/2=s/2.

Misha

I'm afraid I don't understand. Why is the sum of the perpendiculars over 2 equal to the area of the triangle when the side length is 1? I don't really see the logic behind that.

If ABC is the equilateral triangle with sidelength BC = CA = AB = 1, and P is an arbitrary point inside it, call x, y, z the lengths of the perpendiculars from P to the sides BC, CA, AB. Then you ask why $\frac{x+y+z}{2}$ equals the area of triangle ABC. Well, since the area of a triangle equals $\frac12$ $\cdot$ side $\cdot$ corresponding altitude, it follows that the area of triangle BPC equals $\frac12$ $\cdot$ side BC $\cdot$ altitude to side BC. In other words, the area of triangle BPC is $\frac12 \cdot 1 \cdot x = \frac{x}{2}$. Similarly, the area of triangle CPA is $\frac{y}{2}$, and the area of triangle APB is $\frac{z}{2}$. Hence, the area of triangle ABC, being the sum of the areas of triangles BPC, CPA and APB, must equal $\frac{x}{2}+\frac{y}{2}+\frac{z}{2} = \frac{x+y+z}{2}$, and the proof is complete.

Sorry if the above was too sketchy (I am in hurry right now).

  Darij


function
calculus
derivative
blogs
algebra proposed
algebra

Let $ s_p(n)$ denote the sum of the digits of $ n$ when written in base $ p$.  Compute $ \sum_{n \ge 0} s_p(n) x^n$.

Some possibility - $ \prod_{k \equal{} 0}^{\infty} (\sum_{j \equal{} 0}^{p \minus{} 1} y^jx^{j p^k})\equal{}\prod_{k \equal{} 0}^{\infty} \frac{(yx^{p^k})^p\minus{}1}{yx^{p^k}\minus{}1}$
Now we need to differentiate in respect to $ y$ and put $ y \equal{} 1$... Not so sure how it's done.

[EDIT: oh noes. My post count is not a Fibonacci number anymore...]

Applying the product rule formally might give the same answer as mine (which is an infinite sum in terms of $ k$).  I had the same thought, but also have another more indirect solution.

Bambaman, could you explain the idea behind all that? =)

The "unique base-$ p$ representation generating function" is $ \frac{1}{1 \minus{} x} \equal{} 1 \plus{} x \plus{} ... \equal{} (1 \plus{} x \plus{} ... \plus{} x^{p\minus{}1})(1 \plus{} x^p \plus{} x^{2p} \plus{} ... \plus{} x^{(p\minus{}1)p})(1 \plus{} x^{p^2} \plus{} ...)...$.  Each factor accounts for a different digit; the choice of each factor determines the value of the digit, i.e. $ 1$ is a units digit of $ 1$ and $ 2p$ is a tens digit of $ 2$.  Each coefficient is $ 1$ by uniqueness.  Now, the easiest way to keep track of the sum of the digits is to introduce a factor $ y^j$, which gives

$ f(x, y) \equal{} (1 \plus{} yx \plus{} y^2 x^2 \plus{} ... \plus{} y^{p\minus{}1} x^{p\minus{}1})(1 \plus{} yx^p \plus{} y^2 x^{2p} \plus{} ... \plus{} y^{p\minus{}1} x^{p(p\minus{}1)})... \equal{} \sum_{n \ge 0} y^{s_p(n)} x^n$

which is almost what we want:  the original question is recovered by differentiating with respect to $ y$ and substituting $ y \equal{} 1$.

My calculation, which might be wrong, gives
$ \sum_{k \ge 0} \frac{x^{p^k}\minus{}1}{(x^{p^{k\plus{}1}}\minus{}1)(1\minus{}x)}\frac{(px^{p^{k\plus{}1}})(x^{p^k}\minus{}1)\minus{}x^{p^k}(x^{p^{k\plus{}1}}\minus{}1)}{(x^{p^k}\minus{}1)^2}$
How does one justify the use of the product rule here?...
Fermat's Little Turtle - check out the following article for some background and examples for generating functions in 2 variables - [url]http://reflections.awesomemath.org/2006_2/2006_2_zach.pdf[/url] .

[quote="bambaman"]How does one justify the use of the product rule here?...[/quote]
Formally, I think the quickest way is to verify the validity of logarithmic differentiation, at least for a certain class of infinite products that can be evaluated formally.  Anyway, that looks more or less correct.  I'll be writing up my approach as part of a blog post soon.


Ben owns five pairs of socks, and each pair is a different color.  If he loses three socks, what is the probability that he has only two complete pairs left?  Express your answer as a common fraction.

[hide]let the socks be divided into 5 groups where there are two socks in each group such as:
A B C D E
A B C D E       (Fig 1.)

The first sock chosen may be any of them so hypothetically let us pick the first A, leaving

   B C D E       (Fig.2)
A B C D E

The probability of any sock being chosen except the last A from figure two is 8/9

Now let us pick another, but it must not be the A because if we only have three socks to pick, then each sock must be from a different group.

Hypothetically we may pick B so the diagram would look like:

     C D E  (Fig.3)
A B C D E

Now let us take another Sock from a different group that is not A or B
looking at figure three we see that the probability of picking  one under these conditions is 6/8.

Now multiplying our probabilities together:

$ \frac {8}{9}*\frac {6}{8} \Longrightarrow \frac {8}{9}*\frac {3}{4} \equal{} \boxed{\frac {2}{3}}$[/hide]

Or..... you could be smart and realize that there are 20 (5 p 3) Because we are only counting the sock pairs and eliminating them if one is chosen.(Hope you see why). Then, we put that over or regular denominator, 5*4*3 and presto chango, we get 1/3. Just a way to think ahead. Despacito reached 4 bil yesterday. CRI :(


inequalities
geometry
3D geometry
tetrahedron
circumcircle

The circumradius of a tetrahedron $ ABCD$ is $ R$, and the lenghts of the segments connecting the vertices $ A,B,C,D$ with the centroids of the opposite faces are equal to $ m_a,m_b,m_c$ and $ m_d$, respectively. Prove that:

$ m_a\plus{}m_b\plus{}m_c\plus{}m_d\leq \dfrac{16}{3}R$


[quote="Eli"]Eli can solve 15 problems in one hour.  Joe can solve 10 problems in one hour.  Eli and Joe work together to solve some problems for the first 48 minutes.  Joe then stops working and plays Halo for the rest of the time (that lazy fool!).  If there were 100 problems for homework to do and Eli and Joe start working at 8:00 PM, how many minutes will Eli spend slaving into the night finishing the homework? (I made this problem up in case you're wondering.  Any resemblances to names, usernames, people, or homework is purely coincidental.)

[hide="Answer"] Eli can do one problem in 4 minutes.  Joe can do one problem in 6 minutes.  Together, they do 12 problems (Eli) + 8 problems (Joe) = 20 problems before Joe leaves.  This means that Eli has to do 80 more problems.  80 problems * 4 minutes per problem = 160 minutes.[/hide]

Bonus problem: Eli charges Joe 1 dollar for every 160 minutes Joe doesn't work on homework problems with Eli.  If each homework assignment is 100 problems long, after how many homework assignments will Eli become a millionaire? (Pretend that the conditions in the problem are always the same)

[hide="Answer"] 160 minutes * 6250 homework assignments of 100 problems each = 1,000,000.  Therefore, after 6250 homework assignments, Eli will be rich and Joe will be paying through his nose.  Not that Eli would be willing to do so much extra homework (they were [b]supposed[/b] to split the problems equally.[/hide]

I hope I answered my questions correctly. :D[/quote]
[hide="#1"]Everything is correct, but you multiplied incorrectly. $80\cdot4=320$[/hide]

[hide="#2"]Because of your mistake in the first problem, this answer is twice as big as it should be. $1000000\div320=3125$[/hide]

Oops....

20 points to i_like_pie for answering the problems correctly and finding the mistake! :lol:


function
superior algebra
superior algebra unsolved

Hello,

let $k$ be a field.
Consider the field$K$ of all rational functions in $n$ variables $x_{1},\cdots,x_{n}$

Prove that $x_{1},\cdots,x_{n}$ is a transcendence basis of $K$ over $k$

(By transcendence basis I mean : a set of algebraically independent elements over $k$ , such that $K$ is algebraic over the field they generate together with $k$).

This isn't homework  :( , it's just me trying to catch up on some algebra.  This should be straightforward, but it isn't for me. :oops:
Thanks,
fredbel6

use/proof  the following notion of algebraically independentness:

$x_{1}, ... , x_{n}\subseteq L$ are algebraically independant over $K$ if and only if the canonical map $K[T_{1}, ... , T_{n}] \to L, T_{i}\mapsto x_{i}$ is injective, i.e. induces $K(T_{1},...,T_{n}) \cong K(x_{1},...,x_{n})$ (where the $T_{i}$ are indeterminants. this definition manifests that algebraically independence is the concept of indeterminants).

Okay woops,thanks, that was easy but that wasn't my problem. :oops:
This morning I was told that transcendence degree means : the smallest possible size of a set, algebraically independent, and generating a field over which the total field is algebraic.

So my questions is really : why does it have to have at least $n$ elements. :huh:

it's exactly n, because x1,...,xn is a trancencence basis ...

your definition is not very practical :-). one has to prove before that every algebraic independent set can be extended to a transcendence basis.

[quote="-oo-"]it's exactly n, because x1,...,xn is a trancencence basis ...

your definition is not very practical :-)[/quote]
O no no, I'm creating confusion.
You have already proven that that set (which is pretty simple) is a basis, so the minimal size of a basis would have to be at most $n$.

But there seems to a theorem stating that all bases have equal cardinality.....which I haven't seen either in any course? Is the proof of that straightforward (that would solve a lot of problems) :ninja:

it's not easy, but you can find it in a lot of algebra textbooks, i think :-)

[quote="ZetaX"]The proof uses mainly the same ideas like that one for linear bases in linear algebra. Especially that you can replace any element of the basis by others.[/quote]
Well, Wikipedia says :
[url]http://en.wikipedia.org/wiki/Transcendence_degree[/url]
[quote]The fact that transcendence bases always exist (like the fact that bases always exist in linear algebra) requires the axiom of choice. The proof that any two bases have the same cardinality depends, in each setting, on an exchange lemma.[/quote]
 :| What is an exchange lemma? :blush:


calculus
derivative
conics
hyperbola
calculus computations

Given is an hyperbool $xy=16$ so $y=\frac{16}{x}$
and they ask me to find a tangent on the hyperbol so that the distance from p to q will be so small as possible.

So let me try first of all I sketch the problem.

[img]http://img53.imageshack.us/img53/7428/hopeloosbis0xk.jpg[/img]

We know that a tangent will have the equation of the form y=ax+b so that $a=-16*x^{-2}$

Then we know that a good tangent is  $\frac{16}{x}=-16x^{-2}*x+b$

And we have that b is the distance on the y-axis. $\frac{32}{x}=b$
The distance on the x-axis is given by x self.

So we can make a formula for the distance between p and q we get $D=\sqrt{x^{2}+(\frac{32}{x})^{2}}$

And know if I derivate that and set it equal to zero I get 5 as a solution But I need to find 4 Where do I miss?

Thanks.

Bert, first If you set the equation of the tangent line as $y=ax+b,$ you shoud set the tangent point  ,say $\left(p,\ \frac{16}{p}\right)\ (p>0).$

As for me, I usually use $t$ as the abcissa of the tangent pont.

If you dare to use the tangent point as $\left(x,\ \frac{16}{x}\right)\ (x>0),$

You will write down the tangent line as $Y=aX+b$ which you mentioned.

Anyway you are wrong the length of $Op.$

Finally,  hyperbool$\longrightarrow$ hyperbola. :)

oké thanks.

But I will calculate the smallest distance from P to Q is this formula correct ? $d=\sqrt{x^{2}+(\frac{32}{x})^{2}}$

If I derivate this $\frac{D}{dx}=\frac{2x-2048x^{-3}}{2 \ \sqrt{x^{2}+(\frac{32}{x})^{2}}}$

Then I say $0=\frac{2x-2048x^{-3}}{2 \ \sqrt{x^{2}+(\frac{32}{x})^{2}}}$

I this oké? So I'm not looking from the length of Op but the lengt of P to Q.

Thanks.

You are wrong, The length of $OP$ must be $2x.$ :)

yes I see it now.

So I have worked a few hours on this and I don't see it so

new sketch [img]http://img371.imageshack.us/img371/3324/hopeloosbistera9hw.jpg[/img]

Is this the only mistakke?  Cheers Thanks  :lol:

Let the point of intersection of the hyperabola and the line be $\left(p,\frac{16}{p}\right)$, then the equation of the line is given by $y=-\frac{16}{p^{2}}x+\frac{32}{p}$  Then we have $P=(2p,0)$ and $Q=(0,32/p)$, so $PQ = \sqrt{4p^{2}+32^{2}/p^{2}}$ taking the derivative and setting it equal to zero gives $p=\pm 4$, since we are interested in the first quadrant we have $p=4$ and thus $PQ=8\sqrt{2}$.

oké thanks

But why [img]http://www.mathlinks.ro/Forum/latexrender/pictures/a10a99454b14e18ca4e165e3b616ef29.gif[/img] will we have $\frac{32}{p}$ in this equation?

Thanks. :)  You can also use $A.M.-G.M.$  :)

I see that the lenght from O to P is not x but who can you proof that this lenght will be 2x?

[quote="kunny"]The tangent line at the point $\left(p,\ \frac{16}{p}\right)$ is $y=-\frac{16}{p^{2}}(x-p)+\frac{16}{p}.$[/quote]

Substitute $y=0$ for this equation, you will obtain what you want. :)

but who do they come to this factor in this equation?

[img]http://img157.imageshack.us/img157/1906/factor3ph.jpg[/img]

Thanks.

Edit: maybe I find them the lengt of b in the equation $y=ax+b$

will be the y value of the hyperbola plus the point p multiply by $\frac{16}{p^{2}}$

 [img]http://img92.imageshack.us/img92/8404/hopeloosbisf0th.jpg[/img]

thus the lengt from O to q is $\frac{16}{p}+\frac{16}{p}$

But Why told  someone that the constant term in the equation of the tangent will be $\frac{16}{p}$ and not $\frac{32}{p}$

Thanks.


calculus
integration
logarithms
real analysis
real analysis unsolved

$ \int\limits_{ \minus{} 1}^{1} \frac {dx}{\ln (1 \plus{} \sqrt [3]{x})}$

 and $ \int\limits_{2}^{\infty} \frac {x}{\ln x} dx$

The first integral converges, and the second one diverges.

The first integral has singularities at both $ 0$ and $ \minus{}1.$ Near $ 0,$ the integrand resembles $ \frac{1}{\sqrt[3]{x}},$ and that's a slow enough rate of increase for the integral to converge in that neighborhood. The singularity at $ \minus{}1$ is logarithmic, with the logarithm in the denominator. The integrand actually tends to zero there, creating no obstacle to convergence.

For the second integral, the integrand tends to $ \infty$ as $ x\to\infty,$ so the integral diverges by comparison to the integral of $ 1.$


Hello everybody,
                             This time I got stuck up with this problem. It demands me to devise a linear time complexity function to find the majority element in an integer array.
               For an array of dimension n we call an element the majority element if it appears more than n/2 times. Our task is to return the majority element if it exists else return -1 keeping in view that the time complexity of our algorithm is in O(n).

Thanks in advance...[/b]

Well, I have an idea which should work if you think through the details.

Let's have a list of N numbers, let's assume that N is even for now. Pair up the elements in an arbitrary way. Make a new (empty) list of numbers, and for each pair of original elements, if they are equal, insert their value into the new list (only once for each pair!).
Now, let's say x is a majority element and in the new list, there are K elements which are NOT x. Then, there were 2*K elements (K pairs) in the original list that were not x. In the remaining N-2K elements (N/2 - K pairs) there are at least N/2+1 elements equal to x. According to the Dirichlet principle / pigeonhole principle (whatever you call it, it's just common sense), there are at least K+1 pairs where both elements are equal to x (otherwise there are maximum of 2K+(N/2-2K) x's), therefore in the new list, x is also a majority element. It is clear that the size of the new list is not larger than N/2.

Therefore IF the original list has a majority element, it is the majority element of the new list. We can use the same procedure to find the majority element for the new list. If we do this recursively, we end up with a list with constantly many elements (for example 1) and we can check those on time O(1). The time required to solve this for N numbers is then T(N) = O(N) + T(N/2), which, I believe, yields T(N)=O(N).

Of course, the "IF" clause in the last paragraph implies that we must in the end check whether the majority element of the new list is really a majority element of the original list, but it's enough to do this only once - at the end, in time O(N).

Also, there is a nontrivial case of ending up with odd N (which will eventually happen if N is not a power of two). This can be solved for example as follows: pick one element (E), remove it from the list and calculate the majority element of the remaining numbers. If it exists, adding E to the list does not change it being the majority element (remember, N is odd). If it does not exist, adding E to the list might cause E (but no other number) to become the majority element, therefore we should check if E is not a majority element. This all can be done in O(N) time, it doesn't spoil our recurrence relation nor the total O(N) time performance.

I hope my explanation is clear enough ...

I also know an easier solution that has O(N) and uses only O(1) additional memory. Scan the array from left to right and keep at each step the number that is candidate for and a counter. If the next number equals the candidate increase the counter, and if it is different decrese the counter - if the counter reaches 0 then change the candidate to the current number. In the end, the candidate will be the majority element - i think it can be easily proven that , if a majority element exists, the algorithm will finish by finding it. Here's some code:

[code]    res = -1; cnt = 1;
    while (!feof(stdin))
    {
        scanf("%d", &n);
        if (n != res)
           cnt--;
        else
           cnt++;
        if (cnt == 0) res = n, cnt = 1;
    }[/code]

Thank You, Kubus. Infact, your idea is likely to work and, forgive me for what I ask, but can you find some time to post your code for the problem as well? If you do, then kindly post it too.

Hm .. I was hoping my idea would work ...
... but domino's solution is much simpler. I don't have any code to my solution, and I don't feel like producing one;)

Not bad idea kubus but compared to domino's answer just way too complex :) . However, domino's algorithm has one minor flaw and that is that it can't tell whether the majority element really exists. Of course, you can check this in O(N) time but you'll lose the constant memory. Nevertheless, I'd use domino's source code if you need one, Akash.
And by the way, it took me quite a while to prove that your solution really gives the right answer. Is there any simple way to prove this, anybody?


inequalities
special factorizations
inequalities proposed

I proved that $\sum\frac{a}{b+c}\cdot (y+z)\geq\sum\sqrt{(x+y)(x+z)} -(x+y+z)$ ;)

Harazi inequality doesn't help us here. The inequality from the second post is stronger. ;)

what is your proof for the ineq you write in the second post?

Harazi solved these problems (Thuan's problem and Cezar's problem) long time ago  ;)  They were put in Old and New ineqs , problem 101 , p.94  ;)

[quote="cezar lupu"]I proved that $ \sum\frac{a}{b\plus{}c}\cdot (y\plus{}z)\geq\sum\sqrt{(x\plus{}y)(x\plus{}z)}\minus{}(x\plus{}y\plus{}z)$ ;)[/quote]
We rewrite the LHS of ineq given by cezar lupu in the form
\[ (a\plus{}b\plus{}c)\Bigl\{\frac{x\plus{}y}{a\plus{}b}\plus{}\frac{y\plus{}z}{b\plus{}c}\plus{}\frac{z\plus{}x}{c\plus{}a}\Bigl\}\minus{}2(x\plus{}y\plus{}z).\]
By Cauchy-Schwatz inequality, we have
\[ (a\plus{}b\plus{}c)\Bigl\{\frac{x\plus{}y}{a\plus{}b}\plus{}\frac{y\plus{}z}{b\plus{}c}\plus{}\frac{z\plus{}x}{c\plus{}a}\Bigl\}\geq\tfrac12(\sqrt{x\plus{}y}\plus{}\sqrt{y\plus{}z}\plus{}\sqrt{z\plus{}x})^{2}.\]
Completing the square, we get the desired inequality.

[quote=silouan]Also from an ineq which was constructed by Harazi we have $\frac a{b+c}(y+z)+\frac b{c+a}(z+x)+\frac c{a+b}(x+y)\geq\sqrt{3(xy+xy+zx)}$[/quote]
http://www.artofproblemsolving.com/community/c6h591007p3503218

[quote=silouan]Also from an ineq which was constructed by Harazi we have $\frac a{b+c}(y+z)+\frac b{c+a}(z+x)+\frac c{a+b}(x+y)\geq\sqrt{3(xy+xy+zx)}$[/quote]
For positive real numbers $a,b,c $ and real numbers $ x,y,z$ , have
 $$\frac a{b+c}(y^2+z^2)+\frac b{c+a}(z^2+x^2)+\frac c{a+b}(x^2+y^2)\geq\sqrt{3(x^2y^2+x^2y^2+z^2x^2)}\geq\sqrt{|xy+xy+zx|}$$

[quote=pvthuan]Show that for positive real numbers $a,b,c,x,y,z$
\[\frac a{b+c}(y+z)+\frac b{c+a}(z+x)+\frac c{a+b}(x+y)\geq3\frac{xy+yz+zx}{x+y+z}.\]
Prove that
\[\frac u{v+w}\frac{bc}{s-a}+\frac{v}{w+u}\frac{ca}{s-b}+\frac w{u+v}\frac{ab}{s-c}\geq a+b+c\]
where $a,b,c$ are side lengths of a triangle, and $s$ denotes its semiperimeter, $u,v,w$ are positive real numbers.[/quote]

[img]http://s6.sinaimg.cn/mw690/006ptkjAzy7qOPQwbXvf5&690[/img]

[quote=pvthuan]Show that for positive real numbers $a,b,c,x,y,z$
\[\frac a{b+c}(y+z)+\frac b{c+a}(z+x)+\frac c{a+b}(x+y)\geq3\frac{xy+yz+zx}{x+y+z}.\]
[/quote]
Show that for positive real numbers $a,b,c,x,y,z$ $$\frac {a(y+z)}{b+c}+\frac {b(z+x)}{c+a}+\frac {c(x+y)}{a+b}\geq\sqrt{3\left(xy+yz+zx\right)}\geq\frac{3(xy+yz+zx)}{x+y+z}$$
$$\sum\frac {a(y+z)}{b+c}\geq 4\sqrt{\sum \frac {bc}{(c+a)(a+b)}\sum yz}\geq  \sqrt{3\sum xy}$$
[hide=*][quote=luofangxiang]

[img]http://s6.sinaimg.cn/mw690/006ptkjAzy7qOPQwbXvf5&690[/img][/quote]

[/hide]

Let $ x,y,z$ be positive reals  such that $ x+y+z=1.$ Prove  that  $$ yza^2+zxb^2+xyc^2<\frac {1}{6}(a^2+b^2+c^2).$$
where $a,b,c$ are side lengths of a triangle.

For positive real numbers $ x,y,z$  and real numbers $a,b,c $ , have
$$\frac{x}{y+z} (b^2+c^2)+\frac{y}{z+x} (c^2+a^2)+\frac{z}{x+y} (a^2+b^2) \geq ab+bc+ca$$
[url=https://artofproblemsolving.com/community/c6h1141511p12773740]h [/url][url=https://artofproblemsolving.com/community/c6h299899p2314107]h[/url] [url=https://artofproblemsolving.com/community/c6h186636p1025190]h[/url] [url=https://artofproblemsolving.com/community/c4h2821329p25758551]h[/url]


Somewhere I found this inequaity and I can't kill it. for nonnegative$ a,b,c$

$ a^3 \plus{} b^3 \plus{} c^3  \plus{}3abc> \equal{} ab((2a^2 \plus{} 2b^2)^{1/2} \plus{} bc(2b^2 \plus{} 2c^2)^{1/2} \plus{} ca(2c^2 \plus{} 2a^2)^{1/2}$

[quote="matex (L)(L)(L)"]Somewhere I found this inequaity and I can't kill it. for nonnegative$ a,b,c$

$ a^3 \plus{} b^3 \plus{} c^3 > \equal{} ab((2a^2 \plus{} 2b^2)^{1/2} \plus{} bc(2b^2 \plus{} 2c^2)^{1/2} \plus{} ca(2c^2 \plus{} 2a^2)^{1/2}$[/quote]
hm...
$ a\equal{}b\equal{}c\equal{}1$

I missed $ 3abc$ on the left side
now it should be correct.

from schur this is very much clear that

a^3 + b^3 + c^3 +3abc> = ab(a+b) + bc(b+c) + ca(c+a)

now from am, gm we have that cyclic(2a^2+2b^2)^1/2>=2{cyclic(ab)^1/2}

and we have a+b>=2(ab)^1/2

hence

a^3 + b^3 + c^3 +3abc> = ab(a+b) + bc(b+c) + ca(c+a)>=cyclic{ab(2(ab)^1/2))

hence our result is proved

The proof above is wrong.
http://www.mathlinks.ro/viewtopic.php?t=6317


Nastavljam tradiciju otvaranja topica za IMC. Zadatke imate na [url]http://www.mathlinks.ro/index.php?f=79[/url], a rezultate cekamo  :lol:

Posto na facebooku vidim da u Blagoevgradu imate internet, bilo bi fino da se neko javi sa utiscima...

Ahem... Preciznije, na IMCu ucestvuju univerziteti. Da, nijedan bh univerzitet ne ucestvuje (ti znas da je nase omiljeno takmicenje Mediteransko-ne moras nigdje slati ekipu, ne mora se placati... kad bi sva tkmicenja bila takva, ucestovali bi mi  :lol: )

Rezultati su objavljeni ([url]http://www.mathlinks.ro/viewtopic.php?t=218379[/url]), cestitke takmicarima. Kome je mrsko da pregleda cijelu tabelu, povadio sam "nase":

Second Prize:

Marko Jevremovic (Beograd)
Igor Uljarevic (Beograd)
Goran Drazic (Zagreb)

Third Prize:

Aleksandar Pejcev (Beograd)
Spela Spenko (Ljubljana)
Nikola Adzaga (Zagreb)
Nik Stopar (Ljubljana)
Matija Basic (Zagreb)
Petar Sirkovic (Zagreb)
Marijana Smailagic (Beograd)

Honorable Mention:

Rafael Mrden (Zagreb)
Sara Kalisnik (Ljubljana)
Sasa Stanko (Zagreb)
Visnja Polic (Zagreb)
Luka Zunic (Zagreb)
Vili Krainz (Zagreb)
Peter Mursic (Primorska)

Certificate:

Marko Erceg (Zagreb)
Anton Golub (Zagreb)

Ako sam koga preskocio, pogresno napisao... Samo recite :)


integration
limit
real analysis
real analysis unsolved

Let
                 $ a_n\equal{}\frac{n\pi}{4}\minus{}\sum_{j\equal{}1}^{n} \frac{n^2}{n^2\plus{}j^2}$
Find $ lim_{n\rightarrow \infty} a_n$

Let's write this as

$ a_n = n(\frac {\pi}4 - \sum_{j = 1}^n\frac {1}{1 + \frac {j^2}{n^2}} \cdot\frac1n)$

$ = n(\int_0^1\frac {1}{1 + x^2}\,dx - \sum_{j = 1}^n\frac {1} {1 + \frac {j^2}{n^2}} \cdot\frac1n)$

$ = n\sum_{j = 1}^n\int_{\frac {j - 1}n}^{\frac {j}n}( \frac1{1 + x^2} - \frac1{1 + (\frac {j}n)^2})dx$

An exercise in Taylor expansions gets us that

$ \frac1{1 + x^2} - \frac1{1 + (\frac {j}n)^2} = \frac {2(\frac jn - x)\frac jn}{(1 + (\frac jn)^2)^2} + O(n^{ - 2})$

We then note that $ \int_{\frac {j - 1}n}^{\frac {j}n}(\frac jn - x)dx = \frac1{2n^2}.$ That gives us

$ a_n = n\sum_{j = 1}^n(\frac {\frac j{n^3}}{(1 + (\frac jn)^2)^2} + O(n^{ - 3}))$

$ = \sum_{j = 1}^n\frac {\frac j{n}}{(1 + (\frac jn)^2)^2}\cdot\frac1n + O(n^{ - 1})$

We now interpret this as its own Riemann sum and say that

$ \lim_{n\to \infty}a_n = \int_0^1\frac {x}{(1 + x^2)^2}\,dx = \boxed{\frac14}.$

[color=green][Added in edit: what I posted originally was in error by a factor of 2. Subsequent to making the post, I went back and ran this numerically, with an extrapolation step, and discovered the error. I am now very confident of this answer, which agrees with the numerical estimate.][/color]

More on what I said about numerical checking. From my spreadsheet, I have

$ a_4\ \approxeq 0.260416183$
$ a_8\ \approxeq 0.255208318$
$ a_{16}\approxeq 0.252604166$
$ a_{32}\approxeq 0.251302083$
$ a_{64}\approxeq 0.250651042$

With that, I dare say the extrapolation step is apparent to the naked eye.


calculus
integration
algebra
binomial theorem
calculus computations

I don't know if this is the right subforum. I posted this thread on another subforum aswell, but not sure where to put it.

OK, here goes:

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

But what about algrebra of this type. Are there any rules on how to do them. Could probably be simpler than i expect, that wouldn't surprice me, but i have tried several different ways of solving the algrebra problem.

$(a+x)^{-\frac{1}{2}}=...$

And this one:

$(a+x^{\frac{1}{2}})^2=...$

You should have probably posted this under the Algebra subforum under the Olympiad forum. Anyways, for your second question you just replace the $b$ with $x^{\frac{1}{2}}$ and you get you expansion. For your first question you can just write $(a+x)^{-1/2}$  as $\frac{1}{\sqrt{a+x}}$. But if you want to exapand this even further then you would have to use something called Newton's generalization of the Binomial Theorem (this would actually give you an infinite expansion!). You can learn more about this from this website:
http://en.wikipedia.org/wiki/Binomial_theorem

OK, thx for your help. I think i can solve my problem in another way. It was original an integral that i wanted to rewrite before working with it, but i am getting that help on the integral on another subforum. Thx for your help though.


Usually, the easiest way to find the area of triangle is to use the formula $ bh/2$.
This time, we cannot since it will be difficult to find a base and perpendicular height to plug in.

We can use Pick's Theorem to find the area of the triangle.

His theorem states that the area of the triangle is equivalent to the following:

$ A \equal{} I \plus{} B/2 \minus{} 1$

$ I$ is the number of lattice points in the interior and  $ B$ is the number of lattice points on the boundary.

As you can count, there are [b]13[/b] points that are TOTALLY in the interior, and there are [b]6[/b] lattice points on the boundary.

Thus the area is:

$ 13 \plus{} 6/2 \minus{} 1 \equal{} 15$


calculus
integration
trigonometry
calculus computations

Evaluate$\int\ \sqrt{tanx}-\sqrt{cotx}dx$

A single substitution is all that's needed

Reply to Ramanujan
It is possible to evaluate this inegral using the substitution you suggest but there is another substitution that makes very short work of the problem.
My solution using this substitution required only 8 lines of working.

Michael N.

Yes, we have a basic method. Try it, everyone. :lol:

[quote="Ramanujan"]We set $w^{2} = \tan x$ and the integral becomes

$\\ \int \frac{2w^{2}-2}{w^{4}+1} \, dw$

but this is too ugly to do   :P  :)[/quote]

Not that ugly as you think.
$\int {\frac{{2\omega ^2 - 2}}{{\omega ^4 + 1}}} d\omega = 2\int {\frac{{1 - \frac{1}{{\omega ^2 }}}}{{\omega ^2 + \frac{1}{{\omega ^2 }}}}} d\omega = 2\int {\frac{{d(\omega + \frac{1}{\omega })}}{{\left( {\omega + \frac{1}{\omega }} \right)^2 - 2}}} d\omega$
which can be easily integrated by substituting
$\omega + \frac{1}{\omega } = t$


logarithms
function
limit
complex analysis

For which $z \in \mathbb{C}$ the series $\sum_{n \geq 1}{z \choose n}$ converges?

By Raabe theorem the series is absolute convergent when $\Re{z}> 0$. What happens for $\Re{z}\leq 0$?

${z \choose n}$ is combination? then z is a natural number!
Or u wanna say $\frac{n!}{\Gamma(n-z+1)\Gamma(z+1)}$?

${z \choose n}= \frac{z (z-1) \ldots (z-n+1)}{n!}$

For $|w|<1,\ \sum_{n=0}^{\infty}\binom{z}{n}w^{n}=(1+w)^{z}.$

Note that we can choose the principal branch of the logarithm, with $\arg(1+w)\in\left(-\frac{\pi}2,\frac{\pi}2\right),$ so that the function is well-defined.

The question here is about the convergence of this power series at the point $w=1.$

We can write

$\binom{z}{n}=(-1)^{n}\prod_{k=1}^{n}\left(1-\frac{z+1}n\right)$

Taking logarithms, we find that

$\log\left((-1)^{n}\binom{z}{n}\right)= \sum_{k=1}^{n}\log\left(1-\frac{z+1}n\right)$

$=-\sum_{k=1}^{n}\frac{z+1}k+O(1)$

$=-(z+1)\ln n+O(1).$

Taking the exponential again, we find that

$\binom{z}{n}$ is comparable to $(-1)^{n}n^{-(z+1)}$ for large $n.$

(In haven't nailed down all the details but I believe that to be correct.)

In that case, the series converges absolutely for $\text{Re}(z)>0$ and diverges for $\text{Re}(z)\le-1.$

I would conjecture that it converges conditionally in the strip in between, where $-1<\text{Re}(z)\le0.$ Certainly it converges conditionally for real $z$ in that strip, but we need more investigation than I've shown so far to deal with $z$ in that strip but off the real axis.

By Abel's Theorem, if the series converges, its sum is $2^{z}.$

We have $\binom{z}{n}=\frac{z-n+1}{n}\binom{z}{n-1}$, where
\[\frac{|z-n+1|}{n}=\sqrt{(1-(\Re z+1)/n)^{2}+(\Im z/n)^{2}}=1-\frac{\Re z+1}{n}+O(n^{-2}). \]
Taking the product over $n$, find that $\lim_{n\to\infty}\binom{z}{n}=0$ iff $\Re z>-1$, in which case $\left|\binom{z}{n}\right|=\prod_{k=1}^{n}\frac{|z-k+1|}{k}=O(1/n^{\Re z+1})$.

Suppose that $\Re z>-1$. Since $\binom{z}{n}\to 0$, we can group the terms in pairs without affecting convergence.
\[\binom{z}{n}+\binom{z}{n+1}=\binom{z}{n}\frac{z-1}{n+1}=O(1/n^{\Re z+2}). \]
Since $\Re z+2>1$, the series is convergent.


Is there any composite number such that when its prime factors are listed in increasing order and viewed as one number, it's the same number?

To show you what I mean, let's say 123456789.

123456789 = 3*3*3607*3803
Which can be turned into the number 3336073803. So that doesn't work.

...or maybe you can take the prime factorization of THAT and it would equal the original number, or if not then do it again...

Any thoughts on this?

No composite number exists which has the property that if its prime factors are listed in increasing order, the original number is formed. This follows from the more general fact that I will prove: A number $n = \overline{N_1N_2}$, where $N_1$ and $N_2$ are consecutive blocks of digits, is greater than the product $N_1 \cdot N_2$. Suppose that $N_2$ has $k$ digits. Then we need to have $n = N_1\cdot N_2$. But $n = 10^k \cdot N_1 + N_2 > 10^k \cdot N_1 > N_1\cdot N_2$. The result regarding primes follows from the fact that we would have to write $n =  \overline{p_{n_1}p_{n_2}\cdots p_{n_k}} = p_{n_1} \cdot p_{n_2} \cdots p_{n_k}$ which is impossible.

Moreover, we see that the said transformation takes to number $n = p_{n_1}\cdot p_{n_2} \cdots p_{n_k}$ to the number $f(n) = \overline{p_{n_1}p_{n_2}\cdots p_{n_k}}$, with the property that $n < f(n)$. It follows that $n < f(n) < f(f(n)) < \cdots$ so that the procedure never returns to the original number.


Suppose I have

$ \begin{aligned}a\plus{}b&\equal{}c\\2a\plus{}2b&\equal{}2c\end{aligned}$

I'm currently working with

[code]\setlength{\fboxsep}{0.5pt}
\shadowbox{\colorbox{red}{some text}}[/code]

In "some text," I'm try to add that alignment, but I can't.

[code]\setlength{\fboxsep}{0.5pt}
\shadowbox{\colorbox{red}{
$\begin{aligned}a+b&=c\\2a+2b&=2c\end{aligned}$
}}[/code]
But you may like to try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=empheq]empheq package[/url] which does this sort of thing well.


Hello everyone,

I am supposed to find the partial sum by writing each value of k out but I'd like to find the partial sum with this formula. Could anyone please check what I am doing wrong? The answer in the book is different from my answer.

Thanks in advance.

$ \sum_{k=2}^{7}{k(k+2)}$
[u][b]
My work:[/b][/u]

Since the lower limit of the summation is 2, my series formula would be:

$ {s_{n}= \frac{n(a_{2}+a_{n})}{2}}$

$ {s_{7}= \frac{7(8+63)}{2}}$

$ {s_{7}= 231}$

Answer in book: [hide]231[/hide]

[quote="chickendude"]unfortunately, the formula you gave is for arithmetic series
The series you have is quadratic so you will have to do a little more

[hide]
$ \sum_{k=2}^{7}k(k+2) = \sum_{k=2}^{7}k^{2}+\sum_{k=2}^{7}2k$

you can sum the second one with the arithmetic series formula (number of terms * average of terms)

$ \sum_{k=2}^{7}k^{2}+6\frac{4+14}{2}$

$ 54+\sum_{k=2}^{7}k^{2}$

now you have to use the formula

$ \sum_{k=1}^{n}k^{2}= \frac{n(n+1)(2n+1)}{6}$

so that sum is

$ 54+\frac{7(7+1)(14+1)}{6}-\frac{1(1+1)(2+1)}{6}$

$ 54+140-1 = 193$
[/hide][/quote]

Thank you very much for your reply chickendude. I understand everything except the following two concepts and unfortunately, my book does not cover how to solve for these summations so I have two questions:

1. $ \sum_{k=1}^{n}k^{2}= \frac{n(n+1)(2n+1)}{6}$

Could you please tell me how you derived this formula? I know that part of it comes from $ \s_{n}= \frac{n(n+1)}{6}$.

2. With series which have multiple terms, what is the general rule for solving them? Would it be to apply the FOIL rule like you did every time and then solve each term separately?

Thank you again!

USe mathematical induction. Induction in n

Thank you for your replies.

Unfortunately, I am not that familiar with mathematical induction yet. Would there be another way to get the formula?

Pay attention to left side of the screen when you are looking at posts on this forum: sometimes a very nice proof of that formula appears.

I've never seen a proof of the sum of squares, because it has to be proven in three dimensions.

To go ahead and prove this, I'll induce (catalan, see if you can follow along)
I want to prove that the sum of the first $ n$ squares is $ \frac{(n)(n+1)(2n+1)}{6}$.
If $ n=1$, we see that the sum of the first 1 square is 1, and $ \frac{(1)(1+1)(2(1+1)+1)}{6}=\frac{6}{6}=1$. So if $ n=1$, our conjecture holds true.

Assume that the sum of the first $ n$ squares is $ \frac{(n)(n+1)(2n+1)}{6}=\frac{2n^{3}+3n^{2}+n}{6}$.

If I can prove that the sum of the first $ n+1$ squares is $ \frac{(n+1)((n+1)+1)(2(n+1)+1)}{6}=\frac{2n^{3}+9n^{2}+13n+6}{6}$, then I'll have all the necessary proof. This is because if, say I wanted the sum of the first 2007 squares, I could let $ 2007=n+1$ and know that if the formula works for $ n=2006$, then the formula works for $ n+1=2007$. And if I wanted to know if the formula worked for the sum of the first 2006 squares, I know that if the formula works for 2005 squares, then it works for 2006 squares. I could repeat this if-then process to the point where I need to know if the formula works for 1 square, then it works for 2 squares. Since I know that it works for 1 square, then I know it works for 2 squares, and if it works for 2, then it works for 3, and so on until I know it works for 2007 squares.

Maybe you should try to do the algebra yourself, but [hide="here is a hint"]
$ 1^{2}+2^{2}+...+n^{2}+(n+1)^{2}=(1^{2}+2^{2}+...+n^{2})+(n+1)^{2}$

You already know what $ 1^{2}+2^{2}+...+n^{2}$ is, so try to make your sum equal to $ \frac{2n^{3}+9n^{2}+13n+6}{6}$. Then your induction is complete.[/hide]
[hide="and here is a solution"]
$ 1^{2}+2^{2}+...+n^{2}+(n+1)^{2}=(1^{2}+2^{2}+...+n^{2})+(n+1)^{2}$

$ =\frac{2n^{3}+3n^{2}+n}{6}+(n+1)^{2}$

(I can say this because I assumed it is true to prove by induction. This is called the [i]induction assumption[/i])

$ \frac{2n^{3}+3n^{2}+n}{6}+(n+1)^{2}= \frac{2n^{3}+3n^{2}+n}{6}+n^{2}+2n+1$

$ =\frac{2n^{3}+3n^{2}+n+6n^{2}+12n+6}{6}$

$ =\frac{2n^{3}+9n^{2}+13n+6}{6}$

This is what we had to prove to complete the induction, so the induction is complete and we are done.[/hide]

Thank you very much to everyone who replied!


Describe in as many ways possible, the following graph IN POLAR FORM.
The graph is of infinitely many concentric circles, each $1$ unit apart with the center at the origin.

[hide="is this one?"]$\prod^\infty_{k=0}(r-k)=0$
where $r$ is the radius (and coefficient of $\theta$ is $0$).
[/hide]

[quote="cosinator"]$\prod^\infty_{k=0}(r-k)=0$[/quote]
Yes, that one works.  I hadn't thought of that one.  No need to hide your solutions.

[hide="Another"] $\sin (\pi r) = 0$ [/hide]

Remark:  The product that [b]cosinator[/b] gives doesn't converge for any non-integer values of $r$, so if that bothers you then you could rewrite it as

$r \prod_{k=1}^{\infty}\left( 1-\frac{r}{k}\right) = 0$

In fact, this product is closely related to the product form of $\sin x$, which is given by

$\sin x = x \prod_{k=1}^{\infty}\left( 1-\frac{x^{2}}{(k \pi)^{2}}\right)$

In particular,

$\frac{1}{\pi}\sin \pi r = r \prod_{k=1}^{\infty}\left( 1-\frac{r^{2}}{k^{2}}\right)$

Two more:

[hide="Modeled after the Spiral of Archimedes"]$r = \left \lfloor{\frac{\theta}{2\pi}\right \rfloor}$[/hide]
[hide="Modeled after cosinators equation"]$r = \lfloor r \rfloor$[/hide]


geometry
trapezoid
parallelogram
geometry proposed

$ ABCD$ is a rectangular isosceles trapezoid $ (\angle A\equal{}\angle D\equal{}90^{\circ})$ in which the triangle $ ABC$ is rectangular and isosceles $ (AC \perp CB)$. Denote by $ E$ the middle of $ BC$ and $ AC \cap DE\equal{}\{ P \}$.
$ a)$ Prove that $ DP\equal{}PE$
$ b)$ Compute $ \frac{AP}{AE}.$

*rectangular=with a right angle (I don't know if this is the term used in English) :?:

1) let m be the middle of $ AC$ then we have $ MECD$ is a parallelogram so $ DP\equal{}PE$
2) if $ DC\equal{}a$ we have $ AE\equal{} \sqrt{(1/2 a)^{2}\plus{}(3/2 a)^{2}}$
and $ AP\equal{}3/4 AC\equal{} 3/4 \sqrt{2}a$


Hello,

here are more than 700 problems involving inequalities. Enjoy!!

Kantor


Let $ D$ represent a repeating decimal. If $ P$ denotes the $ r$ figures of $ D$ which do not repeat themselves, and $ Q$ denotes the $ s$ figures of $ D$ which do repeat themselves, then the incorrect expression is:

$ \textbf{(A)}\ D \equal{} .PQQQ\ldots \qquad\textbf{(B)}\ 10^rD \equal{} P.QQQ\ldots$
$ \textbf{(C)}\ 10^{r \plus{} s}D \equal{} PQ.QQQ\ldots \qquad\textbf{(D)}\ 10^r(10^s \minus{} 1)D \equal{} Q(P \minus{} 1)$
$ \textbf{(E)}\ 10^r\cdot10^{2s}D \equal{} PQQ.QQQ\ldots$

[hide="Click for solution"]
$ D\equal{}.PQQQ \ldots \equal{} .a_1 \ldots a_rb_1 \ldots b_sb_1 \ldots b_s \ldots$.  So $ \textbf{(A)}, \textbf{(B)}, \textbf{(C)},$ and $ \textbf{(E)}$ are all correct, so $ \boxed{\textbf{(D)}}$ is incorrect.
[/hide]


invariant
linear algebra
matrix
linear algebra unsolved

Let $A,B\in\mathcal{M}_{n}(\mathbb{C})$ s. t. $AB+BA=I_{n}$ and $BA^{2}B=0$.
i) (not difficult) Prove that $n$ is even.
ii) Prove that if $n\leq6$ then $trace(A)=trace(B)=0$.
iii) Does this last result remain true if $n>6$ ?

I thought I'd revive this topic, but it would be a shame if I didn't post anything about the problem. So here's what I got:

From $ AB+BA = I_{n}\ \& \ BA \cdot AB = 0$, we get that $ AB$ is idempotent. Note that the problem is invariant under conjugation. Thus we may WLOG assume that $ AB = \begin{pmatrix}I_{r}& 0 \\ 0 & 0 \end{pmatrix}$. That implies $ BA = \begin{pmatrix}0 & 0 \\ 0 & I_{n-r}\end{pmatrix}$.

Now let's look at this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=86370.
[b]Kent Merryfield[/b] has proved that $ \text{rank}\left( I_{n}-AB \right) = \text{rank}\left( I_{n}-BA \right)$ holds for all matrices $ A,B$. Since in this problem we have that $ AB+BA = I_{n}$, it follows that $ \text{rank}(AB) = \text{rank}(BA).$ In other words, $ r = n-r$, from where $ r = \frac{n}2$, so that $ n$ is even. This settles [b](i)[/b].

(Alternatively we could say that $ \text{tr}(AB) = \text{tr}(BA)$, but I just liked the argument in the other topic :D)

Let $ n = 2k$, so that $ AB = \begin{pmatrix}I_{k}& 0 \\ 0 & 0 \end{pmatrix}, \, BA = \begin{pmatrix}0 & 0 \\ 0 & I_{k}\end{pmatrix}$. My unsuccessful idea was to split both $ A$ and $ B$ into $ k \times k$ blocks $ A = \begin{pmatrix}X & Y \\ Z & T \end{pmatrix}, \, B = \begin{pmatrix}P & Q \\ R & S \end{pmatrix}$, and then to try to solve the equations that follow. When $ k=1$, it's very easy to prove [b](ii)[/b], but already for $ k=2$ I have no idea how to attack it cleanly.

PS: If you were hoping for the harder part, then I'm sorry to disappoint you  :oops:

Hi perfect_radio.
If $ n\leq6$ then the $ A,B$-diagonals are $ 0$; that is to say, with your notations,  $ X=T=P=S=0$. I have proved this result (which is stronger than that required) by writing the algebraic system in the $ a[i,j],b[i,j]$; then I solved this system with the help of Maple and the Gröbner's basis theory. But I can do that only if  $ n\leq6$.


Vieta
quadratics
algebra
quadratic formula

Let $\alpha$ be the root, one of the cubic equation $x^3+3x^2-1=0$.

(1) Express $(2\alpha ^ 2+5\alpha -1)^2$ in the form of $a\alpha ^2+b\alpha+c$, where $a,b,c$ are rational numbers.

(2) Express other two roots except $\alpha$ in the form of $a\alpha ^2+b\alpha+c$.

[hide="1"]The quantity is equal to $(2\alpha^2+5\alpha-1)^2=4\alpha^4+20\alpha^3+21\alpha^2+1$. Furthermore, $0=(\alpha^3+3\alpha^2-1)(4\alpha+8)=4\alpha^4+20\alpha^3+24\alpha^2-4\alpha-8$. Subtract this from the original to get $-3\alpha^2+4\alpha+9$.[/hide]

kunny, it would help if you would say [i]what kind[/i] of mistake people have made -- just saying they are wrong is kind of annoying.

Both of you had errors expanding out the initial expression $(2\alpha ^ 2+5\alpha -1)^2$.  Once you go and fix those, you will probably be all right.

$(2\alpha ^ 2+5\alpha -1)^2=4\alpha^4+20\alpha^3+21\alpha^2-10\alpha+1$. With jb05 method we get $-3\alpha^2-6\alpha-9$.

For 2 we can use Viete formulas :D

Can anyone solve this problem completely? :)

Hello, Pork_Chop. :)

Set $\beta,\gamma$ be the roots except $\alpha$, you can use Vieta's formula.

kunny

$\beta + \gamma = -3 - \alpha$

$\beta\gamma + \alpha(\beta + \gamma) = 0$

$\beta\gamma = \frac{1}{\alpha}$

????

Oops: Maybe we have quadratic in $\beta$ and $\gamma$, by Viete's we get

$x^2 - (\beta + \gamma)x + \beta\gamma = 0$

$\Rightarrow x^2 - (-3 - \alpha)x + \frac{1}{\alpha} = 0$

$\Rightarrow \alpha x^2 - (-3\alpha - \alpha^2)x + 1 = 0$

$\Rightarrow \alpha^2(x) + \alpha(x^2 + 3x) + 1 = 0$

????

using Pork_Chop8's equation: $x^2-(-3-\alpha)x+\frac1{\alpha}=0$ apply quadratic formula

$x=\frac{-3-\alpha}2\pm\frac12\sqrt{(3+\alpha)^2-4\frac1{\alpha}}$

let's work for now the expression inside of square root

$(3+\alpha)^2-4\frac1{\alpha}=(3+\alpha)^2-4\alpha(3+\alpha)=(3+\alpha)(3+\alpha-4\alpha)=3(3+\alpha)(1-\alpha)$

because $\boxed{\frac1{\alpha}=\alpha(3+\alpha)}$

here is one tricky part

from $\alpha^3+3\alpha^2-1=0\Rightarrow 3\alpha^2=1-\alpha^3=(1-\alpha)(\alpha^2+\alpha+1)$

and so $\boxed{1-\alpha=\frac{3\alpha^2}{\alpha^2+\alpha+1}}$

so we have $3(3+\alpha)(1-\alpha)=3\frac1{\alpha^2}\frac{3\alpha^2}{\alpha^2+\alpha+1}=\frac9{\alpha^2+\alpha+1}$

this is another tricky part

from $\alpha^3+3\alpha^2-1=0\Rightarrow (\alpha^3+3\alpha^2-1)(\alpha+1)=0$

$\alpha^4+4\alpha^3+3\alpha^2-\alpha-1=0\Rightarrow \alpha^4+4\alpha^3+4\alpha^2-\alpha^2-\alpha-1=0$

and so $\boxed{\alpha^2+\alpha+1=\alpha^4+4\alpha^3+4\alpha^2=\alpha^2(\alpha+2)^2}$

so now $\frac9{\alpha^2+\alpha+1}=\frac9{\alpha^2(\alpha+2)^2}$

back to $x$ we have

$x=\frac{-3-\alpha}2\pm\frac12\sqrt{(3+\alpha)^2-4\frac1{\alpha}}=\frac{-3-\alpha}2\pm\frac32\frac1{\alpha(\alpha+2)}$

but remember that $\alpha^2+\alpha+1=\alpha^2(\alpha+2)^2$ which means that

$\boxed{\frac1{\alpha(\alpha+2)}=\frac{\alpha(\alpha+2)}{\alpha^2+\alpha+1}}$, also

$\boxed{\frac1{\alpha^2+\alpha+1}=\frac{1-\alpha}{3\alpha^2}}$ and also

$\boxed{\frac1{\alpha^2}=\alpha+3}$ finally!!! so combine them all

$x=\frac{-3-\alpha}2\pm\frac32\frac1{\alpha(\alpha+2)}=\frac{-3-\alpha}2\pm\frac32\frac{\alpha(\alpha+2)(1-\alpha)(\alpha+3)}3$

now multiply out and do a lot of simplification and remembering to use to $\alpha^3+3\alpha^2-1=0$ we should get

$x=\frac{-3-\alpha}2\pm\frac{6\alpha^2+5\alpha-1}2$

choose one sign for one root and another one for another root

$\beta=3\alpha^2+2\alpha-2$, $\gamma=-3\alpha^2-3\alpha-1$

whew, a lot of typing and copy/paste. kunny probably has a slick short solution :D


function
limit
calculus
derivative
algebra
domain
calculus computations

In our text book it is asked to show that the maxima of $ f(x) \equal{} x \minus{} \frac 1x$ is less than the minima of $ f(x)$. They have found that the minima of the function is $ \plus{} 2$ and maxima is $ \minus{} 2$. But as far as I have understood the global maximum should be greater than global minimum.
Moreover we can easily see that $ \lim_{x \to 0} f(x) \to \infty$. So, the maximum value of the function should be $ \infty$.  :|

So, is it correct that with the usual method (I mean at first taking first derivative and finding critical point etc.) we can not always find the global maxima and minima? Then how can we find them?

Could anyone please explain me these things? Probably I have some serious conceptual gap here. Thanks in advance.

That's a local maximum and a local minimum; this function is unbounded both above and below on the real line.
The thing is, you're not remembering the whole method- we also have to check the edges of the domain; in this case, $ \lim_{x\to\infty}f(x)$, $ \lim_{x\to\minus{}\infty}f(x)$, $ \lim_{x\to 0^\plus{}}f(x)$, and $ \lim_{x\to 0^\minus{}}f(x)$.

Except that if the function is really $ f(x)\equal{}x\minus{}\frac1x$ with the minus sign in there, it has no local maxima nor local minima anywhere.

The function $ g(x)\equal{}x\plus{}\frac1x$ does have a local minimum value of $ 2$ at $ x\equal{}1$ and a local maximum value of $ \minus{}2$ at $ x\equal{}\minus{}1.$


Can anyone tell me how to solve this without guess and check?

what is the greatest n where n is a positive integer where 12^n is a divisor of 20!

I know the answer is 8 I just dont know any way to do it other than guess and check.

Thanks

(Note: throughout this post, brackets like [x] denote the greatest integer less than or equal to x.  Thus, [1] = 1, [1.5] = 1, [-12.6] = -13.)

It may be helpful to think of the problem as "how many times is 12 in 20!?"  Since 12 = 2^2*3, we're essentially trying to find the number of factors of 2 and 3 in 20!.

So how many factors of 2 are in 20! = 1*2*3*...*19*20?  Well, for each multiple of 2, there's one...so that's 10 (2, 4, 6, ....18, 20).  But we missed some factors of 2 since some numbers have more than one multiple of 2, like 4.  There's an extra factor of 2 in [20/4] = 5 more numbers (4, 8, 12, 16, 20).  But we've still missed some factors of 2, since some numbers have three factors of 2, like 8.  Thus we need to add in [20/8] = 2.  By similar reasoning, we need to add in [20/16] = 1.  In all, there are 10 + 5 + 2 + 1 = 18 multiples of 2 in 20!

By similar reasoning, there are [20/3] + [20/9] + [20/27] + ... = 6 + 2 + 0 + ... = 8 multiples of 20!

So how many 12's can we make with 18 2's and 8 3's?  Only 8, because we need one 3 for each 12 and we only have 8 3's.  (The other 2's are extras.)  Thus, there are 8 12's in 20!

This is just a variation on the "finding how many factors of a certain number k! has" problem.  Since we are dividing it by 12 which equals 2^2*3 we should find how many factors of 2 and how many factors of 3 it has.

To find the number of factors of 2 that 20! has:

Since 20!=20*19*...*2*1, we will get a 2 for every even number so there are 20/2=10 even numbers.  But what about numbers with 2^2 in them?  So then we have to go back and find how many of them there are: there are 20/4=5 of them.  We still have to find how many factors of 2^3, and 2^4 there are.  There are 2 and 1, respectively.  So there are 18 factors of 2 in this number.

Number of factors of 3 that 20! has:

Using the same procedure we get [20/3]+[20/9]=6+2=8...[x] is the greatest integer less than or equal to x function.

So now we have $\frac{2^{18}\cdot 3^8\cdot k}{2^{2n}\cdot3^n}=2^{18-2n}\cdot 3^{8-n}\cdot k$

We want the right side to be an integer so the largest value is 8.

Sorry but I dont quite get why and how that method works

I solved the problem using similar concept to yours but much more unsophisticated.  I just wrote out all 20 numbers in a form of multiples of prime numbers!!!

Can you specifically explain how [x] method worx?

Thanks!

nvm...  I had a good sleep last nite and LOL I dreamed about the problem!

Now I get why [x] worx...  My brain must've been in a very primitive state when I saw that question.

You can use this...it's easier:
If p is a prime number and n is a pozitive integer:
\[exp(n!)p=[\frac{n}{p^1}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]+... \]


factorial
number theory
number theory unsolved

find all $ x,y \in I$,If $ x!y!\equal{}x!\plus{}y!\plus{}z!$

[quote="apollo"]find all $ x,y \in I$,If $ x!y! \equal{} x! \plus{} y! \plus{} z!$[/quote]
Do you want to solve $ x,y\in N$
Easy to check $ z\geq\text{max}\{x,y\}$
So $ x!|y!,y!|x!$ so $ x\equal{}y$
So
$ (x!)^2\equal{}2x!\plus{}z!$
$ \Leftrightarrow {x!}\equal{}2\plus{}\frac{z!}{x!}$
$ x>2$ and  consider mod 3 we have $ x<z<x\plus{}3$
Case1:$ z\equal{}x\plus{}1$
$ x!\equal{}x\plus{}3$ has $ x\equal{}3$
Case 2: $ z\equal{}x\plus{}2$
$ x!\equal{}2\plus{}(x\plus{}1)(x\plus{}2)$  has no solution.
Case 3:$ z\equal{}x\plus{}3$
Same method ...

I think this is a good question...
the Solution is:     $ x!y! \equal{} x!\plus{}y!\plus{}z!$
                  $ y! \equal{} 1 \plus{} \frac{y!}{x!} \plus{} \frac{z!}{x!}$
which shows that $ x \leqslant y$ because $ LHS$ is a integer.
also we can write $ x! \equal{} 1 \plus{} \frac{x!}{y!} \plus{} \frac{z!}{y!}$
which shows that $ y \leqslant x$ because $ LHS$ is a integer.
hence we can say that $ x\equal{}y$.

substituting this we get
$ (x!)(x!\minus{}2) \equal{} z!$
substituting $ x!$ as $ n$.
we get $ n(n\minus{}2) \equal{} z!$,
and thts how we procede further

[quote="apollo"]why $ x < z < x \plus{} 3$ ???[/quote]
If $ z\geq x\plus{}3$ then $ 3|\frac{z!}{x!}$ and $ 3|x!$

[quote="ashwinrk_jain"]...substituting $ x!$ as $ n$.
we get $ n(n \minus{} 2) \equal{} z!$,
and thts how we procede further[/quote]
Unfortunately, you have reduced this to an unsolved problem in number theory related to a problem Erdos has considered hopeless, namely the Brocard-Ramanujan Diophantine Equation.  Yet, it is interesting that in some sense this problem is related but not as difficult.

Also, you still must check when $ x,y$ are 1,2 (trivial).  Now, you could probably use what you wrote:
\[ (x!)(x!\minus{}2) \equal{} z!~~~(1)\] which implies $ x<2z/3$, for sufficiently large $ x$ (which can be computed).  Then, argue that there exists a prime in the interval $ \big( 2z/3, z \big)$ which divides the RHS but not the LHS of your equation (1).

Have a look here
http://www.mathlinks.ro/Forum/viewtopic.php?t=116937

Pierre.


Let quadrilateral ABCD satisfy $ \angle{ABC}\equal{}\angle{ADC}\equal{}\frac{\pi}{2}$ and $ AC\equal{}AB\plus{}AD$. AC intersect BD at I. Assume that E is a point that in ray IA satisfy quadrilateral EBCD is a tangentical quadrilateral. Let $ I_1; I_2$ are the center of excircle of angle of triangle EBD and the center of excircle of triangle BCD, respectively. Prove that: $ I, I_1, I_2$ are collinear.

Exist $ E$ is a point that in ray $ IA$ satisfy quadrilateral $ EBCD$ is a tangentical quadrilateral
$ \Leftrightarrow AB\equal{}AD$
then $ I,I_1,I_2\in AC$
See my solution:
[url]http://forum.mathscope.org/showthread.php?t=9516[/url]

If $ AB\neq AD$, with out lost of general we can assume that $ AB<AD$.
$ (AB\plus{}AD)^2\equal{}AB^2\plus{}BC^2\equal{}AD^2\plus{}DC^2$
$ \Rightarrow BC^2\equal{}AD^2\plus{}2AB.AD, AB^2\equal{}DC^2\minus{}2AB.AD$
$ \Rightarrow BC^2.AB^2\equal{}(AD^2\plus{}2AB.AD)(DC^2\minus{}2AB.AD)$
We will show that $ (AD^2\plus{}2AB.AD)(DC^2\minus{}2AB.AD)<AD^2.DC^2$
$ \Leftrightarrow DC^2\minus{}AD^2<2AB.AD\equal{}(AB\plus{}AD)^2\minus{}AB^2\minus{}AD^2$
$ \Leftrightarrow DC^2<4R^2\minus{}AB^2\Leftrightarrow DC^2\plus{}AB^2<4R^2$, it's true because $ DC^2<BC^2$
Therefore $ BC^2.AB^2<AD^2.DC^2 \Leftrightarrow 2S_{ABC}\equal{}BC.AB<AD.DC\equal{}2S_{ADC}$
$ \Leftrightarrow IB<ID$
$ \Rightarrow EB<ED$
But $ EB\plus{}DC\equal{}ED\plus{}BC, DC<BC$ then it's contradiction.
Hence $ AB\equal{}AD$ and we are done.

^^! I don't think the proof the same as you did. Now, you trying to prove that perimeter of triangle CBI=perimeter of triangle CDI if not exist point E.^^

It is so easy.
We have:
$ IB \equal{} \frac {a.BC(a.CD \plus{} b.BC)}{(a.BC \plus{} b.BC)(a \plus{} b)}$
$ ID \equal{} \frac {b.CD(a.CD \plus{} b.BC)}{(a.BC \plus{} b.CD)(a \plus{} b)}$
$ IB \minus{} ID \equal{}\frac {(a.CD \plus{} b.BC)(a.BC \minus{} b.CD)}{(a \plus{} b)(a.BC \plus{} b.CD)}$
$ IB \minus{} ID \equal{} CD \minus{} BC$
$ \Leftrightarrow 2ab.BC^2 \equal{} 2ab.CD^2 \minus{} a^2BC^2 \plus{} b^2.CD^2$
$ \Leftrightarrow BC^2(a^2 \plus{} 2ab) \equal{} CD^2(b^2 \plus{} 2ab)$
done.

[quote="Elementaryyy"]It is so easy.
We have:
$ IB \equal{} \frac {a.BC(a.CD \plus{} b.BC)}{(a.BC \plus{} b.BC)(a \plus{} b)}$
$ ID \equal{} \frac {b.CD(a.CD \plus{} b.BC)}{(a.BC \plus{} b.CD)(a \plus{} b)}$
$ IB \minus{} ID \equal{} \frac {(a.CD \plus{} b.BC)(a.BC \minus{} b.CD)}{(a \plus{} b)(a.BC \plus{} b.CD)}$
$ IB \minus{} ID \equal{} CD \minus{} BC$
$ \Leftrightarrow 2ab.BC^2 \equal{} 2ab.CD^2 \minus{} a^2BC^2 \plus{} b^2.CD^2$
$ \Leftrightarrow BC^2(a^2 \plus{} 2ab) \equal{} CD^2(b^2 \plus{} 2ab)$
done.[/quote]
Very good! Thank you so much!:x


OK, here's how the game works.  Post a riddle or thought provoking puzzle, and the next person has to answer the riddle, rate it on a scale from A+ to F(no E's!), and submit another riddle.  I'll start out with one:  What, if made into a book can wrap around the world dozens of times, but is located entirely inside the world?

Sorry, I forgot about posting rules:

1. Do not create a new riddle until all the previous riddles have been solved. (If the riddles are impossible, this rule can be bypassed after a day or so).
2. Don't spam up the forum by complimenting on how "smart you were to think of that" or "what a great riddle".  Instead, PM them about why it was so great and give them a 6 cube rating.
3. Absolutely NO personal insults, harrassment, stalking, or discussion of illegal activities will be tolerated.
4. When you rate the riddle before you, tell them what you thought about it, not just an "F" and "it was horrible".
5. Don't make any non age appropriate riddles.
6. If things become a problem, I will post new rules later.

Thanks:  Lets do everything we can to avoid this getting locked or abandoned.

EDIT: If no one wants to solve this one in the next 10 minutes, please post a new riddle to kick start this marathon.

[quote="sdkudrgn88"]Hold on, don't we already have something like this in Puzzles and Brainteasers? Although it's already two weeks stale...

I don't understand your riddle, what do you mean by "inside the world"? I can tell you that the answer sure isn't "towel"...

Alright, next riddle.

(This puzzle was created by Sam Loyd. I take no credit for this puzzle.)

Two ferries are making trips across a river between New York and New Jersey. The two ships leave at the same time, and meet each other 720 yards away from New York. When they reach their respective destinations, they take exactly the same time to change passengers and start heading back. They meet each other again 400 yards away from New Jersey.

How wide is the river?[/quote]
OK, no one has answered the question.  NOte:  the answer to my question was "The Internet".  New Riddle

If it takes 10 men to dig 1 hole in 1 day, how many days does it take 20 men to dig 1/2 of a hole?

Quarter day. D+...way to simple

Two people are walking. One notices that there is one tree ahead and walks around it. The other sees two trees next to each other, and walks into the tree. What is the difference between the two people?

[quote="Chompy"]Quarter day.[/quote]
I do believe the answer is that "you can't dig half a hole". Either way, that one gets a 38/100, or a D-.
@Chompy: One is drunk, the other one isn't. 56/100 (C+).
The answer to mine: One mile. Figure it out from there.

Two twins are born on Earth, 10 minutes apart. However, on their first birthday, they find that the older twin's birthday occurs [i]three[/i] days after the younger twin's. How is this possible?
And a similar one:
Two twins are born on Earth, 10 minutes apart. However, on their first birthday, they find that the [i]younger[/i] twin's birthday occurs two days after the [i]older[/i] twin's. How is this possible?
(Once again, I cannot take credit for this riddle, it comes from Wu Riddles. However, the first one I can take credit for.)


geometry
trigonometry
trig identities
Law of Cosines

Twelve points $ A_1,A_2,A_3,...,A_{12}$ are equally spaced around a circle of radius 6. Region $ A_1A_6A_8$ is shaded. What is the area of the shaded region?

We need to find the length of $ A_1 A_6$. The angle intercepted is $ \frac{5}{12} \cdot 360^\circ \equal{} 150^\circ$. Using the Law of Cosines, that the square of the length of $ A_1 A_6$ is $ 72\plus{}36\sqrt{3}$. The area of the region is
\[ \frac{1}{2}x^2 \sin 30^\circ \equal{} \frac{1}{2} \cdot \left(72\plus{}36\sqrt{3}\right) \cdot \frac{1}{2} \equal{} \boxed{18\plus{}9\sqrt{3}}.\]

yes the above is for triangle $ A_1A_6A_8$

Instead of using the cosine law a bunch of times we can just use the symetry that $ |A_1A_6|\equal{}|A_1A_8|$
So we have an isoscelese triangle.

[geogebra]e191b6383b2c3c53d91712cf3536bde0c18aed7a[/geogebra]

If we join every other point of the 12-gon, we get a regular hexagon, Since a regular hexagon is made up of six equilateral triangles we know that $ A_6A_8\equal{}6$. We also Know that the height of the equilateral triangle is $ 6\sin 60 \equal{} 3\sqrt{3}$.

$ \therefore [A_1A_6A_8] \equal{} \frac{1}{2}\cdot6\cdot(3\sqrt{3}\plus{}6) \equal{} \boxed{9(2\plus{}\sqrt{3})}$


search
inequalities unsolved
inequalities

[size=150][color=darkblue][u][b]Problem 4[/b][/u]:[/color][/size]

Let  $ x > 1, y > 1, z > 1$  and  $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 2$. Prove that $ \sqrt {x \plus{} y \plus{} z} \ge \sqrt {x \minus{} 1} \plus{} \sqrt {y \minus{} 1} \plus{} \sqrt {z \minus{} 1}$.

$ \equal{}>\frac{x\minus{}1}{x}\plus{}\frac{y\minus{}1}{y}\plus{}\frac{z\minus{}1}{z}\equal{}1$
$ \plus{})\frac{x\minus{}1}{x}\plus{}\frac{y\minus{}1}{y}\plus{}\frac{z\minus{}1}{z} \ge \frac{(\sqrt{x\minus{}1}\plus{}\sqrt{y\minus{}1}\plus{}\sqrt{z\minus{}1})^2}{x\plus{}y\plus{}z}$
$ \equal{}>x\plus{}y\plus{}z \ge \sqrt{x\minus{}1}\plus{}\sqrt{y\minus{}1}\plus{}\sqrt{z\minus{}1}$
=>Done! :P

This is Iran 1998 , and i think it was an IMO problem .

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=210535&search_id=744443372]here[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=744443372&t=144]here[/url]

And you posted same problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=210524&search_id=744443372]here[/url]


calculus
logarithms
derivative
Pythagorean Theorem
geometry
calculus computations

The foot of a 20' feet ladder is being pulled away from a wall at the rate of 1.5ft/sec.  At the instant when the foot is 12ft away from the wall, the angel the ladder makes with the floor is decreasing at the rate (in radians/sec)?

If f(x)=In(the squre root of 1+x^20  then a horizontal tangent line to f(x) exists at x=?

[quote="cdymdcool"]If f(x)=In(...)[/quote]
I see this on students' papers, and it [i]always[/i] bothers me. That letter is not an "I". The letter is an "L". L is for [b]logarithm[/b]. "ln" is for "logarithm, natural."

$ f(x)=\ln\left(\sqrt{1+x^{10}\right)=\frac12\ln(1+x^{10}).}$

Now, you tell me: what is the derivative $ f'(x)?$

And as for the ladder sliding down the wall: label the lengths that matter. Write an equation that remains valid even as things are changing. (The Pythagorean Theorem should be useful.)

What do you have so far? Show me your setup.


My team just took the test today, and I was wondering if it would be ok if I talked about the information on it. I believe I correctly solved all of the subproblems, and I would like to see what other people got.

And if this is in the wrong forum (maybe contests, but I couldn't find a mandlebrot subforum..) please move it.

haha my team did too

how did you do in indivdiual first and second rounds?

duno if we can talk about it

Once again, there are schools taking the Mandelbrot this Saturday!


calculus
integration
function
trigonometry
limit
ratio
calculus computations

For $m=1,2,\cdots,$ let $g_{m}(x)$ be the function of real numbers $x$ as follows.

\[g_{m}(x)=\int_{x}^{x+\frac{\pi}{2}}(\sin \theta)^{2m}\ d\theta .\]

Let $a_{m},\ b_{m}$ be the minimum value and maximum value of $g_{m}(x)$ respectively. Find $\lim_{m\to\infty}\frac{a_{m}}{b_{m}}.$

First, $a_{m}\le g(-\pi/4)\le \frac{\pi}{2}\sin^{2m}\frac{\pi}{4}=\frac{\pi}{2^{m+1}}$.

Second, for any $\epsilon\in(0,\pi/4)$ we have $b_{m}\ge \int_{\pi/2-\epsilon}^{\pi/2+\epsilon}\sin^{2m}\theta \ge 2\epsilon\cos^{2m}\epsilon\ge 2\epsilon(1-\epsilon^{2}/2)^{2m}$. Choose $\epsilon=\frac{1}{\sqrt{m}}$ (assuming $m>1$). This gives $b_{m}\ge \frac{2}{\sqrt{m}}(1-1/(2m))^{2m}\sim \frac{2}{e\sqrt{m}}$ as $m\to\infty$.

$b_{m}$ wins, big time.

we have g'(x) = (sinx)^(2m)
take f(x) = g(x+pi/2) -g(x)
therefore f'(x) = g'(x+pi/2) - g'(x)
                    = [(sinx)^m + (cosx)^m] [(sinx)^m - (cosx)^m]
so what we actually want is f(x)    [sorry for changing notations] :oops:
so maxima and minima is got by f'(x)=0
which gives sinx=cosx
and sinx = -cosx   as two solutions
one will give maxima and other minima
ow find integral of (sinx)^(2m)  using reduction formulae(which is quite easy but lengthy)
and put sinx=cosx and sinx=-cosx conditions to get maxima and minima
take ratio and put m->infinity
to get the answer


The problem as for the reader to evaluate

$\int_{0}^{\frac{\pi}{2}}\frac{x\sin(x)}{1+\cos^{2}(x)}dx$

I have tried $u=\frac{\pi}{2}-x$ but that was not very useful.

Could somone please give he a hint to lead me in the right direction.

I am not sure if this will help:
Note that $\frac{x\sin x}{1+\cos^{2}x}= \sum_{n=0}^{\infty}(-1)^{n}x \sin x \cos^{2n}x$

Then, applying integration by parts to each term of the series, the original integral equals $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}\int_{0}^{\frac{\pi}{2}}\cos^{2n+1}x\, dx$.

Now, we will  have to use reduction formula for the cosine integral terms, but I am not sure how this will help.

Perhaps there is an easier way!


probability
quadratics
complementary counting
AMC 10
algebra
quadratic formula
geometry

Let this thread be a thread for colorful advice for AMC B takers this wednesday.

Yeah, you'd think they would have a more prominent place to put it other than the lower right hand corner of the back of the answer sheet.

---===---

Use scrap paper liberally.

---===---

Post #500!

Obvious one: make sure you don't misbubble, especially when you are skipping problems. There are few things more awful than getting a 37.5 instead of a 139.5 just because you forgot to skip a line. (This is all hypothetical, of course.  :P )

On probability questions, make sure that you have what they want.  I have missed questions before when I used complementary counting and forgot to subtract my answer from one.

[quote="bpms"][quote="Karth"]yea

don't misbubble... my score dropped from like a 145 to like a 118 on the AMC 10 when I was in 10th grade, and I felt like smacking someone.

don't let your mind wander

oh yea, and don't misbubble gender EVEN FOR A JOKE... this kid in my school marked both male and female (though he's not a hermaphrodite) last year and our teacher chewed him out like food[/quote]
What did that show up as on the AMC site?[/quote]

probably in the category for:

(is a male and a female)

Does anyone go into the test with certain Calculator Programs ready? (Number of divisors, quad formula, etc.)

If so, what are some programs I should write before the test so I can save some time during the test? I use TI-83+ Calculator.

You don't really need them, but I always have the quadratic formula on my 83. I don't really have much else, maybe Heron's Formula, but that doesn't really show up all that much on tests.

The only programs I have are the quadratic formula and heron's, and I hardly ever use them.

Yes.
I have all the useful number theory programs (CRT, FLT, Totient Funct.)

And some probability ones I wrote myself.

I tend to come into the test with a symbolic manipulator on the TI-89, and sometimes I'll put a Geo program on there for help with vizualization.

[quote="bpms"][quote="randomdragoon"]Make sure to remember to bubble in your grade, age and gender - don't want to contribute to the (?) category when statistics come out[/quote]
Hah, I think I forgot to bubble gender.[/quote]
Yeah, I am officially one of the ? on the AMC site  :ninja:


Real numbers $ a$ and $ b$ satisfy the equations $ 3^{a} \equal{} 81^{b \plus{} 2}$ and $ 125^{b} \equal{} 5^{a \minus{} 3}$. What is $ ab$?

$ \textbf{(A)} \minus{} \!60 \qquad \textbf{(B)} \minus{} \!17 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 60$

I do believe the first two answer choices are negative.

[quote="now a ranger"]doesn't the combination

12 and 5 work too?[/quote]

No? Plugging them in is a good way to check...

Just equates the base and take the powers to a whole new equation.

I agree to the others answers

The first equation implies $a=4b+8$ and $3b=a-3$.

So $3b=4b+5\implies b=-5$ and $a=-12$.

So these multiply to $\boxed{\textbf{(E)}\ 60}$.


let $ a_{i}> 0$($ i=1\to  100$ )
such that : $ \sum_{i=1}^{100}a_{i}\geq 300$and  $ \sum_{i=1}^{100}a_{i}^{2}\leq 1000.$ prove that : exist 3 real numbers in sequence have sum $ >100.$

Are you sure it's true?
Take $ a_{1}\equal{}a_{2}\equal{}...\equal{}a_{100}\equal{}3$.They satified the condition but there doesnt any 3 numbers which have sum >100. :o
-----------
Are you going to Russia,Hong Quy?


Find continuous function $ {f: (0,\plus{}\infty)}\rightarrow{R}$ such that for any $ x,y\in(0,\plus{}\infty)$, $ f(x\plus{}y)\minus{}f(x)\minus{}f(y)\equal{}f(xy\plus{}1)\minus{}f(xy)\minus{}f(1)$

[quote="ggbg90"]sorry, just 1 question, do we have to find all continuous funcitons or just 1?[/quote]

I think the question is to find all solutions, since finding just one solution is obvious : $ f(x)\equal{}ax\plus{}b$


Very nice proof,arqady :lol:
Solution I knew:we can firstly assume $ \sum a^2=2$ $ \sum x^2=2$(This is because we can subsitute $ x$ by $ tx$ and $ a$ by$ \frac{a}{t}$ similar with $ y,z,b,c$
So our ineq $ %Error. "lLeftrighttarrow" is a bad command.
\sum (a+x)^2 \ge \sum a^2+\sum x^2$ since $ LHS=4$ so we deduce it to  $ \sum (a+x)^2 \ge 4$
after applying Cauchy Schwarz,we are sufficient to prove $ (\sum a+\sum x)^2 \ge 12$ it is obviously true by Am-Gm because $ (\sum a)(\sum x)=3$


function
search
calculus
algebra proposed
algebra

Find all functions $ f: \mathbb{R}\rightarrow: \mathbb{R^{\plus{}}}$ such that
\[f^3\left(\frac{x+y}{2}\right)=f(x)f(y)\sqrt{\frac{f^2(x)+f^2(y)}{2}},\] for all positive reals $ x$ and $ y$.

Trivial : $ y\equal{}x \Longrightarrow f(x)\equal{}|f(x)|$ , $ y\equal{}\minus{}x\Longrightarrow f^3(0)\equal{}f^3(x)\Longrightarrow f(x)\equal{}f(0) \geq 0$
:cool:

[quote="Diogene"]Trivial : $ y \equal{} x \Longrightarrow f(x) \equal{} |f(x)|$ , $ y \equal{} \minus{} x\Longrightarrow f^3(0) \equal{} f^3(x)\Longrightarrow f(x) \equal{} f(0) \geq 0$
:cool:[/quote]

Maybe trivial, but surely not thru your demo :

$ y \equal{} \minus{} x\Longrightarrow f^3(0) \equal{} f(x)f(\minus{}x)\sqrt{\frac{f(x)^2\plus{}f(\minus{}x)^2}{2}}$

You seem to think that $ |f(\minus{}x)|\equal{}|f(x)|$ which is absolutely not obvious.

And, btw, $ f(x) \equal{} |f(x)|$ is an immediate conclusion of $ f(\mathbb{R})\subseteq\mathbb{R}^\plus{}$

[quote="mehdi cherif"]find all functions $ f: \mathbb{R}\rightarrow: \mathbb{R^{ + }}$ such that :

$ f^3(\frac {x + y}{2}) = f(x)f(y)\sqrt {\frac {f^2(x) + f^2(y)}{2}}$ for all $ x$ and $ y$

have fun :)[/quote]

Hello Mehdi Cherif !

I just have one ugly solution. And I hope there is something simpler :)

Obviously $ f(x)=c>0$ is a solution. So, let's search for non constant solutions.

Let $ g(x)=\frac{f^2(x)}{2f^2(0)}$ We have :

$ P(x,y)$ : $ 2g^3(\frac{x+y}{2})=g(x)g(y)(g(x)+g(y))$. and $ g(0)=\frac{1}{2}$

1) If $ g(x)$ is not constant, card($ g(\mathbb{R})$)$ =+\infty$
Demonstration :

Let $ u$ and $ a\neq 0$ such that $ g(u)\neq g(u+a)$
$ P(u,u+2a)$ $ \implies$ $ 2g^3(u+a)=g(u)g(u+2a)(g(u)+g(u+2a))$ which may be written :

$ 2=\frac{g(u)}{g(u+a)}\frac{g(u+2a)}{g(u+a)}(\frac{g(u)}{g(u+a}+\frac{g(u+2a)}{g(u+a)})$

From this, we get :
${ g(u)}>g(u+a)$ $ \implies$ $ g(u+a)>g(u+2a)$
${ g(u)}<g(u+a)$ $ \implies$ $ g(u+a)<g(u+2a)$

And so the sequence $ g(u+na)$ is strictly monotonous and hence all $ g(u+na)$ are different.
Q.E.D;

Now :
Let $ a\neq 0$
Let $ u=g(a)$
Let $ v=g(2a)$
Let $ w=g(-a)$
Let $ t=g(-2a)$

$ P(2a,0)$ $ \implies$ $ 8u^3 = v(2v+1)$
$ P(-2a,0)$ $ \implies$ $ 8w^3 = t(2t+1)$
$ P(-a,a)$ $ \implies$ $ 1 = 4uw(u+w)$
$ P(-2a,2a)$ $ \implies$ $ 1 = 4vt(v+t)$

This system of four equations with four unknown values has a limited set of solutions (demonstration follows).
Hence $ u=g(a)$ can take a limited set of values
Hence card($ g(\mathbb{R})$) is a finite number
Hence $ g(x)$ is a constant function (from point 1) above)

Hence solutions of initial equation are constant functions.
Q.E.D.

=======================
Here is the ugly part of my demo :
$ E1$ : $ 8u^3 = v(2v+1)$
$ E2$ : $ 8w^3 = t(2t+1)$
$ E3$ : $ 1 = 4uw(u+w)$
$ E4$ : $ 1 = 4vt(v+t)$

Showing that this set of equation has a limited set of solutions seems trivial but need some calculus ... :

From $ E3$, we get $ (u+w)^3=u^3+w^3+\frac{3}{4}$ and so $ 1=64u^3w^3(u+w)^3=64u^3w^3(u^3+w^3+\frac{3}{4})$
So, using $ E1$ and $ E2$ :

$ 8=vt(2v+1)(2t+1)(v(2v+1)+t(2t+1)+6)$

$ 8=vt(4vt+2v+2t+1)(2v^2+2t^2+v+t+6)$

$ 8=vt(4vt+2(v+t)+1)(2(v+t)^2 -4vt+(v+t)+6)$

Let then $ v+t=z$ and $ vt=\frac{1}{4z}$ (using $ E4$) :

$ 8=\frac{1}{4z}(\frac{1}{z}+2z+1)(2z^2 -\frac{1}{z}+z+6)$

$ (2z^2+z+1)(2z^3+z^2+6z-1)=32z^3$

And this, obviously, gives a limited set of values for $ z$. And each value of $ z$ gives one value of $ vt$ (thru $ E4$), so at most two values of $ v$ and $ t$, and so values of $ u$ and $ w$.

Q.E.D.


Given a regular $n$-gon, with $n$ odd and vertices $v_1,\cdots,v_n$, and $C$ its circumcircle. At each $v_i$ draw a circle that is internally tangent to $C$ at $v_i$, and suppose all these tangent circles are congruent. Let $P$ be any point on the minor arc from $v_1$ to $v_n$ and let $t_i$ be the length of the tangent from $P$ to the circle tangent to $C$ at $v_i$. Prove that
\[ \sum_{i=1}^n (-1)^it_i=0.  \]

Let angle(v1,O,P)=2x where O is center of C.
Let R be radius of C and r be radius of every little circle.
Let also Oi be the center of the circle at vi.
Let r/R be k.
Then
ti=sqrt(POi^2 - r^2)
=sqrt(R^2 +(R-r)^2 - 2R.(R-r).|cos(2x + (i-1).2.PI/n)| -r^2)=
= sqrt( 2.R^2( 1-k )( 1- |cos(2x+(i-1).2.PI/n)| ) )=
= const. sqrt(1-|cos(2x+(i-1).2.PI/n)|)

from here we need to show that
sum(i=1;n) of (-1)^i.sqrt(1-|cos(2x+(i-1).2.PI/n)|) = 0;  (*)

++ i will write the complete proof tomorow.

Hmm. I don't see how to complete the proof.


Let $S$ be the set of vertices of a unit cube. Find the sum of the areas of all triangles whose vertices are in $S$.

Since $C^8_3 = 56$ there are all together 56 triangles.
There are 24 triangles using just one face of the cube giving an area of $12$
There are 24 triangles with one leg as diagonal of a face and the other leg is an edge of the cube, giving an area of $12\sqrt{2}$
This leaves 8 equilateral triangles each with side $\sqrt{2}$, giving an area of $4\sqrt3$
Adding them up gives $12 + 12\sqrt2 + 4\sqrt3$

[hide]There are $24$ triangles on the faces of the cube which are $12$ complete square faces so $12$.

There are $12$ edges and $24$ triangles between opposite edges so the triangles between opposite edges are each $\sqrt2/2$. The sum of those areas is $12\sqrt2$.

There are $8C3 = 56$ triangles and we've counted 48 so there are 8 left to count. There are 8 equilateral triangles of area $\sqrt3/2$ each. So a sum of $4\sqrt3$.

Add them all up for $\boxed{12 + 12\sqrt2 + 4\sqrt3}$[/hide]


In the figure, line AC is a diameter of the circle with center O. If the circumference of the circle is 32 * pi, and the length of line AB is 16, what is the degree measure of <BAC?

I'm not sure how to approach this one.

Circum = 2 * pi * r ---> the radius = 16

Lines BO, AO and CO all = 8 each

I'm not sure if I need to find the area of the triangle, or of one of the triangles.

It appears to be an Isosceles triangle.

I'm not sure what to do.

Remember: A triangle inscribed in a semicircle is always a right triangle.

[quote="Brut3Forc3"]You've already found that $ \overline{AB} \equal{} \overline{AO} \equal{} \overline{BO}$. So you know that triangle $ ABO$ is equilateral, and that $ \angle BAC \equal{} 60^{\circ}$[/quote]

what?  There's no equalities given in the problem about $ AB$...

The triangles are iscosoles because of their sides (the radii of the circle). Given that these two triangles are congruent (because of SAS), you can assume $ \overline{AB} \equal{} \overline{BC}$.

I need clarification please.

I'm still not sure what the angle degree is.

[quote="7h3.D3m0n.117"][quote="Star00"]Lines BO, AO and CO all = 8 each[/quote]

This piece of information is incorrect.

Because BO, AO, CO, are the radii, and you are given indirectly that the radius is 16, then it makes sense for all three to be 16, and not 8.

Brut3Forc3's answer is correct.[/quote]

Thank you for pointing out my error.


It is quite easy, but I want to post a solution here:

Let $a+b+c=3$, by AM-GM inequality, \[a^{3}+1+1\geq 3a.\] Similarly, we are done.

General power mean inequality has been proven in [url=http://planetmath.org/?op=getobj&from=objects&name=ProofOfGeneralMeansInequality]here[/url].

replace a by a'x where a'+b'+c'=3 and divide by x

Naphthalin


Suppose an equilateral triangle in $R^{3}$ has vertices of the form $(a,b,c)$, where $a,b,c \in \{0,1,2,3\}$. How many such equilateral triangles exist?

This is an extension of a previous[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367586#p367586]AMC problem[/url].

[quote="JBL"]I don't see a nice way to approach this.  Here was how I thought I would go about it:

First, every such triangle can be translated so that it's a triangle with one of its vertices at the origin and its other vertices on the lattice cube $\{-3,-2,-1, 0, 1, 2, 3\}^{3}$ in exactly three ways.  Then we restrict ourselves to considering pairs of vertices at some fixed distance from the origin.  Then for each fixed distance, we have a relatively reduced number of cases to check.  But we still have to do the checking, and I didn't see any reasonable way to go about it.

However, luckily, I don't have to because the sequence already exists in the OEIS: it's sequence [url=http://www.research.att.com/~njas/sequences/A102698]A102698[/url].  There are a number of other similar sequences (acute triangles, obtuse triangles, scalene triangles, isosceles triangles, etc.) there as well, if you do a little searching.

Addendum: there are ways to count that are smarter than what I mentioned above.  In particular, you can consider it recursively: there are first those cases where the triangle fits inside a $2\times2\times2$ cube, then separately the case where it does not (and so where one of the coordinates of the translated triangle is $\pm 3$).  There are also lots of symmetries to expolit to make it more managable.[/quote]

the sloan sequence you found was created by a mathematician that i know at a local university, eugen ionascu.
he's researching the problem and some generalizations. he did in fact get his idea for the project from an AMC problem a few years ago, which paladin8 has referred to.


I probably made a dumb mistake somewhere but anyways ...

Let $ M$ be the mass and $ R$ be the radius of the sphere. Look at a sphere within the big sphere, with mass $ m$ and radius $ r$. Because the sphere has the same density,
\[ \frac{m}{M} \equal{} \frac{r^3}{R^3}\]
\[ m \equal{} \frac{Mr^3}{R^3}\]
\[ dm \equal{} \frac{3Mr^2}{R^3} dr\]
\[ \int r^2 m \, dm \equal{} \int_0^R \frac{3Mr^4}{R^3} dr \equal{} \frac{3M}{R^3} \left[ \frac{r^5}{5} \right]_0^R \equal{} \frac{3}{5} MR^2\]

 :?:  :?:  :?:

It's simply that you're integrating in the radial direction rather than an axial direction (w.r.t. an axis through the center of the sphere).  I would recommend using disks.