Let $ a_1, a_2, \ldots, a_{10}$ be ten distinct positive integers whose sum is 1995. Determine the minimum value of
\[ a_1a_2 \plus{} a_2a_3 \plus{} \cdots \plus{} a_9a_{10} \plus{} a_{10}a_1
\]

Can vieta's formulas be applied here somehow?

But we only have the sum of the roots, and I don't think that a polynomial equation of degree 10 will be much help here. Not sure, but that's my intuition

Maybe we can use the fact that $ ab \plus{} 1\ge a \plus{} b$ for $ a,b\ge 1$. The product $ ab$ increases at a much higher rate than which $ a,b$ increase. Therefore the minimum is attained when nine of the $ a_i$'s are $ 1$ and the remaining one is $ 1984$, which is $ 3976$.

There should be a better argument, I guess..  :maybe:

But the integers must be distinct. So following that logic, the integers must be 1, 2, 3, 4, 5, 6, 7, 8, 9, and 1950, and these can be arranged in some order to give a minimum sum of products. I couldn't find an easy way to prove it though

Since , the question hasnt specified that they arent in an AP or are increasing, We might as well take up both cases and work out a solution.


Best Of luck to this year's Kyrgyzstan IMO team.
Who finds the results first, post em here pls

[quote="ximik"]Who finds the results first, post em here pls[/quote]The results should (and will be) posted in the [url=http://www.mathlinks.ro/Forum/index.php?f=268]IMO 2005 Results Forum[/url]. This includes partial results that will appear over time :)

[quote="Mamat"]My results will be very bad  :([/quote]I am sorry for that, but please continue the discussion about the scores in the deficated forum :) Topic closed.


I just checked and my previously beautiful CSS code :D now malfunctions. Is it just something wrong with my blog or has AoPS changed the blog system?

Can you let us know what it looks like and what it's suppose to look like? When I just visited your blog it looked OK.

Yeah, I changed my code.
It was supposed to have no bar evident on the top of the screen. (The bar that says Home >> Blogs >> Dojo's dojo), and it would come up when I hovered over it. I actually changed that part, although it may be temporary.

Also, when I hover the posting box, the color of the box should have changed, but now the font just becomes more or less dark.

One more thing I noticed is the absence of opacity. Everything looks solid while my code says it shouldn't.

I looked at other blogs that had similar codes as mine, and they seemed to be suffering the same.

That is really odd. It now seems to work. We're all good.

Make sure you use appropriate CSS for Internet Explorer.

See http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=278589 for more info.

Hmmm, if I change the code, will it be changed on other browsers?

I'm not sure what you mean. The CSS will download to all browsers.

I mean, if you change the opacity code in the CSS, will it effect the output on other browsers?

You're not changing code, you are adding code. filter is an IE CSS thing, it is essentially ignored by other browsers. I still don't really get your question since

opacity: 0.5;

and

filter:alpha(opacity=50);

do the [b]exact same thing[/b] but on different browsers. Thus, even if other browsers read it, it wouldn't make a lick of difference.


Find the measures of the angles of a triangle ABC, such that
$ cos\frac{5A}{2}\plus{}cos\frac{5B}{2}\plus{}cos\frac{5C}{2}\equal{}\frac{3\sqrt{3}}{2}$

I think that too, but where's the solution? :D

There's no solution for this easy problem!

Let ${ \frac{5A}2},y=\frac{5B}{2}$, then $ 0<x,y$ and $ x+y<\frac{5\pi }{2}$.
$ f(x,y)=cos x+cos y+cos(\frac{5\pi}{2} -x-y)=cos x+cos y +sin(x+y)$.
We can find $ max(f(x,y))$. It give $ x=y=\frac{pi}{6}$ or $ x=\frac{\pi}{6},y=x+2\pi$ or $ y=\frac{\pi}{6}+2\pi$.
$ max f(x,y)=\frac{3\sqrt 3 }{2}.$
Therefore solutions are
 $ A=\frac{\pi}{15}=B,C=\frac{13\pi}{15}$ and permutations (A,B,C).

$ \cos\frac {5A}{2} \equal{} \cos\frac {5\pi}{6} \equal{} \minus{} \frac {\sqrt {3}}{2}$, yielding $ \minus{} \frac {3\sqrt {3}}{2}$, not $ \frac {3\sqrt {3}}{2}$!!

One possibility is though $ A \equal{} B \equal{} \frac {\pi}{15}$, $ C \equal{} \frac {13\pi}{15}$...

Now, since $ \frac {3\sqrt {3}}{2} > 2$, all three cosines must be positive, or all angles of the triangle must be in $ [0,\frac {pi}{5})$ or in $ (\frac {3\pi}{5},\pi)$.  therefore, two of the angles must be in the first range, one in the second (or the three angles would not add up to $ \pi$).  Assume wlog that $ C\in (\frac {3\pi}{5},\pi)$.  The LHS is written as
\[ 2\cos\frac {5A \plus{} 5B}{4}\cos\frac {5A \minus{} 5B}{4} \plus{} \cos\frac {5C}{2}.
\]
Now, $ \cos\frac {5A \plus{} 5B}{4}$ is directly determined by the value of $ C$, and it is positive.  Or the maximum of the LHS for any given $ C$ is obtained when $ A \equal{} B$, so that $ \cos\frac {5A \minus{} 5B}{4} \equal{} 1$.  Then, for $ B \equal{} A$, we have
\[ \cos\frac {5C}{2} \equal{} \cos\frac {5\pi \minus{} 10A}{2} \equal{} \sin 5A.
\]
In order to find the maximum of the LHS, we need to find thus the maximum of $ \sin 5A \plus{} 2\cos\frac {5A}{2}$, or calling $ \alpha \equal{} \frac {5A}{2}$, we need to find the maximum of $ \sin 2\alpha \plus{} 2\cos\alpha$, which happens when $ \cos 2\alpha \equal{} \sin\alpha$, or when $ \alpha \equal{} \frac {\pi}{6}$ for the given interval where $ A$ may be found.  The maximum (which is found iff $ A \equal{} B \equal{} \frac {\pi}{15}$) is then $ \frac {3\sqrt {3}}{2}$, or the given equation is only true for this maximum, and the given solution is the only solution (except for permutations).

Just when posting saw that Rust beat me to it...  good job Rust!


[i]Prove that at least side of a Pythagorean triple is always prime. [/i]

My coach gave me this one. I couldn't do it on the spot, so I decided to post it on AoPS.

EDIT: Sorry about that! It meant to be at least one side!

You can't prove it, because it's false. A counterexample would be 9,40,41.

Go ahead and try the problem now! :)

Sorry if this is spam! :D

But what about 6,8,10? The "theorem" we're trying to prove is still false.

no as in base triples, like 3-4-5.

6-8-10 is just 2(3,4,5).

By base triple, do you mean primitive Pythagorean triple?

Yeah. I think so.

Like 5-12-13
and so on

So 6-8-10 doesn't count. The farthest I can do on the spot is that they must be relatively prime, or else it would be like 6-8-10.

... I guess I didn't correctly remember the question, or my coach made a mistake... The latter is highly unlikely, but possible...

Proof by contradiction. It doesn't work.

33=3*11
544=2*2*2*2*2*17
545=5*109

This is an original pythagorean triplet.

[quote="nikeballa96"]what about $ 21, 220, 221$?

$ 21 \equal{} 7\cdot3$

$ 220 \equal{} 2^2\cdot5\cdot11$

$ 221 \equal{} 17\cdot13$[/quote]

@batteredbutnotdefeated, both this example and mine are proof of counterexamples. :)

Hope this helps  :)

Yes... It seems there seems to have been a mistake in the question asked... Thanks for all your help guys!


i think you mean to evaluate P in term of a and b  , or  :maybe:

seems really hard, i wonder if anyone have the solution


calculus
integration
function
calculus computations

A function $f(x)$ satisfies $\begin{cases} f(x)=-f''(x)-(4x-2)f'(x)\\ f(0)=a,\ f(1)=b \end{cases}$

Evaluate $\int_0^1 f(x)(x^2-x)\ dx.$

Yes, your answer is correct, Beginner. :)

[quote="kunny"]A function $ f(x)$ satisfies $ \begin{cases} f(x) \equal{} \minus{} f''(x) \minus{} (4x \minus{} 2)f'(x) \\
f(0) \equal{} a,\ f(1) \equal{} b \end{cases}$

Evaluate $ \int_0^1 f(x)(x^2 \minus{} x)\ dx.$[/quote]

I tried using integral by parts but can't get the answer. :(


If $ x,y,z$ are from $ (0,1)$ so that $ x\plus{}y\plus{}z\geq \frac{3}{4}$, then  $ \frac{x}{1\minus{}y\minus{}z}\plus{}\frac{y}{1\minus{}x\minus{}z}\plus{}\frac{z}{1\minus{}x\minus{}y}\geq \frac{3}{2}$

This inequality isn't right. For example $ x\equal{}y\equal{}z\equal{}\frac{2}{3}$, then $ x,y,z \in (0, 1)$ and $ x\plus{}y\plus{}z\ge\frac{3}{4}$ but $ \frac{x}{1\minus{}y\minus{}z}\plus{}\frac{y}{1\minus{}x\minus{}z}\plus{}\frac{z}{1\minus{}x\minus{}y}\equal{}\minus{}6< \frac{3}{2}$

Yeah, consider $ x\plus{}y,y\plus{}z,z\plus{}x<1$

[quote="skylover"]If $ x,y,z$ are from $ (0,1)$ so that $ x \plus{} y \plus{} z\geq \frac {3}{4}$, $ x \plus{} y,y \plus{} z,z \plus{} x < 1$ then  $ \frac {x}{1 \minus{} y \minus{} z} \plus{} \frac {y}{1 \minus{} x \minus{} z} \plus{} \frac {z}{1 \minus{} x \minus{} y}\geq \frac {3}{2}$[/quote]

This inequality does not hold. Try $ x\equal{}\frac 3 8$, $ y\equal{}\frac 1 4$, $ z\equal{}\frac 1 8$. Then $ x\plus{}y\equal{}\frac 5 8<1$, $ x\plus{}z\equal{}\frac 1 2<1$, $ y\plus{}z\equal{}\frac 3 8<1$, $ x\plus{}y\plus{}z\equal{}\frac 3 4$, but

\[ \frac {x}{1 \minus{} y \minus{} z} \plus{} \frac {y}{1 \minus{} x \minus{} z} \plus{} \frac {z}{1 \minus{} x \minus{} y} \equal{} \frac{43}{30} < \frac{45}{30}\equal{}\frac{3}{2}\]

Sergey Markelov


If z is a complex number with |z| = 1, find the value of $|1 + z|^2 + |1 - z|^2$.

[hide="My solution"]Let make $z=a+bi$.  Thus $\sqrt {a^2+b^2}=1$ :arrow: $a^2+b^2=1$.  Substituting that back to the original equation, we get $|a+1+bi|^2+|-a+1-bi|^2=(a+1)^2+b^2+(a-1)^2+b^2=a^2+2a+1+2b^2+a^2-2a+1=2(a^2+b^2)+2=2+2=4$

Thus, the answer is 4.[/hide]

Yeah, that works, and thats how I did it at first... but finding a geometric approach is cooler :P

[hide]
[img]http://img.photobucket.com/albums/v213/joml88/complex.png[/img]

In the picture, $AC=|1+z|$ and $AD=|1-z|$.  We already know a bunch of lengths, $AB=BC=BD=1$.  So, a quick application of Stewart's gives us:

\[|1+z|^2+|1-z|^2=(1)(1)(2)+1^2(2)=\boxed{4}.\] QED[/hide]

Very nice joml88, quite different method to show that problem solving is fun  :coolspeak:   :clap2:

[quote="Psycho"][hide]$|1+z|^2+|1-z|^2$
$=(1+z)^2+(1-z)^2$
$=1+2z+z^2+1-2z+z^2$
$=2(1+z^2)$
$=4$[/hide][/quote]

There is a small problem with your solution.

[hide]$|1+z|^2 \neq (1+z)^2$, because on the left side you have a (necessarily) positive real number, and on the right you have something that may be complex (a complex number squared is still a complex number).

What you really want to use is, $|z|^2$ (of a complex number z) = $z*\overline{z}$, where $\overline{z}$ means the complex conjugate of z. It turns out that the middle terms will still cancel, but they'll be $\pm (z + \overline{z})$ instead of $\pm 2z$.[/hide]


How many integers $ n$ can be written as $ a\plus{}b\plus{}c$, where $ a$, $ b$, and $ c$ are distinct odd integers between 2 and 20?

[hide]Note that a+b+c has to be odd. The minimum you can have is 3+3+3 = 9 and the maximum 19+19+19 = 57. We can have all of the odd integers between these two values, which amounts to 25.[/hide]

for this problem we can use all of the odd integers from 3-19, inclusive.
the minimum sum is 3+5+7=15, and the max is 19+17+15=51. if an odd integer n can be  made by a sum, then n+2 can be made by "brining up"(+2) the largest number in the sume. the only way that this cannot be done is if the largest is 19, in which case you would "bring up the second largest. by similar logic this cannot be done only if the second largest is 17, in which case you would bring up the third largest, which can only be stopped if the numbers are 15,17,19. so if i can get a sum n, i can get n+2 unless the sum is 19+17+15=51.
so i have as possibilities (as sums) all the odds from 15 to 51, which is (51-15)/2+1=19
if i am unclear, im sorry. maybe someone else who understands my logic can explain?.
btw, is 19 the answer?

@davidlu, im pretty sure ur answer is wrong since the integers have to be distinct.

Shh... don't tell anyone my real name! :P
Good point, distinct integers. Your solution is correct.


[hide="Hint for 1"]Each arrangement consists of three different letters.  How many ways are there to put three different letters in alphabetical order?[/hide]

[quote="superwilson"]Q#1
Compute the number of 3 letter arrangements that can be formed if no letter is repeated and the letters are in alphabetical order.

Q#2
Compute the sum of all positive integers greater than 1 and less than 1000, which leave a remainder of one when, divided by either 2, 3, 4, 5, or 6.

Q#3
Three positive integers form an arithmetic progression.  If the smallest number is increased by one, the arithmetic progression becomes a geometric progression.  In the original arithmetic progression if the largest number is increased by two, the arithmetic progression also becomes a geometric progression. Compuite the smallest number of the original arithmetic progression.[/quote]
[hide="1"]24? Probably wrong...[/hide]

[hide="#3"]Let the numbers be $a-d,a,a+d$. Then we have

$\begin{cases}(a-d+1)(a+d)=a^{2}\\(a-d)(a+d+2)=a^{2}\end{cases}\iff\begin{cases}a^{2}-d^{2}+(a+d)=a^{2}\\ a^{2}-d^{2}+2(a-d)=a^{2}\end{cases}$

Subtracting the first equation from the second we get

$2(a-d)-(a+d)=0\iff a-3d=0\iff a=3d$

Plugging that into the first equation we get

$(2d+1)\cdot 4d=9d^{2}$

Obviously $d$ can't be zero, since three equal numbers wouldn't meet the given conditions. Hence the last equation simplifies to

$4(2d+1)=9d\iff 8d+4=9d\iff d=4\iff a=3d=12$

The smallest number is $a-d=8$ (the sequence is $8, 12, 16$)[/hide]

[quote="anirudh"]I probably overcounted but...
$\binom{26}{3}=2600$

EDIT: THis is problem 1[/quote]

I believe that's correct for #1, because it's 26*25*24 divided by 6, as there's only one working arrangement out of six for each combination of letters...

lol no, think about it, i'm just saying there are 26*25*24 ways to arrange letters without repeating, and every set of three letters in alphabetical order has five partners that aren't in alphabetical order, plus you can make a list and count, I did and I'm pretty sure that the answer is 2600

Oops sorry...yes i get it..... :oops:  LoL...

[quote="nebula42"][quote="fizsle"]I think it should be 24*24*24/6 = 2304

The first letter cannot start with Y or Z as then we will not be able to satisfy the condition...same second cannot be started with A or Z....[/quote]
First of all, I think you meant just $24^{3}$.
However, unfortunately, this doesn't work because it greatly overcounts because it also counts arrangements like GRF, which doesn't satisfy the condition. The actual answer is
[hide]${26 \choose 3}=2600$ because for each combination of three distinct letters (out of 26), there is exactly one way to arrange them to satisfy the conditions. (For example, if you chose R, B, and E, the only way you could arrange them would be B, E, R) Also, every arrangement that satisfies the conditions is counted exactly once in this way.[/hide][/quote]

LOL. I was actually right :rotfl:

[quote="mustafa"][hide="Hint for number 2"]Use modular stuff
[/hide]
[hide="#2"]Any number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6  must be exactly 1 more than a number that is divisible my all 5 of those. The smallest such number is 60, so you would find all of the numbers between 1 and 1000 that are one more than a multiple of 60.[/hide][/quote]

[hide]
I think you mean 61, but that's the right idea.[/hide]


While typing a letter in the latex software the command [code]\math[...][/code] will give some different shapes and forms to the letter ,for instance :

$P \longrightarrow \mathcal P \longrightarrow \mathbb P$

I want to know ,what are the possible suffixes for the command [code]\math...[/code]

In the above exmaples ,suffixes are [code] CAL,BB[/code]

There's also \mathbf for bold math font.

Look in [url=http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to $\LaTeXe$[/url] at Table 3.19 page 80 (=page 66 at top of page)  and Table 6.4 page 121 which give
\mathrm
\mathit
\mathnormal
\mathcal
\mathscr
\mathfrak
\mathbb
\mathbf
\mathtt
along with examples.

like this :

$\mathrm {ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ :)
$\mathit {ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ :lol:
$\mathnormal {ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ :D
$\mathcal {ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ :P
$\mathscr {ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ :roll:
$\mathfrak {ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ ;)
$\mathbb {ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ :arrow:
$\mathbf {ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ :idea:
$\mathtt {ABCDEFGHIJKLMNOPQRSTUVWXYZ}$ :ninja:


Ross Mathematics Program
HCSSiM
summer program
PROMYS

I am considering attending the Ross Mathematics Program this upcoming summer, and I was wondering if any former (or current) Ross students would be willing to share their opinion of the program.

Would you recommend the program?
Was the overall experience positive/negative?
How is Ross compared to other programs?

I would recommend it. I heard that it was much more intense than other mathematics programs, so I was a little wary when I chose to go to Ross instead of the Hampshire Studies or PROMYS. But it wasn't bad at all. Sure, they give you lots of math, but everyone goes at their own pace, and there's not too much stress.
You will definitely learn a lot, as long as your attitude is positive. Again, they throw a LOT of math at you in eight weeks, and to some people (like me), it is a completely new way of approaching mathematics. At the same time, there's a lot of fun activities, most notably traditional frisbee games every weekend.
All in all, its a great program; let me know if you have specific questions.

...for the teacher recommendation, can you use a math coach? (i.e somebody you've never formally had as a teacher, so they wouldn't show up on your transcripts.)

Just wondering. Yes, I know, it depends on the other problems as well, but in terms of problems solved, how many did you guys have? I think I can get around 6 of them this year...

I'm not sure about your math coach question, but for the application problems, I solved all of them. That being said, I'm sure you could still get in without solving them all completely, but I don't know if there is a "minimum" number of problems you should solve.

I got into Ross but went to promys, and I also solved all the questions.

how long does it take usually before they reply

A week or two, I think. I was accepted just today.

Will it be too late if I send in my application forms by around April 20th? Do I still get chance then? Do you recommend it? well, I applied to Promys and it just occured to me that i might try ross if I do not get in to promys. I haven't started on the ross problems and if it's not advisable to apply so late, then I won't even try it. Please give me some suggestions.


[quote="warut_suk"]Prove that $(\cos\frac{A}{2}\cdot\frac{2bc}{b+c})^2 = \frac{4bc}{\left(b+c\right)^2}\cdot p\left(p-a\right)$ where $p = \frac{a+b+c}{2}$[/quote]

use double angle theorem and cosine laws ....

[quote="warut_suk"]Actually both of them are angle bisector lengths.  :D
Can you prove it?[/quote]
In triangle ABC, bisector AD intersect BC at D. Let $AD=x$. Use the law of cosine:

$\cos{\frac{A}{2}}=\frac{b^2+x^2-CD^2}{2bx}$

$\cos{\frac{A}{2}}=\frac{c^2+x^2-BD^2}{2cx}$

Solve for CD and BD:

$-BD^2=2xc \cos{\frac{A}{2}} -x^2-c^2$
$-CD^2=2xb \cos{\frac{A}{2}} - x^2-b^2$

Since $\frac{BD}{CD}=\frac{c}{b}$, so

$\frac{c^2}{b^2}=\frac{2xc \cos{\frac{A}{2}} -x^2-c^2}{2xb \cos{\frac{A}{2}} - x^2-b^2}$

Simplify, we get:
$(c^2-b^2)x^2+2bc \cos{\frac{A}{2}} (b-c) x = 0$

Since $x \neq 0$, so $x= \frac{2bc}{b+c} \cdot \cos{\frac{A}{2}}$ is the only root.


When I make a title in $\LaTeX$ without an email address, it works fine.  However, when I insert the \email{} command below the \author{} command I get an error.  I have searched Google and just adding the \email{} command seems like the thing I'm supposed to be doing.  Would someone please tell me how to add an email address to the title page of a $\LaTeX$ document?

It's strange, but try putting this code in the preamble:
[code]\def\maketitle{%
  \thispagestyle{empty}\vbox to \vsize{%
    \haut{}{\@blurb}{}
    \vfill
    \ligne{\Large \@title}
    \vspace{5mm}
    \ligne{\Large \@author}
    \vspace{1mm}\ligne{\texttt{<\@email>}}
    \vspace{1cm}
    \vfill
    \vfill
    \bas{}{\@location, \@date}{}
    }%
  \cleardoublepage
  }
\def\date#1{\def\@date{#1}}
\def\author#1{\def\@author{#1}}
\def\title#1{\def\@title{#1}}
\def\location#1{\def\@location{#1}}
\def\blurb#1{\def\@blurb{#1}}
\def\email#1{\def\@email{#1}}
[/code]
I think that the problem comes from tue use of @ sign...

[b]PS:[/b] I don't have LateX installed in my PC.

[quote="amcavoy"]This isn't working for some reason.  By the way, since you don't have $\LaTeX$ installed, is this just a guess?[/quote]

No, I copied it from a french web site :)
This is the full link: [url]http://zoonek.free.fr/LaTeX/LaTeX_samples_title/0.html[/url]
It's in the second example.

The code define the \email command because it's not defined in the $\LaTeX$ classes.


I know that many more people take the 12A than the 12B, except I'm not sure how well people do in the 12A compared to the 12B. I was just wondering which is more advantageous to score higher on for USAMO? Like if person A only took the 12A and got a 200 index whereas person B only took the 12B and got a 200 index, would it matter in comparison to other people taking the same test?

Well, according to the AMC's criteria for selecting USAMO qualifiers, they will take qualifiers in equal proportion to how many people took the A or B date, respectively. However, I'm willing to bet that, unless you are really really a thin line, it won't even matter.


function
integration
geometry
rectangle
ratio
calculus
real analysis

Fix a positive integer $N \in \mathbb{N}$. Does there exist a Riemann integrable function $F:\mathbb{R}^2 \to \mathbb{R}$ such that
\[
\int_a^b \int_c^d F(x,y) \, \mathrm{d}x \, \mathrm{d}y = 0
\]
if and only if $N\frac{a-b}{c-d} \in \mathbb{Z}$?

I agree with grobber. I saw Vess sometime ago in a problem with tiling :D
 But I don't think such function can exist. Because, as the continuousness of the double integral and the dense of $\mathbb{Q}$ the double integral should be $0$ trivially

[quote="grobber"]Just a question: did this idea, by any chance spring from problems about paving a rectangle with other rectangles? I'm asking because if we had such a function, we would find a quick solution to the following problem (solved in a stronger form on the forum):

Call a rectangle good if the ratio of its sides is rational. Show that a rectangle can be paved with finitely many good rectangles iff it's good.[/quote]

Of course  :) . I know three proofs of the mentioned fact, but I was just wondering whether this standard approach would work as well. Okay, I agree it won't work as stated; see the edited opening post for a revised question that would also imply the assertion.

I don't think this is going to work.

Let $G(x,y)=\int_0^x\int_0^y F(u,v)\,dv\,du$
Then $G$ is continuous. On the curve $y=rx$, where $r$ is an arbitrary rational number, $G$ takes on only rational values. By the Intermediate Value theorem and the fact that the irrational numbers are dense, $G$ is constant and must be equal to its value $0$ at $(0,0)$. By the fact that these rational lines are dense, $G$ is identically zero, and we don't have the uniqueness.

Yes, I realized the question was silly, sorry about this  :blush: . Actually, it is clear that we cannot obtain a solution to the tiling problem by restricting a condition on the integral to be equivalent to $(a-b)/(c-d) \in \mathbb{Q}$, because the condition that the tiling consists of finitely many squares is absolutely essential (the assetion is not true for infinite tilings). Finally, the ammended question which, I believe, is not so trivial.

We might as well set $N=1$. Now, consider rectangles of width $\frac{\alpha}{m}$ and height $\frac{n\alpha}{m}$. Since the ratio height/width is an integer, the integral over any such rectangle is zero. Glue $m$ such rectangles together horizontally; this gives a rectangle of width $\alpha$ and height $\frac nm \alpha$. We're back to where we started, and we already know that doesn't work.

In fact, if $\int_a^{a+c}\int_b^{b+c}F(x,y)\,dy\,dx$ is always rational (integral over squares), the arguments here suffice to show that $\int_a^b\int_c^dF(x,y)\,dy\,dx=0$ always, which implies that $f=0$ almost everywhere.

This idea is dead.

[quote="jmerry"]
This idea is dead.[/quote]

 :(  :(  :(

I see, very stupid of me  :blush: Sorry for wasting your time.


1)The number of ways in which $6$ men can be arranged in a row so that 3 particular men are consecutive is?(My book says my answer for this dumb question is wrong)

2)The total number of words that can be made by writing the letters of the word $PARAMETER$ so that no vowel is between two consonants is?

3)Let $A$ be the set of $4-digit$ numbers $a_1a_2a_3a_4$ where $a_1>a_2>a_3>a_4$ then $n(A)$ is equal to?

4)The number of $5-digit$ even numbers that can be made with the digits $0,1,2,3$ is?

1)

[hide] There are 6 ways to put the designated three men in order.  They can be in slots 123,234,345, or 456 for four possibilities.  There are then 6 ways to place the remaining three men.  6*4*6=144 ways.[/hide]

4)

[hide] The last digit can be 0 or 2 for 2 possibilities.  The first digit can't be 0, for 3 possibilities there.  The other 3 digits have 4 options.  So 3*4*4*4*2=384 numbers.[/hide]

[hide="3"]

If $A_1=9$, we have $9$ possible digits remaining to choose for the other $3$. This provides $\displaystyle\binom{9}{3}$

If $A_1=8$, we have $8$ possible digits to choose from - $\displaystyle\binom{8}{3}$

Continuing this until $A_1=4$, we have $\displaystyle\binom{3}{3}$

The total number of possible numbers is then $\displaystyle\binom{3}{3} + \displaystyle\binom{4}{3}+ ... + \displaystyle\binom{9}{3} = \displaystyle\binom{10}{4}=210$

[/hide]

[quote="shadysaysurspammed"]2)The total number of words that can be made by writing the letters of the word $PARAMETER$ so that no vowel is between two consonants is?[/quote][hide="Maybe"]4 vowels, 2 of them E's, 2 of them A's. 5 consonants, 2 of them R's, the others unique. Group the vowels, group the consonants. Multiply permutations together. $\frac{4!}{2!2!} \times \frac{5!}{2!} = 6 \times 60 = 360$. Multiply by 2 (vowels grouped before or after consonants) so [b]720[/b].[/hide]

hokkage according to my book,that answer is worng,much more than that(i got it different as well)

[quote="236factorial"]I did the same thing as Hokkage, but I multiplied 360 by 5 instead (since cosonants can be between vowels), and got 1800. Is that right?[/quote]
I get 1800 also.

Consider all the consonants as one whole block, then there are $\frac{4!}{2!2!}$ ways to arrange the vowels, and $5$ ways to place the block of consonants within the vowels, so $30$ ways in total.  Then, There are $\frac{5!}{2!}=60$ ways to arrange the consonants, so $30*60=1800$ ways in total

Easy solution for 3...

For any 4 digits you pick, there is only one way to organize them so they are in decreasing order.  Thus the answer is $\binom{10}{4} = 210$

Yeah,1800 is right(thats what i got,not the book)


AMC
USA(J)MO
USAMO
algebra
polynomial
binomial coefficients
algebra proposed

A polynomial p(x) of degree n satisfies p(0) = 0, p(1) = 1/2, p(2) = 2/3, ... , p(n) = n/(n+1). Find $P(n+1)$.

The polynomial $Q(x) : = (x+1)P(x)-x$ is of the degree $n+1$ and has $n+1$ zeroes: $0,1,\dots,n$. Therefore $Q(x)=Cx(x-1)\dots(x-n)$ where $C$ is a constant. That gives

$P(x)=\frac{Cx(x-1)\dots(x-n)+x}{x+1}$

Since the fraction has to be a polynomial, the numerator must vanish for $x=-1$:

$C(-1)^{n+1}(n+1)!-1=0\iff C=\frac{(-1)^{n+1}}{(n+1)!}$

Hence $P(n+1)=\frac{\frac{(-1)^{n+1}}{(n+1)!}(n+1)!+n+1}{n+2}=\frac{n+1+(-1)^{n+1}}{n+2}=\begin{cases}1 & n\ \mathrm{odd}\\{n\over n+2}& n\ \mathrm{even}\end{cases}$

Nice solution, Farenhajt!

The problem doesn't ask to, but we can find a formula for $p(x), \ \forall x \in \mathbb{R}$

[hide]Using B-splines we can find that $p(x) = \sum_{k=2}^{n+1}{(-1)^{k}\cdot \frac{1}{k}\cdot \binom{x}{k-1}}$

[hide="About binomial coefficients"]We'll use the binomial coefficients

$\binom{x}{k}= \frac{x!}{(x-k)!k!}=\frac{x(x-1)(x-2)...(x-k+1)}{k!}$

It's a k-degree polynomial of $x$. So, we can use them even if $x \in \mathbb{R}$.

But notice that, if $x<k$ and $x,k\in \mathbb{N}\Rightarrow \binom{x}{k}=0$
--------------------------------------------------------------[/hide]
We'll see that $p(x)=\frac{x}{x+1}, \ \forall x \in \{0,1,...,n\}$

Indeed,

[hide="calculations..."]
$\frac{1}{k}\cdot \binom{x}{k-1}= \frac{1}{k}\cdot \frac{x(x-1)...(x-k+2)}{(k-1)!}= \frac{x(x-1)...(x-k+2)}{k!}=$

$= \frac{1}{x+1}\cdot \frac{(x+1)x(x-1)...(x-k+2)}{k!}= \frac{1}{x+1}\cdot \binom{x+1}{k}\Rightarrow $
------------------------------------------------------------------------------[/hide]

$\frac{1}{k}\cdot \binom{x}{k-1}= \frac{1}{x+1}\cdot \binom{x+1}{k}$

So, $\boxed{p(x) = \frac{1}{x+1}\sum_{k=2}^{n+1}{(-1)^{k}\binom{x+1}{k}}}\ \ (1)$

Now, we can calculate the values of $p(x)$ for $x \in \{0,1,...,n\}$

$\binom{x+1}{k}=0, \ \forall k>x+1$, so we can simply write $p(x) = \frac{1}{x+1}\sum_{k=2}^{x+1}{(-1)^{k}\binom{x+1}{k}}$

But

$0 = (-1+1)^{x+1}= \sum_{k=0}^{x+1}{(-1)^{k}\binom{x+1}{k}}\Rightarrow $

$\binom{x+1}{0}-\binom{x+1}{1}+(x+1)p(x) = 0$

$(x+1)p(x) = \binom{x+1}{1}-\binom{x+1}{0}= (x+1)-1=x$

So, $p(x) = \frac{x}{x+1}, \ \forall x \in \{0,1,...,n\}$

The polynomial $p(x)$ is well defined from its $n+1$ values, so it is unique. Now we are sure that $p(x)$ is given by the formula $(1)$

[hide="check the answer for p(n+1)"]
Setting $x=n+1$ in $(1)$ we get:

$p(n+1) = \frac{1}{n+2}\sum_{k=2}^{n+1}{(-1)^{k}\binom{n+2}{k}}= \frac{1}{n+2}\cdot S$

$\binom{n+2}{0}-\binom{n+2}{1}+S+(-1)^{n+2}\binom{n+2}{n+2}=0 \Rightarrow $

$1-(n+2)+S+(-1)^{n}=0 \Rightarrow $

$S=(n+1)-(-1)^{n}\Rightarrow $

$S=(n+1)+(-1)^{n+1}$ and $p(n+1)=\frac{S}{n+2}$, same as Farenhajt's[/hide][/hide]


How do you write the math fonts? (like exponents, radical signs, etc.) I don't see any in my keyboard. :huh:  Do you need to download anything? If so, could you folks out there tell me how to do it?

Thanks for the reply, and sorry for my post location. I am a new member, so there are things I don't know. I't try to move it.

Take a look at this: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690[/url]


this if from a mexican tst:

 an integer  N is n-good if it has the following properties:

1. N has at least n distinct prime divisors
2.there exist positive divisors of N: 1,x_2,x_3,...x_n
such that N=1+x_2+x_3+...+x_n

prove that there exists a n-good integer  for n>=6

any solutions

please, moderators, lock this topic!

this problem now is in brazilian training lists, and can't be argued!

thanks!  :)

where can we find all the problems from mexico or can you PM me if they are not on ML...thx


function
integration
calculus
derivative
calculus computations

I am stuck at the point of deciding whether x=0 is point of inflection or point of maximum for the function

$f(x) = \int\limits_0^{x^2 } {\frac{{t^2  - t}}{{e^t  + 1}}} dt$

[hide]First derivative test shows that the function has local maximum at x=0, but second derivative also is becoming zero at x=0.[/hide]

How about some common sense?

$f(x)=\int_0^{x^2}g(t)\,dt$, so $f(x)=f(-x)$. For $0<t<1$, $f(t)<0$, so $g(x)<0$ for $0<x<1$. This makes $x=0$ a local maximum for $f$, and $\pm1$ local minima.

For $|x|$ small, $f(x)=-\frac{x^4}4+ax^6+\cdots$

It's not an inflection point, since the second derivative does not change sign at 0.


calculus
derivative
algebra unsolved
algebra

a)

define: $f(q) = \sum_{i=1}^n q^i = \frac{q^{n+1}-q}{q-1}$

then the derivative of f,

$f'(q) = \sum_{i=1}^n i*q^{i-1} = \frac{(n(q-1)-1)q^n+1}{(q-1)^2}$

b) can be done in a similar fashion by taking the derivative of $q*f'(q)$ since this derivative is just

$\sum_{i=1}^n i^2*q^{i-1}$

Ok! :)
I should had done it! ;)
I thought of using Abel's formula.Would it be resultful?

Not only resultful but also easy.
$1+2q+3q^2+...+nq^n-1=(1-2)+(2-3)\cdot(1+q)+...+n(1+q+q^2+...+q^{n-1})
=-(\frac{q-1}{q-1}+\frac{q^2-1}{q-1}+...+\frac{q^{n-1}-1}{q-1})+n\cdot\frac{{q^n}-1}{q-1}$
and the rest it's obvious. Similarly can be done the b)part!

Oh, sailor, you remembered the Abel identities!

I was thinking over it and I solved it.I just only forgot to post it! :D
...however thanks!
ARTIONA :P

having, \[
\begin{array}{l}
S = 1 + 2q + 3q^2  +  \cdot  \cdot  \cdot  + nq^{n - 1} , \\
 qS = q + 2q^2  +  \cdot  \cdot  \cdot  + (n - 1)q^{n - 1}  + nq^n , \\
 \end{array}
\]
and
\[
(1 - q)S = 1 + q + q^2  +  \cdot  \cdot  \cdot  + q^{n - 1}  - nq^n
\]
Intersting,isn't?


I'm not sure this is exactly geometry, and is really pretty simple, but anyways.

The 4-color theorom states that any map on a two dimensional plane can be colored with only 4 colors so that no two countries (or states, etc.) which share a border (a line, not a point) are the same color. (Ok, I know, I'm oversimplifying it, but that's the gist.)

Consider the 3-d analog, which is much easier to prove. How many colors are needed to color any "map" in three dimensions? Demonstrate why...

.dzhonatan

[quote="liyi"]any number of colours.[/quote]

no, not just "any number"

you'd need an unlimited amount of them, it is not possible thus

(just take some meters of sewing-thread in different colors and try out. You'll see what I mean :) you can connect each color problemlessly to each other color)

A better question is how many colors are needed to color a map on a standard torus? What about on a double-holed torus?

This is a famous problem. It  was actually solved long before the question of coloring maps on the plane was resolved.

The minimum color number which needed to color a surface with Euler Characteristic n is [(7+ \sqrt (49-24n))/2], known as the heawood formula.


Let segement AB be the diameter of a circle with center C and area 36pi. Then P is a point on the circle such that AC=2AP (where AC And AP are line segments). Find the length of segment PB.

How would you do this problem?

[hide]The area of the circle is $36\pi,$ so it's radius, or $AC,$ is 6. We know $P$ is the midpoint of $AC,$ so $PC=3$ and $BC=6.$ Adding, we get $PB=\boxed{9}.$[/hide]

Since
By cosine theorem
$Cos(\angle ACP)=\frac{7}{8}$
$Cos(\angle PCB)=\frac{-7}{8}$
so
$PB=3\sqrt{15}$

I_like_pie, please compare what the problem says:

[quote="mr. math"]Then P is a point [b]on the circle[/b] such that...[/quote]

with what you wrote:

[quote="i_like_pie"]We know $P$ is the midpoint of $AC$[/quote]

and draw the necessary conclusions :)

It also says $AC=2AP,$ but I didn't realize it could be anywhere. :ninja: Conclusion = I messed up.

But then couldn't $P$ be anywhere on the red arc?

[img]http://img140.imageshack.us/img140/6428/locusofpointsfa4.jpg[/img]

Language lesson:

"ON a circle" = on the circular line;

"IN a circle" = within the area bounded by the circular line.


Is there a mathematical way to determine what value I will put in each box below in the matrix/table so that each row and column will have the numbers 1-5 EXACTLY once?

[url=http://img248.imageshack.us/img248/6991/tempbx5.jpg][img]http://img248.imageshack.us/img248/6991/tempbx5.th.jpg[/img][/url]

If so, could someone tell me how I would go about doing a table like the one above, but 30 x 30?...

...so that each row/column has each number 1-29 EXACTLY once.

(You cannot put values into the boxes with Xs in them)

I'm trying to find a mathematical way to do this for a program I'm writing.

Thanks!

I think you're talking about latin squares.  If you are check this out: [url]http://en.wikipedia.org/wiki/Latin_square[/url]

Thanks! You helped me find this website (which was exactly what I needed):

http://www.warebizprogramming.com/text/cpp/section6/part9.htm


Let $x_i \in \mathbb{R^+} , i \in \mathbb{N}, 1 \leq i \leq n$ such that

                                        $\sum_{i=1}^n \frac{1}{1+x_i} \leq 1$

Prove :

$\sum_{i=1}^n \frac{1}{2^{x_i}} \leq 1$

$\sum x_i +1\geq (\sum x_i +1)(\sum \frac{1}{x_i +1})\geq n^2 \rightarrow \sum x_i \geq n^2-n$
now $(1/2)^x$ is concave so
$\sum \frac{1}{2^{x_i}}\le \frac{n}{2^{n-1}}\le 1$if $n> 3$ for the case $n=3$
take $x_i = 2 \rightarrow \sum \frac{1}{2^{x_i}}=\frac{3}{4}$


I'm pretty sure it's safe to talk about this now.  So how'd you do this year?  I got a 33.  I got a 6 on half of them and a 5 on the other half.

I had a disastrous round Tuesday, giving me my lowest score ever since I started doing this in 8th grade: 3 :)

Previously I had 5,6,5,5,6 (and I was hoping for a 6 to pull me up to 33)

Thus I end the season with 30.

36! I'm so happy. I am so sick of OSSM winning in Oklahoma...

34 this year: 6-6-5-5-6-6
Our school has 169.

I was aiming for an arithmetic progression in Math League for my high school years (32, 33, 34, 35), but I accidentally pulled a 36 this year. Since I have another year left, hopefully I can still make an interesting pattern with my scores.

i got  a 34, which seems to be the mode on this forum :)
6-5-6-6-6-5

5-0-6-4-6-6=27

Yeah, somebody posted the questions and the answers for round two before my school administered it and I was discussing them because I thought they were normal math problems that someone just posted. So, I just didn't go that time.


logarithms
limit
real analysis
real analysis solved

Evaluate the sum of the serie

$\frac{\ln 2}{2}-\frac{\ln 3}{3}+\frac{\ln 4}{4}-\frac{\ln 5}{5}+.....=
\sum_{k\geq 1}(-1)^{k}\frac{\ln k}{k}$

The answer, which appears on the page liyi linked to, is $\gamma\ln 2-\frac{\ln^2 2}{2}$, where $\gamma$ is Euler's constant $\lim_{n\to\infty}\sum_{k=1}^n\frac1k-\ln n$.

To get that, introduce a variable by considering $f(s)=\sum_{n=1}^\infty\frac{(-1)^n\ln n}{n^s}$ for $s\ge1.$ This series is absolutely convergent for $s>1$; the value at $s=1$ will be justified by an appeal to Abel's theorem.
We can show that $f(s)=g'(s)$ where $g(s)=1-2^{-s}+3^{-s}-4^{-s}+\dots=\zeta(s)-\frac2{2^s}\zeta(s)= (1-2^{1-s})\zeta(s).$

The problem then becomes finding $\lim_{t\to0^+}g'(1+t).$ We have $\zeta(1+t)=\frac1t+\gamma+O(t)$ as $t\to0$ by a complicated calculation, and we have $1-2^{-t}=\ln 2\cdot t-\frac{\ln^2 2}{2}t^2+O(t^3)$ as $t\to0.$ Therefore, $g(1+t)=\ln 2+\left(\gamma\ln 2-\frac{\ln^2 2}{2}\right)t+O(t^2)$ as $t\to0$ and $\lim_{s\to1^+}g'(s)=\gamma\ln 2-\frac{\ln^2 2}{2}.$


function
algebra
domain
analytic geometry
graphing lines
slope
calculus

[quote="t0rajir0u"]A cross-country racer runs a $10$-mile race in $50$ minutes.  Prove that somewhere along the course the racer ran $2$ miles in exactly $10$ minutes.

[hide="Hint"] The function $f(x)$ that tells you how many minutes it took the racer to run from $x$ to $x+2$ is continuous, or, at least, it is assumed to be so. [/hide][/quote]

Are you asking us to prove that at some point he ran 1 mile in 5 minutes?

[hide]
Consider $f(0)$.  If $f(0)=10$, we're done.  If $f(0)>10$, $f(x)$ must be less than 10 for some $x$, $0<x\leq 8$ (to compensate).  The reverse is true if $f(0)<10$.  Either way, by the Intermediate Value Theorem $f(x)=10$ for some $x$, $0<x\leq 8$.

Probably totally nonrigorous . . .
[/hide]

[quote="E^(pi*i)=-1"]If $f(0)>10$, $f(x)$ must be less than 10 for some $x$, $0<x\leq 8$ (to compensate).[/quote]
You should justify this step further.

If f(0)>10 and $f(x)\ge10$ for all x in the proper domain, then we will end up with more than 50 miles in total.  So, some f(x) has to be less than 10.

Again, I find this statement to be more of an assertion than a rigorous argument.  Where is the justification for the statement?

I would have expected an argument to at least mention the equation
[hide]
\[f(0)+f(2)+f(4)+f(6)+f(8) = 50. \]
[/hide]

Here's a followup question.  Prove or disprove: Somewhere along the course the racer must have ran 3 miles in exactly 15 minutes.

I would say that my argument is self-explanatory; the intermediate steps would be overdoing it:

If $f(0)>10$ gives $f(2)+f(4)+f(6)+f(8)<40$ So, now we have a set of 4 values whose mean is less than 10.  Right away this means at least one of the values is less than ten.  I suppose you could go even further, and further, ... but it would be pointless.

I just used a contradiction-style explanation.

The 3 miles/15 minutes problem has an almost identical solution.  If the first three miles are in 15 minutes, we're done.  If not, some other set of three miles must compensate and in the middle we have some set of three miles that is run in 15 minutes.

Edit:  We can take another approach for this.  Define a function d(t) to be the total distance.  This passes through (0,0) and (50,10).  We want to find some point at which the function has a secant line from t value $t_{1}$ to $t_{1}+15$ and a slope of $\frac{1}{5}$.  So, taking from the difference quotient, we have $\frac{d(t+15)-d(t)}{15}=\frac{1}{5}$.
The LHS is a new, continuous function, say y(t).  Its integral is d.  So, the mean value is $\frac{1}{5}$.  We pget a region bounded below by the t-axis and above by our new function and with an area of 10.  It is also bounded on the left by y=0 and on the right by y=50.  Using contradiction, the function can never have the value of $\frac{1}{5}$.  If, at any point, $y(t)>\frac{1}{5}$, it must be less than $\frac{1}{5}$ somewhere else, and vice versa, or else the area of the region will be more than 10.  By the intermediate value theorem, $y(t)=\frac{1}{5}$ at some t between these two points.  This is a contradiction and we are done.

[quote="dasherm"].The 3 miles/15 minutes problem has an almost identical solution.  If the first three miles are in 15 minutes, we're done.  If not, some other set of three miles must compensate and in the middle we have some set of three miles that is run in 15 minutes.[/quote]
Unfortunately, the answer to the 3 miles/15 minutes problem is negative: There are continuous counterexamples.  That's the advantage of rigorous arguments: We can distinguish true statements from falsehoods.

Then what's the proof of that?

Edit: Do you have a counterexample in which he doesn't run 3 miles in 15 minutes?  What isn't rigorous about my "edit" solution?  Why is the 3 miles one any different from the 2 miles one?

[quote="dasherm"]Why is the 3 miles one any different from the 2 miles one?[/quote]

Simply because $3 \not | 10$, so that there does not exist a corresponding identity to

$f(0)+f(2)+f(4)+f(6)+f(8) = 50$

This identity is paramount to the solution of the original problem.

Constructing a counterexample for the $3$ mile case is more easily done visually than symbolically, so you should try it yourself.

Edit: I wrote this before I saw the reply of t0rajir0u.

[quote="dasherm"]$\frac{d(t+15)-d(t)}{15}=\frac{1}{5}$.  The LHS is a new, continuous function, say y(t).  Its integral is d.[/quote]
Unfortunately, the integral of $y$ is not $d$.  You seem to have confused average speed (difference quotients) with instantaneous speed (derivatives).

Here's a counterexample to the "3 miles in 15 minutes" variant.  Suppose the racer alternately jogs 1 mile (in 8 minutes) and sprints 2 miles (in 6 minutes total).  On the one hand, the 10-mile race takes him 50 minutes.  On the other hand, every 3-mile interval takes him 14 minutes, not 15.

The key difference between the problems is that $\frac{10}{2}$ is an integer, while $\frac{10}{3}$ is not.  Because $\frac{10}{2}$ is an integer, we get an equation like
\[f(0)+f(2)+f(4)+f(6)+f(8) = 50 \, . \]
Because $\frac{10}{3}$ is not an integer, we don't get an analogous equation for the second problem.  That's one reason I viewed the arguments you and E^(pi*i)=-1 gave as insufficient.


probability
function
number theory
relatively prime
algebra solved
algebra

This is really COOL:

What's the probability that 2 randomly chosen integers are coprime?

Let p1,p2,p3,... are prime numbers.
P(n,m are coprime numbers)=\prod P((n is not divisible by pi)or(m is not divisible by pi))=\prod (1-1/pi2)=1/\prod (1+1/pi2+1/pi4+...)=1/\sum 1/n2=6/\pi2.

I remember this question can be asked for 4 numbers as well. I can't remember the probability though. Anyway, it shouldn't be too hard to deduce.

[quote="grobber"]
What's the probability that 2 randomly chosen integers are coprime?[/quote]

What's the probability that 3 randomly chosen integers are corpime ?What's the probability that k randomly chosen integers are corpime ?

Let Z(k,t) be the number of k-tuples of positive integers less than or equal to t which are relatively prime (i.e. gcd(a_1, a_2,...,_an) = 1).

If dz(.) denotes the Riemann's dzeta function, it is known [1] that : Z(2,t) = t 2 /dz(2) + O(t*ln(t)),
and for k  \geq 3, Z(k,t) = t k /dz(k) + O(t k-1 ).

It follows that the probability P(k,n) = Z(k,n)/n k that k randomly chosen (uniform distribution) positive integers in {1,...,n} are relatively prime is P(k,n) = 1/dz(k) + O(1/n) for k \geq 3, and P(2,n) = 1/dz(2) + O(ln(n)/n).
Thus, for each k, we have P(k,n) -> 1/dz(k) as n tends to +oo, which is known as the 'probability' that k randomly chosen positive integers are relatively prime.

Moreover, if A(k) denotes the 'probability' that k randomly chosen positive integers are pairwise relatively prime (i.e. gcd(a_i,a_j) = 1 for every i =/= j), then it is known [2] that :
A(k) = product ( (1 + (k-1)/p)*( 1 - 1/p) k-1 ),
wher the product is over the primes p.

References :
[1] J.E. Nymann, 'On the probability that k positive integers are relatively prime', Journal of number theory, 4 (1972), 469-473.
[2] L. Toth, 'The probability that k positive integers are pairwise relatively prime', Fibonacci Quaterly, 40 (2002), 13-18.

Pierre.


linear algebra
matrix
induction
inequalities
superior algebra
superior algebra solved

A \in Mn(R) invertible
prove that det(((A*))*) + (detA)^((n+1)^2)  \geq 2(detA)^(n^2 +1)

where A* is the matrix verify A.A*=A*.A=detAI_n

Viorel Tudoran si Razvan Tudoran, Arad

Just the 1st idea that hit me!

Induction maybe?

I did not think about this gem
A is invertible=> detA=/=0
with (1) A*.A=detAI we can do thing like
take the determinant detA*.detA= (detA)^n=>

(2) det(A*)=(detA)^(n-1)

substtitute A by A* in (1)
(A*)*.A*=det(A*)I =(detA)^(n-1)I

take the determinant

det((A*)*).det(A*)=(detA*)^(n(n-1))  =>

(3) det((A*)*)= (detA)^(n^2+1) the inequality becomes

if we write x=detA

x^(n^2+1) + x^(n^2+2n+1)  \geq 2x^(2n^2+2n+2) this
is a consequence of AM-GM


[quote="Bloody-Joker"]I'm noobie help me please
$ a,b,c\geq 0$
prove that
$ 3(\sum_{cyc} a) \geq 2(\sum_{cyc} \sqrt {a^2 \plus{} bc})$[/quote]
It's ugly but a proof.
By Cauchy-Schwartz we obtain:
$ \left(\sum_{cyc}\sqrt {a^2 \plus{} bc}\right)^2\leq\sum_{cyc}\frac {a^2 \plus{} bc}{2a^2 \plus{} b^2 \plus{} c^2 \plus{} bc}\sum_{cyc}(4a^2 \plus{} ab).$
Hence, it remains to prove that
$ \sum_{cyc}\frac {a^2 \plus{} bc}{2a^2 \plus{} b^2 \plus{} c^2 \plus{} bc}\sum_{cyc}(4a^2 \plus{} ab)\leq\frac {9(a \plus{} b \plus{} c)^2}{4},$ which is equivalent to
$ \sum_{sym}(a^8 \plus{} 2a^7b \plus{} 9a^6b^2 \minus{} 23a^5b^3 \plus{} 10a^4b^4 \plus{} 5.5a^6bc \plus{} 70a^5b^2c \plus{}$
$ \plus{} 79a^4b^3c \plus{} 131.5a^4b^2c^2 \minus{} 97.5a^3b^3c^2)\geq0,$ which is obvious.

[quote="Bloody-Joker"]I'm noobie help me please
$ a,b,c\geq 0$
prove that
$ 3(\sum_{cyc} a) \geq 2(\sum_{cyc} \sqrt {a^2 + bc})$[/quote]
This is homogeneous, so we normalize to $ a+b+c = 1$
so, we have to proove that :
$ S = \sqrt{a^{2}+bc}+\sqrt{b^{2}+ac}+\sqrt{c^{2}+ab} \leq \frac{3}{2}$
Now by power mean :
$ S \leq \sqrt{3}\sqrt{a^{2}+b^{2}+c^{2}+ab+bc+ca}$
$ a^{2}+b^{2}+c^{2}+ab+bc+ca = \frac{1}{2} ( (a+b+c)^{2}+a^{2}+b^{2}+c^{2})$
now,
${ } a^{2}+b^{2}+c^{2} \leq 1 - (a^{2}+b^{2}+c^{2})$
$ \Rightarrow a^{2}+b^{2}+c^{2} \leq  \frac{1}{2}$
so,
$ a^{2}+b^{2}+c^{2}+ab+bc+ca \leq \frac{3}{4}$
so,
$ S \leq  \frac{3}{2}$

[quote="rajat91"]
$ \Rightarrow a^{2} \plus{} b^{2} \plus{} c^{2} \leq \frac {1}{2}$
[/quote]
try $ a\equal{}1,b\equal{}c\equal{}0$ :lol:

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63645

[quote="Bloody-Joker"]Let $ a,b,c\geq 0,$ prove that

$ 3\sum a\geq 2\sum \sqrt {a^2 \plus{} bc}.$[/quote]$ (a^2 \plus{} bc)(2a \plus{} b \plus{} c)^2 \minus{} (2bc \plus{} ca \plus{} ab)^2$

$ \equal{} 4a^4 \plus{} 4a^3(b \plus{} c) \plus{} 4a^2bc \plus{} bc(b \minus{} c)^2\geq0\Longrightarrow$

$ \sqrt {a^2 \plus{} bc}\geq\frac {2bc \plus{} ca \plus{} ab}{2a \plus{} b \plus{} c}\Longrightarrow$

$ \sum\left(\sqrt {a^2 \plus{} bc} \minus{} a\right) \equal{} \sum\frac {bc}{a \plus{} \sqrt {a^2 \plus{} bc}}\leq\sum\cfrac{bc}{a \plus{} \cfrac{2bc \plus{} ca \plus{} ab}{2a \plus{} b \plus{} c}}$

$ \equal{} \sum\frac {bc(2a \plus{} b \plus{} c)}{2(c \plus{} a)(a \plus{} b)} \equal{} \frac {a \plus{} b \plus{} c}{2}\Longrightarrow$

$ \sum\sqrt {a^2 \plus{} bc}\leq\frac {3}{2}(a \plus{} b \plus{} c).$

See also here : http://www.mathlinks.ro/viewtopic.php?t=196784


A five-digit number is created using the digits 1-5 each once.  What is the probability that the number is odd?  Express your answer as a common fraction.

[hide]there's 5! ways to get a 5 digit number using 1-5, which is 120... and in order for those numbers to be odd, the last digit have to be either 1,3,or 5... so it's 3*4*3*2*1 = 72.            72/120 = 3/5[/hide]

There is a simpler solution.  See if you can find it.

[hide]it would be 3/5 because the end digits have to be 1,3, or 5
so there are 3 of the numbers tat can make it a odd number
so the probability is 3/5![/hide]

mathprincess09, EXACTLY!!!
Lucy, your answer isn't wrong,

it would take up too much time if you did it in the count down round

It can end in 1,2,3,4, or 5. 3 of those numbers would make it odd. there are 5 numbers. $\frac{3}{5}$

The solution is very simple. 1, 3, and 5 can be used to make an odd number, and there are 5 numbers total, so the probability is 3/5.

oooo well, I wasn't sure whether or not 3/5 was right just because 3 out of the 5 numbers is odd... so I did it the long way....  :blush:

[hide]i got 3/5[/hide]
i like your method too, LuCy4EvA

[hide]_ _ _ _ x
0<x>6
3 odds, 2 evens
3/5
This is what should go through your mind naturally.[/hide]

[hide]
I get 3/5 because there are three choices for the last number that make it odd.[/hide]


A blimp, suspended in the air at a height of 600 feet, lies directly over the line from Soldier Field to the Adler Planetarium on Lake Michigan. If the angle of depression from the blimp to the stadium is 42 degrees and from the blimp to the planetarium is 33 degrees, find the distance BETWEEN Soldier Field and the Adler Planetarium.

Draw the diagram.  Using trig, we see that the horizontal distance from the blimp to the stadium is: $ \tan{42}\equal{}\frac{x}{600}\Rightarrow x\equal{}600\tan{42}$.  On the other side, for the planetarium, we have $ \tan{33}\equal{}\frac{x}{600}\Rightarrow x\equal{}600\tan{33}$.  So the combined distance is $ 600(\tan{42}\plus{}\tan{33})$.


$Question 1$ Express $997$ as the sum of natural numbers i.e. $997 = n_1 + n_2 + n_3 + ... + n_x$ so that $P = n_1 * n_2 * n_3 *...* n_x$ is the maximum.

Please, take a note that there is no constraint that you have to divide $997$ in $2$ parts only. If that were the case then the answer would have been $997 = 498 + 499$ and the Productmax would have been $498*499 = 248502$.

$Question 2$Which value is greater of $1999^{1999}$ or $2000^{1998}$ ? You can not use a calculator or a computer. Logic is expected!!!!

[hide]We assign values to the variables in any way.  Then, for each $n_{k}$ such that $n_{k}\ge 5$, replace $n_{k}$ with $3$ and $n_{k}-3$.  Since $2n_{k}>9$, we have $3(n_{k}-3)>n_{k}$, so the product increases.  Therefore, our product consists of no integers greater than $4$.

For convenience, we replace each $4$ with two $2$'s, so that the product remains the same.  Our product now consists only of $2$'s and $3$'s.

Since $2^{3}<3^{2}$, we replace groups of three $2$'s with two $3$'s until we have less than three $2$'s.  Therefore, $n_{1}=n_{2}=\cdots =n_{331}=3$ and $n_{332}=n_{333}=2$.[/hide]


Two triangles are drawn on a plane in such a way that the area covered by their union is an n-gon (not necessarily convex). Find all possible values of the number of vertices n.


An envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 twenties.
Two bills are drawn at random without replacement. What is the probability
that their sum is \$20 or more?

There are three possibilities: we can get a twenty dollar bill and something else, two twenty dollar bills, or two ten dollar bills, for a probability of $ \frac {2\cdot6 \plus{} 1 \plus{} 1}{\binom82} \equal{} \frac {14}{28} \equal{} \boxed{\frac12}$.

Can someone tell me what's wrong with this reasoning:

The wanted probability is the

P(my first draw is a \$20) + P(first draw is a \$10)P(second draw is a \$10) =

$ \frac{2}{8} \plus{} \frac{2}{8}\times \frac{1}{7} \equal{} \frac{2}{7}$

You're missing cases, like when the first draw is not a $ \$20$, but the second is.

[quote="mcbobo"]Are the 20 dollar bills distinct?[/quote]
I would prefer that you didn't revive this....
But yes, the $20$ dollar bills are distinct

No, why are they distinct? Obviously they aren't. If I were to label 1 and 2 on the back of two twenty-dollar bills, then mix them up, could you tell which one's 1 and which one's 2 (without guessing)? Well, unless you're psychic, you can't. Therefore, they are indistinguishible.

Well not necessarily distinct, but you can choose $20$ dollar bill number 1 and say $5$ dollar bill number 1 is the different than $20$ dollar bill number 2 and $5$ dollar bill number 1

What's wrong with $\dfrac{2\cdot7+2}{56}=\dfrac{2}{7}?$

(counting with order)

(wait I did this yesterday?)

I got it again today. :P I've figured it out, $1+6+6+1$ is the numerator and $\dbinom{8}{2}$ is the denominator.

[hide=Cheap Solution]
To omit the trouble of replacements, note that a density of 2 types of 4 bills each choosing 2, is the same as the density of 1 type of 4 bills each choosing 2 without replacement. With this only (1,20), (5,20), and (10,20) work. There are C(4,2)=6 total cases. Therefore, the probability is 3/6=1/2
[/hide]

[quote=centslordm][b]Solution:[/b] We have only one case: The sum of the bills is greater than $20.$ This either happens, or it doesn't happen. Hence, we obtain $\boxed{\tfrac{1}{2}}$ and we are done.[/quote]

Sorry but it is not true.. what you are only right to say is that if you note $A = \text{bill} \geq 20$ then $\mathbb{P}(A) + \mathbb{P}(\bar{A}) = 1$. Which is always true. If you change the value $20$ to be $10$ for instance then your assertion still apply yet the result is not $\frac{1}{2}$.

[hide]
Complementary counting: We don't have a sum of 20, which means we need to have no twenty bills, since having one of those gives us what we want(sum of $20 or greater), which leaves us with 2 ones, 2 tens, and 2 fives. We can not have 2 tens, which there is only 1 way of happening, and thus we have 6C2 ways in total to get 2 bills minus 1, giving us 14 / 8C2, 14/28 = 1/2, 1-1/2=1/2
Ans. 1/2
[/hide]


inequalities
trigonometry
inequalities proposed

[quote="Marius Mainea"]If a,b,c find the sidelengths of a triangle then:

a) $ \frac{1}{2}<\frac{m_a\plus{}m_b}{a\plus{}b}<\frac{3}{2}$

b) $ \frac{3}{10}<\frac{am_a\plus{}bm_b}{a^2\plus{}b^2}<\frac{3}{2}$

c) $ \frac{1}{2}<\frac{am_b\plus{}bm_a}{a^2\plus{}b^2}<\frac{3}{2}$[/quote]
D'you mean medians by $ m_{a}, m_{b}$? :huh:

Hi Marius! So far I remember there is some general expression of $ m_{a}, m_{b}, m_{c}$ using $ a,b,c$ and probably sines of angles. Could you post it here please. I think it might be of some use in proving it. I still got no idea. :maybe:

P.S. Can it be proved in a purely geometric approach?

hello, using the well known formulas $ m_a\equal{}\frac{1}{2}\sqrt{2(b^2\plus{}c^2)\minus{}a^2}$ and $ m_b\equal{}\frac{1}{2}\sqrt{2(a^2\plus{}c^2)\minus{}b^2}$ then the left hand side of our inequality is equivalent with
$ a\plus{}b<\sqrt{2(b^2\plus{}c^2)\minus{}a^2}\plus{}\sqrt{2(a^2\plus{}c^2)\minus{}b^2}$. After squaring two
times we get: $ 2(a\plus{}b)^2(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)>0$ which is true. By the same way we get for the right hand side $ 18(a\plus{}b)^2(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)>0$, which is also
obviously true.
Sonnhard.


vector
analytic geometry
calculus
calculus computations

Can anyone point me to a good website or a free software that I can use to graph equations such as:

$ x^2 \plus{} z^2 \equal{} 3$
$ 4x \plus{} y \plus{} 3z \equal{} 12$

this is a 3 dimensional object.. or can you please describe what it looks like?

hello
http://graph.seriesmathstudy.com/
u can also buy CD of Microsoft student
thank u

If I were to draw that kind of object in 3D in a test, how would I approach the problem first?

Or you can do without the software and just [i]think[/i].

$ 4x\plus{}y\plus{}3z\equal{}12$ is a plane (not through the origin). You know quite a bit about planes. If you're in the course I think you're in, your textbook discusses the normal vector to a plane. You can also find out where this plane intersects each coordinate axis by setting two of the variables equal to zero: $ (3,0,0), (0,12,0),(0,0,4).$ Those three points, plus the fact that it's a plane, should tell you quite a lot about what it looks like.

As for $ x^2\plus{}z^2\equal{}3:$ the very first thing to notice is that $ y$ does not appear in the equation. That means this set is "free" in $ y.$ From any point in the set, you can travel freely in the $ y$ direction and stay in the set. Conversely, from any point not in this set, travel in a direction parallel to the $ y$ axis (that is, change $ y$ while leaving $ x$ and $ z$ the same) and you're still not in the set. From there to understand it, you should look at a simple cross section. As a set in the $ x$-$ z$ plane, what does $ x^2\plus{}z^2\equal{}3$ look like?

For the first one I know it's a plane and for the second one isn't it a surface of a half cylinder along the y-axis?

Please don't quote the entire preceding message, and please don't make any quotes without a specific reason.

[quote="EquinoX"]... isn't it a surface of a half cylinder along the y-axis?[/quote]
For what reason did the word "half" appear in this sentence?


[quote="erdugan"]a,b,c,d are natural numbers such that a^2+ab+b^2=c^2+cd+d^2.prove that a+b+c+d is composite number.[/quote]
Wrong:$ a \equal{} c \equal{} 1,b \equal{} d \equal{} 0$
To t0rajir0u;Sorry,in my country natural numbers includes $ 0$

One should interpret "natural number" to mean $ \mathbb{Z}_{>0}$ here.


geometry
3D geometry
dodecahedron
Gauss
vector

The gas pressure exerts force to the inner surface of the box. But the net force and the net torque of the box should be zero. It would be counterintuitive otherwise. Yet how do we know that they are zero?

yes the net force is zero since all sides get same pressure

what if the shape of the container is arbitrary : ellipsoid, dodecahedron, or any other (non-symmetric) closed surface?

We can use an modification of Gauss Theorem here. Can you see?

Apply Gauss theorem (divergence form) with the constant vector field $ F : \equal{} \mathbf{e}_1$ and one sees that the $ x$-component of the net force is zero.

Apply Gauss theorem (curl form) with the vector field $ F: \equal{} \mathbf{r}$, position vector, and one sees the net torque is zero.


anyone knows any drawing program (preferably free) that can draw 3D stuff???because the one i use can only draw 2D...

If all else fails then I will have to use A-CAD....

well, I use Auto-CAD to draw the picture and it look quite nice.(Just a little blury because of all the converting file type) But are we allowed to distribute pictures(I guess not...)???

[quote="pkerichang"]But are we allowed to distribute pictures(I guess not...)???[/quote]You guess correctly.  Do not distribute any pictures that you create in the course of solving the problem.


Suppose that one constructs open intervals $I(x)$ of center x and length at most $c>0$ (fixed) at each real point. Let $J$ be a compact interval. Prove that one can find a finite set $A$ such that the intervals $I(x)$ with $x\in A$ cover $J$ and the sum of their lengths is at most $c+2\cdot l(J)$ (l(J) is the length of J). Is this value optimal?

I hope I don't misunderstand something, because this seems easy.

The intervals hardly have to overlap at all. But we can construct the covering you wanted by picking each next $x_n$ at the next point outside of the previous $I_{n-1}$. Then $I_{n}$ and $I_{n-1}$ overlap by $\frac{c}{2}$. So if we cover the whole interval this way, we have total length less than or equal to $2l(J) + c$ (equality occurs when $l(J)$ is an integer multiple of $c$).

But we can make the overlap much smaller. We can have each successive interval overlap the previous one on an interval of arbitrarily small length. We can also make the first and last intervals shorter so that they don't extend too far past the edge. I think we can make a covering as close to $l(J)$ as we like (but never equal of course).


[quote="bpms"][quote="7h3.D3m0n.117"]What are the last two digits in the sum of the factorials of the first 100 positive integers?

Please show all work I could not even start this problem  :lol:[/quote]
[hide="hint"]This problem can't be that tedious, so just look at the last two digits of small cases (1-15) and see what you get.[/hide][/quote]

[hide]Notice that all that matters is mod (100), everything abnove 10! has 2 zeros, so do everything below.[/hide]

Adding all factorials 1-9 it would be 13. Right?

yes it's 13 ah previously i accidently added $0!$ before

[quote="7h3.D3m0n.117"]What are the last two digits in the sum of the factorials of the first 100 positive integers?

Please show all work I could not even start this problem  :lol:[/quote]
[hide]
everything 10 and above factorialed would end in 00
9!+8! also ends in 00 because 9!+8!=(9+1)(8!)=10(8!) and 5! and above have one 0 already
7! ends in 40. this can be done fast by multiplying 5 and 2, then 10 by 3, then 30 by 4 (ends in 20), then 20 by 6 (ends in 20 again), then 20 by 7, which ends in 40
5! ends in 20. since 6x20=120, 6! also ends in 20
4! ends in 24
3! ends in 6
2! ends in 2
1! ends in 1
40+40+30+3=80+33=$13$

[/hide]

this might be worth memorizing as i have seen problems like this in countdown.
still.......even if you forget, being able to derive it is important as well... :)


print it out or something and view it in a mirror


Consider a trianglem $ABC$ and a point$P\notin BC,CA,AB$.$AP\cap BC=A';BP\cap CA=B';CP\cap AB=C'$.$O_1,O_2,O-3,O_4,O_5,O_6$ are the circumcircle of $APB',APC',BPA',BPC',CPA',CPB'$.Prove that $O_1,O_2,O_3,O_4,O_5,O_6$are cyclic $\leftrightarrow P$ is the centroid or the heigh point of $ABC$.


geometry
circumcircle
power of a point
radical axis
geometry proposed

A hint:
The problem is equivalent to $MG\cdot MD=MF\cdot ME$, where $M=AP\cap DE$.

[quote="Mathx"]Points $D$ and $E$ are situated on the sides $AB$ and $AC$ of triangle $ABC$
so that $DE\parallel BC$. Let $P$ be an arbitrary point inside $\triangle ABC$. Lines $PB$
and $PC$ meet $DE$ at $F$ and $G$, respectively. Let $O_1$ be the circumcenter
of $\triangle PDG$ and $O_2$ that of $\triangle PEF$. Show that $AP\bot  O_1O_2$.[/quote]

In http://www.mathlinks.ro/Forum/viewtopic.php?t=24889 , it was shown that the points A, P and Q are collinear, where Q is the point of intersection of the circumcircles of triangles PDG and PEF (apart from P). Thus, the line AP coincides with the line PQ. But since the common chord of two circles is perpendicular to the line joining their centers, we have $PQ\perp O_1O_2$ (in fact, the circumcircles of triangles PDG and PEF have the common chord PQ and the centers $O_1$ and $O_2$. Thus, $AP\perp O_1O_2$.

  Darij

I note $M\in DE\cap AP;\ \ DE\parallel BC\Longrightarrow \frac {BD}{BA}=\frac {CE}{CA}$. From the aplication of the Menelaus's theorem to the transversals

$\overline {BFP},\ \overline {CGP}$ in the triangles $ADM,\ AEM$ respectively we obtain the following relations:

$\frac{BD}{BA}\cdot \frac {PA}{PM}\cdot \frac {FM}{FD}=1,\ \frac {CE}{CA}\cdot \frac {PA}{PM}\cdot \frac {GM}{GE}=1 \Longrightarrow \frac {FD}{FM}=\frac{GE}{GM} \Longrightarrow\frac {FD}{FM}+1=$ $\frac {GE}{GM}+1\Longrightarrow$

$\frac {MD}{MF}=\frac {ME}{MG}\Longrightarrow MD\cdot MG=ME\cdot MF$, i.e. the point M has the same power w.r.t. the circumcircles

for the $\triangle ADG,\ \triangle AEF$. Therefore, the line $\overline {APM}$ is the radical axis of the two circles  $\Longrightarrow O_1O_2\perp AP.$


Label the equilateral triangle DEF. (No reason for it to be equilateral). Perpendiculars to AE, AF from at E, F intersect at P. Perpendicular to PD at D cuts AF, AE at B, C. Perpendiculars to EF, FD, DE from A, B, C are concurrent at Q (by Carnot theorem). Q is isogonal conjugate of P. 3 different possibilities to choose vertex D yield 3 possible triangles ABC.


If $xy+yz+zx=1$ Prove that
$(x+y+z)(\frac{2x}{2x+y+z}+\frac{2y}{2y+x+z}+\frac{2z}{2z+x+y})\geq\frac{3\sqrt 3}{2}$ :)

Nice and not so easy.

It maybe has some trig  substitution....??  :?

I created this my friend Thanasis. I think it is good and smart


the point (-2,6) is on the graph of y=f(x) which of the following must be on the graph of the function y=1/3f(2(x-1))
a) (0,2) b) (-6,2) c) (-3,18) d) (-5,18

How come $ f(x) \equal{} y$?? I mean functions should suppose to be like "$ f(x) \equal{} 3x \plus{} 1$" or if sometimes "$ y \equal{} 3x \plus{} 1$" isn't it???Relation of the function of $ x$ to $ y$ can't be established here.

Correct Me if I"m wrong.

EDIT: Ooh.. At least somehow it was clarified.

Look at your own example -- the graph of $ y\equal{}3x\plus{}1$ can be written as the graph of $ y\equal{}f(x)$, where $ f(x) \equal{} 3x\plus{}1$. The point is that we don't [i]know[/i] what $ f$ is for all $ x$, but we know enough information about it to find one point on it.
[hide]
We know that $ f(\minus{}2) \equal{} 6$. We can write $ \minus{}2 \equal{} 2(0\minus{}1)$; that is, $ f(2(0\minus{}1)) \equal{} 6$, and thus $ \frac{1}{3} f(2(0\minus{}1)) \equal{} \frac{6}{3} \equal{} 2$

Thus, the point $ (0,2)$ is on the graph of $ y\equal{}\frac{1}{3} f(2(x\minus{}1))$[/hide]


I have been sent the following question last Sunday but have been away:

"If the radius of convegence of $ \sum c_{n}x^{x}$ is $ a$ and $ b$ for $ \sum d_{n}x^{n}$ what can be said about the radius of convergence of $ \sum(c_{n}\plus{}d_{n})x^{n}$?"

It's obviously at least $ \min (a,b)$ and it can be arbitrarily large (for instance, if $ d_n \equal{} \minus{}c_n$, the series converges everywhere).

Thanks, CommandoGuard. Yes, it seems obvious: that's the reply I sent, but I'm not sure how to prove it.

How to prove it is reasonably simple. Suppose $ |x| < \min(a,b),$ meaning $ |x| < a$ and $ |x| < b.$ Then $ \sum c_nx^n$ and $ \sum d_nx^n$ both converge, and the sum of two convergent series is convergent.

That proof says nothing about what happens for $ |x| > \min(a,b),$ but (as CommandoGuard suggested), that's taken care of by examples showing that several things can happen.


topology
superior algebra
superior algebra unsolved

Post it here on the forum.
And tell us what your problems are, we are not here for simply making your homework.


The length of a pendulum is 20cm. If the tip of the pendulum swings through an angle of 86 degrees, find :

a) The arc length ABC (the distance through which the tip travels)
b) The area of triangle OAC.
c) The area of the sector.
d) The area of the minor segment.
e) The length BX.


Hello,

Let $ A$ part of $ \mathbb{R}$ verifying :

$ \displaystyle A\neq \emptyset ,$

$ \displaystyle \forall x\in A,\exists \varepsilon _{x}>0,]x\minus{}\varepsilon _{x},x\plus{}\varepsilon _{x}[\subset A,$

$ \displaystyle \forall x\in \mathbb{R}: (\forall \varepsilon >0,]x\minus{}\varepsilon ,x\plus{}\varepsilon [\cap A\neq \emptyset )\Rightarrow x\in A.$

[u][b]Prove that[/b][/u]:   $ A \equal{} \mathbb{R}$

Short: An open dense set is the whole space.

But why did you post it as Number Theory¿ Please take that 5 seconds to choose the right forum next time!

This is not a number theory problem but rather a basic topology problem. The first statement implies that A is open; the second statement implies that A is closed. The only nonempty subset of $ \mathbb{R}$ that is both open and closed is $ \mathbb{R}$ itself.

FibonacciFan: No, the second condition is not closed, but dense.

It is apparent that it was meant to be
$ \forall x\in \mathbb{R} ((\forall \varepsilon >0,]x\minus{}\varepsilon ,x\plus{}\varepsilon [\cap A\neq \emptyset )\Rightarrow x\in A)$,
which is a terribly formal way to write down the closedness condition. By the way, there are plenty of open dense proper subsets of $ \mathbb R$.


Daca a,b,c>1 , calculati: \[[ 4(log_{a^{2}bc}a+log_{b^{2}ca}b+log_{c^{2}ab}c ]\]

[quote="Vrajitoarea Andrei"]Daca a,b,c>1 , calculati:
\[[ 4(log_{a^{2}bc}a+log_{b^{2}ca}b+log_{c^{2}ab}c ] \]
[/quote]

Let $A=\log a,\ B=\log b,\ \log c\ (A,\ B,\ C>0)$, we have

$log_{a^{2}bc}a+log_{b^{2}ca}b+log_{c^{2}ab}c$

$=\frac{A}{2A+B+C}+\frac{B}{2B+C+A}+\frac{C}{2C+A+B}: =f(A,\ B,\ C),$

$f(A,\ B,\ C)>\frac{A}{2A+2B+2C}+\frac{B}{2B+2C+2A}+\frac{C}{2C+2A+2B}=\frac{1}{2}$

 and $f(A,\ B,\ C)<\frac{A}{A+B+C}+\frac{B}{B+C+A}+\frac{C}{C+A+B}=1$.

Do you mean that $[\ ]$ is  Gauss sign?

$[x]$ reprezinta partea intreaga al lui $x$ (the floor of $x$). Obtinem:\[2<4f(A,B,C)<4\]
Si cu asta nu putem sa determinam $[4f(A,B,C)]$
Poate daca aratam ca $f(A,B,C)\geq \frac{3}{4}$

hm...mie mi se pare ca
$f(A,0,0)=1/2$.
probabil ca ar fi bine sa aratam ca
$f(A,B,C)<3/4$.

E chiar banala.Folosim dualitate notand $x=a+b , y=b+c , z=c+a$
Astfel: \[%Error. "displaysetup" is a bad command.
\sum_{cyc}\frac{a}{2a+b+c}=%Error. "displaysetup" is a bad command.
\sum_{cyc}\frac{x-y+z}{2(x+z)}= \frac{3}{2}-\frac{1}{2}%Error. "displaysetup" is a bad command.
\sum_{cyc}\frac{x}{y+z}\]
Din inegalitatea lui Nesbitt obtinem in final ca $f(a,b,c)\leq \frac{3}{4}$
Problema e ca noi putem avea egalitate .Daca problema ar fi precizat ca numerele ar fi diferite atunci partea intreaga ar fi 2.Altfel, este 3


Well?  I liked the level 2 test, which I took, but the level 3 test was really hard, even Mr. [url=http://www.math.uncc.edu/~hbreiter/]Reiter[/url] thought so.  So anyways, I figured I'd start the discussion like I always do.  We can start with [url=http://math.cofc.edu/MATHMEET/index.html?pg=winners]results[/url] and a problem:

EDIT:
(Except that I seem to have lost my problem sheet.  But did anyone get the precalculus question on level 2 about the inverse cosine, or something like that?)

Level 3 was a beast of a test. It doesn't make me feel too swell when I leave 1 and 2 blank, but I guess I shouldn't be talking :). It was a lot of messy arithmetic and trivia.

One of our guys had apparently forgotten (?) that his brother's birthday was today, and therefore had to miss. In fact, he didn't even tell us this - we had to search for him around school for an hour before somebody tracked down his phone number. So we couldn't do any team events.

And I know Hoggard did as well as some winning schools, even with only three people. They should have counted us for a team score.

But it was still a lot of fun. I love CofC.


function counting distinguishability calculus

The RBI prints currency notes in denominations of Rs 2,Rs 5,Rs 10,Rs 20,Rs 50, Rs 100 and Rs 500. In how many ways can it display ten currency notes, not necessarily of different denominations?How many of these will have all denominations?

 [hide="click to see my answer"]clearly answer to first part is $ 7^{10}$.
the answer to the second part is given to be (in the book) = $ 7!*7^3$
which i doubt is correct and that should be $ 7^{10} \minus{} (7*6^{10} \minus{} \dbinom{7}{5}*5^{10})$

Explanaton :

if we find number of ways in which not all denominations are present we can get answer to second part by subtracting it to $ 7^{10}$.
if we remove any one of the denomination the remaining denomination can be displayed in $ 6^{10}$ ways.
the denominations to removed can be selected in 7 ways.
So total number of ways of displaying 10 denomination using only 6 currency at a time = $ 7*6^{10}$ ways.
but in any two $ 6^{10}$ ways of display we can find 5 denominations common which are counted twice.

So actual number of ways in which not all denominations are present = $ 7*6^{10} \minus{} \dbinom{7}{5}*5^{10}$

Hence the required answer

tell me if i am wrong or correct[/hide]

I don't know why you are claiming overcounting in the 2nd part...

I think you are right about the book being incorrect, but go about it the wrong way.

I think you start off correctly, but you need to continue on.  You must consider, at every step, that some of the notes may not be used, and adjust accordingly.  So there are $ 7^{10} \minus{} 7*(6^{10} \minus{} 6*(5^{10} \minus{} 5(4^{10} \minus{} 4(3^{10} \minus{} 3(2^{10} \minus{} 1)))))$ ways, or something like that.  I'm probably doing something wrong here (probably should use combinations somewhere...), but the idea is you need to consider that when you do $ 5^{10}$, you in no way insure that all the possibilities you count don't exclude $ 1$ or more notes.

[hide="how about this"]For seven of the 10 notes we have no choice. But the last three notes can be chosen freely:

Case 1: they are all different. Then there are $ 7\choose3$ ways to choose them, and since each of these three are duplicated once from the required seven, the number of ways to arrange all 10 denominations is $ \frac{10!}{2!2!2!}$ (this is like counting how many 'words' can be formed from AABBCCDEFG).

Case 2: two are the same; there are $ 7\choose2$ ways to choose the denominations and $ \frac{10!}{3!2!}$ to arrange them.

Case 3: all three are the same: 7 ways to choose this, and $ \frac{10!}{4!}$ ways to arrange.

I am way too lazy to write the final answer; just multiply the ways to choose with the ways to arrange in each case and add them together :D  [/hide]

236 that sounds good too... your answer is $ 23284800$

Anyway, I think my correct answer would be:

$ 7^{10} \minus{} \binom{7}{1}6^{10} \plus{} \binom{7}{2}5^{10}\minus{}\binom{7}{3}4^{10}\plus{}\binom{7}{4}3^{10} \minus{} \binom{7}{5}2^{10}\plus{}\binom{7}{6} \equal{} 29635200$

First, we count the number of ways to choose $ 7$ notes, then subtract out the ways that have only up to 6 notes.  By doing so, we have over-subtracted because we chose 1 notes not to include but some of these choices exclude more than 1, so we need to adjust by adding back in the choices where up to 5 notes were included (the set which was overcounted in the subtracting), etc.  The idea is principle of inclusion exclusion.

Anyway, one of us is wrong here (probably me).  Can someone please find the error?

Hmmm... tricky. We differ by exactly $ {7\choose2}\cdot\frac {10!}{3!2!}$, which I determined to be the number of combinations for the second case.

EDIT: Aha! I have found my mistake. In my second case, I wrote that "two were the same". Well, either denomination could have satisfied this, so I would have to multiply the number of ways in this case by two, which explains the error.

actually I was just about to point this out; I think I considered it momentarily a while ago then forget about it.

Anyway, I figured out a much better way to formulate my solution: with generating functions.

The generating function that gives the denominator in distinguishable permutations is $ 1 \plus{} x \plus{} x^2/2! \plus{} x^3/3! \plus{} \ldots \equal{} e^x$.  We want to exclude the possibility of having none of a bill, so each bill has the generating function $ (e^x \minus{} 1)$.  We have 7 bills, thus, $ (e^x \minus{} 1)^7 \equal{} e^{7x} \minus{} 7e^{6x} \plus{} 21e^{5x} \minus{} 35e^{4x} \plus{} 35e^{3x} \minus{} 21e^{2x} \plus{} 7e^x \minus{} 1$ describes the indistinguishable permutations of them, when we must have at least one of each kind.

Then this can be rewritten as an infinite maclaurin series (a little calculus - note that $ e^{kx}$ contributes $ k^n$ to the coefficient of $ x^n$): $ \minus{} 1 \plus{} \sum_{n \equal{} 1}^{\infty} \frac {7^n \minus{} 7\cdot6^n \plus{} 21\cdot5^n \minus{} 35\cdot4^n \plus{} 35\cdot 3^n \minus{} 21\cdot2^n \plus{} 7}{n!}x^n$.  Remembering that the coefficients represent only the sum of the denominators in the distinguishable permutations, we have $ 7^n \minus{} 7\cdot6^n \plus{} 21\cdot5^n \minus{} 35\cdot4^n \plus{} 35\cdot 3^n \minus{} 21\cdot2^n \plus{} 7$ ways to choose $ n$ bills, with each chosen at least once.  Plugging in $ n \equal{} 10$ gives the expression I had before.


limit logarithms calculus derivative induction algebra function

Let $ (a_{n})_{n\geq 1}$ be a sequence of real numbers satisfying  $ a_{1}\in\mathbb{R}^{*}$ and
\[ \displaystyle a_{n\plus{}1}\equal{}\sqrt[3]{\frac{a_{1}^{2}}{1}\plus{}\frac{a_{2}^{2}}{2}\plus{}\ldots\plus{}\frac{a_{n}^{2}}{2}}, \forall n\geq 1.\]
Find $ \displaystyle\lim_{n\to\infty}(3a_{n}\minus{}\log n)$.

[quote="Cezar Lupu"]Let $ (a_{n})_{n\geq 1}$ be a sequence of real numbers satisfying  $ a_{1}\in\mathbb{R}^{*}$ and
\[ \displaystyle a_{n \plus{} 1} \equal{} \sqrt [3]{\frac {a_{1}^{2}}{1} \plus{} \frac {a_{2}^{2}}{2} \plus{} \ldots \plus{} \frac {a_{n}^{2}}{2}}, \forall n\geq 1.\]
Find $ \displaystyle\lim_{n\to\infty}(3a_{n} \minus{} \log n)$.[/quote]

Hi :) !!

i think that you mean : $ a_{n \plus{} 1} \equal{} \sqrt [3]{\frac {a_{1}^{2}}{1} \plus{} \frac {a_{2}^{2}}{2} \plus{} \ldots \plus{} \frac {a_{n}^{2}}{\color{red}{n}}}, \forall n\geq 1.$

and thanks ...

I ran it out to a few thousand terms on a spreadsheet. I'm not sure how to do a good extrapolation on this, but I think the limit is between $ 2.42365$ and $ 2.42366$ when $ a_1\equal{}1.$ However, I get different apparent limits for different values of $ a_1.$

For $ a_1\equal{}1,\ 2.424.$
For $ a_1\equal{}2,\ 4.232.$
For $ a_1\equal{}10,\ 13.460.$
For $ a_1\equal{}.5,\ 1.255$
For $ a_1\equal{}.1,\ \minus{}.169.$

(I was using mathema*'s version.)

I don't know how to prove the convergens yet, but it's easy to prove that the sequence $ |3 a_n \minus{} log n |$ is bounded and get some bounds( which depend on $ a_1$, of course) on the limit (if it exists  :)  ) .
edit.- actually, this limit exists, it could be easily proved by considering Taylor expansion of the functon $ (1 \plus{} x)^{\frac {1}{3}}$ up to third term, but I still don't know how to compute it. :)

[quote="leshik"] it could be easily proved by considering Taylor expansion of the functon $ (1 \plus{} x)^{\frac {1}{3}}$ up to third term[/quote]
You mean "by noticing that the derivative of $ x^{1/3}$ is bounded away from $ 0$"?
Indeed, the simple induction shows that $ a_n\le 1\plus{}|a_1|\plus{}\log n$, say and the recurrence relation can be rewritten as $ a_{n\plus{}1}\equal{}\sqrt[3]{a_n^3\plus{}\frac{a_n^2}{n}}$ whence $ \frac 1{3n}\minus{}C\left(\frac{|a_n|}{n^2}\plus{}\frac 1{n^3}\right)\le a_{n\plus{}1}\minus{}a_n\le \frac 1{3n}$, and we conclude that $ a_n\minus{}\left(\frac 13\plus{}\frac 1{3\cdot2}\plus{}\dots\plus{}\frac 1{3(n\minus{}1)}\right)$ is decreasing and bounded.

If you also take into account that the derivative of $ x^{1/3}$ is $ \frac 13x^{\minus{}\frac 23}$, you will see that for large $ a_1>0$, the limit in question is just a little bit below $ 3a_2\minus{}1\plus{}\gamma$, which, for $ a_1\equal{}10$, gives "a bit less than $ 13.502$", agreeing with Kent's answer. So, the answer, indeed, does depend on $ a_1$.

I have no idea whether there is any "closed form expression" for the limit. I surmise not because, if you go to the complex domain, you'll have a singularity at each point where the sequence eventually gets $ 0$ and the set of such points seems to be too wild to be a set of singularities of an elementary function (though I haven't tried to investigate it carefully enough to turn it into a formal proof).

Actually, I meant that one  could just write the estimate $ 1 \plus{} \frac {1}{3} x \minus{} \frac {2}{9} x^2 \le (1 \plus{} x)^{\frac {1}{3}}\le 1 \plus{} \frac {1}{3} x$, and then we can easily prove that $ 3a_n \minus{} log n$ is Cauchy.

$ a(n)$ is equivalent to $ \frac{1}{3}\ln(n)$ when n tend to infty

Certainly, $ \frac {a_n}{\frac {1}{3}\log n}\to 1$ when $ n\to \infty$ (this follows from the fact that $ a_n \minus{} \frac{1}{3}\log n$ has finite limit but converse is not true):)


[quote="cnyd"]Hello I'm new in Muirhead inequality

$ [2,0,0,0] \equal{} \frac {1}{3}$  $ \implies$  find the maximum of $ [3,1,0,0]$[/quote]
$ \sum_{cyc}a^3(b\plus{}c\plus{}d)\leq\frac{3}{4}(a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2\Leftrightarrow\sum_{sym}(a\minus{}b)^4\geq0.$ :wink:

Can we generate this type of inequalities to Muirhead


LaTeX
modular arithmetic
Euler
algebra
polynomial
Vieta
number theory

Εδώ οι ασκήσεις έχουν πιο  γενικό χαρακτήρα. Ας  μαζέψουμε γύρω στα [color=red]50 θέματα [/color] και θα ταξινομήσουμε στο τέλος. Να φροντίσουμε να έχουμε θέματα από :
    [b]α) [/b]Εισαγωγικό μέρος : Διαιρετότητα , ΕΚΠ , ΜΚΔ, Πρώτοι αριθμοί
   [b] β) [/b]Ισοτιμίες
    [b]γ)[/b] Διοφαντικές εξισώσεις κάθε μορφής
    [b]δ) [/b]Θεωρήματα Fermat και Wilson, ή και άλλα θεωρήματα πιο ειδικά.
Φαντάζομαι ότι τα πιο καλά θέματα είναι αυτά που έχουν πέσει σε διάφορους διαγωνισμούς. Πάρτε από τη βιβλιογραφία τα πιο καλά και διδακτικά θέματα , καθώς και ασκήσεις που φτιάξατε μόνοι σας στην προσπάθειά σας να λύσετε κάποιο άλλο!

 Κατά τα άλλα ισχύουν οι κανόνες που βάλαμε και στις άλλες ενότητες.

                     [color=red]Καλή Συγκομιδή! [/color]

[b][color=red]ΑΣΚΗΣΗ 1α[/color][/b]

Δείξτε ότι $\forall n \in \aleph$ υπάρχει $p$ πρώτος με $n<p \le n!+1$ και κατά συνέπεια υπάρχουν άπειροι πρώτοι.

[b][color=red]ΑΣΚΗΣΗ 1β[/color][/b]

Δείξτε ότι $\forall n \in \aleph, n \ge 2$ δεν υπάρχει $p$ πρώτος με $n!+2 \le p \le n!+n$.

Στέλνω τις λύσεις στο pdf. Πείτε μου αν το επίπεδο είναι καλό ή χρειάζεται κάτι άλλο.

λοιπον οι ασκησεις ειναι σωστες
μια αλλη λυση για την ασκηση 1α:
an $n!+1$ ειναι πρωτος τελιωσαμε, διαφορετικα
εστω $p$ ενας πρωτος διαιρετης του  $n!+1$ με $p>2$
αφου ειναι πρωτος $(p,k)=1$ με  $k=1,2,...,n$ αρα $p$ $>n$

[color=darkblue][b]ΑΣΚΗΣΗ 1[/b]-[i]διορθώσεις[/i][/color]

1α) Στην 4η και 5η σειρά της λύσης νομίζω πως όλες οι ανισότητες πρέπει να μετατραπούν σε ανισοϊσότητες, δηλαδή να μετατραπούν τα $>$ σε $\geq $ ώστε να συμπεριλάβουμε και την περίπτωση ο $n$ να είναι ο $2$ ή ο $3$ (...γιατί αναφέρεις ότι $p_{i}$ για $i=1,2,...,k$ , είναι όλοι οι πρώτοι μικρότεροι ή ίσοι του $n$).  :wink:
Φυσικά το ίδιο ισχύει και για την 7η σειρά που στο πρώτο σκέλος της ισοδυναμίας το $n<p<n!+1$ πρέπει να μετατραπεί σε $n<p\leq n!+1$ και συνεπώς να μην υπάρχει καθόλου το δεύτερο σκέλος της ισοδυναμίας  (... :P) .
Επίσης αν θες να είσαι απόλυτα σωστή θα πρέπει να εξετάσεις την περίπτωση $n=1$ ξεχωριστά, μιας και η απόδειξή σου δεν ισχύει σε αυτή την περίπτωση.  :wink:

1β) Σωστή!  :D

Πάντως κατά τη γνώμη μου αν οι λύσεις δεν γράφονται σε pdf, αλλά απευθείας στο forum, θα είναι πιο εύκολες οι διορθώσεις... (γιατί θα μπορούμε να βάζουμε quotes για να δείξουμε σε ποιο σημείο αναφερόμαστε αντί να γράφουμε π.χ. στη μέση της τρίτης σειράς μετά το έβδομο ίσον και πριν τη δεύτερη ισοδυναμία... :P ).

Ευχαριστώ, θα τα συμπληρώσω!
Αλλά αύριο, γιατί τώρα νυστάζω..  :oops:

Και στην επόμενη θα πληκτρολογήσω και τη λύση.  :D

[quote="ambros"]Λήμμα θεωρητικό
Να αποδείξετέ ότι
αν α ρητός  μακάρι να μπορούσα να γενικεύσω κι για παραγματικούς οποίος ξέρει να το κάνει ας το κάνει κι
α/α+1 = κ ακέραιος
τοτε α=1

λύση
εστω α=κ/λ με (κ,λ)=1
τότε μετά από πράξεις
κ/κ+λ άρα κ=λ μόμο άρα α=1[/quote]

 Φίλε Αμβρόσιε !

 Εσύ είσαι  νοήμων νέος  :)  και καλό παιδί. Χωρίς παρεξήγηση θα σε ρωτήσω : Μπορεί μια άσκηση , όπως την έγραψες , να μπει σε βιβλίο ; Θα μας κράξουν !(Sorry! Λίγο τραβηγμένο, αλλά με ξάφνιασες. Περίμενα κάτι άλλο από σένα , γιατί είσαι παλιό μέλος .)  Λέμε να προσπαθήσουμε να γράφουμε τις ασκήσεις όσο πιο περίτεχνα γίνεται. Ένα βιβλίο θα μείνει για χρόνια. Αυτό , αν καταφέρουμε και το βγάλουμε , θα το διαβάζουν και τα παιδιά  σου. Θα σε παρακαλέσω , τελείως  φιλικά , και γνωρίζοντας τη φιλότιμη και ευγενική φύση σου , να ξαναστείλλεις το θέμα διατυπωμένο ωραία , με όλες τις πράξεις , με το σκεπτικό του κλπ. Δε θέλει LATEX .
 Ένας παραπάνω λόγος που σου το λέω αυτό , είναι διότι θέλω, οι ασκήσεις που θα βάλεις να είναι γραμμένες με το δικό σου χέρι, χωρίς να τις τροποποιήσουμε εμείς. Διότι , αν την αλλάξουμε εμείς , παύει να είναι δημιουργία σου. Αλλά όλοι θέλουμε , σε αυτό το βιβλίο , να αφήσουμε κάτι από τον εαυτό μας, από την ψυχή μας! Όχι απλά  μια άσκηση που θα τη συνοδεύει το όνομά μας. Αν με κατάλαβες , έχει καλώς. Αν όχι , τότε θέλω να με συγχωρέσεις για το αυστηρό κάπως ύφος μου και να συνεχίσεις να βοηθάς , όπως εσυ νομίζεις. Γιατί χρειαζόμαστε και τη δική σου βοήθεια , όπως ακριβώς και όλων των άλλων.

    [u]Μπάμπης[/u]

Η [color=red]άσκηση 1α[/color] διορθώθηκε και προστέθηκε και η δεύτερη λύση στο pdf.
Αν θέλει κάποιος, ας κάνει τον κόπο να το ελέγξει άλλη μία ώστε να είμαστε σίγουροι.  :)

[color=red]Άσκηση 1α-διορθώσεις[/color]

Σωστή :D , αλλά αν θες μπορείς να γίνεις ακόμη πιο σωστή  :P . Συγκεκριμένα η πρώτη σειρά της δεύτερης λύσης νομίζω μπορεί να παραλειφθεί και απλά να μπει στη δεύτερη σειρά $p\geq 2$ αντί για $p>2$ και να σβηστεί το "όχι πρώτος".

Btw,εκεί που λες "η ανίσωση ισχύει τετριμμένα" νομίζω θα ήταν σωστότερο "η ανισότητα ισχύει τετριμμένα" ή ακόμη καλύτερα "η ανισοϊσότητα ισχύει κατά τετριμμένο τρόπο". Αν δεν κάνω λάθος, ανίσωση λέμε στην περίπτωση που υπάρχει κάποιος άγνωστος που πρέπει να προσδιορίσουμε.

Sorry αν κούρασα με τις ανούσιες παρατηρήσεις μου. :P

Και το σκεφτόμουν για το "ανίσωση" και "ανισότητα"... Από το σχολείο ακόμα μπερδεύω τα ονόματα!  :blush:

Όχι, καλά κάνεις.  :yup: Το θέμα είναι να τις έχουμε όσο πιο σωστές γίνεται, ακόμα κι αν αυτό έχει να κάνει με τη διατύπωση.  :wink:

Ευχαριστώ που τις ξανακοίταξες!  :lol:

(Αλλά για τη δεύτερη λύση, χρειάζομαι το ok από τον kosta, καθότι αυτός την διατύπωσε έτσι.)

[quote="ambros"]Κύριε Στεργίου ζητώ συγνώμη
θα ξαναγράψω την άσκηση κι θα την δημοσιεύσω σκαναρισμένη σε εικόνα στο φόρουμ πάντωες θα διατηρήσω αυτό το πόστ για να μείνει η ιδέα
ΟΚ???
ευχαριστώ πολύ[/quote]

 [color=blue]Αμβρόσιε , ευχαριστώ για την κατανόηση. Το εκτιμώ ιδιαιτέρως . Θα δεις ότι με πλήρη λύση το θέμα σου θα έχει εξαιρετική μορφή και θα είναι δημοσιεύσιμο ! Έτσι θα μπορούν όλοι να το καταλάβουν και να το απολαύσουν.Αυτό δεν επιδιώκουμε ; [/color]

               [color=red]Καλή Συνέχεια !!![/color]

ola οκ Βικη
παμε για  την επομενη ασκηση,

Επειδή έχουν περάσει 24 ώρες από τη στιγμή που πρωτοδημοσίευσα την άσκηση και δεν μπορώ να επεξεργαστώ το μήνυμά μου, μετά και την τελική διόρθωση την ανεβάζω ξανά εδώ.

[color=red]EDIT:[/color] Ουφ!  :cursing: Πιο πολύ κουράστηκα να διορθώσω το δικό μου παρά τα άλλα... Μου φαίνεται το επόμενο θα το ανεβάσω σε κανά μήνα!  :P

VickiL αν θες σβήσε και το [u]"ότι n!+1 όχι πρώτος και"[/u] από την πρώτη σειρά της δεύτερης λύσης.
[size=75]Lol, δε θα σε αφήσω σε ησυχία μου φαίνεται.  :P[/size]

[quote="silouan"]
$1+k\geq 1+xk(z-1)$  ή $k(z-1)\leq 1$ άρα $k\leq 1$ και $z-1\leq 1$
[/quote]

Εδώ στην πραγματικότητα έχουμε $x(z-1)\leq 1$ κι οπότε $x=1$ και $z=1$ ή $x=1$ και $z=2$
Η πρώτη πράγματι δεν δίνει λύσεις ... η δεύτερη γίνεται $1+k=2^{k}$ και λόγω της Bernoulli έχουμε τη λύση που έγραψες :wink:

[quote="bilot4"]
Εδώ στην πραγματικότητα έχουμε $x(z-1)\leq 1$[/quote]

Thanks για τη διόρθωση  :)

Πολύ καλή η προσέγγιση με την Bernuli !!!

Mπράβο siluan...   την έλυσες πιο γρήγορα και κομψά από τη λύση που έχω εγώ...

Θα σας γράψω τη λύση που έχω σε αρχείο word σε κανα δυο μέρες. Καλό σαβ/κο...

Λοιπον παιδιά έγραψα σε αρχείο word τη λύση που είχα για τη διοφαντική εξίσωση  χ+y = χ*(z^y).  Προσωπικά θεωρώ ότι ο siluan έδωσε πιο κομψή λύση αλλά πολλές φορές το να έχουμε διαφορετικές προσεγγίσεις ενος θέματος βοηθάει.

Όποιος θέλει λοιπόν ας κάνει download  ( Τυχόν επισημάνσεις έίναι χρήσιμες ... ) ( ελπίζω να μην έχω λάθη ... )

ΕΠΙΣΗΣ

αν με το να επισυνάπτω αρχεία σε word  σπαταλάω κάποιο χώρο ( μνήμης ίσως )  παρακαλώ να μου πείτε ώστε να μην επισυνάπτω με το παραμικρό αρχεία word για download. Είμαι πρωτάρης και δεν ξέρω πως λειτουργούν τα Forums

[quote="galois01"]Βάζω μιά εύκολη και ενδιαφέρουσα ασκησούλα-κατά τη γνώμη μου (γιατί με τόση ζέστη δεν είμαστε για ποιό δύσκολα :wink: - τουλάχιστον εγώ :P ):
Βρείτε όλους τους πρώτους αριθμούς n για τους οποίους οι n,n+4,n+8 είναι πρώτοι αριθμοί.[/quote]

Για $n=2$ άτοπο.
Για $n=3$, $n+4=7$, $n+8=11$, άρα μας κάνει
Για $n>3$ κάποιος απ' τους τρεις θα είναι πολαπλάσιο του τρία κι άρα όχι πρώτος.
Άρα γίνεται μόνο για τον $n=3$

Είναι σχεδόν η ίδια με το πρώτο θέμα στον Ευκλείδη Α λυκείου της περσινής χρονιάς :wink:

Αυτή ακριβώς είναι η λύση που σκέφτηκα και εγώ. :idea:
P.S:Δέν ήξερα ότι είναι παρόμοια με αυτή που μπήκε περυσι στον Ευκλείδη της Α΄Λυκείου μου την έδωσε ένας φίλος και είπα να τη βάλω.

Οι προηγούμενες ασκήσεις $ 34$ (bilot4), $ 35$(vangelismak), $ 36$(galois01) δεν είχαν αρίθμηση. Συνεχίζω λοιπόν με την άσκηση $ 37$.
[size=150][b][color=red]
ΑΣΚΗΣΗ 37 [/color][/b][/size]

Να αποδειχθεί ότι για κάθε περιττό πρώτο $ p$, υπάρχει ακολουθία από $ p$ διαδοχικούς σύνθετους αριθμούς. (Προφανώς ισχύει και για $ p=2$ αλλά δεν έχει ενδιαφέρον).

Αλέξανδρος

Αυτό ισχύει γενικά για κάθε αριθμό $ n$ .
Οι αριθμοί $ (p-1)!,\ (p-1)!+1,\ ....,(p-1)!+p-1$
δίνουν τη λύση για p πρώτο

Πολύ σωστά Silouan. Πρέπει όμως να αναφερθεί ότι ο $ (p-1)!+1$ είναι σύνθετος λόγω του θεωρήματος του Wilson (διαιρείται από το $ p$). Οι υπόλοιποι είναι προφανές γιατί είναι σύνθετοι.
Μας κάνουν οι ίδιοι αριθμοί εαν $ n$ όχι πρώτος? Πώς θα εξασφαλίσουμε το ότι ο $ (n-1)!+1$ είναι σύνθετος?

Αν δεν μιλάς για τους ίδιους αριθμούς τότε ποιοι είναι αυτοί ? Ξέρουμε συγκεκριμένα ή είναι απλά ένα θεώρημα ύπαρξης?

Αλέξανδρος

οιπόν δεν το θυμόμουνα ακριβώς . Είναι ν-1 διαδοχικοί σύνθετοι .

Είχες δίκιο Silouan. Η απάντηση είναι πολύ απλή και μου ήρθε μετά.

Για οποιοδήποτε αριθμό $ n$, oι αριθμοί $ (n+1)!+2, (n+1)!+3, \cdots , (n+1)!+n, (n+1)!+(n+1)$ είναι $ n$ σε πλήθος διαδοχικοί σύνθετοι αριθμοί.

Η χρησιμότητα της παραπάνω πρότασης είναι ότι μέσα στο σύνολο των φυσικών αριθμών μπορούμε να βρούμε οσοδήποτε μεγάλα "κενά" θέλουμε στα οποία να μην υπάρχει κανένας πρώτος αριθμός. Μου δίνεις πόσο μεγάλο "κενό" θέλεις και σου βρίσκω σε ποιους αριθμούς ανάμεσα πρέπει να κινηθεις (κοίτα παραπάνω) ώστε να μην υπάρχει κανείς πρώτος ενδιάμεσα.

[size=150][b][color=red]Edit:[/color][/b] [/size]

[color=blue]Ανοικτό παραμένει το ερώτημα αν μπορούμε με κάποιο άλλο τρόπο (εκτός από εξαντλητικό ψάξιμο) να βρούμε τους μικρότερους διαδοχικούς αριθμούς που να έχουν το επιθυμητό κενό. Αυτή είναι εύλογη ερώτηση αν αναλογιστούμε ότι το παραγοντικό μεγαλώνει πολύ πολύ γρήγορα. Να αναφέρω ότι π.χ. με τον παραπάνω τρόπο για να προσδιορίσουμε $ 5$ διαδοχικούς σύνθετους αριθμούς μπορούμε να πάρουμε τους αριθμούς $ 6!+2, 6!+3, 6!+4, 6!+4, 6!+6$ δηλαδή τους αριθμούς $ 722,723,724,725,726$. Όμως οι $ 5$ πρώτοι διαδοχικοί φυσικοί αριθμοί που συναντάμε είναι οι $ 24,25,26,27,28$ (ασφαλώς πολύ πολύ μικρότεροι από εκείνους που προκύπτουν με την παραπάνω μέθοδο).[/color]

Αλέξανδρος

[size=150][b][color=red]ΑΣΚΗΣΗ 38[/color][/b][/size]
Να βρείτε τις ακέραιες λύσεις της εξίσωσης
$ 2x^{6}+3y^{6}=z^{6}$

$ 2x^{6}+3y^{6}=z^{6}$.Eστω $ (x_{0},y_{0},z_{0})$ η λυση της εξισωσης με $ h(x_{0},y_{0},z_{0})=x_{o}y_{o}z_{o}$ να ειναι το ελαχιστο. Εστω $ d=(x_{o},y_{o},z_{o})$ τοτε μια ακομα λυση θα ειναι και η $ (x_{0}/d,y_{o}/d,z_{o}/d)$ ομως τοτε $ h(x_{o}/d,y_{0}/d,z_{0}/d)=x_{o}y_{0}z_{0}/d^{3}<x_{o}y_{0}z_{0}=h(x_{0},y_{0},z_{0})$ .Επομενως βλεπουμε οτι για την λυση $ (x_{0},y_{0},z_{0})$ θα πρεπει να ισχυει οτι $ d=1$
[url]LEMMA[/url]
Για καθε θετ ακεραιο $ x$ ισχυει $ x^{6}=1mod9$,$ x^{6}=0mod9$.Αρα για την λυση μας ισχυει οτι $ 3/z_{0}^{6}-2y_{0}^{6}$.Kατι τετοιο ισχυει μονο αν $ 3/z_{0}$ και $ 3/x_{0}$ αρα και $ 3/y_{0}$ που ειναι ατοπο αρα η εξισωση μας δν εχει ουδεμια θετ ακεραια λυση.

Με τον ιδιο τροπο αποδεικνειω οτι η εξισωση $ 2x^{3}+3y^{3}=z^{3}$ εχει μναδικη λυση την $ (0,0,0)$.Λαβε υποψην οτι $ x^{3}=0mod9$ η $ x^{3}=1mod9$ η         $ x^{3}=-1mod9$  :wink:


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inequalities
inequalities proposed

Let $ a,b,c$ be non-negative real numbers satisfying the condition $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 2$.
Prove that $ a \plus{} b \plus{} c \plus{} a^3b^3c^3\geq 2(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3)$

[quote="kally"]goto search page
[url=http://www.math123xyz.com]Interactive Math Resources[/url][/quote]
Kally is the ownage  :wink:  :rotfl:
Inequalities needs more bots!


If $a,b > 0$ and the triangle in the first quadrant bounded by the coordinate axes and the graph of $ax+by = 6$ has area 6, then $ab =$

$\text{(A)} \ 3 \qquad \text{(B)} \ 6 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 108 \qquad \text{(E)} \ 432$

[hide]The x intercept (and also the base) is $\frac{6}{b}$, while the y-intercept (or height) is $\frac{6}{a}$. The area would be the product or $\frac{36}{2ab}=6$. Solving for $ab$ gives us 3.   [/hide]

[hide]area of triangle formed by ax+by+c=0 with coordinate axes is c^2/2|ab|

with the help of this formula....36/2|ab| = 6

36 = 12|ab|

|ab|=3

therefore,  ab = 3 [/hide]

[quote="sagar_astakala"]area of triangle formed by ax+by+c=0 with coordinate axes is c^2/2|ab|

with the help of this formula....36/2|ab| = 6

36 = 12|ab|

|ab|=3

therefore,  ab = +3 or - 3[/quote]

Okay.. Which one though? (If you weren't given multiple choice)

What do we know about $a$ and $b$ to determine if $ab$ is positive or negative?

In the given it was posted $a, b>0$ so $ab>0$.


The digits 2, 3, 4, 7 and 8 are each used once to form a five-digit number.  What is the probability that the tens digit is odd and the number is divisible by 4?  Express your answer as a common fraction.

The last two digits that satisfy the given conditions are 32 and 72. The number of possible last two digits = 5x4 = 20. So the probability is 1 out of 10.


Let $a,b,c$ be positive reals such that $3(a^2+b^2+c^2) = 4(a+b+c+3)$.  Prove that

\[\frac{1}{\sqrt{a^4+4}} + \frac{1}{\sqrt{b^4+4}}+ \frac{1}{\sqrt{c^4+4}} \geq \frac{3\sqrt{5}}{10},\]

and determine when equality holds.

Well, I'm not 100% sure this is right; I found this by a somewhat contrived method.  So, anyone else have some thoughts on this?

Huh?  I think it still holds with that condition...


vector
algebra
polynomial
induction
linear algebra
matrix
function

More general statement: if an endomorphism $ T$ satisfies any polynomial equation $ P(T)=0$ and if $ \lambda$ is an eigenvalue of $ T,$ then $ P(\lambda)=0.$ Notice in this particular case that $ P(\lambda)=\lambda^{2}-1.$

The proof is straightforward. Suppose that $ x\ne0$ and $ Tx=\lambda x.$ By induction, you can prove that for $ k=0,1,2,\dots,$ we have $ T^{k}x=\lambda^{k}x.$ Then if $ P$ is any polynomial, we have that $ P(T)x=P(\lambda)x.$ But then if $ P(T)=0,$ it follows that $ P(\lambda)x=0,$ and with $ x\ne 0,$ then $ P(\lambda)=0.$

Note that this is [i]not[/i] the celebrated Cayley-Hamilton theorem - this is just an elementary observation.

[quote="kalazealot"]$ E$ is finite dim vector space, $ f$ is an involutary automorphism of $ E$, i.e. $ f^{2}=1_{E}$ (but $ f \neq 1_{E}$).

Prove: The eigenvalues of $ f$ is $ +1$ or $ -1$.[/quote]

$ f^{2}=1_{E}\iff f^{2}-1_{E}=0_{E}$

Then, the Minimal polynomial for $ f$ is $ p(x)=x^{2}-1=(x-1)(x+1)$.

The minimal is product of linear irreductible factors, then the representant matrix for $ f$ , is diagonalizable and the minimal polynominal is equal to characteristic polynominal ( remeber that $ f\neq 1_{E}$).

Then, the eigenvalues for $ f$ are the diagional elemnts of the matrix for $ f$ , and they are $ 1$ and $ -1$

sorry if my english is poor :blush:  :(

[quote="Kent Merryfield"]About that last post ...

We don't know for sure that the minimal polynomial is $ x^{2}-1.$ All we really know is that the minimal polynomial divides $ x^{2}-1.$  For example: suppose our function is $ -I.$ That satisfies our conditions, but the minimal polynomial is $ x+1.$

As for the argument that the matrix is diagonalizable: true, but overkill. We don't need anything that fancy. I think this problem intended the use of elementary tools.

From the conditions of the problem, we can conclude that $ -1$ is an eigenvalue, that $ 1$ may be an eigenvalue, and that there are no other eigenvalues. Notice the "or" in kalazealot's original post.[/quote]

 :o

You have reason.

My english level is very bad and i can't expresate the ideas correctly.

Bye!


Let $ ABC$ be a triangle and let $ D \in \bar{BC}$, $ E \in \bar{CA}$ and $ F \in \bar{AB}$ be points such that $ EA \plus{} AF \equal{} FB \plus{} BD \equal{} DC \plus{} CE$. Prove that

$ DE \plus{} EF \plus{} FD \ge \frac {AB \plus{} BC \plus{} CA}{2}$.

LaTeX question: isn't it possible to put the bar over something bigger than a letter (as I failed to do with $ \bar{AB}$)?

\overline{bigsomething} for your LaTeX question.

The problem itself has been posted before but never answered afaik.

  darij


trigonometry algebra function domain LaTeX

Assume $0\leq\alpha<2\pi$.
1) Show that the identity $\frac{\tan\alpha}{1-\cot\alpha}+\frac{\cot\alpha}{1-\tan\alpha}=\tan\alpha+\cot\alpha+1$ holds for all $\alpha$ in the domain except certain values, and determine those values of $\alpha$ for which this identity is false.
2) Show that the identity is also equal to $\sec\alpha\csc\alpha+1$.

Hello, 3X.lich!

Once again, this requires a lot of LaTeX, so I won't try to hide it.

[quote]Assume $0 \leq x <2\pi$.
1) Show that the identity $\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}=\tan x+\cot x + 1$
holds for all $x$ in the domain except certain values,
and determine those values of $x$ for which this identity is false.

2) Show that the identity is also equal to $\sec x\csc x + 1$.[/quote]
1) We have: $\frac{\tan x}{1 - \cot x} + \frac{\cot x}{1 - \tan x}$

Multiply top and bottom of both fractions by $\tan x:$

$.\qquad \frac{\tan^2x}{\tan x - 1} + \frac{1}{\tan x(1 - \tan x)} \;= \;\frac{\tan^2x}{\tan x - 1} - \frac{1}{\tan x(\tan x - 1)}$

$.\quad= \;\frac{\tan^3x - 1}{\tan x(\tan x - 1)} \;= \;\frac{(\tan x - 1)(\tan^2x +\ tan x + 1)}{\tan x(\tan x - 1)}$

If $\tan x-1 \neq 0$, that is, $x \neq \left\{\frac{\pi}{4},\frac{5\pi}{4}\right\}$, then we cancel

and we have: $\frac{\tan^2x + 1 + \tan x}{\tan x} \;= \;\frac{\tan^2x}{\tan x} + \frac{1}{\tan x} + \frac{\tan x}{\tan x}$

$.\qquad= \;\tan x + \cot x + 1$

2) We had: $\frac{\tan^2x + 1 + \tan x}{\tan x} \;= \;\frac{\sec^2x + \tan x}{\tan x} \;= \;\frac{\sec^2x}{\tan x} + 1$

$.\quad= \;\frac{\frac{1}{\cos^2x}}{\frac{\sin x}{\cos x}} + 1 \;= \;\frac{1}{\cos x\sin x} + 1 \;= \;\sec x\csc x + 1$

Very nice Soroban! Great work and effort. As a reward, you belong to the Pre-Olympiad level and above  :!:  :)


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