Let a and b be distinct, positive integers that satisfy [tex]a < b^2 < a^3 < b^4 < a^5 < ... < b^2^0^0^4 < a^2^0^0^5[/tex]. Find the smallest value of [tex]a + b[/tex].

I forget where I found this problem, but I need help in solving it.  Thanks!

I would think that a and b have to be relatively large numbers that satisfy $a+1=b$.  Just by looking at the problem, my best guess is 2004+2005=4009, but it's unlikely.  You could mabye try solving a simpler version of the problem, like find a and b so that $a<b^2<a^3<b^4<a^5$.  Then, work your way up and try to find a pattern.

Actually, now that ive thought about it a little longer, you can say the following:

$a<b^2$, so $\sqrt a <b$
$b^2<a^3$, so $a>b^{\frac 2 3}$
and so on until
$a^{2003}<b^{2004}$, so $a<b^{\frac{2004}{2003}}$
$b^{2004}<a^{2005}$, so $a>b^{\frac{2004}{2005}}$
From these last two equations, you can see that $b^{\frac{2004}{2005}}<a<b^{\frac{2004}{2003}}$.  All that remains is to find the smallest integers that satisfy this equation.


Prove that
\[ 1{\binom n 1}^2 \plus{} 2{\binom n 2}^2 \plus{}3{\binom n 3}^2 \plus{} ... \plus{} n{\binom n n}^2 \equal{} n \binom{2n\minus{}1}{n\minus{}1} \]

Still no answer !!  :o
Come on guys, this is a really nice problem.

I somehow feel there is a combinatorial proof involved.

[hide]Mainly because you said that its a nice prob.  :rotfl: [/hide]

[hide="Else this is standard problem"]

Consider $ (1\plus{}x)^n \equal{} 1 \plus{} C_1x \plus{} C_2x^2 ... C_nx^n$

Differentiating $ n(1\plus{}x)^{n\minus{}1} \equal{} C_1 \plus{} 2C_2x \plus{} 3C_3x^2 .....nC_nx^{n\minus{}1} \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}1$

Consider$ (x\plus{}1)^n \equal{} x^n \plus{} C_1x^{n\minus{}1} ....\plus{} C_n \minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}2$

Multiply 1 and 2 and coefficient of $ x^{n\minus{}1}$ gives the result. [/hide]

@Conjurer: Yes, there is a combinatorial proof. But your solution is really nice too. Why do you call it a standard problem? Is your method a standard application?

BTW, If anyone wants me to, I will post the combinatorial solution.

Take a look at my post in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=277960]this thread[/url]

Nice one keyree 10 but this sum can also by done by using Lattice Paths by constructing them by moving along by 1 step.
Your proof is very nice.

[quote="keyree10"]Prove that
\[ 1{\binom n 1}^2 \plus{} 2{\binom n 2}^2 \plus{} 3{\binom n 3}^2 \plus{} ... \plus{} n{\binom n n}^2 \equal{} n \binom{2n \minus{} 1}{n \minus{} 1}
\]
[/quote]

I have a purely mathematical method.

First multiply both sides of the equation by 2

then what we have to prove is $ 2{\binom n 1}^2 \plus{} 4{\binom n 2}^2 \plus{} 6{\binom n 3}^2 \plus{} ... \plus{}2 n{\binom n n}^2 \equal{} n \binom{2n}{n}$

Add $ 0{\binom n 0}^2$ to both sides so that it becomes symmetric.

Then pair $ 0{\binom n 0}^2$ to $ 2n{\binom n n}^2$ and so on.................

What we get is $ 2n({\binom n 0}^2\plus{}{\binom n 1}^2.............{\binom n {[n/2]}}^2)\equal{} n \binom{2n}{n}$
So we get $ 2({\binom n 0}^2\plus{}{\binom n 1}^2.............{\binom n {[n/2]}}^2)\equal{} \binom{2n}{n}$

Which is $ {\binom n 1}^2 \plus{} {\binom n 2}^2 \plus{} {\binom n 3}^2 \plus{} ... \plus{}{\binom n n}^2 \equal{}  \binom{2n}{n}$

Consider $ (x\plus{}1)^{2n} \equal{} (x\plus{}1)^n*(x\plus{}1)^n$

Consider the co efficient of $ x^n$ and we are done.

P.S. Beautiful method keyree


function
calculus
integration
calculus computations

find the primitive of the following function:

$ f(x)\equal{}\frac{x^{2}\minus{}x\plus{}1}{(x^{2}\plus{}1)^{2}}$

$ f(x) \equal{} \frac {1}{x^2 \plus{} 1} \minus{} \frac {x}{(x^2 \plus{} 1)^2}$

the integral of the first term is $ \arctan x$, and the second part is done by substitution.


calculus
integration
logarithms
complex analysis
complex analysis unsolved

$ \int _0 ^{ \plus{} \infty}\frac {\ln x}{x^6 \plus{} 1}dx$

I don't know which contour to integrate, using the method of residues...

Well, you need to get back from zero to $ \infty$ to close the loop. You're not going to do it with a negligible integral, so try for an integral that's similar to what you already have. What path can you take that makes $ x^6\plus{}1$ the same, and what does it do to $ \ln x$?


have you seen the problemset of croatian open 5  ?
the last problem was like

given a sequence A , length of A is <=40,000 (or something like that)
find the max length of the sequence that occured in A more than 1;

like if it is

ABCABCDEF

len (ABC) = 3, is the output, it occured twice.

[quote="fahim"]oh!
they published the soln in their site!

what if i am given a seq A, another seq B, and asked how many times B occured in A as a subsequence ?[/quote]

Isn't that just string matching? http://en.wikipedia.org/wiki/String_searching_algorithm

fahim ur new problem is just KMP. Look at CLRS!

Mahbub

yes! the first problem is string matching, i had a look at CLRS, robin-karp seemed easier than KMP.

But isnt it somekind of dp if i am given a sequence A like A="KABUBAABA", and B="ABA", and asked how many subsequence ABA is there in A, like here i have 3 + 2 = 5 "..A..B..A.."

i guess there is an acm problem like that! i am not sure whether recursion with memoization would suffice if len(A) is very big..

[quote]
But isnt it somekind of dp if i am given a sequence A like A="KABUBAABA", and B="ABA", and asked how many subsequence ABA is there in A, like here i have 3 + 2 = 5 "..A..B..A.."
[/quote]

I dont think your example is correct one. dont you think that the answer is: 9 ?

as it is subsequence so i dont think KMP is going to work this time. i dont know the limit, and i have not yet thought on the solution, but i think, u should make a tree and thus count the number of solution.

DP might also come into play, i am not sure!

never mind! i found a dp solution of O(MN) , m = len(A), n = len(B);
acm has this problem at [url=http://acm.uva.es/p/v100/10069.html]10069,[/url] but from the uva eboard, i knew i have to use big int there too!  :mad:

O(MN)  :mad: i thought you wanted linear solution!

:maybe: its confusing me.. my dp soln seemed okay, but its getting wa in uva at 3.762s!  :(
i wrote you a mail, can you check ?


Here's one that I thought wasn't too hard, but kinda fun (from alcumus):

[b]How many 4-digit positive base ten integer palindromes are there?[/b]

If you forgot what a palindrome is, it's a number that when flipped is still the same number (for example, 141).

[hide="Solution"]
All 4-digit- palindromes are in the form of ABBA, where A and B are digits (A and B CAN be the same). Because it is base ten, there are 10 choices for each of A and B. However, a 4-digit number cannot begin with zero, so there are 9 choices for A. $ 9*10\equal{}\boxed{90\text{ palindromes}}$[/hide]

I've seen this problem a bunch of times. :)
Good for beginners to counting.

[quote="Jean-Louis"]We have established that there are 90 palindromic numbers with four digits.

So, here's a problem :

find the 4-digit number abcd such that : abcd * 9 = dcba[/quote]

2178*4=8712

To mz94:

Where on Earth did you dream up that answer?

You can do it with modular arithmetic as well, and bases.

well you realy only need to know the first 2 digits because the second 2 digits will be the mirror of it. so all you need is how many 2 digit integers there are. theres 90. this is to the original problem. its an easier solution

we know a has to be 1. so $ 1bcd\equal{}dcb1$ so $ 999\plus{}100b\plus{}10c\plus{}d\equal{}1000d\plus{}100c\plus{}10b, 999\plus{}90b\equal{}999d\plus{}90c$ so $ 111\plus{}10b\equal{}111d\plus{}10c$. then we have d has to equal 1 otherwise it would be impossible to solve the equation with integers. so b=c

ok i messed up. can someone help

Well, let's see.

For this to be true,
$ 9000a\plus{}900b\plus{}90c\plus{}9d\equal{}1000d\plus{}100c\plus{}10b\plus{}a$
Thus,
$ 9991a\plus{}890b\minus{}10c\minus{}991d\equal{}0$

Um... well, you could use a computer.

or calculator. but my 83+ is in my backpack and im 2 lazy to get it

[quote="Temperal"]With some caseowork, you can discover that it's impossible.  :wink: (Edit: for integers)

You just take various constant integers as two of the vars, then graph on a Cartesian plane.[/quote]

what do you mean edit for integers? you cant have like 5/4 for a digt.

sure lets have five fourths hundred fourtyteen

[quote]ok, so this one you have to use imaginary numbers. you know like eleventy tweleve[/quote]

I mean that there are no integer solutions.

if there are no integer solutions, there ARE no solutions. its asking for a, b, c, and d.

Yes, that's exactly what I meant. There are solutions for the equation, but no solutions for the problem.

Please stop reviving the thread.

[quote="dtlee88"]1089*9 = 9801.

1) We know a = 1 because otherwise the result would have 5 digits.
2) We know d = 9 because d*9 has to end in a 1.
3) We know b = 0 because otherwise there will be carry over and the result will be 5 digits.
4) We know c = 8 because we need c*9 + 8 to end in 0.[/quote]

Nice!
I almost got carried off by previous posts, but phew.. :lol:  good thing I came to page 2.

@pi guy: 90 was the answer for the original problem.. This was the question that [u]Jean-Louis[/u] came up with.

Since a=9, if b>0, then the product would be five digits.


[hide]
This follows a pattern:
The number of 5s in the difference will be the number of 2's in the 2nd expression.
And then there will be one 4 and a bunch of 0's.
$77^2 - 23^2 = 5400$
$777^2 - 223^2 = 554000$
$7777^2 - 2223^2 = 55540000$
So the sum will be $5(14) + 4 = 74$
[/hide]

YAY. There was a problem much like this on the math jam the other night.

It was, in their base 10 representations, the integer a consists of 1985 8's and the integer b consists of 1985 5's. What is the sum of the digits in the base 10 representation of 9ab. And basically, you approach it the same way Snowstorm approached this problem.

That one was WAY TOUGHER.

I didn't quite get it and it's one of mock AMC J question. So, if you can answer it with explanation, I would be grateful.

Or anyone else who knows it.

Thanks.

Well, using SnowStorm's method, if you don't want to proceed with a time consuming constructive approach, we have:

9 * 8 * 5 = 360
9* 88 * 55 = 43560
9 * 888 * 555 = 4435560
9 * 8888 * 5555 = 444355560
.
.
.

As you can see, there is the same pattern here. For n 8's or 5's there are (n-1) 4's and 5's in the product. Since the 3 and 6 are constant, a conjecture could be that for n 8's and 5's, the sum of the digits of the product could be 9n, and this case, n = 1985. 9 * 1985 = 17865.

[quote="Silverfalcon"]The sum of digits of the integer equal to ${777,777,777,777,777}^2 - {222,222,222,222,223}^2$[/quote]Let $d\ldots{d}$ represent a string of $14$ digits, each $d$. Then $7\ldots77^2-2\ldots23^2=10\ldots0\cdot5\ldots54=5\ldots540\ldots0.$ The sum of the digits is thus $5\cdot14+4=74$. I'm usually not comfortable with solutions that rely on "seeing the pattern."

I got 142/150 on that contest back in the day; missed an easy case on the last question   :(


Who is the Winner?
[hide="Somewhat disturbing image"]
[img]http://hagacity.jeeran.com/files/61839.jpg[/img][/hide]
[img]http://antizion.webphoto.ir/photos/an985236.jpg[/img]

Could there ever be a winner in such a war? It's a continuous conflict, in which we haven't seen the winners: but we see the losers every day-innocent civilians.

The ground operations started this evening: is it a step in restoration of 'peace' (peace-always a relative term in the middle East, as the War for Peace's led every couple of years...), or the final act of aggression?

Before it's too late, can anybody fix the poll? At the moment, the six answers are
  	Gazza
  	Gazza
  	Gaza
  	Israel
  	Israel
  	Israel
This may look like someone is trying to skew the statistics - voters for Gaza will always vote the correctly-written (3rd) option, while voters for Israel will choose any of the three Israel's. Now I am pretty sure that this was done without malicious intent - there is a bug in ML that leads to duplicated options in polls, probably it was triggered when Amir tried to correct the spelling of Gaza. So could anyone remove the duplicates, please, so that people don't get suspicious?

(Actually I'm not quite fond of discussing the "winner" before the war has passed, but okay - the round table has worse topics.)

  darij

[quote="darij grinberg"]... This may look like someone is trying to skew the statistics - voters for Gaza will always vote the correctly-written (3rd) option, while voters for Israel will choose any of the three Israel's. Now I am pretty sure that this was done without malicious intent - there is a bug in ML that leads to duplicated options in polls, probably it was triggered when Amir tried to correct the spelling of Gaza. So could anyone remove the duplicates, please, so that people don't get suspicious?

(Actually I'm not quite fond of discussing the "winner" before the war has passed, but okay - the round table has worse topics.)

  darij[/quote]

actually Gazza is not wrong , this is how its pronounced in Arabic and Farsi , we  don't pronounce it as gaza , I don't know can you see the diference or not.
Anyway there is a bug in ML  as you see.

Just out of curiosity, What do you think about the conflict Darij?

man, the first picture is really awful, that would be better if you hide it.

:rotfl: The poll is getting better. Amir, maybe better delete it and open a new one? At the moment, exactly one person has voted in it. I think the majority of us have no idea what you mean by "winner". The war is not over. Should we write whom we [i]want[/i] to be the "winner" or whom we [i]expect[/i] to "win"? And even in this case, the answers could be modified as well - for instance, I wouldn't consider Hamas staying in power to be a win for Gaza... (Malezgerd, here you've got my opinion ;) )

Erken, get used to it, it's the internet, people don't care about wars here if they don't get something to see...

  darij

[quote="darij grinberg"]
Erken, get used to it, it's the internet, people don't care about wars here if they don't get something to see...

darij[/quote]

sure, but concealment of the picture would not make the thread less attractive.

My opinion:
nobody is a winner, the confrontation is so ancient, but nobody has won the war yet and it is quite doubtful that we will ever figure out who is a winner.

I am not moderator hence i can't delete it :wink:
Yes,I translate farsi word by word to english which was wrong , anyway I meant "Who will be the winner?"
hence there are 2 options here besides gaza and Israel, that's 1-administartors solve the problem ,2-Delete the Topic.

In my opinion it is really to early to make a speculation as to who will win, but I feel Israel is overreacting.
You don't need to bomb schools and kill 150+ kids to stop 10-30 militants

In addition to that, I am ashamed of America's unwavering funding and support of Israel

So much for "Never Again"

The UN needs to make an exception in this case and override America's continued vetoes on behalf of Israel

[quote="darij grinberg"]

(Malezgerd, here you've got my opinion ;) )

[/quote]
Quite descriptive, mesmerized  :wink:

This is no war, but a Genocide !
May Allah give that boy and many other victims a place in de djanna !


Out of curiosity - are there any FF music fans here? My favorites so far are FF9's Melodies of Life and FF7's main theme...

i love the ff music. especially the ff7 battle theme. the first time i heard it i fell in love wit it.  the chrono trigger music rules all though. like magus's theme

I have found a kinded spirit... I have every FF8 song in the game on my laptop and I play them all the time. One-winged Angel and Sudeki da Nei are also pretty good.

-Isaac

[quote="PF790b-Jyzzy"]How can you convert .wav files into mp3 files?[/quote]
I think you could use Windows Media Player or the Windows Movie Maker (comes with Windows), or iTunes/Movies/w/e
You could also download some tools for conversion. If you could name the program I might be able to help you out.

Hmm. Question: What're you converting it for?

...don't know whether I should go :lol: or :twisted:

I mean, I'm pretty sure it's for a MP3 player.

your probably right...but most music files should come in MP3...


As I looked today at http://www.mathlinks.ro/Forum/viewtopic.php?t=146020 , most of the $ \text{\LaTeX}$-images there were rather strange (I repaired them now).
The problem seems to be the parser's interpretation of term like a^\frac bc.
Some time ago, this was rendered as a^{\frac{b}{c}}, but now it seems to cause strange effects:
$ a^\frac bc$
This may have caused a lot of other unintended changes in a lot of threads.

That is weird. I will take a look at it. Though again I stress people should totally avoid writing those kind of statements.

Problem is fixed now. Any other similar problems?


calculus
integration
function
calculus computations

Hello,

First of all, a big thank you for taking time to read this.    :)
The problem is how to integrate:

Integral(0,Inf): [Erfc(x)]^2  dx

http://mathworld.wolfram.com/Erfc.html

Equation (10) shows that Integral: Erfc(x) = 1/sqrt(pi)
Equation (11) shows that Integral: Erfc^2(x) = (2-sqrt(2))/sqrt(pi)

http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions

may help too

I have tried my best to prove (10).  Please see bottom of post for it, and please point out any mistakes.  I am unable to use this method on (11).

Any help on how to do (10) differently/correctly, and how to do (11) would be awesome.  I am seeking step by step analytical solution, as I ultimately want to tackle:

Integral(0,Inf): P(x) erfc^2(x)  dx      where P(x) = exp(-x)
Integral(0,Inf): Q(x) erfc^2(x)  dx      where Q(x) = exp(-x^2)

currently, P or Q is 1 for (10) and (11).

So here is my approach to (10):

Integral(0,Inf): erfc(x) dx
=  (2/sqrt(pi) * Integral(0,Inf): Integral (x,Inf): exp(-t^2) dtdx

1/  change limits of outer integral (0,Inf) -> (0,t)
2/  change limits of erfc integral (x,Inf) -> (0,Inf)
3/ Integrate with respect to x first  to get t*exp(-t^2)
4/ Integrate with respect to t to get  1/sqrt(pi)  as per (10)

Even if this is correct, I wouldn't know how to change limits for a triple integral of (11).

Any help is most welcome!    :ninja:

After using the Wolfram Integrator, I found the following.  If anyone can provide the Step- Step for the following, it would be awesome:

http://integrals.wolfram.com/index.jsp?expr=erfc%28x%29&random=false

http://integrals.wolfram.com/index.jsp?expr=erfc%28x%29+*+erfc%28x%29+&random=false

http://integrals.wolfram.com/index.jsp?expr=exp%28-x%29+*+erfc%28sqrt%28x%29%29+&random=false

[quote="weisiguo"]

Integral(0,Inf): [Erfc(x)]^2  dx

http://mathworld.wolfram.com/Erfc.html

[/quote]

all of the integrals listed there are fairly easy to do, including $ (12)$

 you only require high school elementary calculus and that is it. no need for double or triple integrals whatsoever.

anyway, here is one possible solution to $ (11)$ ( i also solved it easily using triple integrals... )

$ \int_0^\infty \text{erfc}^2(x) \,dx \;\;\equal{}\;\; \underbrace{\text{erfc}^2(x)\cdot x\;\bigg|_0^\infty}_{\equal{}\quad 0} \;\;\minus{}\;\; \int_0^\infty 2\;\text{erfc}(x)\;\underbrace{\minus{}\,\dfrac{2}{\sqrt{\pi}}\;e^{\minus{}x^2}}_{\frac{d}{dx} \;\text{erfc}\,(x)}\;  x\,\;dx$

$ \equal{}\dfrac{2}{\sqrt{\pi}}\;\underbrace{\text{erfc}(x)\;\minus{}e^{\minus{}x^2}\;\bigg|_0^\infty}_{\equal{}\quad 1} \;\;\minus{}\;\;\ \dfrac{2}{\sqrt{\pi}} \int_0^\infty \;\underbrace{\minus{}\,\dfrac{2}{\sqrt{\pi}}\;e^{\minus{}x^2}}_{\frac{d}{dx} \;\text{erfc}\,(x)}\;\minus{}e^{\minus{}x^2} \,dx$

$ \equal{} \dfrac{2}{\sqrt{\pi}} \minus{} \dfrac{4}{\pi} \int_0^\infty e^{\minus{}2x^2} \,dx$

$ \equal{} \dfrac{2}{\sqrt{\pi}} \minus{} \dfrac{4}{\pi}\;\left( \frac{1}{2}\;\sqrt{\frac{\pi}{2}}\right)$

$ \equal{} \boxed{\dfrac{2 \minus{} \sqrt{2}}{\sqrt{\pi}}}$


Let $ (A_n)_n$ and $ A$ be sets in $ \mathbb{R}^n$.  What is the default meaning of $ A_n\to A$?

edit: Is it $ A_n\to A$ iff $ \limsup A_n \equal{} \liminf A_n \equal{} A$?

That sounds right. Another way to put it:
For all $ x\in A$, $ x\in A_n$ for all but finitely many $ n$. For all $ x\not\in A$, $ x\not\in A_n$ for all but finitely many $ n$.

There is no relation to the structure of the underlying space.


Dear friends,

while I was solving a problem given by my adviser, I got the result which looks like this \[\mathbf{A}^{H}\mathbf{P}=\mathbf{A}^{H}\mathbf{r}\] where
 $\mathbf{A}^{H}$ is an $m\times n$ complex matrix with full rank ,
$\mathbf{P}$ is an $n\times n$ complex matrix with rank $k<n$
$\mathbf{r}$ is a vector of size $n \times 1$

The above equation can be written as \[\mathbf{A}^{H}( \mathbf{P}-\mathbf{I})\mathbf{r}=\mathbf{0}\] It is found that rank of $( \mathbf{P}-\mathbf{I})$ is found out to be $n-k$. How to solve for real $\mathbf{r}$ from the above equation.

anybody please help me.

[quote="bhupala"]Dear friends,

while I was solving a problem given by my adviser, I got the result which looks like this \[\mathbf{A}^{H}\mathbf{P}=\mathbf{A}^{H}\mathbf{r}\] where
 $\mathbf{A}^{H}$ is an $m\times n$ complex matrix with full rank ,
$\mathbf{P}$ is an $n\times n$ complex matrix with rank $k<n$
$\mathbf{r}$ is a vector of size $n \times 1$

[/quote]

The [b] LHS [/b] is a $n \times n$ - matrix, the [b] RHS [/b] a $m \times 1$ - vector. So they cannot be equal (only if $m=n=1$)

Sorry that the equation should be read as \[\mathbf{A}^{H}\mathbf{P}\mathbf{r}=\mathbf{A}^{H}\mathbf{r}\] regards


[quote="Dida Returns"]Does there exist an integer solution to

$ a^2 \minus{} 3(2b^2 \plus{} 1)^2 \equal{} 1$ ?[/quote]

$ a\equal{}2$ and $ b\equal{}0$


Two fair six-sided dice are rolled.  What is the probability that the product
of the two numbers is a composite number?  Express your answer as a common fraction.

Only non-composite numbers which are product of numbers of 1 to 6 are: 1,2,3,5. As 2,3 and 5 can be gotten in 2 ways, and 1 in one way, and there are 36 possible products, the probability is: $ 1\minus{}\frac{7}{36}\equal{}\frac{29}{36}$.

Possibly they tell students differently at math competitions, but 1 is not composite.

The definition of "composite" is having at least two different prime divisors.

So, I would revise the answer to $ \boxed{\frac {5}{6}}$.

EDIT: Ahh, rofl, sorry,

I thought that 29/36 looked like it was 1/36 away from 5/6, and more plausable then :)  Sure, looks like 29/36 is wrong, sorry for confusion.

Yes, 1 isn't composite, but answer is $ \boxed{\frac{29}{36}}$! I am 100 percent sure!

Both of you are say that 1 is not composite, and I confirmed, the answer is 29/36. If think RightAnnihilator juust thought that Bugi didn't coun't 1,1.

Not possible:
(1,1)
(1,2)
(1,3)
(1,5)
(2,1)
(3,1)
(5,1)


find a triple $ (x,y,z)$ of positive integers that satisfy $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 41^{13}$

[hide="Solution"]$ x\equal{}41^6$
$ y\equal{}6 \cdot 41^6$
$ z\equal{}2 \cdot 41^6$[/hide]


Let $ a$ and $ b$ real numbers distincts such as $ a^{2}\equal{}6b\plus{}5ab$  and $ b^{2}\equal{}6a\plus{}5ab$.

a) Determine the value of a + b.
b) Determine the value of ab.

a) Subtracting, we have $ a^2\minus{}b^2\equal{}6b\minus{}6a \implies (a\minus{}b)(a\plus{}b)\equal{}\minus{}6(a\minus{}b) \implies a\plus{}b\equal{}\minus{}6$ (because a and b are distinct).

b) Note that $ a\equal{}\minus{}6\minus{}b$

$ b^2\plus{}12b\plus{}36\equal{}6b\minus{}5b(b\plus{}6)$

$ b^2\plus{}12b\plus{}36\equal{}6b\minus{}5b^2\minus{}30b$

$ 6b^2\plus{}36b\plus{}36\equal{}0$

$ b^2\plus{}6b\plus{}6\equal{}0$

Also, by a similar process, $ a^2\plus{}6a\plus{}6\equal{}0$, so a and b are the roots of $ x^2\plus{}6x\plus{}6\equal{}0$ so by Vieta's their product is $ 6$

[hide="solution?"]

$ a^2 \equal{} 6b \plus{} 5ab$                 (i)
$ b^2 \equal{} 6a \plus{} 5ab$                 (ii)
(i)-(ii)
$ a^2 \minus{} b^2 \equal{} 6b \minus{} 6a$
$ (a \plus{} b)(a \minus{} b) \equal{} \minus{} 6(a \minus{} b)$

$ a \plus{} b \equal{} \minus{} 6$
(i)+(ii)
$ a^2 \plus{} b^2 \equal{} 6a \plus{} 6b \plus{} 10ab$
$ (a \plus{} b)^2 \equal{} 6(a \plus{} b) \plus{} 12ab$
$ ( \minus{} 6)^2 \equal{} 6*( \minus{} 6) \plus{} 12ab$
$ ab \equal{} 6$

Have I lost anything in the solution?
[/hide]


calculus
real analysis
real analysis theorems

Can i solve a limit problem using the standart part?
Using infinitesimal calculus
http://www.math.wisc.edu/~keisler/calc.html

Is infinitesimal calculus, standart part, etc.  formal?

i think it is good for some problems. but, in keisler book i did not find a transtition of infinitesimal calculus with limite definition, like a demonstration

Abraham Robinson's nonstandard analysis is a rigorous and formal system. However, it has not become a popular tool for teaching introductory level calculus because the development of nonstandard analysis requires some fairly sophisticated techniques and mathematicians seem loathe to teach the methods of nonstandard analysis to impressionable freshmen without knowing whether they'll ever actually see the rigorous development that justifies the witchery.

In my opinion, nonstandard analysis is slightly less unwieldy for introductory analysis, but it's not really amazingly better than using limit definitions. However, it is interesting that nonstandard techniques make the proofs of certain nontrivial theorems much simpler. The idea of nonstandard analysis is to hide some of the difficulties of all of its proofs in the construction (where, I might point out, the Axiom of Choice is required), so that proofs of certain results using nonstandard techniques turns out to be much simpler.

For example, Robinson himself drew attention to his theories when he used nonstandard techniques to settle some open problem about Banach spaces (I'm sure Wikipedia says what the problem was). Also, you can see in Robert Goldblatt's book "Lectures on the Hyperreals" (Springer GTM series) several applications of nonstandard analysis to classical results. In particular I recall that (the infinite version of) Ramsey's theorem is rendered trivial.

thanks for the great text.

I think that using infinitesimal calculus is better for physics problems. The constructuing of a lot of problems begin with infinitesimals, and using nonstandard analysis, like you said, is much more easy to proof and demonstrate some problems.

My first calculus book was that, from Keisler pdf.
And now I think is much more easy to work with inifitesimals calculus, like Leibniz did, that with de full of rigor definition of limit.

But I learnt the definition of limit. I understand much more easy the definition after learnt infinitesimals.

I think that the professor can work with the 2 definitions together.
I know that to define the Real numbers is too difficult, and hyperreal numbers too.

But beggining intuitivilly, like Leibniz, is the better way. Just after we can put more rigor.

I don't know what is a classroom of calculus, because I'm only 18 years. But I can imagine.

Sorry for the wrong english, I'm tired. I will go sleep.

Bye. and Thanks


conics
geometry
parabola
ratio
Asymptote
analytic geometry
projective geometry

[quote="grobber"]Let $P$ be a fixed point on a conic, and let $M,N$ be variable points on that same conic s.t. $PM\perp PN$. Show that $MN$ passes through a fixed point.[/quote]

If the conic is a circle the claim is trivial: the fixed point is the center of circle. Probably is possible to define a projective transformation which map a generic conis with its ortogonal chords in to a circle with its ortogonal chords. I don't have enough teoretical knoledge about Projective Geometry (PG) to state this.

I started to interest to the PG after seeing the following problem of which I have a very short and simple analitical solution but I don't have yet a systhetic solution. [color=red][Moderator edit: An analytic solution can be found at http://www.mathlinks.ro/Forum/viewtopic.php?t=1430 .][/color]

Problem.
Prove that the four point common to two parabolas which have orthogonnal axes are concyclic.

PS
I am sure that some guy of you can quickly solve this question.

[quote="sprmnt21"]
If the conic is a circle the claim is trivial: the fixed point is the center of circle. Probably is possible to define a projective transformation which map a generic conis with its ortogonal chords in to a circle with its ortogonal chords. I don't have enough teoretical knoledge about Projective Geometry (PG) to state this.

[/quote]

I don't know if this is true either. Here's what I did: the map $A\mapsto B$ is an involution on the conic. Take two pairs of corresponding points $(A_1,B_1),(A_2,B_2)$. An involution is determined by these two pairs, and since the map $X\mapsto Y$ where $XY$ passes through $A_1B_1\cap A_2B_2$ is an involution in which the image of $A_i$ is $B_i$, it means that our involution is precisely this one, so a line passing through a point and its image always passes through $A_1B_1\cap A_2B_2$.

[quote="sprmnt21"]Prove that the four point common to two parabolas which have orthogonnal axes are concyclic.[/quote]

I think I have a solution to this problem, but it is so adventurous and strange that it would be better if somebody more familiar with projective geomerty over $\mathbb{C}$ would check.

Let a and b be two parabolas with orthogonal axes, and X, Y, Z, W be the four points of intersection of these two parabolas. We have to prove that the points X, Y, Z, W lie on one circle.

We consider projective geometry in $\mathbb{C}^2$. All circles pass through two fixed points, the so-called circular points at infinity, which we call $I_1$ and $I_2$. These two points lie on the line at infinity. A conic is a circle if and only if it passes through these two points. Any two perpendicular lines meet the line at infinity at two harmonically conjugate points with respect to the segment $I_1I_2$.

A parabola is characterized among all conics by the property that it touches the line at infinity. Let our parabolas a and b touch the line at infinity at the points A and B. Then, these two points A and B are the points of intersection of the axes of these parabolas a and b with the line at infinity, and thus, since these axes are perpendicular, the two points A and B are harmonic conjugates with respect to the segment $I_1I_2$.

Now consider a projective transformation mapping the points X and Y to the circular points $I_1$ and $I_2$ at infinity. Since both parabolas a and b pass through the points X and Y, their images a' and b' under our projective transformation are conics passing through $I_1$ and $I_2$, i. e. circles. Since the points Z and W also lie on the parabolas a and b, the images Z' and W' of these points Z and W lie on the images a' and b' of these parabolas a and b. In other words, we have obtained two circles a' and b' intersecting each other at the two points Z' and W'.

Let g be the line which is the image of the line at infinity under our projective transformation. Since the parabolas a and b touch the line at infinity, their images a' and b' must therefore touch the line g. Since the points $I_1$, $I_2$, A and B lie on the line at infinity, their images $I_1^{\prime}$, $I_2^{\prime}$, A' and B' under our projective transformation must lie on the image of the line at infinity, i. e. on the line g. Since projective transformations preserve cross-ratios, we also see that the points A' and B' are harmonic conjugates with respect to the segment $I_1^{\prime}I_2^{\prime}$ (since the points A and B are harmonic conjugates with respect to the segment $I_1I_2$). In other words, the points $I_1^{\prime}$ and $I_2^{\prime}$ divide the segment A'B' harmonically.

What do we want to prove? We want to show that the points X, Y, Z, W lie on one circle. Speaking projectively, this is equivalent to showing that the points X, Y, Z, W lie on one conic through the circular points $I_1$ and $I_2$; in other words, we want to show that the points X, Y, Z, W, $I_1$ and $I_2$ lie on one conic. This is equivalent to the assertion that the images of these points, namely the points $I_1$, $I_2$, Z', W', $I_1^{\prime}$ and $I_2^{\prime}$, lie on one conic. In other words, it is enough to show that the points Z', W', $I_1^{\prime}$ and $I_2^{\prime}$ lie on one circle.

So it remains to prove the following elementary theorem: If a' and b' are two circles touching a line g at the points A' and B', and intersecting each other at two points Z' and W', and if $I_1^{\prime}$ and $I_2^{\prime}$ are two points on the line A'B' dividing the segment A'B' harmonically, then the points Z', W', $I_1^{\prime}$ and $I_2^{\prime}$ lie on one circle.

In fact, this is an easy application of inversion. Consider the circle k with diameter A'B'; this circle is orthogonal to the circles a' and b', since the circles a' and b' touch the line g at the points A' and B'. Hence, being the two common points of the circles a' and b', the points Z' and W' are mutually inverse with respect to the circle k. The points $I_1$ and $I_2$ are also mutually inverse with respect to k, since they divide the diameter A'B' of k harmonically. Two pairs of mutually inverse points with respect to a circle always lie on one circle; hence, the points Z', W', $I_1^{\prime}$ and $I_2^{\prime}$ lie on one circle. Proof complete.

Any comments or corrections?

Finally, to the original problem and to Sprmnt21's question:

[quote="sprmnt21"]If the conic is a circle the claim is trivial: the fixed point is the center of circle. Probably is possible to define a projective transformation which map a generic conis with its ortogonal chords in to a circle with its ortogonal chords.[/quote]

I don't know whether there exists such a transformation, but here is a solution of the original problem posted by Grobber:

Again, I use $\mathbb{C}^2$ and the two circular points $I_1$ and $I_2$.

Since the lines PM and PN are perpendicular to each other, they intersect the line at infinity at two points harmonically conjugate to each other with respect to the segment $I_1I_2$. Undertake a projective transformation that maps the conic to a circle; then, the problem takes the following form:

Let P be a fixed point on a circle k, and let $I_1$ and $I_2$ be two arbitrary points in the complex projective plane. Let M and N be variable points on k such that the lines PM and PN intersect the line $I_1I_2$ at two points harmonically conjugate to each other with respect to the segment $I_1I_2$. Show that the variable line MN passes through a fixed point.

In fact, if the lines $PI_1$ and $PI_2$ meet the circle k at the points $J_1$ and $J_2$ (apart from P, of course), then, from the fact that the lines PM and PN intersect the line $I_1I_2$ at two points harmonically conjugate to each other with respect to the segment $I_1I_2$, we can conclude that the cyclic quadrilateral $MJ_1NJ_2$ is harmonic. Thus, the line MN passes through the pole of the line $J_1J_2$ with respect to the circle k. This pole is, of course, a fixed point, and thus we are done.

Again, sorry if this was erroneous.

  Darij

[quote="grobber"]You know, I think the problem posted by sprmnt holds for any two conics having a pair of perpendicular axes. This seems to be true from some dynamic sketches I drew, so I'm not too sure.[/quote]

It seems to be true also after my quick calculation following the same way I did for the case of two parabolas. If you are interested in I can post some details.

I don't know how, but it seems that I deleted one of my post  :( .
Later, when I have more time to do it, I will post again the algebraic solution of the problem for the general case of two conics having orthogola axes.

[quote="grobber"]You know, I think the problem posted by sprmnt holds for any two conics having a pair of perpendicular axes.[/quote]

This is standing around here for a rather long time, so let's give a proof.

We have to show the following theorem:

[color=blue][b]Theorem 1.[/b] Let k and k' be two conics such that one axis of the conic k is perpendicular to one axis of the conic k'. Denote by A, B, C and D the four points of intersection of these two conics k and k'. Then, these four points A, B, C and D lie on one circle.[/color]

[i]Proof.[/i] We work in the complex projective plane again. Again, we denote by $I_1$ and $I_2$ the two circular points at infinity.

The conic k has not just one axis, it has two mutually perpendicular axes. Call these axes g and h. Similarly, the conic k' has two mutually perpendicular axes, which we denote by g' and h'. As we assumed, one axis of the conic k is perpendicular to one axis of the conic k'; let's say, WLOG, that $g\perp h^{\prime}$. Then, since $g^{\prime}\perp h^{\prime}$, we get g || g'. Thus, the lines g and g' have a common infinite point. Denote this point by G. Similarly, the lines h and h' have a common infinite point, which we denote by H. Since the lines g and h, being the axes of the conic k, are perpendicular to each other, their infinite points G and H divide the segment $I_1I_2$ harmonically (since this segment is actually the segment joining the two circular points at infinity $I_1$ and $I_2$).

Hence, in turn, the points $I_1$ and $I_2$ divide the segment GH harmonically. Let i be the involution on the line at infinity which maps every point to its harmonic conjugate with respect to the segment GH. Then, the points $I_1$ and $I_2$ are conjugates under the involution i.

Since we are working in the complex plane, every conic intersects every line at two points (which potentially can coincide). So let the conic k intersect the line at infinity at the points X and Y. Similarly, let the conic k' intersect the line at infinity at the points X' and Y'.

In the following, the [i]asymptotes[/i] of a conic will mean the lines joining the center of the conic (i. e. the point of intersection of its two axes) with the points where the conic intersects the line at infinity. The two asymptotes of a conic are symmetric to each other with respect to every of the two axes of the conic; thus, the two axes of the conic are the angle bisectors of the angle formed by the two asymptotes. Hence, the two axes of the conic harmonically divide the angle formed by the two asymptotes. Projecting this harmonic ratio onto the line at infinity (and remembering that the two asympotes meet the line at infinity at the two points where the conic itself intersects the line at infinity), we see that the two points where the axes of the conic meet the line at infinity harmonically divide the segment between the two points where the conic itself intersects the line at infinity. This, applied to the conic k, which intersects the line at infinity at the points X and Y and whose axes g and h intersect the line at infinity at the points G and H, yields that the points G and H divide the segment XY harmonically. Thus, in turn, the points X and Y divide the segment GH harmonically. In other words, the points X and Y are conjugates under the involution i. Similarly, the points X' and Y' are conjugates under the involution i.

So now we have the following picture: We have the line at infinity (which is, projectively speaken, just a line), and six points $I_1$, $I_2$, X, Y, X' and Y' on this line. We also have the points G and H on the line at infinity, but we won't need them anymore, we will only need the involution i on the line at infinity. We know that the points $I_1$ and $I_2$ are conjugates under the involution i, that the points X and Y are conjugates under the involution i, and that the points X' and Y' are conjugates under the involution i. Also, we have four points A, B, C and D somewhere in our projective plane, we have the conic k which passes through the points X, Y, A, B, C and D, and we have the conic k' which passes through the points X', Y', A, B, C and D. We must show that the points A, B, C and D lie on one circle. In terms of projective geometry, a circle is just a conic through the two circular points at infinity $I_1$ and $I_2$; hence, we must show that the points A, B, C and D lie on one conic through the two circular points at infinity $I_1$ and $I_2$. In other words, we must show that the points $I_1$, $I_2$, A, B, C and D lie on one conic.

Some people will already see the light at this point, but for those who don't, here is a final step:

Perform a projective transformation which maps the line CD to the line at infinity and the points C and D to the two circular points at infinity. Then, of course, the line $I_1I_2$ is not the line at infinity anymore, and the points $I_1$ and $I_2$ are just two simple points in the plane. But the configuration has become more familiar: We have a line (the former line at infinity), and six points $I_1$, $I_2$, X, Y, X' and Y' on this line. We have an involution i on this line. We know that the points $I_1$ and $I_2$ are conjugates under the involution i, that the points X and Y are conjugates under the involution i, and that the points X' and Y' are are conjugates under the involution i. Also, we have two points A and B somewhere in our plane, and we have the two circular points at infinity C and D. We have the conic k which passes through the points X, Y, A, B, C and D, and we have the conic k' which passes through the points X', Y', A, B, C and D. We must show that the points $I_1$, $I_2$, A, B, C and D lie on one conic.

But since the points C and D are the circular points at infinity, a conic through the points C and D is simply a circle. Hence, we can describe our situation as follows: We know that the points X, Y, A and B lie on one circle, and that the points X', Y', A and B lie on one circle, and we have to prove that the points $I_1$, $I_2$, A and B lie on one circle. But in fact, since the points $I_1$ and $I_2$ are conjugates under the involution i, the points X and Y are conjugates under the involution i, and the points X' and Y' are conjugates under the involution i, this directly follows from a well-known property of involutions (namely, from the construction of an involution using coaxal circles).

Proof complete.

Okay, I guess the proof using coordinates is easier...

  Darij

The result follows immediately from the fact that all parabolas are
      similar.

     Maj.  Pestich.


Since $ f(x) \equal{} \frac {x}{1 \minus{} x}$, its inverse function is given by $ x \equal{} \frac {f'(x)}{1 \minus{} f'(x)} \Longrightarrow f'(x) \equal{} \frac {x}{x \plus{} 1}$. We want $ a \equal{} f'(f'(\cdots f'(1)\cdots)) \equal{} f'_{1996}(1)$.

We note that if $ m$ is in the form $ \frac {x}{kx \plus{} 1}$, then $ f'(m) \equal{} \frac {\left(\frac {x}{kx \plus{} 1}\right)}{\left(\frac {x}{kx \plus{} 1}\right) \plus{} 1} \equal{} \frac {x}{(k \plus{} 1)x \plus{} 1}$, so by induction,
\[ f'_{1996}\left(\frac {1}{0 \cdot 1 \plus{} 1}\right) \equal{} \frac {1}{1996 \cdot 1 \plus{} 1} \equal{} \frac {1}{1997} \Rightarrow \mathrm{(B)}
\]

The above answer is the most straightforward. If this were on a contest, one might reason it out as follows:

[hide="Solution"]One may notice that the only numbers mapped to positive numbers by f are on the open interval (0,1), and that on this interval, f(x)>1.

We may thus solve f(x)=1 to get x=1/2, and conclude that the target term must be on the open interval (0,1/2).

Hence, the answer is either B or E.

Perform f once on 1/1997 to immediately see the pattern and conclude that B is the answer.[/hide]

[hide="Even more brute force"]

You can simply keep finding $ x_n$.

$ x_0 \equal{} a$
$ x_1 \equal{} \frac{a}{1\minus{}a}$
$ x_2 \equal{} \frac{a}{1\minus{}2a}$
$ x_3 \equal{} \frac{a}{1\minus{}3a}$

That is, $ x_n \equal{} \frac{a}{1\minus{}na}$. So, $ x_{1996} \equal{} \frac{a}{1\minus{}1996a}$. Since that value is 1:

$ 1 \equal{} \frac{a}{1 \minus{} 1996a}$
$ 1\minus{}1996a \equal{} a$
$ 1 \equal{} 1997a$
$ a \equal{} \frac{1}{1997}$ or B. [/hide]


Is there an elegant way to do this.

For example, with any given sides.

If you know 3 altitudes of a triangle.

yeah i figured that out, but i cudnt erase it


vector
linear algebra
linear algebra theorems

Hi all,

can someone prove that the criteria of a set to be a subspace of a vector space (closed under vector addition, and scalar multiplication) are independent?

In particular, I'm trying to answer a problem from my textbook:

Show that the criteria of vector space are independent by constructing a set in R2 which is:

i) closed under vector addition (even subtraction) but not under scalar multiplication

ii) closed under scalar multiplication, but not under vector addition (other than the two opposing quadrant)

Thanks

For (i), how about $\mathbb{Z}^{2},$ the set of integer lattice points in the plane? Closed under addition but not under scalar multiplication.

For (ii), try the intersection of more than one line through the origin. Say, $\{(x,y): x=2y \text{ or }2x=y\}.$

Great!! thanks Kent... you made the problem look so easy.. I was wondering how can I be so blind not to see such an obvious solution... once again thanks a lot!!


i need to solve the series without having to write out each term and adding them.
too many terms

sigma to n terms
k=1 (k is the first term)
equation is x^k (x to the power k.)

sorry i dont have the facility to illustrate the problem.

i would app some assistance please

$\displaystyle\sum_{k=1}^{n} x^k = x + x^2 + x^3 + ... + x^k = \displaystyle\frac{x^{k+1} - x}{x-1}$

Proof:

\begin{eqnarray*} S &=& x + x^2 + x^3 + ... + x^k \\ Sx &=& x^2 + x^3 + ... + x^{k+1} = S - x + x^{k+1} \\ Sx - S &=& S(x - 1) = x^{k+1} - x \\ S &=& \displaystyle\frac{x^{k+1} - x}{x-1} \end{eqnarray*}

Generally, people formulate the question starting from $k = 0$ instead, which makes the series look a little neater:

\begin{eqnarray*} S &=& 1 + x + x^2 + ... + x^k \\ Sx &=& x + x^2 + ... + x^{k+1} \\ S - Sx &=& S(1 - x) = 1 - x^{k+1} \\ S &=& \displaystyle\frac{1 - x^{k+1}}{1 - x} \end{eqnarray*}

This form is particularly useful for $x \leq 1$.


3C/2 where C is the capacitance of each capacitor

Answer is wrong. Perhaps you could tell me your method..... Come on ppl, its a simple prob so try.....

I'm pretty much discouraged if I do post physics problems here and I dont really want pardesi to solve these cause these are too elementary for him..  :rotfl:  :rotfl:

yes that's wrong :o sorry
it's $ \frac{5C}{3}$
and why is pardesi not answering such 'elementary' questions :huh:

Yes I think ur right. :D
and $ C\equal{}\frac{\epsilon _{0}.A}{d}$.

Yeah correct. Post your method.. Or atleast your idea..... just wanna see if there is a better method.

Let the positive plate be A and negative be B and the intermediate plate C.
Redraw the circuit and hence find the effective capacitance. :)


[hide="hint"]The sequence repeats itself...[/hide]

[hide="Answer"]$\cos 45+i^2\cos 225=\cos 45-\cos 225=2\cos 45$, $i\cos 135+i^3\cos 315=i\cos 135-i\cos 315=2i\cos 135$. The sequence repeats every four terms as it cycles around the unit circle, so our total will be $21\cdot \cos 45+20i\cos 135=\frac{\sqrt{2}}{2}(21-20i)\Rightarrow \boxed{D}$.[/hide]

[hide]

another way to do it will be to put it in summation form
$\sum_{k=0}^{40}i^k\cos{45+90n}$

then if you put the $i^k$ in trigonometric form, and do some simplification, you get

$\frac{\sqrt{2}}{2}\sum_{k=0}^{40}\cos^2{90k}-\sin^2{90k}$

which is easily evaluated by periodicity of the sine and cosine functions

Giving us $\frac{\sqrt{2}}{2}(21-20i)=\boxed{D}$

[/hide]

[hide]This sequence seems to go $\frac{\sqrt2}{2}, -\frac{\sqrt2}{2}i, \frac{\sqrt2}{2}, ...$. At the term starting with $i^{39}$, we have gone through 20 cycles of the repeating pattern, which equals $10\sqrt2-10\sqrt2 i$ and adding the last term gives us the answer D. [/hide]

[hide]$\cos {135^\circ}=-\sin {45^\circ}$
$\cos {45^\circ} - i\sin {45^\circ}=e^{i \cdot \pi/4}$

wait, nvm, i didn't notice that the power of $i$ was increasing every term. you could still use this way, but it's not as good as the other ways, i don't think. :(

[/hide]

Can you finish that solution? I don't see how you can use that.  :?

[size=150][b]I have 4000 posts  [/b][/size]:D


The sequence $ a_n$ is defined by $ a_0\equal{}4,a_1\equal{}1$ and the recurrence formula $ a_{n\plus{}1}\equal{}a_n\plus{}6a_{n\minus{}1}.$ The sequence $ b_n$ is given by: $ b_n\equal{}\displaystyle\sum_{k\equal{}0}^{n} \binom {n}{k} a_k.$ Find the coefficients $ \alpha,\beta$ so that $ b_n$ satisfies the recurrence formula $ b_{n\plus{}1}\equal{}\alpha b_n\plus{}\beta b_{n\minus{}1}$. Find the explicite form of $ b_n$.

characteristik equation for $ a_n$ is $ x^2\minus{}x\minus{}6\equal{}(x\minus{}3)(x\plus{}2)\equal{}0$. Therefore $ a_n\equal{}A3^n\plus{}B(\minus{}2)^n$. It give $ b_n\equal{}A(3\plus{}1)^n\plus{}B(\minus{}2\plus{}1)^n$ or characteristik equation for $ b_n$ is $ 0\equal{}(y\minus{}4)(y\plus{}1)\equal{}y^2\minus{}3y_4$. Therefore $ \alpha\equal{}3, \beta\equal{}4.$


bache ha barnameye geometers sketch pad ro az koja mitoonam begiram?(lotfan ye link az site morede nazar bedid.)

[quote="amir2"]bia jensen joon in site GSP: http://www.keypress.com/sketchpad/

inham 2ta link baraye keshidane sheklhaye hende3 begheir az avalli:
http://dynageo.de

http://www.mit.edu/~ibaran/kseg.html[/quote]
mer30. :D

[quote="parinaz.sa"]be nazare man baraye keshidane shekl haye hendesse CABRI kheili rahat tar bashe!!![url]http://www.cabri.com/v2/pages/en/index.php[/url][/quote]
az shoma ham mamnoon. :D


Every one of them can have 1 of 20 values, so 20*20*20=8000

[quote="myyellowducky82"][quote="GameBot"]Daniel, Chris, and Gerry are contestants on the TV game show "The Price is Too High, But I've Got A Coupon." They each get to spin the "big money" wheel, which has 20 values of money on it from as low as 5 cents and up to the grand prize of 1 dollar. If they each give the wheel a spin and get to keep the amount of money they land on, how many different possible outcomes are there?[/quote]
Hi,
I just did this problem and I need a clarification, so I am sorry for reviving an old topic. Wouldn't it be 6840 because there are $ 20\cdot19\cdot18$ outcomes?[/quote]

Actually it should be $ 20\cdot20\cdot20$ because it dosen't say that Daniel, Chris, and Gerry each get a different amount.

But even if they get the same amount of money that still counts as one possibility?

yes
it's possible they get the same amount of money, thus it should be a possibility


Column A

$ 1) Fe^{3\plus{}} \plus{} [Fe(CN)_6]^{4\minus{}}$
$ 2) Fe^{3\plus{}} \plus{} [Fe(CN)_6]^{3\minus{}}$
$ 3) Fe^{2\plus{}} \plus{} [Fe(CN)_6]^{4\minus{}}$
$ 4) Fe^{2\plus{}} \plus{} [Fe(CN)_6]^{3\minus{}}$

Column B

p) Prussian Blue ppt.
q) Brown colouration
r) White ppt. in inert atmosphere
s) No reaction

1-q
2-p
3-s
4-r

shudn't b wrong.... :o  :maybe:

 :rotfl:


geometry
3D geometry
prism
probability
expected value

Find the volume of the prism that is formed when you draw x+y+z=n in the first octant only.

(The volume is a positive finite number)

[hide]
Just look at the intersections with the axes.
\[V= \frac{1}{3}\left( \frac{1}{2}n^{2}\right) \cdot n = \boxed{ \frac{n^{3}}{6}}\] [/hide]

fun problem relating to this (although generalizing to higher dimensions)

Let $x_{i}, i \ge 1$ be random numbers in the interval $[0,1]$.  Let $N$ be the least integer such that $x_{1}+x_{2}+\cdots+x_{N}\ge 1$.  Find the expected value of $N$.

[hide="answer, no solution"]$e$[/hide]


Let $0 <a <b$ be two rational numbers. Let $M$ be a set of positive real numbers with the properties:

(i) $a \in M$ and $b \in M$;

(ii) if $x$ $\in M$ and $y \in M$, then $\sqrt{xy} \in M$.

Let $M^*$denote the set of all irrational numbers in $M$. prove that every $c,d$ such that $a <c <d<b$, $M^*$ contains an element $m$ with property $c<m<d$

Let $S : = \{ \mathrm{ln}(x) | x \in M\}$.
Then the condition gets:
$s=\mathrm{ln}(a) ,t=\mathrm{ln}(b) \in S$ and when $x,y \in S$ then $\frac{x+y}{2} \in S$.
Claim: for any $c,d$ with $s<c<d<t$ there is an $m \in S$ such that $c<m<d$ and $e^m$ is irrational.

Now $S$ is dense in $[a,b]$; this can be seen in various ways:
- $S$ contains all numbers of type $\frac{k}{2^i}a+\frac{2^i-k}{2^i}b$ (the binary fractions of finite length).
- or: take any $u$, then $u$ lies in $[a,\frac{a+b}{2}]$ or $[\frac{a+b}{2},b]$; now continue this till you get near enough to $u$.

So to show that there is such $m \in S$ with $c<m<d$, just take $m$ near enough to $\frac{c+d}{2}$ (distance $\frac{c-d}{2}$ is sufficient).
Now the last part is to get $e^m$ irrational. Well, take another one, lets call it $n$ with $c<m<n<d$ (this can be done since we can use what we already proved to find a number $n$ with $m<n<d$).
Since $x=e^m,y=e^n$ are not equal, there is at least one prime $p$ with $v_p(x) \neq v_p(y)$ (see the formulary for the definition of $v_p$). Now assume that we obtain a rational number with every step of the following, otherwise we would be done: $v_p(\sqrt{xy}) = \frac{v_p(x)+v_p(y)}{2}$ and using $\sqrt{xy}$ instead of $x$ and repeating this we get $v_p(x)$ as near to $v_p(y)$ as we want, but not equal. But in  a small neighbourhood of $v_p(y)$ there is just one integer, thus $v_p(x)$ will not be an integer, thus contradiction.

[hide=Solution]
Suppose that there existed \(x<y\) in \((a,b)\) such that for all irrational numbers \(m\) in \(M\), we had \(m < x < y\) or \(m > y > x\).

So take an irrational m such that \(m > y > x\) (We can use a similar argument if \(m < x < y\)). Then consider the sequence
\[\sqrt{mx}, \sqrt{x\sqrt{mx}}, \sqrt{x\sqrt{x\sqrt{mx}}}, \ldots\]
This is a strictly decreasing sequence which converges to \(x\). So there must be a number \(t\) in this sequence for which \(t < y\). Since \(t\) is in \(M\) and all sufficiently large terms of this sequence are irrational, we have found a contradiction.
[/hide]

Please PM if there's a mistake in my solution.


Hello Everyone. Can any one post problems from IMO shortlist  which were not selected due to being harder than IMO level . Or any other IMO type problem which required far more ingenuinity than routine IMO problems to solve it . Don't forget to comment about that problem.

                                                                     THANKS A LOT

I think that this years imo short list will apear next year.. (like always)
 :?

The following problem was proposed to the 1997 imo.
Find all graph such that any two vertex have exactly 1 common neighbour.
I think this problem wasnt in the shorlist because it was too difficult

This is known as the Friendship Theorem. The solution is the windmill graph.

The difficulty is to prove it combinatorially, this is a notorious example of a problem whose algebraic solution (using the adjacency matrix) is easier.  The result originally appeared in a paper co-authored by Erdos, but I don't know whether the proof was combinatorial.

Yes, it is absurd as an IMO proposal --- famous problem, easy solution if you've seen it (using non-IMO material), and a research-level problem using elementary tools.


limit
probability
ratio
advanced fields
advanced fields unsolved

suppose that $B_t$ is a brownian motion in $\mathbb{R}$. Prove that almost surely, $\lim \frac{B_t}{t}=0$.

How could we fail to have $\lim_{t\to0}\frac{B_t}t=0?$ It would have to be true that there is an $\epsilon>0$ and a sequence $t_n\to\infty$ such that $|B_{t_n}|\ge \epsilon t_n.$

But for any given $t,$ what is $P(|B_t|\ge \epsilon t)?$ $B_t$ is normally distributed with mean $0$ variance $t.$  Let $Z=\frac{B_t}{\sqrt{t}}.$ Then $Z$ is a standard normal distribution. We are asking for the event $|Z|\ge\frac{\epsilon t}{\sqrt{t}}=\epsilon\sqrt{t}.$

For large $t,$ this probability becomes very small. (In fact, we can say that it is asymptotic to $\frac2{\sqrt{2\pi}}\frac{e^{-\epsilon^2t/2}}{\epsilon\sqrt{t}}.$) Hence the probability of having this for a sequence of arbitrarily large $t$ must be zero.

We have actually shown that $P\left(\limsup_{t\to\infty}\left|\frac{B_t}t\right|\ge\epsilon\right)=0.$ Taking a countable union of such sets, we conclude that $P\left(\limsup_{t\to\infty}\left|\frac{B_t}t\right|>0\right)=0$ and hence $\frac{B_t}t$ tends to zero almost surely.

Here's an intuition: replace the Brownian motion by a simple random walk, generated by flipping a fair coin. (The actual Brownian motion is a carefully scaled limit of such a random walk.) Let $X_n$ be the position after $n$ flips. Saying that $\frac{X_n}n\to0$ is saying that in the long run, the ratio of heads to tails tends to $1: 1.$ The claim that that holds almost surely is the strong law of large numbers.

of course this is the idea, but I don't think the proof is very rigorous : this way you prove that for any fixed sequence $t_1,t_2,..,t_n,..$ that tends towards infinity, $\lim_{n \to \infty} \frac{B_{t_k}}{t_k} = 0$. I don't see how you conclude that $\lim_{t \to \infty} \frac{B_t}{t}=0$. One have to show that the brownian motion does not change too much between $t_n$ and $t_{n+1}$ isn't ?


inequalities logarithms function trigonometry calculus integration derivative

Let $ x>\minus{}1,$ prove that

$ (1\plus{}4x)\arctan{x}\leq (1\plus{}5x)\ln{(1\plus{}x)},$

with equality if and only if $ x\equal{}0.$

Can't we do this...?
Consider each to be two separate functions and then find the $ Range$ of each function which we get as:
$ (\minus{} \infty, 0]$ for $ (1 \plus{} 4x) \tan^{\minus{}1}x$ and $ [0, \infty)$ for $ (1 \plus{} 5x) \ln (1 \plus{} x)$ [b]defined for $ [\minus{}1, \infty)$[/b].

[quote="Rajiv"]Can't we do this...?
Consider each to be two separate functions and then find the $ Range$ of each function which we get as:
$ ( \minus{} \infty, 0]$ for $ (1 \plus{} 4x) \tan^{ \minus{} 1}x$ and $ [0, \infty)$ for $ (1 \plus{} 5x) \ln (1 \plus{} x)$ [b]defined for $ [ \minus{} 1, \infty)$[/b].[/quote]
Hello
[hide="Doubt"]
Range of first func is not $ ( \minus{} \infty,0]$.U can easily observe that it will give $ \plus{} ve value$  Kindly Check this.
green is $ (1 \plus{} 4x)tan^{\minus{} 1}x$
blue is$ (5x \plus{} 1)\ln(1 \plus{} x)$

[/hide]
Thank u.

I tried the brute force calculus approach.

Let $ f(x)\equal{}(1\plus{}5x)\ln(1\plus{}x)\minus{}(1\plus{}4x)\arctan x.$

Since $ f(0)\equal{}0,$ if we could show that $ f'(x)>0$ for $ x>0$ and $ f'(x)<0$ for $ x<0,$ that would suffice.

$ f'(x)\equal{}\frac{1\plus{}5x}{1\plus{}x}\plus{}5\ln(1\plus{}x)\minus{}\frac{1\plus{}4x}{1\plus{}x^2}\minus{}4\arctan x.$

We do have that $ f'(0)\equal{}0,$ but it's less obvious that $ f'$ is positive and negative where we need it to be. So we look at $ f''(x).$ If we could show that $ f''(x)>0$ everywhere, that would suffice.

$ f''(x)\equal{}\frac{9\minus{}5x}{(1\plus{}x)^2}\minus{}\frac{8\minus{}2x}{(1\plus{}x^2)^2}.$

There's a problem with trying to show that $ f''(x)>0:$ it's not true. $ f''(0)>0,$ but $ f''(1)<0.$ And $ f''(x)<0$ for all $ x>M$ for some $ M.$

This procedure isn't going well so far.

You can do much better with integration. Let's say $ x>0$. Then $ \arctan x\equal{}\int_0^x\frac{dt}{1\plus{}t^2}$, $ \log(1\plus{}x)\equal{}\int_0^x \frac{dt}{1\plus{}t}$ and, putting the terms without $ x$ on the left and with $ x$ on the right, we see that we need to prove that
\[ \int_0^x \frac{t(1\minus{}t)}{D(t)}\,dt\le x\int_0^x \frac{1\minus{}4t\plus{}5t^2}{D(t)}\,dt\]
where $ D(t)\equal{}(1\plus{}t)(1\plus{}t^2)$ (I do not care much what it is as long as it is increasing and positive). Now, the left hand side decreases for $ x\ge 1$ and the right hand side increases for all $ x>0$. Hence it is enough to assume that $ 0<x\le 1$. Now, $ \frac {1\minus{}t}{D(t)}$ decreases on $ [0,1]$. For any decreasing on $ [0,x]$ function $ g$, we have $ \int_0^x tg(t)\,dt\le\frac x2\int_0^x g(t)\,dt$. Thus, it suffices to show that
\[ \int_0^x \frac{(1\minus{}t)}{D(t)}\,dt\le 2\int_0^x \frac{1\minus{}4t\plus{}5t^2}{D(t)}\,dt\]
or, equivalently, that
\[ \int_0^x \frac{1\minus{}7t\plus{}10t^2}{D(t)}\,dt\ge 0\]
for all $ x\in [0,1]$
Since the denominator increases, it suffices to show that
\[ \int_0^x (1\minus{}7t\plus{}10t^2)\,dt\equal{}x\left(1\minus{}\frac 72x\plus{}\frac {10}3 x^2\right)\ge 0\]
for all $ x\in[0,1]$. But $ \frac{49}4\le 4\cdot \frac{10}3$ and we are done.

I suspect that the case $ x<0$ is even simpler, but I haven't tried it yet...

[quote="fedja"]I suspect that the case $ x < 0$ is even simpler, but I haven't tried it yet...[/quote]
My otherwise unpromising differentiation approach did get that. Assuming I did the algebra right (not a safe assumption), the common denominator version is

$ f''(x) \equal{} \frac {1 \minus{} 19x \plus{} 14x^2 \minus{} 8x^3 \plus{} 9x^4 \minus{} 5x^5}{(1 \plus{} x)^2(1 \plus{} x^2)^2}$

which is clearly positive for $ \minus{} 1 < x < 0.$

[quote="Kent Merryfield"]
$ f'(x) \equal{} \frac {1 \plus{} 5x}{1 \plus{} x} \plus{} 5\ln(1 \plus{} x) \minus{} \frac {1 \plus{} 4x}{1 \plus{} x^2} \minus{} 4\arctan x.$
...
$ f''(x) \equal{} \frac {9 \minus{} 5x}{(1 \plus{} x)^2} \minus{} \frac {8 \minus{} 2x}{(1 \plus{} x^2)^2}.$
[/quote]
Erm, isn't it $ \frac {9\, {\color{red}\plus{}}\, 5x}{(1 \plus{} x)^2}$? I mean $ \frac {10}{1\plus{}x}\minus{}\frac {1\plus{}5x}{(1\plus{}x)^2}\equal{}\frac {10\plus{}10x\minus{}1\minus{}5x}{(1\plus{}x)^2}$.
So the numerator is actually $ 1\minus{}9x\plus{}14x^2\plus{}12x^3\plus{}9x^4\plus{}5x^5$. This still dives below $ 0$ at, say, $ \frac 14$ and is non-negative for $ \minus{}1\le x\le 0$, so both Kent's remarks remain valid.


Find the coefficient of $ x^{21}$ in the expansion of $ (x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^5 \plus{} x^6)^6$ (preferably without complex numbers so stupid me can understand).

eh i think you mean $ x^{21}$.

EDIT: oh oops sry.

hmm wonder if theres an other way besides brute force...

TeX exponent mistake again??!??!?!  :wallbash: sigh

[quote="FantasyLover"]
so its making 21 by adding 6 numbers, which are larger than 0.[/quote]

not quite, FantasyLover, there's more... none of the numbers can be larger than 6.

Hopefully someone knows the answer, I got $ 4332x^{21}$.

TYPO EDITIED :(20 choose 5)-6(14 choose 5) + 15(8 choose 5) is the answer.

Principle of Inclusion Exclusion on the number of ways to get a+b+c+d+e+f = 21 with some of them > 6.

This happens to come out to 4332.

pythag, your solution is giving me $ \minus{}9170$ :(

[quote="pythag011"](14 choose 5)-6(14 choose 5) + 15(8 choose 5) is the answer.

Principle of Inclusion Exclusion on the number of ways to get a+b+c+d+e+f = 21 with some of them > 6.

This happens to come out to 4332.[/quote]

He has a slight typo, it should be (20 choose 5)-6(14 choose 5) + 15(8 choose 5)

Ah, I get it now.
[hide]
$ \binom{20}{5} \equal{}$  number of ways to sum to 21 with 6 positive integers

$ 6\binom{14}{5} \equal{} \binom{6}{1}\binom{14}{5}  \equal{}$ number of ways to sum to 15 with 6 positive integers $ \equal{}$ number of ways to sum to 21 with 6 positive integers with at least one greater than 6

$ 15\binom{8}{5} \equal{} \binom{6}{2}\binom{8}{5}  \equal{}$ number of ways to sum to 9 with 6 positive integers $ \equal{}$ number of ways to sum to 21 with 6 positive integers with two greater than 6[/hide]

Thanks, everyone  :)


function
real analysis
real analysis unsolved

Find a net (a sequence indexed by an upper directed set, ie, $ \mathbb{R}$) in $ E : \equal{} \{ f : \mathbb{R} \in \mathbb{R} : f(x) \equal{} 2 \text{ except finitely often} \}$ with limit the zero function $ z$.

What is a good directed set to use?  One connected to the problem?

I have the feeling you can do it with \mathb{R} but I don't really know much...

No.  The hint for a good directed set is in that phrase "except finitely often"

While your directed set might have the same cardinality as $ \mathbb{R}$, it shouldn't have the same order structure. A convergent net indexed by $ \mathbb{R}$ can have convergent sequences extracted from it.

The appropriate sense of limit here is "pointwise"; this relates to the product topology. We can't find a sequence that converges to zero in this case, just because all but finitely many points give the value $ 2$ for [i]all[/i] elements of the sequence.

So, what structure should we use for the index set? Try a collection of sets, ordered by inclusion. That branches enough to be interesting.

Okay yeah thanks. It took me time, but I figured it out with all those hints... By taking the finite sets of $ \mathbb{R}$ ordered by containment, we have $ \mathbb{R}$ at the end...

Thanks !


integration
trigonometry
calculus
calculus computations

[quote="bgbgbgbg"]integrate?????????????????????????? alpha between 0 and pi/3[/quote]
$ \int_0^{2\alpha} \frac {\sqrt [4]{\sin(3\alpha \minus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}} dx \equal{}$

$ \int_0^{\alpha} \frac {\sqrt [4]{\sin(3\alpha \minus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}} dx \plus{} \int_{\alpha}^{2\alpha} \frac {\sqrt [4]{\sin(3\alpha \minus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}} dx$

But $ \int_{\alpha}^{2\alpha} \frac {\sqrt [4]{\sin(3\alpha \minus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}} dx \equal{} \int_0^{\alpha} \frac {\sqrt [4]{\sin(\alpha \plus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}} dx$

So $ \int_0^{2\alpha} \frac {\sqrt [4]{\sin(3\alpha \minus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}} dx \equal{}$

$ \int_0^{\alpha} \frac {\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}} dx \equal{}$

$ \int_0^{\alpha} 1 dx \equal{} \alpha$

Set $ I \equal{} \int_0^{2\alpha} \frac {\sqrt [4]{\sin(3\alpha \minus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}} dx,$

$ J \equal{} \int_0^{2\alpha} \frac {\sqrt [4]{\sin(\alpha \plus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}} dx$.

Let $ x \equal{} 2\alpha \minus{} t$, we have $ I \equal{} J$ and $ I \plus{} J \equal{} 2\alpha$, yielding $ I \equal{} J \equal{} \alpha$.

Let  $ \ I \equal{} \int_0^{2\alpha} \frac {\sqrt [4]{\sin(3\alpha \minus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}} dx$

$ \equal{}\int_0^{\alpha} \frac {\sqrt [4]{\sin(3\alpha \minus{} x)}}{\sqrt [4]{\sin(3\alpha \minus{} x)} \plus{} \sqrt [4]{\sin(\alpha \plus{} x)}}\plus{}\frac {\sqrt [4]{\sin(3\alpha \minus{}(2\alpha \minus{} x))}}{\sqrt [4]{\sin(3\alpha \minus{}(2\alpha\minus{} x))} \plus{} \sqrt [4]{\sin(\alpha \plus{}(2\alpha \minus{}x))}} dx$

$ \implies \boxed{I\equal{} \int_0^{\alpha} 1\ dx\equal{}\alpha}$


Triangle $ABC$ is isosceles with base $AC$. Points $P$ and $Q$ are respectively in $CB$ and $AB$ and such that $AC=AP=PQ=QB$. The number of degrees in angle $B$ is:
(A) $25\frac{5}{7}$
(B) $26\frac{1}{3}$
(C) $30$
(D) $40$
(E) not determined by the information given

[hide]$\angle C = \theta$

$\angle B = \pi - 2\theta$

The original isosceles triangle is divided into three isoceles triangles.  Working up with substitution, you eventually get:

$2\pi - 3\theta + 2(\pi - 2\theta) = \pi \rightarrow 7\theta = 3\pi \rightarrow \theta = \frac{3\pi}{7}$

$\angle B = \pi - 2\theta = \frac{\pi}{7}$

Which, in degrees, is (A) $25\frac{5}{7}$.[/hide]


abstract algebra
geometry
advanced fields
advanced fields unsolved

a)If $ Y$ is any subset of a topological space $ X$, then $ \dim Y\leq \dim X$.
b)If $ X$ is a topological space which is covered by a family of open subsets $ \{U_i\}$, then $ \dim X\equal{}\sup \dim U_i$.
c)Given an example of a topological space $ X$ and a dense open subset $ U$ with $ \dim U<\dim X$.
d)If $ Y$ is a closed subset of an irreducible finite-dimensional topological space $ X$, and if $ \dim Y\equal{}\dim X$, then $ Y\equal{}X$.
e)Given an example of a noetherian topological space of infinite dimension.

I would rather suspect this problem is about the "combinatorial" dimension used in algebraic geometry, defined as the supremum of the lengths of chains $ Z_0\subset Z_1\subset\ldots\subset Z_n$ with each $ Z_i$ irreducible and closed. In particular, $ \dim\varnothing\equal{}\minus{}\infty$ and $ \dim\mathbb R^n\equal{}0$. owk

[quote="barra"]I am unfamiliar with dimension in topological space. Could you define this term for me, please.
Thanks[/quote]
See in ''Robin Hartshorne, Algebraic Geometry, GTM 52, page 5'' .  :wink:

Can you define the dimension of topological space ? I think there are some interesting problems around this question!


Real numbers $a$ , $b$ , $c$ , $d$ , $e$ , $f$ such that $a \leq b \leq c$ and $d \leq e \leq f$ for which $a + b + c = d + e + f$ and $ab + bc + ca = de + ef + fa$.  If $a \leq d$, prove that $c \leq f$.

let $ P(x) \equal{} (x \minus{} a)(x \minus{} b)(x \minus{} c)$ and $ Q(x) \equal{} (x \minus{} d)(x \minus{} e)(x \minus{} f)$, then from condition we have $ P(x) \minus{} Q(x) \equal{} def \minus{} abc$ and from $ a \le d \le e \le f$ we get $ Q(a) \equal{} (a \minus{} d)(a \minus{} e)(a \minus{} f) \le 0$ so $ def \minus{} abc \equal{} P(a) \minus{} Q(a) \equal{} \minus{} Q(a) \ge 0 (*)$.

and $ P(f) \equal{} (f \minus{} a)(f \minus{} b)(f \minus{} c)$ and since $ a \le d \le f$ we have $ f \minus{} a \ge 0$ and also $ P(f) \equal{} P(f) \minus{} Q(f) \equal{} def \minus{} abc \ge 0$ (from $ (*)$),

so we must have $ (f \minus{} b)(f \minus{} c) \ge 0$ so $ f \le b$ or $ f \ge c$ .

$ \star$ if $ f \ge c$ then we are done.

$ \star$ if $ f \le b$ then $ a \le d \le e \le f \le b \le c$ so $ (c \minus{} d)(c \minus{} e)(c \minus{} f) \ge 0$ but $ \minus{} Q(c) \equal{} P(c) \minus{} Q(c) \equal{} def \minus{} abc \ge 0$ so $ Q (c) \le 0$ therefore  $ (c \minus{} d)(c \minus{} e)(c \minus{} f) \equal{} 0$ so from this and the fact that $ d \le e \le f \le b \le c$ we must have $ b \equal{} c \equal{} f$ and the result: $ f\ge c$ is true.


algebra
polynomial
ratio
trigonometry
invariant
induction
combinatorics unsolved

For two real numbers $ a$, $ b$, with $ ab\neq 1$, define the $ \ast$ operation by
     \[ a\ast b=\frac{a+b-2ab}{1-ab}.\] Start with a list of $ n\geq 2$ real numbers whose entries $ x$ all satisfy $ 0<x<1$. Select any two numbers $ a$ and $ b$ in the list; remove them and put the number $ a\ast b$ at the end of the list, thereby reducing its length by one. Repeat this procedure until a single number remains.

$ a.$ Prove that this single number is the same regardless of the choice of pair at each stage.
$ b.$ Suppose that the condition on the numbers $ x$ is weakened to $ 0<x\leq 1$. What happens if the list contains exactly one $ 1$?

[hide="hint"]Notice that $ \ast$ is associative, $ (a\ast b)\ast c =a\ast( b\ast c)$ [/hide]

In fact, you need both that it's associative [hide="calculation"]\begin{eqnarray*}(a * b)*c &=& \frac{a+b-2ab}{1-ab}* c \\ &= &\frac{\frac{a+b-2ab}{1-ab}+c-2c\frac{a+b-2ab}{1-ab}}{1-c\frac{a+b-2ab}{1-ab}}\\ & = & \frac{a+b+c-2ab-2ac-2bc+3abc}{1-ab-ac-bc+2abc}\\ & = & \frac{a+\frac{b+c-2bc}{1-bc}-2a\frac{b+c-2bc}{1-bc}}{1-a\frac{b+c-2bc}{1-bc}}\\ & = & a* \frac{b+c-2bc}{1-bc}\\ & = & a*(b*c) \end{eqnarray*}[/hide] and that it's commutative (this is obvious) and it follows that any chain can be rearranged into any other, so all give the same value.  For part b, we also note that if $ a \neq 1$ then $ a * 1 = 1$, so the result is 1.

An algebraic interpretation of the operation:
Work in some commutative ring containing a copy of $ \mathbb{R}$, and let $ N$ be a nonzero element with $ N^{2}=0$ (for example, $ A$ is the ring of polynomials mod $ x^{2}$ and $ N$ is $ x$). If $ \alpha=1+a(-1+N)$ and $ \beta=1+b(-1+N)$, then $ \alpha\cdot\beta=(1-ab)+(a+b-2ab)(-1+N)=(1-ab)(1+a*b(-1+N))$.

Since $ a*b$ gives the coefficient ratio of the product of two elements with coefficient ratios $ a$ and $ b$ and the product is associative and commutative, $ *$ is associative and commutative wherever it's defined. Coefficient ratios strictly between $ -1$ and $ 1$ correspond to ring elements of the form $ a+bN$ with $ a$ positive; any products of these must still be of the same form. A coefficient ratio of $ 1$ is an element of the form $ cN$, and those absorb others under multiplication. A second such entry would collapse everything into being undefined, as $ N^{2}=0$, and zero has no coefficient ratio.

Nice one, Jmerry; but how did you come up with that?  :maybe:

For me it was enough to observe that $ \left(1-\frac{1}{x}\right)*\left(1-\frac{1}{y}\right)=1-\frac{1}{x+y-1}$. This, however, has the downside that for part [b]b)[/b] you either need to work in $ \overline{\mathbb{R}}=\mathbb{R}\cup\left\{\infty\right\}$ or a separate argument...

  Darij

It reminded me of the angle addition identity for $ \tan$, and the relation of that identity to complex multiplication. As I was working through it, I realized that the "absorbing" value had to come from a nilpotent.
For the specific  calculations, I wanted $ (1+ak)(1+bk)=(1-ab)+(a+b-2ab)k$; that led to $ k^{2}+2k+1=0$. Setting $ k=-1$ would cause all sorts of problems, so we use $ k=-1+N$ instead.

OK can someone see if this solution makes sense? I thought of it relatively quickly and its very different than above answers (make sure i did not miss anything)
----------------
Obviously, $a*b=b*a$
Lemma: When $n=3$, then any permutation over $a, b, c$ of $a*b*c$ are equivalent. Note: this is the $n=3$ case for part a
Proof: let $g(a,b,c)=(a*b)*c$. Then
$g(a,b,c)=\frac{a+b-2ab}{1-ab} * c
\\ = \frac{ \frac{a+b-2ab}{1-ab} + c - 2c(\frac{a+b-2ab}{1-ab})}{1-c(\frac{a+b-2ab}{1-ab})} (\frac{1-ab}{1-ab})
\\ = \frac{ a+b-2ab+c-abc-2c(a+b-2ab)}{1-ab-c(a+b-2ab)}
\\ = \frac{a+b-2ab+c-abc-2ac-2bc+4abc}{1-ab-ac-bc+2abc}
\\ = \frac{a+b+c-2ab-2bc-2ac+3abc}{1-ab-ac-bc+2abc}$
Which is obviously symmetric over $a,b,c \blacksquare$
Now we shall prove part a) by induction. The process in the problem is equivalent to producing some arrangement of the integers $x_1, x_2, ..., x_n$ with the $*$ operation between each pair of integers. We shall prove that, given an arbitrary arrangement of the integers $x_1, x_2, ..., x_n$ with a $*$ operation between each pair of integers, we can always reorder them to be in the order $x_1*x_2*...*x_n$. This would imply that all orderings result in the same value, proving the problem.
$n=2$ is trivial because $a*b=b*a$, and our proved $n=3$, which is stating that $a*b*c=a*c*b=b*a*c=b*c*a=c*a*b=c*b*a$. Now for the inductive step:

Given we can reorder an arbitrary arrangement $x_1, x_2, ..., x_n$ to be in the order $x_1*x_2*...*x_n$, we shall prove we can do the same with a list of $n+1$ integers.
We start with an arrangement of $x_1, x_2, x_3, ..., x_n, x_{n+1}$.
If the last element in our arrangement is $x_{n+1}$, then the first $n$ elements are some arrangement of $x_1, x_2, ..., x_{n-1}, x_n$, and we can rearrange the first $n$ elements to be in the order $x_1*x_2*...*x_n$. Then our entire value becomes $x_1*x_2*...*x_n*x_{n+1}$, and we are done.
Else, denote our last element to be $x_b$.
if $b=1$, then denote our second to last element $x_c$. We can reorder the final two elements such that they switch from $...*x_c*x_b$ to $...*x_b*x_c$. reassign b=c, so now our last element is $x_b, b \not=1$.
then the first $n$ elements are some arrangement of $x_1,..., x_{n+1}$, for all index $i, 1 \le i \le n+1, i \not= b$. We can rearrange the first $n$ elements to be in ascending order, i. e.  in the order $x_1*...*x_{n+1}$ for all index $i \not=b$. Our entire value is then $x_1*...*x_{n+1}*x_b$. We then rearrange the last $n$ elements to be in ascending order, such that our entire value is then $x_1*x_2*...*x_{n+1} \blacksquare$

[quote=mathbuzz]We will prove that for any number of variables  (say n>=2), the number obtained at last in this process is a symmetric expression in those variables.
we use induction (induction on n)
P(2),P(3) are true (easy to check)
we assume that P(k),P(k-1),...,P(2) is true [I.H.]
P(k+1)--
there are variables a 1,...a k, ,a (k+1)
now we choose any 2 variables a i and a j
then in the next stage , we have k variables a i*a j and the remaining k-1 variables of the previous stage ,i.e. k variables in total.
now , according to our I.H. , the number obtained at last is symmetric in these k variables of this stage ( as stated in the preceding line)
now , a i*a j is also symmetric in a i and a j
so, we can conclude that the expression is symmetric in a 1,..., a (k+1)
then it is easy to say that the last number is independent of the choice of variables in each stage :D[/quote]

If I understand your solution correctly, I think it does not work.

What you are effectively claiming is the following: given a function f(x1,...xn) from R^n to R symmetric in x1,...,xn  and a function g(p,q) from R^2 to R symmetric in p,q, then h(x1,...,xn-1,p,q)=f(x1,...xn-1, g(p,q)) is symmetric in x1,..xn,p,q.

But this is not true, for example take f=x+y+z, g=1/(p+q). Then h=x+y+1/(p+q). Clearly h(x,y,p,q) is identical to h(x,p,y,q).


algebra
polynomial
Putnam
quadratics
algorithm
superior algebra
superior algebra solved

K is a finite field such that the polynomial X^2-5 is irreducible in K[X].  Prove that :

(a) 1+1 != 0
(b) for every a in K, X^5 + a is reducible over K[X]

<i> Romanian Regional Olympiad 2003, 12th grade </i>

Does anyone have an ellegant solution for (b) ?

Z3 is a finite field. X 2-5=X 2 +1 is irreducible in Z3[X] because it has no roots.

We don't have 1+1=0. Maybe 1+1 !=0 means smth else.

1+1 != 0 means actually 1+1 different from 0 (in C++ language notation :D).

it is an easy task, as if 1+1=0 => x^2-5 = x^2+1=(x+1)^2.

the hard part is point b). the official solution is a bit unexpected, however Ana Caraiani (which will probably be a fellow putnam this year - based on the rumours) found that solution, and she was the only one to solve the problem.

[quote="Valentin Vornicu"]1+1 != 0 means actually 1+1 different from 0 (in C++ language notation :D).
[/quote]

I like Turbo Pascal. Even though is less powerful, it is easier to implement in.
In there, the notation for different is "<>" which I saw used on this site too.

Here is a the sketch of a "short" solution. Let p be the characteristic of K.
First, since X^2-5 is irreducible in K[X], 5 is not a quadratic residue
modulo p. By Gauss's law of quadratic reciprocity, it follows that p is
not a quadratic residue modulo 5, and this implies that 5 does not divide p^2-1.

By consequent, there exist integers m and n such that 1 = 5m - (p^2-1)n.
Let us show that x^5 = y^5 (with x,y non-zero in K) implies x=y. Indeed
x^(5m) = y^(5n), so x^(1 + (p^2-1)n) = y^(1 + (p^2-1)n).

But x^((p^2-1)n) = [x^(p-1)]^[(p+1)n] = 1 in K, so x=y.

We have proved that the map
        :K -> K
         x -> x^5
is injective. Since K is finite, this map is also surjective, so
every element a in K is the 5th power of an element b in K and this clearly
implies that X^5-a is reducible (it has X-b as factor).

indeed that is the official solution. well an adaptation on it, but based on the same idea. when I first thought about the problem I didn't expect to factor by an X-b, but rather on a X^2+bX+c. Only after some time of thinking (this was not the only problem on the 3hour paper! - there were other hard problems too!) I have managed to think that it might be possible to try finding a root, and then things went rather naturally. Good for me I was correcting the paper, and not taking it :D:D

For a, 1+1=0 implies that X^2-5 = (X-1)(X+1) which is reducible, a contradiction. Therefore gcd(char(K),2)=1 (*).

It's well known and easy to prove that if a field K is finite and of characteristic p then it contains q=p^n elements for some n. Moreover, x^(q-1) = 1 for all x in K\{0}. Using these notations for our field K, we notice that the polynomial X^(q-1) - 1 has all elements of K\{0} as roots. Since X^2-5 is irreducible over K we have in K[X] that gcd(X^2-5,X^(q-1)-1) = 1 (**).

Let us prove that whenever n is even, (**) is contradicted. If n is even, then (q-1)/2 = (p-1) [(p^(n-1) + p^(n-2) + ... + p + 1)/2], which is a product of two integers, because from (*) we get p!=2. Also we know that X^2-5 divides X^(q-1) - 5^(q-1)/2 , but since p-1 divides (q-1)/2 we have that 5^(q-1)/2 = 1 (mod p), and since char(K)=p we have in fact that 5^(q-1)/2 = 1 in K. Therefore if n is even, X^2-5 is not irreducible over K.

First we assume that p is not 5. Now, let n be even. 5 is not a quadratic residue mod p, or otherwise there exists a in {1,2,..,p-1} such that a^2=5 and therefore X^2-5 = (X-a)(X+a). By the quadratic reciprocity law, (p/5) (5/p) = 1, so 5 is not a quadratic residue mod p iff p is nod a quadratic residue mod 5. But if r is the residue of p mod 5, (p/5)=(r/5) , which is -1 iff r is 2 or 3. Ok, so in fact we have gcd(p^2-1,5)=1 and 5 | p^2 + 1.

Assume that for u!=v in K, u^5=v^5. Then uv^(-1) has order 5 in K, so 5 divides p^n - 1. If n = 4k+1, then since 5 | p^4-1, we have 5|p^4k - 1, so 5|p^(4k+1)-p^4k, whence 5 divides p-1, a contradiction. If n=4k+3, 5|p^(4k+4)-1 and 5|p^(4k+3)-1 implies again 5|p-1 a contradiction.  Therefore f(x)=x^5 , f:K\{0} --> K\{0} is injective. Since K is finite, it's bijective and in particular, there exists b in K such that a=b^5. Therefore X^5 + a = X^5 + b^5 = (X+b)(X^4 - bX^3 + b^2 X^2 - b^3 X + b^4).

All we have to do now is study the case p=5. But this means that X^2-5 = X^2 which is not irreducible, a contradiction.

[quote="iandrei"]It's well known and easy to prove that if a field K is finite and of characteristic p then it contains q=p^n elements for some n.[/quote]

I know the fact, what about the easy proof?

Let a\inK. Then a has order p in (K,+) (or order 1, if a=0), so the only prime which divides |K| is p (from Cauchy's thm).

It is pretty short, isn't it :)?

well if it's finite, then if it would not have the number of elements of the form pn then there would exist two different primes p,q such that pq divides the number of elements.

From Cauchy's theorem there exist an element of order p and an element of order q in the additive group of the field. Using Euclid's algorithm we find that (p,q).1 = 0 => 1=0 contradiction. :D

That's what I meant, but I guess I didn't make it clear enough. Thx, Valentin! :)

nope it wasn't that. We just posted in the same time, I saw your post after I posted mine ;) :P


Is it true that for $a_n,b_n$ 2 strictly increasing sequences of positive integers there are only finitely many couples $(n,m)$ such that $2^{a_n}-1,3^{b_m}-1$ have exactly the same odd prime factors?

Harazi ,i have the genelization problem from your problem and i remamber i have posted it in the forum for a long time.The genelization is :find all a,b,n,m \in Z^{m} such that $a^{n}-b^{n},a^{m}-b^{m}$ have the same prime divisors :D  :D .I found the solution and this result make ma unsatisfactory because it look so unnatural.
HInt:you can find all m,n such that 2^{n}-1,3^{m}-1 have the same prime divisors by the method in IMOSL  :) >good luck


function MATHCOUNTS algorithm probability LaTeX counting distinguishability

# 24, State 2001 Sprint

The number 4 can be written as a sum of one or more natural numbers in exactly five ways: 4, 3+1, 2+1+1, 2+2, and 1+1+1+1; and so 4 is said to have 5 [i]partitions[/i]. What is the number of partitions for the number 7?

[hide]15 (Partitions)[/hide]

My strategy was to look for a pattern in 1-4, but I failed  :rotfl:

The partition function doesn't have an obvious pattern.  It's actually a very complicated function.

Oh, great... The few problems I try to avoid brute force turn out to be ridiculously hard doing it the *Smart* way?

It's not hard to list all the combinations; there's not many of them.

Right, I know I can list the combinations but it's MathCounts, so I was thinking there was some way to do it that was much easier and required less time.

Just out of curiosity--what is the function?

it isn't mathcounts level, so it wouldnt belong in the forum anyways...
for this kind of problem, your best chance is to list them out with a pattern like how many with 7 1's, 5 1's, 4 1's, 3 1's etc. or something like that

The algorithm I used is partitions with no 1s, 1 1, 2 1s, 3 1s, etc.

Chapter 14 in Intermediate Counting and Probability talks about it. It's a really long function. Dont know how to write it in latex though

Erm, but you would have to divide by repeats, and there would be different numbers of repeats depending on the number of repeated digits...
So it would be easier just to list them out.

[quote="Yongyi781"]Couldn't we use balls and urns for this? Since the maximum number of elements in a partition of 7 is 7, anyways, we could have 7 indistinguishable urns and 7 indistinguishable balls?

Explanation:
A partition of $ \{5, 1, 1\}$ is the same as $ \{5, 1, 1, 0, 0, 0, 0\}$ and a partition of $ \{7\}$ is the same as $ \{7, 0, 0, 0, 0, 0, 0\}$, where the rest of the elements up to the seventh element are filled with zeroes.[/quote]

You can use generating functions for partitions, but with balls and urns, you have to be careful.  This was a sprint round question, so just list them out

just a thought: how many topics have the name partitions and are posted in the MC forum. Furthermore, how many of the topics contain the question "how many partitions of 7 are there?"
 :rotfl:  :rotfl:

i havent seen partitions on any test i have taken except that one.(2003? i cant remeber the year). even if you do see it, which i doubt, listing should not be hard. they probably wont go past 7 anyway, because then listing is too hard, and theres really no function for it. it should be a problem if you see it.(i dont think at least)

[quote="Yongyi781"]Couldn't we use balls and urns for this?[/quote]

Balls and urns has a simple formula.  Like I said, the partition function does not.  See [url=http://mathworld.wolfram.com/PartitionFunctionP.html]the Mathworld article[/url] if you really want to know.

Balls and urns does [b]not[/b] have a simple formula where both the balls [b]and[/b] the urns are indistinguishable. As far as I know, they are the same function (Stirling Number of the Second Kind or Catalan Number? I can't remember.).

The Stirling Numbers of the Second Kind and the Catalan Numbers are two different sequences, neither of which are the partition function.  You are correct about the indistinguishable balls and indistinguishable urns, though.  "Balls and urns" without qualification usually refers (in my experience) to distinguishable urns.


Just wondering how people prepared for math competitions such as the AMC, etc.

I voted for "do old tests", because that's what I do most of to prepare, but I also read AoPS books.  :)

I'm too old to compete, of course, but I learn math by reading good expository books about mathematics. I don't have enough peace and quiet in my day to work a lot of math problems, so I gain conceptual knowledge faster than I improve my problem-solving ability.

Studying stuff like AoPS is good, but the best thing to do is to do old tests.  That kind of practice is well geared for the contest, fun, and still teaches you stuff.  What more could you ask for? :)

[quote="EFuzzy"]I'm not very experienced, but here's my philosophy. There's no point in doing old tests until you know all the material and all the theoy behind the tests. So read lots of old books first, and then do some old tests to iron out stupid mistates and get used to the test format. There are only so many old AMC10's, might as well save them  :) .[/quote]-Erm, yes.  The only reason not to pound out theory--and advance your learning in general--is to get ready for a test quickly (against a deadline...).  That said, I chose "tests" mostly because I need to know WHAT theory is on the tests, and since my experience with mathematics is far greater than that with contests, practice problems seem the way to go.  Sometimes, that means I just sit there and stare at a method until I can work something out, then carefully glance through other people's solutions.  Still, the theory is the best route--and will certainly save you from dumb mistakes or an over-reliance on calculators too.  Excepting this, calculator programs, well understood before the test or even engineered quickly can prove an exceptional tool.  Case in point, Zack Abel with the following:

[quote="zabelman"]Writing a program for this problem is very fun to do! So in the last 10 minutes I decided to write one. I came up with the following within a few mins (TI-83 Plus, Basic syntax):

[code]0->K
For(A,2,1000)
Disp A
If fPart(A/2) $\neq$ 0 and fPart(A/3) $\neq$ 0 and fPart(A/5) $\neq$ 0
1+k->k
End
Disp K[/code]

(I put in the line "Disp A" so I could watch it count up to 1000: I *hate* waiting for a program to finish running without knowing if it is going at all.) Anyway, it took about 2 minutes to run, and then it told me 265. So after subtracting off the 165 leftover primes, thats 100. Yay, it worked![/quote]Beautiful.


Jordan buys pens for 3 dollars each and books for 5 dollars each, totaling exactly 2008 dollars.
What is the largest number of items (pens plus books) he could have bought?

We obviously want to maximize the number of pens.  However, we clearly cannot have ONLY pens, because then we wouldn't be able to reach 2008 dollars.  We can, however, add $ 2$ dollars to any total we can reach (substitute a pen for a book).  With only pens, we can reach $ 2010,2007,2004,\ldots$.  Using $ 2004$, we substitute two pens for books, to reach $ 2008$.  We have used $ \frac{2004}{3}\equal{}\boxed{668}$ items.


function
modular arithmetic
number theory unsolved
number theory

determine all pairs of positive integers $ (x,y)$ satisfying the equation
$ 2^x\equal{}3^y\plus{}5$

It's equivalent with :

$ \frac{2}{3}\equal{}\frac{3^{y\minus{}1}\plus{}1}{2^{x\minus{}1}\minus{}1}$

$ \Rightarrow (x,y)\equal{}(3,1) or (x,y)\equal{}(5,3)$

Sorry, but I do not understand how from $ \frac{2}{3}\equal{}\frac{3^{y\minus{}1}\plus{}1}{2^{x\minus{}1}\minus{}1}$, necessarily follows that the [b]only[/b] solutions are $ (3,1)$ or $ (5,3)$. Can you explain?

[quote="alex2008"]It's equivalent with :

$ \frac {2}{3} \equal{} \frac {3^{y \minus{} 1} \plus{} 1}{2^{x \minus{} 1} \minus{} 1}$

$ \Rightarrow (x,y) \equal{} (3,1) or (x,y) \equal{} (5,3)$[/quote]I think you cannot solve it like that because $ \frac {3^{y \minus{} 1} \plus{} 1}{2^{x \minus{} 1} \minus{} 1}$ is function of two variables( if there was one variable you could easily see that the function is increasing but know it doesn't seems true)

Sorry for my delay. I'll motivate my previous post proving that there doesn't exists any other solution except those two i mentioned above.

Taking the equation modulo $64$ (for $x>5$) we'll find that $y$ must have the form $16n+1$. Now taking the equation modulo $17$, and taking account that $3^{16} \equiv 1 \pmod{17}$, we'll find that $LHS$ and $RHS$ give different remainders divided by $17$.


let $ a,b,c$ be positive reals such that $abc=1$ prove that :
$\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{a+c+1}\le 1$

Please, look a little bit in the previous discussions. This one was discussed at least 4 times. Not to mention that his sister from IMO TST 1999 was discussed at least 8 times. And when I say popular problems should have special places...

Here is an idea. We take $a=x^3,b=y^3,c=z^3$ and using the well known inequality $x^3+y^3>=xy(x+y)$ we deduce
 our inequality. ;)


[color=red]ΕΡΩΤΗΣΗ 1.[/color]

 Αγαπητοί φίλοι !

 Θέλω να βρείτε , αν υπάρχει ,  μια ελληνική λέξη( ή ονομα ή τοποθεσία κλπ )  που να έχει τη δομή της  λέξης GIURGIU και να έχει νόημα.Πχ  ΑΝΕΣΑΝΕ ή τερ-σ-τερ αλλά να έχει νόημα. Με τη λέξη αυτή θα φτιάξω μια άσκηση γαι την Α΄Γυμνασίου ή για το διαγωνισμού του Δημοτικού στην Εύβοια. Την άσκηση θα σας τη δείξω.Για όσους δεν κατάλαβαν , η λέξη αποτελείται από δύο όμοια θέματα με ένα γράμμα ανάμεσα: GIU - R - GIU, ΜΟΥ - Ρ - ΜΟΥ κλπ

                 ΕΥΧΑΡΙΣΤΩ

Καλησπέρα κ. Στεργίου,

Μερικά παραδείγματα:

ΜΠΑ-Ρ-ΜΠΑ ( κλητική προσφώνηση / αιτιατική )

ΜΑ-Γ-ΜΑ

ΤΑ-Ρ-ΤΑ (yummy :icecream: !)

ΚΑ-Σ-ΚΑ

Ελπίζω να είναι εντάξει  :D

Στέλιο γεια !

 Το πρώτο μου κάνει , γιατί - δεν το τόνισα δυστυχώς - θέλω λέξη με 7 γράμματα.
Επειδή λιγάκι θα μου κακοφανεί , να με φωνάζουν από ... τώρα ... ΜΠΑΡΜΠΑ  , προσπάθησε (ίσως και με τους φίλους σου ) να βρεις και καμιά άλλη.
 Θέλω να φτιάξω μια πρόσθεση :
--ΜΠΑΡΜ Π Α
----ΠΑΡΜ Π Α
-----ΑΡΜ Π Α      +
------ ΡΜ Π Α
---- ----Μ Π Α
---------- Π Α
------------ Α
---------------------------------- =
1 5 0 6 6 4 1
 Να βρείτε την τιμη των ψηφίων που κρύβουν (παραστάνουν ) τα γράμματα , ώστε η πρόσθεση να είναι σωστή. Διαφορετικά γράμματα παραστάνουν διαφορετικά ψηφία. Για Δημοτικό η άσκηση είναι καλή !

 Μπάμπης

Δεν μπόρεσα να βρω προς το παρόν μία τέτοιου είδους λέξη, αλλά π.χ. :

Έστω η λέξη ΓΑΡ-ΓΑΡ-Α

  Γ Α Ρ Γ Α Ρ Α
  - Α Ρ Γ Α Ρ Α
  - - Ρ Γ Α Ρ Α
  - - - Γ Α Ρ Α
  - - - - Α Ρ Α
  - - - - -  Ρ Α
  - - - - -  - Α +
--------------- =
9 5 0 5 5 8 2

α) Να βρείτε την τιμη των ψηφίων που κρύβουν (παραστάνουν ) τα γράμματα , ώστε η πρόσθεση να είναι σωστή. Διαφορετικά γράμματα παραστάνουν διαφορετικά ψηφία.

β) Να βρείτε για ποιες τιμές των Γ,Α,Ρ η παράσταση αποκτά τη μέγιστη τιμή και στη συνέχεια να βρείτε τη μέγιστη αυτή τιμή. Διαιρείται με το 3;

[u]Απάντηση[/u]: α) Α=6, Ρ=9, Γ=8
                        β) Θέτουμε Α=Ρ=Γ=9

[u]Σημείωση[/u]: Το σκεπτικό στην προκειμένη ίσως είναι λίγο πιο περίπλοκο, αφού το Ρ αν το πάρουμε από τα δεξιά προς τα αριστερά μπορεί να πάρει την τιμή είτε 4 είτε 9. (Πρέπει να εξεταστούν και οι δύο περιπτώσεις).

[quote="mostel"]Δεν μπόρεσα να βρω προς το παρόν μία τέτοιου είδους λέξη, αλλά π.χ. :

Έστω η λέξη ΓΑΡ-ΓΑΡ-Α

  Γ Α Ρ Γ Α Ρ Α
  - Α Ρ Γ Α Ρ Α
  - - Ρ Γ Α Ρ Α
  - - - Γ Α Ρ Α
  - - - - Α Ρ Α
  - - - - -  Ρ Α
  - - - - -  - Α +
--------------- =
9 5 0 5 5 8 2

β) Να βρείτε για ποιες τιμές των Γ,Α,Ρ η παράσταση αποκτά τη μέγιστη τιμή και στη συνέχεια να βρείτε τη μέγιστη αυτή τιμή. Διαιρείται με το 3;

[u]Απάντηση[/u]: α) Α=6, Ρ=9, Γ=8
                        β) Θέτουμε Α=Ρ=Γ=9

[u]Σημείωση[/u]: Το σκεπτικό στην προκειμένη ίσως είναι λίγο πιο περίπλοκο, αφού το Ρ αν το πάρουμε από τα δεξιά προς τα αριστερά μπορεί να πάρει την τιμή είτε 4 είτε 9. (Πρέπει να εξεταστούν και οι δύο περιπτώσεις).[/quote]

 Στέλιο ,  το πρόβλημα που έφτιαξες είναι καλό. Δεν έχω χαρτί μπροστά μου για να το κοιτάξω , αλλά πες μου : Έχει μοναδική λύση στο α) ερώτημα ; Γιατί στο β) βλέπω ένα α= γ = ρ = 9.Αλλά στο α) έχεις α = 6.Αν θέλεις το αναλύεις λίγο , διαφορετικά να το κοιτάξω και εγώ.Εκτός αν γίνεται και κάτι άλλο που δεν το βλέπω.
 Μπάμπης

[quote="mostel"]Από τη στιγμή που Γ, Α, Ρ παριστάνουν ψηφία, Γ,Α,Ρ $\in\{0,1,2,3,4,5,6,7,8,9\}$

Το α) ερώτημα είναι [u]ανεξάρτητο[/u] από το β). Δηλαδή, στο πρώτο ερώτημα η λογική είναι η εξής:

7*Α=κάτι2

Αυτό ισχύει μόνο για [b]Α=6[/b] (7*6=42, άρα 4 τα κρατούμενα).

Στη συνέχεια, 6*Ρ + 4= κάτι8 (πλην 4 που είναι τα κρατούμενα) =>
6*Ρ=κάτι4. Δηλαδή 6*Ρ = 54 ή 6*Ρ=24 => Ρ = 9 ή Ρ=4. Για Ρ=4 στη συνέχεια βγαίνει άτοπο, επομένως [b]Ρ=9[/b].

Κατόπιν, 5*Α+5=5*6+5=35 (3 τα κρατούμενα). Στη συνέχεια, Γ*4+3=κάτι5=> Γ*4=κάτι2 (πλην 3 που είναι τα κρατούμενα). Αυτό ισχύει για 8*4=32 , 3*4=12. Δηλαδή Γ=3 ή Γ=8. Για Γ=3 βγαίνει επίσης άτοπο, οπότε [b]Γ=8[/b].

Αφού βρήκαμε τα απαραίτητα ψηφία, στη συνέχεια είναι θέμα πράξεων για να επαληθευτεί το αποτέλεσμα.

β) Για να αποκτήσει αυτό το άθροισμα τη [u]μέγιστη[/u] τιμή θα πρέπει τα ψηφία να έχουν τη μεγαλύτερη δυνατή τιμή, η οποία είναι το 9. Δηλαδή [b]Α=Ρ=Γ=9[/b]. Στη συνέχεια για να βρούμε το αποτέλεσμα κάνουμε τις πράξεις. Επίσης, προφανές είναι ότι ο αριθμός θα διαιρείται με το 3.

 :)[/quote]
Κάτι τέτοιο φαντάστηκα και εγώ ότι εννοούσες, ότι δηλαδή το β) δεν μπορεί να έχει σχέση με το α) αλλά δεν κατάλαβα τι εννοούσες με τη λέξη ''παράσταση ''.Τώρα όλα είναι εντάξει και η λύση στο πρόβλημα είναι σωστή. Έχουν ωστόσο... πλάκα αυτά τα προβληματάκια και βοηθάνε τους μικρούς μαθητές να καταλάβουν τη δομή των πράξεων και την τεχνική τους.
  [b]Ευχαριστώ[/b] για το χρόνο που αφιέρωσες.

[u] Μπάμπης[/u]

Έχετε δίκαιο κ. Στεργίου, η χρησιμοποίηση της λέξης "παράσταση" ήταν άστοχη. Αν δεν κάνω λάθος θέμα παρόμοιας λογικής (αλλά με πολλαπλασιασμό) είχε τεθεί μία χρονιά στον Ευκλείδη της Β' Γυμνασίου.

Πάντως ακόμη δεν έχω βρει λέξη με τη μορφή που προαναφέρατε...  :|

Αν επιτρέπονται άλλες πτώσεις τότε

ΙΟΥΝΙΟΥ
ΙΟΥΛΙΟΥ

και .. ποιητική αδεία .. ΑΚΙΔΑ -> ΑΚΙΔΑΚΙ

+an epitreponte+ksena onomata apo mexikanika sirial...
Tote....
mar i mar    :rotfl:

 :D

edit:a....episis kai RA K AR.....(ala me 5 gramata..)


calculus
derivative
function
real analysis
real analysis theorems

i am presently studying multivariable analysis.

i am not fully satisfied with my intuitive understanding of total and directional derivatives.
i would be extremely happy if anyone could help me with this doubt.
thanks in advance!!

suppose we have a function f:R^2 -> R. say at the point (1,1) we wish to find the derivative along the direction (2,2).
can I interpret this problem as follows:
I look at the graph of the function in R^3 and restrict the function to the plane perpendicular to the X-Y plane passing through (1,1)
and (2,2). now i look at the tangent to the curve at (1,1) .
Are the two problems same???
secondly
if this view is correct then should'nt  the directional derivative at (1,1) along (2,2) and (3,3)be the  same?
(As the "direction" of the line joining (1,1) and (2,2) is the same as that joining (1,1) and (3,3)). :huh:

could anyone suggest a different geometric interpretation for a more satisfying picture??

it seems that people have gone through this post.
can anyone help?? :?:


Let $ABCD$ an inscribable quadrilateral of sides $a,b,c,d$ and diagonals $p,q$. Then

$ 4(\frac{1}{p^2} +\frac{1}{q^2}) \leq \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}$

It is easy to prove that:  $\displaystyle \sum \frac{1}{a^{2}} \geq 8 \frac{1}{q^{2}}$ and   $\displaystyle \sum \frac{1}{a^{2}} \geq 8 \frac{1}{p^{2}}$


integration
calculus
derivative
function
real analysis
real analysis unsolved

Let $ f \in L^\infty(\Omega)$. Given a sequence $ M_n\to\infty$, can we always find a sequence $ (u_n) \subset W^{1,p}(\Omega)$ such that $ \|u_n\|_{W^{1,p}}\equal{}1$ and
\[ M_n \int_\Omega f|u_n|^\alpha \to 0?
\]
Here $ 1<\alpha<p^*: \equal{}\frac{Np}{N\minus{}p}$, $ 1<p<N$.

I don't quite understand what is the problem: the norm of $ u_n$ in $ W^{1,p}$ involves derivatives and $ |u_n|^\alpha$ does not. Now, we can always find an arbitrarily small in the uniform norm function with arbitrarily large $ L^p$ norm of the gradient. I bet you meant something else but I cannot guess what. :?

Oops... my question is quite stupid.  :blush:  Thanks Fedja.