AMC
AMC 12
quadratics
Vieta
algebra
polynomial
AIME

Hey, a friend from school gave me this problem to work on, but I've been stumped.

Two circle with different radii are drawn in the first quadrant of the Cartesian plane such that they are both tangent to $x=0$ and $y=0$.  One point of intersection between the circles is $(1,3)$.  What is the sum of the radii?

I would really appreciate it if you guys helped me with a hint.  =)

Thanks alot!  And what level problem would this be on AIME/AMC12?

You have to use the pieces of information that it is tangent to the x,y axis and it passes through $(1,3)$. Try writing the equation of the circle in terms of $r$, then plug in the point to solve for $r$.

This would actually be around moderate AMC12, I'd say...
[hide="hint 1"]Either center is equidistant from $x=0$ and $y=0$.[/hide]
[hide="hint 2"]Thus, they both lie on the line $x=y$[/hide]
[hide="hint 3"] So find two points $(x, y)$ such that $x=y=$(the distance from $(3, 1)$)[/hide]

[quote="Bictor717"]The most important thing to notice is
[hide]because the circles are tangent to the x- and y-axes, their centers lie on the line y=x. So they can be expressed as (r,r).[/hide]
Then
[hide]because the circles contain the point (1,3), the distance from the center to (1,3) is the same as the distance from the center to either of the axes.[/hide]
Therefore[hide]
$\sqrt{(r-3)^2+(r-1)^2}=r$, from which you get a quadratic. The roots are the possible radii of circles tangent to both axes and contain (1,3). So the final answer is
[hide][hide="Did you get an answer yet?"]8[/hide][/hide][/hide][/quote]

AAAAAH this makes me so angry!!!!  That's [i]exactly[/i] what I did, but I kept thinking my solution wasn't logical...  Thanks for the posts guys...  I need to think about why I thought that solution was wrong.

Edit:

When I simplified the quadratic, I got

$r^2-8r+10$, and a possible solution is $4\pm\sqrt{6}$  I think that's where I went...woah...

Edit again!!!

Now that I think about it, the possible radii are $4\pm\sqrt{6}$, but the sum of those quantities is 8!

But weren't the circles supposed to be of different radii?

[quote="brady00n"][quote="chess64"]Yeah, this was from math league round 6.

$r^2-8r+10=0$ gives the two possible radii, which you found to be $4+\sqrt{6}\text{ and } 4-\sqrt{6}$. They have sum 8. You can also skip right to the answer, since vieta's formulas give $r_1+r_2=-(-8)=\boxed {8}$.[/quote]

Thanks.

What are vieta's formulas?  Are the the relation between the sums and products of quadratic roots?

Yeah, and I just noticed...$4\pm\sqrt{6}$ are [i]not[/i] the same quantities...

I need to stop getting all nervous and approach the problem calmly...[/quote]

Lol, yeah. Consider the polynomial $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. Let $\sigma_k$ be the sum of the roots taken $k$ at a time. For example, $\sigma_2(1,2,3)=1\cdot 2+2\cdot 3+3\cdot 1$. Vieta's formula states that \[ \boxed {\sigma_k = (-1)^k\frac{a_{n-k}}{a_n}}. \] In this case we want the sum of the roots of $r^2-8r+10=0$, which is just $\sigma_1$. So: \[ \sigma_1=(-1)^1\frac{a_{2-1}}{a_2} = -\frac{-8}{1}=\boxed {8}. \]

Just a coincidence that I had this EXACT problem on a math league this morning? (Alabama-Mississippi-Louisiana Math League)  This problem was posted yesterday...

[quote="pkerichang"]Yeah...next time on any competition when they ask about the sum or product of something, and you somehow got a polynomial, ALWAYS just go for Vieta's formula AND DON'T FORGET THE NEGATIVE SIGN IF IT IS THERE!!! (speed is VERY important in all the competitions we will go to next year)[/quote]

How would you use Vieta's Formula for a product?

Btw, I had this problem on the Ohio Math League this morning too...

Oops, this problem shouldn't have been posted until after today. There are two testing dates for math league; the first one was last week, I guess you could take it today too.

For a product using vieta just take the $n$th symmetric sum of the $n$ roots.

[quote="DanK"]Um, what's a symmetric sum?[/quote]

For a polynomial with $m$ number of roots, the $nth$ symmetric sum is the sum of the products of $n$ of the $m$ roots at a time, ex: the $2nd$ symmetric sum of a polynomial will be the sum of the products of all the possible pairs of roots. Taking the $mth$ symmetric sum of a polynomial with $m$ roots will yield just one product - the product of all of the roots.

At least it was easy...  If this were a Mandelbrot or something...

[quote="DanK"]So to find the product of the $m$ roots you would use $\sigma_m$?[/quote]

Yeah. Here's a problem to try:

$1.\triangleright$ Find the sum and product of the roots of the equation $x^5+2x^4+3x^3+4x^2+5x+6=0$.

$2.\triangleright$ (AIME 2001) Find the sum of the roots, real and non-real, of the equation $x^{2001}+\left(\frac 12-x\right)^{2001}=0,$ given that there are no multiple roots. (see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64586 )

[url=http://www.upal.ca/adeel/final_paper.pdf]Unfinished paper about this stuff[/url]

I'll try to finish it as soon as possible! :)

Well, now that you showed me Vieta's Formula, that first one is pretty easy...

[hide]The sum is -2 and the product is 6.[/hide]

HA! top 3 mistake of the use of Vieta's formula: it's -6 instead of 6.  You take the sign ALTERNATELY, so like -, +, -, +.

For a polynomial with EVEN degree, (for instance if the first problems starts with $x^6$ instead of $x^5$), then the product will be the constant.  But for a polynomial with ODD degree (like stated in the problem), it'll be the NEGATIVE of the constant.  You can try it out by taking alternate signs.

For more clarification: given an equation $2x^3 + 5x^2 + 24x + 7 = 0$, let the roots be $r$, $s$, and $t$. then:

$r + s + t = -\frac{5}{2}$
$rs + rt + st = +12$
$rst = -\frac{7}{2}$

Yeah the degree is 5 so you have to be all like $r_1+r_2+\cdots+r_5=\sigma_1=(-1)^{1}\frac{a_4}{a_5}=-2$ and $r_1r_2\cdots r_5=\sigma_5=(-1)^5\frac{a_0}{a_5}=-6$.

Now find $\sigma_3$.

A relatively simple problem. The equation for this is: ${(x - 1)^2} + {(x - 3)^2} = {x^2}$
Simplifying, we have:
${x^2} - 8x + 10 = 0$
The roots are thus $4 \pm 2\sqrt{6}$, and adding, our answer is $\boxed{8}$

This was in the last VAML, I believe.

[quote="K81o7"]You do understand why Vieta's formula works DanK, right?[/quote]

Yeah.  I can't explain it very well, but it's because of how a polynomial with roots $a,b,c,...$ is $(x-a)(x-b)(x-c)...$, which gives you those coefficients...

If someone can explain it better, please do.

[quote="jrshoch"]This topic's sort of confused me... but noone answered chess's question #2...

x^2001+(1/2-x)^2001=0
x^2001-x^2001+(1/2)x^2000-(1/4)x^1999....+(1/2)^2001=0
(1/2)x^2000-(1/4)x^1999+.....+(1/2)^2001=0

So would the sum be -(-1/4)/(1/2)=1/2?[/quote]

you did the binomial expansion incorrectly, remember the coefficients  ;)

[quote="chess64"][url=http://www.upal.ca/adeel/final_paper.pdf]there's a proof in this article[/url][/quote]

Thanks.  That's a useful paper...

[quote="jrshoch"][quote="Altheman"]

you did the binomial expansion incorrectly, remember the coefficients  ;)[/quote]

Good call..... so instead you'd have

(2000 C 1)(1/2)x^2000-(2000 C 2)(1/4)x^1999.....+(1/2)^2001=0

1000x^2000-499750x^1999+...+(1/2)^2001=0

And the sum would be 499.75?[/quote]

Reviewing the paper (which includes this problem) it appears i made yet another mistake by using 2000 C x instead of 2001 C x  :(
The real answer is 500 check the paper for a solution i dont' want to make another mistake...


Let three circles $ \Gamma_1, \Gamma_2, \Gamma_3$, which are non-overlapping and mutually external, be given in the plane. For each point $ P$ in the plane, outside the three circles, construct six points $ A_1, B_1, A_2, B_2, A_3, B_3$ as follows: For each $ i \equal{} 1, 2, 3$, $ A_i, B_i$ are distinct points on the circle $ \Gamma_i$ such that the lines $ PA_i$ and $ PB_i$ are both tangents to $ \Gamma_i$. Call the point $ P$ exceptional if, from the construction, three lines $ A_1B_1, A_2 B_2, A_3 B_3$ are concurrent. Show that every exceptional point of the plane, if exists, lies on the same circle.

circle's center is radical center O this 3 circles and R=OK, K - is point in circle $ G_i$ OK is both tangent.

Let $ O_i$ and $ r_i$ be the center and the radius of $ \Gamma_i$. Consider an exceptional point $ P$, such that $ A_1B_1, A_2B_2, A_3B_3$ concur at $ Q$. Consider circle $ \Gamma$ with diameter $ PQ$, center $ O$ and radius $ r$. Let $ A_1B_1$ intersect $ PO_1$ at $ X_1$. We have $ r_1^2\equal{}O_1X_1\cdot O_1P\equal{}O_1O^2\minus{}r^2$. So we get $ r^2\equal{}O_1O^2\minus{}r_1^2\equal{}O_2O^2\minus{}r_2^2\equal{}O_3O^2\minus{}r^2$. So $ O$ must be the radical center of $ \Gamma_1,\Gamma_2,\Gamma_3$ and $ r$ does not depend on $ P$. Thus all exceptional points lie on $ \Gamma$.

Let the point of concurancy be $Q$.Dilate with centre $P$ and ratio $1/2$.This maps $A_iB_i$ to the radical axis of point circle $P$ & $\Gamma_i$.Since these radical axes are concurrent,we get them midpt of $PQ$ say $M$  to be radical centre of 3 circles and $P$ point circle.Now since the power of $M$ wrt point circle $P$ is fixed we get $MP^2$ is fixed .Done :)

Wow. If the three polars concur, then the radial axes of the point circle at $P$ with the three circles concur, so $P$ lies on the circle centered at the radical center of $\Gamma_1, \Gamma_2, \Gamma_3$ with radius equal to the square root of the common power.


i have a few questions about IS and subject classes:

1) is there a specific time when we "meet" for class, or do we have a separate forum to discuss hw and be given out the next time's hw and such?

2) do the subject classes have assigned hw, or is it more independent?

3) if we enroll in an IS class, do we separately enroll for a subject class, or wait till the date of the subject class starts?

these questions are of minor importance to me, as i've already signed up for the classes and i have 0 thought of dropping out, but i'm just curious about how the classes will work out. thanks :D

i haven't taken but an IS class, only a subject class, but I think I know enough to be of help.
by taking an IS, you get a credit to enroll in one subject class.  when you use that credit is up to you.  as for the subject class, each week there is an actual class for about an hour an half (varies for each class).  after the class, the instructor posts a problem set on the message board, and students post whatever they can/wish to do on the problems.  in this aspect, it is sort of independent.  but, there is also a midterm challenge and a final challenge that are to be written up and sent in to the instructor by fax or snail mai which you are evaluated on and will receive feedback via email.  the full solutions will also be posted after the due dates.  i think that answers your questions.

ffdbzathf has correctly described the subject classes, but the Independent Study program is considerably different.  Every two weeks you'll receive a reading and homework assignment from us.  During the week we have office hours - you can find the tentative partial schedule of office hours [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35051]here[/url].
At the office hours, you can ask any math questions you like and work 1-1 with AoPS instructors.  If you have no questions at that time, you can join in the small group discussion of a problem if there is one going on.  Usually a few of the students work together with an instructor on a problem during office hours, while other students work 1-1 on specific problems they're having trouble with.

As for the subject of your choice, we will send out an email to all the Independent Study participants when the Independent Study begins asking you for your course selection.

do we have to sign up for the class office hours?? or do we just come and join the classroom?

You simply show up at a scheduled office hour time.  We might switch to a scheduling system in the future if the number of enrollments stretches our ability to staff office hours with experience teachers.

You can also request the addition of a few office hours -- but let us know why you are requesting.  Requests by students who can make it to few of the offered times will take precedence.


calculus
integration
LaTeX
trigonometry
calculus computations

Evaluate the definite integral.

4th root pi (top)     x^3  *  cos(x^4)

0 (bottum)

how would i do this?

There is a blatantly obvious substitution. Try it.

i dont get how to do this, can someone show me the steps

and how do i use latex

I'm not sure what you mean: is it $ \int_{0}^{\frac {\pi}{4}} x^{3} \cos (x^{4}) \ dx$?

Let $ u \equal{} x^{4}$.

Then $ du \equal{} 4x^{3}\ dx$ and we get $ [\frac {\sin (x^{4})}{4}]_{0}^{\frac {\pi}{4}}$ $ \equal{} \frac {1}{4} \sin (\frac {\pi^{4}}{256})$.

If you hover over the stuff written in $ LaTeX$, you can see how it's done.

Hope this helps you.

sory but to be more clear its     pi^(1/4)    not pi/4

hello, in this case (if you meant $ \int_{0}^{\sqrt[4]{\pi}}x^3\cos(x^4)\,dx$) you will get $ 0$ as the result.
Sonnhard.

Thanks again, Sonnhard. (I should have realised what was meant.)

$ \boxed{\cos f(x)*f'(x)\equal{}(\sin f(x))'}$


[hide]First take the square root of 2005 to see the largest prime you can square before going over 2005. Then the following primes are listed raised to the largest power before going over 2005 (since the square root of 2005 is 44.etc the largest prime you can use is 43):

2^10
3^6
5^4
7^3
11^3
13^2
17^2
19^2
23^2
29^2
31^2
37^2
41^2
43^2

Since m >=2
you can subract one from each of the powers and add the differences, meaning:
9+5+3+2+2+1+1+1+1+1+1+1+1+1=30
So there are 30 superprimes
[/hide]

[quote="Hexaditidom"]A superprime can be written in the form $ n^m $ , where n is a prime number and m >= 2. How many superprimes are there that are less than 2005?

Edit - m is an integer[/quote]

Just in case, if you find it reading about math olympiad or number theory in general, what you call superprimes usually are called prime powers!

I called it that because I posted a problem about superpowers and superfactorials in the olympiad section, and I wanted to add another super but this one was easier.

One of these days I'll come up with something completely original...  :?

[quote="Hexaditidom"]One of these days I'll come up with something completely original...  :?[/quote]

A difficult thing to do!  I think you post some very good questions, even if your terminology is sometimes unorthodox :)

oh!  were we meant to use a calculator? lol.

After finding that 43 is the largest prime and all the primes smaller than that, I was wondering for the next 20 minutes how you find out $m$ without using brute force

I agree dangvh; I arrived at the same answer


Prove the following identity for nonnegative integers $n$:

$\sum _{k=0}^{n} {n\choose k} (k+1)^{k-1} (n-k+1)^{n-k-1} = 2(n+2)^{n-1}$

I am very interested where did you get it.

My approach is the following.

We will prove that both parts of this identity means the number of trees of $n+2$ vertices 1, 2, 3, ..., $n+2$, where vertices $n+1$ and $n+2$ are connected!

Indeed, for the LHS. We are to choose $k$ vertices for $n+1$-subtree and $n-k$ vertices for $n+2$-subtree. These subtrees have $k+1$ and $n-k+1$ vertices respectively, so numbers of possible trees on them are $(k+1)^{k-1}$ and $(n-k+1)^{n-k-1}$ due to Cayley's theorem.

For $RHS$ we are just to note that Prüfer's code will necessary ends with $n+1$ or $n+2$, so we have $(n+2)^{n-1}\cdot 2$ such codes!

A bit more generally we have :
$\sum _{k=0}^{n} {n\choose k} (k+a)^{k-1} (n-k+b)^{n-k-1} = (n+a+b)^{n-1}(\frac 1 a + \frac 1 b)$.

This has been proposed in the St-Petersburg Contests (1965-1984, extra problems. I proposed some of them before here).
Moreover, this (last) problem also appears in the excellent book of L. Lovasz, 'Combinatorial problems and exercises', North-Holland.

I have an algebraic solution but it is really long and boring. Your combinatorial approach is really nicer even if it needs more knowledge.

Pierre.

I didn't understand from your remark whether you know how to solve this (last) problem with my approach. So I will write the proof for the general version too.

Consider trees of $n+a+b$ vertices 1, 2, ..., $n+a+b$, for which vertices $n+1$, $n+2$, ..., $n+a+b$ are connected in chain.
First of all, note that the Prüfer's code for such trees ends with $m$, $n+1$, ..., $n+a+b-2$, where $m$ is an arbitrary number from $n+1$, ..., $n+a+b$ (and vice versa). It follows that the number of such trees is $(n+a+b)^{n-1}(a+b)$.

On the other hand, consider subtrees containing ($n+1$, ..., $n+a$) and ($n+a+1$, ..., $n+a+b$) respectively. Suppose the first subtree contains $a+k$ vertices, then we need to choose $k$ vertices from $n$ available vertices for the first subtree and $n-k$ vertices for the second subtree, i.e. $C_n^k$ choices. Further, we are to count the number of subtrees of $k+a$ vertices, where vertices $n+1$, ..., $n+a$ are connected in a chain. Using the same reasoning with Pruffer code we conclude that there are $a(k+a)^{k-1}$ such subtrees. Similar for the second subtree. Hence the result.


algorithm
AoPSwiki
geometry
analytic geometry
rectangle
trigonometry
FTW

What is Pick's Theorem and the Shoelace Algorithm? Also, could someone please post a few examples? Thank you.

Izzy: This may help you:
From AoPSWiki
   Pick's Theorem expresses the area of a polygon, all of whose vertices are lattice points in a coordinate plane, in terms of the number of lattice points inside the polygon and the number of lattice points on the sides of the polygon. The formula is:

 I+B/2-1

where I is the number of lattice points in the interior and  B being the number of lattice points on the boundary. It is similar to the shoestring formula, and though it is less powerful it is a good tool to have in solving problems.

Lattice points are points $ (x,y)$, where $ x$ and $ y$ are integers.

[hide="The Shoelace Algorithm"]Ok, so say you have points of a triangle.  The points are $ (1,4)$, $ (3,1)$, and $ (11,1)$.  Obviously, there are other ways to solve this, but let's use Shoelace.

For $ n$ amount of points, the shoelace algortihm states that the area is:

$ \frac{1}{2} |x_1 y_2 \plus{} x_2 y_3 \plus{} ... \plus{} x_{(n\minus{}1)} y_n \plus{} x_n y_1 \minus{} y_1 x_2 \minus{} y_2 x_3 \minus{} ... \minus{} y_{(n\minus{}1)} x_n \minus{} y_n x_1|$

So the area would be:

$ \frac{1}{2} |1 \plus{} 3 \plus{} 44 \minus{} 12 \minus{} 11 \minus{} 1|$

$ \frac{1}{2} |24|$

$ \boxed{12}$[/hide]

Or, a more visual way
[code]
    1  4
     \/
12   /\    1
    3  1
     \/
11   /\    3
    11 1
     \/
1    /\    44
    1  4
[/code]

Add up columns to get 24 on left hand side and 48 on right hand side.
48-24=24
24/2=12

Personally, I don't think the shoelace formula is needed. Drawing out the polygon, putting it around a rectangle, and subtracting off triangles works just as well.

Or shearing, if you're lucky, can save IMMENSE amounts of time.

[quote="isabella2296"]What are lattice points? :maybe:[/quote]

Hmm...you always mem FTW quesitons with lattice points...

[quote="math154"]MC problems almost always have simple solutions, so it is usually slower to do things like trig bash, shoelace, etc.[/quote]
Every problem in MC can be solved w/o trig.  I don't know bout shoelace though

Shoelace just takes a [b]whole lot of time[/b].

Let's see: Sprint rounds seriously [u]DO NOT NEED SHOELACE[/u] (unless something like #28 Nats)
Target rounds: plausibly. If you can shoelace within the time limits, I suppose it works out.
Count down rounds: If one does shoelace and the other does, say, direct, I think I know who's going to get that point.
Team round: Perhaps.

Program it into your calculator. Never use it otherwise.

I don't think it states anything about programmed calcs in the rules.

Heh, I have Shoelace on my calc.

can someone please post a proof to both theorems?
BTW, i need a programmable calc.

[quote="AIME15"][quote="isabella2296"]What are lattice points? :maybe:[/quote]

Hmm...you always mem FTW quesitons with lattice points...
[/quote]

I knew what lattice points were, just not exactly what. It's kind of easy to infer, I just wanted to check. (This is what happens when you teach yourself Geometry...)

Are there any other quick methods similar to Shoelace but are faster?

[quote="Poincare"]
BTW, i need a programmable calc.[/quote]

TIs from 83 and above are easily programmable (if you really need the power TI 89 is a sure bet)
Not sure about HP's though.

Are calcs allowed in sprint, countdowns, team, or target rounds?

I know they're not allowed in Countdown and Sprint, but they are in Target. I don't know about Team.

[quote="from AOPS WIKI"]Team Round

10 problems in 20 minutes for a team of 4 students. These problems typically include some of the most difficult problems of the competition. Use of a calculator is allowed (and required for some questions). Like all of the other rounds, it gets progressively harder from the School-level competition to the National-level competition. [/quote]

So I guess Team Round calcs are allowed

Wait guys, are you allowed to use a programmed calc with shoelace, herons, and all that stuff on it?  Are you allowed to have like formulas on your calc?  Because...then i need to get a graphing calc.

Well they're only allowed for target and team rounds, so that kind of limits their purpose

[quote="JBoyd"]I have found that it works well with pentagons and triangles that don't have a side that matches up with either thee x-axis or the y-axis.  There other methods that can be used but at times this saves time.  I show the kids several methods and let them chose which method works best for them.

Mathcounts does allow you to have programs in the calc.  You just can't hhave a qwerty keyboard.[/quote]

@AIME- cough cough.


A logician told her daughter: “If you don’t finish your dinner then you will not get to
stay up and watch TV.” That night the girl finished her dinner and was then sent straight
to bed. Was the mother telling the truth?

[hide="is this correct?"] Let "FINISH DINNER" BE F
Let "STAY UP AND WATCH TV" be S

so the statement is $ \equal{} \equal{} > \not{F} \minus{} \minus{} > \not{S}$
since she did eat the dinner $ F \minus{} \minus{} > \not{S}$

so mother was lying? [/hide]

The mother's statement is "If you don't finish your dinner..." which means that if the girl [i]did[/i] finish it, what the mother said has no meaning anymore. The mother was telling the truth, the condition just didn't apply.


linear algebra
matrix
vector
inequalities
ratio
Functional Analysis
real analysis

Suppose we have two norms $ ||- ||_{1}$ and $||-||_{2}$ on $C^n$. They induce two matrix norms $ |||- |||_{1}$ and $|||- |||_{2}$. Prove that if for any matrix $A\in C^{n^2}$ we have $|||A|||_{1}\geq |||A|||_{2}$ then for any matrix $A$ we have $|||A|||_{1}=|||A|||_{2}$. Strange result...

Moreover, in this case the norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are just multiples of each other.  (Consider it as a small hint  ;) )

As for the source, I remember it may be found in Horn and Djonson "Matrix analysis", and some references there. I have the solution (wich works also for infinite-dimensional Banach spaces), but it  requires some time to translate it into English.

So it seems it is not at all easy. Thank you very much, Fedor, I would be very grateful to see a solution for this problem.

[quote="harazi"]I don't like your hints: they are just too subtle for me! :)  :)[/quote]
Well, I just wanted to let other people try to solve the problem first  ;) . If nobody solves it within a week, say, I'll post a complete solution. BTW, I do like most of the problems you post very much!   :lol:  (not that I can solve many of them  :( *sigh* )

Ok, fedja, I will more on the problem and in a week, if nobody comes with a solution, you post it, ok? By the way, you solved practically all problems I posted, so don't worry...

[quote="Myth"]Is an induced matrix norm an operator norm?[/quote]
I think it is (otherwise I completely misunderstood the problem).

I think I have a solution at least when the two norms are equivalent
( in order to say that the corresponding normed spaces have that same bounded
functionals and bounded operators), which is the case when the space is finite
dimensional.
 Suppose that we have two equivalent
norms $ ||-||_{K} $ and $ ||-||_{L} $ on a Banach space
$ X $, which induce norms on bounded operators $ A : X \rightarrow X $ :
$ |||A|||_{K} $ and $ |||A|||_{L} $.
(because the norms are equivalent, the set of bounded operators coinside,
also the set of bounded linear functionals coincide, which I will denote as
$ X^{*} $ ( as a set) , and the induced dual norms $ ||f||_{K^{*}} , ||f||_{L^{*}} $
are also are also equivalent, so we can denote by $ X^{**} $ the dual of $ X^{*} $
(as a set) )
Suppose that $ |||A|||_{K} \leqslant |||A|||_{L} $ for every bounded
$ A : X \rightarrow X $. Take some $ w \in X $, $ f \in X^{*} $ and define
$ A : X \rightarrow X $ as $ Ax := f(x)w $.
Then it is easy to see that $ |||A|||_{K} = ||f||_{K^{*}} ||w||_{K} $ and
$ |||A|||_{L} = ||f||_{L^{*}} ||w||_{L} $.
Therefore for every $ w \in X $, $ f \in X^{*} $ we have
$ ||f||_{K^{*}} ||w||_{K} \leqslant ||f||_{L^{*}} ||w||_{L} $. Therefore
$ \frac{||w||_{K}}{||w||_{L}} \leqslant \frac{||f||_{L^{*}}}{||f||_{K^{*}}} $
for every $ w \in X $, $ f \in X^{*} $ , $ f,w \neq 0 $.
On the other hand, suppose that we have a constant $ C > 0 $ such that
$ ||w||_{K} \leqslant C ||w||_{L} $ for every $ w \in X $. Then for every $ f \in X^{*} $
we have $ |f(x)| \leqslant ||f||_{K^{*}} ||x||_{K} \leqslant ||f||_{K^{*}} C ||x||_{L} $
for every $ x \in X $, so $ ||f||_{L^{*}} \leqslant C ||f||_{K^{*}} $. Now if
$ ||f||_{L^{*}} \leqslant C ||f||_{K^{*}} $ for every $  f \in X^{*} $ , then doing
the same thing we get that $  ||F||_{K^{**}} \leqslant C ||F||_{L^{**}} $
for every $ F \in X^{**} $, but because of the embedding $ X \hookrightarrow X^{**} $
we get that $ ||w||_{K} \leqslant C ||w||_{L} $ for every $ w \in X $. So we see that
the condition $ ||w||_{K} \leqslant C ||w||_{L} $ is equivalent to condition
$ ||f||_{L^{*}} \leqslant C ||f||_{K^{*}} $ for $ C > 0 $, therefore we obtain that
$ sup \frac{||w||_{K}}{||w||_{L}} = sup \frac{||f||_{L^{*}}}{||f||_{K^{*}}} $.
Denote $ C:= sup \frac{||w||_{K}}{||w||_{L}} $. Then
$ sup \frac{||f||_{L^{*}}}{||f||_{K^{*}}} = C $, but for given $ f \in X^{*} $ we have that
$ \frac{||w||_{K}}{||w||_{L}} \leqslant \frac{||f||_{L^{*}}}{||f||_{K^{*}}} $ for each
$ w \in X $, so $ C \leqslant \frac{||f||_{L^{*}}}{||f||_{K^{*}}} $ for every
$ f \in X^{*} $. Therefore we have $ \frac{||f||_{L^{*}}}{||f||_{K^{*}}} = C $
for every $ f \in X^{*} $, so $ ||f||_{L^{*}} = C ||f||_{K^{*}} $ for every
$ f \in X^{*} $. Therefore $ ||F||_{K^{**}} = C ||F||_{L^{**}} $ for every
$ F \in X^{**} $ , and again by the embedding $ X \hookrightarrow X^{**} $
we have that $ ||w||_{K} = C ||w||_{L} $ for every $ w \in X $.

I hope it is correct  :)  .

Bromov is the best! Now, I will have to understand the solution.  :)

It is not hard  :)  .
Only the word "Banach" is scary, but the idea is simple.

Essentially the same solution can be presented in a bit shorter way. Assume that the norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are not multiples of each other (if they are, the induced operator norms coincide and the story is over). Then there exist two vectors $u$ and $v$ such that
$\frac{\|v\|_2}{\|v\|_1}>\frac{\|u\|_2}{\|u\|_1}$. Now use the Hahn-Banach theorem to construct a functional $\psi$ such that $\psi(u)=\|u\|_1$ and $|\psi(x)|\leqslant\|x\|_1$ for all $x$, and consider the linear operator $A: x\mapsto \psi(x)v$. Then $|\|A|\|_1=\|v\|_1$, while $|\|A|\|_2\geqslant \frac{\|Au\|_2}{\|u\|_2}=\frac{\|u\|_1\|v\|_2}{\|u\|_2}>\|v\|_1=|\|A|\|_1$ contradicting the assumption of the problem. The scary word "Banach" still appears, though.

It seems I did the same thing, but in slightly another way.
Suppose $b\in X$ a s.t. non-zero element, that $\frac{||b||_1}{||b||_2}$ is minimal (I don't argue that $b$ exists, but I assume it for simplicity; at least in finite case it is true. The general case can be esily done via minimizing sequence).
Consider an arbitrary element $a$ s.t. $||a||_1=1$, and let $L$ be a hyperplane tangenting unit ball in $||\cdot||_1$ at point $a$. Consider an operator $Ax\equiv A(x_aa+x_L)=x_ab$, $x_a\in \mathbb C$, $x_L\in L$. Then the given inequality means that $\frac{\min_{x_L} ||a+x_L||_1}{\min_{x_L}||a+x_L||_2}\leq \frac{||b||_1}{||b||_2}$, but $\min_{x_L} ||a+x_L||_1=||a||_1$, so $\frac{||a||_1}{||a||_2}\leq \frac{\min_{x_L} ||a+x_L||_1}{\min_{x_L}||a+x_L||_2}\leq \frac{||b||_1}{||b||_2}$. Due to our choice of $b$ we conclude that $\frac{||a||_1}{||a||_2}= \frac{||b||_1}{||b||_2}$.

P.S. I have just realized that I don't need minimality property of $b$. Just assume that this ratio is not a contstant, like fedja did.


vector
induction
algorithm
algebra
polynomial
function
set theory

The theorem "every linear subspace in R^n has a finite basis" can be proved only if the axiom of choice holds. I studied the proof of this theorem, but I can't find where exactly the axiom is applied (it is certainyl not explicit). Can anyone help?

I'm not sure, but I thought that to prove that an arbitary vector space (maybe infinite dimensional) has a basis, we must use [i]Zorn's Lemma[/i] which is equivalent to [i]Axiom of Choice[/i].

One need not invoke the axiom of choice for a finite dimensional space!

Suppose $V$ is a subspace of $\mathbb{R}^{n}$ (or of $F^{n}$ over some field $F.$)

If $V=\{0\},$ then it has the empty basis and we are done. So assume that ${V\ne\{0}.$

Now proceed by induction. Start with $W_{0}=\{0\}.$

Find $v_{1}\in V\setminus\{0\}.$ Let $W_{1}=\text{span}\{v_{1}\}.$

At each stage of the induction, we have an independent set $\{v_{1},v_{2},\dots,v_{k}\}$ which spans $W_{k}\subset V.$

If $W_{k}=V$ we are done. Else find $v_{k+1}\in V\setminus W_{k}.$

Since $v_{k+1}$ is not in the span of the independent set $\{v_{1},\dots,v_{k}\},$ then the set $\{v_{1},v_{2},\dots,v_{k},v_{k+1}\}$ is independent. Continue the induction from here.

At each stage, we have an independent set. But we also have the theorem that any set of $n+1$ or more vectors in $F^{n}$ must be dependent; hence the inductive process must terminate with $k\le n.$

--

That finite dimensional proof gives at least a hint of why we would need Zorn's Lemma or transfinite induction to handle the infinite dimensional case: we need a substitute for the assertion that the induction must terminate.

I don't understand how your proof works formally without having proof by induction established (which depends on well-ordering principle), Kent. The way I work it out is that if we can't find a basis with dimension smaller than $n$, then we can find $n+1$ linearly independent vectors. This proof requires induction over $n$.

Is it really formally ok to just say "if $W_{k}= V$ stop, else find ... repeat this", since we don't know for how long we are repeating it? To me such arguments require induction. But maybe I'm wrong. I don't have a true formal understanding of such proofs even though I will sometimes use them myself and accept them.

Kent Merryfield's algorithm will produce a basis for any finite dimensional space. It is not important that you don't know how long it will take. However, if you are doing the algorithm and you aren't done yet, you can't tell whether your space is finite-dimensional or infinite-dimensional. If you want to be sure your algorithm "terminates" you need one of these stronger assumptions.

If you are interested in this sort of thing, read about computability. :) A guy in my class just gave a presentation about it yesterday. It's really cool. :D

No, I am not convinced that it works without axiom of choice.

I don't really see what distinguishes Kent Merryfield's argument fron the following:

Suppose that we have propositions $P_{1},P_{2}, ...$. Assume that for any k $P_{k}$ implies $P_{k+1}$ and that $P_{1}$ is true.. Then, for any $n$ we can prove that $P_{n}$ is true because since $P_{1}$ is true, $P_{2}$ is true, therefore $P_{3}$ is true "and so on". This process will terminate at $P_{n}$ and therefore $P_{n}$ is true.

This is not a correct argument (although I don't quite understand why). I don't see why Kent Merryfield's argument works better.

Here is how I would formulate Kent Merryfield's proof with induction:

Lemma. Suppose that the linear space $V \ne 0$ does not have a basis with $\le n$ elements. For any integer $1 \le k \le n+1$ we can find a basis with $k$ elements.

Proof. For $k=1$ we can choose any vector $v_{1}$ not equal to zero.
Now assume the statement is true for $k=p$, $1 \le p \le n$. Then it is true for $k=p+1$. Since we can find linearly independent vectors $v_{1},..,v_{p}$ and they don't generate $V$ we can find another vector $v_{p+1}$ so that $v_{1},...v_{p+1}$ are linearly independent. The proof now follows from the induction principle.

Now apply this on the linear subspace in question...

I don't see how to get away from calling the induction principle toward the end.

There is quite a difference between "for some given $n$" and "for all $n$."

If you give Professor Merryfield a finite dimensional vector space, he will give you a basis. That's what his proof says. It requires only finite induction because he knows there exists some $n$ at which algorithm will stop.

It's when you don't know that there exists such $n$, or you want it to work for all $n$ at once, that it doesn't work.

I hope it makes sense... I'm not an expert at this either...

[quote="Xevarion"]There is quite a difference between "for some given $n$" and "for all $n$."
[/quote]
Not really. If we take any given $c \in \mathbb{Z}^{+}$ and show the proposition to be true for this $c$, it will be true for all $n \in \mathbb{Z}^{+}$. On the other hand, if it is true for all $n \in \mathbb{Z}^{+}$ it is true for any given $c \in \mathbb{Z}^{+}$.

It's just proof style that we don't normally distinguish between the "fixed c" and the "general n".

If you apply the algorithm in my post in an infinite dimensional vector space, then (by ordinary induction on the natural numbers), we will obtain a countably infinite independent set of vectors $\{v_{1},v_{n},\dots\}.$ However, there is no guarantee that this set will span the vector space in question - in fact, it is quite likely not to be a spanning set. A basis is a maximal independent set, and the set produced by ordinary induction is very likely not maximal.

Having a vector space with a countable basis wouldn't help. For instance, consider the vector space of all polynomials (over whatever field we're using). I could use the freedom in my proof to select the set $\{1,x^{2},x^{4},\dots\}.$ This will be an infinite independent set but it would not be a basis.

---

Here's a sub-problem. Suppose $U$ is a vector space over $F$ for which you can exhibit a countably infinite basis. (Up to vector space isomorphism, this denotes a unique vector space.) Suppose $V$ is a subspace of $U.$ Can you prove that $V$ has a basis without invoking any equivalent of the Axiom of Choice?

But when you invoke induction in your original reasoning, isn't that invokign the axiom of choice? I thought induction was an implication of the axiom of choice.

Yes, my proofs most definitely use induction (or, equivalently, the well-ordering if the natural numbers.) No, induction (at least ordinary induction, on $\mathbb{N}$) is not a consequence of the axiom of choice. Perhaps you could consult a book on set theory or logic.

There is something called "transfinite induction" which is a consequence of the fact that every set can be well-ordered, which is in turn equivalent to the axiom of choice - but I wasn't using that.

Proofs by ordinary induction may be a little bit mysterious but there's nevertheless something concrete about them. The little bit of mystery is the claim that "$P_{n}$ is true for all $n$" but you can see exactly why $P_{14392}$ is true, even though it would take a long time to get there. More to the point: if I give you any concrete description of a finite-dimensional vector space, I ought to be able, by some specific, concrete algorithm, to explicitly exhibit a basis.

The hallmark of all things produced by the Axiom of Choice is that they are never concretely constructed. Give me the Axiom of Choice and I can prove that every vector space has a basis. This includes $\mathbb{R}$ having a basis considered as a vector space over $\mathbb{Q}.$ Make a completely arbitrary mapping from that basis to the reals, and we can uniquely extend that mapping to a function $f: \mathbb{R}\mapsto\mathbb{R}$ such that $f(x+y)=f(x)+f(y).$ A cardinality argument now shows us that there are more of these functions than there are ordinary linear functions $f(x)=ax.$ But what does one of these monsters - an additive function that is not $\mathbb{R}$-linear - look like? Its graph is unimaginable.

That's what the Axiom of Choice does - it asserts the existence of things we cannot visualize.

Of course you could prove easily graph of all function $f: \mathbb R\longrightarrow\mathbb R$ that are not continuous, is dense in $\mathbb R^{2}$. What a monster !!!

Something strange about Axiom of Choice is that it prove that "[i]Every set $S$ can be ordered with some order $\geq$ that it's well-ordered.[/i]"

This is one of the most strange things about the [i]Axiom of Choice[/i].

In my opinion [url=http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox]the Banach-Tarski paradox[/url] is one of the most strange consequences of the Axiom of Choice... :P

[quote="Kent Merryfield"]Here's a sub-problem. Suppose $U$ is a vector space over $F$ for which you can exhibit a countably infinite basis. (Up to vector space isomorphism, this denotes a unique vector space.) Suppose $V$ is a subspace of $U.$ Can you prove that $V$ has a basis without invoking any equivalent of the Axiom of Choice?[/quote]

My first impression was that we can do it without the A.C. But after a little thinking, it doesn't seem so clear (mainly because $F$ can be "big"). So, what's the answer?

We can do it by ordinary induction. While we have to make choices along the way, we'll be making them one at a time, and that does not require AC.

I'm making the assumption here that the field $F$ has already been constructed; I'm not counting as part of this any machinery needed to construct $F.$

Let $\{e_{1},e_{2},\dots\}$ be a basis for $U.$

Let $U_{n}=\span{e_{1},\dots,e_{n}\}.$ This is a finite dimensional vector space.

Let $V_{n}=V\cap U_{n}.$

Make the inductive assumption that we have already constructed a basis ${f_{1},\dots,f_{k}}$ for $V_{n}.$ (This might be the empty set).

Either $V_{n}=V_{n+1}$ or $V_{n}\ne V_{n+1}.$ If the former, there is no action that we need to take. If the latter, then there exists $f_{k+1}\in V_{n+1}\setminus V_{n}.$  We can uniquely write $f_{k+1}=w+\alpha e_{n+1}$ where $w\in U_{n}.$ Now take any $x\in V_{n+1}\setminus V_{n}.$ We can write $x=u+\beta e_{n+1}.$ Then $x-\frac{\beta}{\alpha}f_{k+1}\in V_{n}.$

This argument allows us to conclude that $\{f_{1},\dots,f_{k},f_{k+1}\}$ is a basis for $V_{n+1}.$

Continuing this process by induction - ordinary induction - allows us to write a potentially infinite independent set $\{f_{1},f_{2}\dots\}.$ But any element of $V$ is contained in $V_{n}$ for some finite $n$ and thus is in the span of this (possibly) infinite independent set. We conclude that this (possibly) infinite independent set is a basis for $V.$

Here's a paraphrase of the construction: given some vector space of polynomials, it is possible to find a basis for this vector space which consists of a list of polynomials of strictly increasing degree.


geometry
circumcircle
inradius
trigonometry
angle bisector
geometry solved

ABC triangle. A' and A" the intersection points with BC  of the interieur and exterieur bissecteur of the angle A.
Same definition of B' , B", C',C".
Consider the six lenghts: AA',AA",BB',BB",CC' and CC"
             the six lenghts: the three sides and the three altitudes
             the five radius of the incercle , circumcerle and the three "excercle"
Construct such a triangle so as the 17 lenghs must be naturel numbers.

Consider the triangle ABC with sidelengths a = 125, b = 154, c = 169. Then, the following lengths are all rational:

$A\text{-altitude }=\frac{3696}{25}$;
$B\text{-altitude }=120$;
$C\text{-altitude }=\frac{18480}{169}$;
$\text{circumradius }=\frac{4225}{48}$;
$\text{semiperimeter }=224$;
$\text{inradius }=\frac{165}{4}$;
$A\text{-exradius }=\frac{280}{3}$;
$B\text{-exradius }=132$;
$C\text{-exradius }=168$;
$A\text{-angle bisector (internal) }=\frac{48048}{323}$;
$B\text{-angle bisector (internal) }=\frac{2600}{21}$;
$C\text{-angle bisector (internal) }=\frac{30800}{279}$;
$A\text{-angle bisector (external) }=\frac{4004}{3}$;
$B\text{-angle bisector (external) }=\frac{975}{2}$;
$C\text{-angle bisector (external) }=\frac{23100}{29}$.

As for now, most of the values are fractions. By multiplying the three sides a, b, c of the triangle with the least common divisor of the denominators of all of these fractions, we get a new triangle where all these lengths are natural numbers. This is the triangle you asked for. Have fun with constructing!

Note that in the above, I have used the formula from http://www.mathlinks.ro/Forum/viewtopic.php?t=805 post #3 for determining the lengths of the angle bisectors.

[b]PS.[/b] [u]Thanks to Yetti[/u] for a correction of a stupid mistake in my computations. He also gave the following additional rational values:

$\text{Area }=9240$;
$\cos A=\frac{119}{169}$;
$\cos B=\frac{2047}{4225}$;
$\cos C=\frac{7}{25}$;
$\sin A=\frac{120}{169}$;
$\sin B=\frac{3696}{4225}$;
$\sin C=\frac{24}{25}$.

  Darij


Princeton
college
real analysis
real analysis solved

If a_1 < a_2 <...< a_k is a set of k distincts positive
integers whose 2^k  subsets all have different sums.
Prove that 1/a_1 + 1/a_2 +...+ 1/a_k < 2

References:
1.Canadian Mathematical Bulletin, 17(1975) 768, problem P.220
2.G.Polya, Mathematics and Plausible Reasoning, Volume 1, Princeton
University Press, New Jersey, 1954, pp. 17-21
3.Ross Honsberger, Mathematical Gem III, MAA p.215

Erdos has the theory that God has a book containing all the theorems
of mathematics with their most beautiful proofs "This one from the book!"

Let b_i=2 i-1 (i=1,2..,k)
We must prove sum{1->k}a_i /leq sum{1->k}2 i-1
This is equivalent to sum(1/b_i - 1/a_i) >=0
               or sum( (a_i-b_i)/a_i*b_i) \geq 0 (1)
Let c_i=1/(ai*bi)
(1) is equivlaent to sum(ai-bi)ci  \geq 0 (2)
Let S_i = sum{1->i} (aj-bj) ,i=1,2,..,k
then (2) equal sum{1->k} (S_i - S_(i-1))c_i = S1c1 + (S2-S1)c2 +....+(S_k-S_(k-1))c_k
               =S1(c1-c2)+...+S_(k-1)(c_(k-1)-c_k) + Skck
Now,we show that S_i  \geq 0 for all i=1,2,..,k (easy!)
We are done! :)

There is a book called 'Proofs from The Book' by Martin Aigner, Gunter M. Ziegler, a tribute to Paul Erdos' enthusiasm.


inequalities
function
logarithms
inequalities proposed

WLOG，
$a \ge b \ge c$
because $f(x)=x^x$ disincrease in $[0,\frac 1 e]$ and increase in $[\frac 1 e ,\infty)$
so one of $b^b \le \max \{a^a,c^c\}$
so we get $b^b(b-c)(b-a)+a^a(a-b)(a-c) \ge 0$ or $b^b(b-c)(b-a)+c^c(c-b)(c-a) \ge 0$
the other term is obvious $\ge 0$
I think it have been proved.
btw,it is really a nice inequality :)
with the menthod I used I think we can get.
$\sum_{cyc}(a-t)^k(a-b)(a-c) \ge 0$
where $k,t$ are real numbers.
I think it is also useful.

[quote="zhaobin"]WLOG，
with the menthod I used I think we can get.
$\sum_{cyc}(a-t)^k(a-b)(a-c) \ge 0$
where $k,t$ are real numbers.
I think it is also useful.[/quote]
Yes!
If $f(x)\geq0,$ where $x\geq0$ and
for $f$ there is not internal maximum when $x>0$, then
$\sum_{cyc}f(a)(a-b)(a-c)\geq0.$ This is new for me.

[quote="arqady"][quote="leepakhin"]This is Schur.[/quote]
I known only that if $a\geq0,b\geq0,c\geq0$ and $f$ is increase function in $[0,+\infty)$ and $f\geq0$ then
$f(a)(a-b)(a-c)+f(b)(b-a)(b-c)+f(c)(c-a)(c-b)\geq0.$ This is Schur! No?[/quote]
I meant the original one.
Your inequality is a generalization of Schur, and the proof is also the same as Schur's.

[quote="zhaobin"]with the menthod I used I think we can get.
$\sum_{cyc}(a-t)^k(a-b)(a-c) \ge 0$
where $k,t$ are real numbers.[/quote]

Hmm, if a - t, b - t and c - t are nonnegative, then this is simply the Schur inequality applied to the numbers a - t, b - t and c - t. However, if a - t, b - t and c - t can be negative, then this inequality is generally wrong...

[quote="arqady"][quote="leepakhin"]This is Schur.[/quote]
I known only that if $a\geq0,b\geq0,c\geq0$ and $f$ is increase function in $[0,+\infty)$ and $f\geq0$ then
$f(a)(a-b)(a-c)+f(b)(b-a)(b-c)+f(c)(c-a)(c-b)\geq0.$ This is Schur! No?[/quote]

Well, there are lots of things we are used to call Schur. In [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=3663#p3663]http://www.mathlinks.ro/Forum/viewtopic.php?t=1166 post #3[/url], Valentin noted that the above inequality

$f\left(a\right)\left(a-b\right)\left(a-c\right)+f\left(b\right)\left(b-c\right)\left(b-a\right)+f\left(c\right)\left(c-a\right)\left(c-b\right)\geq 0$

also holds if f is a positive-valued convex function on the interval containing the numbers a, b, c. Particularly, this generalizes your inequality

$a^a\left(a-b\right)\left(a-c\right)+b^b\left(b-c\right)\left(b-a\right)+c^c\left(c-a\right)\left(c-b\right)\geq 0$,

since the function $f\left(x\right)=x^x$ is convex on $\mathbb{R}^+$ (in fact, $f^{\prime\prime}\left(x\right)=x^{x-1}\left(x\ln x+1+x\ln^2 x\right)\geq 0$, since $\ln^2 x\geq 0$ (as squares are always $\geq 0$) and $x\ln x+1\geq 0$ (as $x\ln x\geq -1$, what follows from $\ln x\geq 1-\frac{1}{x}\geq -\frac{1}{x}$)).

  Darij

[quote="darij grinberg"][quote="zhaobin"]with the menthod I used I think we can get.
$\sum_{cyc}(a-t)^k(a-b)(a-c) \ge 0$
where $k,t$ are real numbers.[/quote]

Hmm, if a - t, b - t and c - t are nonnegative, then this is simply the Schur inequality applied to the numbers a - t, b - t and c - t. However, if a - t, b - t and c - t can be negative, then this inequality is generally wrong...

  Darij[/quote]
sorry,my careless :)

[quote="darij grinberg"][quote="zhaobin"]with the menthod I used I think we can get.
$\sum_{cyc}(a-t)^k(a-b)(a-c) \ge 0$
where $k,t$ are real numbers.[/quote]

Hmm, if a - t, b - t and c - t are nonnegative, then this is simply the Schur inequality applied to the numbers a - t, b - t and c - t. However, if a - t, b - t and c - t can be negative, then this inequality is generally wrong...[/quote]
zhaobin mean $k\in2\mathbb N$
[quote="darij grinberg"]since the function $f\left(x\right)=x^x$ is convex on $\mathbb{R}^+$[/quote]
Enough for $f$   there is not internal maximum when $x>0.$

[quote="arqady"][quote="darij grinberg"][quote="zhaobin"]with the menthod I used I think we can get.
$\sum_{cyc}(a-t)^k(a-b)(a-c) \ge 0$
where $k,t$ are real numbers.[/quote]

Hmm, if a - t, b - t and c - t are nonnegative, then this is simply the Schur inequality applied to the numbers a - t, b - t and c - t. However, if a - t, b - t and c - t can be negative, then this inequality is generally wrong...[/quote]
zhaobin mean $k\in2\mathbb N$[/quote]

Well, in this case, it's right for arbitrary reals a, b, c, t, and follows e. g. from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=271175#p271175]http://www.mathlinks.ro/Forum/viewtopic.php?t=41284 post #4[/url] Theorem 3 [b](b)[/b].

[quote="arqady"]Enough for $f$   there is not internal maximum when $x>0.$[/quote]

Cool! When I first read this assertion in your post #3, I ignored it since I believed it would require some additional condition like continuity. Now I see it doesn't, but it is really useful as it contains the monotonically increasing, the monotonically decreasing and the convex case of Schur as particular cases. Note that it looks like this assertion has already been posted on ML before, namely in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=313522#p313522]http://www.mathlinks.ro/Forum/viewtopic.php?t=49427 post #3[/url], but I am not 100% sure since that post seems to contain some typos. Anyway, I would rewrite this assertion using numbers rather than functions, so that it would look like the following:

[color=blue][b]Theorem 1.[/b] Let a, b, c be three reals and x, y, z three nonnegative reals such that $a\geq b\geq c$, and such that $x\geq y$ or $z\geq y$. Then,

$x\left(a-b\right)\left(a-c\right)+y\left(b-c\right)\left(b-a\right)+z\left(c-a\right)\left(c-b\right)\geq 0$.[/color]

The proof is the same as the proof of the weaker Lemma 1 from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=156212#p156212]http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9[/url]. (Yes, you know this all, I'm just writing it for future reference.)

  Darij

[quote="darij grinberg"][url=http://www.mathlinks.ro/Forum/viewtopic.php?p=156212#p156212]http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9[/url][/quote]
Thank you, Darij

What about the following generalisations of Vornicu-Schur?
$a \ge b \ge c>0$ and either $x \ge y \ge z\ge0$ or $0\le x \le y \le z$. For real (!) $n>0$, prove

(1)  $\sum_{cyc} x(a^2+bc)^n\ge \sum_{cyc} x(ab+ac)^n$

(2)  $\sum_{cyc} x(a^2+bc)^{-n}\le \sum_{cyc} x(ab+ac)^{-n}$.

(The cyclical sums mean of course that $x,y,z$ cycle simultaneously with $a,b,c$.)


Prove that a countable infinite graph is k-colorable if all its finite subgraphs are k-colorable.

It is the compactness principle...

Pierre.

$k$-color the graph on vertices $1$, then $1,2$, then $1,2,3$, and so on. There will be a color which $1$ receives infinitely often, so fix this color and give it to $1$. In the infinitely many cases in which $1$ receives the chosen color, there will be a color that $2$ receives infinitely many times, so give this color to $2$, and so on. We can thus find a color for every vertex, and the resulting graph will clearly be properly colored.

A compactness principle-type proof :).


algebra
polynomial
superior algebra
superior algebra solved

Suppose $a,b,c\in \mathbb{C}$. Let
\[
A = \left(
\begin{array}{ccc}
b & c & a\\
c & a & b\\
a & b & c\\
\end{array}
\right), \quad
B = \left(
\begin{array}{ccc}
c & a & b\\
a & b & c\\
b & c & a\\
\end{array}
\right), \quad
C = \left(
\begin{array}{ccc}
a & b & c\\
b & c & a\\
c & a & b\\
\end{array}
\right)
\]
(i) Prove that $A,B,C$ are similar.
(ii) If $BC=CB$, show that at least two eigenvalues of $A,B,C$ are zero.

Let
$u=\left(\begin{array}{ccc} 0&0&1 \\ 0&1&0\\ 1&0&0\end{array}\right)$
$v=\left(\begin{array}{ccc} 1&0&0 \\ 0&0&1\\ 0&1&0\end{array}\right)$
$w=\left(\begin{array}{ccc} 0&1&0 \\ 1&0&0\\ 0&0&1\end{array}\right)$
$e=\left(\begin{array}{ccc} 0&1&0 \\ 0&0&1\\ 1&0&0\end{array}\right)$
Then $e^2=\left(\begin{array}{ccc} 0&0&1 \\ 1&0&0\\ 0&1&0\end{array}\right)$

We have the following multiplication rules:
\begin{tabular}{|c|c|c|c|c|c|c|}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  * & 1 & e & e^2 & u & v & w \\
  \hline
  1 & 1 & e & e^2 & u & v & w \\
  e & e & e^2 & 1 & w & u & v \\
  e^2 & e^2 & 1 & e & v & w & u \\
  u & u & v & w & 1 & e & e^2 \\
  v & v & w & u & e^2 & 1 & e \\
  w & w & u & v & e & e^2 & 1 \\
  \hline

\end{tabular}
The element (i,j) in the table is the product between the element i and the element j (the element leading line i and element leading column j).

We have
$A=a\cdot u+b\cdot v+c\cdot w$
$B=b\cdot u+c\cdot v+a\cdot w$
$C=c\cdot u+a\cdot v+b\cdot w$.

Also
$e\cdot A\cdot e^2=C$
$e\cdot C\cdot e^2=B$.
These prove that A,B,C are similar.

$BC=CB\Rightarrow (bu+cv+aw)(cu+av+bw)=(cu+av+bw)(bu+cv+aw)$
This is equivalent to
$(bc+ca+ab)1+(ab+bc+ca)e+(a^2+b^2+c^2)e^2=(bc+ca+ab)1+(a^2+b^2+c^2)e+(ca+ab+bc)e^2$
This is equivalent to $a^2+b^2+c^2=ab+bc+ca$.

The characteristic polynomial of A is
$\det(X\cdot 1-A)=X^3-(a+b+c)X^2-(a^2+b^2+c^2-ab-bc-ca)X+(a^3+b^3+c^3-3abc)=(X-(a+b+c))(X^2-(a^2+b^2+c^2-ab-bc-ca))=X^2(X-(a+b+c))$.
This proves that A has 2 of its eigenvalues 0.


linear algebra
matrix
quadratics
combinatorics proposed
combinatorics

$n$ teams take part in a tournament, in which every two teams compete exactly once, and that no draws are possible. It is known that for any two teams, there exists another team which defeated both of the two teams. Find all $n$ for which this is possible.

I hope you have a "smarter" solution.

First notice that if it's possible for $n$, then it's possible for $n+1$: perform the construction for $n$, and add another team which everyone defeates, getting a construction for $n+1$. This means that all we need to find is the minimal $n$ for which it works. Then I showed it's impossible for $n=6$. This isn't hard at all, since the steps one has to make while trying to construct such a graph are pretty limited, but I won't write it here, since it's boring. After that, I just sort of "stumbled upon" a construction for $7$.

I'll just write the adjacency matrix of a tournament with $n=7$ here: a $1$ in position $(i,j)$ means that $i$ defeated $j$.

\[\begin{pmatrix}0&1&0&0&0&1&1\\0&0&0&1&1&0&1\\1&1&0&0&1&0&0\\1&0&1&0&0&0&1\\1&0&0&1&0&1&0\\0&1&1&1&0&0&0\\0&0&1&0&1&1&0\end{pmatrix}\]

This problem was also posted in a topic called [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23065]smallest number[/url].  There we put the n = 7 example in a broader context (quadratic residues).

actually for $n$ odd, $n\geq 7$ there is a simple(elementary) description for such a graph.
label the vertices by $0,1,\ldots,2n$. i shall just write down (some of) the vertices which correspond to defeats alone(i.e. when I write $j$ along the row labelled $i$ i mean that $i$ defeated $j$):

  $0:1,2,\ldots,2n;\\
  1:n,n+2,n+3,\ldots,2n;\text{(n+1 missing)}\\
  2:1,n+1,n+3,\ldots,2n;\text{(n+2 missing)}\\ \\
  n:n-1,n+1,n+2,\ldots,2n-1;\text{(2n missing)}\\ \\
  n+1: 0,1,n+2\\
  n+2: 0,2,n+3\\ \\
  2n: 0,n,n+1$

to explain what is written here, for instance the first row simply means that $0$ defeats $1,2,\ldots,n$ and so on.
 note that this is a proper description of a partial tournament because if $j$ lies on row labelled $i$ then $i$ does not lie on row $j$.

of course this is not the description of a tournament completely but what we need is the following:
    basically notice that  $n+1$ defeats $1$ but all the higher numbered players lose to $1$. similarly $n+2$ defeats $2$ but all the other players among $n+1,\ldots,2n$ lose to $2$, and so on.
once this is the situation, notice that in this partial tournament(i.e at this stage of the tournament), for any unordered pair, they are already in some row. to see this easily notice that any pair of the form $(0,k)$ occurs amonst the last $n$ rows. now any pair $(i,j)$ with both $i,j\in\{1,2,\ldots,n\}$ occurs from the first row.
now if $i\leq n,j>n$ then the pair $(i,j)$ occurs amongst the first $n$ rows unless the pairs are one of $(1,n+2),(2,n+3),\ldots,(n-1,2n),(n,n+1)$ and these occur amongst the last $n$ rows.
      finally if both $i,j>n$ then they occur(actually they occur $n-1$ times) amongst the first $n$ rows and so that completes our requirements.
as remarked before, what remains to be played are some of those games amongst $1,2\ldots,n$ and some amongst $n+1,n+2,\ldots,2n$.but their results are irrelevant.
grobber's solution for $n=7$ is also obtained from this.in that case since such a diagram describes $3$ losses for each player, it turns out that the above described tournament in that case is complete.


If Anna flips 8 coins, what is the probability that she gets more heads than tails?

 :huh:

[hide="Solution with casework"]She can get the following:

5 heads, 3 tails
6 heads, 2 tails
7 heads, 1 tail
8 heads, 0 tails

Let's work this problem with casework:

Case 1: 5 heads, 3 tails
There are C(8,5) combinations of the 8 flips that can be heads/tails. Probability of this case happening=$ C(8,5)*(\frac {1}{2})^8 \equal{} \frac {56}{256} \equal{} \frac {7}{32}$.

Case 2: 6 heads, 2 tails
There are C(8,2) combinations of the 8 flips that can be heads/tails.
Probability=$ C(8,2)*(\frac {1}{2})^8 \equal{} \frac {28}{256} \equal{} \frac {7}{64}$.

Case 3: 7 heads, 1 tail
C(8,1) combinations of flips that can be heads/tails.
Probability=$ C(8,1)*(\frac {1}{2})^8 \equal{} \frac {8}{256} \equal{} \frac {1}{32}$.

Case 4: all 8 heads
This is simply $ (\frac {1}{2})^8 \equal{} \frac {1}{256}$.

So the probabilities add up together. $ \frac {7}{32} \plus{} \frac {7}{64} \plus{} \frac {1}{32} \plus{} \frac {1}{256} \equal{} \boxed{\frac {93}{256}}$.[/hide]

There is 8C4=70 ways to get 4 heads. So there is a $ \frac{70}{2^8}\equal{}\frac{70}{256}\equal{}\frac{35}{128}$ chance that there is 4 heads.
The probability of getting more heads than tails is the same as getting more tails than heads.
Call that probability $ p$.
$ p\plus{}p\plus{}\frac{35}{128}\equal{}1\\
2p\equal{}\frac{93}{128}\\
p\equal{}\frac{93}{256}$

The probability that Anna flips 4 heads and 4 tails is clearly $ \binom{8}{4}\left( \frac {1}{2} \right)^8 \equal{} \frac {35}{128}$. Let the probability that she flips more heads than tails be $ P$. We then have, by symmetry,
\[ 2P \plus{} \frac {35}{128} \equal{} 1
\]
so
\[ P \equal{} \boxed{\frac {93}{256}}.
\]

EDIT: beaten, sorry... :blush:

[quote=First]This is a routine problem, use $\dbinom{8}{5}$ + $\dbinom{8}{6}$ ... $\dbinom{8}{8}$ and divide by 256[/quote]

Hi, there is a much better solution than thus: simply subtract the cases where they both get 4 heads/tails from 128 and put it over 256 by symmetry.

(also please don't use words like "routine problem" to make yourself look good)


[url]http://news.nationalgeographic.com/news/2002/11/1120_021120_GeoRoperSurvey.html[/url]

Edit :Poor geographical knowledge [b]in the USA [/b] has been a source of humor for a long time.

When I visited the USA, I was quite surprised to see that many Americans did know about Belgium, many of them had actually even been there.
So I thought my prejudices were somewhat not justified.
However, this recent study by national geography, conducted in the USA, and several other nations, shows a dramatic lack of knowledge.

11% of young US citizens (18-24) can't find USA on a world map.
69% can't find UK (the crucial ally in the war on terror) and 65% can't find France (which I think is even worse after all the criticism for the criticism on the War on Terror, remember Freedom Fries?, I think judging a country starts with knowing where it is :huh: )

Which is worse, young people in other nations seem better at estimating US population than US youngsters themselves, some even go to 2 billion.

All the 'hot spots' for USA : Iraq, Iran, Afghanistan, Israel, the young people in the USA can't find them.

I would like to hear your thoughts (US-ers and non US-ers) on this.  Is this study being overdramatic?
There are a lot of things I don't get in that article.  First of all 'students taking geography'  :? .  At my school I simply had to do six years of geography, everyone had to.  We spent a full semester on just CIS/USSR.  We had to memorise all Soviet nations and capitals, and all USA states.  I didn't go to any special school though.

I think if this is true, it is definitely a bad thing.  The almost uncontested superpower, a democratic country, has a population that mostly doesn't know about the rest of the world.  Is this a cultural thing, leading to lack of interest?  Or is it deliberate : some sources I found on the internet claim it is part of a political agenda, to not inform people too much.
And who is to blame then?  Teachers?  Parents?  Or the media perhaps.  Whenever the news is about something foreign on the Flemish Public Broadcasting net, a globe is shown, zooming  out of Belgium and zooming in onto that particular country.
The article however doesn't only point the finger at Americans, many people didn't seem to know a lot about nuclear nations, and many French didn't know their own nation was a nuclear nation.

Please don't see this post as an insult.  I am just saying what that article says and requesting opinions.   :roll:

I think this article will sum up my opinion and prove yours:

http://www.msnbc.msn.com/id/12591413/

Stuff like this makes me pray for humanity  :blush:

I have read a book which said that map is the most important book one should read. It can help you learn about the world and let you set ambitions. I personaly love the maps. I think it is a great pleasure just sitting in front of a world map and find countries from every cornors of the earth......

I don't know why there's a lack of geoknowledge among US. But I know that geography is taught in the USA and I think it is the teacher's duty to get the students to become interested in geography.

The link also has a "test yourself" link -- you can try to answer the questions that were given on the survey.  I got 19 out of 20 -- I misguessed what percentage of US population growth is due to immigration.

Interesting, in my link you mean or in yellowmarker's link?
Haven't noticed yet?

I'm pretty sure that if a survey like that was taken in Romania, the results would be very similar (if not the same). I guess most people are not concerned about the whereabouts of another country, unless they have a good education (thus they learn it and remember it from school) or just have been there with work or something like that. Most people here cannot point on the map some of the major counties of Romania, let alone find Israel on the map :)

So i did that test and I got 7/10, I was hoping for more

I got all non US questions right though ;) .

[hide]I thought the Mississippi began in Wisconsin, that the gold rush began in Colorado, and that Pennsylvania was larger than Georgia.[/hide]

Valentin, are you saying that what this study reveals is actually the case everywhere?

Thank you JBL.

How much did people here get at this quiz?
I found those 20 questions still relatively doable (it could have been questions like : what is the top opium exporter, what countries has the highest illiteracy, when did the capital of Tanzania change...)

I like geography quizzes.  All links are welcome in my opinion.

[url]http://www.sheppardsoftware.com/geography.htm[/url] is a good site, especially to learn US States and Capitals

I am particularly looking for a game to learn Chinese provinces, and for a game that tests knowledge of borders around the world (when I ask myself : what countries border Laos, I am sometimes in doubt)

I'm really surprised at this study...

At least now I feel better about not knowing that El Salvador was the smallest country in Central America... :D

[quote="JBL"]Fredbel, I was refering to your link, where there are the actual 20 questions used in the survey.  Right above the start of the article, there is a link which says "Test Yourself" or some such thing.  The quiz you took is from Yellowmarker's post and is a different set of questions.

The URL of the thing I'm talking about is [url=http://geosurvey.nationalgeographic.com/geosurvey/templates/question_1.html]here.[/url][/quote]
I saw stupid question like
"Which of the following organizations endorses the euro as the common currency for its members?"
I call it stupid because I don't think anyone will make mistake here, even if he know nothing about this.

Well uhm, actually that question WAS difficult as it appeared that they meant : of which organisation do the members use this coin, even though the Euro isn't even used in all 15 members BEFORE the 2004 expansion (Denmark, Sweden and UK didn't join in)

I seriously wonder who those few percentages in countries like Sweden were, who didn't know that?

people can simply judge by the key word [b]euro[/b] vs [b]euro[/b]pean union.
Of course, there's a chance to make mistake by using this method, for example, Russian speak Russian, but American do not speak a language called American. But if you know something, you know the answer, otherwise this is the best way to guess.

[quote="Valentin Vornicu"][quote="fredbel6"]Valentin, are you saying that what this study reveals is actually the case everywhere?[/quote]Yup, there's absolutely nothing special about US in that reference I think.[/quote]

While one may debate whether or not lack of geo knowledge/interest is worldwide, the test does demonstrate the US still excels at that.  USA is one of the poorest scorers on questions like its own population or where the Taliban lived.
(And on the other hand, UK is one of the poorest scorers when it comes to finding Argentina  :? , well I dom remember an urban legend about people panicking when Falkland war began, confusing it with Shetlands and stuff, but I thought that was just made up  :huh: )

I see two possible factors here:

One - you have to wonder whether the American results are entirely due to lack of knowledge or possibly to a degree of apathy from Americans towards surveys like these. It may be possible that Americans [i]know[/i] the answers to more of these questions, but that their degree of apathy is so large that they just don't care about putting forward an effort to answer them correctly. I always wonder whether there's any attempt to control for that potential discrepency in these type of surveys.

You also have to remember that it wasn't long ago when America was an isolated country with a very low literacy rate. America has always had a "working man's" culture rather than an intellectual culture. In fact, the primary reason for establishing and mandating public education in the first place was to increase the productivity of American "blue collar" workers -  especially on assembly lines. I believe a lot of young American's still carry this attitude towards education begotten from their parents. History has momentum.

Thank you for your view.

1) Yes, how did they conduct this test.  At the entrance of a supermarket, or exit, or at the exit of a cinema, where people where irked and wanted to leave?  Maybe some people fearing a poor score just boycotted the test all at once.

2) I really did not know that.  I thought it would have been caused mainly by a not very demanding school system (our schools are not that free, we had 33 classes of 50 minutes a week) and by a feeling that USA is the only succeeding country, and the rest of the world is just poorer, more dangerous,...

I had expected that, even though USers aren't in my opinion among the best educated in general, they would still think highly of education, and look down on 'poor ass countries'( as they would call them) like Somalia for its deplorable condition (well there are more problems in Somalia than illiteracy).
I think in USA and in Europe most jobs have become more intellectual (requiring less manual labour), so the situation should be quite similar.  The generation of my grandfather didn't use to go school either until eighteen, they started working at fourteen.

[quote] I misguessed what percentage of US population growth is due to immigration.[/quote]
Don't know which test this question is from,  the link you gave,  did not have this  Q and ALL the questions were ridicously   easy.

[quote]One - you have to wonder whether the American results are entirely due to lack of knowledge or possibly to a degree of apathy from Americans towards surveys like these. It may be possible that Americans know the answers to more of these questions, but that their degree of apathy is so large that they just don't care about putting forward an effort to answer them correctly. I always wonder whether there's any attempt to control for that potential discrepency in these type of surveys. [/quote]
FWIW  I think there is  a real possibility that a significant percentage, as you say,  just didn't care for,  or went out of their way to do "you ask me a silly question you get a  sillier response".  I would be tempted to point to some funny place if some one is silly enough to ask  me  "point out USA on this map"  type   question.

I got the "most widely practised religion" question wrong, hehe.  I recalled seeing a graph that put Islam on top, but then I remembered it was because they seperated Catholics and Protestants.

Edit:  that test was ridiculously easy, for the spot the country questions they could at least have made it a little more difficult, like the Sweden question, the choices werent something close like Norway or Finland, they were all over the place.

Hmm when president Bush was campaigning in 2000, i heard there was a reporter in his cloud of journalists following him , that dared to ask him questions like :'What is the capital of Taiwan?!', and I heard he got punished for that by being ignored most of the time.  ..... He does know the answer I hope?? :?

Edit : actually that question is not that easy, as for a long time the ROC claimed Nanjing (mainland China) to be their capital

I'd be tempted to say that Canada might do even worse on a survey like that.  I've never had to take a geography class in 12 years of schooling.  However, its kinda mixed in with Social/ History class, its kind of assumed that if you're learning about all these countries, you would know where they are on a map.

But...for instance, I know nothing about the USA.  I can barely name states.  I know, its sad, but it was never considered a priority for me.  However, I don't know anyone here who wouldn't be able to find the UK or France on a map.

Well in our schools they don't really teach you where France and UK as they both are almost neighbouring countries :D , but I guess those countries are priority in Canada because of historical ties?

I am still convinced that discipline is the main issue.  I've never chosen any geography class, in fact in Belgium, every two years you choose in what direction you wanna continue (and your options depend on what you did before)
I chose Science-Mathematics.  I simply HAD to take six years of geography, simply had to learn all US states, all Soviet capitals (European capitals are compulsory in 8th grade)

I'm sure Canadian and USA students are smart enough to do that if we could do it, you just have to be willing to do it I guess.

Heh, 8/10 on the first quiz and 20/20 on the second. I can't believe I missed those 2.
I actually have heard about this survey before, it's amazing how much GMA can tell you. And I am not that surprised. The American schooling system is not that good, I have heard of parents teaching kids things in a week that took a couple months in school. I have seen other documentaries on this topic, and they showed many American kids could not tell what the first Amendment was, much less what exactly the Bill of Rights are. Most of them could not tell what the cause of the Civil War was. Also, America's kids are ranked as the 25th smartest in the world. I believe Finland(?) has the smartest. Or was it Belguim? I'm not sure though. I saw one lady on this documentary, where the Kentucky Derby is held, answered "Kansas"  :rotfl:  :rotfl:


1, 11, 111, 1111, 11111, 111111,.....

$U_{n}=\frac{10^{n}-1}{9}$ where does the -1 come from in this formula ?

the same in this sequence
 5, 7, 11, 19, ......

$U_{n}=\frac{2^{n}+3}{1}=2^{n}+3$

where does the +3 come from in this formula ?

i know where other things come from in those formula except those.

[quote="binomial_4eva"]thnx but this doesnt answer my question does it ?[/quote]

How about this?

$1+10+10^{2}+10^{3}+\cdots+10^{n-1}=\frac{1-10^{n}}{1-10}.$

Yeah what he's doing is using the formula for the sum of a finite geometric sequence.

Easily provable by dividing out the first term, then muliplying the sequence by r, subtracting, etc.

Okay.  Let me give this a stab.   :)

$10^{n}$ is a number with $n$ zeros after it.

For example: $10^{3}=1000$

Then we subtract one to get a number with only $9$'s in it.

For example: $1000-1=999$

Then we divide by $9$ to recieve the series that you mentioned.

For example: $\frac{999}{9}=111$

Hope that helps for the first one.  Now let's try the second one...

If you start to list the values in the sequence out: the first term is 5; the second term is 7; the third term is 11; the fourth term is 19; etc.

We can generalize this to $2^{n}+3$ Like you had for the nth term.

The +3 comes into play, because $2^{n}$ is not the number of the nth term, you have to add $3$ first.

I'm not sure if I understood you or if that helped at all, but hopefull it did  :D

[quote]The +3 comes into play, because  $2^{n}$ is not the number of the nth term, you have to add 3 first.[/quote]

so if we think that way then three comes from

first term=a=5

$a-2^{1}$=3 which is true bcoz 5-2=3 i agree. so is it  true for the other sequence?

lets try  a=1
   $a-10^{1}=1-10=-9$ but we have -1 not -9

you can see its not true for the other.

[quote="kunny"][quote="rnwang2"]The second sequence can also be described as:

$\frac{1}{3}\, n^{3}-n^{2}+\frac{8}{3}\, n+3$  :wink:[/quote]
Tha'ts wrong. :(
[/quote]

Well, there could be anything beyond that ellipsis. So, you could use the Lagrange interpolating polynomial to make it any polynomial you want.  :P

[quote="kunny"][quote="cincodemayo5590"][quote="kunny"]I can't understand what you said, but I do know Lagrange's interpolation. :D[/quote]

We are only given a few terms of the sequence. So, we can fill in whatever we want for the rest, and find a polynomial that fits!  :D[/quote]

But isn't $a_{n+1}-a_{n}$ a geometric sequence?[/quote]

The intended solution is that it is geometric. But many argue that we could fill in anything after the last term, because they do not give us these terms explicitly. [b]Rnwang2[/b] did exactly this. (Usually, it is a joke.)   :D

how did u get this
$=5+\sum_{k=1}^{n-1}2^{k}=5+2\cdot\frac{1-2^{n-1}}{1-2}=2^{n}+3$ $(n\geq1)$

xplain  more plz, i didnt get it :?:

kunny how did you get $2^{k}$
 isnt it $2(2)^{k}$

can somebody help? please


Well the world series is here!!!

Who do you think will win?

I don't know, but I was watching the World Series last night and I proved that it was absolutely convergent. :P

But seriously...

[quote="math92"]Go tigers![/quote]
[quote="kheg_k"]Go Tigers!!!!!!!!!!!!!!!!!!!!!!!!!![/quote]
[quote="mustafa"]Go Tigers!!!!!!!!![/quote]
:o :mad: :dry:

Wow. Justin Verlander gave up 7 runs..... :dry:

I am cursed!

To all those people who want Detroit to win: Cardinals are 1-0. :P

pick before game 2's game-Tigers will win game 2. Cardinals will win the series.

But still,...

[b][color=red][size=150]GO TIGERS!!![/size][/color][/b]

[quote="mustafa"]Ok, this time, I will pick game by game, and see how badly I do then.....

[hide="Game 1"]Tigers win

Already got that one wrong, but that is what I thought would happen :rotfl:  0-7
[/hide]
[hide="Game 2"]Tigers win
[/hide]
[hide="Game 3"]Cardinals win
[/hide]
[hide="Game 4"]Cardinals win
[/hide]
[hide="Game 5"]Tigers win
[/hide]
[hide="Game 6"]Tigers win[/hide]
[hide=" Game 7"]Tigers win it all!!!!![/hide]

edit because I can't post again for 20 minutes: Wow. Kenny Rogers looked like he might have had pine tar on his hand illegally. They pointed it out on tv, and then he washed his hand :rotfl:[/quote]
Assuming your predictions are wrong, Cardinals win!

[quote="mathgeniuse^ln(x)"]Go Tigers

Kenny Rogers is gonna pwn.[/quote]
He did. :(

[quote="mustafa"]Looks like I might have a correct pick! Tigers up 3-0 in the bottom of the 7th![/quote]
Congratulations, you are now 1-7.

[hide] LETS GO REDBIRDS![/hide]
[hide] LETS GO REDBIRDS![/hide]
[hide] LETS GO REDBIRDS![/hide]
[hide] LETS GO REDBIRDS![/hide]

The cardinals are going all the way, baby!

I think tigers will win. :)
They will for sure win the fourth gameand the series will end up in game 7 where tigers will win.

[color=red][size=200]CARDINALS WIN!!!!!!!!!!!!!!!!!!!![/size][/color]

I predicted after game one that the Tigers would win game 2 and the Cardinals would win the series. I was right! :lol:  :D

even though I wanted the Tigers to win.  :o  :(  :mad:

[b][/b][b][/b][b][/b][i][b][/b][/i][b][/b][b][/b][b][/b][i][b][/b][/i][b][/b][b][/b][b][/b][i][b][/b][/i][b][/b][b][/b][b][/b][i][b][/b][/i]

[quote="mustafa"]Some might call me......0-16![/quote]
I won't, but actually, you're 0-17. :o

And I remember the [i]last[/i] time Detroit and St. Louis met in the World Series - in 1968. I suppose you guys have heard of Skip Caray and Joe Buck - the Cardinals broadcast team in the mid-60's was Harry Caray and Jack Buck. (They're both dead now.)

In 2006, game 4, Curtis Granderson, who I think is a good defensive centerfielder, fell down chasing a fly ball. In 1968, game 7, Curt Flood, who was certainly an excellent defensive centerfielder, dropped [i]two[/i] fly balls, leading to Detroit winning the game and the series.

Detroit had Denny McLain, who won 30 games as a pitcher - no one has done that since. But McLain (who would later turn out to be a crook) couldn't beat Bob Gibson in the World Series. 1968 was the "year of the pitcher", with the lowest scoring averge in baseball since before 1920, but even in that context, Gibson had an amazing year. In June and July, he pitched 11 consecutive complete games and allowed a total of 3 runs - 8 shutouts, three games in which he allowed 1 run each. He had 13 shutouts. His record was an unlucky 22-9, but his ERA was 1.12.

But Gibson couldn't win Game 7 of the World Series; that honor went to Mickey Lolich, who was the winning pitcher three times in the series.

I've noticed that most of the Cardinals' votes came [b]after[/b] the world series ended.

[quote="futuremontrealian"]Mwuahaha the Cardinals won!  I'm from Missouri, so I'm supposed to be excited, but I'm not.  I'm a redsox fan anyway, but hey!  at least Missouri is famous for something (besides its "karst topography").[/quote]

Really. Where in Missouri?

I'm from Missouri and I'm glad the Cardinals won too. But I used to live in Michigan, so I guess I have mixed loyalties...


geometry
parallelogram
projective geometry
geometry solved

ABCDE is a convex pentagon. F, G, H, I, J are the intersections of BC and DE, CD and EA, DE and AB, EA and BC, AB and CD. Show that the intersection of BD and CE lies on AF iff the intersection of GH and IJ does.

Okay, since this is probably from a projective geometry problem set, I'm solving it using projective geometry:

By appropriate projection, or, equivalently, by a composition of two polar transformations with respect to circles (this is for people like me who don't like $\mathbb{R}^3$ :D ), we can map the line AF to the line at infinity.

Now, of course, we have:

{The point of intersection of the lines BD and CE lies on the line AF}
$\Longleftrightarrow$ {The point of intersection of the lines BD and CE lies on the line at infinity}
$\Longleftrightarrow$ {The lines BD and CE are parallel}.

But since the lines BC and DE are parallel (in fact, they intersect at the point F, which lies on the line at infinity), if the lines BD and CE are parallel, the quadrilateral BCED must be a parallelogram, and thus BC = DE (using directed segments on the parallel lines BC and DE), and conversely, if BC = DE, the quadrilateral BCED is a parallelogram, so that the lines BD and CE are parallel. Hence, we have

{The lines BD and CE are parallel}
$\Longleftrightarrow$ {We have BC = DE}.

Now, remember that the lines BC and DE are parallel. In other words, the lines BI and EH are parallel. On the other hand, the lines BH and EI are parallel (since they intersect at the point A, which lies on the line at infinity). Hence, the quadrilateral BIEH is a parallelogram. This yields BI = HE. Since BC = BI + IC and DE = DH + HE, we thus have BC = DE if and only if IC = DH. In other words:

{We have BC = DE}
$\Longleftrightarrow$ {We have IC = DH}.

Now, think about the two triangles CIG and DHJ. Their respective sidelines are parallel (in fact, at first, CI || DH follows from BC || DE; then, IG || HJ follows from BH || EI; finally, GC || JD is trivial). Hence, these two triangles are homothetic, and we can conclude that $\frac{IC}{IG}=\frac{HD}{HJ}$ (using directed segments on the parallel lines BH and EI). In other words, $\frac{IC}{IG}=\frac{DH}{JH}$. Hence:

{We have IC = DH}
$\Longleftrightarrow$ {We have IG = JH}.

Now, the lines IG and JH are parallel (this is just a rephrase of BH || EI). Hence, if we have IG = JH, then the quadrilateral IGHJ is a parallelogram, and thus the lines GH and IJ are parallel, and conversely, if the lines GH and IJ are parallel, then the quadrilateral IGHJ is a parallelogram, what yields IG = JH. Hence,

{We have IG = JH}
$\Longleftrightarrow$ {The lines GH and IJ are parallel}.

Now we combine everything:

{The point of intersection of the lines BD and CE lies on the line AF}
$\Longleftrightarrow$ {The lines BD and CE are parallel}
$\Longleftrightarrow$ {We have BC = DE}
$\Longleftrightarrow$ {We have IC = DH}
$\Longleftrightarrow$ {We have IG = JH}
$\Longleftrightarrow$ {The lines GH and IJ are parallel}
$\Longleftrightarrow$ {The point of intersection of the lines GH and IJ lies on the line at infinity}
$\Longleftrightarrow$ {The point of intersection of the lines GH and IJ lies on the line AF}.

And this pretty much solves the problem, doesn't it? ;)

  Darij


group theory
abstract algebra
vector
advanced fields
advanced fields theorems

hi all,
Perhaps my doubt stems from some misunderstanding,anyway i am not able to clear my doubt myself,so i would appreciate very much help from u people,here goes my question:
In the notes he difines something called $min(L)$ associated to a Euclidean lattice$L$(discrete subgroup of a eucledian space,withnormal inner product),Now my question is is'nt this value depended on the basis of the lattice,also wo'nt the vectors on which min is attained change with change in basis?.
thanking u all
Rohith

The representation of the minimal vectors will change. The vectors themselves and their lengths won't, since the norm comes from the general Euclidean norm.

thanks,i actually resolved my doubt myself after sometime,then i tried to delete the post bubt i do'nt know why it did'nt get deleted,anyway i am quite stupid sometimes,i forget everything  :blush: ,thank u
Rohith


$a$ and $b$ are positive integers. Prove that,
\[\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{9a^2+b^2}}+\frac{2ab}{\sqrt{a^2+b^2}\sqrt{9a^2+b^2}} \leq \frac{3}{2}\]


abstract algebra
vector
number theory
relatively prime
superior algebra
superior algebra unsolved

A is a ring s.t. 1<>0. It has 2n-1 invertible elements and fewer non-invertible elements. Show that it's a field.

Lemma: Suppose $R$ is a finite ring with an odd number of invertible elements. Then $R$ has characteristic 2, so $x+x=0$ for all $x \in R$.

Proof: Since the additive group of $R$ is finite, there is some minimal positive integer $N$ for which $Nx=0$ (defined by repeated addition) for all $x \in R$. We can also interpret the integers as elements of the ring, and this fact implies that $N=0$ and $k \neq 0$ for all $k<N$. The integers less than $N$ and relatively prime to $N$ are invertible mod $N$, so they are distinct invertible elements in $R$ (since $kx-mx =(k-m)x \neq 0$ for some $x$). We now partition the invertible elements by whether they are integer multiples of each other; each class has $\Phi(N)$ members, so the total number of invertible elements is divisible by $\Phi(N)$. Since $\Phi(N)$ is even for all $N>2$, $N=2$.

Now we know that $A$ has characteristic $2$, making $A$ a module over the integers mod 2. Since the ring of integers mod 2 is also a field, $A$ is in fact (using only the additive structure) a vector space over that field, so the number of elements is a power of $2$. Since there are more elements invertible than not, there must be exactly $2^n$ elements. This means that $0$ is the only noninvertible element, so $A$ is a division ring. By a standard theorem, all finite division rings are fields, and we are done.

I guess it's Ok, but the fact that $0$ is the only non-invertible element characterizes fields, doesn't it? Why do you need to mention division rings? :?

A division ring, by definition, satisfies all the field axioms except commutativity of multiplication (standard example: the quaternions). This includes that all nonzero elements are invertible.

I wasn't assuming that $A$ had commutative multiplication; maybe you were.

Right.. english terminology confused me. In Romanian, we have "corp" (which means "division ring") and "corp comutativ" (guess what this means :)). When I started browsing forums in which English is mandatory, I began to equate "field" with "corp", even though I knew it wasn't correct. I guess it's just a bad habit :). I didn't assume commutativity either.

Sorry for the useless (and pretty stupid :)) remark from my previous post.


geometry
3D geometry
tetrahedron
inradius
sphere
analytic geometry
circumcircle

How do you find the inradius of a tetrahedron, given the side lenghts or the volume or anything else that you need?

$r = 3V/A$ where $V$ is volume, $A$ is surface area, and $r$ is inradius.

Reason: if $A_{i}$ is the area of face $i$, then $V = \sum rA_{i}/3 = r/3 \sum A_{i}= rA/3$

What about the circumscribed sphere? And how are these formulae derived?

I have no clue about the circumradius.

unless $V = abcdef/12R^{3}$ or something random like that... (note: that formula is almost certainly way off)

Edit: and I just showed you my derivation for the insphere one. Just calculate the volume using the inradius! (i.e. divide the inside of the tetrahedron into 4 parts.)

[quote="chickendude"]I know that if you have 4 coordinate points for a tetrahedron, you can average each of the values (x, y, and z) to find the center of the circumsphere

that can be used to calculate the circumradius, but this only works in a coordinate situation[/quote]
i think that gives centroid. or does centroid = circumcenter??

[quote="zeb"]NO

[size=59]The message is too small. Please make the message longer before submitting.[/size][/quote]
is this no to me or to chicken's method?

definitely you, not sure if also chicken's thing
but how you would find the circumradius is drop an altitude and from the circumcenter draw a point then let the radius be x and you can use pythag
too tired to explain right now

but for a triangle, i know average x-coordinates will give x-coordinate of centroid and average of y-coordinates will give y-coordinate of centroid. so i thought this might extend to 3-D stuff too

either you're mistaken or you're misunderstanding

http://www.pballew.net/centroid.html

Averaging the coordinates of the vertices gives the centroid (center of mass).  This will not necessarily (in fact, will usually not) be the circumcenter.  (Indeed, the circumcenter can lie outside the tetrahedron.)  I suspect a formula for the circumradius will be difficult to derive.  One possible approach is as follows: choose one face, choose its circumcenter, and draw a line perpendicular to the face through that point.  The circumcenter of the tetrahedron will lie on this line.  So, for example, we could find two such lines and calculate their intersection, or, we could try and figure out how far along this line we need to go to make the fourth distance equal to those three.

WHAT A COINCIDENCE! I JUST FOUND THE FORMULA FOR THE CIRCUMRADIUS TODAY!!!

$6hT = j$, where $h$ is the circumradius, $T$ is the area of the tetrahedron, and $j$ is the area of the triangle with sides of length $ap, bq, cr$! (and no I don't mean factorial I mean I'm EXCITED!!!!!!! also $a, b, c, p, q, r$ are the lengths of $BC, AC, AB, DA, DB, DC$)

but the proof was longish and involved linear algebra and five by five determinants.


function
number theory unsolved
number theory

Find all functions like $f$ such that:
$f:\mathbb{N} \longrightarrow \mathbb{N}$
$f(f(n))=n+2k+1$ where $k \in \mathbb{N}$ and constant
P.S: i belive there is no one but as usuall i dont know how to prove  :blush:  :blush:

Have a look here :
http://www.kalva.demon.co.uk/imo/isoln/isoln874.html

The only thing used is that $1987$ is odd...

Pierre.


calculus
integration
function
calculus computations

Have anyone any idea about this exercise:

Find the function f where f(x) is

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

hello
$ y \equal{} \int \frac {(x \minus{} 1)^2}{x^2 \plus{} 1}y \minus{} C$
after differentiating and rearranging
$ \frac {dy}{y} \equal{} \frac {(x \minus{} 1)^2}{x^2 \plus{} 1}dx \equal{}[ 1 \minus{} \frac {2x}{x^2 \plus{} 1}]dx$
integrating the equation and get $ y$ or $ f(x)$
thank u

[quote="kabi"]hello
$ y \equal{} \int \frac {(x \minus{} 1)^2}{x^2 \plus{} 1}y \minus{} C$
after differentiating and rearranging
$ \frac {dy}{y} \equal{} \frac {(x \minus{} 1)^2}{x^2 \plus{} 1}dx \equal{} [ 1 \minus{} \frac {2x}{x^2 \plus{} 1}]dx$
integrating the equation and get $ y$ or $ f(x)$
thank u[/quote]

Thank you very much Kabi!!


Find the sum of the roots of \[2^{333z-2}+2^{111z+1}=2^{222z+2}+1,\] where $z\in\mathbf{C}$.

(Sorry, I don't know the source.)

That's from this year's AIME, #7 if I remember correctly.

[hide]

Let

Let's rearrange  to  or

The sums of the roots would therefore be

That's the product of the roots of , which is equal to . Plugging it back in, we see that

Therefore, our answer is [/hide]

AIME 2005I, number 8
[url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=50]Math Jam solution[/url]

For real z, at least. I don't think there are complex solutions.

It upsets me to hear that was from the AIME.. that must have been where he (the guy who told me it today) got it from. I was really out of it when I took the AMC this year and didn't even make the AIME  :blush: even though I easily made it last year. At least I was able to get $\frac{2}{111}$ for this without too much difficulty.

[hide="hint for those who havent done or who couldnt do it"]

look at the exponents, and the coeffieints of x. 111, 222, 333...doesnt that ring the bell for ya?
[/hide]

Expansion on reddevil's hint:[hide]Substitution.[/hide]


Find all real $ x$ that satisfy

\[ \minus{}4x\plus{}7\plus{}3x^2<2x^2\minus{}2x\plus{}7\leq 4\plus{}x^2\plus{}2x.\]

[hide]My solution:
The 1st inequality gives us the interval: $ 0 < x < 2$
The 2nd inequality gives us the interval:$ 1\leq x \leq 3$
Finding the common interval in those 2 inequalities, the final answer is: $ 1\leq x<2$
Btw, I don't see the point of posting it in this box. It should be moved to "High School Basics".

EDIT:Thx for your comment, resurrection. I fixed it![/hide]

Minor error: 2nd inequality is less than or equal to, but it doesn't affect the overall answer.

[quote="resurrection"]it doesn't affect the overall answer.[/quote]
Actually, it does affect the overall result by very little(add 1 to the lower bound) :wink:

oh whoops right... my quick glances fail.


I appear to have misplaced my supply list.
Could someone please scan or type up their sheet and post it?

pillow
twin sheet set
light blanket

soap
toothbrush
toothpaste
shampoo/conditioner
brush/comb
deodorant
sunscreen
beach towel
bath towel
washcloth
tissues

cash (200 is plenty)
stationery/stamps
books
camera/film
umbrella
calling card
sunglasses
waterbottle
athletic gear


geometry
perimeter
rectangle
combinatorics unsolved
combinatorics

A house has the shape of a triangle of perimeter $ P$ meters and area $ A$ square meters. The garden consists of all the land within $ 5$ meters of the house. How much land do the garden and house together occupy?

Draw rectangles off each edge of the house with the house as one side, 5 meters in length. Let x,y,z be the sides of the house. So the area of these rectangles is 5x+5y+5z=5P. The rest of the garden is circular sectors centered at each of the vertices of the house with radii spanning from the 5 meter edge of one rectangle to that of another. But the central angle of each of these sectors is 360-90-90-(an angle of the triangle). Thus, the sum of these angles is 360, so the three sectors together make a circle. So the final answer is $ 5P \plus{} 25 \pi \plus{} A$
This seemed kind of easy for an olympiad, and wasn't really combinatorics.


superior algebra
superior algebra unsolved

how do you apply that formula in this case?