linear algebra
matrix
algebra
function
domain
linear algebra unsolved

S is called a twisting subsituation if $S^2$ is indentically equal .Prove that :All of subsituation can be expressed form product of the twisting situations

... just wondering, do you mean idempotent matrix? every matrix can be written as a product of idempotent matrices? also what domain are you working on?


Let triangle $ ABC$. Prove
$ a^4b^2\plus{}b^4c^2\plus{}c^4a^2\geq a^3c^2b\plus{}b^3a^2c\plus{}c^3b^2a$
Quickly!

[quote="mufc"]Let triangle $ ABC$. Prove
$ a^4b^2 \plus{} b^4c^2 \plus{} c^4a^2\geq a^3c^2b \plus{} b^3a^2c \plus{} c^3b^2a$
Quickly![/quote]
It's true for all reals $ a,$ $ b$ and $ c.$ :wink:

[quote="mufc"]Let triangle $ ABC$. Prove
$ a^4b^2 \plus{} b^4c^2 \plus{} c^4a^2\geq a^3c^2b \plus{} b^3a^2c \plus{} c^3b^2a$
Quickly![/quote]
Inequality follows from these three am-gm inequalities.
$ a^4b^2\plus{}b^4c^2 \geq 2b^3a^2c$
$ a^4b^2\plus{}c^4a^2\geq 2a^3c^2b$
$ b^4c^2\plus{}c^4a^2\geq 2c^3b^2a$.


invariant
combinatorics proposed
combinatorics

A regular $ (5 \times 5)$-array of lights is defective, so that toggling the switch for one light causes each adjacent light in the same row and in the same column as well as the light itself to change state, from on to off, or from off to on. Initially all the lights are switched off. After a certain number of toggles, exactly one light is switched on. Find all the possible positions of this light.

My progress(yet to reach the whole solution :blush: )
Let me define "good" bulb; (1,1), (1,5), (2,2), (2,4), (3,3), (4,2), (4,4), (5,1), (5,5) are good.
Let me call the other "bad".
Then, the number of bad bulb on will be always even.
So, the group of good bulbs contains the solution.
I'm not a contestant, but if I were a contestant, could I get partial credit? :D

[quote="peregrinefalcon88"]The parity in M is not invariant just operate on 2;5. However if u consider a square bound by 1;1, 1;5, 4;1, 4;5 the parity of on lights in that region is invariant and if we reflect this region we conclude that there is no light that can be on while all others are off.[/quote]
Incorrect; operating on any of a lot of lights already tell that it's wrong. Try (3,3).

The invariant is actually the set dhthstn said, only with a few typos there. The parity of the number of turned on lights in the set (1,1), (1,2), (1,4), (1,5), (3,1), (3,2), (3,4), (3,5), (5,1), (5,2), (5,4), (5,5) is an invariant. So the bulbs can only be in the complementary region. Reflecting through the main diagonal, we get that the set (1,1), (1,3), (1,5), (2,1), (2,3), (2,5), (4,1), (4,3), (4,5), (5,1), (5,3), (5,5) also have the same invariant. Thus the lights that are in the complement of both regions are the only candidates; these are (2,2), (2,4), (3,3), (4,2), (4,4). Now for the construction...

Another way of looking at the invariants above is to think of the lights' statuses as a collection of $25$ equations $\pmod 2$, where $a_{i,j} \equiv 1 \pmod 2$ if and only if the light at $(i,j)$ is on (otherwise it is $0$) and $a_{i,j} \equiv a_{i,j} + a_{i-1,j} + a_{i+1,j} + a_{i,j-1} + a_{i,j+1} \pmod 2$ (with some of the RHS terms missing if $(i,j)$ is an edge/corner square). Then if a collection $S$ of squares which, when all toggled (starting from no lights on), leave all lights off, then no square in $S$ can be a lone illuminated square since summing the corresponding equations $\pmod 2$ give a LHS of $0 \pmod 2$ and a RHS of $1 \pmod 2$ (a sum of $0$s along with a single $1$ corresponding to the one illuminated square).

Such sets include, as partly mentioned above:

$\{(1,1), (1,2), (1,4), (1,5), (3,1), (3,2), (3,4), (3,5), (5,1), (5,2), (5,4), (5,5)\}$, implying that the lone illuminated square cannot be a corner square

$\{(1,2), (1,3), (1,4), (2,1), (2,3), (2,5), (3,1), (3,2), (3,4), (3,5), (4,1), (4,3), (4,5), (5,2), (5,3), (5,4)\}$, implying that the lone illuminated square must be on a diagonal.

The remaining five squares are achievable. For the center square, toggle $(1,1), (1,2), (2,3), (3,1), (3,3), (3,4), (4,1), (4,5), (5,2), (5,3), (5,5)$. To achieve only $(4,4)$ illuminated, toggle $(1,4), (2,3), (2,4), (2,5), (3,2), (4,1), (4,2), (4,4), (4,5), (5,2), (5,4)$ (rotate this set to get the other three of $(2,2), (2,4), (4,2)$).

[quote=dhthstn]label the light by its position (i;j) $1\leq\i\leq\ 5;1\leq\j\leq\ 5$
call (i;j) good if there is an algorithm make all lights but light(i;j) off
notice that ;if (i;j) is good then the light symmetric to it through 2 diagonal is also good
now we define an invariant
consider set S (1;1);(1;2);(1;4);(1;5);(3;1);(3;2);(3;4);(3;5);(5;1);(5;2);(5;1);(5;3)
the parity of M,the number of light on in S is invariant.A light not in S has even adjacent light in S(0 or 2);so toggle it will make M unchanged,or reduce 2 or increase 2,so parity is unchanged.A light in S has one adjacent light in S,so toggle it also make the parity of M unchanged
Initially,M is 0,so a good position (i;j) cannot in S
also (1;3);(2;1);(4;1);(2;5);(4;5) is not good since it is symmetric to a position in S
so there are 5 possible good position and we can check that all of it works
easy problem,right?[/quote]

I think(if understands correctly) if (1,2),(3,2) is on, (5,2) is off, then operates on (2,2) make M alternates the parity.

Is there any other invariant? what about bigger fields?

I wonder how to determine a binary matrix(relation matrix) is invertible? Or restrict to some symmetric case like this problem.  Can anyone recommend some theoretical or computational reference ! :)

 The construction took me more time than the solution itself :)
My solution is very similar to dhthstn's, so I'll just post the construction.
In the construction, black squares are the lights which we will toggle:
[hide=First construction] Costruction for the square $(2,4)$: (and $(2,2),(4,4),(4,2)$ which follows by symmetry)
$$\begin{bmatrix}
 \square& \blacksquare  & \square & \blacksquare & \square\\
 \blacksquare& \blacksquare & \square & \blacksquare &\blacksquare \\
 \square& \square & \square & \blacksquare & \square\\
 \blacksquare & \blacksquare & \blacksquare & \square & \square\\
 \square& \blacksquare & \square &\square  & \square
\end{bmatrix}$$[/hide]
[hide=Second construction]Construction for the square $(3,3)$:
$$\begin{bmatrix}
\blacksquare & \square & \blacksquare & \blacksquare & \square\\
\blacksquare & \square & \square & \square & \blacksquare\\
\square & \blacksquare & \blacksquare & \square & \blacksquare\\
\square & \square & \blacksquare & \square & \square\\
\square & \square & \square & \blacksquare & \blacksquare
\end{bmatrix}$$[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be an additive subgroup of $ (R, \plus{})$. Show that either $ G$ is dense in $ R$or it is discrite (дискретная).

Sketch:
If there is a smallest positive element in that subgroup, this element generates the whole group. If such an element doesn't exist, then there are arbitrary small elements. This implies density.


function
limit
calculus
derivative
calculus computations

If $ f, g$ are non negative real valued functions, where $ f$ is continuous,  is it true that $ \inf_{x\in E}f(g(x))\equal{}f\left(\inf_{x\in E}g(x)\right)$?

No, it is not true. Take for example $ f(x)\equal{}|lnx|$, $ g(x)\equal{}x^2\plus{}\frac{1}{2}$, $ E\equal{}\mathbb R^\plus{}$.

The question is same as if $ E$ is set of non negative reals then $ \inf f(E) \equal{} f\left(\inf{E}\right)$.

I hope the result would be true for monotone increasing continuous $ f$.
Here's a proof:

$ f\left(\inf{E}\right)\le f(x)$ for all $ x\in E$ ( since $ f$ is increasing)
Therefore $ f\left(\inf{E}\right)\le \inf f(E)$

On the other hand,
Let $ x_n$ be sequence in $ E$ s.t. $ x_n\to \inf E$

Now $ \inf f(E)\le f(x_n)$ for every $ n$

So, $ \inf f(E)\le \lim_{n\to \infty}f(x_n)$
$ \equal{} f\left(\lim_{n\to \infty}x_n\right)$   (using continuity of $ f$)
$ \equal{} f\left(\inf{E}\right)$

For monotone decreasing continuous $ f$ I think the result would be $ \sup f(E) \equal{} f\left(\sup{E}\right)$.

It seems fine.  Note that if you are assuming that $ f$ is monotone, not only do you get continuity a.e., you get a finite derivative a.e. (see [url=http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem]Lebesgue's Differentiation Theorem[/url] for the second one).  And if you assume $ f$ is surjective, you're guaranteed continuity on the interval.


MIT college Harvard Princeton geometry Stanford email

Hi all!

  I'm going to take 3 SAT subject test in December 2; 2 math and a physics
test.I am taking SAT test fo the first time and I would like someone who
have already taken it to describe the atmospher during the test.I think it
will help me to score batter if I know what to expect.

  So some things I would like to know.Can I choose the order of the exams?
I mean if I want to take Math 1 then Physics and in the and Math
2 can I do that or I have to write the exams in that order in which they will tell me on
the test day?Also, can I make notes on the test form( I mean it may be
helpful to note the questions that you miss).At last I will be gratefull to
hear any advice that you think can help me during the test.

Thank you!

Tigran

Hrmm well in response to your questions, Yes you can choose whatever order you want to take the exams. All the tests will come in a big book and you just flip to a page to take the exams. You can mark up the test as much as you want and it won't be graded.

My advice is to get a good nights sleep because even though it is slightly shorter than the SAT I's, it is still quite demanding on concentration. Cramming usually doesn't work and I don't know about other states but I believe Math II is the only recommended one to take since Math II is like the Math I plus more.

Thanks for the reply and advice.And congratulations with the first post in  Mathlinks  :lol: .

As far as i know, if you are going to take Math 2, i heard that you don't have to take math 1. (my school recommends to take only Math 2...) BTW, I'm taking Physics on Dec. 2 too..(i should've taken last year... :( )

I took all 3 (not at the same time).  Taking Math I and II is good.  When applying to college, I found some schools asked for either, while some asked for one specifically.  They're both easy, though.

On the physics, there's a ton of leeway.  You can get quite a few wrong and still get 800.  One thing to look out for- when I took it there were a couple of questions on topics which weren't part of the AP Physics B curriculum (especially relating to sound).  I specifically remember a question about timbre.  So look out for that, but it's nothing major.

And good luck.  Taking 3 is exhausting.

i took both math 1 and 2 and 2 is a tiny little harder than 1 but has a way better curve so your scores comes out better.

[quote="ShoeFactory"][quote="Tiks"][quote="kimby_102"]As far as i know, if you are going to take Math 2, i heard that you don't have to take math 1. (my school recommends to take only Math 2...) [/quote]
Are you sure about it? :maybe:
I think some of my friends hwo are students already took both math1 ant math 2.So,anyway I 'm going to take them both because I have already registered :) .But I don't think I will have any problems with math,the difficult part for me is physics :( .[/quote]

after you register you can change, it doesnt matter what you registered for. on the test day they give you a big booklet, just do the test that you WANT to do. Dont even do math 1, just do math 2.[/quote]
Look guys,in the colleg application requirements it is written that you should take three subject tests,and I think math 1 and math 2 are different tests.So,in my opinion I need to write both Math 1 and math2,and I 'm not worring about them. :P

[quote="Ravi B"]You may wish to ask the college whether Math 1 and Math 2 count as two different subject tests.[/quote]

That would be my advice too: check the college admission Web pages for what each college recommends.

[quote="Tiks"][Look guys,in the colleg application requirements it is written that you should take three subject tests,and I think math 1 and math 2 are different tests.So,in my opinion I need to write both Math 1 and math2,and I 'm not worring about them. :P[/quote]

Honestly, I doubt a college is going to fall for that. I think if you take math 2, they are not going to look at Math 1, unless you do a lot better on Math 1, in which case, they might be kind enough to ignore your math 2.

Those tests are really designed for everyone to be able to take a math SAT II, yet, still allow different math levels. I can see how some people may end up with both scores, but to take it on the same day may make you look like a Grade Hog, and that is a bad thing to be. A Grade Hog (or someother word) is someone who's actions are only taken to look good on the app, not out of a pursuit of excellence.

I agree with the others. Check with the school before you go too far on this plan.

[quote="Tiks"]I am rather sure that the colleges will admitt math 1 and math 2 as 2 subject tests,because my two friends Iurie Boreico(Iura)(twice perfect scorer at the IMO :P ) and Dimitar Semionov(imortal) (silver medalist from Bulgarian team)took math1,math2 and Physics as three subject tests for their application and now Dimitar is a freshman at MIT,Iura has already applied to Harvard :) .So,I don't think it is a problem to take math1 and math 2 as different subject tests.However,I will try to find out more about that!  :ninja:[/quote]

MIT only requires 2 SAT2s - one in Math 1 or 2 and one in a science. Not sure about Harvard. In any case, standardized test scores are a bit negotiable. I know that Harvey Mudd wants something life 5 or 6 from homeschooled students. I wrote to the admissions and nicely asked them if they are on something. I made it clear my son would not be applying if that was the case. They admitted they would back down on that as long as everything else is good.

some other examples...

Caltech also requires only 2, but wants Math 2, not Math 1, and a science SAT.

Princeton wants 3, and doesnt specify. Presumably they want 3 to see how you have mastered 3 different subjects. If your app is very strong in other areas, they might ignore the fact you havent really shown this, and that you may be trying to do something bogus.

Stanford only wants 2, and with a preference for Math 2.

Here is a cautionary tale told to me by a UCLA prof who used to be at MIT. Some kid has 3 SAT 1 scores. The first one is 1590. He takes it again, and again, gets 1590, but this time it is reversed, so he does have 800s in both M and V. This is not good enough for him. He takes it a 3rd time and gets 1600. Although he looked pretty good in other areas, the 3 SAT scores tipped the balance against him - he appeared highly strung and not beyond doing anything to make his app look good, thus suspect.

Good luck on your apps.

[quote="WindSlicer"]I would never retake a 1590, but I would like to know why an admissions officer would deny a kid who tried to strive for perfection... and accomplished it.

I would hope there were other gaps in the app to justify the decision.. poor kid.[/quote]

SATs are not that important. Not worth the time to retake. Striving for perfection on an SAT shows a lack of maturity. Striving for perfection on your coursework or extracurricular activities looks good.

[quote="WindSlicer"]I would never retake a 1590, but I would like to know why an admissions officer would deny a kid who tried to strive for perfection... and accomplished it.

I would hope there were other gaps in the app to justify the decision.. poor kid.[/quote]
1600 isn't perfection anyway. At my school, when I took SAT, there were 4 1600s. Only one of them was actually a perfect paper. ;)

Anyway I don't have much respect for standardized tests as a way to determine the best from among good students (obviously it is good for determining the good students from the average ones). Just like IQ -- the IQ test gets less and less accurate for people farther from the mean.

@ Tiks: I will try to email an admissions officer and find out what their opinion is about the Math I, Math II thing.

If instead of thinking about it as "getting rejected on account of retaking the SAT to get a 1600" we think of it as "not getting accepted on account of retaking the SAT to get a 1600" I think it seems more reasonable for this to possibly hurt the chances of the applicant.  Admissions officers (at the more selective schools, at least) usually look at applicants holistically.  If they look at your application, see that you spent a lot of time on the SAT (as indicated by the number of times you took it and the progression of scores) then they may very well think that the chances you're just a tool are higher...at least, that's what I would think.  Of course, if you aren't a total tool, and your application shows that, then this impression can be overriden.  Anyway, the whole idea of someone getting rejected for the sole reason that they took the SAT $x$ times, and scored $y_{i}$ on them is dubious.  It's the [i]whole[/i] application that gets someone in a school and, likewise, it's (generally) the whole application that gets someone rejected.

[quote="WatsonLadd"][quote="Tiks"]I am rather sure that the colleges will admitt math 1 and math 2 as 2 subject tests,because my two friends Iurie Boreico(Iura)(twice perfect scorer at the IMO :P ) and Dimitar Semionov(imortal) (silver medalist from Bulgarian team)took math1,math2 and Physics as three subject tests for their application and now Dimitar is a freshman at MIT,Iura has already applied to Harvard :) .So,I don't think it is a problem to take math1 and math 2 as different subject tests.However,I will try to find out more about that!  :ninja:[/quote]
Perhaps going to the IMO had something to do with getting in.[/quote]

Yeah,I went to the IMO too(hmm...4 times :blush: and it is not the end :D  ).


let a, b, c, d  be positive numbers such that abcd=1. prove  that :!:  :!:
\[ \sum{\frac{1}{{1+a+a^{2}+a^{3}}}}\ge 1 \]

[quote="Vorelin Tancu"]$ \sum \frac{1}{1+a+a^{2}+a^{3}}\geq \sum \left(-\frac{3}{8}\left( a-1\right)+\frac{1}{4}\right) =-\frac{3}{8}\left( a+b+c+d\right)+\frac{5}{2}\geq-\frac{3}{8}.4\sqrt[4]{abcd}+\frac{5}{2}=1$[/quote]

but i think the latest inequality is incorrect!!!  :blush:

Another studios... :huh:

http://www.mathlinks.ro/Forum/viewtopic.php?t=17246

  darij


Consider a triangle ABC and two different points P and Q.
The parallel to BC through Q intersects the line AP at A*.
The parallel to AP through Q intersects the line BC at A#.
Define the points B*, B#, C* and C# similarly.
Then prove that the triangles A*B*C* and A#B#C# are perspective.

  Darij

I think the projective generalization is also true: replace the line at infinity with a line $\ell$ and instead of the parallel through $Q$ to $BC$ (for example) draw the line through $Q$ which is concurrent with $BC,\ell$ and so on.

This is important because we might be able to use some projective transformations to make the problem easier. For example, I'm trying to solve it in the case $PQ=\ell_{\infty}$ (I haven't succeeded, though, so maybe this isn't such a slick approach :D).

Apparently, some very interesting projective properties are related to the configuration I mentioned. Let me resume it:

I took $PQ=\ell_{\infty}$, so instead of lines through $P,Q$ we have lines parallel to two given directions $p,q$. Let $\ell$ be the line replacing $\ell_{\infty}$ from the initial problem. We obtain points $A^*,A\#$ and the others in the obvious way, extending the definitions from Darij's post. Then I've noticed that this very interesting thing happens: no matter how we move $\ell$, the point $A^*A\#\wedge B^*B\#$ always stays on the same line parallel to $q$, and the same can be said about the other two points (we want to show that these points are, in fact, one and the same). Then if $A^*A\#,B^*B\#,C^*C\#$ are concurrent for a position of $\ell$, then they're always concurrent, so all we need to do is choose a particular position of $\ell$ and prove the concurrence. We may, for example, choose $\ell||p$. I don't know if it works out Ok, but it looks like we're only dealing with some similarities (we can choose [b]any[/b] of the lines $\ell||p$).

Note that we can't choose $\ell||q$ because in that case the three concurrence points would trivially be one and the same, because the three lines parallel to $q$ which are their loci concur on the infinity line ($\ell_{\infty}$), but we want to show that they concur at a finite point.

Now I have a proof. Admittedly, it uses none of your ideas, Grobber, I haven't managed to complete them; it just uses Ceva and Menelaos (well, Menelaos for quadrilaterals). You can find the proof in the note "On the paracevian perspector" on [url=http://www.cip.ifi.lmu.de/~grinberg/]my geometry website[/url]. (BTW, I have replaced the notations A#, B#, C#, A*, B*, C* by U, V, W, U', V', W' for typographical reasons.)

  Darij

Dear Mathlinkers,
another synthetic proof can be see on
http://perso.orange.fr/jl.ayme   vol. 4  the paracevian perspector (in French)
Sincerely
Jean-Louis

Also, two other solutions have been posted in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=126138]Perpendicular feet[/url], of [b]barasawala[/b].

Kostas Vittas.


The four angles of a quadrilateral are in the ratio of $ 1: 2: 3: 4$. What is the number of degrees in the measure of the smallest angle?

So, you first know that a quadrilateral has $ 360$ degrees in it. Then, you add up the ratios. So, $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \equal{} 10$  So now you can do $ \dfrac{360}{10}$ and get $ 36$ for each unit. The $ 1$ is obviously the smallest so the answer is $ \boxed{36}$

A little clarification:

The reason we added up the ratios and divided, is because we know there exists some number $ x$ such that $ 1x \plus{} 2x \plus{} 3x \plus{} 4x \equal{} 360$. So all we need to do is solve that equation for $ x$ and then multiply by the respective number (1, 2, 3, or 4) to find the angle we wanted. In this case, we wanted the smallest angle, so we took $ x \equal{} 36$ and multiplied it by 1, which is just $ \boxed{36}$.


Why is the number of digits in any integer 1 + log(n)?

Can someone show me a derivation or give me an explanation?

it's not $ 1\plus{}\log n$. It's $ 1\plus{}\lfloor \log n \rfloor$; since we write numbers in base 10, a number with $ n$ digits can be written as $ 10^{n\minus{}1}\plus{}\text{stuff less than }10^n$ (test it for small values of $ n$ to see why). Just rewrite that in logarithmic form.


Let $x,y$ be positive reals such that $\frac{1}{x}+\frac{1}{y}=1$, and let $n$ be a positive integer. Prove that $(x+y)^n + 2^{n+1} \geq x^n + y^n + 2^{2n}$.

[hide="Solution"]
$x + y = xy$
$(x-1)(y-1) = 1$

Let $a = x-1$.  Then $x = a+1, y = \frac{1}{a} + 1$ and the problem condition becomes

$(a + 2 + \frac{1}{a})^n + 2^{n+1} \ge a^n + \frac{1}{a^n} + 2^{2n}$

Let $b = \sqrt{a}$.  We can rewrite the problem as

$\left( b + \frac{1}{b} \right)^{2n} - b^{2n} - \frac{1}{b^{2n}} \ge 2^{2n} - 2^{n+1}$

Let $s_k = b^k + \frac{1}{b^k}$.  It is fairly clear from binomial theorem that

$s_1^{2n} - s_{2n} = {2n \choose 1} s_{2n-2} + {2n \choose 2} s_{2n-4} + ... + {2n \choose n}$

Now, we have the inequality $s_k \ge 2$.  So:

$s_1^{2n} - s_{2n} \ge {2n \choose 1} 2 + {2n \choose 2} 2 + ... + {2n \choose n} = 2^{2n} - 2$

Which seems to be much stronger than the initial statement...  :huh:
[/hide]

[quote="t0rajir0u"]
Which seems to be much stronger than the initial statement...  :huh:
[/quote]

If $x=y=2$, then the equality holds.

[quote="t0rajir0u"]
$x + y = xy$
$(x-1)(y-1) = 1$

Let $a = x-1$.  Then $x = a+1, y = \frac{1}{a} + 1$ and the problem condition becomes

$(a + 2 + \frac{1}{a})^n + 2^{n+1} \ge a^n + \frac{1}{a^n} + 2^{2n}$

[/quote]

Oh, I see where your mistake is. It should be

$(a + 2 + \frac{1}{a})^n + 2^{n+1} \ge (a+1)^n + (\frac{1}{a}+1)^n + 2^{2n}$

Edit: -deleted for mistaken the question-

[quote="t0rajir0u"][hide="Solution"]

[/hide][/quote]

Your solution is not correct. Would you like to try again? :-) Here is a little hint:

[hide="hint"]
Pairing $x^ky^{n-k}$ and $x^{n-k}y^k$ in the binomial expansion of $(x+y)^n$.
[/hide]

[quote="chess64"][hide="AM GM"]
By AM-GM, we have $x+y=xy\ge 2\sqrt{xy}\implies xy\ge 4\implies (x+y)^n \ge 2^{2n}$.

But also by AM-GM, $x^n+y^n\ge 2(\sqrt{xy})^n \ge 2(2)^n = 2^{n+1}$.

Adding these equations, we get $(x+y)^n+(x^n+y^n) \ge 2^{2n}+2^{n+1}$.
[/hide]

Err oops what I did wrong? :P[/quote]

Nice try :-)

You didn't do anything wrong except the thing you proved is a little different from the original inequality :-)


calculus
integration
number theory unsolved
number theory

If x is a root of an equation x^m+C1*x^(m-1)+.....+Cm where C1,....,Cm are integers then x is either integral or irrational

this is a special case of the rational root theorem. the proof of the theorem is simple: seee it at http://planetmath.org/encyclopedia/ProofOfRationalRootTheorem.html


Let $ p$ be a prime number with $ p > 2$ such that $ 3$ divides $ p - 2$. Let $ S$ be a set such that $ S$ = {$ y^2 - x^3 - 1$ and $ x,y$ integers with ${ 0<= x, y<= p - 1}$}
Show that $ p$ divides at most $ p - 1$ members of the set $ S$.

i didnt get it. could you explain it ???? :)

we take $ A \equal{} \{\overline{x^3}|x\in[|0,p \minus{} 1|]\}$ and $ B \equal{} \{\overline{y^2 \minus{} 1}|y\in[|0,p \minus{} 1|]\}$
we have $ a^3\equiv b^3[p]$ ==> $ a^{3}(a^3)^{\frac {p \minus{} 2}{3}}\equiv b^3(b^3)^{\frac {p \minus{} 2}{3}}[p]$ ==> $ a\equiv b[p]$.
then $ A \equal{} \{\overline{x^3}|x\in[|0,p \minus{} 1|]\} \equal{} \{\overline{k}|\ k\in [|0,p \minus{} 1|]\}$
then $ S \equal{} A \plus{} B$
if $ \exists (a_1^3 \plus{} b_1,a_2^3 \plus{} b_2,...,a_{\frac {p \plus{} 1}{2}}^3 \plus{} b_{\frac {p \plus{} 1}{2}})\in S^{\frac {p \plus{} 1}{2}}$ such that for $ i \equal{} 1..p: \ p|a_i^3 \plus{} b_i,\ (a_i^3,b_i)\in A\times B$
then exists $ i\neq j$ such that $ b_i\equiv b_j[p]$ because $ |B| \equal{} \frac {p \minus{} 1}{2} < \frac {p \plus{} 1}{2}$
so $ a_i^3\equiv a_j^3[p]$ ==> $ a_i\equiv b_i$
then $ |\{X\in S: \ p|X\}|\le\boxed{\frac {p \plus{} 1}{2}}\le p$


inequalities
AMC
USA(J)MO
USAMO
inequalities solved

I need a problems, which are solved with Bernoulli's inequality. Can you help me to find them! Bernoulli's inequality says: \[(1+x)^n\geqq 1+nx,\quad x\geqq -1\in\mathbb{R},\quad n\in\mathbb{N}\]
I just want a problems and their decisions.  :)

I believe #4 of 1991 USAMO first introduced Bernoulli to me...

http://www.kalva.demon.co.uk/usa/usa91.html

our imo team has solved many problems using bernulli s ineq.
i will post them later ;)

Here is a problem using Bernoulli's inequality:

   Solve the equation: $2^x+3^x=3x+2$. If you want more I have some hard one's.


Let a,b,c,d \in (0,\infty). Prove that:
(a/a+b)2 + (b/b+c)2 + (c/c+d)2 + (d/d+a)2  \geq 1.

it suffices to prove that \sum a/a+b >2
\sum a/a+b =\sum 1/(1+(b/a))
so we need to show that \sum 1/(1+x) >2 where xyzt=1
which is equivalent to prove that \sum x> \sum 1/x.
consider r.h.s of this last inequality to be proved:
\sum 1/x=xyz+yzt+ztx+txy =xy(z+t)+zt(x+y)
consider (x+y)(1-zt)+(z+t)(1-xy)
obviously one of xy and zt is less than 1. let zt<=1.
consider the change x to root xy and y to root xy.
the new variables still satisfy xyzt=1
but now since x+y>=2 times root xy. therefore the expression is decreased.
hence the minimum occurs when x=y.
expression now equals: (2x)(1-zt)+(z+t)(1-x^2)=(zt-1)(1/z+1/t -2x)
where x^2 z t =1
but 1/z +1/t>= 2/root(zt) =2x hence expression >=0
hence \sum x \geq \sum 1/x.   \

The above solution is wrong. The inequality sum 1/(1+x) >=2 isn't always true: for example, it fails when y,z,t do to infinity and x tends to zero so that xyzt=1. I've seen this inequality before and , if I'm right, it belongs to Vasile Cartoaje. I managed to solve it some time ago, but I've tried know to solve it and I didn't manage.

Just a little help: we have to prove that sum 1/(1+x)^2 is at least 1 when xyzt=1. Make the substitutions 1/(1+x)=a, etc and you will obtain the following inequality proposed by the same author at a GM contest ( in 2000 or 2001) : if a,b,c,d>=0 and abcd=(1-a)(1-b)(1-c)(1-d) , then a^2 +b^2+c^2+d^2>=1.

Harazi, have you a solution for the inequality :

If a,b,c,d>=0 and abcd=(1-a)(1-b)(1-c)(1-d) , then

                    a^2 +b^2+c^2+d^2>=1.

I cannot solve it.

Thank you, very much.

No, I don't have it, but maybe a romanian who has a book with that GM contest would be so kind to post a solution.

well, up to my knowledge, there is no book with the solutions of those problems ( GM contest 2000-2001) and neither the GM presents the solutions, so I guess we have to solve the problem :)

Isn't the problem like this: if a 2 +b 2 +c 2 +d 2 =1 then prove that abcd \geq (1-a)(1-b)(1-c)(1-d)?

The inequality can be obtained from the product of the following inequalities:
  (1 - a)(1 - b)(a2 + b2) \geq ab(c2 + d2)    (1)
  (1 - c)(1 - d)(c2 + d2) \geq cd(a2 +b2)     (2).
Starting with the identity
  2(1 - a)(1 - b) = (a + b - 1)2 +c2 + d2
we have:
  2(1 - a)(1 - b)(a2 + b2)-2ab(c2 +d2) = [(a + b - 1)2 + c2 + d2](a2 + b2 - 2ab(c2 +d2) = (a + b - 1)2 (a 2 + b2) + (c2 + d2)(a - b)2  \geq 0.

Yeah, right. All I can say about that solution is that it is very nice, but not at all natural.

May anyone explain to me why the inequality:

if a,b,c,d>=0 and abcd=(1-a)(1-b)(1-c)(1-d) , then
a^2 +b^2+c^2+d^2>=1.

is equivalent to

if a,b,c,d>=o and  a^2 +b^2 +c^2 +d^2 =1 , then
abcd \geq (1-a)(1-b)(1-c)(1-d)

how pcalin said.
Sorry, but I don't understand.

Sorry, pcalin, I have misunderstood the meaning of the adjective "like".
It means "similar" and not, as I did, "equivalent".

So this beautiful inequality is again an open challenge.

Hi all,

I'm agree with Manlio.

Two problem "if a,b,c,d>=0 and abcd=(1-a)(1-b)(1-c)(1-d) , then
a^2 +b^2+c^2+d^2>=1"

and

"if a,b,c,d>=0 and a^2 +b^2 +c^2 +d^2 =1 , then
abcd  (1-a)(1-b)(1-c)(1-d)".

are not equivalent. (At least I dont know how to reduce one from the other).

So, the problem is open (at least in this forum). I have a solution, not very nice, but contains many interesting ideas. I will post it in future.

Namdung

By dividing all the fractions with a, b, c, d and b/a = x, c/d = y, d/c = z, a/d = t we have:
1/(a+x)2 + 1/(1+y)2 -1/(1+xy) = [(xy-1)2 + xy(x-y)2]/[(1+x)2 (1+y)2(1+xy)2] => 1/(1+x)2 + 1/(1+y)2 \geq 1/(1+xy) = a/(a+c) also 1/(1+z)2 + 1/(1+t)2 \geq 1/(1+zt) = c/(a+c).
From the sum of the two inequalities we get our inequality for a=b=c=d.

[quote="Gil"]By dividing all the fractions with a, b, c, d and b/a = x, c/d = y, d/c = z, a/d = t we have:
1/(1+x)2 + 1/(1+y)2 -1/(1+xy) = [(xy-1)2 + xy(x-y)2]/[(1+x)2 (1+y)2(1+xy)] => 1/(1+x)2 + 1/(1+y)2 \geq 1/(1+xy) = a/(a+c) also 1/(1+z)2 + 1/(1+t)2 \geq 1/(1+zt) = c/(a+c).
From the sum of the two inequalities we get our inequality for a=b=c=d.[/quote]
============
Very nice and neat solution (I corected some misprint). So we can forget my solution.

Namdung


$ a,b,c,x,y,z > 0$such that $ abc \equal{} m^3$ and $ xyz \equal{} n^3$,where $ m,n$are given const.Find the max and the min of the following expression.
\[ \frac {1}{1 \plus{} a \plus{} x} \plus{} \frac {1}{1 \plus{} b \plus{} y} \plus{} \frac {1 }{1\plus{} c \plus{} z}\]

:maybe: one and a half month has passed....still no one have a solution?????? :(

I think there isn't any minimum constant value. We can approach that expression to $ 0$.

If there is a constant $ k>0$ such that the expression $ \geq k$. Let's take $ x,y,c>\frac{3}{k}\minus{}1$ is a contradiction. :wink:


LaTeX
number theory unsolved
number theory

Given two positive integers $m>n$,denote $A=\{a|(a,n)=1,1\le a\le m\}$,show that $\sum_{a \in A}\frac{1}{a}$is not an integer.

[i]moderator edit: no, the problem is not hard; choose better titles next time![/i]

It is not correct. For example $if \ P(m)|n, \ P(m)=p_1*p_2* \dots *p_s \ - \ product \ all \ primes \ \le m.$ $For \ another \ cases \ it \ is \ true, \ and \ may \ be \ proved \ by \ consider \ maximal \ prime \ p \lem: (p,n)=1$.

Rust: please don't write everything in $\LaTeX$ now, especially the text ;)
But I don't see what your condition implies...

Indeed, a full prove (I just need $m$ being greater than the smallest prime not dividing $n$):
Take the smallest $p>1$ with $\gcd(n,p)=1$. $p$ will be prime then since otherwiese every prime divisor of $p$ is smaller and fulfills the conditions, too.
Now to kill this prime $p$ from the denominator, you will need another number having this denominator; let $pq$ be the next multiple of $p$ with $\gcd(q,n)= \gcd(pq,n) =1$.
Since $p$ is minimal, $q \geq p \implies p=q$, thus the next multiple of $p$ (occuring as denominator) is already $p^2$.
Now to kill $p^2$, a $p$ is not enough, you need a denominator that is a multiple of $p^2$ this time, and continuing as before, the next such multiple is $p^3$, the same problems occure again and so on, thus you will never get an integer.


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

(I)Let $G$ be a group of order $2^{m}k$,where $k$ is odd. Prove that if $G$ contains an element of order $2^{m}$,then the set of all odd order in $G$ is a (normal) subgroup of $G$.
(Hint:Consider $G$ as permutations via Cayley’s theorem , and show that it contains an odd permutation.)
(II)Show that a finite simple group of even order must have order divisible by 4.

We have a homomorphism $\varphi: G\to S_{G}$, taking an element $g\in G$ to the permutation $\sigma_{g}$ of $G$, where $\sigma_{g}(x) = gx$.  Suppose $g$ has order $2^{m}$.  Then $\sigma_{g}$ is a permutation of $|G|$ elements, of order $2^{m}$, with no fixed points.  It must therefore consist of $k$ disjoint $2^{m}$-cycles, so it is an odd permutation.  It follows that $sign\circ\varphi : G\to \{\pm 1\}$ is surjective, and that $N=\ker(sign\circ\varphi)$ is a subgroup of $G$ of order $2^{m-1}k$.  Clearly the odd order elements of $N$ and of $G$ are the same.  By induction (since $\sigma^{2}$ has order $2^{m-1}$), these elements form a subgroup, which is obviously characteristic, hence normal.

Now (II) follows from the fact that a group of even order has an element of order 2 (Cauchy).


[quote="Interval"]A recent national survey found that high school students watched an average
(mean) of 6.8 DVDs per month. A random sample of 36 college students
revealed that the mean number of DVDs watched last month was 6.2, with a
standard deviation of 0.5. At the .05 significance level, can we conclude
that college students watch fewer DVDs a month than high school students?[/quote]
You do realize that .5 and .05 are different, right?

You asked:

You do realize that .5 and .05 are different, right?

My reply: Yes, of course.

[hide] Since the survey said that college students watch 6.2 DVDs per month with a standard deviation of .5, we can say that the range of the possible average DVDs per month that are watched by all college students is more than 6.2-.5=5.7 and less than 6.2+.5=6.7.  However, 6.8 is more than the highest possible average, which is 6.7.  This means that the average DVDs watched by the high schoolers is greater than the average DVD watched by the college students. [/hide]


Find A+B, if A and B are positive, rational numbers and

$ A\plus{}B\sqrt{7}\equal{}\sqrt{6\sqrt{28}\plus{}\frac{2}{8\plus{}3\sqrt{7}}}$

[hide] We can simplify some of the terms in the RHS to get $ \sqrt{12\sqrt{7}\plus{}16\minus{}6\sqrt{7}}\equal{}\sqrt{6\sqrt{7}\plus{}16}$. So,

$ A\plus{}B\sqrt{7}\equal{}\sqrt{16\plus{}6\sqrt{7}}$

$ \implies A^2\plus{}7B^2\plus{}2AB\sqrt{7}\equal{}16\plus{}6\sqrt{7}$.

Comparing like terms,

$ A^2\plus{}7B^2\equal{}16$

$ 2AB\equal{}6$

The solution of this is $ A\equal{}\pm3$, $ B\equal{}\pm1$. However, we want the positive values (since square roots are positive), so $ A\equal{}3$, $ B\equal{}1$.

Therefore, $ A\plus{}B\equal{}3\plus{}1\equal{}\boxed{4}$.[/hide]


[quote="joml88"]This is a very popular question.  I have seen it on [i]so[/i] many competitons :D[/quote]

You have? I haven't...but then again the only competitions I have ever participated in are MATHCOUNTS and Math League round 1... :D

This problem is 1980 AHSME #27.

chess64, can you edit your post so it has just the problems and the choices? I want to link this thread to Olympiad Resources page.

Here are choices:

A. $\frac{3}{2}$
B. $\frac{\sqrt[3] {65}}{4}$
C. $\frac{1 + \sqrt[6] {13}}{2}$
D. $\sqrt[3] {2}$
E. None of these

My own solution...
[hide]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=313[/img]
[/hide]

[quote="chess64"]HIDE YOUR ANSWER! And also, how do I know you didn't just use a calculator?? :)[/quote]

In Japan, there is no custom  to use calculator in examination.

Anyway in solving your problem, we are to use the next inequality.

If $a+b+c=0$, then $a^3+b^3+c^3=3abc$. :)

or just cube the whole thing
$(x+y)^3=x^3+y^3+3xy(x+y)$

Which, if you'll notice, is what I did. ;)


I love the MATHCOUNTS competition.
I would like to start a new and fresh round of a [b]Mock Mathcounts Competition[/b]

It will start out with a mock school round.
Everyone will then take the mock chapter round.
Those doing well on the chapter will qualify and take the mock state round.
The top scorers on the state will take the mock national round.
The winners will come to FTW2 for an unrated round of CountDown.
The ranking will be posted.  :)

Those interested are welcome to sign-up!

I would also appreciate a few helpers to write the problems.
[i]Note: Helpers cannot also take the test.[/i]

I hope this works smoothly, have fun!

I'll participate :lol: !!

[hide="Participants"]the cliu[/hide]

Please copy and paste this list, thanks.

No problem.

[hide="Participants(1)"]
the cliu
[/hide]

Thanks!

I would like to help write the problems
(I'm KingSmasher3's sister)

Thanks!

i would like to be a helper also.

since i am in high school now, it would be more helpful to myself to write problems than solve them

I would like to sign up for this mock-mathcounts competition. :D

No Problem.

[hide="Participants(3)"]
the cliu
Street
BarbieRocks
[/hide]

[hide="Helpers(2)"]
15!!!
riggabamboo
[/hide]

I will let everyone know when we are ready.  :)

Thanks!

I will join. I hope i dont fail like i did PoP's :oops:

I'll join. Warning: might forget I'm in this :P

UPDATE:

[hide="Participants(6)"]
the cliu
Street
BarbieRocks
dragon96
TwinDracula
math rulz
[/hide]

[hide="Helpers(2)"]
15!!!
riggabamboo
[/hide]

Thanks all! I will PM all of you regarding the Competition!

About how hard is it going to be? Like, what grade level? :fool:

I'll join.

[hide="Participants (7)"]the cliu
Street
BarbieRocks
dragon96
TwinDracula
math rulz
PowerOfPi[/hide]
[hide="Helpers (2)"]
15!!!
riggabamboo[/hide]

[quote="exmath89"]@Kingsmasher3:
Can I still participate in the State round and pm you the answers but my score will not be counted?[/quote]

Of Course.

can you send me tehh questions ?
of course, im not joinging

Can we sign up? This is getting sort of inactive, because Nats is in 6 days.

Yes, I noticed.  :)

I know that so many people are getting PUMPED for nats.
I'm sorry I havent gotten around to grading all the tests and keeping the tournament going. I'm preparing too.  :P

I am surely going to pick up after nats is over on the 10th.
Thanks!

Yes, it would be great for you to sign up. Copy and paste your name to the most recent list and you may find the School, Chapter, and State Problems on the previous Pages.

Thanks!
I am waiting for a few more answers to be turned in before I grade them
[color=#0040BF]THANKS ALL![/color]

[hide="Participants(A lot)"]AdithyaGanesh
kelleyzhao
the cliu
Street
BarbieRocks
dragon96
TwinDracula
math rulz
PowerOfPi
Maybach
percy314
slayer42
Drepix
bbgun34
cheeseyicecream
CRICKET229
yaofan
AwesomeToad
AIME15USAMO
\LaTeX
basketball9
Minamoto
pi31415926
The Hobbit
Putneni07
TheMan1998
SonyWii
jerry125
origamiguy777
bluecarneal
CDKing129
Just_Beginner
AIME15
biblioman
monkeygirl13
policecap
diger
hahagooman
hi how are you doing toda
dwei1019
barcelona
Iggy Iguana
Demosthenes
141matho
e,i,pi
The Archer
liltobe9
athunder
mathiscool888
matho141
amparvardi
theoneforce
vjnmath
ksun48
hli
colinhy
dwade884
physicsguymark
joshim5
shangdevin
zhanga
jonathanchou711
ksun48[/hide]

[quote="theoneforce"]What's the current deadline for State? I'll try to fit it in my schedule, but the earliest available day is Friday, so...[/quote]

You may turn in your answers for the State round any time before the end of the competition.  :)

Can I still do the state round?  Sorry I've been wrapped up in other things lately.

Ugh.
Me too.
I have had an extremely busy summer.  :dry:

I'll my best to keep this going.
(If not there will be next year)

@Maybach: Of course. The Q's are open to everyone. You are also welcome to Pm me your answers

THANKS ALL!

Ok!
I have set a date for the release of the final test for this year's round. The national round will be posted on July 20th. Feel free to turn in your answers anytime befor the end of August!
Sorry about the delay of the grading of the state answers I will get them back to you as soon as possible.
I hope things will be better next year!

Thanks!

[quote=AwesomeToad][quote="stargroup"]oh oops I didn't realize I posted in a mock MATHCOUNTS competition

nvm I'm out haha[/quote]

Sometimes it would be helpful to read the title of the thread you're posting in maybe? :lol:[/quote]

HI GUYS! I JUST WANTED TO SAY HI BECAUSE  I DON'T KNOW ANYBODY HERE. I'M PRETTY GOOD AT MATH AND I THINK Y'ALL CAN GET PERFECT SCORES. EVERYBODY CAN BE SMART IF THEY ACTUALLY TRY AND DO THEIR BEST. GOOD LUCK!

[color=#f00]Welcome to AoPS!  But please try not to revive ten year old threads to add discussion that isn't relevant to the original topic :) ~dj[/color]


geometry
3D geometry
sphere
parameterization
real analysis
real analysis unsolved

[hide]As you know one can compute the order of contact of two curves at a certain point $ P$ by comparing the terms of their Tailor series, by definition.

there is another way, when we want to compute the order of contact between a curve and it's osculating circle at point $ P$:[/hide]

assume that $ O(s)$ be the origin of osculating circle to curve at point $ s$ for a [b]fixed point[/b] $ s$. define: $ r(t): \equal{}O(s)\minus{}C(t)$ where $ C(t)$ is an arbitrary parametrization for that curve. It is obvious that the first non-zero term of the Tailor series of $ r(t)$ around $ s$ determines the order of contact.

now there is a question: can we extend this method in order to compute the order of contact between a curve and it's osculating [b]sphere[/b]?

That's already stated for spheres, in any number of dimensions.

If you talk about surfaces instead of curves, there isn't an osculating sphere in general.

It's alright. but how can I prove it?
Or this is definition? if yes, can you say where could I find it?


algebra
polynomial
number theory proposed
number theory

Prove that for any integer $ n$, there exists a unique polynomial $ Q$
with coefficients in {$ 0, 1, ...,9$} such that $ Q( \minus{} 2)$ $ \equal{}$ $ Q( \minus{} 5)$ $ \equal{}$ $ n$.

A tough one:
http://www.mathlinks.ro/Forum/viewtopic.php?p=343873#343873


Suppose that the set $ \{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $ \{a_i,b_i\}$ ($ 1\leq i\leq 999$) so that for all $ i$, $ |a_i \minus{} b_i|$ equals $ 1$ or $ 6$. Prove that the sum
\[ |a_1 \minus{} b_1| \plus{} |a_2 \minus{} b_2| \plus{} \cdots \plus{} |a_{999} \minus{} b_{999}|
\]
ends in the digit $ 9$.

Here is one proof. I do not think it's very well written, especially the lemma (did I even need a lemma??) but here it is. (Sorry, I also didn't know how to make the congruence sign, and I was in a hurry, so I just put equal signs.)

[hide="lemma"]
Define $ x$ as the number of pairs where $ \[ |a_{i} \minus{} b_{i}| \equal{} 6$, and similarly define $ y$ as the number of pairs where $ \[ |a_{i} \minus{} b_{i}| \equal{} 1$.

We can prove that $ a \equal{} 0 (mod 2)$.

Assume a was not even. Then, we would have some interval, $ (m, m \plus{} 2n)$, where $ m$ and $ n$ are positive integers, where the middle numbers weren't any $ a_i$ or $ b_i$. We could not assign pairs $ |a_{i} \minus{} b_{i}| \equal{} 1$ to these integers.

So, a must be even.
[/hide]

[hide="main proof"]
Using definitions of $ a$ and $ b$ from the lemma, we know that
$ a \plus{} b \equal{} 999$.
Using our results from the lemma,

$ a \equal{} 0 (mod 2)$
$ \Leftrightarrow 5a \equal{} 0 (mod 10)$
$ \Leftrightarrow 5a \plus{} 999 \equal{} 999 (mod 10)$
$ \Leftrightarrow 5a \plus{} 999 \equal{} 9 (mod 10)$

Using the equation $ a \plus{} b \equal{} 999$, we substitute:

$ 5a \plus{} a \plus{} b \equal{} 9 (mod 10)$
$ \Leftrightarrow 6a \plus{} b \equal{} 9 (mod 10)$

We notice that 6a + b represents the total sum, and it is congruente to 9 modulo 10, i.e. its last digit is nine.

QED

 [/hide]

$ \text{USAMO} \not\subset \text{Intermediate Topics}$.  Choose more appropriate fora in the future.  Unrelatedly, you have some notational errors in your solution you might want to fix.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=343865#343865

Please do not double post.


function
geometry
3D geometry
LaTeX
algebra
domain
logarithms

Show that the cube root of z is an analytic function.
(Sorry about the lack of symbols but I am trying to get Latex up and running)

What do you mean, "the" cube root? You have to choose a branch.
Also, the origin is a bad point- whatever region you choose as a domain will have to exclude it, as well as some curve between zero and $ \infty$.

Some possible approaches:
- Write down the derivative.
- Write down a power series (the binomial series).
- Write it as a composition involving $ \log$ and $ \exp$.


algebra
polynomial
number theory
greatest common divisor

For how many integers $N$ between 1 and 1990 is the improper fraction $\frac{N^2+7}{N+4}$ NOT in lowest terms?

A. 0
B. 86
C. 90
D. 104
E. 105

[hide="Solution"]
$N+4$ and $N^2 + 7$ must have some common factor $p$ in order for the expression not to be in lowest terms.  Thus, $(N+4)^2 - (N^2 + 7) - 8(N+4) = 23$ must also share this common factor $p$, which must therefore be $23$.  Hence we want $N+4 = 23k < 1990$.  We find that there are $86$ values of $N$ for which this is true.  Hence, our answer is $\boxed{B}$
[/hide]

[hide]
Answer: B
Given (N^2+7)/(N-4), we can use synthetic division to get
N-4+23/(N-4)
Since N-4 is always an integer, we only need to ensure that the term 23/(N-4) is in the lowest terms. In this, it is important to consider that 23 is a prime number, rendering it impossible to reduce the numerator. As a result, the only denominator that can be less than 23 so that the resulting fraction can be reduced is 1, which is impossible since then N would equal -3, outside the given range. Also, as a result, N-4 must be a multiple of 23. Between 1 and 1990, there are 86 such multiples of 23. Now, if 4 is added to the greatest multiple in order to solve for N, the greatest value of N remains less than 1990, and thus within the given range. Therefore, the answer is [b]86[/b][/hide]

This seems way too hard for a 19 out of 30. :huh:

[quote="chess64"]Maybe because you guys are making it way too complicated... :roll:

[hide="Solution"]
Use polynomial long division to find $\frac{N^2+7}{N+4}=N-4+\frac{23}{N+4}$. If this is in lowest terms, then $\frac{23}{N+4}$ must be in lowest terms. The only way that $\text {gcd}(N+4,23)\ne 1$ and $1\le N\le 1990$ is if $N\in \{ 19, 42, 65, \ldots, 23\cdot 86-4\}$, so there are $\boxed {86}$ such $N$.
[/hide][/quote]

Ahhhh... that makes me feel a lot better  :P

[quote="236factorial"][quote="t0rajir0u"]Thus, $(N+4)^2 - (N^2 + 7) - 8(N+4) = 23$ must also share this common factor $p$, which must therefore be $23$. [/quote]

Where did that come from?[/quote]

Well, I had considered the synthetic division solution, but thought this one was a little purer, so what I did was to try to show that a constant had to have $p$ as one of its factors, so I considered $(N+4)^2 - (N^2 + 7) = 8N + 9$, which also has to have this factor, but I wanted to eliminate $N$ entirely, so, reasoning that $8(N+4)$ must have this factor, I subtracted and, er, took absolute value.   :D


Hi,
I tried to use latex for my resume. Need to add an article class with .cls..
how to add it (where to save the file I downloaded in my computer)? Thanks!

What do you mean by a new article class? Exactly what did you download and from where? Does it require other packages? What LaTeX distribution are you using? What is the class called? Is it already there? What happens if you use \documentclass{[i]name of class[/i]} in your document?
In other words you must provide much more information about your system and what you've done to answer your question.

I downloaded a resume document in .tex. I found that TexnicCenter could use the usepackage directly. (sorry, somehow, I made a mistake for using "article class" instead of "documentclass" :( ).


In regular pentagon $ PQRST$, $ X$ is the midpoint of segment $ ST$. What is the measure of angle $ XQS$?

[asy]size(101);
picture p;
pair P  = (0,0); pair Q = (-cos(108),sin(108)); pair R = Q + expi(36*pi/180);  pair T = (1,0); pair S1 = T + (cos(108),sin(108));
draw(p,(P--Q--R--S1--T--cycle),linewidth(1)); pair X = (S1+T)/2;
label(p,rotate(20)*"$P$",P,rotate(20)*W,fontsize(10pt)); label(p,rotate(20)*"$Q$",Q,rotate(20)*NW,fontsize(10pt)); label(p,rotate(20)*"$R$",R,rotate(20)*NE,fontsize(10pt)); label(p,rotate(20)*"$S$",S1,rotate(20)*E,fontsize(10pt)); label(p,rotate(20)*"$T$",T,rotate(20)*S,fontsize(10pt));label(p,rotate(20)*"$X$",X,rotate(20)*SE,fontsize(10pt));
dot(p,X);
add(rotate(-20)*p);[/asy]

Triangle RQS is an isosceles triangle, so <RSQ is (180-108)/2=36 degrees. Triangle XQS is a right triangle, so <XQS is 180-<QXS-<XSQ=180-90-(108-36)=90-72=18 degrees.


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let $ ABC$ be a triangle inscribed in circumcircle $ (O)$. Denote $ A_1,B_1,C_1$ respectively be the projections of $ A$,$ B,$ $ C$ onto $ BC,$ $ CA,$ $ AB$. Let $ A_2,B_2,C_2$ respectively be the intersections of $ AO,BO,CO$ with $ BC,$ $ CA,$ and $ AB$. A circle $ \Omega_a$ passes through $ A_1$, $ A_2$ and tangents to the arc of $ BC$ that doesn't contain $ A$ of $ (O)$ at $ T_a$. The same definition holds for $ T_b,T_c$. Prove that $ AT_a$, $ BT_b$ and $ CT_c$ are concurrent.

[i]Best regard,
[b]mathVNpro[/b] [/i]

Since the circle $ \odot(AA_1A_2)$ is tangent to $ (O)$ at $ A,$ then the tangent $ \ell_1$ to $ (O)$ at $ A$ is the radical axis of $ \odot(AA_1A_2)$ and $ (O).$ The sideline $ BC$ is the radical axis of $ \odot(AA_1A_2) , \Omega_a$ and the tangent $ \ell_2$ to $ (O)$ at $ T_a$ is the radical axis of $ \Omega_a$ and $ (O).$ Hence,  lines $ BC ,$ $ \ell_1 ,$ $ \ell_2$ concur at the radical center $ O_a$ of circles $ (O) ,\Omega_a$ and $ \odot(AA_1A_2).$ It's known that $ O_a \equiv \ell_1 \cap BC$ is the center of the A-Apollonius circle of $\triangle ABC,$ which is orthogonal to $ (O)$ $ \Longrightarrow$ $ AT_a$ is the A-symmedian. By similar reasoning we conclude that lines $ AT_a,BT_b,CT_c$ concur at the symmedian point of $ \triangle ABC.$

The generalization of this problem also holds:(instead of $ H$ is orthocenter and $ O$ is circumcenter we have $ H,O$ are arbitary isogonal conjugates points).  :wink:
Proof:Let $ (AA_1A_2)$ cut $ AB$ at $ A_3$ and $ AC$ at $ A_4$.Then $ \angle{A_3A_4A_2} \equal{} \angle{A_4A_3A_1} \rightarrow$ $ A_1A_2A_3A_4$ is isoceles trapezium.From that $ A_3A_4 \| BC \rightarrow$ $ (AA_3A_4)$ is tangent to $ (ABC)$.The back of this proof is same of luisgeometria's proof.


induction
modular arithmetic
number theory unsolved
number theory

Show that every integer greater than $ 89.315$ is the sum of two abundant numbers.

Well, this is not a solution to your problem, but I will sketch a weaker result:

"Every large integer is a sum of two abundant numbers". More precisely, exists a number $ N$ such that every integer larger than $ N$ is a sum of two abundants.

Let $ S(k)$ the sum of all positive divisors of $ k$. If $ S(k)>2k$ then $ k$ is abundant.

[b]Lemma 1:[/b]
If $ k\equal{}{p_1}^{e_1}{p_2}^{e_2}\ldots{p_i}^{e_i}$ is the canonical factoring of $ k$ then
\[ S(k)\equal{}\frac{{p_1}^{e_1\plus{}1}\minus{}1}{p_1\minus{}1} \cdot \frac{{p_2}^{e_2\plus{}1}\minus{}1}{p_2\minus{}1} \ldots \frac{{p_i}^{e_i\plus{}1}\minus{}1}{p_i\minus{}1}\]

This can be proved with double counting techniques or with a simple induction (and even can be generalized : try to calculate the sum of powers of divisors!).

[b]Lemma 1.1:[/b]
\[ \frac{S(k)}{k}\equal{} \prod_{1 \leq j \leq i}\left({\sum_{0 \leq t \leq e_j}{p_j^{\minus{}t}}}\right)\]

[b]Lemma 2:[/b]
For any $ m \geq 1$ and $ k \geq 1$, $ S(km)/km \geq S(m)/m$ with equality if and only if $ k\equal{}1$.
This can be proved by a two-parallel-way induction:

1) If $ k$ has no common prime factors with $ m$ then it adds another primes to the $ km$ factoring, and so it adds other $ {\sum_{0 \leq t \leq e_j}{p_j^{\minus{}t}}}$ factors to the product above;

2) If $ k$ has some common prime factor with $ m$, some of existing factors {\sum_{0 \leq t \leq e_j}{p_j^{-t}}} gain an extra term $ p_i^t$.

In every case, the only way to make none of adds above is $ k\equal{}1$.

[b]Lemma 2.1:[/b]
If $ m$ is perfect or abundant then $ Km$ is abundant for $ K>1$.

[b]Lemma 2.1.1:[/b]
Every multiple of 6, greater than 6, is abundant!

If $ N \equiv t \pmod 6$ then $ N\equal{}6M\plus{}t$, and we can force $ M>1$ just in the case. But we need to guarrantee that $ t$ is abundant. Or, better, we need to arrange a residue class of 6, containing only abundant numbers.
If we find an abundant $ N \equiv 1 \pmod 6$ then our residue class can be just $ \{6N,1N,2N,3N,4N,5N\}$; and if we can find an abundant $ N \equiv \minus{}1 \pmod 6$ it is just fine too (the residue class is reverted!$ \{6N,5N,4N,3N,2N,1N\}$).

Every prime (except 2 and 3) leaves a residue $ 1$ or $ 5 \equiv \minus{}1$ modulo 5. Our idea is to find an abundant number of form $ N\equal{}5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \ldots$. You can verify that $ 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31$ works!

Maybe adapting this idea and using greater exponents instead I can obtain better estimates... But, in fact, as far as I know, the lower limit is very large! 89315=5*17863 appears to be very small...

Obs.: The limit for "even as sum of two abundants" is far less.


two boats A and B lives a port C at the same time on different routes.B travelledon a bearing of 150degrees and A travelled north of B.when A had travelled 8km,B had travelled 10km and the distance between them was 12km.Calculate the bearing of A's route from C.

Those sides call for the Law of Cosines.

We are looking for the center angle at C, right? So use
$ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab \cos C$
and $ 160 \cos C \equal{} 64 \plus{} 100 \minus{} 144$
$ \cos C \equal{} \frac{1}{8}$
Solving with a calculator yields 82.819244218541718676961004373851 degrees. Sadly, I don't remember the way bearings turn anymore.


IMO Shortlist
geometry unsolved
geometry

The circles $ k_1$ and $ k_2$ with respective centers $ O_1$ and $ O_2$ are externally tangent at point $ C$, while the circle $ k$ with center $ O$ is externally tangent to $ k_1$ and $ k_2$. Let $ l$ be the common tangent of $ k_1$ and $ k_2$ at the point $ C$ and let $ AB$ be the diameter of $ k$ perpendicular to $ l$. Assume that $ O$ and $ A$ lie on the same side of $ l$. Show that the lines $ AO_2,BO_1,l$ have a common point.

Well, it's nice to hearing that this problem is from Bulgaria 1996. It was also proposed for IMO Shortlist 2006, you can find some nice solutions here: http://www.mathlinks.ro/viewtopic.php?p=875026#875026


[hide]

If x is relatively prime to 13 then by Fermat's Little Theorem

x^12 = 1 (mod 13)

(x^12)^3 = x^36 = 1 (mod 13)

x^37 = x (mod 13)

Otherwise if 13 | x

x^37 = x = 0 (mod 13)

[/hide]

[quote="Iris Aliaj"]Prove that the congruence x^37≡x(mod 13) is valid for any integer x.[/quote]

I will prove FLT:

Modulo p, [0, 1, .... , p-1] is congruent to [0, a , 2a , ..... a(p-1)]

Hence:

(p-1)! = a^(n-1)(p-1)! mod p

But (p-1)! = -1 mod p -->

1 = a^(n-1) mod p as desired.


Help me please!

What's the maximum number of regions formed by lines in the plane?

This is obviously $ \binom{n\plus{}1}{2}\plus{}1$.  But what if $ k$ are parallel?  Or $ c$ are concurrent?  What about planes in space?

Generally, in the $ n^{th}$ dimension, the maximum number of regions that $ k$ things of $ (n \minus{} 1)^{th}$ dimension can cut it into is the sum of the first $ n \plus{} 1$ numbers in the $ k^{th}$ row of Pascal's Triangle, where the zeroth row is $ 1$, the first row is $ 1,1$, etc., i.e. $ R_n \equal{} \left\{\dbinom nj\right\}_{j \equal{} 0}^n$, or equivalently, $ \sum_{i \equal{} 0}^n\dbinom ki$, where if $ k < i$, we have $ \dbinom ki \equal{} 0$.

[quote="AIME15"]

This is > obviously < $ \binom{n \plus{} 1}{2} \plus{} 1$. [/quote]
This is a silly statement.  There is no "obviously" about it.

You just state "This is $ \binom{n \plus{} 1}{2} \plus{} 1$."

[quote="AIME15"]But what if $ k$ are parallel?  Or $ c$ are concurrent?[/quote]

Then we haven't attained the maximum number of regions formed by lines in the plane.

...I must be missing something.  :maybe:

I think he means the maximum possible given that $ k$ lines are parallel, $ c$ lines are concurrent (probably not too clear, but something of that nature), or both.


No, I am pretty certain that you cannot go as an individual if you do not make your school's team.

I'm pretty sure you can sign up as "homeschooled".  I don't think you actually have to be homeschooled.

Probably the smartest thing to do, instead of asking random AoPSers who have not been in a similar situation before and can do nothing more than make educated guesses, would be to find your chapter coordinater's email and ask him specifically.

[quote="athunder"]If one does not make the school team for whatever reason, is he/she aloud to go and still take the mathcounts test officially?

I am in a situation like this.

I know you can sign up as an individual if your school does not offer it, but what if it does?

Thanks.[/quote]

Do you mean that if you didn't make the top 4? Because, register your school can 4 team members and 4 individuals. Or, do you mean that you didn't make the top 8? If that happens, you can be an alternate in chapter and practice with the team. Thats up to your ranking in the school competition and the discretion of the teacher.


number theory
greatest common divisor
relatively prime
prime factorization

If $ \frac{1}{a}\plus{}\frac{1}{b}\equal{}\frac{1}{c}$, where a, b, c are positive integers with no common factor, prove that $ (a \plus{} b)$ is the square of an integer.

[hide="solution"]$ \dfrac{a \plus{} b}{ab} \equal{} \dfrac{1}{c}$

$ ab \minus{} ca \minus{} cb \equal{} 0$

$ ab \minus{} ca \minus{} cb \plus{} c^2 \equal{} c^2$

$ (a \minus{} c)(b \minus{} c) \equal{} c^2$

Since $ a$, $ b$, and $ c$ have no common factors with each other other than 1, $ a \minus{} c$ and $ b \minus{} c$ may not have any common factors with $ c^2$ other than 1, but that's impossible if $ c^2$ is not equal to 1. Therefore, $ c \equal{} 1$, and $ a \equal{} b \equal{} 2$, but that's impossible.[/hide]

Am I missing something?

I think he means there is no factor which all three have in common.

[hide]
Rearranging and factoring gives $ (a\minus{}c)(b\minus{}c) \equal{} c^2$. If this has integer solutions we must have $ a\minus{}c\equal{}k$ and $ b\minus{}c \equal{} c^2/k$ where $ k$ is some factor of $ c^2$. Then $ a\plus{}b \equal{} \frac{1}{k}(k\plus{}c)^2$. It remains only to show that $ k$ must be a perfect square. Probably something to do with the relatively prime condition.
[/hide]

I think 1=2 makes a good point.

$ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {1}{c}$

Therefore

$ \frac {a \plus{} b}{ab} \equal{} \frac {1}{c}$

Therefore $ a \plus{} b \equal{} \frac {ab}{c}$

But since $ gcd(c,ab) \equal{} 1$ if $ a \plus{} b$ is an integer, $ c \equal{} 1$

Therefore it is as 1=2 suggests

[b]Related Problem[/b]
BMO 2: 1998 Q3

Suppose x, y, and z are integers satisfying
\[ \frac {1}{x} \minus{} \frac {1}{y} \equal{} \frac {1}{c}
\]
Let h be the highest common factor of x, y, and z.

Prove

i) hxyz is a perfect square
ii) h(y-x) is a perfect square

[b]@1=2:[/b] Why is $ (a,b,c) \equal{} (2,2,1)$ impossible? Also consider $ \frac {1}{a} \plus{} \frac {1}{a(a \minus{} 1)} \equal{} \frac {1}{a \minus{} 1}$, where clearly the GCD is 1 since $ a$ and $ c \equal{} a \minus{} 1$ are co-prime.

[b]@SimonM:[/b] I think the thing here is that we don't necessarily have $ \text{GCD}[a,b] \equal{} \text{GCD}[b,c] \equal{} \text{GCD}[c,a] \equal{} 1$, only $ \text{GCD}[a,b,c]$

Note that we must have $ a|bc \; \; \; b|ac \; \; \; c|ab$

[hide="my solution"]
Say $ c \equal{} a \minus{} d$.

$ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {1}{a \minus{} d} \\
\Rightarrow \frac {ab}{a \plus{} b} \equal{} a \minus{} d \\
\Rightarrow d \equal{} a \minus{} \frac {ab}{a \plus{} b} \\
\Rightarrow d \equal{} \frac {a^2}{a \plus{} b} \\
\Rightarrow d(a \plus{} b) \equal{} a^2$

Consider a prime factor $ p$ of $ d$. It is clear that $ p|a$, so $ a$ and $ d$ have p in common, and therefore $ a$ and $ c$ have $ p$ in common. Follows that $ p$ doesn't divide $ b$, so neither doess it divide $ a \plus{} b$. We conclude that $ p$ occurs equally many times in $ d$ as in $ a^2$. We do this for every prime factor of $ d$ and so we see that $ d$ is a perfect square. Therefore also $ (a \plus{} b)$ is a perfect square.
[/hide]

Yeah, that's what I was thinking, [b]Mat(h)s[/b].

Nice proof, [b]Mat(h)s[/b].

Here is my conjecture of a stronger proposition:

Given $ \gcd(a,b,c)\equal{}1$, if $ (a,b,c)$ satisfies $ \frac1a\plus{}\frac 1b\equal{}\frac1c$, then $ (a,b,c)\equal{}\left(p(p\plus{}q),q(p\plus{}q),pq\right)$ for $ \gcd(p,q)\equal{}1$.

True or false?

[quote="The Zuton Force"]
[b]@SimonM:[/b] I think the thing here is that we don't necessarily have $ \text{GCD}[a,b] \equal{} \text{GCD}[b,c] \equal{} \text{GCD}[c,a] \equal{} 1$, only $ \text{GCD}[a,b,c]$
[/quote]

Ah, it didn't read like that when I first read it.

[hide="Solution"]We have $ bc\plus{}ac\equal{}ab$.  Note that if any prime $ p$ divides one of $ a,b,c$, it must divide exactly one of the other two.  Furthermore, suppose for instance $ p^{k}$ strictly divides $ a$.  Then $ p^{k}$ divides $ bc$, so $ p^{k}$ divides either $ b$ or $ c$.  But if $ p^{k\plus{}1}$ divides either $ b$ or $ c$, then $ a$ must be divisible by $ p^{k\plus{}1}$ by the same logic, a contradiction.  Therefore, if $ p^{k}$ strictly divides one of $ a,b,c$, then $ p^{k}$ strictly divides another of $ a,b,c$, and the third is not divisible by $ p$.

Now let $ x\equal{}(a,b)$, $ y\equal{}(b,c)$, and $ z\equal{}(c,a)$.  Every power of a prime $ p_{i}^{a_{i}}$ that is part of the prime factorization of $ a$ is also part of the prime factorization of exactly one of $ b,c$.  Therefore, $ a\equal{}xz$, and similarly $ b\equal{}xy$ and $ c\equal{}yz$.  Substituting, we get $ xy\cdot yz\plus{}xz\cdot yz\equal{}xz\cdot xy$.  Dividing through by $ xyz$, we get $ y\plus{}z\equal{}x$.

Then $ a\plus{}b\equal{}xz\plus{}xy\equal{}x(y\plus{}z)\equal{}x^{2}$ as desired.[/hide]

Oh, thank for all! I don't think we can have many proofs as above.

[quote="NguyenDungTN"]If $ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {1}{c}$, where a, b, c are positive integers with no common factor, prove that $ (a \plus{} b)$ is the square of an integer.[/quote]

Sorry for reviving this thread...but i didnt see a satisfactoryanswer so i decided to post mine.

solving the equation gives us
$ ab\minus{}cb\minus{}ca\equal{}0$
$ \equal{}> (a\minus{}c)(a\minus{}b)\equal{}c^2$  (*)

$ (a,b,c)\equal{}1$ $ \equal{}> (a\minus{}c,b\minus{}c,1)\equal{}1$       (**)
fact * and ** together imply that $ a\minus{}c\equal{}y^2$ and  $ b\minus{}c\equal{}z^2$

this means that $ c \equal{} yz$
$ a\plus{}b \equal{} y^2\plus{}z^2\plus{} 2c\equal{}y^2\plus{}z^2\plus{}2yz\equal{}(y\plus{}z)^2$ . Hence we are done


inequalities
induction
inequalities unsolved

Hey,

We want to prove the inequality: $ (\frac{n\plus{}1}{2})^n(\frac{2n\plus{}1}{3})^n > (n!)^2$.

Notice that the following equality holds for each positive integer $ n$ (it can be proved by induction on n):

     $ 1^2\plus{}2^2\plus{}...\plus{}n^2\equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{6}$

Therefore considering the left part of our desired inequality,we have:

 $ (\frac{n\plus{}1}{2})^n(\frac{2n\plus{}1}{3})^n\equal{}(\frac{(n\plus{}1)(2n\plus{}1)}{6})^n\equal{}(\frac{1^2\plus{}2^2\plus{}...n^2}{n})^n>1^22^2...n^2\equal{}(n!)^2$

because of the AM-GM.Notice that the inequality is strict!!

~Kostas

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=306724[/url]


geometry
3D geometry
vector
LaTeX
calculus
linear algebra
matrix

Find the angle between the diagonal of cube and  another diagonal of the
cube.

I can visualize it, but how do you express the vector for the 2nd diagonal? I'm using a unit cube so $ u \equal{} <1,1,1>$ is my first diagonal but I don't know how to express the other diagonal in component form.

$ \text{\LaTeX}$ hint: use \langle and \rangle for that kind of bracket, not < and >.

Your vector $ \langle 1,1,1\rangle$ is the displacement from the point $ (0,0,0)$ to $ (1,1,1).$ That's two of the 8 vertices of the cube. Now, pick two [i]other[/i] diametrically opposed vertices; it shouldn't matter which. What's another vertex? Say, $ (1,0,0).$ Now find what's opposite to that by just changing ones to zeros and zeros to ones. That gives us $ (0,1,1).$ The displacement vector would then be $ \langle -1,1,1\rangle$ - or maybe that in the other direction, $ \langle 1,-1,-1\rangle.$ You can try both, and report the angle you get as the acute angle.

Vaguely related chemistry question: What is the H-C-H bond angle in methane?

[quote="Kent Merryfield"]$ \text{\LaTeX}$ hint: use \langle and \rangle for that kind of bracket, not < and >.

Your vector $ \langle 1,1,1\rangle$ is the displacement from the point $ (0,0,0)$ to $ (1,1,1).$ That's two of the 8 vertices of the cube. Now, pick two [i]other[/i] diametrically opposed vertices; it shouldn't matter which. What's another vertex? Say, $ (1,0,0).$ Now find what's opposite to that by just changing ones to zeros and zeros to ones. That gives us $ (0,1,1).$ The displacement vector would then be $ \langle - 1,1,1\rangle$ - or maybe that in the other direction, $ \langle 1, - 1, - 1\rangle.$ You can try both, and report the angle you get as the acute angle.

Vaguely related chemistry question: What is the H-C-H bond angle in methane?[/quote]

Ah, thank you. I didn't think of having negative coordinates. I did get the same as the bond angle, which is 109.5. Coincidentally, that happens to be #45 in my book :]

I have a question about the section though. I am doing this problem from 11.1 of Strang's Calculus. I always thought vectors were a part of calculus, so why are they in the Linear Algebra section? I know you can express a vector as a column matrix, but I didn't know they played a big role in matrix algebra. Guess I have a lot more to learn lol!

I'd probably report the answer to this particular problem as $ \cos^{\minus{}1}\left(\frac13\right)\approxeq 70.5^{\circ}$ - the acute angle rather than the obtuse angle. But that's not particularly important.

I'm the one who moved the problem. Yes, I am very much aware that this comes from a calculus course and there's a chapter in most calculus books containing exactly this sort of material. It's just that I wanted to make the point that not everything taught in such courses is actually calculus - if it's actually vector algebra or vector geometry, I think it fits a little better in this forum.

I have one more question.

27 The force F = 3i - 4k acts at the point (1,2,2). How much
force pulls toward the origin? How much force pulls vertically
down? Which direction does a mass move under the force F?

For vertically downward, the answer is stated as 4. How is that computed?

The vertically downward vector should be $ \langle 1,2,2\rangle \minus{} \langle 1,0,2 \rangle$ correct? That gives a vector of $ \langle 0,2,0 \rangle$, but the magnitutde of the projection comes out to be zero.

Not correct.  You're asked to find the downwards component of the force vector, not the $ z$-coordinate of the position.  (Note that in three dimensions, the third coordinate is the vertical coordinate, not the second.)  For the first part of the question, you want to find the component of the force in the direction that joins the origin to the current position.

I got the first and third part, but I'm not sure how to compute the second part.


calculus
integration
trigonometry
Putnam
real analysis
real analysis unsolved

See and the exercise $8.1.$ from http://www.mathlinks.ro/Forum/viewtopic.php?t=66197


vector
abstract algebra
Functional Analysis
advanced fields
advanced fields unsolved

Let $X,Y$ be Banach spaces, and let $T: X\to Y$ be a continuous linear map such that the linear space $Y/T(X)$ is finite-dimensional.

Prove that the range of $T$ (that is, $T(X)$) is a closed subspace of $Y$.

I'll try this:  Suppose not, if necessary by replacing the spaces we can assume that $cl(TX)=Y$.  Let a vector $v$ in $Y$ representing some vector in $Y/X$.  Then we can approximate v in TX by $Tx_{n}$, which I can choose to be a linearly independent set, here I use that v is not in the image of X.  We define $A_{n}\in Y^{*}$ as $A_{n}(v)=1, A_{n}(Tx_{i})=0, \forall i<n$ and extend them each by Hahn-Banach, here I use the linearly independence.  By Banach-Alaoglu(mispelling?) theorem, exist $A\in Y^{*}$ s.t. $A_{n}$ weak* converges to A.  By definition that means $ATx_{n}=0, Av=1$, contradiction.

But I still kinda feel that there's some problem in the proof, I would be very appreciative if someone would like to point it out.  Finite dimension is not used, but I am not concious enough to see where it should be.  I'll sleep first... :lol:

Banach-Alaoglu says that the unit ball of $Y^{*}$ is weak-$*$ compact. In order to apply it, you need a boundedness condition on your sequence of functionals $(A_{n})_{n}$, which you don't have.

P.S.

I just wanted to point out that (a) as you noticed, you did not use the fact that $Y/TX$ is finite-dimensional, and there are plenty of coutnerexamples which show that it's a crucial condition, and (b) you did not use the fact that $TX$ is the range of a continuous operator $T$; that too is crucial, since there are finite-codimensional non-closed subspaces of $Y$ (the kernel of a discontinuous linear functional, for example).

There's nothing redundant in the hypothesis.

First, we can suppose $T$ is injective, otherwise we replace $T$ with $T/\mathrm{ker}T$ wich has the same range.

(i) By the hypotesis, there exist linearly independent $y_{1},...,y_{n}$ s.t. $Y=\langle y_{1},...,y_{n}\rangle+T(X)$. Moreover, using linear independence of the $y_{i}$, it can be shown that the $+$ above is in fact a $\oplus$.

(ii) Define a map $\phi: X\times\mathbb{R}^{n}\to Y$, sending $(x,t_{1},...,t_{n})$ in $t_{1}y_{1}+...t_{n}+y_{n}+T(x)$. This is linear, continuous and surjective by construction (hence an open map) and injective by assumption. So its inverse is continuous.

(iii) Take $\{y_{k}\}$ in $T(X)$ which converge to $y\in Y$. Then $\phi^{-1}(z_{k})=: (x_{k},0,...,0)\to\phi^{-1}(y)$. So $\phi^{-1}(y)=(x,0,...,0)$ for some $x\in X$, that is, $y=T(x)$.


calculus
derivative
integration
logarithms
calculus computations

If $ x^{3}\minus{}3a^{2}x\equal{}a^{2}y$ show that
$ \frac{dx}{\sqrt{x^{2}\minus{}4a^{2}}}\equal{}\frac{dy}{3\sqrt{y^{2}\minus{}4a^{2}}}$
and hence obtain Cardan's formula for the solution of a cubic.

Proving the differential equation is just a straightforward exercise of implicit differentiation, but I can't seem to get the solution of a cubic from it, integrating both sides results in a rather large mess...Any suggestions?

The formula is messy no matter what; you'll need to get the integral in terms of logarithms, and do some complicated algebra.

Actually, this method is a lot nicer if we look at the case where $ y$ is small and there are three real roots- it's an angle trisection.


[quote="surshu"]obviously x=3, y=2[/quote]

You must always show your work surshu...

[hide]
$2^{2x}-2^{2y}=48$
$2^{x}\times2^{y}=32$
Let $X=2^{x}$ and $Y=2^{y}$
$(X-Y)(X+Y)=48$ and $XY=32$
For X and Y to have a natural value in the first equation, there are only 3 possible pairs: (X-Y,X+Y)=(6,8),(2,24),(4,12)

If you solve for X and Y, only (4,12) gives you perfect powers of 2. $X=8$ and $Y=4$. This fits into the second equation well.

Finally, we see that $x=3$ and $x=2$.
$3x-4y=9-8=1$
[/hide]

[quote="Andrew Kim"]
For X and Y to have a natural value in the first equation[/quote]
Although this works, you don't know that they're natural


$p_1,p_2,...,p_{25}$ is primes which is less than  $2004$Find the maximal integer $T$ such that all of positive integer which is not greater than $T$ can be represented as sum of distinct divisor of  $(p_1.p_2...p_{25})^{2004}$


$ \frac{s^2}{2p}\equal{}r$ then is equal to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=215261]this[/url]


limit
logarithms
integration
calculus
calculus computations

Calculate       $\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}$

$\log \left ( \frac{\sqrt[n]{n!}}{n}\right ) = \log \left ( \frac{\prod_{i=1}^{n}i^{\frac{1}{n}}}{n}\right ) = \sum_{i=1}^{n}\log \left ( \frac{i^{\frac{1}{n}}}{n}\right ) = \frac{1}{n}\sum_{i=1}^{n}\log \left ( \frac{i}{n}\right )$

Therefore,
$\lim_{n \to \infty}\log \left (\frac{n}{\sqrt[n]{n!}}\right ) = \lim_{n \to \infty}\log \left (\frac{1}{\frac{\sqrt[n]{n!}}{n}}\right ) =-\lim_{n \to \infty}\log \left ( \frac{\sqrt[n]{n!}}{n}\right )$
$=-\lim_{n \to \infty}\left ( \frac{1}{n}\sum_{i=1}^{n}\log \left ( \frac{i}{n}\right ) \right ) =-\int_{0}^{1}\log x \ dx = 1$

Thus,
$\lim_{n \to \infty}\frac{n}{\sqrt[n]{n!}}= e^{1}= e$

[b]Theorem:[/b] Let $(s_{n})$ be a sequence of non-zero terms and $\lim \left| \frac{s_{n+1}}{s_{n}}\right| = L$ (note $L$ can be finite or infinite) then $\lim |s_{n}|^{1/n}= L$.
----

With above theorem it is simple. Consider the sequence $(s_{n}) = \frac{n^{n}}{n!}$ then $\frac{s_{n+1}}{s_{n}}= \frac{(n+1)^{n+1}}{(n+1)n!}\cdot \frac{n!}{n^{n}}$.
Which is,
$=(1+1/n)^{n}$
So,
$\lim \frac{s_{n+1}}{s_{n}}= e$.

Which means that,
$\lim \left| \frac{n^{n}}{n!}\right|^{1/n}= \lim \frac{n}{\sqrt[n]{n!}}= e$

[quote="Kouichi Nakagawa"]$ \log \left ( \frac {\sqrt [n]{n!}}{n}\right ) = \log \left ( \frac {\prod_{i = 1}^{n}i^{\frac {1}{n}}}{n}\right ) = \sum_{i = 1}^{n}\log \left ( \frac {i^{\frac {1}{n}}}{n}\right ) = \frac {1}{n}\sum_{i = 1}^{n}\log \left ( \frac {i}{n}\right )$[/quote]

Sorry :(
I'm mistake.

\begin{eqnarray*}
\log \left( \frac{\sqrt[n]{n!}}{n} \right)
& = & \log \left( \frac{\prod_{i=1}^n i^{\frac{1}{n}}}{n} \right) \\
& = & \log \left( \frac{\prod_{i=1}^n i^{\frac{1}{n}}}{\prod_{i=1}^n n^{\frac{1}{n}}} \right) \\
& = & \log \left( \prod_{i=1}^n \frac{i^{\frac{1}{n}}}{n^{\frac{1}{n}}} \right) \\
& = & \log \left( \prod_{i=1}^n \left( \frac{i}{n} \right)^{\frac{1}{n}} \right) \\
& = & \sum_{i=1}^n \log \left( \left( \frac{i}{n} \right)^{\frac{1}{n}} \right) \\
& = & \frac{1}{n} \sum_{i=1}^n \log \left( \frac{i}{n} \right) \\
\end{eqnarray*}

$ n! \approx \sqrt{2 \pi n} \frac{n^n}{e^n}$

so

$ \frac{n}{\sqrt[n]{n!}}\approx e \frac{1}{\sqrt[\frac{n}{2}]{2 \pi n} } \Rightarrow \lim_{n \rightarrow \infty}\frac{n}{\sqrt[n]{n!}}=e$


The equation $x^{3}+ax^{2}+bx+c = 0$ has the root $\sqrt[3]{(\sqrt{33}+5)}-\sqrt[3]{(\sqrt{33}-5)}$. Find the ordered triple $(a,b,c)$ if $a,b,$ and $c$ are integers.

hmm thats cool whats the motivation behind this?

well the motivation is that you have seen the problem before...like i suppose you can be like hey lets just cube it, that gets rid of stuff...then i guess you just need to be creative to see the recursive substitution


[quote="xdanielxedgex"]Find all real solutions of the equation [quote]$ x^5 \plus{} 10x^3 \plus{} 20x \minus{} 18 \equal{} 0$[/quote][/quote]

$ f(x)\equal{}x^5 \plus{} 10x^3 \plus{} 20x \minus{} 18$ $ \implies$ $ f'(x)\equal{}5x^4\plus{}30x^2\plus{}20>0$ and so $ f(x)$ is increasing and has exactly one real root.

Then numeric calculus gives $ x\equal{}0.711111084051342...$

[quote="xdanielxedgex"]Hi, Can you explain how you obtain this result?[/quote]

The numeric result ?

There exists a lot of method to numerically solve equations. In this case, I used a spreadsheet and a basic Newton method :

$ u_0\equal{}0$
$ u_{n\plus{}1}\equal{}u_n\minus{}\frac{f(u_n)}{f'(u_n)}$


[quote="MCrawford"][quote="zabelman"]I think there's a typo at the bottom of the screen, when it says there have been 8250144700 users online in the last 24 hours. ([i]has[/i] should be [i]have[/i])[/quote]

There's a mistake in your comment above.  The set of all possible users is less than 8 billion.  And we won't have the attention of all of them for another few years at least.[/quote]

You've worked with zableman enough to know he's always thinking ahead.


conics
ellipse
analytic geometry
graphing lines
slope
geometry
geometric transformation

Let the major axis of an ellipse be $AB$, let $O$ be its center, and let $F$ be one of its foci. $P$ is a point on the ellipse, and $CD$ a chord through $O$, such that $CD$ is parallel to the tangent of the ellipse at $P$. $PF$ and $CD$ intersect at $Q$. Compare the lengths of $PQ$ and $OA$.

$\left(\forall\right)P\in \Sigma\ ,\ PQ=OA\ .$. [b]Very nice and interesting problem ![/b][hide="Here is the proof."][color=darkblue]Let $b^2x^2+a^2y^2=a^2b^2$ be the equation of the ellipse $\Sigma$, where $b^2+c^2=a^2$ and let $F(\epsilon c,0)$ be a focus, where $\epsilon ^2=1$. Let $P(au,bv)\in \Sigma$ be a mobile point, where $u^2+v^2=1$. Then the slope of the tangent $PP$ in the point $P\in \Sigma$ to the ellipse $\Sigma$ is $m_{PP}=-\frac{bu}{av}$. The equation of the line $CD$ is $y=-\frac{bu}{av}\cdot x$ and the equation of the line $CF$ is $y=\frac{bv}{au-\epsilon c}\cdot (x-\epsilon c)$. Thus, for the intersection point $Q\in CD\cap PF$ we have $x_Q=\frac{\epsilon acv^2}{a-\epsilon cu}$, $y_Q=-\frac{\epsilon bcuv}{a-\epsilon cu}$ and $\left(\frac{a-\epsilon cu}{a}\right)^2\cdot PQ^2=$ $\left[u(a-\epsilon cu)-\epsilon cv^2\right]^2+b^2v^2=$ $(au-\epsilon c)^2+b^2v^2=$ $a^2u^2+b^2(1-u^2)+c^2-2\epsilon acu=$ $(a^2-b^2)u^2+a^2-2\epsilon acu=$ $c^2u^2+a^2-2\epsilon acu=$ $(a-\epsilon cu)^2\Longrightarrow PQ=a$, i.e. $PQ=OA$.[/color][/hide]

Let the foci be $F_1$ and $F_2$, and let $Q_1 = PF_1 \cap CD$ and $Q_2 = PF_2 \cap CD$.  Let $M$, $R_1$, and $R_2$ be the projections of $P$, $F_1$, and $F_2$ onto $CD$, respectively.

First, we claim that triangles $PMQ_1$ and $PMQ_2$ are congruent.  We are given that $CD$ is parallel to the tangent at $P$, so by the reflection property of ellipses, $\angle MPQ_1 = \angle MPQ_2$.  Also, $\angle PMQ_1 = \angle PMQ_2 = 90^\circ$, and both triangles share side $PM$, so they are congruent.  Therefore, $PQ_1 = PQ_2$.

Next, we claim that triangles $F_1 R_1 Q_1$ and $F_2 Q_2 R_2$ are congruent.  Since $CD$ passes through the centre of the ellipse, foci $F_1$ and $F_2$ are equidistant from $CD$, i.e. $F_1 R_1 = F_2 R_2$.  By the argument above, $\angle R_1 F_1 Q_1 = \angle R_2 F_2 Q_2$, and $\angle F_1 R_1 Q_1 = \angle F_2 R_2 Q_2 = 90^\circ$, so the two triangles are congruent.  Therefore, $F_1 Q_1 = F_2 Q_2$.

But, $PF_1 + PF_2 = AB$, and
\begin{eqnarray*} PF_1 + PF_2 &=& F_1 Q_1 + Q_1 P + PQ_2 - Q_2 F_2 \\ &=& Q_1 P + PQ_2 \\ &=& 2 PQ_1 = 2PQ_2, \end{eqnarray*}
so $PQ_1 = PQ_2 = AB/2 = OA$.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/f044e9673a5d3b9e6f446ce1be0c9dbb.png[/img]


Euler
algorithm
real analysis
real analysis solved

prove that { \emptyset (n)/n} is dense in [0,1]?
( \emptyset (n) means totient of euler and {x} means the decimal part of x) excuse me I don't know if I send it to right topic or not?

Let 2=p1<p2<... be the sequence of all prime numbers.

 \emptyset (p)/p=1-1/p ->1 for p prime, p-> \infty.

 \emptyset (p1*p2*...*pk)/(p1*...*pk)=(1-1/p1)...(1-1/pk)=1/(1+1/p1+1/p1 2 +...)...(1+1/pk+1/pk 2 +...)=1/(\sum 1/x :x is a number that has only prime divisors int he set p1...pk). It is easy to see that the limit of this, when k ->\infty is 0.

For a number a \in (0,1) use the following algorithm:

(00) for k:=1 to \infty do if  (1-1/pk)>a then goto (01);
(01) n:=pk;
(02) for i:=k+1 to \infty do if  (\emptyset (n)/n)(1-1/pi)>a then goto (03);
(03) k:=i; while ( \emptyset(n)/n)(1-1/pk)>a do {n:=n*pk; k:=k+1;} goto (02);

This alghorithm, chained indefinitely, constructs an increasing sequence of n's such as  \emptyset (n)/n decreases to a.

I hope you don't find it that hard to prove that the sequence actually converges to a. You must use: (1-1/pk) increases to 1 and (1-1/p1)*...*(1-1/pk) decreases to 0. You can a;so use this to prove that the alghorithm doesn't stop at some point.
----------------------------------------------------------------------------------
{} - the decimal part, it is superfluous for the problem because  0 \leq \emptyset(n) <n


Does anyone have a link to a [b]free[/b] geometry rendering program?

I'm using TeXgraph, but there is a lot of programs for that: gnuplot,....

Google is your friend :lol:

try graphmatica, from download.com. If you need geometrical shapes, then do what i did. Use a mixture of ppt and paint. :P


How many ways are there to put 4 distinguishable balls into 2 distinguishable boxes?

Each ball has $ 2$ choices, and there are $ 4$ balls, so this is $ 2 \times 2 \times 2 \times 2\equal{}2^4\equal{}\boxed{16}$.

4 choose 0 + 4 choose 1 + 4 choose 2 + 4 choose 3 + 4 choose 4 = 2^4 = 16  :P

$ 2 \times 2 \times 2 \times 2=2^4=\boxed{16}$

[quote=blijkijker]It appeared that 16 is correct. I am wondering, however, if the correct answer should be 14 instead. It is quite explicitly mentioned that the 4 balls should be put into 2 boxes. Therefore, putting all balls in one box should not be counted as a valid solution. Using the same reasoning, putting all balls in box 2 is also invalid, which results in $16-2=14$ valid ways to put the 4 distinguishable balls into 2 distinguishable boxes.[/quote]

There is no restriction in the problem that said "putting all balls in one box should not be counted as a valid solution" You can imagine it as putting 0 balls in box 1 and 4 balls in box 2.

I know this user probably won't reply as their post was made a few years ago, but can someone else confirm my reasoning.

The generalization is there are $m^n$ ways to put $n$ distinguishable balls in $m$ distinguishable boxes.

[quote=Nutty_Nitty][quote=blijkijker]It appeared that 16 is correct. I am wondering, however, if the correct answer should be 14 instead. It is quite explicitly mentioned that the 4 balls should be put into 2 boxes. Therefore, putting all balls in one box should not be counted as a valid solution. Using the same reasoning, putting all balls in box 2 is also invalid, which results in $16-2=14$ valid ways to put the 4 distinguishable balls into 2 distinguishable boxes.[/quote]

There is no restriction in the problem that said "putting all balls in one box should not be counted as a valid solution" You can imagine it as putting 0 balls in box 1 and 4 balls in box 2.

I know this user probably won't reply as their post was made a few years ago, but Can someone else confirm my reasoning.[/quote]

Yeah that is how I interpreted it too. :)


calculus
integration
function
trigonometry
limit
derivative
real analysis

This one's been eating me up for quite some time. Any help is welcome! :blush:
I found in a paper the following equality:
$ I_0(r,z)\equal{}rz\int_z^r\frac {1}{p\sqrt {r^2 \minus{} p^2}\sqrt {p^2 \minus{} z^2}}\,dp \equal{} \frac {\pi}{2}$
$ 0 < z < r < 1$
But it doesn't say how they got there nor has any helpful reference.
So here I am, asking someone who can solve this and show me how to do it, please!
(There's also $ I_1$ and $ I_n$ in the bag, but I'll be glad to solve this one first  :wink: )

[quote="JRav"]http://en.wikipedia.org/wiki/Elliptic_integral[/quote]
Thanks, but my doubt comes from page 596 of [url=http://books.google.com/books?id=MtU8uP7XMvoC&pg=PA596&lpg=PA587&dq=abramowitz&output=html&sig=i1Wvt7QhvO_-Qf5KjOmcNOLLT7I]Abramowitz and Stegun's handbook of mathematical functions[/url]. They have a table with some integrals similar to mine:
17.4.43
$ a\int_b^x\frac{dt}{\sqrt{(a^2\minus{}t^2)(t^2\minus{}b^2)}}$
and
17.4.44
$ a\int_x^a\frac{dt}{\sqrt{(a^2\minus{}t^2)(t^2\minus{}b^2)}}$

But I don't know how to use this thing. I don't know anything about elliptic integrals. That's why I'm asking for help. How can I use this? What are those 'cn', 'sc', 'nd', etc functions?

Also, I have an extra $ p$ in the denominator, so I can't just apply the formula from the book. There's some extra brain crunching to do, but I feel unprepared for such a momentous task.
On a bright note, I put this integral on Mathematica and, after substituting $ y\equal{}p^2,\,dy\equal{}2p\,dp$ it was able to calculate it and gave a correct result. So my theory is: if the computer can do it, then a human can do it. I just need to learn.
Please, if anyone has any ideas, fell free! :blush:

[quote="pocaracas"]
$ I_0(r,z) \equal{} rz\int_z^r\frac {1}{p\sqrt {r^2 \minus{} p^2}\sqrt {p^2 \minus{} z^2}}\,dp \equal{} \frac {\pi}{2}$
  [/quote]

this is an [b]easy[/b] integral, why on earth r u associating it with elliptic integrals  ?   :?:

  divide both num and den by $ rz\,p^3$ to get

 $ I_0(r,z) \equal{} \int_z^r \frac {\frac {1}{p^3}\,dp } {\sqrt {\frac {1}{p^2} \minus{} \frac {1}{r^2} }\sqrt {\frac {1}{z^2} \minus{} \frac {1}{p^2}} } \equal{} \minus{} \frac {1}{2}\int_z^r \frac {d\left(\frac {1}{p^2}\right) } {\sqrt {\frac {1}{p^2} \minus{} \frac {1}{r^2} }\sqrt {\frac {1}{z^2} \minus{} \frac {1}{p^2}} }$

letting $ x \equal{} \frac {1}{p^2}$ gives (notice $ 1 > r > z > 0 )$

$ I_0(r,z) \equal{}\frac {1}{2}\int_{\frac {1}{r^2}}^{\frac {1}{z^2}} \frac {dx}{\sqrt {x \minus{} \frac {1}{r^2}}\sqrt {\frac {1}{z^2} \minus{} x}} \equal{}\frac {1}{2}\;\left\{ \minus{} \tan^{ \minus{} 1} \left( \; \frac { \frac {1}{2r^2} \plus{} \frac {1}{2z^2} \minus{}x}{\sqrt {x \minus{} \frac {1}{r^2} }\sqrt {\frac {1}{z^2} \minus{} x}} \;\right)\;\bigg|_{\frac {1}{r^2}}^{\frac {1}{z^2}}\right\}$

$ \equal{}  \frac {1}{2}\left\{  \;\lim_{p_0\to z}  \; \minus{} \tan^{ \minus{} 1} \left( \frac{\frac {1}{2r^2} \minus{} \frac {1}{2z^2} }{\sqrt { \frac {1}{p_0^2} \minus{} \frac {1}{r^2} }\sqrt {\frac {1}{z^2}  \minus{} \frac {1}{p_0^2} }} \right) \plus{} \;\lim_{p_1\to r}  \;  \tan^{ \minus{} 1}  \left( \frac{\frac {1}{2z^2} \minus{} \frac {1}{2r^2} }{\sqrt { \frac {1}{p_1^2} \minus{} \frac {1}{r^2} }\sqrt {\frac {1}{z^2}  \minus{} \frac {1}{p_1^2} }} \right) \right\}$

 since $ 1 > r > z > 0,$ we have

 $ I_0(r,z) \equal{} \frac {1}{2} \; \left\{ \minus{}  \tan^{ \minus{} 1}  \left( \frac{ something \quad negative} { something \quad going \quad to \quad 0} \right) \plus{}  \tan^{ \minus{} 1} \left( \frac{ something \quad positive} { something \quad going \quad to \quad 0}\right) \right\}$

$ \equal{}\frac {1}{2} \;\lim_{R\to\infty } \; \left(\minus{} \tan^{ \minus{} 1} ( \minus{} R) \plus{}  \tan^{ \minus{} 1} R  \right)\equal{} \frac {1}{2} \; \left( \minus{} \left( \minus{}\frac {\pi}{2}\right) \plus{} \frac {\pi}{2}\right) \equal{} \frac {\pi}{2}$

$ \equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}\equal{}$

PS: to solve the integral, i used the  fact

$ d(\tan^{ \minus{} 1} \; X) \equal{} \frac {1}{1 \plus{} X^2}$

 can you identify what my $ X$ was ?

[quote="jmerry"]That extra $ p$ is exactly what you need for elementary methods. Substitute $ p^2$, and then combine the square roots into the square root of a quadratic. Trig substitution works.[/quote]
Thanks, jmerry, but it's not so obvious to me... I must have burned some neurons after graduating!  :wink:

[quote="misan"]
... can you identify what my $ X$ was ?[/quote]
Ok,
I got as far as
$ I_0(r,z) \equal{} \minus{} \frac {1}{2}\int_{\frac {1}{z^2}}^{\frac {1}{r^2}} \frac {dx}{\sqrt {x \minus{} \frac {1}{r^2}}\sqrt {\frac {1}{z^2} \minus{} x}}$
no prob!
But I can't see how you did the next step

Using $ d(\tan^{ \minus{} 1} \; X) \equal{} \frac {1}{1 \plus{} X^2}$, one would be led to assume that $ 1\plus{}X^2 \equal{} \sqrt {x \minus{} \frac {1}{r^2}}\sqrt {\frac {1}{z^2} \minus{} x}$, but that can't be right since it leads to $ \sqrt{\minus{}1}$... I think.
The limit afterwards just blew me away, completely. Can you explain how you get there, please?

Ok! I see how the limit works. thanks!
But this leap:
$ I_0(r,z) \equal{}\frac {1}{2}\int_{\frac {1}{r^2}}^{\frac {1}{z^2}} \frac {dx}{\sqrt {x \minus{} \frac {1}{r^2}}\sqrt {\frac {1}{z^2} \minus{} x}} \equal{}\frac {1}{2}\;\left\{ \minus{} \tan^{ \minus{} 1} \left( \; \frac { \frac {1}{2r^2} \plus{} \frac {1}{2z^2} \minus{}x}{\sqrt {x \minus{} \frac {1}{r^2} }\sqrt {\frac {1}{z^2} \minus{} x}} \;\right)\;\bigg|_{\frac {1}{r^2}}^{\frac {1}{z^2}}\right\}$
is still unclear.

How do you go from $ \frac {1}{\sqrt {x \minus{} \frac {1}{r^2}}\sqrt {\frac {1}{z^2} \minus{} x}}$ to $ \frac{1}{1\plus{}X^2}$
where, appearently, $ X$ is $ \frac { \frac {1}{2r^2} \plus{} \frac {1}{2z^2} \minus{}x}{\sqrt {x \minus{} \frac {1}{r^2} }\sqrt {\frac {1}{z^2} \minus{} x}}$?

I found it!
misan, you must have done something wrong to get the right result.
But I found equation 3.3.27 of [url=http://books.google.com/books?id=MtU8uP7XMvoC&pg=PA12&lpg=PA5&dq=abramowitz&output=html&sig=4Z9EC21rwOhDxOsClgtxRaOPM1s]Abramowitz[/url] which gives a ready made formula to find this integral:
$ \int\frac{dx}{\sqrt{(a\plus{}bx)(c\plus{}dx)}} \equal{} \minus{}\frac{1}{\sqrt{\minus{}bd}}\arcsin\left(\frac{2bdx\plus{}ad\plus{}bc}{bc\minus{}ad}\right),\,b>0, d<0$

So,
$ I_0(r,z) \equal{} \frac{1}{2}\int_{\frac{1}{r^2}}^{\frac{1}{z^2}}\frac{dx}{\sqrt{(x\minus{}\frac{1}{r^2})(\frac{1}{z^2}\minus{}x)}} \equal{}\frac{1}{2}\left[\minus{}\frac{1}{\sqrt{\minus{}(\minus{}1)}}\arcsin\left(\frac{2(\minus{}1)x\plus{}(\minus{}\frac{1}{r^2})(\minus{}1)\plus{}\frac{1}{z^2}}{\frac{1}{z^2}\minus{}\frac{1}{r^2}}\right)\right]_{\frac{1}{r^2}}^{\frac{1}{z^2}}$
$ \equal{}\minus{}\frac{1}{2}[\arcsin(\minus{}1)\minus{}\arcsin(1)] \equal{} \minus{}\frac{1}{2}(\minus{}\frac{\pi}{2}\minus{}\frac{\pi}{2}) \equal{} \frac{\pi}{2}$

ok, misan, I double checked your result and deriving the $ \tan^{\minus{}1}$ you present one does get to $ \frac {1}{\sqrt {x \minus{} \frac {1}{r^2}}\sqrt {\frac {1}{z^2} \minus{} x}}$.
On my first check it didn't give that result.
I'm [b]sorry[/b] for saying you were wrong.

One other thing that made me say it was wrong was that $ \frac{1}{1\plus{}X^2}$, with $ X\equal{}\frac { \frac {1}{2r^2} \plus{} \frac {1}{2z^2} \minus{}x}{\sqrt {x \minus{} \frac {1}{r^2} }\sqrt {\frac {1}{z^2} \minus{} x}}$, I arrive at $ \minus{}\frac{4r^2(r^2x\minus{}1)z^2(z^2x\minus{}1)}{(r^2\minus{}z^2)^2}$ which is considerably different from $ \frac {1}{\sqrt {x \minus{} \frac {1}{r^2}}\sqrt {\frac {1}{z^2} \minus{} x}}$ in the degree of $ x$. :ninja:
So... I'm guessing you didn't really use that formula for the derivative of the $ \tan^{\minus{}1}$, did you?

So do it the long way. $ \frac1p\equal{}\frac{p}{p^2}$; now you have $ p\,dp$ and can substitute $ p^2$. There are square roots of linear factors in the new denominator; combine them, complete the square, and perform the appropriate trig substitution. This leaves a rational combination of trigonometric functions, which you may need to use the $ \tan\frac{\theta}{2}$ rationalizing substitution on. It's long, but it's guaranteed to work and find an elementary antiderivative.


function
induction
algebra unsolved
algebra

Let $f: \Bbb{R} \rightarrow \Bbb{R}$ be a function such that
(i) For all $x,y \in \Bbb{R}$, \[ f(x)+f(y)+1 \geq f(x+y) \geq f(x)+f(y) \]
(ii) For all $x \in [0,1)$, $f(0) \geq f(x)$,
(iii) $-f(-1) = f(1) = 1$.

Find all such functions $f$.

by plugging $x=-1$, $y=1$ we get $f(0)\geq0$, by plugging $(0,0)$ we get $f(0) \leq 0$ so $f(0)=0$. You can easy prove that for all $x \in [0,1)$ we have $f(x)=0$ (it's a key step  :) ) Now we'll prove by induction that $f(x)=[x]$. Our inductive sentence $T_n$ is: $f(x)=[x]$ for all $x<n$. Suppose we have proved it for some $n$. Now we have for $n-1 \leq x < n$ and $\epsilon$ such that $x+\epsilon<n$:

$[x+1]=[x]+1=f(x+\epsilon)+f(1-\epsilon)+1 \geq f(x+1) \geq f(x)+1=[x]+1=[x+1]$. So we have proved $T_{n+1}$.

QED

My solution from WOOT:

Let $P(x,y)$ denote the assertion of $(\text i)$.
$P(0,0),P(1,-1)\Rightarrow f(0)=0$
$P(x,1)\Rightarrow f(x+1)\ge f(x)+1$
$P(x+1,-1)\Rightarrow f(x)\ge f(x+1)-1$
Then equality must hold, so $f(x+1)=f(x)+1$.
For $x\in[0,1)$, $f(x)\le0$. Then:
$P(x,1-x)\Rightarrow f(x)+f(1-x)\ge0$, but since $f(x)+f(1-x)\le0+0$ equality holds for each and $f(x)=0$ for $x\in[0,1)$. Then by induction on $f(x+1)=f(x)+1$ we obtain $\boxed{f(x)=\lfloor x\rfloor}$, which fits.

[hide=Solution]
Substituting $x=y=0$, we get $f(0)\geq2f(0)$, so $f(0)\leq0$. Substituting $x=1$ and $y=-1$, we get $f(0)\geq f(1)+f(-1)=0$. Therefore, $f(0)=0$. Substituting $x=1$, we get $f(y+1)\geq f(y)+1$. Substituting $x=-1$, we get $f(y-1)\geq f(y)-1$, so $f(y)\geq f(y+1)-1$. Substituting this into the first inequality, we get $f(y+1)\geq f(y)+1\geq f(y+1)$, so $f(y+1)=f(y)+1$. Substituting $y=1-x$, we get $f(x)+f(1-x)\geq0$. If $0\leq x<1$, then $f(x)\leq0$, so $f(x)=0$. Therefore, we have $f(x)=\lfloor x\rfloor$. We will show that this works. The left inequality is true since $\lfloor x\rfloor+\lfloor y\rfloor+1-\lfloor x+y\rfloor=-\{x\}-\{y\}+1+\{x+y\}>-1$ and it is an integer, so it is nonnegative. The right inequality is equivalent to $\{x\}+\{y\}\geq\{x+y\}$, which is also true. Since $\lfloor -1\rfloor=-1$ and $\lfloor 1\rfloor=1$, this means that the third condition is satisfied. Since $\lfloor x\rfloor=0$ when $0\leq x<1$, this means that the second condition is also true.

Therefore, the only solution is $\boxed{f(x)=\lfloor x\rfloor}$.

As everybody else is posting solutions, I'll also post mine:

We claim that $f(x)=\lfloor x \rfloor$ is the only solution. It clearly works.

[b][color=#f00]Claim:[/color][/b] $f(x+1)=f(x)+1$ for all real $x$.
Notice that plugging in $x=1$ gives $$1+f(y)\le f(1+y)\le 2+f(y).$$ On the other hand, plugging in $x=-1$ gives $$f(y)-1\le f(y-1)\le f(y)\Rightarrow f(y-1)\le f(y) \le f(y-1)+1\Rightarrow f(y)\le f(y+1)\le f(y)+1.$$ The result follows from these two inequalities. $\blacksquare$

Thus it suffices to find the values of $f(x)$ for $0\le x<1$; then we can use the claim to find the rest of the function. Using the claim, we get $f(0)=0$. Let $0<a<1$ and plug $x=a, y=1-a$ to get $$f(a)+f(1-a)\le 1\le 1+f(a)+f(1-a).$$ In particular, we get $0\le f(a)+f(1-a)$. However, from condition (ii) we get $0\ge f(a), f(1-a)$, thus we have $f(a)=f(1-a)=0$. Hence, for all $0\le x<1$, $f(x)=0$. Using the claim, we get $f(x)=\lfloor x \rfloor$, and we are done.

The answer is $f(x) = \lfloor x \rfloor$, which works.

[b][color=blue]Step 1: Extracting $f(0)$.[/b][/color] We claim that $f(0)=0$. First, by setting $x=y=0$, $$2f(0)+1 \geq f(0) \geq 2f(0) \iff -1 \leq f(0)\leq 0.$$ On the other hand, letting $x=1$ and $y=-1$, $$f(-1)+f(1)+1=1\geq f(0) \geq 0.$$ Combining these two inequalities gives $f(0)=0$.

[b][color=blue]Step 2: Getting $f$ over $[0, 1)$.[/b][/color] Now, we claim that $f(x)=0$ for all $x \in [0, 1)$. The idea here is similar to the idea in the first step: by condition (ii), we know that $f(x) \leq 0$ for $x \in [0, 1)$, but on the contrary, letting $y=1-x$ gives us $$f(x)+f(1-x)+1\geq f(1) = 1 \iff f(x)+f(1-x) \geq 0.$$ For $x \in [0, 1)$, this is only possible if $f(x)=f(1-x)=0$, as required.

[b][color=blue]Step 3: Inducting upwards/downwards.[/b][/color] Finally, we are ready to finish. Letting $y=1$ gives us $$f(x+1) \geq f(x)+1$$ for all $x$, and setting $y=-1$ and $x=x+1$ gives us $$f(x+1)+f(-1)\leq f(x) \iff f(x+1)\leq f(x)+1.$$ Therefore, $f(x+1)=f(x)+1$ for all $x$. Inducting upwards and downwards gives us the result.

Let $P(x,y)$ denote the given assertion.

Claim: $f(0)=0$.
Proof: $P(0,0): 2f(0)+1\ge f(0)\ge 2f(0)\implies -1\le f(0)\le 0$.

$P(1,-1): f(0)\ge 0$, so $f(0)=0$.

Claim: $f(k)=0\forall 0\le k<1$.
Proof: Note that $f(k), f(1-k)\le 0$.

$P(k,1-k): f(k)+f(1-k)+1\ge 1\implies f(k)+f(1-k)\ge 0\implies f(k)=f(1-k)=0$.

Claim: $f(x+1)=f(x)+1$.
Proof: $P(x,1): f(x)+2\ge f(x+1)\ge f(x)+1$.

$P(x+1,-1): f(x)\ge f(x+1)-1\implies f(x)+1\ge f(x+1)$.

This proves our claim. We can also find that $f(x)-1=f(x-1)$.

From this claim, $f(x)=\lfloor x \rfloor +\{x\}\implies \boxed{f(x)=\lfloor x\rfloor}$, which works.


The area of the semicircle in Figure A is half the area of the circle in Figure B. The area of a square inscribed in the semicircle, as shown, is what fraction of the area of a square inscribed in the circle? Express your answer as a common fraction.

[asy]defaultpen(linewidth(0.8));

size(5cm,5cm);

draw((0,0)..(1,1)..(2,0)--(0,0));

draw((0.5,0)--(0.5,0.87)--(1.5,0.87)--(1.5,0));

draw(Circle((4,0),1));

pair A,B,C,D;

A=(3.3,0.7);
B=(3.3,-0.7);
D=(4.7,0.7);
C=(4.7,-0.7);

draw(A--B--C--D--A);

label("Figure A",(1,1.3));
label("Figure B",(4,1.3));[/asy]

hello
Area of square in semicircle=$ \frac{4}{5}r^2$
Area of square in circle=$ 2r^2$
so the ratio=$ \frac{2}{5}$
thank u

[quote="covertcognomen"]could you please explain why the area of the square in the semicircle is   (4/5)r^2?

sorry for the fact that I can't type in latex.[/quote]
Letting the radius of the semicircle be $r$, then the line segment from one corner of the the square in Figure A to the center of the semicircle makes a right triangle.  Let the side of the square in Figure A be $x$ units in length, then by the Pythagorean Theorem, we have:
\begin{align*}x^2+\left(\frac{x}{2}\right)^2&=r^2 \\ x^2+\frac{x^2}{4}&=r^2 \\ \frac{5x^2}{4}&=r^2 \\ x^2&=\frac{4r^2}{5} \\ x&=\frac{2r}{\sqrt5}\end{align*}
Since the area of a square is the square of its side, the area of the square in Figure A is:
\[\left(\frac{2r}{\sqrt5}\right)^2 = \frac{4r^2}{5}\]
Which is what kabi got.