Hey, I've got a college interview with MIT coming up and I was just wondering what they're like.  Is it just like the application, only orally?  What type of questions would they ask?

There was a Math Jam a while back about college interviews.  Why not start there?  [url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=71[/url]

Interviews depend on who the interviewer is.  They can really ask you just about anything, but you should at least be prepared to answer the generic and obvious interview questions.  (Why do you want to go to this school?  What are your interests?  Etc.)  You should also try to have at least one question to ask the interviewer about the school.

[quote="cosinator"]Is it just like the application, only orally?[/quote]
Well, no.  That's what the application is for.

My advice for interviews is not to worry about them.  It's easy to tell if you've prepared stock answers; interviewers are looking to see aspects of your identity as an intellectual and as a person that don't come across in the application, so you should be emphasizing the parts of your personality that already come naturally, such as your passion, etc.  Aim for a conversation that flows naturally.

For what it's worth, during my MIT interview we didn't talk about MIT at all.  We talked about  my interviewer's job as an industrial psychologist studying different ways to increase the pay of engineers who work on group projects.

I had my interview just a couple of hours ago.  I didn't worry about it too much, and the conversation just flowed naturally.  The questions were pretty generic, as yeniee said i.e. why do you want to come to MIT, what do you do outside of school, describe your family, etc.

I'm glad to hear that things went well, and also that my powers of ESP have not failed.

I also wanted to add that t0rajir0u's advice is pretty much spot on, except that there's nothing wrong with having a few stock answers.  If you were asked why you want to go to MIT, and you looked like a deer in headlights and stumbled, "Uh... gee, I never quite thought about that.  Let me think about that for a minute," then that might not go over so well.  There's no inherent conflict between being prepared and being genuine.  It's *phony* canned answers that people might not like.

Another point:  While it may be (probably is?) true that MIT interviewers have access to your application, I know for a fact that Harvard interviewers do not, and therefore rehashing your application verbally may or may not be part of the interview process.

Again, every interview is different, so I would avoid making generalizations.

My wife is an interviewer for MIT, and she does not have access to the student's application.

My interviewer for MIT told me at the beginning that she didn't want to hear me give my application - she just wanted to hear about me as a person. Obviously, it varies from person to person.


geometry
incenter
rectangle
angle bisector

Given four concyclic points. For each subset of three points take the incenter. Show that the four incenters from a rectangle

its just some angle chasing,and Im sure it has been posted before several times,so maybe someone could give you the link...

anyway here`s the solution:
[hide="solution"]
name the four points as $ A,B,C,D$ and let $ I_{1},I_{2},I_{3},$ be the incircles of $ \triangle ABD,\triangle ABC,\triangle BCD$ respectively...
now we will show that $ I_{1}I_{2}\perp I_{2}I_{3}$
[b]lemma:in $ \triangle ABC$ we have $ \angle BIC=90+\angle \frac{A}{2}$ where $ I$ is the incenter[/b]

now according to the above lemma,for $ \triangle ABC,\triangle ABD$ we have:

$ \angle AI_{2}B=90+\frac{\angle ACB}{2}$

$ \angle AI_{1}B=90+\frac{\angle ADB}{2}$

but $ ABCD$ is cyclic so we have $ \angle ADB=\angle ACB$ so we get that:

$ \angle AI_{1}B=\angle AI_{2}B$ so the quadrilateral $ AI_{1}I_{2}B$ is cyclic hence

$ \boxed{\angle BI_{2}I_{1}=180-\angle I_{1}AB=180-\frac{\angle BAD}{2}(I)}$

in a similar way we get that:

$ \boxed{\angle BI_{2}I_{3}=180-\angle I_{3}CB=180-\frac{\angle BCD}{2}(II)}$

now from $ (I),(II)$ we get the desired resault...[/hide]

Your lemma isn't making any sense to me. For example, in an equilateral triangle it's false ($ 60 \neq 120$).

Was that supposed to be $ 90+\angle \frac{A}2$?

yeah that was a typo in typing,I have edited it...

Below a nice approach:
Let $M, N, P, Q$ be midpoints of the arcs $\stackrel{\frown}{AB},\stackrel{\frown}{BC}, \stackrel{\frown}{CD},\stackrel{\frown}{DA}$ (arcs which do not contain other vertex of the quadrilateral).
It is well known that $I_1, I_2$ lie on the circle $(M, MA)$, at its intersection with $MD, MC$ respectively; as $MP$ is the angle bisector of $\angle CMD\implies I_1I_2\bot MP$. In a similar way $I_2I_3\bot NQ$, but, again well known, $MP\bot NQ$, a.s.o.
(Here $I_1, I_2, I_3$ are as above).

Note: I think I posted before this proof somewhere around.

Best regards,
sunken rock

Dear Mathlinkers,
you can see
http://jl.ayme.pagesperso-orange.fr/  vol. 4,  le rectangle de Ryokan Maruyama

Sincerely
Jean-Louis

It seems this problem is now called [b]'Japanese theorem[/b]' ...

Best regards,
sunken rock


Let $a_1,a_2,a_3,a_4,a_5$ be distinct real numbers. Consider all sums of the form $a_i + a_j$ where $i,j \in \{1,2,3,4,5\}$ and $i \neq j$. Let $m$ be the number of distinct numbers among these sums. What is the smallest possible value of $m$?

Try (0,1,2,3,4), and we get m=7.  Can we get m = 6?  We can suppose the first number is 0, since adding a constant to each $a_i$ doesnt change anything.  Suppose $a,b,c,d$ are the remaining numbers in increasing order.  Then $a,b,c,d,d+b,d+c$ are distinct.  Then b+c can only be equal to d.  Then d+a can only be b+c.  Contradiction.


Here is a sequence of challenge problems.  Each problem is about the same exponential equation.

(1) Prove that there does not exist a nonzero real solution to the equation ex = 1 + x.

(2) Prove that there does exist a nonzero complex solution to the equation ex = 1 + x.

(3) How many complex solutions does the equation ex = 1 + x have?

Nice problem, I've never thought of this before !

A (not too rigorous ...) try :

[hide]

1-  By Taylor, e^x - 1 - x = x^2/2 e^c > 0 for some c between 0 and x.

2- x=a+ib,  equation is then, after some easy work :

a = b/tanb -1 and f(b) = e^(b/tan(b) -1) - b / sin(b) = 0

f(b) can be defined (continously prolonged on k*pi/2) on ] k*pi, (k+1)pi [ for every k.

Now f(2*k*pi + epsilon) > 0 and f((2*k+1)*pi - epsilon) < 0 (I didn't checked it seriously but it seems to be corrrect) so there is a solution of f(b)=0 in ]2*k*pi .. (2*k+1)[ and a 'a' accordingly.[/hide]

Another proof of the first part (which I wasn't going to post until I solved the whole thing, but then belenos posted, so I changed my mind):

The line 1+x is tangent to ex at x=0. Since ex is convex and nowhere linear, there are no other points of intersection.

Yes, you both have nice proofs for Question 1.  Belenos, nice work on Questions 2 and 3!  So there are infinitely many complex solutions.


function
trigonometry
floor function
real analysis
real analysis solved

Let $f:R->R$ a continous function such that $sin(f)$ is uniform continous.Show that $f$ is uniform continous.

Here is my solution.I hope it is correct.Denote by $G_{k}=\{x|f(x)\in{(-\frac{\pi}{2}+\pi\*k,\frac{\pi}{2}+\pi\*k)}\}$.Since $f$ is continious $G_{k}$ are open sets.Then $f^{*}(x)=\arcsin{(\sin{f(x)})}=(-1)^{k}(f(x)-\pi\*k)$.As $\arcsin{x}$ is uniformly continious function then $f^{*}(x)$ is uniformly continious,too.Now assume that $f$ is continious,but not uniformly.Then there exist two sequences $x_{n}$ and ${x_{n}}^{'}$ and ${\varepsilon}_{0}>0$ such that $f(x_{n})-f(x_{n}^{'})\geq{\varepsilon}_{0}$,but $x_{n}-{x_{n}}^{'}{\rightarrow}0$.If there exist natural number $n_{0}$,such that $\forall n>n_{0} \exists k_{n}|x_{n},{x_{n}}^{'}\in{G_{k_{n}}}$,then we get contradiction,because $\exists \delta>0||x-x^{'}|<\delta$ then $|f^{*}(x)-f^{*}(x^{'})|<{\varepsilon}_{0}$,but if  $x_{n},{x_{n}}^{'}\in{G_{k_{n}}}$ then $|f^{*}(x_{n})-f^{*}({x_{n}}^{'})|=|f(x_{n})-f({x_{n}}^{'})|$,which leads to the contradiction.In the case when such $n_{0}$ doesnt exist
Then for each pair $x_{n},{x_{n}}^{'}$ $x_{n}\in{G_{k_{n}}},{x_{n}}^{'}\in{G_{m_{n}}},m_{n}{\neq}k_{n}.$
Since $f$ is continious we can  assume that $m_{n}=k_{n}+1$(if anyone wants I can explain).Besides
there exist $z_{n}$ between $x_{n}$ and ${x_{n}}^{'}$ such that $f(z_{n})=\frac{\pi}{2}+{\pi}k_{n}$,since
$f({x_{n}}^{'})-f(x_{n})=f({x_{n}}^{'})-f(z_{n})+f(z_{n})-f(x_{n})\geq{\varepsilon}_{0}$,then
$f({x_{n}}^{'})-f(z_{n})\geq{\frac{{\varepsilon}_{0}}{2}}$ or  $f(z_{n})-f(x_{n}) \geq{\frac{{\varepsilon}_{0}}{2}}$,but that means that $|f^{*}({x_{n}}^{'})-f^{*}(z_{n})|\geq{\frac{{\varepsilon}_{0}}{2}}$  or  $|f^{*}(z_{n})-f^{*}(x_{n})| \geq{\frac{{\varepsilon}_{0}}{2}}$.Again contradiction,so we are done.

An informal look at Levon's proof: Because $\sin^{-1}$ is uniformly continuous, then for any $\epsilon>0$, there is a $\delta$ such for any $\delta$-neighborhood $J$, $\sin^{-1}(J)$ consists of a collection of widely separated "islands", each of which is an $\epsilon$-neigborhood.

Now if $x$ and $y$ are close, then $\sin(f(x))$ and $\sin(f(y))$ are close. This means that $f(x)$ and $f(y)$ belong to the same $\sin^{-1}(J)$, but could they be in different "islands"? No, they can't, because by the Intermediate Value Theorem, for $z$ between $x$ and $y$, $f(z)$ would have to take on all of the values in between, forcing you into the deep water between the islands, where $\sin(f(z))$ would then be outside of $J$, and that would violate the uniform continuity of $\sin(f).$

Note that we very much needed to separately assume the continuity of $f$ so we could quote the Intermediate Value Theorem. The need for this hypothesis is shown by the example of $f(x)=2\pi\lfloor x\rfloor.$

Yes,Kent that's exactly that I wanted to say.And I like your interpretation.    :)


The first step in finding the product $ (3x \plus{} 2)(x \minus{} 5)$ by use of the distributive property in the form $ a(b \plus{} c) \equal{} ab \plus{} ac$ is:

$ \textbf{(A)}\ 3x^2 \minus{} 13x \minus{} 10 \qquad \textbf{(B)}\ 3x(x \minus{} 5) \plus{} 2(x \minus{} 5)\qquad \\\textbf{(C)}\ (3x \plus{} 2)x \plus{} (3x \plus{} 2)( \minus{} 5)\qquad \textbf{(D)}\ 3x^2 \minus{} 17x \minus{} 10\qquad \textbf{(E)}\ 3x^2 \plus{} 2x \minus{} 15x \minus{} 10$

[hide]
Using the property, we get
(3x+2)x - (3x+2)5
$ \boxed{\text{C}}$[/hide]


I am intentionally putting this in the Round Table to see views of all AoPSers. The question I want answered is, is Mathcounts "real" problem solving? Or is it a waste of time?

For all you international people out there, Mathcounts is a Middle School competition stressing quick thinking and solving problems in a very short amount of time. There are also alot of tricks involved as well, that show up constantly.

What's your opinion?

Good.  It's fun, and encourages kids to figure out quicker, more elegant shortcuts.  So yes, it DOES provoke thinking.  Not to mention it gets them to meet others.

I think it's great. Before MathCounts, I wasn't in to math a lot, but once I got invovled with MathCounts, I began studying math. It helps your mind work faster and meet new people.

My personal experience with Mathcounts is that I've made ridiculously blatant (and ridiculously numerous) stupid mistakes. Most of these stem from misreading the question, and just plain old rushing. (For example, at Nationals 2006, I did the problem with the cross sums perfectly well, but when it asked for A,BCD,EFG, I put down A+B+C+D+E+F+G instead. :roll:)

I think Mathcounts is more speed-oriented than actual critical thinking.

But they hand out free calculators, so who am I to complain? :D

I became aware of Mathcounts in 1995, when my son first became involved. It has its own style, its own strengths and weaknesses. I think there should definitely be variety in mathematics competitions and that not every competition should work the same way. I'll also add that truly being a middle-school level competition puts some rather strong curriculum-level restrictions on questions that might be asked, and it's not all that surprising that the questions may seem to be mining the same territories over and over.

Mathcounts has the power to sweep a student into the problem-solving world who wouldn't otherwise be there. I will say this: although of course I always knew that my son was good at mathematics, I never had any clear idea just [i]how[/i] good until we knew how he ranked in the state and national levels in this program.

As for the emphasis on quickness, and whether that is different from "true problem-solving"? I've always been surprised at the extent to which the same people bubble to the top in very different contests. In the year we attended the Mathcounts nationals (1996), three of the top four individuals there were Sasha Schwartz, Reid Barton, and Po-Shen Lo. Sasha Schwartz won every round. If I'm remembering correctly, Reid Barton (a 7th grader at the time) climbed quite a few places during the countdown round. Now, are those the names of people who were just quick calculators with no real depth? I don't think so.

[quote="Kent Merryfield"]
Mathcounts has the power to sweep a student into the problem-solving world who wouldn't otherwise be there.[/quote]
I am one of those people. I didn't know what the AMC was until 2 thirds of the way through last year, and wouldn't have found out if it wasn't for MC and this forum. Hopefully, I will make USAMO next year. I am right now at about a 5 level on AIMEs so maybe I'll be up there by then.

As for the question you asked, no I don't believe it's "real" <- (Have no idea what you mean by that) problem solving. However, with the way you phrased your post, you created a false dilemna. Just because something isn't "real" problem solving doesn't mean it's a waste of time. You have to think of the fact that most kids your age aren't concerned with real problem solving and are having a tough enough time with school math. It accomplishes two goals with very different target audiences at once. One, it motivates kids who may be having a tough time with school math and gives them some extra material, etc. to work with. Second, it motivates plenty of extremely strong middle school students by giving them goals to work to and practicing [b]how[/b] to do problems rather than just memorizing them as the same problems don't appear on the same test (well, they do sometimes) but the concepts definitely reappear.

The point isn't to win as it's impossible for most people and some people get caught up too much in trying to win. If it was, all people would be focussing is getting a grasp of the material and then just worrying about stupid mistakes over and over again which some people do. The point is that it introduces some concepts that you probably wouldn't have ordinarily seen in school and gives you chances to apply them, and it sets goals for you which can be applied to many things in life.

In addition to my two years as a competitor, I've volunteered a few times as a grader. I'd like to add another positive point about Mathcounts: the Countdown round is really fun to watch.
At the Seattle chapter this spring, I saw a girl sweep all nine rounds of the Countdown ladder, going from tenth to first. Even with no personal connection to the competitors, I'll remember that for a long time.

I did well in Mathcounts and now that I took the AMC, I was not able to break 135 the last three years on the AMC 12.

Mathcounts is not very in depth problem solving, more like remembering a few quick tips and solving problems accurately.

I agree that you don't need to be really creative to excel at Mathcounts, but I still think that Mathcounts is really good. I was pretty good at math, but I never thought of doing it for fun before Mathcounts. And when they give you a free trip to Washington D.C., and a free calculator, that makes it even better.


geometry
area of a triangle
Heron\u0027s formula
geometry unsolved

3. In the plane there exists a triangle, which sides and one of its altitudes are consecutive positive integers. The altitude is the smaller than all the three sides. Determine this triangle and show, that there are no other triangles satisfying this condition.

$ (a, b, c) \equal{} (n, n\plus{}1, n\plus{}2)$ are the 3 triangle sides and $ h\equal{}n\minus{}1$ is one altitude ($ n > 1$). Triangle area $ \triangle$ is given by Heron's formula

$ 16 \triangle^2 \equal{} (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(c\plus{}a\minus{}b)(b\plus{}c\minus{}a) \equal{} 3(n\minus{}1)(n\plus{}1)^2(n\plus{}3).$

This is equal to $ 4h^2a^2$ or to $ 4h^2b^2$ or to $ 4h^2c^2.$ All these expressions are even, the Heron's formula expression is also even $ \Longrightarrow$ $ n \equal{} 2m\minus{}1$ is odd ($ m > 1$). Then

$ \frac{\triangle^2}{m\minus{}1} \equal{} 3m^2(m\plus{}1) \equal{} (m\minus{}1)(2m\minus{}1)^2$ or $ (m\minus{}1)m^2$ or $ (m\minus{}1)(2m\plus{}1)^2.$

[color=red](a)[/color]  $ 3m^2(m\plus{}1) \equal{} (m\minus{}1)(2m\minus{}1)^2\ \Longrightarrow$ $ m^3 \minus{} 11m^2 \plus{} 5m \minus{} 1 \equal{} 0.$ Possible rational roots are integers $ \pm 1,$ none will do.

[color=red](b)[/color]  $ 3m^2(m\plus{}1) \equal{} 4(m\minus{}1)m^2\ \Longrightarrow$ $ m \equal{} 7, n \equal{} 13, (a, b, c) \equal{} (13, 14, 15)$

[color=red](c)[/color]  $ 3m^2(m\plus{}1) \equal{} (m\minus{}1)(2m\plus{}1)^2\ \Longrightarrow$ $ m^3 \minus{} 3m^2 \minus{}3m \minus{} 1 \equal{} 0.$ Again, possible rational roots are integers $ \pm 1,$ none will do.


inequalities function trigonometry integration geometry trapezoid calculus

If $ a>0$, prove that

$ \frac{1}{\sqrt{a}}>2(\sqrt{a\plus{}1}\minus{}\sqrt{a})$.

hello, your inequality is equivalent with
$ \frac{1}{\sqrt{a}}>\frac{2}{\sqrt{a\plus{}1}\plus{}\sqrt{a}}$
$ \sqrt{a\plus{}1}\plus{}\sqrt{a}>2\sqrt{a}$
$ \sqrt{a\plus{}1}>\sqrt{a}$
and this is true.
Sonnhard.

Hello,
[hide]
$ \frac 1{\sqrt a} > \frac 2 {\sqrt{a+1}+\sqrt{a}$
$ a < \frac {a+1+a+2\sqrt{a^2+a}}2$
$ 0 < \frac 12 + \sqrt{a^2+a}$
from which is true since a>0

[/hide]
Roobiks

Alteratively, since $ a>0$,

$ \frac{1}{\sqrt{a}} >2(\sqrt{a\plus{}1}\minus{}\sqrt{a} \equiv \frac{\sqrt{a\plus{}1}\plus{}\sqrt{a}}{\sqrt{a}}> 2(a\plus{}1\minus{}a)$,

(on multiplying both sides by $ \sqrt{a\plus{}1}\plus{}\sqrt{a}$,)

$ \equiv \sqrt{a\plus{}1}\plus{}\sqrt{a} > 2\sqrt{a}$ which is true.

$ \frac{1}{2 \sqrt{a}}>\sqrt{a\plus{}1}\minus{}\sqrt{a}$

Using $ f(a) \equal{} \sqrt{a}$, this is just:

$ f'(a) > \Delta[f] (a)$

Which must be true since the square-root function is concave everywhere.

[quote="AndrewTom"]If $ a > 0$, prove that

$ \frac {1}{\sqrt {a}} > 2(\sqrt {a + 1} - \sqrt {a})$.[/quote]

Solution 1:

Consider the mid point on real numbers line for $ \sqrt {a},\ \sqrt {a + 1}$, yielding

$ \sqrt {a} < \frac {\sqrt {a} + \sqrt {a + 1}}{2} < \sqrt {a + 1}$

$ \Longleftrightarrow \frac {1}{\sqrt {a+1}} < 2(\sqrt {a + 1} - \sqrt {a}) < \frac {1}{\sqrt {a}}$

[color=red]Last Edited[/color]

Solution 2:

Let $ a = \tan ^ 2 \theta\ (0 < \theta < \frac {\pi}{2})$,

$ \frac {1}{\sqrt {a}} > 2(\sqrt {a + 1} - \sqrt {a})$

$ \Longleftrightarrow \frac {1}{\tan \theta} > 2(\frac {1}{\cos \theta} - \tan \theta )$

$ \Longleftrightarrow \frac {\cos \theta}{\sin \theta} > 2(\frac {1}{\cos \theta} - \frac {\sin \theta}{\cos \theta})$

$ \Longleftrightarrow \cos ^ 2 \theta > 2\sin \theta (1 - \sin \theta) \because 0 < \theta < \frac {\pi}{2}$

$ \Longleftrightarrow 1 - \sin ^ 2 \theta > 2\sin \theta (1 - \sin \theta)$

$ \Longleftrightarrow 1 + \sin \theta > 2\sin \theta \ \because 0 < \sin \theta < 1$

$ \Longleftrightarrow \sin \theta < 1$

Solution 3:

Since $ y = \frac {1}{\sqrt {x}}\ (x > 0)$ is concave down and an decreasing function,

$ \frac {1}{\sqrt {a}} > 2(\sqrt {a + 1} - \sqrt {a})$

$ \Longleftrightarrow \int_a^{a + 1} \frac {1}{\sqrt {x}}\ dx < \frac {1}{\sqrt {a}}$

The right side means the area of the rectangular $ \frac {1}{\sqrt {a}}$ by $ (a + 1 - a)$

Solution 4:

Since $ y = \frac {1}{\sqrt {x}}\ (x > 0)$ is concave down and an decreasing function,

$ \frac {1}{\sqrt {a}} > 2(\sqrt {a + 1} - \sqrt {a})$

$ \Longleftrightarrow \frac {\sqrt {a + 1} + \sqrt {a}}{2} < \frac {1}{\sqrt {a}}$

The left side means the area of the trapezoid with base $ \sqrt {a},\ \sqrt {a + 1}$ and hight $ (a + 1 - a)$.

Comment :

We can also prove the inequality in using  Mean Value Theorem of Integral, Lagrange's Mean Value Theorem.

Nice one, TZF. Omedetoo gozaimasu, kunny-san.

I think kunny left out the intermediate step and made a typo, that is, I suspect he meant to write:

$ \sqrt{a} < \frac{\sqrt{a}\plus{}\sqrt{a\plus{}1}}{2} < \sqrt{a\plus{}1}$

$ \frac{1}{\sqrt{a\plus{}1}}<\frac{2}{\sqrt{a\plus{}1}\plus{}\sqrt{a}}<\frac{1}{\sqrt{a}}$

Then, rationalising the denominator of the middle trerm, this becomes

$ \frac{1}{\sqrt{a\plus{}1}}<\frac{2(\sqrt{a\plus{}1}\minus{}\sqrt{a})}{a\plus{}1\minus{}a}<\frac{1}{\sqrt{a}}$,

That is,

$ \frac{1}{\sqrt{a\plus{}1}}<2(\sqrt{a\plus{}1}\minus{}\sqrt{a})<\frac{1}{\sqrt{a}}$

and the right-hand inequality is the required one.

(However, kunny may have a much neater way of doing this.)

[quote="Arrange your tan"][quote="kunny"][quote="AndrewTom"]If $ a > 0$, prove that

$ \frac {1}{\sqrt {a}} > 2(\sqrt {a \plus{} 1} \minus{} \sqrt {a})$.[/quote]

Solution 1:

Consider the mid point on real numbers line for $ \sqrt {a},\ \sqrt {a \plus{} 1}$, yielding

$ \sqrt {a} < \frac {\sqrt {a} \plus{} \sqrt {a \plus{} 1}}{2} < \sqrt {a \plus{} 1}$ . . . (1)

$ \Longleftrightarrow \sqrt {a} < 2(\sqrt {a \plus{} 1} \minus{} \sqrt {a}) < \sqrt {a \plus{} 1}$ . . . (2)

[/quote]

You did not show/explain the connecting step as to how you got from
the middle quantity in (1) to the middle quantity in (2).[/quote]

Sorry, I made a typo.

kunny

Thank you for the correction, AndrewTom. :)

You're welcome, kunny. I was impressed by all your different solutions. :lol:


limit
LaTeX
calculus
calculus computations

Suppose that $ a_{n}\leq 0$ and that $ \sum_{n\equal{}1}^{\infty}a_{n}<\infty$.
(i) Show that $ \lim\inf_{n\rightarrow\infty}na_{n}\equal{}0$
(ii) Give an example showing $ \lim\sup_{n\rightarrow\infty}na_{n}>0$ is possible.

Presumably you mean $ a_n\geq 0$.

For i), if $ \liminf_{n\to\infty} na_n = \ell>0$, then you could maybe compare $ a_n$ to $ (\ell-\varepsilon)/n$ for some small $ \varepsilon$, and use the divergence of the harmonic series.

For ii), you could maybe have mostly zero terms, but (say) a small subsequence of terms satisfying $ a_k=1/k$...

also in $ \text{\LaTeX}$ $ \liminf_{n\to\infty}$ looks much better than $ \lim\inf_{n\to\infty}$

blahblahblah, for ii wouldn't such a sequence diverge?

For i, you can also choose some non-increasing subsequence of series terms (presuming terms are positive, otherwise let it be non-decreasing).

um, no?  i think what i'm referring to is pretty clear, but i don't want to spell out the answer any more than i've done.

Thanks, I think I figured out at least the first part, and possibly the second.


5 problems will be posted in 3 minutes.

First person to answer the question gets a point.

No solutions necessary, but may be requested by people who don't get how to get an answer.

A total of 5 round, for 25 questions in all.

1. If 3/(2-x/2) = 2, then x equals what?

2. If 3x + 7 = x^2 + k = 7x + 15, then k equals what?

3. (x^-1 - y^-1)^-1 equals what?

4. A metal strip is shaped into a right triangle such that it has legs of x and x + 3 and hypotenuse x + 6 the correct value for x is what?

5.  3x - (4y - (4y - 7y)) equals what?

Make sure you indicate which question you are answering.

1. If 3/(2-x/2) = 2, then x equals what?

[hide]2 -x/2 must equal 1.5, so x must be 1[/hide]

:edited: new answer

JBL wrote:Scrambled wrote:[hide]1/x is x[/hide]

Since when?  As an example, I'd like to present you the number 2.

sorry, typo

BHorseMath incorrect for 2, 4, & 5.  Keep trying.  We'll post solutions to how to do them.

[quote]
Since when? As an example, I'd like to present you the number 2.
[/quote]

Wow.... just wow.  That was great.

Scrambled gets #1.

Please don't edit posts, just make a new post.  Thanks.

2. If 3x + 7 = x^2 + k = 7x + 15, then k equals what?

3x = 7x+6, so 4x+6=0,so x is equal to 3/2.  so 3/2+7 is 17/2, and 9/4+k is 17/2.  so k is equal to 6.25

BHorse, why don't you post a solution instead of just an answer?  Then if you're wrong, it's easier to figure out why.

Darn, came here waaaaay too late :lol: Oh well, when does the next round start?

...it's no competition nowadays anyways :P

Mathfiend incorrect for #5.

BHorseMath incorrect for #2.

2. If 3x + 7 = x^2 + k = 7x + 15, then k equals what?

3x = 7x+8, so 4x+8=0,so x is equal to -2.  so -2+7 is 5, and 4+1 is 5.  so k is equal to 1

topper, Mathfiend was correct as far as the question was written.

My bad, I typed it in wrong.

Point goes to mathfiend.

#2 still out there.

Sorry, Tare gets #2.

Leaderboard:

Scrambled......3
Mathfiend.......1
Tare...............1

More questions in 3 minutes

Scrambled: no, your #2 doesn't work. You made a tiny mistake the step AFTER finding the value of x. (The value of x is right.)


Let the number  $S(n)=1^5+2^5+....+n^5$  with  $n\in N^*$.

Find all the $n$ such that  $(n+1)|S(n)$

If n is even, you can use gaussian pairing to show $n+1$ divides, as follows:

$S(n)=(1^5+n^5)+(2^5+(n-1)^5)+ \cdots$

Where $n+1$ divides each bracket.

For odd n, do exactly the same, but we'll have one term left over, namely, if $n=2m-1$, the term will be $m^5$

But $\gcd(2m+1,m)=1$ therefore no odd number will work.

[quote="seamusoboyle"]For odd n, .[/quote]

In this case you missed something
If $n$ is odd and $n+1$ is a multiple of $4$ then  $n+1|S(n)$

Yes, but he just did some small mistake:
we need that $2m=(n+1)|m^5$, and this happens iff $2|m$, giving the result.


search
inequalities unsolved
inequalities

let$x,y,z$ be three positive  numbers such that :
 $(x+y-z)(1/x+1/y-1/z) =4$
  find the minimal value of expression :
$A =(x^4 +y^4+z^4)(1/x^4 +1/y^4+1/z^4)$

This is a problem in ML contest . Try to search for it !

use a= x/y, b = y/z, c = z/x
then a + 1/a - b - 1/b - c - 1/c = 1 then assume k' = k + 1/k(k =a,b Or c)
In the next one a' = b' + c' +1
a^4 + a^-4 = ((a')^2 - 2) ^2 -2
do this again to b,c, i think i'll be able to find the ans.
(However,my time is up. :ninja: )


group theory
abstract algebra
superior algebra
superior algebra theorems

Let $G$ be an infinite grup, and $K\subset H \subset G$ be a subgrup of $G$ (posibily infinite order too). If the index of $H$ in $G$ and the index of  $K$ in $G$ is both finite prove that they statisfies \[[G: K] = [G: H] [H: K]\]

("group" has an o in it)

There is a natural bijection between $(G/K)/(H/K)$ and $G/H$; if all of these are normal subgroups, it's a group isomorphism. We match$aH$ in $G/H$ with $(aK)H$ in $(G/K)/(H/K)$. The rest is simply a matter of counting the elements on both sides.


Harvard
college
MIT
Putnam
geometry
topology
number theory

Hello Guys,

I am wondering if it is more difficult to get into a graduate school at the place where you did your undergrad (than if you apply from another college). Is this true?

I am not only interested in mathematical graduate programmes, but also medical schools, business schools etc.

For example, If I want to go to Harvard Business School, do I have a better chance to get admitted if I first attend Harvard College?

Thank you for your replies!

Professional schools work differently from graduate schools that way. To get into a good professional school (law school, medical school, etc.) it is ALWAYS a good idea to get into the very best undergraduate university program you can get into. That will continue to be relevant in the job market. I have seen employment advertisements looking for lawyers that specify, "The successful candidate for this position will have an Ivy League undergraduate degree." A LARGE percentage of the professional school admittees at Harvard have Harvard undergraduate degrees, and a person with a Harvard undergraduate degree is sure to get into a good professional school SOMEWHERE, even if not at Harvard.

But for graduate school there is much more of a tendency to encourage applicants to look for a new school. To get into a strong mathematics graduate program, it is mostly necessary to have a solid undergraduate mathematics major, but that need not be at a highly prestigious school. For example, St. Olaf College in Minnesota is a very good school to go to to prepare for mathematics graduate degrees. I learned many of the facts surrounding math graduate study from Steven G. Krantz's book [url=http://www.amazon.com/exec/obidos/tg/detail/-/082183455X/learninfreed]A Mathematician's Survival Guide: Graduate School and Early Career Development[/url]. Harvard specifically encourages its undergraduate math concentrators to apply to other schools for graduate study of mathematics.

Hope this helps! Please ask follow-up questions as needed.

The crucial difference between U.S. graduate and professional
school is that graduate schools pay salary money to
the students whereas professional schools (law, medicine,
business especially) will non-negotiably take your money either
up front or in the form of loans.  This has some consequences:

1. Grad student admission is an EMPLOYMENT INTERVIEW
process.  The department will be paying you, and at
the very least the professors will be seeing you
face-to-face in the hallways, seminars etc  for a
half-decade or so.  They may also be seeing you more
extensively, as an advisor.  So they want to know how
capable you are and "is this somebody we want to have
around".  The hiring is therefore competitive and relatively
meritocratic, and your abilities will be scrutinized more.

2. Professional school admission is more of an AUCTION
process -- selling a scarce resource to the highest
bidder, where the coin of the realm is grades, test scores
and undergraduate institution (not to mention parental
donations, professors' connections, etc).  To some
extent any admission process can be paraphrased that
way, but professional school admission is more cut-and-dried
and impersonal.  For law school especially you can
predict well what the outcomes will be based on LSAT
score and GPA.  Basically a repeat of college undergrad
admissions, but more down-to-business and without
the financial aid.  (Expect to pay, or not go).

3. From the professors' point of view, graduate students
are pseudo-colleagues and a RESOURCE of (at least
potentially, significant) value  --- people who can advance the
professor's research goals or work in their lab, co-author
papers, serve later as collaborators, offer jobs to the
professor's future students, etc.  This means
that the professors of the particular department will
have the primary say in your admission and can over-rule
the university admissions office.  They're more interested
in what you can do and what you've done than the
brand-name of the school you came from (i.e.,
having an MIT physics degree is better than one
from Garden Variety State College, but the credentials
will be scrutinized closely enough that the difference
isn't all-important).

4. At professional schools, on the other hand, students
are a DRAIN ON RESOURCES from the point of view of
a professor.  They require instruction and they will be
out the door once out of the professor's particular class.
Yes, some faculty like teaching or derive other benefits
(e.g. involving students in outside lab or consulting work)
but for the most part the teacher/student interaction
is a somewhat reluctant form of "customer service".
So schools are looking for customers who are problem-free
and docile, i.e. will graduate reliably and without hiccups,
and proceed to pay off their loans.

5.  Grad school admission is dramatically easier if you
are a US permanent resident (thus eligible for various
government monies as well as extremely strong
pro-US affirmative action in the admissions process).
Admission is guaranteed at most places if you bring
your own external scholarship funding (NSF, Fulbright, Hertz
etc).  It is also easier if you are applying for something
relatively obscure where there is a shortage of students,
or in a field that has a lot of money, or both (say, nautical
engineering rather than physics).

Note that graduate school here means doctoral (Ph.D.)
programs. Master's-only programs are just like
professional schools, they are moneymakers for the
university, or else a supply of cheap instructors
for basic courses such as algebra, physics, or writing.

Many of the best grad schools in math for example (eg Harvard) will NOT accept their own math undergraduates unless
a. a professor at the University really really wants to work with such an undergraduate
b. the whole applying pool is of a lower quality (not hapenning)

The reasons for this being that Universities want diversity.

Andrei

[quote="ajorza"]Many of the best grad schools in math for example (eg Harvard) will NOT accept their own math undergraduates unless [/quote]
Harvard has wavered about this over the years. They used to have a rigid policy against it, with a few exceptions such as a couple of Zariski's students (Mumford who started as a freshman, I think also Hartshorne as reported in the notorious back cover paragraph of his book).  When Harvard began monopolizing the pool of US math undergrads via the Putnam, they reversed this and started admitting their own.  It backfired to some extent and they restricted the policy again, though I think it is less hard-line than it once was.

Also, Harvard is compelled to accept any Putnam scholarship winners (I think there is one per
competition) who elect to exercise the award, irrespective of where (or if!) they were undergraduates.

[quote]The reasons for this being that Universities want diversity. [/quote] As you say in the other thread, the reasons are to some extent conjecture by those of us not on the admissions committees.
I don't think there is/was much diversity of any kind in the graduate program.   Rather, the stated reason was always that it's better for the ex-undergraduates to experience different environments.

Hello,

[quote="fleeting_guest"]It is also easier if you are applying for something
relatively obscure where there is a shortage of students,
or in a field that has a lot of money, or both (say, nautical
engineering rather than physics). [/quote]

This use to be a weapon of double blade, at least if you are neither US Citizen or US permanent resident and you plan to study in the US. There are some (actually, a lot) areas that have security checks or are considered "Restricted Technologies". Usually they are not that important when you apply to the school, but when you ask for the visa at the embassy. For example, someone I know was interested in some relevant area in AeroSpace, and he had to loose his first year, because there was a eight months of security check process. Of course, we know that the fact that we would like to be an specialist in a or b branch of sciences does not mean that we are going to use that knowledge to do some terrorism wapons or something like that, but right now, it seems to be a concern for some parts of the goverment.

There are easy ways to avoid this kind of problems, like being very general. If for example you are interested in Cryptography, in your interview in the embassy, you should say that you are interested in Number Theory, and if they ask you about something in especific, well say Elliptic Curves, and finally, if they ask you about applications, give the theorietical part, as group theory, algebraic geometry and topology, or something like that.

YES, I know that this may sound a little paranoic, but I have seen people more than once to have a rejection from the embassy because of their interests.

Best regards,


I have encountered some problems that give you an algebraic expression and in this problem it gives a theory that it (this theory) works in all cases as long as it's related with the algebraic expression.  The problem of this problem...would be...how to prove that the algebraic expression is ALWAYS related to the theory.  Not sure if you understand this....

So I give an example...

Ex.1 In the following expression, f(x) = x^2+3x^2, the result is always a prime number...

[b]It's a fake example[/b]

What sort of ways could we prove these types of problems?

Yeah...I just made that problem out of nowhere...it's not suppose to be true...

I know a problem from Mathcamp 2004...I'm not sure if some people would like it to get solved so I'll put it in spoilers.

[hide]The sequence a0 = 0, a1 = 1, a2 = 4, a3 = 15, ... is defined via the recursion an = 4 an-1 - an-2 for n >= 2. Show that the triple of numbers (an-1, 2 an, an+1) has the property that the product of any two of the numbers is one less than a perfect square.[/hide]


[quote="math92"][hide]
it can't be A becuse 2% of 2 is less than 4, it can't be B because 4% of 4 is less than 4, it can't be C becuase 10% of 10 is 1.  The answer is $\boxed{D}$ becuase 20% of 20 is 4.[/hide][/quote]

Nice way to use elimination but is there any "algebraic" way and perhaps much quicker way? (There is.  :) )

[quote="Silverfalcon"]A positive number $x$ has the property that $x \%$ of $x$ is 4. What is $x$?

A. 2
B. 4
C. 10
D. 20
E. 40[/quote]

Use a, whatchamacalit.

[hide]
$\frac{x}{100} = \frac4x$
$x^2 = 400$, $x=20$ since it's a positive number
D

[/hide]

[hide]
I just ran through all of the numbers up until 20 (which was the answer). $\framebox{D}$[/hide]


This competition will have 15-20 problems each round, the contestants will have 5-7 days to complete each round, and a lot of contestants. There will be enough rounds so that there will be determined a winner.

The sign ups begin now. You can either be a contestants, a regular coach or a head coach. Only 8th graders andbelow can be contestants, any people can be coaches. The contestants do the problems, send them in to the regular coaches, they correct them and hand them in to the head coaches and the head coaches organize them and announce the final scores.

The head coach right now is Treething. You still can be a head coach, you just have to sign up for it. There can be more than one head coach, more than one regular coach, and definately more than one contestant.

So far, the sign ups are

Head coaches:

Treething

regular coaches:

contestants:

Now Treething and the head coaches will be moderating this competition.

I may not take part of this competition, if you didn't know.

I recommend that there be more than one head coach. :)

have fun!

[u]Head Coaches (2):[/u]
Treething
LadyKn1ght

[u]Coaches (3):[/u]
junggi
lotrgreengrapes7926
kheg_k

[u]Contestants (19):[/u]
i_like_pie
13375P34K43V312
bpms
perfect628
daermon
SplashD
anirudh
goldendomer
easyas3.14159...
mgao
kstan013
ProtestanT
Ubemaya
Bluestalk
funcia
meenamathgirl
klutzcherry14
jhredsox
tspa

I think nutz is making the problems... I am going to join...

me too... if I have time.  I'll sign up anyways

I have nothing better to do so I shall do it as well.

I'll do it, just make lots of rounds, and include a countdown round as well  :)

Answer to WHO is making the problems.

Any preson who won't be competing may make the problems. I will probably make some, too.

Also, if the number of contestants number to more than 30, then I strongly suggest that there be more coaches!

Here are the current contestants:

i_like_pie
13375P34K43V312
bpms
perfect628
daermon
SplashD
anirudh
goldendomer
easyas3.14159...
mgao

ragnarok23 didn't say that he wanted to compete!

minor note: Only middle schoolers and below can compete!

I'll be a head coach. I'll be away from Oct 12-15 though, so if the competition is then I can't do it.

i'm in too although i can't guarantee that i'll be able to participate  :huuh:  :?:

(very sorry for retrieving this) you can move this back if you want

-jorian

[quote="jhredsox"]I know this is retrieving a dead post, but is this going anywhere

anyways, i was looking at my post list, and looked in the way back and saw this

-jorian[/quote]

What's the point of spamming the forum with a post like this?

ugh, i just made it so nutz would know why i did it

can this be locked up??? (leet - don't respond to this :P )

-jorian

[quote="nutz_for2.718281828"]PLease, no reviving topcs not posted in since 3 weeks ago. Thank you.

P.S. This competition is officially dead.

P.P.S. I wish this was like AoPS Kings... :([/quote]
What is the point of spamming the forum with a thread like this (not to be offensive)? I have a strong feeling this thread is going to get deleted for not having anything of use whatsoever.

[quote="jhredsox"]wait. do you mean like a post like this or a thread like this...there's a big differencio

-jorian[/quote]

Does it matter??? Just ignore the thread and let it fall to the bottom!

This thread hasn't been posted in in over 2 years....

[quote="topofmath"]where do i sign up?[/quote]
There was a post saying that this competition is dead, so you can't sign up anymore.

I have a bunch of questions.
1) When is this?
2) How do i sign up?
3) How long does this take?

Apparently the last post was about in 2006 so you cannot sign up anymore.

When does the competition start? I'm a 7th grader and would like to be a participant.

Try not to revive an old thread and read the preceding posts above yours.


It can be shown that if $ p : X \to Y$ is a quotient map, and $ Z$ is locally compact
Hausdorff, then $ p \times 1 : (X \times Z) \to (Y \times Z)$ is also a quotient map.
Give an example of a space $ Z$ (obviously not loc. compact Hausdorff) and a quotient map $ p$ such that $ (p \times 1)$ is [i]not[/i] a quotient map.

Take $ {\mathbb Q}\to{\mathbb Q}/{\mathbb Z}$ (collapsing $ {\mathbb Z}$ to a point) and $ Y \equal{} {\mathbb Q}$. Then the map $ {\mathbb Q}\times{\mathbb Q}\to {\mathbb Q}/{\mathbb Z}\times{\mathbb Q}$ is not a quotient map. Took me 2 pages to write this down in detail.

Wow, no wonder it didn't jump out at me :P.

You know, I'd really like texts to give these kinds of counterexamples
whenever they give a proposition like this, because the proof
I've seen is also a bit tricky an unmotivated. Thanks for the
ctr-ex.


If $a,b,c$ are real numbers and $(a-b)^2+(b-c)^2+(c-a)^2=1$ then find the minimum value of $a^2+b^2+c^2.$

the condition is equivalent to $3(a^2+b^2+c^2)-(a+b+c)^2=1$. So $(a^2+b^2+c^2)\ge\frac13$ with equality iff $a+b+c=0$. This can be attained for example if $a=b=\frac1{3\sqrt{2}},c=-\frac2{3\sqrt{2}}$ :)


inequalities
quadratics
calculus
integration
real analysis
real analysis unsolved

Hi, I'd like to ask about the sum of this double series:

\[ \sum_{n\equal{}0}^\infty \sum_{m\equal{}0}^\infty \frac{1}{(n\plus{}m)!},\]

which is very simple-looking, but I don't even know how to begin.

Thank you very much.

Hi sos440, thank you very much. The answer is amazingly simple and excellent.

To mathema*: thank you too, but I think that's not the final answer right?

Now I'd like to ask you: how do you know that the given double series converges (absolutely) before attempting to find
its sum?

Oh, and another question: given this series

\[ \sum_{n\equal{}1}^\infty(\frac{1}{n}\minus{}\frac{1}{n\plus{}x}), \qquad x\in R\minus{}\{\minus{}1,\minus{}2,\minus{}3,\dots,\},\]

how do you show its convergence? I use the limit comparison test, but the answer appears complicated.

Thank you.

[quote="vnca1"]how do you know that the given double series converges (absolutely) before attempting to find its sum?[/quote]
His argument is already an argument that the sum converges absolutely; you have the inequality $ \frac{k\plus{}1}{k!} < \frac{1}{(k\minus{}2)!}$ for $ k \ge 2$, for example.

[quote="vnca1"]given this series
\[ \sum_{n \equal{} 1}^\infty(\frac {1}{n} \minus{} \frac {1}{n \plus{} x}), \qquad x\in R \minus{} \{ \minus{} 1, \minus{} 2, \minus{} 3,\dots,\},
\]
how do you show its convergence? [/quote]
For fixed $ x$ each term is quadratic in $ n$.  It shouldn't be complicated.

[quote="vnca1"]
Oh, and another question: given this series
\[ \sum_{n \equal{} 1}^\infty(\frac {1}{n} \minus{} \frac {1}{n \plus{} x}), \qquad x\in R \minus{} \{ \minus{} 1, \minus{} 2, \minus{} 3,\dots,\},
\]
how do you show its convergence? I use the limit comparison test, but the answer appears complicated.

Thank you.[/quote]

Hi :) !!!

If you look for only the covergence of $ \sum_{n \geq 1} \left( \frac {1}{n} \minus{} \frac {1}{n \minus{} x}\right)$ i think that's conevrge by the comparing series-integrals if $ x < \infty$.
thx
______________________________________
lahoucine

[quote="vnca1"]... how do you know that the given double series converges (absolutely) before attempting to find its sum?[/quote]
There are basically two different hypotheses that can be used to justify all types of rearrangements of sum (including interchanging the order of summation for a double sum and including the the sort of rearrangement done here). One hypothesis is that the overall series converges absolutely (which is what vnca1 seems to be referring to); the other usable hypothesis is that all of the terms are nonnnegative.

If you use the nonnegative terms hypothesis without [i]a priori[/i] verifying absolute convergence, you might wind up with a result of $ \plus{}\infty$ - but then the theorem is that  you would get $ \plus{}\infty$ under any other rearrangement as well.

This manipulations in this problem can be justified because of nonnegative terms. That the final result is finite means we know after the fact that it was absolutely convergent as well.

This same pairing of hypotheses occurs in other (albeit closely related) contexts: Fubini's theorem versus Tonelli's theorem for double integrals; the dominated convergence theorem versus the monotone convergence theorem.

Thank you. What book do you recommend if I want to learn more about series?

Thanks.


I took the AMC 12 Problem Series this year.  The forum for the class is now locked, and has been for a while, and it says "Archive" next to it.  How long will the forum stay there?


Here's a nice one: the projections of a body on 2 planes are 2 disks. Prove that the 2 disks have equal radi.

Not too nice.. It took me a while... :(

Okay:

[b]Solution:[/b]

lets consider that the two planes are not paralel (not the trivial case); lets consider d as the intersection of the two planes(a,b). let C1' and C2' be the projection of circles C1 an C2 on d. For any point on the given body, lets consider it P, we denote now P1 and P2 the projection of P on a, and b respectively.
Now it is obious to see that the three projections of the points on d (P,P1 and P2) corespund; so C1' and C2' have equal radius!!

cheers! :D :D

Thx for the soln. I'll be posting a lot less from now on :( . My school is in a different city (Constanta) from the one I live in (Calarasi), and I don't have internet in Constanta (only at school).


topology
trigonometry
real analysis
real analysis unsolved

Let $ M\subset\mathbb{R}^2$ be a manifold consisting of two circles of radius $ 1/2$, one centered at $ (\minus{}1, 0)$ and the other at $ (1,0)$.

(a) Does the form $ (2x^3\minus{}2x)dy \minus{} ydx$ orient $ M$?

I'll get (b) if I see how to do this. I can come up with orientations, but I'm not sure how to check them. Probably show it is continuous and never zero, but how do you show that is never zero? It seems like that would be difficult.

I normally think of manifolds as connected spaces. Anyway, you indeed should prove that the form is never zero on $ M$. You know how to do that for a form $ f(t)\,dt$ on $ \mathbb R$ - it's zero iff $ f(x)$ is. Differential forms are easy to pull back under smooth maps such as $ t\mapsto \left(1\plus{}\frac12 \cos t,\frac12\sin t\right)$, which is a covering of the right circle. Is the pullback ever zero?

Hmm...that's what I thought.

I get that the simplified version is $ \frac{3}{4}\cos^2(t)\plus{}\frac{3}{4}\cos^3(t)\plus{}\frac{1}{8}\cos^4(t)\plus{}\frac{1}{4}$, which isn't zero for any $ t\in (0,2\pi]$. Now just do it for the other circle $ t \mapsto \left(\minus{}1\plus{}\frac{1}{2}\cos t, \sin t\right)$. That was another question. I figure you can just check each component separately.

If orientation means one of two possible choices of equivalence classes of nonvanishing $ 1$-forms, then each component can be dealt with separately.


$ \sum\frac{a^4\plus{}4a\minus{}4}{(a\minus{}1)^2}\geq54$

$ a,b,c>1$ and $ ab\plus{}bc\plus{}ca\equal{}abc$

[quote="lipako"]$ \sum\frac {a^4 \plus{} 4a \minus{} 4}{(a \minus{} 1)^2}\geq54$

$ a,b,c > 1$ and $ ab \plus{} bc \plus{} ca \equal{} abc$[/quote]

For me it is wrong....
the next is correct (with the same condictions):

$ \sum\frac{c^{4}\plus{}4a\minus{}4}{(a\minus{}1)^{2}}\geq54$


Sign up now!!!

No limit to sign ups, just cut off date by this coming Wednesday, July 1st.

[quote="Thunder365"]What level of questions will the test have? (AMC 10, AIME, etc)[/quote]
AMC 9 =)

Actually, it's hard AMC 8 to easy AMC 10 I guess.

[quote="BOGTRO"]I join. What type of mock test is this?[/quote]

Just a set of 30 questions.

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[quote="Dojo"]
AMC 9 =)

Just a set of 30 questions.
[/quote]

Shouldn't it be AIME 0.5 or AMC 9.5 because it has 30 questions?

click [hide="here"]
i wanna join![/hide]

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I have a feeling there will be some perfects. There's that for my intelligence.

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Okay guys, I'm cutting signups now.

Please download the test.

It is due one week from now.

Is there a time limit like 30 or 60 minutes? Or can we work on it as long as we want?

[quote="Dojo"]It is due one week from now.[/quote]

Reading is a good thing, and posting to increase post count is not.

[quote="AIME15"]Can we change our answers after they are submitted?[/quote]
No. Once you have submitted, you have submitted.

Also, please do not ask me questions about the test. The questions should be clear enough.

I know that I did not give my solutions but who is the winner.


ratio geometry perimeter symmetry inequalities trigonometry inradius

8

First day

8.1 (B.Frenkin) Does a convex quadrilateral without parallel sidelines exist such
that it can be divided into four equal triangles?

8.2 (F.Nilov) Given right triangle $ ABC$ with hypothenuse $ AC$ and
$ \angle A \equal{} 50^{\circ}$. Points $ K$ and $ L$ on the cathetus $ BC$ are
such that $ \angle KAC \equal{} \angle LAB \equal{} 10^{\circ}$. Determine the ratio
$ CK/LB$.

8.3 (D.Shnol) Two opposite angles of a convex quadrilateral with
perpendicular diagonals
are equal. Prove that a circle can be inscribed in this
quadrilateral.

8.4 (F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$;
the perpendicular bisectors
to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A'$, $ B'$; $ C_1$ is
the meet of lines $ AA'$ and $ BB'$. Prove that $ \angle C_1CA \equal{} \angle C_0CB$.

8.5 (A.Zaslavsky) Given two triangles $ ABC$, $ A'B'C'$. Denote by $ \alpha$ the angle between
the altitude and the median from vertex $ A$ of triangle $ ABC$.
Angles $ \beta$, $ \gamma$, $ \alpha'$, $ \beta'$, $ \gamma'$ are
defined similarly. It is known that $ \alpha \equal{} \alpha'$,
$ \beta \equal{} \beta'$, $ \gamma \equal{} \gamma'$. Can we conclude that the
triangles are similar?

Second day

8.6 (B.Frenkin) Consider the triangles such that all their vertices are vertices of a given
regular 2008-gon. What triangles are more numerous among them: acute-angled or
obtuse-angled?

8.7 (F.Nilov) Given isosceles triangle $ ABC$ with base $ AC$ and
$ \angle B \equal{} \alpha$. The arc $ AC$ constructed outside the triangle has angular
measure equal to $ \beta$. Two lines passing through $ B$ divide the
segment and the arc $ AC$ into three equal parts. Find the relation
$ \alpha / \beta$.

8.8 (B.Frenkin, A.Zaslavsky) A convex quadrilateral was drawn on
the blackboard. Boris marked the centers of four excircles each
touching one side of the quadrilateral and the extensions of two
adjacent sides. After this, Alexey erased the quadrilateral. Can
Boris define its perimeter?

9
First day

9.1 (A.Zaslavsky) A convex polygon can be divided into
2008 equal quadrilaterals.
Is it true that this polygon has a center or an axis of symmetry?

9.2 (F.Nilov) Given quadrilateral $ ABCD$. Find the locus of points such
that their projections to the lines $ AB$, $ BC$, $ CD$, $ DA$ form a
quadrilateral with perpendicular diagonals.

9.3 (R.Pirkuliev) Prove the inequality
\[ \frac1{\sqrt {2\sin A}} \plus{} \frac1{\sqrt {2\sin B}} \plus{} \frac1{\sqrt {2\sin C}}\leq\sqrt {\frac {p}{r}},
\]
where $ p$ and $ r$ are the semiperimeter and the inradius of
triangle $ ABC$.

9.4 (F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$; the perpendicular bisectors
to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A_c$, $ B_c$; $ C_1$
is the common point of $ AA_c$ and $ BB_c$. Points $ A_1$, $ B_1$
are defined similarly. Prove that circle $ A_1B_1C_1$ passes
through the circumcenter of triangle $ ABC$.

9.5 (N.Avilov) Can the surface of a regular tetrahedron be glued over with equal
regular hexagons?

Second day

9.6 (B.Frenkin) Construct the triangle, given its centroid and the feet of an altitude and
a bisector from the same vertex.

9.7 (A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$.
Another circle with the same radius passes through the orthocenter
$ H$ of this triangle and intersect its circumcirle in points $ X$,
$ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find
the circumradius of triangle $ ABZ$.

9.8 (J.-L.Aime, France) Points $ P$, $ Q$ lie on the circumcircle
$ \omega$ of triangle $ ABC$. The perpendicular bisector $ l$ to $ PQ$
intersects $ BC$, $ CA$, $ AB$ in points $ A'$, $ B'$, $ C'$. Let $ A"$,
$ B"$, $ C"$ be the second common points of $ l$ with the circles
$ A'PQ$, $ B'PQ$, $ C'PQ$. Prove that $ AA"$, $ BB"$, $ CC"$ concur.

10
First day

10.1 (B.Frenkin) An inscribed and circumscribed $ n$-gon is divided by some
line into two inscribed and circumscribed polygons with different
numbers of sides. Find $ n$.

10.2 (A.Myakishev) Let triangle $ A_1B_1C_1$ be symmetric to $ ABC$
wrt the incenter of its medial triangle. Prove that the
orthocenter of $ A_1B_1C_1$ coincides with the circumcenter of the
triangle formed by the excenters of $ ABC$.

10.3 (V.Yasinsky, Ukraine) Suppose $ X$ and $ Y$ are the common points of two
circles $ \omega_1$ and $ \omega_2$. The third circle $ \omega$
is internally tangent to $ \omega_1$ and $ \omega_2$ in $ P$ and $ Q$
respectively. Segment $ XY$ intersects $ \omega$ in points $ M$ and
$ N$. Rays $ PM$ and $ PN$ intersect $ \omega_1$ in points $ A$ and
$ D$; rays $ QM$ and $ QN$ intersect $ \omega_2$ in points $ B$ and
$ C$ respectively. Prove that $ AB \equal{} CD$.

10.4 (A.Zaslavsky) Given three points $ C_0$, $ C_1$, $ C_2$ on the line
$ l$. Find the locus of incenters of triangles $ ABC$ such that
points $ A$, $ B$ lie on $ l$ and the feet of the median, the
bisector and the altitude from $ C$ coincide with $ C_0$, $ C_1$,
$ C_2$.

10.5 (I.Bogdanov) A section of a regular tetragonal pyramid is a
regular pentagon. Find the ratio of its side to the side of the
base of the pyramid.

Second day

10.6 (B.Frenkin) The product of two sides in a triangle is equal to
$ 8Rr$, where $ R$ and $ r$ are the circumradius and the inradius of
the triangle. Prove that the angle between these sides is less than
$ 60^{\circ}$.

10.7 (F.Nilov) Two arcs with equal angular measure are constructed on
the medians $ AA'$ and $ BB'$ of triangle $ ABC$ towards
vertex $ C$. Prove that the common chord of the respective circles passes
through $ C$.

10.8 (A.Akopyan, V.Dolnikov) Given a set of points inn the plane. It
is known that among any three of its points there are two such that the
distance between them doesn't exceed 1. Prove that this set can
be divided into three parts such that the diameter of each part
does not exceed 1.

Thank you very much for the problems, Doctor A.

By popular demand to translate problems from the Sharygin Contest 2007, the rest of the problems of years 2006 and 2008 have been posted:

Sharygin Olympiad consists of two rounds, Correpondance and Final round. A problems of the Final round can be found here in English:

http://geometry.ru/olimp/sharygin/2007/var-final-e.pdf

[b]Correspondance round:[/b] (Will be available tomorrow, I hope)


Two things cannot exist: topics which are named hard ineq and problems posted 5 times. From now on, I regret, but I will delete every new topic that is called hard ineq or equivalent forms.

Thanks a lot, hope now he will no longer be obsessed by this problem!


inequalities
real analysis
combinatorics unsolved
combinatorics

Let $S$ be the set of points inside and on the boarder of a regular haxagon with side length 1. Find the least constant $r$, such that there exists one way to colour all the points in $S$ with three colous so that the distance between any two points with same colour is less than $r$.

I believe it's $\frac{3}{2}$.

First of all, in order to construct an example, let $ABCDEF$ be the hexagon, and let $O$ be its center. Draw three segments connecting $O$ to the midpoints $M,N,P$ of $AB,CD,EF$ respectively. This divides the hexagon into three regions, which will be the regions we color. On the borders, where the sets overlap, choose an arbitrary color, except in the points $M,N,P$, which we color differently (so that the inequality is strict: every two points having the same color are strictly less than $\frac{3}{2}=MN=NP=PM$ apart).

Conversely, we’ll prove that the intersection of one of the colored regions with the border of the hexagon contains points $X_{n},Y_{n}$ such that $\limsup\angle X_{n}OY_{n}\ge\frac{2\pi}3$ (I’m referring to the angle $\angle X_{n}OY_{n}\le\pi$). Since two points $U,V$ on the border of the hexagon such that $\angle UOV=\frac{2\pi}3$ are clearly at least $\frac{2}{3}$ apart, this will suffice for the converse.

 Since we’re now only dealing with angles, we may as well replace the hexagon with the unit circle. We can also assume that the circumference of the circle is covered with three [i]closed[/i] sets $\mathcal S_{i},\ i=\overline{1,3}$. Our aim will now be to show that in these conditions, there are two points $X,Y$ belonging to the same set $\mathcal S_{i}$ such that $\angle XOY\ge\frac{2\pi}3$. The conclusion will follow.

Assume the contrary, and take an arbitrary $P\in\mathcal S_{i}$. Then $\mathcal S_{i}$ must be contained in the union $\mathcal S$ of the two arcs of length $\frac{2\pi}3$ having $P$ as an endpoint. Now let $Q\in\mathcal S_{i}$ be one of the (at most two) points farthest away from $P$ (there is at least one such point because $\mathcal S_{i}$ are closed). All the other points of $\mathcal S_{i}$ must lie withing the arc of length $\frac{2\pi}3$ starting at $Q$ and contained completely in $\mathcal S$. We’ve shown that each $\mathcal S_{i}$ is contained strictly in an arc of length $\frac{2\pi}3$. Since the $\mathcal S_{i}$ are closed, we conclude that each one has measure (the Lebesgue measure on the circle, with totale size $2\pi$) strictly less than $\frac{2\pi}3$, and this contradicts the fact that the three sets cover the circumference, which has measure $2\pi$. We have our contradiction.


At the end of a professional bowling tournament, the top 5 bowlers have a playoff. First #5 bowls #4. The loser recieves 5th prize and the winner bowls #3 in another game. The loser of this game receives 4th prize and the winner bowls #2. The loser of this game receives 3rd prize and the winner bowls #1. The winner of this game gets 1st prize and the loser gets 2nd prize. In how many orders can bowlers #1 through #5 receieve the prizes?

(A) 10
(B) 16
(C) 24
(D) 120
(E) none of these

I just finished this test... and I didn't get this one either... they had an odd solution which I didn't understand.

I would think it's 120. But then i realize it's a #18.
What does the solution manual say?

[hide]
For 5th place, there are 2 possibilities, 4 and 5.  Then for 4th place, you will either have 3, or (4 or 5) depending on who won the 5th place game, so there are only two possibilities, since it's dependent on the 5th place game.  This continues, and since there are 4 games, there are $2^4=16$ ways.
[/hide]

either the underdog wins the game, or the expected wins, in 4 games, that gives $2^4=16$

[hide]There are two possibilities for the winner of the game #5 v. #4, two for #4 v. #3, ..., two for #2 v. #1, so total $2^4 =16$.

$\therefore \boxed {B}$[/hide]


Let $a, b, c$ be three integrers such that $a+b+c=0$, and let be $d$ such that $d=a^{47}+b^{47}+c^{47}$.

A) Can $d$ be equal to $2$???

B) Can $d$ be a prime number????

[quote="me@home"]D cannot be two, the magnitudes becomes too great and we can easily disprove trivial cases of 1 and -1.[/quote]Can you at least give a more rigorous proof? I'm not convinced by this statement.

That would require time, which I don't have...
But I will try.
[hide="Proof that d cannot equal 2"]
First we see that the difference between consecutive fourth powers is $(a+1)^47 - a^47 = 47a^46 + 47(23)*a^45 + ... + 47a + 1$
We see that this grows quite rapidly and that since $a+b = -c$ (or $c = -a-b$) the crude large differences in the 47th powers can't satisfy a sum of two. Someone could write up a proof of this quite simply.
[/hide]

[quote="scorpius119"][hide]$a^{47}\equiv a\pmod{47}$ by Fermat's Little Theorem (the other variables work similarly), so $a^{47}+b^{47}+c^{47}\equiv a+b+c\equiv 0\pmod{47}$. So $d$ could not possibly be any prime other than 47.

But also $a^{47}$ has the same parity as $a$, so $d$ has the same parity as $a+b+c=0$ making $d$ even. Then $d$ could not equal 47. Therefore $d$ cannot be prime.[/hide][/quote]

Why do you say $a+b+c=0$?

Uh, I didn't ask that question.  Someone must have logged in as me.  :blush:


I couldn't figure this guy out.?

Inside a room of area 5, you place 9 rugs, each of area 1 and an arbitrary shape. Prove that there ar etwo rugs which overlap by at least 1/9.

Any advice / solutions would be awesome, because I am beginning to hate rugs! ha

[hide]

The rugs take up 9 units of area, but only 5 units are needed; so there are 4 units of overlap. With 9 rugs, there are $\binom{9}{2}=36$ possible pairs of rugs. There are 36 different pairs of rugs to measure their overlap.

Assume that each of these areas of overlap is $<\frac{1}{9}$ Then the total of the 36 areas is $<36\left(\frac{1}{9}\right)$, or $<4$. But the total area of overlap must be 4, a contradiction.

[/hide]

[hide]So we have a total overlap of four units. But we only have 36 pairs to overlap. So we must overlap by at least 1/9 by pigeonhole[/hide]


geometry
analytic geometry
graphing lines
slope

Hi..I just became a freshman and I am worrying about my grades for Algebra 1...when i was in 8th grade I took Algebra 1 and I did poorly. So I am trying to see if I can take or fix my grades...can I take it in summer or am I too late? did i had to take algebra in school this year? However, now I am doing really good on Geometry and other classes. Plz answer my question..thanks~!!  :blush:

[quote="mustafa"]How poorly? Did you pass the class? Where I live, if you get a D-, that is good enough to go on to the next class :wink:[/quote]
D-? How does that show that you passed class? The "average" is considered a C, so D- is below average.

honestly im not sure but it might be possible it really all depends if you can pull some strings at the district

well i got two Cs,B, and D....-_- but i want to raise my grades....can i do that in summer school???

If the curriculum has solutions, see how they did the solutions. If you don't know where you went wrong in what problems, where in each problem, and stuff like that, you may not be able to do Algebra. Try finding a book that has solutions.

I would suggest the Art of Problem Solving books:
AoPS Vol. 1 for basic manipulations.
AoPS Vol. 2 for advanced manipulations.

Once you have Volume 1, read the "Possible Curricula" paragraphs and find the Algebra curriculum.

I hpoe that helps!!! :D

but i want to go to good college...:( I heard that people see Algebra 1 grade...or should i just stop worrying and keep going to next level and get good grades? cause i am worrying about Algebras 1...:(

[quote="mustafa"]I'm pretty sure that colleges only look at grades that you got in high school.[/quote]
I believe Algebra is a high school credit course. Or that might just be in my area.

Algebra is a high school credit course....so it counts...noo...:(
my goal is to go to California Institute of Technology...lol well i get straight As but worrying about algebra...i got As whole classes...but...:(

In Ohio, anything that gives you high school credit does.  So like, my Algebra I & II and French I grades all count even though I took them in middle school...

oo...interesting...but i am in maryland...lol >.< it's so confusing...i really wanna go to good college...:( well when people from college see my grades...do they just see only the total 4 years of my GPA? or like each classes grades?

[quote="mustafa"]I am not positive about this, but in wisconsin at least, only classes that you took in high school go onto your GPA.[/quote]

So it is in GA too. Or at least for me.

   Akajoe maybe you could study and take an SAT II for algebra or something because trust me it isn't that hard..to tell the truth the way it is usually split into algebra 1 then geometry then algebra 2 makes it so you forget algebra 1 and have to do lots of review when starting algebra 2 so if you got a passing score for algebra 1 keep going until algebra 2...an AoPS book'd do you good if you try to do an SAT II

okay...danng you guys are really good at this...XD good thing that i joined this forum...lolz...well thanks people...i will write something if i need help...lol well my goal is to get As on math and then try SAT II? yeah...something like that...-_-

Joe,

I would suggest talking to your guidance counselor.  These people are hired to actually help you, not just discipline you (learned that in 8th grade  :rotfl: ).  But seriously, tell your guidance counselor your concern, and see if she can help you.  More often than not, it's better to swallow your pride and talk to someone who can help you.  Learned that in 8th grade too.  XD

yeah i should...well thanks for the advise~!!

[quote="lotrgreengrapes7926"]I'm pretty sure that if you excel at Geometry, Algebra II, Pre-Calc, and Calc (if you're going to take it), the bad grade in Algebra I will not kill you. :D[/quote]

No offense or anything, but you will be unlikely to excel at those higher math courses if you don't have a good base in Algebra I. Grades and colleges aren't everything. You should probably review Algebra I- not just for the grades, but so that you can have a solid base for other classes too.

alg 1 is just like slope and stuff. review that

it wont have your 8th grade stuff on transcript. Taking it in summer just draws attention

but...algebra 1 is a high school course...:/

Often its not.

I recommend you dont waste your time with that, and instead, if you are weak in it, try to review a little on your own. They won't pay attention to it if you have strong grades in math classes ahead.

Plus as I said earlier when doing algebra II most of it is just review of algebra I

thanks to you guys...i went to my counsler to figure this our...XD yeah...she told me that i CANT retake it...but i can just start over and do good from now on...she wants me to go to honors next year so i can just improve and stuff...yupp...THANKS AGAIN~!!!


algebra
polynomial
linear algebra
linear algebra unsolved

[quote="jmerry"]These matrices are already in Jordan form- you can read off the minimal and characteristic polynomials (in factored form) from that.[/quote]

I need the minimal and characteristic polynomials of T , not of B and C.

can someone post a solution?


Q. Assume that two distinct positive odd integers $a$ and $b$ satisfy $(a-b)^{2}\le 8\sqrt{ab}$. If $n=ab$, then show that the roots of equation
\[x^{2}-2(\lfloor \sqrt{n}\rfloor+1)x+n=0 \]
are all positive integers.

(From KMO middle school division)

That floor function makes it difficult, I think :D

[hide="Solution"] Let $k = \lfloor \sqrt{ab}\rfloor$.  Then we have distinct positive odd integers $a, b$ satisfying

$ab < (k+1)^{2}$
$(a-b)^{2}\le 8k$

Together these results imply

$(a+b)^{2}< 4k^{2}+16k+4 \Leftrightarrow$
$a+b < \sqrt{4k^{2}+16k+4}< 2k+4$

But since $a, b$ are odd we cannot have $a+b = 2k+3$, so

$a+b < 2k+3$

But by AM-GM we have

$a+b > 2k$

(The equality case is impossible since $a \neq b$).  Again, since $a, b$ are both odd we must have

$a+b > 2k+1$

Combining our results, we conclude that $a+b = 2k+2$, hence

$x^{2}-2(k+1)x+n = (x-a)(x-b)$ [/hide]


idk if this is the right forum to post this question, but how do you know that $ \binom{1}{0.5}$ is $ \frac{\sqrt{\pi}}{2}$?

how do you derive it?

This probably has to do with the [url=http://en.wikipedia.org/wiki/Gamma_function]Gamma function[/url] extending the definition of factorial, but the value you give doesn't seem to coincide...

By using a generalization of the factorial function so that it is defined for noninteger values, and then using a variation of the factorial formula. The generalized factorial is called the [url=http://en.wikipedia.org/wiki/Gamma_function]gamma function[/url]. Its definition was picked because it gets nice properties in complex analysis.

In other words, there's no elementary proof.

Here's a fascinating identity directly results from the Gamma function: $ \boxed{\binom{x}{y} \cdot \binom{y}{x} \equal{} \frac {\sin((x \minus{} y) \pi)}{(x \minus{} y) \pi} }$

Substitute in $ (x,y) \equal{} (1, \; 1/2)$: you get $ \binom{1}{1/2} \cdot \binom{1/2}{1} \equal{} \frac {\sin (\pi / 2)}{\pi /2}$

We know that $ \binom{x}{1} \equal{} x$, so this becomes $ \frac {1}{2} \binom{1}{1/2} \equal{} \frac {2}{\pi} \implies \binom{1}{1/2} \equal{} \frac {4}{\pi}$

[b]Note:[/b] [i]This is not what the OP claims; what [b]is[/b] true about the value that the OP cited is that $ (1/2)! \equal{} \frac {\sqrt {\pi}}{2}$, if you analytically extend the factorial to the reals using the definition $ x! \equal{} \Gamma (x \plus{} 1)$[/i]

[quote="TZF"][b]Note:[/b] [i]This is not what the OP claims; what [b]is[/b] true about the value that the OP cited is that $ (1/2)! \equal{} \frac {\sqrt {\pi}}{2}$, if you analytically extend the factorial to the reals using the definition $ x! \equal{} \Gamma (x \plus{} 1)$[/i][/quote]
Ah, that's what I thought.
With this extention of factorial, can we calculate $ \binom{1}{.5}\equal{}\frac{1!}{.5!\cdot .5!}\equal{}\dfrac{1}{\frac{\sqrt \pi}{2}\cdot\frac{\sqrt \pi}{2}}\equal{}\frac{4}{\pi}$?

We sure can.

I used the definition of the binomial coefficient to find $ (1/2)!$ in my post, but if you somehow know that value to start off with, you can definitely use it to compute the binomial coefficient.


what do people think of the two slit experiment?

its the experiment where a light is shined on a wall.  there is a barrier in the way though, but its got two slits in it.  the light on the wall isnt two ellipitcal shapes, its an intereference pattern, it alternates between light and dark many times.  it shows that lights a wave (of photons like water's a wave of h2o).

but its gets wierder:  it can also be done with electrons.  but the wierd part is that even if one electon is shot at a time, an interference pattern is created.  one would expect (including me) that the electron would go through one of the slits and the other slit would be irrevelent, but both slits are involved, wierd.

i think it was feynman who theorized that the electon will take every possible path (infinite) from the emitter on one side of the barrier to the sensor on the other side, and the sum of all the paths plus some probability is where it will end up.

does anyone know of any other theroizes or what they think of it?

Don't they say that electron sometimes behave as wave too? You can even calculate it's wavelength using de Broglie's formula.

The same argument could be used for light - imagine a single photon, it must choose one single way to go. But that's not what happens - it will too end up somewhere in the middle of the interference pattern.


ratio
linear algebra
matrix
function
algebra
polynomial
linear algebra unsolved

Suppose $ a,b,c\in C$, $ c\ne 0$. Let $ (x^3 \plus{} ax^2 \plus{} bx \plus{} c)^{ \minus{} 1} \equal{} \sum c_kx^k$.
Show that $ det\left (\begin{array}{ccc}c_k & c_{k \plus{} 1} & c_{k \plus{} 2} \\
c_{k \plus{} 1} & c_{k \plus{} 2} & c_{k \plus{} 3} \\
c_{k \plus{} 2} & c_{k \plus{} 3} & c_{k \plus{} 4}\end{array}\right) /det\left (\begin{array}{ccc}c_{k \plus{} 1} & c_{k \plus{} 2} & c_{k \plus{} 3} \\
c_{k \plus{} 2} & c_{k \plus{} 3} & c_{k \plus{} 4} \\
c_{k \plus{} 3} & c_{k \plus{} 4} & c_{k \plus{} 5}\end{array}\right)$ is independent of $ k$.

To avoid conflicts of notations, let's rename the coefficients $ c_k$ as $ p_k$.

Let $ M_k\equal{}\begin{pmatrix}p_k&p_{k\plus{}1}&p_{k\plus{}2}\\p_{k\plus{}1}&p_{k\plus{}2}&p_{k\plus{}3}\\p_{k\plus{}2}&p_{k\plus{}3}&p_{k\plus{}4}\end{pmatrix}$. We have that $ (x^3\plus{}ax^2\plus{}bx\plus{}c)\cdot\sum p_kx^k\equal{}1\plus{}0\cdot x\plus{}0\cdot x^2\plus{}\cdots$, so the coefficient $ p_k\plus{}ap_{k\plus{}1}\plus{}bp_{k\plus{}2}\plus{}cp_{k\plus{}3}$ of $ x^{k\plus{}3}$ is zero. From this relation, repeated for each entry of the bottom row, $ M_{k\plus{}1}\equal{}\begin{pmatrix}0&1&0\\0&0&1\\ \frac{\minus{}1}c&\frac{\minus{}a}c&\frac{\minus{}b}c\end{pmatrix}\cdot M_k$

The determinant ratio is thus $ \frac{\minus{}b}{c}$. This all makes sense over any field; we're just working with formal power series rather than analytic functions most of the time.

I like companion matrices. In this case, it's really about the mirrored polynomial, with roots the reciprocals of those of $ x^3\plus{}ax^2\plus{}bx\plus{}c$.


Find all functions f from the real numbers to the real numbers which
satisfy
$ f (x^3 ) \plus{} f (y^3 ) \equal{} (x \plus{} y)(f (x^2 ) \plus{} f (y^2 ) \minus{} f (xy))$

for all real numbers x and y.

$ x \equal{} y$ yields $ \boxed{f(x^3) \equal{} xf(x^2)}\quad (1)$. Hence $ \boxed{f( \minus{} x) \equal{} \minus{} f(x)}\quad (2)$. Substituting $ (1)$ into the original equation we deduce that
\[ (x \plus{} y)f(xy) \equal{} xf(y^2) \plus{} yf(x^2).\qquad (3)
\]
Using $ (2)$, $ y\mapsto \minus{} y$ into $ (3)$ yields
\[ \minus{} (x \minus{} y)f(xy) \equal{} xf(y^2) \minus{} yf(x^2).\qquad (4)
\]
Adding $ (3)$ and $ (4)$ we get $ \boxed{xf(y^2) \equal{} yf(xy)}$. Finally, $ y \equal{} 1$ in the last equation yields $ \boxed{f(x) \equal{} xf(1)}$.

So those 4 are the solutions for this problem?

Um... no. The solutions are $ f(x)\equal{}xf(1)$ where $ f(1)$ is any real number. If you read my post carefully you should realize this yourself even though I don't write it specifically. I just boxed those equations to make the reader understand where I was going from where.

But isn't $ f(x)\equal{}x^a$ also a solution?

[quote="malcolm"]But isn't $ f(x) \equal{} x^a$ also a solution?[/quote]
$ f (x^3 ) \plus{} f (y^3 ) \equal{} (x \plus{} y)(f (x^2 ) \plus{} f (y^2 ) \minus{} f (xy))$
$ x^{3a}\plus{}y^{3a}\equal{}(x\plus{}y)(x^{2a}\minus{}x^ay^a\plus{}y^{2a})$
$ 0\equal{}xy^{2a}\plus{}yx^{2a}\minus{}x^ay^a(x\plus{}y)$
Try a=2
Nope :wink:

Let $x=y=0$ in the given functional equation, giving us $f(0)=0$.  Let $y=0$ gives $f(x^3)=xf(x^2)$.  Substituting this in the functional equation gives $$xf(y^2)+yf(x^2)=(x+y)f(xy).$$  Since the range of $x^3$ is all reals, we have $f(-x^3)=-xf(x^2)=-(xf(x^2))=-f(x^3)$, proving $f$ is odd.  Now, let $x=x$ and $y=1$ and $x=-x$ and $y=1$, giving \begin{align*} xf(1)+f(x^2)=xf(x)+f(x)\\-xf(1)+f(x^2)=-xf(-x)+f(-x).\end{align*}  Noting that $f(x)=-f(-x)$, we can subtract the latter from the prior to get $2f(x)=xf(1)$ or $\boxed{f(x)=2xf(1)}$, where $f(1)$ may be $0$. $\blacksquare$


Given a positive integer $ n$, denote by $ \sigma (n)$ the sum of all positive divisors of $ n$. We say that $ n$ is $ abundant$ if $ \sigma (n)>2n.$ (For example, $ 12$ is abundant since $ \sigma (12)\equal{}28>2 \cdot 12$.) Let $ a,b$ be positive integers and suppose that $ a$ is abundant. Prove that $ ab$ is abundant.

An easy problem.

Let $ d_1,d_2\ldots d_k$ be a set of divisors of $ a$ who sum to more than $ 2a$. Then $ bd_1,bd_2\ldots bd_k$ are divisors of $ ab$ which sum to more than $ 2ab$.


topology
group theory
abstract algebra
advanced fields
advanced fields unsolved

Let $ G$ be a connected lie-group. Show that $ \pi_2 (G) \equal{} \{0\}$.

reduce to compact simply-connected. then use classification or the spectral sequence in http://eom.springer.de/C/c023060.htm with $ X \equal{} G, Y \equal{} *$

Another approach:

Again, reduce to the case of a compact group (by a theorem of Iwasawa, every connected Lie group has a compact subgroup as a deformation retract). Using Morse Theory, it can be shown that the [url=http://planetmath.org/encyclopedia/LoopSpace.html]loop space[/url] $ \Omega(G)$ of a compact connected Lie group $ G$ has the homotopy type of a CW-complex with no cells of odd dimension. In particular, its fundamental group vanishes. Since $ \pi_1(\Omega(G))\equal{}\pi_2(G)$, we're done.

Nice! I didn't manage to prove $ \pi_2(G) \equal{} \{0\}$, but only that the rational part $ \pi_2(G)\otimes {\mathbb R}$ must vanish. For this, assume $ \pi_1(G) \equal{} 0$ and $ G$ compact. Then we have $ \pi_2(G)\otimes {\mathbb R}\cong H^2(G;{\mathbb R})$, and this cohomology group coincides with the second lie-algebra cohomology of the semisimple lie-algebra $ L(G)$ which vanishes by the second Whitehead-Theorem.

could you prove this theorem "every connected Lie group has a compact subgroup as a deformation retract."
or is there an online resourse of its proof?

It's proved in a paper of Iwasawa, "On some types of topological groups" (in the Annals of Mathematics, 1949). The proof is neither simple, nor short, and certainly not suitable for posting here; besides, I'm not familiar enough with it :).


As most of us know, the mathematical notations used in different countries are different, and it sometimes causes confusion.  There's a bit of information in [url=http://www.mathlinks.ro/phpBB/viewtopic.php?t=2469]this thread[/url], but it will probably sink to the bottom soon enough, so maybe everybody could clarify the notations used in their country in this thread instead. :)

I'll start with those used in Singapore (and south-east Asia, I think):

- R+ is (0,+oo)
- N is {1,2,3...}
- N0 is {0, 1,2,3,...}, but this notation is very rare, we use N  \cup {0} more often
- tangent and cotangent are tan and cot respectively
- n!/r!(n-r)! is nCr or in 2x1 matrix form with n on top and r below

I suppose this is the British style that we inherited... :)

in China (in current textbooks)

R+ is (0,\infty)
N  is {0,1,2,3,4...}
N+, N*, N+ all means {1,2,3,4...}
Permutation (n choose r) is A_n^r
Combinatorial coefficient is C_n^r (in high school textbooks).


Ok - so I have a problem with a strip containing icons of Calendar, Faq,Search, etc and Advanced Section, College Playground and so - I've attached a screen, look at the bottom (I'd like them to appear at the top). Any solution?

P.S. This problem only appears on Opera 9. On IE as on older versions of Opera it seems to work normally

Bug is confirmed. Will try to fix, but that will come by next week sorry (unless someone gives me a pure CSS solution until then :) )

PS Opera 9 is much faster and cooler than previous versions of Opera. I might end up using it as my main browser if more wedgies appear :)

http://validator.w3.org/check?uri=http%3A%2F%2Fwww.mathlinks.ro%2FForum%2Findex.php&charset=%28detect+automatically%29&doctype=Inline&ss=1

The validator says you're missing a </table> somewhere...maybe that's it?

Good catch. I've added the missing table ;)


|z_1|=1, |z_2|=k , |z_3|=2-k, z_1=cos a+i sin a , z_2=k(cos b+i sin b),
z_3=(2-k)(cos c+i sin c) and z_1+z_2+z_3=0,

(1) fint the minimum of  cos(b-c)=? and  k=?

(2) fint the maximum of   cos(b-c)=? and k=?

Can anybody answer me ?  :maybe:

Please do not double post. :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200401


conics
parabola
analytic geometry
rotation
LaTeX

The focus of a parabola is located at $ (\minus{}1, \minus{}1)$ and its directrix is $ 3x\plus{}4y\equal{}5$. What is the equation of the parabola? Explain how you arrived at your answer.

The directrix is $ y\equal{}\minus{}\frac{3}{4}x\plus{}\frac{5}{4}$, which means our coordinate axes are rotated, which means this problem is ugly :(

I don't know much about rotated conics. Sorry...

A parabola is the locus of points determined by a line (directrix) and point (focus) such that any point on the parabola is equidistant from the directrix and the focus.

Using this definition, call any point on the parabola (x,y) Use the distance between a point and line formula with (x,y) and 3x + 4y = 5 and set this distance equal to the distance between (x,y) and (-1,-1).

I don't know how to use Latex and it looked pretty messy when I started it, but I'm pretty sure this method will work.

So you use the method to obtain three points and then use those points to write the equation?

This was a team round problem at our local math league competition. 3 people and 10 minutes to solve it.

Don't you already have an equation if you equate the distance from the directrix and the distance from the focus?  :maybe:


What is the product of the real roots of the equation

$ x^2\plus{}18x\plus{}30\equal{}2\sqrt{x^2\plus{}18x\plus{}45}?$

Refer to
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60880

Also, AoPS has an archive for most AIME problems. Just go to the following link and click on the year.
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45
You can then click on the number of the problem to view a thread about that problem.

Happy solving! :)


What is alweays ahead of fashion but always out of date?

Fashion Designers  :P ?? They're always ahead of fashion, because they design it, but all of them are the same and thus they're cliche.
Same for models.
 :rotfl:  :rotfl:

yes, but then s, h, i, o, and n would work too

but s, h, i, o, and n isn't [i]ahead[/i] of fashion =)


inequalities
search
inequalities proposed

[quote="Evariste-Galois"]Given a,b,c>0 Prove that :
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq\frac {3}{4} \plus{} \frac {a^2 \plus{} b^2}{2(a \plus{} b)^2} \plus{} \frac {b^2 \plus{} c^2}{2(b \plus{} c)^2} \plus{} \frac {c^2 \plus{} a^2}{2(a \plus{} c)^2}$[/quote]

SOS(Sum of square) can solve it :)

It is equivalent to the following inequality

$ (ab\plus{}bc\plus{}ca)\left(\sum\frac{1}{(a\plus{}b)^2} \right) \ge \frac{9}{4}$

Therefore, needless to say, it is definitely famous Iran 98.

yes!! you're right, I found this result when I was sloving Iran 96 Tst's inequality  ,llool  :)

[quote="dgreenb801"]Yes, it is Iran 1996 (I don't think it is 98), we have
$ \frac {a^2 \plus{} b^2}{2(a \plus{} b)^2} \equal{} \frac {(a \plus{} b)^2 \minus{} 2ab}{2(a \plus{} b)^2} \equal{} \frac {1}{2} \minus{} \frac {ab}{(a \plus{} b)^2}$
So we have to show
$ \sum \frac {c}{a \plus{} b} \plus{} \frac {ab}{(a \plus{} b)^2} \equal{} (ab \plus{} bc \plus{} ca)\sum (\frac {1}{a \plus{} b})^2 \ge \frac {9}{4}$.
Which is Iran 1996.[/quote]

Dear dgreenb801! Please can you show how U solved this problem? I don'y understand how to prove! Thanks in advance! RFZ!

[quote="RFZ"]
Dear dgreenb801! Please can you show how U solved this problem? I don'y understand how to prove! Thanks in advance! RFZ![/quote]
It is easy my friend
You have to prove that:
$ \sum_{cyc}\frac{ab}{(a\plus{}b)^2}\plus{}\frac{c}{a\plus{}b}\ge \frac{9}{4}$

$ \Longleftrightarrow \sum_{cyc}\frac{ab\plus{}bc\plus{}ca}{(a\plus{}b)^2}\ge \frac{9}{4}$

$ \Longleftrightarrow (ab\plus{}bc\plus{}ca)\sum\frac{1}{(a\plus{}b)^2}\ge \frac{9}{4}$,

which is the well known result....

[quote="Dimitris X"][quote="RFZ"]
Dear dgreenb801! Please can you show how U solved this problem? I don'y understand how to prove! Thanks in advance! RFZ![/quote]
It is easy my friend
You have to prove that:
$ \sum_{cyc}\frac {ab}{(a \plus{} b)^2} \plus{} \frac {c}{a \plus{} b}\ge \frac {9}{4}$

$ \Longleftrightarrow \sum_{cyc}\frac {ab \plus{} bc \plus{} ca}{(a \plus{} b)^2}\ge \frac {9}{4}$

$ \Longleftrightarrow (ab \plus{} bc \plus{} ca)\sum\frac {1}{(a \plus{} b)^2}\ge \frac {9}{4}$,

which is the well known result....[/quote]

My dear friend I understand all this! Thak YOU! But I have one question! How to prove the last inequality? Please help!

[quote="Moonmathpi496"][quote="RFZ"]My dear friend I understand all this! Thak YOU! But I have one question! How to prove the last inequality? Please help![/quote]
This is a very well known problem, known as Iran 96. It is so much celebrated that you can find [url=http://www.mathlinks.ro/viewtopic.php?search_id=838963879&t=249265]can_hang2007's 13th solution here.[/url] ;)
Also: http://www.mathlinks.ro/viewtopic.php?search_id=838963879&t=32470[/quote]
Thank YOU my brother!

I have one solution for the last inequaluty! I can post my solution! Will anybody check up my solution?? =)
I want to know my solution is true or not!

$ (ab \plus{} bc \plus{} ca)\sum\frac {1}{(a \plus{} b)^{2}}\ge\frac {9}{4}$

It's my solution:

At first we make some replacements: $ a\equal{}\frac{1}{x}, b\equal{}\frac{1}{y}, c\equal{}\frac{1}{z}$
And our inequality takes such view:

\[ \frac{xy}{xz\plus{}yz}\plus{}\frac{yz}{xy\plus{}xz}\plus{}\frac{xz}{xy\plus{}yz}\geq\frac{3}{4}\plus{}\frac{x^{2}\plus{}y^{2}}{2(x\plus{}y)^{2}}\plus{}\frac{y^{2}\plus{}z^{2}}{2(y\plus{}z)^{2}}\plus{}\frac{x^{2}\plus{}z^{2}}{2(x\plus{}z)^{2}}\]

We make some changes in our inequality:

\[ \frac{x^{2}\plus{}y^{2}}{2(x\plus{}y)^{2}}\equal{}\frac{(x\plus{}y)^{2}\minus{}2xy}{2(x\plus{}y)^{2}}\equal{}\frac{1}{2} \minus{} \frac{xy}{(x\plus{}y)^2}\]

\[ \frac{y^{2}\plus{}z^{2}}{2(y\plus{}z)^{2}}\equal{}\frac{(y\plus{}z)^{2}\minus{}2yz}{2(z\plus{}y)^{2}}\equal{}\frac{1}{2} \minus{} \frac{yz}{(z\plus{}y)^2}\]
\[ \frac{x^{2}\plus{}z^{2}}{2(x\plus{}z)^{2}}\equal{}\frac{(x\plus{}z)^{2}\minus{}2xz}{2(x\plus{}z)^{2}}\equal{}\frac{1}{2} \minus{} \frac{yz}{(z\plus{}y)^2}\]
 We have the following inequality:

\[ \frac{xy}{xz\plus{}yz}\plus{}\frac{yz}{xy\plus{}xz}\plus{}\frac{xz}{xy\plus{}yz}\plus{}\frac{xy}{(x\plus{}y)^2}\plus{}\frac{yz}{(z\plus{}y)^2}\plus{}\frac{xz}{(x\plus{}z)^2}\geq\frac{9}{4}    \Rightarrow\]

\[ \Rightarrow \frac{xy}{z(x\plus{}y)}\plus{}\frac{yz}{x(y\plus{}z)}\plus{}\frac{xz}{y(x\plus{}z)}\plus{}\frac{xy}{(x\plus{}y)^2}\plus{}\frac{yz}{(z\plus{}y)^2}\plus{}\frac{xz}{(x\plus{}z)^2}\geq\frac{9}{4} \Rightarrow\]

\[ \frac{xy(x\plus{}y\plus{}z)}{z(x\plus{}y)^{2}}\plus{}\frac{yz(x\plus{}y\plus{}z)}{x(z\plus{}y)^{2}}\plus{}\frac{xz(x\plus{}y\plus{}z)}{y(x\plus{}z)^{2}}\geq\frac{9}{4} \Rightarrow\]

\[ [x\plus{}y\plus{}z] \left[\frac{xy}{z(x\plus{}y)^{2}}\plus{}\frac{yz}{x(z\plus{}y)^{2}}\plus{}\frac{xz}{y(x\plus{}z)^{2}}\right]\geq\frac{9}{4}\]
Here we use the inequality of AM-GM.
\[ x\plus{}y\geq 2\sqrt{xy}, (x\plus{}y)^{2}\geq 4xy\]

From here we have.

\[ \frac{1}{(x\plus{}y)^2}\leq \frac{1}{4xy}\Rightarrow\]

\[ \frac{1}{4z}\geq \frac{xy}{z(x\plus{}y)^2}\]

Applying our inequality we have:

\[ [x\plus{}y\plus{}z] \left[\frac{1}{4z}\plus{}\frac{1}{4x}\plus{}\frac{1}{y4}\right]\geq [x\plus{}y\plus{}z] \left[\frac{xy}{z(x\plus{}y)^{2}}\plus{}\frac{yz}{x(z\plus{}y)^{2}}\plus{}\frac{xz}{y(x\plus{}z)^{2}}\right]\geq\frac{9}{4}\]

From here:

\[ [x\plus{}y\plus{}z] \left[\frac{1}{4z}\plus{}\frac{1}{4x}\plus{}\frac{1}{4y}\right]\geq \frac{9}{4}\]

We multiply both sides of our inequality to 4.

\[ [x\plus{}y\plus{}z] \left[\frac{1}{z}\plus{}\frac{1}{x}\plus{}\frac{1}{y}\right]\geq 9\]
The last inequality is true by Cauchy inequality.

[quote="great math"]I think you did something wrong because to prove that 4>3 and 4>3.5 doesn't mean 3 > 3.5  :wink:[/quote]

My dear brother ! Look at my solution with attention!!! Please...

\[ \frac{x^{2}\plus{}y^{2}}{2(x\plus{}y)^{2}}\equal{}\frac{(x\plus{}y)^{2}\minus{}2xy}{2(x\plus{}y)^{2}}\equal{}\frac{1}{2} \minus{} \frac{xy}{(x\plus{}y)^2}\]

\[ \frac{y^{2}\plus{}z^{2}}{2(y\plus{}z)^{2}}\equal{}\frac{(y\plus{}z)^{2}\minus{}2yz}{2(z\plus{}y)^{2}}\equal{}\frac{1}{2} \minus{} \frac{yz}{(z\plus{}y)^2}\]
\[ \frac{x^{2}\plus{}z^{2}}{2(x\plus{}z)^{2}}\equal{}\frac{(x\plus{}z)^{2}\minus{}2xz}{2(x\plus{}z)^{2}}\equal{}\frac{1}{2} \minus{} \frac{yz}{(z\plus{}y)^2}\]

I used this ineq! Look with attention...  :lol:

[quote="RFZ"]
From here we have.
\[ \frac {1}{(x \plus{} y)^2}\leq \frac {1}{4xy}\Rightarrow
\]

\[ \frac {1}{4z}\geq \frac {xy}{z(x \plus{} y)^2}
\]
Applying our inequality we have:
\[ [x \plus{} y \plus{} z] \left[\frac {1}{4z} \plus{} \frac {1}{4x} \plus{} \frac {1}{y4}\right]\geq [x \plus{} y \plus{} z] \left[\frac {xy}{z(x \plus{} y)^{2}} \plus{} \frac {yz}{x(z \plus{} y)^{2}} \plus{} \frac {xz}{y(x \plus{} z)^{2}}\right]\geq\frac {9}{4}
\]
[/quote]
You are wrong!!!
You proved nothing....
Look at great math post carefully

Can you help me to prove this inequality, my dear bro (Dmitris)?
I don't understand your proof! Please help me!

Again, I will clear my post with a initial question for RFZ
" Supposed that  a > b and c> b, how can you conclude about the difference between a and c (i,e which number is greater, a or c?"

You are right my dear friend! But can you solve this problem? If you will solve please post solution!


Can somebody please recommend good online texts (down loadable or hyperlinked online) in [b][b]Quantum Mechanics[/b][/b] and [b]Advanced Classical Mechanics[/b] ???

In Quantum Mechanics, I am looking for a book that covers fully the mathematical (rigorous) portion of QM and not just the physical implications and stuff on how to explain QM to a 'chemist' (I already have that!) !! I am looking for a rigorous book for a 'mathematician'!! (Hope you understand !!!)

In Advanced Classical Mechanics, I am looking for a book that covers all Hamiltonian operators and Lagrangian Mechanics etc.  - the theoretical part and NOT the applications!

If you are not aware of online texts but know about printed books meeting the above 'demands' , please recommend those too!

Hi Rushil ,
                 I am Syed Imam from Lucknow. I am a class XII pass out .

 These are two books for learning Quantum Mechanics at deeper level :

 1 ) Quantum Mechanics by Cohen Tannoudji  and

 2 ) Advanced Quantum Mechanics by Sakurai.

    I searched for them online but they are not available online it seems.

  Though there are many graduate level QM lecture notes available online by some top professors. They might be of some help.

                                                   Thanks

Thanks! Can you kindly give some specific sites for the lecture notes you said! Also, do you think the 2 books will be available on some University p2p servers?

Is (a) available in India? Who publishes it? Do yo uknow is price? ( (a) refers to the one by Cohen-....)

Syed,

Doesn't Sakurai have 2 books - Advanced QM and Modern QM? Which one of the two is the more 'phenomenon-explaining' type as opposed to the 'mathematically rigorous and explaining everything through math ' type? Also, do either of the books assume substantial previous knowledge of QM(taking from their names) ? Which one starts the study of QM - but form basics - not from just phenomenon and NOT math!?

also, are you aware of Griffith's QM? How good is it?


posted in epic thread (i know this is destined to become one)

[quote="chuong01"]I have a idea that why we dont write open source math ebook (ebook in format .doc). Since PDF format cant be modify or addition and ebook in PDF format cant be recoverd in Word format (mathtype formula will be error). Word Ebook can be writed and added by a lot of person, its mathtype formula also can be copied and pasted on forums easily. Word ebook is alike Open source Dictionary Wikipedia and open source software Firefox. Word ebooks can be writed by unify font in Times New Roman and unify English language.
We shared free ebook so copyright problem may be not important?[/quote]

Can you be more specific?

If JBL wants to know whom to thank for moving this thread from Calculus Computations and Tutorials to his kingdom, that was I. :P

As to the question itself, I also do not understand what exactly is suggested. Of course, we can write and put online whatever we want but it would be nice to have some idea of what it is going to be first. If you want just some random collection of useful mathematical facts, contribute to AoPSWiki. As far as I can tell, nobody prevents you from writing there (as long as you stick to the rules), and, unlike the "big Wiki", technical math. posts are welcome. As to the idea to write math. books in .doc format, I find it terrible (for many reasons).

It's possible to make .doc files read-only, so that can't be it.  The main reason is that math looks ugly in Word.

Books are publications. Having a book that anyone can edit pretty much defeats the point of a publication. There are many, many other ways to share information, as [b]fedja[/b] has pointed out. Each medium has its own advantages and disadvantages.

Word documents are much slower to open and navigate than pdf's, and in many cases, have huge file sizes (though office 2007 has improved the file size issue). This is for one very simple reason -- Word documents are optimized for processing/editing, while PDF's are optimized for publishing final documents.

[quote="fedja"]... from Calculus Computations and Tutorials ...[/quote]

I somehow found this very funny. But I have to give it to the OP for using capital letters, periods, and attempting to write in complete sentences.  :)

I just see that word ebook can be copied and pasted, so we can linking several related files to one. And sometimes I want to supplement knowledge, examples from many ebooks into one ebook. I also want to solve exercises and extended cases in the ebooks.

$ \text{\LaTeX}$ allows much better ways to link files than "copy-and-paste". ;) Anyway let's start from the beginning. What exactly is the book supposed to be about?

The OP doesn't want to write a book; he's offering a possible explanation for why it seems that eBooks are circulated in .pdf format over .doc.

Well, .doc files are insecure because of the tinkering you can do with macros... whereas .pdf is a secure filetype.

Of course, because Adobe is so eager on delivering security patches. The last one took a month, for example.

  darij

But in fact many person want to convert PDF file to Word file by using convert softwares with purpose editting. Certainly, it cant be executed with math ebooks. So is that unfair with math ebooks? And so softwares that convert PDF to Word are issued for what purpose?

[quote="TZF"]If it does not violate the copyright agreement, then you can copy sections of books and compile them into your own document, in whatever format you please. If you really like Microsoft Word, then go ahead and use that.

Just like, I can take several books, photocopy various pages from them, write comments in the margin, staple it together, and give it to whoever I want. I might decide that I don't want to have the resulting compilation in a hard-cover. But does that mean that actual, published books should not be bound in hard-cover? Certainly not.

Again, the point is, there is a difference between [i]documents[/i] and [i]professional publications[/i]. Make whatever documents you want, as long as you don't break the law. But that has nothing to do with how publishing companies should produce electronic versions of their books.

If it really bothers you, maybe you can open your own publishing company and start the ".docx" e-Book movement.  :)[/quote]

I agree with this. Maybe if we go on with the original proposal, then we are bound to infringe many copyright laws.

OP, I suggest you write a 50 page ebook in .doc format, and then read it.  It should then be immediately apparent why this is a terrible idea (and by extension, a terrible thread).

Bonus point if said ebook is written in 16 pt 'comic sans' font.


Can anyone help with this problem?

What are the imaginary cube roots of 1?

There are three cube roots of 1: 1, $-\frac{1}{2}+\frac{\sqrt 3}{2}i$ and $-\frac{1}{2}-\frac{\sqrt 3}{2}i$, where $i$ is an imaginary unit.

Ahh..I see, so the other root is 1 itself?

In general to find the n roots of unity you can use complex numbers.

The n roots of unity are represented by the equation:
\[x^n=1\]

which can be written as:

\[x=\sqrt[n]{\cos 0+i\sin 0}\]

By De Moivre the solutions are:

\[\cos{\frac{360^\circ \cdot k}{n}}+i\sin{\frac{360^\circ \cdot k}{n}}\]

In your case n=3.  So we have:

$\cos 0+i\sin 0=1 \\
\cos 120^\circ+i\sin 120^\circ=-\frac 12 + \frac{\sqrt{3}}{2}i\\
\cos 240^\circ+i\sin 240^\circ=-\frac 12 - \frac{\sqrt{3}}{2}i
$


algebra
polynomial
combinatorics theorems
combinatorics

Hello all,

In the below chromatic polynomial (see attached picture), I am having trouble justifying the second equal. marked with a (**). Why is that true? The first (*) follows apparently from writing a recurrance relation $ \chi_t (G) \equal{} \chi_t (G \minus{} e) \minus{} X(G/e)$, where $ G$ is the original graph (which is the first expression in the pic), and $ e$ is an edge of the graph. As for (**), my guess is that one applies this recurrance relation yet again. Is that true?

Thanks in advance,

Los

Try applying it twice, once in each terms, in both cases to the "outer" edge that "loops around" most of the graph.