calculus
derivative
function
combinatorics unsolved
combinatorics

Calculate

\[T = \sum \frac{1}{n_1! \cdot n_2! \cdot \cdots n_{1994}! \cdot (n_2 + 2 \cdot n_3 + 3 \cdot n_4 + \ldots + 1993 \cdot n_{1994})!}\]

where the sum is taken over all 1994-tuples of the numbers $n_1, n_2, \ldots, n_{1994} \in \mathbb{N} \cup \{0\}$ satisfying $n_1 + 2 \cdot n_2 + 3 \cdot n_3 + \ldots + 1994 \cdot n_{1994} = 1994.$

uhm..
i think i have another solution, using taylor polynomials...
it's not actually olympic, but problem A3 is not fully olympic itself, so i feel like i can post it...
(moderators can remove it, of course :p).

ok, let's begin.
working in general, $\displaystyle \sum \frac{p!}{n_1!n_2!\dots n_{k}!(p-\sum n_i)}$, once we've fixed $\sum in_i = s$, is the coefficient of the term of degree $s$ in $\displaystyle\left(\frac{1-x^{k+1}}{1-x}\right)^p$.
now, our case is "lucky", so we have $k = p = s$, and we can plug it in: what we have to find is $S_k = \displaystyle \sum_{\sum in_i = k} \frac1{n_1!n_2!\dots n_{k}!(k-\sum n_i)}$, so coefficient of the term of degree $k$ in $\displaystyle\frac{1}{k!}\left(\frac{1-x^{k+1}}{1-x}\right)^k$.
let's work with taylor: $\displaystyle S_k =\left(\frac1{k!}\right)^2\frac{d^k}{dx^k}\left(\frac{1-x^{k+1}}{1-x}\right)^k(0)$.
let me break that fraction and compute the binomial, and evaluate this with a double sum: \[ \displaystyle S_k = \left(\frac1{k!}\right)^2\left(\sum_{i,j=1}^{k} \binom{k}{i}\binom{k}{j}\left[D^i\frac1{(1-x)^k}\right]\left[D^{k-i}x^{(k+1)j}\right]\right)(0) \]
what we have to look for is the term of degree $0$, so we don't have to take any sum for $j>0$; moreover, for $j=0$ sum in $i$ is just a single term ($i=k$), since all further derivatives are $0$. we have now: \[ \displaystyle S_k =\left( \frac1{k!}\right)^2\binom{k}{0}^2D^k\frac1{(1-x)^k}(0) = \left(\frac{1}{k!}\right)^2\frac{(2k-1)!}{(k-1)!} = \frac{\displaystyle\binom{2k-1}{k}}{k!} \]

Who can explain ma_go's solution ?
Sorry for my weak  :(
I have an another solution ( Use  generating function method )

[quote=seshadri]consider the general sum
 $\sum_{(n_1,n_2,\ldots,n_k): \sum_i in_i=k} \frac{1}{n_1!n_2!\cdots n_k!(k-(n_1+n_2+\cdots n_k))!}=\frac{1}{k!} \sum_{(n_1,n_2,\ldots,n_k): \sum_i in_i=k} \sum_{r=n_1+n_2+...n_k} \binom{r}{n_1,n_2,\ldots,n_k}\binom{k}{r}$.
now for each $r$ the tuples $(n_1,n_2,\ldots,n_k)$ cover all partitions of $k$ with $r$ parts,so it follows that the term $\binom{r}{n_1,n_2,\ldots,n_k}$ counts the number of compostios of $k$ with $r$ parts. so the sum on the RHS(without the $k!$ in the denom) counts precisely the following set:
   $\cup_{r}{(\pi, A)|\pi}$ is a composition of $k$ with $r$ parts,${A\subset \{1,2,...,k}$ of size $r\}$. but we know that all compositions of $k$ are in $1-1$ correspondence with the subsets of $\{1,2,\ldots,k-1\}$.(each such set would correspond to the 'halts' we make as we keep adding the consecutive terms of the composition before finally landing on $k$). thus this set is in $1-1$ correspondence with the following set:
 $\{(A,B),A\subset \{1,2,\ldots,k-1\},B\subset\{1,2,\ldots,k\},|A|+1=|B|\}$.
but this is easily evaluated: this is precisely
$ \sum_{r=0}^{k} \binom{k-1}{r}\binom{k}{r+1}=\sum_r \binom{k-1}{k-1-r}\binom{k}{r+1}$ and this is easily seen to be $\binom{2k-1}{k}$(coeff of $x^k$ in the expansion of $(1+x)^{k-1}(1+x)^k$ gives the sum).
thus finally the given sum is $\frac{\binom{2k-1}{k}}{k!}$.[/quote]

Sum goes from $r=1$ and $ \sum_{r=0}^{k} \binom{k-1}{r}\binom{k}{r+1}=\sum_r \binom{k-1}{k-1-r}\binom{k}{r+1}$ is wrong.

[quote=panto]Does anyone have a more elaborate solution that does not use taylor polynomials?
(It seems that seshadri's solution has an error)[/quote]

Does anyone know how to do this?


limit
real analysis
real analysis unsolved

1) Let a_n be the general therm of a sequence. Show that, if $ \lim (a_{n\plus{}1} \minus{} a_n)$ exists then $ \lim \frac{a_n}{n} \equal{} \lim (a_{n\plus{}1} \minus{} a_n)$.

2) Find similar results for $ \lim \frac{a_n}{n^k}$, with k = 2, 3, and 4.

3) Propose a general expression for $ \lim \frac{a_n}{n^k}\,\,, k \in N$.

4) Show that, if $ \lim a_n$ exists then $ \lim \frac{a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} \equal{} \lim a_n$.

5) Find similar results for $ \lim \frac{a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n^k}$, with k = 2, 3, and 4.

6) Propose a general expression for $ \lim \frac{a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n^k} \,\,, k \in N$.

More
$ y_n$ increase and $ \lim_{n\to\infty}y_n \equal{} \plus{} \infty$
We has (Stolz lemma):
$ \lim_{n\to\infty}\frac {x_n}{y_n} \equal{} \lim_{n\to\infty}\frac {x_{n \plus{} 1} \minus{} x_n}{y_{n \plus{} 1} \minus{} y_n}$
Apply to your question.

[hide="Answer for 3"] $ \lim \frac {a_n}{n^k} \equal{} \frac {1}{k!} \lim \left[\sum_{i \equal{} 0}^k ( \minus{} 1)^i \dbinom{k}{i} a_{n \plus{} k \minus{} i}\right]$

as long as the limit on the right exists.[/hide]

[hide="Answer for 6"] $ \lim \frac {a_1 \plus{} a_2 \plus{} a_3 \plus{} ... \plus{} a_n}{n^k} \equal{} \frac {1}{k} \lim \frac {a_n}{n^{k \minus{} 1}}$

provided the limit on the right exists.[/hide]


quadratics
number theory unsolved
number theory

find all positive integers  $ x, y$
$ x^2\equal{}12y^3\minus{} 16y \plus{}1$

$ x^2\plus{}3 \equal{} 4(y\minus{}1)(3y^2\plus{}3y\minus{}1)$

$ 3y^2\plus{}3y\minus{}1 \equiv 5 \bmod{6}$ so if it is positive it must have a prime factor which is $ \equiv 5 \bmod{6}$, but this is not possible because such a prime factor cannot divide $ x^2\plus{}3$ by quadratic reciprocity.

so we need $ 3y^2\plus{}3y\minus{}1 < 0$ which means $ y\equal{}0$ or $ y\equal{}\minus{}1$
so the only solution is $ x\equal{}1$, $ y\equal{}0$


function
calculus
derivative
real analysis
real analysis solved

Find all infinitely differentiable functions f from R to R which satisfy            2f(x+1)=f(x)+f(2x) for any real x.

hey
does anyone here have a teeny-weeny idea of what the function is like :D  :D . the problem is really sticky
[i'm getting f(0)=f(1)=f(2),f'(0)=f'(1),f''(0)=f''(1)=0.bu no further progress :(

I think we can show that $f^{(n)} (x) \rightarrow 0$ as $x \rightarrow \inf$, for all $n \geq 0$.  Does this help anybody?

This is actually Ralph Furmaniak's solution:

let g=f''. Then g''(2x)=(2g'(x+1)-g'(x))/4. Now, it is easy to see that |g| attains its maximum on the interval [-2,2](chosen arbitrairly) because

|g(2x)|<=|g(x)|/4 +|g(x+1)|/2, so |g(2x)|<=MAX(|g(x+1)|,|g(x)|). In particular, |g| achieves its maximum. So suppose |g(y)| is the maximum of |g|. Then we have hat |g(y)|<=|g(x)|/4+|g(x+1)|/2<=3|g(y)|/4. So we have |g(y)|=0. hence, the double derivative is uniformly zero. So the only function is the constant function.

Hi, I just found this forum.

Anyone care to explain what makes this functional equation "polish"?

[quote](why is this a sticky?)[/quote]
Good point.

I guess it has just been stickied up there for so long that nobody could bear to take it off. ;)

[quote="jacob"]This is actually Ralph Furmaniak's solution:

 .[/quote]
Could you explain it once more.I still cant convince myself y it should only be a constant function..


Mike has 5 boxes. The first contains 2 squares and 8 triangles, the second 3 squares and 2 triangles, the third 3 squares and 4 triangles, the fourth 4 squares and 3 triangles and the fifth 5 squares and 4 triangles. The side lengths of all the squares and triangles in all the boxes are the same. Mike wishes to make a few polyhedra by glueing together squares and triangles along their edges. Using all the pieces from one box for each polyhedron, for how many boxes can he succeed?

Source: Australian Math Competition, S30, 2000

Box one wouldn't succeed, since eight triangles make up an 8-sided polyhedron.  You can't do anything with two squares.

Box two does succeed, since you put threes squares in a triangular pattern, and pop on the triangles on the end.

Box three succeeds, because you can build a tetrahedron without the bottom.  put the three squares on the open edges of the triangles, and stick a triangle on the bottom to seal it.

Box four doesn't succeed, since you would repeat the process when experimenting with box 3, but a square would be on the bottom.  This leaves edges sticking out.

Box five succeeds.  ARrange the five squares in a cubical pattern, and four triangles go on top.  This seals off everything.

The answer is $\boxed{3}$.

Hope you like my answer!!


Make a search on this chemistry forum. I have already discussed this with Pardesi, if I remember correctly.

[quote="Reon"]can someone explain hyperconjugation? :D[/quote]

Wikipedia is a great resource: http://en.wikipedia.org/wiki/Hyperconjugation

Have you got a textbook that you can consult? Otherwise how did you come across this term. Anyway, you should be more specific about what you don't understand otherwise we can just simply quote the definition of hyperconjugation to you without doing anything too productive. :)


ratio
number theory theorems
number theory

could anyone prove that the sum of reciprocals of fibonacci numbers $ f_i$ may be written as
\[ \sum_{i \equal{} 0}^{\infty}\left(\frac {f_{4i \plus{} 1} \plus{} l_{2i \plus{} 1} \plus{} 1}{f_{4i \plus{} 2}} \plus{} \frac {1 \minus{} \sqrt {5}}{2}\right)
\]
??

where $ l_i$ are lucas numbers http://mathworld.wolfram.com/LucasNumber.html

[hide="Heres a hint"]
Split the sum $ \sum\frac1{f_i}$ into $ \sum\frac1{f_{2^k}}\plus{}\sum\frac1{f_{3(2^k)}}\plus{}\sum\frac1{f_{5(2^k)}}\cdots$.
They are equal because each positive integer has a unique representation as an odd number times a power of 2. Now, the first sum here is somewhat well known; see for instance, http://mathworld.wolfram.com/MillinSeries.html . Now, sum each series here and each will give you a term in the sum you want to show is equal.
[/hide]

here it is:

we write a telescoping sum for $ \sum\frac{1}{f_{2^k}}$ (the case $ \frac{1}{f_{q2^k}}$ where $ q$ odd, $ q>1$ (here q=1) it will be analogically)

\[ \sum_{k\equal{}1}^{n}\frac{1}{f_{2^k}} \equal{} \frac{1}{f_2}\plus{}\sum_{k\equal{}2}^n\left(\frac{f_{ 2^{k\minus{}1} \minus{} 1}}{f_{2^{k\minus{}1}}} \minus{} \frac{f_{2^k \minus{} 1}}{f_{2^k}}\right)\]
after reductions we obtain $ 2\minus{}\frac{f_{2^n \minus{} 1}}{f_{2^n}}\to 2\minus{}\frac{1}{\phi}$ (where $ \phi$-golden ratio)

$ 2\minus{}\frac{1}{\phi}$ = first term ($ i\equal{}0$) of the series from first post.

analogically for $ q\equal{}3,5,\dots$ (the same idea) what gives next terms :)


Be
$x^4-(3m+5)x^2+(m+1)^2=0$

Find m for this equation have four different roots

[hide="I think..."]
Set $x^2=y$ and rewrite the ecuation $y^2-(3m+5)y+(m+1)^2=0$. To have 4 different $x$, we need to have 2 different $y$. So $\Delta\geq 0\Longleftrightarrow 5m^2+22m+21\geq 0 $ hence $m\in(-\infty,-3]\cup \left[-\frac{7}{5},\infty\right)$.[/hide]

There is one problem that occurs when $y = 0$, or when $m=-1$ ;)

2 roots must be positive!!! Hence, S,P must be positive :lol:


function
algebra
functional equation
algebra proposed

let $ \alpha, \beta \ne 0$ such that $ \alpha \plus{} \beta \equal{} 1$ . Find all continuous function $ f;R \to R$ such that $ f( \alpha \plus{} \beta) \equal{} \alpha f(x) \plus{} \beta f(y), \forall x,y, \in R$

 :idea:

Do you mean $ f(x\plus{}y)\equal{}\alpha f(x)\plus{} \beta f(y)$?

Because then interchanging $ x$ and $ y$, then adding yields $ 2f(x\plus{}y)\equal{}f(x)\plus{}f(y)$, which after plugging in $ x\equal{}y$ yields $ f(x)\equal{}f(y)$, so the only function is the constant function.

[quote="Rofler"]Do you mean $ f(x \plus{} y) \equal{} \alpha f(x) \plus{} \beta f(y)$?

Because then interchanging $ x$ and $ y$, then adding yields $ 2f(x \plus{} y) \equal{} f(x) \plus{} f(y)$, which after plugging in $ x \equal{} y$ yields $ f(x) \equal{} f(y)$, so the only function is the constant function.[/quote]

Plugging in $ x \equal{} y$ yields $ f(2x) \equal{} f(x)$, not $ f(x) \equal{} f(y)$.

But setting $ y\equal{}0$ implies $ f(x)\equal{}f(0)$, hence your good conclusion.

And no continuity is required.

Sorry, i made a big mistake. the problem is

let $ \alpha, \beta \ne 0$. Find all continuous function $ f;R \to R$ such that $ f( \alpha x\plus{} \beta y) \equal{} \alpha f(x) \plus{} \beta f(y), \forall x,y, \in R$

Now you forgot $ \alpha \plus{} \beta \equal{} 1$. Assume it.
Define $ g(x) \equal{} f(x) \minus{} f(0)$, so $ g(0) \equal{} 0$. Then $ g$ satisfies the same functional equation as $ f$, and is as well continuous.
For $ y\equal{}0$ we therefore have $ g(\alpha x) \equal{} \alpha g(x)$, and similarly, for $ x\equal{}0$ we have $ g(\beta y) \equal{} \beta g(y)$.
So we have $ g(\alpha x \plus{} \beta y) \equal{} g(\alpha x) \plus{} g(\beta y)$. Denote $ \alpha x \equal{} X$ and $ \beta y \equal{} Y$. Finally we get $ g(X \plus{} Y) \equal{} g(X) \plus{} g(Y)$, the Cauchy equation (for continuous function), with solution $ g(X) \equal{} \lambda X$. This yields $ f(x) \equal{} \lambda x \plus{} f(0)$, i.e. $ f$ is linear.

i didi'n forget $ \alpha\plus{} \beta \equal{}1$ , so to solve my problem, we have to check : if:$ \alpha \plus{}\beta\equal{}1$ follow $ f(x) \equal{}ax\plus{}b$ and if $ \alpha\plus{} \beta \ne 1$ then $ f(x)\equal{}kx$. if quite easy. Sorry all


Do you think you'll find the best solution... ?     http://mathzone.net.tc   :D

I liked that challenge, it's a lot of fun :P  Did this happen last year too?  Is there somewhere I can see the results?  :?

-Ixeus

You have to wait for the 15th of June :D


Could people please post their ideas about different ways to solve limits, like elimination, plugging in the values, and etc. I'd like to make sure I didn't miss any important ones. Thanks.

Nobody is willing to give me some tips on this? Oh well.

You're asking a very broad question; it's like asking how many different ways are there to solve geometry problems.  The number of different ways of evaluating limits is almost the number of limit problems.

My suggestion is to pick up a good calculus textbook, and do lots of examples.  If you have any more specific questions, just post them here.

Ok, thanks. I've just been introduced to calculus, so I don't really know much about it yet.

I teach my students that if you have a function that's a combination of "normal" elementary functions, then the first thing you do is try to plug the value in. If the function is defined there, then that is your limit (the reason being that "normal" elementary functions are continuous in their domain).

(Note that this does not apply to piecewise defined functions).

If, however, when you try to plug the limiting input value into your function you find that the function is undefined there (because it would require you to do something illegal like, say, divide by zero), then you decide whether the behavior of the function there is infinite, oscillatory, jumps from the left or right, or an indeterminate form.

(If you don't know what I mean by a limit which is infinite, oscillatory, or jumps from the left and the right, then learn about the three commons reasons why a limit fails to exist).

If it is an indeterminate form (make sure you understand what an indeterminate form is), see if you can reduce the function algebraically (either by factoring or multiplying by a conjugate, etc) to a function that is the same at every point except for the point at which you are trying to take the limit. Then you repeat the process. See if you can plug the value in... If not, see if it is infinite... If not, try and reduce some more.. etc, etc.

The most common tricks for reducing are the same ones you learned in Algebra II for simplifying algebraic expressions. Factoring and canceling, multiplying the top and the bottom of an expression by a conjugate, expanding and combining like terms, etc, etc...

If you are in the first few weeks of a Calculus class, these methods will work on all the limits that you encounter. But later on, you will see that even these methods don't work on *every* limit. At that point you will learn some other techniques for evaluating limits (L'Hopital's Rule, Taylor Series expansions, etc).

But for now, the best way to learn how to do limits is to DO lots of them!! The more you do, the better acquainted you will become with the techniques for solving them.


Taking a leaf out of the nsep thread. Which are all the colleges you've applied for?

Me IIT-JEE (obvious), BITS and writing AIEEE. Almost applied for NTU (singapore) but I don't want to do my UG abroad, maybe PG there. ;)

Not yet applied for anything else. I might became eligible for some more if I clear the INOI.

add ISI,CMI,NISER,CBS-DAE,IISER to that -NTU

I am applying for
ISI bangalore
CMI (direct admission)
IISER (direct admission)
NISER,Bhubaneswar and CBS, Mumbai (NEST)
IITJEE

My first option is
ISI bangalore for B. Mat (I am pretty sure I will get through)
2nd option is CMI...
3rd option is IISER (but don't need as 2nd option is available)

I did not apply for US universities for the same reasons as Shreyas..
I gonna do my UG in India...

Me same as Shreyas except NTU , INOI and the PG parts
Btw even as i applied only for 3 ( IIT , BITS , AIEEE ) will be for CMI  :D  my grandparents asked me
" Why are you applying fer sooooo many universities "  :o
 :D

actually we are applying for some basic science courses so

[quote="madness"]no i was just voicing my grandparents opinion
Doubt if my mother would let me even apply for a college like ISI even though i heard its good  :D[/quote]

I think it is a waste of (800+400) rupees that you actually submitted bits and aieee. after all you are assured of a top rank in JEE. Unless you joined under the girls category in which case you would have wasted lesser  :rotfl:

Dei vayavetchikittu chuma iru da  :D   :P   :wink:

[quote="pardesi"]vayu summa dan eriku kay than pesigirudhu[/quote]

Are you a tamilian? How the hell do you know so much tamil??

[quote="pardesi"]as if u don't know  :)[/quote]

Err..Seriously no.. how thee??

i didn't mean to insult someone :huh:
nor was faraday wrong :D
our books were and those people (and i beleive they have sufficient experience) hadn't thought of that before...
what faradya said was interpreted and there was nothing wrong in what he said :P
and when i said nothing is final meant there is always a scope for change and improvement and doesn't matter ur improving upon/discarding  whose theory...if u r right then u r bound to be right....
Einstein(he inevitably comes up in all our aruments  :P ) discarded the established wave theory then...Hugyens the corpuscular theory.... taht doesn' mean we stopped respecting Newton :noo:
it siple gave a new perspective of looking at things
and about Faraday's sucessors i don't know if they were dumb or not but they didn't write down what they saw...since it wasn't there  :rotfl:
well coming back to the subject matter...i said i would still prefer a book(a good one) with a partner for healthy discussion than sit in someone's lecture classes day and night who keeps on explaining things from his perspective(may be right also) but which i am not keen on looking at from :)

[quote="SOUMYASHANT NAYAK"]
I am talking abt productive discussions with the prof...
I believe to properly understand a subject you should have a philosophical debate with an experienced person or any such person who is capable enough...
[/quote]
i said that in my previous post...for that u neddn't sit in someone's lecture day and night :lol:
[quote="SOUMYASHANT NAYAK"]
I don't see what difference maths and physics will have in context of pouring in of new facts...
i think you are making unfair comparisons...
"that's the best part of physics"-- What is that supposed to mean??
None of your arguments show me that the way maths is done is different from the way physics is done...
[/quote]
that means if there is any scope of change a magnificient chang e and remember the stress is upon the word change then it's onl;y possible in physics...yes maths can lead u to great discoveries...old proofs inspiring new ones...but surely they are true u can't change the perspective of looking at the,m
and this is the best part of physics
[quote="SOUMYASHANT NAYAK"]
Although the physics profs hadn't seen that before because they hadn't had a teacher like Feynman...( the supposed fallacy I don't think was a new one ...it was addressed by feynman in his lecture in 1960s...)[/quote]
No it wasn't addressed u interpreted our question wrong nor there is a proof of the fact in the book...
ask aditya ,Sanjay(he had even come up with Feynamann) and what we suppose farday's law is isn't true
[b]He(Prof. Srivatsava) wanted be to send my proof(pretty simple though) and the suppose intricacy  to American Journal Of Physics[/b]  :D
[quote="SOUMYASHANT NAYAK"]You yourself seem to accept my standpoint...[/quote]
about what rotating at $ 2*10^{5} \frac {Km}{s}$
[quote="SOUMYASHANT NAYAK"]There exactly where I want to point out the importance of a good teacher...
Suppose the doubt about faraday's law wouldn't have occurred to you..??
Similarly, there are lot of intricate questions that might not have occurred to you till now and a good guide might be able to open your eyes to them...

And about attending classes, I think it is more important as a measure of discipline than anything else ...

As i said, let us not be idealists..
be practical...[/quote]
i am not being in a hypothetical atmosphere and speaking...i would rather like to think on it before doing anything ...i said it in my previous post :wink:

My take: You need a good set of books, an inspiring peer group and focus/determination more than all so that you will never be diverted from your path. But a guide is a very very important. A person who knows much more than you, with more experience is needed for you to interact with. The human element is a must.

But I do not believe in that much discipline as soumyashant claims because if you are focussed and know WHAT EXACTLY you are doing, your discipline is all that matters to you. You don't need someone to enforce it on you. I believe IIT isn't so strict. not sure.

I think pardesi, your biggest void is that you haven't yet met a physics teacher who can convince you. I don't have that problem, because I have met someone like that. My IIT prof! (:-)

P.S: I think both CMI and ISI have got the same intl recognition no?

[quote="Cyber-Shreyas"]
But I do not believe in that much discipline as soumyashant claims because if you are focussed and know WHAT EXACTLY you are doing, your discipline is all that matters to you. You don't need someone to enforce it on you. I believe IIT isn't so strict. not sure.

[/quote]

Ya i know that a guy who doesn't go to school for half the year, doesn't submit homework at all, doesn't study what he doesn't like,bunks classes when he can, is always late to school doesn't have a right to talk abt discipline...(that's of course me... :rotfl: )

But I am not talking about talking attending classes day and night but attending classes for 3 hours in a day and being in frequent touch with your teacher...

@pardesi
When you yourself become a professor of physics, you will realise what i meant to say...
you want your students to learn from your research experiences...
(not necessarily from academics point of view...)

Faraday gave a law ...and our wrong interpretation doesn't make it wrong...

Can we post the fallacy in a thread...???

no i have never sadi proffesors sitting all around are a crap...i will be soon one :)
but what i said is u should rather learn urself and take other experience...than depending on them to give u their experience...for ex.
a leaf from ur thread suppose u r taking a class on Newton's laws obviusly u  will be asking ur students what does it define mass or force or neither or both...there may be some who would have thought of this  and it may so happen that none of them would have thought about it and won't appreciate the fact and may dismiss it as a crap because they never thought about it and don't feel it's importance..and will feel as if u r wasting their time...i had seen this happening with Nishi Sir he was exceptional but students(we) didn't deserve him!! :)
ye s u need a person who  inspires u because without it there is nothing acheived

Interesting to see how a questions sparked up a nice and big discussion. :D

Personally I'd say that I definitely noticed a difference in the time I was learning Physics by myself and when our current sir started teaching us. I like his insights and chance to look at things from our sir's perspective. Besides I also like the fact that I have someone definite I can rely on when I have some doubts or new ideas.

The same might not be true at other instances. I have never had an impressive Comp. Sci. teacher (besides a 2 week stint at Infosys in 9th grade) but since I enjoy programming, I took it upon myself to learn from various sources. I like to think that I have been pretty successful all by myself.

Clearly a lot depends on our different qualities and the teachers we come across.  :maybe:

On another line, right now I have seriously no idea about what I'll do after I complete my UG in India (if I do decide to go to CMI or ISI). Again if I get to do my PG in the US, research seems good. Incase I'm stuck here, I might prefer having a B Comp. Sci. and keep my options open.

Since I like Comp. Sci. anyway, CMI is an extremely good options (albeit I'm a bit worried about 'Jack of all trades ...' :| ) Maybe I'll ask some more people who are at CMI and try to build up a strong focus and concentrate on a few things.  :)

@Soumyashant: Thanks for linking me to Vipul's blog. There's loads of information there.

@Pardesi: Faraday's Law Debate ... that sounds interesting! What was it all about?

go through it...it's actually a definite mistake that exista in inerpreting it and exists in all our books
and
P.S.: i have more posts than Rushil now,Indian with highest number of posts :first:

first of all,
congrats to pardesi... :P  :P  :P

@Vihang
I would like to keep my options open for going for research in theoretical physics after UG... So I am choosing ISI bangalore..

Also, after my advice, our astro profs have taken keen interest in doing a serious NIUS for us where an annual continuous correspondence followed by exams for B.Sc will be done... (I talked with them this december)
Bachelor's degrees in astrophysics or physics may be awarded..
(just like mathematics nurture programme where you get B. Mat without attending full-fledged bachelor's degree course in mathematics...)

Although recently I have developed a lot and lot of interest in computer programming (earlier i didn't have the remotest idea of what it is..)
, but that will always remain a very good hobby...

thanks soumya :D
i would like to do  a NIUS(if possible with the KVPY) on Maths or Organic Synthesis

When I see people who are good at organic chemistry,
I wonder and think there is sth like 'talent' in chemistry just as in maths, physics...

However hard I try I cannot understand anything in organic chemistry...
(I will have an I.Q. of -100 in organic in apparent contradiction of binet's test...)
how people write mechanisms and all just by looking at the reactants and reagent...
It always was and I think, always will be a mystery beyond my comprehension...

And you reminded me of that 'talent' when i see the confidence with which you say you want to do a research project on Organic Synthesis...

and i can bet a Morrison Boyd i the best book i have ever read :omighty:
and not to forget Carcul & Nishikant  Sir

Congrats pardesi, can we expect a treat next time you are here.

There's only something called talent as Einstein says in "on education" (do you guys have this chapter?). Depending on the way you channelize it I guess it shapes yourself and your own flair.  :D

treat nichyamagha unda :) ....but u never said where u live in thiruvanmyur

[quote="madness"]A question easily answered
His address is :

No. 8 Sincere Street Sincere Sivagami road Thiruvanmiyur Chennai - 600 040  :D

And by the way he calls himself S.Sincere Shreyas but we call him S.Sincere Shreya or aayas  :rotfl:
He is clearly the most unsincere person in the class and he has a punch dialogue :

" Sincerity Blood la odanum "  :rotfl:[/quote]

I am S.(Sincere) Shreyas, and there is no doubt about it. Atleast I'm matured (with a moustache and beard) and not a baby like ahem, ahem.... Vanna da.

~ Yours Sincerely
S.(Sincere)Shreyas


calculus integration trigonometry derivative function logarithms LaTeX

Edited:

grr..yes, i made an error.

Compute the integral from negative infinity to positive infinite of (p+x^2)^-3 dx where p is a positive real number.

Sorry, :(

Isn't the integral infinite? Perhaps you made an error in transcribing the problem?

perhaps it's

[tex]\displaystyle \int_{-\infty}^\infty \frac{1}{p+x^2} dx[/tex]?

My guess is that it really was supposed to be

[tex]

\displaystyle{\int_{-\infty}^{\infty} \frac{dx}{(p + x^{2})^{3}}}

[/tex]

There is a classic trick for solving integrals with powers of [tex]p + x^{2}[/tex].  [hide]Consider the trig substitution x = sqrt(p)tan(y).  The integral can then be rewritten as the integral of an even power of cos(y) that is solvable by standard trig simplification.[/hide]

Mildorf:

That's true, but there's another way of doing it. One which I quite don't understand:

Differentiate the Integral from negative infinity to positive infinity of (p + x^2)^-1 dx with respect to p twice, and that's your answer.

Why???

I think it's an indirect way to evaluate the integral.  Suppose we have

[tex]
\displaystyle{F(p) = \int_{-\infty}^{\infty} \frac{dx}{p + x^{2}}}
[/tex]

This is a reasonably well known arc tangent integral, but, differentiating F with respect to p, we have

[tex]
\displaystyle{\frac{d F(p)}{dp} = \frac{d}{dp}\left(\int_{-\infty}^{\infty} \frac{dx}{p + x^{2}}\right) = \int_{-\infty}^{\infty} \frac{d}{dp} \left(\frac{dx}{p + x^{2}}\right) = \int_{-\infty}^{\infty} \frac{dx}{-(p + x^{2})^{2}}}
[/tex]

and again,

[tex]
\displaystyle{\frac{d^{2} F(p)}{dp^{2}} = \frac{d}{dp}\left( \int_{-\infty}^{\infty} \frac{dx}{-(p + x^{2})^{2}}} \right)} = \int_{-\infty}^{\infty} \left(\frac{2dx}{(p + x^{2})^{3}}}\right)}
[/tex]

This last expression is twice the desired integral - all that is left is to determine a closed form expression for [tex]$F(p)$[/tex], differentiate this twice with respect to [tex]$p$[/tex], and finally divide by 2.

[quote="Mildorf"]There is a classic trick for solving integrals with powers of [tex]p + x^{2}[/tex][/quote]

Actually, it's not really a classical trick... it's called trig substitution... lol... and that's the method I used to solve it before I read any of the below posts... I see no reasons for this method being unacceptable...

[tex]\mbox {\Large{\displaystyle{\frac{3\pi}{8p^{\frac{5}{2}}}}}}[/tex]

I wouldn't call it a "trick", but it is a pretty standard technique. You sometimes have to be careful about interchanging derivation and integration though, they don't always commute. In this case, however, they do.

but notice the integration is in terms of x and the partial differentiation is terms of p... partial integration and partial differentiation of different variables are always interchangable.

I'm not entirely convinced that is true ... couldn't the function (of p) on the inside be discontinuous but become continuous as you integrate over x?

what do you mean on the inside?

and

If we suppose [tex]\displaystyle{F(p) = \int_{-\infty}^{\infty} \frac{dx}{p+x^2}}[/tex]... we STILL must use the same trig substitution method to solve this integral as we would have before, it just becomes a bit simpler... so what is the benefit of this method?

Sorry -- say we have a [tex]\displaystyle{G(p) = \frac{d}{d p}\int_{a}^{b} F(x,p)dx}[/tex].  You claimed that this necessarily implied that [tex]\displaystyle{G(p) = \int_{a}^{b} \frac{\delta F(x,p)}{\delta p}dx}[/tex].  However, that assumes that integration over x cannot change a non-differentiable function of p into a differentiable function of p, and I'm not convinced that that is true.  Can you prove it, or send me to Mathworld to read about it?

JBL, honestly, even though I think it's true just from experience... I cannot prove it and I think you have a legitimate argument... so I'm sticky with the opinion that why on earth would someone want to solve an integral like this when good old trig substitution works fine... it reduces to [tex]\displaystyle{\int \cos{4\theta}\; d\theta}[/tex] which isn't hard to integrate at all.

I don't really have anything to say about the particular question here -- I was more interested in the general rules.  (Theoretical math classes do that to ya ;))

[quote="blahblahblah"]

You refer to these problems as Leibniz's rule problems (which they are) but these are specifically integration under the integral sign problems, correct?[/quote]

Well no, some of them are integration under the integral sign problems but not all of them. These are all just a special case of Leibniz's Rule. I'll post more about this later. I'm too tired to LaTex an explaination right now.

its kind of hard to give spoiler hints (because i need LaTex)

but here goes:

hints in order of how they appear:

1:  consider the function [tex]\displaystyle G(a) = \int_{0}^{\infty}\frac{e^{-x}sin(ax)}{x}dx[/tex], find G'(a), integrate, then use the fact that G(0) = 0 to find G(a)

2:  let [tex]\displaystyle H(a) = \int_{0}^{\infty}\frac{arctan(ax)-arctan(x)}{x}dx[/tex], do the same as above

3:  let [tex]\displaystyle I(a) = \int_{0}^{\infty}\frac{sin^2(ax)}{x^2}dx[/tex]

the rest is up to you, have fun.

Yeah some trick, everyone in BC calc learns trig substitution. Arc Tan is probably the best known. This definitely shouldn't be in an advanced section. I consider this to be HARD questions. Meaning that I would have no shot of solving them. Calc isn't hard, it is the same as posting any other type of math, just some know it, some don't.

OK also, I read through all the posts, and I see now explanation of this "integration by cheating" method. Where is it?

And lol, I had done the arctan question as well, I just made a math error which caused it not to work... we have:

[tex] \displaystyle{F(t)=\int_0^\infty \frac{\arctan(tx)-\arctan(x)}{x}\; dx}
\\ \displaystyle{F'(t)=\int_0^\infty \frac{dt}{t^2x^2+1}\; dx} \\ \displaystyle{F'(t)= \frac{\pi}{2t}\; dt} \\ \displaystyle{F(t)= \int \frac{\pi}{2t}\; dt = \frac{\pi}{2}\ln|t|} \\\displaystyle{F(\pi)= \frac{\pi}{2}\ln(\pi)}[/tex]

The last one took me a bit longer, but I figured out how to do it I think:

[tex]\displaystyle{F(t)=\int_0^\infty \frac{\sin^2(tx)}{x^2}\; dx}\\ \displaystyle{F'(t)=2\int_0^\infty\frac{\sin(2tx)}{x}\; dx\; dt}[/tex]

Now, we'll use the substitutions [tex]\displaystyle{u=2tx, x=\frac{u}{2t},}[/tex] and [tex]\displaystyle{dx=\frac{du}{2t}}[/tex]... We have:

[tex]\displaystyle{F'(t)=2\int_0^\infty\frac{\sin(u)}{u}\; du\; dt}\\ \displaystyle{F'(t)=\pi\; dt}\\ \displaystyle{F(t)=\pi t}\\ \displaystyle{F(1)=\pi}[/tex]

These problems are pretty damn sweet :), and fun to do!

[quote="Spoon"]AHH! I looked at the problem again in my english class at school today and I realized that I had infact done it correctly last night, I merely TeXed it incorrectly! lol... what I meant was...

[tex]\displaystyle{F(t)=\int_0^\infty \frac{\sin(tx)}{x e^x} = \arctan(t), \arctan(1)=\frac{\pi}{4}}[/tex]

I resubmit my "yay" :)[/quote]

Yes, this is correct. You could also introduce the t-variable like you did the first time. But that integral comes out to [tex] \displaystyle {G(t)=\int_0^\infty \frac{\sin(x)}{x e^{tx}} = \arctan(1/t)}[/tex]. So you get the same answer when it's evaluated at t = 1. I think you just made a small calculation mistake or typo the first time.

Your other ones are all right. Nicely done.

Here's the general form of Leibnitz's Rule from MathWorld - http://mathworld.wolfram.com/LeibnizIntegralRule.html

As you can see it's just a combination of the rules for interchange of differentiation and integration, and the Fundamental Theorem of Calculus.

You should also check out Integration Under the Integral Sign. It's a similar technique for computing definite integrals - http://mathworld.wolfram.com/IntegrationUndertheIntegralSign.html

Again, they leave out the conditions under which these are true. My teacher in college was adamant about us knowing the conditions under which things were true.

[quote="gauss202"]They don't always commute. I'll post an example when I think of one. This is called Leibnitz's Rule by the way. There's almost always a question on the GRE Math Subject Test about it.

A nice problem to try: Evaluate [tex] \displaystyle {\int^1_0 \frac {x-1}{\ln x} dx}[/tex][/quote]
hi, can anyone show how to go about this question?  thanks

[quote="gauss202"]To get to my integral, try integrating [tex] \displaystyle {\int^1_0 \frac {x^t - 1}{\ln x} dx}[/tex] by using this differentiation technique, and letting t = 1.[/quote]

[tex]
\displaystyle F(t) = \int\limits_0^1 {\frac{{\left( {\frac{1}{{\ln x}}x^t  - t} \right)}}{{\ln x}}} dx \\
\displaystyle \frac{{dF(t)}}{{dt}} = \int\limits_0^1 {\frac{{x^t  - 1}}{{\ln x}}dx}  \\

[/tex]
I dont know what to do

Here we go....

We wish to solve:

[tex] \displaystyle {\int^1_0 \frac {x - 1}{\ln x} dx}[/tex]

As guass has pointed out, we will write this as a function of [tex]t[/tex].

[tex]\begin{eqnarray*}
f(t) &=& {\int^1_0 \frac {x^t - 1}{\ln x} dx} \\
\frac{\partial}{\partial t}f(t) &=& {\int^1_0 \frac{\partial}{\partial t}\left(\frac {x^t - 1}{\ln x}\right) dx} \\
&=& {\int^1_0 \frac {x^t\ln x - 0}{\ln x} dx} \\
&=& {\int^1_0 x^t \; dx} \\
&=& \frac{x^{t+1}}{t+1}\Bigg{|}_0^1\\
&=& \frac{1}{t+1}\\
f(t) &=& \int \frac{1}{t+1}\; dt\\
&=& \ln |t+1|\\
f(1) &=& \ln 2

\end{eqnarray*}[/tex]

*AHEM*

Might it be important that I note f(t) = ln|t+1| + C, and as it is clear to see that f(0) = 0, C=0. We still must consider the constant of integration when solving problems in this matter.

You should be a bit more careful at the end -- is it necessarily true that if

[tex]\displaystyle f'(t)=\frac{1}{t+1}[/tex]

then

[tex]\displaystyle f(t)=\log|t+1|?[/tex]

simon, you are absolutely right! It doesn't imply log, it's ln :-D.


Does anyone know how to attach a picture in a post? :oops:  :blush:

Scroll down past the "Submit" button to "Attachment Posting Control Panel".

Please direct any further questions to the "Questions, Suggestions, and Announcements" forum.


Find the least $n$ such that whenever the elements of set {1,2,...,n} are coloured red or blue, there always exist $x,y,z,w$ (not necessarily distinct) of the same colour such that $x+y+z=w$


Let m and n be positive integers, and let

S(m,n) = ${n \choose 0} * (2^{n} - 1)^{m} - {n \choose 1} * (2^{n-1} - 1)^{m} + {n \choose 2} * (2^{n-2} - 1 )^{m} - ... + (-1)^{n-1} {n \choose n-1}$

Prove that S(m,n) = S(n,m).

I also interested in the explicit formula for S(m,n) if possible.


probability
function
real analysis
calculus
integration
probability and stats

Let $ X_1, X_2, \ldots, X_n$ be $ n$ discrete uniform random variables in $ [0, A\minus{}1]$ where $ A$ is a large integer and $ n<< A$ (although this does not make a difference). Therefore $ \Pr[X_i\equal{}a]\equal{}1/A$ if $ 0\leq a < A$, and $ 0$ otherwise.

I need an approximation of the probability mass function of $ (X_1\plus{}X_2\plus{}... \plus{}X_n)$

In other words, I want a nice little formula to plot the various PMFs for different $ n$s using a computer.
I know that we need to approximate this using a continuous normal distribution but I could not find any good resources. Can anyone point me to the right direction for the closed form of the approximate normal distribution?

I tried the convolution formula for two variables but I get stuck at that point. Most books stop at $ n\equal{}2$. If someone can explain how to generalize this for any $ n$, that would be great.

[quote="alekk"]if at the end you just want to approximate it by a Gaussian variable, why don't you just put the right mean/variance, which is straightforward.[/quote]

yes I want to approximate it by a Gaussian variable.
However, I am stuck at the definite integral which involves the term $ e^{\minus{}x^2}$

Essentially, I want to evaluate:

\[ \int_{0}^{A}{p(x)dx}\plus{}\int_{A}^{(A\plus{}B/2)}{p(x)dx}\minus{}\int_{A}^{(A\plus{}B)/2}{p(x\minus{}A)dx},\]

where:

\[ p(x)\equal{}\frac{e^{\minus{}\frac{(x\minus{}nA/2)^2}{n(A^2\plus{}2A)/6}}}{\sqrt{\pi n(A^2\plus{}2A)/6)}}\]


algebra
polynomial
number theory unsolved
number theory

Let $ p[X,Y]$ be a polynom with algebraic coefficients, $ p$ is irreducible and both $ X$ and $ Y$ occur in $ p$. Let $ f(z)\equal{}p(z,e^z)$. If $ p(z,e^z)\equal{}0$ and $ z$ is nonzero, then is it true that $ z$ is transcendental over $ \mathbb{Q}$?

Assume $ z$ algebraic and write $ p(x,y)$ as $ a_m(x) y^m \plus{} ... \plus{} a_1(x) \plus{} a_0(x)$. Then $ q(y) \equal{} p(z,y)$ is a polynomial in $ y$ with algebraic coefficients. As $ e^z$ is transzendental, and $ q(e^z) \equal{} 0$, we get that $ q \equal{} 0$. This gives that $ a_k(z) \equal{} 0$ for all $ k$, so $ a_k(x) \equal{} (x \minus{} z) b_k(x)$ for some polynomial $ b$ with algebraic coefficients. But this gives $ p(x,y) \equal{} (x \minus{} z) \cdot r(x,y)$ (with $ r(x,y) \equal{} b_m(x) y^m \plus{} ... \plus{} b_1(x)y \plus{} b_0(x)$), which contradicts irreducibility [I assume you meant in an algebraic closure¿] of $ p$ (as $ r$ cannot be constant: it must contain the variable $ y$ by assumption).

Thanks, ZetaX. In one paper I found another short proof of this fact, but I don't understand it: it says(in conditions of the problem) "If $ z$ is not transcendental over $ \mathbb Q$ then both $ z$, $ e^z$ are algebraic over $ \mathbb Q$ contradicting the Lindemann-Weierstrass Theorem". I don't understand why then $ e^z$ is algebraic over $ \mathbb Q$ ?

For the same reason that ZetaX gave.  If $ z$ is algebraic then $ p(z, *)$ is a polynomial with algebraic coefficients, so $ e^z$ is algebraic - contradiction.

Thanks for answer, t0rajir0u. But will $ e^z$ be algebraic if it is the root of polynom with algebraic coefficients? Should it be the root of the polynom with rational coefficients?

If $ M|L$ and $ L|K$ ar algebraic extensions (of fields), then so is $ M|K$.

The set of algebraic numbers is not just the set of roots of polynomials with rational coefficients - it is, in fact, the algebraic closure of $ \mathbb{Q}$, so any polynomial with algebraic coefficients has algebraic roots.  This is a consequence of the general statement ZetaX gave.


geometry
perimeter
geometric transformation
reflection
trigonometry
geometry proposed

I have been trying hard at this problem for a while, but haven't gotten any leads yet. Was wondering if someone could give a shot at it.

Given an acute triangle ABC, with altitude from A intersecting BC at D, altitude from B intersecting CA at E and altitude from C intersecting AB at F. Prove that the perimeter of triangle DEF < twice any altitude.

Consider $ D'1$ and $ D'2$ the reflected points of $ D$ with respect to line $ AC$ and $ AB$ respectively. Is easy to prove that $ D'1$, $ D'2$, $ E$ and $ F$ are colinear. And $ ED'1\equal{}ED$, $ FD'2\equal{}FD$, so $ D'1D'2\equal{}$ perimter of $ DEF$. Let $ X$ and $ Y$ the intersection points of $ DD'1$ and $ AC$ and $ DD'2$ and AB, then $ XY\equal{}$half perimeter $ DEF$, and $ XY$ $ <\equal{}$ $ AD$ because $ AXDY$ is cyclic and $ AD$ is a diameter, so it is the gratest chord of its circle. This implies the desired result

Hey,
Thanks a lot. Helped me in thinking about other questions.
The reason I was not able to do this question was because we were doing
affine geometry and reflection was not even mentioned so it never crossed
my mind. But thanks it really helps.

trigonometry does the work too  :D we'll prove that the perimeter is less than or equal to $ h_a\equal{}2R\sin B\sin C$. recall that the sides of the orthic triangle are $ a\cos A, b\cos B, c\cos C$, then, the perimeter equals $ \sum 2R\sin A\cos A\equal{}R\sum \sin 2A\equal{}4R\sin A\sin B\sin C$. it follows obviously that $ 4R\sin A\sin B\sin C\leq 2\cdot 2R\sin B\sin C$.

I followed the above geometrical proof, yet I was not able to figure out how you can prove that D'1, D'2, E and F are collinear. It does not really mean that the reflection point will meet EF at D'1. Kindly enlighten me on that.
And campos thanks for the trig proof.

To prove it, it is enough to see that $ <DED'1\plus{}<DEF\equal{}180$, the other points, $ D'2$, $ F$, will be analogous. $ <DEF\equal{}2(90\minus{}<B)\equal{}<FEH\plus{}<HED\equal{}<FAD\plus{}<DCF$ and $ <DED'1\equal{}2<DEC\equal{}2<DHC\equal{}2<B$ so, adding them up you get:
$ <DEF\plus{}<DED'1\equal{}180$


function
real analysis
real analysis unsolved

if $\sum_{n=1}^{\infty}a_{n}$ diverge; prove that $\sum_{n=1}^{\infty}na_{n}$diverge

[hide]Let's suppose that $\sum na_{n}$ converges.
Consider function $f(x)=\sum_{n=1}^{\infty}a_{n}x^{n}$. The radius of convergence of $f'$ and $f$ $\ge1$. $f'(1)$ is convergent, hence $f'(x)$ converges uniformly on $[0;1]$. But this implies that f(x) converges uniformly on [0;1](since f(0) conv.)Particularly f(1) converges. Contradiction.
P.S.Nice problem.[/hide]


Does anyone(I know fredbel does..) know latin? If so, can we chat in this thread with latin to enhance our knowledge? Just so you know I have had two years latin but am just starting HS level.

STARTS OUT:

Quis ad hic addent?

I hope my grammar is correct...

Rewording..

Quis aut quid ad hic addent?

I've learnt latin for two years too. But, I've forgotten a lot (well, not that much).

I'm in latin 4 this year.

Ad hic addidi.

Okay..this one must get off this topic or we'll all be contemplating who has posted.
Funcius in pavimento nummum condebant.

My grammar isn't the best. :lol:

[quote="funcia"]Does anyone(I know fredbel does..) know latin? If so, can we chat in this thread with latin to enhance our knowledge? Just so you know I have had two years latin but am just starting HS level.

Quis ad hic addent?

I hope my grammar is correct...[/quote]
I've had Latin for four years, then I dropped it for the two remaining years in favor of maths and sciences (here in Flanders we can't decide as much about what we take as in the USA).  I can hardly call it a language I know, and if it were one, it would be number five...

Seriously now : "Quis aut quid ad hic addent?"
What do you mean (what is your question in English)?
"Who or what will they add at this?" (by the way "ad hic" doesn't seem right, it should be "ad hoc" I think, "hic" doesn't seem possible because the male singular accusative of that pronoun is "hunc")

Probably it should be "Qui ad hoc addent?" Or "Quis ad hoc addet?"

Well IF funcia means :"Who will add himself here?" then I would propose
"Quis ad hoc se addet?"

Bubka, "quo vadis" means "where are you going", the movie with that title is quite good. :wink:

[quote="bubka"]pro eccelesia et pro patria - that was the motto of my school.[/quote]
You mean "pro ecclesia et pro patria".

"pro patria" means "for the homeland" but does "pro ecclesia" mean "for the church"? :maybe:   I don't know...

Funcia, can you please say what you meant?

Well if we're quoting random stuff, I think my favorite is "nemo me impune lacessit."

I did 4 yrs of latin in secondary school and one semester last year. I still suck at it. I probably won't take any more language classes (well not Latin anyway) because I'm not too interested in translation. The stuff we read was fun, but I was more interested in the content, not the act of translation....

Anyway just quoting the best line from the Cambridge Latin Course (wait actually there are several):
"Multo sanguis ex ore fluit." (From "Venatio" in book 1)
"Nos omnes moribundi summus!" (From "Tremores" (?) in book 1)
"'Vilbia? Barbam habes? Quid tibi accidit? O!'" (From "Vilbia III" [it was something III] in book 2. Modestus et Strythio were hilarious.)

And no, I don't have those books handy. I just had a lot of fun in that class...

I stopped doing Latin six years I ago. :blush:   I'm so cold and boring, I especially liked exploring Julius Caesar's grammatically correct but incredibly long and boring sentences about battles against the Gauls(=us :mad:).

I never refuse a little exercise, could you tell me whether or not I got this straight :

[hide]
"nemo me impune lacessit."="nobody defies me unpunished"
"Multo sanguis ex ore fluit."="A lot of blood flows from the mouth"
"Nos omnes moribundi sumus."="We are all mortal." (I think it must be sumus not summus, summus means "uppermost")
"'Vilbia? Barbam habes? Quid tibi accidit? O!'"="Vilbia? You have a beard? What has happened to you?  O!"[/hide]

And let us not forget "Horum omnium fortissimi sunt Belgae"by Julius Caser in "De bello Gallico IV" :lol:
(even though some experts claim we have no relation at all with those Belgae...)

I had one year of Latin in the ninth grade (that would be three years ago) with a legendary teacher (Scorţescu). She liked discipline and order. How she kept things under control: sheer terror.

All that I can remember is how the Carthagians lost in the third punic war. Grammar - squat. Okay, some words are pretty similar with the ones in Romanian, but those terminations were  :wallbash_red:

[quote="perfect_radio"]
All that I can remember is how the Carthagians lost in the third punic war. Grammar - squat. Okay, some words are pretty similar with the ones in Romanian, but those terminations were  :wallbash_red:[/quote]
Well I wasn't interested in poetry at all (the grammar didn't make sense) but apart from Julius Caesar's boring but correct sentences, I do like the ancient history... but I prefer to read about it in English (or if possible : Dutch).
The Carthagians didn't just lose the third Punic war, they were pretty good at losing all of them :lol: .  They got pretty close during the second one though when Hannibal got to Rome with elephants (he crossed the Alps) but  :huh: instead of going straight for Rome, he went further south in Italy to seek allies
"Vincere scis, Hannibal, victoria uti nescis."="You know how to win, Hannibal, but you don't know how to make use of your victory."

A quite funny, but not equally verifiable,story is that of Roman senator Cato, who kept urging his fellow Romans to start the third Punic war, not being convinced the second Punic war annihilated the threat posed by Carthago.  He would make speeches about anything all the time, but everytime he would finish them with "Ceterum censeo Carthaginem delendam esse"="Moreover, I think Carthago should be destroyed."

i meant who or what will add to this...i forgot to decline hic. sorry :blush: fredbel you know latin...grammar. that is for sure.
mea culpa! i am still reviewing right now forgive my grammar. ani think that the 3rd sentence was right..i remember having trouble with summus and sumus last year.

also i like those sentences the nobody defies me one is like a motto someone had for their nation in nationstates

I think in that case "nos omnes moribundi sumus" meant "we are all going to die!" In that section, Vesuvius was erupting. :D And yes, it's sumus, I wasn't paying attention.

I think the most interesting stuff I read among all of my Latin classes had to be the Seneca letters. He sounded like a crazy old geezer, and he had a good sense of humor. I read a few paragraphs of Caesar since my mom was teaching that to some guy, but I never did it in a class. Too bad: I probably would have liked it. My mom said she had a hard time understanding a lot of the military terminology... :P

@funcia: "nemo me impune lacessit" is the motto of Scotland, I believe. At least, it's on the rim of the Scotland pound coin. It's also the motto of the protagonist of "The Cask of Amontillado" -- a brilliant short story.

Actually the motto thing is my friend's motto for nationstates(my homepage.) did i say friend? i meant other personality. :maybe:

[hide="Slight Off-topic"]
[quote="fredbel6"]The Carthagians didn't just lose the third Punic war, they
were pretty good at losing all of them :lol: .[/quote]
Yes, but in the third one they got pwned, so to speak.
[/hide]

to Xevarion : that quote about the Vesuvius, is it by any chance from Plinius? (I really appreciate his scientific way of describing the eruption..which is still of great importance to scientists studying the Vesuvius)

to Perfectradio : I don't know really know what pwned means (I don't know any l337 speak  :roll: ) but if you mean, totally destroyed, then I agree.

Haha no it's from the first book of the Cambridge Latin Course. The book had a lot of short fictional stories based on actual Roman history. The first book was about a family living in Pompeii.

I did read that letter of Pliny to which you refer, though. (First semester last year.) It was pretty interesting. A lot of Pliny's letters were sort of annoying though. He wrote one about his uncle's books, and then rambled about his uncle's diligence in work and his random personal habits (he pretty much ate, slept, read, and wrote) ... it was bizarre. I really liked Seneca more.

I only am translating sentences but I feel I like the sentence that aren't philosophia(i know it shouldn't be nom.)

[quote="Xevarion"]
I did read that letter of Pliny to which you refer, though. (First semester last year.) It was pretty interesting. A lot of Pliny's letters were sort of annoying though. He wrote one about his uncle's books, and then rambled about his uncle's diligence in work and his random personal habits (he pretty much ate, slept, read, and wrote) ... it was bizarre. I really liked Seneca more.[/quote]
I guess it's the scientist in me.  :D    When in school, we went to Italy in 2002.  When driving to the Vesuvius and Pompei, we watched a documentary about Plinius works and curent findings.  Putting all together, they concluded that during that night, there were four or five times when the volcano lost some power and debris from the edge of the crater fell back in. That is what created a fast powerful storm cloud racing over the town.

Isn't Plinius the one whose uncle even fell asleep during the eruption (and who was snooring too :roll: ).

[quote="Xevarion"]I thought the uncle dies in the eruption while trying to help the townspeople get out....
(He's really fat and not in good physical shape so it is not surprising that he more easily succumbs to the fumes and ash than the people around him)[/quote]
Didn't a couple of slaves who didn't want to wake him die too.

Anyway, I won't deny that the physical shape of his uncle seems to be of utmost importance throughout his writings. :D


Solve for $ x$: $ 2^x\plus{}6\equal{}3\cdot2^x\minus{}26$

\begin{align*}
2^x+6=3\cdot 2^x-26 &\Rightarrow 32=(3-1)\cdot 2^x \\
&\Rightarrow 32=2^{x+1} \\
&\Rightarrow x=\boxed{4} \\
\end{align*}


modular arithmetic
induction
number theory proposed
number theory

For which pairs of positive integers $(m,n)$ there exists a set $A$ such that for all positive integers $x,y$, if $|x-y|=m$, then at least one of the numbers  $x,y$ belongs to the set $A$, and if $|x-y|=n$, then at least one of the numbers $x,y$ does not belong to the set $A$?

[i]Adapted by Dan Schwarz from A.M.M.[/i]

If $m=n$, it is obvious it exists $A$.
Suppose for a pair $(m,n)$ with $m\neq n$ it exists $A$.For $u\in A$, we have $u\pm n\notin A\Rightarrow u\pm n\pm m\in A$. So for $k,l\in\mathbb{Z}$ we have that if $u\in A\Rightarrow u+k(m+n)+l(m-n)=u+(k+l)m+(k-l)n\in A$. The same for $u\notin A$.
Let $d=gcd(m,n)\Rightarrow$ it exists $u,v\in\mathbb{Z}$ such that $um+vn=d$. Chosing $k=u+v$ and $l=u-v$ we have that if $u\in A\Rightarrow u+2d\in A\Rightarrow u+2kd\in A$ and if $u\notin A\Rightarrow u+2d\notin A\Rightarrow u+2kd\notin A$.
If $2d|n\Rightarrow n=2kd\Rightarrow$ if $u\in A\Rightarrow u+n\in A$(false)
If $2d|m\Rightarrow m=2kd\Rightarrow$ if $u\notin A\Rightarrow u+m\notin A$(false)
So, $exp_2(m)=exp_2(n)$. I couldn't find an example of $A$ in this case, but I managed to prove somehow that it exists.

IMHO, this is a very nice problem :D

I found something similar, although I'm pretty late: if $k \in A$, then: $k+an+bm \in A$ for $a+b \equiv 0 \pmod 2$; else $k+an+bm \not\in A$. (to get a certain pair $(a,b)$, we first let $b$ oscillate between $0$ and $1$ until $a$ gets the right value, then we do the same thing with $b$) Thus, the only way to get a contradiction is to have $pn=qm$, where $p \not\equiv q \pmod 2$, which leads to the same answer as the one in the previous.

Now, if $2^t \| m,n$, then we put $1,2,\ldots,2^t$ in $A$, the others will follow by the above observation applied to each of $1,2,\ldots,2^t$. (when applying it to $1$, we will decide whether some number $\equiv 1 \pmod{2^t}$ gets chosen or not). Note that they can't get into any conflict, because $|x-y|=n$, for example, implies $x \equiv y \pmod{2^t}$.

And please tell me if there's something with the previous paragraph.

Just to flesh out the previous two solutions:
I claim it is possible iff $m$ and $n$ have the same maximum powers of $2$ dividing them.
First, if $x$ is in $A$, $x+n$ is not in $A$, $x+n+m$ is in $A$, $x+m$ is not in $A$. Similarly if $x$ is not in $A$, $x+m$ is in $A$, $x+m+n$ is not in $A$, and $x+n$ is in $A$. So basically, if $|x-y|=m$ or $|x-y|=n$, then precisely one of $x,y$ is in $A$.
By induction, if $x$ is in $A$ then $x+(2k+1)m$ is not in $A$, while $x+2ln$ is in $A$, so thus the equation $(2k+1)m = 2ln$ has no solutions, ie the fraction $\frac{m}{n}$ cannot be expressed as an even number divided by an odd number, and by switching variables $m,n$ since it doesn't make a difference, $\frac{m}{n}$ cannot be expressed as an odd number divided by an even number either, ie in lowest terms the fraction $\frac{m}{n}$ is an odd number divided by an odd numbers, so for such a set to possibly exist $m$ and $n$ must have the same maximum powers of $2$ dividing them.
To see this works: if $m$,$n$ are both odd, then consider the set $A$ of odd numbers, which works obviously. Now proceed by induction on the highest power of 2 dividing $m$ and $n$. if $2^{s+1}$ is the highest power dividing $m$ and $n$, consider numbers mod $2^{s+1}$ and for a fixed $t$ $0 \leq t < 2^{s+1}$, $a= r 2^{s+1}+t$, biject that number to $r$. Now let $A$, in this bijected set, consist of all odd numbers, and now biject it back to get a suitable set $A$ for the real thing, and do this for every value of $t$ to get a working construction.


Princeton
college
AMC
AIME
USA(J)MO
USAMO
summer program

Next year, I'll be enrolling as a freshman at Princeton University. As of right now, I'm totally split between majoring in Math or ORFE [Operations Research and Financial Engineering] at Princeton.

In the future, I hope to have get a job on wall street or some other investment/finance related job. What makes me unsure what to major in is I'm unsure of my own abilities. I'm not sure if I'm more a problem solver or a thinker... I do really well on AMC, AIME and other contests, though, I've never really been strong on the USAMO the two times i took it. So I believe I'll do much better academically in ORFE.

In terms of interest, I've attended BU's PROMYS for two summers and I've really enjoyed it there. I did a researhc project which was really confusing, yet at the same time really interesting. ORFE seems to be a lot fo brute force work, whereas majoring in math seems to be more thought provoking.

And a last thing... majoring in math seems so much leet. There are only 18 or so math majors in each class at Princeton, so there's defintely some prestige for math majors. And also... I know it's wrong to look for a high paying job, but, i was wondering if math graduates from Princeton who go into finance on wall street make more than ORFE majors.

And I sort of want to decide before the first semester, because ORFE is engineering and Math is A.B... so i'd have to work out the scheduling differently [for example, A.B. students have a foreign language requirement].

If anyone could help me out on this decision, I'd be really grateful.

Norman!

You have to be kidding me with not being a "thinker." You'll do great as a math major (or any other major).

I would suggest doing math, but I guess that's a little biased?

You're definitely good enough at math, so the only question you should be asking is what you like better.

Btw, I heard you were at the ultimate practice on Saturday. Sorry I couldn't come... This year they have a modular forms seminar and p-sets are about as hard as the ones from Riemann-Zeta function from our first year. It's pretty interesting though.

[quote="yunorman"]And also... I know it's wrong to look for a high paying job, but, i was wondering if math graduates from Princeton who go into finance on wall street make more than ORFE majors.[/quote]

I don't think there's anything wrong with looking for a high paying job. It's nice to be able to live with a little extra money.

As for your primary question, I would be that having attended PROMYS twice you have more experience than most math majors starting out. Plus, it has to say something about your math ability; not everyone can walk into PROMYS and walk out wanting to go again (I say this as a PROMYS alumnus myself in case you are wondering - I almost went back, but had other plans for the last two summers). It sounds like you really want to do math, but you don't want to go into academia. One possibility is to do the ORFE program and just take the math classes you want to, minor if you really feel like it. It seems like a good balance for you, but as the previous poster has said, at this point I think it is a question of your interests and what you want to get out of your college education. Yup, that's about all I have to say.

Good luck at Princeton next year, go class of 2011! :)

Of course, as paladin8 suggested, it's possible to compromise. In fact you can compromise in both directions:

Math major + finance certificate
ORFE major + applied math certificate

I'm doing the first one. I think an important advantage to majoring in math is that it still leaves you the choice of continuing in academia or going into finance. You will get about the same quality job as someone who majored in ORFE, or perhaps a little bit better. You may be surprised, however, at how much your opinions about the subjects change after you're exposed to them at the level of rigor of a Princeton class.

Some other points I should make:

1) You WILL get better grades in the ORFE department than in the math department. It's not even in question. But it's not like your GPA will suck if you major in math, so don't worry too much about that.

2) You have about average experience with math among people majoring in math at Princeton. (You have significantly more experience than the typical freshman considering majoring in math.) I'd say you have a fair amount more experience than I did. So you'll probably be fine, but don't expect to be dominating everyone. You might end up taking classes with IMO gold medalists.

An important lesson I've learned in college is that you don't have to be the best to be successful. There are so many directions to go, so many things to do, that you will find a way to capitalize on your individual interests and abilities. For example, I don't think there are any other undergrad math majors at Princeton who know more about geometric analysis and measure theory than I do. But I am nowhere near the "smartest" or "most advanced" math major. I just got interested in a particular topic and studied it.

3) Princeton's math department is one of the best in the world. But the best finance people are on Wall Street managing billions of dollars, not sitting in a room in central New Jersey trying to pass on their wealth of knowledge to a younger generation. ;)

So, after all this, some actual advice:
Plan on majoring in math and getting a finance certificate. Take classes that are ORFE requirements and also AB distribution requirements freshman year. This way, you will have the option of declaring ORFE at the end of the year if you didn't like your first two semesters of math that much.

You may also want to read the book

The [url=http://www.amazon.com/Black-Swan-Impact-Highly-Improbable/dp/1400063515/]Black Swan: The Impact of the Highly Improbable[/url] by Nassim Nicholas Taleb.

It has an interesting take on careers in finance and in academe.

Xevarion,

How do you plan to fulfill your independent work requirement? I would also like to concentrate in math and do a certificate in finance but I don't necessarily want my math senior thesis to have anything to do with finance.

Also, do you know of anyone at Princeton, or anyone else in this forum, who is doing a certificate or a major in linguistics?

By the way, Princeton doesn't seem to have a physics certificate (since most certificates seem to be quite specialized) and it seems stupid for me to do a certificate in applied math if I am a math major, so what's the best way to study physics along with math (obviously other than voluntarily taking electives in physics)?

paladin8,

which summer were you at PROMYS?

Thanks for all the replies.

Paladin’s got a pretty good gauge on me; I’m really interested in pure math, but I don’t plan on going into academia.

I also have doubts about majoring in mathematics simply because of the “impracticalities” of what I’ll be learning in college with regards to my future profession. Without a question, pure math is definitely the most interesting topic I’ve ever studied, but how will learning it help me in the future? Though, a friend at Princeton tells me major finance companies heavily recruit math majors because they know how smart they are and they will train them for the job… Then again wouldn’t doing really well in the ORFE department make me a good recruit AND teach me some skills?

Also, according to bls.gov [bureau of labor statistics], employees with the title “mathematician” typically have a PhD, unless they work for the government. While a Wall Street employees’ title may not be mathematician, do you really need a PhD to get a good job?

[quote]3) Princeton's math department is one of the best in the world. But the best finance people are on Wall Street managing billions of dollars, not sitting in a room in central New Jersey trying to pass on their wealth of knowledge to a younger generation. Wink [/quote]
Do you mean... majoring in math isn't lucrative?

Right now, I really want to work on Wall Street or some major finance firm in the future. [Actually, my #1 dream/ultimate job is general manager of the Boston Red Sox: P. I’m really into sabermetrics, it’s more stats/applied though.]

Hmmm, so right now… I’m looking at math major plus possibly certificates in finance, management and engineering systems, and applied and computational mathematics [not necessarily all 3]. If I find out I’m really not that into math, I’ll switch to ORFE if I find that a lot more interesting, as you guys suggested.

shbmsc90,

I definitely should have gone to PROMYS again this summer. O well... I'll see you at the PROMYS - RSI alumni again, maybe paladin8 too [but for PROMYS or RSI...?? :P]. How's the research project, modular forms, and being a JC. When I heard you were only doing 1 advanced seminar this time, I thought you were slacking, but then I realized you already finished the other ones :)

[quote="shbmsc90"]paladin8,

which summer were you at PROMYS?[/quote]

Ah, it was ages ago. Summer of '04. We'll see if I'm ever in the Boston area during the summers in college :P.

[quote="shbmsc90"]Xevarion,

How do you plan to fulfill your independent work requirement? I would also like to concentrate in math and do a certificate in finance but I don't necessarily want my math senior thesis to have anything to do with finance.

Also, do you know of anyone at Princeton, or anyone else in this forum, who is doing a certificate or a major in linguistics?

By the way, Princeton doesn't seem to have a physics certificate (since most certificates seem to be quite specialized) and it seems stupid for me to do a certificate in applied math if I am a math major, so what's the best way to study physics along with math (obviously other than voluntarily taking electives in physics)?[/quote]
1) I haven't decided, hehe. I'll probably do something theoretical involving stochastic calculus.

2) No. I do know some people who have taken linguistics classes, but I'm not sure if any of them are actually going for the certificate or anything.

3) Just take physics classes. Do one of your JPs on a physics-related topic. Grad schools/future employers will be able to figure out that you know some physics. (Also, the applied math cert isn't totally pointless; it shows that you know how to do math that would be useful in industry, rather than just purely abstract math, so it makes you more employable.)

[quote="yunorman"] Without a question, pure math is definitely the most interesting topic I’ve ever studied, but how will learning it help me in the future? Though, a friend at Princeton tells me major finance companies heavily recruit math majors because they know how smart they are and they will train them for the job… Then again wouldn’t doing really well in the ORFE department make me a good recruit AND teach me some skills?

While a Wall Street employees’ title may not be mathematician, do you really need a PhD to get a good job?

Do you mean... majoring in math isn't lucrative? [/quote]

1) If you have a good foundation in math, it's much easier to learn anything else that's math-based. It's just like learning Latin and then learning another Romance language. The vocabulary and grammar will be easy, it's only learning how to speak that will be hard.
If you major in ORFE, there's no guarantee that you have a good foundation in math. You personally probably would, because if you like pure math you'd probably take extra math classes, but it's possible to go through the ORFE department doing a minimum of math with minimum understanding. My impression is that employers at the top hedge funds and such really don't want to take risks with their employees; Princeton will guarantee a good foundation in math, and then the hedge fund can personally give you the financial training you need, and there's MUCH less risk for them.

2) You don't need it. An undergrad degree in math (especially at Princeton) already teaches all the pure math you'll ever need in finance. A PhD in math shows personal qualities like the ability to do significant research, dedication, etc., but the math itself isn't really necessary. I had some interviews for finance internships this past year, and I asked them about that, and they mostly seemed to be fairly indifferent. There's no huge advantage for PhD or no PhD.

3) I meant that you should learn math from the best at Princeton and then learn finance from the best on Wall Street. Academia is almost never lucrative, but professors aren't poor either (once they get tenure, at least). Majoring in math, generally speaking, won't guarantee you a high-paying job, but there are a LOT of high-paying jobs available, mainly because of what I said in (1).

[quote="shbmsc90"]... do you know of anyone at Princeton, or anyone else in this forum, who is doing a certificate or a major in linguistics?
[/quote]

I'm leaning towards majoring in math with a certificate in linguistics :)

To be honest, majoring in math is NOT a good idea if you want to make money (its not practical). And in this world today, money IS important. Major in economics/ quantitative finance/ OR to get into financial engineering. I have not taken many math courses in my college, yet I am more knowledgeable than many of the math majors. So you can always self-study math. To make it a career is a risky decision. Whereas, majoring in the OR/economics will be more practical in terms of finding high paying jobs. Nobody will care if you have solved some esoteric problem for your PhD thesis in math (i.e. do math not because of fame and glory). Mathematics is not a lucrative field. So I encourage you to pursue your love of math(but not as a career!). But if you truly love it, then you can do it by yourself. Who the heck cares if math is the 'leet' subject at Princeton. Thats a stupid reason to major in a subject. The PhD degree in general is useless unless you want to be a professor. You waste your twenties for what? Just to get three letters (note I said its useless if you do not want to be a professor). I would advise you to take some time off after college and work/ explore the world. If you go to grad school immediately after, then your life seems like its very one-dimensional. Then afterwards you can go to grad school if you want to. Also I would advise that if you want to become a professor, do it later on in life (i.e become an adjunct professor). Professors today are required to spit out papers to get tenure. As an adjunct professor, you will not have the same responsibilities are regular professors, but still get to teach students. Hope this helps.

Hmm, there have been varying responses with regards to whether or not studying mathematics can be made into a lucrative career. I do plan to get certificates in finance and management and engineering systems, so, I shouldn't be too far behind. And ragingbullrocky, btw, I don't plan to get a PhD, if i do go to grad school, I'll probably go for an MBA... but who knows, i may end up with a PhD one day, and if i do, I probably won't care about money.


Okay, I know mods is very helpful in many math contests, especially MATHCOUNTS.  The only thing is, I've gone over my MATHCOUNTS class transcript for mods, but I still do not understand exactly how they work and when to apply them in a problem.  Could anyone help me, and if you could, give sample problems with solutions?  NOTE: Please do not post the mods section on the AoPS Wiki.  I already know about it!!

All replies are appreciated. Thank you. :lol:

Here's a good mod question from 2004 target:

2 is set $ S$. Also, if $ x$ is in set $ S$ then $ x\plus{}7$ and $ 3x$ are also in set $ S$. What is least integer greater than 2008 that is in set $ S$?

HINT:
[hide]use mod7 [/hide]

Solution:
[hide]Since 2 is in the set, all numbers congruent to 2 in mod7 are in the set. (If x is in the set, x+7 is also in it.)

If 2 is in the set then 6 is also in the set(3x). Thus, using the same logic all numbers greater than 6 that are congruent to 6 in mod7 are also in the set.

2,6,18,54,162,486 are all in the set(using the 3x rule)

Notice that they leave a remainder of 2,6,4,5,1,3 respectively when they are divided by 7.

Thus, all numbers congruent to 1,2,3,4,5, or 6 in mod7 are in the set.

So the only numbers that aren't in the set are numbers that are divisible by 7. (leave remainder of 0).

2009 is divisble by 7 so its the answer.[/hide]


search
combinatorics unsolved
combinatorics

Consider the set [b]U={1,2,3...,6024}.[/b]
Prove that for each partition of $ U$ in $ 3$ subsets of $ 2008$ elements each,we can choose $ 1$ element from each subset s.t.
one element is equal to the sum of the other $ 2$ chosen.

A more general result here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=400307688&t=3781

Pierre.


Four lines are given in the plane such that each three form a non degenerate , non equilateral triangle. Prove that , if it is true that one line is parralel to euler line of the triangle formed by other three lines , then this is true for each of line .


$a=x+y$, $b=y+z$, $c=z+x$, $x,y,z>0$, operando se llega .
MUIRHEAD.

i don`t think it can be solved using muirhead, could you post a complete solution ?

[color=darkblue]I"ll prove that this inequality is stronger than the [b][u]Nesbitt's inequality[/u][/b]

$\sum\frac{b+c}{a}\ge \frac{9}{2}+\sum\frac{a}{b+c}\ .$ Indeed, prove easily that $\sum \frac{a}{b+c}\ge \frac{3}{2}$ and

$\boxed{\ \sum a\cdot \sum\frac{1}{a}\ge 6\cdot\sum \frac{a}{b+c}\ }$ $\Longleftrightarrow$ $3+\sum\frac{b+c}{a}\ge 6\cdot\sum\frac{a}{b+c}$ $\Longleftrightarrow$

$\left\{\begin{array}{c}\sum \frac{b+c}{a}\ge-3+6\cdot\sum \frac{a}{b+c}\\\\ 5\sum\frac{a}{b+c}\ge \frac{15}{2}\end{array}\right\|\bigoplus$ $\Longrightarrow$ $\boxed{\ \sum\frac{b+c}{a}\ge \frac{9}{2}+\sum\frac{a}{b+c}\ }\ .$ [/color]

[quote=pohoatza]Fie $a,b,c$ laturile unui triunghi. Sa se demonstreze inegalitatea:

$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq  6(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})$[/quote]

[b][color=#c00]Clearly[/color][/b], the $LHS$ is [b][color=#c00]decreasing [/color][/b]of $xyz$ while the $RHS$ is [b][color=#c00]increasing [/color][/b]of $xyz$. [b][color=#38761D]So suffice it to prove for[/color][/b] $y=z=1$,$x\in[1,2]$ [b][color=#c00]first [/color][/b]and for $z=x+y$ [b][color=#c00]after[/color][/b].

[b][color=#c00]In the first case[/color][/b] get $(2-x)(x-1)^2 \geq 0$, which is true, while [color=#c00][b]in the second one[/b][/color] we get $\frac{x}{y}+\frac{y}{x} \geq 3(\frac{x}{x+2y}+\frac{y}{2x+y})$ which is obviously [color=#c00][b]true[/b][/color].

[i]Returning to the original problem[/i], [b][color=#c00]equality at[/color][/b] $(1,1,1)$ or $(2,2,0)$ and [b][color=#c00]permutations[/color][/b].

[center][b][color=#38761D]Very beautiful problem.
[/color][/b][/center]

[center][img]https://image.ibb.co/kmHYnz/group.png[/img][/center]


Define "A constant subsequence"

Define $\{x_n\}_n$.

thank you

In Ed5 Round 3 Problem 1, you use these terms

can you please clarify what they mean

What's there to define about a constant subsequence of another sequence?!  :what?:

Pardon me for my poor understanding of math terms, but what does a constant subsequence mean? Can you please give an example? Thanks.

Hmm, I think it's very obvious what a subsequence is, and a constant one ... well anyway if you don't know what a subsequence is, then look [url=http://planetmath.org/encyclopedia/Subsequence.html]here[/url] :)

An example would be: the sequence 1,0,2,0,3,0,4,0,...,n,0,n+1,0,.... contains a constant subsequence with all terms equal with 0.

Hey, i am also unclear about that term!  :blush:

plz, don't think that we all are as great as you :P

So the example given by is not clear enough! plz give much discriptive example. or if anyone else can help us to understand that plz!

Say two seq:
1,2,3,4,5...

and

0,1,-1,3,-3,5,-5,7,-7...
Can i say the common subsequence is 1,3,5,...

Mahbub

I think it clearly states [b]CONSTANT[/b] subsequence. I am sorry, but I think that I have explained enough.

OK :blush:

But well just answer me one thing :blush:

In the example u gave where u said 0 is the constant sub-sequence, if there was finite 0s say only three like this:
0,1,0,2,0,3,4,5,6,...

Then is the seq of 0 in the above series (which i gave) considered as constant-subsequence? :?

[quote="perfect_radio"]i also had problems understanding this term

from what you wrote above i understood that we have to prove that there is an $a$ s.t. it appears infinitely many times in the sequence.

Is this correct?[/quote]
i took the meaning as yours :)

C'mon guys! :) I can't belive you don't know what a subsequence of a sequence is ...  :) Perfect_ratio is right ...

Hey u r wrong this time valentin :D

It is not perfect_ratio it is perfect_radio :P

Ooops :D  :blush:  :blush: Made more sense to me to say perfect_ratio :P Sorry ...  :blush:


calculus
integration
function
derivative
real analysis
real analysis unsolved

I need to show that \[f(\xi,t)=e^{-\frac{\xi^{2}}{4t}}\] is uniformly convergent, preferably using the Weierstrass M-Test.  The professor keeps telling me to use the fact that there exists some constant $t_{0}$ such that $t \ge t_{0}> 0,\ \forall t$, but I can't find a proper $M(\xi)$.  Can anyone help?  Thank you.

I'll prove it for $e^{-t^{2}/2}= \sum_{k=0}^\infty(-1)^{k}\frac{t^{2k}}{2^{k}k!}$, which should give you some idea of how to approach the problem.  On a fixed radius of convergence $R$ Use $M_{n}= \frac{R^{2k}}{R^{k}k!}$, which converges to $e^{R^{2}/2}$.  Since $R$ was arbitrary, the series converges uniformly in any domain.  Does that make sense?

Using the fact that such a $t_{0}$ exists, you simply insert that into the denominator of your power series to get that it converges to $e^{R^{2}/4t_{0}}$

Sorry, I am still confused because I do not understand how you can just pick an arbitrary radius.  Also, I do not know whether a discrete sum representation carries directly over to a continuous integral representation.  The Weierstrass M-Test that I am supposed to use is defined as follows: \[\text{If }e^{-\frac{\xi^{2}}{4t}}\le M(\xi) \text{ and }\int_{-\infty}^{\infty}M(\xi)\,d\xi<\infty , \\ \text{then }e^{-\frac{\xi^{2}}{4t}}\text{ converges uniformly. }\]  The point is that $M(\xi)$ cannot depend on $t$.  Correct me if I'm wrong, but I think you are saying that my $M(\xi)$ should be $e^{-\frac{\xi^{2}}{4R}}$.  I suppose if you are definite that I can introduce such an arbitrary variable, then this satisfies the Weierstrass M-Test.  What are your thoughts on this?

You haven't made clear what the problem is. What "sequence" of functions is converging to what function as what variable tends to what? And is there a sum or integral involved or just a limit? (I associate "Weirstrass M-test" with sums.)

I am trying to show that
\[u(x,t)=\int_{-\infty}^{\infty}f(\xi)e^{-\frac{(\xi-x)^{2}}{4t}}\,d\xi \]
is a solution to the one-dimensional heat equation, simply by plugging it into the heat equation, $u_{xx}=u_{t}$.  During the process however, using the Weierstrass M-Test, I have to justify that I can differentiate under the integral, so that I may write:
\[\frac{\partial^{2}}{\partial x^{2}}\int_{-\infty}^{\infty}f(\xi)e^{-\frac{(\xi-x)^{2}}{4t}}\,d\xi=\int_{-\infty}^{\infty}f(\xi)\frac{\partial^{2}}{\partial x^{2}}e^{-\frac{(\xi-x)^{2}}{4t}}\,d\xi , \] \[\frac{\partial}{\partial t}\int_{-\infty}^{\infty}f(\xi)e^{-\frac{(\xi-x)^{2}}{4t}}\,d\xi=\int_{-\infty}^{\infty}f(\xi)\frac{\partial}{\partial t}e^{-\frac{(\xi-x)^{2}}{4t}}\,d\xi . \]
This justification is brought by proving that the integrand is uniformly convergent.  For simplicity, I was letting $f(\xi)=1$.  Additionally, my professor hinted that it would help to use a change of variables: $\bar{\xi}=\xi-x$.  In the previous posts, I just used $\xi$ instead of $\bar{\xi}$.

You can't do it uniformly in $t$, all at once. If you restrict to $t>t_{0}$, differentiation works on that set; letting $\epsilon\to0$, it works for all $t>0$.
You can't go all the way to $t=0$; there's no reason to expect that the function is even differentiable there.

Differentiation is local; locally nice functions are good enough.

Sorry, more clarifications: the integrand is of class $C^{2}$ and we are only looking at $t>0$.  Thank you for all your responses so far.


analytic geometry
calculus
derivative
function
calculus computations

find  the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane $x+2y+3z=6$.

Lagrange multipliers at their most straightforward. Have you tried that technique?

sorry but i have never heard of that technique.  :(

All right- how about AM-GM? The volume is $\frac16(x\cdot 2y\cdot 3z)$.

ok so that gives us
$\frac{x+2y+3z}{3}\geq\sqrt[3]{6xyz}$
so
$\sqrt[3]{V}=\frac{2}{\sqrt[3]{6}}$
cubing yields
$V=\frac{8}{6}=\frac{4}{3}$
thanks.

how would you do this using the second derivative test that is normally taught in calc lll?

Your function $V(x, y, z) = xyz$.

However, $x = 6-2y-3z$.

So, $V(y, z) = (6-2y-3z)yz = 6yz-2y^{2}z-3yz^{2}$.

Now, $\frac{\partial V}{\partial y}= 6z-4yz-3z^{2}= 0$,
and $\frac{\partial V}{\partial z}= 6y-2y^{2}-6yz = 0$.

So then, $12y-4y^{2}-18z+9z^{2}= 0$.

Then, $9z^{2}-18z+9 = 4y^{2}-12y+9$.

So, $(3z-3)^{2}= (2y-3)^{2}$.

Thus, $3z-3 = 2y-3$ or $3z-3 = 3-2y$.

But in the second case, $2y+3z = 6$, so $x = 0$, obviously not optimum.

Therefore, $2y = 3z$, so $y = \frac{3z}{2}$.

Then we have $6z-6z^{2}-3z^{2}= 0$, by substitution.

So $z = \frac{6}{9}= \frac{2}{3}$.

Then, $y = 1$, and therefore, $x = 2$.

There are other more brute force methods to do this, but I think this method is more creative.

I know maokid7 has never seen Lagrange multipliers, but they're worth showing anyway. In a standard American calculus course and textbook sequence, this is usually placed in the Calculus III or multivariable calculus course; look for it in the section of the book devoted to partial derivatives.

We're trying to optimize $V(x,y,z)=xyz$ subject to the constraint $g(x,y,z)=x+2y+3z=6.$

We invent a new variable, $\lambda,$ which is the "multiplier." The critical points are places where $\nabla V=\lambda\nabla g.$ This gives us the following equations to solve:

$\left(\frac{\partial}{\partial x}\right)$ equation: $\ yz=\lambda$
$\left(\frac{\partial}{\partial y}\right)$ equation:$\ xz=2\lambda$
$\left(\frac{\partial}{\partial z}\right)$ equation: $\ xy=3\lambda$
Constraint equation: $x+2y+3z=6.$

If $\lambda=0,$ we get points at which two coordinates are zero: $(6,0,0),(0,3,0),$ and $(0,0,2).$

If $\lambda\ne0,$ we get that $y=\frac{x}{2}$ and $z=\frac{x}{3}.$ Put those into the constraint equation and we get the point $\left(2,1,\frac23\right).$

By talking about a box, we have ruled out any coordinate being negative. That gives us a compact set - the intersection of the constraint plane with the first octant. Therefore, the continuous function $V$ must assume a maximum. The boudaries of our set (which is a triangle) are lines on which at least one of $x, y,$ or $z$ is zero. $V=0$ at all such points (including the three that turned up as critical points above; in fact, we should downgrade the status of those three points to just being boundary points of the domain.)

But $V\left(2,1,\frac23\right)=\frac43>0.$ Since this is the only critical point at which the function is positive, it must be the maximum.

Of course, this is identical to the solution given by the previous two posters. It should be; all are correct. I've also done the Lagrange multipliers in more detail than would be needed by someone with experience in the technique. It is a genuinely useful technique, and worth knowing.


The pages of a book are numbered consecutively, beginning with 1. If the digit 7 is printed 25 times in the book, what is the largest number of pages that the book can have?

[hide]I got 166
The first 100 numbers have 7 as their units digit 10 times, and then you add the 9 with tens digit. that's 19
then 107, 117, etc. until the sixth one, 157, and then 166 is the maximum [/hide]

I got 156
because from 1-100 there are 20 7's (77 counts as 2-theres 2 sevens) and
5 more sevens later you get 147....so the farthest is 156.

Argh. that's what i got for my first answer, but when i recounted, i got a different answer. Sorry.

[hide]
lets see
for the first 100 it is
0- 1- 2- 3- 4- 5- 6- 8- 9-7
that is 10
in the 70s there are 11
thats already 21
22-107
23-117
24-127
25-137
My answer:$\boxed{146}$[/hide]

[hide]Hopefully everyone knows that 20 of each number except 0 appear in 1-99.
21: 107
22: 117
23: 127
24: 137
25: 147
Hence, $\boxed{156}$[/hide].

[hide]after recounting, i realize my answer is incorrect
ans: $156$[/hide]

[hide]Looking at it in a practical way, we start a list of numbers with the 7 in it:
7 - 17 - 27 - 37 - 47 - 57 - 67 - 70 - 71 - 72 - 73 - 74 - 75 - 76 - 77 - 78 - 79 - 87 - 97 - 107 - 117 - 127 - 137 - 147 ...we've just reached the limit for sevens. Since you can go up until there is another seven, the most pages you can have is therefore [b]156[/b].[/hide]


linear algebra
matrix
algorithm
algebra
polynomial

Let [tex]A,B,C,D \in M_n (\mathbb{R})[/tex] with [tex] CD=DC[/tex]. Prove that

[tex]\displaystyle \det\begin{pmatrix} A & B \\ C & D\end{pmatrix} = \det(AD-BC)[/tex]

[i]Moderator's note: I [tex]\TeX[/tex]ed the problem to make it more readable.[/i]

Maybe I am being stupid, but this problem does not make much sense.  On the LHS you have the determinant of the matrix on the RHS, w/"det" in front of it.  The determinant of a 1x1 matrix is the same as the only element in it, right?  Are you asking to verify the algorithim for finding a determinant?  B/c that's a definition, I think.

No, what he's saying is that A, B, C, and D are each n x n matrices. Then you combine them in the way shown to form a new 2n x 2n matrix. So A, B, C, and D are not "elements" of the matrix on the left; they are matrices themselves.

Let $U_x=xI+D$, then $p(x)=\det(U_x)$ is a monic polynomial in $x$ of degree $n$. So $U_x$ is invertible for all $x\in \mathbb{R}$ except the roots of the polynomial $p(x)$. Clearly $CU_x=U_xC$. It's easy to verify that:

$\displaystyle
\displaystyle \begin{pmatrix} A & B \\ C & U_x\end{pmatrix} \begin{pmatrix} I & O\\ -U_x^{-1}C & I\end{pmatrix}
=\begin{pmatrix} A-BU_x^{-1}C & B \\ O & U_x\end{pmatrix}.
$

Hence

$\displaystyle
\det\begin{pmatrix} A & B \\ C & U_x\end{pmatrix}
=\det\begin{pmatrix} A-BU_x^{-1}C & B \\ O & U_x\end{pmatrix} =\det(A-BU_x^{-1}C)\det(U_x) $

$\displaystyle =\det(AU_x-BU_x^{-1}CU_x)=\det(AU_x-BC)$.

Since this is true for all $x\in \mathbb{R}$ such that $p(x)\not=0$ and both sides are polynomials in $x$, it must be true when $x=0$.

[quote="JBL"]And what if p(0) = 0?[/quote]

Both sides are polynomials and they are equal for infnitely many x. So they are identical. In particular, they are equal when x=0.


vector
geometry
geometric transformation
set theory
superior algebra
superior algebra theorems

It may be introduced by Hilbert, from the word 'Zahlring'.

I ask this because in german the word ring mens a gem what you're wearing on the finger.
The word "Zahlring" is translated word by word "numberring".

Why it has been choosen the word "Ring" and not another word like "Biarithmetic" for instance.

Maybe there is a known orign.

The word "Field" for instance is called so because of the perfection of them.

Field in german means "Körper" and you can also translte it by "body" or "solid".
According to my knowledge the word "Field" comes from the german word what is called "body".

I think you should go to the library and read the paper written by Hilbert at 1896.  :D

I want to reply that if you'll know the reason.

In Korean, 'field' is '체(體)'

If you wish to touch the [b]body[/b] (field) then must pass through the (wedding) [b]ring[/b] (in the Lord's house) !

can i ask?but has this name of the algebraic structure any link with its topological representation ?
i mean i saw somewhere about torus algebra wich implied some superior algebra,
on the other hand a torus resembles to a ring(not the algebraic structure).
i mean...maybe geometrical structures were associated with the names
i dont know too much so...dont bother to read my post...im probably wrong

I am working on my thesis to obtain my masters in educational
mathematics in Dominican Republic. The topic of my thesis is
"teaching algebraic structures in high school". I would like to
know if this topic is included in high school in any state or if
you are aware of any school that includes at least a light introduction. (Some number and set theory, group concept and properties, etc.)
 Also, may you help me with resources, websites or so
where I can analyze methodologies for teaching this topic?

Thanks
 :?:

In the $12th$ grade, in Romania, we are thaught: groups, rings, fields,vector spaces and .... anyway, you get the point.

[quote="luther_driggers"]in spanish i believe it's body (cuerpo)[/quote]

es campo

[quote="Gauß"]Does anybody know where the term "Ring" come from.[/quote]

The name was coined by Hilbert in S31 of "The theory of algebraic number fields"
You can see it in English translation here:
http://books.google.com/books?as_q=zahlring&as_isbn=3540627790


function
geometry
floor function
number theory
prime factorization

Compute the area of the solution set $\lfloor x \rfloor \lfloor y \rfloor =2000.$

[hide]Let's look at the first quadrant's contribution first. Since $\lfloor x \rfloor$ and $\lfloor y \rfloor$ are each integers, we see that each must be a factor of $2000$. In addition, if one is any given factor of $2000$, the other is fixed. Now, $2000$ has prime factorization $2^{4}5^{3}$, so it has a total of $5*4 = 20$ factors, so there are $\frac{20}{2}= 10$ pairs of factors.

Now, we must determine areas. Let's take the factors $1$ and $2000$ for the sake of simplicity. The ways in which $\lfloor x \rfloor$ and $\lfloor y \rfloor$ can be $1$ and $2000$ are when $1 \leq x < 2$ and $2000 \leq y < 2001$ or when $1 \leq y < 2$ and $2000 \leq x < 2001$. Each of these possibilities is a square with side length $1$, so each contributes $1$ square unit of area to the solution set. As a result, the pair of factors $1$ and $2000$ contributes $2$ square units of area. It is easy to see that this holds for each of the $10$ pairs of factors, so the area of the positive part of the solution set is $20$ square units.

Obviously, $\lfloor x \rfloor$ and $\lfloor y \rfloor$ cannot have opposite signs (a negative times a positive equals a positive????). That eliminates the second and fourth quadrants from the solution set. Applying a similar technique to the third quadrant as we did to the first, we see that the third quadrant also contributes $20$ square units of area. Thus, the area of the solution set is $40$ square units.[/hide]


Follow the instructions in the LaTeX page for MiKTeX.  The TeXnicCenter instructions should be updated.  In the meantime, you should follow these instructions:

-When it says "Type the full path to the LaTeX executables," just type "C:\Program Files\MiKTeX 2.5\miktex\bin." (assuming you installed it in C:\Program Files\MiKTeX 2.5)
-When it asks to configure your PostScript options, just click next.
-Click Finish.

You're done.  Check for MiKTeX upgrades often, and don't forget to download your missing packages over the time. You can install the complete MiKTeX set instead if you want to, there is no need to only install the basics.

I do not have a MiKTeX folder.... I just have a TexnicCenter folder. What path should I provide for my executables?

I had TexnicCenter working on my desktop but I got a laptop and now I need to figure out how to do it again.... :(

[quote="mgao"]-When it says "Type the full path to the LaTeX executables," just type "C:\Program Files\MiKTeX 2.5\miktex\bin." (assuming you installed it in C:\Program Files\MiKTeX 2.5)[/quote]
I think it should be MiKTeX 2.8, which is the current version.

EDIT: Oops.

Well, to answer buzzer's question: Where did you install MiKTeX? (if you didn't, you should)

I figured it out: I was trying to use TexnicCenter without MikTex... I downloaded the newest version of MikTex and it worked fine. Thanks!


Find the greatest value of real number c such that for every positive integers m, n, there always exists real number x satisfying
sin(mx) +sin(nx) $\geq$ c.

Dear Truong,

You should not post the problems of recent issues of M&Y (and any), because it will lead to some trouble.

Dear Moderators, please remove this problem (It appeared in M&Y for June 2004)

Namdung


We can write $A=J-B$ where

$J=\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{bmatrix}$ and $B= \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}$.

Some things to note:

Every power of $B$ is a permutation matrix and $B^3=I.$

$J^2=3J.$

$JB^k=B^kJ=J.$

It then follows fairly quickly (and can be proved by induction) that

$A^k=(J-B)^k=a_kJ+(-1)^kB$ where $a_{k+1}=2a_k+(-1)^k.$

That last recursion is quickly solved: $a_k=\frac{2^k+(-1)^{k+1}}3.$

So in general, $A^k=\frac{2^k+(-1)^{k+1}}3\,J+(-1)^kB^k$ and in particular,

$A^{3n}=\frac{2^{3n}+(-1)^{n+1}}3\,J+(-1)^nI.$

For instance, $A^9$ has entries of 170 on the main diagonal and 171 off the diagonal, while $A^{12}$ has entries of 1366 on the main diagonal and 1365 off the diagonal.

Of course, we could have just diagonalized the matrix. All circulants can be diagonalized in the same way (essentially the discrete Fourier transform). The eigenvalues in this case are $2,\,\frac12\pm\frac{\sqrt{3}i}2.$

Needless to say, your answers are correct, Kent Merryfield, Moubinool, thank you!

kunny


The 2008 AMC10 pdf file does not show pictures. Can someone show me how to view pictures in this file?

Thanks.

Please don't post two essentially duplicate topics in the future; if you need to change the old one, use the edit button in the lower right-hand corner.