Yale college Harvard Princeton MIT AMC USA(J)MO

Well, despite 2400 SATs/USAMO, etc, I've been rejected from the majority of my colleges. I'm quite interested in continuing to pursue theoretical math on a high level in college, and I would really like for once to be surrounded by people who are both intelligent and interested in intellectual pursuits (there's a bit of a drought at my school). My "realistic" options are really not even worth mentioning mathematically as far as I can tell, but I've also been accepted into Cornell Engineering and waitlisted at Yale. (Although, Cornell would be somewhere between a severe stretch and totally undoable financially) Does anyone know how the theoretical math options are for Cornell Engineers? I know that the focus seems to be on applied, engineering math. Also, on the off chance that I'm accepted to Yale (which would be quite the salvation, I loved that place even despite the lack of math reputation), does anybody know much about the quality of the math there?

On another note, I'm considering taking a productive gap year and probably getting a job, probably as a programmer (I had multiple offers last time I looked for work, but couldn't afford the time), among other possibilities. Has anyone had that sort of experience or know anything about it? I think it would be a chance to grow as a person before I enter another four years of education, [i]and[/i] to get a better shot at admission to a college I'd feel more at home at than my present safety options, but my parents are fairly convinced I'm never going to go back to college if I don't go now.

Thanks for the help.

The only thing that might prevent you from learning just as much pure math at Cornell as the math majors would be the (probably large number of) required courses for your engineering major.  However, keep in mind that many engineering departments are willing to cut you some slack on the "official" math requirements if you want to replace them with harder, more theoretical math classes.

By the way, Yale and Cornell both have great math departments.

Lots of people take a year away from school before college.  Is there something particular about you that makes your parents think that you won't want to go to college?  You certainly sound like someone who wants to go to college.

Finally, I'm not sure how an extra year will help you get into a college.  Of course, it depends on what you do with that year, but your SAT score certainly isn't going to improve.

[u]Comments, in no particular order:[/u]

- You can try to appeal your financial aid decision if you don't like what you get. Now that colleges are tight on money, many of them are giving lower initial offers, with the idea that those who really need more money will ask for it anyway, and those who don't ask for more money don't really need it.

- I don't understand what has your parents convinced that you won't go to college if you don't go now. Some people enter the workforce and can't imagine going back to spending lots of money instead of earning lots of money, but I don't think that would really apply to you, since a BS would obviously be a worthwhile investment and you seem like an ambitious person.

- Taking a year off to work probably [i]would[/i] increase your chances of getting in to any particular college. You'd have more interesting experience (ie, not the same boring stereotypical math/cs/science study-nerd applicant), and if you do a good job, you can get a killer employer recommendation (especially if you want to major in computer science). Still, there are lots of random factors and fluctuations in college admissions, so there are no guarantees. Make sure you stay in touch with your school community so you can get teacher recommendations as well. Try to stick in some interesting activities, and maybe some community service, along with your job, if you have time.

- There are other options, like going to a "sub-optimal" college where you can get a lot more attention (and are one of the best students), getting a killer GPA, doing some cool research or whatever, and then trying to transfer after a year. You can pick a state school or something cheap, and choose classes that will transfer. This way, you end up paying less, still get the degree from the "good" college, and may even end up having a richer freshman experience (it won't be as hardcore, and you won't be missing out on anything because you'd probably be taking a lot of intro courses).

- You don't have to be at Harvard, Princeton, or MIT to get a good math education. Most of what you'll be learning will be similar no matter where you go. What you should look at is the contents of the various courses that are offered. I know Cornell's course catalog is available on the internet.

- Cornell and Yale both have great math programs. I also know Cornell has a great Computer Science program. Also, Cornell Engineering has a great co-op program by which they can hook you up with a recruiter so you can take a semester off and work (for competitive salaries), if you can fit the rest of your coursework into 7 semesters (or are willing to take summer classes, which isn't bad, since you can work/do research in parallel).

The Yale course list does give me the sense that number theory courses are non-existent there. If you happen to like number theory, you might want to check that.

[quote="sponge008"]Well, the situation's clarified a bit - Cornell did not come through with financial aid and so ended up being unaffordable; Yale's freshman class is over-enrolled so they're not looking to take anyone off the waitlist. I've sent my deposit to Tufts, which on one hand is tuition-free thanks to parent faculty benefits, but on the other hand, I've spent quite a bit of time around campus, taken a class, and gotten to know some people that go there, and Tufts is tied for worst "fit" out of all the colleges I've visited.

Now, the question is, do I go to Tufts and transfer out, or do I take a gap year, orient myself, and then tell Tufts to take me off the reserved-places list if/when I decide to reapply to other places.

Responding to previous points:
-My parents are fairly conservative, even paranoid from my point of view - they don't have any real justification to suspect that I wouldn't go back to college (other than the afore-mentioned that I wouldn't want to go back to spending money after making it, etc), and they're the only ones that I've heard think that way about me, but their perspective is a factor in my decision nonetheless.

-I think I'd have much better admissions results if I reapplied during a gap year, for various reasons. For one, I think I terribly misrepresented myself on my applications - hearing advice to "be honest" and that most people change greatly during college, I decided that it would be unfair to colleges to give them concrete plans. Thus, I painted a rather wishy-washy portrait of myself (checked "undecided" for major although at this point I'm pretty set on mathematics or computer science, etc). Also, I've improved my profile fairly substantially over the course of my senior year; I only qualified for USAMO for the first time this spring, when decisions were probably already made, the math team I founded at my school this year did extremely well, and many other things, not to mention the boost I'd get from gap-year experiences. Also, I'd apply to more schools that suited me, but would probably take me, rather than picking safeties that I'd be horrified to end up at.

I'm leaning towards taking a gap year - am I wrong?[/quote]
i say take a gap year

[quote="jonathanchou711"]Duke is a great math place but it is probably too much money. I need to get a scholarship for Duke. Or MIT. t0rajir0u is in there.[/quote]
[quote="mathemonster"]i say take a gap year[/quote]  Perhaps you two could take self-imposed moratoria on useless posts that could not possibly be helpful or interesting to anyone?

[b]mathemonster[/b], the OP didn't post his concerns for people to vote on what he should do.

Anyway, [b]@ sponge008:[/b]

So I didn't realize you had free tuition to Tufts! Free tuition in this economy is [i]invaluable[/i]. Right now, you've submitted your deposit to Tufts, so I see no benefit in you planning on taking a year off. Most likely, I think, your best course of action would be to take a first year at Tufts and get those intro courses out of the way for free. Why?

1. You'll get a year of college out of the way for free. After that, if you don't like it, you can transfer and do your more substantial degree work somewhere else, for 3/4 the cost. Or, it may turn out that you've found a nice place there and you actually like it! Then you'll get your whole education for free.

2. While Tufts may not seem to be such a good fit for you, there's probably some good opportunities floating around. For one thing, you can still look for a programming job (though probably not full time) to do alongside your intro courses, so you can still get that experience as well. You could also get involved in campus research, or look around for other opportunities in the Boston area (there's no shortage).

3. Solid grades make a strong transfer application. You're probably more likely to get in this way, anyway.

I may or may not have convinced you to take your free "preview" year at Tufts. But either way, it's way too early for you to rule it out (since that won't benefit you to do so). Here are some things you can do over the summer that will be productive.

1. Look around for jobs in the Boston area. If you go to Tufts, maybe you'll have a great part-time job to do alongside your coursework. If you decide to take the gap year, then maybe you'll have a great full/part-time job to do.

2. Look into research opportunities/projects at Tufts. You can get a running start if you do your homework ahead of time, and will be more likely to make the best of your Tufts experience, whether it's 1 year or 4 years.

3. Read ahead on the material for your intro courses. If you go to Tufts, you'll ace your classes with your preparation, which is good whether or not you want to stay on. If you take the year off, you'll have to take those classes later anyway.

4. Anything else that you might find productive.

I concur with most of what TZF said.

I would only take the gap year if the gap year is something you want to do irrespective of where you are matriculating.  Here's a slightly different standard:  If you took the gap year and still ended up at Tufts, would you regret having taken it?  If so, then don't take it.  Sure, if you take a gap year, you might get to go somewhere better, but I think that there is a significant probability that you won't.  More importantly, your experience with Cornell shows that even if you get in somewhere better than Tufts, the financial situation might still keep you out.  So if I were you, I'd just go to Tufts and make the most of it and apply for a transfer.  You never know--you might come to like Tufts.

I'd been seriously considering a gap year even before decision season, and I think I would benefit from it even if I did end up going to Tufts anyways. I skipped a grade, so I would actually end up equalizing in age as well as, in my opinion, becoming more balanced in life overall and ready to benefit from college in other respects. The only problem I see with a gap year, other than my parents' strong resistance, is the mixup in inertia; if I end up working on a startup that takes off, I might end up putting college off by more than one year. (although I'm not sure that that situation would be a bad one, unorthodox as it may be)

In other words, I'm fairly certain my college "experience" would benefit from an extra year before it. The question is, I suppose, whether that's worth that year.

Thanks for the advice, everyone; if anyone has any more thoughts I'd be happy to hear them.

Edit/postscript: Is it true that admissions chances for transfers are better? I was under the impression that the percentages were much lower, and as grades go, my junior and senior years' are almost maxed out,

[quote="sponge008"]Is it true that admissions chances for transfers are better?[/quote]  No: transfer applicants tend to be lower priority to admissions offices than incoming freshmen.

[quote="sponge008"]Is it true that admissions chances for transfers are better?[/quote]  [quote="JBL"]No: transfer applicants tend to be lower priority to admissions offices than incoming freshmen.[/quote]

I've heard that transfer admissions are more straightforward than freshman admissions at high-ranking colleges. While I have no way to verify this claim, it does make sense to me because (1) they've seen you perform in college, so there is less guesswork and (2) they've already inducted an "interesting/balanced" class, so that factor is likely to have less effect.

If this were true, though, it might be considered by some to be "easier".

Yes, but colleges aren't trying to fill an entire class with transfer applicants, they're trying to make up for attrition of students.  The pool of available spots is much smaller and less predictable and the pool of applicants is more highly self-selected.  I don't have a huge amount of data to support my claims, but it's true of [url=http://cornellsun.com/node/26216]Cornell[/url] (the relevant data is in the table under "Arts and Sciences"), [url=http://www.yaledailynews.com/articles/view/23988]Harvard, and Yale[/url].  It seems from sniffing around that Berkeley admits a similar percentage (25-30%) of freshman applicants and transfer students, though the story will be somewhat different for public universities (see most of the other schools at Cornell, for example).  It also appears that transfer admissions to [url=http://admissions.tufts.edu/?pid=115]Tufts[/url] are substantially more competitive than [url=http://admissions.tufts.edu/?pid=121]regular admissions[/url].

(This selection of universities was largely guided by what came up on my search engine.)


The positive integers $ a,b,c$ are pairwise relatives prime. $ a,c$ are odd and the number satisfy the equation :$ a^2\plus{}b^2\equal{}c^2$
Prove that $ b\plus{}c$ is a square of an integer.

$ a^2\equal{}c^2\minus{}b^2\equal{}(c\minus{}b)(c\plus{}b)$. $ (c\plus{}b,c\minus{}b)\equal{}(c\plus{}b,2c)\equal{}(c\plus{}b,c)\equal{}(b,c)\equal{}1$, therefore $ c\minus{}b$ and $ c\plus{}b$ are square.

note that the answers of the equation $ a^2\plus{}b^2\equal{}c^2$ are in the form:

$ a\equal{}d(r^2\minus{}s^2),b\equal{}2rsd,c\equal{}d(r^2\plus{}s^2)$

where $ d\equal{}(a,b)$ and $ r,s$ are natural numbers such that $ (r,s)\equal{}1$

now in your problem we have $ d\equal{}1$ so:

$ b\plus{}c\equal{}2rs\plus{}r^2\plus{}s^2\equal{}(r\plus{}s)^2$

QED


linear algebra
matrix
linear algebra unsolved

Every matrix of the form $ AB - BA$ has trace zero.
The converse (at least over $ \bb{R}$) is also true, so that
$ dim$ $ Span \{ AB - BA : A, B \in \bb{R}^{n \times n} \} = n^2 - 1$.

What about the dimensions of the subspaces $ X^k$, where
$ X^k = Span \{ (AB - BA)^k : A, B \in \bb{R}^{n \times n} \}$ ?

Thanks.

Let $ n\equal{}2$: if $ p$ is even then $ X^p\equal{}\{\lambda{I}: \lambda\in\mathbb{R}\}$ and else $ X^p\equal{}X^1$.
Proof: Let $ U\equal{}\begin{pmatrix}a&b\\c&\minus{}a\end{pmatrix}$; $ U^{2k}\equal{}(a^2\plus{}bc)^k{I}$ and we are done for $ p$ even.
$ U^{2k\plus{}1}\equal{}(a^2\plus{}bc)^k{U}$.  $ X^{2k\plus{}1}$ contains, up to sign, the following matrices: $ \begin{pmatrix}0&1\\1&0\end{pmatrix},\begin{pmatrix}0&1\\\minus{}1&0\end{pmatrix},\begin{pmatrix}1&0\\0&\minus{}1\end{pmatrix}$
Thus $ dim(X^{2k\plus{}1})\geq3$.   QED

Nice, loup blanc.
That settles case $ n\equal{}2$.

Any suggestion for larger matrices?


Another question for you. On page 274 of the Into to Num Theory book,
there is a linear congruence of 5b + 3 = 2a +1.
From this point, it applies modulo 2 to both sides of the equation to isolate
the variables. I am not clear how you apply mod 2 to the left hand side.

Is it as simple as 5*b+3 (mod 2) = b + 1 (mod 2) because 5b/2 leaves a remainder of 1 and 3/2 leaves a remainder of 1?

[quote="baudhack"]Is it as simple as 5*b+3 (mod 2) = b + 1 (mod 2) because 5b/2 leaves a remainder of 1 and 3/2 leaves a remainder of 1?[/quote]

Yes.  :)


anyone has some techniques and tricks for the AMC? share please

This topic is currently being discussed only a few threads down, at

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24821 (formulas and specific problem-solving tools)

and

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24818 (general strategies for doing well on AMC)

Next time, please take a look at the forum or use the [url=http://www.artofproblemsolving.com/Forum/search.php]search[/url] function before posting.

One very important lesson: DO NOT GET BOGGED DOWN IN A SINGLE QUESTION!!! It is better to check easy answers (and I got the 1st question wrong on a recent practice test) than to gamble for a hard answer that may end up wrong.


Prove that for all positive integers $ n$ and all integers $ a_1,a_2,...,a_n,b_1,b_2,...,b_n$ the number $ \prod(a^{2}_{k}\minus{}b^{2}_{k})$ can be written as a difference of two squares.

[u]Proof 1 [/u]
$ (x^2\minus{}y^2)(z^2\minus{}t^2)\equal{}(xz\plus{}yt)^2\minus{}(xt\plus{}yz)^2$
[u]Proof 2[/u]
A number n can represent in the form $ x^2\minus{}y^2$ if and only if it is not in the form $ 4k\plus{}2$
Because $ a^2\minus{}b^2$ is not in the form $ 4k\plus{}2$ so product of them true.
This problem claim.

We have: $ p_n \equal{} \prod(a^{2}_{k} \minus{} b^{2}_{k}) \equal{} \prod(a_{k} \minus{} b_{k})\cdot\prod(a_{k} \plus{} b_{k}) \equal{} (A_{n} \minus{} B_{n})(A_{n} \plus{} B_{n})$
Hence $ p_n \equal{} A^{2}_{n} \minus{} B^{2}_{n}$

Right?? :)


Kind of like the other topic, but a separate discussion.
What calculators have you got throuh mathcounts, and how did you get them?

For me: 2 Ti 83+'s for making state this year and last.
1 Ti 84+ Silver for making countdown at state.

I think Texas gives more prizes than other states.  :P  :D  :rotfl:

I made countdown in the state competitions on both my times I made it...

I have zero calculators from Mathcounts

Edit: I am too slow on the uptake, Texas Instruments are in Texas  :wallbash:

TI-34II-bought in 6th grade
TI-83+SE-bought in 7th grade
TI-84+SE-won at nationals in 8th grade

For no aparant reason, I still use the TI-83+SE. I have the 84 in the package it came in.

So far, my only calculators from mathcounts has been the TI-84+ SE from last year's nationals and a really bad TI-30Xa school edition which was lost and I won in a pi reciting contest at chapter... Anyway, I made nationals again this year and really hope that we get TI-89s.

In CA, they don't give anyone calculators.  However, I got a TI84 Plus Silver for my birthday, which I used in my 7th and 8th grade competitions.

They don't give out calculators in MO either. :(

no calculators in oklahoma, either, just trophies

No calculators in FL, just a $\$$500 scholarship for top four. xD  :lol:  :D

WHAT???? 500$\$$????
In DE we get nothing

...Ohio has only trophies

The chapter competition I went to gives out 5,000 dollar scholarships to the top 2 boys and top 2 girls, but our state only has trophies,

Last year, if you got 7th, all you recieved was a pin

GO TEXAS!!! :lol: i got 2 TI-83's this year and last year. i tried to sell one of them for atleast 50, but no one wanted.
And i also had a bunch of other calculators like TI 30's and stuff... but i lost them.  :furious:

is it the sponsorship in texas that gets us so much?

I feel jealous

Texas Instruments are in Texas, so yeah, that's why

I have no calculators from mathcounts. The best calculator I have is a TI-30xa.

Well you're garunteed a better one.  I hope we don't get an 84 or 83 because I have 2 84's and I might get one later.

I'm using my sister's old TI-83+ for competitions.
It's decorated with about 15 stickers of flowers and stars and things that are impossible to remove especially because she graduated college last year and the calculator is really old  :blush:

CT doesn't give anything at all but trophies so I just want a calculator that's functional and better than this TI-83+ from MathCounts to use.

Well, it's ALMOST definite that we'll will get a 84 or a 89 or a voyager at nats this year because the others are all old or not that good overall or both.

[quote="binonunquineist"]Ti89s are really useful in hard MATHCOUNTS tests, espcially since they have quick solve, factor, and expand functions, and I can't believe we can actually use them. In practices my ti89 has been very useful many times when I can't do the problem manually.

Also, I never knew the voyager was allowed. Doesn't it hav a qwerty keyboard?[/quote]

They give it after the rounds of competition  ;)

[quote="biffanddoc"]Not last year ;)[/quote]

Then you wouldn't be allowed to use it during competition.

Which calculators are you talking about?
If they give out legal calculators (ti-84,83,89) then you can.
If they give out voyagers, then I guess you cant... :huh:
[hide]
i cat believe i am posting on aops at 1:44 in the morning :?:\
[/hide]

I was referring to the Voyagers Kevin was talking about

Off Topic: I can, and I thought my sleeping habits were just starting to become better.

I didn't care what calculator I got before, but if you look up TI-89 on wikipedia, then you can see how it is much better...

en.wikipedia.org/wiki/TI-89

Texas Instruments, please pity us poor Mathletes who work hard and slave to get to nationals, give us each a TI-89 titanium...  :yup:


Let $ABC$ be a triangle right-angled at $A$ and $S$ be its circumcircle. Let $S_1$ be the circle touching the lines $AB$ and $AC$, and the circle $S$ internally. Further, let $S_2$ be the circle touching the lines $AB$ and $AC$ and the circle $S$ externally. If $r_1, r_2$ be the radii of $S_1, S_2$ prove that $r_1 \cdot r_2 = 4 A[ABC]$.

$AB\perp AC\Longrightarrow r=s-a,\ r_a=s,\ S=\frac 12 bc=s(s-a)=rr_a.$

From http://www.mathlinks.ro/Forum/viewtopic.php?t=46418 results:

$\dfrac{r_1}{r}=\dfrac{r_2}{r_a}=\dfrac{bc}{s(s-a)}\Longrightarrow r_1\cdot r_2=rr_a\cdot \left[ \dfrac{bc}{s(s-a)}\right]^2 =4S.$

Use famouse fact to solve this problem
 [geogebra]6924254ec0f294173937496e8155ea972f0c47ef[/geogebra]


superior algebra
superior algebra solved

Clearly f(x) = x and f(x) = -x are automorphisms on Z, but how do you find the rest of them (Aut(Z)) if there are any more?

(Specify structure- you're talking about it as an additive group, not a ring)

That's it. $ 1$ must map to a generator, and once you know where it goes, you know everything.

or if you don't know about generators (although it will go into the same direction), identify for $ n\in\mathbb{N}$ the functional behavior

$ f(n)\equal{}f(\underbrace{1\plus{}\ldots \plus{}1}_n)\equal{}nf(1),f(\minus{}n)\equal{}f(n)^{\minus{}1}\equal{}\minus{}f(n)\equal{}\minus{}nf(1)$

i.e. $ 1\in f(\mathbb{Z})\Longleftrightarrow f(1)\equal{}\pm 1$


integration
inequalities proposed
inequalities

Let $ f:[-1, 1]\to {\mathbb R}  $ such that $ f^{\prime\prime} $ exists on  $ [-1, 1]$  . If    $  \left|f^{\prime\prime}\right| \le 1\;$ on  $  [-1, 1] $, then   prove that
\[\left|\int\limits_{-1}^1(3x^2-1)f(x)\; dx \right|\le \frac{4}{15}\; ,\]
with equality for  $ f_0(x)=\frac{1}{2}x^2+px+q\; ,\;\;p,q\in {\mathbb R}\; .$


quadratics
number theory unsolved
number theory

Find all integer solution to $x^3-y^3=xy+61$,

We have that $x>y$ so $x=y+d$ where $d\geq 1$
The equality rewrites as $(3d-1)y^2+(3d-1)yd+d^3=61$  from where we have that $d=1,2,\ or\ 3$

We take the quadratic which has natural roots only for $d=1$  so $y=5$ and
$x=6$

It says "integer solutions", so you have to cover the cases where $x<y$ and $xy+61<0.$

Sorry, I solved it only in positive integers.

But the remaining case isn't difficult:

From $x<y \land xy<-61<0$ it follows $x<0<y.$ Put $x=-t$ where $t\geqslant 1$ (it's obvious that neither $x$ nor $y$ can be zero). Then the equation becomes $t^3+y^3-ty=-61$, but as $t^3\geqslant t^2$ and $y^3\geqslant y^2$, then $t^3+y^3\geqslant t^2+y^2\stackrel{\scriptscriptstyle\mathrm{AM-GM}}{\displaystyle\geqslant}2ty>ty$, hence $t^3+y^3-ty>0$


This problem is a multiple choice problem with answer choices 0.21, 0.23, 0.25, 0.27, and 0.29. I seem to be solving it wrong, since I always get 0.02 off of any answer choice.

Twelve cards of which 5 are red, 4 are white,a nd 3 are blue are placed in a box. A red card is worth 1 point, a white card is worth 2 points, and a blue card is worth 3 points. Three cards are picked at random from the box. What is the probability that the total point value of the three cards is greater than or equal to 7, rounded to two decimal places?

I would appreciate it if someone would post their solution and not just the answer.

Thanks.

There are four three-card arrangements:
(1 blue, 2 white), (2 blue, 1 white), (2 blue, 1 red), and (3 blue)
The probability for (1 blue, 2 white) is $\frac{{3\choose 1}{4\choose 2}}{{12\choose 3}}$
The probability for (2 blue, 1 white) is $\frac{{3\choose 2}{4\choose 1}}{{12\choose 3}}$
The probability for (2 blue, 1 red) is $\frac{{3\choose 2}{5\choose 1}}{{12\choose 3}}$
The probability for (3 blue) is $\frac{{3\choose 3}}{{12\choose 3}}$
The total probability is then $\frac{23}{110} = 0.2\bar{0}\bar{9} \approx 0.21$


Hi,

This problem came across in a study/problem session with some friends.  Does anybody have any thoughts on it?

Let A,B,C,D be four consecutive points on a quadrilateral that lie on a circle.  Prove:

|A-C||B-D|= |A-B||C-D| + | B-C||A-D|

I'm pretty certain that littlefermat23 means Ptolemy's Theorem, especially if the letters are interpreted as affixes of the points.

A very nice proof comes once we consider inverting about one of the four points, at which point the theorem is just the inequality case for the Triangle Inequality.  Indeed, this is one of the simplest proofs of the stronger Ptolemy's Inequality.


A three-digit number has, from left to right, the digits $ h$, $ t$, and $ u$, with $ h>u$.  When the number with the digits reversed is subtracted from the original number, the units' digit in the difference of $ r$.  The next two digits, from right to left, are:

$ \textbf{(A)}\ \text{5 and 9} \qquad
\textbf{(B)}\ \text{9 and 5} \qquad
\textbf{(C)}\ \text{impossible to tell} \qquad
\textbf{(D)}\ \text{5 and 4} \qquad
\textbf{(E)}\ \text{4 and 5}$

[quote="AIME15"], the units' digit in the difference of $ r$. [/quote]

What does this mean??

GAH!  It should be "the units' digit in the difference is $ 4$".  Moderator please edit.

Set up the subtraction:
[code]  h t u
- u t h
---------
[/code]
Since $ u < h$, to subtract we would add 10 to u and subtract 1 from the upper t. So...
[code]  h t-1 u+10
- u t   h
-------------
[/code]

The units digit is u+10-h=4, so u-h=-6.
The tens digit is t-1+10-t=9.
The hundreds digit is h-1-u=6-1=5.
So the answer is 9 and 5, $ \boxed{B}$.

Also, the difference must be divisible by 99. If the units digit is 4, that means that the difference must be 6*99=594, so the digits are 9 and 5.

You could also guess and check (less advanced but just as fast in this problem).


number theory
modular arithmetic
parni10 used alc sol

How many positive three-digit integers with each digit greater than 4 are divisible by 6?

If the number is divisible by six, then the units digit must be even (6 or 8) and the numbers must add up to a multiple of 3. So, first lets list the possible numbers.

$ 0 \mod 3 \rightarrow \{6,9\}$
$ 1 \mod 3 \rightarrow \{7 \}$
$ 2 \mod 3 \rightarrow \{5, 8 \}$

[b]Case 1:[/b]
$ a \plus{} b \plus{} 6 \equiv a \plus{} b \equiv 0 \mod 3$

$ (0,0), (1,2), (2,1) \mod 3$
$ 2 \cdot 2 \plus{} 1 \cdot 2 \plus{} 2 \cdot 1 \equal{} 8$

[b]Case 2:[/b]
$ a \plus{} b \plus{} 8 \equiv a \plus{} b \plus{} 2 \rightarrow a \plus{} b \equiv 1 \mod 3$

$ (1,0),(0,1),(2,2)$
$ 1 \cdot 2 \plus{} 2 \cdot 1 \plus{} 2 \cdot 2 \equal{} 8$

In total, there are $8+8= \boxed{16 \ numbers}$.

Note: I don't know if this is the fastest way to do this problem, but it is the fastest I can think of right now. If you are comfortable with modular arithmetic, it shouldn't take too long.

This isn't right.  Can someone post a correct solution?

I edited it. It should be 16, not 8, since I forgot to add the results from each case.

The official solution is much simpler; I used a bash method involving selecting the hundreds digit first, but selecting the ones digit first yields an excellent solution!

why is this wrong?

There are $5$ options for the first digit, $6$ options for the second digit, and since we want it to be divisible by $6$ the last digit can only take evens which means there are $3$ options. Now by symmetry exactly a third of these are divisible by $3$. So our answer is $5 \cdot 6 = 30$?

Take this with a grain of salt as I haven't seen this problem in a while:  it's actually not symmetric; if you count out all integers, the number of non-multiples would [b]not[/b] be two times the number of multiples.

It's kind of like if I asked you "How many multiples of 3 are between 0 and 10," you can't use symmetry.

Ok, but how is this problem supposed to be state sprint if it takes this long :(

[quote=618173]ummm who else just literally listed? bad problem imo[/quote]

yeah same i just made a list

[quote=jason.ca]Don’t revive posts from 2009[/quote]

Last post was from 2021 also I clicked the discuss button
If you can’t revive threads started in 2009 then am I not allowed to discuss the problem??


Find all pairs of real numbers $ (x,y)$ such that:

$ \frac{x\minus{}2}{y}\plus{}\frac{5}{xy}\equal{}\frac{4\minus{}y}{x}\minus{}\frac{|y\minus{}2x|}{xy}.$

[hide]$ \frac {x \minus{} 2}{y} \plus{} \frac {5}{xy} \equal{} \frac {4 \minus{} y}{x} \minus{} \frac {|y \minus{} 2x|}{xy}$

$ x^2 \minus{} 2x \plus{} 5 \equal{} 4y \minus{} y^2 \minus{} |y \minus{} 2x|$

$ x^2 \minus{} 2x \plus{} 1 \plus{} y^2 \minus{} 4y \plus{} 4 \equal{} \minus{} |y \minus{} 2x|$

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

Since $ (x \minus{} 1)^2 \plus{} (y \minus{} 2)^2 \ge 0$ and $ \minus{} |y \minus{} 2x| \le 0$, the only solution occurs when:

$ (x \minus{} 1)^2 \plus{} (y \minus{} 2)^2 \equal{} \minus{} |y \minus{} 2x| \equal{} 0 \Rightarrow (x,y) \equal{} (1,2)$.[/hide]


trigonometry
logarithms
real analysis
real analysis unsolved

prove that $ \sum_{n\equal{}2}^{\infty}\frac{\sin n}{\ln n}$ converges, but $ \sum_{n\equal{}2}^{\infty}\frac{|\sin n|}{\ln n}$ not

[quote="rex88"]prove that $ \sum_{n \equal{} 2}^{\infty}\frac {\sin n}{\ln n}$ converges, but $ \sum_{n \equal{} 2}^{\infty}\frac {|\sin n|}{\ln n}$ not[/quote]

For the first series, use Dirichlet's test. For the second, notice that if  |sin(n)| > 1/sqrt(2), then one of |sin(n+3)|, |sin(n+4)| is > 1/sqrt(2) as well (think geometrically!)

For the second, there is a useful trick: $ |\sin n|\ge sin^2 n \equal{} (1 \minus{} \cos 2n)/2$. You will get a divergent series minus a convergent series.


Square BCEF is inscribed in right trangle AGD. B Is on AD (23 units from A). C is on AD (58 units from D). What is the area of square BCFE?

I'll post the picture to save people from having to listen to this. :P
[color=red][size=75][Moderator rcv's note:  Several content-free posts (not by easyaspi) were removed from this thread.  Thanks to easyaspi for posting a picture.][/size][/color]
This is the question: Square BCFE is inscribed in right triangle AGD, as shown below. If AB=28 units and CD=58 units, what is the area of square BCFE?

Honestly i dont have a clue on this kinda stuff...

But please draw a SQUARE... it really confuzles me

[hide="hint"]
Look for similar triangles
[/hide]

[hide]
I will assume that
triangle ABE and triangle FCD are similar.
[color=blue]Moderator kyyuanmathcount's note: Something was wrong with the original LaTeX, I just took it out.[/color]
Since they both have the side of the square as a side, we can call it $x$.
Thus
$\frac{28}{x}=\frac{x}{58}$
$1624=x^2$

That is our answer.

I tried proving that they were similar and failed (probably calculation mistake)
[/hide]

[hide="another hint"]What conclusions can we draw from the fact that lines EF and AD are parellel?  Any equal angles?[/hide]

i did it like biffanddo did it

oops i multiplied wrong    :blush:

Biffanddoc is correct, that's how I did it.

Please don't bump old posts. This one is around 11 years old. If you have a comment or question on the problem, just post the problem again in a new post.

EDIT: Welcome to AoPS! :-D

[quote=BananaPnaplsXie]@icosahedron lol he joined
August 2016
It's just his first post[/quote]

Oh. Well, still congratulations on your first post Sam2005!


I hope mr. mod will make it a sticky...and transfer any irrelevant chatting to the metathread (which will be pretty often :P )

After burning fingers 4 months back by installing ubuntu 7.04 in my desktop (which ultimately forced me to format), I have decided to renew my affair but this time on my laptop...

I installed ubuntu 7.10 yesterday...

As pardesi is a ubuntu veteran I would like to be guided by him and others who have some idea about it...

[b]Question 1:[/b]
How do I change my password?

[b]Question 2:[/b]
Now if I add up my drive memories in Windows Vista, it is 10 GB less than should be....And there is another drive (H:) which needs to be formatted to be able to use it...
But when I formatted it, ubuntu was lost and poor lappy did not start with all kind of weirdo frightening messages and symbols..

I had to use live CD to reinstall it...

So what should I do to get back my 10 GB??

posting from XP...will tell u once i switch back to ubuntu :D

I moved this here to get better views...could have moved this to Computer Science and informatics but don't think it has lots of visitors :D
I think many people come here so ...assuming people use Ubuntu linux u may get quick help :)

There's no reason that this should be limited to ubuntu users. Both of these questions apply to any variation of linux. I use gentoo, personally.

1) open up a terminal, type passwd. There might be a way to do it graphically, but I don't know.

2) The reason you're "missing" 10 GB is called partitioning. Linux uses a different file system than windows. Windows uses NTFS and linux uses EXT3. When you installed linux, you set up a 10GB partition that is EXT3. Well, windows can't use EXT3, so of course it looks like it's missing. The best solutions are to just deal with the smaller amount of disk space or to only run one operating system instead of two. There are ways to have the two operating systems share disk space, but this is not ideal, especially since many things won't work on both sides anyway.

[quote="SOUMYASHANT NAYAK"]Question 3:
I know this sounds stupid but still...
Can I install a software on an external hard drive and use it just by connecting it to the comp?[/quote]

Actually there are operating systems which can be installed on USB Drives *Linux ofc*.... As for the software certain software like Open Office have portable versions just for the purpose you mentioned.

But not all. I'm sure Microsoft word or Outlook would not support such a feature.

[quote="SOUMYASHANT NAYAK"]Ok,
then I suppose loading linux on USB is fine...
will it be very slow...?[/quote]

I believe there is this OS called DSL (D.a.m.n Small Linux) which is supposedly fast.

Other than that I don't have much personal experience.

Flash drives are relatively fast storage devices, the slow part being the USB port. Software you put on the flash drive should work relatively well. (I put firefox 3 on mine)


BackHo[WHITE] has just qualified to the OSL Round of 16, after a stunning disqualification of GoRush in Game 2 of the Best of 3 Series in the first leg of the Round of 36 group C. BackHo swiftly crushed Tester[gm] 2-0 thereafter.

GoRush was playing one of his solid matchups (ZvP) against BackHo's worst (PvZ). The disqualification in the second game occurred after BackHo was up 1-0, and GoRush accidentally typed "a" in the beginning of the second game.

Thoughts?
 Personally I feel that, even though the proceedings were massively unfortunate for GoRush's chances, that there was nothing KesPa could do to save GoRush. After all, he had already violated the rules in the close of game 1, by mistyping GG. With just the "a", I can see how KesPa might've been lenient. But two in one night? That's a tougher sell.

--btw read backho's post-game interview, where he talks about how he got zizi yO and variants banned, and how ironically that leads to gorush's ban just earlier today.

Hmmm I have not been watching recently because my mom is like, SCREW YOUR INTERNET AFTER 10:30 PM.
I should probably practice more, so I can get my APM up from the current pathetic 70...

[quote="not_trig"]zizi yO gorush~ :/

also lol tester[/quote]
Quote from plexa (from memory):
"Tester is way ahead" (4 gates to backho's 3, and equal reavers, tester up an expo)
"And wtf... just happened. Backho has this now."

Do you guys mind explaining what you're talking about?
I have a feeling this is about Starcraft but why are we talking about this on the Bay Area forums?

[quote="Scrambled"][quote="junggi"][quote="Scrambled"]BackHo will make it to / win the semifinals.[/quote]
hey no copying my predictions[/quote]
What? I've predicted this every single starleague since before the Ever.[/quote]
what not in batoo

[quote="junggi"][/quote][quote="Scrambled"][quote="junggi"][quote="Scrambled"]BackHo will make it to / win the semifinals.[/quote]
hey no copying my predictions[/quote]
What? I've predicted this every single starleague since before the Ever.[/quote][quote="junggi"]
what not in batoo[/quote]
Are you serious? Everyone knew in batoo that he would descend from his surgery bed to play in finals against Jaedong.
Which by the way, actually did happen. Fantasy was just a disguise for the Sheer Awesome of the Starleagues.

in other news my fantasy team is getting owned. zero < light what? at least horang won, but then ACE WON AND OOPS REACH -1. (whoa pun I am so cool)

so

oz is failing. jaedong is blearghing, lomo is random, backho SUCKS.
ktf is failing. luxury sucks, forgg sucks, flash is mediocre.
khan is failing. great is now their best player lolwhat?
Stars got 3-0'd last night by MBC.
ESTRO WON.

ACE is not failing.
SKT is definitely not failing, nor is CJ.

what

is

up

they seriously have fantasy leagues for.... Starcraft?

Curling I could imagine.  But STARCRAFT?

how is sc even a sport

no flaming my heresy please


algebra
polynomial
geometry
function
calculus
derivative
complex numbers

Kind of interesting, y=c/x where c are constants ranging from -3 to 3.

Table[
 Plot[
  Table[
   c/x, {c, -3, 3, f}
   ], {x, -3, 3}
  ], {f, {.16, .13, .1, .05, .01}}
 ]

What would it look like if you were to graph the tangent line at every point (or a sufficient number) on an interval?

This image displays x^2, x^3, sin(x), and e^x, respectively.

plot:=Plot[
  Table[
   f'[c] (x - c) + f[c],
   {c, -4, 4, .05}],
  {x, -4, 4},
  AspectRatio -> Automatic,
  PlotRange -> {{-4, 4}, {-5, 5}}
  ]
{
 f[x_] := x^2;
 plot,
 f[x_] := x^3;
 plot,
 f[x_] := Sin[x];
 plot,
 f[x_] := Exp[x];
 plot
 }

Obtained by rotating a figure-8 around and axis and putting a twist in it, three dimensional immersion.

ParametricPlot3D[
 {
  Cos[u] (3 + Cos[u/2] Sin[v] - Sin[u/2] Sin[2 v]),
  Sin[u] (3 + Cos[u/2] Sin[v] - Sin[u/2] Sin[2 v]),
  Sin[u/2] Sin[v] + Cos[u/2] Sin[2 v]
  },
 {u, 0, 2 Pi},
 {v, 0, 2 Pi},
 Mesh -> {{5}},
 MeshFunctions -> {#1^2 + #2^2 &},
 MeshShading -> {Opacity[0.5, Yellow],
   Opacity[0.5, Red]},
 Axes -> False
 ]

Just a simple textured sphere.

ParametricPlot3D[
 {
  Sin[u] Sin[v] + 0.05 Cos[20 v],
   Cos[u] Sin[v] + 0.05 Cos[20 u],
   Cos[v]
  },
 {u, -\[Pi], \[Pi]},
 {v, -\[Pi], \[Pi]},
 MaxRecursion -> 4,
 PlotStyle -> {
   Orange,
   Specularity[White, 10]
   },
 Axes -> None,
 Mesh -> None
 ]

Inspired by the Wolfram Demonstration Project.

Contour plot of a polynomial with zeros at the lattice points m+i*n.  Note that this is the real part, the imaginary part is upcoming.

ContourPlot[
 Evaluate
  [Re[Product[x + I y - (a + I b),
    {a, -2, 2},
    {b, -2, 2}]
   ]
  ], {x, -3, 3}, {y, -3, 3}
 ]

Inspired by the Wolfram Documentation Center

A contour plot of a polynomial with zeros at the lattice points m+i*n.  Note that this shows the imaginary part.

ContourPlot[
 Evaluate
  [Re[Product[x + I y - (a + I b),
    {a, -2, 2},
    {b, -2, 2}]
   ]
  ], {x, -3, 3}, {y, -3, 3}
 ]

Inspired by the Wolfram Demonstration project.

oooo nice picture, Stan.  You're very intelligent.

Stank, you are a genius!  I do not know how you come up with these amazing mathematical ideas.

The area of the region between norm balls, inspired by Wolfram Documentation Center.

RegionPlot3D[
 1 <= Norm[{x, y, z}, 1] && Norm[{x, y, z}] <= 1,
 {x, -1, 1},
 {y, -1, 1},
 {z, -1, 1},
 Mesh -> None, PlotStyle -> Directive[Opacity[0.5],
   Pink,
   Specularity[White, 20]],
 PlotPoints -> 60,
 NormalsFunction -> "Average",
 Axes -> False
 ]

Cool convergence of $ {(1 \plus{} \frac {1}{x})}^x$.
This approaches e as $ x\to\infty$.
It starts looking pretty logarithmic like one would expect.
Then, around 5000000, there is some irregularity.
By x=100000000, the graph is oscillating out control.

Table[
 Plot[
  {
   (1 + 1/x)^x,
   E
   }, {x, 1, 10^n}],
 {n, 6, 9}
 ]

The graph appears to be oscillating but this is due to the lack of sufficient machine precision to evaluate $ (1 \plus{} 1/x)^x$ for very large $ x$.  It is an artifact of the computation, not of the function itself, since we are able to easily show that this function has a positive first derivative for $ x > 1$.  Also observe that the scale on the y-axis becomes increasingly fine, which is misleading when interpreting the scale of these apparent fluctuations.

These plotting artifacts can be removed by increasing the working precision.  If we use the default options, e.g.,

Plot[(1+1/x)^x, {x, 10^7, 2*10^7}]

the plot is jagged.  This is because WorkingPrecision -> $ \$$MachinePrecision by default (the value of MachinePrecision varies by system; on mine, it is roughly 16).  If we instead write

Plot[(1+1/x)^x, {x, 10^7, 2*10^7}, WorkingPrecision -> 100],

the plot is now smooth, as expected.


conics
ellipse
geometry
geometry proposed

Given a and b as the semimajor and semiminor axis of an ellipse, respectly, find the maximum área of a triangle inscribed in this figure.

I made it transforming the ellipse in a circunference, by linear transformations. But i couldn't find any proof using only pure geometry.

Can you think about it???

Thanks!  :)

By pure geometry do you mean by a compass and ruler construction?


function
analytic geometry
inequalities
quadratics
real analysis
advanced fields
advanced fields unsolved

Suppose n, m are non-negative integers, for real-valued function W(n, m) (between 0 and 1), if n<=m(m>0), then W(n, m)=1 and W(n, 0)=0. We have the following recursive equation:
W(n, m)= W(n-2, m)*n/(n+m) + W(n-1, m-1)*m/(n+m).

I conjecture that if f(m)<m^2 (here '<' means smaller in order, e.g., f(m)=2m), then lim_{m->Inf} W(f(m), m)=1 (Inf means Infinity). And if f(m)>m^2 (here '>' means bigger in order, e.g., f(m)=m^3), then lim_{m->Inf} W(f(m), m)=0.
If f(m)~m^2 (~ means with the same order), then lim_{m->Inf} W(f(m), m)=c and 0<c<1?

But I not know how to prove this even if when f(m)=2m. Thanks very much.

Your rough conjectures (about $ n\gg m^2$ and $ n\ll m^2$) are correct and can be proved by comparing $ W$ with $ \varphi\left(\frac {m^2}{n \plus{} Cm}\right)$ with appropriately chosen increasing positive function $ \varphi$ and constant $ C > 0$ (just do the second order Taylor formula to construct the appropriate majorant and minorant). The question about the existence of the limit is more subtle. I am inclined to believe that your conjecture is true but the proof will, probably, require a lot of care (at least, I couldn't prove it in an hour :( )

Thanks to fedja. I have tried your method but failed. Could you show your construction of the majorant and minorant of the simple case W(2m, m)? I think there may exist some problems you have not considered. Looking forward to your reply!

OK. Here goes one half of it. I'll prefer to denote the coordinates by $ x$ and $ y$ rather than $ n$ and $ m$, so we have $ W(x,0)\equal{}0$, $ W(x,y)\equal{}1$ when $ y\ge x>0$, and the recurrence relation is
\[ W(x,y)\equal{}\frac{1}{x\plus{}y}[xW(x\minus{}2,y)\plus{}yW(x\minus{}1,x\minus{}1)]\]
for $ x,y>0$, $ y<x$.
The upper extimate $ W(x,y)\le 1$ is obvious. To get the lower one, it is enough to construct a smooth function $ M(x,y)$ such that $ M(x,y)\le W(x,y)$ on the boundary and
\[ M(x,y)\le \frac{1}{x\plus{}y}[xM(x\minus{}2,y)\plus{}yM(x\minus{}1,x\minus{}1)]\]
for large $ x$ and $ 0<y<x$.
The difference between the right hand side and the left hand one can be written as $ \frac 1{x\plus{}y}$ times
\[ \minus{}(2x\plus{}y)M_x(x,y)\minus{}yM_y(x,y)\plus{}\frac 12[4xM_{xx}(\zeta,y)\plus{}y(M_{xx}(\xi,\eta)
\plus{}2M_{xy}(\xi,\eta)\plus{}M_{yy}(\xi,\eta))]\]
where $ \zeta\in(x\minus{}2,x)$, $ \xi\in(x\minus{}1,x)$, $ \eta\in(y\minus{}1,y)$.

Now let us denote $ t\equal{}t(x,y)\equal{}\frac{y^2}{x}$ and put $ M(x,y)\equal{}\varphi(t(x,y)$. Assume that $ \varphi$ is increasing, $ varphi(0)\equal{}0$ and $ \varphi\le 1$. Then the boundary conditions are clearly satisfied, so the crucial thing is to check the recurrence inequality. Note that
\[ (\minus{}2x\plus{}y)M_x(x,y)\minus{}yM_y(x,y)\equal{}\varphi'(t(x,y))\frac{y^3}{x^2}\]
Also,
\[ d^2 M\equal{}\varphi'(t) d^2 t\plus{}\varphi''(t) (dt)^2\]
Now, $ t(x,y)$ is convex (check it yourself), so the first term in $ d^2M$ only helps us. Restricting our attention to the second term, we see that
\[ dt(\alpha,\beta)\equal{}\frac{2\alpha}{\beta}dy\minus{}\frac{\alpha^2}{\beta^2}dx\]
Since in our Taylor formula we have $ \alpha\le y$ and $ \beta\approx x$ (if needed, we can assume that everything starts with $ x\ge 10$. This will force us to check the inequality $ M\le W$ inside for $ x<10$, but it won't ne a problem), we conclude that
\[ |dt(\zeta,y)(\minus{}2,\minus{}0)|\le A\frac{y^2}{x^2},\qquad |dt(\xi,\eta)(1,1)|\le A\frac{y}x\]
with $ A\equal{}100$, say (here we used that $ y<x$, of course).
Squaring and multiplying by $ x$ and $ y$ respectively, we see that the quadratic term is at least
\[ \minus{}A^2|\varphi''(t')|\left[\frac{y^4}{x^3}\plus{}\frac{y^3}{x^2}\right]\ge
\minus{}2A^2|\varphi''(t')|\frac{y^3}{x^2}\]
The only little problem is that $ t'$ is not exactly $ t$, but, since $ t(x,y)$ is Lipschitz with constant $ 10$ in $ y\le x$, we see that $ t'\ge t\minus{}20$. Now all we need is to find $ \varphi$ such that $ \varphi''$ changes only slightly on each interval of length $ 20$ and $ \varphi'\gg |\varphi''|$ everywhere. The direct inspection shows that $ \varphi(t)\equal{}1\minus{}e^{\minus{}\delta t}$ with very small $ \delta>0$ fits all the requirements.

This takes care of $ y^2\gg x$. The second case is a similar except you should choose $ C$ large and it is the term with $ \varphi'$ in $ d^2M$ that is bad now. Try to do it yourself. If you fail again or if you have any other questions, just ask. :)


calculus integration inequalities function quadratics algebra polynomial

Let $f:[0,1]\to\mathbb{R}$ be an integrable function, such that:

                  $\int_{0}^{1} f(x)dx=\int_{0}^{1} xf(x)dx=\int_{0}^{1} x^{2}f(x)dx=1$.

 Prove that:  $\int_{0}^{1} f^{2} (x)dx\geq 9$ ;)

[hide="Oops."]The conditions $\int_0^1f(x)\,dx=1$ and $\int_0^1x^2f(x)\,dx=1$ are red herrings here.

Of course, we use the Schwarz-Bunyakovsky inequality:

$\left[\int_0^1xf(x)\,dx\right]^2\le\left[\int_0^1x^2\,dx\right]\left[\int_0^1f^2(x)\,dx\right]$

$1\le\frac19\int_0^1f^2(x)\,dx$[/hide]

We should really be working in $L^2[0,1]$ here, although of course the inequality will hold in any subspace, and is trivially true if the integral of the square is infinite.

This is about orthogonal projection: the minimum is achieved when we project 0 onto the translated space defined by the three conditions.

First, we find a point in the space we want to project onto. For quadratic polynomials $p(x)=ax^2+bx+c$, we have $\int_0^1p(x)\,dx=\frac a3+\frac b2+c$, $\int_0^1xp(x)\,dx=\frac a4+\frac b3+\frac c2$, $\int_0^1x^2p(x)=\frac a5+\frac b4+\frac c3$. If we set $q(x)=30x^2-24x+3$, we have $\int_0^1q(x)\,dx=\int_0^1xq(x)\,dx=\int_0^1x^2q(x)\,dx=1$.

Now, we find an appropriate basis for the Hilbert space. The shifted Legendre polynomials $p_n(x)=\left(\frac d{dx}\right)^n(x-x^2)^n$ are orthogonal, and are a basis for the dense subspace of polynomials. Since $p_n$ has degree $n$, $p_n$ is orthogonal to all polynomials of degree less than $n$, and therefore $q$ is orthogonal to $p_n$ for $n\ge3$. Since the space we want to project to is the closure of $q+\mathrm{span }\{p_n: n\ge3\}$, $q$ has minimal norm among all such polynomials, and all that remains is to calculate this norm.

$\int_0^1q(x)^2\,dx=9\int_0^1 100x^4-160x^3+84x^2-16x+1\,dx=9\cdot(20-40+28-8+1)=9$

very nice, jmerry. My solution is the same with yours. ;)

I'm sorry. My solution isn't exactly the same. After, i found the polynomial, i applyed Cauchy to that polynomial and to $f$. jmerry can u explain your solution in detail after you found the polynomial? :?

$f(x)=q(x)+\sum_{n=3}^{\infty}a_np_n(x)$
$\int_0^1f^2(x)\,dx=\int_0^1q^2(x)+\sum_{n=3}^{\infty}a_n^2\int_0^1p_n^2(x)\,dx$ by orthogonality. The second sum is always nonnegative, and we have the desired inequality.

This would have been much uglier with any other orthogonal basis of $L^2[0,1]$.


geometry
3D geometry
group theory
abstract algebra
factorial
number theory proposed
number theory

$ m^2 \plus{} 7$ is a cube for a number $ m\in\mathbb{Z}$.
then $ m\pm 1$ ? or there are other solutions ?

If $ x^3 \equal{} m^2 \plus{} 7$ and $ x,m\in\mathbb{Z}$ then $ x \equal{} 2,m \equal{} \pm 1$ thus $ m\pm 1\in \{ \minus{} 2,0,2\}$. I don't understand the question.

I think he/she is asking whether the equation $ x^3\equal{}m^2\plus{}7$ has any solutions other than $ m\equal{}\pm 1$ in $ \mathbb{Z}$.

Yes, exactly. I asked about nontrivial solutions to this equation.

MeKnowsNothing, could you justify yuor answer?

The curve I wrote has rank $ 1$ and a torsion subgroup isomorphic to trivial group. The group law shows that there is only one solution in integers.

You could also do some arithmetic in $ \mathbb Z [\frac{1\plus{}\sqrt{\minus{}7}}{2}]$.

yes, I was quite sure one can use properties of elliptic curvers
it's very strong theory but quickly gives nice results in NT
thanks :))

ZetaX could you write something more about your method ?

As I was requested to get into detail:

We want to solve $ x^2\plus{}7\equal{}y^3$ for integers. For this we work in the ring $ R\equal{} \mathbb Z \left[ \frac{1\plus{}\sqrt{\minus{}7}}{2}\right] \equal{} \left\{ \frac{a\plus{}b \sqrt{\minus{}7}}{2} | a \equiv b \mod 2 \right\}$ which is factorial (this means: factorisation into prime factors exists and is unique up to permutation and sign). One checks that $ 2$ is prime in $ R$.

Write the equation as $ (x\plus{}\sqrt{\minus{}7})(x\minus{}\sqrt{\minus{}7})\equal{}y^3$ and note that $ \gcd(x\plus{}\sqrt{\minus{}7},x\minus{}\sqrt{\minus{}7}) \equal{} \gcd(x\plus{}\sqrt{\minus{}7},2 \sqrt{\minus{}7})$ which divides $ \gcd(2x\plus{}2\sqrt{\minus{}7},2\sqrt{\minus{}7}) \equal{} \gcd(2x,2\sqrt{\minus{}7})$ which again divides $ 14 \equal{} 2 \cdot 7$. But if the last $ \gcd$ isn't coprime to $ 7$, then $ 7|x$, thus $ 7|y^3$, thus $ 7|y$, finally giving a contradiction $ 7^2|7$ (all these divisibilities being in $ \mathbb Z$). Thus $ \gcd(x\plus{}\sqrt{\minus{}7},x\minus{}\sqrt{\minus{}7})$ divides $ 2$, so it is either $ 1$ or $ 2$.

1. case: $ x$ is even.
Then $ \frac{x \plus{} \sqrt{\minus{}7}}{2}$ is not in $ R$, so $ \gcd(x\plus{}\sqrt{\minus{}7},x\minus{}\sqrt{\minus{}7})\equal{}1$ (as there is no factor $ 2$ on the left term). Using coprimality and unique factorisation on the third power $ (x\plus{}\sqrt{\minus{}7})(x\minus{}\sqrt{\minus{}7})\equal{}y^3$, we get that $ x\plus{}\sqrt{\minus{}7}$ is a third power itself.
This gives us the existence of integers $ a,b$ with $ x\plus{}\sqrt{\minus{}7}\equal{} \left(\frac{a\plus{}b\sqrt{\minus{}7}}{2}\right)^3$. Equivalently, $ 8x\plus{}8\sqrt{\minus{}7} \equal{} a^3\minus{}21ab^2\plus{}\sqrt{\minus{}7}(3a^2b\minus{}7b^3)$. Comparing imaginary parts gives $ 8 \equal{} b(3a^2\minus{}7b^2)$, so $ b|8$. In total, we now just have to check this finitely many cases (for every $ b$ there are at most two possible $ a$); left to the reader (there will be no solution).

2. case: $ x$ is odd.
Then $ \frac{x \plus{} \sqrt{\minus{}7}}{2}$ is in $ R$ and $ \gcd(x\plus{}\sqrt{\minus{}7},x\minus{}\sqrt{\minus{}7})\equal{}2$. So we get $ \gcd\left( \frac{x \plus{} \sqrt{\minus{}7}}{2} , \frac{x \minus{} \sqrt{\minus{}7}}{2} \right)\equal{}1$. The same reasoning as before gives $ \frac{x \plus{} \sqrt{\minus{}7}}{2} \equal{} \left(\frac{a\plus{}b\sqrt{\minus{}7}}{2}\right)^3$, resolving for $ 4\equal{} b(3a^2\minus{}7b^2)$ (so $ b|4$), which this time has two solutions $ b\equal{}1, a\equal{} \pm 1$. Going backwards, this gives $ x\equal{}\pm 1$, as we wanted to show.


Do there exist polynomials $p(x), q(x), r(x)$ whose coefficients are positive integers such that $p(x) = (x^{2}-3x+3) q(x)$ and $q(x) = (\frac{x^{2}}{20}-\frac{x}{15}+\frac{1}{12}) r(x)$?

[quote="Rust"]$r(x)=60r_{1}(x), r_{1}(x)\in Z[x]$ work.[/quote]
Please read carefully problem!

[quote="Rust"]I am sorry. Let $r(x)=60(3x^{2}+4x+5)(x^{2}+3x+3)r_{1}(x)$, were all coefficients of $r_{1}(x)$ are positive integers.
Then $q(x)=(x^{2}+3x+3)q_{1}(x), \ q_{1}(x)=(9x^{4}+14x^{2}+25)r_{1}(x)$,
therefore $p(x)=(9x^{8}-13x^{6}+64x^{4}+51x^{2}+225)r_{1}(x)$.
All coefficients of $p(x),q(x),r(x)$ positive and integer if $r_{x}(x)=(x^{2}+2)(r_{2}(x)+1+x)$, were $r_{2}(x)$ had only not negative integer coefficients.[/quote]
What is $r_{x}(x)$? I will check your solution carefully.

Let $r(x)=60(1+x)^n$, where $n$ is large enough.  This works by this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1219057&sid=3337c041849ded52fc6251ebe519223d#p1219057]theorem[/url] and the fact that $x^2-3x+3$ and $\frac{1}{60}(3x^2-4x+5)$ are both strictly positive for all $x$.


If the value of $ x$ is twice the value of $ y$, what is the value of $ x^2$ in
terms of $ y$?


algebra
function
domain
calculus
calculus computations

hi guys,
i am just confused over this one.

differential equation;
y'+2y=r(t)     y(0) = 0

where r(t) = 1   when t < 1   and   r(t)=0    when   t > 1
so there will be 2 solutions to the equation one for each interval
so far what i did

[b]y'+2y = 1 [/b]             ( when t  < 1 )

solving it i get  [b]1/2 [1 -  e^(-2t) ] [/b] (this is correct in the book) for t < 1

y' + 2y = 0  (when t  > 1 )
solving it i get y = Ke^-2t   now y(0) = 0 so k = 0 ??
i believe i have to make it continuous for t => 1 and find the other solution for t > 1
btw answer is ( 1/2 ( e^2-1)e^-2t for t > 1
thanks!

For the second half of the problem (behavior beyond 1), you should ignore the condition at zero; that's outside the domain of your new solution. Instead, we have the initial condition $y(1)=\frac12\left(1-e^{-2}\right)$, chosen to ensure continuity. That is, of course, simply your solution for $x\le 1$, evaluated at 1.


I know combinatorics is such a general field, but are there any downloads for some of the various subtopics?

Check out Reid Barton's MOP page for some handouts.
 http://web.mit.edu/rwbarton/Public/mop/


limit
logarithms
calculus
calculus computations

Calculate,

$ \lim\limits_{x\rightarrow 0}\frac{\prod\limits_{k\equal{}1}^{n}(e^{k x}\minus{}1)\minus{}\prod\limits_{k\equal{}1}^{n}\ln(1\plus{}kx)}{x^{n\plus{}1}},$

where $ n\geq 1$ is a fixed natural number.

[hide="answer"]Taylor expanding we have,
$ \prod\limits_{k \equal{} 1}^{n}(e^{k x} \minus{} 1) \equal{} n!x^n \plus{} \frac {n(n \plus{} 1)!}{4}x^{n \plus{} 1} \plus{} \mathcal{O}\left(x^{n \plus{} 2}\right)$

$ \prod\limits_{k \equal{} 1}^{n}\log \left(1 \plus{} kx\right) \equal{} n!x^n \minus{}\frac {n(n \plus{} 1)!}{4}x^{n \plus{} 1} \plus{} \mathcal{O}\left(x^{n \plus{} 2}\right)$

Putting everything together,

$ \lim\limits_{x\rightarrow 0}\frac {\prod\limits_{k \equal{} 1}^{n}(e^{k x} \minus{} 1) \minus{} \prod\limits_{k \equal{} 1}^{n}\ln(1 \plus{} kx)}{x^{n \plus{} 1}} \equal{} \frac {n(n \plus{} 1)!}{2}$[/hide]

How can we get that series ?? please help me


calculus
integration
function
Functional Analysis
real analysis
real analysis unsolved

a)Let X be a Banach space and let T be a bdd lin map from X--->X which is 1-1. prove that if {x_n} is a sequence in X such that ||x_n||----->inf, then ||T(x_n)||------>inf

b)Let X=C[0,1] with usual sup norm and define T:X----->X by Tf(t)=integral f(s) ds from 0 to t for all f in X and t in [0,1]. Show that there is a sequence {f_n} in X such that ||f_n||---->inf and ||T(f_n)||=1 for all n. Why doesnt this contradict the result in a)

Problem (a) is not true as stated. A simple counterexample:
Let $C[-1,1]$ be the space of continuous functions on $[-1,1]$, with the uniform norm. Let $T(f)(x)=xf(x)$. $T$ is injective with norm 1, by routine calculations. Let $F(x)=\begin{cases}1-|x|&|x|<1\\0&|x|\ge1\end{cases}$, and let $f_n(x)=nF(nx)$. The norm of $f_n$ is $n$, and the norm of $Tf_n$ is $\frac14$ for all $n$.

Actually, part (b) also gives a counterexample closely related to mine. I can come up with counterexamples in every infinite-dimensional Banach space I know.

Is there something missing from the statement? Is $T$ also meant to be surjective?

If the map is both one-to-one and onto, then that's that: it's invertible. Not only in the algebraic sense (that's obvious), but in the topological sense as well, i.e. its inverse is continuous. This is because on the one hand it's a continuous bijection, and on the other hand it's also an open map by the Open Mapping Theorem (stating that an onto map between Banach spaces is an open map).

is the reason for b) not contradicting a) because it is not a banach space with the given norm? why does the operator have norm 1 for a sequence that goes to inf?

No, it's that the mapping $T$ is not onto. If $g=Tf$, then $g$ is a $C^1$ function with the property that $g(0)=0.$ That's not all of $C[0,1],$ for multiple reasons.


i read a statement written in the book by Morrison& Boyd which goes like this
[b]In E2 dehydrohalogenation of alkyl halides the order of reactivity of Alkyl Halides is : $3^{o}>2^{o}>1^{o}$[/b]
but $3^{o}$ alkyl halides primarily follow E1 mechanism and evne if the cond.s favour the E2 mechanism they react quite slowly.what is then the meaning of the above statement.

That's a true statement, but it also depends on nucleophile/base used and the temperature.

[quote="Pardesi"]but 3º alkyl halides primarily follow E1 mechanism and even if the cond.s favour the E2 mechanism they react quite slowly.[/quote]

If the base/nucleophile is a small one (for example, cyanide, ethoxide), then the less hindered the carbon bearing the halogen the more favoured will be the substitution reaction. So, for primary halides, elimination is only feasible if one uses a strong and hindered base, like tert-butoxide. For secondary halides, substitution is favoured with weak bases (cyanide, carboxylate, halide anions) and elimination is favoured with strong bases (alcoxides). For tertiary halides, weak nucleophiles/bases (like solvolysis) give SN1 at lower temperatures and E1 at higher temperatures. For strong bases E2 is the favoured path, specially at higher temperatures (in this context, low temperature means room temperature, and higher temperature is something like 50ºC or so).

[quote="Carcul"]
If the base/nucleophile is a small one (for example, cyanide, ethoxide), then the less hindered the carbon bearing the halogen the more favoured will be the substitution reaction. [/quote]
if the base is small then how does it matter(it matters very less) that the carbon is hindered or not >but if u take $t-Butoxide$ then may be you can say the primary halides will have osme kinda upper hand in the reaction rate.
[quote="Carcul"]
So, for primary halides, elimination is only feasible if one uses a strong and hindered base, like tert-butoxide.[/quote]
why a hindered base .since this is primary a small base will also do.
[quote="Carcul"]
 For strong bases E2 is the favoured path, specially at higher temperatures (in this context, low temperature means room temperature, and higher temperature is something like 50ºC or so).[/quote]
But then [b]tertiary sulfonates[/b]  will undergo  reaction very slowly if say you use a $t-Butoxide$ ion.

[quote="Pardesi"]if the base is small then how does it matter(it matters very less) that the carbon is hindered or not[/quote]

If the base is small and the carbon is not hindered (primary halide), the SN2 reaction is favoured over the E2 reaction. Why? Both reactions are bimolecular, but the transition state for SN2 is "simpler" to achive: don't forget that in E2 mechanism the substracte molecule must first adopt an antiperiplanar configuration for both atoms that will be eliminated.

[quote="Pardesi"]but if u take t-butoxide then may be you can say the primary halides will have osme kinda upper hand in the reaction rate.[/quote]

[quote="Pardesi"]why a hindered base .since this is primary a small base will also do.[/quote]

For tert-butoxide the SN2 transition state will not be favoured, as the stereochemical hinder is considerable. However, if tert-butoxide is to function as a base (E2) then that path will be favoured because then tert-butoxide will "just" have to approach to very small hydrogen atom. For a small base like ethoxide SN2 is always favoured (in primary halides) because the corresponding transition state is simpler to achieve.

[quote="Pardesi"]But then tertiary sulfonates will undergo reaction very slowly if say you use a t-butoxide ion.[/quote]

For tertiary substractes, if you want E2 to occur then using a small strong base is enough. Use of a bulky base will slower the reaction a little perhaps.

well want i really want is now under conditions favourable for $E2$ tertiary substrates react slowly(because the prefer $E1$ path) then how can this be the fastest

If the conditions are favourable for E2 why do you say that tertiary substractes react slowly? Why do you say that they prefer the E1 path? The preferred path for a chemical reaction depends on the particular conditions employed: if we use conditions that strongly [i]favour[/i] E2 mechanism, then E2 is the preferred path. Also, what is your definition of "preferred path"?

look  [url=http://www.mathlinks.ro/viewtopic.php?t=146977]Here[/url] this thread is some kinda continuation of taht.that is what that confuses me :(

If the following is what confuses you,

[quote="Pardesi"]also the converse for tertiary substrates i.e whether they primarily follow 1st order kinetics or by simply changing the solvent the we can change that to 2nd order kinetics (say i use DMSO)[/quote]

then the outcome of the reaction doesn't depend alone on the choice of solvent. For a tertiary substrate, the conditions that favour 1st order kinetics (SN1) are: low temperature; weak nucleophile; a solvent that helps to stabilize the incoming carbocation, because that is the rate determining step.
On the other hand, 2nd order kinetics (E2) is favoured by: higher temperatures; strong base; a solvent that will not solvate too much the base. So simply changing the solvent alone will not in general make a change from SN1 to E2.


ratio
geometry
3D geometry
sphere
analytic geometry
trigonometry
calculus

Just like weekly contests in most of the other math forums, I am starting a [color=red][size=150]Weekly Physics Contest Forum[/size][/color]. I will propose 5-6 problems at the start of the week (i.e. Monday) which can be solved through Friday. These problems may be standard but are definitely interesting!!!

[size=150][b]Rules[/b][/size]

1) All solutions should be [b]PM-ed to me by Friday[/b]. If a solution is a bit late due to time zone difference, the solution will still be accepted.Points for each problem will be indicated alongside.

2) Please [b]don't post any solutions or discussions of problems here[/b]. I am making a [b]separate forum [/b]for them. Only if you have some queries regarding the problems should you post them here.Please don't mind.I am just trying to keep things manageable.

3) I will be posting problems on [b]Mechanics(all), Thermodynamics (not Statistical Physics), Electodynamics and Magnetism , Waves and Oscillations and Optics [/b]. I am [b]not including Modern Physics and Relativity [/b]since most high school students won't be familiare with them. I would like to keep the level at the high school or introductory undergrad level . If you find the problems are getting too tough or easy, please tell me.

4) I will always [b]accept new problems from the members[/b]. If you would like to suggest any, please make sure they are interesting and not too complicated involving minimal of calculus. Please PM such problems to me.

5) Feel free to [b]post your queries [/b]regarding the contest here. I would also like if someone can volunteer to help me here so that I can distribute the workload. I am a bit busy on Thursdays and Fridays!!!!

Hope the moderator likes it!!!

Anyways, here goes. I hope many people will participate to get this forum going!!! [color=green][size=150]Happy Problem Solving!!![/size][/color]

[color=red][size=150]PROBLEM SET 1 [/size][/color]

[u][b]Problem 1[/b][/u]

Two swimmers leave point $A$ on one bank of the river to reach point $B$ lying right across on the other bank. One of them crosses the river(actually - ie. his actual path of motion) along the straight line $AB$ while the other swims at right angles to the stream and gets drifted and then walks the distance that he has been carried away by the stream to get to point $B$. What was the velocity $u$ of his walking if both swimmers reached the destination simultaneously? The stream velocity is $ v_0 $ and the velocity of each swimmer w.r.t still water is $ v' $.

[color=red](4 points)[/color]

[color=red][size=150]PROBLEM SET 1[/size][/color]

[u][b]Problem 2[/b][/u]
A vertcal cylinder closed from both ends is equpped with an easily moving piston dividng the volume into two parts, each containing one mole of air. In equlibrium at temp. $ T_0$ , the volume of the upper part is $\eta$ times greater than that of the lower part. At what temperature will the ratio of these volumes be equal to $ \eta ' < \eta $

[color=red](4 points)[/color]

[color=red][size=150]PROBLEM SET 1[/size][/color]

[u][b]Problem 4[/b][/u]

A particle of mass $m$ is located in a unidimensional potential field where the potential energy of the particle depends on the coordinate $x$ as $ U(x) =\ U_0 (1 - \cos{ax} ) $ where $ U_0 $ and $a$ are constants. Find the period of small oscillations that the paticle performs about the equilibrium position.

[color=red](3 points)[/color]

[quote="shobber"]Is detailed solution required?[/quote]

Ye , it is. But you can obviously leave out the integration etc. Just the answer might fetch you just one point.

[color=red][size=150]PROBLEM SET 1[/size][/color]

[u][b]Problem 5[/b][/u]

A man standing on the edge of a swimming pool looks at a stone lying on the bottom. The depth of the swimming pool is $h$. At what distance from the surface of the water is the image of the stone formed if the line of vision makes an angle $\theta$ with the normal to the surface?

[color=red](4 points)[/color]

should we assume the index of refraction for air in the last problem to be 1?

Yes of course. I am following all standard conventions in these problems.

[quote="Rushil"][color=red][size=150]PROBLEM SET 1[/size][/color]

[u][b]Problem 3[/b][/u]
Two small equally charges spheres , each of mass $m$ , are suspended from the same pont by silk threads of length $l$. the distance between the spheres $x << l $ . Find the rate $\frac{dq}{dt} $with which the charge leaks off each sphere if their approach velocity varies as $ v =\ \frac{a}{ \sqrt{x} } $ where $a$ is a constant.

[color=red](4 points)[/color][/quote]

problem from Resnick ,Halliday's Physics part 2...

I got nostalgic looking at those problems :lol:

Dunno but I put it from A russian book!!! Post your soln as PM to me to get ponts !!! :lol:

that pool's pure water, right? As a lifeguard, i must say that it is a rather naive assumption  :P

it matters not... just use "n" for the index of refraction. (that's what I did anyway.)

Just wanted to confirm if anyone besides kubus has PM-ed me thier solutions. I have got no one else's solutions!!!

@savy2020
pls mention which integral is for which surface...,atleast that wud make it a bit easier to understand

The first one is for the curved surface. The second term for the top surface (opp to that containing O).
For the surface containing O, flux is zero

[quote="Addarsh"]Q1. Find the rate  with which helium flows out of a thermally insulated vessel into vacuum through a small hole.The flow rate of the gas inside the vessel is asssumed to be negligible under these conditions.The temperature of helium in the vessel is [/quote]
is the answer
[hide]
$\sqrt{\frac{2C_pT}{M}}$[/hide]

[quote="Addarsh"][quote]is the answer
Hidden Text[/quote]
Yes it is! :D  :D
How did u get it?Cud u show a detailed solution?[/quote]

use bernoulli's theorem. i.e conserve energy density
[hide]
$\frac{U}{V}= \frac{nC_VT}{V}= \frac{ \rho C_VT}{M}  $

$P= \rho RT$

$P+\frac{U}{V} = \frac{1}{2} \rho v^{2}$

voila!!  :P [/hide]

Good Idea!!! Contest for every coach and students in the World

[quote="pKa2010"]Can this be a beginner for your ques.
$ Ques.$Find the induced charge density on an infinite plane  at a distance $ x$
above the line joining a charge $ q$(and the plane,perpendicular to the plane) which is situated  at a distance $ d$ in a direction  perpendicular to the plane.[/quote]
I see people from India really love Irodov.I did not read all the problems but steal found 2 from that book,this one and the problem from post #4.
By the law of Gauss we get that $E_{normal}(x)=\dfrac{\sigma(x)}{\epsilon}$ as we have a infinite plane,we shall use the image principle(we put a charge $-q$ in symetry of first charge due to plane),so we get that $E_{normal}=2\cdot \frac{kq\cdot cos\phi}{d^2+x^2}$=$2\cdot \frac{kq\cdot d}{(d^2+x^2)^{3/2}}$ so we can easily derive that $\sigma(x)=\frac{qd}{2\pi\cdot (d^2+x^2)^{3/2}}$.

[i]Let continue forum
Dear people I am  interested in physics so can you give me links where i can participate to contests[/i]

Fixing an $xy$ coordinate system with the central charge at the origin, be $y$ the ordinate of the other two charges.
Thanks to the symmetry of the problem we have that the side charges always have the same ordinate.

For the potential energy of one of the side charges we have:

$U(y) = \dfrac{k q^2}{d} + \dfrac{k q^2}{2 \sqrt{d^2 - y^2}} + cost$

where

$k = \dfrac{1}{4 \pi \epsilon_0}$

and $cost$ can be chosen in arbitrary way.

We have, thanks to Taylor,

$U(y) \approx U(0) + U'(0)y + \dfrac{1}{2}U''(0)y^2$

We choose $cost$ in order to have $U(0) = 0$

Deriving we have

$U'(y) = \dfrac{kq^2}{2} \dfrac{y}{(d^2-y^2)^{\frac{3}{2}}}$

$U''(y) = \dfrac{kq^2}{2} \dfrac{d^2 + 2y^2}{(d^2-y^2)^{\frac{5}{2}}} $

Hence:

$U(y) = \dfrac{1}{2} \dfrac{k q^2}{d^3}$

which is a "harmonic" potential. So we have:

$T = 2 \pi \sqrt{\dfrac{m}{U''(0)}} = 2 \pi \sqrt{\dfrac{2md^3}{kq^2}}$

This is enough to conclude:

$|q| = \sqrt{\dfrac{8 \pi^2 m d^3}{kT^2}}$

I am relatively new in this forum,so please forgive any mistakes.Here is the problem
Calculate the capacitance of a spherical capacitor of inner radius $A$ and outer radius $B$ which is filled with a di-electric varying as $e=c+dcos^2f$ . $c$ and $d$ are constants and $f$ is the angle made with the $z$ axis.(also known as the polar angle)

[quote="luiseduardo"]Very nice problem:

Three small identical coins of mass $m$ each are connected by two light nonconducting strings of length $d$ each. Each coin carries an unknown charge $q$. The coins are placed on a horizontal frictionless nonconducting surface as shown (the angle between the strings is very close to 180°). After the coins are released, they are observed to vibrate with period $T$. Find the charge $q$ on each of the coins.[/quote]
Is the middle coin fixed? Even it it's not, I am getting a different answer... can someone tell me where I am missing a factor of $2$?
[hide]
Fix the coordinate axes just as in the following diagram:
[asy]
import graph;size(10cm);
pair a=(-5,2);pair b=(5,2);pair o=(0,0); dot(o); dot(a); dot(b);
draw((-10,0)--(10,0)); draw((0,5)--(0,-5));
draw(a--o--b); draw(a--b,dashed); draw((5,2)--(8,3.2), EndArrow); draw((5,2)--(8.5,2),EndArrow);
label("$x$",(10,0),S); label("$y$",(0,5),N);
label("$q$",(0,0),SE); label("$q$",(-5,2),NW); label("$q$",(5,2),N);
label("$\textstyle F_1=\frac{q^2}{4\pi\epsilon_0d^2}$",(8,3.2),NE);
label("$\textstyle F_2=\frac{q^2}{16\pi\epsilon_0d^2\cos^2\theta}$",(8.5,2),E);
MA((5,0),o,b,2,black);label("$\theta$",(2,0.56),E);
[/asy]
Then note that the forces act in such a way that the Tension force neutralises the components of $F_1$ and $F_2$ along the string, and the force component perpendicular to it provides the necessary restoring torque. So the path followed by the particles can be assumed to be a circular one, and assume that one of the non-centre particles experiences the restoring torque $\tau_{res}=-F_2d\sin\theta.$ Then observe that
\[\begin{aligned}\tau_{res}=I_{\circ}\frac{d\,^2\theta}{d\,t^2}=-\frac{q^2d\sin\theta}{16\pi\epsilon_0d^2\cos^2\theta}&= -\frac{q^2}{16\pi\epsilon_0d}\cdot\frac{\sin\theta}{\cos^2\theta}\\&\approx-\frac{q^2}{16\pi\epsilon_0d}\cdot\theta.\end{aligned}\]
Thus $\omega^2=\frac{q^2}{16\pi\epsilon_0dI_{\circ}},$ where $\omega$ is the angular frequency of the SHM, and $I_{\circ}$ is the inertia of the small mass about $O,$ the origin, which is $md^2.$ So,
\[q=\sqrt{\frac{64\pi^3\epsilon_0md^3}{T^2}},\]
Which looks like $\sqrt2$ times andrea's answer.
[/hide]
Thanks in advance. :)


Find $ V_o$ and $ i_o$ in the circuit of the figure

[img]http://i35.tinypic.com/2urro5z.jpg[/img]

What is the symbol in the middle?  Just remember, all of the current entering/leaving a node must sum to 0, and the current around a loop must be 0.

its an arrow pointing down

yes I tried that

$ 6\minus{}I_o\minus{}\frac{I_o}{4}$=0?


hello!!
let $ n\in\mathbb{N*}$
prove that:
$ 2^{F_{n}}\equiv2[F_{n}]$
$ F_{n}$ is fermat's number.

I think that's Fibonacci.
By Fermat's numbers, do you mean numbers of the form $ 2^{2^n}\plus{}1$?

[quote="Brut3Forc3"]I think that's Fibonacci.
By Fermat's numbers, do you mean numbers of the form $ 2^{2^n} \plus{} 1$?[/quote]
yes I mean $ 2^{2^{n}}\plus{}1$


has anyone had their chapter competition yet.

Why do you need to know?

Even if anyone has had it, we can't talk about it...


conics
ellipse
geometry
rectangle
circumcircle
symmetry
analytic geometry

P is a point on an ellipse and L is a line through P perpendicular to the tangent to the ellipse at P.F and G are the foci of the ellipse,and E and H are the points on L such that EF and GH are perpendicular to L.Prove that PE.PH is constant

[b]Lemma.[/b] The pedal curve of an ellipse with respect to either focus (i.e., locus of the feet of normals from the focus to the ellipse tangents) is a circle centered at the ellipse center and with radius equal to the ellipse semi-major axis.

Let $AB$ be the ellipse major axis and $O$ be the ellipse center (the midpoint of the segmeny $AB$). Let $P$ an arbitrary point on the given ellipse and $P'$ a point symmetrical to the point $P$ with respect to the center $O$. The point $P'$ also lies on the given ellipse. Let $t, t'$ be the ellipse tangents at the points $P, P'$. Since both the ellipse and the points $P, P'$ are symmetrical with respect to the ellipse center $O$, so are the tangents $t, t'$ and consequently, they are parallel, $t \parallel t'$. Let $Q, R$ be the feet of normals from the foci $F, G$ to the ellipse tangent $t$ at the point $P$ and $Q', R'$ the feet of normals from the foci $F, G$ to the ellipse tangent $t'$ at the point $P'$. The quadrilateral $QRR'Q'$ is a rectangle, because $QR \equiv t \parallel t' \equiv Q'R'$, $QQ' \perp t \equiv QR, RR' \perp t \equiv QR$ and the circumcircle of this rectangle is the ellipse fixed pedal circle $(O)$ with radius $OA = OB = \frac{AB}{2}$. Since $FQ \perp t \equiv QP$, $PE \equiv l \perp t \equiv QP$ and $FE \perp l \equiv PE$, the quadrilateral $FEPQ$ is a retangle and $PE = FQ$. Similarly, since $GR \perp t \equiv PR$, $PH \equiv l \perp t \equiv PR$ and $GH \perp l \equiv PH$, the quadrilateral $GHPR$ is a retangle and $PH = GR$. By symmetry of the ellipse, the points $P, P'$ and the tangents $t, t'$ with respect to the center $O$, $GR' = FQ$. Thus $PE \cdot PF = GR \cdot GR'$ is equal to the power of the focus $G$ to the pedal circle $(O)$, which does not depend on the position of the point $P$ on the ellipse.

[quote="yetti"][b]Lemma.[/b] The pedal curve of an ellipse with respect to either focus (i.e., locus of the feet of normals from the focus to the ellipse tangents) is a circle centered at the ellipse center and with radius equal to the ellipse semi-major axis.[/quote]
Since I do not know a nice proof of the above lemma, here is an ugly one.  Let $a, b$ be the semi-major and semi-minor axes of an ellipse centered at the origin $O$, $a > b$. The equation of the ellipse is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\ \ y^2 = \frac{b^2}{a^2} (a^2 - x^2)$

and the equations of the upper/lower ellipse arcs are

$y = +\frac b a \sqrt{a^2 - x^2},\ \ y = -\frac b a \sqrt{a^2 - x^2}$

The focal points $F, G$ have coordinates $F = \left(-\sqrt{a^2 - b^2}, 0\right)$, $G = \left(+\sqrt{a^2 - b^2}, 0\right)$. Let the pedal point be the left focus $F$. The slope of the tangent to the upper arc of the ellipse at an arbitrary point $P = (x_P, y_P)$ of this arc is the derivative of the upper arc at this point:

$\frac{\text dy}{\text dx}(x_P) = -\frac b a\ \frac{x_P}{\sqrt{a^2 - x_P^2}}$

and the equation of the tangent is

$y - y_P = -\frac b a\ \frac{x_P}{\sqrt{a^2 - x_P^2}}\ (x - x_P)$

Since the tangency point $P$ lies on the upper arc of the ellipse,

$y_P = \frac b a \sqrt{a^2 - x_P^2}$

$y - \frac b a \sqrt{a^2 - x_P^2} = -\frac b a\ \frac{x_P}{\sqrt{a^2 - x_P^2}}\ (x - x_P)$

The product of slopes of this tangent and a normal to it from the focus $F$ is equal to -1. Hence the equation of the normal to the ellipse tangent at the point $P$ dropped frop the pedal point $F = (x_F, y_F) = \left(-\sqrt{a^2 - b^2}, 0\right)$ is

$y - y_F = \frac a b\ \frac{\sqrt{a^2 - x_P^2}}{x_P}\ (x - x_F)$

$y = \frac a b\ \frac{\sqrt{a^2 - x_P^2}}{x_P}\ (x + \sqrt{a^2 - b^2})$

To get the equation of the pedal curve, we have to eliminate the coordinate $x_P$ of the arbitray tangency point $P$. Rewriting the equations of the tangent and the normal as

$y + \frac b a\ \frac{x_P}{\sqrt{a^2 - x_P^2}}\ x = \frac b a \sqrt{a^2 - x_P^2} + \frac b a\ \frac{x_P^2}{\sqrt{a^2 - x_P^2}}$

$\frac b a\ \frac{x_P}{\sqrt{a^2 - x_P^2}}\ y - x  = \sqrt{a^2 - b^2}$

Now, if we square and add both equations, the mixed terms $xy$ cancel out and the coordinate $x_P$ can be factored out:

$(x^2 + y^2) \left(1 + \frac{b^2}{a^2}\ \frac{x_P^2}{a^2 - x_P^2} \right) =$

$= \frac{b^2}{a^2}\left(a^2 - x_P^2 + 2x_P^2 + \frac{x_P^4}{a^2 - x_P^2}\right) + a^2 - b^2 =$

$= \frac{b^2}{a^2}\ \frac{a^2(a^2 - x_P^2) + x_P^2(a^2 - x_P^2) + x_P^4}{a^2 - x_P^2} + a^2 - b^2 =$

$= \frac{a^2b^2}{a^2 - x_P^2} + a^2 - b^2 = a^2 + \frac{a^2b^2 - b^2(a^2 - x_P^2)}{a^2 - x_P^2} =$

$= a^2 + \frac{b^2x_P^2}{a^2 - x_P^2} = a^2 \left(1 + \frac{b^2}{a^2}\ \frac{x_P^2}{a^2 - x_P^2}\right)$

Factoring out $\left(1 + \frac{b^2}{a^2}\ \frac{x_P^2}{a^2 - x_P^2}\right)$ eliminates the coordinate $x_P$ and yields the equation of the pedal curve:

$x^2 + y^2 = a^2$

which is a circle centered at the origin (the ellipse center) and with radius equal to the ellipse semi-major axis $a$.


Theorem : Suppose Y is a subset of X. A subset E of Y is open relative to Y if and only if E =the intersection of Y and G, for some open subset G of X.

Can anyone give some examples?

Thanks A Lott!  =)

Let $ Y$ denote the $ xy$-plane and let $ X$ denote $ \mathbb{R}^3$.  Let $ E$ be an open circle of radius $ 1$ in $ Y$.  Then $ E$ is the intersection of an open [i]ball[/i] of radius $ 1$ and $ Y$.

In other words, the open subsets of $ Y$ can all be described as the "restriction" of open subsets of $ X$ to $ Y$.

Thxx I can totally see the geometric picture now. But can u prove this theorem mathematically?
            Suppose Y is a subset of X. A subset E of Y is open relative to Y if and only if E = the intersection of Y and G for some open subset G of X.

The point is I don't really get how the definition " E is open relative to Y if to each p, which is an element of E, there is associated an r>0 s.t q is an element of E whenever d(p,q)<r and q is an element of Y. " fit with the geometric picture?

This is not a theorem. It's the definition of the subspace topology.

If you're talking about [i]metric[/i] spaces, it's provable as a theorem, assuming you use the inherited metric on the subset. We're using the same distance, so the open and closed sets are the same- except that the universe of points is smaller.


Let $P(x): =x^{n}+\sum\limits_{k=1}^{n}a_kx^{n-k}$  with $0\le a_n\le a_{n-1}\le\ldots a_2\le a_1\le 1$. Suppose that there exists $r\ge 1 ,\; \varphi \in {\mathbb R}$  such that  $P(re^{i\varphi})=0$. Find $r$.

It's a very well-known trick (my polynomial is $P(x)=a_0+a_1x+\ldots+a_{n-1}x^{n-1}+x^n$, so I changed the indexing of the coefficients):

Since $1$ is clearly not a root of $P$, we have $P(x)=0\iff (x-1)P(x)=0$, which can be written as $x^{n+1}=a_0+(a_1-a_0)x+\ldots+(a_{n-1}-a_{n-2})x^{n-1}+(1-a_{n-1})x^n$. For every $x$ s.t. $P(x)=0$ we will thus have $|x|^{n+1}=|a_0+(a_1-a_0)x+\ldots+(a_{n-1}-a_{n-2})x^{n-1}+(1-a_{n-1})x^n|$, which is at most $\sum(a_{i+1}-a_i)|x|^{i+1}\le \max(1,|x|^n)$, i.e. $|x|\le 1$, so all the roots of $P$ must lie in the closed unit disk (which means that the answer to the question is $r=1$).

[quote="grobber"]It's a very well-known trick .......[/quote] Nice solution !/Proposer


algebra
polynomial
calculus
integration
function
domain
Ring Theory

show if $x^{k}\equiv \ 1 mod{p}$ for all nonzero integers $x mod{p}$, then $p-1$ divides $k$:

I saw this problem in [b]Art and Craft of Problem Solving[/b]: [i]Paul Zeitz[/i]

Thats equivalent to the existence of primitive roots (see http://www.mathlinks.ro/Forum/viewtopic.php?t=55473 for a proof that they exist) after some computations.

Another way would be:
we can WLOG assume that $1< k \leq p-1$ by Fermat's little theorem. Then look at the polynomial $x^{k}-1$: since the residues $\mod p$ are a field/an integral domain, it has at most $k$ roots. When $x^{k}\equiv 1 \mod p$ for all nonzero $x$, it has at least $p-1$ roots. Thus $k=p-1$.

I don't see exactly how the problem immediately implies the existence of a primitive root.  The alternative method you showed is actually one of the steps in the proof that I know of the existence of a primitive root.


LaTeX
number theory unsolved
number theory

Let $v_0 = 0, v_1 = 1$ and $v_{n+1} = 8 \cdot v_n - v_{n-1},$ $n = 1,2, ...$. Prove that in the sequence $\{v_n\}$ there aren't terms of the form $3^{\alpha} \cdot 5^{\beta}$ with $\alpha, \beta \in \mathbb{N}.$

Do you mean $\mathbb{N}=\{1,2,...\}$ ?

If yes then the proof is: $3|v_n <=> 3|n$ and $7|v_n <=> 3|n$ - what's easy to check. So if $3|v_n$ then $v_n$ can't be the given form.

Rafal, a tip: the code \Leftrightarrow gives $\Leftrightarrow$.

the latex you wrote also doesn't look nice :)


function
calculus
derivative
integration
real analysis
advanced fields
advanced fields unsolved

Show that $ \chi_{[0,\plus{}\infty)}(x)$ does not have a weak derivative in $ L^1_{loc}(R).$

[quote="yenlee"]Here's the idea:  The weak derivative of the step function must behave like the delta function.  (In fact, it IS the delta function if you allow yourself to work with distributions.)  Prove that no L^1_loc function has such properties.[/quote]

Thanks. I am trying to choose a suitable text function, but I have not found. Please share!

How about this? On $ \mathbb{R},$ suppose distribution $ f$ has derivative $ g\in L^1_{\text{loc}}.$ Now let $ h(x)\equal{}\int_0^xg(t)\,dt.$ That's defined as a Lebesgue integral, so $ h$ is defined. I know two things about $ h:$

(1) $ h$ is continuous on $ \mathbb{R}.$

(2) The weak derivative of $ h$ is $ g.$

But then by (2), $ h$ and $ f$ differ (as distributions) by at most a constant. A constant plus a continuous function is a continuous function, so it follows that $ f$ is continuous.

So a function whose weak derivative is locally integrable is equal as a distribution to a continuous function.

That's what rules out the function with the jump discontinuity.


function
integration
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inequalities
advanced fields
advanced fields unsolved

A function $ u\in H_0^{2}(U)$ is a weak solution of this boundary-value problem for the biharmonic equation

$ \triangle^2u \equal{} f$ in $ U$,  $ u \equal{} \frac {\partial u}{\partial \nu} \equal{} 0$ on $ \partial U$   $ (\star)$

provided $ \int_{U}\triangle u \triangle v dx \equal{} \int_Ufv dx$ for all $ v\in H_{0}^2(U)$.

Given $ f\in L^2(U)$, prove that there exists a unique weak solution of $ (\star)$.

Are you reading Gilbarg und Trudinger?

Los

Sounds more like Evans.

$ math$ sound suspiciously like someone trying to get someone to do his/her homework, but I'll give benefit of the doubt for now.

Hello everyone,

Yes I think that you yenlee are correct, since the first question is exactly problem 1 of chapter 6 in Evans while this question is problem 2 in chapter 6.

tzktzktzk thats a grad class (2nd or 4th semester) . At this level you ought to have developed the ability to collaborate with your classmates!

I would say the #1 skill that you need to do well in first semester grad PDE is, to know Vector Calculus (2. semester, at level of say marsden & tromba).

Yes, yes! This problem is in the Evans's book!

Use Lax-Milgram Theorem and some inequalities!


So i tried the picture in the aops tutorial and it worked but when i try to put in this png that i made from paint it didnt work for some reason no matter what i do can someone please help me see whats wrong and i just put some random scribbles on  the paint document so no one thinks its an error if they saw it blank.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example required[/url]
There are some very clever people on AoPS but none of them can read your mind or know what LaTeX code you've used.

I have put the picture in a latex file and if you try to compile this latex file you will get an error that says no output produced. The file for the picture is in the first post of this thread. Sorry I did not get the minimal example i am just very angry at why its so hard to get a simple picture into a latex document for usamts.

Clicking on your image or USAMTS4.png, Firefox says
[quote]The image “http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=23175&thumb=1” cannot be displayed, because it contains errors.[/quote]so there's clearly something wrong with the image. Have you converted it to png from another file format? Changing the name isn't enough; you need a program like Paint to do that.

i did change a paint picture to a png just by changing its name from usamtsfour.bmp to usamtsfour.png using right click rename. i just got a program called pixillon image converter but that doesnt make a difference.

You cannot change file formats like that. All that does is confuse programs as you've found and is why it doesn't show up in your first post. Load it into Paint then and then [i][b]save[/b][/i] as png.

Yes it does because I tried it. What doesn't work?

:D idk what i did the first time but it worked the second time. AOPS FTW and ty everyone who helped me.

edit: ty stevem XD sorry i phailed to see that you were the one who helped me this whole time.


algebra
polynomial
topology
complex analysis
complex analysis solved

if $f$ in holomorfic in the hole plane and for each $z\in\mathbb{C}$, $f^{(n)}(z)=0$  for some $n$ then $f$ must be  a polynomial.

Well i did it in a non analytical way and got some generalizations, (st we only need that $f^{(n)}(z)=0$  for some $n$ in a closed set with acumulation point), well i thought that it can prove usiing  only "analitical" tools, well i'll aprichiate any commentary.

PD:i saw in a problem list that it's also true for real $C^\infty$ functions.

What do you mean by "non analytical way"?

Anyway, here's what I did for the initial problem:

Let $A_n$ be the set consisting of those $z$ for which $f^{(n)}(z)=0$. The $A_n$'s are closed sets which cover the complex plane, so one of them ($A_m$, say) must have a non-empty interior (Baire's Theorem). However, this means that $f^{(m)}$ vanishes infinitely many times on a non-empty open subset of the complex plane, so it must be zero everywhere, since it's entire.

Suppose $f^{(m)}$ not not identically zero for all $m$ then each $f^{(m)}$ has only countable zeroes on the complex plane because each of them is entire. Thus $\bigcup_{m=0}^\infty \{\mbox{zeroes of }f^{(m)}\}$ is countable. But the assumption tells us $\bigcup_{m=0}^\infty \{\mbox{zeroes of }f^{(m)}\} =\mathbb{C}$... Oops.

I meant just what you did, use baire's theorem, well it looks like a  "topological way"....the hypothesis of $f$ entire is too much, this argument works in a closed set with acumulation point (exactly what it needs to use baire's theorem and the identity principle), so it looks like it can be skipped...... welll i invite you to try the real case, ($f\in C^\infty(\mathbb{R})$, st for all $x$, $f^{(n)}(x)=0$ for some $n$, then $f$ is a polynomial), it's more dificult but very finteresting!.


Which of the sections of the forum, if any of them, would it be appropriate to ask the following?

.............
I've been taking old AMC 10s under timed conditions and have been getting roughly 120s, but I haven't been getting 120 or over consistently. I don't feel like I have a single weak area such as number theory or probability...If you were me, how would you study for the few weeks coming before the AMC 10 after you finished going through all the previous AMC 10s?

Maybe in the AMC forum, under "Contests and Programs."
:)


[quote="nonie"]What is the coeffiecient of $x^{99}$ in $(x+1)(x+2)(x+3)...(x+100)$?[/quote]
[hide] In the polynomial $f(x)=ax^{n}+bx^{n-1}+...$, $-\frac{b}{a}$ is equal to the sum of the roots of the polynomial. In this case, the sum of the roots is $-1-2-3-...-100=-5050$. And $a=1$, so $-\frac{b}{1}=-5050 \Longrightarrow b=5050$ [/hide]


LaTeX
quadratics
trigonometry
geometry
3D geometry
algebra
polynomial

If someone changes this to latex ill give you brownie points :rotfl:

X^6-1 can be factored pretty easily right?

split it into (x^3-1)(X^3+1) and then into
(x-1)(x+1)(x^2+x+1)(X^2-x+1) using (what is that formula called?)

so then the answer is pretty simple if you use quadratic formula.

x= 1,-1, -1+sqrt(-3)/2,-1-sqrt(-3)/2

The problem is if you split x^6-1 into thirds like this:

(x^2-1)(x^4+x^2+1)

(x+1)(x-1)(x^4+x^2+1)

I then used quadratic equation on the trinomial.

x^2 = -1-sqrt(-3)/2,-1+sqrt(-3)/2 is what i get!
So that means x is the square root of that! The first one and the second have diffrent solutions! My entire class including our teacher is confused, can somebody help?

If someone changes this to latex ill give you brownie points :rotfl:

$x^6-1$ can be factored pretty easily right?

split it into $(x^3-1)(x^3+1)$ and then into
$(x-1)(x+1)(x^2+x+1)(x^2-x+1)$ using (what is that formula called?)

so then the answer is pretty simple if you use quadratic formula.

$x= 1,-1, -1+sqrt(-3)/2,-1-sqrt(-3)/2$

The problem is if you split x^6-1 into thirds like this:

$(x^2-1)(x^4+x^2+1)$

$(x+1)(x-1)(x^4+x^2+1)$

I then used quadratic equation on the trinomial.

$x^2 = -1-sqrt(-3)/2,-1+sqrt(-3)/2 $is what i get!
So that means x is the square root of that! The first one and the second have diffrent solutions! My entire class including our teacher is confused, can somebody help?

ALL YOU HAD TO DO WAS PUT IN $!!!

So what is the problem?

... That is assuming that you remembered to use both $(x^2+x+1)=0 $ and $(x^2-x+1) = 0$ and got all the six roots. (in your first part)

For simplicity let $u =  \frac {1 +  \sqrt {-3} } 2 $

You have all the solutions as  $1, u , u^2, u^3, u^4, u^5 $

You do notice that $u^2  =  \frac {-1 +  \sqrt {-3} } 2 $
 and $u^3 =  -1 $
and  $u^4 =  \frac {-1 - \sqrt {-3} } 2 $
and $u^5 =  \frac {1 - \sqrt {-3} } 2 $
and of cource  $u^6 = 1$

(You may have noticed that $u = \cos {(\pi / 6)) + i \sin {(\pi / 6)}}$

Just check out  .. all your answers are rearrangemnt of the above ... right?

Don't be fooled by  just remember that  $(u^4) = u^8 = u^6 u^2 = u^2$ etc..

[quote="pontios"]xdl615x

$(\frac{1 + i\sqrt{3}}{2})^2=$

$\frac{1^2+2i\sqrt{3} + (i\sqrt{3})^2}{4}=$

$\frac{1+2i\sqrt{3} - 3}{4}=$

$\frac{2i\sqrt{3} - 2}{4}=$

$\frac{i\sqrt{3} - 1}{2}=$

$\frac{\sqrt{-3} - 1}{2},$ which is the one solution of $x^2$ you found

I'm confused too!

I mean... What's the problem?[/quote]

Why do you square the answer in the beggining?

$x^2=y$

then

$x=sqrt(y)$ right?

xdl615x, I just noticed that one solution of $x^2=\frac{\sqrt{-3} - 1}{2}\ \ $  is $\ \ x=\frac{1 + i\sqrt{3}}{2}$

I'm pretty sure that the error was that you interpreted the difference of cubes incorrectly, which was stated earlier. If you continue the problem with the factored polynomial, you would get the correct answer.


for all $ x\in[\minus{}1,\ 1]\quad|a\plus{}bx\plus{}cx^2\plus{}dx^3|\leq 1$
prove that also for $ x\in[\minus{}1,\ 1]$ there is $ |c\plus{}2cx\plus{}3dx^2|\leq 9$

Let $ y\equal{}a\plus{}bx\plus{}cx^2\plus{}dx^3$.  Then $ y' \equal{} b\plus{}2cx\plus{}3dx^2$.  For the condition $ |y| \le 1$ to be satisfied, you need only to check the points where $ x\equal{}\minus{}1$, $ x\equal{}1$, and $ y'(x)\equal{}0$ ($ x \in [\minus{}1,1]$).

By the way, is the first $ c$ in $ |c\plus{}2cx\plus{}3dx^2|$ a typo?

[quote="facis"]Let $ y \equal{} a \plus{} bx \plus{} cx^2 \plus{} dx^3$.  Then $ y' \equal{} b \plus{} 2cx \plus{} 3dx^2$.  For the condition $ |y| \le 1$ to be satisfied, you need only to check the points where $ x \equal{} \minus{} 1$, $ x \equal{} 1$, and $ y'(x) \equal{} 0$ ($ x \in [ \minus{} 1,1]$).

By the way, is the first $ c$ in $ |c \plus{} 2cx \plus{} 3dx^2|$ a typo?[/quote]I hardly find this hint helpful. Any other ideas, anyone?


maximization
minimization
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algebra
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IMO Shortlist

Let $ n \in \mathbb{Z}^\plus{}$ and let $ a, b \in \mathbb{R}.$ Determine the range of $ x_0$ for which

\[ \sum^n_{i\equal{}0} x_i \equal{} a \text{ and } \sum^n_{i\equal{}0} x^2_i \equal{} b,\]

where $ x_0, x_1, \ldots , x_n$ are real variables.

The Power-Mean Inequality dictates that \[\frac{b-x_0^2}{n} = \frac{1}{n}\sum_{i=1}^n x_i^2 \geq \left(\frac{1}{n}\sum_{i=1}^n x_i\right)^2 = \left(\frac{a-x_0}{n}\right)^2\,.\]  Therefore, \[(n+1)x_0^2-2ax_0+\left(a^2-nb\right) \leq 0\,.\]  Let $D:=\sqrt{n\left((n+1)b-a^2\right)}$ (and so, we implicitly require $a^2 \leq (n+1)b$, another instance of the Power-Mean Ineq.).  Ergo, the range of $x_0$ is a subset of $\left[\frac{a-D}{n+1},\frac{a+D}{n+1}\right]$.  It is easy to see that this range is exactly the answer.


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Prove that if F is a field, R is a ring, and theta is the mapping from F--> R is a ring homomorphism, then either theta is 1-1 or theta(a)=0 for all a in F

Its kernel is an ideal of $F$, which is then $0$ or $F$, since $F$ is a field.


Prove or disprove that there exists a permutation $(x_{1}, x_{2}, ... , x_{n})$ from ${1,2,...,n}$ such that $x_{1}$ , $x_{1}+x_{2}$ , ... $x_{1}+x_{2}+...+x_{n}$ are distinct modulo $n$

i proved that for odd n, is impossible.

first suppose that is possible.
see that $x_{1}= n$, because the differences of two consecutive terms of our 'sequence' can't be a multiple of n,
so, $x_{2}, x_{3}, ... , x_{n}$ aren't multiples of n.
now, see that $n|x_{1}+x_{2}+...+x_{n}= \frac{n(n-1)}{2}$, because n is odd. impossible, because $n|x_{1}$

_

i have one construction for n even, but is ugly. if no one find one nice way, i'll post!  :)

And it is possible for $n$ even. Just take $(x_{1},x_{2},\ldots,x_{n})=(0,n-1,2,n-3,\ldots)$.


function
calculus
integration
real analysis
real analysis unsolved

Suppose a function $f : R\to R$ satisfies $f(f(x)) = -x$ for every $x$.
Prove that $f$ has infinitely many points of discontinuity.

It is easy to prove that $f$ is injective. Assume that $f$ is continuous on an interval. Then $f$ is monotonic on that interval, then $f(f(x))$ is incresing on that interval. But $-x$ is decreasing on that interval. Contradiction. Hence f cannot be continuous on any interval.

it is not totally correct because if it is monotonous on an interval $I$, maybe $f(I)$ is not included in $I$.

ooops... I think I can construct a such function that have only finitely many dicontinuities.

$f=\begin{cases} \frac 1 x \text { on } (-\infty,-1)\cup (1,\infty)\\ -\frac 1 x \text { on } (-1,0)\cup(0,1)\\ 0 \text { elsewhere }\end{cases}$

this function has only three discontinuities. Am I wrong?

That example is flawed; $F$ needs to be defined on all of $\mathbb{R}$, and there's no way do define $f$ on the three discontinuites that works. As written, $f(f(1))=0\neq -1$.

Ok then.. I try another solution.

Assume the contrary.  Denote by $\pm I_1,\pm I_2,..,\pm I_n$ the intervals on which $f$ is continuous( I used the fact that $f$ is odd.) Since $f$ is a permutation of those integrals of order $4$ ($fofofof=1_R$)it is composed by cycles of length 4. Hence the number of intervals must be divisible by 4.
In the same way one can prove that the number points not included in those intervals (their limits ) mus be divisible by 4. But those points are 1 less than the intervals. Cotradiction.
Hence the nuumber is infinite.
For an example of such function one can split $R$ into sets $[n,n+1)$ make groups of four such intervals and f translates $I_1\to I_2 \to -I_1\to -I_2\to I_1$.

Think it is all right.