inequalities function calculus algebra domain rotation ratio

if a,b,c are sides of a triangle
find the minimum/maximum of a/(a+b)+b/(b+c)+c/(c+a)

this is supposedly given to a grade ten student
maybe this is easier
a/(a+b)+b/(b+c)+c/(c+a)>1.4545

Well, so far I've got (min) + (max) = 3.  But all I've done was lying in bed.  More thoughts later.

yeah, thats why I added in the part about the max.  somebody used numerical analysis techniques but they are way beyond the grade 10 level.  I am sure there exist a crispy method to solve it.

Everything I've looked at so far gives me the wrong directions, but I haven't tried anything more complicated than AM-GM-HM combined with triangle inequalities.  By the way, is that 1.4545 supposed to be a strict lower bound, or an approximation?

By considering the degenerate case, I found a lower bound of approximately 1.4777, whose exact value I'd have to go find again but which involves a square root of 2 and of 5.  That doesn't say anything about the general case, of course.  I'd be interested in knowing one thing (working on the assumption that you don't in fact know the nice solution): do you know what the "answer" triangle is?  Also, I'm thinking that this really isn't a nice problem, although I could be wrong.  I set a former IMO team member on it, so if he gets anything, I'll report that, too.

I think we can reduce the general case to the degenerate case.  Without loss of generality, assume that a :le: b.  Consider the given expression as a function of c.  By calculus, I showed that the expression is increasing from 0 to :sqrt:(ab), and then decreasing from :sqrt:(ab) to infinity.  Of course, c is constrained to lie between b - a and b + a by the triangle inequality.  Thus the minimum value of the expression occurs when c is one of those two endpoints.  In other words, we have a degenerate triangle.

I got the same lower bound as Joel, about 1.4777719.

Without loss of generality, we can set a=1.  Then, using the triangle inequality, we plot the possible domain of b and c on a pair of axes (b on x, c on y).  We find b and c must be bounded by the lines, c=b-1 (b+1>c), c=b-1 (c+1>b), and c=1-b (b+c>1).  Because the function is symmetric if we rotate the sides a, b, c, we can also assume a is the longest side.  We further bound the possible domain of b and c by the lines b=1 (b<=1) and c=1 (c<=1).  The bounded region is a triangle with vertices at the points (0,1), (1,0), (1,1).

The edge from (0,1) to (1,0) represents all aspect ratios of degenerate triangles.  If I numerically trace this edge, I find the function is equal to 1.5 at the endpoints and in the center (b=0, b=0.5, b=1).  The function has a minima of about 1.4777719 near b=0.2038.  The function has a maxima of about 1.5222281 near b=0.7962.

Note that if any two sides are equal, the function value is exactly equal to 1.5.

The edge from (1,0) to (1,1) has a constant function value of 1.5.

The edge from (1,1) to (0,1) has a constant function value of 1.5.

I numerically examined the vicinity of the minima and maxima we encountered along the degenerate edge.  In both cases, more extreme values are located outside our region (and in violation of the triangle inequality).  Less extreme values are located inside our region.

I haven't proven there aren't more extreme local maxima or minima inside our bounded triangle.  But I am confident there are not.  The range of function values is about 1.4777719 to 1.5222281.  And JBL was right in his claim that the sum of the minimum and the maximum was equal to 3.

That 1.477... number seems to be
[tex]\displaystyle{\frac{{5\sqrt 5  - 8\sqrt 2  + 9}}{6}}[/tex].
And the corresponding triangle is
[tex]\displaystyle{a:b:c = \frac{{1 + \sqrt 5  - 2\sqrt 2 }}{2}:\frac{{1 - \sqrt 5  + 2\sqrt 2 }}{2}:1}[/tex]
Considering how "messy" the result is, I'm having doubts on whether a nice proof exists.

Although I left my previous post unproven, I think the method was within reach of some 10th grade students.

I can now complete my proof, but I shall resort to calculus.  We have given a=1, and we have established a triangular region bounding b and c.

We now draw a line through the origin, c=kb, where k is a constant.  We plug a=1 and c=kb into the original expression:

[tex]\displaystyle{\frac{1}{1+b}+\frac{b}{b+kb}+\frac{kb}{kb+1}}[/tex]

Differentiating, and setting to 0, we find minima/maxima occur at [tex]\displaystyle{b=\frac{1}{\sqrt{k}}}[/tex] or [tex]\displaystyle{c=kb=k\frac{1}{\sqrt{k}}=\sqrt{k}}[/tex].

Consider the line c=kb.  For any slope, 0<k<1, the minima/maxima occurs at b>1.  This is outside our bounding triangle.  For any slope, k>1, the minima/maxima occurs at c>1.  This is outside our bounding triangle.  If k=1, we have b=c.  The expression remains constant (1.5), and there is neither a minima nor a maxima.

Having considered all cases that pass through the interior of our bounding triangle, we conclude there are no minima/maxima inside our bounding triangle.  Our previous conclusion is proven.

this was given as a question for grade ten students in vietnam


You have a 4 x 4 grid in a square shape with 16 squares. Just above this grid are attached two more squares, for a total of 18 little squares.

When you flick on a switch, 7 of the 18 squares will light up. The pattern is different and random each and every time you turn on the switch, but there will always only be 7 squares that light up. Let's say that you get to select 8 squares just before you turn the switch on. What is the probability that upon flicking on the switch that the 7 squares that light up will be a part of the 8 that you had selected just before turning on the switch?

The furthest I have gotten is that the denominator (number of combinations in universe) should be C(18,8) x C(18,7) and I can't even guarantee that this is correct. Goodluck!

[quote="K81o7"]Well first, how many ways are there to choose 7 squares?  $\frac{18!}{7!11!}$.  Then, how many ways are there for the 7 to be a subset of your 8?  $8$ ways.  Thus, the probability must be $\frac{1}{13*17*18}$.[/quote]

[quote="blahblahblah"]You choose from $C^{18}_8$ combinations; $8$ of those combinations are good, so the answer is

$\frac {8}{C^{18}_8}$[/quote]

Now I am a little more confused. To me it seems that 7 should be a subset of the ways that eight can be chosen from 18 ($C^{8}_7$, and not that the set C(18,7) should be made as the universal set.
 :?:  :?:  :?:  :|

[quote="K81o7"]Well first, how many ways are there to choose 7 squares?  $\frac{18!}{7!11!}$.  Then, how many ways are there for the 7 to be a subset of your 8?  $8$ ways.  Thus, the probability must be $\frac{1}{13*17*18}$.[/quote]

American Micro Devices K8, you are brilliant!

Can I ask about you?
What is your age?
Level of education?
What line of work are you in?
etc... :?:  :)

American Micro Devices?  ... :D
Age: 14
10th grade student
Essentially, you don't have to count anything about the 8 squares chosen.   We're really only worrying about the 7.  So then, just assume that 8 have been picked.  From there you calculate the probability by $\frac{number of possibilities for the desired situation}{total number of possibilities}$. so yeah, it's $\frac{8}{\binom{18}{7}}$.  I posted the general solution in the other thread in pre-olympiad...


let x,y,z is positive reals and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4$.
Prove that:$\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\leq 1$

Very simple,indeed. Well,we have $1/2x+y+z =1/(x+y)+(x+z)$ but you do know that we have the following
 inequality $1/x+1/y+1/z>=2/x+y +2/y+z +2/z+x $ so the conclusion follows. :D

from the AM-HM inequality (commonly cauchy's inequality)
we have
 $\frac{1}{2x+y+z}=\frac{1}{x+x+y+z}\leq\frac{1}{16}(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
Similarly for the other and we add, exploiting the given relation.
I hope this is true!

Socrates,it is the same thing with what I wrote earlier.


calculus
integration
geometry
rectangle
analytic geometry
real analysis
real analysis unsolved

Big rectangle has been cut into several rectangles, with at least one side of integer length. Prove that big rectangle has at least one side integer length.

I think this is a well-known problem :-)

The Zeitz book, [i]The Art and Craft of Problem Solving[/i], has an extensive discussion of this problem.

There are several very different proofs, all of which work. The slickest one I know: assume the edges of the rectangle are aligned parallel to coordinate axes. Integrate $e^{2\pi i(x+y)}$ over the rectangle.


I wanted to make graphs and stuff so I downloaded Latex, Asy, but I was stopped at Ghostgum. I downloaded it, but it says I have to register. In order to register, I have to pay...Is there other programs out there that I dont have to pay?? thanks

Ghostgum is the software company that publishes GSView and I quote from their website http://www.ghostgum.com.au/
[quote]Registering GSview
If you wish to support the development of GSview or wish to disable the GSview nag screen, please consider registering GSview. There is no requirement for you to register GSview. According to the terms of the GSview licence, you can use GSview without any payment.[/quote]
There's more in the readme file when you install the program.


vector
inequalities
linear algebra
linear algebra unsolved

[may be well known and obvious, but seems nice :)]

$(tr(A \cdot B^*))^2 \leq tr(A \cdot A^*) \cdot tr(B \cdot B^*)$ for $A, B \in M_n(C)$

$(A,B)\mapsto \mbox{tr}(B^*A)$ is a scalar product on the vector space $\mathcal M_n(\mathbb C)$, and that inequality is just the Cauchy-Schwartz inequality for the scalar product of a Hilbert space.

Except, of course, you do need one absolute value in there:

$|tr(A \cdot B^*)|^2 \leq tr(A \cdot A^*) \cdot tr(B \cdot B^*)$


induction
real analysis
real analysis unsolved

Let $ v(x)=x_{1}x_{2}...x_{n}\prod_{1\leq i<j \leq n}{(x_{i}-x_{j})}$.
Show that M{$ x: |v(x)|\leq1$} is bounded, where M is Lebesque measure.

I don't think it's bounded independently of $ n$, if that's what you were asking. Let $ x_{0}=0$. For any $ \epsilon>0$ the supremum ${ \prod_{0\le i<j\le n}|x_{i}-x_{j}|}$ taken over $ x_{1},\dots,x_{n}\in [-2+\epsilon,2-\epsilon]$ is less than $ 1$ if $ n$ is large enough. This follows from the definition of [url=http://eom.springer.de/t/t093670.htm]transfinite diameter[/url] and from the fact that the transfinite diameter of a segment is $ 1/4$ of its length. Therefore, $ M\{x\colon |v(x)|\le 1\}\ge (4-2\epsilon)^{n}$ for large $ n$.

So perhaps you want to show that the measure is merely finite. I tried induction on $ n$, but ended up with a mess. Hopefully someone will have a better idea.

[quote="mlok"]I tried induction on $ n$, but ended up with a mess. [/quote]
Hmm... Induction works just fine if all you need is that the measure is finite. First of all note that $ |v_{n}(x)|$ is the product of $ V_{n,j}(x)=|x_{j}|\sqrt{\prod_{i: i\ne j}|x_{j}-x_{i}|}$ ($ j=1,\dots,n$). So, we can assume that $ V_{n,n}(x)$ is the largest. Now, let $ S\subset \mathbb R^{n-1}$ be the set where $ |v_{n-1}|\asymp 1$.

Let us look at what the possible value of $ x_{n}$ can be if $ (x_{1},\dots,x_{n-1})\in 2^{k}S$ with $ k\ge 0$. Then we must at least have $ \prod_{j<n}V_{n,j}\le 1$. But this product is just $ v_{n-1}\sqrt{\prod_{j<n}|x_{n}-x_{j}|}$. Since $ |v_{n-1}|\ge 2^{k n(n-1)/2}$, we conclude that at least one of the factors $ |x_{n}-x_{j}|$ should be less than $ 2^{-nk}$ (up to a constant factor independent of $ k$), so the length of the cross-section is at most $ 2^{-nk}$. But the volume of $ 2^{k}S$ is only $ 2^{(n-1)k}$ times the volume of $ S$, so we get a convergent series.

Now, assume that $ (x_{1},\dots,x_{n-1})\in 2^{-k}S$, $ k>0$. Then we just write $ |v_{n}|=|v_{n-1}|\cdot|x_{n}|\prod_{j<n}|x_{n}-x_{j}|$. Again, $ |v_{n-1}|\ge 2^{-k n(n-1)/2}$, so one of the factors should be at most $ 2^{k(n-1)/2}$ and again the rough estimate for the length of the cross-section yields a convergent series.


trigonometry
inequalities proposed
inequalities

Let $ ABC$ be a triangle. Prove that:
$ sin \frac{3A}{2}\plus{}sin \frac{3B}{2}\plus{}sin \frac{3C}{2} \leq cos \frac{A\minus{}B}{2}\plus{}cos \frac{B\minus{}C}{2}\plus{}cos \frac{C\minus{}A}{2}$
:)

[quote="now or never"]Let $ ABC$ be a triangle. Prove that:
$ sin \frac {3A}{2} \plus{} sin \frac {3B}{2} \plus{} sin \frac {3C}{2} \leq cos \frac {A \minus{} B}{2} \plus{} cos \frac {B \minus{} C}{2} \plus{} cos \frac {C \minus{} A}{2}$
:)[/quote]
[hide="HINT"]$ cos(\frac{A\minus{}B}{2})\equal{}sin(\frac{\pi\minus{}A\plus{}B}{2})$ and use Karamata ineq [/hide]


MATHCOUNTS AMC AIME USA(J)MO USAMO geometry Duke

Now that the first AMC this year is coming up in just about 2 weeks, how will everyone prepare (if at all)? How many other prospective perfect achievers are out there? What topics and skills will you review? Formulas cracked? Problems done? (and maybe something proven). How hard do you think the problems will be this year and what category? (I think a nice medley of Geo/CP/maybe NT probs). Is anyone gonna take a break from the AoPS books or AMC 10/12/AIME/Above/MATHCOUNTS prep and just get in the mindset for the AMC 8?
(I'm ditching all other math... I want to do well...)

I want to know how everyone's feeling about this.

Well, it's hard for me to ditch MATHCOUNTS prep when I have a MATHCOUNTS minicomp three days before the AMC 8... I'm just practicing MATHCOUNTS, the AMC 8 can't be too much harder than a State test...

Hmm...I don't really care that much about how I do on AMC 8, so I'm just going to take it with absolutely no AMC 8 preparation.

I might also do the same with mathcounts.

Heh, I can't take it this year. That means no perfect AMC 8 score in my life. Oh darn.

holy crap. two weeks?! i haven't even registered. i have done absolutely NO AMC 8 PREPARATION. but, maybe AMC 10/12 prep helps. idk.

i actually think amc 8 is easier than mc state b/c both the content and time limit are generous.

And really, guys, it's not that hard to prepare for AMC 8. If you can normally do well on chapter MC, you should hope for a good score on AMC 8.

Honestly, I think a perfect AMC 8 is easily achieved as long as you review all the formulas beforehand (to make it quicker).

Generally, I like to rush through each of the problems once, then go back for vigorous checking (JUST in case you put an extra 0 at the end or subtracted instead of added, etc (not checking failed me at warmups) ). I have friends going for perfects too, so it's gonna be tight, and I hope they do well.

What checking method works for you guys? If there's a problem you're somewhat unsure of, do you make sure you're positive before you move on? (Does it nag at you when you move on?)

And... HAPPY NOVEMBER EVERYONE ^o^

i don't want to sound bad but AMC 8 really doesn't matter, the people who get perfects on AMC 8 are usually at the level of doing well on AMC 10/12/AIME. so if you want to do well on AMC 8, prepare with material that is alot harder than AMC 8 material.

that is my two cents worth, but i may be way off.

I'm aiming for another perfect...forgot it was coming up though, going to start preparing tommorrow :P

[quote="Textangle"]You're right... i just want to do well because it's my last year doing it. :D[/quote]

no i wasn't implying that you shouldn't try to get a perfect score. in fact, you should try very hard to get a perfect. a perfect on AMC 8 is a great thing to have on your mathematical resume. i was just kind of saying that if you don't get a perfect, its not a big deal. I was just saying, to prepare for a contest, do problems that are harder than those problems. me, i am studying for AMC 10/12 and AIME by doing hard AIME problems, easy USAMO problems, and WOOT...

[quote="AIME15"]Oh dang....2 weeks!?

I have all the old AMC 8's except for 1999 and 2000 and 2007 (but I took 2007 for real), so I'll probably do those *gasp* one a day.[/quote]

you probly know about this, but if it helps

[url=http://www.math.ksu.edu/main/handbook/ProblemSets/AMC8Problems/prb1999amc8.pdf]1999 AMC 8[/url]

[url=http://www.math.ksu.edu/main/handbook/ProblemSets/AMC8Problems/prb2000amc8.pdf]2000 AMC 8[/url]

[url=http://www.math.ksu.edu/main/handbook/ProblemSets/AMC8Solutions/sol1999amc8.pdf]1999 AMC 8 solutions[/url]

[url=http://www.math.ksu.edu/main/handbook/ProblemSets/AMC8Solutions/sol2000amc8.pdf]2000 AMC 8 solutions[/url]

sorry, i don't have 2007. i think i got like a 23 or 24 on the actual test, and i was so mad because i missed #4, that i think i either shredded or burned my copy of the 2007 pamphlet.  :( . i was SOOOOO mad

EDIT: i'm pretty sure i shredded it...

[quote="BOGTRO"]I'd prepare by doing past tests and focusing on areas of which you got wrong.

I wish I could ditch everything else and megaprepare (is that a word?) for AMC 8, but Duke (ARML like) is on the Saturday before (...)

Good luck to all taking it, looking for a 25 this year...

Hmm... Past years:

4th Grade: 17
5th Grade: 22
6th Grade: ? looking for 25.

w/e I'll get 24 at least.

Good luck!!![/quote]

For me:

6th grade: 23

I never got the solutions pamphlet, so I only know that I missed 25, so probly a stupid mistake somewhere...

Looking for 25 also.

im not even taking amc 8 :(
i think its pointless b/c i can just get perfect and no other good prizes...
but surely im taking amc 10

[quote="gauss1181"] i can just get perfect and no other good prizes...[/quote]

first of all, no matter how good you are, you are not guaranteed to get a perfect, you maybe able to get a 42 on USAMO, there maybe just one problem on the AMC 8 that you don't know how to do. and you could make careless mistakes. and what other prizes to do you want. you get a plaque and a very strong item on you resume.

how is an [b]8th grade [/b]award strong on your resume? ur only supposed to put stuff from HS :wink:

we are getting off topic, but are you saying that you cannot put anything you did in mathcounts on your resume?

wow...took a practice AMC 8 today...

I missed number 2.  :rotfl:

:blush:

Blah, im' still going for the 25.  But I'd be surprised if anyone else at my school got higher than 20...we have people taking ALgebra 1 and geometry...class average last year was 16 for geometry, 12 for algebra.

:rotfl:

[quote="Altheman"]I don't think that scores on the AMC 8 or mathcounts mean much in terms of college applications... Those competitions are too focused on speed. I think that later contests like AIME and USAMO can hold more weight. I know that in my application process, they were impressed by the problems that I made in my mock tests.[/quote]

Obviously, AIME and USAMO qualifiers will have a good time applying to colleges :wink:

But AMC 8 and mathcounts are fun, and if you're not at AMC 10/12 level yet, they can serve as "preperation"

I've never prepared for an AMC 8 and Im not doing it this year. 6th grade-24, forgot about 10 when finding numbers whose digit sum was a perfect square, 7th grade-25. JUst dont fail.

Yeah, and you were the only perfect scorer from Missouri, and you DON'T HAVE A PICTURE!!

Darn.

Oh well.

I'm just glad I got a 25 last year (my last chance), so now I'm aiming for at least 120 on the AMC 10 this year.

Knowing my luck I will fail and get 24. Can you take boht the AMC 8 and the AMC 10?

Yeah.  Last year I took AMc 8, amc 10, and amc 12. and aime

[quote="mathemonster"]a) where are you applying
b) u just showed them your mock tests at ur interview or sent them to the colleges?[/quote]

a) I'm going to washu in st. louis
b) haha...i didn't just like bust the problems out...i sent it as a part of my application ( I think that many schools would be unlikely to read a supplement like that, but I was a finalist for a scholarship there so the head of the scholarship committee read my problems :D I also showed a professor a few more on the spot when I was talking to him

[quote="AIME15"]Obviously, AIME and USAMO qualifiers will have a good time applying to colleges :wink:
[/quote]

I'd be careful about that statement...doing well on those tests is not sufficient to get into an upper echelon school. Generally those people do well in other things so thats why their college applications processes go well.

On another note, I don't have solid evidence for this, but I think that people who went to mop and/or RSI tend to get into top schools like MIT (not solely due those achievements, but I feel that is a good measure of the level of the applicants).

Washu St. Louis is good but why not MIT/Harvard u seem very smart

[quote="Altheman"]

[quote="AIME15"]Obviously, AIME and USAMO qualifiers will have a good time applying to colleges :wink:
[/quote]

I'd be careful about that statement...doing well on those tests is not sufficient to get into an upper echelon school. Generally those people do well in other things so thats why their college applications processes go well.

On another note, I don't have solid evidence for this, but I think that people who went to mop and/or RSI tend to get into top schools like MIT (not solely due those achievements, but I feel that is a good measure of the level of the applicants).[/quote]

Sorry, that should be AIME and USAMO qualifiers will have a BETTER time applying to colleges than others.


calculus
integration
algebra
polynomial
inequalities
function
real analysis

Let f be in $ C[0,1]$ such that the integral of $ f(x)x^n$ from 0 to 1 is 0 for every integer n greater than or equal to 0. Prove that f(x)=0 for all x in [0,1].

the hint is to use the fact that f can approximated by a sucession of polynomials in C[0,1].

Any suggestions?

So just use the hint :) (or more precisely, for every $ \varepsilon>0$ you can find a polynomial for which $ |f(x)\minus{}p_n (x)|<\varepsilon$ therefore $ \int\limits_{0}^{1} |f(x)|^2<\int\limits_{0}^{1} |f(x)\minus{}p_n (x)|^2 <\varepsilon^2$ because $ \int\limits_{0}^{1} f(x)p_n (x)\equal{}0$. This observation implies that $ \int\limits_{0}^{1} |f(x)|^2\equal{}0$ or $ f(x)\equal{}0$)

Im sorry ive been  trying to do the algebra but i  cant make it work , how did you get the first inequality? :blush:

Ok, suppose that our function $ f$ is real-valued (otherwise consider $ Re(f)$). Then we have $ \int\limits_{0}^{1}(f(x)\minus{}p_n (x))^2dx\equal{}\int\limits_{0}^{1}(f(x))^2dx \minus{}2 \int\limits_{0}^{1} f(x)p_n (x)dx \plus{}\int\limits_{0}^{1}(p_n (x))^2dx$ $ \equal{}\int\limits_{0}^{1}(f(x))^2dx \plus{}\int\limits_{0}^{1}(p_n (x))^2dx >\int\limits_{0}^{1}(f (x))^2dx$ because $ \int\limits_{0}^{1} f(x)p_n (x)dx\equal{}0$ by the condition of the problem.

ohh  ok  i see it now. thanks for your time.


Αν $ ,b,c>\equal{}0$ και $ a\plus{}b\plus{}c\equal{}1$ να αποδειχθεί ότι $ 5(a^{2}\plus{}b^{2}\plus{}c^{2})\leq 6(a^{3}\plus{}b^{3}\plus{}c^{3})\plus{}1$

Ομογενοποιούμε (πολλαπλασιάζουμε το αριστερό μέλος με $ a\plus{}b\plus{}c$ και αντικαθιστούμε στο δεξί το $ 1$ με $ (a\plus{}b\plus{}c)^3$) ανοίγουμε παρενθέσεις και είναι Schur. :wink:

[quote="Protonios"] ανοίγουμε παρενθέσεις και είναι Schur. :wink:[/quote]

Βασικά είναι μια ειδική περίπτωση της ανισότητας Schur, αφού το τελικό αποτέλεσμα είναι $ a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}3abc\geq a^{2}(b\plus{}c)\plus{}b^{2}(c\plus{}a)\plus{}c^{2}(a\plus{}b)$, αλλά όπως και να έχει, η λύση σου έιναι μία από τις κομψότερες, που μπορεί να δώσει κάνεις για αυτή την άσκηση!


calculus
derivative
function
algebra
polynomial
absolute value
calculus computations

g(x) = sin(x) + cos(.5x) on (-4, 4)

Okay, so for the above expression, I have to:
a) Calculate an expression for the 5th derivative g^(5)(x).
b) Find a bound for the absolute value of g^(5)(x) on the interval indicated.
c) Use Taylor's Theorem to find a cap for the error; that is, a cap for the difference between the function and the Maclaurin polynomial of degree four on the interval.

I did part a and this is what I got:

g'(x) = cos(x)-.5sin(.5x)
g''(x) = -.25cos(.5x) - sin(x)
g'''(x) = .125sin(.5x) - cos(x)
g^(4)(x) = .0625cos(.5x) + sin(x)
g^(5)(x) = cos(x) - .03125sin(.5x)

The last one is the 5th derivative.

I attempted part b and basically what I did was substitute 4 for x in the expression for the 5th derivative of g(x). From this I got approximately .68, which I rounde up to .7. Then for a final answer, I got
the absolute value of cos(x) - .03125sin(.5x) <= .7

I'm not sure if the above is right, but that's what I got.

My real problem lies in doing part c. I guess I'm confused as to the method because my book isn't very clear.

[quote="JBL"][quote="muckymotter"]I attempted part b and basically what I did was substitute 4 for x in the expression for the 5th derivative of g(x).[/quote]

This would only make sense if you happened to know that your function was increasing (and could be extended to the supremum of its domain).

For part c, do you know what Taylor's Theorem says?[/quote]

Taylor's Theorem says that:

Suppose g is a function with derivatives g^(n) of all orders n on an interval (a - r, a + r)(r > 0).  Suppose also that for every natural number n, there's a positive number Max sub n such that g^(n)(x) is less than or equal to Max sub n for all x in (a - r, a + r).  Then for every natural number n, and every x in the interval (a - r, a  + r), the Taylor polynomial T sub n-1(x) of degree n-1 based at a, approximates g(x) with an error no larger than

Max sub n divided by n! * |x - a|^n

I guess I'm confused about how to use this.


rotation
geometry
geometric transformation
LaTeX

Let n be a natural number diffrent from 0 and $ A_1A_2...A_{2n\plus{}1}$ a convex poligon.For $ i\equal{}1,2,3,...,2n\plus{}1$ let $ d_i$ be the distance from $ A_{i\plus{}n\plus{}1}$ to $ A_iA_{i\plus{}1}$, where $ A_{2n\plus{}1\plus{}i}\equal{}A_i$.Prove that \[ \sum_{i\equal{}1}^{2n\plus{}1} \displaystyle\frac{A_iA_{i\plus{}1}}{d_i}>\pi\]

We have to know that in any triangle ABC we have $ \frac{a}{h_a}> <A$ where A is in radians and alsow we have to know that $ \sum_{i\equal{}1}^{2n\plus{}1}<(A_iA_{n\plus{}1\plus{}i}A_{i\plus{}1})\equal{}\pi$.
And now the proof of the problem.We rotate $ A_iA_{n\plus{}1\plus{}i}$ in clockwise with the center $ A_{n\plus{}1\plus{}i}$.Then we rotate $ A_{i\plus{}1}A_{n\plus{}1\plus{}i}$ with center $ A_{i\plus{}1}$ until we get to $ A_{i\plus{}1}A_{n\plus{}1}$.We continue this proces rotating in clockwise $ A_jA_{n\plus{}1\plus{}j}$  with center $ A_{n\plus{}1\plus{}j}$ and we get at $ A_{j\plus{}1}A_{n\plus{}1}$,until we get back at $ A_iA_{n\plus{}1\plus{}i}$>At this moment we've passed thru all the angles $ A_iA_{n\plus{}1\minus{}i}$ only once and we've done a complete rotation for $ A_iA_{n\plus{}1\plus{}i}$,therefore,in total,we've covered $ 2 \pi$ twice,therefore we covered $ \pi$.And it results that \[ \sum_{i\equal{}1}^{2n\plus{}1} A_iA_{n\plus{}1\plus{}i}A_{i\plus{}1} \equal{} \pi\]

Let's prove now that in any triangle ABC we have $ \frac{a}{h_a}><A$
We have 2 cases
[b]Case 1[/b] :<B,<C < 90. Let $ \alpha,\beta$ the angles formed by AB and ha ,AC and ha.Then $ a\equal{}ha(tan\alpha\plus{}tan\beta)$ therefore $ tan\alpha>\alpha$
[b]Case 2[/b] <C >90. $ a\equal{}h_a(tan\alpha\minus{}tan\beta)$ therefore $ \displaystyle\frac{a}{(\alpha\minus{}\beta)h_a}\equal{}\displaystyle\frac{tan\alpha\minus{}tan\beta}{\alpha\minus{}\beta}\equal{}tan\omega$ where $ \omega \in (\alpha,\beta)$.But $ tan\omega\equal{}\frac{1}{cos^2 \omega}>1$ therefore $ \frac{a}{h_a}><A$

Quite possibly a typo.  I think he means $ \frac{a}{h_a} > \angle A$.

@balan razvan: in LaTeX mode, use "\angle" to get the nice little angle sign.

ok, and teki teki is wright about what I was meaning


Harold tosses a nickel four times. The probability that he gets at least as many heads as tails is:

$\text{(A) }\frac{5}{16}\qquad\text{(B) }\frac{3}{8}\qquad\text{(C) }\frac{1}{2}\qquad\text{(D) }\frac{5}{8}\qquad\text{(E) }\frac{11}{16}$

[quote="Klebian"][hide]Well, there are 3 outcomes: He could either have an equal number of heads and tails, or he could have more heads than tails, or he could have less heads than tails. Since there is an equal opportunity for heads or tails, the last 2 outcomes have the same probability. So we only have to find the probability of getting an equal number of heads or tails (HHTT). There's 1/16 chance to have HHTT, and HHTT can be arranged in 6 ways, so there is a 6/16 chance of getting 2 heads and 2 tails. So there is a (1- 6/16)/2 = 5/16 chance of getting LESS heads than tails, so there is a (1 - 5/16) = 11/16 chance of getting at least as many heads as tails. $\boxed{E}$.[/hide]

Btw, I think this is not really HSB level (AMC 8)[/quote]

Yeah, this should be in the middle school section.


synthesise pent-3eneoic acid from acetaldehyde

is this right?  :D
1. aldol with itself to give but-2-ene-1-al
2. HCN and hydrolyse to get pent-3-ene-2-ol-oic acid
3. P/HI to get pent-3-ene-oic acid (According to Carcul P/HI doesnt reduce an acid  :P )

but it does. ive seen the reagent being used in many books. i think carcul said that the reagent was not commonly used during synthesis.
as for reaction 2, you would probably be getting a michael product too, albeit a minor one.

[quote="Carcul"][quote="VARUNARASU"]as for reaction 2, you would probably be getting a michael product too, albeit a minor one.[/quote]

Wrong two times. First, that wouldn't be a Michael addition. Second, cyanide is a weak nucleophile, and 1,4-addition is then the preferred pathway, not carbonyl addition.[/quote]

then what addition would that be?

1.Perform selfaldol with ethanal (base) --> crotonaldehyde
2.Tollen's reagent followed by H+ --> crotonic acid
3.Perform arndt eistert (I am unsure whether it undergoes) --> Pent-3-enoic acid

[quote="Carcul"]A 1,4-addition or conjugate addition of cyanide.[/quote]
my book says that would be an example of michael reaction but again my book could be wrong.

[quote="chemrock"]
Perform arndt eistert (I am unsure whether it undergoes) --> Pent-3-enoic acid
[/quote]
ch2n2 is basically a carbene, so i think it would react with the double bond, wouldnt it, carcul?
mebbe you could protect the double bond by ag+.

[quote="VARUNARASU"]my book says that would be an example of michael reaction but again my book could be wrong.[/quote]

By definition, a Michael addition is the conjugate addition of an enolate anion to an alpha,beta-unsaturated carbonyl system. Yes, your book is wrong.

[quote="VARUNARASU"]ch2n2 is basically a carbene, so i think it would react with the double bond, wouldnt it, carcul?[/quote]

Diazomethane do not react directly as a carbene, but as a base/nucleophile or as a 1,3-dipolar compound. I was also with doubts regarding Chemrock's proposal, but a book I have consulted says that "unsaturation can be present". However, I know from other sources that when we have a beta,gamma-, gamma, delta-, and so on, unsaturated acid, other products are obtained unless we use photolysis in the Wolff rearrangement step. In this case, we have an alpha,beta-unsaturated acid, which on the other hand makes the double bond a very good dipolarophile and so susceptible to 1,3-dipolar cycloaddition with diazomethane. Therefore, as a final opinion, we might get a reasonable yield of the desired product, but always accompanied by significant amounts of other products.

[quote="valeriummaximum"]see if this conversion is right.......

1.Treat with Phosphorous ylide i.e. Ph3P=CH-CH2-Cl
2.treat with KCN and hydrolyse to get the acid...........[/quote]
Well, for starters how would you prepare such a phosphorane. Wittigs reagent is prepared by the foll method.
Ph3P + CH3Br(along with strong base like C6H5Li)->Ph3P=CH2(ylide)
The phosphorane you have used cannot be prepared by this method.


For #1 and #3 of USAMTS, the questions [i]dont[/i] say "find, with proof,_____." My question is that what if you assume something in your solution that may or may not be true, but you dont expand on it and continue with the solution? Will you get full credit if your answer is correct?

Say for part of your solution you write (just an example)"We note that z must be the highest power of 3q in order to fulfill the question requirements of p(z)....[insert rest of solution].....QED"  Will you get full credit for that even if your answer is correct but you dont expand on that important piece of information?

In general, you should prove everything. If you only supply an answer, you definately won't get full credit. If you supply an answer and an explanation that isn't a proof, you probably won't get full credit (3 or 4 is likely, assuming the answer is correct)

HINT: If you're result is nontrivial, then we can assume that

[hide]USAMTS graders are not mindreaders and will expect justification[/hide]

Hope that helps.  Why don't you just try proving your result or assumption?

Dont you mean, if your result is trivial?


There are six cities in a given country. Each two of them are joined by a one way road. Prove that we can leave from a city, visit three other cities and return to visit the initial city, having gone by only $ 4$ roads. Can we guarantee to visit four other cities and go back to the initial one using only $ 5$ roads?

I think I can do the second question: [hide]Denote a city in which all roads lead to it  as a [b] center [/b] , and a city in which all roads lead out of it as a [b] hub [/b].  (What else should I call them?)  First, clearly, there can be at most one center, because if there were two or more, then there could not be a one-way road joining the centers.  Similarly, there can be no one-way road joining two hubs.  Thus, there are at most one each of a hub and a center.
Now, note that no path can pass through a hub, because there are no roads to a hub.  Even if  course starts at a hub, the path will never be able to come back to the beginning.  Similarly, no path can touch a center, or it will never be able to leave.
Hence, we cannot guarantee going to 4 cities and coming back using 5 roads, because if an layout has both a hub and a center, then there are only 4 usable cities, and we need 5. [/hide]


So, I discovered an incredibly useful function in the "toolbox" called "special pages."  It gives me a list of options, such as "double redirects," "dead-end pages," and "wanted pages."  I don't know who has access to this page, but it's incredibly useful -- for instance, you can quickly see what pages don't exist and are being linked to, and then fix that situation (either by creating a stub page with a little bit of content so that someone will stumble upon it and add real content or by noticing bad jobs of linkage ("equations," "fibonacci numbers") and correcting the linking article).  So, for instance, I no longer have to rely on the "Random Page" link to find something interesting to work on.

So, that is just a public-service announcement.  The problem is that when I click on the link to "special pages," I sometimes have to reload or reclick repeatedly (3 seems to be the luck number) before it actually displays this list for me.  This is kind of annoying.

I also don't know who has access to this page, and I don't know how to link to it inside the wiki.  But I do urge anybody who hasn't discovered it yet to check it out.

[quote="JBL"]The problem is that when I click on the link to "special pages," I sometimes have to reload or reclick repeatedly (3 seems to be the luck number) before it actually displays this list for me.  This is kind of annoying.[/quote]Believe me, if I had any idea what was causing this, I'd fix it.  But I'm clueless.

The Special Pages page will come up eventually, if you click on the "Special pages" link in the toolbox enough times.  (I had to click on it 8 times once.)

It should be visible to anyone, whether they're logged in or not.

For permanent links (ie not having to click a random number of times) see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98638]here[/url].


log2(x)^2=/2log2x.
log(2x)^2=2log2x.

P.S.:from now on till an uncertain time i'll be a hodge conjecture.

just to make Cronus' post a bit more readable with $ \text{\LaTeX}$:

If you want to take down the exponent in a logarithm such as this:
\[ \log{(ax)^b}\]
where $ a$ and $ b$ are constants, moving $ b$ down as a coefficient like this:
\[ \log{(ax)^b=b\log{ax}}\]
is valid.

However, in your post, you had
\[ \log{2x^2}\]
which is not equivalent to
\[ \log{(2x)^2}=\log{4x^2}\]
so equating that to $ 2\log{2x}$ was incorrect.

HTH :)

ThNX 4 doing me this huuuuuuuuuuuuuuuuuuuuuuuuuuuge favour.
Undefined:U r welcome.
                                :cool: :pilot:

I thanked u so u were supposed to say that i was welcome.
as u did me a favour so i tried to repay u by saving u from the pain from writing it. :rotfl:

thanks for clearing that up for me mates!


calculus
integration
limit
Putnam
search
real analysis
calculus computations

Evaluate $ \lim_{p\rightarrow 0}\left ( \int_{0}^{1}\left ( bx\plus{}a(1\minus{}x) \right )^{p}dx \right )^{\frac{1}{p}}$ with $ 0<a<b$.

nope! It's from a math contest for undergraduates here in our country. I'm still solving it but I have the answer...

I remember the problem which someone posted as Iran's university problem several years ago.

Anyway, here is the problem.

[url]http://www.mathlinks.ro/viewtopic.php?search_id=1858812326&t=115865[/url]

oh..how remarkable kunny! the problem is really the same from the virginia tech regional math contest...anyway, i think i don't I (or we) need to solve it anymore...c",)...but i'll try giving more problems from our real analysis class...

I remember the problem because I couldn't solve it.
Anyway, good problem repeates. :lol:

I'm looking forward to the problems. :)


geometry
topology
3D geometry
sphere
rotation
geometry proposed

A ring is the area between two circles with the same center, and width of a ring is the difference between the radii of two circles.
[img]http://i18.tinypic.com/6cdmvi8.png[/img]
a) Can we put uncountable disjoint rings of width 1(not necessarily same) in the space such that each two of them can not be separated.
[img]http://i19.tinypic.com/4qgx30j.png[/img]
b) What's the answer if 1 is replaced with 0?

[quote="borislav_mirchev"]I think that Iranian mathematicians have lots ot things to show. There are many problems in Iranian Math Olympiad taken from European countries but math unions people. I'm sorry for offtopic but I think that if you put as more problems in Resources section in this site from Iranian Math Olympiad will be very good. I like this competition. Thank you again![/quote]
It is not so. All of the problems are designed in large sessions and after long sessions. They are not copied from European countries problem. If it is, it's is accidental.

[i]P.S. I believe the same as you : Math unite people together.[/i]

I don't know what is the situation now, but I took in mind problems in Mohammad Katoozian web site. [1997-1999] - it is maybe first round. http://www.geocities.com/CollegePark/Lounge/5284/index.html
1998-1-i saw this in Argentinian math olympiad.
1999-1 - seems to be from a russian olympiad, 5 is a theorem, and maybe 2, 4 - saw something like this in Irish math olympiad, problem 6 is from APMO or Hungary Israel math competition i think, I saw this in mathlinks.

Those problems are the old ones. But I tell you after 2001 or 2002 nothing is copied from other Olympiads.

It is very good! I think it also take lots of time and lots of people are engaged with the competition. I think in future Iran will have very strong team, traditions and success. As I understood you are engaged in the olympiad. I wish success to you and your team!

(a) Let $ R_{1},R_{2}$ be two intertwining rings, and let $ C_{1},C_{2}$ be the outer boundary circles of $ R_{1},R_{2}$ respectively. Finally, let $ \pi$ be the plane in which $ R_{1}$ lies.

$ C_{2}$ cuts the portion of $ \pi$ enclosed by the inner boundary circle of $ R_{1}$ in some point $ P$, whose distance from $ C_{1}$ must thus be at least $ 1$. It follows that the [url=http://planetmath.org/encyclopedia/HausdorffMetric.html]Hausdorff distance[/url] between $ C_{1}$ and $ C_{2}$ is at least $ 1$. This holds for any two rings $ R_{1},R_{2}$ in our set, so the outer circles of such a family form a discrete set in the metric space of all compact subsets of $ \mathbb R^{3}$ with the Hausdorff metric. Since this metric space is separable, we're done: any set of mutually intertwining rings with width $ 1$ is at most countable.

(b) The [url=http://www.mathematik.uni-bielefeld.de/birep/top/hopf/hopf-fib.htm]Hopf map[/url] provides an example. It's a map $ p:\mathbb S^{3}\to\mathbb S^{2}$ such that the preimage of any point in $ \mathbb S^{2}$ is a circle, and such that any two distinct preimages are linked. There are pictures on that website of mutually linked circles on a torus obtained by restricting this map.

Official Solution:
The answer for (a) is no. Assume that two rings $ A,A'$ with centers $ O,O'$ are tied together. We will show that $ |OO'|\geq 1$.
Let $ R,R'$ denote the inner radii of the rings. Assume that $ R'\geq R$. The plane of the ring $ A'$ intersects	the ring $ A$, in two intervals, one inside the ring, and the other outside.
Let $ U$ be a point on the outer border of $ A'$ that is on the plane and inside the ring $ A$.
Then $ R'\plus{}1\equal{}|O'U|\leq |O'O|\plus{}|OU|\leq |OO'|\plus{}R\leq |OO'|\plus{}R'$ which shows that $ |OO'|\geq 1$.
Now for each ring with $ O$ as the center, consider the sphere with radius $ \frac 13$ around $ O$. This sphere contains a point with rational coordinates. So there is an injective mapping from the rings to the rational points. (It is injective because no two such spheres intersect) So the number of rings must be countable.
The answer for (b) is yes. Consider a circle with center $ (0,0,0)$ lying in the xy plane and having radius $ 1$. For each $ 0\leq\theta\leq0.1$, move the circle by $ \theta$ along the positive direction of the x axis, and rotate it along this axis by $ \theta$ radians.
It is obvious that the rings constructed in this way are pairwise tied to each other. They are in one-to-one correspondence to $ [0,0.1]$ which is uncountable.


modular arithmetic
number theory
number theory unsolved

[quote="taozh"]Q89 in Number theory:

if you move the last digit of a number to the front, then it becomes nine times larger, find the smallest such number.

I find there is no solution for this question. The hint in the book is not very clear.

tao zh[/quote]

Consider the number $ A \equal{} 10x \plus{} a$ with $ 10^{n \minus{} 1} > x\geq 10^{n \minus{} 2}$ ($ A$ has $ n$ digits) and $ a\in\{0,1,2,3,4,5,6,7,8,9\}$
We want to have $ a10^{n \minus{} 1} \plus{} x \equal{} 9(10x \plus{} a)$ and so $ 89x \equal{} a(10^{n \minus{} 1} \minus{} 9)$

Since $ x\geq 10^{n \minus{} 2}$, we have $ a(10^{n \minus{} 1} \minus{} 9)\geq 89\cdot 10^{n \minus{} 2}$, we have $ a > 8$ and so $ a \equal{} 9$ and then $ n\geq 4$

So we want $ 9\cdot 10^{n \minus{} 1} \plus{} x \equal{} 9(10x \plus{} 9)$ and so $ x \equal{} 9\frac {10^{n \minus{} 1} \minus{} 9}{89}$

So we are looking for the fewest $ n\geq 4$ such that $ 10^{n \minus{} 1} \equal{} 9\pmod{89}$ or also $ 10^n \equal{} 1\pmod{89}$.

Since $ 89$ is prime, $ n$ must divide $ 88$ and so $ n\in\{1,2,4,8,11,22,44,88\}$.

A quick check gives $ n \equal{} 44$ and so $ x \equal{} 9\frac {10^{43} \minus{} 9}{89}$

And the requested number is $ A \equal{} 90\frac {10^{43} \minus{} 9}{89} \plus{} 9 \equal{} 10112359550561797752808988764044943820224719$


inequalities
logarithms
function
inequalities unsolved

Let $P$ be the product af the positive real numbers $a_{1}, a_{2}, \ldots ,a_{n}$. Prove that
$\prod_{k=1}^{n}({1+\frac{1}{a_{k}^{n}+a_{k}^{2n}})\geq(1+\frac{1}{P+P^{2}})^{n}}$

Let $f(x)=\ln\left(1+\frac{1}{e^{nx}+e^{2nx}}\right)$ for all real $x.$  The function $f$ is convex since

\[f''(x)=\frac{n^{2}e^{2nx}(4e^{2nx}+7e^{nx}+4)}{(e^{2nx}+e^{nx}+1)^{2}(e^{nx}+1)^{2}},\]

which is clearly positive.

Now we apply Jensen’s Inequality to $f$ with the $n$ values $\ln x_{1},$, $\ln x_{2},\ldots,\ln x_{n},$  letting $x_{1}x_{2}\cdots x_{n}=P.$  We get

\[\sum_{i=1}^{n}\ln \left(1+\frac{1}{e^{n\ln x_{i}}+e^{2n\ln x_{i}}}\right)\geq n \cdot \ln \left(1+\frac{1}{e^{n(\frac{\ln x_{1}+\ln x+2+\cdots+\ln x_{n}}{n})}+e^{2n(\frac{\ln x_{1}+\ln x_{2}+\cdots+\ln x_{n}}{n})}}\right).\]

This simplifies to

\[\sum_{i=1}^{n}\ln\left(1+\frac{1}{x_{i}^{n}+x_{i}^{2n}}\right) \geq n \cdot \ln\left(1+\frac{1}{e^{\ln P}+e^{2\ln P}}\right).\]

Exponentiating both sides gives

\[\prod_{i=1}^{n}\left(1+\frac{1}{x_{1}^{n}+x_{1}^{2n}}\right) \geq \left(1+\frac{1}{P+P^{2}}\right)^{n}\]

as desired.


These problems are probably very easy, but I can't remember how to do them.

1. –xy + zy for x = 1, y = 2 and z = 3

2. 5x + 3 – x = 25

3. 3x – 7 + 4x = 10x + 2

4. Solve for n: 8 < n - 6

5. Solve for x: 2x – 10 > 4x + 3

6. Find the direct variation equation for the point (3,4).

Thanks

Post your problems in separate posts next time-I think that still applies. ;)
[hide="1"]Plug the numbers in:-(1)(2)+(3)(2)=-2+6=[b]4[/b].[/hide]
[hide="2"]First, combine the common terms (5x and -x) to get 4x+3=25. Subtract 3 from both sides to get 4x=22, or x=22/4=[b]11/2[/b].The whole idea for these problems is to isolate the variable.[/hide]
[hide="3"]Again, combine common terms to get 7x+7=10x+2. Then, subtract 7x from both sides to get 7=3x+2. Now take 2 away from both sides and you get 5=3x, or x=[b]5/3[/b].[/hide]
[hide="4"]In inequalities, we also want to get the variable by itself. This can be accomplished by adding 6 to both sides, and we get 14<n. Usually this will be written with n first ([b]n>14[/b]).[/hide]
[hide="5"]First, subtract 2x from both sides to get -10>2x+3. Now take away 3 from both sides, and you get -13>2x. Divide both sides by 2 and you get -13/2>x, or [b]x<-13/2[/b].[/hide]
[hide="6"]I'm not so sure about this, but I think its y/x, so $\frac{4}{3}$?[/hide]
Let me know if you need any more help.

Thank you for your help. That basically got me through the problems and I can solve on my own now.

[hide][quote]3
Again, combine common terms to get 7x+7=10x+2. Then, subtract 7x from both sides to get 7=3x+2. Now take 2 away from both sides and you get 5=3x, or x=5/3. [/quote]
combining the terms on the left side yields $7x-7$. solving gives $x=-3$

6. a direct variation equation is in the form $y=kx$ so the y-intercept is 0 and the line goes through the origin. here $x=3$ and $y=4$. substituting gives $4=3k$ and $k=\frac{4}{3}$ so the equation is $y=\frac{4}{3}x$[/hide]

just plug in numbers for variables
1. 4

2. combine terms and solve to get 11/2 or 5.2

3. someone messed up here. x equals -3

4. 14<n. solve like an equation but keep it as an inequality.

5. x<-13/2. the inequality sign switches when you divide or times both sides by a negative number.

6. the equation is y=kx where k is a constant. so when u plug in u get 4=3k which makes k 4/3 and then you plug k into the direct variation equation.


Find five positive integers a,b,c,d,e such that no subset of these integers has their sum divisible by 5.

[hide="Hint"]Look at things mod 5.[/hide]

[quote="4everwise"]Find five positive integers a,b,c,d,e such that no subset of these integers is divisible by 5.

[hide="Hint"]Look at things mod 5.[/hide][/quote]

do you mean the sum of the subset?

[quote="matt276eagles"]I assume you mean the sum of any subset is not divisible by $5$.  Right?[/quote]Yeah, the sum.  :roll:

[hide]If $a_{1},a_{2},a_{3},a_{4},a_{5}$ is a solution, then $a_{1}+5k_{1},a_{2}+5k_{2},a_{3}+5k_{3},a_{4}+5k_{4},a_{5}+5k_{5}$ for integer $k_{i}$ is also a solution.  Therefore, assume WLOG that $0\le a_{i}\le 4$.

Clearly, $a_{i}\ne 0$, or the subset containing only $a_{i}$ is divisible by $5$.  Also, we cannot have $a_{1}=a_{2}=a_{3}=a_{4}=a_{5}$, or the subset containing all elements is divisible by $5$.

We cannot have both a $1$ and a $4$ at the same time, and similarly we cannot have both a $2$ and a $3$ at the same time.  Therefore, we must have exactly two different numbers.

If we have all $1$'s and $2$'s, then we must have a subset of either $\{1,2,2\}$ or $\{1,1,1,2\}$, either of which are divisible by $5$.

If we have all $1$'s and $3$'s, then we must have a subset of either $\{1,1,3\}$ or $\{1,3,3,3\}$, either of which are divisible by $5$.

If we have all $4$'s and $2$'s, then we must have a subset of either $\{2,4,4\}$ or $\{2,2,2,4\}$, either of which are divisible by $5$.

If we have all $4$'s and $3$'s, then we must have a subset of either $\{3,3,4\}$ or $\{3,4,4,4\}$, either of which are divisible by $5$.

Therefore, there are no solutions.[/hide]


geometry
geometric transformation
reflection
USAMTS
email

If I've faxed my entry form to the number stated at the bottom of the form, how long should it take before it reflects on the My USAMTS page?

I can't say, but give it a while; remember they have lots of forms to go through!

It usually does take a while for things to show up on the website though...

We are entering them into a system approximately once a week.  So it will not show up as entered right away, but you should see it within a week or two.

If it has been [b]more than 3 weeks[/b] since you submitted your form, and it still does not appear as "received", please let us know by email at usamts@usamts.org.  When contacting us, it is helpful if you tell us approximately when (what date) and how (mail or fax) you submitted your form.

Also, I will reiterate that if you are planning to submit your Round 1 solutions by mail, then you can simply submit the Entry Form along with your Round 1 solutions.


integration
function
limit
real analysis
real analysis unsolved

$ f\in C(\minus{}\infty,\plus{}\infty), g(x)\equal{}f(x)\int_0^x{f(t)dt}$, $ g(x)$ is monotonous decreasing. Prove that $ f\equiv 0$.

If $ h(x) \equal{} \int_0^x f(f)\;\mbox{d}t$, then $ h \in C^1$ and
$ g(x) \equal{} \frac12 (h^2(x))'$.
Suppose $ g(c) \ne 0$ for a certain $ c$.
Then the graph of $ h^2$ is below the tangent to the graph of $ h^2$ at the point $ c$, because $ g$ is decreasing.
Then part of the graph of $ h^2$ is below the x-axis, which is impossible, because $ h^2 > 0$.
Thus $ g \equal{} h\equal{} 0$ and $ f$ is a constant function (not necessarily 0).

You are right that $ f$ is a constant, but we can also conclude that $ f\equal{}0$.
Since $ g$ is decreasing, we have
$ g'(0)\equal{}\lim_{x\rightarrow 0}{\frac{f(x)\int_0^x{f(t)dt}}{x}}\equal{}f^2(0)\leq 0,$
which implies $ f(0)\equal{}0$, hence $ f\equal{}0$

that means h^2 is a convex(concave) function
h^2((x+y)/2)>=(h^2(x)+h^2(y))/2
and h(0)=0
so,h=0
so,f=0


modular arithmetic
number theory
prime numbers
divisibility tests

What are the greatest and smallest six digit ( all six to be different .... no repeatation ) number divisible by the all the first four prime numbers that is $2$, $3$, $5$, and $7$ ?

no direct calculation....

[hide]A few things to notice:

The number ends in 0, either way.
The sum of digits must be divisible by 3.
This accounts for 2, 3, and 5.

There's a trick for divisibility by 7 that derives from the fact that $7 | 10^{6k}-1$ and $7 | 10^{6k-3}+1$.
So we can group the six digits into ones of three - if the number is $abcdef$, we find that if $7 | (def-abc)$ then $7 | abcdef$.

For the small number, set the first three digits to $123$. Then $de0$ are the last three digits. We know (1) $3 | d+e$. Because $123 \equiv 4\pmod{7}$, (2) $de0 \equiv 4\pmod{7}$. We know that $900$ meets requirements (1) and (2).
So find a number equivalent $900\pmod{7}$ that contains no digits 0 - 3 and have two distinct digits. The lowest number is $480$... so our number is $123480$.

For the big number, set the first three digits to $987$. Still $de0$ are the last three. We know (1), and since $987 \equiv 0\pmod{7}$, we know (3) $de0 \equiv 0\pmod{7}$. We can see that $700$ meets (1) and (3). But there are repeated digits. So the actual largest number is $630$. So our number is $987630$.[/hide]


At $ t\equal{}0,$ a massless bucket contains a mass $ M$ of sand. It is connected to a wall by a massless spring with constant tension $ T,$ (tension is independent of length). The ground is frictionless and the initial distance to the wall is $ L.$ At later times, let $ x$ be the distance from the wall, and let $ m$ be the mass of sand in the bucket. The bucket is released, and on its way to the wall, it leaks sand at a rate $ dm/dx \equal{} M/L.$

a) What is the kinetic energy of the sand in the bucket as a function of $ x$?

Here is my solution:

[hide="My Solution"]We have by Newton's Second Law, that

\[ \sum F \equal{} \frac{d}{dt}(mx') \equal{} \frac{dm}{dt}x' \plus{} \frac{dx'}{dt}m \equal{} T
\]

By the Chain rule,

\[ \frac{dm}{dx} \equal{} \frac{\frac{dm}{dt}}{\frac{dx}{dt}} \equal{} \frac{\frac{dm}{dt}}{x'} \equal{} \frac{M}{L} \Rightarrow \frac{dm}{dt} \equal{} \frac{M}{L}
\]

So, we substitute above to get

\[ T \equal{} \frac{M}{L}x'^2 \plus{} mx'' \Leftrightarrow T \equal{} \frac{M}{L}x'^2 \plus{} \frac{M}{L}xx''
\]

since clearly, $ m(x) \equal{} Mx/L.$ We then have

\[ \frac{TL}{M} \equal{} x'^2 \plus{} xx'' \equal{} (xx')' \Leftrightarrow \frac{TL}{M}t \equal{} xx' \Leftrightarrow \frac{TL}{M} dt \equal{} x dx \Rightarrow TL/M t^2 \equal{} 1/2 x^2 \Rightarrow x \equal{} t \sqrt{2TL/M} \Rightarrow v \equal{} dx/dt \equal{} \sqrt{2TL/M} \Rightarrow mv^2/2 \equal{} Tx
\][/hide]
But this isn't the right answer... :maybe: why?

if i understand the question properly then $ \frac {dm}{dx} \equal{} \minus{} \frac {M}{L}$ similarly $ \frac {dm}{dt} \equal{} \minus{} \frac {Mx^{'}}{L}$

[hide="perhaps"]
you missed a +C in $ m(x)\equal{}\frac{M}{L} x$
or, more likely,
$ \frac{TL}{M}t \equal{} xx' \Leftrightarrow \frac{TL}{M} dt \equal{} x dx \Rightarrow TL/M t^2 \equal{} 1/2 x^2$
you missed a t when typing i think.
should be
$ \frac{TL}{M}t \equal{} xx' \Leftrightarrow \frac{TL}{M}t dt \equal{} x dx \Rightarrow 1/2 TL/M t^2 \equal{} 1/2 x^2$
[/hide]

[quote="genericme"]I could be stupid, but why is it that $ m(x) \equal{} Mx/L$?  The solution is based on this assumption.[/quote]
yes taht's wrong because the sand's leaking so $ \frac {dm}{dx} \equal{} \frac { \minus{} M}{L} \Rightarrow m(x) \equal{} M \minus{} \frac {Mx}{L}$

The problem looks weird, but the mass at position $ x$ should be $ Mx/L$, since $ m$ is linear in $ x$, and blah blah blah you can solve stuff. Then $ x'' \equal{} F/m \equal{} \minus{}kx/(Mx/L) \equal{} \minus{}kL/M$. This means that $ x \equal{} L \minus{} kL/2M \cdot t^2$, so that the velocity at position $ x$ is going to be $ \minus{}kL/M \cdot \sqrt{(L\minus{}x)2M/kL}$.

Then KE is $ 1/2 \cdot Mx/L \cdot (\minus{}kL/M)^2 \cdot (L\minus{}x)2M/kL \equal{} k^2 x (L\minus{}x)$. Hmmm hopefully that's right... I didn't check if it's the same as your answer, but $ T$ is not constant...

[quote="probability1.01"]The problem looks weird, but the mass at position $ x$ should be $ Mx/L$, since $ m$ is linear in $ x$, and blah blah blah you can solve stuff. Then $ x'' \equal{} F/m \equal{} \minus{} kx/(Mx/L) \equal{} \minus{} kL/M$.[/quote]

Ah, but you forgot to use the chain rule like Karth did - you can only say that $ x'' \equal{} F/m$ when $ m$ is constant, but since $ m$ depends on $ t$, you have to use $ F_{net} \equal{} d(mx')/dt$.

But yeah, I can't imagine a spring whose tension is constant.

And I agree with pardesi, $ m(x) \equal{} M \minus{} Mx/L$

Oh yeah oops I first tried a more complicated way and got an answer that looked like that, so I just used it. But actually I had $ kx(L\minus{}x)$ from solving for KE directly. Hmmm... also I just realized that is probably the same as $ Tx$, and the confusion is just in the direction of $ x$.

I'm definitely sure about $ m(x) \equal{} Mx/L$... remember that $ x$ is the distance from the wall to the bucket... so, we have that
\[ dm/dx \equal{} M/L \Leftrightarrow \Delta m \equal{} M \Delta x / L \Leftrightarrow m \minus{} M \equal{} M (x \minus{} L)/L \Leftrightarrow m \minus{} M \equal{} Mx/L \minus{} M \Leftrightarrow m \equal{} Mx/L.
\]

In addition, $ dm/dt < 0, dx/dt < 0,$ so this would mean that $ dm/dx \equal{} M/L> 0.$ The sign for $ dm/dx$ is correct.

Also, I was thinking... is it possible that we must only consider $ F \equal{} ma$ as opposed to $ F \equal{} d/dt(mv)$ since the tension is invariant?

OK, so I was confused about the direction of $ x$.  But it's still necessary to consider $ d(mv)/dt$ because $ m$ varies with $ t$.

I see you guys are having a hard time here. So well here is the solution: :D
Pardesi cannot be right with: $ m(x)\equal{}M\minus{}M\frac{x}{L}$ since, then if we get: $ m(0)\equal{}M$, and $ x$ is measured from the wall.... $ m(L)\equal{}M$ should be..

So the sand leaks from the bucket as:
\[ \dfrac{dm}{dx} \equal{} \dfrac{M}{L}
\]
Now this is a straighforward differential equation we can integrate at once:
\[ m \equal{} \dfrac{M}{L}x \plus{} C
\]
the initial condition is : $ m(L) \equal{} M$, so plugging this in we get: $ C \equal{} 0$.
So the $ m(x)$ function is:
\[ m(x) \equal{} \dfrac{M}{L}x
\]
Now the kinetic energy is $ E_k \equal{} \frac12 mv^2$.
So we need the $ v(x)$ function.
To get this lets solve the equation of motion!
Since the tension in the spring is constant, we have:
\[ m\ddot x \equal{} \minus{} T
\]
We need the kinetic energy as a function of $ x$ so we need only the $ v(x)$ function not the $ v(t)$.
now lets do the usual trick with $ \ddot x$. Remember this trick since its very useful:
\[ \ddot x \equal{} \dfrac{d\dot x}{dx}\dfrac{dx}{dt} \equal{} \dfrac{d\dot x}{dx}\dot x \equal{} \dfrac{vdv}{dx}
\]
Now plugging this into the equation we get:
\[ mv\cdot dv \equal{} \minus{} T\cdot dx
\]
But we know that $ m$ varies with $ x$ too.So:
\[ v\cdot dv \equal{} \minus{} \dfrac{TL}{M}\dfrac{dx}{x}
\]
Now this is a differential equation, with the variables already separated, so we can integrate at once:
\[ \dfrac{v^2}{2} \equal{} \minus{} \dfrac{TL}{M}\ln\left(\dfrac{x}{C}\right)
\]
The initial condition is $ v(L) \equal{} 0$. So from here we get: $ C \equal{} L$.
Hence :
\[ \dfrac{v^2}{2} \equal{} \dfrac{TL}{M}\ln\left(\dfrac{L}{x}\right)
\]
So the kinetic energy:
\[ E_k(x) \equal{} m\dfrac{v^2}{2} \equal{} \dfrac{M}{L}x\dfrac{TL}{M}\ln\left(\dfrac{L}{x}\right) \equal{} Tx\ln\left(\dfrac{L}{x}\right)
\]

[quote="Thaakisfox"]
Since the tension in the spring is constant, we have:
\[ m\ddot x \equal{} \minus{} T
\]
[/quote]

The mass of the bucket is changing, so shouldn't it be
\[ \minus{} T \equal{} \frac {d}{dt}(mv) \equal{} \frac {dm}{dt}v \plus{} \frac {dv}{dt}m \equal{} m\ddot{x} \plus{} \frac {dm}{dt}\dot{x}
\]
We see that by the chain rule
\[ \frac {dm}{dt} \equal{} \frac {dm}{dx} \cdot \frac {dx}{dt} \equal{} \frac {M\dot{x}}{L}
\]
So, we should really have
\[ \minus{} T \equal{} m\ddot{x} \plus{} \frac {dm}{dt}\dot{x} \equal{} m\ddot{x} \plus{} \frac {M}{L}\dot{x}^2
\]
In short, I think you're missing the $ M/L\cdot\dot{x}^2$ term... is this right?  :maybe:

Yes of course you are right..I was in a bit of a hurry there :oops:

Well then here is the correction:
\[ m\ddot x \plus{}\dfrac{dm}{dx}\dot x^2\equal{}\minus{}T\]
So then:
\[ m\cdot v\dfrac{dv}{dx}\plus{}\dfrac{M}{L}v^2\equal{}\minus{}T\]
That is:
\[ m\cdot vv'\plus{}\dfrac{M}{L}v^2\equal{}\minus{}T\]

notice that : $ vv'\equal{}\frac12 (v^2)'$. So lets substitute $ v^2\equal{}\xi$, and divide by $ m$. Then we have:
\[ \xi'\plus{}\dfrac{2}{x}\xi \equal{}\minus{}\dfrac{2TL}{Mx}\]

Now this is a simple nonhomogenous linear differential equation. But its special, because notice, that $ \xi_0\equal{}\minus{}\frac{TL}{M}$ is a particular solution.
So we only need to find the general solution of the homogenous part:
\[ \xi_h'\plus{}\dfrac{2}{x}\xi_h\equal{}0\]
This is straight forward... after separating the variables, and integrating we get:
\[ \xi_h\equal{}\dfrac{C}{x^2}\]
Where $ C$ is some constant.
So the general solution of our original DE, as we know the sum of the general solution of the homogenous part, and a particular solution of the whole DE:
\[ \xi\equal{}\xi_h\plus{}\xi_0\]
THat is:
\[ \xi\equal{}v^2\equal{}\dfrac{C}{x^2}\minus{}\dfrac{TL}{M}\]
But we have initial conditions, that is $ v(L)\equal{}0$. plugging this in we get for the constant: $ C\equal{}\dfrac{TL^3}{M}$
So the square of the $ v(x)$ function is:
\[ v^2\equal{}\dfrac{TL}{M}\left(\dfrac{L^2}{x^2}\minus{}1\right)\]
So the kinetic energy:
\[ E_k(x)\equal{}m\dfrac{v^2}{2}\equal{}\dfrac{M}{L}x\dfrac12 \dfrac{TL}{M}\left(\dfrac{L^2}{x^2}\minus{}1\right)\]
That is:
\[ E_k(x)\equal{}\dfrac{Tx}{2}\left(\dfrac{L^2}{x^2}\minus{}1\right)\]

Now I hope I didnt screw anything, now because I wrote it quite quick. :D

nice problem... i shall come up with a solution too


Well, anxious to get my scores, I go to "My USAMTS" for about the 8th time in two days. Bored, I click over to Profile and notice that I have a reminder saying that I have not yet turned in my entry/permission form! Well, I remember mailing it, but of course it could have gotten lost.

So, three questions:
1. If I e-mail my solutions, must I also e-mail my permission form, or could I also fax/mail it?
2. Should I send my permission form now, or wait til next round?
3. Will I still get my results back/posted on leaders board if I have not sent in my permission form?

I'm pretty sure they have not loged permission forms yet

That is correct; we haven't logged all the forms.


Let the volumes of wine and water in their respective containers be $x$. And let there be a spoon of volume $y$.

We take one spoonful of water from container-1 and empty it into the container-2 (which contains wine). Now we thoroughly shake the liquid in container-2. Then we take a  spoonful of liquid from container-2 and empty it into container--1. Let this be called 1 cycle of mixing.

$Question 1$
Can you deduce and expression for the amount of water remaining in container-2 after $n$ such cycles.

$Question 2$
Also prove that it will take infinite number of cycles to make the liquids contained in the two containers identical in composition.

Let $w=\frac{y}{x}$.
Without loss of generality suppose that [b]x=1 lt[/b]. So the volume of one spoonfull is $y=w$.
Of course we suppose that $w<1$

Let $a$ and $b$ be respectively the amount of water and wine on the container #1 at the start of some cycle
Similarly, define $c,d$ on the container #2

At the start of each cycle #1 contains [b]1 lt[/b].
So, when we take one spoonfull from the container #1, we take an amount  $aw$ of water, and $bw$ of wine

We have:

$\begin{array}{|c|c|c|}\hline \bold{Container\ \#1}&\bold{Container\ \#2}&\bold{SPOON}\\\hline a&c&\\\hline a-aw&c&aw\\\hline a-aw&c+aw&\\\hline \end{array}$

[i](similarly, we can find the amount of the wine but we don't care about that)[/i]

Now the container #2 contains $1+w$ total litters.
So, at the second half of the cycle, the spoon will carry $h=\frac{c+aw}{1+w}\cdot w$ water

Now we have:

$\begin{array}{|c|c|c|}\hline \bold{Container\ \#1}&\bold{Container\ \#2}&\bold{SPOON}\\\hline a-aw&c+aw&\\\hline a-aw&c+aw-h&h \\\hline a-aw+h&c+aw-h& \\\hline \end{array}$

Let $c'$ be the amount of water on the container #2 , after the end of the cycle.

$c' = c+aw-h = (c+aw)-(c+aw)\frac{w}{1+w}\Rightarrow \boxed{c' = \frac{c+aw}{1+w}}\ \ (1)$

At any time (of course when the spoon is empty), the total amount of water in the two containers is [b]1lt[/b].
Setting $a=1-c$ in $(1)$ we get

$c' = \frac{c+(1-c)w}{1+w}= \frac{c(1-w)+w}{1+w}\ \ (2)$

Now let $C = 0.5-c$ and $C'=0.5-c'$.

[hide="some calculations..."]
$(2) \Rightarrow 0.5-C' = \frac{(0.5-C)(1-w)+w}{1+w}\Rightarrow $

$(0.5-C')(1+w) = (0.5-C)(1-w)+w \Rightarrow $

$0.5(1+w)-C'(1+w) = 0.5(1-w)+w-C(1-w) \Rightarrow $

$0.5(1+w)-C'(1+w) = 0.5(1+w)-C(1-w) \Rightarrow $

$C'(1+w) = C(1-w) \Rightarrow $
$\hline$[/hide]$\Rightarrow $
$\boxed{C' = \frac{1-w}{1+w}\cdot C}$

Setting $C_{0}= 0.5-0=0.5$ we get that after the [b]n[/b]-th cycle it is $C_{n}= 0.5\cdot\left(\frac{1-w}{1+w}\right)^{n}\ \ (3),$
which is clearly positive $\forall n,$ so $c_{n}=0.5-C_{n}<0.5$

So $a_{n}= 1-c_{n}>0.5>c_{n}$, which means that they will never be equal

From $(3)$ we get $\boxed{c_{n}= 0.5-0.5\cdot\left(\frac{1-w}{1+w}\right)^{n}}$


superior algebra
superior algebra solved

This is another problem that was given to my brother at the university:

lim(n-->oo) (n^(1/n))^a

where a= sum(k=3,n) tan(pi/k)

Since a actually depends on n I'll call it a_n. By using Cesaro-Stolz we get a_n/(1+1/2+1/3+..+1/n) -> Pi, so all we need is lim(n^(1/n))^(1+1/2+..+1/n). We also have (1+1/2+..+1/n)/ln(n) -> 1, so all we need now is lim(n^(1/n))^ln(n) = n^(ln(n)/n) = e^[(ln(n))^2/n], but it's easy to prove that (ln(n))^2/n -> 0, so the limit we're looking for is 1. Hope there are no mistakes.

There is one thing I don't understand, perhaps it's just me silly, but how come:
we compute a_n/(1+1/2+1/3+..+1/n) -> Pi and then replace a_n by (1+1/2+1/3+..+1/n) ? I don't understand this subtility! Please explain.

thank you very much!

What he means is that if he obtaines the limit l when working with 1+1/2+..., the final limit will be l pi.


$ A,B$ lie on $ PM$ and $ PN$ which $ PMN$ is a triangle. And $ AM\plus{}AN\equal{}BM\plus{}BN$, $ AN$ meets $ BM$ at $ Q$

Prove: $ PAQB$ has a incircle.

Denote by $ C(I)$ the incircle of $ \triangle MBP$ and we draw the second tangent to this circle from $ N$ that touches $ PM$ in $ A'$.

We will show that $ MA'\plus{}A'N\equal{}BM\plus{}BN$. Now, denote by $ P_1,P_2,P_3,P_4$ the intersection points of $ C(I)$ with $ PM,A'N,BM,PN,$ respectively.

$ A'M\plus{}A'N\equal{}A'M\plus{}A'P_2\plus{}P_2N\equal{}A'M\plus{}A'P_1\plus{}NP_2\equal{}MP_1\plus{}NP_2$. Similar $ BM\plus{}BN\equal{}MP_3\plus{}NP_4$.

But $ MP_1\equal{}MP_3$ and $ NP_2\equal{}NP_4 \Rightarrow A'M\plus{}A'N\equal{}BM\plus{}BN\equal{}AM\plus{}AN$

Finally, $ A\equal{}A'$ because otherwise if $ A' \in (AM) \Rightarrow AA'\plus{}AN\equal{}A'N \Rightarrow A\equal{}A'$ which is in contradiction with the hypothesis.

If $ A'\in (PA)$ then we use a similar argument.

Dear Mathlinkers,
I only know that Steiner studied this situation.
Any reference?
Sincerely
Jean-Louis

If in the quadrilateral ABCD denote $ \{E\} \equal{} [AB\cap [DC$ and $ \{F\} \equal{} [AD\cap [BC$, then:
$ ABCD \minus{} circumscriptible \Leftrightarrow |AB| \plus{} |CD| \equal{} |AD| \plus{} |BC| \Leftrightarrow \\
\Leftrightarrow |BE| \plus{} |BF| \equal{} |DE| \plus{} DF| \Leftrightarrow |AE| \plus{} |CF| \equal{} |AF| \plus{} |CE|.$

[b]See:[/b] [i]Sharygin:[/i] "GEOMETRY PROBLEMS"; the problem II.241. [in [u]Russian[/u], [u]Spanish[/u] or English(??)].

take a look at this
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1100789696&t=6559

   M.T.

In Russian, this book Sharagin can be found here:
http://www.math.ru/lib/cat/geom?n=20&page=8&o=0&o_by=author


Suppose that $N$ is a norm on $R^n$ such that $N(x_1,...,x_n)=N(|x_1|,..., |x_n|)$ for all $x_i$. Then $N(a_1+b_1,...,a_n+b_n)$ is at least $N(a_1,...,a_n)$ if $a_i, b_i$ are nonnegative.

Any hint for this one ? I've searched for quite a long time and I still haven't got anything decent ...

It is enough to consider the case when $b_1>0$, $b_2=\dots=b_n=0$. Note that the point $(a_1,a_2,\dots,a_n)$ is a convex combination of $(a_1+b_1, a_2,\dots, a_n)$ and $(-a_1-b_1, a_2,\dots, a_n)$.

I had reduced the problem and I had looked for several similar arguments but I missed this one !!! I feel stupid  :D  :blush:  Thanks fedja !!


Tamame n hayi ra  biabid ke ebarate zir avval shavad:
$ n^10 \plus{} n^5 \plus{} 1$

bayad injuri bashe:

\[ n^{10}\plus{}n^5\plus{}1\]

 :wink:

tamordamo Ddam tu in frumam yeki post kard!!

Salam man ye rah hale sade daram :
\[ n^{10}\plus{}n^5\plus{}1\equal{}(n^2\plus{}n\plus{}1)(n^8\minus{}n^7\plus{}n^6\minus{}n^5\plus{}n^4\minus{}n^3\plus{}n^2\minus{}n\plus{}1)\]
chon in ebarat be hasel zarb 2 parantez tabdil shod pas yeki az in parantez ha bayad barabar ba $ 1$ bashad .
\[ 1:  n^2\plus{}n\plus{}1\equal{}1 \equal{}\equal{}> n^2\plus{}n\equal{}0 \equal{}\equal{}> n\equal{}0, n\equal{} \minus{}1\]
ama ba gharar dadan in maghadir addad mored nazar $ 1$ mishavad ke aval nist
\[ 2 : n^8\minus{}n^7\plus{}n^6\minus{}n^5\plus{}n^4\minus{}n^3\plus{}n^2\minus{}n\plus{}1 \equal{} 1 \equal{}\equal{}> \frac{n^9\plus{}1}{n\plus{}1}\equal{}1 \equal{}\equal{}> n^9\equal{}n \equal{}\equal{}> n\equal{}0 , n\equal{}1\]
ke be ezay $ n\equal{}1$ be addad$ 3$ miresim .

Pas tanha be ezay $ n\equal{}1$ ebarat ma avval ast  :)

Is there a easy way to find this arrangement

n^{10}+n^5+1=(n^2+n+1)(n^8-n^7+n^6-n^5+n^4-n^3+n^2-n+1)

for example using complex numbers or something similar?

[quote="navidj"]
\[ n^{10} + n^5 + 1 = (n^2 + n + 1)(n^8 - n^7 + n^6 - n^5 + n^4 - n^3 + n^2 - n + 1)
\]
[/quote]
fek nemikonam in doros bashe,chon:

\begin{eqnarray*}(n^2 + n + 1)(n^8 - n^7 + n^6 - n^5 + n^4 - n^3 + n^2 - n + 1)=n^{10}+n^8-n^7+n^6-n^5+n^4-n^3+n^2+1\end{eqnarray*}

shayadam man too zarb kardan daram eshteba mikonam... :maybe:

Thank babak . A little mistake !  :oops:
\[ n^{10}\plus{}n^5\plus{}1\equal{}(n^2\plus{}n\plus{}1)(n^8\minus{}n^7\plus{}n^5\minus{}n^4\plus{}n^3\minus{}n\plus{}1)\]

khodetoon baghiyash ro hal konid ( hosele neveshtan nadaram ! )

 :)

[quote="stvs_f"]thanks a lot Navid.
tajziyeye $ n^10 \plus{} n^5 \plus{} 1$ kare sakhty bud.(kheily sakht)[/quote]
Etefaghan tajziye in ebarat kar sakhti nabood .
cheghooneghi tajziye :
Yeki az ravesh haye mamool ( va khoob ) baraye tajziye in joor ebarat ha ezafe va kam kardan jomalat hazf shode ast !
\[ n^{10} \plus{} n^5 \plus{} 1 \equal{} (n^{10} \plus{} n^9 \plus{} n^8) \minus{} (n^9 \plus{} n^8 \plus{} n^7) \plus{} (n^7 \plus{} n^6 \plus{} n^5) \minus{} (n^6 \plus{} n^5 \plus{} n^4) \plus{} (n^5 \plus{} n^4 \plus{} n^3) \minus{} (n^3 \plus{} n^2 \plus{} n) \plus{} (n^2 \plus{} n \plus{} 1) \equal{} (n^2 \plus{} n \plus{} 1)(n^8 \minus{} n^7 \plus{} n^5 \minus{} n^4 \plus{} n^3 \minus{} n \plus{} 1)
\]
agar mikhay dastet to tajziye rah biyofte mitooni ebarat haye zir ro tajziye koni !
\[ a^{5n} \plus{} a^{n} \plus{} 1
\]

\[ (x \minus{} 1)^5 \plus{} x
\]
age to har kodoom moshkel dashti bego ta komaket konam .
  :)

jeddan thanks a  lot Navid.fek konam az in tip soala ziad dide budam.inayi ke gofty ro hatman fek mikonam. :)

tajzie kardanesh moshakhas bud.
agar w rishe 3th vahed bashe.
$ w^{10} \plus{} w^5 \plus{} 1 \equal{} w \plus{} w^2 \plus{} 1 \equal{} 0$ pas $ x^2 \plus{} x \plus{} 1|x^{10} \plus{} x^5 \plus{} 1$

[quote="Aravind Srinivas L"]Why do u people communicate in Arabic or whatever?
This is a global forum so even the national and regional forums must be useful for any person visiting it.
Please make sure u communicate in English.[/quote]
first of all it's not arabic,it's persian,there's a huge difference between them...
and also as you can see from the name of this forum:this forum is a "National & Local Community",so it's for people in Iran,so they should be able to communicate in their own language,which is PERSIAN...
this forum is not built for other nationalities,so it's not our problem that you can't understand persian...
so if you don't understand persian,there's no reason for you to visit this forum...
(respectfully) :)


logarithms
induction
number theory unsolved
number theory

Find all pairs of positive integers $ (x,y)$ such that $ x^y\plus{}y\equal{}y^x\plus{}x$.

- If $ 2<x<y$, we have : $ e \leq x<y$ so $ \frac {\ln x}{x}> \frac {\ln y}{y}$ and $ x^y>y^x$.
Thus $ x^y\minus{}y^x>0>x\minus{}y$, so $ x^y\plus{}y>y^x\plus{}x$.
- By induction it is easy to prove that if $ 3<y$ then $ 2^y\plus{}y>y^2\plus{}2$.
- So the only solutions are $ (x,x)$; $ (1,x)$; $ (x,1)$; $ (2,3)$ and $ (3,2)$.

In fact we can easily prove that if $ x$ and $ y$ are integers such that $ y > x > 2$ then $ x^y \minus{} y^x > y$.
So $ \left| x^y\minus{}y^x \right| \equal{} \left| x\minus{}y \right|$ has no more solutions.


AMC
AIME
geometry
USA(J)MO
USAMO
trigonometry
AMC 10

To this day, I haven't solved an AIME geometry problem while taking a practice AIME, but I've always been really close.

Ok, my goal is next year to make USAMO, and I need to increase my AIME score. I average a 5 right now, so I figure if I can get 2 geometry problems and stop with some stupid mistakes, I should be all right.

My question is, which book is the most helpful for AIME level geometry? Challenging Problems in Geometry or AoPS Volume 1?

Mostly in Volume 1, I can get the easy geometry questions fast. But in Challenging problems, each problem takes at least 5-10 minutes, and are quite challenging for me (I'm getting about 2 out of 3 solved).

//each problem takes at least 5-10 minutes, and are quite challenging for me//

That's not bad. Maybe you're giving up too soon on the ones you aren't getting. I'm kinda surprised you can't solve any, considering some of the 1-5's aren't that bad, I do think maybe you spend too little time thinking about the problems. I'd start with the easiest AIME geometry problems and work your way up. (Or the 20-25 problems on AMC).

Do you have Volume 2? If it's a not know your material problem, like you haven't learned the trig geo, the cyclic quads, collinearity/concurrency, and etc. I'd look at vol 2. Vol 1 does not cover all the topics that show up on the AIME.

Yeah. I definitely spend too much time thinking and investing way too much time in my first attempt (which is usually wrong). Then, I feel slightly uncomfortable finding a totally different way. This is probably why the the questions with drawing an auxiliary line kill me.

I have Volume 2, but I haven't done much geometry. The only somewhat related geometry chapter I've done is Chapter 2, but that's just mainly trig. I'll be sure to work through that as well. But for me, it usually isn't about not knowing anything (at least for the first few chapters in Challenging Problems...just simple stuff that I learned in HS Geo put into a new perspective).

Hello. I have both books and I personally like Challenging better (PS it is a lot harder). I feel as if Challenging poses more problems involving the creative use of having to draw in certain lines. Personally, I would use Challenging. While AIME geometry problems usually involve trig and complex steps rather than difficult choice of lines, Challenging offers good basic introduction and practice for the mindest needed to solve USAMO geometry problems (those not involving trig).

I personally really like the ACOPS section in geometry the best (however it is a little hard for me). Hope that I was a help.  :lol:

Uh...Volume 1 would be best for preparing for AMC 10 and MathCounts. If you want to do well on AIME geo, I suggest looking at Volume 2/Challenging Problems in Geometry. The latter will probably be better for improving the way you think about geometry in general, while the former will be better for honing specific skills.

Definitely challenging problems in geometry! If you're looking for AIME level geometry, challenging beats volume 1 by a long shot! If you really want some info on challenging problems in geometry check [url=http://www.amazon.com/exec/obidos/ASIN/0486691543/artofproblems-20]here[/url]

read the first review.

[quote="veezbo"]Yeah. I definitely spend too much time thinking and investing way too much time in my first attempt (which is usually wrong). Then, I feel slightly uncomfortable finding a totally different way. This is probably why the the questions with drawing an auxiliary line kill me.[/quote]

If you're only spending 10 minutes per problem, then you're not spending too much time thinking (probably you are spending too little time thinking). But it is definitely good to not get stuck on a single idea.


Consider the gambler's ruin problem, with $ N$ the initial amount of money. The player wins or loses $ \$1$ with probability $ \frac{1}2$.
What's the expected duration of game till the player gets ruined?

Infinity.  If not, you could use the martingale convergence theorem to get a contradiction.


Find all functions $f: \mathbb R \to \mathbb R$ such that \[ f( xf(x) + f(y) ) = f(x)^2 + y \] for all $x,y\in \mathbb R$.

[quote="mathmen"]Find all real valued functions defined on the reals such that for every real $ x,y$:

$ f(xf(x) \plus{} f(y)) \equal{} f(x)^2 \plus{} y$[/quote]

Let $ P(x,y)$ be the assertion $ f(xf(x)\plus{}f(y))\equal{}f(x)^2\plus{}y$

$ P(0,x)$ $ \implies$ $ f(f(x))\equal{}x\plus{}f(0)^2$ and so $ f(x)$ is bijective and so $ \exists a$ such that $ f(a)\equal{}0$.
Then $ P(a,x)$ $ \implies$ $ f(f(x))\equal{}x$ and $ f(0)\equal{}0$

$ P(f(x),y)$ $ \implies$ $ f(f(x)f(f(x))\plus{}f(y))\equal{}f(f(x))^2\plus{}y$ and so $ f(xf(x)\plus{}f(y))\equal{}x^2\plus{}y$ and so (comparing with $ P(x,y)$) : $ f(x)^2\equal{}x^2$
So : $ \forall x$, either $ f(x)\equal{}x$, either $ f(x)\equal{}\minus{}x$

Suppose now $ \exists a$ such $ f(a)\equal{}\minus{}a$ and $ \exists b$ such that $ f(b)\equal{}b$

$ P(a,b)$ $ \implies$ $ f(\minus{}a^2\plus{}b)\equal{}a^2\plus{}b$ and so either $ \minus{}a^2\plus{}b\equal{}a^2\plus{}b$, either $ a^2\minus{}b\equal{}a^2\plus{}b$ $ \implies$ either $ a\equal{}0$, either $ b\equal{}0$

So either $ f(x)\equal{}x$ $ \forall x$, either $ f(x)\equal{}\minus{}x$ $ \forall x$ and it is easy to check back that these two solutions both fit the requirements.

Hence the two solutions :
$ f(x)\equal{}x$ $ \forall x$
$ f(x)\equal{}\minus{}x$ $ \forall x$

[quote="pco"]$ f(f(x))\equal{}x\plus{}f(0)^2$ and so $ f(x)$ is bijective ...[/quote]
1. Assume $f(x_1) = f(x_2)$, then $f(f(x_1)) = f(f(x_2))$, and so $x_1 + f(0)^2 = x_2 + f(0)^2$, hence $x_1 = x_2$, hence injectivity.
2. Let $y\in \mathbb{R}$; then take $x = f(y-f(0)^2)$ and get $f(x) = y$, hence surjectivity.

A simpler argument is that from $f\circ f = \textrm{id} + \textrm{constant}$ follows immediately $f$ bijective, since RHS is bijective.

EDIT. This was in answer to some question about bijectivity of $f$, made by user eraydin, who then withdrew his entry, as this was posted.

[quote=pco]Suppose now $ \exists a$ such $ f(a)\equal{}\minus{}a$ and $ \exists b$ such that $ f(b)\equal{}b$

$ P(a,b)$ $ \implies$ $ f(\minus{}a^2\plus{}b)\equal{}a^2\plus{}b$ and so either $ \minus{}a^2\plus{}b\equal{}a^2\plus{}b$, either $ a^2\minus{}b\equal{}a^2\plus{}b$ $ \implies$ either $ a\equal{}0$, either $ b\equal{}0$

[/quote]
Could you please explain why this part was needed? (I mean doesnt the question get solved completely at the point we say $f(x)^2=x^2 \forall x\in \mathbb{R}$ )

[quote=lneis1]Could you please explain why this part was needed? (I mean doesnt the question get solved completely at the point we say $f(x)^2=x^2 \forall x\in \mathbb{R}$ )[/quote]
No, because for example $f(x)=(-1)^{\lfloor 1000\sin x\rfloor}x$ matches the propertyi $f(x)^2=x^2$ too


I'm trying to solve this problem: "Find the quadratic equation, in standard form, for the points (2, -4), (3, -7), (1, -3)". This is what I have but the back of the book says it's wrong :( :

[code]
3 three equations using the points:

-4 = 4a +2b +c
-7 = 9a +3b +c
-3 = a +b +c

so by substitution (3rd equation):

c = -a -b -3

so plug that into the other 2 equations:

-4 = 4a +2b + (-a -b -3)
-1 = 3a +b

-7 = 9a +3b + (-a -b -3)
-4 = 8a -2b

now solve those two with elimination:

3a +b = -1
8a -2b = -4
(multiply first by 2)
6a +2b = -2
8a - 2b = -4
14a = -6
a = -3/7
[/code]

But the back of the book says the answer should be -1. Any help is appreciated, thank you.

[quote][code]3 three equations using the points:

-4 = 4a +2b +c
-7 = 9a +3b +c
-3 = a +b +c

so by substitution (3rd equation):

c = -a -b -3

so plug that into the other 2 equations:

-4 = 4a +2b + (-a -b -3)
-1 = 3a +b

-7 = 9a +3b + (-a -b -3)
-4 = 8a -2b

now solve those two with elimination:

3a +b = -1
8a -2b = -4
(multiply first by 2)
6a +2b = -2
8a - 2b = -4
14a = -6
a = -3/7 [/code][/quote]

Okay, so the issue with your method comes with this step

\begin{eqnarray*}
-7&=&9a+3b+(-a -b -3)\\
-4&=&8a-2b
\end{eqnarray*}

Which should actually be:

\begin{eqnarray*}
-7&=&9a+3b+(-a -b -3)\\
-4&=&8a+2b
\end{eqnarray*}

That should help.

I feel retarded, what a stupid mistake, thanks.


how many ways can i distribute 45 coins among 10 people, if no one can get more than 7 coins?

not sure how to solve this but i think it involves generating funtions.

ok,  i have solved part of the problem.

so far we know that
$ (1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^5 \plus{} x^6 \plus{} x^7)^{10}$  is the function we can set up.  now we must find the coefficient of the $ x^{45}$ term

does anyone know how to do that?

Really?  $ \minus{}126$ ways?  I think you should give this one more thought  :roll:

[quote="Sage!"]$ \mathbf{coeff}[x^{45}] \equal{} {10 \choose 0} \minus{} {10 \choose 1} \plus{} {10 \choose 2} \minus{} {10 \choose 3} \plus{} {10 \choose 4} \minus{} {10 \choose 5}$[/quote]

well i multiplied it on mathmatica and it gave 23392456 as the coefficient. but what was ur reasoning for your method?

[hide="Horrible explanation (somewhat messy)"]You can express $ (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7)^{10}$ as $ \left(\frac {1 - x^8}{1 - x}\right)^{10}.$ Now, that is equal to
\[ (1 - x^8)^{10}\left(\frac {1}{(1 - x)^{10}}\right) = (1 - x^8)^{10}(1 + x + x^2 + x^3 + \dots)^{10}.\]
We only want the coefficients of terms of $ (1 - x^8)^{10}$ that have degrees less than $ 45$ (since we are looking for exponents that sum to $ 45).$ The polynomial is simplified to $ 1 - 10x^8 + 45x^{16} - 120x^{24} + 210x^{32} - 252x^{40}.$ Now we just find the terms in the expanded power series (if you don't know how to do this, just play with it for a while and come up with a conjecture, or just know that the coefficient of $ x^{k}$ in $ (1 + x + x^2 + x^3 + \dots)^{n}$ is $ \binom{n + k - 1}{n - 1}$) we find the terms with degree $ 45, 37,29,21,13,$ and $ 5.$ They are
\begin{align*}\binom{45 + 10 - 1}{9}x^{45} & = 5317936260x^{45} \\
\binom{37 + 10 - 1}{9}x^{37} & = 1101716330x^{37} \\
\binom{29 + 10 - 1}{9}x^{29} & = 163011640x^{29} \\
\binom{21 + 10 - 1}{9}x^{21} & = 14307150x^{21} \\
\binom{13 + 10 - 1}{9}x^{13} & = 497420x^{13} \\
\binom{5 + 10 - 1}{9}x^{5} & = 2002x^5.\end{align*}
Now multiply them with their "matching" term and add the final terms. Since the numbers are huge and messy, I'll leave the multiplication and adding to the OP, and we get a result of $ 23392456x^{45}$ with a final answer of $ \boxed{23392456}.$

BTW, if this were a competition problem, I would expect them to allow you to use calculators, since the numbers are huge.[/hide]

The equivalent alternative is to do PIE based on how many have over $ 7$ coins.


limit
logarithms
real analysis
real analysis unsolved

lim_x-->1     (x^x - x)/[ln(x)  - x + 1]
I'm pretty sure that the limit is -2... but I just can't cancel something in numerator and denominator, by using L'Hospital's rule 2 times
but how can we compute the limit without using L'Hospital's rule ?

[hide]
We have the limit
\[ \lim_{x\rightarrow 1} \frac{\ln x \minus{}x\plus{}1}{(x\minus{}1)^2}\equal{}\minus{}\frac{1}{2}\]
which can be done with L'Hopital, Taylor series, or squeeze theorem (by playing around with integrals).

So you also need to find $ \lim_{x\rightarrow 1} \frac{e^{x\ln x}\minus{}x}{(x\minus{}1)^2}$. Here the limits
\[ \lim_{x\rightarrow 1} \frac{\ln x}{x\minus{}1}\equal{}1, \lim_{u\rightarrow 0} \frac{e^u\minus{}1\minus{}u}{u^2}\equal{}\frac{1}{2}\]
are useful and can be shown just like the limit done above (the first of these is actually weaker than the above limit).
[/hide]


Find all positive integers (x,y) which satisfies the equation: x^2-3y^4=1

There will be infinately many solutions.U put a value of y u will get a value of y.To solve an eqn. of 2 variables u atleast need 2 eqns.


number theory
relatively prime
number theory open

Exercise I: Prove that there are infinitely many primes $ p$ such that $ p|2^n\plus{}3$ with some $ n$..

Exercise II: $ p$ is a prime such that $ p|2^n\plus{}3$ and $ p$ is not Wiefierich prime (i.e. $ 2^{p\minus{}1} \not \equiv 1 \mod{p^2}$). Prove that $ p^2|2^m\plus{}3$ with some $ m$.

Open question: Are there infinitely many primes $ p$ such that $ p^2|2^n\plus{}3$ with some $ n$?
I believe that it is indeed open (because of the strong connection with the Wiefierich primes), does anyone now something more about this?

1. Let p odd prime, suth that $ p|(2^n\plus{}3,2^m\plus{}3)\equal{}(2^n\plus{}3,2^n\minus{}2^m)\to p|2^n\plus{}3, \ p|2^{|n\minus{}m|}\minus{}1$. If $ n\minus{}m|n$, then $ 2^n\plus{}3\equal{}4\mod p$. Therefore prime divisors $ 2^{n!}\plus{}3$ are different.
2. If $ p|2^n\plus{}3$ and $ n\equal{}mp$, then $ p|2^m\plus{}3$, because $ 2^n\equal{}(2^p)^m\equal{}2^m\mod p$. Therefore if exist n, suth that $ p|2^n\plus{}3$, then exist m, suth that $ p|2^m\plus{}3, \ p\not |m$.
Let $ p|2^m\plus{}3,p\not |m, \ p^2\not |2^m\plus{}3$ and p is not Wevirich prime ($ \ p^2\not |2^{p\minus{}1}\minus{}1$). Then $ 2^{m\plus{}(p\minus{}1)k}\plus{}3\equal{}(2^{p\minus{}1}\minus{}1\plus{}1)^k2^m\plus{}3\equal{}(ap\plus{}1)^k(bp\minus{}3)\plus{}3\equal{}(1\plus{}kap)(bp\minus{}3)\plus{}3\mod p^2\equal{}p(b\minus{}3ak)\mod p^2$. Therefore we can find k, suth that $ p|b\minus{}3ak$.

we have a problem :
if we use this problem your problem will be solved!!

 " prove that there are infinity ${ \{n_1}$,${ \{n_2}$,${ \{n_3}$...... such as ${ \{2^{n_i}}$;${ \{2^{n_j}}$ +3 are relatively prime  with i#j

Rust already solved your problem with $ n_k \equal{} k!$.


find all polynomial with real coefficients P(x) such that
P(P(x)+x) =P(x)P(x+1)
For all \[ x \in R
\]

$ p\equiv 1\ or\ 0$
if $ deg(p)=n\ge 1$ then $ p(p(x)+x)=p(x)p(x+1)$ gives $ n^2=2n$ then $ n=2$
we suppos that $ p(x)=ax^2+bx+c$ then $ .....$


At the beginning of the school year, Lisa's goal was to earn an A on at least $80 \%$ of her 50 quizzes for the year. She earned an A on 22 of the first 30 quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?

A. 1
B. 2
C. 3
D. 4
E. 5

80% of 50 = 40

since she got 22 out of 30 correct she must do 18 more to achieve 40

she will have 20 quizzes left in the year

to achieve her goal she can earn less than an A on 2 quizzes

[hide]
$80\%\cdot50=40$
So, out of her 20 remaining quizes, she needs to get 18 right, so the answer is 2 or $B$.[/hide]

That is right. i also got 2 as an answer :lol:

[hide] since there's 20 more quizzes left in the year, if she were to get all quizzes above 90, it would look like this;

42/50

her goal is to get above 80%, so subtract 1 from 42 until you can't get past 80%.

41/50 is 82. you can go further.

40/50 is 80%. You can't go any further. thus, she can only miss 2 quizzes.[/hide]

[hide].80*50=40 should be A. So 10 tests should be B or less.
30-22=8 that should be B or less.
10-8=2 tests is lower than A
Answer=2(B)[/hide]

[hide]If Lisa wants to earn an A on at least 80% of her 50 quizzes, then the maximum number of quizzes that she can score below an A on is 10. She has already scored lower than an A on 8 quizzes, making it impossible to get below an A on more than 2 quizzes if she is to achieve her goal. (answer B)[/hide]


calculus
geometry
parameterization
algebra
function
domain
analytic geometry

If the mass per unit area of a surface is given by p=xy, find the mass

SS xy dS

if S is the part of the cylinder x^2+z^2=25 which is in the first octant and contained within the cylinder x^2+y^2=9..

Looks hard. You need a parametrization, such as $\vec r(x,y)=\langle x,y,\sqrt{25-x^{2}}\rangle$ with the parameter domain $D$ being the quarter of a disk of radius $3$. Since $|\vec r_{x}\times\vec r_{y}|=\sqrt{1+x^{2}/(25-x^{2})}=5/\sqrt{25-x^{2}}$, we have
\[\iint_{S}xy\,dS=\iint_{D}\frac{5xy}{\sqrt{25-x^{2}}}\,dA\]
In polar coordinates this integral becomes
\[\int_{0}^{\pi/2}\int_{0}^{3}\frac{5r^{3}\cos\theta \sin\theta}{\sqrt{25-r^{2}\cos^{2}\theta}}\,dr\,d\theta\]
I think I'll leave this for you to finish.


trigonometry
modular arithmetic
complex numbers

[quote="silouan"][quote="warut_suk"]Find the real part of $(1+i)^{50}$[/quote]
What symbilise the $i$?[/quote]

It is $i^2=-1$, the imaginary unit....

if you put it in exponential notation,
$(\sqrt{2}e^{\frac{i\pi}{4}})^{50}$
$2^{25}*e^{12.5\pi i}$
$2^{25}e^{\frac{\pi}{2} i}$
the real part, $2^{50}* \cos{\frac{\pi}{2} = \boxed{0}}$
but i agree, the other method is somewhat easier

Hides please!
[hide]
$(1+i)(1+i)=1+i+i-1=2i\\(1+i)^50=(2i)^25=2^25*i^25$
$25\equiv1\pmod4$, Thus $2^25*i^25=2^25*i$
Thus the real part is:
[hide="The answer"][hide="is hidden"]0[/hide][/hide]

[/hide]


function
trigonometry
algebra
functional equation

Find all functions $f$  defined on the set of real numbers with   the property
              $f(x+y)f(x-y) = x^2-y^2$ ,
                 for every real number x , y.
[u]Babis[/u]

  [i]Please , send al your nice and short solutions !!! [/i]

[quote="Altheman"]$y=0 \implies f(x)=\pm x$, after checking, both work[/quote]

From $f(x)^2=x^2$ you can't derive $f(x)=\pm x$. For example, all of the following satisfy the first condition:

$1^\circ\quad f(x)=|x|$
$2^\circ\quad f(x)=-|x|$
$\begin{displaymath} 3^\circ\quad f(x) = \left\{ \begin{array}{ll} 2 & x=2\\ -x & x\neq 2 \end{array} \right. \end{displaymath}$
$\begin{displaymath} 4^\circ\quad f(x) = \left\{ \begin{array}{ll} x & x\in\mathbb{Q}\\ -x & x\notin\mathbb{Q} \end{array} \right. \end{displaymath}$

etc. etc.

[hide="Correct solution with the same result"]Put $u=x+y, v=x-y\Longrightarrow x={{u+v}\over 2}, y={{u-v}\over 2}$. Now the given condition becomes

(1) $f(u)f(v)=uv\quad\forall u,v\in\mathbb{R}$

Substituting $v=1$ into (1) we get $f(u)={u\over{f(1)}}$, hence $f(v)={v\over{f(1)}}$. Plugging that back into (1) we get $f(1)^2=1\Longrightarrow f(1)=\pm 1\Longrightarrow f(u)=\pm u$. Checking proves that both work.[/hide]

Can you explain why $f(1)=\pm 1\Longrightarrow f(u)=\pm u$ ? I don't see why all your other examples won't work just because you've shown that $f(1) = \pm 1$.

This then follows directly from $f(u) = \frac{u}{f(1)}$...

Lol, Farenhajt made a very useful remark because I must admit that I would make the same mistake as in the post above :).

Anyway, great solution and when you find that $f(u)=1/f(1)$ then just plug $f(u)=1/f(1)$ for all u,v  and then you obtain that $f(1)=+1$ or $f(1)=-1$ :)

i don't see why this would not work...

$\left(f(x)\right)^2-x^2=0$
$(f(x)-x)(f(x)+x)=0$
$f(x)=x$ or $f(x)=-x$

Obviously one of the two equalities hold for every $x$, but you have no way of knowing that the same equality holds for each one, e.g, $f(x)=|x|$, which satisfies the functional equation $f(x)=x^2$ and is even continuous.

For any fixed number a, your reasoning shows that we must have $f(a) = \pm a$, but that doesn't mean that $f(x) \equiv x$ for all x.

Look at Farenhajt's counterexample
$\begin{displaymath} 4^\circ\quad f(x) = \left\{ \begin{array}{ll} x & x\in\mathbb{Q}\\ -x & x\notin\mathbb{Q} \end{array} \right. \end{displaymath}$

In any point we have either $f(a) = a$ or $f(a) = -a$, but the function isn't just $f(x) = x$ or $f(x) = -x$ over the entire interval of real numbers.

See this !

   << Every person is male or female >> does not mean <<Every person is male >> or << Every person is female >>.

  This is a law in mathematical logic. Many people - me too - have made this mistake. This was the reason , I sent the problem.if you find similar problems , please send them here or in pm.

 [u]Babis[/u]

[quote="stergiu"]<< Every person is male or female >> does not mean <<Every person is male >> or << Every person is female >>[/quote]

Or, in formal quantifier algebra: $(\forall z)(P\lor Q)\nRightarrow\lbrack(\forall z)P\rbrack\lor\lbrack(\forall z)Q\rbrack$

I must also point out another tricky detail, which I haven't explicitly wrote in my solution. When we're substituting the old [i]variables[/i] with the new [i]variables [/i](i.e. NOT taking specific values, but making general substitution), we must always take into account all the restrictions that might induce. In the given example, we have that $(x,y)\mapsto (u,v)$ is a [u]bijection[/u] $\mathbb{R}^2\mapsto\mathbb{R}^2$, hence no solution will be lost. But, for example, if we have $f({{x+y}\over{1-xy}})=A(x,y)$ and we substitute $x=\tan u, y=\tan v$ to obtain $f(\tan(u+v))=A(\tan u,\tan v)$, later we must take into account that $u+v\notin{\pi\over 2}+\mathbb{Z}\pi$, since $\tan u\tan v\neq 1$.

This functional equation helped me cope with many others....
10x to all of you for helping me cope with this


For $ a,b,c$ positive reals prove that

$ \sum  \frac {1}{\sqrt {a^2\plus{}bc}} \geq \frac {6}{ab \plus{} bc \plus{} ca}$

[quote="manlio"]For $ a,b,c$ positive reals prove that

$ \sum \frac {1}{\sqrt {a^2 \plus{} bc}} \geq \frac {6}{ab \plus{} bc \plus{} ca}$[/quote]
Something wrong. Try $ a\equal{}b\equal{}1$ and $ c\rightarrow0.$ :wink:

[quote="manlio"]For $ a,b,c$ positive reals prove that

$ \sum \frac {1}{\sqrt {a^2 \plus{} bc}} \geq \frac {6}{ab \plus{} bc \plus{} ca}$[/quote]
It must be
$ \sum \frac {1}{\sqrt {a^2 \plus{} bc}} \geq \frac {6}{a\plus{}b\plus{}c}$
, which is clearly true by AM-GM inequality and the known inequality
$ (a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab) \le \frac{1}{2^6}(a\plus{}b\plus{}c)^6$.

The stronger one
$ 3(a\plus{}b\plus{}c) \ge 2(\sqrt{a^2\plus{}bc}\plus{}\sqrt{b^2\plus{}ca}\plus{}\sqrt{c^2\plus{}ab})$
is also a problem of hungkhtn, you can see it in Secrets in Inequalities (I forgot the page number).


parameterization
calculus
integration
probability and stats

Let $ B_t$ be a brownian motion and $ T(\omega) \equal{} \inf(t \geq 1 : B_t \equal{} 0)$. Compute the law of $ T$.

1. Note that $ B_1, \ W_t\equal{}B_t\minus{}B_1$ are independent.
2. $ P(\inf\{t|W_t\plus{}B_1\equal{}0\}>x)\equal{}EP(\inf\{t|W_t\plus{}B_1\equal{}0\}>x\parallel{}B_1)$ - conditioning on $ B_1$.
3. Fixed $ B_1\equal{}\minus{}L<0$,  $ P\left(\inf\{t|W_t\minus{}L\equal{}0\}>x\right)\equal{}P\left(\sup\{W_t, \ t\in[0,x]\}<L\right)$, and the last distribution is known as the double normal law(i forgot the parameters, so i dont write the explicit form). The same is when fixed $ B_1\equal{}L>0$, we have the $ \inf$ instead of $ \sup$, so now it remains to calculate the integral in 2 accurately, or just leave it in this form  :lol:

yep, this is the idea, even if I do not understand all what you wrote  :)


modular arithmetic
induction
number theory
relatively prime
number theory unsolved

Given $ \{a_1,a_2,\ldots,a_n\},\{b_1,b_2,\ldots,b_n\},$ and $ \{a_1b_1,a_2b_2,\ldots,a_nb_n\}$ are complete residue in modulo $ n$, show that $ n\equal{}1$ or $ 2$.

lemma- every integer $ a_i$ which is divisible by $ p$ must be with an integer $ b_i$ which is divisible  by $ p$ and every integer $ a_i$ which is not divisible by $ p$ must be with an integer which isn't divisible by $ p$. (to prove it just pay attention to the number of integer which must be divisible by $ p$ in the set $ \{a_1b_1,....,a_nb_n \}$).
see that $ n$ must be a square-free integer,let's suppose that there exists a prime $ p$ such that $ p^2 | n$ then none of the integer which is in the set  $ \{a_1b_1,....,a_nb_n \}$ can  be in the form $ p^2k \plus{} p$ because of the lemma.so we have a contradiction since the set $ \{a_1b_1,....,a_nb_n \}$ must contain all residue modulo $ p^2$.
the remainig case is $ n \equal{} p_1...p_k$.
one of the result of the lemma is that if $ \gcd (a_i,n) \equal{} 1$ then $ \gcd (b_i,n) \equal{} 1$.
note that in this case $ n$ can not have any primitive roots.
without loss of generality,let's suppose that the set $ \{a_1,...,a_{\phi(n)}\}$ is the set consisting of all integers non-congruent modulo $ n$ and relatively prime to $ n$.the same is for $ \{b_1,...,b_{\phi(n)}\}$.
now the set $ \{a_1b_1,...,a_{\phi(n)}b_{\phi(n)}\}$ must have the same property mentioned above.also we must have in this case $ (a_1b_1)...(a_{\phi(n)}b_{\phi(n)}) \equiv a_1a_2...a_{\phi(n)} \equiv (a_1...a_{\phi(n)})^2 \pmod {n}$.
but from the problem D12 in PEN we must have $ P \equal{} a_1a_2...a_{\phi(n)} \equiv \pm 1 \pmod {n}$.so we must have $ P \equal{} a_1a_2...a_{\phi(n)} \equiv 1 \pmod {n}$ (if not then we must have $ n \equal{} 2,1$) which means that $ n$ can not have any primitive root and so can not be in the form $ n \equal{} p$ or  $ n \equal{} 2p$.
i'm too lazy to think about the rest,but i don't think it can be difficult. :D

and here is the rest of the solution,
i proved that $ n \neq p,2p$
suppose that $ n\equal{}p_1...p_k$
we'll use induction on $ k$ to prove the statement.
if $ k\equal{}1$ we know that it is impossible.
now suppose that the problem is proved for $ k\equal{}1,...,t\minus{}1$ i mean we can't have a number whci is the product of $ k$ primes and has the ppropety mentioned
WLOG supposing $ a_i\equal{}i$
we know that each integer which is divisible by $ p$ must be mutiplied by an  integer which is a multiple of $ p$ and also if an integer isn't divisible by $ p$ then it must be multiplied by an integer which isn't a multiple of $ p$.
so by the above lemma we know that $ b_{lp_1}\equal{}p_1s_l$ where $ 1 \le l \le p_2...p_t$
so the number which are divisible by $ p_1$ in the third set are:
$ \{1p \times b_p , 2p \times b_{2p} , .... , (p_2...p_t)p \times b_{(p_2...p_t)p}\}$
also the above set of numbers must be a complete system of residue modulo $ p_2...p_t$ (it is easy to prove )
but since $ \gcd(p_1,p_2...p_t)\equal{}1$ so the set $ \{1 \times b_p , 2 \times b_{2p} , .... , (p_2...p_t) \times b_{(p_2...p_t)p}\}$
must also be a complete system of residue modulo $ p_2..p_t$
we replace $ b_{lp_1}$ by $ p_1s_l$ and we get that the set $ \{1 \times p_1s_1 , 2 \times p_1s_2 , .... , (p_2...p_t) \times p_1s_{p_2...p_t}\}$
is a complete system of residue modulo $ p_2...p_t$ and finally(again!) since $ \gcd(p_1,p_2...p_t)\equal{}1$ the set
$ \{1 \times s_1 , 2 \times s_2 , .... , (p_2...p_t) \times s_{p_2...p_t}\}$
must be a complete system of residue modulo $ p_2...p_t$
but we know that $ \{1,...,p_2...p_t\}$ and $ \{s_1,...s_{p_2...p_t}\}$ are complete system of residue modulo $ p_2...p_t$
so we have the similar property but this time for $ t\minus{}1$ primes which is a contradiction with the assumption that we had at the beginning of the proof(there doesn't exist any number with $ t\minus{}1$ primes for which the statement hold)
so we r done!


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