What is the greatest digit that can be placed in the circle so that the six-digit is divisible by 4?\\
$ 3529\bigcirc2$

For a number to be divisible by $ 4$, the last two digits of a number must be divisible by $ 4$. Since we have a $ 2$ ending the number, then the possible choices are $ 12$, $ 32$, $ 52$, $ 72$, and $ 92$. Out of those the greatest digit possible $ \Rightarrow{9}$


Here's a really simple problem:

Let x1, x2, ..., xn be nonnegative reals with sum at most 1/2.
Prove that (1 - x1)(1 - x2)...(1 - xn) \geq 1/2.

if we view (1 - x_1)(1 - x_2)...(1 - x_n)  as a function in x_i, holding all the other x_j constant, the function is linear. all the x_i fall in [0,1/2]. to minimize the function, we choose each as x_i as one of the endpoints 0 or 1/2. in order to fulfill the sum 1/2 requirement, make all but one of the x_i 0 and a the last 1/2, yielding the minimum 1/2.

That's not correct tetrahedr0n. Do you see why?

I don't think you can "suppose" it's optimal for 1/2,0,0,0... or anything, A1lqdSchool's solution is better (and is same as mine :D)

why cant i assume that? isnt the expression linear in each variable?

[quote="tetrahedr0n"]if we view (1 - x_1)(1 - x_2)...(1 - x_n)  as a function in x_i, holding all the other x_j constant, the function is linear. all the x_i fall in [0,1/2]. to minimize the function, we choose each as x_i as one of the endpoints 0 or 1/2.[/quote]

You fixed all but one of the xi's, so the variable xi can vary only on

[0, 1/2 - (x1 + ... + xi - 1 + xi + 1 + ... + xn)],

as the sum of the xi's is at most 1/2.


$ x$ , $ y$ and $ z$ are real numbers
prove that:
l$ x\plus{}y\plus{}z$l+l$ x$l+l$ y$l+l$ z$l>= l$ x\plus{}y$l+l$ y\plus{}z$l+l$ z\plus{}x$l

[quote="greatestmaths"]how?  :maybe:[/quote]
So $ \left(|x \plus{} y \plus{} z|\plus{}|x|\plus{}|y|\plus{}|z|\right)^2\geq \left(|x \plus{} y|\plus{}|y \plus{} z|\plus{}|z \plus{} x|\right)^2.$ :wink:


Let be given $ a>1$.Find all functions $ f:R^{+} \longrightarrow R$ satisfying:
	i) f is strictly increasing.
	ii) $ f(x+1)=f(x)+\frac1{a^x}$

The solutions are the functions of the form $\dsp f(x) = \alpha - \frac a {a-1} \cdot \frac 1 {a^x}$, where $\alpha$ is an arbitrary constant.

Assume $f$ satisfies the statement of the problem. Then, from 1), $\dsp \lim_{+ \infty} f(x) = L$ does exist and $L$ is either a real number or $+ \infty$.

Let $x$ be fixed. Then for all $n$, we deduce easily from 2) that :
$\dsp f(x+n) = f(x) + \frac 1 {a^x} \cdot \frac { \frac 1 {a^n} - 1} {\frac 1 a  - 1}$.
When $n$ goes to $+\infty$ the RHS has limit $f(x) +  \frac 1 {a^x} \cdot  \frac a {a-1}$ and the LHS has limit $L$, so that $L$ is a real number and $L = f(x) +  \frac 1 {a^x} \cdot  \frac a {a-1}$.
Since this limit must be independant from $x$, it follows that for all $x$, we have
$fx) = \alpha -  \frac 1 {a^x} \cdot  \frac a {a-1}$, where $\alpha$ is some constant.

It is easy to verify that, cnversely, these functions indeed satisfies the statement of the problem.

Pierre.

But if f is not bounded superior, then $ L=+\infty$, how can we get the number $ \alpha$

But this cannot happen since $L$ has to be equal to $\dsp f(x) + \frac a {a-1} \cdot \frac 1 {a^x}$ which is a real number...
I did not assume that $L$ is real. I've just used the relation with $f(x+n)$ and the fact that the limit of the LHS does exist so as the limit of the RHS.

Pierre.


Euler
real analysis
real analysis unsolved

Find the adherence of the set of the numbers of the form $\frac{\phi(n)}{n}$ where $\phi(n)$ is the Euler indicator.

Is it $(0,1]$ (I took $\varphi(1)=1$ to get $\frac{\varphi(n)}n=1$ for $n=1$)?

Those numbers are numbers of the form $\prod_{i=1}^n\left(1-\frac 1{p_{k_i}}\right)$, where $(p_n)_n$ is the sequence of primes. Fix any $a\in(0,1)$, a $t>1$, and take some $n$ for which $\frac{\varphi(n)}n>ta$. Now choose a prime $p_m$ so large that $1-\frac 1{p_m}>\frac 1t$, and start multiplying $\frac{\varphi(n)}n$ with $1-\frac 1{p_m},1-\frac 1{p_{m+1}}$, and so on, until we reach the first $k$ s.t. the product $\alpha(k)=\frac{\varphi(n)}n\cdot\left(1-\frac 1{p_m}\right)\cdot\ldots\cdot\left(1-\frac 1{p_{m+k}}\right)$ is $<ta$. Since $\alpha(k-1)\ge ta$ and $\frac{\alpha(k)}{\alpha(k-1)}\ge 1-\frac 1{p_m}>\frac 1t$, we have $\alpha(k)\in(a,ta)$, and this proves that every $a\in(0,1)$ is a limit point of $\left\{\frac{\varphi(n)}n\right\}_n$.


Prove or disprove by countexample:

If g(t) is continuous on [0,1], then there is subinterval (a,b) $ \subset$ [0,1] with a< b for which g is monotonic on (a,b)

A function monotonic on an interval is differentiable almost everywhere on that interval, but we can find $ g$ that is differentiable nowhere, so no.

Try a classic Weierstrass-style nowhere differentiable function:

$ f(x) \equal{} \sum_{k \equal{} 1}^{\infty}\left(\frac34\right)^k\cos(4^kx)$

Consider it on the interval $ \left[\frac {m\pi}{4^n},\frac {(m \plus{} 1)\pi}{4^n}\right]$ for some integers $ m$ and $ n$. Can you tell whether it (on the net) increases or decreases from the left end to the right end of that interval?

Should that be a $ \left(\frac {3}{4}\right)^{k}$? Or am I misinterpreting it?

[color=green][Fixed now.][/color]


[quote="holmes.roams"]Is speed ever a problem for you guys? I feel like I take too long in my problems. By that I mean it takes a bit too long between the moment I realize how to do the problem to when I actually get the answer.

Is there any way to practice speeding that up? Like what kind of practice should I do?[/quote]The only way to get faster in my experience is to learn computation tricks and practice problems under timed conditions.

I find that using the distributive property for two-digit multiplication is often useful and faster than the standard multiplication algorithm.

Ex.: $ 24\times 57 \equal{} (20 \plus{} 4)(60 \minus{} 3) \equal{} 1200 \plus{} 240 \minus{} 60 \minus{} 12 \equal{} 1380 \minus{} 12 \equal{} 1368$.

(Using this is often good when checking over your work as it is quick and easy to do.)

I am in the same situation as you; what you describe often happens to me. But after participating in a bunch of NSML meets, I started being more aware of time. So I suggest practicing under timed conditions until you eventually become more aware.

[quote="sunehra"]I find that using the distributive property for two-digit multiplication is often useful and faster than the standard multiplication algorithm.[/quote]They're the same thing...

[quote="holmes.roams"]Is speed ever a problem for you guys? I feel like I take too long in my problems. By that I mean it takes a bit too long between the moment I realize how to do the problem to when I actually get the answer.

Is there any way to practice speeding that up? Like what kind of practice should I do?[/quote]

Try to solve problems, being conscious of what is making you take so long. Is it computational? Are you taking long to interpret the problem? If it's geometry, does it take a long time to draw an accurate picture? Do you simply not know how to solve the problem?

The last question is of particular importance. If you already know how to solve most of the problems (this comes from TAKING PAST TESTS; problems repeat), then you are definitely going to work faster than otherwise. So TAKE PAST TESTS!!!

Idk, just basically analyze your thinking while you solve problems to selfdiagnose. Just my two cents :P

I guess so. It's seems faster to me because you don't have to keep track of carrying numbers or anything like that. Note that you only have one number with a units digit so subtraction/addition of all the other numbers is really simply and then you just add/sub on the last number.

Speed is usually not the problem for me on these types of tests; I am pretty good at cheating, I mean seeing the problem, under pressure. (Just kidding about the cheating part).
On tests such as the AMC's, the main problem for me is that I make careless mistakes (usually question interpretation or computational errors) and fail to see my errors. I've gotten better at finding them, but I still tend to make really bad errors once in a while (like number thirteen on the AMC 12).
I guess the problem is different for everyone, really.

I think the best thing to do is practice taking the tests under timed conditions. Once you get used to the time, you will learn to pace yourself better. But really, just do lots of problems.


Here's an interesting problem:

An $8\times8$ chessboard is completely tiled by $2\times1$ dominoes. Prove that we can place positive integers
in all cells of the table in such a way that the sums of numbers in every domino are equal and the numbers placed
in two adjacent cells are coprime if and only if they belong to the same domino. (Two cells are called adjacent if
they have a common side.)

I don't sure this will work,but what about sum of two numbers on a domino is a prime.One is divided by 3 but odd,and one is even. :?

Well, perhaps you are close, but can you explain your construction with more details, please.


find all function$f: R^{+} \to R^{+}$
 and satisfying :
 1/     $f$ decsending in $R^{+}$
 2/   $f(x+y)+f(f(x)+f(y))=f(f(x+f(y))+f(y+f(x))) \forall x \in R^{+}$

ooh, we can show that all of the functions that satisfying these conditions , satisfying in the function $f(f(x))=x$  :?

[quote="tranthanhnam"]but i want to  have a solution completely  .
       can you help me., ehsan2004?[/quote]

It's very hard to find all of the functions that satisfying in these conditions, but one of the function is $f(x)=\frac cx$ (where $c$ is a constant) and as i wrote one of the other is $f(x)=f^{-1}(x)$ .

take a look here :

[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=f%28x%2By%29%2Bf%28f%28x%29%2Bf%28y%29%29%3Df%28f%28x%2Bf%28y%29%29%2Bf%28y%2Bf%28x%29%29%29&t=41361[/url]

PS.Nice URL isnt it :D


I'm debating whether to apply between this and this different program in my area.  When would acceptance letters be out for this camp?  Or does it depend on when the application goes it?

Well since all applications received before the end of April all have equal consideration,it would make good sense that you won't be seeing them before the middle of May.

We may start doing some early acceptances this year for applicants who are very obviously going to get in.  Otherwise, the first general round of acceptances will probably go out by the end of the first week in May.  At that point we'll be a bit conservative about how many spots we have, so there will certainly be people who don't get definite answers until after that.


function
geometry
rectangle
calculus
integration
calculus computations

Hi friends.

Can anyone help me out with problem 6 on this page: http://www.stanford.edu/class/math52h/homework/homework1.pdf

I have been on this problem for some hours now and I can't figure out how the hell to prove it or what the hell the utility of it is. It is also almost 2 in the morning now. I would post the question here but it is too late for me to figure out how the symbolism works in this box.

You might know 'volume zero' as 'content zero' or 'zero content'.

Which direction are you having difficulties with? The if direction is very straightforward, no?

For the only if, use the fact that $1$ is integrable over $S$, so if $\Delta$ is an interval containing $S$, the characteristic function $1_{S}$ is integrable over $\Delta$. It is clear that we may assume that all upper step functions $t$ and all lower step functions $s$ only assume the values $0$ and $1$. In fact, since $\sup s = \iint_\Delta 1_{S}= 0$ we can assume that $s=0$ always (why?). So the condiition that we can find upper and lower step functions with the property that $\iint_\Delta (t-s) < \varepsilon$ in fact becomes  $\iint_\Delta t < \varepsilon$. So the rectangle partition of $\Delta$ on which $t$ is defined gives a collection of rectangles whose union covers the set $S$ and has area less than $\varepsilon$. Some kind of tricky algebra or just some hand-waving now shows that if we can find rectangles like this we can also find balls covering $S$ in the described way.

edit: I am assuming here that a set having volume zero means that every bounded function is riemann integrable on it with value 0. Furthermore, the definition of riemann integral I use is based on the supremum of integrals of lower step functions and infimum of upper step functions. I think you'll find after a while that all proofs regarding riemann integrals are ultimately tedious algebraic manipulations of upper and lower step functions. Riemann integration is boring :)

If a set S has content zero then for every positive epsilon there exists another set V of closed intervals P(1),...,P(n) such that S is contained in the union of the closed intervals in V and , the sum of the 'volumes' of the closed intervals in V is less than epsilon.

'Volume' meaning if P is the closed interval [a(1),b(1)] X ... X [a(n),b(n)] then the volume of P is the product b(k)-a(k), k=1 to n.

Seems like a tedious exercise now, no?

Oh, this is where I suggested hand-waving...

Well is it really hard to do it? Say all rectangles were squares. Then we could choose $x_{1},\ldots,x_{N}$ as the midpoints of these squares, and $\rho_{j}$ as the diameter of the corresponding square so that $\sum_{j=1}^{N}p_{j}^{n}< k\varepsilon$ where $k$ is some constant dependign only on $n$.

I mean, I guess it's a little tedious, but I don't think that it will be hard to complete this argument.

That assumes all rectangles can be partitioned into squares and this is generally not true. I thought of doing that before. I tried partitioning the closed interval into N equal sized rectangles each contained in a ball and then making the content of the rectangles arbitrarily small but that didn't work either.

I think I can get it with a little chiseling though. Can you help me on the last problem in the pdf?


help me

Triangle ABC is any one of the set of triangles having base BC equal to a and height from A to BC equal to h, with h ＜ sqrt(3)/2a. P is a point inside the triangle such that the value of ∠PAB = ∠PBA =∠PCB. Show that the measure of ∠PAB, ∠PBA and ∠PCB is the same for every triangle in the set.

thank you

This was posted (and has been answered) in the Advanced Forum -- can it be cleared out of here please?


i suck at probability so....

i failed this one and I had to miss 10A... and when i took the A at home for practice... i only missed one.. WAH?

oh well i'll get to take 12a/b next year

My school only offered 10B. I tried the 10A (for practice) and was able to get about 135, but I am not sure about my 10B score, definitely lower, much lower than 10A.

There were very few geometry problems, but so many probability problems. Wonder what the cutoff score will be for 10B this year.

Just curious how everyone else did.

umm... well you should have gotten an answer booklet when you finished the test...

i did not receive them, can someone post them?

[quote="codemanz234"]yeah i know what you mean its like they specifically tried to hit my weak point.... ah well i answer a few of a probablity problems but now that i look back on it i got most of them wrong so my score took a dive[/quote]

ditto. it was awful. it made me sad that there were so few geometry questions (24 was beast without compass/protractor/ruler, so i missed it  :( ).

probability and counting are not my strong points either. this test was like a mathcounts test on steroids.

i didn't do well at mathcounts.

I was really glad there wasn't a huge influx of geometry problems, as I am very weak in that area.

I thought this was much easier than the 10A this year. I'm much better at counting/probability problems than geometry. The difficult problems were the last two, but #24 can be done with a ruler/protactor and #25 can be bashed with not much difficulty.

I was so mad when I remembered you could have protractor+ruler(I didn't). I didn't need it, but I liked my solution to #24, and a ton of people got it in a very cheap manner imo.

Cheap, but effective

You're right, finding an actual solution would be better, but in a contest, I do what I need to get the problem solved, and ignore "good practices" if i feel i could do it faster that way than finding a solution.

The minority of geo allowed me to get 150. I was so happy when I actually solved 24.

I took the 12A, 10B, so I'm not sure I can accurately comment, but the B seemed way easier. I am biased though-I'm way better at counting than everything else.

[quote="frost13"]I really thought the B was a lot easier.  My middle school team(I am coach) had 4 kids score 120+ on the B that didn't on the A while only 2 kids scored 120+ on the A( includiing a 139.5 on the B).  Those 4 kids improved on average by 17 points!

I did the 10A problems myself but haven't done all the10B problems yet.  #24 was hard if you didn't use a protractor :lol: and #25 was hard or tedious depending on how you did it.  Those two were harder than anything on 10A, but the rest seemed easier.......to my kids anyway[/quote]
That is a good point, even I scored higher on the B than on the A, but i still found the B too be harder. Maybe you are such a good coach that you were able to pull you students up 17 points. :) . But the reason i did better is I think that 22+ of them were very doable for me while only 20+ were doable for me on the A. I only improved by 4.5 pts though. 121.5 - 126.

Looked at the questions: Difficulty was about the same... In my opinion. xD...

I failed the 12 T.T... I forgot to square root, subtract 1, and then... one more... -_-...

I found the 10B a kit easier. I suck at geometry, so I was glad there wasn't a lot of it, even though I'm taking into Geometry. I've taken intro Counting and Probab. so I got most of the probability problems though 21 stopped me because it LOOKED so hard!!!

OH, well. Anyways, I like the B a lot better than the 10A, but it might not have been a lot easier.


geometry
rectangle
greatest common divisor
search

A 24-foot by 72-foot rectangular dance floor is completely
tiled with 1-foot by 1-foot square tiles. Two opposite corners
of the dance floor are connected by a diagonal. This diagonal
passes through the interior of exactly how many tiles?

the answer is [hide]72[/hide] but i dont kno why.... cud sum1 help? thx a lot

btw this is a 2002 ch sprint...

This is a well-known theorem(?)/technique.
For any $ m*n$ rectangle, the number of unit squares one diagonal passes through is $ m \plus{} n \minus{} (m,n)$, where $ (m,n)$ denotes the GCF function.

Apply that to this: $ 72 \plus{} 24 \minus{} 24 \equal{} 72$

Not sure what this is called.

BTW: Use the search button before posting a problem.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1773065499&t=117544[/url]

ok thx.... i had no idea bout that theorem


trigonometry
geometry
geometric transformation
reflection
trig identities

Let $ABCD$ be a cuadrilateral with $AB \parallel CD$. We know that $AC=BC$, $AB=BD$ and $\angle{ACB}=90^\circ$. Find $\angle{CBD}$.

Let $a=AC=BC$. Since $\triangle ACB$ is an isosceles right, then $AB=BD=a\sqrt{2}$ and $\angle CBA=45^\circ$, which gives $\angle BCD=135^\circ$ (parallel sides). Now let $\phi=\angle CBD$. Then $\angle ABD=\angle BDC=45^\circ-\phi$.

Applying Sine Law to $\triangle BCD$ we get:

$\frac{a}{\sin(45^\circ-\phi)}=\frac{a\sqrt{2}}{\sin 135^\circ}$ or

$\sin(45^\circ-\phi)=\frac{\sin 135^\circ}{\sqrt{2}}={1\over 2}\implies 45^\circ-\phi=30^\circ\implies \phi=15^\circ$

Is there another approach? Like finding similar triangles or stuff like that?

I mean if you don't know trig.

even if you dont know a lot of trig identities and algebraic trig, i think it would be useful to learn some basic geometric trig (extended los, loc, etc.)

Sorry for bumping such an old thread, but a synthetic solution was long overdue here.

See the attached diagram. Reflect $D$ over $BC$ to $E$. Then $\angle pCD=\angle CBA=45^\circ\implies\angle DCE=90^\circ$, hence $CE$ is the extension of $CM$, which is the altitude of $\triangle ABC$. But as $\triangle ABC$ is isosceles-right, $M$ is the midpoint of $AB$. Also, $BE=BD=BA$. Therefore, $\triangle EBM$ is right, with the hypotenuse twice as long as one leg, which means it's $30-60-90$.

Thus $\angle CEB=30^\circ\implies \angle CDB=30^\circ$, hence $\angle CBD=180^\circ-\angle CDB-\angle DCB=180^\circ-30^\circ-135^\circ=15^\circ$


[quote="bryantacademy"]what is 0!=1!=2!=3!=4!=10![/quote]
were those = signs supposed to be + signs?

[quote="bryantacademy"]what is 0!=1!=2!=3!=4!=10![/quote]
um, how are we supposed to do that?

Quit following up every one of noneoftheabove's posts with that. If you're impulse to say it is that big, then just PM him or IM him or just go to a Naruto forum.


Find the number of subsets of $S=\{ 1,...,n\}$ of the form even-odd-even-... or
odd-even-odd-...  ,as elements are sorted in increasing mode.

Let $f(n)$ be the required number and let $g(n)$ be the number of such subsets whose largest element is actually $n$.

By inspection $f(1)=2, f(2)=4, f(3)=7, f(4)=12, f(5)=20$. We conjecture $f(n)=F_{n+3}-1$.

Now we'll find $g(n)$ recursively. The number of good subsets with largest element $n$ is equal to the number of good subsets with largest element one of $n-1, n-3, \ldots, 1$ or $2$ depending on the parity of $n$ (this is obvious if you take away the $n$ element and look at the subset left), unless the subset is just $n$. Thus $g(2k)=g(2k-1)+g(2k-3)+\ldots+g(3)+g(1)+1$ and $g(2k+1)=g(2k)+g(2k-2)+\ldots+g(2)+1$.

Now, since $g(1)=1, g(2)=2$, we have $g(3)=2+1=3, g(4)=3+1+1=5$. Now note that $g(3)=g(2)+g(1), g(4)=g(3)+g(2)$. If $g(2k-2)=g(2k-3)+\ldots+g(1)$ then $g(2k)=g(2k-1)+g(2k-3)+\ldots+g(1)=g(2k-1)+g(2k-2)$, and analogously $g(2k+1)=g(2k)+g(2k-1)$. Thus, since $g(1)=F_2, g(2)=F_3$ we have $g(n)=F_{n+1}$.

Finally $f(n)=g(n)+g(n-1)+\ldots+g(1)+1=F_{n+1}+F_n+\ldots+F_2+1$. Now using the above identities $1+F_3+F_5+\ldots+F_{2k-1}=F_{2k}$, and $1+F_2+\ldots+F_{2k}=F_{2k+1}$, so $f(n)=F_{n+2}+F_{n+1}-1=F_{n+3}-1$, as we wanted.


Find all $x$ such that $[\frac{x}{[\sqrt{x}]}]=100$...

I don't have the answer right now, but I trust you MathLinkers....

Taking $x=z^{2}$, it becomes $100[z]\le z^{2}<101[z]$.
It should suffice studying all the cases $[z]=k$ for every $k$ near enough $100$.


Evaluate $ 2.65\times 10^3 \plus{} 3.5\times 10^4$ in scientific notation. Round your answer to two significant digits.

[hide="Easiest/fastest way"]

By noticing that $ 2.65\times 10^3 $ is one less power of ten than $3.5\times 10^4 $, we can add $0.5 + 0.265 = 0.765$ to get the decimal for $3.$ Thus the answer when rounded to two significant figures is $ \boxed{3.8\times 10^4}. $

[/hide]


calculus
derivative
vector
advanced fields
advanced fields theorems

Hi all,
here is one more problem that I have,if someone know how to solve it, please help me!! :)

this is the problem:

Derive the heat equation:

u=(1/cρ) · (γ/γx) · [k(x)ux] + f(x,t)

for the situation where the thermal conductivity k(x) depends on x.

 [/code]

The left hand side of that should be the derivative with respect to $t,$ right? I think you meant to say this (click on it to see what the code looks like):

$\frac{\partial u}{\partial t}=\frac1{c\rho}\frac{\partial}{\partial x}\left(k(x)\frac{\partial u}{\partial x}\right)+f(x,t)$

In several space dimensions, that would be

$\frac{\partial u}{\partial t}=\frac1{c\rho}\nabla_{x}\cdot\left(k(x)\nabla_{x}u\right)+f(x,t)$

where by $\nabla_{x}g$ i mean the gradient in the space variables of $g$ and by $\nabla_{x}\cdot \vec{G}$ I mean the divergence in the space variables of the vector field $\vec{G}.$

Of course, $f(x,t)$ represents some internal heat source or heat sink. My advice is to find a good derivation of the constant-coefficient (that is, constant heat capacity) version, and just try to copy it step by step, changing only what you have to change.

HI,
yes, the left hand side is the derivative with respect to t.
thanks for a help:) i ll try to do it now.


Fie $a,b,c$ din intervalul $(0;1)$, cu $a+b+c =1$...sa se demonstreze inegalitatea:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq  3(a^{2}+b^{2}+c^{2})+8$

$\, \Longleftrightarrow \, \, \sum \left( \frac1{a}-3 a^{2}\right) \geq 8$

Fie $f : \left( 0,1 \right) \to \mathbb R$, $f(x) = \frac1{x}-3 x^{2}$. Avem $f^{\prime \prime}(x) = \frac{2}{x^{3}}-6$, deci $f$ e strict convexa pe $I = \left( 0,\frac1{\sqrt[3] 3}\right)$.

Cel putin doua numere dintre $a,b,c$ sunt in $\left( 0,\frac12 \right) \subset I$. Folosind inegalitatea lui Jensen, putem presupune ca doua din ele sunt egale.

Daca $a=b$, atunci $a = \frac{1-c}2$. Inegalitatea este echivalenta cu \[\frac{4}{1-c}+\frac1{c}\geq \frac64 (1-c)^{2}+3c^{2}+8\]  \[\, \Longleftrightarrow \, \, 9 c^{4}-15 c^{3}+25 c^{2}-13 c+2 \geq 0\]  \[\, \Longleftrightarrow \, \, \left( c-\frac13 \right)^{2}\left[ \left( c-\frac12 \right)^{2}+\frac74 \right] \geq 0 .\] Avem egalitate pentru $a = b = c = \frac13$.


I do not understand, what do you mean by the right cone. If the base diameter and the lateral height are equal, the cone cross-section through its rotational axis is an equilateral triangle, which means that the cone apex angle is $60^o$, i.e., it is not right.

If the cone is well defined by the base diameter d and the lateral height $h_L$, cut the cone lateral surface along any cone element (a line through the apex) and unroll it. The result is a circular sector with radius $h_L$ and arc length $\pi d$, i.e., with the central angle $\phi = 2 \pi \cdot \frac{\pi d}{2 \pi h_L} = \frac{\pi d}{h_L}$ in radians. The problem then consists in inscribing a rectangle of a given width W (the adhesive tape width) with the maximum possible length L in this circular sector. If $d = h_L = 20\ \text{cm}$, then $\phi = \pi$ is a straight angle and the circular sector is a semicircle. The longest rectangle in this case is clearly placed along the semicircle diameter. Taking into account the tape width $W = 2\ \text{cm}$, the rectangle length L is given by

$L = 2 \sqrt{h_L^2 - W^2} = 2 \sqrt{400 - 4} = 2 \sqrt{396} = 12 \sqrt{11} \doteq 39.80\ \text{cm}$

[quote="yetti"]I do not understand, what do you mean by the right cone. If the base diameter and the lateral height are equal, the cone cross-section through its rotational axis is an equilateral triangle, which means that the cone apex angle is $60^o$, i.e., it is not right.
[/quote]

He means a right circular cone as opposed to a left circular cone.

A right cone is one in which the segment connecting the vertex to the center of the base is perpendicular to the base.  Equivalently (and more generally), the height meets the base at its centriod.

http://mathworld.wolfram.com/RightPyramid.html

[quote="warut_suk"]In a right circular cone, both the diameter of the base and the slant height are 20 cm. Find the maximum length of a 2 cm wide self-adhesive band that can be sticked on the lateral surface of the cone without any creasing, cutting, or overlapping.[/quote]
hey, the band would always crease, wouldnt it? :(  :?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Any help would be more than valuable, thank you.

Let G be a finite group, and r a natural number dividing |G|. Assume that H is the only subgroup of G of order r. Show that H is a normal subgroup of G.

Fix $ g\in G$.  What group is $ g^{\minus{}1} H g$?

I is a subgroup of G. Does this help somehow?
Thanks

And what do we know about subgroups of $ G$ of order r?


[quote="daermon"]$x=0.313131313131..._{5}$  When $x$ is written as a base ten fraction, what is the sum of its numerator and denominator?[/quote]

[hide]$0.313131313131..._{5}=\frac{31}{44}_{5}=\frac{16}{24}=\frac{2}{3}_{10}$[/hide]

correct, but you didn't actually answer the question  :P  :roll:

[quote="kheg_k"]I don t get it,

what does the subscript 5 mean in the question?

does that mean base 5?[/quote]

Yes, it does mean base 5.

[size=150]More Generalized[/size]

If you have a number, $\cdots d_{3}d_{2}d{}_{1}{}_{n}$, where the $d$s are digits with subscript $n$, then the number is $\cdots d_{3}\times n^{2}+d_{2}\times n^{1}+d_{1}\times n^{0}$.

[quote="dodobird"][quote="vishalarul"]$55x = 31.\overline{31}_{5}$
$44x ={31}_{5}$
$x={\frac{31}{44}}_{5}$
$x={\frac{16}{24}}_{10}= \frac{2}{3}$
The answer is $\boxed{5}$.

P.S. In your question, you should of put:
$x=0.\overline{31}_{5}$ When x is written as a base ten fraction([b]in lowest terms[/b]), what is the sum of its numerator and denominator?

Minor error, though.[/quote]

Question, shouldn't it be 100x and not 55x because in base 5 55x is impossible.[/quote]
55=100 in base five
2=10 in base two
It's easier to work with 55 than 100 in this case.

Oh, ok but i still like 100 it looks and sounds bigger :P

[hide]
$\overline{31}_{5}=55x$
$31_{5}=44x$
$x=\frac{31}{44}$
Then, we need to convert to base 10:
$x=\frac{16}{24}_{1}0=\frac{2}{3}$
Then, add the numerator and denominator
$2+3=\boxed{5}$[/hide]

[quote="daermon"]$x=0.\overline{31}_{5}$  When $x$ is written as a base ten fraction, what is the sum of its numerator and denominator?[/quote]

[hide]
$\overline{31}_{5}= 55x$
$x=\frac{31}{44}_{5}$
$\text{You need to convert to base 10.}$
$x=\frac{16}{24}_{10}=\frac{2}{3}_{10}$
$\text{The numerator is 2, and the denominator is 3. You add the numerator and the denominator to get }\boxed{5}$
[/hide]


Given that $x$ is directly proportional to $y$ and to $z$ and is inversely proportional to $w$, why is $\frac{xw}{yz}$ constant?

Edit: sorry; it's on page 38.

[hide="The general idea"] If $x$ is directly proportional to another variable $y$, then when everything else is held constant, $x = cy$, or $\frac{x}{y}$ is constant.

However, there are other variables to take into consideration.  $x$ is directly proportional to both $y$ and $z$.  This means that when $z$ is constant, $x = c_{1}y$ and when $y$ is constant, $x = c_{2}z$.  We can combine this into the statement

$x = c_{3}yz$

When $w$ is held constant; you can see from this expression that it satisfies our requirements.

Now, $x$ is also inversely proportional to $w$.  This means that when $y, z$ are held constant, $x = \frac{c_{4}}{w}$.  We can combine this with the other statement to create the beastly

$x = c_{5}\frac{yz}{w}$

Or

$c_{5}= \frac{xw}{yz}$

Which is a constant.  Can you see why this satisfies every requirement? [/hide]

Something is fishy here.

If $x=ay=bz={c\over w}$, then ${xw\over yz}={abc\over x^{2}}\neq\mathrm{const.}$

That's the mistake I made too.

Of course, you need to realize that $x = ay$, etc. only hold [b]when the other two variables are constant.[/b]

Or they could use an adjective like "simultaneously", which would clear the matter.

is this the logic you're using?

since $xw$ is constant when $y$ and $z$ are constant, we can write $xw=a$. but because $y$ and $z$ are constant $yz$ is constant and we can just divide both sides of $xw=a$ by $yz$ and that still makes $xw$ a constant (but a different constant). this applies to the other cases as well.

in conclusion, we can say that $\frac{xw}{yz}$ is constant.

Er, no.  That's not really the reason.  Refer back to my edited post, and tell me which part you don't understand.


calculus
integration
function
trigonometry
real analysis
real analysis unsolved

Let $ f(x)$ is continuos function .If $ f(1) \equal{} 1$ and

$ \int_0^x tf(2x \minus{} t)dt \equal{} \frac {1}{2}arctan(x^2)$ then Find $ \int_1^2 f(x)dx$

Let $ 2x \minus{} t \equal{} u$, we can rewrite the integral equation as

$ \int_{x}^{2x} (2x \minus{} u)f(u)du \equal{} \frac {1}{2}\tan ^{ \minus{} 1} x^2$

$ \Longleftrightarrow 2x\int_{x}^{2x} f(u)du \minus{} \int_{x}^{2x} uf(u)du \equal{} \frac {1}{2}\tan ^{ \minus{} 1} x^2$

Differentiate with respect to $ x$ both sides,

$ 2\int_{x}^{2x} f(u)du \plus{} 2x\{2f(2x) \minus{} f(x)\} \minus{} \{4xf(2x) \minus{} xf(x)\} \equal{} \frac {1}{2}\cdot \frac {2x}{1 \plus{} x^2}$

$ \Longleftrightarrow 2\int_{x}^{2x} f(u)du \minus{} xf(x) \equal{} \frac {x}{1 \plus{} x^2}.$

Plugg $ x \equal{} 1$ into the equation,  $ 2\int_1^2 f(u)du \equal{} f(1) \plus{} \frac {1}{2} \equal{} \frac {3}{2}$, yielding $ \int_1^2 f(x)dx \equal{} \boxed{\frac {3}{4}}.$


Let $a$, $b$, $c$ be the side lengths of a triangle, and F its area. Prove that

$F\leq \frac{\sqrt 3}{4}(abc)^{\frac{2}{3}}$.

Determine when equality holds.

The statement is $F \leq \frac{\sqrt{3}}{4}(abc)^{2/3}$

[hide="And the solution"]
Since $SinASinBSinC \leq \frac{3 \sqrt{3}}{8}$  $\ldots (*)$we get that $abc=8R^{3}.SinA.SinB.SinC \leq 3 \sqrt{3}R^{3}$ iff $\left( \frac{abc}{4R}\right)^{3}\leq \frac{ 3\sqrt{3}}{64}(abc)^{2}$ iff   $F^{3}\leq \frac{ 3\sqrt{3}}{64}(abc)^{2}$ iff $F \leq \frac{ \sqrt{3}}{4}(abc)^{2/3}$

For (*) you can see the 5th post in   [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15509]www.mathlinks.ro/Forum/viewtopic.php?t=15509[/url] ) note that equality holds iff $SinA=SinB=SinC$ iff $a=b=c$.  So in the statment we have an equality iff the triangle  is equilateral.

[/hide]

$Tipe$

Thank for your proof with it :
I can do stronger :
$F \leq \frac{sinA+sinB+sinC}{6}(abc)^{3/2} \leq \frac{\sqrt{3}}{4}(abc)^{3/2}$

OK??  :P

Sorry, but I don't think that
$F \leq \frac{\sqrt{3}}{4}(abc)^{3/2}$ is stronger than $F \leq \frac{\sqrt{3}}{4}(abc)^{2/3}$

Oh, I see it's just a typo. Seems first bad post.
[]


In how many three-digit integers is a 5 either the tens digit, the units digit, or both?

Let's use complementary counting.

There are $ 9\cdot9\cdot9$ integers that don't satisfy this, so $ 900\minus{}729\equal{}171$ do.

How can there be an odd number of answers if for ever positive number that fits, there exists a negative one?

That queston's wording is confusing.

Another way: In each "hundreds group" ($100-199$, $200-299$, etc.) there are $10+10-1=19$ numbers that satisfy this ($15, 25, 35, 45, 55, 65, 75, 85, 95$ and $50, 51, 52, 53, 54, 55, 56, 57, 58, 59$, minus one since $55$ was double-counted).
Since there are $9$ hundreds groups, there are $9\times19=\boxed{171}$ numbers that satisfy this.


algebra function domain integration topology calculus derivative

Let $ \Omega$ be a bounded smooth domain in $ \mathbf R^n$ and $ \Gamma$ be an open smooth portion of the boundary $ \partial\Omega$, $ \Gamma' : \equal{} \partial\Omega \setminus \Gamma$. Consider the following mixed boundary value problem:
\[ \left\{\begin{aligned} \minus{} \Delta u & \equal{} 0 & & \text{ in } \Omega, \\
u & \equal{} 0 & & \text{ on } \Gamma', \\
\frac {\partial u}{\partial\nu} & \equal{} \lambda u & & \text{ on } \Gamma. \end{aligned} \right.
\]
Set $ X : \equal{} \{u \in H^1(\Omega) \mid u \equal{} 0 \text{ on } \Gamma' \text{ (in the trace sense)}\}$ then the variational formulation of our problem is as follows: find $ u \in H^1(\Omega)$ so that
\[ \int_\Omega \nabla u \cdot \nabla v \minus{} \lambda \int_\Gamma uv \equal{} 0
\]
for any $ v \in X$.

We are interested in $ \lambda > 0$ for which we have $ u\ge0$, $ u\not\equiv0$. Such $ \lambda$ exist. Here's [hide="sketch of the proof."]
Set $ \Phi(v) : \equal{} \frac {\int_{\Omega} |\nabla v|^2}{\int_{\Gamma} v^2}$ and consider the minimization problem
\[ \lambda : \equal{} \inf_{v \in X} \Phi(v) \equal{} \inf_{v \in X, \, \|v\|_{L^2(\Gamma)} \equal{} 1} \int_\Omega |\nabla v|^2.
\]
If $ (v_n) \subset X$ is a minimizing sequence then $ \int_{\Omega} |\nabla v_n|^2$ is bounded, yielding $ (v_n)$ is bounded in $ H^1(\Omega)$. So WLOG one may assume that $ v_n \rightharpoonup v_*$ weakly in $ H^1(\Omega)$ for some $ v_* \in H^1(\Omega)$. It now follows from lower semi-continuity of $ \Phi$ that $ \lambda \equal{} \Phi(v_*)$. Since $ \Phi(|v|) \equal{} \Phi(v)$ for any $ v \in X$, we can replace $ v_*$ with $ |v_*|$. Since $ \|v_*\|_{L^2(\Gamma)} \equal{} 1$, $ v_*\not\equiv0$. Finally, we must have $ \Phi'(v_*) \equal{} 0$, i.e. that $ u \equal{} v_*$ is a solution to our problem.[/hide]

[b]Question:[/b]
1) What can you say about regularity of $ u$? We must have $ u \in C^\infty(\Omega)$ but it is very likely that $ u \notin C^1(\overline\Omega)$. Ok. But is $ u$ a $ C^1$ function at the interior points (in the topology of $ \partial\Omega$) of $ \Gamma$ and $ \Gamma'$?
I need that normal derivative be defined as a classical function both on $ \Gamma$ and $ \Gamma'$ because I want to use Hopf's lemma: $ \frac {\partial u}{\partial\nu} < 0$ on $ \Gamma'$.

2) Also I want to use integration by parts: for any $ v \in H^1(\Omega)$,
\[ \int_\Omega \nabla u \cdot \nabla v \equal{} \minus{} \int_\Omega v\Delta u \plus{} \int_{\partial\Omega} \frac {\partial u}{\partial\nu} v \equal{} \int_{\partial\Omega} \frac {\partial u}{\partial\nu} v \equal{} \lambda \int_\Gamma uv \plus{} \int_{\Gamma'} \frac {\partial u}{\partial\nu}v.
\]
The trouble is in its validity (e.g. we do not have $ \Delta u$ in general, because we only have $ u \in H^1(\Omega)$). Are the identies above valid?

2) I think that
\[ \int\limits_{\Gamma '} {\frac{{\partial u}}
{{\partial \nu }}vdx}  \equal{} 0
\]
is trivial and your calculation is right.

Why $ \int_{\Gamma'} \frac {\partial u}{\partial\nu}v dx \equal{} 0$? Suppose $ \frac {\partial u}{\partial\nu}$ exists as an ordinary function, then by Strong Maximum Principle (or Hopf's lemma) we must have that $ \frac {\partial u}{\partial\nu} < 0$ on $ \Gamma'$.

Why my calculations are right? Recall that we just have $ u \in H^1(\Omega)$, not $ u \in H^2(\Omega)$.

OK, let me check my argument and back to you later.

I need the following inequality:
\[ \int_\Omega \nabla u \cdot \nabla v \le \lambda \int_\Gamma uv\]
for any nonnegative $ v \in H^1(\Omega)$.

If $ v \in X$ then $ \int_\Omega \nabla u \cdot \nabla v \equal{} \lambda \int_\Omega uv$ by the variational formulation of the problem. So all is OK here. But what happens if $ v \notin X$?

If integration by parts is valid and if $ \frac {\partial u}{\partial\nu}$ exists then I will have the desired inequality because
\[ \int_{\Gamma'} \frac {\partial u}{\partial\nu}v \le 0\]
by Hopf's lemma ($ \frac {\partial u}{\partial\nu} < 0$ on $ \Gamma'$).

[quote="volodja"]2) Also I want to use integration by parts: for any $ v \in H^1(\Omega)$,
\[ \int_\Omega \nabla u \cdot \nabla v \equal{} \minus{} \int_\Omega v\Delta u \plus{} \int_{\partial\Omega} \frac {\partial u}{\partial\nu} v \equal{} \int_{\partial\Omega} \frac {\partial u}{\partial\nu} v \equal{} \lambda \int_\Gamma uv \plus{} \int_{\Gamma'} \frac {\partial u}{\partial\nu}v.
\]
The trouble is in its validity (e.g. we do not have $ \Delta u$ in general, because we only have $ u \in H^1(\Omega)$). Are the identies above valid?[/quote]

Something wrongs in your framework, recall that
\[ \int\limits_\Omega  {\left( { \minus{} \Delta u} \right)vdx}  \equal{} \int\limits_\Omega  {\nabla u\nabla vdx}  \minus{} \int\limits_{\partial \Omega } {\frac{{\partial u}}
{{\partial n}}vdS}
\]
holds for every $ u \in H^2 \left( \Omega  \right)$ and $ v \in H^1 \left( \Omega  \right)$.

[quote="bookworm_vn"]
holds for every $ u \in H^2 \left( \Omega \right)$ and $ v \in H^1 \left( \Omega \right)$.[/quote]

We only have that $ u \in H^1(\Omega)$.


Let  $ ABC$ be an acutangle triangle. Prove o Disprove that
 $ \frac{\cot^2(A) \plus{} \cot^2(B) \plus{} \cot^2(C)}{\tan^2(\frac{A}{2}) \plus{} \tan^2(\frac{B}{2}) \plus{} \tan^2(\frac{C}{2})}[\cos(\frac{A}{2})\cos(\frac{B}{2})\cos(\frac{C}{2})]>\frac{\sin(A \plus{} B \plus{} C)}{4}$ .

_____________
 :)

Always true, LHS is positive, but RHS = 0 (0 <A,B,C<90°, sin180°=0), or did you mean...
$ \frac{\cot^2(A) \plus{} \cot^2(B) \plus{} \cot^2(C)}{\tan^2(\frac{A}{2}) \plus{} \tan^2(\frac{B}{2}) \plus{} \tan^2(\frac{C}{2})}[\cos(\frac{A}{2})\cos(\frac{B}{2})\cos(\frac{C}{2})]>sin(\frac{(A \plus{} B \plus{} C)}{4})$

[quote="anderssonsw"]Always true, LHS is positive, but RHS = 0 (0 <A,B,C<90°, sin180°=0), or did you mean...
$ \frac {\cot^2(A) \plus{} \cot^2(B) \plus{} \cot^2(C)}{\tan^2(\frac {A}{2}) \plus{} \tan^2(\frac {B}{2}) \plus{} \tan^2(\frac {C}{2})}[\cos(\frac {A}{2})\cos(\frac {B}{2})\cos(\frac {C}{2})] > sin(\frac {(A \plus{} B \plus{} C)}{4})$[/quote]

yes.... and

$ \frac{\cot^2(A) \plus{} \cot^2(B) \plus{} \cot^2(C)}{\tan^2(\frac{A}{2}) \plus{} \tan^2(\frac{B}{2}) \plus{} \tan^2(\frac{C}{2})}[\cos(\frac{A}{2})\cos(\frac{B}{2})\cos(\frac{C}{2})]>\frac{\sin(\frac{4A \plus{} 2B \plus{} C}{4})}{4}$ ???


pigeonhole principle
combinatorics unsolved
combinatorics

Let $ S$ be a finite set of at least five points in the plane, some of which are colored red and the others are colored blue. No three points of the same color are collinear. Show that there is a triangle with the vertices in $ S$ such that:

$ (1)$ all its vertices are the same color.
$ (2)$ at least one side of the triangle contains no point of the opposite color.

By the Pigeonhole principle, at least one such triangle satifying (1) must exist.
Assume the second condition is false. Then there must be a point on each side of the triangle of the opposite color. But now we have another triangle, for which there must exists three more points on each of its sides of the opposite color, and so on. Thus, we have a contradiction by infinite descent, since there must be a finite number of points.


function
limit
inequalities
conics
parabola
calculus
calculus computations

Let f be the function such that $ f(x) = 1+x^{2}$ if x is irrational and $ f(x) = 1+4x^{4}$ if x is rational.

Use the Squeeze Theorem to find $ \lim_{x\to\\ 0\\ {f(x)}}$

Note: Enter the missing values in their decimal form.

Solution: Clearly, for all real numbers x, f(x) $ \geq 1$. Next, we note that for a real number x,
$ x^{2}-4x^{4}= x^{2}(1-4x^{2})\geq{0}$

if and only if ___ $ \geq$ 0. This means that $ x^{2}\geq{4x^{4}}$ if and only if $ x\in[ , ]$. Particularly, we get that for all (rational or irrational) numbers x in [  ,  ] we have
$ f(x) = 1+4x^{4}\leq 1+x^{2}$.

Therefore, for all $ x\in$ [  ,  ], we have that
$ 1\leq f(x)\leq 1+x^{2}$.

Since $ \lim_{x}\to\0\{1+x^{2}$     ,
by the _______ Theorem,  $ \lim_{x}\to\0\{f(x)}=$

[color=darkblue]You can use the [b]Heine's definition[/b] for the limit of a function in an accumulation point : [/color]

[color=darkred][b][u]Heine's definition.[/u][/b] Let $ f: E\rightarrow\mathcal R$ be a function and the point $ x_{0}\in E'$- the [b]derivate[/b] of the set $ E$.  Then

$ \lim_{x\to x_{0}}f(x) \equal{} L\in R$ if and only if for any sequence $ x_{n}\ ,\ n\in N^{*}$ so that $ \left\{\begin{array}{c}(\forall ) n\in N^{*}\ ,\ x_{n}\in E\\ \ (\forall ) n\in N^{*}\ ,\ x_{n}\ne x_{0}\\ \ x_{n}\rightarrow x_{0}\end{array}\right\|$ $ \implies$ $ \lim_{n\to\infty }f(x_{n}) \equal{} L$.[/color]

I don't think it helps the poster to understand the Squeeze Theorem if you give another Theorem to solve the problem.

ysk1, the solution you've posted so far looks good. If I understand correctly, you are just having trouble finding the values to fill in the blanks? (And a little trouble with LaTex).

You need to solve the inequality $ x^{2}(1\minus{}4x^{2})\geq 0$. First note that $ x^{2}\geq 0$ for all x, so consider only the sign of $ 1\minus{}4x^{2}$. Think about the graph of this expression and where it will be positive?

[hide]Answer: The graph of this expression is a "bowl down" parabola and it is positive between the roots. You can find the values of these roots by setting the expression equal to zero and solving for x.[/hide]


anyone joining narayana aits.......or joind earlier.......????

hey rahul how went ur npt...

mine went ok.......got rank 300 overall

6	2380040	Arnav Mittal	43/873	87/131	32/723	162	300

btw are u giving the grand test on 6th
???


function
integration
calculus
real analysis
calculus computations

How might you prove the following:
$\int_{-\infty}^{\infty}f(x) \delta(g(x)) dx = \sum_{\{x_{i}|g(x_{i})=0\}}\frac{1}{|g'(x_{i})|}f(x_{i})$

where $\delta(x)$ is the dirac delta function.

It seems that the "integral" on the left is not an ordinary Lebesgue (or Riemann) integral. What is it, then?

I guess you can define it as a functional $\Delta (f(x),I)$ s.t

${\Delta (f(x),I) = \begin{cases}f(0) \& \mbox{if }0 \in I\\ 0 \& \mbox{if }0 \notin}I \end{cases}$

where $I \subset \mathbb{R}$.

I don't know how to be any more formal than this.

[quote="V.V."]V.S. Vladimirov, "Generalized functions in mathematical physics" , MIR (1977) (Translated from Russian).[/quote]
I can't find that book at my local library. Would you mind giving me a gist of the proof?

Notice that the transformation property of dirac delta says that \[ \delta\left[g(x)\right]\equal{}\sum_{i\equal{}1}^{\infty}\frac{\delta(x\minus{}x_{i})}{|g'(x_{i})|}\] where $ x_{i}$'s are simple roots of the equation $ g(x)\equal{}0$. This equation ca be found in books related to generalized functions. For example you can refer to Ram P kanwal's book. Substituting this result in $ \int_{\minus{}\infty}^{\infty}f(x)\delta\left[g(x)\right]dx$ we get \[ \int_{\minus{}\infty}^{\infty}f(x)\sum_{i\equal{}1}^{\infty}\frac{\delta(x\minus{}x_{i})}{|g'(x_{i})|}dx\] By interchanging summation and integration and using the relation $ \int_{\minus{}\infty}^{\infty}f(x)\delta(x\minus{}x_{i})dx\equal{}f(x_{i})$ we finally get \[ \int_{\minus{}\infty}^{\infty}f(x)\delta\left[g(x)\right]dx\equal{}\sum_{i\equal{}1}^{\infty}\frac{f(x_{i})}{|g'(x_{i})|}\]


function
advanced fields
advanced fields unsolved

Given a locally Lipschitz continuous function $ f: M \to N$, then the restriction of $ f$ to any compact set $ A \subseteq M$ is Lipschitz continuous.

If not, then there are two sequences $ (a_n)$ and $ (b_n)$ in $ A$ such that $ \frac{d(f(a_n),f(b_n))}{d(a_n,b_n)}\to\infty$. Choose a convergent subsequence of $ (a_n)$. Since $ d(a_n,b_n)\to 0$, the corresponding subsequence of $ (b_n)$ has the same limit. And so on.

it's probably safer to work with "open cover definition" for compact sets,

i.e. for each $ a\in A$ let $ U_{\epsilon_a}(a)$ be an open $ \epsilon_a$-neighbourhood such that there exists $ K_a$ with $ \forall x,y\in U_{\epsilon_a}(a): \,|f(x)\minus{}f(y)|<K_a|x\minus{}y|$. Since $ A$ is compact, we can assume $ A\subseteq\bigcup_{a\in I}U_{\epsilon_a}(a)$ for some finite index set $ I\subseteq M$. Then $ K\equal{}\min\{K_a\,|\,a\in I\}$ "does the job".

[quote="hjbrasch"] $ K \equal{} \min\{K_a\,|\,a\in I\}$ "does the job".[/quote]

I don't think this last step is right. This is were i got stucked with the prove

Indeed, $ \max K_a$ will serve only the pairs of very close points (at best). The trick is to use the compactness of $ K\times K$ rather than that of $ K$. Look at the function
\[ F(x,y)\equal{}\begin{cases}
\frac{|f(x)\minus{}f(y)|}{d(x,y)},& x\ne y
\\
0, &x\equal{}y
\end{cases}\]
It is locally bounded (outside the diagonal it is because we can choose a neighborhood of $ (x_0,y_0)$ in which $ f(x)$ and $ f(y)$ are bounded and $ d(x,y)$ is separated from $ 0$; on the diagonal it is exactly the local Lipshitz condition), so it is bounded on $ K\times K$, which is what we need.

sorry, Salvador, I was very sloppy.

$ \forall a\in A$ let $ K_a$ the local Lipschitz-constant and $ U_{\epsilon_a}(a)$ the corresponding neighbourhood. Choose a finite cover $ A\subseteq \bigcup_{i\in I}U_{\frac{1}{2}\epsilon_a}(a)$ and define $ K_1\equal{}\max\{K_a\,|\,a\in I\}, K_2\equal{}\max\{d_N(f(a),f(b))\,|\,a,b\in I\}$ and $ \epsilon_{\min}\equal{}\min\{\epsilon_a\,|\,a\in I\},\epsilon_{\max}\equal{}\max\{\epsilon_a\,|\,a\in I\}$ then $ K\equal{}\max\left\{K_1,2\frac{K_1\epsilon_{\max}\plus{}K_2}{\epsilon_{\min}}\right\}$ does the job. To see that let $ x,y\in A$ be arbitrary.

If $ d_M(x,y)<\frac{1}{2}\epsilon_{\min}$ choose $ a\in I$ with $ x\in U_{\frac{1}{2}\epsilon_a}(a)$, then $ x,y\in U_{\epsilon}(a)$ which implies $ d_N(f(x),f(y))<K_ad_M(x,y)\leq Kd_M(x,y)$.

If $ d_M(x,y)\geq\frac{1}{2}\epsilon_{\min}$ holds, choose $ a,b\in I$ with $ x\in U_{\frac{1}{2}\epsilon_a}(a),y\in U_{\frac{1}{2}\epsilon_b}(b)$ and calculate:

$ d_N(f(x),f(y))\leq d_N(f(x),f(a))\plus{}d_N(f(a),f(b))\plus{}d_N(f(b),f(y))<\\
K_1(d_M(x,a)\plus{}d_M(b,y))\plus{}K_2\leq K_1\epsilon_{\max}\plus{}K_2\leq 2\frac{K_1\epsilon_{\max}\plus{}K_2}{\epsilon_{\min}}d_M(x,y)\equal{}Kd_M(x,y)$


Find all real-valued functions $ f(x)$ on the reals such that $ f(2002x \minus{} f(0) ) \equal{} 2002 x^2$ for all $ x$.

Just substitute $ t\equal{}\frac{x\plus{}f(0)}{2002}$ to get $ f(t)\equal{}\frac{(t\plus{}f(0))^2}{2002}$. In the case $ t\equal{}0$, we get $ f(0)\equal{}0$ or $ f(0)\equal{}2002$ so the functions are
\[ \boxed{f(t)\equal{}\frac{t^2}{2002},f(t)\equal{}\frac{(t\plus{}2002)^2}{2002}}\]


function
induction
algebra
domain
Diophantine equation
Ring Theory
number theory unsolved

Solve the function:  3^a - 5^b = 2,  a,b are positive integers.

... Harazi, then we just get that $a$ is odd ...

[quote="Arne"]Actually we get $a = 3$ (mod 4) and $b$ is even (looking mod 3 and mod 5) and the problem reduces to the Pell equation \[x^2 - 3y^2 = -2.\]

I just don't remember how to solve an equation of this type (of course I can do $x^2 - dy^2 = 1$ but with $-2$ instead it's a little bit more complicated).

But anyway, I guess it's obvious once you solve the Pell equation.[/quote]

If  Pell equation \[x^2 - 3y^2 = -2.\]  is solved,what to do next? I think only have (a,b)=(3,2) is possible.

I think that (3, 2) is the only solution as well.

Well, Pell equations always have solutions but the number of solutions with powers of 3 and 5 will always be very limited... if you understand what I mean?

There are some nice recurrences for the solutions of Pell equations which might allow you to prove that the only solution for that equation which is of the form $(5^x,3^y)$ is $(1, 1)$ (by induction or so).

I know it sounds pretty vague but I have done and seen similar things before.

How to solve this particular Pell equation? :blush:

$x^2-3y^2=-2$ can be written as $(x-\sqrt 3y)(x+\sqrt 3y)=-2$. Since $-2$ is a prime, it means that both $x-\sqrt 3y$ and $x+\sqrt 3y$ are primes in $\mathbb Z[\sqrt 3]$ (it's a Euclidean domain, so everything works out just fine; irreducible elements are prime elements, and so on).

If $u^2-3v^2=-2$ (i.e. if $(u,v)$ is another solution to the equation), then $-2=(u-\sqrt 3v)(u+\sqrt 3v)$ is another prime decomposition of $-2$, so $u-\sqrt 3v,u+\sqrt 3v$ are the same as $x-\sqrt 3y,x+\sqrt 3y$, up to multiplication with invertible elements in $\mathbb Z[\sqrt 3]$, i.e. we can write $u+\sqrt 3v=(x+\sqrt 3y)(a+\sqrt 3b)\ (*)$, where $a,b$ are integers s.t. $a^2-3b^2=\pm 1$.

That means that, basically, all you need to do is solve the equations $a^2-3b^2=\pm 1$ (actually, just $a^2-3b^2=1$, since the one with $-1$ has no solutions), and find just one solution to $x^2-3y^2=-2$ ($(1,1)$, for example), from which you get all the other ones by $(*)$.

as i know(of course i may be wrong), that can not express all the solutions but some of which.

I have come up with a solution, but I don't know if it was right and it's loaded down with trivial details.
If a>4,
Looking mod 243, we get b=162n+74,
Then looking mod 163(a prime), we get a=162m+103,
Then looking mod 19, we get 2=3^(162m+103) - 5^(162n+74)=3^103 - 5^74=14 - 6=8 (mod 19), which is impossible. Maybe my solution is wrong, I don't know.


If you have an iPod and some bathroom time, this is a good listen if you're interested in a career in science.  [url=http://www.nyas.org/snc/podcasts.asp?PartnerCD=iTunes&TrackCD=pcast]Here[/url] is the podcast description, but [url=http://media.nyas.org/content/podcasts/snc/Doherty.mp3]here[/url] is a direct link to the relevant episode, entitled "How to Win a Nobel Prize."  The episode description reads:

"Immunologist and Nobel Laureate Peter Doherty shares his advice on succeeding in a career in science, and discusses the challenges facing young researchers today."

Something worth checking out.  I haven't heard the whole thing, but so far it's been good.

It's pretty interesting overall.........it's surely worth 13+ minutes of your time....... :)


[quote="grupyorum"]Show that for all positive real numbers a,b and c the following inequality holds:
$ \sum_{sym}\frac {(b \plus{} c)(a^{4} \minus{} b^{2}c^{2})}{ab \plus{} 2bc \plus{} ac}\geq 0$[/quote]

after expanding,

$ \sum_{sym}\frac {(b \plus{} c)(a^{4} \minus{} b^{2}c^{2})}{ab \plus{} 2bc \plus{} ac}\geq 0$

$ \iff \sum_{cyc}\frac {(b \plus{} c)(a^{4} \minus{} b^{2}c^{2})}{ab \plus{} 2bc \plus{} ac}\geq 0$

$ \iff 2[6,3,0] \plus{} 7[6,2,1] \plus{} [5,2,2] \ge [5,4,0] \plus{} 3[4,4,1] \plus{} 6[4,3,2]$

by Muirhead Inequality, we have

$ [6,3,0] \ge [5,4,0]$

$ [6,3,0] \ge [4,4,1]$

$ 2[6,2,1] \ge 2[4,4,1]$

$ 5[6,2,1] \ge 5[4,3,2]$

$ [5,2,2] \ge [4,3,2]$

adding them together, we have done.

[quote="spanferkel"][quote="grupyorum"]Show that for all positive real numbers a,b and c the following inequality holds:
$ \sum_{sym}\frac {(b \plus{} c)(a^{4} \minus{} b^{2}c^{2})}{ab \plus{} 2bc \plus{} ac}\geq 0$[/quote]

Using $ 4bc\le(b \plus{} c)^2$ in both numerator and denominator :D, we get

$ \frac {(b \plus{} c)(16a^{4} \minus{} 16b^{2}c^{2})}{2(ab \plus{} 2bc \plus{} ac)}\ge\frac {(b \plus{} c)(16a^{4} \minus{} (b \plus{} c)^4)}{2a(b \plus{} c) \plus{} (b \plus{} c)^2} \equal{} (2a \minus{} b \minus{} c)(4a^2 \plus{} (b \plus{} c)^2).$

Summing this up cyclically (or symmetrically, if you really want) and using $ \sum(2a \minus{} b \minus{} c)(a^2 \plus{} b^2 \plus{} c^2) \equal{} 0$, we remain with

$ \sum (6a^3 \plus{} 4abc \minus{} 5(a^2b \plus{} a^2c))\ge0$ by AM-GM and Schur. No need for Muirhead, expanding etc.  :P  :lol:[/quote]

very nice...
 :D

But this solution is wrong unfortunately. Because $ a^4\minus{}b^2c^2$ may be negative.  :wink:

[quote="FelixD"]I think spanferkels solution is still true.[/quote]
How can you rescue it? It may be possible that

$ \frac {(b \plus{} c)(16a^{4} \minus{} 16b^{2}c^{2})}{2(ab \plus{} 2bc \plus{} ac)}\ge (2a \minus{} b \minus{} c)(4a^{2} \plus{} (b \plus{} c)^{2})$

always holds, but proving it may need bunching that whole thing. Or can you prove that easily?

I would have liked an [i]elegant [/i]symmetric solution that doesn't need to assume things like $ a\ge b\ge c$.

It's not longer elegant, but if you expand
\[ \frac {16a^4 \minus{} 16b^2c^2}{2(ab \plus{} 2bc \plus{} ca)} \ge \frac {16a^4 \minus{} (b \plus{} c)^4}{2a(b \plus{} c) \plus{} (b \plus{} c)^2},\]
you'll see that it's true. Indeed, the inequality is equivalent to
\[ 16a^4(b \minus{} c)^2 \plus{} 2(ab \plus{} bc \plus{} ca) ((b \plus{} c)^4 \minus{} 16b^2c^2) \plus{} 2bc ( b \minus{} c)^4 \ge 0,\]
which is clearly true. (moreover, this takes about 1 minute^^)

[hide="@spanferkel:"] so unelegant finde ich das ganze dann auch wieder nicht, vorallem weil man eigentlich fast keinen rechenaufwand beim ausmultiplizieren hat und der rest der lösung is eh sehr elegant :P
schon ausgebessert^^[/hide]

:P  Thank you very much, Felix. I think it doesn't happen often that AM-GM can inspire a valid estimation without being actually used in it.  :lol:

[hide="@ FelixD"]kleiner Druckfehler: statt [quote="FelixD"]
\[ \frac {16a^4 \minus{} 16b^2c^2}{2(ab \plus{} 2bc \plus{} ca)} \ge \frac {16a^4 \minus{} (b \plus{} c)^4}{2(a \plus{} c) \plus{} (b \plus{} c)^2},\]
[/quote]sollte es natürlich  $ \frac {16a^4 \minus{} 16b^2c^2}{2(ab \plus{} 2bc \plus{} ca)} \ge \frac {16a^4 \minus{} (b \plus{} c)^4}{2a(b \plus{} c) \plus{} (b \plus{} c)^2}$ heißen. Stimmt, soo furchterregend ist der Ausdruck nicht  :lol: [/hide]

[quote="grupyorum"]Show that for all positive real numbers a,b and c the following inequality holds:
$ \frac {(b \plus{} c)(a^{4} \minus{} b^{2}c^{2})}{ab \plus{} ac \plus{} 2bc} \plus{} \frac {(c \plus{} a)(b^4 \minus{} c^2a^2)}{bc \plus{} ba \plus{} 2ca} \plus{} \frac{(a \plus{} b)(c^4 \minus{} a^2b^2)}{ca \plus{} cb \plus{} 2ab}\geq 0.$[/quote]
Without loss of generality, assume that $ a \ge b \ge c.$ Since $ a^4 \minus{} b^2c^2 \ge 0$ and $ ab \plus{} ac \plus{} 2bc \le bc \plus{} ba \plus{} 2ca,$ we have
$ \frac {(b \plus{} c)(a^4 \minus{} b^2c^2)}{ab \plus{} ac \plus{} 2bc} \ge \frac {(b \plus{} c)(a^4 \minus{} b^2c^2)}{bc \plus{} ba \plus{} 2ca}.$
Similarly, since $ c^4 \minus{} a^2b^2 \le 0$ and $ ca \plus{} cb \plus{} 2ab \ge bc \plus{} ba \plus{} 2ca,$ we have
$ \frac {(a \plus{} b)(c^4 \minus{} a^2b^2)}{ca \plus{} cb \plus{} 2ab} \ge \frac {(a \plus{} b)(c^4 \minus{} a^2b^2)}{bc \plus{} ba \plus{} 2ca}.$
Therefore, it suffices to prove that
$ \frac {(b \plus{} c)(a^4 \minus{} b^2c^2)}{bc \plus{} ba \plus{} 2ca} \plus{} \frac {(c \plus{} a)(b^4 \minus{} c^2a^2)}{bc \plus{} ba \plus{} 2ca} \plus{} \frac {(a \plus{} b)(c^4 \minus{} a^2b^2)}{bc \plus{} ba \plus{} 2ca} \ge 0,$
which reduces to the obvious inequality
$ bc(b \plus{} c)(b \minus{} c)^2 \plus{} ca(c \plus{} a)(c \minus{} a)^2 \plus{} ab(a \plus{} b)(a \minus{} b)^2 \ge 0.$
The proof is completed. Equality occurs if and only if $ a \equal{} b \equal{} c.$

[quote="grupyorum"]Show that for all positive real numbers a,b and c the following inequality holds:
$ \sum_{sym}\frac {(b \plus{} c)(a^{4} \minus{} b^{2}c^{2})}{ab \plus{} 2bc \plus{} ac}\geq 0$[/quote]
It's obviously trues because by Chebychev because with $ a \ge b \ge c$, then
$ (b\plus{}c)(a^4\minus{}b^2c^2) \ge (c\plus{}a)(b^4\minus{}c^2a^2) \ge (a\plus{}b)(c^4\minus{}a^2b^2)$
And $ \frac{1}{ab\plus{}2bc\plus{}ca} \ge \frac{1}{bc\plus{}2ca\plus{}ab} \ge \frac{1}{ca\plus{}2ab\plus{}bc}$  :maybe:

@spanferkel
i almost do the same thing with you in the exam which costed me for 7 points :(

[quote="grupyorum"]Show that for all positive real numbers a,b and c the following inequality holds:
$ \sum_{sym}\frac{(b\plus{}c)(a^{4}\minus{}b^{2}c^{2})}{ab\plus{}2bc\plus{}ac}\geq 0$[/quote]

$ \Longleftrightarrow  \sum_{cyc} \frac{(a^3b^2+2a^3bc+a^2b^3+4a^2b^2c+2b^3ac+ac^4+bc^4)(a-b)^2}{(ab+2bc+ca)(bc+2ca+ab)}\geq 0$.

It's fifth problem of Turkey NMO 2009, also see here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2003335&sid=6ac5f3f753fc1fa1b2df1a8789382121#p2003335


[quote="filomen"]If $ a,b,c > 0$, then

$ \frac {c^2 \plus{} c \minus{} b}{a \plus{} b \plus{} 2c} \plus{} \frac {b^2 \plus{} b \minus{} a}{a \plus{} 2b \plus{} c} \plus{} \frac {a^2 \plus{} a \minus{} c}{2a \plus{} b \plus{} c}\leq\frac {1}{2}(\frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{c \plus{} a} \plus{} \frac {c^2}{a \plus{} b})$[/quote]
It's follows from $ \sum_{cyc}\frac{c\minus{}b}{a\plus{}b\plus{}2c}\leq0$ and $ \sum_{cyc}\frac {c^2  }{a \plus{} b \plus{} 2c}\leq\frac {1}{2}\sum_{cyc}\frac {a^2}{b \plus{} c},$ which are obviously true.


A life raft, set adrift from a sinking ship 150 miles offshore, travels directly toward a Coast Guard station at the rate of 5 mph.  At the same time the raft is set adrift, a rescue hellicopter is dispatched from the Coast Guard station.  If the hellicopter's average speed is 90 mph, how long wiil it take the hellicopter to reach the life raft?

since when do life rafts float directly towards coast guard stations???

the helicopter covers 90 miles in one hour, the raft covers 5, so together they cover 95 miles in 1 hour. distance=rate*time: 150 miles=95 mph*x, x=1.578947368... hours or 1 hour 34 minutes and some.

[quote="Interval"]I did not know the question involved D = rt.

Thanks![/quote]
Usually when you see miles, mph, seconds, etc. (and distance, rate or time unit) you should look for ways to apply $d=rt$


Ok, this one is quite easy :

In an international enterprise, there are $70$ people. For any two of them, say $X$ and $Y$, there is a language spoken by $X$ and not by $Y$, and there is a language spoken by $Y$ and not by $X$.
Determine the minimal number of language spoken in this enterprise.

Pierre.

indeed, if the set of languages spoken is $N,|N|=n$, each individual can be labelled by the subset of $N$ .then this collection of sets has to be an antichain.since $70=\binom{8}{4}$ it follows(Sperner's lemma) that the min number of languages spoken has to be atleast $8$.

Yes, that was my point of view too. :)
In another hand, it seems a bit strange that they are expecting such a solution because I'm quite sure that Sperner's lemma is not assumed to be known. ;)

Pierre.

maybe one thing is that it is necessary to note that $70=\binom{8}{4}$ so that the min number of languages is atmost $8$. also it has to be greater than $6$ since $2^6<70$,which leaves only $7$ as an alternate to be ruled out...


hello
this question is really easy and simple

1) What is the least whole number that can be divisible by all the numbers from one through twelve?

have fun

i m goin to say the answer later
i want others to try it
i just post it online for fun
coz my teacher didnt get it
and i was laughing so hard.
by the way which method did u use

so is any1 else going to try it!
and ya can u tell me which method
u used
coz i think there is more than one way
to solve it :D

[hide]it's basicly the LCM of 1-12... and so I can't show it here but it turns out to be:
2*3*2*5*7*2*3*11 = 27720[/hide]

[hide="something o_o"]
1 and 2 = 2
1 to 3 = 6
1 to 4 = 12
1 to 5 = 60
1 to 6 = 60
1 to 7 = 420
1 to 8 = 840
1 to 9 = 2520
1 to 10 = 2520
1 to 11 = 27720
1 to 12 = 27720
[/hide]

Maybe your teacher should get his/her/its credentials checked o_O .

lol ya he should
but anyways every1 got it right!
by the way i dont need to
worry about my teacher
he cant even solve
1+2+3+4+5+6+7+8+9+...............+100
ha ha ha
lol
 :D  :P


algebra
polynomial
Rational Root Theorem

Prove or disprove:  If a monic polynomial $x^{2}+ax+b$ with integer coefficients has one integer root, then it has two integer roots.

[b]Remark:[/b]  There are a few different approaches to this problem, of varying complexity.

[hide="Approach 1"] Complete the square. [/hide]
[hide="Approach 1b"] Conjugate root theorem. [/hide]
[hide="Approach 2"] Rational root theorem and Vieta's. [/hide]
[hide="Approach 2b"] Division algorithm. [/hide]

[hide]
Viete's formula says that for an equation $x^{2}+ax+b$, the sum of the roots is $a$. However, $a$ is an integer (given). If one of the two roots is rational (by rational roots thm.), then the other must be as well (because we are subtracting two integers, we get an integer)
[/hide]

How do we approach double roots?

Are those considered 2 different roots, or only one?

Without qualification, "two integer roots" means "two (not necessarily distinct) integer roots."


I hope i post it in the right forum.

[b]Problem[/b]
Ole Oil produces three products: heating oil, gasoline, dan jet fuel. The average octane levels must be at least $4.5$ for heating oil, $8.5$ for gas, and $7.0$ for jet fuel. To produce these products Ole purchases two types of oil: Crude 1 (at $\$$12 per barrel) and Crude 2 (at $\$$10 per barrel). Each day, at most 10,000 barrels of each type of oil can be purchased.

Before crude can be used to produce products for sale, it must be distilled. Each day, at most $15,000$ barrels of oil can be distilled. It cost 10¢ to distill a barrel of oil. The result of distillation is as follow:

(1) Each barrel of crude 1 yields $0.6$ barrel naphtha , $0.3$ barrel of distilled 1, and $0.1$ barrel of distilled 2.

(2) Each barrel of crude 2 yields $0.4$ barrel of naphtha, $0.2$ barrel of distilled 1, $0.4$ barrel of distilled 2.

Distilled naphtha can be used only to produce gasoline or jet fuel. Distilled oil can be used to produce heating oil or it can be sent through the catalytic cracker (at cost of 15¢ per barrel). Each day, at most $5,000$ barrels of distilled oil can be sent through the cracker.
Each barrel of of distilled 1 sent through the cracker yields $0.8$ barrel of cracked 1 and $0.2$ barrel of cracked 2. Cracked oil can be used to produce gasoline and jet fuel but not to produce heating oil.

The octane level of each type of oil is as follow: naphtha, $8$; distilled 1, $4$; distilled 2, $5$; cracked 1, $9$; craked 2, $6$.

All heating oil produced can be sold at $\$$14 per barrel; all gasoline produced, $\$$18 per barrell and all jet fuel produced $\$$16 per barrel. Marketing considerations dictate that at least $3000$ barrels of each product must be produced daily. Formulate the Linear programming model to maximize Ole's Daily profit.

First question by me, what is ¢? ,  :blush:  what is it relation with US dollar?

What we have as decision variables here?

¢ is cents=.01$\$$.

"Solved problems" isn't usually the right place to post anything, although I'm not sure where this belongs.

This belongs in some "Computations and Tutorials" section. Matrix Algebra is as good a fit as any.

Ok I'm sorry about that. And thanks for reply.
So anybody have know what is decision variable here?
The requirment of "distilled"  seems to be trouble for me.

I define decision variables to be $x_{1h} = \text{barrels of crude } 1 \text{ used to produce heating oil}$
And define the other $x_{2h}$, $x_{1g}$ and $x_{1g}$ for gasoline.
But how to express the use of naphtha? :?

Can someone give a clue about decision variables?
I thinks this problem is not standard.


If $ a^5b^3 \minus{} a^4b^4$, then what would you do? :)

The distributive property is our friend when solving this problem:

$ (3^5)(2^3)\minus{}(2^4)(3^4)$
$ \equal{}3^4((3^1)(2^3)\minus{}2^4)$
$ \equal{}81(2^3)(3\minus{}2)$
$ \equal{}81(8)$
$ \equal{}\fbox{648}$

you can also use binomial theorem factoring :)


prove that for arbitrary coefficient a,b,c and d the sum (cos32x+acos31x+bcos30x+ccos2x+dcosx) cannot take on only positive values for all x.

by the way, can anyone tell me how i can download latex and how i will use it?

thanks in advance.

You don't need to download $\LaTeX$ to use it on the forums.  You just need to put dollar signs around your math, and then learn a few things.  A large portion of the website is devoted to it; click the relevant link on the left.

[hide]
assume for contradiction that there exist $a,b,c,d$ such that the sum $S$ is always positive.
well, setting $x = \frac{\pi}{2}$ gives $S = 1-(b+c)$, hence $b+c < 1$
setting $x = \frac{\pi}{4}$ gives $S = 1+(a+d)$, hence $a+d >-1$
setting $x = \frac{3\pi}{4}$ gives $S = 1-(a+d)$, hence $a+d < 1$

this isn't getting anywhere fast.  dunno.  maybe use Chebyshev polynomials.[/hide]


quadratics
linear algebra
matrix
linear algebra unsolved

Suppose that ax2 + 2bxy + cy2 is a quadratic form with associated symmetric matrix
A. Let j1+ j2 denote the eigenvalues of A. Prove that j1+j2= a+c!
Please help!

We have $A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}$, so the characteristic equation is $(a - \lambda)(c - \lambda) - b^2 = 0$, or $\lambda^2 - (a + c) \lambda + (ac - b^2) = 0$. The eigenvalues of $A$ are the roots of this equation, and hence the sum of the eigenvalues (sum of the roots) is $a + c$.

thanks, I really appreciate your help :)

Please don't start new problems out in the "solved problems" section. Put them here, or in the "computations and tutorials" forum, depending on the level.

should I  factor this or just leave it as it is?

factor  \lambda^2 - (a + c) \lambda + (ac - b^2) = 0

Why would you do that? It's clear that the sum of the roots is $a + c$.


inequalities
induction
Russia
2004
cyclic inequality
n-variable inequality

Let $n > 3$ be a natural number, and let $x_1$, $x_2$, ..., $x_n$ be $n$ positive real numbers whose product is $1$. Prove the inequality \[ \frac {1}{1 + x_1 + x_1\cdot x_2} + \frac {1}{1 + x_2 + x_2\cdot x_3} + ... + \frac {1}{1 + x_n + x_n\cdot x_1} > 1. \]

Let $f(x_1,x_2,...,x_n)$ be our expression. It's easy to see that for $n=3$, the expression is exactly equal to 1, because it is $\dsp \frac {1}{1+x_2+x_2x_3}(\frac {1}{x_1}+1+x_2)=1$.
Now, prove it by induction over $n$. We will always find one of the $x_i$'s $\geq$ than 1 such that $x_i \cdot x_{i+1} \geq 1$. If there weren't such an $x_i$, then the product of the numbers would obviously be < 1.
So if $x_i>1$ and $x_i \cdot x_{i+1} \geq 1$, it's easy to see that $f(x_1,x_2,...,x_n)>f(x_1,...,x_{i-2},x_{i-1}\cdot x_{i},x_{i+1},...,x_n)$, because

\[\dsp \frac {1}{1+x_{i-2}+x_{i-2}x_{i-1}}+\frac {1}{1+x_{i-1}+x_{i-1}x_i}+\frac {1}{1+x_i+x_ix_{i+1}}>\frac {1}{1+x_{i-2}+x_{i-2}x_{i-1}x_i}+\frac {1}{1+x_{i-1}x_i+x_{i-1}x_ix_{i+1}}\]

This is true because the first two terms in LHS are greater than the first two terms on RHS. So our induction is complete.

Very nice solution Andrei. But the most beautiful solution was that of the problem 8 from the TST. Absolutely incredible!

What are you talking about Harazi?

Pierre.

He is talking about the perfect solution that Andrei Negut gave on this problem:
http://www.mathlinks.ro/viewtopic.php?t=5486

Maybe Andrei can post it for us.

[quote="andrei negut"]Let $ f(x_1,x_2,...,x_n)$ be our expression. It's easy to see that for $ n = 3$, the expression is exactly equal to 1, because it is $ \dsp \frac {1}{1 + x_2 + x_2x_3}(\frac {1}{x_1} + 1 + x_2) = 1$.
Now, prove it by induction over $ n$. We will always find one of the $ x_i$'s $ \geq$ than 1 such that $ x_i \cdot x_{i + 1} \geq 1$. If there weren't such an $ x_i$, then the product of the numbers would obviously be < 1.
So if $ x_i > 1$ and $ x_i \cdot x_{i + 1} \geq 1$, it's easy to see that $ f(x_1,x_2,...,x_n) > f(x_1,...,x_{i - 2},x_{i - 1}\cdot x_{i},x_{i + 1},...,x_n)$, because
\[ \dsp \frac {1}{1 + x_{i - 2} + x_{i - 2}x_{i - 1}} + \frac {1}{1 + x_{i - 1} + x_{i - 1}x_i} + \frac {1}{1 + x_i + x_ix_{i + 1}} > \frac {1}{1 + x_{i - 2} + x_{i - 2}x_{i - 1}x_i} + \frac {1}{1 + x_{i - 1}x_i + x_{i - 1}x_ix_{i + 1}}\]
This is true because the first two terms in LHS are greater than the first two terms on RHS. So our induction is complete.[/quote]

So how did you with $ f(x_1,x_2,...,x_n) > f(x_1,...,x_{i - 2},x_{i - 1}\cdot x_{i},x_{i + 1},...,x_n)$ prove the first inequality $ f(x_1,x_2,...,x_n) > 1$ ?

[b]Edit: This is a completely wrong proof! I claimed a rather embarrassingly absurd statement (observed in the post below.) Please disregard this post. My apologies. :blush:[/b]

[hide]It can in fact be shown that the left-hand side is always larger than $\frac{n}{4}$. We have

\[ \frac{1}{1+ x_1 +x_1x_2} + \frac{1}{1 + x_2 + x_2 x_3}+ \cdots + \frac{1}{1 + x_n + x_nx_1} \\
\&> \frac{1}{1 + x_1 + x_2 + x_1 x_2} + \frac{1}{1 + x_2 + x_3 + x_2 x_3} + \cdots + \frac{1}{1 + x_n + x_1 + x_n x_1} \\
\&= \frac{1}{(1+x_1)(1+x_2)} + \frac{1}{(1+x_2)(1+x_3)} + \cdots + \frac{1}{(1+x_n)(1+x_1)}. \]

Let $y_i = \frac{1}{1+x_i}$. We have $0 < y_i < 1$ and $\frac{1}{y_i} - 1 = x_i$, so $\prod_{i=1}^n \left(\frac{1}{y_i} - 1\right) = 1$, so $\prod_{i=1}^n (1 - y_i) = \prod_{i=1}^n y_i$. Hence, $\prod_{i=1}^n y_i^2 = \prod_{i=1}^n y_i (1 - y_i) \geq \frac{1}{4^n}$ by AM-GM, so $\prod_{i=1}^n y_i \geq \frac{1}{2^n}$. Therefore, by another application of AM-GM, we have
\begin{align*}
\frac{1}{(1+x_1)(1+x_2)} + \frac{1}{(1+x_2)(1+x_3)} + \cdots + \frac{1}{(1+x_n)(1+x_1)}
&= y_1 y_2 + y_2 y_3 + \cdots + y_n y_1 \\
&\geq n (y_1 y_2 \ldots y_n)^{\frac{2}{n}} \\
&\geq \frac{n}{4}.
\end{align*}[/hide]

Sorry, I am very new to inequalities but could you explain to me why:
$Pi[(y_i)(1-y_i)]>=1/4^n$

An excellent question! In retrospect, I wonder how I concluded such an obviously false statement myself. :blush:

Despite the egregious error in my post, I still wonder whether the inequality when 1 is replaced with $\frac{n}{4}$ holds; the inequalities I used in my incorrect proof were all fairly weak.

EDIT: The strengthened inequality $\frac{1}{(1+x_1)(1+x_2)} + \frac{1}{(1+x_2)(1+x_3)} + \ldots + \frac{1}{(1+x_n)(1+x_1)} \geq \frac{n}{4}$ does not hold. For even $n$, let $x_{2i} = \epsilon$ and $x_{2i+1} = \frac{1}{\epsilon}$; for positive $\epsilon$ sufficiently small, each term in the sum will be very close to 0.

I first wasn't able to do this because my substitution was kind of weird. Needed to take hints from the handout I was doing. Now my solution is the same as that of harazi, but I'll post it anyways
Substitute $x_1=\frac{a_2}{a_2}$, $x_2=\frac{a_3}{a_2}$ ... $x_n=\frac{a_1}{a_n}$
Now the given inequality is equivalent to,
$\sum_{i=1}^{n}{\frac{a_i}{a_i+a_{i+1}+a_{i+2}}} > 1$ which is true as $a_i+ax_{i+1}+a_{i+2} < \sum_{i=1}^n{a_i}$.


function
calculus
derivative
calculus computations

Prove rigorously or give a concrete counterexample:Let $f$ be a function with a vertical tangent at $x=a$, then $x=a$ is an inflection point of $f$.

[quote="leibniz"]Is $f$ a differentiable function in $a$?[/quote]
When we speak of a tangent line at a point, we mean that the derivative at that point exists. Note that the converse is not true when infinite derivatives are to be considered, eg check the sign function at the pole.

No idea guys? Let me put it this way; Supposing $f'(a)=\infty$ (ie $f'_{+}(a)=f'_{-}(a)=\infty )$, can we say that $f''$ changes its sign, when passing $x_{0}$?


Given that
\[ \frac{((3!)!)!}{3!} = k \cdot n!, \]
where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k + n$.

[quote=ATGY][hide=Easy Solution]

Note that $1 < 3! < 720$ and:
$$\implies \frac{720!}{3!} = 720\cdot719\cdot718\cdot717...\cdot4$$

Clearly $n = 719$ and $k = 120$. Hence our answer is $719 + 120 = \boxed{839}$.
$\blacksquare$[/quote]

It was pretty much the same as the previous solutions but more detailed

[quote=asbodke][url]https://artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_1[/url][/quote]

AIME problems were so easy back then...

[quote=Possible][quote=AwesomeLife_Math]Well it's an AIME #1. AIME I 2020 #1 was easier than this, in my opinion.[/quote]

IMO, AIME I 2020 #1 was harder than this[/quote]

Well I got #1 wrong because I couldn't read and put down the wrong angle.
I guess they're both really easy for AIME.

yeah i agree! but the easier the better on the hard contests :jump:

amctrivial grind

This is equal to \[\frac{720!}{3!}=\frac{720!}{6}=\frac{719!\cdot 720}{6}=719!\cdot 120\implies \boxed{839}.\]

Obviously $n<720$, because the $\text{LHS }<720!$.

p8: This simplifies to $\frac{720!}{3!}$. It is obvious that $n = 720$ is not possible, but $n = 719$ indeed works, giving $k = \frac{720}{6} = 120 \implies 719 + 120 = \boxed{839}$.


Prove that in any triangle we have the following ineguality :\[ p \geq \sqrt{3r(5R-r)}\]

Actually the stronger inequality holds: $ p^{2}\geq 16Rr-5r^{2}$. This is called the Gerretsen inequality, and you can see it here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=28098[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=15462[/url].

True :D.Without having to know Gerretsen the ineg is equivalent to $ \sum (b+c-2a)^{2}(-a+b+c)\geq 0$


Here's a cryptarhythm problem that I don't know how to solve without writing a program or without doing extensive trials:

$ TROIS$ is a triangular number
$ QUATRE$ is a perfect square
$ SEPT$ cannot be expressed as the sum of three squares.

Find values of $ A, E, I, O, P, Q, R, S, T, U$ which clearly range from $ 0$ to $ 9$, all distinct.

PS: I read somewhere that a number that cannot be expressed as the sum of three squares has to be of the form $ 8n+7$. Please check this out. It could be useful for solving the problem.

[quote="xxxyyyy"]PS: I read somewhere that a number that cannot be expressed as the sum of three squares has to be of the form $ 8n+7$. Please check this out. It could be useful for solving the problem.[/quote]

That's correct, as $ k^{2}\equiv0,1,4\mod8$.

[hide]
Taking all numbers mod 8 and noting that since $ 8k+7$ is odd, $ SEPT$ cannot have $ 0, 2, 4, 6$, or $ 8$ in it. From there it's easy to see that $ SEPT$ is $ (1, 3, 5, 9)$.

Note that if $ R = 0$, we have an instant solution, as both $ TROIS$ and $ QUATRE$ are both equal to $ 0$. There is another solution, though.

$ QUATRE$ has only one solution if $ 0$ is not allowed in it: $ (1, 2, 3, 4, 6, 9)$. However, $ QUATRE$ and $ SEPT$ only share two letters ($ T$ and $ E$), yet $ (1, 3, 5, 9)$ and $ (1, 2, 3, 4, 6, 9)$ have three numbers in common, so $ QUATRE$ cannot be $ (1, 2, 3, 4, 6, 9)$ and must contain a 0 in either $ Q$, $ U$, or $ A$ (assuming $ R$ is not zero). Knowing that the the other letters can be anything, we first solve $ TROIS$.

$ TROIS$ shares two letters with $ SEPT$: $ T$ and $ S$. The other three letters, $ R$, $ O$, and $ I$, must be either $ 2, 4, 6, 7$, or $ 8$. We are looking for a pair of numbers $ n, n+1$ such that $ \frac{n(n+1)}{2}$ can be represented as the product of $ T$, $ R$, $ O$, $ I$, and $ S$. First, we note that the product of $ R$, $ O$, and $ I$ is divisible by 4, so one of the number in our pair must be divisible by 8. We start guessing from the bottom:
$ 7, 8$ and $ 8, 9$ are both too small.
$ 15, 16$ is too small.
$ 24$'s neighbors are $ 23$ and $ 25$, neither of which work.
Same with $ 32$, $ 40$, and $ 48$.
$ 63, 64$ works! $ (8, 4, 7), (9, 1)$ multiply to $ 32 \cdot 63$.

Now that we know where all of the numbers go, we just have to match them up to the letters:
$ S, T = 1, 9$
$ E, P = 3, 5$
$ R, O, I = 4, 7, 8$
$ A, Q, U = 0, 2, 6$
[/hide]

[quote="Ubemaya"][hide]
Taking all numbers mod 8 and noting that since $ 8k\plus{}7$ is odd, $ SEPT$ cannot have $ 0, 2, 4, 6$, or $ 8$ in it. From there it's easy to see that $ SEPT$ is $ (1, 3, 5, 9)$.

Note that if $ R \equal{} 0$, we have an instant solution, as both $ TROIS$ and $ QUATRE$ are both equal to $ 0$. There is another solution, though.

$ QUATRE$ has only one solution if $ 0$ is not allowed in it: $ (1, 2, 3, 4, 6, 9)$. However, $ QUATRE$ and $ SEPT$ only share two letters ($ T$ and $ E$), yet $ (1, 3, 5, 9)$ and $ (1, 2, 3, 4, 6, 9)$ have three numbers in common, so $ QUATRE$ cannot be $ (1, 2, 3, 4, 6, 9)$ and must contain a 0 in either $ Q$, $ U$, or $ A$ (assuming $ R$ is not zero). Knowing that the the other letters can be anything, we first solve $ TROIS$.

$ TROIS$ shares two letters with $ SEPT$: $ T$ and $ S$. The other three letters, $ R$, $ O$, and $ I$, must be either $ 2, 4, 6, 7$, or $ 8$. We are looking for a pair of numbers $ n, n\plus{}1$ such that $ \frac{n(n\plus{}1)}{2}$ can be represented as the product of $ T$, $ R$, $ O$, $ I$, and $ S$. First, we note that the product of $ R$, $ O$, and $ I$ is divisible by 4, so one of the number in our pair must be divisible by 8. We start guessing from the bottom:
$ 7, 8$ and $ 8, 9$ are both too small.
$ 15, 16$ is too small.
$ 24$'s neighbors are $ 23$ and $ 25$, neither of which work.
Same with $ 32$, $ 40$, and $ 48$.
$ 63, 64$ works! $ (8, 4, 7), (9, 1)$ multiply to $ 32\cdot 63$.

Now that we know where all of the numbers go, we just have to match them up to the letters:
$ S, T \equal{} 1, 9$
$ E, P \equal{} 3, 5$
$ R, O, I \equal{} 4, 7, 8$
$ A, Q, U \equal{} 0, 2, 6$
[/hide][/quote]

please check...
[hide]Would u like to check your solution again for we have $ E \equal{} 3$ and $ E$ must be $ 1,4,5,6,9$ for $ QUATRE$ is a square number.[/hide]

But $ Q$, $ U$, or $ A$ is $ 0$, making $ SQUARE$ $ 0$, which is a square number.

Teapot uhh I think by $ TROIS$ he means $ 10000T\plus{}1000R\plus{}100O\plus{}10I\plus{}S$, not $ T\cdot R\cdot O\cdot I\cdot S$.

Yeah... I kinda realized that afterwards. But I don't really think that a problem would make you hunt for 6 digit squares and 5 digit triangular numbers.

[quote="diophantient"][quote="Lazarus"][quote="xxxyyyy"]PS: I read somewhere that a number that cannot be expressed as the sum of three squares has to be of the form $ 8n\plus{}7$. Please check this out. It could be useful for solving the problem.[/quote]

That's correct, as $ k^{2}\equiv0,1,4\mod8$.[/quote]
That's false, the converse of that is true.
How is 22 the sum of three squares?
just because a number is six mod 8 doesn't [b]guarantee[/b] that it is the sum of three squares. A number that is 7 mod 8 is guaranteed not to be the sum of three squares is all we know.[/quote]

$ 22 \equal{} 4\plus{}9\plus{}9$

Alright uhh I meant to prove that all numbers of the form $ 8k\plus{}7$ are prime. Not the converse. No clue why I incorporated it into my bash though. :\

The correct condition is a number cannot be the sum of three squares if and only if it is of the form $ 4^{n}(8k\plus{}7)$ for integers $ n,k$. For example, $ 28$ cannot be written as the sum of three squares, but $ 56 \equal{} 36\plus{}16\plus{}4$


calculus derivative function inequalities trigonometry

Let the function  $f(x)=x\cdot sin\frac{x}{2}$ where x is an angle of a triangle

Solve the inequality  $f''(x)\geq 0$

Appending.

cos(x) * (x^2+16) will always be [u]positive[/u] because both terms are positive in the given interval.

(4-x)(4+x) will be [u]positive[/u] for the same reason.

Therefore, the whole expression is [u]strictly greater[/u] than 0.

[hide="Graphing approach"]Sketch the graph on the interval $(0, \pi)$ and look where it's concave up.[/hide]

[b]Siluan[/b], maybe $f(x)=\sin x\sin\frac x2$ (which is neither convex and nor  concav on $(0,\pi )$), because $a\sin \frac A2=2R\cdot \boxed {\sin A\sin \frac A2}\ !!!!!!!!!!!!!!!$
See the inequality $\sum a\sin \frac A2\ge p$ ([b]Moldova, 2006[/b]).


trigonometry
calculus
integration
calculus computations

Find the integral [b]without Weierstrass substitution[/b]

$ \int \frac{dx}{\sqrt{\cos x \sin^3 x}}$

$ \int \frac{dx}{\sqrt{\cos x\sin ^{3}x}}\equal{}\int \frac{d(\tan x)}{\sqrt{\tan ^{3}x}}\equal{}\minus{}\frac{2}{\sqrt{\tan x}}\plus{}C$

[quote="IndoChina"]$ \int \frac {dx}{\sqrt {\cos x\sin ^{3}x}} \equal{} \int \frac {d(\tan x)}{\sqrt {\tan ^{3}x}} \equal{} \minus{} \frac {2}{\sqrt {\tan x}} \plus{} C$[/quote]

Yes, excellent)


How many different phone numbers are that have exactly two different digits in them, keeping in mind that each phone number has 7 digits, and the first number can't be 0 or 1. For instance, 223-2332 will work, but 112-1221 and 222-2222 won't.

[hide]$28(2^{7}-2)+16(2^{6}-1)=4536$
there are 28 that involve only 2-9, you can have 7C1, 7C1...7C6 of themthere are 16 that involve 0 or 1 (not both), and you can have 6C1..6C6 of them (the first digit is a non 0 or 1 so you can have 6 0s or 1s for the rest of the spots)..add them up.[/hide]


Find all pairs $(a, b)$ of positive integers such that $2a - 1$ and $2b+1$ are
coprime and $a+b$ divides $4ab+1$.


[quote="Carcul"]A car A is at rest on a road. In the same road a car B travels at the constant speed of 80 km/h. In the exact moment that car B passes by car A, car A starts moving with constant acceleration, going from 0 km/h to 100 km/h in 10 seconds, and maintaining that speed of 100 km/h from thereafter.

(a) What is the speed of car A when it reaches car B?

(b) At what instant does that happens, and what is the distance traveled by both cars?[/quote]

Given for constant acceleration, $ s\equal{}ut\plus{}\frac{1}{2}at^2$, lets assume that they meet during the period of constant acceleration.

$ s_b\equal{}80t$
$ s_a\equal{}\frac{1}{2}10t^2$

Hence when they are level

$ s_a\equal{}s_b$
$ 80t\equal{}\frac{1}{2}10t^2$ Ignoring $ t\equal{}0$ when B reaches A
$ t\equal{}16$

Since $ 16>10$ this is after acceleration is complete. Hence the speed of the two cars when the pass is $ A\equal{}100, B\equal{}80$

Now consider the distance travelled for a time after 10s.

$ s_b\equal{}80t$
$ s_a\equal{}500\plus{}100(t\minus{}10)\equal{}100t\minus{}500$

Hence
$ s_a\equal{}s_b$
$ 80t\equal{}100t\minus{}500$
$ 500\equal{}20t$
$ t\equal{}25$

Hence they have travelled for 25 seconds

[quote="Carcul"]Yes, that's correct.[/quote]

I didn't do that last part of the question for the sheer reason that I couldn't be bothered to convert km/h into ms$ ^{\minus{}1}$.

So here goes....

80 km/h is $ \frac{200}{9}$ m/s

Since $ s\equal{}vt$, $ s\equal{}\frac{200}{9} \cdot 25 \equal{} \frac{5000}{9} \approx 555.6 \approx$ half a kilometre


I've just remember a nice inequality from KVANT

Let a, b, c > 0. Show that

(a+b)^c + (b+c)^a + (c+a)^b > 2

Namdung

Very nice, indeed. It remembers me about another memorable inequality: a^b+b^a>1. The technique is the same. If for example a>1, then (a+b)^c>1 and (a+c)^b>1 and we are home. Otherwise, let a,b,c<1. Then we have (a+b)^c>(a+b)/(a+b+c) and the others inequalities. Sum up and you will find exactly the inequality proposed. Very very nice problem.


Hi,

I got a question on the term "convex" when it came to the function. I found this definition in the mathworld:

http://mathworld.wolfram.com/ConvexFunction.html

But I have no clue on what that means. I'm trying to learn some advanced inequalities and develop my algebra skills but I don't seem to understand this term.

Thanks for the help!

[quote="shobber"]One definition of convex is: Take any two points on the curve and construct a segment with the two points. The function is convex if and only if the segment is above the curve.

There is a definition of the convex function in the inequality marathon.[/quote]

Thanks! I never really understood that before either.


Set $ S \equal{} \{a_1,a_2, ..., a_n\}$ that all $ 1 \le i \le n, a_i \in N$ satisfy $ \sum_{a \in A}a \not \equal{} \sum_{b \in B}b$ for two subset $ A, B$ of arbitrary subset $ X$ of $ S$.
prove that $ \sum_{i \equal{} 1}^n a_{i}^2 \ge \frac {4^n \minus{} 1}{3}$ and $ \sum_{i \equal{} 1}^n \frac {1}{a_i} \prec 2$

One solution is like this.
$ WLOG$
$ a_{n} \prec a_{n\minus{}1} \prec ... \prec a_{1}$
Easily we know that

$ a_{1} \ge 2^{0}\equal{}1$, $ a_{1}\plus{}a_{2} \ge 2^{0}\plus{}2^{1}, ..., a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n} \ge 2^{0}\plus{}2^{1}\plus{}...\plus{}2^{n\minus{}1}$
And there exist sufficiently large number M, N such that
$ M\plus{}\sum_{i\equal{}1}^{n}a_{i}\equal{}N\plus{}\sum_{i\equal{}1}^{n}2^{i\minus{}1}$
and then above inequality, $ M \succ N$
Then, By Karamata Inequality (it is known as Hardy-Littlewood-Polya inequality, too)

We can get conclusion.

Anyone can suggest other solutions?

I think order of $ a_{i}$ is reversed;

Sorry


i have a question, plz help me out to solve this question

Question is

I           I
I  3-2x   I
I ------  I    <   4
I  2+x    I     -
I           I

find solution set and draw graph.

need to solve it with the help of inequalities

is it solve $\left | \frac{3-2x}{2+x}\right |\le 4$? Learn how to write with LaTex. If this is the problem put it in Intermidiate.


[code]i see good website: http://www.yourglobaltv.com. Reply favorites channel pls[/code]

[code]i see good banning, if this is ad.[/code]


Prove that

$ \sin 2A \plus{} \sin 2B \plus{} \sin 2C \equal{} 4\sin A \sin B \sin C$

[hide="solution"]It should hold $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi$.

$ \alpha\equal{}(\pi\minus{}(\beta\plus{}\gamma))$

$ \implies \sin \alpha\equal{}\sin (\pi\minus{}(\beta\plus{}\gamma))\equal{}\sin \pi \cdot \cos(\beta\plus{}\gamma)\minus{}\cos \pi \cdot \sin (\beta\plus{}\gamma)\equal{}\sin (\beta\plus{}\gamma)$

$ \implies \cos  \alpha\equal{}\cos (\pi\minus{}(\beta\plus{}\gamma))\equal{}\minus{}\cos(\gamma\plus{}\beta))$

Now by the sum-to-product formulas we obtain:

\[ \sin 2\alpha \plus{} 2 \sin \frac{2(\gamma\plus{}\beta)}{2}\cdot \cos(\frac{2(\beta\minus{}\gamma)}{2}\]

\[ 2\sin \alpha \cdot \cos \alpha \plus{} 2 \sin (\gamma\plus{}\beta) \cdot \cos (\beta\minus{}\gamma)\]

Now we use $ \sin (\gamma\plus{}\beta)\equal{}\sin\alpha$ to obtain:

\[ 2\sin \alpha (\cos \alpha \plus{}\cos (\beta\minus{}\gamma))\]

Now we use $ \cos  \alpha\equal{}\minus{}\cos(\gamma\plus{}\beta)$

\[ 2 \sin \alpha ( \cos (\beta\minus{}\gamma)\minus{}\cos(\gamma\plus{}\beta))\]

Since $ \cos(\beta\minus{}\gamma)\minus{}\cos(\gamma\plus{}\beta) \equal{} 2\sin \beta \cdot \sin \gamma$ the result follows.
[/hide]


logarithms
group theory
abstract algebra
number theory proposed
number theory

Here is a very very nice problem: the largest prime factor of $2^{2^n}+1$ is greater than $2^{n+2}\cdot (n+1)$. Probably this will give an insight for my problem: if $a>1$ then for infinitely many $n$ the largest prime factor of $a^n-1$ is greater than $ n\log_a {n}$.

Let's look at any prime divisor of $2^{2^n}+1$. Let $p$ be such a prime divisor, so $p|2^{2^n}+1$ (1).
Let $u$ be the positive integer in subgroup group $<2>$ in $Z_p$ such that $2^u \equiv -1 (\mod p)$
 - we know that such exists from (1). Because for all odd multiples (and only for them) $t$ of $u$
 we have $2^t \equiv -1 (\mod p)$, thus $u|2^n$ from (1). But since $t$ are odd multiples thus the only possibility is $u=2^n$. Because for $Z_p$ $2^{p-1} \equiv 1 (\mod p)$ and also $2^{2u} \equiv 2^{2^{n+1}} \equiv -1 (\mod p)$ thus ${2^{n+1}}|p-1$ so $p=2^{n+1}m+1$ where $m$ is a positive integer. Now let's put $2^{2^n}+1=p_1^{k_1}p_2^{k_2}...p_i^{k_i}$ where $p_j$ are odd primes. We can prove that
$k_1+k_2+..+k_i < \frac{2^n}{n+1}$. Let's note that if $p_j=2^{n+1}m_j+1$ then $2^{2^n}+1=p_1^{k_1}p_2^{k_2}...p_i^{k_i}=$ $(2^{n+1}m_1+1)^{k_1}(2^{n+1}m_2+1)^{k_2}
...(2^{n+1}m_i+1)^{k_i}$ $ > 2^{(n+1)(k_1+k_2+...+k_i)}$
thus $(n+1)(k_1+k_2+...+k_i) < 2^n$ so indeed $k_1+k_2+..+k_i< \frac{2^n}{n+1}$.
Now let's also note that $(2^{n+1}m_1+1)^{k_1}(2^{n+1}m_2+1)^{k_2}...(2^{n+1}m_i+1)^{k_i}=$ $2^{2(n+2)}A+2^{n+1}(k_1m_1+k_2m_2+...+k_im_i)+1$
where $A$ is a not important :) positive integer.
So we have $2^{2(n+2)}A+2^{n+1}(k_1m_1+k_2m_2+...+k_im_i)=2^{2^n}$ so obviously
$2^{n+1}|k_1m_1+k_2m_2+...+k_im_i$ but $k_1m_1+k_2m_2+...+k_im_i \leq$ $ max\{m_j : 1\leq j \leq i\}(k_1+k_2+..+k_i)$ $< max\{m_j : 1\leq j \leq i\}\frac{2^n}{n+1}$
so $2^{n+1} < max\{m_j : 1\leq j \leq i\}\frac{2^n}{n+1}$ thus $2(n+1) < max\{m_j : 1\leq j \leq i\}$ and that means that if $max\{m_j : 1\leq j \leq i\}=m_k$ then $p_k=2^{n+1}m_k+1 >$ $2^{n+2}(n+1)+1>2^{n+2}(n+1)$ so obviously $p_k$ is the prime we are looking for.

But somehow harazi I don't see any generalization for this in the $a^n-1$ problem.

Intersting note: Landry has proved in $1880$ that $2^{64}+1$ is not a prime number ($2^{64}+1=67280421310721*274177$ when he was 82 years old (!).

Very very nice, that's exactly the solution I also found. I will let you try my problem before giving a hint. A super-smart man told me my problem is too difficult for a TST and I start to give him reason.

Dear harazi, your problem is indeed very hard.
I would like to consider the case where a is even.
Then by considering numbers of the form $a^{2 \cdot a^n} -1$ and the largest prime divisor of $a^{a^n} +1$ we are done. The proof of this fact is a "copy_past" of Rafal's proof.
So it remains to solve it for the case where a is odd.

harazi's solution:

[quote="harazi"]... Ok, here is my solution in case $a$ is not a power of 2 (otherwise it is the chinese problem). Take $p$ an odd prime factor of $a$ and write $ 1+a^{p^{n-1}}+...+a^{(p-1)p^{n-1}}= q_1^{k_1}...q_s^{k_s}$. Then clearly none of $q_i$ is 2 and it is really known that each one of them is $1 (mod p^n)$. Let $q_i=1+p^n r_i$ and reduce the last relation mod $p^{2n}$. We find $\sum{k_i r_i}$ multiple of $p^n$ thus surely it is at least $p^n$. Since $ a^{p^n}> 1+a^{p^{n-1}}+...+a^{(p-1)p^{n-1}}>p^{n(k_1+...+k_s)}$ we can easily conclude that $ max(r_i)> n log_p a$ and thus $max q_i> p^n\cdot n log_a p$ and since all $q_i$ divide $a^{p^n}-1$ it's done.
   [/quote]

I think this problem be true if $n\geq 3$
And I see
If $n\geq 4$ then $2^{2^{n}}+1$ has the largest prime factor be greater than $2^{n+4}(n+2)$.

Rust!Sorry for i don't understand you write what.Is it solution to my problem? :(

He was trying to solve harazi's problem,I think  :wink: .Anyway ,your problem could solved exactly in the same way with a broaden result stating that: All such numbers have prime factors in the form $ 2^{n \plus{} 2}k \plus{} 1$ instead of $ 2^{n \plus{} 1}k \plus{} 1$.
   I have the same solution as harazi's to his own problem!  :lol:


Is it possible to solve $ r\sin{\theta} \equal{} e^{r\cos{\theta}}$ for $ r$? Is so, what is the expression for $ r$?

That would be $ r \equal{} \minus{} \frac {W\left( \minus{} \cot \theta\right)}{\cos\theta}$, where $ W(x)$ is the [url=http://en.wikipedia.org/wiki/Lambert_W_function]Lambert[/url] [url=http://mathworld.wolfram.com/LambertW-Function.html]W-Function[/url], or the function defined to be the inverse of $ f(x) \equal{} xe^{x}$.

you guys are WAY too smart... :arrow:  :?:

The smiley faces represent what i think of this thread


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

An ideal $ I$ of $ A$ is radical iff $ r(I)\equal{}I$.

Show that if $ A$ is Noetherian, then all radical ideals of $ A$ can be expressed as an intersection of finitely many prime ideals. Deduce that if $ A$ is Noetherian, then $ A$ has finitely many minimal prime ideals.

Hint: a quotient of a Noetherian ring is also Noetherian, so it suffices to show that $ \sqrt{(0)}$ is an intersection of finitely many primes.


Ok so I'm going to make a Math Club next year so I can do some math competitions...including this AMC 12. So, to anyone who was ever in the same situation, what do I do? Do I just tell a math teacher I know, tell him'her to procter and register for the exam with me or something?

I'd ask your math teacher and/or the head of the math department, tell them that you're interested in doing this competition, and that this competition is well known and taken by other students all around the country, but needs the school to register, and someone to proctor and administer it. There are a few people here that end up having to pay the fee themselves, but usually that isn't a problem and the school pays. (The fee is like about one AP test or something...).

Then try to get some friends who might be interested to take the test, if only to fill up the extra tests.

I was less than successful starting a math club at my school, so I can't help you there.

I was in same situation. Basically, I started whole AMC program in my school since my freshman year and I think due to my work, my rural high school'll continue to participate (although I'm not sure if there'll be any more AIME qualifier).  :)

Like pianoforte says, you need to talk with a teacher. For me, I talked with my gifted teacher, and because I had a premonition that he was delaying things, I actually went straight to talk with principal himself (I know him too), and also one of my math teachers. They soon registered for it after hearing the school board's approval. So I suggest that you talk to them early, possibly during the summer, since most teachers are in happier mood during the summer (why? They don't have to deal with annoying kids).

As for getting people, I suggest that you don't worry about math club just yet. Is there anything like a gifted club in your school? Or any special club for exceptional kids? I say to talk with that advisor and see if he/she can proctor you. Those people are usually more lenient and understanding because they want to do many projects to satisfy parents.

I wish you luck. AMC is tons of fun anyway.  :)

Just as a warning. Sometimes school administration is not willing to fork over any money or resources (like my school) So decide whether or not you'd be willing to pay for it yourself and also for a proctor, it doesnt have to be a math teacher. Just find any teacher that stays late after school and have them watch you as they do their work.

I was at about the same place you're describing a few months ago, in terms of ability to do AMC problems, but now I feel very confident that I can qualify for AIME this year. The difference is that I've spent a lot of time this summer on problem solving. I'm currently taking the AMC 12 Problem Series class here, and I definitely recommend it. If that's not an option, just keep doing problems, and if you're persistent, your skills will naturally improve.

If you succeed in starting a math club at your school, it's great, but even if you don't, don't give up on taking the AMC. I'm co-president of math club at my school, which has almost no "math culture," but even without the club, math teachers would continue the school's participation. The more people you can convince to take the AMC 12, the better, but don't worry if it's not enough to form a club. Also, you might find it easier to start a math club if you advertise it as "a fun place to discuss math" or something like that, instead of being very serious about competitions. (That's how we keep the club going at my school, but it probably depends on the school.)

Thanks mechanicalpencil. I'll try to do that. Hey, what do you learn in these AMC 12 classes? I'm currently studying out of the AoPS book and doing practice tests, but I don't want to waste more money on a class that might not raise my score above 100.

[quote="mathemonster"]pianoforte btw looking at ur sig wtf didnt u take the amc10 so u could make usamo?[/quote]
No.  :(

Bah, it doesn't matter anymore this year.

[quote="stanford1463"]Thanks guys for the info! Unfortunately, my school doesn't have any gifted/talented groups, so I am going to have to start a Math Club and be President of it :P. Alright, so I'll talk to a math teacher i know and convince her to proctor the AMC 12 test. Thanks! Now...the real problem is how should I attempt to pass the AMC 12...Lol, because I bought the AoPS Volume I and am currently on Chapter 11..but I suck at thinking critically like a mathematician if you know what I mean. [b]So my last question is- how should I train my brain to think like an AMC'er and qualify to AIME? (since that is my only goal-to get above 100 and qualify for AIME).[/b] I can do usually questions 1-10, but after that I just can't think of what steps to take and what to do next, so I get lost :(. I usually get about 13-14 questions right, but then I just can't do the other 11-12... because I don't know where to start or even what they are asking for.[/quote]

Like others said, don't worry about starting a math club, I was less than successful starting one at my own school.

It's cliche, but you just have to practice. Keep doing problems, don't give up too early, understand the solutions on the ones you do give up on, and plan to go back to them. Most of all, have fun, you and you alone is the only rival that matters, strive to improve your abilities.

its ok pianoforte-- i bet youll make it this year!

About the class: I like it because when we do problems in the class, we basically walk through the problem solving process, which is very helpful for learning to problem solve. Now, even when I don't know how to do a problem, I can usually at least find somewhere to start, which is a HUGE improvement. Also, all of the problems are in the harder range, and it's good to have the practice with subjects that aren't covered a lot in school. The books help with some of this, but I think the class is a better way to develop problem solving intuition, especially if you still aren't confident that you can score above 100.

Ok cool..I'm still debating whether I should take it because I am about borderline 100..I get about 14 questions right and omit the rest, which is about 100.5...but sometimes get lower. I will probably just do a lot of practice questions from the AoPS book and finish the entire book, like doing a lot of practice problems like pianoforte said. Thanks..any tips on problem solving techniques would be helpful :)

I don't seem to find a thread on AMC 12A topics, so does anyone know what kind of math is on it? would I have to know base-2, base-10, configurations etc. ?

Frequently Asked Question AMC 10/12 #2 at:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions