I got my 3 USAMTS prizes just today. I was wondering, do we get a small medal along with it? I remember last year when I won silver, i got a small silver medal.

[quote="simon_math"]For those of you who selected Mathematica, do you receive it in the mail? I selected it but I don't have it in the package.[/quote]
I believe it's being sent directly to those who selected it by Wolfram Research. I also didn't receive it with my other prizes.

The Mathematica prizes should come directly from Wolfram and will not be included with your other prize(s).

[quote="123456789"]I was wondering, do we get a small medal along with it? I remember last year when I won silver, i got a small silver medal.[/quote]

I didn't get a medal either... Maybe they didn't do those this year (I think they didn't do medals two years ago, so it may have just been something they did for just one year)?

Also, I spoke today with a representative from Wolfram Research, the makers of Mathematica, who have kindly donated the Mathematica prizes.  They have been running a bit behind, but they should be contacting you in the next week or so, if you selected Mathematica as a prize, with instructions on how to receive the software.

Good job with the t-shirts, guys! They're great.

Its been a while since DPatrick's update, but I still haven't been contacted by Wolfram. I'm wondering, has anyone else who picked Mathematica gotten a message from them?

Update: I just checked with our representative at Wolfram Research, and she confirmed that all of the students who selected Mathematica as a prize have been contacted by email with instructions for how to receive the software.

If you selected Mathematica as a prize and did not yet receive this information via email, please write usamts@usamts.org to let us know, and we will pass on that information to Wolfram.  But please double-check your email first: I believe that the notifications went out the week of June 29-July 3 or thereabouts.


function
real analysis
real analysis unsolved

Let $ f: R \rightarrow R$ be a monotonous function such that $ f \circ f$ is continuous. Prove that $ f$ is continuous

[hide]
We will prove the statement by using contradiction, supposed $ f$ is monotonic, here we assume that $ f$ is monotonic increasing, and $ f(f(x))$ is a continuous function, but $ f$ is discontinuous at, at least one point, name the point $ a \in R$, where $ R$ here denote the set of real numbers, then there exist $ t > 0$ such that for every $ d$ statisfying $ |x \minus{} a| < d$, we have $ |f(x) \minus{} f(a)| > t$, it means $ f(x) \minus{} f(a) > t$ or $ f(x) \minus{} f(a) < \minus{} t$. If $ f(x) \minus{} f(a) > t$, because $ f(x)$ is monotonic increasing, by our assumption above, then $ f(x) > f(a) \plus{} t$, means $ f(f(x)) > f(f(a) \plus{} t)>f(f(a))$, that means $ f(f(x)) \minus{} f(f(a)) > 0$, by Archimedean Property, there exist $ n \in N$, where $ N$ denotes the set of natural numbers, that statisfies $ f(f(x)) \minus{} f(f(a) \plus{} t) > \frac {1}{n}$, that means $ |f(f(x)) \minus{} f(f(a))| > \frac {1}{n}$ (for every $ t > 0$ that statisfies $ |x \minus{} a| < t$, if $ f(x) \minus{} f(a) < \minus{} t$, $ f(x) < f(a) \minus{} t<f(a)$, by our assumption that $ f(x)$ is monotonic increasing, we have $ f(f(x)) < f(f(a) \minus{} t)<f(f(a))$, $ f(f(x)) \minus{} f(f(a)) < 0$, again by Archimedean Property there exist $ n \in N$, such that $ f(f(x)) \minus{} f(f(a)) < \minus{} \frac {1}{n}$, that means $ |f(f(x) \minus{} f(f(a))| < \frac {1}{n}$, both of them contradict the fact that $ f(f(x))$ is continuous (for every $ t > 0$ statisfying $ |x \minus{} a| < t$). Now if $ g(x)$ is monotonic decreasing, we can set $ g(x) \equal{} \minus{} f(x)$, and $ f(x)$ is monotonic increasing in $ R$, then we can do as the proof above.
[/hide]

I am sorry if my answer is false. Thanks in advance.


[hide="little hint"]the measure of angle DBE is 180-60-30=90 degrees.[/hide]

If the areas are equal, then if we split them in half they should also be equal. Draw the heights on each triangle. Now we see that they are similar since all angles are congruent. Then since the areas are congruent, then the triangles are congruent, meaning the sides have the same length. As 236! pointed out, DBE=90. And this triangle has  congruent legs, then EDB=45


calculus
derivative
geometry
3D geometry
sphere
trigonometry
parameterization

I was wondering if there is a method, using partial derivatives maybe, to find the shortest distance between two distinct points going along the curve in 3 dimensions.
More specificially, given the unit sphere in 3d, and given some two random points, what is the easiest method to find the shortest distance between them going on the surface of the sphere?

I haven't tried this so I don't know if it works but perhaps try parametrizing the sphere, perhaps as $ \textbf{X}(s,t)\equal{}(a\sin{s}\cos{t},a\sin{s}\sin{t},a\cos{s}$ where $ a$ is the radius.  Then use the arclength formula with this parametrization and evaluate it.

Please give a specific question with all info so I can try it

This type of curve is called a geodesic, and is the path between the points with minimal curvature. In the case of the sphere, the geodesic lies on the great circle connecting the two points. The distance can be exactly calculated, but it's gross.

I think this is possible to evaluate in case of the sphere, since in this case the geodesics are the great circles, and you can calculate the arclength using  the angle $ \angle POQ$.

OK.
So what about finding the minimal length between two points on a general curve? (eg. between points $ (1,2,5)$ and $ (3,4,25)$ on curve $ z \equal{} x^2 \plus{} y^2$ or something like that)? Is there a method to find this distance?

When you say "curve" you really mean "surface."  Follow Kalle's suggestion to look at [url=http://mathworld.wolfram.com/Curvature.html]curvature[/url].

For general surfaces in $ \mathbb{R}^3$ you need to parameterize the geodesic curves (there is a set of differential equations giving them) and calculate a line integral, take a look here: http://en.wikipedia.org/wiki/Geodesic

First, of course you mean "surface" rather than "curve."

The quick answer is that the shortest path is something called a geodesic.  The root answer for how to find it is the calculus of variations.

In calculus, if you want to minimize a function of, say, three variables, then you look for the point where all the partial derivatives vanish.  In your problem, you want to minimize arclength over the space of all paths along your surface connecting the two points.  The space of all such paths is infinite dimensional, but that doesn't stop you from applying the same idea of demanding that "all the partials vanish."  If you demand that "all the partials vanish," one can show that your path must obey a certain 2nd-order ODE.  In your particular case, this 2nd-order ODE is called the geodesic equation (and more generally in calculus of variations it is called the Euler-Lagrange equation), and any path that solves the equation is called a geodesic.    Geodesics are simply the generalization of what it means to walk "straight" along a curved surface.

So if your two points are p and q, just look for all solutions of the geodesic equation that start at p and go through q, and then compute their arclengths to see which is the shortest.  In practice, this is typically a horrible computation, and typically you can't hope to write down nice solutions of awful nonlinear ODEs, but it all works in theory.

Note that it's not hard to rigorously prove (assuming the surface is smooth) that IF you have a path that minimizes arclength, then it must be a geodesic.  It is far more subtle (and requires some analysis) to prove that a path that minimizes arclength must be *realized*.  (For example, one must use completeness of the surface.)  To see how nontrivial this is, consider the following "easy" question: How do you know that a straight line minimizes the distance between two points in the plane?

I'd also like to clarify that finding geodesics is NOT about "minimizing curvature".

Clarification:  I was confused by Kalle's comment about "minimal curvature," but I guess that comment sort of makes sense because a geodesic is exactly a curve with *zero* (mean) curvature.  (However, the word "curvature" here has to be defined appropriately.  I'm talking about the curvature of the curve considered as being embedded in the surface, which is different from the curvature of the curve considered as a subset of R^3, and also different from the (Gauss) curvature of the surface itself.)


If the following instructions are carried out by a computer, which of $X$ will be printed because of instruction $5$?

$1.$ Start $X$ at $3$ and $S$ at $0$
$2.$ Increase the value of $X$ by $2$.
$3.$ Increase the value of $S$ by the value of $X$.
$4.$ If $S$ is at least $10000$, then go to instsruction $5$; otherwise, go to instruction $2$ and proceed from there.
$5.$ Print the value of $X$.
$6.$ Stop.

$\text{(A)} \ 19 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ 23 \qquad \text{(D)} \ 199 \qquad \text{(E)} \ 201$

[hide]Notice that $X$ is a sequence of odd integers starting at 3: 3,5,7,9,... etc. Therefore, $S=3+5+7+\cdots$. We want to find the smallest value of x such that $S=3+5+7+(2n+1)\le 10000$.

Using the formula for the sum of an arithmetic series, we see that the sum of the first $n$ odd numbers is $n^2$. Therefore, $X$ should be $199$. However, the answer is the next greatest odd integer, or $201.$[/hide]

[quote="Aequilipse"]This is AHSME 1990 #13

[hide="Question"]

I don't see the sequence. I start with $(x,s)=(3,0)$ and then (5,5) and (7,12) and (9,21)

Where is the 3,5,7,... coming from? or am I interpreting the problem wrong?
[/hide][/quote]
I don't see where you can't find the sequence... It is simply starting at 3 and going up by 2 each time because f step 2...

I think he posted it on the AoPS contest page.

And also he probably won't see your comment, since I sent a PM to him last December, and it's still in my outbox :P (lol)

[quote="vinayak-kumar"]@4everwise I don't see the reason of posting a problem, then posting the solution yourself. You could've done that on your blog.[/quote]
If you noticed, this is a really old post, so he shouldn't be active anymore.

[quote="Picroft"]Well, i typed in this code in a c compiler
[hide="Code i wrote"]#include <stdio.h>

int main(int argc, char* argv[])
{
    int x = 3;
    int s = 0;
    while(s<10000) {
    	x = x+2;
    	s = s+x;
    }
    printf("%d",x);
}
[/hide]
And i got 201, so i get it is $\boxed{E}$
But, i also solved it using math.
Making a table
Values of X : 3, 5, 7, 9, 11...
Values of Y : 0, 5, 12, 21, 32...
X is obviously a list of all odd numbers starting at 3, Which means that the nth number in the series is
$2(n+1)-1 = 2n+1$.
Y is a list that has a confusing relation. the list goes:0, 0+5, 0+5+7, 0+5+7+9, 0+5+7+9+11...
So it's like (the sum of the first n+1 odd numbers)-(1+3)
which means that the nth number in the series is basically $(n+1)^2-4$
the first such number greater than 10000 is $101^2-4=10197$
Since n here is 100, Its corresponding X value is $2(100) + 1 = 201$
Which again leads us to the answer $\boxed{E}$[/quote]

Sorry for the revive but I don't see the pattern anymore.
The list for Y goes \[\{ 0,0+5, 0+5+7,0+5+7+9, 0+5+7+9+11, \cdots \}\]

How do we get that that is basically $(n+1)^2-4?$ I don't see that :(

Yes, but isn't the sum $S$ equal to $5+5+7+5+7+9+\cdots \neq 5+7+9+\cdots ?$

No, it's not. You should try listing out the first few values, and maybe you can see something happening after 5 or so terms.


Is there a formula for the pattern:
1,4,10,20,35,......
in terms of n,which represents the term of the humber. E.G., for n=2, it refers to the number,4.

My formura was:
      n
 :Sigma:      i[n-(i-1)] and I was wondering if there is another way to express it.
  i=1

Thanks.

You can find that same forumla if you recognize that those numbers are just part of Pascal's triangle.  They start two rows down(hence the n+2) and always appear 3 spots in from either end of the row(as symmetry applies to the traingle).

Here is a simpler formula, but it is for simpler problems:

$z=a+dn-d$ where $a$ is the first term, $d$ is the common difference, $n$ is the number of terms, and $z$ is the last term.  This might not help you any, but it helped me a lot in MathCounts, and since this is the MathCounts forum...

Hmm... you don't usually see $\Sigma$ in the Mathcounts forum. :lol:

Which is exactly why I posted my other formula.

Well...
1) Take your TI-83+, etc. whatever you have.
2) Press Stat -> Edit.
3) In L1, put 1,2,3,4,5.
4) In L2, put 1,4,10,20,35.
5) Press Stat -> Calc -> CubicReg, ENTER, Y=, VARS -> Stat... ->EQ -> RegEQ
6) Press 2nd -> GRAPH, you'll get the table of values.
 :D The Y= is the equation, it's basically $Y_1=\frac{1}{6}x^3+\frac{3}{6}x^2+\frac{2}{6}x \Rightarrow \frac{x^3+3x^2+2x}{6} \Rightarrow \frac{x(x^2+3x+2)}{6} \Rightarrow \frac{x(x+1)(x+2)}{6}$.

[quote="probability1.01"]$\displaystyle \sum^n_{i=1} i(n-i+1)$

$\displaystyle = \sum^n_{i=1} (n+1)i-i^2$

$\displaystyle = (n+1) \sum^n_{i=1} i - \sum^n_{i=1} i^2$

$\displaystyle = (n+1) \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$

$\displaystyle = \frac{3n^3+6n^2+3n}{6} - \frac{2n^3+3n^2+n}{6}$

$\displaystyle = \frac{n^3+3n^2+2n}{6}$

$\displaystyle = \frac{n(n+2)(n+1)}{6}$

$\displaystyle = {{n+2}\choose 3}$[/quote]

Instead of
$\displaystyle \sum^n_{i=1} i(n-i+1)$
you could just use
$\displaystyle \sum^n_{i=1} \frac{i^2+i}{2}$

Ehh well I was working with the expression xixi found.  I was too lazy to go look at the numbers myself :P.

Also, if $L_1$ is $[1,1,1,1...]$ and $L_x$ is denoted by that the $n$th term of $L_x$ is the sum of the first $n$ elements of $L_{x-1}$ for $x$ is greater than $1$.  So the $n$th term of $L_x$ is
${{n+x-2}\choose {x-1}}$
The way to find out x if you just have a list of numbers, the first two terms are always $[1,x,...]$


When I go into the cool kind of math mode created by \*[ and \*], how do I create new lines? I've tried \\ and \newline but no luck :(

I don't like splitting it up into \*[ math \*] \*[ more math \*] because that puts an extra new line.

P.S. Ignore the *. It's there to help formatting (which got screwed up for some reason).

[b]\\ [/b]and [b]\newline[/b] both work:
\[ \text{line 1} \\
\text{line 2} \newline
\text{line 3}
\]

No, I mean specifically inside the \ [ and \ ] commands. I know how to make a new line :P Just not [i]inside[/i] of those.

It seems picky but it's really frustrating =/

I don't really understand what you're asking for. Can you be a bit more specific (maybe providing an example of what you've tried will clarify what you want to achieve)?

The commands \ [ and \ ] do not work on these forums because of conflict with BBCode or something. But they work with regular LaTeX compilers. They are very nice: they make a new line, they put all the LaTeX inside the brackets in display style, and they center it. It's like dollar signs on steroids.

But I can't create new lines inside of the \ [ and \ ] using the regular methods: like \\ or \newline

P.S. I put a space between \ [ and \ ] b/c without it the forum gets angry.

[quote="ktm"]The commands \ [ and \ ] do not work on these forums because of conflict with BBCode or something. But they work with regular LaTeX compilers. [/quote]You have got to be kidding me. They do work on this forum just like on any TeX compiler:
\[ \frac 12 \equal{} \frac 23 !
\]

Oh, I see. I was putting them in between the dollar signs, my bad. (That's not what I was doing in my LaTeX file, though.)

On this forum, when I do the following:
\[ f(x) \equal{} x \\
g(y) \equal{} y
\]
it works! But on my computer (Mac OS X, TeXShop) it does not make that new line, instead it looks like:
\[ f(x) \equal{} x \; g(y) \equal{} y
\]
So the best I can do on my LaTeX compiler is:
\[ f(x) \equal{} x
\]

\[ g(y) \equal{} y
\]
but this makes an extra new line which is undesirable =/

I didn't know it would work on this forum, so I guess my question is now targeted to Mac users =/

Haha, I realize this question is kinda picky, as I could do it with a \begin{center}. But the other way would be more convenient.

[quote="ktm"]On this forum, when I do the following:
\[ f(x) = x \\
g(y) = y
\]
it works! But on my computer (Mac OS X, TeXShop) it does not make that new line, instead it looks like:
\[ f(x) = x \; g(y) = y
\]
So the best I can do on my LaTeX compiler is:
\[ f(x) = x
\]

\[ g(y) = y
\]
but this makes an extra new line which is undesirable =/

I didn't know it would work on this forum, so I guess my question is now targeted to Mac users =/
[/quote]
Nothing to do with Macs as \ [ and \ ] is only meant for displaying a formula on a single line. To get multiline formulae (with the equals signs lined up) use the align environment and \usepackage{amsmath} in the preamble.
[code]\begin{align*}
f(x) &= x \\
g(y) &= y
\end{align*}[/code]
looks like this
\begin{align*} f(x) & = x \\
g(y) & = y \end{align*}
The * is to stop the equations being numbered and the & lines up the equals signs. No need for dollar signs in the document either.


limit
real analysis
real analysis unsolved

let $ a_n$ be the general term of a positive-term series. Why is not sufficient, to assure that the series diverges, to say that $ a_n \geq 1/n$ for infinite indexes (I mean, why is it necessary to say 'definitely' (which exactly means '$ \exists n_0 : \forall n \geq n_0$ the condition is satisfied')?)

Consider what happens if $ a_n\equal{}1/n$ when $ n$ is a power of $ 2$ and $ a_n\equal{}0$ (or $ a_n\equal{}1/3^n$ if you insist on strict positivity) otherwise.

Now, how about a slightly different question?

Let $ a_n$ be a decreasing (not necessarily strictly) sequence of positive numbers such that $ a_n\ge \frac1n$ for infinitely many $ n$. Does $ \sum a_n$ necessarily diverge?

it has already been posted many times that a decreasing sequence whose sum is finite satisfies $ a_n \equal{} o(\frac{1}{n})$

Then this was probably posted as well: for any sequence $ b_n$ such that $ \lim_{n\to\infty} nb_n\equal{}0$ there exists a convergent series $ \sum a_n$ with positive and decreasing terms such that $ a_n\ge b_n$ for infinitely many $ n$.

Is $ \liminf_{n\to\infty} nb_n\equal{}0$ enough?

[quote="mlok"]
Is $ \liminf_{n\to\infty} nb_n \equal{} 0$ enough?[/quote]
Yes, with the same proof: choose inductively an increasing sequence $ n_k$ such that $ n_kb_{n_k}<\frac 1{k^2}$ and $ b_{n_k}\le b_{n_{k\minus{}1}}$ (we may assume WLOG that all $ b_k>0$) and put $ a_n\equal{}b_{n_k}$ for $ n_{k\minus{}1}<n\le n_k$.


[quote="bpms"]If you submit your sols, VERY early for round 2, can you get your score early? If not, does everyone get their score at the same time?[/quote]
Nope, submitting early will not get your solutions graded early; here's why.  USAMTS grading is sort of like AP grading, from what I've come to understand - after the deadline is passed, all the graders get together in a big room and break up into groups to read over individual problems.  As they look over solutions, they develop a rubric for grading the problems based on the patterns and difficulties they encounter in the solutions they see.  This rubric allows, as nearly as possible, a fair standard to be applied to determine the score for any individual solution to any problem.

This is why your solutions wouldn't be graded early - one, because the graders will not convene in a special early session to grade them, and two, because you would not have the advantage of having a rubric applied to your solutions, which might give you more points than would otherwise be the case.

I hope this answers your questions, and any future questions people ask about whether their solutions would be graded early if they turned them in early.


A={1,2,3,.........150}
B={x*y|x,y € A and x and y aren't equal)
how many elements of B are divisible by 3?

50 members of $A$ are divisible by 3.  A member of $B$ is divisible by 3 if either x or y (or both) are, so either x is divisible and y isn't, y is divisible and x isn't, or they both are.  This gives us $50 \times 100 + 100 \times 50 + 50 \times 49 = 12450$

well , if you counted $3\cdot 4$ and $4\cdot 3$ are two different element in $B$ , then there are $12450$ element in $B$ . But I think they should be the same .

So the number of element in $B$ is just $50\cdot 149=7450$

its smaller than 7450 but i havent found out yet what exactly it is.but as a simple example 9*2=6*3 and there are lots of pairs like this

Answer = $7420$

The biggest element in $B$ is $149\cdot 150=22350$ . Since all element in $B$ must be divisible by $3$ , hence the set is of element is an Arthimetic Progression with first term $3$ and common different $3$ . The last term is $22350$ which have a total of $7450$ elements .Suppose we call that set of element as $C$ . Now we need to determine whether the square numbers in $C$ is formable or not .

If the square number in $C$ is in the pattern $3a \times 3a$ where $a$ is positive integer . Then you can also write ie as $a\times 9a$ with the condition that $3a\leq 50$ .From this we know that all element which is square number and $3a\leq 50$ in $C$ is formable .

If the square number is in the pattern $6b\times 6b$ , we can also write it as $3b\times 12b$ with the condition that $6b\leq 75$ .This means that for every even square number in $C$ with $6b\leq 75$ , is formable . So the odd square number from $51^2$ to $75^2$ (total of 5 elements) cannot be formed . Also , those element greater than $75$ also cannot be form (75 to 147 so total of 25 elements) .

Hence , the number of element in $B$ is $7450-5-25=\boxed{7420}$

I hope this is correct  :roll:

hey, between $1$ and $150 \cdot 150$ are $7500$ numbers divisible by 3. the answer to the problem is significantly smaller, i think $2916$ (that's what i got from a $C++$ program)

You see perfect_radio , the different between $149\cdot 150$ and $150\cdot 150$ is $150$ .  When divide by three gives you $50$ more elements . This is why I take until $149\cdot 150$ . By the way , I doubt there are so few elements in $B$  :o  . Do you think that there are any flaw in my solution  :?

I'm kind of sick right now and I can't concentrate to check your solution :( Sorry if I've written nonsense, but it seems obvious that there are fewer numbers... If i've made a mistake when i wrote that program, well  :blush:

I think I get what you mean . For example , consider the element $151\cdot 147=22197<22350$ . Although it is smaller than $22350$ , it still cannot be an element in $B$ .So probably we have to minus all those combination with prime number $>150$  :o  ....... that doesnt seem easy ......

I think you missed numbers of the form $3n$, where $n$ is an integer having prime factors all being greater than $150$.
[edit] If $n$ has at least two prime factors all greater than $50$, $3n$ also cannot satisfy the condition.

ya ....leepakhin .The problem is , is there an easy way to find those prime factor greater than 150 ? (Not using computer of course) ......


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

I need to sketch this hyperbola. From the equation I have extracted the following information which (I think) is all correct:

The centre of the hyperbola is at $ (\minus{}\frac{1}{2}, \frac{3}{4})$

Its foci are at $ (\frac{\sqrt{5} \minus{} 1}{2}, 0) and (\frac{\minus{}(\sqrt{5}\plus{}1)}{2}, 0)$

The eccentricity is $ \sqrt5$

The directrices are $ y \equal{} (\frac{\minus{}(\sqrt{5}\plus{}1)}{2\sqrt5}$ and $ y \equal{} \frac{1 \minus{} \sqrt5}{2\sqrt5}$

I'm kinda stuck. I thought the centre was the point where the directrices meet? But the two directrices are parallel so they don't meet...can anyone sketch this for me? Thanks

I think
foci are at $ \left(\minus{}\frac{1}{2}, \ \frac{3}{4}\plus{}\frac{\sqrt{5}}{2}\right)$ and $ \left(\minus{}\frac{1}{2}, \ \frac{3}{4}\minus{}\frac{\sqrt{5}}{2}\right)$.

YES! I figured it out and got the same as you too. I was getting my h's and k's a bit mixed up cos it's a north-south hyperbola not an east-west one. I also got my directrices and my vertices mied up. But I ended up with the same you got, thanks a lot.

P.S. japan rules


Express $ 16 \frac{2}{3} \%$ of $ 33 \frac{1}{3}$ as a mixed number.

We can write this as $ \frac{1}{6}\cdot\frac{100}{3}\equal{}\frac{100}{18}\equal{}5\frac{5}{9}$


linear algebra
matrix
number theory proposed
number theory

Given $ k \in [0,1,2,3]$ and a positive integer $ n$, let $ f_k(n)$ be the number of sequences $ x_1,...,x_n,$ where $ x_i \in [\minus{}1,0,1]$ for $ i\equal{}1,...,n,$ and

$ x_1\plus{}...\plus{}x_n \equiv k$ mod 4

a) Prove that $ f_1(n) \equal{} f_3(n)$ for all positive integers $ n$.

(b) Prove that

$ f_0(n) \equal{} [{3^n \plus{} 2 \plus{} [\minus{}1]^n}] / 4$

for all positive integers $ n$.


I was reviewing some problems and came across a couple of ones which gave me trouble. I would really appreciate it if someone could please help me and explain what I should do in order to determine the answer successfully. Thanks!

1. To construct an oscillating LC system, you can choose from a 13 mH inductor, a 4.8 μF capacitor, and a 1.9 μF capacitor. What are the (a) smallest, (b) second smallest, (c) second largest, and (d) largest oscillation frequency that can be set up by these elements in various combinations?

2. In an oscillating LC circuit, when 75.6% of the total energy is stored in the inductor's magnetic field, (a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor?

3. In an oscillating LC circuit in which C = 3.90 μF, the maximum potential difference across the capacitor during the oscillations is 1.44 V and the maximum current through the inductor is 49.3 mA. What are (a) the inductance L and (b) the frequency of the oscillations? (c) How much time is required for the charge on the capacitor to rise from zero to its maximum value?

Thank you very much for your help! Have a great week!

I think we have $\omega = \frac{1}{\sqrt{LC}}$, so since the inductance is pretty much fixed, it just depends on the capacitance. You have four choices... put in the first one, put in the second one, put in both in series, or put in both in parallel. The last two should give  the extreme cases, I think.

Can someone just help me with at least one of the problems? I have tried but I still can't seem to get it right :(

[quote="PhysicsManiac101"]2. In an oscillating LC circuit, when 75.6% of the total energy is stored in the inductor's magnetic field, (a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor?[/quote]
Assume that inductor current is $I = 0$ at time t = 0 (this can always be done for oscillations). Zero energy is then stored in the inductor, all energy of the oscillating circuit is stored in the capacitor, and charge $Q = Q_m$ is maximum. Voltage across the inductor is $V = L \frac{\text dI}{\text dt}$ and energy stored in the inductor at time $t = \tau$ is

$E_L(\tau) = \int_0^\tau V I\ \text dt = \int_0^\tau L \frac{\text dI}{\text dt}\ I\ \text dt = L \int_0^{I(\tau)} I\ \text dI = \frac{L I^2(\tau)}{2}$

When current is maximum, all energy of the oscillating circuit is stored in the inductor, $E = \frac{L I_m^2}{2}$, zero energy is stored in the capacitor, and charge is $Q = 0$.

$\frac{E_L(\tau)}{E} = \frac{I^2(\tau)}{I_m^2} = 0.756\ \Longrightarrow\ \frac{I(\tau)}{I_m} = \sqrt{0.756}$

Voltage across the capacitor is $V = \frac Q C$, current is $I = \frac{\text dQ}{\text dt}$, and energy stored in the capacitor at time $t = \tau$ is

$E_C(\tau) = E + \int_0^\tau V I\ \text dt = E + \int_0^\tau \frac Q C\ \frac{\text dQ}{\text dt}\ \text dt =$

$= E + \frac 1 C \int_{Q_m}^{Q(\tau)} Q\ \text dQ = E + \frac{Q^2(\tau) - Q_m^2}{2C} = \frac{Q^2(\tau)}{2C}$

When charge is maximum, all energy of the oscillating circuit is stored in the capacitor, $E = \frac{Q_m^2}{2C}$, zero energy is stored in the inductor, and current $I = 0$. Since $E_C(\tau) + E_L(\tau) = E,$ $\frac{E_C(\tau)}{E} = 1 - \frac{E_L(\tau)}{E} = 0.244$.

$\frac{E_C(\tau)}{E} = \frac{Q^2(\tau)}{Q_m^2} = 0.244\ \Longrightarrow\ \frac{Q(\tau)}{Q_m} = \sqrt{0.244}$

Thanks. I also managed to figure out the other 2 as well. ;)


prove that the triangle with the following equality must be right triangle :

$ sin^2 \alpha \plus{} sin^2 \beta \plus{}sin^2 \gamma \equal{} 2(cos^2\alpha \plus{} cos^2 \beta \plus{} cos^2 \gamma )$

[quote="TTsphn"]Hint
Let ABC is a triagle then
$ \text{cos}^2A \plus{} \text {cos}^2B \plus{} \text{cos}^2C \plus{} \text{cos}A\text{cos}B\text{cos}C \equal{} 1$
From condition then $ \text{cos}^2A \plus{} \text{cos}B^2 \plus{} \text{cos}C^2 \equal{} 1$
so $ \text{cos}A\text{cos}B\text{cos}C \equal{} 0$[/quote]

but from where did you get :

$ \text{cos}^2A \plus{} \text {cos}^2B \plus{} \text{cos}^2C \plus{} \text{cos}A\text{cos}B\text{cos}C \equal{} 1$

$ \cos2A\equal{}2\cos^2A\minus{}1$
Use this to prove the equality.

hello, from your equation follows
$ 1\equal{}cos^2(\alpha)\plus{}cos^2(\beta)\plus{}cos^2(\gamma)$. Using the theorem of cosine we get after a factorization
$ (a^2\plus{}b^2\minus{}c^2)(a^2\plus{}c^2\minus{}b^2)(b^2\plus{}c^2\minus{}a^2)\equal{}0$.
Sonnhard.


[quote="perfect628"]This sort of reveals people's opinions of states, and should therefore be locked.[/quote]

how?

anyways, I have pretty bad nats scores- on 1999 (my first nats test) I got 17/6=23

on 1992 (my second nats test) I got 23/10=33.

But usually to make countdown is around 30-35. 30 on only really hard years. In nearly all years 30 makes top 50 no matter how easy it is.

usually need 35-36 to make countdown.  I beleive the lowest ever score needed was a 33.

[quote="Ihatepie"][quote="perfect628"]This sort of reveals people's opinions of states, and should therefore be locked.[/quote]

how?[/quote]
If this year's states were harder(hypothetical-I actually haven't seen it), people may be inclined to predict lower scores.

actually, in many years, the national tests have a much different level than states. For example, 2005 State was very easy, yet the nationals was one of the hardest ones ever made. Anyways, I'm basing this off of passed results, which has nothing to do with the state scores.

Also, a 36 made countdown last year.

[quote="Ihatepie"]actually, this was 2005

1st : 39
2nd : 36
3rd - 5th : 35
6th - 8th : 33
9th - 10th : 32
11th : 31
12th - 17th : 30
18th - 28th : 29
29th - 32nd : 28
33rd - 40th : 27
41st - 51st : 26
52nd - 56th : 25
57th - ??? : 24[/quote]

uhh wow? i remember getting 33 on this year for practice.  Are you sure those are the right scores?  i dont think i have much of a shot at making cd...

[quote="infinity4ever"]Well there are always like 50-way ties for getting a score of 29 and 28, etc.  It's still hard to be in good condition on test day, in general.[/quote]

but if you're better, chances are you'll win the tiebreaker(s).

[quote="ra5249"]on the 2007 test i got 25/14
but thats if i even make nats, cause of the fierce competition in texas ( :( )[/quote]

wow...i need to stop considering indiana as a hard state.  :o

[quote="Yongyi781"]On an old Nationals test, I got a 28/16 = 44, but that's because it was a very old one (1989-1990), and old ones are easier, right?[/quote]
dang where do you get 1980's tests?

[quote="Lawrence Wu"][quote="ra5249"]on the 2007 test i got 25/14
but thats if i even make nats, cause of the fierce competition in texas ( :( )[/quote]
u know ur gonna win at states...[/quote]
uhhh yeah, i wish

[quote="Ihatepie"][quote="illusiat"]on 2008 nats (practice), i got 25. 17 sprint, 8 target
on 2003 nats, i got 35. 21 sprint, 14 target
on 1991 nats, i got 44. 28 sprint, 16 target

ya, some of the old nats r easier than recent states.[/quote]

2008 nats??!?!?[/quote]

Apparently, he cheated and hacked the Nationals problems from the website, allowing him to remember the solutions and get a perfect score :D

Really, I think it was just "practice" or "mock" Nationals problems.

[quote="gauss1181"][quote="DonkeyKong"]This is my first time going to Nationals along with many others. I really don't know what to expect. Especially the scores.

What would you say that you would get at Nats?

I took practice tests, and I am embrassed to say my scores. :([/quote]
Huh? What place did you take in Oklahoma? Is it an easy state?

A score of 30 would be material enough for countdown qualification.[/quote]
I took 2nd, but Oklahoma is like the easiest state of all...so, yeah..

[quote="numberdance"]My mother has been trying to convince me that Oklahoma isn't the easiest state.
I'm not sure I believe her. (I got 4th in Oklahoma)

I got a 40 on the 1995 test, but I didn't time myself, and I don't know how long I took.
And then I got a 38 on the 1994 test, but I'd seen at least 3 of the target problems before.[/quote]

Um, no.  Not to be unsupportive of my state team, but Georgia is probably one of the easiest when it comes to MathCounts.  I don't think that's true for high school math tourneys, but MC-wise, I think it's fairly easy to qualify for nats.  But that's just my opinion. :D

[quote="ra5249"][quote="Lawrence Wu"][quote="ra5249"]on the 2007 test i got 25/14
but thats if i even make nats, cause of the fierce competition in texas ( :( )[/quote]
u know ur gonna win at states...[/quote]

uhhh yeah... I WISH!!!
and guys, state department wins by a mile when it comes to easiest state[/quote]

Oh, well I meant the actual 50 states, b/c the state department has like no ppl.  And also, where do they live anyways?

Still, you have Alaska, Wyoming, etc. Georgia may be easy, but not the easiest.

[quote="ra5249"][quote="Lawrence Wu"][quote="ra5249"]on the 2007 test i got 25/14
but thats if i even make nats, cause of the fierce competition in texas ( :( )[/quote]
u know ur gonna win at states...[/quote]

uhhh yeah... I WISH!!!
and guys, state department wins by a mile when it comes to easiest state[/quote]

nonono! well state department isnt THAT bad ! our cutoff was like 34 or 35 last year, and highest was 39... so it isnt that bad right???

[quote="numberdance"]My mother has been trying to convince me that Oklahoma isn't the easiest state.
I'm not sure I believe her. (I got 4th in Oklahoma)

I got a 40 on the 1995 test, but I didn't time myself, and I don't know how long I took.
And then I got a 38 on the 1994 test, but I'd seen at least 3 of the target problems before.[/quote]

hey, I got 38 as well! stupid reading errors...

[quote="kthxbai"]But didn't sprints have 40 questions and targets have 10 questions in the older tests?[/quote]

They did until it was changed to the current format for the 1989-1990 contests.

[quote="DonkeyKong"][quote="Math Geek"]Well, we did come in like 40th last year, so I appreciate the support and understand that we did beat 17 states/regions, but notice that I said [b]one[/b] of the easiest states, not [b]the[/b] easiest state.  So yeah, but still, 39 states infinitely pwn us... :( Ah well...  Anyways, math isn't the only competition/extracurricular activity I do, so I don't feel that bad.[/quote]
From what I heard from my coach who went last year, Oklahoma got 42nd or 43rd. He said that they would have done better, but still... Our state seems to be better than at least 7 states.. That kind of relieves me.

[quote="numberdance"]And then I got a 38 on the 1994 test, but I'd seen at least 3 of the target problems before.[/quote]
Did you time this one? If so, from the information I was able to gather, I think you would have won that year. Wow..

I think this post can be locked now, since I have basic knowledge of what to expect. Thanks.[/quote]

"But I'd seen at least 3 of the target questions before"

Anyways, I usually get from high 30s to low 40s when doing old nats tests.


probability
MATHCOUNTS
number theory
relatively prime

Wait your not back already are you??? :P

[hide]I think basically you start with any combination (like $7\times 9\times 14\times 26$) and then start putting parentheses around them (there are $6$ in this one)

Then you use casework and rearrange the order so that you account for identical values and get the answer.

I don't really feel like doing it now. :D Actually I'm not sure this is the right way. :D [/hide]

Yeah, I think it's much more complicated (in your example, the value would not change no matter where you put the parentheses).

Oh sorry I meant inserting parentheses AND making everything in between them +.

Yeah, I see now that it's more complicated than that.  Forget what I wrote.

[quote="chess64"]Hey does somebody want to post the winning countdown question? :D[/quote]
A jar contains 8 red balls, 6 green balls and 24 yellow balls. In order to make the probability of choosing a yellow ball from the jar on the first selection equal to1/2, Kerry will add x red balls and y green balls. What is the average of x and y?
[hide="Answer"]5[/hide]

[quote="biffanddoc"]Can anyone solve it??

Only 1 team got it (not us or VA)[/quote]

The VA team came up with this thing, they divided it by the ways to do the + and *s and then added them. It's semi brute force.

Oh, and VA was the only team to get a 9 on the team round, everybody else got 8s and below but they didn't get #10 right o_O. Only 5 teams got #4 right.

Here's my solution to #10 on the team if anyone wants it :

[hide]

Since the 4 numbers 7,16,25,27 are relatively prime and a quick check shows no two can add up to a third, I felt pretty comfotable with this approach.

Refer to the 4 numbers as a,b,c,d

List the ways a,b,c,d can be combined by + and x and then find how the numbers can be exchanged

a + b + c + d          1 way
ab + c + d              6 ways (4 choose 2 for the #s that are multiplied)
ab + cd                  3 ways (4 choose 2 again, but divide by 2 since when you pair a &-                                            b , you are also choosing the pair c & d)
abc + d                   4 ways (4 choose 3 or 1)
abcd                       1 way
(a+b)c + d              12 ways (4 choose 2 for the () then x 2 for which one you x by)
(a+b)cd                   6 ways (4 choose 2 again)
(a + b)(c +d)           3 ways( same idea as when we did ab + cd)
(a + b + c)d             4 ways (4 choose 3 or 1)
(a + bc)d                 12 ways(4 choose 2 to multiply, then x 2 for which one you add)

Total    52 ways !!!!

That was definitely a doozy!!  Props to whichever team did get it right.

[/hide]

For those of you I got to meet in DC, nice meeting you, and good job to all!!

[quote="XxHarryxX"][quote="biffanddoc"]I dont see why so many teams got #4 wrong. I mean, it has to be a multiple of 45, and if you brute force you should come out to 315-405 seconds, varying if you counted right and stuff. Since there are only 3 answers that make sense at all, why did 52 teams put the answers that made sense but were wrong? Maybe mathcounts was wrong, but I doubt it.[/quote]

Well, CO missed the fact that the teachers start calling immediately... (Like, when the first teacher finishes her call to another teacher, that other teacher starts calling while the first teacher calls her second person). I was doing #10 very inefficiently, and came up with 20 ways before leaving it alone. We wrote down 24. The girl on the team worked #8 (the palindrome one) couldn't find any that changed in six steps... I also don't remember which other one we missed... Oh, well, I had fun![/quote]

Are there any other competitions in the high school that would also be very fun?

For #4, our team had the right idea but misread and thought that the principal called 2 teachers, not one.

I hope so. I live in a small town, and I need to get my high school involved in everything possible. :)

For #4, our team didn't realize it would take 2*45 seconds per teacher, since each had to make two calls, and the first teacher called starts calling immediately.

I solved it after the competition. But i say that whoever got it guessed. Anyway here's my solution:

[hide="Looks like the one from frost13 which i did not finish reading"]

Note that neither adding nor multiplying any two of 7,16,25,and 27 will give you the same result twice. Therefore, any that will not give an algebraic equivilant (associative, commutative properties) will be distinct.

Split it into many cases:

there are (omitting numbers)
+++ : 1 (trivial)
++x : 6 (choose 2 for the x)
+xx : 4 (choose 1 for the +)
x+x : 6 (choose 2 for the 1st x and the 2nd is determined)
xxx : 1 (trivial)
(+)xx : 6 (choose 2 for the + inside)
(++)x : 4 (choose 1 for outside)
(+x)x : 12 (choose 1 for outside (4) and 1 for the + inside (3 left))
((+)x)+ : 12 (choose 1 for the outermost + (4) and 1 for the inner x (3 left))
total them and you get 52
[/hide]

[quote="biffanddoc"]You are given the integers 9,7,14,26 (different from the real competition but meaningless)

_____ _ _____ _ _____ _ _____

You put the integers without repetition into the longer spaces and the operations x or + in the smaller spaces.
If an unlimited number of parenthesis can be used, and non-congruent expressions have a different value, then how many values can you get?

[hide]Its 52 but Im not sure how to do it[/hide][/quote]

This problem requires systematic case work as follows. I will use $a, b, c, d$ instead of numbers.

Case A: Zero $\times$
We have only $a+b+c+d$, so there is only $\boxed{1}$ way

Case B: One $\times$
We have the following four sub-cases:
(i) Operations of the form $(a \times b) + c + d$. There are $\binom{4}{2} = 6$ ways.

(ii) Operations of the form $(a+b)\times c + d$. There are $\binom{4}{2} = 6$ ways to
select the two numbers to add, then 2 ways to do the single multiplication operation for a total of 12 ways.

(iii) Operations of the form $(a+b) \times (c+d)$. Using the same argument, we get 6 ways to do this.

(iv) Operations of the form $(a + b + c) \times d$. There are 4 ways to select 3 numbers to
add initially. So this case yields 4 ways.

So in Case B we get a total of $6 + 12 + 6 + 4 = \boxed{28}$ ways.

Case C: Two $\times$

We have 3 sub-cases here as follows:

(i) Operations of the form $a \times b \times c + d$. There are 4 ways to do this.
(ii) Operations of the form $(a + b) \times c \times d$. There are 6 ways to do this.
(iii) Operations of the form $((a \times b)) + c ) \times d$. There are 12 ways to do this.
This gives us a total of $4 + 6 + 12 = \boxed{22}$ ways.

Case D: Three $\times$.
We have only $\boxed{1}$ case here: $a \times b \times c \times d$

So adding cases A, B, C and D we get a grand total of $1 + 28 + 22 + 1 = \boxed{\boxed{52}}$.

One way to reduce the number of cases (in say frost13's solution) is to just count the cases when the top-level operation is addition.  Then we can simply double the count to handle the cases when the top-level operation is multiplication.

wait...how come i forgot $(a+b)(c+d)$ and the guy above this forgot $ab + cd$ and yet we both get the same (and correct) answer?

NOTE: they aren't the same: $(a+b)(c+d) = ac+bc+ad+bd \neq ab+cd$

EDIT: checked frost13's solution. The reason is funny. Both of us forgot that the $\binom{4}{2}$ can be on the other side too so we had double for our case but 0 for the other which compensated. Don't you love double errors which work out perfectly?

I did #8, the trick was 19 would be the same as 28, 37, 46, 55 (if you didn't count 55 as a palindrome), 64, etc. Also, if 12 has 1 step, 13 has 2 steps, 14 has 3 steps, etc.

We got #4, (I didn't even get to read it), #9 (I had the answer, started writing the 3 down 12 seconds before time was called, the person that did it called "Wait, why?", I started answering and time was called), and #10 (The same person that did #4 and #9 got did this, what a coincidence)?

CT was the only team to get #10? Amazing :!:

he forgot to put the word "wrong" at the end of his post.

37 would land you in countdown (or a 36 with good tiebreaker)...at least nobody missed the #1 and the #3 easiest question as an 8th grader in the top 25%  One I misread, the other I am sure I got right, but maybe it is my fault as I know my handwriting

[quote="Ignite168"]Yeah I did, and 3 more seconds = 8 points more individually, and 8 points more for me would've landed me in countdown I believe.[/quote]

This is some pretty twisted logic as I understand it...you're saying that if you had 3 more seconds, you would have gotten #9 correct, which adds 2 points to the team score, which is equivalent to getting a total of 8 individual points higher, which would have put you in countdown... :?

How's this.  Indiana had a debate over two different answers :  15 & 37.

Moral of the story??  When you can't decide between two answers?  Add them together :)

[quote="biffanddoc"]I was literally 1 second from writing the answer to the change one I thought with 10 seconds left o yeah if the tens and hundreds add to 9, then it works. So 8! But i couldnt write it in time. I think my calculator malfunctioned on #8. I got 29768 for some reason. THen i tried the exact same thing after and got 30760. Wow thats annoying.[/quote]

Don't you just hate it when that happens to your calculator? Once, during math class, my calculator gave me the same WRONG answer TWICE!!! At least 30760 is somewhat close....

[quote="jclarkemathL314159"]not trying to offend anyone, but did ct just guess?
ironic:
MI got 15
MA got 37

MI+MA= correct answer  :)[/quote]

Guess on what? For #10, we had two people attempt it but they brute forced -_- We ended up putting 29 as that was the number they were on, for #9 we did not guess, we had the answer but one idiot prevented us from writing it down. For #4, I have no idea, the same person that did #9 did #4 and someone else was in the middle of checking him.

The numbers on problem #10 were $7^1,\thinspace 2^4,\thinspace 5^2,\thinspace 3^3.$

Some HP calculators use RPN (Reverse Polish Notation).
[hide]I rewrote the expression in RPN, which eliminates the parentheses.  The RPN must contain 4 operands and 3 operators.  It must start with 2 operands and it must end with 1 operator.  This leaves ${4\choose 2}=6$ possible forms of RPN string to consider:

LLLLOOO
LLLOLOO
LLLOOLO
LLOLLOO
LLOLOLO
LLOOLLO

The last form is invalid, since we don't have enough operands to execute the second operator.

For each of the five valid forms of RPN string, I considered each of the 8 possible choices for the operators.

For each of these 40 items, I wrote the result as a "sum of products".  [So, for example, the RPN string $\hbox{abcd++\times}$ simplifies to $ab+ac+ad.$]

I then combined and categorized the results, looked for duplicates among the combined results, and then counted the number of results of each form.

[code]
Form          Categorized     Count
-----------   -----------     -----
a+b+c+d       1+1+1+1            1
a+b+cd        2+1+1              6
a+bc+d        2+1+1        dup
ab+c+d        2+1+1        dup
ab+cd         2+2                3
a+bd+cd       2+2+1             12
ab+ac+d       2+2+1        dup
ac+bc+d       2+2+1        dup
ab+ac+ad      2+2+2              4
ad+bd+cd      2+2+2        dup
ac+ad+bc+bd   2+2+2+2            3
a+bcd         3+1                4
abc+d         3+1          dup
abc+ad        3+2               12
abd+cd        3+2          dup
ad+bcd        3+2          dup
abc+abd       3+3                6
abd+acd       3+3          dup
acd+bcd       3+3          dup
abcd          4                  1
[/code]

My final tally is 52, which seems to be the accepted answer.

This method was well-organized, and I am pretty confident I didn't omit any cases.  However, this method was quite tedious, and would have consumed most of the time permitted in the Team Round.  In retrospect, I see some shortcuts that might have sped up the process, but frost13's answer is just as well-organized and was far less tedious.[/hide]

lol you used a calculator program?
in one of our practices, george (8th) wrote a calculator program to solve this really tedious problem that another kid on our team made up. So two kids working brute force on an annoying problem, another sitting back and writing a calculator program. not very productive in terms of getting something out of it.

just an fyi. Its mathcounts. not calculatorprogrammingcounts

just fyi.
RPN isn't a calculator program.

GROAN--my team had one person work on that problem the whole time, and we still got it wrong. :( That was an evil problem...

Anyone know the actual values?

BTW, what's Reverse Polish Notation? rcv mentioned it...

[quote="jclarkemathL314159"]But couldn't you have

a(b(c+d))?[/quote]

I'm not exactly good with RPN, but I think you'd do:

d c + b * a *

With an enter between everything.


You're honest opinion.

[youtube]YZnXbdse9oQ[/youtube]

[hide="I think"]Celtic Woman[/hide] sings it better.

I like the sail away part best.  :lol:
Does anyone here know the song "Come Sail Away" by The Styx?


A math teacher was asked how many students she had taught. She replied that the sum of $ \frac15, \frac16$, and $ \frac17$ of the total number of students she had taught was 107. How many students had she taught?

Suppose she taught $ s$ students. We can then say $ \frac{s}{5} \plus{} \frac{s}{6} \plus{} \frac{s}{7} \equal{} 107 \implies 42s \plus{} 35s \plus{} 30s \equal{} 11235 \implies s \equal{} 105$.

I got something different.
Let $ s$ be the total number of students she taught.  We have
\[ \frac {s}{5} \plus{} \frac {s}{6} \plus{} \frac {s}{7} \equal{} 107\]
Therefore
\[ 107 \equal{} \frac {42s}{210} \plus{} \frac {35s}{210} \plus{} \frac {30s}{210} \equal{} \frac {107s}{210}\]
Thus
\[ 107s \equal{} 210(107)\implies s \equal{} \boxed{210}\]


Over the summer I've been doing some Pre-Calc in prep for next year.

I just can't seem to solve it:

Find the maximum value of the function f(x) = 2x^3 - 3ax^2 + 2a^3 on the interval (0 <\= x <\= 2).  Assume that a>0.

The chapter involves using variation tables if that helps.

Here is my solution by hand writing.This is my first Add an Attachment! :lol:
I hope this help you.

kunny

Kunny, you beast, knocking it up with calculus xD

There's a trick to doing this problem without calculus (and without variation tables) that involves observing the shape of the graph.

$ 2x^{3}-3ax^{2}+2a^{3}=2 \left( x^{3}-\frac{3a}{2}x^{2}\right)+2a^{3}$

For a fixed value of $ a$, the maximum value of this can be found by finding the maximum value of $ x^{3}-\frac{3a}{2}x^{2}$

At $ x=0$, this value is $ 0$. Since the quadratic temporarily overwhelms the cubic [i](why? Consider how if $ 0<a<1$, then $ a^{3}< a^{2}\implies a^{3}-a^{2}< 0$)[/i], this begins by decreasing. However, we know that eventually, the cubic will overtake the quadratic, and the value will become positive. As a result, the graph has a U shape. We know the maximum value will occur on one of these two ends.

So at $ x=0$, $ x^{3}-\frac{3a}{2}x^{2}=0$. At $ x=2$, $ x^{3}-\frac{3a}{2}x^{2}=8-6a$

If $ 8-6a>0$, then the right end of the U (x=2) is higher then the left end, and the maximum occurs at $ x=2$. If $ 8-6a<0$, then the left end is higher than the right end, and the maximum is at $ x=0$. If $ 8-6a=0$, then it doesn't matter which we use, since both ends are equally high; we can use 0 for convenience.

$ 8-6a>0 \implies 8 > 6a \implies a < \frac{4}{3}$.

Thus, if $ a < \frac{4}{3}$, the maximum occurs at $ x=2$. Plugging this into our original equation gives a value of $ 16-12a+2a^{3}$.

If $ a \geq \frac{4}{3}$, the maximum occurs at $ x=0$. Plugging this into our original equation gives a value of $ 2a^{3}$.

thanks so much

Nice Calculus there lol
I had to look up some of your functions

anyways thanks
I understand now  :D


What time does ARML happen today? I have a party I have to go to but I really still want to make it. Also how many practices are there before the competition in LV?
-Thanks, MrRubix

ARML practice is from 2-5 PM today, and will be at the same time every Sunday unless you hear otherwise.  The practices are currently scheduled for every Sunday from now right up until the weekend before the competition, except for May 4th.  So those dates are April 13th, 20th, and 27th, and May 11th, 18th, and 25th.  Six total practices.


consider two lenses of focal length $ f_{1}$ and $ f_{2}$ kept close one another
we know equaivalent focal length $ F$ satisfies $ \frac{1}{F}\equal{}\frac{1}{f_{1}} \plus{} \frac{1}{f_{2}}$
now this formule has been derived by applying the lens formulae twic for two lenses where the image due to first lens acts as an object for the second.
but in doing so our formulae for $ f_{1}$ is as if there was air on both sided of the lens whereas there is air on one side and the medium of the other lens on the other side.
how is this justififed??

[quote="pardesi"]consider two lenses of focal length $ f_{1}$ and $ f_{2}$ kept close one another
we know equaivalent focal length $ F$ satisfies $ \frac {1}{F} \equal{} \frac {1}{f_{1}} \plus{} \frac {1}{f_{2}}$
now this formule has been derived by applying the lens formulae twic for two lenses where the image due to first lens acts as an object for the second.
but in doing so our formulae for $ f_{1}$ is as if there was air on both sided of the lens whereas there is air on one side and the medium of the other lens on the other side.
how is this justififed??[/quote]

Now imagine you have a glass slab and a point source.... what do you do?? You first find the devn due to glass slab, erase the glass slab and continue as if object was present at |u-dev| distance from origin.

Similar concept here. first assume that only one lens is there and calculate the deviation due the lens and remove it totally. Now that ray acts as image for 2nd.

Hope you get me:)

[Side Note @ chapplikka: I was asking the same thing before the optics test rem?]

no problem with that
probably u misunderstood me
we use
$ \frac{1}{f}\equal{}(\mu\minus{}1)(\frac{1}{R_{1}}\minus{} \frac{1}{R_{2}})$
but actually the ray after refraction is going from AIR to GLASS then again to GLASS and NOT AIR
but the above formulae holds if the ray were moving from AIR to GLASS to AIR

Hey ,
I think you have misunderstood it. It moves from air to glass in lens 1 and then back to air and then to glass of lens 2 and then again out to air...

@GV
lets have a convex and plano concave lens by side so taht the convex part exactly fits in the concave part of the other lens
now how do u apply the formulae??


Find all ordered pairs (x;y) of integers such that $ x^{3}\minus{}y^{3}\equal{}3(x^{2}\minus{}y^{2})$ .

[hide="Might help.."]We can factor this as

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

$ x^2 \plus{} xy \plus{} y^2 \equal{} 3(x \plus{} y)$

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

*Note: Any ordered pair $ (z,z)$ satisfies the equation, where $ z$ is an integer.[/hide]

$ x^3\minus{}3x^2\equal{}y^3\minus{}3y^2$

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

Clearly all x=y work. WLOG y>x.

We look at the graph of $ x^3\minus{}3x^2$

Using basic calculus the relative extrema occur at $ 3x^2\minus{}6x\equal{}0 \implies 3x(x\minus{}2)\equal{}0$ x=0,2 and are a rel. max of 0 and a rel min of -4.

When -4<f(x)<0 this is a one-to-one function. We now find the domain over which it is not. After some easy math we find that $ \minus{}1 \leq x \le y \le 4$

We now do basic casework which I don't have time for as I have to go

Well, then, x=y ?

x,y E {integers} ?

I don't understand very well the answer.

[hide]Let $ f(x) \equal{} x^3\minus{}3x^2$. We have $ x^3\minus{}y^3\equal{}3(x^2\minus{}y^2) \Leftrightarrow x^3\minus{}3x^2\equal{}y^3\minus{}3y^2 \Leftrightarrow f(x) \equal{} f(y)$

Obviously $ (x,y) \equal{} (n,n)$ is a solution for all $ n \in \mathbb{Z}$.

By graphing $ f(x)$ (or doing some calculus) we see that $ f(x)$ is one-to-one over $ (\minus{}\infty,\minus{}1) \cup (3,\infty)$.

So, if $ x \in (\minus{}\infty,\minus{}1) \cup (3,\infty)$, then $ y\equal{}x$ is the only solution for that value of $ x$.

Now we need to test integers $ x,y \in [\minus{}1,3]$:
$ f(\minus{}1) \equal{} \minus{}4$, $ f(0) \equal{} 0$, $ f(1) \equal{} \minus{}2$, $ f(2) \equal{} \minus{}4$, $ f(3) \equal{} 0$

So, the solutions with $ x \neq y$ are:
$ (\minus{}1,2)$, $ (2,\minus{}1)$, $ (0,3)$, $ (3,0)$.

Therefore, the solutions $ (x,y)$ are:
$ (\minus{}1,2)$, $ (2,\minus{}1)$, $ (0,3)$, $ (3,0)$ and $ (n,n)$ for all $ n \in \mathbb{Z}$.[/hide]

Ah ok. Thanks for your help Xantos C. Guin ;)


Start with a username. The next poster posts a new username that combines elements of both the previously posted username and his own username.

For example, my username is randomdragoon. If Ignite posts after me, he could do randomIgnite or something. Of course, you can probably be a little more creative than that.

----Begin----

randomdragoon

jomack92

The message is too small. Please make the message longer before submitting.

muklermacka

Sounds like a pokemon name :D


Does anyone know how many chapters ther are in PA?  Just wondering

There are 23 chapter competitions
@rihim77.7, if you dont know the answer, you shouldn't post.


quadratics
algebra
polynomial
Rational Root Theorem

[hide="ugly solution"]I really hope there is a more elegant solution than this...

By squaring both sides and combining you get

x^4+4x^3+4x^2+5x-2

(I couldn't find anything by factoring the top into (x-1)(x-2) and the left side into (x+2)(x+1)^2(x)

rational root theorem confirms that this question is evil and has no rational roots...

so the only hope is to factor into two quadratics

so far we know (x^2...+2)(X^2...-1) or (x^2...-2)(x^2...+1)

I went with the positive 2 first because the coefficient of x is positive...

call the x term on both quadratics a and b

since the coefficient of x^3 is 4 and the coefficient of x is 5
a+b=4
2a-b=5

a=3 b=1

(x^2+3x-1)(x^2+x+2)

so yeah...(-1+-sqrt(7)i)/2 or (-3+-sqrt(13))/2

then I applied ti-84 plus' raw calculating power with two random roots I found and it seems to work[/hide]

Are you sure the numerator inside the square root isn't

$x^{2}+3x+2$? That would make this problem a little less nasty.

I'm sure there's no mistake in that problem.


trigonometry
integration
real analysis
real analysis solved

Suppose a tennis ball of diameter 1 has a line connecting its north pole to its south pole. After several games it comes to rest on the ground (in a random position). Looking at the ball from above, what is the expected length of the line as seen by you? That is, take the two ends of the line and project them onto the ground. Measure the line on the ground. What is the average value of this measurement?
For example, if the ball lands in its standard position, the line is vertical and has projected length 0. If the ball lands so that the line is horizontal, its projected length is 1.

The upper endpoint of this line us uniformly distributed on the surface of a hemisphere. From that we can determine the following. Let $\varphi$ be the angle the line makes with the vertical. $\varphi$ is a random variable distributed on $\left[0,\frac{\pi}2\right]$ with density $\sin\varphi\,d\varphi.$ The projected length of the line is $\sin\varphi.$ Thus the expected length of that projection is

$\int_0^{\frac{\pi}2}\sin^2\varphi\,d\varphi=\frac{\pi}4.$

I stand by the answer $\frac{\pi}4.$ The density $dP=\sin \varphi\,d\varphi$ is appropriately normalized, since

$\int_0^{\frac{\pi}2}\sin\varphi\,d\varphi=1.$

The resulting intrgral, $\int_0^{\frac{\pi}2}\sin^2\varphi\,d\varphi,$ works out to "half the box", which is $\frac{\pi}4.$

Why do you think the answer is $\frac12$? Can you explain your reasoning?

One reason to be suspicious: in order for the length of the projected line to be less than 1/2, the line must make an angle with the vertical that is less than 30° - in other words, the line must point into what is nearly the polar regions on the globe. There's not very much polar region on a globe; there's a lot more temperate zone and there's especially a lot more tropics. It's far easier for the line to be nearly horizontal ("tropical"), in which case its projected length will be nearly 1. Expect the average to be much closer to 1 than to 0.


Hey, I need help proving the Pythagorean Theorem.

It needs to be in the format of a 2 Column Proof, a paragraph proof, flow proof, etc.

Im doing a portfolio and i can't seem to understand these proofs.

If someone can please help me out that would be great.

Hello, this link should be pretty helpful: [url]http://en.wikipedia.org/wiki/Pythagorean_theorem[/url]

(As if we all don't use wiki enough already...)

haha yah, i looked there, thanks.

if someone can give me a little more help that would be cool.

The most elementary methods would be using the areas of squares. However, if you know calculus, there is also another simple method. They are all in the link Brut3Forc3 posted.

I think this link should be usefull [url]http://www.mathlinks.ro/viewtopic.php?search_id=1326179976&t=199883[/url]

This is not an exercise on how to prove the theorem. This is an exercise on understanding the logical structure of proofs, using a famous theorem as an example. Choose [b]one[/b] of the many proofs available, and translate it into the various forms. They're all trying to do the same things- only the presentation is different.

Mathematical proofs are a form of writing, much like an essay. The forms you are learning now are analogous to the standard essay forms you learn in school, and they will invariably look stilted to us. You will never see a two-column or flow proof beyond the textbooks you have now- they're tools to learn the logical structure of a proof, but not efficient ways of actually writing one.


linear algebra
matrix
logarithms
complex analysis
function
algebra
polynomial

hi all,
we know that if given a matrix equation $ e^S\equal{}U$,we'll easy to find concrete U when we know S,but conversely you can find concrete S when we know U?(not represent in the form $ S\equal{}\ln U$ )
please help me, and can you give me an example to illustrate?

i want to beg you give me an example with concrete matrix U, then you find concrete matrix S?
for example let $ U\equal{}\left (\begin{array}{cc}
                                  1  & 2 \\
                                   3 & 8 \\
                              \end{array} \right)$
find S?
help me,please!

1) If $ x$ is a complex matrix without  any eigenvalue in $ ]\minus{}\infty,0]$ then there exists an unique complex matrix $ a$ s.t. $ e^a\equal{}x$ and $ \forall\lambda\in{spectrum(a)},Im(\lambda)\in{]\minus{}\pi,\pi[}$.
 2) If $ f: D\minus{}>\mathbb{C}$ is a one to one holomorphic function and if $ spectrum(x)\equal{}\{\lambda_1,\cdots,\lambda_k\}\subset{D}$ and $ x$ is diagonalizable then it's easy to calculate $ f(x)$. (here $ \{\lambda_1,\cdots,\lambda_k\}$ refers to the distinct eigenvalues of $ x$).
3) Let $ p$ be the lagrange's polynomial of degree $ k\minus{}1$, the graph of which, contains the points $ (\lambda_i,f(\lambda_i))$. Then $ f(x)\equal{}p(x)$.
4) Here $ x$ has 2 positive distinct eigenvalues and thus $ k\equal{}2$.
$ log(x)\equal{}[584.0357125, 1509.788122, 2264.682184, 5868.294142]$ row by row.
   The following is a Maple program which calculates $ ln(a)$ if $ a$ is a diagonalizable $ (n,n)$ complex matrix with non negative eigenvalue.
with(linalg):
n:=2:
> a:=matrix(n,n,[1,2,3,8]):s:=[evalf(eigenvalues(a))]:
> d:=nops(s):
> ei:=NULL:lo:=NULL:
> for i from 1 to d do
> ei:=ei,s[i]:lo:=lo,ln(s[i]):
> od:
> ei:=[ei]:lo:=[lo]:
> p:=interp(ei,lo, z):p:=unapply(p,z):
> lg:=evalm(p(a)):
>

thanks to loup blanc !
i'll check your good idea!but will we use maple program?can you give me an well example of matrix U to find S which not odd number as you find log(x)=[-1.148620062, .8543963730, 1.281594560, 1.841767244] or log(x)=[584.0357125, 1509.788122, 2264.682184, 5868.294142] , please!
best,Luan.


inequalities
calculus
integration
logarithms
geometry
rectangle
calculus computations

Prove that
$ \displaystyle \int\limits_0^1 {{e^{{x^2}}}dx}  > \int\limits_1^e {\sqrt {\ln x} dx}$

Lets draw the graph of $ e^{x^2}$.  It is easy to see that $ \int\limits_{0}^{1}{{e^{{x^{2}}}}dx} \plus{} \int\limits_{1}^{e}{\sqrt{\ln x}dx}$ is equal to the square of rectangle with vertices $ (0,0),(1,0),(1,e),(0,e).$ So we need to prove that $ \int\limits_{0}^{1}{{e^{{x^{2}}}}dx}>\frac{e}{2}$.
$ \forall x>0\ e^x>1\plus{}x\plus{}\frac{x^2}{2}$, so $ e^{x^2}>1\plus{}x^2\plus{}\frac{x^4}{2}$, and $ \int\limits_{0}^{1}{{e^{{x^{2}}}}dx} > 1 \plus{} \frac{1}{3} \plus{} \frac{1}{10} > 1.43 > \frac{2.8}{2} > \frac{e}{2}$


algebra
function
domain
Ring Theory
calculus
integration
polynomial

This is Exercise 11.2 from Eisenbud's Commutative Algebra book. Let $ R$ be any domain, and let $ P\subset R$ be a prime ideal. Show (by Zorn's lemma) that there exists a subring $ R'\subset K(R)$ containing $ R$ and maximal among those rings such that $ PR'\neq R'$ (I know how to do this) Then

a) $ R'$ is local, integrally closed in $ K(R)$, $ R_P\subset R'$ and if $ \mathfrak{m}\subset R'$ is the maximal ideal, then it contracts to $ P$ in $ R$ (ie it intersects $ R$ at $ P$).
b) For any domain $ S$ and any element $ s\in K(S)$ write $ S[s]$ as the subring of $ K(S)$ generated by $ S$ and $ s$. Show that $ sS[s]\equal{}S[s]$ iff $ s^{\minus{}1}$ is integral over $ S$. Use this remark to prove:
c) $ R'$ is a valuation ring.
d) The integral closure of $ R$ is the intersection of the valuation rings containing $ R$.

So s a hint for part $ a)$, Eisenbud suggests localizing at $ P$ by replacing $ R$ with $ R_P$ I'm confused as to why you're allowed to do this, but given that you are allowed to do it, I know how to do the rest of part $ a$. Im basically now totally stuck on $ b\minus{}c\minus{}d$, so any tips are appreciated. Eisenbud also makes some argument in the hint which makes no sense to me: He said this is a hint for 11.2a (which I don't really see how it applies, i worked it out without this hint though, so I think im ok for that part): "Reduce to the case $ R\equal{}R_P$ If $ x\in K(R)$ suppose neither $ x$ nor $ x^{\minus{}1}$ were contained in $ R$. By maximality of $ R$ (he probably means $ P$ here) $ PR[x]\equal{}R[x]$ so we may write $ 1\equal{}r_0\plus{}r_1x\plus{}...r_nx^n$. Multiplying by $ x^{\minus{}n}$ we see that $ x^{\minus{}1}$ is integral over $ R$ and thus in $ R$."
Im thinking he wants us to use this for part c... but... I'm basically totally confused.

(a) Notice that if $ pR' \neq R'$, then $ pR'_p \neq R'_p$. (Suppose otherwise, then $ 1 \equal{} \sum p_i \frac{r_i}{s_i}$, where $ p_i \in p$, $ r_i \in R$, $ s_i \in R \minus{} p$. But then if we clear the denominator and multiply the denominator to the other side, then LHS is something not in $ p$, but RHS is something in $ p$, contradiction)

Therefore $ R' \equal{} R'_{p}$, in particular, $ R_p$ is contained in $ R'_p \equal{} R'$, and of course in this setting $ R'$ is also a maximal subring containing $ R_p$ with the desired property. So in the first step one can replace $ R$ by $ R_p$.

(b)Note that $ sS[s] \equal{} S[s]$ iff there exists a polynomial $ p(s)$ such that $ sp(s) \equal{} 1$ iff $ s^{\minus{}1}$ is integral over $ S$.

(c)SUppose that $ x \notin R'$, we want to show that $ x^{\minus{}1}$ is in $ R'$. By (a), it suffices to show that $ x^{\minus{}1}$ is integral over $ R'$. Then basically you can just use the hint (I guess it's for (c)). Since $ R'$ is maximal, $ pR'[x] \equal{} R'[x]$. Then $ p_1f_1(x) \plus{} ... \plus{} p_nf_n(x) \equal{} 1$, where $ p_i \in p$, $ f_i$ are polynomials with coefficients in $ R'$. We can of course simplify the left hand side and write it as $ r_0 \plus{} ... \plus{} r_nx^n \equal{} 1$. We claim that $ 1\minus{}r_0$ is a unit in $ R'$. This is because $ r_0 \in pR'$, and $ pR' \subset m$, the maximal ideal of $ R'$. Since $ R'$ is local, $ 1\minus{}r_0$ is a unit.

(d)I haven't figured it out yet.

Once again thank you very much. I was talking to a friend and he helped me for part d, here's what I have

We know first of all, that any valuation
ring containing $ R$ is integrally closed in $ K(R)$. Therefore the
integral closure of $ R$ is contained in each valuation ring
containing $ R$ and hence is in their intersection. For the converse,
let $ x\in K(R)$ be non integral. Then $ x^{\minus{}1}R[x^{\minus{}1}]\equal{}R[x^{\minus{}1}]$ so
there is a nontrivial maximal ideal containing $ (x^{\minus{}1})$ in
$ R[x^{\minus{}1}]$. Then consider the valuation ring $ R'$ containing this
ring, its maximal ideal will contain $ x^{\minus{}1}$, and so $ x^{\minus{}1}$ is
not a unit in that valuation ring. Hence $ x$ is not in that
valuation ring and is not in any valuation ring containing $ R$.

It looks right, so I'm just posting it for completion


graph theory
combinatorics unsolved
combinatorics

Hi, everybody. I'm new to this forum.  :)
I got a really difficult problem when i was doing mathmatical modeling. Since I'm not soooooo good with math,(so good like you guys) it took me many days to think and I still can not figure it out. But for you guys this may not be a problem. I would appreciate if someone would help me with this, even it is too late for the mathmatical modeling paper.

Given arbitrary n uniform points, what is the maximum number of triangles you can draw from the n points without any two triangles sharing the same edge?

For example:
If given n=5, the maximum number of triangles you can get is 2, and their vertices are:
1 2 3
1 4 5
If given n=6, the maximum number of triangles you can get is 4, and their vertices are:
1 2 3
1 4 5
2 4 6
3 5 6
Of course those are not the only way to get 2\4 triangles, just the maximum number of triangles is 2\4.

If given n = 7,you get 7 triangles, forming a complete graph, with their vertices:
1 2 3
1 4 5
1 6 7
2 4 6
2 5 7
3 4 7
3 5 6

But what about an arbitrary n points?(n>3)

Since this is a problem from mathmatical modeling, computer programing is allowed. But I find no way to calculate even using computer, not even knowing how to burst. :(

It's sequence [url=http://www.research.att.com/~njas/sequences/A001839]A001839[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


Let $a_0$ be a positive integer. Define the sequence $\{a_n\}_{n \geq 0}$ as follows: if \[ a_n = \sum_{i = 0}^jc_i10^i \] where $c_i \in \{0,1,2,\cdots,9\}$, then \[ a_{n + 1} = c_0^{2005} + c_1^{2005} + \cdots + c_j^{2005}. \] Is it possible to choose $a_0$ such that all terms in the sequence are distinct?

No. Denote $m$ the number of digits in $a_n$.

We have that $a_{n+1}<\lceil _{10}\log(a_n)\rceil.9^{2005}$ (replacing all digits with 9), so if $a_n\ge10^{2005}$ then $a_{n+1}<a_n$.
Now, $a_k\ge10^{2005}$, is $a_{k+1}\ge10^{2005}$? If yes, go on until you're below $10^{2005}$. (which will always occur as it's decreasing) If not, you also have another number below $10^{2005}$. So there are infinitely many numbers below $10^{2005}$, so they cannot all be distinct by pigeonhole.


A man is runnung through a train tunnel. When he is $\frac{2}{5}$ of the way through, he hears a train that is approaching the tunnel from behind him at a speed of 60 mph. Whether he runs ahead or back, he will reach the end of the  tunnel at the same time the train reaches the end. At what rate, in miles per hour is he running assuming he runs at a constant rate?

if you need help with this problem i think it was posted a few weeks ago. just search the problem in the search engine for this website and you will find it :lol: .
-jorian

yes. it was

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119545[/url]

ok, i remember there was a long discussion about this problem so i figured it didn't need to happen again.

-jorian

So a moderator should lock this thread since the problem has already been discussed.  :police:

[hide="solution"]When the man runs through 2/5 of the tunnel, the train, no matter what, will be at the first end of the tunnel. Then the man will run 1/5 of thwe tunnel when the trian goes all of the way through. [b][i][u]12[/u][/i][/b][/hide]

Wow! That's a nice problem!! :lol:  I kinda feel sorry for the guy though.


trigonometry
induction
search
real analysis
real analysis unsolved

n is positive integer,n>1,Calculate
$ \prod_{i\equal{}1}^{n\minus{}1} \sin \frac{i\pi }{n}$

hello,prove by induction that $ \prod_{i\equal{}1}^{n\minus{}1}\sin\left(\frac{i\pi}{n}\right)\equal{}n2^{1\minus{}n}$.
Sonnhard.


Can someone help me with this:

What is the sum of the infinite series: 1/1-3/3+5/9-7/27+9/81-11/243...?

so what we have to find is
$ \boxed{\sum_{k \equal{} 0}^{\infty}\frac{(\minus{}1)^{k}(2k\plus{}1)}{3^{k}}}$
undoubtedly this converges
$ \rightarrow\sum_{k \equal{} 0}^{\infty}\frac{(\minus{}1)^{k}(2k)}{3^{k}}\plus{}\sum_{k \equal{} 0}^{\infty}\frac{(\minus{}1)^{k}}{3^{k}}$
use the identity
$ \boxed{\sum_{k \equal{} 0}^{\infty}(\minus{}1)^{k}x^{k}\equal{}\frac{1}{1\plus{}x}}$ provided $ |x| < 1$ ******
putting $ x \equal{}\frac{1}{3}$ we get the second sum to be $ \frac{3}{4}$
also differentitating ****** once gives
$ \sum_{k \equal{} 0}^{\infty}(\minus{}1)^{k}kx^{k\minus{}1}\equal{}\frac{\minus{}1}{(x\plus{}1)^{2}}$ multiplying $ x$ both sides gives
$ \boxed{\sum_{k \equal{} 0}^{\infty}(\minus{}1)^{k}kx^{k}\equal{}\frac{\minus{}x}{(x\plus{}1)^{2}}}$
putting $ x \equal{}\frac{1}{3}$ we get the first sum as $ 2(\frac{\minus{}3}{16}) \equal{}\frac{\minus{}3}{8}$
adding we get,

$ \boxed{\sum_{k \equal{} 0}^{\infty}\frac{(\minus{}1)^{k}(2k\plus{}1)}{3^{k}}\equal{}\frac{3}{8}}$

EDIT:$ 4\rightarrow 2$
due to [b]t0rajir0u[/b] thanks!

I think the summation is in fact $ \sum_{k\equal{}0}^{\infty}\frac{(\minus{}1)^{k}(2k\plus{}1)}{3^{k}}$.  Of course, the process is identical.

thanks, do you know where I can find some of these identities?

A lot of these can be found in 104 Number theory Problems from the Preparation of the USA IMO team.


Let $ \{a_n\}$ be sequence of positive numbers and the sequence $ \{x_n\}$ is defined by:
$ x_1\equal{}1,x_2\equal{}2$ and $ x_{n\plus{}2}\equal{}\minus{}x_n\minus{}a_n x_{n\plus{}1}$ for $ n\geq 1$.
Show that among the terms of $ x_n$ there exist infinitely many positive and infinitely many negative numbers.

[quote="Ibrogimov"]Let $ \{a_n\}$ be sequence of positive numbers and the sequence $ \{x_n\}$ is defined by:
$ x_1 \equal{} 1,x_2 \equal{} 2$ and $ x_{n \plus{} 2} \equal{} \minus{} x_n \minus{} a_n x_{n \plus{} 1}$ for $ n\geq 1$.
Show that among the terms of $ x_n$ there exist infinitely many positive and infinitely many negative numbers.[/quote]

$ 1)$ It is impossible to have two successive elements of  $ \{x_n\}$ equal to zeroes (else it would be easy to show $ x_2\equal{}0$)
$ 2)$ If $ x_n\equal{}0$, $ x_{n\plus{}1}\neq 0$ (see 1) and $ x_{n\plus{}1}$ and $ x_{n\plus{}2}$ are non-zero and has opposite signs
$ 3)$ If $ x_n\neq 0$ and $ x_{n\plus{}1}\equal{}0$, $ x_n$ and $ x_{n\plus{}2}$ are non-zero and has opposite signs
$ 4)$ If $ x_n\neq 0$ and $ x_{n\plus{}1}\neq 0$ and $ x_n$ and $ x_{n\plus{}1}$ has opposite signs, then $ x_n$ and $ x_{n\plus{}1}$ are non-zero and has opposite signs
$ 5)$ If $ x_n\neq 0$ and $ x_{n\plus{}1}\neq 0$ and $ x_n$ and $ x_{n\plus{}1}$ has same signs, then $ x_n$ and $ x_{n\plus{}2}$ are non-zero and has opposite signs

So, in $ \{x_n,x_{n\plus{}1},x_{n\plus{}2}\}$, we always have at least two non-zero elements with opposite signs.

Q.E.D.


Determine $ Q \in \mathbb{R}$ which is so big that a sequence with non-negative reals elements $ a_1 ,a_2, \ldots$ which satisfies the following two conditions:

[b](i)[/b] $ \forall m,n \geq 1$ we have $ a_{m \plus{} n} \leq 2 \left(a_m \plus{} a_n \right)$
[b](ii)[/b] $ \forall k \geq 0$ we have $ a_{2^k} \leq \frac {1}{(k \plus{} 1)^{2008}}$

such that for each sequence element we have the inequality $ a_n \leq Q.$

A slight correction: It should be 2.008 instead of 2008 in the power.

But I think this is the number An will converge,the exact answer may be a little higher.Right?

This is also a special case of IMO 2007 shortlist problem A5 by Vjekoslav Kovač (Croatia):
http://www.artofproblemsolving.com/community/c6h214704p1187165

But can we figure out the smallest bound for An?


11 students scored 8 on the individual round - 10 from U.S. teams and one from an international team. I was trying to copy down the winners, but Paul D. was reading the names faster than I could write:

Here's what I have - can anyone fill it in?

1. ?, Montgomery A
2. ?, Eastern Mass.
3. ?, T. Jefferson
4. Jeff Manning, Southern California A
5. ?, Vermont
6. Zhuo Ming "Aubrey" Yang, Southern California A
7. ??
8. ??
9. Thomas Davids, Colorado Granite
10. ?, Vermont

The fourth person to get an 8 at the Las Vegas site was Yat Wui Cheung from Hong Kong.

In addition, the following students got 7's at Las Vegas:

Joshua Wang, San Diego Surf - the site winner
Edward Zu, Sam Diego Surf
Jonathan Hung, Washington Gold
Sam Keller, Washington Gold (who also played a key role in WWTBAM)
H.Y. Ku, Taiwan
Lee Chun Lu (?), Hong Kong
Michael ?, Hong Kong
Patrick Chan, Southern California A
Zhifan "Ivan" Zhang, Southern California A
Jiunn Song, Southern California A
David Yang, Southern California B (a 6th grader)

[color=green][Added in edit: deferred to the spelling of the Hong Kong participant from gtf's post below; identified two people as being from San Diego.[/color]

Allen Huang from Michigan reels was either 8th or 7th and then Kevin Chen from Texas Gold was the other, 7th or 8th

There were 11 sevens from Iowa and Jeffrey Chu was the site winner, and I'm pretty sure I got second in the tiebreaker to him

i believe

1st was Qinxuan Pan. (me thinks)
3rd was Daniel Li

I have a scribble next to #8 that might be "Kevin" - so that matches what nat mc said about Kevin Chen of Texas being 8th.

I [i]think[/i] the two people from Vermont were David Rolnick and Colin Sandon, in some order (though I'm not sure of that).

right, Kevin Chen = 8th and Allen Yuan = 7th.

That matches another scribble I have: David Rolnick was 10th.

So far, it looks like we have:

1. Qinxuan Pan, Montgomery A
2. Vlad Firoiu, Eastern Mass A
3. Daniel Li, TJA
4. Jeff Manning, SoCal A
5. Colin Sandon, VT
6. Aubrey Yang, SoCal A
7. Allen Yuan, Mich A
8. Kevin Chen, Texas Gold
9. Thomas Davids, Col. Granite
10. David Rolnick, VT

[quote="Fanatic"]Correction to above -- it was Allen Yuan who got 7th I believe, not Allen Huang, although I can't blame you... several people around me thought the announcer was announcing the same person twice when he called up "Allen Yuan" and "Allen Huang" to the tiebreakers.[/quote]

i believe that it is Allen Yuan and [b]Alan[/b] Huang, not Allen Huang, but whatever.

2. Vlad Firoiu, Eastern Mass A

The rest are correct except I have Yat Wui Cheung from Hong Kong listed as 8th. I assume that he is not listed since he is on an international team.

Site winners were:
Las Vegas: Joshua Wang, San Diego
Iowa: Jeffrey Chen, Texas Gold
Georgia: John Berman, North Carolina A
Penn State: Meng Li, Eastern Mass E

All received 7's

I wish they would've said the times on the tiebreakers, I would've liked to have known how close I was to site winner and the times of those with 8s

Somebody elsewhere mentioned that one of the site winners won by virtue of tiebreaker #1, possibly making him the [i]only[/i] person to solve it correctly.

I was monitoring the tiebreakers at UNLV and saw many, many answers that were oh-so-close to being correct  :P

Among those with 8's, 4th place was 13:52 and 6th place 15:40.

Allen Yuan was sitting next to me and he missed 1 and 2 and then got the third in about 3 minutes, so his time was like 18:40, and he was 7th.  My time was around 17 minutes. I missed 1 and got 2 quickly (fastest at the site) but three took me almost 5 minutes which sucked

EDIT: Now I get how it works :) And actually, Allen got it right just at the end, so his time then was about 16:00, since the third tiebreaker doesn't matter. My time then was about 12:40, since the third isn't counted after I got the second one right, and I guess that wasn't fastest at our site among the 7s because Jeffrey was the site winner. Kinda cool that I beat the time of at least 7 of the 10 people with 8s

Also, what did everyone else get for getting a 7. All I got was the choice of a frisbee and a book (I took the firsbee), but then the 6s on our team got medals for being the team high score, since they weren't "recognized nationally". So getting a frisbee is better than a medal....?

[quote="nat mc"]So getting a frisbee is better than a medal....?[/quote]

For sure. Frisbees are practical.

[quote="nat mc"]Allen Yuan was sitting next to me and he missed 1 and 2 and then got the third in about 2:40, so his time was like 18:40, and he was 7th.  My time was around 17 minutes. I missed 1 and got 2 quickly (fastest at the site) but three took me almost 5 minutes which sucked

Also, what did everyone else get for getting a 7. All I got was the choice of a frisbee and a book (I took the firsbee), but then the 6s on our team got medals for being the team high score, since they weren't "recognized nationally". So getting a frisbee is better than a medal....?[/quote]

Lol...I actually got #3 wrong, which was embarrassing; I still don't understand how the tiebreaker works - do they add up your times for ALL the problems?  Because I thought it was something to do with which problem you get right first so since they announced my name for getting #2 right, I just subconsciously shut off my mind for #3 (yes, I used coordinates).

I think Kevin Chen was #9 because I remember one person getting announced between me and him.

First person to solve a problem wins the tiebreaker. If you solved problem #2 in one minute, it took you eleven minutes overall to solve [i]a[/i] problem.

The whole adding-times business is done so that a person who solves tiebreaker #1 in 9:59 is ranked above ALL people who miss tiebreaker #1 (regardless of how quickly they may solve tiebreaker #2).

Ok, so what happens if two people miss the first tiebreaker. Then they both get the second tiebreaker, but person A gets it 1 minute faster than person B. Then, they both get the third tiebreaker, but person B is 2 minutes faster than person A. Who wins?

nat mc, the event you describe ends with the second tiebreaker. Neither of those people should do the third tiebreaker. What kyuanmathcounts describes sounds like a minor (and correctable) error in administering the round.

[quote="kyyuanmathcount"]I think Kevin Chen was #9 because I remember one person getting announced between me and him.[/quote]

Hey, I was at Iowa and I thought so too...

Ah thanks for posting these; at the Georgia site, the awards were just skimmed over, since so few people there won. They didn't even say the individual winners; felt like any other little contest in the south, weather included.

In Georgia, there were no 8s and just four 7s:

John Berman (North Carolina)
Jeremy Hahn (North Carolina)
Marcel Nunez (Florida)
Oliver Wang??? (Georgia)

We got a fairly easy substitute tiebreaker problem.

[quote="generating"][quote="pythag011"][quote="gtf"]It wasn't a substitute. It was the 2nd tiebreaker. We were told (by the author of the tiebreakers I believe) that since we were just looking for the site winner, the 2nd question would be better. I think they were pretty sure that the 1st question was near impossible.

Tom[/quote]

It was near impossible ( about 1 or 2 people in the whole country got it right.)[/quote]

What, can't people draw perfectly proportioned triangles and straight lines freehand anymore?  :P[/quote]
hehe i hear a couple people on our b team counted the triangles right

[quote="jb05"]Oliver Wang??? (Georgia)[/quote]

Huang  :)

Where can I get a solution to tiebreaker 1? I got it from my seat, but I'm not sure that my solution was legit...

I believe the 7 is Patrick Chan, not Patrick Daly.
Guess the first name got caught right, but not the last.

[color=green][Oops! I'll go back and fix that. - K.M.][/color]

[quote="Allan Z"][quote="E^(pi*i)=-1"][quote="generating"]Somebody elsewhere mentioned that one of the site winners won by virtue of tiebreaker #1, possibly making him the [i]only[/i] person to solve it correctly.[/quote]

That was Meng Li at Penn State, who is actually a girl.

The rest of us 7's were pretty mad at her (although I wouldn't have gotten #2 in the time limit anyway).[/quote]

Lol, I was like "yes" until someone announced that someone from the back got 91..  The second tiebreaker was pretty straight forward and quick if you draw the graph =/[/quote]

She goes to my school. Apparently, she just drew out the triangle accurately and counted the regions  :D

Wait, so were there some number above but close to 10 8's, and the site winners the people that didn't make the top 10 but were the best in their site?

I think there were exactly 10 8's, not counting internationals.  The site winners were the 7's at each site who won the tiebreaker.

That 11 includes one from Hong Kong, who was ineligible for the main awards.

Wasn't the person from Hong Kong the fastest tiebreaker, with ~45 seconds? That was the only indiv. result mentioned at the GA site.

[quote="JAM"]Um, no he wasn't :maybe: . If I remember correctly, he missed the first two tiebreaker questions and ended up being third at Las Vegas.[/quote]
I'm pretty sure he was behind the 7th place official finisher and ahead of the 8th place official finisher - which would make him 3rd at Las Vegas, behind Jeff and Aubrey and ahead of Thomas. There's some confusion above about whether Kevin Chen was 8th or 9th: that would be 8th among the U.S. official list, 9th if you include everyone.

[quote="nat mc"][quote="rrusczyk"]First problem you answered correct was all that mattered.  Not sure why they still had you up there after getting #2...[/quote]
Hmm.. if that's right, then a lot of people must not have seen Jeffrey turn in his answer for #2.

Everyone at our site missed #1, then for number two, I got it right in about 2 and a half minutes and my coach, my parents and some people from my team all thought that I turned my answer in second to Neil Gurram, who was sitting next to me, and before everyone else and he got it wrong. So I thought I was the fastest on #2, but they announced Jeffrey as the site winner

Jeffrey had to have turned it in before me, or else there'd be a mistake. I guess we all just missed it

Also, now that I know how it works: yeah I forgot that Allen got the second one right at about 6 minutes so his time then was about 16:00, since the third tiebreaker doesn't matter. My time then was about 12:40, Kinda cool that I beat the time of at least 7 of the 10 people with 8s[/quote]

Yeah, well, I did turn it in like a whole minute before you, so anything could have happened.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group and let $ H,K<G$  be s.t. $ \gcd(|H|,[G: K])\equal{}1$ and one of them is a normal subgroup of $ G$.
Prove that $ H<K$ or $ K<H$.

we can derive $ H \subseteq K$ in each case.

let $ K$ be normal, and $ h \in H$. if $ p : G \to G/K$ denotes the projection, then $ ord(p(h))$ divides $ ord(h)$ and $ |G/K|$, so that this is a common divisor of $ |H|$ and $ |G/K|$. thus $ p(h)\equal{}1$, i.e. $ h \in K$. this shows $ H \subseteq K$.

now let $ H$ be normal. then $ HK$ is a subgroup of $ G$ where $ |HK|$ divides $ |H| |K|$ and $ |G|$, i.e. $ |HK : K|$ divides $ |H|$ and $ [G: K]$. thus $ [HK : K]\equal{}1$, i.,e. $ H \subseteq K$.

[b]-oo-[/b], you are the best, man! Keep up the good work!


geometry
3D geometry
algebra unsolved
algebra

Two numbers are such that the sum of their cubes is 5 and the sum of their squares is 3. Find the sum of the numbers.

No matter I tried it from whatever method, I always got x+y=5/(3-x.y). Anybody, who can tell how to get the product of two numbers?

Let $ A\equal{}x\plus{}y$ and $ B\equal{}xy$
$ A^3\equal{}x^3\plus{}y^3\plus{}3xy(x\plus{}y)\equal{}5\plus{}3AB$
$ A^2\equal{}x^2\plus{}y^2\plus{}2xy\equal{}3\plus{}2B \Rightarrow A^3\equal{}3A\plus{}2AB$
$ 6AB\equal{}2A^3\minus{}10\equal{}3A^3\minus{}9A$
$ \Rightarrow A^3\minus{}9A\plus{}10\equal{}0$
$ A\equal{}2, \minus{}1\plus{}\sqrt{6},\minus{}1\minus{}\sqrt{6}$


If $n>m$, prove:
\[ \sqrt[m]{x^m+y^m} > \sqrt[n]{x^n+y^n}  \]
Edit: $x,y \geq 0$.

It seems that $x,y$ must be positive, since $\sqrt[3]{2^3+(-1)^3}\ngtr \sqrt[4]{2^4+(-1)^4}$, and for $y=0$ the inequality obviously doesn't hold.


[hide]The number's actually didn't matter since it's in the form
$x^{2}-(x+1)(x-1)$

Anytime you see this, the value of the expression is $1$. (Multiply the (x+1)(x-1) part out)[/hide]


$ x$ and $ y$ are two given reals.
$ n$ an integer :$ n>\equal{}0$
If $ cos(2^nx)>cos(2^ny)$ for all $ n$ then $ x\equal{}2k\pi$

Put $ U_n \equal{} cos(2^nx) ,V_n \equal{} cos(2^ny)$=>$ U_{n \plus{} 1} \equal{} 2U^2_n \minus{} 1;V_{n \plus{} 1} \equal{} 2V^2_n \minus{} 1$=>$ U_{n \plus{} 1} \minus{} V_{n \plus{} 1} \equal{} (U_n \minus{} V_n)(U_n \plus{} V_n) > 0$=>$ U_n \plus{} V_n > 0$=>$ U_n > 0$ for all$ n$ =>$ U_{n \plus{} 1} \equal{} 2U^2_n \minus{} 1 > 0$=> $ U_n > \frac {1}{\sqrt {2}}$ for all $ n$.
Now it's easy to continu


modular arithmetic
quadratics
number theory unsolved
number theory

Can you find an $ n$ such that $ n^{2}\equiv \minus{}1\pmod{p}$ when $ p\equal{}3$?  when $ p\equal{}5$?  Can you characterize the primes for which this congruence holds?

[hide="Some Solutions"]
When $ p\equal{}5$, we can choose $ n\equal{}2,3,7$ and so on.  When $ p\equal{}13$, we can choose $ n\equal{}5$.  When $ p\equal{}17$, we can choose $ n\equal{}4$.  When $ p\equal{}29$, we can choose $ n\equal{}12$.[/hide]

The question$ \Leftrightarrow \minus{}1$ is a quadratic residue $ mod \;\; p$  $ \Leftrightarrow (\minus{}1)^{\frac{p\minus{}1}{2}}\equal{}1(mod \;\; p) \Leftrightarrow p\equal{}1(mod \;\; 4)$

This is just the law of quadratic reciprocity supplement one.
If $ p$ is odd prime,
$ n^2\equiv \minus{}1\pmod{p}$
has a solution if and only if $ p\equiv 1\pmod{4}$.

The question of how one actually goes about computing $ n$ is interesting.  If you've got your hands on a generator $ g$, you can just take $ g^{\frac {p \minus{} 1}{2} }$, but it's not practical to go so far as to compute a generator.  One way to proceed is to run the first part of the [url=http://en.wikipedia.org/wiki/Miller-Rabin_primality_test]Miller-Rabin test[/url].


[quote="slogger"]since i seem to have become active here  :lol:  , i think this forum deserves my stupid qns as well   :D

1st ->
What role does ether play in[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/d/f/6/df631c4b14c8796f87204a6b09a265bb46925492.gif[/img]
(during reaction of phenol with SOCl_2 , i have this doubt because ether doesn't seem to come anywhere in the mechanism :D )

2nd->
Whats the main motive of ring contraction ,
in better words , what drives a 6 membered ring to 5 membered ring ?[/quote]

First of all, dont act too humble da!  :D

1) And what do you mean by $ SN_{i}^2$ reaction  :P
(am asking this because both $ SN_i$ and $ SN_2$ mechanisms are possible with $ SOCl_2$  :wink: )
2) the stability of the resulting carbocation?  :D
( like in pinacol-pinacolone and benzil-benzilic acid rearrangement where the resulting carbocation is stabilised by the +R effect of oxygen  :) )

[quote="Da Vinci"]

First of all, dont act too humble da!  :D

1) And what do you mean by $ SN_{i}^2$ reaction  :P
(am asking this because both $ SN_i$ and $ SN_2$ mechanisms are possible with $ SOCl_2$  :wink: )
[/quote]

arey , if reagent is ether it is SN i no?
 else when its pyridine , its SN 2 right ?

in 2nd qn , i encountered a carbocation in which ,->
consider a cyclohexane ring with positive charge at one carbon , and to the adjacent carbon  0H  is attached .
Now why does the carbon next adjacent to OH attach to the original carbon containing positive charge.

i dont get ur 1st qn
as for 2 its a plain experimental fact...but te reasoning being in the 5 ringed stucture it;s no more the carbon that has the positive charge rather it is the oxygen that bears the brunt and in that structure since each atom is able to fulfill it's octet it is considerably stable(check out morrison boyd solns where in one qn which i think was why in a certain reaction methyl alcohol is 2000 times more reactive that methane the explaination being all atoms had acheived octet)

[quote="slogger"]
in 2nd qn , i encountered a carbocation in which ,->
consider a cyclohexane ring with positive charge at one carbon , and to the adjacent carbon  0H  is attached .
Now why does the carbon next adjacent to OH attach to the original carbon containing positive charge.[/quote]

This is because of back-bonding. Sigma electrons of oxygen are delocalized to empty carbocation orbital, leading to a resonating structure which is highly stable, as resonance from adjacent positions is far more more effective than from separated positions, as in the case of an enolate, say.(moreover, in an enolate, pi electrons are delocalized.)

[quote="slogger"]
Why is ether used in alkyl chloride formation?[/quote]

Ethers, as is their most fundamental and common purpose in most reactions, is to behave as a non-polar solvent. Alkyl chlorides, furthermore, are readily soluble in ethers, It provides an excellent base for the reaction.

 :)


In how many ways you can express $ 2^{n}$ as a sum of perfect squares?  :)

[hide]Let $ 2^n\equal{}x_0^2\plus{}y_0^2$.  If $ n<2$ leave the equation as is.  If not, then modulo $ 4$ the equation is $ 0\equiv x_0^2 \plus{}y_0^2\pmod{4}$.  Since squares are either $ 0$ or $ 1 \pmod{4}$ then we find that both $ x_0$ and $ y_0$ must be even.  Let $ x_0\equal{}2x_1$ and $ y_0\equal{}2y_1$.  Then the equation becomes $ 2^{n\minus{}2}\equal{}x_1^2\plus{}y_1^2$.
Repeat this process until $ n<2$.  When we are done we will have one of these equations:

$ 1\equal{}x_k^2\plus{}y_k^2$ where $ k\equal{}\frac{n}{2}$ if $ n\equiv0\pmod{2}$

$ 2\equal{}x_k^2\plus{}y_k^2$ where $ k\equal{}\frac{n\minus{}1}{2}$ if $ n\equiv1\pmod{2}$

For the first equation, the only solution is one of $ x_k,y_k$ is $ 0$ while the other is $ 1$.  Thus, $ 2^n\equal{}0^2\plus{}(2^{\frac{n}{2}})^2$ is the only solution.  For the second equation, the only solution is $ x_k\equal{}y_k\equal{}1$.  Thus $ 2^n\equal{}(2^{\frac{n\minus{}1}{2}})^2\plus{}(2^{\frac{n\minus{}1}{2}})^2$ is the only solution.

In conclusion, there is only $ 1$ solution for all $ n$.[/hide]


inequalities
factorial
function
inequalities unsolved

Prove that for any non-negative integers $n$ and $p\geq 1$ and any non-negative real numbers $a,b$,

(i)
\[{1\over p^2}\sum_{j=0}^n{a^{j\over p}b^{n-j\over p}\over
\left(j\over p\right)!\left(n-j\over p\right)!}\leq
{(a+b)^n\over\left(n\over p\right)!}\]

(ii)
\[{1\over p}\sum_{j=0}^n{a^{j\over p}b^{n-j\over p}\over
\left(j\over p\right)!\left(n-j\over p\right)!}\leq
{(a+b)^n\over\left(n\over p\right)!}\]

(iii)
\[{1\over p^2}\sum_{j=0}^n{a^{j\over p}b^{n-j\over p}\over
\left(j\over p\right)!\left(n-j\over p\right)!}\leq
{(a+b)^{n\over p}\over\left(n\over p\right)!}\]

(iv)
\[{1\over p}\sum_{j=0}^n{a^{j\over p}b^{n-j\over p}\over
\left(j\over p\right)!\left(n-j\over p\right)!}\leq
{(a+b)^{n\over p}\over\left(n\over p\right)!}\]

P.S. Factorial of non-integer numbers is defined via gamma function: $x!=\Gamma(x+1)$.


Inside an angle with vertex $ P$ is given a point $ A$. Points $ X,Y$ lie on different rays of the angles so that $ PX\equal{}PY$ and the sum $ AX\plus{}AY$ is minimal. Prove that $ \angle XAP\equal{}\angle YAP$

Wowo, another problem from PMO (problem 2).
This one had many proofs,  by eliptic curves, by Ptolemy theorem, by Ptolemy inequality, by constructing symmetric segments etc. Just look into the link that I gave you for the inequality :wink:


hello guys...
im harold from manila, philippines.
im a computer science student.

i would like to ask help from u guys.
we our doing our thesis this semester.
we call our thesis "SP" (special problem).
its not a group project. we have to do it individually.
its not just paper work but mostly programming.
we are required to create a system (e.g. information system).
until now i havent found a good topic (and kinda feeling desperate already).
some of my classmates have proposed some topics but they have been rejected because the teachers thought they were "too easy". personally i dont really understand what they meant by "too easy".

maybe u could suggest a good topic for my SP.
im not really an expert programmer (meaning im not really that good in programming) :-)
maybe u could suggest a topic that is not too easy nor too hard.
im gonna be using mysql, apache, html and PHP. its ok if i also incorporate SMS and email features to my system, as long as its just PHP code.

i hope u could help me guys. im planning to finish my SP before october. so, there's only little time..
i would appreciate any suggestion.

u could post your reply here or u could email me at h_del_castillo@yahoo.com

tnx! :-)

what kind of topics did the teachers reject?

It would be easier if we know the difficulty level that teachers reject for us to come up with something.

Create a script that asks for the users City and fetches newsheadlines and weather information for that area.  PHP was the first language I learned and it has some really cool functioning to do this, such as CURL.

Are you required to propose a completely new idea or just some programming project?

What is your specialization? Are you doing your project under for example, DB topics or AI, IS etc...?


Is it true that if I randomly choose $ n\plus{}2$ integers from the set $ \{1,\ldots,2n\}$, there will be at least two consecutive ones, the first one of which is even. I don't know how to use the fact that if I choose $ n\plus{}1$ integers from the same set, there will be two consecutive ones.

I proved that among the $ n\plus{}2$ integers there will be three couples of consecutive integers, but don't know how to continue :blush:.

[hide]
Consider a different set of pigeonholes.

{1}
{2, 3}
{4, 5}
...
{2n-2, 2n-1}
{2n}[/hide]

[quote="Stewie"]Ok, so I just have to say that, since those sets are $ n\plus{}1$ and I have $ n\plus{}2$ integers, at least one of those subsets will be covered.[/quote]

Yes...moreover, there must be a 2-element set which is covered (i mean, this is obvious, but jsut to specify)

Yes, of course. Thank you very much :lol:


Determine the largest mathematically unreachable number of points scored by a team in
an (infinitely long) game. can anyone tell me the answer