Let $\epsilon\in(0,1)$. Prove that there are infinitely many integers $n$ for which $\cos(n)\ge{1-\epsilon}$

Let $\tau = \cos^{-1}{\1-\epsilon}$.

Then it suffices to show that there exist infinitely many integers $a_{i}$ in the intervals $\left(2k\pi-\tau, 2k\pi+\tau\right)$ which is a theorem of Jacobi that can easily be shown using the pigeonhole principle and letting $N > \frac{1}{\tau}$.

Sorry - should be $\tau = \cos^{-1}{\left(1-\epsilon\right)}$ (won't let me edit)

Teki-teki said, roughtly this:
Plot the cos(x) function.
Now plot the line y=1-e.
The two plots intersect at infinitely many points.
The distance between two points consecutive of intersection is fixed (the points should be such that for every point x between them cos(x)-(1-e)>0). And your question follows immediately form the irrationality of pi (the period between the intervals)

so basically you are using the fact that between any two numbers $a,b\in\mathbb{R}$ we know $\exists{c}\in\mathbb{Q}$ such that $a<c<b$


Επειδή δε βλέπω να μπαίνουν related θέματα από I.M.Os σκέφτηκα να βάλω εγώ ένα...

Καθορίστε τον ελάχιστο πραγματικό αριθμό $ M$ τέτοιον ώστε η ανισότητα $ \left| ab(a^{2} \minus{} b^{2}) \plus{} bc(b^{2} \minus{} c^{2}) \plus{} ca(c^{2} \minus{} a^{2})\right|\leq M(a^{2} \plus{} b^{2} \plus{} c^{2})^{2}$ να ισχύει για όλους τους πραγματικούς αριθμούς $ a,b,c$.

$ //S(cyclic)(ab(a^2 \minus{} b^2 )//$
$ < \equal{} // S(cylic){(a^2 \plus{} b^2)(a^2 \minus{} b^2)}/{2}//$
$ \equal{} //{(a^4 \minus{} b^4 \plus{} b^4 \minus{} c^4 \plus{} c^4 \minus{} a^4)}/{2}// \equal{} 0$
ara arkei na vrume to elaxisto M wste
$ M(a^2 \plus{} b^2 \plus{} c^2)^2 > \equal{} 0$pou profanws einai to $ M \equal{} 0$.
P.S. opou dio payles ennow to apolito. :oops:

[quote="Athinaios"]$ \left|\left|\sum_{cyc}ab(a^{2} - b^{2})\right|\leq \left|\sum_{cyc}\frac {\left(a^{2} - b^{2}\right)\left(a^{2} + b^{2}\right)}{2\right|} = \left|\sum_{cyc}\frac {a^{4} - b^{4}}{2}\right| = 0$ αρα αρκεί να βρούμε το ελάχιστο $ M$ τέτοιο ώστε $ M(a^{2} + b^{2} + c^{2})^{2}\geq 0\Longrightarrow M = 0$ [/quote]

Εάν θέσεις όμως στην αρχική ανισότητα όπου $ M = 0$ τότε θα έχουμε ότι $ \left| ab(a^{2} - b^{2}) + bc(b^{2} - c^{2}) + ca(c^{2} - a^{2})\right|\leq 0$ όπου εδώ τώρα ισχύει μόνο η ισότητα και όχι η ανίσωση. :!:

[hide]Η σωστή απάντηση είναι $ M = \frac {9}{32}\sqrt {2}$[/hide]

[quote="Γιώργος"]Επειδή δε βλέπω να μπαίνουν related θέματα από I.M.Os σκέφτηκα να βάλω εγώ ένα...

Καθορίστε τον ελάχιστο πραγματικό αριθμό $ M$ τέτοιον ώστε η ανισότητα $ \left| ab(a^{2} \minus{} b^{2}) \plus{} bc(b^{2} \minus{} c^{2}) \plus{} ca(c^{2} \plus{} a^{2})\right|\leq M(a^{2} \plus{} b^{2} \plus{} c^{2})^{2}$ να ισχύει για όλους τους πραγματικούς αριθμούς $ a,b,c$.[/quote]

Εχεις ενα μικρο typo... το σωστο ειναι

$ \left| ab(a^{2} \minus{} b^{2}) \plus{} bc(b^{2} \minus{} c^{2}) \plus{} ca(c^{2} \minus{} a^{2})\right|\leq M(a^{2} \plus{} b^{2} \plus{} c^{2})^{2}$

Τι έγινε ρε παιδιά? κανένας? :o

Γιώργο δεν ασχολήθηκε κάποιος με την άσκηση γιατί είχε συζητηθεί και παλαιότερα στο αγγλικό Forum (στο section με τις ανισότητες) αλλά και στο Ελληνικό. Σε παραπέμπω λοιπόν στις εξής σελίδες:

http://www.mathlinks.ro/Forum/viewtopic.php?p=571945#571945

καθώς και στην http://www.mathlinks.ro/Forum/viewtopic.php?t=101316 όπου ο Σιλουανός είχε βάλει μία πολύ καλή λύση.

Αλέξανδρος


For $-1\leq x\leq 1$ and $n\in\mathbb N$ define $T_{n}(x)=\frac{1}{2^{n}}[(x+\sqrt{1-x^{2}})^{n}+(x-\sqrt{1-x^{2}})^{n}]$.
a)Prove that $T_{n}$ is a monic polynomial of degree $n$ in $x$ and that the maximum value of $|T_{n}(x)|$ is $\frac{1}{2^{n-1}}$.
b)Suppose that $p(x)=x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}\in\mathbb{R}[x]$ is a monic polynomial of degree $n$ such that $p(x)>-\frac{1}{2^{n-1}}$ forall $x$, $-1\leq x\leq 1$. Prove that there exists $x_{0}$, $-1\leq x_{0}\leq 1$ such that $p(x_{0})\geq\frac{1}{2^{n-1}}$.

But then $x^{2}-1\leq 0$ with $-1\leq x\leq 1$.

Let $-1\le x\le 1$, then exist y, suth that $x=cos y, \ \sqrt{1-x^{2}}=siny$.
It give $T_{n}(x)=\frac{1}{2^{n}}[(cosy+isiny)^{n}+(cosy-isiny )^{n}]=\frac{cos(ny)}{2^{n-1}}.$ Therefore $T_{n}(x)=\frac{1}{2^{n-1}}cos(n arccos x).$


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations
number theory unsolved

Find all pair integer numbers $ (x,y)$ satisfy equation: $ x^3\minus{}y^3\equal{}2xy\plus{}8$

Multiply through by 27 and use $ a^3 \plus{} b^3 \plus{} c^3 \minus{}3abc \equal{} (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2 \minus{}ab\minus{}bc\minus{}ca)$

[quote="KevinB"]Multiply through by 27 and use $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$[/quote]

I don't suppose the result from multiplying by 27 would yield the form $ a^3\plus{}b^3\plus{}c^3\minus{}3abc$...

[quote="KevinB"]$ x^3 \minus{} y^3 \equal{} 2xy \plus{} 8 \implies x^3 \minus{} y^3 \minus{} 1 \minus{} 2xy \equal{} 7 \implies (3x)^3 \plus{} ( \minus{} 3y)^3 \plus{} ( \minus{} 3)^3 \minus{} 2(3x)( \minus{} 3y)( \minus{} 3) \equal{} 7\cdot3^3$[/quote] Sorry ignore this, its wrong.

Seems like you can limit the range of values for x,y by analyzing the difference of cubes on the left with the expression on the right.

I.e. Consider the case when the cubes on the left are consecutive, x=y+1
(y+1)^3-y^3 = 2y(y+1)+8

This has roots around y=-3.1926 and y=2.1926 which suggests for y>2.192 and y<-3.1926 the left side is greater than the right.

With that knowledge, you should only have to test y=2,1,0,-1,-2,-3 and evaluate in the original equation if x can be integer.

This occurs for
y=0, x=2
y=-2, x=0


I am finding many probems that are using some geometric terminology I do not yet know, especially related to triangles.  So I'm deciding to start a thread asking about the definition of these terms, such as: orthocenter, circumcenter, centroid, incenter,... there might be others I don't know.

Here are some of the common ones:

$\text{centroid}$ - $\text{where the three medians of a }\triangle\text{ meet}$
$\text{circumcenter}$ - $\text{ where the three }\perp\text{ bisectors of a }\triangle\text{ meet}$
$\text{incenter}$ - $\text{ where the three }\angle\text{ bisectors of a }\triangle\text{ meet}$
$\text{orthocenter}$ - $\text{ where the three altitudes of a }\triangle\text{ meet}$

Note that the orthocenter and the circumcenter may occur outside of the triangle.

Also note that the centroid is sometimes referred to the 'center of mass', I believe.

The incenter is also the center of the inscribed circle; similarly, the circumcenter is the center of the circumscribed circle.

[quote="mna851"]Does the perpendicular referred to in the definition of the circumcenter have to come from the opposite vertex.[/quote]
No.


probability
function
probability and stats

Suppose someone put popcorn in a microwave.
The minimum time to a popcorn explode is $ a$ and the maximum time is $ b$. This can be found empirically.

I wanted to build a density of probability function for this model. After $ t$ times what is the probability of a popcorn explode?

There is $ n$ ways to make this function. I thought of one that I think is simple.

You put the popcorn in the microwave and record the sound. :)

But the next step; how can I make this function knowing only the Sound?

I wanted to build a simple program to compute this function empirically. So this program computes the maximums of intensity of the sound and return the time of this peak.

And in the final I have the time of each explosion (to make it by hand would be problematical and imprecise). This is simple.

Knowing this final result, we can say that this is a function. At each time it return 1 for explosion and 0 for non-explosion.

But the Time is continuous but the number of explosions is discrete. This is my problem.

If someone has another method to find this density of probability, I would be happy to learn.

you should google "maximum likelihood". This is a usual way to deal with such kind of problems, once you know a little more what kind of density you are looking for.

Er... I saind I wanted to "convert" the sound in probability and asked a method to do this.

And I asked how to make a simple program that do that.

I had a idea of how to do this, but is fairly complicated.

On internet I doubt that there is something like what I wanted.

If I had the DATA I wouldn't be asking here.


[b]If $ A,B,C$ are the angles of any triangle, prove that
$ 3(\sin^2A\plus{}\sin^2B\plus{}\sin^2C)\minus{}2(\cos^3A\plus{}\cos^3B\plus{}\cos^3C)\leq 6$[/b]

^The LHS will be maximized when $ A\equal{}B\equal{}C$?


calculus
integration
function
trigonometry
calculus computations

$ g(x)$ is a differentiable function for $ 0\leq x\leq \pi$ and $ g'(x)$ is a continuous function for $ 0\leq x\leq \pi$.
Let $ f(x) \equal{} g(x)\sin x.$ Find $ g(x)$ such that $ \int_0^{\pi} \{f(x)\}^2dx \equal{} \int_0^{\pi}\{f'(x)\}^2dx$.

Your answer is correct, Yuriy Solovyov. :)


Well, as the thread's title says, I'm planning to make the sixth installation of The Mole in AoPS!
Although my first attempt failed, I think that this time I'll make it better. I would like to rather have a co/backup/replacement mod, given that I'll be insanely busy due to university... offers to the charge can be PM'd to me.
About the setup... I've already made about what, 20% of the setup. So, if I'm lucky (and university isn't so much work, and I get a good backup mod), the plot will be complete before end of January, signups would start at early or mid-february, and the game would be underway around March.
So, comments come here, and co/backup mod offers go to my PM box.

[size=75]If any mod thinks this is spam, they can lock the thread if wished... this is just for information, and as I thought this would rather belong to FF, I posted here.[/size]

Judging by my game, the mole has become relatively unpopular now.

I'll play, but co-modding is out of my time... :\

I'll play, with a possibility of co-modding.

I'd like in.
No co-modding for me.
I'm lazy at the moment.

[quote="pacman2812"][b] I would like to rather have a co/backup/replacement mod, given that I'll be insanely busy due to university... offers to the charge can be PM'd to me.
 (and university isn't so much work,if I get a good backup mod), the plot will be complete before end of January, signups would start at early or mid-february, and the game would be underway around March.
So, comments come here, and co/backup mod offers go to my PM box. [/b][/quote]

I'll play, but can't mod due to creating a game that's also The Mole plus a bunch of other elements :O

I'll play, but please let it have an end :)

[quote="pacman2812"]Well, maybe given that the inmediately previous game imploded due to me losing internet access, that wasn't much popular.
And, if I'm not wrong, the Mole 5 exploded due to [i]you[/i] being inactive :wink:
And, after all, I posted this in order to get a co-mod (in case I don't get computer acces on time, or in case the university makes my brain explode due to excessive homework >.<)[/quote]

ME being inactive? I tried to restart it, prodded everyone at least 6 times, and waited for something good to happen. Which it didn't.

Maybe I can co-mod, I don't know yet.

well, it seems that I'll be OK with access from February-...
The problem is, the plot I have is HUGE, and the game would end around October or November :what?: :wacko:

@musicnmath  One way to find the rules would be to wait till the game starts, another is to study a previous game 2-3 game pages back.

Plot status: 55% complete.
Estimated start for signups: February 1.

Note: the signups are made by applications, via PM. At the date above, I'll create a thread where all the regulations are exposed, then you can start signing up.
Anyone interested should look at the prior The Mole games (preferred games 1-3, as 4-5 were abandoned)

Finally, a Mole game run by someone who will be active and knows how to make real games without throwing in random dumb twists b/c they got tired of modding!

Now how far are you into the storyline?

2 weeks ago=55%

just wondering.....

Around 90%. Only one task (the final task), details on other tasks, some quizzes and some flavor to go.
Whenever I get a computer (currently from internet cafe), I'll start the applications.
I'm still accepting offers for co-mod.


For distinct natural numbers $a,b$, prove that $gcd(a,b) \leq \frac{a+b}{3}$

Ooh...I'm not good with these kind of problems...can someone please help me with these.... :blush:

We know that if $a>b$ then the gcd. is not more than $\frac{a}{2}$

I sort of have an idea, but I'm not sure if it's rigorous enough. It follows what you said. I'm sorry if it's too wordy or if it's wrong. I'm not really good at proofs

[hide]
I thought that since the numbers are distinct, we can assume without loss of generality that a>b. The maximum the gcd could possibly be would then be b. But any common denominator would be a factor of both numbers, so if b is a factor of a then so is a/b. Since factors must be integers, the smallest a/b can be is 2 (1 wouldn't work because then a=b), which would give us the maximum value of b as a/2. So gcd(a,b) has to be less than or equal to a/2, or (a+b)/3.
[/hide]

Ewww.... I definitely need latex.

hopefully i got it...
[hide]we can let $g=\gcd(a,b)$ and so we can write $a=m*g$ and $b=n*g$.
im going to pick $a<b$.
it is easy to see $g\leq a$.
but $a,b$ are distinct, and so $2\leq n$ and $2g\leq ng=b$.
add them together and get $3g\leq a+b$ and we done.[/hide]

Yeah, that makes a load of sense. And it's less wordy compared to mine  :blush:


1 from a faculty of six professors,six associate profresors,ten assistant professors,and
twelve instructors, a commitee of size six is formed randomly.
what is the  probability that there is at least one person from each rank on the committee?

2from the set of integers {1,2,3,....100000} a number is selected at random
what is the probability that the sum of its digits is 8?

3 the chairperson if the industry academic partnership of a town invites all 12 members of the board and their
spouses to his house for a christmas party.If a board member may attend without his spouse , but not vice
versa, how many different groups can the chairperson get?

4 bill and jogn play in a background tournament. A player is the winner if he wins three games in a row or
four games altogether.In what percent of all possible cases does the tournmant end because john wins four games without winning three in a row?

1.
[hide]There are 34 people in total, so there are 34 choose 6 ways to choose a committee of 6.  There are 6 ways to choose one professor, 6 for one associate, 10 for one assistant, and 12 for one instructor, and 30 choose 2 ways to pick the remaining 2 people for the committee.  So there are 6*6*10*12*30C2/34C6 ways to choose such a committee. [/hide]
3.
[hide]Either a board member may come with their spouse, or the spouse may be left behind, so there are 2 possibilities for each board member, and since there are 12 board members, there are 2^12 different groups that can come.  If in the problem it was meant that a board member may choose to not come, then this would be a third possibility, and there would be 3^12 different groups that may come.[/hide]

2.  [hide="This answer is probably wrong"] I get 87/20000, but that doesn't sound right. [/hide]

4.  [hide="Another guess"] 26%? [/hide]

Do you know the answers to these?


algebra
function
domain
factorial
calculus
integration
Ring Theory

1. Is being a principal ideal ring a local property?

2. Prove that every finite ring is isomorphic to a product of local rings.

1. What do you mean by local property?
Is it: A property $P$ of rings is local if:
$R$ has $P$ if and only if $R_{q}$ has $P$ for all primes $q\in Spec R$.

If this is the case, then the answer is no:
A Dedekind domain that is not factorial has the property that all localizations at primes are Discrete Valuation Rings, hence Principal Ideal Domains, but it is not itself Principal.

2. Because of the title, I assume that we are in the commutative world.
Let $R$ be a finite commutative ring  with $1$.
Every prime ideal of $R$ is maximal because $R/p$ is a finite integral domain, hence also a field.
There are only finitely many prime ideals in $R$ because $R$ is finte. Let these ideals be $p_{1},\ldots, p_{k}$, and let $I=\prod_{i=1}^{k}p_{i}$. Then $I$ is contained in all the prime ideals of $R$, hence it consists of nilpotents. Since it is finite, there exists $n\in\mathbb N$ such that $I^{n}=0$.
Choose $n_{1},\ldots,n_{k}$ all minimal such that $\prod_{i=1}^{k}p_{i}^{n_{i}}=0$.
It is easy to see that all $p_{i}^{n_{i}}$ are comaximal (if a prime $q$ contains $p_{i}^{n_{i}}+p_{j}^{n^{j}}$, then it contains $p_{i}^{n_{i}}$ and $p_{j}^{n_{j}}$. Because it is a radical ideal, it contains $p_{i}$ and $p_{j}$, hence their sum also. But these are distinct maximal ideals, so their sum is the unit ideal).
It follows by the Chinese Remainder Theorem that \[R=R/0=R/\prod_{i=1}^{k}p_{i}^{n_{i}}\simeq\prod_{i=1}^{k}R/p_{i}^{n^{i}}.\]
$R/m^{n}$ is a local ring for any maximal ideal $m$ of $R$, because prime ideals of $R/m^{n}$ correspond to prime ideals of $R$ that contain $m^{n}$ which are the prime ideals of $R$ that contain $m$. Since $m$ is maximal, there aren't many of these.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a group of order $2m$ for some positive integer $m$, and let $H$ be the subset of $G$ consisting of those $x$ for which $x^m=1$.

Prove that $H$ is a subroup of $G$, and either $H=G$ or $\left[G: H\right]=2$.

By a theorem I've proved elsewhere on this forum, the number of elements of order dividing $m$ is divisible by $m$. If there are $2m$, we are done. If there are $m$, we need only to prove that $H$ is a subgroup.

I'll come back later- I don't have the time for this right now.

That theorem needs some more conditions.  For example, when $m=2$, the cyclic group $C_4$ has three elements of order dividing 2.

It doesn't need more conditions; the only condition is that $m$ divides the order of the group. My proof is in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22490]this thread[/url].

Your example is simply wrong; there are only two such elements in $C_4$.

Consider G acting on G by left multiplication, this gives a representation of G as a set of permutations on n elements. (n is the order of G).  It isn´t hard to see that H is exactly the subgroup of even permutations. From where both things follow.


chand kore be sho'a'e 1 mitavan hamzaman bar yek kore ba hamin sho'a momas kard? :)


geometry limit logarithms integration ratio calculus derivative

$x=3$ is almost harmonic series and $x=-3$ is almost alternating harmonic series...

(Use \infty for that upper limit)

You have found that the radius of convergence is 3. This means that the series converges absolutely when $|x|<3$ by comparison to a geometric series, and diverges when $|x|>3$ since the terms tend to $\infty$. The only question that remains is whether the series converges or diverges at each point $x$ with $|x|=3$.

If $x$ is a real variable, the only points to check are $x=3$ and $x=-3$. When $x=3$, the series becomes $\sum_{n=1}^\infty \frac1{3n}$, which is a multiple of the harmonic series and divergent. When $x=-3$, the series becomes $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{3n}$ which is a conditionally convergent alternating series.

If $x$ is a complex variable, we need more complicated methods to deal with the rest of the circle. The series does converge for all $x$ with $|x|=3$ except $3$ itself, and we can calculate the exact values.

oké I see but for calculating the convergentia area I use always x=1 is this oké? why can I don't use for example 2 or 4 ect...?

[quote](Use \infty for that upper limit)
[/quote]

What is the different?

Greets thanks a lot.

the difference is that $\propto$ is a completely different sign than $\infty$, it's like using $\lhd$ for $<$ because it looks like it...

But I don't understand your question on $|x|=1$... on what mathematical grounds would you [i]choose[/i] the convergence radius? :? Usually it has to follow from the problem, and you don't have the freedom to 'take' it $|x|=1$ or something.

[quote]But I don't understand your question on ... on what mathematical grounds would you choose the convergence radius?  Usually it has to follow from the problem, and you don't have the freedom to 'take' it  or something.
[/quote]

I know but the question is just calculate the convergence radius I don't know who to do that if I use x=1 then I find the rigth answer but I don't know why I need to choice x=1?

Greets.

I don't understand what you mean by "if I use x=1 then I find the rigth answer"? :?

first of all I need to find the convergence radius but I don't know who so  I go look into my solutions in the back of my book and then I see that I do the next calculation that I get 3 as the answer.

So $\sum^{%Error. "intify" is a bad command.
}_{n=1}\frac{x^{n-1}}{n3^n}$

then I put 1 in my exprsession so $\sum^{%Error. "intify" is a bad command.
}_{n=1}\frac{1^{n-1}}{ n3^n}$ and if n become infinite then we get ooh sorry I can't get the right value

But can someone tell me who I need to do that? Greets.

[quote]In fact, I still can't follow you... what's the problem with jmerry's solution?
[/quote]

Nothing but I can not find  the convergention radius on a correct way that's the problem.

Let $S_n(p)=\sum_{k=1}^n \frac{1}{k\cdot p^k}\ (p>2),$ we have $\lim_{n\to\infty} S_n(p)=\ln \frac{p}{p-1}.$

Here is my proof, but this may be a famous.

For $x\neq 1,$ let $\ f_n(x)=\sum_{k=1}^n x^{k-1},$ we have $\frac{1}{1-x}-f_n(x)=\frac{x^n}{1-x}.$ Thus for $0\leq x\leq \frac{1}{p}\ (p>2),$ we have $1\leq \frac{1}{1-x}\leq \frac{p}{p-1},$ yielding
$0\leq \frac{x^n}{1-x}\leq \frac{p}{p-1}x^n.$ Integrate with respect to $x$ for the interval $0\leq x\leq \frac{1}{p},$ we have $0<\int_0^{\frac{1}{p}} \left\{\frac{1}{1-x}-f_n(x)\right\}dx<\frac{p}{p-1}\int_0^{\frac{1}{p}} x^ndx.$

$\therefore 0<\ln \frac{p}{p-1}-\sum_{k=1}^n \frac{1}{k\cdot p^k}<\frac{p}{p-1}\cdot \frac{1}{(n+1)p^{n+1}}.$

Let $n$ approach infinity, $\lim_{n\to\infty} \left(\ln \frac{p}{p-1}-\sum_{k=1}^n \frac{1}{k\cdot p^k}\right)=0.$

Therefore $\sum_{n=1}^{\infty} \frac{1}{n\cdot p^n}=\ln \frac{p}{p-1}.$

briljant proof but can I stil find the convergention radius?

[quote="Bert"]Hello,

Given is $\sum^{\propto}_{n=1}\frac{x^{n-1}}{n3^n}$ I can find that the convergention region is 3 but what can I say about the points 3 and -3 ?

Greets.[/quote]

  You can try ratio test and find that the radius of convergence is 3. About the end point, jmeery has answered you already.
  Ratio test stated that : If $lim_{n \to \infty } abs ({\frac{a_{n+1}}{a_n})=L <1}$ then the series $\sum_{n=1}^{\infty } a_n$ is absolutely converge and therefore it is converge.

  BTW, Kunny sols for $x=1$ is very nice.

In fact, it can be done much easier than Kunny's proof...

Termswise differentiation gives $\sum -\frac{1}{p^{k+1}} = -\frac{1}{p(p-1)}$, integrating again gives $\ln(p) - \ln(p-1)$, which is the same. ;)

Just as trivial we have that the original sum is $x^{n-1} \ln\left(\frac{p}{p-1}\right)$.

About finding that radius of convergence:

The ratio test and/or the root test compare the series to a geometric series. In practice, they are used to find what it would take to balance out everything that grows or shrinks geometrically fast or faster, while rendering invisible everything that grows or shrinks slower than geometrically fast.

The powers of the variable in a power series are always geometric, which is why the root and/or ratio tests are always in order. On the other hand, once you're used to this, there are many cases (including this one) in which you can determine the radius of convergence "by inspection."

Look at what you have: you have $x^{n-1},$ which is geometric, but whose exact behavior depends on $x.$ You have $\frac1{3^n}$ which is geometric - and you can see that setting $|x|=3$ would exactly balance it out. And you have $\frac1n$ which shrinks and a slower-than-geometric race, and hence has no influence on the radius of convergence. If you go ahead and use the root or ratio test, then the $\frac1n$ will contribute a factor of $1$ to the limiting root or limiting ratio - that's what I meant above by being "rendered invisible." So, for the radius of convergence, once we see that $|x|=3$ will balance the geometric things we have and there's nothing else that matters, then the radius of convergence has to be $3.$

Determining the radius of convergence is quite different than determining the behavior at the boundary points. If $|x|=3$ then all the geomtric things balance out perfectly (at least in size) and all that's left is that "slower" object, the $\frac1n,$ along with whatever $\pm$ signs we have. Follow up on that thinking and you have jmerry's first post in this thread.

[Note: the limsup version of the root test is theoretically superior to any version of the ratio test - you will see it used to give a formula for the radius of convergence. That doesn't matter in this problem.]


vector
advanced fields
advanced fields unsolved

Hello folks! Here another one that I could not handle!
Thankl for any help!

Let $E$ and $F$ normed vector spaces. Suppose that $T : E \to  F$ is a closed linear mapping :

a) Let $K$ be a compact subset of $E$. SHow that $T(K)$ is a closed subset of $F$.

b) If $K$ is compact subset of $F$, show that $T^-{1}(K)$ (the inverse image) is a closed subset of $E$

a) Assume $(T(x_{n}))$ is a sequence of $T(K)$ converging to a point $e \in F$. By compacity of $K$, there is a subsequence $(x'_{n})$ of $(x_{n})$ which converges to a point $x' \in K$. But then $(x'_{n},T(x'_{n}))$ is converging to $(x',e)$, and since the graph of $T$ is closed, $e=T(x') \in T(K)$. This proves that $T(K)$ is closed in $F$.

b) Assume $(x_{n})$ is a sequence of $T^{-1}(K)$ converging to $x \in E$. Then $(T(x_{n}))$ is a sequence of $K$ from which you can extract a subsequence $(T(x'_{n}))$ converging to $y \in K$. Now since $T$ is closed, $(x'_{n},T(x'_{n}))$ converges to $(x,y)$ and $y=T(x)$, meaning $x \in T^{-1}(K)$. Whence $T^{-1}(K)$ is closed.


quadratics
LaTeX
algebra
quadratic formula
calculus
calculus computations

I have problems finding the limits of this fraction:

$f(x)=\frac{x^2+3x}{x+4}$

$x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$

$x=0;-3$

$f(x)=\frac{(x+0)(x+3)}{x+4}$

$f(x)=\frac{x(x+3)}{x+4}$

But my calculator doesn't agree. What am i doing wrong?

[u]EDIT:[/u] My calculator says there's one more around -3,9. How do i find that one?

(For my own usage: 322,1)

what do you mean with

EDIT: My calculator says there's one more around -3,9. How do i find that one?

Anyhow

if  x->-4(-) [respectively x->-4(+)]    then f->-inf  [respectively +inf]

if x -> +inf  [-inf] then f ->+ inf    [-inf]

Well, on my calculator it looked like it was $-3,9$, but i figured it was exactly $-4$.

Ok, so i have to take each factor i get:

$(x+0);(x+3);(x+4)$

and from there i can find the limits..

Of course, i knew that. How am i stupid :lol:

Welcome to the forum BTW

And here are two other ones. I only need help to find the full fraction from which i can do the factorizing.

1) $\frac{1}{x}+\frac{5}{x-2}=0$

2) $\frac{x+6}{x+1}=2$

I've read this earlier:

$f(x)=x+\frac{x-4}{x+2}$
$f(x)=\frac{x(x+2)}{x+2}+\frac{x-4}{x+2}$
$f(x)=\frac{x^2+3x-4}{x+2}$

I can follow the direction of the book, but now that we have a fraction + another fraction i get confused.

Thx in advance. BTW, thx to you, Pic, with the first problem.

[quote="Sporally"]I have problems finding the limits of this fraction:

$f(x)=\frac{x^2+3x}{x+4}$

$x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$

$x=0;-3$[/quote]

Hehe, you don't need the quadratic formula there :) Just take out a common factor of $x$ from $x^2+3x$, and you get $(x)(x+3)$, which is the same thing you had, but a thousand times easier. As my math teacher tells me almost daily, "Common factoring is the #1 skill in math."

Ah, of course. I have done that several times now, i don't know why i didn't do it this time though. Can you help me with my number 1 and 2 in the last post?

[quote="Sporally"]And here are two other ones. I only need help to find the full fraction from which i can do the factorizing.

1) $\frac{1}{x}+\frac{5}{x-2}=0$

2) $\frac{x+6}{x+1}=2$[/quote]
Basically, you need to take the common denominator of the two fractions, just like when you calculate $\frac{1}{2}+\frac{1}{3}$.

[hide="Number 1"]
$\frac{1}{x}+\frac{5}{x-2}=0$

$\frac{x-2}{x(x-2)}+\frac{5x}{x(x-2)}=0$

$\frac{x-2+5x}{x(x-2)}=0$

$\frac{6x-2}{x(x-2)}=0$[/hide]

For #2, move 2 (on the right side) to the left side of the equation and then take the common denominator.

I have a suggestion: Every time you don't know how to do a problem, try manipulating the equation a little and see what you can come up with. Just because it's not exactly like what you've seen before doesn't mean you can't figure it out.

I figured i had to find a common denominator, it was more like i didn't know how to get it like that. But it seems like i can do it now, thx to you. Regarding nr. 2 i knew that i could move 2 over to the other side as -2, but didn't know what to do from there.

I usally try and manipulating with the fractions, but just couldn't find a way that my calculator would show the graph the same way everytime.

BTW: Regarind LaTex, how do i make the +/- symbol so that i can write

$x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$

properly?

Yeah, got the same info from the LaTex-subforum. Found it just a little after i've posted the last post in this thread. But thx for the info anyway.

And just as you think you understand it all in this chapter you encounter another problem.

Ok, i have this fraction which i first of rewrite, from which i must find the numbers that can not be used (can't remember the english word for it, i hope you get it anyway once you see the fraction):

$\frac{1}{x}+\frac{5}{x-2}=0$

$\frac{6x-2}{x(x-2}=0$

If i want to find the numbers that i can't used for this fraction i have to find the numbers for x that makes the denominator equal 0 as you can't divide with 0. But on the other hand you used the denominator when you want to find the limits of the fraction, where you will see what numbers makes either the nominator and the denominator equal 0.

$x=0$

and

$x-2=0$

$x=2$

So these numbers are both used as the limits for the denominator and the numbers which can't be used due to the law that you can't divide by 0.

I believe i have misunderstood something about the law of dividing with 0, so can anyone guide me, plz?


From my homework, I have no idea:

Boron occurs naturally as 10B and 11B. Which isotope is most abundant? (Hint: Refer to the Periodic Table.)
A) boron-5  	B) boron-6  	C) boron-10  	D) boron-11  	E) none of the above

[quote="Gibbenergy"](Hint: Refer to the Periodic Table.)  [/quote]

Might not be a bad idea.

Look up the atomic weight of Boron.  Is it closer to $10$ or $11$?


I'm not sure if this problem is supposed to be stated in this subforum, but here it goes:

As you probably all know, exponential functions of the form $ a^{x}$ (whereby $ a$ is a constant), all have inverse functions of the form $ log_{a}(x)$. When graphed, it becomes evident that some of these functions and their inverses (those with large values of $ a$) don't intersect with one another, while others do so twice (with smaller $ a$ values);

Try getting a graphics calculator and graphing $ e^{x}$ and $ ln(x)$. It's evident that these don't intersect, and if you graph something like $ 1.1^{x}$ and $ log_{1.1}(x)$, you can see that they do twice.

Therefore, there must be a value for $ a$ in $ a^{x}$ which only intersects once with its inverse,  $ log_{a}(x)$ . The problem is, what is it?

[hide="a little calculus"]
At any place where the graphs of $ a^{x}$ and $ \log_{a}x$ intersect, $ x$ does too.  So we want the value of $ a$ such that $ a^{x}=x$ has a unique solution.
Notice that if this is the case, then at the solution $ x$ to this equation, the derivative wrt $ x$ of $ a^{x}$ must be $ 1$.  The derivative wrt $ x$ of this function is $ a^{x}\ln{a}$.  So we have

\[ a^{x}=x\]
\[ a^{x}\ln{a}=1\]

Okay, so clearly $ x=\frac{1}{\ln{a}}$.  So then

$ a^{\frac{1}{\ln{a}}}=\frac{1}{\ln{a}}$.
But $ a^{\frac{1}{\ln{a}}}=e^{\frac{\ln{a}}{\ln{a}}}=e$
$ \implies e=\frac{1}{\ln{a}}\implies a=\boxed{e^{\frac{1}{e}}}$[/hide]


Find every pair of positive real numbes $ (\alpha,\beta)$ such that for all sequence $ (u_n)$ which satisfies $ \exists K: |u_n|\leg K,\forall n$ and $ \alpha u_{n + 1} + \beta u_n \geq u_{n - 1}$ then $ (u_n)$ converges.

[quote]Find every pair of positive real numbes  such that for all sequence  which satisfies $ \exists K: |u_n|\leg K,\forall n$ and $ \alpha u_{n + 1} + \beta u_n \geq u_{n - 1}$ then  converges.
[/quote]
Do you mean $ K > 0$ and $ |u_n|\le K$?

If $ \alpha +\beta \ge 1$ exist sequence ${ u_{3n}=1,u_{3n+1}=a,u_{3n+2}=b, \{a,b\}\not=\{1,1}$, tytrefore nonconverged, satisfyed $ \alpha b +\beta a\ge 1,\alpha a+\beta \ge b, \alpha +\beta b\ge a$.
If $ \alpha +\beta <1$, then $ \alpha (u_{n+1}-u_n)+(\alpha+\beta )(u_n-u_{n-1})\ge (1-\alpha +\beta )u_{n-1}$ give, that $ u_n$ converged.


[quote="zaizai-hoang"]$ \bigstar$ [b]Problem:[/b]
Let $ a,b,c,d$ be positive real numbers. Prove that:
\[ \frac {a^3}{b^3 \plus{} a^2b} \plus{} \frac {b^3}{c^3 \plus{} b^2c} \plus{} \frac {c^3}{d^3 \plus{} c^2d} \plus{} \frac {d^3}{a^3 \plus{} d^2a}\ge 2
\]
[/quote]

Put $ x\equal{}\frac ba$ etc. We have to prove $ \sum_{cyc}\frac1{x^3\plus{}x}\ge2$ for $ xyzt\equal{}1$.

Now, $ \frac1{u^6\plus{}u^2}\plus{}\frac1{v^6\plus{}v^2}\ge\frac2{u^3v^3\plus{}uv}$ (easy by AM-GM), so we can prove it by mixing variables.  :lol:

Can you explain clearly ? :) The result of n-variables is also correct  :D

What is not clear? My proof applies to $ n$ variables, too. :)

$ LHS\equal{}\sum\frac{\frac{a}{b}(a^2b\plus{}b^3)\minus{}ab^2}{b^3\plus{}a^2b}\equal{}\sum\frac{a}{b}\minus{}\sum\frac{ab^2}{b^3\plus{}a^2b}\geq 2$
I think this proof can solve n-variables :)


geometry
inequalities unsolved
inequalities

:oops: a,b,c,A,B,C are nonnegative numbers satisfying a+A=b+B=c+C=k
So prove that aB+bC+cA$ \le k^2$

brian_gold, why do you need to post the same problem again?

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=219077[/url]

great,kunny!but use geometry is much nicer and I think have a ineq tronger than bri_an gold 's problem!
I'm trying it! :blush:


I'm a college freshman right now, trying to figure out what to do this summer. I've identified some REU's (NSF summer research programs) that I plan to apply for, but the problem is that many of them are geared towards more advanced students (sophomores or juniors), so I definitely need a backup plan. I am several years advanced in math courses, so it's not prereqs holding me back so much as

Long term, I'm basically interested in pure mathematics, most likely as a research mathematician. Barring that, I basically want to be thinking and solving problems. So short term I'm looking for something that will both allow me to solve problems, keep me from doing nothing all summer, and maybe give me some useful experience and money on the side.

Where would be a good place to look for interesting internships or other opportunities? I imagine I will need to start applying very soon.

I'm in a similar situation and would appreciate any advice.  Like eugevir, the internships I've seen are geared towards older students (sophomores, juniors), but I would like to get some experience this year.

REUs (NSF or otherwise) generally don't mind if you're a frosh as long as you've had the prereqs.  Don't let that hold you back---apply to at least 6 - 8 programs, though, to increase your chances of getting in.  (I'm saying that because my students have recently been getting into one or two of the 6 - 8 they apply for... they're often alternates at a couple other REUs.)

I worked for a professor the summer after my first year of college, doing very low-level research.  (I looked for papers in the library and ran data analyses against some of their theories.)  You might ask around to see if anyone at your institution has that kind of funding.


What number should be added to both the numerator and the denominator of $ \frac{1}{5}$ to get a fraction equivalent to $ \frac{4}{5}$?

$ \frac{4}{5}\equal{}\frac{8}{10},\frac{12}{15},\frac{16}{20}$

$ \frac{16\minus{}1}{20\minus{}5}\equal{}\frac{15}{15}$

If you add 15 to the numerator and denominator of $ \frac{1}{5},$ you get $ \frac{16}{20}\equal{}\frac{4}{5}$

instead of guessing and checking, which may take a while if the numbers were larger, we use an equation. Let $ x\equal{}\text{the number added}.$ We have \[ \frac{1\plus{}x}{5\plus{}x}\equal{}\frac{4}{5}\implies 5x\plus{}5\equal{}4x\plus{}20\implies x\equal{}\boxed{15}\]


how does one thermodynamically define freezing point and in particu;ar why is there 'depression of freezing point' in a 'impure' solution

In the freezing point of a pure solvent, the rate at which the solvent molecules stick together to form the solid is equal to the rate at which solvent molecules come out of the solid to return to the liquid phase. However, when a solute is present (volatile or not) its molecules "get in the way" of the solvent molecules and so there are fewer solvent molecules in contact with the solid solvent. However, the rate at which solvent molecules leave the solid to go into the solution is unaffected, and so there is a net flow of solvent molecules from the solid to the liquid, and solidification in therefore not possible. Only if the temperature is lowered, in order for the solid -> liquid flow of solvent molecules to be slowed, will there be a net flow of molecules from the liquid to the solid and so solidification can now occur.


Triangle $ ABC$ is isosceles with angle $ A$ congruent to angle $ B$.
The measure of angle $ C$ is 30 degrees more than the measure of
angle $ A$. What is the number of degrees in the measure of angle $ C$?

call angle A x

then angle B is also x

and angle C is x+30

since they must be equal to 180;

x+x+x+30=180

3x+30=180

3x=150

x=50; angle C=50+30=80


geometry
circumcircle
power of a point
radical axis
geometry proposed

11.4 Let $AA_1$ and $BB_1$ are altitudes of an acute non-isosceles triangle $ABC$, $A'$ is a midpoint of $BC$ and $B'$ is a midpoint of $AC$. A segement $A_1B_1$ intersects $A'B'$ at point $C'$. Prove that $CC'\perp HO$, where $H$ is a orthocenter and $O$ is a circumcenter of $ABC$.
([i]L. Emel'yanov[/i])

Denote $H'$ intersection of $CH$ and $A'B'$, $O'$ intersection of $CO$ and $A_1B_1$

It is well known that $ CH'\perp A'B'$ and $CO'\perp A_1B_1$. So $CC'O'H'$  is cyclic.

Consider inversion with th pole $C$ and power $k = CA'*CA_1=CB'*CB_1$.

Line $A'H'B'$ transforms in $(A_1HB_1C)$
Line $A_1O'B_1$ transforms in $(A'OB'C)$
It can easily to see $I(O')= O, I(H')= H$

So circle $(CC'O'H')$ transforms in line $OH$ which is perpendicular to diameter $(CC'O'H')$ - $CC'$.

So $OH\perp CC'$.

Here's another solution:

Since $A_1A'B'B_1$ is cyclic, $C'A'\cdot C'B'=C'A_1\cdot C'B_1$, so $C'$ lies on the radical axis of the circles $(CA_1B_1),(CA'B')$. This means that $CC'$ is perpendicular to the line of centers of these two circles (call this $(*)$). However, the center of $(CA_1B_1)$ is the midpoint of $CH$, while the center of $(CA'B')$ is the midpoint of $OH$, so $OH$ is parallel to the line of centers, and thus perpendicular to $CC'$, according to $(*)$.

[quote="Myth"]The official solution is the same as grobber's one.
It is quiate unusual, since they needed to present a notion of radical axe which is out of school education.[/quote]
Maybe L. Emel'yanov was expecting the contestants to deduce the radical axe idea by themselves :D

[quote="Myth"]11.4 Let $AA_1$ and $BB_1$ are altitudes of an acute non-isosceles triangle $ABC$, $A'$ is a midpoint of $BC$ and $B'$ is a midpoint of $AC$. A segement $A_1B_1$ intersects $A'B'$ at point $C'$. Prove that $CC'\perp HO$, where $H$ is a orthocenter and $O$ is a circumcenter of $ABC$.[/quote]

More general (but equally trivial):

Let P and Q be two isogonal conjugate points with respect to triangle ABC. Let X and Y be the orthogonal projections of the point P on the sides BC and CA of triangle ABC, and let X' and Y' be the orthogonal projections of the point Q on these sides. The lines XY and X'Y' intersect each other at the point C'. Prove that $CC^{\prime}\perp PQ$.

The above problem 11.4 is a particular case of this if P = H and Q = O.

Both Prowler's and Grobber's solutions of problem 11.4 work similarly for the general version (one just has to apply the well-known result that the points X, Y, X', Y' are concyclic).

  darij

[quote="darij grinberg"][quote="Myth"]11.4 Let $AA_1$ and $BB_1$ are altitudes of an acute non-isosceles triangle $ABC$, $A'$ is a midpoint of $BC$ and $B'$ is a midpoint of $AC$. A segement $A_1B_1$ intersects $A'B'$ at point $C'$. Prove that $CC'\perp HO$, where $H$ is a orthocenter and $O$ is a circumcenter of $ABC$.[/quote]

More general (but equally trivial)...[/quote]
Say it only for yourself. If this problem appeared as 11.4 it means that the jury had some reasons for that.

[quote="grobber"]The center of $(CA'B')$ is the midpoint of $OH$.[/quote]

Small typo here: it should be the midpoint of $OC$. A very nice solution, grobber!


Let P(x) be a polynomial with real coefficients. Prove that there exists a nonzero polynomial Q(x) with real coefficients such that P(x)Q(x) has terms which are all of a degree divisible by 10^9.

Try $Q(x)=P(z_1x)P(z_2x)\dots P(z_{n-1}x)$, where $z_k$ are the roots of $z^m=1$ different from $1$. Here $m=10^9$.


P is in Triangle ABC and makes <PBC=<PCA=<PAB
Prove this ineq:
a^2+b^2+c^2>=(PA+PB+PC)^2

Nice problem!

[color=blue][b]Problem.[/b] Let P be a point inside a triangle ABC such that < PBC = < PCA = < PAB. Prove that $a^2+b^2+c^2\geq\left(PA+PB+PC\right)^2$.[/color]

[i]Solution.[/i] Let the lines AP, BP, CP meet the sides BC, CA, AB of triangle ABC at the points X, Y, Z. Then, by the Gergonne theorem (see http://cut-the-knot.com/triangle/Gergonne.shtml or http://www.mathlinks.ro/Forum/viewtopic.php?t=31501 property ii) ), we have $\frac{PX}{AX}+\frac{PY}{BY}+\frac{PZ}{CZ}=1$.

Now, remember that < PBC = < PCA; this can be rewritten as < YBC = < YCP. Also, trivially, < BYC = < CYP. Thus, the triangles YBC and YCP are similar; hence, $\frac{CY}{BY}=\frac{PC}{CB}$ and $\frac{PY}{CY}=\frac{PC}{CB}$. Hence, $\frac{PY}{BY}=\frac{PY}{CY}\cdot\frac{CY}{BY}=\frac{PC}{CB}\cdot\frac{PC}{CB}=\left(\frac{PC}{CB}\right)^2=\left(\frac{PC}{a}\right)^2$. Similarly, $\frac{PZ}{CZ}=\left(\frac{PA}{b}\right)^2$ and $\frac{PX}{AX}=\left(\frac{PB}{c}\right)^2$. Hence, the equation $\frac{PX}{AX}+\frac{PY}{BY}+\frac{PZ}{CZ}=1$ becomes $\left(\frac{PB}{c}\right)^2+\left(\frac{PC}{a}\right)^2+\left(\frac{PA}{b}\right)^2=1$. Now, by the Cauchy-Schwarz inequality,

$\left(a^2+b^2+c^2\right)\left(\underbrace{\left(\frac{PC}{a}\right)^2+\left(\frac{PA}{b}\right)^2+\left(\frac{PB}{c}\right)^2}_{=\left(\frac{PB}{c}\right)^2+\left(\frac{PC}{a}\right)^2+\left(\frac{PA}{b}\right)^2=1}\right)\geq\left(\underbrace{a\cdot\frac{PC}{a}+b\cdot\frac{PA}{b}+c\cdot\frac{PB}{c}}_{=PA+PB+PC}\right)^2$;

in other words,

$a^2+b^2+c^2\geq\left(PA+PB+PC\right)^2$,

and the problem is solved.

[i]Note.[/i] The point P satisfying < PBC = < PCA = < PAB is called the [i]first Brocard point[/i] of triangle ABC, and the angle < PBC = < PCA = < PAB is referred to as the [i]Brocard angle[/i] of triangle ABC.

  darij


For how many integers $ n$ is $ \frac{n}{20\minus{}n}$ the square of an integer?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 10$

[quote="worthawholebean"]For how many integers $ n$ is $ \frac {n}{20 \minus{} n}$ the square of an integer?

$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 10$[/quote]

Through brute force, one can easily tell it is 3.
$ N>10$
N works when $ N\equal{}10, 16, 18$
C

Doesn't n=0 also work? Because it doesn't say it has to be positive.
So the answer is D.

[hide]Let the square be $ k^2$. Then, $ \frac{n}{20\minus{}n}\equal{}k^2$, and solving this for $ n$, we have
\[ n\equal{}\frac{20k^2}{k^2\plus{}1}\]Because $ n$ is an integer, the fraction must be reducible, and since $ k^2$ and $ k^2\plus{}1$ are relatively prime, $ k^2\plus{}1$ must divide $ 20$, and that is true only when $ k\equal{}0, 1, 2, 3$. The corresponding values of $ n$ are $ 0, 10, 16, 18$, and the answer is $ 4 \Rightarrow \boxed{D}$.[/hide]


solve equation :$x^2+y^2+1=xyz$ where $x,y,z \in Z$

I didn't you can post it here...in wrong part...


real analysis
function
integration
real analysis unsolved

Prove or disprove:

Let $ f: \mathbb{R}\to\mathbb{R}$ be L-measurable.

$ \int f\,dx \equal{}\sup\{\int s \,dx |0\le s\le f\}$

where s is a step function

False.

On $ [0,1]$ let $ f(x)\equal{}\begin{cases}1,&x\in\mathbb{R}\setminus \mathbb{Q}\\ 0,&x\in\mathbb{Q}\end{cases}.$

Then $ \sup\int_{[0,1]}s\equal{}0$ where the supremum is taken over step functions $ s\le f,$ while $ \int_{[0,1]}f\equal{}1.$


My school is offering the AMC 12 A, but I want to take both tests... I was wondering if there is anywhere in Maryland where I can take the AMC 12 B test this year. Thanks in advance!

Yes, dunno how close it is to you.

http://www.unl.edu/amc/b-registration/b1-archive/2009-2010/2010-HS-CU-list.shtml

thank you for the link... i will check it out!


geometry
3D geometry
sphere
power of a point
radical axis
geometry solved

"A circle is the set of points in a plane that are equidistant from a given point $O$"

For any two points $A,B$, their distance is $\geq 0$
(it is $0$ iff $A \equiv B$)

So, a zero-distance is allowed.

So, we have "The set of points in the plane that have distance $0$ from the point $O$ " is the point $O$ itself

Further more, $O$ is a sphere with a zero-radius, or even an n-dimensional sphere

[quote="jensen"]Point is a circle or not?[/quote]

Well, this depends on how you interpret the word "circle". The same holds for your question whether a line is a circle, which you have posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=65374 . Mathematicians have noted that under some circumstances, points and lines behave similarly to circles, so they have found it useful to consider points and lines as "degenerate" cases of circles. The exact definitions are the following:

- A point P can be identified with a "degenerate" circle with center P and radius 0; this circle is called the [b]point-circle[/b] P. This point-circle passes through one single point, namely P.
- A line g can also be identified with a "degenerate" circle, a so called [b]line-circle[/b]. The center of this line-circle is the infinite point where all lines perpendicular to g intersect. The radius of this circle is undefined (you could say it is $\infty$, but this is misleading, since usually, a circle is uniquely determined by its center and its radius, but in this case this would not be true, as there would be lots of circles all centered at one infinite point and having the same radius $\infty$).

Actually, both notions of point-circles and line-circles are suggested by limiting observations: Imagine a fixed line t, a fixed point P on the line t, and a variable point M on the line t. Consider the circle with center M and radius MP. When the point M comes closer and closer to P, the radius MP tends to zero, and the circle resembles more and more to the single point P (point-circle). When the point M moves farther and farther from P, the radius MP tends to infinity, and the circle resembles more and more to a line, namely the perpendicular to the line t at P (line-circle).

Of course, in order to ensure that these "definitions" are more than empty words, we have to extend some possible properties of circles to these "degenerate" circles. For instance, when are two degenerate circles, or a non-degenerate and a degenerate circle orthogonal? Well, the usual definition of orthogonality of two circles is: Two circles $k_1$ and $k_2$ are called orthogonal if $O_1O_2\ ^2=r_1^2+r_2^2$, where $O_1$ and $O_2$ are the centers and $r_1$ and $r_2$ are the radii of the circles $k_1$ and $k_2$. This definition works for non-degenerate circles, and it also works for point-circles. However, since a point-circle has radius 0, the equation $O_1O_2\ ^2=r_1^2+r_2^2$ simplifies to $O_1O_2\ ^2=r_1^2$ when the circle $k_2$ is a point-circle, and this obviously means that the point $O_2$ lies on the circle $k_1$. Hence, we see that a point-circle P is orthogonal to a circle k if and only if the point P lies on the circle k. As a consequence of this, two point-circles are orthogonal if and only if they coincide. Well, so far we have defined orthogonality for point-circles. Defining orthogonality for line-circles is a bit harder, since they have no radii. But actually, line-circles are lines; so it is suggesting to say that two line-circles are orthogonal when they are perpendicular lines. Furthermore, a line-circle is said to be orthogonal to a point-circle when it passes through the point, and a line-circle is said to be orthogonal to a non-degenerate circle if the center of the non-degenerate circle lies on the line. (Both of these postulations are not arbitrary, but suggested by a limiting argument.)

Now we can note that lots of theorems about circles continue to hold if some of the circles "degenerate" to point-circles or line-circles. Particularly in inversive geometry, you will find lines and circles having more or less equal rights. You can also define the radical axis of a circle and a point-circle, for instance,[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=249398#p249398]and use it in some problems[/url] (however, as usual, degenerate cases should be treated with care - e. g., you cannot define radical axes for line-circles). This is the main reason why the notions of point-circles and line-circles are used.

  Darij

point is a circle with a very very small radius. ;)

Darij grinberg, thanks a lot for your answer, I enjoyed reading it! :)
That was really interesting.


The sequence $(a_n)$ is given by $a_1=x\in R$ and $3a_{n+1}=a_n+1$ for $n\geq 1$. Set
$A=\sum_{n=1}^\infty [a_n-\frac 16]$ and $B=\sum_{n=1}^\infty [a_n+\frac 16]$. (where $[x]$ denote the greatest integer less that or equal to $x$).Compute the sum $A+B$ in terms of $x$. :lol:


number theory
relatively prime
Divisibility Theory

Prove that among any ten consecutive positive integers at least one is relatively prime to the product of the others.

Clearly the only common prime factors amongst 10 consecutive positive integers will be 2,3,5,7.

5 of them will be divisible by 2, at least one of which must also be divisible by 3 and at least one of which must also be divisible by 5, and, if two of the numbers are divisible by 7, one of them will be even.

This leaves two multiples of 3, one multiple of 5, and one multiple of 7 unaccounted for, enough factors to dish out to 4 more of our integers.

But this still leaves one integer not divisible by 2,3,5,7, and therefore co-prime with all other integers of the set, and so coprime with their product.

Can we replace $ 10$ by $ n$?

I have found that this fails for $ 25$, although I am not sure of the other values. Can anyone give anything?

[quote="peter"]Can you post your example for 25? Maybe that gives us a clue.[/quote]

Well my example is the natural numbers from $ 1$ to $ 25$.

[quote="MayankM"]
I have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon.
[/quote]

The source of the problem said [b]IHH pp.211[/b] and so I took that out and I found what [b]MayankM[/b] had siad to be true.

[quote="manjil"][quote="peter"]Can you post your example for 25? Maybe that gives us a clue.[/quote]

Well my example is the natural numbers from $ 1$ to $ 25$.[/quote]Really? I think $ 23$ is relatively prime to the product of the others though...  :wink:

[quote="MayankM"]I have read some where that this result is true for n up to 16 consecutive integers. I'll look up the source and post soon.[/quote]
it's true. 16 can be done and 17 can't. the counterexample for $ n\equal{}17$ are consecutive numbers of which the first equals $ 27830$.
(in fact the first can be $ 27830\plus{}30030k$ for a natural $ k$, mybe there exists a smaller example that i missed but the point is that there are infinitely many)
u can proove it for $ n\equal{}16$ basicaly same as ilthigore has done for $ n\equal{}10$ above only a bit more complicated.

$ n\equal{}17$
let $ a_1,...a_{17}$ be these consecutive numbers. if we set $ a_1$ to be devisable by $ 2,5,11$, $ a_2$ devisable by $ 3$, $ a_3$ by $ 7$ and $ a_4$ by $ 13$. it's easy to check thet none of them is than relatively prime to all others. we can calculate $ 27830$ with china reminder theorem now.

what is left to do is to show that if there is a counter wxample for $ n$ than we can also find a counter example for $ n\plus{}1$. im working on this now.

[hide=drawn-out solution]Let any set of $10$ numbers be defined as $S=\{n,n+1,n+2,n+3, \dots ,n+9\}$.

[b]Lemma $1$[/b]: The only prime factors that the numbers in $S$ can have in common are $2,3,5$ and $7$.
We can take the set $S$ modulo $2,3,5,$ and $7$, and since there must contain at least two $0$'s in $S$ (by spacing of the modulo), there are two numbers with the modulo as their factors. However, if we take the modulo of any prime greater than $7$, there is at most one $0$ contained in the set. (Also, due to spacing)
Also, note that this argument makes it so that the only divisors that the elements in $S$ can have in common are the ones described above, due to euclidian algorithm (For example, $\gcd(n,n+5)=\gcd(n,5)$) $\square$

There can be a maximum of $5$ factors of $2$ contained in $S$, $4$ factors of $3$, $2$ factors of $5$, and $2$ factors of $7$. Using a spacing argument, there must exist $2$ factors of $2$ and $3$; $1$ factor of $2$ and $5$; and $1$ factor of $2$ and $7$.Thus, to account for the overcounting, we have to subtract $4$, and we can now simply sum the numbers and subtract our overcounting factor to get $5+4+2+2-4=9$, and so there must exist some number that is relatively prime to the rest of the numbers.[/hide]

As mentioned above, the statement is true when 'ten' is replaced by any positive integer $m \le 16$. It is false, however, if one replaces 'ten' by any integer $m > 16$. References for this fact:

In a paper that you can find here: [url]https://www.ias.ac.in/article/fulltext/seca/011/01/0006-0012[/url], Pillai proved that it is true for $m \le 16$ and it is false for all $m$ with $17 \le m \le 430$.

In a paper that you can find here: [url]https://www.ams.org/journals/bull/1941-47-04/S0002-9904-1941-07455-0/S0002-9904-1941-07455-0.pdf[/url], Brauer proves that it is false for all $m \ge 300$, thereby finishing the problem.


geometry
3D geometry
number theory proposed
number theory

These questions are almost classical rigth now but...

Which is the smallest positive integer that could be written as the sum of two (positive) squares in two different ways?
Which is the smallest positive integer that could be written as the sum of two (positive) squares in three different ways (three distinct pairs - any two elements are not equal)?
Which is the smallest positive integer that could be written as the sum of two (positive) cubes in two different ways?
There is an integer such that is the smallest positive integer that could be written as the sum of two (positive) cubes in three different ways?
Which is the smallest positive integer that could be written as the sum of two (positive) fourth powers in two different ways?

[quote="t0rajir0u"]3.  Ramanujan's number![/quote]

Also called taxicab number ;)

fOR THE LAST also Ramanajuan  reply to this at  5 minutes after the question without using a computer and being on hospital that the smallest is

$635318657=133^4+134^4=158^4+59^4$ :)

[quote="silouan"]without using a computer[/quote]
There was no computers at that time ;)

Sorry for spamming but Ramanujan was a computer himself :D
i think noone has doubt for this.


function
integration
trigonometry
real analysis
real analysis unsolved

Let $ m\equal{}2a\minus{}1$ and $ n\equal{}2b\minus{}1$ and so,
$ \int_{0}^{\pi/2}\sin^{2a\minus{}1}x\cos^{2b\minus{}1}x dx \equal{}\int_{0}^{\pi/2}\sin^{2a\minus{}1}x\cos^{2b\minus{}2}x\cos xdx$
Let $ y\equal{}\sin x$ to get,
$ \int_{0}^{1}y^{2a\minus{}1}(1\minus{}y^{2})^{b\minus{}1}dy$
Let $ z\equal{}y^{2}$ to get,
$ \frac{1}{2}\int_{0}^{1}z^{a\minus{}1}(1\minus{}z)^{b\minus{}1}dz \equal{}\frac{1}{2}\mbox{B}(a,b) \equal{}\frac{1}{2}\cdot\frac{\Gamma (a)\Gamma(b)}{\Gamma(a\plus{}b)}$
So, subsitute back,
$ \frac{1}{2}\frac{\Gamma\left(\frac{n\plus{}1}{2}\right)\cdot\Gamma\left(\frac{m\plus{}1}{2}\right)}{\Gamma\left(\frac{n\plus{}m\plus{}2}{2}\right)}$
Apply the infinite product formulas for Gamma functions ... and it should work out.  :maybe:


Do all squares differ by odd integers (e.g. 4, 9, 16, 25, etc.)?

Are all squares perfect squares?

[quote="erik-the-red"]Are all squares perfect squares?[/quote]
Uhh, what do you mean?

well let x be an integer, then x+1 is the next integer...so the difference is
$(x+1)^{2} - x^{2}$
$2x + 1$
so yes ...since $2x +1$ for any integer x is odd

[quote="Elemennop"]No, because $n^2\equiv 0,1\pmod 4$, so that two non-adjacent ones can be $1-1=0\pmod 4$, and therefore not an odd integer.[/quote]But that's assuming that there are two adjacent squares that are both $\equiv 1\pmod 4$...it turnes out they alternate between $0$ and $1$.  If the square root is even then the square is $\equiv 0\pmod 4$, if the square root is odd then the square is $\equiv 1\pmod 4$.

[quote="Elemennop"]He never said anywhere they had to be adjacent, plus I specified specifically that they have an even difference only if they are not adjacent.[/quote]Oh good point, I probably just assumed he meant consecutive squares.

No, not all squares are perfect squares.  A perfect square is a positive integer which is the square of some other positive integer.  A square can be one of many things.  It can be any number.  Or, it could be a regular quadrilateral.

Perfect squares are the square of an integer.

Only squares of integers differing by an odd number will differ by an odd number. For example: $n^2$ and $(n-3)^2=n^2-6n+9$ differ by an odd number. However, squares of integers differing by an even number will differ by an even number.


We make n² equal squares inside of a square and we write on each square a real number.

$ \forall 1 \le i \le n$ we denote $ r_i$ the sum of the numbers on the [u]row[/u] i and $ c_i$ the sum of the numbers on the [u]column[/u] i
(Thank you JBL for correcting my english)

prove that there exists i and j such that:
$ (r_i \minus{} c_i)(r_j \minus{} c_j) \le 0$

Where was this problem proposed please?

Thanks !

[quote="FOURRIER"]... [u]line[/u] ... [u]array[/u] ...[/quote]

"Row" and "column," respectively.  (Easy to remember in this instance, because of the $ r$ and the $ c$ ;).)

Indeed very easy:
If $ (r_i \minus{} c_i)(r_j \minus{} c_j) > 0$ for all $ i$ and $ j$ then
$ r_i > c_i\ \ \forall i$           (or $ (r_i < c_i\ \ \forall i$ wich gives the same)
then $ r_1 \plus{} r_2 \plus{} ... \plus{} r_n > c_1 \plus{} c_2 \plus{} ... \plus{} c_n$ which contradicts $ r_1 \plus{} r_2 \plus{} ... \plus{} r_n \equal{} c_1 \plus{} c_2 \plus{} ... \plus{} c_n$

btw no idea where it was proposed...


Let $ a,b,c>0$ such that $ abc\equal{}1$. Find the minimum value of the following inequality:
\[ \frac{1}{{\sqrt {1 \plus{} a^2 } }} \plus{} \frac{1}{{\sqrt {1 \plus{} b^2 } }} \plus{} \frac{1}{{\sqrt {1 \plus{} c^2 } }}
\]

I think it is correct :lol:
Inequality holds for a->0 and b,c--->infinity.
I think for these kind of problem(only one inflection point in f(e^x)) it is easy because
you can put a=b=t, c=1/t^2 and differentiate.

How to solve it? This is the most important things in this topic  :) Can anyone post solution?

[quote="ywl"]
you can put a=b=t, c=1/t^2[/quote]

I think you can't

only if you have shown : $ f(a,b,c)\geq f(\sqrt{ab},\sqrt{ab},c)$
 :wink:

[quote="FOURRIER"][quote="ywl"]
you can put a=b=t, c=1/t^2[/quote]

I think you can't

only if you have shown : $ f(a,b,c)\geq f(\sqrt {ab},\sqrt {ab},c)$
 :wink:[/quote]

I think I have shown. Have you tried the convexity of f(e^x) with f(x)=1/(root(1+x^2)? It has one inflection point
so that you can put a=b=t and c=1/t^2 This follows from applications of Karamata and Jensen
(See post in Theorems and Formulas sections by Zhaobin regarding the "Half convex, half concave theorem) :)

$ S(a,b,c)\equal{}\frac{1}{\sqrt{1\plus{}a^{2}}}\plus{}\frac{1}{\sqrt{1\plus{}b^{2}}}\plus{}\frac{1}{\sqrt{1\plus{}c^{2}}}$. Let's prove that $ \inf{S(a,b,c)}\equal{}1$ (It hasn't got a minimum value) with $ abc\equal{}1$. It's clear that if $ a>\equal{}1,b<\equal{}1,c<\equal{}1$ , we have $ S(a,b,c)>\frac{1}{\sqrt{1\plus{}b^{2}}}\plus{}\frac{1}{\sqrt{1\plus{}c^{2}}}>\equal{}2\cdot\frac{1}{2}\equal{}1$ , so $ a>\equal{}1, b>\equal{}1, c<\equal{}1$. Then $ b<\equal{}\frac{1}{c}$. So we have $ S(a,b,c)>\frac{1}{\sqrt{1\plus{}b^{2}}}\plus{}\frac{1}{\sqrt{1\plus{}c^{2}}}>\equal{}\frac{c\plus{}1}{\sqrt{1\plus{}c^{2}}}>1$
So $ S(a,b,c)>1$. But choosing $ a\equal{}b\equal{}x$, $ c\equal{}\frac{1}{x^{2}}$ we have $ S(a,b,c)\equal{}\frac{x\plus{}2}{\sqrt{1\plus{}x^{2}}}\rightarrow 1$ with $ x\rightarrow \infty$. So $ \inf{S(a,b,c)}\equal{}1$

[i]I am a member of LeHongPhong highschool so i can say that  the problem is " find maximum of the expression"[/i]
[i]very beautiful problem

    But  it's origin is very , very similiar to erveryone :)
   [/i] Denote $ x = 1 , y = \frac {1}{a^2} , z = \frac {1}{(ab)^2}$
 => $ a^2 = \frac {x}{y} , b^2 = \frac {y}{z}, c^2 = \frac {z}{x}$
  [i]We have to prove with x,y,z  be positive real numbers we have[/i]  B=$ \sum\limits_{cyc} \sqrt {\frac {2x}{x + y}}$  $ \leq$  3
 Deduce B= $ \sum\limits_{cyc}\sqrt {\frac {2x(y + z)}{(x + y)(y + z)}}$

   => $ \ B^{2}$          $ \leq$(BCS) A= 4( $ xy + yz + zx$).($ \frac {1}{(x + y)(y + z)} + \frac {1}{(y + z)(z + x)} + \frac {1}{(z + x)(x + y)}$)
      = 8$ \frac {(xy + yz + zx)(x + y + z)}{(x + y)(y + z)(z + x)}$

[i]    From a known ineq[/i]
    $ (x + y)(y + z)(z + x)$$ \geq$ $ \frac {8}{9}$ $ (x + y + z)(xy + yz + zx)$ :lol:
[i]  ( Cauchy with 6 positive real numbers)[/i]
           [i] We have[/i] [size=150]A[/size] $ \leq$ 9
    => LHS $ \leq$ 3 ( QED)
   [i] Hero TVƠ Yêu An mãi mãi[/i]

[quote="zaizai-hoang"]Let $ a,b,c > 0$ such that $ abc \equal{} 1$. Find the minimum value of the following inequality:
\[ \frac {1}{{\sqrt {1 \plus{} a^2 } }} \plus{} \frac {1}{{\sqrt {1 \plus{} b^2 } }} \plus{} \frac {1}{{\sqrt {1 \plus{} c^2 } }}
\]
[/quote]
Assume $ ab \le 1$, then
$ \frac {1}{{\sqrt {1 \plus{} a^2 } }} \plus{} \frac {1}{{\sqrt {1 \plus{} b^2 } }} \plus{} \frac {1}{{\sqrt {1 \plus{} c^2 } }} \\ \ge \frac {1}{{\sqrt {1 \plus{} a^2 } }} \plus{} \frac {1}{{\sqrt {1 \plus{} b^2 } }} \\ \ge \sqrt{\frac{1}{1\plus{}a^2}\plus{}\frac{1}{1\plus{}b^2}} \\
\ge 1$.

We know the general inequality
\[ \min \left(1;\frac{n}{2^k} \right) \le \frac{1}{(1\plus{}a_1)^k}\plus{}\frac{1}{(1\plus{}a_1)^k}\plus{}...\plus{}\frac{1}{(1\plus{}a_1)^k} \le \max \left(n\minus{}1;\frac{n}{2^k} \right)
\]
where $ a_1,a_2,...,a_n > 0$ such that $ a_1a_2...a_n \equal{}1$ and $ k$ is a positive real number.


Let $ a,b,c\ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$,prove that\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\ge \frac {1}{4}\]

[quote="skywalkerJ.L."]Let $ a,b,c\ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$,prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\ge \frac {1}{4}\]
[/quote]
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\ge \frac {1}{4}\Leftrightarrow4(a^3\plus{}b^3\plus{}c^3)\plus{}24abc\geq(a\plus{}b\plus{}c)^3,$ which is obviously true by Schur.

[quote="skywalkerJ.L."]Let $ a,b,c\ge 0$ and $ a \plus{} b \plus{} c \equal{} 1$,prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\ge \frac {1}{4}\]
[/quote]

$ \Longleftrightarrow a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\ge \frac {1}{4}(a\plus{}b\plus{}c)^3$

$ \Longleftrightarrow \frac{3}{4}\sum{a(a\minus{}b)(a\minus{}c)}\plus{}\frac{9}{4}abc\geq 0$


absolute value
number theory unsolved
number theory

Prove that the number $[(5+\sqrt{35})^{2n-1}] $ is divisible by $ 10^n$ for each $n\in N$
(where $ [x]$ denote the greatest integer less than or equal to $x$). :huh: :)

Define $x_0:=2 ; x_1:=10 ; x_{n+2} := 10( x_{n+1} + x_{n} )$, then every odd term of the sequence will be identically what your term gives (simply solve the recursion, you get some term $5-\sqrt{35}$ that has absolute value $<1$ and so can be neglected when taking the integer part) and conclusion (even more) follows.


I guess the first term in the denominator should be $ 5x^4$?

$ f\left( x \right) \equal{} \frac {x^5}{5x^4 \minus{} 10x^3 \plus{} 10x^2 \minus{} 5x \plus{} 1} \equal{} \frac {x^5}{5(x^3 \minus{} x^2 \plus{} x)(x \minus{} 1) \plus{} 1}$.

And it's 1005, but I don't know how to get there from this...

$ f(x) \equal{} \frac {x^5}{x^5 \plus{} (1 \minus{} x)^5}$
$ f(x) \plus{} f(1 \minus{} x) \equal{} \frac {x^5}{x^5 \plus{} (1 \minus{} x)^5} \plus{} \frac {(1 \minus{} x)^5}{x^5 \plus{} (1 \minus{} x)^5} \equal{} 1$

Hence, all terms pair up except $ f(\frac {2009}{2009}) \equal{} f(1) \equal{} \frac {1}{1 \plus{} 0} \equal{} 1$, giving the sum to have value $ 1004 \plus{} 1 \equal{} 1005$

So B is the correct answer.

The trick was seeing that you could write the denominator like that.

[quote="rofler"]The trick was seeing that you could write the denominator like that.[/quote]

Ah... I factored it wrong.

Bleh >_<. Mind's not working. Thank you!


[quote="hanschinaman"]a,b,c,d are postive real numbers whose sum is 4, prove
$ \prod_{cyc}{(a^{3}+1)}\geq 16$[/quote]
The following inequality is also true.
Let $a$, $b$, $c$, $d$, $e$, $f$ and $g$ be positive numbers such that $a+b+c+d+e+f+g=7$. Prove that:
\[(a^3+1)(b^3+1)(c^3+1)(d^3+1)(e^3+1)(f^3+1)(g^3+1)\geq128\]


inequalities
calculus
inequalities proposed

:D   i realized my mistake in calculus, so i erased it.


probability
geometry
3D geometry
calculus
integration
quadratics

I have n sticks, of length 1, 2, 3, ..., n.
I choose three sticks at random. Whats the probability that they can form a triangle?

I have an answer, hopefully it agrees with your ones.

Haven't tried it yet, but a possibly interesting variation:  Given three real numbers between 0 and 1, with all triples equally likely to be chosen, what is the probability that a triangle can be formed with sides of those lengths?

Well, Joel's variant can be solved by considering a 1*1*1 cube and finding the volume that makes it work. Stephen's one seems like a very difficult problem.

Hmm, hadn't thought about that.  Makes sense.  It should also be the limiting case of the solution to Steven's problem, which I now attempt to give.

Hmm, I realize that I did a different problem than the one you gave -- you question didn't allow repeats, while my solution allowed for them.  I'll try and do it your way later, but here's what I did first:

[hide]

Okay, so, denominator first:  C(n, 3) ways to pick all different, C(n, 1) ways to pick all the same, 2C(n, 2) ways to pick two the same and one different, so (C(n, 3) + 2C(n, 2) + C(n, 1)) = (n3 + 3n2 + 2n)/6 total picks.

Now, for the numerator.  Let M(i) be the number of triangles with integral sides between 1 and i with largest side i.  Then the numerator should be  :Sigma:M(i) from i = 1 to n.

I wish to generate M(i) recursively.  I claim M(i) = M(i - 1) + [i/2], where [x] is the greatest integer less than or equal to x.

Proof:

Consider any triple (a, b, i) that makes a triangle, with b > a.  Then (a, b - 1, i - 1) is a triple that makes a triangle, with b - 1 :ge: a.  That gives us M(i - 1) such triples.  The only ones we've missed are those in which a = b, of which we have [i/2] which make triangles.    Thus, we have the relation I claimed.

So, if we just look at what that gives us, we have M(1) = 1, M(2) = 2, M(3) = 3, M(4) = 5, M(5) = 7, M(6) = 10, M(7) = 13, and so on.  This is roughly quadratic -- for odd i, it gives us (i2 + 3)/4, and for even i it gives us (i2/4 + 1), which can be encapsulated nicely in the formula ([i2/4] + 1).

Then we have to sum that up from i = 1 to n.

[/hide]

I have to go, so I can't finish this now -- I will try to finish my way and adjust for the way I was supposed to do it later.  Anyone, feel free to jump off of this.

My solution to Joel's problem:

[hide]

Well, I couldn't figure out how to get integral signs, so S_a^b means the integral from a to b:

S_0^1 S_0^1 (min(1,x+y)-|x-y|) dy dx =

S_0^1 S_0^1 min(1,x+y) - S_0^1 S_0^1 |x-y| dy dx =

S_0^1 S_0^(1-x) x+y dy dx + S_0^1 S_0^(1-x) 1 dy dx - S_0^1 S_0^x (x-y) dy dx + S_0^1 S_x^1 (y-x) dy dx =

S_0^1 (-1/2 x^2 + 1/2) dx + S_0^1 x dx - S_0^1 1/2 x^2 dx - S_0^1 (1/2x^2-x+1/2) dx =

S_0^1 (-3/2 x^2 + 2x) dx =

1/2

[/hide]

The fact that the answer has such a simple form makes me think there must be a more elegant solution.  Any ideas for say [hide]a bijection between triples that form a triangle and triples that don't[/hide]?

--Dan

Well heres an outline of how I did it anyway.

[hide]

We count the possible number of groups of 3 that work. First we consider the ones with largest stick of the 3 being the one length n:

Now if we take n-1, we can choose one of (2,3,...,n-2) ie n-3 ways.

If we take n-2, we choose one of (3,4,..,n-3) ie n-5 ways.

etc etc

So the number of ways in this case is 0+2+4+..+(n-3) if n is odd, 1+3+5+..+(n-3) if n is even.

Work out those sums, (the formulas end up differing by 1/4). So to find the total number of ways, you sum the (higher) formula from 3 to n, subtract off [(n-1)/2] quarters, and get an answer (after dividing by nC3).

[/hide]

Okay, yeah, that was exactly what I was trying to do, except that I was working on it when I wasn't at the computer so I got that one part (about repeatability) wrong.


inequalities
calculus
calculus computations

Does any one have good problems involving greatest integer and fractional part ?

do it by yourself.

go and use google, c'mon, you can't be that lazy.

you can even look at the forum.

Hi. A book which has problems on this is "Equations and Inequalities", pages 251 to 257. See here: http://www.muni.cz/research/publications/313831/


S is the complexe of the numbers in form p+a^p which a is a fix integer and p is a prime number and suppose that k is also an integer.is it true that we can find two member from the complexe like x and y so that   (x,y)>k
thank you very much.

Take some prime $q>k$ not dividing $a$. Then lets set $s \equiv 1 \mod (q-1)$ and $\equiv -a \mod q$ since this gives $s+a^s \equiv s+a \equiv 0 \mod q$.
By Dirichlets theorem on primes in arithmetic progressions, there are two different primes $x,y \equiv s \mod q\cdot(q-1)$, fulfilling $\gcd(x+a^x , y+a^y) =q \cdot \mathrm{SPAM} \geq q >k$.


abstract algebra
group theory
superior algebra
superior algebra unsolved

just checking if i did this right
$\phi:H\rightarrow H$, where $H$ is any group, and $\phi(x)=x^{-1}$.  Show that $\phi$ is a homomorphism.

$\phi(x+y)=(x+y)^{-1}$, $\phi(x) + \phi(y)=x^{-1} + y^{-1}$ ,   $\phi(x+y) \neq \phi(x) + \phi(y)$ , so $\phi$ wouldnt be a homomorphism

Are you sure that's + and not $\times$ ?

If we set * the "law" (is it the good word in english ?) on H, we get :
$\phi(x) * \phi(y)=x^{-1} * y^{-1}$, $\phi(x*y)=(xy)^{-1}=y^{-1} * x^{-1}$
$\phi$ is an homomorphism if H is an abelian group

[quote="Rodolphe2005"]If we set * the "law" (is it the good word in english ?) on H, we get :
[/quote]
operation.

No, for example take $( \mathbb{Z}_2 ; + )$

[quote="chenwb"]$\phi:H\rightarrow H$, where $H$ is any group, and $\phi(x)=x^{-1}$. Show that $\phi$ is a homomorphism.

This holds if and only if $H$ is abelian? Thus statement is not true  :?[/quote]

Totally right, because by hypothesis, for every $x,y\in H$ you'll have $\phi(xy)=\phi(x)\phi(y)$, but therefore $(xy)^{-1}=y^{-1}x^{-1}=x^{-1}y^{-1}$, hence, $H$ is abelian!


geometry
angle bisector
Pythagorean Theorem

Right triangle ABC is divided, as shown, such that angle BAC is bisected by AD.  What is the length of AD?  Express your answer as a decimal to the nearest hundredth.

--> Click on PDF attachment for diagram.

[hide="ANSWER"]10.06[/hide]

[hide]Recognizing Pythagorean Triplets tells us that BC=12.
Let BD=x so CD=12-x.
The Angle Bisector Theorem tells us that $ \frac{9}{x}=\frac{15}{12-x}$ or
$ 108-9x=15x$
$ x=\frac{9}{2}$ Applying the Pythagorean Theorem to triangle ABD gives that AD is approximately 10.06[/hide]


category theory
abstract algebra
superior algebra
superior algebra unsolved

Let $R$ a ring, $I$ ideal of $R$, $S$ a multiplicatively closed subset of $R$. Prove that $S^{-1}I=S^{-1}R$ iff $I\cap S\neq\phi$.

Is $R$ commutative and what does the notation $S^{-1}$ stand for ?

$S^{-1}R$ is as always the localization of $R$ at $S$.

Here is my proof:

$M \mapsto S^{-1}M$ is an exact functor because $S^{-1}M = S^{-1}R \otimes_{R}M$ and $S^{-1}R$ is a flat R-module.

We have the short exact sequence $0 \to I \to R \to R/I \to 0$, and therefore the short exact sequence $0 \to S^{-1}I \to S^{-1}R \to S^{-1}(R/I) \to 0$.

Here $S^{-1}I \to S^{-1}R$ is surjective if and only if the kernel of $S^{-1}R \to S^{-1}(R/I)$ is all of $S^{-1}R$. But this map is also surjective, so this happens if and only if $S^{-1}(R/I) = 0$.

Now $S^{-1}(R/I) \simeq \bar{S}^{-1}(R/I)$ where $\bar{S}$ is the image of S in R/I. And this is 0 if and only if $0 \in \bar{S}$ which happens when and only when $S \cap I \neq \emptyset$.

[quote="henrik"]$M \mapsto S^{-1}M$ is an exact functor because $S^{-1}M = S^{-1}R \otimes_{R}M$ and $S^{-1}R$ is a flat R-module[/quote]

Usually, this implication is vice-versa ;)

I'm not quite sure about that! May I see a proof? :maybe:

you've already got a proof (although, as always, you did not show any cooperation, also in the case of these trivial exercises ...)! what's unclear about it?


A woman wants to open a lock. She has 10 keys. She tries them one by one, setting them aside if the lock does not open. Find the chance
that the seventh key opens the lock.

For the seventh key to open the lock, she must have picked six wrong keys first.

I think that there is a probability of $ \frac{9}{10}$ of picking an incorrect key the first time, then $ \frac{8}{9}$ and so on, giving

$ \frac{9}{10} \cdot \frac{8}{9} \cdot \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{4}{5} \cdot \frac{1}{4} \equal{} \frac{1}{10}$.

Or you could think of it as "Of the letters A B C D E F G H I J, what is the probability that A is 7th?" In this case, A is the key that unlocks the lock and B C D E..... are all the other keys. A has an equal probability of being in any of the 10 spots: aka 1/10 probability of being first, 1/10 probability of being 2nd, 1/10th probability of being 3rd and so on. So A has a 1/10 probability of being 7th.

Yet another way to do this is find the total number of ways to arrange the keys (10!) and then the number of ways to arrange the keys with A being 7th (9!) and divide the two to get 1/10


articles MIT college Princeton Harvard Yale Stanford

I've read some articles saying that the gap between the quality of education at schools ranked say, #1-20 is very minute.  Meaning, top schools (MIT, Princeton, Harvard, etc.) compared with good, but not elite schools (Michigan, Minnesota, Rice, Texas, etc.).

What do you guys think about this?  How big really is the gap between these schools?  And if this gap isn't that big, do you think it's really worth it to travel long distances and pay big bucks for schools like Harvard? A lot of people have told me its much more important to go to a good school for a Graduate/Professional degree than for undergraduate.

As someone living in Austin, I will most likely go to Texas (UT Austin), or possibly Rice/A&M, which are among those second-tier schools I mentioned.

The general definition of "second tier" is very different from yours. Rice and Michigan are in no way second tier schools. What's your definition of first and second tier?

Sorry if my meaning is unclear.  I mean to compare the very elite, ivy league schools, (ie. Harvard) with good but not elite schools (as I said before, schools like Michigan and Rice).

Just wondering, what is the general definition of first and second tier?

'Lots' of people consider Harvard, Princeton, Yale, MIT, Stanford, and Caltech to be the definitive first tier schools (inasmuch as undergrad is concerned).  I personally concur with this, although I'm sure others do not.


group theory
abstract algebra
modular arithmetic
linear algebra
matrix
algebra
polynomial

This is a very beautiful problem:
  a) Let $K$ be a finite field and let $G$ be the set of linear transformations of the form $f(x,y)=(ax+by, y)$ with $a,b\in K$, $b\ne 0$. Prove that $G$ has a  subgroup of order $n$ if and only if $n$ can be written $NM$ with $N\parallel K|$, $M\parallel K|-1$ and $M=1\pmod N$.
  b) If $n$ has the property that any group of order divisible by $n$ has a subgroup of order $n$, then $n$ is a power of a prime. Is the converse true?

There some things I don't quite get:

First of all, I think $a$ should be non-zero, not $b$, since $G$ is just the group of matrices of the form $\begin{pmatrix}a&b\\0&1\end{pmatrix}$.

That's is not so important though. What I'm having trouble with is this: the order of $G$ is $|K|\cdot(|K|-1)$. If we take $n=|K|\cdot(|K|-1)$ then of course, $G$ has subgroups of order $n$. The only candidates for $N,M$, however, are $N=|K|$ and $M=|K|-1$, and clearly, $M\ne 1\pmod N$..:? Should it be the other way around (that is, $N=1\pmod M$)?

Sorry if I'm missing something obvious.

I'm sorry, you are right for the two remarks.

Let $K$ be a field with $q=p^{k}$ elements for some prime $p$ and some positive integer $k$.

(a)

Let $H\le G$ be a subgroup of $G$. Since $G$ has $q(q-1)$ elements, the order $n$ of $H$ is indeed of the form $n=uv,\ u|q,\ v|q-1$. We want to prove that $v|u-1$. Notice that the subgroup $P$ of $H$ consisting of matrices of the form $\begin{pmatrix}1&*\\0&1\end{pmatrix}$ is the unique Sylow $p$-subgroup of $H$ (and so it has cardinality $u$). The action of $\begin{pmatrix}a&*\\0&1\end{pmatrix}\ (\#)$ by conjugation on $P$ depends only on $a$, as does the equivalence class of $(\#)$ modulo $P$. Moreover, conjugation by $(\#)$ sends $\begin{pmatrix}1&b\\0&1\end{pmatrix}$ to $\begin{pmatrix}1&ab\\0&1\end{pmatrix}$. This tells us what the orbits of this action look like: the orbit of the identity is, of course, trivial, and all the other orbits have precisely $\frac{|H|}{|P|}=v$ elements. This shows that $P\setminus\{1\}$ is partitioned into $v$-element subsets, so $v|u-1$, as desired.

Conversely, let $n=uv,\ u|q,\ v|q-1$ and $v|u-1$. Suppose $u=p^{l},\ l\le k$. Since $v$ divides both $p^{k}-1$ and $p^{l}-1$, it divides $p^{d}-1$, where $d=(k,l)$. The roots of the polynomial $x^{p^{d}}-x$ form a subfield $K'$ of $K$ with $p^{d}$ elements. Let $V$ be a vector subspace of $K$ over $K'$ of dimension $\frac{l}{d}$ (and hence of cardinality $u=p^{l}$), and let $a$ be a primitive $v$’th root of $1$ contained in $K'$. The subgroup of $\mbox{Gl}_{2}(K)$ generated by the groups $\left\{\begin{pmatrix}a^{n}&0\\0&1\end{pmatrix}\ |\ n\in\mathbb Z\right\}$ and $\left\{\begin{pmatrix}1&b\\0&1\end{pmatrix}\ |\ b\in V\right\}$ has the properties we’re looking for.

(b)

This is a direct consequence of (a). Let $n=p_{1}^{a_{1}}\cdot\ldots\cdot p_{t}^{a_{t}},\ t\ge 2$, and assume that $p_{1}^{a_{1}}$ is the smallest among all $p_{i}^{a_{i}},\ i=\overline{1,t}$. Then we can find a field $K$ with $p^{k}=p_{1}^{k}$ elements for some large $k$ such that $n$ divides the order of the group $G$ defined as above, but $n$ does not satisfy the condition proved necessary for $G$ to have a subgroup of order $n$ (this condition is $\frac n{p_{1}^{a_{1}}}\ |\ p_{1}^{a_{1}}-1$, which is clearly impossible because of the minimality of $p_{1}^{a_{1}}$ among $p_{i}^{a_{i}}$).

P.S.

The converse of (b) is true and well-known :).


Support
abstract algebra
calculus
integration

Me and my mate have been trying to solve this question for 2 weeks now and we haven't made any progress. We were given it for a challenge by our maths teacher and he knows we won't be able to solve it, he just likes toying with us, but i want to rpove him wrong...with your help :P

The question is as follows:

The pilot of a small plane is attempting to land on a small airstrip, and the wheels hit the runway only 1000 metres short of a large group of trees at the end of the runway.  The plane’s speed on the tarmac is modelled by the equation v = A – 10^kt, where v the plane’s velocity in m/s, t is the time taken in seconds to come to a complete halt, and A and k are constants.

b)  Investigate at least 3 different values of k between (but NOT including) 0.01 and 0.2 and for each value determine whether the plane comes to a complete stop before the end of the runway.  Use any graphs and/or diagrams, any numerical techniques and any technology to support your arguments, justify your procedures.  What physical conditions could make k vary?  List any assumptions you have made.

Would there be any values of k (not just the values between 0.01 and 0.2) for which this equation could not be used to model the velocity of a plane landing on a tarmac?

My friend does the hardest maths module at my high-school and yet still cannot get this question out, neither can i. I know it is a lost of physics as well but i think there may be some integration needed to solve. Any help would be appreciated.

Cheers

Hi, welcome to the forum.
I've got two questions for you.
1. Do you know the value of $A$? I don't think you can solve the problem without it.
2. Can you integrate simple expressions like $10^{kt}$? Is it at your level?

"t is the time taken in seconds to come to a complete halt"
so, is t a variable? I'll assume it is.

a) find the t value when v is zero, that would be the time it takes for a complete stop. integrate v from 0 to this value, that is the stopping distance (in terms of k). Then see when the distance is less than 1000m

b) k cannot be negative or zero, since the plane is supposed to slow down, not speed up. If you want to be more precise, the model isn't a very good model at all, since either case, (k>0 or not), the limit of v as t goes to infinity is non-zero, unless A is zero and k is negative.


function
induction
algebra proposed
algebra

Find all $ f: \mathbb{R}\rightarrow\mathbb{R}$ which satisfies
$ f(xy \plus{} f(x \plus{} y)) \equal{} xy \plus{} f(x) \plus{} f(y)$
for all $ x,y \in \mathbb{R}$.

Here's my progress (Still not coming up with the solution.  :(  )

Substitute $ y \leftarrow 0$, $ f(f(x)) \equal{} f(x) \plus{} f(0)$. Denote by $ a$ the term $ f(0)$. This means $ f(y) \equal{} y \plus{} a \forall y \in Im(f)$.

Consider any $ y \in Im(f)$, substituting $ x \rightarrow a$ yields $ f(ay \plus{} f(a \plus{} y)) \equal{} ay \plus{} f(a) \plus{} f(y)$ (1).

Note that $ a \plus{} y \equal{} f(y) \in Im(f)$ and $ f(a) \equal{} 2a$, $ (1) \Rightarrow f(ay \plus{} y \plus{} 2a) \equal{} ay \plus{} y \plus{} 2a$
$ \Rightarrow ay \plus{} y \plus{} 2a \in Im(f) \Rightarrow f(ay \plus{} y \plus{} 2a) \equal{} ay \plus{} y \plus{} 3a$
So, $ ay \plus{} y \plus{} 2a \equal{} ay \plus{} y \plus{} 3a$ or $ a \equal{} 0$.

For now, we have: $ f(0) \equal{} 0$ and $ f(x) \equal{} x \forall x \in Im(f)$.

How do we proceed ? Could we prove that $ f$ is surjective ?

[quote="mathangel"]
Note that $ a \plus{} y \equal{} f(y) \in Im(f)$ and $ f(a) \equal{} 2a$, $ (1) \Rightarrow f(ay \plus{} y \plus{} 2a) \equal{} ay \plus{} y \plus{} 2a$[/quote]

I think there is a $ a$ missing in the final RHS, since $ f(a)\plus{}f(y) \equal{} 2a\plus{}y\plus{}a$.
It follows that we don't have $ a\equal{}0$, which is confirmed by the fact that any function of the form $ f(x)\equal{}x\plus{}c$, where $ c$ is a constant, is a solution of the problem.

Pierre.

[quote=Ryoma.Echizen]What is $Im(f)$?[/quote]
image/range of $f$, $f(\mathbb{R})$


hi guys

Im in the toronto/scarborough area, my school just told me that there have been a little problem in ordering AIME II --they were asked to send in money before AIME but our school doesnt allow that, somehow. so the math department emailed AIME but no reply yet.

So i'm wondering if any of your schools, if, in that area, offers AIME to students who took AMC in their own school?

Thanks!

[quote="sansan"]they were asked to send in money before AIME but our school doesnt allow that, somehow. [/quote]

Doesn't allow?
What a [b][color=red]-censored-[/color][/b] school policy

(censored on my own)

LOl I don't know
i think because the money has to go through our school board, that's why
the school board has some policies regarding prepayments...they somehow want to receive the exams before they pay.....can i somehow write it though?

First of all, you should ask someone from your school to [b]call[/b] the AMC to figure out what's going on (the number is on the inside cover of the AIME Teachers' Manual).

The fee is only 25 dollars US.  (Maybe more for Canada.)  It's absurd to me that a school won't fork over such a small amount of money (for a school) without having to go through layers of bureaucracy.  The AMC does accept credit cards so maybe your parents can just pay the fee directly (if they are so inclined).

The problem with taking it at a different school is that for the AIME II, each school generally only has exactly enough copies of the contest for their own students, and no extras.  You might be able to arrange to take it at a different school if their AIME exams haven't been shipped yet.

But I would start by having your teacher call the AMC and seeing what your options are.  Good luck!

HI

My teacher has emailed the coordinator but I asked this morning, he said there's still no reply yet.

does anyone know if the email they provide on the manual is active? people do use that right?

thanks!

The most obvious solution is to ask your parents, or some other sympathic adult, to write the check. It isnt that much money.

well thanks everyone

I've tried to ask my teacher about it, but he said he would take care of it, so i don't want to sound/act as if I can't trust him.

even though i really want to pay the 25 dollars on my own, i dont know how to ask him...

should i keep waiting or what?

thanks!

I would just recommend that you talk to whoever runs the competitions and pay.

Last year, I had to pay for the exam, as my school would never imagine paying for anything like that. (In fact, we have to pay 3 dollars each for the AIME.) Their philosophy is that if they didn't force you to do the competition, they won't pay for it. Anyway, all I'm saying is that it's certainly possible for you to pay, as I did last year.

[quote="sansan"]hi calc

how did you pay last year?
did you write a checque and mail it to AMC organizers?
when did you pay, did you pay before your AIME? do you still know the deadline?

thanks![/quote]

I think all I did was give 25 dollars to the math department head at my school, though I don't exactly remember.

25 dollars in checque right?

also, did you pay before AIME ?

do you think i still have enough time...? i should pay it tomorrow then!

Yes, the email is active.  It is checked and replied to daily, provided the mail was sent to the address on the contest booklets.  (We have other addresses used for other purposes, and those may not be answered daily as workflow increases or decreases.)

This is a simple problem that can be easily resolved between the school and the AMC office.   We have business procedures that allow a teacher to order with out prepayment in cash.  The best procedure is for the teacher to call the AMC office and make arrangements.

 It does not make sense of a number of non-involved students to attempt to resolve the issue here without the teacher and without the AMC office.  Please allow teachers and the AMC office to conduct regular business with our regular procedures.  Some trust in your teachers would also be well-placed.

Steve Dunbar
Director, American Mathematics Competitions


AMC AIME USA(J)MO USAMO MATHCOUNTS search

i wasnt sure where to put this post, sorry if its in the wrong place. i was wondering for all you amc, aime, & usamo people out there, did you attend any kind of math program or participate in some workshop, or read some book to improve your scores? thanks in advance  :lol:

You've come to the right place :D

Math Programs: Tons of them out there, you can find info about them in the "Other US Contests & Programs" forum.
Classes: AoPS offers classes ranging from MathCounts to Olympiad level. Go to the home page and click on the "Classes" link.
Books: AoPS v1 and v2 are very useful for AMC/AIME and perhaps even USAMO. There's a lot of talk about other books randomly in the forum, so you can search them up.


Yann and Camille go to a restaurant.  If there are 10 items on the menu, and each orders one dish, how many different combinations of meals can Yann and Camille order if they refuse to order the same dish?

Yann can order 1 of 10 dishes and Camille can order 1 of 9 dishes (since they don't order the same one)

therefore, there are 90 possibilities

Note that vallon got this by doing $ 10 \times 9 \equal{} 90$.

Since I hace 10 things and the friends has 9 the total is just 90. (remember you don't eivide by two since we have distinct people.


function
limit
real analysis
real analysis unsolved

Hey I have a question that I can't seem to figure out.  I have to find the limit as n goes to infinity of:

lim (X^n/(1+X^n))

And this should equal 0, but I keep getting 1.  Any help would be greatly appreaciated!

Thanks

Set $f_n(x)=\frac{x^n}{1+x^n}$. Then \[ \lim_{n\rightarrow\infty}f_n(x)=\begin{cases}1,& x>1\\ \frac{1}{2},& x=1\\ 0,&1\ge x\ge 0\end{cases} \] I think the limit does not exist for $x<0$.


trigonometry
calculus
conics
parabola
analytic geometry
graphing lines
slope

A certain football quarterback can throw a football a maximum range of 80 meters on level ground. What is the highest point reached by the football if thrown this maximum range? Ignore air friction.

a) 10m
b) 20m
c) 30m
d) 40m
e) 50m

Here's my long-winded [hide="solution"]

Let $ v_{o}$ be the initial speed, $ t$ the total time in flight and $ h$ the maximum height.

We have 3 equations:

1) Height and time: $ h \equal{} v_{o}sin\theta \frac{t}{2} \minus{} \frac{1}{2}a(\frac{t}{2})^2$

2) Range and time: $ R \equal{} v_{o}cos\theta t$ or $ t \equal{} \frac{R}{v_{o}cos\theta}$

3) Conservation of energy: $ \frac{1}{2}mv_{o}^2 \equal{} \frac{1}{2}m(v_{o}cos\theta)^2 \plus{} mgh$ or equivalently $ v_{o}^2 \equal{} \frac{gh}{\frac{1}{2} \minus{} \frac{1}{2}cos^2\theta}$

Now to plug E2 into E1:

$ h \equal{} v_{o}sin\theta \frac{R}{2v_{o}cos\theta} \minus{} \frac{1}{2}a(\frac{R}{2v_{o}cos\theta})^2 \equal{} \frac{rtan\theta}{2} \minus{} \frac{aR^2}{8v_{o}^2cos^2\theta}$

Plugging E3 into the above equation:

$ h \equal{} \frac{rtan\theta}{2} \minus{} \frac{aR^2(\frac{1}{2} \minus{} \frac{1}{2}cos^2\theta)}{8ghcos^2\theta}$

Since the maximum range is achieved only when $ \theta \equal{} \frac{\pi}{4}$:

$ h \equal{} \frac{r}{2} \minus{} \frac{aR^2(\frac{1}{2} \minus{} \frac{1}{2}(\frac {\sqrt{2}}{2})^2)}{8gh(\frac {\sqrt{2}}{2})^2} \equal{} \frac{R}{2} \minus{} \frac{aR^2}{16gh}$

Thus, $ h \equal{} 20$ [/hide]

With that ugly mess written, can someone suggest an easier method?

This one can be done without using energy.

[hide="solution"]$ v_0 \sin(\theta) t \equal{} 80$ [i][b](1)[/b][/i]
$ v \equal{} v_0 \plus{} at \implies 2v_0 \cos\theta \equal{} gt \implies t \equal{} \frac {2v_0\cos\theta}{g}$ [i][b](2)[/b][/i]

Plug [i][b](2)[/b][/i] back in to [i][b](1)[/b][/i] to get $ \frac {2v_0^2\sin\theta\cos\theta}{g}$. And since $ \sin\theta \equal{} \cos\theta$ when $ \theta \equal{} \frac {pi}{4}$, this simplifies to $ \frac {2v_0^2\cos^2\theta}{g}$[i][b] (3)[/b][/i]

So, for the vertical motion, $ v \equal{} v_0^2 \plus{} 2at$ and plugging in values will get $ 0 \equal{} (v_0\cos\theta)^2 \minus{} 2gx \implies x \equal{} \frac {(v_0\cos\theta)^2}{2g}$. [i][b](4)[/b][/i]

Plug the above equation [i][b](3) [/b][/i]into [i][b](4)[/b][/i] to get $ x \equal{} \frac {40g}{2g} \equal{} \boxed{20m}$.[/hide]

[quote="pascal12"]This one can be done without using energy.

[hide="solution"]$ v_0 \sin(\theta) t \equal{} 80$
$ v \equal{} v_0 \plus{} at \implies 2v_0 \cos\theta \equal{} gt \implies t \equal{} \frac {2v_0\cos\theta}{g}$

Plug it back in to get $ \frac {2v_0^2\sin\theta\cos\theta}{g}$. And since $ \sin\theta \equal{} \cos\theta$ when $ \theta \equal{} \frac {pi}{4}$, this simplifies to $ \frac {2v_0^2\cos^2\theta}{g}$

So, for the vertical motion, $ v \equal{} v_0^2 \plus{} 2at$ and plugging in values will get $ 0 \equal{} (v_0\cos\theta)^2 \minus{} 2gx \implies x \equal{} \frac {(v_0\cos\theta)^2}{2g}$.

Plug the above equation to get $ x \equal{} \frac {40g}{2g} \equal{} \boxed{20m}$.[/hide][/quote]

Thanks. Your "it"s and "plug the equation" without references to the target were kinda confusing, but I understand that the three main equations that you used are:

1) $ v_{o}sin\theta t \equal{} 80$

2) $ v \equal{} v_{o} \plus{} at$

3) $ 0 \equal{} (v_{o}sin\theta)^2 \minus{} 2gx$

I guess your method is shorter and involves less algebra because you used different equations from what I used.

Oh yeah, sorry, I should have made that clearer. Ok, I edited it and hopefully it should make more sense.

Also, I generally try to avoid the equation $ x \equal{} vt \plus{} \frac {1}{2}at^2$ because unless $ v \equal{} 0$, it usually leads to something really messy.

Have you taken calculus? If so, you will know some facts (unrelated to physics) that will make this problem doable with almost no calculation.

[hide="Comments"][b]Facts:[/b]
- The ball follows a parabolic path
- All parabolas are similar
- The distance is maximized at an angle of 45 degrees, which corresponds to a slope of magnitude 1.

I just want the ratio of the height to the distance for this throw. I can use any parabola I want, since they are all similar! I'll use $ y \equal{} x^2$. The slope has magnitude one at $ x \equal{} \pm 1/2$. The height of the portion of the parabola within these bounds is $ \frac{1}{4}$, and he width is just the difference in $ x$ values, or $ 1$, so the height is 1/4 of the width.[/hide]

While I certainly am not encouraging anyone to use methods like that to solve problems, it can be helpful when there is little time left, or for checking answers in a different way.

That's a really ingenious way to solve the problem. I'll definitely look out for these mathematical methods from now on. Thanks!

It's useful to know the range and height formulas for projectile motion:

$ R \equal{} \frac{v_0^2 \sin(2\theta)}{g}$

$ H \equal{} \frac{v_0^2 \sin^2(\theta)}{2g}$

let: y=-(x/a)^2+(x/b)

from dy/dx = 1 at x=0 we get b = 1

hence range is a (the other solution -(x/a)^2+(x) = 0)

height is y at a/2.


[quote="RDeepMath91"]Mykolka the numismatist possesses $241$ coins, each worth an integer number of turgiks. The total value of the coins is $360$ turgiks. Is is necessarily true that the coins can be divided into three groups of equal total value ?  :maybe:[/quote]

Let $a_{1}$ the number of coins whose value is exactly $1$ turgik. Considering that all the other coins have a value of at least $2$, we have $a_{1}+2(241-a_{1})\leq 360$, and so $a_{1}\geq 122$
Since we have $241$ coins whose total value is $360$, the biggest coin value possible is $120$

Now split all the $241-a_{1}$ coins whose values are greater than $1$ turgik in two groups A and B whose difference of values is minimum, say $B=A+\Delta$. Take then one coin in B (any coin). Let $m$ its value. Obviously, moving this coin from group B to group A may not reduce the difference between A and B (since $\Delta$ is supposed the minimum one) and so $m\geq \Delta$.
Since $A+\Delta+A+a_{1}=360$ and since $a_{1}\geq 122$, we have $A=180-\frac{a_{1}}{2}-\frac{\Delta}{2}< 120$

Consider now the 3 groups :
$F$ is the group of $a_{1}$ $1$-turgik coins plus the selected $m$-turgiks coin : Total is $a_{1}+m$
$G$ is the group $A$ : Total is $A=180-\frac{a_{1}}{2}-\frac{\Delta}{2}<120$
$H$ is the group $B$ minus the selected $m$-turgiks coin. Total is $A+\Delta-m=180-\frac{a_{1}}{2}+\frac{\Delta}{2}-m\leq A<120$

In order to complete group $G$ up to $120$, we need $C_{G}=120-A=\frac{a_{1}}{2}+\frac{\Delta}{2}-60$
In order to complete group $H$ up to $120$, we need $C_{H}=\frac{a_{1}}{2}-\frac{\Delta}{2}+m-60$

And $C_{G}+C_{H}=a_{1}+m-120$ and, since no coin may have a value $>120$, $C_{G}+C_{H}\leq a_{1}$

Hence, we have enough $1$-turgik coins to move from $F$ to groups $G$ and $H$ to value them at $120$ (and group $F$ will then also have value of $120$).

And it is true that the coins can be divided into three groups of equal total value.

--
Patrick


inequalities
trigonometry
geometry
3D geometry

I'm afraid the left side is only true for $n > \frac{25}{12}$...
Take for example $n=1$ or $n=0$, and the left side doesn't hold.

Now if we assume that $n > \frac{25}{12}$, it is easy to prove:

By AM-GM:

$3(\sqrt[3]{n^2(n+2)} + \sqrt[3]{n(n+2)^2}) \geq 6 \sqrt{n(n+2)}$

Now, because $n > \frac{25}{12}$, it's true that \[ \frac{n}{3} > \frac{25}{36} \]
So \[ n^2+2n > n^2 + \frac{5n}{3} + \frac{25}{36} \]
And taking the square root of both sides yields \[ \sqrt{n(n+2)} > n + \frac{5}{6} \]

Plugging this in the original inequality gives us \[ 3(\sqrt[3]{n^2(n+2)} + \sqrt[3]{n(n+2)^2}) \geq 6 \sqrt{n(n+2)} > 6n + 5 \]

So it works in case $n > \frac{25}{12}$

Notice that between the inequality sines, we have something in the form $3a^2b + 3ab^2$, though cube roots are involved.  Does that bring any methods ideas to mind?


If a curve with a radius of 88m is perfectly banked for a car traveling 75 km/hr, what must be the coefficient of static friction for a car not to skid when traveling at 95 km/hr.

I got [hide="ans"]0.304.[/hide]

clarification:  when the road is "perfectly banked," does that mean no static friction is needed to keep the car on the road?

yes, and your answer is wrong : (

I'll post a solution tomorrow if you need it.

whoops, forgot to base my $ \mu_s$ on the normal force.

[hide]
.352?
[/hide]

noope =[

hmm i don't feel like posting the solution
maybe this will get you to the answer
http://en.wikipedia.org/wiki/Banked_curve#Banked_turns_with_friction

actually, that basically gives the answer to the solution... :|

bleh, now I'm too lazy.

had my centripetal force going the wrong direction.
it's been too long...

Yeah, the most difficult part (for me at least) was realizing that the frictional force does not go towards the center of the circle but rather vertically below that center and down the incline.

Friction force is parallel to contact surfaces.