HMMT
summer program
PROMYS
number theory
prime numbers

Well, I finally have full solutions to all 9 of this years questions (let's just hope they are right ;) ), but I have a few questions about the application questions. For the second question, it asks us:

[quote]Tell us about any other mathematical experiences you have had. For example, have you ever done a special mathematics project, or entered a math competition?[/quote]

I was just wondering, does entering a math competition include stuff like the AMC's or the HMMT? I have been doing all of that kind of stuff (like math team, etc.) but I wasn't sure if that was what the question is asking. As far as a "special math project" goes, the only thing I really ever did was did some reasearch on the partitioning of integers and prime numbers, but I didn't write a paper or anything. What else do you think should be put as an answer to this question?

My next question is regarding the "essay" question that you are supposed to write telling them about your mathematical development. What exactly does it mean by mathematicial "development"? Also, how long should it be?

[quote="fanaticsm"]I was just wondering, does entering a math competition include stuff like the AMC's or the HMMT? I have been doing all of that kind of stuff (like math team, etc.) but I wasn't sure if that was what the question is asking. . . .

My next question is regarding the "essay" question that you are supposed to write telling them about your mathematical development. What exactly does it mean by mathematicial "development"? Also, how long should it be?[/quote]

Of course the AMCs and HMMT are math competitions. There are high school kids who have never done any of those, so you definitely want to mention experiences like those if you have them on your PROMYS application.

Your next question is an English writing question. Not to write it for you, I will just say that certainly you would want to describe how you got more interested in math, and more capable at math, over the course of your childhood until now.

That's enough from me. Good luck on your application. Please tell us how PROMYS was after you attend.

Ahh I see, thanks for the help tokenadult!

Well, I finally mailed in my application today...just have to wait to hear a decision now :D


[quote="enndb0x"][quote="Farenhajt"]Solve the following equations:

1. $ \left\lfloor x\{x\}\right\rfloor \plus{} \left\{x\lfloor x\rfloor\right\} \equal{} x$

2. $ \left\lfloor x\{x\}\right\rfloor \minus{} \left\{x\lfloor x\rfloor\right\} \equal{} x$[/quote]

Hi,
 1. The fractional part of $ x$ is the fractional part of number $ \left\lfloor x\{x\}\right\rfloor \plus{} \left\{x\lfloor x\rfloor\right\}$, so $ \left\{x\lfloor x\rfloor\right\} \equal{} \left\{x\right\}$ $ \Longrightarrow \lfloor x \rfloor \equal{} 1$.Also $ \left\lfloor x\{x\}\right\rfloor \equal{} \lfloor x\rfloor \equal{} 1$
Now we just have to solve the inequality $ x(x \minus{} 1)\geq 1$ and $ x < 2$ since $ \lfloor x \rfloor \equal{} 1$

 If I'm correct to 1 :blush:  the solution to the second equation will look like the first.[/quote]

Your solution to $ 1$ is not complete,  $ \left\{x\lfloor x\rfloor\right\} \equal{} \{x \}$ just means $ \{x(\lfloor x \rfloor \minus{} 1)\} \equal{} 0$ ,so there are some more steps .


parameterization function calculus derivative logarithms

Let be the equation :

$ a^x \equal{} \left| x\plus{} 2 \right| \minus{} \left| 2x \plus{} 8 \right|$ where $ a > 0$ . Find $ a$ such that the given equation has only one solution .

[hide="hint"]
Break down the right hand side by cases and look for extrema
[/hide]

[hide]
If you break down the RHS into cases split by -4 and -2 and analyze (either by graph or sign checks), you will see that there is a maximum at -4. (The function is increasing for $ x<\minus{}4$, and it is decreasing for $ x>\minus{}4$.

If the exponential is to hit the function exactly once, it must hit it only at that one point, the maximum.
plugging x=-4, the equation becomes.

$ a^{\minus{}4} \equal{} 2$

$ a \equal{} \frac{1}{\sqrt[4]{2}}$
[/hide]

Note $ a^x > 0$. Let $ f(x)$ be the RHS. We can determine that $ f(x)$ is positive only in $ ( \minus{} 6, \minus{} 10/3)$ and that

$ f(x) \equal{} x \plus{} 6$ when $ x \in ( \minus{} 6, \minus{} 4)$
$ f(x) \equal{} \minus{} 3x \minus{} 10$ when $ x \in ( \minus{} 4, \minus{} 10/3)$

Suppose $ a^x < f(x)$ for some value of $ x$. This must occur in $ ( \minus{} 6, \minus{} 10/3)$ Note that $ a^x > f(x)$ at $ x \equal{} \minus{} 6, \minus{} 10/3$. Since $ a^x$ and $ f(x)$ are continuous, there must be two intersections, contradicting the condition. So when $ a^x$ and $ f(x)$ intersect, one of the following must be true:

1. $ a^x$ intersects $ f(x)$ at its cusp.
2. $ a^x$ and $ f(x)$ are tangent.

Case 1: The cusp of $ f(x)$ is at $ ( \minus{} 4,2)$, so $ a^{ \minus{} 4} \equal{} 2 \Rightarrow a \equal{} \sqrt [4]{1/2}$. The derivative of $ a$ at $ \minus{} 4$ is $ \ln \sqrt {1/2} > \minus{} 3$, so $ a^x$ and $ f(x)$ do not intersect again. So $ \boxed{\sqrt [4]{1/2}}$

Case 2 is messy. Calculators show that there are no solutions this way, but I think you have to do some tricky bounding that I didn't see at first glance.


Prove that for all $ a,b,c$ be nonnegative real numbers, we have
\[ a^3\plus{}b^3\plus{}c^3\plus{}6abc\ge a^2(b\plus{}c)\plus{}b^2(c\plus{}a)\plus{}c^2(a\plus{}b)\plus{}2abc\left(\frac a{b\plus{}c}\plus{}\frac b{c\plus{}a}\plus{}\frac c{a\plus{}b}\right)
\]

[quote="dduclam"]Prove that for all $ a,b,c$ be nonnegative real numbers, we have
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\ge a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \plus{} 2abc\left(\frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b}\right)
\]
[/quote]
It's equivalent to $ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\plus{}\sum_{cyc}ab(a^2\plus{}ab\plus{}b^2\minus{}c^2)(a\minus{}b)^2\geq0,$ which is obviously true.

I found out the general problem follows:

Consider the following Inequality
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(k \plus{} 1)abc\ge \sum_{cyc} a^2(b \plus{} c) \plus{} 2kabc\left(\frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b}\right)
\]
$ (a)$ Prove above Inequality when $ k\le1$ and $ a,b,c$ be nonnegative real numbers.
$ (b)$ Prove above Inequality when $ 1\le k\le 2$ and $ a,b,c\in[k \minus{} 1; 1].$

When the equality occous?


Does anyone who took USAPHO last year have the 2005 tests on the local level?

Thank you!

I don't think we were allowed to keep the tests-you do mean the qualifying test, right?

I mean the very first test that you take to see if you make it to semifinals.

yeah, that'd be the qualifiers.

I kinda remember the problems, but I'm pretty sure we weren't allowed to keep them.

Of course we're allowed to keep them...
they are avalible online ... http://www.compadre.org/psrc/evals/olympiad.cfm
but 2005 is not on there

I have it though so PM me if you want it...

And I'd appreciate it if someone gave me solutions to previous years


Now, when the hard one has been solved, let me post an easy one:

Let $f,g:[0,1]\to[0,1]$ be continuous. Assume that $f$ is non-decreasing. Then $\int_0^1 f(g(x))\,dx\leq \int_0^1 f(x)\,dx+\int_0^1 g(x)\,dx$.

This one was posted by Valentin, it is also from Leningrad olympiad. Is it is easy?  ;)

Solution by Pliviu.

Suppose for some $c$ we have $f(c)-c> \int_0^1 f(x)dx$. But \[\int_0^1 f(x)dx\geq \int_c^1 f(x)dx\geq (1-c)f(c).\] So $f(c)-c>(1-c)f(c)$. It follows that $cf(c)>c$ -- contradiction!
Therefore $f(c)-c \leq \int_0^1 f(x)dx.$
Take $n$ a positive integer and $a_{i,n}=\sup\{g(x)\mid (i-1)/n\leq x\leq i/n\}$.
We have $\frac{1}{n}(f(a_{i,n})-a_{i,n})\leq \frac{1}{n} \int_0^1 f(x)dx$ for all $i$.
Sum it up for all $i=\overline{1,n}$ and get \[\frac{1}{n} \sum_{i=1}^n f(a_{i,n}) \leq \int_0^1 f(x)dx + \frac{1}{n} \sum_{i=1}^{n} a_{i,n}.\]
But $a_{i,n}\geq g(i/n)$, so $f(g(i/n))\leq f(a_{i,n})$. It implies that \[\frac{1}{n}\sum_{i=1}^{n} f(g(i/n)) \leq \int f(x)dx + \frac{1}{n} \sum_{i=1}^n a_{i,n}.\]
Let now $n\to \infty$ and obtain $\int_0^1 f(g(x))dx\leq \int_0^1 f(x)dx + \int_0^1 g(x)dx$.


graphe sadeye $G$ ba $n$ raas va $m$ yal dar nazar begirid.. sabet konid $G$ ,$2^{m-n+1}$ zirderakhte faragir darad ke darajeye tamame rOose an zoj ast.

age ye dore n rasi ro dar nazar begiri kheyli bishtar az 2^(n-n+1) zir derakhte faragir dare ke darajeye tamamiye rOose an dar graphe asly zoj bashe!(be vozooh nemitoone darajeye tamame rOos too khode zirderakht zoj bashe).

mishe yekam bishtar tozih bedi?

vaghan sorry zir derakht na !!!!! zir graphe faragir ....


might I ask WHAT IS THIS TOPIC
at least put it in the right forum....

First of all this shouldn't be in this forum.

Second, What! No Em (Eminem in case if you don't know who he is), no Lloyd Banks, no Ludacris? What is this This is outrageous! Ok. In this case, dre is good...

EDIT: Em, Banks, and Ludacris are way better than rappers like Jibbs...

ok now its in the right forum.

rap sucks.
boo

rape
tr.v. raped, rap·ing, rapes
1. To force (another person) to submit to sex acts, especially sexual intercourse; commit rape on.
2. To seize and carry off by force.
3. To plunder or pillage.

[b]raper[/b] n.

Did you mean [i]rapper[/i]?

Dude where's Eminem yo. He's my favorite rapper!
say word

I'm sorry for the missing choices, put it just didn't allow me to put anymore. By the way Em is nice, but none could mess with the B.I.G.

By the way if you didn't kno, today is 10 years since biggie died. I went to Bed Stuy, cause there was supposed to be some huge undertground party, but i left cause people started gettin high and I didn't want to get in trouble.

the fact that this thread is on mathlinks of all places is brilliant.


calculus
integration
logarithms
symmetry
trigonometry
calculus computations

Hello
[hide="hint 1"]
Use partial fraction.
[hide="hint 2"]
$ \frac {1}{x(x^6 \plus{} 1)^2} \equal{} \frac {1}{x} \minus{} \frac {x^5}{x^6 \plus{} 1} \minus{} \frac {x^5}{(x^6 \plus{} 1)^2}$[/hide][/hide]
This u can integrate easily.
Thank u.

I'll do you one better:

$ \frac{1}{x(1\plus{}x^k)^2} \equal{} \frac{1}{x} \minus{} \frac{x^{k\minus{}1}}{(1\plus{}x^k)^2} \minus{} \frac{x^{k\minus{}1}}{1\plus{}x^k}$.

[color=blue][b]Thanks All,
And How can u solve it if $ I\equal{}\int {\frac {1}{{x^6 \plus{} 1}}} dx$[/b][/color]

[color=blue][b]We have: :lol:
$ \int {\frac {1}{{x(x^6 \plus{} 1)^2}}dx \equal{} } \frac {1}{3}\int {\frac {{3x^2 }}{{x^3 (x^6 \plus{} 1)^2}}dx}$

Let: $ t \equal{} x^3 \Rightarrow dt \equal{} 3x^2 dx$
Then:
$ \frac {1}{3}\int {\frac {1}{{t(t^2 \plus{} 1)^2}}dt}$
The problem becomes easier.... :rotfl: [/b][/color]

[quote="quocbao153"][color=blue][b]We have: :lol:
$ \int {\frac {1}{{x(x^6 + 1)^2}}dx = } \frac {1}{3}\int {\frac {{3x^2 }}{{x^3 (x^6 + 1)^2}}dx}$

Let: $ t = x^3 \Rightarrow dt = 3x^2 dx$
Then:
$ \frac {1}{3}\int {\frac {1}{{t(t^2 + 1)^2}}dt}$
The problem becomes easier.... :rotfl: [/b][/color][/quote]

\begin{eqnarray*}
\frac{1}{x(x^n+1)}
& = & \frac{A}{x} + \frac{B}{x^n+1} \\
& = & \frac{A x^n+B x+A}{x(x^n+1)}
\end{eqnarray*}
So, $ (A, B) = (1, -x^{n-1})$.

Therefore, $ \frac{1}{x(x^n+1)} = \frac{1}{x} - \frac{x^{n-1}}{x^n+1}$.

Thus,
\begin{eqnarray*}
\int \frac{1}{x(x^n+1)} dx
& = & \int \left(\frac{1}{x} - \frac{x^{n-1}}{x^n+1}\right) dx\\
& = & \int \frac{1}{x} dx - \int \left(\frac{1}{n}\right)\frac{(x^n+1)'}{x^n+1} dx \\
& = & \ln x - \frac{1}{n} \ln (x^n+1)
\end{eqnarray*}

[color=blue][b]Find Integral  :D
$ I \equal{} \int {\frac {1}{{x^6 \plus{} 1}}} dx$[/b][/color]

hello, note that $ \frac{1}{x^6\plus{}1}\equal{}\frac{1}{3(x^2\plus{}1)}\plus{}\frac{\minus{}x^2\plus{}2}{3(x^4\minus{}x^2\plus{}1)}$.
Sonnhard.

If you change your upper limits in both integrals to $ \frac{\pi}{2}$, then using symmetry we obtain quickly $ \minus{}\frac{\pi\ln 2}{2}$ for both integrals (solved many times here on ML).

If you insist on using $ \frac{\pi}{4}$ then the answer will include Catalan constant (that solution is also somewhere on ML, if I recall correctly).

[hide]$ \int_0^{\pi/4} \ln(\sin x)\,dx \equal{} \minus{}\frac{\pi}{4} \ln 2 \minus{} \frac{K}{2}$ (where K is Catalan's constant), which you can see after calculate the sum of the series $ \sum_{n\equal{}1}^{\infty} \frac{\cos(nx)}{n}$. For $ I \equal{} \int_0^{\pi/4} \ln(\cos x)\,dx$ we just have to see that

$ \int_0^{\pi/4} \ln(\sin x)\,dx \equal{} \minus{} \int_{\pi /2}^{\pi /4} \ln\left[\sin \left(\frac{\pi}{2} \minus{} t\right)\right]\,dt \equal{} \int_{\pi /4}^{\pi /2} \ln(\cos x)\,dx$

$ \equal{} \int_0^{\pi /2}\ln(\cos x)\,dx \minus{} \int_0^{\pi /4}\ln(\cos x)\,dx$

and so

$ \minus{}\frac{\pi}{4} \ln 2 \minus{} \frac{K}{2} \equal{} \minus{} \frac{\pi}{2} \ln 2 \minus{} I$

$ I \equal{} \minus{}\frac{\pi}{4} \ln 2 \plus{} \frac{K}{2}$[/hide]

[color=blue][b][quote="quocbao153"]1. $ I \equal{} \int_{0}^{\sqrt {7}} \sqrt {x^3 \plus{} 1}dx$

2. $ j \equal{} \int\limits_1^e {\sqrt {\ln x} } dx$[/quote]
May 2 problems have no solution????????????[/b][/color]

At first sight they don't seem to have a nice closed form. May I ask what is the source of these integrals?


[hide]Change denominators to $6x$ so multiply the first fraction's denominator/numerator by $6$, the second ones by $3$ and the third by $2$. That gives us $\frac{6}{6x} + \frac{3}{6x} + \frac{2}{6x}=\frac{6+3+2}{6x}=\boxed{\frac{11}{6x}}$[/hide]

[hide]
Ruduce to the LCD, which is 6x.
$\frac{6+2+3}{6x}=\frac{11}{6x}$
$D$[/hide]


A triangle has vertices A=(-3,1), B=(-2,-3), C=(x,0)
What is x? The triangle has a area of 7.5.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=282647]Here[/url] is a link with a discussion of the shoelace theorem. It also has very easy to understand explanation of what it is and how to use the theorem, as well as an example. (heh Im a bit biased since I wrote some of this :D )

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=287748]This[/url] has a proof outline for the Shoelace Theorem and has a diagram that actually shows how to use it graphically.

Oh, and to answer your problem:

[hide="Solution"]
We have $ \frac{1}{2} | (\minus{}3 \times \minus{}3 \plus{} \minus{}2 \times 0 \plus{} x \times 1) \minus{} (1 \times \minus{}2 \plus{} \minus{}3 \times x \plus{} \minus{}3 \times 0) | \equal{} \frac{1}{2} | (x\minus{}9)\minus{}(\minus{}2\minus{}3x) | \equal{} \frac{1}{2} |4x\minus{}7|\equal{}7.5$.  You should be able to finish from here.
[/hide]

I think you made a slight calculation error. The solution should look something like [hide="this."] $ 7.5 \equal{} \frac{1}{2}|(x\plus{}9)\minus{}(\minus{}2\minus{}3x)| \equal{} \frac{1}{2}|4x\plus{}11|$. You forgot the double negative with $ \minus{}3 * \minus{}3$. Solving the equation for its double values, we get the [hide="solution."] $ x \equal{} 1, \minus{}6.5$ [/hide] [/hide]

Then, perhaps, since our help is obviously not good enough, you could try the problem yourself...we already told you how.


For any positive real numbers $a,b$ and $c$,
\[\frac{a(b+c)}{a^{2}+bc}+\frac{b(c+a)}{b^{2}+ca}+\frac{c(a+b)}{c^{2}+ab}\le \frac{1}{2}\sqrt{27+(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\]

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac {a(b \plus{} c)}{a^{2} \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^{2} \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^{2} \plus{} ab}\le \frac {1}{2}\sqrt {27 \plus{} (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)}
\]
[/quote]

The Inequality is equivalent to
$ \sum\frac {a^2(b \plus{} c)^2}{(a^2 \plus{} bc)^2} \plus{} 2\sum\frac {ab(b \plus{} c)(c \plus{} a)}{(a^2 \plus{} bc)(b^2 \plus{} ca)}\le\frac {15}2 \plus{} \frac1{4}\left(\sum\frac {b \plus{} c}a\right)$
Notice that  $ (a^2 \plus{} bc)(b^2 \plus{} ca) \minus{} ab(b \plus{} c)(c \plus{} a) \equal{} c(a \plus{} b)(a \minus{} b)^2$
then
$ 2\sum\frac {ab(b \plus{} c)(c \plus{} a)}{(a^2 \plus{} bc)(b^2 \plus{} ca)}\le6$    (1)
Other hand,
$ \sum\frac {a^2(b \plus{} c)^2}{(a^2 \plus{} bc)^2}\le\sum\frac {a^2(b \plus{} c)^2}{4a^2bc} \equal{} \frac1{4}\sum\left(\frac {b}c \plus{} \frac {c}b \plus{} 2\right)$  (2)
From (1) and (2) we have done!

Besides, by the sam ways, we have a nice solution for an old problem:
$ \sum\sqrt{\frac {a(b \plus{} c)}{a^2 \plus{} bc}}\le\sqrt{\left(\sum\sqrt a\right)\left(\sum\frac1{\sqrt a}\right)}$
:)

Dear, Lam. This is exactly my solution too. This inequality is created as an example for proving a nice inequality without any theorem but pure algebraic transformation.

For the inequality you cited, see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=119080[/url]. For additional similar inequalities, see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=78204[/url], post [b]#11[/b].


Suppose that $p>5$ be the prime number and $X=\{p-n^2 \mid n\in \mathbb{N} , n^3<p\}$ .Show that there exists two distinct elements like $x,y$ s.t. $x$ is not equal $1$ and $x\mid y$

[quote]bebakhshid ke saal ro engilisi ferestadam [/quote]

 :P


function
limit
algebra
domain
calculus
topology
calculus computations

Give an example of some real functions $ f, g$ with the following conditions, or prove that such functions do not exist.

$ f, g$ are differentiable over an interval $ (a, b)$ containing zero and \[ \lim_{x\to 0}f(x)\equal{}0\equal{}\lim_{x\to 0}g(x), \ L\equal{}\lim_{x\to 0}\frac{f(x)}{g(x)}, \ L'\equal{}\lim_{x\to 0}\frac{f'(x)}{g'(x)},\]
where $ L, L'\in \mathbb{R}\cup \{\pm\infty\}$, but $ L\neq L'$.

[b]Note[/b] that when defining $ \lim_{x\to t}h(x)$ we do not require $ h$ to be defined in a punctured neighborhood of $ x\equal{}t$. It suffices for our purposes that $ x\equal{}t$ be a limit point of the domain of $ h$.

Diogene, before applying any theorem you should make sure that all of its conditions are satisfied, and L'Hospital's Rule is no exception. Now, as an exercise, state that rule and compare it with my question. Tell me if you still think that L'Hospital's Rule is applicable with some supporting reasons.

Of course such functions as you require , dont exist :  still l'Hôpital rule. Go to http://planetmath.org/encyclopedia/LHpitalsRule.html
 :cool:

Diogene, have you read the 'Note' in my first post? The version of L'Hospital's Rule that you are referring to does not accept this assumption for limits, that is, in this case you are implicitly assuming that $ g'(x)\equal{}0$ has no root over an interval containing zero, but the original question makes a weaker assumption. See 'Note.'

[quote="The Captain"][b]Note[/b] that when defining $ \lim_{x\to t}h(x)$ we do not require $ h$ to be defined in a punctured neighborhood of $ x \equal{} t$. It suffices for our purposes that $ x \equal{} t$ be a limit point of the domain of $ h$.[/quote]

How is this relevant to your original question? because you are given that $ f$ and $ g$ are defined on $ (a,b)$.

$ \frac{f'(x)}{g'(x)}$ need no be defined for each individual $ x\in (a, b)$, that is, in our case $ x\equal{}0$ might be a limit point for $ Z(g'(x))$, the set of zeros of $ g'(x)\equal{}0$, as well as for the domain of $ \frac{f'}{g'}$.

[quote="The Captain"][b]Note[/b] that when defining $ \lim_{x\to t}h(x)$ we do not require $ h$ to be defined in a punctured neighborhood of $ x \equal{} t$. It suffices for our purposes that $ x \equal{} t$ be a limit point of the domain of $ h$.[/quote]
Are you claiming that these two definitions are same? The way I learnt limit was that we define $ \lim_{x\to t}h(x)$ only when $ h$ is defined atleast in a punctured neighborhood of $ t$
Anyway your question seems to be interesting because in L'Hospital's rule we want $ g'(x)\neq 0$ for all $ x\in (a, b)$. So we cannot use L'Hospital's rule and hence claim that $ L\equal{}L'$.

The way we define the limit of a function at a point depends on our context. My definition is a generalization of what you have learned in calculus, and it still can be generalized to topological spaces with the target set usually a Hausdorff space to imply the uniqueness of limits.

[quote="The Captain"]The way we define the limit of a function at a point depends on our context.[/quote]

What's $ \lim_{x\to 0}\frac{f'(x)}{g'(x)}$ according to your definition if $ g'(x)\equal{}0$ for all $ x\in (a,b)$ (Remember $ (a,b)$ contains $ 0$)

When we talk about the limit of $ h(x)$ at the point $ x \equal{} t$, it is assumed that $ x \equal{} t$ is a limit point of the domain of $ h$. So, in this question, it follows that $ x \equal{} 0$ is a limit point of the domain of $ \frac {f'}{g'}$, and hence $ g'$ cannot be zero over the entire interval $ (a,b)$ containing zero.

Once again I ask you to read the question and its assumptions more carefully.


[size=150][color=darkblue][u][b]Problem 2[/b][/u]:[/color][/size]

Let $ a,b,c>0$. Prove that $ \frac{a}{1\plus{}a}\plus{}\frac{2b}{2\plus{}b}\plus{}\frac{3c}{3\plus{}c}\le \frac{6(a\plus{}b\plus{}c)}{6\plus{}a\plus{}b\plus{}c}$

$ <\equal{}>\frac {1}{a \plus{} 1} \plus{} \frac {4}{b \plus{} 2} \plus{} \frac {9}{c \plus{} 3} \ge \frac {36}{a \plus{} b \plus{} c \plus{} 6}$
It's true by cauchy-swarchz

[quote="thanhnam2902"][quote="thanhnam2902"][size=150][color=darkblue][u][b]Problem 2[/b][/u]:[/color][/size]

Let $ a,b,c > 0$. Prove that $ \frac {a}{1 \plus{} a} \plus{} \frac {2b}{2 \plus{} b} \plus{} \frac {3c}{3 \plus{} c}\le \frac {6(a \plus{} b \plus{} c)}{6 \plus{} a \plus{} b \plus{} c}$[/quote]

Yes!

$ \frac {a}{1 \plus{} a} \plus{} \frac {2b}{2 \plus{} b} \plus{} \frac {3c}{3 \plus{} c}\le \frac {6(a \plus{} b \plus{} c)}{6 \plus{} a \plus{} b \plus{} c}$

$ \Longleftrightarrow \left( \frac {a}{1 \plus{} a} \minus{} 1 \right) \plus{} \left( \frac {2b}{2 \plus{} b} \minus{} 2 \right) \plus{} \left( \frac {3c}{3 \plus{} c} \minus{} 3 \right) \le \left[\frac {6(a \plus{} b \plus{} c)}{6 \plus{} a \plus{} b \plus{} c} \minus{} 6 \right]$

$ \Longleftrightarrow \frac { \minus{} 1}{1 \plus{} a} \plus{} \frac { \minus{} 4}{2 \plus{} b} \plus{} \frac { \minus{} 9}{3 \plus{} c}\le \frac { \minus{} 36}{6 \plus{} a \plus{} b \plus{} c}$

$ \Longleftrightarrow \frac {1}{1 \plus{} a} \plus{} \frac {4}{2 \plus{} b} \plus{} \frac {9}{3 \plus{} c}\ge \frac {36}{6 \plus{} a \plus{} b \plus{} c}$

Then it's true by use [i][b]Cauchy-Swarchz[/b][/i].[/quote]

This is old problem  :D


Could anyone post the problems of Northen,South East, and Western Mathematical Olympiad contest?Thank you very much. :roll:

Some problems are in the Resources section.

There are just a few problems in the Resources section now.I hope some Chinese Mathlinkers will translate more.Thanks! :blush:

Western Olympiad has been recently posted. Take a look at my recent posts.

Western 2008: http://www.mathlinks.ro/viewtopic.php?t=238324
Southeast 2008: http://www.mathlinks.ro/viewtopic.php?t=217830
Western 2007: http://www.mathlinks.ro/viewtopic.php?search_id=1563807731&t=175295
Western 2005: http://www.mathlinks.ro/viewtopic.php?search_id=1563807731&t=63032
Southeast 2005: http://www.mathlinks.ro/viewtopic.php?search_id=1563807731&t=44857
Western 2003: http://www.mathlinks.ro/viewtopic.php?search_id=1563807731&t=18378

Use the search engine to find more!


function
algebra
rational function
combinatorics proposed
combinatorics

Let $ r < s$ be two rational numbers.  Show that the generating function

$ \sum_{ r \le \frac {b}{a} \le s } x^a y^b$

is a rational function of $ x$ and $ y$, where the sum is taken over all lattice points $ (a, b), a, b \ge 0$ in the first quadrant.  What can you say about the case that at least one of $ r, s$ isn't rational?

[hide="Hint"] In the rational case, write $ r \equal{} \frac{d}{c}, s \equal{} \frac{f}{e}$ and impose the lattice generated by $ (e, f)$ and $ (c, d)$ on the lattice points in question.  A good diagram will help. [/hide]


Does anyone have any good resources to learn (beginner-level) coloring?

Thanks.

[quote="JBL"]How about [url=http://www.allcraftsupplies.com/p-48352/paint-by-number-kit-tall-17-1-4x6-1-4-fruit.html]this[/url] or [url=http://www.coloringbook.com/]this[/url]?

(Seriously, though, we're not mind-readers out here.  What are you actually trying to ask?)[/quote]

Rofl!!! :rotfl: I'm talking about coloring like in combinatorics.

Coloring is used in lots of different ways in combinatorics (e.g., as an invariant-type object to show that certain kinds of tilings don't exist, or as in graph coloring (chromatic number, etc.), or in Ramsey theory, etc.) -- really, if you want people to help you, you need to make your questions sufficiently precise as to be comprehensible.  Links to a couple of examples of what you're interested in might help.

He's talking about coloring problems like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274249]this[/url] one. Or for example, proving that you cannot cover an $ 8$ by $ 8$ board with the top right corner taken off, with tiles that are $ 2$ by $ 1$ without overlapping.

[quote="geniusbliss"][quote="modularmarc101"]I really don't remember any specific problems...Oh and I also want to learn more about problems about tilings. Do you know any good ones?[/quote]
Try Proofs that Count by Jennifer Quinn and Arthur Benjamin[/quote]

Thanks!


1)Princple Of vinylogy
2)Retrograde Aldol Reaction

See whether these help?
http://www.technion.ac.il/~keinanj/pub/106.pdf

http://www.arkat-usa.org/?VIEW=MANUSCRIPT&MSID=368
 :D

[i]Vinylology[/i] refers to the acidity of $ \gamma$ hydrogens in $ \alpha ,\beta\minus{}$unsaturated carbonyls and nitriles.

[i]Retrograde aldol reactions[/i] are simply the reverse of an usual aldol reaction: cleavage of an alpha,beta unsaturated or alpha-hydroxy aldehyde or ketone to enolates and carbonyl compounds.

A bit of trivia: [i]retro[/i] is derived from the latin (I think) prefix meaning 'backwards' hence retroaldol would mean the reverse of aldol. Remember [i]Retrosynthesis[/i]? Thinking about a synthetic problem in reverse, a common technique used in formulating total synthetic paths.

Sorry for the slight irrelevance.

On the contrary, very good observation. Another example is the retro-Diels-Alder reaction. And yes, "retro" is derived from Latin.

is the $ \gamma$ $ \ce{H}$  of the conjugated system more acidic than the $ \alpha$  $ \ce{H}$ of a carbonyl roup.
what about the $ \ce{H}$ of carboxylic acid ,$ \beta$ keto ester and phenol.

The gamma and alpha H have approximately the same acidity.

[quote="Pardesi"]what about the $ \ce{H}$ of carboxylic acid, $ \beta$ keto ester and phenol.[/quote]

What about them?

i wanted the comparison of their acidic hydrogen with the $ \gamma$ hydrogen

That should be obvious to you by now:

carboxylic acid > phenol > beta-keto esther > gamma- or alpha hydrogen.

yes that was since u asked me their meanings i posted it again :)

i have  a doubt....
we say that Cinnamaldehye doe not undergo aldol because of absence of gamma hydrogen....
Is the presence of gamma hydrogen an essential thing for aldol???

that's because it is an $ \alpha,\beta$ unsaturated compound hence in order for it to undergo aldol it would require a $ \gamma$ hydrogen...actuallyt as u know the basic requiremnet for an aldol or even aldol kind condensation is an acidic hydrigen to generate a carbanion...for ex.perkin,Claissen ester condensation...


problem 1
\[ {\rm{Given that one root of the quadratic equation }}ax^2 {\rm{ }} \plus{} {\rm{ }}bx{\rm{ }} \plus{} {\rm{ }}c{\rm{ }} \equal{} {\rm{ }}0{\rm{ isthree times the other}}.{\rm{ Show that }}3b^2 {\rm{ }} \equal{} {\rm{ }}16ac
\]
problem2

$ 6\left( {x \plus{} \frac{1}{x}} \right)^2  \minus{} 25\left( {x \minus{} \frac{1}{x}} \right) \equal{} 0$

members please submit more sums of same type with solutions
STUDENTS FROM INDIA
ONLINE COACHING FOR SSC CBSE ICSE BOARD MATHS SITTING AT YOUR HOUSE
FOR DETAILS MAIL sanbittu@hotmail.com

[quote="sanbittu"]problem 1
\[ {\mbox{Given that one root of the quadratic equation }}ax^2 {\rm{ }} + {\rm{ }}bx{\rm{ }} + {\rm{ }}c{\rm{ }} = {\rm{ }}0{\rm{ isthree times the other}}.{\rm{ Show that }}3b^2 {\rm{ }} = {\rm{ }}16ac
\]
[/quote]

There are two cases, either

$ \frac { - b + \sqrt {b^2 - 4ac}}{2a} = 3 \left ( \frac { - b - \sqrt {b^2 - 4ac}}{2a} \right )$
or
$ \frac { - b - \sqrt {b^2 - 4ac}}{2a} = 3 \left ( \frac { - b + \sqrt {b^2 - 4ac}}{2a} \right )$

I will consider case $ 1$ though it will demonstrate why case $ 2$ works anyway

\begin{align*} \frac { - b + \sqrt {b^2 - 4ac}}{2a} & = 3 \left ( \frac { - b - \sqrt {b^2 - 4ac}}{2a} \right ) \\
- b + \sqrt {b^2 - 4ac} & = - 3b - 3\sqrt {b^2 - 4ac} \\
4\sqrt {b^2 - 4ac} & = - 2b \\
16b^2 - 64ac & = - 4b^2 \\
12b^2 & = 64ac \\
3b^2 & = 16ac \\
\end{align*}

For question 2, seeing as the sheer number of times I've seen you post that question, I'm going to create a very simple generalization.

\begin{align*} a \left ( x + \frac {1}{x} \right ) ^2 + b \left ( x - \frac {1}{x} \right ) + c & = 0 \\
ax^2 + 2a + \frac {a}{x^2} + bx - \frac {b}{x} + c & = 0 \\
\frac {a - bx}{x^2} + ax^2 + 2a + bx + c & = 0 \\
a - bx + ax^4 + 2ax^2 + bx^3 + cx^2 & = 0 \\
ax^4 + bx^3 + (2a + c)x^2 - bx + a & = 0 \end{align*}

Using the given $ a = 6, b = - 25, c = 0$

$ 6x^4 - 25x^3 + 12x^2 + 25x + 6 = 0$
$ (x - 3)(x - 2)(2x - 1)(3x + 1) = 0$

Therefore, $ x = 3$ or $ x = 2$ or $ x = - \frac {1}{3}$ or $ x = - \frac {1}{2}$


vector
geometry
geometric transformation
rotation

A(a), B(b), C(c) are forming an equilateral triangle. Show it's equal to a+b+c-ab-ac-bc=0  .

a,b and c are complex numbers.

I can't solve this problem...
Thank you for your help!

I dont think i understand the question. Could you paraphrase it?

Consider a,b,c complex numbers. A(a), B(b), C(c) are the points in the complex plane given by a,b,c. Show a^2+b^2+c^2=ab+ac+bc.

let $\omega^3=1$ be a primitive 3rd root of unity. then one can factor $a^2+b^2+c^2-ab-ac-bc=(a+\omega b+\omega^2 c)(a+\omega^2 b+\omega c)$. now, assume wlog $a+\omega b+\omega^2 c=0$ (else exchange $b$ and $c$). this is equivalent to $a-b=(1+\omega)(c-b)$ since $1+\omega=-\omega^2$. but $1+\omega$ is a primitive 6th root of unity, thus the vector $BA$ is the vector $BC$ turned by 60 giving the result.

Peter

I apologize for my poor english :(.

Thank you very much for helping me solving the problem.

Thank you another time!

Best regards,
Charles.

Hi
Peter scholze method was good.
Here is mine.My concept works on vectors.
As a,b,c are complex nos. and represent sides of a triangle,fisrt find the lengths of the triangle by taking a,b,c as position vectors. by rotation theorem one side can be written in terms of other sides(u get the other side by rotating this vector anticlockwise by 60 degrees).repeat the same process for the other side and when u compare u get the answer. ;)


This inequality was given to be my friend(own) - Agr_94_Math(user id)
[u]Q[/u]. For positive reals $ a,b,c$ such that $ a \plus{} b \plus{} c \equal{} 1$ , Prove that,
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3abc \ge \frac {4}{9}$
waiting to see your proofs!

[hide]Homogenising, we have to prove that $ (a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c) \plus{} 3abc \ge \frac{4}{9} (a\plus{}b\plus{}c)^3$ which after simplification yields

$ 5 \sum a^3 \plus{} 3abc \ge 3 \sum ab(a\plus{}b)$

From Schur, $ \sum a^3 \plus{} 3abc \ge \sum ab(a\plus{}b)$

From Rearrangement $ \sum a^3 \ge a^2b\plus{}b^2c \plus{} c^2 a$ and $ \sum a^3 \ge ab^2\plus{}bc ^2\plus{} ca^2$

So that $ 2 \sum a^3 \ge \sum ab(a\plus{}b)$ or $ 4 \sum a^3 \ge 2 \sum ab(a\plus{}b)$

Adding the three inequalities we get the desired result.[/hide]


Harvard college Yale Princeton Stanford MIT

Just a quick question regarding college application process-

For the schools that I have already get into, what GPA/Grades shall I maintain to be not 'kicked' out of them? It seems like I would have 5 As and 1 Bs or GPA 3.8; will it be a critical problem/cause/factor for the change of decision that has been already made? (I am talking primarily about top schools like Harvard, Yale, Princeton, Stanford, MIT, CalTech, etc..)

It is very urgent/critical problem that I have at this moment... I certainly do not wish for the decision to be altered just because I screwed up in one course that I am not intending to major...

Please leave your remarks... Although it seems highly unlikely that my final grade will be changed for the course, I will, if it is very necessary so, study as hard as I can for the the final...

If by "screwed up," you mean the B, you can stop worrying.  Go do something fun.

If you are already admitted, one B among several A's will make no difference at all.


topology
linear algebra
matrix
LaTeX
function
analytic geometry
real analysis

Let $f: R^{n+k}\to R^{n}$ be of class $C^{r}$. Let M be the set of all x such that f(x) = 0. Assume that M is non-empty and that Df(x) has rank n for x in M.
Then M is a k-manifold without boundary in $R^{n+k}$. Furthermore, if N is the set of all x for which $f_{1}(x) = ... = f_{n-1}(x)$ and$f_{n}(x) \ge 0$, and if the matrix $\partial (f_{1},...,f_{n-1})/\partial x$ has rank n-1 at each point of N, then N is a k+1 manifold, and $\partial N = M$

I am not sure how to prove this theorem. I have something which I am not confident of but would like to see another's solution to it. I will try to post my solution later since putting it in Latex will not be so facile.

The [url=http://en.wikipedia.org/wiki/Implicit_function]implicit function theorem[/url] is not so elementary. Its proof is, like, hard. And boring.

Here is my solution for the first portion. I made a few liberal assumptions which seemed plausible in the proof of which I am unsure but I pointed them out in case any of you would be so kind as to explain them for me (if they are right), or otherwise how to correct them. My proof for the second part comes next.

Let $p \in M$. Define $F: R^{n+k}\to R^{n+k}$by $F(x) = (x_{1},...,x_{k},f(x))$.

Then DF is invertible (can someone assure me of why this follows) and by the inverse function theorem, there is an open set W containing p such that $F: W \to F(W)$ is a diffeomorphism.
Let $\pi_{k}: R^{n+k}\to R^{k}$ project $R^{n+k}$ to its first k coordinates. Take g to be the inverse of $\pi_{k}\circ F$ and set $U = \pi_{k}\circ F(M \cap W)$. (How do we prove that U is open in $R^{k}$?).
$M \cap W$ is open in M and $g: U \to M \cap W$ is a $C^{r}$bijective function (why is it a $C^{r}$bijective function?).

Dg(x) has rank k on U (why?). Therefore g is a coordinate patch about M and M is a k-manifold without boundary in $R^{n+k}$.

What does it have to do with the implicit function theorem (which I don't even know)? I used the inverse function theorem (which is a special case) in my solution but I didn't have to prove it.

Here is the second part of my solution:

Take $p \in N$ and define $G: R^{n+k}\to R^{n+k}$by $G(x) = (x_{1},...,x_{k+1},f_{1}(x),...,f_{n-1}(x))$. Then DG is invertible and by the inverse function theorem we can find an open set $X \subset R^{n+k}$ containing p such that $G: X \to G(X)$ is a surjective diffeomorphism. $\pi_{k+1}\circ G$ is bijective on $N \cap X$. Set $V = \pi_{k+1}\circ G(N \cap X)$ and let h be the inverse of $\pi_{k+1}\circ G$. As above, h is a bijection of class $C^{r}$because $\pi_{k+1}\circ G$ is (and this follows since G is). $h: V \to N \cap X$ is the required coordinate patch on N about p and p is an interior point of N, for V is open in $R^{k+1}$ (proof?).

But it doesn't make sense that V is open in $R^{k+1}$ otherwise N would be a k+1-manifold without boundary.

Does no one know this?  :(

My solution is certainly wrong.

your problem is certainly interesting, I'll try to find a solution today.

No one knows how to solve this?  :maybe:

$ f_n (x_0 ) > 0$ implies $ f_n$ is positive in a neighborhood U of $ x_0$

If we assume that at $ x_0$the last n-1 columns of the Jacobi matrix of the following transformation are independent then the inverse function shows it is a diffeomorphism in a neighborhood V of $ x_0$. Restrict this diffeomorphism to the intersection of U and V.

$ t_i \equal{} x_i ,i \in \{ 1,...,k \plus{} 1\}$
$ t_{i \plus{} k \plus{} 1} \equal{} f_i (x_1 ,...,x_{n \plus{} k} )i \in \{ 1,...,n \minus{} 1\}$

If $ f_n (x_0 ) \equal{} 0$ the following transformation is again a diffeomorphism in a neighborhood U of $ x_0$ by the inverse function theorem

$ t_i \equal{} x_i ,i \in \{ 1,...,k\}$
$ t_{k \plus{} 1} \equal{} f_n (x_1 ,...,x_{n \plus{} k} )$
$ t_{i \plus{} k \plus{} 1} \equal{} f_i (x_1 ,...,x_{n \plus{} k} )i \in \{ 1,...,n \minus{} 1\}$

and it is the requisite local chart.


Uhhh dude let them be p and q, respectively; if we have p + q^2 = t^2, then p = (t-q)(t+q) >= t+q > p... their being prime is totally random...

[quote="diophantient"]Ok, so you have two primes $4x-1$ and $4x+3$.
Show that
$4x-1+(4x+3)^{2}$ can't be a perfect square.[/quote]
[hide]If $p_{1}+p_{2}^{2}=t^{2}$, then $p_{1}=t^{2}-p_{2}^{2}=(t-p_{2})(t+p_{2})$, which makes $p_{1}$ composite.[/hide]

i think that Mr. i_like_pie leaved one condition that we have to show in the solution that if $t-p_{2}= 1$then this may be true....
for proving that this is also nt possible we do following
$t=p_{2}+1$
$p_{1}= t+p_{2}$
$p_{1}= 2p_{2}+1$
$4x-1 = 2(4x+3)+1$
$4x =-8$
$x=-2$
bt for $x=-2$given numbers are not prime. hence we can say that this is not the perfect square

oops, this is the wrong question, it's easier than the one I made up.
Unfortunately, I don't remember the intended problem, although I wrote it down somewhere.


A number of $n > m \geq 1$ soccer teams play a full tournament, each team meeting (once) each other. Points are awarded: $2$ for a victory, $1$ for a tie and $0$ for a loss. At the end, each team has won half of its points against the $m$ teams placed last (including each of these teams, who won half of its points against the other $m-1$).

Find all possible values for $n$ and $m$, supported with examples of such tournaments.

I believe the answer is $(n, m) = (4a^{2}, 2a^{2}-a)$ or $(n, m) = ((2a+1)^{2}, 2a^{2}+a)$. It's not too hard to prove those are the only possible, but it's annoying to do the construction.

yes, i remember that the construction was a nasty part, but the first part was ok, i will sketch the official solution, i think:

let us call the last $m$ teams the losers, and the other $n-m$ teams the winners.
now we have that the losers together have won $4 \cdot{{m}\choose{2}}$ points, while the winners together have won $4 \cdot{{n-m}\choose{2}}$ points, but in all there are $2 \cdot{{n}\choose{2}}$ points awarded, so $4 \cdot{{m}\choose{2}}+4 \cdot{{n-m}\choose{2}}= 2 \cdot{{n}\choose{2}}$, therefore $n^{2}-(4m+1)n+4m^{2}= 0$ or $(n-2m)^{2}= n$, so by this wee need to have $n=(k+1)^{2}$ , $k \geq 1$.
but at least one loser won not less than $2(m-1)$ points, while at least one winner has won at most $2(n-m-1)$ points, thus $n \geq 2m \Rightarrow n-2m= k+1$, so $m=\frac{k(k+1)}{2}$ and $n-m=\frac{(k+1)(k+2)}{2}$.


Solve $ x\frac{d^2y}{dx^2}\plus{}(x\minus{}1)\frac{dy}{dx}\minus{}y\equal{}0$ with $ y(0)\equal{}1,\ \lim_{x\to\infty} y(x)\equal{}0$.

hello, one can easily show that $ y_1(x) \equal{} x \minus{} 1$ and $ y_2(x) \equal{} e^{ \minus{} x}$ are solutions of our equation, so we have
the gerneral solution as $ y(x) \equal{} C_1(x \minus{} 1) \plus{} C_2e^{ \minus{} x}$. Because of the condition $ y(0) \equal{} 1$ we get $ \minus{} C_1 \plus{} C_2 \equal{} 1$ and from $ \lim_{x \to \infty}y(x) \equal{} 0$ we get $ C_1 \equal{} 0$ so our solution is $ y(x) \equal{} e^{ \minus{} x}$.
Sonnhard.


What is the attachment for?

what does it mean if a thread has an attachment?

It's for if you need to post a picture, like when reporting a bug, or a word document like in a tournament. The little symbol indicates that some post in the thread has an attachment.


If $ A$ and $ B$ are matrices, show that if $ AB$ and $ BA$ is defined, then A and B are square matrices.

That is certainly false. Consider, for example, $ A \equal{} A_{m \times n}$ and $ B \equal{} B_{n \times m}$, with $ m \neq n$: then both products AB and BA are defined although A and B are not square matrices.


How would you do this problem?

If the perimeter of a square is equal to the circumfrence of a circle, what is the ratio of the square's area to that of the circle?

[hide]If the perimeter of a square is $c,$ its area is $\frac{c^{2}}{16}.$

If the circumference of a circle is $c,$ its area is $\frac{c^{2}}{4\pi}.$

The ratio of the areas of a square and circle with equal perimeters is $4: \pi.$[/hide]


function
geometry
geometric transformation
integration
real analysis
Support
topology

Hi,

I am looking for an elegant approach to this quite obviously correct theorem :

let $f: \mathbb{R}^n\rightarrow \mathbb{C}$ a measurable and integrable function,so
\[ \int_{\mathbb{R}^n} |f(x)| \]
exists

Prove that for each $\epsilon>0$, there is a $\delta>0$ such that $||h||<\delta$ implies :

 \[ (\int_{\mathbb{R}^n}( |f(x+h)-f(x)|)<\epsilon \]

I was thinking of a Lebesgue theorem of dominated convergence....but I did not assume the function is continuous!!

This is probably not the most elegant approach, but here goes...

Since $f$ is integrable, we know there exists $R$ so that $\int_{|x|>R} 2|f(x)|dx < \frac{\epsilon}{3}$.

Consider the ball of radius $R$ around the origin. We can apply the theorem of Lusin, which says that for any $\eta$, there is some closed set $E$ with measure $m(E) > m(B(R, 0)) - \eta$ and $f|_E$ ($f$ restricted to $E$) is continuous.

It is a well-known fact about integrable functions that if $\eta$ is small enough, then
$\int_A |g(x)|dx < \epsilon$ for any $A$ with $m(A) < \eta$. $|f(x+h) - f(x)| \le |f(x+h)| + |f(x)|$, so we pick $\eta$ small enough that $\int_A 2|f(x)|dx < \frac{\epsilon}{3}$.

Now with that $\eta$ we use Lusin's theorem to get our set $E$. The set $B(R, 0) \backslash E$ is open, so a countable union of connected open sets, and we just continuously complete $f$ in those holes. This modified $f^*$ is continuous, so we may pick small enough $h$ that $\int_{|x| \le R} |f^*(x+h) - f^*(x)|dx < \frac{\epsilon}{3}$.

This is our $h$. We have
$\int_{R^d} |f(x+h) - f(x)|dx \le \int_{|x| > R} 2|f(x)|dx + \int_{B(R, 0)\backslash E} 2|f(x)|dx + \int_{B(R, 0)}|f^*(x+h) - f^*(x)|dx < \epsilon$

Thanks everyone.
In light of the path my syllabus takes, alekk, would just continuous with compact support do the trick just as well?

of course  :D  it was just a reflex to take $C^{\infty}$ ..

[quote="alekk"]The $C^{\infty}$ functions with compact support are dense in $L^1(R^n)$. Suppose that one knows that the result is true for these regular functions. Then take such a regular function $g$ such that $|f-g|_{L^1} < \epsilon$ and call for simplicity $F_h(x)=F(x+h)$. Then $|f_h-f|_1 \leq |f_h-g_h|_1+|g_h-g|_1+|g-f|_1 \leq 2 \epsilon + |g_h-g|_1$. So the conclusion follows from the case of $C^{\infty}$ function with compact support. And for these functions it is quite easy since they are uniformly continuous.[/quote]

Actually aren't there some pitfalls here?  I slept one night over it and got in doubt.

a) Are you doing this : $f$ is continous on$\mathbb{R}^n$ and has compact support$\Longrightarrow f$ is uniform continuous on the total $\mathbb{R}^n$

b) so you have your continuous compact support function f, let us call the support $K$
Isn't there a little danger here, I mean $\int_{\mathbb{R}^n} |f(x+h)-f(x) | = \int_{K} |f(x+h)-f(x) |$ is too fast right?

Is there a problem ? Just take a compact a little larger than $K$ that contains the support of $f$ and $f_h$.

If we agree we don't consider $||h||>1$ it suffices to just take the compactum of all points of distance at most 1 from $K$ right?

The nice thing about this  :)  is that it actually works for every $L_p$ with $1\leq p<\infty$


Let $ a_n \equal{} 3^n \minus{} 2^n$ for every $ n\in\mathbb{N}$.
Prove that the sequence $ \left(a_1,a_2,a_3,...\right)$ does not contain three terms in arithmetic progression.

This is just a split off the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=64838 (see in this topic for a solution).


calculus
derivative
real analysis
real analysis unsolved

if: $ f^{2}$ is differentiable and $ f$ is continuous, then: $ f$ is differentiable and $ f\left( x\right) \neq0$

The statement should read: if $ f^{2}$ is differentiable, $ f$ is continuous, and $ f(x)\ne 0$, then $ f$ is differentiable at $ x$.
If, for instance, $ f(x) > 0$, then (by continuity) $ f > 0$ in some neighborhood of $ x$, and you can use the chain rule for $ \sqrt {f^2}$. The case $ f(x) < 0$ is similar.


Suppose we flip four coins simultaneously: a penny, a nickel, a dime, and a quarter.  What is the probability that at least 15 cents worth of coins come up heads?

When doing this problem, you don't have to consider the penny at all. It will not make any difference in accumulating at least 15 cents whether you include the value of the penny or not. Without the penny, there are 8 total combinations. 4 of these have 25 cents and one has the 5, 10 and no 25. So the probability is 5/8.

There are $ 2^{4}\equal{}16$ possible outcomes, since each of the 4 coins can land 2 different ways (heads or tails). There are 2 possibilities for the penny and the dime: either they're both heads or they're both tails. There are also 2 possibilities for the nickel and 2 possibilities for the quarter. So there are $ 2\times 2\times 2 \equal{} 8$ successful outcomes, and the probability of success is $ \frac{8}{16}\equal{}\boxed{\frac{1}{2}}$ .

[quote="IcePhoenix"][hide]I have a rather involved solution here:
So, there are 16 possible outcomes (by $2*2*2*2$).  The ones we don't want are the ones that have only pennies, dimes, and nickels (except the ND, nickel and dime).  So then the ones you don't want are D, N, PN, PD, and P.  This is 5 UNWANTED outcomes.  Out of the total outcomes, that gives us 11 WANTED outcomes.
Then for each sequence, the probability you will have that sequence is $1/16$ (you must count the probability for tails, too:  for PND, you have to count that the quarter is tails).  So if you multiply it by 11 (by adding $1/16$ to itself eleven times) then you get 11/16.
However, since the answer was 10/16, I'm wondering if I forgot an unwanted outcome.  [/hide][/quote]
Sorry for this ever so big bump, but you are missing not getting any of the coins at all.

[hide=Casework]
To get less than 15, we can have:

p
n
d
pn
pd
nothing

So 6 possibilities.

So $1-\frac{6}{16}=\boxed{\frac{10}{16}}$
[/hide]

how is this level 24?

don't really want to list out the $16$ cases but it honestly doesn't take that long. $10$ working cases, total of $16$, answer is $\boxed{\frac{5}{8}}$.

[hide=spoiler]use casework on # of heads flipped
should only take like 3 mins max there are 4 cases: 1 head, 2 heads, 3 heads, 4 heads[/hide]

[hide=sol]let the penny be P and nickel be N and dime D and quarter Q.
if the nickel heads then we have PND, PNQ, and PNDQ
if the nickel tails then we have PDQ and PQ
five ways if penny is heads
now for the nickel
if the dime heads we have ND and NDQ
if the dime tails we have NQ
now for the dime
there is only one way just DQ dairy queen
now for the quarter
there is only one way just Q
adding the cases up we have 10 cases
there are 2^4 ways to have four coin flips so it is 10/16=5/8[/hide]

I have a question: why are a lot of the problems from the C&P textbook in here?

[hide=Solution]Notice that whenever a quarters comes up heads, then it is always greater than $15$. Thus, we want to count the possibilities when the quarter doesn't come up heads. Notice this only happens if and only if there is a dime and nickel or a dime, nickel, and penny. Thus $$\frac{8+2}{2^4}=\boxed{\frac{5}{8}}$$


algebra
polynomial
sum of roots
product of roots

The roots of the equation $x^5-40x^4+Px^3+Qx^2+Rx+S=0$ are in geometric progression.  The sum of their reciprocals is 10.  Compute the numerical value of $|S|$.

[hide]let the number decompose into the following factorization:

(x+a)(x+a^2)(x+a^3)(x+a^4)(x+a^5).

that means that a+a^2+a^3+a^4+a^5=-40

and the sum of the reciprocals of the roots are:

1/a+1/(a^2)+...+1/(a^5) which is (a+a^2+...+a^5)/a^15.  We know that the numerator is 40.  So that means that a is the -15th root of 10.  and after multiplying out we get the -15th root of 10^15.  And that is -10.  So |s|=10[/hide]

Can anyone see a use of reverse factorization that could make this problem a bit easier (think about the ways we could express the sum of a finite geometric series).

There might be some error  inabove post .....

I[hide]f the numbers are a, ar, ar^2, ar^3 ar^4

The sum of reciprocals = 10  = (1/(a*r4)) (1+r+r^2+r^4)

The sum  = 40 = a(1+r+r^2+r^3+r^4)

This gives a^2*r^4 = 4   and this will make  |s|=|a^5*r^10| = 32.[/hide]


geometry
analytic geometry
graphing lines
slope
trigonometry
MATHCOUNTS
AMC

really need some help right now...im learning everything, but my foundation is falling while learning new stuff. i need something to help me understand!

im learning geometry right now..the reason why i put algebra up there was because im losing memory for algebra and start to forget everything like slopes and stuff. and for trig and cal, i put it up there because i want to start up early and not get confused during future classes. i mean if im struggling geometry, ill prob stuggle on later courses.

i need a good overall software for each courses. dont want just basic stuff on one course and the other..

a lot of the algebra stuff is reviewed at the beginning of pre calc.  im not sure about any softwares though.  maybe try the aops geometry book by mr. ruszyk.  the first like 5 chapters go over triangle similareites and stuff.  are you struggling with proofs? (i always hated those two collumn proofs. :dry: )

can anyone else back me up?

All i remember from my geo classes (yes, I took three different geo classes, not because I didn't get it, but because I got *too* much, and wanted harder stuff...)

is this:

CPCTC

(corresponding parts of congruent triangles are congruent.)

And OMG yes I hated two column proofs/

if you're trying to do it all at once, that might be the source of your problems. I'm not saying you can't handle it, but make sure you don't just look over material, say 'o I know that!' and move on. do all practice problems, even if they seem too easy, because that builds up your foundation. If you spent an hour sitting there calculating the slopes of a hundred different lines, you wouldn't be likely to forget it, if you glanced over a formula that said delta y over delta x, there's a chance you would forget it.

its not those problems im having..when i learn geo, i get confused yet still find a way with a hard time getting it. ive been doin that for last 10weeks or so. then when we review for algebra, i forgot most of em..i almost have to start all over cuz i didnt get any of them..they looked so different from geometry. now from doin proofs(yea we're done) and theorem stuff for almost whole semester, not doin any single algebra, i forget ><.

there's a program called Acces that has like all the mathcounts problems up to 2000.  it improved my math ability from  16 on the AMC8 to a 138 on the AMC10

graphing calculators are alawys fun to mess with (the ones on the computers)
those can "improve" graphing skills and you can figure out what graphs sort of look like


function
number theory unsolved
number theory

Prove that $\tau(n) \leq 2\sqrt{n}$ for all $n \in \mathbb{Z}$
Note: $\tau(n)$ is number of divisors of $n$

See here : http://www.mathlinks.ro/Forum/viewtopic.php?t=70289 :)


What is the positive difference between the degree measures of the complement and supplement of the same angle if the sum of their degree measures is $ 184^\circ$?

Let the angle be $ x$.

Therefore, $ 270\minus{}2x\equal{}184$, or $ x\equal{}43$.

Therefore, $ 137\minus{}47\equal{}90$.

Note that this will always be $ 90$.


calculus
integration
function
real analysis
real analysis unsolved

Let f be an non negative extended real valued Lebesgue measurable function on a set D with m(D) is finite. Suppose f >0 a.e. on D.
show that for every a >0 there exists b>0 such that for every  Lebesgue measurable subset E of D with m(E)>= a,
we have $ \int_{E}f\,dm$ >= b.

As far as $ f>0$ and $ m(D)$ is finite, we have $ m(\{f<\delta\})\to 0,\delta\to 0$. Now if $ m(E)\ge a$, then for any $ \delta>0$ we have $ \int_E f(x) m(dx) \ge \delta (m(E) \minus{} m(\{f<\delta\}))\ge \delta (a\minus{}m(\{f<\delta\}))$. Now take $ \delta$ s.t. $ m(\{f<\delta\})<a/2$. qed


calculus
integration
real analysis
complex analysis
complex analysis unsolved

Let $ u(z) \equal{} Im[({\frac {1 \plus{} z}{1 \minus{} z})^2}]$ , this defined on the  unit disc. Show that u is not the poisson integral of any measure nor is it the difference of two positive harmonic functions. I think i started ok but i got stuck. Can someone help? This is what i did...

For the first one, i assumed that it is. Let $ m_{1}$ denote such measure. Then by radon nikodym we can write $ dm_{1} \equal{} fdx \plus{} dm_{2}$, where f is integrable. By well known theorem the poisson integral of $ m_{1}$ ( assumed to be u) has non tangential limit $ f(e^{i\theta})$. But we can make a sketch of what $ u$ does to the unit disc- it maps disk to the cut plane, so radial limits exists and is 0. Since non tangential limit exists it also has to be 0. So we get f=0 a.e. So $ m_{1}$ is singular with lebesgue measure. This means it is supported on set of measure 0. I want to look for a contradiction somewhere but i cant figure where to get a contradiction from. Hints will be fine. Thanks

Just look at the rate of growth near $ 1$ ;).


point p is 9 away from center of circle with radius 15
how many chords thru p have integer lengths?

I'm not sure if this is right, but here goes:
If we draw the diameter AB through P, we can divide it into 2 segments, say AOP and PB. AO = radius = 15, so AOP=24, which means PB = 6. Now draw a chord perpendicular to AB through P, and denote it by CD. We know that CP = PD, and let's set the length of CP = x. So we have 144=x^2. x = 12. This will the smallest value possible for any chord, and it happens to be an integer =) The upperbound is 30 because of the diameter through P.
Then, every integer between 12 and 30 is a possible chord length, and there are two possible chords for each chord length. Thus,
30-12-2=16, and 16*2+2=34.
So there are 34 possible chords with integer values.

Actually u forgot that x is only half of CD, ie. CD = 24 is the lower bound...


A classroom has some students in it. The ratio of boys to girls is 3:2. After a minute, a boy enters the classroom and two girls leave. The ratio of boys to girls is now 2:1. How many students are in the classroom at the point?

Set up a equation

$ \frac{b}{g} \equal{} \frac{3}{2}$

$ \frac{b\plus{}1}{g\minus{}2} \equal{} \frac{2}{1}$

Cross multiply

$ 2b \equal{} 3g$

$ b\plus{}1 \equal{} 2g \minus{} 4$

$ 2b\plus{}2 \equal{} 4g \minus{} 8$

Subsitute

$ 3g \plus{} 2 \equal{} 4g\minus{}8$

$ \minus{}g \equal{} \minus{}10$

$ g \equal{} 10$

So

$ 2b \equal{} 3(10)$

$ b \equal{} 15$

$ 15\plus{}10 \equal{} 25$

But a boy enters and two girls leave so

$ 25 \plus{}1 \minus{} 2 \equal{} \boxed{24}$

Couldn't the answer to this also be 6? Because if you start of with 3 boys and 2 girls, the ratio of boys to girls would be 3:2 and then and one boy and the ratio would be 4:2 which is equivalent to 2:1. So 4+2 would equal 6.

I think you forgot about the part about the two girls leaving the room.


geometry
parallelogram
vector
area of a triangle

Please help me figure this one out...

If X is the area of any triangle-PQR and you make a hole in the center of this triangle by first joining the midpoints of each side to create smaller triangle-STU and then again joining the midpoints of triangle-STU to create the hole in the center of this triangle, it would create a donut.

What is the area of the triangle donut in terms of X?

Thanks!!

Trying it with an equilateral triangle gives $\frac{13}{16}$ but I'm not sure if this is a general solution.

thanks for responding... but the question is for any triangle and not for one particular triangle. an equilateral triangle, in this sense, would be a particular triangle.

the solution would have the variable x in it, representing the unknown area of ANY triangle.

Yeah, my answer was $\frac{13}{16}X$. I left out the X because it was implied that the $\frac{13}{16}$ was the fraction of the triangle still left.

hi. thanks for getting back to me once again....
... i actually tried to figure it out myself and got 15/16x.

the first larger triangle PQR has area X
the smaller one made by joining mid points STU has an area of 1/4X. this can be proved by calculating the area of a triangle using vectors.

1. The area of a parallelogram can be calculated by the use of vectors. If PQ and PR are vectors pointing from P to Q and from P to R respectively, the area of parallelogram PQZR is |PQ × PR|, the magnitude of the cross product of vectors PQ and PR. |PQ × PR| is also equal to |h × PR|, where h represents the altitude h as a vector.

The area of triangle PQR is half of this, or X = ½|PQ × PR|.

2. when connecting midpoints STU and creating a smaller triangle inside of the larger triangle, there are actually three other small triangles as a result: PST, SQU and TUR.

3. by using the same method of calculating the area of a triangle, i can now attempt to calculate the area of triangle PST.

the area of PST = 1/2 PSxPT

PS=1/2 PQ and PT=1/2PR
therefore:
area of PST=1/2x1/2PQx1/2PR or in other words: 1/4x ½|PQ × PR| which is exactly a 1/4 of X, the area of PQR.

4. i can apply the same method on the other two triangles: SQU and TUR. Finding that the area of all three triangles equal 1/4 the area of the larger triangle PQR. knowing this, i can easily deduce that the area of the fourth triangle within PQR - named STU, is 1/4 the area of PQR as well.

5. since X is the area of ANY triangle, this becomes a rule. the area of a small triangle made by connecting midpoints of a larger triangle is 1/4 the area of the larger one.

6. realizing this, the area of the smallest triangle, or the "donut hole" as it is refered to in the question, made by connecting midpoints of triangle STU, will be 1/4 of the area of STU..... or in other words 1/4 of 1/4X, which equals 1/16X.

7. now the rest is easy. The area of the "donut" is the area X minus the area of the "donut hole" 1/16X. so: X-1/16X=

15/16X

that's what i fugured out. what do you think?
now i can create the equation:


limit
real analysis
real analysis unsolved

Let $ \alpha > 1$, and take $ x_1 > \sqrt \alpha$.  Define
$ x_{n\plus{}1} \equal{} \frac{\alpha \plus{} x_n}{1\plus{}x_n}$.
Show that
(i) $ x_1 > x_3 > x_5\cdots$
(ii) $ x_2 < x_4 < x_6 \cdots$
(iii) $ \lim x_n \equal{} \sqrt \alpha$
(iv) estimate the error in the approximation.

i), ii) and iii) use $ f'(x)\equal{}\minus{}\frac{\alpha\minus{}1}{(x\plus{}1)^2}<0\forall x>0$, here $ f(x)\equal{}\frac{\alpha \plus{}x}{1\plus{}x}, x>0$.


bob wnats to write down all n digit integers using digits 1,2,3 os that there are no two 1s next to each other. How many integers can bob write?

[hide="Well..."] It depends on what base he's using.

If he's using Fibonacci base, he can write all of them  :rotfl: [/hide]

[quote="mna851"][hide]Let $\ R_{n}$ be the number of $\ n$ digit numbers made from the digits $\ 1,2,3$ such that no two $\ 1$'s are next to each other.  If one of these numbers begins with $\ 2$ or $\ 3$ then there are $\ R_{n-1}$ such numbers.  If one of these numbers begins $\ 12$ or $\ 13$ then there are $\ R_{n-2}$.  Therefore we have $\ R_{n}= 2R_{n-1}+2R_{n-2}$, however I don't know how to find the explicit form of this recurrence.

EDIT: I'm actually semi-sure this recursion has no explicit form.

[/hide][/quote]

Every recursion of that type has an explicit form ;)

[hide="Assuming yours is correct..."] The recursion has characteristic polynomial $x^{2}-2x-2 = 0$, which has solutions

$r = 1+\sqrt{3}, s = 1-\sqrt{3}$

Hence our final form is given by

$R_{n}= ar^{n}+bs^{n}$

For some constants $a, b$, where $R_{1}= 3, R_{2}= 8$ by inspection.  Solving for our constants,

$a(1+\sqrt{3})+b(1-\sqrt{3}) = 3$
$a(4+2 \sqrt{3})+b(4-2 \sqrt{3}) = 8$
$a(2+\sqrt{3})+b(2-\sqrt{3}) = 4$
$a+b = 1$
$a \sqrt{3}-b \sqrt{3}= 2$
$a-b = \frac{ 2 \sqrt{3}}{3}$

This gives

$a = \frac{3+2 \sqrt{3}}{6}$
$b = \frac{3-2 \sqrt{3}}{6}$

So, finally, we have

$R_{n}= \frac{3+2 \sqrt{3}}{6}(1+\sqrt{3})^{n}+\frac{3-2 \sqrt{3}}{6}(1-\sqrt{3})^{n}$

As a check, by this formula, $R_{3}= 22$.  Those $22$ strings are

$121, 122, 123, 131, 132, 133, 212, 213, 221, 222, 223, 231, 231, 233, 312, 313, 321, 322, 323, 331, 332, 333$
[/hide]

[quote="mna851"][hide]Let $\ R_{n}$ be the number of $\ n$ digit numbers made from the digits $\ 1,2,3$ such that no two $\ 1$'s are next to each other.  If one of these numbers begins with $\ 2$ or $\ 3$ then there are $\ R_{n-1}$ such numbers.  If one of these numbers begins $\ 12$ or $\ 13$ then there are $\ R_{n-2}$.  Therefore we have $\ R_{n}= 2R_{n-1}+2R_{n-2}$, however I don't know how to find the explicit form of this recurrence.

EDIT: I'm actually semi-sure this recursion has no explicit form.

[/hide][/quote]

Could you explain your resonings on why If one of these numbers begins with $\ 2$ or $\ 3$ then there are $\ R_{n-1}$ such numbers.  If one of these numbers begins $\ 12$ or $\ 13$ then there are $\ R_{n-2}$.

[quote="D.P.L"]Could you explain your resonings on why If one of these numbers begins with $\ 2$ or $\ 3$ then there are $\ R_{n-1}$ such numbers.  If one of these numbers begins $\ 12$ or $\ 13$ then there are $\ R_{n-2}$.[/quote]

[hide="Explanation"] Let $\mathbb{S}_{n}$ be the set of possible $n$-length strings formed by $1, 2, 3$ such that there do not exist consecutive $1$s.  Of course, $| \mathbb{S}_{n}| = R_{n}$.

Consider all the ways in which an element of $\mathbb{S}_{n}$ can start.

[b]Case:[/b]  It starts with $1$.  If it starts with $1$, then the next digit cannot be $1$, so it must be $2$ or $3$.  Hence it must start with either $12$ or $13$, and the rest of the string can be any string of length $n-2$ that has no consecutive $1$s - in other words, it must be a string in $\mathbb{S}_{n-2}$, so there are $2R_{n-2}$ of these strings.

[b]Case:[/b]  It starts with $2$.  Then the rest of the string can be any string of length $n-1$ that has no consecutive $1$s - in other words, it must be a string in $\mathbb{S}_{n-1}$, so there are $R_{n-1}$ of these strings.

[b]Case:[/b]  It starts with $3$.  Identical to the other case; there are again $R_{n-1}$ of these strings.

There are a total of $2R_{n-2}+2R_{n-1}$ strings in $\mathbb{S}_{n}$, and therefore

$R_{n}= 2R_{n-2}+2R_{n-1}$ [/hide]

Yep, that was my reasoning.  Ahh, I see how to get the explicit solution.  I actually tried writing the characteristic equation but I messed up the constants.


if two circles with radius r have combined area of (3 :pi: +2)(r^2)/2, find how far apart the circles are

[hide]Let the circles have centers A and B. Since they obviously intersect, let X and Y be the intersections and let  :theta: =  :ang: XAY. We find that the area = r^2 (2 :pi: - :theta: + sin :theta: ). Since this is equal to r^2 (3 :pi: + 2)/2, we guess that 2 :pi: - :theta: = 3 :pi: /2 and sin :theta: = 1 by comparing these two values. It works!  :theta: = :pi: /2. So AXBY is a square and AB = r :rt2: [/hide]

Assuming you want the distance between the centers....

(3pi+2)(r^2)/2 is ugly.  We split it up into 3/2*pi*r^2 + r^2.  Ignore the second term for a second.  If the two circles didn't intersect, their total area would be 2pi*r^2.  We only have 3/2*pi*r^2, or 3/4 of that area.  From this we guess that 3/4 of each circle (i.e. 270o) is involved, although we don't know this for sure.  Now we look at the r^2 part.  This could represent a bunch of things, but the simplest is a square with side r.

Now we try and fit two 3/4-of-a-circle pieces and one square piece together.  Voila!  (See diagram.)  Since the two centers are opposite corners of a square, the distance between them is the same as the length of a 45-45-90 right triangle with leg r, or r*sqrt(2).

Hmm...any ideas on how to do this problem had AXBY not been a square (that is, had the initial area given in the problem been just A instead of an arbitrarily chosen number that worked out nicely)?

If you define theta as i did in my solution, you're gonna get an equation involving theta and sin(theta), which is probably not easy to solve algebraically (if even possible) unless theta is some special value (for example pi/2 as in this example). So, in a problem like this, the theta will probably be easy to guess.

Yeah, pretty sure such questions aren't solvable algebraicly.  Ditto for questions with exponentials and polynomials or logs and polynomials -- they don't mix well.


integration
calculus
calculus computations

$ \int\frac{  (x^{4} \plus{} 1 ) dx}{x^{6} \plus{} 1 }$ =
                                                     $ \int\frac{ ((x^{4} \minus{} x^{2} \plus{}1 ) \plus{} (x^{2} \plus{} 1 )\minus{}1)dx}{ (x^{2} \plus{}1) ( x^{4} \minus{} x^{2} \plus{}1 ) }$
                                                     =$ \int\frac{dx}{x^{2} \plus{} 1 }$  + $ \int\frac{dx}{x^{4} \minus{} x^{2} \plus{}1}$  -  $ \int\frac{dx}{1\plus{}x^{6}}$

I think the first two integarls can be evaluated easily . Now for the 3rd integral.

 $ \int\frac{dx}{1\plus{}x^{6}}$ = $ \int\frac{x^{2} \plus{} 1 \minus{} x^{2} }{1\plus{}x^{6} }$

then evaluate .

ihope u got if not please leave a reply i put down the full solution..

While $ \int \frac{x^{2}\,dx}{1\plus{}x^2} \equal{} \frac{1}{3}\int \frac{3x^2\,dx}{x^{6}\plus{}1} \equal{}\frac{1}{3} \arctan (x^{3}) \plus{}c$, the other two integrals, $ \int \frac{dx}{x^{4}\minus{}x^{2}\plus{}1}$ and $ \int\frac{x^{2}\plus{}1\, dx}{x^{6}\plus{}1}$ don't seem to be that easy.

I see what you're doing, except that the numerator in your second integral on the first line should be $ (1\plus{}x^{2}) \plus{}(1\minus{}x^{2})$.

My problem is integrating $ 1/(x^{4}\minus{}x^{2}\plus{}1)$.

Now i have edited everything and have posted the solution . i hope its clear now and sorry for the typing error

vijaymenon, I've noticed that we get some cancellation which simplfies the work:

From your line $ \int\frac {dx}{1 \plus{} x^{2}} \plus{} \int\frac {dx}{x^{4} \minus{} x^{2} \plus{} 1} \minus{} \int\frac {dx}{x^{6} \plus{} 1}$, we have

(well you have):

$ \int\frac {dx}{1 \plus{} x^{2}} \plus{} \int\frac {dx}{x^{4} \minus{} x^{2} \plus{} 1} \minus{} \int\frac {x^{2} \plus{} 1 \minus{} x^{2}\,dx}{x^{6} \plus{} 1}$

$ \equal{} \arctan x \plus{} \int\frac {dx}{x^{4} \minus{} x^{2} \plus{} 1} \minus{} \int\frac {x^{2} \plus{} 1\, dx}{(x^{2} \plus{} 1)(x^{4} \minus{} x^{2} \plus{} 1)} \plus{} \int \frac {x^{2}\,dx}{x^{6} \plus{} 1}$

$ \equal{} \arctan x \plus{} \int\frac {dx}{x^{4} \minus{} x^{2} \plus{} 1} \minus{} \int\frac {dx}{x^{4} \minus{} x^{2} \plus{} 1} \plus{} \frac {1}{3}\arctan x^{3}$

$ \equal{} \arctan x \plus{} \frac {1}{3} \arctan x^{3} \plus{} c$.

Checking by differentiating, we get

$ \frac {1}{1 \plus{} x^{2}} \plus{} \frac {x^{2}}{1 \plus{} x^{6}}$

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

$ \equal{} \frac {1 \minus{} x^{2} \plus{} x^{4} \plus{} x^{2}}{1 \plus{} x^{6}}$

$ \equal{} \frac {x^{4} \plus{} 1}{x^{6} \plus{} 1}$.

Oh ya i never bothered abt the minus sign. Sorry Andrew :(


modular arithmetic
number theory proposed
number theory

find all prime p such that
$ \frac{2^{p\minus{}1}\minus{}1}{p}$ is a square number.

this was a problem from the [url=http://www.mathlinks.ro/viewtopic.php?p=341880#341880]Indian National Mathematics Olympiad,1995[/url]

[quote="quangpbc"]But, no have solution on this link. :D Please post your solution.I have a solution but it is not well :P[/quote]
[u][b]Solution[/b][/u]
observe that $ p\equal{}2$ can't satisfy the condition in fact it is not even an integer.
[u][b]CASE-1[/b][/u]
 let $ p\equal{}4k\plus{}1$
let $ \frac{2^{p\minus{}1}\minus{}1}{p}\equal{}m^{2}$ clearly $ m$ is odd.
this gives $ 2^{4k}\minus{}1\equal{}(4k\plus{}1)m^{2}$
observe that $ 2^{4k}\minus{}1\equiv 3\pmod 4$ but $ (4k\plus{}1)m^{2}\equiv 1\pmod 4$ so $ p\neq 4k\plus{}1$
[b][u]CASE-2[/u][/b]
hence let $ p\equal{}4k\plus{}3$
clearly if $ k\equal{}0$ then thsi holds let $ k >0$
thsi gives $ 2^{4k\plus{}2}\minus{}1\equal{}(2^{2k\plus{}1}\minus{}1)(2^{2k\plus{}1}\plus{}1)$ but observe that $ (2^{2k\plus{}1}\minus{}1,2^{2k\plus{}1}\plus{}1)\equal{}1$
hence one must be $ pm^{2}$ other $ v^{2}$
clearly by same reasoning as teh first case we  have
$ 2^{2k\plus{}1}\minus{}1\equal{}pu^{2}$
$ 2^{2k\plus{}1}\plus{}1\equal{}v^{2}\implies 2^{2k\plus{}1}\equal{}(v\minus{}1)(v\plus{}1)$
thus $ v\minus{}1\equal{}2^{a}$
$ v\plus{}1\equal{}2^{b}$
thus $ 2^{a}\minus{}2^{b}\equal{}2$ and $ a\plus{}b\equal{}2k\plus{}1$
$ 2^{b}(2^{a\minus{}b}\minus{}1)\equal{}2$ thus $ b\equal{}1$
$ 2^{a\minus{}b}\minus{}1\equal{}1$ so $ a\equal{}2$
thus $ v^{2}\equal{}9$ and $ k\equal{}1$ so $ p\equal{}7$
so only possible primes are $ 3,7$

[quote="pardesi"]
[u][b]Solution[/b][/u]
observe that $ p \equal{} 2$ can't satisfy the condition in fact it is not even an integer.
[u][b]CASE-1[/b][/u]
 let $ p \equal{} 4k\plus{}1$
let $ \frac{2^{p\minus{}1}\minus{}1}{p}\equal{} m^{2}$ clearly $ m$ is odd.
this gives $ 2^{4k}\minus{}1 \equal{} (4k\plus{}1)m^{2}$
observe that $ 2^{4k}\minus{}1\equiv 3\pmod 4$ but $ (4k\plus{}1)m^{2}\equiv 1\pmod 4$ so $ p\neq 4k\plus{}1$
[b][u]CASE-2[/u][/b]
hence let $ p \equal{} 4k\plus{}3$
clearly if $ k \equal{} 0$ then thsi holds let $ k > 0$
thsi gives $ 2^{4k\plus{}2}\minus{}1 \equal{} (2^{2k\plus{}1}\minus{}1)(2^{2k\plus{}1}\plus{}1)$ but observe that $ (2^{2k\plus{}1}\minus{}1,2^{2k\plus{}1}\plus{}1) \equal{} 1$
hence one must be $ pm^{2}$ other $ v^{2}$
clearly by same reasoning as teh first case we  have
$ 2^{2k\plus{}1}\minus{}1 \equal{} pu^{2}$
$ 2^{2k\plus{}1}\plus{}1 \equal{} v^{2}\implies 2^{2k\plus{}1}\equal{} (v\minus{}1)(v\plus{}1)$
thus $ v\minus{}1 \equal{} 2^{a}$
$ v\plus{}1 \equal{} 2^{b}$
thus $ 2^{a}\minus{}2^{b}\equal{} 2$ and $ a\plus{}b \equal{} 2k\plus{}1$
$ 2^{b}(2^{a\minus{}b}\minus{}1) \equal{} 2$ thus $ b \equal{} 1$
$ 2^{a\minus{}b}\minus{}1 \equal{} 1$ so $ a \equal{} 2$
thus $ v^{2}\equal{} 9$ and $ k \equal{} 1$ so $ p \equal{} 7$
so only possible primes are $ 3,7$[/quote]

It also is my solution.I think have a greater solution  :P


geometry
circumcircle
trigonometry
geometric transformation
homothety
rotation
geometry solved

Given a real number $\lambda > 1$, let $P$ be a point on the arc $BAC$ of the circumcircle of $\bigtriangleup ABC$.  Extend $BP$ and $CP$ to $U$ and $V$ respectively such that $BU = \lambda BA$, $CV = \lambda CA$. Then extend $UV$ to $Q$ such that $UQ = \lambda UV$. Find the locus of point $Q$.

$\frac{BU}{AB}=\frac{CV}{AC}=\lambda$ and $\angle UBA=\angle ACV$ there for $\Delta ABU \cong \Delta ACV$
it means that $\Delta AUV \cong \Delta ABC$then $\frac{AV}{AC}=\frac{AU}{AB}=\frac{UV}{BC}$
we can get from this: $UV,UQ$ are constant now

$AQ=\frac{\sin \angle B.QU}{\sin \angle QAY}$ and it's constant so the locus of Q is a circle by center A.

Amir

Hello Amir,

I do not think that you are right.

[quote="amir2"]$\frac{BU}{AB}=\frac{CV}{AC}=\lambda$ and $\angle UBA=\angle ACV$, therefore $\Delta ABU \cong \Delta ACV$
it means that $\Delta AUV \cong \Delta ABC$...[/quote]

This is good stuff, but I do not get the rest. In fact, the triangles $\triangle ABU \sim \triangle ACV$ and the triangles $\triangle AUV \sim \triangle ABC$ are just similar and not congruent. I would add that the last pair is similar, because from similarity of the triangles $\triangle ABU \sim \triangle ACV$, we have

$\frac{AB}{AC} = \frac{AU}{AV}$

$\angle UAV = \angle UAB - (\angle CAV - \angle CAB) = \angle CAB$

The triangles $\triangle AUV, \triangle ABC$ have a common vertex $A$ and they can be made centrally similar with positive coefficient of similarity by rotating the triangle $\triangle AUV$ counter-clockwise by the angle $\angle UAB$ or centrally similar with negative coefficient of similarity by rotating the triangle $\triangle AUV$ clockwise by the angle $180^o - \angle UAB$. Hence, they are spirally similar with ceneter of similarity $A$. Let $D$ be a point on the extended line $BC$, such that $BD = \lambda BC$ and the points $B, C, D$ follow on the line $BC$ in this order. Since $\frac{BD}{BC} = \frac{UQ}{UV} = \lambda$ and since the triangles $\triangle ABC \sim \triangle AUV$ are similar, the quadrilateral $ABCD \sim AUVQ$ (each with one internal angle equal to 180^o) are also spirally similar with the same center of similarity $A$ and the same coefficient of similarity. As the point $P$ moves on the arc $BAC$ of the circumcircle $(O)$ of the triangle $\triangle ABC$, the locus of points $U$ is an arc of a circle $(B)$ centered at the vertex $B$ and radius $r_B = \lambda BA$. Similarly, the locus of points $V$ is an arc of a circle $(C)$ centered at the vertex $C$ and radius $r_C = \lambda CA$. When the point $P$ coincides with the vertex $B$, the locus arc $(B)$ of the points $U$ is limited by the intersection $S$ of the circumcircle tangent at the vertex $B$ with the circle $(B)$ and the locus arc $(C)$ of the points $V$ by the intersection $B'$ of the ray $CB$ with the circle $(C)$. When the point $P$ coincides with the vertex $C$, the locus arc $(C)$ of the points $V$ is limited by the intersection $T$ of the circumcircle tangent at the vertex $C$ with the circle $(C)$ and the locus arc $(B)$ of the points $U$ by the intersection $C'$ of the ray $BC$ with the circle $(B)$. Thus the central angles of the locus arcs $C'S \equiv (B)$, $B'T \equiv (C)$ of the points $U, V$ are

$\angle C'BS = 90^o + \angle OBC = 90^o + 90^o - \angle A  = \angle B + \angle C$

$\angle B'CT = 90^o + \angle OCB = 90^o + 90^o - \angle A  = \angle B + \angle C$

[color=blue][b]Theorem. [/b][/color] Let the figure $\Phi$ move in such a way that it is always directly similar to a figure $\Phi_0$ and one common point $A$ of the similar figures $\Phi \sim \Phi_0$ does not move at all. Let the locus of a point $B$ of the figure $\Phi$ be a curve $\Gamma$. Then the locus of any other point $C$ (except $A$) of the figure $\Phi$ is a curve $\Gamma'$ similar to the curve $\Gamma$.

For a simple proof, see Yaglom, Geometric Transformations II (chapter 1, \1, pp. 68-69, theorem without a number between theorems 2 and 3). It follows that the locus of points $Q$ is an arc similar to the arcs $C'S \equiv (B)$, $B'T \equiv (C)$ of the points $U, V$. Since these 2 arcs are centered at the vertices $B, C$ of the quadrilateral $ABCD$, the locus arc $(D)$ of the points $Q$ is centered at the point $D$ corresponding to the point $Q$ in the spiral similarity of the quadrilaterals $ABCD \sim AUVQ$ and with the same central angle $\angle B + \angle C$. Radius $r_D$ of the circle $(D)$ is easily found when the point $P$ coincides with the vertex $A$ and the triangles $\triangle AUV \sim \triangle ABC$ and the quadrilaterals $ABCD \sim AUVQ$ become centrally similar with homothety center $A$ and homothety coefficient $\frac{AU}{AB} = \frac{\lambda - 1}{\lambda}$. Denote $a = BC, b = CA, c = AB$. Using the cosine theorem for the triangle $\triangle ABD$,
$DA = \sqrt{AB^2 + BD^2 - 2AB \cdot BD \cos{\widehat B}} = \sqrt{c^2 + \lambda^2 a^2 - 2\lambda ca \cos{\widehat B}}$
$r_D = DQ = DA + AQ = DA \left(1 + \frac{\lambda - 1}{\lambda}\right) = \lambda DA$
Substituting $\cos{\widehat B} = \frac{c^2 + a^2 - b^2}{2ca}$, we get
$r_D = \sqrt{c^2 + \lambda^2 a^2 - \lambda(c^2 + a^2 - b^2)} = \sqrt{\lambda(\lambda - 1) a^2 + \lambda b^2 - (\lambda - 1)c^2}$
The locus arc $(D)$ of the points $Q$ is limited by the appropriate intersection $S'$ of the line $C'S$ with the circle $(D)$, when the point $P$ becomes identical with the vertex $B$ and by the appropriate intersection $T'$ of the line $B'T$ with the circle $(D)$, when the point $P$ becomes identical with the vertex $C$.

ooops,sorry really I had a bad day, I wasn't OK.

[quote="pestich"]Oh,  c'm on Yetti,  it is so uncharacteristic of you!
   Do not drop the ball half-court,  take it to the hoop.

   Describe the bunches of circles centered at B and C,
   reveal the envelopes of  the UV  line for LAM=1,  LAM>1, LAM<1,
   disclose the role of vertex A, pinpoint  the exact location of  center D,
   identify and explain  the locus of intersection of two UV lines for different LAMs,
   and I hope you can elaborate (but not too much, just a bit!) on the connection
   (if there is any?) of this problem to the Simson line of Poncelet rotation problem.

   That would be a worth while reply, provided you keep it under 4 pages

   Maj.  Pestich.[/quote]

 :rotfl: :rotfl: :rotfl: Any geometric ankers (sic) ?

What Mathlinks really should do is to set up a new section [b]Psychology[/b] or better yet, a new section [b]Psychology of Yetti[/b]. That way, your life would become organized, because you could keep all your messages in one place.

Yetti


quadratics
number theory unsolved
number theory

We have $y_{n+1}=y_{n}^{2}+y_{n}$, with $y_{1}=2$. Do these $y_{i}$ have a closed form?


Fie $a, b, c>0$ astfel incat $abc=1.$ Sa se demonstreze ca:

   $\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\geq\frac{3}{4}$ ;)

(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1=ab+bc+ca+a+b+c+2=a(c+1)+b(a+1)+c(b+1)+2
Se aduce la acelasi numitor, se separa fractia si se foloseste inegalitatea $(a+1)(b+1)(c+1)\geq8$

Intr-adevar, e mult prea banala problema. :oops:

Si totusi a fost data la un test de selectie pentru OIM.

Ehei Claudiu, cunosti cintecul [b]"Si totusi exista iubire, si totusi exista blestem"[/b]?

Franta e tara EGALITATII , nu a INEGALITATILOR  :rotfl:[youtube]k1RSuGWkreQ[/youtube]

Nu o stiam. Buna faza cu Franta-tara egalitatii ...:))

In orice caz vad ca post-ul lui Cezar Lupu e din 2005. Eu am aflat ca a fost data la un test de selectie din articolul pe care l-a publicat in Gazeta impreuna cu dl. Lascu... . Mi se pare ca problema a fost data in 2006.


graph theory
combinatorics unsolved
combinatorics

Let f(n) = no. of labelled graphs of n vertices.

Let these graphs be "2-regular"

Also you cannot have cycles that have length that are a multiple of 10.

Find an expression for f(n)

Thankyou

If by "2-regular" graph you mean  each vertex has degree precisely 2.
Then such a graph IF connected is necessarely a cycle of length >=3.
More generally this graph is a set of "disjoint" cycles.
So now you have to enumerate these graphs under the constraint "avoid 0 (mod10) length"
This is not a graph theory problem but mainly an enumeration problem.
It looks a bit messy, you should try with small examples replacing "10" by "4" , "5" and "6".
Look also at graphs labeled or not.
I hope recasting this as an enumeration problem clarifies things a bit.


find the number of integral solutions to \[ x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}20\] given the conditions that \[ 1\le x_1\le 6, 0\le x_2\le 7, 4\le x_3\le 8, 2\le x_4\le 6\]

i don't know how usefull this may be, but one way that might work is to start at 27(the maximum possible sum under the constraints), and see how many ways you can subtract 7 total from the variables.