geometry calculus integration rectangle limit logarithms function

Prove that the area bounded between the curve $y = \frac{1}{x}$ and the $x$ and $y$ axes is infinite.

This is false.  However, it is possible to prove that the area is [i]not[/i] finite without calculus (although using ideas which are very important in calculus).

So, without using the machinery or terminology of integral calculus:

1) Prove that the region "bounded" by the line $x = 1$, the $x$-axis and the curve $xy = 1$, does not have finite area.

2) Using a similar method, prove that the region "bounded" by the line $x=1$, the $x$-axis and the curve $x^{2}y = 1$ [i]does[/i] have bounded area.

(I put "bounded" in quotation marks because these regions are, in fact, unbounded.  But whatever.)

Sorry, let me edit the original post. :blush:

[hide="1"]Looking at the diagram, it is obvious that between the lines $x=k-1$ and $x=k$, we can place a rectangle with length 1 and height $\frac{1}{k}$ underneath $y=\frac{1}{x}$ in the first quadrant.  Hence the area bounded by the axes and the graph $y=\frac{1}{x}$ is at least $1+\frac{1}{2}+\frac{1}{3}+\cdots$, which is infinite.

I used this diagram for a previous problem, but you can just flip it along the line $y=x$.[/hide]

Of course non-calculus is always preferred, but using calculus:

[hide]
Let the area A be the area in the first quadrant only. Then:

$A=\lim_{c\to{\infty}}\int_{0}^{c}\frac{1}{x}dx$

$A=\lim_{c\to{\infty}}{(\ln{c}-\ln{0})}$

Which is clearly a non-finite calculation.[/hide]

Just a minor nit-pick: that integral is improper at both sides, so you should have been taking two limits simultaneously.

[quote="DiscreetFourierTransform"]For the $x^{2}y=1$, can we go like this?
Bear in mind I haven't taken calculus yet, so this is probably incorrect, but based off of what I know about integrals.
We can see that the function is indeed $f(x)=\frac{1}{x^{2}}$.
Then we can say that $\sum_{n=1}^{\infty}\frac{1}{x^{2}}*f(x) = \frac{\pi^{2}}{6}*f(x)$, which is clearly finite?

That probably had a bunch of holes in it, but do I need to use an integral, or is this crude definition of Riemann sums correct?[/quote]

I don't really understand what you're trying to do.  Your summation doesn't make sense; what is the $\frac{1}{x^{2}}$ for if you already have a $f(x)$ term, and why does your expression not involve $n$?

A good "non-calculus" way to do it would be to bound the given area between $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$ and $\sum_{k=2}^{\infty}\frac{1}{k^{2}}$ (the quotation marks are necessary because any infinite sum is calculus in some sense).

[quote="DiscreetFourierTransform"]Can you show me how that would work?
I'm going into calculus next year, and I want to have a good idea of what it is before I start.[/quote]

Well do what I did, pick up the textbook for the next year and learn all of it during the last week of summer.

Simpler approach to the original problem (does not involve harmonic series):
[hide][img]http://img.photobucket.com/albums/v194/mukmasterxps/altpic.gif[/img]
Consider the series of rectangles above obtained by taking various $y$ intervals with lengths negative powers of two, and projecting them onto the curve to create rectangles.

The area is greater than $(1-1/2)(1)+(1/2-1/4)(2)+(1/4-1/8)(4)+...=1/2+1/2+1/2+...$, which is a diverging sum. Therefore, the area is not finite.[/hide]

A similar breakdown can be used for the second:
[hide]Consider:
[img]http://img.photobucket.com/albums/v194/mukmasterxps/altpic2.gif[/img]

We're doing the same $y$ scale breakdown as for the first problem, except we're doing an outer Reimann sum instead of an inner Reimann sum.

Area from $x=1$ to infinity is less than $(1-1/2)(\sqrt{2})+(1/2-1/4)(2)+(1/4-1/8)(2\sqrt{2})+...+2^{-n/2}+...$
This is a converging geometric series, and converges to $\frac{1/ \sqrt{2}}{1-1/ \sqrt{2}}=1+\sqrt{2}$. Subtracting $1$ for that extra square in the beginning, the area must be less than $\sqrt{2}$, and is therefore finite.

(Using calculus, the area is in fact $1$)[/hide]


how do you get a big bracket on the left side of several equations/ expressions that are on different lines (maybe eqnarray) that encompasses all the lines?

Manually size the brackets using \big, \bigg, \Big or \Bigg (see [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math mode[/url] section 8 page 23). Examples follow in 8.1 on page 25.
Please don't use eqnarray as it is obsolete and has been replaced by align - see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ 5[/url].


Which number is bigger
$\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{x+y}}$ or $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$

Isn`t the sign reversed?
I don`t know what is rearangement and how to prove it.. Maybe that`s why it was hard for me. But i hope you can show me, please.
Thank you.

x=(x1,x2,......,xn)
y=(y1,y2,....,yn)
if x and y are ordered  equally then sigma(xiyi)>=sigma(xiyw(i)) where w is a permutation of (1,2,...,n)

:oops_sign:
Perhaps my mind wasn't clear.

[quote="leepakhin"][quote="ZetaX"]As hard as $x^2 > 0$ for $x \neq 0$ ;)
First: I think it is $x,y,z>0$.
Second:
Just use Rearrangement on the equally ordered sequences $x,y,z,$ and $\frac{1}{\sqrt{y+z}} , \frac{1}{\sqrt{x+z}} , \frac{1}{\sqrt{x+y}}$ (to show that ordering: assume $x \geq y \geq z$, then $x+y \geq x+z \geq y+z$ and the rest follows). Thus $\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{x+y}} \geq \frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ follows.[/quote]
I don't think you can assume the ordering of $x,y,z$ since $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ isn't symmetric. :?[/quote]
But in any case we have $(x,y,z)$ and $(\frac{1}{\sqrt{y+z}},\frac{1}{\sqrt{z+x}},\frac{1}{\sqrt{x+y}})$ is equal sorted.So we can apply the Rearangement inequality,

[quote="leepakhin"][quote="ZetaX"]As hard as $x^2 > 0$ for $x \neq 0$ ;)
First: I think it is $x,y,z>0$.
Second:
Just use Rearrangement on the equally ordered sequences $x,y,z,$ and $\frac{1}{\sqrt{y+z}} , \frac{1}{\sqrt{x+z}} , \frac{1}{\sqrt{x+y}}$ (to show that ordering: assume $x \geq y \geq z$, then $x+y \geq x+z \geq y+z$ and the rest follows). Thus $\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{x+y}} \geq \frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ follows.[/quote]
I don't think you can assume the ordering of $x,y,z$ since $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ isn't symmetric. :?[/quote]
No, we can. Because if $x\ge z\ge y$  it is similar


which is me.  I found out just by searching for math websites on google.

I found out from the link on the AMC web page, http://www.unl.edu/amc

Well , I was searching for something to do with maths in google , and this site just showed up.Ever since then , there's only one place for queries relating to maths....:D

[quote="xxreddevilzxx"]a topic like this have already been discussed. please search before posting.[/quote]

Actually, since this is not a technical question or anything it's okay to make a topic like one that was posted before. This is actually the logic behind our "dont revive old threads" rule.


In the figure shown, a circle passes through two adjacent vertices of a square and
is tangent to the opposite side of the square. If the side length of the square is 3,
what is the area of the circle?

[hide]

First, raw radii from the center of the circle to the point of tangency and two vertices. Then, if you drop a perpendicular from the center of the circle to the square, you get a right triangle with sides $ 3\minus{}r$, $ r$, and $ \frac{3}{2}$, with $ r$ as the hypotenuse. By Pythagorean theorem, $ r^2\equal{}r^2\minus{}6r\plus{}9\plus{}\frac{9}{4}$. Solving gives $ r\equal{}\frac{15}{8}$.

[/hide]


function
modular arithmetic
number theory

(Also posted on the number theory proposed/own problems.)

Define f(x) to be what we will call the "eventual sum" of x, which is an iterating function until it reaches a single digit. For example, f(4435672) goes first to 4+4+3+5+6+7+2 = 31, and since 31 is not a single digit we f(31) which gives 3+1 = 4, which is the single digit we're looking for, so f(4435672)=4.

My idea: prove that for any integer a, there exists some b such that a-100<b<a+100 and f(1),f(2),...,f(b) yields the exact same number of 1's, 2's, ..., 9's. Else: provide a counterexample for a.

Don't really know where to go with this one, but I'm pretty sure it's true. You can sort of see it in the graph of f(x).

$f(x) = (x - 1) \bmod 9 + 1$

So, for some $a-100 < b < a+100$ we have $b \equiv 0 \bmod 9$ and the desired result.

So would that mean this is true for $a-6<b<a+6$?

In fact, it's true for $a - 5 < b < a + 5$.  There are nine numbers in that range, so one of them has to be divisible by 9.

does one of them being divisble by 9 guarantee the result? we know for sure that $b \equiv 0 \mod 9$, because of how there must be an equal number of 1's, 2's, ... 9's, but this does not guarantee the result that there will be an equal number of 1's, 2's, etc

also, i think your $f(x) = (x-1) \mod 9 + 1$ works but id like to see a proof to that


Ok the question is: A boy has a unlimited supply of 7 different types of candy. Given that he does not need each type of candy, how many different ways can he pick 12? If you could give a solution with your anwser that would be appreciated.

I might be wrong but I think that this is the solution.

Since he doesn't need every kind of candy, there are 7 choices for the first candy, 7 for the next, and so on. So I think our answer is $ 7^{12}$ or $ \boxed{1.38412872 \times 10^{10}}$.

Well, you see, it's just an 01 at the end, so I don't see why you need to express it in scientific notation... and I don't think it's correct anyways, since ABCDEFGGGGGG is the same as GCGBGAGFGEGD (just order reversed, but you have the same candies). It is correct if you consider the positions distinguishable, however.

By the way, $ 7^{12} \equal{} 13421287201$. :)

isabella2296 has overcounted the number of ways, as Yongyi781 has said.

[hide="Consider a slightly different problem"]
In this slightly different problem, the boy must select at least one of each type of candy.

Pretend these asterisks are the $ 12$ candies:
************
We can denote any selection by using $ 7 \minus{} 1 \equal{} 6$ dividers, like so
*/*/***/*/****/*/*
That would mean he took 1 of the first kind, 1 of the second kind, 3 of the third kind, 1 of the fourth kind, 4 of the fifth kind, 1 of the sixth kind, and 1 of the seventh kind.

Since he takes at least one of each type of candy (in this problem), there are $ 12 \minus{} 1 \equal{} 11$ places to put the $ 6$ dividers. Therefore the solution to this problem is the number of ways to pick 6 things out of 11, or $ \binom{11}{6}$. (In general, for $ n$ candies and $ k$ types, the answer is $ \binom{n \minus{} 1}{k \minus{} 1}$.)
[/hide]

[hide="And now the solution to the original problem"]
We run into problems if we try to use the above strategy on the original problem, because two of the dividers could potentially be next to each other, in which case we have complications....

So instead note that for every sequence of $ 12$ asterisks and $ 6$ dividers, like
/*/*****///***/***
we can add an $ 7$ more asterisks in each section, like so:
*/**/******/*/*/****/****
[i]___Confuzzlement starts here.___[/i]
Every such sequence has at least one asterisk before the first divider, between any two dividers, and after the last divider. And for every sequence of $ 12\plus{}7\equal{}19$ asterisks where each section has at least one divider, we can take out an asterisk from each section and we get $ 12$ asterisks with dividers placed without restrictions. Therefore the number of ways to place $ 12$ asterisks and $ 6$ dividers without restrictions is the same as the number of ways to place $ 19$ asterisks and $ 6$ dividers with the restriction that each section must have one asterisk. This is just the first problem we solved above, so the answer is
$ \binom{19\minus{}1}{7\minus{}1}\equal{}\binom{18}{6}\equal{}\boxed{18564}$.

Uh-oh... now someone will tell me this is wrong.  :maybe:

The general case for $ n$ candies and $ k$ types is $ \binom{n\plus{}k\minus{}1}{k\minus{}1}$.
[/hide]

yea, forgot to mention order does not matter or else i probably would have seen 7^12 but thanks anyway Isabella.

I think leoxnlin is correct. What distinguishes the choices from one another is the amounts he picks from each type of candy, regardless of order. So this could be thought of as giving 12 pieces of candy to 7 kids, or a+b+c+d+e+f+g=12.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=367062880&t=197582]This[/url] may help.

Wow leoxnlin. Thanks, that helped. And your solution was correct.


So I know it's true that if $ n$ is composite, then $ 2^n\minus{}1$ is also composite, but Engel makes the claim that this fact "follows from the fact that $ a\minus{}b|a^n\minus{}b^n$... how?? (Problem is pp. 131, "solution" pp. 138).

That gives $ a\minus{}1|a^{xy}\minus{}1$... If we try to set $ a \equal{} 2$ we get $ 1|2^n\minus{}1$ which is obviously true and doesn't say anything. Setting a = anything else, I don't see how that helps.

Take the other hint

$ 2^n\minus{}1 \equal{} 2^{xy}\minus{}1 \equal{} (2^x)^y\minus{}1$


function
topology
real analysis
real analysis unsolved

Let $f_n$ be a sequence of functions defined on $[0,1]$ that converges pointwise to 0. Does it follow that there exists a dense subset of $[0,1]$ on which the convergence is uniforme? What about such an open set?

I believe the answer to both questions is "no". For the first one, take $f_n$ to be the characteristic function of $\left(0,\frac 1n\right]$. For the second one, let $(q_n)_n$ be an enumeration of the rationals in $[0,1]$, and then take $f_n$ to be the characteristic function of $\{q_n\}$.

(I've edited it according to pleurestique's remark below; for some reason I thought we had to have some sort of special enumeration of the rationals, when any such enumeration works)

i really like this prob. and actually any enumeration works...

any open set must contain in it some interval $(\frac{i}{n}, \frac{i+1}{n})$ for some $i$ and $n$. now, wherever we are in our enumerated list, we've only listed up to that point a finite number of denominators, say those $\leq M$. so past that point in our list must lie *all* fractions of denominators $>M$. these include all those fractions of denominator $m = \min\{M+1, n+1\}$; because $m>n$, among these fractions there is one, $\frac{k}{m}$, which is contained in $(\frac{i}{n},\frac{i+1}{n})$.

And if we restrict to infinite sets of uniform convergence?  ;)

Then the answer becomes affirmative: there always exists an infinite set on which $(f_n)_n$ converges uniformly.

To each $x\in[0,1]$ associate the sequence $(S^x_m)_{m\ge 1}$, where $S^x_m$ is the number of terms of $(f_n(x))_n$ lying outside $\left(-\frac 1m,\frac 1m\right)$. What we want to prove translates to showing that there is an infinite set $X\subseteq[0,1]$ such that for all $m\ge 1$, the set $\{S^x_m\}_{x\in X}$ is bounded. We do this as follows:

Let $c$ be the continuum cardinality. $X_0=[0,1]$ has $c$ elements, and as $x$ varies in $X_0,\ S^x_1$ takes only non-negative integers as values, so there must be some $n_1$ such that $X_1=\{x\in X_0\ |\ S^x_1=n_1\}$ has cardinality $c$. Then, by the same argument, we can find $n_2$ such that $X_2=\{x\in X_1\ |\ S^x_2=n_2\}$ has cardinality $c$. After that, we find $n_3$ and $X_3$, and so on. $X_0\supseteq X_1\supseteq X_2\supseteq X_3\supseteq\ldots$ is a nested sequence of sets of cardinality $c$. In order to construct our set $X$, pick an element in each $X_k,\ k\ge 0$.

Edit:

Harazi pointed out to me that this is wrong. However, I think it can be repaired by changing the definition of $S_m^x$: take it to be the minimal $n_0$ such that $n\ge n_0$ implies $f_n(x)\in\left(-\frac 1m,\frac 1m\right)$.

i believe the answer might be yes if the every function in the sequence is continous

No, none of the answers change if $f_n$ are continuous.

For the first question, take $f_n,\ n\ge 1$ to be $1$ on $\left[\frac 1{n^2},\frac 1n\right]$, and $0$ outside $\left(\frac 1{n^2+1},\frac 1{n-1}\right)$ (a slight modification of the example given in post #2, so as to make the functions continuous).

For the second one, my example is a bit more complicated. There may be simpler ones, of course.

Let $(I_m)_{m\ge 1}$ be an enumeration of the open subintervals of $I=[0,1]$ which have rational endpoints (set $I_m=(a_m,b_m)$), and partition $\mathbb N=\{1,2,\ldots\}$ into sets $\mathbb N_m$, each of which contains arbitrarily large positive integers. For each $m\ge 1$ let $(I_m^k)_{k\ge 1}$ be a sequence of disjoint open subintervals of $I_m$, and let $n_m^1<n_m^2<\ldots$ be the natural ordering of the elements of $\mathbb N_m$. Make $f_{n_m^i}$ equal to $\frac 1m$ in some point of $I_m$, but zero outside $I_m^i$.

And of course, the answer to the third question stays positive.


An approach using newton sums:
[hide]Let $ a$, $ b$, and $ c$ be the roots of $ x^3\plus{}mx^2\plus{}nx\plus{}p$.  If $ S_n\equal{}a^n\plus{}b^n\plus{}c^n$:

$ S_1\plus{}m\equal{}0 \Rightarrow 1\plus{}m\equal{}0 \Rightarrow m\equal{}\minus{}1$

$ S_2\plus{}(\minus{}1)(S_1)\plus{}2n\equal{}0 \Rightarrow 2\minus{}1\plus{}2n\equal{}0 \Rightarrow n\equal{}\minus{}\frac{1}{2}$

$ S_3\plus{}(\minus{}1)(S_2)\plus{}(\minus{}\frac{1}{2})(S_1)\plus{}3p\equal{}0$
$ \Rightarrow 3\minus{}2\minus{}\frac{1}{2}\plus{}3p\equal{}0 \Rightarrow d\equal{}\minus{}\frac{1}{6}$

$ S_4\plus{}(\minus{}1)(S_3)\plus{}(\minus{}\frac{1}{2})(S_2)\plus{}(\minus{}\frac{1}{6})(S_1)\equal{}0$
$ S_4\minus{}3\minus{}1\minus{}\frac{1}{6}\equal{}0$
$ \boxed{S_4\equal{}\frac{25}{6}}$[/hide]


geometry
geometric transformation
reflection
linear algebra
matrix
linear algebra unsolved

Prove that every bijective linear transformation (Over $ R^2$) has either one or infinite fixed points.

I think that in fact a bijective linear transformation is either the identity, has a fixed line or only one fixed point.

Obviously false. The identity is a bijective linear transformation.

Dear jmerry you are right, I've corrected my initial post  :)

Still false.  Bijective linear transformations on $ \mathbb{R}^3$ can preserve planes and still not be the identity (for example, reflections through a plane).

In general, the fixed points of a transformation with matrix $ \mathbf{A} \in \mathcal{M}_n$ are precisely the eigenspace corresponding to the eigenvalue $ \lambda \equal{} 1$, and this eigenspace can have any dimension (the geometric multiplicity of $ 1$) from $ 0$ to $ n$.

Of course, dimension zero is a point.

This is not a particularly interesting question; the real answer is that the fixed points are a subspace.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove or disprove that
If $ N$ is a normal subgroup, and $ N \cap Z(G) \equal{}$ {$ e$},then $ N\equal{}${$ e$}.

Can you think of a group that has trivial center and contains one or more non-trivial, proper normal subgroups?

Counterexample:
The center is always normal. Take $ G \equal{} S_{5}$. Then $ N \equal{} A_{5}$ is the only normal subgroup of $ G$ and $ Z(G)$ is trivial.


[quote="Mathx"]if a positive integer $n$ is a square $\mod p$ for every prime $p$,Must $n$ be square number?[/quote]

Yes; this was proven by Vess in http://www.mathlinks.ro/Forum/viewtopic.php?t=5997 post #14. Anyway, an elementary proof of this fact would be really interesting...

  darij


Let two sequence number $ (a_n)$ and $ (b_n)$ we have:
  $ a_{i \plus{} 1} \equal{} a_i \minus{} \frac {a^3_i}{4} ;b_i \equal{} |a_i|$
Prove that:Exist value of $ a_1$ so that $ limb_n\neq 0$

[quote="karis"]Let two sequence number $ (a_n)$ and $ (b_n)$ we have:
  $ a_{i \plus{} 1} \equal{} a_i \minus{} \frac {a^3_i}{4} ;b_i \equal{} |a_i|$
Prove that:Exist value of $ a_i$ so that $ limb_n\neq 0$[/quote]

Should it be $ a_1$ instead of $ a_i$?

Try $ a_1 \equal{} 2\sqrt{2}$, you will get a constant $ b_i$.

I presume you mean that it has a limit.

Suppose the sequence itself without the absolute value is convergent.
Limit boths sides to infinity, you will get that the limit has to be 0.

Suppose the sequence only converges with the absolute value, so I GUESS that the sequence will alternatively being positive and negative.
Then I put this guess into the equation to get that the result.


$a, b$ are two fixed lines through $O$. Variable lines $x, y$ are parallel. $x$ intersects a at $A$ and $b$ at $C$, $y$ intersects $a$ at $B$ and $b$at $D$. The lines $AD$ and $BC$ meet at $M$. The line through $M$ parallel to $x$ meets $a$ at $L$ and $b$ at $N$. What can you say about $L, M, N$? Find the locus $M$.

The variable lines $x$ and $y$ should be parallel to a [b][u]fixed[/u][/b] direction.

Best regards,
sunken rock


In right $ \Delta ABC$, $ \angle CAB$ is a right angle. Point $ M$ is the midpoint of $ \overline{BC}$. What is the number of centimeters in the length of median $ \overline{AM}$? Express your answer as a decimal to the nearest tenth.
[asy]pair A,B,C,M;
A = (0,0); B = (4,0); C = (0,3); M = (B+C)/2;
draw(M--A--B--C--A);
label("A",A,W); label("B",B,E);
label("C",C,W); label("M",M,NE);
label("3 cm",A--C,W); label("4 cm",A--B,S);[/asy]

It is a right triangle, so the other side is 5. The median is half of the hypotenuse, so it is 2.5.
To show the median is half of the hypotenuse, extend it until it forms a rectangle ABCD, where AD is the new line. The two diagonals are equal, and they bisect each other.

Actually, $ \overline{AM}$ is an altitude.

You can turn the triangle so that the hypotenuse is the base.

The area of the whole triangle is $ 6$, using $ 3$ as the altitude and $ 4$ as the base.

Now, you can see the area in a new way, $ \frac {5a}{2}$, where $ a$ is the altitude.

[asy]draw((0,0)--(1.8,2.4)--(5,0)--cycle);
label("$3$",(0.9,1.2),W);
label("$4$",(3.4,1.2),E);
label("$5$",(2.5,0),S);[/asy]

Then, solve for $ a$.

$ \frac {5a}{2} \equal{} 6$

$ 5a \equal{} 12$

$ a \equal{} \boxed{2.4}$

EDIT: I should learn how to read...

That would mean ABM is isoceles. Is that the case?

And ernie, no it isnt an altitude, its a median!

[quote="GameBot"]What is the number of centimeters in the length of [b]median[/b] $ \overline{AM}$?
[/quote]
It says median, ernie.
@mewto: yes, it's isosceles because the four triangles that form from the diagonals of a rectangle are always isosceles.

[geogebra]2de8fd3ca5254f1f66aa201783d7964b54e4c587[/geogebra]

Visual--the median is always equal to the circumradius of a right triangle, which is half the hypotenuse. This explains the isosceles triangles, too.


floor function
logarithms
number theory unsolved
number theory

Let $\{a_{n}\}$ be an infinite sequence of positive integers such that $a_{n}<a_{n+1}\le 10 a_{n}$ for all $n\geq 1$.  Consider the number $\alpha=0.a_{1}a_{2}a_{3}\cdots$ formed by concatenating the numbers $a_{n}$ (for e.g. if $a_{n}=2^{n}$ then $\alpha=0.248163264128\ldots$).  Prove $\alpha$ is irrational.

What I am stuck on:
Assume $\alpha$ is rational then it has a repeating part with say length $r$.  I was able to show that for every sufficiently large integer $k$, there exists an $a_{n}$ with exactly $kr$ digits.  So, by the property $a_{n}<a_{n+1}\le 10 a_{n}$, we have that $a_{n+1}$ must have $kr$ or $kr+1$ digits.  The latter results in a contradiction (which we want).  Now, I need to obtain a contradiction with $a_{n}$ and $a_{n+1}$ both having $kr$ digits.  Please help with this part of the argument.  Thanks!

[quote="baz"]Let $\{a_{n}\}$ be an infinite sequence of positive integers such that $a_{n}<a_{n+1}\le 10 a_{n}$ for all $n\geq 1$.  Consider the number $\alpha=0.a_{1}a_{2}a_{3}\cdots$ formed by concatenating the numbers $a_{n}$ (for e.g. if $a_{n}=2^{n}$ then $\alpha=0.248163264128\ldots$).  Prove $\alpha$ is irrational.[/quote]

If $\alpha$ is rational, consider $n_{0}$ great enough for $a_{n{0}}$ be in the periodic part of $\alpha$.
Consider this period is $T$-digits long.
Let $l(x)=1+\lfloor \log_{10}(x)\rfloor$ be the length (in decimal digits) of $x$.
We have $l(a_{n+1})=l(a_{n})$ or $l(a_{n+1})=l(a_{n})+1$
So, it exists $n_{1}\geq n_{0}$ such that $l(a_{n_{1}})=kT$
Then the only possibility for $a_{(n_{1}+1)}$ is $a_{(n_{1}+1)}=10a_{n_{1}}$
But this means that the first digit of $a_{n_{1}}$ or of $a_{(n_{1}+1)}$ is $0$, which is impossible.

And $\alpha$ can't be rational.

[quote="baz"][quote="pco"]
Then the only possibility for $a_{(n_{1}+1)}$ is $a_{(n_{1}+1)}=10a_{n_{1}}$.[/quote]

This is what I am not seeing.  Could you please explain why this is the only possibility.  Thanks.[/quote]

Since $l(a_{n_{1}})=kT$, the $kT$ next digits are exactly the same as the $kT$ digits of $a_{n_{1}}$.

So $a_{(n_{1}+1)}$ can't be made of $kT$ digits, else it would be equal to $a_{n_{1}}$.

So $a_{(n_{1}+1)}$ is $(kT+1)$-long and $a_{(n_{1}+1)}=10a_{n_{1}}+0..9$

But, since $a_{(n_{1}+1)}$ must be $\leq 10a_{n_{1}}$, the only solution must be $a_{(n_{1}+1)}=10a_{n_{1}}$


Do you know how to obtain them? thanks..

The 2006 contest is available on the ARML website (http://www.arml.com).  There are also books of past contests you can order.  Try Amazon.  Alternatively, the website directs you to email J. Bryan Sullivan about ordering them.


inequalities
number theory unsolved
number theory

Prove that if $a,b,c,k$ are integer $k>1$then
$(-1+ab+bc+ca)^k+(a+b+c-abc)^k$ ($k$ even, not in form $2^n$) can be factor.

I think he said that $(ab+bc+ca-1)^k+(a+b+c-abc)^k$ for $k=2r+1$ form is not prime.
and I know that $(ab+bc+ca-1)^2+(a+b+c-abc)^2=(1+a^2)(1+b^2)(1+c^2)$

[quote="jin"]I'm sorry I can't understand.
What do you mean by "factor"?[/quote]

Sorry because I didn't state clear my problems. To be honest with you, as solving 1 inequality problem, I found it but unable to prove. You can find it as http://www.mathlinks.ro/Forum/topic-82579.html post #11 :(


abstract algebra
group theory
superior algebra
superior algebra unsolved

(i) Prove: The abelian group $ A$ is a direct sum of subgroups $ B$ and $ C$ iff $ A\equal{}B\plus{}C$ and $ B\cap C\equal{}{0}$. This is the case iff the map
$ BxC\to A$ given by $ (b,c)\to b\plus{}c$
is an isomorphism.

(ii) Prove that an abelian group $ A$ is a direct sum of subgroups $ B_1,B_2,...,B_r$ iff the map
$ (b_1,...,b_r)\to b_1\plus{}b_2\plus{}...\plus{}b_r$ of $ \Pi B_i\to A$
is an isomorphism

what have you done so far? you don't need any ideas here to check the definitions ..


Where could a rising 9th grader take high school, AP, or intro college level physics over the summer?

Aside from this options:
- local college or summer college program (age requirement problems)
- CTY (hoping to find a physics class that requires calculus)

I vaguely recall reading about places where you can just go and take a high school or AP level course during the summer.
[hide="The Idea"]
The idea is that I want a certain will-be- rising 9th grader who knows calculus now to be able to learn some intro physics at least enough to pass both AP physics C exams the spring after the coming spring.

Frankly, I don't trust this clever will-be-rising 9th grader to study on his own extremely diligently. [/hide]

If you want to take it in a local school, you should see if this student can register for the AP course.  Otherwise, try a local college and see if they will allow the student to take the course.

k. I'll try that.

In case that doesn't work out, which it might not since the local colleges require you to be a junior for dual enrollment, is there any place I could send him to take a physics class?

Self-study.  Anyway, I don't think it's that important to take AP physics as a freshman in high school.  In my opinion, unless one has already displayed some sort of genius-type skill in physical sciences, there is no reason to push it.  If this is for college, I can tell you that most adcoms don't care when you take physics.  In fact, some might look at this as simply an effort to get into college and do too much.  In other words, not for passion, just for application padding.


function
integration
ceiling function
calculus
limit
logarithms
calculus computations

Need some help here, Please!!!

Define $G_n= \sum_{i=1}^{n} \frac{1}{i}$

1. Show that $G_n - log(n)= 1+\int_{1}^{n} \frac{t-\lceil t\rceil}{t \lceil t\rceil}dt$ and concude that there is a constant C such that $G_n=log(n)+C+R_n$, where $|R_n|\leq\frac{1}{n}$

2. Use (1) to show that $\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=log(2)$

3. Show that $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...=log(2\sqrt{2})$

In 3. willn't your series equal to  $log(2)$, what is in 2 ?

That's the question asks for, but I dont know how to start. Could anyone here to help me with this? Thx alot.

$\sum_{j=1}^{2k}\frac{(-1)^{1+j}}{j}=2\left(\sum_{j\le2k, j\text{ is even}}\frac1j\right)-\sum_{j=1}^{2k}\frac1{j}= \sum_{j=1}^{k}\frac1{j}-\sum_{j=1}^{2k}\frac1{j}=G_{2j}-G_j$.
Taking limits, we can do part 2. Part 3 ($1+\frac02+\frac13-\frac24+\frac15+\frac06+\frac17-\frac28\dots$) should be a similar exercise in setting up linear combinations of partial sums of the harmonic series.

That leaves part 1. Here, we want $C=1+\int_1^\infty \frac{t-\left\lceil t\right\rceil}{t \left\lceil t\right\rceil}\,dt$, and you need to show that $\int_n^\infty \frac{\left\lceil t\right\rceil-t}{t \left\lceil t\right\rceil}\,dt\le\frac1n$. Use the general fact that $0\le\left\lceil t\right\rceil-t\le1$ and $\frac1{t\left\lceil t\right\rceil}\le\frac1{t^2}$ for this.

Don't try to actually compute $C$ beyond defining it as an integral.

Did you use the part 1 to show part 2?

I still dont know understand what you meant by setting up linear combinations of partial sums of the harmonic series to show the part 3.

And I totally lost when you say about the " Don't try to actually compute $C$ beyond defining it as an integral for the part 1.

Need more explanation, Thanks.

Could anyone help more to explain how to get the part (2) by (1) and (3)? Thx in advance.

Maybe the notation confuse you. But Jmerry did it  :) . he got
$\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = G_{2n} - G_{n}$. And from the condition $|R_n| \leq \frac{1}{n}$, you'll get $\lim_{n \rightarrow \infty} R_n = 0$. Now take a look $G_{2n}-G{n}$. use part 1. and taking limit you'll get part 2.More precisely $G_{2n}=\ln 2n + C +R_{2n}$ and $G_{n}=\ln n + C +R_{n}$ substract them and taking limit.

[quote="koola"]Did you use the part 1 to show part 2?

And I totally lost when you say about the " Don't try to actually compute $C$ beyond defining it as an integral for the part 1.

Need more explanation, Thanks.[/quote]

If you try to compute the numerical value of $C$, it is equivalent to  compute $\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \left( \frac{1}{k} \right)- \ln n$ that is known as euler's constant. It seem that the excact value of C can not determined by that integral.


calculus
integration
calculus computations

[quote="b555"]the  integral  on the right  hand  side  can be seen  as   k/(x+a)^2.    and left  side  integral is an obvious  one, so where  is the difficulty? :?:[/quote]

The LHS is easy. In the RHS, $ h$ is not a constant. What do we do there?

thats  what i am saying,  u can  see that  as  [b]k/(x+a)^2  [/b]           where  k=GM  and  a=R.

now  to evaluate  the integral of the  bolded part above  is  an easy task......... :)

[quote="b555"]now  to evaluate  the integral of the  bolded part above  is  an easy task......... :)[/quote]

O.k. All I wanted to know is if variation of $ h$ with respect to $ t$ causes any change in the value of the integral or if we can simply consider it to be constant and integrate as we do normally.

[quote="b555"]thats  what i am saying,  u can  see that  as  $ \int \frac{k}{(x\plus{}a)^2}$  [/b]           where  k=GM  and  a=R.
[/quote]

And integrate w.r.t what? How does that give me the time $ t$?

i guess u need to  write h as a variable  of  t  there.  then the job is done. :)

[quote="b555"]i guess u need to  write h as a variable  of  t  there[/quote]

Don't you think that would complicate this?  :maybe:

[quote="Rajiv"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1426871#1426871]Preface[/url]

I want the value of:
$ \int_{0}^{v_f}dv_h \equal{} \int_{0}^{t}\frac {GM}{(R \plus{} h)^2}dt$

Where $ v_f \equal{} \sqrt {\frac {2GM_E}{R \plus{} H}}$

$ H$ is a constant whereas $ h$ is not (for those who are too busy to read the preface :) )[/quote]

In your preface.
see the step in which you went wrong was writing the basic equation. you forgot to include final P.E. which is $ \frac { \minus{} GM}{R}$
also you forgot (-) sign of PE.

As for integarting write g as vdv/dh. then you integrate one side wrt h another side wrt v. then again dh/dt=v which gives t.

Oops! Thanks a ton for pointing it out! I never thought of that, and that mistake was so inconspicuous, I suppose no one else noticed it.


1.  Two rats, one male and one female, scampered on board a ship that was anchored at a local dock. The ship set sail across the ocean. When it anchored at a deserted island in late December, the two rats abandoned the ship to make their home on the island.

Under these ideal conditions, it might be interesting to estimate the number of offspring produced by this pair in one year. you should make the following assumptions:
_ The number of young produced in every litter is 6, and 3 of those 6 are females.
_ The original female gives birth to 6 young on January 1, and produces another     litter of six 40 days later and every 40 days thereafter as long as she lives.
_ Each female born on the island will produce her first litter 120 days after her birth, and then produce a new litter every 40 days thereafter.
_The rats are on an island with no natural enemies and plenty of food. Therefore, in this first year, there are only births, and no rats die.

What will be the total number of rats by the next January 1, including the original pair?

[hide="hint"]Leap years don't matter[/hide]

[hide="second hint"]Do you need to know the date when the next litter is born, or just the day?(By day, I mean like the 41st day, the 243rd day, so on)[/hide]

[hide="third hint"]This problem isn't "hard", it's just very tedious.[/hide]

First of all, is the problem asking you to assume that during that current year, there is no leap year? :?:

To get the answer, you must first figure out how many rats will be alive on january 1. This answer is 8. Then, for the next 120 days, the number of rats will only increase by 6. 8+18=26

(there is an 8 there because of the two parental rats).

120 days later...........
Finally, the 12 female babies, and the mother will be able to give birth. There will then be  98 rats. Just keep track of the number of babies born at that time, , the number of female rats, and the total number of rats.

(sorry if this is a little confusing)

rihim

A litter is usually a piece of trash thrown on the floor...No, but in this case it's a group of babies born for animals. Like, a litter of kittens, puppies, etc.

Wait, so if I say a litter of cats that means a group of baby cats?

That was just born, yes. They are usually less than a month old.

@ rihim, you might want to check your solution. Every 40 days [b][i][u]all[/u][/i][/b] females give birth (unless they are under 120 days old. So for example, in those first 120 days, the mother has given birth 3 times. So what you are doing is correct (brute force), but with the wrong numbers.  :)


Is there any equivalent to the gather environment usable on the site? \[\] by itself yields right-aligned lines, and eqnarray only works if you are aligning equations.

Please don't use eqnarray as it is obsolete - see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ 5[/url].
The gather environment can be made to work on the forum in the same way as align works, but in the meantime the workaround to get what gather does, ie centre each line of equations, is to use more than one set of displayed equations, one per line. For example
\[ \frac{1}{2} \plus{} \frac{3}{4}\]\[ 1 \plus{} 2 \plus{} 3 \plus{} \ldots \plus{} n \equal{} \frac{n(n\plus{}1)}{2}
\]

Then you end up with giant spaces between each line.

I did say it's a workaround and the line spacing is not that much greater than in the gather environment itself
[img]http://sixthform.info/gblog/pictures/e4ee851ab7de55f0a6c9871ac510e7ad.gif[/img]

\[ 1 \plus{} 2 \plus{} 3 \plus{} \ldots \plus{} n \equal{} \frac {n(n \plus{} 1)}{2}\]


Let $ A\in \mathcal{M}_n(\mathbb{C})$ .Does the following equality hold ?

 $ rank(A)\equal{}rank(A\cdot \overline{^tA})$

Yes, because the inner product it's based on is positive definite. You don't have to make $ A$ square, either.

If $ Av\equal{}0$, $ A^*Av\equal{}0$. If $ A^*Av\equal{}0,$ $ v^*A^*Av\equal{}\|Av\|^2\equal{}0$ and $ Av\equal{}0$. (I interchanged the roles of $ A$ and the conjugate transpose; it doesn't change the content)
This means that the two matrices have the same null space; they must therefore have the same rank.


The March 10 Math Jam that discussed the 2005 AIME used a technique called Power of a Point to solve problem number 15. Would anyone mind describing the Power of a Point technique or pointing me to a resource where I can learn about them? Are they related to Mass Points?

Thanks!

I don't think there's any relation to mass points. The power of a point P is denoted omega ($\omega_P$).
The tangent from $P$ to circle $O$ is $\overline{PT}$. A secant to circle $O$ through $P$ is $\overline{PAB}$.
$\omega_P=\left(PT\right)^2=\left(PA\right)\left(PB\right)$

If I recall correctly, the proof involves drawing $\overline{TB}$ and $\overline{TA}$. Then use similar triangles.

Power of a Point states that
a. CQ*QB=DQ*QE
b. FR*FG=FS*FH=FP^2
and can be proved using similar triangles


Princeton
college
trigonometry
HMMT
email
ARML
Siemens Competition

[quote="chess64"][quote="Iversonfan2007"][quote="NeverOddOrEven"]anyone going to go to the princeton math comp for north carolina?[/quote]

 :wink:  :wink:[/quote]

 :huh:  :huh: [size=0][color=white]there are too many smilies you can have at most three smilies in this message please remove some smilies before submitting this message[/color][/size][/quote]

yes there is a Carolina Team  :wink:

[quote="NeverOddOrEven"]:huh:  :huh:[/quote]  :arrow:  :idea:

[quote="Iversonfan2007"][/quote][quote="chess64"][quote="Iversonfan2007"][quote="NeverOddOrEven"]anyone going to go to the princeton math comp for north carolina?[/quote]

 :wink:  :wink:[/quote]

 :huh:  :huh: [size=0][color=white]there are too many smilies you can have at most three smilies in this message please remove some smilies before submitting this message[/color][/size][/quote][quote="Iversonfan2007"]

yes there is a Carolina Team  :wink:[/quote]

 :)  :) [size=0][color=white]there are too many smilies you can have at most three smilies in this message please remove some smilies before submitting this message[/color][/size]

[quote="jb05"]What an enlightening discussion! We have a NC/SC team. :wink:  :wink:[/quote]

indeed  :pilot:  :spidy:

any leftovers that wanna form another one?

 :wink:  :huh:  (just so we can continue the enlightening discussion)

Me?  :D  :D

And what's with the double smileys?

k so we currently have me and not trig as a team. i think we're gonna need more people soon though hehe. how bout you bflat? :lol:  :P

You guys should get together a team for Princeton Math contest. It's December something. 16th. You can get ncssm people too if you don't have enough.

Excellent observation! The entire last page has been devoted to it.

If you want another team, I think Jeremy Hahn was interested, and you may try Misha or John Pardon or someone like that. Yasha, Arnav, and I are going on the original Carolina team.

wtf is this princeton math contest...i wanna come! :what?:  :o

Maybe those of us in the triangle could actually meet somewhere to work on the problems over the weekend?

Thanks guys for your hard work! Hopefully we'll win! :D We have a pretty good chance. GO NC

dang good job guys. almost got the perfect score. yea you guys have a really good chance of winning.

[quote="n^4+4"]Alright guys!   Team NC is going to win the itest, Princeton, and ARML without breaking a sweat.  This was just a warmup[/quote]

You forgot the IMO.

And a Fields Medal too, and let's throw in a Wolf Prize or an Abel Prize in case we don't get a team organized until we're past 40.

Man we are just too good... no one has a chance. Everyone else should just automatically capitulate to our overpowering skill. :wink:

GUYS WE WIN ITEST

$200 GIFT CARD

SWEET

 :D

dude yes north carolina is clearly going to win IMO

Of course - how could North Carolina not secede from the Union?

By the way, is that one 200-dollar gift card for each of us or one 200-dollar gift card for us as a group?

When do we get this?

For each of us. I imagine they shipped it out around the time they put out results, but let's just be patient. Unless you're just REALLY REALLY eager for some shopping at Bust Buy.

[quote]Of course... how could North Carolina not secede from the Union?[/quote]

That would be South Carolina. Like 5 times. Hmm... we'll all move to South Carolina, win the IMO, and Stephen Colbert will be our president. Right?

hmmm...

i don't remember filling out my address at the itest website.. i wonder how they're going to mail it out

@blueboat: if you gave your school, your city, and your state they can send it to you at school.  That's what happened to me last year with an AMC plaque.  Or they might just find out your address through your school.

you could probably email bradley about it if you want to check

also it's probably 200 / person

they give good prizes

also dude NC

Yes - NC is doing well.  for example, three North Carolinians will happen to be representing the Southeast Region in the team division of the National Finals of the Siemens Competition in early December in New York.

Okay, yes, one of them is me - I had gotten questions about what I did, so here's [url=http://www.newsobserver.com/print/sunday/city_state/story/760151.html]how and what we did[/url].


calculus
derivative
limit
calculus computations

Show that the derivative of $x^{-4}=\frac{-4}{x^5}$.

Just by the exponent rule it works, unless the problem is to actually use the definition of the derivative to evaluate the limit. Anyways, this shouldn't be under getting started.

Using the definition of the derivative, let $f(x)=x^{-4}$. Then you just need to show that $\lim_{\Delta x \to 0} \frac{f(x+ \Delta x)-f(x)}{\Delta x} = -4x^{-5}$ which is trivial.

i put it in getting started and it got moved.


Find three consecutive integers such that the sum of 2 times the first and 3 times the second equals 10 more than the third.
I did post this in the beginner section and nobody gave me the right answer can somebody help me

Let $x$ be the middle integer

$2(x-1)+3x=(x+1)+10$

Solve for $x$ and you will be done.  But there are no such integers which will work.

Yeah, there is no such integer.  If you solve it, you can get that 9/4, 13/4 and 17/4 work, but there are no integer solutions.

Hey you posted this same problem in getting started, anyway, you don't get an integer so there is no solution.


Just an hour ago, I upgraded to Firefox 3.0 and Fedora 9. Now, the the little animation things on the side of AoPS don't really work on my computer. Sometimes they work like normal, but a lot of the time they just freeze or are blank. Is this my computer or AoPS?

It's probably just your computer. It works fine on mine. So it's not AoPS.

I agree with Gauss. Local issue, nothing to do with AoPS. Check your flash instalation and/or your Firefox extensions.


[quote="leepakhin"][quote="pvthuan"]Given three non-negative numbers $x,y,z$ such that $xy+yz+zx=1$, prove that
\[ \frac1{1+x^2+y^2}+\frac1{1+y^2+z^2}+\frac1{1+x^2+z^2}\leq\frac95.  \][/quote]
An obvious and but ugly solution:  :P
Homogenize the inequality,
$\frac{1}{xy+yz+zx+x^2+y^2}+\frac{1}{xy+yz+zx+y^2+z^2}+\frac{1}{xy+yz+zx+x^2+z^2}\leq\frac{9}{5(xy+yz+zx)}$
Expanding everything, this is equivalent to
$4\sum_{sym}x^5y+7\sum_{sym}x^4y^2+3\sum_{sym}x^3y^3\geq 10\sum_{sym}x^3y^2z+4\sum_{sym}x^2y^2z^2$
which is true by Muirhead :)[/quote]

my solution is the same.  :)

[quote="socrates"]It would be interesting to find a solution without Muirhead
(that a lot of people hate, including me... :P )
 :D[/quote]
I have just solved an easy part of the problem: Suppose that one of the number is zero, say, $x=0$, so that the desired inequality takes the form
\[ \frac1{1+y^2}+\frac1{1+y^2+z^2}+\frac1{1+z^2}\leq\frac95.  \]
Note that when $x=0$, the given restriction gives $yz=1$. Therefore, we need to show that
\[ \frac1{1+y^2}+\frac1{1+y^2+\frac1{y^2}}+\frac1{1+\frac1{y^2}}\leq\frac95,  \]
which can be written equivalently as
\[ \frac1{1+y^2}+\frac{y^2}{1+y^2}+\frac{y^2}{y^4+y^2+1}\leq\frac95.  \]
And this is to $4y^4-y^2+4\geq0$. Q.E.D

Have you got any idea for the other case?  :lol:

[quote="hungkhtn"]DO you complete it, PVThuan?[/quote]
Of course, not.

I wait for your solution.

[quote="hungkhtn"]here is it ( I don't sure because It's found out in a short minute)

We need to prove:

$\frac{x^2+y^2}{1+x^2+y^2}+\frac{y^2+z^2}{1+y^2+z^2}+\frac{z^2+x^2}{1+z^2+x^2} \ge \frac{6}{5}$

Notice that:

$\frac{x^2+y^2}{1+x^2+y^2} \ge \frac{(x+y)^2}{2+(x+y)^2}$

Then :
 $\sum \frac{x^2+y^2}{1+x^2+y^2} \ge \sum\frac{(x+y)^2}{2+(x+y)^2}$
$\ge \frac{4(x+y+z)^2}{6+(x+y)^2+(y+z)^2+(z+x)^2}=\frac{4(x+y+z)^2}{6(xy+yz+zx)+(x+y)^2+(y+z)^2+(z+x)^2}$

And now all we need to do is:

$20(x+y+z)^2 \ge 36(xy+yz+zx)+6(x+y)^2+6(y+z)^2+6(z+x)^2$

Which is very easy because $x^2+y^2+z^2 \ge xy+yz+zx$.[/quote]

Very nice and correct solution. Thank you Hung.

I think this problem is really easy.
By some computations,we have the given inequal is equivalent to:
$6+2(\sum x^2)+9 x^2y^2z^2\leq 4\sum x^4+12\sum x^2y^2+9(\sum x^2)^2.(\sum y^2z^2)$
Since the initial relation:$xyz\leq \frac{1}{3\sqrt3}$
And using some result:$(\sum x^2)^2\geq \sum x^2$( because $\sum x^2\geq 1)$
We have q.e.d immediately..

The following is true and stronger.

Given three non-negative numbers $x,y,z$ such that $xy+yz+zx=1$, prove that \[\frac1{1+zx+y^{2}}+\frac1{1+xy+z^{2}}+\frac1{1+yz+x^{2}}\leq\frac95.\]

Seems to be similar to: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30719

[quote="pvthuan"]Given three non-negative numbers $x,y,z$ such that $xy+yz+zx=1$, prove that \[\frac1{1+x^{2}+y^{2}}+\frac1{1+y^{2}+z^{2}}+\frac1{1+x^{2}+z^{2}}\leq\frac95.\] [/quote]
There's a simple solution by MV for this inequality. But it's not very short.


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

A thought that I don't have time to chase down:

First, the function is even so the integral might as well be taken over the whole real line.

Looked at under the Fourier transform, we are looking for the following: Evaluate at zero the value of the convolution of a finite collection of normalized box functions. (By "normalized box function", I mean a function of the form $f(x)=\frac1b$ on the interval $[-b/2,b/2]$ and zero elsewhere; the normalization is that they all have the same integral.)

[quote="Buffalo"]I can't understand what you said.
Related problem: Let $a_1=1$, $a_2=t$, $0<t<1$, let $a_n$ be a positive sequence s.t. $\forall n\ I_n=\int_{-\infty}^{\infty}\prod_{i=1}^{n}\frac{\sin a_i x}{a_i x}dx\equiv\pi$, How long could this sequence be?
An example: $1,\ \frac{1}{3},\ \frac{1}{5},\ \frac{1}{7},\ \frac{1}{9},\ \frac{1}{11},\ \frac{1}{13},$ and next, $\frac{2021}{45045}$, and, the end.[/quote]
What a silly question! We can always get an infinite sequence.


Prove that \[\sum_\textrm{cyc}\frac{a+b}c \geq 4 \sum_\textrm{cyc}\frac{c}{a+b}, \, \forall a,b,c > 0\] I'm sure that it was discussed, but I can't find the topics. Please help me with that :blush:
Also, I think I found a
[hide="solution."]
It's equivalent to $\sum S_{a}(b-c)^{2}\geq 0$, where $S_{a}= \frac{a^{2}+ab+ac-bc}{bc(a+b)(a+c)}$. If we suppose that $a \geq b \geq c$, then $S_{a}, S_{b}\geq 0$ and $S_{b}+S_{c}\geq 0$, which solves it, since $(c-a)^{2}\geq (c-b)^{2}+(b-a)^{2}$ (SOS method).
[/hide]
Is it correct?

I think I found a more elementary proof. \[\sum_{cyc}{\frac{\frac{a}{c}+\frac{a}{b}}{2}\geq \sum_{cyc}{\frac{2}{\frac{c}{a}+\frac{b}{a}}}}\] By AM-HM. Do some manipulation to obtain the original inequality  :)

$\sum_{\sigma}{\frac{a+b}{c}}\geq4\sum_{\sigma}{\frac{c}{a+b}}\geq6$ by Nesbitt's .  Can someone finish.

${1 \over x}+{1 \over y}\ge{4 \over x+y}.$

$LHS=\sum\left({a \over b}+{a \over c}\right) \ge \sum{4a \over b+c}.$


Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangements of the 8 coins.

[hide="Solution"]Disregarding faces, there are $\frac{8!}{4!4!}=70$ ways to arrange all of the coins.

Now let's see what happens when we consider the faces. If one of the coins is face up, all the others must be face up, because we can't have any faces facing each other. Thus, there are 8 different ways here.

However, we need to consider one more arrangment; the case when they're all face down. There's 1 way here.

Therefore, the answer is $9*70=630$ arrangments.[/hide]

[hide="Answer"]There are $\binom{8}{4}=70$ ways to place the eight coins independently of their faces. Considering the faces: every time a coin is face-up, all coins after it must be face-up. Therefore, we have eight cases of each of the different coins being the first coin face-up, and one case with all coins face-down, so we have $9(70)=\boxed{630}$.[/hide]

where do u get 8 coins facing up? there r only 4 coins...nd y is it "1" tat for the cases facing down? isnt it facing down just flipping over the facing up? so facing up should equal facing down??? idk im confused

when he says there are eight choices for the first coin face up the case that all the coins are face up is included because if the FIRST coin is face up then all of them are...
there are 9 choices however, to include the case where NONE of the coins are face up

There are eight coins. The gold and silver coins are in the same stack. Also, do you understand that if any coin (starting from the bottom) is facing up, all of the coins after that must also be facing up, or else two faces will be touching each other? So, there is one combination for each of no coins facing up, the first coin and on facing up, the second and on, etc., so we have $9$ possibilities there, after $70$ possible ways to place the coins.

Also, please try to type with proper punctuation and grammar, it makes your posts much easier to understand.

[quote="NeverOddOrEven"]are there faces on both sides or on just one?[/quote]

[quote="4everwise"]Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. [b]Each coin has an engraving of one face on one side, but not on the other.[/b] He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distunguishable arrangements of the 8 coins.[/quote]

I think that's clear enough.

[quote="breez"][hide="Question"]If there's 70 unique arrangements, why do you multiply by 9? How are there 8 ways they can all be facing up?  There's 70 ways of them all facing up, and 70 all facing down. So it should be 2 x 70 =  140.[/hide][/quote]

There are also the cases:

DUUUUUUU
DDUUUUUU
DDDUUUUU
DDDDUUUU
DDDDDUUU
DDDDDDUU
DDDDDDDU

[b]DU[/b]UUUUUU

D[b]DU[/b]UUUUU

DD[b]DU[/b]UUUU

DDD[b]DU[/b]UUU

DDDD[b]DU[/b]UU

DDDDD[b]DU[/b]U

DDDDDD[b]DU[/b]

I may be misundestanding the question, but the bold adjacent coins are face to face.

In Paladin's notation, left is down and right is up.  So two coins DU are not facing each other, while two coins UD are.  If you want to use the other possible orientation (so left is up and right is down), then you'll still have 9 possibilities, which you would get by flipping the order of each of Paladin's possibilities.

[hide]You can have 9 possibilities, depending on where  the heads start.

So there are 70 ways of arranging 4 gold and 4 silver, so it is 70*9=630

$630$[/hide]

[quote="vishalarul"]The problem would be easier if there are engravings on both sides. That's what I did. I got an answer of 140.

Suppose we have a generalization of this problem:
Suppose there are $a_{1}$ gold coins, $a_{2}$ silver coins,... $a_{n}$ of whatever. Find the number of arrangements you can do if no coin can be face to face.

Correct me if I am wrong...

[hide="Am I right?"]$9\frac{(a_{1}+a_{2}+a_{3}+...+a_{n})!}{a_{1}!\cdot a_{2}! \cdot a_{3}! \cdot ...\cdot a_{n}!}$[/hide][/quote]

I don't think it would be $9$ anymore... more like $a_{1}+a_{2}+\cdots+a_{n}+1$ so you could add that to the factorial to simplify the expression.

I belive the way we did this in AIME class was something along the line of:

[hide]If you think about the coins in a stack, starting with them all facing down, as soon as you turn any coin in the stack upward, all of the ones on top of it must also face up.  So, you're just looking at the number of possible locations of the bottom-most upward facing coin.  There are 9 (one where none face up, and once for each location in the stack of 8).  The number of ways to arrange 4 gold and 4 silver coins in a stack is 8!/(4!4!), or 70.  Multiply this by the 9 facing arrangemetns and you have 630.[/hide]

Sorry to bump this one.

We're allowed to tear off little pieces of paper to visualize the coins for problems like this, correct?

We can think of it as 8 circles up-down, and split each circle as the "north pole" and the "south pole."
What we need to consider is at which step the coin "ruins" it for everyone else. Basically, a down forces others to face down as well.

[hide]Before we choose which coins are gold and which coins are silver, we can choose the orientation of the faces on the coins in the stack. The key of the problem is to notice that after we have one coin that is face up, all other coins above it will also be face up. Therefore, we are basically choosing where we want to put the bar in a group of $8$ stars. In this stars and bars problem, the bar can be at the ends of the star. Therefore, there are $9$ possibilities. Now we can choose which coins are gold and which coins are silver. Therefore, the solution is $\boxed{\binom{8}{4}\cdot 9=630}$.[/hide]

[hide=Recursive Approach](In this solution we will disregard the fact that gold and silver are indistinguishable and multiply by $\tbinom84$ at the end.) Build a stack of $n$ coins, and consider the bottom coin. Then, let $a_n$ be the number of ways to arrange these coins following the conditions given in the problem. If the bottom coin is facing up, there is exactly $1$ way, because the orientation of every single coin is forced. However, if the bottom coin is facing down, there are $a_{n-1}$ ways, because we can simply disregard the first coin that we place; it doesn't affect the count. Thus, it is the equivalent of the number of ways to place $n-1$ coins. We have the recursion $$a_n=a_{n-1}+1,$$ with $a_1=2$ (there are $2$ faces of a coin). Thus, we get $$a_1=2,~a_2=3,~a_3=4,~a_4=5,~a_5=6,~a_6=7,~a_7=8,~a_8=9.$$ Then, due to the fact that we have to now account for the ordering of the coins, the answer is $9\cdot\tbinom84=\boxed{630}$.[/hide]

@below it's a different solution

hi pi_is_3.14! :)

hi does anyone have practice problems with recursion in combinatorics

thanks

Hello, can someone please confirm I'm doing this right? I don't think I am. Sorry, I am not good at recursion.

Let $a_k$ be the amount of ways to have a stack with $k$ coins with the top face facing up and let $b_k$ be the amount of ways to have a stack with $k$ coins with the top being a non-face facing up, and let $c_k$ be the total. If a face is facing up, then the next coin also has a face facing up. Otherwise, the next coin can be either way. Therefore, $c_{k+1} = a_k + 2b_k$. Also, $a_{k+1} = a_k + b_k$ because we can in both cases ($a_k, b_k$) just add a coin with the top face facing up, and $b_{k+1} = a_k$ because we can only have the $a_k$ case, where we add a coin with the top face facing up. (in the $b_k$ case, we'd get two "heads" facing each other)


How many four digit numbers less than $ 5000$ are possible with:
a. no repeatation of digits in the four digits
b. digit $ 1$ compulsorily appearing as one of the four digits
c. the four digit number being divisible by $ 11$
d. All ten digits between $ 0$ to $ 9$ (both including) are qualified to appear

Since no one has responded to this i will post my very, very unelegant and confusing solution
[hide="solution...sort of"]
remember from primary school that if a number is divisible by $ 11$ then the alternating subtraction and addition of the digits is also divisible by $ 11$. In this case, if a four digit number in the form $ abcd$ is divisible by $ 11$ then $ a\minus{}b\plus{}c\minus{}d$ is also divisible by 11, or $ a\plus{}c \equiv b\plus{}d \mod 11$

Now one digit has to be $ 1$ so for now we let $ a\equal{}1$ and $ a\plus{}c \equal{} {1,2...10}$
So we have to find the partions of {1,2...10} in the form $ b\plus{}d$ so that $ a\minus{}b\plus{}c\minus{}d$ is divisible by 11

To do this consider a function, $ f(x)\equal{}x^0 \plus{} x^1 \plus{} x^2...x^9$,
the expansion of $ f(x)^2$ will show us all the possible sums of two digits, but instead of actually expanding this we just realise that the exponents will be $ {0,1,2,3...18}$ and the respective coefficients are $ {1,2,3...9,10,9...3,2,1}$
Now since we are only concerned with numbers that sum to $ {1,2...10}$ we find that there are $ {1,2,3,4,5,6,7,8,9,10,9}$ such pair. (now it gets a little messy)
From the first condition there can be no repeating digits so we have to subtract 1 from every even sum to get
$ {2,2,4,4,6,6,8,8,10,8}$
Also, since we have counted the each pair twice (e.g. 8=5+3 and 3+5) we divide by 2 and get
$ {1,1,2,2,3,3,4,4,5,4}$
And again; one of each solution will be the form (1+c) so we must subtract because (a+c) is already in that form...$ {0,0,1,1,2,2,3,3,4,3}$

So the sum of these is $ 1\plus{}1\plus{}2\plus{}2\plus{}3\plus{}3\plus{}4\plus{}4\plus{}5\plus{}4 \equal{} 29$
These groups of four can be organised 8 ways and still be divisible by 11 but 8 will begin with 0

So the answer is $ 29\times 8 \minus{} 8 \equal{} 224$
 :oops:

[/hide]


[quote="linj"]你们知道不知道哪里可以买 TI-nSpire CAS?[/quote]

不知道這是什麽東西。。。有錢人啊


geometry
analytic geometry
graphing lines
slope

I don't know what this question is asking me to do. I can easily figure out which one is the right answer, but there has to be a way of getting the answer without having to look at the answers and figure it out.
[img]http://i47.tinypic.com/8x6j5h.jpg[/img]

Thanks.

And by the way, the picture may say review, but this book has not went over this at all. This is the first time it's mentioned anything like this.

[hide]This problem is basically just asking for the equation of the line that goes through the three points given: $ (10,45)$, $ (20,60)$, $ (30,75)$. Choose two points and find the slope between these two points. I'll choose the first and the last, so the slope is $ \frac{30}{20}\equal{}\frac{3}{2}$. Also using point-slope formula $ y\minus{}y_1\equal{}m(x\minus{}x_1)$ where $ (x_1,y_1)$ is a point on the line and $ m$ is the slope. Since we used two points to form this line you can choose either one to put into this equation. Choose the first point for simplicity sake. So we have that the equation of the line is $ y\minus{}45\equal{}1.5(x\minus{}10)$
This simplifies to $ y\equal{}1.5x\plus{}30$, or since it's with cost and money, $ c\equal{}1.5m\plus{}30$ or $ F$.
[/hide]

Thanks. I was trying to figure out how to use the slope intercept form for this one. I forgot about the point slope form. How did you know to use that?

[quote="gaussintraining"]A good indication to use point slope form in a problem is if two (or more) points are given, and you are asked to find the equation of the line. Here's a practice problem, for example:

Given that a line passes through the point $ (3,2)$ and has an $ x$-intercept of $ 7$, find the equation of the line.[/quote]
y=2x-4? I used (3,2) as one point and (7,0) as the other. And using the point slope form, I got my answer.

Thanks for the help guys.

One more quick 2 questions that I can't figure out.
It says "Find the value of the variable and MP, if P is between M and N."

MP = 7x, PN = 3x, PN = 24.

And the other one is:

MP = 2c, PN = 9c, PN = 63.

What I can't figure out is how to get the MN value when I don't know if P is halfway between M and N. If it can be anywhere on the line between M and N, I don't know how to figure it out. I got the other ones where I was given the value of MN.
Thanks.

[quote="ckck"]Well for this problem it helps to draw out the line. You have that P is between M and N. Remember due to the variables given it doesn't matter whether P is halfway or not...because we're trying to find out where it is :P. We have that MP is 7x and PN is 3x. Because MP+PN=PN(just look at your drawing) we have the equation 10x=24 or x=2.4. Try the other one :)[/quote]
Oh, I'm stupid. I was racking my brain trying to figure out why they gave me two values for PN, when in fact, they did that to show me the value of X.
Thanks.


geometry
parallelogram
geometry unsolved

Let a parallelogram $ABCD: AD=DB$.Let a point $M$ such that:$MA\perp AC;MD\perp DC;N=DB\cap MC;I=AC\cap MD$.Find $\angle MNI$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88917]View all problem for VN IMO team 2006[/url]

[quote="nthd"]Let a paralell $ABCD: AD=DB$.Let a point $M$ such that:$MA\perp AC;MD\perp DC;N=DB\cap MC;I=AC\cap MD$.Find $\angle MNI$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88917]View all problem for VN IMO team 2006[/url][/quote]

What is a parallel $ABCD: AB=DB$ ??

Daniel

I think he meant a [i]parallelogram[/i].

One can easily observe that $AMBI$ is a kite, thus $\angle{MBI}=90^{\cdot}$.
Since $I$ is the centroid of $\triangle{ABD}$ and  $AMCD$ is cyclic we get: \[ \angle{IMC}=\angle{CAD}=\angle{IBD}  \] consequently $MBNI$ is cyclic. Therefore $\angle{MNI}=90^{\cdot}$.


induction? it works for 1 and 2 so.... :D

[hide]We define the sequence $ \{a_{k}\}$ in $ \mod n$ such that $ a_{k}\equiv 7\cdot 10^{k}\mod n$

We can define $ \{a_{k}\}$ with the reccursive formulas:

$ a_{0}=7 \mod n$ and $ a_{k+1}=10\cdot a_{k}\mod n$

$ a_{k}$ cannot be all different in modulo n, because they are infinite

Accurately, there is some positive $ m$ which repeats after some steps, say for example $ a_{m+T}\equiv a_{m}$.
Then of course $ a_{k+T}\equiv a_{k}, \ \forall k\geq m$

Then we can choose $ n$ copies of $ a_{m}$, namely the terms $ a_{m}, a_{m+T}, ..., a_{m+(n-1)T}$

If we add these $ n$ equal values, then of course we get a multiple of $ n$,

and the corresponding number  $ x=\sum_{i=0}^{n-1}7\cdot 10^{m+iT}$ is a multiple of $ n$

[color=brown][u][hide="Note"][/u]
The number $ x$ is not necessarily the smaller with that property.

For example, if $ n=11$ we get $ \{a_{k}\}= \{7,4,7,4,7,4,...\}$

$ 4+7\equiv 0\mod 11$, so the number $ 77$ is a multiple of $ 11$

The number $ x$ becomes $ 707070707070707070707$, much larger, but the problem asks only for the existence of those numbers

Another example, for $ n=13$ we have $ \{a_{k}\}=\{7,5,11,6,8,2,7,5,11,6,8,2,...\}$

We can choose each of those elements we like to make a multiple of $ 13$

For example one could take the numbers $ 7+5+11+6+8+2=39$, which corresponds to the number $ 777777$

But someone else could take just the numbers $ 7+6=13$, so it's the number $ 7007$

$ \star$ If there is some positive $ m$ with $ a_{m}\equiv 0 \mod n$ then the number $ 7\cdot 10^{m}$ is a multiple of $ n$[/color][/hide][/hide]

[quote="mdk"]Show that for any positive integer n, there exists a positive multiple of n that contains only the digits 7 and 0 .[/quote]

[hide="Solution"]
I give a solution with an example, say $ n=11$.
Now create the following set:
$ 7,77,777,....,777777777777$
Now there are $ 12$ numbers.
So by pigenhole two when subtracted are divisible by $ 11$.
When you subtract those two you end up with a number $ 77..70...0$.

Now generalize!
[/hide]

[quote="0714446459923"]if this is true for 7, woulnd't it be true for any digit 1-9?[/quote]

In particular, it's true for the digit $ 1$, which implies its truth for all the other digits trivially.


geometry
perimeter
rectangle
calculus
trigonometry
3D geometry
prism

Hi ya'll, just had a question from a math competition . . . [moderator edit:] [i]Which math competition was that? Are there other people in other places who will still be taking the same competition with the same question? [/i]

I'll use a little lemma that's pretty easy to prove;

If $a,b\geq 0$, then $a+b\geq 2\sqrt{ab}$

Proof: $(\sqrt {a}-\sqrt{b})^2\geq 0$

Okay, so let $a,b$ be the dimensions of the sides parallel to the ground.

We have that $V=4ab+\frac {4}{2}ab=6ab$

Similarly, because the perimeter of the uppermost rectangle is $2a+2b=360\Rightarrow a+b=180$, we have that:

$\frac {3}{2} (a+b)^2=\frac {3}{2}\cdot 180^2\geq \frac {3}{2}\cdot 4ab=6ab$

So $V\leq \frac {3}{2}\cdot 180^2$ (I don't have a calculator handy)  And yes, I'm blah_blah from ibscrewed.

[quote="WindSlicer"]Hey blah! (figured that was you)

Hm, I understand your solution, and I do believe you arrived at the correct conclusion. The thing is, it's all based on the fact that a + b  >=  2 :sqrt: ab   I don't really see how you just pulled that out of thin air, and I kind of need to learn how to do these types of problems rather than learn how to do this one. Is there a more geometry way to do it, I just don't see the writers of that test writing it and having that as the way to get the solution as it's a geometry test. Or perhaps that a + b  >=  2 :sqrt: ab is some geometric/trig thing, and they just wanted us to know that *shrugs*

Could you, or another, explain a general way to approach a problem like this? Thanks![/quote]

Well, you know how to do calculus, so you should know that the area of any rectangle with constant perimeter is maximized when it's a square.  This is just a way of showing it without calculus.

Blah, I don't know calc :p I'm taking geometry/partial trig in highschool. I'll be dual enrolling precalc/trig in the summer, but atm I don't know calc. Anyways...

Hm, so detect the flaw in my reasoning.

I know that the maximum area can be obtained by setting a,b equal to 90 each, creating a square. In order to find the volume of the rectangular prism (subtracting the pyramid) I would multiply (length)(width)(height) = (90)(90)(4) to find the volume, which equals 32,400. Now, we must find the area of the square pyramid. We know the formula for a pyramid is the (area of base)(height)(1/3)  =  (4x90)(90)(1/3) = 10,800.  Addint the two values we get 32,400 +10,800 = 43,2000  which is the wrong answer.

I'm guessing it has to do with assuming the maximum area is when a,b is 90? But I don't see why it is not..

Unless I'm looking at your diagram wrong, that's not a square pyramid - it's just half of a box, and hence has volume $\frac {1}{2}bh$ instead of $\frac {1}{3}bh$

I thought I remembered you saying that you were taking multivariable calc or something; I guess I just remembered wrong.

Ah, no, wasn't me.. maybe I said I wish I could *shrugs*

And ohhhh.. I've been so used to computing pyramids in math calss that when I see that shape I just assume it's one. Now it makes sense, so my logic isn't wrong, I weas just using a wrong formula. Thanks for the help, would have been thinking about it the whole weekend and not getting anywhere!

why cant u say the top is a rectangular prism 90x90x4 and the bottom is a triangular prism (i dont know what u call it) is  90x90x4 (making its area 90x90x2).

Yeah, that's what I ended up doing, I just misclassified the bottom part as a square pyramid rather than half of a rectangular prism, therefore giving me a wrong answer.


For some reason, when you put an apostrophe in a hide tag it doesn't work. For example, I could be saying [code][hide=this user's problem]Solution[/hide][/code] and it wouldn't show up.

Is there any way to fix this?

[hide="This hide tag won't be hidden"]Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=284089]here[/url][/hide]

[quote="RoFlLoLcOpT"]Is there any way to fix this?[/quote]
No. (Unless you don't put it in at all).


Let [x] be defined as the greatest integer which is less than or equal to x.  If the postal rate for first class mail is 35 cents for the first ounce or fraction thereof and is 25 cents for each additional ounce or fraction thereof, then the cost in cents of first class postage on a letter weighing w ounces where w>0, is always k+f[pw+g].  Find the value of k+f+p+g.

The answer key says 10, but I don't see it.   Any ideas?  Wouldn't the formula be 35 + 25[w] ??  This would give 61 for the answer.

Actually it would be
35 + 25[w-1]
k=35
f=25
p=1
g=-1
$35+25+1+(-1)=60$
It probably meant something like k+f*p*g
That would give you 10

Actually, it's 35-25[-w+1]
35-25-1+1=10

35 + 25[w] doesn't work with w=1
35 + 25[w-1] doesn't work with w=1/2 or something like that

Great!  Thanks.   This problem was trickier than I originally thought.


Find the specific heat capacities $ c_{V}$ and $ c_{p}$ for a gaseous mixture consisting of $ 7g$ of nitrogen and $ 20g$ of argon. The gases are assumed to be ideal.

I've got 0,5 and 0,73, but the solution says 0,42 and 0,65$ \frac{J}{g\cdot K}$

Got it, book is right.

[hide="Solution"]

The heat transferred to the gas is spent on heating both ingredients. $ (m_1\plus{}m_2)c\Delta T\equal{}m_1c_1\Delta T\plus{}m_2c_2\Delta T$, so $ c\equal{}\frac{m_1c_1\plus{}m_2c_2}{m_1\plus{}m_2}$

That holds for isochoric and isobaric specific heat capacities also, and as it's easy to find them using that nitrogen is diatomic, and argon monoatomic, the solutions are 0,5 and 0,73...

[/hide]


In the diagram, $ ABCD$ is a square. The area of rectangle $ NFMD$ is half the area of $ ABCD$, and $ ND \equal{} \frac{1}{2} CN$. If the area of $ ABCD$ is 36 square centimeters, what is the number of centimeters in the perimeter of rectangle $ ABEM$?

[asy]pair A,B,C,D,I,F,M,N;
A=(0,0); B=(0,5); C=(5,5);
D=(5,0); I=(12,5);M=(12,0);
N=(5,1.8); F=(12,1.8);
draw(A--B--I--M--cycle,linewidth(0.7));
draw(C--D,linewidth(0.7));
draw(N--F,linewidth(0.7));
label("$A$",A,SW);
label("$B$",B,NW);
label("$C$",C,N);
label("$D$",D,S);
label("$E$",I,NE);
label("$F$",F,E);
label("$M$",M,SE);
label("$N$",N,W);[/asy]

AB = 6, CD = 6
Reason: Square of area 36 => sides of length sqrt(36)=6
ND = 2
Reason: ND=CN/2, CD=6 => 3 ND = CD = 6, ND = 2
NF = 9
Reason: 36/2=2*NF => NF = 9
BE = 15
Reason: BC+CE=BE

9+15+9+15 = 48.

Answer: [b]48[/b]


For any positive integer $n$ the sum $\displaystyle 1+\frac 12+ \cdots + \frac 1n$ is written in the form $\displaystyle \frac{P(n)}{Q(n)}$, where $P(n)$ and $Q(n)$ are relatively prime.

a) Prove that $P(67)$ is not divisible by 3;

b) Find all possible $n$, for which $P(n)$ is divisible by 3.

I think I know what the answer is, but I have no proof. By checking a lot of numbers with Mathematica (I couldn't have done that during the contest, of course :D), it looks like the only numbers $n$ s.t. $3|P(n)$ are $2,\ 7,\ 22$. Strange set of solutions, huh? Maybe the fact that there are no more has something to do with (a) and with the fact that $7=3\cdot 2+1,\ 22=3\cdot 7+1,\ 67=3\cdot 22+1$.

I have no clue about this, but maybe these are important observations.

What is Mathematica?

:):):)

_______________
"Our days are never coming back... " (Highway Song -> System of a down)

I agree with you, grobber.
Define S[n] the residues mod 3^n and coprime to it.
then for every a there is a' s.t.aa'/3^n=1(mod 3^n)
Also if 3^k|a, define a' s.t. aa'/3^(n-k)=1(mod 3^(n-k)).
Suppose 3^k<n<3^(k+1) we must have
sum a' (a=1..n) divisible by 3^(k+1) (*)
 We group 1..n into A[1]..A[k] by the power of 3 dividing them
By (*) we must have 3|sum(a')  ai n A[k].
Hence both 3^k and 2*3^k belong to A[k]. So 2*3^k<=n.(1)
If we have k=1 then this yields n=2.
Now consider A[k-1] By (1) we can show 1,2,4,5 belong to A[k-1] and maybe 7 or 7, 8.
But we use (*) and (1) to get 3|sum(a') a' in A[k-1] + 1.
Checking we get just 7 belong to A[k-1].
So 7*3^(k-2)<=n<8*3^k-2. If k=0 we get n=7.
Continuing like this we get 22/27<=n<23/27 yielding n=22 for k=3.
But this procedure gives no solution further (we have 2 possible residues to choose, and the desired residue in luckily not among them).
Hence we stuck and we get grobber's result.
I know my solution is hard to understand. Sorry I can't explain it.
To solve it you must draw the tables of a' for S[1],S[2],S[3],S[4]. 81 numbers.
Elegant, huh?
If this is the only solution they've got, this is a horrible problem.
Maybe it involves a beatiful idea, but counting 81 numbers..

[quote="iura"]I agree with you, grobber.
[...]
If this is the only solution they've got, this is a horrible problem.
Maybe it involves a beatcan someone explain it because I can't understand it?please write the thought and the conclusion. iful idea, but counting 81 numbers..[/quote]
can someone explain it because I can't understand it?please write the thought and the conclusion.
I also have a thought.for P(k) is not divisiable by 3,we can prove P(3k),P(3k+1),P(3k+2) is not ,either.

I think I have found a much easier solution:
$\displaystyle 1+\frac 12+ \cdots + \frac 1n$ = $\displaystyle \frac { \sum^{n}_{j=1} \frac{n!}{j}} {n!} $
Take the greatest dividing exponent $a$ such that $3^a$ || $n!$, which means that $3^a$ divides $n!$, but $3^{a+1}$ does not divide it.
For $n \geq 27$ the greatest dividing exponent of $\sum^{n}_{j=1} \frac{n!}{j}$ is less than $a$.
1) $\displaystyle 3^b \leq n \leq 2*3^b-1$
For $1 \leq j \leq n$, $j$ $\neq$ $3^b$, then $3^{a-b+1}$ | $\frac{n!}{j}$.
But $3^{a-b}$ || $\frac{n!}{3^b}$, so $3^{a-b}$ || $\sum^{n}_{j=1} \frac{n!}{j}$, done.
2) $\displaystyle 2*3^b \leq n \leq 3^{b+1}-1$
For $1 \leq j \leq n$, $j$ $\neq$ $3^{b-1}$, $2*3^{b-1}$, $3^b$, $4*3^{b-1}$, $5*3^{b-1}$, $2*3^b$, and eventually $7*3^{b-1}$, $8*3^{b-1}$, then $3^{a-b+2}$ | $\frac{n!}{j}$.
But  $\displaystyle 3^{a-b+1}$ || $\displaystyle \frac{n!}{3^{b-1}} + \frac{n!}{2*3^{b-1}} + \frac{n!}{3^b} + \frac{n!}{4*3^{b-1}} + \frac{n!}{5*3^{b-1}} + \frac{n!}{2*3^b}$ = $\displaystyle \frac{49n!}{20*3^{b-1}} $.
Consider each subcase:
2.1) $\displaystyle 2*3^b \leq 7*3^{b-1} - 1 $
$3^{a-b+1}$ || $\displaystyle \frac{49n!}{20*3^{b-1}} $ , done.
2.2) $\displaystyle  7*3^{b-1} \leq 8*3^{b-1} - 1$
$3^{a-b+2}$ || $\displaystyle \frac{363n!}{140*3^{b-1}} $
This case is again subdivided and consider $j$ such that $3^{a-b+3}$ | $\frac{n!}{j}$ while for the others give:
$3^{a-b+2}$ || $\sum^{n}_{j=1} \frac{n!}{j}$, that's why we must have $ b \geq 3 $.
2.3) $\displaystyle  8*3^{b-1} \leq 3^{b+1} - 1$
$3^{a-b+1}$ || $\displaystyle \frac{761n!}{280*3^{b-1}} $ , done.
I have overviewed it and I think it is quite wrong, but I will not delete it anyway.

well,I think your solution are right and beautiful.but we have to check n from 1 to 26.I think it will also be a huge work.if anyone in the test could do this work? :)

@heartwork: I like your approach very much. Your response is a bit hard to understand, but if I understand correctly,  the only case we have to worry about is when $7*2^{b-1}\le n<8*2^{b-1}$. As you noted, we again need to take subcases, i.e by considering the numbers of the form $3^{b-2}, 2*3^{b-2},...26*3^{b-2}$. But this seems tedious, and how do you know you won't have to take subcases again.

Can anyone fill in the details here?

Using $3$-adic valuation is good! For a rational number $x$, as usual with $\nu_3(x)$ we denote the largest (possibly negative) integer $k$ such that $\frac{x}{3^k}$ is an integer. Note the key property $\nu_3(A+B) \geq \min(\nu_3(A),\nu_3(B))$ with equality if $\nu_3(A) \neq \nu_3(B)$. Let $S_n = \sum_{i=1}^n \frac{1}{i}$. Then $S_{3n} = \frac{1}{3}S_n + \sum_{i=1, 3\nmid i}^{3n} \frac{1}{i} = \frac{S_n}{3} + \frac{a_n}{b_n} = \frac{S_nb_n + 3a_n}{3b_n}$ where $b_n$ is an integer which is not a multiple of $3$ and $a_n$ is a multiple of $3$ (the latter since $1/i \equiv i \pmod 3$ for $3\nmid i$ since $i^2 \equiv 1 \pmod 3$). Hence $1 + \nu_3(S_{3n}) = \nu_3(3S_nb_n) = \nu_3(S_nb_n + 3a_n)$ and so if  the numerator of $S_n$ is not divisible by $9$ (meaning $\nu_3(S_n) \leq 1$), then $9\mid 3a_n$ and the above key property yield $1+\nu_3(S_{3n}) = \nu_3(S_{n})$; in particular, the numerator of $S_{3n}$ would not be divisible by $3$. Analogously, if the numerator of $S_n$ is not divisible by $9$ (meaning $\nu_3(S_n) \leq 1$), then $1+\nu_3(S_{3n+2}) = \nu_3(S_{n})$ and so the numerator of $S_{3n+2}$ would not be divisible by $3$; and if the numerator of $S_n$ is not divisible by $3$ (meaning $\nu_3(S_n) \leq 0$), then $1+\nu(S_{3n+1}) = \nu_3(S_{n})$ and so the numerator of $S_{3n+1}$ would also not be divisible by $3$.

Now all that is remaining is annoying numerical calculations. For simplicity, call a number $n$ "bad" if the numerator of $S_n$ is not divisible by $3$ and "good" otherwise. By above we have the key inductive statement that if $n$ is bad, then so are $3n$, $3n+1$ and $3n+2$. Since $S_1 = 1$, $S_2 = \frac{3}{2}$, $S_3 = \frac{11}{6}$, $S_4 = \frac{25}{12}$, $S_5 = \frac{137}{60}$, $S_6 = \frac{147}{60} $, $S_7 = \frac{363}{140}$, $S_8 = \frac{761}{280}$ we see that $2$ and $7$ are good, $1$, $3$, $4$, $5$, $6$ and $8$ are bad and now we easily obtain that $9,\ldots,20$ are also bad. An unpleasant calculation (by using the above recursion between $S_{3n+1}$ and $S_n$) shows that $22$ is good (with numerator not divisible by $9$, hence $66$ and $68$ are bad) but $67$ is bad (and similar calculations yield that $21$ and $23$ are bad) and now a straightforward induction confirms that all other numbers are bad. Therefore the only solutions are $n=2,7,22$.


THe first one:

[hide]Multiply 6 by -3. It's -18. Look for two numbers that numbers that add up to 7 and multiply to -18. 9 and -2 work.
Rewriting the equation,

$(6x^2-2x)+(9x-3)=2x(3x-1)+3(3x-1)=(2x+3)(3x-1)$[/hide]

The second one:

[hide]$x^4-y^4=(x^2+y^2)(x^2-y^2)=(x^2+y^2)(x+y)(x-y)$[/hide]


my first topic. I'm so proud.

found this cartoon:

[url]http://www.xpmath.com/videos/donald.php[/url]

it's walt disney's cartoon with donald the duck

I watched that in 6th grade. Our math teacher made us watch it.
It was kinda creepy....

After doing a unit on mini-golf in Geometry, our teacher showed the pool section in that video.


integration
conics
hyperbola
probability and stats

Suppose U and V are independent and uniform on the interval (0,1).

a) find the pdf of S = UV

b) find the pdf of T = U/V

c) find the joint pdf of S and T

Solution for (a):
$ P(UV\le x)\equal{}\int_{0}^{x} \int_{0}^{1}1.du\, dv \plus{}\int_{x}^{1} \int_{0}^{\frac{x}{v}}1.du\, dv$ (You have a hyperbola intersecting the unit square)


Let $a \in \mathbb{N}$ and $p$ be a prime greater than $3$. Prove that there are no numbers $n$ which are congruent to $0, 3, 4, 18, 22 \bmod{24}$ of the form $a^{2}+p$.

[hide="Comment"]

I'm not sure if this problem is correct; I was coming up with a problem and happened to prove this to be true, but am not sure if my proof is completely watertight.

[/hide]

[quote="scorpius119"]Look at
$a=1,\ p=17,\ n=18$
$a=1,\ p=23,\ n=24$
$a=2,\ p=23,\ n=24+3$
$a=3,\ p=13,\ n=22$
$a=3,\ p=19,\ n=24+4$
:wink:[/quote]

Hmm... I'll go revise it, then  :D I'm sure it's right, except I just probably overlooked a restriction.

Ah, I see the mistake I made... I incorrectly assumed that all primes are equivalent to $1, 5, 7, 11 \bmod{24}.$ I forgot to count the second half of the residues, or $13, 17, 19, 23.$


function
floor function
algebra proposed
algebra

For every natural number $n$ we define \[f(n)=\left\lfloor n+\sqrt{n}+\frac{1}{2}\right\rfloor\] Show that for every integer $k \geq 1$ the equation \[f(f(n))-f(n)=k\] has exactly $2k-1$ solutions.

It is easy to chek, that if $f(n)\not =k^{2}\ \forall k\in N$ and if $m\not =k^{2}$ then exist unique n, suth that $f(n)=m$, $n=m-[\sqrt{m-1}]$.
Therefore $f(n)-n=[\sqrt{f(n)-1}]=[\sqrt{f(n)]}$, and equation equavalent to $k^{2}<f(f(n)))=m<(k+1)^{2}$.
It give $f(n)=m-k$, t $m\not =k^{2}+k$.
All solutions are $n=g(g(m)), m=k^{2}+1,...,k^{2}+k-1,k^{2}+k+1,....k^{2}+2k, \ g(m)=m-[\sqrt m ].$


[quote="moldovan"]Let $ a,b,c > 0$ and $ \frac {1} {a} \plus{} \frac {1} {b} \plus{} \frac {1} {c} \equal{} 1$

Prove that: $ a^2 \plus{} b^2 \plus{} c^2\ge 3(a \plus{} b \plus{} c)$[/quote]We must prove $ (\frac {1} {a} \plus{} \frac {1} {b} \plus{} \frac {1} {c})(a^2 \plus{} b^2 \plus{} c^2)\ge 3(a \plus{} b \plus{} c)$ open brac and $ (2: 0;\minus{}1)\succ (1;0;0)$

By $ \textrm{QM}\ge \textrm{AM}$ we get: $ \displaystyle \textrm{LHS}\ge \frac{(a\plus{}b\plus{}c)^2}{3}$.

We now have to prove $ \displaystyle \frac{(a\plus{}b\plus{}c)^2}{3}\ge 3(a\plus{}b\plus{}c)$

which is $ \frac{a\plus{}b\plus{}c}{3}\ge 3\equal{}\textrm{HM}$ from the conditions.


[quote="mdk"]If x and y are real numbers with (x+y)^4 = x-y, what is the maximum possible value of y?[/quote]
If $ x=-a$ and $ y=a$ where $ a>0$ we have:
$ (-a+a)^{4}=-(-a)-a\implies 0 = 0$
Which works.

But sonce $ a$ can be anything positive, there is no maximum value.

Um, I think that use misread the problem.  The RHS is $ x-y$, not $ -x-y$.  Anyways, let $ a=x+y$ and $ b=x-y$.  Then we know $ a^{4}=b$, and we want the maximum value of $ y=\frac{1}{2}(a-b)=\frac{a-a^{4}}{2}$.  It is easy to show that $ y\le max \{f(a)\}=3\cdot 2^{\frac{-11}{3}}$ and this happens where $ a=2^{\frac{-2}{3}}$.


I just noticed it said I had 208 posts, but I was pretty sure I had more than that so I clicked on view all posts by sailingdog and it said I had 710 there. Is there something wrong and are other people having the same problem?
It happened to my brother too, it says he has 1 post but he actually has more than 5.
Does it not count the ones in private forums like my math team's? Or the new topics?
Please help!

Basically, only "serious" forums count, for example posts in "Games and Fun Factory" don't.

OK, but I logged out then looked at my posts and it listed 512...
and that included ones in games and fun factory, etc. but not ones in my math team's forum or message boards of classes I've taken.
Also, it included posts I made to the FTW forum but not ones I made to that forum in December during testing of the earlier version, but I guess that forum was deleted or something...

so the number of posts thing attached to my posts says 208 or so but I posted 512 that should count...

This is the list of places your post does not go towards your post count.
Games and Fun Factory
Polls
Any of the Community forums.


algebra
polynomial
superior algebra
superior algebra unsolved

Let F be a field of non-zero characteristic, L be the splitting field of some polynomial over F, and K be the set of separable elements x of L over F. Show that K is a subfield of L and find |Aut(L/F)|.

this should be found in every textbook on algebra. in particular, |Aut(L/F)| = [K : F].

Can you explain me why you answered a couple of post more or less the same way¿
I don't have the impression that you've thought a lot about the problems or answers given.


geometry
parallelogram
AoPS Books
similar triangles

my basic question is that how do you know that a certain thereom can be stated without proof, some theorems are standard whereas some others are less known
for example if you said ptolemy's theorem, there would be no confusion

but if you took a theorem that is true but not a standard theorem, the test graders would not be sure you statement is true (ex. the sum of the squares of the sides of a parallelogram equal the sum of the squares of the diagonals of that parallelogram), (this is from one of the AoPS books)

so if you wanted to use the parallelogram theorem stated above, would you have to prove it separately?

well my guess is that the non-heard theorems should and have to be proved.

then that raises another question about what is known

in my geometry book there is a theorem called the midline theorem

it basically states that connecting the midpoints of a triangle makes a line parallel to the 3rd side of the triangle, also it is half the length of the 3rd side

this theorem is useful in a bunch of the AoPS problems but it is not that common, (searching on google gives a few results though)

also the proof for this is pretty obvious (similar triangles)

Midline theorem is well known through out the world. You can use it in any competitions.

When in competitions , it's best to prove everything to a basic level
so that the judges dont have a chance of accusing your answer.

As far as the midline theorem is concerned , it's one of the most widely used throught the world....Here in India , I learnt it as the "Midpoint" theorem.By the way , most reputed math competitions have questions that you really need to think and answer , and I doubt you'll find a problem which is easy to solve with an unconventional theorem.

I still of course , agree with you on one aspect.- I doubt there is any central authority in the world which defines a specific theorem as "Standard" or not.


Tara rolls three standard dice once.  What is the probability that the sum of the numbers rolled will be three or more?  Express your answer as a percent.

Each dice must roll at least 1, so the total is at least 3, making the probability 100%.

[hide="ANSWER"]1.we know the following facts

 I.there are 3 die

 II.every roll it's either:1,2,3,4,5,6

2.on every roll it's a 100% guarantee that the # will be 1,2,3,4,5,6

3.the #'s can be 1 and still be 3 or more

4.so the answer is 100%

5.[hide]100%[/hide][/hide]

[= I HOPE MY ANSWER IS GOOD

There is no possibility of rolling a sum less than $3$ when rolling three dice. So the answer is 100%

that was totally not an unnecessary revive. @Pi_


inequalities
algebra
polynomial
function
system of equations

wow i was playing with some numbers and i noticed that if

$x_{1}=x_{2}=x_{3}=...=x_{n}=1$ then:

The first equation would be

$1+2+3...+n=\frac{n(n+1)}{2}$ (sum of the integers from 1 to n)

The second one would be

$1+1+1+1...$ n times$=n$

so I dunno if thats the onyl solution but I just noticed that...

Well, that has to be the only solution.  Look at the second equation, there are n numbers and they all sum to n.  The only way to make n [b]positive integers[/b] sum to n is to have them all equal to one.

[hide="Sketch"] Let $a_{i}= ix_{i}$.  Then $\sum a_{i}$ is fixed by the first equation, so we can use Karamata on the second to deduce that $a_{i}= i$ is the only solution.

I think. [/hide]

[hide="Solution"]It's easy to notice that $x_{1}=\dots=x_{n}=1$ is a solution. Hence substitute $x_{1}=1+y_{i}$ where $y_{i}\geqslant-1$. Then the equations become

$y_{1}+2y_{2}+\dots+ny_{n}=0$
$(1+y_{1})+(1+y_{2})^{2}+\dots+(1+y_{n})^{n}=n$

Assume at least one of $y_{i}$ is $\neq-1,0$. Then by Bernoulli's inequality we have that the LHS of the last equation is strictly greater than $1+y_{1}+1+2y_{2}+\dots+1+ny_{n}$, which yields $y_{1}+2y_{2}+\dots+ny_{n}<0$, and that contradicts the first equation.

(Note: If $y_{1}\neq-1,0$ and all the other $y_{i}$ are $=-1,0$, Bernoulli can't be applied directly, but then it's easy to see that the system has no solution.)

Therefore, $(\forall i)y_{i}\in\{-1,0\}\implies(\forall i)x_{i}\in\{0,1\}$. Now it's obvious that $x_{1}+2x_{2}+\dots+nx_{n}$ can reach the desired value of ${n(n+1)\over 2}$ iff $x_{i}=1$. Hence that's the only solution.[/hide]

[quote="chess64"]Let $n$ be a positive integer. Solve the system of equations
\begin{align*}x_{1}+2x_{2}+\cdots+nx_{n}&= \frac{n(n+1)}{2}\\ x_{1}+x_{2}^{2}+\cdots+x_{n}^{n}&= n\end{align*}
for $n$-tuples $(x_{1},x_{2},\ldots,x_{n})$ of nonnegative real numbers.[/quote]

The second equation minus the first equation equals

$(x_{2}-1)^{2}+(x_{3}-1)^{2}(x_{3}+2)+\cdots+(x_{n}-1)^{2}(x^{n-2}+2x^{n-3}+3x^{n-4}+\cdots+(n-1))=0$

and the polynomials that have all postive integer coefficients clearly can't have postive real roots, and thus the only solution is $n$  $1s$

Haha, those are both much cooler than what I did.   :)

What function would you have used for your karamata solution?

Are you talking about the Karamata/Majorization Inequality?

Huh.  Yeah, I was, but I don't think it works anymore.  Intuitively, what I was thinking of amounts to Farenhajt's solution, so...

[url=https://artofproblemsolving.com/community/c6h593044p3516391]another link[/url] and a source ([url=https://artofproblemsolving.com/community/c6h531729p3038032]Czech-Polish-Slovak 2005[/url]), as I  am creating the [url=https://artofproblemsolving.com/community/c1118889_2006_mop_homework]2006 MOP Homework [/url]post collection


B and C are points on the segment AD. If AB=CD, prove that PA is greater than or equal to PB + PC for any point P.

In the original statement it's $PA+PD\ge PB+PC$.
Conversely, if the above is fulfilled for all $P$, then $B,C$ lie on the segment $AD$ and $AB=CD$.


On page 182 of Volume 2, the book finds $ \sum{i^2}$ using a nice method. They then say "We can, in principle at least, extend this type of argument to find $ \sim i^r$ for any integer r."

In other words, it's stated that you can write $ i^n\equal{}\sum_{k\equal{}1}^n a_k\binom{i}{k}$.

Is there any elementary proof? Or am I merely overlooking something obvious?

Thanks.

In the process, we let $ a_n \equal{} n!$ so that the expanded form of your RHS expression has coefficient $ 1$ for $ i^n$, so the expanded form so far is $ n!\dbinom in \equal{} i^n \plus{} c_{n \minus{} 1}i^{n \minus{} 1} \plus{} \cdots$. Then, for every other $ 1\le k\le n \minus{} 1$, we make the coefficient $ a_k$ such that we "cancel out" the $ i^k$ term in the expanded form (for example, for $ k \equal{} n \minus{} 1$, we let the coefficient $ a_{n \minus{} 1} \equal{} \minus{} c_n(n \minus{} 1)!$ so $ n!\dbinom in \minus{} c_{n \minus{} 1}(n \minus{} 1)!\dbinom i{n \minus{} 1} \equal{} i^n \plus{} 0i^{n \minus{} 1} \plus{} c_{n \minus{} 2}i^{n \minus{} 2} \plus{} \cdots$, and we then continue in a similar manner). Since the degree of the term with coefficient $ a_k$ is $ k$, this process works, since we eliminate the $ i^k$ term as we go from $ k \equal{} n \minus{} 1$ to $ k \equal{} 1$.


inequalities
calculus
ratio
function
inequalities open

Let $ a,b>0$ such that $ 3\minus{}2\sqrt{2}\le\frac ab\le3\plus{}2\sqrt{2}$. Prove or disprove for all $ x,y>0$
\[ \frac1{x^{a}\plus{}x^{b}}\plus{}\frac1{y^{a}\plus{}y^{b}}\ge\frac2{\sqrt{xy}^a\plus{}\sqrt{xy}^b}.\]
This has been inspired by the little auxiliary inequality [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1385055]here[/url].
[hide]
The problem has some similarity with [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=716555]Vasc’s sharpening of Zeikii’s one (still unproved :huh: )[/url].
Note that it is “homogenous” in $ a$ and $ b$. If we put $ a\equal{}\sqrt{2}\minus{}1,b\equal{}\sqrt{2}\plus{}1$, the ‘critical point’ is $ x\equal{}y\equal{}\sqrt{2}\minus{}1$, as my computer tells me that the inequality is false for $ (a,b,x,y)\mapsto (a\minus{}\epsilon, b\plus{}\epsilon, x\minus{}\epsilon, y\plus{}\epsilon)$ and small $ \epsilon>0$. Even proving only this is really ugly :wallbash: [/hide]
I think this problem gets a good position in the quest for the toughest inequality ever posted here ;)

It is not as tricky as I thought. :blush: I’ve found a proof, using a bit of calculus, but nice.

Now for the weighted version:
[b]Conjecture[/b]. Let $ a,b$ as above and let positive weights $ \lambda \plus{}\mu\equal{}1$. Then for all $ x,y>0$
\[ \frac{\lambda}{x^{a}\plus{}x^{b}}\plus{}\frac\mu{y^{a}\plus{}y^{b}}\ge\frac1{ (x^\lambda y^\mu)^a\plus{}(x^\lambda y^\mu)^ b}.\]
Moreover, for each pair of weights, the extremal ratio of $ a$ and $ b$ seems to remain always the same, $ 3\pm2\sqrt{2}$.  :roll:
I think my proof can be extended to suit that one, but hardly without a computer algebra system  :(

Note that if this conjecture is true, we would get directly by iteration the weighted version for $ n$ variables, which is exactly the same thing: :lol:

[b]Let $ a,b>0$ such that $ 3\minus{}2\sqrt{2}\le\frac ab\le3\plus{}2\sqrt{2}$. Let $ \lambda_1,\dots, \lambda_n\ge0$ and $ \sum\lambda_i\equal{}1$ and for $ x_1,\dots,x_n>0$ put $ r: \equal{}\prod x_i^{\lambda_i }$. Then
\[ \sum\frac{\lambda_i}{x_i^{a}\plus{}x_i^{b}} \ge\frac1{ r^a\plus{}r^ b}.\] [/b]
In particular the unweighted one would follow, which can [i]not at all[/i] be inferred directly from its $ n\equal{}2$ case:
\[ \sum\frac1{x_i^{a}\plus{}x_i^{b}} \ge\frac n{ r^a\plus{}r^ b}\] where $ r\equal{}\sqrt[n]{x_1\cdots x_n}.$ :roll:

:lol: … and nobody out there has noticed that all this is just an application of Jensen for the function $ f(t) \equal{} \frac1{e^{at} \plus{} e^{bt}}$ :blush: ? This function is convex iff $ a^2 \minus{} 6ab \plus{} b^2\le0$, i.e. the above condition.


Hi, I'm kinda confused on functional equation. May I know how is it different from other types of equation that we do normally in high school algebra?

Functional equations are usually much different than the equations you solve in high-school.  For functional equations, you look for (a) generalized solution(s).  Search for functional equations on the web (e.g. this website) and you'll see the difference immediately.

I see. I looked through some of the examples of functional equation in this website. However, it looks complicated to me especially all sorts of defining(s).
Nevertheless it seems caught my attention and I really keen to learn this topic. If i'm to, what pre-topic do I need to get hand on before really starting to get into functional equations?


[url=http://mathworld.wolfram.com/ConnellSequence.html]Connell Sequence[/url]
[url=http://www.research.att.com/~njas/sequences/A001614]A001614 in OEIS[/url]

This sum was incidentally in Arthur Engel :)
But how does one solve such a sum?? :maybe:

[b]Hint[/b]
1st group : $ 1$
2nd group : $ 2, 4$
3rd group : $ 5, 7, 9$
4th group : $ 10, 12, 14, 16$
$ \vdots$
k-th group : $ k^2\minus{}2k\plus{}2(\equal{}(k\minus{}1)^2\plus{}1), k^2\minus{}2k\plus{}4, \cdots, k^2$ ($ k$ terms)

Such a sequence is called [b]group sequence (群数列)[/b] in Japan.
([url=http://search.yahoo.co.jp/search?p=%E7%BE%A4%E6%95%B0%E5%88%97&ei=UTF-8&fr=moz2&x=wrt]Yahoo! Search in Japanese[/url])

Possibly group sequence may say double sequence.


calculus
integration
trigonometry
search
calculus computations

Evaluate $ \int e^{\minus{}\cos \left( 2x\plus{}\frac{\pi }{3}\right) }\cos 3x\,dx$

I = $ \int  e^{ \minus{} \cos (2x \plus{} \frac{\pi}{3})} \ \cos 3x \ dx$

$ 2x \plus{} \frac{\pi}{3} \equal{} 2z \ \Rightarrow \ dx \equal{} dz \\ \\ 3x \equal{} 3z \minus{} \frac{\pi}{2}$

This gives

I = $ \int \ e^{ \minus{} \cos 2z } \ \cos ( 3z \minus{} \frac{\pi}{2} ) \ dz$

$ \equal{} \int \ e^{ \minus{} \cos 2z } \ \sin 3z \ dz$

$ \equal{} \int \ e^{ \minus{} \cos 2z } \ ( 3 \sin z \minus{} 4 \sin ^3 z ) \ dz$

$ \equal{} \int \ e^{ \minus{} \cos 2z } \ \sin z ( 3 \minus{} 4 ( 1 \minus{} \cos ^2 z ) \ dz$

$ \equal{} \int \ e^{ \minus{} \cos 2z } \ \sin z ( 4 \cos ^ z \minus{} 1 ) \ dz$

Integrate $ e^{ \minus{} \cos 2z } \ 4 \sin z \cos ^2 z$  by parts taking $ \cos z$ as 1 function.

$ I \equal{} \int \ e^{ \minus{} \cos 2z } \ ( 2 \sin 2z \ \cos z \minus{} \sin z ) \ dz$

$ \equal{} \cos z \int \ e^{ \minus{} \cos 2z } \ 2 \sin 2z \ dz \minus{} \int ( \frac{d}{dz}( \cos z )  \int  e^{ \minus{} \cos 2z } \ 2 \sin 2z \ dz ) \ dz  \minus{} \int \ \sin z \ e^{ \minus{} \cos 2z } \ dz$

$ \boxed { \int \ e^{ \minus{} \cos 2z } \ 2 \sin 2z \ dz \equal{} e^{ \minus{} \cos 2z } }$

So,

$ I \equal{} \cos z \ e^{ \minus{} \cos 2z } \minus{} \int ( \minus{} \sin z \ e^{ \minus{} \cos 2z }) \ dz  \minus{} \int \ \sin z \ e^{ \minus{} \cos 2z } \ dz  \ \plus{} C$

= $ e^{ \minus{} \cos 2z } \ \cos z \plus{} C$

= $ \  e^{ \minus{} \cos (2x \plus{} \frac{\pi}{3}) } \ \cos (x \plus{} \frac{\pi}{6}) \plus{} C$

 :D

[hide="Even Lengthier Method"]

same substitution as above :  $ 2x + \frac{\pi}{3} = 2z$

I = $ \int \ e^{- \cos 2z } \sin 3z \ dz$

$ = \frac{1}{2} \ \int e^{- \cos 2z} \ \sin 2z ( - \sec z + 4 \cos z ) \ dz$

Integrate by parts taking $ ( - \sec z + 4 \cos z )$ as 1 function.

you get

[b]2I[/b] = $ {(-secz+4cosz) \int e^{-cos2z}.sin2z.dz - \int ((-secz.tanz - 4sinz) \int e^{-cos2z}.sin2z.dz )dz}$

$ \int e^{- \cos 2z}. \sin 2z. dz = \frac{1}{2} e^{- \cos 2z}$

[b]2I[/b] = $ { (-secz+4cosz) \frac{1}{2} e^{- \cos 2z} + \int ((secz.tanz + 4sinz) \frac{1}{2} e^{- \cos 2z} . dz}$

[b]4I[/b] = $ { (-secz+4cosz) . e^{- \cos 2z} + \int 4 sinz . e^{- \cos 2z} . dz + \int e^{- \cos 2z}.secz.tanz.dz }$

= $ { (-secz+4cosz) . e^{- \cos 2z} +  \int 2 sin 2z . e^{- \cos 2z} . secz. dz + \int e^{- \cos 2z}.secz.tanz.dz }$

Integrating the 2nd part by Parts :

= $ { (-secz+4cosz) . e^{- \cos 2z} +  secz. \int 2 sin 2z . e^{- \cos 2z} . dz - \int (secz.tanz.\int 2 sin 2z . e^{- \cos 2z} . dz )dz + \int e^{- \cos 2z}.secz.tanz.dz }$

= ${ { (-secz+4cosz) . e^{- \cos 2z} +  secz. e^{- \cos 2z} - \int e^{- \cos 2z}.secz.tanz.dz } + \int e^{- \cos 2z}.secz.tanz.dz }$

All but one term cancel out leaving you with

$ 4I = 4 \cos z. e^{- \cos 2z} + C$

$ I = \cos z. e^{- \cos 2z} + c$

$ \Rightarrow \ I = e^{- \cos (2x+\frac{\pi}{3})}. \cos (x + \frac{\pi}{6}) + c$

[/hide]

Well done Nora!!

.... I truely apprciate your "never give up" spirit :)

Now left with one problem to go :P ... Great work!

Thanks IndoChina

but for your [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1782961647&t=279746]first Question[/url], im sorry i don't have a clue as I haven't done Double Integrals in high school yet !!

 :yoda:


vector
linear algebra
linear algebra unsolved

Are the elements t(1,1,-1,-1)?
can you find a basis?
I know that dimension is number of vectors in a basis

:what?:
Your problem contains $\mathbb{Z}_2^4$ and you don't know what it means!
I am affraid I am unable to explain it in a clear way. Ask someone else, please.

$\mathbb{Z}_2^4$ basically means that you are working in a number system where $2 = 0$.


Alright, so I got this question:
"what is the greatest possible GCD of $n^{3}+2007$ and $(n+1)^{3}+2007$

I got it down to $k|((3011)(9030)+1)\implies k|(13)(2091487)$
($2091487$ is prime)

I then got that there are infinite cases for which $k=13$ (such as $n=5$), but haven't been able to find any for $k=2091487$ or $k=(13)(2091487)$.

So is the largest value of $k$ $13$, or are there values of $n$ for which $k=2091487$ or $k=(13)(2091487)$?

Thanks  :)

EDIT: Wow I'm an idiot, $k|n+3011 \to n+3011=0 \implies n=-3011$ gives the highest GCD...