Find a solution of the simultaneous equations
\[ xy \plus{} yz \plus{} zx \equal{} 12 \\
xyz \equal{} 2\plus{}x\plus{}y\plus{}z
\]in which all of $ x,y,z$ are positive, and prove that it is the only such solution.

Show that a solution exists in which $ x,y,z$ are real and distinct.

[hide="First Part"] Let $ t\equal{}x\plus{}y\plus{}z$.  Then $ xyz\equal{}t\plus{}2$, and by AM-GM we have $ \frac{t}{3}\ge(t\plus{}2)^\frac{1}{3}$ with equality when $ x\equal{}y\equal{}z$.  We cube both sides to get $ t^3\minus{}27t\minus{}54\ge0$.  This gives $ (t\minus{}6)(t\plus{}3)^2\ge0$.  Thus, $ t\ge6$.

But, from AM-GM on $ xy\plus{}yz\plus{}zx$ we get $ \frac{xy\plus{}yz\plus{}zx}{3}\ge(xyz)^\frac{2}{3}$. This gives $ 8\ge xyz$ and thus $ 6\ge t$.

So we get that equality must happen in both cases.  That is, $ x\equal{}y\equal{}z$.  From $ xy\plus{}yz\plus{}zx\equal{}12$ we find $ x\equal{}y\equal{}z\equal{}2$ is the only solution for $ x,y,z>0$.[/hide]

Second part
[hide]
$ \left(\minus{}2,\frac{\minus{}12\plus{}2\sqrt{21}}{5}, \frac{\minus{}12\minus{}2\sqrt{21}}{5}\right)$
[/hide]

for the first part
we know from am-gm that
$ xy\plus{}yz\plus{}zx\geq3\sqrt[3]{(xyz)^2}<\equal{}>xyz\leq8<\equal{}>x\plus{}y\plus{}z\leq6$(1)
but we know $ (x\plus{}y\plus{}z)^2\geq3(xy\plus{}yz\plus{}zx)\equal{}36<\equal{}>x\plus{}y\plus{}z\geq6$(2)
$ (1),(2)\equal{}>x\plus{}y\plus{}z\equal{}6$
and because $ (x\plus{}y\plus{}z)^2\equal{}3(xy\plus{}yz\plus{}zx)\equal{}>x\equal{}y\equal{}z\equal{}2$the only solution for x,y,z>0


function
geometry
geometric transformation
rotation
analytic geometry
graphing lines
slope

If $ x$ and $ y$ are real numbers with $ (x\plus{}y)^4 \equal{} x \minus{} y$, what is the maximum possible value of $ y$?

And why is this graph a 135 degree clockwise rotation and scaling of the function $ y \equal{} x^4$? Would someone explain this.. thanks

The maximum value of $ y$ is $ \dfrac{3}{8 \sqrt[3]{4}}$ and is attained at $ x\equal{}\dfrac{5}{8 \sqrt[3]{4}}$.
First, the answer to the second part. If we look at the point $ (x,\,y)$ on the $ xy$-plane, the $ x$ coordinate is basically the (signed) distance of that point from the $ y$-axis, while the $ y$ coordinate is the (signed) distance from the $ x$ axis. Transform the given equation as follows:
$ (x\plus{}y)^4\equal{}x\minus{}y \quad\Rightarrow \ (\sqrt{2})^3 \left(\dfrac{x\plus{}y}{\sqrt{2}}\right)^4 \equal{} \dfrac{x\minus{}y}{\sqrt{2}}\qquad[1]$
we see that $ \dfrac{x\minus{}y}{\sqrt{2}}$ is the distance of the point $ P(x,\,y)$ from the line $ x\equal{}y$ while $ \dfrac{x\plus{}y}{\sqrt{2}}$ is the distance of $ P$ from the line $ x\plus{}y\equal{}0$. So setting $ X\equal{}\dfrac{x\plus{}y}{\sqrt{2}}$ and $ Y\equal{}\dfrac{x\minus{}y}{\sqrt{2}}$, the equation becomes
$ 2^{3/2}\ X^4 \equal{} Y$
Further, since $ (x\plus{}y)^4 \geq 0$ for all real $ x,\,y$, the graph of $ y$ vs. $ x$ is confined to the region $ x\geq y$ i.e. to the right-side of the line $ y\equal{}x$.  The graph is shown in the fig, from where it easy to see that it is a rotated and scaled version of $ y\equal{}x^4$: first rotate by 135$ ^\circ$, then scale the (rotated) $ y$ axis by $ (\sqrt{2})^3$ and the (rotated) $ x$ axis by $ \minus{}1$.

Again, the maximum of $ y$ as given by the original line is obtained when the tangent line is horizontal (i.e. its slope is zero). Accordingly, in the $ XY$-plane the slope of the corresponding tangent will be $ 1$ (see the dashed-line in the fig). It happens when
$ 1\equal{}\dfrac{\mathrm{d}Y}{\mathrm{d}X} \equal{} 4 (2^{3/2}) X^3$
i.e. at $ X\equal{}2^{\minus{}7/6}$ and the corresponding value of $ Y\equal{}2^{\minus{}19/6}$
Using the relations $ X\equal{}\dfrac{x\plus{}y}{\sqrt{2}}$ and $ Y\equal{}\dfrac{x\minus{}y}{\sqrt{2}}$, we get the required maximum value  of $ \boxed{y\equal{}\dfrac{3}{8 \sqrt[3]{4}}}$

Thanks for the solution and also for producing a neat diagram in $ \text{\LaTeX}$ :P I was thinking in terms of "the x-axis is $ y = 0$, so I need to get the new x-axis to be $ x-y = 0$". So I substituted $ x-y$ for $ x$ but I didn't get the given equation. So we must think in terms of the x-coordinate being a signed distance from the y-axis, and likewise for the y-coordinate!


[hide]For odd $ n$:

$ [n^2 \minus{} (n \minus{} 1)^2 ] \plus{} \ldots \plus{} [3^2 \minus{} 2^2] \plus{} [1^2 \minus{} 0^2]$
$ \equal{} (2n \minus{} 1) \plus{} \ldots \plus{} (2(3) \minus{} 1) \plus{} (2(1) \minus{} 0)$
$ \equal{} [n \plus{} (n \minus{} 1)] \plus{} \ldots \plus{} [3 \plus{} 2] \plus{} [1 \plus{} 0]$
$ \equal{} \frac {n(n \plus{} 1)}{2}$

For even $ n$:

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

We can easily combine the two:

$ \sum \equal{} \frac {( \minus{} 1)^{n \minus{} 1}}{2} n (n \plus{} 1)$[/hide]

ugh...it looks so easy when i see it get done..how about this:

51. Let Sk = 1^k + 2^k + · · · + n^k. Prove that
n =S0
n2 =2S1 − S0
n3 =3S2 − 3S1 + S0
n4 =4S3 − 6S2 + 4S1 − S0

the numbers in front of the S's are subscipts...and did a couple of substitutions here and there but reached a dead end.

BTW thanks.

[hide="Generalization"]We start with your definition and subtract $ n^k$:
$ S_k \equal{} \sum _{h \equal{} 1}^n h^k$
$ S_k \minus{} n^k \equal{} \sum _{h \equal{} 1}^{n \minus{} 1} h^k$

Now, we perform an index shift on this. Note that we can add in $ 0 \equal{} 0^k$ to help us out:
$ S_k \minus{} n^k \equal{} \sum _{h \equal{} 0}^{n\minus{}1} h^k \equal{} \sum _{h \equal{} 1}^{n} (h \minus{} 1)^k$

Now, expand this with the binomial theorem:
$ S_k \minus{} n^k \equal{} \sum _{h \equal{} 1}^{n} \sum_{j \equal{} 0}^{k} ( \minus{} 1)^{j} \binom {k}{j} h^{k \minus{} j}$

We can change the order of finite summations by the commutative property:
$ S_k \minus{} n^k \equal{} \sum_{j \equal{} 0}^{k} \sum _{h \equal{} 1}^{n} ( \minus{} 1)^{j} \binom {k}{j} h^{k \minus{} j}$

Now, we factor out the terms in the second summation that are independent of $ h$:
$ S_k \minus{} n^k \equal{} \sum_{j \equal{} 0}^{k} ( \minus{} 1)^{j} \binom {k}{j} \sum _{h \equal{} 1}^{n} h^{k \minus{} j}$

We use our definition of $ S_k$:
$ S_k \minus{} n^k \equal{} \sum_{j \equal{} 0}^{k} ( \minus{} 1)^{j} \binom {k}{j} S_{k \minus{} j}$

We can then negate this:
$ n^k \minus{} S_k \equal{} \sum_{j \equal{} 0}^{k} ( \minus{} 1)^{j \plus{} 1} \binom {k}{j} S_{k \minus{} j}$

Finally, add $ S_k$ to each side, to achieve the general form of your pattern:
$ n^k \equal{} \sum_{j \equal{} 1}^{k} ( \minus{} 1)^{j \plus{} 1} \binom {k}{j} S_{k \minus{} j}$[/hide]

that makes a lot of sense.

Prove that any n-gon can be cut into triangles by non-intersecting diagonals

Prove that the number of triangles obtained by cutting an n-gon by non-intersecting diagonals is equal to n-2

Your question has _nothing_ to do with the original problem. And bumping as early as after 24 hours is disliked, too.
Please start a new thread and ask there.

i thought it would be best to post all my proof questions in this thread.And i just wanted people to see the thread thats why i bumped it. Can you help with the problems? Thanks in advance.

[quote="The Zuton Force"][hide="Generalization"]We start with your definition and subtract $ n^k$:
$ S_k \equal{} \sum _{h \equal{} 1}^n h^k$
$ S_k \minus{} n^k \equal{} \sum _{h \equal{} 1}^{n \minus{} 1} h^k$

Now, we perform an index shift on this. Note that we can add in $ 0 \equal{} 0^k$ to help us out:
$ S_k \minus{} n^k \equal{} \sum _{h \equal{} 0}^{n \minus{} 1} h^k \equal{} \sum _{h \equal{} 1}^{n} (h \minus{} 1)^k$

Now, expand this with the binomial theorem:
$ S_k \minus{} n^k \equal{} \sum _{h \equal{} 1}^{n} \sum_{j \equal{} 0}^{k} ( \minus{} 1)^{j} \binom {k}{j} h^{k \minus{} j}$

We can change the order of finite summations by the commutative property:
$ S_k \minus{} n^k \equal{} \sum_{j \equal{} 0}^{k} \sum _{h \equal{} 1}^{n} ( \minus{} 1)^{j} \binom {k}{j} h^{k \minus{} j}$

Now, we factor out the terms in the second summation that are independent of $ h$:
$ S_k \minus{} n^k \equal{} \sum_{j \equal{} 0}^{k} ( \minus{} 1)^{j} \binom {k}{j} \sum _{h \equal{} 1}^{n} h^{k \minus{} j}$

We use our definition of $ S_k$:
$ S_k \minus{} n^k \equal{} \sum_{j \equal{} 0}^{k} ( \minus{} 1)^{j} \binom {k}{j} S_{k \minus{} j}$

We can then negate this:
$ n^k \minus{} S_k \equal{} \sum_{j \equal{} 0}^{k} ( \minus{} 1)^{j \plus{} 1} \binom {k}{j} S_{k \minus{} j}$

Finally, add $ S_k$ to each side, to achieve the general form of your pattern:
$ n^k \equal{} \sum_{j \equal{} 1}^{k} ( \minus{} 1)^{j \plus{} 1} \binom {k}{j} S_{k \minus{} j}$[/hide][/quote]

I just started like you to prove that the formula for $ n^k$ but I need to know what is the simple formula for $ S_k$.

Could you answer it please , I asked about here :

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1021139

thanks. :)


Here's a fun probability problem:
A fair coin is tossed, if a head is tossed a balanced die is rolled, if a tail is tossed then two balanced dice are rolled.  Let $y$ be the total number rolled on the die/dice for one trial. Find the probability that $y=6$.

[hide="casework"]Case 1:  Heads is flipped
The probability of flipping heads is $\frac12$.  The probability of rolling 6 on the die is $\frac16$, for a total probability of having a sum of 6 to be $\left(\frac12\right)\left(\frac16\right)=\frac1{12}$.

Case 2:  Tails is flipped
The probability of flipping tails is $\frac12$.  The ways to roll a sum of 6 on two die are:
(1,5)
(2,4)
(3,3)
(4,2)
(5,1)
Thus, the probability of rolling a sum of 6 on two die is $\frac{5}{36}$.

Total probability:

$\frac1{12}+\frac5{72}=\frac{11}{72}$[/hide]


[quote="redcomet46"]Given $3x+2y=33$, find $x$ when $xy+\frac{x^{2}\sqrt{3}}{4}$ is maximized.[/quote]

[hide]$2y=33-3x$, so $y=\frac{33-3x}{2}$.
Then
$xy+\frac{x^{2}\sqrt3}{4}$ $= x \cdot \frac{33-3x}{2}+\frac{x^{2}\sqrt{3}}{4}$ $=\frac{-3}{2}x^{2}+\frac{33}{2}x+\frac{\sqrt{3}}{4}x^{2}$ $=\frac{-6+\sqrt{3}}{4}x^{2}+\frac{33}{2}x$
which is a quadratic function that is maximized at $x= \frac{-\frac{33}{2}}{\frac{-6+\sqrt{3}}{2}}= \frac{33}{6-\sqrt{3}}= \boxed{6+\sqrt{3}}$.[/hide]

I guess it is maximizing the area of a figure that looks like an equilateral triangle attached to a rectangle.


Let $ a,b,c$ be positive reals such that $ abc \ge 1$. Prove that

$ a^3\plus{}b^3\plus{}c^3 \ge ab\plus{}bc\plus{}ca$

Of course we could just use Muirhead.
[hide]By Muirhead,
$ a^3 \plus{} b^3 \plus{} c^3 \ge a^{4/3}b^{4/3}c^{1/3} \plus{} a^{1/3}b^{4/3}c^{4/3} \plus{} a^{4/3}b^{1/3}c^{4/3} \equal{} (abc)^{1/3}(ab \plus{} bc \plus{} ca) \ge ab \plus{} bc \plus{} ca$[/hide]
Here is a nicer solution.
[hide]
By AM-GM,
$ a^3 \plus{} b^3 \plus{} 1 \ge 3ab$
$ b^3 \plus{} c^3 \plus{} 1 \ge 3bc$
$ c^3 \plus{} a^3 \plus{} 1 \ge 3ac$
Adding them,
$ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3 \ge 3(ab \plus{} bc \plus{} ca)$
But by AM-GM, $ a^3 \plus{} b^3 \plus{} c^3 \ge 3abc \ge 3$. Thus,
$ 3(a^3 \plus{} b^3 \plus{} c^3) \ge 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3 \ge 3(ab \plus{} bc \plus{} ca)$[/hide]
And here is another solution
[hide]By Holder's inequality,
$ (a^3 \plus{} b^3 \plus{} c^3)(b^3 \plus{} c^3 \plus{} a^3)(1 \plus{} 1 \plus{} 1) \ge (ab \plus{} bc \plus{} ca)^3$
But by AM-GM, $ a^3 \plus{} b^3 \plus{} c^3 \ge 3abc \ge 3$ (since $ abc \ge 1$). So
$ (a^3 \plus{} b^3 \plus{} c^3)^3 \ge (a^3 \plus{} b^3 \plus{} c^3)(b^3 \plus{} c^3 \plus{} a^3)(1 \plus{} 1 \plus{} 1) \ge (ab \plus{} bc \plus{} ca)^3$
So $ a^3 \plus{} b^3 \plus{} c^3 \ge ab \plus{} bc \plus{} ca$[/hide]

[hide]OK .Inequality is symmetric in its variables , so without loss of generality let $ a \geq b \geq c$ .Then by Chebyshev :
\[ a^3 \plus{}b^3 \plus{}c^3 \geq \frac{1}{3} (a\plus{}b\plus{}c)(a^2 \plus{}b^2 \plus{}c^2)\]
 By AM-GM $ a\plus{}b\plus{}c \geq 3\sqrt[3]{abc} \equal{}3$ , then we only need to prove that :
 \[ a^2 \plus{}b^2 \plus{}c^2 \geq ab \plus{}bc \plus{}ca\]
\[ (a\minus{}b)^2 \plus{}(b\minus{}c)^2 \plus{}(c\minus{}a)^2 \geq 0\]

nice solutions dgreenb801  :)
[/hide]


vector
linear algebra
linear algebra unsolved

If U1 and U2 are subspaces of a vector space V, prove that U1 +U2 is the smallest subspace of V containing U1 and U2..

How would you go about trying to prove this??

This breaks down into two tasks:

1. Prove that $ U_1\plus{}U_2$ is a subspace.

2. Prove that if $ W$ is a subspace of $ V$ that contains both $ U_1$ and $ U_2,$ then $ U_1\plus{}U_2\subseteq W.$

Can you do either one of these?


AMC AIME geometry trigonometry function circumcircle inradius

1) There is a set of 1000 switches, each of which has four positions, called A, B, C, and D. When the position of any switch changes, it is only from A to B, from B to C, from C to D, or from D to A. Initially each switch is in position A. The switches are labeled with the 1000 different integers (2^x)(3^y)(5^z), where x, y, and z take on the values 0, 1, ..., 9. At step i of a 1000-step process, the ith switch is advanced one step, and so are all the other switches whose labels divide the label on the ith switch. After step 1000 has been completed, how many switches will be in position A?

2) Point P is located inside triangle ABC so that angles PAB, PBC, and PCA are all congruent. The sides of the triangle have lengths AB = 13, BC = 14, and CA = 15, and the tangent of angle PAB is m/n, where m and n are relatively prime positive integers. Find m + n.

Hint:

[hide]Look for tangent by looking for Cosine and Sine using their respective Laws.

Then, see what you can do with the Trig form of Ceva's Theorem.[/hide]

Response:

[hide]Hmm...Im a bit slow, couldn't find a nice relationship between the sin/cos of the angle and the sides used in Ceva's. Perhaps you could give a bit more detailed explanation?[/hide]

Hmm...

What I think of using Ceva:

[hide] First I took the 13-14-15 triangle and split it into a 9-12-15 and a 5-12-13. Using those right triangles I came up with the sine and cosine of the three angles in the original triangle.

Using the trig variant of Ceva's theorem as well as the sin(a-b) formula (since I knew those original angles), I was able to set up an equation in which the only variable was the angle we were looking for.

However, it was a very messy long equation that involved trig functions to the third power and multidigit fractions. Probably not the best solution. I couldn't see any other way to apply Ceva.

[/hide]

A brief hint to the elegant solution... (I cheated)...

[hide]... it's a lot easier than it seems. Try Law of Cosines and a couple of area formulas.[/hide]

O yea!

Rite. Stupid me.

Fierytycoon, try that.

And it's really easy.

Hint:

[hide]

Express Cosine as (13^2+14^2+15^2)/(2*(13z+14y+15x))[/hide]

I remember when I first did that AIME, and found that solution. I was soooo proud of myself.

[quote]Using the trig variant of Ceva's theorem [/quote]

Waiit, what's this?

A somewhat incoherent explanation of the theorem:

In the given problem, each one of the three angles of the original 13-14-15 are split into two smaller angles; let's call them the three "equal" angles (the three which are the same in the problem) and the three "other" angles.  the product of the  sines of the three equal angles is equal to the product of the sines of the other angles by Ceva's theorem. The same is true for any six angles formed in a triangle by three lines which all pass through one point.

edit: I did mean the sines of the angles. Whoops.

[color=cyan]You mean the product of the sines of the angles.  It's just Ceva + Law of Sines, really.[/color]

Thanks for the feedback guys. Hmm, now I see what you mean by the trig form of Ceva's--it was even in AoPS 2, but I hadn't read that chapter yet :/ Anyways, I first tried the long solution, and got an equation in theta, but, as you folk mentioned, expanding out all the sin(x-y) was quite painful, and I didn't manage to solve the equation in the end.

Now, for the supposedly more elgant solution...

Well, I took the hint from 314159265358979323846, and looked at each of the common area formulas:

(1/2)ab sin(c) : Well, we use the sin of an angle, but we can already figure out the sin of each angle after splitting the triangle into 5-12-13 and 9-12-15 right triangles

abc/(4R) : the circumradius isn't really part of the problem, so it doesn't look too promising

(1/2) b*h : No use, only involves sides.

Sqrt(s(s-a)(s-b)(s-c)): Similarly, no use.

rs: no mention of inradius in the problem

So..I am stuck (as usual)  :(

By the way, darkquantum, what were you refering to by the variables x, y, and z that you used in your equation?

O sorry, fierytycoon, let me give u the solution.

We let x, y, z be the segments PA, PB, PC. Then, if we use the Law of Cosine on the three congruent angles, add them together, and solve for cosine, you should get cosA (A is the angle we need)
as (13^2+14^2+15^2)/(2*(13z+14y+15x))

Now, we know the area of the triangle, by Heron's formula. Then, we use the 1/2*absinC formula on the three separate triangles and sum up that area, which should equal the number u get from heron's. From that, you will also get a 13z+14y+15x term. THus, when u do sin/cos, the variables cancel out.

DO you get it now?

[quote]We let x, y, z be the segments PA, PB, PC. Then, if we use the Law of Cosine on the three congruent angles, add them together, and solve for cosine, you should get cosA (A is the angle we need)
as (13^2+14^2+15^2)/(2*(13z+14y+15x)) [/quote]

Hmm, I might just be really tired (since its getting a bit late over here), but when we add fractions with different denominators, don't we have to multiply each fraction to get a common denominator, then add the numerators? After calculating the three expressions for cos(theta), it seems (at a glance) that you are just adding the numerators and denominators without doing the above.

nono, fierytycoon

add them first
 in the a^2=b^2+c^2-2bcCosA form

When u add them in that form
its nice
cuz all the x^2, y^2, z^2 cancel out

[quote]Hmm, I might just be really tired [/quote]

See, I told you.  :D

Ah, ok, I get it now. Thanks for putting up with my occasional brain clogs.  8-)

hmmm
no prob

happens to me all the time :D

BTW, has anyone solved the first one??

Yeah, anyone solve the first counting problem?

Ok well I may as well try..

[hide]

Basically the switch with label 2^x 3^y 5^z is switched over (x+1)(y+1)(z+1), as we switch it over for each one with the power of 2 bit being 0, 1, .., x, the power of 3 bit being 0, 1, .., y, etc.

We want to count how many have (x+1)(y+1)(z+1) a multiple of 4.. oh, this may take a little while.. probably have to use PIE but I'll just leave it here. Though I'm probably leaving out the main bit related to the title..

[/hide]

Ok. That was enough to help me count all of the possible switches. Thanks for the help.


Show that there is a surjective function f : N  to N*N. Use this to
show that there is a surjective function N to Q.

For a function f : A to A denote by f^(n) the composition of f(f(f(...))) (n times). Show that if for each x in A there is an n(x)in N (the natural numbers) such that f^(n(x))(x) = x then f is bijective.

[hide="1"]$ f(1) \equal{} (1,1)$
$ f(2) \equal{} (2,1)$
$ f(3) \equal{} (1,2)$
$ f(4) \equal{} (3,1)$
$ f(5) \equal{} (2,2)$
$ f(6) \equal{} (1,3)$
$ f(7) \equal{} (4,1)$
and continue moving in a triangular pattern.
Let $ g : \mathbb{N}^2 \rightarrow \mathbb{Q}$ be $ g(m,n) \equal{} m/n$. Then $ f(g(n))$ is a surjective function from $ \mathbb{N}$ to $ \mathbb{Q}$.[/hide]

[hide="2"]Suppose $ a \in A$ is not taken by $ f$. Then $ f^{(n(a))}(a) \equal{} a$ is impossible for any $ n(a)$. So $ f$ is surjective.

Suppose $ f(a) \equal{} f(b) \equal{} c$. Consider the sequence $ f^{(i)}(c)$ for $ 0 \leq i < n(c)$. We get
$ c, f(c), f^{(2)}(c), \ldots f^{(n(c)\minus{}1)}(c)$
None of these terms are equal to $ c$ except the first. However, we must have both $ a$ and $ b$ in this sequence since repeated applications of $ f$ on $ a$ and $ b$ (which go into this sequence) must return to $ a$ and $ b$. But if $ f^{(k)}(c) \equal{} a$, $ f^{(k\plus{}1)}(c) \equal{} c$. So $ k \equal{} n(c) \minus{} 1$. Similarly, if $ f^{(m)}(c) \equal{} b$, $ f^{(m\plus{}1)}(c) \equal{} c$. So $ m \equal{} n(c) \minus{} 1$. Then $ m \equal{} k$, so $ f^{(k)}(c) \equal{} f^{(m)}(c)$ and thus $ a \equal{} b$. So $ f$ is injective.[/hide]


function integration inequalities calculus algebra polynomial quadratics

Prove that for any continuous function $f: [0,1]\to\mathbb{R}$, are loc inegalitatea:

                        \[ \int_{0}^{1}f(x)dx\cdot\int_{0}^{1}x^{4}f(x)dx\leq\frac{4}{15}\int_{0}^{1}f^{2}(x)dx. \]

   [i] Dorian Popa, Ioan Rasa [/i]

It seems that we can use Cauchy-Schwalz's inequality for integral version.

Yes, Kunny, you are right. I think it would be great if you could post a full solution. ;)

I think that one must find first the function for which the equality holds and then apply Cauchy-Schwartz, because every time we apply C-S, we need to have equality for the previously found function . The desired function seems to be a polynomial of small degree (at most 5).  I cannot see it, and I'm to lazy to find it, or maybe my claim is false  :D . Can someone post solution?

A I expected. Now the problem becomes simpler. The ineq is equiv with:

$\int_0^1f(x)dx\int_0^1(3x^4+1)f(x)dx-\int_0^1f(x)dx\int_0^1f(x)dx\le \frac 4 5 \int_0^1f^2(x)dx$

But aplying Cauchy-Schwartz for $\int_0^1(3x^4+1)f(x)dx$ we get.

$\int_0^1f(x)dx\int_0^1(3x^4+1)f(x)dx-\int_0^1f(x)dx\int_0^1f(x)dx\le \int_0^1f(x)dx\sqrt{\int_0^1f^2(x)dx \frac{16} 5 }$.

I we denote $a=\int_0^1f(x)dx$ and $b=\int_0^1f(x)dx$ we only need to prove that:

$a^2-a\sqrt{b \frac{16} 5 }+\frac{4b} 5 \ge 0$

Treating it as a quadratic ecuation in $a$ we can eaily get that this ineq holds.

The method Kent Merryfield used [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=orthogonal&t=77089]here[/url] kills this inequality.

More specifically, let $\left< f, 1 \right> = a, \, \left< f, x^{4}\right> = b$.
The functions are $u(x) = a, \, v(x) = \frac{45}{16}\left( 5b-a \right) \left( x^{4}-\frac15 \right)$ and $g(x) = f(x)-u(x)-v(x)$.
We see that $\left< u, v \right> = \left< g, u \right> = \left< g, v \right> = 0$, so
\[\left< f, f \right> = \left< g, g \right>+\left< u, u \right>+\left< v, v \right> . \]
We have to prove that it's $\geq \frac{15ab}4$.
$\left< g, g \right>$ is $\geq 0$, so $\left< f, f \right> \geq \left< u, u \right>+\left< v, v \right>$.
The inequality $\left< u, u \right>+\left< v, v \right> \geq \frac{15ab}4$ reduces after many computations to $\left( a-3b \right)^{2}\geq 0$.
For equality to hold we must have that $g \equiv 0$ and that $a = 3b$, which, after some more computations, gives $k \left( 3 x^{4}+1 \right)$.

The question I want to put is pretty stupid: [i]Why does such a method work?[/i]


I think :
   Let x=a^2+b^2+c^2
        y=ab+bc+ca
        z=abc

nice arqady :)
another solution
it is easily now two of $a-1,b-1,c-1$,have the same symbol
so we can assume $(a-1)(b-1) \ge 0$
then the because it euqal to $a^2+b^2+c^2+abc+2 \geq a+b+c+bc+ca+ab$
we can prove it by $a^2+b^2+1 \ge 2ab+1 =ab+(a-1)(b-1)+a+b \ge ab+a+b$
and $c^2+abc+1 =c^2+1-c+c(a-1)(b-1)+ac+bc \ge c^2+1-c+ac+bc \ge c+ac+bc$
is it nicer ;)

[quote="zhaobin"]
it is easily now two of $a-1,b-1,c-1$,have the same symbol
so we can assume $(a-1)(b-1) \ge 0$
[/quote]
This is nice!

Today is  really day of Schur's inequality. :D .
If (a-1)(b-1)(c-1)\geq 0 it's trivial.
Else then(1-a)(1-b)(1-c)\geq 0.
It's easily to prove the following inequality:(Schur's one conjugate of Am-Gm)
a^2+b^2+c^2+2abc+1\geq 2(ab+bc+ca)
And we remain to prove that:1-abc\ge \sum a - \sum bc<->(1-a)(1-b)(1-c)\geq 0 which's obvios.

Today is  really day of Schur's inequality. :D .
If $(a-1)(b-1)(c-1)\geq 0$ it's trivial.
Else then $(1-a)(1-b)(1-c)\geq 0.$
It's easily to prove the following inequality:(Schur's one conjugate of Am-Gm)
$a^2+b^2+c^2+2abc+1\geq 2(ab+bc+ca)$
And we remain to prove that:$1-abc\ge \sum a - \sum bc<\to (1-a)(1-b)(1-c)\geq 0$ which's obvios.


geometry
geometric transformation
reflection
circumcircle
geometry unsolved

[quote="Leonhard Euler"]Let $ ABC$ be a trianlge and $ P,Q$ be the point such that $ \measuredangle BPC \equal{} \minus{} \measuredangle BQC,\measuredangle CPA \equal{} \minus{} \measuredangle CQA,\measuredangle APB \equal{} \minus{} \measuredangle AQB$ $ \mod 180$( $ \measuredangle BPC \equal{} \minus{} \measuredangle BQC$ $ \mod 180$ is equivalent  with fact that $ B,P_{A},C,Q$ are cyclic where $ P_A$ is reflection of $ P$ on $ BC$ and it is obvious that $ Q$ is well define for each every point $ P$). Let $ X,Y,Z$ be a circumcenter of triangle $ BPC,CPA,APB$ and $ O$ be the circumcenter of triangle $ XYZ.$ Let $ Q'$ be the a isogonal conjugate point of $ Q$ wrt triangle $ ABC$. Prove that point $ P,O,Q'$ are collinear.[/quote]

$ Lemma1$.Q' is the inversion of the point P'wrt circumcircle of ABC.

$ Lemma2$.Let O'be the circumcentre of the triangle $ ABC$.Then O'M//PP', where P' is the isogonal conjugate of the point P wrt $ ABC$.

$ Lemma3$.Let N be the midpoint of the side PP'. Then P'N/OO' = R² - O'P'²/2R², where R is the circumradi of the tiangle $ ABC$.

$ Prove$.Let $ D,E,F$ be the projectives of the point P' wrt $ ABC$. As [ABC]/[XYZ] = 4[DEF]/[ABC], and [DEF]/[ABC] = R² - O'P'²/4R², hence [DEF]/[XYZ] = (R² - O'P'²/4R²)². Also △XYZ∽△DEF , we can conclude that P`N/OO` = R² - O'P'²/2R²

Now note that P'Q'/O'Q'=R²-O'P'²/R², and PP'/OO'=R²-O'P'²/R², and OO'//PP', hence ,O,P,Q' are collinear.


superior algebra
superior algebra solved

Let E be an elliptic curve, y^2 = x^3 + bx + c (mod p), p>=5 is an odd prime. Assume that if x is a root of x^3 + bx + c (mod p), then 3x^2 + b is not 0 mod p. Prove that if the order of the group of the curve is odd, then x^3 + bx + c = 0 (mod p) has no solutions.

1. $b,c\in\mathbb Z/p\mathbb Z$?
2. If $x$ is a root of $x^{3}+bx+c\ (mod\ p)$ then.... You mean $x\in Z/p\mathbb Z$ or in its algebraic closure?
3. The group of the curve is the group of $Z/p\mathbb Z$ points?

A point of order $2$ on $E$ has the form $(x,0)$ with $x^{3}+bx+c=0\ (mod\ p)$.
If the group of $\mathbb Z/p\mathbb Z$-points on $E$ has odd order, then $E$ has no element of order $2$ which translates into the required condition.

For my question $3$ you don't really need to limit your attention to $\mathbb Z/p\mathbb Z$ points. Points over any finite extension of $\mathbb Z/p\mathbb Z$ also work. The only condition is that $E(k)$ is a finite group which is by default if $k$ is a finite extension of $\mathbb Z/p\mathbb Z$.


inequalities
function
induction
inequalities unsolved

an easy one for the beggining :

$x_1,x_2,...,x_n$ and $y_1,y_2,...,y_n$ are given real numbers such that
$x_1 \geq x_2 \geq ... \geq x_n >0$ and $y_1 \geq y_2 \geq ... \geq y_n >0$
and also $\prod_{i=1}^{k}x_i \geq \prod_{i=1}^{k}y_i$ for all $k$ such that $1 \leq k \leq n$.

Proove that $\sum_{i=1}^{n}x_i \geq \sum_{i=1}^{n}y_i$.

see i tried this inequality for some time now and didnt come to any conclusion. i dont have any experience how to use karamata i only know some about the theoretical part. thats why i first wanted to read some sultions done with it.
id be glad to see the solution to your prob.
also im still looking for links

thx alot though peeta

umm.. i a bit suspicious about the correctness of this problem. it is sure the log should be applied. but log is concave function. then the sign will be opposite in majorization inequality. so is the problem correct?

ok, note that putting: $a_i=lnx_i$, $b_i=lny_i$ for $1 \leq i \leq n$ and $f(x)=e^x$ - convex and increasing funcion, then apllying Hardy-Littlewood-Polya (or equivalently Karamata's) inequality we get that:
$e^{a_1}+e^{a_2}+...+e^{a_n} \geq e^{b_1}+e^{b_2}+...+e^{b_n}$, so
$x_1+x_2+...+x_n \geq y_1+y_2+...+y_n$, q.e.d.

[Edit] we take $e^x$ not $lnx$ mahbub.

well, it does not say about another condition: the equality of the last condition. What about that?

well, can it be inequality as well?

hmm...I see I have to quote the Karamata's theorem:

If real numbers $x_1,x_2,...,x_n$ and $y_1,y_2,...,y_n$ satisfie :
$\sum_{i=0}^{k}x_i \geq \sum_{i=0}^{k}y_i$ for all $k$ such that $1 \leq k \leq n$
then if $f(x):R->R$ is a concave and non-decreasing function then :
$f(x_1)+f(x_2)+...+f(x_n) \geq f(y_1)+f(y_2)+...+f(y_n)$

ps. Mahbub, please formulate your ideas in a more precise way, cause I have problems with understanding your concise replies.

[quote="peeta"]hi
i just learned about majorization and the karamata theorom. and i'd really like to know how to apply this theorom. might somebody be so friendly to share some threads where it has been applied successfully?
thx peeta[/quote]

Karamata's paper was later than that of Hardy-Littlewood-Polya, and in any case the majorization concept was independently discovered many times before and after their papers.
It is simpler and more accurate to refer to it as simply majorization or the majorization inequality.

[quote="Rafal"]hmm...I see I have to quote the Karamata's theorem:

If real numbers $x_1,x_2,...,x_n$ and $y_1,y_2,...,y_n$ satisfie :
$\sum_{i=0}^{k}x_i \geq \sum_{i=0}^{k}y_i$ for all $k$ such that $1 \leq k \leq n$
then if $f(x):R->R$ is a concave and non-decreasing function then :
$f(x_1)+f(x_2)+...+f(x_n) \geq f(y_1)+f(y_2)+...+f(y_n)$
[/quote]

Well as far i know the majorization inequality is:

If real numbers $x_1,x_2,...,x_n$ and $y_1,y_2,...,y_n$ satisfie :
$\sum_{i=0}^{k}x_i \geq \sum_{i=0}^{k}y_i$ for all $k$ such that $1 \leq k < n$ and $\sum_{i=0}^{k}x_i = \sum_{i=0}^{k}y_i$ for $k=n$
then if $f(x):R->R$ is a convex
$f(x_1)+f(x_2)+...+f(x_n) \geq f(y_1)+f(y_2)+...+f(y_n)$

hey i got it!!
thx alot for this first problem. i guess i kinda got the idea.
i would think it'd be really useful though if somebody could write down some more problems or links to threads where there are some

thx peeta

[quote="Rafal"]If you knew the proof of the generalised inequality then you wouldn't be confused, here's my proof on a Polish forum
of the Karamata's inequality [url=http://matematyka.org/forum/viewtopic.php?p=13524&highlight=#13524]matematyka.org/forum/viewtopic.php?p=13524&highlight=#13524[/url]
- as you can see I don't use the assumption about the equality: $\sum_{i=1}^{n}x_i=\sum_{i=1}^{n}y_i$. But even by intuition it's quite obvious that for convex and increasing functions if the theorem holds for $\sum_{i=1}^{n}x_i=\sum_{i=1}^{n}y_i$ then especially for $\sum_{i=1}^{n}x_i > \sum_{i=1}^{n}y_i$[/quote]
but you wrote concave and non-decreasing in the previous post!

ok, I apologise for the typo , I've written concave in one of those posts but of course it should be "convex" - but that doesn't change the situation - I was thinking about convex and writting concave :) (in the proof I used a convex function of course).

Now let's clarify something:

The Karamata's inequality - also named alternatively H-L-P (Hardy-Littlewood-Polya)
has two possible variants :

a) everything is the same as I wrote while quoting the theorem some posts before but(!) if
$\sum_{i=1}^{n}x_i=\sum_{i=1}^{n}y_i$ - then the inequality is true even(!) when the convex function is
not monotonimous.

b) the same as I wrote in posts before especially $\sum_{i=1}^{n}x_i \geq \sum_{i=1}^{n}y_i$ - then it's true for all
convex and non-decreasing functions.

I don't know why the link with my proof of the second variant of H-L-P doesn't work, so in near future I promise to translate this proof from Polish to English and show it at mathlinks.

[Edit] More details about this inequality can be found in : "Analityc Inequalities"; D.S.Mitrinovic; Berlin-Heidelberg-New York 1970

ok, as I promised :

Here is the proof:

Assumptions :,$ x_1 \geq x_2 \geq x_3 ... \geq x_n $ and $ y_1 \geq y_2 \geq y_3 ... \geq y_n $ and $x_1+x_2+...+x_i \geq y_1+y_2+...+y_i$ for $ i=1,2,...,n $ and obviously $x_i , y_i \in S$ dla $i=1,2,...,n$ where $f(x)$ is convex and non-decreasing in $S$ .

Thesis : $ f(x_1)+f(x_2)+...+f(x_n) \geq f(y_1)+f(y_2)+...+f(y_n)$ .

Let's use induction :

1.For $n=1$ it's easy to check , for $n=2$ we have :

$x_1 \geq y_1$ and $x_1+x_2 \geq y_1+y_2$ , We know that $f(x_1) \geq f(y_1)$ , let's put $x_1$ as  $y_1 + r$ , thus $x_2 \geq y_2-r$ , let's then take the minimal value of $x_1$ so that $f(x_1)$ would be possibly the smallest - so we have $x_2=y_2-r$ .Let's note that $f(y_1+r)-f(y_1) \geq f(y_2)-f(y_2-r)$ , but because $f(x)$ is convex so $g(x)=f(x+r)-f(x)$ is non-decreasing . So we have the proof for $n=2$ . Let's use this method in the next induction step.

2. Let's assume that for $i=1,2,...,k$ the thesis is true . We will proove that it's also true for $k+1$ .
We have a sequence ofinequalities :
\[x_1 \geq y_1\]
\[x_1+x_2 \geq y_1+y_2\]
\[...\]
\[x_1+x_2+...+x_{k+1} \geq y_1+y_2+...+y_{k+1}\]

Let's assume first that there exists such $i$ , that  $k+1 \geq i>1$ and $x_i \geq y_i$ , so using the induction assumption (for $i=1,2,...,k$ the thesis is true) we see that :
 $ f(x_1)+f(x_2)+...+f(x_{i-1}) \geq f(y_1)+f(y_2)+...+f(y_{i-1})$ and
 $ f(x_i)+f(x_{i+1})+...+f(x_{k+1}) \geq f(y_i)+f(y_{i+1})+...+f(y_{k+1})$ , so summing up we get the thesis.

So let's assume that for $i=2,3,...,k+1$ is $x_i < y_i $.
From the inequality $x_1+x_2+...+x_{k+1} \geq y_1+y_2+...+y_{k+1}$ concludes that
$x_1-y_1 \geq (y_2-x_2)+(y_3-x_3)+...+(y_{k+1}-x_{k+1})$ .
Thus we can divide the segment $x_1-y_1$ into smaller ones $(y_2-x_2),...,(y_{k+1}-x_{k+1})$.
Let's put for $i=2,3,...,k+1$ $y_i-x_i = r_{i-1}$ , so $y_2-x_2=r_1,...,y_{k+1}-x_{k+1}=r_k$.
From the observation in the proof for $n=2$ we know that :
\[f(y_1+r_1)-f(y_1) \geq f(y_2)-f(y_2-r_1)=f(y_2)-f(x_2)\]
\[f(y_1+r_1+r_2)-f(y_1+r_1) \geq f(y_3)-f(y_3-r_2)=f(y_3)-f(x_3)\]
\[...\]
\[f(y_1+r_1+r_2+...+r_k)-f(y_1+r_1+r_2+...+r_{k-1}) \geq f(y_{k+1})-f(y_{k+1}-r_k)=f(y_{k+1})-f(x_{k+1})\]

Summing up we get :
\[f(y_1+r_1+r_2+...+r_k)-f(y_1) \geq (f(y_2)-f(x_2))+(f(y_3)-f(x_3))+...+(f(y_{k+1}-f(y_{k+1}))\], but we know that $f(x_1) \geq f(y_1+r_1+r_2+...+r_k)$, cause $x_1-y_1 \geq (y_2-x_2)+(y_3-x_3)+...+(y_{k+1}-x_{k+1})$ , so finally :

\[ f(x_1)+f(x_2)+...+f(x_{k+1}) \geq f(y_1)+f(y_2)+...+f(y_{k+1})\] , q.e.d.

Analogically we could write a proof for $f(x)$ non-increasing and concave, but then the direction of the inequality is opposite.

Hi everybody,

I don't understand one thing in the proof at this level :
[quote]Let's assume first that there exists such $ i$ , that  $ k \plus{} 1 \geq i > 1$ and $ x_i \geq y_i$ , so using the induction assumption (for $ i \equal{} 1,2,...,k$ the thesis is true) we see that :
 $ f(x_1) \plus{} f(x_2) \plus{} ... \plus{} f(x_{i \minus{} 1}) \geq f(y_1) \plus{} f(y_2) \plus{} ... \plus{} f(y_{i \minus{} 1})$ and
 $ f(x_i) \plus{} f(x_{i \plus{} 1}) \plus{} ... \plus{} f(x_{k \plus{} 1}) \geq f(y_i) \plus{} f(y_{i \plus{} 1}) \plus{} ... \plus{} f(y_{k \plus{} 1})$ , so summing up we get the thesis.[/quote]

I don't see why the last inequality is true. $ x_i, ..., x_{k \plus{} 1}$ don't verify the assumptions for me. Somebody can explain me ?

Last thing, the case "for $ i \equal{} 2,3,...,k \plus{} 1$ is $ x_i < y_i$" doesn't use induction. No ?

Thank you.

http://www.sftw.umac.mo/~fstitl/10mmo/2000-majorization.html
定理(Majorization 不等式)：
設函數 f 在閉區間 I=[a, b]為凸的，且
(x1, x2, ...., xn)(y1, y2, ...., yn)其中實數xi, yj在I。則有 f(x1)+ f(x2)+ .... + f(xn)≧f(y1)+ f(y2)+ .... + f(yn)。
對於嚴格凸的函數f，等號成立當且僅當
這兩個n元組相等，即(x1, x2, ...., xn) = (y1, y2, ...., yn)。
對於下凹的函數，只須要將結論中的不等式的方向換過來。
證：已知(x1, x2, ...., xn)(y1, y2, ...., yn)，所以對所有的i=1,2,..,n，有xi+1≦ xi及yi+1≦ yi。令mi=[f(xi)-f(yi)]/(xi-yi)，由引理，可以證明：mi≧ mi+1如下：

如果 yi+1≦yi≦xi+1≦xi：應用引理，可以有
mi+1=[f(xi+1)-f(yi+1)]/(xi+1-yi+1)
　≦[f(xi+1)-f(yi)]/(xi+1-yi)　　第二個不等式：yi+1≦yi≦xi+1
　≦[f(xi)-f(yi)]/(xi-yi)　　　第一個不等式：yi≦xi+1≦xi
　=mi。
對於 yi+1≦yi、xi+1≦xi的其他可能分配，可以運月用類似的方法，略去。
記Xk=x1+x2+ ...+xk、Yk=y1+y2+ ...+yk，此外為了以後方便，令X0=Y0=mn+1=0。由已知條件得於Xk≧Yk當k=1,2,..,n-1，及Xn=Yn。所以有
0≦Σ1≦k≦n(Xk-Yk)(mk- mk+1)
　=Σ1≦k≦n[(Xk-Xk-1)-(Yk-Yk)]mk
　=Σ1≦k≦n(xk-yk)mk
　=Σ1≦k≦n[ f(xk)-f(yk)]
即Σ1≦k≦nf(xk)≧Σ1≦k≦nf(yk)。
註：代數解釋：
(x1, x2, ...., xn)(y1, y2, ...., yn)可以看成凸性的組合條件。

http://episte.math.ntu.edu.tw/articles/mm/mm_06_4_02/page3.html

You can use Google Translate to view these pages in English.


Q: Find the radii of convergence of the Taylor expansion at i of
f(z)=1/[(cosz)-0.5]

All I can think of is f(z)=(-2) * [1/(1-2cosz)]

|2cosz|<1
|cosz|<(1/2)
|z|<|arccos(1/2)| =  |-i Log[(1/2)+sqrt((1/4)-1)] |     <---formula for arccos (1/2)
 = 5pi/3

 Is my concept totally wrong???  Cuz I didn't use the fact that it is centred at i...and I have no idea on how to find the Taylor expansion of f(z).

If I'm wrong, how to do this question???? Thanks

Hmm It would be nice if a more experienced comples analyst would help
but I though it was as simple as this : your radius of convergence defines in what circle around the point the function is analytic
well $cos(z)=1/2$ happens if and only if $z=\frac{\pi}{3}+k\pi$ so I would go for $\frac{\pi}{3}$ ?

That would be $\sqrt{\left(\frac\pi3\right)^2+1}$. $\cos z$ is $\frac12$ only on the real axis, at $\pm\frac\pi3+2n\pi$. The radius of convergence is the distance to the nearest singularity.

Yes I was calculating from the origin.
But for the rest I am correct?


Estimate the time between each photoflash of the apple in the figure. Assume the apple is 10 cm in diameter.

Bad picture :P

[hide="Possible HINT"]
$ d(t)\equal{} \minus{}v_{n}t \minus{}\frac{gt^{2}}{2}$.
[/hide]

Questio

1)IS there the same apple in each photograpy?

If it is. I think the high friction with the atmosphere diminished the apple     :P


Suppose you have an unlimited number of A cent postage stamps and B cent postage stamps,
where A and B have no common divisor greater than 1. What is the largest total that you
can not make using these stamps?

Is this a homework question? The solution is $ AB \minus{} A \minus{} B$. And exactly half of the solutions less than that (and including) cannot be done. That is rather neat huh =)
Are you looking for a derivation?

no, it's not a homework problem haha
a derivation would be nice =)


function
number theory unsolved
number theory

Find the total number of different integer values the function \[ f(x) = [x] + [2x] + [\frac{5x}{3}] + [3x] + [4x] \] takes for real numbers $x$ with $0 \leq x \leq 100$.

consider $x\in[0,3)$ we find 22,$x\in[0,1)$,we find 8.
And $x\in[k,l)$ is the same number with $x\in[k+3,l+3)$
So $22\times33+8=734$

[quote="wangsacl"]Excuse me, but what does a [x] mean here?
Does it mean the nearest integer to x?[/quote]

$ [x]$ is meant as the biggest integer smaller than $ x$. So $ [x]\leq x<[x]\plus{}1$

[quote="jin"]
And $x\in[k,l)$ is the same number with $x\in[k+3,l+3)$
[/quote]

How did you get this?

Bumping this thread to see if anybody knows?

Jin's solution is correct, but let me formalize it.  Let $ g(a, b) $ be the number of distinct integer values $ f $ takes on the interval $ [a, b) $.

Lemma 1: $ g(a, b) = g(a + 3, b + 3) $ for all $ a, b \in \mathbb{R} $.

[i]Proof:[/i] Let $ S = \{z \in \mathbb{Z} : \exists r \in [a, b)\:s.t.\:f(r) = z\} $ and let $ T = \{z \in \mathbb{Z} : \exists r \in [a + 3, b + 3) \:s.t. \:f(r) = z\} $.  Letting $ p(x) = x + 35 $ I shall show that $ p: S \longrightarrow T $ is a bijection which will imply the desired result.  Clearly if $ p(j) = p(k) $ then $ j = k $ so we have an injection.  Now to show a surjection it suffices to show that for all $ t \in T $ we have that $ t - 35 \in S $ which is clear since if $ f(r) = t $ then $ f(r - 3) = t - 35 $.  Therefore $ |S| = |T| $ which implies the lemma.

Lemma 2: $ g(0, 3) = 22 $.

[i]Proof:[/i] By counting we get that letting $ x \in [0, 3) $ the integers that are reachable (from $ f $) are $ 0, 1, 2, 4, 5, 6, 7, 11, 12, 13, 14, 16, 17, 18, 19, 23, 24, 25, 26, 28, 29, 30 $

Lemma 3: $ g(0, 1) = 7 $

[i]Proof:[/i] By counting we get that letting $ x \in [0, 1) $ the integers that are reachable (from $ f $) are $ 0, 1, 2, 4, 5, 6, 7 $

Now the answer to the APMO question itself is given by $ \sum_{i = 0}^{32}g(3i, 3i + 3) + g(99, 100) + 1 = 33g(0, 3) + g(0, 1) + 1 = 734 $ as desired.


invariant
linear algebra
matrix
linear algebra unsolved

assume that $A$ and $B$ are two n*n complex matrix such that $A^2=I=B^2$.prove that there exsit one dimensional or two dimensional invariant subspace under $A$ and $B$.

We can find $\alpha\ne 0$ s.t. $B(I_n-\alpha A)$ has a non-zero eigenvalue $\beta$. Take $x$ to be an eigenvector corresponding to $\beta$. We get $Bx-\alpha BAx=\beta x$, and, after left multiplication with $B$, this becomes $x=\alpha Ax+\beta Bx$.

If any two of $x,Ax,Bx$ are collinear, then all three of them are, so $x$ is a common eigenvector of $A,B$, meaning that we can find a one-dimensional $A,B-$invariant subspace, i.e. the one generated by $x$. Otherwise, if the three are mutually non-collinear, then take the plane generated by them (by any two of them). It's generated by $x,Ax$, so it's $A-$invariant, and it's also generated by $B$, so it's also $B-$invariant.

Is it Ok? :?


calculus
Pythagorean Theorem
geometry
calculus computations

Boats A and B leave the same place at the same time. Boat A heads due north at 12 km/hr. Boat B heads due east at 18 km/hr. After 2.5 hours, how fast is the distance between the boats increasing (in km/hr)?

A) 21.63
B) 31.20
C) 75.00
D) 09.84
E) 54.08

[hide](A) is the answer. ((E) is the distance between A and B after 2.5 hours, but if the question is about the rate of change of the distance between them, then the answer is not (E).)

If the boats have been moving for time $ t$ hours, then Boat A has moved $ 12t$ km while Boat B has moved $ 18t$; by Pythagoras, the distance between them is $ \sqrt{(12t)^2\plus{}(18t)^2}\equal{}6\sqrt{13}t$. Hence the rate of change of the distance between the boats is $ \frac{6\sqrt{13}t}{t}\equal{}6\sqrt{13}\ \textrm{km}\,\textrm{h}^{\minus{}1}$. This rate is constant, however long the boats have been travelling.[/hide]

[hide]At $ t \equal{} 2.5$ you can see that Boat A will be $ 30$ miles north and Boat B will be $ 45$ miles east.
The boats form a right triangle so by pythagorean theorem we see that their distance apart is $ \sqrt {2925}$

Now some calculus:
$ a^2 \plus{} b^2 \equal{} c^2$
Differitiating with respect to time gives us:
$ 2a\frac {da}{dt} \plus{} 2b\frac {db}{dt} \equal{} 2c\frac {dc}{dt}$

$ \frac {da}{dt} \equal{}$speed of a=$ 12$

$ \frac {db}{dt} \equal{}$speed of b=$ 18$

$ \frac {dc}{dt} \equal{}$speed of c=$ ??$

$ 2\cdot 30\cdot 12 \plus{} 2\cdot 45\cdot 18 \equal{} 2\cdot \sqrt {2925}\cdot \frac {dc}{dt}$

$ \frac {dc}{dt} \equal{} 6\sqrt {13} \equal{} \boxed{21.633}$[/hide]


algebra
polynomial
limit
number theory unsolved
number theory

let w be a polynominal of degree two with integer coefficients. suppose that for each integer x the value w(x) is the square of an integer. prove that w the square of a polynominal.

This answer is wrong, see below.
[hide]
Let $w(x)=c(x-a)(x-b)$.
WLOG, let $a \leq b$.
If $x=a+1$, then $w(x)=c(a-b+1)$.
If $x=b+1$, then $w(x)=c(b-a+1)$.

Now we have two possibilities. Either (1)$a=b$ or (2)$b=a+1$. Otherwise, there are values of $x$ for which $w(x)$ is negative, and so not a square.

(1)
$w(x)=c{(x-a)}^{2}$.
$w(a+1) = c = k^{2}$ for some $k$.
Hence $w(x)=(k(x-a))^{2}$ and we're done.

(2)
$w(x) = c(x-a)(x-a-1)$.
$w(a+2) = 2c = m^{2}$, $m$ is an integer.
$w(a+3) = 6c = 3m^{2}= n^{2}$, $n$ is an integer.
$n=\sqrt{3}m$. But this isn't an integer, contradiction.
So (1) is the only possibility.
[/hide]

[quote="TomciO"]Why do you suppose that $w$ has integer roots?[/quote]
:blush:

I was wondering why it seemed so straightforward...

$w(x)=ax^{2}+bx+c$.We have $\sqrt{w(k)}=n$for each integer k.

$\sqrt{w(k+1)}-\sqrt{w(k)}= \frac{w(k+1)-w(k)}{\sqrt{w(k+1)}+\sqrt{w(k)}}=n_{1}-n_{2}$.
So

$n_{1}-n_{2}=\lim_{k \to \infty}{\frac{(2k+1)a+b}{\sqrt{a(k+1)^{2}+b(k+1)+c}+\sqrt{ak^{2}+bk+c}}=\sqrt{a}}$.

So $\sqrt{a}$ is an integer and seqence $\sqrt{w(k+1)}-\sqrt{w(k)}$ converges to an integer.Therefore, there exists the smallest integer $k_{0}$ for which $\sqrt{w(k_{0}+1)}=\sqrt{w(k_{0})}+\sqrt{a}$.So we have $p(m)=w(k_{0}+m)-(\sqrt{w(k_{0})}+m\sqrt{a})^{2}=0$ for all positive integers m.Therefore polynomial $p(x)$ has infinately many roots.Therefore $p(x)=0$ for all x.So $w(x)=(\sqrt{w(k_{0})}+(x-k_{0})\sqrt{a})^{2}$.


trigonometry
analytic geometry
graphing lines
slope
geometry
trapezoid
perpendicular bisector

1) In triangle ABC, D is on AC and F is on BC.  Also, AB perpendicular to AC, AF is perp. to BC and BD=DC=FC=1. Find AC.

2) Two circles of radii 8 and 6 are drawn with their centers 12 units apart.  at P, one of the points of intersection, a line is drawn in such a way that QP = PR, where Q is on the circle with radius 8, and R on the circle of radius 6. Find PQ.

[b][color=blue](1)[/color][/b] Let $M$ be the midpoint of the hypotenuse $BC$ The triangle $\triangle BCD$ is isosceles with $BD = CD = 1$. Hence, $DM$ is the perpendicular bisector of $BC$.

$\cos{\widehat C} = \frac{CA}{BC} = \frac{CF}{CA} = \frac{CM}{CD} = \frac{BC}{2CD}$

Since $CF = CD = 1$,

$\frac{BC}{2} = \frac{1}{AC} = \frac{AC}{BC}$

$\frac{1}{AC} = \frac{AC^2}{2},\ \ AC^3 = 2,\ \ AC =\ ^3\sqrt 2$

[b][color=blue](2)[/color][/b] Let $O_1, O_1$ be the centers of the circles with radii $r_1 = 6, r_2 = 8$. Denote $d = O_1O_2$ the distance between the circle centers. Put the coordinate origin at the point $O_1$ and the positive x-axis along the ray $O_1O_2$. The circles $(O_1), (O_2)$ have equations

$x^2 + y^2 = r_1^2,\ \ (x - d)^2 + y^2 = r_2^2$

Solving these 2 equations, we get the coordinates of the point $P$:

$x_P = O_1N = \frac{d^2 + r_1^2 - r_2^2}{2d} = \frac{144 + 36 - 64}{24} = \frac{29}{6}$

$y_P = PN = \sqrt{r_1^2 - x_P^2} = \sqrt{36 - \frac{841}{36}} = \frac{\sqrt{7 \cdot 65}}{6}$

Let the equation of the line $QPR$ passing through the point $P$, such that $PQ = PR$, be

$y - y_P = m(x - x_P)$

The normals to this line through the circle centers $O_1, O_2$ cut the chords $PQ, PR$ at their midpoints $M, N$ and $MN = \frac{QR}{2} = PQ$. These normals have equations

$O_1M:\ \ y = -\frac x m$

$O_2N:\ \ y = -\frac{x - d}{m}$

Solving for the coordinates of the points $M, N$, we get

$-\frac{x_M}{m} - y_P = m(x_M - x_P),\ \ x_M = \frac{m^2x_P - my_P}{m^2 + 1}$

$y_M = -\frac{x_M}{m} = -\frac{mx_P - y_P}{m^2 + 1}$

$-\frac{x_N - d}{m} - y_P = m(x_N - x_P),\ \ x_N = \frac{m^2x_P - my_P + d}{m^2 + 1}$

$y_N = -\frac{x_N - d}{m} = -\frac{mx_P - y_P}{m^2 + 1} - \frac{d}{m(m^2 + 1)} + \frac d m = -\frac{mx_P - y_P}{m^2 + 1} + \frac{md}{m^2 + 1}$

$MN = \sqrt{(x_N - x_M)^2 + (y_N - y_M)^2} = \sqrt{\left(\frac{d}{m^2 + 1}\right)^2 + \left(\frac{md}{m^2 + 1}\right)^2} =$

$= \frac{d}{m^2 + 1} \sqrt{1 + m^2} = \frac{d}{\sqrt{m^2 + 1}}$

Since $P$ is the midpoint of the segment $MN$, we have

$x_P = \frac{x_M + x_N}{2} = \frac{m^2x_P - my_P + \frac d 2}{m^2 + 1}$

$x_P = \frac d 2 - my_P,\ \ m = \frac{\frac d 2 - x_P}{y_P} = \frac{36 - 29}{\sqrt{7 \cdot 65}} = \sqrt{\frac{7}{65}}$

Substituting the slope $m$ into the expression for the segment $MN = PQ$, we obtain

$PQ = MN = 12 \sqrt{\frac{65}{7 + 65}} = \sqrt{130}$

The equation of the line $QPR \equiv MN$ is

$y - \frac{\sqrt{7 \cdot 65}}{6} = \sqrt{\frac{7}{65}} \left(x - \frac{29}{6}\right)$

and it intersects the x-axis at a point with the x-coordinate

$x = \frac{29}{6} -  \frac{\sqrt{7 \cdot 65}}{6} \cdot \sqrt{\frac{65}{7}} = \frac{29 - 65}{6} = -6$

Since this is the intersection of the x-axis with the circle $(O_1)$, i.e., it lies on this circle, this point is identical with the point $Q$.

There is also a synthetic (this doesn't mean short) way to solve problem [b](2)[/b], based on a generalization of this problem.

In the following, for a given point Z and a given real number r, we denote by Z(r) the circle with center Z and radius r. Then, we will show the following generalization of problem [b](2)[/b]:

[color=blue][b]Theorem 1.[/b] Consider two points U and V and two reals u and v such that UV = 2u. Assume that the circles U(u) and V(v) have two points of intersection, and let P be one of these two points. Also, let a line through the point P intersect the circles U(u) and V(v) again at two points R and Q such that QP = PR. Then, $QP=PR=\sqrt{\frac12\left(\left(3u\right)^2-v^2\right)}$.[/color]

[i]Proof of Theorem 1.[/i] If M is the midpoint of the segment UV, then $UM=\frac{UV}{2}=\frac{2u}{2}=u$, so that the point M lies on the circle U(u).

The center of a circle always lies on the perpendicular bisector of any chord of the circle. Thus, the point U, being the center of the circle U(u), lies on the perpendicular bisector of its chord PR, and the point V, as the center of the circle V(v), lies on the perpendicular bisector of its chord QP. In other words, if X and Y are the midpoints of the segments PR and QP, we have $UX\perp PR$ and $VY\perp QP$. Equivalently, $UX\perp QR$ and $VY\perp QR$. Thus, UX || VY, so that the quadrilateral UXYV is a trapezoid.

Now, since X and Y are the midpoints of the segments RP and QP, we have $QY=YP=\frac{QP}{2}$ and $PX=XR=\frac{PR}{2}$. Since QP = PR, this becomes QY = YP = PX = XR. Denote QY = YP = PX = XR = t. Then, YR = YP + PX + XR = t + t + t = 3t.

Since YP = PX, the point P is the midpoint of the segment YX. Also, we know that M is the midpoint of the segment UV. Thus, the points M and P are the midpoints of the legs UV and YX of the trapezoid UXYV; hence, the line MP is the midline of this trapezoid, so that it is parallel to its two bases: MP || UX || VY. Since $UX\perp QR$, this yields $MP\perp QR$. In other words, < MPR = 90°. Therefore, the segment MR is a diameter of the circle U(u). Hence, it passes through the center U of this circle. So, the points M, R and U lie on one line; of course, this line must also pass through the point V (since M is the midpoint of the segment UV). Also, since the points M and R lie on the circle U(u), we have RU = u and UM = u, and since M is the midpoint of the segment UV, we have UM = MV, so that also MV = u. Thus, RV = RU + UM + MV = u + u + u = 3u.

Now, since $VY\perp QR$, the triangle RYV is right-angled, so that Pythagoras yields $RV^2=YR^2+YV^2$. On the other hand, for the same reason, the triangle PYV is right-angled, so that Pythagoras yields $PV^2=YP^2+YV^2$. Hence,

$RV^2-PV^2=\left(YR^2+YV^2\right)-\left(YP^2+YV^2\right)=YR^2-YP^2$.

But since the point P lies on the circle V(v), we have PV = v. Also, RV = 3u, YR = 3t and YP = t. Thus, this equation becomes $\left(3u\right)^2-v^2=\left(3t\right)^2-t^2$. Thus,

$2\cdot\left(2t\right)^2=8t^2=\left(3t\right)^2-t^2=\left(3u\right)^2-v^2$,

so that $2t=\sqrt{\frac12\left(\left(3u\right)^2-v^2\right)}$. Since QP = PR = PX + XR = t + t = 2t, this becomes $QP=PR=\sqrt{\frac12\left(\left(3u\right)^2-v^2\right)}$, and Theorem 1 is proven.

Your problem [b](2)[/b] is a particular case of Theorem 1 for u = 6 and v = 8. Nice problem, I must say!

  darij

More generally, if UV = d and QP = PR = L then
(d^2-L^2)(2u^2+2v^2-d^2) = (u^2-v^2)^2


Cashews cost\ \$1.60 a pound and pecans cost\ \$1.85 per pound. Joe bought cashews but the clerk thought they were pecans and charged him\ \$35.15. How much money was Joe overcharged?

We divide 35.15 by 1.85, and we know that Joe bought 19 pounds of nuts.

Since every pound of nut was overcharged by 15 cents, Joe was overcharged $ \boxed{\$ 2.85}$

He was actually overcharged by 25 cents so the answer is $4.75


Find a number whose double exceeds its half by exactly 99.

Assume that number is = n
Double of the number = 2n
Half of the number = n/2

So 2n = n/2 + 99
==> 4n = n + 198
==> 3n = 198
==> n = 66

Thanks,
http://www.teacherone.com


Four politicians and three lawyers attend a party. Each politician shakes hands exactly once with everyone, and each lawyer shakes hands exactly once with each politician. How many handshakes take place?

The first politician shakes hands with $ (4 \plus{} 3) \minus{} 1 \equal{} 6$ other people. The second one, thus shakes hands with $ 6 \minus{} 1 \equal{} 5$ other people, since the handshake with the first politician was already counted. The third thus adds $ 5 \minus{} 1 \equal{} 4$ more handshakes, and the fourth $ 4 \minus{} 1 \equal{} 3$ more. In this, the handshakes the lawyers participated in were already included, since the politicians shook hands with [i]everybody[/i]. Thus, $ 6 \plus{} 5 \plus{} 4 \plus{} 3 \equal{} \fbox{18}$ handshakes take place.

Each politician shakes hands with $ 6$ people and each lawyer shakes hands with $ 4$ people.  $ \frac{4 \times 6 \plus{} 3 \times 4}{2}\equal{}\boxed{18}$.


Hi everyone  :lol: ,
i m naveenkr

[b]Start[/b] Posting any (whatever it may be inc. cutoffz)  about INPhO2009.  :roll:

What was the score of last rank in INPHO09? (that u call cutoff)

Any Strategy???

post answers of INPho 2008 ques [url=http://www.mathlinks.ro/viewtopic.php?t=249483]here[/url]

u can find the paper HBCSE


inequalities
vector
algebra
polynomial
linear algebra
inequalities theorems

First I will quote two posts:

[quote="Xevarion"]To avoid confusion I will use $[n, n, n]$ to represent symmetric sums and $(n, n, n)$ for cyclic in this post. Wikipedia uses square brackets for symmetric.[/quote]
from
http://www.mathlinks.ro/Forum/viewtopic.php?t=103761

[quote="conceive"][size=150]I suggest this notation:[/size]
$[r,s,t]$ will denote CYCLIC sum while
$(r,s,t)$ will denote SYMMETRIC sum (EDITED)[/quote]
from
http://www.mathlinks.ro/Forum/viewtopic.php?p=583292

I am sure there is necessary to have this anywhere in the forum the same. Because write "I understand here () as a symmetric sum" in every new topics is quite ridiculous.

I suggest () as a symmetric and [] as a cyclic sum.

[quote="Peter VDD"]I've not seen $(n,n,n)$ too much, but $[n,n,n]$ is just too commonly used for symmetric to start using it for cyclic all at once...[/quote]

First note that it is very important to have this everywhere in common.

Second note that it is also very useful to [b]have a sign for cyclic[/b] ones. Because you couldn't write $a^{4}+b^{4}+c^{4}\ge a^{3}b+b^{3}c+c^{3}a$, which is easy by $[4,0,0]\ge[3,1,0]$.

Therefore I think my suggestion is a good one, because it contains cyclic ones too.

Then I suggest you switching them like Xevarion did. The notation $[3,1,0]$ is pretty much the standard I've seen here for $\sum_{sym}x^{3}y$.

[quote="Peter VDD"]The notation $[3,1,0]$ is pretty much the standard I've seen here for $\sum_{sym}x^{3}y$.[/quote]

Not true. Muirhead inequality states that $(a_{1},a_{2},a_{3})\ge(b_{1},b_{2},b_{3})$ with this notation.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=15896 posts 4,9,10.

If you read carefully, you will see that it denotes the triplet $(a,b,c)$, not the symmetric sum. Majorization is defined on n-tuples of real numbers, not on symmetric sums.

I haven't claimed $(a,b,c)$ for cyclic sum would be a good idea either (for the confusion you mention there), I'm just saying $[a,b,c]$ would in my opinion be far worse.

[quote="Peter VDD"]If you read carefully, you will see that it denotes the triplet $(a,b,c)$, not the symmetric sum.[/quote]

You are wrong, in that topic () denotes symmetric sum. See:

[quote]$(3, 3, 1)$ for $2x^{3}y^{3}z+2x^{3}y z^{3}+2x y^{3}z^{3}$[/quote]

On the other hand see:
http://www.mathlinks.ro/Forum/viewtopic.php?p=281573

post 13.

Everything I want to point out is that my suggestion is a more natural one.

[b]EDIT:[/b]

[quote="Xevarion"]Wikipedia uses square brackets for symmetric.[/quote]

He teels downrigt lies!! Wikepedia uses () for symmetric.

See:
http://en.wikipedia.org/wiki/Muirhead_inequality

As for what I can see here, wikipedia uses $[a] \ge [b]$, which is square. And uses $(1,0,..,0)$ for the $n$-tuples.

[quote="conceive"][quote]$(3, 3, 1)$ for $2x^{3}y^{3}z+2x^{3}y z^{3}+2x y^{3}z^{3}$[/quote][/quote]where exactly does that stand? I fail to see it on the page you mention... The only post about that is jhaussmann5's, and he is clearly posting that the triple of real numbers $(a_{1},a_{2},a_{3})$ is filled in with the triple of real numbers $(3,3,1)$, and this in the formula mentioned some posts above that.

This confusion between the triplet $(a,b,c)$ and the symmetric sum $[a,b,c]$ you already seem to have now, so I don't think it's wise to mix up their notations... :?

[quote="Peter VDD"]where exactly?[/quote]
in
http://www.mathlinks.ro/Forum/viewtopic.php?t=15896

[quote="jhaussmann5"]Bruno,
you have left out the condition that $a_{1}\ge a_{2}\ge \dots \ge a_{n}$ and $b_{1}\ge b_{2}\ge \dots \ge b_{n}$.

JSRosen3,
shuffling your numbers around, your example would now be $(a_{1}, a_{2}, a_{3}) = (3, 3, 1)$ and $(b_{1}, b_{2}, b_{3}) = (3, 2, 2)$, and the resulting inequality would be \[2x^{3}y^{3}z+2x^{3}y z^{3}+2x y^{3}z^{3}\ge 2x^{3}y^{2}z^{2}+2x^{2}y^{3}z^{2}+2x^{2}y^{2}z^{3}.\] [/quote]

Hope it's clear now.

[quote="Peter VDD"]As for what I can see here, wikipedia use $(1,0,..,0)$ for the $n$-tuples.[/quote]

Well, that's right. So, we can use it for eg. $(4,0,0),(3,2,0),(5,4,0)$ for symmetric sums.

[quote="Peter VDD"]You're doing for this, right?  :lol:[/quote]

what?

I just don't want to spend here the whole day posting every 20 minutes a reply.

The only one reason for me to make this topic was, that this denotion was not the same everywhere. I'm sure it will be better to have everywhere the same.

I just think my suggestion is more natural one.

$\left( a_{1},a_{2},\ldots,a_{n}\right)$ means $n$-uple, not a bloody symmetric sum. Or, in other words, a vector with $n$ components. It's a convention used in linear algebra for example. Do you know anything about vectors? This may cause your HUGE confusion.

On the other hand, $\left[ a_{1}, a_{2},\ldots,a_{n}\right]$ reffers to symmetric sums $\left( \sum_\textrm{sym}x_{1}^{a_{1}}\ldots x_{n}^{a_{n}}\right)$, a convention used on ML and on Wikipedia (although in a slightly different form).

Vector $\neq$ Sum (i.e. Vector $\neq$ Polynomial in several variables).

Muirhead's inequality says that $\left( a_{1},a_{2},\ldots,a_{n}\right) \succ \left( b_{1},b_{2},\ldots,b_{n}\right)$ implies $\left[ a_{1},a_{2},\ldots,a_{n}\right] \geq \left[ b_{1},b_{2},\ldots,b_{n}\right]$ (for positive variables $x_{1},\ldots,x_{n}$).

I hope it's clear. It's really annoying to argue about such small things, regarded as conventions.

I have to admit that I didn't fully understand fully the definition on Wikipedis. OK

But still as well as

http://www.mathlinks.ro/Forum/viewtopic.php?p=281573
post 13

as well as in
http://www.mathlinks.ro/Forum/viewtopic.php?t=56369

I still suggest to compare mutiindexes; I mean that I [b]unite[/b] the [i]definition of multiindex [/i]and a [i]definition of sym/cyc polynomial[/i].

like that ones:
$(3,0,0)\ge(2,1,0)$

$(3,0,0)=2(a^{3}+b^{3}+c^{3})$
$(2,1,0)=a^{2}b+ab^{2}+b^{2}c+bc^{2}+c^{2}a+ca^{2}$

and

$[3,0,0]=a^{3}+b^{3}+c^{3}$
$[2,1,0]=a^{2}b+b^{2}c+c^{2}a$

I think the main idea is clear.

[quote="conceive"] http://www.mathlinks.ro/Forum/viewtopic.php?t=56369[/quote] Everyone is writing it correct there? :?

[quote="conceive"]$(3,0,0)=2(a^{3}+b^{3}+c^{3})$
$(2,1,0)=a^{2}b+ab^{2}+b^{2}c+bc^{2}+c^{2}a+ca^{2}$

and

$[3,0,0]=a^{3}+b^{3}+c^{3}$
$[2,1,0]=a^{2}b+b^{2}c+c^{2}a$

I think the main idea is clear.[/quote]You are clear. Buy why don't we denote open intervals as $[a,b]$ and closed intervals as $a^{b}$ from now on? Your proposal seems quite similar...

[quote="Peter VDD"][quote="conceive"] http://www.mathlinks.ro/Forum/viewtopic.php?t=56369[/quote] Everyone is writing it correct there? :?[/quote]

post number 27 and the following ones

[quote="Peter VDD"]You are clear. Buy why don't we denote open intervals as $[a,b]$ and closed intervals as $a^{b}$ from now on? Your proposal seems quite similar...[/quote]

Nothing in common with intervals.

[quote="Peter VDD"]I suggest $\sum_{sym}$ and $\sum_{cyc}$ which appear in half the topics. Then there can be no confusion. ;)[/quote]

Well, good idea, but I am sure in future there will very useful to work just with multiindexes. I am pretty sure that many inequalities will be about comparing them.. and $\sum$ are not so useful here.. :maybe:

[quote="conceive"][quote="Soarer"]in fact the topics you posted are good examples to see people using [] for symmetric sums.

For the post 13 you quoted again and again, what I see is

[quote][10, 1, 1] + [6, 3, 3) >= 2 [8, 2, 2] by AG mean inequality, and the rest are by murhead:
[8, 2, 2]>= [5, 5, 2] =>
[10, 1, 1] + [6, 3, 3] >= [8, 2, 2] + [5, 5, 2]
[7, 5, 0] >= [6, 5, 1]
[7, 5, 0] >= [6, 4, 2]
collecting the last three gives the desired result. [/quote]

unless you are talking about [6,3,3), which is an obvious typo. :)[/quote]

This is a falsehood!!!!! I am sure that one weak before it was written using these (). If any mod had rewrited it, just to show that I am wrong in my arguments, I am really aghas. :mad:[/quote]

sorry.... but i don't really think any person will dare to modify a post one year ago for just a minor argument in inequalities forum. I guess you might have remembered wrongly.

Maybe a mod corrected the wrong notation? But [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=91319]here[/url]'s another round-bracketed example if you think one example makes you right. ;)

[quote="Peter VDD"]Maybe a mod corrected the wrong notation?[/quote]

Sorry, MY NOTATION IS NOT WRONG, it is just a suggestion. The whole point is to choice to right notation, no matter if it will be the mine or the your. To choice the right for the future of Forum.

[quote="Peter VDD"]
But [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=91319]here[/url]'s another round-bracketed example if you think one example makes you right. ;)[/quote]

OK, one for you.

But the one I wrote I am totally sure it was rewritted. Shame on the one who did it, because we still didn't come to any conclusion at all. If someone did it just to make me believe that I am absolutelly the only one uses this notation, that's not fair.

[quote="Rzeszut"]One of the problems with denotion can be easily solved: $n$-tuples of numbers will not be mistaken with polynomials if we will use [b]darij grinberg[/b]'s notation with [b];[/b] for number arrays, for example $(2;1;0)$ (and, of course, still use [b],[/b] for denoting cyclic and symmetric sums).[/quote]

Good idea, but there is no need for that. Many people will not obey this.

When you write

$(4,2,0)\ge(3,3,0)$

nobody will think you are talking about n-tuples..

On the other hand if you write

$(a,b,c)=(1,5,2)$

everybody will get that you are not talking about inequalities...

[quote="Ilthigore"]Perhaps anyone who wishes to use the opposite notation could put a note in their signatures telling people they are using a different notation.[/quote]

I disagree with that... It would become confusing. After all, it's best to use $\sum_\textrm{cyc}\, \, \textrm{or}\, \, \sum_\textrm{sym}$.

PS: I agree with the fact that we should use the "inverse" notation rather than the one [i]conceive[/i] proposes. But I'll try to avoid it and use the symbols mentioned in the paragraph above.

[quote="Ilthigore"][quote]Sorry, MY NOTATION IS NOT WRONG, it is just a suggestion. The whole point is to choice to right notation, no matter if it will be the mine or the your. To choice the right for the future of Forum. [/quote]
I agree that notation be standardised, but I would suggest the converse to your notation be the standard, with () cyclic and [] symmetric. Until this thread appeared I had assumed this was the actual standard. Perhaps anyone who wishes to use the opposite notation could put a note in their signatures telling people they are using a different notation.[/quote]

No reason. In every post I do I write there what do I mean by $(r,s,t)$.

I agree that notation shoud be the same everywhere.

[quote="perfect_radio"]I disagree with that... It would become confusing. After all, it's best to use $\sum_\textrm{cyc}\, \, \textrm{or}\, \, \sum_\textrm{sym}$.[/quote]

It is not convenient and in addition it is not so good readable.
See:
http://www.mathlinks.ro/Forum/viewtopic.php?t=56369&start=20

[quote="Rzeszut"]Maybe you should add a poll to this topic, [b]conceive[/b].[/quote]

Well, I can do that but it won't change anything.

I think that it is important on how many threads are both notations used. That is important.

[quote="Peter VDD"]I think it is important you just stop spamming here. You have proposed, everyone claims against, so please accept it.[/quote]

Spamming is the last thing I want to do. But I have to rewrite every topic I made using my notation?  :maybe:

My personal opinion is that it doesn't matter, as long as it's clear. So if you keep your notation and add "(note that [...] stands for cyclic sum)" or anything that is no problem for me personally. It might be to others, I don't know.

With spamming I just meant: it has no sense to keep urging for it in this topic, looking at people's comments it looks quite obvious that people are not eager to put your notation as "the standard notation".

I don't see what is inconvenient about $\sum_{cyc}$, but as long as you make clear what you mean I see no problem. [note that using your notation without any word of explaining makes many people assume you're talking about symmetric instead of cyclic, that's not "clear" in my opinion]

How to simplify the cyclic or symmetric sums as soon as possible?

In cyclic ways,it is a bit easier.

For example,we easily know that $ \sum \equal{} \sum_{cyc}$

$ (\sum{x})^3 \equal{} \sum{x^3} \plus{} 3\sum{x^2y} \plus{} 3\sum{x^2z} \plus{} 6xyz$.

$ \sum{x^2}\sum{yz} \equal{} \sum{x^2yz} \plus{} \sum{y^3z} \plus{} \sum{yz^3}$.

But in symmetric ways, it's a bit more difficult.

For example, how to simplify it quickly?

$ (\sum_{sym}{x^2y})^3$


Hi!

$f: [0,1]\to{\mathbb R}$ is a continuous function such that $f(0)=f(1)=0$. We assume that $f\left(x+{3\over{10}}\right)\neq f(x)$ for all $x\in \left[0,{7\over{10}}\right]$.

Prove that $f$ has at least 7 roots on $[0,1]$.

[quote="Rust"]Your condition is not correct. For example $f(x)=ax,0<x<0.7$ and $f(x)=0.7a-\frac{7}{3}(x-0.7)a, 0.7\le x\le 1$.
I think correct condition is $sign f(x+0.3)\not =sign(f(x))$.[/quote]

You're right Rust!

Do you have an idea how to solve the problem with this new condition?


Slove equation:  $ x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \equal{} 2xyzt$, with $ x, y, z, t \in Z$.

[quote="Bugi"][hide="Solution"]
Those numbers can be all even, 2 even or all odd. Trying mod 8 we see that all are even: $ x \equal{} 2x_1$ ... $ t \equal{} 2t_1$.
Dividing by 4, we get: $ {x_1}^2 \plus{} y_1^2 \plus{} z_1^2 \plus{} t_1^2 \equal{} 8x_1y_1z_1t_1$. Then do this again and again and again ...
The only number which is divisible by $ 2^{\infty}$ is $ 0$, so $ x \equal{} y \equal{} z \equal{} t \equal{} 0$.
[/hide][/quote]

[hide]Thank you very much![/hide]


If $ |x|\plus{}x\plus{}y \equal{} 10$ and $ x\plus{}|y|\minus{}y \equal{} 12$, find $ x\plus{}y$.

Answer

Lets take $ x\geq 0$ and $ y\geq 0$

Equations become: $ 2x\plus{}y \equal{} 10$ and $ x \equal{} 12$

$ \Rightarrow 24\plus{}y \equal{} 10$

$ \Rightarrow y \equal{}\minus{}14$

$ y < 0$ which contradicts the choice that $ y\geq 0$. Hence no solutions in this range.

Lets take $ x\geq 0$ and $ y < 0$

Equations become: $ 2x\plus{}y \equal{} 10\,\, (\Rightarrow 4x\plus{}2y \equal{} 20)$ and $ x\minus{}2y \equal{} 12$

$ \Rightarrow 5x \equal{} 32$

$ \Rightarrow x \equal{}\frac{32}{5}$

$ \Rightarrow\frac{64}{5}\plus{}y \equal{} 10$

$ \Rightarrow y \equal{}\minus{}\frac{14}{5}$

These values of $ x$ and $ y$ lie in the chosen range. Hence $ x\plus{}y \equal{}\frac{18}{5}$ is a solution.

Lets take $ x < 0$ and $ y\geq 0$

$ \Rightarrow x \equal{} 12$ which contradicts $ x < 0$

Lets take $ x < 0$ and $ y < 0$

$ \Rightarrow y \equal{} 10$ which contradicts $ y < 0$

Hence $ x\plus{}y \equal{}\frac{18}{5}$ is the only solution.

I'm confused? Where does it say any where that y>0.

It doesn't say $ y\geq 0$ anywhere. I have treated the cases where $ y\geq 0$ and $ y < 0$ seperately. This is because when $ y\geq 0$, $ |y| \equal{} y$ and when $ y < 0$, $ |y| \equal{}\minus{}y$. So by treating the cases seperately, the awkward modulus signs are removed. If you look at my solution, you will see I treated the 4 cases depending on the signs of $ x$ and $ y$ seperately.


[quote="BanishedTraitor"]In a triangle $ ABC$ the points $ A',B',C'$ lie on the sides $ BC,CA,AB$ respectively. It is known that $ \angle{AC'B'} \equal{} \angle{B'A'C}$, $ \angle{CB'A'} \equal{} \angle{A'C'B}$ and $ \angle{BA'C'} \equal{} \angle{C'B'A}$. Prove that $ A',B',C'$ are the midpoints of the corresponding sides.[/quote]

[hide="Hint"]Prove that $ \angle{AC'B'} \plus{} \angle{CB'A'} \plus{} \angle{BA'C'} \equal{} 180$
So $ C'B'CA'$ and $ C'B'A'B$ are parallelograms and the result follows[/hide]


If $r,s,t$ are the complex roots of the equation

\[x^3+2x^2+5x+7 = 0\]

Find $r^2+s^2+t^2$.

A. -6
B. 6
C. -4
D. 4
E. 2

[hide]Method 1:
Hmm...using Newton's Sums:
$s_i$ represents the sum of the $ith$ power of the roots.

$s_1=-2$
$(1)(s_2)+(2)(-2)+(2)(5)=0$, solving for $s_2=-6$. So the answer is $\text{A}$.

Method 2:
$r+s+t=-2$ and $rs+rt+st=5$ from the coefficients.  Squaring the first equation we have $r^2+s^2+t^2+2rs+2rt+2st=4$.  Subtracting twice the second equation from this we have $r^2+s^2+t^2=4-10=-6$.  So the answer is $\text{A}$.[/hide]

Newton Sum will do it.

EDIT: didnt see eryaman. I should look at what others post next time :blush:

So why does it say "complex roots" when the equation clearly does not have all complex roots.

Uh, darktreb???  I'm pretty sure that numbers like $1+0i$ are complex even though they lie on the real line.  If it had said they were imaginary numbers then you would have a point.

The problem could have said "$r$, $s$, and $t$ are complex roots of the equation with one of them being real also...'  but that's a bit unnecessary and trivial.

Strictly complex I guess.

And joml, you're just making it more ridiculous. The problem could have said, r, s, t are the ROOTS of the equation and been just fine. Using the word complex is just a little awkward here.

I believe the word "complex" was added so the negative answers would not be ignored.  When I solved the problem, I got $-6$ and thought, "Wait, that can't be right, the answer has to be positive!" until I looked back up and saw that the roots were [u]complex[/u], thus allowing for $r^2+s^2+t^2$ to be negative.  If it wasn't denoted that the roots were "complex", I probably would have reattempted the problem, which would have only been frustrating when I came out again with the same negative answer.

Mathworld also states that numbers such as $1+0i$ (as joml stated) are complex.  A complex number just takes the form $a+bi$; with $a$,$b$ real.  Personally, I think the addition of the word "complex" makes the problem nicer on slower people, such as myself. ;)

Very good points, no_one_2000. I do see the logic now.

I guess at this point it should go without saying that cubics always have at least one real root, and the complex conjugates are very nice when you add them or add their squares :).

And yes, $\mathbb{R} \in \mathbb{C}$

[quote="darktreb"]$\mathbb{R} \in \mathbb{C}$[/quote]

Actually, $\mathbb{R} \subset \mathbb{C}$  :D

Ya I knew that, but now I finally learned \subset by looking at Yufei's javascript :).

Darktreb, I'll agree that the problem would be fine without the word 'complex' in there since it's a cubic and they are specifying that they want something about all three of them.  If they were asking for, say, the sum of the roots then it would be ambiguous to not specify complex or real.

Just for one note:

I got this problem word by word from a real contest. So, it's not me who did it.  :P


linear algebra
matrix
function
linear algebra unsolved

Let $ A$ be a real symmetric matrix. Let $ M_{A}$ be the matrix of minors of $ A$ (i.e, the $ (i,j)$th element of $ M_{A}$ is the $ (i,j)$th minor of $ A$). What is the differential of $ \mathrm{tr}\: M_{A}$?

The hypothesis "symmetric" seems useless.
1) Let $ spectrum(A)=\{\lambda_{1},\cdots,\lambda_{n}\}$.
If $ A$ is invertible then $ trace(M_{A})=det(A)trace(A^{-1})=\sum\lambda_{i_{1}}\cdots\lambda_{i_{n-1}}=\sigma_{n-1}$, the $ \{\lambda_{i}\}_{i}$-elementary symmetric function of order $ n-1$. It's, up to signum, the coefficient of $ x$ in the $ A$-charpoly. By continuity, this formula is yet valid if $ A$ isn't invertible.
2) We assume that $ A$ is invertible.
Let $ w(A)=trace(M_{A})=uv(A)$ where $ u(A)=det(A)$ and $ v(A)=trace(A^{-1})$. $ Du_{A}(H)=det(A)trace(HA^{-1})$ and $ Dv_{A}(H)=-trace(A^{-1}HA^{-1})$. Thus $ Dw_{A}(H)=det(A)[trace(HA^{-1})trace(A^{-1})-trace(A^{-1}HA^{-1})]$.


As in my avatar is chocolate and my signiture is Homer Simpson's way of saying "I gotta get some!"

How do your avatar and signature entwine?

my avatar and signature entwine in the fact they both have absolutely nothing to do with each other

my avatar and signature AND username intertwine

figure out how

I don t see what u mean.

 :blush:

[size=9]i am kidding[/size]

[quote="Xantos C. Guin"]It is fairly obvious that my avatar and signature are somewhat related.[/quote]
really? how?  :rotfl:

my avatar is cool, and my signature is cool.  see, they are related.  :D

My avatar is a lollerskater.

I have no signature. I should go make one.

my signature is dull and my lack of avatar shows how non-creative i am. that's entwined.  :D

i love that thing in my avatar.....

but i have to think about a signature as welll.........

 :D

My avatar is just too cool for my signature.

I never changed my avatar, but I change my signature frequently. They do not have any connections at all!

Before I posted this post, my avatar and sig had nothing to do with each other.  I will have probably changed that by the time you read this.

no avatar..
but my signature shows some curiosity...
should i or should i not never give up?

yeah my avatar is a thing and my signature never shows up. Can u guess how they're entwined?

My avatar is about obvious things.  My signature is about the dumb mistakes people like me make.  Captain obvious is my favorite superhero--- W000t.


1.
$ \sum{i=1]^k \frac{x_i}{i}\le\sum_{i=1}^n \frac{x_n}{i}}$ :huh:

2.
Set $ a=b=c$ and see that equality holds, so the inequality isn't true :D

[quote="bertram"]
2.
Set $ a \equal{} b \equal{} c$ and see that equality holds, so the inequality isn't true :D[/quote]
$ a \equal{} b \equal{} c$ gives that the left hand side is greater than $ 0$ and the right hand side is $ 0$... so it holds  :wink:
[hide="Solution"]
Without loss of generality, say that $ a\ge b\ge c\implies \max (|a^2 \minus{} b^2|, |b^2 \minus{} c^2|, |c^2 \minus{} a^2|) \equal{} a^2 \minus{} c^2$.  So, we must show that
\[ a^2 \plus{} b^2 \plus{} c^2 > \sqrt {3}(a^2 \minus{} c^2)\implies b^2 \plus{} c^2(1 \plus{} \sqrt {3}) > a^2(\sqrt {3} \minus{} 1)
\]

\[ 2a^2 < b^2(1 \plus{} \sqrt {3}) \plus{} c^2(4 \plus{} 2\sqrt {3})
\]
Now, we will attempt to show that
\[ 2(b \plus{} c)^2\le b^2(1 \plus{} \sqrt {3}) \plus{} c^2(4 \plus{} 2\sqrt {3})
\]

\[ b^2(\sqrt {3} \minus{} 1) \plus{} c^2(2 \plus{} 2\sqrt {3})\ge 4bc
\]

\[ b^2(\sqrt {3} \minus{} 1) \plus{} c^2(2 \plus{} 2\sqrt {3})\ge 2bc\sqrt {2(\sqrt {3} \minus{} 1)(\sqrt {3} \plus{} 1)} \equal{} 4bc
\]
by AM-GM and so $ b^2(1 \plus{} \sqrt {3}) \plus{} c^2(4 \plus{} 2\sqrt {3}) \ge 2(b \plus{} c)^2 > 2a^2$ (notice that the last part of the inequality follows from the triangle inequality) and thus we are done.  [/hide]
EDIT: To bertram: I don't have an equality case... the triangle inequality is strictly without an equality (I am assuming a non-degenerate triangle here.)

[quote="The QuattoMaster 6000"][quote="bertram"]
2.
Set $ a \equal{} b \equal{} c$ and see that equality holds, so the inequality isn't true :D[/quote]
$ a \equal{} b \equal{} c$ gives that the left hand side is greater than $ 0$ and the right hand side is $ 0$... so it holds  :wink:
[hide="Solution"]
Without loss of generality, say that $ a\ge b\ge c\implies \max (|a^2 \minus{} b^2|, |b^2 \minus{} c^2|, |c^2 \minus{} a^2|) \equal{} a^2 \minus{} c^2$.  So, we must show that
\[ a^2 \plus{} b^2 \plus{} c^2 > \sqrt {3}(a^2 \minus{} c^2)\implies b^2 \plus{} c^2(1 \plus{} \sqrt {3}) > a^2(\sqrt {3} \minus{} 1)
\]

\[ 2a^2 < b^2(1 \plus{} \sqrt {3}) \plus{} c^2(4 \plus{} 2\sqrt {3})
\]
Now, we will attempt to show that
\[ 2(b \plus{} c)^2\le b^2(1 \plus{} \sqrt {3}) \plus{} c^2(4 \plus{} 2\sqrt {3})
\]

\[ b^2(\sqrt {3} \minus{} 1) \plus{} c^2(2 \plus{} 2\sqrt {3})\ge 4bc
\]

\[ b^2(\sqrt {3} \minus{} 1) \plus{} c^2(2 \plus{} 2\sqrt {3})\ge 2bc\sqrt {2(\sqrt {3} \minus{} 1)(\sqrt {3} \plus{} 1)} \equal{} 4bc
\]
by AM-GM and so $ b^2(1 \plus{} \sqrt {3}) \plus{} c^2(4 \plus{} 2\sqrt {3}) \ge 2(b \plus{} c)^2 > 2a^2$ (notice that the last part of the inequality follows from the triangle inequality) and thus we are done.  [/hide][/quote]

:( The $ a^2,\sqrt{3}$ and $ a^2\minus{}b^2$ made me think of the Hardwiger-Finslinger Inequality, but you have equality there... reading comprehension for the win.

[quote="hell_ever"]q 1) bertram i can't understand what ur trying to imply can u explain?

(and the last term on the lhs of the first expression is $ x_{n^{2}}$)[/quote]

I used a ] instead of a }. :(

How is $ x_{n^2}$ defined? After all $ n^2>n$ for almost all integers...


limit
function
real analysis
real analysis unsolved

Let a sequence $(x_{n})_{n\in\mathbb{N}}$ with $x_{0}\geq 0$ defined by   $\sqrt{x_{n}}=\frac{x_{n}-x_{n+1}+1}{x_{n+1}-x_{n}}$. Evaluate $\lim_{n\to\infty}{\frac{x_{n}^{6}}{n^{4}}}$.

[hide="Here is an idea,"]We have $x_{n+1}-x_{n}=\frac{1}{1+\sqrt{x_{n}}}$. I declare $n$ to be a continuous variable and replace $x_{n+1}-x_{n}$ with $\frac{dx}{dn}$ with a blatant disregard for mathematical rigour. The inverse function $n(x)$ satisfies $\frac{dn}{dx}=1+\sqrt{x}$. This integrates to $n(x)=x+\frac23 x^{3/2}+C$, hence $\frac{x_{n}^{6}}{n^{4}}\to \frac{81}{16}$. [/hide] but does it work?

First of all, $x_{n}\to \infty$ (easy to prove).

Next, by Stolz-Cesaro, $\lim_{n \to \infty}\frac{x_{n}^{3/2}}{n}= \lim_{n \to \infty}\left( x_{n+1}^{3/2}-x_{n}^{3/2}\right)$.

Now put $x_{n+1}= x_{n}+\ldots$ (see mlok's post), then compute $\lim_{x \to \infty}\left[ \left( x+\frac1{1+\sqrt x}\right)^{3/2}-x^{3/2}\right]$ using power series :) You'll get $\frac32$, so the answer is $\frac{81}{16}$.

Problems of similar nature have been proposed on Romanian olympiads.

After rationalise it works without power series..

[hide="only for perfect_radio"][quote="perfect_radio"]
Problems of similar nature have been proposed on Romanian olympiads.[/quote]

..sau la ASE  :mad:
[/hide]


are we allowed to use calculators in AIME and AMC 10/12

Not anymore (although you coulnd't use them for AIME anyway).

AIME never allowed calculator usage. But AMC 10/12 is beginning to ban calculator use beginning in the 2008 competitions.

But I find that I never really use calculators anyway, except for long calculations, which I could do by hand but I'm too lazy. Plus, now they probably won't have those kinds of problems anymore.

Frquently Asked Question AMC 10/12 #7, on
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


Prove that for real nos. $ a,b,c>2$ the relation holds good:
$ a\plus{}b\plus{}c<abc$
 :P

[quote="Sunkern_sunflora"]Prove that for real nos. $ a,b,c > 2$ the relation holds good:
$ a \plus{} b \plus{} c < abc$
 :P[/quote]
Let $ c\equal{}max (a,b,c)$ then $ RHS>4c>3c$ done! :maybe:

I edit my post, sorry!

Please clarify how you conclude that $ R.H.S. > 4c$?? So far I can conclude is $ R.H.S. > 4a$.....where $ a\equal{}min(a,b,c)$ :maybe:

P/s: What if the same problem is asked if $ a,b,c > 1$? Does it hold? :maybe:

Well $ abc>2\cdot 2\cdot c\equal{}4c$ because $ a>2$ and $ b>2$ :wink:

For P.S.: Try $ a\equal{}b\equal{}c\equal{}1.5$

Ooh! I was so stupid! Sorry Allnames! Thank u Bugi! :oops:

Can you modify the problem Bugi?


If $ a,b,c>0,$ such that $ a\plus{}b\plus{}c\equal{}1,$ prove that $ \displaystyle\sum\frac{ab}{(1\minus{}a)(1\minus{}b)}\geq\frac{3}{4}$

From $ a\plus{}b\plus{}c\equal{}1$, the problem changes to $ \sum \frac{ab}{(b\plus{}c)(a\plus{}c)}\equal{}\sum \frac{a^2b^2}{ab(b\plus{}c)(a\plus{}c)}\ge \frac{3}{4}$

From Cauchy Schwarz

$ \sum ab(b\plus{}c)(a\plus{}c).\sum \frac{a^2b^2}{ab(b\plus{}c)(a\plus{}c)} \ge (ab\plus{}bc\plus{}ac)^2$
$ \sum \frac{a^2b^2}{ab(b\plus{}c)(a\plus{}c)}\ge\frac{(ab\plus{}bc\plus{}ac)^2}{\sum ab(b\plus{}c)(a\plus{}c)}$

So, we need to prove
$ \frac{(ab\plus{}bc\plus{}ac)^2}{\sum ab(b\plus{}c)(a\plus{}c)}\ge \frac{3}{4}$
$ 4\sum (ab)^2\plus{}8abc(a\plus{}b\plus{}c)\ge 3\sum (ab)^2\plus{}6abc(a\plus{}b\plus{}c)$
$ (ab\plus{}bc\plus{}ac)^2\ge 0$, which is true

$ QED$

[quote="nickolas"]If $ a,b,c > 0,$ such that $ a \plus{} b \plus{} c \equal{} 1,$ prove that $ \displaystyle\sum\frac {ab}{(1 \minus{} a)(1 \minus{} b)}\geq\frac {3}{4}$[/quote]
We got to prove that
$ \displaystyle\sum ab(1\minus{}c)\geq\frac{3}{4}(1\minus{}a)(1\minus{}b)(1\minus{}c)$
Or
$ 4[(ab\plus{}bc\plus{}ca)\minus{}3abc]\geq3(ab\plus{}bc\plus{}ca\minus{}abc)$
Which is
$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\geq9$
But we can easily prove that
$ (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\geq9$
But $ a\plus{}b\plus{}c\equal{}1$ so we have a Q.E.D. :P


If a year had 364 days, then the same calendar could be used every year by changing only the year.  A "regular" year has 365 days and a leap year has 366.  1980 was a leap year and leap years occur every 4 years.  Steffie has a calendar of 1981.  What would be the next year that she could use this calendar by merely changing the year?

[hide="answer"]
Let's say that 1981 started with a monday, and each following year the first day is one more.
So 1982 starts with a tuesday.
1983 starts with a wednesday, and 1984(leap year) starts with a friday, 1985 starts with a saturday, 1986 starts with sunday and finally 1987 starts with a monday.

so our answer is $ \boxed{1987}$
[/hide]


Putnam
function
vector
topology
advanced fields
advanced fields unsolved

I had no idea where to post this. It's B3 from Putnam 1962.

Show that a convex and open set which contains the point $P$ but doesn't contain any ray from $P$ must be bounded. Does it remain true if we drop the openness condition?

$S = {(x,y) | 0<x<1} \cup {(0,0)}$ it is convex, it does not contain any ray from the origin, however, it is unbounded.

Yeah, this is very similar to the counterexample I gave to the question. What about the proof of the first assertion?

Let $P=0$ and $\Omega\in\mathbb{R}^n$ is mentioned open set.
Consider Minkowski function $f(x)\equiv \inf\{\frac{1}{r}\ :\ rx\in\Omega\}$, $f(0)=0$.
We have $f(ax)=af(x)$ for $\forall a\geq 0$. And it is easy to show that $f(x+y)\leq f(x)+f(y)$. It follows that $f(x)$ is convex function in $\mathbb{R}^{n}$. It implies that $f(x)$ is continuous $\Rightarrow$ there exists $k=\inf_{|x|=1}f(x)>0$, since $f(x)\neq 0$ for $x\neq 0$ ($\Omega$ doesn't contain any ray from $P$). Hence $\Omega\in B_{1/k}(0)$.

I was knowing the result on its contraposed (?) form, that is :
If $C$ is a open convex and unbounded set of a normed vector space of finite dimension, then from each point $P \in C$ we may find a ray contained in $C$.

The result is no longer true if the set $C$ is not supposed to be open, but when I looked at the problem I proved that, in $|R^n$, the set $C$ contains at least one half-line.

Pierre.

[quote="the game"]What does the term "ray" mean?[/quote]
Half of a line, that is, a straight line extending from a point.

[quote="Myth"]
Consider Minkowski function $f(x)\equiv \inf\{\frac{1}{r}\ :\ f(rx)\in\Omega\}$, $f(0)=0$.
[/quote]
I think that Myth meant  $f(x)\equiv \inf\{\frac{1}{r}\ :\ (rx)\in\Omega\}$, $f(0)=0$.
and i'd like to ask whether you used the openness condition
in addition: very nice solution,Myth

Thank you for correction. Of course, it was a typo.
The opennes is required to well define $f$. Otherwise, it is possible that there is no such $r$ that $rx\in\Omega$, so $\inf$ is undefined.


modular arithmetic
number theory
relatively prime
number theory proposed

Determine the largest integer $ n$ with the following property:
Whenever $ p\equiv q\bmod n$ and $ s\equiv t\bmod n$, then also $ p^s\equiv q^t\bmod n$.

well...
if $ t,s$ can be zero then the problem is trivial(choose $ p \equal{} q \equal{} n,s \equal{} 0,t \equal{} n$ to get $ n^n \equiv 1 \pmod{n}$...)
but if $ t,s$ must be positive then the problem is trivial for odd $ n$.
(choose $ p \equal{} \minus{} 1,q \equal{} n \minus{} 1,s \equal{} 1,t \equal{} n \plus{} 1$ to get $ \minus{} 1 \equiv 1 \pmod{n}$)
if $ n$ is even then we can easily prove that $ n$ is square free.
(let $ l$ be a prime dividing $ n$ and suppose that $ l^2 |n$ then choose $ p \equal{} l,q \equal{} l,s \equal{} 1,t \equal{} n \plus{} 1$ to get $ l \equiv l^{n \plus{} 1} \pmod{n}$ which implies $ l \equiv l^{n \plus{} 1} \pmod{l^2}$ which is impossible)
so if we don't consider the case $ n \equal{} 2$ then $ n$ must be of the form $ 2n'$ where $ n' > 1$ is odd.
now obviously for every integer $ s$ relatively prime to $ n$ we have :
$ s \equiv s \pmod{n}$
$ 1 \equiv n \plus{} 1 \pmod{n}$
it implies
$ s^n \equiv 1 \pmod{n}$
also we can consider a primitive root like $ r$ modulo $ p$ which is a divisor of $ n$ and then since we can easily find an integer like $ r'$ which is congruent to $ r$ modulo $ p$ and is relatively prime to $ n$ then:
$ r'^n \equiv 1 \pmod{n}$ which implies $ r'^n \equiv 1 \pmod{p}$ or $ r^n \equiv 1 \pmod{p}$ so we must have $ p \minus{} 1 | n$ since $ r$ is a primitive root modulo $ p$.
and since $ 2 | p \minus{} 1$ and also $ 2 \parallel{} n$ we must have $ \forall p$ which divides $ n$ :
$ p \equiv \minus{} 1 \pmod{4}$
so for example $ 5$ isn't a divisor of $ n$ and we must have $ 5^n \equiv 1 \pmod{n}$.
but as i mentioned for all prime $ p$ we must have $ p \minus{} 1 |n$ and since $ n$ has no prime divisors of the form $ 4k \plus{} 1$ so all odd prime divisors of $ p \minus{} 1$ must be of the form $ 4k \plus{} 3$.
which means that $ 11$ can't divide $ n$ since if it divides it then we must have $ 5 |n$.
also we can easily prove that your problem(for the case: $ n$ is even) and this problem are equivalent :
Detemine the largest square free integer $ n$ for which if $ s$ is an integer relatively prime to $ n$ then $ s^n \equiv1 \pmod{n}$.
here is one of them $ n \equal{} 2 \times 3 \times 7 \times 43$.
u can easily see that for each prime $ p$ which divides $ n$ we have the following property : $ p \minus{} 1 | n$.
now we want to prove that this is our answer.
let $ p$ be the smallest odd prime factor of $ n$ we want to prove that $ 3 \equal{} p$
we know that $ p \minus{} 1 | n$ and since $ p$ was the smallest prime factor of $ n$ so the number $ p \minus{} 1$ doesn't have a prime divisor other than $ 2$ so $ p \minus{} 1 \equal{} 2^a$ but we know that $ 2 \parallel{}n$ which means that $ a \equal{} 1$ and so $ p \equal{} 3$.
until now we have proven that $ 6 | n$
let $ p$ be the prime factor of $ n$ which is greater then $ 3$ but not greater than the other prime factor.
since $ p \minus{} 1 | n$ and $ p \minus{} 1$ hasn't any other prime factor greater than $ 3$ so we get $ p \minus{} 1 | 6$ which will imply $ p \equal{} 7$.
repeat the same story for the prime $ p$ greater than $ 7$ but lesser than the other prime we get $ p \minus{} 1 | 42$
then by checking all cases we obtain that $ p$ must be equal to $ 43$.
suppose that there exists another prime which divides $ n$ and different from $ 2,3,7,43$
let this prime be $ p$ then we must have
$ p \minus{} 1 | 2 \times 3 \times 7 \times 43$
by checking all cases u can see that such a prime(different from $ 2,3,7,43$) doesn't exist and so we r dOnE!


[quote="anirudh"][quote="Padmavathi"]I like this one...
A cuckoo clock chimes the number of each hour (eg. 8 o clock, 8 chimes). It also chimes 2 times per half an hour and 1 time per quarter hour.  How many times does it chime in
a) one day
b)one month[/quote]

[hide]
For 12 hours: 78
78*2 for a day so:
$78\cdot2+23\cdot2+46$ for the quarter hours and half hours which equals $7440$[/hide][/quote]
Where'd you get the 78 from?

I know that is very confusing...I'll try to fix it...

the 78 is from:
$1+2+3+4+5+6+7+8++10+11+12$

and the 7440 is after multiplying by 30 (which i hope is the number of days in the month Padmavathi wants)

and then i will post answer

[hide="day"]248[/hide]
[hide="month"]7440[/hide]

I don't think they make these, but if the cuckoo was programmed for a 24-hour day, then it would be...

[hide]Day: $392$

Month: $11760$[/hide]

[quote="parm12"]How many days in your "month"?  :P[/quote]
30 days.

it may be 28, 29, 30, or 31 :lol:
the author should make it clear :D

Hi!
They DO make such cuckoo clocks, as I have one in my house direct from Germany and hence I was quite surprised when my math teacher(If u can call her that) said there was some "special formula" to find out the number of times it chimes. I don't think so...is there?? I mean I just add everything by Trial and Error...but she claims there s a formula...have been searching but no success...
Re: month, 30 days only!  :D


logarithms
limit
calculus
Euler
real analysis
real analysis unsolved

Let $x_n$ be the minimum value of $x$ which gives extreme value for the following function.
\[y=x(x-1)(x-2)\cdots(x-n)\ (n=1,2,\cdots)\ \]
Find $\lim_{n\to\infty} x_n\ln n$.

An interesting question.

We are looking for the smallest $x$ where $f(x)=\prod_{k=0}^{n}(x-k)$ has an extreme. Alternatively we may answer the same question for $g(x)=\sum_{k=0}^n \ln(x-k)$ or, applying calculus, the following: what is the smallest root of $1/x=\sum_{k=1}^{n}\frac{1}{k-n}$. Now assume, that $x_n$ fullfilles this equation, then clearly $\frac{1}{2}+\ldots+\frac{1}{n}\leq \frac{1}{x_n}-\frac{1}{1-x_n}\leq 1+\ldots+\frac{1}{n-1}$. By the well known convergence of the difference of the harmonic sum up to n  with $\log(n)$ to the Euler constant $\gamma$, we have for large $n$ (avoiding the use of $\delta$s and $\epsilon$s) that $\log n+\gamma-1\leq \frac{1}{x_n}-\frac{1}{1-x_n}\leq \log n+\gamma$. Since $x_n\to 0$ as $n\to\infty$ and  $\frac{1}{1-x_n}\to 0$ it follows, that $\log n \cdot x_n\to 1$.

Is this right?

If yes, can we discover second order behaviour of $x_n\log n$ by this method?

Leonard


I saw many students on this website use AMC B as their part of index score.
I do not think this is a good idea.
#6 of selection process for USAMO says:
[[u]u][quote]6. To adjust for variations in contest diffi culty, the number of
students selected from A & B contests will be proportional
to the number of students who took the A & B Contests.[/quote][/u]In general, there are fewer students taking AMC B, but relatively more higher score(strong) students. In this case, when being selected for USAMO, it will be more competitive for students who took AMC B, because there are proportionally fewer students.
For example 100000 students took AMC A, and there will be 200 for USAMO(fewer higher score students, like 1000 to compete 200, 5---1).
However, 20000 students took AMC B, there will be 40 for USAMO(more higher score students, like 300 to compete 40, 7.5---1).
So I think it's harder for students who took AMC B to get into USAMO.
Do my words make sense? Am I right? :?  :?:

Unless I am quite mistaken, that quote refers to the regular and alternate AIME's, rather than AMC A and B.

And you can't take both AIME's.

Since the new qualifying rules were introduced in 2002, the cutoffs for the 12A and 12B have always been the same.  The only time there have been separate cutoffs were last year, when there was an AIME-I cutoff and an AIME-II cutoff.  That happened when there were pronounced differences in the difficulty of the two AIMEs.  Thus unless there is a big, big difference between the two AMC12's, this trend will likely continue, and your index will be the same regardless of which AMC you took.

Plus, imagine the headache that would ensue for the AMC folks if they had to sort through everyone's 12A and 12B scores, with some people qualifying through only one test, some people through both, etc.

I think those rules that they list are rather bendable; they seem to try to strike a balance between simplicity and fairness.  Thus far, I think they've been very successful with this.  As an example, in 2002, the rules originally stated that the floor applied only to people who took the AMC10.  After the original list of qualifiers came out, though, it became apparent that some people had been denied qualification simply because they took the 12, and they would have made the floor if they had taken the 10.  As a result, they retroactively instituted the floor rule as it stands now.

I know that is OK, but I am wondering if the committee will take the higher score you made to be your part of index, and not both count. If they do so, then it is bad news.

[quote]Since the new qualifying rules were introduced in 2002, the cutoffs for the 12A and 12B have always been the same.  The only time there have been separate cutoffs were last year, when there was an AIME-I cutoff and an AIME-II cutoff.  That happened when there were pronounced differences in the difficulty of the two AIMEs.  Thus unless there is a big, big difference between the two AMC12's, this trend will likely continue, and your index will be the same regardless of which AMC you took.[/quote]
I do agree with you. Thank you for your idea, Sly Si. But I somehow feel there is a little unfair---Students who take a relatively easier AMC will be a little easier to roll into USAMO than those who take a harder one. :oops:
Then :idea:  :arrow: take both is the best idea.

the # of AMC B takers this year was only slightly lower than A takers

in my school    only like 20 students took AMC A    and about 100 students took AMC B .......................

Perhaps I'm missing something, but it doesn't seem the least bit unfair to me. Everyone is allowed to take both AMCs and the higher score is taken. Furthermore, generally the difficulty level between the two AMCs is not dissimilar enough to cause a major shift in the would-be qualifiers. You can use either AMC 12 score for purposes of index calculation. I'm not sure it's a matter of the CAMC's willingness to do more work to determine qualified individuals for USAMO selection. To me it seems like there's absolutely no reason to take proportional numbers since a vast majority of USAMO quals take both in the first place, and everyone that wishes to do so has the opportunity to try A and B.

The AIME is completely different: in two consequtive scores (especially near the qualifying floor) there are thousands of test takers. You are allowed to take only one AIME, and the reputation of the AMC has stemmed partially from its ability to distinguish the best problem solvers despite incongruency in the difficulty of its exams. Because USAMO qualification is a prestigous honor, it is, arguably, the CAMC's responsibility to choose qualifiers which have best shown their problem solving abilities and to give each aspirant a fair chance. That, I think, is why "proportional number of qualifiers" policy has never been applied to the AIME and not the AMC.


function
trigonometry
integration
limit
complex analysis
complex analysis theorems

[b]Theorem[/b] Every function $f(x) = x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_n$ has a real or complex root.

[b]Sketch of Proof[/b]
Suppose $x=r(\cos\theta+i\sin\theta)$ ($x^k=r^k(\cos k\theta+i\sin k\theta)$) such that
$f(x) = P+Qi$
where
\[P = r^n \cos n\theta + \cdots,\quad Q=r^n \sin\theta + \cdots\]

If there is a pair of $(r,\theta)$ such that $P^2+Q^2=0$, the theorem is proved.

Consider the function $U = \arctan \frac{P}{Q}$
then
\[\frac{\partial U}{\partial r}=\frac{\frac{\partial P}{\partial r}Q-P\frac{\partial Q}{\partial r}}{P^2+Q^2}\quad \frac{\partial U}{\partial \theta}=\frac{\frac{\partial P}{\partial\theta}Q-P\frac{\partial Q}{\partial\theta}}{P^2+Q^2}\]
and
\[\frac{\partial^2 U}{\partial r\partial \theta} = \frac{H(r,\theta)}{(P^2+Q^2)^2}\]
where $H(r,\theta)$ is a continuous function of $r,\theta$.

Finally, let us consider
$ I_1 = \int_0^R dr \int_0^{2\pi} \frac{\partial^2 U}{\partial r\partial \theta}d\theta $ and $ I_2 = \int_0^{2\pi} d\theta \int_0^R \frac{\partial^2 U}{\partial r\partial \theta}dr $
where $R$ is positive.

If $P^2+Q^2\neq 0$, then the integrand is continuous and it must holds that $I_1=I_2$. Then we prove that the equality fails for $R$ large enough.

Anyone who likes to complete the proof?

Hint:

It is clear that $I_1 = 0$
Now try to prove that $\lim_{r\to\infty} \frac{\partial U}{\partial \theta} = -n$ and it tends [u]uniformly[/u] to $-n$ about $\theta$.
Then show that $\lim_{R\to\infty}I_2 = -2\pi n$