Let $ m$ and $ n$ be positive integers. Prove that $ 6m | (2m \plus{} 3)^n \plus{} 1$ if and only if $ 4m | 3^n \plus{} 1$

$ 6m|(2m \plus{} 3)^n \plus{} 1 \Leftrightarrow 6m|2^nm^n \plus{} 3^n \plus{} 1 \Leftrightarrow 3|2^nm^n \plus{} 1$ and $ 2m|3^n \plus{} 1$. So we are proving $ 3|2^nm^n \plus{} 1$ and $ 2m|3^n \plus{} 1 \Leftrightarrow 4m|3^n \plus{} 1$.
Suppose that $ m$ is even, then $ 4m\nmid 3^n \plus{} 1$, because $ 8\nmid 3^n \plus{} 1$. Also, $ 2m|3^n \plus{} 1$ implies that $ n$ is odd, and with this http://www.mathlinks.ro/viewtopic.php?t=199361 in mind  $ m \equal{} 2(3k \plus{} 1)$, but then $ 3\nmid 2^nm^n \plus{} 1$, so that is also imposible.

Hence, $ m$ is odd.
Then from $ 4m|3^n \plus{} 1$ we get $ n$ is odd so $ m\equiv_3 1$, so $ 2m|3^n \plus{} 1$ and $ 3|2^nm^n \plus{} 1$.
From $ 3|2^nm^n \plus{} 1$ and $ 2m|3^n \plus{} 1$, we get that $ n$ can't be even (because then $ 3|m^n \plus{} 1$, which is imposible when $ n$ is even), so $ n$ is odd, and $ 4m|3^n \plus{} 1$. With this, we concluded the equivalence.

Bye

$ 3|(2m+3)^n+1$ so we have $ m \equiv 1(mod3)$ and also n is odd.
we know that $ m| 3^n+1$ but odd prime diviser of $ 3^n+1$ has the form $ 3k+1$ so if m is even we have $ 4|m$
so we have $ 8|(2m+3)^n+1$ so we have $ 8|3^n+1$ which contradicts(n is odd)
so m is odd. but we have $ 2m|(2m+3)^n+1$ so $ 2m|3^n+1$ so $ 4m|3^n+1$

I really don't understand SpongeBob 's solution so Can anybody post another solution ? :(

My idea:
Lemma: if $ p\in \mathbb P$ satisfies $ p|3^n \plus{} 1$ then $ p$ does not have the form $ 3k \plus{} 2$
Applying the lemma:
 $ 6m | (2m \plus{} 3)^{n} \plus{} 1 \leftrightarrow 6m|(2m)^n \plus{} A \plus{} 3^n \plus{} 1$
(we can find out $ A$ from Newton expansion, and noting that $ 6m|A$)
$ \leftrightarrow 6m|(2m)^n \plus{} 3^n \plus{} 1$ (1)
$ \rightarrow$
(1) implies $ m|3^n \plus{} 1$ and $ 3|(2m)^n \plus{} 1$ thus $ n$ is an odd hence $ 4|3^n \plus{} 1$.We have done!
$ \leftarrow$
if $ 4m|3^n \plus{} 1 \leftarrow m \equal{} 3k \plus{} 1$ (the lemma),and $ n$ is an odd.
$ \rightarrow 6m|(2m)^n \plus{} 3^n \plus{} 1$
Completing my idea :lol: .
P/s:I think my idea  is true but I can't make a complete solution. Sorry  :(

[quote="broniran"][quote="Allnames"]My idea:
Lemma: if $ p\in \mathbb P$ satisfies $ p|3^n \plus{} 1$ then $ p$ does not have the form $ 3k \plus{} 2$[/quote] What about $ n \equal{} 2$?[/quote]

and also for $ n\equal{}10$....
I think that n must be odd.

Yes n must be odd ,but it is easy to prove !

$ 6m | (2m\plus{}3)^{n}\plus{}1$ implies $ 2 | 3^{n} \plus{}1$, $ 3 | (2m)^{n} \plus{}1$, and $ m | 3^{n} \plus{}1$.
First one is trivial and from the second we can easily verify $ n$ is odd and $ m \equiv 1(mod 3)$.
From this $ 3^{n} \plus{}1 \equiv  4 (mod 8)$, and therefore $ 4m | 3^{n}\plus{}1$.

$ 4m | 3^{n}\plus{}1$ implies $ 4 | 3^{n} \plus{}1$ and $ m | 3^{n} \plus{}1$.
First one implies $ n$ is odd and $ 3^{n}\plus{}1 \equiv 4(mod 8)$. With this and given condition $ m$ must be odd.
If $ p$ is an odd prime number which divide $ m$, then $ p | 3^{n} \plus{}1$.
$ \left( \frac{\minus{}3}{p} \right) \equal{}1$
$ p \equiv 1 (mod 3)$
$ m \equiv 1 (mod 3)$
Therefore  $ 3 | (2m)^{n} \plus{}1$ , $ 6m |(2m\plus{}3)^{n}\plus{}1$ and we are done.

[quote=imfeyn]  $ m | 3^{n} +1$.
From this $ 3^{n} \plus{}1 \equiv  4 (mod 8)$, and therefore $ 4m | 3^{n}\plus{}1$.
[/quote]
If $m$ is even?


algebra
polynomial
search
algebra unsolved

Consider the polynomial
$ T(x)\equal{}x^3\plus{}14x^2\minus{}2x\plus{}1$
Show that there exists a natural number $ n>1$ such that $ T^{(n)}(x)\minus{}x$ is divisible by 101 for all integers x
where $ T^{(n)}(x)\equal{}T(T(...T(x)...))$ --> we have n $ T(x)$
I need your help  :blush:

See the general problem .
http://www.mathlinks.ro/viewtopic.php?search_id=1100931740&t=209092


function
analytic geometry
algebra
polynomial
geometry
geometric transformation
reflection

Given the points $ A(0,5)$, $ B(3,7)$, and $ C(6,2)$, find the point $ P$ on the x-axis that minimizes $ (PA)^{2}+(PB)^{2}+(PC)^{2}$.

[hide]I think it should be equidistant from the two points that are closest to the $ x$-axis, $ (0,5)$ and $ (6,2)$.

The equation of the line through those two points is $ y=-\frac{x}{2}+5$, and the perpendicular bisector is $ y=2x-\frac{5}{2}$.

Set $ y=0$...

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

$ x=\frac{5}{4}$.

I think $ P\left(\frac{5}{4},0\right)$, but I'm not entirely sure.[/hide]

Well the first poster wasn't really sure what he was doing, so I think you're right, DFT. Although I was close... :)

[quote="da ban man"]umm no actually both answers are wrong...I think $ (3,0)$ is right..[/quote]

Yeah, you are right... DFT made an error. The polynomial should be
$ 3x^{2}-18x+123$

EDIT: Could someone solve this follow-up? I thought that the original problem was this:
Follow-up: Given the points $ A(0,5)$, $ B(3,7)$, and $ C(6,2)$, find the point P on the x-axis that minimizes $ PA+PB+PC$.

I tried using [hide="this"]reflections[/hide] but that didn't work so I tried using [hide="this"]coordinates[/hide] but that made it worse.

EDIT EDIT:
Another follow-up
Given the points $ A(0,5)$, $ B(3,7)$, and $ C(6,2)$, find the point P on the y-axis that minimizes $ (PA)^{2}+(PB)^{2}+(PC)^{2}$.
What if P can be any point on the x y plane?

I am able to solve these, but I have no idea how to start if we wanted to minimize $ PA+PB+PC$.

[quote="DiscreetFourierTransform"]Well, we can use the distance formula and some algebra.

[hide="Solution"]
Let point P be (X, 0).
The distances from points A, B, and C to point P are (by the distance formula)
PA: $ \sqrt{(0-x)^{2}+(5-0)^{2}}$
PB: $ \sqrt{(2-x)^{2}+(7-0)^{2}}$
PC: $ \sqrt{(6-x)^{2}+(2-0)^{2}}$

We want $ (PA)^{2}+(PB)^{2}+(PC)^{2}$, and we want to minimize this function.
Well, this function is just

$ f(x) = \sqrt{(0-x)^{2}+(5-0)^{2}}^{2}+\sqrt{(2-x)^{2}+(7-0)^{2}}^{2}+\sqrt{(6-x)^{2}+(2-0)^{2}}^{2}$
or just
$ f(x) = 3x^{2}-16x+118$,
if I did my algebra right.

The x coordinate of the minimum is $ \frac{-b}{2a}$, or $ \frac{16}{6}= \frac{8}{3}$
Isn't that right?
That's a different answer than the first poster..
[/hide][/quote]

So it should have been:

[hide="click for answer"]

Let point P be (X, 0).
The distances from points A, B, and C to point P are (by the distance formula)
PA: $ \sqrt{(0-x)^{2}+(5-0)^{2}}$
PB: $ \sqrt{(3-x)^{2}+(7-0)^{2}}$
PC: $ \sqrt{(6-x)^{2}+(2-0)^{2}}$

and the function is:

$ f(x) = \sqrt{(0-x)^{2}+(5-0)^{2}}^{2}+\sqrt{(3-x)^{2}+(7-0)^{2}}^{2}+\sqrt{(6-x)^{2}+(2-0)^{2}}^{2}$
= $ x^{2}+25+x^{2}-6x+9+49+x^{2}-12x+36+4$
= $ 3x^{2}+6x+123$

so:

$ -b/2a =--18/(2*3) = 3$

$ (3, 0)$

[/hide]

Interesting afterthoughts:
Given N points (A, B) , (C, D) ... (Y, Z), find a point P on the X axis such that the sum of the squares of the distances from point P to each of the N points is minimized.

Well, we know that by the distance formula, the distance from P (X, 0) to the first of the points is $ \sqrt{(A-X)^{2}+B^{2}}$

We can repeat this for the other points:
$ PB = \sqrt{(C-X)^{2}+D^{2}}$

and so on.

Because we want the squares of these distances, we can just do this:
$ f(x) = (A-X)^{2}+B^{2}+(C-X)^{2}+D^{2}+(E-X)^{2}+F^{2}$
expanding each row, we get
$ f(x) = a^{2}-2ax+x^{2}+b^{2}+c^{2}-2cx+x^{2}+d^{2}+e^{2}-2ex+x^{2}+f^{2}$

Now, adding vertically, we'll get a polynomial like this:
$ f(x) = nx^{2}-2(a+c+e+g+\ldots)\cdot x+b^{2}+d^{2}+...$
but nothing but our first two coefficients matter.

the X coord is at $ -\frac{b}{2a}$, or $ \frac{2(a+c+e+g+\ldots)}{2n}$ = $ \boxed{\frac{a+c+e+g+\ldots}{n}}$

This is interesting, because it implies that the Y values of the points don't matter in the slightest. It's just the average of the X values..

Can this be right?

[quote="DiscreetFourierTransform"]This is interesting, because it implies that the Y values of the points don't matter in the slightest. It's just the average of the X values..

Can this be right?[/quote]

Yes, because you are finding a point on the x-axis, so y=0 always.

The more interesting problem arises when P is an arbitrary point on the xy plane, in which both x- and y- values matter. (It's also the average, by the way).

Right.
Another possible approach: Construct a convex polygon through the points in consideration (is this always possible?) and find the centroid. You can use mass points to do this.

Because the points are weighted equally, this should be the point where the distance to all the vertices is minimized, right?

Or am I COMPLETELY wrong?


Six people are going to be seated in a row of 8 chairs. How many different ways may they be seated?

Hello
[hide="solution"]
$ \frac{8!}{2!}$=20160
[/hide]
Thank u.

There are 8 choices for the first person, 7 for the second, 6 for the third, and so on until there are 3 choices for the 6th, so there are $ 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\equal{}20160$ ways.

Extension: How many different ways can $ r$ people be seated in a row of $ n$ chairs?

I have a question for the first solution (well they're basically the same but) why do you divide by 2!?

-vdjejhehsw


If $a,\, b,\, c$ are the sides of a triangle, prove that
\[\frac{(c+a-b)^{4}}{a(a+b-c)}+\frac{(a+b-c)^{4}}{b(b+c-a)}+\frac{(b+c-a)^{4}}{c(c+a-b)}\ge ab+bc+ca \]


LaTeX
inequalities
rearrangement inequality
inequalities unsolved

[quote="Detective"]i think i'll change the word $prove$ into the phrase $prove or disprove$.[/quote]Detective, welcome to the AoPS - MathLinks forums. Please note that

1) You have an edit button with which you can edit your post in case you made a mistake in the statement
2) Please do not use normal text inside $\LaTeX$ tags. That loads up the server unnecesarily. Thanks.

[quote="Detective"]For any positive real numbers $x,y,z$,
prove or disprove

$\frac{x^4 - x^3}{x^3 + y^2 + z^2} + \frac{y^4 - y^3}{y^3 + x^2 + z^2} + \frac{z^4 - z^3}{z^3 + x^2 + y^2} \geq \frac{x + y + z}{3} - 1$

And determine when does equality occur.[/quote]
Elementary transformation turn the given iequality into  SUM(x4+y2+z2)/x3+y2+z2)>=((x+y+z)/3)+2 where the summation on LHS is cyclic.If we apply Cauchy-Schwarz inequality we get (x4+y2+z2)(x2+y2+z2)>=(x3+y2+z2)2 so (x4+y2+z2)/(x3+y2+z2)>=(x3+y2+z2)/(x2+y2+z2)Summing up all such inequalities yields LHS>=2+((x3+y3+z3)/(x2+y2+z2))According to rearrangement inequality (x3+y3+z3)/(x2+y2+z2)>=(x+y+z)/3 Thus the prooof is ccoplete.Equality holdsiff x=y=z=1(herex4 means x to the power of 4 etc
SUM (x4+y2+z2)/(x3+y2+z2) stands for (x4+y2+z2)/(x3+y2+z2)+(x2+y4+z2)/(x2+y3+z2)+(x2+y2+4)/(x2+y2+z4)

[quote="Arne"][quote="Makaveli"]Elementary transformation turn the given iequality into  SUMx4+y2+z2/x3+y2+z2)>=((x+y+z)/3)+2 where the summation on LHS is cyclic.If we apply Cauchy-Schwarz inequality we get (x4+y2+z2)(x2+y2+z2)>=(x3+y2+z2)2 so (x4+y2+z2)/(x3+y2+z2)>=(x3+y2+z2)/(x2+y2+z2)Summing up all such inequalities yields LHS>=2+((x3+y3+z3)/(x2+y2+z2))According to rearrangement inequality (x3+y3+z3)/(x2+y2+z2)>=(x+y+z)/3 Thus the prooof is ccoplete.Equality holdsiff x=y=z=1(herex4 means x to the power of 4 etc[/quote]

Are you sure you can't make your posts a bit "neater"? :|[/quote]
That SUM stands for sigma sighn where, as I said, the summation is cyclic.Is some particular part that is not understandable?

[quote="Valentin Vornicu"][quote="Makaveli"]That SUM stands for sigma sighn where, as I said, the summation is cyclic.Is some particular part that is not understandable?[/quote]Yes, your whole post!

Try to

- leave a space between . and the next word
- use $\LaTeX$ (you can find a quick guide by following the link in my signature).[/quote]
Is it any better now?

[quote="Makaveli"]Elementary transformation turn the given iequality into  SUM(x4+y2+z2)/x3+y2+z2)>=((x+y+z)/3)+2 where the summation on LHS is cyclic.If we apply Cauchy-Schwarz inequality we get (x4+y2+z2)(x2+y2+z2)>=(x3+y2+z2)2 so (x4+y2+z2)/(x3+y2+z2)>=(x3+y2+z2)/(x2+y2+z2)Summing up all such inequalities yields LHS>=2+((x3+y3+z3)/(x2+y2+z2))According to rearrangement inequality (x3+y3+z3)/(x2+y2+z2)>=(x+y+z)/3 Thus the prooof is ccoplete.Equality holdsiff x=y=z=1(herex4 means x to the power of 4 etc
SUM (x4+y2+z2)/(x3+y2+z2) stands for (x4+y2+z2)/(x3+y2+z2)+(x2+y4+z2)/(x2+y3+z2)+(x2+y2+4)/(x2+y2+z4)

Is it any better now?[/quote]

No, it's not actually. It looks exactly the same to me. You can either make it

[quote="Makaveli"]Elementary transformation turn the given iequality into  SUM (x^4+y^2+z^2)/(x^3+y^2+z^2)>=((x+y+z)/3)+2 where the summation on LHS is cyclic.

If we apply Cauchy-Schwarz inequality we get (x^4+y^2+z^2)(x^2+y^2+z^2)>=(x^3+y^2+z^2)^2 so (x^4+y^2+z^2)/(x^3+y^2+z^2) >= (x^3+y^2+z^2)/(x^2+y^2+z^2).
Summing up all such inequalities yields LHS >= 2+((x^3+y^3+z^3)/(x^2+y^2+z^2)).

According to rearrangement inequality (x^3+y^3+z^3)/(x^2+y^2+z^2) >= (x+y+z)/3, thus the prooof is complete. Equality holds iff x=y=z=1.
[/quote]

or (much better) into

[quote="Makaveli"]Elementary transformation turn the given iequality into $\sum \dfrac { x^4+y^2+z^2}{x^3+y^2+z^2} \geq \frac {x+y+z}3+2$ where the summation on LHS is cyclic.

If we apply Cauchy-Schwarz inequality we get $(x^4+y^2+z^2)(x^2+y^2+z^2) \geq (x^3+y^2+z^2)^2$ so $\dfrac {x^4+y^2+z^2}{x^3+y^2+z^2} \geq \dfrac {x^3+y^2+z^2}{x^2+y^2+z^2}$.
Summing up all such inequalities yields $\textrm{LHS} \geq 2+\dfrac {x^3+y^3+z^3}{x^2+y^2+z^2}$.

According to rearrangement inequality $\dfrac {x^3+y^3+z^3}{x^2+y^2+z^2} \geq \dfrac {x+y+z}3$, thus the prooof is complete. Equality holds iff $x=y=z=1$.
[/quote]


floor function
number theory proposed
number theory

For all positive integer n,define$b(n)=min\{(2k+\frac{n}{k^2}),k\in Z+\}$Find the closest fomula for$[b(n)]$

I found $b ( n ) = 2 \left\lfloor \sqrt[3]{n} + \frac{1}{2} \right\rfloor + \frac{n}{\lfloor \sqrt[3]{n} + \frac{1}{2} \rfloor^2}$. Correct ?

your formula isn't simple,can you simply your fomula?I'll post my formula.It's very nice

Feel free to post your formula.
I'm impatient to see it :)

your formula is true,but my require is$[b_n]$,can you solve for my require?

I tought that the brackets $[$ in $[b(n)]$ were just parantheses ! :oops: I'll post the formula for $\lfloor b(n) \rfloor$ as soon as I can (I'm going on hollydays so in the worst case that will be in 10 days...). I'm sorry for all this confusion :(

Rust,your formula is very false.The true formula is
$\lfloor\ b_n\rfloor$=$\left\lfloor\sqrt[3]{27n+9\sqrt[3]{n}}\right\rfloor if n=k^3+r\left(0\leq r<\frac{k^2(3k+2)}{2k+1}\right)\\ \left\lfloor\sqrt[3]{27n+9\sqrt[3]{n}+3}\right\rfloor otherwise$
From my formula,i suggested another problem:
Problem 2 Find all positive integer n such that$\left\lfloor\sqrt[3]{27n+9\sqrt[3]{n}}\right\rfloor+1=\left\lfloor\sqrt[3]{27n+9\sqrt[3]{n}+3}\right\rfloor$


function
quadratics
algebra unsolved
algebra

Find all functions $ f : \mathbb{R} \to \mathbb{R}$ such that
\[f( x^2 + y + f(y)) \ = \ (f(x))^2, \quad \forall x , y  \in \mathbb{R}.\]

[quote="mai quoc thang"]Pco , where are you ? :D[/quote]
I have an ugly solution and I'm writing it. I'll soon post.

[quote="nguyenvuthanhha"][i]Find all function $ f : \mathbb{R} \to \mathbb{R} \#$ such that :

   $ f( x^2 + y + f(y)) \ = \ (f(x))^2 \ \forall \ x ;y \ \in \ \mathbb{R}$[/i][/quote]

I really think there is a shorter demo, but here is the best I can do up to now :

Let $ P(x,y)$ be the assertion $ f(x^2+y+f(y))=f(x)^2$
Let $ U=\{x+f(x),\forall x\in\mathbb R\}$
Let $ Q(x,u)$ be the assertion $ f(x^2+u)=f(x)^2$ where $ x\in\mathbb R$ and $ u\in U$

1) [b]$ f(-x)=f(x)\geq 0$ $ \forall x$[/b]
=====================
Suppose $ \exists x$ such that $ f(x)<0$. Then $ P(\sqrt{-f(x)},x)$ $ \implies$ $ f(x)=f(\sqrt{-f(x)})^2\geq 0$ and so contradiction, and so $ f(x)\geq 0$ $ \forall x$

Comparing then $ P(x,y)$ and $ P(-x,y)$ and using the fact that $ f(x)\geq 0$, we get $ f(-x)=f(x)$ $ \forall x$

2) $ \exists z$ as great as we want such that either $ f(z)=0$, either $ f(z)=1$
======================================================
Let then $ u,v\in U$ and $ w=\max(u+v,u^2+v)+f(\max(u+v,u^2+v))$ such that $ w\in U$ and $ w>u+v$ and $ w>u^2+v$

$ Q(x+u,v)$ $ \implies$ $ f((x+u)^2+v)=f(x+u)^2$
If $ x\geq 0$ and comparing $ Q(\sqrt x,u)$ and $ Q(\sqrt x,w)$ : $ f(x+u)=f(x+w)$ and so $ f((x+u)^2+v)=f(x+w)^2$

Consider then the quadratic $ (x+u)^2+v=x+w$ $ \iff$ $ x^2+(2u-1)x+u^2+v-w$ whose discriminant is $ \Delta=1-4u-4v+4w$
Since $ w>u+v$, $ \Delta>0$ and this quadratic has two real roots $ a\geq b$
Since $ w>u^2+v$, $ ab=u^2+v-w<0$ and so $ a>0$

So, since $ a>0$, we have $ f((a+u)^2+v)=f(a+w)^2$ and, since $ (a+u)^2+v=a+w$, we get $ f(a+w)=f(a+w)^2$ and so $ f(a+w)
\in\{0,1\}$
And since $ a>0$ and $ w$ as great as we want, $ \exists z$ as great as we want such that either $ f(z)=0$, either $ f(z)=1$
Q.E.D.

3) The only two solutions are $ f(x)=0$ $ \forall x$ and $ f(x)=1$ $ \forall x$
=================================================

3.1) Suppose $ \exists z$ such that $ f(z)=0$
$ P(z,y)$ $ \implies$ $ f(z^2+y+f(y))=0$ and so we can find a $ z$ as great as we want (just choosing $ y$ great enough).
Let then $ x\in \mathbb R$ and $ z>u=x+f(x)$ such that $ f(z)=0$ and so $ f(-z)=0$ too.

$ P(\sqrt{z-x-f(x)},-z)$ $ \implies$ $ f(-x-f(x))=f(\sqrt{z-x-f(x)})^2$
$ P(\sqrt{z-x-f(x)},x)$ $ \implies$ $ 0=f(z)=f(\sqrt{z-x-f(x)})^2$
So $ f(-x-f(x))=0$ $ \forall x$ and so $ f(x+f(x))=0$ $ \forall u\in U$. Then :

$ P(0,x)$ $ \implies$ $ f(0)=0$

$ P(\sqrt{|x+f(x)|},0)$ $ \implies$ $ f(\sqrt{|x+f(x)|})=0$

$ P(\sqrt{|x+f(x)|},y+f(y))$ $ \implies$ $ f(|x+f(x)|+y+f(y))=0$
$ P(\sqrt{|x+f(x)|},-y-f(y))$ $ \implies$ $ f(|x+f(x)|-y-f(y))=0$
So $ f(x+f(x)-y-f(y))=0$

And now, just set $ x=\frac a2$ and $ y=-\frac a2$ and we get $ f(a)=0$ $ \forall a$

3.2) Suppose $ \not\exists z$ such that $ f(z)=0$
Using paragraph 2 result, we know that $ \exists z$ such that $ f(z)=1$

$ P(z,y)$ $ \implies$ $ f(z^2+y+f(y))=1$ and so we can find a $ z$ as great as we want (just choosing $ y$ great enough).
Let then $ x\in \mathbb R$ and $ z>u=x+f(x)$ such that $ f(z)=1$ and so $ f(-z)=1$ too.

$ P(\sqrt{z-x-f(x)},-z)$ $ \implies$ $ f(1-x-f(x))=f(\sqrt{z-x-f(x)})^2$
$ P(\sqrt{z-x-f(x)},x)$ $ \implies$ $ 1=f(z)=f(\sqrt{z-x-f(x)})^2$
So $ f(1-u)=1$ $ \forall u\in U$ and so $ 2-u\in U$ $ \forall u\in U$ (since $ 2-u=1-u+f(1-u)$)

Then, comparing $ Q(\sqrt x,2-u)$ and $ Q(\sqrt x, 2-v)$, we get $ f(x+2-u)=f(x+2-v)$ $ \forall x\geq 0$ $ \forall u,v\in U$

So $ f(-x-2+u)=f(-x-2+v)$ $ \forall x\geq 0$ $ \forall u,v\in U$

So $ f(x+u)=f(x+v)$ $ \forall x\leq -2$ $ \forall u,v\in U$

So $ f(x)=f(x+u-v)$ $ \forall x\leq v-2$ $ \forall u,v\in U$

Let then $ u=a+f(a)$ and $ v=-a+f(-a)$ : $ f(x)=f(x+2a)$ $ \forall x\leq -a+f(-a)-2$ $ \forall x, a\in \mathbb R$

So $ f(x)=f(x+2a)$ $ \forall a$ $ \forall x\leq -a-2$

Let then $ a=-\frac{x+y}2$ : $ f(x)=f(-y)$ $ \forall y$ $ \forall x\leq \frac{x+y}2-2$

So $ f(x)=f(y)$ $ \forall y$ $ \forall x\leq y-4$

And so $ f(x)=$ constant $ =1$
Q.E.D.

Here is a short solution of my friend :

Step 1: $ f(x) \geq$ 0 for all $ x$.
Proof: Suppose $ f(y) \equal{} \minus{} t^2$ with positive $ t$.
We have $ f( t ^ 2 \plus{} y \plus{} f (y ) ) \equal{} f(y) \equal{} \minus{} t^2 < 0$, false.

From step 1 ,since $ f^ 2 ( x ) \equal{} f^2 ( \minus{} x )$ we have $ f(x) \equal{} f( \minus{} x)$.

Step 2: For all $ a \geq f(y) \minus{} y$ we have $ f(a) \equal{} f(a \plus{} 2y)$

Proof: For all $ x,y$ :
$ f ( x^ 2 \plus{} y \plus{} f ( y ) ) \equal{} f ( x^ 2 \minus{} y \plus{} f ( y ) )$ , so step 2 follows.

Step 3: For all $ b > a > f(1) \minus{} 1, f(a) \equal{} f(b)$

Proof: Since $ b > a > f(1) \minus{} 1$, from step 2 we have $ f(a \plus{} 2n) \equal{} f(a)$ and $ f(b \plus{} 2n) \equal{} f(b)$ for all natural $ n$.

Choose n sufficently large so that $ a \plus{} 2n > f( \frac {b \minus{} a} {2} ) \minus{} \frac {b \minus{} a} {2}$ , with this n we have $ f(a \plus{} 2n) \equal{} f(a \plus{} 2n \plus{} b \minus{} a) \equal{} f(b \plus{} 2n)$, so
$ f(a) \equal{} f(b)$.

Note that in this step we can replace 1 by an abritary positive real $ r$ and we still have the result.

Step 4: $ f(x) \equal{} 0$ for all $ x$ or $ f(x) \equal{} 1$ for all $ x$.

Proof: From step 3 and our initial equation, clearly $ f ^ 2 ( a ) \equal{} f(a)$ for all

$ a > f(1) \minus{} 1$, so $ f(a) \equal{} 1$ or $ 0$. But from step 3, it follows that $ f(a) \equal{} 0$ for all $ a > f(1) \minus{} 1$ or $ f(a) \equal{} 1$ for all $ a > f(1) \minus{} 1.$

Choose sufficently large a such that $ f(a) \minus{} a < 0$.

Now we repeat the reasoning of step 3 to get $ f(x) \equal{} f(y)$ for all $ x,y > f(a) \minus{} a$

In particular, $ f$ is constant for all non - negative reals. But since $ f$ is an even function, $ f$ must be constant and we obtain the desired result.


[quote="outback"]Find all functions $ f: \mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}$ such that for all reals $ x,y,z$,

$ f(f(x,z),f(z,y))\equal{}f(x,y)\plus{}z$[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=364515


number theory
greatest common divisor
arithmetic sequence
number theory open

The problem is actually quite interesting. It is equivalent to the following:

If $p>5$ is a prime, and $G=\mathbb{Z}_p^*$, then, $|\langle2\rangle\cap\langle3\rangle|<\min\{|\langle2\rangle|,|\langle3\rangle|\}$.

I haven't been able to solve it yet, but it seems to be right!, at least it is true for $p=7,11,13,17,19,23,29,31$.

[quote="pat.costello"]It appears that
gcd(3^n-2,2^n-3) = 5 if n=3(mod 4) and
gcd(3^n-2,2^n-3) = 1 otherwise
Does this hold in general?[/quote]

No, the first counterexemple is n=3783:
gcd(2^3783-3,3^3783-2) = 26665 = 5*5333

For n<100000 it only fails for the terms of the
arithmetic progression 3783+k*5332;
for all of them the gcd is 26665.

Best regards,

j. h. nieto

Jeje, Jose, I was just going to post this answer (Jose is member in another list and I proposed the problem there, he found that solution above) :D .

Well, It is very nice, we already know that there are examples where it is not true. It would be nice to prove that there are either just finitely many prime that may divide suc expresion, or to prove that, on the other hand, there are infinitelly many.

BTW, I think (but it is just a guess) that $\gcd(2^n-3, 3^n-2)$ is always square free!

Best,

[quote="pat.costello"]It looks like it is still the case that
gcd(2^n-3,3^n-2) = 0 (mod 5)   when n=3 (mod 4)   and
gcd(2^n-3,3^n-2) = 1                otherwise
Is this true in general?[/quote]

Well, the first statement: $gcd(2^n-3,3^n-2) = 0 \mod 5$   when $n=3 \mod 4$ is definitely true and easy to proove, nevertheless the second one I am not sure, but I think it may be.

Best,


Number triangles with different numbers in the bottom line or with the same numbers but in a different order are regarded as different.

a)A number triangle has six consective integers in some order in the bottom line and top number 2006.
Give one example of this with 61,62,63,64,65,66 in some order in the bottom line.

Show that this is the only possible set of six consecutive numbers for the bottom line of a number triangle with top number 2006.

How many different number triangles are there with six consecutive numbers in the bottom line , in some order and with top number 2006?

Umm.. this is straight out of the Maths Challenge. Talk about lazy.

Could a mod lock this, please? And the other topics?


geometry
rectangle
combinatorics unsolved
combinatorics

Find all $m \times n$ rectangles that can be tiled with the T-tetromino.

It has been proved in the A.M.M. (1965, p.986-988) that an $m \times n$ rectangle can be tiled with T-tetrominos if and only if 4 divides both $m$ and $n$.

The proof is more than two pages long, so I will not write it here.

Pierre.

Well, we could start solving it, and who knows? maybe we'll even finish it.

First of all, let's show that $4$ divides at least one of $m,n$ (this is easy, but I'll still do it :)). It's clear that $4|mn$, so let's assume $m=4a+2,n=4b+2$. Color the squares of the board chessboard-style. A tetromino either covers $3$ black squares or $3$ white squares, so, since there are as many white squares as there are black squares on the board, it means that we have an even number of tetrominoes, so $8|mn$, contradiction our hypothesis. This means that $4$ divides at least one of $m,n$ and, we have also shown that $8|mn$.

And Pierre, if you do find the time to post the solution, would you mind postponing that? I'd like to give it some thought :).

can you post the rest of solution?pierre


algebra
polynomial
algorithm
superior algebra
superior algebra theorems

Does there exist some way to do polynomial division when you have $ f,g\in R[X,Y]$?
Say $ f\equal{}Y^4\minus{}X^2$ and $ g\equal{}Y^4\minus{}X^2Y^2\plus{}Y^3X\plus{}X^3$. How would one proceed?

when $ f,g$ are polynomials over some commutative ring $ R$ with one variable, and the leading coefficient of $ g$ is a unit, then you can divide $ f: g$ with remainder. (the proof and the algorithm works as in the familiar case where $ R$ is a field). here we have the ring $ R[X]$ resp. $ R[Y]$.


function
inequalities
algebra unsolved
algebra

show that thee is no   function f definedonthe positive real numbes such that :
                 $f(y) > (y-x)f(x)^2$

Am I right? :huh:
Let $x$ and $y$ be positive real numbers such that $y>x$. By the given condition,
$f(y)>(y-x)f(x)^2$
Interchange the positions of $x$ and $y$, we have
$f(x)>(x-y)f(y)^2$
From the first inequality,
$f(x)>(x-y)f(y)^2>(x-y)[(y-x)f(x)^2]^2=(x-y)^3f(x)^4$
$\Rightarrow f(x)<\frac{1}{x-y}<0$
However, $f(x)>(x-x)f(x)^2=0$ so there is a contradiction.

P.S. Why are you posting the same question again in the same forum? :? [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65810]Take a look here.[/url]

No....I think you're not right... cos' the fact that \[ f(y)>(y-x)f(x)^2 \] doesn't mean that \[ f(y)^2>[(y-x)f(x)^2]^2 \] Look..for example... if $f(y)=1$ and $f(y+1)=10$ then of course $1=f(y)>0>(-1)(10)^2$ so our first inequality is true...but $1^2<10^2$ so the second one is false


inequalities
calculus
inequalities proposed

Let $a$, $b$, $c$ be positive real numbers with $abc < 512$. Prove that
$\displaystyle \frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}>1$

The original problem, posted by  cn2_71828182846 (in the "Useful inequalities" thread), asked about this inequality for $abc=8$. When I decided to extend it, I was surprised I could get this far.

This was discussed here:

http://artofproblemsolving.com/Forum/viewtopic.php?t=5635

All right, I don't read every thread, and that dates from well before I joined. I had an entertaining calculus-based solution- for $abc=M$, I calculated the values at the critical points- the ones that matter are of the form $(a,b,b)$ or a permutation thereof with $a \le 2$. The value at that point is $\frac{\sqrt{4a+1}}{\sqrt{a+1}} > 1$, and the value on the boundary is > 1, so the inequality can only be violated if $\frac{3}{\sqrt{1+a}} < 1$ for $a^3=M$.

Also discussed on mathlinks forum. I remember to have posted a solution.


Three points are always coplanar.

For four points, use the scalar triple product to find the volume of the parallelepiped determined by your points. It's 0 iff the points are coplanar.

[quote="guile"]How do you know that if three points are coplanar? For example
    $ a \equal{} (2, 3, 5) b \equal{} (1,6,2)$ and $ d \equal{} ( \minus{} 3, 1, 4)$[/quote]

Remember that lines are solutions to systems of 2 equations with 3 unknowns, planes are solutions to systems of 3 equations with 4 unknowns, and so on. It's obvious that such systems always have solutions. In particular, the equation of a plane looks like \[ ax\plus{}by\plus{}cz\plus{}d\equal{}0\], specified by the values of $ a,b,c,d$. If someone gives you 3 distinct points ([i]any[/i] 3 distinct points), then you will now have 3 equations (one occurence of the above for each $ (x_0,y_0,z_0)$), and 4 unknowns ($ a,b,c,d$). A solution always exists for this, so [i]some[/i] plane will always work.

If someone gives you [i]four[/i] points, however, you have a system of four equations, four unknowns, which may or may not have a solution, and that's when you can either solve the system (if possible), or use Bictor717's suggestion if all you care about is the co-planarness.

Hope this rather long explanation helps clear up any confusion... :)

[quote="AdrianP"]planes are solutions to systems of 3 equations with 4 unknowns[/quote]

No: planes in 3-space are the solution set of one (linear) equation in three unknowns, and lines in 4-space are the solutions of 3 equations with 4 unknowns.  In general, a $ k$-dimensional (affine) subspace of $ n$-space is the solution of $ n \minus{} k$ equations in $ n$ unknowns.


There is a place with 1,10,100,and 1000 dollar bills.  can one have exactly half a million bills with a value of 1 million dollars?

[hide]Say that you have 500,000 $1 bills - that's $500,000. Replacing a $1 bill with a $10, $100, or $1,000 bill will cause the amount of money you have to increase by a multiple of 9. However, $500,000 is not evenly divisible by 9, so you cannot have exactly $1,000,000.[/hide]


[hide]21???
1 goes with 13
2 goes with 14
so 9 goes with 21[/hide]

[quote="joejia"]Twenty-six people, numbered consecutively 1 through 26, are seated equally spaced around a circular table. Which numbered person is seated directly across from person number 9?[/quote]

[hide]
diamtrically opposite increments of $\frac{26}{2} = 13 $

Thus, the person who is sitting opposite person 9 is person $9+13=22$[/hide]

[hide]
The corresponding numbers, in ordered pairs, are $(1,14), (2,15), (3,16)\cdots (9,22)$. $22$ is seated directly across from $9$.[/hide]


trigonometry
geometry
geometric transformation
reflection
geometry unsolved

Show that there is a point inside any triangle such that the feet of the perpendiculars to the three sides form an equilateral triangle.

See here http://www.mathlinks.ro/Forum/viewtopic.php?t=6489 and here http://www.mathlinks.ro/Forum/viewtopic.php?t=208618 .

But the 1st isodynamic point $ Q_1$ is isogonal conjugate of the 1st Fermat point $ F_1,$ therefore $ Q_1$ is inside of $ \triangle ABC$ as long as $ F_1$ is inside, i.e., only when all triangle angles are less then $ 120^\circ.$

Hmmmm... But assume that point $ P$ lies inside triagnle $ ABC$ and $ D,E,F$ are foots of perpendiculars from $ P$ respectively on $ BC,CA,AB$ and $ DEF$ is equilateral. Then \[ \frac{\sin \angle DCP}{\sin \angle DCE}\equal{}\frac{DP}{DE}\equal{}\frac{DP}{FD}\equal{}\frac{\sin \angle DBP}{\sin \angle DBF}\iff \frac{BP}{CP}\equal{}\frac{\sin \angle DCP}{\sin \angle DBP}\equal{}\frac{\sin \angle BCA}{\sin\angle ABC}\equal{}\frac{AB}{AC}\] which means that $ P$ lies on the A-Apollonius circle and analogously on the B and C... So where is the mistake?

The mistake is in the problem, because it claims that a point with equilateral pedal triangle exists [color=red]inside[/color] of every triangle. The 1st and 2nd isodynamic points $ Q_1, Q_2$ (the 2 intersections of the 3 Apollonius circles) are the only 2 points with equilateral pedal triangles. They are isogonal conjugates of the 1st and 2nd Fermat points $ F_1, F_2.$

For the 1st Fermat point inside a triangle $ \triangle ABC,$ $ \angle BF_1C \equal{} \angle CF_1A \equal{} \angle AF_1B \equal{} 120^\circ,$ therefore $ F_1$ is inside of a triangle only when all angles are less than $ 120^\circ.$ If one angle is greater, say $ \angle CAB > 120^\circ,$ $ F_1$ is outside in the angle opposite to $ \angle CAB,$ lines $ BF_1, CF_1$ are outside of the triangle, lines $ BQ_1, CQ_1$ are reflections of $ BF_1, CF_1$ in internal bisectors of $ \angle B, \angle C,$ also outside of the triangle, $ Q_1$ is outside.

Similar reasons put the 2nd Fermat point $ F_2$ always outside (unless one angle is equal to $ 60^\circ,$ in which case $ F_2$ is identical with this vertex). $ Q_2$ is therefore always outside (the special case easy to handle) and it is out of consideration.


Source: AIME 1994

[img]http://www.kalva.demon.co.uk/aime/soln/diag942.jpg[/img]

The large circle has diameter $40$ and the small circle diameter $10$. They touch at $P$. $PQ$ is a diameter of the small circle. $ABCD$ is a square touching the small circle at $Q$. Find $AB$.

[hide]Let the center be $O$, and let $AB=n$.  Then,
$OA=OB-20$
Extend the diameter of the long circle passing through $P$ and through $O$.  Let the point at which is intersects the square again be $M$.  Then
$QO=10 \implies OM=n-10$.  We also know that $MA=MB=\frac{n}{2}$.  Therefore,
$(n-10)^2+(\frac{n}{2})^2=400 \implies n^2-16n-240=0 \implies n=\frac{16 \pm \sqrt{256+960}}{2}=8 \pm \sqrt{304}$
So that should be the answer...I really like using the $\implies$ sign today for some reason  :D[/hide]

forgot to reduce $\sqrt{304}$ :P and remember that the answer can't be negative

[hide]Oh yeah.  Does $\sqrt{304}$ have to be reduced though?  I'm guessing it asks for either the form $a+b\sqrt{c}$ or $m+\sqrt{n}$.  So then it's either $8+\sqrt{304}$ or it's $8+4\sqrt{19}$.[/hide]
Edit: Whoops, forgot to hide solutions  :D


calculus
integration
function
calculus computations

Please help with a double integral:

0<y<1
y<x<(sqrt(y))

The function is y / (sqrt(x^2 +y^2)) dx dy

As, the condition is:
0<y<1 and y<x<(sqrt(y))

I have changed the order of the integration to:

0 < y < x < sqrt(y) <1

Therefore, the limits for outer integral are 0<x<1 and for inner integral: x^2 < y < 1

I guess I have to use substitution for y / (sqrt(x^2 +y^2)), but I am not sure how to do it. I could not find anything in my books about integration with 2 variables in a fraction.

Any help would be appreciated!

$ \begin{array}{ll}\int\limits_0^1\int\limits_y^{\sqrt{y}}\frac{y}{\sqrt{x^2+y^2}}dxdy & = \int\limits_0^1\int\limits_{x^2}^x\frac{y}{\sqrt{x^2+y^2}}dydx= \int\limits_0^1\left[\sqrt{x^2+y^2}\right]_{x^2}^xdx\\[0.2cm]
 & = \int\limits_0^1\left[\sqrt{2x^2}-\sqrt{x^2+x^4}\right]dx= \int\limits_0^1\left[x\sqrt{2}-x\sqrt{1+x^2}\right]dx
\\[0.2cm]
 & =  \left[\frac{x^2}{\sqrt{2}}-\frac{(1+x^2)^{\frac{3}{2}}}{3}\right]_0^1=\frac{2-\sqrt{2}}{6}\end{array}$


probability
geometry
parallelogram
3D geometry
sphere
Asymptote
probability and stats

1) Lines are laid down independently at random on the plane, dividing it into polygons. Show that the average number of sides of this set of polygons is 4.

2) Prove that four point picked at random in a parallelogram form a convex quadrilateral with probability $ \frac{25}{36}$.

For (1), what do you mean "at random on the plane"? There's no canonical probability density here.

My book does not specify. There is an hint: consider n random great circles of a sphere of radius R; then let R and n increase.

Hint for the second:
choose P,Q,R independently at random in the square S(a) of side a. Show that $ E(AreaPQR)\equal{}\frac{11a^2}{144}$, and from this deduce the requires result.

#1 isn't a probability question at all.

Let $ n$ lines ($ n>2$) be placed in general position in the plane; any two intersect, and no three meet at a single point. This divides the plane into $ \binom{n\plus{}1}{2}\plus{}1$ regions, of which $ 2n$ are unbounded (The $ n$th line cuts each of $ n$ old regions into two parts). There are $ \binom{n}{2}$ intersection points, and $ n(n\minus{}2)$ segments between adjacent intersection points on a line.
Now, look at the bounded polygons. If we cut off all of the rays that go to $ \infty$, there are $ \frac{n(n\plus{}1)}{2}\minus{}2n\plus{}1\equal{}\frac{n^2\minus{}3n\plus{}2}{2}$ faces (and the outside), $ \frac{n^2\minus{}n}{2}$ vertices, and $ n^2\minus{}2n$ edges.

Let the number of edges of the "outside" be $ m$. The average number of edges per face is $ \frac{2(n^2\minus{}2n)\minus{}m}{(n^2\minus{}3n\plus{}2)/2}\equal{}4\frac{n^2\minus{}2n\minus{}m/2}{n^2\minus{}3n\plus{}2}$
I conjecture that $ m\le 2n$ for any $ n$ and any configuration. If this is true, or even if $ m\equal{}o(n^2)$, the average goes to $ 4$ regardless of configuration as $ n\to\infty$.

This would all be simpler on the sphere- there, the average is exactly $ 4\minus{}\frac{8}{n^2\minus{}n\plus{}2}$.

If $ n$ great circles meet each other, not more than two at any given point, then there are $ 2\binom{n}{2}$ intersections. It follows that there are $ 4\binom{n}{2}$ segments between vertices, and Euler's formula gives the number of regions as $ n(n \minus{} 1) \plus{} 2$. We may think of the plane as obtained by taking the limit as $ R \minus{} > \infty$ and 'stretching out' the sphere. Each segment is a side of two polygons, so the average number of sides satisfies
$ \frac {4n(n \minus{} 1)}{2 \plus{} n(n \minus{} 1)} \minus{} > 4$ as $ n \minus{} > \infty$.

[quote="jmerry"]
I conjecture that $ m\le 2n$ for any $ n$ and any configuration.
[/quote]
This is false:
[asy]import cse5; size(100); D(dir(90)--dir(-80)--dir(110)--dir(-40)--dir(-140)--dir(70)--dir(-100)--cycle);[/asy]
$ o(n^2)$ may still be true though. Let's think more of that :).


function
trigonometry
analytic geometry
graphing lines
slope
integration
calculus

I'm wondering about the function $ f: R \rightarrow R$ defined so that $ f(x) \equal{} y$ is the solution to $ \sin{(y \plus{} x)} \equal{} y$, i.e. the fixed point of sine with a shift of $ x$ units leftward.  The function is well-defined - there is a unique solution to that equation for each $ x$.  You can prove this fairly easily, or convince yourself by drawing lines of slope one through a sine wave.

I've graphed the function with Matlab; unfortunately I did not save the image.  It looks similar to $ \sin{x}$, but it has vertical tangents at at the zeroes $ x \equal{} 2\pi k$.  Specifically I'm curious to know if there is a nice closed form for $ \int_0^\pi f(x)dx$.  I don't believe there is a closed form for the function since $ f(\frac {\pi}{2})$ is the [url=http://mathworld.wolfram.com/DottieNumber.html]Dottie Number[/url].  If anybody has heard of this or similar functions I'd be glad to know about them. :)

Yes, the integral works out nicely. $ \int_0^\pi f(x)\,dx\equal{}2$.

Try the parametric form $ x\equal{}t\minus{}\sin t$, $ y\equal{}\sin t$, and calculate $ \int y\,dx$. Ending at $ x\equal{}t\equal{}\pi$ works nicely, unlike other points.

Ah-hah, very nice.  Thank you.  So simple, but not obvious for me.  Interesting that it's the same as $ \int_0^\pi \sin{(x)}dx$.


[quote="aktyw19"]$ a,b,c$ - real positive .  It prove, that
$ \left( ab \plus{} bc \plus{} ac \right) \cdot \left( \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \right) \ge \frac {9}{4}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3547

[quote="caubetoanhoc94"][quote="aktyw19"]$ a,b,c$ - real positive .  It prove, that
$ \left( ab \plus{} bc \plus{} ac \right) \cdot \left( \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \right) \ge \frac {9}{4}$[/quote]

It's Iran 96  :P[/quote]

Yeah , and it 's not very hard ;)


geometry
circumcircle
trigonometry
analytic geometry
complex numbers
trig identities
Law of Cosines

$ P$ is a point on the circumcircle of the square $ ABCD$ between C and D. Show that $ PA^{2}-PB^{2}= PB \cdot PD-PA \cdot PC$.

Kalva has a trig solution, I tried to use cyclic quads but didn't get anywhere.

[hide]Place the square on the complex plane and WLOG let the coordinates of $ A,B,C,D,P$ be $ -1,-i,1,i,z$ repectively. Let $ \arg (z)=\theta$

$ PA^{2}-PB^{2}= PB \cdot PD-PA \cdot PC$
$ \Leftrightarrow |z+1|^{2}-|z+i|^{2}=|z^{2}+1|-|z^{2}-1|$

but we have

$ |z+1|^{2}-|z+i|^{2}=(z+1)(\overline{z}+1)-(z+i)(\overline{z}-i)=(2+z+\overline{z})-(2+i\overline{z}-zi)=(z+\overline{z})+i(z-\overline{z})=2\cos\theta-2\sin \theta$

$ |z^{2}+1|-|z^{2}-1|=|1+\cos 2\theta+i\sin 2\theta|-|-1+\cos 2\theta+i\sin 2\theta|=\sqrt{2+2\cos 2\theta}-\sqrt{2-2\cos 2\theta}=2\cos\theta-2\sin \theta\ \blacksquare .$[/hide]

This is a bit more direct $ |z^{2}+1|=|z|*|z+z^{-1}|$ [and circumvents the positive negative squareroot bs].

Oh snap not only is [insert name] stalking my posts, but he is reading that complex numbers book.

[quote="Altheman"]This is a bit more direct $ |z^{2}+1|=|z|*|z+z^{-1}|$ [and circumvents the positive negative squareroot bs].[/quote]

Nice, but doesn't my choice of $ C$ and $ D$ force $ 0\le \theta\le \frac{\pi}{2}$ which also avoids the positive negative squareroot bs.

[hide="A more geometric solution"]Let the center of the circumcircle of $ ABCD$ be $ O$ and let its radius be $ r$.

By the Law of Cosines on $ AOP$, $ AP^{2}=r^{2}+r^{2}-2r^{2}\cos \angle AOP =2r^{2}(1+\sin\angle DOP )$ Similarly, $ BP^{2}=2r^{2}(1+\sin\angle COP)$

$ PB\cdot PD=(2r\cos\angle PBD)(2r\sin \angle PBD)=2r^{2}\sin 2\angle PBD=2r^{2}\sin\angle DOP$ Similarly, $ PA\cdot PC=2r^{2}\sin \angle COP$ and the identity follows. [/hide]


geometry
circumcircle
combinatorics proposed
combinatorics

There is a frog in every vertex of a regular $ 2n$ gon with circumcirle($ n \ge 2$) At certain time, all the frogs jump to the neighborhood vertices simultaneosly ( maybe there are more than one frog in one vertex), we call it it as "[i]a way of jump[/i]" . It turned out that there exists  "[i]a way of jump[/i]"  with respect to $ 2n$ gon, such that the line connecting any two distinct vertices having frogs on it, after the jump, does not pass through the circumcircle center of [b]2n [/b] gon. Find all the possible value of $ n$.

Solution: There exists a way of jump for a 2n-gon if 4|n with the exception of n=2. If such a jump exists for a 2n-gon then, exactly n of the vertices have to be empty ( do not contain a frog) as there are exactly n pairs of vertices so that line segment joining a pair passes through the centre. A vertex with 0 frog implies the existence of a vertex (different one) with 2 frogs(after the jump a vertex cannot have more than 2 frogs). Hence, there are exactly n vertices with 2 frogs and n vertices with 0 frog.

Let us look at the possible assignment(s) of exactly n 2's and n 0's to each of the vertices so as to satisfy the required condition. Let us call it the vertex sequence.
1. We cannot have three consecutive 2's ('222')and equivalently we cannot have three consecutive 0's ('000').
2.We cannot '202' in the vertex sequence.

So we are left with one way of building the vertex sequence that is 00220022....

If the vertex sequence starts with 00 (22) it has to end with 22 (00). Hence 2|n . Suppose (wlog) the sequence started with '00'( corresponding to vertex numbers 1 and 2), then vertices $ 1\plus{}\frac{n}{2}$ and $ 2\plus{}\frac{n}{2}$ have to be '22'. Hence, $ \frac{n}{2}$ is even.

When n=2, we have the way of jump 1->2,2->1,3->2,4->1


function
real analysis
real analysis unsolved

Show that if f is weakly rectifiable on [a, b] then f is integrable on [a, b].

What is a weakly rectifiable function? :?

Let f be bounded on [a,b].
f is weakly rectifiable on [a,b] when l(f) is finite, for l(f) being sup{l(f,X)|X is a partition on [a,b]} (l(f,X) is the arclength of f on X).


The aqueous solutions of these are acidic or basic?
1) $NaHSO_{4}$
2) $Na_{2}HPO_{4}$
3) $NaHCO_{3}$
4) $NaHSO_{3}$

The first one is acid, because the $HSO_{4}^{-}$ ion is a weak acid (Ka = 0.012). For the other three solutions, try to find the Ka's of the relevant ions and then compare with the corresponding Kb's. The equlibrium with the biggest K will predominate in solution.

Hmm... in my textbook it says $NaH_{2}PO_{4}$ gives an acidic solution, because $H_{2}PO_{4}^{-}+H_{2}O\longrightarrow HPO_{4}^{2-}+H_{3}O^{+}$, but then I think what about $H_{2}PO_{4}^{-}+H_{2}O\longrightarrow H_{3}PO_{4}+OH^{-}$...?

$H_{3}PO_{4}$ is a stronger acid than $H_{2}PO_{4}^{-}$, and hydroxide is a stronger base than water, and so the equilibrium

$H_{2}PO_{4}^{-}+H_{2}O \rightleftharpoons H_{3}PO_{4}+HO^{-}$

will be considerably shifted to the left. On the other hand, $H_{2}PO_{4}^{-}$ is a stronger acid than $HPO_{4}^{2-}$.

OK, so can we, in the same way, predict the solutions of 2),3),4)?

Can we predict that 2),3),4) will be acidic?  :maybe:

For solution number 2:

$Na_{2}HPO_{4}(s)+aq \longrightarrow 2Na^{+}(aq)+HPO_{4}^{2-}(aq)$

$HPO_{4}^{2-}(aq)+H_{2}O(l) \rightleftharpoons PO_{4}^{3-}(aq)+H_{3}O^{+}(aq), K = 2.1 \times 10^{-13}$

$HPO_{4}^{2-}(aq)+H_{2}O(l) \rightleftharpoons H_{2}PO_{4}^{-}(aq)+HO^{-}(aq), K = 1.6 \times 10^{-7}$.

As the second reaction have a greater equilibrium constant, then a solution of hydrogenphosphate ion should be alcaline.

For number 3, I know by experience that a bicarbonate solution is alcaline. Let's check:

$HCO_{3}^{-}(aq)+H_{2}O(l) \rightleftharpoons CO_{3}^{2-}(aq)+H_{3}O^{+}(aq), K = 5.6 \times 10^{-11}$

$HCO_{3}^{-}(aq)+H_{2}O(l) \rightleftharpoons H_{2}CO_{3}(aq)+HO^{-}(aq), K = 2.3 \times 10^{-8}$,

and so a bicarbonate solution is slightly alcaline.

Proceed the same way for solution number 4:

$HSO_{3}^{-}(aq)+H_{2}O(l) \rightleftharpoons SO_{3}^{2-}(aq)+H_{3}O^{+}(aq), K = 1.2 \times 10^{-7}$

$HSO_{3}^{-}(aq)+H_{2}O(l) \rightleftharpoons H_{2}SO_{3}(aq)+HO^{-}(aq), K = 6.7 \times 10^{-13}$.

So, a bisulphite ion solution is expected to be slightly acidic.

Well, why do you call it "bisulphite" and "bicarbonate". Where does "bi" come from?

The "bi" is the common name for the ion of a polyprotic acid with one hydrogen.  The IUPAC naming conventions use the word "hydrogen " instead.  I can't recall where the prefix "bi" itself originated.

One should also note that it is possible to calculate the pH of a solution of a polyprotic acid salt, even when it has one or more hydrogens still attached.  The formula for such a salt with one or more hydrogens still attached, say $\text{HA}^{-}$, is $\text{pH}= \frac{\text{pK}_\text{a1}+\text{pK}_\text{a2}}{2}$, where $\text{K}_\text{a1}, \text{K}_\text{a2}$ are the acid dissociation constants for $\text{H}_{2}\text{A}$ and $\text{HA}^{-}$, respectively.  Note that the pH is independent of the concentration of the salt.


Consider the six by six grid of unit squares below. How many
rectangles of area 8 square units can be formed using only the
line segments of the grid as the sides of the rectangles?

[asy]unitsize(0.2inch);
draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);
draw((1,0)--(1,6));
draw((2,0)--(2,6));
draw((3,0)--(3,6));
draw((4,0)--(4,6));
draw((5,0)--(5,6));
draw((0,1)--(6,1));
draw((0,2)--(6,2));
draw((0,3)--(6,3));
draw((0,4)--(6,4));
draw((0,5)--(6,5));[/asy]

Hello
$ 4*4 \equal{} 16$
16 squares are of area $ 8$ square units
Thank u.

i got a different answer

there are 4 ways to make rectangles of 8 square units
1 by 8
2 by 4
4 by 2
8 by 1

the 1 by 8 and 8 by 1 are impossible since we do not have so many squares, so those can be eliminated

now we think about the 2 by 4 rectangles

the grid is 6 by 6, so we can have a total of $ 3\times5$ rectangles
if you don't believe me, try putting a 2 by 4 rectangle in the top left corner
now move it to the right, and record how many ways it can be put
now go back and move it down, and record how many ways
the numbers should be 3 and 5

now we think about 4 by 2 rectangles
for this, there are the same number

so we have a total of 30 rectangles

Hello
Vallon ur answer is right.By mistake I read no. of squares with area $ 8$ sq. units.
Thank u.

[quote="NQ123cute"]Sorry but both of ya'll are incorrect... it's 46.... I saw the solution just to verify it...[/quote]

Um, I don't think $46$ is correct.  There are only two possible cases--a $2\times4$ rectangle and a $4\times2$ rectangle, and each appear $3 \cdot 5 = 15$ times.  If $46$ is the correct answer on Alcumus, then maybe there is an error.

If you don't believe me, click [url=http://www.imlem.org/sch/winchester/mc-2003-state-team-with-answers.pdf]here[/url].  It's #2.  Answers are shown below.

lol i wish mathcounts state was this easy...


conics
parabola
analytic geometry
geometry unsolved
geometry

Let $ A,B,C,D$ be the four distinct points. Prove that $ A,B,C,D$ are cyclic iff there exist two parabola passing $ A,B,C,D$ which directrix are perpendicular.

Let's suppose there exists two parabola passing both passing through $ A,B,C,D$, and its directrixes are perpendicular to each other. The two parabolas can be denoted by the equation; $ (y \minus{} a)^2 \equal{} 4p(x \minus{} p), (x \minus{} b)^2 \equal{} 4q(y \minus{} q)$.  (I setted the $ xy$ coordinate system that $ x,y$ axises are the two directrixes.)

The curve system passes through $ A,B,C,D$ is:

$ \left((x \minus{} a)^2 \minus{} 4p(y \minus{} p)\right)h \plus{} \left((y \minus{} b)^2 \minus{} 4q(x \minus{} q)\right) \equal{} 0$, $ h$ is an arbitrary real number. This curve always passes through the four distinct points $ A,B,C,D$. We have to prove that there exists a circle that passes through $ A,B,C,D$, and it can be done by putting $ h$ to $ 1$.

 I want to see the proof of necessary condition. (I'm trying, but it doesn't work....)


algebra
polynomial
Rational Root Theorem

consider the following expression:
$ 4f^3 \plus{} 9f^2 \minus{} 4f$ where $ f$ is a reduced fraction.

for which value of $ f$, the above expression is an integer ?

[hide]
From the rational root theorem, if $ f \equal{} \frac {p}{q} \longrightarrow q \equal{} \{1,2,4\}$
For now we let $ q \equal{} 4$ because that covers everything.

$ \Rightarrow \frac {p^3}{16} \plus{} \frac {p^2}{16} \minus{} p \equal{} k$ for some intiger $ k$
$ \Rightarrow p^3 \plus{} 9p^2 \minus{} 16p \equiv 0 \mod 16$

Since $ f$ is a reduced fraction $ \gcd(p,q) \equal{} (p,4) \equal{} (p,16) \equal{} 1$, so
$ p^2 \plus{} 9p \minus{} 16 \equiv 0 \mod 16$
$ p \plus{} 9 \equiv 0 \mod 16$

$ p \equal{} 7 \plus{} 16k$

Checking we see that $ f\left(\frac {7}{4}\right) \equal{} 42$ which is an intiger so

$ f \equal{} \dfrac{7 \plus{} 16k}{4}$ where $ k$ is a positive intiger
[/hide]


geometry
circumcircle
trigonometry
geometry proposed

Let $ \gamma$ be circumcircle of $ \triangle ABC$.Let $ R_a$ be radius of circle touching $ AB,AC$&$ \gamma$ internally.Define $ R_b,R_c$ similarly.
Prove That $ \frac {1}{aR_a} \plus{} \frac {1}{bR_b} \plus{} \frac {1}{cR_c} \equal{} \frac {s^2}{rabc}$.

[color=darkblue]Let $ B'$ and $ C'$ b tangency points of $ \gamma$ with $ AB$ and $ AC$. It' s wellknown that $ I$ is midpoint of $ B'C'$. Let $ O_A$ be the center of $ \gamma$. Then

$ IB' \equal{} AI \tan \dfrac{A}{2} \equal{} \dfrac{AI \sin \dfrac{A}{2} }{\cos \dfrac{A}{2}} \equal{} \dfrac{r }{\cos \dfrac{A}{2}}$

$ R_a \equal{} \dfrac{IB'}{\cos \dfrac{A}{2}} \equal{} \dfrac{r }{\cos^2 \dfrac{A}{2}} \equal{} \dfrac{r}{\frac {p(p \minus{} a)}{bc}} \equal{} \dfrac{r \cdot bc}{p(p \minus{} a)}$

Then

$ \sum \dfrac{1}{aR_a} \equal{} \sum\dfrac{p(p \minus{} a)}{r abc} \equal{} \dfrac{p}{rabc}\sum (p \minus{} a) \equal{} \dfrac{p^2}{rabc}$

Hm.... Am I right?[/color]

[quote="Ahiles"]It' s wellknown that $ I$ is midpoint of $ B'C'$[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=31739&start=0

The radius of the $A-$ mixtilinear incircle $R_a=r\sec^2\frac{A}{2}=\frac{r}{\frac{1+\cos A}{2}}=\frac{2r}{1+\cos A}$

So, $\frac{1}{aR_a}=\frac{1+\cos A}{2ra}$

and hence

${\sum\frac{1}{aR_a}=\frac{1}{2r}\sum\frac{1+\cos A}{a}=\frac{1}{2r}\sum\frac{1+\frac{b^2+c^2-a^2}{2bc}}{a}}$

$\Longrightarrow \sum\frac{1}{aR_a}=\frac{1}{4r}\sum\frac{(b+c)^2-a^2}{abc}=\frac{1}{4r}\sum\frac{4s(s-a)}{abc}$

$\Longrightarrow \sum\frac{1}{aR_a}=\frac{s}{rabc}\sum(s-a)=\frac{s^2}{rabc}$ as required.

Using the Casey theorem for A,B,C and circle internally tangent to arc BC(not containing A) we get
$AX=\frac{2bc}{a+b+c}$ and analogous expressions for others.
So$ LHS=\frac{a+b+c}{2abc}(\sum\frac{1}{tan\frac{A}{2}})=\frac{a+b+c}{2abc}\frac{1}{tan\frac{A}{2} tan \frac{B}{2} tan \frac{C}{2}}=\frac{r^2}{sabc}$.

Please point out if I am wrong anywhere....

Bye..

Sayantan....

The right hand side of the equation should be
\[\frac{s^2}{rabc}\]
Mods, please fix [s]I couldn't solve this because I was trying to prove the other way around...[/s]

[color=#f00]Fixed, thanks!  ~dj[/color]


A rectangle has length $ 15$ inches and width $ 10$ inches. If these dimensions are both increased by $ 20%
$, by how many square inches will the area of the rectangle be increased?


I had a question on a previous Mathematics UIL Test (If you live in Texas, you'll be familiar).  The question asked about finding the amount of vertices on a dodecahedron, with only the knowledge of the number of edges and the number of sides per face (which I believe is 6).  Is their a specific equation for finding the relationship?

Hey there, I just graduated but I did UIL in High School. I think this is one of the state release tests,(lettered from A-I) and the test writer most of the times includes a question that involves the formula $ F \plus{} V \equal{} E \plus{} 2$ which is also known as Euler's Formula.

yea.. thanks for the reply, and.. how long ago did you graduate, if you mind me asking?

I will definitely look over Euler's Formula.

I graduated this past year (class of 08) from La Joya HS(5A).

The formula is also written commonly as F+V-E=2, and can be manipulated as necessary.


calculus integration trigonometry limit logarithms function derivative

Dear everyone.

please answer my shoddy question  :blush:

calculate $ \int_{0}^{\infty}{ \frac{1\minus{}\cos{x}}{x(e^{x}\minus{}1)}dx }$

thanks.

Let's do a side calculation first:

Let $ g(t) \equal{} \int_0^{\infty}\frac {1 \minus{} \cos x}{x}\cdot e^{ \minus{} tx}\,dx$

Differentiating, we get $ g'(t) \equal{} \minus{}\int_0^{\infty}(1 \minus{} \cos x)e^{ \minus{} tx}\,dx \equal{} \minus{}\frac1t \plus{}\frac {t}{t^2 \plus{} 1}.$

Since it's also clear that $ \lim_{t\to\infty}g(t) \equal{} 0,$ we can integrate back to get that

$ g(t) \equal{} \frac12\ln\left(\frac {t^2 \plus{} 1}{t^2}\right).$

Now write $ \frac1{e^x \minus{} 1} \equal{} \frac {e^{ \minus{} x}}{1 \minus{} e^{ \minus{} x}} \equal{} \sum_{k \equal{} 1}^{\infty}e^{ \minus{} kx}.$

Then (justifying the interchange of sum and integral by the nonnegativity of the integrands)

$ \int_0^{\infty}\frac {1 \minus{} \cos x}{x(e^x \minus{} 1)}\,dx \equal{} \sum_{k \equal{} 1}^{\infty}\int_0^{\infty}\frac {1 \minus{} \cos x}{x}\cdot e^{ \minus{} kx}\,dx$

$ \equal{} \frac12\sum_{k \equal{} 1}^{\infty}\ln\left(\frac {k^2 \plus{} 1}{k^2}\right).$

I don't yet see what I'm going to do with that sum.

I did take the time to run a numerical estimate on the sum, using an extrapolation technique on the partial sums. The number I got, $ 0.650923199302,$ agrees with milin's answer to at least that many digits.

How do we know when we can differentiate an improper integral?

That particular step can be justified by the Dominated Convergence Theorem. Use some multiple of $ e^{\minus{}t_0x}$ as the dominating function, and only do this for $ t>t_0>0.$ The claim that $ g(t)\to 0$ as $ t\to\infty$ can also be justified by the DCT.

Yes, DCT makes it easier, but how'd you justify it by not using analysis tools?

Oh, I can make it more elementary, albeit less elegant:

$ \frac{g(t\plus{}h)\minus{}g(t)}{h}\equal{}\int_0^{\infty}\frac{1\minus{}\cos x}{x}\cdot\frac{e^{\minus{}(t\plus{}h)x}\minus{}e^{\minus{}tx}}{h}\,dx$

$ \equal{}\int_0^{\infty}\frac{1\minus{}\cos x}{x}\cdot\frac{e^{\minus{}hx}\minus{}1}{h}\cdot e^{\minus{}tx}\,dx$

Now we try estimating $ \frac{e^{\minus{}hx}\minus{}1}{h}$ using Taylor's theorem with remainder. If $ f(h)\equal{}e^{\minus{}hx}$ then $ f'(0)\equal{}\minus{}x$ and $ f''(\xi)\equal{}x^2e^{\minus{}\xi x}$ for some $ |\xi|<|h|$ We estimate that as $ |f''(\xi)|<x^2e^{|h|x}$ so that

$ \left|\frac{e^{\minus{}hx}\minus{}1}{h}\plus{}x\right|\le \frac12x^2|h|e^{|h|x}.$

Let $ D(t)\equal{}\minus{}\int_0^{\infty}(1\minus{}\cos x)e^{\minus{}tx}\,dx,$ which is the thing I claimed is the derivative of $ g(t).$ (Note: I've got to go back to my original post to clean up a sign error - one that was reversed a step or two later, so the final result is sound.) Then

$ \left|\frac{g(t\plus{}h)\minus{}g(t)}{h}\minus{}D(t)\right|\le\int_0^{\infty}\frac{1\minus{}\cos x}{x}\cdot e^{\minus{}tx}\cdot \frac12x^2|h|e^{|h|x}\,dx$

$ \le C|h|\int_0^{\infty}x^2e^{\minus{}(t\minus{}|h|)x}\,dx\equal{}\frac{2C|h|}{(t\minus{}|h|)^3}$

And that goes to zero as $ h\to 0,$ verifying our result.

Somewhere in there, I used that $ \frac{1\minus{}\cos x}{x}$ is (if you fill in the removable discontinuity) a continuous function on $ [0,\infty)$ that tends to zero at $ \infty$ and is therefore bounded. I didn't feel motivated to compute its actual maximum.


A number's persistence is the number of steps required to reduce it to a single digit by multiplying all its digits to obtain a second number, then multiplying all the digits of that number to obtain a third number, and so on until a one-digit number is obtained.  For example, 77 has a persistence of four because four steps are required to reduce it to one digit: 77 --> 49 --> 36 --> 18 --> 8.

The smallest number of persistence one is 10, the smallest of persistence two is 25, the smallest of persistence three is 39, the smallest of persistence four is 77.  What is the smallest number of persistence five?

Posted by MrMcGeek

[hide]

679 ?  I did it by guessing and trying but is there an astute way to find it ?

[/hide]

[hide]you could try figuring out waht number in a number make sure that it is a 2, or a 3, or a 1. for example a 0 means it's 1, a 1 means it the number of digits. etc..[/hide]

[hide]Another way to solve this is to find the smallest number that has persistence of four, and has no prime factors over 10.  Using those factors, you can find the number of persistence five that gives that number.  That might not work though.[/hide]

belenos got it right, but i want to see ml2's and bookworm's way of doing it.

remember... show your work

Posted by MrMcGeek


Konig's theorem:

Let $G=(A,B,E)$ a bipartite graph. Then the number of edges in a maximum matching equals the number of vertices in a minimum vertex cover.

Use Hall's theorem to prove the above.

Clue: If $C = X \cup Y$ such that $X \subseteq A$ and $Y \subseteq B$ then there is a matching of $X$ in the sub-graph on the vertices $X \cup (N(X) \setminus Y)$ and there is a matching of $Y$ in the sub-graph on the vertices $(N(Y) \setminus X) \cup Y$.
($N(X)$ denotes the set of neighbors of $X$, $N(Y)$ denotes the set of neighbors of $Y$)

denote $\alpha' (G)$ as the number of edges in a maximum matching,and also let $\beta (G)$ denote the number of vertices in a minimum vertex cover...
now we want to prove that $\beta (G)=\alpha' (G)$...
first trivilay we have that $\boxed{\beta (G)\geq \alpha' (G)}(I)$,now assume that $M$ is a matching with $\alpha'$ edge in $G$,and let $U$ be a minimum vertex cover in $G$,we show that $|U|\leq \alpha'$
let $X,Y$ be the bipartite of $G$ and denote $T=U \cap X$ and $S=U \cap Y$
now let $G_{1}$ be the graph with vertices $S,X-T$ and let $G_{2}$ be the graph with vertices $T,Y-S$,now we claim that there is a matching in $G_{1}$ which saturates $S$,and there is a matching in $G_{2}$ which saturates $T$...
now according to $Hall$ its sufficient to prove that $\forall A\subset S$ we have $|N(A)|\geq |A|$,wwe`ll prove this with contradiction:
assunme the contrary $|N(A)|<|A|$,in this case if we put $N(A)$ instead of $A$,in $U$ we`ll get a vertex cover which is smaller than $U$ which is a contradiction...
so the claim is proved,there for according to $Hall$ there exists a matching with $\beta (G_{1})$ edges therefor we get that:
$\boxed{\alpha '\geq \beta}(II)$
$(I),(II)\Rightarrow \alpha'=\beta$ and we are done...


calculus
integration
derivative
function
advanced fields
advanced fields theorems

The problem statement:

Solve the initial value problem:
\begin{align}&yu_x + u_y = x  \\ &u(x,0) = x^2\\&\frac{dx}{y}=\frac{dy}{1} \end{align}

Solving (3) for a first integral gives:

$ x- \frac{y^2}{2} = c_1= u_1$

Shouldn't first integrals satisfy the P.D.E. ?

$ u_x = 1$
$ u_y = -y$

Substitute into the (1)
$ y(1) -y \neq x$ ??

I'm not sure about the terminology here, but the solutions of (3) should satisfy the PDE given by (3), i.e., the [b]homogeneous[/b] PDE with the right hand side $ 0$ instead of $ x$, which they do. Remember that they just give you the curves such that the left hand side (the one with the derivatives) is exactly the derivative of the unknown function in the direction of the curve. After that you have to solve the ODE to which (1) reduces on each such curve.

I think I understand what you are saying. The text does not explicitly make that statement (unless of course, I've missed it).

Are you saying that a first integral, i.e. solution to (3), results in a curve that is a directional derivative in the direction of the curve that equation (1) comes from?

Charact equation:
$ \frac{dx}{y}=\frac{dy}{1}$

Then $ C=x-\frac{y^2}{2}$.

Time $ \tau$ alone characteristic we find from
$ \frac{dy}{1}=d\tau$, $ y|_{\tau=0}=0$.

Then $ y=\tau$.

Hence $ x=C+\frac{\tau^2}{2}$, $ y=\tau$,

$ \frac{du}{d\tau}=C+\frac{\tau^2}{2}$, $ u|_{\tau=0}=(C+\frac{\tau^2}{2})|_{\tau=0}=C^2$.

We receive $ u=C\tau+\frac{\tau^3}{6}+C^2$ and

in terms $ x$, $ y$
$ u=(x-y^2/2)y+y^3/6+(x-y^2/2)^2$.

[quote="V.V."]
in terms $ x$, $ y$
$ u \equal{} (x \minus{} y^2/2)y \plus{} y^3/6 \plus{} (x \minus{} y^2/2)^2$.[/quote]

Text book solution...

[quote="V.V."][quote="vector2"]Text book solution...[/quote]

My textbook (in Russian) on [url]http://joker.botik.ru/~trushkov[/url] :)[/quote]

I think they're all in Russian...

[quote="V.V."]
Time $ \tau$ alone characteristic we find from
$ \frac {dy}{1} = d\tau$, $ y|_{\tau = 0} = 0$.

Then $ y = \tau$.

Hence $ x = C + \frac {\tau^2}{2}$, $ y = \tau$,

$ \frac {du}{d\tau} = C + \frac {\tau^2}{2}$, $ u|_{\tau = 0} = (C + \frac {\tau^2}{2})|_{\tau = 0} = C^2$.
[/quote]

Rereading your post, I realized that I didn't realize considering 1 characteristic $ \longrightarrow \frac {dy}{1} = d\tau$, $ y|_{\tau = 0} = 0$ , as you have, was valid.  Excellent...


SALAMALEYKO????????????????????????????????? :rotfl:
turkey=lung(in ARMENIAN language) :rotfl:  :D

You are so funny  :thumbup:

But  I am afraid that urgently you need some medical  :help:

Thanks a lot for your magnanimous, but I needn't any help. :wink:

[quote]Thanks a lot for your magnanimous, but I needn't any help. [/quote]

OK But (according to your writings) it doesn't seem like that
Please don't write these stupid messages again and please don't write [b]selamaleyko[/b] any more or if you want to write it please write it in correct form as [b]Esselamu-alejkum[/b]
This incorrect form insult me personally.


[quote]Prove that among sixteen consecutive integers it is always possible to find one that is relatively prime to all the rest.
[/quote]

Maybe casework?

I think a mod should delete one or the other topic, because they are the same questions being asked.

[quote="lotrgreengrapes7926"][quote="sapphyre571"]I really want to say that in every set of sixteen integers, there is a prime, but that hasn't been proven yet[/quote] That's not true, though.  :maybe:[/quote]

 :D

[hide="Say no to caseworking"]Clearly, the $\gcd$ of two of those integers is smaller than 16. So if we could prove that there's an integer that isn't divisible by 2, 3, 5, 7 and 11, we're done. There are 8 even numbers. Forget about them. There are at most 3 odd multiples of 3 and 2 multiples of 5 that aren't divisible by 2 and 3. Forget about them too. There is at most one odd multiple of 7 and 11. These are 8+3+2+1+1 = 15 numbers. The one who's left, is the one we're looking for.[/hide]

By the way, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=104101]1[/url], [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=104103]2[/url] and this one's the third. Triplepost  :dry:

[quote="sapphyre571"]I really want to say that in every set of sixteen integers, there is a prime, but that hasn't been proven yet

this takes lots of casework[/quote]

You can have as many consecutives composite numbers as you want, but always a finite quantity.

[quote="Kurt Gödel"][quote="lotrgreengrapes7926"][quote="sapphyre571"]I really want to say that in every set of sixteen integers, there is a prime, but that hasn't been proven yet[/quote] That's not true, though.  :maybe:[/quote][/quote]

 :D

[hide="Say no to caseworking"]Clearly, the $\gcd$ of two of those integers is smaller than 16. So if we could prove that there's an integer that isn't divisible by 2, 3, 5, 7 and 11, we're done. There are 8 even numbers. Forget about them. There are at most 3 odd multiples of 3 and 2 multiples of 5 that aren't divisible by 2 and 3. Forget about them too. There is at most one odd multiple of 7 and 11. These are 8+3+2+1+1 = 15 numbers. The one who's left, is the one we're looking for.[/hide]

what about 13?

Hmmm, you're right. Let's find sixteen consecutive integers with 8 even numbers, 5 multiples of 3, 3 multiples of 5, 2 multiples of 7, 2 multiples of 11 and two multiples of 13  :lol:

In fact, by chinese remainder theorem you can construct arbitrarily large sets of consecutive integers without primes.

CRT is not something that a Getting Started user should be expected to know.  If you're going to use it you should offer an explanation.

A much easier and understandable way to make the same point would be to note the following sequence of numbers: $\{ n!+2, n!+3, n!+4, \ldots , n!+n\}.$  Each term is of the form $n!+k$ where $1<k\le n$ so we can factor as $k\left(\frac{n!}k+1\right)$.  We know that $\frac{n!}k$ is an integer since $n!=1\cdot2\cdots k\cdots n.$

Indeed, I forgot about that example. :)

Either way just googling chinese remainder theorem gives a very good explanation on number 1, that's why I pointed to google, not because I assumed it known.

( and I again wonder how the heck I got into getting started... this forum is supposed to be hidden for me :huh: )


calculus
integration
trigonometry
limit
calculus computations

By trigonometric substitution, let $ x\equal{}\tan\theta$, then $ dx\equal{}\sec^2\theta d\theta$...Solving, the integral, I got $ (4k\minus{}3)\pi/2$, where $ k\in \mathbb{Z}^\plus{}$.

Um ... this is a specific definite integral, and it equals a single specific number. A cheap estimate: it's no more than
\[ \int_0^1x^2\,dx\plus{}\int_1^{\infty}\frac1{x^2}\,dx\equal{}\frac12\plus{}1\equal{}\frac43.\]
You can probably use that to throw out most of your array of possible values.

\begin{eqnarray*}
\int_{0}^{+\infty}\frac{x^{2}}{\left(x^{2}+1\right)^{2}}dx
&=& \displaystyle\lim_{M \rightarrow \infty} \int_{0}^{M}\left(\dfrac{1}{x^2 + 1} - \frac{1}{\left(x^{2}+1\right)^{2}}\right)dx
\\&=& \dfrac{\pi}{2} - \displaystyle\lim_{M \rightarrow \infty} \int_{0}^{M} \frac{1}{\left(x^{2}+1\right)^{2}}dx
\\&=& \dfrac{\pi}{2} - \displaystyle\lim_{M \rightarrow \frac{\pi}{2}} \int_{0}^{M} \frac{1}{\left(\tan^{2}\theta+1\right)^{2}}\sec^2\theta \, d\theta
\\&=& \dfrac{\pi}{2} - \displaystyle\lim_{M \rightarrow \frac{\pi}{2}} \int_{0}^{M}\cos^2\theta \, d\theta
\\&=& \dfrac{\pi}{2} - \displaystyle\lim_{M \rightarrow \frac{\pi}{2}} \dfrac{1}{2}\int_{0}^{M}(1 + \cos 2\theta)\, d\theta
\\&=& \dfrac{\pi}{2} - \displaystyle\lim_{M \rightarrow \frac{\pi}{2}} \dfrac{1}{2}\left[\theta + \dfrac{1}{2}\sin 2\theta\right]_{0}^{M}
\\&=& \dfrac{\pi}{4}
\end{eqnarray*}

Used the substitution $ x = \tan \theta$ in the middle.

Setting $ x\equal{}\frac{1}{y}$

$ I\equal{}\int_{0}^{\infty }\frac{x^{2}}{\left( x^{2}\plus{}1\right) ^{2}}dx\equal{}\allowbreak \int_{0}^{\infty }\frac{1}{\left( 1\plus{}y^{2}\right) ^{2}}\,dy$

$ 2I\equal{}\int_{0}^{\infty }\frac{1}{1\plus{}x^{2}}\,dx\equal{}\frac{\pi }{2}$

Hence $ \int_{0}^{\infty }\frac{x^{2}}{\left( x^{2}\plus{}1\right) ^{2}}dx\equal{}\frac{\pi }{4}$

$ \int_{0}^{\plus{}\infty}\frac{x^{2}}{\left(x^{2}\plus{}1\right)^{2}}dx \equal{} \frac12 \int_{\minus{}\infty}^{\plus{}\infty}\frac{x^{2}}{\left(x^{2}\plus{}1\right)^{2}}dx \equal{} \left. \pi i Res_{z\equal{}i} \frac{d}{dz}\left((z\minus{}i)^2 \frac{z^2}{(z^2\plus{}1)^2}\right)\right|_{z\equal{}i} \equal{} \left. \pi i Res_{z\equal{}i} \frac{d}{dz}\left( \frac{z^2}{(z\plus{}i)^2}\right)\right|_{z\equal{}i} \equal{} \pi i Res_{z\equal{}i} \left( \frac{2z(z\plus{}i)^2\minus{}2(z\plus{}i)z^2}{(z\plus{}i)^4}\right) \equal{} \pi i \frac{\minus{}8i\plus{}4i}{(2i)^4} \equal{} \frac{\pi}{4}$

oopss..yeah! sorry I misunderstood the problem...I know it's a definite integral, an improper one, so it must diverge to a single value only...BTW, my answer to my last post is supposed to $ (4k\minus{}3)/4$...What I did was to look for values of $ \theta$ for which $ \tan \theta$ goes to positive infinity, & that's how I did my solution..Anyway, the solution should be only for $ k\equal{}1$, hence, $ \pi /4$. :)


Hi everyone. I am coming after a long time to this forum, thanks to a friend of mine, who gave me a problem which I am unable to figure out.

He suggested a 3D sudoku. It is very difficult for me to explain what it actually means. I just understood it as he said this to me. And I hope you too will understand it. You can ask some options as to what you are understanding by the word "3d sudoku". I can only say that there will be 81 rows, 81 columns, in the form of 9 rows and 9 columns back to back...

Well the problem he has put before me is that, whether there exists such a thing. I am unable to figure out will a 3d sudoku be possible or not.

2D Sudokus are already challenging and interesting, I don't see what is the purpose of creating such things as 3D Sudokus.

What about, "Every xy, yz, and zx plane, and the 27 3x3x3 cubes have the numbers 1-81 exactly once"?

kde's ksudoku comes with a 3d option. Its version is a 3x3x3 cube where each cublet has one number (the same on each face) and every plane parallel to a side of the cube has the numbers 1-9.


Given are triangle $ABC$ with in radius $r$ and three exradius $r_{a},r_{b},r_{c}$ prove that
$\sqrt{\frac{1}{rr_{a}}+\frac{1}{r_{b}^{2}}}+\sqrt{\frac{1}{rr_{b}}+\frac{1}{r_{c}^{2}}}+\sqrt{\frac{1}{rr_{c}}+\frac{1}{r_{a}^{2}}}\ge\frac{2}{r}$
 :D

Very nice inequality, but not so difficult.
If we put $p-a=x,p-b=y,p-c=z$ and normalize $x+y+z=1$ we change it in the (well known ?  :P ) inequality
$\sum_{cyc}\sqrt{x+y^{2}}\geq 2$
which can be proved by using
$\sqrt{a}+\sqrt{b}\geq \sqrt{c}+\sqrt{a+b-c}$ where $c=min\{a,b,c\}$

ok that's right Albanian Eagle, you known its source,  :D  but is there a geometric or trigonometric solution ?  :)


For the list of awards page, is it limited to high-school awards?

Are you talking about [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPS_Community_Awards]this[/url] page?  If you look closely you'll see MathCounts and Putnam on there, so...

[quote="anirudh"]I mean that can we add other stuff, like MOEMS?[/quote]
I'm no admin, but I don't think that is advised because it would be near impossible to get everyone whose done well on MOEMS' names listed. Also, doing well on MOEMS is no big deal (no offense meant). It isn't very difficult. The awards compilation doesn't even list every national mathcounts participant, or every USAMO qualifier. These are difficult things to accomplish. Again, I'm no admin or mod, but I strongly suggest that you don't for these reasons. Then again, maybe I have no clue what I'm saying.

[quote="joml88"]It's preferable to keep the list to a reasonable size.  So I would say no.  However, one thing you can do is make a page for AoPSers who score high on MOEMS.  Then on the other big page you can say something like "there currently $n$ AoPS/ML members who have done [such and such] well on MOEMS."  I don't really know much about MOEMS though...

It would be pretty cool to do something like this for USAMO.[/quote]
Let's just say that it isn't that hard to master Moems problems.


geometry
Gauss
vector
linear algebra
matrix

Let a triangle $ABC$ with $\angle A$ acute,  a point  $D$ on side $AB$ , a point $E$ on side $AC$ , the orthocenter $Z$ of triangle $ADE$ and the orthocenter $H$ of triangle $ABC$. If $K, L$ are the midpoints of segments $BE , CD$, prove that
   $KL \perp ZH$.

 [b] NOTE[/b]

 [color=blue]On my paper I have noted that the problem is  from a local competition in Bucarest 1991 , but I have not a source to ensure this. Probably I’m mistaken.[/color]

[color=darkblue][b][u]Very nice and well-known problem ! Here is an original proof ![/u][/b]

[b][u]Lemma.[/u][/b] $\boxed{\ MN\perp XY\Longleftrightarrow MX^{2}-MY^{2}=NX^{2}-NY^{2}\ }\ .$

[size=134][b][u]The general enunciation.[/u][/b][/size] $\triangle ABC\ : |\begin{array}{c}\ D\in AB\ ,\ E\in AC\ .\\\ \mathrm{orthocenter}\ Z\ \mathrm{of}\ \triangle ADE\ .\\\ \mathrm{orthocenter}\ H\ \mathrm{of}\ \triangle ABC\ .\\ \ K\in BE\ ,\ KB=KE\ .\\ L\in CD\ ,\ LC=LD\ .\end{array}\Longrightarrow HZ\perp KL\ .$

[b]Proof[/b] ([u]a very interesting use of the above remarkable lemma[/u]).

$\boxed{\mathrm{I}}\blacktriangleright\ AH\perp BC\Longrightarrow \boxed{\ HC^{2}-HB^{2}=AC^{2}-AB^{2}\ }\ \ (1)\ .$

$|\begin{array}{ccc}CH\perp AD & \Longrightarrow & HD^{2}-HA^{2}=CD^{2}-CA^{2}\\ BH\perp AE & \Longrightarrow & HA^{2}-HE^{2}=BA^{2}-BE^{2}\end{array}|$ $\Longrightarrow$ $\boxed{\ HD^{2}-HE^{2}=CD^{2}+BA^{2}-CA^{2}-BE^{2}\ }\ \ (2)\ .$

Apply the theorem of the $H$- median in the triangles $HCD\ ,\ HBE$ and the previous relations $(1)\ ,\ (2)\ :$

$|\begin{array}{c}4\cdot HL^{2}=2\cdot(HD^{2}+HC^{2})-DC^{2}\\ 4\cdot HK^{2}=2\cdot(HE^{2}+HB^{2})-BE^{2}\end{array}|$ $\Longrightarrow$ $4\cdot(HL^{2}-HK^{2})=2\cdot(HC^{2}-HB^{2})+2\cdot(HD^{2}-HE^{2})+(BE^{2}-DC^{2})$ $\Longrightarrow$

$4\cdot(HL^{2}-HK^{2})=$ $2\cdot (AC^{2}-AB^{2})+2\cdot(CD^{2}+BA^{2}-CA^{2}-BE^{2})+(BE^{2}-DC^{2})$ $\Longrightarrow$ $\boxed{\ 4\cdot(HL^{2}-HK^{2})=CD^{2}-BE^{2}\ }\ \ \ (3)\ .$

$\boxed{\mathrm{II}}\blacktriangleright\ AZ\perp DE$ $\Longrightarrow$ $\boxed{\ ZD^{2}-ZE^{2}=AD^{2}-AE^{2}\ }\ \ \ (1')\ .$

$|\begin{array}{ccc}EZ\perp AB & \Longrightarrow & ZA^{2}-ZB^{2}=EA^{2}-EB^{2}\\ DZ\perp AC & \Longrightarrow & ZC^{2}-ZA^{2}=DC^{2}-DA^{2}\end{array}|$ $\Longrightarrow$ $\boxed{\ ZC^{2}-ZB^{2}=DC^{2}+EA^{2}-DA^{2}-EB^{2}\ }\ \ \ (2')\ .$

Apply the theorem of the $Z$- median in the triangles $ZCD\ ,\ ZBE$ and the previous relatons $(1')\ ,\ (2')\ :$

$|\begin{array}{c}4\cdot ZL^{2}=2\cdot(ZD^{2}+ZC^{2})-DC^{2}\\ 4\cdot ZK^{2}=2\cdot(ZE^{2}+ZB^{2})-BE^{2}\end{array}|$ $\Longrightarrow$ $4\cdot(ZL^{2}-ZK^{2})=2\cdot (ZC^{2}-ZB^{2})+2\cdot(ZD^{2}-ZE^{2})+(BE^{2}-DC^{2})$ $\Longrightarrow$

$4\cdot(ZL^{2}-ZK^{2})=$ $2\cdot(DC^{2}+EA^{2}-DA^{2}-EB^{2})+2\cdot(AD^{2}-AE^{2})+(BE^{2}-DC^{2})$ $\Longrightarrow$ $\boxed{\ 4\cdot(ZL^{2}-ZK^{2})=CD^{2}-BE^{2}\ }\ \ \ (3')\ .$

From the relations $(3)$ and $(3')$ obtain $HL^{2}-HK^{2}=ZL^{2}-ZK^{2}$, i.e. $HZ\perp KL\ .$

[b]Remark.[/b] Let $ABCD$ be a convex quadrilateral for which denote the intersections $E\in AB\cap CD$ and $F\in AD\cap BC\ .$
Then the middlepoints $M$, $N$, $P$ of the "diagonals" $[AC]$, $[BD]$, $[EF]$ are collineary (the [b][u]Gauss' line[/u][/b]) ,
the orthocenters $X$, $Y$, $Z$, $T$ of the triangles $CDF$, $BCE$, $ABF$, $ADE$ are collineary (the [b][u]Aubert's line[/u][/b]) and $\overline{XYZT}\perp\overline{MNP}\ .$[/color]

[quote="Virgil Nicula"][b]Remark.[/b] Let $ABCD$ be a convex quadrilateral for which denote the intersections $E\in AB\cap CD$ and $F\in AD\cap BC\ .$
Then the middlepoints $M$, $N$, $P$ of the "diagonals" $[AC]$, $[BD]$, $[EF]$ are concurrently (the [b][u]Gauss' line[/u][/b]) ,
the orthocenters $X,Y,Z,T$ of the triangles $CDF$, $BCE$, $ABF$, $ADE$ are collineary (the [b][u]Aubert's line[/u][/b]) and $\overline{XYZT}\perp \overline{MNP}$. [/quote]

I'll prove the Gauss' line result in a classical way (note that the midpoints are not concurrent, they are collinear).

I accidentally interchanged the notations: $\left\{ E \right\}= AD \cap BC$ and $\left\{ F \right\}= AB \cap CD$, so I'll leave it this way.

[hide="Here is the solution using skew axis"]
The vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$ are not collinear, so they form a basis, with respect to which we define (skew) coordinates.

We have that $A(0,0), \, B(b,0), \, D(0,d), \, C(u,v)$;

- $AD$: $x = 0$;

- $BC$: $xv+y(b-u)-bv = 0$;

- $AB$: $y = 0$;

- $CD$: $x(v-d)-yu+du = 0$.

Thus, $E \left( 0, \frac{bv}{b-u}\right)$ and $F \left( \frac{du}{d-v}, 0 \right)$, so $M \left( \frac{u}2, \frac{v}2 \right)$, $N \left( \frac{b}2, \frac{d}2 \right)$ and $P \left( \frac{du}{2(d-v)}, \frac{bv}{2(b-u)}\right)$.

We have that
\begin{eqnarray*}\left| \begin{matrix}x_{M}& y_{M}& 1 \\ x_{N}& y_{N}& 1 \\ x_{P}& y_{P}& 1 \end{matrix}\right| &=& \left| \begin{matrix}\frac{u}2 & \frac{v}2 & 1 \\ \\ \frac{b}2 & \frac{d}2 & 1 \\ \\ \frac{du}{2(d-v)}& \frac{bv}{2(b-u)}& 1 \end{matrix}\right| \\ \ &=& \left| \begin{matrix}0 & 0 & 1 \\ \frac{b-u}2 & \frac{d-v}2 & 1 \\ \\ \frac{du}{2(d-v)}-\frac{u}2 & \frac{bv}{2(b-u)}-\frac{v}2 & 1 \end{matrix}\right| \\ \ &=& \left| \begin{matrix}\frac{b-u}2 & \frac{d-v}2 \\ \\ \frac{uv}{2(d-v)}& \frac{uv}{2(b-u)}\end{matrix}\right| \\ \ &=& \frac{uv}4-\frac{uv}4 \\ \ &=& 0 ,\end{eqnarray*}
so $M,N,P$ are collinear.

$\blacksquare$
[/hide]

For synthetic proofs of the first result, see [url=http://www.cut-the-knot.org/Curriculum/Geometry/Quadri.shtml]Theorem of Complete Quadrilateral[/url].
For the second result, see [url=http://www.cut-the-knot.org/Curriculum/Geometry/CircleOnCevian.shtml]Circles on Cevians[/url] (at the end of the page).
For additional results, see [url=http://www.cut-the-knot.org/ctk/CompleteQuadrilateral.shtml]The Complete Quadrilateral[/url].

I think that those are among the easiest proofs one can find.

[b]PS[/b]: [i]Skew axis[/i] are sometimes helpful in problem which do not involve angles (including perpendicularity).


I read somewhere that there is a fundamental equation of fluid mechanics jst like Newton's laws and Maxwell's eqns. Does anyone know of such an equation???

Do you mean Navier-Stokes equation? It's something like Newton's 3rd law for liquids, but I don't see very far into it...

... if you dare, just type it in google and you'll get a bunch of quite friendly sites about the equations.

No not such differntial eqns... Just a simple eqn involving pressure gradient etc...

I think I got it!!!

[b]Euler's Equation of Inviscid Motion[/b]
The equation of motion which applies to high Reynolds number regimes in which the viscous force is small compared to the inertia force, so it can be neglected in the Navier-Stokes equations. For nonsteady flow, this becomes

 $ \varrho \frac{D \vec{u}}{Dt} = \varrho ( \frac{\partial \vec{u} }{ \partial t} + \vec{u}\times \vec{ \bigtriangledown } . \vec{u})  = - \vec {\bigtriangledown} P $

where $ \frac{D \vec{u}}{Dt} $ is a substantive derivative, $ \varrho $ is the density, $u$ is the fluid velocity, and $P$ is the pressure. For steady flow, it simplifies to Fourier's law

$ \varrho \vec{u} \bigtriangledown . \vec{u} = - \bigtriangledown P $

Also,   [b]Newton's Law for Fluids[/b]

[url=http://scienceworld.wolfram.com/physics/NewtonsLawforFluids.html]Newton's law for fluids[/url]


Not sure where to ask this or find this...

Why is it that problems in most of the larger high school mathematics competitions do not require calculus? From my understanding, there are many problems in these same competitions that require other skills and knowledge that, typically, are taught at the same time or even later than calculus.

De-emphasizing calculus levels the playing field between younger and older students.  Younger students are clearly less likely to have taken calculus in a formal setting even though young and old alike benefit similarly from coaching and learning other kinds of math outside the classroom.

I also find calculus boring in the high school setting.  Finding the extrema of a function by setting its derivative to zero rarely gets you any actual understanding of the behavior of that function; this is even truer of the method of Lagrange multipliers.  These methods are very general but tell you very little besides the answer.


Sarah's parents decide to reward her (with money!) for her semester grades. Sarah has 5 classes. She will get $\$ 15$ for each A, $\$ 10.5$ for each B, $\$ 7.5$ for each C, and $\$ 1.5$ for each D. She will receive nothing for an F. The last 3 semesters, Sarah has avoided F's, and received the same amount of money ([i]a whole number of dollars[/i]) each semester, although the grade distribution of A's, B's, C's, and D's was different each time. How much money did Sarah earn each semester? What were her grades each time?

This problem is from regular Algebra II class, so it shouldn't be that hard, but I don't see a nice way to approach...

A [i]regular[/i] class? Well, some problems are nasty too at classroom level. I guess you can list out all the possible ways, and find the satisfactory conditions.


vector
algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $ V$ be $ n$ dimensional vector space and $ f: V \longrightarrow V$ be a linear map such that maps $ id, f, f^2, \ldots, f^{n \minus{} 1}$ are linear independent (of course as vectors of space $ End(V)$). Prove that there exists $ v \in V$ such that $ v, f(v), f^2(v), \ldots, f^{n \minus{} 1}(v)$ are linear independent.

This result is a special case of Aupetit's Theorem in Linear Algebra.  See Theorem $ 7.3$ in Prasolov's Problems and Theorems in Linear Algebra for a proof over $ \mathbb{C}$.

The characteristic and minimal polynomials of $ f$ coincide, so the rational canonical form of $ f$ is the [url=http://en.wikipedia.org/wiki/Companion_matrix]companion matrix[/url] of its characteristic polynomial.  Verify that $ v \equal{} \left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ \vdots \end{array} \right]$ works.


By Cauchy-Schwarz Inequality $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$.

[hide="Solution"]
AM-GM gives
\begin{eqnarray*} x+\frac 1x\ge 2\\ y+\frac 1y\ge 2\\ z+\frac 1z\ge 2\\ \Rightarrow x+y+z+\frac 1x+\frac 1y+\frac 1z\ge 6\\ \Rightarrow 3+\frac 1x+\frac 1y+\frac 1z\ge 6\\ \Rightarrow \frac 1x+\frac 1y+\frac 1z\ge 3\end{eqnarray*}
[/hide]

[quote="yassin88"]If \[ x+y+z\leq 3  \]prove that:\[ \frac1{x}+\frac1{y}+\frac1{z}\geq 3  \][/quote]

By Jensen's inequality,

Let $F(x)=\frac{1}{x}$
F is convex.

so, $F(x)+F(y)+F(z)\geq3f(\frac{x+y+z}{3})$

F is decreasing function.  so, when $x+y+3=3$, we have the smallest amount of $f(\frac{x+y+z}{3})$

By AM-HM Inequality $\frac1x+\frac1y+\frac1z\geq\frac{3}{\frac{1}{\frac1x}+\frac{1}{\frac1y}+\frac{1}{\frac1z}}\geq 1$.

Can we just apply AM-HM without stating that x,y,z must be positive ? :?

And obviously it cant hold for all real x,y,z ..... pick x=y=z=0 ....

[quote="maokid7"]good point. it did not say that x,y,z were all non-negative. so i dont think that we can use AM-GM.
so using cauchy's is prob the best way to go.  :)[/quote]

Cauchy doesn't hold for negative numbers either... the inequality itself doesn't hold for negative numbers, just take $a,b,c$ all negative.

[quote="maokid7"]good point. it did not say that x,y,z were all non-negative. so i dont think that we can use AM-GM.
so using cauchy's is prob the best way to go.  :)[/quote]

It doesn't hold for negative numbers, period.

oops didnt know that b4. thanks guys. :blush:  :oops:

Cuachy's inequality holds for any real numbers.
Just in our prolem it doesnot work since the trples to $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 9$  are $\sqrt x$ etc.....

I meant that this inequality does not hold for negative numbers.  If $x,y,z0$, then $1/x,1/y,1/z<0$ also.


[hide] Firstly, let's calculate the probability of getting all reds.  Since there are currently 8 out of 15 reds, the probabilty of taking a red is 8/15.  If you take out one red, there will be 7 out of 14 reds; thus 7/14=1/2.  When you take out another red, there will be 6 out of 13 reds; thus 6/13.  Thus, the probability of taking all reds is 8/15*7/14*6/13=8/65.  Now, let's look at blacks.  Using the same concept, we have the probabilty for blacks is 7/15*6/14*5/13=1/13.  Add the probabilty of black and of red to get 8/65+1/13=(8/65)+(5/65)=13/65=$\frac{1}{5}$; which is our answer. [/hide]

Just get the probability of getting 3 of black: 8/15*7/15*6/15 or 8/65.  Now get the probability of red: 7/15*6/15*5/15 or 1/13.  Add the two together (8/65+5/65) and you get 13/65, or 1/5.

[quote="#H34N1"]A set of cards consists of 8 red and 7 black cards. Three cards are dealt at random. What is the probability of obtaining three cards of the same color?[/quote]
[hide]$\frac{8}{15}\cdot\frac{7}{14}\cdot\frac{6}{13}+\frac{7}{15}\cdot\frac{6}{14}\cdot\frac{5}{13}=\boxed{\frac{1}{5}}$[/hide]


If the roots of ax^2 + bx + c = 0 are real, rational, and equal, what is true about the graph of the function y = ax^2 + bx + c?

(1) It intersects the x-axis in two distinct points.
(2) It lies entirely below the x-axis.
(3) It lies entirely above the x-axis.
(4) It is tangent to the x-axis.

[hide="The Answer"]Since the roots are equal, the equation can be written in the form $y=a(x-h)^2$ This means that the graph is tangent to the x-axis.[/hide]

[hide]
If the roots are equal, it is tangent to the x axis , (y=x^2 for instance).
Since this is true, the answer is 4.
[/hide]


algebra
function
domain
calculus
derivative
trigonometry
calculus computations

is f(x)=tan x continuous and differentiable ? ( I think it is, because it is continuous and differentiable at every points in its domain)
if it is not, please elaborate to me,

if it is true, please go on reading,

there comes Role's Theorem,
if f(x) is continuous in [a,b] and differentiable for a<x<b and if , further more, f(a)=f(b), then there exists c, where a<c<b such that f'(c)=0

consider the function f(x)= tan x for 0<x<pie
obviously this part of function fulfils the condition of the theorem, but there is no such point in this interval with a derivative equal to 0,

so where goes wrong? can you tell me :(

Function $\tan x$ is differentiable on intervals $(k\pi-\pi/2,k\pi+\pi/2)$, $k\in\mathbb{Z}$.
You apply Role's theorem on interval $(0,\pi)$, which doesn't satisfy condition that $f$ is differentiable everywhere on interval, since $\tan x$ isn't differentiable at point $x_0=\pi/2$.


calculus
integration
limit
logarithms
inequalities
calculus computations

$ \int_{0}^{\sqrt{3}}\frac{1}{1\plus{}x}dx\equal{}\allowbreak \ln \left( 1\plus{}\sqrt{3}\right)$

$ \int_{0}^{\sqrt{3}}\frac{1}{1\plus{}x^{2}}dx\equal{}\allowbreak \frac{\pi }{3}$

$ \lim_{n\rightarrow \infty }\int_{0}^{\sqrt{3}}\frac{1}{1\plus{}x^{n}}dx$

$ \equal{}\lim_{n\rightarrow \infty }\left( \int_{0}^{1}\frac{1}{1\plus{}x^{n}}dx\plus{}\int_{1}^{\sqrt{3}}\frac{1}{1\plus{}x^{n}}dx\right)$

$ \equal{}\int_{0}^{1}1dx\plus{}\int_{1}^{\sqrt{3}}0dx\equal{}1$

[quote="IndoChina"]
$ \lim_{n\rightarrow \infty }\int_{0}^{\sqrt {3}}\frac {1}{1 \plus{} x^{n}}dx$

$ \equal{} \lim_{n\rightarrow \infty }\left( \int_{0}^{1}\frac {1}{1 \plus{} x^{n}}dx \plus{} \int_{1}^{\sqrt {3}}\frac {1}{1 \plus{} x^{n}}dx\right)$

$ \equal{} \int_{0}^{1}1dx \plus{} \int_{1}^{\sqrt {3}}0dx \equal{} 1$[/quote]

I can't understand it.

Kunny, a more general argument is that for all $ n,\alpha > 1,$

$ \int_{0}^{1}1dx < \int_{0}^{\alpha }\frac {1}{1 \plus{} x^{n}}dx < \int_{0}^{1}1dx \plus{} \int_{1}^{\alpha }\frac {1}{\left( 1 \plus{} x\right) ^{n}}dx \equal{} 1 \plus{} \frac {\left( 1 \plus{} \alpha \right) ^{1 \minus{} n} \minus{} 2^{1 \minus{} n}}{n \minus{} 1}$

By squeeze theorem, $ \lim_{n\rightarrow \infty }\int_{0}^{\alpha }\frac {1}{1 \plus{} x^{n}}dx \equal{} 1$

O.K.

You can also prove that $ \boxed{1\plus{}\frac{1}{n\plus{}1}< \int_0^1 \frac{1}{1\plus{}x^n} \ dx<1}$.

Kunny how $ \boxed{1\plus{}\frac{1}{n\plus{}1}<1}$ :!: ?
Indochina why $ \int_{0}^{1}1dx <\int_{0}^{\alpha }\frac{1}{1\plus{}x^{n}}dx  : ?:$

[quote="kunny"]O.K.

You can also prove that $ \boxed{1 \plus{} \frac {1}{n \plus{} 1} < \int_0^1 \frac {1}{1 \plus{} x^n} \ dx < 1}$.[/quote]

[quote="mathxl"]Kunny how $ \boxed{1 \plus{} \frac {1}{n \plus{} 1} < 1}$ :!: ?:[/quote]
Sorry, it's a typo,

Correction: $ 1 \minus{} \frac {1}{n \plus{} 2}\leq \int_0^1 \frac {1}{1 \plus{} x^{n}}\ dx\leq 1$.

Mhmhm...IndoChina and Kunny, still i cant [b]prove[/b] why
$ \int_{0}^{1}1dx <\int_{0}^{\alpha }\frac{1}{1\plus{}x^{n}}dx$
also why $ 1\minus{}\frac{1}{n\plus{}2}\leq\int_{0}^{1}\frac{1}{1\plus{}x^{n}}\ dx$
So how can we prove the above :huh:  ?

@ mathxl

see, since

$ \int_{0}^{\alpha }\frac{1}{1\plus{}x^{n}}dx\equal{}\allowbreak\int_{0}^{1}\frac{1}{1\plus{}x^{n}}dx\plus{}\int_{1}^{\alpha }\frac{1}{1\plus{}x^{n}}dx$ and $ n \rightarrow \infty$ therefore $ \int_{0}^{1}\frac{1}{1\plus{}x^{n}}dx \equal{}  \int_{0}^{1}1dx <\int_{0}^{\alpha }\frac{1}{1\plus{}x^{n}}dx$ for $ \alpha > 1$ because as  $ n \rightarrow \infty$  and $ 0 < x <  1   , x^{n} \rightarrow \infty$and $ \frac{1}{1\plus{}x^n} \rightarrow 1.$

[quote="vijaymenon"]@ mathxl

see, since

$ \int_{0}^{\alpha }\frac {1}{1 \plus{} x^{n}}dx \equal{} \allowbreak\int_{0}^{1}\frac {1}{1 \plus{} x^{n}}dx \plus{} \int_{1}^{\alpha }\frac {1}{1 \plus{} x^{n}}dx$ and $ n \rightarrow \infty$ therefore $ \int_{0}^{1}\frac {1}{1 \plus{} x^{n}}dx \equal{} \int_{0}^{1}1dx < \int_{0}^{\alpha }\frac {1}{1 \plus{} x^{n}}dx$ for $ \alpha > 1$ because as  $ n \rightarrow \infty$  and $ 0 < x < 1 , x^{n} \rightarrow \infty$and $ \frac {1}{1 \plus{} x^n} \rightarrow 1.$[/quote]
Yup, but this is using limits
The above inequalities are true without limits. I think Kunny's is true for big n but Indochinas is true for n=1 and also for the rest values of n. How can we prove the inequality of Indochina without using limits?
i.e imagine n=8 (a proof for not only big values of n)
An exercise could be: show that $ \int_{0}^{1}1dx < \int_{0}^{\alpha }\frac {1}{1 \plus{} x^{8}}dx$?. How can we prove it?
Thank you

$ \int_0^1 (1 \minus{} x^n) dx < \int_0^1 \frac {1}{1 \plus{} x^n} dx < \int_0^1 1 dx$
$ \iff \frac {n}{1 \plus{} n} < \int_0^1 \frac {1}{1 \plus{} x^n} dx < 1$
So, $ \lim_{n \to \infty} \int_0^1 \frac {1}{1 \plus{} x^n} dx \equal{} 1$.

$ \int_1^{\sqrt {3}} 0 dx < \int_1^{\sqrt {3}} \frac {1}{1 \plus{} x^n} dx < \int_1^{\sqrt {3}} \frac {1}{x^n} dx$
$ \iff 0 < \int_1^{\sqrt {3}} \frac {1}{1 \plus{} x^n} dx < \frac {\left(\sqrt {3}\right)^{1 \minus{} n} \minus{} 1}{1 \minus{} n}$
So, $ \lim_{n \to \infty} \int_1^{\sqrt {3}} \frac {1}{1 \plus{} x^n} dx \equal{} 0$.

Thus, $ \lim_{n \to \infty} \int_0^{\sqrt {3}} \frac {1}{1 \plus{} x^n} dx \equal{} \lim_{n \to \infty} \int_0^1 \frac {1}{1 \plus{} x^n} dx \plus{} \lim_{n \to \infty} \int_1^{\sqrt {3}} \frac {1}{1 \plus{} x^n} dx \equal{} 1 \plus{} 0 \equal{} \boxed{1}$.

[quote="mathxl"]Thank you very much Kouichi Nakagawa, for your nice solution and explanation[/quote]
Thank you.
But my solution may be difficult for a Japanese high school student.
Because Japanese high school student don't learn the series expansion of the function.

Oh i thought that  you used \[ 1 \minus{} {x^n} \le \frac{1}{{1 \plus{} {x^n}}}\] for x at [0,1]
and then you integrate (no series expansion) :maybe:

I inspire $ \frac{1}{1\plus{}x^n}\equal{}\sum_{i\equal{}0}^\infty (\minus{}x^n)^i\equal{}1\minus{}x^n\plus{}x^{2n}\minus{}x^{3n}\plus{}\cdots$
By Maclaurin


calculus
integration
calculus computations

[quote="Kibu"]Is that correct?
$ \int_{f(0)}^{f(k)} g(f(x)) df \equal{} \int_0^k g(f(x)) dx$[/quote]

It should be $ \int_{f(0)}^{f(k)} g(f(x)) df \equal{} \int_0^k g(x) dx$.

So e.x. $ \int_{f(0)}^{f(2)} (1 \plus{} f(x)) df \equal{} \int_0^2 (1 \plus{} x) dx \equal{} 2 \plus{} 0.5\cdot 2^2 \equal{} 4$and it doesn't matter how does f(x) look?

It should really be $ \int_0^k g(f(x))\,df\equal{}\int_{f(0)}^{f(k)}g(x)\,dx$.  Also, be aware that there may be cases where the right hand side makes sense but the left hand side does not. As written, it is true if $ f,g$ are defined on the entire line and are reasonably nice (say, $ g$ continuous and $ f$ continuously differentiable).


I'm not sure about this please check...sorry if it's wrong
homogenize $ a \plus{} b \plus{} c \equal{} 1$
for a = max{a,b,c} and $ f(a,b,c) \equal{} \sum_{cyc} \frac {1}{a^2 \plus{} b^2} \minus{} 10$
i will prove that $ f(a,b,c)\geq f(a,b \plus{} c,0)$ or
$ (\frac {1}{b^2 \plus{} c^2} \minus{} \frac {1}{(b \plus{} c)^2}) \plus{} (\frac {1}{a^2 \plus{} b^2} \plus{} \frac {1}{a^2 \plus{} c^2} \minus{} \frac {1}{a^2} \minus{} \frac {1}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 2bc})\geq 0$
it's
$ \frac {2bc}{(b^2 \plus{} c^2)(b \plus{} c)^2} \plus{} \frac {2a^4bc \minus{} (2a^2b^2c^2 \plus{} b^4c^2 \plus{} b^2c^4 \plus{} 2b^3c^3)}{a^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(a^2 \plus{} (b \plus{} c)^2)}\geq 0$
or
$ 2a^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(a^2 \plus{} (b \plus{} c)^2) \plus{} 2a^4(b^2 \plus{} c^2)(b \plus{} c)^2 \minus{} 2a^2bc(b^2 \plus{} c^2)(b \plus{} c)^2 \minus{} bc(b^2 \plus{} c^2)(b \plus{} c)^4 \geq 0$
since a = max{a,b,c} we have
$ 2a^4(b^2 \plus{} c^2)(b \plus{} c)^2 \minus{} 2a^2bc(b^2 \plus{} c^2)(b \plus{} c)^2\geq 0$
and from
$ 2a^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(a^2 \plus{} (b \plus{} c)^2) \equal{} 2a^4(a^2 \plus{} b^2)(a^2 \plus{} c^2) \plus{} 2a^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b \plus{} c)^2$
together with $ 2a^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b \plus{} c)^2\geq 2bc(b^2 \plus{} c^2)(b^2 \plus{} c^2)(b \plus{} c)^2\geq bc(b^2 \plus{} c^2)(b \plus{} c)^4$
so we have $ f(a,b,c)\geq f(a,b \plus{} c,0)$
it remains to prove that
$ \frac {1}{a^2 \plus{} b^2} \plus{} \frac {1}{a^2} \plus{} \frac {1}{b^2}\geq 10$
or
$ (1 \minus{} 4q)(1 \minus{} 5q^2)\geq 0$
which is obviously from $ a \plus{} b \equal{} 1$ where $ q \equal{} ab$

[quote="SUPERMAN2"]I try to solve this problem but fail.Can anyone post a solution?Thanks.[/quote]
I have solved it, but I will not post the solution right now :) Where did you find it?

[quote="SUPERMAN2"]Let $ a,b,c$≥0.Prove that:
$ \frac {1}{a^2 \plus{} b^2} \plus{} \frac {1}{b^2 \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} a^2}$≥$ \frac {10}{(a \plus{} b \plus{} c)^2}$[/quote]
Assume that $ c\equal{}\min\{a,b,c\},$ then it is clear that
$ a^2\plus{}b^2 \le x^2\plus{}y^2,b^2\plus{}c^2 \le y^2,c^2\plus{}a^2 \le x^2$
where $ x\equal{}a\plus{}c/2,y\equal{}b\plus{}c/2$. So, we only need to prove that
$ \frac{1}{x^2}\plus{}\frac{1}{y^2}\plus{}\frac{1}{x^2\plus{}y^2} \ge \frac{10}{(x\plus{}y)^2},$
which is very simple. :)

[quote="SUPERMAN2"]Let $ a,b,c$≥0.Prove that:
$ \frac {1}{a^2 \plus{} b^2} \plus{} \frac {1}{b^2 \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} a^2}$≥$ \frac {10}{(a \plus{} b \plus{} c)^2}$[/quote]

Let $ a,b,c$ be nonnegative real numbers . Prove that :

$ \sum\frac{1}{a^4\plus{}b^4} \geq \frac{20}{(a\plus{}b\plus{}c)^4}$