Otherwise, someone would send some large amount of spam to everyone.

No ability to mass PM makes sending Mafia Role PMs a serious headache  :stretcher:. Maybe if the mods or authorized people could do it, sending them would be a lot easier.
Same with big games, FTW tournaments, etc...

There was an issue with this a while back during the AoPS Team Chess Tournament. If you're going to mass message someone, do it on email. The problems with mass PMing outweigh the benefits.


Is there a positive integer $m$ such that the equation \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}= \frac{m}{a+b+c}\] has infinitely many solutions in positive integers $a, b, c \;$?

For $ m\equal{}12$ then it has many solution.
Prove with the Pell's equation and has discuss .

Can you please prove it has infinitely many for 12?

$ x\equal{}a\minus{}k,y\equal{}a,z\equal{}a\plus{}k$
Then the equation equivalent:
$ a^2\minus{}3k^2\equal{}1$
It is the Pell equation.
The root of equation is :
$ a_0\equal{}1,k_0\equal{}0$
$ a_1\equal{}2,k_1\equal{}1$
$ a_{n\plus{}2}\equal{}4a_{n\plus{}1}\minus{}a_n$
$ k_{n\plus{}2}\equal{}4k_{n\plus{}1}\minus{}k_n$
Easy to check $ a>k$ so the equation has positive solution.
More consider the equation:
$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\plus{}\frac{c}{xyz}\equal{}\frac{3c\plus{}9}{x\plus{}y\plus{}z}$
With the same method we can prove this equation has infinite natural solution.


Here are the rules:
Keep naming songs as long as you can!
Note: name 1 song per post!

I'll start the chain:
Holiday (green day)

"Epitaph, including March For No Reason and Tomorrow And Tomorrow" - King Crimson.


Points $ A_1,B_1,C_1$ are chosen on the sides $ BC,CA,AB$ of and isosceles triangle $ ABC(AB\equal{}BC)$.It is known that $ \widehat{BC_1A_1}\equal{}\widehat{CA_1B_1}\equal{}\widehat{A}$ Let $ BB_1$ and $ CC_1$ meet at $ P$.Prove that $ AB_1PC_1$ is cyclic.

It's easy. $ \angle{A}\equal{}\angle{C}$ the conditions imply that]. $ C_1A_1CA$ and $ BA_1B_1A$ are cyclic. (Directly from the given angles) Then, $ \angle{B_1BA_1}\equal{}\angle{A_1AB_1}$ and $ \angle{C_1CB}\equal{}\angle{C_1AA_1}$ this means $ \angle{BPC}\equal{}180^o\minus{}\angle{A}$ and hence the result.

There are another solution:
Construct P' is the intersection of CC1 and cicumcircle of BA1C1, then simple to relize that C1P'B1A is cyclic. So <BP'C1=<BA1C1=<A, <C1P'B1=180-<A, then P' is on BB1. So P'=P and AC1PB1 is cyclic

Point $ A_1, B_1, C_1$ chosen on the sides [BC], [CA] and [AB], respectively; that quadrangles $ ABA_1B_1, ACA_1C_1$ cyclic. Let $ \{P\}\equal{}BB_1\cap CC_1$, prove that $ AB_1PC_1$ cyclic.

It use the same idea with the problem before. :roll:


geometry
inradius
inequalities
inequalities unsolved

Prove that in any triangle

$ \frac{(m_am_b)^2}{ab} +\frac{(m_bm_c)^2}{bc}+\frac{(m_cm_a)^2}{ca}\geq \frac{81}{4}r^2 $

$m_a,m_b,m_c = $ medians

$ r = $ inradius

Perhaps I have a solution but I must check it so I have posted this one in Unsolved Section Problems.

Nice inequality!!
We  know  that  [tex]  m_a , m_b , m_c [/tex]  are  sides  of triangle  with  area  [tex] S' = \frac{3}{4} S [/tex]
In  any triangle  ABC we have  [tex]  ab+bc+ca \geq 4\sqrt{3} S  [/tex]  so we  aply  this  inequality  to  the  triangle  with sides  [tex] m_a , m_b , m_c [/tex]   and we have  [tex] m_a\cdot m_b + m_b\cdot m_c + m_c\cdot m_a \geq 3\sqrt{3}S [/tex] .
Now  we  use  BCS :

[tex]\Sigma \frac{m_a^2m_b^2}{ab} \geq \frac{[\Sigma m_am_b]^2}{\Sigma ab} \geq \frac{[3\sqrt{3}S]^2}{\Sigma ab}= \frac{81p^2r^2}{3\Sigma ab} \geq\frac{81p^2r^2}{(2p)^2} = \frac{81r^2}{4} [/tex]

My solution is less simple. I hope it is correct.

It is based on these three inequality not difficult to prove so I leave them as exercises.

$2s^2 \geq 27Rr$---------------------$R \geq2r$-----------------------$m_am_bm_c \geq r_ar_br_c$

We have $s^3 \geq (27/2)^{3/2}R^{3/2}r^{3/2} \geq \sqrt{2}(27/2)^{3/2}Rr^2=81(\sqrt{3}/2)Rr^2 $

Hence $S^3=r^3s^3 \geq 81(\sqrt{3}/2)Rr^5$

so $(rr_ar_br_c)^2=S^4 \geq 81(\sqrt{3}/8)4SRr^5=81(\sqrt{3}/8)abcr^5$  and   $(r_ar_br_c)^2 \geq 81(\sqrt{3}/2)r^3$

Furthermore $(m_am_bm_c)^2 \geq (r_ar_br_c)^2 \geq 81(\sqrt{3}/8)abcr^3 $

By AM-GM inequality we obtain

$\frac{1}{3}\sum \frac{m_a^2m_b^2}{ab} \geq \frac{(m_am_bm_c)^(\frac{4}{3})}{(abc)^(2/3)} \geq 81(\frac{\sqrt{3}}{8})^(2/3)r^2 =\frac{27}{4}r^2 $


Let  A={1,2...n}. How many solutions $(X,Y,Z)$ with $2\leq|X|<|Y|<|Z|$ does the equation $X\cup Y\cup Z=A$ have?

Very nice problem!!

Let's consider all the possible sets containing
a) all numbers at least once together.
b) each a different amount of numbers.

It is clear that all these fit in the definition of X,Y,Z. exactly once; And it is also clear that all not fitting in this definition don't fit in the X,Y,Z. That means, they're equivalent.  (I just find that wording simpler)
____

Now I'll try to solve that problem:
For each number n, there are 7 options of how to appear: $7^n$ in total

If we now consider the amount of ways where 2 out of 3 sets have the same amount of elements: (double this)
- $x=y\not=z$: ...
- $x\not=y=z$: ...
And all 3 the same: (triple this)
- $x=y=z$: ...

$7^n$ minus these last amounts results the final answer after some calculation.
But I guess there is a way with less calculation :P

Have you tried to make these calculations? Believe me Peter there is no way on Earth to make these calculations, unless you find a trick or a nice recurrency formula or something. I couldn't find none of these as far as now. If you get something, anything, please post.


If 2 vertices of a Cube and a point in its interior are randomly chosen, what is the probability that they form an obtuse triangle?

[hide="Solution (I think)."]Pick any vertex with which to begin.  The probability of having the second vertex be adjacent to it is $ \frac {3}{7}$.  For the triangle formed by these two points and an interior point to be obtuse, you can draw a sphere through the two points (with the two points determining a diameter).  If the interior point lies on the sphere, its measure is $ 90$, while if it lies inside it, it is obtuse.  If the side length of the cube is $ 2x$, the volume of the portion of the sphere within the cube is $ \frac {\pi x^3}{3}$.  The probability of the point being chosen in the interior of the sphere is $ \frac {\pi}{24}$.  If the two vertices chosen are opposite to each other, which also happens with probability $ \frac {3}{7}$, the volume of the sphere is $ \frac {4\pi x^3\sqrt {2}}{3}$ and the probability is $ \frac {\pi\sqrt {2}}{6}$.  If the two points are doubly opposite from each other, which happens with probability $ \frac {1}{7}$, then all possible points chosen will satisfy the condition.  Consequently, the probability is $ \frac {3\pi \plus{} 12\pi\sqrt {2} \plus{} 24}{168}$.[/hide]

[quote="The Zuton Force"][b]@ Begoner:[/b] Are you sure that you've taken the right section of the sphere in your second case (ie. with the face diagonal as a diameter)?[/quote]

Yeah, I took the right section.  Unfortunately, multiplying and dividing by $ 2$ proved to difficult for me.  Fixed, I believe.

I believe the second volume is mistaken (for the opposites).

I noticed that since if I the hemisphere is equal to half the cube plus 6 side bits.
So i got:

Edit: WRONG

[quote="The Zuton Force"]Draw your section of the sphere -- I'm not 100% sure I interpreted your wording correctly, but if I did, I think part of your sphere's section should spill out of the sphere, and that needs to be accounted for (consider a square, and a circle such that a particular diagonal of the square is a diameter of the circle).[/quote]

You're completely correct -- I missed that entirely.  I'm not sure how to find that volume, though, other than by calculating the volume of the slices of the sphere lying outside the cube and subtracting from it the pyramidal portions lying inside the cube.  That seems to be an overly messy way to solve the problem.

Edit: Iniquitus, that's certainly a far better approach than mine, but how can you determine that the hemisphere is equal to half the cube plus $ 6$ side bits?  Isn't it just $ 3$ side bits (well, I guess it depends on your definition of "side bit," but if you shifted the sphere to the center of the circle, you'd have $ 6$ portions sticking out; by moving it back to a given face, half of those are removed)?

Yes  :blush:  I was thinking of a cube inscribed in a sphere...


If $ a,b,c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}3$, then

$ (3a\plus{}\frac 2{b})(3b\plus{}\frac 2{c})(3c\plus{}\frac 2{a}) \ge 125$

[quote="Vasc"]If $ a,b,c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ (3a \plus{} \frac 2{b})(3b \plus{} \frac 2{c})(3c \plus{} \frac 2{a}) \ge 125$[/quote]
I think, the Mixing Variables method kills it.

[quote="Vasc"]If $ a,b,c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ (3a \plus{} \frac 2{b})(3b \plus{} \frac 2{c})(3c \plus{} \frac 2{a}) \ge 125$[/quote]
Here is my solution:
Suppose that $ c \equal{} max\{a;b;c\}$ then $ c \geq 1;ab \leq abc \leq 1$
We can rewrite this inequality:
$ 27abc \plus{} \frac {12(ab \plus{} bc \plus{} ca)}{abc} \plus{} \frac {8}{abc} \geq 71$
<=>$ 27abc \minus{} (a \plus{} b \plus{} c)^3 \plus{} \frac {12(ab \plus{} bc \plus{} ca) \minus{} 36abc}{abc} \plus{} \frac {8 \minus{} 8abc}{abc} \geq 0$
We have:
$ \frac {12(ab \plus{} bc \plus{} ca) \minus{} 36abc}{abc} \equal{} \frac {8c(a \minus{} b)^2 \plus{} (4a \plus{} 4b)(a \minus{} c)(b \minus{} c)}{abc}$
$ 27abc \minus{} (a \plus{} b \plus{} c)^3 \equal{} \minus{} (a \plus{} b \plus{} 7c)(a \minus{} b)^2 \minus{} (a \minus{} c)(b \minus{} c)(4a \plus{} 4b \plus{} c)$
$ \frac {8 \minus{} 8abc}{abc} \equal{} \frac {8(a \plus{} b \plus{} 7c)(a \minus{} b)^2 \plus{} 8(4a \plus{} 4b \plus{} c)(a \minus{} c)(b \minus{} c)}{27abc}$
So this inequality becomes:
$ M(a \minus{} b)^2 \plus{} N(a \minus{} c)(b \minus{} c) \geq 0$
with $ M \equal{} \frac {8c}{abc} \plus{} \frac {8(a \plus{} b \plus{} 7c)}{27abc} \minus{} a \minus{} b \minus{} 7c$
and $ N\equal{}\frac{4a\plus{}4b}{abc}\plus{}\frac{8(4a\plus{}4b\plus{}c)}{27abc}\minus{}4a\minus{}4b\minus{}c$
We will prove that $ M \geq 0;N \geq 0$
$ M \geq 0$<=>$ \frac {8c}{abc} \plus{} \frac {8(3 \plus{} 6c)}{27abc} \geq 6c \plus{} 3$
<=>$ \frac {88}{9ab} \plus{} \frac {24}{27abc} \geq 6c \plus{} 3$
We need prove that $ \frac {88}{9ab} \geq 6c \plus{} \frac {19}{9}$
<=>$ 88 \geq 54abc \plus{} 19ab$
This inequality true because $ ab \leq abc \leq 1$
 We need prove that $ N \geq 0$
or $ \frac {24}{27abc} \plus{} \frac {100(a \plus{} b)}{9abc} \geq 3a \plus{} 3b \plus{} 3$
But we have:
$ \frac {24}{27abc} \plus{} \frac {100(a \plus{} b)}{9abc} \geq \frac {24}{27} \plus{} \frac {200}{9c}$
We will prove $ \frac {200}{9c} \geq 3a \plus{} 3b \plus{} \frac {19}{9}$
<=>$ 200 \geq 27c(a \plus{} b) \plus{} 19c$
But by AM-GM $ 27c(a \plus{} b) \leq \frac {243}{4}$
And we need prove $ \frac {557}{4} \geq 19c$
<=>$ \frac {557}{76} \geq c$
Done!

Nice, Chien Than.
But, since $ 8(abc\plus{}\frac 1{abc}) \ge 16$, it suffices to show that
$ 19abc\plus{}12(\frac 1{a}\plus{}\frac 1{b}\plus{}\frac 1{c}) \ge 55$.

Other problem:
Let $ a;b;c$ be positive real numbers.Find min of:
$ S\equal{}(1\plus{}\frac{a}{3b})(1\plus{}\frac{b}{3c})(1\plus{}\frac{c}{3a})$


limit
algebra
polynomial
algebra unsolved

1. Prove that exists some number n, such that in decimal representation of $ 5^n$ there is block of at least 100 zeros.

2. Find limit

$ \mathop {\lim }\limits_{n \to infinity} \{ { (5 \plus{} \sqrt {24} )^n } \}$

$ n\to \plus{}\infty\Rightarrow L\to \plus{}\infty
        n\to \minus{}\infty\Rightarrow L\to 0$:wink:

[quote="karis"]$ n\to \plus{} \infty\Rightarrow L\to \plus{} \infty n\to \minus{} \infty\Rightarrow L\to 0$:wink:[/quote]

The mistake is corrected.

I am new here, so I am already not good in Latex...

[quote="t0rajir0u"]$ (5 \plus{} \sqrt {24})^n \plus{} (5 \minus{} \sqrt {24})^n \in \mathbb{N}$, so the fractional part is $ 1 \minus{} (5 \minus{} \sqrt {24})^n$ and approaches $ 1$.[/quote]

That's beautiful! Thank you very much. And could somebody say me the main idea of the second problem? I seen many problems of that type, but I cannot imagine how to start with it.

One of the ideas is that $ s_n \equal{} (5 \plus{} \sqrt {24})^n \plus{} (5 \minus{} \sqrt {24})^n$ satisfies the second-order linear recurrence
$ s_{n \plus{} 2} \equal{} 10s_{n \plus{} 1} \minus{} s_n$

so every term in the sequence is an integer.  There is a very general theory about how to handle sequences of this type; see [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=215833]here[/url] and [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url].  There are two other ways to look at this:

- When you add $ (5 \minus{} \sqrt {24})^n$, the terms in the binomial expansion containing the odd powers of $ \sqrt {24}$ cancel, so you are left with a sum of integers.  This is a good elementary perspective but does not readily generalize without more work.

- The sum of the powers of the roots of $ x^2 \minus{} 10x \plus{} 1$ are symmetric polynomials in the roots, hence polynomials in the coefficients.  This is closely related to the linear recurrence idea; see [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums]Newton sums[/url] and the [url=http://en.wikipedia.org/wiki/Fundamental_theorem_of_symmetric_polynomials#The_fundamental_theorem_of_symmetric_polynomials]fundamental theorem of symmetric polynomials[/url].


function
arithmetic sequence
number theory proposed
number theory

I only have two things to say :):

$F_n+L_n=2F_{n+1}$

$(F_n,F_{n+1})=1$.

[quote="Collins"]$(F_n)$ denote Fibonacci numbers and $(L_n)$ Lucas numbers.
Prove that $F_{2005}$ and $L_{2005}$ are coprime.[/quote]

Use [i]soustraction formula[/i]: $2F_{m-n}=(-1)^n(F_mL_n-L_mF_n)$ with $(m,n)=(2006,2005)$.
In fact $\gcd (F_n,L_n)=1$ if $n\equiv 1 or 2[3]$ and $\gcd (F_n,L_n)=2$ if $n\equiv 0[3]$.

 :( You are too fast for me, Grobber!!!

Actually, there's still the possibility that $2|F_n,L_n$, so what I wrote wasn't complete. We still have to show that $F_{2005},L_{2005}$ are not both even, but that's not hard. :)

loocking both sequences modulo 2 and finding a period will do the job.

Here's a nice fact: if $k$ is odd, $\gcd(F_k,L_m)|2$.

I bet some nice and slick forumla solves this one, but here's a solution which doesn't use any slick formulae :).

Let $u=\frac{1+\sqrt 5}2$ and $\bar u=\frac{1-\sqrt 5}2$. Assume an odd prime $p|F_k=\frac 1{\sqrt 5}\cdot(u^k-\bar u^k)$ and $p|L_m=u^m+\bar u^m$. If $p$ is $5$, we're done: $5$ never divides an $L_m$. Otherwise, work in $\mathbb Z_p[\sqrt 5]$. We have $u^{2k}+1=0$ and $u^{2m}+(-1)^m=0$.

If $m$ is even, this is impossible, because the order of $u$ in the multiplicative group of our field divides $4k$, and, on the other hand, we find that it must divide $4m$ and it must contain all $2$'s in the decomposition of $4m$, and there are more than $2$ (because $m$ is even).

If $m$ is odd, the order of $u$ divides $2m$, but it also divides $4k$ and has all the $2$'s in the decomposition of $4k$, again a contradiction.

jaja, thats what i had in mind!!!!!!!!!!! Nice Grobber!

[quote="grobber"]I bet some nice and slick formula solves this one[/quote]
Why bother with anything so crude as a formula?

(Let $p$ be a prime in the following)
In any sequence obeying the Fibonacci recurrence (or any second-order recurrence which is reversible mod $p$), the zeros mod $p$, if they exist, must form a single arithmetic sequence with difference independent of the initial conditions; if the series have zeros mod $p$, they differ mod $p$ only by a constant multiple and a horizontal translation.
The absolute values of the Lucas sequence are an even function, so an element divisible by $p$ must be mirrored by another with the negative of its index. For $p>2$, this means that if the Lucas sequence is ever zero mod $p$, the difference between zeros mod $p$ is even. This difference must carry over to the Fibonacci sequence as well; since $p|F_0$, $p$ must not divide any Fibonacci number with odd index.

For powers of 2, we directly show that no Lucas numbers are divisible by 4.


trigonometry
Euler
algebra unsolved
algebra

Following equation i've to reform it like "a+bi" :

cos(2t-1)+i sin(2t+1)

Can someone help me solving this prob?

THX

It's already in the form $ a\plus{}bi$ with $ a\equal{}\cos (2t\minus{}1)$, $ b\equal{}\sin (2t\plus{}1)$.  Can you make your question more specific?

[quote="JoeBlow"]It's already in the form $ a \plus{} bi$ with $ a \equal{} \cos (2t \minus{} 1)$, $ b \equal{} \sin (2t \plus{} 1)$. [/quote]

This was my first thought also but i was also thinking that it was too easy, so i started using Euler, but i got stuck.

Thx JoeBlow for the reply


1. How would you rearange the letters in the word "new door" to make one word?

2. Yes, I am heavy. but backwards I am not. Who am I?

[hide="1"]new door$\rightarrow$one word[/hide]
[hide="2"]ton[/hide]

[hide]1. New door is one word rearanged

2.ton not[/hide]


I am SO FRUSTRATED with this problem. I have spent hours and hours on it checking it and re-checking it, reworking it, and I cannot get the correct solution. I am willing to bet it is some minor oversight, but it completely eludes me. Someone please show me the error of my ways before I blow a gasket.

This is my feeble attempt to work the problem.  What is really bizzarre is that this problem consistently gives me the same wrong answer!

Solve the equation. Check the result.

[(3 + p) / 3] - 4p = 1 - [(p + 7) / 2]

----------------------------------------------------------------------------------------------------------

The first thing we need to do is eliminate the fractions by multiplying both sides of the equation by the LCD. In this case it is 6.

(6){[(3 + p) / 3] - 4p} = (6){1 - [(p + 7) / 2]}

Use the distributive property:

(6)[(3 + p) / 3] - (6)4p = (6)1 - (6)[(p + 7) / 2]

Simplify non-fractional terms:

(6)[(3 + p) / 3] - 24p = 6 - (6)[(p + 7) / 2]

Use the commutative property:

(6/3)(3 + p) - 24p = 6 - (6/2)(p + 7)

Simplify:

(2)(3 + p) - 24p = 6 - (3)(p + 7)

Distribute:

(2)3 + (2)p - 24p = 6 - (3)p + (3)7

6 + 2p - 24p = 6 - 3p + 21

Combine like terms and place in conventional order:

(-22p) + 6 = (-3p) + 27

Add 3p to both sides:

(-22p) + 6 + 3p = (-3p) + 27 + 3p

(-19p) + 6 = 27

Subtract 6 from both sides:

(-19p) + 6 - 6 = 27 - 6

(-19p) = 21

Use the associative property and divide both sides by -19:

(-19)(p) / -19 = 21 / -19

p = -21/19

----------------------------------------------------------------

To check our solution, we plug it back into the original equation:

[(3 + p) / 3] - 4p = 1 - [(p + 7) / 2]

[(3 + (-21/19)) / 3] - 4(-21/19) = 1 - [((-21/19) + 7) / 2]

----------------------------------------------------------------

I will begin by solving the FIRST expression:

[(3 + (-21/19)) / 3] - 4(-21/19)

Solve the numerator first:

[((57/19) + (-21/19)) / 3] - 4(-21/19)

[(36/19) / 3] - 4(-21/19)

Then the fraction itself:

(12/19) - 4(-21/19)

Multiply and simplify:

(12/19) - (-84/19)

(12/19) + (84/19) = 96/19

The first expression simplifies to 96/19.

----------------------------------------------------------------

Now we check the SECOND expression:

1 - [((-21/19) + 7) / 2]

Do the numerator first:

1 - [((-21/19) + (133/19)) / 2]

1 - [(112/19) / 2]

Then the / 2 in the fraction:

1 - (56/19)

(19/19) - (56/19) = -37/19

----------------------------------------------------------------

96/19 =? -37/19 FALSE!!!

And false again, and again

[quote="kamikaze762"]I am SO FRUSTRATED with this problem. I have spent hours and hours on it checking it and re-checking it, reworking it, and I cannot get the correct solution. I am willing to bet it is some minor oversight, but it completely eludes me. Someone please show me the error of my ways before I blow a gasket.

This is my feeble attempt to work the problem.  What is really bizzarre is that this problem consistently gives me the same wrong answer!

Solve the equation. Check the result.

[(3 + p) / 3] - 4p = 1 - [(p + 7) / 2]

----------------------------------------------------------------------------------------------------------

The first thing we need to do is eliminate the fractions by multiplying both sides of the equation by the LCD. In this case it is 6.

(6){[(3 + p) / 3] - 4p} = (6){1 - [(p + 7) / 2]}

Use the distributive property:

(6)[(3 + p) / 3] - (6)4p = (6)1 - (6)[(p + 7) / 2]

Simplify non-fractional terms:

(6)[(3 + p) / 3] - 24p = 6 - (6)[(p + 7) / 2]

Use the commutative property:

(6/3)(3 + p) - 24p = 6 - (6/2)(p + 7)

Simplify:

(2)(3 + p) - 24p = 6 - (3)(p + 7)

Distribute:

(2)3 + (2)p - 24p = 6 - (3)p + (3)7
[/quote]

Your error is right here.  You distributed incorrectly, there should be a minus in front of $ (3)7$.  If you repeat the calculation from here with the minus there, you should get the correct answer $ p\equal{}\frac{21}{19}$

He's right. Here's what should have happened:

$ \frac{3\plus{}p}{3}\minus{}4p\equal{}1\minus{}\frac{p\plus{}7}{2}$

$ 6\left(\frac{3\plus{}p}{3}\minus{}4p\right)\equal{}6\left(1\minus{}\frac{p\plus{}7}{2}\right)$

$ 2(3\plus{}p)\minus{}6(4p)\equal{}6(1)\minus{}3(p\plus{}7)$

$ 6\plus{}2p\minus{}24p\equal{}6\minus{}3p\minus{}21$ (Your mistake was here. When you distributed, you didn't distribute the minus sign to the 21.)

$ 21\equal{}19p$

$ \frac{21}{19}\equal{}p$

[u][b]Check:[/b][/u]

$ \frac{3\plus{}\frac{21}{19}}{3}\minus{}4\left(\frac{21}{19}\right)\equal{}1\minus{}\frac{\frac{21}{19}\plus{}7}{2}$

$ \frac{\frac{57}{19}\plus{}\frac{21}{19}}{3}\minus{}\frac{84}{19}\equal{}1\minus{}\frac{\frac{21}{19}\plus{}\frac{133}{19}}{2}$

$ \frac{\frac{78}{19}}{3}\minus{}\frac{84}{19}\equal{}1\minus{}\frac{\frac{154}{19}}{2}$

$ \frac{26}{19}\minus{}\frac{84}{19}\equal{}\frac{19}{19}\minus{}\frac{77}{19}$

$ \minus{}\frac{58}{19}\equal{}\minus{}\frac{58}{19}$

First off, I really appreciate your responses, and I see where I messed up, but I am still having trouble grasping the concept of negative distibution and/or distribution after a minus sign.

Let me make certain I understand this rule.

6 - (3)(p + 7)

Ok, we have a minus just hanging around there ready to perform subtraction.

Since (3) comes after it, (3) becomes negative once it is distributed?

If that were the case, why does the problem not end up like this since p and 7 are both positive?

6 - (-3p) - 7
= 6 + 3p - 7

My question is, why does the sign of the first term remain the same while the sign of the second term changes?  After all, both terms receive the same treatment here, so it seems counter-intuitive to change only the last term.

For additonal clarity:
This is the correct distribution, but how?

6 - (3)(p + 7)

6 - (3)p + (-3)7

That blows my mind.

Can someone lay this out for me in a logical way so I can see how this works?  Any help is greatly appreciated.   :)

Wait a sec.  I think I understand.

As a matter of logic, it would actually be done in this way?

6 - (3)(p + 7)

6 - (3)p + - (3)7

Simply saying that since the (p + 7) is grouped and appears after a minus sign, it must be interpreted that that operation to be performed on the group must be performed on all terms within the group.  In this case, we are multiplying all terms within the group, and also subtracting all terms within the group.

Is that the correct logical analysis of this rule?

yes, that is correct.

$ 6\minus{}3(p\plus{}7)\equal{}6\plus{}(\minus{}3(p\plus{}7))$, so we must distibute the negative.

$ 6\minus{}3(p\plus{}7)\equal{}6\minus{}(3p\plus{}21)\equal{}6\plus{}(\minus{}1(3p\plus{}21))$. We must distibute the $ \minus{}1$.

[quote="fishythefish"]That solution has already been stated. If you saw the 58/19, then you were probably looking at the checking portion.

BTW: Your second gold line should probably say "(+((p+[b]7[/b])/2))[/quote]

thanks. Yes it should say (+((p+7)/2)))

You dont need to use all the terms like commutative , distributive...

its a simple format question and you do not need divide by 19 and all.
Just take the LCM both the sides treating p + 7 as a whole.

pretty easy.

I don't quite understand what you're saying. I understand that there might be an alternate method of solving, I was just showing this one step by step.

Feel free to post your solution if this bothers you so much.


function
integration
calculus
complex analysis
complex analysis unsolved

Hello,

I was trying to do a problem of the forum and I am stuck on the following: does the fact that $\forall z \in \mathbb{C}, \int_{\gamma_{z}}{\frac{Re(f(\omega))}{w-z}}=0$ imply that $f$ is constant ? Here $\gamma_{z}$ denotes any simple piecewise smooth path such that $Ind(\gamma,z)=1$, and $f$ is an entire function.

Note that it would be enough to show that $Re(f)$ is entire ( I say that in case we could apply Morera's theorem ). Anyway, I'm note even sure the result holds.

This is a bit strange question: take $\gamma_z$ to be a very small circle around $z$. Then the integral is almost $Re(f(z))$ and we conclude that $Re(f(z))=0$ for all $z$, which, of course, implies that $f$ is a (purely imaginary) constant. Are you sure that you stated the question correctly? :?

No, my question actually does not make any sense at all. Sorry for that.


geometry
geometric transformation
reflection
circumcircle
projective geometry
cyclic quadrilateral
geometry proposed

[quote="yetti"]Let K, L, M be the feet of the internal bisectors of the angles $\angle A,\ \angle B,\ \angle C$ of a triangle $\triangle ABC.$ Let D, E, F be the tangency points of the triangle incircle (I) with the sides BC, CA, AB and D', E', F' reflections of D, E, F in the angle bisectors AK, BL, CM. Show that the points $X \equiv LM \cap EF',\ Y \equiv MK \cap FD',\ Z \equiv KL \cap DE'$ lie on the line IO, where O is the triangle circumcenter.[/quote]

First we will prove that the lines $FD'$,$F'D$ and $MK$ are collinear.

Let's use this well know fact.

[color=blue]Theorem 1.Let $ABCD$ is an inscribed quadrilateral.Let $X\in{AB\cap{CD}}$,$Z\in{AC\cap{BD}}$ and the tangents lines to the circumscribed circle at the point $A$ and $B$ intesecting at $Y$,then $X$,$Y$ and $Z$ are collinear.[/color]

Really this theorem is special case of the Pascal's theorem about hexagon!

It's easy to see that the point's $F$,$D$,$D'$ and $F'$ are on the

incircle,and lines $MF$,$MF'$,$KD$ and $KD'$ are touching the incircle.

So if $FD$ intersecting $F'D'$ at $T$ and $FD'$ intersecting $DF'$ at $R$,then from the theorem we have that the points $T$,$R$,$M$ and $T$,$R$,$K$ are collinear=>the points $M$,$R$,$K$ are collinear too,=>$Y\equiv{R}$=> the lines $FD'$,$DF'$ and $MK$ are collinear.

Now let's continiu the lines $FF'$,$DD'$ and $EE'$,they will construct a triangle $A'B'C'$,but us we have that $FF'\parallel DL$ and $F'\in(I)$,we can easely see that (I) is the Euler's circle for the triangle $A'B'C'$;$F'$,$D'$ and $E'$ are feet of altituds and $D,E,F$ are middelpoint of it's sides.
From another well-know fact we know that the line $IO$ is same with Euler's line of the triangle $DEF$,and Euler's line of the triangle $DEF$ is same with the Euler's line of the triangle $A'B'C'$.

So finaly we have to prove that the points $X,Y,Z$ are collinear with Euler's line of the triangle $A'B'C'$,let's write this as another problem.

[color=blue]Problem 2.In the triangle $ABC$,$A_{1}$ and $B_{1}$ are fetes of altitudes from  $A$ and $B$,and $A_{2}$ and $B_{2}$ are middelpoints of the sides $BC$ and $AC$ respectively.Let $H$ is the orthocentr and $M$ median's intersection point of $ABC$,also $X\in{A_{1}B_{2}\cap{A_{2}B_{1}}}$.Prove that $H$ ,$M$ and $X$ are collinear. [/color]

But if we remember the Papus' theorem we can see that the [color=blue]Problem 2       [/color] is the special case of that theorem! :wink:

Very nice, thank you  :!:

Let KD', LE' meet at P. KPLC is a (concave) tangential quadrilateral, with diagonals KL, PC and DEE'D' its contact cyclic quadrilateral with diagonals DE', E'D. These 2 quadrilaterals have the same diagonal intersection $Z \equiv KL \cap PC \equiv DE' \cap ED'.$ Similarly, $X \equiv LM \cap EF' \cap FE'$ and $Y \equiv MK \cap FD' \cap DF'.$

Let $U \equiv KD' \cap CA,\ V \equiv LE' \cap BC.$ From the kites ABKU and BVLA, $AU = AB = BV$ According to [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=105048[/url] etc., $UV \perp IO.$ ED' is a polar of U and DE' is a polar of V with respect to the incircle (I). The poles U, V of ED', DE' are collinear with the direction UV perpendicular to IO, i.e., with the pole W of IO at infinity, hence the polars ED', DE', IO of the collinear points U, V, W (the last one at infinity) are concurrent at Z. Similarly, X, Y also lie on IO.

Thank you for corrections and for very nice problem [b]yetti[/b] :!:


Συγχαρητηρια σε οσους ξεκινησαν και συντηρουν αυτη εδω την κοινοτητα αλλα και στους υπολοιπους που ασχολιουνται ενεργα με αυτη!
Ας βαλω και εγω το πρωτο μου προβλημα

 Να εξεταστει αν συγκλινει η σειρα

$ \sum_{n=1}^{\infty}{\int_0^1\cos(nt^2)dt}$

[quote="Durandal"][quote="mathsforfun!"]Να εξεταστει αν συγκλινει η σειρα

$ \sum_{n = 1}^{\infty}{\int_0^1\cos(nt^2)dt}$[/quote]

First off, welcome :welcomeani:

Αν δεν κάνω λάθος, η σειρά αποκλίνει... με την αντικατάσταση $ u = t\sqrt {n}$ ο γενικός όρος γράφεται $ \frac {1}{\sqrt {n}} \int_0^{\sqrt {n}} \cos(u^2) du$ και το ολοκλήρωμα έχει θετικό όριο ($ \sqrt {\pi}/4$? κάτι τέτοιο) καθώς $ n\to\infty$,  οπότε...

Βέβαια, για να δείξει κανείς ότι το ολοκλήρωμα αυτό συγκλίνει σε θετική τιμή θέλει κάποια δουλίτσα, η οποία είναι όμως γενικά γνωστή (η λύση που ξέρω είναι με μιγαδική ολοκλήρωση). Αν θέλετε περισσότερες λεπτομέρειες, let me know.

Cheerio,

Durandal 1707
PS: Tα ολοκληρώματα έχουν σίγουρα κάποιο φτηνό lower bound, αλλά τώρα είναι αργά και βαρίεμαι να σκεφτώ... από αύριο :)[/quote]
Δες στο http://en.wikipedia.org/wiki/Cornu_spiral#Cornu_spiral
Ολοκλήρωμα Fresnel χρήσιμο στην γεωμετρική οπτική


Here is the IMO shortlist problems 1992, I got it from internet.

I notice that problem 45 of the [b]1992[/b] short list appears on the [b]1993[/b] IMO.  Does this strike anyone else as strange?

[b][size=160]IMO LongList 1992 Problems (Posted In The Forums)[/size][/b]

Problem 1 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004282

Problem 2 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004289

Problem 3 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004807

Problem 4 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004811

Problem 5 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004306

Problem 6 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004818

Problem 7 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005940

Problem 8 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004820

Problem 9 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004823

Problem 10 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366404

Problem 11 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004308

Problem 12 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004829

Problem 13 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004837

Problem 14 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004841

Problem 15 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004846

Problem 16 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004310

Problem 17 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366410

Problem 18 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004855

Problem 19 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004863

Problem 20 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004866

Problem 21 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004869

Problem 22 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366420

Problem 23 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004314

Problem 24 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004925

Problem 25 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004935

Problem 26 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366399

Problem 27 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004965

Problem 28 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=126017

Problem 29 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222516

Problem 30 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2002366

Problem 31 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222819

Problem 32 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004995

Problem 33 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2002365

Problem 34 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005037

Problem 35 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222519

Problem 36 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004327

Problem 37 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005065

Problem 38 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366415

Problem 39 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005071

Problem 40 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005073

Problem 41 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005088

Problem 42 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222521

Problem 43 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005096

Problem 44 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=849468

Problem 45 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005112

Problem 46 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005114

Problem 47 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004331

Problem 48 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005214

Problem 49 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005480

Problem 50 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005699

Problem 51 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222525

Problem 52 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005715

Problem 53 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=154338

Problem 54 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005719

Problem 55 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222823

Problem 56 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005726

Problem 57 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005730

Problem 58 : False

Problem 59 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005737

Problem 60 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222532

Problem 61 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005739

Problem 62 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005745

Problem 63 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005749

Problem 64 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005753

Problem 65 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005754

Problem 66 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005759

Problem 67 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005765

Problem 68 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2004333

Problem 69 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222814

Problem 70 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005771

Problem 71 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005889

Problem 72 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005893

Problem 73 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005906

Problem 74 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005912

Problem 75 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005916

Problem 76 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005919

Problem 77 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005922

Problem 78 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005925

Problem 79 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1222822

Problem 80 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005926

Problem 81 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005928

Problem 82 : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2005932

Here is typed form of IMO 1992 ShortList.


For $ n$ a positive integer, if $ \frac {1}{(4n^2 \minus{} 2n)^2}$ is the general term,
find the exact value of the sum of the following infinite series:

$ \frac{1}{2^2} \plus{} \frac{1}{12^2} \plus{} \frac{1}{30^2} \plus{} ...$

[quote="Arrange your tan"]For $ n$ a positive integer, if $ \frac {1}{(4n^2 \minus{} 2n)^2}$ is the general term,
find the exact value of the sum of the following infinite series:

$ \frac {1}{2^2} \plus{} \frac {1}{12^2} \plus{} \frac {1}{30^2} \plus{} ...$[/quote]
[hide]We want to find $ \sum_{n\equal{}1}^{\infty} \frac{1}{(4n^2\minus{}2n)^2}$. This splits into $ \sum_{n\equal{}1}^{\infty} (\frac{1}{n}\plus{}\frac{1}{4n^2}\minus{}\frac{2}{2n\minus{}1}\plus{}\frac{1}{(2n\minus{}1)^{2}})$[/hide]
This is what I have so far...unfortunately I have to go...:| (I'll work on this later)

[quote="Arrange your tan"]For $ n$ a positive integer, if $ \frac {1}{(4n^2 \minus{} 2n)^2}$ is the general term,
find the exact value of the sum of the following infinite series:

$ \frac {1}{2^2} \plus{} \frac {1}{12^2} \plus{} \frac {1}{30^2} \plus{} ...$[/quote]

Suggested hint for starting:
[hide]
The general term can be written as $ \frac {1}{[(2n \minus{} 1)(2n)]^2}$.

And the first few terms can be written as:

$ \frac {1}{(1 \cdot 2)^2} \plus{} \frac {1}{(3 \cdot 4)^2} \plus{} \frac {1}{(5 \cdot 6)^2} \plus{} ...$
[/hide]

Additional hint:
[hide]
The general term can be rewritten as $ (\frac {1}{2n \minus{} 1} \minus{} \frac{1}{2n})^2$.

And the first few terms can be rewritten as:

$ ( \frac 11 \minus{} \frac12)^2 \plus{} (\frac13 \minus{} \frac14)^2 \plus{} (\frac15 \minus{} \frac16)^2 \plus{} ...$

[/hide]

[hide="Solution"]
As mentioned in the tip, the general term can be rewritten as $ (\frac{1}{2n\minus{}1}\minus{}\frac{1}{2n})^{2}$.
Thus, we have:
$ (\frac{1}{1}\minus{}\frac{1}2)^{2}\plus{}(\frac{1}3\minus{}\frac{1}4)^{2}\plus{}(\frac{1}5\minus{}\frac{1}6)^{2}\plus{}...$.
If we expand the terms and reorganize, we get:
$ \left ( \frac{1}{1^{2}} \plus{}\frac{1}{2^{2}}\plus{}\frac{1}{3^{2}}\plus{}...\right )\minus{}2\left ( \frac{1}{1*2}\plus{}\frac{1}{3*4}\plus{}\frac{1}{5*6}\plus{}...\right )$

We have now expressed the series as the sum of two other series.
The first series is the $ \frac{1}{n^{2}}$ series, whose sum Euler proved to be $ \frac{\pi ^{2}}{6}$.
The second series is not as obvious. Disregarding the factor of $ 2$, the general term can be expressed as $ \frac{1}{2n(2n\minus{}1)}$, which can be rewritten as $ \frac{1}{2n\minus{}1}\minus{}\frac{1}{2n}$.  Thus, we have $ \left (\frac{1}{1}\minus{}\frac{1}{2}  \right )\plus{}\left (\frac{1}{3}\minus{}\frac{1}{4}  \right )\plus{}\left (\frac{1}{5}\minus{}\frac{1}{6}  \right )\plus{}...$.  This is merely the alternating harmonic series, whose sum is $ \ln 2$.

So, the answer is $ \boxed{\frac{\pi ^{2}}{6}\minus{}2\ln 2}$.
[/hide]

I liked the problem, may I ask how you came up with it?


function
algebra
polynomial
trigonometry
linear algebra
matrix
calculus

Of course the solutions of $y''+y=0$ differential equation are well known.I need a particular solution for the differential equation $y''+y=4cosx$.

One of the method of finding the particilar solution is using [b]Differential Operator.[/b]

Let $D=\frac{d}{dx},$ we can rewrite the given differential equation as $(D^{2}+1)y=Re 4e^{ix}.$ Thus $y=\frac{4}{D^{2}+1}e^{ix}$

$\therefore y=\frac{4}{(D+i)(D-i)}e^{ix}.$

This is a constant-coefficient linear differential equation for which the nonhomogeneity lies within the class of functions (linear combinations of exponentials times polynomials) that are themselves annihilated by some such operator.

Notation: define the operator $L(y)=y''+y.$ The characteristic polynomial associated to this operator is $\lambda^{2}+1,$ which has roots $\pm i.$ What expontial $e^{\lambda x}$ is $4\cos x$ associated with? That's also $\pm i,$ and that's a complication, since it means that $L(4\cos x)=0.$

The situation is essentially the same as having a double root of the characteristic polynomial; we get ourselves out of it by multiplying by $x.$

So: Assume $y_{p}(x)=Ax\cos x+Bx\sin x.$

Differentiate that at put it into the differential equation - can you now solve for $A$ and $B?$

Thank u very much. I found $y_{p}(x)=2x\sin x$ is a particular solution. Thanks again.

What I showed you above was the method of undetermined coefficients. One could also use the method of variation of parameters.

Start by assuming that $y=u\cos x+v\sin x.$ We're going to take advantage of the fact that the solution space of the homogeneous equation is spanned by $\{\cos x,\sin x\}.$ Since we've introduced two degrees of freedom, we will impose one extra condition when it is convenient to do so.

$y'=u'\cos x+v'\sin x-u\sin x+v\cos x.$

We impose the condition $u'\cos x+v'\sin x=0.$ That gives us

$y'=-u\sin x+v\cos x$

$y''=-u'\sin x+v'\cos x-u\cos x-v\sin x$

$y''+y=-u'\sin x+v'\cos x=4\cos x$ by the differential equation.

That gives us two linear equations for the two unknown functions $u'$ and $v'.$ In matrix form:

$\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}\begin{bmatrix}u'\\ v'\end{bmatrix}=\begin{bmatrix}0\\ 4\cos x\end{bmatrix}.$

This has the solutions $u'=-4\sin x\cos x$ and $v'=4\cos^{2}x.$ There's still quite a bit of integration and algebraic manipulation to go, but with patience, you will arrive at the same answer as above.

In this particular instance, the technique of undetermined coefficients proved far simpler; but the technique of variation of parameters can reach cases that are beyond the reach of undetermined coefficients.


Consider the integers $2$, $3$, and $6$. They are distinct, and their reciprocals add up to $1$.

Can you think of four distinct integers whose reciprocals add up to $1$?

How about five such integers?

And six?

And $n$ such integers?

This is related to perfect numbers.

If n = sum-of-proper-divisors-of-n, then dividing both sides by n gives $1 = 1/d_{1}+1/d_{2}+...$.

28 = 1 + 2 + 4 + 7 + 14
1 = 1/28 + 1/14 + 1/7 + 1/4 + 1/2

You know that this works for $n=3$, then you can show that it works for $n=4$:
[hide]for example
$1=2^{-1}+4^{-1}+6^{-1}+12^{-1}$[/hide]
 and then do the easy induction step from $n$ to $n+2$
[hide]simply replace the reciprocal of the largest integer $n_{max}^{-1}$ by the sum
$n_{max}^{-1}(2^{-1}+3^{-1}+6^{-1}).$[/hide]

And then do the even easier step from $n$ to $n+1$ - halve it!


oh my god, the first time, argady have made mistake :P I hope you will correct it and you will have another nice inequality :D If it is symetric, maybe Schur will kill it ...


[quote="shobber"]Find all nagetive $a$ which makes the quadratic inequality $\sin^2{x}+a\cos{x}+a^2 \geq 1+ \cos{x}$ true for all $x \in R$.[/quote]

Is it this way ?
We have $cos^2 x+(1-a)cos x - a^2\leq 0$ after simplying . So taking discriminant ,

$(1-a)^2+4a^2=5(a-\frac{1}{5})^2+\frac{4}{5}>0$ which is always true for all $a$ . Since $a$ is non-negative,so $a\geq 0$ .  :?

[quote="shobber"]Be more careful.
(1) $a$ is negative.
(2) $|\cos{x}| \leq 1$[/quote]

$\cos ^2 x \le \cos x\,\,\,\,\forall x$ such that cosx positive

so, $\cos ^2 x + (1 - a)\cos x\, - a^2  \le 0\,\,\,$

$ \Rightarrow  cos x + (1 - a)\cos x\, - a^2 \, \le 0$

$ \Rightarrow (2 - a)\cos x\, - a^2 \, \le 0$

for this to be true for all x,

$\frac{{a^2 }}{{2 - a}} \ge 1$ and as a is given negative, this inequality gives $a \le  - 2$

[quote="shobber"]Be more careful.
(1) $a$ is negative.
(2) $|\cos{x}| \leq 1$[/quote]

Opps .... misread as $a$ is non-negative  :blush: But I cant figure out how $|\cos{x}| \leq 1$ would have affected the roots .  :o

[quote="vidyamanohar"][quote="shobber"]Be more careful.
(1) $a$ is negative.
(2) $|\cos{x}| \leq 1$[/quote]

$\cos ^2 x \le \cos x\,\,\,\,\forall x$

so, $\cos ^2 x + (1 - a)\cos x\, - a^2  \le 0\,\,\,$

$ \Rightarrow  cos x + (1 - a)\cos x\, - a^2 \, \le 0$

$ \Rightarrow (2 - a)\cos x\, - a^2 \, \le 0$

for this to be true for all x,

$\frac{{a^2 }}{{2 - a}} \ge 1$ and as a is given negative, this inequality gives $a \le  - 2$[/quote]

How this $ \Rightarrow  cos x + (1 - a)\cos x\, - a^2 \, \le 0$ can happen ?
And why this is true ? $\cos ^2 x \le \cos x\,\,\,\,\forall x$
If $x$ is $120^0$ i.e  , that cant be true ...

Im abit confused for what you have write . Can you explain in word ?

[quote="shyong"][quote="vidyamanohar"][quote="shobber"]Be more careful.
(1) $a$ is negative.
(2) $|\cos{x}| \leq 1$[/quote]

$\cos ^2 x \le \cos x\,\,\,\,\forall x$

so, $\cos ^2 x + (1 - a)\cos x\, - a^2  \le 0\,\,\,$

$ \Rightarrow  cos x + (1 - a)\cos x\, - a^2 \, \le 0$

$ \Rightarrow (2 - a)\cos x\, - a^2 \, \le 0$

for this to be true for all x,

$\frac{{a^2 }}{{2 - a}} \ge 1$ and as a is given negative, this inequality gives $a \le  - 2$[/quote]

How this $ \Rightarrow  cos x + (1 - a)\cos x\, - a^2 \, \le 0$ can happen ?

Im abit confused for what you have write . Can you explain in word ?[/quote]

we are looking at the smallest value that can be bigger than  $\cos ^2 x + (1 - a)\cos x\, - a^2$.

When positive, cosx is bigger than  $\cos ^2 x$.

I have compared the $cos x + (1 - a)\cos x\, - a^2$ with zero,  in place of  $\cos ^2 x + (1 - a)\cos x\, - a^2$.
and $|\cos{x}| \leq 1$ is reflected in taking $\frac{{a^2 }}{{2 - a}} \ge 1$

hope this clears your doubt

[quote="vidyamanohar"][/quote][quote="shyong"][quote="vidyamanohar"][quote="shobber"]Be more careful.
(1) $a$ is negative.
(2) $|\cos{x}| \leq 1$[/quote]

$\cos ^2 x \le \cos x\,\,\,\,\forall x$

so, $\cos ^2 x + (1 - a)\cos x\, - a^2  \le 0\,\,\,$

$ \Rightarrow  cos x + (1 - a)\cos x\, - a^2 \, \le 0$

$ \Rightarrow (2 - a)\cos x\, - a^2 \, \le 0$

for this to be true for all x,

$\frac{{a^2 }}{{2 - a}} \ge 1$ and as a is given negative, this inequality gives $a \le  - 2$[/quote]

How this $ \Rightarrow  cos x + (1 - a)\cos x\, - a^2 \, \le 0$ can happen ?

Im abit confused for what you have write . Can you explain in word ?[/quote][quote="vidyamanohar"]

we are looking for the smallest value that can be bigger than  $\cos ^2 x + (1 - a)\cos x\, - a^2$.

When positive, cosx is bigger than  $\cos ^2 x$.

I have compared the $cos x + (1 - a)\cos x\, - a^2$ with zero,  in place of  $\cos ^2 x + (1 - a)\cos x\, - a^2$.
and $|\cos{x}| \leq 1$ is reflected in taking $\frac{{a^2 }}{{2 - a}} \ge 1$

hope this clears your doubt[/quote]

Why can we compare $cos x + (1 - a)\cos x\, - a^2$ with zero in place of $\cos ^2 x + (1 - a)\cos x\, - a^2$ ? I cant get it  :(


vector
geometry
geometric transformation
reflection
advanced fields
advanced fields unsolved

Sketch the vector field on the real line, find all the fixed points, classify their stability, and sketch the graph of x(t) for different initial conditions.

(a) x* = 1 - (x^14)
(b) x* = sinx/(e^x)
(c) x* = x - (x^3)

The roots for (a) and (b) are x=+/- 1 but I don't know how to draw these or where to take it from here.
I'm not sure about (c).  I think sinx=0 so that corresponds to x=0, 2pi, pi.....???

For drawing, all you need to do is to figure out when the vector points to the right and when it points to the left (i.e., when the expression is positive and when it is negative). Once you figured that out, imagine (or try to draw) an arrow at each point of the real line in the direction corresponding to that point. Now just follow the arrows and see whether they lead you toward a fixed point (stability) or away from it (instability). Your typical drawing will look like this:

[asy]size(250);
real f(real x){return 0.14*sgn(x);}
D((0,0.2),rgb(0.95,0.99,0.99));
D((-2,0)--(2,0));
D("-1",(-1,0));
D("0",(0,0));
D("1",(1,0));
for(real x=-1.8;x<1.9;x+=0.2)
if(abs(x-x^3)>0.0001)
{
real y=x+f(x-x^3);
D((x,0)--(y,0),black+linewidth(1), EndArrow);
}[/asy]

If you want to be more accurate, you should make the lengths of the arrows reflect the size of the right hand side (the larger the size, the longer the arrow) but it is not so important.

But don't you need to know what the graph looks like in order to know what the vector field is?  How do you graph x-x^3? Or sinx/e^x, or x^14?


function
integration
analytic geometry
topology
complex analysis
complex analysis unsolved

Let $ \Omega$ be an open set of the complex plane $ F$ be a family of analytic functions $ f: \Omega \rightarrow \mathbb{C}$ .
Suppose that there is an $ M>0$ such that
$ \int\int_{\Omega} |f(x,y)| dxdy<M$ for all $ f\in F$

Prove that $ F$ is a normal family

Use polar coordinates + mean value theorem, or green's theorem or whatever to express $ \iint f(x,y) dA$ over discs in $ \Omega$


[hide]$8\equiv 1\pmod 7$

so you just see which numbers leave a remainder of 1 when divided by 7 and the answers are A and E[/hide]

15 gives the same remainder as 8 when divided by seven.  I hope that helps. :)

[quote="mr. math"]but why is 8= 1(mod 7)[/quote]

it's like the remainder when you divide by 7

Look at the formal definition:
If $n\equiv 1\pmod 7$, then $n=1+7q$, where $q$ is any integer. So you see if $q=1$, then $n=8$. If $q=2$, then $n=15$.

The formal definition is:
If $a\equiv b \pmod{m}$, then $a-b$ is divisible by $m$.

Personally, I like it in simple terms.  Modular division is just like regular division, except the result isn't the quotient, but the remainder.

For example: 14 divided by 3 Remainder 2, whereas modular division just results in 2, the remainder left over.

Thusly, 8 in Mod 7 would be 1, since 8 divided 7 is 1 Remainder 1.  That means you need to find another number that divided by 7 leaves a remainder of 1, and 15 divided by 7 is 2 Remainder 1, so E's the answer.

[quote="Micalix"] Modular division is just like regular division, except the result isn't the quotient, but the remainder.[/quote]

Uh, no.

Modular division is defined as follows:

$\frac{a}{b}= q \bmod m : = a \equiv bq \bmod m$

The concept to which you refer is modular [b]equality / congruence[/b].

[quote="mr. math"]I need help with this problem, and if there are any good sites to learn about this topic, i would like to know about them. thanks.

which of the following is/are congruent to 8 in mod 7.  (u may have more than one answer)

A. 1    B. 3  C. 6    D. 10   E. 15[/quote]

Buy the introduction to number theory book if you can!
[hide]
The answer is A and E.

You have to look at the difference of the two numbers and see if that is divisible by the mod. Also, they have to have the same remainder when divided by the mod. [/hide]

You can learn modular arithmetic from either the Introduction to Number Theory book or class.  I would highly recommend getting the book, taking the class, or doing both, if you can.

The definition for mods is

If $a\equiv b\pmod{n}$, then $n|a-b$.

Example:
$12\equiv5\pmod{7}$ is a true statement, because $7|12-5$.

It is not just the remainder, because not only is $12\equiv5\pmod{7}$, $12\equiv-2\pmod7$, $12\equiv698\pmod7$, etc.

Learn more about the Introduction to Number Theory class [url=http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php#begnum]here[/url]

Learn more about the Introduction to Number Theory text book [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=10]here[/url]


Mike draws five cards from a standard 52-card deck. What is the probability that he draws a card from at least three of the four suits?

If he gets exactly one suit, there are $ 4\binom{13}5$ ways. If he gets exactly two suits, there are $ \binom42\left(2\binom{13}1\binom{13}4\plus{}2\binom{13}2\binom{13}3\right)$ ways. Thus, there are $ 384384$ ways to get less than three suits, for a probability of $ 1\minus{}\frac{384384}{\binom{52}5}\equal{}\boxed{\frac{507}{595}}$.

Why do we have $ \binom42\left(2\binom{13}1\binom{13}4+2\binom{13}2\binom{13}3\right) $ ways?

EDIT: 1234th post!

You can pick two suits out of four in $\binom42$ ways.

You can pick one out of the two suits to have one card in $2\binom{13}1$ ways and 4 cards out of the other suit $\binom{13}4$ ways. Multiply these two together to get $2\binom{13}1\binom{13}4$.

Do the same thing for the 2nd part.

WHAT IS IT WITH THREAD REVIVALS THE LAS POST WAS 6 YEARS AGO!!!!!!!!!

Thread revivals are encouraged in the Alcumus forum (I believe).

Each problem has a thread to go with it, and when you click the discuss button at the solution of a problem, it links you to the problem thread. If people revive the threads, that is fine in this forum! :)

[quote=Historybuff]Thankyou lion 11202 [hide]You want to be friends?[/quote]

Stop asking random people to be "friends" with you. You can discuss that in a PM.

Have a quick question, is there a gamebot question for EVERY question on alcumus?

I get this problem wrong every time I get it

This can be done with a little more casework (and no complementary counting).

Here is my solution:
[hide]
[b]Case 1: 3 suits [/b]
For this case, there are two sub cases. In the first subcase, one card gets 3 cards and the two suits get one card each. In the second subcase, two suits get two cards each and another suit just gets one card.
[b]First subcase (3,1,1):[/b]
There are $\binom{4}{3}$ ways to choose which 3 suits to choose. There are also $\binom{13}{1}$ ways to choose one card from each of the two suits that get one card each. There are $\binom{13}{3}$ ways to choose 3 cards from the suit that gets 3 cards. Finally, there are $\binom{3}{1}$ ways to choose which of the three suits selected gets 3 cards.
Thus, the probability for this case is:
 $\frac{\binom{13}{1}\binom{13}{1}\binom{13}{3}\binom{4}{3}\binom{3}{1}}{\binom{52}{5}}$

[b]Second subcase (2,2,1):[/b]
There are $\binom{4}{3}$ ways to choose which three suits out of the four are chosen. Then, there are $\binom{13}{2}$ ways to choose cards for each of the 2 suits selected that give 2 cards each. There are $\binom{13}{1}$ ways to choose the single card to be selected out of the final suit. Finally, there are $\binom{3}{1}$ ways to choose the one out of 3 suits selected that will not have 2 cards chosen (it will have 1).
Thus, the probability of this case is:
 $\frac{\binom{13}{2}\binom{13}{2}\binom{13}{1}\binom{4}{3}\binom{3}{1}}{\binom{52}{5}}$

[b]Case 2: 4 suits[/b]
In this case, 3 suits must get one card each and 1 suit must get two cards. Thus, there are $\binom{4}{1}$ ways to choose which suit gets the 2 cards. For each of the 3 suits that get one card each, there are $\binom{13}{1}$ ways to choose a card. Likewise, for the final suit, there are $\binom{13}{2}$ ways to choose two cards. Thus, the probability of this case is:
 $\frac{\binom{4}{1}\binom{13}{1}\binom{13}{1}\binom{13}{1}\binom{13}{2}}{\binom{52}{5}}$

If you sum up all of the probabilities of these cases, you will obtain $\frac{507}{595}$.
[/hide]

[quote=to_chicken][quote=apolloheo]I always get confused in deck of cards counting/probablility problem. Is there a nice article I can read to help me know what to do for what?[/quote][/quote]

can someone please answer this question? i think it would be helpful for a lot of people (including me)

thats kinda sad

got the right answer but proceeded to simplify using $585$ instead of $595$ :(

[quote=GoondaMATH]There is no way this would be asked in a competition, or perhaps I just did it in a poor way. ultimately had to simplify 265749120/311875200[/quote]

Next time, factor out numbers before multiplying them out  :)


probability
geometry
geometric transformation

In a game, each match consists of the launching of a honest coin till 10 times. If the number of gotten heads reach value five, you lose; contrary case you win.

Find the probability to win a match of this game.

I hope my translation wasn't too bad.

IF I understand the problem correctly, to my opinion it is $1/2$

Well, I think its:
 $\frac{1}{2^5}*\binom{5}{5}+\frac{1}{2^6}*\binom{6}{5}+\frac{1}{2^7}*\binom{7}{5}+\frac{1}{2^8}*\binom{8}{5}+\frac{1}{2^9}*\binom{9}{5}+\frac{1}{2^{10}}*\binom{10}{5}$
.....which is too sickening to evaluate, but hopefully I got LoboSolitario's answer.

[hide]

If i understand, if number of heads exceeds 5, you lose?

Well, the whole thing is symmetric around $2^5$
so you're looking for $\frac{1}{2^{10}} * (\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4})$

So subtract $\frac{1}{2^10}\binom{10}{5}$ from 1 and divide it by two...
so you have $\frac{1}{2}(1-\frac{10*9*8*7*6){2^10*5*4*3*2}$
which is

$\frac{1}{2}(\frac{193}{256})$

which is
$\frac{193}{512}$

yes?
[/hide]

[quote="daveed"]

If i understand, if number of heads exceeds 5, you lose?

[/quote]

[hide="Hint"]
If you get 5 heads you lose. Launching 10 times the coin you win for 0, 1, 2, 3 or 4 heads.

Hence
[hide="Monster Hint"]
$P = \frac{1}{2^{10}} \cdot \left(\binom{10}{0} + \binom{10}{1} + \binom{10}{2}+\binom{10}{3}+\binom{10}{4}\right)$[/hide]
[/hide]

All possibilities can be grouped into one of three cases: 0-4 heads, 5 heads/5 tails, 0-4 tails.
Since the coin is fair, the probability of getting 0-4 heads is the same as the probability of getting 0-4 tails.
The probability of getting 5 heads/5 tails is ${10 \choose 5}0.5^{10}$.
So the probability of getting 0-4 heads is $\frac12 \left(1-{10 \choose 5}0.5^{10}\right)=\frac{193}{512}$.


superior algebra
superior algebra unsolved

Show that if k is not= 23, there are infinitely many primes that are k mod 24. If anyone could post the case when k= 23, I would be grateful.

[url=http://www.math.uconn.edu/~kconrad/blurbs/dirichleteuclid.pdf]Here's[/url] something that might interest you.


a,b,c positive real numbers prove

$ \frac{3a^2\plus{}ab}{(a\plus{}b)^2}\plus{}\frac{3b^2\plus{}bc}{(a\plus{}b)^2}\plus{}\frac{3c^2\plus{}ac}{(b\plus{}c)^2}\geq3$

[quote="gret"]a,b,c positive real numbers prove

$ \frac {3a^2 \plus{} ab}{(a \plus{} b)^2} \plus{} \frac {3b^2 \plus{} bc}{(a \plus{} b)^2} \plus{} \frac {3c^2 \plus{} ac}{(b \plus{} c)^2}\geq3$[/quote]

I think you have typo, becouse it's not work:
try $ a\equal{}b\equal{}\frac{1}{2}, c\equal{}\frac{1}{4}$

$ \frac{3a^2\plus{}ab}{(a\plus{}b)^2}\plus{}\frac{3b^2\plus{}bc}{(b\plus{}c)^2}\plus{}\frac{3c^2\plus{}ac}{(c\plus{}a)^2}\geq3$
I am sorry again


The archeologists are working on an excavation site of a Roman marketplace. They know that the marketplace is square shaped, was surrounded by 4 walls, there was a well in the middle of the marketplace in which the merchants poured their gold in case of a barbarian attack. In two opposite corners of the market place stood the statues of Jupiter and Juno. The archeologists found the statue of Juno and the corner of the marketplace marked by a 1 meter long section of both walls running from this corner behind Juno, and a marking pole (showing the different directions towards the two statues) that was placed outside of the walls. How can the archeologists locate the well with all the gold in it?

I'm bad at visualising; picture please :P ?

I'll probably die reading this, popping my right brain because of excessive mind visualization

Sorry no picture.

If there was one, it wouldn't be as challenging. The no picture thing is the part that is confusing me.


John and a group of friends took a bus trip. Each person paid the bus driver with the same 9 coins. If the bus driver recieved $8.41 from the group, how many dimes did he recieve?

well 841=29^2

So there were 29 cents in the 9 coins for 29 people

And that is 5(5)+4(1)=29 so the bus driver got

5 nickes and 4 pennies from everyone, no dimes

so 0(29)=0 dimes


logarithms inequalities integration calculus function derivative limit

Let f(x)=x+x^2+x^4+x^8+... and g(x)=log (base 2) 1/(1-x).
 a) Prove that f-g is bounded.
 b) Does there exists the limit of f-g when x approaches 1 but is smaller than 1?
  I solved a).

How did you guys prove (a)?

Here's what I did: We can see that $1/(1-x)=(1+x)(1+x^2)(1+x^4)\ldots$ (we have all the powers of 2 as exponents), so $f(x)-g(x)=\displaystyle \sum_{i=0}^{\infty}\left(x^{2^i}-\log_2{(1+x^{2^i})}\right)$. If we use the inequalities $x\le \log_2{(1+x)}\le \frac x{\ln 2}$, we're done (these inequalities are true for all $x\in [0,1]$).

I tried to investigate point (b) with Mathematica, but since I'm not that skilled in using it, I didn't manage to do much.

I was expecting that you found a super-interesting and beautiful solution, but I can't believe it's so nice. My solution was the following:
   We have  \sum (n>=0) x^2^n \geq  \sum (n>=0)  \int (n-n+1) x^2^t dt and also  \sum (n>=0) x^2^n=x+ \sum (n>=1) x^2^n<=x+ \sum (n>=1)  \int (n-1-n) x^2^t dt. Thus,  \sum (n>=0) x 2 2 2 n -  \int (0 to infinity) x^2^t dt . Make the change of variable in the integral and find that it is 1/ln2*  \int (1 to infinity) x^u/u du and make the same trick: write the last integral  as an infinite sum of integrals and use the same inequality. I think you've got the idea.

Thanks for showing your solution to me. There was a tiny mistake in mine (a typo), so I edited the post.

Regarding (b), to be honest, I don't think the limit exists. I tried to use Mathematica to make a plot of f'-g', but I only did this for some values of n, not the whole series, and it's really crazy-looking.

Anyway, we definitely can't say somthing like "f-g is increasing" (or "decreasing").

I have ploted f-g in MathCad. It looks like decreasing continuos concave (maybe, approximately) function.

My solution of a) is similar to grobber's one.

Well, I didn't know how to plot x+x2+.. in Mathematica and I only worked with approximations of f'-g', but I think there's a region "realy close" to 1 where f-g becomes "very decreasing" :).

I was trying to make some progress on this problem, when I realized that my solution for (a) doesn't show squat :). The first inequality shows that $(f-g)(x)\le 0$, but the second inequality shows that $(f-g)(x)\ge (1-\frac 1{\ln 2})(x+x^2+x^4+\ldots)$, and this isn't bounded from below on $(0,1)$. It tends to $-\infty$ when $x\to 1,\ x<1$.

Any progress anyone? (I mean for point (b))

I can't manage to show that when you sum members like the last one for $x^{2^n}$ instead of $x$ you get a bounded sum, even with your hints. Anyway, I think Harazi's solution gives the best idea about what's going on. I think we get a tight bound by using that method: $(f-g)(x)\ge\frac {-x}{\ln 2}$.

Yours might be just as good, but I'm sort of tired right now, and this problem is giving me a really hard time. I can't try to solve other problems because I'm obsessed, but, on the other hand, I have no idea how to continue :(.

Thanks! Pretty stupid of me not to see that :).

I think that I have finally managed to prove that the limit exists. Anyway, there are so many computations that I am sure it is wrong.
   Let us take a number $0<a<1$ and take the function $ f(x)=a^{2^x}$, continuous decreasing function. Let the function $ s(x)=\sum_{k=1,[x]}{f(k)}$ for $x\geq 1$ and 0 for the other values of x. Now, define $ a_k=\int_{k-1}^{k}{f(t)dt}-f(k)$. The reader will prove without great difficulty that there exists a bounded function T and a number $C=\sum_{k\geq 1} {a_k}$ such that $ s(x)=\int_{0}^{x}{f(t)dt}-C+f(x)T(x)$. (**)Now, define the numbers $x(a)$ by $2^{x(a)}=\frac{1}{(1-a)^2}$. It is easy to prove that $ a^{2^{x(a)}}$ and $\frac{a^{2^{x(a)}}}{lna}$ tend to 0 when a tends to 1.
  Next, observe that $ 0\leq\sum_{n\geq 0}{f(k)}-s(x(a))\leq\frac{a^{2^{x(a)}}}{1-a^{2^{x(a)}}}$, which proves that it is enough to show that the limit of $ s(x(a))-\frac{ln(1-a)}{ln2}$ exists when a goes to 1. Now, we use (**) to reduce this existence to the existence of the limit $\int_{0}^{x(a)}{f(t)dt}-\frac{ln(1-a)}{ln2}$. Changing the variable in the integral and the use of the fact that the integral $\int_{1}^{\inf}{\frac{a^u}{u}du}$ and the fact that $\int_{2^{x(a)}}^{\inf}{\frac{a^u}{u}du}$ goes to 0 as a goes to 1 ( it suffices to observe that this integral is between 0 and $\frac{a^{2^{x(a)}}}{-lna}$) reduces us to proving that the limit of $\int_{1}^{\inf}{\frac{a^u}{u}du}+ln(1-a)$ exists when a goes to 1. If I am not making a stupid mistake, then the derivative of the function $ h(a)=\int_{1}^{\inf}{\frac{a^u}{u}}+ln(1-a)$ is $\frac{-1}{lna}-\frac{1}{1-a}$ which tends to $-\frac{1}{2}$ when a goes to 1. Thus the derivative becomes negative in a certain zone close to 1 and there the function is decreasing and so has a limit.
 Sorry if I made idiot mistakes, but this problem already gets me mad.

[quote="harazi"]Now, define $ a_k=\int_{k-1}^{k}{f(t)dt}-f(k)$. The reader will prove without great difficulty that there exists a bounded function T and a number $C=\sum_{k\geq 1} {a_k}$ such that $ s(x)=\int_{0}^{x}{f(t)dt}-C+f(x)T(x)$.[/quote]
Maybe I am very stupid, but $C=C(a)$, but you ignore it in further proof.
Am I missing something?

Ups, sorry, you are right.  :(  Back to the table.  :(  :blush:

;) I believe that there is no limit. If you want to know why, [hide="click here"]

Here is an idea why. If the limit exists, then $g(x)=f(x)-f(x^{\sqrt 2})$ must tend to $\frac12$ (because this is true for the logarithm). Now, $g(x)$ can be written as
\[
g(x)=\sum_{k=1}^\infty \left(x^{2^k}-x^{2^{k+1/2}}\right)
\]
Now take any $x\in(0,1)$ and consider $x^{2^{-n}}$. Letting $n\to+\infty$, we get
\[
\frac12=\lim_{n\to\infty}g(x^{2^{-n}})=\sum_{k=-\infty}^\infty \left(x^{2^k}-x^{2^{k+1/2}}\right)\,.
\]
And this must be true for any $x\in(0,1)$! Such an identity is just unbelievable.
Of course, if it is false, it will be easy to show it using a computer, but the problem is too nice to resort to such brute force methods. If, instead of powers of $2$, we had powers of some big number $M$, the story would be pretty much the same but we would be able to say that there exists $x$ with $x-x^{\sqrt M}>\frac12$, period. And it is hard to believe that $2$ is special here. I understand that this is not a solution, but I hope the argument is fairly convincing even if we agree to outrule using computers for this problem...
[/hide]

The equality from http://www.mathlinks.ro/Forum/viewtopic.php?t=28814 seems unbelievable at first sight too. :)

OK. I figured the one I need out. If you apply $x\frac d{dx}$ to it many times and choose $x$ appropriately, one term will dominate the sum of all others. :lol:

[quote="fedja"][quote="Myth"]Sorry. What are talking about? Do you speak russian?[/quote]
The identity I declared "unbelievable" in my previous post. Da, ja govorju po russki luchshe chem po anglijski... ;)[/quote]
Я про?то не пон?л, к чему ты ?казал про дифференцирование. Ты учишь??/работаешь в СШ??

To Valentin. I have own opinion when I am allowed to write in Russian ;)

Why don't we switch to PM's not to spam the forum? ;)

Aww well guys I didnt get what you are trying to say & the link is not opening in my browser.
So that leaves with the million dollar question whether the limit exists or not.According to my "wrong"calculations it doesnot... :(

[quote="fedja"][quote="the game"]Aww well guys I didnt get what you are trying to say & the link is not opening in my browser.[/quote]
Sorry, the game! :blush: Do you understand what was written in my posts #16 and #18? Together they are assumed to prove that there is no limit...[/quote]

Sorry fedja the link didnt open .Maybe some kind of browser problem.Thats what I meant :blush:


Let p(x) be a polynomial with integer coefficients. Prove that if k is an even integer, we have p(k)=p(0)(mod 2) and if k is odd, then we have p(k)=p(1)(mod 2)

[hide="First Problem"]
When k is a even integer(2b):
$ P(k) \equal{} (a_n)(2b)^n \plus{} (a_{n \minus{} 1})(2b)^{n \minus{} 1} \plus{} ... \plus{} a_0.$
And
$ p(0) \equal{} a_0$

We know that $ a_0\equiv a_0(mod 2)$ and that if $ a\equiv b(mod z)$ and $ c\equiv d(mod z)$ then $ a \plus{} c\equiv b \plus{} d(mod z)$.

Since every term that contains (2b) is even and congruent to 0 mod 2, when we add the terms in $ p(k)$ in (mod 2) we are left with $ a_0$ and that equals the value of $ p(0)$.
[/hide]
[hide="Second Problem"]
To Prove that when k is odd, $ p(k)\equiv p(1)(mod 2)$ we in other words have to prove that $ p(k)\minus{}p(1)\equiv 0(mod 2)$

Let's say that $ k\equal{}2b\plus{}1$
So we have
$ p(k)\equal{}(a_n)(2b\plus{}1)^n\plus{}(a_{n\minus{}1})(2b\plus{}1)^{n\minus{}1}\plus{}\cdots\plus{}a_0$
$ p(1)\equal{}(a_n)(1)^n\plus{}(a_{n\minus{}1})(1)^{n\minus{}1}\plus{}\cdots\plus{}a_0$

$ (a_n)(2b\plus{}1)^n\minus{}(a_n)(1)^n\equiv (a_n)((2b)^n\plus{}(2b)^{n\minus{}1}\plus{}\cdots\plus{}1)\minus{}(a_n)(1)^n(mod 2)$

Since we know that every term that contains $ 2b$ is congruent to 0 (mod 2) we are left with:

$ (a_n)(2b\plus{}1)^n\minus{}(a_n)(1)^n\equiv 0(mod 2)$
We repeat this process for every n:
$ (a_{n\minus{}1})(2b\plus{}1)^{n\minus{}1}\minus{}(a_{n\minus{}1})(1)^{n\minus{}1}\equiv 0(mod 2)$
$ (a_{n\minus{}2})(2b\plus{}1)^{n\minus{}2}\minus{}(a_{n\minus{}2})(1)^{n\minus{}2}\equiv 0(mod 2)$
$ \vdots$
$ (a_{0})(2b\plus{}1)^{0}\minus{}(a_{0})(1)^{0}\equiv 0(mod 2)$
So this proves that $ p(k)\minus{}p(1)\equiv 0(mod 2)$ or $ p(k)\equiv p(1)(mod 2)$
[/hide]


The path of a ball is given by y=(-1/20)x^2 + 3x + 5 , where y is the height in feet and x is the horizontal distance in feet.

(a) Find the maximum height of the ball

(b) Which term determines the height at which the ball was thrown? Does changing this term change the coordinates of the maximym height of the ball? Explain.

have fun

Complete the square for (a). You know squares cannot be less than 0, so replacing the square with 0 gives the maximum height.

The height at which the ball is thrown is the height at x=0.

haha
yeah
i know that i have to complete the square
but the prob is that wen i do

i get a really weird fraction

i dont know what im getting wrong

and the initial height should be 5
not 0
because of the value of c

Zuton Force said it occurs [i]at[/i] x=0, so the initial height is 5.  :wink:

For future reference, this problem would be more appropriate in High School Basics.

Ya i was wondering y it was in the pre-olympiad section when it can be easily solved..


$P(x)=ax^{3}+bx^{2}+cx+d$ is a polynomial with integer coefficients and $a$ is non-zero .We have $xP(x)=yP(y)$ for many pairs $(x,y)$ of unequal integers. Show that $P(x)$ has an integer root.

This problem is A3 in IMO2002 Shortlist.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118702#p118702]here[/url] is the respective thread.
While [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=2002]here[/url] you can see all problems of IMO2002 Shortlist.


[quote="Beat"]Let $a, b, c$ be positive real numbers such that
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$.
Prove that
$\frac{1}{4a+1}+\frac{1}{4b+1}+\frac{1}{4c+1}\ge 1$[/quote]
Nicer:
$\frac{1}{4a+1}+\frac{1}{4b+1}+\frac{1}{4c+1}\ge 1\Leftrightarrow\sum_{cyc}\left(\frac{1}{4a+1}-\frac{1}{a+1}+\frac{1}{3}\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(2a-1)^{2}}{(4a+1)(a+1)}\geq0.$ :)

I found even more nicer:
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$, equivalently
$ab+bc+ca+2abc=1$.
By the Cauchy-Schwarz inequality it follows that:
$((a+b+c)+\frac{1}{2})(bc+ca+ab+2abc)\ge (4\sqrt{abc})^{2}=16abc$
Or equivalently: $2(a+b+c)+1\ge 32abc$, $\frac{1}{1+4a}+\frac{1}{1+4b}+\frac{1}{1+4c}\ge1$

[quote="Beat"]
By the Cauchy-Schwarz inequality it follows that:
$((a+b+c)+\frac{1}{2})(bc+ca+ab+2abc)\ge (4\sqrt{abc})^{2}=16abc$
[/quote]
Nice! :lol:

See also here .
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%2A32abc%2A&t=79554&sid=7a3f16552285148fbb15b3b510b1c3d9

Campos posted a very nice solution using Cauchy . :)

[quote=Beat]Let $a, b, c$ be positive real numbers such that
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$.
Prove that
$\frac{1}{4a+1}+\frac{1}{4b+1}+\frac{1}{4c+1}\ge 1$[/quote]

$\frac{1}{{4a + 1}} \ge \frac{1}{{a + 1}} - \frac{1}{3} \Leftrightarrow \left( {1 - 2a} \right)^2  \ge 0$
$ \Rightarrow \sum {\frac{1}{{4a + 1}}}  \ge \sum {\frac{1}{{a + 1}}}  - 1 = 1$

[quote=scpajmb][quote=Beat]Let $a, b, c$ be positive real numbers such that
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$.
Prove that
$\frac{1}{4a+1}+\frac{1}{4b+1}+\frac{1}{4c+1}\ge 1$[/quote]

$\frac{1}{{4a + 1}} \ge \frac{1}{{a + 2}} - \frac{1}{3} \Leftrightarrow \left( {1 - 2a} \right)^2  \ge 0$
$ \Rightarrow \sum {\frac{1}{{4a + 1}}}  \ge \sum {\frac{1}{{a + 2}}}  - 1 = 1$[/quote]
$\sum {\frac{1}{{4a + 1}}}  \ge \sum \left({\frac{1}{{a + 1}}- \frac{1}{3}}\right)= 1$


floor function
fractional part
algebra
IMO Shortlist

Let $ a, b, c$ be positive integers satisfying the conditions $ b > 2a$ and $ c > 2b.$ Show that there exists a real number $ \lambda$ with the property that all the three numbers $ \lambda a, \lambda b, \lambda c$ have their fractional parts lying in the interval $ \left(\frac {1}{3}, \frac {2}{3} \right].$

Why are some other files written that the interval is $ \left(\frac{1}{3},\frac{2}{3}\right]$ not $ \left(\frac{1}{2},\frac{2}{3}\right]$?
And what is the correct statement?

I checked the IMO Compendium, and it says that (1/3, 2/3] is the correct interval.

[quote="JSGandora"]That is equivalent to finding some integer $x$ and $y$ such that
\[\frac{3x+1}{3y+1}\leq \frac{a}{b}\leq \frac{3x+2}{3y+2}.\][/quote]
Alternatively, note that this is equivalent to $\frac{1}{3} \le \frac{ya - xb}{b-a} \le \frac{2}{3}$. If $a=gu$, $b=gv$ with $(u,v)=1$, we then just need to find $k\ge1$ with $\frac{1}{3} \le \frac{kg}{b-a} = \frac{k}{v-u} \le \frac{2}{3}$ (since $g = na-mb$ for some positive integers $n,m$). But $b>2a\implies v>2u$, so $v-u \ge u+1 \ge 2$, and we can simply take $k = \lfloor{(v-u)/2}\rfloor$ to finish (the "hardest" case is when $v-u=3$).

I'm not sure if we can use rational approximation when $a/b$ is irrational (the problem still holds in this case by the previous solution), but at least the two-dimensional version of Kronecker's theorem doesn't seem to work directly.

Consider a "good $x$ interval" a interval of continuous $\lambda$s that work for $x$. It measures $1/3x$, and by inequality, any good-b interval intersects some good-c interval, so is enough to prove some good-b interval is completely inside a good-a interval. Is trivial to see that the $(\lambda_1+1)/2$'th good-b interval and the $(\lambda_2+1)/2$'th good-a interval satisfy this iff

$ \frac{ | \lambda_1b - \lambda_2a | }{2} \le \frac{b-a}{3}$.

By bezout lemma, is easy to find odd $\lambda$s that satisfy this (one needs to divide in cases $a|b$ and $a$ doesnt divide $b$), so we finish.

Solved with [b]Smileyklaws[/b]

Consider all intervals of the form $\left(\frac{3k+1}{3b}, \frac{3k+2}{3b}\right]$ for $k \ge 0$. Each of these intervals has length $\frac{1}{3b}$. Now consider all intervals of the form $\left(\frac{3k+1}{3c}, \frac{3k+2}{3c}\right]$ for $k \ge 0$. Each of these intervals has length $\frac{1}{3c}$, and the space between two consecutive intervals is $\frac{2}{3c}$. But $c > 2b \implies \frac{2}{3c} < \frac{1}{3b}$, so it follows that each $c$-interval overlaps with a $b$-interval. We wish to show that there exists some $\lambda$ in an $a$-interval, a $b$-interval, and a $c$-interval.

Now we claim that there exists a $c$-interval that is contained within an $a$-interval. Note that each $c$-interval has length $\frac{1}{3c}$, while the space between two consecutive intervals is $\frac{2}{3c}$. Then $$c>4a \implies \frac{4}{3c}<\frac{1}{3a} \implies \frac{1}{3c}+\frac{2}{3c}+\frac{1}{3c} < \frac{1}{3a}$$so with some effort it follows that there exists a $c$-interval contained within an $a$-interval.

Since each $c$-interval overlaps with a $b$-interval, by our work above, there exists an $a$-interval, a $b$-interval, and a $c$-interval that overlap. Consider some real $d$ in the union. Then for some positive integer $m$, $$\frac{3m+1}{3a} < \lambda \le \frac{3m+2}{3a} \implies m+\frac{1}{3} < \lambda a \le m+\frac{2}{3} \implies \frac{1}{3} < \{\lambda a\} \le \frac{2}{3}$$Similarly, $\frac{1}{3} < \{\lambda b\} \le \frac{2}{3}$, $\frac{1}{3} < \{\lambda c\} \le \frac{2}{3}$. This is what we needed, so we are done. $\blacksquare$

[quote=JSGandora]If $\lambda\in\left(\frac{3k+1}{3a}, \frac{3k+2}{3a}\right]$, then $\frac{1}{3}<\{\lambda a\}\leq \frac{2}{3}$.  Define the intervals similarly for $b$ and $c$.  Call the interval for $a$ - $L_k$, the interval for $b$ - $M_k$ and the interval for $c$ - $N_k$.  Now the length of each $M_k$ is $\frac{1}{3b}$ which are spaced $\frac{2}{3b}$ apart from each other.  The interval for each $N_k$ is $\frac{1}{3c}$ and they are spaced $\frac{2}{3c}<\frac{1}{3b}$ apart.  Thus no $M_k$ can not contain an interval $N_k$ because the intervals not containing $N_k$ are all of length less than $\frac{1}{3b}$.  Thus each interval $M_k$ must intersect some interval $N_k$.

Now what remains to prove find is some interval $M_k$ that is contained entirely within some interval $L_k$.  If that is true, then we want some $x$ and $y$ such that
\[\frac{2x+1}{3a}\leq \frac{3y+1}{3b}<\frac{3y+2}{3b}\leq \frac{3x+2}{3a}.\]
That is equivalent to finding some integer $x$ and $y$ such that
\[\frac{3x+1}{3y+1}\leq \frac{a}{b}\leq \frac{3x+2}{3y+2}.\]
Now if $\frac{a}{b}\leq \frac14$, then let $y=2^{e}$ where $e$ is the maximal $e$ such that $\frac{1}{3\cdot 2^e+1}\leq \frac{a}{b}$.  The minimal $e$ is $0$ and thus the minimum $\frac{a}{b}$ is $\frac{1}{4}$ which explains the bound.  Now then $\frac{1}{3\cdot 2^{e-1}+1}\geq \frac{a}{b}$ by definition so $\frac{2}{3\cdot 2^e+2}\geq \frac{a}{b}$.  Therefore, if we let $x=0$ and $y={2^e}$, we have
\[\frac{2x+1}{3y+1}=\frac{1}{3\cdot 2^e+1}\leq \frac{a}{b}\leq \frac{2}{3\cdot 2^e+2}=\frac{3x+2}{3y+2}.\]
Now if $\frac{a}{b}\geq \frac{1}{4}$.  Note that $\frac{a}{b}<\frac12$ because $b>2a$.  Now the sequence $\frac{3\cdot2^e-2}{3\cdot2^{e+1}-2}$ approaches $\frac12$ as $e$ approaches infinity.  Now choose the maximal $e$ such that $\frac{3\cdot2^e-2}{3\cdot2^{e+1}-2}\leq \frac{a}{b}$.  Then $\frac{3\cdot 2^{e+1}-2}{3\cdot 2^{e+2}-2}=\frac{3\cdot 2^e-1}{3\cdot 2^{e+1}-1}\geq \frac{a}{b}$.  Thus let $x=2^{e}-1$ and $y=2^{e+1}-1$ and thus that satisfies
\[\frac{3x+1}{3y+1}\leq \frac{a}{b}\leq \frac{3x+2}{3y+2}.\]
Therefore it is always possible to find some integer $x,y$ such that $M_k$ is contained entirely within some $L_k$, and the $M_k$ will always intersect some $N_k$ which proves that there exists some real number $\lambda$ that is in all three intervals so the fractional parts of $\lambda a, \lambda b,$ and $\lambda c$ are all in the interval $\left(\frac13, \frac23\right]$. $\blacksquare$[/quote]

I think you meant $3x+1$ in LHS of the inequality instead of $2x+1$.

Note that $\{\lambda x\}\in \left(\tfrac {1}{3}, \tfrac {2}{3} \right]$ is equivalent to $\lambda \in \left(\tfrac{3k+1}{3x},\tfrac{3k+2}{3x}\right]$ for some integer $k$. Call these the $x$-intervals, for $x=a,b,c$. Each $x$-interval has a starting point of $\tfrac{3k+1}{3x}$, a length of $\tfrac{1}{3x}$ and are $\tfrac{2}{3x}$ apart. Since $\tfrac{2}{3c}<\tfrac{1}{3b}$, each $b$-interval intersects some $c$-interval at least somewhere.
\\$~$\\
It remains to show that some $b$-interval is contained in another $a$-interval. To do this, some $b$-interval's starting point must be in \[\left(\frac{k}{a}+\frac{1}{3a},\frac{k}{a}+\frac{2}{3a}-\frac{1}{3b}\right] = \left(\frac{3bk+b}{3ab},\frac{3bk+2b-a}{3ab}\right]\]
Meanwhile, the starting points of the $b$-intervals are
\[\frac{1}{3b}+\frac{l}{b}=\frac{3la+a}{3ab}\]
Clearly, if $b>4a$ then $2b-a-b>3a$ so there must exist some $l$ that makes $3la+a$ fall into the interval $(3bk+b,3bk+2b-a]$. If $b=4a$ then $7a$ and $2b-a$ coincide.
\\$~$\\
Now, we claim that no matter what $r=\tfrac{b}{a}$ is, we will be able to find a value $k$ and $l$. We have already done $k\ge 4$. Consider $l=2k+1$. We have $3la+a=(6k+4)a$. As long as
\[b(3k+1) < a(6k+4) \le b(3k+2)-a \implies 2+\frac{1}{3k+2}\le r < 2+\frac{2}{3k+1}\]
Thus, it is true for $r\in[2+\tfrac1{3k+2},2+\tfrac2{3k+1})$. Note that the lower bound can get arbitrarily close to $2$. It remains to show that the upper bound is at least equal to the lower bound of the previous, that for all $k\ge 1$, \[\frac2{3k+1}\ge \frac1{3k-1}\]
Which is true. Note that when $k=0$ the upper bound is $4$. Therefore, if $r\in (2,4)$ we can find a value of $k,l$ that fit. We are done.

onk

b-interval intersects c-interval

prove b-interval is contained by a-interval

now do spam stuff

(oops i actually goofed and got to @above for $k=0$ and didn't generalize L me)

wait just do what gandora did trust

okay yeah cool my brain just farted but this works. hopefully i don't forget this but the moral of the story is to just do it smh my head


geometry
geometric transformation
reflection
modular arithmetic
national olympiad

16th Hungary-Israel olympiad
Israel, 28.3.2006

day 1

1. let x,y,p,n,k be naturals satisfying
x^n + y^n = p^k
if n>1 is odd and p is odd prime prove that n is power of p.

2. Box with dimensions aXbXc is paved by 3D domino blocks 1X1X2
that have sides parallel to the box sides. any side of the box have
equal number of blocks paralleling. what is the possible dimensions of the box?
(domino block is paralleling to side of the box if it's side of length 2 is paralleling to
the box side).

3. Let H be convex polygon with n sides. let A(1), A(2), ..., A(n) be his vertexes and
A'(1), A'(2), ..., A'(n) be the reflection by the middle section of the neighboring diagonal
A(i-1)A(i+1) (here A(n+1)=A(1)). The vertex A(i) is good if A'(i) is inside of H.
prove that H have at least n-3 good vertexes.

day 2

4. P is point inside a given circle. there are three cords passing through P
with equal length. prove that P is the center of the circle.

5. For any non-negative x, y, z let:
S= x^2y + y^2z + z^2x + xy^2 + yz^2 + zx^2
C= x^2y + y^2z + z^2x
find the maximal values of S and C when x + y + z = 1.

6. Group of 100 contestants numbered 1, 2, 3, ..., 100 playing the following game:
the judge writing the numbers 1, 2, 3, ..., 100 on 100 cards, turn them over
and place them on a table in arbitrary order.
for any j, contestant number j entering the room and flip 50 cards.
he gain one point if the number j is on one of the flipped cards.
is there a strategy giving more than 1%  to win?
(contestants can't talk to each other when the game begin).

1. assume that $ p^m|x$, and $ p^m$ is the highest power doing so. write $ x \equal{} p^mt$, $ t\in \mathbf{N}$. clearly $ mn < k$, and $ y^n \equal{} p^k \minus{} p^{mn}t \equal{} p^{mn}(p^{k \minus{} mn} \minus{} t)$, so the highest power of $ p$ dividing $ y$ is $ p^m$. dividing out the highest power of $ p$ we reduce the problem to the case where $ p$ doesn't divide $ x$ or $ y$. now since $ n$ is odd, we may write the equation as

$ p^k \equal{} (x \plus{} y)(x^{n \minus{} 1} \minus{} x^{n \minus{} 2}y \plus{} x^{n \minus{} 3}y^2 \minus{} \cdots \plus{} y^{n \minus{} 1})$ (*)

clearly $ (x \plus{} y) > 1$, so $ p|x \plus{} y$, i.e. $ y\equiv \minus{} x \pmod{p}$.

denote by $ E$ the second factor on the RHS of (*). we then have

$ E \equiv x^{n \minus{} 1} \plus{} x^{n \minus{} 1} \plus{} \cdots \plus{} x^{n \minus{} 1} \equal{} nx^{n \minus{} 1} \pmod{p}$ (**)

now if both $ x$ and $ y$ were 1, we would have $ p^k \equal{} 2$, but $ p$ is odd. so one of $ x,y$ is $ > 1$. thus, since $ n > 1$, $ x \plus{} y < x^n \plus{} y^n \equal{} p^k$, so $ E > 1$. so $ p|E$, and in light of (**) it follows that $ p|n$.

now we have $ (x^{p})^{n/p} \plus{} (y^{p})^{n/p} \equal{} p^k$. if $ n/p \equal{} 1$, we're done. otherwise, $ n/p > 1$, so we may apply the above argument. continuing like this, we must eventually reach $ n/p^i \equal{} 1$ for some $ i$, and so we end.

4. if $ P$ weren't the center, and the point $ O$ instead were the center, we would get three congruent right triangles all with the same hypotenuse, $ OP$, so that the leg ending at point $ P$ is the same length for all three of them. but there are at most two such triangles.

6. i've heard this problem somewhere else. you forgot to say that the goal of the game is that at least one of the contestants turn over the correct card.

Problem 6 is a very classic problem.... (I have heard of a version with prisoners and boxes... same idea...)

the strategy is that the nth contestant first goes to the nth card on the table, and then if the card said k, he would go to the kth card, and so on until they either find their card, or exhaust all 50 tries. This would increase the chances of them winning by a great deal, as the card numbers will cycle (i.e. they will lead to each other through a set repetition of values). Thus, each competitor will eventually find the winning card through the maximum cycle length (in the worst case scenario).

So, for every single competitor to win, all cycles must have a length that is less than 50... Thus, since there are 100! permutations, and of these, there are 100!/N that are bad for every N>50... we can see that the number of good permutations is: 100! - 100! / 51 - 100! / 52 - ... - 100!/100

This yields an approximate winning rate of: 31.1%... not bad! :P  :P

In 6 i forgot to mention that the aim is to get 100 points.
Nawahd Werdna can you expalin how you get the method to win?
Can someone give a link to similar problems?

Well... to find this method, you just need to play around with the problem for a while... and you'll get it...

To give you an example why this solution works, lets take an example with 10 cards...
Lets say the cards are in the order
10 2 4 5 7 1 9 6 3 8

Now, contestanat 1 goes to the table, and he goes to card position 1, which is 10... then, he goes to card position 10, with value 8... he then goes to card position 8, which is 6, and to the 6th position, which has value 1.... and so on

We can write this as a cycle, namely:

10, 8, 6, 1, 10 , ...
and so, since this cycle has length that less than 5, we can see that every one of the numbers  10, 8, 6, and 1 will be guaranteed to find their card using this method... :D


calculus Princeton college Stanford articles topology

I use a book written by Paul A. Foerster for Calculus, and it's pretty good.

As far as physics goes... I used a Princeton Review AP Physics book a bunch, and I've read about half of Richard Feynman's Lectures on Physics Volume 1. That's a tough read; though. You need to pretty much know the stuff before hand.

Feynman's books are fantastic, but confuted is exactly right.  According to legend, when Feynman actually gave the course (a freshman physics course), he quickly ran off all the students - but all the later year students and grad students (and some profs) started coming to solidify their knowledge.

The Feynman lectures are excellent for exactly that purpose.  If you find yourself really into physics, get them.

[quote="hello"]This might not be appropriate for this category but those of you taking single variable Calculus and algebra based physics do any of you use the Osteebe, Zorn Calculus book and Giancoli for physics. If yes, do you think the books are good?[/quote]

Hmmm, I think below I'll put my VERY rough draft of a FAQ on studying calculus. To answer your basic question, I have the preliminary edition of the Ostebee, Zorn textbook and it looks all right as a high school textbook for an AP calculus class. It has a very strong "reform" flavor. I have several other calculus textbooks at home. By far my favorite, among all the calculus books I hope to slog through in the next few years (I've never had a calculus class) is Michael Spivak's textbook Calculus.

We haven't had an algebra-based physics textbook around the house yet. I imagine next year we will buy whatever EPGY specifies for its computer-based beginning physics course. I recommend LOTS of reading in adult popular books about physics to supplement what you find in your textbook, as I do for studying any science course.

Hope this helps!

VERY rough draft of calculus study FAQ follows, and I'd love to hear comments from participants here with suggestions on how to improve the FAQ. [FAQ begins here:] The best calculus textbook, at least in English, is Michael Spivak's book Calculus (Publish or Perish Press, third edition). This textbook is not specifically written to the AP Calculus syllabus and thus is unknown to many AP calculus teachers, but it is unanimously recommended by mathematicians.

In answer to a question in the Usenet sci.math newsgroup, I wrote

> What comes after calculus?

What comes after a first course in calculus is a second try at learning calculus to really UNDERSTAND calculus. Hardly any universities use Michael Spivak's book Calculus (third edition, Publish or Perish Press) as a textbook for a first calculus course, but it is positively SCARY how unanimous mathematicians are in recommending that book for people who really want to understand calculus. Spivak's book also has good humor, as most good mathematical books do.

A little while earlier, someone else on the same newsgroup asked a series of detailed questions about studying calculus.

> Anyone got any idea as to good book on calculus?

The calculus book that I have seen most consistently get glowing reviews from professional mathematicians, with many calling it the "best" calculus book, is Calculus by Michael Spivak (3rd edition, Publish or Perish Press). I own Spivak's book, and know several people locally who think it is wonderful. However, Spivak's Calculus is not an EASY introduction to calculus--it's very useful to have for clear explanations and working through it would probably give you a deep understanding of calculus that would be difficult to forget, but it wouldn't be the fastest way to learn, for example, the calculus tested on the AP calculus exams.

I second-source (actually, quintuple-source or more) calculus books, as I am also trying to learn calculus by self-study. In the end, the only way to learn calculus is to do A LOT of problems that build progressively on earlier understanding--on that, all teachers of calculus agree, whatever else they disagree on. And calculus is, indeed, one of the math courses for which there is the most varied assortment of textbooks, sometimes taking wildly differing approaches.

> Can I learn calculus in the internet?

There is a distance learning course on calculus for which you can get college credit (if you PAY for the college credit and keep up with your required homework) offered by the University of Illinois Urbana-Champaign

http://www-cm.math.uiuc.edu/

called Calculus and Mathematica which some math professors like a lot, and others decry. I MAY just buy the book for that course through Amazon.com and work through it myself with a student copy of Mathematica, forgoing the college credit, but I'm still trying to decide.

There is another distance learning course about calculus offered by the Education Program for Gifted Youth (EPGY) at Stanford University

http://epgy.standford.edu/

but that course, as its name implies, is for gifted young people who haven't yet graduated from high school. That course uses a combination of the Anton textbook and innovative course software developed by Stanford. My oldest son is taking EPGY courses that are prerequisites for the calculus course right now, and may be taking calculus within two years.

> Is there a web site to learn calculus?

There are a lot of great Web sites with articles ABOUT calculus, including what amounts to a complete, FREE textbook by a famous mathematician in at least one case. Those include

http://www.math.gatech.edu/~morley/1507/MeaninglessSymbolicManipulation.html

(approaches to studying Calculus, including a recommendation of some of the books I just mentioned)

http://www.math.washington.edu/~duchamp/124notes/k124-main.pdf

http://www.math.washington.edu/~duchamp/125notes/k125-main.pdf

(Two lengthy Adobe Acrobat .PDF files with a college calculus course, free for the downloading, by Professor Neal Koblitz of the University of Washington)

http://www.math.hawaii.edu/~lee/calculus/

(Web page with links to many articles by Professor E. Lee Lady of the University of Hawaii)

There is a lot more where these links came from; you could find good stuff about calculus on the Web for hours. Try Google

http://www.google.com/

to run more searches of your own.

That's as far as I wrote for the Usenet inquirer. Most parents of homeschooled kids are looking for something like a high school calculus book. The first place to look is in your friendly local public library, where you may find several different calculus books aimed at high school students. Stewart's book is used here in Minnesota by the University of Minnesota Talented Youth Mathematics Program (UMTYMP) and the Anton book is used by EPGY; I can get both in my local library systems. How to Learn Calculus: A Streetwise Guide looks pretty good to me, although I would have your local library request it rather than buying it yourself. Calculus Made Easy is NOT a complete calculus course, but it is good for motivation and keeping the learner from being psyched out. Spivak's Hitchhiker's Guide to Calculus is a good introductory text; there are lots of high school cram books on calculus in good public libraries. There are many good calculus preparation sites for university students, for example,

http://www.math.mun.ca/~apics/calculus/

The person I replied to on Usenet asked another useful question:

> What about other books that would serve as backgrounders
> along the same vein as the above point?

There are LOTS of books for "transition courses" that bridge the gap between undergraduate calculus and math courses beyond calculus for math and science majors. Those books often have titles like "Introduction to Proof," or "Problem Solving" or "Introduction to Advanced Mathematics" or "Mathematical Thinking" or phrases more or less like that. It is a good idea to start reading a book like that, and working through its exercises, BEFORE finishing a calculus course. Probably your local university has several book like this in the campus bookstore, and the math department could recommend whatever title is used where you go to school.

[End of calculus FAQ]

What do the rest of you have to say about studying calculus?

Is Spivak's [i]Calculus on Manifolds[/i] the "Volume II" of his [i]Calculus[/i]?

Please don't revive old threads.  Also, I'm sending this off to another forum, where it belongs.

[quote="confuted"]I use a book written by Paul A. Foerster for Calculus, and it's pretty good.
[/quote]
We use the same book. It has good problems, and clearly outlines topics. However, the author's sense of humor (i.e. the name puns) I could do without.


1. Denote by $p_{k}$ the $k$th prime number. Show that $p_{1}p_{2}\cdots p_{n}+1$ cannot be the perfect square of an integer.

2. Prove that it is impossible for three consecutive squares to sum to another perfect square.

[hide="Hint for 1"] $\bmod 4$. [/hide] [hide="Hint for 2"] $\bmod 3$. [/hide]

[hide="Solution to #2"]
Let the squares be $(x-1)^{2}$, $x^{2}$, and $(x+1)^{2}$. Their sum is $3x^{2}+2\equiv 2\pmod 3$.

Since $0^{2}\equiv 0\pmod 3$, $1^{2}\equiv 1\pmod 3$, and $2^{2}\equiv 1\pmod 3$, it follows that a perfect square cannot be congruent to 2 mod 3. [/hide]

[hide="Solution to #1"]
Assume $p_{1}p_{2}\cdots p_{n}+1=m^{2}\implies p_{1}p_{2}\cdots p_{n}=(m+1)(m-1)$.

Since $p_{1}=2$, the LHS is even, so $m$ must be odd. Then $m+1$ and $m-1$ are both even and the RHS is divisible by 4. Contradiction.[/hide]

[quote="junggi"][quote="computer_nuke"][quote="junggi"][hide="1"]Since all primes except 2 is in the form $4n\pm1,$
$p_{1}p_{2}\cdots p_{n}+1=2\cdot\pm1+1=2+1=3\mod 4$
but a square cannot be $3\mod4$[/hide][/quote]

How do you know that it is 2+1 = 3 instead of 2-1 = 1?

(because $p_{1}p_{2}\cdots p_{n}+1=2\cdot\pm1+1=2+1=3\mod 4$ )[/quote]
Sorry, I texed it weird...I meant it to be like this:
$p_{1}p_{2}\cdots p_{n}+1=2\cdot1+1=2+1=3\mod 4$
or
$p_{1}p_{2}\cdots p_{n}+1=2\cdot(-1)+1=-2+1=-1=3\mod 4$[/quote]

Yeah that makes more sense, thanks.

[hide="1"]Since 2 is a prime,  $p_{1}p_{2}\cdots p_{n}$ is always 2mod4. You add 1 to get 3mod4, but a square can only be 0 or 1 mod 4.[/hide]

[hide="2"]$n^{2}+n^{2}+2n+1+n^{2}+4n+4=3n^{2}+6n+5$

Since that will be congruent to 2 mod 3, and squares can only be 0 or 1 mod 3, that equation can't be another square.[/hide]


calculus
integration
algebra unsolved
algebra

find all integeral solution pf the equation
        $[2x -y -2 )^2 =7 * ( x-2y-y^2-1)$


Evaluate
\[ \sum_{n=-\infty}^\infty{(x-n)}^{-k} \] where k>1 is an integer

Does anyone know how to do this? It's related to the Huruwitz Zeta function. Can it even be done by reasonably elementary methods?


Hi all,

Just wondering, is it possible to have multiple headers per page? Like for example if I had two different half-page tests with the same header, and wanted to print them on the same page to save paper. Then one header would be at the top of the page, and one would be halfway down the page.

I'm using the fancyhdr package

Try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=twoinone]twoinone[/url] or [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=twoupltx]twoupltx[/url] packages to see how well they work.

You know, you could just manually make a header by using the center function.

hmmmm... alright I'll check those things out

thanks for the replies everyone!