[quote="tridecalogism_lover"]help me on  this one....i just can't find the area of square ABCD... i know it's gotta be simple!  :blush:[/quote]
[hide]Since $\overline{AE}\cong\overline{BE}\cong\overline{BF}\cong\overline{CF}\cong\overline{CG}\cong\overline{DG}\cong\overline{DH}\cong\overline{AH}\cong$, the four triangles become 45-45-90 triangles in which the sides are $x, x, x\sqrt{2}$. Since $x\sqrt{2}=10$, solve for x:
$x\sqrt{2}=10$
$x=\frac{10}{\sqrt{2}}$
Rationalize (multiply by $\frac{\sqrt{2}}{\sqrt{2}}$): $x=\frac{10\sqrt{2}}{2}$
Finally, cancel: $x=5\sqrt{2}$
Double and then square to get $A=\boxed{200}$[/hide]
But ckck's way is probably easier... :P

[quote="bpms"][hide="a more intuitive method"]Draw both the diagonals of the smaller square. Notice that for each of the four triangles that make up the smaller square there is another triangle outside that is congruent to it. Clearly its area is half the area of the larger square. Doubling 100, we get 200.[/hide][/quote]

Wow that is a very interesting way of solving the problem....

i think bpms's way is clearly better cause its faster and your less likely to make mistakes.
those problems might be worth memorizing

[quote="tridecalogism_lover"]help me on  this one....i just can't find the area of square ABCD... i know it's gotta be simple!  :blush:[/quote]

[hide]
the inner square has half the area of the outer square. the inner square has $100 units^{2}$, so the outer square has $200 units^{2}$

why the inner square has half the area of the outer square:
if you drew a diagonal down the inner square, shifted the outer square right to meet the left edge with the diagonal of the inner square, and moved the left half of the inner square right, you will see that the outer square is divided into four equal pieces. two of which are the two halves of the inner square.
(in other words, you divide the inner square into two pieces and flip them so that theyre backwards to see that they are half of the full square)

[/hide]

[quote="jhredsox"][quote="ckck"][hide]Square EFGH has a side length of 10. Notice that the diaganol of the square is also the length of square ABCD. So the diaganol length is $10\sqrt{2}$. So squaring this will get the area of ABCD. So the answer is $\boxed{200}$.[/hide][/quote]
same here :)
jorian[/quote]
by saying "same here" doesn't help the person understand the problem better. post your solution next time.

ok, fine, i just didn't want to repost :P (and wanted to look like i knew what i was doing)

-jorian


calculus
integration
geometry
3D geometry
sphere
analytic geometry
limit

let [tex]B_n(r) = \{x \epsilon R^n| |x| \le r\} [/tex] be the sphere around the origin of radius r in [tex] R^n. [/tex]  let [tex] V_n(r) = \int_{B_n(r)} dV [/tex] be the volume of [tex]B_n(r)[/tex].

a)show that [tex] V_n(r) = r^n * V_n(1) [/tex]
b)write [tex] B_n(1) [/tex] as [tex] I*J(x) * B_{n-2}(x,y), [/tex]  where I is a fixed interval for the variable x, J an interval for y dependent on x, and [tex]B_{n-2}(x,y) [/tex] a ball in [tex] R^{n-2} [/tex] with a radius dependent on x and y.  this decomposition should allow for use of fubini's theorem in part c)

c)find [tex]V_n(1) [/tex] in terms of [tex] V_{n-2}(1)[/tex]

d)find [tex] V_n(1) [/tex] in terms of only n (eg. find a closed form for [tex] V_n(1) [/tex])

for a), i just used the fact that [tex] V_n(r) [/tex] is proportional to [tex]\pi r^n [/tex] . (not sure how this relationship is derived though - i just got that from the formula for 3 dimensions).

for b), is the answer
[tex] B_n(1) = \{I, J \epsilon R^n, B_{n-2}(x,y) \epsilon R^{n-2} | I \epsilon [0,1], J \epsilon [-\sqrt{1-x^2}}, \sqrt{1-x^2}], B_{n-2}(x,y) \epsilon [0,1]*[0,1]*[0,1]\} [/tex]
not sure about the last part...how do i show that radius of B_{n-2}(x,y) is dependent on x and y?

for c), i found the answer to be [tex] V_n(1) = V_{n-2}(1) * \int_{0}^{2 \pi} \int_{0}^{1} r dr d{\theta} [/tex] (using polar coordinates)

i'm stuck on d).  what does it mean by "closed form"

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=ball&t=26191]This topic[/url] has some ideas.

If you want a single formula for all $n,$ you can write it as

$V_n(r)=\frac{\pi^{\frac n2}}{\Gamma\left(\frac n2+1\right)}\cdot r^n$

Or you could split it into two formulas, for odd and even $n.$

Shocking fact: for any fixed radius, no matter how large,

$\lim_{n\to\infty}V_n(r)=0.$


On each tile in a set of dominoes, each integer 0 through 12 is paired exactly once with each integer 0 through 12. One number of the pair is written on one half of the domino, and the other number is written on the other half. If one domino is randomly selected from all the dominoes in the set, what is the probability that the same number will be on both halves? Express your answer as a common fraction.

There are $ 1\plus{}2\plus{}3\plus{}...\plus{}13$ $ \equal{}$ $ \frac{13(14)}{2}$ $ \equal{}$ $ 91$ dominoes in the set, with each integer being paired with another.
Of those, there are 13 integers from 0 to 12, so there are 13 dominoes with the same number on both halves.
Thus the probability is $ \frac{13}{91}$ $ \equal{}$ $ \boxed{\frac{1}{7}}$.

why is it 1+2+3+...14 is there a restriction in dominoes the way the numbers are combined?

I'm confused why do you add all the integers from 1 through 13 to get the total number of dominoes? Isn't it $13^2$? Because you have 13 choices for the first half and then thirteen choices for the second half, so you multiply them...

Remember that $13^2$ includes both dominos $(6,5)$ and $(5,6)$ (replace the numbers with anything else). Usually in domino problems order doesn't count, so the domino with a 5 and a 6 is only counted once.

The sum of the first 13 numbers counts

$(0,0), (0,1), (0,2), \ldots, (0,12)$ (13 terms)

$(1,1), (1,2), \ldots, (1,12)$ (12 terms)
.
.
.
$(11,12), (12,12)$ (2 terms)

$(12,12)$ (1 term)

Thus, the number of dominos is $13+12+\cdots+2+1= \frac{13\times14}{2} = 91$.

Hope this helped.


The digits 2, 3, 4, 7, and 8 form a five-digit number. The tens digit is odd and the number is divisible by 4. How many numbers are possible?

There are two odd numbers, $ 3$ and $ 7$, to choose from for the tens digit. If the number is to be divisible by $ 4$, therefore, the ones digit must be $ 2$. (Consult a set of divisibility rules if you don't know are are not clearly sure why.) Now the remaining three numbers can be put in $ 3!\equal{}6$ orders for the first three digits of this five-digit number. Thus, there are $ 2\times6\equal{}\fbox{12}$ such numbers.


function
search
real analysis
real analysis unsolved

Show also that for real valued $ f \not \in L^p(E)$, there exists a function $ g \in L^{p'}(E)$, $ 1/p \plus{} 1/{p'} \equal{} 1$, such that $ fg \not \in L^1(E)$.

Any idea?


There are twelve buttons in an elevator numbered from 1 to 12.  Seven buttons are pressed, and as a result, they light up.

Possible arrangement:

12
[color=orange]11[/color]
10
[color=orange]9[/color]
[color=orange]8[/color]
[color=orange]7[/color]
6
5
[color=orange]4[/color]
[color=orange]3[/color]
2
[color=orange]1[/color]

If no button is pressed more than once, prove that there is a pair of numbers lit that have a difference of 6.

[hide] Let's group the numbers into 6 sets $\{1,7\}$, $\{2,8\}$, $\{3,9\}$, etc. We must choose 7 of the numbers. By Pigeonhole, there is at least 1 set with 2 numbers chosen, and therefore, their difference is 6. [/hide]


Twenty tiles are numbered 1 through 20 and are placed into box A. Twenty other tiles numbered 11 through 30 are placed into box B. One tile is randomly drawn from each box. What is the probability that the tile from box A is less than 15 and the tile from box B is even or greater than 25? Express your answer as a common fraction.

The probability for the first condition is $ \frac {15 \minus{} 1}{20} \equal{} \frac7{10}$, and $ \frac {\left(1 \plus{} \frac {26 \minus{} 12}2\right) \plus{} (30 \minus{} 27 \plus{} 1)}{30 \minus{} 11 \plus{} 1} \equal{} \frac {12}{20}\equal{}\frac35$ for the second condition, so it's

$ \frac7{10}\cdot\frac35 \equal{} \boxed{\frac {21}{50}}$ for both.

[b]Oh I get it, I thought that it meant 'Even to or greater than 25' i.e.25,26,27,28...
I understand now.
 :idea: [/b]

Agreed with above.  Wording is slightly misleading if doing it quickly.

[quote=ThriftyPiano][quote=OlympusHero]how is this level 18?

amount of ways to get a tile less than $15$ from box $A$ is $\frac{14}{20}$. amount of ways to get a tile that is either even or greater than $25$ from box $B$ is $\frac{7+5}{20}=\frac{12}{20}$.

multiplying these gives $\frac{14}{20} \cdot \frac{12}{20} = \boxed{\frac{21}{50}}$.[/quote]

Good job on finding a 2 digit one!
Only 56![/quote]

only 56 what

[quote=ThriftyPiano][quote=ilovepizza2020][quote=awesomeming327.]problem number[/quote]

o nice I've seen 26[/quote]

Wait WHAT?[/quote]

Yeah it has been here and there for the past 8 months...


Given $m\geq u$,$m+n\geq u+v$,$w=m+n-u-v$,$p=o$$m\geq n\geq p$,$u\geq v= w$
$a,b,c\geq 0$
prove that:
$\sum_{cyc}a^mb^n\geq \sum_{cyc}a^ub^vc^w$
anyone have a nice solution ?
I think this Inequality can be generalized.but I don't have a nice form of the constraint of $m,n,p,u,v,w$.
does anyone have the interest?

Allow me to assk a stupid question:
What do you mean by $\sum_{cyc}a^m b^n$?

I feel really sorry about my mistakes of type.
the condition of the Inequality should be $u\geq v=w$ but not $u\geq v\geq w$.
the original Inequality I want to prove is with the condition $u\geq v\geq w$
but I found I can prove it .then I changed the constrait to $v=w$ ,and managed to prove the easier one.
But when I wrote the topic,I forgot I had  changed the condition of the Inquality,so I wrote the wrong one.
Forgive me.
Now the Ineq come easier ,right?
I post my solution here :
let$x=\frac{(v-m)n+un}{m^{2}-mn+n^{2}}$
$y=\frac{(m-v)m-(m-n)u}{m^{2}-mn+n^{2}}$
$z=\frac{vn+vm}{m^{2}-mn+n^{2}}$
notice that $x+y+z=1$
we can check that $x>0,y>0,z>0$
then we have $\sum a^{m}b^{n}=\sum (xa^{m}b^{n}+yb^{m}c^{n}+zc^{m}a^{n})$
by holder Inequality we have :
$\sum{xa^{m}b^{n}+yb^{m}c^{n}+zc^{m}a^{n}}\geq \sum a^{xm}b^{xn}b^{ym}c^{yn}c^{xm}a^{xn}$
we can check that
x,y,z,be the solution of the system of the equation:
$xm+yn=u$
$xn+zm=v$
$ym+zn=v$
so $xa^{m}b^{n}+yb^{m}c^{n}+zc^{m}a^{n}\geq a^{xm}b^{xn}+b^{ym}c^{yn}+c^{xm}a^{xn}=\sum a^{u}b^{v}c^{w}$


find the least $ M$ such that following inequality holds for positive reals $ a,b,c$
  \[ |(a-b)(b-c)|^{3}+|(b-c)(c-a)|^{3}+|(c-a)(a-b)|^{3}\le M(a^{3}+b^{3}+c^{3})^{2}\]
i have no solution and no comments about this inequality .

[quote="anuj kumar"]find the least $ M$ such that following inequality holds for positive reals $ a,b,c$
\[ |(a \minus{} b)(b \minus{} c)|^{3} \plus{} |(b \minus{} c)(c \minus{} a)|^{3} \plus{} |(c \minus{} a)(a \minus{} b)|^{3}\le M(a^{3} \plus{} b^{3} \plus{} c^{3})^{2}
\]
i have no solution and no comments about this inequality .[/quote]
If $ b \equal{} c\rightarrow0^ \plus{}$ then $ M\geq1.$
Let $ a\leq b\leq c$ and $ b \equal{} a \plus{} v,$ $ c \equal{} a \plus{} u.$
Hence, $ u\geq v$ and for $ M \equal{} 1$ your inequality equivalent to the following:
$ \left(3a^3 \plus{} 3(u \plus{} v)a^2 \plus{} 3(u^2 \plus{} v^2)a \plus{} u^3 \plus{} v^3\right)^2\geq u^3v^3 \plus{} (u^3 \plus{} v^3)(u \minus{} v)^3.$
Let $ u \equal{} tv.$ Then it remains to prove that for $ t\geq1$ holds:
$ \left(t^3 \plus{} 1\right)^2\geq t^3 \plus{} (t \minus{} 1)^3(t^3 \plus{}1),$ which obviously true.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ G$ be a finite group and $ \mathbb{Z}[G] \equal{} \{\sum \limits _{g \in G}n_gg, n_g \in \mathbb{Z} \}$,

 let $ A$ be an abelian group, $ B \equal{} Hom_{\mathbb{Z}}(\mathbb{Z}[G],A)$ and $ C \equal{} \mathbb{Z}[G] \otimes_{\mathbb{Z}}A$.

Show that the morphism $ \psi : B \rightarrow C$ with $ f \mapsto \sum \limits_{g \in G}g^{ \minus{} 1} \otimes_{\mathbb{Z}}f(g)$ is an isomorphism. Do you please help me.

You need to construct an inverse map $ \theta: C\to B$ sending each $ g\otimes a$ to the map $ g^{\minus{}1} \to a$.  To do this use the universal property of tensor products.


An unsolved BMO question:

Determine the least possible value of the largest term in an arithematic progression of 7 distince primes. :wink:

EDIT
The difference must be even, clearly...
So.. If the difference ends in 2, 4, 6 or 8
I think there must be a number ended with 5 in one of the 7 number.. And that number won't be a prime unless it's 5..

Otherwise, the difference must ends in 0... And this is the only possibility...
I can't find one ex. Anyone ??  :|

Please tell me the reason. Incidentally, it should be 'distinct primes'.

His reasoning is faulty. No reason for any of the numbers to end in $5$.

Well, by a mod 3 argument, the difference is also a multiple of 3, then by a mod 5 argument and mod 7 argument, also of 5 and 7.  So the common difference between the terms is a multiple of 210...
To be clearer...
If the common difference between the terms were not a multiple of 2, then the terms would alternate even, odd, even, odd.
If the common difference were not a multiple of 3, then because 3 is prime and has no zero divisors, we would have a multiple of 3.
If the common difference were not a multiple of 5, we would have a multiple of 5
and so on for 7.
Don't know where to go from there.

Come on, who can solve it? Is it that hard?

I think, the answer is: 907

K81o7 is right that difference must be divisible by 2,3,5 but it's not true for 7!! More precisely - it's true unless the first term of the sequence equals 7.

If not and difference is $\geq 210$ then seventh term is $>6\cdot210 = 1260 > 907$.

If the first term is 7, then the difference is of the form $30k$, where $k$ is natural.  We can't have $k\in\{1,2,3\}$ because then $187=11\cdot17$ would belong to our sequence ($187=7+6\cdot30$). If $k=4$ than $247=13\cdot19$ would be one of the sequence's terms. So $k\geq5$. And we can check that indeed it is the solution, as all of the numbers 7, 157, 307, 457, 607, 757, 907 are primes.


Sorry if the arithmetic is off, but i've only got 2 minutes, and I'm doing this in my head:
[hide] Ok, so draw the perpendicular from the smaller radius to the larger radius, so that the larger is divided into segments of 1 and 2. Then, draw the segment containing the centers. There's a right triangle, with sides $5,1,\sqrt{24}$. But $\sqrt{24}$ is the same length as the external tangent, and the quadrilateral is a right trapezoid with bases $3$ and $2$, with height $\sqrt{24}$, thus the area is $5\sqrt{6}$.[/hide]

[quote="i_like_pie"]Circle $O$ with radius $3$ is externally tangent to circle $P$ with radius $2$. Their common external tangent is tangent to circle $O$ at $A$ and to circle $P$ at $B$. Find the area of quadrilateral $ABPO$.[/quote]
[hide]
the vertex at A and at B are right angles
this is a trapezoid, with 1 paralel set of sides (OA and PB)
OA is 3, and PB is 2.
the height would be AB, which is $2\sqrt{6}$ (there are various ways of figuring this out)
so it is $\frac{1}{2}(3+2)(2\sqrt{6})$
simplified, it is $5\sqrt{6}$
[/hide]


By placing a "1" at either end of a number, the value is increased by 14,789. What was the original number?

let x be the number when 1 is added to the end of that number that number is multiplied by 10 and added to 1
10x+1-x=14789
9x+1=14789
that's impossible!?

when a number adds a one in front of it usually increases by a power of ten for 23 123 adds 100.

ok lets see....

let us assume that x is a three digit number (a reasonable assumption)

$10x + 10,000 + 1 - x = 14789$
$9x + 10,001 = 14789$
$9x = 4788$
$\boxed{x = 532}$

checking:

$15321-532 = 14789$

And it's good! :jump:

oh i thought you meant that
1532 is 14879 more than 532 and so is 5321

either should be changed to both ends of the number

has to be 3 digit number otherwise the increase will be too big or too small if fewer digits.  call the number x = 100a + 10b + c.  Make an equation based on the given information and its simple from there, just move stuff around and you'll see that 100a + 10b + c = 532, so the number is x = 532.

Oh, at either end of the number. My first thought was that it wasn't possible because you had to add it to either side.  :oops:


[quote="hjbrownies"]There is not too many Koreans here... (I shall type in english, cuz this is faster for me)

But like.. I thought this place would be FILLED with Koreans...

 XD Where is everyone hiding?? maybe too busy studing??[/quote]

Hi, Koreans friends,
I am here to have a look. :)

I'm a Korean. I don't come to this part of the forum because it's always empty(no new post). :(

Y don't many Korean people come to the Korean forum when there are quite a few people in the Chineese forum?! Wat's with this. I mean shouldn't  people who want to be better at math have some skill increase in forums?


Does anyone know when is the AMC12 B?

According to the website, it is "Wednesday, February 26, 2003" (next to the words "Registration Form B"). Could someone please clarify this for me?

[quote="Fierytycoon"]25th, i believe[/quote]

its the 25th

[quote="mathnovice"]Does anyone know when is the AMC12 B?[/quote]

Wednesday 25 February 2004. The old information on the Web site was also a Wednesday, but last year.

My son takes AMC 10B that day.

Is it too late to register? (I assume it is)

The 'official' registration date is this coming Sunday:

[url]http://www.unl.edu/amc/whatswhen.html[/url]

If you want to take it at another school the registration part doesn't matter, right?  You just have to ask them if you can come on the testdate?

I have another question about taking AMC tests at another school:  If one were to qualify for AIME taking the AMC-12 at a different school, would that person be counted at the school he/she attends or the one at which the test is taken?

your school.  On the visiting student registration form, it says something like 'if you qualify for the AIME, where would you like us to send the materials to?'

Can i ask at what school you're planning to take it?

I actually only asked because someone from my school took the AMC 12-A at a different school and got a better score than I did (even though he failed out of my math class). I'm not actually planning to take the AMC 12-B.

interesting...  but why did he take it at another school if yours offered it?

He moved to my school after the deadline to sign up at the school and therefore had to take it at a different school, although not his old school.

where do you get the list of schools that administer amc???

[quote="**********"]it's never too late to register (unless its, like, the day of the test or something)  just call up the people at AMC...they allow late registrations, u have to pay more though...but you still have 2 weeks, plenty of time...[/quote]

Even after the "official" deadline?

yea, i signed up as a visitor for the AMC 12A exactly 4 days before the day of the test...this one other person signed up like 2 or 3 days before, i think...

all the AMC needs is like 2 or 3 days to Fedex the test materials over to the test center

Can anyone post that e-mail address you write to in order to find out which school in your area administer the B tests again?


MATHCOUNTS probability geometry rectangle ratio inequalities

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}

are the answers shown anywhere for these problems?  :?:

*shrugs*
but anyways... I don't feel like doing anymore math today... Mathcounts...

1.
[hide]
The total area is 4^2pi=16pi.
The possible area is 3^2pi-2^2pi=5pi.
5/pi/16pi= 5/16.
[/hide]

10.

[hide]
Solve and get x is less than/equal to 6, x is greater than/equal to 2.5.
The difference is 2.5 to 6, so 3 1/2.
[/hide]

5.
[hide]
She is buying 3/4 dozen of roses, 1/2 dozen of daisies, and 3/2 dozen of chrysanthemums.   (20*3/4)+(5*1/2)+(10*3/2)=32.5 dollars.

[/hide]

2.
[hide]
Notice that the height of the cylinder is 6 lengths of a ball, and the radius is equal to the radius of the ball.
Thus, the cylinder volume is 4^2*24=384pi.
The 3 tennis ball areas are 3*(4/3*64)=216pi.
384pi-216pi=168pi.[/hide]

lol this thread is in MO forum too

why would you come to NJ forum and practice mathcounts :rotfl:

I think it's because there are mods in MO forum now. :o


Let N={0,1,...}.
Determine all functions $ {f(N)}=N$ such that $ {x.f(y)+y.f(x)=(x+y).f(x^2+y^2)}$.

Uh sorry but what are all the dots for in

$ {x.f(y)+y.f(x)=(x+y).f(x^{2}+y^{2})}$?

And this is not High School Basics. THis should be [b]at least[/b] Intermediate Topic.

[quote="darkdieuguerre"]Those denote multiplication.[/quote]

Oh allright, that's what I thought but I wasn't sure since it wasn't $ {x\cdot f(y)+y\cdot f(x)=(x+y) \cdot f(x^{2}+y^{2})}$

[hide="Hint"]Let $ y\equal{}0$ while $ x$ is some nonzero positive integer.[/hide]

Let $ y\equal{}0$. Our equation becomes

$ xf(0) \equal{} x f(x^2)$

$ f(0) \equal{} f(x^2)$

We divided by $ x$ so $ x$ is nonzero.

But this restricts its domain to perfect squares. So I don't really know what happens now.. lol. Help?

[quote="modularmarc101"]Let $ y \equal{} 0$. Our equation becomes

$ xf(0) \equal{} x f(x^2)$

$ f(0) \equal{} f(x^2)$

We divided by $ x$ so $ x$ is nonzero.

But this restricts its domain to perfect squares. So I don't really know what happens now.. lol. Help?[/quote]So what happens when you plug in [u]any[/u] positive integer value for $ x$? It still equals $ f(0)$. But, $ f(0)$ is a constant *idea idea* :)

Just a note: The notation written in the problem in post #1 should be $ f: \mathbb{N}\rightarrow\mathbb{N}$ instead of $ f(N)\equal{}N$.

[quote="modularmarc101"]But just because we plug in a number for $ x$ doesn't mean it is in it's domain (in this case). Notice that $ 2,3,5,6,7...$ are not in the domain.[/quote]Ooooh, yea... $ x\equal{}y\equal{}1\Rightarrow f(1)\equal{}f(2)\equal{}f(0)$, $ x\equal{}1,y\equal{}2\Rightarrow 2f(1)\plus{}f(2)\equal{}3f(5)$, $ x\equal{}1,y\equal{}7\Rightarrow 7f(1)\plus{}f(7)\equal{}8f(50)$ and $ x\equal{}5,y\equal{}5\Rightarrow 5f(5)\plus{}f(5)\equal{}10f(50)$  :what?: . Is there a way to find the other integers? This is getting too hard for High School Basics and hope our moderators DO check this :P (I suggest Pre-Olympiad).

Only the most basic proofs are allowed in HSB, so this is going to be moved to Pre-Olympiad.  I highly doubt anything even close to this difficult would arise in a typical high school classroom.

Easy problem ^^

Let $ y \equal{} 0 \Rightarrow f(x^{2}) \equal{} f(0)$ this means that for all $ c$ a positifs integers wa have $ f(c) \equal{} f(0)$
So $ f(x^{2} \plus{} y^{2}) \equal{} f(0)$ $ \Rightarrow$  $ xf(y) \plus{} yf(x) \equal{} f(0)$

And $ f(x^{2}) \equal{} f(0)$ implies $ f(1) \equal{} f(0)$ , and now , let $ y \equal{} 1$ , $ x \equal{} 0$ $ \Rightarrow$  $ xf(1) \plus{} f(x) \equal{} xf(0) \plus{} f(0)$

and since we have $ f(0) \equal{} f(1)$ so : $ f(x) \equal{} f(0) \equal{} b$ where $ b$ is a constant

reciprocally the constants fonctions are a solutions.

[quote="R.Maths"]

Let $ y \equal{} 0 \Rightarrow f(x^{2}) \equal{} f(0)$ this means that for all $ c$ a positifs integers wa have $ f(c) \equal{} f(0)$
[/quote]

Not really. Notice that it said $ f: N \rightarrow N$. So it doesn't work for [b]all[/b] positive integers, just perfect squares. That makes the rest of the solution false.

This problem is also solvable for the set $ \mathbb{N} \equal{} \{1,2,3,\ldots\}$ (i.e., no 0).
[hide="Hint"]This problem doesn't use any number theoretic things about $ x^2\plus{}y^2$.
[hide="Bigger Hint"]In fact, you could put many different expressions in place of $ x^2\plus{}y^2$ for an equivalent problem...[/hide][/hide]

lol , sorry , yes you're all right modularmarc101 ,but i think that the rest is true ... :)

@R.Maths, no the rest doesn't hold true.

[hide="less vague hint"] Replace $ x^2 \plus{} y^2$ with any function $ g(x,y)$ such that $ g(x,y) > \text{max}(x,y)$ (for $ x,y \neq 0$). If $ f(x) < f(y)$, what can you say about $ f(g(x,y))$?  [/hide]

sorry , i wasn't make attention for the topic ,
and what is about did you say azjps , you're al right , just you can delet my mistakes ;)  and the rest will be abvious


Consider a 100 by 100 square grid of dots, consisting of the lattice points in the square going from (0,0) to (99,99).

Each point that is not (0,0) or (99,99) is marked with an integer from 0 to 100. A robot starts on the point (0,0), and walks up, down, left, or right on the grid in order to get to (99,99).

This robot, as it passes a point, keeps a running total of all the numbers it has passed, and is trying to find an optimal path.

My question is, what is the optimal path with the greatest total weight you can have? (For example, for a grid where every square is marked 100, the optimal weight is always 19700.)

this questions is sort of vague.. what is an optimal path?
by optimal path.. you mean a path of length 197?

If you define optimal as greatest weight, then as long as there exists one point with a positive integer value, then you can have an infinitely high value using cycles.

Therefore, I will assume that optimal is equivalent to least weight.

If you want a shortest path as in number of edges traveled in conjunction with smallest weight, make all edges point either "right" or "up", and then DP is actually easier to implement than Dijkstra's.

If not, then simply construct the regular graph.

I hope I'm understanding this correctly.


Find the smallest positive integer $n$ for which the expansion of $(xy - 3x +7y - 21)^n,$ after like terms have been collected, has at least 1996 terms.

[hide="Solution"]Rewrite as $((x+7)(y-3))^n$.
The expansion will have $(n+1)^2$ terms for a specific $n$, since each monomial contributes $n+1$ terms. So we must meow find the smallest $n$ such that
$(n+1)^2 \ge 1996$
So $n=\boxed{44}$.[/hide]

Unless the problem was typed incorrectly, you can't rewrite it like that, because the final term is negative, not positive.

Hmmm.... I think it should be $+21$, since otherwise it is impossible to factor.

The problem was typed incorrectly.  It should be \[ (xy-3x+7y-21)^n. \]

Sorry guys. I got this from Kalva and tha't the way it was written there. No wonder I was so frustrated about this problem!

Clicked wrong link and got this, but hey I have no problem with solving it :)

[hide=Solution]Rewrite as $(x+7)^n(y-3)^n$. From the Binomial Theorem, each of these has $n+1$ terms, so we must have $(n+1)^2 \geq 1996 \implies \boxed{44}$.[/hide]


function algebra domain Columbia polynomial logarithms AMC

I was looking over some old tests in AMC/AIME/and other national contests,and I saw how often function appeared. In AoPS book (at least in volume I), it looks easy but I know how hard function can get. So I like to set up similar to what towersfreak2006 is doing, except in AMC forum.

You can add problems or answer the ones that are posted.

1. (NCIML) Let $f$ be a function with domain $x>1$, such that $f(x) + 2f \left( \frac{x+2002}{x-1}\right) = 4009-x$. Find $f(2004)$.

2. (Canada Open Math Challenge 2004) A function $f(x)$ has the following properties:

i) $f(1) = 1$
ii) $f(2x) = 4 f(x) + 6$
iii) $f(x+2) = f(x) + 12x + 12$

Compute $f(6)$.

3. (Canada Fermat 1997) If $f(x) = px+q$ and $f(f(f(x))) = 8x+21$, and if $p$ and $q$ are real numbers, then $p+q$ equals what?

4. (Canada Fermat 2002) The function $f(x)$ has the property that $f(x+y) = f(x) + f(y) + 2xy$, for all positive integers $x$ and $y$. If $f(1) = 4$, then the numerical value of $f(8)$ is what?

5. (British Columbia Colleges Senior Round) Let $f_0 (x) = \frac{1}{1-x}$ and $f_{n+1} (x) = f_0 (f_n (x))$. Then compute the value of $f_{100} (3)$.

I'll post more soon. Anyone can add some problems.  :D  :)

looking over these. #2 is very simple.
[hide]
$f(2)=10,f(4)=46, f(6)=106$[/hide]

#1 is quite easy:

[hide]Let $x = 2$.  Then

$f(2) + 2f(2004) = 4007$

Now let $x = 2004$.  Then

$f(2004) + 2f(2) = 2005$

Subtracting the second equation from twice the first equation gives

$3 f(2004) = 4007 \cdot 2 - 2005 = 6009$

Therefore $f(2004) = 2003$[/hide]

[quote="zanttrang"][quote="amirhtlusa"][hide="4"]
$f(f(f(x)))=p^3x+p^2q+pq+q\Rightarrow p=2,q=3$[/hide][/quote]

That's #3.[/quote]
 :blush:  fixed

6. (University of South Carolina 1987) A certain function $f$ satisfies $f(x) + 2f(6-x) = x$ for all real numbers $x$. Compute the value of $f(1)$.

7. (University of South Carolina 1987) If $f(x)$ is a function that satisfies $f(2x+1) = 2f(x) + 1$ for all $x$, and if $f(0) = 2$, then $f(3)$ equals what?

8. (University of South Carolina 1990) For all real numbers $x$, the function $f(x)$ satisfies

\[2f(x) + f(1-x) = x^2\].

Find $f(5)$.

9. (University of South Carolina 1996) Given that $f(x) = (x^5-1)(x^3+1), g(x) = (x^2-1)(x^2-x+1)$, and $h(x)$ is a polynomial such that $f(x) = g(x) h(x)$, what is the value of $h(1)$?

10. (University of South Carolina 1996) Suppose that $f(n) = \log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdots \log_{n-1} n$. Then compute the value of $\sum^{10}_{k=2} f(2^k)$.

Hope you enjoy them!  :lol:

#6:

[hide]Note that

$f(1) + 2f(5) = 1$

$f(5) + 2f(1) = 5$

Subtracting twice the second equation from the first gives

$-3f(1) = -9$

$f(1) = 3$[/hide]

#7:

[hide]$f(1) = 2f(0) + 1 = 5$

$f(3) = f(2 \cdot 1 + 1) = 2f(1) + 1 = 11$[/hide]

#8:

[hide]$2f(5) + f(-4) = 25$

$2f(-4) + f(5) = 16$

Subtracting twice the first equation from the second gives

$-3f(5) = -34$

$f(5) = \frac{34}{3}$[/hide]

#9:

[hide]Note that $f(x)$ has a real zero at $x = 1$ from the first factor and $x = -1$ from the second factor.  Factoring these terms out gives

$f(x) = (x - 1)(x^4 + x^3 + x^2 + x + 1)(x + 1)(x^2 - x + 1)$

Rearranging terms gives

$f(x) = (x^2 - 1)(x^2 - x + 1)(x^4 + x^3 + x^2 + x + 1)$

Noting that $f(x) = g(x)(x^4 + x^3 + x^2 + x + 1)$, we find that

$h(x) = x^4 + x^3 + x^2 + x + 1$

Plugging in x = 1, we find

$h(1) = 5$[/hide]

[hide="10"]
notice that $f(x)$ could be simplified to $\frac{log_n}{log_2}$ because we can put every logs into fractions then cancel out. so $\displaystyle \sum^{10}_{k=2} f(2^k)$ would be  same as $\displaystyle \sum^{10}_{k=2} k$, which is $55-1 = 54$
[/hide]

10 is more a log problem than a function problem, but whatever.
Here's a really simple function/polynomial problem I wrote:
If $f(x)=x^2-\frac{23}{6}x+\frac 72$, prove that the roots of f(x) are also roots of $f^{2005}(x)$.
Note that by $f^{2005}(x)$ I mean f(f(f(f...f(x)))) with 2005 f's, NOT (f(x))^2005.

[quote="LynnelleYe"]
Here's a really simple function/polynomial problem I wrote:
If $f(x)=x^2-\frac{23}{6}x+\frac 72$, prove that the roots of f(x) are also roots of $f^{2005}(x)$.
Note that by $f^{2005}(x)$ I mean f(f(f(f...f(x)))) with 2005 f's, NOT (f(x))^2005.[/quote]

Here's the solution:

[hide]
Let $x_i$ be an arbitrary root of $f(x)$.  Now list the first several $ f^{(n)}(x_i)$.

$f^1(x_i) = 0$

$f^2(x_i) = f^1(0) = \frac{7}{2}$

$f^3(x_i) = f^1\left(\frac{7}{2}\right) = \frac{7}{3}$

$f^4(x_i) = f^1\left(\frac{7}{3}\right) = 0$

Therefore, we know that for any integer $n \geq -1$,

$f^{4+3n}(x_i) = 0$

As $4 + 3 \cdot 667 = 2005$, we have proven that

$f^{2005}(x_i) = 0$

[/hide]

11. (MA$\theta$ National Convention 2004) If $f(x) = \frac{x+1}{2}$, find $f^{-1} (4)$.

12. (MA$\theta$ National Convention 2004) Given that $f(x-1) = 2x^2+3x-10$, find $f(5) - f(3)$.

13. (MA$\theta$ National Convention 2004) Find $u(w(v(-1)))$ given that $u(x) = 2x-1, v(x) = 2x^2, w(x) = x^{-2}$.

14. (MA$\theta$ National Convention 2004) It is known that $F(x) = 2x^2-x+5$ and $G(x) = -x^2+15x+1$. Furthermore, $H(x)$ is defined as:

$F(x) - G(x)$ for $x>0$
$G(x) - F(x)$ for $x<0$
Max of $G(x)$ or $F(x)$ for $x=0$

What is the maximum value of $H(x)$ on the interval $(-\infty, 5)$?

15. (MA$\theta$ National Convention 2003) Let $f(x) = 3x+2$ and $g(x) = 2x^2-3$. Evaluate $(g+f)(-3)$.

Above problems aren't really type of problems you'll see on AIME or possibly AMC but they are going to be good review (especially if you can't remember how to do inverse) and I think I'll post the problems about 5 more and open new thread for new topic. I like functions but we have to prepare for other type as well. :)

I'm surprised #11 was on a math contest; it seems far too easy.  Oh well.

#11:

[hide]$2f(x) - 1 = x$

$f^{-1}(x) = 2x - 1$

$f^{-1}(4) = 7$[/hide]

#12:

[hide]$f(6 - 1) = f(5) = 2 \cdot 6^2 + 3 \cdot 6 - 10 = 80$

$f(4 - 1) = f(3) = 2 \cdot 4^2 + 3 \cdot 4 - 10 = 34$

$f(5) - f(3) = 80 - 34 = 46$[/hide]

#13:[hide]

$v(-1) = 2 \rightarrow u(2) = \frac{1}{4} \rightarrow u\left(\frac{1}{4}\right) = -\frac{1}{2}$[/hide]

#14:

[hide]Turn H(x) into a split function:

$H(x) = 3\left(x-\frac{8}{3}\right)^2 - \frac{52}{3}, x > 0$

$H(x) = 5, x = 0$

$H(x) = -3\left(x-\frac{8}{3}\right)^2 + \frac{52}{3}, x < 0$

The minimum value of $H(x)$ for $x > 0$ is on the interval $(0, 5)$, so test the limits of the endpoints to get

$\lim_{x \rightarrow 0^+}H(x) = 4$

$\lim_{x \rightarrow 5} H(x) = -1$

For $x < 0$ the maximum value of $H(x)$ is

$\lim_{ x \rightarrow 0^-}H(x) = -4$

Therefore, the maximum value of $H(x)$ on the interval $(- \infty, 5)$ is $H(0) = 5$[/hide]

#15:

[hide]Assuming I remember my notation, $(f + g)(x) = f(x) + g(x)$.  Then

$(f+g)(-3) = f(-3) + g(-3) = 8$[/hide]

[quote="zanttrang"]I'm surprised #11 was on a math contest; it seems far too easy.  Oh well. [/quote]

it comes from $MA\theta$ MAO problems are multiple-choice, which makes it all the more easy.

#14 certainly do not need caculus to find the answer.

[hide="Hint"]
What does $F(x) - G(x)$ and $G(x)-F(x)$ look like? And remember, their $x$ are exactly opposite (positive and negative). [/hide]

#12 I solved liked you did, without finding the function. Can anyone find the function? (I know how to do it but I'm just posting it as challenge problem)

To find the function for 12, just let $x=y+1$.  Then we have $f(y)=2(y+1)^2+3(y+1)$

Oh and for 11, there's another method to find the inverse.  I call it the socks and shoes method because when you get dressed, you put on your socks and then your shoes, but when you get undressed, you take off your shoes and then your socks.

so if $f(x)=\frac{x+1}2$, say "f(x) adds one, then divides by 2.  Therefore, the inverse multiples by 2 and subtracts 1, i.e. it is $2x-1$" There's nothing to write down!

Here are last five problems.. on function.. :)

16. (OCTM 2000) The graph of the cubic function $f(x) = x^3+ax^2+bx+c$ crosses x-axis at the points (-1,0), (2,0), and (-3,0). Find the value of $1+a+b+c$.

17. (Thunder Bay High School Math Competition 2002 Senior) The function $f(x)$ has the property that $f(x+y) = f(x) + f(y) + xy$, for all positive integers $x$ and $y$. If $f(1) = 1$, then what is the numerical value of $f(7)$?

18. (Texas A&M University Math Contest CD Exam 2004) For $f(x)$, $f(1) = 4, f(2) = 1, f(3) = 3, f(4) = 5, f(5) = 2$. If $u_0 = 4$ and $u_{n+1} = f(u_n)$ for $n \geq 0$, what is the value of $u_{1234}$?

19. (University of Vermont High School Math Contest 2004) A function $f$ satisfies $f(x+y) = 4f(x)f(y)$ for all real numbers $x$ and $y$. If $f(3) = 32$, find $f(1)$.

20. (University of Tennesse Fermat I 2003) Let the function $f(x)$ be defined recursively by the fact that $f(x+1) = f(x)+3$ if $f(x) \leq 100$, and $f(x+1) = f(x)-2$ if $f(x) > 100$. If $f(1) = 50$, what is the function value at $x=205$?

That's it.. I hope you enjoyed them.

I think during this 20 questioned function lesson, I learn that the best way to solve function problems is two ways: either plug in the values and see what you get or use algebra to see if anything repeats.

[hide="16"]We're basically told that -1, 2, and -3 are the roots of the function, so it can be written as (x+1)(x-2)(x+3). 1+a+b+c is sum of the coefficients, so it's the function evaluated at x=1, or 2*(-1)*4=-8. Once I saw a similar problem on an AMC-12 and decided that the sum of the coefficients is just a+b+c. Won't make that mistake again haha...[/hide]
[hide="17"]Substituting x=1, y=1, we find f(2)=3. Substituting x=2, y=2, we find f(4)=10. Substituting x=1, y=2, we find f(3)=6. Finally, substituting x=4, y=3, we find f(7)=28. I did that pretty much in my head so it's probably all wrong.[/hide]
[hide="18"]$u_0=4$
$u_1=5$
$u_2=2$
$u_3=1$
$u_4=4$
And it repeats 5, 1, 2, 4 forever. 1234 is 2 mod 4, and therefore is equal to u_2, or 2.[/hide]
[hide="19"]We have f(1+2)=4f(1)f(2)=32, so f(1)f(2)=8. We also have f(2)=f(1+1)=4f(1)f(1). So $4(f(1))^3=8$, or $f(1)=\sqrt[3]{2}$.[/hide]
[hide="20"]f(1)=50
f(2)=53
.
.
.
We continue adding 3 until we get to f(18)=101 (since 50+3*17=101 and we start at 50+3*0). Then f(19)=99, f(20)=102, etc. The function applied to the even numbers begins incrementing by 1 while the function applied to the odd numbers is simply 3 less than the function of the next even number. f(206) would then be 101+188=289 since it's the 188th even number after 18, and f(205) is 289-3=286.[/hide]

who can post some basic functional equations here?

Here is neat one (very similar to one on AMC Practices. Actually almost twin) from British Math Olympiad.

A function $f$ is defined over the set of all positive integers and satisfies:

\[f(1) = 1996\]

and

\[f(1) + f(2) + f(3) + \cdots f(n) = n^2 f(n)\]

for all $n>1$.

Calculate the exact value of $f(1996)$.

I believe it's

[hide]$f(1996) = \frac{2}{1997}$

I won't go through the details of my derivation (I will if somebody asks me when I get up tomorrow, but it's too late tonight and it would take a long time to type out), but I can prove this result by induction.

I assert that

$f(n) = \frac{2}{n(n + 1)}f(1)$

for $n \geq 1$.  Note that this formula holds for $n=1$, as

$f(1) = \frac{2}{2}f(1) = f(1)$

Now assume that

$f(n) = \frac{2}{n(n + 1)}f(1)$

This must imply that

$f(n+1) = \frac{2}{(n+1)(n+2)}f(1)$

We know by the formula given in the problem that

$f(n + 1) = \frac{f(1) + ... + f(n)}{(n+1)^2 - 1}$

$= \frac{f(1) + ... + f(n)}{n(n + 2)}$

So we must have

$ \frac{f(1) + ... + f(n)}{n(n + 2)} = \frac{2}{(n+1)(n+2)}f(1)$

$ f(1) + ... + f(n) = \frac{2n}{n+1}f(1)$

However, we know that

$f(1) + ... + f(n) = n^2 f(n)$

Therefore we must have

$n^2 f(n) = \frac{2n}{n+1}f(1)$

$f(n) = \frac{2}{n(n+1)}f(1)$

which is our original assumption.  Therefore, we have shown that

$f(n) = \frac{2}{n(n+1)} f(1)$

implies

$f(n + 1) = \frac{2}{(n+1)(n+2)}f(1)$

and our assertion is proven.

Now simply let $n = 1996$, and we have

$f(1996) = \frac{2}{1996 \cdot 1997} \cdot 1996$

$f(1996) = \frac{2}{1997}$

$QED$

Edit: I realized that given my assertion, you can plug it into the original formula and verify that the sum on the left is equal to the right-hand side, but induction was what occured to me first (and I'm trying to practice my induction skills
:lol: )[/hide]


Euler
function
algebra
functional equation
number theory unsolved
number theory

Hello:

I have been reading the book "Prime Obsession". I find it very interesting and informative. I did not want something at the graduate level in complex analysis. I do not understand how zeta(0)=-1/2. Can anyone explain that without too much trouble?. It can not be plugged into the zeta(1-s) equation because zeta(1) is undefined. I found a formula where the Kronecker delta is used but I was wondering if there is another way.

Thanks,
Cody

you can use the functional equation, since you can just compute the coefficient of the pole of both sides - they have to be equal since it's an equality of functions.

Thank you.  I used:  zeta(s)=((1-2^(1-s))^-1)*(1/2+1/2*sum((-1)^(n-1)*(n^(-s)-(n+1)^(-s))),n=1 to infinity).
When 0 is subbed in we get -1/2. This is a variation on the functional equation theme from the Euler transformation and some analytic continuation.

Cody

just a readable version of the equality before:

$\zeta(s)=\frac 12 \frac{1}{1-2^{1-s}}\left(1+\sum_{n\geq 1} \left( (-1)^{n-1}(\frac 1{n^s}-\frac 1{(n+1)^s})\right)\right)$.

More generally, $\zeta (1-n) = - \frac{B_n}{n}$, where $B_n$ is the $n$th [url=http://mathworld.wolfram.com/BernoulliNumber.html]Bernoulli number[/url].

shall i state the result for general dirichlet L-functions, complex and p-adic ;)

Please state it for Hilbert/Poincare series. ";)"


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Let $ n\in \mathbb{N^*}$ and $ f: [0,1]\rightarrow \mathbb{R}$ a continuos function with the prop. $ \int_{0}^{1}(1\minus{}x^n)f(x)dx\equal{}0$.
Prove that $ \int_{0}^{1}f^2(x)dx \geq 2(n\plus{}1)\left(\int_{0}^{1}f(x)dx\right)^2$

Are you sure that is not supposed to be 2n+1?

I wish it was $ 2n\plus{}1$ also but I checked again the source of the problem (a regional contest in my country) and it is $ 2(n\plus{}1)$. Could be a mistake...

[quote="c_adryan"][quote="Aalexandru"]I wish it was $ 2n \plus{} 1$ also but I checked again the source of the problem (a regional contest in my country) and it is $ 2(n \plus{} 1)$. Could be a mistake...[/quote]

what contest?[/quote]

Grigore C. Moisil , Cluj, 2008, 23rd edition   :)

[quote="mareleG"]This problem has somehow a standard approach.
Try first to find out a function that satisfies initial conditions and for which left hand equals right hand.

Function $ f_0(x) \equal{} x^n \minus{} \frac {1}{2n \plus{} 1}$ satisfies initial conditions and $ \int_0^1 f_0^2 (x)dx \equal{} 2(n \plus{} 1) (\int_0^1 f_0(x) dx)^2$, so LHM = RHM

Apply then Cauchy-Schwartz inequality for $ f$ and $ f_0$

and be aware that $ \int_0^1f(x)f_0(x)dx \equal{} \int_0^1f(x)x^n dx \minus{} \frac {1}{2n \plus{} 1}\int_0^1f(x)dx \equal{} \frac {2n}{2n \plus{} 1}\int_0^1f(x)dx$

Equality holds for functions of type: $ f(x) \equal{} k(x^n \minus{} \frac {1}{2n \plus{} 1})$, where k is a real constant[/quote]

Thank you mareleG for your post. It was very useful!


there are 10 signposts each a mile apart.  They go in a straight line numbered 1-10. You order a man to nail the first sign at a sign 1-10.  Then you tell him where to go next but it can't be the same place.  You do this until he can't nail any more signs.  Find the longest path you can make him take.

Hint:

[hide]
The most common route first said is 1,10,2,9,3,8,4,7,5,6 which covers
45 miles

5,10,1,9,2,8,3,7,4,6 covers 5+9+8+7+6+5+4+3+2 or 49 miles.

Can you find more?
[/hide]

[quote="236factorial"]I think the hints are supposed to be answered 45 and 49.[/quote]
oops i'll correct those.

[quote="Devendra Patel"]Jimlee could you find some way more than 59 ,I am stuck up at 59 too.Ill give it to my students lets see if they can find a new way[/quote]

I'm actually Jim Li

Let's see.
I found one of equal route which was 5,10,2,9,1,8,3,7,4,6.  I guess that the maximum is dependent on the sign you start on.
Then I found 5,10,4,9,3,8,2,7,1,6 which totaled 5+6+5+6+5+6+5+6+5, 49.  I guess I'll work some more.
Starting on 7 (lucky number?)
7,1,10,2,9,3,8,4,6,5 totals 49 still. wierd.   7,1,9,2,10,3,8,4,6,5 is worse though.  What if we tried a formula approach?  We could find the worst for 1-5 signs easily.

with 1:you go to 1,0 miles
with 2:you go to 1-2,1 mile
with 3:you go to 1-3-2,3 miles
with 4:you go to 2-4-1-3,7 miles
with 5:you go to 3-5-1-4-2,11 miles

Maybe the best way for an odd number of signs is starting middle and going back and forth.  Sort of makes sense.  I think i might have given the answer away.

Hi Mr. Devendra Patel,
I'm interested in knowin ghow you got 59.  I wrote a computer program just now that I think works correctly to try all possible permutations (I barely did any bug -checking but the code seems right).  It says the best is 49, and one of the paths it gave fitting the 49 description is:

5,7,1,8,2,9,3,10,4,6

Perhaps another way to approach this problem is to look at average moving distances and whether there is some system limit to the most.  for example, there are 9 steps to go from start to finish.  49 / 9 ~ 5.45 average miles per step.  is that number particularly significant to anyone?

Very good problem  :D

[quote="4everwise"][hide]I got 5.45 per step.[/hide][/quote]
solution?


For any $ a,b,c$ positive reals such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$, prove that
$ \frac {1}{2 \minus{} a} \plus{} \frac {1}{2 \minus{} b} \plus{} \frac {1}{2 \minus{} c} \ge 3$

[quote="Sunkern_sunflora"]Expanding leads us to:
$ \sum (2 \minus{} b)(2 \minus{} c) \geq 3(2 \minus{} a)(2 \minus{} b)(2 \minus{} c)$
$ \Rightarrow \sum (4 \minus{} 2b \minus{} 2c \plus{} bc) \geq 3(8 \minus{} 4a \minus{} 4b \minus{} 4c \plus{} 2ab \plus{} 2bc \plus{} 2ca \minus{} abc)$
$ \Rightarrow 12 \minus{} 4(a \plus{} b \plus{} c) \plus{} (ab \plus{} bc \plus{} ca) \geq 24 \minus{} 12(a \plus{} b \plus{} c) \plus{} 6(ab \plus{} bc \plus{} ca) \minus{} 3abc$
$ \Rightarrow 8(a \plus{} b \plus{} c) \plus{} 3abc \geq 12 \plus{} 5(ab \plus{} bc \plus{} ca)$

Now can't we solve it using "[b]pqr[/b]" technique?
 :)[/quote]
Yes we can. But I like the $ uvw$ technique.
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Then $ t\geq1,$ $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3\Leftrightarrow3u^2 \minus{} 2v^2 \equal{} 1$ and
$ 8(a \plus{} b \plus{} c) \plus{} 3abc \geq 12 \plus{} 5(ab \plus{} bc \plus{} ca)\Leftrightarrow$
$ \Leftrightarrow24u(3u^2 \minus{} 2v^2) \plus{} 3w^3\geq(12(3u^2 \minus{} 2v^2) \plus{} 15v^2)\sqrt {3u^2 \minus{} 2v^2}\Leftrightarrow$
$ \Leftrightarrow24u^3 \minus{} 16uv^2 \plus{} w^3\geq(12u^2 \minus{} 3v^2)\sqrt {3u^2 \minus{} 2v^2}.$
But by Schur we obtain $ w^3\geq4uv^2 \minus{} 3u^3.$
Id est, it remains to prove that $ 7u^3 \minus{} 4uv^2\geq(4u^2 \minus{} v^2)\sqrt {3u^2 \minus{} 2v^2}.$
But $ 7u^3 \minus{} 4uv^2\geq(4u^2 \minus{} v^2)\sqrt {3u^2 \minus{} 2v^2}\Leftrightarrow t(7t \minus{} 4)^2\geq(4t \minus{} 1)^2(3t \minus{} 2)\Leftrightarrow$
$ \Leftrightarrow(t \minus{} 1)^2(t \plus{} 2)\geq0.$

I think, the following solution is better.
$ \sum_{cyc}\frac{1}{2\minus{}a}\geq3\Leftrightarrow\sum_{cyc}\left(\frac{1}{2\minus{}a}\minus{}1\minus{}\frac{a^2\minus{}1}{2}\right)\geq0\Leftrightarrow\sum_{cyc}\frac{a(a\minus{}1)^2}{2\minus{}a}\geq0.$

Fantastic arqady! :coolspeak:
Your solution is much much better! Thank u for that nice solution.
 :)


Given: line segment AB is congruent toline segment BC  ;   line segment  AD   is congruent to line segment  CD

Prove: BD is perpindicular bisector of AC

the figure looks like a kite. point a is on the left, bis on the to, c is on the left, d is on the bottom. E is in the middle

Let angle BAE= x =angle BCE, &  angle ABC=z
so,
2x+z=180 (1)
z=90+x
We sub the value for z into (1):
2x+90+x=180
3x=90
x=30
z=120
Lets backtrack a bit,
In triangle ABE & CBE,
AB=CB (given)
BE=BE(common side)
AE=EC(pythagorean theorem)
Thus triangle ABE is congruent to triangle CBE by SSS congruency.
Since they are congruent,
angle ABE=angle CBE
since z=120
angle ABE=angle CBE=60
Therefore:
angle AEB=angle CEB=90
So we know that BE is the perpendicular bisector of AC. Similarly, we can prove DE is the perpendicular bisector of AC. And so we're done.  :D

[quote="bitternlutz"] The figure looks like a kite. point A is on the left, B is on the [b]top[/b], c is on the [b]right[/b], & D is on the bottom. E is in the middle.[/quote]
I assumed this is the case.


vector
linear algebra
linear algebra unsolved

let $ V_1,V_2$ be two subspace of $ E$ with the same codimension ,prove that they have a commun supplement .
HINT:use the fact that $ H$ U $ G\equal{}E$ then $ H\equal{}E$ or $ G\equal{}E$

[quote="drapt@"]let $ V_1,V_2$ be two subspace of $ E$ with the same codimension ,prove that they have a commun supplement .
HINT:use the fact that $ H$ U $ G \equal{} E$ then $ H \equal{} E$ or $ G \equal{} E$[/quote]
If I'm not mistaken, by "commun supplement" you mean simultaneous complement.
Here is a counterexample to your assertion: Using your notation, let $ E\equal{}\mathbb{R}[x], V_1\equal{}\langle x \rangle, V_2\equal{}\langle 1, x \rangle$. Clearly, $ \text{codim}(V_1)\equal{}\text{codim}(V_2).$ But $ V_1$ and $ V_2$ do not have a simultaneous complement.

but the codimension of $ V_1$ is infinit , when we talk about codimension ,it's implicitly a finite codimension

Codimension of a vector space can be finite or infinite, just as dimension. Whenever you mean finite codimension, you should explicitly mention that. :wink:
With that extra assumption, it is clear that $ V_1$ and $ V_2$ have the same dimension. Define $ \Sigma\equal{}\{X\subseteq E|\langle X\rangle \cap V_1\equal{}0\equal{}\langle X\rangle \cap V_2\}$. Choose a maximal element, say $ Y$, for $ (\Sigma,\subseteq)$. Show that $ \langle Y\rangle$ is the desired subspace, using your hint.


geometry
conics
function
algebra
polynomial
absolute value
geometry unsolved

If $ x, y \in R$ then the equation:
$ 3x^4 \plus{} 2(19y \plus{} 8)x^2 \plus{} (361y^2 \plus{} 2(100 \plus{} y^4) \plus{} 64) \equal{} 2(190y \plus{} 2y^2)$
represents in the rectangular cartesian system a,
$ A)$ A pair of circles.
$ B)$ A circle and a hyperbola.
$ C)$ Four distinct points.
$ D)$ two distinct points.

The given equation $ F(x,y) = 3x^4 + 2(19y + 8)x^2 + 2y^4 + 357y^2 - 380y + 264 = 0$ may represents a general 4th degree curve. The equation may be factorable, for example into two 2nd degree equations representing two real conic sections. Or it can represent an even number of up to 4 distinct points (0 or 2 or 4), because only even powers of $ x$ are present.

Assuming the equation is factorable into two conic equations and one conic is a circle, from absence of the $ x^2y^2$ term, the other conic would have to be a rectangular hyperbola. Then the coefficients at $ x^4, y^4$ would have to be equal in absolute value and have opposite signs, which is not the case.

Assuming the equation represents a curve, then by the implicit function theorem, a point $ (x_0, y_0)$ would have to exist, such that $ F(x_0, y_0) = 0$ and gradient $  \nabla F(x, y)|_{(x_0, y_0)} \neq 0.$ Solving the given equation for $ x^2$ in terms of $ y,$

$ x^2 = \frac { - (19y + 8) \pm \sqrt {D(y)}}{3},$

where the discriminant cannot be negative:

$ D(y) = (19y + 8)^2 - 3(2y^4 + 357y^2 - 380y + 264) \ge 0,$

$ 3y^4 + 355y^2 - 722y + 364 \le 0.$

Comparing the polynomial coefficients $ a_4 = 3, a_3 = 0, a_2 = 355, a_1 = - 722, a_0 = 364$ yields $ \sum_{i = 0}^4 a_i = 0,$ which means that the last polynomial has a root $ y = 1.$ Substituting $ y = z + 1$ to expose it, we get

$ z^2(3z^2 + 12z + 373) = z^2 (3(z + 2)^2 + 19^2) \le 0.$

This is possible only if $ z = 0$ or $ y = 1.$ Thus the discriminant polynomial has double root $ y = 1$ and no other real root, which means $ D(1) = 0$ and $ D(y) < 0$ otherwise. The only real $ x^2$ is then $ x^2 = - \frac {_{27}}{^3} = - 9,$ which is impossible. It follows that the original equation represents zero points, which is not on the original list of answers (A), (B), (C), (D).

[quote="kabi"]Hello
The equation represent four distinct point.
Thank u.[/quote]
The solution set says the answer is two distinc points.

For this, the original equation would have to be $ 3x^4 \minus{} 2(19y \plus{} 8)x^2 \plus{} (361y^2 \plus{} 2(100 \plus{} y^4) \plus{} 64) \equal{} 2(190y \plus{} 2y^2),$ representing 2 points $ (\pm 3, 1).$ Either you made a typo copying it, or the typo is in your book.


The state income tax where Kristin lives levied at the rate of $ p$% of the first of $ \$28000$ of annual income plus $ (p \plus{} 2)$% of any amount above $ \$28000$.  Kristin noticed that the state income tax she paid amounted to $ (p \plus{} 0.25)$% of her annual income.  What was her annual income?

[b](A)[/b] $ \$28000$       [b](B)[/b]$ \$32000$    [b](C)[/b]$ \$35000$     [b](D)[/b]$ \$42000$    [b](E)[/b]$ \$56000$

[hide]B[/hide]

A charity sells 140 benefit tickets for a total of $ \$2001$.  Some tickets sell for full price (whole dollar amount), and the rest sell for half price.  How much money is raised y the full-price tickets?

(A)782  (B) 986    (C)1158    (D) 1219    (E) 1449

[hide]a[/hide]

When you explain it, please don't use the solutions from the AMC 10 packet, I didn't understand those...

[quote="mathemagician1729"]Might I ask which year and what version these came from?[/quote]

2001

Ok for number two basically u just relate the expression with what you have and then simplify the equation as much as possible. You'll end up with an equation in two variables. From there you just factorize to find answers for the respective domain $ 0\le x\le140$.

For number 1:
I'm too lazy to really do it now but you could just use the answer choices and plug in. I think back then you could have used a calculator anyway.

[quote="mathemagician1729"]Ok for number two basically u just relate the expression with what you have and then simplify the equation as much as possible. You'll end up with an equation in two variables. From there you just factorize to find answers for the respective domain $ 0\le x\le140$.[/quote]

Would equation be this if $ x \equal{} \text{ price of full ticket}$ and $ y \equal{} \text{ number of people who paid full ticket}$:

$ xy \plus{} \frac12x(140 \minus{} y) \implies xy \plus{} 70x \minus{} \frac12xy \equal{} \implies \frac32xy \plus{} 70x \equal{} 2001$

Do I just guess $ x$ and $ y$ now?

[quote="mathemagician1729"]'m too lazy to really do it now but you could just use the answer choices and plug in. I think back then you could have used a calculator anyway.[/quote]

1. We can't use a calculator now
2. But how can we guess and check if $ p$ is unknown.....I can solve # 24 but can't solve #9....sad...

[hide="1"] Let it be x. We have $ \frac{p(28000)\plus{}(p\plus{}2)(x\minus{}28000)}{100}\equal{}\frac{(p\plus{}.25)(x)}{100}$. Multiplying by 100 and expanding gives you $ 28000p\plus{}px\minus{}28000p\plus{}2x\minus{}56000\equal{}px\plus{}.25x$ so $ 1.75x\equal{}56000$ and $ x\equal{}32000$, so B.[/hide]

[hide="2"] Let x be the number of full price, and let them cost y each. We have $ xy\plus{}\frac{y(140\minus{}x)}{2}\equal{}2001$. Multiplying by 2 and expanding gives $ xy\plus{}140y\equal{}4002$ so $ y(x\plus{}140)\equal{}4002$. 4002=2*3*23*29, and (x+140) must be greater than 140, x+140 must be 174, so x is 34. y is then 23, so our answer is 23*34=782.[/hide]


Find all positive integer solutions

$2b^3 = a^3 +c^3$

 :D

[quote="Chang Woo-JIn"]thank you for information  :D[/quote]

More you can find some studies of this type equation in the book : [b]Elementary Theory of Numbers[/b] deu to Waclaw Siearpinski, published in 1963, in Warsaw, Poland.


$\cup_{i=1}^{\infty} A_{i}$
On wikipedia, this works out fine, but here the indices aren't where they should be: under and above the union sign, what did I do wrong?

Second of all : there used to be some links I believe on the latex help page allowing you to download tex files of real professional math texts, and I found them very useful

As I will be starting to think of my thesis of next year soon, I have to get a little better at latex eventually, that's why.

Thans to all the people who have helped me to learn latex!

[quote="fredbel6"]$\cup_{i=1}^{\infty} A_{i}$
On wikipedia, this works out fine, but here the indices aren't where they should be: under and above the union sign, what did I do wrong?[/quote]
You should be using \bigcup for this to look good $\bigcup_{i=1}^{\infty} A_{i}$ but if you insist on using the smaller \cup you need to convert it into an operator with \mathop so that you can use \limits (which is normally used with sums and integrals to put the limits above and below) $\mathop{\cup}\limits_{i=1}^{\infty} A_{i}$

\cup looks a bit small with those limits, hence the preferred use of \bigcup with limits above and below.

okay but what about

$\Sigma_{i=1}^{\infty} x_{i}$

\mathop would work here as well but it's best not to use \Sigma which is just a Greek letter - you want \sum which can have limits. This forum uses \displaystyle by default so you get $\sum_{i=1}^{\infty} x_{i}$ here but in a document this would give $\textstyle \sum_{i=1}^{\infty} x_{i}$ so you must either add \displaystyle or \sum\limits $\displaystyle\sum_{i=1}^{\infty} x_{i}$ or $\sum\limits_{i=1}^{\infty} x_{i}$

The same system applies to integrals.


calculus
integration
inequalities
function
calculus computations

WLOG, we may consider the following equivalent problem.
Let such an f, but defined in [-1,1] and f(0)=0. Prove that $\int_{-1}^{1} f(x)dx\geq 0 $.
But this is obvious since $ f(x) \geq x\cdot f'(0)$ for all x in [-1,1] .

[quote="kunny"]A function $f(x)$ satisfies $f''(x)>0$ for all real numbers $x$. Prove the following inequality.

\[\int_{2003}^{2005} f(x)dx\geq 2f(2004)\][/quote]

it can be

$\begin{array}{l}
\int\limits_{ - 1}^0 {\left[ {f(x + 2004) - f(2004)} \right]} dx \ge \int\limits_0^1 {\left[ {f(2004) - f(x + 2004)} \right]} dx\\
 \Leftrightarrow \int\limits_{ - 1}^0 {xf'(\theta (x))} dx \ge \int\limits_0^1 { - xf'(\theta (x))} dx \Leftrightarrow f'({c_1})\int\limits_{ - 1}^0 x dx \ge f'({c_2})\int\limits_0^1 { - x} dx,{c_1} \in \left( { - 1,0} \right),{c_2} \in \left( {0,1} \right)
\end{array}$ (*)

but, f''(x)>0, there for (*) allaways true. it's the result

Um, isn't that just a special case of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=473225]this[/url]?


algebra polynomial function calculus integration trigonometry limit

Maybe I'm just on crack right now, perhaps this is pretty obvious.

How do we know that the Taylor Polynomial the best approximation of an appropriately differentiable function about a certain point?

$ \blacktriangleright$ What is meant by the "best approximation" of a continuous function, anyway?

For example, suppose I define $ f_t (x)$ as the cubic in $ x$ that minimizes the integral $ \int_{\minus{}t}^{t} (\sin x \minus{} f_t (x) )^2 dx$

$ \blacktriangleright$ Does the fact that $ x \minus{} \frac{x^3}{6}$ is the cubic Taylor Polynomial at $ 0$ for $ \sin x$ [i]imply[/i] that $ \lim_{t \to 0} f_t (x) \equal{} x \minus{} \frac{x^3}{6}$?

$ \blacktriangleright$ What if the bounds of that integral were from $ 0$ to $ t$ instead of from $ \minus{}t$ to $ t$?

$ \blacktriangleright$ What if the difference was raised to the 4th power, or some other even power, instead of the 2nd power? Would the limit still be the Taylor Polynomial?

"Best approximation" can mean various things. If you just fixed the value of $ t$ (rather than let it go to $ 0$), you'd probably have another minimizer.

But as $ t\to 0$, the minimizing cubic indeed approaches the Taylor polynomial (call it $ T(x)$). It does not matter which integral norm you use, or whether the interval is symmetric or not. A very rough sketch: $ \int_{\minus{}t}^t |\sin x\minus{}T(x)|^p\,dx$ is $ O(t^{5p\plus{}1})$ as $ t\to 0$. Let $ q_t(x)\equal{}c_0\plus{}c_1x\plus{}c_2x^2\plus{}c_3x^3: \equal{}f_t(x)\minus{}T(x)$ (the coefficients depend on $ t$, of course). The minimizing property of $ f_t$ implies $ \int_{\minus{}t}^t |\sin x\minus{}T(x)\minus{}q_t(x)|^p\,dx\equal{}O(t^{5p\plus{}1})$, hence $ \int_{\minus{}t}^t |q_t(x)|^p\,dx\equal{}O(t^{5p\plus{}1})$. This forces all coefficients of $ q_t$ to go to $ 0$ as $ t\to 0$.

The question you raise on a [i]fixed[/i] interval is actually very interesting since the answer is almost never the Taylor polynomial; in particular, for $ p \equal{} 2$ you are minimizing a norm induced by a certain inner product and the answer you want can be given in terms of some system of [url=http://en.wikipedia.org/wiki/Orthogonal_polynomials]orthogonal polynomials[/url].

That's actually where my question came from  :) I was revisiting a chapter on the Graham-Schmidt process which showed a very effective quintic approximation to $ \sin$ over $ [\minus{}\pi, \pi]$. So then I was wondering how the approximations obtained in this way related to the Taylor polynomials.


[quote="chien than"][quote="phuc-90"]
$\frac{6}{abc+3}\geq \frac{3}{abc+3}$[/quote][/quote]

wrong! with $a,b,c \geq 1$

if should be $\frac{6}{abc+3}\geq \frac{3}{abc+1}$


function
real analysis
real analysis unsolved

[quote="knoppix"]$ x_0 \equal{} a$ and $ x_{n \plus{} 1} \equal{} x_n \minus{} (1 \plus{} x_n^2)\arctan x_n$

find all reals $ a$ such that this sequence converges[/quote]This is essentially the question for which $ x_0\equal{}a$ Newton's method for solving $ \arctan x \equal{} 0$ will converge. The answer is the interval $ (\minus{}M,M)$, where $ M$ is the unique positive solution of the equation $ x\minus{} (1\plus{}x^2)\arctan x \equal{}\minus{}x$.

[quote="knoppix"]please justify this answer

I know it's Newton's method, but how did you obtain this $ M$ ? I know only that there exists such interval[/quote]This is the largest interval that the iteration function $ \varphi(x) \equal{} x \minus{} (1\plus{}x^2)\arctan x$ maps to itself. In fact, $ M$ is the unique positive solution of $ \varphi(x) \equal{} \minus{}x$ (Cal;culus exercise).

In the interior of the interval, $ |\varphi'(x)| < 1$,  each interval $ [\minus{}K,K] \subset (\minus{}M,M)$ is mapped to itself by $ \varphi$, and $ |\varphi'(x)| \le \epsilon_K < 1$ on $ [\minus{}K,K]$.  That's sufficient for convergence, see e.g. the corresponding chapter in Burden/Faires.


inequalities
absolute value
inequalities unsolved

Is this hard or easy.

Bomb :omighty: only to BOMB.

Bomb - :starwars:

[quote="bomb"]For each distinct reals x,y,z let
E(x,y,z)=(f(x)+f(y)+f(z))/f((x-y)(y-z)(z-y)) where f(x) denoted the absolute value or modulus of x.
Determine the smallest possible value of E.[/quote]
You probably typed your question wrongly. Anyway, I think the answer -> 0.

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

bomb wrote:
For each distinct reals x,y,z let
E(x,y,z)=(f(x)+f(y)+f(z))/f((x-y)(y-z)(z-y)) where f(x) denoted the absolute value or modulus of x.
Determine the smallest possible value of E.

Yeah sorry, Its supposed to be (f(x)+f(y)+f(z))^3/etc....

Bomb

We want to minimize the expression:

$(|x| + |y| + |z|)^3(|x-y||y-z||z-x|)^{-1}$

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

Now, $|x-y| \leq |x| + |y|$, whereby $(|x-y||y-z||z-x|)^{-1} \geq (\prod |x| + |y|)^{-1}$.

Observe that in the expression,

The numerator is $\sum_{cyc} |x|^3 + 3\sum_{sym} |x|^2|y| + 6|x||y||z|$
The denominator is $(\prod |x| + |y|) = 2|x||y||z| + \sum_{sym}|x|^2|y|$

whereby the expression is now $3 + \frac{\sum_{cyc} |x|^3}{2|x||y||z| + \sum_{sym}|x|^2|y|}$.

It can be shown that the last bit is $\geq 0$, so final answer 3.

so if the last inequality holds!( i believe it's true i didn't follow the demo :blush: )
but you have to minimize the bit!
wich sequence did y take to proove that it's infimum is 0? :?:


What is your favorite board game? Mine is chess :)

:blush: i must admit that sometimes i am the one who eats the cookies and watermelon :lol:

[quote] i must admit that sometimes i am the one who eats the cookies and watermelon [/quote]

i don't understand. what do you mean?

anyway, i like chess, but, my favorite game is backgammon.

I don't like chess at all.  My favorite is Risk.

And I think jhollenback means that he's just an observers some of the time.

Risk is awesome  :P

I do play all those other games though, except checkers.. I cant take that.

you spelled 'stratego' wrong.

and there's plenty of board games not on that list..  you even forgot monopoly (not that I like monopoly) but out of that list, I think I like Stratego and Backgammon the best. But I tend to pick games for more than 2 people more often.. Like Pictionary or Clue.

how can you edit polls? I remembered other and monopoly right when i posted it

what i mean by the cookies and watermelon choise is that you just watch people play games and eat all the food :P

:o you like life? I thought it was OK until our box fell apart.

seems like checkers isn't very popular.....

Would that be the "classic" board game Life or the game by John Conway called Life? Conway's life is a good 1person board game ;) Stratego's probably my favorite right now; Monopoly and Risk are good. Civilization looks like fun.

i like chess alot, but i'm not that good at it.

Sheesh, no go, chinese chess, or shougi.

Oh, and I absolutely hate stratego. It's just the exact same thing as a chinese board game called "zhun qi" except watered down.

I like chess. I also like playing crazy versions of chess where the pieces do different things.

yay for chess-starcraft hybrid games...where the pieces do insane things :D

but risk all the way

what about Balderdash which is so funny and is old school since it was made in the 80s

risk is the best.
i might like chess more if im any good at it  :?

Hey, I just thought of something - does anyone know of a place online where you can play multiplayer RISK - thats good- ? It'd be cool to have an AoPS RISK tourney.. then again it would take 4 hours per game so i dunno.

chess, then stratego, but ches is unbeatable.  :)  :)

:D yay, chess caught up in the polls! :)

I votedl so now chess is beating stratego!!!. Guys vote for chess.

i remember some game called Crazy Bones.  You had to flick your piece and try to hit your opponents piece.  This was about 6 years ago though.

[quote="math92"]Risk takes too long to play...I don't have the time to play a full game :(
(i have 2500 posts)[/quote]

Yea, but it's still fun while you play it.  Especially if there's a lot of people (4-6).  It's not as much fun when you only have 2 people.

there is only me and my sister so that's two people. my parents are too busy to play and we rarely have friends over.

has any1 heard of the game Diplomacy before?? it is basically a better version of risk which requires less luck and more skill.

[quote="Super Naterul"]My brothers and I have played "flicking" chess. We would flick our pieces at the other persons and try to knock down all of theirs before they knock down all of ours. :lol:  :lol:  :lol:[/quote]I always thought i made that game up! :o now i hear that it is becoming popular!  I like atomic chess, play it a lot on USCL.

i like to play chess since its stratigic and simple.
im not good at it tho... :P

Keep playing! The more you play, the better you get at chess, the best board game ever! :)

I think chess is boring..'flicking chess' sounds like fun though :)

[quote="M"]Does anybody of you know [url=http://www.gipf.com]Gipf[/url]? It's a whole family of games and the are all very strategic without any presence of probability (like chess). Of this family I lile yinsh very much.

Oh, and I like chess (even if I'm not too good a player) and don't like risk so much because of the probability[/quote]

I heard of Gipf! I haven't got the chance to play them yet though.

chess is more interesting than flicking chess. :)

[quote="zeb"][quote="nonie"]What's 'go'?? Never heard of it..[/quote]

AAAAUGH what is the world comng to?! It is the best board game - it has like three rules, but the strategy is insane. The rules are basically 1) every turn you put a piece on the board 2) surrounding a group of pieces kills it and 3) whoever surrounds the most territory at the end of the game wins. The strategy is so complicated that we can make computers that beat masters at chess but no computer can even beat an intermediate go player.[/quote]

Sounds like fun!! :lol:


Some sick dude made this modifications  :D

Edit:969.6

But,its pretty bloody,i will switch to the old version.The mines idea,is not bad,however the blood thingie is pretty sick  :?

I don't care bout the blood, it's a great game!

And it IS all about mathematics :D

[quote="Valiowk"]This is utterly morbid...
And I can't judge where the club is supposed to be now so (thankfully) I don't hit the penguin most of the time.[/quote]

Hey Valiowlk it's just a game, no pinguins had to suffer for it ;)

[quote="Peter VDD"] Hey Valiowlk it's just a game, no pinguins had to suffer for it ;)[/quote] that's what they said about Counterstrike, untill they began to play Paintball :D:D

whahaha!! :D:D:D

that's probably the best one I ever heared :P

so new signature food :)

[quote="Valiowk"]It's nice to see my name appearing in Peter's signature but could you please get my name spelled correctly! ;)[/quote]
hahaha :D:D sorry but I couldn't help it :D.

Bloody, but fun  :D
too erratic, though
=====================
Pingu Fan Club 2004
[url]http://obelix.jol.hu/pingu/[/url]

[quote="PiDeltaPhi"][url]http://www.dirtytrain.com/dmx/pingu/[/url][/quote]
this site needs username and password now.. doesn't it?

i found it at
[url]http://www.jengajam.com/r/bloody-pingu[/url]
and i met an 1072.3

too much blood slows down the speed of the game...

strange, didn't I see one with a darts game as well?


function
real analysis
limit
calculus
integration
real analysis unsolved

Examining the function $f(x)=\begin{cases}0~ \text{for}~x\in\mathbb{R}\setminus \mathbb{Q}\\ \frac{1}{q}~ \text{for}~x=\frac{p}{q}\in\mathbb{Q}\\ \end{cases}$ we get that $f$ is continuous on the irrationals and discontinuous on the rationals. This is an example of a function that is defined on an interval continuous at a point in the interval, but not continuous on any interval around that point.

I was wondering if it is possible to have a function $\textit{differentiable}$ at a point, but not continuous on any interval around that point.

Also, since $f$ is only discontinuous on a countable number of points, does this mean that it is Riemann Integrable? This seems hard to believe.

$f(x)=\begin{cases}0&\text{ for }x\in\mathbb R\setminus\mathbb Q \\ x^{2}&\text{ for }x\in\mathbb Q\end{cases}$
is differentiable at $0$.

Yes, any bounded function whose set of discontinuity has Lebesgue measure zero is Riemann integrable. The converse is true as well.

My attempted proof at why it is differentiable at $0$:

$f'(x)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{f(x)}{x}$, so as long as this limit exists it is differentiable at $0$. Let $\{a_{n}\}$ be a sequence of rationals such that $a_{n}\to 0$ as $n\to\infty$. Then $\lim_{x\to 0}\frac{f(x)}{x}=\lim_{n\to\infty}\frac{(a_{n})^{2}}{a_{n}}= \lim_{n\to\infty}a_{n}=0$. Now let $\{b_{n}\}$ be a sequence of irrationals such that $b_{n}\to 0$ as $n\to\infty$, then $\lim_{x\to 0}\frac{f(x)}{x}=\lim_{x\to\infty}\frac{0}{b_{n}}=0$. Therefore $f'(0)=0$.

I'm not so sure about jumping from if any rational series and any irrational series, then any series...

Easier with the squeeze theorem: $0\le |f(x)/x|\le |x|$.

The Riemann integrability of the Dirichlet functions can be shown directly and explicitly, without quoting the theorem about measure-zero discontinuities.

Use the upper and lower sum (Darboux) version of the Riemann integral. Just for specificity, assume we're trying to integrate the function over $[0,1].$

Every lower sum for this integral, no matter what the partition, is zero, since there are irrational numbers in any interval.

Choose an arbitrary $\epsilon>0.$ There are only $n,$ (a finite number) of points $x_{j},1\le j\le n$ in $[0,1].$ for which $f(x)>\frac{\epsilon}2.$ $n$ depends on $\epsilon.$ (If we were to be careful, we could estimate that dependence - but for this proof, we don't need to be that careful.) Choose a partition such that each point $x_{j}$ is isolated within an interval of length no greater than $\frac{\epsilon}{2n}.$ Compute upper sum on this partition. There are two kinds of interval in the partition - those that contain one of the $x_{j}$ and those that do not. Each interval containing an $x_{j}$ contributes no more than $\frac{\epsilon}{2n}$ to the upper sum. Since there are no more than $n$ of those intervals, the total contribution of them is no more than $\frac{\epsilon}2.$ In the other intervals, $f(x)<\frac{\epsilon}2,$ so each such other interval contributes no more than $\frac{\epsilon}2$ times the length of the interval to the sum. Add them up, and with the total lengths not exceeding 1, we get no more than $\frac{\epsilon}2.$ Thus we have an upper sum that is less than $\epsilon.$ Since we can do this for any $\epsilon,$ the infinum of the upper sums equals the supremum of the lower sums equals zero. The function is Riemann integrable, and its integral is zero.

[Lebesgue integral attitude: why are you wasting this much effort on a function that might as well be the zero function, since it's zero almost everywhere?]


MATHCOUNTS
geometry
probability
AMC
AIME
ARML
geometric transformation

Since most of you guys are in the mathcounts tournament, I was wondering something. Right now I have a crappy calculator and am hoping to get a good one for Christmas. Should I get the TI-89, or is there something better that I could possibly look into?

Thank You,
-Jorian Hoover

NO!!! Waste of a present. Get AoPS books. I've said this before. A graphing calculator is NOT NECCESSARY. I believe I got one of the top 30 scores in the country at state without a graphing calculator. They're overrated. They don't help that much.

lol, very funny answer 1337 :rotfl: ...so i should ask for what books if i already am getting vol 1 and 2 within the next few days....since you know im horrible at geometry, should i get the intro to geom.?

and btw, what calculator SHOULD i get?
-jorian

I don't think 13375P34K43V312 was kidding at all.  If you are going to get a calculator, check first with your high school.  Some schools require kids to have a Ti83/84 and if so get a TI84+SE.  If not get a TI89 Titanium.

Since Counting and Probability are the topucs MC loves to focus on and are not taught much in class, I'd go with that book first.

Mathcounts has steered away from doing as much Geometry as they once did.  There is still a fair amount, but if you know your area formulas and Pythag you could be fine.

I would order them in this order :C & P, N.T, Geo.

They are all worth getting though!!!

thank you frost 13

----------
what do u mean they dont let u use the ti 89 on some tests?
-jorian

[quote="frost13"]
I would order them in this order :C & P, N.T, Geo.
[/quote]

Well, it depends on your strengths and stuff. I am finding NT incredibly easy, and Geo a challenge. I finished CP, and I think it was the most fun to work with. But hey, maybe I like Combinatorics, am good at Number Theory, and am bad at geometry! I'm pretty sure the third is correct.

[quote="jhredsox"]
so i should ask for what books if i already am getting vol 1 and 2 within the next few days[/quote]

Based on your current level, you should put volume 2 aside for at least another six months, there's no way you're going to understand it if you aren't comfortable with stuff like ${n\choose r}$ and modular arithmetic.

Work through volume 1 (or at least skim) and see what your weaknesses and strengths seem to be. Believe me, you're going to end up being surprised. I always thought I sucked at combinatorics, but I went through CP no problem.

at the time, whe i ordered the books i thought vol 2 was an addition to vol1, so now i know to hold it off for a year or so
ty,
-jorian

Yeah, I thought that too.

Volume 2 helps pretty much zero with Mathcounts, and I understand that's what you're focusing on this year. Volume 2 is very helpful with the AIME, and the easier parts of olympiads.

Then again, volume 1 is only about 1/3 mathcounts level.

...as long as it helps:)

basically my goal for this year is to get to nats...i don't really care from there as i REALLY wanna go there

...if i go, i want to play (and hopefully beat) carl in tennis:)
-jorian

[quote="pianoforte"]Keep in mind that the TI-89 isn't allowed on some tests.[/quote]

Like? I have a TI-89 titanium and I use it on every test I take (that allows calculators...). It doesn't have a QWERTY keyboard.

Well, if you want a calculator for Christmas (your parents would buy you the books without an excuse such as holidays anyways, right?), I'd recommend the TI-83 if you're looking into math contests, because the TI-89 isn't allowed on some tests. The TI-84 is weirdly curved, IMO.

Although a good calculator obviously isn't necessary, I've found that although I used to make a lot of calculation errors on small calculators with small keys, I never make them anymore on my graphing calculator (hand me down from my brother that he got from Mathcounts, it's a TI-83 SE).

Get a TI-84 Plus Silver Edition.
If you bid carefully, you'll get it for 60 bucks/dollars/molars like me.

TI-84 SE, definitely

all programs available on the website are predownloaded, and its not too clunky.  you can get the hang of it easier, and the screens nice.  just make sure to use whatever calculator you get tons before contests, so you know all the shortcuts.

[quote="Fanatic"]I'd recommend the TI-83 if you're looking into math contests, because the TI-89 isn't allowed on some tests.[/quote]

LIKE????????????

[quote="Phelpedo"][quote="13375P34K43V312"]Oh another point I would like to make: which team do you think would win at nationals if they practiced exactly the same way, and if there was a very difficult team round?:

State: XX

46
45
45
14

State: YY

39
39
39
39[/quote]

Hard to say because scores don't always reflect skill. Ask yourself this: who would win if the situation was

State: XX
46
0
0
0
State: YY
16
15
15

XX I think.[/quote]

that's why the Mathcounts team round doesn't really encourage team work, but something like the relay round of arml does.

[quote="Phelpedo"]The ARML relay involves no teamwork; it just requires you to have a balance of strong individuals. There's no actual interaction and working together.[/quote]

That's what I sorta meant by teamwork, though it was kinda vague, w/e. I'm just giving an example of when state YY would have the advantage.

[quote="nebula42"][quote="bpms"]Did a mod delete a post of mine? I was pretty sure I made the post, but maybe I didn't...[/quote]
Yes, I deleted it. It's better to avoid making comments about specific people (including "inequalities" of people  :wink: ), so please try to avoid doing this in the future.
Thanks.[/quote]
Sorry, I was just pointing out something with the scores... I'll try not to degrade anyone in future.

I have had posts deleted before, but sometimes I don't know what I posted. Is there a way to know if a mod's delted it?

This makes it very confusing!
-jorian

[quote="funcia"]i am really wondering right now what is this :
[quote]State: XX

46
45
45
14

State: YY

39
39
39
39[/quote]
?????????????????????????????????????????
It seems these diagrams are very popular...yet I haven't seen one until now.[/quote]
I don't think so because I haven't seen one until now either.

no i meant they are popular on this thread!!! ahh but i haven't seen them till this thread either too(as i said and you noticed) :D

[quote="Fanatic"]I want an ipod for christmas.

I'm an anomaly on this board :(.[/quote]

I got an ipod nano last year. :D

[quote="Ubemaya"]I want a megaminx, a square-1, a 5x5x5 rubik's cube, and a bunch of math books for christmas. And a Wii. But my parents hate video games.[/quote]

This is turning into a 'what are you gettting for christmas topic'

I think a mod should move it to G&FF.

I find that this topic has served its purpose, suggesting things for he whom first posted, and is now off-topic, this this topic beist locked.


How exactly does one rigorously prove that the ratio of any circle's circumference to its diameter is constant.

It's a direct consequence of the similarity of all circles. If the circumference of one circle is $ C$ and the diameter is $ d$, then all other circles have circumference $ kC$ and diameter $ kd$ for some constant $ k$.

I've actually thought about this many times before.  It certainly isn't so obvious, since in some geometries it doesn't seem to even be true.  I didn't figure it out until someone asked me how to prove that the limit of regular n-gons is a circle.

Essentially, it comes down to the similarity of isosceles triangles with common peak angles.
Consider two regular n-gons, one of radius 1 and perimeter $ p_1$, and one of radius r and perimeter $ p_r$.  Now by drawing lines from the centroid to the vertices of the same n-gon, divide each into isosceles triangles.

Since they are both regular n-gons, these isosceles triangles share peak angles, so they are all similar (for a given n).  Thus $ \frac {baselength}{sidelength}$ is constant among all of them.  From here, we get that $ \frac {p_1}{2} \equal{} \frac {p_r}{2r}$.  Let $ n\rightarrow\infty$ and use the fact that the limit of regular n-gons is a circle to show pi is constant (at least when isosceles triangles with a common peak angle are similar).

I seem to recall that Archimedes addressed this very issue.

A very rough outline of the approach Archimedes took:

1. Argue that the circumference of a circle is greater than the perimeter of an inscribed polygon and less than the perimeter of a circumscribed polygon.

2. Construct regular inscribed polygons and regular circumscribed polygons. The similarity you need comes from the similarity of the triangles that these polygons can be decomposed into.

3. By successive bisection of angles, construct sequences of such polygons with increasing numbers of sides. Trap the value of $ \pi$ in between these as a matter of Eudoxus's "method of exhaustion."

Edit: I started this before I saw what Mathnerdmo wrote. He is talking about the same thing.


parameterization
integration
real analysis
real analysis unsolved

Hi all

I was helping a friend with a physics problem and I got it down to solving the following ODE:

y'' = 1/y^2

I don't have a ton of experience with ODEs and I have been unable to find the general solution to this, I would appreciate it if someone could show me how to do this.

Thanks

Letting $ v = y'$ yields an ODE for $ v$ w.r.t. independent variable $ y$:
\[ vv' = \frac {1}{y^2} \\
\Rightarrow v = \pm \sqrt {\frac {c_1y - 2}{y}
}\]
which yields two-parameter families of solutions for $ y$ in implicit form:
\[ \int^y{\sqrt {\frac {\xi }{c_1\xi - 2}}d\xi } = c_2 \pm x
\]
where $ c_1,c_2$ are arbitrary.


[hide="Solution"]
Let the 3-digit integer be a and b

Then $ 1000a\plus{}b\equal{}kab$

The only part which is maybe not divisible by $ a$ is $ b$, so $ a\mid b$. Let $ b\equal{}xa$

$ 1000a\plus{}xa\equal{}kxa^2$
$ 1000\plus{}x\equal{}kxa$

Observe $ x<10$, and $ x\mid 1000$ from the previous equation.

1) x=1 $ 1001\equal{}ka$, $ (a,b)\equal{}(143,143)$

2) x=2 $ 501\equal{}ka$, $ (a,b)\equal{}(167,334)$

3) x=5 $ 201\equal{}ka$, No solutions

Finnaly, 2 pairs...

[/hide]

[quote="admire9898"]A issue of number theory
for a digit problem
need a complete solution and a solving outline:

Two three-digit figures formed a six-digit figure(such as: Given 263,458,formed a 263458), and the six-digit figure equals to the Integer multiples of the product of the two three-digit integer figures. Find the six-digit figure.
I have found only one answer:167334=3*(167*334)
Anybody can get the complete answers? :)  :?:[/quote]

Let $ x$ be the first number and $ y$ be the second. The equation is $ kxy\equal{}1000x\plus{}y$

Let $ u\equal{}\gcd(x,y)$ and $ x\equal{}x_1u$ and $ y\equal{}y_1u$. The equation becomes $ kx_1y_1u\equal{}1000x_1\plus{}y_1$ And so $ x_1 | y_1$ and $ x_1\equal{}1$ and the equation is $ ky_1u\equal{}1000\plus{}y_1$

So $ y_1 | 1000$ and so $ y_1\in\{1,2,4,5,8,10,20,25,40,50,100,125,200,250,500,1000\}$. Each value gives an equation $ ku\equal{}\frac{1000}{y_1}\plus{}1$ and different values for $ u$ and we just have to cjeck.

$ y_1\equal{}1$ $ \implies$ $ ku\equal{}1001\equal{}7*143$ and $ u\in\{1, 7, 143, 1001\}$ and $ y\equal{}x\in\{1, 7, 143, 1001\}$ and so 3 solutions with 3 digits :
$ 1001*001*001 \equal{} 001001$
$ 143*007*007\equal{}007007$
$ 7*143*143\equal{}143143$

$ y_1\equal{}2$ $ \implies$ $ ku\equal{}501\equal{}3*167$ and $ x\equal{}u\in\{1, 3, 167, 501\}$ and $ y\equal{}2u\in\{2, 6, 334, 1002\}$ and so 3 solutions with 3 digits :
$ 501*001*002\equal{}001002$,
$ 167*003*006\equal{}003006$
$ 3*167*334\equal{}167334$

$ y_1\equal{}4$ $ \implies$ $ ku\equal{}251$ and $ x\equal{}u\in\{1, 251\}$ and $ y\equal{}4u\in\{4, 1004\}$ and so 1 solutions with 3 digits :
$ 251*001*004\equal{}001004$

$ y_1\equal{}5$ $ \implies$ $ ku\equal{}201\equal{}3*67$ and $ x\equal{}u\in\{1, 3, 67, 201\}$ and $ x\equal{}5u\in\{5, 15, 335, 1005\}$ and so 3 solutions with 3 digits :
$ 201*001*005\equal{}001005$
$ 67*003*015\equal{}003015$
$ 3*067*335\equal{}067335$

$ y_1\equal{}8$ $ \implies$ $ ku\equal{}126\equal{}2*3*3*7$ and $ x\equal{}u\in\{1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126 \}$ and $ y\equal{}8x\equal{}u\in\{8, 16, 24, 48, 56, 72, 112, 144, 168, 336, 504, 1008 \}$ and so 11 solutions with 3 digits :
$ 126*001*008\equal{}001008$
$ 63*002*016\equal{}002016$
$ 42*003*024\equal{}003024$
$ 21*006*048\equal{}006048$
$ 18*007*056\equal{}007056$
$ 14*009*072\equal{}009072$
$ 9*014*112\equal{}014112$
$ 7*018*144\equal{}018144$
$ 6*021*168\equal{}021168$
$ 3*042*336\equal{}042336$
$ 2*063*504\equal{}063504$

$ y_1\equal{}10$ $ \implies$ $ ku\equal{}101\equal{}1*101$ and $ x\equal{}u\in\{1,101\}$ and $ y\equal{}10u\in\{10,1010\}$ and so 1 solution with 3 digits :
$ 10*001*010\equal{}001010$

$ y_1\equal{}20$ $ \implies$ $ ku\equal{}51\equal{}3*17$ and $ x\equal{}u\in\{1, 3, 17, 51\}$ and $ y\equal{}20u\in\{20, 60, 340, 1020\}$ and so 3 solutions with 3 digits :
$ 51*001*020\equal{}001020$
$ 17*003*060\equal{}003060$
$ 3*017*340\equal{}017340$

$ y_1\equal{}25$ $ \implies$ $ ku\equal{}41\equal{}1*41$ and $ x\equal{}u\in\{1, 41\}$ and $ y\equal{}25u\in\{25, 1025\}$ and so 1 solution with 3 digits :
$ 41*001*025\equal{}001025$

$ y_1\equal{}40$ $ \implies$ $ ku\equal{}26\equal{}2*13$ and $ x\equal{}u\in\{1, 2, 13, 26\}$ and $ y\equal{}40u\in\{40, 80, 520, 1040\}$ and so 3 solutions with 3 digits :
$ 26*001*040\equal{}001040$
$ 13*002*080\equal{}002080$
$ 2*013*520\equal{}013520$

$ y_1\equal{}50$ $ \implies$ $ ku\equal{}21\equal{}3*7$ and $ x\equal{}u\in\{1, 3, 7, 21\}$ and $ y\equal{}50u\in\{50, 150, 350, 1050\}$ and so 3 solutions with 3 digits :
$ 21*001*050\equal{}001050$
$ 7*003*150\equal{}003150$
$ 3*007*350\equal{}007350$

$ y_1\equal{}100$ $ \implies$ $ ku\equal{}11$ and $ x\equal{}u\in\{1, 11\}$ and $ y\equal{}100u\in\{100, 1100\}$ and so 1 solution with 3 digits :
$ 11*001*100\equal{}001100$

$ y_1\equal{}125$ $ \implies$ $ ku\equal{}9\equal{}3*3$ and $ x\equal{}u\in\{1, 3, 9\}$ and $ y\equal{}125u\in\{125, 375, 1125\}$ and so 2 solutions with 3 digits :
$ 9*001*125\equal{}001125$
$ 3*003*375\equal{}003375$

$ y_1\equal{}200$ $ \implies$ $ ku\equal{}6\equal{}2*3$ and $ x\equal{}u\in\{1, 2, 3, 6\}$ and $ y\equal{}200u\in\{200, 400, 600, 1200\}$ and so 3 solutions with 3 digits :
$ 6*001*200\equal{}001200$
$ 3*002*400\equal{}002400$
$ 2*003*600\equal{}003600$

$ y_1\equal{}250$ $ \implies$ $ ku\equal{}5$ and $ x\equal{}u\in\{1, 5\}$ and $ y\equal{}250u\in\{250, 1250\}$ and so 1 solution with 3 digits :
$ 5*001*250\equal{}001250$

$ y_1\equal{}500$ $ \implies$ $ ku\equal{}3$ and $ x\equal{}u\in\{1, 3\}$ and $ y\equal{}500u\in\{500, 1500\}$ and so 1 solution with 3 digits :
$ 3*001*500\equal{}001500$

$ y_1\equal{}1000$ $ \implies$ $ ku\equal{}2$ and $ x\equal{}u\in\{1, 2\}$ and $ y\equal{}500u\in\{1000, 2000\}$ and so no solution with 3 digits :

[quote="pco"][quote="Farenhajt"]@pco: Not at all sure things like $ \overline{00a}$ and $ \overline{0ab}$ are considered legitimate three-digit numbers.[/quote]

OP used "three digits figures", not "three digits numbers", so ... .[/quote]

Which could very well be a language issue. (The OP is not a native english speaker, that much can be gathered from the post.)

Farnehajt is right...

Issue in number theory isn't a common expression, as well as figures instead of numbers...

[quote="admire9898"]I'm a asian,so not a native english speaker.Sorry for the expression.
the problem(not a issue :P ) is not considered for the number(not the figure :P ) such as 001 or 010,just 【100，999】
So，thanks to all of you to give me the answer in time and help me immediately[/quote]

So I did useless work ...  :(

No,your work does not useless.If next time,the problem change,your work can be the standard answer.


What is the least number of coins you can have and still pay exactly the cost of any purchase less than one dollar?

I do not understand what he is saying in this problem. Am I looking at it the wrong way, or is it worded wrong? :?:  :?:

It's saying what is the least number of coins you can have so you can form any money value through 99 cents. I hope this makes sense...

So far, I'm getting 1 nickel, 2 dimes, 3 quarters, and 4 pennies, for a total of 10 coins. Can anyone find a more efficient combination? Are half-dollars allowed?

The question is saying that if you pick a number of coins out of your total coins, you can make any total less than 1.00. So we would definitely need 4 pennies because if there was a price $ .44$, then we would need those 4 pennies. We would also need a nickel for $ .49$. And we would also need 3 quarters. And last we need 2 dimes.

@Thinkflow: Good job!

As a final note, if half-dollars are allowed, we can reduce our coin count by one, since we can then replace our three quarters with one half-dollar and one quarter.

Umm yes the answer is 9 because we are allowed to use half dollars (according to FTW)


Let $ n\in\mathbb N,$ $ n\geq2$ and $ a_1,$ $ a_2,...$  $ a_n$ are non-negative numbers, $ a_{n \plus{} 1} \equal{} a_1.$ Prove that:
\[ \prod_{i \equal{} 1}^n(a_i \plus{} a_{i \plus{} 1})(1 \plus{} a_i)\geq \prod_{i \equal{} 1}^na_i\left(2 \plus{} a_i \plus{} a_{i \plus{} 1}\right)
\]
It's a generalization of
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=220496

Multiplying the inequalities
$ (1\plus{}\frac 1{a_i})(1\plus{}\frac 1{a_{i\plus{}1}}) \ge (1\plus{}\frac 2 {a_i\plus{}a_{i\plus{}1}})^2$
for $ i\equal{}1,2,...,n$, we get the desired inequality.

[quote="Vasc"]Multiplying the inequalities
$ (1 \plus{} \frac 1{a_i})(1 \plus{} \frac 1{a_{i \plus{} 1}}) \ge (1 \plus{} \frac 2 {a_i \plus{} a_{i \plus{} 1}})^2$
for $ i \equal{} 1,2,...,n$, we get the desired inequality.[/quote]
Yes!  :lol: It's also my proof.
The following proves the Mongolian inequality for $ n\equal{}5.$
Let $ a$ and $ b$ are positive numbers such that $ a\plus{}b\leq1.$ Prove that:
$ \left(1\minus{}\frac{1}{a}\right)\left(1\minus{}\frac{1}{b}\right)\geq\left(1\minus{}\frac{2}{a\plus{}b}\right)^2.$

[quote="Sunkern_sunflora"]I think arqady your last ineq follows from expanding.

since $ a \plus{} b \leq 1$, we have,

$ \Leftrightarrow \frac {1}{ab} \minus{} \left(\frac {1}{a} \plus{} \frac {1}{b}\right) \geq \frac {4}{(a \plus{} b)^2} \minus{} \frac {4}{a \plus{} b}$

$ \Leftrightarrow (a \plus{} b)^2 \minus{} (a \plus{} b)^3 \geq{ 4ab \minus{} 4ab(a \plus{} b)}$

$ \Leftrightarrow (a \plus{} b)^2 \geq {4ab}$...which is obvious!

Am I right?
:)[/quote]
You are right! Now you can  prove the privet case of Mongolian inequality:

Let $ a,$ $ b,$ $ c,$ $ d$ and $ e$ are positive numbers. Prove that:
\[ \prod_{cyc}\frac {a \plus{} b \plus{} c}{3}\geq \prod_{cyc}\frac {a \plus{} b}{2}
\]

Hello arqady! I will try the privet part u posted here.

Tell me one thing. Let there be two set of positive real numbers $ a_{i}$'s and $ b_{i}$'s, $ i \equal{} 1(1)n$. A.M. of the first set is $ A_{1}$ and G.M. is $ G_{1}$ and that for the second set is $ A_{2}$ and $ G_{2}$. If we see that $ A_{1} \geq A_{2}$ holds always, can we conclude that $ G_{1} \geq G_{2}$ will also hold always, or there will be some conditions? Thank u.
 :)

[quote="Sunkern_sunflora"]Let there be two set of positive real numbers $ a_{i}$'s and $ b_{i}$'s, $ i \equal{} 1(1)n$. A.M. of the first set is $ A_{1}$ and G.M. is $ G_{1}$ and that for the second set is $ A_{2}$ and $ G_{2}$. If we see that $ A_{1} \geq A_{2}$ holds always, can we conclude that $ G_{1} \geq G_{2}$ will also hold always, or there will be some conditions? Thank u.
 :)[/quote]
It's wrong. $ \frac{1\plus{}3}{2}\geq\frac{2\plus{}2}{2}$ is true but $ \sqrt{1\cdot3}\geq\sqrt{2\cdot2}$ is wrong.
By the way $ \frac{2\plus{}2}{2}\geq\frac{1\plus{}3}{2}$ is true but $ \sqrt{2\cdot2}\geq \sqrt{1\cdot3}$ is true.

Ok, let us apply some conditions now. All the $ a_{i}$'s and $ b_{i}$'s can be arranged in a cyclic sequence. Does it hold now?
I am suffering from high fever, so plz don't mind if I post something foolish, Because it seems to me that this things will hold under some conditions.


function
logarithms
complex analysis
algebra
domain
complex analysis unsolved

I was thinking about the functions

$f_n: \frac{1}{x}, \frac{1}{x\log(x)}, \frac{1}{x\log(x)\log(\log(x))}, ...$

and

$g_n: \frac{1}{x^2}, \frac{1}{x\log(x)^2}, \frac{1}{x\log(x)\log(\log(x))^2}, ...$

Each $f_n$ is not integrable on the real line where it is defined, and each $g_n$ is. But obviously, neither series tends to a limit that is defined from $R$ to $R$. But what about the complex? What do these functions look like as functions from $R$ to $C$? I guess that's a pretty vague question. I mean, do they converge to the same limit? Is this limit holomorphic? Analytic? What else might you say about it?

I'm sorry for the imprecision and noobery; I don't really know complex analysis but for a few basic things. But I was very interested in these two sequences this afternoon and I'd like to see what people think of them.

Thanks!

[quote="Xevarion"]I was thinking about the functions

$f_n: \frac{1}{x}, \frac{1}{x\log(x)}, \frac{1}{x\log(x)\log(\log(x))}, ...$
[/quote]
These functions have domains that are contained in $(0,\infty), \, (1,\infty), \, (e,\infty), \, (e^e,\infty), \, (e^{e^e},\infty)$ and so on. Since the intersection of their domains is empty.we can't talk about their limit.

Well on the real line, sure. But you can extend the logarithm to the slit plane (let's do the normal one where we exclude the negative real axis), and then the functions should all be defined everywhere but for finitely many points (the power towers of $e$ that aren't tall enough). Or am I wrong?


calculus
integration
derivative
function
calculus computations

Well for part a, we can just plug in 0 so the problem is $ \int_{\minus{}3}^{0}f(t)\mbox{ d}t$, which is whatever it is (sorry, too lazy to evaluate).  For b, take the derivative of the function at 0, which means $ \frac{d}{dt}\int_{\minus{}3}^{x}f(t)\mbox{ d}t$, which, by the Fundamental Theorem, is just $ f(x)$ for $ x\equal{}0$, so $ f(0)\equal{}1$.

are you saying that

g(0) = 0

and

g'(0) = 1

?

I don't think $ g(o)\equal{}1$.  I think it equals that integral which is 4.5 (check me on that).  And yes, I do believe $ g'(0)\equal{}1$.  Do you understand what I did in solving the problems?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Show that the multiplicative group of (Z/2^nZ) is not cyclic for n >2. Hint: find two distinct subgroups of order 2.

there are exactly three elements of order $2$:

$-1 , 2^{n-1}\pm 1$