Let a_1 \leq a_2 \leq a_3 \leq ... \leq a_n be n positive reals(n \geq2) such that

  sum(i=1,..,n)a_ia_(i+1)=1   (a_(n+1)=a_1)

Determine the minimum of sum(i=1,...,n)a_i

This problem is very hard and I have never seen a solution.

Just a quick thought, since I have to leave for three days.I know that it is not very difficult (but not trivial at all) to prove that for all x_1,...,x_n>=0 we have (x_1+x_2+...+x_n)^2>=4(x_1*x_2+x_2*x_3+...+x_n*x_1). So we must have sum a_i*a_i+1<=1/4. Moreover , if we take a_n=a_n-1=1/2 and the rest of them very close to 0, sum a_i*a_i+1 tends to 1/4. So I believe that if a_i>0 there doesn't exist a maximum value, but a supremmum , which is 1/4. I really don't have the enough time to check myself.


geometry
3D geometry
LaTeX
number theory unsolved
number theory

find all the pairs of integers$\left( x,y\right) $ satisfiing :$y^{2}=x^{3}+16$

Well factorizing we have: (y+4)(y-4) is the cube of an integer. but mcd of (y+4,y-4) is 1,2,4 or 8 so we divide the problem in 4 cases:
mcd=1 wich has no solution becuse the diference of two cubes is at its smallest 7 ie (1,8) but then the next diference is 19 so no solution .
and so on i have a solution but i dont know latex and to write it would be impossible so i put the part i kan in hopes for somone to kontinue.

[quote="marko avila"]so the solution is not so impossible to write anyway divide the problem in four cases:
case1
mcd= 1 we saw before implies the diference of two cubes is 8 which is a contradiction.
case2
mcd=2 wich implies the diference of two cubes is 4 which is a contradiction.
case3
mcd=4 wich implies the diference of two cubes is wich is a contradiction
case4
mcd=8 wich implies the diference of two cubes is 1 and so 0 and 1 are the only ones with that diference and hence only solution y=4 and x=0.

so there we you have it. :lol:[/quote]

Uh no. How about: $y-4 = 2a^3$, $y+4 = 4b^3$, where $a$ is odd?

See  no because y+4=4b3 and y-4=2a3 (the 3 at the end is the exponent) implies that :

4b3-2a3=8 hence dividing by 2 we have: 2b3-a3=4 this forces a to be even wich contradicts your afirmation. so i think im right. :blush:

[quote="marko avila"]another way too see your wrong in kase you didnt understand the last one is the following: since you say y+4=4b3(again the three is the exponent ) this implies that y is a multiple of 4. hence y-4=2a3 is also a multiple of 4 and hence a has to be even.[/quote]

Now that you said it, so what if a is even? Note I am not saying that what you are saying is wrong, I am just saying that it is incomplete and you need some more things to complete your proof (=

ok i see what your saying. so i have kompleted it. here goes:
well since we have that a is even well this forces y=4(mod16) hence 4b3=y+4=8(mod16) but then this forces b to be even and hence y+4 is a multiple of 64  but then mcd(y+4,y-4)=16 wich is a contradiction¡¡¡ so there you go¡¡¡ :lol:


Let $a,b,c,d \in R$ so that $a+c=1$ and $b+d=1$.
Show that $a^{2}b+c^{2}d \geq bd$ and $ab^{2}+cd^{2} \geq ac$.


function
calculus
derivative
calculus computations

consider the family of functions defined by f(x)= bxe^(bx), where b is a non zero constant. show that the absolute minimum value of bxe^(bx) is the same for all nonzero values of b

Note that $ f'(x) \equal{} be^{bx} [1 \plus{} bx]$, which is 0 when $ x \equal{} \frac{\minus{}1}{b}$ and this value is a minima since the second derivative is positive at this value. Thus the minimum is given by:
$ f(\frac{\minus{}1}{b}) \equal{} \minus{} 1$, for any nonzero $ b$.

can you just post the second derivative i'm a bit confused still


Given: a,b,c>0
2a+4b+7c<=2abc
*:<= less or equal to
what is the minimum value of a+b+c?

I'm probably wrong, but here goes:

[hide]

[/hide]

Very nice application of AM-GM.

However, I am not sure it is a minimum when $ 2abc \equal{} 2*\sqrt [3] (189abc)$.

Hmm. I figured that would minimize $ abc$, which would in turn minimize $ a\plus{}b\plus{}c$. It's not rigorous, but it feels right lol.


function
real analysis
real analysis unsolved

Let $f$ be a bounded function on $(a,b)$ such that
$f(\frac{x+y}2)\leq \frac{f(x)+f(y)}2.$
Show that $f$ is continuous.

suppose that $f$ is not continuous at $x_{0}\in (a,b)$. One can find a sequence $(x_{n}) \in (a,b)$ and $\epsilon > 0$ such that $x_{n}$ tends to $x_{0}$ and $|f(x_{n})-f(x_{0})| > \epsilon$. WLOG, $(a,b)=(-1;1)$, $x_{0}=0$, $|x_{n}| < \frac{1}{n}$, $f(x_{n})>f(0)+\epsilon$. This is why
$f(n x_{n}) \geq n f(x_{n}) > n \epsilon$. This shows that $f$ is unbounded on $(-1;+1)$.

More general , let $J$ be an open interval at left. I will suppose only finite superior bound for $f: J\rightarrow R$ . Let $x_{0}$ be an element of $J$
For a convergent sequence $X=\{x_{n}\}\in J,\ x_{n}\rightarrow x_{0}$ , let $F_{X}$ be the adherance (set of the finite or infinite limit points) of $f(x_{n})$.
Let $F_{x_{0}}$ be the set $F_{x_{0}}= \cup F_{X}$ . Because $f$ is superior bounded, then there is finite $S=\sup (F_{x_{0}})$ , so there is $u_{n}\rightarrow x_{0}$ such that $f(u_{n})\rightarrow S$
The initial relation ships gives : $f(u(n))\leq \frac{f(x_{0})+f(2u_{n}-x0)}2\Longrightarrow 2f(u_{n})-f(2u_{n}-x0)\leq f(x_{0})$, and that gives  $S \leq f(x_{0})$
Let $I$ be the  inferior (finite or infinite) bound $I = inf (F_{x_{0}})$, so there is $v_{n}\rightarrow x_{0}$ such that $f(v_{n})\rightarrow I\leq S\leq f(x_{0})$
The initial relation ships gives : $f(x_{0})\leq \frac{f(v_{n})+f(2x0-v_{n})}2\Longrightarrow 2f(x_{0})-f(2x0-v_{n})\leq f(v_{n})$, and that gives  $f(x_{0})\leq I$
Finally $S=I=f(x_{0})$, so $f$ is continue .
Do you see when I needed the choice axiom ?
 :cool:

[quote="Diogene"]
The initial relation ships gives : $f(x_{0})\leq \frac{f(v_{n})+f(2x0-v_{n})}2\Longrightarrow 2f(x_{0})-f(2x0-v_{n})\leq f(v_{n})$, and that gives  $f(x_{0})\leq I$

 :cool:[/quote]
how can you get:$f(x_{0})\leq I$?
I think we can only get:$f(x_{0})\leq \frac{S+I}{2}$
But this doesn't matter.


Determine the minimum value of $ \frac{x^4\plus{}x^2\plus{}1}{x^2\plus{}x\plus{}1}$ for $ x\in \mathbb{R}$ .

[hide="Using Truffles hint"]$ x^4\plus{}x^2\plus{}1 \equal{} (x^2\minus{}x\plus{}1)(x^2\plus{}x\plus{}1)$
So we simply need to find the minimum value of:
$ x^2\minus{}x\plus{}1 \equal{} \left(x\minus{}\frac{1}{2}\right)^2 \plus{} \frac{3}{4} \geq \frac{3}{4}$.
[/hide]


logarithms
real analysis
calculus
calculus computations

Fix $ b > 1, y > 0$, and prove that there is a unique real $ x$ such that $ b^x = y$. I don't follow Rudin's argument on p.22 problem 7 (Principles of Mathematical Analysis).

I don't have the book.  Could you by any chance post his argument or the part you don't understand so we can help you with it?

(a) For any positive integer $ n$, $ b^{n}\minus{}1 \geq n(b\minus{}1)$.

(b) Hence $ b\minus{}1 \geq n(b^{1/n}\minus{}1)$.

(c) If $ t >1$ and $ n > (b\minus{}1)/(t\minus{}1)$, then $ b^{1/n} < t$.

(d) If $ w$ is such that $ b^w < y$, then $ b^{w\plus{} (1/n)} < y$ for sufficiently large $ n$.

(e) If $ b^w > y$, then $ b^{w\minus{}(1/n)} > y$ for sufficiently large $ n$.

(f) Let $ A$ be the set of all $ w$ such that $ b^w < y$, and show that $ x \equal{} \sup A$ satisfies $ b^x \equal{} y$.

(g) Prove that $ x$ is unique.

I don't understand the point of steps (a) and (b). Is it to show that it is monotonically increasing? Is this the only way to prove this?


Let $ d$ be a positive integer.Integers $ t_1,t_2,...,t_d$ and real numbers that $ a_1,a_2,...,a_d$ are given such that $ a_{1}t_{1}^{j}\plus{}a_{2}t_{2}^{j}\plus{}...\plus{}a_{d}t_{d}^{j}$ is an integer for all integers $ j$ with $ 0 \leq j < d$.Prove that $ a_{1}t_{1}^{d}\plus{}a_{2}t_{2}^{d}\plus{}...\plus{}a_{d}t_{d}^{d}$ is also an integer.


What percent of this circle is shaded? Express your answer as a mixed number.

i got that question on FTW....  IT DOESN"T SHOW THE PICTURE SO YOU CANT DO IT!!!!!!
[hide="don't"] [hide="look"] THe answer was 331/3 [/hide] [/hide]

Well, if the answer is $33\frac{1}{3}\%$, then $\frac{1}{3}$ of the circle was shaded.


advanced fields
advanced fields unsolved

let $ (X,\sigma)$ be a measurable space $ (\{0,1\}^{\aleph_0},2^X)$.
let $ (Y,\Sigma)$ be a product of $ \aleph_0$ spaces $ (X,\sigma)$.

obviously $ (X,\sigma)$ is finitely generated.  what about $ (Y,\Sigma)$ ?


trigonometry
calculus
LaTeX
calculus computations

hello

well I basically need problems in differential equations, I already have trouble with these...

1)
for which positive values of the constant $k$ does the differential equation $y''+ky=0$ have solutions which satisfy $y(0)=y(\pi)=0$. Determine all such solutions.

2)
Solve $f'(0)+f(1)$ when you know that $f'(1)+f(0)=1$ and that for every value of $x$ you have $f'(x)+f(x)=0$}

3)
A thumb rule  in population approximations is that if the growth is $p \%$ a year the time when the population is dubbeld is $\frac{70}{p}$ years.

a) the rising of the population can be described with a mathematical model in the form of a differential equation. Produce such a model.

b) use your model and show that rule is true

Any ideas about these problems Salah?  I'd like to hear how you've tried to solve them.

1) Find a general solution in terms of $k$, and then see what you need $k$ to be to satisfy those conditions.

2) What particular values of $x$ might be useful in $f'(x) + f(x) = 0$ to help you figure out what you want? How can you use those?

3) Hint: More people can have more children. :D

well

on the first one I suppose I would transform the general statement $y''+ky=0$  into $x''=-k$ hence:
$x=\pm i\sqrt{k}$ this would imply that the equation would be of a complex, dubble root  and so we have;$y=Acos\sqrt{k}x+Bsin\sqrt{k}x$.

Then; $y(0)=0 \Rightarrow y(0)=Acos(0)+Bsin(0) \Rightarrow A=0$; since $cos(0)=1$ and then we have the other one,

$y(\pi)=0 \Rightarrow y(\pi)=Bsin \sqrt{k}\pi=0$ This cant be right since one could just set $B=0$ and be done with it, so_? I suppose It might be $y(0)=y'(\pi)=0$ ill check on it later..

on the second one I have no clue however on the third one I suppose that one should somehow form an idea of what the formula must be able to do and be compatible with the rule which they describe. I hade a discussion with my teacher about this particular problem and even though he went about it almost an hour or so he could give me a straight answer. Well he said that its intendend to be a little more demanding problem where the student would have the opportunity to demonstrate his analytic abillities.

I remeber something about that one should first understand that its an exponential growth of the type similar to
$y=C()^{kx}$ and that $\frac{70}{p} \Rightarrow$ (somehow dont remember) $\Rightarrow \frac{p}{100}$ since its a procentual increase and by some logic the factor of 70 would be done away with because its an differential equation and so it changes by input of the other variables.

I need help  :(

1)  You have a good start on this problem.

Just because $B\sin(\sqrt{k}\pi)=0$, that doesn't necessarily mean that $B$ must be equal to $0$.  There are certain values of k for which $\sin(\sqrt{k}\pi)=0$.

2)  $f'(x) + f(x)=0$ is a first-order linear ODE, arguably easier to solve than the second-order linear ODE that you managed to solve in your partial solution of problem # 1.  Can you find a general solution to the differential equation $f'(x) + f(x)=0$?  That would be a good start.

3) To elaborate on Xevarion's hint from earlier... If a population were twice as large, then its growth rate would double.  If a population were three times as large, then its growth rate would triple.  (Okay, these statements probably are not true in the real world.  In the real world, if a population doubles, then famine may occur, which may keep the growth rate from doubling.  But if we are trying to make a simple model of population growth, then we can start by accepting these assumptions for the time being.)  Perhaps you could let $N(t)$ be the population at time t, and then write down an equation that expresses the idea that I stated above.

1) What Aunt Sally said is right. You're almost done.

2) Aunt Sally's suggestion of solving the differential equation will work. It is a pretty easy one. But you can actually solve this as an algebra problem, not as a calculus problem. It's probably good to see both solutions. Try to do it her way, and then see if you can figure out what I'm talking about. Consider $x=0$ and $x=1$ in the differential equation they give you...

3) It isn't actually exactly 70. 70 is rounded. Hint: $100\ln(2) \approx 69.3$. Where does that come from? Follow Aunt Sally's suggestion to make a differential equation for population growth. Solve it, and then you'll be able to see how to get that relation.

hi

1)  $B\sin(\sqrt{k}\pi)=0 \Rightarrow B=0$ or
$k_1=0\vee k_2=1$
which would make it to
$B\sin(0 )=0 \vee B\sin(\pi)=0$ I guess I got stuck here because I thought that I had to find a pair of solutions for every k,B

2)$f'(x)+f(x)=0 \Rightarrow f(x)=Ce^{kx}$and $f'(x)=kCe^{kx}$
so k has to be negative?...$k=-1$
$funct: Ce^{-x}-Ce^{-x}=0$
and so $f'(1)+f(0)=1 \Rightarrow Ce^{0}-Ce^{-1}=1 \Rightarrow C- \frac{C}{e}=1$ which would give  $C=\frac{e}{e-1}$
and then to calculate $f'(0)+f(1)=\frac{e^2}{e-1}-\frac{e}{e-1}= \frac{e^2-e}{e-1} = \frac{e(e-1)}{e-1}= e$
Is that right?

Ill try the third one some other time, this Latex stuff is kind of tricky, thanks for the help though, I would appreciate one thing ([i]how does one hide the solution in Latex?[/i] )

(2) $f'(0) + f(0) = 0$ and $f'(1) + f(1) = 0$ and $f'(1) + f(0) = 1$ are given. Take the second, subtract the third, and add the first, and we get $f(1) + f'(0) = -1$.
Either you or I did something wrong, but I can't see any mistake in either your work or mine... Probably it is something obvious....

It's that calculation of $f(1)$. $f(1)=\frac{1}{e-1}$, not $\frac{e^2}{e-1}$.

There are actually infinitely many positive values of $k$ such that $\sin(\sqrt{k}\pi)=0$.

(I was posting as Aunt Sally before, but nobody seemed to catch the reference...)

yeah I see my err, I wrote $Ce^{1} \Rightarrow Ce$  instead of  $Ce^{-1} \Rightarrow \frac{C}{e}$
anyways thats correct Xevarion, nice method too, thats the algebra you were talking about I guess.

please someone. show me how to hide the latex :huh:

It is just as follows

[ hide = short text ] Your code and whatever [ / hide ]

But with no spaces.

Example:

[hide="Example"] I hid this text! [/hide]

I hope that's what you were wondering. :-)


vector
geometry
3D geometry
sphere
analytic geometry
parameterization
calculus

I have an exam in vector analysis in January and I have some questions about this subject.

When I have to calculate the flux through a surface it's often useful to use the divergence theorem and then subtract the flux through the surfaces not included in the area I'm looking at. However it's quite often hard for me to find a way to parametrize these surfaces. For example I have this one problem where I'm supposed to find the flux through the boundary surface of the solid given by x^2+y^2+z^2=4a^2 and with the restriction x^2+y^2>a^2 (that is: It's a sphere with a cylindrical hole in it). To find the flux through the boundary surface I apply the divergence theorem to the whole sphere, but when I have to subtract the curved ends of the cylinder I'm lost. How can I find an equation for these curved ends? In my case the vectorfield is F(x,y,z)= (x+yz)i+(y-zx)j+(z^2-2z+e^x*sin(y))k, but I'm most interested in a way to parametrize the surface. (it would be wise to change to cylinder coordinates I'm sure)
In general: What is the magic of finding the easiest way to parametrize a surface?
Anyone know a good website/book about parametrization of surfaces in R^3 or with a lot of examples on vector analysis?

Don't use the curved ends- use something else with the same boundary, like a pair of disks.

If you do this (and the divergence isn't zero), use cylindrical coordinates for the divergence integral.

The other possibility is parametrizing the sphere and the part we're working with; the limits at least work out nicely in the standard coordinates $ \theta$ and $ \phi$. ($ \frac {\pi}6\le \phi\le \frac {5\pi}{6}$)

[Edit- this is based on a different interpretation of the question, leaving out the inner cylindrical part]

Parametrization of surfaces 101: (1) write the equation of the surface in an appropriate system of coordinates; (2) solve this equation for one of three coordinates; (3) call the other two coordinates "parameters"; (4) identify the possible values of those parameters at the points of the surface ("parameter domain").

However, I don't know if you really need to parametrize. If you are "supposed to find the flux through the boundary surface of the solid ... ", does not that mean the entire surface of the solid? It's not clear to me what "and with the restriction..." actually means: it could either define the solid or separate a part of its boundary.

[quote]Don't use the curved ends- use something else with the same boundary, like a pair of disks. [/quote]

But how can I be sure that the flux through the two disks is the same as through the curved ends? They do have the same boundary surface, but according to stokes theorem that only means that the flux of the curl of F is equal.

[quote]
However, I don't know if you really need to parametrize. If you are "supposed to find the flux through the boundary surface of the solid ... ", does not that mean the entire surface of the solid? It's not clear to me what "and with the restriction..." actually means: it could either define the solid or separate a part of its boundary.[/quote]

Sorry for not being specific enough. Problem is to find the flux of the sphere (radius 2a) with a cylindrical hole through it's center (radius a))

It isn't the same; that choice also comes with a different region to apply the divergence theorem to. Using spherical caps is just as hard as doing the original integral directly.

Thanks for all the good answers. I managed to do it by shifting to polar coordinates and integrating up from the bottom of the cylinder to the top. Wasn't that hard once you got it fully visualized in your head :) It all summed up to zero for the whole sphere with hole.

Yet I now how a new problem (still the same exerscise). I now need to find the flux through the cylinder part. It's terrible simple but I just can't seem to get it right:
Just to eliminate doubt I'll try to be more specific about the exercise:
Let E be solid defined by x^2+y^2+z^2<4a^2 and x^2+y^2>a^2
Let F be the vectorfield defined by:
F(x,y,z)= (x+yz)î+(y-zx)j+(z^2-2z+e^x*sin(y))k
Find the flux through the cylindrical part of the surface of E.
My idea is this:
First I find the parametrization of the cylindrical surface, r(t)=(a*cos(t),a*sin(t),z). Then I take the cross product between r differentiated with respect to t and z - this gives me r'(t)=(-a*sin(t),a*cos(t),0).
Then I calculate the double integral over F(r(t)) dot r('t) dA where dA is dt*dz. This seems to give me some constants multiplied by a^3. The limits on t is 0 to 2 Pi and the limits on z is -sqrt(3)*a to sqrt(3)*a. I know that the correct answer is the constants multiplied by a^2. Where is my error? (forgive the sloppy notation but I haven't learned LaTeX yet, it's on my to do list)

I think the flux should be proportional to $ a^3$. Indeed, the area of the surface is proportional to $ a^2$, and the field contributes another $ a$ from the $ x\vec i\plus{}y\vec j$ term.

That sounds reasonable but being a physics student I just figured that since flux is always measuered in square meters it should be a^2. It's also in the solution manual, but maybe my instructor just got it wrong?

I'm not a physics student, but I think that the flux of $ \vec F$ is measured in $ (\textrm{unit of }\vec F) \cdot (\textrm{unit of length})^2$.


inequalities
number theory
Diophantine equation
number theory proposed

Find $n$ and $m$ positive integers such that \[n!+1 = (m!-1)^{2}.\] :wink:

Rearrange to obtain $n!=m!(m!-2)$ so $m\geq 3$

We can spot the case $(n,m)=(4,3)$ relatively easily.

Now note that $3|m!$, so $3\not| m!-2$, so if $3^{c}\parallel n!$, $3^{c}\parallel m!$.

Thus $m \geq (n-2)$, so $n!=m!(m!-2)\geq (n-2)!((n-2)!-2)$
So $n(n-1) \geq (n-2)!-2$, which implies $n<7$.

Checking cases reveals no more solutions.

Thanks for the nice problem :).

This equation is equivalent to

$(1)\;\; n! \;=\; m!(m!-2)$

Hence $m! > 2 = 2!,$ so $m \geq 3$. Equation (1) implies $m \neq n$ and $m!\,|\,n!,$ giving $n > m$. Thus $n = m+k$ where

$k$ is positive integer. In other words,

$(2) \;\; \prod_{i=1}^{k}m+i \;=\; m!-2.$

according to (1). Assume $k \geq 3$. Then ${\textstyle \frac{n!}{m!}}$ is divisible by $(m+1)(m+2)(m+3)$, which itself is

divisible by 3, hence ${\textstyle 3 \,|\, \frac{n!}{m!}}$. Furthermore $3\,|\,m!$ since $m \geq 3,$ hence $3\,|\,2$ by (2).

This contradiction gives $k \leq 2,$ so

$(3) \;\; m!-2 \; \leq \; m(m+1)(m+2)$.

acording to (2). The inequality (3) imply $m \leq 5$. Trying with $m = 3, 4, 5$ in (1), we find only one solution of the

Diophantine equation (1), namely $(m,n) = (3,4).$

Find n and m positive integers such that n! + 1 = m^2


AMC
USA(J)MO
USAMO
floor function
inequalities
induction
strong induction

[b]Problem)[/b] Prove that \[\sum_{k=1}^{n}\frac{\lfloor kx\rfloor}{k}\leq\lfloor nx\rfloor\] [i]Remark: The solutions are as nice as the statment.  The solution at Kalva starts off by noting that it suffices to prove the inequality only for rational $x=\frac{a}{b}$ where $b\leq n$.  Then it finishes off with some number theorey and the rearrangement inequality.  Though induction seems futile, one of my professors came up with the following slick inductive solution.[/i]

[b]Solution)[/b] [hide]Let $G_{n}(x)$ denote the sum.  Add the equations

\begin{eqnarray*}G_{1}(x)&=&\lfloor x\rfloor\\ \ 2(G_{2}(x)-G_{1}(x))&=&\lfloor 2x\rfloor\\ \ \vdots~~~~~~~~~~&=&~~~\vdots\\ \ n(G_{n}(x)-G_{n-1}(x))&=&\lfloor nx\rfloor \end{eqnarray*}

to get

\begin{eqnarray*}nG_{n}(x)&=&\sum_{k=1}^{n-1}G_{k}(x)+\sum_{k=1}^{n}\lfloor kx\rfloor\\ \ &\leq&\sum_{k=1}^{n-1}\lfloor kx\rfloor+\sum_{k=1}^{n}\lfloor kx\rfloor~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(\mbox{strong induction})\\ \ &=&\sum_{k=1}^{n}\Bigl(\lfloor kx\rfloor+\lfloor (n-k)x\rfloor\Bigr)\\ \ &\leq&n\lfloor nx\rfloor~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(\mbox{since }\lfloor a\rfloor+\lfloor b\rfloor\leq\lfloor a+b\rfloor\mbox{)}\end{eqnarray*}

which gives the desired result.[/hide]

That's a really amazing inductive solution :clap2:


Let $ N$ be a number composed of 1998 consecutive 1's. Find the 1000th  digit after the period in the decimal expansion of $ \sqrt{N}$.

No, it is not 3. Perhaps it is better for all of us If you post the whole solution.

we're looking for $ \sqrt{\frac {10^{1988} \minus{} 1}{9}} \equal{} \frac13 \sqrt {10^{1988} \minus{} 1} \equal{} \frac {10^{999}}{3} \minus{} \frac {1}{3(10^{999} \plus{} \sqrt {10^{1988} \minus{} 1})}$
multiply everything by 1000 and take mod 10
we want the unit digit of $ \frac {10}{3} \minus{} 1,6666... \equal{} 1,666$  so the answer is 1 (the first non-3 digit.)
:P i was off by a factor of 10 in the second fraction


Use AM-GM with $2a,2a,2a,3b,3b$.  :wink:

[quote="N.T.TUAN"][quote="Sunjee"][quote="mayhem"]$a,b$ are positive reel numbers and $a+b=5$. Find maximum value of $a^{3}.b^{2}$.[/quote]
i think $max a^{3}b^{2}=3^{5}$[/quote]
If so, $6=5$ :P[/quote]
sorry mistyped


inequalities
floor function
linear algebra
matrix
real analysis
real analysis unsolved

Let $ m\equal{}\Big\lfloor\frac{1}{2}\left(\frac{2}{\sqrt{3}}\right)^n\Big\rfloor$.

Is this inequality completely obvious to anyone? It went without mention...

$ \left(\begin{matrix} 2m\\ 3\end{matrix}\right)3\left(\frac{3}{4}\right)^n \leq m$

Here was my guess approach:

$ \left(\begin{matrix} 2m\\ 3\end{matrix}\right)3\left(\frac{3}{4}\right)^n \leq
\frac{(2m)^3}{2}\left(\frac{3}{4}\right)^n \leq \frac{4}{2^3}\left(\frac{2}{\sqrt{3}}\right)^{3n}\left(\frac{3}{4}\right)^n \leq \cdots \leq m$

Where I ended didn't seem to be simplifying in the right direction. I think it went too far. Also, the first step I did came from $ \left(\begin{matrix} n \\ k\end{matrix}\right) \leq \frac{n^k}{k!}$

The inequality is equivalent to $ (m\minus{}1)(m\minus{}2) \leq \left(\frac{4}{3}\right)^n$
and estimating left hand side by $ m^2$ is good enough.

Should that not be $ (2m\minus{}1)(2m\minus{}2)\leq(4/3)^n$?

[quote="ggbg90"]Should that not be $ (2m \minus{} 1)(2m \minus{} 2)\leq(4/3)^n$?[/quote]

Yes, sorry.  :blush:


vector
geometry
complex numbers
geometry theorems

Dear friends,

I am preparing some geometry problems in which we could using
vector method to solve them ie... Really i got about 30 simple problems and very few nice problems or hard problems[Special i do not have much such problems which appear in Olympiads].If you have any exprience about this please let me know!
 Thanks in advanced!
 Yours,

Here are two problems:

1. Prove that if in convex quadrilateral ABCD holds the equality
|AB|^2 + |CD|^2 = |BC|^2 + |DA|^2
then AC and BD are perpendicular.

2. Given a triangle ABC, over the sides AC and BC we construct squares. Denote the centres of these squares K and L, and D the centre of AB. Prove that the triangle KDL is  right-angled and isosceles.

here is a link where you may find some problems like that :
 http://popacalin.tripod.com/calinpopa/geometry.pdf

1. Let A1, A2, A3, A4 be distinct points on a circle. Let Wi be Euler's nine points circle of triangle AjAkAl where {i,j,k,l}={1,2,3,4}. Prove that this circles meet at a point.
(you don't realy nead the 4 points to be on a circle, but you do need them on a circle if you want a nice vector solution).

2. It's odds on you know this problem:
Let A,B,C,D be distinct points on a circle centered on O, so that AC and BD are perpendicular, and let T=AC \cap BD.
Prove that 2*OT= OA+OB+OC+OD (vectors not segments).

vecters are not so power full . but in some problems if u use them ,theproblems will solved very nice and soon. :) .

for example:three trisngles ,in wich each one has equal sides(OAB,OCD,OEF).
X is the midpoint of FA.Y is the midpoint of BC.and Z is the mid point of ED.
prove that :triangle XYZ has equal sides.

try to solve it by using vectors ;) .

[quote="Arne"]Yes, but it's even easier to prove that using complex numbers, so I think it is better to master complex numbers, rather than vectors.[/quote]
This is good book ''Complex Numbers from A to ...Z by Titu Andreescu , Dorin Andrica'' .  :wink:

[quote="Arne"]I am not fond of vectors... I don't think you can apply it very often. (Maybe it's a good idea to master complex numbers instead: they are more powerful than vectors but not much harder to work with.)[/quote]

But complex numbers applied to geometry are, in fact, generalizations of vectors which it's origin is the same origin of the graphic.

See, for example, when you have a complex number and want to rotationate it using the De Moivre theorem, you are in fact rotationating a vector that has it's origin on the graphic's origin and it's extremitiy in the complex number's afix.

The bad thing about complex numbers is that you have to have it in the origin and you can only work with it in 2D figures; vectors are useful in 3D space too.

Cheers

well i don't like solving geometry problems with vectors or complex numbers.....
but if you want, the first problem of the last imo con be solved with vectors..
it's a good example... i didn't solved it like that in spain but it could be done with simple vectors' manipulations ..

Can anyone recommend a good book which teaches how to use vectors in geometry with lots of exercises and examples?

[quote="apratimdefermat"][i][size=150][b][color=#F778A1]Can anyone recommend a good book which teaches how to use vectors in geometry with a lot of Olympiad standard examples and exercises.

I think barycentrics are also a very good technique in solving many problems[/color][/b][/size] [/i][/quote]
I think this artcle may be very interesting to you  :wink:
[url]http://www.math.ust.hk/excalibur/v6_n5.pdf[/url]

Dear Mathlinkers,
I think that the tool "vector" can be used for solving every geometric problem involving lines and circles.
The knot is that the relevant technics are not clearly teach and named like the most important theorems in Geometry.
In my honest point of view, this space is very open for those who want to developp a personnal approach. I have investigated one time the marvellous space...
Sincerely
Jean-Louis

I have some experience with vectors and vector analysis, resultant vectors, etc., but most of that is from my physics class. If you want vector problems made by me, either PM me or post on the thread.

vectors are interchangeable with matrices, so there fore you can do 2 or 3 d vector operations, using matrices
some people find working with numbers easier, so all laws that apply to matrices(numbers) , apply to vectors.

the vector method will be more powerful if we use it in barycentric coordinates.
We can use this method to prove geometric properties. And research the properties of geometric objects, too...


logarithms
calculus
derivative
trigonometry
calculus computations

Any reason why?

Can be simplifed to

$\frac{2(x-\sqrt{a^2-x^2}a}{\left(-\sqrt{a^2-x^2}-x+a\right)^2\sqrt{a^2-x^2}}$

Another solution:

put $x = a \sin \theta$. Then it turns into

$\frac{a+a \sin \theta + a \cos \theta}{-a+a \sin \theta + a \cos \theta}
= \displaystyle \frac {\sin \theta + \cos \theta + 1} {\sin \theta + \cos \theta - 1}
= \displaystyle 1 + \frac{2}{\sin \theta + \cos \theta - 1}$

which should be easy by chain rule, if my shaky calculus doesn't fail me.

[quote="vidyamanohar"][quote="indybar"]Differentiate $f(x)=\frac{a+\sqrt{a^2-x^2}+x}{\sqrt{a^2-x^2}+x-a} (0 < x < a) $[/quote]

Given $f(x)=\frac{a+\sqrt{a^2-x^2}+x}{\sqrt{a^2-x^2}+x-a} (0 < x < a) $

$\Rightarrow f(x) = \frac{{\sqrt {a + x} (\sqrt {a + x}  + \sqrt {a - x} )}}{{\sqrt {a - x} (\sqrt {a + x}  - \sqrt {a - x} )}}$

Taking logarithms on both sides, we get

$f(x) = \frac{1}{2}\log (a + x) + \log (\sqrt {a + x}  + \sqrt {a - x} ) - \frac{1}{2}\log (a - x) - \log (\sqrt {a + x}  - \sqrt {a - x} )$

Then differentiation is easy.[/quote]

Yes! I did also like this way. :)

you say $f(x)=$ that big thing, then take natural logs of both sides to simplify things, then differentiate implicity with respect to x and solve for $f ' (x)$.  So without writing the whole expressions, after you take natural logs of both sides you have on the left hand side $\ln{f(x)}$, then you wanna differentiate both sides of equation to get $1/f(x) \cdot f ' (x)$=stuff, so then solve for $f ' (x)$.

[quote="indybar"]How can you take logarithms on both sides and the derivative doesn't change? (I'm sorry. I'm just in high school.)[/quote]

This technique is called logarithmic differentiation.  You can refer to any standard text book on differential calculus.

why not simply write $f(x)=1+2a\frac{1}{\sqrt{a^2-x^2}+x-a}$? The derivative is namely $f'(x)=-2a\frac{1}{(\sqrt{a^2-x^2}+x-a)^2}(1-\frac{x}{\sqrt{a^2-x^2}})$

[quote="galeos"]why not simply write $f(x)=1+2a\frac{1}{\sqrt{a^2-x^2}+x-a}$? The derivative is namely $f'(x)=-2a\frac{1}{(\sqrt{a^2-x^2}+x-a)^2}(1+\frac{x}{\sqrt{a^2-x^2}})$[/quote] :?

i see :o  . there must be a minus in the parenthesis. i have edited it!


search
combinatorics proposed
combinatorics

Let $ S$ be the set of all odd positive integers less than $ 30m$ which are not multiples of $ 5$, where $ m$ is a given positive integer. Find the smallest positive integer $ k$ such that each $ k$-element subset of $ S$ contains two distinct numbers, one of which divides the other.

Why doesn't anyone do a search?  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200223 was posted only a week ago!  (Also, the correct spelling is "integers," not "intergers.")


Hi there

I would like to selfstudy this and I got the AP Computer Science book from Barrons but I would like a course or something (preferaably online) A couple of threads down, Ive seen one from ICT.org or something like that but that's too expensive for me :/

http://www.bluepelicanjava.com/

It's not a course, but it clearly and thoroughly goes through a lot of the information you'll need.


In how many ways can you misspell the word WOOTHCOUNTS?

WOOTHCOUNTS?! Never heard of that...

Anyway, there are infinitely many ways to construct an arbitrary string of letters that don't spell "WOOTHCOUNTS" (which I don't think is a real word). However, if we're restricted to the letters in the original "word", we get the following solution:

[hide]
Since there are 3 O's and 2 T's, there are $\frac{11!}{3!2!}=3326400$ spellings. Of these, only one is correct, so there are $3326399$ misspellings.
[/hide]

[hide]$\frac{11!}{3!2!}$ ways to arrange it -1=$\boxed{3326399}$
[/hide]
WOOT Met Mathcounts. Hm.
by the way, it could be infinity, because I could spell it "a" and it would be wrong.

There is a total of

(11!/3!2!) = 3326400 spellings. Since only one of these is a correct spelling, there are a total of 3326399 mispellings!

Ragnarok is right, actually. All the question is asking is how many ways can we mispell it, not necessarily using all or any of the letters given.
Answer:
[hide]$\infty$[/hide]

assuming that you can only use the given letters, [hide]$\frac{11!}{2!3!}-1=\boxed{3326399}$[/hide]

[quote="mathgeniuse^ln(x)"][quote="bpms"][quote="b-flat"]All of this is actually assuming that WOOTHCOUNTS is spelled correctly and you must use those letters.

[hide]There are a total of $\frac{11!}{2!\cdot3!}=3,326,400$ arrangements possible (3 Os and 2 Ts).  Of these, only one is WOOTHCOUNTS, so there are $3,326,400-1=3,326,399$ misspellings.[/hide]

Assuming that you can use whatever letters you want, and as many of them as you want
[hide]There are an infinite number of "words" to be created, and only one is correct, so there are $\infty$ misspellings.[/hide][/quote]
Well since one [b]is[/b] correct there are $\infty-1$ misspelings.[/quote]

There is no such number as infinity-1, right?[/quote]

No. Quote somewhere 'infinity is a value in which any nonempty subset of it is itself' or something like that.

[hide]
$\frac{11!}{2!3!}-1=3326400$
We subtract one because WOOTHCOUNTS is the correct spelling.[/hide]

[hide]Assuming you must use only those given letters, there are $\frac{11!}{2!3!}-1=3326399$ misspellings.[/hide]


How many zeros appear at the end of the whole-number representation of $ \frac{20!}{10^4}$?

$ 20!$ is divisible by $ 5^4$, but not by $ 5^5$
$ 10^4$ is also divisible by $ 5^4$, so there isn't a zero at the end of the representation.

[color=#f00][/color] thank you very much

that is q hard problem lokking at it but if you know the trick it's simple

[quote=nt811581147]If you were me in your mathcounts days, you memorized many factorials, and know that

20!=2,432,902,008,176,640,000 so when you remove the 4 0s, there are $ 0$ zeroes left.[/quote]

why you flexing tho


function
logarithms
analytic geometry
calculus
calculus computations

(i) Using the definitions of $ \sinh x$ and $ \cosh x$ in terms of $ e^{x}$ and $ e^{\minus{}x}$, show that

(a) $ \cosh(\ln a) \equal{} \frac{a^{2} \plus{} 1}{2a}$, where $ a>0$,

(b) $ \cosh x \cosh y \minus{} \sinh x \sinh y \equiv \cosh (x\minus{}y)$.

(ii) Use part (i)(b) to show that $ \cosh^{2}x\minus{}\sinh^{2}x\equiv 1$.

(iii) Given that $ R > 0$ and $ a > 1$, find $ R$ and $ a$ such that

$ 13\cosh x \minus{} 5\sinh x \equiv R\cosh(x\minus{}\ln a)$.

(iv) Hence write down the coordinates of the minimum point on the curve with equation $ y\equal{}13\cosh x \minus{} 5\sinh x$.

hello, for a) we have $ \cosh(x)\equal{}\frac{1}{2}(e^x\plus{}e^{\minus{}x})$ and so we get $ \cosh(\ln(a))\equal{}\frac{1}{2}(e^{\ln(a)}\plus{}e^{\minus{}\ln(a)})\equal{}\frac{1}{2}(a\plus{}\frac{1}{a})\equal{}\frac{a^2\plus{}1}{2a}$.
Sonnhard.

Any takers for the rest of this question?

For (ii) put $ y \equal{} x$ in (b) to get $ \cosh 0\equal{} 1$.

Thanks, lordmahone123.

For part (iii), I have, using the previous two parts, $ R\equal{}12$ and $ \cosh y \equal{} \frac{13}{12}$.

So $ y\equal{}\ln(\frac{13}{12} \plus{} \sqrt{(\frac{13}{12})^{2}\minus{}1}\,)$

$ \equal{}\ln (\frac{3}{2})$

$ \Rightarrow\, a \equal{} \frac{3}{2}$

For part (iv), differentiating and setting to zero, I have at a glance, $ (x, y) \equal{} (\ln \frac{3}{2}, 13)$, which I need to check.

Thanks, Sonnhard. Your post arrived while I was (painfully slowly) typing mine (and feeling guilty about being indoors in such fine sunny weather). Is mine correct? AndrewTom.

P.S. Have you done the log(cos x), Newton-Raphson (I only have a very basic calculator) and reduction formula questions?

hello, i have also got $ R\equal{}12$ and $ a\equal{}\frac{3}{2}$, for (iv) i have found the point $ \ln(\frac{3}{2});12$.
Sonnhard.

Thanks Sonnhard.

I made a silly mistake (yet again): of course the $ y$-coordinate is $ R \times \cosh (0) \equal{} 12 \times 1 \equal{} 12$ and not $ 13$.

Are there any other ways of doing the various parts of this question?

I'm still hoping to see other ways of looking at, or solving, the various parts of this question.


integration
calculus
function
number theory
prime factorization
calculus computations

[url]http://www.mathlinks.ro/viewtopic.php?t=193915[/url]

You could've just bumped the first topic.

Double posting is strongly discouraged  :spam: I removed your identical post. As to the integrals themselves, the antiderivatives are not elementary functions, period.

Sory [b]fedja[/b] and [b]Prime factorization[/b] very much. I will careful for next post. Thank for your reply!

 :(   :(   :(   .

both of them are odd functions so both of them have value equal to zero....both these functions are symmetric abt the origin...

now replace x by -x and verify. now there are many integral properties that help us to prove this as zero.


inequalities
induction
inequalities unsolved

[quote]Prove that the inequality $\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2} \geq \frac{3\sqrt{2}}{2}$ holds for arbitrary real numbers $a,b,c$.[/quote]

The solution I have for this problem says that it applies Minkowski's Inequality to come up with
$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2} \geq \sqrt{(a+b+c)^2 + (3-a-b-c)^2}.$$

and from there the final solution can be found (and I understand the rest of the proof from this step).  However, I can't find any way to apply Minkowski's inequality as shown above.  In order to make it work, Minkowski's would have to be extended to include three sets of numbers ([tex]a_i, b_i, c_i[/tex]) rather than just the normal two sets.  Can anybody explain to me how Minkowski's is being applied in this instance?

Just check Kiran Kedlaya's $a<b$ for a more general statement of Minkowski (that is to say that yes, it does hold for three sets of numbers)

Otherwise, Jensen's gives us

$\sqrt{\frac {a^2+(1-b)^2}{2}}\geq \frac {a+(1-b)}{2}$

That inequality needs some explanation.  Certainly it's true if $0\leq b\leq 1$; if $b>1$, we get

$\sqrt{\frac {a^2+(1-b)^2}{2}}\geq \frac {a+(b-1)}{2}\geq\frac {a+(1-b)}{2}$

Summing, we get

$\sum \sqrt{a^2+(1-b)^2}\geq \sqrt {2}\cdot \sum \frac {a+(1-b)}{2}=\frac{3\sqrt{2}}{2}$

which is the desired result.

[quote="blahblahblah"]Just check Kiran Kedlaya's $a<b$ for a more general statement of Minkowski (that is to say that yes, it does hold for three sets of numbers)
[/quote]
Ah, thanks a lot!  What is Kedlaya's $a<b$, exactly?  I've never heard of it before.  Also, can Minkowski's be extended indefinitely, i.e.

$\left(\sum(a_i + b_i + ... + z_i)^r \right)^{1/r} \leq \left(\sum a_i^r \right)^{1/r} +  \left(\sum b_i^r \right)^{1/r} + ... +  \left(\sum z_i^r \right)^{1/r}$

(I apologize if this is explained in the reference you gave me)

[quote]Also, can Minkowski's be extended indefinitely, i.e.

$\left(\sum(a_i + b_i + ... + z_i)^r \right)^{1/r} \leq \left(\sum a_i^r \right)^{1/r} + \left(\sum b_i^r \right)^{1/r} + ... + \left(\sum z_i^r \right)^{1/r}$[/quote]

Yes. Proof follows from induction with respect to the number of letters.


Hey this is to inform you about Frugal,i am just copy pasting the mail i got from the coords of the event so what u see here is the description of the event on their words.
[quote="Frugal Coords"]
Want to take a crack at beating all those awesome "programmers" at their own game? Read on.

Frugal is a new event at Shaastra, the technical fest of IIT Madras that hopes to be different. It aims to test your ingenuity whilst not being dependant on various algorithms. If you know the basics of programming in C, then you just might have a good chance of winning Frugal. Frugal is a chance for you to prove that your programming skills are enough to put yourself right up there with the best.

Frugal is about minimizing code to maximize your score and win. The problem statements will require a fair bit of thinking and creativity to solve, but we'll make sure that that's all they will require. You won't be required to have super amounts of knowledge of algorithms.

To register and for more details, go to http://frugal.shaastra.org .
[/quote]

When is the event? Looks perfect for a newbie like me with little algorithmic knowledge :P


abstract algebra
superior algebra
superior algebra solved

How do you prove that $\ds M\otimes_R\frac{R}{I}\cong\frac{M}{IM}$ where $M$ is an $R$-module and $I$ is an ideal of $R$?

The part that I need help with is how should one define a bilinear map $\psi: M\times \frac{R}{I}\longrightarrow \frac{M}{IM}$?

[quote="baz"]The part that I need help with is how should one define a bilinear map $\psi: M\times \frac{R}{I}\longrightarrow \frac{M}{IM}$?[/quote]

$\psi(m,r+I)=(r\cdot m+IM)$. $r+I$ denotes the equivalence class of $r$ modulo $I$ and $rm+IM$ denotes the equivalence class of $rm$ modulo $IM$.

The map I defined is obviously bilinear. To see that it is well defined, let $r,r'\in R$ such that $r+I=r'+I$. Then $r-r'\in I$, so $(r-r')m\in IM$, hence $rm+IM=r'm+IM$.

Since you don't need help with the rest of the problem I'm sure you wouldn't mind sharing the rest of the proof with us.

[u]Wouldn't this be another way?[/u]

Consider $g: M\rightarrow M\otimes R/I$ defined by $m\mapsto m\otimes\overline{1}.$  This is clearly well-defined.  Notice that $\ker(g)=IM$ since
$m\otimes \overline{1}=m\otimes\overline{i}=im\otimes\overline{1}$ for any $i\in I$
and
$g(im)=im\otimes \overline{1}=m\otimes \overline{i}=m\otimes \overline{1}$ implying that $g(i_1m_1+\cdots+i_sm_s)=m_1+\cdots+m_s\otimes\overline{1}.$
Thus, the First Isomorphism Theorem (or if you call it the Fundamental Isomorphism Theorem or whatever) yields the result.

See this for a solution: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76587[/url]


Find the bounds of
$\cos(A-B)+\cos(B-C)+\cos(C-A)$
when
(a) $A,B,C$ are angles in a triangle
(b) $A,B,C$ are real numbers

the maximum bound is 3 for both cause the maximum of the cosine function is 1 (for x=0) and it can be achieved in both (a) and (b). In (a) with a triangle with the three angles of 60º:
cos(60-60)+cos(60-60)+cos(60-60) = 3

And in (b) with A=B=C.

Say, $A-B=x, B-C=y, C-A=z \Rightarrow x+y+z=0$
So, $\displaystyle\sum_{cyc}\cos(A-B) = \displaystyle\sum_{cyc}\cos x = S$.
Now, $2S =  \displaystyle\sum_{cyc}(\cos x + \cos y) = \displaystyle\sum_{cyc}(2\cos\frac{x+y}{2}\cos\left(\frac{x}{2}-\frac{y}{2}\right))$

$ = \displaystyle\sum_{cyc}2\cos\frac{-z}{2}\left(\cos\frac{x}{2}\cos\frac{y}{2}+\sin\frac{x}{2}\sin\frac{y}{2}\right) = \displaystyle\sum_{cyc}2\left(\cos\frac{z}{2}\cos\frac{x}{2}\cos\frac{y}{2}+\cos\frac{z}{2}\sin\frac{x}{2}\sin\frac{y}{2}\right)$

$\Rightarrow S = \displaystyle\sum_{cyc}\left(\cos\frac{z}{2}\sin\frac{x}{2}\sin\frac{y}{2}\right) + 3\cos\frac{x}{2}\cos\frac{y}{2}\cos\frac{z}{2}$

Now, say $T = \displaystyle\sum_{cyc}\left(\cos\frac{z}{2}\sin\frac{x}{2}\sin\frac{y}{2}\right)$ and then we can say:

$2T = \displaystyle\sum_{cyc}\left(\cos\frac{z}{2}\sin\frac{x}{2}\sin\frac{y}{2}+\cos\frac{x}{2}\sin\frac{y}{2}\sin\frac{z}{2}\right)$

$= \displaystyle\sum_{cyc}\sin\frac{y}{2}\left(\cos\frac{z}{2}\sin\frac{x}{2}+\cos\frac{x}{2}\sin\frac{z}{2}\right) = \displaystyle\sum_{cyc}\sin\frac{y}{2}\left(\sin\frac{x+z}{2}\right)$

$= \displaystyle\sum_{cyc}\sin\frac{y}{2}\left(\sin\frac{-y}{2}\right)= \displaystyle\sum_{cyc}-\sin^2\frac{y}{2}$

$\Rightarrow 4T = \displaystyle\sum_{cyc}-2\sin^2\frac{y}{2}$

$\Rightarrow S = 4T + 3 = \displaystyle\sum_{cyc}\left(1-2\sin^2\frac{y}{2}\right) = \displaystyle\sum_{cyc}\cos y =S$

$\Rightarrow   4T + 3 = T + 3\cos\frac{x}{2}\cos\frac{y}{2}\cos\frac{z}{2}$

$\Rightarrow   T = \cos\frac{x}{2}\cos\frac{y}{2}\cos\frac{z}{2} - 1$

$\Rightarrow S = 4\cos\frac{x}{2}\cos\frac{y}{2}\cos\frac{z}{2} - 1$

so assuming Pork_Chop8 is correct
[hide="bounds"](a) $-1\leq S\leq\frac{3\sqrt3}2-1$
(b) $-2\leq S\leq 3$
right? :)[/hide]

Thanks, Pal (Palytoxin). I went to eat dinner after posting my little identity and hadn't had a chance to do that. Could you post a solution instead of just an answer?

Well, Palytoxin, I think you make a little mistake for the upper bond in part (a).As Joao Guerreiro already pointed out, 3 can be reached when A=B=C. Beside, from the indentity  Pork_Chop8 has worked out,$\displaystyle \cos\frac{x}{2}\cos\frac{y}{2}\cos\frac{z}{2}$can reach 1 when x=y=z=0.
I'm not quite sure how you get the lower bond for part (b). Care to explain, Pal? [/quote]


Let $ k$ be a fixed positive integer. For all positive integers $ n$, prove that there exists a sequence of positive integers $ a_{1},\ a_{2},\ \ldots,\ a_{n}$ such that $ (j,a_{j})\equal{}1$, for all $ 1\leq j \leq n$, and satisfying $ \sum_{j\equal{}1}^{n}{\frac{j^{k}}{a_{j}}}\equal{}1.$

[quote="pohoatza"]Let $ k$ be a fixed positive integer. For all positive integers $ n$, prove that there exists a sequence of positive integers $ a_{1},\ a_{2},\ \ldots,\ a_{n}$ such that $ (j,a_{j}) \equal{} 1$, for all $ 1\leq j \leq n$, and satisfying $ \sum_{j \equal{} 1}^{n}{\frac {j^{k}}{a_{j}}} \equal{} 1.$[/quote]

If $ n\equal{}1$, $ a_1\equal{}1$ is the solution.

If $ n\geq 2$
Let the sequence $ u_p$, for $ p\in[2,n]$ defined in reverse order as :
$ u_n\equal{}1$
If $ n>2$ : $ u_p\equal{}u_{p\plus{}1}(u_{p\plus{}1}(p\plus{}1)^k\plus{}1)$ $ \forall p\in[2,n\minus{}1]$

Let then $ a_p$ defined as :
$ a_1\equal{}u_2(u_22^k\plus{}1)$
$ a_p\equal{}u_pp^k\plus{}1$ $ \forall p\in[2,n]$
We obviously have $ (p,a_p)\equal{}1$

$ \forall p\geq 2$ : $ \frac{p^k}{a_p}\equal{}\frac{p^k}{u_pp^k\plus{}1}$ $ \equal{}\frac{1}{u_p}\minus{}\frac{1}{u_p(u_pp^k\plus{}1)}$

$ \implies$ $ \sum_{i\equal{}2}^n\frac{i^k}{a_i}$ $ \equal{}\frac{1}{u_n}\minus{}\frac{1}{u_2(u_22^k\plus{}1)}$ $ \equal{}1\minus{}\frac{1}{a_1}$

$ \implies$ $ \sum_{i\equal{}1}^n\frac{i^k}{a_i}\equal{}1$

Q.E.D.


Determine all functions $f: \mathbb{R}\to\mathbb{R}$ satisfying:

$i)$ $f(1)=1, f(-1)=-1$
$ii)$ $f(x) \le 0, \forall x \in (0,1)$
$iii)$ $f(x)+f(y)+1 \ge f(x+y) \ge f(x)+f(y), \forall x,y\in\mathbb{R}$

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=apmo&t=78729

ii) is different a bit but it comes to the same.


1.Let $ a,b,c\in R \plus{}$ which none pair of them are equal prove that
$ \sum_{cyc} \frac {4 \plus{} (a \plus{} b)^2}{c(c \minus{} a)(c \minus{} b)} > \frac {24}{\sqrt {abc(a \plus{} b \plus{} c)}}$

Could you please expand the sigma?I’m not familiar with such notation .

This is not the complete solution but I think that be useful.
We can easily prove that :
1/(b-c)(b-a) = -1/((c-a)(c-b))-1/((a-b)(a-c))
  Substituting the 3rd  term in left side by ( -1/((c-a)(c-b))-1/((a-b)(a-c)) )(4+ (a+c)^2)/b yields :
{   (4+(b+c)^2) / (a(a-b)(a-c)) +   (4+(a+c)^2) / (b(a-c)(b-a))}  + { (4+(a+b)^2) / (c(c-a)(c-b))  +(4+(a+c)^2) / (b(a-c)(c-b))}
  Using this arrangement and after a little long calculations we get to the following algebraic equality:
Left side = (4+(a+b+c)^2)/(abc)
This result was checked for some examples by MATLAB software .We already know both two sides are positive then we can compare square of them so:
 16(a+b+c) +(a+b+c)^5+ 8(a+b+c)^3 should be compared by  576 abc .
I hope we be able to continue this solution .I think yensen inequality may be useful .

Let me change the 24 by 14 .Let’s see the result of selecting   a=1.1  , b = 1,c = 0.9 by MATLAB software .
>> a=1.1,b=1,c=0.9
a =
    1.1000
b =
     1
c =
    0.9000
>> (4+(a+b)^2)/(c*(c-a)*(c-b))+(4+(b+c)^2)/(a*(a-b)*(a-c))+(4+(a+c)^2)/(b*(b-c)*(b-a))-24/(a*b*c*(a+b+c))^0.5
ans =
   -0.7949
>>
As you see the inequality seem to be not true for 24 .By 14 instead of 24 a result of yensen inequality for mean values yields the following :
(a+b+c)/3 > (abc)^(1/3)
(a+b+c)^3 > 27abc
8(a+b+c)^3 > 216abc> 196abc = (14^2)abc
16(a+b+c) +(a+b+c)^5+ 8(a+b+c)^3 > (14^2)abc
The last inequality proves the intended  inequality .


Each of eight houses on a street is painted brown, yellow, or white.  Each house is painted only one color and each color is used on at least one house.  No two colors are used to paint the same number of houses.  In how many ways could the eight houses on the street be painted? One such way is shown here.
Brown |    :    |White
White  |   :     |Yellow
Yellow |    :    |Yellow
Yellow |    :    |Yellow

[hide]First count the number of ways in which we can assign a number of houses to each color. We have $ a+b+c=8$, WLOG let $ a>b>c\ge1$, and there are only two possible solutions, $ (5,2,1)$ and $ (4,3,1)$. If $ (5,2,1)$, there are 6 ways to match up colors with numbers, then $ \binom{8}{5}\binom{3}{2}$ ways to color the houses. Similarly, if $ (4,3,1)$, there are 6 ways to match up colors with numbers, then $ \binom{8}{4}\binom{4}{3}$. Hence our answer is

$ 18\binom{8}{5}+24\binom{8}{4}=2688$[/hide]


Find an analytic function $ f(x)$ such that $ f(f(x)) \equal{} e^x$.

I don't think you can find an explicit expression for such a function. [url=http://www.mathpages.com/home/kmath507.htm]Here[/url] is some information.
But, for arbitrary $ f$ one can find some variants.


Murtha has decided to start a pebble collection. She collects one pebble the first day and two pebbles on the second day. On each subsequent day she collects one more pebble than the previous day. How many pebbles will she have collected at the end of the twelfth day?

[quote="Kouichi Nakagawa"]$ \sum_{k\equal{}1}^{11} k \equal{} \frac{11 \times 12}{2} \equal{} \boxed{66}$.[/quote]
That's not even correct.. It's 78

$\frac{12*13}{2}$

Basically, its an arithmetic sequence
$1+2+3...12=?$
The sum of the first $n$ numbers is $\frac{n(n+1)}{2}$
Plugging in $12$, we get $\boxed{78}$


function
integration
real analysis
real analysis unsolved

Suppose $ f$ is a measurable function on some measure space, $ 1<p<\infty,$ and $ \frac1p\plus{}\frac1q\equal{}1.$ Suppose that $ \int |f|^p\equal{}\infty.$

Prove that there exists $ g$ such that $ \int|g|^q$ is finite but $ \int fg\equal{}\infty.$

[Proofs that concretely construct $ g$ would be appreciated.]

Something must be assumed about the measure space to prevent it from being a singleton with infinite measure.

Assume something: OK, let's say $ \sigma$-finite.

This is somewhat constructive. :maybe: There is a countable family of disjoint sets $ E_n$ of finite measure such that $ \int_{E_n}|f|^p > 2^{np}$ for every $ n$. Choose $ M_n > 0$ so that $ \int_{E_n}(|f|\wedge M_n)^p > 2^{np}$. Let $ g_n \equal{} \mathrm{sign}(f)(|f|\wedge M_n)^{p \minus{} 1}\chi_{E_n}$. Since $ \int fg_n > 2^n\|g_n\|_{q}$, the function $ g \equal{} \sum_{n}\frac {g_n}{2^n \|g_n\|_q}$ works.

$ \wedge$ stands for minimum, $ \vee$ for maximum.


Show that given any integer $0 < n \leq 2^{1000000}$ we can find at set $S$ of at most $1100000$ positive integers such that $S$ includes $1 $and $n$ and every element of $S$ except $1$ is a sum of two (possibly equal) smaller elements of $S$.

[quote="DusT"]I hope I understand the problem corectly, because I think it should be very easy to take all powers of $2$ less than $n$, and the number $n$. Obviously, $n$ has a binary representation and every power of $2$ is the double of the previous one  :?[/quote]

how does that ensure that $n$ is a sum of $2$ smaller numbers?if $n$ in binary form has more than 2 ones then just using the powers of $2$ would not suffice to be able to write $n$...
in fact i have the following formulation:if $f(n)$ denotes the smallest possible size of a set satisfying the above property with $n$ as its last element, then if $n=\sum_{i=1}^{k} 2^{a_i}$ with $a_1>a_2>\cdots>a_k$ then $f(n)\leq a_1+k$(easy to prove by induction).
actually one can prove that $f(mn)\leq f(m)+f(n)-1$ and $f(m+n)\leq f(m)+f(n)-1$.
we need a uniform bound: $\forall\ 1\leq n\leq 2^{10^6},f(n)\leq 10^6+10^5$.

I was wondering , if it can be shown that for any positive integer n, that f(2n+1) >= f(2n).


$21)$ Find the volume in cubic inches of the solid generated by revolving a region bounded by an equilateral triangle of side $4$ inches about one side.

Could someone draw a picture to go with it? I'm having trouble with these kinds of problems.  :blush:

The solid generated is a double cone with volume $V= 16\pi$ cubic inches.

Carcul

can you draw a picture or explain further? Because I don't follow what you said....  :maybe:

okay. so the radius is $2\sqrt{3}$ since it is revolved around a $SIDE$. There are two circles, and the heights are both $2.$

so it's a long cone with small height.  :lol:

therefore, use the cone formula to find that both cone areas are both $8$$\pi$

That adds up to $16 \pi$, which is the answer.

[quote="Carcul"]The solid generated is a double cone with volume $V=\frac{48\pi}{3}$ cubic inches.

Carcul[/quote]

last time i checked $\frac{48}{3}=16$  :P

each cone has a volume of $\frac{({2\sqrt3})^{2}\times2\times\pi}{3}=8\pi$

[quote="Walk Around the River"]last time i checked $\frac{48}{3}=16$[/quote]

No comment...

Thanks, Carcul


hello i dont know if anyone remembers me but i am one of the assistants in the imo. i was in  the room for codes 1 and 2, i was looking for pascual, darij , and severius but im not sure if anyone of you guys has code one or two so do you??? also extraoficially at the end of the test please be sure to have all your papers ready and in your folders before the exam is finished because it takes too much time for us assistants to pick them up and tomorrow we have to get everything picked up including the desks and chairs by 2 o clock also please remain seated until the assistants say you can leave because we had a little problem with the participants leaving or staying and contestants still writing after the test wich we wish to completly eliminate on the second day. anyway if you have a code one or two then i will be your assistant for tomorrow look for my name i will have on my tag !!!!! good luck everybody!!!!!!!

Hmm, for some reason, I hadn't noted this post until the end of the exams  :(  . Anyway, I was in that room (my code was DEU1), so I have probably seen you without noting who you are...

  darij

I met Marko. His father and me coordinated problem 3  :lol:

COOL¡¡¡
 YES I  heard your name in the final ceremony you got silver right? congrats¡¡
  very nice¡¡
 yes  i probably took you to the bathroom one time  :P

hi valentin¡¡ i definitely will go to woot¡¡¡¡ my dad already said he will pay  :D  :rotfl:

For the record, I was ARG2.  :P Unfortunately I hadn't read this post earlier.

Cool, I was in Room 1/2 too. I wonder if you were the person whom I constantly bothered with the query form and asked for more water from. :oops: :P


For $ 1\leq i\leq 2005$, $ a_{i}$ is any real number greater than or equal to 1. Determine the minimum value of

$ \frac{\prod_{i\equal{}1}^{2005}{a_{i}}^{2}\plus{}1}{\prod_{i\equal{}1}^{2005}({a_{i}}^{2}\plus{}1)}$.

First, this is an inequality problem, not an algebra problem.  Second, you've already posted this problem here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168172[/url]

Do not double post.


modular arithmetic
number theory
least common multiple
AoPSwiki
AMC
AIME
search

For the first one, I don't know any particular equation.  For the second one, split up the two expressions.

$ 6^{45} \equiv ( \minus{} 1)^{45} \pmod{7}$

This is $ 6$.

$ 8^{31} \equiv 1^{31} \pmod{7}$

This is $ 1$.

So $ 6^{45} \plus{} 8^{31} \pmod{7} \equal{} 0$.

[color=darkred][size=75]Fixed the exponents.
-nebula42[/size][/color]

:|  :|
idk wut ur talking about. i'm kinda new to mod and remainder stuff, so could u provide some more detailed solution?  :)

Okay, well, 8 has a remainder of 1 when divided by 7, so we write $ 8\equiv1\pmod{7}$. Similarly, 6 is 1 less than a multiple of 7, so we write $ 6\equiv\minus{}1\pmod{7}$ (we can also write it as 6 (mod 7). What's useful about a number being $ \pm1\pmod {m}$ is that when we raise it to a power, it is either 1 or -1.

In this case, $ 8^{31}\equiv(1)^{31}\equiv1\pmod{7}$. It has a remainder of 1 when divided by 7.
Similarly, $ 6^{45}\equiv(\minus{}1)^{45}\equiv\minus{}1\pmod{7}$. The remainders cancel off, so the remainder is 0.

[quote="toadoncart"]i understand modular notations and stuff, but i don't really know how to actually solve equations/questions that deal w/ modular stuff. like for instance "some number gets a remainder of 5 when divided by 7 and it also gets a remainder of 8 when divided by 11. what is the number?"

so u would do
$ x \equiv 5 \pmod{7}$
$ x \equiv 8 \pmod {11}$.

then wut?  :huh:

edit: oh yeah, (yes, i rly suck at number theory) how would u solve something like "find the remainder of 8^31+6^45 when divided by 7." thx!  :)[/quote]

About you first question, how to finish solving "Some number gets a remainder of 5 when divided by seven.  The same number gets a remainder of 8 when divided by 11.  What is the number?"

You the setup right I believe:

$ x \equiv 5 \pmod{7}$
$ x \equiv 8 \pmod{11}$

Now we add 58 to both sides to equal everything out.  (Add 58 to 5 and you get 63, 63/7=0, and 58+8=66, 66/11=0, you have to "zero" it out)

$ x+58 \equiv 0 \pmod{7}$
$ x+58 \equiv 0 \pmod{11}$

Since the left and right parts are the same, we can make a new equation with x-58 on one side, and the LCM of 7 and 1 on the other.

$ x+58=LCM(7,11)$
$ x+58=77$
$ x=19$

We can test 19 put and see it works.  So our answer is $ \fbox\fbox\fbox{19}}}$

[quote="ajain"][quote="toadoncart"]i understand modular notations and stuff, but i don't really know how to actually solve equations/questions that deal w/ modular stuff. like for instance "some number gets a remainder of 5 when divided by 7 and it also gets a remainder of 8 when divided by 11. what is the number?"

so u would do
$ x \equiv 5 \pmod{7}$
$ x \equiv 8 \pmod {11}$.

then wut?  :huh:

edit: oh yeah, (yes, i rly suck at number theory) how would u solve something like "find the remainder of 8^31+6^45 when divided by 7." thx!  :)[/quote]

About you first question, how to finish solving "Some number gets a remainder of 5 when divided by seven.  The same number gets a remainder of 8 when divided by 11.  What is the number?"

You the setup right I believe:

$ x \equiv 5 \pmod{7}$
$ x \equiv 8 \pmod{11}$

Now we add 58 to both sides to equal everything out.  (Add 58 to 5 and you get 63, 63/7=0, and 58+8=66, 66/11=0, you have to "zero" it out)

$ x + 58 \equiv 0 \pmod{7}$
$ x + 58 \equiv 0 \pmod{11}$

Since the left and right parts are the same, we can make a new equation with x-58 on one side, and the LCM of 7 and 1 on the other.

$ x + 58 = LCM(7,11)$
$ x + 58 = 77$
$ x = 19$

We can test 19 put and see it works.  So our answer is $ \fbox\fbox\fbox{19}}}$[/quote]

o. i c. huh, that's pretty interesting. second problem?

o yeah, how do u find out that u were supposed to add 58 to both sides?  :huh:
like if next time i were to do a similar problem, how would i find out what to add to both sides?  :)

edit: o yeah, isn't that the way to find the smallest possible value of x? like if the question were "wut is the second smallest possible value of x" would u just multiply the smallest value by 2 or something?  :huh:

[hide][quote="toadoncart"]i understand modular notations and stuff, but i don't really know how to actually solve equations/questions that deal w/ modular stuff. like for instance "some number gets a remainder of 5 when divided by 7 and it also gets a remainder of 8 when divided by 11. what is the number?"

so u would do
$ x \equiv 5 \pmod{7}$
$ x \equiv 8 \pmod {11}$.

then wut?  :huh:

edit: oh yeah, (yes, i rly suck at number theory) how would u solve something like "find the remainder of 8^31+6^45 when divided by 7." thx!  :)[/quote]

first, notice that if x is a solution, x+lcm(7,11)=x+77 is also a solution, since it has the same remainder as x when divided by 7 and 11. so there are an infinite number of solutions. usually the problem will put a limit on this. to solve this notice that x=5 mod7 means x=5+7k for some integer k. plug this into the second equation so 5+7k=8 mod 11, so 7k=3 mod 11, so here you start from 3 and count up by 11's to find the inverse of 7 in mod 11.the first number divisible by 7 is the inverse of 7.(inverses multiply to 1 mod something-3 and 2 are inverses in mod 5. so, 3,14. wow that was fast. so 7k=3=14 mod 11, so k=2 mod 11.(we can divide because gcd(11,7)=1) so,k=2 mod 11, so x=5+7k=5+14 (mod 77=7*11), so x=19. checking, this works. so a general solution is 19+77y for some integer y. just to prove my point, you can see that 19+77=96 is a solution.

for more information on these problems, i think this has to do with the Chinese Remainder Theorem. (probably in the wiki)

for another example to a problem like this...

a guy has some number a candies less than 20 on halloween. he hears his doorbell. at the door he sees 2 kids dressed as a ghost and a lion. he realizes that if he distributes his candies evenly,  one is left over. at this point dracula jumps out from behind that ghost. now he again has 1 left over. now some other kid comes out, and he still has 1 left over. how many candies did he start with.(he has more than 1 candy)

don't be oblivious to creative solutions. this problem says
x=1 mod 3
x=1 mod 4 (x=1 mod 2 is a more specific case of this)
so x-1=0 mod 4 and mod 3, so x-1=0 mod 12, so x=13+12k, for integers k. so any k other than 0 would not be within the limits of the problem so the answer is 12.[/hide]

if the hide doesn't work, can a mod hide it(i forgot how to)

[quote="Yongyi781"]Ajain's method, simply put, just begged the question again. How DO you find the number 58 so that it will zero out the mods? You have to do the same process...

If a formula is not desired, a speedy method is to start with 8, increment by 11's until you get a number that leaves a remainder of 5 when divided by 7.

$ 8 \equiv 1 \pmod 7$
$ 19 \equiv 5 \pmod 7$

Look! We found it already![/quote]

Yeah, thats probably the fastest way to find it unless the number is far off in the thousands.

what math154 said and three words: chinese remainder theorem.

Eh...what exactly is the Chinese Remainder Theorem. I never followed it--is it the same thing, or a different but related thing?

[quote="mathking123"]so,k=2 mod 11, so x=5+7k=5+14 (mod 77=7*11), so x=19. checking, this works. so a general solution is 19+77y for some integer y. just to prove my point, you can see that 19+77=96 is a solution.

so x-1=0 mod 4 and mod 3, so x-1=0 mod 12, so x=13+12k, for integers k. so any k other than 0 would not be within the limits of the problem so the answer is 12.[/quote]

srry for digging up an old post, but don't get these two parts... :(

2 days is not an old post :D

look it up on AoPsWiki, the best way i found to be comfy with mod equations is to make up two numbers that satisfy two mod equations, then solve them and see if you get the same number :)

[quote="math154"]If you want a general method,
[hide]
For our example, note that the equations simply mean that
i. When x is divided by 7, the remainder is 5.
ii. When x is divided by 11, the remainder is 8.
Thus, we can express this as $ x \equal{} 7a \plus{} 5 \equal{} 11b \plus{} 8$ for some integers $ a,b$.
Now, if we take the equation $ \bmod7$, we have $ 5\equiv4b \plus{} 1\pmod7\implies4b\equiv4\pmod7$.
Since $ \gcd(4,7) \equal{} 1$, we have $ b\equiv1\pmod7$. As above, this means that $ b \equal{} 7c \plus{} 1$ for some integer c. Substituting this back in $ x \equal{} 11b \plus{} 8$, we get $ x \equal{} 77c \plus{} 19$, so our solutions are $ \boxed{19,96,...}$.
I think you should remember this method, because it's generally pretty fast. Of course, for small numbers like these, the method Yongyi posted is faster.
[/hide]
I think this is CRT or something...not sure.[/quote]

i'm trying to solve the mod problems like this, but it somehow doesn't work for me:
ex:
$ x\equiv 2\pmod{4}$
$ x\equiv 4\pmod{7}$

k, so 7k+4 substitute into first equation. u get $ 3k\equiv 2\pmod{4}$. so we could write 3k=4c+2, or $ k \equal{} \frac {4c \plus{} 2}{3}$ for some integer c. substitute back into 7k+3 to get $ 7(\frac {4c \plus{} 2}{3}) \plus{} 3 \equal{} \frac {28c \plus{} 14}{3} \plus{} 3$, so da solution to "what is the smallest possible value for x" is 17 (c=1). but if u substitute this back into the original two equations, it doesn't work.... :huh:

o yeah, could u put c as a negative number if it works out? and do u usually assume the problem is asking for a nonnegative solution? :|

[quote="ajain"][quote="toadoncart"]i understand modular notations and stuff, but i don't really know how to actually solve equations/questions that deal w/ modular stuff. like for instance "some number gets a remainder of 5 when divided by 7 and it also gets a remainder of 8 when divided by 11. what is the number?"

so u would do
$ x \equiv 5 \pmod{7}$
$ x \equiv 8 \pmod {11}$.

then wut?  :huh:

edit: oh yeah, (yes, i rly suck at number theory) how would u solve something like "find the remainder of 8^31+6^45 when divided by 7." thx!  :)[/quote]

About you first question, how to finish solving "Some number gets a remainder of 5 when divided by seven.  The same number gets a remainder of 8 when divided by 11.  What is the number?"

You the setup right I believe:

$ x \equiv 5 \pmod{7}$
$ x \equiv 8 \pmod{11}$

Now we add [b]58[/b] to both sides to equal everything out.  (Add 58 to 5 and you get 63, 63/7=0, and 58+8=66, 66/11=0, you have to "zero" it out)

$ x + 58 \equiv 0 \pmod{7}$
$ x + 58 \equiv 0 \pmod{11}$

Since the left and right parts are the same, we can make a new equation with x-58 on one side, and the LCM of 7 and 1 on the other.

$ x + 58 = LCM(7,11)$
$ x + 58 = 77$
$ x = 19$

We can test 19 put and see it works.  So our answer is $ \fbox\fbox\fbox{19}}}$[/quote]

how did you get the 58??

He either used guess and check or my method posted above.

@toadoncart:
We have $ x \equal{} 4a \plus{} 2 \equal{} 7b \plus{} 4$ for INTEGERS $ a,b$. Take this $ \bmod4$ to get $ 3b\equiv2\pmod4$, which is essentially what you had. But if we simply said, for all integers $ c$, $ 3b \equal{} 4c \plus{} 2\implies{b \equal{} \frac {4c \plus{} 2}3}$, we wouldn't always have an integer value for $ b$, which we already assumed above. You can't just plug in any value of $ c$...you have to plug in a value of $ c$ such that $ b$ is an integer. So you need to first solve the congruence $ 3b\equiv2\pmod4\implies{3b\equiv6\pmod4}$, and since $ \gcd(3,4) \equal{} 1$, $ b\equiv2\pmod4\implies{b \equal{} 4c \plus{} 2}$ for some integer $ c$ (notice how this holds for ALL values of $ c$). THEN you can plug it back in and solve. I hope this helps!

it sure does! that was pretty clever.  :)

edit: wat, i just thot of something. u just increased c in c(mod x) (c wuz this word that started with a "r" but i forgot it) by x until it wuz divisible by k in kb (where b is the variable). is it possible for c+ax (where c is increased by x "a" times) to never be divisible by k?  :|

Yes, it is possible for that to happen. For example, say we have $ 2a\equiv3\pmod4$. Well, it is clear that for all integers, we have $ 2a\equiv{0,2}\pmod4$ (0 if a is even and 2 if a is odd).

for all integers $ 2a \equiv 0, 2 \pmod{4}$ (0 if a|2 and 1 otherwise)? idk where u got that...

so wut would u do if that happens?  :huh:

yes, i suck at number theory... :|

Sorry, I typoed in the parentheses...it's edited. So basically, there would be no solutions to that.

sigh...

??? i still don't understand. how could 2a be 3 and 0,2 (mod 4) at the same time (regardless to if a is even/odd)?

yes, i REALLY suck at number theory.  :|  :|  :|

Oh, so that's what you didn't understand. I meant that if you were trying to solve $ 2a\equiv3\pmod4$, there would be no solutions. Sorry! :oops:

division is not defined in modular arithmetic

Yes it is. There are just a few extra rules. To divide $ ax\equiv{b}\pmod{c}$ by $ a$, you have $ x\equiv\frac {b}{a}\pmod{\frac {c}{\gcd(a,c)}}$. Do you see why?

yes, but for MC/AMC 10/12/medium AIME ignoring that is fine.

Um...ok??? It is not exactly fine. Some of the tougher problems do need this fact. The fact is not as complicated as it looks.

[quote]About you first question, how to finish solving "Some number gets a remainder of 5 when divided by seven. The same number gets a remainder of 8 when divided by 11. What is the number?"

You the setup right I believe:

Now we add 58 to both sides to equal everything out. (Add 58 to 5 and you get 63, 63/7=0, and 58+8=66, 66/11=0, you have to "zero" it out)

Since the left and right parts are the same, we can make a new equation with x-58 on one side, and the LCM of 7 and 1 on the other.

We can test 19 put and see it works. So our answer is
[/quote]

its not that hard. they already answered your questions a couple of times.i didn't know about equaling it out and writing a new equation with the LCM.thanks[/quote]

how do solve a system of three or more modular equations?  :huh:

I'll give you a problem...

$ x\equiv1\pmod2$
$ x\equiv3\pmod5$
$ x\equiv2\pmod7$

[hide="Hint"]
You know how to solve systems with $ 2$ equations. Can you reduce the number of equations by solving $ 2$ of them first?
[/hide]

k, i'll first plug in the first equation into the second: $ 2k\plus{}1 \equiv 3 \pmod{5} \longrightarrow 2k \equiv 2\pmod{5} \longrightarrow 2k \equiv 12\pmod{5} \longrightarrow k \equiv 6\pmod{5} \longrightarrow k \equiv 1 \pmod{5}$, so k=5c+1 for some integer c. then plug into next equation? 2(5c+1)=10c+2=x, so $ 10c\plus{}2\equiv 2 \pmod{7} \longrightarrow 10c \equiv 0 \pmod{7} \longrightarrow 10c \equiv 70 \pmod{7} \longrightarrow c \equiv 7 \pmod{7} \longrightarrow c \equiv 0 \pmod{7}$, so c=7a, for some integer a. so 10(7a)+2=70a+2, so erm, i think i got this wrong. ??? idk how to do it.

$ x\equal{}2a\plus{}1$, so we can plug it in to the second equation to get $ 2a\plus{}1\equiv3\pmod5\implies2a\equiv2\pmod5$. We can divide both sides by $ 2$ to get $ a\equiv1\pmod5\implies{a\equal{}5b\plus{}1}$. Plugging this into the first equation, we have $ x\equal{}10b\plus{}3$. You got most of this correct, but you made a mistake when you said that $ x\equal{}10b\plus{}2$. Now try it again. :)

*slams head on desk* tells self: "y did u screw up on ur algebra!!!!!!!!!!!"

$ 10c \plus{} 3\equiv 2 \pmod{7} \longrightarrow 10c\equiv \minus{} 1 \pmod{7} \longrightarrow 10c \equiv 6\pmod{7} \longrightarrow 10c \equiv 20 \pmod{7} \longrightarrow c \equiv 2 \pmod{7}$

so c=7a+2 for some integer a. so 7a+2 plugs into 5c+1, so 5(7a+2)+1=35a+11. now substitute this back into the original x=2k+1. 2(35a+11)+1=70a+23. so x would be 23. :)

so for four or more, u just keep on substituting and stuff, over and over again, right? will they usually ask u 2-3 equations at the max in middle school comps? :|

Actually, $ \bmod$ is usually not mandatory for middle school competitions. I think it may be useful on some problems, but I've never seen one that required it.

there wuz one on this year's mc state comp. (i had to do it the "increase one by one method until u c a number that satisfies it" since i didn't know it back then) :|

Are you talking about State Target #8? That doesn't necessarily need mods. Search it, if you need a solution.

no, i'm talking about the one on the sprint (i forgot which one). do u usually assume x is greater than or equal to 0? if u get like 4c at the end after u substituted the value of x in terms of some variable in the first equation and substituted x=blahblahblah into the next equation to get that variable=something(mod something) and something in the (mod something)=4, do u just put 0 as the answer? :| (that was a bit confusing, but i wuz just really meaning like when u get 4c+3 at the end, for instance, since u substituted and all that and just put the smallest integer for c, which is 0, u would get 3 to be x. but if it had 4c at the end, would u put 0 or 4?)

(i wuz editing some confusing stuff awhile ago)
edit: @math154: #20 :)

Sorry...I don't understand. :(
Could you find the problem from [url=http://mathcounts.org/Document.Doc?id=212]here[/url]?

EDIT: I DIDN'T REALIZE THERE ARE 2 PAGES TO THIS THREAD


[quote="jensen"]if $\frac{sinx+2cosx}{sinx-3cosx}=2$ then $\frac{1}{sinx \times cosx}=?$[/quote]
[hide="see"]we have
$\frac{sinx+2cosx}{sinx-3cosx}=2$
$1+\frac{5cosx}{sinx-3cosx}=2$
$8cosx=sinx$
then$\frac{1}{sinx \times cosx}=\frac{1}{8 cos^{2}x}$=$\frac{65}{8}$[/hide]


[quote="ninni"]$ f(x) \equal{} (a \minus{} x)^{1/n}$[/quote]This function is not compositely periodic of any order. I think you meant $ f(x)\equal{}(a\minus{}x^n)^\frac1{n}$.

[quote="10000th User"][quote="ninni"]$ f(x) \equal{} (a \minus{} x)^{1/n}$[/quote]This function is not compositely periodic of any order. I think you meant $ f(x) \equal{} (a \minus{} x^n)^\frac1{n}$.[/quote]

oh yeah sorry.

[quote="ninni"]If $ f(x) \equal{} (a \minus{} x^n)^{1/n},a > 0$ and $ n\epsilon N$, then prove that $ f(f(x)) \equal{} x$ for all x. Please help. Thank you.[/quote]
[url=http://www.servimg.com/image_preview.php?i=266&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk14.jpg[/img][/url]
 :pilot:


superior algebra
superior algebra solved

Let $(G,\cdot)$ be a group. There exists and element $a\in G$ s.t.  $x^4=axa^{-1}$ for all $x\in G$. Prove that $x^9=1$ for all $x\in G$.
What about the general case $x^p=axa^{-1},\forall x\in G$, $ p\in\mathbb{N}^*$?

Well, it's not the result asked for, but I can show that $x^{63}=1.$ Maybe that's a start.

The identity applies to $a$ in the place of $x$, so $a^3=1.$

Write the identity as $ax=x^4a$. Moving an element across $a$ raises it to the 4th power.

$x=a^3x=a^2x^4a=ax^{16}a^2=x^{64}a^3=x^{64}$ so $x^{63}=1.$

I also managed to prove this.. but now I don't know what to do next... If the group $G$ would be finite it is suffices to prove that there are no elements $x\in G$ with $x^7=1$ (I couldn't prove this either). But... we have only that each element of $G$ has a finite order...

Actually, it suffices to show that the only element with $x^7=1$ is $x=1$, because $(x^9)^7=1,\ \forall x\in G$, so if we show that, we have $x^9=1,\ \forall x\in G$. I tried to prove it for a while, but failed and let it go. It's not really a nice, inspiring problem, is it? :)

neah... it's not... it's a buggy one.. but is was at the local olympiad...  :( and at the answers it writes to look in the manual where is treated the more general form.. $x^p=axa^{-1}$  :(

:blush:
Because $axa^{-1}$ is an isomorphism of $G$ we have that $x^4$ is an isomorphism:
so $(xy)^4=x^4y^4$ from which results that $(yx)^3=x^3y^3$ (1) (we should not forget that $a^3=1$ which follows from the hypothesis for $x=a$)
Now $x^4=axa^{-1}\Leftrightarrow x^3a=x^{-1}ax\mid ^3\Leftrightarrow (ax^3)^3=x^{-1}a^3x=x^{-1}x=1$ so we have that $(ax^3)^3=1$ but from (1) this is equivalent with $1=(ax^3)^3=(x^3)^3a^3=x^9\Rightarrow x^9=1$

It seems to me I found the solution. Suppose $G$ is a group and
there exists some $ p \in \mathbb{N} $ and $ a \in G $ such that
for every $ x \in G $, $ x^{p}= axa^{-1}$. Denote now $ \varphi :
G \rightarrow G $, $ \varphi(x):= x^{p} = axa^{-1} $. From $
\varphi(x) = axa^{-1} $ we see that $\varphi$ is an endomorphism
of $ G $. From $ \varphi(x):= x^{p} $ we see that for every $ x,y
\in G $, $ \varphi (y^{-1}xy) = y^{-1}\varphi(x)y $. So if we
denote $ H:= \{ x\in G | \varphi(x)=x \} $ then $ H \subset G $ is
a normal subgroup. Also evidently $ a \in H $ and for every $ x
\in H $, $ x^{p-1} = x $. Now , if we take quotient $ K:= G/H $,
then because of $ a \in H $ and $ x^{p}= axa^{-1}$ for every $ x
\in G $, we get that $ z^{p}=z $ for every $ z \in K $, hence $
z^{p-1} = 1 $ for every $ z \in K $, so for every $ x \in G $, $
y=x^{p-1} \in H $, therefore , as we have seen, $ x^{(p-1)^{2}} =
y^{p-1} = 1 $ in $ G $. So for every $ x \in G $, $ x^{(p-1)^{2}}
= 1 $.


induction
inequalities
superior algebra
superior algebra theorems

given a set with the symbols :

" + " for addition

" - " for inverse

the constants :

0

AND the axioms:

for all a,b,c : a+(b+c) = (a+b) + c

for all a : a+0 = a

for all a : a +(-a) =0

for all a,b : a+b = b+a

PROVE using CONTRADICTION

1) THE uniqness of zero

2) THE uniqness of the inverse

Just assume there's two and show their difference is equal to zero.

[quote="ZetaX"]Then just do so  :wink:
Sorry, but we will not do all of your homework. You may post what you did so far.[/quote]

[b]THIS IS NOT HOMEWORK[/b]

I HAVE looked uppon nearly all books of linear and abstract algebra [b]and none of them gives a proof by contradiction[/b]

That inspired the question in concern.

I tried my self but i got mixed up along the way .

SO if anybody can do it i will be pleased to learn

Ok, a start: Assume that $ a,b$ both work as zero but are different. Then $ a\equal{}a\plus{}b\equal{}b$, contradiction.

But that kind of proof is somewhat a fake contradiction.

[quote="ZetaX"]Ok, a start: Assume that $ a,b$ both work as zero but are different. Then $ a \equal{} a \plus{} b \equal{} b$, contradiction.

But that kind of proof is somewhat a fake contradiction.[/quote]

But this proof is not even fake contradiction.

This is a direct proof coming from the definition of uniqness of zero

This is the proof produced in linear and abstract algebra books

[quote="ZetaX"]It's the standard proof if you delete "assume" and "contradiction". Nonetheless, it's rather useless to look for proof by contradiction: every proof can made to contain one, but some just don't do canonically (and especially, you loose some information, namely constructivity, if doing it by contradiction).[/quote]

CAN you do,please the contradiction proof.

I am very interested to see how is done

[quote="ZetaX"]I already mentioned one. You can also do it the stupid way by just writing $ \neg \neg  A \implies  A$ (being the main ingredient for contradictions) somewhere into your proof without any further reason.
But you better learn that mathematics doesn't work that way  :wink:[/quote]

CAN YOU PLEASE write [b] A COMPLETE CONTRADICTORY PROOF[/b] instead of confusing me with hints??

The one that you suggested ,as i said, [b] IS NOT A CONTRADICTORY PROOF[/b]

[quote="ZetaX"]Can you please just stop shouting¿ You better define what you mean by a "contradictory proof" instead of just telling people that their way is wrong.
I didn't give you just hints, I gave you a full explaination. And I repeat myself: this is not the way mathematics works.[/quote]

So we want to prove that:

[b]There exists a unique y for all x such that: x+y =x[/b]

And now using contradiction we must negate the above statement

[b]AND the question  now is what is the negation of the above statement??[/b]

In quantifier form the above statement is:

$ \exists y!\forall x[ x\plus{}y\equal{}x]$...... ($ \exists !y$ means there exists a unique y) , which by definition is equivalent to:

$ \exists y\forall x[x\plus{}y \equal{} x]$ [b]AND[/b] $ \forall y\forall z[\forall x( x\plus{}y\equal{}x)\wedge\forall x(x\plus{}z\equal{}x)\Longrightarrow y\equal{}z]$,

which in words is:

there exists a, y for all x such that :  x+y =x

AND

FOR all y,z :if for all x ( x+y=x) and for all x(x+z=x) ,then y=z

Now to negate the above is not an easy task.

can you curry on from here?

The negation of the original statement
[quote]
There exists a unique y for all x such that: x+y =x
[/quote]
is quite easy to make as long as you do not try to write it in formal language and confuse yourself and others. It is
(for all $ y$ there exists $ x$ such that $ x \plus{} y\ne x$) [b]or[/b] (there exist $ y'\ne y''$ such that for all $ x$, we have $ x \plus{} y' \equal{} x \plus{} y'' \equal{} x$)

Anyway, what's the point in insisting on a proof by contradiction? :? I mean, you could also ask us to derive this particular statement using mathematical induction or whatever else that might come into your head. My natural reaction would be "What for?". Normally, you just use the most suitable tools when solving the problem, not the ones artificially prescribed by somebody else. What you ask us to do is more or less equivalent to climbing a tree in snow shoes. OK, this can be done, but the easiest way to do it would be just to hold the snow shoes in your teeth to have the arms and the legs free. :P.

if every time in mathematics you asked "what for" mathematics would not exist.

unless you do not know how to finish the proof by contradiction ,and this is an excuse.

EUCLID was asked the same question by one of his pupils in trying to prove a geometrical theorem.

Then EUCLID ordered one of his servants to pay his pupil every time ,he (the pupil that is) proved a theorem

MANY times in mathematics one is asked to prove a theorem a particular way ,although there is another perhaps easier.

Apart from their [b] practical application [/b] mathematics is a game of arguments

I still don't know why exactly you doubt the proof I gave doesn't fit (except that from a very formal point of view, this proof is not using the axiom of "tertium non datur"). Uniqueness (ignoring existence, which is already given) of zero just says: $ \forall a,b: \mathrm{zero}(a) \wedge \mathrm{zero}(b) \implies a\equal{}b$; here $ \mathrm{zero}(a)$ is just $ \forall x: x\plus{}a\equal{}x$. The contraposition is what I assumed.

[quote="τρικλινο"]MANY times in mathematics one is asked to prove a theorem a particular way ,although there is another perhaps easier.[/quote]
In fact, if no reason is given, this almost everytime is just a very stupid exercise then. It's often nothing more than restriction of free mind.
And even more, I never saw any formal definition of "prove X using Y"; from a theoretical point of view, you could just write down $ Y$ without further reason, followed by a $ Y$-less proof of $ X$.

This becomes too philosophical for me. OK, the OP wanted to play a certain mathematical game with us, so be it. One question is more than enough for this game, so I'll choose the one about the uniqueness of zero.

Assume that there exists a zero $ a$ different from $ 0$. Consider any $ x\in X$. If $ x\ne 0$, then $ x\plus{}a\equal{}x\ne 0$. If $ x\equal{}0$, then $ x\plus{}a\equal{}0\plus{}a\equal{}a\plus{}0\equal{}a\ne 0$. Thus, for all $ x\in X$, we have $ x\plus{}a\ne 0$, which contradicts the existence of $ \minus{}a$ in $ X$. Note that this proof is different because it also works under the assumption that $ a$ is a "partial" zero, i.e., the identity $ x\plus{}a\equal{}x$ needs to hold only for $ x\ne 0$, which destroys the "standard" argument completely.

So, τρικλινο, if you are still not satisfied with it, tell us what exactly is wrong now. If it is OK with you, the turn passes to you. Your task is to prove the uniqueness of zero using math. induction.

[quote="fedja"]This becomes too philosophical for me. OK, the OP wanted to play a certain mathematical game with us, so be it. One question is more than enough for this game, so I'll choose the one about the uniqueness of zero.

Assume that there exists a zero $ a$ different from $ 0$. Consider any $ x\in X$. If $ x\ne 0$, then $ x \plus{} a \equal{} x\ne 0$. If $ x \equal{} 0$, then $ x \plus{} a \equal{} 0 \plus{} a \equal{} a \plus{} 0 \equal{} a\ne 0$. Thus, for all $ x\in X$, we have $ x \plus{} a\ne 0$, which contradicts the existence of $ \minus{} a$ in $ X$. Note that this proof is different because it also works under the assumption that $ a$ is a "partial" zero, i.e., the identity $ x \plus{} a \equal{} x$ needs to hold only for $ x\ne 0$, which destroys the "standard" argument completely.

So, τρικλινο, if you are still not satisfied with it, tell us what exactly is wrong now. If it is OK with you, the turn passes to you. Your task is to prove the uniqueness of zero using math. induction.[/quote]

If, a is another identity element then certainly [b]we do not have[/b] : for all ,x $ ( x \plus{} a\neq 0)$ ,because for all x we do have ( x + a = x ) and for x =0 , ( o +a = 0 ).

If a is a "partial identity element" there is no argument to follow whatsoever.

FURTHER MORE to prove indirectly that there  exist a unique zero one has to deny the unique existence of zero partially or completely i.e:

either deny the existence part and end up with a contradiction (this is a partial denial )

or deny the uniqueness part and end up with a contradiction (this is a partial denial again)

or deny the existence and the uniqueness and end up with a contradiction ( this the complete denial )

[quote="ZetaX"]I still don't know why exactly you doubt the proof I gave doesn't fit (except that from a very formal point of view, this proof is not using the axiom of "tertium non datur"). Uniqueness (ignoring existence, which is already given) of zero just says: $ \forall a,b: \mathrm{zero}(a) \wedge \mathrm{zero}(b) \implies a \equal{} b$; here $ \mathrm{zero}(a)$ is just $ \forall x: x \plus{} a \equal{} x$. The contraposition is what I assumed.

.[/quote]

Uniqueness is [b]defined[/b] as follows :

$ \forall a\forall b [(\forall x ( x \plus{} a \equal{} x )\wedge\forall x ( x \plus{} b \equal{}x ))\Longrightarrow a\equal{}b]$

and using contraposition the above formula is equivalent to:

$ \forall a\forall b [ a\neq b\Longrightarrow (\exists x ( x \plus{} a\neq x)\vee\exists x ( x \plus{} b\neq x))]$

But contraposition is not contradiction

What is "tertium non datur" ??

[quote="τρικλινο"]
If, a is another identity element then certainly [b]we do not have[/b] : for all ,x $ ( x \plus{} a\neq 0)$ ,because for all x we do have ( x + a = x ) and for x =0 , ( o +a = 0 ).
[/quote]
Objection rejected. This merely means that we can get the contradiction earlier. Also, any attempt to use the identity $ 0\plus{}a\equal{}0$ will result in just a repetition of the standard proof, which, by the rules of the current game, should not be repeated. So, no, you cannot use $ 0\plus{}a\equal{}0$ in either the proof, or your objections to it (used in the proof, it will mean repetition of the standard approach, used in an objection, it will just mean that there exists another proof, which we know perfectly well anyway).
[quote]
If a is a "partial identity element" there is no argument to follow whatsoever.
[/quote]
Objection rejected. The proof I presented works in this case perfectly well.
[quote]
or deny the uniqueness part and end up with a contradiction (this is a partial denial again)
[/quote]
This is exactly what has been done. So, this comment is considered empty.

Is there anything else you want to say before embarking upon the proof by math. induction challenge?

嗯哦 屗嗯大屗亽撒他啊嗯大

[quote="JBL"]Well, that was no less meaningless than most of your other comments on this thread  :roll:[/quote]

I like forums with a high sense of humor and I am glad you are watching the discussion

[quote="Fermat's Little Turtle"][url=http://en.wikipedia.org/wiki/Law_of_excluded_middle]Google is not hard to use[/url]

How does that make you feel? The word choices I have used for the link. When you state something like "But contraposition is not contradiction" bare in mind how childish and condescending you sound. Both of these people are trying to sort this out with you so appreciate their efforts. Also they are well developed mathematicians, try not to take yourself for granted so much. And your definitions, where do you get them from? A book? Online source? etc..[/quote]

The author is: ANGELO MARGARIS

The book is : FIRST ORDER MATHEMATICAL LOGIC

The pages are :

1) PAGE 109 .for the definition of uniqueness and

2)page 121 ,where he produces a formal proof for the uniqueness of the identity element , which  is non contradictory ,and which is the ordinary mathematical proof that many books follow and a lot of people wrongly consider as contradictory

Other ,books explaning quite well contradiction are:

1) THEORY AND PROBLEMS OF LOGIC ,in Schaum's OUTLINE SERIES

2)INTRODUCTION TO LOGIC ,BY IRVING COPI

3)SYMBOLIC LOGIC ,BY IRVING .M. COPI again

of course there is a dozen of other excellent books ,but to start with, those are the books recommended by University professors.

And again other professors may recommend a different package

[quote="fedja"][quote="τρικλινο"]
If, a is another identity element then certainly [b]we do not have[/b] : for all ,x $ ( x \plus{} a\neq 0)$ ,because for all x we do have ( x + a = x ) and for x =0 , ( o +a = 0 ).
[/quote]
Objection rejected. This merely means that we can get the contradiction earlier. Also, any attempt to use the identity $ 0 \plus{} a \equal{} 0$ will result in just a repetition of the standard proof, which, by the rules of the current game, should not be repeated. So, no, you cannot use $ 0 \plus{} a \equal{} 0$ in either the proof, or your objections to it (used in the proof, it will mean repetition of the standard approach, used in an objection, it will just mean that there exists another proof, which we know perfectly well anyway).
[quote]
If a is a "partial identity element" there is no argument to follow whatsoever.
[/quote]
Objection rejected. The proof I presented works in this case perfectly well.
[quote]
or deny the uniqueness part and end up with a contradiction (this is a partial denial again)

[/quote]
This is exactly what has been done. So, this comment is considered empty.

Is there anything else you want to say before embarking upon the proof by math. induction challenge?[/quote]

Before I venture into the unknown fields of discovering the uniqueness of the identity element by using induction,I will kindly ask you to rewrite the proof you wrote in post #16 ,in steps ,justifying each step if possible.

This will help us to really examine your proof and throw away any misunderstanding

[quote="τρικλινο"]
Before I venture into the unknown fields of discovering the uniqueness of the identity element by using induction,I will kindly ask you to rewrite the proof you wrote in post #16 ,in steps ,justifying each step if possible.[/quote]
This, certainly, can be done :).

[b]Step 1[/b]. Assume that there exists $ a\ne 0$ in $ X$ such that $ x\plus{}a\equal{}x$ for all $ x\in X$ (or even just for all $ x\ne 0$). This is just the negation of uniqueness.

[b]Step 2[/b] Let $ x\in X$. We want to show that $ x\plus{}a\ne 0$. There are two cases to consider:

 [b]Case 2.1[/b] $ x\ne 0$. Then $ x\plus{}a\equal{}x\ne 0$ by the assumption that $ a$ is a (partial) zero.
 [b]Case 2.2[/b] $ x\equal{}0$. Here we do [i]not[/i] want to use that $ a$ is a zero at all but want to use only that $ 0$ is a zero. So, $ x\plus{}a\stackrel 1\equal{}0\plus{}a\stackrel 2\equal{}a\plus{}0\stackrel 3\equal{}a\ne 0$. The first equality holds because $ x\equal{}0$, the second is the commutativity axiom, the third holds because $ 0$ is a zero, and the last inequality is our assumption that $ a\ne 0$ made in Step 1.

[b]Step 3[/b]. Since we found that $ x\plus{}a\ne 0$ in both cases and since for each $ x$, either Case 2.1, or Case 2.2 holds (the law of excluded middle is used here, of course, but all logic systems that allow the proofs by contradiction have to contain some form of it anyway), we conclude that for every $ x\in X$, one has $ x\plus{}a\ne 0$.

[b]Step 4[/b]. On the other hand, by the axiom about $ \minus{}a$, we have $ (\minus{}a)\plus{}a\equal{}0$. Thus, there does exist $ x\in X$ (namely, $ \minus{}a$) satisfying $ x\plus{}a\equal{}0$. This is in direct contradiction with the result of Step 3. The end.

Feel free to ask as many further [b]meaningful[/b] questions about this argument as you need. ;)

OBJECTION:

Each step must be :

1) an axiom ,or

2) The conclusion of the application of a law of logic in one or more previous steps

In that way we be able to say : that law or that axiom is wrong.

For example , for the first step we want to negate the uniqueness ,so that we can start a contradiction.

SO.

STEP 1: $ \neg\forall a\forall b[(\forall x ( x \plus{} a \equal{} x)\wedge\forall x ( x \plus{} b \equal{} x))\Longrightarrow a \equal{} b]$..............................................................................................this is an assumption for the contradiction law ( a law in logic)

STEP 2: $ \exists a\exists b[\neg[(\forall x ( x \plus{} a \equal{} x)\wedge\forall x ( x \plus{} b \equal{} x))\Longrightarrow a \equal{} b]]$..............................................................................................this is the result by applying the following law of logic in step (1) : $ \neg\forall x[ Q(x)]\Longrightarrow\exists x[\neg Q(x)]$

STEP 3: $ \neg[(\forall x ( x \plus{} a \equal{} x)\wedge\forall x ( x \plus{} b \equal{} x))\Longrightarrow a \equal{} b]$..................................................................................................this is the result by applying the law of Existential Elimination in step (2)

STEP 4: $ (\forall x ( x \plus{} a \equal{} x)\wedge\forall x ( x \plus{} b \equal{} x))\wedge a\neq b$.........................................................................................................this is the result by applying the following law in step (3): $ \neg[ (p\wedge q)\Longrightarrow r]\Longrightarrow [(p\wedge q)\wedge \neg r]$

STEP 5 : $ (\forall x ( x \plus{} a \equal{} x)\wedge\forall x ( x \plus{} b \equal{} x))$........................................................................................................this is the result by applying the law called addition elimination in step (4)

e.t.c  .............e.t.c

I see what you mean but I'm too lazy to do that. ;) This would just take about an hour of my time and, while I generally do not value my time too much, I still have to draw a line somewhere as far as wasting it is concerned :P. Converting verbal proofs into symbolic ones is certainly beyond that line. So, we'll have to patiently wait now until somebody else does that for us.  :D


trigonometry
calculus
calculus computations

Since $ \sin x\equal{}\sum_{n\equal{}0}^{\infty}(\minus{}1)^n\frac{x^{2n\plus{}1}}{(2n\plus{}1)!}$, it follows that $ x^2\sin x\equal{}\sum_{n\equal{}0}^{\infty}(\minus{}1)^n\frac{x^{2n\plus{}3}}{(2n\plus{}1)!}$. You should find the coefficient of $ x^{12}$ in this series, because the coefficient of $ x^{12}$ in the Maclaurin series of $ f$ is equal to $ \frac{f^{(12)}(0)}{12!}$.


[quote="redkimchi"]Let $A_{1}, A_{2}, A_{3}, \ldots$ be a sequence of sets such that
\[i \neq j \implies A_{i}\neq A_{j}\]
Let $P$ be the set of all sets that can be obtained from $A_{1}, A_{2}, ...$ via a finite number of set operations : union, intersection, and set-minus.  (not including infinite union nor infinite intersection).
For example, $(A_{3}-A_{2}) \cup A_{9}\in P$.

Prove that $P$ contains infinitely many disjoint sets, i.e. there exists non-empty sets $B_{1}, B_{2}, ...$ in $P$ with
\[i \neq j \implies B_{i}\cap B_{j}= \emptyset \]
[/quote]

Given a finite number of sets, their union provides a certain universe of elements, and by forming all possible Boolean combinations of these sets the universe is partitioned into a finite number of disjoint pieces, call them "molecules".  Each time a new set $A$ is given, it can add a new molecule consisting of elements (if any) not in the earlier sets, and it could split some of the existing molecules (that is, if for molecule $X$ the partition into $X \cap A$ and $X-A$ is nontrivial, it is split into two new molecules).

If there are infinitely many stages where new elements are added, the problem is trivial.   If not, consider the splitting process if the universe stays fixed after some point.

If the splitting process stabilizes at a population of $n$ molecules, then all the sets  $A_{i}$ are finite unions of some of these molecules, so there are at most $2^{n}$ different possible sets on the list.    So there are infinitely many stages where molecules are split.    Koenig's tree lemma shows that the tree of molecules has an infinite branch.   If one descends this branch then the halves of the molecules that are pruned away from the branch form an infinite collection of distinct sets.