modular arithmetic
ratio
number theory unsolved
number theory

For a positive integer, we define it's [i]set of exponents[/i] the unordered list of all the exponents of the primes, in it`s decomposition. For example, $18=2\cdot 3^{2}$ has it`s set of exponents $1,2$ and $300=2^{2}\cdot 3\cdot 5^{2}$ has it`s set of exponents $1,2,2$. There are given two arithmetical progressions $\big(a_{n}\big)_{n}$ and $\big(b_{n}\big)_{n}$, such that for any positive integer $n$, $a_{n}$ and $b_{n}$ have the same set of exponents. Prove that the progressions are proportional (that is, there is $k$ such that $a_{n}=kb_{n}$ for any $n$).

[i]Proposed by A. Golovanov[/i]

I cannot assure everything is right; however...
Let $a_{n}=qn+A$ and $b_{n}=rn+A$ with $n\geq 0$ . Then taking $n=m(AB)^{2}$ we have that  $a_{m(AB)^{2}}=A(mAB^{2}q+1)$ has the same set of exponents as $b_{m(AB)^{2}}=B(mA^{2}Br+1)$ for all $m\geq 0$. Since $A=a_{0}$ and $B=b_{0}$ have the same set of exponents and no prime dividing  $A$ ($B$) divides  $mAB^{2}q+1$  ($mA^{2}Br+1$)  this means that also  $mAB^{2}q+1$ and $mA^{2}Br+1$ have the same set of exponents for all $m\geq 0$.
Let  $AB^{2}q=Q$ and  $A^{2}Br=R$. Suppose  $Q$ and  $R$ are different and without loss of generality, suppose  $Q<R$. $Q=1$ leads to  $A=B=q=1$, which is easy to verify that doesn't work with $Q<R$ (take  $m=R+2$ and  $m=R-2$ and see that the condition implies the existance of two positive squares at distance 4 from each other), so suppose $Q>1$. Take a $x>1$ such that  $hcf(x,\phi(R))=1$ which means that modulo $Q$ and modulo $R$ the only solution to $a^{x}\equiv1$ is $a\equiv1$. When  $m=\frac{(Q+1)^{x}-1}{Q}$, $a_{m}=(Q+1)^{x}$, so in order to have the same exponent set, $b_{m}$ should be an $x$th power too. However:  $b_{m}=\frac{(Q+1)^{x}-1}{Q}R+1<\frac{(R+1)^{x}-1}{R}R+1=(R+1)^{x}$ and $(R+1)^{x}$ is the least $x$th power $>1$ and $\equiv1$ modulo R, so we must admit $Q=R$, which means $AB^{2}q=A^{2}Br \rightarrow \; \frac{A}{B}=\frac{q}{r}=\frac{a_{n}}{b_{n}}$, for all $n$.

[quote="barasawala"][quote="Mimoide"]When  $m=\frac{(Q+1)^{x}-1}{Q}$, $a_{m}=(Q+1)^{x}$,[/quote]

I don't think this is right!  :([/quote]

So how will you fix it?

Perhaps this is simpler:

Let $a_{n}=qn+A$, $b_{n}=rn+B$, and take $n=m(AB)^{2}$ for all $m$. Then we get the sequences $c_{m}=qmA^{2}B^{2}+A=A(qAB^{2}m+1), d_{m}=B(rA^{2}Bm+1)$. Now let $p$ be a prime which is coprime to $q, r, A, B$. Then of course we can find arbitrarily large $k$s such that $p^{k}\equiv 1 \pmod{qAB^{2}}$, and for each such $k$ there exists $m$ such that $c_{m}=Ap^{k}$. Now, since $A$ and $B$ have the same set of exponents and they are coprime to $qAB^{2}m+1$ and $rA^{2}Bm+1$, this means that $p^{k}$ and $rA^{2}Bm+1$ must have the same set of exponents too. However, the set of exponents of $p^{k}$ is just $\{k \}$, so $rA^{2}Bm+1=s^{k}$ for some prime $s$. However, if we take a large enough $k$, then for any $s$ not equal to $p$ the ratio between $Ap^{k}$ and $Bs^{k}$ will be too large to be asymptotically equal to the ratio between $q$ and $r$, since it will get bigger the bigger $k$ is. Then for infinitely many $m$ we have $c_{m}=Ap^{k}$ and $d_{m}=Bp^{k}$. This implies that $q/r=A/B$, and we're done.

[quote="Severius"]then for any $s$ not equal to $p$ the ratio between $Ap^{k}$ and $Bs^{k}$ will be too large to be asymptotically equal to the ratio between $q$ and $r$, [/quote]

What does this mean?

Okay... suppose $c_{m}=Ap^{k}$ and $d_{m}=Bs^{k}$ where $s$ is a prime not equal to $p$. Then $\frac{c_{m}}{d_{m}}=\frac{A}{B}( \frac{p}{s})^{k}$ (*).

Let $p_{1}< p_{2}< \ldots$ be the sequence of all primes and $p=p_{l}$; then if $s<p$, we have $s \leq p_{l-1}$, while if $s>p$, we have $s \geq p_{l+1}$. Now, look at $\frac{c_{m}}{d_{m}}$ again. Evidently, that ratio should tend to $\frac{q}{r}$ as $m$ tends to infinity. However, for those $m$ such that $c_{m}=Ap^{k}$ and $d_{m}=Bs^{k}$, if $p \neq s$ then $( \frac{p}{s})^{k}$ will either be $\geq ( \frac{p}{p_{l-1}})^{k}$ or $\leq ( \frac{p}{p_{l+1}})^{k}$. If we take $k$ large enough, these numbers will be really big and really small respectively, so that they can't tend to a nonzero constant as $m$ tends to infinity.

A bit overpowered but the problem is anyway not an easy one. Very nice! :)

Let us write $$a_n=g_1(c_1n+d_1)$$ and $$b_n=g_2(c_2n+d_2)$$ where $g_1,g_2$ are non zero integers and $c_1,d_1$ and $c_2,d_2$ are relatively prime integers, with $c_1$ and $c_2$ non zero. Let $M>0$ be any natural number and $q$ be any sufficiently large prime number. Define the prime number $p$ (using the notion of inverse congruence) as follows: $$p \equiv \, d_1 \, (\bmod \, c_1)$$ and $$p \equiv \, d_1-\frac{c_1}{c_2}\cdot d_2 \, (\bmod \,q^M)$$ and observe that such a prime number $p$ exists by the Dirichlet's Theorem and the Chinese Remainder Theorem. Let $p=c_1n+d_1$ and note that $a_n=g_1p$ and $$b_n=\frac{g_2}{c_1}\cdot \left(c_2(p-d_1)+c_1d_2\right)$$ and we get that $q^M \mid b_n$. In particular, a number $v_q(b_n) \ge M$ belongs in the set of exponents of $b_n$. However, the largest element in the set of exponents of $a_n$ in this case, is bounded by the largest exponent of a prime dividing $g_1$ and we get a contradiction as $M$ sufficiently large violates our hypothesis. Thus, either $d_1=0$ or $\frac{c_1}{c_2}=\frac{d_1}{d_2}=l$; in the former case, set $n=p$ satisfying $$p \equiv \, -\frac{c_1}{c_2}\cdot d_2 \, (\bmod \, q^M)$$ and repeat the same argument. In the latter case, our conclusion follows by setting $k=\frac{g_2}{g_1l}$, $b_n=ka_n$ for all $n$. Other trivial possibilities, where either of the sequences is constant are ignored and can be resolved in a fairly easy way following the same argument.

Suppose that $a_n=a_0+nr,b_n=b_0+ns$ for all positive integer $n$
Consider $a_m,b_m$ where $m=la_0^2b_0^2$ we get $a_m=a_0(1+lb_0^2a_0r),b_m=b_0(1+la_0^2b_0s)$
Let $f(n)$ denote "set of exponent" of $n$ for all $n\in \mathbb{Z}^+$
Since $gcd(a_0,1+lb_0^2a_0r)=1$ we get that $|f(a_0)\cap f(1+lb_0^2a_0r)|=0$
WLOG that $a_0s\geq b_0r$, if $a_0s=b_0r$ we are done, so suppose that $a_0s>b_0r$
Since there exist infinitely many prime $p$ that $b_0^2a_0r\mid p-1$
Take $l=\frac{p-1}{b_0^2a_0r}$ we get that $f(a_m)=f(a_0)\cup \{ 1\} =f(b_0)\cup f(1+la_0^2b_0s) \Rightarrow 1+|f(a_0)|\geq |f(b_0)|+|f(1+la_0^2b_0s)|\geq |f(b_0)|+1$
So $|f(a_0)|\geq |f(b_0)|$, similarly  $|f(b_0)|\geq |f(a_0)|$, so $|f(a_0)|=|f(b_0)|$
Take $l=\frac{p^z-1}{b_0^2a_0r},z>\max \{ t\mid t\in f(a_0)\cup f(b_0)\}$ we get that $f(a_m)=f(a_0)\cup \{ z\} =f(b_0)\cup f(1+la_0^2b_0s)$
So $f(a_0)=f(b_0)$, so $\{ z\} =f(1+la_0^2b_0s)$
This give us for each $z\in \mathbb{Z}^+$ there exist prime number $q$ such that $q^z=1+\frac{p^z-1}{b_0^2a_0r}\times a_0^2b_0s$
Consider $z>\log_{\frac{p+1}{p}}\left( \frac{a_0s}{b_0r}\right)$ we get $\frac{a_.s}{b_0r}<\left( \frac{p+1}{p}\right)^z\rightarrow \sqrt[z]{\frac{a_0s}{b_0r}} <\frac{p+1}{p}$
Since $q^z=1+\frac{p^z-1}{b_0^2a_0r}\times a_0^2b_0s$ we get $1< \frac{a_0s}{b_0r}=\frac{q^z-1}{p^z-1}$, so $q>p$
Since $q>p$ we easily get that $\frac{q^z-1}{p^z-1} \geq \frac{q^z}{p^z}$, so $\frac{a_0s}{b_0r} \geq \frac{q^z}{p^z}\rightarrow \sqrt[z]{\frac{a_0s}{b_0r}} \geq \frac{q}{p}$ but we have $ \sqrt[z]{\frac{a_0s}{b_0r}} <\frac{p+1}{p}$, so $p+1>q$ impossible

Finally, we get $a_0s=b_0r$ and so the sequence $(a_n),(b_n)$ satisfy $\frac{a_n}{b_n}=\frac{a_0}{b_0}=\frac{r}{s}$ for all $n\in \mathbb{Z}^+$ and we are done.

Let $a_n=d(a_0+xn),\ b_n=e(b_0+yn)$ with $(a_0,x)=(b_0,y)=1$. By list I mean set of exponents.

Note that the arithmetic progressions $\alpha_n=d(a_0+xa_0^2b_0^2den)$ and $\beta_n=e(b_0+ya_0^2b_0^2den)$ also have the same list for any $n$ as they are just corresponding subsequences of $(a_n)_{n\ge 0}$ and $(b_n)_{n\ge 0}$.

Write $\alpha_n=da_0(1+kn),\ \beta_n=eb_0(1+\ell n)$ with $da_0|k$ and $eb_0|\ell$. We thus have that $(da_0,1+kn)=1$ for any $n$, so the list of of $\alpha_n$ is the reunion of the list of $da_0$ and the list of $1+kn$ (and the similar for $\beta_n$). Picking $n_0=\dfrac{(k+1)^{\varphi (k) t}-1}{k}$ with $t$ greater than anything in the list of $eb_0$, we have that the list of $\alpha_{n_0}$ is the reunion of the list of $da_0$ with $\varphi(k)t$. The list of $\beta_{n_0}$ is the reunion of the list of $eb_0$ and the list of $1+\ell n$. As the two lists have to be equal, we infer that the list of $da_0$ is greater or equal in length than the list of $eb_0$. Similarly we can get the reversed inequality, so the two lists actually have the same length. Nonetheless, $\varphi(k)t$ is greater than anything in the list of $eb_0$, so actually the lists of $eb_0$ and $da_0$ are the same.

By picking $n=kt^2+2t$ with variable $t$, we have $1+kn=(kt+1)^2$, so the list of $\alpha_n$ is the reunion of the list of $da_0$ with an array of even numbers. As $eb_0$ and $da_0$ have the same list and $\alpha_n$ and $\beta_n$ have the same list, we infer that the list of $1+\ell n$ is an array of even numbers, i.e. $1+\ell n$ is a perfect square for any $n=kt^2+2t$. Thus, $\ell k t^2+2tl+1$ is a perfect square for any $t$. It is easy [hide="to see"]If $an^2+bn+c$ with $a,b,c\in \mathbb{N}$ is a perfect square for any $n\in \mathbb{N}$, then $a=x^2,b=2xy,c=y^2$ with $x,y\in \mathbb{N}$. Note that there is an infinite number of primes $p$ dividing $(an^2+bn+c)_{n\ge 1}$ (unless $a=b=0$ in which case the claim is obvious). Pick $p>a$ a prime dividing $an^2+bn+c$. If $p$ divides $an^2+bn+c$, then $p^2$ divides $an^2+bn+c$ because it is a square. However, $p$ also divides $a(n+p)^2+b(n+p)+c$, so $p^2$ divides $a(n+p)^2+b(n+p)+c$. Substracting, we get that $p$ divides $2an+b$. Substituting in $p|an^2+bn+c$, we get $p|b^2-4ac$. This happens for an infinite number of primes, whence $b^2-4ac=0$, so $an^2+bn+c=a \left ( n+\dfrac{b}{2a} \right )^2$. From here it is easy to see that $a$ has to be a square and that $\dfrac{b}{\sqrt{a}}\in \mathbb{N}$ and the conclusion follows.[/hide] that $\ell k=a^2, 2l=2ab, 1=b^2$, so $\ell=k$, or equivalently $xb_0=ya_0$.

We are now done: $\dfrac{a_n}{b_n}=\dfrac{ d \left ( a_0+ \frac{ya_0}{b_0}n \right ) }{e(b_0+yn)}=\dfrac{da_0}{eb_0}$ for any $n$.


Since there are many interested people in olympiads (including me), I thought we could post problems and solve them to help others who may need it.

The first question i give is from chemistry. I deliberately give a rather simple one so that the 11thers can also try. If any 12ther wants to give a solution, then please hide them.

Chemical bonding (Source: USA chemistry olympiad 1st round):

Xenon forms several compounds including XeF2, XeF4 and XeO3.

(1) Draw Lewis structures for the above compunds.
(2) Describe the geometry of each molecule including bond angles.
(3) State whether the above molecules are polar or non polar.
(4) Account for the fact that these molecules are highly reactive.

same here !
i think i would do it with the help of geogabra !
a prob might be there with the geometry!
 :)

[quote="rituraj007"]these r tooooooooooo easy
as even i cud solve them......
 :rotfl:[/quote]

He mentioned in the beginning only that its a simple one!  :P
Why dont you ppl post your solutions?  :huh:

hell,....
Should i post some of my sir's "reaction sequences"?  :P

of course but please they should be at the olympiad level only!

Yes, please post them as they usually involve reagents unknown to me as i experienced in chem forum.  :lol:

Post them please

Good to see hell_ever online :)
good thing,I will be extremely benefitted by these(I am not that smart enough)

[hide="4"]
I am trying this as I am not sure in it(Though for brilliant ones like you all it may be a piece of cake)
Is it due to fact that orbital for bonding in Xe is 5d that is much bigger in comparison to respective orbitals of Oxygen and Fluorine????

please answer.We have not reached inert gases in studies.I am just guessing by the concepts of bonding only(that starts with,we don't know what is right or wrong,just we try to explain.........)

Geometry is trigonal bipyramidal in XeF2(l.p at equatorial),octahedral in XeF4(l.p at axial) and pyramidal in XeO3.

XeF2,and XeF4 are non-polar but XeO3 is polar(not !00% sure,chem is not my cup of tea :( )
Edit:mistakes in typing :oops:
[/hide]

XeF4 is square planar. Other answers are right. What about geometry of XeF2? Think about excited state electronic configuration!

The reason for the high reactivity is probably due to the oxidation states of Xe. In XeO3, XeF6, it is +6 and hence a very powerful oxidising agent. However, the answer you gave can also be deemed appropriate i think.

no no i mistaked in typing

i was telling that Xef2 has trigonal pyramidal geometry(with equatorial positions occupied by l.p)
and XeF4 has octahedral(or square bipyramidal with l.p at axial positions)
I committed error in typing  :oops:

Your answer is right but when asked about the geometry of a molecule we only consider the atoms, not the lone pairs. Hence in case of XeF2, the answer should be linear, but your answer is correct, nonetheless.

one more thing I'ld like to ask

if we say geometry,it is as i have given or just by telling XeF2 to be linear,etc(not considering l.p.s)

thanks

Edit:got the answers,thanks again  :)  :)

[quote="rituraj007"]i can't challenge hell in chem
but i think we do consider lone pairs in geometry and only bond pairs in "structure/shape :?:  :!: "
so the geometry is trigonal bipyramidal and shape is linear[/quote]

i think u meant "the hybridisation is sp3d and shape is linear"

(geometry = shape i guess )

between can the answer of d ) part be that ->
since Xe is inert , the compounds it form would be highly reactive to give back Xe its inert status ??

[quote="slogger"]
between can the answer of d ) part be that ->
since Xe is inert , the compounds it form would be highly reactive to give back Xe its inert status ??[/quote]

that's the best answer ,i wud say :P

but i m not sure geometry = shape  :|


A function $ f$ from $ [0,1]$ to $ \mathbb{R}$ has the following properties:
$ (i)$ $ f(1)\equal{}1;$
$ (ii)$ $ f(x) \ge 0$ for all $ x \in [0,1]$;
$ (iii)$ If $ x,y,x\plus{}y \in [0,1]$, then $ f(x\plus{}y) \ge f(x)\plus{}f(y)$.
Prove that $ f(x) \le 2x$ for all $ x \in [0,1]$.

$ 0 \le f(0) \le f(1 \plus{} 0) \minus{} f(1)\equal{} 0$. By PMI: $ f(2^{ \minus{} n}) \le 2^{ \minus{} n} \forall n \in \mathbb{N}$. $ \forall x \in (0,1] \exists n \in \mathbb{N}$ s.t. $ x \in [2^{ \minus{} n \minus{} 1}, 2^{ \minus{} n})$, so $ f(x) \le f(x) \plus{} f(2^{ \minus{} n} \minus{} x) \le f(2^{ \minus{} n}) \le 2^{ \minus{} n} \le 2x$.


How do you make a horizontal line in the normal text, in order to divide content? I've tried \hline, but an error appeared when I tried to compile it.

It depends on what you want to achieve. For example
[code]some text \\
\noindent \rule{\linewidth}{0.4pt}\\
more text[/code]looks like this

$ \text{some text} \\
\noindent \rule{\linewidth}{0.4pt}\\
\text{more text}$


Four identical particles of mass 0.50 kg each are placed at the vertices of a 2.0m X 2.0m square and held together by four massless rods, which form the sides of a square. What is the rotatioal inertia of this rigid body about an axis that a) passes through the midpoints of opposite sides and lies in the plane of the square b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square and c) lines in the plane of the square and passes through two diagonally opposite particles? HEHEHE IF YOU THINK WHAT IT IS REPLY!

Just using the definition formula $J = \sum_i{m_ir_i^2}$, we get

a) $J = 4\cdot0.5\cdot1^2 = 2\ \textrm{kg.m}^2$

b) $J = 2\cdot0.5\cdot1^2 + 2\cdot0.5\cdot(\sqrt{1^2+2^2})^2 = 6\ \textrm{kg.m}^2$

c) $J = 2\cdot0.5\cdot0^2 + 2\cdot0.5\cdot(\sqrt{1^2+1^2})^2 = 2\ \textrm{kg.m}^2$

finally someone replyed thanks now can someone reply to " a wave "


quadratics
geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

If a and b are integers with a,b >0 and (a-b)(a 2 +ab+b 2 ) = 37

well after using guess and check ...I found the answer ....but it took too long so I am wondering if there's a logical way to solve it without using guess and check..please help me thank you

Since $37$ is a prime, and the second factor $a^2+ab+b^2$ is greater than $1$, the first factor must be $1$ itself, which gives $a=b+1$. Now set this result back into the original equation and solve the quadratic you get

[quote="towersfreak2006"]Better yet, $(a-b)(a^2+ab+b^2)=a^3-b^3=37$, so we have a difference of cubes.
Play guess and check, and you should notice that the difference of cubes of integers gets big really quickly.
Starting small then, the nice values of $a=4, b=3$ should jump out.[/quote]

I think his solution was guess and check and he want's to see one working 'more' with numbertea.
How do you proof your 'cubes of integers gets big really quickly' in a shorter way¿

Well, like you said, $a-b=1$, so it's the difference of consecutive cubes.
$(a+1)^3-a^3=3a^2+3a+1$, which is increasing and concave up for all positive integers.


function calculus integration algebra domain Euler

hi everyone

i was thinking if anybody could solve the numerical integral(below) in terms of wellknown functions and constants (like sin , pi , e ,log ,tan, root,..) gamma and zeta are not allowed !

of course it is possible that it is impossible, a proof of that would be nice too (but as good as impossible i guess)
i however believe in a solution at first sight.

you might have a deja vu feeling  :P

y and z are positive integers which you may choose!
y is minimum 3.

INTEGRAL FROM 0 TO INFINITY OF [exp(-(x^y)/z] dx = ???

for clarity you are supposed to pick a valid y ( >2) and z. both integer and positive. and then replacing ??? bye an analitic identity (no gamma, zeta , limits , infinite series or integrals)

if you find such a solution you have my respect !!!

oh yes i like making conjectures :p

->for any numerical integral considered at integers or infinity(for one variable) of a continous function( at the domain of integration), lets say int(f(x) dx) , where f(x) contains only wellknown functions (gamma and zeta now allowed) and f(x) contains no  multiplications , or "additions above the fraction sign",....
air  :P
let A be the number of powers, and adittions under the fraction sign , the the integral can not equal a power tower of degree A+2 with algebraic numbers or pi or e or eulers gamma.

so  e.g.   integral from 0 to infinity of exp( -exp (x²)) can not equal root (2)^(root(3)^(root(5)^(root(7)^(root(31)^pi))))

a counterexample proof or disproof also gets my respect (if i didnt define wrong or dont misinterpret what i said :p)
if you dont know what a power tower is its just a^b^c^d,..( ignoring the brackets ! ! )
with degree i mean the number of exponents (or variables or whatever i should call them :p )look it up on mathworld.com if you dont get it.

srr if im making long texts wich are complicated
thats just me  :D
greetz
tommy1729

ps looking for a name for the conjecture :p     tommys farfetched integral conjecture ? :p

i think you have posted it in the wrong place ;)


A goat is tied to the corner of a rectangular barn which has dimensions 40 feet by 80 feet. Assuming that the rope is 50 feet long, what is the maximum area that the goat can graze in squared?

how would u do this?

[hide="hint"]fractions of circles[/hide]
[hide="solution"]call the rectangle ABCD, with AB=40 and AD=80.  the goat is tied at A.  so he can graze anywhere in the circle of radius 50, except that the barn takes up a forth of it.  so he really only can graze in $\frac{3}{4}$ of the area, which is $\frac{3}{4}\pi (50)^{2}=1875\pi$

also, when the goat gets to point B, he still has 10 feet of rope, so he can go that far along side BC.  the area is a forth of a circle of radius 10, so the area is $\frac{1}{4}\pi (10)^{2}=25\pi$

so we have a total area of $1875\pi+25\pi=1900\pi$[/hide]

[hide]Draw a simple picture. You should note that there are $2$ sectors, one with central angle $270$ degrees and radius $50$, and another with central angle $90$ degrees and radius $10$.

It's easy too see that the area is $1875\pi+25\pi = 1900\pi$[/hide]


inequalities
calculus
calculus computations

What condition must hold to satisfy equality in the triangle inequality for sums of complex numbers?

$ |z_1 \plus{} z_2|\leq |z_1| \plus{} |z_2|$

Obviously equality holds if $ z_1 \equal{} 0$ or if $ z_2 \equal{} 0$ (or both). Can you see whether these are the only cases?

Equality happens if origin,$ z_{1}$ and $ z_{2}$ are collinear with origin at one end.This leads to
$ |z_{1} \plus{} z_{2}| \equal{} |z_{1}| \plus{} |z_{2}|$
$ |z_{1} \minus{} z_{2}| \equal{} \parallel{}z_{1}| \minus{} |z_{2}\parallel{}$
Also,
$ |z_{1} \minus{} z_{2}| \equal{} |z_{1}| \plus{} |z_{2}|$
$ |z_{1} \plus{} z_{2}| \equal{} \parallel{}z_{1}| \minus{} |z_{2}\parallel{}$
occurs when
origin,$ z_{1}$ and $ z_{2}$ are collinear with origin lying between $ z_{1}$ and $ z_{2}$.

Further,we have the most general statement

$ \parallel{}z_{1}| \minus{} |z_{2}\parallel{} \leq |z_{1} \pm z_{2}| \leq |z_{1}| \plus{} |z_{2}|$


parameterization
calculus
calculus computations

How to solve Differential Equation with Constaints?

Let's say we have an ODE that is: dy / dx = f(x) subject to g(x, y) = 0. How to solve this type of ODE?

It is not quite clear what you mean by "solve" here: the ODE itself determines the solution uniquely up to a few constant parameters and it can easily happen that no solution of the ODE satisfies the given constraint :?. So, your "solution" will have to violate either the ODE or the constraint. Or, perhaps, I misunderstand the question completely and you impose your constraints just at the endpoints, not throughout the entire interval. In this case proceed as usual: find the general solution and determine the parameters from the equations given by the constraints at the endpoints.


$ (cos(x) \plus{} isin(x))^n \equal{} cos(nx) \plus{} isin(nx)$ for natural n, real x.

Derivation (sort of): [url=http://en.wikipedia.org/wiki/Euler%27s_formula]Euler's wonderful formula[/url] tells us that

$ e^{i\theta} \equal{} \cos \theta \plus{} i \sin \theta$, for $ \theta \in \mathbb{R}$.

Now,

$ (\cos \theta \plus{} i \sin \theta)^{n} \equal{} (e^{i\theta})^{n} \equal{} e^{i(n\theta)} \equal{} \cos n\theta \plus{} i \sin n\theta$,

for any $ \theta, n \in \mathbb{R}$, not just natural numbers.

$ \cos\theta \plus{} i\sin\theta$ is also commonly abbreviated as $ cis\theta$.

Hmm this stuff is useful since

$ e^{ni\theta} \equal{} \cos{n\theta} \plus{} i\sin{n\theta}$

$ e^{ \minus{} ni\theta} \equal{} \cos{( \minus{} n\theta)} \plus{} i\sin{( \minus{} n\theta)} \equal{} \cos{n\theta} \minus{} i\sin{n\theta}$

so you can see that

$ \cfrac{e^{ni\theta} \plus{} e^{ \minus{} ni\theta}}{2} \equal{} \cfrac{\cos{n\theta} \plus{} i\sin{n\theta} \plus{} \cos{n\theta} \minus{} i\sin{n\theta}}{2} \equal{} \cos{n\theta}$

do something similar for $ \sin{n\theta}$.

Show off to your friends by deriving all the trig identities with this stuff XD

I'm pretty sure I read that off of a wikipedia page too.. but eh okey is this high school basics?

Take a look at the Wikipedia article on it.

http://en.wikipedia.org/wiki/De_Moivre's_formula

EDIT: The link's not showing up right, just copy and paste the URL into the address bar.


Diophantine equation
number theory solved
number theory

Find all pairs $(n,x)$ of positive integers such that $x^2=\frac{n(n+1)}{2}$.

Rewrite it as $2(2x)^2 +1= 4n^2+4n+1=(2n+1)^2$, so we are looking for solutions to the pell equation $a^2-2b^2=1$ with odd $a=2n+1$ (and thus even $b=2x$). The smallest solution is $(a,b)=(3,2)$ and the further solutions by the mapping $(a,b) \mapsto (3a+4b,2a+3b)$. So $\mod 2$ we see that $(a,b) \mapsto (a,b)$, so parity is preserved.


linear algebra
matrix
vector
linear algebra unsolved

[quote="djimenez"][quote="alexrait1"]Let $A \in C^{nxn}$ be nonsingular. Find an expression for the induced matrix norm $S_{\phi \psi}(A^{-1})$ in terms of the matrix A.[/quote]

alexrait1,

Could you please define that "operator" that you define: $S_{\phi \psi}(A^{-1})$. The induced matrix norm by a norm $\|\cdot\|_*$, as long as I know, is $\|A\|=\max_{\|z\|_*=1}\|Az\|_*$, but I don't understand your notation.[/quote]

The definition is this: $S_{\phi\psi}(A) = sup_{x \neq 0} \frac{\phi(Ax)}{\psi(x)}$

Thanx for you help.

[quote="amfulger"]THE QUESTION: Who are $\phi$ and $\psi$?[/quote]
these are the appropriate vector norms..


calculus
integration
parameterization
LaTeX
calculus computations

Hello again...

Meanwhile i have another question.
During a proof in solid state, i need to show that Cv(heat capacity) propotional to T/@ (T is the temp', @ is debye temp').
while writing it down i need to solve a complicated integral:

it's complicate to write it, so I wrote it down nice and preety in equation editor, and saved in doc file.

Thanks a lot!

[quote="fedja"]So, do I understand it right that the integral is
\[
\int_0^\infty\frac{2L}{\Pi a}\frac{h\omega}{(\omega_{max}^2-\omega^2)^{1/2}} \frac 1{ e^{ {\textstyle \frac { h\omega}{ k_B T}}}-1}\,d\omega
\]
and the task is to show that it is proportional to $\frac{k_BT^2}{h\omega_{max}}$ assuming that $L$, $\Pi$ and $a$ are constants? If not, please correct me.

Assuming that my interpretation is correct, all I can say is that it is false. The dependence on the parameters is more complicated unless you want an asymptotic relation for very small or very large $T$. But even in this case, since the integral depends only on the product $k_B T$, any formula for it should depend also only on the product $k_B T$, not on $k_B$ and $T$ separately. So, something is wrong here. Please, check your integral and your suggested formula for it.

At last, as a new person to the forum, you might not know that posting the same thing in two different places is considered a spam here. So, I am deleting your identical post in Calculus Unsolved Problems. Be sure, it won't result in fewer people seeing your post  :)[/quote]

First of all, thanks for your answer.
I thought some things are clear, so i didn't write them. sorry.
so, i rewrite the file, and it's updated, and i am SURE that now it can be solved. there are NO too many parameters as before....
thanks!!!!

[quote="fedja"]Oops, my version of Microsoft Word cannot read one of the symbols in your file! Do you want $T$ to be much less or much larger than $\theta$ (Or is it something else there? I mean the phrase $C_\nu\approx \frac T\theta$ for $T$ "something" $\theta$ in the beginning of the file). Also, was my understanding that the integral is taken from $0$ to $\infty$ correct?[/quote]

Hey!

The integral is on dw, and as you can see there is a const named w(max), so te integral is from 0 to w(max). (the meaning is that for this particular system there are finite alloud states, each has it's own frequency w)
one more thing: T is much smaller the eta (debye temp').

Thanks fedja!
Hope now you'll work it out....

[quote="Hen"]Thanks fedja!
Hope now you'll work it out....[/quote]What about you learning $\LaTeX$ ? :) It would be easier to express yourself, and you can write math stuff much easier and quicker. Read the link in my signature for a quick $\LaTeX$ guide, and take a look (click on the formulaes) at other people's formulaes to see how they are created (it's very easy to learn).

Yes, now it is not hard to figure out. We have $\int_0^{\omega_{max}}\frac{2L}{\Pi a}\frac{h\omega}{(\omega_{max}^2-\omega^2)^{1/2}} \frac 1{ e^{ {\textstyle \frac { h\omega}{ k_B T}}}-1}\,d\omega \approx \frac{\pi^2}3\frac {Lk_B}{\Pi a}\frac {T^2}\theta$ (note that an extra $k_B$ appears now in the proportionality constant). The (informally put) reason is that the exponential factor is huge if $\omega$ is noticable compared to $\omega_{max}$, so the main part of the integral is concentrated in the region where $\omega$ is much smaller than $\omega_{max}$. This allows to replace the complicated expression $(\omega_{max}^2-\omega^2)^{1/2}$ by $\omega_{max}$ and extend the integration to the entire interval $(0,+\infty)$. After that, just make the change of variable $\omega=\frac {k_B T}{h}\omega'$ and observe that $\int_0^\infty \frac {\omega'}{e^{\omega'}-1}\,d\omega'=\frac{\pi^2}{6}$. So, you are right, with everything else fixed, $u$ is approximately proportional to $T^2$ for $T$ small compared to $\theta$. Try to make this informal reasoning rigorous yourself. If you face any unsurmountable difficulty in the process, you are welcome to ask your questions any time :)


geometry
geometric transformation
rotation
reflection

In how many ways can 4 purple balls and 4 green balls be placed into a 4 x 4 grid such that every row and column contains one purple ball and one green ball? Only one ball may be placed in each box, and rotations and reflections of a single configuration are considered different.

are all the individual purple or green balls distinct form eachother or not.

Im assuming they are all the same?

If all the balls were the same colour, we would have $ \binom{4}{2}= 6$ ways to place the first 2 balls in the first row. The balls in the second row can likewise be placed in several ways:

If the 2 balls in the second row coincide with the columns of the balls in the first row, then the placement of balls in the 3rd row and 4th row are uniquely placed. There are 6 ways for this to happen.

If exactly one ball in the second row coincides in columns with a ball in the first row (there are $ 6 \cdot 4 = 24$ ways for this to happen), then the third row can be selected in 2 ways and the fourth row is uniquely determined.. So 48 ways.

If the balls in the second row are in different columns from the balls in the first row (6 ways again), then the 3rd row can be placed in $ 3! = 6$ ways and the fourth row is uniquely determined. So $ 6 \cdot 6 = 36$ ways.

The sum is $ 48+42 = 90$.

Now we place the colours: If the upper leftmost ball is coloured purple, the colours of all other balls are uniquely determined. Likewise if it is green, hence the answer is 180.

Hope there are no mistakes there

I got something different from that. :maybe:

[hide="My solution"]
First, place the four green balls. It's easy to see there are $ 4! = 24$ ways to do this. Now place the four purple balls. The one in the first column can go in any of the three open spaces. The row in which it is placed also has a green ball in another column. Consider the column that green ball is in. We can place the purple ball in any of the three open spaces in that column. We also find that each placement determines uniquely the location of the last two purple balls. So we have $ 24 \cdot 3^{2}= 216$ ways.
[/hide]


Can the cells of a 2002 x 2002 table be filled with the numbers from $1$to $2002^2$ (one per cell) so that for any cell we can find three numbers $a, b, c$ in the same row or column (or the cell itself) with$a = bc$?

assume that one have done this
now delete the numbers $1,...,2001$ from the table now we have a row and a column that are full with numbers $2002,...,(2002)^2$ but the subtract of any two of these numbers is greater than $(2002)^2$ so it cant be equal to another number and its a contradiction
so we cant do this


Hello!

Here's my problem:

The pressure of a closed system is $ P_{sys}$. Let the external pressure and temperature be $ P_{ext}$ and $ T$. There's a hole in the wall of the system of area $ A$. Supose that every molecule that "touchs" the hole breaks into the system. (a) How many molecules gets into the system at a time $ t$? (b) If the system's volume is $ V_0$, what's the variation of pressure? (c) Show that the number of molecules that gets out of the system are despreciable, for a small hole area, room external pressure and temperatures, and low pressure on the system.

The only part of your question that I understand is (a).  Here are my thoughts (they may be completely wrong, or answering this question may require more knowledge of kinetic molecular theory than I have):

I'll assume that the gas is ideal.  Consider an imaginary cylinder that covers the hole and juts out into the environment, with base area $ A$ and length $ v_{rms}t$, where $ v_{rms}$ is the rms speed of the external molecules.  The cylinder contains $ \rho Av_{rms}t/M$ moles of gas (where $ \rho$ is the mass density and $ M$ is the molar mass), or $ N_A\rho Av_{rms}t/M$ molecules, where $ N_A$ is Avogadro's number.  About half of the molecules in this cylinder are going toward the hole, and about half are going away from it.  Hence, in a time $ t$, the number of molecules going into the closed system is about
\[ \frac{N_A\rho Av_{rms}t}{2M}.
\]
We can substitute the relations $ v_{rms} \equal{} \sqrt{\frac{3RT}{M}}$, $ \rho \equal{} \frac{PM}{RT}$, and $ R \equal{} N_A k$ ($ k$ being Boltzmann's constant) to get
\[ \mbox{number going in} \equal{} \frac{P_{ext}At}{2k}\sqrt{\frac{3R}{MT}}.
\]
Unfortunately, we're not given $ M$ for the gas.

In addition, we'd have to account for the molecules going out of the closed system during a time $ t$, using a similar analysis, and we're not given the temperature of the closed system.  But then we would have
\[ \mbox{net number of molecules going in} \equal{} \mbox{number going in} \minus{} \mbox{number going out}.
\]

Part (b) states that if there's a given volume for the system with the lowest pressure (the pressure on the system in this case), what's the change on pressure.

Part (c) asks you to show that, at external room temperatures and pressure (let's statessay about 300 K and 1 atm) and at low temperatures inside the system (let's say like $ P_{sys}<0.2 [Pa]$), the number of molecules that gets out of the system are despreciable.

This can be solved very similarly to the problem in the marathon section, you posted.. :wink:


Seven distinct points of a circle are chosen. How many line segments are determined by these points?

A line segment is determined by $ 2$ points. We have to choose $ 2$ out of the $ 7$ points. This is just $ \binom{7}{2} \equal{} \frac{7\times6}{2\times1} \equal{} \frac{42}{2} \equal{} \boxed{28}$.

RoFl did everything right until the end.  $ \frac {42}{2}$=$ \boxed {21}$, not $ 28$


function
limit
trigonometry
calculus
calculus computations

Levi,can you repost it.I don't understand everything you have written. ;)

Levi,I think the conclusion of your statement follows imediately from Cesaro-Stolz criterion. :?

      What do you think?

Certainly, but you present a solution here.

you know why I told you that? Because I proved with Cesaro-Stolz a similar result witch states that:

  Let $\mathbb{D}$ a set in $\mathbb{R^*}$ (avand originea ca punct de acumulare)(i don't know how to write this in english,sorry :oops: ).Let $f:\mathbb{D}\to\mathbb{D}$ and the sequence $(x_n)_{n\geq1}$ such that $x_1\in\mathbb{D}$ and $x_{n+1}=f(x_n)$.We assume that the following conditions are fullfiled:

  1) $x_n\rightarrow0$ and 2) $\frac{x-f(x)}{x\cdot f(x)}\rightarrow l $ when $x\rightarrow0$.Then:

  $ n\cdot x_n\rightarrow\frac{1}{l}$ ;)


induction
geometry
3D geometry
number theory unsolved
number theory

''For every b>0 the 48th power of b can be written as the sum of 3 cubes''. I think it's difficult but math's mysterious...

Come on guys iff you see the solution in mathworld you'll be amazed!

maybe check for some small cases of b.
then use induction by b
say that P(b) is $\exists \ m,n,p$ so that $b^{48}=m^3+n^3+p^3$
we have to proove P(b+1) namely : $\exists m'(b),n'(b),p'(b) \ so \ that \ (b+1)^{48}=m'(b)^3+n'(b)^3 +p'(b)^3$
the difference between $(b+1)^{48}$ and $b^{48}$ is a sum of products of
combinations and $b^k$,how do we add those terms to $m^3 \ and \ n^3 \ and \ p^3$
to form new cubes $m',n',p'$ *
do we have anything to help us here ?

also fermat case n=3 says here that $m^3+n^3,m^3+p^3,n^3+p^3$ can never be
cubes.
that is also true for $m',n',p'$
maybe that can offer some constraints for our main goal *


limit
logarithms
integration
calculus
LaTeX
calculus computations

Let $ m$ be a positive integer and let $ n$ be a non-negative integer.

(i)  Use the result $ \lim_{t \to \infty} e^{\minus{}mt}t^{n} \equal{} 0$ to show that

$ \lim_{x \to 0}x^{m}(\ln x)^{n} \equal{} 0$.

By writing $ x^{x}$ as $ e^{x \ln x}$ show that

$ \lim_{x \to0} x^{x} \equal{} 1$.

(ii) Let $ I_{n} \equal{} \int_{0}^{1}\, x^{m}(\ln x)^{n}\,dx$. Show that

$ I_{n\plus{}1} \equal{} \minus{}\frac{n\plus{}1}{m\plus{}1}I_{n}$

and hence evaluate $ I_{n}$.

(iii) Show that

$ \int_{0}^{1} x^{x} dx \equal{} 1\minus{}(\frac{1}{2})^{2} \plus{} (\frac{1}{3})^{3} \minus{} (\frac{1}{4})^{4} \plus{} ... .$

[quote="AndrewTom"]Thanks kabi: unfortunately I have a number of problems with my computer and one is that if I click on the blue writing where something is hidden it brings me back to the top of the page and I never get to see what's been hidden.[/quote]
sorry ,
I would n't be using hide in posting replies to u any more .

(3) can be done by expanding $ x^x \equal{} e^{x \log x}$ into a series, the integral can be evaluated by (2) or by other recursion methods like substituting $ e^{ \minus{} u}$ and using $ \Gamma(n\plus{}1)$, resulting in the RHS of (3).

For your comp.problems you could also temp. quote the message if you can read plain Latex :).

How would you evaluate the integral with the recursion? I am referring to ii)

We fixed $ m$ and saw that $ I_{n\plus{}1} \equal{} \minus{} \frac{n\plus{}1}{m\plus{}1} I_n$.

But this means that $ I_n \equal{} \minus{} \frac{n}{m\plus{}1} I_{n\minus{}1} \equal{} \frac{n(n\minus{}1)}{(m\plus{}1)^2} I_{n\minus{}2} \equal{} (\minus{}1)^n \frac{n!}{(m\plus{}1)^n} I_0$.

However, $ I_0 \equal{} \int\limits_0^1 x^m (\ln x)^0 \, dx \equal{} \int\limits_0^1 x^m \, dx \equal{} \frac1{m\plus{}1}$.

So we get $ I_n \equal{} (\minus{}1)^n \frac{n!}{(m\plus{}1)^{n\plus{}1}}$


limit
real analysis
probability and stats

trying to show if a process $ W_t$ satisfies these three conditions:
(i) $ t\ne s \to W_t$ independent of $ W_s$
(ii) $ W_t$ stationary
(iii) $ E(W_t)\equal{}0, \forall t\ge 0$

cannot have continuous paths.  The hint says to consider
$ W_t^{(N)}\equal{}\min(\max(W_t, \minus{}N),N)$ the truncated process, and look at
$ E((W_t^{(N)}\minus{}W_s^{(N)})^2)$.. but i dont see how this helps me.

I tried by defining
$ A_k\equal{}\{w: W_t(w)$ continuous on $ t\in [k,k\plus{}1) \}$
and then saying
$ P($W_t continuous$ )\equal{}P(\cup_k A_k)\equal{}\sum_k P(A_k)\equal{}\lim_{k\to \infty} P(A_0)^k$
and so it suffices to show $ P(A_o)<1$, but I dont know how to show that..

any help would be great.

Something is missing. What if $ W$ is identically $ 0$?

Hi kenn4000

where do you get $ P(A_0)^k=P(A_k)$ from ?

Personnaly I would look at the limit when $ t\to s$ of $ E((W_{t}^{(N)}-W_{s}^{(N)})^{2})$ (supposing first that the process is continuous),

We get from independence assumption :
$ E((W_{t}^{(N)}-W_{s}^{(N)})^{2})= E((W_{t}^{(N)}^2) +E((W_{s}^{(N)}^2)$

this would imply (I clearly lack of rigor here) that it converges to $ 0=2.E((W_{s}^{(N)}^2)$ which imply that $ W_{s}^{(N)}=0$a.s.for every s. Passing to the limit with $ N$ we get that $ W$ is not continuous or is a.s. null.

why do we even need the truncated processes anyway?

$ E((W_s\minus{}W_t)^2)\equal{}2E(W_t^2)\minus{}2E(W_t)^2\equal{}2E(W_t^2)\to 0$ means $ W_t\equal{}0$
seems much simpler to say?  but the book clearly gives the truncated things as a hint

maybe i just dont get the wording.. i think maybe its asking to show that continuous paths are a measure zero set?  this problem sucks, but i want to finish them all before i start writing up the next chapter, so any more help would be great.

[quote="kenn4000"]
$ E((W_s \minus{} W_t)^2) \to 0$ [/quote]
You cannot really say this without using dominated convergence theorem or something similar. ;)


Where can I find problems in .pdf format?

[url]http://www.google.com/advanced_search?hl=en[/url]

Choose .pdf file type, type in "math problems," or whatever you want, and search. :)

Look above:  :roll:

Click on the Contests section. We have over 4000 Problems or so.


geometry
circumcircle
trapezoid
geometry unsolved

Let  $A_1$  and $B_1$   be two  points  on the base  AB  of the isoceles  triangle ABC ( angleC>60)  such  that  angle$A_1CB_1= angleABC$.  A   circunference tangent externally to  the  circuncircle of $A_1B_1C$  is also  tangent to  the rays  CA   and  CB in the points $A_2$ and $B_2$ respectively.  prove that  $A_2B_2 = 2AB$

 Please  solve with  inversion  :)

http://www.mathlinks.ro/Forum/viewtopic.php?t=23118
http://www.mathlinks.ro/Forum/viewtopic.php?t=33745

Ask Ashegh for an inversive proof :D .

  Darij

ok here is an inversive proof...

i dont know wether this problem is easy or hard...

but what should i use this power full tool to solve easy problems...

i think its not the way of nice problem solving...

i decided not to use inversion unnesserily...

inversion will make this problem very easy... :D

the two circles are tangent to each other at point $G$.

circumcircle of triangle $CA1B1$ intersects $CA,CB$ at points $E,F$.

consider a n inversion with center$(C)$,and arbitrary radius.

$AB$ changes to circum circle of triangle $CA'B'$

circumcircle of triangle $CA1B1$ changes to line $A1'B1'$.

$E'$ and $F'$ are the intersections of line $A1'B1'$ with $CA,CB$

the other circle will change to incircle of triangle $CE'F'$.(externaly tangent changes to internaly tangent)

and is tangent to $CA,CB,E'F'$ at points $A2',B2',H$ respectively.

now in this shape,we should prove:$A2,B2$,are the midpoints of $CA',CB'$.

lets prove it:

in the new shape we have this facts:

$1)$$\angle A1'CB1'=\angle CB1'B' =\angle CA1'A'$

$2)$then: $C,A1',B1',B'$,is an isoceles trapezoid

also $C,A',A1',B1'$ is an isoceles trapezoid.

...............................................................

then:in $C,A1',B1',B'$ we have$B'B2=B1'H , B2C=HA1'$

and in $C,A',A1',B1'$ we have $A'A2=A1'H , A2C=B1'H$

the conclusion is $A2C=A2A' , CB2=B2B$

and it shows us that,$A2,B2$,are the midpoints of $CA',CB'$.

marko is the solution clear...

if not,i would explain u each part u want ;) .

i  understood completely  many  thanks  ashegh!!  :)


calculus
integration
inequalities
calculus computations

Prove the following inequalities
1)
$ \frac{\pi}{4}\leq \int^1_0 \frac{1}{1\plus{}x^4}dx \leq 1 \quad\text{aplying}  \quad m(b\minus{}a)\leq \int^b_a f(x)dx\leq M(b\minus{}a) \\
\Rightarrow\frac{1}{2}\leq \int^1_0 \frac{1}{1\plus{}x^4}dx \leq 1$ but how do i get $ \frac{\pi}{4}$
2)

$ 1\leq \int^1_0 \sqrt{1\plus{}x^4}dx\leq \frac{6}{5}$

3)

$ \sqrt{2}\leq \int^9_5 \frac{1}{\sqrt{x\minus{}1}}dx\leq 2$

4)

$ \frac{1}{9\sqrt{2}}\leq \int^1_0 \frac{x^8}{\sqrt{1\plus{}x}}dx\leq \frac{1}{9}$

thanks beforehand

1) Hint:
Compute $ \int_0^1 \frac {1}{1 \plus{} x^2}\ dx$

#2, $ 1 \le \sqrt {1 \plus{} x^4 } \le 1 \plus{} x^2$

#3, Substitution of x+1

#4, on the interval, $ \frac{{x^8 }}{{\sqrt 2 }} \le \frac{{x^8 }}{{\sqrt {1 \plus{} x} }} \le x^8$


[quote="Soccer_Chica_018"]I really need the lcm of this by tonight!  Please help me!!!!!!!!!!!!!!!!!!![/quote]
I'm assuming you mean the lcm of all four of the numbers.

You must factorize each of the numbers and take the highest powers you find and multipy them together.

include all factors regardless of whether they are in  more than one number.

Ex. the lcm of 2, 3 and 4  is 12 because $2^2$*3=12

I think you should be able to do it from here[/hide]

There is a upside down division method that is nearly impossible to explain.

But you can go [url=http://www.dmacc.cc.ia.us/instructors/meshedenhelm/Class%20Notes%203%20&%204.htm]here[/url].

To classroom it goes.

if you need it by tonight you should ask a teacher not make a aops account to have other people solve your problems

[hide]
Since 2 is a multiple of 86, all you need to do is 86*179*5.
I hope this helped, but I agree with apusey, you should ask the teacher instead.[/hide]


search
inequalities
real analysis
real analysis unsolved

Let $ a_1,a_2,...,a_n$ be nonnegative real numbers. Prove that \[ (  \sum_{i=1}^na_i)(  \sum_{i=1}^na_i^{n-1}) \leq  n \prod_{i=1}^na_i+ (n-1) (  \sum_{i=1}^na_i^n).\]

[i]J. Suranyi[/i]

a solution is [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1351715794&t=14008]here[/url]
an old discussion mentions a [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1351715794&t=89757]generalization[/url].
Also there's a bunch of these types of inequalities [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1351715794&t=110149]here[/url]


linear algebra
matrix
calculus
real analysis
integration
trigonometry
geometry

I think I have come up with the derivation of the Jacobian matrix (in 2 variable), but I want to make sure that it is correct.  (It wasn't in my book and the teacher didn't know).

So here goes:
By the chain rule we have
$dx=\frac{\partial x}{\partial u} du + \frac{\partial x}{\partial v} dv$.
$dy=\frac{\partial y}{\partial u} du + \frac{\partial y}{\partial v} dv$.
Since $dx$ and $dy$ are perpendicular to each other, $dxdy$ is really $dx \otimes dy$.
$\left |dx \otimes dy\right | =\left | \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} dudv - \frac{\partial x}{\partial v} \cdot \frac{\partial y}{\partial u} dudv \right |$.
Therefore $J= \left | \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \cdot \frac{\partial y}{\partial u}\right |$.

Well, I'm not completely sure what you mean by "derivation" here, i.e., what you assume known. It seems like you rely on the calculus of differential forms but usually the change of variable theorem is used to derive the properties of differential forms, so you risk to get a vicious circle in your argument. Maybe Kent can say more about this...

Can you explain this more, and/ or give a complete proof?
Thanks.

My question is "What do you assume known when deriving the formula for the Jacobian?" I mean you manipulate with $dx$, $dy$, $du$, $dv$. What are those in your opinion? And why are you allowed to manipulate them the way you do?

As to posting the complete proof, do you mean the proof of the change of variable formula for the Lebesgue integral? I would, but it is a bit too long to post it here   :( Also you can find it in all standard analysis textbooks.

Or, maybe, I just do not understand what you are asking?  :?  Then, please, explain again and I'll try to answer in a better way.

You're right, I don't know the derivation of the chain rule, $dx=\frac{\partial x}{\partial u} du +\frac{\partial x}{\partial v}dv$.
Can you show me why this is true?  And as for the Lebesgue integral, I'm not too sure what you're talking about.
Thanks.

Never mind, I found a site which proves the chain rule, but assuming that the chain rule is known, then could I have done what I did to get the Jacobian?

[quote="Jimmy"]...assuming that the chain rule is known, then could I have done what I did to get the Jacobian?[/quote]
And how do you define the Jacobian? I mean, to derive a formula for something, you must first have some definition of that something. I thought you were talking about the change of variable theorem but apparently this is not true. So I'm still a little bit too confused to answer properly  :?

I was referring to the change of variable theorem, and my definition of the Jacobian would be the factor by which the unit square is morphed by the integral transformation.

[quote="Jimmy"]my definition of the Jacobian would be the factor by which the unit square is morphed by the integral transformation.[/quote]
Can you be more precise? I do not quite understand the words you wrote. I apologize for my stupidity, but to answer whether you did your derivation correctly I first need to understand a few things, two most important of which are "[b]How do you define the Jacobian?[/b]", and "[b]What are $dx$, $dy$, $du$, $dv$ in your opinion?[/b]". The point is that your derivation can be made certain sense of, but I am very unsure about whether you really understood what you were doing when writing it. That is probably the reason why your teacher couldn't answer you whether it was correct or not too. So, if you want to get my opinion, please, answer the two questions above as clearly as you can. Maybe, once you answer them, you'll be even able to see whether your derivation is correct yourself. The general rule is that it is correct if you can explain each symbol, notion, and step in it to somebody else who sees it for the first time (I'm trying to play the role of such a person now) and incorrect otherwise. As I said, it can be made some sense of and is not something that is really wrong. The question is whether [b]you[/b] can make any sense of it.

When my teacher showed us the Jacobian, I asked where it came from, and she said she didn't know.  Then I told her about taking the cross product between $dx$ and $dy$ and she said it made sense, but I understand what you're saying.
I think I can explian what I think the Jacobian is by using a basic calc example $\int \sin\left(\frac{x}{2}\right) \ dx$.  We make the substitution $x=2u$, and then $dx=2du$, so our integral becomes $2\int \sin(u) \ du$.  The Jacobian is the three dimensional (or more) analogue of the $2$ in the above example.  When we take area strips under a curve, we say they have a width $dx$, but now that we are integtrating with respect to $u$, we have to find a relationship between $dx$ and $du$, namely $dx=2du$, and this relationship shows how a change in $x$ affects $u$, or how in a sense the strip is morphed.  I hope this makes sense.

Yes, this makes sense though it is not quite a formal definition yet. Seems like the integral you are dealing with is the so-called Riemann integral (but I may be confused again). The idea of the Riemann integral in 2 dimensions is, roughly speaking, the following. Suppose we have a domain $\Omega$ and want to integrate some function $f:\Omega\to\mathbb R$. Then we split $\Omega$ into many pieces of small diameter $\Omega=\bigcup_j\Omega_j$, choose some points $p_j\in\Omega_j$ and form the so-called Riemann sum $\sum_j f(p_j)\text{Area}(\Omega_j)$ (we assume that some notion of area is available to us and that all $\Omega_j$ are good enough to talk about their areas). As the diameters of $\Omega_j$ tend to $0$, it may happen that the Riemann sums tend to some number $I$ that is then called the integral of $f$ over $\Omega$. Does this sound familiar? Now assume that we have a one-to-one smooth mapping $T$ from $\Omega$ to some $\Omega'$ and a function $f:\Omega'\to\mathbb R$. Then, instead of splitting $\Omega'$ directly, we may split $\Omega$ and consider the induced partition $\Omega'=T(\Omega)=\bigcup_jT(\Omega_j)$. Also, we can choose points $p_j\in\Omega_j$ and consider the corresponding points $T(p_j)\in T(\Omega_j)$. Then the Riemann sum for $f$ becomes $\sum_j f(T(p_j))\text{Area}(T(\Omega_j))=\sum_j f(T(p_j))\frac{\text{Area}(T(\Omega_j))} {\text{Area}(\Omega_j)}\,\text{Area}(\Omega_j)$. This looks like a Riemann sum for the function $f\circ T$ integrated over $\Omega$ except we have an extra factor $\frac{\text{Area}(T(\Omega_j))} {\text{Area}(\Omega_j)}$. We may think that the smooth mapping is closed to a linear one near each point and, therefore, changes the area of any small region around a point $p\in\Omega$ by the same factor depending on the point only, say, $J(p)$, so  $\frac{\text{Area}(T(\Omega_j))} {\text{Area}(\Omega_j)}\approx J(p_j)$ Then the sum we had becomes just a Riemann sum for the function $(f\circ T)\cdot J$ integrated over $\Omega$. So we arrive to the following definition of $J(p)$:
\[
J(p)=\lim_{p\in Q, \text{diam}(Q)\to 0}\frac{\text{Area}(T(Q))}{\text{Area}(Q)}.
\]
As usually happens with the limit concept, it means that to approximately find the Jacobian $J(p)$ at the point $p$, we should take some small region $Q$ around $p$, say a square, or a circle, or whatever your favorite geometric shape is, and to find the ratio of the area of the image to the area of the region. Is this the definition you had in mind? (of course, the derivation is not really rigorous, but to make it completely rigorous one would need a few pages of $\varepsilon-\delta$ arguments). If so, let us accept this definition and go to the second question I asked. You have the nice "chain rule" formula $dx=\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv$. I assume that you consider here some mapping $T: (u,v)\mapsto(x(u,v),y(u,v))$. Now, every object in mathematics belongs to some class: you may have numbers, functions, vectors, etc. If I understand things right, your chain rule is written at some point $p$ for which you want to find the Jacobian, so the partial derivatives in the formula are just numbers (the values of the corresponding functions at $p$). Now comes the main question we should answer if we want to make sense of your derivation: what kind of objects are $dx, dy, du, dv$?

[quote="Jimmy"]Well I thought of them as vectors which was why I was able to take the cross product of $dx$ and $dy$.  And yes, you're right I was talking about the Riemann integral.[/quote] But then what vectors are they? How do they relate to the formula with the quotient of areas? You see, if we agree on the definition in my previous post, to find the Jacobian, we have to start with some small region $Q$ of your choice and so far we have no vectors to talk about.  :?

Let's say we have a rectangular area with its sides parallel to the $x$ and $y$ axes.  I think of there being infinitely many infinetely small $dx$ vectors in the positive $x$ direction of the rectangle and infinitely many infinetely small $dy$ vectors in the positive $y$ direction, so that $dx \otimes dy$ represents a rectangular area similar to the larger area.  Just so I'm on the right track, $dx, dy, du, \text{and } dv$ are vectors right?

[quote="Jimmy"]  Just so I'm on the right track, $dx, dy, du, \text{and } dv$ are vectors right?[/quote]
That you'll have to decide for yourself: just see if you can answer the questions I'll ask assuming that they are vectors. So let me repeat: if we accept your point of view, we must have some region $Q$, its image $T(Q)$, some vectors $dx,dy,du,dv$ associated in some reasonable way with $Q$ and $T(Q)$ so that the area of $T(Q)$ is about $|dx\times dy|$, the area of $Q$ is $|du\times dv|$ and the vectors satisfy the chain rule formula (by the way, it would be interesting to see how you understand it: are $du$ and $dv$ just arbitrary vectors and if they are, what are $dx$ and $dy$ on the left?).
I know that I'm trying to confuse you here instead of just presenting a ready solution but it is a good way to see whether somebody understands something and next time you'll be able to try a similar exercise yourself ;)

[quote="fedja"]I know that I'm trying to confuse you here[/quote]
Well, mission accomplished. :lol:

But I do appreciate your time in helping me understand this.

From my understanding of what you wrote, $Q$ is the original image and $T(Q)$ is the "transformed" image right?  So then wouldn't $Q=|dx \otimes dy|$ and $T(Q)=|du \otimes dv|$?

Well here is what I think I understand:
$dx$ is a vector parallel to the $x$-axis.
$dy$ is a vector parallel to the $y$-axis.

Now, these are ideas:
The chain rule relates the magnitudes of $dx,dy,du, \text{and} dv$.
When the $xy$ plane is transformed into a $uv$ plane:
$du$ is a vector parallel to the $u$-axis.
$dv$ is a vector parallel to the $v$-axis.

OK, let me show you the standard derivation. First of all, how to interpret the chain rule. One way to think of it is the following. Let $p=(u,v)$ be some point and $\widetilde p=(u+du,v+dv)$ be a very close point (so $du$ and $dv$ are actually [b]scalars[/b] in this interpretation). Let $T(p)=(x,y)$ be the image of $p$ and $T(\widetilde p)=(x+dx, y+dy)$ be the image of $\widetilde p$. Then $dx\approx \frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv$ and similarly for $y$.
More precisely, $dx= \frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv+o(|du|+|dv|)$ as $du,dv\to 0$. This is the "engineering" interpretation of the chain rule formula. As you can easily see, this interpretation will not validate your derivation. On the other hand, it can be used to derive the formula for the Jacobian in the following way. Let us consider a rectangle $Q$ with vertices $(u,v)$, $(u+du,v)$, $(u+du,v+dv)$ and $(u,v+dv)$. This rectangle will be mapped by $T$ approximately to the parallelogram, whose vertices will be $(x,y)$, $(x+\frac{\partial x}{\partial u}du, y+\frac{\partial y}{\partial u}du)$, $(x+\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv,y+\frac{\partial y}{\partial u}du+\frac{\partial y}{\partial v}dv)$, $(x+\frac{\partial x}{\partial v}dv, y+\frac{\partial y}{\partial v}dv)$. Finding the area of this parallelogram from the coordinates of its vertices according to the standard formula, we get the expression for the Jacobian. As to your derivation, it will require a completely different interpretation of $dx$, $dy$, $du$ and $dv$. Just keep it in mind now and return to it when you start to learn differential forms. That is as much as I can say now. As I said, maybe Kent has something to add.

Thanks again, I really have a much better understanding of the Jacobian and chain rule now.


Why are there more males in math?
Are boys smarter? Are girls less likely to achieve mathematic excellence?
Why are there so few girls on this site?

It's probably difficult to answer honestly without being accused of sexism  :D

Usually women use language to carry feelings, men to carry informations (see the nice book "women come from Venus, men from Mars" ...).  Maths is language without any feelings. I think that's why they are less interested -and so less active/good/smart - in maths.

Why are there more women in litterature, socology, .... ??

I think that the mathematics is an hermetic field.
We mathematicians "learn" to like math. You cannot say that math attracted you from the first grade when you learned to add positive integers.
What's attractive about math is the wonderful way ideas gather to form a proof.
There is also something wrong about the way math is perceived by the outside world. People usualy think math is about computations and following routines.
I think math is a dream world. Is the only model built by the human mind for a perfect world, where everything is known. I mean everything is either true or false. There is only black and white and there are no grays.

This only coveres why most people don't appreciate maths.
I think that this general mentality about what women think and what they are able of doing is the cause for them not liking math.
Ask a girl what she wanted to be when she was a kid and she will give you all kind of artistic answers: a ballerina... But she only tells you what her parents like her to say (it's the only thing she knows).
Boys on the other hand are less bounded in their options. Their parents know that a boy can do anything. So they will tell you they want to be astronauts, construction workers, auto mechanists, or even garbadge collectors.

My conclusion is that girls use language to carry feelings because of the wolrd's mentality about what girls do and they don't choose math because of the world's mentality about maths.

[quote="amfulger"]My conclusion is that girls use language to carry feelings because of the wolrd's mentality about what girls do and they don't choose math because of the world's mentality about maths.[/quote]

It was also my opinion for a long time. Until I have two children. One girl and one boy.

Boys and girls minds ARE different. Different doesn't mean one is superior to the other. Not everything is cultural.

Okay I too admit that there are some differences in the way boys and girls think.
Some of then are in our DNA, inherited for many generations from the stoneage, when the man (because of his superior force) required the abbilites of a good provider and the woman's life was focused on the safety of the group and ...
What I mean is that these differences are instinctual and I think these are the only differences that should exist. This means that there should be no difference in the way man and woman feel.
I think that by instinct, women are more imaginative and because of this, they should be better in math. There is a chance I'm wrong about this but you said it yourself: there are more women in literature,... Maybe there is another characteristic that makes their superiority in the imagination useless.

One more thing: when someone says he/she loves maths, what do they realy mean?

the education in 18th-19th century in Europe was mainly for men not women. There was a good French female mathematician. She was Sophie Germian. But after she died, the report of her death said that her identity is a "rentire-annuitant" not a "mathmaticienne". And the twenty-two names of experts inscribed on Eiffel Tower don't include hers -- Sophie Germian, who made great contributions to vibrations of elastic plates.
It seems that we also have a prejudice on women...

Mathematics is very much a cultural endeavor. This is why even some countries produce more mathematicians than others. And since cultures have traditionally excluded women from mathematics, that stigma has remained in people's minds of math being a more "masculine" endeavor. To change that only comes through very slow change in cultures. You can see it is already happening, there are now more great female mathematicians than ever and slowly growing. It will probably never come close to even though, because cultural stigmas never completely go away.

Harold Reiter, a teacher here in America has done some studies of the differences between female and male students betweent he ages of 12 and 14 wohen it comes to mathematical ability - and it indicates that females on whole actually have better spacial reasoning and are better at geometry at that age than boys. But few of them pursue an interest in mathematics into high school. No surprise that this happens around puberty when social pressures come into factor.

[quote="gauss202"]female and male students between the ages of 12 and 14 wohen it comes to mathematical ability - and it indicates that females on whole actually have better spacial reasoning and are better at geometry at that age than boys.[/quote]

is this true?
When I was in junior middle school, I always get a high score in a plane geometry examination, but i never seen a girl who got such a high score. they even failed the exams many times, but they can get a good result in algebra calculations.
And In high school, we have solid geometry courses. we can use space vectors to solve the problems, and this makes many problems much easier. indeed, many of them have difficulties on finding a pure geometrical solution.

I don't think individual experience can tell us one way or the other, because the sample size of people that we know personally is too small. His study was based on the AHSME/AMC-12 exam, which is the first step to the USA IMO Team, and is given to 300,000 American students each year.

But if it is really a culturally influenced result, then that could even possibly account for differences between countries in the way that males and females perform in math.

girls differ in attitude to boys.

For ex, a lot more boys go into computer science because a lot more boys play video games.

There is a correlation between computer usage/videogaming and majoring in computer-related majors.

In my secondary school (a girls' school) there was no way you could learn computer programming even if you wanted to...there was no club, no lessons...I learnt Basic and VB only because I attended some programme organised by a local university in lieu of a boring school programme :D (and that was because there were boys attending that programme!).  I always thought that it's kind of sad because I would really love to learn (and I was good at Basic, but the course was so short :( ), and even if I join my high school programming club now I'll be way behind the others in terms of skills.

A (worrying) trend that I've noted is that (at least in Singapore), there's a lot less support for girls to excel in mathematics or programming.  It's like: if you do really well in class, that's good, but that's the end of the story.  Some of my schoolmates really had potential in mathematics, but they always seemed to have something "more important" than maths olympiad training, so after some time their standard goes down... :(

[quote="PiDeltaPhi"][quote="amfulger"]I think that the mathematics is an hermetic field.
We mathematicians "learn" to like math. You cannot say that math attracted you from the first grade when you learned to add positive integers.[/quote]

I loved to add positive integers,even before going to school  :D  :D[/quote]

Well, good for you! :D
Do you still like it?

Heh, reminds me of a scanned paper of Godel when he was a kid and trained addition, 4 + 2 = 6 or something like that. He had written it like 40 times on the paper (including 2 + 4 = 6), and an error was visible, 4 + 2 = 5 crossed over. :-)

[quote]Do you still like it?[/quote]

I still train it today  :D (though many told me i should've understood it by now  :D  :D )

[quote="Alucard"]Djimenez.
Does it mean that you agree with my idea or point of view??[/quote]

Actually I don't. I do consider that it is possible that as well as there is a big differenciation on muscle strength and flexibility from women to men, then maybe there can be a similar biological correlation between gender and math skills, nevertheless, in the first case, it is mainly because of hormones production, and it is not quite different between the humans and other mamals (females have children while males do not, then, the hormonal charges are different), but even when I do not discard it as a possibility, I do not think it is the main reason. I honestly think that it is more because of a social issue than a biological issue (even when the second option still remain!).

[quote="Alucard"]Do you think that the primates and our direct ancestor had some social rules to distribute these "thinking-required" tasks?[/quote]

Well, they distribute them, but that does not mean that just one of the gender took all of them. I think that most of the primates live in a gregarian way, and usually the males are the ones who fight or chase the predators and, for the few of them who are actually carniborus, the ones who haunt, while the females are the collectors and the ones who take care of the small members of the group.

[quote="Alucard"]...If so....what do you think they took into account to make this distribution?[/quote]

As well as in our case, probably phisical strength. Primates are not that different from us biologically. They have the basic same set of hormones, therefore males produce a lot more testosteron, therefore they develop more muscle, therefore they are stronger, therefore they are a little more prone to win a fight with a predator or a prey than a female. Of course, during the execution of such duties, they would be training different skills, and therefore, potentially different parts of their brain, but I honestly don't think that it can be genetically passed from generation to generation, but trained and tought from one generation to the following.

[quote="Alucard"]In fact there are a lot of studies and experiments that show that these kind of skills developed by males and females must be passed genetically. It must be that way, because that is the same way every living being evolves.[/quote]

Do you have reference for at least one of such studies that we can corroborate? Because I think that most of the human beens skills are learned, those skills are trainable and not necessarly genetical

[quote="Alucard"]I mean, the evolution is REAL, and the only way this can happen is by passing these skills or info from one being to another genetically[/quote]

Information yes, skills, not sure. And yes, the evolution is REAL

[quote="Alucard"]that's the only way, because if it would be as you say, there would had been no evolution at all.[/quote]

Why is that the only way?

women didnt play that big of a role back then so the famous mathematicians were male

Well I think Alucard is saying that our skills are hereditary (to some extent).  And this must be true, since we are innately superior to monkeys intellectually.  This is the basis for his argument.  But we don't know just what parts of our minds are determined genetically.

Yes, he is saying that skills are hereditary.  But he is also saying that skills are hereditary along gender lines.  I would like him to give reports corroborating this claim but he insists that he is "tired of trying to explain this" and doesn't seem to be willing to discuss this any further.

[quote="liyi"]It has been reported that men has more gray matter (brownish-gray nerve tissue) in the brain, which makes them better at thinking abstract things while women has more white matter (Whitish nerve tissue), which makes them better in communications and speaking languages.[/quote]

From medical point of view (I am speaking with medical background), the above statements are not accurate. Gray matter is where most neurons are located and white matter is the where the supporting tissue (blood vessels and glial cells, in other words, connective tissue) is located. Simply using the size of the brain to measure a person’s intellectual ability is no way to be accurate. Einstein’s brain is not in any sense larger than an average sized brain. Research of his brain did suggest that there were more complicated connections. And those connections happened to be in the white matter of his brain. The average size of male’s brain is larger than a female’s brain. However, the brain size is generally proportional to a person’s general physical size; in other words, you do need more brain mass to mobilize your muscle and other masses in your body. The brain mass is not composed of  all thinking cells.

Nature assigns different tasks for males and females, but not superiority of one over the other.

Alucard,

I think you've been reading some outdated anthropology and/or evolutionary biology.  The theories you espouse are not, as far as I know, currently held as fact by the scientific establishment.  Perhaps you could cite sources to the contrary?

[quote="polymorphic"]Yes, he is saying that skills are hereditary.  But he is also saying that skills are hereditary along gender lines.  I would like him to give reports corroborating this claim [b]but he insists that he is "tired of trying to explain this" and doesn't seem to be willing to discuss this any further.[/b][/quote]

Probably because of the lack of such sources. I mean, that is my guess, there is no reports and it may be just a personal opinion expressed as a scientific fact!

It's due to history and gender norms. In the past, women were expected to stay home. Later, the 1920s saw a revolution for women, and later during the 1960s, equal pay was enforced. 90% of it is due to these gender norms.


uh... not sure wat u mean by "analytically" but...

[hide]

the two equations imply:

$x^2 = 2x \implies x(x-2) = 0$

so, $x = 0$ or $x=2$ are possible

checking we see only when $x=2$ does $2^2 = 2(2) = 4$

[/hide]

doesn't a system include 2 variables and 2 equations rather than 1 variable and
2 equations?

I think that it is not a system, because he would have put "solve the system". So, I think that there are two different equations....................

[quote="cjqsg"]solve analytically

2^x = 2x

and

x^2 = 2^x[/quote]

1>The graphs of y=2^x and y = 2x intersect each other only at two points viz. x=1 and x= 2

For x < 0 , 2^x > 0 but 2x < 0

For  0=< x < 1 , 2^x > 2x

For x = 1 , 2^x =  2x

For  1 < x < 2 , 2^x < 2x

For x = 2 , 2^x =  2x

For  2 < x , 2^x >  2x

[quote="cjqsg"]they are two different equations and i know the answers to both of them.

what i mean by an analytic solution is one that shows how these answers can be gotten systematically, i.e. not by observation and showing that they are the only solutions[/quote]

Well in that case u need to employ a little bit of Calculus ,which is not allowed in this forum,to supplement ur reasoning.

[quote]what i mean by an analytic solution is one that shows how these answers can be gotten systematically, i.e. not by observation and showing that they are the only solutions[/quote]

For the second:  http://www.qc.edu.hk/math/Advanced%20Level/y%5Ex_x%5Ey.htm


Find the current $ I_b$
[img]http://i38.tinypic.com/28809lc.jpg[/img]

[hide="So far"]I tried applying KCL at the node the middle
$ \frac {24 \minus{} V_1}{250} \plus{} \frac {60I_b \minus{} V_1}{150} \equal{} \frac {V_1}{50}$

How to proceed from here?[/hide]
Why cannot I apply $ V \equal{} IR$ i.e. $ I_b \equal{} \frac {24}{250}$ ??

First correctly write the 3 loop equations.
Use atmost 2 variables: Ib and I, which is a branch current.

First of all...tell me, is the current being shared only by the 250 ohm resistor?? I think then everything will be clear to you, because you can write that Joule's equation if and only if the entire circuit can be simplified to a battery and the equivalent resistance of the circuit in closed loop.

It is better to solve problems involving meshes by using the Kirchoff's method....
 :)

I used KCL at the node, but I get two unknowns in 1 eqn.

What other moethod can I use??


Assume that $ a$, $ b$, $ c$ and $ d$ are the sides of a quadrilateral inscribed in a given circle. Prove that the product $ (ab \plus{} cd)(ac \plus{} bd)(ad \plus{} bc)$ acquires its maximum when the quadrilateral is a square.

We have that
$ (ab \plus{} cd)(ac \plus{} bd)(ad \plus{} bc)\leq(\frac {ab \plus{} cd \plus{} ac \plus{} bd \plus{} ad \plus{} bc}{3})^3\leq$
$ \frac {1}{2^9}(a \plus{} b \plus{} c \plus{} d)^6\leq 4^3\cdot r^6\,.$
The equality holds if and only if $ a \equal{} b \equal{} c \equal{} d$.

Let $ e,f$ diagonals of this quadrilateral,$ R$ radius of a given circle,$ S$area of this quadrilateral so by Ptolemey $ T\equal{}(ab \plus{} cd)(ac \plus{} bd)(ad \plus{} bc) \equal{} ef(ab \plus{} cd)(ad \plus{} bc) \equal{} (abe \plus{} cde)(adf \plus{} cdf) \equal{} 16R^2(\frac {abe \plus{} cde}{4R})(\frac {abe \plus{} cde}{4R}) \equal{} 16R^2S^2$
S max when the quadrilateral is a square, so T max the quadrilateral is a square

When the product $ (ab \plus{} cd)(ac \plus{} bd)(ad \plus{} bc)$ acquires its maximum, we can show that $c=d.$


A種水管一支，注水於桶５min. 可滿，若用B種水管一支，則要８min. 才滿，現同以A管3枝，B管5枝注水於桶7個，問全部注滿要幾min. ?


Your family traveled 260 miles to the beach at an average speed of 65 mph. Traffic on the return trip slowed your speed to an average of 40 mph. To the nearest whole number, what was the mean speed for the entire trip? Express your answer in miles per hour.

We use the harmonice mean here, $ \frac{2\cdot{40}\cdot{65}}{40\plus{}65}\equal{}\frac{2\cdot40\cdot65}{105}$

Which to the nearest whole number is 50.

Harmonic mean of two numbers $ x$ and $ y$ is:

$ \frac{2}{\frac{1}{x} \plus{} \frac{1}{y}}$


function
algebra
polynomial
number theory proposed
number theory

Let $ n$ be a positive integer and denote the Mobius function by $ \mu$. Define a polynomial $ f$ over $ \mathbb{Z}$ by

\[ f(X) \equal{} \sum_{d\mid n} \mu(d) (1 \plus{} X^d)^{n/d}\]
where the sum is over all the positive divisors $ d$ of $ n$. Show that all the coefficients of $ f$ are divisible by $ n$.

A sketch of what this is about:

For each positive interger $ n$ denote by $ Z_n$ the cyclic additive group of integers mod $ n$.
Let us say that a subset $ S$ of $ Z_n$ has [i]period[/i] $ p$ if the conditions $ x\in S$ and $ x \plus{} p\in S$ are equivalent.
Any subset $ S$ of $ Z_n$ has period $ n$ and therefore some [i]smallest[/i] positive period $ d$.
We have $ d\mid n$ (not hard) and, if $ S$ has $ m$ elements, $ n\mid dm$.

1. (corrected :blush:) The number of subsets of $ Z_n$ having $ m$ elements and period $ d$ (or smaller) is $ \binom{d}{dm/n}$ if $ d\mid n$.

2. The number of subsets of $ Z_n$ having $ m$ elements and [i]least[/i] period $ n$ is
\[ \sum_{d\mid n,d\mid m}\mu(d)\binom{n/d}{m/d}
\]
3. That number is the coefficient of $ X^m$ in $ f(X)$.

4. That number is divisible by $ n$, because the $ m$-element subsets of $ Z_n$ having no period $ < n$, fall into equivalence classes each of which has $ n$
elements.


The cube has edge length $ 9$, so the sphere's diameter is $ 9\sqrt {3}$.

Then, divide this by $ \sqrt2$, to get $ \boxed{\frac {9\sqrt6}{2}}$

I'm not so sure...

EDIT: Read it wrong. :P  And I made a stupid mistake.  :P

Look at BOGTRO's solution.

[quote="GameBot"]A sphere is inscribed in a cube with edge length 9 inches. Then a smaller cube in inscribed in the sphere. How many cubic inches are in the volume of the inscribed cube? Express your answer in simplest radical form.[/quote]

The cube has edge length $ 9$, so the sphere has a diameter of $ 9$.

Therefore, since all sides of the cube are equal, $ s\sqrt{3}\equal{}9$.

Thus, $ s\equal{}\frac{9}{\sqrt{3}}$, which is equal to $ \frac{9}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\equal{}\frac{9\sqrt{3}}{3}\equal{}\boxed{3\sqrt{3}}$.

Note that $ s\sqrt{3}$ was used due to the fact that it is the hypotenuse of a right triangle with legs $ s$, $ s\sqrt{2}$.

Thus, $ \sqrt{s^2\plus{}2s^2}\equal{}s\sqrt{3}$.

@ernie: The sphere is [b]inscribed[/b] in the cube. Also, you don't divide by $ \sqrt{2}$, that would be for a square. However, we are talking about a cube.

it asks for volume, so it's side lenght cubed.

81sqrt3.


At time t=0, 4 insects A,B,C and D stand at the corners of a square of side a.
For time t>0 each insect crawls with constant speed v in the direction of the next insect in cyclic order(that is A crawls towards B, B towards C and so on)
Show that the insects meet after time a/v and that they go round the centre of the square an infinite number of times.

[quote="Michael Niland"]
Show that the insects meet after time a/v...[/quote]

Why would they meet if they all have speed v?


P.S.  I put it into my own words.  Just forgot to cite the source! :(  :(  :(  :(  :(

[quote="mli"]AAAAAAAAARRRRGGGGGGG!!!! I think I might be one of those people![/quote]

Thanks for coming forward, Chenchuan.  We will add the source to your quiz, and continue to consider your application.

Note that it's not necessary to have posted anything publicly.  Private message would have been fine.


Let a and b be given positive real numbers with a < b. If two points are selected at random from a straight line segment of length b, what is the probability that the distance between them is at least a?

It's an old Putnam problem and it's been posted before:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=341172544&t=11013


Does anyone knows where I can find the official solutions?

i think they go up at http://sms.math.nus.edu.sg/simo/simo.aspx but they aren't up yet.

last year, they went up at the official imo website at http://www.imo2004.gr/fimo , so maybe this year they'll put them up at http://erdos.fciencias.unam.mx/competition.htm but i don't know.

valentin has also said he'd put them up.


limit
trigonometry
geometry
circumcircle
geometry open

Prove:
\[ \pi = \displaystyle\lim_{n\to\infty}n\sin(\frac{\pi}{n}) \]

I don't really use the L' Hospital Rule to solve it, because there's a geometric way to do it..........

Hi !

We have $\lim_{x \to 0}{\frac{sinx}{x}= 1}$ and with $x=\frac{\pi}{n}$ it's O.K. ! ;)

I've forgotten about this, and i've constructed a regular n-gon with its circumradius is 1. When n tends to infinity, it'll be a circle.......


geometry
incenter
trapezoid
geometry unsolved

circles $ S_1$ and  $ S_2$ with centers $ O_1$ and $ O_2$ respectively intersect each other at points $ A$ and $ B$ .ray $ O_1B$ intersects $ S_1$ at point $ F$ and ray $ O_2B$ intersects $ S_1$ at point $ E$.the line parallel to $ EF$ and passing through $ B$ intersects $ S_1$ and $ S_2$ at points $ M$ and $ N$ .prove that $ B$ is incenter of $ EAF$ and $ MN\equal{}AE\plus{}AF$

It is easy to acquire O1,F,E,O2 are concyclic
∠O1EF=∠O1O2F=∠AMB=∠AEB   because we have MN∥EF  =〉 ∠BEF=∠O1EF=∠EBM
AB=EM=BE1=BF1=NF     BEMA is an isosceles trapezoid  =〉 AE=BM
Analogously  BFNA is an isosceles trapezoid  =〉 AF=BN
=〉MN=BM+BN=AE+AF
It is obvious that B is the incenter of △AEF


Proof that closed subspace of locally compact space is locally compact.

[quote="rmfgcampos"]Let $ X$ be the locally compact space and $ Y$ a subspace.

let $ y \in Y$

As $ y\in X$ and $ X$ is locally compact there is a compact neighborhood containing $ y$. So $ Y$ is locally compact.

What is wrong with this proof? I'm not using $ X$ is closed.[/quote]
See the definition of being locally compact means there is a open set(open in $ X$) say $ U$ around the point $ y$ whose closure is compact.Here since $ Y$ is closed subset of $ X$ so $ Y \cap \tilde{U}$ is closed in $ Y$ and also $ U \cap Y$ is open in $ Y$ where $ \tilde{U}$ is closure of $ U$.


So the ARML runoff for New Haven County (and possibly other areas) was today, if anyone wants to compare answers, just drop me a PM.

[quote="worthawholebean"]Has anyone gotten anything from the ARML people?[/quote]

Yes. Check your PMs.

So who's on the team this year?
Couple kids from Avon, I hear.
At least that's what's in my ear
I think I'll go have a beer.


Given $ f(x)\equal{}x^n\minus{}e^x$ with $ 1<x<n$
I am to prove $ f'(x)>0$ with $ 1<x<n$

[quote="bibobeo"]Given $ f(x) \equal{} x^n \minus{} e^x$ with $ 1 < x < n$
I am to prove $ f'(x) > 0$ with $ 1 < x < n$[/quote]
What about $ n\equal{}2$ ?

I have no idea what it is. I tried wikipedia, but in vain. It is too hard to understand, and I don't know how to apply it.

Could you please be more specific?

[hide="Taylor Series"]
Defintion
The Taylor Series of a function is $ \sum_{i\equal{}0}^{\infty} \frac{f^{i} \left( a \right) \cdot \left( x \minus{} a \right)^{i}}{i!}$, where $ f^{i} \left( x \right)$ is the $ i$-th derivative of the function. The value $ a$ says about what number the series is centered.

The Taylor Series for $ e^{x}$ centered at zero is $ 1 \plus{} x \plus{} \frac{x^{2}}{2!} \plus{} \frac{x^{3}}{3!} \plus{} \dots$.
So the problem statement becomes,
Proof that $ n x^{n\minus{}1} > 1 \plus{} x \plus{} \frac{x^{2}}{2!} \plus{} \frac{x^{3}}{3!} \plus{} \dots$ for all $ x \in \left( 1 , n \right) \ , \ n \ge 3$.

That should be easier to solve.
[/hide]

Leave it as $ nx^{n\minus{}1} > e^x$.  Then we need only show that $ (n\minus{}1)\ln{x} \plus{} \ln{n} > x$.  Then use Taylor series for $ \ln{x}$.  We might want to center the series at $ x\equal{}1$ and then substitute $ y \equal{} x\minus{}1$ on both sides.

Follow you, and we have to prove
$ (n\minus{}1)((x\minus{}1)\minus{}\frac{(x\minus{}1)^2}{2}\plus{}...)\plus{}lnn>x$
and I still don't know what to do next
(Sorry if I am so noob)


\[ \prod_{k=1}^{n} \frac{(2k-1)^4+\frac{1}{4}}{(2k)^4+\frac{1}{4}}  \]
Could you give me some hints about evaluating sums and products. What methods should one use to do it?


Yep, it's the time of the year again where everyone chows down on lotus paste and egg yolks. Happy moon festival!

sounds fun!

And yes, I [i]do[/i] know what the Moon Festival is.

愉快的中秋节!

yay!! moon cakes are sooooooooooo yummy =]

[hide="And guess what today is???"]Tuesday!!![/hide]


trigonometry
calculus
integration
calculus computations

a. $ \int_{0}^\infty\;\frac {\cos\,x}{x}\;\textbf dx$

b. $ \int_{0}^\infty\;\frac {{e}^{\minus{} x}}{x}\;\textbf dx$

Are you sure what you want is that and not $ \int_0^{\plus{} \infty} \frac{1 \minus{} \cos x}{x}\,dx$ and $ \int_0^{\plus{} \infty} \frac{1\minus{}e^{\minus{}x}}{x}\,dx$?

What have you done so far? For example, did you already proved the nature (convergence/divergence) of both integrals?

No.....,,, I don't know how to do it... i tried to integrate it by parts...

but still, not effective

For example, let's take a look at the first integral. We have

$ \int_0^{\plus{} \infty} \frac{\cos x}{x}\,dx \equal{} \left[\frac{\sin x}{x}\right]_0^{\plus{} \infty} \plus{} \int_0^{\plus{} \infty} \frac{\sin x}{x^2}\,dx \equal{} \minus{}1 \plus{} \int_0^1 \frac{\sin x}{x^2}\,dx \plus{} \int_1^{\plus{} \infty} \frac{\sin x}{x^2}\,dx$.

Now, what can you say about the two integrals on the right?


Find the sum of $ \frac{4!}{6!3!} \plus{} \frac{3!}{5!}$. Express your answer as a common fraction.


Find the sides of the right triangle if they are positive integers and their product is three times bigger than the perimeter.

I already have a solution which is there are no solutions. :maybe:  Maybe Ive done any terrible error. :blush:

abc=3(a+b+c) =>
3│a or 3│b or 3│c.

If 3│c than we have a 2  + b 2 ≡0(mod 9) The residues of X 2  modulo 9 are 2, 0, 1, 4 hence we see that a 2  + b 2 ≡0(mod 9) if and only if both a 2 ≡0(mod 9) and b 2 ≡0(mod 9).
Hence if 3│c then 3│b and 3│a.

abc=3(a+b+c) =>
c=3(a+b)/(ab-3) Its obvious that ab-3 cant be neither 1 nor 3 because wed have
c=3(a+b) or c=a+b which is impossible hence ab-3│a+b that is
a+b=kab-3k
Therefore if 3│a from this equality well have 3│b and therefore 3│c since c 2  = a 2  + b 2 .
Similarly if 3│b well have 3│a and 3│c.
So we showed that in any case we have 3│a and 3│b and 3│c.
That is a=3m, b=3n, c=3q where m, n, q are positive integers.
Since cc 2  = a 2  + b 2
 then q 2 =m 2  + n 2
From  abc=3(a+b+c) we have
3mnq=m+n+q =>
q(3mn-1)=m+n=>
q=(m+n)/(3mn-1) =>
m+n≥3mn-1 =>
m+1≥n(3m-1) =>
n≤(m+1)/(3m-1)=1- 2(m-1)/(3m-1) ≤ 1 With equality when m=1.
Since n is a positive integer we have to have n=1, thats m=1.
So m=n which gives a=b which doesnt give an integer for c. :maybe:  :maybe:

[quote="Iris Aliaj"]
abc=3(a+b+c) =>
c=3(a+b)/(ab-3) Its obvious that ab-3 cant be neither 1 nor 3 because wed have
c=3(a+b) or c=a+b which is impossible hence ab-3│a+b
[/quote]

I don't quite understand this part. Just because ab-3 | 3(a+b) and ab-3 doesn't divide 3 does not mean that ab-3 | (a+b)..
Quick example: (not using these exact as and bs): 6|3*8, but 6 doesn't divide either 3 or 8.

We have to find positive integers $a,b,c$ such that $c^2 = a^2 +b ^2$ and $abc = 3(a+b+c)$.
Dividing the latter by $abc$, it leads to $\frac 1 3 = \frac 1 {ab} + \frac 1 {bc} + \frac 1 {ca}$.
Note that $a \leq b < c$, so that $ab = \min \{ab,bc,ca \}$. It follows that $\frac 1 3 \leq \frac 3 {ab}$ which leads to $ab \leq 9$.

In another hand, we cannot have $a=b$ otherwise $c$ would not be an integer. Then, $1 \leq a < b < c$ from which we deduce that $c \geq 3$. Therefore $a^2 = c^2 - b^2 \geq 9 - 4$ (this is the least possible difference between two squares), which gives $a \geq 3$.
Then $b \geq 4$ and $ab \geq 12$. A contradiction.

Pierre.

One of the old tricks, and works well with this kind of problemss is:
 Sides of a right angel triangle  can be takes as:

k(2mn) , k(m^2-n^2), k(m^2+n^2)

(m, n are prime to each other, of different parity, - that is one even and one odd))

So our problem reduces to:

( k^3)*(2*m*n)(m^2-n^2)(m^2+n^2) = 3*k*(2*m*n+2*m^2)

or (k^2)*n*(m-n)(m^2+n^2) = 3

which is impossible since minimimum walue (even with k=1,n=1,(m-n)=1) of LHS is 5)


a,b,c £ IR
prove that:
http://www.monsterup.com/image.php?url=upload/123887715257.gif

Hello, do you mean $ (a\plus{}b\minus{}c)^2\plus{}(a\minus{}b\plus{}c)^2\plus{}(\minus{}a\plus{}b\plus{}c)^2\plus{}\frac{3}{4}\geq a\plus{}b\plus{}c$  for all real numbers $ a,b,c$?
Sonnhard.

It is so easy .

After expanding we get that :

$ 3a^2\plus{}3b^2\plus{}3c^2\minus{}2ab\minus{}2bc\minus{}2ca\plus{}\frac{3}{4}\ge a\plus{}b\plus{}c$

$ \Leftrightarrow (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\plus{}(a\minus{}\frac{1}{2})^2\plus{}(b\minus{}\frac{1}{2})^2\plus{}(c\minus{}\frac{1}{2})^2\ge 0$

hello, or alternatively, setting $ x\equal{}a\plus{}b\minus{}c$,$ y\equal{}a\minus{}b\plus{}c$,$ z\equal{}\minus{}a\plus{}b\plus{}c$, so we have $ x\plus{}y\plus{}z\equal{}a\plus{}b\plus{}c$ and our inequality is equivalent with $ (x\plus{}\frac{1}{2})^2\plus{}(y\plus{}\frac{1}{2})^2\plus{}(z\plus{}\frac{1}{2})^2\geq0$.
Sonnhard.

we have, $ (a\minus{}b\plus{}c)^2\plus{}\frac{1}{4} geq a\minus{}b\plus{}c$
and $ (a\plus{}b\minus{}c)^2\plus{}\frac{1}{4} geq a\plus{}b\minus{}c$
and $ (\minus{}a\plus{}b\plus{}c)^2\plus{}\frac{1}{4} geq \minus{}a\plus{}b\plus{}c$
after summation we get, $ (a\plus{}b\minus{}c)^2\plus{}(a\minus{}b\plus{}c)^2\plus{}(\minus{}a\plus{}b\plus{}c)^2\plus{}\frac{3}{4}\geq a\plus{}b\plus{}c$


Given six numbers which satisfy the relations:

$ x^2\plus{}xy\plus{}y^2\equal{}a^2$
$ y^2\plus{}yz\plus{}z^2\equal{}b^2$
$ z^2\plus{}zx\plus{}x^2\equal{}c^2$

Determine the sum $ x\plus{}y\plus{}z$ in terms of $ a, b, c$.  Give a geometric interpretation if the numbers are all positive.

[hide="hint"]
Use the Cosine law: $ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab\cos(\theta)$
[/hide]

[hide="solution"]
Summing up the equations gives;
$ 2(x^2\plus{}y^2\plus{}z^2) \plus{} (xy\plus{}yz\plus{}zx) \equal{} a^2\plus{}b^2\plus{}c^2$

$ \therefore 2(x\plus{}y\plus{}z)^2 \minus{} 3(xy\plus{}yz\plus{}zx) \equal{} a^2\plus{}b^2\plus{}c^2 \qquad(1)$

From the cosine law, we get the geo interpretation of three triangles with sides $ \{x,y,a\}$, $ \{y,z,b\}$ and $ \{z,x,c\}$. If we put all these triangles together we get a larger triangle with sides $ a,b,c$.

So the area of each smaller triangles are:
 $ \frac{1}{2}xy\sin(120),\frac{1}{2}yz\sin(120)$, and $ \frac{1}{2}zx\sin(120)$

The area of the larger triangle is given by heron's $ \sqrt{\frac{1}{8}(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(c\plus{}a\minus{}b)(b\plus{}c\minus{}a)}$

This gives $ xy\plus{}yz\plus{}zx \equal{} \frac{4}{\sqrt{3}}\cdot \sqrt{\frac{1}{8}(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(c\plus{}a\minus{}b)(b\plus{}c\minus{}a)} \qquad (2)$

(1) and (2) gives

$ 2(x\plus{}y\plus{}z)^2 \equal{} a^2\plus{}b^2\plus{}c^2 \plus{}\sqrt{6(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(c\plus{}a\minus{}b)(b\plus{}c\minus{}a)}$

$ x\plus{}y\plus{}z \equal{} \sqrt{\frac{1}{2}(a^2\plus{}b^2\plus{}c^2 \plus{}\sqrt{6(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(c\plus{}a\minus{}b)(b\plus{}c\minus{}a)})}$

You may be able to simplify this..I can't

[/hide]

[hide="Geometric Interpretation"]
The picture is a triangle with sides a, b, and c.
Line segments coming from the vertices (A, B, and C) meet at a point inside in the triangle, point O.
given that angles AOB, BOC, and COA are all 120 degrees, find $ x\plus{}y\plus{}z$ in terms of $ a,b,c$.
[/hide]

As for the proof... it's 11:34 where I live, better get to sleep  :(

[hide="hint"]It's a well-known result that $ EC, FB$ and $ AG$ are concurrent and $ x\plus{}y\plus{}z\equal{}EC\equal{}FB\equal{}AG$
[geogebra]dbe0eee201f1761f9388467a35e48aec4087535b[/geogebra][/hide]

[quote="ocha"][hide="hint"]
Use the Cosine law: $ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab\cos(\theta)$
[/hide]

[hide="solution"]
Summing up the equations gives;
$ 2(x^2 \plus{} y^2 \plus{} z^2) \plus{} (xy \plus{} yz \plus{} zx) \equal{} a^2 \plus{} b^2 \plus{} c^2$

$ \therefore 2(x \plus{} y \plus{} z)^2 \minus{} 3(xy \plus{} yz \plus{} zx) \equal{} a^2 \plus{} b^2 \plus{} c^2 \qquad(1)$

From the cosine law, we get the geo interpretation of three triangles with sides $ \{x,y,a\}$, $ \{y,z,b\}$ and $ \{z,x,c\}$. If we put all these triangles together we get a larger triangle with sides $ a,b,c$.

So the area of each smaller triangles are:
 $ \frac {1}{2}xy\sin(120),\frac {1}{2}yz\sin(120)$, and $ \frac {1}{2}zx\sin(120)$

The area of the larger triangle is given by heron's $ \sqrt {\frac {1}{8}(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(c \plus{} a \minus{} b)(b \plus{} c \minus{} a)}$

This gives $ xy \plus{} yz \plus{} zx \equal{} \frac {4}{\sqrt {3}}\cdot \sqrt {\frac {1}{8}(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(c \plus{} a \minus{} b)(b \plus{} c \minus{} a)} \qquad (2)$

(1) and (2) gives

$ 2(x \plus{} y \plus{} z)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} \sqrt {6(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(c \plus{} a \minus{} b)(b \plus{} c \minus{} a)}$

$ x \plus{} y \plus{} z \equal{} \sqrt {\frac {1}{2}(a^2 \plus{} b^2 \plus{} c^2 \plus{} \sqrt {6(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(c \plus{} a \minus{} b)(b \plus{} c \minus{} a)})}$

You may be able to simplify this..I can't

[/hide][/quote]
The Heron formula must be
$ \sqrt {\frac {1}{16}(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(c \plus{} a \minus{} b)(b \plus{} c \minus{} a)}$
 :P

Thanks for the solution Ocha you just made a small mistake with Heron's formula.  The final expression should be:

$ x\plus{}y\plus{}z \equal{}\sqrt{\frac{1}{2}(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}\sqrt{3(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(c\plus{}a\minus{}b)(b\plus{}c\minus{}a)})}$

What about when the numbers are NOT all positive?

[quote="Tomekk"][quote="Machodog"]
What about when the numbers are NOT all positive?[/quote]

Maybe it's only me, but I belive that a,b,c,x,y,z are sidelenghts of triangles which cannot be negative :)

Sidelenght a cannot be -1 for instance, it would be hard to make that kind of triangle picture on paper ^^.[/quote]
$ x^2\plus{}xy\plus{}y^2\equal{}a^2 \implies (\minus{}x)^2\plus{}(\minus{}x)(\minus{}y)\plus{}(\minus{}y)^2\equal{}a^2$ or $ (\minus{}a)^2$ :roll:

[quote="Tomekk"][quote="Machodog"]
What about when the numbers are NOT all positive?[/quote]

Maybe it's only me, but I belive that a,b,c,x,y,z are sidelenghts of triangles which cannot be negative :)

Sidelenght a cannot be -1 for instance, it would be hard to make that kind of triangle picture on paper ^^.[/quote]

Sorry perhaps I was unclear.  The "first part" of the original problem is to find $ x\plus{}y\plus{}z$ when $ x, y, z, a, b, c$ are not necessarily all positive.  The "second part" is the geometric interpretation when they are all positive (that is when we can say that they are all side lengths  of triangles).  ANyway, here is the solution for when the numbers are not all positive:

Adding all three equations together we get:
$ a^2\plus{}b^2\plus{}c^2 \equal{} (x\plus{}y\plus{}z)^2 \plus{} x^2\plus{}y^2\plus{}z^2 \minus{} xy \minus{}yz\minus{}zx$
$ a^2\plus{}b^2\plus{}c^2 \equal{} (x\plus{}y\plus{}z)^2 \plus{} \frac{1}{2} \left((x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2 \right)$

Subtracting the first equation from the second gives:
$ x^2\plus{}y(x\minus{}z)\minus{}z^2\equal{}a^2\minus{}b^2 \implies (x\minus{}z)\equal{}\frac{a^2\minus{}b^2}{x\plus{}y\plus{}z}$
Doing the same for all three pairs, and substituting in we get:

$ a^2\plus{}b^2\plus{}c^2 \equal{} (x\plus{}y\plus{}z)^2 \plus{}  \frac{(b^2\minus{}c^2)^2\plus{}(c^2\minus{}a^2)^2\plus{}(a^2\minus{}b^2)^2}{2(x\plus{}y\plus{}z)^2}$

This is a quadratic in $ (x\plus{}y\plus{}z)^2$ which we can solve, and simplify to get:

$ 2(x\plus{}y\plus{}z)^2\equal{}a^2\plus{}b^2\plus{}c^2 \pm \sqrt{6(a^2b^2 \plus{} b^2c^2\plus{}c^2a^2)\minus{}3(a^4\plus{}b^4\plus{}c^4)}$


ratio
geometry
similar triangles
Pythagorean Theorem

[img]http://img217.imageshack.us/img217/3410/trianglezlw3.png[/img]

AC, BD, and m are all perpendicular to BC.
CD = 20; AB = 15; BC = d

Find and prove $ \frac{1}{m}$ in terms of d.

You are allowed to have radicals in denominators.

hint.
[hide]1/m is the sum of the reciprocals of ac and bd.
using that fact, one can determine the length in terms of d[/hide]

[hide]Label the intersection of $ AB$ and $ CD$ $ E$ and the intersection of $ BC$ and $ m$ $ F$.
$ \triangle{ABC} \sim \triangle{EBF}$ by Angle-Angle because $ \angle{ACB} \cong \angle{EFB}$ because both are right angles, and both triangles share $ \angle{ABC}$.
By similar reasoning, $ \triangle{DCB} \sim \triangle{ECF}$.

Using ratios of sides with each set of similar triangles, one gets $ \frac {AC}{m} \equal{} \frac {d}{BF}$ and $ \frac {BD}{m} \equal{} \frac {d}{d \minus{} BF}$.
Rewrite these to get $ md \equal{} AC \cdot BF$ and $ md \equal{} BD \cdot d \minus{} BD \cdot BF$.
Then,
$ AC \cdot BF \equal{} BD \cdot d \minus{} BD \cdot BF$.
Solve for $ BF$ to get $ BF \equal{} \frac {BD \cdot d}{AC \plus{} BD}$.

$ md \equal{} AC \cdot BF$
$ md \equal{} AC \cdot \frac {BD \cdot d}{AC \plus{} BD}$
$ m \equal{} \frac {AC \cdot BD}{AC \plus{} BD}$
$ \frac {1}{m} \equal{} \frac {AC \plus{} BD}{AC \cdot BD} \equal{} \frac {1}{BD} \plus{} \frac {1}{AC}$

By the Pythagorean Theorem, $ AC \equal{} \sqrt {15^2 \minus{} d^2}$ and $ BD \equal{} \sqrt {20^2 \minus{} d^2}$.

$ \frac {1}{m} \equal{} \boxed{\frac {1}{\sqrt {400 \minus{} d^2}} \plus{} \frac {1}{\sqrt {225 \minus{} d^2}}}$
[/hide]


trigonometry
inequalities
inequalities unsolved

For any triangle $ABC$, \[{ \prod \sin A \leq ( \frac{3 \sqrt{3}}{2 \pi}} )^3 ABC  \] holds.

Incredibly enough, $log \left (\frac{\sin x}{x} \right)$ is concave, and ineq comes directly from Jensen  :D