Two candles are the same height. The first takes 10 hours to burn completely whilst the second takes 8 hours to burn completely.
Both candles are lit at midday. At what time is the height of the first candle twice the height of the second candle?

[hide="My Solution"]
Let the height of both candles be $ h$. The first candle burns at a rate of $ \frac {h}{10}$ per hour and $ \frac {h}{8}$ per hour for the second. We can set up the following equation in terms of $ x$ where $ x$ is the time desired.

$ h \minus{} \frac {hx}{10} \equal{} 2(h \minus{} \frac {hx}{8}) \\
\frac {(10 \minus{} x)h}{10} \equal{} \frac {(8 \minus{} x)h}{4} \\
\frac {10 \minus{} x}{10} \equal{} \frac {8 \minus{} x}{4} \\
40 \minus{} 4x \equal{} 80 \minus{} 10x \\
6x \equal{} 40 \\
x \equal{} \frac {20}{3}\text{ hours}$

So the time at which the desired condition occurs is $ \boxed{6: 40\text{ P.M.}}$.
[/hide]

[hide="Solution"]
Without loss of generality, let the candles have height $ 1$.
The first candle's height decreases at a rate of $ \minus{}\frac{1}{10}$ units/sec, so its height at a given time $ t$ is simply: $ \minus{}\int\frac{dt}{10}\equal{}C\minus{}\frac{t}{10}$.

 :rotfl: Just kidding (though that works too).

The height of the first candle at time $ t$ is $ 1\minus{}\frac{t}{10}$. (From that integral up there, or by simply noting that after $ t$ minutes, $ t/10$ units have melted, so $ 1\minus{}t/10$ units are left).
The height of the second candle at time $ t$ is $ 1\minus{}\frac{t}{8}$.

Solve $ 1\minus{}\frac{t}{10}\equal{}2\minus{}\frac{t}{4}$ for $ t$ to get $ t\equal{}\frac{20}{3}\equal{}6\plus{}\frac{2}{3}$.

$ \frac{2}{3}\cdot 60\equal{}40$, so it takes 6 hours and 40 minutes.

The answer, then, is $ \boxed{\text{6: 40 PM}}$.
[/hide]

Heh. I just solved an IMO problem. (pun intended)


On what interval does \[ \frac{|x-|x||}{x} > x \]?

(A) (-2,0)
(B) (-infinity, -2)
(C) (2,infinity)
(D) (-infinity,0)
(E) (0,infinity)

I answered (wrongly) [hide]A[/hide]

It's clear that $x$ must be negative.  So we can reduce the problem to

$|x-|x||<x^2$ and then to

$|x-(-x)|<x^2$ and then to

$2|x|<x^2$ and then to

$-2x<x^2$ and then to

$x^2+2x>0$ which is equivalent to $x(x+2)>0$; e.g, for $x<-2$ or $x>0$.  But the last possibility was excluded, so...

On a multiple choice test like this it is always a good idea to plug in numbers if you can do so in a timely manner (as is the case with this problem).

Otherwise, I probably would have done this:
[hide]
We start from the inside and work out.  We first examine the $|x|$.  We have two cases, $x\ge 0$ and $x<0.$

[b]Case 1: $x\ge 0$[/b]

The problem simplifies to $0>x$ which contradicts our assumption that $x\ge 0.$

[b]Case 2: $x<0$[/b]

This simplifies to $\frac{|2x|}x>x\Rightarrow \frac{-2x}x>x\Rightarrow -2>x.$  Therefore the answer is $(-\infty,-2).$ [/hide]


group theory
abstract algebra
search
superior algebra
superior algebra solved

Let $ p$ be a prime number. If a group $ G$ has $ p\cdot p$ elements, prove that $ G$ is a commutative group.

Use the class equation:
|G|=|Z(G)|+ \sum|G: C(x)| [1] where
G is the group
Z(G) is the center of G, is {x in G| xg=gx for all g in G}. It is easy to see that Z(G) is a subgroup of G.
C(x)={g in G| xg=gx} this is also a subgroup of G
|M| is the number of elements of a set M
|G:H| is the number of distinct elements of the set {xH| x is in G}
xH={xh| h is in H}
The sum is taken for a system of representands such that x is not in Z(G) and each G:C(x) is distinct from the others.
Or you could search for it in a first year algebra course and correct me if I made some mistakes (hope not  :D ).

Back to [1] we have p 2 =|Z(G)|+ \sum|G: C(x)|
Z(G) is a subgroup of G, so by Lagrange |Z(G)| is 1,p or p 2 . If it is p 2 , we are done Z(G)=G and G is commutative.
If Z(G)=1 then since C(x) is a subgroup different from G (remember x is not in Z(G)) then by Lagrange, each |G:C(x)|=p and we obtain a contradiction mod p.
If Z(G)=p, then G/Z(G) is izomorphic to Z_p and use the following result:
[b]If H is a subgroup of G contained in Z(G) and such that G/H is cyclic then G is commutative.[/b]
Proof: First note the the factorization G/H makes sense. This is because H<Z(G) and since it is normal G.
Take x^ to be a generator of G/H. Then G/H=<xH>={x k H} and so each element in G has a writing g=x ih.
Take g'=x jh' another element of G
gg'=x ihx jh'=x ix jhh' (because H is in Z(G))=x jx ih'h (H is commutative and powers of x commute)=x jh'x ih (and again H comm...)=g'g.
So G is commutative.

Apply this result to G and H=Z(G).

For more on group theory search for a first year algebra or search for J.S. Milne's notes on the NET. :D


linear algebra
matrix
analytic geometry
superior algebra
superior algebra unsolved

prove:
If D i si division algebra, then $ M_{n}(D)$ is a prime algebra for every $ n \in N$

You may want to show that it is a "simple algebra" (don't know your definition of "prime algebra", but indecomposable would follow from simple, too).
Assume $ I$ is a two-sided ideal containing some nonzero element $ A$. Show by elementary matrix computations that for any $ i,j$ there must be a matrix in the ideal that is $ 1$ at coordinates $ (i,j)$ and $ 0$ otherwise. Conclude that $ I$ is the full algebra.


linear algebra
matrix
linear algebra unsolved

Let $A \in {\mathbb C}^{n \times n}$ such that $rank[I|A|A^2| ... |A^k]<n$ where $k \in {\mathbb N}^*$ is fixed. Show that the matrix $I+A+...+A^k$ is singular (non-inverteble).

$B$ is not a square matrix, hence it does not have an inverse. What do you actually mean when you say "singular"?


geometry
geometric transformation
reflection
\/closed

Yes, the price has gone up, in part because Google declined to renew their sponsorship.

Also, the 2005-06 price was an introductory price to build interest in the new program.  The current price more accurately reflects the cost of running the program.

We also hope to involve more past Olympiad students (college students) to supplement instructional time.

Remember that the program lasts for many months and is less than a quarter of the cost of many summer programs.  Obviously, there's no room and board in WOOT, but we try to stretch the resources as well as we can.

In general, we started all our online programs underpriced with the plan of letting the prices rise a little bit at a time, along with interest in the programs, until a good system is in place.  We will take in [appropriate] corporate sponsorship when we can to defray costs.

[quote="DPatrick"][quote="probability1.01"]May I ask what the primary expenses are for running WOOT?[/quote]Labor -- both in the curriculum-development stage (e.g. preparing a syllabus) and the actual execution-stage (e.g. lecturing and grading problem sets and tests).

Also, the site has generic overhead -- to cite our most obvious example, the forum and Math Jams are free -- that have to be supported through our revenue-producing activities.[/quote]

It's also probably not clear to most students that selecting problems and writing solutions is a process that passes through several hands for a variety of reasons -- including the desire to use problems that most students will not have seen before.

However, I wouldn't say that we explicitly take site overhead into account (in the broad sense) when planning WOOT.  But we do take into account the time spent by instructors managing the WOOT forum, which is very large with hundreds of problems selected to go with the lecture material.


function
limit
real analysis
real analysis theorems

Dear everyone.

$ \forall a,b(\in R), f(x)$ satisfies $ \left|f(a) \minus{} f(b)\right|\leq (a \minus{} b)^{2}$.

Prove $ f(x)$ is constant.

Thanks in advance. :)

Suppose $ a < b$. For any positive integer n take $ x_k \equal{} a \plus{} \frac {k(b\minus{}a)} {n}$ for $ 0 \leq k \leq n$. Then $ |f(a) \minus{} f(b)| \equal{} | \sum_{k\equal{}0}^{n\minus{}1} (f(x_k) \minus{} f(x_{k\plus{}1})) | \leq \sum_{k\equal{}0}^{n\minus{}1} (| f(x_k) \minus{} f(x_{k\plus{}1} |) \leq$ $ \sum_{k\equal{}0}^{n\minus{}1} (x_k \minus{} x_{k\plus{}1})^2 \equal{} n\left (\frac {b\minus{}a} {n}\right )^2 \equal{} \frac {(b\minus{}a)^2} {n}$.
Since $ n$ can be taken as large as wanted, it follows $ |f(a) \minus{} f(b)| \equal{} 0$, therefore $ f(a) \equal{} f(b)$.

Or you can just prove that $ f'(x)\equal{}0$ everywhere.

Following Kent Merryfield:
$ |\frac {f(a) - f(b)}{a - b}|\leq |a - b| \to 0$ as $ a\to b$. Therefore, $ f'(a) = \lim_{b\to a}\frac {f(a) - f(b)}{a - b} = 0$ for every $ a$. It follows that $ f$ is constant.

The same task:
http://www.mathlinks.ro/viewtopic.php?t=265086

It is enough to have $ \left|f(a) \minus{} f(b)\right|\leq |a \minus{} b|^{1 \plus{} \varepsilon}$, for some given $ \varepsilon > 0$. In this problem $ \varepsilon \equal{} 1$; in Kirill's link $ \varepsilon \equal{} 1/2$.


function
Functional Analysis
real analysis
real analysis solved

Let $f:X\to \mathbb{R}$ is a continuous functional defined on banach space $X$, $a\in X$.
It is known that neighbourhood $B_{r}(a)$ is free from local maxima and local minima.

Is it true that there is continuous path $p:[0,1]\to X$ s.t. $p(0)=a$, $p(1)\neq a$ and $f(p(t))$ is a decreasing function on $[0,1]$.

1) $X=\mathbb{R}^n$.
2) $X$ is an infinite-dimensional space.

[quote]Myth is out of here. Bye[/quote]
But now you're back?

I think the answer is no. My example is in $\mathbb{R}^2.$ Let $f(x,y)=x(y-x^{-1/3}\sin(1/x))$ for $x\ne 0$ and zero for $x=0.$ Let your point $a$ be the origin. The curve in question would have to lie in the set $f^{-1}([-b,0])$ for some $b$ but it appears that that set is not path-connected.


Some object is let fall from 1000 kilometers above the ground. How long will it take to reach the ground? Don't consider aerial friction.

Your equation is only valid for uniform acceleration. Are you sure $ g$ will remain constant through the 1000 kilometers? :wink:

look we in general cannot use law of conservation of energy since 1000 km is a huge distance it is comparable to the earth's radius so we use it in a different form as
Gravitational potential energy(replacement for mgh)=KE gained....
we then are able to find its velocity then neglecting air drag....
we use simple equation ... time=distance/speed

[quote="Anvay"]time=distance/speed[/quote]

Isn't that valid only when $ v$ is constant?

You found the $ KE$, all right but that is the $ KE_{final}$, not the KE throughout the fall. This problem is not easy, and it involves a bit of calculus work.

[quote="Anvay"]SHW UR CALCULUS WRK[/quote]

I didn't get the answer. But I will dump here my rough work which might help someone get a lead.

$ \frac{d}{dt}(KE\plus{}PE) \equal{} 0
\\
\\
\\
\equal{}\frac{d}{dt}KE\plus{}\frac{d}{dt}PE
\\
\\
mv_tg_h \plus{} \frac{GMm}{R\plus{}h}v_t\equal{}0
\\
\\
m(R\plus{}h)\frac{d^2h}{dt^2} \equal{} (GMm)dt$

i dont understand why we need to use calculus.the velocity increases by g metres per second for every second .g is a constant .it is the velocity that is increasing by g.not acceleration

[quote="aadil"]the velocity increases by g metres per second for every second .g is a constant .it is the velocity that is increasing by g.not acceleration[/quote]

The $ g$ that you are talking about is with $ h \equal{} 0$. $ g$ is a function of $ h$, $ R$ being a constant.

$ g_h \equal{} \frac {GM_E}{(R \plus{} h)^2} \\
 \\
 \\
\frac {d}{dh}g \equal{} \frac { \minus{} GM_E}{R \plus{} h} \neq0$

Which means accleration due to gravity is [b]not[/b] constant throughout and decreases with increase in height (see that minus sign there?) and that the velocity does [b]not[/b] change by $ g\text{ m/s}^2$.

[quote="Immanuel Bonfils"]About 510 seconds[/quote]

If we [b]consider[/b] gravity to be uniform throughout the fall,
$ 1000000m \equal{} \frac {1}{2}gt^2 \\
 \\
t \equal{} 447.213595s.$

And our answer should be > the time we found above. How did you get it?

so considering earths gravitational field the value of g varies

[quote="Conjurer"]510 > 447  :wink:[/quote]

Oh! Thanks.

[quote="Immanuel Bonfils"]Obviously, the idea is to consider the variation of acceleration...[/quote]

Could you please show your work? I'm stuck!

Where I'm stuck:

$ \text{Note: }M_E \equal{} \text{Mass of Earth, } H \equal{} \text{1000km, }h \equal{} \text{height at any instant, }g_h \equal{} \text{acceleration due to gravity at a height }h \\
 \\
PE_i \equal{} KE_f \\
 \\
\therefore \frac {1}{2}mv_f^2 \equal{} \frac {GM_E}{R \plus{} H} \\
 \\
v_f \equal{} \sqrt {\frac {2GM_E}{R \plus{} H}} .....(1) \\
 \\
 \\
 \\
g_h \equal{} \frac {GM_E}{(R \plus{} h)^2} \\
 \\
 \\
\frac {dv_h}{dt} \equal{} \frac {GM_E}{(R \plus{} h)^2} \\
 \\
 \\
 \\
\int_{0}^{v_f}dv_h \equal{} \int_{0}^{t}\frac {GM}{(R \plus{} h)^2}dt$

Let r be the distance to earth center, $ M = 5.9742\; .{10}^{24} kg \; and \; R = 6378\; km$ mass and earth radius and $ \;G = 6.674.{10}^{ - 11} {{m^3}\over {kg.s^2}}$ the gravitational constant.

From COME (apud Rajiv), at distance r and instant t(r) ,

$ v^2 = 2GM.\;\; {{R + h - r}\over{r.(R + h)}} \;\;$, where h=1000 km is the total falling height.

So we get to the DE
$ {{dr}\over{dt}} = - A \sqrt {{{R + h - r}\over r}}\;\;$ where $ A = \sqrt {{{2GM}\over{R + h}}}$, which can be solved as
\[ t = {1 \over A}\left [\left({\pi\over 2} - \arctan {{\sqrt {{R\over h}}}\right) .(R + h) + \sqrt {Rh}\right]
}\]
.

How does it match for small h?

So you considered variation of displacement with time rather than velocity with time. Good trick! Thanks.

[quote="castigioni"]I found t=1217 seconds[/quote]
Could you, please, show your work?


Wait, until they make twice in a row, or they just shoot twice?

And do we want the expected #s?

doesn't specify, but i don't think it matters.

[quote="dingzhou"]Alex and Brian take turns shoting free throws until they each shoot twice. Alex and Brian have 80% and 60% of making their free throws, respectively. What is the probability that after each free throw they take, Alex hs made at least as many as Brian if Brian shoots first.[/quote]

[hide]
Brian must miss his first shot, or else Alex has less than him after that free throw.  Chance: 40%
Now we have two paths:
Alex makes his next shot.
80%
It doesn't matter what Brian does, then it doesn't matter what Alex does.
100%

Alex misses his next shot
20%
Now Brian has to miss.
40%
Now it doesn't matter what Alex does.
100%

So it's $ \frac{2}{5}(\frac{4}{5}\plus{}\frac{2}{25})\equal{}\frac{44}{125}$

[/hide]

I do not think your answer is correct. I think the answer is 508/625

The probability that Brian makes more free throws than Alex at any point is the probability that Brian
makes the first and Alex misses the first plus the probability that they made the same number of free
throws after each taking one and then Brian makes the second and Alex misses the second.

So i guess (1/5)*(3/5) + (1/5)*3/5* (3/5*4/5 + 1/5*2/5) = 117/625

So the answer is : 1- 117/625= 508/625

I am pretty sure this is correct

Umm, it's Alex has made at least as many as Brian, not the other way around, unless I'm being stupid.

You are not being stupid but I don`t think you read the last part.

[i]the probability that Alex has made at least as many free throws as Brian after each one taken is
1 − 117/625 = 508/625[/i]. If you subtract 1 form it you are finding the probability that Alex has made at least as many free throws as Brian

dingzhou do you know what the right answer is? Is this correct...

[quote="dingzhou"]Alex and Brian take turns shoting free throws until they each shoot twice. Alex and Brian have 80% and 60% of making their free throws, respectively. What is the probability that after each free throw they take, Alex hs made at least as many as Brian if Brian shoots first.[/quote]

[hide] I got 196/625

 so the chance that Alex made it is 8/10, that Brian made it is 6/10. So the chance they both made it was 48/100. Then the chance that Alex didn't make it was 2/10 and the chance that Brian didn't make it was 4/10. The chance they both didn't make it was 8/100. So add it together, and 56/100. It is exactly the same with the next shot, so multiply the two days and you get 196/625.[/hide]

Also, I think your solution is wrong.

So i guess (1/5)*(3/5) + (1/5)*3/5* (3/5*4/5 + 1/5*2/5) = 117/625
That doesn't equal 117/625. It equals 107/625.


Let n be a positive integer and $ a_1,a_2,..,a_n$ are integer.Let $ f: Z\rightarrow R$ such that:
$ \sum_{i\equal{}1}^{n} f(a_il\plus{}k)\equal{}0$ for all $ k,l$  integer and $ l\not\equal{} 0$

I need a solution elementery.
this problem is IMC 2007 but I don't like offical solution.

can anyone post ideas problem and solution?


calculus
derivative
function
integration
probability
conditional probability
probability and stats

[quote="mathlova"]I will try to write this up to get some feedback, verifying that I understood it correct.

If f is a nonnegative function on a measure space $ (\Omega, \mathcal{F}, \mu)$, then $ \nu(A) \equal{} \int_Af d\mu$ defines another measure on $ \mathcal{F}$. $ \nu$ has density f with respect to $ \mu$.
Theorem 1.
If $ \mu$ and $ \nu$ are the $ \sigma$=finite measures such that $ \nu < < \mu$, then there exists a nonnegative f, a density, such that $ \nu(A) \equal{} \int_A f du$ for all $ A \in \mathcal{F}$.
For two such densities f and g, $ \mu[f \neq g] \equal{} 0 \blacksquare$
and
$ \int_A h dr \equal{} \int_A h f du$
$ \int_A h dr \equal{} \int_A h \frac {dr}{du}du$
where $ \frac {dr}{du}$ is often denoted the Radon-Nikym derivative of r with respect to u.

I understand that the Radon-Nikodym derivative is the equivalent to a derivative in calculus in the sense that it describes the rate of change of density of one measure with respect to another.

It can be used on discrete sets and measures.

This theory is used in conditional probability, martingales and stochastic processes.
This includes change of numeraie and change of measure for a stochastic process.[/quote]

Your understanding is right.I have just learned this theorem.I gave a lecture to my classmates about it this Tuesday .

[quote="mathlova"]Is there anything else to add?

What would be some good examples of the theorem in use?[/quote]

I have not encounterd axamples.what do you major in?probability theory?


If a quadrilateral is formed by the equation xy-3x-4y=-12 and the x-axis and y-axis, then what is the area of the quadrilateral?


ok, this works in the AoPS TeXer, but it won't work in TeXnicCenter...
[code] 2+\cfrac{1}{3+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{2+\cdots}}}}=x [/code]
it comes out as 2+13+12+13+12+...=x
help?

You need to add \usepackage{amsmath} in your preamble

$\cfrac{a}{b+\cfrac{c}{d+\cfrac{e}{f+\cdots}}}$

$\frac{a}{b+\frac{c}{d+\frac{e}{f+\cdots}}}$

So all the c part does is make it bigger?

\cfrac has a couple of extra options [l] or [r] which left- or right- justify the numerators
\[2+\cfrac[l]{1}{3+\cfrac[r]{1}{2+\cfrac[l]{1}{3+\cfrac[r]{1}{2+\cdots}}}}=x\]


The incorrect statement amongst the following is:

a) The order of reduction reaction by $ LiAlH_4$ and that of the reduction by $ NaBH_4$ each is two.

b) $ Al\minus{}H$ bond is more ionic than $ B\minus{}H$ bond and hence $ LiAlH_4$ can produce larger concentration of hydride ion that that of $ NaBH_4$.

c) $ LiAlH_4$ cannot reduce $ \minus{}NO_2$ group.

d) Inspite of very small rate constant of reduction of any carbonyl function other than aldehyde and ketones, the rate of reduction with $ LiAlH_4$ becomes appreciable due to large concentration of hydride ion.

No. (C) is correct, I have never seen using $ LiAlH_4$ to reduce $ \minus{}NO_2$

that's what even I'm telling . LiAlH4 can't reduce . NO2 grp.

Read the first paragraph given [url=http://en.wikipedia.org/wiki/Lithium_aluminium_hydride]here[/url] very carefully.[/url]

The incorrect statement if (C) of course, without looking at the link in the previous post. The nitrogen in -NO2 is strongly positively polarised, and hydride is a very good nucleophile/base. LiAlH4 redudes nitro groups all the way into amine groups.


abstract algebra
vector
group theory
superior algebra
superior algebra solved

Show that any group with at least 3 elements has a non-trivial
  automorphism.

Let $G$ be a group with at least three elements.

Case 1: $G$ is nonabelian: there exist $x,y$ such that $x^{-1}yx\neq y$. Let $\phi(x)=x^{-1}yx$.
Case 2: $G$ is abelian, and some element $y$ has order greater than 2. Let $\phi(x)=x^{-1}$.
Case 3: $G$ is abelian, and every element has order 1 or 2. Then $G$ is the additive group of a vector space over $\mathbb{Z}_2$. Find a basis (which has at least two elements) and exchange the first two elements. This generates an automorphism.

Actually, if $G$ is infinite and Case 3 applies, we need the Axiom of Choice, unless we already know a basis.

If the group is not abelian, then we can consider an inner automorphism obtained by conjugation with a non-central element, so we can restrict our attention to abelian groups.

After that, if there is at leaast one non-trivial element of order different from $2$, then $x\mapsto x^{-1}$ is a non-trivial automorphism.

Finally, if all the non-trivial elemetns have order $2$, then the group can be regarded as a vector space over $\mathbb Z_2$ (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=224391#p224391]this topic[/url]). Let $(a_i)_{i\in I}$ be a basis for this vector space. The subspace generated by $(a_i)_{i\in I\setminus\{i_0\}}$ for some $i_0$ is a subgroup $H$ of index $2$ in our group. Take an element $a\in H$, and define the automorphism to be $f(x)=x$ if $x\in H$ and $ax$ if $x\not\in H$ (it can easily be checked to be an isomorphism).

I remember posting this a long time ago, but I couldn't find it.

Edit: darn! too late :)


Find the unit vector having the same direction as v.

(1) v = - 3j

(2) v = 2i - j + k

In general for $ \mathbf{v}$, the unit vector $ \hat{\mathbf{v}} \equal{} \frac {\mathbf{v}}{\| \mathbf{v} \| } \equal{} \frac {\mathbf{v}}{\sqrt {\mathbf{v} \cdot \mathbf{v}} }$


In a certain group of people:
a) any two who are acquainted do not have a common acquaintance
b) any two who are not acquainted have exactly 2 common acquaintances.

Prove that everyone has the same number of acquaintances.

Here's a prb which is related to this one, and maybe it can help (I haven't solved the prb, MMM):

A set M contains n people s.t.:

a)every person knows exactly k other people from M;
b)every 2 persons who know each other have exactly l (small L) common acquaintances;
c)every 2 people who don't know each-other have exactly m common acquaintances.

Prove that m(n-k)-k(k-l)+k-m=0.

This was given at a Romanian selection test (4'th test in 1990).

Here's what I thought of:
Consider a person a. Then all the people a doesn't know can be put in bijection with the pairs of people he knows:

For every 2 people he knows it means that they do not know each-other, so they must have another common acquaintance besides a. It follows that a doesn't know this other common acquaintance. So to every pair of people a knows we associate this other person that knows that specific pair of people. This function is surjective because for any person a doesn't know there are 2 people which know both that person and a, and it's also injective because otherwise we would have 2 people who don't know each-other and have >2 common acquaintances. This means that the function is bijective, so the number of people a doesn't know=number of pairs of people a knows. Assume a knows k people and we have n people in total. Then k(k-1)/2= n-1-k, iff k^2+k+2-2n=0, which only has one positive soln, and that soln is the common number of acquaintances of all n people.


geometry
3D geometry
geometric sequence
algebra unsolved
algebra

The sum of the first few terms in a geometric progression is 11, the sum of their squares is 341, and sum of their cubes is 3641. Find the terms of the sequence

Let sequense is $ b,ba,...ba^{n\minus{}1}$, then
$ b\frac{a^n\minus{}1}{a\minus{}1}\equal{}11,b^2\frac{a^{2n}\minus{}1}{a^2\minus{}1}\equal{}341,b^3\frac{a^{3n}\minus{}1}{a^3\minus{}1}\equal{}3641$.
Therefore $ b\frac{a^n\plus{}1}{a\plus{}1}\equal{}31,b^2\frac{a^{2n}\plus{}a^n\plus{}1}{a^2\plus{}a\plus{}1}\equal{}331$.
$ 11(a^n\plus{}1)(a\minus{}1)\equal{}31(a^n\minus{}1)(a\plus{}1)$,
$ 331(a^{2n}\plus{}a^n\plus{}1)(a^2\minus{}1)\equal{}341(a^{2n}\minus{}1)(a^2\plus{}a\plus{}1)$
Solution is $ b\equal{}1,a\equal{}\minus{}2,n\equal{}5$.


hey.. i've been going to the gym lately and they sell those protein shakes there... what do they do? any side effects? my friends tell me that they help build muscle.. but im mainly trying to burn some stomach fat.. so you think those would help?

If you're just trying to burn fat, they won't really help. If you want to get some muscles, they do.

And there are no *known* side effects...


Let $A=\sum ab\sum\frac{1}{ka^{2}+bc}$.Prove that:
 a. A$\geq \frac{9}{k+1}when k\geq\sqrt{7}+3$
 b. A$\geq1+\frac{2}{k}when \frac{2}{3}\leq k\leq 3+\sqrt{7}$
 c.A$\geq 4 when0\leq k\leq \frac{2}{3}$

Is it very hard?
I saw
$\frac{1}{a^{2}+2bc}+\frac{1}{b^{2}+2ca}+\frac{1}{c^{2}+2ab}\geqslant\frac{2}{ab+bc+ca}$
which is the case $k=1/2$

I found out this problem in a topic of hungkhtn .I'll post my solution on tuesday(it's very complicated)

OK
I'll look at yours after I try to solve it

Today,I'll post my solution for case $k\geq 3+\sqrt{7}$
$\sum \frac{1}{ka^{2}+bc}\geq \frac{9}{\sum ab}$(1)
 I have:
$(1)\leftrightarrow k^{2}(k-8)\sum(ab)^{3}+(k^{2}+k+1)(K+1)abc\sum ab(a+b)+k(k+1)\sum(ac)^{2}(a^{2}+c^{2})+k(2k-7)abc\sum a^{3}\geq (9(k^{3}+1)-3(K+1))a^{2}b^{2}c^{2}$(2)
 note that:$2(k+1)\geq k(8-k)$ (3)
        and $k\geq 3+\sqrt{7}\geq 3,5(4)$
From (3)(4),it's easy to prove (2)
With the remains of two case,I use sos but very complicated


probability
AMC
AIME
geometry
3D geometry
algebra
polynomial

Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac{1}{3}$. The probability that Club Truncator will finish the season with more wins than losses is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[hide]Well, since ties basically make it into a lower number of game season we need to consider all of the seasons with less games possible. 5 ties/1 win= $\frac{6}{729}$. 4 ties/2 win= $\frac{15}{729}$. 3 ties/3 wins=$\frac{20}{729}$. 3 ties/ 2 wins/ 1 loss=$\frac{60}{729}$. 2 ties/ 4 wins = $\frac{15}{729}$.
2 ties/ 3 wins/ 1 loss = $\frac{60}{729}$. 1 tie/ 5 wins =$\frac{6}{729}$. 1 tie/ 4 wins/ 1 loss = $\frac{30}{729}$. 1 tie/3 wins/ 2 loss = $\frac{60}{729}$. 6 wins/ 0 losses= $\frac{1}{729}$. 5 wins/ 1 loss = $\frac{6}{729}$. 4 wins/ 2 loss =$\frac{15}{729}$ Adding these all up we get $\frac{98}{243}$ so our answer is 341.[/hide]
I'm sure there is a better way but I did this in about 8 minutes so...

[quote="mathgeniuse^ln(x)"]Yeah, I hope there is because if they had 7 matches or 8 matches, it would be much longer.[/quote]
Hey, I solved an AIME problem number 11 in 8 minutes, that's good :). However, I think I found a faster way... I'm not sure if it'll work but it might
[hide="an idea"]the expansion method of a trinomial.[/hide]

[hide]You just need to evaluate the sum \[\sum_{i=0}^{6}\sum_{j=0}^{i-1}\frac{6!}{i!j!(6-i-j)!}\] which is like \[\sum_{i=0}^{6}\binom{6}{i}\sum_{j=0}^{i-1}\binom{6-i}{j}\] if you cleverly rewrite it but it's faster just to compute the sum by hand since there's only 6 terms anyway[/hide]

[hide="hint"]i think you should consider a complement argument[/hide]

[hide="solution"]the probability of more wins than losses is the same as more losses than wins, so we need to find probability of same wins and losses, and subtract that from 1 and divide by 2

so (W,L,T)=(0,0,6),(1,1,4),...(3,3,0)

sum of those is $x=\frac{1}{3^{6}}(\frac{6!}{6!0!0!}+\frac{6!}{4!1!1!}+\frac{6!}{2!2!2!}+\frac{6!}{0!3!3!})=\frac{1+30+90+20}{3^{6}}=\frac{47}{243}$

$\frac{1-x}{2}=\frac{98}{243}\implies m+n=\boxed{341}$[/hide]

[hide="Another approach"]
Let losses be $ \minus{}1$, ties $ 0$, and wins $ 1$, and notice that any configuration of wins, losses, and ties must be greater than $ 0$ for it to count.

So we need the sum of the coefficients of $ x^6,\ldots,x$ in $ \left(\frac 1x\plus{}1\plus{}x\right)^6$.

Now you just expand carefully and get the cube of a larger polynomial, which you then square again and drop all terms with degree less than or equal to 0, and then sum the coefficients.

This ends up being $ 294$, and then $ \frac{294}{3^6}\equal{}\frac{98}{243}\implies m\plus{}n\equal{}\boxed{341}$.
[/hide]

Still pretty ugly, but it didn't take long.

Personally I thought this was very easy for a #11. I listed all possible (W,L,T)'s.

The complement solution does provide some insight though. It teaches us to look for useful symmetries.

We find the probability that Club Truncator has the same amounts of wins and losses. Then, we subtract it from 1 and divide by 2 to get our answer.
Case 1(0 ties): Club Truncator needs to win 3 times and lose 3 times. There are $\binom{6}{3}$ ways to choose the team that Club Truncator wins against. Thus, the probability is $\frac{\binom{6}{3}}{3^6}=\frac{20}{729}$. \\ Case 2(2 ties): Club Truncator needs to win twice and lose twice. There are $\binom{6}{2}$ ways to choose the teams in which Club Truncator ties with and $\binom{4}{2}$ ways to choose the team that Club Truncator wins against. Thus, the probability is $\frac{\binom{6}{2}\cdot\binom{4}{2}}{3^6}=\frac{90}{729}$. \\ Case 3(4 ties): Club Truncator needs to win once and lose once. There are $\binom{6}{4}$ ways to choose which teams Club Truncator ties with. Then, there are $2$ ways to choose the team Club Truncator wins against. The probability for this case is $\frac{\binom{6}{4}\cdot 2}{3^6}=\frac{30}{729}$. \\ Case 4(6 ties): Only one way so it's $\frac{1}{729}$. \\ Thus, the probability that Club Truncator wins is $\frac{20+90+30+1}{729}=\frac{141}{729}$. Thus, our desired probability is $\frac{\frac{729-141}{2}}{729}=\frac{98}{243}$. Hence, our answer is $98+243=341$.

We proceed by casework:
Case 1 (1 win): $6$
Case 2 (2 wins): $\binom{6}{2} \cdot (1+4)$
Case 3 (3 wins): $\binom{6}{3} \cdot (2^3-1)$
Case 4 (4 wins): $\binom{6}{4} \cdot 2^2$
Case 5 (5 wins): $6 \cdot 2$
Case 6 (6 wins): $1$
Adding them up, we get $294.$ Thus the probability is $\frac{294}{3^6} = \frac{98}{243}.$ Hence, our desired answer is $\boxed{341}.$

[b]Solution:[/b] We exploit symmetry. For now, let us find the probability that the number of wins will equal the number of losses.
Case 1: $WWWLLL$ can happen in $20$ ways.
Case 2: $WWLLTT$ can happen in $90$ ways.
Case 3: $WLTTTT$ can happen in $30$ ways.
Case 4: $TTTTTT$ can happen in $1$ way.
So, overall, the probability for getting the same number of wins and losses is $\frac{20+90+30+1}{3^6}=\frac{47}{243}$. Now using the symmetry we first mentioned, our answer is $\frac{1-\frac{47}{243}}{2} = \frac{98}{243} \Rightarrow \boxed{341}$.


function
rotation
analytic geometry
logarithms
ratio
complex numbers
topology

[quote]
Let $ G$ be a region and suppose that $ f: G \to \bf{C}$ is analytic such that $ f(G)$ is a subset of a circle. Show that $ f$ is constant
[/quote]

Is there an elementary way to show this (i.e. not using the open mapping theorem)

Assume that the circle is centered at $ 0$, and show that $ \frac{f'}{f}$ is pure imaginary. (can you figure out the rest?)

I will assume you know the fact that if $ f$ is analytic in a region $ G$ and $ \mid f(z) \mid$ is constant for all $ z \in G$ then $ f$ is constant.

Now $ G$ is a region which means that $ G$ is a connected open subset of the complex plane, and since $ f$ is analytic, hence continuous, $ f(G)$ is connected.

Therefore for some fixed $ w \in \mathbb{C}$ we have that $ f(G) \equal{} \{ w \plus{} re^{i\theta}: \theta_1 < \theta < \theta_2 \}$

So $ \mid f(z) \minus{} w \mid \equal{} g(\theta) \equal{} \mid re^{i\theta} \mid \equal{} r$ a constant. Hence $ f(z) \minus{} w$ is constant which means that $ f$ must be constant.

[quote="msaid"]I will assume you know the fact that if $ f$ is analytic in a region $ G$ and $ \mid f(z) \mid$ is constant for all $ z \in G$ then $ f$ is constant.[/quote]

?!?

Why on Earth would you solve a problem by assuming the result?

[quote="kouboy"]in other words, f is analytic function and |f| is constant... it follows from just the cauchy riemann equations that f itself is constant.[/quote]

how?

if $ f \equal{} u(x,y) \plus{} iv(x,y)$

then

$ |f|^2 \equal{} c^2 \equal{} u^2 \plus{} v^2$

and

$ \frac {\partial u}{\partial x} u \equal{} \minus{} \frac {\partial v}{\partial x} v \plus{} 0$

or

$ \frac {\partial v}{\partial y} u \equal{} \minus{} \frac {\partial v}{\partial x} v$

How to concude?

What you've written implies that $ \frac {f'}{f}$ is pure imaginary.  Say, for example, that we are looking near a point with positive $ x$- and $ y$-coordinates. (if not, we can rotate the picture easily)  Then $ f \equal{} u \plus{} i\sqrt {1 \minus{} u^2}$, $ f' \equal{} ig (u \plus{} i\sqrt {1 \minus{} u^2}) \equal{} \minus{} g\sqrt {1 \minus{} u^2} \plus{} igu \equal{} u_x \minus{} i \frac {u_x}{\sqrt {1 \minus{} u^2}}$, for some $ g$.

So $ u_x \equal{} \minus{} g\sqrt {1 \minus{} u^2}$ and $ gu \sqrt {1 \minus{} u^2} \equal{} \minus{} u_x$.  We conclude that $ u_x u \equal{} u_x$.  But $ u\neq 1$ (we assumed the $ y$-coordinate was positive) so $ u_x \equal{} 0$ in our neighborhood.  Similarly, $ u_y \equal{} 0$, so $ u$ (hence $ v$, hence $ f$) is locally constant.  Since $ G$ is connected this implies that $ f$ is constant.

It is probably nicer to solve for $ f$ more directly from $ \frac {f'}{f} \equal{} ig$ using a branch of the logarithm--purely real analytic functions are constant--but you asked how to do it with Cauchy-Riemann...

[quote="LydianRain"]What you've written implies that $ \frac {f'}{f}$ is pure imaginary.
[/quote]
I don't understand that. Here, already, you say that $ f/f' \equal{} ig$? Why is that? What is $ g$?

[quote]
 $ f \equal{} u \plus{} i\sqrt {1 \minus{} u^2}$, $ f' \equal{} ig (u \plus{} i\sqrt {1 \minus{} u^2}) \equal{} \minus{} g\sqrt {1 \minus{} u^2} \plus{} igu \equal{} u_x \minus{} i \frac {u_x}{\sqrt {1 \minus{} u^2}}$, for some $ g$.
[/quote]
Why does the last equality hold?

[quote]
-but you asked how to do it with Cauchy-Riemann...
[/quote]
Not necessarily. I was looking for the most elementary way (not open mapping theorem). Since Conway, which you are reading too, lists the exercise in a whole chapter of Mobius transformations, we maybe missed another approach?

[quote="Jacobson"][quote="LydianRain"]What you've written implies that $ \frac {f'}{f}$ is pure imaginary.
[/quote]
I don't understand that. Here, already, you say that $ f/f' \equal{} ig$? Why is that? What is $ g$?[/quote]
You wrote down an equation which states that the inner product of $ f$ with $ f'$ is zero.  Hence the angle between them is $ \frac{\pi}{2}$, so their ratio is pure imaginary.  $ g$ is some analytic function which takes only real values.

[quote="Jacobson"][quote]
 $ f \equal{} u \plus{} i\sqrt {1 \minus{} u^2}$, $ f' \equal{} ig (u \plus{} i\sqrt {1 \minus{} u^2}) \equal{} \minus{} g\sqrt {1 \minus{} u^2} \plus{} igu \equal{} u_x \minus{} i \frac {u_x}{\sqrt {1 \minus{} u^2}}$, for some $ g$.
[/quote]
Why does the last equality hold?[/quote]
$ f' \equal{} u_x \plus{} i v_x \equal{} u_x \plus{} i (\sqrt{1\minus{}u^2})_x$.

[quote="Jacobson"]Since Conway, which you are reading too, lists the exercise in a whole chapter of Mobius transformations, we maybe missed another approach?[/quote]
You are absolutely right.  Consider $ f$ in some neighborhood in which its image is not the whole circle.  Let $ m$ be a Möbius transformation that takes the circle in question to the real line, mapping a point not in the image of $ f$ to $ \infty$.  Then $ m \circ f$ is an analytic function that takes only real values, hence it is constant.  Since $ m$ is injective, $ f$ is constant as well.  This is a much better proof.

[quote="LydianRain"]  Consider $ f$ in some neighborhood in which its image is not the whole circle.  Let $ m$ be a Möbius transformation that takes the circle in question to the real line, mapping a point not in the image of $ f$ to $ \infty$.  Then $ m \circ f$ is an analytic function that takes only real values, hence it is constant.  Since $ m$ is injective, $ f$ is constant as well.  This is a much better proof.[/quote]

That's the one  :D

Thanks a lot.

When looking at your proof more closely, I came across the last problem :( sorry :)

so $ F \equal{} m\circ f$ only takes real values on a circle (say $ |z| \equal{} 1$)

$ F(z) \equal{} u(z) \plus{} iv(z)$

by maximum principle for harmonic functions

$ v(z) \equal{} 0$ on the unit disc $ B(0,1)$, so

$ F(z) \equal{} u(z)$ on the unit disc, and is real valued.

but open mapping theorem says that the image of an open set under a non constant function is open. But the unit disc is open and $ F(B(0,1)) \subset \mathbb{R}$, so $ F$ must be constant..

again I had to rely on a heavy theorem. Is there a more elementary way to show that $ F$ is constant, after all, the exercise is listed before the open mapping theorem

[quote="Jacobson"]Is there a more elementary way to show that $ F$ is constant, after all, the exercise is listed before the open mapping theorem[/quote]
I thought that Conway proved this in the text.  Anyway, it can be seen directly from the Cauchy-Riemann equations.  $ v(z)\equal{}0$, so $ \frac{\partial u}{\partial x} \equal{} \frac{\partial v}{\partial y}\equal{}0$, and similarly $ \frac{\partial u}{\partial y} \equal{} 0$, so $ u$, hence $ F$, is constant.


In $ \triangle{ABC},\hat{ABC}\equal{}60$ and $ \hat{ACB}\equal{}70$.Point $ D$ is on the line segment $ BC$ such that $ \hat{BAD}\equal{}20$.Prove that $ AB\plus{}BD \equal{} AD\plus{}DC$


Dissect a $ 4$ by $ 9$ square into two pieces which can be assembled to form a square.

[hide]let the lower-left corner be the origin, and the points on grid points.
draw a line from (3,0) to (3,2)
and from (3,2) to (6,2)
and from (6.2) to (6,4).
now shift the right piece down two and left three to form a square.[/hide]

[hide]Good. This problem is a 'proof by example' problem. Surely if this were a written proof, it would be easy to draw. [/hide]

See the rather fascinating [url=http://mathworld.wolfram.com/Wallace-Bolyai-GerwienTheorem.html]Wallace-Bolyai-Gerwien Theorem[/url].

If I remember correctly, the proof for convex polygons is doable. The main parts of a proof are showing that: any convex polygon can be triangluated. Then any triangle can be made into a rectangle. Any two rectangles can be made into a rectangle. And any rectangle can be made into a square.

So I didn't have to show how I got that?

If I did show how I got that, would that be extra credit?

[quote="1=2"]
If I did show how I got that, would that be extra credit?[/quote]

I doubt it because that would be equivalent to penalising those people who did the question correctly but did not give an explanation (which is not required)

All that is necessary is to prove that your answer is correct or whatever. You usually don't need to explain where everything came from. It depends where you are sending the solution. I feel like for formal contests like the usamo, one should remove how you found your answer. If you are discussing problems with others, it is nice to explain how you got your solution so others can learn better from what you did.


A mixture consisting only of KBr and NaBr has a mass of 0.4820 grams. After treating this sample with Ag+ (aq), all of the Br- in the original sample is recovered as 0.8251 g of pure AgBr. What is the percent by mass of NaBr in the original sample?

Molar mass AgBr =107.8+79.9=187.7g/mol

moles Br-=0.8251/187.7=0.004396 mol

let x and y be amount of K+ and Na+ in moles respectively.

then x+y=0.004396
and x * molar mass KBr + y * molar mass NaBr =.4820g

solve for x and y simultaneously


[quote="spix"]Prove that if $0\leq a_{1}<a_{2}<...<a_{s}\leq 1$ and $0\leq b_{1}<b_{2}<...<b_{s}\leq 1$, where $s,t\in N$ and
$[x+a_{1}]+[x+a_{2}]+...+[x+a_{s}]=[x+b_{1}]+[x+b_{2}]+...+[x+b_{t}],\forall x \in R$ then $s=t$ and $(a_{1},a_{2},...,a_{s})=(b_{1},b_{2},...,b_{t})$.[/quote]
I'm sure you mean $s, t \in \mathbb{N}^{+}$ and $0 \le b_{1}< b_{2}< \ldots < b_{t}\le 1$, don't you? Moreover, I guess that $[\,\cdot\,]$ is here to indicate the floor function. Is it right?

You are right, sorry for the typo's, it's correct now.


I'll be at National MATHCOUNTS this year - please come introduce yourself if you are there!  I'll be around the ESPN crew a fair amount, and hanging out with AoPSers the rest of the time.

Any chance you can do the interesting problems this time rather than the "add these numbers" problems? I suppose not. See you (and everybody else) there.

We'll see!  I can't even remember how the problems I did last time were chosen.  I think I got to choose them the first time around, but the second time around I don't remember being as involved.

I do hope the problems are more interesting and less reading than they were at State this year...

Also, if any of you have ideas for little math tidbits to use in the broadcast, post them here.

Yes SIR Mr. Rusczyk SIR!

My mother was so afraid that I would say that to Mr. Roker instead of Yes SIR Mr. Roker SIR! If I'm in the countdown round, not likely, I'll be saying that and having fun!

Billy

Penguin! is flying up there tomorrow.
I doubt I'll make it in the countdown round, but if I do I'll be the first penguin to do so.

I think that you should say how good the seventh graders did this year, with Neal winnning and Nathan gettiing 2nd written.

[quote="09husbri"]Mr. Rusczyk - dunno if you remember me, we talked a [i]bit[/i] during the GM trip on the bus. I'm the tall kid from Ohio. It was great to meet you![/quote]
I agree. It was great to meet you.  I came close to the countdown but lost the tiebreaker. (I did really well on the target round.) But I had fun.

Mr. Rusczyk, do you remember me from last year? I was the white guy with the goatee on the CA team. :lol:

[quote="jb05"]You aren't the first penguin to [b]get[/b] there? Hooray! I'm flying up at 9:30 and landing at 4:00 (or something like that). That's 6 and a half hours of undisturbed math![/quote]
I meant in the CDR
Oh well, I'm the first penguin! at nationals.

Gaah! You're that psychopathic penguin kid!

Are you an eighth grader? 'Cuz if not...

Yes, I, um...[i]noticed[/i] Mr. Batman Villain over there in the yearbook. Hint: It was in his hobbies.

I saw you there at nationals, Mr. Rusczyk, but I didn't know who you were and I wasn't in Art of Problem Solving then.  I am the tallish boy with the long pony tail, bow tie, vest, and jacket who commented on your Polish name.  Do you remember me?

You are Miles from GA aren't you?  You were talking to Mr Fish and David and Nathan and me about how your parents taught math at Honey Creek (where Mr Fish teaches math) and how you were born in Terre Haute and that Mr Fish wanted to talk to your parents

That's exactly who I am—except you must have mis-heard me.  My parents taught at Indiana State University, not Honey Creek.
In that case, who would you be?

One of the people standing by Mr Fischer, not David or Nathan


Let $\ a_{i}\geq 0$ such that $\sum_{i=1}^{n}a_{i}=1$
 Prove that $\sum_{i=1}^{n}\frac{a_{i}}{2-a_{i}}\geq\frac{n}{2n-1}$
         good luck  :D


calculus
integration
trigonometry
limit
calculus computations

Few students will be able to solve this completely out of our expectation.

it's equal to :
$ \frac{(2k\minus{}1)!!}{(2k)!!}\frac{\pi}{2}$ if $ n\equal{}2k$
and equal to
$ \frac{(2k\minus{}1)!!}{(2k)!!}$
if $ n\equal{}2k\plus{}1$ which calculated by standart method.

after is simpl...

[quote="kunny"]Hum, we don't need to find $ x_n$ directly.[/quote]
Yes I agree with you :)

This problem seems to be easy for Experinced solver in this forum, but few students could solve the problem completely.

[quote="kunny"]Let $ x_n \equal{} \int_0^{\frac {\pi}{2}} \sin ^ n \theta \ d\theta \ (n \equal{} 0,\ 1,\ 2,\ \cdots)$.

(1) Show that $ x_n \equal{} \frac {n \minus{} 1}{n}x_{n \minus{} 2}$.

(2) Find the value of $ nx_nx_{n \minus{} 1}$.

(3) Show that a sequence $ \{x_n\}$ is monotone decreasing.

(4) Find $ \lim_{n\to\infty} nx_n^2$.[/quote]
(1) using reduction formula on $ \int\sin ^ n \theta \, d\theta \equal{}\minus{}\int\sin ^ {n\minus{}1} \theta \cdot(\minus{}\sin\theta)\, d\theta$ will give us
$ \int\sin ^ n \theta \, d\theta \equal{}\minus{}\frac{1}{n}\sin ^ {n\minus{}1} \theta\cos\theta\plus{}\frac{n\minus{}1}{n}\int\sin ^ {n\minus{}2} \theta\, d\theta$

putting the limits then gives us
$ x_n\equal{}\frac{n\minus{}1}{n}\,x_{n\minus{}2}$

(2) when $ n\;(\equal{}2k)$, we get
$ x_{2k}\equal{}\frac{\pi}{2}\;\;\frac{1\cdot 3\cdot 5\ldots (2k\minus{}1)}{2\cdot 4\cdot 6\ldots (2k)}$

and when $ n\;(\equal{}2k\minus{}1)$, we get
$ x_{2k\minus{}1}\equal{}\frac{2\cdot 4\cdot 6\ldots (2k\minus{}2)}{3\cdot 5\ldots 7\ldots (2k\minus{}1)}$

now, $ nx_nx_{n \minus{} 1}$ involves the product of an even and an odd term either way if we choose $ n$ as even or odd, and so using even $ n$ here, we get
$ 2k\;x_{2k}\;x_{2k \minus{} 1}\equal{}2k\;\frac{\pi}{2}\;\;\frac{1\cdot 3\cdot 5\ldots (2k\minus{}1)}{2\cdot 4\cdot 6\ldots (2k\minus{}2)\cdot (2k)}\;\;\frac{2\cdot 4\cdot 6\ldots (2k\minus{}2)}{3\cdot 5\ldots 7\ldots (2k\minus{}1)}\equal{}\frac{\pi}{2}$

(3) a monotone decreasing sequence requires us to show $ \frac{x_n}{x_{n\minus{}1}}\leq 1$.
we can pick either of $ \frac{x_{2k}}{x_{2k\minus{}1}}$ or $ \frac{x_{2k\plus{}1}}{x_{2k}}$ to demonstrate it, so i will pick the first one.

$ \frac{x_{2k}}{x_{2k\minus{}1}}\equal{}\frac{\pi}{2}\;\frac{1}{2k}\;\left(\dfrac{3\cdot 5\cdot 7\ldots (2k\minus{}1)}{2\cdot 4\cdot 6\ldots (2k\minus{}2)}\right)^2$
$ \equal{}\frac{\pi}{4k}\;\left(\dfrac{(2k\minus{}1)!}{2^{2k\minus{}2}\cdot((k\minus{}1)!)^2}\right)^2$

now $ \left\{\frac{x_{2k}}{x_{2k\minus{}1}}\right\}$ is a sequence that is bounded (below) by $ \frac{\pi}{4}\;(\approx 0.7853)$ and gradually increases to $ 1$ as $ {k\to\infty}$, since $ \lim_{k\to\infty} \frac{x_{2k}}{x_{2k\minus{}1}}\equal{}\pi\;\left(\dfrac{1}{\sqrt\pi}\right)^2\equal{}1$

this therefore shows monotone decreasing.

note: $ \lim_{k\to\infty} \left(\frac{\sqrt k\cdot(2k)!}{2^{2k}\cdot(k!)^2}\right)\equal{}\frac{1}{\sqrt\pi}$ has been shown several times before on the forum (by myself and others), so i am not going to repeat it

(4) using the expression for $ n$ even [or $ n$ odd] (shown in $ (2)$) will again work equally well here, so i will pick the even case.
$ \lim_{n\to\infty} nx_n^2\equal{}\lim_{k\to\infty} 2k\cdot x_{2k}^2\equal{}\lim_{k\to\infty} 2k\;\left(\frac{\pi}{2}\;\;\frac{1\cdot 3\cdot 5\ldots (2k\minus{}1)}{2\cdot 4\cdot 6\ldots (2k)}\right)^2$
$ \equal{}\frac{\pi^2}{2}\;\lim_{k\to\infty} k\cdot\left(\frac{1\cdot 3\cdot 5\ldots (2k\minus{}1)}{2\cdot 4\cdot 6\ldots (2k)}\right)^2$
$ \equal{}\frac{\pi^2}{2}\;\lim_{k\to\infty} \left(\frac{\sqrt k\cdot(2k)!}{2^{2k}\cdot(k!)^2}\right)^2$
$ \equal{}\frac{\pi^2}{2}\;\left(\frac{1}{\sqrt\pi}\right)^2$
$ \equal{}\frac{\pi}{2}$

Hum, your solutions seem to be very special,guys. Can anyone solve this problem naturally or more simply by following the given questions?

[quote="bookworm_vn"]Nothing special...  :idea:[/quote]

How do you solve this problem,bookworm_vn? :D

I think [b]kunny[/b] refers to sth like, e.g., for (2) using (1)

$ (n\plus{}1)x_{n\plus{}1}x_n\equal{}(n\plus{}1)\frac{n}{n\plus{}1}x_{n\minus{}1}x_n\equal{}nx_nx_{n\minus{}1}$, i.e. is a constant sequence, hence $ nx_nx_{n\minus{}1}\equal{}x_1x_0\equal{}\frac{\pi}{2}$

or as for (4) using (1,2,3)

$ \frac{\pi}{2}\equal{}\lim_{n\rightarrow\infty}\frac{n^2x_nx_{n\minus{}1}}{n\plus{}1}\equal{}\lim_{n\rightarrow\infty}nx_{n\plus{}1}x_n\leq \lim_{n\rightarrow\infty}nx_n^2\leq\lim_{n\rightarrow\infty}nx_nx_{n\minus{}1}\equal{}\frac{\pi}{2}$

P.S. (3) is a no-brainer due to $ \sin^n\theta\geq\sin^{n\plus{}1}\theta$ on the interval $ \left[0,\frac{\pi}{2}\right]$

or just subtracting gives $ x_{n\minus{}1}\minus{}x_{n}\equal{}\int_0^{\frac{\pi}{2}} \sin ^ {n\minus{}1} \theta (1\minus{}\sin \theta )\ d\theta >0.$


Is there any way to simplify:

$ \displaystyle \sum^{n}_{i = 1}x_{i}^{2}$

Into something more easily computable if I know what ${ \displaystyle \sum^{n}_{i = 1}x_{i}}$ is?

Thanks in Advance

What restrictions exist on the $ x_i$ term?

For example, how would you differentiate when $ x_1 \equal{} 1$ and $ x_2 \equal{} \minus{}1$ versus having $ x_1 \equal{} 2009$ and $ x_2 \equal{} \minus{}2009$?

Well I'm just trying to see if there is an easier way to estimate standard deviation. I guess the only restriction which exists on $ x_{i}$ is that it is positive.

[quote="RobRoobiks"]$ \sum^{n}_{i \equal{} 1}x_{i}^{2}*n\ge\sum^{n}_{i \equal{} 1}x_{i}$
is the only thing that popped up in my mind  :roll: cauchy-schwartz-bunawinksy or something like that[/quote]

Actually, it would be

$ \sum^{n}_{i \equal{} 1}x_{i}^{2}*n\ge (\sum^{n}_{i \equal{} 1}x_{i})^2$


trigonometry
inequalities
geometry
rectangle
geometry unsolved

Let ABCD be a convex quadrilateral. Prove that if exists a point P inside ABCD such that $\angle PAB=\angle PBC=\angle PCD=\angle PDA=\frac{\pi}{4}$ then ABCD is a square.

See the diagram for additional notation.

By Sine Law:

\begin{eqnarray*}{a\over\sin\beta}&=& b\sqrt{2}\\{b\over\sin\gamma}&=& c\sqrt{2}\\{c\over\sin\delta}&=& d\sqrt{2}\\{d\over\sin\alpha}&=& a\sqrt{2}\end{eqnarray*}

Multiplying those up, we get

$\sin\alpha\sin\beta\sin\gamma\sin\delta={1\over 4}$

Since $\alpha+\beta+\gamma+\delta=\pi$, all of those angles are between $0$ and $\pi$, hence their sines are positive and $f(x)=\sin x$ is concave on the interval. That gives

\begin{eqnarray*}\sin\alpha\sin\beta\sin\gamma\sin\delta &\leqslant& \left({1\over 4}\sin\alpha+{1\over 4}\sin\beta+{1\over 4}\sin\gamma+{1\over 4}\sin\delta\right)^{4}\quad\text{[AM-GM]}\\ &\leqslant& \sin^{4}{\alpha+\beta+\gamma+\delta\over 4}\quad\text{[Jensen]}\\ &=&{1\over 4}\end{eqnarray*}

Hence the equality holds iff $\alpha=\beta=\gamma=\delta={\pi\over 4}$, which gives us a rectangle, but then the first four equations yield $a=b=c=d$, which gives us a square ($AB=BC=CD=DA=a\sqrt{2}$).


hi,
Prove that for every odd $ n$ the set
\[ \{\binom{n}{1}, \binom{n}{2}, ..., \binom{n}{\frac{n\minus{}1}{2}}\}\]
contains an odd number of odd numbers.

$ (1\plus{}1)^n\equal{}1\plus{}(\binom{n}{1}\plus{} \binom{n}{2}\plus{} ...\plus{} \binom{n}{\frac{n\minus{}1}{2}})\plus{}(\binom{n}{\frac{n\minus{}1}{2}}\plus{} ...\plus{}\binom{n}{2}\plus{}\binom{n}{1})\plus{}1\equal{}2^n$, so $ (\binom{n}{1}\plus{} \binom{n}{2}\plus{} ...\plus{} \binom{n}{\frac{n\minus{}1}{2}})\equal{}2^{n\minus{}1}\minus{}1$ which is odd , so done.


A coin is flipped 10 times in a row. Given that exactly 7 heads appeared, find the probability that no two consecutive flips were both tails.

There are $ \dbinom{10}7\equal{}120$ ways to get 7 heads and 3 tails.
Then, say we have
" H H H H H H H "
Then there are 8 spots that we can have the tails, and in each spot, we are limited to one tail. So this is like choosing 3 spots, which has $ \dbinom83\equal{}56$ ways.
Our final probability is $ \boxed{\frac7{15}}$.

Another way to approach this problem is by finding the number of ways you can have at least 2 tails together.
So, there are 7 heads and 10 spots:
HHHHHHH
Therefore, there are 8 places to put the first tail, and 8 places to put the other two tails (because there have to be at least 2 tails together, we just count all the ways in which 1 tail can go and the other two tails can go; additionally, all 3 heads can be together, so there aren't just 8*7 ways).

There are C(10,7) ways to choose the places where the heads go, so our answer is:
1-(64/120)=56/120=[u]7/15[/u]


Prove that

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\geq\frac{\sqrt{25rR\minus{}2r^2}}{4rR}$

[quote="Ligouras"]Prove that

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\geq\frac {\sqrt {25rR \minus{} 2r^2}}{4rR}$[/quote]

$ \Longleftrightarrow  \frac{s^2\plus{}4Rr\plus{}r^2}{4Rrs}\geq\frac {\sqrt {25rR \minus{} 2r^2}}{4rR}$

$ \Longleftrightarrow  \frac{s^2\plus{}4Rr\plus{}r^2}{s}\geq \sqrt {25rR \minus{} 2r^2}$

$ \Longleftrightarrow  s^4\minus{}17rRs^2\plus{}4r^2s^2\plus{}16r^2R^2\plus{}8r^3R\plus{}r^4\geq 0$

$ s^4\minus{}17rRs^2\plus{}4r^2s^2\plus{}16r^2R^2\plus{}8r^3R\plus{}r^4\equal{}(s^4\minus{}20Rrs^2\plus{}r^2s^2\plus{}4r^2(4R\plus{}r)^2)$ $ \plus{}3r(R\plus{}r)(s^2\minus{}16Rr\plus{}5r^2)\plus{}9r^3(R\minus{}2r)\geq 0$


Galois Theory
superior algebra
superior algebra unsolved

Hilbert Theorem 94: Let $ L/K$ be an unramified cyclic extension of prime degree $ p > 2$, then there is an element $ \mathfrak{a}$ of the class group of $ K$ of order $ p$ which becomes trivial in the class group of $ L$.

Hilbert Theorem 92: $ \#\hat{H}^{ - 1}(G,U_K) \neq \{1\}$ for cyclic extensions $ K/k$ of prime order of number fields with Galois group $ G$. Here $ U_K$ is the unit group of $ K$ and $ \#\hat{H}^{-1}(G,U_K) = \ker(N_{K/k}: U_K \to U_k)/U_K^{1-G}$ is Tate cohomology.


i can't figure this out...maybe i didn't learn it yet..

$H_{3}O^{+}$ :::::: the hybrid orbital for this should be $sp^{3}$?

If so, I tried doing the promoting thing with the up and down arrows and can't seem to figure it out.

Maybe it has to do with antibonding (which I haven't learned yet) because $H_{3}O^{+}$ is the same as $H^+$

First, for its Lewis Structure, I got $O$ as the center atom with three $H$'s around it and a pair of electrons.  The total electron count is eight.

Now I try doing the arrow thing for $O$ since it's the center atom.

$O$ is

$2s^2 : \upharpoonleft\downharpoonright$
${2p^4 : \upharpoonleft\downharpoonright}$  ${\upharpoonleft\ \upharpoonleft}$

Then when I do the hybrid orbitals of $sp^3$, I need to promote one of the paired electrons in the $2p$ orbital to somewhere (I was guessing d-orbital, but oxygen is only on the n=2 level and doesn't have a d-orbital) to get 3 sigma bonds and an electron pair to form the $sp^3$ hybrids...

H3O has a +1 charge, so there is one less electron in the entire thing.

yup i already factored that in when i calculated 8 electrons total...

1 : H
1 : H
1 : H
6 : O

equals 9

minus 1

equals 8

Ooh... nice $\LaTeX$.

Oxygen on its own has the valence electron configuration

$\underbrace{\underline{\upharpoonleft\downharpoonright}}_{2s}\underbrace{\underline{\upharpoonleft\downharpoonright}\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }}_{2p}$

Since $\text{O}^+$ has a positive charge, $\text{O}^+$ has the following configuration:  $\underbrace{\underline{\upharpoonleft\downharpoonright}}_{2s}\underbrace{\underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }}_{2p}$

We now hybridize into $sp^3$:

$\underbrace{\underline{\upharpoonleft\downharpoonright}\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }}_{2sp^3}$

You would be perfectly justified moving the paired electron from the $2s$ orbital into one of the $2p$ orbitals, but that is unnecessary in this case (try it; the result is exactly the same except that the order of the orbitals has changed, but this doesn't matter since all the hybridized orbitals are equivalent).

We're really done at this point; I'll include, for completion, the electrons in the oxygen atom's hybridized orbitals during bonding.

$\underbrace{\underline{\upharpoonleft^\text{O}\downharpoonright^\text{O}}\ \underline{\upharpoonleft^\text{O}\downharpoonright^{\text{H}_1}}\ \underline{\upharpoonleft^\text{O}\downharpoonright^{\text{H}_2}}\ \underline{\upharpoonleft^\text{O}\downharpoonright^{\text{H}_3}}}_{2sp^3}$

The superscripts on each electron show, in a vague sort of way, where each of the electrons traces its origin, but remember that the electrons shift around haphazardly, and might find themselves in any orbital.  In other words: don't take that last diagram too literally.

Hope that helps.

ohh i get it...thanks a lot!!!

i didn't know to include the $^+$ for the $O$ when doing the hybrid orbitals!!

ohh and nice $LaTeX$

Oh...here's another one i don't understand...

$SO_{4}^{-2}$

I drew the Lewis Structure with best formal charges...

and I got $S$ in the middle with four $O$'s around it...

two $O$'s with double bonds and two $O$'s with single bonds...

and there are two extra electrons so, 5 x 6 = 30...then 30 +2 = 32 total electrons

I got $sp^3$ hybrid orbitals, but I don't know where to put the two electrons that pi bond..

So for $S$...I start with...

$\underbrace{\underline{\upharpoonleft\downharpoonright}}_{3s}\underbrace{\underline{\upharpoonleft\downharpoonright}\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }}_{3p}$

then promoting one of the paired electrons in the s orbital, and one of the paired electrons in the p orbital, they go to the d orbital..since $S$ is in the n=3 energy level, it can do that...

so I now have...

$\underbrace{\underline{\upharpoonleft\ }}_{3s}\underbrace{\underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }}_{3p}\underbrace{\underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }\ \underline{\ \ }\ \underline{\ \ }\ \underline{\ \ }}_{3d}$

Here's where I'm unsure....

Can the d-orbital account for the two pi bonds when I do the hybrids for $S$?

$\underbrace{\underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }}_{sp^{3}}\underbrace{\underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }\ \underline{\ \ }\ \underline{\ \ }\ \underline{\ \ }}_{3d???}$

Disclaimer: I'm not completely sure I know what I'm doing.

I believe the hybridization is $spd^2$

We initially have the electron configuration $\underbrace{\underline{\upharpoonleft\downharpoonright}}_{3s}\underbrace{\underline{\upharpoonleft\downharpoonright}\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }}_{3p}\underbrace{\underline{\ \ }\ \underline{\ \ }\ \underline{\ \ }\ \underline{\ \ }\ \underline{\ \ }}_{3d}$

We hybridize together the $s$ orbital, a $p$ orbital, and two $d$ orbitals (promoting as appropriate):

$\underbrace{\underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }}_{3spd^2}\underbrace{\underline{\upharpoonleft\ }\ \underline{\upharpoonleft\ }}_{3p}$

(the unused $d$ orbitals are omitted; there are three of them).  Now we can show the bonds:

$\underbrace{\underline{\upharpoonleft^\text{S}\downharpoonright^{\sigma\text{O}_1}}\ \underline{\upharpoonleft^\text{S}\downharpoonright^{\sigma\text{O}_2}}\ \underline{\upharpoonleft^\text{S}\downharpoonright^{\sigma\text{O}_3}}\ \underline{\upharpoonleft^\text{S}\downharpoonright^{\sigma\text{O}_4}}}_{3spd^2}\underbrace{\underline{\upharpoonleft^\text{S}\downharpoonright^{\pi\text{O}_1}}\ \underline{\upharpoonleft^\text{S}\downharpoonright^{\pi\text{O}_2}}}_{3p}$

(The same disclaimer applies to the above diagram as in my last post).

What is the motivation for this unusual hybridization?  Well, $\pi$ bonds must always occur in unhybridized $p$ orbitals (to the best of my knowledge; "always" is a tricky word).  So we must leave two $p$ orbitals unhybridized.

Hope that helps.

It's an interesting thought I haven't been introduced to yet...thanks A LOT!!!

I'll keep this in mind!!

OK, so as I feared... $\text{The above is FALSE!}$

What actually happens is this:

The hybridization is $sp^3$.  The $\pi$ bonds are not actual $\pi$ bonds, but they exhibit $\pi$ bonding, meaning they are $\pi$ bond-like in that they are not along the central axis between the two nuclei (that would make them $\sigma$ bonds).  The electrons responsible for the $\pi$ bonding are found in unhybridized $d$ orbitals.  Basically, electrons get promoted to the $d$ orbitals for $\pi$ bonding.

Hope that helps.

oh wow....yea i told my ap chem teacher bout the $spd^2$ hybrid and he had no clue...

but i also got the $sp^3$ hybrid and i just didn't know if the electrons responsible for the $\pi$ bond came from the $d\ orbital$

thanks a lot for the correction!!


calculus
derivative
induction
calculus computations

let f(x)=ln[(1+x)/(1-x)]. find a general formula for [f^(n)] (x) the n-th derivative of f(x) -and prove the assertion by induction.

Step 1: $ f(x)\equal{}\ln(1\plus{}x)\minus{}\ln(1\minus{}x).$ You really need to do that much.

I'm not entirely sure where the "by induction" is coming from here, but I'd suggest taking a few derivatives and looking for patterns.


In a convex quad $ABCD$ with $\measuredangle ADB = 2\measuredangle ACB$ and  $\measuredangle BDC = 2\measuredangle BAC$, prove that $AD=CD$

Draw a circle with center at $D$ with radius AD.  $\measuredangle ADB=2\measuredangle ACB$, so B and C lie on the circle, and AB is the intercepted arc of ACB.  $\measuredangle BDC=2\measuredangle BAC$, so BC is the intercepted arc of A.  Then, $AD$, $BD$, and $CD$ are radii of the circle, and so $AD=CD$.
I believe that proves it...
In order to further demonstrate that such a circle exists, inscribe triangle ABC in a circle.  With $\measuredangle BDC=2\measuredangle BAC$, there exists a curve on which D may lie to keep $\measuredangle BDC$ constant.  With $\measuredangle ADB=2\measuredangle ACB$, there exists a curve on which D may lie to keep $\measuredangle ADB$ constant.  With the constraints that ABCD is convex, there is only one intersection point for these two curves.  You could extend AD, BD and CD into lines and show that as they rotate at the same rate around A, B, and C, there is only one point at which they are concurrent.

Let DM, DN  bisectors of the angles$A\hat DB\mathop {}\limits^{} ,\mathop {}\limits^{} B\hat DC$, respectively.

 Then : $B\hat AN = B\hat DN$  and $M\hat BD = B\hat CM$

  So the quadrilateral ABND and the BMDC is writable, so we have:

  triangles AMD ,  MBD    are equal  .Similarly  triangles BND, CND  are  equal.

     So: $M\hat AD = M\hat BD = N\hat CD$ .

Therefore the triangle ADC  is isosceles , with  DA=DC


A population is described by a N(25,16). If you take a sample of size 36, what is the probability that the sample mean is between 22 and 27?

It's more likely to be N(mean, standard deviation).
The population of sample means would be N(25, 16/6). Use the calculator.

Hmm, k.

And if you're learning this, there should be formulas in your book on how to do it. It's just pretty much look at the table and plug in the numbers.


1)  Find the smallest natural numbers which can be written as
$\frac{a^2}{24}+\frac{b^2}{25}+\frac{c^2}{26}+\frac{d^2}{27}+\frac{e^2}{28}$, $\ a,b,c,d,e$ naturals.
2)  Prove that the equation $\ (x+2y)(y+2x)=2^z$ has no solutions in $\ N$.

Yes, you're right.This is the answer.But how can you prove this?

2) Both factors must be powers of two.
Let $x+2y=2^a\Rightarrow x=2^a-2y$.
Let $y+2x=2^b\Rightarrow y+2(2^a-2y)=2^b\Rightarrow 2^{a+1}-3y=2^b$.  $y$ clearly cannot be natural as long as $a$ and $b$ are integers, so it is impossible.

A second method to solve #2:
$x$ and $y$ must both be even in order for both factors to be even.
So, let $2x_1=x$ and $2y_1=y$.
Then, $(2x_1+4y_1)(2y_1+4x_1)=2^z \Rightarrow (x_1+2y_1)(y_1+2x_1)=2^{z-2}$.
$x_1$ and $y_1$ must also be even.
So, let $2x_2=x_1$ and $2y_2=y_1$.
We continue the procedure until $(x_n+2y_n)(y_n+2x_n)=2^0$ or $(x_n+2y_n)(y_n+2x_n)=2^1$.  Either is impossible for natural numbers.


Find all real numbers  $ x$  such that
$ \frac{\sqrt{x}}{x\sqrt{x}\minus{}3\sqrt{x}\plus{}3}$ is integers number

[quote="aktyw19"]Find all real numbers  $ x$  such that
$ \frac {\sqrt {x}}{x\sqrt {x} \minus{} 3\sqrt {x} \plus{} 3}$ is integers number[/quote]

[hide="solution"]
obviously $ x\geq0$ and $ \sqrt{x}\geq0$
let $ \sqrt{x}\equal{}t$ which $ t\geq0$ and $ f(t)\equal{}x\sqrt {x} \minus{} 3\sqrt {x} \plus{} 3\equal{}t^3\minus{}3t\plus{}3$
then $ f'(t)\equal{}3t^2\minus{}3$
let $ f'(t)\equal{}0$ then $ t\equal{}1$
so $ f_{min}(t)\equal{}f(1)\equal{}1>0$
$ x\sqrt {x} \minus{} 3\sqrt {x} \plus{} 3>0$
so if $ \frac {\sqrt {x}}{x\sqrt {x} \minus{} 3\sqrt {x} \plus{} 3}\equal{}\frac{t}{f(t)}$ is an integer then
$ t\equal{}0$ or $ t\geq f(t)$
$ t^3\minus{}4t\plus{}3\leq0$
$ (t\minus{}1)(t^2\plus{}t\minus{}3)\leq0$
$ (t\minus{}1)(t\minus{}\frac{\minus{}1\minus{}\sqrt{13}}{2})(t\minus{}\frac{\minus{}1\plus{}\sqrt{13}}{2})\leq0$
$ 1\leq t \leq\frac{\minus{}1\plus{}\sqrt{13}}{2}$
let $ g(t)\equal{}2f(t)\minus{}t$
then $ g'(t)\equal{}2f'(t)\minus{}1\equal{}6t^2\minus{}7$
let $ g'(t)\equal{}0$ then $ t\equal{}\sqrt{\frac76}$
so $ g_{min}(t)\equal{}g(\sqrt{\frac76})$ for $ 1\leq t \leq\frac{\minus{}1\plus{}\sqrt{13}}{2}$
notice that $ g(\sqrt{\frac76})>0$
so $ 2f(t)>t$ for $ 1\leq t \leq\frac{\minus{}1\plus{}\sqrt{13}}{2}$
$ 1\leq\frac{t}{f(t)}<2$
so only when $ \frac{t}{f(t)}\equal{}1$ is $ \frac{t}{f(t)}$ an integer
so $ t\equal{}0,1$ or $ \frac{\minus{}1\plus{}\sqrt{13}}{2}$
that is $ x\equal{}0,1$ or $ \frac{7\minus{}\sqrt{13}}{2}$

I'm not sure whether I've lost something in the solution
[/hide]

[quote="aktyw19"]Find all real numbers  $ x$  such that
$ \frac {\sqrt {x}}{x\sqrt {x} \minus{} 3\sqrt {x} \plus{} 3}$ is integers number[/quote]

Obviously $ x\equal{}0$ is a solution.

If $ x>0$, we can write the equation as $ x\sqrt x\minus{}(3\plus{}\frac 1n)\sqrt x \plus{}3\equal{}0$ with $ n\in\mathbb Z^*$ (set of non zero integers).

So we need to find values of $ n$ for which the equation $ P(t)\equal{}x^3\minus{}(3\plus{}\frac 1n)x\plus{}3\equal{}0$ has at least one positive real root.
Since this equation has a negative root, we need $ 3$ real roots, and so we need $ 27\cdot 3^2\minus{}4(3\plus{}\frac 1n)^3\le 0$

And so $ 243\le 4(3\plus{}\frac 1n)^3$ and so $ n\equal{}1$ and then $ P(t)\equal{}t^3\minus{}4t\plus{}3\equal{}(t\minus{}1)(t^2\plus{}t\minus{}3)\equal{}0$ and so $ t\equal{}\sqrt x\in\{1,\frac{\sqrt{13}\minus{}1}2\}$

Hence the final result : $ \boxed{x\in\{0,1,\frac{7\minus{}\sqrt{13}}2\}}$


Given $ a\plus{}b > 0$ where $ a \neq b$, prove that $ 2^{n\minus{}1}(a^{n} \plus{} b^{n}) > (a\plus{}b)^{n}$, for any natural number $ n > 1$

[quote="AndrewTom"]Given $ a \plus{} b > 0$ where $ a \neq b$, prove that $ 2^{n \minus{} 1}(a^{n} \plus{} b^{n}) > (a \plus{} b)^{n}$, for any natural number $ n > 1$[/quote]

[hide]I'll prove this inequality $ (\frac{a\plus{}b}{2})^n \leq \frac{a^n\plus{}b^n}{2}$ where $ a,b\ge0 , n\in N$
By induction:
For $ n\equal{}1$ it's true.
Suppose that it's true for $ n\equal{}k>1$ we have: $ (\frac{a\plus{}b}{2})^k \leq \frac{a^k\plus{}b^k}{2}$

Since $ (\frac{a\plus{}b}{2})^{k\plus{}1}\equal{}(\frac{a\plus{}b}{2})^k \cdot \frac{a\plus{}b}{2} \leq \frac{a^k\plus{}b^k}{2}\cdot\frac{a\plus{}b}{2}$

and

$ (\frac{a\plus{}b}{2})^{k\plus{}1}\minus{}(\frac{a\plus{}b}{2})^k \cdot \frac{a\plus{}b}{2}\equal{}\frac{1}{4}(a^{k\plus{}1}\plus{}b^{k\plus{}1}\minus{}a^kb\minus{}ab^k)\equal{}\frac{1}{4}(a\minus{}b)(a^k\minus{}b^k) \geq 0$

therefore it's true for $ k\plus{}1$, thus it's true $ \forall n$[/hide]

Nice. Alternatively,

Since $ a\neq b$, $ (a\minus{}b)^{2} > 0$.

Adding $ (a\plus{}b)^{2}$ to both sides, $ 2(a^{n} \plus{} b^{n}) > (a\plus{}b)^{n}$.

Therefore true for $ n\equal{}2$.

Assuming $ 2^{k\minus{}1}(a^{k} \plus{} b^{k}) > (a\plus{}b)^{k}$, multiply both sides by $ a\plus{}b$.

$ 2^{k\minus{}1}(a^{k}\plus{}b^{k})(a\plus{}b)> (a\plus{}b)^{k\plus{}1}$

So we need to show that $ 2^{k}(a^{k\plus{}1}\plus{}b^{k\plus{}1}) > 2^{k\minus{}1}(a^{k} \plus{} b^{k})(a\plus{}b)$,

that is,

$ a^{k\plus{}1} \plus{} b^{k\plus{}1} > a^{k}b \plus{} ab^{k}$,

that is,

$ (a^{k} \minus{} b^{k})(a\minus{}b)>0$ which is true (for $ a>b$ and for $ a<b$).


A number of points are given on a unit circle so that the product of the distances from any point on the circle to the given points does not exceed $ 2$.
Prove that the points are the vertices of a regular polygon.

they dont necesarily have to be the vertices of a regular polygon  :huh:


inequalities
algebra
polynomial
inequalities unsolved

Let $ a,b,c \geq 0$ .Prove that :
$ \sum_{cyc} \frac{\sqrt{bc\plus{}4a(b\plus{}c)}}{b\plus{}c} \geq \frac{9}{2}$

[quote="conan_naruto236"]Let $ a,b,c \geq 0$ .Prove that :
$ \sum_{cyc} \frac {\sqrt {bc \plus{} 4a(b \plus{} c)}}{b \plus{} c} \geq \frac {9}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3769
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28175


Let $a>b>c$ be real numbers. Prove that $a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)>0$

Okie, inspire from your problem,
Prove that if $a \geq b \geq c \geq 0$ and for any integer n then:
$a^{n}(b-c)+b^{n}(c-a)+c^{n}(a-b) \geq 0$

for $n>1$, $a \ge b \ge c$ we have
$f(x)=x^{n}$ is convex
Lemma for $c < b < a$
$\frac{f(b)-f(c)}{b-c}\le \frac{f(a)-f(c)}{a-c}\le \frac{f(a)-f(b)}{a-b}$
(if I am right Yulia must know this Lemma...)
$\frac{f(a)-f(b)}{a-b}$ is angel coefficient of segment AB, where  A(a,f(a)),B(b,f(b))
then remember that f(x) is convex.


If $x,y>0$ and $x^5+y^5=x-y \geq 0$, then $x^4+y^4<1$.

made by Ghenghiu R.

cheers! :D :D :cool:

This is similar to C146
Suppose that $ x^4+y^4 \geq 1 $;so
  [tex] x^5+y^5 \leq (x^4+y^4)(x-y)=x^5-x^4y+y^4x-y^5 [/tex]
  [tex] \Leftrightarrow y^5 \leq -x^4y+y^4x-y^5<0 [/tex].This is unreasonable.
So  $x^4+y^4<1 $:)

Sorry, how do you got that  \[ x^4y+y^4x-y^5<0\]
??? :blush:

we  have  2y^5 <=xy^4 - x^4y = -x(x^3-y^3)<0  which  is  not  true , so  we must  have  x^4 + y^4<1.
By  the way...can we  find x,y >0  such  x^5+y^5=x-y>0 ??

Put x = t*y, and the solutions x,y>0 for x5 + y5 = x - y >= 0  are  y = ((t-1)/(t5+1))1/4 , x=t*((t-1)/t5+1))1/4 , with t > 1
By the way, the asked inequality becomes t4*(t-1)/(t5+1) + (t-1)/(t5+1) < 1  and that's trivial for t >= 1
:lol:


Find the least n such that 1+4+9+...+n^2 is divisible by 200.


function
induction
absolute value
algebra unsolved
algebra

Given a function$f$ from the integers to the integers ($f: Z \rightarrow Z$ is this how you represent that?)

$f(x-y+f(y))=f(x)+f(y)$

Find all functions that fit the above criteria.

I'm not quite sure this is ok.

First putting $x=y$ we get $f(f(x))=2(f(x))$. On the other hand, letting $g(y)=f(y)-y$ we get $f(x+g(y))$=$f(x)+f(y)$, this allows us to prove by induction  $f(x+kg(y))=f(x)+kf(y)\;\forall x,y,k\in\mathbb Z$. Putting $k=g(y)$ we get
$f(f(x+g(y)^{2})=f(f(x)+f(y)g(y))=f(f(x))+f(y)^{2}=2f(x)+f(y)^{2}$
But also $f(f(x+g(y)^{2}))=2f(x+g(y)^{2})=2f(x)+2g(y)f(y)$, so comparing both ecuations we obtain
$f(y)^{2}=2g(y)f(y)\rightarrow f(y)(f(y)-2g(y))=0\rightarrow$
$(f(y)=0 \vee f(y)=2y\;\;\forall y\in \mathbb Z$. Note that this means that in any case $f(0)=0$. Suppose that exists some $x$ such that $f(x)=0$ other than $0$ and call $z$ one with minimal absolute value; then putting $y=z$ we get $f(x-z)=f(x)$, which allows us to prove by induction $f(x+kz)=f(x)\; \forall k,x\in \mathbb Z$. Because, since $f(0)=0$, $f(kz)=0 \; \forall k \in\mathbb Z$ we can assume without loss of generality $z>0$. If $z=1$, $f(x)=0 \;\; \forall x \in \mathbb Z$; if $z>1$, $x=z-1, \;y=z-1$ yields $0=f(z)=f(z-1-1+2)=f((z-1)-1+f(1))=f(z-1)+f(1)=2(z-1)+2=2z$, which is a contraddiction. Therefore, if $f$ is not the 0-function, $f(x)=2x \;\;\forall x \in \mathbb Z$.  I suppose the last is actually the same as a proof by induction.

So we have: $\ x+f(0)=f(x)-x$ $\ \to f(x)=2x+f(0)$

Question:  Why is f an injection?

can you prove that? :D

Well we might be able to argue and find that if $f(x)$ isn't injective it will lead to a contradiction somehow. I am not sure. :maybe:

[quote="El fisico"]Let $\ f(a)=f(b)$ such that $\ a\neq b\to a-b\neq 0$. Replace $\ x=a$ and $\ y=b$ in the equation.

We have $\ f(a-b+f(a))=2f(a)$ but the right hand side of the equation depends only on $\ a$ so it's impossible. then $\ a=b$.[/quote]

Probably it's me; I cannot sleep very well this times, but; why is it impossible?  $a$ and $b$ are not free variables all over $\mathbb Z$, and they are mutually dependent. What do I miss?  :huh:  :oops:

Another proof.
(1) $ P(y,y): f(f(y))\equal{}2f(y)$

(2) $ P(x,f(y)): f(x\plus{}f(y))\equal{}f(x)\plus{}2f(y)$.
Replace $ x$ with $ x\minus{}y$ above, we get
$ f(x\minus{}y\plus{}f(y))\equal{}f(x\minus{}y)\plus{}2f(y)\equal{}f(x)\plus{}f(y)\Rightarrow f(x\minus{}y)\plus{}f(y)\equal{}f(x)$.
Now we can easily get $ f(x)\equal{}xf(1)$. Let $ f(1)\equal{}a$ and put in (1),
$ f(f(y))\equal{}a^2y\equal{}2ay\Rightarrow a^2\equal{}2a$.

Hence, $ f(x)\equal{}2x$ or $ f(x)\equal{}0$.


trigonometry
real analysis
real analysis unsolved

Study the convergence of

$\sum_{n\geq 1}\frac {(\frac{2+\sin {n}}{3})^{n}}{n}$

This has been posted before. See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093[/url]


Any last minute (in my head time goes 14400 times as fast) ideas for tricks, tips, fomulas, types of questions, what to know and etc? (for aime) and also some hard examples of Aime questions!
I would greatly appreciate any help

So you really are in 4th grade? If so, wow.
Erm...lemesee...Stewart's Theorem, Rational Roots Theorem, Viete's Formula, Ptolemy's Theorem...trig identities, make sure you know your combinatorics (balls-and-urns), and sometimes complex numbers (i.e. DeMoivre's, etc.), but I haven't seen that many problems with them.
EDIT: By the way, we already have a thread like this, but oh well.

Make sure you know A LOT of number theory stuff.

[quote]So you really are in 4th grade? If so, wow. [/quote]

and you know this how?

didn't you post that in another thread?...

Lol.. you cant be a 4th grader-you also said in the mathcounts forum that you were 2nd in chapter countdown..4th graders cant do mathcounts...

Ah, exposed, then.  :roll:
He must be a middle schooler, then. Unless he's lying about MATHCOUNTS as well. I don't check that forum at all.

"i got 2nd place [in chapter]"
"i'm in 4th grade and i got a 120, ezzz"
 :rotfl:


Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac{a^3}{(2a^2\plus{}b^2)(2a^2\plus{}c^2)}\plus{}\frac{b^3}{(2b^2\plus{}a^2)(2b^2\plus{}c^2)}\plus{}\frac{c^3}{(2c^2\plus{}a^2)(2c^2\plus{}b^2)}\leq \frac{1}{a\plus{}b\plus{}c}$

By CBS we have $ \frac{a^3}{(a^2\plus{}a^2\plus{}b^2)(c^2\plus{}a^2\plus{}a^2)}\leq\frac{a^3}{(ac\plus{}a^2\plus{}ba)^2}\equal{}\frac{a}{(a\plus{}b\plus{}c)^2}$

and similarly.


Let $ x,y,z>0$ and $ \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z} \equal{} 1$. Prove that $ \sqrt {x \plus{} yz}  \plus{} \sqrt {y \plus{} xz}  \plus{} \sqrt {z \plus{} xy}  \ge \sqrt {xyz}  \plus{} \sqrt x  \plus{} \sqrt y  \plus{} \sqrt z$

[quote="thanhnam2902"]Let $ x,y,z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that $ \sqrt {x \plus{} yz} \plus{} \sqrt {y \plus{} xz} \plus{} \sqrt {z \plus{} xy} \ge \sqrt {xyz} \plus{} \sqrt x \plus{} \sqrt y \plus{} \sqrt z$[/quote]
Let : $ \frac{1}{x} \equal{} a,\frac{1}{y} \equal{} b$ and $ \frac{1}{z} \equal{} c$.With $ a,b,c > 0$ and $ a\plus{}b\plus{}c\equal{}1$,we have to prove that :

$ \sqrt{a\plus{}bc}\plus{}\sqrt{b\plus{}ac}\plus{}\sqrt{c\plus{}ab} \geq 1 \plus{} \sqrt{ab}\plus{} \sqrt{bc}\plus{} \sqrt{ca}$

Which follows from Caushy-Scwartz inequality :

$ \sqrt{a\plus{}bc} \equal{} \sqrt{(a\plus{}b)(a\plus{}c)} \geq a \plus{} \sqrt{bc}$ ,...

[quote="rachid"][quote="thanhnam2902"]Let $ x,y,z > 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 1$. Prove that $ \sqrt {x \plus{} yz} \plus{} \sqrt {y \plus{} xz} \plus{} \sqrt {z \plus{} xy} \ge \sqrt {xyz} \plus{} \sqrt x \plus{} \sqrt y \plus{} \sqrt z$[/quote]
Let : $ \frac {1}{x} \equal{} a,\frac {1}{y} \equal{} b$ and $ \frac {1}{z} \equal{} c$.With $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$,we have to prove that :

$ \sqrt {a \plus{} bc} \plus{} \sqrt {b \plus{} ac} \plus{} \sqrt {c \plus{} ab} \geq 1 \plus{} \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}$

Which follows from Caushy-Scwartz inequality :

$ \sqrt {a \plus{} bc} \equal{} \sqrt {(a \plus{} b)(a \plus{} c)} \geq a \plus{} \sqrt {bc}$ ,...[/quote]

agree with you

that's very easy...


geometry
3D geometry
pyramid
perimeter
geometry unsolved

Let $ S.ABCD$ is a pyramid and let $ SA \equal{} SB \equal{} SC \equal{} SD \equal{} a$, $ ABCD$ is a square with $ AB \equal{} BC \equal{} CD \equal{} DA \equal{} b$ and $ a > b$. Let $ AC$ meet $ BD$ at point $ O$. Let $ M,N$ are two points lie on two planes $ (SAB)$ and $ (SCD)$ respectively. Let $ p \equal{} MO \plus{} ON \plus{} MN$. Find place of points $ M$ and $ N$ for $ p$ be minimum value.

$ M$   and $ N$ lies in the medians, from vertex S form triangles $ SAB$ and $ SCD$ whose angle wrt to the basis we'll designate by $ \alpha$ (constant).

Let the angle (variable) of $ OM$ equal to that of $ ON$ , also wrt to the basis,  be $ \beta$.

Then, the perimeter of triangle $ OMN$ can be written as $ p \equal{} b{{sin \alpha} \over {sin (\alpha \plus{} \beta)}}.( cos \beta \plus{} 1)$.

Minimum $ p$ occurs for $ cos ( \alpha \plus{} \beta ) \equal{} \minus{} cos \alpha  \Longrightarrow  \beta \equal{} \pi \minus{} 2\alpha$.

So, the distance , along the medians , of $ M$ and $ N$ to the pyramids basis is $ {b\over 2}$.


inequalities
limit
inequalities proposed

[u][b]The author of this posting is : tranminhhoang[/b][/u]
____________________________________________________________________

Consider an infinite sequence {$x_n$} consisting of positive reals such that $x_0=1$ and $x_0 \geq x_1 \geq x_2 \geq ...$.

Let be an integer, $t>1$. Prove that

[b](a)[/b] For every $s>0$, there is a positive integer $n$ such that:

$\sum\frac{x_i^t}{x_{i+1}} \geq\frac{t^\frac{t}{t-1}}{t-1} - s$

[b](b)[/b] Find all {$x_n$} such that: for all $n$, the inequality

$\sum\frac{x_i^t}{x_{i+1}} \leq\frac{t^\frac{t}{t-1}}{t-1}$

holds.

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

I have a solution.

[b](a)[/b] I will only show key idea without boring calculations.

We can write finite sum as follows

\[ \frac{1}{x_1}+\frac{x_1^t}{x_2}+...+\frac{x_n^t}{x_{n+1}}=t^n\cdot\frac{1}{t^nx_1}+t^{n-1}\cdot\frac{x_1^t}{t^{n-1}x_2}+...+1\cdot\frac{x_n^t}{x_{n+1}}\].

Applying AM-GM we obtain

\[ \sum_{i=0}^n\geq(t^n+t^{n-1}+...+1)\cdot\left(\frac{1}{t^{nt^n}x_1^{t^n}}\cdot\frac{x_1^{t^n}}{t^{(n-1)t^{n-1}}x_2^{t^{n-2}}}\cdot...\cdot\frac{x_n^{t}}{x_{n+1}}\right)^\frac{1}{t^n+t^{n-1}+...+1}= \]
\[ =(t^n+t^{n-1}+...+1)\cdot\left(\frac{1}{t^{nt^n+(n-1)t^{n-1}+...+t}x_{n+1}}\right)^\frac{1}{t^n+t^{n-1}+...+1}\].

Taking into consideration $x_{n+1}<1$ and calculating limit $n\to \infty$ we obtain $ \lim_{n\to\infty}\sum_{i=0}^n\geq \frac{t^{t/(t-1)}}{t-1} $.

[b](b)[/b] Using described technique it is possible to show that required sequence must satisfy the condition $ \frac{x_k^t}{x_{k+1}}=\frac{1}{t^k}\cdot t^{1/(t-1)} $.


Yes, or perhaps it will be on their website.

[quote="alexmcavoy@gmail.com"]I will be a senior this fall.  My high school doesn't offer the AMC exam., so I'm trying to find a school in my area that does (if it is even possible to take it at another school).  Could someone please send me a link or tell me where AMC's are offered in the Western Massachusetts area?

Much appreciated.[/quote]

I take the AMC's and AIME's through this accelerated math class called ICAE. (sorry, it's only in Michigan)

Maybe you can take it through things like Kumon or other extra classes... Maybe places like Sylvan Learning Center (or other tutor places) can give you info. You really need to take the AMC. It's the main test for high school students (mathstudents)

Thanks for the suggestions, but I probably won't end up being able to take it.  I talked to my GC about it and I won't be able to take it at my school.  To my knowledge, it isn't offered anywhere a feasible distance away either.

Thanks again.

you can write it yourself (by registering as a homeschooler or something for some nominal fee - $15?), so don't give up. The AMC people want you to take it, after all.

where do you live?  near worcester?  if you're in the area i might be able to help.

I actually live near Springfield, but depending on the time(s) the tests are administered, I am willing to drive (Worcester isn't that far away).

Thanks a lot for your help.

i dont know really where springfield is, but there's a lot of schools in central MA who give the tests, mostly ~8 am.  the only school i know giving it later, at like 8pm i think is phillips academy.  but that's probablly a pretty big hike for you.  i dont know if you've seen a list of the schools that give the test, but if you aks hstran@amc.unl.edu for a list of schools in MA who do it they'll help you out.  if none of those schools are too close to you but you're willing to drive a little bit to take it at 8am PM me and i cangive you some ppl to try emailing.  by the way you should probablly try to do arml this year.

[quote="Drunken_Math"]I take the AMC's and AIME's through this accelerated math class called ICAE. (sorry, it's only in Michigan)[/quote]

Does anyone know if you can take it through CTY? (http://www.cty.jhu.edu)

[quote="alexmcavoy@gmail.com"]Thanks for the suggestions, but I probably won't end up being able to take it.  I talked to my GC about it and I won't be able to take it at my school.  To my knowledge, it isn't offered anywhere a feasible distance away either.
[/quote]Simon's Rock College in Great Barrington and Longmeadow HS both offer the AMC 12 -- you or your GC might be able to work something out with them.  Press your GC hard to do this -- it's their job!  Better yet, ask the head of your school's math department to have it at your school.  It only costs 54 dollars for your school to register and receive 10 copies of the AMC 12.

As a first resort, we always recommend what DPatrick just suggested in the previous post.  We suggest you form a math club at school, find about 10 interested students, and talk to the Math Department head or supervisor.  They can hardly say no at that point.  To raise money, do the usual things:  have a car wash, sell snacks at sports events, sell T-shirts, etc.   In the process, you not only achieve your goal of having the AMC administered at your school, you organize a group to study and learn with, you learn some organizational skills, and you impress your counselors and teachers with your entrepeneuership.

It is much, much easier to organize, register and hold the AMC competition at your school than it is to be a visitor at another school for all concerned.  Being a visitor at another school is intended to be an emergency, last-resort procedure for unavoidable conflicts.

Steve Dunbar
AMC Director

Last year some kid at my school did what you said AMCDirector, and he got in trouble for it.  So I don't think that'd work in my case.

I did that and got 10 signatures for the AMC 10A. Except some people "chickened out" and said that they never said they were interested :?. So probably we won't get AMC 10 next year. Anyway, as a bonus, we got AMC 12A too. Then some ambitious juniors paid for the 10 AMC 12B tests (rich people) and people who were interested paid them. I reserved my place, but this senior wanted my place... :mad: so yeah...
I got to take the AIME though.  :)

My son 7th grade like to take the AMC 8 exam but the math coordinator does not like to register him. We offer to pay the registration fee and our son's GC agree to be a proctor at the exam but the math coordinator said no. I think she does not understand what should I do? Can you give me some official letter to give to the principal to convince the school to register. We have a math team and the same math coordinator lead the team. What would be the reason for not willing to register?

Go to the AMC website, you should be able to find a place. If you are willing to travel a little, you can go to somewhere in Eastern Mass. I usually go to an RSM; there are about 5 in Eastern Mass.

Try taking it at a local community college. That is what I did in 8th grade. As far as I am aware of, there are plenty of high-performing schools who offer the AMCs around the Western Massachusetts area. If you go check on the school websites, there is sometimes a section about the AMCs where you can register to take it at their schools. Better yet, email the math coordinator/coach at the school.

Before last year, my school didn't offer the AMC. However, I asked them to register for it, and they did. My school did pay for the test, but if you say that you would be willing to pay, your school would have a hard time saying no.

My school usually has people pay a small fee ($5 or so) to cover the cost of ordering the tests.  This seems to work out pretty well overall.

This year, regular A date registration ends on January 4th and Expedited Shipping for A ends on January 19th.  Regular shipping for the B date ends on this same date, and expedited shipping for B ends on February 3rd.  You've got plenty of time to either convince your school to take the test or find an alternate site to do so.


search
function
calculus
calculus computations

I tried a search to see if this has been asked before and was overwhelmed with just how many functional equations have been posed on this site.  Have fun guys!

Find all real-valued, continuous functions $ f$ on $ \mathbb{R}$ which satisfy $ f(x)f(y)=f(x_{1})f(y_{1})$ for all $ x,y,x_{1},y_{1}$ such that $ x^{2}+y^{2}=x_{1}^{2}+y_{1}^{2}$.

(I think the functions are easy to guess, so I am really interested in seeing what different approaches you may take)   :)

it's easy to see that if $ f(0)=0$ then $ f(x)=0$, so lets suppose $ f(0)\neq0$.

make $ g(x)=\frac{f(x)}{f(0)f(1)^{x^{2}}}$, we have that $ g(x)$ respects the same condition than $ f$ and $ g(0)=g(1)=1$.

now we prove that if $ \sum_{i=1}^{n}x_{i}^{2}=\sum_{i=1}^{n}y_{i}^{2}$ then $ \prod g(x_{i})=\prod g(y_{i})$ because
$ \prod g(x_{i})=g(\sqrt{\sum x_{i}^{2}})g(0)^{n-1}=g(\sqrt{\sum x_{i}^{2}})=\prod g(y_{i})$.

setting $ x_{i}=x$ for $ 1\leq i\leq n^{2}$,
$ y_{1}=nx$ and $ y_{i}=0$ for $ 1<i\leq n^{2}$ we get

$ g(x)^{n^{2}}=g(nx)$, putting $ x=1$ we have $ g(n)=1$ and $ x=\frac{m}{n}$ we conclude $ g\left(\frac{m}{n}\right)=1$

as the set of the rational numbers is dense in $ \mathbb{R}$ and $ g$ is continuous we have $ g(x)=1$ so $ f(x)=f(0)f(1)^{x^{2}}$.

[quote="Bajoukx"]make $ g(x)=\frac{f(x)}{f(0)f(1)^{x^{2}}}$, we have that $ g(x)$ respects the same condition than $ f$ and $ g(0)=g(1)=1$.[/quote]

First of all, how do you know that $ f(1) \neq 0$?  Second of all, where did $ g(1) = 1$ come from?

sorry, make first $ h(x)=\frac{f(x)}{f(0)}$, $ h$ respects the same conditions than $ f$ and $ h(0)=1$.

we have $ h(x)=h(x)h(0)=h\left(\frac{x}{\sqrt(2)}\right)^{2}\geq0$

suppose that for some $ y$ we have $ h(y)=0$, as $ h$ is continuous the set $ \{x, h(x)=0\}\0$ is closed and like $ h(0)\neq 0$ the set $ \{x, h(x)=0\}$ as a minimum, lets say $ y_{0}$.

$ h(y_{0})h(0)=h\left(\frac{y_{0}}{\sqrt 2}\right)^{2}\Leftrightarrow h\left(\frac{y_{0}}{\sqrt 2}\right)=0$ contradiction.

so we have $ h(1)\neq 0$ and now we can make $ g(x)=\frac{h(x)}{h(1)^{x^{2}}}$

and now we have $ g(0)=g(1)=0$ and if the rest of the proof is correct in the end we get

$ f(x)=\left(\frac{f(1)}{f(0)}\right)^{x^{2}}f(0)$

sorry for the mistakes...


calculus integration function logarithms algebra polynomial geometry

Let x be the number of the prime and f(x) = 2x^1.177 be the

expression that gives the approximate value of the prime at position

x.  Estimate the sum of 30 consecutive prime numbers, beginning

with the 5th prime number, inclusive.  You must show your use of

integral calculus in the steps in this thread to get your answer.  Then

round your answer to an integer.  In lacking of special

symbols, you may use something similar to the following general

set-up:

Integral from x1 to x2 of f(x) dx

Note:  For reference to compare your answer, I am including the actual sum of these prime numbers several lines below:  >>>>
>
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2022

Beginning at the 5th prime, where 30 is the total # of primes,  the last to be summed, then, is the 34th prime.

Integral from 4.5 to 34.5 of 2x^1.177 dx

Evaluated at (2/2.177)[x^2.177] from 4.5 to 34.5

= 2022.18 which is correct to two decimal places

This rounds to 2022, which matches the problem's sum given at the bottom of the original post.

The real prime estimator function:

The number of primes less than $n$ is approximately $\int_2^n \frac1{\ln x}\,dx$; the errors are asymptotically small compared to this. A slightly worse estimate that can actually be calculated: $\frac{n}{\ln n}$ primes less than $n$, and the $n$th prime at $n\ln n$.

The estimate you're using here is fitted to this, and won't work at a million.
See [url=http://mathworld.wolfram.com/PrimeNumberTheorem.html]here[/url] for more detail.

You didn't have anything to add to the solution to the problem at hand, jmerry.

Stating that it won't work for a million is a given.  You chose poorly in going that route, because I never stated it was good for all integral values of x.

I wasn't talking to you, pleaseread- my post was directed at the original poster.

I object to the highly artificial setup- it was chosen to make exactly this question work, and fails miserably in any other context.

[Typo fixed]

The word "highly" is subjective.  And "artificial" is the spelling you meant I'm sure?  College Algebra and calculus texts regularly have word problems with polynomials in them that do not fit real life.  That makes them artificial, yes.  The point of those problems, and this one, is to test the pertinent knowledge at hand.  This formula is an oversimplification that fits these stretch of x's.  The challenge is to see if a calculus student/instructor can apply it to get the solution.

"miserably" is another subjective word.  Your comment about failing
is moot because the problem NEVER pretended that it was applicable for other contexts.

jmerry, you are the one using repeated subjectivity and displaced anger in your tone.  I could demand more from you academically and in forming better arguments,.............. but I won't.

The problem is reasonable -- it tested my calc that I used in college.

The objections here are emotion-driven.

Creating multiple usernames now, are we?  And the function used in this problem is clearly not a polynomial.

Anger? Now you've provoked me.

The 30 primes in question:
[hide="List"]11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139[/hide]
Their sum is 2110, not 2022.

Were you by any chance using 51 as a prime (and leaving off 139) when you calculated the sum and generated this estimator?

The estimate is high for $2\le n\le 15$, except $n=13$ for which it is nearly exact (40.9 versus 41).
The estimate is low for $16\le n\le 34$. I didn't test it for any larger $n$.

It's not a good estimate, even in this restricted range. A far better problem could be built from the estimate that comes from the Prime Number Theorem, especially if we focus on approximations rather than exact values.

Make it be the

Integral from 4.5 to 34.5 of (2x^1.19068) dx

It comes out to 2109.99+  which rounds to 2110.

So what if this problem is artificial?  So is the the objection raised to it; it teaches

how to work with midpoints of rectangles, integrating, and a familiarity with prime

numbers.  Go for it!

Is someone trying to get around a banning?

You tell me.  You're in the best position to know.

Ever since you joined this thread, there's been nothing but insults and anger.

Take a long hard look in the mirror before asking others if "someone is trying to get

around a banning."  Don't take us for fools, jmerry.


integration Euler calculus function algebra polynomial search

how do you integrate $\int_{a}^{b}e^{ix},dx$ ?
do you use euler identity to convert to cos and sin or can leave it as i and integrate using the rule of exponential integration regularly?
thanks.

You can integrate as you would with a real number.

A justification: the function $F(t): =\int_{a}^{b}e^{tx}\,dx$ is analytic for all $t\in\mathbb C$. Since we know that $F(t)=(e^{tb}-e^{ta})/t$ for $t\in\mathbb R$, the same holds for all $t\in\mathbb C$.

i tried to integrate this, a is a constant
$\int e^{i*t-t^{2}+i*a*t}dt\$$
I got
$\exp^{i*t-t^{2}+i*a*t}/(i-2*t+a*i )\$$ is that correct?

No, that's not correct. The addition of $t^{2}$ changes everything, and your integral is no longer an elementary function. Did you know that you can check your work with [url=http://integrals.wolfram.com/index.jsp]The Integrator[/url]?

then how can I solve that? I wanna know how to do by hand
i could use matlab too but it gives weird answers and i don't understand the answer

That's because there is no answer that we can understand without knowing special functions like $\mathrm{Erf}$.

What you may be able to do by hand is to find $\int_{-\infty}^{\infty}\dots$. And even that requires a working knowledge of complex analysis.

ok so let's take out the i, does it make the problem simpler?
say F(x) = $\int_{0}^\infty\ 1/x*e^{-a/x}*e^{-x}dx\$$
how do you solve exponential problems like that? I tried substitution but it doesn't seem to work

$i$ was not a problem: $t^{2}$ was. There is no general method to "solve exponential problems like that". Substitution does not help much.

Also, you don't need to put that \ at the end of a formula. It adds an unnecessary dollar sign.

Normally, when you have $\int_{a}^{b}f(x) \ dx$, you find the antiderivative of $f(x)$, usually denoted as $F(x)$, and then you have $\int_{a}^{b}f(x) \ dx = F(b)-F(a)$.

Here's the problem:  The function $f(x) = e^{x^{2}}\ dx$ does not have an elementary antiderivative.  By elementary I mean any function that can be made out of a finite number of compositions of polynomial, exponential, logarithmic, trigonometric, hyperbolic, ect functions.  (Example $\frac{\sin(x^{2}+2)}{\ln(2\cosh(e^{x}))}$ would be elementary).

When we are given the bounds for integration, we can sometimes evaluate the integral without knowing the antiderivative:

$\int_{0}^\infty e^{-x^{2}}\ dx = \frac{\sqrt{\pi}}{2}$.

$\int_{0}^\infty \frac{\sin(x)}{x(x^{2}+1)}\ dx = \frac{\pi}{2}\left(1-\frac{1}{e}\right)$

There is no general method to do this, but search the forum and you'll find some cool tricks.

this one does not have the t^2 in it tho
$\int_{0}^\infty\ 1/x * e^{-a/x}*e^{-x}dx$
however i still have that erf function when integrating this, why?

The reciprocal of $x$ under the exponential function is not good either.


inequalities
integration
advanced fields
advanced fields unsolved

Let $f\in C^{1}[a,b]$ and $\parallel f\parallel _{\infty}: =\max_{t\in [a,b]}|f(t)| .$ Prove that \[\parallel f\parallel _{\infty}\le \frac{1}{b-a}\int_{a}^{b}|f(t)|\; dt+\int_{a}^{b}\left|f^{\prime}(t)\right|\; dt\; \; .\]

well know

$|f(c)|=\parallel f\parallel _{\infty}$ and $|f(d)|=min_{[a,b]}|f|$

we have $|f(c)-f(d)|\leq \int_{a}^{b}|f'|$

so

(1)  $|f(c)|\leq |f(d)|+\int_{a}^{b}|f'|$

on the other hand

(2) $\int_{a}^{b}|f| \geq (b-a)|f(d)|$

(1) and (2) gives the result


[quote="dnkywin"]What were the final results. (I didn't get to go because either :furious: :furious: [u] [i] [b] Carnage cheated [/b] [/i] [/u]  :furious: :furious: or a teacher at my school screwed up.)[/quote]

What were the final results. (I didn't get to go because either :furious: :furious: [u] [i] [b] Carnage cheated [/b] [/i] [/u]  :furious: :furious: or a teacher at my school screwed up.)

Dude... thats just obvious kid

Carnage and I think one of Martin's team got first in team round (10/10), but in individual scores we got like last...

:noo:  :bomb:  :furious: Carnage was fair.


From Merriam-Webster:
[url]http://www.merriam-webster.com/cgi-bin/dictionary?book=Dictionary&va=life[/url]
 :D

[quote="KBabe"]sry, this post was a mistake.....[/quote]
Now that's life.

Life is also:
The property or quality that distinguishes living organisms from dead organisms and inanimate matter, manifested in functions such as metabolism, growth, reproduction, and response to stimuli or adaptation to the environment originating from within the organism.

I'll tell you what the life is all about...it's a suggested or gifted opportunity to eat food

you live by eating food

that's the main stuff you need in your life..


Let $ ABCD$ is a parallelogram. Let $ P$ is a point and $ P$ lie on $ BD$, let $ I$ is a point lie on the line pass two points $ C$ and $ P$ such that $ P$ is midpoint of $ IC$. Let $ (d)$ is a line pass point $ I$ and $ (d) // BC$ and $ (d)$ meet $ AB$ at point $ E$. Let $ (d')$ is a line pass point $ I$ and $ (d) // AB$ and $ (d)$ meet $ AD$ at point $ F$. Prove that three points $ P, E, F$ are collinear

Let $ IE\cap DB\equal{}L, IF\cap DB\equal{}J$.
Use Thales's theorem and Menelaus's theorem for triangle ADB with line (PEF)