geometry
3D geometry
sphere
analytic geometry
calculus
calculus computations

Say I have the following sphere equation:

$ (x\minus{}4)^2 \plus{} (y\minus{}3)^2 \plus{} (z\plus{}2)^2 \equal{} 20$

If I am asked how the sphere intersect with the xy-plane, xz plane, and yz plane, how would I find it? do I just set z, y, and x to 0 respectively?

[quote="mlok"]That's right.[/quote]

ok so the intersection with the xy-plane is:

(x-4)^2 + (y-3)2 = 20, in this case it's a circle..

so the intersection in all plane is a circle?

[quote="mlok"]Is that surprising? Imagine a sphere being cut by a plane.[/quote]

Not really as the question was like this:

How does the sphere intersect each of the following planes?
a. In a circle
b. At one point.
c. They do not intersect.
d. At two points.

so I assume for all those 3 planes the answer is in a circle, correct?

By the way, your equation for the intersection with the $ xy$ plane is not correct.

It is true in this case that the intersection of the sphere with the coordinate planes is a circle. But that is not true about its intersection with the plane $ x\equal{}\minus{}1.$

I am not sure I understand what you mean, can you please clarify? so the answer is not circle for all those 3?

Kent merely means that when finding the intersection with the $ xy$ plane, you should plug $ z\equal{}0$ into the equation of the sphere, not just to drop the entire expression in parentheses containing $ z$.

what if the question now asks the intersection with the y-axis, do I just set x and z as 0 here and calculate what ever is left?


Definetely [i]do-able[/i] one could find the 6th, 5th, ... 1st Newton's sum of the original function.  Perhaps there is a nicer way though.


calculus
integration
derivative
calculus computations

compute this integral by setting $ t\equal{}x\minus{}\frac{1}{x}$

$ \int_{1}^{e} \frac{x^2 \plus{} 1}{x \sqrt{x^4 \minus{} x^2 \plus{}1} } dx$

I don't see how...

"Derivative."

misan claims that by reconstructing a derivative the result follows easily. Note that you'll end up with an integral like $ \int_b^a\frac1{\sqrt{1\plus{}x^2}}\,dx,$ this can be solved via substitution $ z\equal{}x\plus{}\sqrt{1\plus{}x^2}.$

[quote="misan"]$ I \equal{} \int_1^e\;\dfrac{\textbf d\left(x \minus{} \frac {1}{x}\right)}{\sqrt {\;1 \plus{} \left(x \minus{} \frac {1}{x}\right)^2\;}}$[/quote]

I think you are wrong somewhere.

No, it's not wrong, it's correct, check it.

[quote="misan"]$ I \equal{} \int_1^e\;\dfrac{\textbf d\left(x \minus{} \frac {1}{x}\right)}{\sqrt {\;1 \plus{} \left(x \minus{} \frac {1}{x}\right)^2\;}}$[/quote]
We have $ dt \equal{} (x \minus{} \frac {1}{x})'dx \equal{} (1 \plus{} \frac {1}{x^2})dx$
So $ I \equal{} \int_{0}^{e \minus{} \frac {1}{e}} \frac {dt}{\sqrt {t^2 \plus{} 1}} \equal{}ln \left| x\plus{}\sqrt{x^2\plus{}1}\right| |_{0}^{e \minus{} \frac {1}{e}} \equal{} ln \left|e\minus{}\frac{1}{e}\plus{}\sqrt{e^2\plus{}\frac{1}{e^2}\minus{}1}\right|$


probability
geometry
3D geometry
quadratics
modular arithmetic

We throw normal playing cube three times.
"Count" the probability of:
A - on each cube will be odd number of dots
B - sum of squares dots is divisible by 3.

I hope you understand :)

[hide="Part A"]Since each cube has $ 3$ odd numbers out of $ 6$ total numbers, the probability is:
\[{ {\left(\frac {3}{6}\right)}^3 = \left(\frac {1}{2}\right)}^3 = \boxed{\frac {1}{8}}
\]
[/hide][hide="Part B"]Denoting $ a, b, c$ as the numbers from the die, we have that $ 3 \mid a^2+b^2+c^2$.
The quadratic residues in $ \pmod{3}$ are $ 0$ and $ 1$. Thus, for $ a^2+b^2+c^2$ to be divisible by $ 3$, we must have:
\[ a^2 \equiv b^2 \equiv c^2 \equiv 0 \pmod{3}\]Then:
\[ a \equiv b \equiv c \equiv 0 \pmod{3}\]
Only the numbers $ 3$ and $ 6$ on a die are divisible by $ 3$, so the request probability is:
\[ {\left(\frac{2}{6}\right)}^3={\left(\frac{1}{3}\right)}^3=\boxed{\frac{1}{27}}\][/hide]

Heh, I'll give this a shot.

[hide="A"]
The number of dots is odd if you roll a 1, 3, or 5. Since there are 6 equally likely outcomes, the probability for each individual roll is 1/2. Thus, the probability is $ (\frac{1}{2})^{3} \equal{}$

$ \boxed{\frac{1}{8}}$[/hide]

[hide="B"]
Observe is modulo 3. The chance to get each individual modular residue is 1/3.
Let's count the total number of ways it can sum to a number divisible by 3. Obviously, if all the numbers have the same remainder, the sum is divisible by 3. Since there are 3 modular residues and there are 2 numbers for each, there are 2^3 * 3 = 24 ways. There is also 0, 1, 2. There are 6 arrangements, so we now have 6 * 2^3 = 48 ways for this case. There are 24 + 48 = 72.

We know that the total number of combinations is 6^3 = 216. Thus, the probability is 72/216, which simplifies to

$ \boxed{1/3}$

Uh oh, since the answer was that simple, this tells me there was probably a much simpler way. (stabs self with a knife)[/hide]

123456789, I'm not sure that my B is right, but I'm pretty sure yours is wrong. For example, a can be congruent to 0, b congruent to 1, and c congruent to 2. Just an example....

i agree with above post for B

we can have (0,0,0) (1,1,1) and thats it becuase all numbers squared are either 0 or 1 mod 3.

(3m+1)^2 has a 1 at the end everything else is divisible by 3
(3m+2)^2 has a 4 at the end everything else is divislbe by 3, 4 is 1 mod 3
(3m)^2 obviously is 1 mod 3

so its (1/3)^3+(2/3)^3=1/27+8/27=9/27=1/3

[quote="Nerd_of_the_Ages"]
123456789, I'm not sure that my B is right, but I'm pretty sure yours is wrong. For example, a can be congruent to 0, b congruent to 1, and c congruent to 2. Just an example....[/quote]

You read the question wrong.  :wink: It says squares.

In fact, for your interpretation of the problem, for any possible result of the first two dice rolls, there is a $ \frac {2}{6}$ chance that you'll get a third dice roll which will make the sum divisible by 3, so the answer is just $ \frac{1}{3}$.

[hide="Answers"]
[b](a)[/b] Since all three have to be odd, the only possiblities for all three are 1, 3, or 5. This means that the total for this part is $ \left(\dfrac{1}{2}\right)^3\equal{}\boxed{\dfrac{1}{8}}$.

[b](b)[/b] In order for the sum to be divisible by 3, we have $ a^2\plus{}b^2\plus{}c^2|3$. In order for this to occur, we need $ a^2\equiv b^2\equiv c^2\equiv 0\pmod 3$. This simplifies to $ a\equiv b\equiv c\equiv 0\pmod 3$. Since between 1-6, only the numbers 3 and 6 are divisible by the number 3, the answer is $ \left(\dfrac{1}{3}\right)^3\equal{}\boxed{\dfrac{1}{27}}$.
[/hide]

Whoops! Sorry, I misinterpreted the problem.  :blush:

The answer for number 2 (B) should still be 1/3 though

Its not just all 3 in 0 mod 3

It can be all 3 in 1 mod 3 (which is 8/27) 1mod 3+1 mod3+ 1mod3=3mod3=0mod 3 divisible by 3.

4^2+2^2+6^2=84, which is divisible by 3...

So its 8/27+1/27=9/27=$ \frac{1}{3}$

Forget the other cases since we cant have 2 mod 3 as squares

So you still got the right answer


geometry
angle bisector
geometry unsolved

In olympiad problems from around the world, it has the following problem:
In convex quadrilateral ABCD, $ AD \equal{} CD$ and $ \angle DAB \equal{} \angle ABC$. The line through D and the midpoint of BC (M) intersect AB at E. Prove that $ \angle BEC \equal{} \angle DAC$.

[hide="Solution"]Let G be the intersection point of AD and BC, and H be the intersection point of AB and CD. Let E' be a point on AB such that B is inside AE' and $ \angle AE'C \equal{} \angle CAD$. Let F be the intersection point of DE' and BC. From Menelaus theorem $ \frac {GA\cdot DH\cdot CB}{DA\cdot CH\cdot GB} \equal{} 1 \equal{} \frac {HD\cdot BE' \cdot CF}{CD\cdot BE'\cdot BF}\iff \frac {CB}{CH} \equal{} \frac {BE'\cdot CF}{HE'\cdot BF}$ (1). ACH is similar to HCE' so $ \frac {HC}{HE'} \equal{} \frac {AC}{CE'}$ (2). $ \angle ACE' \equal{} 180^{\circ} \minus{} \angle ABC$. Hence ACE' is similar to BCE' => $ \frac {AC}{CE'} \equal{} \frac {BC}{BE'}$ (3). From (1),(2) and (3) => BF=CF.[/hide]

let AB intersect CD in J
AD intersect BC in I
DE intersect BC in N
EC intersect AI in F

by theorem of Menelai for triangle BCJ and points N, D, E

BN/NC.CD/DJ.JE/EB=1, but BN=NC, then JD/DC=JE/EB;

by theorem of Menelai for triangle IDC and points A, J, B

IA/AD.DJ/JC.CB/BI=1, but IA=IB, then JD/DC=JC/CB;

JC/CB=JD/DC=JE/EB, then JC/CB=JE/EB then CE is the bisector of angle DCI
Let angle DCI=2a; angle AIB=2b, then angle ABI=90-b, then angle EBC=90+b, angle BCE=angle FCI=a, then angle BEC=90-a-b
angle CAD= angle DCA=angle CDI/2=(180-2b-2a)/2=90-a-b=angle AEC, then

angle CAD= angle DCA=angle AEC, then (AEC) touthes AI and DC at points A and C.

[b]To limes123 [/b](to shorten little bit the proof):

By the isosceles triangles ABG and ACD, from triangle ACG we easily get <DCG = 2*<BAC, then draw CE' the angle bisector of <BCH, see that the triangles CE'H and ACH are similar; further, continue your way, but already got HE'/BE' = HC/BC.

Best regards,
sunken rock


Hi!

Let ${\cal P}$ be an $n$-gon $A_1A_2\cdots A_n$ such that:
\[
A_1A_2=A_2A_3=\cdots=A_nA_1.
\]

(1) what is the maximum area of ${\cal P}$.

(2) compute
\[
\max{{\hbox{area}({\cal P})}\over{\hbox{perimetere}^2 ({\cal P})}}
\]

First of all, you shouldn't have posted it in the "Solved Problems" section. This is for problems which have already been discussed :).

As for the problem itself, it (something more general than this, actually) has been discussed before. The maximal area occurs for cyclic polygons, and, in the case of equilateral polygons, this means they must be cyclic. Try searching for it, using keywords such as "cyclic polygon", or "isoperimetric inequality", stuff like that.

[quote="grobber"]]First of all, you shouldn't have posted it in the "Solved Problems" section. This is for problems which have already been discussed :).
[/quote]
OK!! :P  :P

[quote="grobber"]

As for the problem itself, it (something more general than this, actually) has been discussed before. The maximal area occurs for cyclic polygons, and, in the case of equilateral polygons, this means they must be cyclic. Try searching for it, using keywords such as "cyclic polygon", or "isoperimetric inequality", stuff like that.[/quote]

All right :P :P

[quote="ddziabenko"]Actually you can find it right [url=http://mathworld.wolfram.com/IsoperimetricTheorem.html]here[/url]
How do you prove it though?[/quote]

Thank you very much ddziabenko :)


$ABC$ is an acute-angled triangle with $\measuredangle A = 30$. $H$ is the orthocenter and $M$ is the midpoint of $BC$. $T$ is a point on $HM$ such that $HM = MT$. Show that $AT = 2 BC$.

Of course you mean that $T$ is such that $H \neq T$.

We use the following lemma: let $ABC$ be a triangle, $H$ its orthocenter, $O$ the circumcenter and $A'$ the point on the circumcircle opposite $A$. Then $HA'$ and $BC$ bisect each other, that is, $HBA'C$ is a parallelogram. To prove it, note that $BH, CH$ are perpendicular to $AC, AB$ respectively, and $A'C$ is perpendicular to $AC$ because $AA'$ is a diameter; similarly $A'B$ is perpendicular to $AB$. Then $BH, A'C$ and $CH, A'B$ are parallel.

The converse is clearly true, because if we take $M$ such that $BM=MC, HM=MT$, then $T$ coincides with $A'$. Then $T$ is opposite $A$ through $O$, that is, $AT$ is a diameter. Also, since $ \measuredangle A = 30$, we have $ \measuredangle BOC=60$ and $BO=OC$, from which $BOC$ is equilateral and $BC$ is a radius. We're done.


If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?

$\text{(A)}\ 32 \qquad \text{(B)}\ 34 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 36\qquad \text{(E)}\ 38$

My solution for this problem isn't rigorous, but I guess it works.  :)

[hide="Solution"]First, let's examine the prime factorizations of the number of divisors of $2n$ and $3n$.

$2n$ has $28=7(2^2)$ divisors, so it must be in the form $(x)(y)(z^6)$ or $y^3(z^6)$. The form $x^{27}$ obviously doesn't work. $3n$ has $30=2(3)(5)$ divisors, so it must be in the form $x(y^2)(z^4)$, $y^5z^4$, or some other forms that don't make sense.

Some quick experimentation with each form shows us that $n=2^53^3$, so that $6n=2^63^4$ and has $7\cdot 5=35$ divisors.
[/hide]

Can somebody provide a more elegant solution for this problem? Thanks.  :)

I think your solution is pretty quick. Sometimes there isn't a faster solution than just working with a couple of cases.  :lol:

I solved the same way as you, but I have a slightly more rigorous reasoning:
we see that n must be made up of twos and threes, because other wise by taking 2n and 3n respctively you are adding a new prime factor, which doubles the number of divisors. Since 28 and 30 are so close, this can't be the case. Now we see that 4*7=28 and 5*6=30, which naturally leads to the answer (we see n must have 4*6 divisors and 6n has 5*7 divisors).
Its really the same as what you did.

[hide]For both statements to be true, the prime factorization of $n$ must be ${2^5}{3^3}$ (Feel free to try and form a counterexample, you can't.) Thus the factorization of $6n$ is ${2^6}{3^4}$ and the number of factors is $(6 + 1)(4 + 1) = 35 \implies \boxed{C}$[/hide]

[hide="another way"]

Sorry, this is somewhat hard for me to explain, but I'll have a go at it.  In general, the number of divisors of a number $n = a_1^{p_1}a_2^{p_2}\dotsc a_n^{p_n}$ is given by $(p_1 + 1)(p_2 + 1)\dotsc (p_n+1)$ (A good explaination for this is in the AoPS books).  Let $f(n)$ be the number of factors $n$ has.

Now, $f(2n) = f(n)\frac{b+2}{b+1} = 28$ where $b$ is the number of 2s in the prime factorization of $n$.

And similarly, $f(3n) = f(n)\frac{c+2}{c+1} = 30$ where $c$ is the number of 3s in the prime factorization of $n$.

To find $f(6n)$ we need to find the value of $f(n)\frac{b+2}{b+1}\frac{c+2}{c+1}$.  It follows from multiplying the first two equations together that $f(n)^2\frac{b+2}{b+1}\frac{c+2}{c+1} = 840$.  Therefore, 840 must be divisible by $f(n)\frac{b+2}{b+1}\frac{c+2}{c+1}$, so 840 must be divisible by the answer.  By checking divisibility, it's clear that 35 (C) is the only possible solution of the 5.  Unfortunately, this isn't very sound because it depended on the fact that the question was multiple choice.

It's interesting to note that we can determine from this that $\frac{b+2}{b+1} = \frac76$ and that $\frac{c+2}{c+1} = \frac54$, which confirms LordOfTheMorons' argument that the prime factorization of $n$ must be $2^5*3^3$ (since using higher powers would cause more factors and so would adding more prime factors).

Can anyone come up with a way to show that $2^5*3^3$ is the only possible prime factorization without knowing the solution to the problem? Or a more elegant solution in general?

[/hide]

[hide]Let $n=p_{1}^{a_{1}-1}p_{2}^{a_{2}-1}p_{3}^{a_{3}-1}\cdots$, where $p_{k}$ is the $k$th prime number.  Then $n$ has $a_{1}a_{2}a_{3}\cdots$ divisors.  Let $x=a_{3}a_{4}a_{5}\cdots$ for simplicity, so that $n$ has $a_{1}a_{2}x$ divisors.

$2n$ has $(a_{1}+1)a_{2}x$ divisors, so $(a_{1}+1)a_{2}x=28$.  Also, $3n$ has $a_{1}(a_{2}+1)x$ divisors, so $a_{1}(a_{2}+1)x=30$.

Subtracting the two equations above, we have $a_{1}x-a_{2}x=2$, or $x(a_{1}-a_{2})=2$.

If $x=2$, then our two equations become $a_{2}(a_{1}+1)=14$ and $a_{1}(a_{2}+1)=15$.  There are no solutions to these equations in positive integers $a_{1}$ and $a_{2}$, so $x$ must equal $1$.

Our two equations are $a_{2}(a_{1}+1)=28$ and $a_{1}(a_{2}+1)=30$.  Therefore, $a_{1}=6$ and $a_{2}=4$, and $6n$ has $(a_{1}+1)(a_{2}+1)x=\boxed{35}$ divisors.[/hide]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38673&postorder=asc&start=400]Welcome to the Basics Forum[/url]

[quote="matt276eagles"]Wow... that problem seems a little hard for the basics forum, don't you think?[/quote]

Not really; you just have to understand how to find the number of divisors, and how to solve equations.

Some of the people here overcomplicated the problem.

236factorial, I read your solution in that marathon.  I don't think you can just assume that the number contains no prime factors other than $2$ and $3$, because that affects the total number of divisors.


Hình học sơ cấp:
Câu 1 : Sử dụg tiên đề Hilbe của hình học Ơclit để chứng minh rằng : có những hình đòng dạng với nhau mà không phải là bằng nhau.
Câu 2 :
CMR có thể chọn được 6 điểm trên mặt phẳng sao cho khôg có bộ 3 nào tạo thành tam giác có góc lớn hơn $\frac{2\pi}{3}$
Câu 3 : cho tam diện vuông $Oxyz$ GỌi $A,B,C$là 3 điểm theo thứ tự trên cạnh $Ox, Oy, Oz$ . giả sử $A$cố định ,$B$và $C$lưu động sao cho : $OB+OC = const$. Tìm quỹ tích trọng tâm của tam giác $ABC$
 Với $OA=1, OB+OC=2 .$ Dựng tam giác $ABC$ với đường trung tuyến hạ từ $A$của tam giác$ABC$ có độ dài bằng m cho trước.
Câu 5: cho vòng tròn $C$ tâm $O$ đường kính $AB=2R$. Gọi $S$là hình cầu nhận $C$ làm 1 vòng lớn .$M$ là điẻm trong không gian $AM$ cắt $S$ tại 1 điểm thứ nhì$N$.
Gọi $\delta$ là 1 đường thẳng ở trong $P$ Khoảng cách từ$B$ đến$\delta$ là $d$.
a)	giả sử $M$chạy trên $\delta$. CMR: N vạch nên 1 vòng tròn tâm I bán kính $r$, tính $r$ theo $R$ và $d$. Tìm quĩ tích của I khi $\delta$ thay đổi ở trong mặt phẳng$P$mà $d$ không đổi.
b)	 GỌi$T$là hnìh trụ tròn xoay mà C là 1 thiết diện thẳng. Khi M vạch nên 1 đường sinh G và T thì N vạch nên 1 vòng tròn$\gamma$ Tìm quĩ tích tâm của $\gamma$ khi G thay đổi

Mấy cái bài này chú cứ post ra mục Olympiad ngoài kia là anh em họ làm thịt giúp ngay ấy mà. Chịu khó dịch nó ra English đi.

thế thì ai dịch giùm em với?  Cái khoản hình học em dịch không giỏi đâu ?
   :(  :(  :(


What is the area of an equilateral triangle when it's inscribed in a circle which has radius $10$?

[quote="matt276eagles"]Do you mean inscribed or circumscribed?[/quote]

Sorry, fixed it :blush:

[hide="Answer"]Given circle with center $O$, $\triangle ABC$, and radii $OA$, $OB$, $OC$. Drawing the perpendicular bisectors from $O$ to each side of the equilateral triangle, we form six 30-60-90 triangles, with the radius as the longer side, meaning that the middle side has length $5\sqrt{3}$, and each side of the equilateral triangle has length $10\sqrt{3}$. Therefore, the area of $\triangle ABC=\frac{(10\sqrt{3})^2\sqrt{3}}{4}=75\sqrt{3}$.[/hide]


I was wondering whether it would be a good idea or not.

I see [url=http://www.nytimes.com/1999/09/19/opinion/l-should-we-raise-the-minimum-wage-573795.html]this[/url] and then [url=http://www.helium.com/items/360265-why-we-should-not-raise-the-minimum-wage]that[/url]

What's right?

[color=green]Moderator says: I've fixed broken BBCode for you -- use the preview button![/color]

Increasing the minimum wage hurts people who earn more than minimum wage. Suppose a company pays has $ k$ workers and the company pays its workers at minimum wage. By increasing minimum wage, the company has to lay off workers increasing unemployment. Or, if the company wants to keep the workers, it increases the cost of its goods. There is also a relative decrease in wages of workers who earn more than minimum wage. For the workers who are earning minimum wage, since the cost of living is increased, they are not affected at all by the change. Another effect is that there is increased inflation which decreases the value of the dollar. Consequently it hurts the economy by reducing jobs, decreasing the value of the dollar, and does not help people earning minimum wage.

It causes unemployment, decreases profit margins for companies, and has a negative effect on GDP. However, the argument presented about "relative wages" and inflation looks confused. The wages are paid in money taken from a fixed pool of money, so that your co-worker gets paid more only affects you if it freshly printed money. If raised minimum wages increase inflation in the long run, it's only because economic growth is impeded.

There's also a moral point here: why should someone not be allowed to accept a job offer for whatever salary they please? Although some people turn this argument around to make it sound like something bad: "why should companies be allowed to pay whatever they want?". Those damned companies.

There are really two questions:  Should there be a minimum wage, and if so, what should that minimum wage be?

The standard argument against the minimum wage (and increases in the minimum wage) is that it (1) increases unemployment, and (2) raises the prices that consumers must pay.  Both of these assertions are backed by standard microeconomics, and imho it is difficult to argue that these things do not happen.

But the real question is:  Do the benefits of a minimum wage (or increase in minimum wage) outweigh the costs?  In order to answer this question, a reasonable person (with his or her own set of normative values) would want to know *how many* people will lose their jobs, *how much* prices will increase, and *how many* people’s wages will increase and by *how much* (among other things).  These questions are nearly impossible to answer, but there is scant evidence that minimum wage increases produce the crushing effect on unemployment and prices that is often asserted by minimum wage opponents (typically citing Economics 101, which theoretically makes a qualitative but not quantitative prediction here).

@Erdong171, you said, "For the workers who are earning minimum wage, since the cost of living is increased, they are not affected at all by the change."  This is an excellent example of how ridiculous the anti-minimum wage argument can get.  This statement is illogical on its face and is not supported by any valid economic argument.

@Kalle, the key "moral" issue that I see here is the one I described above.  We have to decide what policy we think is "better for society" using our own normative values.  You want to enshrine "Freedom to choose to toil away for wages that are insufficient to satisfy basic needs," as a basic human right, but others would argue for "Freedom to choose to toil away for wages that are sufficient to satisfy basic needs," as a basic human right.

As I tried to explain in another thread, I think it is an absurd idea that it is okay to deny some people work opportunities because some calculation shows that "the benefits outweigh the costs" (a calculation made at a societal level, I might add). Do you really think that an expected value calculation is the solution to all moral issues? I also cannot relate to how you on an emotional level feel that you or some group of people have the authority (perhaps via the democratic process) to decide over other people's lives just because they happen to live in the united states.

It seems to me that the obvious solution to this problem is for the government to agree to hire everyone for make-work jobs.  The default wage can be set at a living wage, with an optional form that can be filled out by those who wish to exercise their (apparently) fundamental human right to be paid at lower-than-subsistence wages.  But I'm sure that you can explain to me how the good that comes from this is clearly outweighed by the horrible human rights violation that is higher marginal tax rates.

We should also pause here to celebrate "the majestic equality of the laws, which forbids rich and poor alike to sleep under bridges."  [url=http://answers.google.com/answers/threadview?id=58976](cite)[/url]


The ratio of boys to girls in a room is $ 6: 5$. If three boys leave the room, the ratio is $ 1: 1$. How many girls are in the room?

We can express the number of boys as $ 6x$, and the number of girls as $ 5x$. Thus, we can say $ 6x \minus{} 3 \equal{} 5x \implies \minus{}3 \equal{} \minus{}x   \implies x \equal{} 3$. That means we have $ 5 \times 3 \equal{} 15$ girls.

Hi;

My answer is:

b = # of boys
g = # of girls
r = # number in the room

$ b \plus{} g \equal{} r \\
b \minus{} 3 \equal{} g \\
g \equal{} \frac {5}{6}b$

$ b \equal{} 18, g \equal{} 15, r \equal{} 33$


OMG guys its snowing, and its almost May!!!

OMG, arent you excited?

What sucks is that I'm delivering mulch the whole day, so I'm going to be dirty, cold, and wet by the time I get home  :(

But its snowing, so that makes everything up!!

It's freezing in here in Romania, when we usually wear shorts, or don't know how to install the air conditioner quicker :) It seems we'll also have some snow ...

[quote="sky9073"]No signs of snow here in Mexico City. :(
I haven't seen snow in 3 and a half years.[/quote]Mexico City :) So close to Merida :D

[quote="sky9073"][quote="Valentin Vornicu"]Mexico City :) So close to Merida :D[/quote]
Close? It takes 10+ hours to drive there.. (went two summers ago)
But I guess it's close compared to most other parts of the globe....[/quote]It takes 9 lives to swim there from here :lol:

lol, it isn't snowing in North Carolina. It barely snowed here... where is it snowing, hspotter2002?

It was hailing yesterday in central indiana.

Well, here in Atlanta it is a little cooler than a few days ago (two days ago it was 30 C = 86 F) and the weather channel says that tonight it is going to be 0 C = 37 F.

By the way, the weather is crazy. Usually here in Georgia when it rains, it rains strong for twenty minutes or so, and then it may rain very soft (that kind of rain that you can run a hundred yards under the rain and be still almost dry), but yesterday, it rained strong for eight hours (at least in Fulton County!) :?

It's been snowing here in Ohio.  The bad thing is is that my school has exactly 1 snow day every year, which already happened in December.

[quote="byrdgyrl"]Well, it isn't snowing here in Wyoming yet, although we got about 6" a couple days ago.  But I guess that isn't really news since it's snowed in the middle of July before...[/quote] :what?:  :?  :wacko: I thought that only happened in the South of Chile, Argentina and New Zeland!

It's snowing where I am, in northern Indiana. It's a big change from the 80 degree weather on monday and tuesday! Now it's 31, and the snow is sticking to the ground! BRR!

Es muy frio. :)

Yay!
Snow!
Penguins like me love snow.
I wish i weren't stuck in Georgia. :(

I'm in Ohio.  It was only snowing for a while though.  But it's still weird.

Here in Austin, Texas it's like 80 degrees.  I've only seen snow [b]once[/b] my entire life, a few years ago.  Living in Arizona and then moving to Texas does that to you. :P

I heard somewhere that Colorado had a big snowstorm not too long ago.  I thought it was always hot down there...

There's no snow here-and there probably won't be any more for a while. We had a couple of snow days back in winter because they couldn't get the roads cleared and the streets were really icy. Hmmm... I think yesterday it was at about 32 degrees F. Cold, but tolerable. Not ideal for a track meet though. Everyone was wearing sweatshirts and stuff... but we can't wear then when we're racing.  :(

 :)  :)  :)

Near Detroit, MI - horrible weather, been snowing for most of the night and still snowing now. Expected 4-9 inches by the end of Sunday.

[quote="djimenez"]Aughh!!

This morning I took a friend to the airport, and the temperature was 37 F, what shouldn't be that bad, but it was too windy... It felt like 20 or so![/quote]

we've had some really strong gusts up here. i was driving home from small group, and the wind nearly blew the car into the other lane. i suspect they were gusts of up to 50mph or thereabouts. maybe not, but it was pretty bad!

Cleveland, Ohio - Let go of winter!!!! We had like 1 1/2 weeks of phenomenal weather, and now this...this...[i]augh![/i] Well, at least today is a snow day. It's nice to have a day off.

[quote="juicybooty911"]don't get too much snow in Ga(southern part)
it is cold today[/quote]
Well, a little more to the north (Atlanta) it is not snowing either, but it was 32F this morning.


here's another nice problem from spoj:)...i am getting time limit exceeded..

As you are probably well aware, in Byteland it is always the military officer's main worry to order his soldiers on parade correctly. In Bitland ordering soldiers is not really such a problem. If a platoon consists of n men, all of them have different rank (from 1 - lowest to n - highest) and on parade they should be lined up from left to right in increasing order of rank.

Sounds simple, doesn't it? Well, Msgt Johnny thought the same, until one day he was faced with a new command. He soon discovered that his elite commandos preferred to do the fighting, and leave the thinking to their superiors. So, when at the first rollcall the soldiers lined up in fairly random order it was not because of their lack of discipline, but simply because they couldn't work out how to form a line in correct order of ranks. Msgt Johnny was not at all amused, particularly as he soon found that none of the soldiers even remembered his own rank. Over the years of service every soldier had only learned which of the other soldiers were his superiors. But Msgt Johnny was not a man to give up easily when faced with a true military challenge. After a moment's thought a solution of brilliant simplicity struck him and he issued the following order: "men, starting from the left, one by one, do: (step forward; go left until there is no superior to the left of you; get back in line).". This did indeed get the men sorted in a few minutes. The problem was solved... for the time being.

The next day, the soldiers came in exactly the same order as the day before, and had to be rearranged using the same method. History repeated. After some weeks, Msgt Johnny managed to force each of his soldiers to remember how many men he passed when going left, and thus make the sorting process even faster.

If you know how many positions each man has to walk to the left, can you try to find out what order of ranks the soldiers initially line up in?

Input
The first line of input contains an integer t<=50, the number of test cases. It is followed by t test cases, each consisting of 2 lines. The first line contains a single integer n (1<=n<=200000). The second line contains n space separated integers wi, denoting how far the i-th soldier in line must walk to the left when applying Msgt Johnny's algorithm.

Output
For each test case, output a single line consisting of n space separated integers - the ranks of the soldiers, given from left to right in their initial arrangement.

Example
Input:
2
3
0 1 0
5
0 1 2 0 1

Output:
2 1 3
3 2 1 5 4

time limit is 13 secs..please help
best regards
vasu

Well, I suppose an interval tree would work... (Try to figure out how to do it.)

Well... I think you should work from the end.

mahbub..what do u mean by working from the end...?
i already used a stack based approach fixing the maximum number ...and decrementing the maximum..it times out

I think you need to use BST. The idea is:

take the second example:
0 1 2 0 1

{1 to 5} See the last number, 1, move 1 to left. so it must be 4.
{1 to 3, 5}See the second last, 0, no move. so it is 5
{1 to 3}then 2, move 2 to left, 1
{2,3}then 1, move 1 to left, 2
{3}then 0, move 0, 3

move to left means: moving in array of the number that is left.

Now use BST to store which r visited and which r not. then move and blahblah... :D

thanx a lot mahbub...but one more thing...i got the idea basically...but i cant understand the thing abt the binary search tree...how dou exactly use it for this problem...sorry abt the incompetence
thanks agen
best regards
vasu

well, first make a BST with: 1 to n.

and make all the nodes active.

Now, begin from the last, by that u will understand which'th (i don't know what would have been suitable word) number soldier is he. then search it. and deactivate him.

u can probably understand by now that, u have to keep size field in every node. by activating and deactivating node means u update it's all parents siuze field.

what exactly is the running time of this algorithm using the BST.. is it nlgn..?


I'm having some confusion with net dipole moments. For example, does $ cis$ HBrC=CBrH have a net dipole moment? If so, can someone draw a visual about how to add the vectors up?

THere is a practice question:

Find the least acidic of the following compounds:

$ CH_{3}CO_{2}H$, $ H_{2}O$, $ HCl$, $ C_{2}H_{6}$, and $ NH_{3}$. We are not given $ pKa$ values, so how does one approach these problems? Drawing lewis structures?

Also, Can someone clarify what a lewis acid or base means? What exactly is a proton? Simply a $ H^{\plus{}}$ ion without the electron and just the proton?

Thanks.

The dipolar moment vector is directed from the least to the most electronegative atom on a bond. So, in cis-1,2-dibromoethene we have two vectors with equal intensity, each directed from C to Br: their "horizontal" components cancel, but their "vertical" components add up to yield a non-zero dipolar moment for the molecule.

[quote="Bythecliff"]so how does one approach these problems? Drawing lewis structures?[/quote]

Compare the strenght of the conjugate bases.

[quote="Bythecliff"]Can someone clarify what a lewis acid or base means[/quote]

A Lewis acid is any species capable of accepting a pair of electrons. A Lewis base is any species capable of making a bond with another species by providing a pair of electrons.

The hydrogen atom is made up of a proton around which there is an electron. An H+ ion is obviously only a proton.


calculus
integration
trigonometry
calculus computations

hello
[hide="solution"]
 if $ I_{n} \equal{} \int \frac {dx}{(x^2 \plus{} a^2)^n}$
now by applying the integration of parts ::
 $ I_{n \plus{} 1} \equal{} \frac {x}{2na^2(x^2 \plus{} a^2)^n} \plus{} \frac {2n \minus{} 1}{2na^2}I_{n}$
$ \int_0^{2(2 \plus{} \sqrt {3})} \frac {16}{(x^2 \plus{} 4)^2}\ dx$
$ \equal{} [\frac {2x}{x^2 \plus{} 4} \plus{} tan^{ \minus{} 1}(\frac {x}{2}]_0^{2(2 \plus{} \sqrt {3})}$
$ \equal{} \frac {1}{4}\plus{} tan^{ \minus{} 1}(2 \plus{} \sqrt3) \plus{} C$
[/hide]
thank u

What's value of $ \tan ^{\minus{}1}(2\plus{}\sqrt{3})?$

[quote="kunny"]What's value of $ \tan ^{ \minus{} 1}(2 \plus{} \sqrt {3})?$[/quote]
$ \frac{5\pi}{12}$

[quote="∫FaiLurE∮"]the answer is $ \frac {3 \plus{} 5\pi}{12}$
let $ x \equal{} 2tan (k)$,then$ \frac {16}{(x^2 \plus{} 4)^2} \equal{} cos^{4}k$
$ dx \equal{} 2sec^{2}kdk$,
when $ x \equal{} 2(2 \plus{} \sqrt {3})$, $ k \equal{} \frac {5\pi}{12},$
when$ x \equal{} 0,k \equal{} 0$
$ \int_0^{2(2 \plus{} \sqrt {3})} \frac {16}{(x^2 \plus{} 4)^2}\ dx$
$ \equal{} 2\int_0^{\frac {5\pi}{12}} cos^{2}kdk$
$ \equal{} \int_0^{\frac {5\pi}{12}} cos2k \plus{} 1 dk$
$ \equal{} [\frac {sin2k}{2} \plus{} k]_{0}^{\frac {5\pi}{12}}$
$ \equal{} \frac {sin \frac {5\pi}{6}}{2} \plus{} \frac {5\pi}{12} \equal{} \frac {3 \plus{} 5\pi}{12}$[/quote]

Yes, that's a typical solution for Japanese high school students.

Note that $ \boxed{\tan 22.5^\circ \equal{} \sqrt {2} \minus{} 1,\ \tan 15^\circ \equal{} 2 \minus{} \sqrt {3}}$

hello, note that $ \frac{1}{x^2\plus{}4}\equal{}\frac{1}{4}\left(\frac{1}{x^2\plus{}4}\minus{}\frac{x^2}{(x^2\plus{}4)^2}\right)$, so we have $ \int\frac{x^2}{(x^2\plus{}4)^2}\,dx\equal{}\minus{}\frac{x}{2(x^2\plus{}4)}\plus{}\frac{1}{2}\int \frac{1}{x^2\plus{}4}\,dx$, putting all things together we got $ \frac{1}{16}\arctan(\frac{x}{2})\plus{}\frac{x}{8(x^2\plus{}4)}$.
After some algebra we get the result $ \frac{1}{12}(3\plus{}5\pi)$.
Sonnhard.


For a number n , we have --

    omega(n) --> Number of Distinct Prime factors of n.
big-omega(n) --> Number of Prime Factors (Counting Multiplicity.)

So for example ,

    omega(18) = 2.
big-omega(18) = 3.

Clearly, omega(n) <= big-omega(n).

Now we have been given the two series that is we have been given omega(n) and also big-omega(n), and we need to calculate
the series given here -- http://www.research.att.com/~njas/sequences/A143519.

I am listing out some terms here --

                 n              omega                big-omega               Required term.

                  2                  1                         1                             1
                  3                  1                         1                             1
                  4                    1                         2                             -1
                  5                   1                         1                              1
                  6                   2                         2                              -2
                  8                   1                         3                             0
                  10                    2                          2                             -2

Clearly there exists a relation between the required term and the two series omega and bigomega but I am not able to figure that out.

Please suggest what the relation can be ?

Thanks.

What do you mean by a "relation"?  Neither value determines the other for a particular choice of $ n$.  Do you mean an asymptotic relation?


should a chromate ion be written as $ \ce{CrO4^{2\minus{}}}$ or $ \ce{CrO4^{\minus{}2}}$

Doesn't really matter.  Not sure what the official IUPAC rules are on this, but either one should be accepted.  I usually write \ce{CrO4^{2-}} because the numbers line up nicely.

actually what i meant was the former case normally means the Oxidation number of the central atom which  here isn't obviuosly $ 2\minus{}$ .
that is $ 2\minus{}$ represents oxidation state  and $ \minus{}2$ net charge

That number represents the charge of the ion and not the oxidation state of the metal. To give information about the oxidation state of a (transition) metal we use the Shock number: the oxidation state of the metal written in Roman and between brackets (). Also, it is common practice to write charges as "2-" and oxidation numbers as "-2".


abstract algebra
group theory
superior algebra
superior algebra unsolved

Can someone help me explain why the only simple, cyclic groups are those of prime order?

Cyclic groups are abelian, so a cyclic group of order $ pq$ has normal subgroups of orders $ p, q$.

subgroups of a cyclic group of order n correspond to the divisors of n.


superior algebra
superior algebra unsolved

Is $ \pi$ transcendental over $ \bar{\mathbb{Q}}$?

Yes. If it is then it is algebraic over $ \mathbb Q$.

N.T.TUAN means the right thing but his second sentence can be misunderstood.
Algebraicness is a tranistive property, thus if $ C|B|A$ are field extensions, then $ C|A$ is algebraic if and only if $ C|B$ and $ B|A$ are algebraic. Especially, being algebraic over $ \bar{\mathbb Q}$ is the same as being algebraic over $ \mathbb Q$ (so iff it is already in $ \bar{\mathbb Q}$).


Call a positive real number [i]special[/i] if it has a decimal representation that consists entirely of digits 0 and 7. For example, $ \frac{700}{99}\equal{}7.0707070707...$ and 77.007 are special numbers. What is the smallest [i]n[/i] such that 1 can be written as a sum of [i]n[/i] special numbers?

(A)7 (B)8 (C)9 (D)10 (E)1 cannot be represented as a sum of finitely many special numbers

[hide="Hint"] Phrase it this way:  write $ \frac{1}{7}$ as a sum of numbers that consist only of the digits $ 0$ and $ 1$.  What is the decimal expansion of $ \frac{1}{7}$? [/hide]


In a class of 28 students, the teacher selects four people at random to participate in a geography contest. What is the probability that this group of four students includes at least two of the top three geography students in the class? Express your answer as a common fraction.

[hide="Solution"] $\frac{_{25}C_{1}\times_{3}C_{3}+_{25}C_{2}\times_{3}C_{2}}{_{28}C_{4}}=\frac{37}{819}$.  :? I remember this question. I don't know if this is right, though. [/hide]

It's right, my answer key of truth says so.

I missed the problem because I made the mistake of going

3(ways to select at least 2)*26*25/2=975

but that double counts 50 of them as for example selecting 1,2 and then 3 is the same as selecting 2,3 and then 1

[quote="pgpatel"]it's wrong[/quote]
I looked, and answer key says it's right.

Never mind :D. I for some reason recall the answer being distinctly higher.

[hide]
Umm...
I think that the easiest way to do it is
$1-\frac{3\cdot C_{26,4}}{C_{28.4}}$. I don't know how to calculate that, though.
The fraction is the probability that there won't be two top students.[/hide]

[quote="ajai"]what exactly was used to solve the problem (the formula above)?[/quote]

The reasoning is as follows:

There are 3C2 ways to have exactly 2 of the top 3. Then we have to choose another 2 from the 25 people that are not in the top 3, so 25C2 * 3C2.

Likewise, there are 3C3 ways to have all 3 of the top 3, leaving room for us to choose 1 from the other 25, so 25 C 1.

There are 28C4 ways to pick 4 students, so that is the denominator.

We count the number of ways to choose a group of four students including at least two of the top three geography students. This is just $\binom{3}{2}\cdot \binom{25}{2} + \binom{3}{3}\cdot\binom{25}{1} = 925$, since we can choose either 2 or 3 of the top students to be in this group. In all, there are $\binom{28}{4} = 20475$ groups of four students. Thus our desired probability is $\frac{925}{20475} = \boxed{\frac{37}{819}}$.


Using the distance formula, we get:
$\cot{\frac{\sqrt{325}}{\sqrt{850}}}$
$\cot{\frac{\sqrt{850}}{\sqrt{153}}}$
$\cot{\frac{\sqrt{153}}{\sqrt{325}}}$

[quote="croatian_dan"]Find the angles of a triangle with vertices at (0,0), (17,6), and (-12,3)

thanks,
pierre[/quote]
This should go in HSb not classroom math.

Whoops, I'm wrong. It's not a right triangle, never mind.  :blush:

[quote="philB"]Try the shoelace formula...[/quote]

Shoelace is for side measurements...not angles...but u can use the Law of Sines/Cosines?

[quote="xpmath"]The Law of Sines should work...[/quote]

 :D  I keep forgetting which ones the laws work for  :)

Rule of Cosines would work too, but I'm too lazy to do it.

or one could just do the following, which should go in high school basics.

[hide]using determinants, one finds that the area is 123/4.  but we have that $[ABC] = \frac{ab\sin{\angle C}}{2}$, hence $\sin{\angle C}=$ whatever.i'm lazy.  you do the calculations.[/hide]

[quote="Teki-Teki"]...whatever.i'm lazy.  you do the calculations.[/hide][/quote]
Exactly.  :lol:


Is there a way to get some of the features (like centroid, orthocenter...ect) that is available on Geogebra on AoPS to your personal Geogebra...

I'm not a user of GeoGebra but that of Asymptote. But since they're similar, try looking for packages?

GeoGebra is not similar to Asymptote.  GeoGebra has a more user-friendly interface, meaning you can just plot points, with your mouse, it's much more visual.

With Asymptote, you type in code.

To Nandubetax:  Can't you just construct them?


counting
distinguishability
combinatorics unsolved
combinatorics

1) How many ways can I make a collection of 20 bottles from 8 different types?

2) Repeat the same problem with the condition: there must be between 4 and 7 bottles of EACH type in the collection.

1) the answer is the number of solutions in non-negative integers to $ x_1\plus{}x_2\plus{}\cdots\plus{}x_8\equal{}20$, which is $ \binom{20\plus{}8\minus{}1}{8\minus{}1}\equal{}\binom{27}{7}$ (the number of ways to arrange 20 indistinguishable $ 1$'s and 8-1 indistinguishable bars in a row, the bars being there to divide the 1's into 8 parts).

2) this is impossible, as there would have to be at least $ 8\cdot 4 \equal{} 32$ bottles. maybe the question was something else...

Thank you. I should have typed 40 bottles instead. I have reposted the problem.


Find the number of $ (a,b,c) \in N$ such that
$ abc|(a\plus{}1)(b\plus{}1)(c\plus{}1)$

(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1
From the equation above, we deduce:
ab+bc+ca+a+b+c+1=kabc with k is a natural number
=> 1/a+1/b+1/c+1/ab+1/bc+1/ca +1/abc=k
By assumming that a>=b>=c, we have k is less than or equal to 7/c
 :blush:  I can't think further at the moment, anybody helps ?  :blush:

[hide]
$ k \le 7/c \Rightarrow k \le 7$...i think you could just bash from there
[/hide]

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=250991[/url]


$ \sum_{r\equal{}0}^{q}{(^qC_r)(^NC_{n\minus{}r})(a^{N\minus{}n\plus{}r})}$
guys please tell me how to split the T_r to get a  telescoping sum
other methods are also invited
 :)


vector
geometry
3D geometry
sphere
geometry unsolved

If $P_{i}, (i=1,...,n)$ are points on a unit sphere, then $\sum_{i\leq j}|P_{i}P_{j}|^{2}\leq n^{2}.$

If $O$ is center of that sphere, then your ineq is $(OP_{1}+OP_{2}+...+OP_{n})^{2}\geq 0$. Done! :D


MATHCOUNTS
geometry
3D geometry
algebra
polynomial

Well, I wasn't quite sure if this problem was MATHCOUNTS level, but it's from a "sprint round" from a local competition. So, it should be. I doubt it's too easy--and I also doubt it's too hard. Here goes:

Given these two equations:
$ a \plus{} b \equal{} 7$
$ a^3 \plus{} b^3 \equal{} 42$

Solve for
$ \frac{1}{a} \plus{} \frac {1}{b}$

[hide="Answer"]$ \frac{21}{43}$[/hide]

We tried substituting in $ 7 \minus{} b$ in for $ a$, but that didn't work.
...I think you factor this somehow.
Our coach says that "Since it's a sprint round, it wouldn't be this messy. There must be some sort of trick or something."
>.< Any ideas?

$ \frac{1}{a}\plus{}\frac{1}{b}\equal{}\frac{a\plus{}b}{ab}$
Use that. It's helpful.

cube the first equation $ a^3\plus{}3a^2b\plus{}3ab^2\plus{}b^3\equal{}343$ factor the middle terms and substitute the second in. $ 42\plus{}3ab(a\plus{}b)\equal{}343$ solve for ab. $ ab\equal{}\frac{43} {3}$. divide the first equation by ab to get $ \frac{1} {a} \plus{} \frac{1} {b}\equal{}\frac{7} {ab}\equal{}\frac{7} {\frac{43} {3}}\equal{}\frac{21} {43}$

Hmm...cool. :D Thanks.

Is there a certain way to know when to factor or cube, or do you just mess around with the problem, or do you just know by experience? Just curious. :D

This is a type of problem you learn to recognize by experience.  Here's the key:  [b]you don't have to find $ a$ and $ b$ to find $ \frac {1}{a} \plus{} \frac {1}{b}$.[/b]

A specific piece of technique:  any symmetric polynomial $ f(a, b)$ in two variables (a polynomial such that $ f(a, b) \equal{} f(b, a)$) can be written as a polynomial in the variables $ u \equal{} a \plus{} b, v \equal{} ab$.  These are known as the [i]elementary symmetric polynomials in two variables[/i] (and the generalization to three variables is given by $ a \plus{} b \plus{} c, ab \plus{} bc \plus{} ca, abc$).  Here we have

$ u \equal{} 7$
$ u(u^2 \minus{} 3v) \equal{} 42$

and $ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {u}{v}$.

[b]Exercise:[/b]  Given that

$ x \plus{} y \equal{} 5$
$ x^2 \plus{} y^2 \equal{} 13$

find $ x^3 \plus{} y^3$.

[hide="Hint"] $ x^2 \plus{} y^2 \equal{} u^2 \minus{} 2v$. [/hide]


let f an automorphism of C from C

Do we have for any z in C , $f(e^{z})=e^{f(z)}$ ?

If $f$ is continuous, we do; the only continuous automorphisms are the identity and conjugation.

If $f$ isn't continuous, it doesn't work. We don't have to send $e$ to a real number, after all.

What you mean by automorphism?
Do you mean a comform automorphism,  is it $\frac{\partial f}{\partial \overline{z}}=0$ and bijective?


I'm confused about Hamiltonians.  What is an example of a system where the Hamiltonian is NOT equal to the energy of the system?

Possibly I'm misunderstanding your question, but the Hamiltonian is always an energy function for a system: $ H\equal{}T\plus{}U$, where $ T$ is kinetic energy and $ U$ the potential energy.

I think you really want to know is when the Hamiltonian equates total energy $ E$ of the system. A necessary condition for that is that the equations of motion must be independent of the time (not in a strictly sense) but time must not explicity appear in $ H$, i.e.: $ \frac{\partial H}{\partial t}\equal{}0$.
This is equivalent to the kinetic energy is an homogeneous quadratic function of the generalized coordinates $ \dot{q}_{j}$. In this case $  T\equal{}\sum_{j,k}a_{jk}\dot{q}_{j}\dot{q}_{k}$

For an example when $ H$ is not equal to $ E$, consider ths system generated by a pendulum of lenght $ l$ and mass $ m$, where the point supporting the pendulum is moves on a circle of radius $ a$ with constant angular velocity $ \omega$: $ x\equal{}acos(\omega t)\plus{}bsin(\phi)$ and $ y\equal{}asin(\omega t)\minus{}bcos(\phi)$.


[quote="Cezar Lupu"]Let $a, b, c$ three non-negative real numbers. Prove that the following inequality holds:

     \[ \frac{a^{2}+2bc}{b+c}+\frac{b^{2}+2ca}{c+a}+\frac{c^{2}+2ab}{a+b}\geq\frac{3}{2}(a+b+c).  \][/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=b%2Bc&t=85938
 :)

[quote="Cezar Lupu"]Let $a, b, c$ three non-negative real numbers. Prove that the following inequality holds:

     \[ \frac{a^{2}+2bc}{b+c}+\frac{b^{2}+2ca}{c+a}+\frac{c^{2}+2ab}{a+b}\geq\frac{3}{2}(a+b+c).  \][/quote]

  My proof is here. I don't know if it is as good as SOS of Nameless or not. But anyway, I found it:
After expanding, it is equivalent to:
$2\sum a^2(a-b)(a-c)+ \sum ac (a-c)^2 \geq 0$
 :D


limit
real analysis
real analysis unsolved

Let $(a_{n})_{n\geq 1}$ be reals sequence such that $e^{a_{n}}+na_{n}=2(\forall n\geq 1)$. Evaluate $\lim_{n\to\infty}n(1-na_{n})$.

For any $n$ , denote $f_{n}(x)=e^{x}+nx$. We have $f'_{n}(x)=e^{x}+n>0\forall x\in\mathbb{R}$ and $f_{n}(0)=1<2,f_{n}(\frac{2}{n})=e^{\frac{2}{n}}+2>2$ therefore exists $a_{n}$ and
$0<a_{n}<\frac{2}{n}$(1).

Now, since (1) we have $\lim_{n\to\infty}a_{n}=0$. Therefore $\lim_{n\to\infty}n(1-na)=\lim_{n\to\infty}\frac{(e^{a_{n}}-1)(2-e^{a_{n}})}{a_{n}}=1$.


What is the number of sides of a regular polygon for which the number of diagonals is 44 or 45?

[hide]
The number of diagonals in a regular polygon is $\frac{n(n-3)}{2}$ (prove it).
$\frac{n(n-3)}{2} = 44 \Rightarrow n=11$.
$\frac{n(n-3)}{2} = 45$ has no solutions.[/hide]


HMMT function logarithms algebra polynomial

In general,
[hide]$f: \mathbb{Q}-\{0,1\}\rightarrow\mathbb{R}\ni f(x)=\log\left|\frac{x}{x-1}\right|.$[/hide]

[quote="Phelpedo"]Let ${A : = Q\{0,1}$ denote the set of all rationals other than $0$ and $1$. A function $f: A-\to R$ has the property that for all $x \in A$,
\[f(x)+f(1-1/x) = \log|x| \]
Compute the value of $f(2007)$[/quote]

[hide]Notice that

$f(x)+f(\frac{x-1}{x})=\log|x|$

$f(\frac{x-1}{x})+f(\frac{1}{1-x})=\log|\frac{x-1}{x}|$

$f(\frac{1}{1-x})+f(x)=\log|\frac{1}{1-x}|$

So we can solve the equation that

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

So we get that this is just

$f(x)=\log|\frac{x}{x-1}|$

So $f(2007)=\log|\frac{2007}{2006}|=\log{\frac{2007}{2006}}$[/hide]

man you got to know your cyclic functions

those are awesome

at first you are like OMG THIS IS SO RANDOM

but then you are like woa, it simplifies really nicely

this one was kinda easy for a #7 or whatever it was...it was obviously cyclic something or else it'd be IMPOSSIBLE so then it was like a #2 difficulty.

[quote="Elemennop"][quote="Altheman"]i agree that it was pretty easy in that sense...but by that definition, all the problems are easy...almost all problems are just like one or two tricks[/quote]

yea i guess but cyclic functional eq's are one of the most basic tricks, as opposed to symmetric sum polynomials (#9) or mass products of complex polynomials with w factoring (#10).[/quote]

i guess...but i ono...#9 there was another way to do it that resembled another problme, like if you have a poly, find another poly with roots ab, etc, then make u=x+2/x, so that is old

10, same as hmmt 2004 #10

so like...i ono...all of the tricks on the test were "standard"...i just timed out...

[quote="Elemennop"]yea i did think it was kinda cheesy that #10 was the same as 2004

too bad that even konwing that I couldn't figure out how to do it[/quote]

lol yea...i was like OK I HAVE SEEN THIS; and i have posed a similar problem

wtf its not factoring...i just did not think to mulitply by r_i/r_i...which was stupid

yea you have to factor it so you get like f(-w)f(-w^2) on top

[quote="chess64"]which reminds me... #1 combinatorics was EXACTLY the same (not even different numbers) as a problem from a previous year, don't remember which year.[/quote]
Yeah, I know.
I thought I was hallucinating, or that it wasn m't really a problem I was just being uninteligent.


Find all sequences $x_{1},...,x_{n}$ of distinct positive integers such that
$\frac{1}{2}=\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}+...+\frac{1}{x_{n}^{2}}$

Are you sure of this statement?

Until recently it was not known if there exists such a decomposition with more than $12$ terms in the sum...

A solution with $12$ terms is $(x_{1},\cdots,x_{12})=(2,3,4,6,7,9,12,14,21,36,45,60).$

Pierre.


We know that the jumbo jet can travel $ 710$ extra feet in $ 50$ extra seconds.

This means that in $ 20$ seconds, it traveled $ 710 \cdot \frac{2}{5} \equal{} 284$ feet.
Since it traveled the distance of the jumbo jet in $ 20$ seconds, we know that the jumbo jet is $ \boxed{284}$ feet long.

Sorry for not making this clear, but basically….

It takes $ 20$ seconds to travel the length of the plane.  Since it takes $ 70$ seconds to travel the length of the plane + the $ 710$ feet.  This means that it takes $ 50$ seconds to travel $ 710$ feet.  Then we multiply $ 710$ by $ \frac{20}{50}$ in order to find the length of the plane.

One trick to this problem is understanding that to achieve whats on the diagram, the plane travels the length of the hangar plus the length of the jet (x). So,
(710+x)/70 seconds=x/20
then, solve to get x=284


abstract algebra
topology
vector
group theory
advanced fields
advanced fields unsolved

I was wondering, are there any examples of a set $ X$ with
\[ \pi_1(X,x_0) \equal{} \mathbb{R}\]
for some fixed point $ x_0\in X$?

Here $ \mathbb{R}$ is treated as an additive abelian group (or $ \mathbb{Z}$-module)

If there is such an example, can $ X$ be realized in a finite dimensional space?

From Wikipedia:
Every group can be realized as the fundamental group of a connected CW-complex of dimension 2.
As a divisible group, infinite-dimensional over $ \mathbb{Q}$, this one comes out absurdly complicated. I don't actually know how to build even something with $ Q$ as its fundamental group.

Oh, and "addictive abelian group"? Fun typo.

[quote="jmerry"]Oh, and "addictive abelian group"? Fun typo.[/quote]
LOL!

Here's how you do it in general.

Take a free group of generators $ F$ for your group $ G$. Let $ X_1 \equal{} \bigvee_{f \in F} S^1_f$. Then $ \pi_1(X_1) \to G$ is a surjection.

Now, for every element in the kernel $ \alpha$ of this surjection, write attach a $ 2\minus{}$cell to $ X_1$ with attaching map $ \alpha$ to get a 2-skeleton $ X_2$. Then $ \alpha$ becomes zero in $ \pi_1(X_2)$. You can check that $ \pi_1(X_2) \equal{} G$ -- i.e. that only elements $ \alpha$ are killed by the attaching of these cells.

This is a dirty little trick that allows you to replace any space $ X$ with a CW-complex $ K$ and a weak homotopy equivalence $ K \to X$. (You can do a little better than this, even... but it's too late for me to remember it offhand).

Nukular:
Since this post is for the particular example $ G \equal{} \mathbb{R}$, can you write explicitly in this case, what is $ F$ and what are the element in the kernel?

Also, I have trouble making sense of an uncountable wedging of the $ S_f^1$'s...

It seems that for this example, we can only talk about it abstractly, but impossible to actually construct explicitly the set $ X$.

you can build $ F$ in a very "expensive" way, defining just $ F\equal{}\mathbb{R}$ and forcing all relations of the form $ x*y*z$* whenever $ x\plus{}y\plus{}z\equal{}0$, but that's quite "redundand" (if it has any sense to say it).

you can do it in a slightly "cheaper" way, assuming the axiom of choice: you can find a basis for $ \mathbb{R}$ as a $ \mathbb{Q}$-vector space, call it $ B_0$. now consider $ B\equal{}\{x/n|x\in B_0, n\in\mathbb{N}_0\}$. you then have the presentation $ \mathbb{R}\equal{}\langle B | (x/n)^n*x\rangle$*. what we're really doing here is a wedge of uncountably many CW-complexes having $ \mathbb{Q}$ as a fundamental group.

* i'm using the $ *$ to denote the operation on the free module, and i use multiplicative notation, just because it gets more clear.

I think a simple wedge just gets you a free product; you have to build in commutativity relations as well.

sure, you're right. i was just confusing "free" things... thanks jmerry.

This could be a considerably easy construction of a pointed space $ (X,x_0)$ such that $ \pi_1(X,x_0)\cong{\mathbb Q}$. Take an countably infinite disjoint sum of cylinders $ {\mathbb S}^1\times [0,1]$ and glue the bottom of the $ n$-th cylinder along the top of the $ n\plus{}1$-th cylinder via tha map $ z\mapsto z^n$. Finally, collaps all line segments $ \{1\}\times [0,1]$ in the cylinders to a point and take this as a basepoint.

This could be (?) a pointed space with the desired property, though I haven't checked the details.  :roll:  :roll:


Let $ ABC$ be a triangle for which $ a\ge b$ and $ a\ge c$ . Prove that exists the equivalence

$ \left(\sqrt {a\plus{}b}\plus{}\sqrt {a\minus{}b}\right)\cdot \left(\sqrt{a\plus{}c}\plus{}\sqrt{a\minus{}c}\right)\equal{}(a\plus{}b\plus{}c)\sqrt 2\ \Longleftrightarrow\ A\equal{}90^{\circ}$ .

Both sides of the equation are positive, so we can square them and the equation is still equivalent. The if part is easy. The only if part can be proved by contradiction. Consider two cases, $ a^2<b^2\plus{}c^2$ or $ a^2>b^2\plus{}c^2$.

Please, can present your proof of [b]only if part[/b] ? Thanks.

squaring 2 sides of the equality we get
$ (a^2\minus{}b^2\minus{}c^2)(1\plus{}\frac{2a}{\sqrt{a^2\minus{}c^2}\plus{}b}\plus{}\frac{2a}{\sqrt{a^2\minus{}b^2}\plus{}c}\plus{}\frac{2a^2}{\sqrt{a^4\minus{}a^2c^2\minus{}a^2b^2\plus{}b^2c^2}\plus{}bc})\equal{}0$ so we complete the solution

[quote="Virgil Nicula"]Let $ ABC$ be a triangle for which $ a\ge b$ and $ a\ge c$ . Prove that exists the equivalence

$ \left(\sqrt {a \plus{} b} \plus{} \sqrt {a \minus{} b}\right)\cdot \left(\sqrt {a \plus{} c} \plus{} \sqrt {a \minus{} c}\right) \equal{} (a \plus{} b \plus{} c)\sqrt 2\ \Longleftrightarrow\ A \equal{} 90^{\circ}$ .[/quote]
$ A \equal{} 90^\circ \Rightarrow a^2\equal{}b^2\plus{}c^2$. After squaring we get $ 2(a\plus{}\sqrt{a^2\minus{}b^2})(a\plus{}\sqrt{a^2\minus{}c^2}) \equal{}$ $ (a\plus{}b\plus{}c)^2 \iff 2(a\plus{}c)(a\plus{}b) \equal{} (a\plus{}b\plus{}c)^2 \iff a^2\equal{}b^2\plus{}c^2$. So $ A \equal{} 90^{\circ} \Longrightarrow \left(\sqrt{a\plus{}b}\plus{}\sqrt{a\minus{}b}\right)\cdot\left(\sqrt{a\plus{}c}\plus{}\sqrt{a\minus{}c}\right) \equal{} (a\plus{}b\plus{}c)\sqrt 2$
The other direction:
Let $ f(x) \equal{} \sqrt{a\plus{}x}\plus{}\sqrt{a\minus{}x}$. Then $ f'(x) \equal{} \frac{1}{2\sqrt{a\plus{}x}} \minus{} \frac{1}{2\sqrt{a\minus{}x}} < 0$.
Fix $ a,b$. Then $ LHS$ is decreasing as function of $ c$, since $ LHS \equal{} f(b)f(c), f'(x)<0$, and $ RHS$ is increasing as function of $ c$. Hence there is at most one solution. $ a^2\equal{}b^2\plus{}c^2$ is a solution by the argument given above. And we are done.

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle for which  $ a\ge b$ si $ a\ge c$ . Prove that

$ \left(\sqrt {a \plus{} b} \plus{} \sqrt {a \minus{} b}\right)\cdot \left(\sqrt {a \plus{} c} \plus{} \sqrt {a \minus{} c}\right) \equal{} (a \plus{} b \plus{} c)\sqrt 2\ \Longleftrightarrow\ A \equal{} 90^{\circ}$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ X \equal{} (a \plus{} b)(a \plus{} c)$ , $ Y \equal{} (a \minus{} b)(a \minus{} c)$ , $ Z \equal{} (a \minus{} b)(a \plus{} c)$ , $ T \equal{} (a \plus{} b)(a \minus{} c)$ .

Prove easily that  $ 2p^2 \minus{} X \equal{} 2(p \minus{} a)^2 \minus{} Y \equal{} 2(p \minus{} b)^2 \minus{} Z \equal{} 2(p \minus{} c)^2 \minus{} T \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2} .$ Thus

$ \left(p\sqrt 2 \minus{} \sqrt X\right)\ .s.s.\ \left[(p \minus{} a)\sqrt 2 \minus{} \sqrt Y\right]\ .s.s.\ \left[(p \minus{} b)\sqrt 2 \minus{} \sqrt Z\right]\ .s.s.\ \left[(p \minus{} c)\sqrt 2 \minus{} \sqrt T\right]\ .s.s.\ \frac {b^2 \plus{} c^2 \minus{} a^2}{2}$ .

Therefore, $ \left(p\sqrt 2 \minus{} \sqrt X\right) \plus{} \left[(p \minus{} a)\sqrt 2 \minus{} \sqrt Y\right] \plus{}$ $ \left[(p \minus{} b)\sqrt 2 \minus{} \sqrt Z\right] \plus{} \left[(p \minus{} c)\sqrt 2 \minus{} \sqrt T\right]\ .s.s.\ \frac {b^2 \plus{} c^2 \minus{} a^2}{2}$ ,

i.e. $ (a \plus{} b \plus{} c)\sqrt 2 \minus{} \left(\sqrt {a \plus{} b} \plus{} \sqrt {a \minus{} b}\right)\cdot \left(\sqrt {a \plus{} c} \plus{} \sqrt {a \minus{} c}\right)\ .s.s.\ \frac {b^2 \plus{} c^2 \minus{} a^2}{2}$ . In conclusion,

$ A\ \begin{array}{c} < \\
\ \equal{} \\
\ > \end{array}\ 90^{\circ}\ \Longleftrightarrow\ b^2 \plus{} c^2\ \begin{array}{c} > \\
\ \equal{} \\
\ < \end{array}\ a^2\ \Longleftrightarrow$ $ \underline{\overline{\left\|\ \left(\sqrt {a \plus{} b} \plus{} \sqrt {a \minus{} b}\right)\cdot \left(\sqrt {a \plus{} c} \plus{} \sqrt {a \minus{} c}\right)\ \begin{array}{c} < \\
\ \equal{} \\
\ > \end{array}\ (a \plus{} b \plus{} c)\sqrt 2\ \right\|}}$ .

[b]Remark.[/b] I used the notation $ X\ \mathrm{.s.s.}\ Y\ \Longleftrightarrow\ XY > 0$ or $ X \equal{} Y \equal{} 0$ ([b]s[/b]ame [b]s[/b]ign).[/color]

\[ \left(\sqrt{a\plus{}b}\plus{}\sqrt{a\minus{}b}\right)\cdot\left(\sqrt{a\plus{}c}\plus{}\sqrt{a\minus{}c}\right)\equal{}(a\plus{}b\plus{}c)\sqrt 2\Leftrightarrow2(a\plus{}\sqrt{a^2\minus{}b^2})(a\plus{}\sqrt{a^2\minus{}c^2})\equal{}(a\plus{}b\plus{}c)^2\]

If $ A\equal{}90^{\circ}$, then \[ 2(a\plus{}\sqrt{a^2\minus{}b^2})(a\plus{}\sqrt{a^2\minus{}c^2})\equal{}2(a\plus{}c)(a\plus{}b)\equal{}2a^2\plus{}2ab\plus{}2bc\plus{}2ca\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2bc\plus{}2ca\equal{}(a\plus{}b\plus{}c)^2\]

If $ A\ne90^{\circ}$, consider two cases. (i) $ a^2>b^2\plus{}c^2$, then \[ 2(a\plus{}\sqrt{a^2\minus{}b^2})(a\plus{}\sqrt{a^2\minus{}c^2})>2(a\plus{}c)(a\plus{}b)\equal{}2a^2\plus{}2ab\plus{}2bc\plus{}2ca>a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2bc\plus{}2ca\equal{}(a\plus{}b\plus{}c)^2\] (ii) $ a^2<b^2\plus{}c^2$, then \[ 2(a\plus{}\sqrt{a^2\minus{}b^2})(a\plus{}\sqrt{a^2\minus{}c^2})<2(a\plus{}c)(a\plus{}b)\equal{}2a^2\plus{}2ab\plus{}2bc\plus{}2ca<a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2bc\plus{}2ca\equal{}(a\plus{}b\plus{}c)^2\] So if $ A\ne 90^{\circ}$, then $ 2(a\plus{}\sqrt{a^2\minus{}b^2})(a\plus{}\sqrt{a^2\minus{}c^2})\ne(a\plus{}b\plus{}c)^2$.


linear algebra
matrix
function
vector
parameterization
logarithms
analytic geometry

Consider the parametrizations $ (U,V)\mapsto (U\cosh{V},u\sinh{V},U^{2})$ of the surface $ S: \equal{}\{(x,y,z)\in\mathbb{R}^{3}\mid z\equal{}x^{2}\minus{}y^{2}\}$ and $ (u,v)\mapsto (u,v,uv)$ of the surface $ \tilde{S}: \equal{}\{(x,y,z)\in\mathbb{R}^{3}\mid z\equal{}xy\}$.  Show that left multiplication by the matrix $ T\equal{}\begin{pmatrix}1&\minus{}1&0\\
1&1&0\\
0&0&1
\end{pmatrix}$ defines a linear transformation from $ \mathbb{R}^{3}$ to itself, mapping $ \tilde{S}$ one-to-one onto $ S$.  Use the diagram $ (u,v)\rightarrow\tilde{S}\rightarrow S\leftarrow (U,V)$ to evaluate $ U$, $ V$, as functions of $ u$, $ v$.

Now if you actually multiply $ T$ by $ (u,v,uv)$, you get $ (u\minus{}v,u\plus{}v,uv)$, which is not quite in $ S$ (off by a factor of 4).  Is it possible get rid of these factors (I'm not sure it is)?  And then how would one get $ U$ and $ V$ as functions of $ u,v$?  Would we just invert the matrix or something?

To get rid of those factors, you need to change something; I'd probably switch $ T$ to either $ \begin{pmatrix}\frac12&\frac12&0\\ \frac12&\minus{}\frac12&0\\0&0&1\end{pmatrix}$ or $ \begin{pmatrix}1&1&0\\1&\minus{}1&0\\0&0&4\end{pmatrix}$. I'm also transposing; if you're feeling generous, that sign mistake might just be about using row vectors on the left rather than column vectors on the right.

Getting $ U$ and $ V$ is not a linear operation; we have to invert the parametrization, which involves logarithms. If we use the version of $ T$ with $ \frac12$ coefficients, we get $ U^2\equal{}uv$, so $ U\equal{}\pm\sqrt{uv}$... uh, we have a problem. We can choose the sign as $ U\equal{}\text{sgn}(u)\sqrt{uv}$, but we miss the points on $ S$ with $ z$ zero or negative entirely; that's not a complete parametrization of $ S$. On the part it does cover, we then have $ Ue^V\equal{}U\cosh V\plus{}U\sinh V\equal{}\left(\frac12u\plus{}\frac12v\right)\plus{}\left(\frac12u\minus{}\frac12v\right)\equal{}u$, and $ e^V\equal{}\sqrt{\frac{u}{v}}$, making $ V\equal{}\frac12\ln\frac{u}{v}$.

Whoever wrote this problem should have done a lot more proofreading. There is a linear bijection between the two surfaces, but it's not exactly the one given by that matrix and these parametrizations aren't the way to go after it.

Yes, there are problems with it and the professor admitted he didn't think it was completely correct.  The goal is to compute to the Jacobian for $ U$ and $ V$ (w.r.t. $ u$ and $ v$); are you suggesting that it would not be possible?

The Jacobian makes sense; just keep in mind that $ U,V$ parametrize only half of their surface.

Alas, it seems that the mistakes have been (hopefully) corrected.  Here is how it now reads:

Consider the parametrizations $ (u,v)\mapsto (u\cosh{v},u\sinh{v},\frac{u^{2}}{2})$ of the surface $ S: \equal{}\{(x,y,z)\in\mathbb{R}^{3}\mid 2z\equal{}x^{2}\minus{}y^{2}\}$ and $ (U,V)\mapsto (U,V,UV)$ of the surface $ \tilde{S}: \equal{}\{(x,y,z)\in\mathbb{R}^{3}\mid z\equal{}xy\}$. Show that left multiplication by the matrix $ T\equal{}\begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\ \minus{}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\0&0&1 \end{pmatrix}$ defines an orthogonal linear transformation from $ \mathbb{R}^{3}$ to itself, mapping $ \tilde{S}$ one-to-one onto $ S$. Use the diagram $ (u,v)\rightarrow S\rightarrow\tilde{S} \leftarrow (U,V)$ to evaluate $ U$, $ V$, as functions of $ u$, $ v$, and use this to calculate the Jacobian matrix $ \textbf{U}'(\textbf{u})\equal{}\begin{pmatrix}U_{u}&U_{v}\\ V_{u}&V_{v} \end{pmatrix}$.

My main question is how do we recover $ U$ and $ V$ as functions of $ u$ and $ v$ so that we can compute the Jacobian?

You read $ U$ and $ V$ off from the first and second coordinates after applying $ T$. The new version is the easy direction.

I'm sorry, but could you clarify that last statement?  After applying $ T$ to $ (u\cosh{v},v\sinh{v},\frac{u^{2}}{2})$ and reading off the first two terms in the resulting vector?


Your notation is incomprehensible to me.  How does the outer sum in that double sum interact with the inner one?  (This is not the only reason I don't understand your post -- it would be really helpful if you chose intelligible notations and provided sufficient information to make your post self-contained.)

fair enough and i apologize

let me try again

$ n$ is the order of a connected graph
$ m$ is the edge number
$ m \ge n \minus{} 1$
let $ d_1,\ldots,d_n\neq 0$ be the vertex degrees
we know that $ \sum_{i \equal{} 1}^n d_i \equal{} 2m$
let $ I \equal{} \{\{d_1,\ldots,d_n\}\vert \sum_{i \equal{} 1}^n d_i \equal{} 2m\quad \mbox{and}\quad d_1,\ldots,d_n\neq 0\}$ be the set of partitions of $ 2m$ in $ n$ integers

how to write $ \sum_{\{d_1,\ldots, d_n\}\in I}\sum_{i \equal{} 1}^n 2^{d_i}$ in terms of $ m$ and $ n$?

does it make sense now :maybe:

It still lacks precision, and has a probable typo: with $ n > m \minus{} 1$ there are few connected graphs, since any connected graph contains at least $ m \geq n \minus{} 1$ edges (when a tree).
Also, the fact that the $ d_i$'s represent the degree structure of a graph adds restrictions to just having $ \sum_{i \equal{} 1}^n d_i \equal{} 2m$; later on, in your double sum, should we assume the $ d_i$'s are degrees of some connected graph? If not, why even speak about graphs?

yes i meant $ m\ge n \minus{} 1$ sorry. it's corrected already

 [quote="mavropnevma"]
Also, the fact that the $ d_i$'s represent the degree structure of a graph adds restrictions to just having $ \sum_{i \equal{} 1}^n d_i \equal{} 2m$; later on, in your double sum, should we assume the $ d_i$'s are degrees of some connected graph? [/quote]

the sum of the degrees of a grph without loops is twice the edge number. Since the graphs are connected, there are no vertices of zero degree

 [quote="mavropnevma"]
If not, why even speak about graphs?[/quote]
of course, one can formulate the problem without speaking about graphs


Let $ a,b,c$ be the positive real numbers.Prove that:
$ \frac{ab^3}{c}\plus{}\frac{bc^3}{a}\plus{}\frac{ca^3}{b}\plus{}3abc$≥$ 2(a^2b\plus{}b^2c\plus{}c^2a)$

If you divide both sides by $ abc$, and after substituting $ x \equal{} \dfrac{a}{b}, y \equal{} \dfrac{b}{c}$ and $ z \equal{} \dfrac{c}{a}$ the inequality becomes equivalent to:

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} 3\ge 2(xy \plus{} yz \plus{} zx)$,  for $ xyz \equal{} 1$.

But this is an application of the following one:

[quote]Let $ a,b,c\ge 0$, then

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \plus{} 1\ge 2(ab \plus{} bc \plus{} ca)$[/quote]

Try to prove it. It's easy enough.

[quote="SUPERMAN2"]Let $ a,b,c$ be the positive real numbers.Prove that:
$ \frac {ab^3}{c} \plus{} \frac {bc^3}{a} \plus{} \frac {ca^3}{b} \plus{} 3abc$≥$ 2(a^2b \plus{} b^2c \plus{} c^2a)$[/quote]

It's equivolent to:
$ a^2b^4\plus{}b^2c^4\plus{}a^4c^2 \plus{}3a^2b^2c^2 \ge 2a^3b^2c\plus{}2ab^3c^2\plus{}2a^2bc^3$
Let's get $ x\equal{}(a^2b^4)^{\frac{1}{3}}$, $ y\equal{}(b^2c^4)^{\frac{1}{3}}$, $ z\equal{}(a^4c^2)^{\frac{1}{3}}$
Then inequality equiovolent to:
$ x^3\plus{}y^3\plus{}z^3\plus{}3xyz \ge 2(xy)^{\frac{3}{2}}\plus{}2(yz)^{\frac{3}{2}}\plus{}2(xz)^{\frac{3}{2}}$
wich is right becouse by Schur ineqality we have:
$ x^3\plus{}y^3\plus{}z^3\plus{}3xyz \ge \sum_{sym}{x^2y} \ge  2(xy)^{\frac{3}{2}}\plus{}2(yz)^{\frac{3}{2}}\plus{}2(xz)^{\frac{3}{2}}$


a,b,c > 0

prove that :

http://www.monsterup.com/image.php?url=upload/1238886362528.gif

[quote]$ \sqrt {\frac {{2a}}{{a \plus{} b}}} \plus{} \sqrt {\frac {{2b}}{{b \plus{} c}}} \plus{} \sqrt {\frac {{2c}}{{c \plus{} a}}} \le 3$[/quote]

$ S^2 \equal{} ( {\sqrt {\frac {{2a}}{{a \plus{} b}}} \plus{} \sqrt {\frac {{2b}}{{b \plus{} c}}} \plus{} \sqrt {\frac {{2c}}{{c \plus{} a}}} })^2 \le (2 \plus{} 2 \plus{} 2)( {\frac {a}{{a \plus{} b}} \plus{} \frac {b}{{b \plus{} c}} \plus{} \frac {c}{{c \plus{} a}}}) \le 9$

$ \Leftrightarrow \frac {a}{{a \plus{} b}} \plus{} \frac {b}{{b \plus{} c}} \plus{} \frac {c}{{c \plus{} a}} \le \frac {3}{2}$

[quote="L_Euler"]
$ \Leftrightarrow \frac {a}{{a \plus{} b}} \plus{} \frac {b}{{b \plus{} c}} \plus{} \frac {c}{{c \plus{} a}} \le \frac {3}{2}$[/quote]
which is wrong. Try $ a \equal{} b \equal{} 1$ and $ c\rightarrow0^ \plus{} .$ :wink:
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17549


Let $ a,b,c \in [1,2]$. Prove or disprove that
\[ \sqrt{\frac{(2a\minus{}b)(2a\minus{}c)}{b^2\plus{}c^2}} \plus{}\sqrt{\frac{(2b\minus{}c)(2b\minus{}a)}{c^2\plus{}a^2}} \plus{}\sqrt{\frac{(2c\minus{}a)(2c\minus{}b)}{a^2\plus{}b^2}} \ge \frac{3\sqrt{2}}{2}\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ (a,b,c)\equal{}(2,1,1)$.

I think this is a very hard inequality (at least with me :(). I have tried to prove it many times but I failed. :(

I do not dare to try it even.....maybe arqady or vasc can help us..... :roll:


linear algebra
matrix
trigonometry
algebra unsolved
algebra

Solve sytem equation with x,y,z>0: $ \left\{\begin{matrix}
z^2\plus{}2xyz\equal{}1 & \\
3x^2y^2\plus{}3y^2x\equal{}1\plus{}x^3y^4 & \\
z\plus{}zy^4\plus{}4y^3\equal{}4y\plus{}6y^2z &
\end{matrix}\right.$

i have a solution for your problem
setting $ y\equal{}tana (0 < a <\frac{\pi}{2})$
from the third equation we get $ z\equal{}tan4a$, from the first equation we get $ xy\equal{}\frac{1}{tan8a}$, substituting to the second we get $ tana\equal{}tan24a$ and we can easily solve this equation

The last equation is the easiest one to solve for one variable (it is linear in $ z$), so
\[ z(1 \plus{} y^4 \minus{} 6y^2) \equal{} 4y \minus{} 4y^3\implies z \equal{} \frac {4y \minus{} 4y^3}{1 \minus{} 6y^2 \plus{} y^4} \equal{} \frac {\binom41y \minus{} \binom43y^3}{\binom40 \minus{} \binom42y^2 \plus{} \binom44y^4}.\]
This should remind you of [url=http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Tangents]this tangent sum identity[/url].

It might also remind you of $ \cos{4x} \plus{} i\sin{4x} \equal{} (\cos{x} \plus{} i\sin{x})^4$.


[quote="MCrawford"]In a group of cows and chickens, the number of legs was 14 more than twice the number of heads.  The number of cows was:
\[ \text{(A)}\ 5 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 14 \][/quote]
[hide]
$ x $ is the number of cows, $ y $ is the number of chickens.

$ 4x+2y=14+2(x+y) $

$ 4x+2y=14+2x+2y $

$ 2x=14 $

$ x=7 $ cows.

So the answer is $ (B) $[/hide]

[hide]Let the number of cows be $ x $ and the number of chickens be $ y $.

$ (4x+2y)-14=2(x+y) $

$ 4x+2y-14-2x-2y=0 $

$ 2x=14 $

$ x=7 $

Answer: $ \boxed{\text{(C)}} $[/hide]

[quote="nat mc"][quote="YooSam"][hide]Let the number of cows be $x$ and the number of chickens be $y$.

$(4x+2y)-14=2(x+y)$

$4x+2y-14-2x-2y=0$

$2x=14$

$x=7$

Answer: $\boxed{\text{(C)}}$[/hide][/quote]
I hope you mean $\text{(B)}$[/quote]

 :blush: Oops, I was going to typed B since I got the right answer, but don't know why I typed C  :blush: .


integration
logarithms
real analysis
real analysis unsolved

By considering $\int\frac{x^{n}}{1-x^{8}}dx$ or otherwise prove that

 \[ \pi=\sum_{i=0}^{\infty}\frac{1}{16^{i}}\left(\frac{4}{8i+1}-\frac{2}{8i+4}-\frac{1}{8i+5}-\frac{1}{8i+6}\right).  \]


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

Let G be a group and g,h in G. Define the commutator of g and h as [g,h]= gh(hg)^-1. Then define the commutator subgroup, denoted [G,G], of G as the subgroup generated by all the commutators of elements of G, i.e. [G,G]=<{[g,h]: g,h in G}>.

(a) What does an element of [G,G] look like?
Hint: it is not enough to only consider elements of the form [g,h].
 *DONE*

(b) Prove that [G,G] is a normal subgroup of G.
*DONE*

(c) Prove that G/[G,G] is an Abelian group.

I can't get part (c)....is it asking to show that ab=ba for some a,b in [G,G] where a and b are commutators?

What you want to prove is that $ gh[G,G] \equal{} hg[G,G]$ equivalent to $ hgh^{\minus{}1}g^{\minus{}1}[G,G] \equal{} [G,G]$ equivalent to $ hgh^{\minus{}1}g^{\minus{}1} \in [G,G]$

For your first expression, are g and h just elements of G?

Is the product: $ g[G,G]h[G,G]g^{\minus{}1}[G,G]h^{\minus{}1}[G,G]$ not a commutator?
And if it is then call it $ [u,v]$ with $ u$ and $ v$ in G so that we have
$ gh[G,G] \equal{} [u,v](hg[G,G]) \equal{} ([u,v][G,G])hg \equal{} [G,G]hg \equal{} hg[G,G]$.

Where is your first equality from? (it implies gh[G,G]=hg[G,G])

Prove that the union of intervals [1,n] from n=1 to n=infinity is all of N.

Do I use induction on this? Archimedes? (This question is before the section of Archimedes though). I need help on how to start it!

Woops! Sorry...new to the site and didn't realize I did that..thanks!


Start with four congruent triangles. In one step, you may take any triangle and cut it in two with the altitude from the right angle. Prove that you can never get rid of congruent triangles

I don't know what you mean by 'get rid of.' Obviously if you cut one, then they are different. Perhaps the point of the problem is that the cut makes two triangles similar to the first...

it basically means that you will always have congruent triangles. Think invariance.

[quote="mdk"]you may take any triangle and cut it in two with the altitude from the right angle. [/quote]

So are all four triangles right?

[hide="hint"]
Remember the problem with the amoeba which starts at (0, 0), and splits to create one amoeba at (1, 0) and one at (0, 1)?  And then every amoeba thereafter can split to put one amoeba above it and one to the right?  Okay...now how does this apply to our problem?[/hide]

[hide="bigger hint"]If you start with four amoebas in the bottom left corner, show that you'll always have some two amoebas on the same lattice point.[/hide]


Propanol(C3H7OH) melts at -126.5 degree Celsius and boils at 97.4 degree Celsius. Draw a qualitative sketch of how the entropy changes as propanol at 150.0 degree Celsius is cooled to a solid at -150.0 degree Celsius. The entire process occurs at a constant pressure of 1 atm

Quantitatively, since the pressure is constant and considering a reversible process between phase transitions,

$ \Delta S \equal{} \frac{\Delta H}{T}$.

Quantitatively, I think it's obvious that $ \Delta S$ must decrease.

[quote="Carcul"]Quantitatively, since the pressure is constant and considering a reversible process between phase transitions,

$ \Delta S \equal{} \frac {\Delta H}{T}$.

Quantitatively, I think it's obvious that $ \Delta S$ must decrease.[/quote]
Can you explain a little bit more detailed in the final conclusion?I think delta H is a constant and T increases, so delta S decreases? By the way, the problem asks for the sketch, so does that mean we have a graph of the function shown above?

$ \Delta H$ can only be considered constant at phase transitions. The liquid state is more ordered than the gaseous state, and the solid state is more ordered than the liquid state, so, given the interpretation of Entropy of a system as its degree of order, on this process $ S_{syst}$ must decrease.


LaTeX
function
integration
calculus
inequalities
derivative
real analysis

A significant fraction of the posters here - myself included - will refuse to click through to any problem posted as an external link. So I don't know what your problem is or whether or not it is of any interest.

If you want to speak to the people here, type your problem out in this forum. Learning a little $ \text{\LaTeX}$ would help.

suppose $ f(t)$ is a nonnegative continuous concave function on[a,b] not identically zero.

define $ F \equal{} \frac {2}{b \minus{} a}\int_a^b f(x)dx$. and $ \phi (x)$ is a convex function on [0,F]

prove $ \frac {1}{F}\int_0^F\phi(x)dx \ge \frac {1}{b \minus{} a}\int_a^b \phi(f(t)) dt$

so i suppose the first thing to notice is that F is twice the average of a convex function over the interval,
but is it possible that a point exists such that f(x)>F in [a,b]? if so.. isnt the integral not really defined?

Well, Kent told you why, didn't he? ;)

The problem is not difficult at all (it is next to impossible to ask a difficult question about an arbitrary convex function of one variable :)). Without loss of generality, $ [a,b] \equal{} [0,1]$. Note that adding a linear function to $ \psi$ doesn't change anything. Now, every convex function is a linear function plus a weighted average of $ (x \minus{} t)_ \plus{}$, so it will suffice to prove the result for $ \psi(x) \equal{} (x \minus{} t)_ \plus{}$ with $ t\in[0,2\bar f]$. The inequality in question then reduces to $ \int_0^1 (f \minus{} t)_ \plus{} \le \frac {(2\bar f \minus{} t)^2}{4\bar f}$. Now, if you look at this picture,
[asy]size(200,200,IgnoreAspect);
D((-0.3,0)--(1.3,0),black);
D((0,0)--(0,2),black);
real a=0.7;
pair[]Q=IPs(D(MP("t",(0,a),W)--(1,a)),D((0,0)--(0.3,2)--(1,0),red));
D((0,0.5)..Q[0]..(0.5,0.95)..Q[1]..(1,0.2),blue+linewidth(1));
D((1,0)--(1,0.2),black);
D(Q[0]--Q[1],orange+linewidth(1));
MP("s",(Q[0]+Q[1])/2);
MP("0",(0,0)); MP("1",(1,0));[/asy]
you'll realize after a minite of thinking that $ \int_0^1(f \minus{} t)_ \plus{}$ can be estimated by both $ s^2\bar f$ and $ \bar f \minus{} \frac {s \plus{} 1}2 t$. The first estimate increases with $ s$ and the second decreases with $ s$. These estimates are equal for $ s \equal{} s_0 \equal{} 1 \minus{} \frac t{2\bar f}$, which gives exactly the value we want. That's it (I leave the details to you to figure out).

thank you Kent Merryfield and fedja, sorry, I am learning mathmaitcal analysis by myself and I am not familiar with latex and i will follow your advice to learn latex.

i still have a question: every convex function is a linear function plus a weighted average of ???  did you mean it can be approximated?

Either approximated, or represented as  $ \psi(x) \equal{} ax \plus{} b \plus{} \int_I (x \minus{} t)_ \plus{} \,d\mu(t)$ with respect to some non-negative measure $ \mu$ on $ I$: either interpretation is fine (at least, inside the interval $ I$: you may have some stupid problems with the endpoints if $ \psi$ is discontinuous or has infinite derivatives there, but they do not really matter here)

$ \int  _{0} ^{ \infty }  f(x)dx$

I am leaning write fomular with latex.

if  is discontinuous or has infinite derivatives there, but they do not really matter here ????????????????????

I still have trouble with this point?????????????

dear fedja, would you please give an explanation about the case when if the function  is discontinuous or has infinite derivatives

Sorry for the delay. Note that $ \psi$ can be represented as a limit of an [i]increasing[/i] sequence of good (say, piecewise linear) convex functions on $ (0,2\bar f)$ and that, since $ f$ is not identically $ 0$, the sets $ \{f\equal{}0\}$ and $ \{f\equal{}2\bar f\}$ have measure $ 0$. Thus, the Monotone Convergence Theorem does the trick.

how can we construct such and [b]increasing[/b] sequencesse.  why the second set  has measure 0  ???


Let {$ a_n$} be a strictly increasing sequence of positive integers. Can the series

$ (1 \minus{} a_1/a_2) \plus{} (1 \minus{} a_2/a_3) \plus{} (1 \minus{} a_3/a_4) \plus{} ...$

ever converge?

Please provide a proof.

yes. see for example
$ a_1\equal{}3$ and $ a_{i\plus{}1}\equal{}\frac{i^2}{i^2\minus{}1}a_i$

Well that doesn't really work. Say you wanted a_2, the next term. You would need to set i = 1. But... that would be 1/0 * 3... bad :P

Yes there is! Take any function $ f$ for which we have:

$ \sum^\infty_{n = 1} f(n) < \infty$

and

$ f(n) > 0$

Now, we can set:

$ 1 - \frac {a_{2n - 1}}{a_{2n}} = f(n) \, \Rightarrow \, a_n = a_0 \prod ^n _{i = 1}\frac {1}{1 - f\left(\frac {n}{2}\right)}$

and there you go. You can try a simple example:

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

and then we get

$ a_n = a_0 \prod ^n _{i = 1}\frac {1}{1 - \frac {1}{(i + 1)^}} = a_0 \prod ^n _{i = 1}\frac {(i + 1)^2}{(i + 1)^2 - 1} = 2 a_0 \frac {(n + 1)!^2}{n!(n + 2)!} = 2 a_0 \frac {n + 1}{n + 2}$

From this we have

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

and I suppose no further explanation is needed! ;)

I know it seems I'm playing advocatus diaboli here, but one simple question: how can you guarantee the "integerness" of these sequences, milin? As the problem states, the sequence is made of positive integers. For instance, in your example $ a_0$ has to be divisible with every prime form $ \mathbb{N}$... Maybe I'm overseeing something?

It's obvious that a bounded sequence will never work, so the examples posted thus far are ridiculous.

On the other hand, it's enough to find any sequence of positive real numbers that tends to infinity and satisfies the given condition. This can certainly be arranged.

edit: Ohps, I forgot what the condition was and thought of a different one.

Theorem: A series $ \sum x_n$ with no term equal to $ \pm 1$ converges absolutely if and only if $ \prod (1\plus{}x_n)$ (or $ \prod (1\minus{}x_n)$ converges to a nonzero value.

In our case, we use the latter form: $ \frac{a_1}{a_2}\cdot\frac{a_2}{a_3}\cdot\frac{a_3}{a_4}\cdots$.

The product telescopes. If $ a_n\to\infty$, it diverges to zero and the series diverges to $ \infty$. If $ a_n$ stays bounded and converges to some limit $ L$, the product converges to $ \frac{a_1}{L}$ and the series converges.
The second case is impossible with a strictly increasing sequence of positive integers, so the series diverges.

would it be possible for you to give a proof of that theorem or provide a link to where i can find one. thanks.


geometry
3D geometry
sphere
tetrahedron
pyramid
Pythagorean Theorem

Three mutually tangent spheres of radius $ 1$ rest on a horizontal plane. A sphere of radius $ 2$ rests on them. What is the distance from the plane to the top of the larger sphere?

$ \textbf{(A)}\  3 \plus{} \frac {\sqrt {30}}{2} \qquad \textbf{(B)}\  3 \plus{} \frac {\sqrt {69}}{3} \qquad \textbf{(C)}\  3 \plus{} \frac {\sqrt {123}}{4}\qquad \textbf{(D)}\  \frac {52}{9}\qquad \textbf{(E)}\ 3 \plus{} 2\sqrt2$

Wow. Interesting problem, but very subtle solution.
[hide]
Let the centers of the smaller spheres be $A$, $B$ and $C$. Let the center of the bigger sphere be $D$. $s=AD=BD=CD=2+1=3$. $AB=BC=AC=1+1=2$. This forms tetrahedron $ABCD$. We need to find the height $h$ of tetrahedron $ABCD$. Using $s$ as hypotenuse of a right triangle, and the distance $d$ from a vertex of the base $ABC$ to the center of $ABC$, we have $d=2\sqrt{3}/3$. Using pythagorean theorem we find that $h=\frac{\sqrt{69}}{3}$. Therefore we conclude that the distance from the bottom to top is $h+1+2=\boxed{\frac{\sqrt{69}}{3}+3}$, which is $\boxed{B}$.[/hide]

Interesting, but I did that a different way, using orthogonal views

[hide]
If you were to draw the "top view" of the configuation [looking from the top straight down to the bottom] you would draw the 3 radius 1 circles with their centers making an equalateral triangle and a big circle with radius 2 concentric with the "triangle."

The "front view" of the configuration [looking from the front through the shape], you would draw 2 circles of radius 1 and a circle radius 2 on top [becuase they are mutually tangent]

Looking at are top view, we can calculate the distance between the center of the radius-1 circles to the center of the radius-2 circle, or the centroid of the formed triangle.  Becuase the triangle is formed by connecting the circles of radius 1, it is equalateral with side length 2.  The distance to the centroid would thus be $\frac{2\sqrt3}{3}$.

Now we use the front view.  Lets find the distance between the center of the radius-1 circle and the center of the radius-2 circle.  Since they are tangent to each other, the distance is 3.

Becuase the views are orthogonal, we notice that the center of the radius-2 circle is directly above the centroid of the equalateral triangle.  Since they form a right triangle [in 3-d space], we can find the length from the centroid of the triangle to the center of the radius-2 circle.  Through phythagorean theorem, we have the the required distance to be $=\sqrt{3^2-(\frac{2\sqrt3}{3})^2}=\frac{\sqrt{69}}{3}$

By knowing that distance, to make it fit the requried distance, we add 1 for the radius 1 circle, and add 2 for the radius 2 circle to find the distance from the top of the configuration to the absolute bottom of the confuguration.  Therefore the answer is $\frac{\sqrt{69}}{3} +3$[/hide]

Although my method is rather complicated, I think it can be extended to solve similar problems, say a radius-2 sphere resting on 4 tangent radius-1 spheres.

[hide="hint:"]connect the centers of the spheres to create a pyramid that you can find the height of[/hide]

[hide]Connecting the centers of the spheres, we get a pyramid with an equilateral base of side $2$ and the other $3$ edges of side $3.$ Finding the height of this is trivial... it yields $\frac{\sqrt{69}}{3}$. Add this to the radius of one of the lower spheres and the radius of the upper sphere to get $3+\frac{\sqrt{69}}{3}.$[/hide]

[b]Solution:[/b] Ugh. These problems. 3D Geo = Take cross sections. Take the cross section joining the center of the big sphere and two of the smaller spheres. Let those centers be $X, Y, Z$ respectively. Let the foot of the altitude from $X$ be $M$. Notice that $XY = 2 + 1 = 3$ and that $YM = 1$. So we have that $XM= \sqrt{8}$. Now, let the bottom of the large sphere be called $O$. Call the distance from $O$ to the plane $h$. Notice that we have $(1+h)^2 + (\frac{\sqrt{3}}{3})^2 = {\sqrt{8}}^2$ (we get the $\frac{\sqrt{3}}{3}$ value as that is the distance from $O$ to $M$ and we can exploit the equilateral triangle formed from the centers of all three congruent smaller spheres). Solving, we have that $1+h = \sqrt{\frac{23}{3}}$. However, notice that the question is asking for the distance from the plane to the top of the sphere, so we want to find $4+h$. Hence our answer is $\boxed{B}$.


How can I solve the following problem:

$ \sum_{n \equal{} 0}^\infty \sum_{k \equal{} 0}^{n} [{2n\choose2k} {(2k\plus{}1)}^{\minus{}x}] / 2^{n\plus{}1}$

Note: The denominator, 2^(n+1) is inside of the infinite series and outside of the inner series, though come to think of it, I don't think it matters.

Note: x is a constant.

I don't know what to do for $ x \neq 0$...

But for $ x\equal{}0$:

$ (1\minus{}1)^{2n} \equal{} \binom{2n}{0} \minus{}\binom{2n}{1} \plus{} \binom{2n}{2} \minus{} \cdots \plus{} \binom{2n}{2n}$ so $ \binom{2n}{0} \plus{} \binom{2n}{2} \plus{} \cdots \plus{} \binom{2n}{2n} \equal{} \binom{2n}{1} \plus{} \binom{2n}{3} \plus{} \cdots \plus{} \binom{2n}{2n\minus{}1}$.  Since $ (1\plus{}1)^{2n} \equal{} 2^{2n} \equal{} \binom{2n}{0} \plus{} \binom{2n}{1} \plus{}\binom{2n}{2}  \plus{} \cdots \plus{} \binom{2n}{2n}$, we have $ 2^{2n} \equal{} 2(\binom{2n}{0} \plus{} \binom{2n}{2} \plus{} \cdots \plus{} \binom{2n}{2n})$ or $ \binom{2n}{0} \plus{} \binom{2n}{2} \plus{} \cdots \plus{} \binom{2n}{2n} \equal{} 2^{2n\minus{}1}$.

So for $ x\equal{}0$ your summation becomes $ \sum_{n\equal{}0}^\infty \frac{2^{2n\minus{}1}}{2^{n\plus{}1}} \equal{} \sum_{n\equal{}0}^\infty 2^{n\minus{}2}$ which clearly diverges.

This looks suspiciously similar to the right-hand side of the identity

 $ [\sum_{n \equal{} 0}^\infty f(n)x^n]*[\sum_{n \equal{} 0}^\infty g(n)x^n] \equal{} \sum_{n \equal{} 0}^\infty [\sum_{k \equal{} 0}^n f(k)g(n \minus{} k)]x^n$

[quote="epkid08"]x is a complex number with a real part greater than zero.

Is there a way to simplify the sums to an easy function of x.

Is there a way to simplify the sums at all?[/quote]

Oh I think I got it!

What if you split up the binomial coefficient, and factor the (2n)! out of the inner series (can you even do this?), so the inner series is left with:

$ \frac {1} {k!(n \minus{} k)!(2k \plus{} 1)^x}$

So...

$ g(a) \equal{} \frac {1}{a!}$
$ f(a) \equal{} \frac {1}{a!(2a \plus{} 1)^x}$

but the problem is the function outside of the inner sum would not be x^n, it would be $ (2n!)(2^{ \minus{} (2n \plus{} 1)})$. I'm sure it's fine that it's -(2n+1) and not n, but what about (2n!)?

Any ideas?

I don't think there's any chance this is going to be nice.  $ (2k \plus{} 1)^{\minus{}x}$ is just not a nicely-behaved expression.

[quote="JBL"]I don't think there's any chance this is going to be nice.  $ (2k \plus{} 1)^{ \minus{} x}$ is just not a nicely-behaved expression.[/quote]

Probably not later on, but I want to split it up so I can simplify the other part further, so I need to find out how to simplify $ \frac{(2n)!}{2^{n\plus{}1}}$ into something like $ x^n$.

Well I found this handy identity:

$ cosh(z) \equal{} \sum_{n\equal{}0}^{\infty} \frac{z^{2n}}{(2n)!}$

As close as I am, I'm off by one number.

Does anyone have ideas on how to find...

$ \sum_{n\equal{}0}^{\infty} \frac{z^{2n\plus{}1}}{(2n)!}$

...in terms of...

$ cosh(z)$

?

[quote="t0rajir0u"]Try $ z \text{ cosh } z$.  But that doesn't help with the summation you gave at all.[/quote]

Let's say

$ \sum_{n \equal{} 0}^{\infty} \frac {z^{2n \plus{} 1}}{(2n)!} \equal{} zcosh(z)$

Couldn't we then say

$ \sum^{\infty}_{n \equal{} 0} \frac {(2n)!}{2^{2n \plus{} 1}} \sum^n_{k \equal{} 0} \frac {1} {k!(n \minus{} k)!(2k \plus{} 1)^x} \equal{} \sum^{\infty}_{n \equal{} 0} \frac {1}{2cosh(2)} \sum^n_{k \equal{} 0} \frac {1} {k!(n \minus{} k)!(2k \plus{} 1)^x}$

And further more

$ \sum^{\infty}_{n \equal{} 0} \frac {(2n)!}{2^{2n \plus{} 1}} \sum^n_{k \equal{} 0} \frac {1} {k!(n \minus{} k)!(2k \plus{} 1)^x} \equal{} \frac {\sum^{\infty}_{n \equal{} 0} \sum^n_{k \equal{} 0} \frac {1} {k!(n \minus{} k)!(2k \plus{} 1)^x}}{2cosh(2)}$

?

Edit: On second thought, I don't think it's "legal" to do that. Or is it okay since it's an infinite sum?

- First of all, information about a sum $ \sum a_n$ doesn't tell you anything about $ \sum \frac{1}{a_n}$.  You don't know anything about the form of $ \sum \frac{(2n)!}{2^{2n\plus{}1}}$, and for good reason - this sum diverges horribly and is totally unrelated to $ \text{cosh } 2$.

- The original factorization is invalid.  In your product you'll have terms where the $ (2n)!$ doesn't match up with the factorials on the bottom (which aren't correct in the factorization either).

[quote="t0rajir0u"]- First of all, information about a sum $ \sum a_n$ doesn't tell you anything about $ \sum \frac {1}{a_n}$.  You don't know anything about the form of $ \sum \frac {(2n)!}{2^{2n \plus{} 1}}$, and for good reason - this sum diverges horribly and is totally unrelated to $ \text{cosh } 2$.

- The original factorization is invalid.  In your product you'll have terms where the $ (2n)!$ doesn't match up with the factorials on the bottom (which aren't correct in the factorization either).[/quote]

Do you think summation by parts will get me anywhere?

[quote="t0rajir0u"]The standard trick for manipulating double summations is to change the order of summation, which is essentially a generalization of summation by parts.  What is the source of this problem?  Knowing whether a solution exists is crucial to finding it (if it exists).[/quote]

Well actually I don't know if a solution exists. I do know that it diverges, but I wanted to simplify it to find the rate at which it diverges. I also know that the finite differences between terms n+1 and n, approach zero as n approaches infinity.

Finding asymptotics and bounds is a very different (and much simpler) question than computing a closed form.  You should probably have asked that question to start with.


[quote="Phoenix257"]If n is a prime number, what is the smallest composite number produced by $n^{2}-n-1$?[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=prime&t=134338

[hide]n=13, 169-13-1=155=composite

[/hide]-jorian