The sum of five different positive integers is 320. The sum of the greatest three integers in this set is 283. The sum of the greatest and least integers is 119. If x is the greatest integer in the set, what is the positive difference between the greatest possible value and least possible value for x?

[hide="Solution"]Let the integers be $ x_3,x_2,x_1,x_0,x$ where $ x>x_0>x_1>x_2>x_3$.
$ x\plus{}x_0\plus{}x_1\equal{}283,x\plus{}x_3\equal{}119,$ and $ x\plus{}x_0\plus{}x_1\plus{}x_2\plus{}x_3\equal{}320$
Substitute: $ 283\plus{}x_2\plus{}119\minus{}x\equal{}320$
$ x\equal{}82\plus{}x_2$
Another way: $ x_0\plus{}x_1\plus{}x_2\equal{}201$
$ 283\minus{}x\plus{}x_2\equal{}201$
$ x\equal{}82\plus{}x_2$

Now, given that $ x\plus{}x_3\equal{}119$ and that all $ x_i>0$, we can see that for $ x_2>x_3$, we need $ x\minus{}82>119\minus{}x\Rightarrow x>100$
Now, $ 100<x<119$. The smallest possible value is $ 101,$ and the largest is $ 118$.
The difference is thus $ 17$.

Check: $ 18, 19, 90, 92, 101$- $ x\equal{}101$ works. There are many values of $ x_0$ and $ x_1$ that satisfy $ x_0\plus{}x_1\equal{}182$

$ 1, 36, 82, 83, 118$.  Once again, there are many such sets.
[/hide]


How many permutations of the numbers 1, 2, 3, 4, 5, 6, 7 have 1 somewhere to the left of 2, 2 somewhere to the left of 3, and 3 somewhere to the left of 4?

[hide="Wrong Solution"]
There are $ 7!$ permutations total. There is a $ \frac12$ probability for each of the conditions, so $ \frac1{2^4}\cdot7! \equal{} \boxed{315}$ is our answer.
[/hide]

Edit: Sorry, I wasn't thinking. :oops:

[hide="Solution"]
Start with $ 1234$, so that we have the digits $ 5,6,7$ left to place.

This is equivalent to $ 3!$ times the number of ways to distribute $ 3$ indistinguishable objects to $ 5$ distinguishable objects (before the $ 1$, between the $ 1$ and the $ 2$, etc.), so our answer is

$ 3!\dbinom{3\plus{}5\minus{}1}{5\minus{}1}\equal{}\boxed{210}$.
[/hide]

Are you sure?  If we extended the result to $ 6$ conditions in the obvious way, $ \frac{1}{2^6} \cdot 7!$ isn't an integer and the correct answer is $ 1$.

(The conditions aren't independent.)


topology
real analysis
real analysis unsolved

$ \begin{gathered}
  {\text{Suppose that }}\left( {S,d} \right){\text{ is a metric space}}{\text{.  Show that }}\left( {S,d'} \right){\text{ is a metric space where }} \\
   \\
  d'\left( {x,y} \right) \equal{} \frac{{d\left( {x,y} \right)}}
{{1 \plus{} d\left( {x,y} \right)}}. \\
\end{gathered}$

[quote="dubito"]$ \frac {{d\left( {x,y} \right)}} {{1 \plus{} d\left( {x,y} \right)}}$[/quote]
Its easy.
$ d'(x,z) \equal{} \frac{ d(x,z)}{1\plus{}d(x,z)}$, $ d'(z,y) \equal{} \frac{ d(z,y) }{1\plus{}d(z,y)}$, and $ d'(x,y) \equal{} \frac{d(x,y)}{1\plus{}d(x,y)}$.
We wanna prove that,
$ d'(x,y) \leq d'(x,z) \plus{} d'(z,y)$
Equivalently,
$ \frac{d(x,y)}{1\plus{}d(x,y)} \leq \frac{ d(x,z)}{1\plus{}d(x,z)} \plus{} \frac{ d(z,y) }{1\plus{}d(z,y)}$
Thus,
$ \frac{a}{a\plus{}1} \leq \frac{b}{b\plus{}1}\plus{}\frac{c}{c\plus{}1}$ where $ a\equal{}d(x,y),b\equal{}d(x,z),c\equal{}d(z,y)$.
Multiply out the with commond denominator and use fact $ a \leq b \plus{} c$ in the problem.

[quote="sylow_theory"][quote="dubito"]$ \frac {{d\left( {x,y} \right)}} {{1 \plus{} d\left( {x,y} \right)}}$[/quote]
Its easy.
$ d'(x,z) \equal{} \frac { d(x,z)}{1 \plus{} d(x,z)}$, $ d'(z,y) \equal{} \frac { d(z,y) }{1 \plus{} d(z,y)}$, and $ d'(x,y) \equal{} \frac {d(x,y)}{1 \plus{} d(x,y)}$.
We wanna prove that,
$ d'(x,y) \leq d'(x,z) \plus{} d'(z,y)$
Equivalently,
$ \frac {d(x,y)}{1 \plus{} d(x,y)} \leq \frac { d(x,z)}{1 \plus{} d(x,z)} \plus{} \frac { d(z,y) }{1 \plus{} d(z,y)}$
Thus,
$ \frac {a}{a \plus{} 1} \leq \frac {b}{b \plus{} 1} \plus{} \frac {c}{c \plus{} 1}$ where $ a \equal{} d(x,y),b \equal{} d(x,z),c \equal{} d(z,y)$.
Multiply out the with commond denominator and use fact $ a \leq b \plus{} c$ in the problem.[/quote]

I never said it was hard. Your answer is correct.  :lol:


Let $ a,b,c > 0$ .

Prove that

$ ab \plus{} bc \plus{} ca \leq a^4 \plus{} b^4 \plus{} c^4\plus{}\frac {3}{4}.$

That leads to $ a^4 \plus{} b^4 \plus{} c^4 \plus{} \frac {1}{4} \geq \sqrt {2} \cdot (ab \plus{} bc \plus{} ca)$.   :roll:
[hide="What you should have done"]If you'd started with $ a^4 \plus{} b^4 \plus{} \frac {1}{4} \plus{} \frac {1}{4}$, you'd get the desired inequality.

Alternatively, do two steps: first one is $ a^4 \plus{} \frac{1}{4} \geq a^2$.[/hide]

Wait; what? $ \sqrt [4]{a^4b^4/64} \equal{} ab/4$; multiplying both sides by 4 we get $ a^4 \plus{} b^4 \plus{} 1/4\ge ab$.

EDIT: Oh whoops. Adding $ a^4 \plus{} b^4$ cyclically gets you $ 2(a^4 \plus{} b^4 \plus{} c^4)$, not $ a^4 \plus{} b^4 \plus{} c^4$.

EDIT: That too.  :oops:

[quote="4865550150"]Let $ a,b,c > 0$ .

Prove that

$ ab \plus{} bc \plus{} ca \leq a^4 \plus{} b^4 \plus{} c^4 \plus{} \frac {3}{4}.$[/quote]
$ a^4 \plus{} \frac {1}{4} \ge a^2$ and cyclic. So $ a^4 \plus{} b^4 \plus{} c^4 \plus{} \frac{3}{4}\ge a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$.

[quote="4865550150"]Let $ a,b,c > 0$ .

Prove that

$ ab \plus{} bc \plus{} ca \leq a^4 \plus{} b^4 \plus{} c^4 \plus{} \frac {3}{4}.$[/quote]
$ a^4\plus{}b^4\plus{}c^4 \ge a^2b^2\plus{}b^2c^2\plus{}c^2a^2$
So we have to prove that:
$ 4a^2b^2\plus{}4b^2c^2\plus{}4c^2a^2\minus{}4ab\minus{}4bc\minus{}4ca\plus{}3 \ge 0 \Leftrightarrow
(2ab\minus{}1)^2\plus{}(2bc\minus{}1)^2\plus{}(2ca\minus{}1)^2 \ge 0$,which is true


geometry
ratio
conics
projective geometry
geometry unsolved

Some years ago my father introduced me to a construction problem that was associated with Papus.

Given a circle and a line with three points on it, construct the triangle inside the circle those sides can be extended to intersect the line at the three points.

Can anyone tell me more and indeed the solution?

Thank you. I shall work on it over the weekend.

[quote="grobber"]Let $A,B,C$ be the points on the line and let $\mathcal C$ be the circle.

Take any point $P_1\in\mathcal C$, let $P_2$ be the second intersection of $AP_1$ with $\mathcal C$, then let $P_3$ be the second intersection of $BP_2$ with $\mathcal C,\ P_4$ - the second intersection of $CP_3$ with $\mathcal C,\ P_5$ - the second intersection of $AP_4$ with $\mathcal C$, and, finally, $P_6$ - the second intersection of $BP_5$ with $\mathcal C$.

$P_1P_2\ldots P_6$ is a cyclic hexagon, so its pairs of opposite sides intersect on a line. Since $A=P_1P_2\cap P_4P_5$ and $B=P_2P_3\cap P_5P_6$, the mentioned line must be $ABC$, and we must have $C=P_3P_4\cap P_6P_1$.

The above result means this, basically: the projective map $P_1\mapsto P_4$ from $\mathcal C$ to itself is an involution. This means that $P_1P_4$ passes through a fixed point $X$ (it can be shown that $X$ actually belongs to $ABC$, but that's not really important here).

Now, for the construction, simply do this: take two points $P_1,P_1'$ as above, draw their corresponding $P_4,P_4'$, and you have $X$: it's $P_1P_4\cap P_1'P_4'$. Let $T$ be any of the at most two points where a tangent from $X$ to $\mathcal C$ touches the circle (note that it's possible that there are no solutions, if $X$ lies inside the circle, for instance). $T$ is a vertex of the triangle we're looking for, namely the one whose adjacent sides cut the line $ABC$ in $A$ and $C$.[/quote]

this is a nice soln, but i didn't understand some things that well

1) you say the map $P_1 \mapsto P_4$ is projective; i have seen maps from a plane to another and from a pencil of lines to another (in your solutions); what about this?

btw, can you say in a contest that some map is projective with no explanations, or you have to give some kind of a proof?

2) i understand why it's an involution; but why does this mean that $P_1 P_4$ passes through a fixed point? (also, why it is on $ABC$?)

3) (not related to the problem) in a projective theorem we can project one line to infinity (i saw this in a solution of darij); what kind of projection is this?

i'm new to projective geometry and i'm trying to learn. it is seems hard  :blush:

$P_1\mapsto P_4$ simply means that $P_1$ is ampped onto $P_4$. The transformation is from the circle to itself, and projective here means that it preserves cross-ratios, since we can define the cross-ratio of four points on the circle as well (or on any conic): just take any other point on the circle, connect it to the four points, and take the cross-ratio of the points to be the cross-ratio of the four lines, which turns out to be independent of the fifth point you chose on the circle.

There's a theorem stating that every involution on a conic is of this form: $P\mapsto Q$, s.t. $PQ$ passes through a fixed point, called the center of the involution.

thanks for the clarifications

do you know where i may find the proof of that theorem? (you know, i have problems taking things for granted)

Books on the subject, or, if taking things for granted is so much of a problem, you could try to prove it yourself :). It's very easy.


[hide]Inscribed angle A cuts off major arc BDE, but this equals arcBC + arcCD + arcDE + arcEF.
Inscribed angle C cuts off major arc BFE, but this equals arcAB + arcAF + arcFE + arcED
Inscribed angle E cuts off arc FBD, but this equals arcAF + arcAB + arcBC + arcCD.

Since inscribed angles = half the arc angles, A = arcBDE/2, C = arcBFE/2, and E = arcFBD/2.

subsitute for the arcs:

A + C + E = (arcs BC + CD + DE + EF + AF + AB + FE + ED + AF + AB + BC + CD)/2 = arcs AB + BC + CD + DE + EF + FA. This equals a whole circle, so...

A + C + E = 360
E + 220 = 360
E = 140.[/hide]

[quote="The QuattoMaster 6000"][quote="ngurnani"]Could someone help me on this problem, I guessed the answer on my test and got it right however I am not able to understand why.

"There is a Hexagon ABCDEF that is inscribed within a circle. Angle A is 100º and Angle C is 120º what is the size of Angle E?"

Thanks for your help.[/quote]
[hide] Draw AD to disect the hexagon into two cyclic quaderilaterals.  Now, we have that $\angle DAB=180-\angle C=180-120=60$.  Thus, we have that $\angle FAD=\angle FAB-\angle DAB=100-60=40$.  Now, we can say that $\angle E=180-\angle FAD=180-40=\boxed{140}$.  [/hide][/quote]

[hide]
$\angle DAB=180-\angle C=180-120=60$
why is this true?[/hide]

Ish: in cyclic quadrilaterals, opposite angles add to 180.

This is because the arc that <C cuts off + the arc that <DAB cuts off = 360 degrees, so <DAB + <C = 180.


integration
trigonometry
calculus
calculus computations

[quote="suweijie"]yeah,it's really easy.the first is $3\ln(2+\sqrt{3})$,and the second is 2. :D[/quote]

Exactly, in both of them just switch the order of integration.


Solve the equation $ (z\plus{}1)^5\equal{}z^5$ for $ z\in\mathbb C$.

My answer:
$ z_k\equal{}\minus{}\frac1{2\sin\frac{k\pi}5} e^{\frac{5\minus{}2k}{10}\pi i}$, where $ k\equal{}1,2,3,4$.

$ (z\plus{}1)^5 \equal{} z^5 \iff \left(1 \plus{} \frac{1}{z}\right)^5 \equal{} 1 \iff z_k \equal{} \frac{1}{e^\frac{2k\pi i}{5} \minus{} 1} \ \text{for} \ k\equal{}1, 2, 3, 4$


algebra
polynomial
function
induction
algebra proposed

Suppose that  $ f $ is a polynomial of exact degree  $ p.$ Find a rigurous proof that $ S(n) $,  where  $ S(n)= \sum\limits_{k=0}^{n}f(k) , $  is a polynomial function  of (exact) degree  $ p+1 $ in varable $n $ .

[quote="flip2004"][quote="Diogene"]The degree in $n$ of $\displaystyle \sum_{k=1}^{n}k^m$  is exactly $m+1$
 :cool:[/quote]
O.K, but find an elemenatry proof of your assertion (without using Bernoulli polynomials).  It's possible ![/quote]

General notation : $\displaystyle P_m(n) = \sum_{k=1}^nk^m $
$P_{m+1}(n+1)=P_{m+1}(n)+(n+1)^{m+1} = P_{m+1}(n)+\sum_{i=0}^{m+1}C^i_{m+1}n^i$
$ (k+1)^{m+1}= \sum_{i=0}^{m+1}C^i_{m+1}k^i \Longrightarrow P_{m+1}(n+1) =1+ \sum_{i=0}^{m+1}C^i_{m+1}P_i(n)$
So  $(m+1)P_m(n)=n^{m+1}+\sum_{i=0}^{m-1}(C^{i+1}_{m+1}n^{i+1}-C^i_{m+1}P_i(n))$
Finally, by using induction on $m$, the degree in $n$ of $P_m(n)$ is exactly $m+1 $
Is the proof  more elementary  with  Bernoulli polynomials ?! :D
 :cool:

[quote="Diogene"][quote="flip2004"][quote="Diogene"]The degree in $n$ of $\displaystyle \sum_{k=1}^{n}k^m$  is exactly $m+1$
 :cool:[/quote]
O.K, but find an elemenatry proof of your assertion (without using Bernoulli polynomials).  It's possible ![/quote][/quote]
Suppose that you are a teacher and propose following question : [i] to prove that for all [/i] $ n\in {\mathbb N} $ [i] the following equality  is satisfied [/i]
\[
 4\sum\limits_{k=0}^{n}k(k+1)(k+2)= n(n+1)(n+2)(n+3) \; .
\]
Someone ,  present a solution the verification of above equality for  $ n\in \{0,1,2,3,4\}. $ How do you appreciate this solution ??


calculus
derivative
function
vector
integration
real analysis
limit

I wnated to know the relation between $ \delta^{(n)}(at)$ and $ \delta^{(n)}(t)$. Can anybody help me out??

Here $ \delta^{(n)}(t)$ stands for $ n^{th}$ derivative of $ \delta(t)$.

regards,
bhupala.

Because $ \delta$ is not a function, neither the notion of its derivative nor the notion of $ \delta(ax)$ are obviously defined. So how do we define things like that? By analogy.

To start with, what is a "distribution" or "generalized function"? It's a (continuous) linear function on a vector space of "test functions." Let's work for now with the test functions being the $ C^{\infty},$ compactly supported functions on $ \mathbb{R}.$ In other words, a distribution is defined by its action on all of the test functions.

Any locally integrable function $ f$ defines a distribution by the action $ \int_{\minus{}\infty}^{\infty}f(x)\varphi(x)\,dx.$

Of course, there are plenty of distributions that are not locally integrable functions. Here's a notational convention: write the action of the arbitrary distribution $ f$ on the test function $ \varphi$ by $ \int_{\minus{}\infty}^{\infty}f(x)\varphi(x)\,dx.$

Just understand this: that is NOT a Riemann integral or Lebesgue integral or any other well-known type of integral. It's not even an integral at all - it's just a metaphor, a suggestive (and possibly dangerously misleading) notation for the action of a linear functional.

Now, what is the Dirac delta? $ \int_{\infty}^{\infty}\delta(x)\varphi(x)\,dx\equal{}\varphi(0).$

Note that that "integral" (which isn't an integral at all) isn't something we can derive or prove: it's the very definition of the Dirac delta. It defines the action (point evaluation) of a particular linear functional.

[color=green][To be continued.][/color]

I have a question on the meaning of "Generalized function". I happen to know perfectly well that Dirac's Delta isn't a function, even though that's how engineering math books present it. And I know, as Dr. Kent was explaining, that we need to use generalized functions, test functions, ... here. The question I have is a historical one. Who introduced a rigorous approach to Dirac's delta function?

[color=green][Part 2 of 3: derivatives of distributions.][/color]

How can we define some operation or another on distributions? By analogy. In particular, what do we mean by the derivative of a distribution?

Suppose the distribution in question were actually given by a nice, smooth function. Then $ \int_{\minus{}\infty}^{\infty}f(x)\varphi(x)\,dx$ would be an integral rather than a metaphor.

Then, $ \int_{\minus{}\infty}^{\infty}f'(x)\varphi(x)\,dx\equal{}\left.f(x)\varphi(x)\right|_{\minus{}\infty}^{\infty}\minus{}\int_{\minus{}\infty}^{\infty}f(x)\varphi'(x)\,dx$

Since $ \varphi$ is compactly supported, the boundary terms evaluate to zero and we get

$ \int_{\minus{}\infty}^{\infty}f'(x)\varphi(x)\,dx\equal{}\minus{}\int_{\minus{}\infty}^{\infty}f(x)\varphi'(x)\,dx$

The next step: adopt this formula, true by integration by parts, as the [b]definition[/b] of the derivative of an arbitrary distribution.

Example: Let $ H(x)\equal{}1$ for $ x>0$ and zero otherwise: the Heaviside unit step function. Since $ H$ is a locally integrable function, it is a distribution in a natural way. What is its derivative?

$ \int_{\minus{}\infty}^{\infty}H'(x)\varphi(x)\,dx\equal{}\minus{}\int_{\minus{}\infty}^{\infty}H(x)\varphi'(x)\,dx$

$ \equal{}\minus{}\int_{0}^{\infty}\varphi'(x)\,dx\equal{}\minus{}\lim_{x\to\infty}\varphi(x)\plus{}\varphi(0)\equal{}\varphi(0).$

Hence, $ \int_{\minus{}\infty}^{\infty}H'(x)\varphi(x)\,dx\equal{}\varphi(0)\equal{}\int_{\minus{}\infty}^{\infty}\delta(x)\varphi(x)\,dx,$

which can be summarized as $ H'(x)\equal{}\delta(x).$ They have the same action, so they're the same distribution.

The Dirac delta has its own derivatives.

$ \int_{\minus{}\infty}^{\infty}\delta'(x)\varphi(x)\,dx\equal{}\minus{}\varphi'(0).$

More generally, $ \int_{\minus{}\infty}^{\infty}\delta^{[n]}(x)\varphi(x)\,dx\equal{}(\minus{}1)^{n}\varphi^{[n]}(0).$

I think Dirac himself introduced the notion of a "test function." He was a physicist at heart, and that language is really a metaphor inspired by the notion of a "test charge" or "test mass" used, hypothetically, to determine or define an electric or gravitational field.

To really obtain rigor, you must specify the topology of the test function space, and hence specify what I meant when I said a distribution was a [i]continuous[/i] linear functional. That work - the topological underpinnings - as far as I know are mostly due to Laurent Schwartz, and the word "distribution" (well, that word in French) comes from Schwartz. This is mostly happening in the 1930's and 1940's. I do mention to my undergraduate Fourier analysis class when I present this to them that this is, unlike much of the curriculum, genuine 20th-century mathematics.

Spelling:

H.A. Schwarz - 19th century, German. Cauchy-Schwarz inequality. Schwarz lemma for complex functions.

L. Schwartz - 20th century, French. Schwartz distribution. The Schwartz space (the test function space for tempered distributions.)

"Distribution" and "generalized function" mean essentially the same thing; I think that "distribution" is the French vocabulary choice, after Schwartz, while "generalized function" is more of a Russian vocabulary choice.

[color=green][Part 3 of 3: scaling distribution.][/color]

Suppose $ a\ne0$ and $ f(x)$ is a distribution. What do we mean by $ f(ax)?$

Again, we seek our answer by analogy. Suppose $ f$ is a nice function. Then

$ \int_{\minus{}\infty}^{\infty}f(ax)\varphi(x)\,dx \equal{}\frac{1}{|a|}\int_{\minus{}\infty}^{\infty}f(u)\varphi\left(\frac{u}{a}\right)\,du$

So once again, we make this a definition: if $ f$ is an arbitrary definition, then we define

$ \int_{\minus{}\infty}^{\infty}f(ax)\varphi(x)\,dx \equal{}\frac{1}{|a|}\int_{\minus{}\infty}^{\infty}f(x)\varphi\left(\frac{x}{a}\right)\,dx$

In particular, what is $ \delta(ax)?$

$ \int_{\minus{}\infty}^{\infty}\delta(ax)\varphi(x)\,dx \equal{}\frac{1}{|a|}\int_{\minus{}\infty}^{\infty}\delta(x)\varphi\left(\frac{x}{a}\right)\,dx \equal{}\frac{1}{|a|}\,\varphi(0).$

Hence, $ \delta(ax) \equal{}\frac{1}{|a|}\,\delta(x).$

Now, suppose $ g(x) \equal{} f(ax),$ where $ f$ is an arbitrary distribution. What is $ g'?$

$ \int_{\minus{}\infty}^{\infty}g'(x)\varphi(x)\,dx \equal{}\minus{}\int_{\minus{}\infty}^{\infty}g(x)\varphi'(x)\,dx$

$ \equal{}\minus{}\frac{1}{|a|}\int_{\minus{}\infty}^{\infty}f(x)\varphi'\left(\frac{x}{a}\right).$

While at the same time, consider $ h(x) \equal{} af'(ax)$

$ \int_{\minus{}\infty}^{\infty}h(x)\varphi(x)\,dx \equal{} a\int_{\minus{}\infty}^{\infty}f'(ax)\varphi(x)\,dx$

$ \equal{}\frac{a}{|a|}\int_{\minus{}\infty}^{\infty}f'(x)\varphi\left(\frac{x}{a}\right)\,dx \equal{}\frac{1}{|a|}\int_{\minus{}\infty}^{\infty}f(x)\varphi'\left(\frac{x}{a}\right)\,dx.$

We conclude that $ \frac{d}{dx}\,[f(ax)] \equal{} af'(ax).$

---

[color=green][Incorrect answer to bhupala's original question edited out. Figure that our for yourself.][/color]

Thank you for a very useful work, Dr. Kent. BTW we sometime call Dirac measure, so how should we understand this exactly ?

Calling it a measure is emphasizing its action on test functions that are merely continuous and de-emphasizing the fact that you can (weakly) differentiate it. "Point masses" are a staple of measure theory and probability theory.

and something relates to Dirac measure it's interesting to know that $ \minus{}\Delta\Phi\equal{}\delta$ where $ \Phi$ is a fundamental solution of Laplace equation $ \Delta u\equal{}0$ say $ \Phi(x)\equal{}\minus{}\frac{1}{2\pi}\log|x|$ for $ x\in{\mathbb R}^{2}$ and $ C\frac{1}{|x|^{n\minus{}2}}$ for $ n\geq3$.
The equation $ div (ub)\equal{}\delta$ where $ b$ us a snooth vector field in $ {\mathbb R}^{2}$  also show flow of dye without diffusion and implies that the dye density is transported  with the fluid  motion at points far from zero.

What are more appearence of delta in fields and its role in this field ?

If $ P(D)$ is any constant coefficient partial differential operator, it always has a fundamental solution - that is a solution $ E$ to $ P(D)(E)\equal{}\delta.$ You just supplied the specific case of the Newtonian potential for the Laplace operator.

The thing about fundamental solutions is this: if we want to solve the nonhomogeneous PDE $ P(E)(u)\equal{}g,$ the one solution will be $ u\equal{}E*g.$ Thus we can find the electric potential due to a charge distribution $ g$ by simply convolving the Newtonian potential with that charge density. This works because $ \delta$ is the identity element for convolution, and because any derivative of a convolution can be assigned to whichever factor we find convenient.

Hence, if $ u\equal{}E*g,$ then $ P(D)(u)\equal{}[P(D)(E)]*g\equal{}\delta*g\equal{}g.$

---

Here's a question: if we have a sequence of functions $ f_{n}$ such that $ \lim f_{n}\equal{}f,$ then is it true that $ \lim f_{n}'\equal{}f'?$

Now if these are functions and we're talking about classical derivatives and pointwise - or maybe uniform - convergence, then up jump the exceptions and the counterexamples and the extra conditions needed to make it true. In classical analysis, this is a very delicate question.

If $ f_{n}$ and $ f$ are distributions, if the sense of convergence of the sequences is weak convergence (the only kind of convergence there is for distributions), and the derivatives are weak derivatives (the kind I defined above), then this is true. Period. Always. And the proof is easy.

But you don't have to like the answer.

Example: let $ f_{n}(x)\equal{}\frac{1}n\sin(n^{2}x).$ This tends uniformly to zero, hence weakly to zero.  We have $ f_{n}'(x)\equal{}n\cos(n^{2}x).$ By what I said above, $ f_{n}'(x)$ tends weakly to zero. This is all true.

Dear Kent,

do you mean to say that $ \delta^{(n)}(at)\equal{}(\minus{}a)^{n}\delta^{(n)}(at)$?

regards
bhupala

[quote="bhupala"]Dear Kent,

do you mean to say that $ \delta^{(n)}(at) \equal{} (\minus{}a)^{n}\delta^{(n)}(at)$?

regards
bhupala[/quote]

There's one thing I wonder: you meaned $ \delta^{(n)}(at)$ or $ (\delta(a.t))^{(n)}$

So I started looking up things about this and came across the following equivalent way of defining the Dirac delta function:

$ \pi\delta (x)\equal{}\lim_{\eta\to\infty}\frac{\sin(\eta x)}{x}$. To show these are equal you need an arbitrary test function say $ f(x)$, then show the integrals are the same, so LHS easy:

$ \int_{\minus{}\infty}^{\infty}f(x)\pi\delta(x)dx\equal{}\pi f(0)$, but how do you get the RHS to be this. It makes some intuitive sense if you could pull the limit out of the integral, but that limit doesn't seem to converge uniformly, so I don't think you can do that.

Any ideas?

The sense of convergence native to the distribution spaces is called [b]weak[/b] convergence. (If you want to be really technical, since we're talking about convergence in a dual space when tested by elements of the original space, then it's weak* convergence.) But weak convergence, is, well, very weak. It's much weaker than uniform convergence, or convergence in $ L^{1}.$ I wouldn't call that sequence a way of "defining" the Dirac delta, but it is true that it converges weakly to the Dirac delta.

Now, how do you show that? What you need to prove is that for any test function $ \varphi$, then

$ \lim_{\eta\to0}\int_{\minus{}\infty}^{\infty}\frac{\sin(\eta x)}{x}\,\varphi(x)\,dx \equal{}\pi\varphi(0).$

My advice for doing that: Split it into two integrals, one on $ (\minus{}\infty,0),$ the other on $ (0,\infty).$ In each case integrate by parts, integrating the $ \frac{\sin(\eta x)}{x}$ and differentiating $ \varphi(x).$ Yes, we know that integral isn't an elementary function, but give it a name, and note that you do know how it behaves as $ x\to\infty.$

Use the fact that $ \int_{0}^{\infty}\frac{\sin x}{x}\,dx \equal{}\frac{\pi}{2}.$


induction
inequalities
real analysis
real analysis unsolved

problem: f:[0,+infinity)->reals is continuous and |f(x+y)+f(x)+f(y)|<1 for every x,y in[0,+infinity). prove the existence of the limit lim(x->+infinity)f(x)/x.

we only must prove that in infinity $f(x)$ is bounded.
really. if we substitute instead $y$ in our condition $y=0$ then we take
$|2f(x)+f(0)|<1$ $\Rightarrow$
$\frac{|f(0)|}{2}-\frac{1}{2}<|f(x)|<\frac{|f(0)|}{2}+\frac{1}{2}$
or
$\frac{-|\frac{|f(0)|}{2}-\frac{1}{2}|}{x}<\frac{f(x)}{x}<\frac{\frac{|f(0)|}{2}+\frac{1}{2}}{x}$
and consider limit in infinity.

[quote="martin1985"]the problem is the same but |f(x+y)-f(x)-f(y)|<1[/quote]
some idea: show that :
$\frac{f(2^{k}x)}{2^{k}}-\frac{1}{2}<f(x)<\frac{f(2^{k}x)}{2^{k}}+\frac{1}{2}$
Where $k\to+oo$

Using induction we can show $|\sum_{i=0}^{n}f(x_{i})-f(\sum_{i=0}^{n}x_{i})| < n$. Define $M(x) = \max\{ |f(y)| : y \in [0,x] \}< \infty$. For each $a$ and $b$ with $0<a<b$ there is $n\in \mathbb{N}$ and $c \in [0,a)$ with $b = na+c$. Since $b$ is the sum of $n$ $a$'s and one $c$, we have $|f(b)-nf(a)-f(c)| < n$. Then $|f(b)-nf(a)| < n+f(c)$. Divide by $na$ to get
\[|\frac{b}{na}\frac{f(b)}{b}-\frac{f(a)}{a}| < \frac1a+\frac{M(a)}{na}\]
. This inequality implies that the limsup and liminf of $f(x)/x$ as $x\to\infty$ are finite. To see that, fix a value of $a$ while $b$ goes to $\infty$. If $b$ is sufficiently large, then $\frac{b}{na}$ is close to 1 and $\frac{M(a)}{na}$ is close to 0.

Note that the last inequality achieves a comparison of two values $\frac{f(b)}{b}$ and $\frac{f(a)}{a}$. Let $L$ be the liminf of $f(x)/x$ as $x\to\infty$. We want to show that $L$ is indeed the limit. Again fix $a$ and this time let $b$ takes values of a sequence $(b_{i})_{i}$ such that $b_{i}\to \infty$ and $f(b_{i})/b_{i}\to L$. Then it follows from the last inequality that $|L-\frac{f(a)}{a}| \le \frac1a$.  Now let $a\to\infty$ to see that $L$ is the limit.

1. if f(x) is bounded for big x then the limit f(x)/x exists and it's zero.
2. if lim(x->infinity)f(x)=infinity then

we use f((2^k)x)/2^k-1/2<f(x)<f((2^k)x)/2^k +1/2
we have two cases

2.1)lim(k->infinity)f((2^k)x)/2^k<infinity, let say M=>for x=1 and using that f is continuous f(2^k)=2^k=>f(x)=x and the limit f(x)/x=1 and exists

2.2)lim(k->infinity)f((2^k)x)/2^k=infinity=>for even for small x we get that f(x) is very big. take x=1/2^k then f(1)/2^k-1/2<f(1/2^k)<f(1)/2^k-1/2. can we say now that for some very big k LHS  and RHS are -1/2 and 1/2 correspondingly. then f(1/2^k) is not so big(in(-1/2,1/2)) as noted above. contradiction.

does this works?


If $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$, then

       $ 2(a^2 \plus{} b^2 \plus{} c^2) \plus{} 3abc\geq a \plus{} b \plus{} c \plus{} 6$

[quote="Ligouras"][quote="Pain rinnegan"][quote="Ligouras"]If $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$, then

       $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3abc\geq a \plus{} b \plus{} c \plus{} 6$[/quote]

It's false, Ligouras.[/quote]

The correct is:

        $ 2(a^2 \plus{} b^2 \plus{} c^2) \plus{} 3abc\geq a \plus{} b \plus{} c \plus{} 6$

Sorry.....[/quote]

For $ ab\plus{}bc\plus{}ca\equal{}3$ the following inequalities holds:

1)$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3abc\ge 6$(Vasc's inequality)

2)$ a^2 \plus{} b^2 \plus{} c^2\ge a \plus{} b \plus{} c$

[quote="Soarer"]Hi Pain rinnegan, can you give the reference to ?(Vasc's ineq)?[/quote]

dear siu ho chung, see http://www.mathlinks.ro/Forum/viewtopic.php?t=62537

[quote="Pain rinnegan"][/quote][quote="Ligouras"][quote="Pain rinnegan"][quote="Ligouras"]If $ a,b,c > 0$ and $ ab + bc + ca = 3$, then

       $ a^2 + b^2 + c^2 + 3abc\geq a + b + c + 6$[/quote]

It's false, Ligouras.[/quote]

The correct is:

        $ 2(a^2 + b^2 + c^2) + 3abc\geq a + b + c + 6$

Sorry.....[/quote][quote="Pain rinnegan"]

For $ ab + bc + ca = 3$ the following inequalities holds:

1)$ a^2 + b^2 + c^2 + 3abc\ge 6$(Vasc's inequality)

2)$ a^2 + b^2 + c^2\ge a + b + c$[/quote]

Thanks PAIN   :oops:


An aquarium has a rectangular base that measures 100 cm by 40 cm
and has a height of 50 cm. The aquarium is filled with water to a
depth of 37 cm. A rock with volume $ 1000 \text{cm}^3$ is then
placed in the aquarium and completely submerged. By how many
centimeters does the water level rise?  Express your answer as a decimal to the nearest 100th.

The area of the base is 4000 centimeters squared.

The volume of the base solid that is 1 centimeter tall is 4000 centimeters cubed.

1000/4000 = 1/4 = 0.25

So, the water level rises .25 cm

The rock displaces the water by a total volume of 1000 cubic cm.  The area of the base 4000 cubic cm so the water rises by .25cm.


Given $ a,b,c$ positive reals that satisfy $ abc \equal{} 1$, prove that:
\[ ab \plus{} bc \plus{} ca \geq \sqrt [4]{7(a \plus{} b \plus{} c)^2 \plus{} 6\sqrt{3(a \plus{} b \plus{} c)}}
\]
:)

[quote="shaaam"]Given $ a,b,c$ positive reals that satisfy $ abc \equal{} 1$, prove that:
\[ ab \plus{} bc \plus{} ca \geq \sqrt [4]{7(a \plus{} b \plus{} c)^2 \plus{} 6\sqrt {3(a \plus{} b \plus{} c)}}
\]
:)[/quote]
The above inequality is equivalent to
\[ (ab\plus{}bc\plus{}ca)^4 \ge 7(a\plus{}b\plus{}c)^2 \plus{}6\sqrt{3(a\plus{}b\plus{}c)}\]
Since $ (ab\plus{}bc\plus{}ca)^4 \ge [3abc(a\plus{}b\plus{}c)]^2 \equal{}9(a\plus{}b\plus{}c)^2$, it suffices to prove that
\[ 9(a\plus{}b\plus{}c)^2\ge 7(a\plus{}b\plus{}c)^2 \plus{}6\sqrt{3(a\plus{}b\plus{}c)}\]
\[ (a\plus{}b\plus{}c)^2 \ge 3\sqrt{3(a\plus{}b\plus{}c)}\]
\[ a\plus{}b\plus{}c \ge 3\]
which is obviously true by AM-GM. :)

Yes, Can. That is it! Sorry it was too easy.
The following inequality is true too.
Given that $ a,b,c \geq 0$, and $ abc\equal{}1$, then we have:
\[ ab\plus{}bc\plus{}ca \ge \sqrt[3]{\frac{7}{2}(a\plus{}b)(b\plus{}c)(c\plus{}a) \minus{} 1}
\]


Given the equation of an ellipse, what is the eccentricity of it?

Eq: 16x^2 + 7y^2 =112

1/7
23/112
7/16
3/4
7/4

Thanks a lot.

[hide]
Formula for eccentricity is
$ e \equal{} \sqrt{ \frac {a^2 \minus{} b^2}{a^2}}$ i think  :maybe:   :)
[/hide]

what are a and b here?  :)

[quote="madness
what are a and b here?  :)[/quote]

Aren't they the coeffecients to the x and y terms?
In this case, a=16 and b=7?

Is that right?

Thank you

No it isnt right!
First what is the genl form of an ellipse?  :)

Oh, my bad
I realize it now. a^2 is the denominator of the x term and b^2 is the denominator of the y term to make the constant 1.

So, in this case,
a= sqrt 7
b= 4

Now is it right?

If it is, how is the eccentricity imaginary    e---> sqrt  -9/7?

the eccentricity of an ellipse is $ c/a$.,where $ a>b$
the form of the ellipse is:$ x^2/(sqrt7)^2  \plus{} y^2/4^2 \equal{} 1$
so we have $ a\equal{}4$ and $ b \equal{}sqr7$
we know that $ c^2\equal{} a^2 \minus{} b^2\equal{}9$
<=>
$ c\equal{}3$,because $ c>0$
so the eccentricity is $ c/a\equal{}3/4$ :wink:

Where $ c\equal{}\sqrt{a^2\plus{}b^2}$, so bascially the same thing as madness's post.


What lower bound can be placed on $ \frac{\varphi n}{n}$, $ n \geq 10$

If the prime factors of $ n$ are $ p_{1}, p_{2}, \cdots, p_{k}$, then $ \frac{\varphi n}{n}=(1-\frac{1}{p_{1}})\cdots (1-\frac{1}{p_{k}})$ which clearly tends to 0 as $ k \rightarrow \infty$ since all factors are less than 1.

Well $ \frac{\varphi(n)}{n}=\prod_{p|n}^{}\left(1-\frac{1}{p}\right)$, where $ p$ are primes, so the best lower bound we can set is

$ \frac{\varphi(n)}{n}>\prod_{p}^{}\left(1-\frac{1}{p}\right)$
Well you may recognize this as $ \frac{1}{\zeta(1)}$...therefore, the best possible lower bound is, sadly, $ 0$...

edit: SplashD beat me to it...but note that knowing all factors in that product are less than one is not enough to know is converges...consider, for example, $ \prod_{k=1}^{\infty}\frac{\frac{1}{2}+2^{-k}}{\frac{1}{2}+2^{-k+1}}$

The following is what I did last night.

[b]Proposition.[/b] $ 0<\frac{\varphi(n)}{n}<1$ for $ n\geq 2$ ($ n\geq 10$ does give us any extra condition other than $ n\neq 1$).

[b]Proof.[/b] Notice that by Euler's Totient Formula (trivial to prove by for example probability),
\[ \frac{\varphi(n)}{n}=\prod_{i=1}^{m}\left(1-\frac{1}{p_{i}}\right) \]
when $ p_{i}|n$ are primes.  Notice that as $ m\rightarrow\infty$ (there are infinitely many primes),
\begin{eqnarray*}\frac{\varphi(n)}{n}&=& \lim_{m\rightarrow\infty}\prod_{i=1}^{m}\frac{1}{\frac{1}{1-\frac{1}{p_{i}}}}\\ &=& \lim_{m\rightarrow\infty}\prod_{i=1}^{m}\frac{1}{1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\hdots}\\ &=&\lim_{m\rightarrow\infty}\frac{1}{\sum_{k=1}^{m}\frac{1}{k}}= 0. \end{eqnarray*}
$ S=\lim_{m\rightarrow\infty}\sum_{k=1}^{m}\frac{1}{k}=\infty$ since otherwise, $ S=1+\frac{1}{2}+\frac{1}{3}+\hdots>\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+\frac{1}{6}+\hdots=S$.    Contradiction!

As a result, $ \frac{\varphi(n)}{n}>0$.  Also, since $ p_{k}\rightarrow\infty$ as $ k\rightarrow\infty$, $ \frac{\varphi(n)}{n}<1$, and so $ 0<\frac{\varphi(n)}{n}<1$. $ \Box$

[quote="K81o7"]SplashD beat me to it...but note that knowing all factors in that product are less than one is not enough to know is converges...consider, for example, $ \prod_{k=1}^{\infty}\frac{\frac{1}{2}+2^{-k}}{\frac{1}{2}+2^{-k+1}}$[/quote]

What he meant to say is that $ 0< a_{j}< 1$ does not imply that $ \prod_{j \in \mathbb{N}}a_{j}=0$; the example that he gave does in fact converge (to $ 1/3$?).


Prove that in any regular polygon with even number of sides no three diagonals are concurrent except at the center. Does the same also hold for a regular polygon with odd number of sides?

The diagonals of a regular polygon with an even number of sides can concur at a point other than the centre.  Here is a link to a counterexample:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60521[/url]

No three diagonals of a regular polygon with an odd number of sides can concur, but this is difficult to prove.


Nice inequality  :)
By Cauchy Swartz inequality we have that
$ \sum\frac{a^{4}}{a^{2}b(a\plus{}2b)}\geq\frac{(a^{2}\plus{}b^{2}\plus{}c^{2})^{2}}{\sum\ a^{2}b(a\plus{}2b)}$

So we have to prove that
$ \frac{(a^{2}\plus{}b^{2}\plus{}c^{2})^{2}}{\sum\ a^{2}b(a\plus{}2b)}\geq 1$

which is equivalent to $ a^{4}\plus{}b^{4}\plus{}c^{4}\geq a^{3}b\plus{}b^{3}c\plus{}c^{3}a$  which true by Rearengement .

[quote="arqady"][quote="nayel"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{a^{2}}{b(a\plus{}2b)}\plus{}\frac{b^{2}}{c(b\plus{}2c)}\plus{}\frac{c^{2}}{a(c\plus{}2a)}\ge 1\]
[/quote]
$ \frac{a^{2}}{b(a\plus{}kb)}\plus{}\frac{b^{2}}{c(b\plus{}kc)}\plus{}\frac{c^{2}}{a(c\plus{}ka)}\ge\frac{3}{1\plus{}k}$ true too for all positive $ a,$ $ b,$ $ c$ and $ k.$[/quote]

From CBS we have that
$ \sum\frac{a^{4}}{a^{3}b\plus{}ka^{2}b^{2}}\geq\frac{\left(\sum a^{2}\right)^{2}}{\sum a^{3}b\plus{}k\sum a^{2}b^{2}}\geq\frac{3}{k\plus{}1}$
$ \Leftrightarrow$
$ (k\minus{}2)\left(\sum a^{4}\minus{}\sum a^{2}b^{2}\right)\plus{}3\left(\sum a^{4}\minus{}\sum a^{3}b\right)\geq 0$ ,true for $ k\geq 2$

Nice solution silouan! :)

[quote="arqady"][quote="nayel"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{a^{2}}{b(a\plus{}2b)}\plus{}\frac{b^{2}}{c(b\plus{}2c)}\plus{}\frac{c^{2}}{a(c\plus{}2a)}\ge 1\]
[/quote]
$ \frac{a^{2}}{b(a\plus{}kb)}\plus{}\frac{b^{2}}{c(b\plus{}kc)}\plus{}\frac{c^{2}}{a(c\plus{}ka)}\ge\frac{3}{1\plus{}k}$ true too for all positive $ a,$ $ b,$ $ c$ and $ k.$[/quote]

Yes I found that too. But $ k\equal{}2$ makes it kind of special. Actually I proved that \[ \frac{a^{2}}{b(ma\plus{}nb)}\plus{}\frac{b^{2}}{c(mb\plus{}nc)}\plus{}\frac{c^{2}}{a(mc\plus{}na)}\ge\frac{3}{m\plus{}n}\]
Anyway for my proposed inequality I have two different solutions, I will post them tomorrow. :wink:

By the same way :)

By Cauchy-Schwarz we have that

$ \sum\frac{a^{4}}{ma^{3}b+na^{2}b^{2}}\geq\frac{\left(\sum a^{2}\right)^{2}}{\sum_{cyclic}m.a^{3}b+n\sum a^{2}b^{2}}\geq\frac{3}{m+n}$

$ \Leftrightarrow ({m+n)(a^{2}+b^{2}+c^{2})^{2}\ge 3({m\sum_{cyclic}a^{3}.b+n\sum a^{2}.b^{2})}}$

Then easy to prove that :

$ n(a^{2}+b^{2}+c^{2})^{2}\ge 3n\sum a^{2}b^{2}$

And by Vasc'inequality, we have :

$ m(a^{2}+b^{2}+c^{2})^{2}\ge 3m\sum_{cyclic}a^{3}b$

Very nice solution quanpbc! :lol:  But I do not use Vasc's inequality to solve this problem. So people who want to solve it may give it another try. :P

The following is my first solution. As I mentioned before, I have another solution using a nice inequality but I'm not going to post it now. Because I want to create more inequalities using it.

[hide="My first solution"](All sums are cyclic)
First asume that $ 2n\ge m$. By AM-GM inequality,
\[ \sum\frac{a^{2}}{mab+nb^{2}}\ge\sum\frac{2a^{2}}{ma^{2}+(m+2n)b^{2}}\]
Now it is well known (and can be proven by Cauchy's Inequality), that
\[ \sum\frac{a}{a+kb}\ge\frac{3}{1+k}\ \forall k\ge 2\]
Using this we get
\begin{eqnarray*}\sum\frac{2a^{2}}{ma^{2}+(m+2n)b^{2}}&=&\frac{2}{m}\sum\frac{a^{2}}{a^{2}+(1+\frac{2n}{m})b^{2}}\\ &\geq &\frac{2}{m}\cdot\frac{3}{1+1+\frac{2n}{m}}\\ &=&\frac{3}{m+n}\end{eqnarray*}
As $ 1+\frac{2n}{m}\ge 2$.

Now let $ 2n\le m$. In this case we have by Cauchy's inequality,
\begin{eqnarray*}\sum\frac{a^{2}}{mab+nb^{2}}&\ge&\frac{(a+b+c)^{2}}{n\sum b^{2}+m\sum ab}\\ &\ge&\frac{(a+b+c)^{2}}{\frac{m}{2}\sum b^{2}+m\sum ab}\\ &=&\frac{1}{\frac{m}{2}}\ge\frac{3}{m+n}\end{eqnarray*}
Where the last inequality follows from $ m\ge 2n$. :) [/hide]

[quote="nayel"]Very nice solution quangpbc! :lol:  But I do not use Vasc's inequality to solve this problem. So people who want to solve it may give it another try. :P
[/quote]

Suppose $ a \equal{}\min(a,b,c)$
Let $ b \equal{} a\plus{}x,c \equal{} a\plus{}y$ for $ x,y\ge 0$

The inequality is equivalent to :

$ (x^{2}\minus{}pq\plus{}p^{2}).a^{2}\plus{}(x\plus{}3\minus{}5x^{2}y\plus{}4xy^{2}\plus{}q^{3})a\plus{}x^{4}\minus{}3x^{3}y\plus{}2x^{2}y^{2}\plus{}y^{4}\ge 0$

$ \delta \equal{}\minus{}3(x^{3}\minus{}x^{2}y\minus{}2xy^{2}\plus{}y^{3})^{2}\le 0$

Sorry for reviving this topic. But I had to do it as I couldn't edit my post anymore.

[quote="nayel"]Now let $ 2n\le m$. In this case we have by Cauchy's inequality,
\begin{eqnarray*}\sum\frac {a^{2}}{mab + nb^{2}} & \ge & \frac {(a + b + c)^{2}}{n\sum b^{2} + m\sum ab} \\
& \ge & \frac {(a + b + c)^{2}}{\frac {m}{2}\sum b^{2} + m\sum ab} \\
& = & \frac {1}{\frac {m}{2}}\ge\frac {3}{m + n}\end{eqnarray*}
Where the last inequality follows from $ m\ge 2n$. [/quote]

The above part of the solution is incorect. It should be the following:
\begin{eqnarray*}\sum\frac {a^{2}}{mab + nb^{2}} & \ge & \frac {(a + b + c)^{2}}{n\sum b^{2} + m\sum ab}\end{eqnarray*}
So we need to show that
\[ (m + n)(a + b + c)^2\ge 3\left(m\sum ab + n\sum b^2\right)
\]
which is equivalent to

$ (m + n)\sum a^2 + 2(m + n)\sum ab\ge 3m\sum ab + 3n\sum b^2$
$ \iff (m - 2n)\sum a^2\ge (m - 2n)\sum ab$.

This is true because $ m - 2n\ge 0$.

What is the the most easy way to prove Vasc inequality? Who knows?

[quote=luofangxiang][quote=arqady][quote="nayel"]Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{a^{2}}{b(a\plus{}2b)}\plus{}\frac{b^{2}}{c(b\plus{}2c)}\plus{}\frac{c^{2}}{a(c\plus{}2a)}\ge 1\]
[/quote]
$ \frac{a^{2}}{b(a\plus{}kb)}\plus{}\frac{b^{2}}{c(b\plus{}kc)}\plus{}\frac{c^{2}}{a(c\plus{}ka)}\ge\frac{3}{1\plus{}k}$ true too for all positive $ a,$ $ b,$ $ c$ and $ k.$[/quote]

x,y,z=a/b,b/c,c/a[/quote]

indeed very nice

[quote=nayel]Let $ a,b,c$ be positive real numbers. Prove that \[ \frac{a^{2}}{b(a\plus{}2b)}\plus{}\frac{b^{2}}{c(b\plus{}2c)}\plus{}\frac{c^{2}}{a(c\plus{}2a)}\ge 1\][/quote]

By Cauchy-Schwarz we have:

 \[ \frac{a^{2}}{b(a+2b)}+\frac{b^{2}}{c(b+2c)}+\frac{c^{2}}{a(c+2a)}\]

$$= \sum \frac{\frac{a^2}{b^2}}{\frac{a}{b}+2}$$

$$\ge \frac{(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^2}{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+6}$$

Let $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=t \ge 3$, we need:

$$t^2 \ge t+6$$ equivalent to $$(t-3)(t+2) \ge 0$$.

It is true because $t \ge 3$ and we are done.


How many composite factors does 60 have?

[hide] We see that $ 60$ has $ 12$ positive factors. But of these $ 1$, $ 2$, $ 3$, and $ 5$ are not composite. So we have a total of $ \boxed{8}$ composite factors.[/hide]


HI!
 I`m Lucas Molina.I`m Brazilian. I never didn´t paticiped a olimpy´math.Sorry, but I am learning English :lol: . I`m 14 years old. I love math very much!!! :P See you tomorrow!!

Obs.: Is there Brazilian`s forum here?

Welcome to the forums.

I don't think there is a Brazil forum, but maybe you could help create one or something.

Welcome!  We don't have a Brazilian forum as far as I know.  We all love math here, so you're in the right place.  Enjoy! :D

Hi Lucas,

You can try the Spanish Forums (look my signature for a direct link) most of my brazilian friends understand pretty well Spanish!! :D !

hello!
I think this forum is not only a good place to learn math, but also to study English! I myself do it...


group theory
abstract algebra
superior algebra
superior algebra solved

I have problem with this task and
I will be very grateful for all help. :blush:

Let [b]R [/b]be a ring and let [b]H[/b] be subset to [b]R[/b].
Show that [b]H [/b]is ideal in [b]R[/b] [u][b]if and only if [/b][/u][b]H [/b]fulfil
such condition :
1. If [i]r [/i] in R and [i]h[/i] in [b]H[/b] then both [i]rh[/i] and [i]hr[/i] in [b]H[/b].
2. If [i]a [/i]and [i]b [/i] in [b]H[/b] then difference [i]a-b [/i] in [b]H[/b].

 :ewpu:
Thank You very much
Tanja

From 2 we have $0\in H$. From 1 we have $x\in H \implies -x\in H$ thus $x+x=x-(-x)\in H$. This together with 1 is enough to prove that $H$ is an ideal of $R$.

The other way, i.e. every ideal satisfies the conditions you gave, follows directly from the fact that an ideal is an additive subgroup of $R$.

Thank You very much. :lol:

Best regards
Tanja


trigonometry
inequalities
inequalities unsolved

prove or disprove:
let $0<x<\pi$ then
$\frac{\sin x}{x} \geq \frac{\pi^3-x^3}{\pi^3+x^3}$

Could I have a question :what is the "redheffer"?

the redheffer's inequality is:
for  $x\neq 0$ then
$\frac{\sin x}{x} \geq \frac{\pi^2-x^2}{\pi^2+x^2}$

This seems to be wrong: Take $x : = \pi / 2$.


function
inequalities
inequalities unsolved

let $ 0<p<1$ and $ x>0, y>0$
prove that :

$ |x^{p}\minus{}y^{p}|\leq |x\minus{}y|^{p}$

many thanks

Use the fact that for $ 0 < p < 1$ and $ a > 0, b > 0$

$ (a \plus{} b)^p < a^p \plus{} b^p$

(To prove that, consider the function $ f(x) \equal{} (1 \plus{} x)^{p} \minus{} 1 \minus{} x^p$ and it's derivative.)

[quote]$ (a\plus{}b)^{p}< a^{p}\plus{}b^{p}$[/quote]

it is not what I need.
What I need is $ |a^{p}\minus{}b^{p}|\leq |a\minus{}b|^{p}$

Why reject it outright? Did you even try to think about it? It was meant as a sketch, not a complete proof.

Anyway...

Assume $ x < y$

Put $ a \equal{} y\minus{}x, b \equal{} x$ and apply the inequality I stated.


AMC
AMC 12
AMC 12 A
trigonometry
AIME
MATHCOUNTS
factorial

So, what did you get?  Harder or easier than last year?  And all the other usual topics of discussion?  Why is the previous sentence a question?

I got 132, answering all 25 and missing #17 (really really stupid mistake), #24 (kind of stupid mistake), and #25 (missed the triangles that have side length $\sqrt{6}$.  However, despite the fact that I got 137 last year, I did find this year's test easier, so either that's the case or I improved a bit since last year (I'm guessing some of both).  My personal favorite problem was #24, and I was quite proud of myself for solving it (except for a little mistake in the solution).

My team score (we think) is 357.

I need to get to class now.  Bye!

I got a 143 answering all but #24 and #25

[quote="joml88"]My question is: WHERE IS THE TRIG!!!????[/quote]

I'd look for it on the AIME  ;)

140.5 here. I answered all except #24 and got #25 wrong (didn't count triangles with side of $\sqrt6$.

I got 140.5 as well- left 24 blank and made a stupid mistake on number 14- read the question wrong.  I'm really happy with my score, especially because it seemed to me like the problems were harder than last year.

I got 104.5. I answered 1 - 11 and then 18. I got none wrong. I think I could have answered more, but I played it safe this year.

Hey, Sly Si, are you going to be at the City of Lakes South MATHCOUNTS chapter meet? My dad (tokenadult) would like meet you.

It's funny how all the mock tests made extensive use of factorials and trig...

[quote="Eugevir"]Can someone post the answer key for the AMC 12A? I have my answers but cannot find the teacher who would have the solutions book.[/quote]

I believe that there is a board about that.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25587 ?

I'm not sure how reliable that is though...

[quote="joml88"]124 Answered 21 and missed 2 (#'s 9 and 14).  Those were very careless errors and I'm kicking myself for them.  I think difficulty wise it was average.  My question is: WHERE IS THE TRIG!!!????[/quote]

Let me guess, you didn't see the $x$ in $8x$, so you answered $0$?  That's what I did, blah.

What bugs me is that I know someone who found the equation via brute force then found the roots ... they shouldn't permit 89s on the AMC ....

Oh, could someone explain 14, 16, 19, and 20?  The head of the math department said that he didn't get a solutions pamphlet.

I got 116, and although I am not very ahppy with it, it was still the best in my school.  I answered 17, and missed 1.

[quote]Oh, could someone explain 14, 16, 19, and 20? The head of the math department said that he didn't get a solutions pamphlet.[/quote]

I think I can answer #19 off the top of my head.  It's the odometer one, I think.  Basically, what I did was see that there were 9 numbers you could put into any digits place.  Therefore, up to 1000, there were 9^3 = 729 numbers which did not contain a 4.  Since you wanted it to be 2005, you could multiply 729 by 2, since there were no fours in the additional thounsands place.  You now have 1458 working numbers.
You now have up to 2000.  The rest of the numbers are: 2001, 2002, 2003, 2005.  Add 4, and you get 1462.  I think that was answer B.

I could be wrong, since I missed one question, though.

Also, I looked at the AMC 10A, and I think I could have scored in the 140s with it.  Should I take the 10B or the 12B?  Does it make a difference?  I am new to this test, so any more help would be nice.

Thanks.

Well i just completed my first AMC 12 yesterday. I was wondering whether any website shows the solutions along with the problems.

This doesnt make any sense, but on the official amc website they have the answers for the amc-12 b also? Is that a typo, or did they accidentl ygive them out?

Attend the math jam tonight (referring to boblard's question).

Ha, I was so close to expanding the expression on that one.  But right before I began doing that I realized that I could just factor 2005!  It saved me a lot of time and pain.

Oh, wow.

I thought the A, B, and C were digits of the the numbers, so what I did was

$100A$
  $10B$
$+$      $C$
______
$110(A+B+C)$

thus

$(110(A+B+C))(A+B+C)=2005$

and since 2005 didn't have a factor of more than 1000, this confused me immensely. jeez :blush:

Umm... AMC were digits, but 100A + 10M + C != 110(A+M+C)...

[quote="white_horse_king88"]Made a bunch of stupid stupid mistakes...

Spent too much time on 18 since I was stupid enough not to notice the inclusion-exclusion. Eh...

Man.. 119.5... I swear... I should have done much much better.[/quote]Tisk, tisk; same here.  18 was a killer.  I worked it out THREE times fully w/out seeing it (or checking my mistakes on 8,12,16) and left myself too little time to get all the way through 22.  123 good, but I would've felt a lot better if I hadn't known so much of the math that well and [i]still[/i] not answered.  Also, while I didn't miss 17, spending 10 minutes trying to come up with a proof to make myself feel comfortable (I integrated in the end, OK) wasn't -- Good.

I still hold that 19,23,25 should have been in the ten-section; they were just too easy (you were right, codeblue):
[quote="codeblue87"]...significantly easier, at least in the difficulty of the last 5 problems. compared to the 2004 12B, the last 5 in this one are [i]much[/i] easier. our school's team score is 423.[/quote]And I'll hold out for that trig...  I couldn't have felt dumber for it not coming up because that's what I gave the people at my school to look over the afternoon of the test.  (We took it late cause of our State's core testing.)  Scary AIME :?   :?:

Joe, I think your diagram is sideways.  Nevertheless, that's a good one.  I've seen it before (still made a mistake looking at my diagram--which I copied out of the booklet to preserve its usability), and I think:  There is one we should definitely just memorize (the method, not the ratio).

[quote="Simfish"]-. This is the first community where I am below-average. I feel so... worthless.[/quote] $\text {%Error. "endnotesize" is a bad command.
\hfill Don't worry, you'll get over it (and it makes you smarter, too: 30 points or so in a year for me). \hfill%whatever to the centering
}$
[quote="Simfish"]It is very likely to be either 104.5 or 98.5 (if 98.5, I'm hoping that more people taking the AMC = more worse students and a lowering of the curve).

Some very smart person obtained a perfect score of 0! This person must be literally very intelligent.[/quote]He he to the last (but such a waste).  I saw that too in the world stats book for last year.  But for the score:  Don't count on it.  I've never seen the cutoff break below 100.  Usually, it's actually around a 103 or 104 at least (do I remember??).  But there shouldn't be any circumstance to keep you from getting back your book...

i think number 24 was the hardest question on the test, the one about the polynomial.  hands down.

did you get it? I didn't have time to fool with it by the time I got there.

[quote="vera"]Can anyone explain to me number 19 again? I still don't really get it...
I did really bad on this contest, only 13 questions... I spent way too much time on number 18 and  I still forgot the exclusion thing. *sigh* that's life..
I also don't really get the dice question[/quote]

19)  The odometer skips every "4" so instead of 10 numbers per place holder, there are only 9.

Thus, 2005 is in base 9, so $2\cdot 9^3 + 0\cdot 9^2 + 0\cdot 9^3 +5\cdot 9^3 = 1463$, but, the last number is $2005$, so a number was just skipped, therefore it is $1463 - 1  =1462$.

14) There are 21 total dots on the die, so the probability of making aside of the die odd is $\frac{\left(2 + 4 + 6\right)}{21} = \frac{12}{21}$ and the probability of making a side of the die even is $\frac{\left(1 + 3 + 5\right)}{21} = \frac{9}{21}$.

If we made the side odd, then the probability of rolling an odd has increased to $\frac{4}{6}$, but if we made the side even, the probability has fallen to $\frac{2}{6}$.

So, we then multiply and add the respective probabilities.

$\frac{12}{21}\cdot\frac{4}{6} + \frac{9}{21}\cdot\frac{2}{6} = \frac{11}{21}$

[quote]There are 10 greedy (and perfectly logical) bandits with 100 indivisible pieces of gold. The system for dividing the gold is suggested by the bandit with the most seniority (bandit #1), and all of the bandits vote on it. If his plan gets at least half of the votes, it passes. If not, he is killed and bandit #2 suggests a plan, and so on down to bandit #10. You are bandit #1. The following are true:

1) Each bandit is completely logical (and therefore can predict the actions of every other bandit).
2) Each bandit's first priotiry is to stay alive.
3) Each bandit wants to maximize the amount of gold he/she recieves.

What system do you suggest so that you don't die and you receive as much gold as possible? (remember, you need at least 4 votes plus your own) [/quote]

this problem is very interesting but wouldn't you have to understand the logic of each of the bandits?
I mean if you suggest giving each bandit 10 pieces how do you know how each of them are going to vote?
Will 2 say yes because he is afraid of dying or no because he can get more once 1 dies?

qweretyq: the point of the problem is that because every pirate is perfectly logical, we can deduce what each pirate will do in any situation.  For example, you asked

[quote]
this problem is very interesting but wouldn't you have to understand the logic of each of the bandits?
I mean if you suggest giving each bandit 10 pieces how do you know how each of them are going to vote?
Will 2 say yes because he is afraid of dying or no because he can get more once 1 dies?[/quote]

Pirate 2 will say no to a proposal made by pirate 1, only if 2 can think of some proposal that he can make if 1 dies that 2 is sure at least three other pirates will support and where 2 will get more gold than in the proposal that 1 makes.

The point is that because all the pirates are perfectly logical, 2 can accurately predict whether he can come up with a plan where he gets more gold and at least three other pirates will support him.

Does that make sense?

Petra

Actually, number 2 would still need four others to vote for his proposal.

[hide="A huge hint to this question"](and the way I actually figured it out) was the context in which it was referenced.[/hide]

Each pirate is supposed to have perfect logic, and will vote in whatever way most benefits him, wanting 1) to stay alive and 2) as many coins as he can get. [hide="The key"]All pirates think alike, so they know how all the others think (and will vote)[/hide]

The point is for you, pirate #1, to keep as many gold coins for yourself as possible and survive.

Way to catch "subtle" hints, origamimasterjared :D

Has anyone else figured it out?

[quote="Elemennop"]I still have no idea how to do #16; I [b]always[/b] get the identity property.

Edit> Wow, I finally saw it; it was right in front of me the whole time and I never saw it. How embarrasing.

About #18 for the last two posts; the LCM of 2, 3, and 5 is 30, and there are 22 numbers between 1 and 30 that are divisible by 2, 3, or 5. Since 30 is the LCM, there are 22 non-prime looking composite #'s for every set of 30 counting numbers; thus the ratio of 22:30 = 726:990, plus 992, 993, 994, 995, 996, 998, and 999, for a total of 733 non-PM composites.

Add the 168 primes for 901, subtract 2, 3, and 5 because you counted them twice (remember, they are prime when not adding the 30n to it). This gives 101 prime-looking, but 1 is neither prime or composite, so 100 is the answer.[/quote]

You peeps did 18 the HARD way. The easy way is to do it the back door aproach. You find the number not divisable by 2. Than subtract the ones not divisble by 3. Than subtract the ones not divisble by 5. Than subtract the primes. DUHHHHHH

I think you mean subtract the ones divisible by 2, 3, and 5...

I actually found it easier, less tricky, and almost as fast to just go through them one by one.: 7*7, 7*11...11*11, 11*13... Noting that there are 25 primes below 100, and then adding on the ones remaining before 143 (1000/7=142.9) then removing 2, 3, and 5.  So a list of primes from 7-139.  Then finding all the 7*(prime >= 7). 11*(prime >= 11) (because 7*11 is already used) and so on (after the 7-sequence it becomes really simple to count, so long as the numbers are listed).  There should be 94. Then there are six remaining, where there are three such primes multiplied: 7*7*7, 7*7*11, 7*7*13, 7*7*17, 7*7*19, 7*11*11. 94+6=100.


topology
function
advanced fields
advanced fields unsolved

Is it true that a Hausdorff separable space can't have cardinality greater than $2^{c}$?

A topological space $X$ with a countable dense subset $A$ cannot have cardinality $>c$, because for each point $x\in X$ there is a sequence of $a_{n}\in A$ that converges to $x$.

Edit: see jmerry's correction below.

Caveat: sequences must determine the topology for mlok's post to work.

For example (not Hausdorff), consider an arbitrary set $S$ with the cofinite topology. Any infinite subset of $S$ is dense in the space.

See fedja's last post in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=28436]this thread[/url] for a correct proof.
That thread also contains an example of a separable Hausdorff space with $2^{c}$ elements: $\mathbb{R}^{\mathbb{R}}$ with the product topology. We could also use $\{0,1\}^\mathbb{R}$ in the same way, which is slightly less confusing. One countable dense set in $\{0,1\}^{\mathbb{R}}$ is the set of characteristic functions of finite unions of intervals with rational endpoints.


MATHCOUNTS
probability
rotation
analytic geometry
geometry
geometric transformation
function

1. how many pairs of vertical angles are formed by 5 distinct lines that have a common point of intersection?

2. three fair standard six faced dice of different colors are rolled. in how many wyas can the dice be rolled such that the sum of the numbers rolled is 10?

3. container I holds 8 balls and 4 green balls; containers II and III each hold 2 red balls and 4 green balls. a container is selected at random and then a ball is randomly selected from that container. what is the probability that the ball selected is green? express your answer as a common fraction.

4. how many full seven day weeks are there in seven consecutive years? assume the first day of the first year is the first day of the week.

5. the triangle with vertices A (6,1), B(4,1), and C(4,4) is rotated 90 degrees counterclockwise about B. what are the coordinates of the image of C (the point where C is located after the rotation)? express your answer as an ordered pair.

6. seventy two 50 cent pieces are lined side by side on a table. joe replaces every second 50 cent piece with a quarter and leaves. jane replaces every third remaining 50 cent piece with a dime and leaves. jim replaces every fourth remaining 50 cent piece witha nickel and leaves. what is the number of dollar in the amt of money left on the table? express your answer to the nearest cent.

Here's what I think for number 5:

[hide] The answer is (1,1). I'm not sure but I think it is because the phrase "counterclockwise about B" is that it is rotated to counterclockwise by point B staying in the same coordinates.. So, before the rotation, the coordinates of points were:  A (6,1), B(4,1), and C(4,4). But after the rotation, it would be A (4,3), B (4,1), and C(1,1).. I'm not sure whether it's right or not but it's kind of way you need to do with this kind of problem.. Post back if it's incorrect answer...[/hide]

2)
631 - 6 ways
622 - 3
541 - 6
532 - 6
442 - 3
433 - 3

27 ways (or so I hope)

If you had the time you could use generating functions to find the total number of ways to get 3-18 on 3 6 sided di, but yeah.

3)

(I assume that for container 1 you meant 8 red balls and 4 green balls)

I don't believe the answer is 1/2, Yiffie.

container 1:
1/3 x 1/3=1/9

container 2:
1/3 x 2/3=2/9

container 3:
1/3 x 2/3=2/9

1/9+2/9+2/9=5/9


Consider a cube with edges of length s . In simplest terms, what is the ratio of the number of cubic inches in the volume of a cube to the number of square inches in its surface area?

[hide="Solution"]Volume of a cube with side length $ s$: $ s^3$.

Surface area of a cube with side length $ s$: $ 6s^2$.

Ratio: $ \boxed{s: 6}$[/hide]


calculus
integration
logarithms
function
calculus computations

Find the minimum value of $ \int_0^1 \frac{2}{|2t\minus{}x|\plus{}3}\ dt$　for $ 0\leq x\leq 4.$

We can just grind through this. Let $ g(x)$ be the integral in question.

For $ 2\le x\le4,$ we have $ g(x)\equal{}\int_0^1\frac{2}{x\plus{}3\minus{}2t}\,dt$

For $ 0\le x\le 2,$ we have $ g(x)\equal{}\int_0^{\frac{x}{2}}\frac{2}{x\plus{}3\minus{}2t}\,dt\plus{} \int_{\frac{x}{2}}^1\frac{2}{2t\minus{}(x\minus{}3)}\,dt$

Working out the details, we get

$ g(x)\equal{}\begin{cases}\ln\left(\frac{(x\plus{}3)(5\minus{}x)}{9}\right)&x\in[0,2]\\ \ln\left(\frac{x\plus{}3}{x\plus{}1}\right)&x\in[2,4]\end{cases}$

This is clearly decreasing on $ [2,4].$

$ g(0)\equal{}\ln\left(\frac{5}{3}\right),\ g(2)\equal{}\ln\left(\frac{5}{3}\right),$ and $ g(4)\equal{}\ln\left(\frac{7}{5}\right)$

We need to explore the interval $ [0,2]$ further. In that interval, we find that

$ g'(x)\equal{}\frac{1}{x\plus{}3}\minus{}\frac{1}{5\minus{}x}\equal{}\frac{2\minus{}2x}{(x\plus{}3)(5\minus{}x)}.$

This is positive for $ x\in[0,1]$ and negative for $ x\in[1,2].$

$ g(1)\equal{}\ln\left(\frac{16}{9}\right).$ This is the [b]maximum[/b] value of $ g.$

The minimum value that was asked for is $ g(4)\equal{}\ln\left(\frac{7}{5}\right).$

For $ 2\leq x \leq 4$

$ I \equal{} \int_0^1 \frac{2}{x\minus{}2t\plus{}3}dt$

$ \minus{}\ln |x\minus{}2t\plus{}3| \Big | _0^1$

$ \ln\left|\frac{x\plus{}3}{x\plus{}1}\right|$

Min at $ \ln\frac{7}{5}$ for x=4

For $ 0\leq x \leq 2$

$ I \equal{} \int_0^\frac{x}{2} \frac{2}{x\minus{}2t\plus{}3}dt \plus{} \int_\frac{x}{2}^1 \frac{2}{2t\minus{}x\plus{}3}dt$

$ I \equal{} \ln\left|\frac{x\plus{}3}{3}\right| \plus{} \ln\left|\frac{5\minus{}x}{3}\right| \equal{} \ln\left|\frac{(x\plus{}3)(5\minus{}x)}{9}\right|$

This has min at $ \ln\frac{15}{9}$ for x=0,2   (and also max at $ \ln\frac{16}{9}$ for x = 1)

Since the min on the [2,4] interval is smaller than the min on the [0,2] interval, the min is

$ \ln\frac{7}{5}$

The three solutions appear to be in complete agreement, and use very similar methods.


analytic geometry
geometry
perpendicular bisector

Circle $C_1$ goes through points $A(0,1)$ and $B(4,-1)$ and its center is on the line $y=x-1$. Circle $C_2$ is created by reflecting $C_1$ over the line $AB$. If $P$ is a point on $C_1$ and $Q$ is a point on $C_2$, find the maximum possible length of $PQ$.

i'm not entirely sure, but is it ${2}{\sqrt{5}}$ ?

Something is wrong here, I think, because there can be but one circle passing through two given points and having its center on a line not parallel and not congruent with the bisector of the segment between the given points. Hence there's only one possible value for $PQ$ and that's $2\sqrt{5}.$

Perhaps the problem needs some editing.

Yes, you're right :blush: Sorry, I fixed the problem.

hmm..then is it just $2\sqrt{5}(\sqrt2+1)$ ?

just adding the length of two radii from the distance between the centers..

[quote="arkmastermind"]hmm..then is it just $2\sqrt{5}(\sqrt2+1)$ ?[/quote]
:yup:

Same here.  I got the center at $(3, 2)$, and from there it's simple...twice the sum of the radius and the distance from the center of $C_1$ to $PQ$.

[hide]
After finding the center $(3,2)$ by the intersection of $y=x-1$ and $y=2x-4$, the perpendicular bisector of $AB$, we see that the radius of $C_1$ and $C_2$ is $\sqrt{3^2+1^2}=\sqrt{10}$. The distance from either origin to the midpoint of $AB$ is $\sqrt{2^2+1^2}=\sqrt{5}$. Thus, the maximum value of $PQ$ is $\boxed{2(\sqrt{5}+\sqrt{10})}$
[/hide]


AMC AIME MATHCOUNTS trigonometry conics geometry calculus

I'm just curious as to how you guys adapted to the different style competition of the AIME. To me, its an entirely different look at a math contest from what came before it (MATHCOUNTS and AMC), and represents a seperate sort of challenge. Any suggestions as to how to crack it?

I didn't do so well myself( I got a 4)  However, I think that it is everything in high school math competitions.  You need to know trigonometry, conics, advanced geometry, a bit of calculus,etc.  But, not proofs. Besides that,  they give you problems of 15 at a time of 3 hours.  You need to be extremely careful in order to do well.  You have to look at many possibilites for every problem. 8-)

On AIME, it is best to work slowly and carefully. It is the mistakes that will cost you so much if you are careless. As for preparation, you don't need to do only AIME problems. Any problems you solve will help you to do well on all competitions. Incidentally, there's a Math Jam on AIME problems this Thursday, so it would be a good idea to watch that.

Hmm, well, I am personally more worries about the AMC12. I mean, if I study a lot and do many practice problems to get a very comfortable feel for what I am expecting, I can be somewhat confident that my work will pay off on the test, since, with 12 minutes per problem, I have plenty of time to think. The AMC12, however, is a lot more hectic. 3 minutes per problem does not give a lot of time to think about each problem and carefully check over each answer. And furthermore, there is that notion that if you do not score at least 140 or so, it will be very difficult to make USAMO, with the qualifying score increasing by considerbale amounts every year. I think that it will be very important to maintain your confidence, don't break down during the test, and stick to your typical methods for solving math problems instead of panicking on the actual exam.

Fierytycoon

One of the most important factors in AIME is to avoid careless mistakes as mentioned earlier.  Also, if you want to score really high, you have to have a high level of mathematics.  Here's what I mean:

After the first hour, I finish the first 7 on one of the previous years of AIME, but didn't manage to get any of the other questions right.  This probably indicates that there is some sort of level barrier between the 7th and 8th problem that I need to cross in order to get more questions correct.

Good luck on your preparation!  :)

guangweicao said;

    I didn't do so well myself( I got a 4) However, I think that it is everything in high school math competitions. You need to know trigonometry, conics, advanced geometry, a bit of calculus,etc.

  You don't need to know any calculus at all to make the IMO team, and not even the IMO questions have to be solved using calculus. All the problems in the math olmpiad can be solved using pre-calculus techniques.

Please don't revive five-year-old nonmathematical threads.


Let $ f$ be a function with the following properties:
(i) $f(1) \equal{} 1$, and
(ii) $ f(2n) \equal{} n\times f(n)$, for any positive integer $ n$.
What is the value of $ f(2^{100})$?

$ \textbf{(A)}\  1 \qquad \textbf{(B)}\  2^{99} \qquad \textbf{(C)}\  2^{100} \qquad \textbf{(D)}\  2^{4950} \qquad \textbf{(E)}\ 2^{9999}$

Such a function exists by defining $f(n)=n$ for odd $n$.


calculus
integration
trigonometry
logarithms
calculus computations

Calculate the following indefinite integrals.

[1] $\int \sqrt{x}(\sqrt{x}+1)^2 dx$

[2] $\int (e^x+2e^{x+1}-3e^{x+2})dx$

[3] $\int (\sin ^2 x+\cos x)\sin x dx$

[4] $\int x\sqrt{2-x} dx$

[5] $\int x\ln x dx$

[1], just expand it
[2] $e^{x+a}dx = d(e^{x+a})$
[3] $(\sin^x + \cos x)\sin xdx = -(1-\cos^2x+\cos x)d(\cos x)$
[4] variable substitution
[5] integrate by parts

[quote="liyi"][1], just expand it
[2] $e^{x+a}dx = d(e^{x+a})$
[3] $(\sin^2 x + \cos x)\sin xdx = -(1-\cos^2x+\cos x)d(\cos x)$
[4] variable substitution
[5] integrate by parts[/quote]

edited by kunny

hmm.. for #2, it seems that the integral is the same thing. because you can rewrite $e^{x+1}=e.e^x$ so if you integrate you get:
$e^x+2e.e^x-3e^2.e^x$

[quote="kunny"][quote="amirhtlusa"]1
$\frac{2\sqrt{x^5}}{3}+x^2+\frac{2\sqrt{x^3}}{3}$[/quote]

That's wrong.[/quote]
what? i did it again and i got the same thing. expand it you get:
$x\sqrt{x}+2x+\sqrt{x}$ now if you integrate this, you'll get what i got... ;)

[quote="kunny"][quote="amirhtlusa"]3
$\cosh-\frac{\cos^3x}{3}+\frac{cos^2x}{2}$[/quote]

That's wrong. :?[/quote]
i fixed it, it was a typo...

[quote="amirhtlusa"][quote="kunny"][quote="amirhtlusa"]1
$\frac{2\sqrt{x^5}}{3}+x^2+\frac{2\sqrt{x^3}}{3}$[/quote]

That's wrong.[/quote]
aaaakhhh...it was another typo
it should be:
$\frac{2\sqrt{x^5}}{5}+x^2+\frac{2\sqrt{x^3}}{3}$[/quote]

Yes. :)


if a,b,c,d are real numbers then the inequality holds:

(a+b+c-d)(b+c+d-a)(c+d+a-b)(d+a+b-c)<=8(a^2d^2+b^2c^2)

[i]unknown source[/i]

It's pretty easy, isn't it?

We have
(a+b+c-d)(-a+b+c+d)=(b+c) 2 -(a-d) 2
(a-b+c+d)(a+b-c+d)=(a+d) 2 -(b-c) 2

We then multiply these two (it's boring, I know :D ) and we have to prove
E=4(ad+bc) 2 - (a 2 +d 2 -b 2 -c 2 ) 2 <= 8((ad) 2 +(bc) 2 ), but E<=4(ad+bc) 2 <= 8((ad) 2 +(bc) 2 ), and we're done.


[quote="Naib Stilgar"]So...in the heat of SATs and APs and all that what not, I forgot to select a prize for USAMTS:(. Now I don't get any new books!

:'([/quote]

You'll still get something, but we'll be picking them more or less at random from our stash :)

They're going out this week.

Ah I got them. and two happened to be ones I wanted. but now I have two copies of countdown....so....looks like christmas shoppoing for some relative will be easy ;)


I have a realy simple question, but still I don't know the answer to it.... :lol:
I have a JPEG picture from wich I would like to remove the background (meaning I whant to have an image only with the object) ..
I used Photoshop and it can only save my picture the way I want it with the PNG extension...
What should I do? Cause I have to make the picture a JPEG or a GIF.. and I cannot convert it afterwards....
Can anyone help me with my problem? Thank you...

On Photoshop, when you save you can select which file type you'd like to save as. Go hunt for it. If you can't get it, post again and I'll try to give more details.

You mean the quality of the JPEG or GIF isn't as good? Well, theres reason, because JPEG and GIF arent as high quality as PNG is. But when you save it under JPEG type in photoshop, atleast in the newer versions, it allows you to adjust quality. So adjust the quality to the top, and make all compression settings as best quality as you can, it should come out very very very close to PNG quality, if not the quality.

And, if you need to remove the background.. you can make the background a transparency or just fill it with white..

I don't have enough experience with Adobe Photoshop so I can't say that I can help you in this case.

As far as erasing a background goes, if you simply just use the eraser tool and get rid of it, on Photoshop it'll appear transparent but it'll appear white once it's converted into a JPG and used elsewhere.

And like ToBe said, you can always just save it as a high quality JPG. I'm not sure why you absolutely need it as a JPG or GIF and not PNG, because most softwares and websites do accept PNG files...

If you need a jpeg/gif file, a window should pop up at the end of saving asking for the quality you want (depends on what version of photoshop).
And for a less tedious way to get rid of a background, just make a new layer, use the magnetic lasso tool to outline the unwanted backgound, and color it all white.
Hope this helped.


Given $ a,b,c,d>0$ and $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$.Prove that:
$ \frac{a^4\plus{}b^4\plus{}c^4\plus{}d^4}{abcd}\plus{}\frac{16}{3}abcd \ge \frac{28}{3}$ :)

How can we prove that $ (a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2\le 3(a^4\plus{}b^4\plus{}c^4\plus{}d^4)\plus{}4abcd$?

:)

quykhtn-qa1, thank you for your gift.
ElChapin, see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6397
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204175

[quote="secrets"]$ \frac {3(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4}{abcd} \equal{} \frac {3(a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4) \plus{} 4abcd}{abcd} \minus{} 4 \ge \frac {(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2}{abcd} \minus{} 4 \equal{} \frac {16}{abcd} \minus{} 4$
Finally,we will to show that
$ \frac {16}{abcd} \plus{} 16abcd \ge 32$
This ineq is true by AM-GM.This proof is completed.Equality occurs if and only a=b=c=d=1.[/quote]
Nice proof,secrets :lol:


:? How should I simplify
Sqrt(6-Sqrt(6-Sqrt(6-Sqrt(6-Sqrt(6-... (to infinity) :?
Sqrt is the Square Root function (I'm unaccustomed to LaTeX)

[quote="Aryth"]:? How should I simplify
$\sqrt{6-\sqrt{6-\sqrt{6-\sqrt{6-\sqrt{6-...}}}}}$ :?
[/quote]

Let all that be x.

Then $x^2=6-\sqrt{6-\sqrt{6-\sqrt{6-\sqrt{6-\sqrt{6-...}}}}}$
$x^2=6-x$
$x^2+x-6=0$
$x=2$
$x=-3$

It obviously isn't negative, so $\sqrt{6-\sqrt{6-\sqrt{6-\sqrt{6-\sqrt{6-...}}}}}=2$

[quote="nat mc"][quote="anirudh"]I also was wondering what that was except with 2[/quote]
$x=\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}}$
$x=\sqrt{2-x}$
$x^2=2-x$
$x^2+x-2=0$
$x=1$[/quote]

thanks


HMMT
number theory
prime numbers
puzzles

This is one of the problems on an HMMT competition:
[quote="HMMT"]
Write down at least one, and up to ten, different 3-digit prime numbers. If you somehow fail
to do this, we will ignore your submission for this problem. Otherwise, you’re entered into a
game with other teams. In this game, you start with 10 points, and each number you write
down is like a bet: if no one else writes that number, you gain 1 point, but if anyone else
writes that number, you lose 1 point. [/quote]

I think we should start a game like this on AoPS. We'll start with [b]two-digit[/b] primes, then in each round increase the number of digits by one. The rules are the same; you can submit from one to ten prime numbers of the specified amount of digits in each round, in a PM to me. If you are the only one to submit a specific prime number, you get a point; if two or more people submit the same prime, they all lose one point each from their total score. Every person begins with 10 points.

[b]Anyone can join at any time, and will join everyone else, starting with 10 points. Everyone who signs up [u]before[/u] Monday, Feb. 19, will begin in the first round.[/b]  :)

[u]These are the people who have signed up for the game so far:[/u]
[b]
calculuslover800 :coolspeak:
ckck
King Paul
now a ranger
moogra
junggi :coolspeak:
funcia
tjhance
Hamster1800 :coolspeak:
perfect628 :coolspeak:
rjt  :coolspeak:
[/b]
(Those with this little smiley :coolspeak: have their numbers submitted for the current round.)

Please, DON'T use any outside resources, i.e. prime number lists or websites, on this.... I think the fourth or fifth round will serve as an end to the game.

Note that [u]any[/u] digit in any of the prime numbers you submit can be a zero, i.e. 008 is valid for the second round; but keep in mind that there are only four of these numbers (2,3,5,7) available, so they are risky to choose.

Enjoy the game!   :)

Ok, great :P  We've got 8 people playing so far.

I was thinking about changing the rules a bit: instead of each person losing only one point if they all have the same number, I think that each person should lose a number of points based on how many had the same number. (Like, if 3 people had the number 1337, then they all would lose 3 points.)  Do you think we should change it?  :D

I'd like to play.

As for the rules change, I don't think it'd be a good idea to change it since losing so many points due to one prime would be disastrous to everyone who chose it. +/- 1 point is probably better (seems likely for us to lose more points than we gain anyway).

Ok, so for some reason the forum isn't letting me edit my original post. So... here's a list of the people who are currently in the game:
[b]
calculuslover800 *
ckck *
now a ranger *
moogra
junggi *
funcia *
tjhance *
Hamster1800 *
perfect628 *
rjt  *
kstan013 *
mustafa *
mathmom *
bpms
AiW *
[/b]
(Oh no! I've been smiley-limited! :( Those with an asterisk, then, have their numbers submitted.)

Remember, sign-up for the first round ends today, at midnight (or whenever I go to sleep, whichever comes first.)  :)

[quote="ZetaX"]This forum is for Puzzles and Brainteasers as it's title says. Games belong to the Games and Fun Factory, also as the name says.[/quote]

Ok, sorry. Can a mod move it, then?

[quote="bpms"]I'll do it, I think you should change it back to 3 digit though, 2-digit doesn't work to well[/quote]

You're probably right; but let's see what happens for this first round. I might change the rules a bit so this round doesn't destory anyone's chances of winning -- the rules for losing points in this round might be different, or I might only take randomly two numbers from each person's list to (hopefully) create less overlap. But I'll wait to see which rule change is best.  :)

Ok guys, this first round wasn't too good. I think there were just too many people for the first round, and not enough primes; so I'm considering either only giving points to those who have a prime number solely in their list, or just dropping this round's results completely. Actually, I will do the former; this round will only go towards your benefit. If anyone has a problem with that (and I don't know why anyone would  :P ), just PM me or post here. Here are the results:

[b] [size=117] 13 -- Hamster1800 [/size] [size=109]

11 -- junggi, rjt, perfect628 [/size]

10 -- calculuslover800, ckck, now a ranger, moogra, funcia, tjhance, kstan013, mustafa, mathmom, bpms, AiW. [/b]

[size=150]Round Two (3-digit primes) will be open for submissions until Thursday or Friday.[/size] :)

Enjoy!

[quote="kstan013"]is round 2 3-digit numbers? Or is it still 2-digit?[/quote]

i just pmed him a 2 digit and a 3 digit one just in case

The digit count increases by one each round, so this round is [b]3-digit[/b] primes. :)

Oy, I'm really sorry that I haven't kept up with this game; I've been waaay too busy with school work for my own good.

I'll give whoever wants to join the chance to submit answers until [b]Saturday[/b], at which point we'll see how everyone did in this round, and continue to the next one (which might be the last one, depending on how it works out).  :)

uhh it's saturday (or sunday depending on where you are). do we have our answers yet?

It seems, cincodemayo5590 is really too busy to run this game right now  :(  , so I thought maybe someone else should take it over. I would be glad to try, if you guys have no objections.

Please let me know, what do you think?  :roll:

dude, i believe so this is what happens to aops games :wink:
 it was interesting


linear algebra
matrix
linear algebra unsolved

Suppose that $A\in\mathcal M_{n}(\mathbb R)$ with $\text{Rank}(A)=k$. Prove that $A$ is sum of $k$ matrices $X_{1},\dots,X_{k}$ with $\text{Rank}(X_{i})=1$.

Isn't this clear? Let $v_{1},...,v_{k}$ a basis of the image of the linear map. Each $Ax$ is of the form $A_{1}(x)v_{1}+...+A_{k}(x)v_{k}$. This gives the desired linear maps: $B_{i}(x)=A_{i}(x)v_{i}$ which have rank 1 and which add up to $A$.

Hello.

Let $ A$ be the $ n \times n$ matrix assosiated with the linear map. Now $ A$ having rank $ k$ will mean that $ A$ will have $ k$ linearly independent rows. So if we perform elementary row operations on the rows of $ A$ we can form a matrix $ A'$ with the first $ k$ rows linearly independent, and the other rows all with zero entries. So we may express $ A'$ as the sum of $ k$ matrices, $ A'_i$, with rank $ 1$ - namely $ A'_i$ is the matrix which shares the same same $ i$'th row as $ A'$, but all the other entries are $ 0$.

Now $ B$ be the "elementary row operation" matrix which transforms $ A$ to $ A'$.

Thus $ BA \equal{} A' \Rightarrow A \equal{} B^{ \minus{} 1}A'$.

$ \Rightarrow A \equal{} \sum_{i \equal{} 1}^{k} B^{ \minus{} 1}A'_i$

$ B^{ \minus{} 1}$ is also an "elementary row operation" matrix. But performing row operations on the matrix $ A'_i$, with only one (non-zero) row will still leave it with one lineraly independent row! Hence $ A_i \equal{} B^{ \minus{} 1}A'_i$ is a one rank matrix, which when summed over $ i$ will give us $ A$.

I may include a further short post for those that do not no how to construct $ B$, and also show why the existance of $ B^{ \minus{} 1}$ is trivial.

As I promised:

Elementary row operations can be broken into two different classes. Those which exchange two rows, and those which add a multiple of one row onto another.

Say we have a square $ n \times n$ matrix $ A$ and we wish to exchange the $ a$'th and $ b$'th rows of this matrix where $ a \neq b$. Premultiplying by the following matrix does the trick:

$ B_{ii} = 1$ whenever $ i \neq a,b%
$
$ B_{ab} = 1$
$ B_{ba} = 1$
And all other entries are 0.

We can verify this with suffix notation (I hope it is clear where summation convention does or does not apply):

When $ i \neq a,b$: $ (BA)_{ij} = B_{ik}A_{kj} = B_{ii}A_{ij} = A_{ij}$
When $ i = a$: $ (BA)_{aj} = B_{ak}A_{kj} = B_{ab}A_{bj} = A_{bj}$
When $ i = b$: $ (BA)_{bj} = B_{bk}A_{kj} = B_{ba}A_{aj} = A_{aj}$

Thus it does the required job. Note that $ B$ is it's own inverse. Muliplying by $ B$ swaps the rows back into their intial position

Now we find the matrix which adds a multiple, $ m$, of the $ b$'th row onto the $ a$'th row. This matrix is:

$ B_{ii} = 1  \, \forall i$
$ B_{ab} = m$
And all the other entries are zero.

Again we check by suffix notation...

When $ i \neq a$: $ (BA)_{ij} = B_{ik}A_{kj} = B_{ii}A_{ij} = A_{ij}$
When $ i = a$: $ (BA)_{aj} = B_{ak}A_{kj} = B_{aa}A_{aj} + B_{ab}A_{bj} = A_{aj} + mA_{bj}$

As required. The inverse is merely the matrix which adds a multiple $ -m$ of the $ b$'th row to the $ a$'th row.

Since each row operation has an inverse, the full composition of row operation matrix is also invertible. Hence we have shown the existance of $ B$ and $ B^{-1}$. it should also be clear why $ B^{-1}$ is a row operation matrix. It is because the inverse of each row operation is just a row operation.

Now swapping two rows  in a matrix which has only one non-zero row, will leave it with one row. Adding a multiple of another row will only change the matrix if "the other" row is the non-zero row. So there can be two rows or more...but this doesn't cause a problem. Since the rows will still be linearly dependent (they are just multiples of each other).

I hope this clarifies any difficulties with the explaination


[quote="arkhammedos"]prove that  $\sum_{cycl}({\frac{a+2b}{a+2c}})^{3}\geq3$ where $a$ $b$ and $c$ are positive :lol:[/quote]
My ugly proof:
$\sum_{cyc}\left({\frac{a+2b}{a+2c}}\right)^{3}=\sum_{cyc}\frac{(a+2b)^{4}}{(a+2c)^{3}(a+2b)}\geq\frac{\left(\sum(a+2b)^{2}\right)^{2}}{\sum(a+2c)^{3}(a+2b)}\geq3.$
The last inequality is very ugly but true.

I think, for $a,$ $b$ and $c$ are positive numbers the following inequality true too:
$\sum_{cyc}\left({\frac{a+4b}{a+4c}}\right)^{3}\geq3$ :wink:

prove  (or disprove) that

        sum_cyc[(a+kb)/(a+kc)]^3>=3,

 where a,b,c>0 ,and k>=1/5.

After sometimes  that is my solution  since $x^{3}$ is convexe for all positive numbers  we aplly $\sum_{cyc}(\frac{a+2b}{a+2c})^{3}\geq3(\frac{\sum_{cyc}\frac{a+2b}{a+2c}}{3})^{3}$
In other hand we have $\sum_{cyc}\frac{a+2b}{a+2c}+3=2(a+b+c)(\sum_{cyc}\frac{1}{a+2c})$
Now via $AM-HM$ we get $\sum_{cyc}\frac{1}{a+2c}\geq\frac{3}{a+b+c}$ then it comes
$\sum_{cyc}\frac{a+2b}{a+2c}+3\geq6$ so $\sum_{cyc}\frac{a+2b}{a+2c}\geq3$ we return to the first
We get $\sum_{cyc}(\frac{a+2b}{a+2c})^{3}\geq3(\frac{3}{3})^{3}$ and we are done.


calculus
integration
geometry
parameterization
calculus computations

Let $a$ be a positive number. Assume that the parameterized curve $C: \ x=t+e^{at},\ y=-t+e^{at}\ (-\infty <t< \infty)$ is touched to $x$ axis.

(1) Find the value of $a.$
(2) Find the area of the part which is surrounded by two straight lines $y=0, y=x$ and the curve $C.$

[hide="Part 1"]
$\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{-1+ae^{at}}{1+ae^{at}}$
Find $t_{1}$ such that this is $0$, i.e. $-1+ae^{at_{1}}=0$... $ae^{at_{1}}=1$
at this point $t_{1}$ y value must be zero, so $t_{1}=e^{at_{1}}$
we easily solve system of two equations for $a, t_{1}$ $at_{1}=1$ $t_{1}=e^{at_{1}}=e, \boxed{a=\frac{1}{e}}$[/hide]

I know it is really easy to make stupid mistakes in these kind of problems, can anyone tell me if I was even on the right path? Kunny?

Sorry for my reply.

Regrettably Your answer is incorect.

The curve of $C$ is decreasing and cocave down on $0\leq x\leq 2e.$

kunny

[quote="me@home"][hide="Part B)"]
$y=0$ when $t=e^{t/e}$ which is only at $t=e$, the point is $y=0,x=2e$
$y=x$ only when $t=0$, at $y=x=1$.
[img]6861[/img]
We try to find the area of the shaded region. It is the integral from 0 to 1 of the x parametrization from 0 to e (corresponds to green line), minus the triangle with area 1/2.
So: $\left(\int_{t=0}^{e}t+e^{t/e}dt \right)-\frac12 = \frac32 e^{2}-e-\frac12$ if I didn't make computational or stupid mistakes.[/hide][/quote]

I don't think $\int x dt$ gives the area... maybe you need to do $\int y dx$ instead.


calculus
integration
trigonometry
analytic geometry
geometry
calculus computations

[quote="kunny"]Let $ H$ be the piont of midpoint of the cord $ PQ$ that is on the circle centered the origin $ O$ with radius $ 1.$
Suppose the length of the cord $ PQ$ is $ 2\sin \frac {t}{2}$ for the angle $ t\ (0\leq t\leq \pi)$ that is formed by half-ray $ OH$ and the positive direction of the $ x$ axis. Answer the following questions.

(1) Express the coordiante of $ H$ in terms of $ t$.

(2) When $ t$ moves in the range of $ 0\leq t\leq \pi$, find the minimum value of $ x$ coordinate of $ H$.

(3) When $ t$ moves in the range of $ 0\leq t\leq \frac {\pi}{2}$, find the area $ S$ of the region bounded by the curve drawn by the point $ H$ and the $ x$ axis and the $ y$ axis.[/quote]

For Part (1)

$ PQ \equal{} 2\sin\frac {t}{2}\implies PH \equal{} \sin\frac {t}{2}$

$ OH \equal{} \sqrt {OP^2 \minus{} PH^2} \equal{} \sqrt {1 \minus{} \sin^2\frac {t}{2}}\implies OH \equal{} \left|\cos \frac {t}{2}\right| \equal{} \cos \frac {t}{2}$

$ \implies\boxed{\boxed{ H\equiv \left(\cos \frac {t}{2}\cos t,\ \cos \frac {t}{2}\sin t\right)}}$

For Part (2)

$ x \equal{} \cos \frac {t}{2}\cos t$

$ \implies \frac {dx}{dt} \equal{} \minus{} \frac 12 \sin\frac {t}{2} \cos t \minus{} \cos \frac {t}{2}\sin t \equal{} \minus{} \frac 12 \sin\frac {t}{2} \left(1 \minus{} 2\sin^2\frac {t}{2}\right) \minus{} 2\sin \frac {t}{2}\cos^2 \frac {t}{2}$

$ \implies \frac {dx}{dt} \equal{} 3\sin^3\frac {t}{2} \minus{} \frac 52 \sin\frac {t}{2} \equal{} 3\sin\frac {t}{2}\left(\sin\frac {t}{2} \plus{} \sqrt {\frac 56}\right)\left(\sin\frac {t}{2} \minus{} \sqrt {\frac 56}\right)$

$ \implies \min\ x$ will occur at $ \ t \equal{} 2\arcsin \sqrt {\frac 56}$

$ \implies\boxed{\boxed{\min x \equal{} \minus{} \frac {2}{3\sqrt {6}}}}$

For Part (3)

Required area $ \ \equal{}\frac 12\int_0^{\frac {\pi}{2}}\left(x(t)y'(t)\minus{}y(t)x'(t)\right)dt$

$ \equal{}\frac 12\int_0^{\frac {\pi}{2}}\left\{ \cos \frac {t}{2}\cos t \left(\minus{}\frac 12\sin \frac {t}{2}\sin t\plus{}\cos \frac {t}{2}\cos t\right)\plus{}cos \frac {t}{2}\sin t \left(\frac 12 \sin\frac {t}{2} \cos t \plus{} \cos \frac {t}{2}\sin t\right) \right\} dt$

$ \equal{}\frac 12\int_0^{\frac {\pi}{2}}\left\{ \cos^2 \frac {t}{2}\cos^2 t \plus{}cos^2 \frac {t}{2}\sin^2 t \right\} dt$

$ \equal{}\frac 14\int_0^{\frac {\pi}{2}}\left( 1\plus{}\cos t \right) dt\equal{}\frac 14\left( x\plus{}\sin t \right)_0^{\frac {\pi}{2}}$

$ \boxed{\boxed{\text{Required area} \ \equal{}\frac 14\left( \frac {\pi}{2}\plus{}1 \right)\equal{}\frac {\pi\plus{}2}{8}\ \text{square units}}}$

what is the meaning of [b]half-ray[/b] $ OH$ ?

or, [b]ray[/b] ; a line which extends from the origin $ O$ toward $ H$ infinitely.


$ 1\blacktriangleright \{x,y,z\}\subset (0,\infty )\implies 0\le \boxed{(x+y+z)^{3}-27xyz\le 11\cdot\left(x^{3}+y^{3}+z^{3}-3xyz\right)}$.

$ 2\blacktriangleright \left\{\begin{array}{c}\{x,y,z\}\subset (0,\infty )\\\ x+y+z=1\end{array}\right\|\implies 1+6xyz\le 11\cdot \left(x^{3}+y^{3}+z^{3}\right)$.

$ 1.\blacktriangleright$ Let $ S=x+y+z,\,P=xy+yz+zx,\,Q=xyz$
We have
\begin{eqnarray*}\left(\sum x\right)^{3}-27\cdot xyz&\le& 11\cdot\left(x^{3}+y^{3}+z^{3}-3\cdot xyz\right)\\ \iff S^{3}-27\cdot Q&\le& 11\cdot S\cdot \left(S^{2}-3P\right)\\ \iff 33\cdot SP&\le& 10\cdot S^{3}+27\cdot Q\\ \end{eqnarray*}
On the other hand,
\begin{eqnarray*}27\cdot\prod\left(y+z\right)&\le& 8\cdot\left(\sum x\right)^{3}\\ \implies 27\cdot\left[\sum yz(y+z)+2xyz\right]&\le& 8\cdot\left(\sum x\right)^{3}\\ \implies 27\cdot\left(PS-Q \right)&\le& 8\cdot S^{3}\\ \iff 27\cdot PS&\le& 8\cdot S^{3}+27\cdot Q \end{eqnarray*}
So, we have to prove that $ S^{2}\ge 3\cdot P\iff \sum\left(y-z\right)^{2}\ge 0,$ what is truly

\[ 2.\blacktriangleright\quad 1+6\cdot xyz\le 11\cdot\left(x^{3}+y^{3}+z^{3}\right)\\ \iff \left(\sum x\right)^{3}-27\cdot xyz\le 11\cdot\left(\sum x^{3}-3\cdot xyz\right)\]


Is it possible to approximate the x in the given expression $2^x=x^2$ with series? If so, what is the series? If not, how else would we approximate it other than using calculators?

i'm not really sure how to solve this, but here's as much as i got:

[hide] $2^x=x^2$

$x\log{2}=2\log{x}$

$\frac{\log{2}}{2}=\frac{\log{x}}{x}$

and then i guess the only thing you can do with that is graph $y=\frac{\log{x}}{x}$, which has a max at e.  and you find it equals $\frac{\log{2}}{2}$ when x=2 or 4.  but, actually from the way the graph of $y=\frac{\log{x}}{x}$ looks, you know there will only be 2 solutions.  but i don't know how to do it without the calculator  :?
[/hide]

well u should be able to see that $x=2$ and $x=4$ work but for the other solution(s) u prob. will have to use newton's method.

[hide]basically if you have $a^x=x^a$, then you end up with $\frac{\log{a}}{a}=\frac{\log{x}}{x}$, so i don't really know what to do with that except graph it.  but sometimes it won't have real solutions, like when $\frac{\log{x}}{x}$ is greater than (don't know the latex symbol for greater than) e.  but how would you figure out those imaginary answers??  hmm.  maybe you could use calculus, somehow, as in, for getting exact values for the real answers.  i don't know....[/hide]

$2^x=x^2$  This was discussed previously, search and try to find it......

[quote="Hokkage"]Whoops, my question is sort of lame. I was thinking something more along the lines of if you have $x^a=a^x$ (where $x \neq a$ then is there a way to use series to approximate it. For example, try $x^3=3^x$ find the solution where $x \neq 3$.[/quote]

I have repeatedly posted the link to the general solution of $a^b = b^a$.  Try to find it.

how do we solve this?

((( a ^ b ) ^ c ) ^ d ) = ((( d ^ c ) ^ b ) ^ a ) when all a, b, c and d are different.

any takers on this?

[/code]

[quote="t0rajir0u"]It simplifies to $a^{bcd} = d^{abc} \implies a^{d} = d^{a}$ which is the previous problem anyway.[/quote]

Thanks !, I can solve it now.


Consider the sequence of positive real numbers

\[ 4,\; 1/3,\; 4/3,\; 4/9,\; 16/27,\; 64/243,\;\ldots,\]

in which each term (after the first two) is the product of the two previous terms. Note that for this particular sequence, the first and third terms are greater than one while the second and fourth terms are less than one. However, after that the "alternating" pattern fails: the fifth and all subsequent terms are less than one. Do there exist sequences of positive real numbers in which each term is the product of the two previous ones and for which [i]all [/i]odd-numbered terms are greater than one, while all even numbered terms are less than one? If so, find, with proof, all such sequences.  If not, prove that no such sequences exist.

[hide]Yes, such sequences exist. We claim that we can characterize all such sequences as follows: let $ a$, the first term, be an arbitrary real number greater than 1. Then, take the second term $ b$, such that $ b\equal{}a^{\minus{}1/\phi}$, where $ \phi\equal{}\dfrac{1\plus{}\sqrt{5}}{2}$.

[b]Lemma 1.[/b] Let $ x_{n}$ denote the $ n$-th term of the sequence, and let $ x_{1}\equal{}a,x_{2}\equal{}b$. Then, for $ n\ge2$, we have $ x_{n}\equal{}a^{F_{n\minus{}2}}b^{F_{n\minus{}1}}$, where $ F_{k}$ denotes the $ k$-th Fibonacci number, where we take $ F_{0}\equal{}0,F_{1}\equal{}1$ and $ F_{k}\equal{}F_{k\minus{}1}\plus{}F_{k\minus{}2}$ for $ k\ge2$.
[b]Proof.[/b] We induct on $ n$. First, we prove the base cases, $ n\equal{}2,3$. For $ n\equal{}2$, we have $ x_{2}\equal{}b\equal{}a^{F_{0}}b^{F_{1}}$. For $ n\equal{}3$, we have $ x_{3}\equal{}x_{1}x_{2}\equal{}ab\equal{}a^{F_{1}}b^{F_{2}}$. This establishes the base cases. Now, assume the claim holds for $ n\equal{}i$ and $ n\equal{}i\plus{}1$, where $ i\ge2$. We have $ x_{i\plus{}2}\equal{}x_{i}x_{i\plus{}1}\equal{}a^{F_{i\minus{}2}}b^{F_{i\minus{}1}}\cdot a^{F_{i\minus{}1}}b^{F_{i}}\equal{}a^{F_{i}}b^{F_{i\plus{}1}}$ by the definition of the Fibonacci numbers. This completes the proof of the lemma.
---end lemma---

[b]Lemma 2.[/b] Consider the ratio $ \dfrac{F_{n\plus{}1}}{F_{n}}$, where $ F_{n}$ again denotes the $ n$-th Fibonacci number. Then, if $ n$ is odd, $ \dfrac{F_{n\plus{}1}}{F_{n}}<\phi$, and when $ n$ is even, $ \dfrac{F_{n\plus{}1}}{F_{n}}>\phi$. Furthermore, as $ n$ goes to infinity, $ \dfrac{F_{n\plus{}1}}{F_{n}}$ converges to $ \phi$.
[b]Proof.[/b] By Binet's Formula, we have

$ \dfrac{F_{n\plus{}1}}{F_{n}}\equal{}\dfrac{\dfrac{1}{\sqrt{5}}(\phi^{n\plus{}1}\minus{}(\minus{}1/\phi)^{n\plus{}1})}{\dfrac{1}{\sqrt{5}}(\phi^{n}\minus{}(\minus{}1/\phi)^{n})}\equal{}\dfrac{\phi^{n\plus{}1}\minus{}(\minus{}1/\phi)^{n\plus{}1}}{\phi^{n}\minus{}(\minus{}1/\phi)^{n})}$.

Observe that when $ n$ is odd, the numerator is greater than $ \phi^{n\plus{}1}$, since $ \phi>0$, and furthermore the denominator is less than $ \phi^{n}$. Thus, the fraction is more than $ \phi$. Similarly, when $ n$ is even, we find that $ \dfrac{F_{n\plus{}1}}{F_{n}}<\phi$. Now, since $ 1/\phi<1$, as $ n$ goes to infinity, $ (1/\phi)^{n}$ goes to zero. Thus, these terms vanish, and we are left with $ \dfrac{\phi^{n\plus{}1}}{\phi^{n}}\equal{}\phi$. Thus, the ratios of consecutive terms converge to $ \phi$.
---end lemma---

We are now ready to prove the main claim. Let $ \{x_{n}\}$ be such a sequence and let $ x_{1}\equal{}a>1,x_{2}\equal{}b<1$. Now, for $ n$ odd greater than 1, we need $ a^{F_{n\minus{}2}}b^{F_{n\minus{}1}}<1$, from which we get $ a<b^{\minus{}F_{n\minus{}1}/F_{n\minus{}2}}$ for all odd $ n>1$. By Lemma 2, $ \minus{}F_{n\minus{}1}/F_{n\minus{}2}>\minus{}\phi$ for all odd $ n>1$, and furthermore the left hand side converges to $ \phi$. Therefore, $ a\le b^{\minus{}\phi}$. Doing a similar analysis for $ n$ even, we find that $ a\ge b^{\minus{}\phi}$. Combining these, we get the sufficient condition $ a\equal{}b^{\minus{}\phi}$, or $ b\equal{}a^{\minus{}1/\phi}$.

Clearly, it is also sufficient that $ a>1$. Thus, all such sequences must have this form. It suffices to prove that sequences of this form alternate between terms greater than 1 and less than 1. We can do this by reversing our steps; If $ n$ is odd, $ a^{F_{n\minus{}2}}b^{F_{n\minus{}1}}\equal{}a^{F_{n\minus{}2}}a^{\minus{}F_{n\minus{}1}/\phi}>a^{F_{n\minus{}2}\minus{}F_{n\minus{}2}}\equal{}1$, and if $ n$ is even, $ a^{F_{n\minus{}2}}b^{F_{n\minus{}1}}\equal{}a^{F_{n\minus{}2}}a^{\minus{}F_{n\minus{}1}/\phi}<a^{F_{n\minus{}2}\minus{}F_{n\minus{}2}}\equal{}1$, as desired.[/hide]

[hide]Let $ \{ u_n \}_{n \ge 1}$ be a sequence that works, satisfying $ u_{n + 2} = u_{n + 1}u_n$, $ u_{2n - 1} > 1$, and $ u_{2n} < 1$ for all positive integers $ n \ge 1$.

If we form a new sequence $ \{ s_n \}_{n \ge 1}$ so that $ s_n = \ln{u_n}$ for all positive integers $ n$, then $ s_n$ satisfies $ s_{n + 2} = s_{n + 1} + s_n$, $ s_{2n - 1} > 0$, and $ s_{2n} < 0$.  By a trivial induction on $ n$ using our recurrence relation, we have that $ s_n = F_{n - 2} s_1 + F_{n - 1} s_2$, where $ F_n$ denotes the $ n$th Fibonacci number.

Now, by Binet's formula, we have that $ F_n = \frac {1}{\sqrt {5}}(\phi ^n - \varphi ^n)$, where $ \phi = \frac {1 + \sqrt {5}}{2}$ and $ \varphi = \frac {1 - \sqrt {5}}{2}$.
Thus, for odd $ n$, we have
\begin{eqnarray*} (\phi ^{n - 2} - \varphi ^{n - 2}) s_1 + (\phi ^{n - 1} - \varphi ^{n - 1}) s_2 & > & 0 \\
\phi ^{n - 2} s_1 + \phi ^{n - 1} s_2 & > & \varphi ^{n - 2} s_1 + \varphi ^{n - 1} s_2 \\
\phi ^{n - 2}( s_1 + \phi s_2) & > & \varphi ^{n - 2}( s_1 + \varphi s_2) \end{eqnarray*}
Since $ s_1 > 0$ , $ \varphi < 0$ and $ s_2 < 0$, we have $ s_1 + \varphi s_2 > 0$.  In addition, $ \phi > 0$, so we must have
\[ \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2} > \left( \frac {\varphi }{\phi } \right)^{n - 2} \qquad (*)
\]
for all odd $ n$.  Similarly, we must have
\[ \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2} < \left( \frac {\varphi }{\phi } \right)^{n - 2} \qquad (**)
\]
for all even $ n$.

Let us prove the intuitive fact that this cannot be the case for all $ n$ unless the LHS is 0:  Note that the RHS of both our inequalities goes to 0 as $ n$ approaches positive infinity, because $ |\frac {\varphi }{\phi }| < 1$.  This means that for any  $ \epsilon$, there exists some $ k$ such that $ - \epsilon < (\frac {\varphi }{\phi })^{N - 2} < \epsilon$ for all $ N > k$.

Thus, if $ \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2}$ is positive, then we can choose $ 0 < \epsilon < \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2}$ so that $ (\frac {\varphi }{\phi })^{N - 2} < \epsilon < \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2}$ for all $ N$ greater than some $ k$.  This is a contradiction, as we must have $ \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2} < ( \frac {\varphi }{\phi })^{n - 2}$ for all even $ n$, which is clearly violated for $ N$ even.

Similarly, if $ \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2}$ is negative, then we can choose $ \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2} < - \epsilon < 0$ so that $ \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2} < - \epsilon < (\frac {\varphi }{\phi })^{N - 2}$ for all $ N$ greater than some $ k$.  This is a contradiction, as we must have $ \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2} > (\frac {\varphi }{\phi })^{n - 2}$ for all odd $ n$, which is clearly violated for $ N$ odd.

Therefore, we must have $ \frac { s_1 + \phi s_2}{ s_1 + \varphi s_2} = 0$, so $ s_1 + \phi s_2 = 0$, or equivalently, $ s_2 = \varphi s_1$.  We can quickly check that all sequences with $ s_1 > 0$ and $ s_2 = \varphi s_1$ work using $ (*)$ and $ (**)$.

[/hide]


if abc=1 then  prove that:

             2(a+b+c)=<a^2(b+c)+b^2(a+c)+c^2(a+b)

darim $3 \le \sum{a}$ va $3 \le \sum{ab} \longleftrightarrow 1 \le (\sum{ab} -2)$
pas agar do ebarate bala ra dar ham zarb konim va ba tavajoh be in ke $0 \le 1,3$
darim
$3 \le (\sum{a})(\sum{ab} -2) =(\sum{a})(\sum{a^2} +\sum{ab} -\sum{a^2} -2)$ pas
$3 +(\sum{a})(\sum{a^2} -\sum{ab}) +2(\sum{a}) \le (\sum{a})(\sum{a^2})$
amma
$(\sum{a})(\sum{a^2} -\sum{ab}) +3 =\sum{a^3}$ pas ba in jaigozari va sadesazi be dast miaiad ke $2(\sum{a}) \le \sum{a^2b}$