abstract algebra
function
superior algebra
superior algebra theorems

Hi

let us work in a random category C

suppose $(A,f_1,f_2)$is a categorical product of $A_1$ and $A_2$ with $f_i: A \rightarrow A_i$

do the $f_i$ have to be epic morphisms?

suppose $(A,f_1,f_2)$is a categorical coproduct of $A_1$ and $A_2$ with $f_i: Ai\rightarrow A$
do the $f_i$ have to be monic morphisms?

When I do this in $C=R-mod$ the answer seems to be yes?

But is it true in general??

Both is true for $R$-mod, and probably even for abelian categories. But already in Set it's not true: Consider the second projection of the empty product $\emptyset \times \mathbb{N}$.

Hmm, I must say I am not so familiar with Set.
I mean, with the empty set as object.  What morphisms leave and arrive there :?
I feel a little insecure with that empty set actually :oops:

Basics are useful before starting with category theory ;).

For every set $M$, there's exactly one function $\emptyset \to M$. It is surjective if and only if $M = \emptyset$. For every set $M$, there's a function $M \to \emptyset$ if and only if $M = \emptyset$. In this case, there's also only one function. To see this, you have to know the definition of a function as a special relation, which must be empty here, being a subset of $\emptyset \times M=\emptyset$ or $M \times \emptyset = \emptyset$.

I haven't found a non-monic coproduct-injection, yet.

In additive categories, the projections of a product are always epic and the injections of a coproduct are always monic:

http://www.math.washington.edu/~smith/Teaching/513nag/hwksols.ps , Exercise 1.6.

The proof even shows, that the projections $pr_i : \prod\limits_{i \in I} M_i \to M_i$ are epic and the injections $\iota_i : M_i \to \coprod\limits_{i \in I} M_i$ are monic, if there is some morphism $M_i \to M_j$, for all $i,j$.

Just use your set-example in the opposite category, to get a coproduct with nonmonic injections.

LiMa

But sorry for me being stubborn  :roll:  but what if I 'oppose' your set example?

What if I claim that in the category of sets, no morphisms leave or arrive here?

Is an example still possible?

The example is correct, for sure. Where's your problem?

I just think that it is fairly reasonable to see Set as a category where the zero set doesn't have any morphisms leaving or entering

So actually I am changing the category Set

I was just wondering, is there another example,  :roll:

another example: in the category of (commutative unitary) rings, the coproduct is given by the tensor product over $\mathbb{Z}$. especially, we have $\mathbb{Z}/2 \coprod \mathbb{Z}/3 = 0$ so that the inclusions are not monomorphisms.


Let a and b numbers be nonconsective numbers.Let say "close numbers" to a and b numbers if   $a|b^2-1$ and $b|a^2-1$.(For example 3 and  8 close numbers.)Show that there will be a pair of close number in interval [n,8n-17] for every n >=4 positive number.

n and n+1 are always close numbers since $n^2-1=(n+1)(n-1)$ and $(n+1)^2-1=n^2+2n=n(n+2)$.

Read the first sentence -- consecutive integers don't count.

[hide]Actually, alternating Fibanocci numbers work.
1,1,2,3,5,8,13,21,34,55,89,144,..
3,8 works.
8,21 works
21,55 works
55, 144 works...  Not only that, $a^2-1/b$ is also an Fibanocci number.  The only exception to this is 4,15.
When the number is 4 the highest number allowed is 15 and those work.  When that's set to five, the highest allowed is 23 so we can have 8,21 until we get to 9.  Then, we have 55 as a maximum but that's the number we're using.  So that's allowable until 22.  By then, we have a max of 159.  The next pair is 144 and 377.  But at 56, we have 431 as our maximum.  It seems that the difference between the maximum and the largest number in the pair is diverging so the largest number in the pair shouldn't exceed the maximum.[/hide]


[quote="Vasc"]
If $a,b,c,d$ are non-negative real numbers such that $a+b+c+d=4$, then

${\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}}+\frac{d}{a+3}\geq 1$.

In addition, I conjecture the following statement:
If $a,b,c,d,e$ are non-negative real numbers such that $a+b+c+d+e=5$, then

${\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{d+1}}+\frac{d}{e+1}+\frac{e}{a+1}\geq \frac{5}{2}$.

Can somebody to prove this one?[/quote]

[quote="manlio"][quote="Vasc"]
If $a,b,c,d$ are non-negative real numbers such that $a+b+c+d=4$, then

${\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}}+\frac{d}{a+3}\geq 1$.
[/quote][/quote]
By Cauchy,
\[16\displaystyle\sum_{cyc}\frac{a}{b+3} = (\frac{1}{4}(\displaystyle\sum_{cyc}a)^2+12)(\displaystyle\sum_{cyc}\frac{a}{b+3}) \ge (\displaystyle\sum_{cyc}ab + 3\displaystyle\sum_{cyc}a)(\displaystyle\sum_{cyc}\frac{a}{b+3}) \ge (\displaystyle\sum_{cyc}a)^2 = 16 \Rightarrow \displaystyle\sum_{cyc}\frac{a}{b+3} \ge 1.\]

But the seond one is much much much harder!  :?


Each edge and each diagonal of the convex $ n$-gon $ (n\geq 3)$ is colored in red or blue. Prove that the vertices of the $ n$-gon can be labeled as $ A_1,A_2,...,A_n$ in such a way that one of the following two conditions is satisfied:

(a) all segments $ A_1A_2,A_2A_3,...,A_{n\minus{}1}A_n,A_nA_1$ are of the same colour.
(b) there exists a number $ k, 1<k<n$ such that the segments $ A_1A_2,A_2A_3,...,A_{k\minus{}1}A_k$ are blue, and the segments $ A_kA_{k\plus{}1},...,A_{n\minus{}1}A_n,A_nA_1$ are red.

http://www.mathlinks.ro/Forum/viewtopic.php?t=111249

  darij


If 20% of a school was made up of out-of-district students last year, but the number of out-of-district students increased by 50% this year, without any change in total enrollment, what percent of the students is out-of-district this year?


calculus integration function limit factorial Euler Gauss

Is it true?

${\int_0^\infty{\frac{x^k}{e^x}}dx}=k!$

See  [size=150][i]$\gamma$[/i][/size]  (gamma) function for details.....

Use integration by parts.
Let \[ \begin{gathered} u = x^k \hfill \\ \end{gathered}  \] and \[ \begin{gathered} dv = e^{-x} dx \hfill \\ \end{gathered}  \].
Then, \[ \int_0^\infty {x^k e^{ - x} dx = } \int_{}^{} {(u)dv = } \left. {\left( {x^k } \right)\left( {e^{ - x} } \right)} \right|_0^\infty - \int_0^\infty {\left( {kx^{k - 1} } \right)\left( { - e^{ - x} } \right)dx = } \left( {\mathop {\lim }\limits_{x \to \infty } \left( {x^k e^{ - x} } \right) - (0)^k e^{ - (0)} } \right) + \int_0^\infty {kx^{k - 1} e^{ - x} dx = k} \int_0^\infty {x^{k - 1} e^{ - x} dx}  \].

Prove that the original integral is true when k=0.  Then, you can use mathmatical induction.

Write a big $\Gamma$, it's more common I think ;)

To prove the formula, show that it works for $k=0$ (trivial) and take integration by parts to show that you have the same recurrence as for the faculty.

Yes, the gamma function is an extension of the factorial to all complex numbers ([url=http://mathworld.wolfram.com/GammaFunction.html]http://www.mathworld.wolfram.com/GammaFunction.html[/url]).

Be careful, though!  Know that \[ x! = \Gamma(x+1)  \].  (This can get easily confused.)

Even though the gamma function is often defined (I think this is due to Euler) using an improper integral similar to the one above (except k is k-1 for the gamma function), it was also defined by Gauss as \[ \Gamma (x) = \mathop {\lim }\limits_{n \to \infty } \frac{{n!n^x }}{{x(x + 1)(x + 2)...(x + n)}}  \].

I'm normally not very good with integration by parts, but I just recently worked this one out as a practice problem.
I found Gauss' definition in a book of function used for engineering.  the gamma function is a very common function and there has been much written about it.   I've only read an extremely small fraction of it.

Yes, capital gamma is used for the gamma function.  Small gamma is used for other things (like the Euler-Mascheroni constant [url=http://mathworld.wolfram.com/Euler-MascheroniConstant.html]http://mathworld.wolfram.com/Euler-MascheroniConstant.html[/url]).

draperp: please don't post three times in a row, especially in such short intervalls of time.

Indeed, the $\Gamma$-function has some nice properties, like the gaussian multiplication formula or the functional equation of the $\zeta$-function.

Thanks everybody  :roll:

I have used this to solve that one:
Show that the exponential generating function $\widetilde{G}(z)$ of a sequence is related to the ordinary generating function $G(z)$ by the formula
$\int_0^\infty \widetilde{G}(zt)e^{-t} dt = G(z)$
if the integral exists.

I am unfamiliar with either of the functions you mentioned.  What are they?

Take a sequence $a_0,a_1,a_2,...$. Then consider the functions $G(x) = \sum_{n=0}^\infty a_n x^n$ and $\widetilde{G}(x) = \sum_{n=0}^\infty \frac{a_n}{n!} x^n$.


Tyler's age is half Mary's age. In four years, Tyler's age will be two-thirds Mary's
age. How old is Tyler now?

$ T\equal{}\frac{M}{2}$.  Then $ T\plus{}4\equal{}\frac{2M}{3}$.  Now you have two equations in two unknowns, which is quite easy to solve.

Yes, forgot that little sum in there.  Should read
$ T\plus{}4\equal{}\frac{2}{3}(M\plus{}4)$.


prove that there is exist one positive integer n such that
$2^8+2^{11}+2^n$ is a square.

it equals
2^8 (9+2^(n-8))
Obviosuly, 9+16=25 is a perfect square.
Therefore when n=12 it's a perfect square.

for n >= 8, we have that 2^8(1+2^3 + 2^k) is a square, which implies that

1+2^3+2^k must be a square.  Now if k is odd, then 1+2^3+2^k = 2 mod 3,

implying that it cannot be a square.  Otherwise if k is even, and k > 4, we

have that:

2^k = (2^(k/2))^2 < 1+2^3+2^k < (1+2^(k/2))^2 = 1 + 2^2^(k/2+1)+2^k

so we have reduced n to a finite number of cases, and manually checking we

find that as siuhochung posted, n = 12.


Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?

What do you mean by [quote]When we pick point C, we can pick it in the 60 degree sector or 30 degree in a different sector.[/quote]?

[quote=basketball8][hide="Solution"]
We can draw a circle with a hexagon inscribed in it. We pick a edge of the hexagon and name it A. When we pick point B, we have a 1/3 probability that it will be less than the radius. When we pick point C, we can pick it in the 60 degree sector or 30 degree in a different sector. This means that the probability is 1/3 * 1/4 = $\boxed{1/12}$
[/hide][/quote]

THAT IS SOOOOOOOOO SMART!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!  :omighty:


calculus
integration
inequalities
trigonometry
triangle inequality
real analysis
real analysis unsolved

For  $ 0 < \delta < \frac {2}{\pi}$ show that:

$ \left|\int^{\frac {2}{\pi}}_{\delta}sin(\frac {1}{x})dx\right|\leq\frac {4}{\pi^{2}}$

Hint:

$ \int_{\delta}^{\frac{2}{\pi}} \sin \frac{1}{x}\ dx\equal{}\int_{\delta}^{\frac{2}{\pi}} \minus{}\frac{1}{x^2}\cdot \sin \frac{1}{x}\cdot (\minus{}x^2)\ dx$.

Integral by parts

Triangle inequality

Make the substitution $ t\equal{}\frac1x.$ Then integrate by parts as kunny suggested.

$ \left|\int_{\delta}^{\frac{2}{\pi}}\sin\frac1x\,dx\right|\equal{} \left|\int_{\frac{\pi}2}^{\frac{1}{\delta}}\frac{\sin t}{t^2}\,dt\right|$

$ \equal{}\left|\left.\minus{}\frac{\cos t}{t^2}\right|_{\frac{\pi}2}^{\frac1{\delta}} \minus{}\int_{\frac{\pi}2}^{\frac{1}{\delta}}\frac{2\cos t}{t^3}\,dt\right|$

$ \le \delta^2\plus{}\int_{\frac{\pi}2}^{\frac{1}{\delta}}\frac{2}{t^3}\,dt$

$ \equal{}\delta^2\plus{}\frac4{\pi^2}\minus{}\delta^2\equal{}\frac{4}{\pi^2}$

This is essentially the same argument kunny was advocating. I just found it easier to see what to do after I made that change of variables.

[quote="kunny"]Triangle inequality[/quote]
About what? I don't get it. Can you explain?

The triangle inequality, in simplest terms, says that $ |a\plus{}b|\le|a|\plus{}|b|.$ An idea that is related is that $ \left|\int_Ef\right|\le\int_E|f|.$ If you look at what I did from the second line to the third line of my argument above, you'll see that I used both of these things. It is quite likely that one or both of those are what kunny meant with his remark.


geometry
perimeter
3D geometry
blogs
analytic geometry
algebra
polynomial

A good use of this formula would be this question:

What is the area of a triangle with sides of 13, 14, 15?

Either the pythagorean theorem with variables or heron's formula.

I don't think so. I've never seen it like that.

[quote="236factorial"]try it for a 3-4-5 right triangle. It isn't the same.[/quote]

Oh see why. thanx for clearing up the doubt

You usually use $s$ for semiperimeter - less confusion that way ;)

Anyways, you're not supposed to create a topic just to find out what Heron's Formula is. That's what MathWorld's for: mathworld.wolfram.com/HeronsFormula.html

[quote="shobber"][quote="anirudh"][quote="socrates"]It's a formula that gives the area of a triangle given its sides. Here it is :

If $a,b,c$ are the sides of a tringle and $p$ its semiperimeter then its area is equal to

$E=\sqrt{p(p-a)(p-b)(p-c)}$

Very interesting and of course very useful !!! :lol:

You can try to prove it, too!! :)[/quote]

Couldn't you also write it as:

$\sqrt{\frac{p(p-a)(p-b)(p-c)}}{2}$

where p is the perimeter?[/quote]
Your identity is false. There are $p$ in every term. If you prove it your self, you will know that $p$ is much better than $C$.[/quote]

MY IDENTITY IS NOT FALSE

ANd chess64, the word is anyway

[quote="shobber"]If $C$ stand for the perimeters, then:
\[S=\frac{\sqrt{C(C-2a)(C-2b)(C-2c)}}{4}\][/quote]

You should have used p

[quote="236factorial"]it's the same thing  :roll:

shobber, what did you mean why you said that the identity was false  :?:[/quote]

Yes, that's my question

also, for more qeustions, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46626]click here[/url]

[quote="anirudh"][quote="236factorial"]Please don't quote all that and shout.

I don't get what the identity part means though  :?[/quote]

neither do i[/quote]
if you dont understand the identity part and yelling
[quote="anirudh"]MY IDENTITY IS NOT FALSE[/quote]
what's the point of doing that? i dont think you are fair to shobber, who was pointing out that your previous identity is not true.
you want to know what is Heron's formula? here: for any triangle with sides $a,b,c$, area $A$, semiperimeter $s=\frac{a+b+c}2$ and perimeter $p=2s=a+b+c$

\[A=\sqrt{s(s-a)(s-b)(s-c)}\]
\[A=\sqrt{\frac{p(p-2a)(p-2b)(p-2c)}{16}}\]

[quote="chess64"]:rotfl:  :rotfl:  :rotfl: OMG I can't believe you guys misinterpreted him! :rotfl:

He meant "identity" as in mathematical statement.

anirudh's statement was false.

 :rotfl:[/quote]

That's what I thought  :?

[quote="chess64"]:rotfl:  :rotfl:  :rotfl: OMG I can't believe you guys misinterpreted him! :rotfl:

He meant "identity" as in mathematical statement.

anirudh's statement was false.

 :rotfl:[/quote]

What is an identity

[b]PLEASE STOP SPAMMING!!![/b], especially you anirudh, like the post in "Cubes and 1's" you answered
[quote="anirudh"]yes,
but i don't really understand why

[/quote]

im saying no more...

Where can I find the derivation of this formula?

*Braces for an answer that makes him look like a fool*

Check this out:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40603

Jacob's Geometry says that high-school students are not supposed to be able to comprehend any part of a proof of Heron's ... However, I have a proof for isosceles triangles and Heron's on my blog [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=545]here[/url]

i have heard of Jacob's Geometry two times today. Is that a book or something?

I am, in some way, a bit fond of that formula because I found it by my own. I was in my last days of the 8th grade and one day, I suddenly came up with an idea: If the three sides of a triangle are given then the triangle is fixed. And then we shall be able to find the area of the triangle just with the length of the three sides. Then I started working and, I first used coordinates and ended up with an ugly equation (later i found out that i only needed to factor it a bit  :( ). Then I choosed to use the most famours one: $a^2+b^2=c^2$, and quickly got this statement:
\[S=\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}}{4}\]
I thouht it was an extremely beautiful one, just symmetry. (maybe it was because I found it!  :P ) Then, several months later, I found Heron's formula in a book and it used a little substitution to make the statement neater. Anyway, I was still very happy about my result!

that's pretty smart of you  :lol:

What grade are you in now?

I am entering my 11th grade soon. I lost my geometry feelings when I entered high school (10th grade). And I just get it back recently. So I made few discovers after that......

Jacob's Geometry is a textbook by Harold Jacobs on geometry, much used in homeschooling circles. It is generally regarded as a high-school book. I have not used it - I used Saxon - but I've looked through it. It's all right. The AoPS books are better ;)

Looking, I think just knowing the formula is enough  :D

OK, now I am getting irritated. CAN PEOPLE JUST STOP SPAMMING. It is quite annoying to find out that you click on a topic and navigate to
a second page just to find out something that is not germane[pertinent, relevant, appropriate, fitting with the standards of propriety] to the topic. In this case, it is heron's Formula, so it is not pertinent to have
things about Jacob's formula.  :spam:

Also, i think that a mod should lock this topic

[quote="anirudh"]OK, now I am getting irritated. CAN PEOPLE JUST STOP SPAMMING. It is quite annoying to find out that you click on a topic and navigate to
a second page just to find out something that is not germane to the topic. In this case, it is heron's Formula, so it is not pertinent to have
things about Jacob's formula.  :spam:[/quote]
how is a prrof of herons formula spam? To be honest, you are the one that is spamming the most, so please stop.
[quote="chess64"]Good job. :) How long did that take you?[/quote]
Probably around 5 minutes, maybe less.  Its not really too hard to derive.  Just get some eqquations and try to knock out any extra variables besides a,b,c. I drew it out on paper and then I saw how the x,y,and h would cancel out by using the above equations, and then solved.  The hardest part is factoring that big polynomial

[quote="anirudh"][quote="chess64"][quote="anirudh"]It is quite annoying to find out that you click on a topic [i]["Heron's Formula"][/i] and navigate to a second page just to find out something that is not germane to the topic. ... it is not pertinent to have things about Jacob's formula.  :spam:[/quote]

(italic emphasis mine) Wtf is Jacob's formula? What are you talking about? Where on this page do you see something about ... "Jacob's formula"???????????????????[/quote]

Look at post #31[/quote]

We're talking about the book and the Heron's formula's difficulty, not Jacob's formula.

It is quite annoying to find out that you click on a topic ["Heron's Formula"] and [b]navigate to a second page just to find out something that is not germane to the topic[/b]. ... it is not pertinent to have things about Jacob's formula.

Did you look up the definition of germane? If so, I think you're right that you have to navigate to a second page to find out something that is not germane to the topic. What's more, there's a little bug so I can't delete the malapropos posts that you speak of.

Let's stop spamming, okay?

Billy

Identity: look it up in AoPS Book 1 somewhere in the Functions chapter (Chp 21?)
  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:   :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:   :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:   :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:   :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:   :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:  :idea:

Can you just explain here. And also, i think that a mod should lock this topic up before everyone gets off topifc


Assume that ABC is a triangle with the angles [tex]\alpha[/tex], [tex]\beta[/tex], [tex]\gamma[/tex]. Assume that there are 3 points X, Y, Z inside the triangle, so that the lines (AY, AZ), (BZ, BX), (CX, CY) trisect the angles [tex]\alpha[/tex], [tex]\beta[/tex], [tex]\gamma[/tex]. Prove that the triangle XYZ is equilateral.

This is Morley's Theorem. Proofs (which aren't simple) can be found online.

The same well-known problem posted two times in one day! Actually a number of proofs can be found on http://www.cut-the-knot.org/triangle/Morley/index.shtml .

  Darij


[hide]Assuming it is
[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/3/8/c/38c67c75d812435dee2b268ed45906d8c62e3eef.gif[/img]

The xy terms simplify to:
7xy-6xy=xy

[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/c/5/1/c51e17c830821b0f19682c72215e622dba103e36.gif[/img] is the part in parentheses

So the equation becomes:
[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/2/d/d/2ddfa11734d97391b08f2e5176d64d1c73a39098.gif[/img] or
xy(1-y)[/hide]

is it possible he could have meant 2x as the exponent? Although that is rather unlikely...

I'm not sure if this is of any help. On my computer I cannot read what Rocket95 has written: clicking on the squares with red crosses in them and followed by links produces no response on my machine. However I presume it is readable on other machines. Anyway, just in case,

$ 4xy\plus{}3xy\minus{}(6xy\plus{}y^{2}x)$

$ \equal{}7xy\minus{}(6xy\plus{} xy^{2})$

$ \equal{}7xy\minus{}6xy\minus{}xy^{2}$

$ \equal{}xy\minus{}xy^{2}$

$ \equal{}xy(1\minus{}y)$

hello, your answer $ xy(1\minus{}y)$ is right.
Sonnhard.


We have $\frac {2k^2 + 3k+ 2 } {k(k+1)^22^kk!} = \frac 1 {2^{k-1}k \cdot k!} - \frac 1 {2^k(k+1) \cdot (k+1)!}$.
Thus the given sum $S_n$ simplifies by dominos, and leads to :
$S_n = 1 - \frac 1 {2^n(n+1) \cdot (n+1)!}$.

Pierre.

To Pierre: Exactly you are right!

kunny


geometry
incenter
cyclic quadrilateral
geometric transformation
geometry proposed

Let be given a cyclic quadrilateral $ ABCD$. Let $ K$ be a point on $ BD$ such that $ \angle AKB \equal{} \angle ADC$. Denote by $ I$, $ I'$ the incenters of triangles $ ACD$ and $ ABK$, respectively. Let $ X$ be the intersection of $ II'$ and $ BD$. Prove that points $ A$, $ X$, $ I'$ and $ K$ are cyclic.

From two triangle $ ABK, ACD$ are similar then:
$ \widehat{AI'X}\plus{}\widehat{AKX}\equal{}\widehat{AI'I}\plus{}\widehat{AKB}\equal{}\widehat{ABC}\plus{}\widehat{ADC}\equal{}180^0$

[b]Here is my solution[/b] (Using [b]Spiral Similarities[/b], no [b]angle chasing[/b]):
It is easy to see that $ \triangle AKB\sim \triangle ADC$, further, these $ 2$ triangles have the same [b]direction[/b]. Therefore, there exist an unique spiral similarity with center $ A$ which turns $ \triangle AKB\mapsto \triangle ADC$. Denote $ f$ is that spiral similarity of center $ A$. It is easy to notice, $ f: I'\mapsto I$, $ K\mapsto D$, therefore, $ f: \triangle AI'I\mapsto \triangle AKD$, which implies the fact that $ \angle AI'I\equiv \angle AI'E \equal{} \angle AKD$, which leads to the result that $ A,I',E,K$ are concyclic.
Our proof is completed :lol:
[b]p/s[/b]: In my solution, I consider $ E\equiv X$ in [b]April's[/b] problem statement


calculus
integration
LaTeX
absolute value

Am I the only person who thinks that the \displaystyle command should not be automatically enabled?  Displaystyle is, well, just huge, and the small sums, integrals, and whatnots are much more managable.  Plus it makes it necessary to be constantly typing \left and \right to get the appropriate brackets or absolute value signs or whatever to match.

You can always disable it with \textstyle: $\textstyle\int_0^1x^2\, dx=\frac13$

Most of the time, I want displaystyle; I'd prefer leaving it alone.

[quote="Valentin Vornicu"][quote="blahblahblah"]Am I the only person who thinks that the \displaystyle command should not be automatically enabled?[/quote]Yes. :D[/quote]Agreed  :D I honestly prefer in the forum the displaystyle comand automatically enabled!

\vert or \mid.  They produce different results though.

[quote="xxreddevilzxx"]just a question popping out about latex:

how do we do absolute values??? I tried to find the straight upanddown bar on the computer key, but i couldnt find one.[/quote]

Hold Shift and press the backslash \ key.

[quote="xxreddevilzxx"]just a question popping out about latex:

how do we do absolute values??? I tried to find the straight upanddown bar on the computer key, but i couldnt find one.[/quote]It depends. you can do either [code]|x|[/code] that looks like $|x|$ or if the expression is a little bigger, say, the absolute value of a sumatory, you can do [code]\left|\sum_{k=-n}^{n}\frac{a_k}{k^2+1}\right|[/code] that looks like $\left|\sum_{k=-n}^{n}\frac{a_k}{k^2+1}\right|$. I hope it helps!

I forgot, you may do also norm signs, easily. Similarly

[code]\|\mathbf{x}\|

\left\|\sum \frac{e^{ix}}{n!} \right\|_{L^2}[/code]

that looks like $\|x\|$ and $\left\|\sum \frac{e^{ix}}{n!} \right\|_{L^2}$


pigeonhole principle
combinatorics proposed
combinatorics

Let $n_1<n_2<n_3........<n_{2000}<10^{100}$. Prove there exist two disjoint subsets of $\left\{n_1,n_2,n_3.....,n_{2000}\right\}$, $\textbf{\emph{A}}$ and $\emph{\textbf{B}}$ such that $\left|\textbf{\emph{A}}\right|=\left|\textbf{\emph{B}}\right|$, $\displaystyle\sum_{x\in A}x=\sum_{x\in B}x$ and $\displaystyle\sum_{x\in A}x^2=\sum_{x\in B}x^2$.

I haven't actually worked this out completely, but I bet that counting the triples $(|A|,\sum_{a\in A}a,\sum_{a\in A}a^2)$ reveals the fact that there are too many of them, and thus two of them must be equal (by pigeonhole principle).

Is this the right track? :D

you are right as usual, but try to finish it, it is the standard method

Assuming all of those triples are distinct for our subsets, we get at least $2^{2000}-1$ of them. On the other hand, there can be at most $2000\cdot (2000\cdot 10^{100})\cdot (2000\cdot 10^{200})$, and it really is not that hard to see that the second number is much smaller than the first one.


[hide="answer"]

A+B = 10
C+1 = A
C does not equal to 9

soon can easily find out that B = 1
since A + B = 10, A= 9
C + 1 = A
so C = 8

therefore
BAC = 198[/hide]

[hide="Clarification"]
[code] AA
 BB
+CC
---
BAC[/code]

$A+B+C=10+C\Rightarrow A+B=10$
We carry a $1$ over from the units column, so $C+1=A$, and we carry another $1$ over from the tens column. Therefore, $B=1$, so $A=9$ and $C=8$, and $BAC=198$.[/hide]

ummm... little clearer please? I am just startiing algebra and don't quite understand.

Each of $A,B,C$ represents a distinct digit from $0$ to $9$. In the units column, we have $A+B+C=C$. This means that $A+B=10$, since $A+B$ cannot be equal to $20$ even with maximum value. Then, in the next column, $A+B+C+1=A$, $C+11=A$. Again, we must carry a $10$ (we have to to make $B$, anyway), so $C+1=A$. Then, in the final column, the only possible value for $B$ is $1$, since we can't have carried $2$. Then, we see $A+1=10\Rightarrow A=9$, and $C+1=9\Rightarrow C=8$. Checking: $99+11+88=198$, so it works.


Jason posted the A problems, so if you want to see those, look back a few pages.

B1: A certain mechanical oscillator can be modeled as an ideal massless spring connected to a movable plate on an incline.  The spring has spring constant $k$, the plate has mass $m$, and the incline makes and angle $\theta$ with the horizontal.  When the system is operating correctly, the plate oscillates between points $A$ and $B$ in the figure, located a distance $L$ apart.  When the plate reaches point $A$ it has zero kinetic energy, but then trips a small lever that instantaneously loads a block of mass $M$ to the plate.  The block and plate then move down the incline to point $B$, where the force from the spring stops the plate.  At this point, the block falls through a hole in the incline, allowing the plate to move back up under the force of the spring.  Upon returning to point $A$ it collects another block, and the cycle repeats.  Both the plate and the block have a coefficient of friction $\mu$ with the incline for both kinetic and static friction.  It is reasonable that the motion in either direction is simple harmonic in nature.

a. (10 points) Let $\mu_{c}$ be the critical value of the coefficient of friction where the block will just start to slide under the force of gravity on an incline (without the spring acting on it.  Then let $\mu = \frac{\mu_{c}}{2}$.  Find $\mu$ in terms of $g$, the acceleration of free fall, and any or all of the following variables: $\theta$ and $M$.

b. (14 points) In order for this system to work correctly, it is necessary to have the correct ratio between the mass of the block and the mass of the plate.  These masses are chosen so that the downward moving block and plate just stop at point $B$ while the upward moving plate just stops at point $A$. Find the ratio $R = \frac{M}{m}$.

c. (13 points) The system delivers blocks to point $B$ with period $T_{0}$, until the blocks run out.  After that, the plate alone oscillates with a period $T'$. Find the ratio $R = \frac{T_{0}}{T'}$.

d. (13 points) The plate only oscillates a few times after delivering the last block.  At what distance up the incline, measured from point $B$, does the plate come to a permanent stop?

[hide="a"]
Simple relationship between forces along the ramp:
\[ Mgsin\theta \equal{}\mu_c Mg cos \theta\]
\[ \mu_c \equal{} tan\theta\]
\[ \mu \equal{} \frac{tan\theta}{2}\]
[/hide]

[hide="b"]
From A to B, using conservation of energy:
\[ \Delta U_{spring} \plus{} \Delta U_g \equal{} \minus{} F_{friction} L\]
\[ \frac{1}{2}k(\frac{L}{2})^2 \minus{} (M \plus{} m)g(L sin \theta) \equal{} \minus{} \mu (M\plus{}m) g cos \theta L\]
Substituting $ \mu \equal{} \frac{tan\theta}{2}$, we have
\[ M \plus{} m \equal{} \frac{kL}{4gsin\theta}\]
\[ \frac{M}{m} \equal{} \frac{kL}{4mgsin\theta} \minus{} 1\]
Similarly, using conservation of energy for B to A:
\[ \minus{}\frac{1}{2}k(\frac{L}{2})^2 \plus{} mg(L sin \theta) \equal{} \minus{} \mu m g cos \theta L\]
With the same subsitution, this yields
\[ m \equal{} \frac{kL}{12gsin\theta}\]
Substituting this relationship into the equation for $ M/m$ gives
\[ R \equal{} \frac{M}{m} \equal{} 3 \minus{} 1 \equal{} 2\]
[/hide]

[hide="c"]
\[ \frac{T_0}{T'} \equal{} \frac{\frac{T_{AB}}{2} \plus{} \frac{T_{BA}}{2}}{T_{AB}}\]
Using the energy equations from part b and solving for $ k$,
\[ T_{AB} \equal{} 2\pi\sqrt{\frac{(M\plus{}m)L}{4(M\plus{}m)gsin\theta}}\]
\[ T_{AB} \equal{} 2\pi\sqrt{\frac{mL}{12mgsin\theta}}\]

Therefore,
\[ \frac{T_0}{T'} \equal{} \frac{1\plus{}\sqrt{3}}{2}\]
[/hide]

I hope "deliver" means dropping the block through $B$. If the plate comes to a halt, at equilibrium we have:

$mg\sin\theta = kx \pm \mu mg\cos\theta$

\[\therefore\boxed{\frac{L}{2}-\frac{mg\cos\theta(\mu + \tan\theta)}{k} \le d \le \frac{L}{2}+\frac{mg\cos\theta(\mu - \tan\theta)}{k}}\]

...where $x$ is the compression in spring and $d$ is the distance of the plate from $B$. I guess you can also calculate the number of oscillations using energy conservation taking into account the energy lost to friction and the knowledge of the final position of the block!


geometry
circumcircle
geometric transformation
homothety
angle bisector
geometry proposed

Let triangle  $ ABC$, $ DE \parallel BC$ and the circle $ (A,D,E)$. Segments $ BF,CZ$ are tangent to the circle. If $ K$ is the common point of $ DZ,EF$  prove that $ AK$ besects angle $ A$.

Babis

*** How can I send the second  figure (with geogebra ) in a right way ?The first figure is taken with printscreen.

As $ BF$ is tangent to the circle $ (ADE)$, $ \left( \frac{AF}{DF} \right)^2 \equal{} \frac{AB}{BD}$; similarly, $ \left( \frac{AZ}{ZE} \right)^2 \equal{} \frac{AC}{CE}$. If $ AK$ intersects the second time the circle $ (ADE)$, we shall get from Ceva trigonometric:
$ \frac{AF}{FD} \cdot \frac{DT}{TE} \cdot \frac{ZE}{AZ}\equal{}1$ which, with $ DE \parallel BC$, we get $ DT\equal{}TE$, i.e. $ AK$ is the angle bisector of $ \angle BAC$.

Best regards,
sunken rock

Thank you for this nice solution. I thought this problem should be easier  :( , but as I see, it is really difficult. I have not the source.

Babis

[quote="stergiu"] [color=darkred]Let $ ABC$ be a triangle and let $ D\in AB$ , $ E\in AC$ be two points so that  $ DE \parallel BC$ . Consider the tangents  $ BM$ , $ CN$

to the circumcircle $ w$ of $ \triangle ADE$ , where $ \{M,N\}\subset w$ . Denote $ K\in ME\cap ND$ . Prove that $ AK$ bisects the angle $ \widehat {BAC}$ .[/color] [/quote]
[color=darkblue][b]Remark.[/b] Inspirational idea of [b]Sunken Rock[/b] to apply this well-known remarkable property ![/color][quote][color=darkred]Let $ ABCDEF$ be a hexagon ([u]convex or not convex[/u] !) which is inscribed in a circle $ w$ .

Then exists the equivalence $ AD\cap BE\cap CF\ne\emptyset\Longleftrightarrow AB\cdot CD\cdot EF \equal{} BC\cdot DE\cdot FA$ .[/color][/quote]
[color=darkblue][b]Example.[/b]  If $ ABCDEF$ is convex, then $ ACDFBE$ isn't convex and in

this case $ AF\cap CB\cap DE\ne\emptyset\Longleftrightarrow AC\cdot DF\cdot BE \equal{} CD\cdot FB\cdot EA$ .

[b]Proof of the proposed problem.[/b] Denote the second intersections $ U$ , $ V$ of the circumcircle $ \theta$ of $ \triangle ABC$ with $ AM$ , $ AN$ respectively.

From an well-known property obtain that the rays $ [CU$ , $ [BV$ are  bisectors of $ \angle ACB$ , $ \angle ABC$ respectively. Since $ DE\parallel BC$ obtain

that the circles $ w$ , $ \theta$ have a common tangent $ AA$ in $ A$ (the circles $ w$ , $ \theta$ are tangent in a point $ A$). Therefore, the rays $ [DN$ , $ [EM$

are the bisectors of $ \widehat {ADE}$ , $ \widehat {AED}$ respectively and the ray $ [AK$ is the bisector of $ \widehat {DAE}$ , i.e. $ \widehat {BAC}$ (here is a homothety with vertex $ A$ ).[/color]


We have a finite alphabet $ A$ from which we can form words.  The channel over which we send these words is faulty, and certain letters can be replaced by certain other letters in transmission.  In this case, we call a pair of such letters "confusable."  Two words are distinguishable with 0 error if at some position the letters they contain are not confusable.  The zero-error capacity of the channel measures the largest set of words we can use such that every pair of words is distinguishable.

My question is the following: if we require a pair of words to have 2 positions at which they contain nonconfusable letters before we call them distinguishable, does the resulting analogue of zero-error capacity have a name/has it been studied?

Thanks.

There are several papers that study similar problems:
"Channel Capacity for a Given Decoding Metric" by Imre Csiszar and Prakash Narayan
http://ieeexplore.ieee.org/iel3/18/8485/00370120.pdf?tp=&isnumber=&arnumber=370120

"Erasure, List and Detection Zero Error Capacity for Low Noise and a Relation to Identification"
by Rudolf Ahlswede, Ning Cai, Zhen Zhang
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.50.7761

On the other hand, quantitatively your definition is similar to the standard one. Let $ C_n$ be the zero error capacity and $ \tilde C_n$ the capacity w.r.t your stronger definition. Then
\[ C_n \geq \tilde C_n \geq  C_n/(|A| n).\]
Thus $ C_n$ and $ \tilde C_n$ have the same asymptotic behavior:
\[ \lim C_n^{1/n} \equal{} \lim \tilde C_n^{1/n}.\]


Solve the equation:
 $ 4x^2 - 4x - 10 = \sqrt{8x^2-6x-10} $

Squaring, it leads to $16x^4-32x^3-72x^2+86x+110=0$
which is equivalent to $(x+1)(2x-5)(4x^2-2x-11)=0$
The solutions of this last equation are $-1, \frac 5 2 , \frac {1+3\sqrt 5} 4, \frac {1-3\sqrt 5} 4$.

Conversely, for $x=-1$ and $x=\frac {1+3\sqrt 5} 4$ we have $4x^2-4x-10 <0$ so that they are not solutions of the problem, whereas the two other values are.

Pierre.

O_O [b]pbornsztein[/b] fighting to solve [b]Pre-Olympiad[/b] problems???
Did you have a tough day? :D  :D  :D O_-

Well, since my Alzheimer is beginning (see the combi section), I'm going back to basics  ;)

Pierre.

P.S. Since it seems that you did not make any remark to Myth, I presume you consider him as a pre-olympiad member  :D  ;)

What are you talking about, Pierre?
Couple of problems, which I solved in Pre-Olympiad, were not so easy ;) :D

BTW, 3X.lich considers me as intermediate member:
http://www.mathlinks.ro/Forum/viewtopic.php?p=125160  :lol:  :lol:


an infinite plane is completely filled with congruent circles in the pattern shown.  What percentage of the plane is covered y the circles?  Round your answer to the nearest percent.

1. Picture?
2. This was already posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76187]here[/url].
3. Didn't you (kchande) post this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=57395]here[/url]?

It's 91%.  And I do not remember posting that problem.  And I DO NOT forget how to do problems so that is really weird.

wow some people are really slow sometimes .....
[hide="i remember this problem"]
you divide the circles into equilateral triangles with the two of the circle's radii as the sides.
Thus if the radius is $x$, the side is $2x$ and the area of the triangles is $\frac{4x^2\sqrt{3}}{4}=x^2\sqrt{3}$
The area of the three sixth circles is half of a circle = $\frac{x^2\pi}{2}$
Thus the total percent is
$\frac{\frac{x^2\pi}{2}}{x^2\sqrt{3}}=\frac{x^2\pi}{2x^2\sqrt{3}}=\frac{\pi}{2\sqrt{3}}\approx.91\approx91\%$[/hide]

[hide]Divide the plane into equilateral triangles that tessellate the plane, with the vertices at the centers of the circles of each group of 3 mutually tangent circles. We can just find the percentage covered in that one equilateral triangle.

Let s be the radius of the circles. 2s is the side length of the equilateral triangle. The area of the equilateral triangle is therefore $s^2\sqrt{3}$. Each of the 3 sectors in the equilateral triangle is 1/6 the area of a whole circle, or 1/2 the circle all together. The area covered is therefore $\frac{s^2\pi}{2}$.

$\frac{s^2\pi}{2\cdot s^2\sqrt{3}} \approx 91$%[/hide]

There is a complete solution in the Mathcounts Greatest Problems book, you can and SHOULD buy it from the AoPS bookstore, not too late for nats!

I suggest going through all the problems there.


inequalities
number theory
relatively prime
inequalities unsolved

[quote="Sung-yoon Kim"][quote="M4RI0"]Let $a$, $b$, $c$ be positive integers such that $a^{2}+b^{2}=c^{2}(1+ab)$,
prove that $a \geq c$ and $b \geq c$.[/quote]

Counterexample: $a=b=\sqrt{\frac{1}{2}},c=\sqrt{\frac{2}{3}}$[/quote]

[quote="M4RI0"]Let a, b, c be positive integers[/quote]

[hide]Let > denote the greater than or equal to sign

a>c
a^2>c^2
c^2(1+ab)-b^2>c^2
1+ab>b^2/c^2+1
ab>b^2/c^2
ac^2>b
a(a^2+b^2)/(1+ab)>b
a(a^2+b^2)>b(1+ab)
a^3+ab^2>b+ab^2
a^3>b

Let < denote the less than sign.  Let n an integer by greater than 3.  Now if instead a^3<b then

(a^2+a^(2n))/(1+a^(n+1))=c^2
a^2(1+a^(2n-2))/(1+a^(n+1))=c^2

Assuming a does not equal 1 I think each factored term in the numerator is relatively prime with the denominator...

Similarly, we can prove the same with b[/hide]

[quote="ehehheehee"][hide]Let > denote the greater than or equal to sign

a>c
a^2>c^2
c^2(1+ab)-b^2>c^2
1+ab>b^2/c^2+1
ab>b^2/c^2
ac^2>b
a(a^2+b^2)/(1+ab)>b
a(a^2+b^2)>b(1+ab)
a^3+ab^2>b+ab^2
a^3>b

Let < denote the less than sign.  Let n an integer by greater than 3.  Now if instead a^3<b then
[b](&&&)[/b]
(a^2+a^(2n))/(1+a^(n+1))=c^2
a^2(1+a^(2n-2))/(1+a^(n+1))=c^2

Assuming a does not equal 1 I think each factored term in the numerator is relatively prime with the denominator...

Similarly, we can prove the same with b[/hide][/quote]
ehehheehee,  i dont follow after where i marked (&&&) in the quote. could u explain?

[hide=" simple solution"][b]Statement:[/b]
if $a, b, c \in N^{*}$ are such that $a^{2}+b^{2}= c^{2}(1+ab)$, then $a \ge c$ and $b \ge c$
--------------------------------------
[b]Contrapositive statement:[/b]
if $a < c$ or $b < c$ then  $a^{2}+b^{2}\neq c^{2}(1+ab)$
Now, WLOG, we have two cases:
[b]1)[/b] $a \le b < c$
[b]2)[/b] $a < c < b$

[b]Case 1[/b]
$\Longrightarrow c^{2}> a^{2}$ and $c^{2}> b^{2}$
$\Longrightarrow a^{2}+b^{2}< 2c^{2}$

[b]Also[/b], $2c^{2}= c^{2}+c^{2}< c^{2}+(ab)c^{2}$ for $a\cdot b \neq 1---\to{Result (*)}$

$\Longrightarrow a^{2}+b^{2}< c^{2}+(ab)c^{2}$
$\Longrightarrow a^{2}+b^{2}\neq c^{2}(1+ab);$ Since the above inequality shows that $a^{2}+b^{2}$ is strictly less than $c^{2}+(ab)\cdot c^{2}$

[b](*)[/b]Now, in the event that $a = b = 1$, $\exists$  no $c: \{c \in N^{*}\mid c > a ;c>b\}$ that satisfies $a^{2}+b^{2}= c^{2}(1+ab)\Longrightarrow$ the implication of the contrapositive statement holds for this case.

[b]Case 2[/b]
Now, let us assume that given  $a<c<b$ that
$a^{2}+b^{2}= c^{2}(1+ab)$;

$\Longrightarrow c^{2}= \frac{a^{2}+b^{2}}{1+ab}\Longrightarrow c^{2}> \frac{a^{2}+b^{2}}{ab}\Longrightarrow \boxed{ c^{2}> \frac{a}{b}+\frac{b}{a}}$ where $b > a$

Case 2 also $\Longrightarrow c^{2}+b^{2}> c^{2}+(ab)c^{2}\Longrightarrow b^{2}> c^{2}(ab) \Longrightarrow b^{2}> c^{2}a \Longrightarrow c^{2}< \frac{b}{a}$

[b]BUT[/b]
$c^{2}> \frac{a}{b}+\frac{b}{a}\Longrightarrow c^{2}> \frac{b}{a}$
(where $\frac{a}{b}< 1$)
Hence, we obtain
$c^{2}> \frac{b}{a}$ and $c^{2}< \frac{b}{a}$ which leads to a contradiction (especially since $c \in N^{*}$).
Therefore the assumption must be wrong (since the logic is undeniable),
$\Longrightarrow a^{2}+b^{2}\neq c^{2}(1+ab)$ for $a, b, c$ define by Case 2.

Conclusion:
Since our case1 and case2 conclusions hold, this implies the contrapositive argument is true in all cases, hence implying truth of the original statement (logical equivalence) for all $a, b, c$ with the originally defined restrictions.[/hide]


La Olimpiada Nacional Escolar de Matematica  ($ONEM$) consta de 4 fases y de tres niveles (I, II y III) . La fase final se llevó a cabo el 19 de noviembre. A diferencia de otros años , ahora hemos tratado que los problemas de la fase final se han nuevos ( i.e no los copiamos de otras olimpiadas!) . La mayoría de los problemas han sido propuestos por ex-olímpicos peruanos ( un problema del Nivel II es mío :P ) ... estos son los exámenes:

Excellente informacion ..
recien los estaba editando ..

pero  me haz ganado.....

estan bien lindas las preguntas ..
veamos si alguien las traduce al ingles pra colgarlas a Resource ..
se podra??
..

Gracias !!

Carlos Bravo :)
Lima -PERU

Yo podría hacerlo, pero sinceramente no sé como hacer con las imágenes en $\LaTeX$ , si me explicas.. puede ser...

$Tipe$

Mira .. si tenes problemas con imagenes ..
solo traducelo sobre el mismo documento en Word ..

y yo genero el pdf a partir de este documento...
estoy un poco atareado y no podria editar estos doc estos dias ...

Carlos Bravo :)
Lima -PERU


Find all real polynomials $ p(x)$ such that $ p(x)^2 \plus{} 2p(x)p(\frac {1}{x}) \plus{} p(\frac {1}{x})^2 \equal{} p(x^2)p(\frac {1}{x^2})$ for all non-zero $ x$.

[hide="possible hint"]
Set $ x\equal{}1$
[/hide]

 :)

With $ x\equal{}1$ we have $ p(1)\equal{}0$ then what do you do to continue?


Compute the number of solutions to $x = 2007\cos(x)$ if $x$ is in radians.

[quote="mna851"]Is $x$ in degrees or radians?[/quote]Try both  :D

[hide="Radians"]
Note that our bounds of observation are $[-2007,2007]$

$\frac{2007}{\pi }\approx 319.4$ complete cycles. Note that each complete 2-pi cycle gives two solutions, so this adds $319(2)(2)$ solutions from complete cycles.

On the positive side, there is one additional solution (the rightmost) from the additional $0.4$ cycle. We know it has one because $x$ is near 2007, so it will occur very early on in the 320th cycle, but it is less than a half-cycle, so there will not be 2 solutions. Thus we must add one more solution here.

On the negative side, it's the opposite: 0.4 of a cycle isn't quite long enough to get to the first solution of the cycle, so we don't add any solutions here.

In total, there are $319(2)(2)+1=1277$ solutions.

[/hide]

[hide="Degrees"]
$2007/360=5.575$

This produces $5(2)(2)$ solutions from complete cycles. There is one additional on the right side and two additional on the left side, for a total of $5(2)(2)+1+2=23$ solutions.
[/hide]


I've been looking at the math program at Georgia Tech, and I noticed that the only math degrees they offer are:

Applied Mathematics
Discrete Mathematics

This disappointed me, in that I wanted to major in pure math.

How different could an applied mathematics degree be from a pure math degree? How would a company look at it when put aside a pure math degree?

Depends what company you're talking about.  If you want to work for "the man," the applied math would actually likely be better.  If you want to teach and do research, the pure math would be better.

The differences will be in the level of abstraction.  I don't imagine you'll touch abstract algebra, for example, as an applied math major.  You won't need analysis, with all it's deltas and epsilons and the such.

whoa, a discrete math degree? that seems sort of cool, not quite too sure what it's use would be other than in conjunction with computer science.

pure math will only get u a teaching job.. but i think generally, majoring in math wont make much money... unless your one of those crazy people who does math crap to manipulate/predict stocks..

Well....it could get you a lot of money.  Just not as a professor.  Working as an engineer or some other job of the sort, one could probably make >$1 mil a year (though that would be relatively rare).  Actuaries make a load of money.  I'm not sure how the NSA pays.

The only thing is that most people who are smart enough to get math degrees aren't stupid enough to care about money  :lol:

[quote="kentliu"]pure math will only get u a teaching job[/quote]

This actually isn't true.  Many math, applied math and physics majors end up working in finance or doing i-banking -- any industry that values an ability to handle numbers well, basically.  (Also, anywhere that values abstract thinking probably wants its share of mathematicians as well.)  In general, your college major will not make that much of a difference in your life circumstances -- if you're a capable person, you're likely to do well regardless of your major.  At most schools, a major only occupies around one third of your classes, anyhow.  Harvard's math major requires only 12 courses out of a total of 32 to graduate, for example.

In terms of coursework, an applied math degree will probably require you to take a bunch of courses from physics, engineering, computer science and less-theoretical analysis-related courses.  For example, I doubt there are any applied math programs that [i]require[/i] you to take abstract algebra, while it is a standard for pure math majors.


algebra
polynomial
search
number theory proposed
number theory

Find all monic polynomials $f(x)$  in $\mathbb Z[x]$ such that $f(\mathbb Z)$ is closed under multiplication.
[i]By Mohsen Jamali[/i]

Nice problem, but observe that it is an immediate consequence of some problems already discussed on the forum: what are the polynomials whose image contains all numbers $2^{n}$, or $a^{n^{2}}$ and so on. Here, it is clear that once you take some $a\geq 2$ in the image of $f$, all powers of $a$ are in this image and thus you reduced the problem to the mentioned ones. I think however that this should have an absolutely trivial solution, since the constraints are really excessive here.

Harazi can you give some links to the problems you mentioned?

Here you go for the one with $2^{n}$. http://www.mathlinks.ro/Forum/viewtopic.php?search_id=834109250&t=26076
  For my problem with $a^{n^{2}}$ see the files of mathlinks contest.

[hide="Here is another solution"]Clearly, the only constant polynomials satisfying this are $f(x) = 0$, $f(x) = -1$, and $f(x) = 1$. We assume from now on that $f(x)$ is nonconstant. Note that if $f(x)$ is a polynomial satisfying the requirements then $f(x+a)$ must also satisfy the requirements for any integer $a$. Therefore, we can assume WLOG that if $f$ has degree $k$, then the coefficient of the $x^{k-1}$ term for $f$ is $0$. It follows that for sufficiently large $|x|$, $(x-1)^k < f(x) < (x+1)^k$, so if $f(x)$ is a power of $k$ for sufficiently large $k$, $f(x) = x^k$. Now since $f$ is nonconstant, $f(\mathbb Z)$ is an infinite set, and since it is closed under multiplication, then for any $x \in f(\mathbb Z)$, we know that $x^k \in f(\mathbb Z)$ so $f(x) = x^k$ for infinitely many values of $x$. Therefore, $f(x) = x^k$ for all $x$, so it follows that all nonconstant solutions are of the form $f(x) = (x+a)^k$ for some $a$ and $k$. $\Box$ [/hide]

The following is a classic (known result):
[rule]
If $f \in \mathbb Z[x]$ is non-constant such that $f(\mathbb Z) := \{f(x) : x \in \mathbb Z \}$ contains an infinite geometric progression, then
$$ f(x) = (ax + b)^n $$
for some $a,b,n \in \mathbb Z_{\ge 0}$ with $a,n \ge 1$ (converse is also true directly).
[rule]
In fact, it came as a problem in [url=https://artofproblemsolving.com/community/c5390_2014_iran_team_selection_test]Iran TST 2014[/url].

We return to our main problem now. Only constant solutions are clearly $f(x) \equiv -1,0,1$. Assume $f$ is non-constant henceforth. As $f(\mathbb Z)$ is closed under multiplication, it is easy to see that it must contain a infinite geometric progression. Using our above result we get
$$ f(x) \equiv (ax + b)^n $$
for some $a,b,n \in \mathbb Z_{\ge 0}$ with $a,n \ge 1$. As $f$ is monic, hence $a=1$. Thus,
$$ f(x) \equiv (x+b)^n $$
and it is easy to see all such $f$ indeed work.


find all $m,n$ in naturam number such that $m^{2}+1$ divisible by $n$ and $n^{3}+1$ divisible by $m$ .

It has been posted before by  [b]harazi[/b] and the solution is still a mystery.


There are $2005$ young people sitting around a large circular table. Of these, at most $668$ are boys. We say that a girl $G$ has a strong position, if, counting from $G$ in either direction, the number of girls is always strictly larger than the number of boys ($G$ is herself included in the count). Prove that there is always a girl in a strong position.

It's clear that if the result is true for $668$ boys, then it is true for any smaller number of boys (just switch some girls into boys till you have $668$ boys, then the result is true and by switching back the boys into girls it stays true). We'll prove the result for $2n+1$ girls and $n$ boys.

This can be done by induction. Suppose the result is true for $1, 2, ..., n-1$. Now take $2n+1$ girls and $n$ boys. Take any girl $G$ which has a boy for a neighbor (it's clear there is one), and then go round the table till you find another $G$, so you'll have something like $GBBB...BBBG$. Now take away $G, B, G$ where $B$ is any of the boys in that interval. Among the $2n-1$ girls and $n-1$ boys left there's a girl in a strong position by hypothesis; it's easy to see she's still strong when we put back $G, B, G$.


geometry
vector
analytic geometry
geometric transformation
rotation
complex numbers
geometry proposed

Let an isosceles triangle ABC and its height AM.Let MN perpendicular to AC.If P is  the midpoint of  MN, show that AP is perpendicular to BN.

Really one of the best preolympiad (for today standards)problems !

   [b]The best solution is this:[/b]

 If $K$ is the midpoint of  $NC$  , then $P$ is the orthocenter fo triangle $AMK$ , since $PK // MC , KP \perp AM , MK // BN, MN \perp AC$.

 But we can find solution with vectors or coordinates as well.

[u] babis[/u]

Mr Stergiu thank you for the solution!I have to say that i knew three solutions:
1)my solution using some very interesting identities..
2)one friend's solution (about the same with mine)
3)the most synthetic solution using rotation over point M.
As you said your solution is the fourth and the best of all.Congratulations and thanks!

Robpal as you can see problems in olympiads are about the same................................................................

I know that really well:D Both problems have also really beautiful solutions using vector dot prodcut :)

[quote="robpal"]This is problem 1.62 in prasolov book;). Last year, on our polish olympiad there was similiar problem:

Height CD is dropped from vertex C of acute triangle ABC. Point E is a point od segment BC and <CED=90. Let's choose point F on segmet DE so:

EF/FD=AD/DB

Show that AE is perpendicular to CF.[/quote]

  DEJA VU ?

  T.Y.

   M.T.

It's not exactly the same problem, but quite similliar;) and idea is the same

To Nick Rapanos.

  Nick , this solution is not mine. This is the classical solution. I have seen one more  solution in << Γεωργακάκης  "Μεθοδική Γεωμετρία ">>, in Greek , using similar triangles. I would very happy , if I could make a group of olympiad contestants or olympial level students in Greece  to make a good book with a collection of the best proof -geometry problems .  preolympiad and olympiad level.I have allready a colection of 500 solved problems but I have not time this period ! We' ll see.
   Keep sending good problems to the forum !

 [u]Babis[/u]

Stergiu please explain your solution if you have a time ...

Davron

:)  no no thank you  i got the solution thank you very much thats good soluiton

Davron

Mr Stergiu, this sounds very promissing and i hope that you will make it!I am very happy with your idea (it would be really good for Greek students..), but it's a pitty that i 'll not be an active contestant of mathematical olympiads when your book will be published!!
But nevermiind, lets hope that the <<new>>  students will use all the materials,which we don't have now, and maybe someone could do sth better for Greece (in olympiads i mean!) !
Also i want to say that i would be more active member of the forum, if i had the time....(so i can understand you..).But  this period i get prepered for the Mediterenean Competition, which as you probably know better than me is pretty hard!So it is not very easy to keep sending good problems (because i don't have the time!).We'll see....

With best wishes for publishing your [u]books
Nick Rapanos

Very well !!!

 I wish you personaly and to the whole  greek team GOOD LUCK to all  competitions !!!!!!!!!
 I'm sure you 'll solve the most problems.

 Read carefully << Engel , problem solving strategies >> and IMO problems from the last ten years(shortlists as well!)

 [u]babis[/u]

[color=darkred]Indeed, this problem is easily, but is interesting in view of the its didactic character. In the my book [b]"Complex numbers. Exercises and Problems, Ed. Scorpion7, Bucuresti, 1996"[/b] it appears with four solutions (syntetical, analytical, using complex numbers and using vectorial calculus). I shall offer here the its syntetical proof.[/color]

[color=darkblue][b]A syntetical proof.[/b] Define: $S\in BN\cap AP$ ; the point $R\in AC$, $BR\perp AC$. From the similitude $\triangle AMN\sim \triangle BCR$ results: the side $[AM]$ of the triangle $AMN$ is similarly with the side $[BC]$ of the triangle $BCR$ and the median $AP$ in the triangle $AMN$ is similarly with the median $BN$ in the triangle $BCR$. Thus, $\widehat {MAS}\equiv \widehat {MAP}\equiv \widehat {CBN}\equiv \widehat {MBS}$, i.e. $\widehat {MAS}\equiv \widehat {MBS}$, i.e. the quadrilateral $ABMS$ is inscribed. Therefore, $MA\perp MB$ $\Longrightarrow$ $SA\perp SB$, i.e. $AP\perp BN$.[/color]

Nice and what does synthetical proof: mean ?

Davron

It means, that in solution you use basic geometry facts (no complex plain nor analitic sulotion) =)

[hide="Only for Robpal."][color=darkred][b]Dear Robpal[/b], you must know the all instruments ([u]the first level[/u]) and you know to choose the best, efficient and suitable instrument (from the four means) for the solving a problem ([u]the second level[/u]) ! There are and another stronger instruments ! But the nicest solution is syntetically ! If you didn't find the syntetical solution of a problem, then you don't abandon it ! I wish you every success ![/color][/hide]


[quote="amcavoy"]Let $\left\{a_{k}\right\}$ and $\left\{b_{k}\right\}$ be positive real sequences.  Prove that if $\left\{a_{k}/b_{k}\right\}$ converges and $\sum b_{k}$ converges, then $\sum a_{k}$ converges as well.[/quote]

[b]Theorem:[/b] If $\sum_{n=1}^{\infty}a_{n}$ is a convergent series and $\{b_{n}\}$ is bounded then $\sum_{n=1}^{\infty}a_{n}b_{n}$ is a convergent series.

[b]Proof:[/b] We can write $|b_{n}|<K$ for $K>0$, then for any $\epsilon >0$ consider $\epsilon / K>0$ that means there exists $N>0$ so that,
$\left| \sum_{k=m}^{n}a_{n}\right|<\epsilon / K$ for all $n\geq m>N$.
But then,
$\left| \sum_{k=m}^{n}a_{n}b_{n}\right| \leq K\cdot \left| \sum_{k=m}^{n}a_{n}\right| < K \cdot \epsilon/K = \epsilon$.
Thus, $\sum_{n=1}^{\infty}a_{n}b_{n}$ satisfies the Cauchy criterion and therefore must converge.Q.E.D.

[b]Theorem:[/b] Every convergent sequence is bounded.

[b]Proof:[/b] Trivial. Q.E.D.

[b]Answer to Question:[/b] If $\sum_{n=1}^{\infty}b_{n}$ converges and $\lim a_{n}/b_{n}$ coverges then $|a_{n}/b_{n}|<K$. But then $|a_{n}| = |a_{n}/b_{n}|\cdot |b_{n}|$. Q.E.D.

If $\sum b_{k}$ converges then $\lim_{k\to\infty}b_{k}= 0$. Let's now compare the series $\sum a_{k}$ and $\sum \frac{a_{k}}{b_{k}}$:

$\lim_{k\to\infty}\frac{a_{k}}{\frac{a_{k}}{b_{k}}}= \lim_{k\to\infty}b_{k}= 0$.

This limit tells us that the $a_{k}$ is bounded superiorly by a convergent series, and so the series $\sum a_{k}$ is convergent.


calculus
integration
derivative
function
real analysis
real analysis unsolved

f(x,y) is defined on [0,1]x[0,1]. For each x, f(x,y) is Leb integrable w.r.t y.
df/dx (partial derivative) is a bounded function of (x,y). Show:
Let g(x)=Int(from 0 to 1){f(x,y)}dy.

show:
dg/dx=Int(from 0 to 1)(df/dx)dy

My question is:
1. How do we know dg/dx exists?
2. I know df/dx is limit of a sequence of differencial quotient, so probably I am supposed to use Leb dominated conv Thm, but I failed to find an integrable function dominating all the differencial quotient.

Can someone give help? thanks.

If $\frac{\partial f}{\partial x}$ is bounded by $M$, then $\frac{|f(x+h,y)-f(x)|}{h}\le M$ by the mean value theorem.


[hide]tan{(n+1)a\2}={sin a+sin 2a+...+sin na}\{cos a+...+cos na}
sin{(n+1)a\2}/cos{(n+1)a\2}={sin a+sin 2a+...+sin na}\{cos a+...+cos na}
sin{(n+1)a\2}{cos a+...+cos na}={sin a+sin 2a+...+sin na}cos{(n+1)a\2}
sin{(n+1)a\2}cosa+sin{(n+1)a\2}cos2a...+sin{(n+1)a\2}cosna=cos{(n+1)a\2}sina+cos{(n+1)a\2}sin2a...cos{(n+1)a\2}sinna
sin{(n+1)a\2}cosa-sinacos{(n+1)a\2}...+sin{(n+1)a\2}cosna-sinnacos{(n+1)a\2}=0
sin{(n-1)a\2}+sin{(n-3)a\2}...+sin{(-n+1)a\2}
Since sin(-x)=-sinx, the beginning term and the end term cancel out.  The second term and the second to last term cancel out and so on.
If n is odd, then there is still one term left, that term is sins{(-n+1)a\2+(-n+1)a\2}=sin(0)=0[/hide]

You are to find $\frac{\Im\sum_{k=1}^{n}e^{ika}}{\Re\sum_{k=1}^{n}e^{ika}}$.  After finding an explicit formula for the summation using geometric sums and simplifying, you obtain \[\frac{\Im\frac{e^{i\frac{a}{2}(n+1)}\sin\left(\frac{an}{2}\right)}{\sin\left(\frac{a}{2}\right)}}{\Re\frac{e^{i\frac{a}{2}(n+1)}\sin\left(\frac{an}{2}\right)}{\sin\left(\frac{a}{2}\right)}}=\frac{\sin\left(\frac{a}{2}(n+1)\right)}{\cos\left(\frac{a}{2}(n+1)\right)}=\tan\left(\frac{a}{2}(n+1)\right)\] which is as required.

Masoud Zargar


for all  positive real numers xand y let
$f(x,y)= min (x,\frac{y}{(x^2+y^2)})$
 show that there exist $x_0 , y_0$ such that $f(x,y) \leq f(x_0,y_0)$ for all positive x and y , and find $f(x_0,y_0)$

$x\cdot \frac{y}{x^2+y^2}=\frac{xy}{x^2+y^2} \leq \frac{1}{2}(x^2+y^2 \geq 2xy)$ so $f(x,y)\leq \frac{1}{\sqrt{2}}$(because $x$ and $y$ are positive) but for $x=y=\frac{1}{\sqrt{2}} \Rightarrow f(x.y)=\frac{1}{\sqrt{2}}$ so $x_0=y_0=\frac{1}{\sqrt{2}}$ and $f(x_0,y_0)=\frac{1}{\sqrt{2}}$

It's obvious that $x$ must be egual to $\frac{y}{x^2+y^2}$ because if it isn't, one of the two elements get higher, but the other one get smaller.
Now: $\frac{x}{y}=\frac{1}{x^2+y^2}$.
Let's multiply by $y^2$: in fact $f(kx;ky)=k\cdot f(x;y)$. So $xy=1-\frac{x^2}{y^2+x^2}$ while $f(y;x)$ arrive to the maximum if $xy=\frac{x^2}{x^2+y^2}$
. So $1=\frac{x^2+x^2}{x^2+y^2}=2\cdot \frac{x^2}{x^2+y^2}\rightarrow x^2=y^2$ and $x=y$ because they're positive. Solving $x^2=\frac{x^2}{2x^2}$ we get $x=y=\frac{1}{\sqrt{2}}$


function
combinatorics unsolved
combinatorics

For each x ∈ $ R$ we are given an interval $ I_{x}\equal{} [x\minus{} \delta_{x}, x\plus{} \delta_{x}]$ with $ \delta_{x}$>= 0. Moreover, for each $ x, y$∈ $ R$ with $ y$∈$ I_{x}$, we have $ \delta_{y} < \delta_{x}$ if $ x$ is not equal to $ y$. Show that $ \delta_{x} \equal{} 0$ for uncountably many $ x$.

Hint: For every $ n\equal{}1,2,3,\ldots$ consider the set $ A_n : \equal{} \{x \in R: \delta_x > \frac 1 n\}$, and show that it is countable.

While thinking about the problem, I also thought about the following generalization: Let $ f: R \to R$ be any function. Let $ A$ be the set of points, where $ f$ has a strict local maximum. (In the given setting the function $ f(x) \equal{} \delta_x$ has a strict local maximum at every $ x$ such that $ \delta_x>0$.) What can you say about the size of $ A$ (or $ A^c$)? Is it necessarily true that $ A^c$ is uncountable (or even that $ A$ is countable)? I am not sure.


The sum of the numerical coefficients in the complete expansion of $ (x^2 \minus{} 2xy \plus{} y^2)^7$ is:

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 128 \qquad \textbf{(E)}\ 128^2$

[hide="Answer"]
\[ x^2 \minus{} 2xy \plus{} y^2 \equal{} (x \minus{} y)^2
\]

\[ \therefore (x^2 \minus{} 2xy \plus{} y^2)^7 \equal{} (x \minus{} y)^{14} \equal{} \sum_{k \equal{} 0}^{14}\binom{14}{k}(\minus{}1)^{k}x^{14 \minus{} k}y^{k}
\]
Letting $ x \equal{} y \equal{} 1$, we get that the sum of the coefficients is
\[ (1 \minus{} 1)^{14} \equal{} \sum_{k \equal{} 0}^{n}(\minus{}1)^{k}\binom{14}{k} \equal{} \boxed{0}\ \blacksquare
\]
[/hide]

[hide="Solution"]$ (x^2 \minus{} 2xy \plus{} y^2)^7 \equal{} (x \minus{} y)^{14}$

If $ x \equal{} y \equal{} 1$, we have $ 0^{14} \equal{} \boxed{\textbf{(A)}\ 0}$.[/hide]


Let real numbers $ a,b,c$ such that $ a\minus{}2b\plus{}3c\equal{}24$
Find the minimum of $ C\equal{}a^2\plus{}b^2\plus{}c^2\minus{}a\minus{}b\minus{}c\plus{}2424$

hello, with the method of Lagrange multpliers we get $ 4a^2 \plus{} b^2 \plus{} c^2 \minus{} a \minus{} b \minus{} c \plus{} 2424\geq\frac {68909}{28}$
and the equal sign holds if $ a \equal{} \frac {15}{7},b \equal{} \minus{} \frac {39}{14},c \equal{} \frac {38}{7}$.
Sonnhard.


[img]http://img.photobucket.com/albums/v259/drspyder/Honor.gif[/img]
Yes, I know it's too quick. I gotsta find a way to slow it down.

[img]http://img.photobucket.com/albums/v259/drspyder/Eye_Big_Sig.gif[/img]
It's supposed to be an eye :P .

[img]http://img.photobucket.com/albums/v259/drspyder/SeizureSplat.gif[/img]
Stare at it long enough and your brain will rot...

[img]http://img.photobucket.com/albums/v259/drspyder/BombAsplode.gif[/img]
Boredom reigns supreme :P .

[img]http://img.photobucket.com/albums/v259/drspyder/DrspyderGun.gif[/img]
Again, I was bored :P .

[img]http://img.photobucket.com/albums/v259/drspyder/FlashySkullThingy.gif[/img]
I was dying of boredom when I made this.

I made some slideshow animations using powerpoint during keyboarding class.  Enjoy.

This is one I made in sixth grade.  Sorry about the multiple posts, I couldn't find a way to add more than one attachment.

something i did
[img]http://img.photobucket.com/albums/v475/commanderkewl/Copy2ofStickfight.gif[/img]

something else
[img]http://img.photobucket.com/albums/v475/commanderkewl/backflip.gif[/img]

done with Megaman sprites
[img]http://img.photobucket.com/albums/v475/commanderkewl/megaman.gif[/img]


For some reason, the text on my user page is vertically aligned in the center on my user page no matter what I do: [[User:Temperal]]

Is there a method to fix this?

Have no idea. Use firefox, and play with the CSS yourself :)

Hmmm using the !important declaration on my Web Developer>>Edit CSS works it out.

Ok. I tried it out, but it doesn't work very well when actually doing it... :wink:

I reverted your edits; they moved the header to an incorrect location.

I didn't mean the header; I meant the text in the tables.

They're not aligned to the top in Firefox or any other browser I have (Gran Paradiso, IE, Safari, Opera, Netscape, Netcaptor, Communicator, Lynx, and mBrowser GX at last count  :P )

Can you post a screenshot with what you want fixed?


Let $ x,y,z \in R$ satisfying
$ x$≥$ y$≥$ z$≥$ 1$
$ 2y\plus{}3z$≥$ 6$
$ 11x\plus{}27z$≥$ 54$
Find the maximum value of $ A\equal{}\frac{1}{x^2}\plus{}\frac{2008}{y^2}\plus{}\frac{2009}{z^2}$

hello, the maximum of $ A$ is $ \frac{2115274}{729}$ which is obtained at $ x\equal{}\frac{27}{11}$,$ y\equal{}\frac{3}{2}$ and $ z\equal{}1$.
Sonnhard.

Can you post the full solution,please,Dr Sonnhard Graubner?


geometry
circumcircle
trapezoid
cyclic quadrilateral

I have an idea on this problem, but I don't think I was allowed to use it.

Could you please help?

Let ABC be a triangle with $AB \neq AC$.  Let D be the foot of the altitude from A to BC.  Let P be a point on AD and let E and F be the intersections of rays BP and CP with AC and AB respectively.  Show that if CEFB is a cyclic quadrilateral, then P is the orthocenter of triangle ABC.

Very good question, the source was MOP 2003 apparently.

[hide="Funny idea."]Outline of a solution. It is well known fact (and very easy to prove) that if $H$ is an orthocenter of triangle $ABC$ and $O$ is it's circumcenter then $\angle HAB = \angle OAC$. Because $CEFB$ is cyclic it's clear that $ABC$ and $AEF$ are oppositely similar, so from our fact we know that the line $AD$ is passing by the circumcenter of $AEF$. With very easy angle chasing it can be proved in the same manner that circumcenter of triangle $PEF$ also lies on line $AD$. Suppose that these circumcenters are different. Then, the line containing the centers of these two circles, namely the line $AD$, is perpendicular to the common chord of these two circles, the line $EF$. But $AD$ is perpendicular to $BC$ as well, so $BC \parallel EF$, but since $CEFB$ is cyclic it must be isosceles trapezoid and we must have $|AB|=|AC|$, contradiction. Therefore, points $AEPF$ lie on one circle as well. Now it's only trivial angle chasing for finishing it off.[/hide]


calculus
integration
function
derivative
real analysis
complex analysis
topology

My local university has Real Variables as a prerequisite to Complex Variables, so I was wondering if it is possible to take Complex Variables without having knowledge of real variables. I'm mostly interested in the applications to physics, and as far as I know real variables has none (?). There is also a course called Applied Complex Variables that does not have Real Variables as a prereq, but I think any of the "applied" courses will be dumbed down too much. Any advice would be appreciated!

Here are the descriptions of the courses:

Applied Complex Variables
For students who desire a working knowledge of complex variables; covers the standard topics and gives an introduction to integration by residues, the argument principle, conformal maps, and potential fields. Students desiring a systematic development of the foundations of the subject should take Complex Variables.
Text: Churchill and Brown, Complex Variables and Applications, 7th Edition, McGraw Hill, 2004

Complex Variables
For students who desire a rigorous introduction to the theory of functions of a complex variable; topics include Cauchy's theorem, the residue theorem, the maximum modulus theorem, Laurent series, the fundamental theorem of algebra, and the argument principle.
Text: S.D. Fisher, Complex Variables, Dover, 1999.

I don't think real analysis is a necessary prerequisite for the "regular" complex variables course as long as you have had exposure to the ideas of partial differentiation, uniform continuity, infinite series and convergence (as taught in, say, an advanced calculus course). In addition you should be comfortable following proofs and reading books in the "definition-theorem-proof" format.

But if you're really interested in learning complex analysis to help you with physics, you should take the applied complex variables course. A lot of the stuff in the "regular" complex variables course will be mainly of interest to pure mathematicians and how things can be applied to physical situations may not be made apparent. The Churchill and Brown book is good, and even if some sections are not covered in your class you can read them in your book.

On my campus, we have just one such course at that level, and we do list it as having the first semester of our two-semester "introduction to real analysis" course as prerequisite. That first semester would have basic epsilon mechanics, continuity, and completeness in it, but not yet either the Riemann integral or uniform continuity, both of which we put off to the second semester of the intro real analysis.

I taught it once a number of years ago, and I'm pretty sure I used the Fisher book. That book is at a compromise level - more attention to theory and proof than Churchill, but less than Ahlfors or than a graduate text such as Conway. I think most of my students had had both semesters of the real analysis (meaning they had seen Riemann integration and uniform convergence, along with some metric space stuff), but I think one or two hadn't had the even first semester, and they did survive. (I just tried to open my gradebook from that semester and discovered that my current version of Excel won't open a file created that many versions ago. Oh, well.)

There should be a considerable overlap in subject between the two courses. Both should deal at considerable length with the Cauchy Integral Theorem; both should compute a wide variety of integrals by using the Residue Theorem.

interesting ....

In Chinese universities,complex analysis is always taught before real analysis


algebra
polynomial
superior algebra
superior algebra unsolved

Find those real numbers  $x$ such that $\mathbb{C}$ contains a proper subfield $K$ such that adjoining $x$ to $K$ gives all complex numbers.

Sorry, I am in a hurry now (economy course), but I don't have the same answer. I don't have the time now to look on the proof.

Sorry again for my stupid question the other night. I had just woken up and I was a bit confused :). Now for the problem.

Denote the property of $x$ in the statement by $(P)$. I think that the reals $x$ which do not exhibit $(P)$ are precisely those algebraic numbers all of whose conjugates are real. I’ll call these numbers [i]totally real[/i].

$x$ is not totally real $\Rightarrow\ x$ has the property $(P)$:

First assume that $x$ is transcendental. Consider a transcendence basis $\Lambda$ of $\mathbb R$ over $\mathbb Q$ which contains $x$, and a transcendence basis $\Omega$ of $\mathbb C$ over $\mathbb Q$ which contains $x+i$. There is an isomorphism of $\mathbb Q(\Lambda)$ onto $\mathbb Q(\Omega)$ obtained via a bijection of $\Lambda$ onto $\Omega$ which sends $x$ to $x+i$. This isomorphism extends to an automorphism $f$ of $\mathbb C$. Let $K=f(\mathbb R)$. We then have $\mathbb C=K(i)$, which equals $K(x)$ because $x+i\in K$.

Now assume that $x$ is algebraic, with at least one non-real conjugate $\tilde x$. Just as in the previous paragraph, we can extend the isomorphism $\mathbb Q(\tilde x)\to\mathbb Q(x)$ to an automorphism $f$ of $\mathbb C$. Once again, we take $K=f(\mathbb R)$ and we have $\mathbb C=K(i)$, so in particular, $[\mathbb C: K]=2$. This means that $\mathbb C$ is the only strict extension of $K$. Because $\tilde x\not\in\mathbb R$ and $f$ sends $\mathbb R$ onto $K$ and $\tilde x$ to $x$, we get $x\not\in K$. By the previous sentence this implies $\mathbb C=K(x)$, as desired.

$x$ exhibits $(P)\Rightarrow\ x$ is not totally real:

If $x$ is transcendental there’s nothing to prove, so assume it’s algebraic. $\mathbb C$ is then a finite extension of $K$ so, by the Artin-Schreier theorem (see the first paragraph on [url=http://eom.springer.de/R/r110050.htm]this[/url] page), $\mathbb C=K(i)$ (we don’t need this part :)) and $K$ is a real closed field. $K$ contains $\mathbb Q$, of course, so, being real closed, it contains a real closure of $\mathbb Q$. The real closure of an ordered field is unique up to isomorphism, so $K$ contains an isomorphic copy of the field of real algebraic numbers. If $x$ were to be totally real, its minimal polynomial over $\mathbb Q$ would split completely in the field of real algebraic numbers, hence also in the isomorphic embedding of this field in $K$. In particular, we would have $x\in K$, contradicting $K(x)=\mathbb C$.

I realize it may be too long to read through, but I just wanted to know: did you get the same result, harazi?


Let (O) be the circumcircle of a triangle ABC. A circle passing through points A and C meets the sides BC and BA at D and E, respectively. The lines AD and CE meet (O) again at G and H, respectively. The tangent lines of (O) at A and C meet the line DE at L and M, respectively. Prove that the lines LH and MG meet at (O).

This is from APMO 2008: http://www.mathlinks.ro/viewtopic.php?p=1073985#1073985


Credits to iLord for posting this game a few months ago.

PM me a number. Everybody gets a number of points equal to the number you PM'd me. However, the person who PM's me the highest number loses that number of points. It is possible to go into negatives. The winner is the first person to a googol (which basically means it will go on for a while.) Drop to a negative googol and you are out. The number MUST have commas or your number is valued at 0.

Round 1 deadline is Sunday at noon central time.

People who have sent me numbers:

mewto55555
pacman2812
ProtestanT
ernie
andersonw
james4l

[quote="Yongyi781"]Technically, everyone can avoid getting out by saying 0 (or any negative number such that if you add them up each round, it would take over 100 years to reach googol - way more than the average lifetime of the person, much less this topic :D)[/quote]
It's not about avoiding dropping out. It's about winning, and to do that you need to reach a googol.

1=2 YOUR NUMBER DIDN'T HAVE COMMAS HA HA HA

First, you have to PM me the number, otherwise your number is 0.

Second, you have to put the commas in.

woot 2nd place but only by 9
whens next round end?

Aww, are pacman2812 and I the only ones who posted "interesting" numbers? :P

WTH how am I not on the list

I chose

314,159,264

if you wanted to know

a small number  :(

Is this a sign up thing or is it join when ever you feel like it?

I think you can join whenever you feel like it.
I did not choose 1, in case any of you guys are wondering.  :)

I am sorry james4l. Your number will be included in the next list, added to your second round number.

good luck to everyone
i think brute3forc3 and i will have a psychology showdown now

Yes there is absolutely no way we can beat stevenmeow:

He just has to submit 20.

that will only hurt me 20 and heal you at most 19
so you only catch up at most 39
ownt
\

stevenmeow wins!!! (he sent me 20 and ppl sent me more than 20)

He could have not won if several other people sent in bigger numbers.

For instance, if cf249 sent in 9.1x10^99 and some one else sent in 9.2x10^99 (with commas of course), cf249 would have won.

[quote="stevenmeow"]lol
score
stevenmeow 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000[/quote]

Lol stevenmeow, of all people, you were the least likely to miscopy your number.

You put tritrigintillion ($ 10^{99}$) which is just short of googol by a factor of 10.
I knew that because you started with 1,000... where googol starts with 10,000...

:D :D

no i put dotrigintillion
b/c u divide the power by 3 and subtract 1, giving 32


$ \blacktriangleright\ \{a,b,c\}\subset(0,\infty)\ \Longrightarrow\ \{\begin{array}{c}
8\cdot\prod(b^2 + c^2 - bc)\ \ge\ \prod[a(b + c) + |a - b|\cdot |a - c|]\\\\
8\cdot\prod(b^2 + c^2 + bc)\ \ge\ \prod[a(b + c) + (a + b)\cdot (a +c)]\end{array}$

$ 8\prod(b^2\plus{}c^2\plus{}bc) \equal{} \sqrt {\prod[(b\plus{}c)^2\plus{}b^2\plus{}c^2][(a \plus{} b)^2 \plus{} a^2 \plus{} b^2]}\geq\prod[(b \plus{} c)(a \plus{} b) \plus{} ba \plus{} cb]$


Let $G$ be the centroid of $\triangle ABC$. Let $\angle GAB = \angle GBC = \angle GCA = x$, prove that $x \leq 30$

Here's a much stronger problem (and not that hard to prove either :)):

Given a point $P$ inside the triangle $ABC$, then at least one of the angles $\angle PBC,\angle PCA,\angle PAB$ measures at most $30^{\circ}$.

Much nicer, isn't it? :)

[quote="warut_suk"]Let $G$ be the centroid [/quote]

I am reading this word $centroid$ for a long time........

I JUST CAN'T GET IT WHAT IT MEANS!!!!!!!!

If it is what i think it is then when one calculate angles in this problem he gets that $G$ is (becide centroid) an orthocenter of $\triangle ABC$ and from that $\triangle ABC$ is equalliteral

Or did you mean that some of those three angles must be smaller or equal to $30^o$ and not that they are all equal...

[quote="M@re"]

If it is what i think it is then when one calculate angles in this problem he gets that $G$ is (becide centroid) an orthocenter of $\triangle ABC$ and from that $\triangle ABC$ is equalliteral

[/quote]

Is it? If it is, I'm sorry then.

Well, first "centroid" is it crossing point of medians, if isn't sorry my mistake but if is keep reading...

Let $D,E,F$ be midpoints of sides $BC,CA,AB$ respectivly. We have $DE||AB, FD||AC, EF||BC$ from that $\angle GBD=\angle ACG=\angle GFD$ so from that we se that points $G,F,B,D$ are concyclic, similary we obtain $\angle GAF= \angle GBD= \angle GEF$, so the points $A,F,G,E$ are also concyclic. Now we have $\angle B=\angle AFE=\angle AGE$ (because of cyclic quadraliteral $AFGE$)$=\angle BGD=\angle BFD$ (because of cyclic quadraliteral $FBDG$)$=\angle A$ So we obtain $\angle A=\angle B$ similary we can obtain $\angle A=\angle C$ and from that $\triangle ABC$ is equalliteral.

I hope you get it, and i hope i guess what centroid is...

m@re

[quote="Rafal"]grobber's problem was at IMO in the 90's...I know a very stupid solution to it  :)(with manipulating with the lengths of some segments) so I won't post it.[/quote]

There's a very nice and elegant solution using [url=http://mathworld.wolfram.com/Erdos-MordellTheorem.html]the Erdos-Mordell inequality[/url].

um, "centroid" should be replaced with "Brocard point", right?


inequalities
calculus
integration
geometry
rectangle
real analysis
real analysis unsolved

Draw a picture, use the notion of approximating an integral by summing the areas of rectangles whose upper righthand corner is on the curve...

[quote="kahsi"]Edit... Stupid post. It was so obvious.[/quote]
No such thing! Maybe to you it is obvious now but maybe it is helpful to someone else who is reading the forum later. Anyway a thread which has an empty first post and an answer that refers to something that no longer exists is not useful to anyone! So please don't delete your post even after you figure out the answer to your question.


Find all the solutions to the equation: $(x+1)^{y}-x^{z}=1$; $x,y,z$ are integers greater than 1.


For which rational $ x$ is the number $ 1 \plus{} 105 \cdot 2^x$ the square of a rational number?

Obviously x is an integer as $ 1\plus{}105*2^x$ is rational.
For $ x\equal{}0$,    $ 106$ is not a square.
Let $ 1\plus{}105*2^x\equal{}n^2$, then
$ (n\minus{}1)(n\plus{}1)\equal{}105*2^x$ , we get $ 105\equal{}n\minus{}1$  or  $ n\plus{}1$ has no solution.As $ (n\minus{}1) \neq (n\plus{}1)\;(mod\;4)$,
$ n\minus{}1$  or  $ n\plus{}1\equal{}2k$ , where $ k$ is odd .
Substitute  $ 2,6,10,14,30,42,70,210\equal{}n\minus{}1 \;\;\; or \;\;\; n\plus{}1$ we get $ x\equal{}3, \;4$ are the positive solutions.

If $ x\equal{}\minus{}k$ is negative.So $ \dfrac{2^k\plus{}105}{2^k}$ , So $ k\equal{}2m$ even and $ ((2^m)^2\plus{}105)\equal{}n^2$ is a square. $ (n\plus{}2^m)(n\minus{}2^m)\equal{}105$, substitute $ 2\minus{}2^m\equal{}1,3,5,7$ and we get $ x\equal{}\minus{}4, \minus{}6 \;\;or \;\;\minus{}8$ are the negative solutions.


Does the Countdown Round do part of the tiebreakers to determine the final winners of a competition?

The countdown round is actually for the top ten (in county and state) or top twelve (nationals). Tiebreakers are to determine who gets to the countdown round.

well actually, county and state can be any format. but otherwise what u said is correct.

Well but then whats the purpose of the Countdown Round? Does it determine who wins the competition or something like that?

Countdown is for fun at school or chapter level.  Neither are required to do the round, bust most chapters will do it.  The format of the countdown round varies from chapter to chapter, although they generally use the official State and National rules.  However, this is not required.

At the state and national levels, the countdown is used to determine your place overall.

Countdown is actually unofficial in most states. This means that it does not determine official placings, as in who gets to go to nationals and who doesn't. However, some states, notably Texas, New Jersey, and Florida, have official countdowns.

Countdown's use varies from state to state.  Some use it to decide the top, while others use the written round results.  In states where countdown is used, there are no tiebreaks, b/c its single elimination.  In states where written is used, it is based on the tiebreaking rules, first going to last 10 sprint problems or something like that.  Also, style used varies, some use ladder and some use the nats type, with 12 ppl and top 4 getting seeded, with byes


inequalities
function
Functional Analysis
real analysis
real analysis unsolved

Let $H$ be a Hilbert space. Prove that for any $x, y, z\in H$ the following inequality is valid

     \[\parallel x\parallel +\parallel y\parallel +\parallel z\parallel +\parallel x+y+z\parallel \geq \parallel x+y\parallel +\parallel y+z\parallel +\parallel z+x\parallel .\]

I have never understood one thing: what is the difference with the complex case? I mean, isn't it clear that the proof is identical? Is there a nice counter-example in Banach spaces?

Real vs. complex spaces does not matter here: a Hilbert space over $\mathbb C$ becomes a Hilbert space over $\mathbb R$ if one forgets about the complex structure (which is not involved in the inequality).

Counterexample in $\mathbb R^{3}$ with $\ell_{\infty}$ norm: $x=(1,1,0)$, $y=(1,0,1)$, $z=(0,1,1)$. Nice?  :cool:

My approach to the inequality was to rewrite it in the form
\[f(x)+f(y)+f(z)+3f\left(\frac{x+y+z}{3}\right)\ge 2f\left(\frac{x+y}{2}\right)+2f\left(\frac{y+z}{2}\right)+2f\left(\frac{x+z}{2}\right)\quad (*)\]
and try to prove (*) for an arbitrary convex function $f\colon\mathcal H\to\mathbb R$ rather than just for $f(x)=\parallel x\parallel $. :noo:
However, a majorization argument implies (*) for "one-dimensional" convex functions, namely $f=g\circ \varphi$, where $\varphi\colon\mathcal H\to\mathbb R$ is linear and $g\colon\mathbb R\to\mathbb R$ is convex. Consequenly, (*) holds when $f$ is a sum/integral of "one-dimensional" convex functions. I don't know if there is a name for this class, but $f(x)=\parallel x\parallel $ (Euclidean norm) is in it. :yup:


inequalities
vector
linear algebra
matrix
linear algebra unsolved

Let $ T_1,T_2,\ldots,T_m$ be linear transformations of an $ n$-dimensional vector space $ V$. Suppose that:
(i) $ \dim\mathrm{Im}(T_i) \equal{} 1$, for all $ i \equal{} 1,2,\ldots,m$.
(ii) $ T_i^2\ne0$ and $ T_iT_j \equal{} 0$ for each $ i\ne j$.
Prove that $ m\le n$.

From first part of (ii): $ \mathrm{Im}\, T_{i} \not\subset \ker T_{i}$, and hence, by (i): $ \mathrm{Im}\, T_{i} \cap \ker T_{i} \equal{} \{0\}, \ \forall i \equal{} 1, \ldots, m$.
Now second part of (ii) yields $ \mathrm{Im}\, T_{j} \subset \ker T_{i},\ \forall i \neq j$, so $ \mathrm{Im}\, T_{i}\cap \mathrm{Im}\, T_{j} \equal{} \{0\}, \ \forall i \neq j$, and finally $ \mathrm{Im}\, T_{1}\oplus\ldots\oplus \mathrm{Im}\, T_{m}\subset V$ implies $ m \leqslant n$.

We have $ T_{i}^2\equal{}a(i)T_{i}$  with $ a(i) \neq 0$

$ p_{i}\equal{}\frac{T_{i}}{a(i)}$ is a projector each rank 1

$ p_{i}p_{j}\equal{}0$ for each $ i\neq j$

there exist a basis $ (e_{1},...,e_{n})$ such that

for any  i integer from 1 to m

$ p_{i}(e_{i}) \equal{} 1$

$ p_{i}(e_{k})\equal{}0$ for $ k\neq i$

Consider the matrix of $ p_{1}\plus{}...\plus{}p_{m}$ in the basis  $ (e_{1},...,e_{n})$
this matrix is of the form

$ diag(I_{m},0,...0)$ then $ m\leq n$


probability
expected value
probability and stats

One fair coin is flipped many times. Player A gets one point for each head, player B gets one point for each tail. We stop when at least one of players has reached $ n$ points.
What is expected duration?

Warning: i am not sure about existence of closed form

if $ \tau_n$ denotes the first time that $ A$ or $ B$ has reached $ n$ point, then for $ k \leq n\minus{}1$,
\[ P(\tau_n \equal{} n\plus{}k) \equal{} 2 \cdot \frac{1}{2^{n\plus{}k}} \binom{k}{n\plus{}k\minus{}1} \equal{} 2 \cdot \frac{1}{2^{n\plus{}k}} \binom{k}{n\plus{}k} \frac{n}{n\plus{}k}\]
since there are $ \binom{k}{n\plus{}k\minus{}1}$ ways for this to happen if one assumes that $ A$ wins. Hence
\[ E[\tau_n] \equal{} \frac{n}{2^{n\minus{}1}} \sum_{k\equal{}0}^{n\minus{}1} \frac{1}{2^k} \binom{k}{n\plus{}k} \]
is it what you obtained ?

(Alekk posted while I was typing.)

Let $ e(a, b)$ be the expected number of moves if A needs $ a$ more points and B needs $ b$ more points.  Then $ e(0, b) \equal{} e(a, 0) \equal{} 0$ and $ e(a, b) \equal{} 1 \plus{} \frac {1}{2}(e(a \minus{} 1, b) \plus{} e(a, b \minus{} 1))$.  This gives $ 2^{a \plus{} b} e(a, b) \equal{} 2^{a \plus{} b} \plus{} 2^{a \minus{} 1 \plus{} b}e(a \minus{} 1, b) \plus{} 2^{a \plus{} b \minus{} 1}e(a, b \minus{} 1)$, so we get an integer sequence and we can plug it into the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  It's [url=http://www.research.att.com/~njas/sequences/A033504]A033504[/url], which gives the closed formula $ n\cdot 2^{n \minus{} 1} \plus{} \frac {n}{2}\binom{2n}{n}$ and several summation formulas.


If three blots equal four bleets, and five bleets equal six blits, then what is the ratio of one blit to one blot? Express your answer as a common fraction.

1. Three blots equal four bleets. Multiplying both sides by 5, 15 blots equal 20 bleets.

2. Five bleets equal six blits. Multiplying both sides by 4, 20 bleets equal 24 blits.

3. Since 15 blots and 24 blits are both equal to the same thing, they are equal to each other.

4. Dividing on both sides, the ratio of a blit to a blot is 15/24.

5. We need to express the answer as a common fraction. Simplifying, we find that the answer is 5/8.


Are the following true?

(a) $\arccos x = \arcsin (\sqrt{1-x^{2}})$
(b) $x = \cos (\arcsin(\sqrt{1-x^{2}})$

i don't believe either are except for the first one where $0 \leq x \leq 1$

(a) $\arccos$  has its values in $[0,\pi]$, $\arcsin$ has its values in $[-\pi/2,\pi/2]$.  So we must have $\arccos x\in[0,\pi/2]$, hence $0\le x\le1$. On the other hand for $x\in[0,1]$ set $x=\cos\alpha, \alpha\in[0,\pi/2]$: $\arccos x=\alpha, \arcsin\sqrt{1-x^{2}}=\arcsin|\sin\alpha|=\arcsin\sin\alpha=\alpha$.
(b) Since $\sqrt{1-x^{2}}\ge0$, we have $\arcsin(\sqrt{1-x^{2}})\in[0,\pi/2]$ and $\cos(\arcsin(\sqrt{1-x^{2}}))\ge0$. Hence we must have $x\in[0,1]$. On the other hand for $x\in[0,1]$ the claim follows from (a).

$\arcsin \sqrt{1-x^{2}}= \{\begin{array}{ccc}\arccos x & , & 0\le x\le 1\\ \pi-\arccos x & , &-1\le x<0\end{array}\ .$
$\cos(\arcsin\sqrt{1-x^{2}})=|x|\ .$

So, what are the following equal to

(a) $\sin (\arccos x)$
(b) $\cos (\arcsin x)$
(c) $\sin (\arctan x)$
(d) $\cos (\arctan x)$
(e) $\tan (\arcsin x)$
(f) $\tan (\arccos x)$


linear algebra
matrix
function
linear algebra unsolved

Where can I find a proof of the Sylvester's matrix theorem?

If A has distinct eigenvalues $ \lambda_{1}, \dots, \lambda_{n}$ and $ f$ is a matrix function, then $ f(A)\equal{}\sum{f(\lambda_{i})c_{i}r_{i}}$, where $ c_{i}$ and $ r_{i}$ are normed column and row eigenvectors.

I tried to google it, but found nothing. Also, the theorem has been generalized by Buchheim to the case when eigenvalues of A are not distinct, but I couldn't find anything about that.

Hi neworder; your formula is false if $ A\equal{}\begin{pmatrix}1&1\\0&2\end{pmatrix}$ and $ f\equal{}id$. But it's true if $ A$ is a hermitian matrix and if you choose $ \forall{i},r_i\equal{}c_i^*$ (there exist 2 choices for $ r_i$ and $ c_i$).
Proof: There exists an unitary matrix $ P$ s.t. $ A\equal{}Pdiag(\lambda_i)P^*$; then $ f(A)\equal{}Pdiag(f(\lambda_i))P^*$ where $ P\equal{}[c_1,\cdots,c_n]$. $ Pdiag(f(\lambda_i))\equal{}[f(\lambda_1)c_1,\cdots,f(\lambda_n)c_n]$. $ f(A)\equal{}[f(\lambda_1)c_1,\cdots,f(\lambda_n)c_n][r_1,\cdots,r_n]^T\equal{}f(\lambda_1)c_1r_1\plus{}\cdots\plus{}f(\lambda_n)c_nr_n$. QED
 :roll: Sylvester rolling eyes.

The exact theorem is the following: If the $ \lambda_i$ are pairwise distinct then $ f(A)=\sum\dfrac{f(\lambda_i)}{r{}_ic{}_i}c{}_ir{}_i$ where $ c_i,r_i$ are right-hand and left-hand $ A$-eigenvectors associated to $ \lambda_i$.
Proof: there exists an invertible matrix s.t. $ A=Pdiag(\lambda_1,\cdots,\lambda_n)P^{-1}$. $ f(A)=Pdiag(f(\lambda_1),\cdots,f(\lambda_n))P^{-1}=\sum{f(\lambda_i)Q_i}$ where $ Q_i$ is the one rank projector $ Pdiag(0,\cdots,1,\cdots,0)P^{-1}$.
$ Ac_i=\lambda{}_ic{}_i$ and we may assume that the $ c_i$ are the $ P$-columns; $ r_iA=\lambda_i{r_i}$ or $ A^Tr_i^T=\lambda{}_ir{}_i^T=P^{-1}^Tdiag(\lambda_1,\cdots,\lambda_n)P^Tr_i^T$ and  $ r_i=\alpha{}_il{}_i$ where $ l_i$ is the $ i^{th}$ $ P^{-1}$-row and $ \alpha_i\in\mathbb{R}$; $ r{}_ic{}_i=\alpha_i$.
 Thus $ Q_i=c{}_il{}_i=\dfrac{1}{r{}_ic{}_i}c{}_ir{}_i$.   QED.


Solve the following for $ x$: $ \tan^{\minus{}1}(x) \minus{} \sin^{\minus{}1}(x) \equal{} 0$

[quote="sunehra"]Solve the following for $ x$: $ \tan^{ - 1}(x) - \sin^{ - 1}(x) = 0$[/quote]
[hide]Hint:
Using the formula: ${ \tan^{-1}(x)=sin^{-1}(x/sqrt(x^2+1})$ with 0=<x=<1. Now solving the equation:
x/sqrt(x^2+1)=x, which gives the unique solution x=0.[/hide]

[hide]Note that the if the solutions are (x,y) for the inverses, the solutions are (y,x) for these functions. Clearly, if we consider the inverses of the functions the only solution in ${ ( - \frac {\pi}{2},\frac {\pi}{2}})$ is (0,0). x=0 is the only solution; we can just uses periods of the tangent and sine to get the other values for the output, all of them are in the form $ n\pi$. [/hide]

hello, let $ f(x)=arctan(x)-arcsin(x)$, then we have $ \lim_{x \to -1+0}f(x)=+\infty$ ,$ \lim_{x \to 1-0}f(x)=-\infty$ and $ f^'(x)=\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}$, now we have $ f^'(x)\le 0 \Leftrightarrow 0\le x^4+x^2$ so our function is strictly monotonously decreasing, hence the only solution of our
equation is $ x=0$.
Sonnhard.

I'm not sure if this is right

[hide]
$ \tan^{ \minus{} 1}(x)\equal{}\sin^{ \minus{} 1}(x)\equal{}y \implies \tan(y)\equal{}\sin(y)\equal{}x$

$ \tan(y)\equal{}\sin(y)$

$ \frac{\sin(y)}{\cos(y)}\equal{}\sin(y)$

The solutions are either $ \sin(y)\equal{}0$ or $ \cos(y)\equal{}1$ which are equivalent, so $ x\equal{}\boxed{0}$ is the only solution.[/hide]


Prove that the product of the first $1000$ positive even integers differs from the product of the first $1000$ positive odd integers by a multiple of $2001$.

This is trivial: $2001=3.23.29$, all factors apear in both the first 1000 odds and the first 1000 evens, so both expressions are a multiple of 2001, and so is the difference.

I have another solution ..........
2 cong. -1999(mod 2001)
4 cong. -1997 mod 2001)
.
.
.
2000 cong. -1(mod 2001)
As we have even number of congruence relations , by multiplying .........
2.4. ..... 2000 cong. 1.3. ...1999(mod 2001)
QED.....

Just for fun:
Can we generalise it?
Does the product of the first $2k$ positive even integers differ from the product of the first $2k$ positive odd integers by a multiple of $4k+1$?  :wink:

Yes, we can. If $4k+1$ is not a prime and not the square of a prime, then it can be factored as $p = ab$, with $a\neq b$, $1< a,b < 2k+1$, so both $A=\prod_{j=1}^{2k} (2j-1)$ and $B=\prod_{j=1}^{2k} 2j$ are divisible by $ab$.
If $4k+1 = 3^2$, then $B-A = 3^2\cdot 31$, while if $4k+1 = p^2$, $p>3$ prime, then again $A$ and $B$ are both divisible by $p^2$.
Finally, if $4k+1 = p$ prime, then by Wilson's theorem $AB = (p-1)! \equiv -1 \pmod{p}$, and also $B= 2^{2k} (2k)!$, so $B^2 = 2^{p-1} (((p-1)/2)!)^2 \equiv 1\cdot (-1) = -1 \pmod{p}$ (by Fermat's little and Wilson's again), so $-A \equiv AB^2 \equiv -B \pmod{p}$.

There should be a correction in the problem

The original question stated
"Prove that the product of the first $200$ positive even integers differs from the product of first $200$ odd integers by a multiple of $401$"

ONE MORE SOLUTIONS U HAVE
[color=#BF9000]product of 1000 even number =2.4.6.8.10.........2000
[/color] Now 2001=3*667 which implies that difference should be of form 3*667
So if we take differ of odd and even we can common out 3 from even and odd product and also 667 from both of so difference would be multiple of 2001


algebra
polynomial
function
inequalities
Rational Root Theorem

Is this a homework problem?  Well here's a hint...
[hide]In that equation, suppose the three roots are $a$,$b$ and $c$.  Then \[ 0=(x-a)(x-b)(x-c)=x^3-(a+b+c)x+(ab+bc+ac)x-(abc)x \] \[ \implies a+b+c=6 \] \[ ab+bc+ac=12 \] \[ abc=8 \]
[/hide]
Now solve for a, b and c.  I'll tell you now that a=b=c.

This is memorized *hopefully*  :P  as $(x-2)^3$, but somehow I smell homework help...

K81o7, how would you solve that inequality?

How about testing rational roots and dividing by one that works. Then use synthetic division and solve the resulting depressed polynomial.

nope, i'm studying for a quiz..
it used to be a hw question..
this was written on the board for the answer:

(x-2) (x^2-4x+4)
(x-2)(x-2)^2
(x-2)^3

and that is the answer..

but i don't get how you get x^3 - 6x^2 + 12x-8 to (x-2) (x^2-4x+4)..

(i would ask my teacher, but.. every time i ask him a question, he always looks at me like i'm stupid.. and then goes like.. "c'mon.. this is easy" UGH lol.. so i'm asking you guys for help.. so i can pass my quiz and be :D lol thanks

Have you learned about the Rational Root Theorem? It says that all roots must be a factor of the constant term, divided by the leading coefficient.

In "non-mathy" language: For your polynomial $y = x^3 - 6x^2 + 12x -8$, the roots must be a factor of $\frac{-8}{1}$. (last number divided by leading coefficient)

You can then test the factors to see if they work. Generally in school math classes, the factors are almost always whole numbers. :P

Hope this helps. :)

[quote="serenityx33"]hi, i need help with this quesiton:

Graph each of the following. (HInt: Consider the expansion of (a +/- b) ^n for appropriate values of n)

factor:
y = x^3 - 6x^2 + 12x -8

(I DO NOT NEED THE GRAPH, I ONLY NEED TO KNOW HOW TO... FACTOR IT.. hehe, THANK YOU SO MUCH IN ADVANCE)[/quote]

this is easily done by grouping (when you see factoring problems, you should first look for grouping)
so group the terms 1 and 4 then group terms 2 and 3

$(x^3-8)+(-6x^2+12x)$
$(x-2)(x^2+2x+4)+(-6x)(x-2)$

now it is pretty trivial...