function
trigonometry
complex analysis
complex analysis unsolved

1.If $ f$ is entire and $ \Im f > 0$, then $ f$ is constant.
2.If $ f$ is entire and $ f(\mathbb{C}) \subset \mathbb{C} \minus{} [0,1]$, then $ f$ is constant.

Any tips? Thanks in advance.

2) [hide="small hint"] if you substitute the interval $ [0,1]$ with a segment $ L$, the assertion will be still true. Find a suitable Mobius transformation $ g$ from $ \mathbb{C}\minus{}L$ to a suitable region like the unit disc. Can you show that $ g \circ f$ is constant?  :maybe:  [/hide]

[quote="wupanruo"]1.If $ f$ is entire and $ \Im f > 0$, then $ f$ is constant.
.[/quote]

let $ g\equal{}i*f$ then g is entire funtion

we have :$ Re(f)\equal{}1/2[f\plus{}g]$ sine $ f\plus{}g$ is entire and analytic, using "the Open mapping theorem" we have $ f\plus{}g$ is constant, and f is constant

one may find that $ e^{  i*f(z)}$ is bounded,due to Picard Theory,it is proven.

Well, if you're allowed to use Picard, then you don't even need to exponentiate. You have in both cases that the function omits uncountably many points. So it must be constant (since if it omitted even two points then it would be constant).

Well,I mean the Theorem of Liouville,which is somekind equal to Picard.

$ \mathbb{C}^* \setminus [0,1]$ is simply connected and can by the riemann theorem be conformally mapped to the unit disc by some map $ \phi$. $ \phi \circ f$ is then entire and bounded. ehsan suggests $ \phi$ can be made explicit, but I don't see it.

To make $ \phi$ explicit, note that $ z\plus{}z^{\minus{}1}$ maps the unit disk onto $ \mathbb C^*\setminus [\minus{}2,2]$.


Solve $ (x\minus{}5)^{x^{2}\minus{}4}\equal{}1$, for real values of $ x$.

Take logarithm of both sides, you get:

1) $ x \minus{} 5 \equal{} 1$

2) $ x^2 \minus{} 4 \equal{} 0$

Solutions: $ x\in\{6,2, \minus{} 2\}$

hello, here it can be x-5=1, hence $ x\equal{}6$  or $ x^2\minus{}4\equal{}0$ hence $ x\equal{}2$ or $ x\equal{}\minus{}2$ so the solutions are
$ x_1\equal{}6,x_2\equal{}2,x_3\equal{}\minus{}2$.
Sonnhard.

Don't forget $ x \minus{} 5 \equal{} \minus{} 1$ AND $ x^2 \minus{} 4$ is an even integer also yields a solution: $ x \equal{} 4$.

I make $ x\minus{}5 \equal{} \pm1$ and so $ x \equal{} 4$ or $ 6$.

So I get $ x\equal{} 2, \minus{}2, 4$ or $ 6$.


ll A X B ll = llall llbll sin ( theta)

can anyone help me to prove the magnitude of cross product?  I try to use some property of cross product but I couldn't do it.  Can you show me how to use the following property to prove it :

A X B X C = ( A . C )B - (A . B)C
and also the property  A. ( B X C )= ( A X B ) . C

the first is a cross b cross c equal a dot c times b minus a dot b times c

can someone please help me out with this?  I try to use those two properties to make both side equal each other but I can't do it.  I can only do it up to certain step but can't continue anymore.


geometry
geometric transformation
reflection

let L be a line .and A,B are two points wich arent on L.

find point  P on L  inwich we have : PA-PB=max

[quote="ashegh"]let L be a line .and A,B are two points wich arent on L.

find point  P on L  inwich we have : PA+PB=max[/quote]

interesting but i think that it may be wrong... :?

I think he means PA+PB = min... because if its max you can just take P off to infinity. For the min, reflect point B over L and see what you get.

thanks for u information.

i edited it...


parameterization
geometry
3D geometry
trigonometry
analytic geometry
inequalities
calculus

[quote="color"]How do you parametrize a solid cone $x^{2}+y^{2}\leq z^{2}, \,\,\, 0\leq z\leq b$[/quote]

 :| the equation of your cone surface in cylindrical co-ordinates is just $z = r$,
so you can then take as your parameters $r\;\;and\;\;\theta$, such that

$x\;=\;r\;\cos\theta,\;\;\;y\;=\;r\;\sin\theta,\;\;\;z\;=\;r$

where
$0\;\leq\;r\;\leq\;b,\;\;\;0\;\leq\;\theta\;\leq\;2\pi$

for your solid cone, you need to introduce a 3rd co-ordinate,

$x\;=\;\sqrt{2}\;\;r\;\cos\theta\;\cos\phi,\;\;\;y\;=\;\sqrt{2}\;\;r\;\sin\theta\;\cos\phi,\;\;\;z\;=\;\sqrt{2}\;\;r\;\sin\phi$

where
$0\;\leq\;r\;\leq\;b,\;\;\;0\;\leq\;\theta\;\leq\;2\pi,\;\;\;0\;\leq\;\phi\;\leq\;\frac{\pi}{4}$

That cone will have a spherical base instead of the plane $z=b$??

Indeed. We want cylindrical coordinates rather than spherical coordinates: $(r\cos\theta,r\sin\theta,z)$, with $0\le r\le z\le b$.

Yes that should work.
Thanks  :lol:

So it is not possible to parametrize without connecting the parameters as you did with the inequality?

It's possible- it's just more natural that way. If we set $w=\frac{r}{z}$, our region becomes $0\le w\le 1,0\le z\le b,0\le \theta\le 2\pi$.

[quote="jmerry"]It's possible- [color=red]it's just more natural that way[/color]. .[/quote]
Youre right, and I assume multiple integrals over this cone becomes easier, dont they?

Thanks a lot anyway  :) .


Asymptote
function
calculus
calculus computations

Find the locations of all absolute extrema if they exist.
$ f(x) \equal{} (x \minus{} 1) / (x^2 \plus{} 3x \plus{} 5)$

We've got to understand the shape of the graph just so we'll know how to interpret what we find.

The single most important issue: is that denominator ever zero? If it is, then we have a vertical asymptote as a dominant feature of the graph. If not, then the function is continuous and differentiable on the whole line.

After you've decided that, then answer these questions: where is the function zero, and (based on that), where is it positive and where is it negative?

Finally, what happens to this function in the limit as $ x\to\infty$ or $ x\to\infty?$

[hide="Hidden hint"]If you do all of that correctly, you should realize that the function must have an absolute maximum somewhere to the right of $ x\equal{}1$ and an absolute minimum somewhere to the left of $ x\equal{}1.$ I'll let you do the work needed to find exactly where these two points are.[/hide]


function
geometry
3D geometry
sphere
topology
algebra
polynomial

Let $ S^1$ be the unit sphere in the plane an let $ f: S^1\rightarrow S^1$ be a continuous function not homotopic to the identity. Prove that there exist $ x\in S^1$ such that $ f(x)\equal{}\minus{}x$

Suppose there was a function $ f: S^1\rightarrow S^1$ such that $ f(x)\neq \minus{}x$ for all $ x$. Then up to homotopy, $ f(z)\equal{}z^n$ for some integer $ n$. Consider the map $ z\plus{}f(z): S^1\rightarrow \mathbb{C}$, this function would never zero. But $ z\plus{}z^n\equal{}z(1\plus{}z^{n\minus{}1})$, so one of the roots of this polynomial is a root of unity. That means it lies on the unit circle.

So $ z \plus{} f(z)$, a function which is never zero, is homotopic to a map that has a zero; I don't see why this is a contradiction.

Solution:
If $ f(x) \ne \minus{} x$ for all $ x$, then the line segment from $ f(x)$ to $ x$, for each $ x \in S^1$, that does not go through the origin. We have a linear
homotopy $ H : S^1 \times [0,1] \to \mathbb{C} \setminus \{0\}$ such that $ H(x,0) \equal{} x$ and $ H(x,1) \equal{} f(x)$. Composing with the retract $ r : \mathbb{C} \setminus \{0\} \to S^1$ gives the desired homotopy $ rH : S^1 \times [0,1] \to S^1$.

EDIT: Fixed a typo.

Oops, good catch Nukular, I totally forgot i said "up to homotopy" >.< My bad, my solution is wrong.

I think that you mean $ H(x,0)\equal{}x$, thanks.


calculus
derivative
LaTeX
calculus computations

Given 3(x^2)(y^2)+5(y^3)x=128, what is the value of dy/dx when y=2 and x=2?

The answer they gave me is -21/11,
but I got -11/21,
I don't know what I did wrong,
PLEASE HELP ME![/list]

Hey, try some $ \text{\LaTeX}$ in your posts please. Use $ \$\$$'s to outline Mathmode (see my code):

$ 3 x^2 y^2 + 5 y^3 x = 128$.

Could you show what you have done so far?


For any positive real numbers a, b and c, prove that the expression

     [u] a^2 + bc[/u]    +   [u] b^2 + ac[/u]    +    [u]c^2 + ab[/u]
            a(b + c)                      b(a + c)                     c(a + b)

                  is greater than or equal to 3.

The inequality can be manipulated as follows:
$ {\sum }_{cyc}\frac{{a}^{2}\plus{}bc}{a(b\plus{}c)} \equal{}{\sum }_{cyc}(\frac{a(a\plus{}b\plus{}c)\plus{}bc}{a(b\plus{}c)}\minus{}1) \equal{} {\sum }_{cyc}(\frac{a}{b\plus{}c}) \plus{} {\sum }_{cyc}(\frac{bc}{ab\plus{}ac})$
According to Nesbitts inequality $ {\sum }_{cyc}(\frac{a}{b\plus{}c}) \geq\frac{3}{2}$
By substituting x = ac, y = bc and z = ab it also turns out that the second sum also is Nesbitts inequality;
$ {\sum }_{cyc}(\frac{bc}{ab\plus{}ac}) \geq\frac{3}{2}$
, which proves the statement.


This is kind of a silly question but which of the following is true?

$ A^{K}=AA^{k-1}$ or $ A^{k}=A^{k-1}A$ where $ A$ is a nonsingular $ n$x$ n$ matrix.

Also is there anyway to easily remember the correct relation?

Both are true. Multiplication of matrices (singular or not) is associative.

It is true because every matrix commutes under the opreation of multiplication with itself nad in particular with high powers of itself.
What does associativity has to do with this?

Example:

$ A^{4}=AAAA$ by definition, so it is how you group these that you get the different exponents.

$ A^{4}=(AAA)A=A^{3}A=A(AAA)=AA^{3}$ and all the other combinations. I believe in arbitrary structures (that need not even be associative), but do display this property are called [i]power associative[/i].

Thanks for the explanation HilbertThm90 and others.


geometry
geometric transformation
rotation
probability
algebra
function
domain

I spent some time typing this up to help people at my school with the basics of combinatorics. Most of it is fairly elementary, but there are some nice concepts included. Also, it does not focus on combinatorial identities; that will be for the next guide/lecture I write. Let me know of any mistakes.
Enjoy.

Right-click on the link and select "save target as" to download the pdf file.

when will solutions to the exercises be posted?

Not sure. I need to finish the other lectures first, and then I'll give out solutions. I didn't want to attach solutions directly to that document because I know that people get lazy.

[quote="nateharman1234"]err... number 22 says: how many positive integers are relatively prime to 2007?  are you sure you dont want to restrict that anymore?[/quote]

Yeah, thanks. i will change that. I'm going to wait a little longer, and then make an updated version. The integers need to be less than 2007.

[quote="anirudh"]So you are going to make more of these :)[/quote]

I believe so. Number theory, some geometry, and sequences/series.

[quote="drunner2007"][quote="jhredsox"]would this EVER help w/ mcounts?
-jorian :huh:[/quote]

Yeah. Many middle schoolers have difficulty grasping the concept of $\binom{n}{k}$. I think that some of the material covered will help you with the harder problems.[/quote]

some do...nicely done...is there a chance you'd make one on prob.? JW I think it is very nice what you are making....not to mention how people will benefit and how you will become better at explaining things...that is personally my weak point-from the looks of this I don't think it is your's  :D

[quote="funcia"]
some do...nicely done...is there a chance you'd make one on prob.? JW I think it is very nice what you are making....not to mention how people will benefit and how you will become better at explaining things...that is personally my weak point-from the looks of this I don't think it is your's  :D[/quote]

Thanks.

So has anyone worked through the whole thing?

This is very impressive. You explain things very clearly. Great job!

Comments on the problem set:

[hide]9. You might want to specify that there are two $20$s on that row. Your wording implies that there is only one.

10. It might help to say: "Rotations of a table are indistinguishable."

13. This is nice. I don't see a "not casework" solution. Take each center of a side. You have a 6 vertex graph. Connect centers adjacent faces. Upon coloring the verticies with two colors, what is the probability there is no 3 cycle with the same colors. I don't exactly know how to go from here.

15. Is there really an easy way to do this? There seems like there are many ways to overcount due to collinearity.

16. I know what you mean here, but it could be phrased better. Something like: the domain and range of a function are: $\{1,2,...,n\}$ and $\{1,2,...,m\}$, respectively, where $n\ge m$. The issue with your wording is that if the elements of your domain could be $\{1,2\}$ or $\{2,3\}$, and that would make a different function.

22. Positive integers less than 2007...[/hide]


ratio
geometry
symmetry
trigonometry
trig identities
Law of Cosines

In equilateral triangle $ \triangle ABC$, point P is on AB such that $ AP \equal{} \frac{1}{3}AB$, point Q is on BC such that $ BQ \equal{} \frac{1}{3}BC$ and point R is on CA such that $ CR \equal{} \frac{1}{3}CA$. The lines AQ, BR and CP are drawn, and the intersection of these lines forms another triangle. Find the ratio of the area of the new triangle to the area of $ \triangle ABC$

We can find the areas of each of the outside triangles using $ absinC/2$.  Letting each side have length $ 3s$, the combined areas of these triangles are $ 3\sqrt{3}s^{2}/2$.  Hence, the inside triangle has area $ 3\sqrt{3}s^{2}/4$, and our answer $ 1/3$.

[quote="J.Mayers"]We can find the areas of each of the outside triangles using $ absinC/2$.  Letting each side have length $ 3s$, the combined areas of these triangles are $ 3\sqrt {3}s^{2}/2$.  Hence, the inside triangle has area $ 3\sqrt {3}s^{2}/4$, and our answer $ 1/3$.[/quote]

No, but please try again :)

Oh wow...I just connected $ P,Q$ and $ R$ and found that area  :wink:

For the ACTUAL problem, it's easily done by noting that, by symmetry, the middle triangle is equilateral.  Therefore, the smaller outside triangles are similar to the larger outside triangles by $ AA$.  Using the Law of Cosines and ratios, the problem is solved.

Let each side have length 3x. Let BR and PC intersect at E. Let G be the perp from P to AC, D the perp from B to AC, F the perp from E to AC. We then have $ \frac {EF}{BD} \equal{} \frac {FR}{DR} and \frac {EF}{PG} \equal{} \frac {FC}{GC}$. In other words, $ \frac {EF}{3\sqrt {3}x/2} \equal{} \frac {FR}{x/2}$ and $ \frac {EF}{\sqrt {3}x/2} \equal{} \frac {FC}{5x/2}$. In other words,
$ EF \equal{} \frac {\sqrt {3}FC}{5} \equal{} 3\sqrt {3}FR$, so $ FC \equal{} 15FR$. But $ FC \minus{} FR \equal{} RC \equal{} x$, from here we have $ FC \equal{} \frac {15x}{14}$ and $ FR \equal{} \frac {x}{14}$. But the area of the inner triangle=Total area-$ 3 \cdot [BRC] \plus{} 3[ERC] \equal{} 3[ERC] \equal{} 3 \cdot EF \cdot RC/2 \equal{} \frac {9\sqrt {3}x^2}{28}$ out of a total $ \frac {9\sqrt {3}x^2}{4}$, so the answer is $ \frac {1}{7}$?

I remember there is an easier way, but forgot it.

Yes, correct, and your solution is much faster than mine :)

Another solution
Let D be the intersection of AQ and CP. By Menelaus, $ DQ\equal{}\frac{4}{3} AD$ and $ CD\equal{}6DP$. So $ [CDQ]\equal{}8[APD]$. So $ [CEFQ]\equal{}[CDQ]$-area of inner triangle$ \equal{}5[APD]$. So the area of the inner triangle/ the total area$ \equal{}3/(3\plus{}1\plus{}1\plus{}1\plus{}5\plus{}5\plus{}5)\equal{}1/7$.
I still think there's an even better solution using symmetry.


geometry
symmetry
inradius
geometry proposed

Dear everyone.

For an isosceles triangle ABC (CA=CB), let CD be perpendicular to AB and P be on CD.
inradius(ABP)=inradius(PECF) where E=AP$ \cap$ BC, F=BP$ \cap$ CA.
prove that inradius(ADP)=inradius(BCP).

I think it is interesting and nice.
May I ask you beautiful answers? :)

Thanks.

[hide="Simple proof"]Denote $ (I_1,r_1),(I_2,r_2),(M,r_M),(N,r_N)$ be the incircles of $ PECF, ABP, BCP, BDP$ respectively. See that $ I_1,I_2\in CD$ and $ PI_1\equal{}PI_2$.
We get $ \frac{MI_1}{MB}\equal{}\frac{PI_1}{PB}\equal{}\frac{PI_2}{PB}\equal{}\frac{NI_1}{NB}\Rightarrow MN\parallel I_1I_2$, which implies that $ r_M\equal{}r_N$. By the symmetry, the inradius of triangle $ APD$ is equal to $ r_N$. Hence result. [/hide]

Dear mr.dahn

Thanks for nice answers.
I am so glad to see you again, not having seen you for a long time. :)


Find x: $ f(x)\equal{}\frac{x(x\plus{}1)(x\plus{}2)(x\plus{}3)\plus{}1}{x(x\plus{}1)}\equal{}0$ and
Find $ R_n\equal{}f(1)\plus{}f(2)\plus{}..\plus{}f(n)$

[quote="quangpbc"]$ f(x) = 0\Leftrightarrow x(x + 1)(x + 2)(x + 3) + 1 = 0 \\
\Leftrightarrow (x^2 + 3x + 1)^2 = 0 \\$[/quote]

Could you please tell me how you were able to factor that?


algebra
polynomial
function
calculus
calculus computations

Find (preferably by Lagrange multiplier method) all the extremes of the function:
$\boxed {F(x,y,z)=(x^2+15yz)(z^2+15xy)(y^2+15xz)}$

given the condition: $\boxed {xyz=1}$ .

($x,y,z$ are nonegative reals).

There's a minimum when $x=y=z=1$, and the function has no maximum (it's unbounded above).

Just multiply through by $xyz$; we're looking for the extremes of $(x^3+15)(y^3+15)(z^3+15)$ given $xyz=x^3y^3z^3=1$. Let $u=x^3,v=y^3,w=z^3$.

$F(\sqrt[3]u,\sqrt[3]v,\sqrt[3]w)=uvw+15(uv+vw+wu)+15^2(u+v+w)+15^3$
$=1+15^3+15(\frac1u+\frac1v+\frac1w)+15^2(u+v+w)$.
By AM-GM, $\frac1u+\frac1v+\frac1w\ge 3$ and $u+v+w\ge 3$, with equality in both cases when $u=v=w=1$.
For the upper bound, note that $x$ can be arbitrarily large, and $f(x,y,z)>15^2x^3$.

If you really want calculus in there, it's easy to verify that $G(r,s,t)=F(e^{\frac r3},e^{\frac s3},e^{\frac t3})$ is convex. Restricting to the plane $r+s+t=0$, only the global minimum can be a critical point.


What grade will you guys be in next year?

[quote="EUCLA"]excuse me.....where's the  13th?? :? I'm not im the 13th..but just wondering :huh:[/quote]

13th? Is that some grade I don't know about? I only know up to 12th  :)
here, 13th is college  :lol:

Hm do you have a kindergarten? Or is your first grade in elementary/primary school called "1st grade"? (to EUCLA)

I'll be going into 11th.. oh how time flies.

yeah during the year it seems like it takes forever to get to the next grade but when i look back it was really fast. :huh: anyway im going to 8th

I'm going to 8th also, but it seems like I just finished 5th grade. Nothing happened? Time is going too fast!

[quote="furious"]SENI0R. yea. actually, it's not that great (so far). i'm just waiting another year for college.[/quote]

ALWAYS a pessimist...tsk tsk tsk

For me, I'm a $\mathbb{SENIOR.}$

And, uh yeah, there is a 13th grade...just look at the AIME results.  Certain other countries have a grade 13.

Is the grade 13th just because the other countries don't have a grade labeled "kindergarten" or is it 13th because they really have an extra year?

Like do their systems go "1,2,3...13" or "K,1,2,3...13"?

i'm going into 10th. YAY!!! i'm not the youngest at my school now :D

$8th!$   $YEAH!$ ... one more year till I'm out of this school... then to high school  :lol:

Wow, a lot of 8th graders so far.

I'm going to 9th.

[quote="furious"][quote="joml88"][quote="furious"]SENI0R. yea. actually, it's not that great (so far). i'm just waiting another year for college.[/quote]
ALWAYS a pessimist...tsk tsk tsk
[/quote]
how is being a senior gonna be any different from being a junior? well, besides the senior breakfasts  :lol:[/quote]

You gotta be active, not passive.  If you keep saying it's no different from any other year, it won't be any different.  But if you try to go out with a bang, it should be a memorable year.

they don't give me time in school to do homework :(

13th grade is an extra grade before college for those who do a extra year in their first 12 years.  Statistics is a 13th grade class that i am hoping to complete.

I would normally be going into the 8 for this semester (aug of 05). But I've aways been weird like that, I started school at 2 (even though I wasn't talking) I was reading at 3, so I was in kindergarden by then (didn't do preschool) and skipped 2 grades, blah blah blah...  :blush:  And now I am doing geometry (almost done so I will soon be doing alg 2), bio, english 10 2nd semester, world history ( I think, it may be U.S., I do things out of order because I'm a lil behind in history, well sort of), geography.  :mad: Er, geography. Well I guess I need it. I thought that georgia was next to louisiana *cough cough*.  :oops:

Next aug I'll probably be going into 11. Depends on how I do this year. OH yeah! I'm homeschooled...  :lol:  I did some private school and I was miserable!!!

[quote]I almost never have homework. I finish it in class.[/quote]

i agree. i have so much free time in math, french, and social studies class that i finish the homework then. the advantages of easy classes......... :D

[quote="h78reg"]it seems that there are few new middle schoolers this year. only 4 new 7th graders. :([/quote]

Well, we have plenty of 8th graders.  :)

gee. there is so little going to 7th grade

i'd be surprised if we have 5th graders or below on AOPS

I think colts18, math92's sister, is in like 6th I think...
chess64's sister, Diophantine27 (sp?), is young too.

I think school is boring.... They're gonig to start algebra I, and I've finished a major portion of Algebra II... going onto trig...

it seems to me that algebra 2 is almost the exact same thing as 1 except you deal with equations of larger degrees and with matrices. isn't precalc just algebra 2 and trig combined?

Im looking forward to my last year of middle school


Let the sequence $(a_n)$ define by:

$a_1=1, a_2=3$
$a_{n+2}=3a_{n+1}-2 a_n, \forall m,n \in \mathbb{N}^*$

Prove that:

$a_{n+2}=|\frac{7a_{n+1}^2-8.a_{n}.a_{n+1}}{1+a_n+a_{n+1}}|$,

where |x| denotes the integer part of the real number $x$.
_____________________________

I have a problem in typing "[x]". Sorry.


limit
integration
real analysis
real analysis solved

Let  $a_n=\sum_{k=1}^n \frac{1}{\sqrt{k}},b_n=\sum_{k=1}^n \frac{1}{\sqrt{2k+1}}$.Evaluate $\lim_{n\to\infty} a_n,\lim_{n\to\infty} \frac{b_n}{a_n}$ .

$a_n\geq \sum_{i=1}^n \frac{1}{i}$ so $lim\; a_n=\infty$.

Misha

Suppose $\sum x_n$ diverges, $x_n>0$, $\lim_{n\to\infty}\frac{y_n}{x_n}=a$, then $\lim_{n\to\infty}\frac{y_1+y_2+...+y_n}{x_1+x_2+...+x_n}=a$.

Thus $\lim\frac{b_n}{a_n}=\frac{1}{\sqrt{2}}$.

Thank you for your replies, Misha,Myth.

You are right.Here is my solution.

$a_n=\sum_{k=1}^n \frac{1}{\sqrt{k}}>\int^n_1 \frac{1}{\sqrt{x}} dx=2(\sqrt{n}-1)$.Thus $\lim_{n\to\infty} a_n=\infty$

$b_n=\sum_{k=1}^n \frac{1}{\sqrt{2k+1}}>\sum_{k=1}^n \frac{1}{\sqrt{2k+2}}=\frac{1}{\sqrt{2}}\sum_{k=1}^n \frac{1}{\sqrt{k+1}}$

${=\frac{1}{\sqrt{2}}\left(a_n-1-\frac{1}{\sqrt{n+1}}\right)>\frac{1}{\sqrt{2}}(a_n-1)} \cdots[1]$ .on the other hand

$b_n=\sum_{k=1}^n\frac{1}{\sqrt{2k+1}}<\sum_{k=1}^n\frac{1}{\sqrt{2k}}=\frac{1}{\sqrt{2}}a_n \cdots[2]$

From $[1],[2]$,  $\frac{1}{\sqrt2}\left(1-\frac{1}{a_n}\right)<\frac{b_n}{a_n}<\frac{1}{\sqrt{2}}$,yielding $\lim_{n\to\infty} \frac{b_n}{a_n}=\frac{1}{\sqrt{2}}$

kunny

It was clear that we can solve that problem in many ways :D


ratio
analytic geometry
invariant
geometry
geometric transformation
Euler
trigonometry

In triangle $ ABC$, squares $ ABDE, ACGH$ are externally drawn. Let $ M,N$ be midpoints of $ DH,GE$ respectively. Find all possible values of $ MN: BC$.

This problem becomes very easy once you employ complex numbers.  Let $ O$ be the origin of the complex plane.  Then obtain the complex coordinates of D,E,G, and H in terms of A,B, and C.  Then you can obtain M and N.  By doing so, compute $ \left|\frac {m \minus{} n}{b \minus{} c}\right|$.  This sketch is satisfactory enough for one who is familiar with complex numbers in geometry.  Due to the fact that it is 3:20 a.m. here, I may post the complete solution later.

The direction and length of the segment $ MN$ are invariant under any translation of the segment $ ED.$ That is, translating $ ED$ into $ E'D'$ and taking midpoints $ M', N'$ of $ D'H, E'G,$ then $ M'N' \equal{} MN, M'N' \parallel MN.$ Likewise, the direction and length of the segment $ MN$ are invariant under any translation of the segment $ HG.$ Translating $ ED$ to coincide with $ AB$ and $ HG$ to coincide with $ AC,$ $ M'N'$ becomes the midline of the $ \triangle ABC$, $ M'N' \parallel BC,$ $ M'N' \equal{} \frac {BC}{2},$ hence $ \frac {MN}{BC} \equal{} \frac {1}{2}.$

w.l.o.g $ c \geb$. Apply the Euler Theorem on the quad DEHG using: $ y=c \cos C$, $ z=c \cos B$,  $ x=z-y$ and $ h$ the A-altitude, clearly $ y+z=a$, so we have:

[img]http://rogercortesi.com/eqn/tempimagedir/eqn9056.png[/img]


Steve Olson, the author of Countdown, was on "The Screen Savers" on TechTV about ten minutes ago. Did anybody else see him?

Yeah I did.  It was much the same as when he was on the Connection though.


function
inequalities unsolved
inequalities

In one post, Namdung suggested the following estimation
\[\frac{a(b+c)}{a^{2}+(b+c)^{2}}\leq\frac{28a+b+c}{25(a+b+c)}.\]
Of course you know what immediately follows from this and how to verify that relation. I would like to know how to come up with this bound. Does there exist a strategy to determine $\alpha,\beta,\gamma$ such that
\[\frac{a(b+c)}{a^{2}+(b+c)^{2}}\leq \frac{\alpha a+\beta b+\gamma c}{a+b+c}.\]

I will try to explain it . (it has a small gap somewhere)
Well I think that he considered the function

$f(a)=\frac{a(b+c)}{a^{2}+(b+c)^{2}}-\frac{ka+lb+mc}{a+b+c}$

Now take $f'(a)$  and at $f'(a)$ put $a=1$ , $b=1$ , $c=1$
and take this expression equal to 0
We will have
$\frac{54-50k+25l+25m}{225}=0$  (1)

Now try to find solution to this (here is the gap)
But it easy to verify that the numbers that  mr Namdung took ,ie
$k=\frac{28}{25}$ , $l=\frac{1}{25}$ and $m=\frac{1}{25}$
satisfay (1)
Hope that I helped

Thank you. It would be simpler if we suppose without loss of generality that $b=c=1$. Then we investigate the function of $a$. This is just a necessary part.


I was wondering that since there are AP and college level courses available in high school, do some groups of classes prepare people better for certain professions or for the classes leading up to more specific topics? For example, if someone wanted to become a computer engineer or be premed in undergraduate, what should they take in high school? [hide="List of Classes in my High School"]
Not Language or Humanities Classes:
http://academics.tjhsst.edu/scitech.html
http://academics.tjhsst.edu/mathcs.html
[/hide]

I didn't read your course list, but unless you know you are going to get college credit for more advanced courses, there is no reason to rush it because you'll probably have to re-take the classes in college anyway.  Your eventual college academic advisor will be a better source of info.

Does everyone get a college academic adviser? I've never heard of a college academic adviser before.

Is there any advantage to take advanced courses that you will repeat in undergraduate college?

Is it more important to focus on classes that I can get college credit for even if they are not towards a certain area or not?

Supposing you gather as many college credits as you can during high school, will you have more flexibility in choosing your classes in college so that they can be towards a specific subject?

[quote="aufha"]Does everyone get a college academic adviser? I've never heard of a college academic adviser before.[/quote]

Almost everyone in an undergraduate program has to get a signature or approval from someone before registering for classes. That person is often called an adviser, but often doesn't give very much advice.

Unless you can take courses that you KNOW you will get credit for, it is not necessarily worth it.  Maybe it will make the course easier when you take in college, maybe not.  I know people that took Calc III in HS but they're still re-taking it.  I had nothing above AP Calc AB after HS, but I jumped right into Honors Multivariable Calculus (and did well!!).


Hockey Stick Identity Part 1

We start our investigation of the Hockey Stick Identity by trying to count how many ways we can distribute 18 dollars among 6 people.

[i]according to my calculations  I got 108 different ways as well [/i]

:):(:):(:):(:):(

Amazing i cant see how many combination there would be !

Actually there is only one way to give out the money...
You give it all to me.

[hide="Homework Hint"]If we look at the table more carefully it looks like pascal's triangle in a different way(I think (I didnt watch the next one))[/hide]

I believe that [quote="basketball9"]The answer is $\binom{23}{5}$[/quote] comes from stars and bars.
[quote="fdas"][hide="Spoiler Alert"]He doesnt give out the money[/hide][/quote]
Yeah... I wonder why he didn't give it out... :)

That did come from Stars and Bars because he explains it later on.


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

It follows from the equality $ \int_{0}^{\infty}\frac{\tan^{\minus{}1}sx\minus{}\tan^{\minus{}1}x}{x}\; dx\; \equal{}\;\frac{\pi}{2}\ln s$.

You can verify it by differentiating both side.

But, Sos440, how can that equality be deduced?

First I let $ I(s) \equal{}\int_{0}^{\infty}\frac{\tan^{\minus{}1}sx\minus{}\tan^{\minus{}1}x}{x}\; dx$, and noticed that $ I(0) \equal{}\minus{}\infty$, $ I(1) \equal{} 0$.

I also have seen integrals of similar forms which give logarithm (for example, $ \int_{0}^{\infty}\frac{e^{\minus{}x}\minus{}e^{\minus{}sx}}{x}\; dx \equal{}\ln x$), so I thought that $ I(s)$ would be so.

It was easy to verify that it really does, since $ I'(s)\; \equal{}\;\int_{0}^{\infty}\frac{1}{1\plus{}s^{2}x^{2}}\; dx\; \equal{}\;\frac{\pi}{2s}$.

More generally:

[color=blue][b]Frullani's Integral.[/b] If $ f$ is continuous (or maybe even weaker), $ f\left( 0^{\plus{}}\right) \equal{} A$ and $ f\left(\infty\right) \equal{} B$, then
\[ \int_{0}^{\infty}\frac{f(ax)\minus{}f(bx)}{x}\, dx\equal{}(A\minus{}B)\ln\frac{b}{a},\ \text{where}\ a,b > 0 .\]
[/color]

Also see http://www.mathlinks.ro/Forum/viewtopic.php?p=288711&highlight=specific#288711

Ok, thanks for the explanation Perfect_radio.

Oh, that's cool perfect_radio. Then it was not a coincidence that logarithm appears in such integrals.

[quote="misan"]show :D

$ \int_{0}^{\infty}\;\frac {\tan^{ - 1}\;(\pi x)\; - \;\tan^{ - 1}\;(x)}{x}\;\;\textbf dx\;\; = \;\;\boxed{\frac {\ln\left(\pi^{\pi}\right)}{2}}$[/quote]
$ \int_0^\infty\frac {\arctan ax - \arctan bx}x\,dx = \frac\pi2\ln\frac ab,\,\forall a,b > 0.$

$ \frac {\arctan ax - \arctan bx}x = \int_b^a\frac1{1 + u^2x^2}\,du.$ So,

\begin{eqnarray*}\int_0^\infty\frac {\arctan ax - \arctan bx}x\,dx & = & \int_0^\infty\int_b^a\frac1{1 + u^2x^2}\,du\,dx \\
& = & \int_b^a\int_0^\infty\frac1{1 + u^2x^2}\,dx\,du \\
& = & \int_b^a\left(\left.\frac {\arctan ux}u\right|_0^\infty\right)\,du \\
& = & \int_b^a\frac\pi{2u}\,du \\
& = & \frac\pi2\ln\frac ab\,\blacksquare\end{eqnarray*}

Plug $ a = \pi,\,b = 1$ and yields the desired answer.


Asymptote
AoPSwiki
abstract algebra
articles
LaTeX

I followed the instruction on AOPSwiki for setting up asymptote and the document I created as part of the setup tutorial works correctly. However, when I created a new document, in a different folder, it no longer worked saying "cannot find module."

I had that problem when creating the example document in the tutorial, and then after I completed the steps in problem 4 it worked. The file paths and stuff are still the same. It doesn't quite make sense to me why it works in one document but not the other.

That may happen if the folder containing the document that works also contains required files that are not in the folder containing the other document. Or perhaps the new document requires a different module.
Try the new document in the first folder. If that doesn't work then you will need to post a minimal example on the [url=http://www.artofproblemsolving.com/Forum/index.php?f=519]Asymptote forum[/url]and saying [i]exactly[/i] what the error is - you haven't for example said which module can't be found. It could be a missing import line in your document hence the importance of giving more information.

I tried the document in the folder that did work. This time I didn't get any compile/build errors, but asymptote threw the same error. It says cannot find module 'Section' The name of my document is Section 0.1

So where is the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198207]minimal example[/url]? How can we analyse the document without knowing what it contains? Since it is likely to be a syntax error in Asymptote you should post this minimal example in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=519]Asymptote forum[/url]. This example must be complete enough to be compilable but without unnnecessary lines of code.

\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
\begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{document}

I was doing something wrong before, and now it compiles just fine in both folders. Its just that asymptote says cannot find module "section" in one folder but it doesnt say that in the other folder.

Anyone have any ideas? Is there any more information I should include?

I thought it was solved. I tried your document and it was fine providing that I gave it a sensible name. [i]Section 0.1[/i] is not sensible.

Great! It works now under a different name. What are the guidelines for a "sensible" name?

Thanks again

Don't use words that Asymptote thinks might mean something else (see for example Problem 5 at [[Asymptote:_Help]]) and it's a good idea not to use spaces (use _ instead) or dots which are best left for use for file extensions eg .tex .asy


Let $x; y$ and $z$ be positive real numbers such that $\sqrt{x}+\sqrt{y}+\sqrt{z}= 1$. Prove that $\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}+\frac{y^{2}+zx}{\sqrt{2y^{2}(z+x)}}+\frac{z^{2}+xy}{\sqrt{2z^{2}(x+y)}}\geq 1.$

In APMO, I solved this problem by following way.
$\sum_{cyc}\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}=\sum_{cyc}\frac{(x-y)(x-z)}{\sqrt{2x^{2}(y+z)}}+\sum_{cyc}\sqrt{\frac{y+z}{2}}$.
But $\sum_{cyc}\sqrt{\frac{y+z}{2}}\ge \sum_{cyc}\frac{\sqrt{y}+\sqrt{z}}{2}=1$,
WTS : $\sum_{cyc}\frac{(x-y)(x-z)}{\sqrt{2x^{2}(y+z)}}\ge0$
WLOG $x\ge y\ge z$ -> $\frac{1}{y\sqrt{z+x}}\le \frac{1}{z\sqrt{x+y}}$.

Done.

$x,y,z$ are positive reals such that $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$
then
\[sum_{\substack{cyc}}\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}=\sum_{\substack{cyc}}\frac{x}{\sqrt{2(y+z)}}+\sum_{\substack{cyc}}\frac{yz}{x\sqrt{2(y+z)}}\]
now we have :
lemma 1:$\sum_{\substack{}}\frac{x}{\sqrt{2(y+z)}}\ge\frac{1}{2}.$
proof:we apply jensen to get above cyclic sum as
\[\ge \sqrt{\frac{(x+z+y)^{3}}{4(xy+zx+zy)}}\ge \frac{\sqrt{3}}{2}.\sqrt{x+y+z}\ge 1/2 \]

\[as\ x+y+z \ge 1/3.(\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}=1/3 \ and\ (x+z+y)^{2}\ge 3(xz+zy+xy) \]
now lemma 2:$\sum_{\substack{sym}}\frac{yz}{x\sqrt{2(y+z)}}\ge 1/2$
proof : again applying jensen we have above cyclic expression as
\[\ge \frac{1}{xyz}.\sqrt{\frac{(x^{2}z^{2}+x^{2}y^{2}+y^{2}z^{2})^{3}}{2.\sum_{\substack{sym}}x^{3}y^{2}}}\ge \frac{\sqrt{3(x+z+y)}}{2}\]
$iff \sum_{\substack{sym}}x^{6}y^{6}+6\sum_{\substack{sym}}x^{6}y^{4}z^{2}+12x^{4}y^{4}z^{4}\ge 3\sum_{\substack{sym}}x^{6}y^{4}z^{2}+3\sum_{\substack{sym}}x^{5}y^{5}z^{2}+3\sum_{\substack{sym}}x^{5}y^{4}z^{3}$
 from schur inequality
\[\sum_{\substack{sym}}x^{6}y^{6}+6x^{4}y^{4}z^{4}\ge 2\sum_{\substack{sym}}x^{6}y^{4}z^{2}\]
so we have to prove $3(\sum_{\substack{sym}}x^{6}y^{4}z^{2}-x^{5}y^{5}z^{2})+2(\sum_{\substack{sym}}x^{6}y^{4}z^{2}+1/2(x^{4}z^{4}y^{4}))-3\sum_{\substack{sym}}x^{5}y^{4}z^{3}\ge 0$by AM-GM and MUIRHEAD inequality
adding two lemmas we have the result.

i made a different solution... it's key is... faith  :D

from the condition and AM-GM we have that

$\sum_{cyc}\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}=\sum_{cyc}\frac{x^{2}+yz}{x(\sqrt{2x(y+z)}+\sqrt{2y(y+z)}+\sqrt{2z(y+z)})}\geq \sum_{cyc}\frac{2(x^{2}+yz)}{x(2x+5y+5z)}$...

the rest is plain muirhead. after clearing the denominators, it's exagerated how so much terms cancell and the thing to prove at last falls so easy from muirhead...

We have
\[
\begin {align*}
\sum \frac {x^2 + yz} {\sqrt {2x^2 (y + z)}} = \sum \frac {x^2 + yz} {x(y + z)} \cdot \sqrt {\frac {y + z} {2}} \ge \sum \frac {x^2 + yz} {x(y + z)} \left (\frac {\sqrt y + \sqrt z} {2} \right)
\end {align*}
\]
Now
\[
\begin {align*}
\sum \frac {x^2 + yz} {x(y + z)} \left (\frac {\sqrt y + \sqrt z} {2} \right) &\ge 1 \iff\\
\sum \frac {(x + y) (x + z)} {x (y + z)} \left (\frac {\sqrt y + \sqrt z} {2}\right) & \ge 2 \iff \\
\sum \frac {\sqrt y} {2} \left (\frac {(x + y)(x + z)} {x(y + z)} + \frac {(z + y) (z + x)} {z (x + y)}\right) & \ge 2
\end {align*}
\]
From AM-GM,
\[
\begin {align*}
\frac {\sqrt y} {2} \left (\frac {(x + y)(x + z)} {x(y + z)} + \frac {(z + y) (z + x)} {z (x + y)}\right) \ge (x + z) \sqrt \frac {y} {xz} \ge 2 \sqrt y
\end {align*}
\]
Summing cyclically gives us the desired result.

[hide]\[\sum_{cyc}\frac{x^2+yz}{\sqrt{2x^2(y+z)}}\geq 1\]
\[\Rightarrow \frac{1}{\sqrt{2}}\sum_{cyc}\frac{x^2+yz}{x\sqrt{y+z}}\geq 1\]\[\Rightarrow \left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )+\left ( \sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )\geq \sqrt{2}....(*)\]
Before going through the proof we'll prove some lemmas.
[color=#00BF00]Lemma 1:[/color]
$3(x+y+z)\ge 1$
It is a direct consequence of Cauchy applying to the sequences $1,1,1$ and $\sqrt {x},\sqrt {y},\sqrt {z}$.
[color=#00BF00]Lemma 2 :[/color]
$\sum_{cyc}\frac {xy}{z}\ge x+y+z$
Proof:From AM-GM,
\[2\left ( \sum_{cyc}\frac {xy}{z} \right )=\sum_{cyc}\frac {xy}{z}+\frac {yz}{x}\geq 2(x+y+z)\]
And also this is an obvious case of Muirhead:\[[1,1,-1]\ge [1,0,0]\]
[color=#00BF00]Lemma 3 :[/color]
$\frac {1}{3(xy+yz+zx)}\ge \frac {1}{(x+y+z)^2}$
It is a direct consequence of AM-GM.So left for you.

Now we'll start the proof:
From Holder's inequality,\[\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )\left ( \sum _{cyc} x(y+z)\right )\geq (x+y+z)^3\]
\[\Rightarrow 2\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )^2\geq \frac{(x+y+z)^3}{xy+yz+zx}\]
\[\Rightarrow 2\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )^2\geq \frac{3(x+y+z)^3}{3(xy+yz+zx)}\geq \frac{3(x+y+z)^3}{(x+y+z)^2}\]
\[= 3(x+y+z)\geq 1\]\[\Rightarrow 2\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )^2\geq 1\]\[\Rightarrow \sum_{cyc}\frac{x}{\sqrt{y+z}} \geq \frac{1}{\sqrt{2}}.....(i)\]

Again from Holder,\[\left ( \sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )\left ( \sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )\left ( \sum_{cyc}yzx^2(y+z) \right )\geq \left ( xy+yz+zx \right )^3\]\[\Rightarrow 2\left ( \sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )^2\geq \frac{\left ( xy+yz+zx \right )^3}{xyz(xy+yz+zx)}=\frac{\left ( xy+yz+zx \right )^2}{xyz}\]\[=\frac{x^2y^2+y^2z^2+z^2x ^2}{xyz}+2\frac{xy^2z+yz^2x+zx ^2y}{xyz}\]\[=\sum_{cyc}\frac {xy}{z}+2(x+y+z)\geq 3(x+y+z)\geq 1\]\[\Rightarrow 2\left (\sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )^2\geq 1\]\[\Rightarrow \sum_{cyc}\frac{yz}{x\sqrt{y+z}}\geq \frac{1}{\sqrt{2}}....(ii)\]
Adding $(i)$ and $(ii)$ we get $(*)$.\[\left ( \sum_{cyc}\frac{x}{\sqrt{y+z}} \right )+\left ( \sum_{cyc}\frac{yz}{x\sqrt{y+z}} \right )\geq \frac {1}{\sqrt{2}}+\frac {1}{\sqrt{2}}=\sqrt {2}\][/hide]

I have a generalization for this in cases of :
$$\lambda =2,n=2,k=1,p=1,\sqrt{x}+\sqrt{y}+\sqrt{z}=1$$

[b]Generalization 1[/b]
Let $x,y,z$ be positive reals. Then prove that

$$\sum_{cyc}{\left(\dfrac{x^{k+p}+\left(yz\right)^{p}}{\sqrt[n]{\lambda.x^{np}\left(y+z\right)}}\right)}\geq \dfrac{\sqrt[n]{\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^{nk-1}}{2.3^{nk-2}}}+\sqrt[n]{\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^{np-1}}{2}}}{\sqrt[n]{\lambda }}$$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that a group with 300 elements is not simple  :lol:

It would need to have $ 6$ Sylow $ 5$-subgroups, action by conjugation gives an injective map into $ S_6$, but $ 300$ doesn't divide $ 6!$.


quadratics
calculus
integration
modular arithmetic
pigeonhole principle
Ring Theory
number theory unsolved

Prove that if $p$ is a prime, then $-2$ is quadratic residue mod $p$ iff there exist $a,b\in \mathbb{N}$ such that $a^2+2b^2=p$.

I know that it is also true for $-1$:

 If $p$ is a prime, then $-1$ is quadratic residue mod $p$ iff there exist $a,b\in \mathbb{N}$ such that $a^2+b^2=p$.

Can this be generalized for other numbers? For what other numbers is it true?

Can add onething:
If $p=a^2+2b^2$ for p is prime. Then it has the form of $8n+1$ or $8n+3$.

[u][b]Proof: [/b][/u]
It is obvious that p is odd ,since it is greater than two the smallest prime (which is even) number.

[b][i]1.Case[/i][/b] Let both a and b be an even numbers, this implies to contradiction of the statement.

[b][i]2.Case[/i][/b] Let b an even number. Then we have $b=2k$, $a^2+8k^2=p$ thus we have p congruent to 1 in mod8, since a is odd from. Let's check it: $a=2r+1$ ... $a^2=4r^2+4r+1$ and we know that $8|4r(r+1)$. So we have a congruent to 1 up on the division with 8.
Resulting, case 2 impliest that p is of the form $8n+1$.

[b][i]3.Case[/i][/b] Let b be an odd number. So we know that $b^2$ is congruent to 1 in mod8 and $2b^2$ is congruent to -6. also $a^2$ is congruent to a in mod8 so summing up we have $a^2+2b^2$ congruent to -5 which is 3 in mod8.
Resulting, case 3 implies that p is of the form $8n+3$

Davron

No,like $a^2+11b^2$,and $-11$ is quadratic residue of $3$ but we can't find $a,b$ such that $a^2+11b^2=3$.

It's true for $1,2,3$ (where we exclude $p=2$ for $d=3$; it also works for $d=4$ when $p \neq 2$, but thats nonsense) only.
1: http://www.mathlinks.ro/Forum/viewtopic.php?t=46300
2: http://www.mathlinks.ro/Forum/viewtopic.php?t=55462
3: http://www.mathlinks.ro/Forum/viewtopic.php?t=55714
For all those, Thue was used.

Assume $d$ as squarefree.
Let's work in $R=\mathbb{Z}[\sqrt{-d}]$.
The condition that all the $p$ with $\left( \frac{-d}{p} \right)$ can be written as $p=a^2+db^2=(a+b \sqrt{-d})(a-b \sqrt{-d})$ means that all prime ideals that split totally in the integral closure of $R$ already splitt in $R$ and are principal ideals there. That means that $R$ is a principal ideal domain. But that only happens for $d=1,2$ (the case $d=3$ commes from the fact that the only exception is $p=2$).

[quote="Davron"]Can add onething:
If $p=a^2+2b^2$ for p is prime. Then it has the form of $8n+1$ or $8n+3$. [/quote]
Well, that's what xirti stated in his post ;)

let $t^2=-2 \pmod p$
Consider integers $x-ty$ , $0\le x,y < \sqrt{p}$ from pigenhole pronciple
we have that there exist $a-tb$ and $c-td$ satisfy "
$a-tb=c-td \pmod p$
$(a-c)=t(d-b) \pmod p$ let $a-c=k$ and $d-b=l$
$k^2=-2l^2\pmod p$
$k^2+2l^2=0 \pmod p$
now $|k|<\sqrt{p}$ and $|l|<\sqrt{p}$ so
$k^2+2l^2<3p$
if $k^2+2l^2=p$ we are done else
$k^2+2l^2=2p$ and $k$ must be even
so $4k_0^2+2l^2=2p$
$2k_0^2+l^2=p$

sorry i'm late


function
integration
calculus
inequalities
algebra
polynomial
linear algebra

This calls for an orthogonal decomposition. Note that $1$ and $x$ are not orthogonal functions on $[0,1]$ but $1$ and $x-\frac12$ are.

Note also that $\int_0^1\left(x-\frac12\right)^2\,dx=\frac1{12}.$

Let $g(x)=f(x)-1-6\left(x-\frac12\right).$

Then $\int_0^1g(x)\,dx=\int_0^1f(x)\,dx-\int_0^1dx-6\int_0^1\left(x-\frac12\right)\,ex$

$=1-1-0=0$ and

$\int_0^1\left(x-\frac12\right)g(x)\,dx=\int_0^1\left(x-\frac12\right)f(x)\,dx-\int_0^1 \left(x-\frac12\right)^2-6\int_0^1\left(x-\frac12\right)^2\,dx$

$=\left(1-\frac12\right)-0-6\cdot\frac1{12}=0.$

Then $\int_0^1[f(x)]^2\,dx=\int_0^1\left(1+6\left(x-\frac12\right)+g(x)\right)^2\,dx$

[use orthogonality to eliminate cross terms]

$=\int_0^11^2\,dx+36\int_0^1\left(x-\frac12\right)^2\,dx+\int_0^1[g(x)]^2\,dx$

$=1+36\cdot\frac1{12}+\int_0^1[g(x)]^2\,dx$

$\ge 4.$

It's also clear that the equality condition for this inequality is precisely that $f(x)=1+6\left(x-\frac12\right)=6x-2.$ (a.e.)

Note that the equality condition implicit in Ramanujan's post is that $f$ is a constant multiple of $\left(x-\frac13\right).$ This will quickly lead to $f(x)=6x-2,$ the same as in my proof. His proof is more elegant; mine is basically brute force.

A completely abstract version of this problem: Suppose that $u,v,w\in V$ where $V$ is a real inner product space. Suppose that:

$\|u\|^2=1,\,\|v\|^2=\frac13,\, \langle u,v\rangle = \frac12,\,\langle w,u\rangle = 1,\, \langle w,v\rangle=1$

Prove that $\|w\|^2>4.$

[quote="Kent Merryfield"]This calls for an orthogonal decomposition. [/quote]

What is ortogonal decomposition of $f$ , and what calls for it in this problem.

Guessing here... orthogonal decomposition means writing $f = g_1+g_2+\ldots$, where $g_i$ is orthogonal to $g_j$, for all $i \neq j$ ??

Yes perfect_radio, that's what I meant, and that's why I used $x-\frac12$ instead of $x.$

I looked in that direction because the problem clearly involved inner products, but with more than one function.

I'll try to generalize Kent's proof for the first problem.  :)
I will use the [color=blue]shifted Legendre polynomials[/color] which are polynomials given by

$P_{n}(x)=(-1)^n \sum_{k=0}^{n}\binom{n}{k}\binom{n+k}{k}(-x)^k$ , they are orthogonal in $[0,1]$ and

$\int_{0}^{1}P_{n}(x)P_{m}(x)dx=\frac{1}{2n+1}\delta_{mn}$ where $\delta_{mn}$ is the Kronecker delta.

We consider the function

$g(x)=f(x)-\sum_{i=0}^{n-1}c_{i}P_{i}(x)$ with  $c_{i}=\frac{\int_{0}^{1}f(x) P_{i}(x) dx}{\int_{0}^{1}P_{i}^2(x) dx} , \forall i$

because we want $\int_{0}^{1}g(x)P_{i}(x)dx =0 , \forall i$.

From the condition given we have that $\int_{0}^{1}f(x) P_{i}(x) dx=P_{i}(1)$ and then  $c_{i}=(2i+1) P_{i}(1)$

Now $\int_{0}^{1}f^2(x) dx =\int_{0}^{1}[g(x)+\sum_{i=0}^{n-1}c_{i}P_{i}(x)]^2 dx$  and since $\int_{0}^{1}g(x)P_{i}(x)dx =0 , \forall i$ we have

$\int_{0}^{1}f^2(x) dx =\int_{0}^{1}g^{2}(x)dx +\sum_{i=0}^{n-1}c_{i}^{2} \int_{0}^{1}P_{i}^{2}(x)dx\Longrightarrow$

$\int_{0}^{1}f^2(x) dx \geq \sum_{i=0}^{n-1}(2i+1) P_{i}^{2}(1)$

Can anyone simplify the last sum ???  :D  :D
Kent Merryfield i can't understand why you prove that $\int_0^1g(x)\,dx=0$ Where we need that ?
Also explain a bit the connection between this problem and the linear algebra problem you posted  :)

[quote]Kent Merryfield i can't understand why you prove that $\int_0^1g(x)\,dx=0$ Where we need that ? [/quote]
That's actually $\int_0^11\cdot g(x),dx$ - that is, the inner product of $g$ and $1.$

What I need to know about $g$ is that it is orthogonal to $\text{span}\{1,x\}.$

Essentially, I found the othogonal projection of $f$ onto that span. $g$ was what was left over, hence othogonal.

The linear algebra formulation? That matrix is a "Grammian." That is, we have a collection of vectors $\mathbf{v}_k$ and we form the matrix whose $(j,k)$ entry is $\langle\mathbf{v}_j,\mathbf{v}_k\rangle.$

A Grammian is always at least positive semidefinite; positive definite if the vectors are independent.

[quote]That's actually $\int_0^11\cdot g(x)\,dx$ - that is, the inner product of $g$ and $1$.[/quote]

Of course .....what a stupid question ... :blush:

[quote="Kent Merryfield"]A completely abstract version of this problem: Suppose that $ u,v,w\in V$ where $ V$ is a real inner product space. Suppose that:

$ \|u\|^2 \equal{} 1,\,\|v\|^2 \equal{} \frac13,\, \langle u,v\rangle \equal{} \frac12,\,\langle w,u\rangle \equal{} 1,\, \langle w,v\rangle \equal{} 1$

Prove that $ \|w\|^2 > 4.$[/quote]

Let $ d: \equal{} 2u \minus{} 3b \plus{} w$. $ < d,u >  \equal{}  < d,v >  \equal{} 0$. Expand $ < w,w >$.

there is more standard way  which help as to solve such problems .  One can say that, it's in exact about  Lagrange multipliers.
.
Consider two functional
\[ G_{1}(f) \equal{} \int_0^1 f(x) \, dx \equal{} 1
\]
and
\[ G_{2}(f) \equal{} \int_0^1 x f(x) \, \equal{} 1
\]
and we need  to find
\[ G(f) \equal{} \int_0^1 f^2 (x) \, dx \to \min
\]
Now we have isoperimetric problem .
we Consider new functional say :
\[ F(f) \equal{} G(f) \minus{} C_{1}G_{1}(f) \minus{} C_{2}G_{2}(f)
\]
isoperimetric problem says : there must be :
$ \frac {d}{dt} F(f \plus{} th)|_{t \equal{} 0} \equal{} 0$ for all $ h \in C_{0}^{\infty}[0,1]$
thus we have  simplest Euler equation :
$ 2f \minus{} C_{1} \minus{} C_{2}x \equal{} 0$ but since :$ G_{1}(f) \equal{} G_{2}(f) \equal{} 1$ we get
$ f \equal{} 6x \minus{} 2$
similar way works for Kent's generalization case

[quote]Let $f : \left[ 0,1 \right] \to \mathbb R$ be an integrable function such that \[ \int_0^1 f(x) \, dx = \int_0^1 x f(x) \, dx = 1 .  \] Prove that \[ \int_0^1 f^2 (x) \, dx \geq 4 .  \]
[i]Ion Rasa[/i][/quote]

Construct a linear polynomial $p(x)=ax+b$ so that $$\int_0^1p(x)dx=\int_0^1xp(x)dx=1$$
$p(x)=6x-2$ is the unique such polynomial.
From the given condition we have,
$$\int_0^1(p(x)-f(x))dx=0=\int_0^1x(p(x)-f(x))dx$$
So we have,
$$\int_0^1p(x)(p(x)-f(x) dx=0$$
But then,
\begin{align*}
0\le\int_0^1 (p(x)-f(x))^2dx=\int_0^1f(x)^2-f(x)p(x)dx\\
\implies \int_0^1 f(x)^2dx\ge \int_0^1 f(x)p(x)dx=6\int_0^1 xf(x)-2\int_0^1f(x)dx=4
\end{align*}
We are done. $\blacksquare$

Just read through the thread, and I had a small question: could there be a solution using Chebyshev’s Sum Inequality here? From it we have
$$\int_0^1 x(f(x))^2\ge \int_0^1 f(x)\,dx\int_0^1 xf(x)\,dx\implies \int_0^1 x(f(x))^2\ge 1.$$

For Chebyshev inequality you need monotonic condition on the functions
[url=https://mathworld.wolfram.com/ChebyshevIntegralInequality.html] see here  [/url]

[quote=rassvet][quote="Ramanujan"]From

$\int_0^1 f(x) \, dx = \int_0^1 x f(x) \, dx = 1$ we have  $\int_0^1 f(x)(x-\frac{1}{3}) \, dx = \frac{2}{3}$

Now from Schwarz's integral inequality

$\int_0^1 f^2(x) \, dx \cdot \int_0^1 (x-\frac{1}{3})^2 \, dx \geq [\int_0^1 f(x)(x-\frac{1}{3}) \, dx)]^2 = \frac{4}{9}$

and because $\int_0^1 (x-\frac{1}{3})^2 \, dx = \frac{1}{9}$ it follows that

$\int_0^1 f^2(x) \, dx \geq 4$

It is correct ???  :roll:[/quote]

A little bit shorter

Using Holder inequality

$\int_0^1 f^2(x) \, dx \cdot \int_0^1 f(x) \, dx \cdot \int_0^1 x^3 \, dx \geq [\int_0^1 xf(x) \, dx)]^3 = 1$

Thus,
$\frac{1}{4} \cdot \int_0^1 f^2(x) \, dx \geq 1 $[/quote]

cant understand what you did there, could you write a few more step in the solution

Generalization’s Hölder inequality

[url= https://www.emis.de/journals/HOA/IJMMS/26/17.pdf] see [/url]

https://www.emis.de/journals/HOA/IJMMS/26/17.pdf

[url= https://en.wikipedia.org/wiki/Hölder%27s_inequality] see [/url]

https://en.wikipedia.org/wiki/Hölder%27s_inequality

p=q=r=3 ,  1/p + 1/q +1/r =1
$xf(x)=xf^{1/3}(x)f^{2/3}(x)$

[quote=rassvet][quote="Ramanujan"]From

$\int_0^1 f(x) \, dx = \int_0^1 x f(x) \, dx = 1$ we have  $\int_0^1 f(x)(x-\frac{1}{3}) \, dx = \frac{2}{3}$

Now from Schwarz's integral inequality

$\int_0^1 f^2(x) \, dx \cdot \int_0^1 (x-\frac{1}{3})^2 \, dx \geq [\int_0^1 f(x)(x-\frac{1}{3}) \, dx)]^2 = \frac{4}{9}$

and because $\int_0^1 (x-\frac{1}{3})^2 \, dx = \frac{1}{9}$ it follows that

$\int_0^1 f^2(x) \, dx \geq 4$

It is correct ???  :roll:[/quote]

A little bit shorter

Using Holder inequality

$\int_0^1 f^2(x) \, dx \cdot \int_0^1 f(x) \, dx \cdot \int_0^1 x^3 \, dx \geq [\int_0^1 xf(x) \, dx)]^3 = 1$

Thus,
$\frac{1}{4} \cdot \int_0^1 f^2(x) \, dx \geq 1 $[/quote]
I dont think we can use holders inequality here as holders inequality requires functions to be positive and $f$ is not nesaccary positive

For some g(x), by Cauchy Schwartz, we have $$\int_0^1 f(x)^2 d x \geq \frac{\left[\int_0^1 f(x) g(x) d x\right]^2}{\int_0^1 g(x)^2 d x}$$ we would want the expression in LHS to be equal to $4$. Taking a linear function $g(x)=ax+b$, we would get $$\frac{\left[a \int_0^1x f(x)dx+b \int_0^1 f(x)dx\right]^2}{\int_0^1(a x+b)^2 d x}=4$$ Solving it, we get the relation $a=-3b$. Now putting any such $g(x)$ in C-S would yield the desired result.

[quote=Kent Merryfield]How about this problem: Suppose $f\in L^2[0,1]$ (real-valued) and

$\int_0^1f(x)\,dx=1,\,\int_0^1xf(x)\,dx=1,\dots,\int_0^1x^{n-2}f(x)\,dx=1.$

There is a number $C_n$ such that $\int_0^1[f(x)]^2\,dx\ge C_n.$

From above, we can take $C_3=4.$

Compute $C_4,C_5,\dots.$

I have actually double-posted this problem. The other location is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77352]here.[/url][/quote]

From some engineer's induction, it looks like the answer is $C_n=(n-1)^2$. Using the same approach used in post #17, we want to find a polynomial $P_n(x)$ of degree at most $n-2$ that satisfies $\int_0^1 P_n(x)Q(x)\, dx=Q(1)$ for any polynomial $Q(x)$ of degree at most $n-2$.
Once we find this polynomial, the same approach shows that the lower bound is $\int_0^1 P_n(x)P_n(x)\, dx=P(1)$.
Recognizing that this is an inner product, finding $P_n(x)$ resolves to computing
$$H_n^{-1}\begin{pmatrix}1\\\vdots\\1\end{pmatrix}$$
Where $H_n$ is the $(n-1)\times (n-1)$ Hilbert matrix. In particular, the coefficients of $P_n(x)$ are the entries of this vector. We want $P_n(1)$ which is just
$$\begin{pmatrix}1&\cdots&1\end{pmatrix}H_n^{-1}\begin{pmatrix}1\\\vdots\\1\end{pmatrix}$$
Here I am stuck, but with some searching I find [url=https://en.wikipedia.org/wiki/Matrix_determinant_lemma]this lemma[/url]. This seems like it could be a more doable approach then just using the explicit formula for $H_n^{-1}$.


In Vegetableville, an election took place.  Everyone who voted for the Broccoli Party had eaten broccoli at least once in a lifetime.  Among those who voted for other party's, 90% had never eaten broccoli.  What percent of the voters did the Broccoli Party get n the election if exactly 46% of all voters taking part in the election had eaten broccoli at least once in their lifetime?

[quote="myspacesucks"]In Vegetableville, an election took place.  Everyone who voted for the Broccoli Party had eaten broccoli at least once in a lifetime.  Among those who voted for other party's, 90% had never eaten broccoli.  What percent of the voters did the Broccoli Party get in the election if exactly 46% of all voters taking part in the election had eaten broccoli at least once in their lifetime?[/quote][hide]

46% of all voters has eaten broccoli.
10% of the 46%, or 4.6% of all voters with broccoli, voted for the other party.
The answer is 46-4.6=41.4%.[/hide]

[hide="Solution 1"]let $x$ rep. the percentage of people who ate broccoli and voted for the "other" party then:

Percentage of votes received by each party:

Broccoli Party: $46-x$
Non-Broccoli Pary: $54+x$

Number of people who ate broccoli and voted for the other party:

$(0.1)(54+x)=5.4+0.1x$ so
$5.4+0.1x+46-x=46$
$0.9x=5.4$
$x=6$

Therefore $6$ percent of all broccoli eaters voted for the non-broccoli party thus the broccoli party got $46-6=\boxed{40}$[/hide]

[hide="Solution 2"]
Clearly the $54$ percent that dont eat broccoli voted for the non-broccoli party, they rep. 90% of the total vote that, that party got so

$0.9x=54\implies x=60 \therefore \boxed{40}$ percent voted for the broccoli party
[/hide]

[hide="solution"]Let's say that there are 100 people!

Now there are x people who are voting for brocolli. Then there are  100-x people who aren't voting for brocolli. So

$10+\frac{9x}{10}$ is how many people who've eaten brocolli after simplification. But that equals 46. So

$\frac{9x}{10}=36$

$\frac{x}{10}=4$

$x=\boxed{40}$[/hide]


[i]The integers from $1$ to $n$ are written on a board.  On a turn, a player (there are two) erases some number $k$ and all the remaining divisors of $k$.  Then the other player takes a turn.   The player who erases the last number wins the game.[/i]

It isn't hard to show that the first player always has a winning strategy, but it seems to be very hard to actually find that strategy.  Has anybody tried out this game and found a good method of play?

I make a mental note of the parity of the number of primes and numbers that are relatively prime to everything else on the list. That is usually enough for me to win against the kinds of people that I see at school... but a winning strategy seems to be very tough for general n. I'm sure I could devise one for many specific n though, and that should be enough to knock out most laypeople  ;)


Given are many squares, each joining at the midpoint of each others sides. The resistance of the side of the big square is $R$. The resistance of the other sides is proportional to their lengths. Find the resistance between points $A$ and $B$ of the infinite fractile.

By using symetry we can change the figure in the way shown below.
$R_{x}$ is the resistance between $A$ and $B$, $R_{1}=R\frac{\sqrt{2}}{4}$ (from geometry) and $R_{2}=R(\sqrt{2}-1)$ (parallel connection of resistance $R$ and $R\frac{\sqrt{2}}{2}$).

From Kirchoff's laws we get:
$\frac{I}{2}=I_{1}+I_{2}$
and $2I_{2}R\frac{\sqrt{2}}{4}+2I_{2}\frac{R_{x}}{2}=I_{1}R(\sqrt{2}-1)$
Solving for $I_{1}$ we get:
$I_{1}=\frac{I(R\sqrt{2}+2R_{x})}{2(R(3\sqrt{2}-2)+2R_{x})}$

By definition: $R_{x}=\frac{U_{AB}}{I}$ and $U_{AB}=2\frac{I}{2}\frac{R}{2}+I_{1}R(\sqrt{2}-1)$
Putting into this $I_{1}$ we get an equation for $R_{x}$. By solving it we get $R_{x}=\frac{R(1-\sqrt{2}\pm\sqrt{3})}{2}$.
$R_{x}>0$ implies the answer with +:

$R_{x}=\frac{R(1-\sqrt{2}+\sqrt{3})}{2}$

Yes very nice golduck :D
it can be solved without using kirchoffs laws also :wink:

[quote="Thaakisfox"]it can be solved without using kirchoffs laws also :wink:[/quote]
Could you tell how :?:

golduck, how did you simplify the resistors? I couldn't follow that part...

I think he used that the whole system is symmetrical on the diagonals of the square.so all the points on the diagonals are equipotential. hence you can simplify it the way he did. :D

yes, because of the symmetry all points on the red line (as below) are equipotential so we can split the circuit in the marked points.


[quote="siuhochung"]f has to be continuous for this to be true.[/quote]except for when $n$ is a power of 2 :roll:


geometry
3D geometry
modular arithmetic
induction
algebra
factorization
sum of cubes

[hide]
We can change this to $n^3 + (n+1)^3 + (n+2)^3 \equiv 0 \pmod 9 \forall n > 1$
Simplifying, the expression is equal to
$3n^3 + 9n^2 + 15n + 9 = 3(n^3 + 3n^3 + 5n + 3)$
Then we just have to show $n^3 + 3n^3 + 5n + 3 \equiv n(n^2 + 5) \equiv 0 \pmod 3$
There are only three possibilities, $n \equiv 0,1,2 \pmod 3$, and testing these three, we find the above equivalence is satisfied for all $n$.
[/hide]

[hide="my solution"]
$(n+1)^3+(n+2)^3+(n+3)^3$ can be seen as $(k-1)^3+k^3+(k+1)^3$ where $k=n-2$ and $k\geq 3$.
Taking the sum of cubes for $(k-1)^3 + (k+1)^3$ yields $(2k)(k^2-2k+1-(k^2-1)+k^2+2k+1)$.
This simplifies to $(2k)(k^2+3)$. Combining this with the original equation, we get:
$3k^3+6k$, and factoring gives $3k(k^2+2)$. Now, we try cases where $k=3p, k=3p-1$ and $k=3p-2$.
$k=3p$ obviously works. If $k=3p-1$, we have $3(3p-1)(9p^2-6p+3) = 9(3p-1)(3p^2-2p+1)$, which is divisble by $9$.
For $k=3p-2$, we have $3(3p-2)(9p^2-12p+6) = 9(3p-2)(3p^2-4p+2)$ which is also divisble by $9$, so we are done.
[/hide]

[quote="Gyan"][quote]Since the terms are consecutive, $n+1$, $n+2$, and $n+3$ are $\equiv {0,1,2}\mod{3}$. So then the cubes are $\equiv {0,1,8}\mod 9$ so then the sum is $\equiv {0}\mod {9}$? Does this work? [/quote]
FWIW The part $n \equiv a \pmod 3 \implies n^3 \equiv a^3 \pmod 9$ though true,  perhaps, may not be so obvious to all so
it may require an additional step.[/quote]

Well actually I don't know how i transferred steps, but I had to provide some explanation as to how I got there...
Care to explain?  :)

You can say something like:  if $n \equiv a \pmod 3$ you can wrtire   $n$ as $3k+a$  and thus you have:

$n^3=(3k+a)^3 = 27k^3+ 27k^2a+9ka^2+a^3 = 9\cdot(something) + a^3$
$\implies n^3 \equiv a^3 \pmod 9$

(In general if $a \equiv b \pmod n \implies a^n \equiv b^n \pmod {n^2}$

[quote="srulikbd"][quote="jli"][quote="Leao"]Prove that $(n+1)^3+(n+2)^3+(n+3)^3$ is divisible by $9$ for all positive integers $n$.[/quote]

[hide]by expansion we get $3x^3+18x^2+42x+36$

$3x^3+18x^2+42x+36\equiv 3x^3+42x \bmod 9\equiv x^3+14x\bmod 3\equiv x(x^2+14)\bmod 3 \equiv x(x^2-1)\bmod 3\equiv x(x+1)(x-1)\bmod 3$[/hide][/quote]
I didn't understand why does $x(x^2+14)\bmod 3 \equiv x(x^2-1)\bmod 3$
tnx!
[quote="ch1n353ch3s54a1l"][quote="Leao"]Prove that $(n+1)^3+(n+2)^3+(n+3)^3$ is divisible by $9$ for all positive integers $n$.[/quote]

[hide]Since the terms are consecutive, $n+1$, $n+2$, and $n+3$ are $\equiv {0,1,2}\mod{3}$. So then the cubes are $\equiv {0,1,8}\mod 9$ so then the sum is $\equiv {0}\mod {9}$? Does this work?[/hide][/quote]
amazing!  :)[/quote]

$x^2+14 \bmod 3 \equiv x^2-1 \bmod 3$

For future reference, proofs shouldn't really be in getting started.

[quote="Franz Joseph"]Why not just use Mathematical Induction?[/quote]
Could someone please show a proof using mathematical induction?  :roll:

[quote="boxedexe"]Using sum of cubes, we have
$\\(n+1)^3+(n+2)^3+(n+3)^3=\left(\frac{(n+3)(n+4)}{2}\right)^2-\left(\frac{n(n+1)}{2}\right)^2\\=\left(\frac{(n+3)(n+4)-n(n+1)}{2}\right)\left(\frac{(n+3)(n+4)+n(n+1)}{2}\right)\hdots$
Sorry, the rest is easy, so I'll leave it...I must go work on some late assignments due to math.  :P

Masoud Zargar[/quote]
amm look ns but I don't know how to continue so u will need to continue  :)  tnx!
with induction it's easy but very ugly:
1$(x - 1)^3 + x^3 + (x + 1)^3=6x + 3 x^3$
2$x^3 + (x + 1)^3 + (x + 2)^3=9 + 15x + 9 x^2 + 3 x^3$
$2-1=(1?)9 + 15x + 9 x^2 + 3 x^3-(6x + 3x^3)=9+9x+9x^2$

[quote="jli"][/quote][quote="srulikbd"][quote="jli"][quote="Leao"]Prove that $(n+1)^3+(n+2)^3+(n+3)^3$ is divisible by $9$ for all positive integers $n$.[/quote]

[hide]by expansion we get $3x^3+18x^2+42x+36$

$3x^3+18x^2+42x+36\equiv 3x^3+42x \bmod 9\equiv x^3+14x\bmod 3\equiv x(x^2+14)\bmod 3 \equiv x(x^2-1)\bmod 3\equiv x(x+1)(x-1)\bmod 3$[/hide][/quote]
I didn't understand why does $x(x^2+14)\bmod 3 \equiv x(x^2-1)\bmod 3$
tnx!
[quote="ch1n353ch3s54a1l"][quote="Leao"]Prove that $(n+1)^3+(n+2)^3+(n+3)^3$ is divisible by $9$ for all positive integers $n$.[/quote]

[hide]Since the terms are consecutive, $n+1$, $n+2$, and $n+3$ are $\equiv {0,1,2}\mod{3}$. So then the cubes are $\equiv {0,1,8}\mod 9$ so then the sum is $\equiv {0}\mod {9}$? Does this work?[/hide][/quote]
amazing!  :)[/quote][quote="jli"]

$x^2+14 \bmod 3 \equiv x^2-1 \bmod 3$[/quote]
tnx now I understand  :)

My solution

[hide]
If we work in mod 3, then for $n \in \mathbb{Z}^{+}$ we can say that $(n+1),(n+2),(n+3)$ are homologous.  Thus, we can show that the statement is true for any $n$ and it will be true for all $n$.

Let $n = 1$.  $2^3 + 3^3 + 4^3 = 8 + 27 + 64 = 99 = 9*11$, and we're done.

Basically, if there is homology, the variables are interchangeable.  This works in this problem because as long as $n$ is an integer, then the quantities are interchangeable in mod 3.  We show it for one case, and then we have shown it for all cases - similar to induction, but homology is kinda hard to prove.
[/hide]


algebra
polynomial
integration
trigonometry
calculus
calculus computations

Show that if n is a natural number and x > 1 then

$ \frac{1}{\pi }\int\limits_0^\pi  {\left( {x \plus{} \sqrt {x^2  \minus{} 1} \cos \varphi } \right)^n } d\varphi  \equal{} \frac{1}{\pi }\int\limits_0^\pi  {\frac{{d\varphi }}{{\left( {x \minus{} \sqrt {x^2  \minus{} 1} \cos \varphi } \right)^n }}}$

Thought about this one for some weeks but can't figure it out.

Something is certainly wrong. The integrand on the right is [i]pointwise[/i] less than the integrand on the left.

Hmm. I actually realized that the moment I posted this. I suppose there is an error in the book (wouldn't be the first). Is a transformation of that sort known for Legendre polynomials or is that completely wrong?


AMC
AIME
graph theory
combinatorics proposed
combinatorics

A subset $ S$ of $ {(1,2,....,2008)}$ is given such that no two elements from $ S$ differ for $ 4$ or $ 7$. Find the maximal number of elements in $ S$.
[hide]I have generalized it.[/hide]

Does anybody have a solution for this problem?

In my solution, maximal number of elements in S is 732. :?:

EDIT: This is wrong.  $ G$ does have odd cycles...

[hide]I think it can be as large as $ 1004$.  Let $ G$ be a graph with the set of vertices $ V: \equal{} \{1,2,\dots,2008\}$ such that there is an edge joining between two vertices in $ G$ iff their difference is $ 4$ or $ 7$.  You should be able to show that there are no odd cycles in $ G$ (I'm not mistaken, right?).  Hence, $ G$ is a bipartite graph; let say $ \left(V_1,V_2\right)$  is a bipartition of $ G$.  Now, either $ \left|V_1\right| \geq 1004$ or $ \left|V_2\right| \geq 1004$.  Choose $ S$ to be the bigger of the $ V_i$'s, and thus, $ 1004$ can be achieved.

It is left to be shown whether this is the maximum.  (Sorry, I have no idea yet.)

Hope this would work.  Define $ A_i : \equal{} \{8i \minus{} 7,8i \minus{} 3\}$, $ B_i: \equal{} \{8i \minus{} 6,8i \minus{} 2\}$, $ C_i: \equal{} \{8i \minus{} 5,8i \minus{} 1\}$, and $ D_i: \equal{} \{8i \minus{} 4,8i\}$, where $ i \equal{} 1,2,\dots,251$.  Thus, $ V$ can be partitioned in to $ A_i$'s, $ B_i$'s, $ C_i$'s, and $ D_i$'s, and there are $ 1004$ of them.  If $ S$ is bigger than $ 1004$, then obviously, two elements of $ S$ must lie in the same $ A_i$, $ B_i$, $ C_i$,or $ D_i$, and we are done.[/hide]

Here is the original AIME problem solution:

[url]http://www.artofproblemsolving.com/Wiki/index.php/1989_AIME_Problems/Problem_13[/url]


FTW
MATHCOUNTS
trigonometry
geometry
trig identities
Law of Sines

In triangle ABC, the measure of A=30 and the measure of C=105. If AC=8 units, what is the length of segment BC? Express your answer in simplest radical form.

How do you solve this prob??

or you can drop a perpendicular down from C to AB :)

[quote="JRav"][hide]Hint: You can easily find the third angle since you know the other two.  Then just use the law of sines.[/hide][/quote]

Isn't FTW supposed to be MATHCOUNTS level? If it isn't that's why I keep losing... I never use methods above MATHCOUNTS in these types of problems.

The MC level solution is a lot easier to see with the diagram.

[hide]The law of sines can be computed without a calculator for this problem: \[\frac{\sin(30)}{x}=\frac{\sin(45)}{8}\] The $\sin(45)$ is the third angle. So after some cross multiplication, the answer becomes \[\frac{8(\sin(30))}{\sin(45)}\]. Which is just $4\sqrt2$.[/hide]


inequalities
algebra
polynomial
inequalities proposed

[quote="bodom"]this is true for any 3 positive numbers,not only for a,b,c=sides of a triangle ;)[/quote]
I think you are mistake.
Try $a=7,b=1,c=2.$ ;)

of course i am.i was tired and i thought i found a solution for a,b,c postive reals but i forgot to look more carrefully.what a stupid i am.thanks arqady for pointing this out

[quote="Sasha"]One can check that $\sum a^3bc(\frac bc-1)=\sum a^2(b-c)^2(a+c-b)\ge0$.  :P[/quote]

Please, tell me, what's the motivation behind this beautiful solution?

Another hint for another solution .
Make the Ravi subtitution and then the munipulation .You will have to prove something which is a simple AM-GM . :)

A beautiful solution that I have read in "Old and new inequalities . Vol 2"
First notice that we need to prove

$\sum \frac{a^{2}b}{c} \geq \sum a^2$
By Cauchy- Schwartz $(\sum \frac{a^{2}b}{c})(\sum a^{2}bc) \geq (\sum a^{2}b) ^2$
Thus it suffices to prove
$(\sum a^{2}b) ^2 \geq abc(\sum a)(\sum a^2)$
For this consider the trinomial
$f(x) = (\sum a^2)x^2 - (\sum a^{2}b)x+abc( a+b+c)$
Assume wlog that $(a-b)(a-c)\leq 0$
Notice that $f(a)=a(a-b)(a-c)(a-b+c)\leq 0$ (we can see this by considering the polynomial$ f $in variable $a$ with roots $0,b-c,b,c$
Since $x->\infty$ gives us a positive value for $f(x)$ we deduce that It has a real root and that its discriminant is non-negative .
..........................
Nice solution , though I think Ravi substitution is more obvious ;)

[quote=sqing]Let $a,b,c$ be sides of the triangle. Prove that
\[ a^3\left(\frac{b^2}{c^2}-1\right)+b^3\left(\frac{c^2}{a^2}-1\right)+c^3\left(\frac{a^2}{b^2}-1\right)\geq 0 . \][/quote]

i know you'll post it here ,i have proved it

[quote=prowler]Let $a,b,c$ be sides of the triangle. Prove that
\[ a^2\left(\frac{b}{c}-1\right)+b^2\left(\frac{c}{a}-1\right)+c^2\left(\frac{a}{b}-1\right)\geq 0 . \][/quote]

[img]http://s16.sinaimg.cn/middle/006ptkjAzy74SfXQw9h6f&690[/img]

[quote=prowler]Let $a,b,c$ be sides of the triangle. Prove that
\[ F = a^2\left(\frac{b}{c}-1\right)+b^2\left(\frac{c}{a}-1\right)+c^2\left(\frac{a}{b}-1\right)\geq 0 . \][/quote]
Replace $(a,b,c) = (x+y,y+z,z+x)$ then
\[F = \frac{\displaystyle \sum x(x^2+xy+yz)(x-y)^2+\sum z(x+y-z)^2(x-y)^2}{(x+y)(y+z)(z+x)} \geqslant 0.\]

[quote=sqing][quote=prowler]Let $a,b,c$ be sides of the triangle. Prove that
\[ a^2\left(\frac{b}{c}-1\right)+b^2\left(\frac{c}{a}-1\right)+c^2\left(\frac{a}{b}-1\right)\geq 0 . \][/quote]
Let $a,b,c$ be sides of the triangle. [url=http://www.artofproblemsolving.com/community/c6h1306310p6981977]Prove that[/url]
\[a^2\left(\frac{b}{c}-1\right)+b^2\left(\frac{c}{a}-1\right)+c^2\left(\frac{a}{b}-1\right)\geq \frac{(a-b)^2(b-c)^2(c-a)^2}{abc(a+b+c)}.\][/quote]

still not difficult to prove

[quote=sqing][quote=prowler]Let $a,b,c$ be sides of the triangle. Prove that
\[ a^2\left(\frac{b}{c}-1\right)+b^2\left(\frac{c}{a}-1\right)+c^2\left(\frac{a}{b}-1\right)\geq 0 . \][/quote]
Let $a,b,c$ be sides of the triangle. [url=http://www.artofproblemsolving.com/community/c6h1306310p6981977]Prove that[/url]
\[a^2\left(\frac{b}{c}-1\right)+b^2\left(\frac{c}{a}-1\right)+c^2\left(\frac{a}{b}-1\right)\geq \frac{(a-b)^2(b-c)^2(c-a)^2}{abc(a+b+c)}.\][/quote]
Let $a,b,c$ be sides of the triangle. [url=https://artofproblemsolving.com/community/c6h1064117p8845467]Prove that [/url]
$$a^2\left(\frac{b}{c}-1\right)+b^2\left(\frac{c}{a}-1\right)+c^2\left(\frac{a}{b}-1\right)\geq\frac{a^2+b^2+c^2-ab-bc-ca}{a^2+b^2+c^2}. $$


Let $p, q$ and $r$ be the roots of the equation $x^3-14x^2+26x-4=0$. Compute the value of the expression $\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)\left(1-\frac{1}{r}\right)$.

A stick is broken at two random points into three peices. What is the probability that the three peices can be used to form a triangle?

Problem 2: Let the stick be 1 foot long. If we make an imaginary line that divides the stick in half, we can see that if the two points selected are on the same half, it is impossible to make a triangle because the length of the large piece is greater than the sum of the lengths of the two short pieces. So there is a 1/2 probability that the two points lie on the same half, and 1/2 that they are on different halves, enabling a triangle.
But there is also a 1/2 probability that if they are on different halves, the points are 6 or more inches apart. This would also make a triangle impossible. So there is a 1/4 chance of being able to make a triangle. Hope thats right.


logarithms
inequalities
algebra
polynomial
algebra unsolved

[quote="ikap"]Solve the following: \[2003^{2-x^{2}}+2003^{2004x-4\sqrt x-2000}=2004\] [/quote]
It is obviosly $x=1$ is solution. Equation give $x^{2}>2-r, \ r=\ln_{2003}(2004)$ and $2004x-4\sqrt x-2000<r$. Therefore $0.999<x<1.001$. In this interval $f(x)=2003^{2-x^{2}}+2003^{2004x-4\sqrt x-2000}$ decrease. Therefore we have not another solution.

Yes. I check only $f'(1)<0$ and interval is small.

Rewrite it thus:

$2003\cdot 2003^{(1-\sqrt{x})(1+\sqrt{x})(1+x)}+2003^{(\sqrt{x}-1)(2004\sqrt{x}+2000)}=2004$

Put

$u=2003^{1-\sqrt{x}}$

to get

$2003(u^{1+\sqrt{x}+x+x\sqrt{x}}-1)+(u^{-(2004\sqrt{x}+2000)}-1)=0\qquad(*)$

Assume $u>1$. Since $1+\sqrt{x}+x+x\sqrt{x}\geqslant 1$ and $-(2004\sqrt{x}+2000)<0$, we can apply generalized Bernoulli's inequality to get

\begin{eqnarray*}&& 2003(u^{1+\sqrt{x}+x+x\sqrt{x}}-1)+(u^{-(2004\sqrt{x}+2000)}-1)\\ &>& 2003(1+\sqrt{x}+x+x\sqrt{x})(u-1)-(2004\sqrt{x}+2000)(u-1)\\ &=& (3-\sqrt{x}+2003x+2003x\sqrt{x})(u-1)\end{eqnarray*}

Polynomial $P(t) : = 2003t^{3}+2003t^{2}-t+3$ is always positive for $t\geqslant 0$, since it can be written as $2003t^{3}+2003\left(t-{1\over 4006}\right)^{2}+3-{1\over 8012}$. Therefore $(*)$ can't have solution if $u>1$.

Similar argument if $u<1$. Hence the only possibility is $u=1$, which indeed satisfies the equation and gives $x=1$.

[quote="Myth"]But for $f(x)=2003^{2-x^{2}}+2003^{2004x-4\sqrt x-2000}$ we have $f(0)>2004$, $f(2)>2004$, $f(1)=2004$ and it seems $f'(1)\neq 0$. So there is another root.[/quote]
I thought I clearly showed that thre is another root!

[quote="Farenhajt"]Then be so kind to point out the fault in my proof. Equally clearly.[/quote]You know I reviewed more than 100 topics during some last days. And surely, THIS problem doesn't deserve as much attentation as I have already spent on it.

Just forget this problem. Something is wrong with it.


function
real analysis
real analysis unsolved

Is there a $C^{\infty}$ real function whose Taylor series diverges in any point of the real axes? Is the set of these functions dense in the set of $C^{\infty}$ functions with the uniform norm?

How do you define the uniform norm for functions defined on the real line, which can thus be unbounded? :?

As for the initial question, I believe $e^{e^x}$ should work, or something in the same spirit :) (if I understand the question, that is: do we expand the Taylor series around a fixed point, such as $0$, and then want it to diverge for all real $x$, or do we want it to diverge for all real $x$, no matter around which point we expand it?).

The second question was a joke, of course! The answer to the second question is:there is $x_0$ such that  for  non-zero $x$ the series ${\sum_{k\geq 0}{\frac{ f^{(k)}(x_0)}{k!} x^k}}$ diverges.


YAY!!!! (^_^) is in the lead! I voted for you too!

aw darn, i lost the lead, stupid lakers....

LOL I just gave you the lead =D =D =D =D!!!


Let $ x,y,z$ be non-negative real numbers. Prove that
$ x\plus{}y\plus{}z\plus{}\sqrt{x^2\plus{}y^2\plus{}z^2}\geq\sqrt{x^2\plus{}y^2}\plus{}\sqrt{y^2\plus{}z^2}\plus{}\sqrt{z^2\plus{}x^2}$

[quote="Sudoku"]Let $ x,y,z$ be non-negative real numbers. Prove that
$ x \plus{} y \plus{} z \plus{} \sqrt {x^2 \plus{} y^2 \plus{} z^2}\geq\sqrt {x^2 \plus{} y^2} \plus{} \sqrt {y^2 \plus{} z^2} \plus{} \sqrt {z^2 \plus{} x^2}$[/quote]
Assume $ z\equal{}\min\{x,y,z\}$, then
\[ \sqrt{x^2\plus{}y^2\plus{}z^2} \ge \sqrt{x^2\plus{}y^2}\]
\[ \sqrt{y^2\plus{}z^2} \le \sqrt{y^2\plus{}yz} \le \sqrt{y^2\plus{}yz\plus{}\frac{z^2}{4}}\equal{}y\plus{}\frac{z}{2}\]
Similar,
\[ \sqrt{x^2\plus{}z^2} \le x\plus{}\frac{z}{2}\]
Adding up these inequalities, we get the result. :)
Equality holds if and only if $ xyz\equal{}0.$
:)

My solution:

The inequality is equivalent to

$ \sum_{cyc} {x \plus{} y \minus{} \sqrt {x^2 \plus{} y^2}} \geq \sqrt {x^2 \plus{} y^2 \plus{} z^2} \minus{} (x \plus{} y \plus{} z)$

$ \Leftrightarrow \sum_{cyc} \frac {2xy}{\sqrt {x^2 \plus{} y^2} \plus{} x \plus{} y} \geq \frac {2xy \plus{} 2yz \plus{} 2zx}{\sqrt {x^2 \plus{} y^2 \plus{} z^2} \plus{} x \plus{} y \plus{} z}$

$ \Leftrightarrow \sum_{cyc} 2xy ( \frac {1}{\sqrt {x^2 \plus{} y^2} \plus{} x \plus{} y} \minus{} \frac {1}{\sqrt {x^2 \plus{} y^2 \plus{} z^2} \plus{} x \plus{} y \plus{} z} ) \geq 0$

The last inequality is obviously true.

[quote="mathangel"]
The inequality is equivalent to

$ \sum_{cyc} {x \plus{} y \minus{} \sqrt {x^2 \plus{} y^2}} \geq \sqrt {x^2 \plus{} y^2 \plus{} z^2} \minus{} (x \plus{} y \plus{} z)$

[/quote]
I think the RHS should be $ (x\plus{}y\plus{}z)\minus{}\sqrt{x^2\plus{}y^2\plus{}z^2}$.

anyway,nice solution. :)

Uh oh, thanks BaBaK for pointing out my mistake.

Nice Solution, Can
Your solution is same as mine  :)

See here the similar one:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164798


In a right triangle with points L, R, A.
AL=13
RA=13.84
RL=4.75
Angle L is 90 degrees(this is a right triangle)
So find $ \ cos^{\minus{}1}$ of measure of angle A.
Don't have pic sorry. I dunno how.
[hide="Question"]
What is the code for mod?
[/hide]

No. If you are looking for the angles, then you find the $ \cos^{ \minus{} 1} (\cos A)$, in the case when you wanna find A. Also, since you have all the sides of the triangle, you can use any trig identidy (sine, cosine,tangent,etc.).

*Note: You need a calculater (TI 83 or better).

Use a calculator, not a series ... cos A = adjacent/hypotenuse = 13/13.84 = .9393, so A = arccos(.9393) = 20.06 degrees.

hmm.. yes i noticed a flaw in the problem i calculated the other unknown angle and got something like 77.
hmmm 90+20+77=187.

$ \sqrt{4.75^2\plus{}13^2}\approx13.8406105357\neq13.84$ so the stated triangle isn't exactly right.

ya this came from Edhelper.com so i gess they rounded it.

Do not use these types of websites for learning anything. They're full of typos and consumption.

And annoying worksheets that all you need to is plug in a value or equations where you need to do the same thing over and over and over again.

Hmm... thx guys but my mom will still make me do them.
Anyone now how to do mod in latex?

As in (example) \[ 227\equiv11\pmod{18}?\]
[code]227\equiv11\pmod{18}[/code]

Aww that sucks.  :(

11 \equiv 2 \pmod 9 gives us

$ 11 \equiv 2 \pmod 9$

Edit: Beaten :P


Bob starts to climb a mountain at 9 AM one day. He walks up the mountain at a rate of 6 kilometers per hour. Once at the top, he rested for 1.5 hours, and started to come down at a rate of 7.5 kilometers per hour. When he got to the bottom, it was 1:30 PM. How many kilometers did he walk that day climbing up and down the mountain?

[hide="Solution"]
There are $ 4.5$ hours between 9:00 AM and  1:30 PM. Only $ 4.5\minus{}1.5\equal{}3$ hours were spent actually climbing the mountain, though.  Let $ t_{up}$ be the time it took to climb up the mountain and $ t_{down}$ be the time it took to climb down the mountain.

Then,
$ t_{down}\equal{}3\minus{}t_{up}$
$ 6t_{up}\equal{}7.5(3\minus{}t_{up})$
$ t_{up}\equal{}\frac{5}{3}$
$ t_{down}\equal{}3\minus{}\frac{5}{3}\equal{}\frac{4}{3}$

So,
$ d_{total}\equal{}r_{up}t_{up}\plus{}r_{down}t_{down}\equal{}2*r_{up}t_{up}\equal{}2*6*\frac{5}{3}\equal{}20$ km.
[/hide]


Okay so lets says i have to divide 10 cookies among 4 people. Isn't that like saying there are 11 spaces to choose 3 dividers to put in? Apparently my reasoning is wrong? Can someone explain why?  :( thanks

Nope, there are 13 objects - 10 cookies and 3 dividers.  Where did you get eleven from?


AMC
USA(J)MO
USAMO
Olimpiada de matematicas

[color=blue]
[b]PROB 01.[/b]
Sean $n$ un número natural y $p$ un número primo tales que $p|n^{6}-1$ y $n|p-1$.
Pruebe que $4p-3$ es un cuadrado perfecto o $p=n+1$.[/color]

vemos  que $p = nk+1$ ahora tambien que $\frac{(n-1)(n+1)(n^{2}+n+1)(n^{2}-n+1)}{p}$ es entero entonces $p=n+1$ para $k=1$ , ahora veamos para $k\geq 2$ ,
vemos q $p$ o bien divide a $n^{2}+n+1$ o a $n^{2}-n+1$ ya que si este dividira a los dos entonces $nk+1$ dividiria a $2n$ lo cual es imposible , ahora $\frac{(n^{2}+n+1)-(nk+1)}{nk+1}$ es entero entonces $\frac{n(n-k+1)}{nk+1}$ es entero ahora como $nk+1$ es primo este divide a $(n+1-k)$ pero $1<k\leq n+1$ ya q $\frac{(n^{2}+n+1)}{nk+1}\geq 1$ entonces como $p$ divide a $n-k+1$ entonces supongamos q $\frac{(n-k+1)}{nk+1}$ sea mayor q 1 , luego de ello se llega a una contradiccion entonces ello deberia ser cero por ello $k=n+1$ , para el caso en q divida a $n^{2}-n+1$ es analogo en el cual se obtendra q $k=n-1$ , ahora como $p=n(n\pm1\))$$+1$ luego se ve q $4p-3$  es un cuadrado perfecto
QED

[color=blue]Prob 02

hallar los $n$ enteros positivos, talque
$(5^{n}-3^{n})/n$ sea entero.[/color]

[color=blue]He demostrado ya varias cosas. Más tarde podría enviar la justificación, si lo desean. Un ejemplo es que el único primo que cumple la condición es el $2$.

Mi pregunta es: ¿Qué puedo aplicar teniendo que $5^{\varphi(n)}\equiv3^{\varphi(n)}\equiv1(modn)$ y $5^{n}\equiv3^{n}(modn)$? Solicito la asistencia de ustedes.

Saludos, y muchas gracias.[/color]

Bueno, evidentemente ni $3$, ni $5$ dividen a $n$.
Hasta ahora sólo he podido demostrar que $n$ no es impar (quitando el $1$, claro).
Usando el hecho de que, fijado $n$, si $a$ es el menor entero positivo que satisface la propiedad
$n|5^{a}-3^{a}$ $\dots$ $(\alpha)$
Entonces cualquier otro entero $b$ que satisface la propiedad $(\alpha)$, es divisible por $a$.
En cuanto a los pares todavia no he podido probar todo.
Aunque me han salido algunas soluciones medias raras.
Por ejemplo:
Es fácil ver que $n$ puede ser cualquier potencia de $2$.
Si $p|5^{2^{m}}-3^{2^{m}}$, con $p$ primo, entonces $n=2^{m}p^{k}$ cumple la condición pedida, para cualquier $k$ entero positivo.

R.G.A.M., el hecho que mencione arriba, implica que si $b$ y $c$ satisfacen la propiedad $(\alpha)$, entonces tambien la satisface $(b,c)$.

[color=blue]Al parecer, falta algo a su explicación. ¿Podría revisarla nuevamente, por favor? Entre una de mis dudas está el uso que otorga a la variable $c$, que no la hallo en ningún otro lado.

Saludos.[/color]

[color=blue]Prob. 03
Mostrar que no existen enteros positivos $m$ y $k$ tal que:
$k!+48 = 48(k+1)^{m}$[/color]

Emvien sus soluciones..

Carlos

[quote="carlosbr"]Prob.
Mostrar que no existen enteros positivos $m$ y $k$ tal que:
$k!+48 = 48(k+1)^{m}$
[/quote]
SOLUCION.
-Analisando $mod 48$, tenemos que : $k\geq 6$.
I CASO.-
*k+1, que sea un numero compuesto, entonces notamos que $k+1>6$.
Si $n=(a.b)$, y $n>5$, donde: $a,b>1$,notamos que:
si $a=2$ y $b>2$ se cumple que:
$n$ divide $(n-1)!$,
dem:
1.-si $a=b$, entonces $a,b>2$.
$(n-1)!=1x2x...xax(a+1)x..x(2a)x...xn$, donde $n=a^{2}\Rightarrow n>2a$, por lo tanto $n$ divide a $(n-1)!$.
2.-si $a,b$ son distintos es claro que sin perdida de generalidad le damos un orden: $a<b<n-1$entonces el producto de $a$ y $b$ estara dentro de $(n-1)!$. $Q.E.D.$
Usando $mod(k+1)$, tenemos que:
$(k+1)$ divide a $48$, entonces los posibles valores de $k+1$, solo pueden ser los divisores de $48$, de ahi que los posibles valores de $k$, son:${1,2,3,5,7,11,15,23,47}$,como $k\geq 6$, se descartan los valores, ${1,2,3,5}$,
si: $k=7$.
$7!+48=48(8)^{m}\to 48(106)=3(2)^{3m+4}, (= >< =)$.
si: $k=11$.
$11!+48=48(12)^{m}$->analisemos $mod9$, en el lado izquierdo tenemos que es congruente a $3$ en $mod9$ y en el derecho a $0$ en $mod9(= >< =)$
si: $k=15$.
$15!+48=48(16)^{m}$, analisando $mod2^{8}$, tenemos que el lado izquierdo es congruente a 48 y el derecho a cero, $(= >< =)$.
si: $k=23$.
$23!+48=48(24)^{m}$, entonces analisando $mod2^{7}$, tenemos que en el lado izquierdo es congruente a 48 y en el derecho a cero $(= >< =)$
si: $k=47$.
$47!+48=48^{m+1}$, entonces analisando $mod2^{6}$, en el lado izquierdo es congruente a 48 en el derecho a cero $(= >< =)$.
por lo tanto si $k+1$, es un numero compuesto entonces no hay solucion.
II CASO.
**$k+1=a$, si $a$ es primo.
 reemplesemos.
$(a-1)!+48=48(a^{m})$, analisando $mod$ $a$, y utilisando el teorema de Wilson.
"si $p$ es primo entonces $(p-1)!$ congruente a $-1$ en modulo $p$."
entonces, $a$ divide a 47, como $a$ es primo, $a=47$, reemplazando.
$46!+48=48(47^{m}).$
entonces vastaria con demostrar que no existe el $"m"$, en esta ultima ecuacion.

[quote="jonfv"]Para la última ecuación de aev5peru :
Se puede probar por inducción que para todo entero positivo $n$ vale
$47^{2^{n}}\cong 1+2^{n+4}mod (2^{n+5})$
En particular; si $n$ es entero positivo, el menor entero positivo $a$ que satisface
$47^{a}\cong 1 mod 2^{n+4}$
es $a=2^{n}$.
Volviendo a la ecuación, si existiera tal $m$, tendriamos que $46!=48(47^{m}-1)$ es múltiplo de $2^{1}4$ y, luego, que
$47^{m}\cong 1 mod 2^{1}0$
Así, por lo visto anteriormente, deberiamos tener que $m$ es múltiplo de $2^{6}$ y, luego, que $m\geq 2^{6}$.
Finalmente, tendriamos
$47^{4}6>46!=48(47^{m}-1)>47^{6}4$
lo cual es absurdo.[/quote]
BUENA FORMA jonfv, PERO AHORA TERMINO MI SOLUCION.
$46!=48(47^{m}-1)$, notamos que en el lado izquierdo es congruente a $0$ en $mod 48^{3}$, entonces $48^{m}-1$, debe ser divisible por $48^{2}$. Analisando $mod 48$, $m$ debera ser par.
entonces sea: $m=b(2^{a})$, donde $b$ es impar.
$= > 47^{m}-1=(47^{b}+1)(47^{b}-1)(47^{2b}+1)...(47^{2^{a-1}b}+1)$, analisando $mod48$, notamos que el unico factor que es multiplo de $48$ es : $(47^{b}+1)$, entonces este factor debe ser multiplo de $48^{2}$.
 $(47^{b}+1)/(47+1)=47^{b-1}-47^{b-2}+..-47+1$,analisando $mod48$
  tenemos que $b$ es multiplo de $48$. pero $b$ por hipotesis es impar.
$(= > < = )$.
por lo tanto no existe $n,k$. que cumplen la condicion.

[quote="carlosbr"][color=blue]Prob. 04

Encontrar todos los naturales $n$, tal que existan los primos relativos $x$ e $y$, y un entero $k>1$ que satisfagan la ecuacion:

$3^{n}= x^{k}+y^{k}$.[/color]

Carlos Bravo
Lima -PERU[/quote]

Respuesta:
Los unicos valores de n son 0 y 2 (contando a 0 como natural)

Solucion:
Notemos que $(x,3)=(y,3)=1$, esto es inmediato pues son primos relativos y que $3$ divida a alguno implicaría que los divide a los dos.

Primero supongamos que $x>y$. Veamos pues que siempre tenemos que $x>|y| \ge 0$. Si $y=0$ tomamos $x=1$ y $k=2$ y tenemos que  $n=0$ sirve. Como $(0,k)=|k|$ tenemos que soluciones con $y=0$ no puede haber más pues el unico otro primo relativo con $0$ es $-1$ que nos da la misma solución. Ahora mirando módulo $3$ nos damos cuenta que $k$ tiene que ser impar en los otros casos, pues $(3,x)=1 \implies x^{2}\equiv 1 (\mod 3)$. Ahora como $3^{n}=x^{k}+y^{k}> 0$ y como $a^{k}$ y $a$ tienen el mismo signo tenemos que $a >0$ y además que $a > |b|$.

Además como $k$ es impar tenemos que $x+y | x^{k}+y^{k}$, por lo que tenemos que $x+y = 3^{n-t}$ para algún $t < n$ entero. Ahora usemos el lema Lifting the exponent con $p=3$. Tenemos que:
$\\ 3^{n-t}\parallel x-(-y) \\ 3^{s}\parallel n \\ \implies 3^{n-t+s}\parallel x^{n}-(-y)^{n}$
Entonces para que sea $3^{n}$ la máxima potencia de $3$ que divide $x^{k}-(-y)^{k}$ debe pasar que $3^{t}=k$ para que además sea el menor exponente que cumple $3^{n}\parallel x^{k}+y^{k}$

Entonces ya tenemos el problema reducido a $3^{n}= x^{3^{t}}+y^{3^{t}}$ para $t<n$.

Separaremos el problema en 2 casos.
$y>0$ y $y<0$.

En el primer caso probaremos primero que para $t \ge 2$ no hay soluciones. Para eso basta probar que $m= \frac{x^{3^{t}}+y^{3^{t}}}{x+y}> 3^{t}$ pues esta fracción debería ser igual que $3^{t}$ por el lifting. Como $x>y\ge 1$ y $(x-y) \ge 1$ usamos que:
$\\ m = \sum_{i=0}^{3^{t}-1}(-1)^{i}x^{3^{t}-1-i}y^{i}\\ = \sum_{i=0}^{\frac{3^{t}-1}{2}}x^{3^{t}-1-2i}y^{2i}-x^{3^{t}-1-2i-1}y^{2i+1}\\ = \sum_{i=0}^{\frac{3^{t}-1}{2}}(x-y)x^{3^{t}-1-2i-1}y^{2i}\\ \ge \sum_{i=0}^{\frac{3^{t}-1}{2}}(2-1){2^{3^{t}-1-2i-1}1^{2i}\\ = \frac{2^{3^{t}}+1}{3}}$

Por lo que es suficiente probar que:
$\\ \frac{2^{3^{t}}+1}{3}> 3^{t}\iff \\ 2^{3^{t}}+1 > 3^{t+1}$
Que se puede probar fácilmente por inducción de la siguiente forma:
Caso base:
$t=2$
$2^{9}+1 = 513 > 27 = 3^{3}$ y eso es cierto.

Paso inductivo:
Supongamos que para $t=s$ se cumple que:
$2^{3^{s}}+1 > 3^{s+1}$
Probaremos que para $t=s+1$ se cumple que:
$2^{3^{s+1}}+1 > 3^{s+2}$
Para eso tenemos que:
$3^{s+2}< 3\times2^{3^{s}}+3$  por lo que es suficiente probar que:
$3\times2^{3^{s}}+3 < 2^{3^{s+1}}+1$ que es claro pues:
$3\times2^{3^{s}}+3 < 2^{3^{s}+2}< 2^{3^{s+1}}< 2^{3^{s+1}}+1$ pues $s \ge 2$.

Para terminar este caso solo nos falta ver el caso en que:
$3^{n}= x^{3}+y^{3}= (x+y)(x^{2}-xy+y^{2})$
Por lo del lifting tenemos que:
$3 = x^{2}-xy+y^{2}$ y por otro lado tenemos que:
$x(x-y)+y^{2}\ge 2(2-1)+1 = 3$
Entonces $x=2$ y $y=1$ es una solución y tenemos que en este caso $n=2$ y no hay mas soluciones en este caso.

Ahora solo falta ver que pasa en el segundo caso. Cuando $y<0$. Para eso suponemos que $y>0$ y cambiamos la ecuación por:
$3^{n}= x^{k}-y^{k}= (x-y)(\sum_{i=0}^{3^{t}-1}x^{i}y^{3^{t}-1-i})$
Pero tenemos que por el lifting se tiene que cumplir que:
$\sum_{i=0}^{3^{t}-1}x^{i}y^{3^{t}-1-i}= 3^{t}$ que es imposible porque hay $3^{t}-1$ terminos a la derecha que estan sumando y son mayores o iguales que $2$ y el término restante es mayor o igual que $1$ pues $x>y \ge 1$

Con esto queda probado que los únicos valores posibles de $n$ son $0$ y $2$

QED

PD: Carlos podrias poner una¨solucion mas corta?

Nota: el lifting the exponent esta explicado en el Reto 2.

$n=k=2$, $a=3$, $b=4$ y $p=5$ satisfacen la ecuación, pero $n=2$ no es potencia de $p=5$.

[quote="jasamper88"]Bueno para seguir con problemas de la misma linea ahi va otro de la misma olimpiada que el anterior:

$a^{n}+b^{n}= p^{k}$ para $a$,$b$,$k$, $n$ enteros positivos y $p$ un primo impar. Pruebe que $n$ es una potencia de $p$.[/quote]
como dijo jonfv, el problema no te pedira probar que $k$ es una potencia de $n$ ¿?, podrias corregir el problema.

perdon... un error de copia.... hagan el problema que es parcido al anterior.. despues pongan otro

JAS

Sean $a$ y $b$ enteros positivos. Se define la secuencia $\{x_{i}\}_{i=1}^{\infty}$ como:
$\\ x_{0}= 0 , x_{1}= 1 \\ x_{2n}= ax_{2n-1}-x_{2n-2}\\ x_{2n+1}= bx_{2n}-x_{2n-1}$

Demostrar que para toda pareja de enteros no negativos $m,n$ con $m\ge 2$ se cumple que:
$x_{m}\cdot x_{m-1}| \prod_{i=1}^{m}x_{n+i}$

[quote="jasamper88"]$\{x_{i}\}_{i=1}^{\infty}$[/quote]
[color=blue]¿Cómo se lee esta nomenclatura? ¿Qué significa?[/color]

Significa la sucesión infinita $x_{1}, x_{2}, x_{3}, x_{4}, \ldots$.

$Tipe$

dados dos enteros positivos $n$ y $a$ se forma una lista de 2001 numeros como sigue : el primer numero es $a$ ; a partir del segundo , cada numero es el residuo que se obtiene al dividir el cuadrado del anterior entre $n$ . A los numeros de la lista se les ponen los signos $+$ y $-$ , alternadamente , empezando con $+$ . los numeros con signo , asi obtenidos, se suman y a esa suma se le llama suma final para $n$ y $a$ . ¿ para que enteros $n\geq 5$ existe alguna $a$ tal que $2\leq a\leq  \frac{n}{2}$ y la suma final para $n$ y $a$ es positiva ?

jeje hace rato me encontre por primera vez el post de las desigualdades y ahora me encuentro este... jeje, esta tuanis... propongo dos prob muy bonitos q aparecieron en tst's de estados unidos, 2002 y 2003 creo.. "...y dice:"

1. (TST EEUU 2002) Sea $p$ un primo mayor que 5. Para cada entero $x$ se define

$ f_{p}(x)=\sum_{k=1}^{p-1}\frac{1}{(px+k)^{2}}$.

Probar que para todos los enteros positivos $x,y$ el numerador de $f_{p}(x)-f_{p}(y)$, cuando la fracción esta simplificada ("written in lowest terms"), es divisible por $p^{3}$.

2. (TST EEUU 2003) Hallar todas las tripletas de primos $(p,q,r)$ tal que

$p|q^{r}+1, \ q|r^{p}+1, \ r|p^{q}+1.$

...ojala los disfruten tanto como yo  :lol:

Para ir "cerrando" problemas abiertos:

[quote="jasamper88"]Sean $ a$ y $ b$ enteros positivos. Se define la secuencia $ \{x_{i}\}_{i \equal{} 1}^{\infty}$ como:
$ \\
x_{0} \equal{} 0 , x_{1} \equal{} 1 \\
x_{2n} \equal{} ax_{2n \minus{} 1} \minus{} x_{2n \minus{} 2} \\
x_{2n \plus{} 1} \equal{} bx_{2n} \minus{} x_{2n \minus{} 1}$

Demostrar que para toda pareja de enteros no negativos $ m,n$ con $ m\ge 2$ se cumple que:
$ x_{m}\cdot x_{m \minus{} 1}| \prod_{i \equal{} 1}^{m}x_{n \plus{} i}$[/quote]

Para la solución de este problema solo es necesario ver que dos términos consecutios de la secuencia son primos relativos y que $ X_m$ divide a todos los términos de la forma $ X_{km}$. Lo primero es muy fácil de provar por contradicción. Para probar lo segundo basta definir la secuencia para los negativos. Como
$ \\
x_{2n} \equal{} ax_{2n \minus{} 1} \minus{} x_{2n \minus{} 2} \\
x_{2n \plus{} 1} \equal{} bx_{2n} \minus{} x_{2n \minus{} 1}$
Implica:
$ \\
x_{2n \minus{} 2} \equal{} ax_{2n \minus{} 1} \minus{} x_{2n } \\
x_{2n \minus{} 1} \equal{} bx_{2n} \minus{} x_{2n \plus{} 1}$
Tenemos que $ X_j \equal{} \minus{} X_{ \minus{} j}$ Entonces la secuencia, módulo $ X_m$, es como simétrica con respecto a caulquier $ X_{km}$ y así vemos que la secuencia tiene 0's modulo cualquier $ x_i$ cada i términos

[quote="campos"]
1. (TST EEUU 2002) Sea $ p$ un primo mayor que 5. Para cada entero $ x$ se define

$ f_{p}(x) \equal{} \sum_{k \equal{} 1}^{p \minus{} 1}\frac {1}{(px \plus{} k)^{2}}$.

Probar que para todos los enteros positivos $ x,y$ el numerador de $ f_{p}(x) \minus{} f_{p}(y)$, cuando la fracción esta simplificada ("written in lowest terms"), es divisible por $ p^{3}$.
[/quote]

Este ya se probó en otro post del foro pero aprovecho para poner mi solución (y para mostrar un teorema interesante):

[b]Teorema[/b] ([i]de Wolstenholme Revisited[/i]):
Sea $ p$ un primo y $ k$ un entero no negativo tal que $ k < p \minus{} 2$. Entonces el numerador de la fracción:
                                                   $ \sum_{1\leq i \leq p^n}^{(i,p) \equal{} 1}\frac {1}{i^{k}}$
Es divisible por $ p^n$ si k es par o por $ p^{n\plus{}1}$ si $ k$ es impar.

Solución:
$ \frac {1}{ (px \plus{} k)^{2} } \minus{} \frac {1}{(py \plus{} k)^{2}} \equal{} p(y \minus{} x) \frac {p(y \plus{} x) \plus{} 2k}{ (px \plus{} k)^{2}(py \plus{} k)^{2} }$
Mirando módulo $ p^2$ la parte de abajo:
$ \frac {1}{(px \plus{} k)^{2}(py \plus{} k)^{2}} \equiv \frac {1}{2k^3p(x \plus{} y) \plus{} k^4} \equiv \frac {1 \minus{} 2p(x \plus{} y)}{k^4} \ (\bmod \ p^2)$
Entonces módulo $ p^3$
$ p(y \minus{} x) \frac {p(y \plus{} x) \plus{} 2k}{ (px \plus{} k)^{2}(py \plus{} k)^{2} } \equiv p(y \minus{} x)\frac {(p(y \plus{} x) \plus{} 2k)(1 \minus{} 2p(x \plus{} y))}{ (k)^{4} }$
$ \equiv p(y \minus{} x)(1 \minus{} 2p(x \plus{} y)) [\frac {p(y \plus{} x)}{ (k)^{4} } \plus{} \frac {2}{ (k)^{3} }] \ (\bmod \ p^3)$

Entonces, factorizando adecuadamente y usando el teorema enunciado:
$ f_{p}(x) \minus{} f_{p}(y) \equiv p(y \minus{} x)(1 \minus{} 2p(x \plus{} y)) [p(y \plus{} x)\sum_{k \equal{} 1}^{p \minus{} 1}\frac {1}{ (k)^{4} } \plus{} 2\sum_{k \equal{} 1}^{p \minus{} 1}\frac {1}{ (k)^{3} }] \equiv 0 \ (\bmod \ p^3)$

Sean $ p\ge5$ un primo y $ r$ el número de formas en que se pueden poner $ p$ fichas en un tablero $ p\times p$ de manera que no todas las fichas estén en la misma fila (pero pueden estar todas en la misma columna). Muestre que $ r$ es divisible por $ p^5$.

r es el numero de formas de ubicar p fichas en p^2 posiciones excepto por p posiciones que son los cubrimientos de una fila entera entonces r es $ \frac {(p^2)(p^2 \minus{} 1)(....)(p^2 \minus{} p \plus{} 1)}{(p)(p \minus{} 1)(....)(1)} \minus{} p$, habria q demostrar que es sivisible por p^5, que es lo mismo q probar: $ \frac {(p^2 \minus{} 1)(p^2 \minus{} 2).....(p^2 \minus{} p \plus{} 1)}{(p \minus{} 1)!} \minus{} 1$, es multiplo de p^4, pero al pasar 1 al numerador de la fraccion, el numerador queda como:
$ (p^2 \minus{} 1)(....)(p^2 \minus{} p \plus{} 1) \minus{} (p \minus{} 1)!$ y el denominador es primo relativo con p.
escribiendo de otra forma el numerador: $ x^p \plus{} t_1 x^(p \minus{} 1) \plus{} ..... \plus{} t_(p \minus{} 1) x \plus{} (p \minus{} 1)! \minus{} (p \minus{} 1)!$, donde t_1, t_2,....,t_(p-1) son las sumas de los productos simetricos de 1 en 1, de 2 en 2, ..... de p-1 en p-1 y solo por abreviar puse x=p^2.
basta probar que $ t_(p \minus{} 1)p^2$ es congruente a 0 mod p^4, que es lo mismo que   t_(p-1)=0 mod p^2, t_(p-1) se puede escribir como $ (p \minus{} 1)!(1 \plus{} 1/2 \plus{} 1/3 \plus{} ... \plus{} 1/p \minus{} 1)$
tengo que probar que el numerador del segundo factor es multiplo de p^2, agrupandolo queda:
$ p(1/(1)(p \minus{} 1)) \plus{} 1/(2)(p \minus{} 2) \plus{} ...... \plus{} 1/(\frac {p \minus{} 1}{2})(\frac {p \plus{} 1}{2})$, sea $ i_j$ el inverso de j en mod p
es facil probar que  i_1,i_2,.....i_p-1 es un sistema reducido de residuos mod p, y ademas $ i_j \equal{} \minus{} i_(p \minus{} j)$, para terminar el problema entonces: $ 1/j(p \minus{} j) \equal{} i_j.i_(p \minus{} j)modp$, solo queda probar que
$ (i_1)^2 \plus{} (i_2)^2 \plus{} ....... \plus{} i_(p \minus{} 1)^2$ es congruente a 0 mod p, pero dicha expresion es la mitad de:
$ 1^2 \plus{} 2^2 \plus{} ..... \plus{} (p \minus{} 1)^2 \equal{} S_p$, esto ya es cosa teorica nomas. sea g raiz primitiva mod p, entonces g, 2g, 3g,......,(p-1)g es el sistema reducido de residuos mod p, entonces $ g^2S_p \equal{} S_p(modp)$, de donde p divide a S_p y ya esta.

6: Sean a>b>c>d>0 enteros tales que

[size=150]ac + bd = (b + d + a - c)(b + d - a + c)[/size]

probar que ab + cd NO es primo