This isn't really for a debate, but do you guys have any idea on how to find tutoring opportunities? To be specific, college student tutoring younger students? I've been looking in Google, and many of them require Bachelor's Degree.. So, any idea?

I can just tell you that here (germany) it wasn't any problem to get one, even without having a "vordiplom" (something you normally get around your 4th or 5th semester).
But I find it rather strange to do this at any place you don't stidy or work yourself. And in general, you can always ask, if you are very keen and qualified, they may make an exception.

When I was in college, I found them through the job listing section of my school's financial aid webpage.

My school also had a peer tutoring program. It was free for the tutees, but the university paid the tutors. There was always demand for help in calculus and other large courses, so you could work as much as you want. Even if there is not something exactly like this at your school, they must provide some services to students struggling in introductory courses, so you can get involved with that.

If you're interested in tutoring non-college students, I would suggest contacting high schools near your college. You can do things like:
1. Try to advertise through the school. This may or may not be difficult, some schools might not want to help you before you build some good will
2. Approach/get involved with a local math team/whatever you're interested in, and work from there
3. Talk to any of your peers that might be from the area, and see if they can help spread a good word

[u]Some steps, if you're starting from scratch:[/u]

1. Identify what kind of tutoring you're interested in

2a. Find out how students currently receive that kind of tutoring
3a. Figure out the best way to weasel your way into the system and intercept students that would otherwise take up the kind of tutoring from 2a

2b. Find out why some target students aren't receiving that kind of tutoring
3b. Adapt your tutoring so that these 2b students will be willing to take it

[quote="JBL"]I do think Bictor is right that a great many schools (especially large ones with big intro courses) make subsidized tutoring available through the school/math department.[/quote]

Yes, I'm aware of that. But, that's during a school year, not during summer. My institution has TAs as well, but I didn't apply for it because of my busy schedule during semesters.

TZF, that's a really good advice. How should I approach other schools though? Email? Phone (which would be REALLY random I think?)? Any idea?


function
logarithms
geometry
inequalities
number theory unsolved
number theory

Which is the minimum positive integer f(b) which cannot be expressed by the arithmetic mean of different nonnegative powers of b?

f(2)=13 from [url]http://www.mathlinks.ro/viewtopic.php?t=177161[/url]

Is there any ideas?

Should I post my solution?

Claim: $ f(3)\equal{} 4$ and $ \forall b>3: f(b)\equal{}2$

Proof: Trivially, we have $ f(b)>1$, since $ 1\equal{}b^0$ holds. Now, consider the function $ h(s,t)\equal{}s^t\minus{}1\minus{}2t(s\minus{}1)$ for $ s\geq 3,t\geq 2$. We observe $ h(s,2)\equal{}(s\minus{}2)^2\minus{}1\geq 0$ and $ \frac{\partial h(s,t)}{\partial t}\equal{}s^t\ln s\minus{}2(s\minus{}1)>(s\minus{}1)^2\plus{}1>0$, i.e. $ h$ is non-negative for the chosen area and $ h(s,t)\equal{}0\Longleftrightarrow s\equal{}3,t\equal{}2$.

Now assume $ f(b)>2$, then $ 2$ can be represented as the arithmetic mean of $ n$ different powers of $ b$. Because of $ b>2$ we must have $ n\geq 2$ and we get the inequality:

$ 2\geq \frac{1}{n}\,\sum_{i\equal{}0}^{n\minus{}1}b^i\equal{}\frac{b^n\minus{}1}{n(b\minus{}1)}\equal{}\frac{h(b,n)\plus{}2n(b\minus{}1)}{n(b\minus{}1)}\equal{}2\plus{}\frac{h(b,n)}{n(b\minus{}1)}$. This is only possible for $ b\equal{}3,n\equal{}2$ proving the second part of the claim.

From now on we can assume $ b\equal{}3$. Because of $ 2\equal{}\frac{3^0\plus{}3^1}{2}\,,\,3\equal{}\frac{3^1}{1}$ we must have $ f(3)>3$. Now assume $ f(3)>4$, then similar to the earlier reasoning we get $ n\geq 2$ and $ 4\geq \frac{3^n\minus{}1}{2n}\Longrightarrow 3^n\minus{}1\minus{}8n\leq 0\Longrightarrow n\equal{}2$. But the representation $ 4\equal{}\frac{3^x\plus{}3^y}{2}$ implies $ x,y< 2$, which can be checked by hand that this cannot be solved.


Support
rotation
trigonometry
geometry
congruent triangles
trig identities
Law of Cosines

Can you figure out what is wrong with my bogus proof.

Consider triangle $ABC$ with $AB=AC$. Draw an abitrary line through $A$ intersectiong $BC$ at $D$. We have $AB=AC$, $\measuredangle ABC=\measuredangle ACB$, $AD=DA$. Implying $\measuredangle BAD=\measuredangle DAC$. :rotfl:  :ninja:  :huh:

for BAD to equal DAC, d would need to be on the midpoint of BC. You stated earlier that AD was an arbitrary line, meaning it could intersect anywhere

I think $AD$ must be $\perp$ to $\overline{BC}.$

[hide="why"] Because $\triangle ABD\cong\triangle ADC$[/hide]

there was not enough information in your proof to support that statement

You're using SSA (side-side-angle) to "prove" a congruence, but you actually can't use SSA for proving congruence  :wink: , because given the two sides and an angle of a triangle (in that order), the triangle could actually be one of two possible triangles. The two triangles $ABD$ and $ADC$ actually aren't congruent.

you do have SAS but the angle is not the included angle if that's what you're trying to get at

One of the triangle congruency criteria, known as $SSA$ (Side-Side-Angle), states that two triangles are congruent if two pairs of their sides and the angle opposite the LONGER of them are equal. But here you have an angle opposite the shorter side, hence the triangles aren't congruent.

What I don't understand is why you think you have a proof with an SSA. The only ways that you can have congruent triangles are with SSS, SAS, and the scenario which Farenhajt mentioned. Explain where you are getting your facts from.

[quote="kstan013"]The only ways... are with SSS, SAS, and the scenario which Farenhajt mentioned.[/quote]

Not so. There's also ASA.

[quote="Phelpedo"]Side-side-angle does NOT mean congruent triangles. Why not? Draw the two sides that include the angle you have. You know the length of one of them but not the other, so it is represented by a an infinite line. Now the other side is attached to the segment that you do know. This side can be rotated to intersect the line TWO times (use law of cosines to find the possible lengths).[/quote]

That's the case only if the angle is opposite the shorter side. If it's opposite the longer side, you always have exactly one point of intersection.

To sum up:

1. SSS holds always
2. SAS holds always
3. ASA holds always
4. SSA holds only if the angle is opposite the longer side.

yes, true. I forgot about anything along the lines of 2 similar triangles with any sides congruent

[quote="piperson"][quote="nebula42"]You're using SSA (side-side-angle) to [b]"prove" a congruence[/b], but you actually can't use SSA for [b]proving congruence  [/b]:wink: , because given the two sides and an angle of a triangle (in that order), the triangle could actually be one of two possible triangles. The two triangles $ABD$ and $ADC$ actually aren't congruent.[/quote]

I dont understand what you are getting to[/quote]
What I mean is that SSA doesn't necessarily imply congruence, which is why his "proof" is incorrect.
(Notice the quotation marks in "prove"  :wink: )


logarithms MATHCOUNTS function algebra domain

What real number $ x$ satisfies the equation $ (3^x)(4^x)\equal{}7^x$?

12^x=7^x which clearly only works when x=0

To be completely rigorous, we take the log of 7 to both sides:
$ \log_7{12}*x\equal{}x$
$ (\log_7{12}\minus{}1)*x\equal{}0$
$ \log_7{12}\minus{}1\neq0$
so, x=0

Of course, logs aren't mathcounts level.

I suppose you just look for the obvious answer, 0, like mew did.

Plotting the graphs, it becomes even more obvious.

hello, you can write $ \left(\frac{12}{7}\right)^x\equal{}1$, from here we get $ x\equal{}0$.
Sonnhard.

$ \left(\frac {12}{7}\right)^x$ is bijective function mapping all reals to all positive reals. Hence there exists only one solution for $ \left(\frac {12}{7}\right)^x\equal{}1$, which is x=0.

Just trying to be rigorous due to the fact that "obvious" is not an acceptable answer in Olympiad math...

This means that there is a one-to-one correspondence from domain to range. (The function is reversible)


if u play game runescape, then use this worldchanger.. its greaat :D

http://rsworldchanger.has.it

i think runescape is boring and people become too obsessed with it for some reason

yeah i'm getting pretty bored with it :)


Which of these numbers is less than its reciprocal? \\ \\
$ \text{(A)}\ \minus{}2 \qquad \text{(B)}\ \minus{}1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$ \\ \\ Select the most appropriate letter.


trigonometry
number theory
least common multiple
greatest common divisor

Find the period of $ f(x)\equal{}\sin(\frac{ax}{b})\plus{}\cos(\frac{cx}{d})$, where a, b, c, and d are integers and $ \frac{a}{b}$ and $ \frac{c}{d}$ are in lowest terms.

Period of sin(ax/b) =(2$ \pi$.b)/a ;
Similarly period of cos(cx/d) = (2$ \pi$.d)/c ;
Therefore period of the two together is [lcm of ((2$ \pi$.d),(2$ \pi$.b))]/[gcd of (c,a)]


geometry
calculus
analytic geometry
integration
LaTeX

A semicircle of diameter $ 1$ sits at the top of a semicircle of diameter $ 2$, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.
[asy]unitsize(2.5cm);
defaultpen(fontsize(10pt)+linewidth(.8pt));

filldraw(Circle((0,.866),.5),grey,black);
label("1",(0,.866),S);
filldraw(Circle((0,0),1),white,black);
draw((-.5,.866)--(.5,.866),linetype("4 4"));
clip((-1,0)--(1,0)--(1,2)--(-1,2)--cycle);
draw((-1,0)--(1,0));
label("2",(0,0),S);[/asy]$ \textbf{(A)}\ \frac {1}{6}\pi \minus{} \frac {\sqrt {3}}{4} \qquad \textbf{(B)}\ \frac {\sqrt {3}}{4} \minus{} \frac {1}{12}\pi \qquad \textbf{(C)}\ \frac {\sqrt {3}}{4} \minus{} \frac {1}{24}\pi\qquad\textbf{(D)}\ \frac {\sqrt {3}}{4} \plus{} \frac {1}{24}\pi$
$ \textbf{(E)}\ \frac {\sqrt {3}}{4} \plus{} \frac {1}{12}\pi$

[hide]Okay, the approach to this problem will be to add and subtract areas as necessary.  We draw two radii from the center of the large semicircle to the intersection points of the two semicircles.  This forms an equilateral triangle with side length 1.  Note that the central angle of the sector is thus equal to sixty degrees.  The area of the lune is the area of the small semicircle minus the area of the sector that we drew plus the area of the triangle (it sounds confusing without a diagram).

The area of the semicircle is $ \frac{1}{2}(\frac{1}{2})^2\pi \equal{} \frac{1}{8}\pi$

The area of the sector is $ \frac{60}{360}(1)^2\pi\equal{}\frac{\pi}{6}$

The area of the triangle is $ \frac{1^2\sqrt{3}}{4}\equal{}\frac{\sqrt{3}}{4}$

The area, as described above, is $ \frac{1}{8}\pi\minus{}\frac{\pi}{6}\plus{}\frac{\sqrt{3}}{4}\equal{}\minus{}\frac{\pi}{24}\plus{}\frac{\sqrt{3}}{4}$

This is answer $ C$[/hide]

Thats the solution I used, but you can use calculus if you are so inclined (even though it is unnecessarily long, it looks cool on paper  :P )

[hide]

First off, plot this on the coordinate plane so that the center of the diameter of the larger semicircle is at the origin.  Draw the equilateral triangle as detailed in the previous solution and determine that the vertical distance from the center of the diameter of the large semicircle to the center of the diameter of the small semicircle is $ \frac {\sqrt3}{2}$ .  So here are the equations of the two semicircles on the coordinate plane:

Smaller-
$ y \equal{} \sqrt {\frac {1}{4} \minus{} x^2} \plus{} \frac {\sqrt3}{2}$

Larger-

$ y \equal{} \sqrt {1 \minus{} x^2}$

So the area of this region will be the integral from $ \frac { \minus{} 1}{2}$ to $ \frac {1}{2}$
So:

$ \displaystyle\int^\frac {1}{2}_\frac { \minus{} 1}{2} \sqrt {\frac {1}{4} \minus{} x^2} \plus{} \frac {\sqrt3}{2} \minus{} \sqrt {1 \minus{} x^2}\,dx \equal{} .302113$

(I plugged it into my calculator at this point to integrate it  :rotfl: )

And you find that choice $ C$ is correct.

And there you have it, my needlessly long exercise in LaTeX  :P .
[/hide]


Can anyone post probability problems similar to the spider one?

(
A spider has 8 legs, and has to wear 8 socks and 8 shoes. How many ways can he put them on, given that the sock must go before the shoe on a leg?
)

I haven't seen any other probabilty questions like this...

http://www.kalva.demon.co.uk/aime/aime90.html

Look at #8. It's a cool problem.

That's a nice problem, but its not similar to the spider one.
For that one, you simply have to know that there is only ONE way to break the targets in any given column so you can label the columns 1,2,3. Then you just have to find all arrangements of 11122333.

However, for the spider one, I don't quite understand how to do it. I guess you find the number of ways you can put them on without restrictions, and then keep eliminating half of them (??)
The answer was $\frac{16!}{2^8}$

This is why I was looking for more problems of this type. (Problems that have $n!/2^{(n/2)}$ as their answer)


Anyone here take the USABO open? It was a tough exam this year, the highest was a 44, I got a 37 blegh, but then again I haven't touched my textbook since I left the program last year.

Good luck to everyone who made the semifinals!

Semifinals are this tuesday/wednesday! Is anyone else taking it?

Also, has anyone heard anything about the USAPhO semifinalists? I'm dying to get some closure!


limit
integration
trigonometry
calculus
real analysis
real analysis unsolved

Prove that $ \lim_{n\rightarrow\infty}\int_{0}^{\frac{\pi}{2}}\left | x\minus{}\sin^n(\frac{\pi}{2}x) \right |dx\equal{}\lim_{n\rightarrow\infty}\int_{0}^{\frac{\pi}{2}}\left(x\minus{}\sin^n(\frac{\pi}{2}x)\right)dx$.

 Thanks.

Well, just use  LDT and you will get that $ RHS\equal{}LHS\equal{}\int\limits_{0} ^{\frac{\pi}{2}} x dx$ (you have that $ sin^n (\frac{\pi}{2}x)$ converges to $ 0$ a.e)

What is LDT? It must be something I know....

i think it stands for "Lebesgue's Dominated Theorem."

[quote="J.Y.Choi"]Prove that $ \lim_{n\rightarrow\infty}\int_{0}^{\frac {\pi}{2}}\left | x \minus{} \sin^n(\frac {\pi}{2}x) \right |dx \equal{} \lim_{n\rightarrow\infty}\int_{0}^{\frac {\pi}{2}}\left(x \minus{} \sin^n(\frac {\pi}{2}x)\right)dx$.

 Thanks.[/quote]

Old fashioned proof: Show the integral of $ sin^n(\frac {\pi}{2}x)$ over $ [0,\frac {\pi}{2}]$  tends to 0 . Then use $ |a| \minus{} |b|\leq |a\minus{} b|\leq|a| \plus{} |b|$  to finish.


Let $x_1 \le x_2 \le \dots \le x_n$ be positive real numbers such that $x_1 + x_2 + \dots + x_k \ge 2^k - 1$ for all $1 \le k \le n$.  Prove that
\[\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n} < 2.\]


Consider p, a prime.
Prove that the residue class of ${a^{i}}\bmod p$ with $1\leq i \leq (p-1)$ is some permutation of $(1, 2, 3 \cdots p-1)$

Perhaps you mean $a\cdot i$, instead of $a^{i}$?

Or do you mean to show the existence of such $a$?


limit
logarithms
real analysis
real analysis unsolved

Does

converge?

In particular, does anyone have a nicer proof than [hide]guessing what the asymptotics are, then applying l'hopital/taylor expanding a bunch of times?[/hide]

The series diverges since

$ \frac{e}{2n\plus{}2}<e\minus{}e_{n}<\frac{e}{2n\plus{}1}$

and hence $ \lim\limits_{n\rightarrow \infty}n(e\minus{}e_{n})\equal{}e/2$.

However the following series

$ \sum\limits_{n\equal{}1}^{\infty}(\minus{}1)^{n}(e\minus{}e_{n})$ converges, but I don't know what is the value of its sum.

It can be represented as:

$ \sum^{\infty}_{n \equal{} 1}( \minus{} 1)^n\left(e \minus{} \left(1 \plus{} \frac {1}{n}\right)^n\right) \equal{} \sum^{\infty}_{n \equal{} 3}\frac {1}{k!}\sum^{n \minus{} 1}_{k \equal{} 2}S_1(n,n \minus{} k)\zeta(k)\left(1 \minus{} 2^{1 \minus{} k}\right) \minus{} \ln(2)\frac {e}{2}$

Where $ S_1(n,k)$ is the Stirling's number of the first kind and $ \zeta(n)$ is the Riemann's zeta function. Also,

$ \sum^{\infty}_{n \equal{} 1}( \minus{} 1)^n\left(e \minus{} \left(1 \plus{} \frac {1}{n}\right)^n\right) \approx \minus{} 0.445622$

diverges because $ (1 \plus{} 1/n)^{n}\approx e(1 \minus{} \frac {1}{2n})$
Hence $ \sum ..your serie .. \approx \frac{e\ln n }{2}$


Let be $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that :

$ f(x\plus{}2009)\le x\le f(x)\plus{}2009\ ,\ (\forall)x \in \mathbb{R}$ .

Calculate : $ f(\minus{}2009)\plus{}f(\minus{}2007)\plus{}...\plus{}f(1)\plus{}f(3)\plus{}...\plus{}f(2009)$

$ x \le f(x)\plus{}2009 \le (x\minus{}2009)\plus{}2009\equal{}x$, so $ f(x)\equal{}x\minus{}2009$ for all $ x$

There are $ 2010$ terms, so the sum $ \equal{}(\minus{}2009) \times 2010\equal{}4038090$


geometry
geometric transformation
reflection
circumcircle
geometry unsolved

In acute triangle $ ABC$, $ \angle A < 45$. Point $ D$ lies in the interior of triangle $ ABC$ such that
$ BD = CD$ and $ \angle BDC = 4\angle A$. Point $ E$ is the reflection of $ C$ across line $ AB$, and point $ F$ is
the reflection of $ B$ across line $ AC$. Prove that $ AD \perp EF$.

hint

Lema
O is a circumcenter of triangle ABC and I is incenter. P,Q is on AC,BC satisfying AP=AB=BQ then OI is perpendicular to PQ. Similarly this is true about excenter

I cant see a way  :blush:
How to solve it?

this problem has been posted by [b]april[/b] and is solved here:
[url]http://mathlinks.ro/Forum/viewtopic.php?t=154043[/url]


geometry
articles
percent
AMC
USA(J)MO
USAMO
number theory

I'm just curious, does anyone know what their IQ is?

ha! i cant even spell IQ! :lol: lol.

i think its 125

Welcome to AoPS... pretty much everyone (with a few notable exceptions) will be over 100, so the poll won't help  :P
You should add a 131+ category too, cause there are handful of those dudes here
[i](I'm sure there's an 8th grader doing differential geometry somewhere on this forum)[/i]

Yes, why don't you have a 131+ category?

Umm, what is the point of this poll?  No one who has the ability to use a computer is going to be $ \leq50$, and 100 is the average intelligence in America...  Just about everyone on AoPS is going to be smarter than average...

Also there are different IQ tests available. For one instance I recieved a 143 on one test but a 120 on another  :huh: . To test the IQs of AoPSers why not have them take the same test?

I have 148 IQ, so I can't vote in the poll.

You should definitely add a 131+ category, I'm pretty sure I'm at least 135.

Hey you can't edit polls.

If you could, also add a less than 20 category :P

[quote="#H34N1"]Hey you can't edit polls.[/quote]

Hehe I just did.

Well, I haven't actually had my IQ officially tested (so I don't think I should vote), but I took an online test for 18-year-olds (which I'm not yet) and got a 137. Of course, online tests aren't really accurate and are just for fun, so I still don't really know what my IQ is.

i've never gotten it tested officially, and online tests vary considerably (125-208)

[quote="ZeroFive1"]
IQ's aren't very reliable, though.[/quote]

Yes, true. And there's no point in having a super high IQ if you lack basic life skills.

148.

About Bush having 91, I got 90 when I chose all A's on that website.

Hmm, lol

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I think mine was 135 last time I checked, and I'm 14 years old.

i feel really bad...
mine was 126, and im twelve. :(

126 is still way above the average IQ level, 98, in the US. And considering you're 12, that's even better. :)

I just realized: Bush's IQ (91) is lower than the national average! :rotfl:

Bush's IQ is 91? Come on, where do you get this information?

There is no solid proof (or any evidence at all, for that matter), but some rumors are circulating that Bush's IQ is 91. Despite being a Bush-hater, I would like to defend him (for once in my life), and say that his IQ should be at least 95. Some random study concluded that it was between 111 and 132, but I find that a bit high to believe. I would estimate 110 max, and no lower than 95. However, my guess is as good as yours, and I don't think Bush plans on revealing his true IQ (if he knows what it is).

I tested a real one out of pure curiousity in the 7th grade. I believe I scored a 131 or 134, can't remember which. I remember seeing this Mensa book for sale at Barnes and Noble.

In the future do you think IQ testing will be looked upon as genoism?

http://www.csbsju.edu/uspp/Bush/Bush-IQ-Myth.html
bush iq myth

however...
http://en.wikipedia.org/wiki/U.S._Presidents_IQ_hoax

where do you go to find official IQ?
$ \mathfrak{PLEASE}$ tell me!

serach iq test on google... none are that great though.

hey this has bothered me for a while, but I've heard that east asians (mongoloids),
all they take a large part of the USAMO and other math things, that they have
a narrow IQ range, although very high on average, caucasians (including Indians and
middle easterners and europeans), have a wide IQ range, meaning lots of geniuses and lots of idiots,
I being East Asian am really not trying to be racist, but this severely bothers me
to the point that I'm getting very depressed, if you notice, there used to be
a stereotype that all asians were smart and even more that they are good at math,
however as you see nearly all the famous mathematicians are caucasian and
the ones that do extremely well are generally white and caucasian, however,
I feel really bad that I've seen the pattern of so many white geniuses, verses
lots of east asians/orientals that are very smart but not as smart as those
really genius white people, can anybody explain, an Asian (oriental) who knows he
has a really genius IQ like 180+ or even 160+????!!!! Someone, please help
explain this phenomenon, like Albert Einstein, Isaac Newton, they were all of the
caucasian race!!! and on this website I've also noticed the pattern, asians tend to
have more complicated solutions, while another random dude posts the solution,
and it's extremely clever and simple, and every time I check on who the heck
that person is, it is either some eastern european dude, russian dude, indian dude,
white american, somebody of the caucasoid stock owning these problems!!! Please
forgive, this is very racist, and I truly apologize, I am an Asian person and
this REALLY is driving me crazy, my life has been so upset and painful every since
I've seen things like http://infoproc.blogspot.com/2008/06/asian-white-iq-variance-from-pisa.html AND THIS !!!!:
http://www.google.com/imgres?imgurl=http://www.lagriffedulion.f2s.com/sftfi1%257Bimage3%257D.gif&imgrefurl=http://www.lagriffedulion.f2s.com/sft.htm&usg=__ct7JDJg9zgIP4Ko7k_Q3W9ytkWs=&h=347&w=462&sz=5&hl=en&start=0&zoom=1&tbnid=VC0Nv4bfF9bJmM:&tbnh=134&tbnw=179&ei=veXjTYT2JpTUgQfqpK3WBg&prev=/search%3Fq%3Deast%2Basian%2Bcaucasian%2Bnarrow%2BIQ%26um%3D1%26hl%3Den%26sa%3DN%26biw%3D1106%26bih%3D622%26tbm%3Disch&um=1&itbs=1&iact=rc&dur=379&page=1&ndsp=15&ved=1t:429,r:1,s:0&tx=133&ty=116&biw=1106&bih=622

4-year revival..
please don't do that
next time make a new thing
and call it What's your IQ? Revival!


vector
algebra
polynomial
Ring Theory
abstract algebra
calculus
integration

Let I=(x-1)(x-2)(x-3)Q[x] be the set of all polynomials with rational coefficients that are multiples of the cubic polynomial (x-1)(x-2)(x-3) so that I is an ideal of the ring Q[x]. Moreover, R:=Q[x]/I (the factor group) has a natural structure as a ring and the natural surjection h: Q[x]-->R is a ring homomorphism. R contains Q as a subring as well, so V:=Q[x]/I is a Q-vector space.

Let T, an endomorphism on V, be the linear operator defined by:

T(f(x)+I)=xf(x)+I, where for every f(x)+I in R := Q[x]/I.

Show the dimension of V is 3.

I was thinking of using the dimension formula here and that the kernel has inputs x=1, x=2, x=3 s.t. T(f(x)+I)=xf(x), and I disappears because those are the roots of the cubic polynomials, so the dimension of the kernel is 3 (and the dimension of the image is 0). But to be honest, I'm not sure if that actually is the kernel -- basically I don't really know how to think of this problem. Any help is much appreciated!

This might be kind of an overkill proof using the theory of integral relations from commutative algebra (the theorem I am referring to can be found in the first few pages of chapter 4 of Eisenbud's Commutative Algebra with a view toward Algebraic Geometry)

Proof: Consider $ \mathbb{Q}/((x\minus{}1)(x\minus{}2)(x\minus{}3))$ as a $ \mathbb{Q}$-module. Notice that $ I$ is generated by a monic polynomial (multiplying the expression out, we find that the highest power of $ x$ has unity as its coefficient). Therefore, we conclude $ \mathbb{Q}/((x\minus{}1)(x\minus{}2)(x\minus{}3))$ is a finitely generated free $ \mathbb{Q}$-module with free basis of the form $ 1, h(x), h(x^2)$ where $ h$ is the quotient map.

The theorem I am citing says the following: Let $ R$ be a ring and $ I$ and ideal. Let $ S\equal{}R/I$ considered as an $ R$-module. Then if $ I$ is generated by a monic polynomial of degree $ n$, $ S$ is free over $ R$ of rank $ n\minus{}1$ and is generated by the images of $ 1$, $ x$,... $ x^{n\minus{}1}$ through the quotient map $ R\rightarrow R/I$. Its proof is essentially an application of the Cayley Hamilton Theorem.

Thanks, but to be honest I don't understand any of that -- my knowledge doesn't extend beyond an undergrad algebra course...any way to deal with it with a more limited set of tools?

I'd rather skip $ T$ entirely for looking at dimension.

Lemma: Given any triple of values $ a,b,c$, there is exactly one coset $ p\plus{}I$ for which $ p(1)\equal{}a,p(2)\equal{}b,p(3)\equal{}c$.
Proof: For existence, construct it by polynomial interpolation: $ p(x)\equal{}a\frac{(x\minus{}2)(x\minus{}3)}{(1\minus{}2)(1\minus{}3)}\plus{}b\frac{(x\minus{}1)(x\minus{}3)}{(2\minus{}1)(2\minus{}3)}\plus{}c\frac{(x\minus{}1)(x\minus{}2)}{(3\minus{}1)(3\minus{}2)}$. For uniqueness, note that $ p(x_i)\equal{}p'(x_i)$ for each $ x_i\in \{1,2,3\}$ if and only if $ (p\minus{}p')$ is zero at those three points. By elementary theorems of polynomial algebra, this means that $ x\minus{}1,x\minus{}2,$ and $ x\minus{}3$ are all factors of $ p\minus{}p'$. Their product must also be a factor, so $ p\minus{}p'\in I$ and these two polynomials with the same values are in the same coset.

The interpolation basis is also very convenient for the algebra; we've essentially diagonalized it.

I hate to be annoying but it seems like what you just wrote came out of nowhere...I've been trying to make sense of it for a while and just can't.

You know what? Let's be simple.
A coset $ f(x)\plus{}I$ has exactly one representative of degree $ \le 2$, by long division. The polynomials of degree $ <n$ are a vector space of dimension $ n$, with standard basis $ 1,x,x^2,\cdots,x^{n\minus{}1}$. They're also one of the first examples you'll ever run into.

That's it.

Now, what's going on with the rest of what we wrote? We were trying to understand the multiplication operator inherited from the polynomial ring. Since $ x$ generates everything, it suffices to understand the action of multiplication by $ x$, which we call $ T$.
In the usual basis $ 1,x,x^2$, $ T$ has matrix $ \begin{bmatrix}0&0&6\\1&0&\minus{}11\\0&1&6\end{bmatrix}$ (That last column comes from the coefficients of $ (x\minus{}1)(x\minus{}2)(x\minus{}3)\equal{}x^3\minus{}6x^2\plus{}11x\plus{}6$). That's a companion matrix, and we've put $ T$ in rational canonical form. (LkNsngth's post)
In the interpolation basis $ \frac{(x\minus{}2)(x\minus{}3)}{(1\minus{}2)(1\minus{}3)},\frac{(x\minus{}1)(x\minus{}3)}{(2\minus{}1)(2\minus{}3)},\frac{(x\minus{}1)(x\minus{}2)}{(3\minus{}1)(3\minus{}2)}$, $ T$ has matrix $ \begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}$. This change of basis diagonalized it. The technique generalizes to become partial fractions, and leads to Jordan form there. (my first post)

Of course, neither of these complications is necessary for merely finding the dimension. They're both about answering the more general questions you didn't actually ask.


algebra
polynomial
number theory theorems
number theory

Hello!

Does anybody have a complete proof on Wolstenholmes theorem?

Wolstenholme theorem:
(2p-1,p-1)==1 (mod p^3) where p is prime and p>=5.

Note: (n,k)=n!/(k!(n-k)!).

Cheers! 10x!

I have a solution (found on a website) but can someone explain it much more detalied?  :?

If p is a prime >3 , then the numerator of
1+1/2+1/3+..+1/(p-1) is divizible with p^2 and the numerator of
1+1/2^2+1/3^3+...+1/(p-1)^2 is divizible with p.   These imply that if p>=5 is prime, then  we have Wolstenholme's theorem!!

cheers!  :D

Sorry for the previous message it was a mistake with my key board
-------------------------------------------------------------------------------

Ok Lagrangia i will try to explain

Let p prime number, m be integer such that p does not divide m, we will
write m* an integer such that m.m* # 1 (mod p 2), as you  guess # is congruence notation ( a#b mod c means c divides a-b).

Lemma: If p>2 prime, then 1*+2*+...+(p-1)* # 0 (mod p)

Proof: In the field Z/pZ,  m* is the inverse of m so

1*+2*+..+(p-1)* # 1+2+...+(p-1) =p(p-1)/2 # (mod p)

Theorem Wolstenhome 1862: If p>3 prime,
1*+2*+...+(p-1)* # 0 (mod p 2 )

Proof: Let  \alpha =1*+2*+...+(p-1)* then
\alpha.(p-1)! # \sum _{i=1^(p-1)}(p-1)!/i (mod p 2 )

Let A1 = \sum_{i=1^(p-1)}(p-1)!/i belong to N, A1 is the coefficient of
X in the polynomial

Q(X)=(X-(p-1))...(X-2)(x-1) = X^(p-1)-A_(p-2)X^(p-2)+...-A1X +(p-1)!

Modulo p,  Q(X) = X^(p-1)-1 so p divides all A_i for 1 \leq i \leq p-2.

Q(0)= =Q(p)= (p-1)! so

p^(p-1) +A_(p-2)p^(p-2)+...+A_2p^2 +A1p = 0 we divide by p

p^(p-2) +A_(p-2)p^(p-3)+...+A2p+A1 = 0

but p>3 , p 2  divides A_2p,..., A_(p-2)p^(p-3) and p^(p-2)

So we have A1 # 0 (mod p 2 )

And \alpha # 0 (mod p 2 )  because gcd((p-1)!;p^2)= 1

QED

 :D oubi :D

Well ,on a generized version you can see on AMM ,where {1,...,p-1} is replaced by {r1,...,rt} a reduction reduse of an odd integer m>1[so t= \emptyset (m)].We call p is an odd prime which divides by m then we have
 THEOREM: The numerator of 1/r1+...+1/rt is divisible by p

How do we deduce this, i.e.

[quote="Lagrangia"]Wolstenholme theorem:
(2p-1,p-1)==1 (mod p^3) where p is prime and p>=5.

Note: (n,k)=n!/(k!(n-k)!).[/quote],

from the results proven above?


[b]Case 1:[/b] $ y$ and $ z$ are even
$ y \equal{} 2a$
$ z\equal{} 2b$

$ 2x \plus{} 2a \plus{} 2b \equal{} 2n \rightarrow x \plus{} a \plus{} b \equal{} n$

There are $ n$ balls and $ 2$ dividers, so there are $ \binom{n\minus{}1}{2}$ solutions.

[b]Case 2:[/b] $ y$ and $ z$ are odd.
$ y \equal{} 2a \plus{} 1$
$ z \equal{} 2b \plus{} 1$

$ 2x \plus{} 2a \plus{} 2b \plus{} 2 \equal{} 2n \rightarrow x \plus{} a \plus{} (b \plus{} 1) \equal{} n$

Here you do the same as before, except you can have that $ a$ and $ b$ can be $ 0$.

I have to go now, so if no one finishes this, I will when I get back.

Hi modularmarc101.

I make it $ (n\minus{}1)^{2}$ as follows:

Since $ 2x \plus{} y \plus{} z \equal{} 2n$, $ y\plus{}z$ must be even, say $ y\plus{}z \equal{} 2k$.

So $ x\plus{}k\equal{}n$.

Now $ k$ can take any value from $ 1$ to $ n\minus{}1$, to get positive integer solutions.

Then $ y$ can take any value from $ 1$ to $ n\minus{}1$ and $ z$ is determined by $ y\plus{}z\equal{}2k$ and $ x$ is determined by $ 2x\plus{}y\plus{}z\equal{}2n$, giving $ (n\minus{}1)^{2}$ positive integer solutions.

[quote="AndrewTom"]How many positive integer solutions does the equation 2x + y + z = 2n have, where n is a given positive integer?[/quote]
Using this
$ a_0  \plus{} a_1  \plus{} ... \plus{} a_n  \equal{} k$
Have $ C_{n \plus{} k}^n$ positive integer solutions  :lol:


[quote="jli"]The median of the list

$n, n+3, n+4, n+5, n+6, n+8, n+10, n+12, n+15$

is 10.  What is the mean?[/quote]

[hide="solution"]
The mean is the middle number when all of the numbers are listed in numerical order, so the mean is n+6.

n + 6 = 10
n = 4

That means the numbers are:

4, 7, 8, 9, 10, 12, 14, 16, 19, so the arithmetic mean is [b]11[/b].
[/hide]

[hide]the median is n+6 so n is 4.
u add up the numbers and u get 9n + 63, divide that by 9 and u get n+7.
which leads to 11[/hide]

[hide="Solution"]Same here... I got 11. Because the median was 10, and n+6 was the median, N= 4, then you add up all the terms and numbers to get 9n = 36 + 63, = 99/9 = 11.[/hide]


Solve the inequality in the set of real numbers

$ \frac{8x^3 \minus{}1}{4x^2 \plus{} 2x \plus{} 1} \leq 0.$

[hide="I did"]
a)
NUM: positive
$ 8(x^3 \minus{} 1/8) \equal{} 8(x\minus{}1/2)(x^2 \plus{} 1/2 x \plus{} 1/4)>0$
so $ x>1/2$ and the quadratic one does not have any real solutions because its discriminant is less than $ 0$.

DENOM: Negative
Never?

b)
NUM: NEGATIVE
$ x<1/2$
DENOM: Positive
All real numbers

c)
NUM: zero
$ x\equal{}1/2$

d) THE DENOMINATOR = 0
Since the discriminant is less than zero as well, so there's no real answer to this.

I don't know where I went wrong...
[/hide]

the denominator is always positive so we can ignore it, then we have $ (2x)^3\le 1$, or $ 2x\le 1$.

Since the leading term has degree 2, subbing any number will yield a positive result

Hence you can take the denominator to the other side, and solve. But sometimes you dont know whether the denominator is positive or negative. In this case you can't take it to the ohter side.


WHAT'S 1.68499667e+66 in decimal notation!!??!! Need Help!!!

we are not here for homework...but all that problem involves is moving the decimal point over 66 times

i think this problem is not nice and even it is nice it is not good for here. :D


hello mathlinkers,

because some people flooded the old thread i will give you some true information about the german IMO team for the 48th IMO in vietnam.

another year is over, so there is a new team.
and as usually there were some surprising changes in the German team.
First the team and after this a few words.

Peter Scholze (i think you know him)
Friedrich Feuerstein (was at the IMO a few years ago but not in the last 3 years)
Georg Schröter (aka Naphthalin, first participation)
Stephan Neupert (also first one)
Lisa Sauermann (the first girl after about 3 years, her first participation)
Malte Lackmann (first one again)

Lisa Sauermann is just 14 years old and can particpate till 2011 so you can await something Razz

i think without peter and me there is no one of the team here at ml.

for further misinformation see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150116]here[/url]
and [url=http://wmls.no-ip.org/aimo]here[/url] are the results of the 7 TST-exams of germany.

Naphthalin

[i]Edited by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=1991]orl[/url].[/i]

Congrats! What happened to Scellier, Benjamin ?

He's an exchange from France, so he didn't participate for the german team. I don't know if he goes to IMO for France.


Let $ x \equal{} \sqrt[3]{\frac{4}{25}}\,$.  There is a unique value of $ y$ such that $ 0 < y < x$ and $ x^x \equal{} y^y$.  What is the value of $ y$?  Express your answer in the form $ \sqrt[c]{\frac{a}{b}}\,$, where $ a$ and $ b$ are relatively prime positive integers and $ c$ is a prime number.

This is kinda random but basically we want to parameterize solutions so look at:

$ y=kx$ so

$ x^x=(kx)^{kx}$ take the $ x$th root...

$ x=k^k\cdot x^k$

$ k^{-k}=x^{k-1}=(x^s)^{\frac{k-1}{s}}$

So equate the bases and exponents (we know we can do this because we choose such an $ s$).

$ -k=\frac{k-1}{s}\implies \frac{1}{s}=\frac{k}{1-k}$  $ k=x^s\implies x=k^{\frac{k}{1-k}}$.

Now in the problem statement, $ x=\sqrt[3]{\frac{4}{25}}=\left(\frac{2}{5}\right)^{\frac{2}{3}}$ which corresponds to $ k=\frac{2}{5}$   (yay!)

So $ y=kx$ gives us the answer :)

That's fine, but once you had the equation $ k^{\minus{}k} \equal{} x^{k \minus{} 1}$, couldn't you deduce $ x \equal{} k^{k/(1 \minus{} k)}$ in one step, without introducing $ s$?


Find the radius of convergence of $\sum a_{k}x^{k}$ where $k^{-1}\leqslant a_{k}\leqslant k$ for all $k\in\mathbb{N}$.

Use the root test, and the fact that $k^{1/k}\to 1.$ The radius of convergence is 1. We'd draw the same conclusion sandwiching $a_{k}$ between any pair of rational functions of $k.$

In this very particular case, we do in addition have that the series diverges for $x=1$ by comparison to the harmonic series.


inequalities
advanced fields
advanced fields unsolved

Given
$a>b>c$
$a+b+c=1$
$a^{2}+b^{2}+c^{2}=1$
Prove that
$1<a+b<\frac{4}{3}$
$\frac{8}{9}<a^{2}+b^{2}<1$

so far I got
$1 < a+b < \sqrt{2}\quad\mbox{and}\quad a^{2}+b^{2}<1$
and $0<a<1,\quad 0<b<1,\quad-1<c<0$

maybe you can use NLP to find max and min ??

This looks like a Lagrange multipliers problem, but it's simple enough to be solved without them.

The inequalities you are supposed to prove are equivalent to $-\frac13<c<0$ and $0<c^{2}<\frac19$. Thus, the second inequality is a consequence of the first. You already got $c<0$. To prove $c>-\frac13$, notice that
\[(a-2/3)^{2}+(b-2/3)^{2}=a^{2}+b^{2}-\frac43(a+b)+\frac89 =1-c^{2}-\frac43(1-c)+\frac89=1-(c-2/3)^{2}\]
hence $(c-2/3)^{2}\le 1$. Equality is impossible because $a$ and $b$ cannot be both equal to $2/3$.


geometry
3D geometry
sphere
frustum
algebra
system of equations

A sphere with radius 3 is inscribed in a frustum of slant height 10.  What is the volume of the frustum? (A frustum is a non-conical region formed by cutting a cone with a plane parallel to its base.)

[hide="Solution"]
OK, so since the sphere is inscribed, I'm assuming that means it touches the top and the bottom of the frustum, so the height of the frustum should be 6. By that reasoning, the bottom radius is 8 bigger than the top radius. Their average is in the middle, which is again, twice the sphere's radius. So, the top radius is 2 and the bottom one is 10. Using the frustum volume formula (below), we get:
$\frac{pi}{3}$*$top radius^2+Top radius*bottom radius+bottom radius^2=$

$\frac{pi}{3}$*$10^2+20+2^2=$

$\frac{124pi}{3}$
However, I probably made some mistake, as it is hard for me to believe that I actually solved something from Intermediate.

[/hide]

Hmm, In response to sponge, couldn't the top radius be 1 and the bottom be 11, or the top 5 and the bottom 7, etc. etc.?

But you see, if you make a right triangle with the slant height as the hypotenuse and the height as a leg, you get that the difference in radius between the top and the bottom is 8. The 6 in the middle must be the average, so, to solve the system of equations where R is the top radius and r is the bottom radius,
$\frac{R+r}{2} = 6$
and
$R-r=8$

and by solving that we get r=2 and R=10.

[quote="sky9073"][quote="sponge008"][hide="Solution"]Their average is in the middle, which is again, twice the sphere's radius.
[/hide][/quote]
I wouldn't quite agree with this... The sphere meets the frustrum's sides a little closer to the top.
You might want to try:
[hide]Power of a point.. top radius + bottom radius = slant height =10.
Since the difference of the two radii is 8, top radius = 1 and bottom radius = 9.
I got 705pi/4 for the answer.
[/hide][/quote]

hmm i got something different. :?
[hide]extend the frustrum to complete the cone
it would have a height of 27/4 and radius of 9.
$\frac{9^2\pi*\frac{27}{4}}{3}=\frac{729\pi}{4}$.
the missing top cone has radius of 1 and a height of 3/4.
$\frac{1^2\pi*\frac{3}{4}}{3}=\frac{\pi}{4}.$
$\frac{729\pi-\pi}{4}=182\pi$[/hide]

Oops.. calculation error.
(I hate fractions)


Prove that if $ \{a_n\}$ converges then the limit of $ {a_n}$ as $ n$ approaches infinity equals the limit of $ {a_{n\plus{}1}}$ as $ n$ approaches infinity.

Because for all $ \varepsilon>0$ exist $ N$ such that for all $ m,n>N$ then
$ |a_m\minus{}a_n|<\varepsilon$


The functions $ f(x) \equal{} x^2\minus{}2x \plus{} m$ and $ g(x) \equal{} x^2\minus{}2x \plus{} 4m$ are
evaluated when $ x \equal{} 4$. What is the value of $ m$ if $ 2f(4) \equal{} g(4)$?

$ 2f(4)\equal{}2(4)^2\minus{}2(2)(4)\plus{}2m\equal{}32\minus{}16\plus{}m\equal{}16\plus{}m$

$ g(4)\equal{}4^2\minus{}2(4)\plus{}4m\equal{}8\plus{}4m$

$ 16\plus{}m\equal{}8\plus{}4m\implies8\equal{}3m\implies m\equal{}\boxed{\frac{8}{3}}$


What is the $ 11^{\text{th}}$ term in the geometric sequence $ 3, \frac{\minus{}3}{2}, \frac{3}{4}, \frac{\minus{}3}{8}, \ldots$?  Express your answer as a common fraction.

Geometric sequences have what's known as a common ratio. That means that each term is multiplied by the same number. In this sequence, we can easily see that the common ratio is $ \minus{} \frac {1}{2}$. Let's continue the sequence:

$ 3, \frac { \minus{} 3}{2}, \frac {3}{4}, \frac { \minus{} 3}{8}, \frac {3}{16}, \frac { \minus{} 3}{32}, \frac {3}{64}, \frac { \minus{} 3}{128}, \frac {3}{256}, \frac { \minus{} 3}{512}, \frac {3}{1024}$

Thus, our answer is $ \boxed{\frac {3}{1024}}$.

Or we could just multiply our first term, $ 3$, by $ \minus{} \frac {1}{2}^{10}$, to get $ \boxed{\frac {3}{1024}}$.

EDIT 2: Sorry fishy, I was typing up the first edit before I saw your post.

You really don't have to do that.

$ A_n \equal{} A_1\cdot r^{n \minus{} 1}$
$ A_{11} \equal{} 3\cdot (\frac { \minus{} 1}{2})^{11 \minus{} 1}$
$ A_{11} \equal{} \frac {3}{1024}$


[quote="GameBot"]``I weigh 40 pounds more than I did when I graduated from college,'' laments the math teacher. ``I weigh 25$ \%$ more now than I did then.'' How much did the math teacher weigh, in pounds, when he graduated from college?[/quote]

So $ 1.25x\equal{}x\plus{}40 \implies \frac{x}{4}\equal{}40 \implies x\equal{}160$.


algebra
partial fractions
complex analysis

This one seems, dare I say the word, obvious!  :D

Use Mittag-Leffler to establish the partial fractions expansion of $ \coth z$ as:

$ \coth z= \frac{1}{z}+2z \left( \frac{1}{z^{2}+\pi^{2}}+\frac{1}{z^{2}+4 \pi^{2}}+\frac{1}{z^{2}+9 \pi^{2}}+\cdots \right)$.

This can be rewritten to establish that:

$ \sum_{j=-\infty}^{\infty}\frac{1}{j^{2}+a^{2}}= \frac{\pi \coth(\pi a)}{a}$

Since your double sum separates, we get automatically:

$ \sum_{j=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\frac{1}{(j^{2}+a^{2})\cdot (k^{2}+b^{2})}= \frac{\pi^{2}\coth(\pi a) \coth(\pi b)}{ab}$


Let $a_n$ be a sequence of complex numbers.
Show that if $\sum_{n\in{\mathbb{Z}}}\left|a_n\right| \in {\mathbb {R}}$ then $\sum_{n\in{\mathbb{Z}}}\left|a_n\right|^2 \in {\mathbb {R}}$ . Show that the reciproc is not true.

because $|a|^2 < |a|$ if $|a|<1$ and there are only finite many $n$ such that $|a_n| \geq 1$.
The converse is obviously false. (harmonic series for example)

Since $|a_n|$ is a convergent series, we know by Cauchy's Criterion that for $|n| > N$ (for some fixed $N$), $|a_n| \le 1.$ If $|a_n| \le 1$, clearly $|a_n|^2 \le |a_n|$. Then you have convergence of $|a_n|^2$ just by direct comparison of the two series. (The sum over all $|n| > N$ of $|a_n|^2$ is less than that of $|a_n|$, and then you're just adding finitely many more terms, so the total must still be a real number.)

I don't see how his counterexample can be made any more "rigorous" than it already is.

The quantitative version:

Suppose that $\sum_n|a_n|$ converges. Prove that

$\sum_n|a_n|^2\le\left(\sum_n|a_n|\right)^2.$

[hide="hint"]Consider a partial sum and think about what happens when you expand the square on the RHS.  :)  Then, how can you use the fact that $\sum_n|a_n|$ converges?[/hide]

Not the hint I would have given.

[hide="My hint."]This is a homogeneous inequality. We may WLOG assume that something useful equals 1.[/hide]

$|a_k| \leq \sum_n |a_n|$ for all k so QED?


Triangle $ABC$ has sides $a,b,c$ and opposite angles are in order $\alpha, \beta, \gamma$. Prove that:

$a(\sin \beta-\sin \gamma)+b(\sin \gamma- \sin \alpha) + c(\sin \alpha - \sin \beta)=0$

This is very simple problem
Use the sine law

take sinα = ak
        sinß = bk
        sinΓ = ck

put in Left side of the equation and get 0.


function
inequalities
algebra proposed
algebra

Let $ f,g: \mathbb N \minus{} > \mathbb N^*$ ,be two functions with the following properties :
1) $ g$ is surjective
2)$ \exists a \in \mathbb N$ such that $ f(a)\leq a$
3)$ f(g(n) \plus{} m)\geq f(n)\cdot f(m)$ $ \forall m,n \in \mathbb N$

Prove that $ \exists$ $ n_{0}\in \mathbb N$ such that $ f(n_{0}) \equal{} g(n_{0})$

I did not completely solve the problem although I have found some things that can be helpful to others
1.f is increasing set g(n)=1 (g surjective) and get by the inequality that $ f(1\plus{}a)>\equal{}f(n)f(a)>\equal{}f(a)$ since f(n) is in N*
2.If a is the least a for which the first inequality holds, then f(a)=f(a-1)=a.

I hope this will help anyone who wants to solve it

Remark: your problem description introduced $ f,g: \mathbb{N}\rightarrow\mathbb{N}^*$. From that I conclude your definition $ 0\in\mathbb{N}$.

[i]Proof:[/i]

Using (1) choose $ n_1$ such that $ g(n_1)=1$ holds. If $ f(n_1)\geq 2$ holds, the substitution into (3) gives: $ f(1+m)\geq f(n_1) f(m)\geq 2f(m)$ or more generally: $ \forall m\in\mathbb{N}^*: f(m)\geq 2^mf(0)>mf(0)\geq m$. This contradicts (2), therefore we have $ f(n_1)=1=g(n_1)$ and we are finished.


use the existence and uniqueness theorem to find all solutions of the equation $ x''\plus{}4x\equal{}0$.

The method:

Step 1: Find a two-dimensional space of solutions, using standard differential equations methods.
Step 2: Show that this this space contains solutions with all possible initial conditions $ (x(a),x'(a))$.
Step 3: Use the uniqueness side of the existence-uniqueness theorem to show that there are no other solutions.

The existence side is irrelevant here- it's not something that you can reasonably use for computations.


geometry
3D geometry
tetrahedron
geometric transformation
reflection
geometry proposed

A tetrahedron is called a [i]MEMO-tetrahedron[/i] if all six sidelengths are different positive integers where one of them is $ 2$ and one of them is $ 3$. Let $ l(T)$ be the sum of the sidelengths of the tetrahedron $ T$.
(a) Find all positive integers $ n$ so that there exists a MEMO-Tetrahedron $ T$ with $ l(T)\equal{}n$.
(b) How many pairwise non-congruent MEMO-tetrahedrons $ T$ satisfying $ l(T)\equal{}2007$ exist? Two tetrahedrons are said to be non-congruent if one cannot be obtained from the other by a composition of reflections in planes, translations and rotations. (It is not neccessary to prove that the tetrahedrons are not degenerate, i.e. that they have a positive volume).

No response for ten years. I have found a solution in https://skmo.sk/dokumenty.php.
But this is written in Slovak, can someone help to translate it into English?


How many factors of $ 2^5\cdot3^6$ are perfect squares?

A perfect square that is a factor of $ 2^5\cdot3^6$ must be in the form of $ 2^a\cdot3^b$ where a and b are both even and $ a\le5$ and $ b\le6$. Thus, there are 3 possibilities for a and 4 for b. (0 counts) $ \boxed{12}$ possibilities.


algorithm
Stanford
college
least common multiple
greatest common divisor
number theory unsolved
number theory

(a) Find all triples $(x,y,z)$ of positive integers such that $xy \equiv 2 (\bmod{z})$ , $yz \equiv 2 (\bmod{x})$ and $zx \equiv 2 (\bmod{y} )$
(b) Let $n \geq 1$ be an integer. Give an algoritm to determine all triples $(x,y,z)$ such that '2' in part (a) is replaced by 'n' in all three congruences.

it might have already appeared in AMM journal. check back issues say 02', 03'. it was posed by s.o from stanford univ.

[quote="Rust"]$xy \equiv n (\bmod{z})$ mean z>n or not?
For example $3 \equiv 5 (\bmod{2})$.[/quote]

(Sorry I'm poor at English)
I think it means 0<z≤n.
I found [color=white](1,1,1)[/color] and [color=white](2,2,2)[/color](Sorry I don't know the method that blinds that) but could not found the triple that x,y or z > n. :huh:

It is easy to show, that if $xy=n(mod \ z),xz=n (mod \ y), yz=n(mod \ x)$, then at least one of them (x,y,z) less than n.

CASE I) Suppose two of the numbers, let's say $x$, $y$ satisfy $xy=n$, then it's easy to prove that
             $z$ can be any positive multiple of $\frac{[x,y]}{(x,y)}$ where $[x,y]$ is the least common
             multiple and $(x,y)$ is the greatest common divisor of $x$ and $y$.

CASE II) Suppose two of the numbers, let's say $x$, $y$ satisfy $xy<n$, then $z\vert n-xy$ so there
              are a finite number of posibilities to check.

CASE III) Suppose $xy\geq n$ and $zx\geq n$, then there are non-negative integers $b$ and $c$
               satisfying
\[ xy-cz=n  \]
\[ -by+xz=n  \]
this implies that $(x+b-z)y-(x+c-y)z=0$ where $-z<x+b-z<z$ and $-y<x+c-y<y$ and this implies
that, if $d=(y,z)|n$, there exists and integer $k$, $-d<k<d$, such that
\[ x+b-z=\frac z d k  \]
\[ x+c-y=\frac y d k  \]
So we have $c=\frac y d k+y-y$ that replaced in $xy-cz=n$ and after some operations we arrive
at the following conclusion: to obtain the remaining solutions of the initial system it suffices to find
the positive solutions $(x,y',z')$ to the finite number of systems (and discard the strange solutions)
\[ xy'-ly'z'+z'x=\frac n d  \]
\[ d^2y'z'\equiv n (\mod x)  \]
\[ xy'\neq\frac n d  \]
\[ xz'\neq\frac n d  \]
where $d$ runs over the positive divisors of $n$ and $l$ is and integer $1\leq l\leq 2d-1$.
NOTE: Of course, for every solution $(x,y',z')$ to one of the systems, $(x,dy',dz')$ is a candidate
          solution to the original system. Also, we have discarded the solutions yet covered in CASE I.

Now it suffices to prove that every system has only a finite number of solutions that can be easily
checked.
Effectively, multiplying the first equation by $d^2$ and adding the congruence multiplied by $l$ we
obtain $x\vert n(d+l)$, so there are only a finite number of posibilities for $x$ to check.
Now, for every $d$, $l$ and $x$ given, being
\[ z'=\frac x l + \frac{\frac{x^2}l-\frac n d}{ly'-x}  \]
we should have $(ly'-x)\vert(x^2-l\frac n d)$, so there remain only a finite number of posibilities
for $y'$ to check.
NOTE: Observe that we should restrict ourselves to $x^2-l\frac n d$ being an integer distinct from
          zero, else the solutions satisfy $xy'=\frac n d$ or $z'x=\frac n d$, that were covered by
          CASE I.

ociretsih can you please exlain how did you thought of making 3 cases


[quote="manlio"]Prove that for $ a,b,c$ positive reals

$ a\sqrt {a^2 \plus{} 2bc} \plus{} b\sqrt {b^2 \plus{} 2ac} \plus{} c\sqrt {c^2 \plus{} 2ba}$ ≥ $ \sqrt {3(ab \plus{} bc \plus{} ca)}$[/quote]

Manlio, it's not homogeneus.

[quote="manlio"]Prove that for $ a,b,c$ positive reals

$ a\sqrt {a^2 \plus{} 2bc} \plus{} b\sqrt {b^2 \plus{} 2ac} \plus{} c\sqrt {c^2 \plus{} 2ba}$ ≥ $ \sqrt {3}(ab \plus{} bc \plus{} ca)$[/quote]

$ \Longleftrightarrow (a\sqrt {a^2 \plus{} 2bc} \plus{} b\sqrt {b^2 \plus{} 2ac} \plus{} c\sqrt {c^2 \plus{} 2ba})^2 \minus{} 3(ab \plus{} bc \plus{} ca)^2\geq 0$

$ \Longleftrightarrow \sum_{cyc} a^2(a^2 \plus{} 2bc) \plus{} 2\sum_{cyc}ab\sqrt {(a^2 \plus{} 2bc)(b^2 \plus{} 2ac)}$ $ \minus{} 3(ab \plus{} bc \plus{} ca)^2\geq 0$

Note

$ \sum_{cyc} a^2(a^2 \plus{} 2bc) \plus{} 2\sum_{cyc}ab\sqrt {(a^2 \plus{} 2bc)(b^2 \plus{} 2ac)}\geq \sum_{cyc} a^2(a^2 \plus{} 2bc) \plus{}$ $ 2\sum_{cyc}ab(ab \plus{} 2c\sqrt {ab})$

we need only prove

$ \sum_{cyc} a^2(a^2 \plus{} 2bc) \plus{}$ $ 2\sum_{cyc}ab(ab \plus{} 2c\sqrt {ab}) \minus{} 3(ab \plus{} bc \plus{} ca)^2\geq 0..........(*)$

substitution $ a \equal{} x^2,b \equal{} y^2,c \equal{} z^2$,($ x,y,z\geq 0$)

$ (*) \Longleftrightarrow x^8 \minus{} 4x^4y^2z^2 \plus{} y^8 \minus{} 4y^4z^2x^2 \plus{} z^8 \minus{} 4z^4y^2x^2 \minus{} y^4x^4 \plus{} 4y^3x^3z^2 \minus{} y^4z^4$ $ \plus{} 4y^3z^3x^2 \minus{} z^4x^4 \plus{} 4z^3x^3y^2\geq 0$

$ \equal{}\frac{1}{2}\sum_{cyc} (x^3\plus{}x^2y\plus{}2zxy\plus{}xy^2\plus{}y^3)(x^3\plus{}x^2y\minus{}2zxy\plus{}xy^2\plus{}y^3)(x\minus{}y)^2\geq 0$

assume $ x\geq y\geq z$

let

$ S_y\equal{}(z^3\plus{}z^2x\plus{}2zxy\plus{}x^2z\plus{}x^3)(z^3\plus{}z^2x\minus{}2zxy\plus{}x^2z\plus{}x^3)$

$ S_x\equal{}(y^3\plus{}y^2z\plus{}2zxy\plus{}yz^2\plus{}z^3)(y^3\plus{}y^2z\minus{}2zxy\plus{}yz^2\plus{}z^3)$

$ S_z\equal{}(x^3\plus{}x^2y\plus{}2zxy\plus{}xy^2\plus{}y^3)(x^3\plus{}x^2y\minus{}2zxy\plus{}xy^2\plus{}y^3)$

easy prove that

$ S_y\equal{}(z^3\plus{}z^2x\plus{}2zxy\plus{}x^2z\plus{}x^3)(z^3\plus{}z^2x\minus{}2zxy\plus{}x^2z\plus{}x^3)\geq 0$

$ S_y\plus{}S_x\equal{}(z^3\plus{}z^2x\plus{}2zxy\plus{}x^2z\plus{}x^3)(z^3\plus{}z^2x\minus{}2zxy\plus{}x^2z\plus{}x^3)$ $ \plus{}(y^3\plus{}y^2z\plus{}2zxy\plus{}yz^2\plus{}z^3)(y^3\plus{}y^2z\minus{}2zxy\plus{}yz^2\plus{}z^3)\geq 0$

$ S_z\equal{}(x^3\plus{}x^2y\plus{}2zxy\plus{}xy^2\plus{}y^3)(x^3\plus{}x^2y\minus{}2zxy\plus{}xy^2\plus{}y^3)\geq 0$

so.............

 :lol:

[quote="manlio"]Prove that for $ a,b,c$ positive reals
$ a\sqrt {a^2 \plus{} 2bc} \plus{} b\sqrt {b^2 \plus{} 2ac} \plus{} c\sqrt {c^2 \plus{} 2ba}\ge \sqrt {3}(ab \plus{} bc \plus{} ca).$[/quote]
Using Holder's Inequality, we have
$ \left( \sum a\sqrt{a^2\plus{}2bc}\right)^2 \left( \sum \frac{a}{a^2\plus{}2bc}\right) \ge \left( \sum a\right)^3.$
Thus, it suffices to prove that
$ \frac{a}{a^2\plus{}2bc}\plus{}\frac{b}{b^2\plus{}2ca}\plus{}\frac{c}{c^2\plus{}2ab} \le \frac{(a\plus{}b\plus{}c)^3}{3(ab\plus{}bc\plus{}ca)}.$
Using now the known inequalities
$ \frac{a}{a^2\plus{}2bc}\plus{}\frac{b}{b^2\plus{}2ca}\plus{}\frac{c}{c^2\plus{}2ab} \le \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca},$
we see that it is enough to check that
$ \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca} \le  \frac{(a\plus{}b\plus{}c)^3}{3(ab\plus{}bc\plus{}ca)},$
which is equivalent to the obvious inequality
$ 3(ab\plus{}bc\plus{}ca) \le (a\plus{}b\plus{}c)^2.$

[quote="can_hang2007"]$ \frac {a}{a^2 \plus{} 2bc} \plus{} \frac {b}{b^2 \plus{} 2ca} \plus{} \frac {c}{c^2 \plus{} 2ab} \le \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca},$[/quote]
See a Cauchy-Schwarz proof for this inequality here: http://can-hang2007.blogspot.com/2010/03/inequality-122-unknown-author.html

[quote="manlio"]Prove that for $ a,b,c$ positive reals

$ a\sqrt {a^2 \plus{} 2bc} \plus{} b\sqrt {b^2 \plus{} 2ac} \plus{} c\sqrt {c^2 \plus{} 2ba}$ ≥ $ \sqrt {3}(ab \plus{} bc \plus{} ca)$[/quote]
Another solution :
W.O.L.G $ a \ge b \ge c$
[b]Case 1 :[/b]
If $ a \ge 2(b \plus{} c)$ then
$ \begin{array}{l} LHS \ge {a^2} \plus{} {b^2} \plus{} {c^2} \equal{} \frac {{{a^2}}}{2} \plus{} \frac {{{a^2}}}{2} \plus{} {b^2} \plus{} {c^2} \ge \frac {{{a^2}}}{2} \plus{} 2{(b \plus{} c)^2} \plus{} {b^2} \plus{} {c^2} \\
\equal{} \frac {{{a^2}}}{4} \plus{} 3{b^2} \plus{} \frac {{{a^2}}}{4} \plus{} 3{c^2} \plus{} 4bc \ge \sqrt 3 (ab \plus{} bc \plus{} ca) \\
\end{array}$  ( By AM-GM ineq )
[b]Case 2 :[/b]
If   $ a < 2(b \plus{} c)$ then
The inequality equivalent to :
$ {a^4} \plus{} {b^4} \plus{} {c^4} \plus{} 2abc(a \plus{} b \plus{} c) \plus{} 2\sum {ab\sqrt {({a^2} \plus{} 2bc)({b^2} \plus{} 2ac)} } \ge 3{(ab \plus{} bc \plus{} ca)^2}$
We prove the ineq :
 $ \begin{array}{l} ab\sqrt {({a^2} \plus{} 2bc)({b^2} \plus{} 2ac)} \ge ab\left( {ab \plus{} c(a \plus{} b)} \right) \\
\Leftrightarrow {(a \minus{} b)^2}(2a \plus{} 2b \minus{} c) \ge 0 \\
\end{array}$  ( which is true )
Similarly, adding up these ineqs, we have :
$ LHS \ge {a^4} \plus{} {b^4} \plus{} {c^4} \plus{} 2abc(a \plus{} b \plus{} c) \plus{} 2\sum {ab(ab \plus{} c(a \plus{} b))} \ge 3\sum {{a^2}{b^2} \plus{} 6abc(a \plus{} b \plus{} c)} \equal{} 3{(ab \plus{} bc \plus{} ca)^2}$
We are done !

[quote="can_hang2007"][quote="manlio"]Prove that for $ a,b,c$ positive reals
$ a\sqrt {a^2 \plus{} 2bc} \plus{} b\sqrt {b^2 \plus{} 2ac} \plus{} c\sqrt {c^2 \plus{} 2ba}\ge \sqrt {3}(ab \plus{} bc \plus{} ca).$[/quote]
Using Holder's Inequality, we have
$ \left( \sum a\sqrt {a^2 \plus{} 2bc}\right)^2 \left( \sum \frac {a}{a^2 \plus{} 2bc}\right) \ge \left( \sum a\right)^3.$
Thus, it suffices to prove that
$ \frac {a}{a^2 \plus{} 2bc} \plus{} \frac {b}{b^2 \plus{} 2ca} \plus{} \frac {c}{c^2 \plus{} 2ab} \le \frac {(a \plus{} b \plus{} c)^3}{3(ab \plus{} bc \plus{} ca)}.$
Using now the known inequalities
$ \frac {a}{a^2 \plus{} 2bc} \plus{} \frac {b}{b^2 \plus{} 2ca} \plus{} \frac {c}{c^2 \plus{} 2ab} \le \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca},$
we see that it is enough to check that
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \le \frac {(a \plus{} b \plus{} c)^3}{3(ab \plus{} bc \plus{} ca)},$
which is equivalent to the obvious inequality
$ 3(ab \plus{} bc \plus{} ca) \le (a \plus{} b \plus{} c)^2.$[/quote]
I made a little mistake while typing my proof. Very sorry.

The line after "Thus, it suffices ..." should be
$ \frac {a}{a^2 \plus{} 2bc} \plus{} \frac {b}{b^2 \plus{} 2ca} \plus{} \frac {c}{c^2 \plus{} 2ab} \le \frac {(a \plus{} b \plus{} c)^3}{3\color{red}{(ab \plus{} bc \plus{} ca)^2}}.$

The line after "we see that it is..." should be
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \le \frac {(a \plus{} b \plus{} c)^3}{3\color{red}{(ab \plus{} bc \plus{} ca)^2}},$

[quote="manlio"]Prove that for $ a,b,c$ positive reals

$ a\sqrt {a^2 \plus{} 2bc} \plus{} b\sqrt {b^2 \plus{} 2ac} \plus{} c\sqrt {c^2 \plus{} 2ba}$ ≥ $ \sqrt {3}(ab \plus{} bc \plus{} ca)$[/quote]
The following one also holds
\[ a\sqrt {a^2 \plus{} bc} \plus{} b\sqrt {b^2 \plus{} ac} \plus{} c\sqrt {c^2 \plus{} ba}\geq \sqrt {2}(ab \plus{} bc \plus{} ca)\]
So I just want to know what's the best $ k$ such that
\[ a\sqrt {a^2 \plus{} kbc} \plus{} b\sqrt {b^2 \plus{} kac} \plus{} c\sqrt {c^2 \plus{} kba}\geq \sqrt {k\plus{}1}(ab \plus{} bc \plus{} ca)\]

[quote="can_hang2007"]$ \frac {a}{a^2 \plus{} 2bc} \plus{} \frac {b}{b^2 \plus{} 2ca} \plus{} \frac {c}{c^2 \plus{} 2ab} \le \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca},$[/quote]
In fact, the more general statement holds:

Let $ a,$ $ b,$ $ c$ be nonnegative real numbers, no two of which are zero. If $ k \ge 2,$ then
$ \frac{a}{a^2\plus{}kbc}\plus{}\frac{b}{b^2\plus{}kca}\plus{}\frac{c}{c^2\plus{}kab} \le \frac{\left(\sum a\right) \left[ (k\minus{}2)\sum a^2\plus{}(5\minus{}k)\sum ab\right]}{(k\plus{}1)\left( \sum ab\right)^2}.$

[quote="Tourish"]So I just want to know what's the best $ k$ such that
\[ a\sqrt {a^2 \plus{} kbc} \plus{} b\sqrt {b^2 \plus{} kac} \plus{} c\sqrt {c^2 \plus{} kba}\geq \sqrt {k \plus{} 1}(ab \plus{} bc \plus{} ca)\]
[/quote]
All value of $ k$ such that this inequality holds is $ 0 \le k \le 3.$ This problem has been posted by Vasc here: http://www.mathlinks.ro/viewtopic.php?t=33515

[quote="can_hang2007"]In fact, the more general statement holds:

Let $ a,$ $ b,$ $ c$ be nonnegative real numbers, no two of which are zero. If $ k \ge 2,$ then
$ \frac {a}{a^2 \plus{} kbc} \plus{} \frac {b}{b^2 \plus{} kca} \plus{} \frac {c}{c^2 \plus{} kab} \le \frac {\left(\sum a\right) \left[ (k \minus{} 2)\sum a^2 \plus{} (5 \minus{} k)\sum ab\right]}{(k \plus{} 1)\left( \sum ab\right)^2}.$[/quote]
See here: http://can-hang2007.blogspot.com/2010/03/inequality-136-v-q-b-can.html


quadratics
LaTeX
algebra
quadratic formula

The golden mean, $ g$, is a positive number that is one more than its reciprocal. Find $ g$ to the nearest hundredth.

We have $ g \equal{} 1 \plus{} \frac {1}{g} \implies g^2 \minus{} g \minus{} 1 \equal{} 0$.  By the Quadratic Formula, $ g \equal{} \frac {1 \pm \sqrt {5}}{2}$.  We look for the positive root, which is $ \frac {1 \plus{} \sqrt {5}}{2}$.  Since $ \sqrt {5}\approx 2.236$, we have $ g \approx \frac {3.236}{2} \approx \boxed{1.62}$.

LaTeX-ified for you. :)

For $ ax^2\plus{}bx\plus{}c\equal{}0$...

$ x\equal{}\frac{\minus{}b\plus{}\pm\sqrt{b^2\minus{}4ac}}{2a}$

So you just plug in the numbers that are in $ a$, $ b$, and $ c$ into the formula in order to find the roots of it.

You mean:

$ x\equal{}\frac{\minus{}b\pm\sqrt{b^{2}\minus{}4ac}}{2a}$

ergh yes izzy. I just checked and yes it is wat u syd.

Srry, about the latex thing, I was using a public computer (and still am) at the library, and they didn't have latex... :blush:


Prove that if $y^3x+1<x+y^3$ , then $x^3y+1<y+x^3$.

the given ineq is equevalent to $y^3(x-1)<(x-1)$
or $(y^3-1)(x-1)<0$ or $(y-1)(x-1)<0$  (1)

The ineq we want to prove is equevalent to $x^3(y-1)<(y-1)$ or
$(x^3-1)(y-1)<0$ or$(y-1)(x-1)<0$ which is true due to result (1)

We have the equavalence
$\displaystyle (y-1)(y^2+y+1)(x-1)<0$  <=>  $(x-1)(y-1)<0$
because $x^2+x+1=(x+\frac{1}{2})^2+\frac{3}{4}>0$ for all real x.[/hide]


trigonometry
trig identities
Law of Cosines

[img]http://img452.imageshack.us/img452/4879/triangle4ft.jpg[/img]

I have been trying to solve this triangle using trigonometry, but it is getting me no where. Can Anyone help me with this question, to save me fustration and a failing grade?

[hide="hint:"]do some angle chasing, let $\angle APQ=x$, then find all the other angles of the triangle in terms of x[/hide]

I'm doing this assignment too, and all i know is that triangle APQ is equal to triangle PQD. But i'm not sure what you mean by APQ = x and what to do with it.

[hide="a step i did"] $\overline{PQ}$ is the perpendicular bisector of $\overline{AD}$
[/hide]

ahh...very similar to an old ahsme problem from the 80s I believe...and yes angle chasing is definitely the way to go :)

[hide="another hint"]$\triangle PBD \sim \triangle DCQ$[/hide]

[hide="additional hint"]

use the law of cosines

let $x = AQ$ and let $3 - x = QC$

Since $QD = AQ$ let $QD = x$

You can then find $QD$ or $x$ by deriving an equation using the law of cosines where $x$ would be your unknown.

$AP$ would be solved in the same way. $PQ$ could then be found since $PD$ and $QD$ would then be known.

[/hide]

[hide="...solution"]Let $AP=a$, then $PD=a$, and $PB=3-a$, using $\triangle PBD \sim \triangle DCQ$ and proportion properties, $\frac{3-a}{2}=\frac{a}{DQ}=\frac{1}{CQ}=\frac{a+1}{DQ+CQ}=\frac{a+1}{3}$ then $a=AP=\frac{7}{5}$, $CQ=\frac{5}{4}$, $AQ=\frac{7}{4}$, and $DQ=\frac{7}{4}$
Using LoC, $PQ^2=\frac{49}{25}+\frac{49}{16}-\frac{49}{20}=\frac{49*21}{400}\implies PQ=\frac{7}{20}\sqrt{21}$[/hide]

P.S. what kind of class did you get this problem? this seems difficult for school


geometry
3D geometry
pyramid
algebra
polynomial
calculus
topology

Hi. I'm new here. I've really enjoyed reading past forum posts.

Why is the Fundamental Theorem of Algebra so, um, fundamental? And, if it's so great, why is there only a couple sentences about it in my alg. 2 book?

Is there something it's necessary for? Does it sort of "tie the knot" on the entire field of algebra?

Thanks.

The Fundamental Theorem of Algebra is of central importance to the study of polynomials. Unfortunately, it's not [i]really[/i] an "algebraic" theorem - meaning that it can't be proven with just algebra. It's actually a theorem of calculus. That's why it's not talked about much, or proven in your algebra 2 book.

But it's important and useful enough to be introduced before you can see exactly why it's true right now. You shouldn't be surprised by this when you think about it though. There are lots of examples of this happening in math. Do you really know why the area of a circle is pi*r2? Or why the volume of a pyramid is 1/3 its height times the area of its base? These are things that you're just told on faith at first; and then (hopefully) you learn why they're true later in your education.

But I suppose the FTA is important enough to deserve the name "fundamental". It allows you to tell when polynomials can be factored, and what kinds of roots they might have in certain situations.. and lots of other interesting things about polynomials. Whether it should be called a theorem of algebra or not is debatable. But the name is stuck now!

Without the FTA, you are really, really limited in what you can do with real or complex polynomials.  It also tells you something very deep about the real and complex numbers themselves.

I've seen it proved in ways that aren't exactly calculus -- I mean, I guess they are, but they're more topological.  You just have to know what a polynomial would look like in the complex plane over a circle or radius r about the origin.

It seems to me that all the theorems in algebraic number theory are essentially just corollaries of the Fundamental Theorem of Algebra. My topology prof showed us a neat topology proof of it.

Even the topological proofs are based on calculus in a subtle way. They rely on the fact that the real numbers don't have any "gaps" in them. That's why the FTA doesn't hold for, say, rational fields like Q or Q[i]. This seemingly "self-evident" fact about the real numbers is actually a result from analysis. That's why it's generally said that the FTA is a calculus (analysis) result. It can't be proven rigorously without appealing to this property of the real numbers.

Note: Don't worry about the rest of this until you take Galois Theory, but...
You can state a more general form of the FTA, over more general fields - which sort of skirts the issue of the real numbers not having any "gaps" head on. But this result still boils down to an analytic result in most of those other fields.

All of this is really pedantic, though, about where algebra ends and where analysis begins. The only important thing to realize is how important this fact about the real numbers not having any "gaps" is. We use it so often without even realizing it. Even Euclid used it in his first Theorem in The Elements. But it wasn't until calculus was made rigorous in the 1800s that we really understood why it was true.

The proof we did in class went something like this:

Suppose there's no z in C so that p(z)=a0+a1z+...+anzn=0. Define f:C to S1 by f(z)=p(z)/|p(z)|, which is continuous. (Is this what you mean when you say it involves calculus?) Let R be large enough so that |a0+a1z+...+an-1zn-1|<|anzn| for |z|>=R. Let g: [0,1] to S1 with g(t)=f(Re2 :pi: it).

He didn't finish the proof, but I think that since :pi: 1(S1)=Z, |z|=R is not homotopic to a point. But if f has no roots in C in |z|<=R, then we can homotope f along g to the origin. But the only way we can do this is if n=0 since the loops corresponding to n=0 is just a point in the first place. Other loops aren't contractible. Thus the only polynomials without roots in C are constant nonzero polynomials.

I'm don't know how much analysis is used in this proof (compared to, say, the proof with Liouville's Theorem, which makes FTA a pretty trivial corollary).

fundamental therorem of arithmetic (facotring numbers)
fundamental throrem of claclus (integrating rules)


topology
advanced fields
advanced fields unsolved

To me it's not completely unreasonable to ask when $ \widehat{A} \cong \widehat{B}$ implies $ A \cong B$ (where $ \widehat{}$ denotes the completion of the ring, or group or whatever), but in the specific case of the local ring of a point on a variety this seems to be a very rare occurance.  In other words, to have $ \widehat{\mathcal{O}_{X,P}} \cong \widehat{\mathcal{O}_{Y,Q}}$ (this is the definition of an analytic isomorphism of the points $ P$ and $ Q$) is a much weaker statement than $ \mathcal{O}_{X,P} \cong \mathcal{O}_{Y,Q}$.  A problem in Hartshorne asserts that if $ \mathcal{O}_{X,P} \cong \mathcal{O}_{Y,Q}$, then there are open sets $ P \in U \subseteq X$ and $ Q \in V \subseteq Y$ and an isomorphism of $ U$ to $ V$ sending $ P$ to $ Q$, so that $ X$ and $ Y$ are birational.  So another way to phrase it is, "when does analytic isomorphism imply birational equivalence?"  Analytic isomorphism means that the two points have very small isomorphic neighborhoods, while birational equivalence means they have isomorphic dense open sets.  First, a question: when, if ever, in the general case do isomorphic completions have isomorphic rings (or groups, or whatever)?  And secondly, what does this mean for varieties?

Well, the Cohen Structure Theorem gives us a horrifying starting point. This basically says that the completion of any (local) rings (containing the same field) of the same dimension are isomorphic.

For varieties, I think this basically means that locally (in the standard topology) all varieties look the same. Well, at least at non-singular points. But non-singular points are basically your whole variety, since they are the complement of a dense set.

I don't think I can contribute much more.

Right, it's sort of the algebro-geometric analogue of manifolds of the same dimension being locally isomorphic.  I was just wondering if there were any special cases where isomorphic completions have isomorphic rings.


How come on the boards for everyone (like getting started, mc, amc forums), i can edit my post, but in the AMC/AIME class, i can't?

I can edit my posts in Intermediate Counting's forum, but not in the AMC/AIME, is it a problem? or is it on purpose?


integration
trigonometry
real analysis
real analysis unsolved

Find $ I\equal{}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1\plus{}\cos^3x.\cot^3x}{\cos^3x\plus{}\cot^3x}dx$

[quote="thanhnam2902"][quote="kunny"]$ I \equal{} \int_{\frac {\pi}{4}}^{\frac {\pi}{3}} \frac {1 \plus{} \cos^3x.\cot^3x}{\cos^3x \plus{} \cot^3x}dx$

$ \equal{} \int_{\frac {\pi}{4}}^{\frac {\pi}{3}} \frac {\sin ^ 3 x \plus{} \cos ^ 6 x}{\cos ^ 4 x(\sin ^ 3 x \plus{} 1)}\ dx$

$ \equal{} \int_{\frac {1}{2}}^{\frac {1}{\sqrt {2}}} \frac {t^3 \plus{} (1 \minus{} t^2)^3}{(1 \minus{} t^2)^2(t^3 \plus{} 1)}\ dt\ [t \equal{} \sin x]$

$ \equal{} \minus{} \frac {1}{12}\int_{\frac {1}{2}}^{\frac {1}{\sqrt {2}}} \frac {1}{(t \plus{} 1)^3}\ dt \plus{} \frac {1}{8}\int_{\frac {1}{2}}^{\frac {1}{\sqrt {2}}} \frac {1}{(t \minus{} 1)^2}\ dt \plus{} \frac {13}{24}\int_{\frac {1}{2}}^{\frac {1}{\sqrt {2}}} \frac {1}{t^2 \minus{} t \plus{} 1}\ dt$[/quote][/quote]

[color=darkblue]Oh. Dear Kunny, you are wrong.

[b][u]First[/u][/b] $ t \equal{} \sin{x} \Longrightarrow \int_{\frac {\sqrt {2}}{2}}^{\frac {\sqrt {3}}{2}}$.
[b][u]Two[/u] [/b]$ t \equal{} \sin{x} \Longrightarrow dt \equal{} \cos{x}.dx$. You are wrong at $ I\equal{} \int_{\frac {\pi}{4}}^{\frac {\pi}{3}} \frac {\sin ^ 3 x \plus{} \cos ^ 6 x}{\cos ^ 4 x(\sin ^ 3 x \plus{} 1)}\ dx$.
[/color]

Sorry, my head seemed to have been tired such that I made silly mistakes.

kunny


A finite sequence of three-digit integers has the property that the tens and units digits of each terms are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with terms $ 247$, $ 475$, and $ 756$ and end with the term $ 824$. Let $ S$ be the sum of all the terms in the sequence. What is the largest prime number that always divides $ S$?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 37 \qquad \textbf{(E)}\ 43$

[hide="Solution"]Each digit appears once in the tens, the hundreds, and the units place. Therefore the sequence sum, $(100a+10b+c)+(100b+10c+d)+\cdots+(100y+10z+a)+(100z+10a+b)$, can be rewritten as $111(a)+111(b)+111(c) \cdots+111(z)$, or $111(a+b+c \cdots+z)$. Thus the sum is divisible by $111$. But that is not prime; $111=3 \cdot 37$, so the answer is $37$ or $\boxed{\text{D}}$.[/hide]

[hide]Each digit will appear in each place value once, so summing gives $100+10+1=111$ times the sum of all the digits. Since $111=3\times37$, $\boxed{(D)37}$ is the largest prime that divides S.[/hide]
[quote="Totally Zealous"]Just make a random sequence and divide it by the answers.[/quote]
[hide]Ooo, {111} would work, yay.[/hide]

[quote="chess64"]I just guessed 37 because I figured it would be divisible by 111 :D[/quote]
So I did something stupid when I took the test...  I figured that instead of missing a #3 or something like last year, I would warm up by starting on 24 and 25.

It was okay, but I was still on 25 when the teacher called halfway at 37 minutes :!:
Talk about a warm up.
Anyway, I managed to finish all the problems in time but it was tough... 2 minutes left and I hadn't checked yet any answers, and this was the last one (along with 23) left to do.

I was so happy that the AMC did that - two problems with a glance - at solution - that actually gave me to time to correct 2 other answer I would have missed.

Sorry for the narrative.
Basically, you see 37 as a choice, and think, that's gotta be it. No other way.
For fun, I consctructed 123,234,345,456,671,712 on my calculator summed up to a multiple of 111 (just as a self doubt check)

Just try the first answer to find the solution from the given sequence
 and 37 seems to be correct


ratio
factorial
calculus
integration
trigonometry
geometric series
calculus computations

Hi everyone,

When you are trying to determine whether or not an alternating series (-1^n) converges, when can you ignore the "-1^n" part?

For example,

When you are doing the ratio test you can leave the -1^n out. Can you do the same thing when using the test for divergence?

For example,

the series (-1^n)2n^2/4n^2 does this diverge because the limit is equal to 1/2?

Thank you very much

If you know that the limit converges to something (in this case 1/2), then the series converges.  I don't think you can ignore the $ (\minus{}1)^{n}$ because that determines the signs, which help to increase/decrease your limit.

chocolatelover:

In your post, you haven't clearly distinguished between a series and a sequence. A series is a sum. You really shouldn't ever say "the series $ a_n$" for some formula. If you really mean the series, then you should say "the series $ \sum a_n$" or at least include the word "sum" in what you say. If you really mean the sequence, then you should use the word "sequence." Because of that imprecise use of language on your part, it's quite possible that JRav responded to a question that you didn't mean to ask.

So if you're faced with the series $ \sum_n ( \minus{} 1)^nb_n,$ how do you proceed?

I would take it up in this order:

1. Do the terms go to zero? If the answer is "no", then the series diverges. For instance, $ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^n2n^2}{4n^2}$ diverges because the terms don't go to zero. It is reasonable in some cases to record the answer here as "hard to tell" and move on - that mostly applies to series whose terms are full of factorials and factorial-like objects which are hard to balance at first glance. But yes, to answer your original question, we can ignore an alternating sign if all we're asking is whether the terms go to zero. That $ ( \minus{} 1)^n$ won't change whether or not the terms go to zero.

2. Assuming the terms go to zero (or it's hard to tell), we next address the issue of whether the series converges absolutely. That is, does $ \sum |b_n|$ converge? It is at this stage that we ignore the $ ( \minus{} 1)^n.$ We decide absolute convergence - or the convergence of a positive term series, which is the same thing - by some comparison test. Mainstream calculus texts include five comparison tests: simple (that is, one-sided) comparison; limit comparison; the integral test; the ratio test; the root test. The latter two, ratio and root, [i]are[/i] comparisons - they're comparisons to geometric series. They may return the result "inconclusive" but all that means is that the series in question doesn't usefully compare to a geometric series (and we won't get that result very often if we dom't even try to apply these tests in unsuitable or unpromising cases.) We ignore $ (\minus{}1)^n$ for the purposes of the root and ratio tests because they're comparison tests and we're testing for absolute convergence.

Browsing around this forum, you may encounter the "condensation test." That would be a sixth comparison test; its utility is roughly similar to the integral test.

3. Finally, if we don't have absolute convergence but the terms do go to zero, then we can use the alternating series test, or more generally, Dirichlet's test or other results that follow from summation by parts. So $ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^n}n$ converges conditionally by the alternating series test. And $ \sum_{n \equal{} 1}^{\infty}\frac {\sin n}{n}$ converges conditionally by one of those fancier generalizations of the AST.


geometry
geometric transformation
rotation
3D geometry
sphere

If the voyager is between planet saturn,and Jupiter how would an astronomy ascertain the voyager's present direction wether such is in the nort[b],east[/b],south[b],and north?

                                                                   Respectfully,
                                                                   A.Eddington[/b]

Can you restate the question?  There are no cardinal directions in space, unless you mean on one of those planets themselves.  With the difference in both their rotation and magnetic field, there is no guarantee that any of thier cardinal directions have a single star or constellation as a celestial marker.

As for Earth is only really one "marker" for a cardinal direction, Polaris for North.  Even this star is not a completely accurate marker since it is moving slowly across the sky and does not exactly coincide with true magnetic North.  Then of course I guess you could say Ursa Minor is the marker for North.  Again the South marker on Earth is not really accurate either, only the faint $ \sigma$ Octantis.  Though Crux in the Southern Hemisphere does point in the South direction.
Wikipedia Picture
[img]http://upload.wikimedia.org/wikipedia/en/c/cd/Pole01-eng.jpg[/img]
As for the other two cardinal directions, there is not single direction marker since it would vary day to day, hour to hour with the rotation of the celestial sphere.
While observing only mount that really takes the celestial poles into direct consideration is the equatorial mount, which needs to be pointed at the celestial pole to function.

[b]Links[/b]
[url=http://en.wikipedia.org/wiki/North_Star]Wikipedia - North Star[\url]
[url=http://en.wikipedia.org/wiki/South_Star]Wikipedia - South Star[\url]
[url=http://en.wikipedia.org/wiki/Pole_star]Wikipedia - Pole Star[\url]
[url=http://en.wikipedia.org/wiki/Celestial_sphere]Celestial Sphere[\url].


limit
trigonometry
calculus
calculus computations

Compute $\lim_{n \to \infty}\cos(\pi e (2n+1)!)$

Use the formula $e=2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+\frac{\theta_{n}}{n!n}$ and try to prove that
$\lim_{n\to\infty}\theta_{n}=1$


[quote="Anonymous52111"]I need your step-by-step here and need answer, of course. Thanks.

If 6a - 4 = - 2, what is the value of 12a - 4?[/quote]
Please don't post questions in two different forums. I know you're new here so it's okay. Or if you thought that you put itin the high school basics by accident, just ask a mod to move it.

The answer is $0$.

Because you need to add 4 to both sides making the equation:

$6a=2$ .

then, you need to divide both sides by 6 making \[a=\frac{2}{6}=\frac{1}{3}\]  so we know that \[a=\frac{1}{3}\]

Now, plug the \[\frac{1}{3}\] into the other equation$12a-4$ .

Hence, \[\frac{12}{1}\cdot\frac{1}{3}=4\] and $4-4=\boxed{0}$

Does that help?

[quote="Anonymous52111"]I need your step-by-step here and need answer, of course. Thanks.

If 6a - 4 = - 2, what is the value of 12a - 4?[/quote]

Sorry about that. I will do it next time. By the way, yes, it helps. Thanks.

ok...good...as long as it helps... :lol:

[quote="DaReaper"]The answer is $0$.

Because you need to add 4 to both sides making the equation:

$6a=2$ .

then, you need to divide both sides by 6 making
\[a=\frac{2}{6}=\frac{1}{3}\]
so we know that
\[a=\frac{1}{3}\]
Now, plug the
\[\frac{1}{3}\]
into the other equation$12a-4$ .

Hence,
\[\frac{12}{1}\cdot\frac{1}{3}=4 \]
and $4-4=\boxed{0}$

Does that help?[/quote]
There is a way to do it with less fractions, but it may take the same amount of time.
[hide] Take the original equation: 6a-4=-2.  Add 4 to both sides to get 6a=2.  Now, add the original equation and the new equation.  We have that 12a-4=0. [/hide]

[quote="Anonymous52111"]I need your step-by-step here and need answer, of course. Thanks.

If 6a - 4 = - 2, what is the value of 12a - 4?[/quote]
[hide]$6a-4=-2$

$6a=2$

$12a=4$

$12a-4=\boxed{0}$[/hide]


Aldebaran has $\$$5.00 in quarters, nickels, dimes, and pennies. If Aldebaran has 10 quarters, 10 nickels, and 10 pennies, how many dimes does Aldebaran have?

[hide]Well, 10 quarters means 2.50, 10 nickels means 0.50, and 10 pennies, is, of course, 0.10. Adding these up, they total to 3.10. Well, since the total is 5.00, the dimes add up to 1.90, so there are [b]19 dimes[/b].[/hide]

[hide="The no decimal way!"]You add up the number of cents he has not including the dimes and get 250+10+50=310. Subtract that from 500 to get 190. [b][i][u]19[/u][/i][/b][/hide]


Gerald Carter and Marysia Wokawski bought a property by making semi-annual
payments of 2,500 dollars for seven years. If the first payment is due on the date
of purchase and interest is 9 percent compounded quarterly, what is the purchase
price of the property?

Do I use Quarterly = P (1 + r/4)^4 = (quarterly compounding)?

P is the principal (the initial amount you borrow or deposit)

r is the annual rate of interest (percentage)

Am I close?

Close.

$P(1+\frac{r}{4})^{4}$ = the amount of money in the bank after three months. Quarterly compounding is $P((1+\frac{r}{4})^{4}-1)$.

Are you saying that I can find the answer using the formula as you suggested?

Pretty much, unless I made a typo or misremembered something.


function
inequalities
limit
trigonometry
calculus
calculus computations

please help me with these questions I have been struggling with

Find a nonzero value for the constant k that makes

F(x) = {  tan kx/ x,     x<0
            3x+2k^2       X=>0
continuous at x=0

-----------------

Find lim         sin(pi x)/ x-1
      X ->2

------------------

sketch the grpahs of y = 1 - x^2, y = cosx, and y = f(x), where f is any continuous function that satisfies the inequalities
1-x^2 <= f(x) <= cos x

for all x in the interval (-pi/2, pi/2). What can you say about the limit of f(x) as x -> 0?

[quote="furia"]please help me with these questions I have been struggling with

Find a nonzero value for the constant k that makes

F(x) = {  tan kx/ x,     x<0
            3x+2k^2       X=>0
continuous at x=0 [/quote]

$\lim_{x < 0, x \to 0}F(x) = \lim_{x<0, x \to 0}\frac{\tan kx}{x}= k$

$\lim_{x \geq 0, x \to 0}F(x) = 2k^{2}$.

Hence, $F$ is continuous at $0$ if and only if $k = 2k^{2}\Leftrightarrow k=0 \vee k = \frac{1}{2}$

[quote="furia"]
Find lim         sin(pi x)/ x-1
      X ->2    [/quote]

$x \to \frac{\sin \pi x}{x-1}$ is continuous at $2$, hence

$\lim_{x \to 2}\frac{ \sin \pi x}{x-1}= \frac{ \sin 2 \pi}{2-1}= 0$

[quote="furia"]
sketch the grpahs of y = 1 - x^2, y = cosx, and y = f(x), where f is any continuous function that satisfies the inequalities
1-x^2 <= f(x) <= cos x

for all x in the interval (-pi/2, pi/2). What can you say about the limit of f(x) as x -> 0?[/quote]

$\forall \epsilon > 0: \exists \delta > 0: (|x| < \delta \Rightarrow |1-x^{2}-1 |, |\cos x-1|< \epsilon)$

Because of $1-x^{2}-1 \leq f(x)-1 \leq \cos x-1$, either $|f(x)-1| \leq |1-x^{2}-1 |$ or $|f(x)-1| \leq | \cos x-1|$. In both cases, $|x| < \delta \Rightarrow |f(x)-1| < \epsilon$


At what exact time must I be under/over thirteen for me to submit the correct forms? I am turning 13 in 13 days.

Federal Law says that we if collect "identifying information" from you (such as your name, address, etc.) and you are under 13, then we must get a permission form.  So it is based on the time that you register.


Find three different polynomials $P(x)$ with real coefficients such that $P\left(x^2 + 1\right) = P(x)^2 + 1$ for all real $x$.

eum... the polynomials $x, x^2+1, (x^2+1)^2+1, ((x^2+1)^2+1)^2+1$... etc all fulfil that condition?

Yes, they do... The Belgian team didn't solve this problem! :(

Could you post a full solution Peter VDD ?This problem is also Serbia TST 2005.

So how many questions did they not copy from serbia 2005? :?

Anyway silouan, what is there to solve? just fill them in... ;)


find all the continuous function f: R -> R so that:
               f(x+y). (1 - f(x).f(y)) = f(x)+f(y)

[quote="Rust"]Solution is $f(x)=tg(ax)$. But continiosly function give only a=0 or $f(x)\equiv 0$.[/quote]
Rust, for every member of mathlink  they so can post as. Please don't post as. Post concrete!

I think it is obviosly. Let $f(x)=tg(ax)$, then $f(2x)=tg(2ax)$, and $f(nx)=tg(nax)$ for $\frac{x}{2^{k}}$ we get $f(\frac{x}{2^{k}})=tg\frac{bx}{2^{k}}\to a=b$. Because f(x) is continiosly $f(x)=tg(ax) \ \forall x$, therefore a=0.

i dont understand, so  b =........ i dont know. i hope you can post again that it is clear. is that ok?/

For all x we can choose a=a(x), suth that $f(x)=tg(ax)$.
If $a(x)=b(y)=a$, then from $f(x+y)(1-f(x)f(y))=f(x)+f(y)$ we get $a(x+y)=a$. It give $a(nx)=a(x)$, therefore we can choose $a(\frac{x}{n})=a(x)$. Because f(x) is continiosly $a(x)=const=a$ and a=0.


find all functions $f: C\to C$ such that $f(0)=0$ and for all $z\in C$ and $w \in 0,1,i$ we have:
             $\left| f(z)-f(w) \right|=\left| z-w\right|$

(and as everyone knows $i^{2}=-1$ :) )

what i have found:

[hide]if we put $w=0$ then we have for all complex $z$: $\left|f(z)\right|=\left|z\right|$

now if we replace $z$ by $zw$ we will have:
$\left|f(zw)-f(w)\right|=\left|zw-w\right|=\left|f(zw-w)\right| \Longrightarrow \left|f(w(z-1))\right|=\left|f(w)\right|\left|f(z-1)\right|$

which says that for all $z \in C$ and $w\in 0,1,i$ we have:
$\left|f(zw)\right|=\left|f(z)\right|\left|f(w)\right|$[/hide]
Can this help?!!!!!!!!!!!!!!!!!!

Let $f(a+bi) = x+yi$.

Well, then we have:
$a^{2}+b^{2}= x^{2}+y^{2}$ ($w = 0$),
$(a-1)^{2}+b^{2}= (x-1)^{2}+y^{2}$ ($w = 1$),
and $a^{2}+(b-1)^{2}= x^{2}+(y-1)^{2}$ ($w = i$).

Subtracting EQ 2 from EQ 1, we have $2a-1 = 2x-1$.
Therefore, $a = x$.

Similarly, subtracting EQ 3 from EQ 1, we have $2b-1 = 2y-1$,
and thereofre $b = y$.

Thus, the only function to satisfy the given criteria is $f(z) = z$.

PTynan89: that's what I though too, but we have $|f(z)-f(w)|$, and not $|f(z)-w|$ in the formulas.
The function $f(z)=az$ with $|a|=1$ works, too.

This is really a geometry problem; it has nothing to do with complex analysis. Given the distances between an unknown point in the plane and three fixed points which aren't collinear, we can triangulate and find the unknown point. Once we know $f(0),f(1),f(i)$, we know $f(z)$ for all $z$.
Applying the relation to $z=0,1,i$, the triangle with those vertices is sent to another with the same side lengths, which is congruent to the original. With that congruence, we find that $f$ must be an isometry of the plane. It fixes the origin as well.
All isometries $f$ fixing the origin are of the form $f(z)=\alpha z$ or $f(z)=\alpha \overline{z}$ with $|\alpha|=1$.


Let the $ABCA'B'C'$ be a regular prism. The points $M$ and $N$ are the midpoints of the sides $BB'$, respectively $BC$, and the angle between the lines $AB'$ and $BC'$ is of $60^\circ$. Let $O$ and $P$ be the intersection of the lines $A'C$ and $AC'$, with respectively $B'C$ and $C'N$.

a) Prove that $AC' \perp (OPM)$;
b) Find the measure of the angle between the line $AP$ and the plane $(OPM)$.

[i]Mircea Fianu[/i]

does anyone know how to do this problem?

i was bad at geometry (solid or otherwise) back in 8th grade and this hasn't changed much


calculus
integration
real analysis
real analysis unsolved

Is it a joke !!! That's trivial,  with only increasing non constant $f$  !!:D
 :cool:

Indeed, it was a joke. Well, i've edited the problem. :)


Fie $a,b,c$ laturi ale unui triunghi astfel incat $2\sum{ab}-\sum{a^{2}}=4$.Sa se arate ca:

$9(a+b+c)\geq 10(ab+bc+ca)$

Dupa cateva substitutii de baza imi da ca inegalitatea nu e buna... :P Cand a,b,c is egale e adevarat sau am gresit eu la calcule?

Cred ca ai gresit undeva coeficientii, pt ca daca luam $a=b=c=\frac{2}{\sqrt{3}}$, inegalitatea din enunt devine echivalenta cu $9\sqrt{3}\geq 20$.Fals.Totusi, cred ca e doar o chestie de coeficienti..
Uite o problemuta similara(inspirata de asta):
Daca $a,b,c$ sunt lungimile laturilor unui triunghi si $2\cdot \sum ab-\sum a^{2}=12$, sa se arate ca
\[a+b+c+18\leq 2(ab+bc+ca) \]

a, da. scuze. am calculat eu prost cand am bagat un schur, si n-am mai verificat.

[quote="andyciup"]
Daca $a,b,c$ sunt lungimile laturilor unui triunghi si $2\cdot \sum ab-\sum a^{2}=12$, sa se arate ca
\[a+b+c+18\leq 2(ab+bc+ca) \]
[/quote]

Nu raspunde nimeni la inegalitatea asta?

raspund eu.

facem substitutile $a=x+y$, $b=y+z$, $c=z+x$. si conditia aia devine:

$\sum{xy}=4$, si ne ramane sa demonstram ca $4\sum{x^{2}}+7\sum{xy}\geq 4\sum{x}$ , cu $x,y,z$ pozitive

se ridica la patrat + omogenizam in dreapta dupaia (inmultind cu $\sum{xy}$), si obtinem o inegalitate in gradu 4, tre sa fie un muirhead, schur pacolo.

Not nice.. :P solutia mea e muult mai simpla..nu trebe` calcule decat foarte putin..Mai incearca :P

A,am si editat problema..scrisesem aiurea prima data.

nu prea am cum sa redactez acum o solutie completa ca nu prea am unde sa scriu (nu-s la birou meu) & no foi, dar cumva scrii conditia aia cu : $2\sum{ab}-\sum{a^{2}}= 12 \Longleftrightarrow r^{2}+4Rr =3$, si ce trebuie sa demonstram e echivalent cu $p^{2}\geq p+6$ ?

Eu nu am facut-o asa..
Mai mult,inegalitatea are loc si daca $a,b,c$ nu sunt laturi de triunghi :)


how many ways can you tile a $2$ by $n$ board with $1$ by $2$ and/or $2$ by $2$ tiles

n.b. a recurrence relation solution would be preferable...

Let the number of ways for $n$ be $f(n)$. We see that $f(1)=1, f(2)=3$. Now take $n>2$. Consider the right end; if the rightmost column is tiled with a 1x2 tile, there are $f(n-1)$ ways of covering the rest; if the two rightmost columns are tiled with a 2x2 tiles, there are $f(n-2)$ ways of covering the rest; if the two rightmost columns are tiled with two horizontal 1x2 tiles, there are $f(n-2)$ ways of covering the rest, so the solutions are given by $f(n)=f(n-1)+2f(n-2), f(1)=1, f(2)=3$. Of course, if you meant that the 1x2 tiles are only vertical, the answer is the Fibonacci numbers.


The first digit of a 6-digit number is 1. If the 1 is shifted to the other end, the new 6-digit number is 3 times bigger than the old one. What was the old number?

It's simply a matter of what I guess you can call "working backwards".  The first number: 1ABCDE.  Second number: ABCDE1.

$3E \equiv 1 \pmod{10} \rightarrow E = 7$

$3E = 21 \rightarrow 3D + 2\equiv 7 \pmod{10} \rightarrow D = 5$

$3D = 15 \rightarrow 3C + 1\equiv 5 \pmod{10} \rightarrow C = 8$

$3C = 24 \rightarrow 3B + 2\equiv 8 \pmod{10} \rightarrow B = 2$

$3A\equiv 2 \pmod{10} \rightarrow A = 4$

So the first number was 142857:

$142857 \cdot 3 = 428571$

Ill wait for other solutions. I have an algebraic solution for this but I guess Ill wait for the others.

[quote="gunes"]1a*3=a1
$3*10^5-1=7a$
$a=42857$
the number is 142857[/quote]

Could you explain why ?  :?


Determine the maximum value attained by:$\frac{x^{4}-x^{2}}{x^{6}+2x^{3}-1}$ over the real numbers $x>1$

[color=darkblue]$x>1\Longrightarrow x-\frac{1}{x}>0$ and $f(x)\equiv \frac{x^{4}-x^{2}}{x^{6}+2x^{3}-1}>0$. Thus,
$f(x)$ - max $\Longleftrightarrow$ $\frac{1}{f(x)}$ - min $\Longleftrightarrow$ $\frac{x^{4}+x^{2}+1}{x^{2}}+\frac{2x}{x^{2}-1}$ - min $\Longleftrightarrow$ $1+\left(x^{2}+\frac{1}{x^{2}}\right)+\frac{2}{x-\frac{1}{x}}$ - min $\Longleftrightarrow$ $1+\left(x-\frac{1}{x}\right)^{2}+1+1+\frac{1}{x-\frac{1}{x}}+\frac{1}{x-\frac{1}{x}}$ - min. But $1\cdot \left(x-\frac{1}{x}\right)^{2}\cdot 1\cdot 1\cdot \frac{1}{x-\frac{1}{x}}\cdot\frac{1}{x-\frac{1}{x}}=1$. Therefore, $1=\left(x-\frac{1}{x}\right)^{2}=\frac{1}{x-\frac{1}{x}}$ $\Longleftrightarrow$ $x^{2}-x-1=0$, i.e. $x_{\mathrm{max}}=\frac{1+\sqrt 5}{2}$ and $\mathrm{max}\ f(x)=f\left(x_{\mathrm{max}}\right)=\frac{1}{6}\ .$

[b]Remark.[/b] $n\in N,\ n\ge 2$, $a_{k}>0$ for any $k\in \overline{1,n}$  and $\prod_{k=1}^{n}a_{k}=a$ (constant).
Then the sum $\sum_{k=1}^{n}a_{k}$ is minimum iff for any $k\in \overline{1,n}$, $a_{k}=\sqrt [n] a\ .$[/color]

Let $t=x-\frac{1}{x}\ (t>1),$
Rewrite the given equation as $\frac{t}{t^{3}+3t+2}.$

Since $t^{3}\geq 3t-2\Longleftrightarrow (t-1)^{2}(t+2)\geq 0\ (t>0),$ the given equation is minimized when $t=1,$ yielding the desired maximum value is $\frac{1}{6}$.