topology geometry trigonometry calculus Gauss probability function

In your opinion, what would be the best order to study math subjects? Also, if possible, include more advanced course (abstract algebra, topology,...)

Is this at the high school or college level?  At high school, I would think geometry, algebra, trig is an okay path, although I guess the way most schools do it is okay.  In college, take Calc I, Calc II, Calc III, Linear Algebra and if you want Differential Equations.  Then you'll have a solid background, from there, you can pretty much go in whichever order you want.  I'm personally taking a Real/Complex Analysis path, and then going back to Abstract Algebra in a year or two.  Then you can go on to topology, differential geometry, or whatever else interests you.

Theoretically, I suggest taking Linear Algebra before Multivariable Calculus or Differential Equations, and Topology before Real Analysis. But in college it really depends more on what professor is teaching the course. If a good professor is teaching a course in a certain semester, and possibly not later and you have the necessary prerequisites to do well in the course, it's a good idea to take it then regardless of any theoretical sequence you have in mind.

I agree with gauss that linear algebra should be taken as soon as possible, but not that topology should come before analysis.

i tend to think that topology is not that important -- you pick up most of the necessary material in real analysis.  same goes for differential equations -- while this is important to engineers, a lower division course on ODEs is almost completely computational and not particularly useful to math majors.  i actually have never taken a course on ODEs despite taking courses in dynamical systems, PDE, graduate PDE, etc.

linear algebra and real analysis are probably the two most important introductory courses for a math major and should be taken as soon as possible.

after you have taken them, the most important courses are abstract algebra and complex analysis (for example, if you have taken these, you can pass your qualifying exams at a lot of schools).

the most important upper division courses are probably a course in measure theory and an advanced course in algebra.

from here, there's a lot of flexibility in other courses to take.  PDE, topology, discrete math, fourier analysis, probability, differential geometry, algebraic geometry -- it's really up to you!

I took multivariable calculus before linear algebra (both proof-based).  I don't think that hurt me at all.  I also took an ODE course but honestly, I don't think it taught me anything that I couldn't do without.  It was mostly an engineering course so there were very few proofs but if you need another course to take, I guess it's okay to take.

Many colleges require pre-requisites for upper level courses, so once you're done with calc III, linear algebra, and sometimes differential equations, you're free to take whatever you want.

I have one thing to add about the order of Real Analysis and Topology.

My first analysis professor was horrible, and he jammed topology squarely in the middle of his first class of the 3 part sequence.  I can safely say I didn't bring very much away from it, and we hardly applied anything from it in that sequence.
Then I took Topology from Alexander Arhangelskii, and man, Analysis became a lot clearer.  (In fact, all of my math got better.)

My theory is that working my way through proving the lattices of implications between various topological properties helped me get a grip on other subjects.  So, I mean to say that Topology is a great way to refine students' mathematical thought, so I would recommend a thorough course after Real Analysis.

However, after calculus and before Analysis, I would recommend that students be exposed to:
--equivalent definitions of "continuous" for real valued functions on R
--convergence, uniform convergence and Cauchy sequences in metric spaces
--open sets, closed sets and the topology of R
--compactness and the nature of compact sets in R

Especially equivalent forms of continuity.  The epsilon/delta definition of equivalence is so often used, and I think is a common source of confusion for those who first meet it.


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac{1}{(a\plus{}2b)^2}\plus{}\frac{1}{(b\plus{}2c)^2}\plus{}\frac{1}{(c\plus{}2a)^2} \ge \frac{1}{ab\plus{}bc\plus{}ca}\]
:)

P/s: There is a very nice solution by Cauchy Schwarz for it. :)

Sorry ,i can find the proof that be used CS,But i found an other ideal
Let $ a\plus{}2b\equal{}X,b\plus{}2c\equal{}Y,c\plus{}2a\equal{}Z$,And $ \sum X\equal{}1$,the ineq is
$ \sum \frac{1}{X^2}\geq \frac{21}{4\sum X^2\minus{}\sum {XY}}$
The last we can easyly prove by pqr,Who can finish it ?

[quote="Allnames"]Sorry ,i can find the proof that be used CS,But i found an other ideal
Let $ a \plus{} 2b \equal{} X,b \plus{} 2c \equal{} Y,c \plus{} 2a \equal{} Z$,And $ \sum X \equal{} 1$,the ineq is
$ \sum \frac {1}{X^2}\geq \frac {21}{4\sum X^2 \minus{} \sum {XY}}$
The last we can easyly prove by pqr,Who can finish it ?[/quote]
I dont think it can be proved by $ pqr$ technique, Thien. Since it is not symmetric for $ X,Y,Z$ (take a look at the way we denoted $ X,Y,Z$) and it is hard to forces $ q$ and $ r$ after we fixed $ p$. But after using Cauchy Schwarz Inequality, $ pqr$ technique works well. :)
By the way, I think there is something wrong in your calculation after we changed the inequality into $ X,Y,Z$.

[quote="Allnames"]Sorry ,i can find the proof that be used CS,But i found an other ideal
Let $ a \plus{} 2b \equal{} X,b \plus{} 2c \equal{} Y,c \plus{} 2a \equal{} Z$,And $ \sum X \equal{} 1$,the ineq is
$ \sum \frac {1}{X^2}\geq \frac {27}{4\sum X^2 \minus{} \sum {XY}}$
The last we can easyly prove by pqr,Who can finish it ?[/quote]
Sorry,i have some mistake in my calculation ,with Can_hang 2007 's help i have edited that
$ \sum \frac {1}{X^2}\geq \frac {27}{ \minus{} 2\sum X^2 \plus{} 5\sum {XY}}$
Any one can continue?

[quote="shfdfzhjj"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {1}{(a \plus{} 2b)^2} \plus{} \frac {1}{(b \plus{} 2c)^2} \plus{} \frac {1}{(c \plus{} 2a)^2} \ge \frac {1}{ab \plus{} bc \plus{} ca}
\]
:)

P/s: There is a very nice solution by Cauchy Schwarz for it. :)[/quote]

Yes,and the follwoing is also true. :)[/quote]
Nice :)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {1}{(a \plus{} 2b)^2} \plus{} \frac {1}{(b \plus{} 2c)^2} \plus{} \frac {1}{(c \plus{} 2a)^2} \ge \frac {1}{ab \plus{} bc \plus{} ca}
\]
:)

P/s: There is a very nice solution by Cauchy Schwarz for it. :)[/quote]
Here is the Cauchy Schwarz solution for this :)
We will consider 2 cases
Case 1. If $ 4(ab\plus{}bc\plus{}ca) \ge a^2\plus{}b^2\plus{}c^2$, then by Cauchy Schwarz Inequality, we have
\[ \left( \sum \frac{1}{(a\plus{}2b)^2}\right)\left( \sum (a\plus{}2b)^2(a\plus{}2c)^2\right) \ge 9\left( \sum a\right)^2\]
It suffices to prove that
\[ 9\left( \sum a\right)^2 \left( \sum ab\right) \ge \sum (a\plus{}2b)^2(a\plus{}2c)^2\]
\[ \Leftrightarrow 9\left( \sum a\right)^2 \left( \sum ab\right) \ge \left( \sum a\right)^4 \plus{}18\left( \sum ab\right)^2\]
\[ \left( \sum a^2\minus{}\sum ab\right) \left( 4\sum ab \minus{}\sum a^2\right) \ge 0\]
which is true.
Case 2. If $ a^2\plus{}b^2\plus{}c^2 > 4(ab\plus{}bc\plus{}ca)$, then assume that $ a\equal{} \max\{a,b,c\}$. We will show that $ a \ge 2(b\plus{}c)$, indeed, if $ a<2(b\plus{}c)$, then we have
\[ a^2\plus{}b^2\plus{}c^2 \minus{}4(ab\plus{}bc\plus{}ca) \equal{}a(a\minus{}2b\minus{}2c) \plus{}b(b\minus{}a) \plus{}c(c\minus{}a) \minus{}4bc \le 0\]
which is contradict.
Now, with $ a \ge 2(b\plus{}c)$, using AM-GM Inequality, we have
\[ \frac{1}{(a\plus{}2b)^2}\plus{}\frac{1}{(b\plus{}2c)^2} \ge \frac{2}{(a\plus{}2b)(b\plus{}2c)}\]
It suffices to prove that
\[ \frac{2}{(a\plus{}2b)(b\plus{}2c)} \ge \frac{1}{ab\plus{}bc\plus{}ca}\]
\[ \Leftrightarrow b(a\minus{}2b\minus{}2c) \ge 0\]
which is obviouly true.
We have done.
Equality holds if and only if $ a\equal{}b\equal{}c$.
:)

Let $ a \plus{} 2b \equal{} x \ ;\ b \plus{} 2c \equal{} y \ ;\ c \plus{} 2a \equal{} z$

$ \equal{} > a \equal{} x \minus{} 2y \plus{} 4z \ ;\ b \equal{} ....$

$ Ine < \equal{} > \sum \frac {1}{27x^2} \ge \frac {1}{5(xy \plus{} yz \plus{} zx) \minus{} 2(x^2 \plus{} y^2 \plus{} z^2)} (1)$

$ (\ and \ 5(xy \plus{} yz \plus{} zx)\ > \ 2(x^2 \plus{} y^2 \plus{} z^2)\ )$

$ < \equal{} > \sum (5(xy \plus{} yz \plus{} zx) \minus{} 2(x^2 \plus{} y^2 \plus{} z^2)).\frac {1}{x^2} \ge 27$

$ < \equal{} > \sum(x \minus{} y)^2.(\frac {5}{yx} \plus{} \frac {5(xy \plus{} yz \plus{} zx)}{x^2y^2} \minus{} \frac {2(x \plus{} y)^2}{x^2y^2}) \ge 0$

suppose $ x \ge y \ge c \equal{}> S(y) ; S(x) \ge 0$ and $ S(y)\plus{}S(z) >0$

DONE
 $ " \equal{} " \ < \equal{} > \ x \equal{} y \equal{} z \ < \equal{} > \ a \equal{} b \equal{} c$

[quote="can_hang2007"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \frac {1}{(a \plus{} 2b)^2} \plus{} \frac {1}{(b \plus{} 2c)^2} \plus{} \frac {1}{(c \plus{} 2a)^2} \ge \frac {1}{ab \plus{} bc \plus{} ca}
\]
:)

P/s: There is a very nice solution by Cauchy Schwarz for it. :)[/quote]
Here is the Cauchy Schwarz solution for this :)
We will consider 2 cases
Case 1. If $ 4(ab \plus{} bc \plus{} ca) \ge a^2 \plus{} b^2 \plus{} c^2$, then by Cauchy Schwarz Inequality, we have
\[ \left( \sum \frac {1}{(a \plus{} 2b)^2}\right)\left( \sum (a \plus{} 2b)^2(a \plus{} 2c)^2\right) \ge 9\left( \sum a\right)^2
\]
It suffices to prove that
\[ 9\left( \sum a\right)^2 \left( \sum ab\right) \ge \sum (a \plus{} 2b)^2(a \plus{} 2c)^2
\]

\[ \Leftrightarrow 9\left( \sum a\right)^2 \left( \sum ab\right) \ge \left( \sum a\right)^4 \plus{} 18\left( \sum ab\right)^2
\]

\[ \left( \sum a^2 \minus{} \sum ab\right) \left( 4\sum ab \minus{} \sum a^2\right) \ge 0
\]
which is true.
Case 2. If $ a^2 \plus{} b^2 \plus{} c^2 > 4(ab \plus{} bc \plus{} ca)$, then assume that $ a \equal{} \max\{a,b,c\}$. We will show that $ a \ge 2(b \plus{} c)$, indeed, if $ a < 2(b \plus{} c)$, then we have
\[ a^2 \plus{} b^2 \plus{} c^2 \minus{} 4(ab \plus{} bc \plus{} ca) \equal{} a(a \minus{} 2b \minus{} 2c) \plus{} b(b \minus{} a) \plus{} c(c \minus{} a) \minus{} 4bc \le 0
\]
which is contradict.
Now, with $ a \ge 2(b \plus{} c)$, using AM-GM Inequality, we have
\[ \frac {1}{(a \plus{} 2b)^2} \plus{} \frac {1}{(b \plus{} 2c)^2} \ge \frac {2}{(a \plus{} 2b)(b \plus{} 2c)}
\]
It suffices to prove that
\[ \frac {2}{(a \plus{} 2b)(b \plus{} 2c)} \ge \frac {1}{ab \plus{} bc \plus{} ca}
\]

\[ \Leftrightarrow b(a \minus{} 2b \minus{} 2c) \ge 0
\]
which is obviouly true.
We have done.
Equality holds if and only if $ a \equal{} b \equal{} c$.
:)[/quote]

Very nice proof, can_hang2007.:)
But can you prove by a very nice way the following:

$ \bigstar$
Given non-negative real numbers $ a,b,c$. For $ k\ge 2$, prove that
\[ \frac {1}{(ka \plus{} b)^{2}} \plus{} \frac {1}{(kb \plus{} c)^{2}} \plus{} \frac {1}{(kc \plus{} a)^{2}}\ge \frac {9}{(k \plus{} 1)^{2}(ab \plus{} bc \plus{} ca)}.
\]

[quote=Inequalities Master]
Given non-negative real numbers $ a,b,c$. For $ k\ge 2$, prove that
\[ \frac {1}{(ka \plus{} b)^{2}} \plus{} \frac {1}{(kb \plus{} c)^{2}} \plus{} \frac {1}{(kc \plus{} a)^{2}}\ge \frac {9}{(k \plus{} 1)^{2}(ab \plus{} bc \plus{} ca)}.
\][/quote]
Given non-negative real numbers $ a,b,c$. For $ k>0$, prove that
\[ \frac {1}{(ka + b)^{2}} + \frac {1}{(kb + c)^{2}} + \frac {1}{(kc + a)^{2}}\ge \frac {9}{(k + 1)^{2}(ab + bc + ca)}.
\]
Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. [url=https://artofproblemsolving.com/community/c6h257563p1404150]Consider the following inequality[/url]
\[ \frac{1}{(a+b)(a+kb)}+\frac{1}{(b+c)(b+kc)}+\frac{1}{(c+a)(c+ka)} \ge \frac{9}{2(k+1)(ab+bc+ca)}.\]
a) Prove that the inequality holds for $ 0 \le k \le \frac{4}{11}$, $ k=1$ and $ k \ge \frac{11}{4}.$

b) Prove that the inequality holds for all $ k \ge 0$
[url]https://artofproblemsolving.com/community/c6h141336p799232[/url]

I am very happy to notice this inequality is again considered very hard and very nice

Let $a,\,b,\,c$ be nonnegative real numbers such that $ab+bc+ca>0.$ If $p>0,$ then \[(ab+bc+ca)\left[\frac{1}{(a+pb)^2}+\frac{1}{(b+pc)^2}+\frac{1}{(c+pa)^2}\right] \ge \frac{9}{(p+1)^2}.\]

Can_hang2007 said to me 10 years ago :

"It is a really hard problem but now I have completely proved it with a not-hard proof, using elementary inequality like AM-GM and Cauchy,.... but I don't want to reveal this solution."

He never published his solution in anyone of his books on inequalities :huh:

It is a problem I have always in my mind :maybe:

[quote=manlio]

Can_hang2007 said to me 10 years ago :

"It is a really hard problem but now I have completely proved it with a not-hard proof, using elementary inequality like AM-GM and Cauchy,.... but I don't want to reveal this solution."

He never published his solution in anyone of his books on inequalities :huh:

It is a problem I have always in my mind :maybe:[/quote]
With Pierre Fermat was a similar story. We know how it ended. ;)  :-D

[quote=arqady]
With Pierre Fermat was a similar story. We know how it ended. ;)  :-D[/quote]

Very funny :lol:


[url]http://www.cmc.uwaterloo.ca/results/2003-2004/contest_results.html[/url]

Comment #1: Congrats to Massey again! Beating the second place school by a whole 10 points.

Comment #2: Jacob I feel your pain. ;D

Haha, Jacob wasn't even first at his own school.

[quote="kool_dudy"][quote="blahblahblah"][quote="Osiris"]Well, Janos is just as good. But isn't it nerve-wracking when you discover you did all the part C questions right only to flop on some part A question?

Jacob said regarding last year's Fermat, "I'll remember this from now on - x doesn't equal 1/x."[/quote]

A kid from my school (Steve Kim) also missed a part A question.  But at least he was first in British Columbia.[/quote]

I think I heard Steve Kim before...which province/territory u in?  which school?[/quote]

Port Moody Secondary School, Port Moody, British Columbia.

darn, we should have 3 perfect papers on Fermat...  too bad!  :(

there was only one perfect fermat right?

i don't remember quite well...

btw...i got 110 on pascal...ewww

there are two from our school.. there are like five or six in total

[quote="billzhao"]Bah, this year's Fermat was mess. I solved everything but made two stupid arithmetic errors on #24, 25, both times forgot to divide by two.

Anyways, Pascal/Cayley/Fermat contests are way too easy to do a good judgement of one's skills. Most ppl just complete the first 23 questions and leave the last two blank. Too many ppl get near perfect marks. I think Waterloo should make a new super-tough multiple choice (like AMC12) for the grade 12. And I fear that there'd be nothing for me to do once I reach that grade.   :?[/quote]

Yeah, that was pretty much the way the fermat was for me last year, except that I missed question #10 or something.  It's dumb because the questions get so much harder in the last two.

The year I did the Cayley, it was really straightforward until question #25, where I forgot to check one case.  Unfortunately, because it was easy that year, 142 --> 50th in Canada, or something like that.

[quote="bugzpodder"]I am the best, I missed #1 and #25 on the Fermat last yr[/quote]

Wow...how do you miss #1???

thats a good question.  I think i got #1 wrong on all of CMO and APMO and USAMO... so thats another record...

i was -3 years old when this topic was made : D


Let $ f: R \minus{}> R$ be an infinitely differentiable function such that $ f'(x) \equal{} f(1\minus{}x)$ for all x. Given $ f(0)\equal{}1$. Find $ f(1)$.

[quote="Conjurer"]Let $ f: R \to R$ be an infinitely differentiable function such that $ f'(x) \equal{} f(1 \minus{} x) \;\; \forall x$. Given $ f(0) \equal{} 1$. Find $ f(1)$.[/quote]

$ \int_0^1 f'(x)dx \equal{} f(1) \minus{} f(0)$

Where is it that you find the problems of JEE '09. I'd also like to have a look at a few* of them ;)

*quantity may vary from person to person.

You didnt do much and did the obvious step.

[hide="HINT"]

Read the first 5 words of the question carefully.[/hide]

PS:These questions' level would be matching the hardest ones in the JEE paper.

EDITED: Thnx for the bolds.

Do you have a solution to [b]your[/b] question?

The only infinitely differentiable function i can think of is e to the power something. :oops:

[quote="abhitayal"]how did you manage to get a question of JEE 2009 :o
If its true i want the link too.[/quote]

It's a joke, my dear friend. How could you possibly get questions of an exam like JEE before the exam? Do you know why the JEE '97 was cancelled?

Why nobody :wink:

[hide="HINT"]Differentiate the given relation[/hide]

the function is
sin(x+(pi/4)-1/2)/sin((pi/4)-1/2)
or the one in which x is replaced by -x ,-1/2 by 1/2[/code]

WOW!!
Could you explain how you got this??
Was it just KO(Keen Observation)?

differentiate the given realation on both sides wrt x
replace x by 1-x
then we see that the function satisfies differential equation of simple harmonic motion

I am new to differential equations

Could you please explain the solution more elaborately?

P.S. I learnt about differential equations like $ ay'' \plus{} by' \plus{} cy \equal{} 0$ but not such ones with both x and 1-x :oops:

Guys please explain the method u used to solve the differential equation involving x and 1-x.

Befuddlers

Dude you think I posted a question with a "method"? I think you are not familiar with calculus so why not just leave it for now.


limit
logarithms
calculus
calculus computations

Let $ a_n$ be a sequence such that:
$ a_1\equal{}1$ and $ \frac{n}{a_{n\plus{}1}}\minus{}\frac{n\plus{}1}{a_n}\equal{}n(n\plus{}1)$, for all $ n \geq 1$.

Evaluate the limit:
\[ \lim_{n \to \infty}{\frac{1}{n^3}\cdot \sum_{k\equal{}1}^{n}{\frac{1}{\ln(1\plus{}a_k)}} }.\]

Rewrite your equation as:

$ \frac{1}{n+1} \frac{1}{a_{n+1}}-\frac{1}{n} \frac{1}{a_n} = 1$

Now, define:

$ b_n = \frac{1}{n a_n}$

Then you have:

$ b_1= 1$

$ b_{n+1}-b_n= 1 \quad / \sum^{i}_{n=1}$

$ b_{i+1} - b_1 = i \Rightarrow b_i = i$

Then:

$ a_n = \frac{1}{n b_n}=\frac{1}{n^2}$

We can evaluate the limit now:

$ \lim_{n\rightarrow \infty} \frac{1}{n^3} \cdot \sum^n_{k=1} \frac{1}{\ln(1+\frac{1}{k^2})}$

We have

$ \sum^n_{k=1} \frac{1}{\frac{1}{k^2}}<\sum^n_{k=1} \frac{1}{\ln(1+\frac{1}{k^2})}<\sum^n_{k=1}( \frac{1}{\frac{1}{k^2}}+1)$

$ \frac{n^3}{3} + \frac{n^2}{2}+\frac{n}{6}<\sum^n_{k=1} \frac{1}{\ln(1+\frac{1}{k^2})}<\frac{n^3}{3} + \frac{n^2}{2}+\frac{7 n}{6}$

From this, the limit is obvious and we have:

$ \lim_{n\rightarrow \infty} \frac{1}{n^3} \cdot \sum^n_{k=1} \frac{1}{\ln(1+\frac{1}{k^2})} = \frac{1}{3}$


Factor out the two and the sum becomes $ 2(1\plus{}2\plus{}3\plus{}...\plus{}50)$.  Now, we can use the formula on the sum in the parentheses to get $ 2\left(\frac{50(51)}{2}\right)$.  The two's cancel, leaving you with $ 50(51)$ which can be easily evaluated to $ \boxed{2550}$

If you mean factorize as in the factoring out the 2 there to get a sum that we know how to compute, yes, because all the numbers in the sum are divisible by 2

Or, we could say that let's call the sum $ S$, so \[ S\equal{}2\plus{}4\plus{}6\plus{}...\plus{}100\]Thus, \[ S\equal{}100\plus{}98\plus{}96\plus{}...\plus{}2\]Adding these two equations together, we get \[ 2S\equal{}(100\plus{}2)\plus{}(98\plus{}4)\plus{}(96\plus{}6)\plus{}...\plus{}(2\plus{}100)\]Now, note that \[ 100\plus{}2\equal{}98\plus{}4\equal{}96\plus{}6\equal{}94\plus{}8\equal{}...\equal{}2\plus{}100\]Our whole sum can then be rewritten as \[ 2S\equal{}n(100\plus{}2)\]where $ n$ is the number of terms in the sequence 2,4,6,...,100.  It's pretty clear that $ n\equal{}50$, so \[ 2S\equal{}50(102)\] \[ S\equal{}25\times 102\equal{}2550\]


[color=darkblue]Let $ \begin {cases} ax\plus{}by\equal{}c\\bx\plus{}cy\equal{}a\\cx\plus{}ay\equal{}b\end{cases}$. Prove that $ a^3\plus{}b^3\plus{}c^3\equal{}3abc$.

I want you give me a nice solution and short.[/color]

[quote="rofler"]Sum them up

We have either x+y=1 or a+b+c=0

If x+y=1, then y=1-x

(-x)(a-b)=b-c
(-x)(b-c)=c-a
(-x)(c-a)=a-b

Now, if 2 are equal, WLOG a=b, we see b=c, so a=b=c

If all three distinct, (-x)^3(a-b)=(a-b) so (-x)^3=1 so -x=1

Then we get
a+b+c=3a
a+b+c=3b
a+b+c=3c

So a=b=c, contradiction!

Now, what is to be proven turns into $ (a \plus{} b \plus{} c)((a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2) \equal{} 0$ so if a=b=c or a+b+c=0, the equation to be proven becomes true, so we are done.[/quote]

Thank you!  :P

Here's another approach: the homogenous linear system
\[ \left\{
\begin{array}
[c]{c}
a\alpha\plus{}b\beta\plus{}c\gamma\equal{}0\\
b\alpha\plus{}c\beta\plus{}a\gamma\equal{}0\\
c\alpha\plus{}a\beta\plus{}b\gamma\equal{}0
\end{array}
\right.
\]
with the unknowns $ \alpha,\beta,\gamma$ has a nonzero solution $ \left(
x,y,\minus{}1\right)  ,$ hence its determinant must be zero. But
\[ \left\vert
\begin{array}
[c]{ccc}
a & b & c\\
b & c & a\\
c & a & b
\end{array}
\right\vert \equal{}3abc\minus{}a^{3}\minus{}b^{3}\minus{}c^{3}.
\]

It's nice. Thank you very much. :lol:  :P


trigonometry geometry circumcircle inradius incenter

[quote="warut_suk"]In $\triangle ABC$, is it true that
$4Rsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2} = r$
where $R,r$ is the circumradius and the inradius of the triangle?[/quote]

It is true.

If I is incentre of $\triangle ABC$, then in $\triangle IBC$, if ID is perpendicular from I on to BC, then ID=r and IB, IC are bisectors of angles ABC, BCA.

We have $a=BC=BD+DC=r(cot\frac{B}{2}+cot\frac{C}{2})$

using $a=2RsinA$ and simplifying the right hand side,  we get the required result

Right.

You can also use $\sum \cos A = 1+\frac{r}{R} = 1+4\prod \sin \frac{A}{2}$


[hide]$645=15x+\frac{14\cdot15}{2}$, because $645=x+(x+1)+(x+2)+...+(x+14)$. So, $645=15x+105$. Solve for x, which is $36$. We are looking for the largest of the addends, $x+14$. Therefore, our answer is $50$.
[/hide]

[hide]

Let the number we are looking for be $x$.

Then the sum is $x+(x-1)+\cdots+(x-14)=15x-105=645$

$15x=750$

$x=50$[/hide]

[quote="rcv"]If 645 is written as the sum of fifteen consecutive integers, what is the largest of the addends?[/quote]
[hide]
645/15=43
43+7=50
50[/hide]

jorian

How can there be a #40 on sprint?

EDIT: Never mind, there used to be 40 questions on sprint.


I am thinnking of buying an Educaide database of mathcounts problems for $\$125$. I am going to buy the MATHCOUNTS Handbooks database. If I buy that, would it have all of the handbooks from 1984? Also, would the mathcounts competition have all the competitions from '84? Thanks.

Yes, you would have every single problem period.

[quote="#H34N1"]Would it be orgainzed by year?

I'm sorry but I do not know about this.[/quote]
why don't you use their site instead of this one?
It must be possible because it's almost impossible to prove it's impossible.

Remember that in addition to purchasing the MathCounts database, you need Educaide's Acces program to view the database.  Your school might have that program already, but if not, I believe Acces costs more than 400 dollars.  Also, Educaide might have a policy of selling only to educators.

Also, does it contain the national, state, and chapters? What year range? 1984-2001?

I responded to your pm, but here is the info.  The database cost a ton I think Ravi B might be under estimating, but I bought it 5 years ago, so maybe it has come down.  The program though includes NO problems.  Then you have to buy the problem sets.  I believe the choices are Handbooks 1983-1995 or Contests 1983-1995(not sure if includes Nats).  Each of those cost 100 + dollars.  Not worth it for 1 kid!!!  Unless you are insanely in the funds in which case maybe not :D

EDIT: It has nationals and now goes thorugh 2001


geometry
perimeter
inequalities
calculus
triangle inequality
calculus computations

[b]Here are three questions from my paper ,,, since I don't know how to solve them ,,, will someone please help me ,,, ?

They are 5 marks each ,,,

The second questions seems very easy ,,, but is 5 marks worth for it ,,,  ? so I guess it need some tricky solution to beat the marks ,,,[/b]

[u]Questions :[/u]

1. Trace the following curves
$ y\equal{}x/(1\plus{}x^2)$

2. Prove that $ |4x\minus{}5y|<\equal{} 4|x| \plus{} 5|y|$

3. Find the perimeter of the cord $ r \equal{} a(1\plus{}cos\theta)$

[b]I think I got the third questions solution ,,,
is the below one correct ,,, ? what say you guys ,,,[/b]

[u]My Third questions solution :[/u]

$ (|4x\minus{}5y|)^2 \equal{} (4x\minus{}5y)^2$                                 (since, $ |x|^2 \equal{} x^2$  $ \forall x$ belongs to $ R$)
$ \equal{} 16^2 \minus{} 40xy \plus{} 25y^2$
$ <\equal{} 16 |x^2| \plus{} 40 |x\parallel{}\minus{}y| \plus{} 25 |y^2|$                     (since $ x<\equal{}|x|$  $ \forall x$ belongs to $ R$)
$ <\equal{} 16 |x^2| \plus{} |x\parallel{}y| \plus{} 25 |y^2|$
$ <\equal{} (4|x| \plus{} 5|y|)^2$

Removing squares on both sides we get:
$ |4x \minus{} 5y| <\equal{} 4|x| \plus{} 5|y|$

The 2nd Question is proved by Cauchy-Schwarz . You can wiki it.

[quote="mchoi815"]Isn't the question just triangle inequality?  Because |a+b| <= |a|+|b| (you could represent numbers for 1D vectors).  And then you get |4x+-5y| <= |4x|+|-5y| = 4|x| + 5|y|.[/quote]
Yeah, pretty much.

well, for your first problem, it is a rational equation.  So you need to find some asymtotes.  There are no verticle ones (since x^2+1 is always positive) but the horz. asymtote is y=0.  knowing that, you can just plot some points.

Stop it.These problems are Calculus except Q2.

Should be moved in Calculus Tutorials Section.

Thanks in advance.

Japanese Communities Moderator

kunny

[b]Yea stop it now ,,, as I got all the answers ,,, thanks guys for the tips ,,, !

,,, mod ,,, plz feel free to move this topic to the desired location ,,, I can't find the move option myself ,,,[/b]


a) We call [i]admissible sequence[/i] a sequence of 4 even digits in which no digits appears more than two times. Find the number of admissible sequences.

b) For each integer $ n\geq 2$ we denote $ d_n$ the number of possibilities of completing with even digits an array with $ n$ rows and 4 columns, such that

(1) any row is an admissible sequence; (2) the sequence 2, 0, 0, 8 appears exactly ones in the array.

Find the values of $ n$ for which the number $ \frac {d_{n\plus{}1}}{d_n}$ is an integer.

[b]a)[/b] The sequence has elements from the set $ \{ 0,2,4,8\}=A . $ The number of sequences of four distinct elements of $ A $ are $ 120.\quad\text{(i)}  $
Justification: the first slot has $ 5 $ possibilities, the second, $ 4 $ (because we're not allowed to put the same digit as that of the first slot), the third, $ 3 $ (we're not allowed to put the digits of the first and second slot) the fourth, $ 2 $ (same argument), in total, $ 5\times 4\times 3\times 2 . $
The number of sequences of four elements of $ A $ such that they have exactly two same digits are $ 360.\quad\text{(ii)} $
Justification: $ (\alpha ) $ Pick an element of $ A.\quad(\beta ) $ Put that digit in the first and second slots of a sequence. The third slot can occupy $ 4 $ elements (all, but the element chosen in the beginning). The fourth slot can occupy $ 3 $ elements (all, but the element chosen at the beginning and that of the third slot). In total, $ 4\times 3. $ The number of ways in which the step $ (\beta ) $ can be made, is $ \binom{4}{2}, $ and the ways in which step $ (\alpha ) $ can be made is $ 5. $ In total, $ 4\times 3\times\binom{4}{2}\times 5. $
Adding $ \text{(i)} $ and  $ \text{(ii)}, $ we obtain $ 480. $

[b]b)[/b] $ (\gamma ) $ Fix the $ 2,0,0,8 $ sequence in the first row. The other $ n-1 $ rows have, according to [b]a),[/b] $ 480 $ possibilities of forming [i]admissible sequences,[/i] forming a total number of $ 480\times\overbrace{\cdots}^{n-1\text{ times}}\times 480. $ The numbers of ways in which the step $ (\gamma ) $ can be done is $ n $ times. So, $ d_n=n\times\left( 480\times\overbrace{\cdots}^{n-1\text{ times}}\times 480\right) . $ Now,
$$ \frac{d_{n+1}}{d_n}=\frac{d_n=(n+1)\times\left( 480\times\overbrace{\cdots}^{n\text{ times}}\times 480\right)}{d_n=n\times\left( 480\times\overbrace{\cdots}^{n-1\text{ times}}\times 480\right)} =480+\frac{480}{n}\in\mathbb{N}\iff n|480\iff $$
$$ \iff n\in\{ 2,3,4,5,6,8,10,12,16,20,24,30,32,40,48,60,80,96,120,160,240,480\} . $$


Prove that from a set of ten distinct two-digit numbers, it is possible to select two disjoint nonempty subsets whose members have the same sum.

There are $2^{10}-2=1022$ subsets of 10 elements (exclude the emtpy and the whole set).
Now, a sum of the numbers in such a subset is always smaller than $99\cdot10=990$.
So at least 2 of these subsets have the same sum. If they are disjoint, we're done. If they have numbers in common, just delete the common elements. The 2 new set will still have the same sum, and are disjoint.


trigonometry
trig identities
Law of Cosines

From a smooth $30 cm$ diameter circular plywood disk, two smaller circular disks of diameters $20 cm$ and $10 cm$ are cut. What is the largest circular disk that can be cut from one (either) piece of the remaining plywood?

Ignore width of sawblade when cutting.

im not sure how to figure it exactly but try graphing the circle and see if that helps

Construct the circle, and connect the centers of the three inner circles to make a triangle with sides $r+10, r+20, 15$.  Then connect the center of the new circle to the center of the large circle to split side of length $15$ into $10$ and $5$.  Then apply law of cosines twice, using $\cos \theta = - \cos (\pi - \theta)$ to get an equation which should be easily solvable.

[quote="xxxyyyy"][quote="drunner2007"]deej21. you forgot that we were given the diameters so the triangle would have side lengths $5+r$, $15$, and $r+10$. I applied the law of cosines twice, once on the the angle $\theta$ and the other time on the angle $180-\theta$. Then, knowing that $\cos(180-\theta)=-\cos\theta$, I arrived at $\frac{-40r+300}{20(15-r)}=\frac{50r-100}{10(15-r)}$. This implies $r=\frac{25}{7}$. Let me know if you find any mistakes.[/quote]

The correct answer = $30/7$.

I am told that there is an "exotic" formulae by Descrates that gives the answer directly...

anyone?[/quote]

why yes there is!

now basically if you have 4 circles all tangent to eachother then the curvatures $k_i$ satisfy:

$2[ (k_1)^2+(k_2)^2+(k_3)^2+(k_4)^2]=[k_1+k_2+k_3+k_4]^2$

now defining curvature for this problem is a little tricky

if the circle is externally tangent, then the curvature= $\frac{1}{r}$

if the circle is internally tangent, then the curvature is $-\frac{1}{r}$

as for why this is true, i don't know

then the problem becomes (ignoring the 10's)

$2(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{r^2})=(-\frac{1}{3}+\frac{1}{2}+1+\frac{1}{r})^2$


superior algebra
superior algebra unsolved

Prove that in every ring $A$, and for every $n\in \mathbb{N}^*$, we have:
\[ x^{2(2n+1)}=x,\forall x\in A\Leftrightarrow x^{2n}=x,\forall x\in A  \]

the condition " for EVERY $n \in \mathbb{N}^*$ " is the key of the problem.
so, we have :

" $\Leftarrow$ "
$x^{2n}=x$ means also $x^2=x$ multiply by $x$ and you will be able to prove that $x^n=x$ $\forall n \in \mathbb{N}^*$ .

" $\Rightarrow$ "
your condition is $x^{4n+2}=x$ (1). Put n+1 instead of n and you get $x^{4n+6}=x$.
multiply relation (1) by $x^4$ and you will get $x^{4n+6}=x^5$.
hence $x^5=x$ (2)
but, for n=1, we have $x^6=x$. now, multiply relation (2) by x and you get $x^2=x$ and from here $x^n=x$ $\forall n \in \mathbb{N}^*$

in conclusion, we have managed to prove an even stronger relation. :)
hope I haven't disregarded anything in my proof...  :?

Your proof works,...as I see. I thought that it was harder :?  Thank you anyway!

[quote="DiaDi"]the condition " for EVERY $n \in \mathbb{N}^*$ " is the key of the problem.
[/quote]

How about the condition "for EVERY ring $A$"?

I think you have solved this problem:
Let $A$ be a ring. Prove:
\[ (x^{4n+2}=x\ \forall\ x\in A,\ \forall\ n\in\mathbb N^*)\Leftrightarrow(x^{4n}=x\ \forall\ x\in A,\ \forall\ n\in\mathbb N^*). \]

Do you think this problem and the problem posted by xirti are the same?

I thought it was too easy to be true. I'm working at the [i] proper [/i] solution as we speak...it's just that I feel like I'm going circles... :?


Anna divides her collection of marbles into two equal piles.
Her little sister then takes three marbles from one of the
piles, leaving 28 marbles in that pile. How many marbles were
in Anna's original collection of marbles?

28+3=31, so there was 31 marbles in one pile.

thus there was total 62 marbles in two piles.

answer : 62


AMC
AIME
MATHCOUNTS
number theory
relatively prime
special factorizations

Next question

Problem 56.

If $x^{2}+y^{2}-30x-40y+576=0$, then the largest value for $\frac{y}{x}$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime, positive integers.  Determine $m+n$.

this seems rather aimelike

completing the square gets a circle centered at (15, 20), largest value thingy is 20/8 so the answer is 7.

idk, i just did it in my head lol

Yes this is an AIME Like question and your answer is wrong.

Assume that $y=mx$.  Then plug it in.

Then set up an equation using discriminant and you see that you get an answer not equal to $\frac{5}{2}$

Yes it is from a mock AIME, now you have to solve it.

Draw the graph, a circle and a line $y=kx$. to maximize $k$, the line has to be the "steeper" tangent to the circle. (noticed there are two tangent lines, the steeper one has a higher value of $k$)

so plug in $y=kx$ to the original equation, you get:
$x^{2}+(kx)^{2}-30x-40kx+576=0$
$(k^{2}+1)x^{2}-(40k+30)x+576=0$

take the discriminant and set it to 0.
$(-40k-30)^{2}-4*(k^{2}+1)*576=0$
solve for k, you get $k=\frac34$ or $k=\frac{117}{44}$

so you take the bigger value of k, which is $k=\frac{117}{44}$

$m+n=117+44=\boxed{161}$

I think that problems like this may be too advanced for the mathcounts forum...

i think this should be in intermediate forum..

Deracinated from the MC marathon for being much too hard.


[i]Taken from MathWorld:[/i]
This is a classic arithmetical problem probably first posed by Euclid  and investigated by various authors in the Middle Ages.

The problem is formulated as a dialogue between the two animals after which it is called. The mule says to the donkey, "If you gave me one of your sacks, I would have as many as you." The donkey replies, "If you gave one of your sacks, I would have twice as many as you." The question, of course, is "How many sacks do they have?"

[hide]mule = 5 sacks

donkey= 7 sacks

total= 12 sacks

is this a trick question?[/hide]

[quote="MithsApprentice"]
is this a trick question?[/quote]

No. It's just a regular question. But it's good to ask yourself why Euclid would have bothered to pose a question like this, and why people up through the Middle Ages might have found it interesting. Think about when algebra, as we know it now, was invented.

A lot of things that we take for granted now, like symbols for variables, arabic numerals, the concept of negative numbers and zero, were a long time in coming historically. For an example, imagine trying to do multiplication or long division with Roman Numbers.

[color=white]mule=2
donkey=3

i doubt it's right[/color]

Be careful, guys -- it's not "if I had one more sack," it's "if you gave me one more sack."

[hide]

x=mule

x-2=donkey

2(x-3)=x+1

2x-6=x+1

x=7(mule)

7-2=5(donkey)

am i rite?

[/hide]

who would've thought donkeys and mules could talk.

[hide]x=# of bags on donkey, so mule must have x-2. also, x+1=2x-6 which means x=7.

Donkey-7

Mule- 5[/hide]

[quote="yif man12"]who would've thought donkeys and mules could talk.
[/quote]

haha...that's what i was thinking...

but this is a long time ago...who knows....

you must remember...all great mathematicians are crazy

[quote]For an example, imagine trying to do multiplication or long division with Roman Numbers.[/quote]

actually, the romans didn't really use roman numerals to multiply and divide--they used an abacus--with the roman numeral place values.

[quote="ComplexZeta"]Actually, even if you try extremely hard, I doubt you could come up with more than three great mathematicians who went crazy.[/quote]

I guess it's what you call crazy. Looking at some of those perfect AMC12ers, I could question their sanity. . . I mean shit, learn to smile.

I was referring to those who ended up in asyla, but you're right.

Well, I don't think anybody's posted an algebraically correct interpretation of the problem yet with two variables and two equations, so if you're still wondering, here you go.

[hide]Let donkey have x, mule have y sacks.

y+1= x-1

x+1=2(y-1)

Solve, get x=7, y=5.[/hide]

[quote="ComplexZeta"]Actually, even if you try extremely hard, I doubt you could come up with more than three great mathematicians who went crazy.[/quote]

Blaise Pascal, Gregor Cantor, Bertrand Russell, John Nash, Alexander Grothendieck... just to name a few.

Bleh. I forgot about Grothendieck. What happened to Pascat and Russell? The other one I was thinking of was Gdel.

After a near death experience, Pascal became somewhat of a religious fanatac (Jansenist). He gave up mathematics for a while, thinking it was sinful, and got into self-mutilation to teach himself piety.

Oddly enough, he took up mathematics again later in life to distract himself from the pains of a toothache and stomach cancer. He died at the age of 39. He may not have been certifiably "crazy", but his bahaviour was definitely.. abnormal.

Bertrand Russell wrote extensively about his battles with depression. He probably suffered from manic depression; He claims that his desire to discover new mathematics was the only thing that kept him alive.

Ok I knew that Bertrand Russell said that when he was 16 he only didn't commit suicide because he wanted to learn more math. But I really don't think that that constitutes going crazy. I used that quote by Russell (without the quote) as the first sentence of my one of my college applications, by the way.

man, i didn't know there were so many crazy mathematicians. almost as many of them as sadistic authors.

The only mathematician I know who was crazy is John Nash(through a movie)

[quote="gauss202"]
Oddly enough, he took up mathematics again later in life to distract himself from the pains of a toothache and stomach cancer. He died at the age of 39. He may not have been certifiably "crazy", but his bahaviour was definitely.. abnormal.[/quote]

Actually, he had a toothache, got distracted by the math he had given up for years, and forgot about his pain.  He figured that was a sign from God that he was meant to do math.  It wasn't for himself, at least, in his mind, was for God.

Newton and Gauss, two of the three greats, were both a little nutso by the end.  Newton moreso.  And Archimedes, the third great, died (at least the legend says) when a soldier made him stop doing math, and he wouldn't, so the soldier killed him.  That's crazy.  And anything Pythagoras did, ever, like killing people because they proved that  :rt2: was irrational, was insane.

Many of those things are myths (or at best are unsubstantiated). In particular the stories about Archimedes and Pythagoras are part of the mathematical folklore.

[quote]And Archimedes, the third great, died (at least the legend says) when a soldier made him stop doing math, and he wouldn't, so the soldier killed him. That's crazy. And anything Pythagoras did, ever, like killing people because they proved that  was irrational, was insane.[/quote]

That isn't exactly the legend. After the Romans invaded (they were held of for a while by Archimedes' machines), one day Archimedes was doing math and he was writing in the sand. A soldier came up to him and told him to get out of his way. Archimedes asked him to wait until he finished. Then the soldier killed him. The soldier didn't care really whether Archimedes was doing math or not. As for Pythagoras, that is just a rumor. While it is true that he denied the existence of irrationals, killing people over it is ridiculous. I'm curious though, how were Newton and Gauss crazy toward the end of their lives? I read "The Stories of the Great Mathematicians" by Jane Muir, and while it mentions the Pascal story, it says little of Newton and Gauss' endings.

mule has 5
donkey has 7

here's how i did it
 m+1=d-1
 d+1=2(m-1)
rearranging these you get:
  m - d = -2
-2m+d = -3
add: m=5
substitute: d=7


For positive reals [tex]x,y,z,[/tex] prove that

[tex]\displaystyle\frac{x^2+1}{(x+y)(x+z)}+\displaystyle\frac{y^2+1}{(y+z)(y+x)}+\displaystyle\frac{z^2+1}{(z+x)(z+y)}[/tex]  [tex]\geq
\displaystyle\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{2(x^2+y^2+z^2)}[/tex]

Nice solution.  It's a bit different than mine though (I made this problem  :D ).  Well, here it is...

We will be using the following two inequalities, which rearrange to become the trivial inequality:

[tex]2x^2+2 \geq (x+1)^2[/tex]

[tex]x^2+y^2+z^2 \geq xy+yz+zx[/tex]

Thus,

[tex]\left(\displaystyle\sum_{cyc} \displaystyle\frac{(2x^2+2)}{(x+y)(x+z)}\right)
\left(4x^2+4y^2+4z^2\right)[/tex]

[tex]\geq \left(\displaystyle\sum_{cyc} \displaystyle\frac{(x+1)^2}{(x+y)(x+z)}\right)
\left(x^2+y^2+z^2+3xy+3yz+3zx\right)[/tex]

[tex]= \left(\displaystyle\sum_{cyc} \displaystyle\frac{(x+1)^2}{(x+y)(x+z)}\right)
\left(\displaystyle\sum_{cyc} (x+y)(x+z) \right)[/tex]

[tex]\geq
(x+y+z+3)^2=(x+y+z+1+1+1)(1+1+1+x+y+z)[/tex]

[tex]\geq 4(\sqrt{x}+\sqrt{y}+\sqrt{z})^2[/tex],

where the last two inequalities come from Cauchy-Schwartz.  So,

[tex]\left(\displaystyle\sum_{cyc} \displaystyle\frac{(2x^2+2)}{(x+y)(x+z)}\right)
\left(4x^2+4y^2+4z^2\right)
\geq
4(\sqrt{x}+\sqrt{y}+\sqrt{z})^2[/tex]

[tex]\left(\displaystyle\sum_{cyc} \displaystyle\frac{(x^2+1)}{(x+y)(x+z)}\right)
\geq
\displaystyle\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{2(x^2+y^2+z^2)}[/tex],

as desired.

[quote="ThAzN1"]Nice solution.  It's a bit different than mine though (I made this problem  :D ).  [/quote]

Nice problem. Both solutions are pretty much the same. I guess it's a lot easier to make a problem than to discover it's proof. :-)


find the smallest number with 24 divisors

If $ n \equal{} p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$, then number of divisors is $ (\alpha_1 \plus{} 1)(\alpha_2 \plus{} 1)...(\alpha_k \plus{} 1)$.
So, $ 24 \equal{} (\alpha_1 \plus{} 1)(\alpha_2 \plus{} 1)...(\alpha_k \plus{} 1)$, and it's easy to check that $ 2^33^25\equal{}360$ is the smalest number with 24 divisors.

[quote="Kirill"]If $ n \equal{} p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$, then number of divisors is $ (\alpha_1 \plus{} 1)(\alpha_2 \plus{} 1)...(\alpha_k \plus{} 1)$.
So, $ 24 \equal{} (\alpha_1 \plus{} 1)(\alpha_2 \plus{} 1)...(\alpha_k \plus{} 1)$, and it's easy to check that $ 2^33^25 \equal{} 360$ is the smalest number with 24 divisors.[/quote]
How can you find that the solution is 360, there must be many cases. I think that's not very easy to check all cases.

No need for checking the cases, the formula Kirill used is common in Number theory.

Further reading:
http://mathworld.wolfram.com/Divisor.html
http://www.jstor.org/pss/2315183

P. S. Doublepost:

http://www.mathlinks.ro/viewtopic.php?t=235774

He asked, given the condition on $ \alpha_1$, how do you compute the list $ (\alpha_1, \alpha_2, \ldots)$ that will minimize $ \prod p_j ^{\alpha_j}$? Wouldn't there be several cases to consider?

There are some easy ways to get it down to a small number of cases.

Note that $ \alpha_j$ should be decreasing -- clearly, the largest exponents should be on the smallest primes.

The longest factorization of $ 24$ is $ 24 \equal{} 3 \cdot 2 \cdot 2 \cdot 2$, with 4 factors greater than $ 1$, so we will need at most $ 4$ primes; that is, $ N \equal{} 2^{\alpha_1} \cdot 3^{\alpha_2} \cdot 5^{\alpha_3} \cdot 7^{\alpha_4}$.

Now solve $ (\alpha_1 \plus{} 1)(\alpha_2 \plus{} 1)(\alpha_3 \plus{} 1)(\alpha_4 \plus{} 1) \equal{} 24$ with $ \{ \alpha _j \}$ decreasing. The only solutions are:

$ (2,1,1,1) \qquad (3,2,1,0) \qquad (5,1,1,0) \qquad (5,3,0,0)$
$ (7,2,0,0) \qquad (11,1,0,0) \qquad (23,0,0,0)$

After trying the first, which results in $ 420$, we can immediately rule out the last 3 solutions. It is fairly easy to check between the other 4 by comparing factors. For example, $ (5,1,1,0)$ is better than $ (5,3,0,0)$ because we'd rather have an extra factor of $ 5$ than an extra factor of $ 3^2$.

It turns out the best is $ (3,2,1,0)$, which results in $ 2^3 3^2 5^1 7^0 \equal{} 360$.


[img]http://img426.imageshack.us/img426/1984/math0ct.jpg[/img]

and these:

a) mary takes at least 10 min and Joanna at least15 min to speak on phone. What is fewest min to speak 2 bro if they together wen they call? (not a three way call)

b)  mary takes 10 min & and Joanna 15 min to read paper. if they share diff. sections of paper wat is fewest min to read the paper together?

For the first one, the perimeter remains the same.


integration
calculus
calculus computations

$ \sum_{n\equal{}1}^{\infty }(\minus{}1)^{n\plus{}1}\left(\prod_{i\equal{}1}^{n\minus{}1}\frac{2 i\minus{}1}{2 i}\right)\frac{1}{2 n (2 n\minus{}1) 4^{n}}$

$ \frac{(\minus{}1)^{n\plus{}1}\prod_{i\equal{}1}^{n\minus{}1}\frac{2 i\minus{}1}{2 i}}{2 n (2 n\minus{}1) 4^{n}}\equal{}\frac{(\minus{}1)^{n\plus{}1}2^{1\minus{}4 n}\Gamma (2 n)}{(1\minus{}2 n)^{2}n\Gamma (n)^{2}}$

The problem is:

$ \sum_{n\equal{}1}^{\infty }\frac{(\minus{}1)^{n\plus{}1}2^{1\minus{}4 n}\Gamma (2 n)}{(1\minus{}2 n)^{2}n\Gamma (n)^{2}}$

and. this sum i dunno how to solve

$ f(z): \equal{}\text{arsinh}(z)\equal{}\int_0^z (1\plus{}t^2)^{\minus{}\frac12} \, dt\equal{}\sum_{k\equal{}0}^\infty {\minus{}\frac12 \choose k} \frac{z^{2k\plus{}1}}{2k\plus{}1}$

$ g(z): \equal{}(1\plus{}z)^\frac12\minus{}1\equal{}\frac12 \int_0^z (1\plus{}t)^{\minus{}\frac12} \, dt \equal{}\sum_{k\equal{}0}^\infty {\minus{}\frac12 \choose k} \frac{z^{k\plus{}1}}{2k\plus{}2}$

$ z\, f(z)\minus{}g(z^2)\equal{}\sum_{k\equal{}0}^\infty {\minus{}\frac12 \choose k} \, \left(\frac{1}{2k\plus{}1}\minus{}\frac{1}{2k\plus{}2}\right) z^{2k\plus{}2}$

In particular $ \sum_{k\equal{}0}^\infty {\minus{}\frac12 \choose k} \, \frac{1}{(2k\plus{}1)(2k\plus{}2)}  \frac{1}{2^{2k\plus{}2}}\equal{}\frac12\, \text{arsinh}\left(\frac12\right)\minus{}\left(\frac{\sqrt{5}}{2}\minus{}1\right)$


function
quadratics
algebra
polynomial
calculus
calculus computations

Another homework problem that I have no clue:

Find the quadratic polynomial $ g(x) \equal{} ax^2 \plus{} bx \plus{} c$ which best fits the function $ f(x) \equal{} e^x$ at $ x\equal{}0$, in the sense that

$ g(0) \equal{} f(0)$, and $ g'(0) \equal{} f'(0)$ and $ g''(0) \equal{} f''(0)$.

I only got the part where if $ x\equal{}0$, $ f(x)\equal{}1$.

Note that $ f(x) \equal{} f'(x) \equal{} f''(x)$ so that $ f(0) \equal{} f'(0) \equal{} f''(0) \equal{} 1$
$ g(x) \equal{} ax^{2} \plus{} bx \plus{}c$
$ g'(x) \equal{} 2ax \plus{} b$
$ g''(x) \equal{} 2a$
We get:
$ g(0) \equal{} c$
$ g'(0) \equal{} b$
$ g''(x) \equal{} g''(0) \equal{} 2a$

$ g(0) \equal{} f(0)$ implies that $ c \equal{} 1$
$ g'(0) \equal{} f'(0)$ implies that $ b \equal{} 1$
$ g''(0) \equal{} f''(0)$ implies $ 2a \equal{} 1$, so that $ a \equal{} \frac{1}{2}$

Thus, $ g(x) \equal{} \frac{1}{2}x^{2} \plus{} x \plus{} 1$ is such a polynomial

Now try taking this and applying it to get a cubic polynomial that best matches $ e^x$. After doing this, go for a quartic, and see if you can generalize.

Let $ g(x) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} a_1x \plus{} a_0$
$ g^{(r)}(x) \equal{} 1\implies a_r \equal{} \frac {1}{r!}$
Hence $ g(x) \equal{} 1 \plus{} \frac {x}{1!} \plus{} \frac {x^2}{2!} \plus{} ... \plus{} \frac {x^n}{n!}$
:wow:


Given the matrix $\left(\begin{matrix}0&1\\-2&-2\end{matrix}\right)$.
a.) Determine its eigenvalues and a pair of eigenvectors

b.) Solve the system of differential equation :  $x'=Ax$

a) We have $\det(A-\lambda I) = \left|\begin{matrix}-\lambda &1\\-2&-2-\lambda\end{matrix}\right| = \lambda^{2}+2\lambda+2=0\Rightarrow \lambda =-1\pm i$.

For $\lambda =-1+i$ we have $\left(\begin{matrix}1-i & 1\\-2 &-1-i\end{matrix}\right)\left(\begin{matrix}n_{1}\\n_{2}\end{matrix}\right) = \left(\begin{matrix}0\\0\end{matrix}\right)\Rightarrow n_{1}= n_{2}\left(-\frac{1}{2}-\frac{1}{2}i\right)$.  Thus $\left(\begin{matrix}-1-i\\2\end{matrix}\right)$ is an eigenvector.

For $\lambda =-1-i$ we have $\left(\begin{matrix}1+i&1\\-2&-1+i\end{matrix}\right)\left(\begin{matrix}n_{1}\\n_{2}\end{matrix}\right) = \left(\begin{matrix}0\\0\end{matrix}\right)\Rightarrow n_{1}= n_{2}\left(-\frac{1}{2}+\frac{1}{2}i\right)$.  Thus $\left(\begin{matrix}-1+i\\2\end{matrix}\right)$ is an eigenvector.

b) The solution to the given differential equation is thus:
\[x = C_{1}e^{(-1+i)t}\left(\begin{matrix}-1-i\\ 2\end{matrix}\right)+C_{2}e^{(-1-i)t}\left(\begin{matrix}-1+i\\ 2\end{matrix}\right) \]
Your professor may want only only real values in the answer, in which case you'll have to break this up into real and imaginary parts.

It's a nice solution. What my profesor wanted? Even him does not know. But who cares? I took the exam:)


abstract algebra
group theory
superior algebra
superior algebra unsolved

If we for the moment consider $<R,+>$ as a finite abelian group, we see by the structure theorem that the squarefreeness of the group's order implies that $<R,+>$ is cyclic.

Let $a$ be a generator. We can write $r_{1},r_{2}\in R$ as $r_{1}= ba$, $r_{2}= ca$ for $b,c \in \mathbb{Z}$. Then $r_{1}r_{2}= (ba)(ca) = (bc)a^{2}=(ca)(ba) = r_{2}r_{1}$

[quote="The Captain"]Prove that if $R$ is a finite ring then $|R|$'s squarefreeness implies commutativity.
PS. Nice result. Isn't it?[/quote]

*Language Problem*

$|R|$ is squarefree means that there does not exist an $a \in \mathbb{N}$ such that $|R|=a^{2}$? Or does it mean that $|R|=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdot p_{n}^{a_{n}}$ implies that $a_{i}=1$?

It means that no square divides the order, which is equivalent to what I think you mean by the second statement.


function
inequalities
linear algebra
matrix
quadratics
symmetry
conics

im testing out of algebra 2 but i need some help....questions below...reply plz.

1. How do you know whether a function has an inverse?
2. How to you add, subtract, multiply, and divide functions?
3. What is function notation?
4. How to you operate with functions? (addition, subtraction, inverse, and composition)
5. What is the parent graph of typical functions such as f(x)=x,absolute value of x,               1/x,x squared, x cubed, x to 1/2 power, x to 1/3 power, and log(x)
6. How do you graph functions and transformations of the parent function?
7. How do you write equations of functions from graph.
8. How to write equations for and graph piece-wise functions?
9. How do you recognize a linear equation by transformation from y=x?
10. How do you write an equation given transformations from the parent equation y=x?
11. How do you demostrate proficiency with a graphing calculator?
12. How to make graphical connections to each type of equation and inequality?
13. Find inverse of 2 X 2 matrix?
14. How do you find, use, and apply inverse of matrix, sometimes by calculator?
15.How to solve linear system by matrices?
16. How to use technology to manipulate matrices?
17. How to check solution to quadratic equation graphically?
18. Use discriminant to determine the nature of roots?
19. Find quadratic equation given roots?
20. Make graph of a quadratic equation with plotted points or sketch with vertex, axis of symmetry, and intercepts of parabola?
21. How to graph quadratic functions in either vertex form or standard form and transform one equation form to another?
22. Approximate roots of an equation using technology?
23. Graph quadratic inequalities?
24. Use quadratic equations to model real-life situations?
25.Write an equation and sketch graph of conics(parabola, circle, ellipse, hyperbola?)
26. How to use directrix, focus, focal width, asymptotes, etc. of the parabola, parabola and ellipse to both graph and write an equation?
27. Solve systems of quadratic equations graphically and algebraically?
28. Use conics to solve real-life problems?
29. Solve problems and related applications involving direct, inverse and joint variations?

THANKS!!!!!!!!!

There is really too much here to help with.  If you have this many questions about functions, maybe you shouldn't be placing out because from the look of these questions, you have no idea what a function even is.

Function notation is usually $ f(x)$ when we use a single variable.  Adding, subtracting, and multiplying functions is done the way you would expect.  Division is a bit tricker.

With respect to some of the matrix questions: The inverse of a 2x2 matrix is given by
$ \frac{1}{\det(A)}\begin{pmatrix}
d&\minus{}b\\
\minus{}c&a
\end{pmatrix}$
assuming the determinant is nonzero.  To solve systems of equations using matrices, one uses row-reduction.

Quadratics: If you have roots $ x\equal{}a$ and $ x\equal{}b$, then the original quadratic is $ (x\minus{}a)(x\minus{}b)\equal{}x^{2}\minus{}(a\plus{}b)x\plus{}ab$.

I think you should re-think what you're doing if you have this many questions.

I don't think posting 29 questions in a single thread will get them all answered. I would suggest you go back, and re-read the lessons in your math book.

Why might I ask are you attempting to test out? That might not be such a good idea consider that might large list of yours.

[hide="EDIT"]Consider the monic quadratic given by the square of a linear term $ x=h\Rightarrow x-h=0$

$ f(x)=(x-h)^2=x^2-2hx+h^2$

where $ h$ is some constant. Next allow these terms to be denoted as $ a,b,c$ respectively. Thus we have

\begin{eqnarray*}f(x)&=&ax^2+bx+c\\&=&x^2-2hx+h^2\\\end{eqnarray*}

We can see here, that perfect squares will have a certain property. Namely, that $ b^2=4c$.  Take a moment here note that $ b=\pm2\sqrt{c}$. This means that for any value of $ h$, this distinctive property holds. Keeping that in mind, lets consider a [i]non[/i]-monic perfect square. We will do this by multiplying our previous polynomial by some constant $ a$.

$ f(x)=a(x-h)^2=a(x^2-2hx+h^2)$

Now lets try to show that \[ b^2=4c\]
\[ (-2ah)^2=4a^2h^2\neq 4ah^2\]
Clearly, doesn't work out so well. However, note that[i] if we factor[/i] $ a$ [i]out of both terms again, we can produce the proper condition for a perfect square.
[/i]
Right now, we have the graph of a parabola, or U-shaped function which may be pointing up or down (This can be determined by a leading coefficient test: if the term is negative then the function has a maximum value; a positive term implies the function has a minimum value.). But to continue, consider that the function may be shifted vertically up or down by the addition of some constant $ k$. Now,
\[ f(x)=a(x-h)^2+k\]
and setting it equal to zero to find the roots, \[ x=h\pm\sqrt{-\frac{k}{a}}\]
This is, in fact, the quadratic formula, just not its traditional form. To find this form, we'll go back to the definition of a quadratic polynomial with coefficients $ a, b,$ and $ c$, where $ k$ has been added to $ c$. Now,
\[ f(x)=ax^2+bx+c=0\]
First things first: we can't do anything until we get rid of $ a$,
\[ x^2+\frac{b}{a}=-\frac{c}{a}\]
Now recall the relation that $ b^2=4c\Longrightarrow c=\frac{b^2}{4}$. In this case, $ b$ is actually $ \frac{b}{a}$, so by substitution, $ c=\frac{b^2}{4a^2}$. Now we can add the $ c$ term to both sides, which completes the square on the LHS (left hand side.):
\[ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2}\]
Ring a bell? Compressing the LHS and subtracting the RHS from both sides we realize the general form a of a completed square,
\[ \displaystyle \left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}=0\]
or, we could have compressed the LHS, then square-rooted both sides and performed some subtraction to yield the quadratic formula:
\[ x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\]
The easiest way to complete the square is probably to simplify the quadratic equation after you've plugged in the values, then isolate the square root. You would then square both sides and be done.[/hide]
Hope that helps ya.

I think it's easier to just memorize the simple formula than to re-derive it any time one needs it.

Undoubtedly true JRav. It's far more efficient that way. You may have noticed at the end that I suggested the option of reversing the quadratic formula into a completed square.

On the other hand, some problems require that you find (just as an example) the standard from for the equation of a circle. To do that, you're [i]probably[/i] going to need to the complete the square. So, to that end, a nice balance between memorizing things and deriving things is best. Besides, if you don't understand where things come from and maintain a purely formulaic comprehension of math, you aren't going to be able to do many problems anyways.

I completely agree with you that one should know how to get these things as some of the methods are important but I think some of the lengthier derivations are worth skipping.

If you don't know this entire list of stuff, why are you testing out of Algebra 2?
Ever consider an honors or advanced Algebra 2 class?

22. Use a calculator. Use the square root button. If you mean roots of an equation, get a graphing calculator (preferable TI) and use calc-intersect. [/hide]

Many of these are basic Algebra 2 things. If you don't know them, don't try to test out. If you are trying to test out for next year, get an Algebra 2 book, work through the sections over the summer, and try to test out at the beginning of the school year.


abstract algebra
category theory
algebra
function
domain
superior algebra
superior algebra unsolved

Give an example of an $R$-submodule of an flat $R$-module which is not flat as an $R$-module.

[hide="hint 1"]long exact Tor sequence[/hide]
[hide="hint 2"]free modules are flat, so any example of non-vanishing $\mathrm{Tor}_{2}$ will do[/hide]
[hide="hint 3"]$\mathrm{Tor}_{2}^{k[X,Y]}(k,k)\ne0$[/hide]
owk

What flat/free $k[X,Y]$-module is $k$ a $k[X,Y]$-submodule of?

So your reply didn't answer my question?

Can you give an example? Is there a classification or "nontrivial" necessary/sufficient conditions for rings having the property that a submodule of a flat module is flat?

Consider a short exact sequence $0\to K\to F\to Q\to 0$ with $F$ flat. Then, $\mathrm{Tor}_{2}(Q,N)=\mathrm{Tor}_{1}(K,N)$ for every $N$. Hence, $K$ is flat iff $\mathrm{Tor}_{2}(Q,N)=0$ for all $N$. So, a necessary and sufficient condition for submodules of flat modules being flat is the vanishing of the functor $\mathrm{Tor}_{2}$. This is true for Dedekind domains, but I do not know whether this may be true in more general rings.

In the example, $K=(X,Y),F=R,Q=k$. owk


Find the $\text{maximum value}$ of $3x + 4y$ when,

$x^2+y^2=14x +6y +6$

blahblahblah's awesome solution

[quote="blahblahblah"]Using Cauchy Schwarz always makes me happy

[hide]$8^2\cdot 5^2=((x-7)^2+(y-3)^2)(3^2+4^2)\geq (3(x-7)+4(y-3))^2$

This is the same as $40\geq 3x-21+4y-12\implies 73\geq 3x+4y$[/hide][/quote]

Alternatively,
the question is nothing but five times the maximum distance of a point
on the circle with centre $(7, 3)$ with radius $8$ units ,
from the line $3x + 4y =0$ , since the reqd point clearly lies
in the first quadrant. This can be found by adding the distance of the centre
from the line to the radius.
 i.e. reqd dist is $5*(33/5 + 8)$ = $73$

[hide="Somebody will kill me for presenting a solution this lame"]
Define $g(x,y) = x^2 + y^2 -14x-6y-6=0$.
Define $f(x,y) = 3x + 4y$

Based on Lagrange multipliers, we know that all extremia for $f(x,y)$ with constraint $g(x,y) = 0$ satisfy the system:

$\nabla f(x,y) = \lambda \nabla g(x,y)$ and $g(x,y) = 0$.

Evaluation the partial derivatives and calculating $\nabla f(x,y)$ and $\nabla g(x,y)$, we have:

$2x - 14 = 3\lambda$

$y-3 = 2\lambda$

$x^2 + y^2 -14x - 6y - 6=0$

It's a lot of messy algebra but it's doable (solve $x$ and $y$ in terms of $\lambda$ and plug into the 3rd equation).  You get two solutions, the important one being $x = 59/5, y = 47/5, \lambda = 16/5$.

This must be the absolute maxima, so $3x + 4y = 73$.
[/hide]


Show that one can find $50$ distinct positive integers such that the sum of each number and its digits is the same.

Well, I found it this way: we build a number $a_1$ accordingly to a certain rule. It goes like this: its first digit to the right is $1$. Then goes $9$ and $0$, so that the ending is $091$. The next digit is again $1$ and then there have to be a little more nines - exactly a thousand and finely again $0$, so now we have a longer ending: $0\underbrace{9...9}_{1000}1091$. Then the story goes from the begining, that means we add $1$, an appropriate number of $9$'s and finely $0$. The question is: what is the "appropriate" number of $9$'s? Well, that depends on what position is the $1$ before the $9$'s. If it stands for $1\cdot 10^k$ in a decimal system, then we add $10^k$ nines. We perform this operation (at least) 49 times, so that the number has (at least) 49 one's. Well, it's a bit long number to write...  :oops:
Now, let's say this is $a_1$ and let the sum of the digits of $a_1$ will be $S(a_1)$. Let's also write: $f(a_1)=a_1+S(a_1)$. Note that: $f(a_1)=f(a_1+9)$, because the ending of $a_1+9$ is $100$ and that means that $S(a_1+9)=S(a_1)-9$.
Similarly, $f(a_1+9000)=f(a_1)$, because the ending of $a_1+9000$ is $1\underbrace{0...00}_{1000}0091$ and so $S(a_1+9000)=S(a_1)-9000$.
Now its obvious how to continue this argument. We see that $f(a_1+9\cdot 10^k)=f(a_1)$, where $k+1$ is a number of a position for digit $1$ in $a_1$.
Of course, the solution above means that there are not only $50$ numbers, but there are $n$ distinct numbers such that the sum of each number and its digits is the same, for $n\in \mathbb{N}$.

how did you come up with the method to select such digits?

Let $S$ be the digit-counting function. The idea is that $S(91)=S(100)$. We will construct sets of $2^k$ elements such that $e+S(e)$ is the same for all $e \in S$ inductively, with the base case of $k=0$ being trivial. Now, if some set $S_0$ contains $2^k$ elements, all with at most $M$ digits, form $S$ with $2^k$ elements as follows: let $N_1$ be the number whose decimal representation contains $10^M$ nines, then $1$ one, then $M$ zeroes, and let $N_2=10^{10^M+M+1}$. Then for every element $e \in S_0$, add $e+N_1$ and $e+N_2$ to $S$. One can see that $(e+N_1)+S(e+N_1)=(e+N_2)+S(e+N_2)=e+S(e)+N_2+1$, so we are done. $\blacksquare$


Consider in the interior of a convex quadrilateral four circles such that each one of them is tangent to two sides of the quadrilateral and tangent to two of the other circles.
suppose that the qudrilateral has in inscribed circle.
Prove that two of the four circles have the same radius.

Assume that radius of $O_1=r_1$ and $O_2=r_2$ and $\ldots$.
We have $AB+CD=AD+BC$.
$AP_1=AP_8$ and $BP_2=BP_3$ and $\ldots$.
Subtracting these from them main one gives $P_1P_2+P_5P_6=P_3P_4+P_7P_8$.
$P_1P_2=2\sqrt{r_1r_2}$
$P_3P_4=2\sqrt{r_2r_3}$
$P_5P_6=2\sqrt{r_3r_4}$
$P_7P_8=2\sqrt{r_4r_1}$
So we have $\sqrt{r_1r_2}+\sqrt{r_3r_4}=\sqrt{r_2r_3}\sqrt{r_4r_1}$.
But write it as $(\sqrt{r_1}-\sqrt{r_3})(\sqrt{r_2}-\sqrt{r_4})=0$.
So we must have $r_1=r_3$ or $r_2=r_4$.

The problem has been discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=24332 . Note a very interesting comment by Yetti on a possible extension of the problem:

[quote="yetti"]It does not seem possible that such circles exist, unless the diagonals $AC, BD$ of the tangential quadrilateral $ABCD$ are perpendicular to each other, i.e., unless $ABCD$ is at least a kite, or unless at least two of the circles ($a, c$ or $b, d$) coincide with the quadrilateral incircle.[/quote]

I actully have a "half-proof" of this, but it is missing some important and nontrivial arrangement considerations, so the extended problem remains open. Any ideas?

  Darij


Hello,

I know the answer to the following question but I don't exactly understand how you come about getting it. If anyone could try and explain that would be great.

Define the functions $ f: \mathbb{R}\minus{} \left \{\minus{}2\right \}\to\mathbb{R}\minus{}\left \{1\right \}$ by $ f(x)\equal{}\frac{x\minus{}3}{x\plus{}2}$ and $ g: \mathbb{R}\to\mathbb{R}^\plus{}$ by $ g(x)\equal{}x^2\plus{}1$.

Find $ f^{\minus{}1}(x)$.

[hide]$ \boxed{f^{\minus{}1}(x)\equal{}\frac{2x\plus{}3}{1\minus{}x}}$[/hide]

$ g$ is making no sense there.

As for finding the inverse, first prove injection and then surjection, thus the function will be invertible.

Given a function $ f(x)$, the inverse function $ i(x)$ must satisfy $ f(i(x)) \equal{} x$.  When $ f$ has a nice simple form like the one you've shown us, this gives an equation that you can explicitly solve for $ i$ (treating $ i(x)$ as the variable to be solved for and $ x$ as a constant, you get a linear equation).

you have $ f(f^{\minus{}1}{x})\equal{}x$ right? let make it easy $ f^{\minus{}1}{x}\equal{}t(x)$ So by $ f(f^{\minus{}1}{x})\equal{}x$ we have.
$ f(t(x))\equal{}x$
$ \frac{t(x)\minus{}3}{t(x)\plus{}2}\equal{}x$
$ t(x)\minus{}3\equal{}xt(x)\plus{}2x$
$ t(x)(1\minus{}x)\equal{}2x\plus{}3$
$ t(x)\equal{}\frac{2x\plus{}3}{1\minus{}x}$
May that help you.

To find the inverse of a function $ f(x)$ isolate the $ x$. Only one-one functions are invertible.

So, here:
$ f(x)\equal{}y\equal{}\frac{x\minus{}3}{x\plus{}2} \\ \\
y(x\plus{}2)\equal{}x\minus{}3 \\ \\
x(1\minus{}y)\equal{}3\plus{}2y \\ \\ \\
\therefore x\equal{}f^{\minus{}1}(y)\equal{}\frac{3\plus{}2y}{1\minus{}y}$

[quote="Rajiv"]Only one-one functions are invertible.[/quote]That's necessary, but not sufficient. Also need to be [i]onto [/i](surjective).


OK now it's just getting creepy...I tried to post the following:
"Possibilities (from the most likely to least likely):
a1.) Server Trouble
a2.) Server Maintainance
a3.) Switching server
b1.) Switching to independent server
b2.) Forest fire disrupted the server :) (Although I'm not sure if there [b]is[/b] any forest fire left :lol:)
b3.) Construction in/around server
c.) major change somewhere on AoPS
d.) Human error on the site/in the server whether it's turning the server off accidentally or tripping over the cord :) or whatever
e.) phpBB programming problem

j.) Problems on the user side

Pick your reason :)"
under the "Slowness" thread in the Support forum and the following never-before-seen (well at least by me) error came up:
"Could not insert new word matches

DEBUG MODE

SQL Error : 1030 Got error -1 from table handler

INSERT INTO phpbb_search_wordmatch (post_id, word_id, title_match) SELECT 25518, word_id, 0 FROM phpbb_search_wordlist WHERE word_text IN ('possibilities', 'likely', 'least', 'server', 'trouble', 'maintainance', 'switching', 'independent', 'forest', 'fire', 'disrupted', 'although', 'sure', 'left', 'lol', 'construction', 'major', 'change', 'aops', 'human', 'error', 'site', 'whether', 'turning', 'accidentally', 'tripping', 'cord', 'whatever', 'phpbb', 'programming', 'problem', 'problems', 'user', 'side', 'pick', 'reason')

Line : 242
File : /usr/local/4admin/apache/vhosts/artofproblemsolving.com/httpdocs/wforum/includes/functions_search.php"

So odd, it's like the phpBB itself is angry at the criticism I wrote, it's like it has a mind of it's own...da dum dum da-dum, da dum dum da-dum (theme from Terminator :lol: )
Sorry, I had to joke because I have goosebumps, seriously.

Doubtful.

I know virtually nothing about how this forum operates, but it seems to me like you ran into troubles with the function used to add individual words to the database for when you use[url=http://www.artofproblemsolving.com/wforum/search.php]Search[/url].

That is to say, the program picked out uncommon words (basically, all words other than "the", "from", "to", etc.).  The fact that it failed to pick up pseudo-words like "a1.)" is interesting.  It's quite smart in that respect.

So the program picked out all the non-common words, such as forest, fire, phpbb, and all the others, but failed to insert them into the search database.  I have an inkling that when you try to search for this post, say, by typing in several of the words you used in that post, nothing will come up.

[Edit] Oh, the fact that the debug line points to something called function_search.php makes it even more obvious.  I didn't even notice that.  I was just operating on the knowledge that there was something wrong with the "word matches", and a lot of non-common words were picked out in a list.

Mmm, how come this happened only to my post? I mean I'm sure someone already posted a word not in the list before...

All that scared-ness for nothing...

Maybe the phpbb DOES have a mind of it's own.....  :lol:


conics
parabola
geometry
limit
perimeter
college contests

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137054]Problem 1[/url]   50 points

Let $p$ be prime number and $n$ be non negative integers.

(1) Let $m$ be integer such that $0\leq m\leq n.$ How many numbers are there among integers 1 through $p^{n+1}$ which can be divided by $p^{m}$ not but by $p^{m+1}?$
(2) For two integers $x,\ y$ 1 through $p^{n+1},$ how many pairs of $(x, y)$ such that the product $xy$ can be divided by $p^{n+1}?$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=138036]Problem 2[/url]   60 points

For $a>0$, let $l$ be the line created by rotating the tangent line to parabola $y=x^{2}$, which is tangent at point $A(a,a^{2})$, around $A$ by $-\frac{\pi}{6}$.
Let $B$ be the other intersection of $l$ and $y=x^{2}$. Also, let $C$ be $(a,0)$ and let $O$ be the origin.

(1) Find the equation of $l$.
(2) Let $S(a)$ be the area of the region bounded by $OC$, $CA$ and $y=x^{2}$. Let $T(a)$ be the area of the region bounded by $AB$ and $y=x^{2}$. Find $\lim_{a \to \infty}\frac{T(a)}{S(a)}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=136343]Problem 3[/url]   70 points

Three points $P,\ Q,\ R$ move on the perimeter of the regular octagon $A_{1}A_{2}\cdots A_{8}$ of side-length $1.$

(1) Find the maximum value of the area of $\triangle{PQR}.$
(2) If $Q$ coincides with the vertex $A_{1}$ of the regular octagon and $\angle{PQR}=90^\circ,$ then find the maximum value of the area of $\triangle{PQR}.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137061]Problem 4[/url]   70 points

(1) For integer $n=0,\ 1,\ 2,\ \cdots$ and positive number $a_{n},$ let $f_{n}(x)=a_{n}(x-n)(n+1-x).$ Find $a_{n}$ such that the curve $y=f_{n}(x)$ touches to the curve $y=e^{-x}.$

(2) For $f_{n}(x)$ defined in (1), denote the area of the figure bounded by $y=f_{0}(x), y=e^{-x}$ and the $y$-axis by $S_{0},$ for $n\geq 1,$ the area of the figure bounded by $y=f_{n-1}(x),\ y=f_{n}(x)$ and $y=e^{-x}$ by $S_{n}.$ Find $\lim_{n\to\infty}(S_{0}+S_{1}+\cdots+S_{n}).$

Working Time: 2.5 hours

[hide="Solution for number one"]
The number of integers that is divisible by $ p^m$ is $ (n\plus{}1\minus{}m)\plus{}1$. And the number of integers that is divisible by $ {p}^{m(m\plus{}1)}$ is $ (n\plus{}1\minus{} (m^2\plus{}m))\plus{}1$, then by principle of inclusive exclusive we obtain that the number is $ m^2$
[/hide]


limit
trigonometry
real analysis
real analysis unsolved

hello, the result is $ \frac{4}{9\pi^2}$.
Sonnhard.


1. find the number of sets A,B,C suc that AUBUC ={1,2,.......................2008}

2. 324.	Every possible triplet is taken from the finite G.P.
        { 1,2,4,8,16,32,64,128,256,512,1024} and multiplied. What is the sum of these
        Product results?
Try generalizing the problem to n terms. :lol:

[hide]
i think the second prob can be done without combi by atterns
[/hide]

2)[hide="Hint"] $ \Sigma \Sigma \Sigma 2^{\alpha} 2^{\beta} 2^{\gamma}$

where $ 0 <\equal{} \alpha < \beta < \gamma <\equal{}10$[/hide]

guys could somebody post the answer with their methods.
@ conjurer: could u plz post ur final answer and method  :)


trigonometry
geometry
analytic geometry
calculus
integration
algebra
polynomial

Mathematical puns are the first sines of madness ...

[quote="nutz_for2.718281828"]Let e be less than 0.[/quote]

Um, that's a math joke, not a pun. A pun:

You hear the one about the geometer who went to the beach to catch some rays and came back a tangent?

Anirudh, A post of

[quote]So?

:spam:[/quote] is actually itself spam. It has been deleted.

[quote="nutz_for2.718281828"]A math teacher was arrested this morning for bording a plane with a calculator, a protractor, a ruler, an Analytic Geometry book and a laptop. She was charged for boarding a plane with weapons of math instruction.

There's more to it, but I forgot it.[/quote]

She is believed to be part of the Al-Jabr group, an extremist organization that believes in solving equations with radicals.

[quote="bubka"]And is the square root of ab absurd?[/quote]

XD nice

[quote="nutz_for2.718281828"]Let e be less than 0.[/quote]

i believe it's supposed to be epsilon, the thingy that looks as if E and e had an inbred child

It's epsilon : $\epsilon$
It's not a joke at all, but if you say something like that, you can immediately find out who's a mathematician and who isn't.

How to catch a lion in the Sahara desert?
The lion, having spent his life under the Sahara sun, will surely have a tan. Induce him to lie on his back; he can then, by virtue of his reciprocal tan, be cot.

[quote="mathnerd314"]"What's new?"
"Oh, $\frac{c}{\lambda}$"

I don't really get it :blush:

I thought $\sin$ was when you did something wrong, $\tan$ was what you got after a long day at the beach, and $\cot$ was an uncomfortable bed. :roll: (Is it a sin to get a tan on a cot?)[/quote]

Do you seriously don't get it?

$\nu=\frac{c}{\lambda}$

$\nu$ = frequency of an electro magnetic wave
$\lambda$ = wavelength of an electro magnetic wave
$c$ = speed of an electromagnetic wave ($3.0\times10^{8}m/s$ in free space)

And did you hear the one about the boy who broke up with his girlfriend, a student of maths, accusing here of cheating after she admitted to have been in bed struggling with three unknowns? (not really a pun, but has a double meaning word, and is funny anyway)

well I don't really get why she was in bed but ok

and frequency is also written $f$ sometimes, so that may explain it, eh

[quote="bubka"]And is the square root of ab absurd?[/quote]

That's a pretty good one, but unfortunately, most people here probably wouldn't know that radical=surds... though maybe i'm wrong.

[quote="junggi"][quote="bubka"][quote="mathnerd314"]"What's new?"
"Oh, $\frac{c}{\lambda}$"

I don't really get it :blush:

I thought $\sin$ was when you did something wrong, $\tan$ was what you got after a long day at the beach, and $\cot$ was an uncomfortable bed. :roll: (Is it a sin to get a tan on a cot?)[/quote]

Do you seriously don't get it?

$\nu=\frac{c}{\lambda}$

$\nu$ = frequency of an electro magnetic wave
$\lambda$ = wavelength of an electro magnetic wave
$c$ = speed of an electromagnetic wave ($3.0\times10^{8}m/s$ in free space)[/quote]
I dont get it.[/quote]

Ok I'll add this: nu($\nu$) is pronounced new. Now do you get it?

Just a sec, why do you say that if you have a tan on your back, you lie down, then you have a cot? Well, if you sin near a sec. minister, you certainly will have a tan on your back!

Here's another one:

"See the secant?"
"If I see can't, then how do I see it?"

Math jokes are never funny: Do Greek cows say $\mu$?

at amsp, there was this one anonymous kid who was pretty much the master of sick mathematical puns, a guy who would've been worshipped at mathpath05

sadly odds are nobody here would get any of the puns

"amsp" is the stupid acronym

and try us

Yeah try us.

Meanwhile, here's one to keep us going:
No horse can understand rectangualr coordinates, for its no use placing Descartes before the horse.

A polar bear is a rectangular bear after a coordinate transform. :roll:

Which reminds me:

What is the contour integral around Western Europe?
[hide="Answer"]Zero, as all the Poles are concentrated in the Eastern Europe![/hide]

phelpedo doesnt go in this forum much, so i guess he didnt see this topic

however, he started this topic over here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=108404

this was too funny though i think it is a joke...

http://naturalmath.com/jokes/joke8.html

that makes no sense

you can get a pretty hefty polynomial tacked on or vanish completely! pretty major changes what

like I'm pretty sure you could approximate any continuous function arbitrarily closely


Hmm, it seems that my signature will not show up, I have re-submitted it quite a few times. It shows up in my profile, but not on the board.

Do a search, so I don't have to write the same thing that has been written a dozen times before:

When posting, check off the box that says "attach signature."  Or, go to your preferences and set it to always display signature.


Mafia game 22 has begun! It is now the first day in the... well... slightly dangerous town of Murdorsia. The inhabitants of this town are feeling a bit murderous (not for any particular reason), and it quite likely a lynch mob will kill a citizen before the day is out.

[b]Players:[/b]

1.  bubka
2.  Klebian
3.  Ars
4.  mustafa
5.  kheg_k
6.  Ignite168
7.  junggi
8.  wobster109
9.  tim1234133
10.  Peter VDD
11.  BoesFX
12.  akalra1
13.  ragnarok23
14.  dts
15.  LadyKn1ght
16.  daine
17.  perfect628
18.  laughinghead505
19.  mad_skillz_aops
20.  pianoforte

[b]Roles:[/b]

--- 4 mafia ---
No day communication allowed, except during day 1. The mafia also get the corpseripping ability, which means that nobody but them get to know who they killed.

--- 2 angels ---
No day communication allowed, except during day 1.

--- 1 mathematician ---
This innocent role can, once during the game, during the day, reveal him/herself and pick a group of 2 or more people (not including him/herself) to "inspect". I will then tell everybody how many mafia there are in the group, but not which of them is mafia. It is illegal to claim mathematician if you are not really mathematician. Once the mathematician has used his/her power, (s)he becomes the same as a normal civilian.

--- 1 magician ---
This innocent role gets to choose 2 people each night and whatever would have happened to them, is swapped round. For example, if A would have been killed by the mafia and B would have been inspected, B would be killed and A would be inspected. I will tell the mafia that they have killed B and the role of B, and the inspector will be told that he inspected A, and if A is guilty or not.

--- 1 inspector ---
--- 1 assassin ---
--- 1 vigilante ---
--- 1 fairy ---
--- 1 prosecutor ---
--- 1 priest ---
--- 1 devil ---
--- 2 masons ---
--- 2 civilians ---
--- 1 miller (who thinks (s)he is a civilian) ---

[b]Other info:[/b]

- Everybody has to post at least once every 72 hours (unless they said they would be away), or else they might be replaced.

- If a day lasts longer than a week, the town will go to sleep without a lynch.

- One person has a fake ID. If this person is killed, it will only be revealed that (s)he had a fake ID.

Looks like a good game.

But then again... The last game that Melissa modded, I was the singular reason innocents lost, BECAUSE I WAS GROUNDED.

So she modded before, did she? Ok try not to be grounded this tiem round, whatever that means.

OMG three mafia games on at once! Actually four if you count 18.
None will end ...  :(

no they will end 20 should end in a couple of days which leaves 2 games which should be fine

Well, 21 is going really fast, the nights are longer than the days.

I don't think mafia game 21 is that fast, but it is really fast. [Days are still longer than nights]   :wink:

So right now we're just checking in, or what?

Ok, what we need to do right now is mass role claim. That way we can know what to go after instantly as there are not that many safe outs for mafia.

Would you like to begin this roleclaim, then, Ignite?

Yeah I forgot to ask about that what will that do in the game

[quote="Ars"]A couple of Questions will the masons be able to induct or just talk at night also from what I understand the maifa and angels will have night communication or is that wrong for the angels (for mafia I understand it has to be right)[/quote]
Yes, sorry, I forgot to mention that the masons get the ability to induct once.
Mafia and angels both have only night-time communication (out of the thread, that is), except for during the first day, which is now.

[quote="bubka"]I did not get the fake ID part[/quote]
One person has a fake ID. Nobody but me knows who this is - it might be an innocent or a guilty. If this person is killed by the mafia, the mafia only get to know that the person they killed, had had a fake ID. If this person is killed by the vigilante or the assassin, the town only gets to know that (s)he had a fake ID. If the person gets inspected, I will also only reply that (s)he has a fake ID.

This is stupid. Lets just wait for Melissa.

[quote="ragnarok23"]The guilties are pianoforte....Is that all of them?[/quote]
I wasn't guilty- I was miller (apparently).

[quote="junggi"][quote="Klebian"]This is idiocy. Especially when the dead people are revealing their roles.[/quote]
yea ive noticed[/quote]
yea ive noticed too

Yeah.  While we're on the subject, I have something to say too.

[hide="I was..."]alive, before that blasted assassin killed me.[/hide]

OK, it seems all are in favour of restarting the game... I will do so shortly. Sorry guys for a very boring game, the new one will be better :)

Sorry, but my Internet connection is not working very well at the moment. I wanted to send out new role PM's, but it might stop working alltogether while I'm busy, and that would result in some people having new roles and some not.

I will start the new game as soon as my Internet is fixed, or else I will start it when I go to school on Monday.

There are many fixes. I'm sure Melissa will come up with something interesting. :)


Alright, I stole this from Pei, but anyway...

So, is anyone interested in setting up a match in Starcraft Battle.Net?

Several things we need to decide:
Should we play on a rescourse map?
How many players?
What mode (free for all? 2 v 2? etc.)

Also, my B-net unfortunately does not work. It says that my CD key is not intended for B - Net (no, I did NOT pirate the game). So I'm currently looking for my Starcraft case. I found the Brood War, but not the original. So unless someone PM's me a Cd key...  :wink:

Lastly, for certain reasons (no, I did NOT get grounded or anything like that), I will not be able to access the computer with SC until 2 weeks. Unless I can use my brother's computer, and I'm not sure if B-net works for that one. AND I probably won't get a chance to. He's usually using it 24/7.

Btw, I'm not hosting the game. I'm not taking charge. Too lazy. Just asking if anyone is interested.

I love Capture the Flag on BZFlag, is there the same mode on Starcraft?

4v4 Hunters
Or we could do 4v4c hunters.
Or something like that.

hunters... hmm... anyone wanna play 4v4 on one of those hacked up fastest map ever things (the ones where your hatchery is right next to the mineral, which holds 50000)?
Hunters is good also though.
Rofl, the team I'm on should seriously just have everybody zerg rush =).


For which real value of x is (1 - x)5(1 + x)(1 + 2x)2 maximum?
 Determine its maximum value.

Please don't use derivatives ;)

I hope this argument is correct...:D

The function takes positive values for -1 \leq x \leq 1 for these values we have
X 5YZ where X=1-x,Y=1+x,Z=(1+2x) 2 ...where X,Y,Z are  \geq 0 now assuming the function to be the GM it can attain a maximum iff X=Y=Z...

Giving us x=0 to be the only feasible solution....hence maximum value is 1

For which real value of x is (1 - x)5 ( 1 + x)(1 + 2x)2  maximum?
Determine its maximum value.

Expand it to get (1-x)(1-x)(1-x)(1-x)(1-x)(1+x)(1+2x)(1+2x).

Using AM-GM, we'll get, 1  $\geq$  [(1-x)5 (1+x)(1+2x) 2]1/8 so 1  $\geq$ (1-x)5 (1+x)(1+2x) 2.  Indeed 1 is the maximum value of the expression and that will be attained when x equals 0.

[quote="WarpedKlown1335"]For which real value of x is (1 - x)5 ( 1 + x)(1 + 2x)2  maximum?
Determine its maximum value.

Expand it to get (1-x)(1-x)(1-x)(1-x)(1-x)(1+x)(1+2x)(1+2x).

Using AM-GM, we'll get, 1  $ \geq$  [(1-x)5 (1+x)(1+2x) 2]1/8 so 1  $ \geq$ (1-x)5 (1+x)(1+2x) 2.  Indeed 1 is the maximum value of the expression and that will be attained when x equals 0.[/quote]

nice solution..

hmmm... isnt am-gm only guaranteed to work when all the terms are positive?

If $ x < \minus{} 1$, then the expression is negative and clearly can't reach the maximum there, and if $ x\geqslant \minus{} 1$, the argument stands.


Let $ M \equal{} \{ n^{2}\plus{}n | n \in \mathbb{Z}^{\plus{}} \}$ and prove that for all $ 2 \leq k \in \mathbb{Z}^{\plus{}}$ there exist distinct elements $ a_1 , ..... , a_k , b_k \in M$ such that $ a_1 \plus{} ... \plus{} a_k \equal{} b_k$.

[quote="BanishedTraitor"]Let $ M \equal{} \{ n^{2} \plus{} n | n \in \mathbb{Z}^{ \plus{} } \}$ and prove that for all $ 2 \leq k \in \mathbb{Z}^{ \plus{} }$ there exist distinct elements $ a_1 , ..... , a_k , b_k \in M$ such that $ a_1 \plus{} ... \plus{} a_k \equal{} b_k$.[/quote]

Given that $ a_1 \ge 3$, then

$ (b_m^2\plus{}b_m)\plus{}(a_{m\plus{}1}^2\plus{}a_{m\plus{}1})\equal{}(b_{m\plus{}1}^2\plus{}b_{m\plus{}1})$
$ \Leftrightarrow  b_m^2\plus{}b_m\equal{}(b_{m\plus{}1}\minus{}a_{m\plus{}1})(b_{m\plus{}1}\plus{}a_{m\plus{}1}\plus{}1)$

Clearly $ b_{m\plus{}1}\equal{}a_{m\plus{}1}\plus{}1$, $ a_{m\plus{}1}\equal{}\frac{b_m^2\plus{}b_m\minus{}2}{2}>b_m>a_m,a_{m\minus{}1},...,a_2,a_1$ satisfies the condition.

Now it's clear that for $ k\equal{}2$ the statement is true. And assume that the statement is true for $ k\equal{}m$ (i.e. such $ a_1, a_2,....,a_m,b_m$ exists) then the statement is also true for $ k\equal{}m\plus{}1$ (as from the above result we can find $ a_{m\plus{}1}$ and $ b_{m\plus{}1}$ if $ b_m$ is given).

By principle of mathematical induction the statement is true for all $ k \ge 2$


geometry
geometric transformation
reflection
number theory unsolved
number theory

Let $p$ be a prime number of the form $4k+1$ such that $2^{p} \equiv {2} (mod p^2)$. Prove that there exists a prime number $q$,divisor of $2^{p}-1$,such that $2^{q}>(6p)^{p}$   :(  :(

Please don't post problems from current issue of Mathematical Reflections till 1st March.


Take log: $\lg f(mn)=\lg f(m)+\lg f(n)$. Let $g(x)=\lg f(x)$, then $g(xy)=g(x)+g(y)$, so $g(x) = \log_a x$
Since: $f(2)=2$, so $g(2)=\lg 2 = \log_a 2$. Hence $a=10$. Therefore: $g(x)=\lg x = \lg f(x)$. Thus $f(x)=x$.

Is that correct?

shobber, the function is not real-valued, and you have to use the hypothesis $f(n) < f(n+1)$ otherwise function $f(2^n (2*m+1)) = 2^n$ (for instance) would work.

$f(1) = 1$, $f(2^n) = 2^n$, $f(2) < f(3) < f(4)$ so $f(3) = 3$
It's also clear that $f(n) \geq n$

If n is the smallest integer such as $n < f(n)$.
If $n=2N$ then $f(n) = 2f(N) = 2N$ contradiction.
If $n=2N+1$ then $f(2N)=2f(N)=2N < f(n) < f(2N+2) = 2(N+2)$ and $f(n) = 2N+1=n$ contradiction.

What is real valued? Why my proof was wrong because it is real valued?

[quote="seamusoboyle"]We can use Cauchys equations[/quote]
That is what I did.

shoober, seamusoboyle, do you mean that :

$f(mn) = f(n)f(m), \forall m, n \in N_0 \Rightarrow f(n) = n^a$ ?

If yes, this is false :

Take $f(2n) = f(n)$ and $f(2n+1) = 2n+1$ for instance.

[quote="K.W.M.A.N"]I've never heard of Cauchyequations but this might just do it.

Because of (1) and (2) we get $f(0)=0$ and $f(1)=1$. Now from (3) we get $f(4)=f(2\cdot2)=4$, which leaves only one possibility for $f(3)$, that is $f(3)=3$. Continuing this way, we get that $f(n)=n$ is the only possible answer, unless i made a mistake somewhere.[/quote]Hi Karsten, welcome to MathLinks! :)

Your solution is correct (or at least it's the same as mine) but I have a small style remark: for making mathematical proofs more clear, it's better not to write "and continue like that" or something alike. (on a competition, the graders cannot know if you really saw the trick and will not give you full marks)

When possible, try to word it somehow like this:
[quote]First, we will show that $f(2^k)=2^k$ for all $k\in\mathbb{N}$. Proof by induction: $f(1)=1, f(2)=2, f\left(2^k\right)=2f\left(2^{k-1}\right)$.

Now, we have that $f(n)=n$ for all $n\in\mathbb{N}$.
Between $2^{k-1}$ and $2^k$ we have $2^{k-1}-1$ numbers and $2^{k-1}-1$ different values the function can attain, thus we have a unique way to sort them ascending. As $f(n)=n$ is a valid way, this is our unique solution.[/quote]

:D KWMAN are my initials in case you were wondering.

It appears that I was not so clear indeed because your proof is slightly different from mine: I'll try to explain what I meant with "continue like this", I use induction of course:
Suppose that i have shown already that $f(n)=n$ for $n<2k+1$. Then I will proof that $f(n)=n$ for $n<2(k+1)+1=2k+3$.  For $n=2k+2$ is $f(n)=2f(k+1)=2(k+1)$, because $k+1< 2k+1$ ($k \neq0$). But now is $2k=f(2k)<f(2k+1)<f(2k+2)=2k+2$  so
$f(2k+1)=2k+1$.

I think the difference is that I used this order
 2,(0,1),4,3,6,5,8,7,&c
while you used this one
 2,(0,1),4,3,8,(5,6,7),16,(9,10,11,12,13,14,15),32,&c,

[quote="jensen"][quote="shobber"]Let $N_0=\{0, 1, 2 \cdots \}$. Find all functions: $N_0 \to N_0$ such that:

(1) $f(n) < f(n+1)$, all $n \in N_0$;

(2) $f(2)=2$;

(3) $f(mn)=f(m)f(n)$, all $m, n \in N_0$.[/quote]
we can prove that $f(2^k)=2^k,k \in N_0$ and $f(n+1)>f(n)$ $\rightarrow$ we have
$f(2^k)<f(2^k+1<...<2^{k+1}-1<2^{k+1} \rightarrow$
we know that $f(n) \in N_0$ $\rightarrow$ the numbers between $f(2^k)$ and $f(2^{k+1}$ should be the numbers between $2^k$ and $2^{k+1}$ $\rightarrow$
$f(n)=n$(?) :lol:[/quote]isn't that exactly what I wrote? ;)

oh yes,i didn't see your solution,sorry. :D


limit
number theory unsolved
number theory

Prove that: $ \lim_{n\to \infty} \ s(3^{n})\equal{}\plus{}\infty$, where $ s(x)$ means sum of the digits of x.

more general:[url]http://www.mathlinks.ro/viewtopic.php?t=118580[/url]
but the problem didn't solve there...


Given are two concentric spheres with radii 8 and 10.  A sphere with radius 1 is stained in ink and slides around in the space between these spheres such that it paints a region of area 2 on the larger sphere.  What is the area of the region painted on the smaller sphere?

In spolier

[hide]

Ink circle must always be tangent to both spheres, so the ink area on the big sphere divided by total surface area of big sphere equals the ink area on the small sphere divided by total surface area of small sphere.  Call ink area of small sphere x, then:

So:

so x=1.28

[/hide]


find this limit:  :D

$1\;+\;\cfrac{1}{2\;+\;\cfrac{2}{3\;+\;\cfrac{3}{4\;+\;\cfrac{4}{5\;+\;\cfrac{5}{6\;+\;\cfrac{6}{\cdots}}}}}}$

answer is given below, but give it a try first before looking:

[hide]

haha....... :P

 i said try it out first !   :rotfl:

[hide]
$answer: \;\;\;\quad \boxed{\frac{1}{e-2}}$[/hide][/hide]

$\textbf{I have more}$  :rotfl:

[img]http://bbs.pep.com.cn/attachments/Pi-Continued%20fractions_H9T7JRnbF62F.jpg[/img]

[img]http://bbs.pep.com.cn/attachments/E-Continued%20fraction_4f7P6NSuEsYB.jpg[/img]


What's the difference between the outbox and sentbox?

If they've read it, it's in the sentbox.  If they haven't, it's in the outbox.

ok thanks people.  BTW, if I delete a message in the outbox, will the originally intended recipient still receive the message?

[quote="probability1.01"]BTW, if I delete a message in the outbox, will the originally intended recipient still receive the message?[/quote]

There was one time when I COUNTED ON deleting a message from my outbox to keep the recipient from receiving it. (I was bringing up an issue that was resolved soon after I composed the message, and wasn't worth bothering the person about after that.) That's the expected behavior; it's like fishing a letter out of a mailbox before the postal service picks up the letter to send it through the mail system.


conics
hyperbola
geometry
geometry unsolved

i need 1 more point of this hyperbola

 :(

the 1 more point is one of the orthocenter of $ \triangle ABC$, $ \triangle ABD$, $ \triangle ACD$, $ \triangle BCD$ as well-know.


calculus
derivative
function
calculus computations

Let f(x) be a function continuous in [a,b] and differentiable in (a,b).Also     f(a)=a and f(b)=b.Show that there exist distinct 'c' and 'd' such that       f'(c)+f'(d)=2.

Since f is continuous on $[a, b]$ and also differentiable on $(a,b)$, the Mean Value Theorem says that a point c exists with $a< c < b$ such that $\frac{f(b)-f(a)}{b-a}= f'(c)$. Note that the derivative must be the same throughout the open interval $(a, b)$, which means $f'(d) = f'(c)$ as long as both c and d are in $(a, b)$. Using the fact that $f(b) = b$ and $f(a) = a$, we get $f'(c) = f'(d) = 1$.

[quote="akech"] Note that the derivative must be the same throughout the open interval $(a, b)$, which means $f'(d) = f'(c)$ as long as both c and d are in $(a, b)$. .[/quote]

How did you get that?

akech, you just proved that every differentiable function passing through the points $(a, a)$ and $(b, b)$ is linear.  Mayhap you want to reconsider?

akech's proof is clearly invalid. Let's try something else.

Define $g(x)=f(x)-f(a+b-x)-2x+a+b.$

(Adding the constants on the end was optional.) $g$ is continuous on $[a,b]$ and differentiable on $(a,b).$ By direct calculation, $g(a)=g(b)=0.$ By Rolle's Theorem, there exists $c\in(a,b)$ such that $g'(c)=0.$

That is, $f'(c)+f'(a+b-c)-2=0.$

This is what we want. Well, it's almost what we want. What about the requirement that $c$ and $d$ be distinct? In this proof, it was possible for $c=d=\frac{a+b}2.$ That still needs some fixing.

Oops! Sorry folks, I don't know what I was thinking. Thanks for pointing out my error.

HINT:
Let k=(a+b)/2

apply LMVT on [a,k] and again on [k,b]