limit
function
real analysis
real analysis theorems

What does the notation $x^{+}$ mean ? Would it be the fractional part of $x$ ? I have one in my course but I cannot figure out ...

That requires some context, but it's definitely not a fractional part.

The most common version I've seen is $\lim_{t\to x^{+}}f(t)$, which is the limit from above. In some situations, this might be abbreviated $f(x^{+})$.
The other version I know is $g^{+}(x)=\begin{cases}g(x)&g(x)>0\\0&g(x)\le0\end{cases}$; the positive part of a function.

Thank you. I eventually figured out it means $max(0,x)$ (since usually I see this notation only for functions, I was misleaded ...)


The "Log me on automatically each visit: " option does not seem to work. It hasn't been working for the past few months now.

This may be a cookies problem.  Check your Tools->Internet Options->Privacy if you're on IE.

I had that same problem, the cooke for aops seemed to be corrupt, and i was able to auto login to other websites.  What i did was goto tools->internet options then i clicked delete cookies, and the next time i logged in and checked off remember password it worked.


Find the greatest constant $ k$ such that the following inequality holds for $ a,b,c \in [1;k]$:
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{2a}{b+c}+\frac{2b}{c+a}+\frac{2c}{a+b}$

we know a/(b+c)+b/(a+c)+c/(a+b)<3/2 (when a,b,c>0)
the problem reduce to prove
a/b+b/c+c/a>3
we can use AM>GM inequality to prove above  inequality
so k is infinite.

[quote="The soul of rock"]Where is k??[/quote]

$ a,b,c \in [1;k]$

No. It is not true for all a,b,c. In some post in mathlinks, maybe Am-GM vesus Cauchy, I proved this inequality for $ a,b,c$ are three side-lengths of a triangle. But I forgot the links at the moment.

$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-2\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)$
$ =(a-b)^{2}\left(\frac{1}{ab}-\frac{2}{(a+c)(b+c)}\right)+(a-c)(b-c)\left(\frac{1}{ac}-\frac{(a+b+2c)}{(a+b)(a+c)(b+c)}\right)$
Now, assume that $ c=max\left\{a,b,c\right\}$, it's all right.
Hoping  that my solution is wight.

[quote="The soul of rock"]$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}-2\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)$
$ =(a-b)^{2}\left(\frac{1}{ab}-\frac{2}{(a+c)(b+c)}\right)+(a-c)(b-c)\left(\frac{1}{ac}-\frac{(a+b-2c)}{(a+b)(a+c)(b+c)}\right)$
Now, assume that $ c=max\left\{a,b,c\right\}$, it's all right.
Hoping  that my solution is wight.[/quote]

Sorry to say that it is wrong.  :lol:

Sorry, I just proved in case $ a\le b\le c$

I found $ k \approx 6.041584888$
Hope that it's not wrong. :lol:

sorry for my wrong solution, :blush:
a/(b+c)+b/(a+c)+c/(a+b)>3/2 :blush:
I think k is 1.


Let $a,b,c$ be three positive numbers. Prove that:
$\frac{bc}{a^{2}+2bc}+\frac{ca}{b^{2}+2ca}+\frac{ab}{c^{2}+2ab}\leq 1$

[hide]
Multiplying by two and then subtracting each term from 1, we find that this is equivalent to
\[\frac{a^{2}}{a^{2}+2bc}+\frac{b^{2}}{b^{2}+2ca}+\frac{c^{2}}{c^{2}+2ab}\geq 1\]
This follows easily from
\[2bc\leq b^{2}+c^{2}\Rightarrow \frac{a^{2}}{a^{2}+2bc}\geq \frac{a^{2}}{a^{2}+b^{2}+c^{2}}\]
because after using similar inequalities on the other terms and adding them up,
\[\frac{a^{2}}{a^{2}+2bc}+\frac{b^{2}}{b^{2}+2ca}+\frac{c^{2}}{c^{2}+2ab}\geq \frac{a^{2}}{a^{2}+b^{2}+c^{2}}+\frac{b^{2}}{a^{2}+b^{2}+c^{2}}+\frac{c^{2}}{a^{2}+b^{2}+c^{2}}\]
\[=1\]
[/hide]

bc/a*a+2bc = 1/ a*a/bc + 2
u can solve also making following substitution:
x=a*a/bc
y=b*b/ac
z=c*c/ab
xyz=1
starting from this the problem is quite simple

[quote="scorpius119"][hide]
Multiplying by two and then subtracting each term from 1, we find that this is equivalent to
\[\frac{a^{2}}{a^{2}+2bc}+\frac{b^{2}}{b^{2}+2ca}+\frac{c^{2}}{c^{2}+2ab}\geq 1 \]
This follows easily from
\[2bc\leq b^{2}+c^{2}\Rightarrow \frac{a^{2}}{a^{2}+2bc}\geq \frac{a^{2}}{a^{2}+b^{2}+c^{2}}\]
because after using similar inequalities on the other terms and adding them up,
\[\frac{a^{2}}{a^{2}+2bc}+\frac{b^{2}}{b^{2}+2ca}+\frac{c^{2}}{c^{2}+2ab}\geq \frac{a^{2}}{a^{2}+b^{2}+c^{2}}+\frac{b^{2}}{a^{2}+b^{2}+c^{2}}+\frac{c^{2}}{a^{2}+b^{2}+c^{2}}\]

\[=1 \]
[/hide][/quote]

[hide="Or just"]\[\sum_{\textrm{cyc}}\frac{a^{2}}{a^{2}+2bc}\geq \frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+2ab+2bc+2ca}= 1\]according to Cauchy in Engel form.[/hide]

Let $a,b,c$ be  positive real numbers. Prove that
$$\frac{a^{2}}{ka^{2}+bc}+\frac{b^{2}}{kb^{2}+ca}+\frac{c^{2}}{kc^{2}+ab}\leq \frac{3}{k+1}.$$
Where $k\geq 2.$

[quote=sqing]Let $a,b,c$ be  positive real numbers. Prove that
$$ \frac{a^{2}}{ka^{2}+bc}+\frac{b^{2}}{kb^{2}+ca}+\frac{c^{2}}{kc^{2}+ab}\leq \frac{3}{k+1}.$$
Where $k\geq 2.$[/quote]
$$\iff$$
$$\frac{bc}{ka^{2}+bc}+\frac{ca}{kb^{2}+ca}+\frac{ab}{kc^{2}+ab}\geq \frac{3}{k+1}$$
$$\sum_{cyc}\frac{bc}{ka^{2}+bc}= \sum_{cyc}\frac{b^2c^2}{ka^{2}bc+b^2c^2}\geq\frac{\sum_{cyc}b^2c^2}{kabc\sum_{cyc} a+\sum_{cyc}b^2c^2}$$
$$\frac{\sum_{cyc}b^2c^2}{kabc\sum_{cyc} a+\sum_{cyc}b^2c^2}\geq \frac{3}{k+1}\iff \sum_{cyc}b^2c^2\geq abc\sum_{cyc} a $$


Convert $101010_2$ to a number in base 4.

Or,

Each 2 numbers in base 2 is a number in base 4 (11 in base 2 is 3 in base 4)

So, 101010 in base 2 can be taken as:

10=2, 10=2, 10=2

222.

(Not sure if that's clear enough, but I couldn't think of a better way to put it)


Support
real analysis
real analysis unsolved

I cooked up the following questions but couldn't solve them, need help. The space is $R$,

1.$f_{n}\in L^{2}$, $f_{n}\to f$ uniformly on every compact set, then can we conclude that:
$f \in L^{2}$ and $f_{n}\to f$ in $L^{2}$?

2.$f_{n},f\in L^{1}\cap L^{2}$, $f_{n}\to f$ in $L^{2}$,
can we conclude that $f_{n}\to f$ in $L^{1}$? If not, what if $f_{n}, f$ are "nice": e.g. $f$ is bounded
and has compact support and $f_{n}$ is in, say, schwartz space?

1) No. $f_{n}(x)=(1-|x-n|)^{+}$, $f=0$

2) No. $f_{n}(x)=n^{-1}(1-|x|/n)^{+}$, $f=0$

Smoothness (Schwartz space, etc) would not help in (2), as it is irrelevant to such questions.


AMC
AIME
geometry
3D geometry
pyramid
AMC 10
number theory

I thought it was easier, but then again it was only my second official AMC 10 ever. I got a 131-- and was jubilant. Then upon looking back, I realized i could've easily gotten 150 (and I came into it expecting a 120 at most)...

My first 10  :blush:

You took the 10 when you were in 6th grade?

Yeah.

Seems like a lot of people thought it was easy...

138

I thought it was easy but i missed some careless problems...

143

I completely forgot to do the license plate one  :oops:
I thought it was easier than last year's but then again, that was last year.

I can't believe it I had a perfect score and then genious me I got number 16 wrong b/c I flipped the hypotenuse and leg... wow I'll never forget to mark my right angles on the AIMEs  :D 144 is not good enough...

Oh by the way I thought this was a really easy test. There were no brutal problems geometrically; and the hardest questions were number theory questions... which really favors me. I hope the AIMEs are this easy!

i thought the end was sorta hard and then the last problem was really simple.
I got approx 130 (not sure of real score)

You think flipping the hypotenuse and leg was a stupid mistake? Get a load of this one--on number 5, I did 7-3=3 !!!??? WTF lol.

But I felt I did alright considering I abhor counting/probability.

[quote="dogseatcheese"]129......First year in AMC10....8th grade....not particularly bright....am I.... :blush:

Difficult test.....whats next....AMC12 or AIME......If you've past the 10....[/quote]

126.5, 8th grade, first year as well  :roll:

I'm not sure if I'm taking the 12 or not.

WAAA I FORGOT TO DIVIDE BY THREE ON A PYRAMID!!!

I f'd up 15 with a stupid mistake and I made the wrong similar triangles on 16. I blanked 17 and 23 but upon further review they were really easy... darnit!!

There were 3 problems that sorta slowed me down. I don't really see how, but it was I believe 15, 16, and 17. It was just kinda hard to sort through them, but once I got through those the rest were all considerably easy. So I got a 150.


trigonometry
induction
function
complex numbers
algebra proposed
algebra

Solve
\[ \begin{gathered}
  \sum\limits_{k \equal{} 1}^n {\left( {\begin{array}{*{20}c}
   n  \\
   k  \\

 \end{array} } \right)} \cos \left( {k\phi } \right) \hfill \\
  \phi  \in \mathbb{R} \hfill \\
\end{gathered}\]

[quote="transseries"]I'd use complex numbers.[/quote]

Could you post your solution, please?
thanks

hello, you can prove by induction
$ \sum_{k\equal{}1}^n\binom{n}{k}\cos(k\phi)\equal{}2^n\cos\left(\frac{\phi}{2}\right)^n\cos\left(\frac{n\phi}{2}\right)\minus{}1$
Sonnhard.

[hide]$ \sum_{k\equal{}0}^{n}\binom{n}{k}cis(\phi)^k 1^{n\minus{}k} \equal{} (1\plus{}cis(\phi))^n$

$ \sum_{k\equal{}0}^{n}\binom{n}{k}cis(k\phi) \equal{} (1\plus{}\cos(\phi)\plus{}i\sin(\phi))^n$

$ \binom{n}{0}cis(0) \plus{} \sum_{k\equal{}1}^{n}\binom{n}{k}cis(k\phi) \equal{} \left(2\cos(\frac{\phi}{2})cis(\dfrac{\phi}{2})\right)^n$

$ 1 \plus{} \sum_{k\equal{}1}^{n}\binom{n}{k}cis(k\phi) \equal{} 2^n\cos^n(\frac{\phi}{2})cis(\dfrac{n\phi}{2})$

Take the real part of both sides:

$ 1 \plus{} \sum_{k\equal{}1}^{n}\binom{n}{k}\cos(k\phi) \equal{} 2^n\cos^n(\frac{\phi}{2})\cos(\dfrac{n\phi}{2})$

$ \sum_{k\equal{}1}^{n}\binom{n}{k}\cos(k\phi) \equal{} 2^n\cos^n(\frac{\phi}{2})\cos(\dfrac{n\phi}{2}) \minus{} 1$[/hide]

More generally, if you know either the ordinary or the exponential generating function of a sequence $ a_k$, you automatically know the ordinary or the exponential generating function for $ \sum_{k\equal{}0}^{n} {n \choose k} a_k$.

In the first case, the transformation is $ A(x) \mapsto \frac{1}{1 \minus{} x} A \left( \frac{1}{1 \minus{} x} \right)$.  In the second case, the transformation is $ A(x) \mapsto e^x A(x)$.  A useful special case is $ A(x) \equal{} \frac{1}{1 \minus{} x^m}$.

Just use the binomial development of:

$ [(cos\phi \plus{} isen\phi) \plus{} 1]^n \equal{}$
$ \equal{} \sum_{k \equal{} 0}^n (cos(k\phi) \plus{} isen(k\phi)) \equal{}$
$ \equal{} \sum_{k \equal{} 0}^n cos(k\phi) \plus{} i\sum_{k \equal{} 0}^n sen(k\phi)$

Now, using the expression:

$ [(cos\phi \plus{} isen\phi) \plus{} 1]^n \equal{}$
$ \equal{} 2^n(cos(\phi/2))^n(cos(n\phi/2) \plus{} isen(n\phi/2)) \equal{}$
$ \equal{} 2^n(cos(\phi/2))^n(cos(n\phi/2)) \plus{} 2^n(cos(n\phi/2))^n(sen(n\phi/2))i$

Igualating the real parts we obtain the final result:

$ \sum_{k \equal{} 1}^n (cos(k\phi)) \equal{} 2^n(cos(\phi/2))^n(cos(n\phi/2)) \minus{} 1$

And, just as curiosity:

$ \sum_{k \equal{} 1}^n (sen(k\phi)) \equal{} 2^n(cos(n\phi/2))^n(sen(n\phi/2))$

Hope I've helped!

Valle

Thanks everyone, I understood at all.
:D


I think you can use the fact that $ (\sqrt{3}\plus{}\sqrt{2})^{2008}\plus{}(\sqrt{3}\minus{}\sqrt{2})^{2008}$ is a integer number to solve your problem!

I tried it previously, that's why I posted this problem on this forum...

[quote="Invariant"]What is the first digit after 'coma' in its expansion:?? (sorry for english, I hope you understand it.)

$ (\sqrt2 \plus{} \sqrt3)^{2008}$[/quote]

$ (\sqrt2 \plus{} \sqrt3)^{2008}\equal{}a^{1004}$ where $ a\equal{}5\plus{}2\sqrt{6}$

$ a\equal{}5\plus{}2\sqrt{6}$ implies $ \frac{1}{a}\equal{}5\minus{}2\sqrt{6}$ and so $ a^{2004}\plus{}\frac{1}{a^{2004}}\equal{}m\in\mathbb{N}$

So $ (\sqrt2 \plus{} \sqrt3)^{2008}\equal{}a^{1004}$ $ \equal{}m\minus{}\frac{1}{a^{2004}}$

And, since $ a^{1004}>10$, we get  $ m\minus{}\frac{1}{10}<\sqrt2 \plus{} \sqrt3)^{2008}<m$ and so the required digit is $ 9$.

[quote="pco"]
$ (\sqrt2 \plus{} \sqrt3)^{2008} \equal{} a^{1004}$ where $ a \equal{} 5 \plus{} 2\sqrt {6}$

$ a \equal{} 5 \plus{} 2\sqrt {6}$ implies $ \frac {1}{a} \equal{} 5 \minus{} 2\sqrt {6}$ and so $ a^{2004} \plus{} \frac {1}{a^{2004}} \equal{} m\in\mathbb{N}$

So $ (\sqrt2 \plus{} \sqrt3)^{2008} \equal{} a^{1004}$ $ \equal{} m \minus{} \frac {1}{a^{2004}}$

And, since $ a^{1004} > 10$, we get  $ m \minus{} \frac {1}{10} < \sqrt2 \plus{} \sqrt3)^{2008} < m$ and so the required digit is $ 9$.[/quote]

Hi! You don't need to use $ a \equal{} 5 \plus{} 2\sqrt {6},\frac1a \equal{} 5 \minus{} 2\sqrt {6}$. We can use $ b \equal{} \sqrt3 \plus{} \sqrt2,\frac1b \equal{} \sqrt3 \minus{} \sqrt2$

[quote="tdl"]
Hi! You don't need to use $ a \equal{} 5 \plus{} 2\sqrt {6},\frac1a \equal{} 5 \minus{} 2\sqrt {6}$. We can use $ b \equal{} \sqrt3 \plus{} \sqrt2,\frac1b \equal{} \sqrt3 \minus{} \sqrt2$[/quote]

 :oops:  Indeed. Quite strange path in my brain ...   :blush:

And you said it before.

Another question:

Prove, that for every natural number $ n$

$ \lfloor (5\plus{}\sqrt{19})^{n}\rfloor$    is odd number.

[quote="Invariant"]eee....

How?? Can you explain??[/quote]

$ 1>5\minus{}\sqrt{19}>0$ and so $ 1>(5\minus{}\sqrt{19})^n>0$

$ (5\plus{}\sqrt{19})^n \plus{} (5\minus{}\sqrt{19})^n$ is a positive integer since it is $ a_n$ with $ a_1\equal{}10$, $ a_2\equal{}88$ and $ a_{n\plus{}2}\equal{}10a_{n\plus{}1}\minus{}6a_n$

So $ (5\plus{}\sqrt{19})^n \equal{} a_n \minus{} (5\minus{}\sqrt{19})^n$ and so $ [(5\plus{}\sqrt{19})^n] \equal{} a_n \minus{} 1$

And, since $ a_n$ is even, .... .


function
limit
real analysis
real analysis solved

compute the limit of the following sequence:

[a^n*n!]/n^n, for a>0

cheers! :D

I don't have a complete soln, just one for the easy cases  :D :
let a_n=[a^n*n!]/n^n. a_n+1/a_n=a*n^n/(n+1)^n. This has limit a/e as n->infinity. If a<e we get a_n->0, if a>e we get a_n->infinity. I'm still working on the case a=e.

I think that Stirling's Formula works fine :

n! =  \sqrt (2n* \pi) * (n/e)^n * e^(a_n) , where  \lim (n->\infty) a_n = 0

hmm...the solution in my book goes like so...

for a<>e you do a_(n+1)/a_n which grobber solved too :D

the difficult part comes to a=e so the solution they gave is like so:

a_(n+1)/a_n=e/(1+1/n)^n>1, n is natural. So a_n is strictly growing. we can see that (this is the part I don't like):
(a_n)^2/a_(2n)=(e^(2n)*(n!)^2)/n^(2n) * (2n)^(2n)/(e^(2n)*((2n)!))=
(after simplifications)=(2*4*...*(2n))/(1*3*..*(2n-1))=b_n. So (a_n)^2=a_(2n)*b_n, n is natural. Because a_n is strictly growing we have that a_n -->a which is from R U { \pm oo}. But b_n -->oo. if a belongs to R\{0} then a=oo, which is absurd. so a_n -->oo  (n^n/(e^n*(n!))-->0)

I really don't like this solution.... cause it's kinda "author like" solution... who would come with this solution?! :)

cheers! :D

What "the author" did is a part from the prof of Stirling`s formula

Here?s what I did:

a_n=e^n*n!/n^n. a_(n+1)/a_n = e/(1+1/n)^n=b_n. In order to prove that a_n->infinity it?s enough to prove that ln(b_1)+..+ln(b_n)->infinity as n->infinity. ln(b_n)=1-n*ln(1+1/n). If we denote 1/n by x_n ln(b_n)=1-(ln(1+x_n)/x_n). Try to prove this: 1-(ln(1+x)/x) >= x/3 for any x in the interval [0,1/2]. We apply this for x_n for n>=2 and we get ln(b_2)+..+ln(b_n) >= (1/3)*(1/2+..+1/n) which tends to infinity as n->infinity, so we proved what we wanted.

Were you talking seriously when you said it was easy? It took me quite some time. At the beginning I wanted to use Stirling?s formula to approximate n! and then apply L?Hospital, and all kinds of scary stuff J. Bye!

Find the limite of the function R(x) = (a^x.Gamma(x+1))/x^x
when x-->+oo

where Gamma(x) = int_{0^+oo} exp(-t).t^(x-1)dt with x > 0.

Remark: When x = n positive integer, Gamma(n+1) = n.Gamma(n)= n!
and R(n) = a^n.n!/n^n


trigonometry function calculus algebra polynomial conics parabola

Hey, (first post here by the way)

I'm a grade 10 student in Canada, but I've found that my math class goes waay too slow for me. It's not that the material is too easy, it's just that it takes the class several hours to go over one concept that I could've understood fully after an hour or less of reading it over and doing exercises. This whole month I've kind of been idling in class since I'd gone through the rest of the year's material :/

Long story short, this summer I'd like to look over what it is they teach in grade 11. However, although I can find plenty of online tutoring sites, I can't find anything that just teaches the concepts by way of text and independent study.

Is there such a thing? Maybe something geared toward adults who want to re-learn their high school math?

Thanks!

(also, since I really have no clue where this site originates (and sorry about that lol), here's a page listing math concepts for the grade level I'm looking at: http://math.about.com/library/bleleven.htm . I don't want to google those concepts individually, because they're pretty vague, and I doubt I'd find anything of the skill level I want..)

What math class are you currently in? I know that they use a different system in Canada, but what is the rough U.S. equivalent?

It would be much more worthwhile if you looked at AoPS Volume I or II instead of studying 11th grade math concepts - chances are, the 11th math material will be equally boring. Competition math in general will truly enhance your understanding of mathematics - something that school math, no matter how "advanced" [as they say], cannot do. The AoPS volumes are ideal for independent study.

Maybe you should buy some AoPS books and read them in class instead.  :P

If you insist on studying 11th grade material, I suggest that you look through a good trigonometry book. Other very important concepts include polar and parametric functions, and to a lesser degree, series and sequences. For me, [u]Schaum's Outline of Precalculus[/u] was good. If you have already covered this, it's time to order Apostol's Calculus Volumes I and II.  :yup:

Well, it should be the equivalent of U.S. Grade 11, the two school systems are very similar. The course code is MPM-3D, if that's what you're looking for, dunno how U.S. codes compare.

I suppose I could just list off what we did this year. I would want to expand on this knowledge: (hopefully I'm saying this right, since all of my education to date has been in french :P)
- Applying tangent, sine, cosine, and the laws of sine and cosine, to solve triangles
- Solving equations by substitution or elimination
- Solving polynomials
- using equations of the type y = ax^2+bx+c (parabolas)

etc. that's just off the top of my head, but it's the gist of what we learned this year.

I'll look into the books you suggested, that may be what I'm looking for. I'm just the type who thinks the internet holds all knowledge :P

Alright thanks, I'll check those out too.

I also read an excerpt from AoPS, it was about different bases and it blew my mind :P I'd thought about it when it came up in some Larry Niven books i read (sci-fi), but my head asploded when i realized binary was base 2, lmao..

So I'll definitely look into that. Anyway thanks for the help :D

The U.S. system of math is very different from ours, so giving them the course code won't help much. As somebody who took MPM3D (although it was quite a while ago now that I think about it  :P) trust me when I say that if you work AoPS vol. I and maybe the beginning of II you will be more than prepared to tackle the course material. Another good book would be intro. to Algebra, which would cover much more thoroughly most of the curriculum. The only thing it would lack is the trigonometry, the intro. to trigonometric functions, and the chapter on finance. It would cover whatever you need in exponents, transformations of functions (except trig. functions), solving equations, and sequences and series. It would also make up for the sections in the curriculum which they've taken out, like an introduction to complex numbers

Yeah, now that I worked it through my head, vol.1 and intro. to algebra would be perfect (and also ch 2 from vol. 2). Then do volume 2 and any other intro. books that will come to interest you while you're in class  :P. If you want to get ahead in math, learning ahead in the Ontario curriculum is not the way to go (I learned way too late...)

hah, I didn't realize the systems were so different..

Anyway, I checked at the biggest bookstore nearby, and the library just in case, but couldn't find the book (tons of books with really similar names though). Is it a hard book to come by? Would I have more luck at a used book store, or the university bookstore nearby, or is it easiest to just order online?

Not really sure how best to get my hands on it :P It's not exactly a bestseller, all I could find were a couple results on amazon and 2 ebay auctions to the U.S. only..

Edit: oh and also, is it worth getting the solutions book too?

These books are all available at this website, it is their books afterall. I'd get the solution manuals because the problems will be of a style that you've never seen before (that is, they're not just formulaic exercises). That being said, you will have to learn how to have enough self-control not to jump to the solutions the second you get caught on a problem, struggling with a problem is a good thing. Oh, and use this forum to talk about stuff you have trouble with, it's awesome.

Hmm, cause clicking the books button gives a 404 not found. It did that yesterday too. I guess something's down?

Edit: nvm found the right link with a little help from google. Thanks!

[quote="ChaosTheory"]It seems that you're in an Algebra II class. Therefore, I'd highly suggest [url=http://www.amazon.com/Schaums-Outline-Precalculus-Fred-Safier/dp/0070572615/ref=pd_bbs_sr_1/105-6862544-0534843?ie=UTF8&s=books&qid=1181418519&sr=8-1]Schaum's Outline of Precalculus[/url], or if you want something more advanced, [url=http://www.amazon.com/Schaums-Outline-Beginning-Calculus-Mendelson/dp/0070417334/ref=sr_1_2/105-6862544-0534843?ie=UTF8&s=books&qid=1181418561&sr=8-2]Schaum's Outline of Beginning Calculus.[/url] Also, you can look through [url=http://tutorial.math.lamar.edu/]Paul's Online Math Notes[/url] to learn some basic Trigonometry concepts.

Here are some websites I found:
http://jwbales.home.mindspring.com/precal/
http://www.analyzemath.com/precalculus.html <-- Polar Coordinates
http://quiz.uprm.edu/tutorials/ <-- Probably a review for you.
http://euphrates.wpunj.edu/faculty/goldsteine/precal/plainprecal.htm <-- You can get an idea of what a Precalculus test looks like.
http://www.math.hmc.edu/calculus/tutorials/
http://www.sci.uidaho.edu/polya/math143/video_instruction/video_instruction.htm <-- Video lectures.

Hope I've helped.  :jump:[/quote]

Are you sure?  the things he said were all covered in our 7th grade algebra 1 class...
and a little of it (law of sines and cosines) in 8th grade geometry.

Grade 11 math in Ontario gets into trig functions and identities, conic sections (at least, it used to) and a couple other things you wouldn't see in Algebra I. That being said, it's not as comprehensive as Algebra II (we have a new course called "Advanced Functions" that covers what that course misses in the Algebra II/ Pre-calc couses).

NetworkTutors has online tutors for school and college level Math courses and you may want to try there. They have thousands of problems in online worksheets too so try your luck.
[url]http://www.networktutors.com[/url]

[quote="dheerajsl"]Hi I found this site with a lot of info
http://www.iitdreams.com
Prbably the largest database of study material on the net

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http://www.iitdreams.com/getting-into-engineering[/hide][/quote]

Please don't revive old topics.

Thanks, Locked.


They will be [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2009-12a/09-2-1012-Answers.shtml]here[/url] before tomorrow.

What did you guys get on the cube problem?

The answer on the cube problem (assuming you mean the one with the stripes) is $ \frac{3}{16}$.

The answer is 3/16 for that cube problem.

Edit: Oops, beaten.

Well I got the answer key when I left the testing area so here you go:
1. B
2. C
3. C
4. C
5. D
6. D
7. C
8. B
9. A
10. E
11. A
12. A
13. C
14. D
15. E
16. B
17. C
18. D
19. A
20. B
21. D
22. B
23. C
24. A
25. B

cube one: there are 64 ways you can permute those strips.

there are only 3 ways a full strip can be made. For each full-go-around strip, you can adjust the side strips in 4 ways. 3*4 = 12

Thus, 12/64 = 3/16.

I didn't get this problem (didn't attempt) because I thought it was hard. After the test, I read the last line to the problem, then I smacked myself on the desk. Then I solved the question under 30 seconds.

126 <- EPIC FAIL. (missed 21... forgot to add 3 and 1)

btw congrats to solitxnj (or w.e. his account is here) for perfect score on AMC 10B

alekim: You think you failed? I got 25 wrong because I divided 64/4 = 8 so I chose D. 24, I divided 1440 by 10 instead of 9. 23, I really don't know. 14, I skipped to do later because I didn't want to do all the math, and 5 I'm just going to not talk about.

120... even bigger FAIL.

[quote="athunder"]omg 23 was C???? what answer fraction is that...[/quote]

C was 3/16. I could swear it was B (1/8)...

are u sure that 4 was 1/2

everyone said it was 1/5

Yes, I made a mistake.  It should have said 1/5.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=261085

Both runners have to be in the photo at the same time... I multiplied the individual probabilities but sometimes they're both not in the shot at the same time even though one runner is.


algebra
polynomial
Ring Theory
superior algebra
superior algebra solved

Let f,g polynomials in A[X], (A is a ring). A polynomial is said to be primitive iff the ideal generated in A by its coefficients is A. Prove that fg is primitive iff f and g are primitive.

It seems very easy. $\Rightarrow$ was easy, but $\Leftarrow$ wasn't friendly at all :)

Is it true for all rings ? Even for ring with zero divisors ? Doesn't A need to be commutative as you mentioned in the title ?

$\Rightarrow$ Suppose fg is not primitive. Then its coefficients lie in a maximal ideal M. Take A/M. Then fg=0 in A/M, so f=0 or g=0. Thus f or g is not primitive.
Indeed, an easy one :P

Suppose $f$ or $g$ are not primitive. WLOG suppose $f$ is not primitive. Then its coefficients lie in a maximal ideal $M$. Then in $A/M[X]$, $f=0$, so $fg=0$, so $fg$ is not primitive in $A[X]$.


inequalities
complex analysis
complex analysis unsolved

Let f be analytic in the annulus $1\leq \mid z\mid \leq2$ .                            let $\mid f\mid\ \leq  4$ on $\mid z\mid$ =$2$ and $\mid f\mid\ \leq 1$ on $\mid z\mid =1$ . show that $\mid f(z)\mid\leq \mid z^{2}\mid$ on the entire annulus.

Maximum principle applied to $\frac{f(z)}{z^{2}}$. That's all.


geometry
trigonometry
geometric transformation
reflection
incenter
rhombus
circumcircle

This is just what is says in the title; a geometry marathon.

NOTE: This is a geometry marathon, not a trig marathon, pure trig problems (as in trig sequences, proving trig equations) aren't allowed, but you are very welcome to use trig in your solutions. Also, please number the problems since every 15 problems, I'm going to make a PDF out of em.

If a problem is clogging up the marathon for more than 2 days, and the poster hasn't posted a solution despite a PM (which will be sent out after the first day), I'll post a new problem.

1. In a triangle ABC, median BD is such that angles A and DBC are equal, and angle  ADB = 45°, prove that angle  A = 30°.

[hide="Teensy hint"]
Use trig.
[/hide]

[quote="mathwizarddude"]Man I suck at Euclidean geometry... I stared at the diagram trying to do some construction work but failed...
[hide="trig bash"]$ Let \angle A \equal{} x$. $ DB/{DA} \equal{} \sin x/\sin(135 \minus{} x) \equal{} \sin(45 \minus{} x)/\sin x$
$ \implies \sin^2x \equal{} \sin(45 \minus{} x)\cos(45 \minus{} x)$
$ \equal{} 1/2(\cos2x) \equal{} 1/2 \minus{} \sin^2x$ So $ \sin x \equal{} 1/2$ (acute) and hence the result. [/hide]
A synthetic solution is appreciated
Problem 2: Try the last two unsolved problems in this thread [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310548[/url][/quote]
One problem please. (those look a bit hard, are you sure they fit into intermediate? One may think they're the ones clogging your marathon!)

Alternate solution to problem 1.

Because of $\triangle BDC \sim \triangle ABC,$ we have $\frac {BC}{DC} \equal{} \frac {AC}{BC}$ $\Longrightarrow$ $BC^2 \equal{} DC \cdot AC \equal{} 2 \cdot DC^2$ $\Longrightarrow$ $BC \equal{} \sqrt {2}DC.$ If $X,Y$ denote the orthogonal projections of $D$ and $ B$ onto $BC$ and $AC,$ we have $\frac {BC}{DC} \equal{} \frac {BY}{DX} \equal{}\sqrt {2}.$ Since $\triangle BYD$ is isosceles with apex $Y,$ then $BY \equal{} \frac {\sqrt {2}}{2}BD$ $\Longrightarrow$ $DX \equal{} \frac {1}{\sqrt {2}} \cdot \frac {\sqrt {2}}{2}BD \equal{} \frac {1}{2}BD$  $\Longrightarrow$ $\angle CBD \equal{} \angle BAC \equal{} 30^{\circ}.$

New problem:

http://www.gogeometry.com/problem/problem002.htm

[quote="gogeometry"]Prove the following: if, in a triangle $\triangle ABC,$ $\angle BAO=\angle OAC=20^{\circ},$ $\angle ACO=10^{\circ},$ and $\angle OCB =30^{\circ},$ then $\angle OBC=80^{\circ}.$ Try to use elementary geometry (Euclid's Elements.)[/quote]

Let the reflection of $CA$ across $CO$ cut $BA$ at $P$ and let $Q \equiv AO \cap BC.$ Then we have $\angle OCP = \angle OCA = 10^{\circ}$ and $\angle PCB = \angle QAB = 20^{\circ}.$ Since $\triangle BAC$ is isosceles with apex $B,$ it's clear that $PQ \parallel AC$ $\Longrightarrow$ $\angle QPC = \angle PCA = 20^{\circ}.$ On the other hand, since $CO$ and $AO$ are bisectors of $\angle PCA$ and $\angle PAC,$ it follows that $O$ is the incenter of $\triangle APC$ $\Longrightarrow$ $PO$ bisects $\angle APC$ $\Longrightarrow$ $\angle APO = \frac {_1}{^2}(180^{\circ} - 40^{\circ} - 20^{\circ}) = 60^{\circ}.$ But $\angle BQA = 40^{\circ} + 20^{\circ} = 60^{\circ},$  thus the quadrilateral $PBQO$ is cyclic $\Longrightarrow$ $\angle OBC = \angle OPQ = 60^{\circ} + 20^{\circ} = 80^{\circ}.$ $\blacksquare$

[b]Problem 3:[/b] $ABCD$ is a rhombus such that $\triangle ABD$ is equilateral. An arbitrary line $\ell$ through $C$ cuts the extension of its sides $AB$ and $ AD$ at $M$ and $N,$ respectively. Show that the lines $BN$ and $DM$ meet on the circumcircle of $\triangle ABD.$

[hide="Hint for problem 3"]
Try to show that $\triangle BDM \sim \triangle DNB,$ then the result follows by simple angle chasing. [/hide]

Okay. I promise we'll come back to your problem, I don't have time to solve it, for now. I'll solve it as soon as I get home from school tomorrow.

New no. 3:
in right triangle ABC, BF is a median from B (that is where the right angle is) to the hypotenuse, BE bisects angle ABC and BD is an altitude from B to the hypotenuse (AC). Prove angle DBE equals angle EBF.

[hide="Hint"]
$ BF\equal{}AF\equal{}CF$
[/hide]

[hide="solution"]We must prove that $ \angle ABD\equal{}\angle FBC$
So let $ \angle DAB\equal{}\alpha \equal{}> \angle ABD\equal{}90^\circ \minus{}\alpha$
Now since $ AF\equal{}BF \equal{}> \angle DAB\equal{}\angle ABF\equal{}\alpha   \equal{}>  \angle CFB\equal{}2\alpha$
Now $ CF\equal{}BF  \equal{}> \angle FBC\equal{}\angle FCB\equal{}90^\circ \minus{}\alpha$
Hence $ \angle DBE\equal{}\angle EBF$.[/hide]

Problem 4.:
The incircle of a triangle ABC is tangent to BC,CA,AB at M,N and P, respectively. Show that
the circumcenter and incenter of tr.ABC and the orthocenter of tr.MNP are collinear.

[quote="Yavor Tsvetanov 52"]Problem 4.: The incircle of a triangle ABC is tangent to BC,CA,AB at M,N and P, respectively. Show that the circumcenter and incenter of tr.ABC and the orthocenter of tr. MNP are collinear.[/quote]
The inversion with respect to the incircle $(I)$ of $\triangle ABC$ takes the vertices $A,B,C$ into the midpoints $A',B',C'$ of $MP,PM,MN.$ Hence the circumcircle $(O)$  is taken into the circle $\odot(A'B'C'),$ which is the 9-point circle of $\triangle MNP.$ Since the center of the inversion is collinear with the centers of the circle and its inverse image, it follows that the 9-point center of $\triangle MNP$ lies on $IO$ $ \Longrightarrow$ $IO$ is  Euler line of $\triangle MNP,$ thus the orthocenter of $\triangle MNP$ lies on $IO.$

Next problem is the unsolved problem 3. A hint can be seen in the post number #6


Hi everyone,

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For those of you which have no clue as to what I am taking about, please read the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16275]FAQ topic[/url]. All those that know what WAP is can use the above link to access the forum directly on your preffered mobile device, be it a mobile phone, a pda, or a i-pod.

Wap version supported is WAP 1.1, available since 2000, so basically almost any mobile phone younger than 4 years can access it.

Also there is a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16274]bug / support thread.[/url] Please report there any such instances.

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This is the first "official" post to the AoPS - MathLinks forum from a mobile device. This WAPgateway was throughly tested with the Sony-Ericsson K700i phone.

[i]Posted from a Sony Ericsson K700i [/i]


Δείτε μια εξαιρετικά όμορφη άσκηση . Νδο

Αν $ a,b,c$ θετικοί πραγματικοί με $ abc \equal{} 1$, τότε

$ \frac 1{1 \plus{} a \plus{} b^2} \plus{} \frac 1{1 \plus{} b \plus{} c^2} \plus{} \frac 1{1 \plus{} c \plus{} a^2}\le 1$


Determine $ 2^x$ if

$ x \equal{} \frac{\log_5 {63}}{\log_5 {2}}$

[quote="Silverfalcon"]Determine $ 2^x$ if

$ x \equal{} \frac {\log_5 {63}}{\log_5 {2}}$[/quote]

[hide]$ x \equal{} \frac {\log_5 {63}}{\log_5 {2}} \equal{} \log_2 {63}$

So $ 2^x \equal{} 2^{\log_2 {63}} \equal{} 63$.[/hide]


[quote="kunny"]Let $a,b,c$ be positive reals.

Prove that $\frac{a}{a^4+a^2+1}+\frac{b}{b^4+b^2+1}+\frac{c}{c^4+c^2+1}\leqq \frac{1}{3}\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)$.[/quote]

Isn't this a bit trivial? I mean, since AM-GM yields $a^4+1\geq 2\sqrt{a^4\cdot 1}=2a^2$, so that $a^4+a^2+1=\left(a^4+1\right)+a^2\geq 2a^2+a^2=3a^2$, and similarly $b^4+b^2+1\geq 3b^2$ and $c^4+c^2+1\geq 3c^2$, we have

$\frac{a}{a^4+a^2+1}+\frac{b}{b^4+b^2+1}+\frac{c}{c^4+c^2+1}\leq\frac{a}{3a^2}+\frac{b}{3b^2}+\frac{c}{3c^2}$
$=\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}=\frac{bc+ca+ab}{3abc}\leq\frac{a^2+b^2+c^2}{3abc}$ (since $bc+ca+ab\leq a^2+b^2+c^2$ by a well-known inequality)
$=\frac13\cdot\frac{a^2+b^2+c^2}{abc}=\frac13\cdot\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)$.

  darij

Yes. darij's solution is the back ground of making this problem.


function
inequalities
inequalities proposed

For a sequence of positive reals $\{a_i\}_{i=1}^{k+1}$, with $a_{k+1}=a_1$ and $\sum_{i=1}^ka_i=1$, find the minimum value (as a function of $n$ and $k$) of

\[\sum_{i=1}^k\left(\frac{a_i^{n+1}}{\sum_{j=0}^na_i^ja_{i+1}^{n-j}}\right).\]
This is a generalisation of an old Irish MO problem, which had $n=2$ and $k=4$.

Bump, bump.

I know it looks a bit messy, but it's not that bad.

Just by way of encouragement, this was the original problem: Given $a,b,c,d\in\mathbf{R}$ with $a+b+c+d=1$, prove that
\[\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}\ge\frac{1}{2}.\]

[hide="Hint"]
Consider the relationship between $S_1=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a}$ and $S_2=\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{d^2}{c+d}+\frac{a^2}{d+a}$.
[/hide]

The original problem is immediate with Cauchy.  That implies that you may be able to solve the generalization using Holder.  It's 3AM here, so I'll work on it when I wake up.

And just so Arne doesn't get mad at me  :D, I mean the original one is just

$\sum \frac {a^2}{a+b}\geq \frac {(a+b+c+d)^2}{2(a+b+c+d)}=\frac {1}{2}$

Cool problem... let
\[A = \displaystyle \sum_{i=1}^k\left(\frac{a_i^{n+1}}{\sum_{j=0}^na_i^ja_{i+1}^{n-j}}\right) \qquad B = \displaystyle \sum_{i=1}^k\left(\frac{a_{i+1}^{n+1}}{\sum_{j=0}^na_i^ja_{i+1}^{n-j}}\right).\]
We have $A - B = \sum_{i=1}^k (a_i-a_{i+1}) = 0 \Longrightarrow A = B.$  Thus,

$A = \frac{1}{2}(A+B) = \frac{1}{2} \displaystyle \sum_{i=1}^k\left(\frac{a_i^{n+1}+a_{i+1}^{n+1}}{\sum_{j=0}^na_i^ja_{i+1}^{n-j}}\right)$
$\geq \frac{1}{2(n+1)}\sum_{i=1}^k (a_i+a_{i+1}) = \frac{1}{n+1}$

because $a_i^{n+1}+a_{i+1}^{n+1} \geq \frac{1}{2}(a_i+a_{i+1})(a_i^ja_{i+1}^{n-j} + a_i^{n-j}a_{i+1}^j)$ follows from

$(a_{i}^{j+1} - a_{i+1}^{j+1})(a_{i}^{n-j}-a_{i+1}^{n-j}) + (a_{i}^{j} - a_{i+1}^{j})(a_{i}^{n-j+1}-a_{i+1}^{n-j+1}) \geq 0,$

and we just have to sum over $j = 0,1,\dots , n$...

(Actually, I recall seeing the k-variable inequality with n = 3 in some Chinese Olympiad book...)

[quote="cezar lupu"]I'm sorry blahblahblah, but you can't apply Cauchy because $a,b,c,d\in\mathbb{R}$ or you can if Diarmuid edits his third post then you can :D[/quote]

Well, the original problem stipulates positive reals.

http://www.kalva.demon.co.uk/irish/soln/sol998.html

Great solution ThAzN1, and yeah, the original problem should have $a,b,c,d>0$.


A circle has a radius of three inches. The distance from the center of the circle to chord $ CD$ is two inches. How many inches long is chord $ CD$? Express your answer in simplest radical form.

The distance from the center to chord $ CD$ is a perpendicular distance.  In circles, a line perpendicular to the chord that is the diameter or on the line of the diameter bisects the chord.  When you draw in that distance, and then draw in the radius from the center to an endpoint of the chord, you get a right triangle.  The hypotenuse is $ 3$, the legs are $ 2$ and $ \frac{1}{2}CD$.  So we get $ 3^2\minus{}2^2\equal{}(\frac{1}{2}CD)^2$.  We get $ 5\equal{}\frac{1}{4}CD^2$.  $ CD^2\equal{}20$, so $ CD\equal{}\sqrt{20}\equal{}\boxed{2\sqrt{5}}$

On FTW, I typed in the write answer, but I wrote it like this, 2sqrt(5). When I reviewed the question, I saw that I had to use latex. Will it always be like this?

You can just write 2sqrt5 without parentheses.


group theory
abstract algebra
superior algebra
superior algebra unsolved

A group is generated by elements $ x,y$ s.t. $ x^2\equal{}y^3$ and $ x^4\equal{}y^6\equal{}e$.
Find the number of elements of this group.

this description does not make the group unique. such a group is a quotient of $ <x,y : x^2 \equal{} y^3 , x^4 \equal{} y^6 \equal{} 1> \equal{} \mathbb{Z}/4 *_{\mathbb{Z}/2} \mathbb{Z}/6$, which is known to be isomorphic to $ SL(2,\mathbb{Z})$. thus you have to determine the normal subgroups of $ SL(2,\mathbb{Z})$ and then take the quotients. examples are  $ SL(2,\mathbb{Z})$ and $ PSL(2,\mathbb{Z})$.

... there are infinitely many other normal subgroups like $ ker(SL_2(\mathbb Z)\rightarrow SL_2(\mathbb Z/n\mathbb Z))$ for every natural $ n$.


geometry
circumcircle
calculus
real analysis
real analysis unsolved

The figure* illustrates circumscribing regular polygons.  The number of sides of the polygon successively increases.  Will the radii grow without limit? If not, can you estimate the limit, assuming the smallest circle to have a radius of 1 inch?
*the figure starts with a central incircle , a triangle and its circumcircle, a square and its circumcircle; a pentagon and its circumcircle, a hexagon and its circumcircle, and so on with the number of sides of the polygon increases by one each time.  The circumcircle of the smaller polygon is the incircle of the larger polygon.
(the answer given is the increasing radii will approach a limit that is about 12 times that of the original circle.  Can anyone post the solution?)

It seems more like a kind of calculus problem, there is where I am sending it!!


A student spins the spinner twice and records the number that
the spinner lands on each time. What is the probability that the
sum of the two recorded numbers is less than 4, if the
probability of spinning each number is $ \frac 14$? Express your answer as a common fraction.

[asy]size(100);
draw(Circle((0,0),20));
draw((20,0)--(-20,0));
draw((0,20)--(0,-20));
label("1",(-8,11),S);
label("3",(-8,-11),N);
label("2",(8,-11),N);
label("4",(8,11),S);
draw((0,0)--(5,11),Arrow);[/asy]

[hide]We have a total of $ 4\cdot4\equal{}26$. The ways to get a sum less than $ 4$ is \[ 1,2\]\[ 2,1\]\[ 1,1\] Thus, our answer is $ \frac{3}{16}$.[/hide]

[quote="pytheagle"]We have a total of $ 4\cdot4\equal{}26$.[/quote]
$4\cdot 4\equal{}16$, not $26$.

[quote=Ryan2010]i accidently wrote 3/17[/quote]

:skull:

[hide=sol]Three ways to obtain a valid spin-off, (1,1 1,2 2,1) out of 4*4=16 total ways so $\boxed{\dfrac{3}{16}}$[/hide]

[quote=Marcus_Zhang][quote=Ryan2010]i accidently wrote 3/17[/quote]

:skull:

[hide=sol]Three ways to obtain a valid spin-off, (1,1 1,2 2,1) out of 4*4=16 total ways so $\boxed{\frac{3}{16}}$[/hide][/quote]

ftfy


Let a,b are two natural numbers such that p = b/4  \sqrt ((2a-b)/(2a+b)) is a prime number. How large can p be ?

We have a=[b*(b 2 +16p 2 )]/[2(b 2 -16p 2 )]. Thus, b is even. Let b=2c. Then a=[c*(c 2 +4p 2 )]/(c 2 -4p 2 ). Thus, we must find the maximal value of p such that there exists c such that 8p 2 c is divisible by c 2 -4p 2 . First, suppose p|c and let c=rp. Then r 2 -4| 8pr. So, r 2 -4| 8pr 2 and thus r 2 -4| 32p.It follows immediately that r 2 -4 cannot be coprime woth p, so it is divisible by p. In this case we must have r 2 -4 is p,2p,4p,8p,16p,32p,-p,-2p.... and these cases are trivial to analyse.
  Now suppose that (c,p)=1. Then (p 2 ,c 2 -4p 2 )=1 and so c 2 -4p 2 |8c. Suppose p is not 2. It follows that any prime dividing c 2-4p 2  is 2. So, c 2 -4p 2 = \pm 2^k. Of course, k is not 1. Suppose k>2. Then c=2d and d 2 -p 2 =2^(k-2) and d 2 -p 2  |4d. So, 2^(k-2)|4d and since d is not even (otherwise p is even), we must have k \leq 4. Anyway, we have that k is 0,2,3 or 4. Again these cases are not dificult.


Let $ a_1,a_2,...$ be strictly increasing sequence of positive integers such that $ gcd(a_m,a_n)\equal{}a_{gcd(m,n)}$ for all positive integers $ m$ and $ n$. There exists a least positive integer $ k$ for which there exist positive integers $ r<k$ and $ s>k$ such that $ a_k^2\equal{}a_ra_s$. Prove that $ r|k$ and $ k|s$

[quote="forevercontinue"]There exists a least positive integer $ k$ for which there exist positive integers $ r < k$ and $ s > k$[/quote]
Do you mean that there are 3 numbers r<k<s such that ....?  :huh:

ive seen this problem somewhere.
where did you get it?


quadratics logarithms inequalities function conics hyperbola ellipse

I am looking for an algebra book/ebook (the latter is preferred) that emphasises on the following topics: [hide="topics"] quadratic equations, remainder and factor theorem, exponential equations, logarithms, inequalities (only quadratic, linear, rational functions, and modulus; no AM-GM-HM whatsoever), composite functions, inverse functions, absolute value functions, linear law, and finally graphs of linear, quadratic, reciprocal, logarithmic, piecewise, hyperbolas, ellipses, rational functions, circles, and geometrical transformations of graphs to obtain other graphs. [/hide]

I am looking for a problem book as I am quite familiar with these topics and would like to have a test of my skills.

can't go wrong with intermidiate algebra. or you could go with 101 Problems in Algebra from the Training of the US IMO Team by Titu Andreescu and Zuming Feng

I've already finished the necessary portions of Intermediate Algebra.

If you need problems, AoPSWiki has a lot of categorized problems.

For example, [url=http://www.artofproblemsolving.com/Wiki/index.php/Category:Intermediate_Algebra_Problems]Intermediate Algebra Problems.[/url]

As for books, I guess [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts_ProbSolve.php]AoPS Volumes 1 and 2[/url] (mainly 2) have several sections on the stuff you mentioned.


I've found that apart from pdf and dvi, TeXnicCenter can also generate output file in .ps format, but how to open it????

Download Ghostscript and GSview from this site
http://www.cs.wisc.edu/~ghost/


[quote="gracexcor3"]Can anyone explain how to solve this problem?
"The measure of one angle of a triangle is 70. The other two angles have a ratio of 4:7. Find each angle."
thanks!
 :)[/quote]

One angle is 70°. What can you then say about the sum of the other two? How do you solve an equation of te form $ x\plus{}y\equal{}a,\frac{x}{y}\equal{}b$?

[hide="Hint"]$ 70 \plus{} 4x \plus{} 7x \equal{} 180$[/hide]

[hide="Solution"]Well since we know that the sum of the internal angles of a triangle are 180 we know that the other angles have to sum to 110. Since the other two angles are in ratio 4:7 they can be written as $ 4x$ and $ 7x$ respectively.

So we get $ 4x \plus{} 7x \equal{} 110$

and $ x \equal{} 10$

From there we substitute that back in to get $ 4x \equal{} 40$

and $ 7x \equal{} 70$

So the three angles have measure $ 40, 70, 70$[/hide]

oh. i got it before you posted the answer.
CHYEAH!
thanks anyways!
:]


calculus
integration
calculus computations

Your integral equals

$\sum_{l\geq 0}\frac{1}{(2m+1+2nl)^{3}}$.

This result can be obtained by expanding the integrand and then integrating term by term.


The sum of three numbers $ a$, $ b$ and $ c$ is 77. If $ a$ decreased  by 5 is equal to $ b$ increased by 5 and is also equal to $ c$ multiplied by 5, what is the value of $ b$?

$ a\plus{}b\plus{}c\equal{}77$, $ a\minus{}5\equal{}b\plus{}5\equal{}5c$.

$ a\equal{}b\plus{}10$
$ b\equal{}5c\minus{}5$
$ 2b\plus{}c\equal{}67$.
$ 10c\minus{}10\plus{}c\equal{}67 \implies 11c\equal{}77 \implies c\equal{}7$.

This gives $ b\equal{}30$, and $ a\equal{}40$.


geometry
angle bisector
geometry unsolved

I put this problem in an exam in the state of Chihuahua in a test to select the state team for the Mexican Mathematical Olympiad. I later found my solution to be wrong and the problem is tougher than I anticipated, I have worked on it about 10 hours with no luck, hopefully someone here can help out.
The problem is:

Led AD be the altitude of triangle ABC (D in BC). Let BM be and angle bisector (M on AC). Let C' be the midpoint of AB. If AD,BM and CC' intersect in one point O. Show that AO = BO.

[quote="borislav_mirchev"]May we use that: tgB=sinA/cosC ?
(Proposed as problem in Bulgaria and India)[/quote]

What was proposed as problem in Bulgaria and India? The problem I stated or the equality you stated?. Is that equality you state true in any triangle? If so, I would like to see a proof.

The problem he is referring to is http://www.mathlinks.ro/Forum/viewtopic.php?t=33634

Now, your problem is definitely wrong. As if $ AO\equal{}BO$, then triangles $ AOC'$ and $ BOC'$ would be congruent SSS. That would mean that $ \angle AC'O \equal{}\angle BC'O \equal{} 90^{\circ}$, so $ O$ would be on the bisector of the segment $ AB$. But $ O$ is also on the median, so the triangle would be isosceles. Furthermore as $ \angle C'AO \equal{}\angle C'BO \equal{}\frac{\angle B}{2}$, it follows that $ \angle B \equal{} 60^{\circ}$, so the triangle is in fact equilateral. But by the above observations of Darij, not all triangles having those 3 lines concurrent are equilateral, therefore your problem is stated wrong. That's probably why you weren't able to solve it in the first place.

[quote="Valentin Vornicu"]The problem he is referring to is http://www.mathlinks.ro/Forum/viewtopic.php?t=33634

Now, your problem is definitely wrong. As if $ AO \equal{} BO$, then triangles $ AOC'$ and $ BOC'$ would be congruent SSS. That would mean that $ \angle AC'O \equal{}\angle BC'O \equal{} 90^{\circ}$, so $ O$ would be on the bisector of the segment $ AB$. But $ O$ is also on the median, so the triangle would be isosceles. Furthermore as $ \angle C'AO \equal{}\angle C'BO \equal{}\frac{\angle B}{2}$, it follows that $ \angle B \equal{} 60^{\circ}$, so the triangle is in fact equilateral. But by the above observations of Darij, not all triangles having those 3 lines concurrent are equilateral, therefore your problem is stated wrong. That's probably why you weren't able to solve it in the first place.[/quote]

Thanks, I guess I copied the problem incorrectly from the book I found it in. Oh well, it is not the first time I waste 10 hours trying a problem that wasn't stated correctly

Original formulation of the problem is: If AL, BM, CH are angle bisector, median and altitude from vertices A, B, C of triangle ABC and they meets at common point then the following equality is true: tgA=sinC/cosB. /Bulgarian Math Olympiad 1969 - III round/
It is true also: If for a triangle ABC is true that: tgA=sinC/cosB  - then AL, BM, CH - meets at a common point.
You may try to prove that: If in triangle ABC AB=15, BC=13, CA=12 then AL, BM, CH - meets at a common point. /Proposed many times in II round - Bulgaria - district Vratza/.
Another interesting problem about AL, BM, CH is:  If AL=BM=CH - prove that triangle ABC is equaliteral. /Russia 1968 I think... 8 grade/. The list may be continued to infinty...


MATHCOUNTS
AMC
AIME
USA(J)MO
USAMO
USAMTS
AMC 8

just wondering what grade u guys r in on this forum

I'm in 7th grade. 1 more year in MATHCOUNTS! :D

lucky its so much cooler to be a 7th grader   u got 1 more year ahead of u

last year one of the 8th graders was saying that to me and i didn't get what she meant. now i do!

7th... I feel so old! I remember in 6th grade I was one of a few people that voted "6th" in this poll and it felt great. Now... :(

not as bad as being in 8th :stink: :stink:

I'm eight. waaaaaaaaaa
quote from freaky friday: "I'm like the crypt keeper!!"

I am in 8th.....and its my first as well as last year in MATHCOUNTS :wallbash_red:  :ewpu:

I'm a Freshman from Dallas and just to let u guys know, eighth grade isnt the end, its just the beginnning. you guys have so much more to look forward to like the AMC 10, 12 and AIME and USAMO and USAMTS and Mandelbrot and everything else... its hard to find yourself at the end of MathCounts but you'll realize there are only bigger and better things ahead. Keep After It!!

You know....you can do those test as well as mathcounts in middle school...so in high school etc...you have 1 less comp......

I'm glad I'm in seventh so I have one more year for Mathcounts.  I still do the AMCs though, so that won't be something new.

[quote="robinhe"][quote="math92"]8TH GRADE ALL THE WAY!!!!
I hope an 8th grader wins mathcounts this year...[/quote]

NEAL WU  :wow:[/quote]
or TOMMY GILL :wow: :D

[quote="biffanddoc"]i like to be credited as biffanddoc
(then some XXX rXpist doesn't try to track me down and kill me) :chief:  [/quote]

XD

http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2005-8a/05Perfect/DE-Gill,T.jpg

[quote="easyas3.14159..."]SCRUBS all the way!!I still have another year for MathCounts and the AMC 8.[/quote]
So do I but I probably won't do the AMC 8 again because I got a perfect score this year.

you should anyway, you get a cool medal  :w00tb:  :w00tb:

Heh, I have a friend who got all of the AMC8 right except #1.
When he was checking it, he said:
"Hm, I got number 1 wrong, so I can't get a perfect. Oh well. (starts to check the rest of it) Right... Right... Right... (checks the last one) WHAT? NO!!! SON OF A #$%#
$%#%#$%#
$%# !!!
So funny...

Aarg, im in 8th grade, last year :wallbash_red: Next year im :censored: from mathcounts


Let $ABC$ be a triangle, and let $P$ be a point inside it such that $\angle PAC = \angle PBC$. The perpendiculars from $P$ to $BC$ and $CA$ meet these lines at $L$ and $M$, respectively, and $D$ is the midpoint of $AB$. Prove that $DL = DM$.

Let AP meet BC at U and let BP meet CA at V. The quadrilateral ABUV is cyclic and P is its diagonal intersection. Let E be the midpoint of UV. According to the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55836]Quadrilateral[/url] in the Pre-Olympiad section, the quadrilateral DLEM is a kite with DL = DM and EL = EM.

[color=red][b]Generalization.[/b] Given are the triangle $ABC$, the midpoint $D$ of the side $[AB]\ ,$ a inside point $P$ and the points $L\in (BC$, $M\in (AC$ so that $m(\widehat {BLP})+m(\widehat {AMP})=180^{\circ}$ and $\frac{PM}{PL}=\frac{AM}{BL}\ .$ Prove that $\boxed {\frac{DL}{DM}=\frac{BL}{BP}\cdot \frac{AP}{AM}}$ and $\boxed {m(\widehat {LDM})=m(\widehat {LBP})+m(\widehat {MAP})}\ .$[/color][color=blue]

[b]Remark 1.[/b] Can add to the conclusion of the proposed problem and $\boxed {m(\widehat {LDM})=2\cdot m(\widehat {MAP})}\ !$

[b]Remark 2.[/b] The generalization and the proposed problem are
some [u]nice particular cases[/u] of the [u]my problem[/u] which was posted at
http://www.mathlinks.ro/Forum/viewtopic.php?t=75581 ,
where is and a nice [u]Enescu's proof[/u].\[ \boxed {\mathrm {\ A\ Happy\ Easter\ Day\ !\ }}  \][/color]

Virgil please, prove you generalization!

Davron


For a natural number $ n$, let $ w(n)$ denote the number of (positive) prime divisors of $ n$. Find the smallest positive integer $ k$ such that:

$ 2^{w(n)} \le k \sqrt[4]{n}$ for each $ n \in \mathbb{N}$.

Let $ k$ be smallest possitive integer saitisfy condition.
Assume $ p_1 < p_2 < . . .$ is squence of all primes, then $ n > p_1p_2 . . . p_{w(n)}$.
$ \rightarrow \frac {2^{w(n)}}{\sqrt [4]{n}} \leq \frac {2^{w(n)}}{\sqrt [4]{p_1p_2 . . . p_{w(n)}}} \equal{} f(n)$.
$ f(n) < 5$ for $ n \equal{} 1,2,3,4,5,6$ and $ f(n \plus{} 1) < f(n)$ for any $ n\geq 6$, so $ f(n) < 5$ for any $ n$.
$ \rightarrow k\leq 5$.
We have $ \frac {2^{w(2.3.5.7)}}{\sqrt [4]{2.3.5.7}} > 4$
$ \rightarrow k \equal{} 5$


geometry
incenter
parallelogram
trapezoid
geometric transformation
reflection
geometry proposed

Let triangle $ ABC$ with $ H,I,O$ be orthocentre ,incenter and cỉcumcircle respectively. $ (I)$ tangent to $ BC$ at $ D$. Prove that: If we have $ OI//BC$ then $ AO//HD$

[quote="Victory.US"][color=darkred]Let $ ABC$ be a triangle with orthocentre $ H$ , incircle $ w\equal{}C(I,r)$ and cỉcumcircle $ C(O,R)$ .

Denote $ D\in BC$ for which $ ID\perp BC$ . Prove that $ \boxed {\ OI\parallel BC\  \Longleftrightarrow\ AO\parallel HD\ }$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ L\in BC\cap AH$ , the midpoint $ M$ of $ [BC]$ , $ S\in MI\cap AH$ , the diameter $ [DE]$ of $ w$ and $ F\in BC$ for which $ MD\equal{}MF$

($ F$ is the tangent point with $ BC$ of the  $ A$-exincircle). Is well-known that $ HA\equal{}2\cdot OM$ , $ AS\equal{}r$ and $ E\in AF$ .  Using these last mentioned

properties obtain that $ OI\parallel BC\Longleftrightarrow ID\equal{}OM$ $ \Longleftrightarrow AS\equal{}OM$ $ \Longleftrightarrow$ $ SA\equal{}SH$ $ \Longleftrightarrow$ $ ASMO$ is parallelogram $ \Longleftrightarrow$ $ E\in AO$ $ \Longleftrightarrow$

$ O\in AF$ $ \Longleftrightarrow$ $ \overline {MIS}\parallel \overline {AEOF}$ $ \Longleftrightarrow$ $ AHDF$ is trapezoid $ \Longleftrightarrow$ $ HD\parallel AO$  . Maybe somebody draws an figure for this proof. [b]Many thanks[/b] ![/color]

Obviously we can prove that $ S \in (I)$

[quote="Virgil Nicula"][quote="Victory.US"][color=darkred]Let $ ABC$ be a triangle with orthocentre $ H$ , incircle $ w \equal{} C(I,r)$ and cỉcumcircle $ C(O,R)$ .

Denote $ D\in BC$ for which $ ID\perp BC$ . Prove that $ \boxed {\ OI\parallel BC\ \Longleftrightarrow\ AO\parallel HD\ }$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ L\in BC\cap AH$ , the midpoint $ M$ of $ [BC]$ , $ S\in MI\cap AH$ , the diameter $ [DE]$ of $ w$ and $ F\in BC$ for which $ MD \equal{} MF$

($ F$ is the tangent point with $ BC$ of the  $ A$-exincircle). Is well-known that $ HA \equal{} 2\cdot OM$ , $ AS \equal{} r$ and $ E\in AF$ .  Using these last mentioned

properties obtain that $ OI\parallel BC\Longleftrightarrow ID \equal{} OM$ $ \Longleftrightarrow AS \equal{} OM$ $ \Longleftrightarrow$ $ SA \equal{} SH$ $ \Longleftrightarrow$ $ ASMO$ is parallelogram $ \Longleftrightarrow$ $ E\in AO$ $ \Longleftrightarrow$

$ O\in AF$ $ \Longleftrightarrow$ $ \overline {MIS}\parallel \overline {AEOF}$ $ \Longleftrightarrow$ $ AHDF$ is trapezoid $ \Longleftrightarrow$ $ HD\parallel AO$.[/color][/quote]
Kostas Vittas.

Thank you very much, Vittasko ! Is very nice and clear your draw.

in this problem, if we denote $ E,G \in AB,AC$ respectively for which $ IE\perp AB$ and $ IG\perp AC$. And $ D' \in EG$ for which $ DD'\perp EG$.
$ B_1,C_1$ be midpoint of $ AC.AB$ respectively.
Prove that $ B_1C_1,DD',EG$ are concurrent

Let $ K$ be the intersection of $ EG$ with $ BC$. It;s easy to see division $ (KDBC)$ is harnomic =>$ DD'$ is interior bisector of $ \angle BD'C$. It's easy to see $ \triangle ED'B$ \similar $ triangle GD'C$ => $ \ratio ED'/D'G \equal{} \ratio BE/CG$. Applying Menelaus's theorm for $ \triangle AEG$ we get $ B1,C1,D'$ collinear (QED)
 sORRY IF i'M WRONG :)  :) [/code][/hide][/url]

[quote="livetolovemath030894"]Dear victory. Seem your problem isn't really correct for any case. $ D'$ could be a interior point of $ \triangle ABC$ but $ B1C1$ could intersect $ EG$ at a external point of $ \triangle ABC$.My solution above only consider the case They intersect other at a interior point. In this case,your problem is wrong ( see the figure) :)
 Sorry for my bad english  :)[/quote]

you have to draw $ OI//BC$  :)

Why????? The Problem wasn't mentioned that thing  :)  :)

[quote="Victory.US"][size=150]in this problem[/size], if we denote $ E,G \in AB,AC$ respectively for which $ IE\perp AB$ and $ IG\perp AC$. And $ D' \in EG$ for which $ DD'\perp EG$.
$ B_1,C_1$ be midpoint of $ AC.AB$ respectively.
Prove that $ B_1C_1,DD',EG$ are concurrent[/quote]

So i think if don't have the condition which $ OI//BC$ the problem is wrong :)

Dear thuyanh158.
I don't think so.Can you prove for your idea??? :)  :)

Yes we must have $ OI//BC$.
If the projection $ D'$ of $ D$ on $ EG$ lies on the midline $ B_1C_1$, we get the reflection $ D''$ of $ D$ wrt $ EG$ lies on the line through $ A$ and parallel to $ BC$. But this result is well-known: $ AD", OI, BC$ are concurrent then they must concur at infinity point, or $ OI//BC$.

it's a result of http://www.mathlinks.ro/viewtopic.php?t=323922  :)
[img]http://www.shrinkpictures.com/processed/phps5TkxzAM.jpg[/img]


analytic geometry
graphing lines
slope
algebra
system of equations

Positive integers a,b, and c are chosen such that a<b<c, and the the system of equations
$2x+y=2003$ and $y=|x-a|+|x-b|+|x-c|$
has exactly one solution. What is the minimum of c?

(a) 668, (b) 669, (c) 1002, (d) 2003, (e) 2004

I need help on this question. Sorry if this has been posted before, I could not find it. Thanks

[hide]If you look at the graph of , then it must intersection one of , , .  is clearly out of the question by looking at the graph, but  is also not possible, because  has a slope of , so if it intersected at  then  it would be above the graph of  immediatley to the left of , but since for ,  has a slope of ,  would eventually be above  and thus they would have two intersection points.

So

 intersects  at . , so , but  so , so b is at most 1001 and thus  is at least .

[/hide]


calculus
integration
logarithms
trigonometry
calculus computations

solve this integral

$\int_{0}^{\frac{\pi}{2}}\;\left( \frac{\ln{(\cos{x})}}{\sin{x}}\right)^{2}\;dx$

I have a good relation

${\int_{0}^{\frac{\pi }{2}}\left(\frac{\log (\cos (x))}{\sin (x)}\right)^{2}\, dx=\int_{0}^{\frac{\pi }{2}}\left(\frac{\log (\sin (x))}{\cos (x)}\right)^{2}\, dx}$

but i couldn't find a good way to solve it. :)

$\int_{0}^{\pi/2}\frac{\ln^{2}\cos x}{\sin^{2}x}dx=\frac{1}{4}\int_{0}^{\pi/2}\frac{\ln^{2}(1-\sin^{2}x)}{\sin^{2}x}dx=\frac{1}{4}\sum_{n,m\geq 1}\frac{1}{nm}\int_{0}^{\pi/2}\sin^{2n+2m-2}xdx$

$=\frac{\sqrt{\pi}}{8}\sum_{n,m\geq 1}\frac{1}{mn}\frac{\Gamma(n+m-1/2)}{(n+m-1)!}=\frac{\pi}{8}\sum_{n,m\geq 1}\frac{(2n+2m-2)!}{nm[(n+m-1)!]^{2}4^{n+m-1}}$.

I don't know whether this integral can be expressed in terms of well-known constants, maybe it is possible. I hope that there are not too many mistakes in my calculations.

[quote="didilica"]It seems to me that

$I=\int_{0}^{\pi/2}\frac{\ln^{2}\cos x}{\sin^{2}x}dx=\pi\ln 2$.

Integrate by parts and you get that

$I=-\frac{\cos x \ln^{2}\cos x}{\sin x}\bigm|_{0}^{\pi/2}-2\int_{0}^{\pi/2}\ln \cos x dx =\pi\ln 2$.[/quote]

cool.   :jump:


Start with the fact that the units digit of A*7 is A.

Did you delete my post? I'm sure i posted a solution here....  :(

Yes, I deleted your post, because bryantacademy asked for a hint, not a solution.

[quote="bryantacademy"]BA
   7
-----
HAA

A,B,H are different digits. What is the sum of A, B, H?
Once again, do not tell me, but lead on MacBeth.[/quote]

[hide]So A =1 so B=2 and H=2 but that doesn't work since they are all different
so then A=5 B=6 H=4 The answer is 15[/hide]

Would that mean that A is one? I don't get this because that would mean that B was also one, but they have to be different! Is there a formula for this?

no a cant be one. i figured out answer, but here is hint:

the unit's digit of $a\times7=a$ from there you can figure out what $a$ is, then what $b$ and $h$ are

[hide="A solution if you are looking for one"]
You know that $7A=10n+A\implies 6A=10n\implies A=5$.
Now, $7B=10H+\frac{7A-\text{R}7A/10}{10}=10H+3\implies B=6$
Therefore, $A=5,B=6, H=4\implies A+B+H=15$
Note: R in this case means the remainder.[/hide]
Masoud Zargar


Prove that $ \prod_{k\equal{}1}^{n\minus{}1}(2\cot \frac{\pi}{n}\minus{}\cot \frac{k\pi}{n}\plus{}i)$ is a pure imaginary.


linear algebra
matrix
superior algebra
superior algebra solved

I have another matrix and determinantes problem this time! :)

If x and y are two pozitive reals and X and Y two matrices with real entris, X and Y are in M_n.
i)hypotesis:
X^2=x/(x+y)XY+y/(x+y)YX and Y^2=y/(x+y)XY+x/(x+y)YX
prove that: det(X-Y)=0
ii)hypotesis:
X^2=x/(x+y)XY-y/(x+y)YX and Y^2=y/(x+y)XY-x/(x+y)YX
prove that: det(X-Y)=0 or det(X+Y)=0

(italiano contest 1997)

was this a  regional / national contest? how did you perform? in which year of school do you study matrices? :D (hope you won't mind me asking these questions but I'm curious about the system).

Easy (i). Sum the 2 equalities and get
 X 2 +Y 2 =XY+YX, so (X-Y) 2=0 and hence det(X-Y)=0.

Hey.. huw about ii)? it's been given me a big head ache! :?

some one can help?
thanks! :D

[quote="Vialli"]

If x and y are two pozitive reals and X and Y two matrices with real entris, X and Y are in M_n.

ii)hypotesis:
X^2=x/(x+y)XY-y/(x+y)YX and Y^2=y/(x+y)XY-x/(x+y)YX
prove that: det(X-Y)=0 or det(X+Y)=0

[/quote]

I wonder if the sign - is on the right place for Y, i have the result for

 X =x/(x+y)XY-y/(x+y)YX and Y= -y/(x+y)XY+x/(x+y)YX

X-Y = x/(x+y)(XY-YX) + y/(x+y)(XY-YX)

X-Y = XY-YX

X-XY+YX -Y=0

(X+Y)(X-Y)=0

det(X+Y)det(X-Y)=0

det(X+Y)=0 or det(X-Y)=0

Mr Omarjee


logarithms
limit
calculus
calculus computations

Find derivate of that continous funtions: $ f(x)\equal{}2^x$ and $ f(x)\equal{}3^x$

hello, the answer is
$ f^'(x)=2^x\ln(2)$
$ g^'(x)=3^x\ln(3)$
Sonnhard.

As a general rule,

$ \frac{d}{dx} a^x \equal{} a^x \ln{a}$  :)

Can a mod move this to the Calc forum? Thanks.

[quote="L_Euler"][quote="Dr Sonnhard Graubner"]hello, the answer is
$ f^'(x) = 2^x\ln(2)$
$ g^'(x) = 3^x\ln(3)$
Sonnhard.[/quote]
Can you prove that? Thanks very much! :wink:[/quote]

$ f'(x) = \lim_{h\rightarrow 0} \frac {f(x + h) - f(x)}{h}$

and use the fact that $ \lim_{x\rightarrow 0} \frac {a^x - 1}{x} = \ln a, a\ne 1$


Let $m$ and $n$ be integers such that they can be expressed as the sum of two perfect squares. Show that $mn$ has this property as well.

For all you guys who try to do it the easy way, click below:
[hide="A very useful hint"]A simple and elegant proof can be found by using complex numbers.[/hide]

[quote="rem"]It's still easy without complex numbers. we have:
$m=a^{2}+b^{2}, n= c^{2}+d^{2}$ where $a,b,c,d$ are positive integers.
We have:
$mn=(a^{2}+b^{2})(c^{2}+d^{2})=(ac)^{2}+(ad)^{2}+(bc)^{2}+(bd)^{2}$
$=(ac)^{2}-2abcd+(bd)^{2}+(ad)^{2}+2abcd+(bc)^{2}$
$= (ac-bd)^{2}+(ad+bc)^{2}$
QED[/quote]

This is Lagrange's identity for perfect squares isn't it?

For the 4 square case:

$(a^{2}+b^{2}+c^{2}+d^{2})(x^{2}+y^{z}+w^{2}+z^{2}) = (ax+by+cw+dz)^{2}+(ay-bx+cz-dw)^{2}+(aw-bz-cx+dy)^{2}+(az-dx+bw-cy)^{2}.$

Or, you could apply quaternions, which introduces a system such that i^2=j^2=k^2=ijk=-1 and do expansion on a general quaternion number

[quote="maokid7"]so how would one do the original problem using complex numbers?[/quote]
[hide="hint"]what are the real and the  imaginary parts when you multiply  two complex numbers:
say $(a+ib)(c+id)$?
Also you know that if one factors in Gausian primes..eg
$a^{2}+b^{2}=(a+ib)(a-ib)$
$c^{2}+d^{2}=(c+id)(c-id)$
......
[/hide]

[quote="maokid7"]so how would one do the original problem using complex numbers?[/quote]

[b]Definition:[/b]  Let $z = a+bi \in \mathbb{C}$ where $a, b \in \mathbb{R}$ be a complex number.  Then $\bar{z}= a-bi$, the conjugate.

[b]Definition:[/b]  Let $N(z) = z \cdot \bar{z}$.  This is the square of the absolute value; number theorists call it the norm.

[b]Lemma:[/b]  $z, w \in \mathbb{Z}[i]$:  $\overline{zw}= \bar{z}\bar{w}$

Proof:  Let $z = a+bi, w = c+di$.  Then $\bar{z}= a-bi, \bar{w}= c-di, \bar{z}\cdot \bar{w}= (ac-bd)-(ad+bc) i = \overline{zw}$.  QED.

[b]Lemma:[/b]  $N(zw) = N(z) N(w)$

Proof:  $N(zw) = (zw) ( \overline{zw}) = (zw)( \bar{z}\cdot \bar{w}) = (z \bar{z})(w \bar{w}) = N(z) N(w)$.  QED.

Explicitly, this says

$(a^{2}+b^{2})(c^{2}+d^{2}) = (ac-bd)^{2}+(bc+ad)^{2}$

In other words, every product of a sum of two squares is another sum of two squares.

The original problem is actually simply the statement that the product of two complex numbers which have a norm is a complex number which has a norm; in other words, that $\mathbb{C}$ is closed under multiplication.  Extremely simple.

Haha...the first thing I found in Exploration Lab in PROMYS...

[quote="t0rajir0u"][b]Follow-up:[/b]  A similar proof exists for four squares.  Can you find it?[/quote]

[b]Follow-follow-up:[/b]  A similar proof exists for eight squares (not using that any positive integer is already sum of four squares).

If i'm not mistaken, there is no explicit formula for a sum of 2^n squares* sum of 2^n squares where n>2 so don't bother looking for a general formula

There is for $n=3$...
And there is one with a condition on one of the numbers in general.


I am writing a research paper and I want to leave a one page blank buffer between my paper and my references. How do I do this? Thanks.


MATHCOUNTS
factorial
quadratics
analytic geometry
graphing lines
slope
number theory

Mrs. Weatherbee spent $325 dollars on hats and scarves. If each hat cost $20 and each scarf cost $25, what is the square of the sum of all the possible numbers of hats plus the square of the sum of all the possible numbers of scarves?

Keep in mind that eash number of accessories must be positive.

Excellent problem for mathcounts level

Hint: [hide]mods[/hide]

anyone??? it's really easy if you set up a table on a graphing calc...but it isn't necessary...

I'm currently stuck on trying to understand what "Diophantine Equation" is  , but I got [hide]13 (325/25)[/hide]

diophantine refers to an equation with 2 or more variables where the solutions are all integral. in other words, an number theory equation.

Oooooooh.

I don't get why you have to use those hard mumbo-jumbo for this problem :lol: (includes modular arithmetic)

your answer is incorrect...explain your reasoning and maybe we can find out what you did wrong

*We're doing Diophantine Equations in Algebra right now*

*What grade/level of math are you in?*  </jk, except for the question which I'm curious to know since I thought you were in 8th>

Anyways, my answer wasn't what the problem asked for...13 isn't even the total number of possibilities, 4 is, and besides the question asked for something else...*actually reads the question* er, "the square of the sum of all the possible numbers of hats plus the square of the sum of all the possible numbers of scarves"  </jk>

So [hide](0+5+10+15):^2:+(1+5+9+13):^2=(5(1+2+3)):^2:+28:^2:=(5*6):^2:+784=900+784=1684[/hide]

There's probably a more smarter way to do this...

I'm in Algebra 1 and in 8th grade...am I ahead of the "game" or something???

Anyways...you're sooooooo close.

MathFiend wrote:Mrs. Weatherbee spent $325 dollars on hats and scarves.

and

MathFiend wrote:Keep in mind that each number of accessories must be positive.

[hide]So how can you have 0 & 13 as answers? In Diophantine equations you normally respond with all whole number answers, but in this case, that isn't so.[/hide]

*sigh* this is why people are against skipping grades...I skipped two, meaning graphing, factorials, basic number theory, basic algebra. Surprisingly I found out this doesn't have anything to do with me not knowing the Diophantine Equation since I'm supposed to learn it in precalc or something, I don't know. Here in N***** we tend to not teach material unless you can prove that it's correct, for example the Quadratic Formula we learned at the end of 8th since apparently we needed to prove it's correct before we use them

Anyways so is the answer [hide]1125[/hide]?

I think it's 1684.

If you make the problem into the line 20H+25S=325 (H=hats, S=scarves), the slope is -4/5. So you take the y (or S) intercept, which is (0,13) and go from there. At each step, you subtract 4 from the number of scarves and add 5 to the number of hats until you get negatives, then you add things up.

This gets you the answer without mods or diophantine equations, whatever they are.

Yes Tare, you're right now.

Topper...you did the same thing that Tare did earlier...0,13 isn't a possibility because of the reasons I posted above...

1) 0 isn't positive
2) Mrs. Weatherbee bought both hats and scarves...so she can't have 0 of one

But yes, you could graph the line...that's what we did in class, although I prefer to set up a table on my graphing calc. but that doesn't work too well when you can't use a calc. ie. Sprint round...Tests (like the one I have tomorrow AHHHHHHHHHH) etc.


calculus derivative parameterization trigonometry logarithms integration function

how would one evaluate

$\sum_{i=0}^\infty(x^{i}/(i^{2}+1))$.

You can try to combine the derivatives of $f(x)=\sum_{i=0}^{\infty}\frac{x^{i}}{i^{2}+1}$ into a power series that you can recognize. Say, you want to multiply the $i$th term by $i^{2}+1$. Differentiation in $x$ amounts to multiplication by $i$, but it also reduces the power of $x$ by one. So you multiply by $x$ to restore the exponent to $i$. Then do it again...
\[x(xf'(x))'+f(x)=\sum_{i=0}^{\infty}x^{i}=\frac{1}{1-x}\]
The differential equation should not be too hard to solve, although who knows..

That differential equation is pretty nasty; it requires something like variation of parameters to solve it, and that's not even enough.

To solve $x^{2}f''+xf'+f=\frac1{1-x}$, we first solve the homogeneous equation $x^{2}y''+xy'+y=0$. Standard methods (looking for solutions $y=x^{r}$) give solutions $x^{\pm i}$, or $\cos(\ln x)$ and $\sin(\ln x)$ in real terms.

Using variation of parameters from here, I found
$f(x)=\cos(\ln x)\int \frac{-1}{x-x^{2}}\sin(\ln x)\,dx+\sin(\ln x)\int \frac{1}{x-x^{2}}\cos(\ln x)\,dx$
Those are indefinite integrals, of course; the constants of integration generate solutions of the homogeneous equation.

Those integrals are almost certainly impossible in terms of elementary functions. There's also a nasty singularity at $x=0,$ exactly where our function should be well-behaved. This series solution is the only solution to the differential equation in which that singularity is actually removable.

It's possible to express everything in terms of definite integrals, but horribly complicated.

ya the DE is the easy part, its recognizing how to set it up thats challenging...that brings me to a general question that u bring up, how does one express an infinite series as a definite integral?

Yes, the point $x=0$ is special because of vanishing coefficients. This is called [url=http://mathworld.wolfram.com/EulerDifferentialEquation.html]Euler's equation[/url]. It reduces to a constant coefficients equation by substitution $x=\pm e^{t}$. (This also avoids the singularity at $0$). With $x=e^{t}$ I get $f_{tt}+f=\frac{1}{1-e^{t}}$.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

$ R$ is a commutative ring  and $ M$ is a R-module .$ \forall  x  \in  R$  the homomorphism  $ f: M \rightarrow  M$  so that  $ f(y)\equal{}xy$    $ \forall  y   \in  M$   is onto or nilpotent.show that if $ M   \not\equal{}  {0}$  then $ r(0: M)$  is a prime ideal  of $ R$

What do you mean by $ f: M\rightarrow M$ is nilpotent and what does $ r(0: M)$ mean? thanks.

To above: nilpotent should mean $ f^n \equal{} 0$ for some $ n$. $ r(0: M)$ should mean the radical of the annihilator of $ M$. ((0:M) means all the elements in R that sends M to 0, I think in general it's something called ideal quotient for the case of ideals)

To OP:
The given condition means that for any $ x \in R$, either $ xM \equal{} M$ or $ x \in r(0: M)$.
So suppose $ xy \in r((0: M))$. If $ y \in r((0: M))$ we are done. If not, by given condition we have $ yM \equal{} M$, and $ xy \in r((0: M))$ implies $ x^ny^n \in (0: M)$ for some n, meaning $ x^ny^nM \equal{} 0$. But $ yM \equal{} M$, so it means $ x^nM \equal{} 0$, i.e. $ x \in r((0: M))$, done.


geometry
geometric transformation
rotation
reflection

The number of ways in which we can tile a $16 \text{ x }16$ square board with two types of tiles, $2\text{ x }1$ "sticks" and $2 \text{ x }2$ "squares", is $N$. Rotations and reflections of one board are considered distinct, as long as the board looks different. Find the remainder when $N$ is divided by $1000$.

[i](Is there a nice recursion we can set up for this problem?)[/i]

[quote="JBL"][quote="cincodemayo5590"][i](Is there a nice recursion we can set up for this problem?)[/i][/quote]

I don't think so.  You could set up a really, really ugly recursion for a $16 \times n$ square based on all possible ways to tile the leftmost edge.  (You can get a sense of this, minus the massive ugliness, by considering such a recursion for $2 \times n$, $3 \times n$, $4\times n$ and $5 \times n$ boards.  (The odd cases behave a little differently from the even cases.)  Did this problem come from somewhere (in the sense of, is there any reason that you expect it to have a nice solution?)?[/quote]

The problem didn't come from anywhere; but I had thought that there might be a similarity between this problem and the one-dimensional tilings (i.e. $1 \times n$) which are related to the Fibonacci numbers. But I guess not, considering now the ugliness of the recursion.  :P  Can you think of a way to make the problem nicer? Would just the smaller dimensions work, or should the unit tiles be different also?

Thanks.  :)


[quote="jli"]Let $a_{n}=\frac{n^{3}}{n^{2}-15n+75}$ where $n$ is an integer between 1 and 15 inclusive.

What is $\sum_{n=1}^{15}a_{n}$?[/quote]

You should have made the question more difficult by changing the limits to something that would be tedious to compute via brute force.  But as it is, it's still rather easy.  Note the denominator $q(n) = n^{2}-15n+75$ is a quadratic.  Therefore there exists a number $m$ for which $q(m+x) = q(m-x)$.  What is this number, and why is it relevant?

[i]([b]atomicwedgie[/b]: On the test, there was a restriction that solutions could not involve direct computation.)[/i]

[hide="Hint"]
Add $a_{0}= 0$.

[hide=" Bigger Hint"]
Pair up the terms.  :wink:

[hide="Really Big Hint"]
Find $a_{k}+a_{15-k}$.
[/hide]
[/hide]
[/hide]

$\frac{n^{3}}{(n)(n-15)+75}$...

$n\leftarrow 15-n$...

$\frac{ (15-n)^{3}}{(15-n)(-n)+75}$

if we add those...we are like WOA

stuff cancels to give 45


In triangle ABC, angle A is twice angle B, angle C is obtuse, and the three side lengths a, b, c are integers. Determine, with proof, the minimum possible perimeter.

I think is $ 77$. I will post my solution if my result is correct and/or if no one else posts a simpler proof.


linear algebra
matrix
group theory
linear algebra unsolved

Take the order of the matrix $ A \equal{} (a_{i,j})$ to be $ m$.
Build a bijection between such matrices and the symmetric group $ \mathcal{S}_m$ of the permutations on $ m$ elements by $ \sigma(i) \equal{} \textrm{ that } j \textrm{ such that } a_{i,j} \equal{} 1$. Then the powers of $ A$ correspond to compositions of $ \sigma$. There is therefore some $ n$ such that $ \sigma^{n\plus{}1} \equal{} \textrm{id}$, so $ A^{n\plus{}1} \equal{} I_m$. Then $ A^n \equal{} \; ^t  \hspace{\minus{}4pt} A$.


Let $ a,b,c$ be positive numbers. Prove the following inequality :
      $ \large \sum\limits \frac{ab\plus{}ac\minus{}2bc}{b^2\plus{}c^2\minus{}bc} \geq 0$

Dtrung,
Why title of this topic is '...''?  :maybe:

[quote="Dtrung"]Let $ a,b,c$ be positive numbers. Prove the following inequality :
      $ \large \sum\limits \frac {ab \plus{} ac \minus{} 2bc}{b^2 \plus{} c^2 \minus{} bc} \geq 0$[/quote]
The SOS method kills it easy enough.  :wink:


Iata o reformulare a mea a acestei probleme, care este echivalenta cu problema initiala:

Fie ABCD un parallelogram şi O punctual de intersecţie al diagonalelor sale.
Notăm cu M, N şi P proiecţiile vârfului A pe dreptele BC, CD şi respective DB.
Să se arate că punctele M, N, O şi P sunt patru puncte conciclice.
(sub aceasta forma problema apare si in cartea noastra: "ANALOGII TRIUNGHI-TETRAEDRU":
problema 2, din Cap.3)

In ATTACHMENT gasiti solutia analitica (aparuta in "Matematika v skole") si o foarte frumoasa solutie
sintetica a prof.univ. Dan Branzei!

[quote="mihai miculita"][color=darkred][b]Fie $ ABCD$ un paralelogram şi $ O$ punctul de intersecţie al diagonalelor sale.
Notăm cu $ M$, $ N$ şi $ P$ proiecţiile vârfului $ A$ pe dreptele $ BC$, $ CD$ şi $ DB$ respectiv. Să se arate că patrulaterul $ MNOP$ este ciclic.[/b][/color][/quote]
[color=darkblue][b]Demonstratie.[/b] Presupunem fara a restrange generalitatea ca $ A\equal{}m(\widehat{BAD})<90^{\circ}$.

$ AMBP$, $ AMCN$ sunt ciclice [b]si[/b] $ \left\{\begin{array}{c}AB\parallel CD\\\ AD\parallel BC\end{array}\right\|$ $ \implies$ $ \widehat{BPM}\equal{}\widehat{BAM}\equal{}\widehat{DAN}\equal{}$ $ \widehat{OAN}\minus{}\widehat{OAD}\equal{}$

$ \widehat{ONA}\minus{}\widehat{ACM}\equal{}$ $ \widehat{ONA}\minus{}\widehat{ANM}\equal{}\widehat{ONM}$ $ \implies$ $ \widehat{BPM}\equal{}\widehat{ONM}$, adica patrulaterul $ MNOP$ este ciclic.[/color]


I was just wondering....how many of you AoPSers dance? Any kind of dance. I thought it would be pretty awesome to have a little place to trade advice and chat....just for dancers. Since I have no idea how many people are dancers here, this topic may die quickly. So if you are a dancer, please don't hesitate to post and introduce yourself!

I'm royalflower.
I do ballet, been on pointe for almost 3 months now.
I like to help people strengthen for pointe with cool exercises.

heh.
You know io dance.
I do bharatnatyam, My favorite item SO far. is Bho Shambo..it's awesome, and the music is SO cool

@FlyAgaric:

Ikr? What toepads do you use? Do you have tapered toes? I do.  :o

@k00lperson:

That's really cool too!
Maybe I'll look on youtube and learn some!  :P

@royalflower: Sorry, but you can't learn from youtube :P
And they do REALLY REALLY badly on youtube. no offense to the performers, but no one's AWESOME. with lot's of aramandi and good poses.

Aw... It sounded cool.
Maybe you can teach me orally.

rofl
Well. it's not possible to teach orally :P
Nice idea though....
u need a teacher

Oh, well.
Wish I could learn though.  :P

haha :P
umm
i've been dancing for 4 years, and it's pretty hard still

I can dance! I danced ballet and tap dance when I was 3 years old. Then I quit because I was too busy.

Now, I do free dance. When I hear music, I dance whatever I want to do.

My sister dances. Pretty well. But then again what do I know right?

I do Bharatanatyam like K00lperson! In 1 year, I'll have to go to India and write this extensive test (Theory and dance) to get a certificate as an accomplished dancer. I have been doing dance for maybe 4 years now! But, my teacher teahces very fast, like VERY fast. It is hard sometimes if your teacher goes too fast. I sometimes get cobfused but after I practice it a while, I learn it by heart. The weird thing: I can't dance without music, I dont remeber any movements at all, but the minute you turn on music, I remeber all these movements withouit even thinknig about them! Does anyone else feel like that?

Hi. I'm IDKorean.
I do a bit of popping/locking ( just started 2 months ago) on and off.

hmm, not very in tune with the other dances

I do some windmills, L-kicks, freezes, that kind of thing

started... a bit more than a year ago

haha yes, music helps doesnt it@nammuchms!!
mm...
when r u doing ur arangetram?!

Probably next summer or the summer after. I might not go to India next year (lot of camps to attend and comps to prepare for). I hope I go the summer after because then I can be certain that I will pass the test (I will probably finish all my dances June, and it will take around 2-3 months to leanr all the teory and mantras which will be September). K00lperson, when are you going to do ur arangetram? Do you plan on doing one, I mean everyone doesn't have to. I might not...but it depends which mood I am in, I might. You just need to do a couple of dnaces in front of the judges (without any audience) and wirte the extensive theory test. The arangetram is just for everyone to know that you are an accomplished dancer....but from those arangetrams you can earn awards which might be good for college and for your self ego :P

Mhm. Yes, I am doing an arangetram. My teacher told me when I'm doing..Not tellingg ;)!

I dance only when no other people are watching.

Baby I don't need dollar bills to have fun tonight
(I love cheap thrills)
Baby I don't need dollar bills to have fun tonight
(I love cheap thrills)
But I don't need no money
As long as I can feel the beat
I don't need no money
As long as I keep dancing

Hi dancers, I think you'd like this thread I made, check it out!

[url]https://www.artofproblemsolving.com/community/c252065_calling_all_aops_dancers[/url]

[color=#0B5394]Your AoPS friend,[/color]
[hide=~MathMania11] [img]http://cdn.artofproblemsolving.com/images/e/1/f/e1f42eac6c50593d6d7f3654cd06fde7098029c0.gif[/img]

[i][color=#38761D]"Pledge to Always be a Mathlete." ~ MathMania11[/color][/i] [/hide]

[quote=MathMania11][quote=FlyAgaric]Hi I am a dancer. I take ballet. I started pointe last year. It is severely painful. :([/quote]

ME TOO! I am a dancer and i do many types, including ballet, pointe, jazz, modern, hip hop, lyrical, musical theater, and tap. :-D By the way, is this thread still alive? :P :heli:

[color=#0B5394]Your AoPS friend,[/color]
[hide=~MathMania11] [img]http://cdn.artofproblemsolving.com/images/e/1/f/e1f42eac6c50593d6d7f3654cd06fde7098029c0.gif[/img]

[i][color=#38761D]"Pledge to Always be a Mathlete." ~ MathMania11[/color][/i] [/hide][/quote]

No, mathmania. See, It was posted 6 years ago. Please don't recive old posts.


Functional Analysis
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For given cardinal number $ \kappa > \mathfrak{c}$ does there exists Banach Space of cardinality $ \kappa$?
Remark: We are not assuming GCH.

$ \ell^1(I)$ where $ I$ is a set of cardinality $ \kappa$ should work just fine.

I revive this topic since the problem seems to be a little bit harder.

We have $ \text{card}(\ell^{1}(I)) \equal{} \text{card}(I^{\mathbb{N}}) \equal{} \kappa^{\aleph_{0}},$ but using [url=http://en.wikipedia.org/wiki/K%C3%B6nig%27s_theorem_%28set_theory%29]König's theorem[/url] it is not hard to construct cardinal number $ \kappa > \mathfrak{c}$ satisfying $ \kappa < \kappa^{\aleph_{0}},$ so the example above doesn't work for all cardinals $ > \mathfrak{c}.$

Right you are. :blush: OK, let's show that the condition $ \kappa\equal{}\kappa^{\aleph_0}$ is necessary and sufficient. We already know sufficiency. To show necessity, note first that the cardinality of a Banach space is the same as the cardinality of its unit ball $ B$. Now, for each $ k\ge 1$, choose a maximal $ \frac{1}{2^k}$-separated subset $ X_k$ in $ B$ so that $ X_{k\plus{}1}\supset X_k$. Note that $ X_k$ is an $ \frac 1{2^k}$-net in $ B$. Let $ \lambda_k$ be the cardinality of $ X_k$. On one hand, since every point $ x\in B$ is a limit of some sequence of elements $ x_k\in X_k$ and the sequences corresponding to different $ x$ are different, we have $ \kappa\le \prod_k \lambda_k$. On the other hand, if we choose a sequence $ \delta_k>0$ decaying so fast that $ \sum_{n>k}\delta_n<\frac{\delta_k}{2^{k\plus{}1}}$, then for every sequence of elements $ x_k\in X_k$, the series $ \sum_k \delta_k x_k$ will converge and the corresponding points will be different. Thus, $ \kappa\ge \prod_k \lambda_k$, so, by Bernstein, $ \kappa\equal{} \prod_k \lambda_k$. But every countable product $ \kappa$ of a non-decreasing sequence of cardinals satisfies the identity $ \kappa^{\aleph_0}\equal{}\kappa$.