Consider the sequence of numbers $(a_n)$ ($n = 1, 2, \ldots$) defined as follows: $ a_1\in (1, 2)$, $ a_{k + 1} = a_k + \frac{k}{a_k}$ ($k = 1, 2, \ldots$). Prove that there exists at most one pair of distinct positive integers $(i, j)$ such that $a_i + a_j$ is an integer.

[quote="April"]Consider the sequence of numbers $ (a_n)$ $ (n \equal{} 1, 2, \ldots)$ defined as follows: $ a_1\in (1, 2)$, $ a_{k \plus{} 1} \equal{} a_k \plus{} \frac {k}{a_k}$ $ (k \equal{} 1, 2, \ldots)$. Prove that there is at most any two terms of this sequence whose sum is an integer.[/quote]

1) $ a_k>k$ $ \forall k\geq 1$ and $ a_k\minus{}k$ is a strictly decreasing sequence.
$ a_1>1$ and, if it's true for $ a_k$, we have $ a_k>k\geq \sqrt k$ and, since $ x\plus{}\frac{k}{x}$ is an increasing function for $ x\geq \sqrt k$, $ a_{k\plus{}1}\equal{}a_k\plus{}\frac{k}{a_k}\geq k\plus{}\frac{k}{k}\equal{}k\plus{}1$
The last point is an immediate consequence of $ a_{k\plus{}1}\minus{}(k\plus{}1)\equal{}a_k\minus{}k\minus{}1\plus{}\frac{k}{a_k}$ $ \equal{}(a_k\minus{}k)\minus{}\frac{a_k\minus{}k}{a_k}$

2) $ a_k<k\plus{}\delta_k$ where $ \delta_n$ is defined as $ \delta_1\equal{}1$ and $ \delta_{n\plus{}1}\equal{}\delta_n\minus{}\frac{\delta_n}{n\plus{}\delta_n}$
$ a_1<1\plus{}1$ and so it's true for $ n\equal{}1$
If $ a_k<k\plus{}\delta_k$, and since $ x\plus{}\frac{k}{x}$ is an increasing function for $ x\geq \sqrt k$, $ a_{k\plus{}1}<k\plus{}\delta_k\plus{}\frac{k}{k\plus{}\delta_k}$ $ \equal{}k\plus{}1\plus{}\delta_k\minus{}\frac{\delta_k}{k\plus{}\delta_k}$ $ \equal{}k\plus{}1\plus{}\delta_{k\plus{}1}$

3) $ a_i\plus{}a_j$ can never be an integer if $ i>1$ and $ j>1$
Since $ \delta_k$ is a decreasing sequence, we have, since $ i>1$ and $ j>1$ :  $ \delta_i\leq \delta_2\equal{}\frac{1}{2}$ and $ \delta_j\leq \frac{1}{2}$
So $ i\plus{}j<a_i\plus{}a_j<i\plus{}j\plus{}1$

4) There is at most one couple $ (i,j)$ such that $ a_i\plus{}a_j$ is an integer (the requested assertion) :
From 3 above, we know that if such a couple exist, we have wlog $ i\equal{}1$
So, if we have two couples, it means we have $ j>k$ such that :
$ a_1\plus{}a_j$ and $ a_1\plus{}a_k$ both are integers and so $ a_j\minus{}a_k$ is an integer too.

If $ k\equal{}1$, we have $ a_1\plus{}a_1$ integer, so $ a_1\equal{}\frac{3}{2}$ and $ j<a_j<j\plus{}\frac{1}{2}$, so $ j\minus{}\frac{3}{2}<a_j\minus{}a_k<j\minus{}1$ and so cant be an integer.

If $ j>k>1$, we know that $ j<a_j<j\plus{}\frac{1}{2}$ and that $ k<a_k<k\plus{}\frac{1}{2}$ and so $ j\minus{}k\minus{}\frac{1}{2}<a_j\minus{}a_k<j\minus{}k\plus{}\frac{1}{2}$ and so can be an integer only if $ a_j\minus{}a_k\equal{}j\minus{}k$, so if $ a_j\minus{}j\equal{}a_k\minus{}k$. But this is impossible since (point 1 above), we know that $ a_n\minus{}n$ is strictly decreasing.
.

Hence the result.

[quote="April"]Consider the sequence of numbers $ (a_n)$ $ (n \equal{} 1, 2, \ldots)$ defined as follows: $ a_1\in (1, 2)$, $ a_{k \plus{} 1} \equal{} a_k \plus{} \frac {k}{a_k}$ $ (k \equal{} 1, 2, \ldots)$. Prove that there is at most two terms of this sequence whose sum is an integer.[/quote]

I still have some doubts: plug in $ a_1\equal{}1$ gives $ a_1\plus{}a_2\plus{}a_3\equal{}6$, plug in $ a_1\equal{}2$ gives $ a_1\plus{}a_2\plus{}a_3\equal{}7.8$. So there must exist such value for $ a_1$ strictly between 1 and 2 that $ a_1\plus{}a_2\plus{}a_3\equal{}7$. But this will cause the sum of three terms equals an integer. Why?

[quote="greentreeroad"][quote="April"]Consider the sequence of numbers $ (a_n)$ $ (n \equal{} 1, 2, \ldots)$ defined as follows: $ a_1\in (1, 2)$, $ a_{k \plus{} 1} \equal{} a_k \plus{} \frac {k}{a_k}$ $ (k \equal{} 1, 2, \ldots)$. Prove that there is at most two terms of this sequence whose sum is an integer.[/quote]

I still have some doubts: plug in $ a_1 \equal{} 1$ gives $ a_1 \plus{} a_2 \plus{} a_3 \equal{} 6$, plug in $ a_1 \equal{} 2$ gives $ a_1 \plus{} a_2 \plus{} a_3 \equal{} 7.8$. So there must exist such value for $ a_1$ strictly between 1 and 2 that $ a_1 \plus{} a_2 \plus{} a_3 \equal{} 7$. But this will cause the sum of three terms equals an integer. Why?[/quote]

I think - and this is what I demonstrated in point 4 - that the statement is "there is at most one couple $ (a_i,a_j)$ such that $ a_i\plus{}a_j\in\mathbb N$


When can we discuss the second CNML...math league?

seeing how I'm only taking it in 2 hours...
I'd say either tomorrow, or the end of the week

Please do not discuss this as of now. I'm pretty sure people in other time zones are taking it, so I'd postpone discussion until Thursday to be safe. Thanks.

[quote="boxedexe"]:( Even though teh contest is quite useless, I want to know the answers. :P  Anyway, I shall wait.[/quote]

Don't they let you see the answers?  We grade ours on the spot.

Yup, no discussion until Thursday, please. My school has permission to take it today (Wednesday).


find all $ x \in R$ that :
\[ \frac{x^2\plus{}12x\plus{}4}{x\plus{}2}\equal{}6\sqrt{x}\]

[hide]Square both sides and factor to get $ x^2(x^2 \minus{} 12x\plus{}4) \plus{} 4(x^2\minus{}12x\plus{}4)\equal{} 0$

or $ (x^2\plus{}4)(x^2\minus{}12x\plus{}4)\equal{}0$

$ x \equal{} 6\pm 4\sqrt{2}$[/hide]

[hide="Involves squaring trinomials :("]Multiply both sides by $ x\plus{}2$ and simplify a little to get
\[ x^2\plus{}12x\plus{}4\equal{}6x^\frac{3}{2}\plus{}12x^\frac{1}{2}
\]
Now square both sides  :help:
\[ x^4 \plus{} 24 x^3 \plus{} 152 x^2 \plus{}96 x\plus{}16 \equal{} 36 x^3 \plus{} 144 x^2 \plus{} 144 x\\
x^4 \minus{} 12 x^3\plus{} 8 x^2 \minus{} 48 x \plus{}16 \equal{} 0\\
(x^2\plus{}4) (x^2\minus{}12x\plus{}4)\equal{}0
\]

If $ x^2\plus{}4\equal{}0$ then our solutions are nonreal, so $ x^2\minus{}12x\plus{}4\equal{}0\implies x\equal{}\frac{12\pm8\sqrt{2}}{2}\equal{}\boxed{6\plus{}4\sqrt{2},\;6\minus{}4\sqrt{2}}$[/hide]

[quote="grn_trtle"][hide="Involves squaring trinomials :("]Multiply both sides by $ x \plus{} 2$ and simplify a little to get
\[ x^2 \plus{} 12x \plus{} 4 \equal{} 6x^\frac {3}{2} \plus{} 12x^\frac {1}{2}
\]
Now square both sides  :help:
\[ x^4 \plus{} 24 x^3 \plus{} 152 x^2 \plus{} 96 x \plus{} 16 \equal{} 36 x^3 \plus{} 144 x^2 \plus{} 144 x \\
x^4 \minus{} 12 x^3 \plus{} 8 x^2 \minus{} 48 x \plus{} 16 \equal{} 0 \\
(x^2 \plus{} 4) (x^2 \minus{} 12x \plus{} 4) \equal{} 0
\]
If $ x^2 \plus{} 4 \equal{} 0$ then our solutions are nonreal, so $ x^2 \minus{} 12x \plus{} 4 \equal{} 0\implies x \equal{} \frac {12\pm8\sqrt {2}}{2} \equal{} \boxed{6 \plus{} 4\sqrt {2},\;6 \minus{} 4\sqrt {2}}$[/hide][/quote]
Now square both sides  :help:
\[ x^4 \plus{} 24 x^3 \plus{} 152 x^2 \plus{} 96 x \plus{} 16 \equal{} 36 x^3 \plus{} 144 x^2 \plus{} 144 x \\
x^4 \minus{} 12 x^3 \plus{} 8 x^2 \minus{} 48 x \plus{} 16 \equal{} 0 \\
(x^2 \plus{} 4) (x^2 \minus{} 12x \plus{} 4) \equal{} 0
\]

HOw did you factor the polynomial?

Substitute $ \sqrt x \equal{} u$ to get $ u^4\minus{}6u^3\plus{}12u^2\minus{}12u\plus{}4\equal{}0$. Divide by u^2 and group:

$ (u^2\plus{}\frac{4}{u^2})\minus{}6(u\plus{}\frac{2}{u})\plus{}12\equal{}0$. Substitute $ u\plus{}\frac{2}{u}\equal{}v$, as $ u \ge 0 \; \equal{}>v \ge 2\sqrt 2$ to obtain

$ v^2\minus{}6v\plus{}8\equal{}0$. Now $ v\equal{}4$, and go back!


How did you do the final step riddler?
I think i'm getting 1 as well...i know its wrong.....

I've got another way of doing it, but i must admit, its the most inelegant method ever...

diffrentiating,
dm/da=4a^3-42a^2+134a-126=0
Using Newton's approximation several times gets
a=1.697224362....

and then substitution gets m=1 as the min value. (of course after checking the second differential)

[i]Edit:  looking at stephenyamqc's solution verifies that 1.697224362...=(7-sqroot13)/2=a [/i]

yup stephenyamqc, your answer's perfect :coolspeak:  :first: . Coool , you did by EXACTLY the same method as me ! :lol:  :lol:

hey riddler :rotfl:  :rotfl:  , $-80$??? is that supposed to be a joke??i thjink you got the wrong factors .... just do it again

come on ,zeus ............., Differentiaton  :wacko:  :w00t: ! :(  :(  in intermediate?? ok , no probs , atleast you got it , even though i couldnt understand a word of your soln( who's going to learn differentiaton now... :P )

Please refrain from using calculus in the Intermediate forum.

Minor points :
[hide][quote="stephenyamqc"]
m=(a^2-7a+6)(a^2-7a+12)+10
 [/quote]
A little  easier, perhaps,   if you let $y=a^2-7a+9$ and you get $(y-3)(y+3)+10=y^2 -9+10 = y^2+1$
Also
zeus-three  , if you use calculus,
 . There are three roots (and not one) (others are 7/2 and (7+sqrt(13))/2.)  .. and x=7/2 willl NOT give a minimum (but a rather  a "local" maximum)  .. so you have to be careful.
BTW the equation you had becomes $(7x-2) (x^2-7x+9)$   = 0
  [/hide]

yes, thats why you need the second diffrential...

[quote="zeus_three"]sorry, some countries teach calculus in the higher intermediate school years so i thought...[/quote]

Have you seen the announcement at the top of the Intermediate forum that says PLEASE do not post calculus problems in this section?   This also extends not only to posting problems but posting calculus solutions.

[quote="dondigo"][hide]erer[/hide][/quote]

Is that your final answer?  :P


I have a suggestion. Often one has a posting with much expressions like $\displaystyle \frac{a^2}{b^2+c^2+\frac14 a^2}$. In order to have such expressions look nice one has to use \displaystyle all the time (or [ tex ]). Is there a possibility to make a command (something like \alldisplaystyle) one has to put in the beginning of such a message in order to have [i]all[/i] dollar-dollar inline formulas compiled as if they had a \displaystyle automatically?

  Darij

Unfortunately, there's no easy way to do this given the structure of the code; you can type it once and copy it, so you just hit Ctrl-V at the beginning of each bit of TeX.

This trick doesn't really answer your question, but it makes using displaystyle a lot easier:

Put

\newcommand{\D}{\displaystyle}

at the beginning of your document and you can you \D instead of \displaystyle.

Using two dollar signs instead of one also puts you into displaystyle I think.

Problem solved. From now on using $ signs to write tex stuff up, will automatically include a \displaystyle command.

For example $ \frac 12 $ is indeed large.

another question... sometimes I really want to post it in small... how do I do this now? :)

Perhaps make it a checkbox below the post: auto-displaystyle or not? :)

[quote="darij grinberg"]Test: $\frac12$ $ \frac12 $... THANKS Valentin!

Just one thing (I don't know whether there is a good way to realize it): Could you please make the \displaystyle command invisible when one holds the mouse above the LaTeX formula?

Thanks a lot again!

  Darij[/quote] Yes I could, but why?

To Peter: to many checkboxes and options, that people already don't use. I do not imagine a situation in which you want to post "small" :) Please provide me with one.

[quote="Peter VDD"]if I want to type a matrix whose rows and columns should not be a screen broad/high... ;)[/quote]by default the matrix array is considered as if \displaystyle has been written in front. This does not affect that.

Oh well doesn't matter I'll write 1/2 if I need so :)

[quote="darij grinberg"][quote="Valentin Vornicu"][quote="darij grinberg"][...]
Just one thing (I don't know whether there is a good way to realize it): Could you please make the \displaystyle command invisible when one holds the mouse above the LaTeX formula? [...][/quote] Yes I could, but why?[/quote]

Well, I think these \displaystyle's are just needless information... but anyway, it's nothing really important...

@ Peter: Why on the earth do you want to have something displayed "small"? I don't see any advantages... Matrices are indeed \displaystyle'ed by default, and I think it's good so (at least I don't want to see a 5x5 matrix compressed to the height of a letter  :D )...

  Darij[/quote]Indeed, they show up once too many times :) it's done now.

[quote="Peter VDD"]another question... sometimes I really want to post it in small... how do I do this now? :)[/quote]
The opposite of \displaystyle is \textstyle
$\textstyle \int_0^1 f(x)\;dx$ instead of $\int_0^1 f(x)\;dx$

For fractions it easier to use \tfrac
$\tfrac{1}{2}$ instead of $\frac{1}{2}$

Thanks Valentin! And now one final thing...

Some people use the apostroph ' in their TeX fields. Though it seems this is actually not good, in most cases the compiler understands the apostroph and renders it as an apostroph: $B'C' = C'A'$. This is a LaTeX version of the formula B'C' = C'A'. Now, when one holds the mouse over this formula, one sees only the B. In other words, all the text behind the apostroph is cut off.

(I generally don't use apostrophs, so I write $B^{\prime}C^{\prime} = C^{\prime}A^{\prime}$, but some other people do it.)

  Thanks a lot,
  Darij

Yes, it's probably because of the php language. I will take a look into it.

Oh you're there yet :)

Sent you a PM about it.

Ok, problem is fixed. You can use the normal ' character, but on the alt_text the equivalent in \prime will appear.

using stevem's advice, I have devised a different solution for this problem. Now you can use the ' character, and it will be correctly rendered in the alternative text.


Find all $ f: R^ \plus{} \rightarrow R^ \plus{}$ such that $ : (x \plus{} y)f(f(x)y) \equal{} x^2f(f(x) \plus{} f(y))$ $ \forall x,y\in R^ \plus{}$

this problem was appear in iran olympiad and so hard!! :o  i will write this offical solution

this is posted before. here is the link.
[url]http://www.mathlinks.ro/viewtopic.php?t=266483[/url]


With $ a,b,c >0$ such that $ a\plus{}b\plus{}c\equal{}3$.  Prove that
   $ \frac{a^2}{4a\plus{}b^2}\plus{} \frac{b^2}{4b\plus{}c^2}\plus{} \frac{c^2}{4c\plus{}a^2} \geq 3/5$

Using Titu's Lemma

LHS $ \geq \frac{(a\plus{}b\plus{}c)^2}{\sum{a^2}\plus{}4\sum{a}} \equal{} \frac{9}{12\plus{}\sum{a^2}}$

But $ \sum{a^2} \geq \frac{(a\plus{}b\plus{}c)^2}{3} \equal{} 3$ (Titu's Lemma)

Am I doing something wrong?

[quote="madness"]Am I doing something wrong?[/quote]

$ \sum{a^2} \geq 3\Rightarrow \frac {9}{12 \plus{} \sum{a^2}}\le \frac{9}{15}$

[quote="alex2008"][quote="madness"]Am I doing something wrong?[/quote]

$ \sum{a^2} \geq 3\Rightarrow \frac {9}{12 \plus{} \sum{a^2}}\le \frac {9}{15}$[/quote]
But this doesn't prove the required inequality does it? Does it mean Titu's lemma is not a strong enough inequality?  :maybe:

[quote="madness"][quote="alex2008"][quote="madness"]Does it mean Titu's lemma is not a strong enough inequality?  :maybe:[/quote]

yes  :)[/quote]

Can you please tell me how inequalities are classified as strong and weak (or is it only relative)? Thanks :)[/quote]
I guess that "relatively" speaking: the more "classical methods" fail to prove it, the more a given ineq is strong   :lol:
Interesting question: does an ineq cease to be strong the day somebody finds an elegant (=genius type) solution?

[quote="arqady"][quote="2424"]With $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 3$.  Prove that
   $ \frac {a^2}{4a \plus{} b^2} \plus{} \frac {b^2}{4b \plus{} c^2} \plus{} \frac {c^2}{4c \plus{} a^2} \geq 3/5$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201362[/quote]
Hi araqady,I am sorry.
I don't no It is your ineq! :blush:


ratio
geometry
parallelogram
geometry proposed

(V.Protasov, 10-11) Given parallelogram $ ABCD$ such that $ AB \equal{} a$, $ AD \equal{} b$. The first circle has its center at vertex $ A$ and passes through $ D$, and the second circle has its center at $ C$ and passes through $ D$. A circle with center $ B$ meets the first circle at points $ M_1$, $ N_1$, and the second circle at points $ M_2$, $ N_2$. Determine the ratio $ M_1N_1/M_2N_2$.

We will prove that $ M_{1}N_{1} / M_{2}N_{2} \equal{} b/a$. Indeed, since $ M_{1}N_{1} \cdot a \equal{} 4\left[AM_{1}B\right]$, we have that $ M_{1}N_{1}\equal{}\frac{4\left[AM_{1}B\right]}{a}.$ Similarly, $ M_{2}N_{2}\equal{}\frac{4\left[BM_{2}C\right]}{b},$ and thus
\[ \frac{M_{1}N_{1}}{M_{2}N_{2}}\equal{}\frac{b}{a} \cdot \frac{\left[AM_{1}B\right]}{\left[BM_{2}C\right]}.\]
On the other hand, since $ AM_{1}\equal{}AD\equal{}BC$, $ M_{1}B\equal{}BM_{2}$, and $ AB\equal{}CD\equal{}CM_{2}$, the triangles $ AM_{1}B$ and $ BCM_{2}$ are congruent, and therefore $ \left[AM_{1}B\right]\equal{}\left[BM_{2}C\right]$, from which we conclude that $ M_{1}N_{1} / M_{2}N_{2} \equal{} b/a$.


floor function
integration
logarithms
calculus
limit
factorial
ratio

List special equations e is in or ways to derive it
Here's two.

e^pi*i = -1
1/1! + 1/2! + 1/3! + ...

given [tex]\lfloor en! \rfloor + 1 [/tex] points in which each pair of points is connected with a line colored one of [tex]n[/tex] colors, there must exist a monochromatic triangle.  This result is the best possible, there need not exist a monochromatic triangle for fewer points.

[quote="zscool"]given [tex]\lfloor en! \rfloor + 1 [/tex] points in which each pair of points is connected with a line colored one of [tex]n[/tex] colors, there must exist a monochromatic triangle.  This result is the best possible, there need not exist a monochromatic triangle for more points.[/quote]

This is a nice result. What is it called, or who was it found by?

[quote]This is a nice result. What is it called, or who was it found by?[/quote]

first off, I had to edit my post; a monochromatic triangle need not exist for fewer points not more.  I do not know who is credited with that result.  I first saw it in Engel's Problem Solving Strategies.

[quote="JBL"]Given the following game:

You have a given range of integers, with one of these numbers the target (this is secret), and there are two players. Each player takes a turn to guess a number in that range and then will be told whether he is too high or too low. Then its the other players turn.  The first player picks his guesses according to a "best strategy," while the second player guesses randomly.

As the size of the game goes to infinity, the chances of the first player winning approach  [tex]\displaystyle\frac{1+e^{-2}}{2}[/tex].[/quote]

Did you guys ever figure out if this result was original? If it is, you all should consider writing it up collectively and submitting it somewhere. It is a nice result.

[tex]
\displaystyle\int e^x \, dx = e^x + C

$\displaystyle\frac{d^n y}{dx^n} e^x = e^x$
[/tex]

[quote="nr1337"][tex]
\displaystyle\int \!\!\! \int \ldots \int \text{ ($n$ integrals) } e^x \, dx = e^x + C [/tex][/quote]

I see what you're getting at, but this is not quite right. You should just leave one integral to say what you're trying to say. Performing multiple integrals uses a slightly different notation, and works a bit differently than how you're using it.

[tex]
(\pi^4 + \pi^5)^{\frac{1}{6}} \approx e
[/tex]

Sadly, after the first 7 digits digits after the decimal point, this isn't true anymore.

[quote="gauss202"]
I see what you're getting at, but this is not quite right. You should just leave one integral to say what you're trying to say. Performing multiple integrals uses a slightly different notation, and works a bit differently than how you're using it.[/quote]

This is also awkward because the constant C needs to be integrated.

[quote="nr1337"]I wish I had a real calculus course.  Sadly, I doubt I'll get accepted into my local college in 7th grade.  :([/quote]

EPGY is happy to let seventh graders take calculus. You can get college credit for it as well if you aren't too lazy to submit a transcript when you get to college, as I was.

[tex]\frac{1 kilometer}{1 mile} \approx \ln2[/tex]

You could use the compound interest formula too:

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
[tex]A=P[(1+\frac{r}{n})^n]^t[/tex]

[tex]\displaystyle\lim_{n\to\infty}{(1+\frac{r}{n})^n}=e^r[/tex]

And that's where we get the continuously compounded interest formula:

[tex]A=Pe^{rt}[/tex]

You can also do it using a number of limits:

[tex]\displaystyle\lim_{n\to\infty}{(1+\frac{1}{x})^x}=e[/tex]

(I used this one before w/the compound interest formula)

And there is also one w/n!, but I don't remember it.

You can do most of these limits by using L'Hopital after rewriting it as [tex]x=e^{\ln{x}}[/tex].

Take an sequence of independent random variables, each one uniformly distributed on the interval [0, 1]. Add them until the first time the sum exceeds 1. What is the expected number of terms in this sum? If you've been following the theme of this thread ...

[quote="JS1527"]You could you the compound interest formula too:

[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
[tex]A=P[(1+\frac{r}{n})^n]^t[/tex]

[tex]\displaystyle\lim_{n\to\infty}{(1+\frac{r}{n})^n}=e[/tex]

And that's where we get the continuously compounded interest formula:

[tex]A=Pe^{rt}[/tex]

You can also do it using a number of limits:

[tex]\displaystyle\lim_{n\to\infty}{(1+\frac{1}{x})^x}=e[/tex]

(I used this one before w/the compound interest formula)

And there is also one w/n!, but I don't remember it.

You can do most of these limits by using L'Hopital after rewriting it as [tex]x=e^{\ln{x}}[/tex].[/quote]

There are several errors in your posting, Julie.  See if you can spot them (some are typos, some are mathematical).

The factorial one you are referring to is probably:

Sum(1/i!,i,0, :inf: ) = e which is a special case of:

Sum(x^i/i!,0,  :inf: ) = e^x (a taylor series which is convergent everywhere because by the ratio test)

lim n-->  :inf:  x^(i+1)/(n+1)!/x^(i)/(n!)

= lim n-->  :inf: x/(n+1)

= 0

I fixed one thing and a grammatical error, but I don't know what the others are.  Darn, I make so many mistakes.

I was referring to:

[tex]\displaystyle\sum_{k=0}^{\infty}{\frac{1}{k!}}=e[/tex]

Which is the same thing as you said before.

http://mathworld.wolfram.com/FactorialSums.html

Oh and you would not actually put x as e^ln(x), you would put some expression as e^ln(x).  Is this another error you were talking about?

For example, to evaluate the limit:

[tex]\displaystyle\lim_{x\to\infty}(1+\frac{1}{x})^x[/tex]

Change (1+1/x)^x into e^(ln(1+1/x)*x).

Now you just have to find the limit:

[tex]\displaystyle\lim_{x\to\infty}{ln(1+\frac{1}{x})*x}[/tex]

Using L'Hopital:

[tex]\displaystyle\lim_{x\to\infty}{ln(1+\frac{1}{x})*x}=\displaystyle\lim_{x\to\infty}{\frac{\frac{d}{dx}ln(1+\frac{1}{x})}{\frac{d}{dx} \frac{1}{x}}}[/tex]

And the rest is just derivatives.  I think?


I have a math problem from the game I play. I thought yall might have fun trying to solve it. Not to mention it would help me out :p

While sitting around waiting for interior decorators to finish up in the scroll repository, a bunch of bored and mischievous workers decided that they'd build a catapult. When they finally finished, it was able to shoot a projectile straight up into the air. So. in a move which they later regretted, they decided to launch one of the pyramid blocks, which had a density of 2400 kilograms per cubic meter.

When the block crashed back into the catapult, smashing it to pieces, it was determined to have a velocity of 20 meters per second. Assuming that gravity in Neopia is exactly 9.90 meters per second squared, the block hit the catapult at the same elevation as it was launched from the catapult, and there was, miraculously, no air resistance acting on the block, how high, in centimeters, did the block fly above the catapult? Please round to the nearest centimeter and please, do not try this at home.

Looks more like physics to me. But definitely not Fun and Games. *prods*

Odd, this problem was just posted in another [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=342306#p342306]part of the forum[/url] by another user...


geometry
rectangle
group theory
superior algebra
superior algebra solved

Find out all the symmetries of a square, rombe and rectangle

The groups of symmetries are up to an izomorphism:
For the square D_4 the dihedral group defined by the relations x 4=y 2 =e and yxy=x 3.
The symmetries generate a group, but not all the elements of the group are symmetries.
For the romb and the square it is Z_2*Z_2.

The geometric descriptions for the symmetries are quite easy to find


Define an operation * on Z by

[color=green]a*b = ab - a - b +2 for a, b in Z.  We know that * is commutative and associative and that * has an identity element, namely 2.  Determine which elements in Z have inverses relative to *.  Justify your answer.[/color]

Thanks very much :lol:

[color=blue]Let b is the inverse of a.
Then a*b=2 since 2 is the identity of the operation *.
But  a*b=ab-a-b+2=2  and you must solve the equation  ab-a-b+2=2 in integers.
This equation gives (a,b) = (0,0) or (a,b) = (2,2).

Hope it's correct  :)[/color]


search function integration trigonometry calculus limit derivative

How do you find

Int sin(x)/x dx

over the real line?

There are many methods - lots of different things work.

Here's an outline of my favorite method. I know I've posted it, but I haven't found the right way to search for it. It's a real variable method rather than a complex variable method.

(1) Since the function is even, we might as well talk about $ \int_0^{\infty}\frac{\sin x}x\,dx.$

(2) Integrate by parts, using $ u\equal{}\frac1x,dv\equal{}\sin x\,dx,$ and $ v\equal{}1\minus{}\cos x$ to get

$ \int_0^{\infty}\frac{\sin x}x\,dx\equal{}\int_0^{\infty}\frac{1\minus{}\cos x}{x^2}\,dx$

(3) Set $ \frac1{x^2}\equal{}\int_0^{\infty}te^{\minus{}tx}\,dt.$ Put that in the integral to make a double integral.

(4) Switch the order of integration in the double integral (justified for nonnegative functions).

(5) Compute.

Or just recall whose Fourier transform it is ;)

Yes, indeed - if I already have some version of a Fourier inversion theorem. It might be useful to have this particular integral computed as a lemma to be used in a proof of such a theorem.

Usually I do Fourier inversion using $ e^{\minus{}x^2}$. But this integral can be used too. It is enough to have the inversion formula for just one reasonable function at $ x\equal{}0$ to recover the full theorem and the characteristic function of an interval is as good as any :)

Since this was posted in the complex analysis section, we might as well look at this argument.

For $ 0 < \delta < R,$ let $ \Gamma_{\delta,R}$ be the simple closed curve consisting of these segments:

$ \Gamma_1:$ the interval $ [ \minus{} R, \minus{} \delta]$ on the real line,
$ \Gamma_2:$ the upper semicircle of $ |z| \equal{} \delta,$ parameterized clockwise,
$ \Gamma_3:$ the interval $ [\delta,R]$ on the real line, and
$ \Gamma_4:$ the upper semicircle of $ |z| \equal{} R,$ parameterized counterclockwise.

By Cauchy's Thoerem, $ \int_{\Gamma_{\delta,R}}\frac {e^{iz}}{z}\,dz \equal{} 0.$

We can show that $ \lim_{R\to\infty}\int_{\Gamma_4}\frac {e^{iz}}{z}\,dz \equal{} 0.$

The key is that $ \left|e^{iz}\right| \equal{} e^{ \minus{} y}$ is very small away from the real axis.

$ \int_{\Gamma_2}\frac {e^{iz}}{z}\,dz \equal{} \minus{} \int_0^{\pi}\frac {e^{i\delta e^{i\theta}}}{e^{i\theta}}\,ie^{i\theta}\,d\theta.$

As $ \delta\to0,$ this tends to $ \minus{} \int_0^{\pi}i\,d\theta \equal{} \minus{} i\pi.$

$ \int_{\Gamma_1\cup\Gamma_3}\frac {e^{iz}}{z}\,dz \equal{} \left(\int_{ \minus{} R}^{ \minus{} \delta} \plus{} \int_{\delta}^R\right) \frac {\cos x}{x} \plus{} \frac {i\sin x}{x}\,dx$

$ \equal{} i\left(\int_{ \minus{} R}^{ \minus{} \delta} \plus{} \int_{\delta}^R\right)\frac {\sin x}{x}\,dx$

since $ \frac {\cos x}{x}$ is an odd function and integrates to zero over that set.

Take the limit as $ \delta\to0$ and $ R\to\infty$ to conclude that

$ \lim_{\stackrel{\delta\to0}{R\to\infty}}i\left(\int_{ \minus{} R}^{ \minus{} \delta} \plus{} \int_{\delta}^R\right)\frac {\sin x}{x}\,dx \minus{} i\pi \equal{} 0$ or

$ \int_{ \minus{} \infty}^{\infty}\frac {\sin x}{x}\,dx \equal{} \pi.$

Numerical power series computation:

We know that $ \frac{2}{\pi}\int_0^{\infty}\frac{\sin x}{x}\,dx\equal{}1.$

It follows (essentially from the Alternating Series Test) that the "first bump" is bigger:

$ \frac{2}{\pi}\int_0^{\pi}\frac{\sin x}{x}\,dx>1.$

How much bigger than $ 1?$ I claim that it's roughly $ 18\%$ bigger. Use partial sums of power series to verify this.

Key words: Gibbs phenomenon.

[quote="Kent Merryfield"]
We can show that $ \lim_{R\to\infty}\int_{\Gamma_4}\frac {e^{iz}}{z}\,dz = 0.$
[/quote]

See plot below:

Letting $ z=Re^{it}$, we have:

\begin{align*}
\left|\,\mathop\int\limits_{\Gamma_4}\frac{e^{iz}}{z}dz\right|&=
\left|\int_0^{\pi} \frac{Exp[iRe^{it}]}{Re^{it}}iRe^{it}dt\right| \\
&=\left|i\int_0^{\pi} Exp[-RSin(t)+iRCos(t)]dt\right| \\
&\leq \int_0^{\pi} e^{-RSin(t)}dt=2\int_0^{\pi/2} e^{-RSin(t)}dt \\
&< 2\int_0^{\pi/2} e^{-R(2t/{\pi})}dt
\end{align*}
and:
\[ \lim_{R\to\infty}\int_0^{\pi/2} e^{-R(2t/\pi)}dt=0\]

[quote="Kent Merryfield"]There are many methods - lots of different things work.

Here's an outline of my [color=red][b]favorite[/b][/color] method.[/quote]

wanna see mine ? LOL :D

let $ I(a)\equal{}\int_0^\infty\frac{\sin\,x}{x}\;e^{\minus{}ax}\;dx$

$ \Rightarrow \frac{d}{da}I(a)\equal{}\minus{}\int_0^\infty\sin\,x\cdot e^{\minus{}ax}\;dx\equal{}\minus{}\frac{1}{1\plus{}a^2}$

$ \Rightarrow I(a)\bigg|_I^0\equal{}\minus{}\int_0^\infty\frac{1}{1\plus{}a^2}\;da$

giving

$ I\equal{}\frac{\pi}{2}$

That's really the same idea- except that justifying the interchange is tricky without absolute convergence. In integral form, it's
$ \int_0^{\infty}\frac{\sin x}{x}\,dx\equal{}\int_0^{\infty}\int_0^{\infty} \sin x\, e^{\minus{}ax}\,da\,dx \equal{}\int_0^{\infty}\int_0^{\infty} \sin x\, e^{\minus{}ax}\,dx\,da$.
That last interchange is nontrivial- the Fubini hypothesis of absolute convergence doesn't hold.

In misan's computation, there's no trouble justifying the differentiation under the integral, and the fact that $ \lim_{a\to\infty}I(a)\equal{}0.$ Hence, we know that for all $ a>0,\ I(a)\equal{}\frac{\pi}2\minus{}\arctan(a).$

Now comes the trickiest part analytically: We would like to show the following:

$ \lim_{a\to0^\plus{}}I(a)\equal{}\lim_{a\to0^\plus{}}\int_0^{\infty}f(x)e^{\minus{}ax}\,dx \stackrel{?}{\equal{}}\int_0^{\infty}f(x)\,dx$

as long as that last integral converges as an improper integral.

Do you know what that is? It's Abel's theorem, or an integral analogue of Abel's theorem. Here's the original of Abel's theorem:

$ \lim_{r\to1^\minus{}}\sum_{n\equal{}0}^{\infty}a_nr^n\equal{}\sum_{n\equal{}0}^{\infty}a_n,$ provided the latter series converges.

Compare that with what we want. It really is the same thing, isn't it? And we should be able to adapt any proof that works for the original theorem so that it works for the integral version.

Hmmm. If I were to study on my own and try to learn complex analysis, what book might you suggest? And to that extent, are there any prerequisites I should absolutely get a hold on? Currently partway through Calc III.

Thankya!

Complex analysis is generally an advanced undergraduate or early graduate course. Real analysis is generally a prerequisite.

I think one book is not enough. Lars Ahlfors book is really awesome, but he also does some things a little bit too complex, and likes to use homology in a lot of places where there are simpler classical proofs. I also don't think he has a deep enough discussion of conformal maps between surfaces, but I don't know if any book does.

Pre-requisites are basically a good understanding of basic metric topology, in particular compactness and connectedness. But Ahlfors book is pretty self-contained, so it spends a lot of pages on defining all those things for the inexperienced reader.

I don't agree with jmerry that it is an advanced undergraduate topic. Is everything past the first year advanced? The theory in basic complex analysis is not that hard since everything behaves so nicely, easier than for real analysis in my opinion.

At the university I took it at, complex analysis was the last of the core upper-division courses for the major, and the only one with another of those courses as a prerequisite. That metric topology is part of the real analysis course, of course.

i always have this irresistible tendency to ask this set of questions on an exam, of-course, very few will get all 5 right. :D btw/ i don't ask these in complex variables, only in an integral calculus class.

(1) find $ \int_0^\infty\;\dfrac{\sin\,x}{x}\;\textbf dx$ [hide]$ \text{answer: }\qquad\boxed{\dfrac{\pi}{2}}$[/hide]

(2)  find $ \int_0^\infty\;\dfrac{\sin\,x}{\sqrt x}\;\textbf dx$ [hide]$ \text{answer: }\qquad\boxed{\sqrt {\;\dfrac{\pi}{2}\;}}$[/hide]

(3) find $ \int_0^\infty\;x^{\alpha \minus{} 1}\;\sin\,x\;\textbf dx$ (where $ 0<\alpha<1$) [hide]$ \text{answer: }\qquad\boxed{\sin\left(\dfrac{\alpha\pi}{2}\right)\;\Gamma(\alpha)}$[/hide]

(4) find $ \int_0^\infty\;\left(\dfrac{\sin\,x}{x}\right)^2\;\textbf dx$ [hide]$ \text{answer: }\qquad\boxed{\dfrac{\pi}{2}}$[/hide]

(5) find $ \int_0^\infty\;\dfrac{(1 \minus{} \cos\,\beta\,x)}{x^2}\;\textbf dx$ [hide]$ \text{answer: }\qquad\boxed{\dfrac{\pi}{2}\;|\beta|}$[/hide]

as a textbook, i think marsden and hoffman's one is pretty cool. and marsden is a great teacher !  :coolspeak:

Well since you're asking them here, how about $ \displaystyle \mathop\int\limits_{C} \frac{e^{iz}}{\sqrt{z}}dz$ over a quarter turn
for the second one?  The contour $ C$ is then:

$ \Gamma_1$: zero (or $ \delta$ to be more precise) to infinity on real line
$ \Gamma_2$: Large quarter circle in first quadrant
$ \Gamma_3$: Down the imaginary axis
$ \Gamma_4$: small quarter-circle around branch point of sqrt function.

We can show the integral tends to zero over $ \Gamma_2$ and $ \Gamma_4$. However down the imaginary axis we get:

\[ \mathop\int\limits_{\Gamma_3}\frac{e^{iz}}{\sqrt{z}}dz=-i e^{-\pi i/4}\int_0^{\infty}\frac{dr}{e^r r^{1/2}}dr\]

So that:

\begin{align*}
\int_0^{\infty} \frac{Cos(r)}{r^{1/2}}dr+i\int_0^{\infty} \frac{Sin(r)}{r^{1/2}}dr&=i\left[\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right]\int_0^{\infty}\frac{1}{e^{r}r^{1/2}}dr \\
&=i\left[\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right] \Gamma[1/2] \\
&=\sqrt{\frac{\pi}{2}}+i\sqrt{\frac{\pi}{2}}
\end{align*}

to find $ \int_{0}^ {\infty}{\frac{sinx}{x}}$ we can also use laplace transform. Indeed one can cpmpute the value of $ \mathcal{L} \left\{\frac {sinx}{x} \right\} \equal{} F(p)$ and then find $ F(1)$

Maldo, you probably mean $ F(0).$ Go look at misan's post #9 in this topic, and my response to that in #11 - that's what is being discussed.

consider integrals :
$ \int_{0}^{\infty}t^{\alpha \minus{} 1}\cos t dt$ and  $ \int_{0}^{\infty}t^{\alpha \minus{} 1}\sin t dt$
it's proved similary:
$ \Gamma _{1} \equal{} (\epsilon, R)$
$ \Gamma_{2} \equal{} circle (R ,iR)$
$ \Gamma_{3} \equal{} (iR,i\epsilon)$
$ \Gamma_{4} \equal{} smallcircle(i\epsilon,\epsilon)$
$ \Gamma \equal{} \oplus_{i \equal{} 1}^{4}\Gamma_{i}$
$ \int_{\Gamma}z^{\alpha \minus{} 1}e^{iz}dz$ where $ z^{\alpha \minus{} 1}$ define as $ (1 \plus{} i0)^{\alpha \minus{} 1} > 0$
then easly seen that :
$ \int_{\Gamma}z^{\alpha \minus{} 1}e^{iz}dz \equal{} 0$
on other hand we have
$ \int_{\Gamma}z^{\alpha \minus{} 1}e^{iz}dz \equal{} \sum_{i \equal{} 1}^{4}\int_{\Gamma_{i}}z^{\alpha \minus{} 1}e^{iz}dz$
note that as$ \epsilon \to 0$ and $ R \to \infty$then  $ \int_{\Gamma_{2,4}}z^{\alpha \minus{} 1}e^{iz}dz \to 0$
hence we have :
$ \int_{\Gamma_{1}}z^{\alpha \minus{} 1}e^{iz}dz \equal{} \minus{} \int_{\Gamma_{3}}z^{\alpha \minus{} 1}e^{iz}dz$

Consider :
$ \int_{\Gamma_{3}}z^{\alpha \minus{} 1}e^{iz}dz$
but now after $ \Gamma_{2}$ our function $ z^{\alpha \minus{} 1}$ is $ |z|^{\alpha \minus{} 1}e^{\frac {\pi}{2}(\alpha \minus{} 1)}$
or we have :
$ \int_{\Gamma_{3}}|z|^{\alpha \minus{} 1}e^{\frac {\pi}{2}(\alpha \minus{} 1)}e^{iz}dz$
now by parametrization of $ OY$ say $ z \equal{} ix$ we have : (just susbtitute)
$ i\int_{R}^{\epsilon}|x|^{\alpha \minus{} 1}e^{i\frac {\pi}{2}(\alpha \minus{} 1)}e^{ \minus{} x}dx \equal{} \minus{} ie^{\frac {i\pi}{2}}e^{i\frac {\pi}{2}(\alpha)}\int_{\epsilon}^{R}|x|^{\alpha \minus{} 1}e^{ \minus{} x}dx \equal{} \minus{} e^{i\frac {\pi}{2}(\alpha)}\Gamma(\alpha)$ (as $ \epsilon \to 0$ and $ R\to\infty$)
hence:
$ \int_{\Gamma_{1}}z^{\alpha \minus{} 1}e^{iz}dz \equal{} \int_{0}^{\infty}x^{\alpha \minus{} 1}e^{ix}dx \equal{} \minus{} ( \minus{} e^{i\frac {\pi}{2}(\alpha)}\Gamma(\alpha))$ after is simple.
Note that this is true :
$ \alpha \neq 0, ( \minus{} 1 < Re\alpha < 1)$
and if $ \alpha \equal{} 0$, then answer will be $ \frac {\pi}{2}$
but for $ \int ... cos$ it's true only when $ (0 < Re\alpha < 1)$

now consider integrals of the form :
$ \int_{0}^{\infty} (\frac {\sin x}{x})^{n}dx$ for $ n \in \mathbb N$
we need to represent $ \sin^{n}x$ as $ Re$ or $ Im$ of something $ R(e^{iz})$ where $ |R(e^{iz})| < something$ for $ Im(z)>0$  :lol:  say $ < C$
or see (Jordan lemma ), because contour will be
$ \Gamma _{1} \equal{} ( \minus{} R, \minus{} \epsilon)$
$ \Gamma_{2} \equal{} smallcircle ( \minus{} \epsilon , \plus{} \epsilon)$
$ \Gamma_{3} \equal{} (\epsilon, \plus{} \infty)$
$ \Gamma_{4} \equal{} bigcircle( \plus{} \infty, \minus{} \infty)$
$ \Gamma \equal{} \oplus_{i \equal{} 1}^{4}\Gamma_{i}$
for exapmle:
$ \sin^{2}x \equal{} \frac {1}{2}(1 \minus{} \cos(2x)) \equal{} Re (\frac {1}{2}(1 \minus{} e^{2ix}))$ :lol:
then :
consider $ f(z) \equal{} \frac {\frac {1}{2}(1 \minus{} e^{iz})}{z^{2}}$
similary :
$ 0 \equal{} \int_{\Gamma}f(z)dz \equal{} \sum_{i \equal{} 1}^{4}\int_{\Gamma_{i}}f(z)dz$
easly seen that $ \int_{\Gamma_{4}}f(z)dz \to 0$ as $ R \to \infty$
function $ f(z)$ is even, hence $ \int_{ \minus{} \infty}^{\infty}f(z)dz \equal{} 2\int_{0}^{\infty}f(z)dz$
then  it's enough to find $ \int_{\Gamma_{2}}f(z)dz$
let's find Laurent serie for $ f(z)$ az $ z \to 0$
$ f(z) \equal{} \frac {1}{2z^{2}}(1 \minus{} e^{2iz})\approx$ ( :blush: )$ \frac {1}{2z^{2}}(1 \minus{} (1 \plus{} 2iz \plus{} z(.)))$
and after integration

$ \approx \int_{\Gamma_{2}} \frac {1}{2z^{2}}(1 \minus{} (1 \plus{} 2iz \plus{} z(.))) dz \approx \minus{} i\ln(z)|_{ \minus{} \epsilon}^{ \plus{} \epsilon} \plus{} \epsilon (..)\approx \minus{} i(\ln(\epsilon) \minus{} (\ln(\epsilon) \plus{} \pi i)) \plus{} \epsilon(...)\approx \minus{} \pi$ as $ \epsilon \to 0$

hence answer will be $ \frac {\pi}{2}$

similary we can find $ \int \frac {1 \minus{} \cos ()}{x^{2}}dx$.., and  e.t.c
for example (own   :P  ): try to find
$ \int_{0}^{\infty}\frac {\sum_{i \equal{} 0}^{m}( \minus{} 1)^{i}\frac {x^{2i}}{(2i)!} \minus{} \cos x}{x^{2m \plus{} 2}}dx$
similary for $ .... \sin ....$(but serie will be $ \sum ( \minus{} 1)^{...}x^{2k \plus{} 1}/(2k \plus{} 1)!$)
also about
$ \int_{0}^{\infty}\frac {\sin x}{x}dx$ I know some theorem about :
if $ f$ has period $ \equal{} \pi$ then we have
$ \int_{0}^{\infty}f(x)\frac {\sin x}{x}dx \equal{} \int_{0}^{\pi/2}f(x)dx$ if we put $ f(x) \equal{} 1$ then we are done.
as I know this theorem proved by Fourier ...
 :sleeping:  I go to sleep.

Extremal- compare what you just wrote to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1087087#1087087]#6[/url]. It's the same argument, and I'd say Kent's version is much easier to read. You tend to be a very sloppy writer- try to work on it.

[quote="jmerry"]Extremal- compare what you just wrote to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1087087#1087087]#6[/url]. It's the same argument[/quote]
Yes, I know, but $ \int_{0}^{\infty}\frac{\sin x}{x}dx$ and $ \int_{0}^{\infty}\frac{\sin^{2}x}{x^{2}}dx$ is different.
Maybe fore someone it's same idea  (form me yes, I agree with you  :) ), but maybe there exist people, who know How to solve this integral $ \int_{0}^{\infty}\frac{\sin x}{x}dx$ (by Kent's argument), but they can't to solve this : $ \int_{0}^{\infty}\frac{\sin^{2}x}{x^{2}}dx$.
Much different problems can be solved by same argument, and many people can't to see it. for example:
$ \int_{0}^{\infty}t^{\minus{}1/2}\sin t dt$ and $ \int_{0}^{\infty}t^{\alpha\minus{}1}\sin t dt$ is same , There is no difference between shawtoos and My post  :blush:  it's same argument.

[quote="jmerry"]There's nothing wrong with the content. This is about style, and the fact that it is the same idea just makes it easier to compare the two versions.[/quote]

you   understood the contents of his post ?   :huh:  i gave up before even reading half-way through it....   :(  still somewhat agree with what was said in the later part of post # 22.

[quote="Extremal"]There is no difference between shawtoos and My post  :blush:  it's same argument.[/quote]

Hi Extremal.  Thanks for posting that work above.  You talk of people like me.  No surprise there I know.  It's difficult to read but I attribute that to your need to learn a second language.  I plan to work through it all.  How about this:  what is the absolute best way to present your work in the clearest way for beginners that barely know the subject so that after seeing it for the first time, they come away completely understanding it?  For me, that is a matter of expository style and represents one quality of a gifted teacher.

No offense JMerry and Misan.  You guys are tops in my book.


$ a,b,c$ are réel non negative réel numbers. put: $ a \plus{} b \plus{} c \equal{} p$, $ ab \plus{} bc \plus{} ca \equal{} q$ , $ abc \equal{} r$
prove that:
$ p^5 \plus{} 27rq \plus{} 6pq^2 \geq 9rp^2 \plus{} 5p^3q$
with equality if $ a \equal{} b \equal{} c$

[quote="Albanian Eagle"]Wrong. try a=b=c=1[/quote]

Well $ a\equal{}b\equal{}c\equal{}1 \rightarrow p\equal{}3, q\equal{}1, r\equal{}1$

$ \Rightarrow (3)^5 \plus{} 27(1)(1) \ge 9(1)(1)^2 \plus{} 5(3)^3(1) \plus{} 6(3)(1)^2$
$ \Rightarrow 270 \ge 162$

Looks right to me :P ...but still very hard

[quote="peine"]$ a,b,c$ are réel non negative réel numbers. put: $ a \plus{} b \plus{} c \equal{} p$, $ ab \plus{} bc \plus{} ca \equal{} pq$ , $ abc \equal{} r$
prove that:
$ p^5 \plus{} 27rq \geq 9rp^2 \plus{} 5p^3q \plus{} 6pq^2$[/quote]

If you try $ a\equal{}b\equal{}c\equal{} \frac{1}{3}$, then $ LHS\equal{}\frac{4}{3}$ and $ RHS\equal{}\frac{8}{3}$. :wink:

It is then true and very easy if you mean $ ab\plus{}bc\plus{}ca\equal{}q$ instead of $ pq$  :wink:

very well, it is the same resolution which I have make, it is not difficult, but I used it to prove that:
$ \frac{2(a^3\plus{}b^3\plus{}c^3)}{abc}\plus{}\frac{9(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2} \geq 33$

which you can also prove by using Am-Gm nicely!

Since you asked me, here it goes:
We want to prove $ \frac{a^3\plus{}b^3\plus{}c^3}{abc}\plus{}\frac{9(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2}\geq 12$

Now this is equivalent to $ \frac{a^3\plus{}b^3\plus{}c^3}{abc}\minus{}3\geq 9\minus{}9\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}$

or $ \frac{a\plus{}b\plus{}c}{abc}(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)\geq \frac{9}{a^2\plus{}b^2\plus{}c^2}(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)$

Since $ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$ all we need to show is $ (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\geq 9abc$ which is a trivial application of Am-Gm


vector
function
rotation
analytic geometry
calculus
derivative
real analysis

A disc rotates back and forth with angular velocity (cos t) rad/sec. An insect starting 10 cm from the center of the disc at time t=0 crawls outward at a rate of 5t cm/sec. Find the position, velocity and speed of the insect after 2pi seconds. Give the components of position and velocity relative to a fixed Cartesian coordinate system.

I'm not sure where to begin on this one..any help would be appreciated =)

Let's start by writing down what we know. The disk has $ \dot{\theta}\equal{}\omega\equal{}\mbox{cos}(t)\rightarrow \theta \equal{}\mbox{sin}(t)$. The bug moves with $ \dot{r} \equal{}v\equal{}.05t\rightarrow r\equal{}.1\plus{}.025t^2$. Thus, the parametric equations of motion are:

$ \theta \equal{}\mbox{sin}(t)$
$ r\equal{}.1\plus{}.025t^2$

Converting these into Cartesian coordinates, we get:

$ x\equal{}r\cdot \mbox{cos}(\theta )\equal{}(.1\plus{}.025t^2)\cdot\mbox{cos}(\mbox{sin} (t))$
$ y\equal{}r\cdot \mbox{sin}(\theta )\equal{}(.1\plus{}.025t^2)\cdot\mbox{sin}(\mbox{sin} (t))$

That answers the last question. Taking the first and second derivatives and substituting in $ t\equal{}2\pi$ gives you the rest of the answers. Note that the work I did was in meters, not centimeters.


geometry
3D geometry
prism
ceiling function

Carlos wishes to paint the five faces of the solid shown with one coat of paint. The solid is a right prism with triangular bases. A one-ounce container of paint costs $ \$1.50$ and covers 30 square inches. This is the only size container he can purchase. How much will it cost to purchase the total number of containers of paint he needs?

[asy]import three;
unitsize(1inch);
currentprojection=orthographic(1,1,1);
draw((0,0,0)--(-30,-26,0)--(-30,-26,30)--(0,0,30)--cycle3);
draw((0,0,0)--(0,0,30)--(54,-26,30)--(54,-26,0)--cycle3);
draw((-30,-26,30)--(54,-26,30));
draw((-30,-26,0)--(54,-26,0),dashed);
draw((-2.5,-2.2,0)--(0,-3.4,0)--(1.9,-0.9,0));
draw((-2.5,-2.2,30)--(0,-3.4,30)--(1.9,-0.9,30));
label("4''",(-15,-13,0),E);
label("6''",(27,-13,0),SE);
label("3''",(54,-26,15),E);[/asy]

$ \lceil x \rceil$ refers to the least integer greater than or equal to $ x$, so yes.


prove that sqrt(a)+sqrt(b)=c is impossible assuming a, b, and c are integers and a isnt a perfect square.

CAUTION: this is probably easier for beginners than the more advanced moderators

It's actually not as hard as you may think for advanced math people:

[hide]sqrt(a)+sqrt(b)=c

(sqrt(a)+sqrt(b))^2=c^2

a+2*sqrt(a)*sqrt(b)+b=c^2

Since a is not a perfect square, sqrt(a) is irrational (this is fairly easy to prove, but I won't go through it).  Therefore, 2*sqrt(a)*sqrt(b) is irrational, since an integer times an irrational number is still irrational.  This makes the entire left side of the equation irrational, becuase the sum of an integer and an irrational is also an irrational.

Since c^2 is an integer, the right side is an integer.

With the above two paragraphs, we have

(Irrational Number)=(Integer)

Which cannot be true because an integer is never irrational and vice versa.[/hide]

[quote="towbomb"]Since a is not a perfect square, sqrt(a) is irrational (this is fairly easy to prove, but I won't go through it).  Therefore, 2*sqrt(a)*sqrt(b) is irrational.  [/quote]
who said b is a perfect square?

Not so fast ...
[quote="towbomb"]
Since a is not a perfect square, sqrt(a) is irrational (this is fairly easy to prove, but I won't go through it).  Therefore, 2*sqrt(a)*sqrt(b) is irrational,  [/quote]

As someone already has pointed out  ... if a = 2 and b = 8 , neither sqrt(a) nor  sqrt(b) is rational, yet 2*sqrt(a)*sqrt(b) is an integer.

Just  be a  little careful. .. Hope this helps.


How can one get better at predicting reactions? For example, what is the thought process in determining the reactions of the following (or do you just have to memorize a lot of reactions):

1) sulfuric acid and ammonia
2) sulfuric acid and sodium carbonate
3) potassium chromate and sulfuric acid
4) sulfuric acid and copper

[hide="Answers"]
1) $ H_2SO_4\plus{}2NH_3\rightarrow (NH_4)_2SO_4$

2) $ H_2SO_4\plus{}Na_2CO_3\rightarrow  Na_2SO_4\plus{}H_2O\plus{}CO_2$

3) $ 2K_2CrO_4\plus{}H_2SO_4\rightarrow  K_2Cr_2O_7\plus{}K_2SO_4\plus{}H_2O$

4) $ 2H_2SO_4\plus{}Cu\rightarrow  CuSO_4\plus{}2H_2O\plus{}SO_2$[/hide]

1) and 2) are common acid-base reactions.

3) requires you to know that $ 2\ce{CrO4}^{ 2 \minus{} } \plus{} \ce{H \plus{} } \rightarrow \ce{Cr2O7}^{2 \minus{} } \plus{} \ce{H2O}$.

4) is a special reaction. Because normally textbooks tell you that $ \ce{Cu}$ won't react with sulphuric acid. I don't know how to reason this one out but I just know it because I have seen it before. :P

Experience plays a role here. Besides that, and regarding reactions 3 and 4, you should have a good list of reduction half-reactions and corresponding potentials at hand: then you would be in a better position to predict the outcome of redox reactions.

Another interesting example is provided by the reactions of nitric acid with metals like Cu and Zn: the outcome depends not only on the nature of the metal but also on the acid concentration and temperature.

Could we maybe start an (inorganic) reactions marathon or something, or just post lots of problems? I'm preparing for USNCO, and #5 on part 2 of the national exam is always predicting reactions.

AP Chemistry Exams run by the College Board have always involved a question in the free-response section in which one has to predict a reaction described by substances being brought together under certain conditions.

Here is a link to the College Board's website about the AP Chemistry Exam.  They have some of the more recent free-response sections up; look at question #4 for each of them.

http://www.collegeboard.com/student/testing/ap/chemistry/samp.html?chem


I'm going to the program as a first year student.
Before, I never thought that I would enjoy writing proofs,
but the application problems itself made me discover my passion:)
(Really, writing the eleven-page answer wasn't painful at all.
If it were a history paper, I would rather weep.)

This would be a nice opportunity in which I could meet American kids who feel really passionate about math.
Since I came over to America (one and half years ago), I never met such a kid.

ps. I have never been Ohio, so I don't know how weather would be there
ps2. Is it possible for me to bring a musical instrument? Although music is not my primary passion and I certainly do not want to let anything hinder me from perusing math, I do not want to lose my musical abilities by forgoing practice during the two months altogether.

The weather in Ohio is temperate, or at least it should be. And I'm definitely bringing a musical instrument, I'm sure it won't be a problem. I have to play for like three hours a day, haha. What do you play? (I play violin. and saxophone, but I won't be bringing a saxophone with me.)


o there:  :spam:  :wacko:  :rotfl:

                       Trig/Calc, difficulty level 3. Prove that the are root of 2 is irrational, and explain how to use a straightedge and compass to construct a segment that is the square root of 2 in length.

        P.S. u have to prove it

This isn't a trig/calc problem at all. I think he was joking...
For the first
[hide="1"]Proof by contradiction.
Assume that it can be written as a fraction $\frac{p}{q}$ in lowest terms
$p^{2}=2q^{2}$
So p is even. Plugging in p=2k, we see that
$2k^{2}=q^{2}$
So q is even too.
But if both p and q are even, then it can't be written as a fraction in lowest terms.
[/hide]
number two
[hide="2"]
Take a segment of length 1.
Construct a perpendicular to its endpoint (just pick one).
Copy a segment of length 1 on this perpendicular.
Connect the intersection points to have a 1-1-rt2 triangle[/hide]


calculus
integration
logarithms
trigonometry
complex numbers
real analysis
real analysis unsolved

hello :D exams are now finally over, so i am back...  :D

show :D

$ \textbf{(i)}\;\;\int_{0}^{\infty}\;\;\frac{e^{\minus{}a\,x}\;\minus{}\;e^{\minus{}b\,x}}{x\cdot\sec\,(p\,x)}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\frac{1}{2}\;\ln\left(\frac{b^{2}\plus{}p^{2}}{a^{2}\plus{}p^{2}}\right)}$

$ \textbf{(ii)}\;\;\int_{0}^{\infty}\;\;\frac{e^{\minus{}a\,x}\;\minus{}\;e^{\minus{}b\,x}}{x\cdot\text{cosec}\,(p\,x)}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\tan^{\minus{}1}\left(\frac{b}{p}\right)\;\minus{}\;\tan^{\minus{}1}\left(\frac{a}{p}\right)}$

For all complex numbers $ z$ with a positive real part $ \int_{0}^{\infty}\frac{\cos x\minus{}e^{\minus{}zx}}{x}dx \equal{}\ln z$

Hence $ \int_{0}^{\infty}\frac{e^{\minus{}(a\minus{}ip)x}\minus{}e^{\minus{}(b\minus{}ip)x}}{x}dx \equal{}\ln (b\minus{}ip)\minus{}\ln(a\minus{}ip)$

$ \int_{0}^{\infty}\frac{(e^{\minus{}ax}\minus{}e^{\minus{}bx})\, e^{ipx}}{x}dx$ $ \equal{}\ln\sqrt{b^{2}\plus{}p^{2}}\plus{}i\arctan\left(\minus{}\frac{p}{b}\right)\minus{}\ln\sqrt{a^{2}\plus{}p^{2}}\minus{}i\arctan\left(\minus{}\frac{p}{a}\right)$

$ \equal{}\ln\sqrt{\frac{b^{2}\plus{}p^{2}}{a^{2}\plus{}p^{2}}}\plus{}i\left[\left(\minus{}\frac{\pi}{2}\plus{}\arctan\frac{b}{p}\right)\minus{}\left(\minus{}\frac{\pi}{2}\plus{}\arctan\frac{a}{p}\right)\right]$

Use $ e^{ipx}\equal{}\frac{1}{\sec px}\plus{}i\frac{1}{\csc px}$ and compare real and imaginary part.


function algebra polynomial calculus derivative

Hi, so I just started reading the generatingfunctionology book found at:
http://www.math.wisc.edu/~propp/491/gfologyLinked.pdf

And I have a question. On page 11, he finds a sequence explicitly under an assumption that $r_{+}^{N}\neq r_{-}^{N}$, and leaves the case where $r_{+}^{N}=r_{-}^{N}$ to the reader.  So I tried to find out how to do it in this case, trying to use the fact that $b^{2}-4ac=0$, but that didn't really get me anywhere...Could anyone tell me how to do it, or point me in the right direction?

Sorry if this post is kind of unclear, I wasn't really sure how to explain it.

Here's a thought: if a polynomial has a double root at $r$, what can you say about the value of its derivative at $r$?  (And note that if $D(x)$ is considered a known function, $D'(x)$ must be as well.)

(Also, as a timesaver for anyone else: page 11 of the document is page 16 of the .pdf file, and it's part of an example which begins on the previous page.)


Αυτή η συναρτησιακή μου άρεσε πολύ :)

Εστω συνάρτηση $ s: N\rightarrow N$, όπου $ N$ οι φυσικοί, τετοια ώστε $ s(s(n))\equal{}3n$ και η $ s$ είναι άυξουσα. Nα βρεθούν οι πιθανές τιμές τoυ $ s(1997)$.

Λοιπόν,

1. Η $ s$ είναι $ 1\minus{}1$: Έστω $ s(a)\equal{}s(b)$ τότε εφαρμόζοντας την $ s$ και στα δύο μέλη της ισότητας παίρνουμε $ 3a\equal{}3b \implies a\equal{}b$. Αυτό σε συνδυασμό με το ότι η $ s$ είναι αύξουσα σημαίνει ότι είναι γνησίως αύξουσα.

2.$ s(a)>a$: Έστω $ s(a)\leq a$ τότε αφού η $ s$ είναι γνησίως αύξουσα έχουμε ότι $ 3a\equal{}s(s(a))\leq s(a)\leq a$, άτοπο.

3.$ s(1)\equal{}2$: Έστω $ s(1)\equal{}a$. Τότε έχουμε $ a>1$ και εφαρμόζοντας την $ s$ έχουμε λόγω της δοσμένης ότι $ s(a)\equal{}3$ και άρα $ 3>a$, συνεπώς $ a\equal{}2$. Αυτό επίσης συνεπάγεται ότι $ s(2)\equal{}3$ και ότι $ s(3)\equal{}6$

4. $ s(3^n\plus{}k)\equal{}2*3^n\plus{}k$ και $ s(2*3^n\plus{}k)\equal{}3^{n\plus{}1}\plus{}3k$ για $ 0\leq k\leq 3^n$:
Επειδή έχουμε $ s(3^0)\equal{}2*3^0$, $ s(2*3^0)\equal{}3^{0\plus{}1}$ και $ s(3^1)\equal{}2*3^1$ μπορούμε να εφαρμόσουμε επαγωγή. Έστω λοιπόν ότι για κάποιο $ k$ έχουμε $ s(3^k)\equal{}2*3^k$, οπότε εφαρμόζοντας την $ s$ έχουμε $ s(2*3^k)\equal{}3^{k\plus{}1}$ και ξαναεφαρμόζοντας την, έχουμε $ s(3^{k\plus{}1})\equal{}2*3^{k\plus{}1}$. Όμως έχουμε ότι $ s(3^n\plus{}k)\equal{}2*3^n\plus{}k$ για $ 0\leq k\leq 3^n$  (επειδή $ s(3^k)\equal{}2*3^k$, $ s(2*3^k)\equal{}3^{k\plus{}1}$ και λόγω του ότι η $ s$ είναι γνησίως αύξουσα) και εφαρμόζοντας ξανά την $ s$ παίρνουμε $ s(2*3^n\plus{}k)\equal{}3^{n\plus{}1}\plus{}3k$ για $ 0\leq k\leq 3^n$.

5. Τέλος, υποθέσαμε ότι στους φυσικούς ο Mim δεν περιείχε το 0. Αν βάλουμε μέσα και το μηδέν, τότε επειδή $ s(1)\equal{}2$, $ s(0)\equal{}1$ ή $ s(0)\equal{}0$. Αλλά αν $ s(0)\equal{}1$, τότε $ s(1)\equal{}0$, άτοπο

Συνεπώς, επειδή $ 1997\equal{}2*3^6\plus{}539$ και $ 0<539<3^6$ έχουμε ότι $ s(1997)\equal{}3^7\plus{}3*539\equal{}3804$

Ηλίας

:10: Μπράβο ρε Ηλία!!! Πολύ σωστός  :omighty:


inequalities
calculus
derivative
function
inequalities proposed

This ineq is look like it:
$x,y,z>0, x+y+z=1$.Prove
$\sum\frac{1}{1+x^2}\leq \frac{27}{10}$

I think two problem are totally different .
 Nameless , your problem is easy . Just define m,n such that :
 $\frac{1}{1+x^2} \leq mx+n$

[quote="pvthuan"][quote="arqady"][quote="pvthuan"]Given three positive real numbers $x,y,z$ so that $x+y+z=xyz$, prove that
\[ \frac1{2+x^2}+\frac1{2+y^2}+\frac1{2+z^2}\leq\frac35.  \][/quote]
Try $x=\frac{220}{21},y=z=1.1$ ;)[/quote]
You mean that is a counterexample. How did you come up with it? Maybe, Maple?
Anyone has a proof for my problem?[/quote]
$1.1+1.1+\frac{220}{21}=1.1\cdot1.1\cdot\frac{220}{21},$ but
$\frac{1}{2+1.1^2}+\frac{1}{2+1.1^2}+\frac{1}{2+(\frac{220}{21})^2}>\frac{3}{5}.$ :)

I created this inequality from trigon ideas. I like the numbers $2,3$ and $5$ in it, but unfortunately it is not true. So I may pose it this way: Let $x,y,z$ be positve real numbers such that $x+y+z=xyz$, find the smallest possible number $\lambda$ such that
\[ \frac1{\lambda+x^2}+\frac1{\lambda+y^2}+\frac1{\lambda+z^2}\leq\frac3{\lambda+3}.  \]

IT is not so difficult.

first, f(x)=1/(2+x^2)

it is concave.

second, we can find the range of xyz by using AM-GM.

last, we can prove this.

If x+y+z=xyz then we can assume that x=tan(a), y=tan(b) and z=tan(c), and a, b e c are the angles of a triangle. And we have look to the second derivate of f(x)=1/{c+[tg(x)]^2} . I´m doing this now. (c is lâmbida)

[quote="Kimjiman"]
first, f(x)=1/(2+x^2)

it is concave.
[/quote]

dream seldom comes true.

I analyzed the second derivative of this function and I proved that it changes of sign. Thus, I can´t apllye Jensen´s Inequalitie.

[quote="pvthuan"]HUNGKHTN, Where are you?[/quote]

your inequailty is wrong. argady has given you a counterexample.

your inequailty is wrong. argady has given you a counterexample.[/quote]
Can you find $\lambda$?

Because x+y+z=xyz so we should take: x=tgA, y=tgB,z=tgC. Then, we obtain a triangle inequality and it's easy to prove that.


Fie $ABCD$ un patrulater inscriptibil si $I_{1},I_{2}$ respectiv incentrele triunghiurilor $AOB$, $COD$, unde $O \in AC \cap BD$. Sa se arate ca perpendicularele duse respectiv din $I_{1}, O, I_{2}$ pe $BD, AD, AC$ sunt concurente.

Fie $E%Error. "inBD" is a bad command.
$, $F%Error. "inAC" is a bad command.
$ astfel incat $I_{1}E \perp BD$ si $I_{2}F \perp AC$. Este suficient sa se demonstreze ca $AO \cdot OF=DO \cdot OE$.


quadratics
trigonometry
geometry
angle bisector
geometry unsolved

prove that cos 36 = [u]-1 + sqrt5[/u]
                                        4

[hide="A pure geometric approach"]Let $ ABC$ be triangle such that $ A\equal{}36,B\equal{}C\equal{}72$. Draw angle bisector $ BD$. Easy to see that $ BD\equal{}BC\equal{}a$, then $ AD\equal{}BD\equal{}a$. So $ CD\equal{}AC\minus{}AD\equal{}b\minus{}a\equal{}c\minus{}a$. Now, $ BDC$ and $ ABC$ is similar, $ \frac{c}{a}\equal{}\frac{a}{c\minus{}a}$. You'll get a quadratic equation. From here, you can get the value of $ \sin18$, and hence $ \cos 18$. So $ \cos 36$ can be computed easily.[/hide]

$ \sin{18} \equal{} \textonequarter{(\sqrt{5} \minus{} 1)}$ see: [url]http://www.mathlinks.ro/viewtopic.php?t=246587[/url]
$ \cos{36}\equal{}1\minus{}2\sin^2{18}$
 :rotfl:  :rotfl:  :rotfl:

hello, see here
http://www.cut-the-knot.org/pythagoras/cos36.shtml
Sonnhard.


function
algebra
polynomial
superior algebra
superior algebra unsolved

Consider the power sums $ p_1\equal{}x_1\plus{}...\plus{}x_n, p_2\equal{}x_1^2\plus{}...\plus{}x^2_n, ..., p_n\equal{}x_1^n\plus{}...\plus{}x_n^n.$
These are polynomials in the elementary symmetric functions which Artin denotes by $ a_1,...,a_n$ they must therefore be expressible as polynomials in $ a_1,...,a_n.$ What is the formula for arbitrary n?


calculus
integration
function
algebra
domain
complex numbers
complex analysis

Can someone tell me why $\int_{0}^{1}t^{p+z-1}dt=\frac{1}{z+p}$ ? ($Re(z)>-p, p=0,1,2...$)
I know $|t^{p+z-1}|$ is integrable when $Re(z)>-p$.
But why can we use antiderivative to evaluate integral here?

Is it from the following corrolary of Cauchy integral Thm?i.e: If $f(z)$ is analytic  in a simply
connected domain and  $z_{0}, z$ are two points in this domain. And if $DF=f$, then
$\int_{z_{0}}^{z}f dz=F(z)-F(z_{0})$ (integral is along any path connecting $z_{0}$ and $z$ in the domain).
But what if $-1<p+z-1<0$, then $f(t)=t^{p+z-1}$ is not continuious at $0$. So $0$ is not in the domain.

When there is a singularity at $0$, you can replace $0$ by $\epsilon$, apply the theorem, and let $\epsilon\to 0$.

Without Cauchy theorem: the identity $F'(z)=f(z)$ can be written as $\frac12\left(\frac{\partial F}{\partial x}-i\frac{\partial F}{\partial y}\right)=f$. The Cauchy-Riemann equations for $F$ give $\frac{\partial F}{\partial x}+i\frac{\partial F}{\partial y}=0$, hence $\frac{\partial F}{\partial x}=f$. Separate the real and imaginary parts of $F$ and $f$ and apply FTC.

[quote="spx2"]FTC beeing ?[/quote]
Fundamental Theorem of Calculus


I'm stuck on a question (WOCOMAL, March 30, 2005, round 4 #2).  Although technically it's a two-point question, my recollection is that students found it much harder than the three point question, and the league statistics bear that out (quite a few students earned 4/6 on the round; and because the answer is an integer, those with 6's could have been lucky guessers).  Moreover, the question is poorly phrased.

So let's change the question to...

[quote]
Find the real solution of \[ x\equal{}\sqrt{x\minus{}\frac{1}{x}}\plus{}\sqrt{1\minus{}\frac{1}{x}}\] in simplified form.
[/quote]

Ideas?

Rick Yanco
Worcester Academy

P.S. Original question at [url]http://www.matharchive.org/wocomal/varsity/2004-05/20050330_wocomal_v.pdf[/url]

[hide="Solution"]Start by squaring both sides.
\[ x^2 \equal{} x \minus{} \frac1x \plus{} 2\sqrt{x \minus{} 1 \minus{} \frac1x \plus{} \frac1{x^2}} \plus{} 1 \minus{} \frac1x\]
\[ x^2 \minus{} x \minus{} 1 \plus{} \frac2x \equal{} 2\sqrt{x \minus{} 1 \minus{} \frac1x \plus{} \frac1{x^2}}\]
\[ x^3 \minus{} x^2 \minus{} x \plus{} 2 \equal{} 2\sqrt{x^3 \minus{} x^2 \minus{} x \plus{} 1}\]
\[ (x\minus{}1)^2(x\plus{}1) \plus{} 1 \equal{} 2\sqrt{(x\minus{}1)^2(x\plus{}1)}\]
Substitute $ a \equal{} (x\minus{}1)^2(x\plus{}1)$.
\[ a \plus{} 1 \equal{} 2\sqrt{a}\]
\[ a^2 \plus{} 2a \plus{} 1 \equal{} 4a\]
\[ (a\minus{}1)^2 \equal{} 0\]
So $ a \equal{} 1$. Thus
\[ (x\minus{}1)^2(x\plus{}1) \equal{} 1\]
\[ x^3 \minus{} x^2 \minus{} x \plus{} 1 \equal{} 1\]
\[ x^2 \minus{} x \minus{} 1 \equal{} 0\]
(note 0 doesn't work in the original equation, so that root is ignored)
Then do the quadratic formula and take the positive root since the square roots in the original equation don't allow negative ones. This gets you $ \frac{1\plus{}\sqrt{5}}{2} \equal{} \phi$, the golden ratio. Using the formula $ \phi^k \plus{} \phi^{k\plus{}1} \equal{} \phi^{k\plus{}2}$, you can see it works in the original equation.[/hide]

Thanks.  Still not something I could do in three minutes, but I see how to do it. :-)

Rick


Let $ABC$ be triangle and $T$ his center. Let  $P,Q$ be points on sides $BC,AC$. Prove that $T \in PQ\leftrightarrow\frac{BP}{CP}+\frac{AQ}{CQ}=1$.

This theorem will help:
$ABC$ be a triangle. $M$ is the mid-point of $BC$. $R \in AM$. $P \in AB$ and $Q \in AC$. If $R \in PQ$, then:
\[ \frac{1}{2}(\frac{AC}{AQ}+\frac{AB}{AP})=\frac{AM}{AR} \]

1)	Apply Menelaos (${R}=AB\cap PQ$)
2)	Apply the ‘transversal theorem’:  ${M}=AB\cap CT$, then $P, Q$ and $T$ collinear iff $\frac{AQ}{CQ} \cdot MB+\frac{PB}{PC}\cdot AM=\frac{TM}{AT}\cdot AB$ , straight forward!

Best regards,
sunken rock


calculus
derivative
real analysis
real analysis unsolved

let f be differntiable fun.show derivative of f has intermediate value property.


geometry
circumcircle
Euler
ratio
power of a point
radical axis
geometry unsolved

Let $ ABC$ be an acute triangle and the points $ A_1,B_1,C_1$ are the feets of the altitudes from $ A,B,C$ respectively.
A circle passes through $ A_1,B_1$ and touches the smaller arc $ AB$ of the circumcenter of $ \triangle ABC$ in point $ C_2$.
Points $ A_2,B_2$ are defined analogously.
Prove that $ A_1A_2,B_1B_2,C_1C_2$ have a common point,which lies on the Euler line of $ \triangle ABC$

Let $ \mathcal{T}_c$ be tangent of the circumcircle $ (O)$ at $ C_2.$ $ \mathcal{T}_c$ is radical axis of $ (K_c) \equiv \odot(A_1B_1C_2)$ and $ (O),$ $ A_1B_1$ is radical axis of $ (K_c)$ and $ \odot(A_1B_1AB),$  $ AB$ is radical axis of $ (O)$ and $ \odot(A_1B_1AB).$ Therefore $ \mathcal{T}_c,$ $ AB$ and $ A_1B_1$ concur at the radical center $ P_c$ of $ (O),$ $ (K_c),$ $\odot(A_1B_1AB).$ Since cross ratio $ (P_c,C_1,A,B)$ is harmonic, it follows that $ C_2C_1$ is the polar of $ P_c$ with respect to $ (O).$ Mutatis mutandis, $ B_2B_1$ and $ A_2A_1$ are the polars of $ P_b$ and $ P_a.$ Since $ P_a,P_b,P_c$ are collinear on the orthic axis $\tau$ of $ \triangle ABC,$ then $ A_1A_2,$ $B_1B_2,$ $C_1C_2$ concur at the pole of $\tau$ WRT $ (O),$ which lies on the perpendicular to $\tau$ passing through $ O,$ i.e. the Euler line of $ \triangle ABC$ and the proof is completed.


[quote="bpms"]http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/2006-2007,RegAMC8.pdf
Here it says that you can get an AMC 8 "Math Club" Package. Can anyone tell me what it is?[/quote]
Um, I kinda need to know kinda badly. Can someone help me with this?

Check out this link:

http://www.math.unl.edu/~amc/registration/start8.php

in the future, its probably easier to direct this kind of question directly at the AMC, as they probably know more about their products, especially after there is no reply from the forum


With one skein of wool yarn, Sharon can knit three inches of a scarf that is nine inches wide. What is the least whole number of skeins with which she can knit a scarf that is six inches wide and two feet long?

[hide]One skein of yarn can make 27 square inches of scarf. Sharon wants to make a 6 in x 24 in scarf, which is 144 in^2. 144/27 is 5 with some remainder, so round up to get 6.[/hide]


Let [b]ABC[/b] be a triangle and [b]D,E,F[/b] the points where the [i]incircle[/i] touches the sides [b]BC,CA,AB,[/b] respectively.  [b]Show that the lines AM,BN,CP intersect [i]if and only if[/i]the lines DM,EN,FP intersect.[/b]

Davron

I guess $M$, $N$, $P$ are points on $EF$, $FD$, $DE$.
And isn't the fact that $AD$, $BE$, $CF$ are concurrent enough? Don't think you really need the incircle stuff.

If the points M, N, P are on the lines EF, FD, DE, then your statement follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=6579 , using the fact that the lines AD, BE, CF are concurrent (in fact, they concur at the Gergonne point of triangle ABC).

I can also imagine that the points M, N, P are supposed to lie on the incircle of triangle ABC. Then, the problem is not correct, but the following is true: The lines AM, BN, CP concur at one point if and only if either the lines DM, EN, FP concur at one point or the points $DM\cap EF$, $EN\cap FD$, $FP\cap DE$ are collinear. This is Theorem 2 in my note "Variations of the Steinbart Theorem" on http://de.geocities.com/darij_grinberg .

If the points M, N, P can be arbitrary points in the plane, then your statement is just wrong.

  Darij

I assume $M, N, P$ are on $EF, FD, DE$ respectively, in the configuration of the original problem ($D, E, F$ are the contact points of the incircle with the sides).

$DM, EN, FP$ concur iff $\frac{EM}{MF}\cdot \frac{FN}{ND}\cdot \frac{DP}{PE}= 1$ by Ceva.

By the Theorem of Sines $\frac{EM}{\sin \angle CAM}= \frac{AM}{\sin \angle MEA}$ and $\frac{MF}{\sin \angle BAM}= \frac{AM}{\sin \angle MFA}$ and so on.

Since $\angle MEA = \angle MFA$ and so on, they concur iff $\frac{\sin \angle CAM}{\sin \angle BAM}\cdot \frac{\sin \angle ABN}{\sin \angle CBN}\cdot \frac{\sin \angle BCP}{\sin \angle ACP}= 1$ hence, by the Trig Ceva, iff $AM, BN, CP$ concur.


[hide=" click to reveal hidden content"]

Not sure about this but:

For the first six flips, we can have $ 6$ tails, $ 5$ tails and $ 1$ head, $ 4$ tails and $ 2$ heads (but not $ 2$ consecutive heads) or $ 3$ tails and $ 3$ heads (alternating). This has a probability of $ (1\plus{}5\plus{}10\plus{}2)(\frac{1}{2})^{6} \equal{} \frac{9}{32}$.

Then, multiplying by $ (\frac{1}{2})^{2} \equal{} \frac{1}{4}$ for the next two being heads, we get $ \frac{9}{128}$.

[/hide]

[quote="mr. math"]A fair coin is repeatedly flipped. What is the probability that the first time heads is flipped twice in a row is on the seventh and eighth flips?[/quote]

[hide]
$ T,T,T,T,T,T,H,H$

--------------------------------
$ H,T,T,T,T,T,H,H$

$ T,H,T,T,T,T,H,H$

$ T,T,H,T,T,T,H,H$

$ T,T,T,H,T,T,H,H$

$ T,T,T,T,H,T,H,H$

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

$ H,T,H,T,T,T,H,H$

$ H,T,T,H,T,T,H,H$

$ H,T,T,T,H,T,H,H$

$ T,H,T,H,T,T,H,H$

$ T,H,T,T,H,T,H,H$

$ T,T,H,T,H,T,H,H$
---------------------------------

$ H,T,H,T,H,T,H,H$

$ \frac {13}{256}$[/hide]

[hide="AndrewTom"]Actually, AndrewTom, your count is off by one. For the case where you have 3 heads and 3 tails, there is only 1 possibility. Because there cannot be a heads on the sixth position. So it has to be HTHTHTHH. Because otherwise the first consecutive heads is going to be on the sixth and seventh. So, it should have been 17. And the answer, 17/256, I think.
[/hide]

[hide="EDIT"]EDIT: actually Arrangeyourtan has it right, because when choosing the two heads in the first six slots we have to take into consideration the fact that there cannot be two consecutive heads...[/hide]

[hide=" limac"]

Thanks limac.

AndrewTom.

[/hide]

Ok so #6 has to be T, 7/8 are both H.

Now for the first 5, we just sort by what it could be:

5 T: 1 way

4 T 1 H: 5

3T 2H: There are 10 ways, however, if we treat HH as a unit, we see there are 4 ways so that 2 heads are in a row. Thus, 6.

2T 3H: 1 way (HTHTH) so we then have 17 total, and 8 flips so 17/256.


How to not make careless mistakes on AMC/AIME? I (pretty sure many other people) made many mistakes on AMC/AIME cause us to do horribly on them. Does anyone have any suggestions(beside checking since we don't have much time)?

I'm waiting for Tokenadult to post the link to the article on this website about just that question.  It's pretty comprehensive -- poke around and look for it if he doesn't get here first :)

no need to do so tokenadult, here's the link: [url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php[/url]

On the AMC I spent more time checking my answers than actually solving.  That might help.

i usually look at my watch halfway through, and think: at this time, i will stop what i'm doing and check over all the other problems. i've caughts tons of errors in the process.. and then i work on more later.

I know checking is a great way, but if you spend so much time on checking, won't you not solve enough problems and getting actually a lower score than you would expect??

I did get a lower score than I should've, but I don't mind at all.  For me, all I need is the aime floor and I'm happy - index doesn't really apply for 10th and below, remember?

time's not really a problem on the amc8 for me though.  i just go really slow and after im done ive got time to check through twice.  working at a snail's pace seems to work for me-- but it might not work for you.  heh i got a 3 on the aime. by all rights i shouldve gotten a 5 but i made 2 dumb mistakes and some more

one thing that's mentioned but a lot of people (including myself) still mess up on is misreading the question. for a great example of this, see number 5 on the 2004 AIME-2. practice reading questions correctly, it really got me this year.

[quote="captcha000"]According to one of the Georgia ARML coaches:

"The best way to not make careless is to make them. Just make them the week before and not on the test."[/quote]

LOL! Yah, I've heard that before...

Um, one strategy to take (though probably not by itself) is to look at the problem beforehand and try to decide which answer choices make the most sense. Then, if you work out the problem and get something completely different, check back over your work again (with a different method, optimally.) I've used a variation of this on other multiple-choice tests...

Of course, if you make a REALLY stupid mistake and the answer you get isn't one of the choices... :D

For a few years when I did contests I almost never made a careless error.  At some point I started making them again (I did just as well or better on most of the AMC/AIME/ARML-like competitions early in high school than I did later on).

Thinking back I think I have a good understanding of what helped me avoid a lot of careless errors.  The more I practiced solving problems in my head the more likely I was to catch an error [i]in the making[/i].

I suppose this is somewhat similar to the way chess players play entire games of chess on a board mentally and can therefore recite entire games by memory -- when you're so comfortable with a process that you can follow the entire process mentally then you're far less likely to stumble during that process.

I would suggest that students practice solving problems without pencil or paper.  Write a problem down and walk around school with it until you've solved it in your head.  See if you can solve the problems while playing basketball or something that engages other areas of the mind.  At some point you might just develop a greater control over the thinking process.

LOL!  This is exactly what I do!  I look at a problem on AoPS and then pretty soon I can be found pacing around the house everywhere looking at the ground.  Surprised no one has reported me to an insane asylum yet :-).


[quote="nutz_for2.718281828"]Prove that $\sin{(\cos{x})}=\cos{(\cos{x}-90)}$[/quote] If the second argument is $\cos(x)-90^\circ$, then set $y=\cos(x)$. This gives $\sin(y)=\cos(y-90^\circ)$, which is true.

[quote="Treething"]but sin(x) = cos(90-x) :o[/quote]
Yes, but $\cos{a}=\cos{(-a)}$, so $\sin{y}=\cos{(90^\circ-y)}=\cos{(y-90^\circ)}$.


AMC
AMC 12
AIME
MATHCOUNTS
USA(J)MO
USAMO
AMC 8

The AMC forum is the forum in which the most of the questions about what should a student do with that score on AMC12, AIME etc.

The work of some people leaded by [b]orl[/b], tried to introduce the rating system to the AoPS/ML. As only a few look at the Wiki, I'll post some shortened ratings of problems here, so the students can know where they are, and which competitions to try.

Here is the scale, everything is approximate:

1

MOEMS, easy MATHCOUNTS problems, first half of AMC 8, first 5 on AMC 10

2

Second half of AMC 8, 5-15 on AMC 10, 1-5 on AMC 12

3

Last problems of AMC 10, middle AMC 12, easiest AIME problems (1,2,3)

4

Last problems of AMC 12, middle AIME problems, simplest proof problems (#1 on JBMO)

5

10-12 on AIME, a little bit more challenging proof problems (#2,3 JBMO)

6

13-15 on AIME, hard non-proof problems, introductory olympiad proofs (#4 JBMO, #1/4 USAMO, IMO, and #1 BMO)

7

#1/4 which require a little bit more work, introductory #2/5, and also BMO #2,3, USA TST #1/4/7

8

Harder #2,5, BMO #4, USA TST 2/5/8, easier #3,6 on USAMO and IMO

9

#3,6 on IMO, USA TST #3/6/9

10

Extremely hard problems for high school students, mostly suitable for undergraduate students.

Difficulties 1-3 can be considered as beginner level, 4-6 as intermediate level, 7-10 as olympiad level.

The link to the Wiki Page: http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Competition_ratings

Could someone also post popular books and where they rate on the scale?

Books are hard to rate, because almost all have a variety of problems, from easy to hard. If, for example, 3-10 for a book could help you, I don't know...

And you have a lot of topics about which books to use...

Maybe its just me, but I think easy AIME problems (1-3) are a lot harder than "hard" AMC 10 problems (16+) or "medium" AMC 12 problems (6-15)

All rating are approximate. And, last AMC 10= 21+, middle AMC 12=6 - 20.

Compare:
http://www.mathlinks.ro/viewtopic.php?p=1439392#1439392
http://www.mathlinks.ro/viewtopic.php?p=1403057#1403057

I think the geometric series was easier, don't you?

I am trying again to say, everything is approximate.

For more competitions, look at Wiki.


Sa se determine doua progresii, una aritmetica si cealalta geometrica, fiecare formata din cate patru termeni , astfel incat adunand termenii de acelasi rang din cele doua progresii , sa obtinem drept suma, respectiv, numerele 18,18,26,58 .

am luat-o dintr-o carte,dar nu va zic din care ca nu ar mai avea farmec daca v-as zice

haideti!!!!!nu o incearca nimeni??hai ca nu-i grea!!!!!

Parerea mea e ca e prea simpla...Trebuie doar rezolvat sistemul
$a_{1}+b_{1}=18$
$a_{1}+r_{1}+b_{1}r_{2}=18$
$a_{1}+2r_{1}+b_{1}r_{2}^{2}=26$
$a_{1}+3r_{1}+b_{1}r_{2}^{3}=58$
Unde $a_{1},\ b_{1}$ sunt primele numere din fiecare progresie, iar $r_{1},\ r_{2}$ ratiile celor doua progresii.
Deci, e doar calcul...

[hide="solutie"]$a,r$ - primul termen respectiv ratia progresiei aritmetice
$x,q$ - primul termen respectiv ratia progresiei geometrice
$a+x=18$. (1)
$a+r+xq=18$. (2)
$a+2r+xq^{2}=26$. (3)
$a+3r+xq^{3}=58$. (4)
scadem (1) din (2)
$r+x(q-1)=0$.
scadem (2) din (3)
$r+xq(q-1)=8$, deci $xq(q-1)-x(q-1)=8$, adica $xq(q-1)^{2}=8$.
scadem (3) din (4) si facem la fel ca mai sus
$32=xq(q-1)^{2}(q+1)$.
deci $q=3$ si de aici $x=2$, $a=16$, $r=-4$.
obtinem
$16,12,8,4$ si $2,6,18,54$.[/hide]


floor function
algebra
polynomial
number theory proposed
number theory

$ m$ is a positive integer.Prove that   $ \left. {{2^{m \plus{} 1}}} \right|\left\lfloor {{{\left( {\sqrt 3  \plus{} 1} \right)}^{2m \plus{} 1}}} \right\rfloor$

Since $ \sqrt {3} \minus{} 1 < 1$, the floor is equal to $ (1 \plus{} \sqrt {3})^{2m \plus{} 1} \plus{} (1 \minus{} \sqrt {3})^{2m \plus{} 1}$. This gives the odd terms of the second order linear recursion with characteristic polynomial $ x^2 \minus{} (1 \plus{} \sqrt {3} \plus{} 1 \minus{} \sqrt {3})x \plus{} (1 \plus{} \sqrt {3})(1 \minus{} \sqrt {3}) \equal{} x^2 \minus{} 2x \minus{} 2$ so that the recurrence relation is $ a_{n \plus{} 1} \equal{} 2a_n \plus{} 2a_{n \minus{} 1}$, with $ a_1 \equal{} 2,a_2 \equal{} 8$. The result then easily follows by induction.

[quote="Rust"]It is easy problem:
\[ |\lfloor {{{( {\sqrt 3 + 1} )}^{2m + 1}}} \rfloor = (1 + \sqrt 3)^{2m + 1} + (1 - \sqrt 3)^{2m + 1} = 2^{m + 1}Int[(1 + \sqrt 3) (2 + \sqrt 3)^m].\]
Were $ Int(a + b\sqrt 3) = a.$[/quote]
Thank you very much, Rust. Nice Proof!!


Ok now i will tell you all about something which i was pondering  for 4 years and i still try to find something out about this and put it in mathematical sense.
i dont know if i call it a conjecture or....

Conjecture:
Whenever I see something New and it amazez me and i think about it for like 5 sec. i see that thing again soon, you might say well that is obvious you see things everyday, i will explain
it is so unbelievable,shocking and interesting.
[color=green]
examples of my experience:[/color]
1:oneday i saw the number 77 in a maths book, i said 77 it looks nice and thought about it for a second, then tommorow i was downstairs my brother  gave me an equation to solve the answer was surprisingly 777.

2:another day i was reading a book, i saw the word, hamlet, this word sounded kinda new so i looked at it again,then i went outside that day i think i saw the word helmet, not sure i forgot the word but it was the word hamlet exacept one letter was different in someones T-shirt.

3:
4:
5:
and many many more i have seen but i am sleepy now cant remember them all.

I said there must be reason , i asked my friends if this is true, they all said it is true, but it is because when you see something new and you pay more attention to it, you see it around soon because u payed attention this time,other times it was there but u didnt realise.the second time when u see it, it is different by one or two or more things.

what they said is not true, because i saw many many more of this that was opposite to what they said.

i wonder if this is true,  which i think it is true and this has a mathematical relation.
i will write the maths for it soon.

You are insane. I'm not even sure if this is supposed to be an elaborate joke are not.

I know what you mean binomial4eva. One time I learned the word "vehemently" that's seldom used in a conversation and the next day I heard the word on television. Also, after reading "The Great Gatzby", I heard about it a lot (outside of school). However, had I not read the book or learned that word, their reference wouldn't have meant anything to me and thus I wouldn't have noticed them in the same way.

I'm not sure if there's a mathematical relationship for this, because this deals with neurology (in particular, brain patterns and memory).

I'm not sure this is the right place to post this topic, though, but thanks for sharing  :lol:

There is a place for personal coincidences with so-called "mathematical explanations." Its called your blog.

http://en.wikipedia.org/wiki/List_of_memory_biases
http://en.wikipedia.org/wiki/List_of_cognitive_biases

See, in particular: http://en.wikipedia.org/wiki/Confirmation_bias

As far as mathematics is relevant to your ramble, all it will tell you is that coincidences, mathematically speaking, have to happen.  The rest of what you seem to be talking about is cognitive.

Edit:  See also http://en.wikipedia.org/wiki/List_of_fallacies .  A particular example relating to coincidences: http://skepdic.com/lawofnumbers.html

Yea, you have a few examples of that happening...And you've been alive for how long?  Statistically, it's supposed to happen every once and a while...

[quote="binomial_4eva"]2:another day i was reading a book, i saw the word, hamlet, this word sounded kinda new so i looked at it again,then i went outside that day i think i saw the word helmet, not sure i forgot the word but it was the word hamlet exacept one letter was different in someones T-shirt.[/quote]

Wow... definitely didn't see that one coming.

you know wahts even weirder?

like, i'm driving, and i see the word STOP on a red sign or whatever

and then I see it AGAIN  almost immediatley afterwards

AND AGAIN AND AGAIN

its ridiculous I tell you


rotation
combinatorics unsolved
combinatorics

in how many ways can put $n$ number $a_1,a_2,...,a_n \in {0,1}$ around a circle.

Just some observations, partial results and the case where $n$ is prime:

Let $f(n)$ be the number of ways. Obviously $\frac{2^n}{n} \leq f(n) \leq 2^n$. Now, let $a_i=a_{n+i}$ for all $i$. We say that the minimal period of $a_1, ..., a_n$ is $r$ if $a_i=a_{i+r}$ for all $i$, and there's no $r_2<r$ such that $a_i=a_{i+r_2}$ for all $i$.

We'll show that we always have $r \mid n$. Indeed, let $(r, n)=d, n=dp, r=dq, (p, q)=1$. Suppose the sequence has minimal period of length $r$. Consider two different numbers $i, j$ in $1, 2, ..., d$. Clearly, $a_i$ and $a_j$ aren't necessarily equal, because if that were true then $i=j+mr$ for some $m$, which is impossible because then $r \mid j-i \Rightarrow d \mid j-i$. Now suppose $i$ is in $d+1, ..., n$. Write $i=dm+s$. Then since the sequence has period $r$, $a_i=a_{dm+s}=a_{d(m+q)+s}=...=a_{d(m+(n-1)q)+s}$. Clearly if two of those indices are congruent mod $n$ iff $n \mid d(a-b)q$, which happens iff $p \mid a-b$. Then there are $p$ noncongruent indices. Since they are congruent mod $d$, there must be one in each of the intervals $(1, ..., d), (d+1, ..., 2d),$ etc. So any $a_i$ with $i>d$ has some $a_j$ with $j \leq d$ such that it is forced $a_i=a_j$. Then it's easy to see the period must be $d$. But $d \mid n$, so the claim is true.

Now it's easy to see what happens when $n$ is prime. The period of a sequence is either $1$ or $n$, but if it is $1$ the sequence is $1, 1, ..., 1$ or $0, 0, ..., 0$. Out of the total $2^n$ sequences, those two are discarded. Since the rest have period $n$, we can't rotate one of them around the circle $k$ places to get the same sequence unless $k \mid n$, so among the $2^n-2$ left each one is counted $n$ times. Then for $n$ prime, $f(n)= \frac{2^n+2n-2}{n}$.

I'm not sure this method can be generalized for $n$ composite, as it looks rather tedious, but maybe it's more obvious than it seems to me.

Let $f(n)$ be the number of ways, and $M(d)$ be the number of sequences $a_1,a_2,...,a_d$ $\in$ $\{0,1\}$ then we have:
$f(n)=\sum_{d|n}M(d)$ ,

but  $\sum_{d|n}dM(d)=2^n$ ,because $\sum_{d|n}dM(d)$ is the number of all sequences $a_i$ $\in$ $\{0,1\}$ ,  $i=1,2,...,n$
then from mobius refletion  we have:

$nM(n)=\sum_{d|n}\mu(d)^{2^{\frac{n}{d}}}$
therefore:

$f(n)=\sum_{d|n}M(d)=\sum_{d|n}\frac 1{d} \sum_{c|d}\mu (\frac{d}{c})2^c
=\sum_{c|n}2^c\sum_{d|\frac{n}{c}} \frac {\mu(d)}{d}=\frac 1{n}\sum_{c|n} \varphi (\frac{n}{c})2^c$

-Amir


vector
algorithm
calculus
real analysis
real analysis unsolved

I didn't really know where to post this, but this forum seemed the most appropriate.

A cool, instructive math problem was posted on a forum I visit. A friend of the poster invented the riddle and told him about it. The link to the thread I found it posted in is at the bottom, but the entire question is reproduced so no need to redirect your web browser.

Now enough for crediting the sources! Here's the problem:
[i]
"Requires some math knowledge.

There's a boat, that has an initial starting point on the plane and a constant velocity vector.
The starting point and the velocity vector are integers; i.e., the starting point is (a,b) and the boat's position is advanced by (c,d) every second, and a,b,c,d are integers.
You have no knowledge of the boat's position or velocity, but you are allowed, once per second, to bomb a point of your choice in the plane. (After each bombing attempt you know only whether you failed or succeeded.)

Give an algorithm to find and bomb the boat."[/i]

link: http://forum.valhallalegends.com/index.php?topic=17079.0.

It's not calculus- this is entirely a discrete problem, despite the infinite sets. I'm not sure where it belongs, but this forum isn't it.

A tangentially related problem from a Russian competition in the 1990s (paraphrased from memory, and hidden for length):

[hide="Problem"]God and the Devil play a game as follows:
A robot can move around a maze by following its program: it attempts to move one square in the direction indicated. If there's a wall in the way, it fails to move. The program consists simply of a finite sequence of directions: up, down, left, or right. The maze is a $ 100\times 100$ grid, with some paths open and some blocked by walls. There is an exit somewhere in one of the outer walls, and it can be reached from a starting point anywhere inside the maze.
God writes a program. The Devil examines the program, and then builds a maze. God wins if the robot escapes the maze, and the Devil wins if it does not. Who wins?

The wording (God and the Devil) is the reason I remember it, of course. The Russian problems from that time have a lot of good stuff.[/hide]


Let ABC be a triangle and take any D point on [AC] line segment.Let lengths of |AB|=x, |BD|=y, |AC|=x+y  and angle <ABD=18,  <CBD=9  then find angle <ACB.Syntetic and analitical solution desired.Thank you.

Let $\triangle ABD$ be an isosceles triangle with the base BD = y and the shoulder sides AB = AD = x. Denote $\angle BDA = \angle DBA = \phi$. Let $\triangle DBC$ be another isosceles triangle with shoulder sides DB = DC = y and the angles $\angle DBC = \angle DCB = \frac \phi 2$. Then the angle $\angle CDB = 180^\circ - 2\ \cdot \frac \phi 2 = 180^\circ - \phi$. Consequently, the angle $\angle CDA = \angle CDB + \angle DBA = 180^\circ - \phi + \phi = 180^\circ$ is straight, so that the 2 isosceles triangles $\triangle ABD, \triangle DBC$ form a triangle $\triangle ABC$ with the sides AB = x, AC = x + y and the cevian BD = y, as required. If $\phi = 18^\circ$, then $\angle ACB \equiv \angle DCB = \frac \phi 2 = 9^\circ$.


If $a, b, c$ are the sides and $R$ the circumradius of a triangle $ABC$, prove
that:

$\frac{a^{2}}{b+c-a}+\frac{b^{2}}{c+a-b}+\frac{c^{2}}{a+b-c}\geq 3R\sqrt{3}$

hello, setting $ R=\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$ with $ s=\frac{a+b+c}{2}$ we have to prove that
$ \frac{a^2}{b+c-a}+\frac{b^2}{c+a-b}+\frac{c^2}{a+b-c}\geq\frac{3\sqrt{3}abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$ after some algebraic computaton we get
$ {(a+b+c)^3(a^3+b^3+c^3-ab^2-cb^2-bc^2-a^2b-c^2a-a^2c+4abc)^2\geq27(abc)^2(-a+b+c)(c+a-b)(a+b-c)}$ since
$ (a+b+c)^3\geq27abc$ it remains to show that
$ %Error. "dispplaymath" is a bad command.
{(a^3+b^3+c^3-ab^2-cb^2-bc^2-a^2b-c^2a-a^2c+4abc)^2 \geq abc(-a+b+c)(c+a-b)(a+b-c)}$ this is equivalent with
$ {(b^3-b^2a-a^2b+a^3-b^2c+6abc-a^2c-bc^2-c^2a+c^3)(b^2-b^2a-a^2b+a^3-b^2c+3abc-a^2c-bc^2-c^2a+c^3)\geq0}$
setting $ a=y+z,b=x+z,c=x+y$ in the inequality above we get
$ 4(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2)(-6xyz+x^2y+xy^2+x^2z+xz^2+y^2z+yz^2)\geq0$
the first factor is positive because of the definition and the second one because of the AM-GM inequality.
Sonnhard.


linear algebra
combinatorics proposed
combinatorics

Let $X=\{1,2,\ldots,n\}$, and let $S$ be any subset of $\mathcal P(X)$, the power set of $X$. We denote by $S'$ the subset of $\mathcal P(X)$ consisting of those subsets of $X$ which are contained in an odd number of elements of $S$ (we call $S'$ the [i]dual[/i] of $S$). Prove that $(S')'=S$ (i.e. the double dual of a subset $S$ of $\mathcal P(X)$ coincides with $S$).

Isn't it nice? :)

Linear algebra strikes back  :)

Pierre.

[quote="grobber"]$\mathcal P(X)$, the power set of $X$[/quote]

What do you mean by the power set of $X$? :oops:

[quote="pbornsztein"]Linear algebra strikes back  :)

Pierre.[/quote]

Not necessarily :). I didn't use linear algebra, for example.

[quote="Arne"]What do you mean by the power set of $X$? :oops:[/quote]

It's the set of all the subsets of $X$. Here's another good question: what is this Google I keep hearing about, and how do you use it? :)

[url]http://www.google.ro/search?hl=ro&q=power+set&meta=[/url]

one way of seeing this is to get a 'computable' definition of the dual of a set.
we formally write $S: =\sum_{A\in \mathcal{P}} 1_{A\in S}$; then for a set $A$, we have $A\in S'\Leftrightarrow n_S(A)=1$ where $n_S(A)=\sum_{B\in S} 1_{A\subseteq B}$(this sum is computed over $\mathbb{F}_2$).
 Then $A\in S''\Leftrightarrow n_{S'}(B)=1$.
Now,$n_{S'}(B)=\sum_{A\in\mathcal{P}} 1_{B\subseteq A}\cdot 1_{A\in S'}$. But $A\in S'\Leftrightarrow n_S(A)=1$. So we have
$n_{S'}(B)=\sum_{A\in\mathcal{P}} 1_{B\subseteq A}\cdot n_S(A)=\sum_{A\in\mathcal{P},C\in S} 1_{B\subseteq A\subseteq C}=\sum_{C\in S}\sum_{A\in\mathcal{P}}1_{B\subseteq A\subseteq C}$.
now the last sum is e even unless $B=A=C$, which gives us what we want.


Cauchy it!

The following is also true
\[\frac a{a+2b+c}+\frac b{b+2c+d}+\frac c{c+2d+a}+\frac d{d+2a+b}\geq1.\]

You're right...That's how I've did it(One of the solutions)

[quote="luke999"]Consider a,b,c pozitove real numbers.Proove that
$\sum\frac{a}{b+2c+d}\ge 1$[/quote]

        $\sum\frac{a}{b+2c+d}=\sum\frac{a^{2}}{ab+2ac+ad}\ge\frac{[\sum a]^{2}}{4[ab+ac+ad+bc+bd+cd]}$                                                    which is trivially true.

you can use convexite.
WLOG  assume $a+b+c+d=1$
then we have $\sum\frac{a}{b+2c+d}\geq{\frac{1}{4(ab+ca+ad+bc+bd+cd)}\geq{\frac{1}{(a+b+c+d)^{2}}}=1}$

[quote="hitek"]Cauchy it!

The following is also true
\[\frac a{a+2b+c}+\frac b{b+2c+d}+\frac c{c+2d+a}+\frac d{d+2a+b}\geq1. \]
[/quote]
It's nice, hitek! :lol:
But $\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3d}+\frac{d}{d+3a}\geq1$ true too. :wink:

By the way, $\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3d}+\frac{d}{d+3a}<3.$ :wink:


When a biased coin is tossed the probability of a head is p
Two players A and B alternately toss a coin until one of the sequences HHH, HTH occurs.
A wins if HHH occurs first .B win s if HTH occurs first.
For what values of p is the game fair that is  such that Probability A wins = probability B wins

Lets forget p = 0, 1 - answer is obvious.

Obv. one of the two people win.  So we just want chance of A winning to be 1/2.  Say A wins a dollar by winning the game.  Then, we want the expected value of the game to be 1/2.

The four states W,X,Y,Z correspond to the last two flips being HH HT TH TT.

W = p + (1-p)X
X = 0 + (1-p)Z
Y = pW + (1-p)X
Z = pY + (1-p)Z

Then, $Z=Y$ from eqn 4, $X = (1-p)Y$ from eqn 2, $Y = pW + (1-p)^2Y \Rightarrow Y(2-p) = W$ from eqn 3,
and $Y(2-p) = p + (1-p)^2Y$ implying $Y = \frac{p}{1+p-p^2}$

Finally, we want $\frac{W+X+Y+Z}{4} = \frac{1}{2}$.

So $2 = W+X+Y+Z = Y( (2-p) + (1-p) + 1 + 1) = Y(5-2p) = \frac{(5-2p)p}{1+p-p^2}$

It comes to  $-2p^2 + 5p = -2p^2 + 2p + 2$ implying $p = 2/3$.


linear algebra
matrix
function
calculus
derivative
real analysis

Call a real matrix $A$ hyperbolic if it has two real eigenvalues, one inside the unit disc and another one outside. Suppose that $A$ is hyperbolic and in $GL_2(Z)$. Prove that one can find $k>0$ such that for any function $f$ in a set $H$ that will be described a little bit later, one can find a continuous function $g: R^2\to R^2$ such that $g(x+n)=g(x)$ for all $x\in R^2$ and $n\in Z^2$ and $(A+f)*(Id+g)=(Id+g)*A$. Here $*$ means composition of applications of $R^2$. Finally, the set $H$ is the set of $C^1$ class functions $h$ such that $h(x+n)=h(x)$ if $x\in R^2$ and $n\in Z^2$ and also the absolute values of $h_1(x), h_2(x), D_1 h_1(x), D_2 h_2(x)$ ( $D_i$ is the partial derivative) is smaller than $k$ for all $x$.  Why are these old ens problems so impossible?  :(

Ecole Normale Supérieure 1992 seconde composition

Harazi there were many questions before this one  it appear in
the part III.2

Dear Moubinool, first of all I don't consider solving problems is just looking at previous questions on an exam. I mean, I want to see the essence of this problem. Next, it is true there were many questions before, but they weren't particularly useful. Obviously, the way to proceed is the study of a linear application and the use of fixed point theorems. But in this question this is what is asked: to do all this, and this wasn't in previous questions.

The authors V.MAILLOT and Y.FAUCHEUX of this problem ENS Paris 1992 said
the aim is to prove  a particular case
of ANOSOV theorem following MOSER approach.

see the article by MOSER, "theorem of ANOSOV "
Journal of Differential Equations n°5, p.411- 440

Dear Moubinool, thanks for the information. At least, they could have specified that, I really had the impression they just put many hard questions in the exam with no specific purpose. I understand now, but I still have some problems in proving the above assertion. By the way, do you happen to have that article? If so, could you be so kind to post it? Thanks.


After four Algebra tests Stephanie had a mean of 90. After 5 tests she had a mean of 91. What did she score on the fifth test?

[hide]4*90=360

5*91 =455

455-360=95 [/hide]

[hide]Let's just say that she 90 on each test

[/hide]

[hide=Solution]
We can solve this by finding the total sum of the test scores and comparing them. First, the sum of the four tests is $4 \cdot 90 = 360$.
We can do the same with the $5$ tests which is $5 \cdot 91 = 455$. Subtracting $360$ from $455$ gives us that teh score for the fifth test is $\boxed{95}$.
[/hide]

[quote=hashtagmath][hide=Solution]
We can solve this by finding the total sum of the test scores and comparing them. First, the sum of the four tests is $4 \cdot 90 = 360$.
We can do the same with the $5$ tests which is $5 \cdot 91 = 455$. Subtracting $360$ from $455$ gives us that [b]the[/b] score for the fifth test is $\boxed{95}$.
[/hide][/quote]

@above FFTY :)

[quote=judgefan99][quote=hashtagmath][hide=Solution]
We can solve this by finding the total sum of the test scores and comparing them. First, the sum of the four tests is $4 \cdot 90 = 360$.
We can do the same with the $5$ tests which is $5 \cdot 91 = 455$. Subtracting $360$ from $455$ gives us that [b]the[/b] score for the fifth test is $\boxed{95}$.
[/hide][/quote]

@above [b]FTFY[/b] :)[/quote]

FTFY :P

[quote name="franzliszt" url="/community/p19260748"]
[quote=judgefan99][quote=hashtagmath][hide=Solution]
We can solve this by finding the total sum of the test scores and comparing them. First, the sum of the four tests is $4 \cdot 90 = 360$.
We can do the same with the $5$ tests which is $5 \cdot 91 = 455$. Subtracting $360$ from $455$ gives us that [b]the[/b] score for the fifth test is $\boxed{95}$.
[/hide][/quote]

@above FTFY :) :gleam:[/quote]

FTFY :P
[/quote]

FTFY :agent:

can we stop post farming? You can't say "FTFY" when you haven't fixed anything


hello everyone,

i have a document with a list of tables and figures in the beginning.
the \listoftables.

creates something like:

1.1 Caption..........20
1.2 Caption..........22

and so on

but i am required to have the following format

Table 1.1 Caption..........20
Table 1.2 Caption..........22

and so on.
Anyone know how to do this modification?
Thanks a lot...

This was posted on comp.text.tex and seems to do what you want:

\usepackage{tocloft}
\renewcommand{\cfttabpresnum}{Table } %Prefix table numbers with Table
\newlength{\mylen} % a "scratch" length
\settowidth{\mylen}{\bfseries\cfttabpresnum\cfttabaftersnum} % extra space
\addtolength{\cfttabnumwidth}{\mylen} % add the extra space