function
algebra
domain
real analysis
real analysis theorems

The statement : a sequence diverges to infinity .

is equivalent to the statement:

The sequence is unbounded from above??

If yes how do we prove it?

If no give a counter example

No, it's not: the sequence $ 0, 1, 0, 2, 0, 3, \ldots$ (i.e., $ (a_n)_{n \geq 1}$ defined by $ a_{2n} \equal{} n$ and $ a_{2n \minus{} 1} \equal{} 0$) is unbounded above and does not diverge to infinity.  I refer you to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1712881#1712881]my comment on your earlier post[/url]: these are basic definitions that can be found in any textbook.  Why don't you just look them up?  Is there some point here?  (These comments also apply, for example, to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=326067]this thread[/url].)

[quote="JBL"]No, it's not: the sequence $ 0, 1, 0, 2, 0, 3, \ldots$ (i.e., $ (a_n)_{n \geq 1}$ defined by $ a_{2n} \equal{} n$ and $ a_{2n \plus{} 1} \equal{} 0$) [/quote]

So according to your definition we have:

for n=1 .....$ a_{2}  \equal{} 1$  [b]and[/b]  $ a_{3} \equal{} 0$

for n=2 ........$ a_{4}\equal{}2$  [b]and[/b] $ a_{5} \equal{} 0$

e.t.c  e.t.c

[b]Does the above definition define a sequence?[/b]

Where is $ a_{1}$ ??

How do we prove that the sequence : 0,1,0,2,0,3.................................[b]does not[/b] diverge to infinity??

[quote="BlueVelvet"]You're right, a_1 isn't defined in this way. Obviously (since he wrote 0,1,0,2,0,3) JBL meant the first term to be 0, i.e.

$ (a_n)_{n \geq 1}$ defined by $ a_{2n} \equal{} n$ and $ a_{2n \minus{} 1} \equal{} 0$.

But as JBL has said a few times already, these are basic definitions found in any textbook. Write out the definition of "diverging to infinity" (which someone already did for you in another thread), and apply it.[/quote]

First of all the way that JBL  wrote  the definition [b]does not[/b]define a function,because for each value for n there correspond two values :

$ a_{2n}$ [b]and[/b] a_{2n+1}.

Supporting nonsenses it will lead you nowhere.

The definition i was given in another thread  [b]it is wrong[/b].

When i asked :

How can can we prove that the sequence o,1,0,2,0,3...........does not diverge to infinity

[b]it means i do not know how to do it[/b]

Do you??

Oh dear, I don't think I have the energy to do remedial lessons in definitions and notations.  So, I'll just assume that you are feigning incompetence and actually are able to understand the definition of the sequence that I've suggested.  Then we can move on: why don't you write down the definition of the property you are interested in (and perhaps its negation)?  After that, you can proceed to attempt the proof that this sequence does not possess that property.  After you pass this version of the Turing test and contribute something to the thread, then perhaps someone will be sufficiently impressed that they'll be willing to help you along with any difficulties.

Do you mean the sequence:

for n=1 ,$ a_{1} \equal{}0$

for  n=2 ,$ a_{2} \equal{} 2$

for n=3 ,$ a_{3} \equal{} 0$

for n=4 ,$ a_{4} \equal{} 4$

for n=5 ,$ a_{5} \equal{} 0$ ............and so on ...........

Where the general term $ a_{n}$ is n [b]if[/b] n is even and 0 [b]if[/b] n is odd??

[quote="stauros"]Do you mean the sequence:

for n=1 ,$ a_{1} \equal{} 0$

for  n=2 ,$ a_{2} \equal{} 2$

for n=3 ,$ a_{3} \equal{} 0$

for n=4 ,$ a_{4} \equal{} 4$

for n=5 ,$ a_{5} \equal{} 0$ ............and so on ...........
[/quote]
Do you actually read replies?
[quote="BlueVelvet"]You're right, a_1 isn't defined in this way. Obviously (since he wrote 0,1,0,2,0,3) JBL meant the first term to be 0, i.e.

$ (a_n)_{n \geq 1}$ defined by $ a_{2n} \equal{} n$ and $ a_{2n \minus{} 1} \equal{} 0$.[/quote]
The above definition is as clear as it can get.
[quote]
Where the general term $ a_{n}$ is n [b]if[/b] n is even and 0 [b]if[/b] n is odd??[/quote]
No, $ a_n$ equals $ \frac {n}{2}$ if $ n$ is even, since I just defined $ a_{2n} \equal{} n$.

Whether you post questions here or at [url=http://www.physicsforums.com/showthread.php?t=371238]physicsforums[/url], people will expect you to show some effort, otherwise one gets the impression of you letting your homework being done for you.

[quote="JBL"]Oh dear, I don't think I have the energy to do remedial lessons in definitions and notations.  .[/quote]

Particularly if you are so good on definitions

So, I see you still haven't managed to state the definition of the relevant term, nor begin the process of trying to prove anything.  No wonder your rating is 1.2  :wink:  Let's see if you can try for a post with some mathematical content, eh?

In case it's not clear from BlueVelvet's response, yes, that is the sequence I was suggesting, though your description of it is odd.  Here are two ways alternative ways you could have written it:

Way 1:
We have $ a_1 \equal{} 0, a_2 \equal{} 1, a_3 \equal{} 0, a_4 \equal{} 2, a_5 \equal{} 0, a_6 \equal{} 3$, and so on.

Way 2:
When $ n \equal{} 1$, $ a_n \equal{} 0$.  When $ n \equal{} 2$, $ a_n \equal{} 1$.  When $ n \equal{} 3$, $ a_n \equal{} 0$.  When $ n \equal{} 4$, $ a_n \equal{} 2$.  And so on.

Why am I saying this?  There's no $ n$ anywhere in the statement "$ a_1 \equal{} 0$," so this makes the statement "when $ n \equal{} 1$, $ a_1 \equal{} 0$" somewhat odd -- it's as if I said, "When $ x$ is a positive integer, $ 3^2 \plus{} 4^2 \equal{} 5^2$."  This is true, but there's an extra quantifier hanging around for no reason whatsoever.
Of course, this discussion, like most of your posts, relates to mathematics in a similar way that a discussion of font choices relates to literature.

Also, the details of this sequence are almost totally irrelevant, but since you've so far devoted zero words to things recognizable as "mathematical thought," we haven't gone anywhere near the point of assessing what the properties of this sequence are that make it a good example.  Do try to make your next post an improvement, won't you?

[quote="JBL"]

Way 1:
We have $ a_1 \equal{} 0, a_2 \equal{} 1, a_3 \equal{} 0, a_4 \equal{} 2, a_5 \equal{} 0, a_6 \equal{} 3$, and so on.

Way 2:
When $ n \equal{} 1$, $ a_n \equal{} 0$.  When $ n \equal{} 2$, $ a_n \equal{} 1$.  When $ n \equal{} 3$, $ a_n \equal{} 0$.  When $ n \equal{} 4$, $ a_n \equal{} 2$.  And so on.

[/quote]

So the definition now of your sequence is:

For all,n : (if n is odd ,then $ a_n \equal{} 0$) [b]and[/b] ( if n is even,then $ a_n \equal{}\frac{n}{2}$).

Now let us see if logic allows for that definition.

If we put :

 n is odd as statement p............(n is even) = q

($ a_n \equal{} 0$) = r.....................($ a_n \equal{} \frac{n}{2}$) = s

The definition becomes:

($ p\Longrightarrow r$) [b] and [/b] ($ q\Longrightarrow s$) ..........................................................................................1

Logic tell us that (1)  logically implies :

$ p\wedge q\Longrightarrow r\wedge s$ ,or otherwise

p and q implies r and s.

Which means:

[b] if n is odd and even ,then $ a_n \equal{}0$ and $ a_n \equal{}\frac{n}{2}$[/b]

BUT

Can you have [b] n being odd and even at the same time [/b]???

So your 2nd definition is not [b] very successful as well [/b]

@ stauros:

If it rains, I won't play basketball.
If it doesn't rain, I will play basketball.

Following your intention to put complementary statements together:

If it rains and it doesn't rain, I won't and I will play basketball.

BUT

Can it rain and not rain at the same time?

Oh god, what can I do?  :rotfl:
***
Enough joking. I just don't understand what's so unclear about JBL's post, or why are you insisting on using logic rules to deduce stuff.

[quote="Soarer"]@ stauros:

If it rains, I won't play basketball.
If it doesn't rain, I will play basketball.

Following your intention to put complementary statements together:

If it rains and it doesn't rain, I won't and I will play basketball.

.[/quote]

[b] The conclusion is correct[/b]

That shows if you feed logic with nonsenses you will get nonsenses.

Look at the following example:

if you are not stupid then i am a monkey .But since i am not a monkey , logic tell us, that you are stupid

BUT

Are you stupid???

By the way the law that tell us that you are stupid is called : M.Tollens

[quote="stauros"]
Can you have [b] n being odd and even at the same time [/b]???

So your 2nd definition is not [b] very successful as well [/b][/quote]

1. Given that you love going back to rules so much, I can't see why you would say "feed logic with nonsenses you will get nonsenses". In fact, if A is a false statement, then whatever statement B is, we have $ A \Rightarrow B$ is true.

2. Back to your second-to-last post, so can you explain why JBL's second definition is "not very successful"? We don't have a natural number $ n$ being odd and even at the same time, so this is a false statement, so "if n is odd and even ,then $ a_n \equal{} 0$ and $ a_n \equal{} \frac {n}{2}$" is fine. Then what's your possible objection to JBL's second definition, assuming that a reasonable objection could exist?

Right .

But in a mathematical proof we use true statements ,and that definition has both  its supposition and conclusion false.

Can we use it in a proof??

If you are so intent in creating contradiction, I have no idea what mathematics you can do at all.

For example, if I want to define a sequence $ a_1 \equal{} 0$, $ a_n \equal{} 1$ for $ n \ge 2$, then using your argument, I can oppose this definition by intentionally placing $ n \equal{} 1$ and $ n \ge 2$ together, making it a false statement, and you would argue that this definition does not work.

Anyway, I think it is better to post something with mathematical content. The last few posts (yours and mine) are really meaningless.

Soarer, I don't think it's worth it -- s/he has just "proved" that the sequence we've all written down and given formulas for, can't exist :roll:  (Actually, I think specifically s/he has "proved" that every sequence is a constant sequence.  Or maybe that every function must have domain of size 1.  :rotfl: )  Maybe after your post s/he will understand that there is a difference between $ (A \Rightarrow B) \wedge (C \Rightarrow D)$ and $ (A \wedge C) \Rightarrow (B \wedge D)$ (namely that the former implies the latter, but not vice-versa), but I'm somewhat skeptical.

Also, it seems pretty clear that s/he is incapable of actually writing down a sensible mathematical statement and also may not understand in even a vague sense what a "sensible mathematical statement" is; certainly, s/he appears not to have understood that from your very nice example it is not a valid conclusion that it is both raining and not raining, and my several attempts (on several threads) to get him/her to make any substantive statement (even to write down a basic definition) have been utterly futile.

The only thing I'm left wondering is what s/he is getting out of this (unless irritating people who understand mathematics is a source of pleasure for him/her).

[quote="JBL"] Maybe after your post s/he will understand that there is a difference between $ (A \Rightarrow B) \wedge (C \Rightarrow D)$ and $ (A \wedge C) \Rightarrow (B \wedge D)$ (namely that the former implies the latter, but not vice-versa), but I'm somewhat skeptical.[/quote]

Let that be the statement of the year

[quote="JBL"]Also, it seems pretty clear that s/he is incapable of actually writing down a sensible mathematical statement and also may not understand in even a vague sense what a "sensible mathematical statement" is; certainly, s/he appears not to have understood that from your very nice example it is not a valid conclusion that it is both raining and not raining, and my several attempts (on several threads) to get him/her to make any substantive statement (even to write down a basic definition) have been utterly futile. [/quote]

What is a  "sensible mathematical statement",any book to refer me ,for a definition??
Thanks for your efforts and sorry that i am not a good pupil

[quote="JBL"]The only thing I'm left wondering is what s/he is getting out of this (unless irritating people who understand mathematics is a source of pleasure for him/her).[/quote]

That sentence should be addressed to LOGIC.

Finally take my advise ( I know you won't) open a book and learn all the mysteries of logic

Besides you cannot have [b] any mathematics at all without logic[/b]


Who's coming?

I've gathered that the director has a talent for building tension and excitement through unusual formatting at this tournament.

I'm coming.

Yes, I think the director does have such a knack... Who else sends out daily updates to the registered schools?! And who has ciphering before written?! (Ok, so the Cobb fall contest[I think] had ciphering before written for JV, but I wasn't in JV there.)

Actually, I think that that's normal for JV.  Then again, I haven't been to many tournaments this year, and of course I didn't go to many in middle school, and I've never been to one out of state.

Our teacher, however, talked of a "head-to-head" competition about two years ago, in which each competitor was paired with another competitor, either randomly or through seeding (methinks the latter).  Although each of the competitors answers the same question at the same time, a given competitor's only goal is to beat his/her partner.  Thus in each round (roughly) half the competitors are eliminated.  It made me wish I was still doing Mathcounts countdown round practice.  (Of course, she said that he didn't do it last year.)  And then there's the relay.  I imagine that they announce the top fifteen or so at the awards ceremony, too, instead of the top three.  They did that at GACS, and I noticed that it made the awards ceremony much more intense.

I'm in.

(Have you ever noticed that if you make a message as short as the one above, it won't let you post it?)

Can anyone send me the questions for practice? Thanks.

[quote="captcha000"]what school did you go with?[/quote]

I went for Northview.

[quote="solafidefarms"]Ah, so you are Bill Qaio? You beat me by .4 point :D.[/quote]

I'm Karthik Narayan. Who are you?  :)

[quote="Xantos C. Guin"][quote="Boy Soprano II"]Actually, I think that that's normal for JV.  Then again, I haven't been to many tournaments this year, and of course I didn't go to many in middle school, and I've never been to one out of state.[/quote]

Yes, Ciphering is normally 1st for JV students.

What do you mean by not that many? You have been to way more than I even knew existed.[/quote]
I meant not many compared to other people at my school.  I went to two tournaments in the first semester, one of which was Lassiter Invitational (so I couldn't participate), and in second semester I have missed Rockdale and GT, and our school didn't send a team to Milton.  All of these are because of an elaborate conspiracy in which there is a math tournament scheduled during every All-State Chorus/All-State Orchestra/Honor Orchestra audition/event (or at least, 90% of them).  Thus at the time of my post, I had participated in a total of three tournaments this year, excepting Lassiter Invitational.

While we're sharing names, I'm Miles Edwards, by the way.


The mean probability of 100 observations is .0422. If you are not given the data for a sample size of 200, how do you find the mean probability of this data using the mean probability you found from .0422?

The Mean will stay approximately the same no matter how large the size of the sample. However, the standard deviation will decrease if the sample gets larger


Let $ x \equal{} a \plus{} \frac {1}{b} ;y \equal{} b \plus{} \frac {1}{c} ;z \equal{} c \plus{} \frac {1}{a}$. ($ a,b,c > 0$) .Prove that:

$ xy \plus{} yz \plus{} zx \ge 2(x \plus{} y \plus{} z)$

I remember this problem has already been in another topics.
Assume:
(x - 2)(y - 2) ≥ 0
<=> xy + 4 ≥ 2(x + y)

Now we only have to prove that:
yz + zx ≥ 2z + 4
This is quivalent to
$ \left(c\plus{}\frac1a\right)\left(a\plus{}\frac1c\plus{}b\plus{}\frac1b\right)\ge2\left(c\plus{}\frac1a\right)\plus{}4$
which is true because:
$ \\\left(c\plus{}\frac1a\right)\left(a\plus{}\frac1c\right)\ge4\\\left(c\plus{}\frac1a\right)\left(b\plus{}\frac1b\right)\ge2\left(c\plus{}\frac1a\right)$


Le $ S$ be the set of all pairs of real numbers $ (\alpha,\beta)$ such that the equation $ x^3\minus{}6x^2\plus{}\alpha x\minus{}\beta\equal{}0$ has three ral roots (not necessarily disctint) and they are all greater than $ 1$. Find the maximum value of $ T\equal{}8\alpha\minus{}3\beta$ for $ (\alpha,\beta)\in S$

[quote="Algadin"]Le $ S$ be the set of all pairs of real numbers $ (\alpha,\beta)$ such that the equation $ x^3 \minus{} 6x^2 \plus{} \alpha x \minus{} \beta \equal{} 0$ has three ral roots (not necessarily disctint) and they are all greater than $ 1$. Find the maximum value of $ T \equal{} 8\alpha \minus{} 3\beta$ for $ (\alpha,\beta)\in S$[/quote]

the third degree Schur's inequality   $ (a\plus{}b\plus{}c)^{3}\plus{}9abc\geq 4(a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc)$ -----> $ abc\geq \frac{4}{9}(a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc)\minus{}\frac{1}{9}(a\plus{}b\plus{}c)^{3}$
$ a,b,c\geq 1$ ----->

$ a\plus{}b\plus{}c\equal{}6$
$ ab\plus{}ac\plus{}bc\equal{}\alpha$
$ abc\equal{}\beta$ ------>

$ T \equal{} 8\alpha \minus{} 3\beta\equal{}8(ab\plus{}ac\plus{}bc)\minus{}3abc\leq8(ab\plus{}ac\plus{}bc)\minus{}3(\frac{4}{9}(a\plus{}b\plus{}c)(ab\plus{}ac\plus{}bc)\minus{}\frac{1}{9}(a\plus{}b\plus{}c)^{3})\equal{}8(ab\plus{}ac\plus{}bc)\minus{}8(ab\plus{}ac\plus{}bc)\plus{}72\equal{}72$ ---->

$ a\equal{}b\equal{}c\equal{}2$ ---> $ \beta\equal{}8$ ,$ \alpha\equal{}12$ ----> $ max(T)\equal{}72$


Find all functions $f: \Re \rightarrow \Re$ such that,for all real numbers $x$ and $y$:
$f(x\cdot f(y))=f(x.y)+x$

[hide]Put $x=1$ into the function to get $f(f(y))=f(y)+1\Rightarrow f(x)=x+1$ is the only possibility.
Testing the identity, $f(xf(y))=f(xy+x)=xy+x+1=f(xy)+x$ which checks.
Therefore, $f(x)=x+1$.[/hide]

[quote="scorpius119"][hide]Put $x=1$ into the function to get $f(f(y))=f(y)+1\Rightarrow f(x)=x+1$ is the only possibility.
[/hide][/quote]

Why?

($f: \Re \rightarrow \Re \Rightarrow$ the domain and range of $f$ includes ALL reals, right?)
Let $z=f(y)$. From $f(f(y))=f(y)+1$ for all real numbers $y$, since the range of $f$ can include any real number $z$, then $f(z)=z+1$ for all real $z$.

Ok, thanks. Didn't see the substitution of f(y).

[quote="scorpius119"]($f: \Re \rightarrow \Re \Rightarrow$ the domain and range of $f$ includes ALL reals[/quote]

That's true but it should but proved (by $x = f((x-f(0)) \cdot f(0))$ for example), doesn't it ?


trigonometry
AMC
AIME
AMC 12
floor function
logarithms
function

OK, since marathon is the flavor of the month, its time.
.....Im not going to mention that after we get a few problems down, we can compile them into a pdf for practise use. :D

So everyone knows how it works? One person posts a problem, and the one who solves that problem gets to post the next, and so on.

This first one is from the Michigan MAA.....

$Problem 1:$
$The$ $ expression$ $ cos(2a)(sec^2a)$  $is$ $ equivalent$ $ to$ $ what$ $trigonomic$ $ expression?$
$a)1$
$b)sec$ $ a$
$c)1+tan^2 a$
$b)1-tan^2 a$
$e)none$ $ of$ $ these$

[quote="PenguinIntegral"]$\text {Problem 1}$
The expression $\cos{2a} \sec^2{a}$ is equivalent to what trigonometric expression?

a) $1$
b) $\sec{a}$
c) $1+\tan^2{a}$
b) $\1-\tan^2{a}$
e)none of these[/quote]

[quote="JesusFreak197"]$\cos 2a=\cos ^2a-\sin ^2a$
$\cos 2a\sec ^2a=(\cos ^2a-\sin ^2a)\cdot \sec ^2a=1-\tan ^2a=\boxed{E}$

As for another problem... umm... what level of problem should I post? AMC 10/12? I don't know where to find problems like that, only AIME. :?[/quote]

OK, some REALLY EASY AIME then.

But you can find AMC level probs here:http://www.delta.edu/math/mmpc/

Because he isnt posting, and he is offline, I will post another one:
$Problem 2$
How many of the first 1000 integerts have a remainder of 1 when divided by 5, 2 divided by 6, and 5 when divided by 7.

$LCM(5,6,7)=210$

The first number for which this is the case is $26$, and just add $210$ for each number after that. Therefore, there are $\boxed{5}$ numbers between 1 and 1000 that satisfy the condition (those being $26$, $236$, $446$, $656$, $866$).

[b]Problem 3 - 1985 AIME #1[/b]

$x_1=97$, $x_2=\frac{2}{x_1}$, $x_3=\frac{3}{x_2}$, $x_4=\frac{4}{x_3}$, etc. Find $x_1x_2x_3...x_7x_8$.

[quote="amirhtlusa"]just group them, $(x_1,x_2),(x_3,x_4)...$
it is easily seen that $x_1\cdot x_2 ... x_8=2*4*6*8=\boxed{384}$

 [color=blue]Problem 4[/color]
How many Positive even integers less than 2004 are divisble by 3?

source: 2004 Mock AMC12, by Joml88[/quote]
2*3=6
less than would be 333, because 333*6=1818+180=1998, one more 6 would make 2004, so that 6 isn't less than 2004.

 Find the sum of the solutions to x^1/4 = 12/(7 - x^1/4).

[quote="PenguinIntegral"]
 Find the sum of the solutions to x^1/4 = 12/(7 - x^1/4).[/quote]

$x^{\frac{1}{4}} = k \Rightarrow k(7-k) = 12 \Rightarrow k = 4, 3 \Rightarrow x=256, 81 \Rightarrow x_1 + x_2 = \fbox{337}$

Solve $\sum_{k=1}^n \lfloor \log_2 k \rfloor = 1994$ for $n$

[quote="zanttrang"][quote="PenguinIntegral"]What do those brackets mean?[/quote]

It's the floor function, or "greatest integer function," i.e. the greatest integer less than or equal to the number.  It's sometimes represented as $[x]$, but that can also mean the nearest integer function, so I used the floor brackets to avoid confusion.[/quote]

I wouldn't know how to do it other than BF-ing it. :(

[quote="zanttrang"][quote="PenguinIntegral"]
I wouldn't know how to do it other than BF-ing it. :([/quote]

Try looking for a pattern - how many terms are 2, how many are 3, etc.[/quote]

0+2*1+4*2+8*3+16*4........=1994

Find a closed expression for the left and solve? :(

[quote="JesusFreak197"]I figured there was probably a way to prove that, but I don't know how to do that... the values that you're given are kind of odd. :?

[b]1993 AIME #4[/b]

How many 4-tuples $(a,b,c,d)$ satisfy $0<a<b<c<d<500$, $a+d=b+c$, and $bc-ad=93$?[/quote]

Let $b=a+x$, $c=a+y$. From the first equation, that makes $d=a+x+y$. Plugging into the second equation, we get

$(a+x)(a+y)-a(a+x+y)=93\Longrightarrow xy=93$

Since $x<y$, then the two possibilities $(x,y)$ are $(1,93)$ and $(3,31)$. Since our variables always satisfy the first equation, we merely have to make sure it satisfies the second. The second is a function only of x and y, and thus adding any integer value to $(a,b,c,d)$ will satisfy the second equation, so long as the original four-tuple satisfied it.

Thus, letting $a=1$ as our first ordered pair, then one set of them is when $x=1$ and $y=93$, which then makes $x+y=94$. Since $d<500$, $a+x+y=a+94<500$ then $a<406$, and thus $a$ is any integer from 1 to 405, which is 405 possibilities.

Looking at the other ordered pair, we can do the same thing where $x+y=34$, and then $a+x+y=a+34<500$ becomes $a<466$, which means $a$ is any integer from 1 to 465, so 465 possibilities.

The total four-tuples is thus $405+465=870$.

My Problem (Mandlebrot):

Given the sequence $F_0=F_1=1$ and $F_n=F_{n-1}+F_{n-2}$, then express $\sum_{n=0}^{\infty}\frac{F_n}{3^n}$ as a real number.

[quote="Elemennop"]
My Problem (Mandlebrot):

Given the sequence $F_0=F_1=1$ and $F_n=F_{n-1}+F_{n-2}$, then express $\sum_{n=0}^{\infty}\frac{F_n}{3^n}$ as a real number.[/quote]
Consider the solution to the Fibonacci sequence: $F_n = (\alpha^n - \beta^n) / \sqrt{5}$, where $\alpha, \beta$ are the roots of the equation $x^2 - x - 1 = 0$.  So, our sum becomes

$\frac{1}{\sqrt{5}} \sum_{n = 0}^{\infty} \left(\frac{\alpha}{3}\right)^n - \left( \frac{\beta}{3}\right)^n = \frac{1}{\sqrt{5}}\left(\frac{1}{1 - \frac{\alpha}{3}} - \frac{1}{1 - \frac{\beta}{3}}\right) = \frac{1}{\sqrt{5}} \left(\frac{3(1 - \beta) - 3(1 - \alpha)}{(3 - \alpha)(3 - \beta)}\right) = \frac{1}{\sqrt{5}}\left(\frac{3(\alpha - \beta)}{9 - 3(\alpha + \beta) + \alpha \beta}\right) = \frac{3}{9 - 3 - 1} = \frac{3}{5}$

You can also use the fact that the generating function for the Fibonacci sequence is $\sum_{n=1}^{\infty}F_n x^n = \frac{x}{1 - x - x^2}$, so that $\sum_{n=1}^{\infty} F_n \left(\frac{1}{3}\right)^n = \frac{3}{5}$  $F_0 = 0$, so the fact that the sum begins at $n = 0$ rather than $n = 1$ can be ignored, and we have our solution.

[quote="frt"]Yes, that's correct (although you got the answer by eliminating other choices) so you can post the next problem :)

Can anyone prove that the answer is indeed $3\pi$?[/quote]
It involves the fact that if you let the two roots of the equation $x^2 - 9x + 1 = 0$ be $r_1, r_2$, and then you let $\tan a = r_1$ and $\tan b = r_2$, then $\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} = \frac{r_1 + r_2}{1 - r_1 r_2}$.  Noting that $r_1 r_2 = 1$, we find that the expression is undefined, so $a + b = \pi / 2 + k \pi$.  I don't really want to think about the extra ranglings you have to do to prove this rigorously, but you'll probably find that out of the four possible solutions in the domain: $a_1, b_1, a_2, b_2$, two pairs will add up to $3 \pi / 2$, giving you a final solution of $3 \pi$.  Anybody who wants to work that out in detail is welcome to.

Since nobody's posting the next problem...

${\large \textbf{Problem x}}$

Let $\overline{AB}$ be a diameter of a circle and $C$ be a point on $\overline{AB}$ with $2 \cdot AC = BC$. Let $D$ and $E$ be points on the circle such that $\overline {DC} \perp \overline {AB}$ and $\overline{DE}$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$? (AMC 12 2005A #15)

You like AMC12 2005A... :P

[hide="Answer"]Both triangles have $\frac{1}{2}(d)h=A$ where $d$ is the diameter of the circle. All we need to find is the ratio of the two heights $\frac{CF}{CD}$. Because they are similar triangles, we have $\frac{CF}{CD}=\frac{OC}{r}$. Since $2AC=BC$, we have $OC=\frac{1}{3}r$, so $\frac{OC}{r}=\frac{\frac{1}{3}r}{r}=\frac{1}{3}\Rightarrow \frac{[DCE]}{[ABD]}=\boxed{\frac{1}{3}}$.[/hide]

Next problem:

Let R be a rectangle. How many circles in the plane of R have a diameter both of whose endpoints are vertices of R? (AMC12 1998 #11)

[quote="JesusFreak197"]You like AMC12 2005A... :P

[hide="Answer"]Both triangles have $\frac{1}{2}(d)h=A$ where $d$ is the diameter of the circle. All we need to find is the ratio of the two heights $\frac{CF}{CD}$. Because they are similar triangles, we have $\frac{CF}{CD}=\frac{OC}{r}$. Since $2AC=BC$, we have $OC=\frac{1}{3}r$.[/hide][/quote]

Ah, I got everything up to the last step. :mad:

The part with $OC=\frac{1}{3}r$, etc.? I used that ratio, then went on and tried to find other ratios, until I noticed that I actually know what $\frac{OC}{r}$ is. :P

I am sad that this marathon died out, so here is a problem from Niels Henrik Abel Contest, Second Round, 1994-1995 (about AMC 12 level):

How many natural numbers are equal to three times the sum of their digits?

[hide="solution"]
For a two digit number, let it equal
$10a+b$.

Then,
$10a+b=3a+3b$
$7a=2b$

The only number that satisfies this ratio is 27.

For numbers of three digits or more, the sum of their digits can never be great enough to work.  The equation will always be something like this:

$\dots+97x+7y=2z$

Since z can at maximum equal 9, and x>0, the RHS is never large enough.

So, 27 is the only solution, making the answer [b]1[/b]

[/hide]

My question was never answered, though.

As for this question, yes, $27$ is the only answer. Once you realize that three-digit numbers are impossible, you can easily just quickly guess-and-check.

I want to say
[hide]
6.

One on each edge and one for each diagonal.
[/hide]

But I'm not sure.  On the other hand, it's only a problem 11, so it can't have too many subtle tricks.

[quote="dakyru"]Yes, you're simply choosing two of the four vertices: ${4\choose2} = 6$[/quote]
Not the case.
Two of them, namely, those formed by choosing vertices on diagonals, form same circles.

I'd say: $5$ if the rectangle is not degenerated. Otherwise, $1$.

Oops, you're right, the two diagonals are just diameters of the same circle.   :oops:  Cyclic quad.


Alice and Bob play a game with a baseball. On each turn, if Alice has the ball, there is a 1/2 chance that she will toss it to Bob and a 1/2 chance that she will keep the ball. If Bob has the ball, there is a 2/5 chance that he will toss it to Alice, and if he doesn't toss it to Alice, he keeps it. Alice starts with the ball. What is the probability that Alice has the ball again after two turns?

Case 1: A -> A -> A

Probability: $ \frac12 \times \frac12 \equal{} \frac14$

Case 2: A -> B -> A

Probability: $ \frac12 \times \frac25 \equal{} \frac15$

Total Probability = $ \frac14 \plus{} \frac15 \equal{} \boxed{\frac9{20}}$


A cubical box with $ 10$"-long sides contains a sphere with diameter $ 10$". What is the maximum number of inches that any point on the sphere could be from any point on the surface of the cube? Express your answer as a decimal to the nearest hundredth.

Draw a space diagonal in the cube. Clearly, this contains the longest possible length of $ 10\plus{}\frac{10\sqrt3\minus{}10}2\equal{}\boxed{5\plus{}5\sqrt3}$.

in decimal rounded to nearest hundredth, that answer is equal to
13.66

Just draw the space diagonal which is $\sqrt{300}=10\sqrt{3}.$ Then we have to subtract the diameter of the sphere which is 10 and divide by two to make sure we only count once. And thus we have $\frac{10\sqrt{3}-10}{2}=5\sqrt{3}-5$ Subtracting this from the entire space diagonal we get $\boxed{5\sqrt{3}+5}$

[hide="Solution"]
Note that if you pick a point P on the cube, then the maximum distance PQ where Q is a point on the sphere is achieved when P, Q, and the center of the sphere are collinear (with the center of the sphere between P and Q). Thus the problem reduces to finding the point P which maximizes the length of PQ.

Let the center of the sphere be O. Then PQ = PO + OQ = PO + 5. Now we want to maximize PO, which is achieved by letting P be one of the corners of the cube. When P is there, then PO = $5\sqrt{3}$ and $OQ = 5$ so then $PQ = 5 + 5\sqrt{3}$ and this is the maximum.

[/hide]


[quote="vuthanhtu_hd"][b]Problem 2[/b] (Vu Thanh Tu).
Let a,b,c be [b]real[/b] numbers. Prove that

$ \sqrt {\left( {12a^2 \plus{} (b \plus{} c)^2 } \right)\left( {12b^2 \plus{} (c \plus{} a)^2 } \right)\left( {12c^2 \plus{} (a \plus{} b)^2 } \right)} \plus{} 8(a \plus{} b)(b \plus{} c)(c \plus{} a) \ge 0$

When does equality hold?  :)[/quote]
See here: http://can-hang2007.blogspot.com/2009/12/inequality-12-v-t-tu.html


Find all function $f: \mathbb R \to \mathbb R$ such that for all real $x$, we have
\[f(f(f(x)))-3f(f(x))+6f(x)= 4x+3.\]

no. i thing it's easy your problem.its classical:
f is increasing.and use the classical method of iteration .and you will find solution:$ f(x) = x+1$  :P

[quote="greatestmaths"]find all function f: R----->R such that for all x in R.

f(f(f(x)))-3f(f(x))+6f(x)= 4x+3[/quote]

The difficulty is to study the non-trivial solutions (the trivial one is $ f(x)\equal{}x\plus{}1$).

I think I have shown there is no other bijective solution with an Abel form $ f(x)\equal{}g^{\minus{}1}(1\plus{}g(x))$
But it remains job to do.

I'm glad you posted this meassage in this forum, greatestmaths, since it means you have the solution.

Thanks to post it
or just a hint.

here you are a hint:let be $ x_{n}\equal{}f^{n}\minus{}n$ then its easy to see that $ x_{n\plus{}3}\minus{}3x_{n\plus{}2}\plus{}6x_{n\plus{}1}\minus{}4x_{n}\equal{}0$ then study this sequence via the fact that $ f$ is increasing
 :wink:

[quote="fermat3"]here you are a hint:let be $ x_{n}\equal{} f^{n}\minus{}n$ then its easy to see that $ x_{n\plus{}3}\minus{}3x_{n\plus{}2}\plus{}6x_{n\plus{}1}\minus{}4x_{n}\equal{} 0$ then study this sequence via the fact that $ f$ is increasing
 :wink:[/quote]

Why is $ f(x)$ increasing ?

supose that $ f is decreasing$ then $ f0f0f\minus{}3f0f\plus{}6f$ is deacreasing wich absurd because $ f^{3}\minus{}3f^{2}\plus{}6f\equal{}4x\plus{}3$ :wink:

[quote="pco"][quote="fermat3"]supose that $ f is decreasing$ then $ f0f0f\minus{}3f0f\plus{}6f$ is deacreasing wich absurd because $ f^{3}\minus{}3f^{2}\plus{}6f \equal{} 4x\plus{}3$ :wink:[/quote]

But f can be neither decreasing, neither increasing ! (increasing on some parts, decreasing on others), so your demo seems wrong.[/quote]

Yep, I agree with pco; we know that $ f$ is injective (it is enough to assume that $ x,y$ exist such that $ f(x) \equal{} f(y)$ and see that this leads to $ 4x\plus{}3 \equal{} 4y\plus{}3$), but I do not know how you can go from there to saying that $ f$ needs to be monotonous.  For all that we know (unless proved otherwise), it could even be not continuous at any $ x$!

My own efforts go more in the direction of defining $ g(x) \equal{} f(f(x))\minus{}2f(x)\plus{}4x$ so that obviously $ g(f(x))\minus{}g(x) \equal{} 3$ for all $ x$, and $ h(x) \equal{} f(x)\minus{}x$ so that $ h(f(f(x)))\minus{}2h(f(x))\plus{}4h(x) \equal{} 3$, but I have not been able to crack it yet (the idea is showing that $ h(x)$ is constant and/or $ g(x) \equal{} 3x$).

[quote="greatestmaths"]find all function f: R----->R such that for all x in R.

f(f(f(x)))-3f(f(x))+6f(x)= 4x+3[/quote]

Here is a partial result : $ f(x) \equal{} x\plus{}1$ is the unique continuous solution.

Demo :
Let $ x_{0}\in\mathbb{R}$ and the sequence $ a_{1}\equal{} x_{0}$, $ a_{n\plus{}1}\equal{} f(a_{n})$.
We have $ a_{n\plus{}3}\equal{} 3a_{n\plus{}2}\minus{}6a_{n\plus{}1}\plus{}4a_{n}\plus{}3$
Let then $ b_{n}\equal{} a_{n\plus{}1}\minus{}a_{n}$. We have $ b_{n\plus{}2}\equal{} 2b_{n\plus{}1}\minus{}4b_{n}\plus{}3$ and so $ b_{n\plus{}3}\minus{}1 \equal{}\minus{}8(b_{n}\minus{}1)$

So, if $ b_{n}\neq 1$ for some $ n$, we are sure that the sequence $ b_{n}$ will have some $ b_{i}> 0$ and some $ b_{j}< 0$. Or $ a_{i\plus{}1}> a_{i}$ and $ a_{j\plus{}1}< a_{j}$ or again $ f(a_{i}) < a_{i}$ and $ f(a_{j}) > a_{j}$.
Alternatively, $ b_{1}\equal{} 1$ implies $ a_{2}\equal{} a_{1}\plus{}1$ and so $ f(x_{0}) \equal{} x_{0}\plus{}1$

So : either $ f(x_{0}) \equal{} x_{0}\plus{}1$, either it exists reals $ u$ and $ v$ such that $ f(u) < u$ and $ f(v) > v$. Then, if $ f(x)$ is continous, it exists $ w$ such that $ f(w) \equal{} w$. But then $ f(f(f(w))) \equal{} w$ and $ 3f(f(w))\minus{}6f(w)\plus{}4w\plus{}3 \equal{} 3w\minus{}6w\plus{}4w\plus{}3 \equal{} w\plus{}3$ and so $ f(f(f(w)))\neq 3f(f(w))\minus{}6f(w)\plus{}4w\plus{}3$

So, either $ f(x) \equal{} x\plus{}1$ $ \forall x$, either $ f(x)$ is not continuous.

And, since $ x\plus{}1$ is a solution, the only continuous soluition is $ f(x) \equal{} x\plus{}1$
Q.E.D.

It remains to study non continuous solutions.

And, once again, greatestmaths, I would be happy to have some hint (or the complete solution).

[quote="greatestmaths"]let f is a solution.
[u]i begin by f is increasing :[/u]
if f is decreasing so :          f0f0f-3f0f +6f is decreasing.  so it isn't equal for  the function  x---->4x+3  because this function is  increasing.
i take the sequence (Xn)  such that X0=X  and for all  n>=0 , X(n+1) = f(Xn)

so   X(n+3)- 3X(n+2)+ 6 (Xn+1) = 4Xn +3.

now i take   for all n>= 0  , Yn= Xn-n . so it 's easy . we find:

Y(n+3)-3Y(n+2) + 6Y(n+1) - 4 Yn = 0

we find  the equation   X^3 - 3X^2  + 6X - 4 = 0  or  (X - 1)(X^2 - 2 X + 4) = 0.
so there is three solutions . they are  x= 1 or x = e^i/3  or e^-i/3

that's tha first part of solution think about the second part if you can't i 'll help you  and  the solution is  f(x)= x+1

i'm sorry because i don 't know write on LATEX but i' ll do .[/quote]

Have you read previous posts ?
Dichotomy "f is increasing" or "f is decreasing" is wrong.

[quote="mj93"]i got -3/5f LOL.

because couldn't you just simplify the original equage to

3fx - 6fx + 6fx = 4x + 3
3fx = 4x + 3
3x = 4fx + 3f
3fx = 8fx - 3
-5fx = 3
x = -3/5f

...? im sure this is totally wrong but i'd like to know where i messed up[/quote]

Hello,

$ f$ is not a kind of number. it's a function or a real variable : $ f(x)$
So $ f(f(f(x)))$ is $ f(y)$ where $ y\equal{}f(z)$ where $ z\equal{}f(x)$

So you can't do simplifications like $ f(f(f(x)))\equal{}3fx$ (which BTW would be wrong even if $ f$ was a real number and not a function).


Να δειχτεί στους θετικούς ότι:
$ \frac{x^3}{yz} \plus{} \frac{y^3}{zx} \plus{} \frac{z^3}{xy} \ge x \plus{} y \plus{} z$

Είναι κάπως εύκολη αλλά προυσιάζει ενδιαφέρον.

Βάζω κάτι σε Hide

[hide]Είναι ομογενής αρα εχομε δικαώμα να θεσουμε xyz=1 αρα η αχική μεταφράζεται σε $ x^4\plus{}y^4\plus{}z^4\ge{x\plus{}y\plus{}z}$ .Δεν ξέρω αν βοηθάει. [/hide]

Πολλαπλασιάζοντας με $ xyz$ και τα δυο μέλη έχουμε:
$ x^{4}\plus{}y^{4}\plus{}z^{4}\geq x^{2}yz\plus{}y^{2}xz\plus{}z^{2}yx$
Όμως   $ x^{4}\plus{}y^{4}\plus{}z^{4}\geq x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}z^{2}x^{2}$
Η οποία είναι εφαρμογή της βασικής $ a^{2}\plus{}b^{2}\plus{}c^{2}\geq ab\plus{}bc\plus{}ac$
Αρκεί λοιπόν $ x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}z^{2}x^{2}\geq x^{2}yz\plus{}y^{2}xz\plus{}z^{2}yx$

Η οποία όμως είναι επίσης εφαρμογή της βασικής $ a^{2}\plus{}b^{2}\plus{}c^{2}\geq ab\plus{}bc\plus{}ac$
για $ a\equal{}xy, b\equal{}xz, c\equal{}yz$

 :lol: QED

Συγχαρητήρια κ για το Θαλή! Τα λέμε αύριο στο μάθημα ρε Μανιέ :ninja:

Απόδειξη

Μπορούμε να θεωρήσουμε (πιστεύω) $ \dfrac{x}{y}\le{\dfrac{y}{z}}\le{\dfrac{z}{x}}$ και $ \dfrac{x^2}{z}\le{\dfrac{y^2}{x}}\le{\dfrac{z^2}{y}}$ και σύμφωνα με την ανισοτητα της αναδιάταξης καταλήγουμε στο $ \frac {x^3}{yz} \plus{} \frac {y^3}{zx} \plus{} \frac {z^3}{xy} \ge x \plus{} y \plus{} z$, με τους κατάλληλους πολλαπλασιασμούς .

[quote="maniopas"]Να δειχτεί στους θετικούς ότι:
$ \frac {x^3}{yz} \plus{} \frac {y^3}{zx} \plus{} \frac {z^3}{xy} \ge x \plus{} y \plus{} z$

Είναι κάπως εύκολη αλλά προυσιάζει ενδιαφέρον.[/quote]

$ \frac {x^3}{yz} \plus{} \frac {y^3}{zx} \plus{} \frac {z^3}{xy} \geq \frac {(x \plus{} y \plus{} z)^3}{3(xy \plus{} z \plus{} zx)}$

Αρα αρκεί

$ \frac {(x \plus{} y \plus{} z)^3}{3(xy \plus{} z \plus{} zx)}\geq x \plus{} y \plus{} z\Longleftrightarrow (x \plus{} y \plus{} z)^2\geq 3(xy \plus{} yz \plus{} zx)$ που ισχύει

[quote="maniopas"]Να δειχτεί στους θετικούς ότι:
$ \frac {x^3}{yz} \plus{} \frac {y^3}{zx} \plus{} \frac {z^3}{xy} \ge x \plus{} y \plus{} z$

Είναι κάπως εύκολη αλλά προυσιάζει ενδιαφέρον.[/quote]

Μετά την απλοποίηση έχουμε να δείξουμε ότι

$ \sum_{cyc} x^4 \geq \sum_{cyc}x^2yz$ δηλαδή ισοδύναμα

$ 2\sum_{cyc} x^4 \geq 2\sum_{cyc}x^2yz$ δηλαδή ισοδύναμα

$ \sum_{sym} x^4 \geq \sum_{sym}x^2yz$ που ισχύει από την ανισότητα Muirhead αφού

$ (4,0,0) \succ (2,1,1)$

Αλέξανδρος

[quote="maniopas"]Να δειχτεί στους θετικούς ότι : $ \frac{x^3}{yz} \plus{} \frac{y^3}{zx} \plus{} \frac{z^3}{xy} \ge x \plus{} y \plus{} z$[/quote]πάλι [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=742235#p742235]εδώ[/url]


Prove that by connecting, in order, the midpoints of the sides of any quadrilateral, a parallelogram is formed.

Hint: [hide]Think about drawing a diagonal of the quadrilateral created, and then using the diagonal.[/hide]

Anybody want to give this problem a try?

Hmm, I'll move this to the Mathcounts forum.

You have to have takena little geometry to figure this one out.  Draw each diagonal, then we can easily see that the opposite sides of this new quad are parallel, which makes it a prarallelorgram.

Theres a thereom I used here that states in triangle ABC, if points D,E on AB and AC respectiviely such that DB/AB=EC/AC, then DE||BC


I don't really understand how bases work. I don't get how you are supposed to add, subract, multiply, or divide bases. I also don't know how to change from one base to another. Can anyone help me???

Is this similar to the binary, ternary etc systems?
If so I might be able to help a bit.

Well for base 2, there are only 2 numbers: 0 and 1
You count like
0, 1, 10 (since there is no 2, we go to the next possible value), 11, 100, etc.
Changing it into base 10 works like this:
Ex: 100
0*2^0+0*2^1+1*2^2
Ex: 1101
1*2^0+0*2^1+1*2^2+1*2^3

:huh: I don't get a single thing. Can you explain more clear please and with other numbers :?:

take 34 (for example) normally we would write 34 because 34 consists of 3 x 10^1
and 4 x 10^0

in binary 34 would be

100010

because 34 consists of
1 x 2^5
0 x 2^4
0 x 2^3
0 x 2^2
1 x 2^1
0 x 2^0

I hope that that makes some sense.   :)

Suppose you are working in base "n". The first digit(this is from the left of the base ten-decimal point) would be the value of $ n^0$, the second digit would be $ n^1$, and so on. Also, all digits in base "n" has to be less than "n".

Let's say you have the number $ {\overline{a_ma_{m-1}\ldots a_3a_2a_1a_0}}_n$.
The number in base 10 would be $ a_0 \cdot n^0 + a_1 \cdot n^1 + a_2 \cdot n^2 + a_3 \cdot n^3 + \ldots$

For example, if you have $ 3025_3$, in base 10, it's:
\begin{align*}
3025_3 &= \left(5 \cdot 3^0 + 2 \cdot 3^1 + 0 \cdot 3^2 + 3 \cdot 3^3\right)_{10} \\
&= \left(5+6+81\right)_{10} \\
&=92_{10}
\end{align*}


Find all positive integers $n$ such that $n^n+1$ is divisible by 3.

Is this a joke or something? :)

Did you mean $n^n+1$? (you wrote $n^n$, and I assumed it was $n^n+1$ because that's what you wrote in the title). First of all, $n\equiv 2\pmod 3$, and since $2^2\equiv 1\pmod 3$, $n$ must also be odd, so the answer is all $n$ of the forms $6k+5$? :?

To moderator:Please move this problem to pre olympiad section or intermidate section.

In Japan we aren't taught $mod$ sign in senior high school.Almost students are poor at Number Theory.That's why I sometimes confusing the level of problem.

kunny

I think, for most places in the world, mod is not taught as a normal high school math syllabus.

[quote="Myth"]Anyway the problem is solved.
If you wish I will move it to pre-olympiad...[/quote]

Thank you,Myth.

kunny


When I do some physics derivation, I find that on a seesaw, if the object is farther away from the fulcrum, the angular acceleration decreases. Is this true? If not, where did I go wrong?
$ \tau \equal{} I \cdot \alpha$
$ \tau\equal{}F \cdot r$
Then, $ \alpha \equal{} \frac{F \cdot r}{I} \equal{} \frac{F_g \cdot r}{mr^2}\equal{}\frac{g \cdot m \cdot r}{mr^2}\equal{}\frac{g}{r}$, so if $ r$ increases, angular acceleration decreases...did I do something wrong?

I think the problem lies in $ I$.  $ I\equal{}\int r^{2}\ dm$, which is not necessarily equal to $ mr^{2}$.  You need to use the parallel-axis theorem to get an accurate measure of $ I$.

Oh, ok. So in this case, if I apply that, angular acceleration should increase with length, right?

For this particular instance, could you derive me the proof?

Can't derive anything because there is no concrete information.

Angular acceleration does indeed decrease as the object is moved further from the
fulcrum.  For simplicity, consider a case of a point mass on a horizontal, massless seesaw.
As you've derived (the value of $ I$ is valid when everything except the point mass
in question is massless), the angular acceleration decreases with $ r$.

Consider what it would mean if it didn't.  Linear acceleration in this case is $ r\alpha$.
If $ \alpha$ increased with $ r$ (or even stayed constant, or even decreased as,
say, $ \sqrt{r}$), then linear acceleration would also increase with $ r$.  However,
the only force acting on the mass is gravity, which gives it linear acceleration $ a\equal{}g$,
constant with respect to $ r$.

In the case of a non-massless seesaw, $ I$ is no longer equal to $ mr^2$, but we
still see that $ I\equal{}I_0\plus{}mr^2$ for some $ I_0$ that represents the moment of the
seesaw.  In this case, $ \alpha\equal{}\frac{mgr}{I_0\plus{}mr^2}\equal{}\frac{mg}{\frac{I_0}{r}\plus{}mr}$.
This achieves a maximum for some $ r_0$ (specifically, where $ \frac{I_0}{r}\plus{}mr$ has a
minimum, which in turn occurs at $ r_0\equal{}\sqrt{\frac{I_0}{m}}$, and has decreasing
angular acceleration as you move away from $ r_0$.

But how come this doesn't make logical sense?

What part of it doesn't make sense?  (Or, what part of the conclusion doesn't make sense?)

Remember that, the further something is from the center of angular motion,
the faster is has to be traveling to achieve the same angular velocity.


LaTeX
analytic geometry
parameterization

Hi,

I am using WinEdt for typing in Latex. I have inserted picture file using
\includegraphics{filename}.

It is working but a space is getting created just below the picture in the pdf file.

plz help

[quote="stevem"]I think you need to give us much more information than that. What sort of space, what sort of graphics file, is the space part of the image, what is the code nearby, what graphics package are you using, are you compiling via PS etc etc?

What happens if you use \includegraphics* which clips the graphics outside the bounding box of an eps file? What about using the coordinates of the bounding box?

It would all be so much simpler to solve if you gave us a minimal document with a sample graphics file which featured the problem.[/quote]

I drew a circle using Paint(Windows) and then converted it into .eps file.
I enclosed it like
\begin{figure}
\includegraphics[width=3.0in]{file.eps}
\end{figure}

I have added \userpackage{graphicx}.

After the figure some white space are being displayed. I am using dvitopdf. I am attaching both the files.

[quote="stevem"]You forgot to include the files so I can't see your document or graphics.

Is the figure environment necessary ie what happens if you don't use \begin{figure}...\end{figure} or is floating graphics necessary for your document? Can the text go in the figure environment? In other words need to know more about the surrounding text.

There are other graphics optional parameters which may help once you've attached the file.[/quote]

I tried to attach files but .eps and .bmp files are not allowed here it seems.
It also works without the \begin{} \end{figure}.

I'm not surprised that eps file doesn't work:
1. there isn't a bounding box and/or it is corrupt since the log says [i]! Expected to see %%EndBinary at end of data[/i]
2. the bmp consists of a circle diagram in the corner of the page - the figure should just be the circle itself without the blank space to the right and below it
3. Fig 1.3 should not be part of the image but should be added by LaTeX to keep track of the numbering.

You don't say how you created the eps image but I would recommend OpenOffice Drawing, which exports very good quality eps images, to recreate the image. Or, of course, you could use Paint to save it as a png then use pdflatex to compile (not dvi2pdf).


limit
topology
calculus
integration
advanced fields
advanced fields unsolved

Suppose that $E \subset \mathbb{R}^{2}$ is a 1-set, i.e. $0 < \mathcal{H}^{1}(E) < \infty$.

We say that $x \in E$ is a [i]regular point[/i] of $E$ if
$\lim_{r \to 0}\frac{\mathcal{H}^{1}(B_{r}(x) \cap E)}{2r}= 1$
and $x \in E$ is an [i]irregular point[/i] of $E$ otherwise.

Let $R(E)$ be the set of regular points of $E$. It is not hard to show that almost every point of $R(E)$ is a regular point of $R(E)$. How large can the exceptional set be? Is it possible to construct a set $E$ such that $R(E) \setminus R(R(E))$ has Hausdorff dimension 1?

Now, denote $R^{n}(E) = R(R(\cdots R(E)\cdots ))$ ($n$ times). Can we construct a set $E$ such that $R^{n}(E) \supsetneq R^{n+1}(E)$ for all $n \ge 0$?

One way to make $R(E)\setminus R(R(E))$ nonempty is to arrange tiny copies of an irregular set (such as the $(1/2)\times (1/2)$ Cantor product) converging to a point $x$ so that $x$ is regular for the union of the irregular sets. Then $x$ is an isolated point of $R(E)$. The construction extends in an obvious way to a countable $R(E)\setminus R(R(E))$.

Do you have an example with uncountable $R(E)\setminus R(R(E))$?

[quote="mlok"]
Do you have an example with uncountable $R(E)\setminus R(R(E))$?[/quote]
Well, the complement of every closed set $F$ on the line is a union of intervals, so just construct an appropriate necklace of purely irregular sets for each interval to get an example where $R(E)=F$. The next logical step would be to construct a set $E$ with purely irregular $R(E)$ of positive 1-dimensional Hausdorff measure. Can we do it?

[quote="fedja"]The next logical step would be to construct a set $E$ with purely irregular $R(E)$ of positive 1-dimensional Hausdorff measure. Can we do it?[/quote]
This shouldn't be possible. By definition, $E \setminus R(E)$ would be purely irregular. Then $E$ is the union of two purely irregular sets, so it would still be irregular*. But then $R(E)$ was actually measure 0....

*Here I am using the fact that any irregular 1-set in the plane intersects with each rectifiable curve in a set of measure 0. There are weaker generalizations in higher dimensions, but I am not familiar enough with them to know if this would still prevent what you're trying to do. If the dimension of $E$ is not an integer, then the set of regular points of $E$ has to be of measure 0. (I didn't study sets of non-integral dimension in detail so I am not sure the details of the obstacle here or how big that set can be. But it can't be very big.)

Correct (actually, almost :P: $E\setminus R(E)$ is irregular not quite by definition; by definition, it consists of irregular points of $E$ but some extra work is needed to show that almost all of them are irregular points of $E\setminus R(E)$ as well). Now, since, as you mentioned from the beginning, $R(E)=R(R(E))$ up to a set measure $0$ and adding a set of measure $0$ to $X$ doesn't change the set $R(X)$, we must have $R(R(R(E)))=R(R(E))$ and so on, which answers your last question. :)

[quote="fedja"]Correct (actually, almost :P: $E\setminus R(E)$ is irregular not quite by definition; by definition, it consists of irregular points of $E$ but some extra work is needed to show that almost all of them are irregular points of $E\setminus R(E)$ as well). [/quote]
Well I should think if $F \subset E$ then $\mathcal{H}^{1}(B_{r}(x) \cap F) \le \mathcal{H}^{1}(B_{r}(x) \cap E)$ and that's all the extra work we need...  :wink: But yes I guess I should be a bit more precise with my words. :P


trigonometry
ceiling function
logarithms
function
floor function
puzzles

Here are two quickies:

1. Given 3 4, put any one standard mathematical symbol between them so that the value is between 3 and 4.

-----

2. Given two 2's and any standard mathematical symbols make 5.

[quote="NoSoupForYou"]heh heh.

$[2\cdot\sqrt{\tan(\tan(\tan(2)))}]=5$ This is in radians.[/quote]That would be, I think $[2\cdot\sqrt{\tan(\tan(\tan(2)))}]\approx5$, and with very low pressition, because according to a calculator, you have
$[2\cdot\sqrt{\tan(\tan(\tan(2)))}]=5.09180873144569597806011126838$.

Ohhhh, OK, you mean
$\left\lfloor2\cdot\sqrt{\tan(\tan(\tan(2)))}\right\rfloor=5$

I was thinking $\log_2 \sqrt{\sqrt{\sqrt{2}}}$

[quote="randomdragoon"]There are probably several solutions using the round/up down functions...

Can you get exactly 5 with two 2's without using any of the rounding functions?[/quote]

No. I've been trying for the past several minutes and am convinced that if there is a solution it will require the use of functions so esoteric that the number of AoPSers who have even heard of them is in the low single-digits.

2+2 = 5 according to some t-shirts. perhaps they have defined '+' in an unconventional way :D

[quote="djimenez"][quote="NoSoupForYou"]$\frac{\lceil sin(2) \rceil}{\lceil ln(tan(tan(tan(2)))!) \rceil}=\frac{1}{8}$
Not pretty and probably not valid, but it's the shortest method I could find.[/quote]NoSoupForYou,

Do you by any chance work (or study) on either Approximation or in Computer Science? It is just that is funny, you tend to use truncations a lot! :D.[/quote]

No, I don't work in Approximation or in Computer Science. I just use $\lceil\text{  }\rceil$ and $\lfloor\text{  }\rfloor$ because I'm too lazy to actually think about the problem.  :D

[quote="ThAzN1"][quote="randomdragoon"]There are probably several solutions using the round/up down functions...

Can you get exactly 5 with two 2's without using any of the rounding functions?[/quote]

$2+\psi(2) = 5$.  See http://mathworld.wolfram.com/DedekindFunction.html[/quote]

Wow. How is it that I get the feeling that nobody really knows about this function... ;) Good work, though.

Oh yeah, well how about this: !2+2$=5

http://mathworld.wolfram.com/Superfactorial.html
http://mathworld.wolfram.com/Subfactorial.html

How about $F_{2\cdot 2}$, where $F_n$ is the nth Fibonacci number?  (You have to start $F_0 = F_1 = 1$ for this to work, of course.)

[quote="randomdragoon"]hmm... I was actually thinking of

[hide] sqrt(.2^-2)[/hide][/quote]

Oh. That's pretty cool.


Real numbers $ a, b, c$ satisfy $ |a|\geq 2$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} abc \plus{} 4$. Show that
there exist real numbers $ x$ and $ y$ such that

\[ a\equal{}x\plus{}\frac{1}{x},b\equal{}y\plus{}\frac{1}{y},c\equal{}xy\plus{}\frac{1}{xy}.\]

It's false.
Take $ a\equal{}2,b\equal{}0,c\equal{}0$.
The additional conditions $ |b|\geq{2},|c|\geq{2}$ are also necessary. We have a look on the converse:
then $ a\equal{}x\plus{}1/x,b\equal{}y\plus{}1/y,c\equal{}z\plus{}1/z$ with $ xyz\not\equal{}0$. $ a^2\plus{}b^2\plus{}c^2\equal{}abc\plus{}4$ is equivalent to $ (x\minus{}yz)(xy\minus{}z)(zx\minus{}y)(xyz\minus{}1)\equal{}0$. Thus we have not necessarily $ z\equal{}xy$ and we can throw this exercise.


Find all ordered pairs of positive integers $ (m,n)$ such that $ 1<m<n<10$ and
\[ S_{mn}\equal{}\frac{1}{m}\plus{}\frac{1}{m\plus{}1}\plus{}\frac{1}{m\plus{}2}\plus{}\cdots \plus{}\frac{1}{n} > 1
\]
but $ S_{mn}\minus{}\frac{1}{n}<1$.

[hide="Answer"]$ (2,4), (3,7)$[/hide]

Thanks in advance!

*Bump* Could someone please help me? Thanks!

If $ m\ge 4$ then $ \frac{1}{4}\plus{}\frac{1}{5}\plus{}\frac{1}{6}\plus{}\frac{1}{7}\plus{}\frac{1}{8}\plus{}\frac{1}{9}<1$. Hence there are no solutions for $ m\ge 4$.

Then just check $ m\equal{}2$ and $ m\equal{}3$ to get your solutions.


Would anyone be interested in making a fantasy baseball league this year? Right now, I think we should use this league http://www.mlb.com/mlb/fantasy/wsfb_c/info/index.jsp because it's probably the easiest to use and keep track of, and it's free. We would need twelve teams, and I'd like to screen entrants, just to make sure that we don't get people signing up and then not participating at all, so I don't really want anybody who just signed up last week and has 13 posts. If you want to participate, please post here, and suggestions for different websites or whatever are welcome.

Sounds like lots of fun..
I don't think I would be able to do it, though. How many people do you need?

What [i]is[/i] this exactly? Because I don't want to be the guy who screws the whole game up just because I don't know what to do...

@sailingdog- We would need twelve people

@1=2- Fantasy baseball is kind of like being a manager of a baseball team, you draft major league players, and then gain points based on how well they do during the season. This is a highly simplified version, because I doubt that there are enough hardcore baseball fans here to pay for a more complicated version.

This sounds great. We should do a fantasy football in the fall though.

[quote="mustafa"][quote="1=2"][quote="Klebian"]sure
don't follow baseball much but that's ok[/quote]

Ditto

In, but I won't be here Saturday.[/quote]

That's ok, the season doesn't start for a couple of weeks[/quote]

Correction: Next Saturday

If I make Nationals, I won't be very active.

[quote="mustafa"]Ok, the season starts fairly soon, so I made a league, hopefully a few more people will join up when everything is ready to go.

For anyone that wants to join, go here http://www.mlb.com/mlb/fantasy/wsfb/registration/step1.jsp
The league name is aops, and the password is mustafarocks

The draft is going to be a list draft, meaning that you rank all the players and then when everyone is ready, the computer drafts based on your rankings. The players are preranked for you, so that you don't have to spend an hour ranking everyone, but you can switch players around if you like some people more than others.[/quote]

Well, I made it. I will try to be active....

And I entered my real DoB, and it said I'm not eligible.

Edit: I joined on my own computer, and is this a public or private league?

[quote="distracted523"]Hmm:
[quote="MLB site"]

Sorry, no league with those credentials could be found. Please return to the previous step and try again.

[/quote]

Is it already full?[/quote]

that's the same message I got.

Sorry, I think you need the league id number, 52685, instead of aops. Does that work?

me and my brother signed up ...
anywhere else you could post this to advertise?
we need to start soon, lol
like, this week

does anyone in the league know people that would be willing to play? not just aops people, but anyone.

if you do, go ahead and invite them ...


[color=darkblue]$ (a_n)$ is sequence, satisfying:

$ \bullet \ a_0 \equal{} 1$
$ \bullet \ a_{n \plus{} 1} \equal{} \sum_{i \equal{} 0}^n a_i a_{n \minus{} i}$

Is it possible to find the general term of the sequence (I mean not recursive)?[/color]

Let $ P(x) \equal{} \sum_i a_ix^i$, then $ P^2(x) \equal{} \sum_k x^k\sum_{i \equal{} 0}^ka_ia_{k\minus{}i} \equal{} \sum_k a_{k \plus{} 1}x^k \equal{} \frac {P(x) \minus{} 1}{x}$.
Therefore $ P(x) \equal{} \frac {1 \minus{} \sqrt {1 \minus{} 4x}}{2x} \equal{} 1 \plus{} x \plus{} ...$ and $ a_n \equal{} Ca_n \equal{} \frac {C_{2n}^n}{n \plus{} 1}$.


Determine the number of ways of constructing a $ 4 \times 4 \times 4$ block from 64 unit cubes, exactly 16 of which are red, so that there is exactly one red cube within each $ 1 \times 1 \times 4$  sub-block in any orientation.

This is easily equivalent to the number of $ 4 \times 4$ [url=http://mathworld.wolfram.com/LatinSquare.html]Latin squares[/url].  There is no nice formula known in general, and the [url=http://www.research.att.com/~njas/sequences/A002860]OEIS[/url] has only 11 terms.  This is also the number of "pairwise deranged" permutations, i.e. the number of tuples $ (\pi_1, \pi_2, \ldots, \pi_n)$ of permutations of length $ n$ such that $ \pi_i(k) \neq \pi_j(k)$ for all $ i \neq j$, $ i, j, k \in \{1, \ldots, n\}$.

This question (although sufficiently hard on its own terms) suggests the following generalization:

How many ways are there to mark boxes of a $ d$-dimensional cube of side $ n$ in such a way that every line parallel to the axes contains exactly $ k$ marked boxes?

Thanks for the Latin Squares website.  :)  I took 20 minutes to count all possibilities in this problem, it was actually the last problem of a Chinese Mathematical Olympiad Paper II (forgot which year).


Is there another way to do this without graph paper?  Please download the pdf. for the problem.  Thanks.

The distance from B to C is 3 units.  Therefore the after the triangle is rotated 90 counterclockwise the new point has to be 3 units to the left of B,  ((1,1)


function
integration
calculus
topology
real analysis
real analysis unsolved

How do I know there is a $C^\infty$ function f: R->[a,b], such that f(t)=a for t<1/3, and f(t)=b for t>2/3?

Let's start with $f(x)=\begin{cases}e^{-1/x},&x>0\\0,&x\le0\end{cases}.$

Now build $\varphi(x)=cf(x+1)f(1-x),$ with $c$ chosen so that

$\int_{-\infty}^{\infty}\varphi(x)\,dx= \int_{-1}^{1}\varphi(x)\,dx=1.$

We've built ourselves a smooth, nonnegative, compactly supported "bump" function, normalized to integral 1. This can be further manipulated in a number of ways. To get the example you want, I'd take

$g(x)=a+6(b-a)\int_{-\infty}^{x}\varphi(6t-3)\,dt$

That should do it.

A phrase to look for in the index of any of a variety of analysis or topology books: "partition of unity" or "smooth partition of unity."


integration
real analysis
real analysis unsolved

$ I \equal{} \int_{0}^{ \plus{} \infty}x e^{ \minus{} x^2}dx$

$ I \equal{} \left[\minus{} \frac {1}{2}e^{ \minus{} x^2}\right]_0^{\infty} \equal{} \frac {1}{2}$.


What is the value of $ 5 \times 5 \times 0 \plus{} 5 \plus{} 5 \plus{} 5$?

We see that $ 5\cdot5\cdot0\plus{}5\plus{}5\plus{}5\equal{}0\plus{}3\cdot5\equal{}\boxed{15}$.


We should create a classroom math marathon i think. Anyone can post a question

Marathons in this forum are temporarily banned.

Read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23215]this[/url] for more info.


Prove that, for any positive real numbers $a,b$ and $c$,
\[ \frac{b^2+c^2}{a(b+c)}+\frac{c^2+a^2}{b(c+a)}+\frac{a^2+b^2}{c(a+b)}\ge(a^2+b^2+c^2)\sqrt{\frac{3}{abc(a+b+c)}} \]


How many real numbers $x$ satisfay  the equation

$\frac{1}{5}log_2x=sin\left( 5\pi x\right)$


trigonometry
ratio
calculus
integration
quadratics
algebra
number theory

Find all integer values of $x$, where $0^\circ < x < 90^\circ$ which satisfy the equation $\sin{x} = \sin{(x^2)}$

In $\triangle{ABC}$, the ratio of $\sin{(A)} : \sin{(B)} : \sin{(C)} = 4 : 5 : 6$, while the ratio of $\cos{(A)} : \cos{(B)} : \cos{(C)} = x : y : 4$. Compute the value of $x$ and $y$.

for 1:
[hide]
$x=x^2 \rightarrow x=0,1$
or
$180-x=x^2 \rightarrow x^2+x-180=0$
so $x \in (0,1,\frac{1 \pm \sqrt{721}}{2})$

Edit--it said integer values, so not the last one.
[/hide]

I think $x$ is supposed to be in degrees...

well out of the 4 ansewrs i got, 3 of them don't work since 2 aren't integer and 0 is not in 0<x<90.
so is 1 degree the only answer?

[quote="matematikator"]1. There are two conditions:
          i) sina=sin(180-a):    x+x^2=180 no integer solution.
         ii) x=x^2 :                x=1 is the only solution.

2. Let the sides corresponding to the angles A,B and C be a,b,c;
  *    Using the proportion: a=4k, b=5k, c=6k
 **   Using the law of cosine: x=3/8,   y=9/32[/quote]

How did you get $a=4k$, etc? Could you write it out with a little more detail?

*Use the law of sine: a/sinA = b/sinB = c/sinC and by the proportion  a=4k,b=5k,c=6k
**The law of cosine : cosA=3/4, cosB=9/16 and cosC=1/8 and by the proportion
    x=24, y=18

A question for solvers of number 1:  What is $\sin{405^\circ}$?

For those interested, I find [hide="positive_integer"]4[/hide] distinct solutions strictly between 0 and 90.  There are [hide="larger_positive_integer"]16[/hide] solutions in the interval $[0^\circ, 360^\circ)$, and all other integral solutions are a multiple of 360 degrees from one of these.

Matematikator, you have the right values for the cosine of the angles. However, you did not answer the question that was asked. It said the ratio was of the form $x:y:4$. Therefore, you have to multiply all fractions by 32 to achieve this ratio, from which you find $x=24,y=18$.

i think, if you cannot find any other ways to solve it, then try the law of sine and cosine.

[color=blue][b](1)[/b][/color] If $\sin x = \sin{x^2}$, where x is in degrees, then either $x^2 = x + 360k$ or $x^2 = 180 - (x - 360k)$, where $k$ is an integer.

[color=red][b](a)[/b][/color] Consider the 1st case. We get a quadratic equation for x:

$x^2 - x - 360k = 0$

which has roots $x = \frac{1 + \sqrt{1 + 1440k}}{2}$. For x to be an integer, $1 + 1440k = n^2$ must be positive and a full square. Since $0 < x < 90$, $0 < x^2 < 8100$ and $0 \le 360k < 8100$, $0 \le k < \frac{8100}{360} = \frac{45}{2}$. Hence, $0 \le k \le 22$. Let's factor $1440k$:

$1440k = 12^2 \cdot 10 = 2^5 \cdot 3^2 \cdot 5 \cdot k$

Thus $n^2 = 1 + 1440k$ cannot not have factors 2, 3, 5. Since the unit digit of its square is 1, the unit digit of n must be 1 or 9. In addition, $n \le \sqrt{1 + 1440 \cdot 22} = \sqrt{31681} < \sqrt{31684} = 178$. Thus n must either be 1 or it must have a prime factor greater than 5 and end with the digit 1 or 9. As $7^2 \cdot 11 = 539, 7 \cdot 43 = 301, 7 \cdot 43 = 301, 11 \cdot 19 = 209$ are all greater than 178, the possibilities for n are

$7^2 = 49$, $7 \cdot 13 = 91$, $7 \cdot 17 = 119$, $7 \cdot 23 = 161$, $11^2 = 121$, $13^2 = 169$

and the primes between $5 < p < 178$ ending with 1 or 9, i.e., 11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, 131, 139, 149, 151. This is too many possibilities, so it is better to try all integers $0 \le k \le 22$ to see if we get a full square:

$1 + 1440 \cdot 0 = 1$
$1 + 1440 \cdot 1 = 1441 = 11 \cdot 131$
$1 + 1440 \cdot 2 = 2881 = 43 \cdot 67$
$1 + 1440 \cdot 3 = 4321 = 29 \cdot 149$
$1 + 1440 \cdot 4 = 5761 = 7 \cdot 823$
$1 + 1440 \cdot 5 = 7201 = 19 \cdot 379$
$1 + 1440 \cdot 6 = 8641$ [color=white].[/color] is a prime
$1 + 1440 \cdot 7 = 10081 = 17 \cdot 593$
$1 + 1440 \cdot 8 = 11521 = 41 \cdot 281$
$1 + 1440 \cdot 9 = 12961 = 13 \cdot 997$
$1 + 1440 \cdot 10 = 14401$[color=white].[/color] is a prime
$1 + 1440 \cdot 11 = 15841 = 7 \cdot 31 \cdot 73$
$1 + 1440 \cdot 12 = 17281 = 11 \cdot 1571$
$1 + 1440 \cdot 13 = 18721 = 97 \cdot 193$
$1 + 1440 \cdot 14 = 20161$ [color=white].[/color] is a prime
$1 + 1440 \cdot 15 = 21601$ [color=white].[/color] is a prime
$1 + 1440 \cdot 16 = 23041$ [color=white].[/color] is a prime
$1 + 1440 \cdot 17 = 24481$ [color=white].[/color] is a prime
$1 + 1440 \cdot 18 = 25921 = 7^2 \cdot 23^2$
$1 + 1440 \cdot 19 = 27361$ [color=white].[/color] is a prime
$1 + 1440 \cdot 20 = 28801 = 83 \cdot 347$
$1 + 1440 \cdot 21 = 30241$ [color=white].[/color] is a prime
$1 + 1440 \cdot 22 = 31681 = 13 \cdot 2437$

The only full squares we got are for $k = 0$ and $k = 18$. Hence

$x^2 - x = 0$
or
$x^2 - x - 25920 = 0$

Since by the problem condition $x > 0$, the non-positive roots are not acceptable. The 1st quadratic equation has the positive root 1 and the second one

$x = \frac{1 + \sqrt{1 + 1440 \cdot 18}}{2} = \frac{1 + \sqrt{25921}}{2} = \frac{1 + 7 \cdot 23}{2} = \frac{1 + 161}{2} = 81$

[color=red][b](b)[/b][/color] The 2nd case is handled in an entirely similar way. We get a quadratic equation for x:

$x^2 + x - 180 - 360k = 0$

which has roots $x = \frac{-1 + \sqrt{1 + 720 + 1440k}}{2}$. For x to be an integer, $1 + 720 + 1440k = n^2$ must be positive and a full square. Similarly as in the 1st case, we can limit k to $0 \le k \le 22$. Let's factor $720 + 1440k$:

$720 + 1440k = 2^4 \cdot 3^2 \cdot 5 \cdot (1 + 2k)$

Again, $n^2 = 1 + 720 + 1440k$ does not have factors 2, 3, 5 and we can make similar conclusions about n. Trying all integers $0 \le k \le 22$ to see if we get a full square:

$1 + 720 + 1440 \cdot 0 = 721 = 7 \cdot 103$
$1 + 720 + 1440 \cdot 1 = 2161$ [color=white].[/color] is a prime
$1 + 720 + 1440 \cdot 2 = 3601 = 13 \cdot 277$
$1 + 720 + 1440 \cdot 3 = 5041 = 71^2$
$1 + 720 + 1440 \cdot 4 = 6481$ [color=white].[/color] is a prime
$1 + 720 + 1440 \cdot 5 = 7921 = 89^2$
$1 + 720 + 1440 \cdot 6 = 9361 = 11 \cdot 23 \cdot 37$
$1 + 720 + 1440 \cdot 7 = 10801 = 7 \cdot 1543$
$1 + 720 + 1440 \cdot 8 = 12241$ [color=white].[/color] is a prime
$1 + 720 + 1440 \cdot 9 = 13681$ [color=white].[/color] is a prime
$1 + 720 + 1440 \cdot 10 = 15121$ [color=white].[/color] is a prime
$1 + 720 + 1440 \cdot 11 = 16561$ [color=white].[/color] is a prime
$1 + 720 + 1440 \cdot 12 = 18001 = 47 \cdot 383$
$1 + 720 + 1440 \cdot 13 = 19441$ [color=white].[/color] is a prime
$1 + 720 + 1440 \cdot 14 = 20881 = 7 \cdot 19 \cdot 157$
$1 + 720 + 1440 \cdot 15 = 22321 = 13 \cdot 17 \ 101$
$1 + 720 + 1440 \cdot 16 = 23761$ [color=white].[/color] is a prime
$1 + 720 + 1440 \cdot 17 = 25201 = 11 \cdot 29 \cdot 79$
$1 + 720 + 1440 \cdot 18 = 26641$ [color=white].[/color] is a prime
$1 + 720 + 1440 \cdot 19 = 28081$ [color=white].[/color] is a prime
$1 + 720 + 1440 \cdot 20 = 29521 = 53 \cdot 557$
$1 + 720 + 1440 \cdot 21 = 30961 = 7 \cdot 4423$
$1 + 720 + 1440 \cdot 22 = 32401$ [color=white].[/color] is a prime

The only full squares we got are for $k = 3$ and $k = 5$. Hence

$x^2 + x - 5040 = 0$
or
$x^2 + x - 7920  = 0$

The 1st quadratic equation has the positive root

$x = \frac{1 + \sqrt{1 + 720 + 1440 \cdot 3}}{2} = \frac{-1 + \sqrt{5041}}{2} = \frac{-1 + 71}{2} = 35$

and the second one

$x = \frac{1 + \sqrt{1 + 720 + 1440 \cdot 5}}{2} = \frac{-1 + \sqrt{7921}}{2} = \frac{-1 + 89}{2} = 44$

As a result, the original equation $\sin x = \sin{x^2}$ has 4 positive integer roots: $x = 1^o$ or $35^o$ or $44^o$ or $81^o$. Do not believe anybody in the whole world when faced with a problem like this! :lol:

[b]Note:[/b] I certainly did not do the prime factorizations by hand, there are various prime factorization utililities on the web, for example [url=http://www.rsok.com/~jrm/factor.html]Prime Factors of Integers[/url] by John Moyer.

[quote="PeNtaGoNaL"]In $\triangle{ABC}$, the ratio of $\sin{(A)} : \sin{(B)} : \sin{(C)} = 4 : 5 : 6$, while the ratio of $\cos{(A)} : \cos{(B)} : \cos{(C)} = x : y : 4$. Compute the value of $x$ and $y$.[/quote]
Let $a = BC, b = CA, c = AB$ be the triangle sides, $\alpha = \measuredangle A, \beta = \measuredangle B, \gamma = \measuredangle C$ and $R$ the circumradius. By the extended law of sines,

$2R = \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$

Consequently, the sides $a, b, c$ are in the ratios of sines of the opposite angles:

$a : b : c = \sin \alpha : \sin \beta : \sin \gamma$

Using cosine theorem for the triangle $\triangle ABC$ three times, each time with a different angle,

$a^2 = b^2 + c^2 - 2bc \cos \alpha$

$b^2 = c^2 + a^2 - 2ca \cos \beta$

$c^2 = a^2 + b^2 - 2ab \cos \gamma$

Calculating the cosines, which obviously depend only on the ratios of sides,

$\cos \alpha = \frac{b^2 + c^2  - a^2}{2bc} = \frac{25 + 36 - 16}{60} = \frac{45}{60} = \frac{3}{4}$

$\cos \beta = \frac{c^2 + a^2  - b^2}{2ca} = \frac{36 + 16 - 25}{48} = \frac{27}{48} = \frac{9}{16}$

$\cos \gamma = \frac{a^2 + b^2  - c^2}{2ab} = \frac{16 + 25 - 36}{40} = \frac{5}{40} = \frac{1}{8}$

Thus, after expanding by 32, in order to make $z = 4$ in the ratio $\cos \alpha : \cos \beta : \cos \gamma = x : y : z$,

$\cos \alpha : \cos \beta : \cos \gamma = \frac{3}{4} : \frac{9}{16} : \frac{1}{8} = 24 : 18 : 4$

[quote="RedSickle&Hammer"]The values of sine never repeat from 0 to 90, so we can say x=x^2, where the only answers are 1 and 0.  Since it has to be greater than 0, the only answer is 1.[/quote]

Please go read yetti's response above.


What is the sum of the digits of the decimal representation of $\frac{10^{22}+8}{9}$?

[quote="mstoenescu"][hide]$23$[/hide][/quote]

Solution?

How will anyone who doesn't know how to do this problem find out unless someone posts a solution? (this isn't supposed to mean what it literally means)

[quote="mstoenescu"]FINITE INDUCTION: we begin with $10^2+8$ and we get $12$ as result so the answer is $1+2=3$
then $1008: 9=112$
        $10008: 9=1112$ etc.
we see that we have [b]the same[/b] number of zeros in $10008$ and of $1$ in $1112$(three times) and so on...
But $10^{22}+8$ has $21$ zeros then$\frac{10^{22}+8}{9}$ has $21$ of $1$ too !
Finally we get $21+2=23$[/quote]

[hide]my way is sad. I did 1000000000....8/9 actually great when you are bored :) :P[/hide]


inequalities
Triangle
three variable inequality
geometric inequality

[quote="hungvuong"]please help me to solve this problem
Let a,b,c be sides of triangle with a :le: b :le: c
prove that a/b+b/c+c/a+3 :le: 2*(a/c+c/b+b/a)
thanks :)[/quote]

It is obvious that a/c+c/b+b/a >= 3 (*) by AM-GM

Then consider a^2b + b^2c + c^2a >= a^2c + c^2b + b^2a which follows from (a-c)(b-c)(a-b) >= 0

Upon dividing both sides of the latter inequality by abc and adding it to (*), the result follows.

[quote="Ji Chen"]

$F(a,b,c)=F(x+y+z,x+z,x+y)$

$=(k-1)(y^{2}-yz+z^{2})x+(k-1)y^{3}-ky^{2}z+yz^{2}+(k-1)z^{3}$

$=(k-1)(y^{2}-yz+z^{2})x+\frac{u(322y+vz)(322y-wz)^{2}}{1535767408}\geq 0$,
[/quote] Wow. Looks impressive,  :cool: but where is $k$ in your last big fraction? It seems like $u=k-1$, but $k-1$ doesn't factor out  :?: Should it be something like $k-1-u$ instead of $u$, and $v$ and $w$ should also depending on $k$?  :lol:
And how did you find the factors?

[quote="hungvuong"]please help me to solve this problem
Let a,b,c be sides of triangle with a :le: b :le: c
prove that a/b+b/c+c/a+3 :le: 2*(a/c+c/b+b/a)
thanks :)[/quote]

Vasile Critoaje's inequality!


An art teacher buys  exactly 100 items at this sale, and the total cost before tax is 100. She buys at least one of each item. How many of each item does she buy?

Markers= 0.50
Sketch pads= 3.00
Paint sets= 10.00

can some1 think of any other ways to solve it other than using guess and check?
if u dont its k
u can just answer the question!
coz i no the answer using guess and check.

Let m equal the number of markers, p the number of paint sets, and s the number of sketchpads. then
$p+m+s=100$  (:blush:  no pun intended) which implies
$10p+10m+10s=1000$
$10p+.5m+3s=100$ .  Subtracting gives
$9.5m+7s=900$ which is the same as
$m=\frac{1800-14s}{19}=94-\frac{14(s-1)}{19}$ Therefore for m to be an integer, s=20. Then,
$m=94-\frac{14(20-1)}{19}=80$ and
$10p=100-.5*80-3*20=0$ gives $p=0$
[b]The teacher bought 80 markers, 20 sketchpads, and no paintsets.[/b]
 :D

Note that dollar signs can mess up a post due to the fact that dollar signs trigger TeX in the forums.  i simply removed the dollars signs from this post.

With the numbers I was given, this is the only valid solution.  There is no way she could have bought one of each.  The next possible value of s is 39 which would cost 117.

ummmmmmmmmmm.........
sry bout the dollar signs
by the way she
has to buy one of each
things

Sorry, I overlooked a solution. If s=1, then s=1, m=94, and p=5 is a valid solution as well.

[quote="diophantus"]Let m equal the number of markers, p the number of paint sets, and s the number of sketchpads. then
$p+m+s=100$  (:blush:  no pun intended) which implies
$10p+10m+10s=1000$
$10p+.5m+3s=100$ .  Subtracting gives
$9.5m+7s=900$ which is the same as
$m=\frac{1800-14s}{19}=94-\frac{14(s-1)}{19}$ Therefore for m to be an integer, s=20 or 39.  :D[quote="mathprincess09"]did u use guess and check?[/quote]
[/quote]
Did you read my first post? :blush:
The trick is $\frac{s-1}{19}$ must be an integer for m to be an integer and since you can't buy half of a marker....

i dont get wat u r trying to say! :blush:

I think this may be too difficult to be solved analytically at a middle school level.  Perhaps your teacher meant for you to guess and check.  To explain my problem, I solved the equation for m in terms of one other variable.  Since m must be an integer (1,2,3...), since you can't buy $.1247$ markers, I did whatever I had to do to make the right hand side of the equation an integer. In this case, it was making $\frac{s-1}{19}$ an integer. So s could only be 20 or 39. (20-1=19), (39-1=38)

The way I solve most problems is by educated guess and check. You wouldn't know how well it worked at Nats last year  :blush: .

My definition of educated guess and check is using analytical methods to get as far as you can, and if the problem isn't completed yet, use guess and check :) .

actually thats wat i do when i cant solve a problem! anyways the only thing i can do for this problem is guess and check! coz i dont get that method :blush:


Two students attempted to solve a quadratic equation $x^2+bx+c = 0$. Although both students did the work correctly, one miscopied the middle term and obtained the solution set {2,3}, while the other miscopied the constant term and obtained the solution set {2,5}. What is the correct solution set?

[hide]The first person solved the quadratic $(x-2)(x-3)=x^2-5x+6=0$. The second person solved the equaton $(x-2)(x-5)=x^2-7x+10$. Thus the true equation must be $x^2-7x+6=0$, $(x-1)(x-6)=0$

the correct solution set is {1,6}.[/hide]

I thought I already posted this in the Mathcounts Forum. Oh well.  :roll:

[hide="Another approach"]
You can use Vieta's formulas.

The first student copied down the constant term correctly, so the constant is $2\cdot 3=6$.

The second student copied down the middle term correctly, so its coefficient is $-(2+5)=-7$.

So the correct equation is $x^2-7x+6=0$, which gives us $x=1,6$.
[/hide]


quadratics
summer program
PROMYS
LaTeX
induction

Additional problem:
$\forall n \in \mathbb{N}, \forall p \in \mathbb{P}: p \equiv 3 \bmod 4 \implies \not\exists x \in \mathbb{Z}: p^{n}| (x^{2}+1)$

$\exists$: see the $\LaTeX$-code  :P
It means: "exists".

[quote="t0rajir0u"]$\forall n \in \mathbb{N}, \forall p \in \mathbb{P}: p \equiv 1 \bmod 4 \implies \exists x \in \mathbb{Z}: p^{n}| (x^{2}+1)$[/quote]

hmm...all the basic mod stuff i know does not really seem to help here, but i remember seeing something with quadratic residues (quadratic reciprocity theorem, etc) so those might help...i am looking at it though

[quote="t0rajir0u"]$\forall n \in \mathbb{N}, \forall p \in \mathbb{P}: p \equiv 1 \bmod 4 \implies \exists x \in \mathbb{Z}: p^{n}| (x^{2}+1)$[/quote]

Some variant of QR can probably bash this out? Also we can still do that sum of squares thing you put on the WOOT forum. But here's a low-tech way:

If -1 is not a square residue mod p, then we can group the residues by

$\{ x,-x, \frac{1}{x},-\frac{1}{x}\}$

It's clear that these come in sets of four, except {1, -1}. But this would mean that p has $2 \mod 4$ residues, which is contradiction, so -1 is indeed a residue.

From here it should be easy by induction and probably various other methods.

[hide="Hint for one possible elementary method"] Induction. ;) [/hide]

[hide="Hint for my preferred method"] Gaussian integers work well again here. :D [/hide]


geometry
incenter
circumcircle
trigonometry
trig identities
Law of Sines

$\triangle$ABC is not equilaterial of area 1. For which points P can teh area of $\triangle$PAB be determinied?
A. P is centroid of ABC
B. P is incenter of ABC
C. P is circumcenter of ABC
D. More than 1 of the above
E. None of the above.

What's a centroid?

I'm pretty sure B can be determined...

[quote="nateharman1234"]assuming you know everything about the original triangle
[hide]
centroid - 1/3 of total area

incenter- half the angles, then law of sines for side values, and then find the area

circumcenter - find the area of the original then use to find the circumradius, then you know the sides of ABP and can find the area[/hide][/quote]
So the answer is?...?... :huh: ?

[quote="Ubemaya"]What's a centroid?[/quote]
The centroid of a triangle is the point at which the medians intersect.

[quote="13375P34K43V312"]Yes are you given the original area and the sidelengths? Because if you aren't, then this problem is pointless.[/quote]
Well i dunno but that's all the problem says.

YOu can only determine the area if P was centroid.

B.

The centroid and medians divide the triangle into smaller triangles of equal area, so A?


calculus conics ellipse hyperbola integration trigonometry function

a $ 90\;\text{min}$ calculus quiz i gave yesterday to students
preparing for the upcoming famous competition... :D  :lol:
can you solve them ?

$ \textbf{(i)}$ an ellipse $ x^2 \plus{} \frac {y^2}{4} \equal{} 1$ with it's 2 foci aligned
horizontally rests on the inside of the upper branch of
the hyperbola $ \frac {y^2}{4} \minus{} x^2 \equal{} 1$. find the [i]centroid [/i]of the triangle
formed by the foci of the ellipse and the focus of the upper
branch of the hyperbola.

$ \textbf{(ii)}$ solve $ \int\;\dfrac{1}{\sqrt {\;1 \minus{} \sqrt {\;2 \minus{} \sqrt {\;3 \minus{} \sqrt {\;4 \minus{} \sqrt x\;}\;}\;}\;}}\;\;\;dx$

$ \textbf{(iii)}$ find the condition that $ ax^2 \plus{} by^2 \plus{} cz^2 \plus{} dxy \plus{} eyz \plus{} fzx \equal{} 0$
represents a [i]pair [/i]of planes. show that the angle between those planes is
$ \tan^{ \minus{} 1}\left(\dfrac{\sqrt {\;d^2 \plus{} e^2 \plus{} f^2 \minus{} 4(ab \plus{} bc \plus{} ca)\;}}{a \plus{} b \plus{} c}\right)$

$ \textbf{(iv)}$ solve $ \int_0^1\;\left(\sqrt [5]{\;1 \minus{} x^9\;}\; \minus{} \;\sqrt [9]{\;1 \minus{} x^5\;}\right)\;dx$

$ \textbf{(v)}$ find all functions $ f(x)$ that satisfy the equation
$ \left(\int\;f(x)\;dx\right)\cdot\left(\int\;\dfrac{1}{f(x)}\;dx\right)\; \equal{} \; \minus{} 1$
justify your steps properly.

$ \textbf{(vi)}$ if $ 0 < a < b$, show
$ \lim_{x\to 0}\;\left(\int_0^1\;\left(b\,y \plus{} a\,(1 \minus{} y)\right)^x\;dy\right)^{1/x}\; \equal{} \;\frac {1}{e}\cdot\left(\dfrac{b^b}{a^a}\right)^{1/(b \minus{} a)}$

$ \textbf{(vii)}$ find $ \frac {d^4}{dz^4}\;\int_3^{e^z}\;\int_2^{\ln y}\;\int_1^{\cos x}\;\int_0^{\sin w}\;\sqrt {\;1 \plus{} v^3\;}\;dv\;dw\;dx\;dy$

$ \textbf{(viii)}$ evaluate $ \lim_{x\to 0}\;\frac {1}{\sin x}\;\int_0^{\ln\,(1 \plus{} x)}\;\left(1 \minus{} \tan\,2y\right)^{1/y}\;dy$

$ \textbf{(ix)}$ evaluate
$ \lim_{n\to\infty}\;\left\{\;\left(1 \plus{} \frac {1}{n^6}\right)\cdot\left(1 \plus{} \frac {2^6}{n^6}\right)^{1/2}\cdot\left(1 \plus{} \frac {3^6}{n^6}\right)^{1/3}\cdot\;\;\;\ldots\;\;\;\cdot\left(1 \plus{} \frac {n^6}{n^6}\right)^{1/n}\;\right\}$

$ \textbf{(x)}$ solve $ \frac {d^2y}{dx^2} \minus{} \frac {dy}{dx} \minus{} 2y\; \equal{} \;\sec x\cdot\tan x$

All right, some of these problems actually look pretty nice. I've seen a few before, of course.

I object to the notation of #5: the indefinite integral is not a function.

For (ii), I'd let:

$ u \equal{} \sqrt {1 \minus{} \sqrt {2 \minus{} \sqrt {3 \minus{} \sqrt {4 \minus{} \sqrt {x}}}}}$

solve for x, and then solve:

$ \int\frac {1}{u}dx$

It's just a polynomial in u never mind the fact it would probably take me 90 minutes to write down the answer in that form and I double Misan would accept that mixed-variable expression above.

[quote="misan"]$ \textbf{(ix)}$ evaluate
$ \lim_{n\to\infty}\;\left\{\;\left(1 \plus{} \frac {1}{n^6}\right)\cdot\left(1 \plus{} \frac {2^6}{n^6}\right)^{1/2}\cdot\left(1 \plus{} \frac {3^6}{n^6}\right)^{1/3}\cdot\;\;\;\ldots\;\;\;\cdot\left(1 \plus{} \frac {n^6}{n^6}\right)^{1/n}\;\right\}$[/quote]
\[ \Large e^{\int\limits_0^1 {\frac {{\ln \left( {1 \plus{} x^6 } \right)}}{x}dx} } .
\]

Is [hide="iv"]
$ \frac{5\cdot B(\frac{1}{5}, \frac{10}{9}) \minus{} 9\cdot B(\frac{1}{9}, \frac{6}{5})}{45}$
[/hide]
Where $ B(a, b)$ is the beta function defined as $ \frac{\Gamma(a)\Gamma(b)}{\Gamma(a\plus{}b)}$
Sorry for not posting a solution. I want to make sure the answer is right before I go ahead and post one.
 :blush:  Should save some embarrassment down the line if it's wrong.

There is a [b]much[/b] simpler answer to #4. (I've seen it before)
Trying to calculate integrals is definitely the hard way.

[quote="Extremal"]in 6)...
$ ... = \frac {(b - a)}{e}((\frac {b}{b - a})^{x + 1} - (\frac {a}{b - a})^{x + 1})^{\frac {1}{x}}$=
$ \frac {b - a}{e}\exp(\frac {1}{x} \n (\frac {b}{b - a}\exp x\ln \frac {b}{b - a} - \frac {a}{b - a}\exp x\ln \frac {a}{b - a} ))$=
$ \frac {b - a}{e}\exp (\frac {b}{b - a}\ln{\frac {b}{b - a} - \frac {a}{b - a}\ln \frac {a}{b - a})}$=
$ \frac {1}{e}(\frac {b^{b}}{a^{a}})^{\frac {1}{b - a}}$
it's really good limit.  :)[/quote]

Isn't there a $ \ln$ after the $ \frac{1}{x}$ of the remaining terms in the 2nd line? Also can you help me understand how you eliminate the x in the next line. Thanks.

[quote="apatch"]Extremal in your soln to (ix) above after the integral step shouldn't you have k^2 terms in the denominator?[/quote]
I don't uderstand your first question, can you write where must be $ k^{2}$  by latex
and in seccond question:
I use only the fact that :
$ \exp(xa)$ when $ x\minus{}>0$ is equivalent to $ 1\plus{}ax\plus{}o(.)$

For (v), why not solve it just like a differential equation but instead use antiderivatives.  Define the integral operator as $ I$ and let:

$ If \equal{} \int f dx$  Then:

$ (If)(I\frac {1}{f}) \equal{} \minus{} 1$

$ I\frac {1}{f} \equal{} \minus{} \frac {1}{If}$

Differentiating:

$ \frac {1}{f} \equal{} \minus{} \frac {f}{(If)^2}$

$ (If)^2 \equal{} \minus{} f^2$

$ If \equal{} \pm f$

Differentiating again:

$ f' \equal{} \pm f$

$ f \equal{} \pm e^x$

[quote="misan"]$ \textbf{(iv)}$ solve $ \int_0^1\;\left(\sqrt [5]{\;1 \minus{} x^9\;}\; \minus{} \;\sqrt [9]{\;1 \minus{} x^5\;}\right)\;dx$[/quote]
[hide="Solution"]Let $ S$ be the set of points $ (x, y)$ in the first quadrant such that $ x^9 \plus{} y^5 \le 1$.  We can interpret $ \int_0^1 \sqrt[5]{1 \minus{} x^9} \, dx$ as the area of $ S$.  Similarly, we can interpret $ \int_0^1 \sqrt[9]{1 \minus{} y^5} \, dy$ as the area of $ S$.  So the answer is $ \boxed{0}$.[/hide]

[quote="Extremal"]
I use only the fact that :
$ \exp(xa)$ when $ x \minus{} > 0$ is equivalent to $ 1 \plus{} ax \plus{} o(.)$[/quote]

Thanks very much I understand now.

[quote="apatch"]Please see below:

$ \lim_{n \minus{} > \infty}\exp( \frac {1}{n}\sum_{k \equal{} 1}^{n}\frac {\ln (1 \plus{} (\frac {k}{n})^{6})}{\frac {k}{n}}) \equal{} \exp(\int_{0}^{1}\frac {\ln (1 \plus{} x^{6})}{x}dx) \equal{} \exp(\int_{0}^{1}\frac {\sum_{k \equal{} 1}^{\infty}\frac {x^{6k}( \minus{} 1)^{k \plus{} 1}}{k}}{x}dx)$=
$ \exp(\int_{0}^{1}\sum_{k \equal{} 1}^{\infty}\frac {x^{6k \minus{} 1}( \minus{} 1)^{k \plus{} 1}}{k}dx) \equal{} \exp(\sum_{k \equal{} 1}^{\infty}\frac {( \minus{} 1)^{k \plus{} 1}}{6k^2}) \equal{} \exp(\frac {1}{6}.\frac {\pi^2}{12})$[/quote]
You are right.  :)  thanks

[quote="misan"]
$ \textbf{(viii)}$ evaluate $ \lim_{x\to 0}\;\frac {1}{\sin x}\;\int_0^{\ln\,(1 \plus{} x)}\;\left(1 \minus{} \tan\,2y\right)^{1/y}\;dy$[/quote]

Let $ \int (1 \minus{} \tan 2y)^{\frac {1}{y}}\ dy \equal{} F(y) \plus{} Const,$

$ \lim_{x\rightarrow 0} \frac {1}{\sin x}\int_0^{\ln (1 \plus{} x)} (1 \minus{} \tan 2y)^{\frac {1}{y}}\ dy$

$ \equal{} \lim_{x\rightarrow 0} \frac {1}{\sin x}\{F(\ln (1 \plus{} x)) \minus{} F(0)\}$

$ \equal{} \lim_{x\rightarrow 0} \frac {x}{\sin x}\cdot \frac {F(\ln (1 \plus{} x)) \minus{} F(\ln (1 \plus{} 0))}{x \minus{} 0}$

$ \equal{} \frac {1}{1}\cdot \frac {d}{dx}\left[F(\ln (1 \plus{} x))\right]_{x \equal{} 0}$

$ \equal{} \left[\frac {1}{1 \plus{} x}F'(\ln (1 \plus{} x))\right]_{x \equal{} 0}$

$ \equal{} F'(0)$

$ \equal{} \lim_{y\rightarrow 0} (1 \minus{} \tan 2y)^{\frac {1}{y}}$

$ \equal{} \lim_{y\rightarrow 0} \left\{(1 \minus{} \tan 2y)^{ \minus{} \frac {1}{\tan 2y}}\right\}^{ \minus{} \frac {\tan 2y}{y}}$

Now $ \lim_{y\rightarrow 0}\ln \left\{(1 \minus{} \tan 2y)^{ \minus{} \frac {1}{\tan 2y}}\right\}^{ \minus{} \frac {\tan 2y}{y}}$

$ \equal{} \minus{} \lim_{y\rightarrow 0} \frac {\tan 2y}{y}\left\{(1 \minus{} \tan 2y)^{ \minus{} \frac {1}{\tan 2y}}\right\}^{ \minus{} \frac {\tan 2y}{y}}$

$ \equal{} \minus{} 2\lim_{y\rightarrow 0} \frac {\tan 2y}{2y}\ln \left\{(1 \minus{} \tan 2y)^{ \minus{} \frac {1}{\tan 2y}}\right\}^{ \minus{} \frac {\tan 2y}{y}}$

$ \equal{} \minus{} 2\cdot \ln e^{ \minus{} 1} \equal{} 2$, yielding the answer is $ \boxed{e^2}.$

So (i), (iii) and (vii) are still open. (Vii) is easy but complicated as if
\[ F\left( z \right) \equal{} \int_a^{f\left( z \right)} {\left( {\int_b^{g\left( y \right)} {h\left( t \right)dt} } \right)dy}
\]
then
\[ F'\left( z \right) \equal{} f'\left( z \right)\int_b^{g\left( {f\left( z \right)} \right)} {h\left( t \right)dt} .
\]
Wow...  :idea:

[quote="misan"]

$ \textbf{(x)}$ solve $ \frac {d^2y}{dx^2} \minus{} \frac {dy}{dx} \minus{} 2y\; \equal{} \;\sec x\cdot\tan x$[/quote]

Let $ g(x)\equal{}\sec x\cdot\tan x\equal{}\minus{}\left(\frac{1}{\cos x}\right)'$, we can the differential equation rewrite in two ways as follows.

$ y''\plus{}y'\minus{}2(y'\plus{}y'')\equal{}g(x)\ \cdots [1]$

$ y''\minus{}2y'\plus{}y'\minus{}2y\equal{}g(x)\ \cdots [2]$

Let $ p(x)\equal{}y'(x)\plus{}y(x),\ q(x)\equal{}y'(x)\minus{}2y(x)$, we have $ p'(x)\minus{}2p(x)\equal{}g(x)\ \cdots [3],\ q'(x)\plus{}q(x)\equal{}g(x)\ \cdots [4]$

$ e^{\minus{}2x}*[3]\Longrightarrow (e^{\minus{}2x}p(x))'\equal{}e^{\minus{}2x}g(x)\Longrightarrow p(x)\equal{}e^{2x}\int e^{\minus{}2x}g(x)\ dx\plus{}Ae^{2x}$,

$ e^x*[4]\Longrightarrow (e^{x}q(x))'\equal{}e^xg(x)\Longrightarrow q(x)\equal{}e^{\minus{}x}\int e^xg(x)\ dx\plus{}Be^{\minus{}x}$.

$ \therefore y(x)\equal{}\frac{1}{3}\{p(x)\minus{}q(x)\}$

$ \equal{}\frac{1}{3}e^{2x}\int e^{\minus{}2x}g(x)\ dx\minus{}\frac{1}{3}e^{\minus{}x}\int e^xg(x) dx\plus{}C_1e^{2x}\plus{}C_2e^{\minus{}x}$.

Another approach:

Use Heaviside's differential operator,

$ (D\plus{}1)(D\minus{}2)y\equal{}g(x)\Longrightarrow y\equal{}\frac{1}{3}\left(\frac{1}{D\minus{}2}\minus{}\frac{1}{D\plus{}1}\right) \sec x\cdot\tan x.$

or Laplace transform will be solved the problem.

Bookworm no doubt has a more sophisticated approach to (v).  I'd just write them out as:

$ f(v)\equal{}\sqrt{1\minus{}v^3}$

$ g(w)\equal{}\int_0^{Sin[w]} f(v)dv$

$ h(x)\equal{}\int_1^{Cos[x]} g(w)dw$

$ u(y)\equal{}\int_2^{ln(y)} h(x)dx$

$ v(z)\equal{}\int_3^{e^z}u(y)dy$

Ok then, so:

$ \frac{d}{dz} v(z)\equal{}u[e^z]e^z$

Differentiating again:

$ \frac{d^2}{dz^2}\left\{u[e^z]e^z\right\}\equal{}
u[e^z]e^z\plus{}e^z\frac{d}{dz}u[e^z]$

but:

$ u[e^z]\equal{}\int_2^z h(x)dx$

so:

$ \frac{d}{dz}\int_2^z h(x)dx\equal{}h(z)$

Now, two more times.  I end up with:

$ \frac{d^4}{dz^4} v(z)\equal{}u[e^z]e^z\plus{}3e^zh[z]\minus{}2e^zSin[z]g[Cos(z)]\plus{}e^zSin^2[z]Cos[z]f[sin[Cos(z)]]
\minus{}e^{2z}Sin[z]g[Cos[z]][Cos[z]\plus{}Sin[z]]$

[quote="misan"]
$ \textbf{(iii)}$ find the condition that $ ax^2 \plus{} by^2 \plus{} cz^2 \plus{} dxy \plus{} eyz \plus{} fzx \equal{} 0$
represents a [i]pair [/i]of planes. show that the angle between those planes is
$ \tan^{ \minus{} 1}\left(\dfrac{\sqrt {\;d^2 \plus{} e^2 \plus{} f^2 \minus{} 4(ab \plus{} bc \plus{} ca)\;}}{a \plus{} b \plus{} c}\right)$[/quote]

We are to find the condition of existing real numbers $ (p,\ q,\ r,\ l,\ m,\ n)$ such that $ ax^2 \plus{} by^2 \plus{} cz^2 \plus{} dxy \plus{} eyz \plus{} fzx \equal{} (px \plus{} qy \plus{} rz)(lx \plus{} my \plus{} nz)$, that is to say this equation becomes equality.

Expanding right sides of the equation and comparing the coefficients both sides, we have

$ a \equal{} pl,\ b \equal{} qm,\ c \equal{} rn,\ d \equal{} pm \plus{} ql,\ e \equal{} qn \plus{} rm,\ f \equal{} rl \plus{} pn$.

By Cauchy-Schwarz's inequality, $ (p^2 \plus{} q^2 \plus{} r^2)(l^2 \plus{} m^2 \plus{} n^2)\geq (pl \plus{} qm \plus{} rn)^2$

$ \Longleftrightarrow (pl)^2 \plus{} (pm)^2 \plus{} (pn)^2 \plus{} (ql)^2 \plus{} (qm)^2 \plus{} (qn)^2 \plus{} (rl)^2 \plus{} (rm)^2 \plus{} (rn)^2\geq (pl \plus{} qm \plus{} rn)^2$

$ \Longleftrightarrow a^2 \plus{} b^2 \plus{} c^2 \plus{} (pm \plus{} ql)^2 \minus{} 2pl\cdot qm \plus{} (qn \plus{} rm)^2 \minus{} 2qm\cdot rn \plus{} (pn \plus{} rl)^2 \minus{} 2rn\cdot pl\geq (a \plus{} b \plus{} c)^2$

$ \Longleftrightarrow a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2 \plus{} f^2 \minus{} 2(ab \plus{} bc \plus{} ca)\geq a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca)$, yielding the desired condition is $ \boxed{d^2 \plus{} e^2 \plus{} f^2\geq 4(ab \plus{} bc \plus{} ca)}.$ Under this condition, denote the angle two planes $ px \plus{} qy \plus{} rz \equal{} 0,\ lx \plus{} my \plus{} nz \equal{} 0$ by $ \theta$

By the scalar product two normal vectors $ (p,\ q,\ r),\ (l,\ m,\ n)$, we have $ \cos \theta \equal{} \frac {pl \plus{} qm \plus{} rn}{\sqrt {p^2 \plus{} q^2 \plus{} r^2}\sqrt {l^2 \plus{} m^2 \plus{} n^2}}$,

$ 1 \plus{} \tan ^ 2 \theta \equal{} \frac {1}{\cos ^2 \theta } \equal{} \frac {(p^2 \plus{} q^2 \plus{} r^2)(l^2 \plus{} m^2 \plus{} n^2)}{{pl \plus{} qm \plus{} rn}^2}$

$ \tan ^ 2 \theta \equal{} \frac {(p^2 \plus{} q^2 \plus{} r^2)(l^2 \plus{} m^2 \plus{} n^2) \minus{} (pl \plus{} qm \plus{} rn)^2}{(pl \plus{} qm \plus{} rn)^2}$

using the result above,

$ \therefore \tan ^ 2 \theta \equal{} \frac {d^2 \plus{} e^2 \plus{} f^2 \minus{} 4(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2}$, yielding

$ \boxed{\theta \equal{} \tan ^ { \minus{} 1} \left(\frac {\sqrt {d^2 \plus{} e^2 \plus{} f^2 \minus{} 4(ab \plus{} bc \plus{} ca)}}{|a \plus{} b \plus{} c|}\right)}$

[quote="misan"]

$ \textbf{(i)}$ an ellipse $ x^2 \plus{} \frac {y^2}{4} \equal{} 1$ with it's 2 foci aligned
horizontally rests on the inside of the upper branch of
the hyperbola $ \frac {y^2}{4} \minus{} x^2 \equal{} 1$. find the [i]centroid [/i]of the triangle
formed by the foci of the ellipse and the focus of the upper
branch of the hyperbola.[/quote]

How can we construct the triangle?


Each principal of Lincoln High School serves exactly one 3-year
term. What is the maximum number of principals this school could
have during an 8-year period?

4, cause one could be ending with 1st year.

I would just like to add that the four principals (A,B,C,D) have the following terms:

A- 1 year
B- 3 years
C- 3 years
D- 1 year


How many divisors of $ 1996^{1996}$ are perfect squares?
[hide="what i tried"]First I factored 1996 into $ 2^2 * 499$. Then $ 1996^{1996} \equal{} 2^{3992} * 499^{1996}$. From here, I do not know how to find the number of perfect squares.[/hide]

You can further factor that expression into $ (2^2)^{1996}\times{(499^2)^{998}}$. Your answer will then be $ 1997(999)$.
Edited to fix $ LaTeX$.

[quote="gaussintraining"]You can further factor that expression into $ (2^2)^1996\times{(499^2)^998)}$. Your answer will then be $ 1997(999)$.[/quote]

What is the final answer? Is it 1997*999= 1,995,003?
I don't understand the way you wrote the factor of the expression. Did you mean 2 square to the first power time 996 time 499 square to the 9th and then time 98.
I'm not a genius so can you explain more?

Sorry, I had some $ \text{\LaTeX}$ errors which I fixed. As was mentioned on a recent thread, the number of divisors of $ a_{1}^ba_{2}^ca_{3}^d\dots a_{n}^z$ is $ (b+1)(c+1)\dots(z+1)$. This time, we're looking for perfect square factors. Therefore, we write our values as perfect squares raised to some power. Since our factorization is now perfect squares raised to powers, we can do the same thing as before (try a few examples if you have to).

@ the cliu You're welcome!  :D


How many positive perfect cube factors does $ 3^65^{10}$ have?

The cubes are:

$ 3^0$
$ 3^3$
$ 3^6$
$ 5^3$
$ 5^6$
$ 5^9$

So the answer is $ 6$.

Oh, wait...

No its not.

I think I forgot cubes like $ 3^3 \cdot 5^6$ and etc.

Continuing Ernie's cuubes list of:
$ 3^0, 3^3, 3^6$
and
$ 5^0, 5^3, 5^6, 5^9$

We can multiply the number of cubes in each factor, because $ \sqrt[3]{5^9 \cdot 3^3}$ is also a cube. Therefore, there are $ 4\cdot 3\equal{}12$.

Oh wait, on the last sentence, take out  the cube root. I don't know what I was thinking...


[quote="kevin gong"]If $abcde*4=edcba$, and each letter represents a distinct digit, and $a$ is not equal to zero, then find the number $abcde$.[/quote]

I don't think we can answer this here because it is almost the exact same problem as the USAMTS problem 1/1/18. I solved this problem and I'll post an answer once the First Round ends just to be sure I don't violate any rules or stuff.

oops! I didn't notice that before. :(
Thanks for letting me know.

I'll ask a moderator to lock this until Round 1 is over.

Ok, i'll just lock this.

Careful Kevin...  :D


Find all functions $g: \mathbb R \to \mathbb R$ such that for all real $x$ and $y$,
 \[g(x\plus{}y)\plus{}g(xy)\equal{}g(x)g(y)\plus{}g(x)\plus{}g(y).\]

[quote="Algadin"]Find all functions $ g: R\to R$ such that for arbitrary real number $ x$ and $ y$:
$ g(x \plus{} y) \plus{} g(xy) \equal{} g(x)g(y) \plus{} g(x) \plus{} g(y)$[/quote]
[i]Already posted, I think[/i]

Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\plus{}f(xy)\equal{}f(x)f(y)\plus{}f(x)\plus{}f(y)$

$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$
If $ f(1)\equal{}0$, then $ P(x\minus{}1,1)$ $ \implies$ $ f(x)\equal{}0$ $ \forall x$ which indeed is a solution. So we'll consider from now that $ f(1)\ne 0$

$ P(\minus{}1,1)$ $ \implies$ $ f(\minus{}1)\equal{}\minus{}1$

$ P(\minus{}1,\minus{}1)$ $ \implies$ $ f(\minus{}2)\equal{}\minus{}1\minus{}f(1)$
$ P(\minus{}2,1)$ $ \implies$ $ f(\minus{}2)f(1)\equal{}\minus{}1\minus{}f(1)$
Comparing these two lines, we get $ f(1)^2\equal{}1$

If $ f(1)\equal{}\minus{}1$, then $ P(x,1)$ $ \implies$ $ f(x\plus{}1)\equal{}\minus{}f(x)\minus{}1$ and so $ f(x\plus{}2)\equal{}f(x)$ and so $ f(2)\equal{}f(0)\equal{}0$, but then $ P(\frac 12,2)$ $ \implies$ $ f(1)\equal{}0$, contradiction

So $ f(1)\equal{}1$ and $ P(x,1)$ $ \implies$ $ f(x\plus{}1)\equal{}f(x)\plus{}1$

Let then $ x\ne 0$ : $ P(x,\frac yx\plus{}1)$ $ \implies$ $ f(x\plus{}\frac yx)\plus{}f(x\plus{}y)\equal{}f(x)f(\frac yx)\plus{}2f(x)\plus{}f(\frac yx)$
Subtracting $ P(x,\frac yx)$ from this equality, we get : $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$, still true for $ x\equal{}0$

Then $ P(x,y)$ becomes $ f(xy)\equal{}f(x)f(y)$ and we have the very classical system :

$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$
$ f(xy)\equal{}f(x)f(y)$

Whose unique non constant solution is $ f(x)\equal{}x$ which, indeed, is a solution of your equation.

Hence the two solutions :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}x$ $ \forall x$

[quote=pco][quote="Algadin"]Find all functions $ g: R\to R$ such that for arbitrary real number $ x$ and $ y$:
$ g(x \plus{} y) \plus{} g(xy) \equal{} g(x)g(y) \plus{} g(x) \plus{} g(y)$[/quote]
[i]Already posted, I think[/i]

Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\plus{}f(xy)\equal{}f(x)f(y)\plus{}f(x)\plus{}f(y)$

$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$
If $ f(1)\equal{}0$, then $ P(x\minus{}1,1)$ $ \implies$ $ f(x)\equal{}0$ $ \forall x$ which indeed is a solution. So we'll consider from now that $ f(1)\ne 0$

$ P(\minus{}1,1)$ $ \implies$ $ f(\minus{}1)\equal{}\minus{}1$

$ P(\minus{}1,\minus{}1)$ $ \implies$ $ f(\minus{}2)\equal{}\minus{}1\minus{}f(1)$
$ P(\minus{}2,1)$ $ \implies$ $ f(\minus{}2)f(1)\equal{}\minus{}1\minus{}f(1)$
Comparing these two lines, we get $ f(1)^2\equal{}1$

How??? if we compare the two lines we will get $f(-2)=f(-2)f(1)$ ??? how we will get $f(1)^2=1$

[quote=Understandingmathematics]How??? if we compare the two lines we will get $f(-2)=f(-2)f(1)$ ??? how we will get $f(1)^2=1$[/quote]
If we compare the two lines, we have $f(-2)=f(-2)f(1)$ and $f(-2)=-1-f(1)$ and so :
Either $f(1)=1$
Either $f(-2)=0$ and so $f(1)=-1$

Which indeed implies $f(1)^2=1$

[quote=Understandingmathematics]and also can we apply same method if we are given the same function from $Q$ to $Q$ ...???
thankyou[/quote]
Why not ?
Obviously each step can be used in the same way if function is from $\mathbb Q\to\mathbb Q$

We just could shorten a bit since from $f(1)=1$ and $f(x+y)=f(x)+f(y)$ we could then immediately conclude $f(x)=x$ when on $\mathbb Q$ (no need for the supplementary step $f(xy)=f(x)f(y)$)

thankyou so much...pco , i don't know enough about you but till now i come to a realization that you are master in functional equations....