Each day Walter gets $\$3$ for doing his chores or $\$5$ for doing them exceptionally well. After 10 days of doing his chores daily, Walter has received a total of $\$36$. On how many days did Walter do them exceptionally well?

$\textbf{(A)}\ 3 \qquad
\textbf{(B)}\ 4 \qquad
\textbf{(C)}\ 5 \qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 7$

[hide]
$\\ 3x+5y=36\\ x+y=10$

Multiply the bottom equation by 3 and subtract it from the top equation.

$\\ 2y=6\\ y=3$

Plugging that back into the first equation we get:
$\\ 3x+15=36\\ 3x=21\\ x=7$
Then to answer the question he did exceptionally well on 3 days.[/hide]

[quote="pascal_1623"][hide]
If he did exceptional for all ten days, he would get 40 dollars.
But he got 36.
40-2-2=36, so he did his chores exceptionally well 7 days
$E$[/hide][/quote]

Not sure about that pascal, please see your mistake. The answer is $3$.

[hide]We can write a systems of equations:
$x+y=10$
$3x+5y=36$
Where x is the amount of chores done ok and y is the amount of chores done exceptionally well.
Solving we get:
$2y=6\Longrightarrow y=3$
Therefore, he did his chores exceptionally well on $\boxed{\boxed{3}}$ days. [/hide]

It is deffinitley not 7. 7x5 is 35 and the boy made 36. In order to achieve the 36 you would need to add one dollar. But since 1 isnt an option, you can cancel out E. Cancel out D because 6 times 5 30. Which mean she only worked for 6 days. If he does 2 lousy cleanings that will make 36 but only in 8 days. Same thign for 4 and 5 days it wont satisdy the time given. Therfore it must be 3! :)

Ok I think i will tell you how I got these...

[hide]$x+y=10$, and $3x+5y=30$. I used elimination, and I got $x=7$ and $y=3$ (x represents the number of lousy days and y is good days). I got A.[/hide]

 :thumbup: Cheese is cool!!!!!!!!!!!!!!!!!!!! Avocados are cool too.

Sorry I did not include my work earlier I forgot :oops_sign:

[hide]a[/hide] I got confused b/c I thought it was 7 days! :roll:

[hide="quick algebraless method"]Walter gets 3 bucks a day but if he does a good job he gets 2 bucks bonus. He has worked 10 days so he earned 3X10=30 bucks. Since there's 6 bucks extra, he got a bonus on 6/2=3 days.

A. 3[/hide]
Sorry to revive this old topic.

you can also eyeball and get the answer in 2 seconds

jorian

sys of equations....
hmmm.
the answer should be 3


n is a natural number>=3

prove that (n^n^n^n)-(n^n^n) is a multiple of 1989

this is a quite common question.try solving it in as simple manner as possible

You mean $ n^{n^{n^{n}}}\minus{}n^{n^{n}}$
You can find this problem in Hojoo Lee's file about Number Theory.Also he said that it is Unused problem for some BMO.
Look here  http://www.mathlinks.ro/Forum/viewtopic.php?t=150378


Hello,

I'm doing like a small paper on a scientific subject I used \documentclass{report} which I thought it's the best for such purposes  :) . I have a nice table of contents I think but the problem that I used \chapter{Thing 1} so it appears a half begin of the page is blank and it's written chapter 1 for example but I don't want this what I want is like in the beginning of the page it's
1 Something
1.1 Another thing (I used \section{Another thing})
.
.
.
2 A lot of thing

I don't want the layout as previous.
I searched for the nearest example like what I want and I found it here:  http://www.artofproblemsolving.com/Resources/Papers/FracBase.pdf

Thanking you

In the book and report document classes, chapters start on a new page as you would expect. So if you don't want this don't use chapters; use sections, subsections etc. Also why not use the article document class as you have no need for chapters?

I want it in a new page but I don't want it to leave half a page in the beginning and write in the big size Chapter 1

Use the [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/titlesec/titlesec.pdf]titlesec package[/url] with \titlespacing and \titleformat (sections 3.1, 3.2 of the manual). Use something like
[code]\usepackage{titlesec}
...
\titlespacing*{\chapter} {0pt}{-20mm}{6mm}
\titleformat{\chapter}[block]
  {\bfseries\LARGE}
  {\LARGE\thechapter}
  {0.8ex}
  {} [/code]though you'll need to adjust the variables to suit you.

Hello,

Sorry if this irrelevant to the title but because I want to decrease the topic numbers and spam.

I wanted to make a Title Page in a format that is in the center of the page. I searched for a near example and I found this as a good example http://www.artofproblemsolving.com/Resources/Papers/YiSunIntel.pdf

Thanks a lot for help

\maketitle after you've completed \title{...}, \author{...}, plus, if required, \date{...},\thanks{...}

There is a problem when I do this.
The date automatically appear in the first page which is good.
But this "\title{Hello} \author{World}" even though when I put \maketitle appears in a second page alone as "Hello World"  :(

[quote="stevem"]\maketitle [i][b]after[/b][/i] you've completed \title{...}, \author{...}, plus, if required, \date{...},\thanks{...}[/quote]
[code]\title{Hello}
\author{World}
\maketitle[/code]


Prove that for every x,y,w > 0 , the following inequality holds:

cyclic sum of ((2w(2y-x)-2xy)/(x(y+w))) >= cyclic sum of (((y-x)/(y+x))^2)

Could u please be more specific?!..i didn't really understand the content of the page , maybe because of the language..and as a consequence your solution wasn't very clear to me!


Prove that 0.110001000000000000000001000...1000000..1...
is transcendental. It is all zeroes except for ones in the the n!th places, where n is an integer.

Thats a tough problem you got there. This was the first number known to be transandental, im pretty sure. You have to use Louvilles criteria, of which a proof can be found at:

http://en.wikipedia.org/wiki/Transcendental_number

Its there somewhere

(Edit: here is the exact link: http://en.wikipedia.org/wiki/Liouville_number)


Let $ (F_{n})$ is Fibonaxi sequence $ (F_{0}\equal{}F_{1}\equal{}1;F_{n\plus{}2}\equal{}F_{n\plus{}1}\plus{}F_{n} \ \forall n)$
Prove that $ \forall n \geq 2 : 4F_{n\minus{}2}F_{n}F_{n\plus{}2}F_{n\plus{}4}\plus{}9$  is  a square number

[hide="Hint"]You need only prove, that number is equal to $ (2F_nF_{n \plus{} 2}\pm 3)^2$.[/hide]

We can (or no ;p) see that $ (F_{n\minus{}2}F_{n\plus{}4}\plus{}F_{n}F_{n\plus{}2})^2\equal{}4F_{n\minus{}2}F_{n}F_{n\plus{}2}F_{n\plus{}4}\plus{}9$
Tranforming, we get $ (F_{n\minus{}2}F_{n\plus{}4}\minus{}F_{n}F_{n\plus{}2})^2\equal{}9$.
On the other hand, we have $ F_{n\minus{}2}F_{n\plus{}4}\minus{}F_{n}F_{n\plus{}2}\equal{}3$ for odd $ n$ and $ F_{n\minus{}2}F_{n\plus{}4}\minus{}F_{n}F_{n\plus{}2}\equal{}\minus{}3$ for even $ n$, what we can, with some difficulties, prove by induction.


[quote="b.s.o"]$ a,b,c \ge 0$            and           $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} 8$

Prove that :

$ \frac {a \plus{} b \plus{} c}{3} \ge(\frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{3})^{\frac {1}{27}}$[/quote]
A hint: $ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} 8 \iff \frac{a^3 \plus{} b^3 \plus{} c^3}{3} \plus{} 8 \equal{} 9 \left ( \frac{a \plus{} b \plus{} c}{3} \right )^3$


inequalities
AMC
USA(J)MO
USAMO
function
inequalities unsolved

Some old USAMO question.

It is sufficient to notice that the function \[ f(x,\;y,\;z) = \sum_{cyclic}\frac{x}{y + z + 1} + \prod_{cyclic}(1 - x) \] is convex in each of its variables, so you only have to check the cases where $x,\;y,\;z \in \{0,\;1\}$.


[quote="math92"]How many perfect square factors does the number 46,656 have?[/quote]

[hide="Hint"]
Find the prime factorization of $46656$ first.
[/hide]

This problem is beyond the scope of the Basics forum.

Um. Maybe I'm thinking about this wrong, but aren't there only three perfect squares in $2^6$ and three in $3^6$?

$2^2*2^2*2^2*3^2*3^2*3^2$, which would make the total number of factors 9 instead of 16, right?

No, $4^a*9^b$  can be a  possiblle factror  and  $a$ and $b$ each can take four possible values ($0,1,2,3)$)


floor function
ceiling function
combinatorics proposed
combinatorics

Determine the largest integer n for which there exists a set of n positive integers, none of which exceeds 2002, such that if a and b are numbers in S (not necessarily distinct) then ab is not in S.

Is the answer $2003-\lfloor \sqrt{2002} \rfloor$ ?

You got the main idea. How about a full proof?

I will show that the largest n is 1958.  As an example of a large set S, consider S = 45..2002.  It has 1958 elements, and has the desired property.  We will next show that any set S with the given property has at most 1958 elements.

Clearly 1 is not in S.

Consider the pairs $\{k, k^2 \}$ for $k$ from 7 to 44.  These pairs are disjoint subsets of 7..2002.  At most one number from each pair can be in set S.  Thus at least 38 numbers from 7..2002 are not in S.

If the numbers 2, 3, 4, 5, and 6 are not in S, then we would be done, because we would have a total of 44 excluded numbers.

If a small number such as 2 to 6 is in S, then it is easy to show that S is missing more than 100 numbers.  Specifically, suppose $m$ is a number from 2 to 6 that is in S.  Consider the pairs $\{ k, mk \}$ for k from 167 to 333.  These pairs are disjoint subsets of 167..2002.  At most one number from each pair is in S.  Thus we have excluded 167 numbers from S.

my proof: if we cancel the first $\lfloor \sqrt{2002} \rfloor = 44$ elements, then we are clearly ok. Is it possble with 43? Suppose it was, then we had at least one number in [1,44]. Say there are $k$ numbers in this interval. then every square of a number in that interval, plus all their products, becomes 'forbidden'. So the number of numbers that get 'forbidden' is at least $k+1$, as
-$44 < \sqrt{2002}$ (that gives +k already)
-for k=1 the number multiplied by 45 is still below 2002, so +1
-for k>1 there is at least 1 product, so at least +1

As there are at only $k$ numbers missing in [45,2002], this yields a contradiction.

Hmm, I didn't quite get that.  You let $k$ be the number of elements of S in [1, 44].  You have to show that at least $k$ numbers from [45, 2002] are missing.  But how do you know that the numbers you are forbidding are in the range [45, 2002]?

Because if it happens one is not, so is in [1,44] then one of the numbers must be below $\sqrt{44}$, so in $\{1,...,6\}$, and I guess you've figured out that this cannot work.

If one of the numbers is indeed $k\leq6$, that means any number from $\lceil \frac{45}{k}\rceil$ up to $\lfloor{\frac{2002}{k}\rfloor}$ excludes it's k-double, and that are far above 44 numbers on their own :)

[quote="Ravi B"]Got it.

In hindsight, I think I would choose the following proof as a little simpler than my first.  Consider the sets $\{44 - k, 44 + k, (44-k)(44+k) \}$ for $k$ from 0 to 43.  These 44 sets are disjoint subsets of 1..2002.  Each of these sets must have at least one number not in S.  Thus we have at least 44 numbers missing from S, and we are done.[/quote] :coolspeak:


Given two non-negative integers $m>n$, let's say that $m$ [i]ends in[/i] $n$ if we can get $n$ by erasing some digits (from left to right) in the decimal representation of $m$. For example, 329 ends in 29, and also in 9.

Determine how many three-digit numbers end in the product of their digits.

I recommend checking my casework.

[hide]

Consider the one-digit ends: For a*b*c=c, a=b=1, so the possible numbers abc are 110, 111, 112, ..., 119; as well as 100.

Consider the two-digit ends: For a*b*c=bc, we have a=(10b+c)/(b*c)=10/c+1/b=int. 1/b is never an integer so 10/c must not be; c cannot be 1, 2 or 5. Examining the other cases:
c=3 - 1/b= 2/3,5/3,... (b not int)
c=4 - 1/b= 2/4,6/4,... (b=2 in 1/2) [u]324[/u] works
c=6 - 1/b= 2/6,8/6,... (b=3 in 1/3) [u]236[/u] works
c=7 - 1/b= 4/7,13/7,... (b not int)
c=8 - 1/b= 6/8, 14/8, ... (b not int)
c=9 - 1/b= 8/9, 17/9, ... (b not int)

The only possible numbers are 100, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 236, 324 - 13 in total.
[/hide]

You missed something.
[hide]Actually, there's 101.
You observed that 111-119 are possible, but you ignored most of the ones ending in 0.
There's 9 third digit choices.
There's 10 second digit choices.
There's only 1 third digit choice since that's 0.
Thus, there's 99 one-digit numbers' products that are the ends of the three-digit numbers.
You found 2 two-digit numbers where the products are the two-digit end numbers.
Thus, we have 101.[/hide]

[quote="Jutaro"]Given two non-negative integers $ m>n$, let's say that $ m$ [i]ends in[/i] $ n$ if we can get $ n$ by erasing some digits (from left to right) in the decimal representation of $ m$. For example, 329 ends in 29, and also in 9.

Determine how many three-digit numbers end in the product of their digits.[/quote]

Let $ a,b,c$ be digits of a number.
If $ n$ is one-digit number then we have: $ abc=c$ or $ c(ab-1)=0$. If $ c=0$ then we 90 numbers that satisfy conditions. If $ ab=1$ then where are 9 such numbers.

If $ n$ is two-digit number then we have: $ c(ab-1)=10b$ or $ k(ab-1)=10$ where $ c=kb$. After checking three cases $ k=1, k=2$ and $ k=5$ we get 4 solutions: $ 236, 324, 612$ and $ 315$.

All in all, there are 103 such numbers.

Hello. I'm new here and I'm from Chile. This forum will give me the possibility to practise English and to learn MATH.

First it's very easy to notice that the product of the digits of all 3-digits numbers that ends in 0 is cero. (0)

The number abc can ends in bc or c. That's why that now we have 2 possibilities:
(1)  c=a*b*c  (c>0) --> ab=1  --> a=b=1  (because they must be positive integers)

(2) bc=a*b*c --> 10*b + c=a*b*c --> 10*b = c*(a*b -1)   so now he have that c can be 1,2,5,10,b,2*b,5*b,10*b . But c<=9 so the possibilities now are 1,2,5,b,2b,5b.
c=1 --> ab-1=10b --> b(a-10)=1 Impossible (a<=9)
c=2 --> ab-1=5b --> b(a-5)=1 --> a=6, b=1
c=5 --> ab-1=2b --> b(a-2)=1 --> a=3, b=1
c=b --> ab-1=10 --> ab=11 (impossible, they are integer and <=9)
c=2b --> ab-1=5 --> ab=6 --> a=2,3,6 and b=3,2,1 --> c=6,4,2  (in that order)
c=5b --> ab-1=2 --> ab=3 --> a=3, b=1, c=5

So the numbers are: (0) :100,110,120...200,210,...300,......990  (1): 111, 112, 113,....,119  (2) 236, 315, 324, 612.

Finally from (0) we get 90 numbers, from (1) 9 numbers and from (2) 4. So there are 90+9+4=103 numbers.


The set of rational numbers is countable, the irrationals are not.

[quote="conceive"]Is there any function mapping the set of iratinal numbers bijectively to rational numbers?

Does at least one such function exist?[/quote]

Come on, 'open questions' is not the place for a problem as simple as this.

[quote="blahblahblah"]
Come on, 'open questions' is not the place for a problem as simple as this.[/quote]

sorry..  :blush:

EDIT: I was not aware of the real difficuilty of the problem.


Mark sets fence posts for a living. He takes a job to enclose a rectangular field 200 feet by 500 feet. Posts are set in every 10 feet. He pays his helper $ \$8.50$ for setting each post. The cost of materials for the job is $ \$4.30$ per post plus $ 7.2\%$ tax on the purchase, and his overhead for the job is $ \$50.00$. How much profit will Mark make if he charges $ \$25.00$ per fence post?

There are $ 1400$ posts.  For each post he makes $ 25\minus{}8.5\minus{}(4.3*1.072)\equal{}11.8904$, so he makes $ 16646.56\plus{}50\equal{}\boxed{16696.56}$ dollars...

However, this seems quite a lot for setting fence posts...am I off by a factor of ten or something?

Yep, you forgot to account for the fact that posts are set every [b]10[/b] feet.

And cheating with my calculator says that $ 11.8904\cdot 140\plus{}50\equal{}\boxed{1714.66}$ dollars

Heh, I just had this question on OPLET 10 minutes ago, and it says the answer is 1614.66...

I dunno how you count the overhead for this.  Apparently you subtract it...

[hide=Solution]
There's $140$ posts, and with each post he earns $25 - 8.5 - 4.3*1.072 = 11.8904$. The overhead means business expenses not related to labor or materials, so you'd subtract it. So Mark's total profit is $140 \cdot 11.8904 - 50 = 1614.656 \approx \boxed{1614.66}$. [/hide]


floor function
number theory
prime factorization

Prove that ,if $ [n^2x]$ is a perfect square for any natural number $ n$ then $ x$ must be a perfect square.

do the brackets [] around n^2x mean anything? if not the problem isn't hard

[hide="My horrible attempt"]If it's a perfect square, the prime factorization is in the form $ e_1^{a_1} \plus{} e_2^{a_2} \plus{} ... \plus{} e_x^{a_y}$, where all the $ a_i$ are even.
Case 1-$ GCD(x,n) \equal{} 1$.
Then, for the sake of contradiction,  assume x is not a perfect square, so atleast 1 exponent in the prime facotrization is odd.In $ [N^2x]$, we know that all the exponents are even, but the factor of x introduces at least one odd exponent, and we have a contradiction.

Case 2-$ GCD(n,x)\ne{1}$, so they share common factors.
Assume, once again that x is not a perfect square. Then, $ [N^2x]$ will have atleast one odd exponent (so one of the $ a_i$ is odd), since even+odd=odd. But we have that all the $ a_i$ are even, and we reach a contradiction.

So, we have proven x must have all even exponents in its prime factorisation, which means it is a perfect square.[/hide].

[hide]
$ n^{2}x$
If $ x$ is a perfect square, then let $ x\equal{}y^{2}$.
$ n^{2}y^{2} \equal{} \left( ny\right)^{2}$ which is a perfect square.
[/hide]

[b]AstroPhys:[/b]  You haven't done anything to show that $ x$ [i]must[/i] be a perfect square, only that it [i]can[/i] be a perfect square.

[b]xpmath:[/b]  You've misinterpreted the problem.  $ x$ is a real number, and $ [x]$ is the floor function.

[b]Problem statement:[/b]  Given a real number $ x$ such that $ \lfloor n^2 x \rfloor$ is a the square of an integer for [b]every natural number $ n$[/b], prove that $ x$ must be the square of an integer.

Then the statement is not true, a contradiction is easily found as

$ n\equal{}1, x\equal{}0.1,$ and $ \lfloor n^2x \rfloor \equal{} \lfloor 0.1\rfloor \equal{} 0$ which is a perfect square while $ x$ is not.

For [b]every[/b] natural number $ n$. You showed that it is a square for one $ n$, but not for every $ n$.


analytic geometry
conics
ellipse
algebra
system of equations

Kevin is playing with his Solar System Creator game. The game is meant for little kids, so the solar systems are two dimensional. He creates a system with suns at the points (6,1) and (-2,1). He creates a planet with an elliptical orbit with the two suns as foci and with a major axis of 10. He then creates a new system. The new sun is at the midpoint of the two suns in the original system. The planet in the new system has a circular orbit aroudn the new sun with a radius of 4. If the two planets' orbits are plotted on a sheet of graph paper, find the coordinates of the point of intersection of the two orbits in the fourth quadrant.

[hide="Solution"]The length of the semimajor axis is $ a \equal{} 5$, and the value of $ c$ is $ 4$. Then, $ a^{2} \equal{} b^{2} \plus{} c^{2}$, so $ b \equal{} 3$. The center of the ellipse is at $ (2, 1)$.

Then, the equation of the ellipse is $ \frac {(x \minus{} 2)^{2}}{25} \plus{} \frac {(y \minus{} 1)^{2}}{9} \equal{} 1$.
The equation of the circle is $ (x \minus{} 2)^{2} \plus{} (y \minus{} 1)^{2} \equal{} 16$.

Solving our system of equations, the only solution we get that's in the fourth quadrant is $ (\frac {5\sqrt {7} \plus{} 8}{4}, \minus{} \frac {5}{4})$[/hide]


This is a question from this site.
For how many of the first 500 natural numbers, n, does n=[2x]+[4x]+[8x]+[20x] has a solution?
Note that [2x] means the largest integer less than 2x i.e. [14.67]=14

The method I tried involves counting how many "skips" there are, due to overlaps in the coefficients, and then subtracting that from 500, but I consistantly got the wrong answer.

Sorry.  My answer was wrong.  I blame the crack.

Hopefully this is clearer, and hopefully my answer matches that site  :D

[2x] increases as x reaches a multiple of 0.5
[4x] increases as x reaches a multiple of 0.25
[8x] increases as x reaches a multiple of 0.125
[20x] increases as x reaches a multiple of 0.05

First  we look for places where more than one of these functions increases:
At multiples of 0.25 for x our sum increases by at least 3 (4 if x is a multiple of 0.5, 3 otherwise)

Therefore, if x is a multiple of 0.5 we "skip" 3 values of n.
Otherwise, if x is a multiple of 0.25 we "skip" 2 values of n.

So from this we see that x=0 gives n=0, x=0.5 gives n=17 and we skipped 5 values of n between these.  Notice that the numbers skipped are congruent to 6,7,14,15,16 (mod 17).

So we now see that this is true for 12 of the 17 numbers from 17k+1 to 17(k+1) for all k>0, meaning that from 1 to 493 there are 29*12=348 such values of n.

However, since we want from 1 to 500 we check 494 to 500 and see that 499 and 500 are skipped (6,7 mod 17), meaning 494 to 500 contribute 5 more values of n, making our total values of n equal to 353.

Also, if you were consistantly getting 295 then it's because you were counting the skips as 4 and 3 instead of 3 and 2.


advanced fields
advanced fields unsolved

We have that for every algebraic k-prevariety $ (V;\mathcal{O}_V )$ such that $ \Gamma(V;\mathcal{O}_V )$ is an affine k-algebra, there is a canonical morphism of prevarieties $ f: (V;\mathcal{O}_V )\to Spm\Gamma(V;\mathcal{O}_V )$, such that $ f(P) \equal{} \{f\in \Gamma(V;\mathcal{O}_V ) : f(P) \equal{} 0\} \equal{} : \mathfrak{M}_P$
Show that $ (V;\mathcal{O}_V )$ is affine if and only if $ \Gamma(V;\mathcal{O}_V )$ is an affine k-algebra and $ f$ is an isomorphism.

Just to make sure that we use the same definitions which might differ from book to book:
[quote="lyly"]We have that for every algebraic k-prevariety $ (V;\mathcal{O}_V )$[/quote] This is a locally ringed space which locally looks like the maximal spectrum of a finitely generated k-algebra, right?[quote] such that $ \Gamma(V;\mathcal{O}_V )$ is an affine k-algebra, [/quote]This is a finitely generated k-algebra, right?[quote]there is a canonical morphism of prevarieties $ f: (V;\mathcal{O}_V )\to Spm\Gamma(V;\mathcal{O}_V )$, such that $ f(P) \equal{} \{f\in \Gamma(V;\mathcal{O}_V ) : f(P) \equal{} 0\} \equal{} : \mathfrak{M}_P$[/quote]I think $ f$ is used twice here. I suspect that there is a morphism $ F: (V;\mathcal{O}_V )\to Spm\Gamma(V;\mathcal{O}_V )$ with $ F(P) \equal{} \mathfrak{M}_P$ with the latter definition of $ \mathfrak{M}_P$.[quote]
Show that $ (V;\mathcal{O}_V )$ is affine if and only if $ \Gamma(V;\mathcal{O}_V )$ is an affine k-algebra and $ f$ is an isomorphism.[/quote]Anyway, the basic tool for the proof is the Hilbert Nullstellensatz.


calculus
derivative
limit
function
complex numbers
complex analysis

Show that
(a) directly
(b) using the Cauchy Riemann equations
that if $ f(z) \in \mathbb{R}$ $ \forall z \in \mathbb{C}$, then for any $ z \in \mathbb{C}$ if $ f'(z)$ exists then $ f'(z) \equal{} 0$.

I don't know how to do this. For part (a) I have tried using the definition of the derivative and wasn't sure how to do it. I am not sure how to use the Cauchy Riemann equations, we have just started on them. Thanks for any help.

For using the definition of the derivative, you're looking at $ \lim_{w\to0}\frac{f(z\plus{}w)\minus{}f(z)}{w}.$

Try looking at that limit along two specific path (among many other possibilities) for $ w\to 0:$ one path along the real axis, and one path along the imaginary axis. Note that a real divided by a real is real, and a real divided by a pure imaginary is pure imaginary. What can you say about a number that is both real and pure imaginary?

One common notation for the C-R equations is to let $ u$ be the real part of $ f$ and $ v$ be the imaginary part of $ f.$ The statement that $ f$ is real-valued is the the same as the statement that $ v\equal{}0.$

One part of the C-R equation formality is that for differentiable functions, $ f'\equal{}u_x\plus{}iv_x\equal{}v_y\minus{}iu_y.$

Using Cauchy-Riemann:

Let $ f(x,y)\equal{}u(x,y)\plus{}iv(x,y)$. Any function $ f(z)$ can be written like this letting $ z\equal{}x\plus{}iy$. The CR equations state: $ \partial_x u \equal{} \partial_y v$ and $ \partial_y u \equal{} \minus{}\partial_x v$ where $ \partial_x$ means differentiation with respect to $ x$ and $ \partial_y$ means differentation with respect to $ y$. Now the condition that $ f$ is real valued simply means $ v\equal{}0$. Therefore our diff. equations reduce to:

$ \partial_x u \equal{} 0$ and $ \partial_y u \equal{} 0$.

Now: we know $ \partial_z \equal{} \frac{1}{2}(\partial_x \plus{} i \partial_y)$ So applying this to $ f(x,y)\equal{}u(x,y)$ (we drop $ v$ because its equal to zero) we get:

$ \partial_z u(x,y)\equal{}\frac{1}{2}(\partial_x u \plus{} i \partial_y u) \equal{} \frac{1}{2}( 0 \plus{} 0 ) \equal{} 0$

q.e.d.

[quote="LkNsngth"]Now: we know $ \partial_z \equal{} \frac {1}{2}(\partial_x \plus{} i \partial_y)$[/quote]
That should actually be $ \partial_z \equal{} \frac {1}{2}(\partial_x \minus{} i \partial_y)$

Just apply the operator to the function $ f(z)\equal{}z\equal{}x\plus{}iy$ to verify.

$ \frac {1}{2}(\partial_x \plus{} i \partial_y)$ would be $ \partial_{\overline{z}}.$

(None of which changes the answer to this problem.)


$ a,b,c>0$
Prove that: $ \sum (\frac{a}{b})^2\plus{}3(\frac{ab\plus{}cb\plus{}ca}{a^2\plus{}b^2\plus{}c^2})^2 \ge 6$
 :)

[quote="talaaiha"]$ a,b,c > 0$
Prove that: $ \sum (\frac {a}{b})^2 \plus{} 3(\frac {ab \plus{} cb \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2})^2 \ge 6$
 :)[/quote]

I think we use SOS
Prove that
$ \sum (a \minus{} b)^2 .\left[\frac {1}{b^2} \minus{} \frac {3}{2(a^2 \plus{} b^2 \plus{} c^2)}\right] \geq 0$


modular arithmetic
vector
function
linear algebra
combinatorics proposed
combinatorics

Let $B$ be a subset of $\mathbb{Z}_{3}^{n}$ with the property that for every two distinct members $(a_{1},\ldots,a_{n})$ and $(b_{1},\ldots,b_{n})$ of $B$ there exist $1\leq i\leq n$ such that $a_{i}\equiv{b_{i}+1}\pmod{3}$. Prove that $|B| \leq 2^{n}$.

Use the linear algebra method:
 For each $b \in B$, define $f_{b}: Z_{3}^{n}\rightarrow \mathbb{Z}_{3}$ as follows: \[f_{b}(x_{1},\ldots,x_{n})=\prod_{i=1}^{n}(x_{i}-b_{i}-1).\] From the hypothesis we have that for all $b \neq b' \in B$, $f_{b}(b')=0$. Further $f_{b}(b)\neq 0$. By  a simple linear algebra argument $\{ f_{b}$, $b \in B\}$ are all independent in the vector space of functions from $\mathbb{Z}_{3}^{n}$ to $\mathbb{Z}_{3}$. Let $\mathcal{U}$ be the linear span of $\{ f_{b}, b \in B\}$, so that $dim(\mathcal{U})=|B|$. Now, since each $f_{b}$ is a linear combination of monomials of the form $\prod_{i \in S}x_{i}$, $S \subset [n]$ and there are $2^{n}$ such monomials, $dim(\mathcal{U}) \leq 2^{n}$. Therefore $|B|\leq 2^{n}$.

[quote="hhaller"]$dim(\mathcal{U})=|B|$.[/quote]
What is this ? :) I don't understand :)


The existence of photons as quanta of energy can be experimentally verified only when they strike matter and are fully absorbed, loose some of their energy, or are anihilated - the photoelectric effect, Compton effect and the electron-positron pair formation, the last one only if the photon energy is at least 1022 keV (2 times the electron rest mass). As long as the light just propagates, it is described as a wave (spherical for a point source, plain for parallel rays), which is a solution of Maxwell's equations. The measurable electric and magnetic fields in these equations, as well as the current and charge densities in the 1st pair of these equations, are the mean values $\left<\right.\vec E(\vec r, t)\left.\right>$, $\left<\right.\vec B(\vec r, t)\left.\right>$, $\left<\right.\vec j(\vec r, t)\left.\right>$, $\left<\rho(\vec r, t)\right>$ in some small spherical volume around the point $\vec r$. The difficulty is in calculating the mean current and charge densities, which are different from the conduction current and free charge densities due to the polarization and induction effects of the electromagnetic wave on the medium consisting of charged particles with spin. Recall that the light velocity in a medium is $c' = \frac c n$, where $n = \sqrt{\varepsilon \mu}$ is the refraction index and $\varepsilon, \mu$ the relative electric permitivity and magnetic permeability of the medium. The mean values have to be taken over a volume containing many atoms. On the other hand, the volume diameter must be much smaller than the light wavelength $\lambda$. For visible light, the wavelength is really much larger (about $\approx 10^4 \times$) than the atom dimensions, so this averaging is possible. The photons can be imagined as being repeatedly absorbed and emitted by atoms or collections of atoms, which takes time and slows down the wave propagation. If the wavelength is too short (i.e., for $\gamma$-rays), this averaging is not possible and the wave will propagate with the same velocity c as in vacuum, regardless of what the medium is.

excellent info. Yoy mean speed of light in vacuum also depends on its frequencey?!!!

I believe you are crystal clear in this > Where did you get to know all this. Can you tell me your resource.

Also , in most of your post , you have told me about fields and maxwell's eqns. CAn you tell me the significance of photon in maxwell's eqns. Don't they refer just to waves?!

Do you mean that there is no photon which moves unhindered?!!

[quote="Rushil"]You mean speed of light in vacuum also depends on its frequencey?!!![/quote]
By no means. The speed of light in vacuum is constant, $c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$, where $\varepsilon = 0.885 \times 10^{-11}\ \text{F/m},\ \mu_0 = 1.257 \times 10^{-6}\ \text{H/m}$ are the electric permitivity and magnetic permeability of vacuum. This is the speed of plain electromagnetic waves in vacuum that comes out of Maxwell's equations when the current and charge densities are zero. But the speed of light in a medium depends on frequency, because the refraction index depends of frequency, which is demonstrated by chromatic aberration of lenses, dispersion of white light by a prism or by raindrops (rainbow), etc.

[quote="Rushil"]
I believe you are crystal clear in this > Where did you get to know all this. Can you tell me your resource.

Also, in most of your post , you have told me about fields and maxwell's eqns. CAn you tell me the significance of photon in maxwell's eqns. Don't they refer just to waves?![/quote]
The 1st part about the particle-wave duality comes from "philosophically minded" textbooks on quantum mechanics. But I looked into the quantum mechanics textboks, which I have, and they do not mention it. A free "particle-wave" is completely unlocalized, because for a definite value of momentum, we need to count an infinite number of wavelengths (see also the uncertainty principle). Hence, it lacks the main particle-like property, so we call it a wave. On the other hand, if the particle-wave strikes matter, it becomes localized up to the resolution of the detector (the size of  a photographic emulsion grains, the size of bubbles in liquid hydrogen in a bubble chamber, etc) and particle-like. Once we decide that the light propagating through a medium behaves mainly like a wave, we can invoke Maxwell's equations for this wave. A completely satisfactory answer would require using the quantum field theory and I am simply not good at it.

The 2nd source is the textbook "Principles of the Special Theory of Relativity" by V. Votruba, 1969 - the best relativity textbook that I have ever seen, but it is written in Czech, so it will not help you. Votruba was a professor of theoretical physics at the Charles University of Prague, Czechoslovakia.  In the book. he was able to put on the mindset of an experimental physicist, who is not willing to accept the new concepts about the space, time and matter, based on the results of just Michelson-Morley's experiment. Hence, he develops Lorentz's electron theory and the theory of ether, trying to explain (with detailed calculations) just about every relevant phenomenon or experinent (even the recent ones) from these viewpoints, having some success, but eventually demonstrating their inadequacy. Only then he introduces the 2 relativity principles, develops the theory and applies it to the entire classical physics, mainly mechanics and electrodynamics, again with detailed calculations and constantly comparing the theoretical deductions with the available experimental results.

Yetti

So basically(since I don't know much about detailed Quantum mechanics and absolutely nothing about QFT) , you want to say that while light is propagating in space, there are just so many paticles and wavelenghts , that we can completely forget(effectively) about the granular nature of light and consider it to be a wave satisfying maxwell's eqns. iT's only when that photon strikes an atom etc, thatt its existence becomes significant. Rest of the time we may totally negelct it and consider light to be a wave!!!

So I interpret it as this :- when light is moving in the tiny vacuum between 2 atoms in matter , it consists of innumerable photons travelling at speed of light(as required by special relativity). Since the no. of paticles is large, we invoke Bose- einstein Statistics and consider light to be a wave travelling at speed of light in vacuum between those 2 atoms. Only when light (photon ) strikes the atom does its influence become significant. But then one question - why is speed of light reduced in matter? Since light only travels in vacuum between any 2 atoms at speed of light(in vacuum), and when it strikes an atom , photons are absorbed and reemitted in arbitrary directions, why should it's speed be reduced? Maxwell's eqns are for continous matter. What if we analyse the atoms discretely?? Does QFT come in???

Also, are you an undergrad or grad student? just for knowing. do you know of any place on net where I can get info (in english) about the book you are recommending

[quote="Rushil"](since I don't know much about detailed Quantum mechanics and absolutely nothing about QFT) [/quote]
I would place this discussion under classical electrodynamics rather than quantum mechanics. Also, what is QFT?

QFT is quantum Field theory which is basically an experiment at combining quantum mechanics and relativity. Am I right?!!!

We've hopelessly wandered off from the original question . We've reached to QFT which is a matter of PhD and I don't have any interest right now. Also I don't see it's significance in my question(plz correct me if I'm wrong).
[size=150]I would like to ask again....[/size]
The speed of light is reduced in some medium(like water) . Maxwell's eqns dont hypothesise photons and the EM wave equation and all other tool of Maxwell's thoery work perfectly well for light travelling in a medium.

[b]Now if we consider the quantum theory of light. It says that light is made up of photons. At this point forget completely about Maxwell's eqns and just keep in mind that the photon has wave-like character , nothing else. Now if we want to explain that light travels slower in some medium , how do we do that. Effectively, that is my question. Relativity seems to contradict because it says that photons exist ONLY at speed of light in vacuum. Also in a medium , when photons travel between 2 atoms(say) , there is vacuum.... So how does light travle slowly in a medium acc. to Quantum Theory. Also, if it does, how are photon's able to exist at speeds lower than $ 3 \times 10^8 $????[/b]

One way out could be that if we are on the microscopic level in a medium, all of the lorentz transformations etc. involve $c$ as NOT $ 3 \times 10^8 $ but as the speed of light in THAT medium. At this point we forget about even the existence of free spcae, and formulate all our eqns with that speed in THAT medium. We know nothing about vacuum.... Is this correct???([hide]I've asked  a deeply related question seperately ... PLZ SEE!!!   :lol: )[/hide]

Plz answer this question or guide me!!!!

Yes indeed...

Actaully, I figured the right thing is this. When photons travel in a medium , they do move at $ 3 \times 10^8 \frac{m}{s} $ .They are absorbed by atoms and immediately emmited in random directions due to excitaion and emiision. It so happens, that because of the propogation of photns etc, and other reasons, to us who are ovserving, it [i]appears[/i] as if light is moving slowly. The light is deviated a bit and it [i]apperas[/i] to us as such. It's not that the speed of photons has changed. It's just that, they continue to move at speed of light(as demanded by relativity) , but because of scattering by atoms, the speed seems to be lowered.
I think this is the best explanation for my question. If somebody thinks otherwise or would like to add, please post. PLEASE post your opinion!!!

My opinion is [b]stop triple posting[/b]. Please use the edit button. Not fair to other people's post just to bump this topic to the top. Also, you are the one who started wandering off topic and you are the one who started mentioning QFT

yeah, I always told myself it was because there's a time delay between when atoms absorb/emit photons.

[quote="Rushil"]According to Special Relativity, A particle with rest mass = 0 can exist if and only if it is moving at the speed of light. A photon is a particle which comes into this category. However, when light travels in a medium it's speed is always reduced.
[b]Does this contradict relativity? What I want to ask is that when light travels at a lower speed in some medium. do photons exist and why? Why do photons 'seem' to exist at speeds less than that of the speed of light.[/b]

[hide]If I am not clear , feel free to ask!![/hide][/quote]

Mass is a scalar quantity and thus rotationally invariant. The concept that mass is velocity dependent comes from a misunderstanding of how to apply the Lorentz transform.

The concept of rest mass being distinct from mass in a moving reference frame is a useful phantom created by particle physicists to simplify their bookkeeping.  Mass does not change with increasing velocity; momentum changes with increasing velocity according to Relativity, in a nonlinear way.

In general for relativistic particles: $E^2 = p^2c^2 + m_0^2c^4$

Letting $m_0 = 0$ as for a photon, $E^2 = p^2c^2$

Therefore, $p = \frac{E}{c}$; the answer is, "Yes, photons exist in media."

For a given photon energy: $E =h\nu$, multiplying both sides by $\lambda$ yields $E\lambda =hc$.

Thus, assuming $E = \text{const.}$ (this is not a particularly valid assumption as all media have absorption spectra), as $c$ drops [i]in situ[/i] $\lambda$ also drops. Photons in media are different colors than they are  in vacuum.


geometry
trigonometry
rotation
geometric transformation
invariant

Note: The test comes with a diagram, so for clarification, check [url=http://www.delta.edu/math/mmpc/graphics/exams/1975.II.problems.pdf]the fourth problem here.[/url]

A,B,C are three equally spaced points on a horizontal hoop. Prove that as the hoop rolls along a circular line l, the sum of the distances of the points A,B,C  above line l is constant.

Oh, and if you solve it,

[hide]find another way :)[/hide]

I believe you didn't mean it when you said "circular line l" :D.

Well, sometimes I believe

circular $\equiv$ horizontal :)

Here's a big hint for one way to do it:

[hide]$\sin x + \sin (x+120) + \sin (x-120)$
$= \sin x + \sin x \cos 120 + \sin x \cos 120$
$= \sin x (1 + 2\cos 120)$
$= 0$[/hide]

is this solvable without trig (want to know whether or not i should spend time trying)?

Hint hint  :)

[hide]Complex Numbers?[/hide]

Hmmm didn't think of that.  That's cool.

ive got another
if A moves to A1 and B moves to B1 and C moves to C1, draw in the lines AA1, BB1, CC1.
if you extend them, they connect to an equilateral triangle and are a center of it. this means we can shift these imwards equally to make another equilateral equidistant from each of AA1, BB1, and CC1.
call this tirangle XYZ, and it remains to prove that if we rotate X Y and Z around, we get the same distance from the line L
continue the above mentioned step so that it converges on a dot. which is a constant

edit: after looking it over, my method is very similar to seamus boyels

[quote="seamusoboyle"]Since the triangle is equilateral, the median lies on the center. if you thought of the three points as having co-ordinates, the x, co-ordinate being how high they are up from L, the median of this triangle is,

$\frac{x_1+x_2+x_3}{3}$, which we know is constant. (Defination of a circle), the rest is obvious. Look for the beautiful proof![/quote]

Ahh. This is a very nice, simple proof.

Ahem... someone has yet to post the complex numbers method.

seamus's method essentially uses complex numbers.

WLOG we can write

$z_j=e^{2\pi i j/3}$.  It's clear that $z_j^3-1=0$ and hence $\sum_j z_j=0$ by the Viete formulas.  Under rotation, we get $z_j\to cz_j$ where $\vert c\vert=1$.  But then they satisfy the equation $(cz)^3=c^3$ and hence $\sum_j cz_j=0$ again.  Another rotation leaves the sum of the roots invariant and we are done.

You can, of course, apply some sort of linear transformation to the roots to move them to an arbitrary circle; it's clear that this can be done with scaling and translation; translation moves up all roots by an equal amount, and scaling still leaves the sum of all of the roots at the center of the circle.

Let a, b, and c be third roots of a magnitude 1 complex number z.  They are solutions to $x^3 - z = 0$.  Therefore their sum is 0 by Viete's formulae.


Prove that p>=3/2  :sqrt: 6Rr

pr=S+abc/4R≤{[(a+b+c)/3]^3}/4R so pr≤2 p^3/27R thus p^2≥27/2 R r
and the result follows

Dear Iris, you are too kind with the moderators  :D  : you post a problem, give a solution and directly put it into the solved section. All that with only one post. Wow!

Pierre.

Greetings Pierre!

I guess that was some irony in that  :cool: but what am i supossed to do if i want to show the others the solution of a problem that i have liked?


Prove that in a triangle ABC we have
$\large \frac{1}{-a+b+c}+\frac{1}{a-b+c}+\frac{1}{a+b-c}\geq \frac{\sqrt{3}}{2r}$

It 's very easy to see that r/(p-a)=tg(A/2), and we have done. :)


find all positive integers $n$ that $n+1,n+11,n+111$ are all primes.

Thank's Arne, I solve it. They are all primes only when $n=2$


Find min
\[ A\equal{}\sqrt {x^2  \minus{} 2x \plus{}58}  \plus{} \sqrt {x^2  \plus{} 2x \plus{} 10}
\]
This is solution in my book
\[ A \equal{} \sqrt {\left( {x \minus{} 1} \right)^2  \plus{} 4}  \plus{} \sqrt {\left( {x \plus{} 1} \right)^2  \plus{} 9}
\]
Choose
$ \vec{u}\equal{}(x\minus{}1;2),\vec{v}\equal{}(\minus{}x\minus{}1;3)$
So we have
$ |\vec{u}|\plus{}|\vec{v}|\ge |\vec{u}\plus{}\vec{v}|\equal{}\sqrt{29}$

But if we choose $ \vec{v}\equal{}(\minus{}x\minus{}1;\minus{}3)$ we have  $ |\vec{u}\plus{}\vec{v}|\equal{}\sqrt{5}<\sqrt{29}$
Why we don't choose it. I know if we choose it so the equality is not occur.

But my problem : How to choose the first vector which is accurate in a fast way.

[quote="Altheman"]You want u and v to be in the same direction [as opposed to opposite directions].[/quote]
?????????????????????????
What are u saying ??

If $ u\equal{}(1,2)$ and $ v\equal{}(3,6)$, then $ |u\plus{}v|\equal{}|u|\plus{}|v|$, but if $ u\equal{}(\minus{}1,\minus{}2)$ and $ v\equal{}(3,6)$, then $ |u|\plus{}|v|>|u\plus{}v|$.

It is not easy to determine if the vectors in the given problem are the same direction or not because you don't know the sign of the first component until you solve the equation.

In this case, you can tell because the second term $ \sqrt{(x\plus{}1)^2\plus{}9} \geq \sqrt{9} > \sqrt{5}$, so $ \sqrt{5}$ is likely not the bound you are looking for. Rather, you know that $ \sqrt{(x\minus{}1)^2 \plus{} 4} \plus{} \sqrt{(x\plus{}1)^2\plus{}9} > \sqrt{4} \plus{} \sqrt{9} \equal{} 5 \equal{} \sqrt{25}$, so $ \sqrt{29}$ is the more reasonable bound.

So we have that $ A \equal{} \sqrt{(x\minus{}1)^2\plus{}(\minus{}2)^2} \plus{} \sqrt{(x\plus{}1)^2 \plus{} 3^2}$ is the distance from $ (x\minus{}1,\minus{}2)$ to $ (0,0)$, plus the distance from $ (0,0)$ to $ (x\plus{}1,3)$. If we pick $ x$ such that $ (x\minus{}1,\minus{}2)$, $ (0,0)$, and $ (x\plus{}1,3)$ are collinear, then our sum is minimized, and $ A \equal{} \sqrt{((x\plus{}1)\minus{}(x\minus{}1))^2\plus{}(3\minus{}(\minus{}2))^2} \equal{} \sqrt{29}$.

Notice that the sum is minimized when the points are collinear, which is analogous to when $ \vec{u}$ and $ \vec{v}$ are aligned.

It's best to select signs in a way that you can [u]see [/u]something being minimized - in this case, the two distances and the net distance form a triangle, which gives direct, geometric interpretation to the triangle inequality of vectors. We want to make this triangle into a degenerate triangle.


That is not always possible. If $ P$ is the center of the circle then $ PA/PB\equal{}1$.

how is the \[ \minus{}\frac{PA}{PB}\]  construction?

 :(


inequalities
triangle inequality
inequalities unsolved

[quote="IOO"]For all $ a_1,a_2,b_1,b_2,c_1,c_2\in R$ prove that
$ \sqrt {a_1^2 \plus{} a_2^2} \plus{} \sqrt {b_1^2 \plus{} b_2^2} \plus{} \sqrt {c_1^2 \plus{} c_2^2} \plus{} \sqrt {(a_1 \plus{} b_1 \plus{} c_1)^2 \plus{} (a_2 \plus{} b_2 \plus{} c_2)^2}\geq \sqrt {(a_1 \plus{} b_1)^2 \plus{} (a_2 \plus{} b_2)^2} \plus{} \sqrt {(b_1 \plus{} c_1)^2 \plus{} (b_2 \plus{} c_2)^2} \plus{} \sqrt {(c_1 \plus{} a_1)^2 \plus{} (c_2 \plus{} a_2)^2}$[/quote]
I found a very nice proof (and new?) in my opinion! :)

Let $ a \equal{} a_1\plus{}ia_2, b \equal{} b_1\plus{}ib_2, c \equal{} c_1 \plus{} ic_2$. Then we have to prove:

$ |a\plus{}b\plus{}c| \plus{} |a| \plus{} |b| \plus{} |c| \ge |a\plus{}b| \plus{} |b\plus{}c| \plus{} |c\plus{}a| \iff$

$ |a\plus{}b\plus{}c| \minus{} (|a| \plus{} |b| \plus{} |c|) \ge \sum_{cyc} |a\plus{}b| \minus{} |a| \minus{} |b|$

But $ |a\plus{}b\plus{}c| \minus{} (|a|\plus{}|b|\plus{}|c|) \equal{} \frac{|a\plus{}b\plus{}c|^2 \minus{} (|a|\plus{}|b|\plus{}|c|)^2}{|a\plus{}b\plus{}c| \plus{} |a|\plus{}|b|\plus{}|c|} \equal{}$ $ \minus{}2 \frac{\sum_{cyc} (|a| |b| \minus{} a \cdot b)}{|a\plus{}b\plus{}c| \plus{} |a|\plus{}|b|\plus{}|c|}$. Note that $ a \cdot b$ is the real product. It is obvious that $ |a\parallel{}b| \ge a \cdot b$..

Likewise $ |a\plus{}b| \minus{} |a| \minus{} |b| \equal{} \frac{|a\plus{}b|^2 \minus{} (|a|\plus{}|b|)^2}{|a\plus{}b|\plus{}|a|\plus{}|b|} \equal{} \minus{}2\frac{|a| |b| \minus{} a \cdot b}{|a\plus{}b|\plus{}|a|\plus{}|b|}$

Now we are left to prove:

$ \sum_{cyc} \frac{|a| |b| \minus{} a \cdot b}{|a\plus{}b|\plus{}|a|\plus{}|b|} \ge \sum_{cyc} (|a| |b| \minus{} a \cdot b) \left ( \frac{1}{|a\plus{}b\plus{}c|\plus{}|a|\plus{}|b|\plus{}|c|} \right ) \iff$

$ \sum_{cyc} (|a| |b| \minus{} a \cdot b) \left ( \frac{|a\plus{}b\plus{}c|\plus{}|c|\minus{}|a\plus{}b|}{(|a\plus{}b\plus{}c|\plus{}|a|\plus{}|b|\plus{}|c|)(|a\plus{}b|\plus{}|a|\plus{}|b|)} \right ) \ge 0$

But that is obvious since $ |a\plus{}b\plus{}c| \plus{} |c| \equal{} |a\plus{}b\plus{}c| \plus{} |\minus{}c| \ge |a\plus{}b\plus{}c\plus{}(\minus{}c)| \equal{} |a\plus{}b|$ by the triangle inequality, and the fact that $ |a| |b| \ge a \cdot b$.

[quote="dima ukraine"][quote="Mathias_DK"][quote="IOO"]For all $ a_1,a_2,b_1,b_2,c_1,c_2\in R$ prove that
$ \sqrt {a_1^2 \plus{} a_2^2} \plus{} \sqrt {b_1^2 \plus{} b_2^2} \plus{} \sqrt {c_1^2 \plus{} c_2^2} \plus{} \sqrt {(a_1 \plus{} b_1 \plus{} c_1)^2 \plus{} (a_2 \plus{} b_2 \plus{} c_2)^2}\geq \sqrt {(a_1 \plus{} b_1)^2 \plus{} (a_2 \plus{} b_2)^2} \plus{} \sqrt {(b_1 \plus{} c_1)^2 \plus{} (b_2 \plus{} c_2)^2} \plus{} \sqrt {(c_1 \plus{} a_1)^2 \plus{} (c_2 \plus{} a_2)^2}$[/quote]
I found a very nice proof (and new?) in my opinion! :)

[/quote]

I dont think so :wink:  Inequality That you use - special case of Hlawka ineq :)  But cool solution :)[/quote]
Can you tell me more about the Hlawka ineq ? I have no idea what it is  :oops:

[quote="dima ukraine"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16600  :)[/quote]
Ok, thanks :)

I was actually pretty sure before hand that the ineq $ |x| \plus{} |y| \plus{} |z| \plus{} |x\plus{}y\plus{}z| \ge |x\plus{}y| \plus{} |y\plus{}z| \plus{} |z\plus{}x|$ was well-known. I was only thinking, that the proof for Hlawka's ineq was new :)


function
logarithms
algebra
floor function
number theory
prime factorization

Find the number of zeros in which $ 2009!$ ends.

The answer will be 500.

Use $ [\frac{2009}{5}]\plus{}[\frac{2009}{25}]\plus{}[\frac{2009}{125}]\plus{}[\frac{2009}{625}]$ where [.] denotes the greatest integer function
Hoping no calculation mistake is done.

[quote="21 7 15"]My anwser is 7820[/quote]How plz tell :?:

Ok, so how do you add a 0 in the end of a number? Any number? Multiply it by 10. So, we have to find all of the factors of 10 in 2009!

But how do we get factors of 10? Well, think of the prime factorization of 10. What is it? 2x5. So, if we just find the number of 2x5's in 2009!, we can get the number of 0's.

Since all of the even numbers contain 2, we have plenty of them already, so well look for 5's. Which numbers have 5 in them?

5, 10, 15, 20, ...

So, one in every 5 numbers has a 5 factor in it. So we divide 2009 by 5 and get 401. So, you say that there are 401 0's In 2009. But are we done? NO! Some factors of 5 have 2, or even 3 5's in them.

25, 50, 75

125, 250, ...

625, ...

So, we count how much of each factor is in 2009 and add it to 401.

2009/25 = 80
2009/125 = 16
2009/625 = 3

Add them all up and our answer is 401 + 80 + 16 + 3 = 500

500 is our answer!
(I hope I didn't make any mistakes. :D)

Hmm.. just a note if you want to learn more about these problems you should read Intro to NT

[quote="RAGHAV GROVER"]But how will you calculate $ \log_{10}2009!$  :?:[/quote]

You don't. Thats the point. To solve the problem you don't actually multiply it out. Instead you find the factors of ten. $ \log_{10}2009!$ is irrational.

[quote="21 7 15"]If x=2009! Then log[2009!]+1 is the required anwser which is not 7820 sorry[/quote]

That would be the number of digits in 2009!


Hey everybody!!! LEt's play chess?

It's the following, I make a move, and the next person to post here make another move, and etc...

I Start

1- d4 ...

darn you i love queen's gambit =)

2.) nf3

hey.

My play
1- d4 ...

you need to move the black pieces. not the white. hehe

you made 1- ...d5    and after  2- nf3   ?

i did not understant. the queen'd gambit is 2- c4

ahm!?

i will suppose and made

1- d4   d5
2- Nf3  Bf5

[quote="7h3.D3m0n.117"]soz my bad nf6 =)[/quote]

1. d4 Nf6
2. c4

1. d4 Nf6
2. c4 e5

There is a strange thing.

I searched in http://www.chesslive.de/  the opening 1. d4 Nf6  and i didn't found nothing!!!

In chesspartner this opening is very common. strange...

I wonder why chesslive dont have a data basis of this...

because its called the Nimzo-Indian...

1. d4 Nf6
2. c4 e5

eww who did e5? ugh

I did e5 :)

i know this oppening

the strange thing is there is no databasis of this on the site.

[quote="Zarark"]i know this oppening
[/quote]

Correction: you know white's parts. You want black to do e5. the real counter (and possibly the best one) involves Bb4.

Nimzo-Indian Defence
1.d4 Nf6 2.c4 e6 3.Nc3 Bb4

Correction

the nimzo-indian defence is e6 not Bb4.

i made e6 to become more interesting.

but, if you prefer a """Traditional""" (read this like more usual) game i made e6.

no problem

EDIT: The name

Budapest Gambit
1.d4 Nf6 2.c4 e5

http://chess.about.com/library/weekly/aa04g31.htm

[quote="probability1.01"][quote="Zarark"]1. d4 Nf6
2. c4 e5
[/quote]
3. dxe5[/quote]
3. ...Ng4.

[img]http://server6.theimagehosting.com/image.php?img=game.8e4.png[/img]

This is not an easy opening to play for white, because you screw up your pawn structure right from the start, and black plays on that

i go for 4.Bf4

need some development

My play

1. d4 Nf6
2. c4 e5
3. dxe5 Ng4
4. Bf4 Nc6

http://s6.fileho.com/download3/f9ecfe839655_c19m5t4re809nsgu/Chess.JPG

[quote="Zarark"]My play

1. d4 Nf6
2. c4 e5
3. dxe5 Ng4
4. Bf4 Nc6
[/quote]
5. Nf3

1. d4 Nf6
2. c4 e5
3. dxe5 Ng4
4. Bf4 Nc6
5. Nf3

...5. Bc5

No opening knowledge  :roll:

[quote="ch1n353ch3s54a1l"]1. d4 Nf6
2. c4 e5
3. dxe5 Ng4
4. Bf4 Nc6
5. Nf3 Bc5

No opening knowledge  :roll:[/quote]
same here... 6. e3

[quote="probability1.01"]1. d4 Nf6
2. c4 e5
3. dxe5 Ng4
4. Bf4 Nc6
5. Nf3 Bc5
6. e3[/quote]
My play: 6. o-o

[quote]1. d4 Nf6
2. c4 e5
3. dxe5 Ng4
4. Bf4 Nc6
5. Nf3 Bc5
6. e3 O-O[/quote]

7. Bd3

[quote]1. d4 Nf6
2. c4 e5
3. dxe5 Ng4
4. Bf4 Nc6
5. Nf3 Bc5
6. e3 O-O
7. Bd3 [/quote]

7. ...Bb4+

check

[quote="Zarark"]1. d4 Nf6
2. c4 e5
3. dxe5 Ng4
4. Bf4 Nc6
5. Nf3 Bc5
6. e3 O-O
7. Bd3 Bb4+
[/quote]
8. Nbd2

[quote]1. d4 Nf6
2. c4 e5
3. dxe5 Ng4
4. Bf4 Nc6
5. Nf3 Bc5
6. e3 O-O
7. Bd3 Bb4+
8. Nbd2 Re8
[/quote]

9. Qc2


[quote="April"]For $\ a,b,c>0$ prove that:
$\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq\frac{3}{\sqrt[3]{abc}(\sqrt[3]{abc}+1)}$.[/quote]
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=ineq+balkan&t=109888[/url].  :)


search
inequalities unsolved
inequalities

Thiswas proposed and solved by Zhaobin. Search the web, silouan.


function
modular arithmetic
ceiling function
algebra
binomial theorem

What are the last three digits of the smallest integer greater than $ (\sqrt{11}\plus{}\sqrt{10})^{100}$?

I don't know if you're allowed to use a calculator for this problem (and this is not high school basics level) if it's a competition problem. Take 2 cases: rational no.s, and then irrationals. Through incorporation of binomials we get 001 as the sum of the rationals (I might've done this wrong since I'm rushing a bit), and the sum of the irrationals would therefore give us $ 2^{99}\sqrt{110}$ (Again, make use of binomials, and could someone please check because I'm rushing). Now use the ceiling function on the irrational result, and it will be time-consuming to figure out what that is unless we memorize the digits of $ \sqrt{110}$. So is there any other way than this way? I'm probably wrong but...

You would probably want to add $ (\sqrt {11} \minus{} \sqrt {10})^{100}$ and use the binomial theorem on the even exponents to get rid of the messy radicals. I am too lazy to work out the details, however.

[hide="Hints"]Binomial Theorem is your friend here. $ (\sqrt{11}\plus{}\sqrt{10})^{100}\plus{}(\sqrt{11}\minus{}\sqrt{10})^{100}$ is an integer and $ (\sqrt{11}\minus{}\sqrt{10})^{100}<1$. Notice that there is a multiple of 10 inside that square root. Take advantage of that factor to reduce a bunch of terms on mod 1000.[/hide]

[quote="Caelestor"]What are the last three digits of the smallest integer greater than $ (\sqrt {11} \plus{} \sqrt {10})^{100}$?[/quote]
I think it is HSB level.

[hide="The answer"] Add $ (\sqrt {11} \minus{} \sqrt {10})^{100}$ to get rid of all the radicals. This gives something like $ (2)(11)^{50} \plus{} (100C2)(11)^{49}(10) \plus{} (100C4)(11)^{48}(10)^{2}$... (all of the ... doesn't matter because they all end in 3 or more zeroes) through the binomial theorem. I don't have time to finish it, but you get the idea.[/hide]

EDIT: beaten by 10000th and xpmath

Oh sorry, I was really careless  :( . Well then according to my earlier post, the answer should be 002.

[hide="Solution"]
The binomial theorem gives that the expansion is equal to:

$ 11^{50} \plus{} \binom{100}1 11^{49}\sqrt {11\cdot10} \plus{} \cdots \plus{} \binom{100}1 10^{49}\sqrt {10\cdot11} \plus{} \cdots \plus{} 10^{50}$

Unfortunately, all the radicals makes this considerably messy. Fortunately, adding $ (\sqrt {11} \minus{} \sqrt {10})^{100}$ will not make a difference, considering how a number less than one raised to teh 100th power is extremely small. That expansion is:

$ 11^{50} \minus{} \binom{100}1 11^{49}\sqrt {11\cdot10} \plus{} \cdots \minus{} \binom{100}1 10^{49}\sqrt {10\cdot11} \plus{} \cdots \plus{} 10^{50}$

And adding these two expansions will give us approximately the same answer, or when rounded, the exact answer. Now the expression is 2 times the integer parts of $ (\sqrt {11} \plus{} \sqrt {10})^{100}$. Since the problem asks for last three digits, Most of these terms will end in 000, beginning with the term having $ 10^{3}$. So we must find the last three digits of:

$ 2\left(11^{50} \plus{} \binom{100}2 11^{49}10 \plus{} \binom{100}411^{48}10^2\right)$

Finding the last thee digits of $ 11^{50}$ must be a pain without a calculator, right? Actually, you can use the binomial theorem to find this again.

$ (10 \plus{} 1)^{50} \equal{} \text{ ( lots of messy algebra here ) } \plus{} \binom{50}{2}100 \plus{} \binom{50}1 10 \plus{} 1^{50} \equal{} 122500 \plus{} 500 \plus{} 1 \equiv 001 \pmod{1000}$

As for the second term, 100C2 would be easy to calculate, and find the modular residue.

For the 3rd, you can bash this out as well, but you might notice that $ \frac {100\cdot99\cdot98\cdot97}{4!}$ has 0 factors of 2's, but some factors of 5. Therefore, the units digit of only that part is 5, so the modular residue is 5*100=500.

The modular residue of each of these three terms are 001, 500, and 500, so the last three digits is equal to 2(1+500+500) (mod 1000) = 002.
[/hide]

EDIT: Thanks Ihp, fo rletting me know what to put into my solution.

Is there any quick way to find the last 3 digits of $ (11)^{50}$?

you can use pascals triangle
in this case its the 50th row
so the ones digit is 50C0, tens digit is 50C1 and hundreds digit is 50C2. If the digit is a two or more digit number then just carry that digit over to the next place so it would be like
------1
----50
1225
so it would be 001

$ 11^{50}\equal{}(10\plus{}1)^50.$ When you take this modulo 1000 all the terms with $ 10^3$ or higher are 0. That means the only terms left are
$ \binom{50}{48}10^2\cdot1^{48}\plus{}\binom{50}{49}10^1\cdot1^{49}\plus{}\binom{50}{50}1^{50}\equal{}122500\plus{}500\plus{}1\equal{}123001\equiv\boxed{001\pmod{1000}}.$

That was specifically $ 11^50$, so don't box the answer; overall it would be $ 002$.


Prove that product 5 successive integers it is not perfect square.


Right this idea all started from:
suma_milli: lol I like the new avatar izzy! :D
guineapiggies: lol izzy.  i  :love: your new avatar! :D
isabella2296: Thanks Suma, piggies! :)
suma_milli: Your welcome :P
I can never change my avatar because I already know that I would be forced to change back by guassy.. :P
gauss1181: kill razz :D
isabella2296: See, :D is encumbering and horrid, like a dictator. Join  :P, freedom and equality for all!
dragon96: I will join 3rd party:  :yoda:
suma_milli: lol Freedom and equality sounds pretty good...hmmm...

That being said welcome to the 1st annual Smiley Presidential Election!  Join a party today!  Convince the public to vote for your party!

1) Parties must be based on one smiley.
2) You can switch parties in the middle but that says something about your charachter. *cough*[size=59]traitor[/size]*cough*
3) We may have a preliminaries in which a certain number of parties get cut off and then the real thing.

I am neutral at this point
(please disregard the avatar and signature..they are for my safety..)

Wow.. nice choice

For me it's a tie between
 :D  :rotfl:  :P

:huh: The message is too small. Please make the message longer before submitting.

unless someone decides to keep score, this is a really dumb idea.
hmm, there should be a smiley tug-o-war or hurt/heal

[quote="suma_milli"]Er you're right.

Gauss created a smiley hurt and heal before.. I don't know what happened to that.

(P.S I intended to keep score)

 :rotfl: hands out chocolates and gets +5 votes.

:D = 1
:P = 1
 :cursing: = 1
:rotfl: = 5
:huh: = 1

[hide]Too many smilies in your post: 6.
You can have at most 4 smilies in this message. Please remove some smilies before submitting this message.Too many smilies in your post: 6.
You can have at most 4 smilies in this message. Please remove some smilies before submitting this message.[/hide][/quote]
LAWLZ SUMA U TRAITOR
MAYBE YOU SHOULD HAVE PICKED  :D !!!! :D  :D

I support  :starwars: . We require work to save our economy!

[quote="isabella2296"]See, :P represents democracy, freedom, happiness, and JOY. Can I represent :P ? I was the original promoter of it.[/quote]

Idk, talk to your party about it.

[quote="suma_milli"]
:D = 1
:P = 12
 :cursing: = 1
:rotfl: = 5
:huh: = 1
:fool: = 1
 :starwars: = 1

[hide]Too many smilies in your post: 6.
You can have at most 4 smilies in this message. Please remove some smilies before submitting this message.Too many smilies in your post: 6.
You can have at most 4 smilies in this message. Please remove some smilies before submitting this message.[/hide][/quote]

lol, gaussy I'm a freelance promoter.

:P gets 10 more votes for izzy's little speech.

You guys can do it if you want to.  But keep the + or - # of votes to $ \le 10$.

:D = 1
:P = 12
 :cursing: = 1
:rotfl: = 5
:huh: = 1
:fool: = 1
 :starwars: = 1
 :idea: = 3

 :idea: has an idea! +3
[hide=" "]this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler BTM3! this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler this is filler BTM3!!![/hide]

this isnt much of a [b]debate[/b] is it? Its more like just voting.

Er yeah I figured afterwards that nobody wants to "debate" why people should vote for [i]smileys[/i].

So it's now more like a smiley popularity contest.

Plus just go crazy with it cuz' there are no rules against bribing, blackmailing, etc. :P

xD xD I don't know what the heck this topic is for anymore, so whatever everyone just go with the flow!

yeah I think we should have a smiley showdown in song writing

Hi Razzy!
Hi Greeny!
Do you want to go for a ride?
Sure Greeny!
Jump in...

I'm a Razzy girl in a razzy world
Life in aops, it's fantastic!.....

Maybe a mod should delete this lame topic :|

You are turning into mew...

[quote="Mewto55555"]Hi izzy!
Hi Mew!
Do you wanna play a cd?
Sure mew!
Join mine!
[/quote]

:rotfl:

Let's have some plot to overthrow :D

VOTE :rotfl: EVERYBODY

Fav 3 (besides Mr. Green):

 :wallbash_red:
 :yoda:
 :pilot:

+1 Mr. Green
+1 wallbash_red
+2 yoda for not being a smiley
+1 pilot
-1 rofl for being the most annoying smiley

[img]http://www.smiliegenerator.de/s36/smilies-18406.png[/img]

Obv best smiley.

[quote="eggylv999"]:spam: will rule all!!![/quote]
DUDE YO YOU CAN'T SPAM ON AOPS

@pythag: Ha ha ha ha ha.

@AIME15: wtf.

 :D  :D  :D PEOPLE SEE MY BLOG BACKGROUND

Sorry I haven't seen this earlier. :D
[votes as many times as possible for  :D ]
[joins :D party]

[img]http://www.smiliegenerator.de/s36/smilies-18406.png[/img]

+pythag's index score (for those of you that dont knwo thats AMC 10/12+10*AIME)

which is probably 300 but...

[quote="eggylv999"][quote="coolts"]Hmm...I gotta say... :spam: is king.[/quote]

yay!!!  :spam: has votes!!![/quote]

It's only 2! :D has 11 votes!!! :rotfl:

thanks for voting for  :D , people.

i highly appreciate it.

Are you kidding me?

VOTE FOR  :yoda:!!!  now, now, don't be discriminatin' against ugly ppl.  so.... VOTE FOR :yoda:!!!

:spam: is a good practice! (not) but it is still a good way to annoy people.

:rotfl: hands out chocolates and gets 9001 votes

:P hands out $9999999 cars to the whole multiverse and gets TREE(3) votes


Find min and max of $ A\equal{}x^2\plus{}y^2$ with $ x;y \in [\minus{}2009;2009]$ and satisfying
$ (x^2\minus{}2xy\minus{}y^2)^2\equal{}4$

hello, with the method of Lagrange Multipliers i have got
$ x^2\plus{}y^2\le 16144326\minus{}8036\sqrt{2018041}$
which is obtained by $ x\equal{}2009$ and $ y\equal{}\minus{}2009\plus{}2\sqrt{2018041}$.
Sonnhard.

For the minimum value, I make it $ A_{\min}\equal{}\sqrt2,$ attained at $ x\equal{}\sqrt{\frac{\sqrt2\plus{}1}2},\ y\equal{}\sqrt{\sqrt2\minus{}1}\minus{}\sqrt{\frac{\sqrt2\plus{}1}2}.$ There may be an easy way to verify this. My solution is too cumbersome to be worth posting here. :oops:


I just switched to a Dvorak layout because it is optimized for efficiency, unlike Qwerty, which was made inefficient on purpose so that people wouldn't type so fast that their typewriters broke. I was wondering if anyone else here uses this layout.

I've thought about using Dvorak a few times, but never have stuck to it.

i use QWERTY.
i don't even know where they sell Dvorak ones.

I use a QWERTY keyboard. I don't know anyone who doens't use one.

[quote="236factorial"]What's a Dvorak keyboard?  :?[/quote]
keyboard with a different layout to improve typing speed. i had never heard of it before; had to google.

You don't need a special keyboard to use the Dvorak layout. In Windows XP, go to Control Panel and choose Regional and Language Options. On the Languages tab, click Details. Then select US Dvorak and click Add. I don't know how to change this setting in other operating systems, but I know it is possible.

i have a QWERTY keyboard. where can you buy the other type of keyboard?

[quote="solafidefarms"]One way to do it is to set the arrangement on the OS to Dvorak, push all the keys in order, record what they do, and then with some pliers pull the keys off and replace them.  Somewhat drastic though.[/quote]
wow, i did not know you can take off the keys.

yea, keys are frequently found missing from laptops because the keys are very delicate there.

It's relativly easy to replace the keys to rearrange them :)

whats dvorak? :?:  i no what frogpad is tho...

[quote="math92"]i wish i had this key board
http://www.artlebedev.com/portfolio/optimus/[/quote]

Woah.  I want one of those...

OMG!! I want one of those!!

How much does one of those cost? It doesn't say on the webpage... I guess they don't want to give anyone a heart attack  :D

[quote]How much does one of those cost? [/quote]
According to the website, they're not for sale.

I use a QWERTY, although I want to switch to a Dvorak or asetniop keyboard (more effiecient).  However, that'd require learning to use it, and readjusting whenever I use a different computer/keyboard... :?

[EDIT] OK, switched my old computer to Dvorak, typing slowly now...

i switched to dvorak a few days ago after reading this thread (am now slowly typing this reply at about 20 wpm).  hopefully i'll be decently up to speed by the time school starts.

QWRty is the faster one, according to me, but it's the users preference. I can type up to 70wpm with qwrty.

QWERTY for me! Never even hear of other types of keyboards really..

[quote="anirudh"]. I can type up to 70wpm with qwrty.[/quote]
wow, i can only type 40 wpm

i'm now over 30wpm with dvorak!!! (and 70 w/ qwerty)
Instant messengers really help typing speed

yay for being almost exactly the same as L_Li...

lol dvorak is totally weird! but fun!

but only with words. I cant do that with like qweer stuff like xyeiosh eils ieh.  :)

I've heard of Dvorak, but use QWERTY.
qwertyuiop[]\asdfghjkl;'zxcvbnm,./
 :rotfl:

nrn ,day ap. frg yantcbi axrgy ege.ZZZ
(lol what are you talking about dude???)  :D

umm he is refering to the letters on the keyboard for a qwerty and what is the keyboard layout for the the dvorak?? that makes it so efficient? i didnt know there were any other keyboards i just thought a group of people decide this is where we put this now everyone is going to use it or die

the problem with dvorak is that you are suppiosed to use only 1 hand, but people are used to 2, and what will you do at school?  :oops:

err actually...you're supposed to use two hands w/ dvorak...?
the only dif is the key placement which "should" be easier cuz more words are more evenly split b/w the two hands and the more common letters are in easier to reach places lol
school is for WCIII and calculator games and hearts  :D

[quote="anirudh"]but are you going to do the lengthy procedure to change to dvorak including the reboot twice everytime you use the computer? :D[/quote]

Have you ever taken apart a keyboard?  If you have, you would see that the circuits go into a chip and then out to the PS/2 or USB plug.  PS/2 is a standard, no matter what keyboard you use.  The same letters are inputted into the computer, so you wouldn't really need to reboot.

WEll, would they even allow you to access the control panel at school?  ;)  ;)  ;)  ;)  ;)

there's no trouble for me to switch back and forth between dvorak and qwerty. i just added the dvorak configuration to my keyboard settings control panel, and fiddled around w/ some settings. now switching between the two is as easy as alt-shift-1 and alt-shift-2  :D

dvorak is supposed to help you type faster by placing letters so that it alternates between hands more (all vowels on one hand, most common consonants on the other)

well, but you will get so used to dvorak, that you will forget qwerty, and at school you might suffer(unless you do the alt ctrl 1 and 2 at school).

nah i now type decently with dvorak (about 40wpm), and...
*switches mode*
my qwerty typing speed is thus far unaffected, as far as i can tell

and no...it's no problem changing things on school computers even tho the admins try to make it hard...too bad the students have like 5 different admins to choose to use  :D

and yeah...once you know qwerty to a certain point...you'll never really forget it...and altho you might lose a bit of speed if you don't use it for a long time...it'll come back in at most a coupla minutes

do u use the same keyboard for both types because if you do then i would like to try it although
can you show me the exact keyboard configuration by running your finger on top of the rows of the computer

[quote="noneoftheabove"]can you show me the exact keyboard configuration by running your finger on top of the rows of the computer[/quote]
for which type of keyboard, qwerty or dvorak?


calculus
derivative
geometry
parallelogram
rectangle
calculus computations

Can anyone explain why the derative and differential coefficient are the same?

(p.s. I have a geometry problem: answer always, sometimes, or never.
If two chords bisect each other in the same circle, at least one must be a diameter. the answer is in the next post by me

Do 2 chords of the same length bisect each other by the intersecting chords theorem? Two diameters bisect each other. Therefore sometimes.

[quote="sairam"]the answer is sometimes, can someone explain this too.

Thanks[/quote]

This was mistakenly posted in another thread, now removed.

AndrewTom, sorry, I still don't understand what you mean.

It's wrong. If the two diagonals of a quadrilateral bisect each other, it's a parallelogram. The only parallelograms that can be inscribed in circles are rectangles. The diagonals of a rectangle are diameters of the circumcircle.


I found this ineq on an italian Contest.. it's really nice according to my opinion.. here you are

let $a,b,c$ positive reals. Prove that

$\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{3}\leq \sqrt[3]{\frac{a+b}{2}\cdot \frac{b+c}{2}\cdot \frac{c+a}{2}}$

 :)  Good job!

P.S if this has already been posted, you are welcomed to tell me..

This was problem 2 of the 3rd MathLinks contest, 1st round...

http://www.mathlinks.ro/Forum/viewtopic.php?t=4526
http://www.mathlinks.ro/Forum/viewtopic.php?t=4569 for an equivalent version

  darij


A sample of chlorine-38 is needed for an experiment. The radioactive material must be generated in a lab and then shipped six hours to the location where the experiment is being conducted. What is the minimum mass of the isotope that must be generated in order to have 2.00 mg at the start of the experiment? ($t_{1/2}=37.2$ minutes.)

I think I solved it.

We are given the half-life is 37.2 min = 0.62 hr. The sample is shipped 6 hr, which is

$\frac{6}{0.62}\approx 9.6774$ half-lives.

In order for there to be 2.00 mg left after 6 hr,

$N\cdot\left(\frac{1}{2}\right)^{9.6774}=2$ and therefore $N=1637.637$.

Thus, $\boxed{1.64\,\,\text{g}}$ is the minimum mass that must be generated.


AMC
USA(J)MO
USAMO
AIME
summer program
Mathcamp
geometry

Did you ever wonder which high schools produce the most USAMO qualifiers?  Here is a list of the top schools, ranked from high to low, according to the number of USAMO qualifiers since 1996.

Edit: See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31725]Top USAMO High Schools[/url] for the latest, most accurate list.

[code]
SCHOOL  STATE 96 97 98 99 00 01 02 03 04 TOT
T. Jeff.   VA  5  4  3  4 11 14 17 14 14  86
IMSA	    IL  6  5  3  3  5  8  9  6  6  51
Stuyvesant NY  7  5  1  4  5  9  8  8  4  51
AAST       NJ  5  2 15  2  2  3  8  5  6  48
Homeschool --  0  2  1  4  9  6  7  5  7  41
Blair      MD  2  3  2  0  5  2  9  8  5  36
Exeter     NH  1  2  2  1  1  1  4  9 11  32
Lexington  MA  2  2  1  1  5  6  4  4  3  28
Gunn       CA  4  3  2  2  2  3  3  3  5  27
Naperv. N. IL  1  0  1  3  2  6  2  3  3  21
Lynbrook   CA  1  2  1  0  1  4  3  4  3  19
Hunter     NY  0  1  2  1  3  4  4  2  2  19
Andover    MA  2  2  1  0  2  3  4  4  0  18
W. Windsor NJ  0  0  1  1  4  2  2  4  2  16
Madison W. WI  0  0  2  2  4  3  2  2  1  16
J. Madison WI  0  1  1  2  3  2  2  2  2  15
St. Mark's TX  0  0  1  2  2  4  3  1  1  14
NCSSM      NC  3  3  0  2  1  1  1  1  1  13
Saratoga   CA  0  0  0  0  3  2  2  3  1  11
LBJ Acad.  TX  4  2  1  1  1  1  0  0  1  11
Rox. Latin MA  0  0  0  2  1  3  2  3  0  11
[/code]
Let me know if you see any errors in this list.

Source:  A parent at AAST posted an initial version of this list at [url]http://www.nynjmq.org/math/usamo.htm[/url].  That parent in turn credits the AMC as his or her source.  I updated the list for the years 2003 and 2004, using information on the AMC USAMO page.

Here's an interesting related question -- how many unique qualifiers does each school produce?  For example, some people will be counted in that list 3, 4 or 5 times -- if we only count them once, what happens?  Just as an example, James Madison Memorial High School, in Wisconsin, has 15 qualifiers over that period -- however, half or so of those 15 come from two or three siblings, which is somewhat distorting, I'd say.

I agree with your numbers for Stuyvesant for 2000-2003.

I know some of the schools are wrong in the list of USAMO quals right now. Wait a few days, and hopefully they'll fix them. They did last year.

Joel, yes, that's an interesting question too, but I don't have the energy to figure it out.

Simon, thanks for the tip.  I'll check their list again in a week or two.  I already had to downgrade TJ's number this year by 1, because (as someone messaged me) one of this year's qualifiers is listed as a TJ student, but is actually a home-schooler.  TJ's numbers are too high anyway.  :)

[quote="yif man12"]i don't think anyone from my high school's ever made usamo[/quote]

Good luck becoming the first =)

I'm ashamed.. AAST.. 4th?

too bad i won't be going there so there goes that idea. well, they're bound to have an usamo qualifier someday.

[quote="MithsApprentice"][quote="Caroline"]
I'm ashamed.. AAST.. 4th?[/quote]
at least ur on the list...[/quote]
amen to that

Not only that, but if you discount 1998, you drop to 5th, a full 17 behind Stuy :)

1998 was an amazing year for AAST.  Actually, I couldn't double-check that number on the AMC page, because they don't have the 1998 qualifiers listed there.  How did AAST get 15 qualifiers in 1998, but only 2 in 1997 and 1999?!

[quote="JBL"]I know there was a year where there was an issue with AAST cheating on one of these exams ... I don't know much else about it, though.[/quote]

Hmm... I'm a freshman now, so I don't know much about years past.  That's surprising, though; AAST is always really good about no cheating..

[quote="JBL"]I know there was a year where there was an issue with AAST cheating on one of these exams ... I don't know much else about it, though.[/quote]

cheating???...how?

[quote="JBL"]I know there was a year where there was an issue with AAST cheating on one of these exams ... I don't know much else about it, though.[/quote]

Sounds like Stuy sour grapes to me :D

[quote="rrusczyk"]Not inexpensive ones.  Exeter had 3 IMO team members last year, and has both an excellent peer group and Zuming Feng (IMO Team leader).

Summer programs also offer outstanding peer groups.

As a self-serving aside, I'll note that it's tough to beat AoPS for an excellent peer group.  It's not in-person, to be sure, but students in our school and on our site in general find a very challenging and rewarding peer group.  If we could build an AoPS math team for something like ARML or the IMO, I think it would be a pretty good one :) .[/quote]

The only downside to AoPS is that you aren't face-to-face with the other people.  Other than that, AoPS is the perfect community, imo.

It's kinda weird...I was just thinking what would happen if AoPS could make an ARML team.  But that wouldn't work too well since it would make a lot of the local, state, and school teams furious because they were losing their best competitiors.  ;)

[quote="joml88"][quote="rrusczyk"]

It's kinda weird...I was just thinking what would happen if AoPS could make an ARML team.  But that wouldn't work too well since it would make a lot of the local, state, and school teams furious because they were losing their best competitiors.  ;)[/quote][/quote]

Yeh, good point.  But it'll be nice to have an all-star team at the competition.  :)

[quote="yif man12"]Edit: Also, Andover has more people than Exeter?  :huh:  :what?:[/quote]

Eew, over my dead body.  I don't understand how they even have as many as they do -- I guess '02 and '03 were good to them, but none in '04!

Heh, my bad. I was looking at the link and not the posted stats on the first page.

[quote="Ravi B"]Well, they might have 5 if you add up all those years, but you need 11 or more to make the list.[/quote]

Yeah, but the link in the OP's post has a chart going all the way down to 5.

There are no Southern California schools on that list. For one thing, there are no state-level or even metropolitan area-level magnet schools. Nothing like IMSA, or AAST, or some of the others. Even though there are no such magnet schools, several San Francisco Bay area schools did make the list: Gunn, Lynbrook, Saratoga. Southern California in general doesn't have the depth of participation of Northern California.

The one school I'm personally familiar with is Long Beach Poly. In the years indicated (and in the entire history of the school) the total number of USAMO qualifiers is 6. That's just two individuals - one who took it 4 times, and one who has taken it twice (so far). Is there anyone familiar with another school (North Hollywood, Harvard-Westlake, Irvine University, Arcadia) who can top that?

My school (Troy High School, Troy, MI) currently sits at 8.

It is also not included in the OP's lists (despite 7 qualifications thru 2002.)

A breakdown of our qualifications so far:
1999 - M. Asmar (10th)
2000 - M. Asmar (11th), S. Agarwal (10th)
2001 - M. Asmar (12th), A. Gupta (10th)
2002 - A. Gupta (11th), S. Agarwal (12th)
2003 - J. Yang (10th)
2004 - none.

I wish I can earn some honor for my school by becoming an USAMO qualifier this year...

Same, and the first in my school district i think =) We've been close, but none that i know of yet.

We had this list elsewhere, and I mentioned that you had forgotten my school, which had at least 20 qualifiers (I think it might be as many as 24, I don't have the older records though) in the time period mentioned.

This isn't reflected in this list.

Specifically, going by the records on the AMC site, (counting Stephen Guo who always attended our school but took the AMC at the local (not a school) Olympia Institute), we had:
2005 - 2
2004 - 0
2003 - 2
2002 - 3
2001 - 4
2000 - 5
1999 - 2

Li-Chung Chen I believe qualified at least twice as well (he made the IMO team in his last year).

I wrote this thread in 2004, and am not planning to update the list here.  Instead, I have edited my first post in this thread to refer to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31725]2005 list[/url].

The 2005 list already includes your school, Monta Vista.

Sorry, didn't notice that, someone linked to it in another thread and I thought it was a newer one.  :blush:

[quote="joml88"]It's kinda weird...I was just thinking what would happen if AoPS could make an ARML team.  But that wouldn't work too well since it would make a lot of the local, state, and school teams furious because they were losing their best competitiors.[/quote]
Well, USACO camp will be going on at the same time as ARML.  I looked at the rankings for USACO to get an idea of who will be there, and there are 7 MOSP qualifiers in the top 20 for USACO.  That would be more than twice what any school could bring, because TJ and Exeter will lose a MOSP qualifier to the camp.

On the topic of peer group, I definitely think that it makes a huge difference.  Being the only USAMO qualifier at my old school, I thought that I achieved enough just by qualifying, so I didn't work at all for USAMO (1pt in 9th grade, 2pts in 10th).  This year, at Exeter, I've been driven by the other students (and Mr. Feng too, of course) and did much better.
I think that middleschool matters, but not really how good the math program is.  I didn't start doing competition math until this summer.  I think the more important issue is that a good middleschool can challenge you, and force you to learn how to work hard.

[quote="MathChauffeur"][quote="Kent Merryfield"]...  (North Hollywood ... [/quote]Not NH. It has perhaps 3 to 4: 1 to 2 over 10 years ago, 1 about 5 years ago, and 1 last year. They have a Science Bowl and AP culture, but not a math culture as yet.[/quote]

I go to NH; while we don't have a group of outstanding math people, we seem to generally have a couple at the school at any one time. We had a USAMO qualifier this year, Tedrick Leung (10th), who apparently scored about in the middle. I expect him to qualify in 11th and 12th, as well.

Based on my performance on the AMC + AIME this year, I'm hoping to make it next year, and if I do, I can probably make it both years after that, but I will not be going to NH after this year, since I'm moving to Fullerton, so I won't be contributing to that.


geometry
parallelogram
geometric transformation
rotation
geometry proposed

Given are A, B, C and D in that order on the same circle centered O. A rotation of angle a around O gives the new quadrilateral A'B'C'D'. Let's denote M the intersection of AB and A'B', N, P and Q being similarly obtained.
Prove that MNPQ is a parallelogram.

Best regards,
sunken rock

We have:
$ \widehat{MNP}\equal{}180^o\minus{}\widehat{MNB'}\minus{}\widehat{PNC'}$ and $ \widehat{MQP}\equal{}180^o\minus{}\widehat{MQA}\minus{}\widehat{PQD}$

But $ \widehat{MNB'}\equal{}180^o\minus{}\widehat{ABB'}\equal{}180^o\minus{}\widehat{AA'B}\equal{}\widehat{MQA}$ and $ \widehat{PNC'}\equal{}\widehat{PCC'}\equal{}\widehat{PD'D}\equal{}\widehat{PQD}$ so $ \widehat{MNP}\equal{}\widehat{MQP}$.

Similarly, we also have $ \widehat{NMQ}\equal{}\widehat{NPQ}$.

Notice that a quadrilateral with two opposite angles equal to each other is a parallelogram, we got what desired.


Please join my competition. Info is in my blog and in my sig. I have extended the deadline for round one til midnight today.

Thanks

I've only received one answer (guess who :P )

-jorian

I want to join!

OK my answers for this round are:
1) Lake Michigan is the only one entirely in the US.
2) Maine is the northernmost state in the contiguous US.
3) Florida is the southernmost state.

Is this going to be a weekly thing? Or is it on some other schedule?

1. dunno (sees lake michigan)
2. I thought it was alaska
3. Hawaii? or florida

ahhhhhh!!!!!!!! you have to pm me these

u can't post them in public!!!!!!

i guess i'll count them but next time pm me

jorian


devide  $2x^{3}-3x^{2}+3x-3$ by $x-3$

i know how to do it in long division method, but is there any easier way than that ?

[hide]I'm not sure how to say it but i just learnt in a text book.
One way is as you said; long division.
There is another way where you work with co-efficients.
First list the co-efficients; 2,-3,3,-3.
Then as you are dividing by $x-3$, write it like this

3  l  2  -3  3  -3
...l____________

then take down 2(write it under the line under2)  Then multiply it by 3 and write the value under -3.  Then add the value with -3 and put it under the line again.  Do the same again.  You'll get this:

3  l  2  -3  3  -3
...l____________
.....2   3  12  33

These are one less then the original in terms of powers of $x$
So when it is divided, the value is $2x^{2}+3x+12$ with a remainder of $33$[/hide]
I'm not sure if I was clear enough.  If someone can find one put a link to a website explaining this method please.  And my diagram of equation is quite poor

factor the polynomial and see if x-3 is a factor?

and I believe the method you tried to describe is synthetic division

You can use Factor  and Remainder Theorem and see if it's a factor...then find the other 2 factors...

synthetic division is the best way.  almost always.

go here:

http://en.wikipedia.org/wiki/Synthetic_division#Method_1

also, ask your alg. 2 teacher.  its really easy and it takes 2 seconds to do

Since the remainder is $2\cdot 3^{3}-3^{3}+3^{2}-3 = 54-27+6 = 33,$ you can proceed easily.

Remember that synthetic division only works when you divide by a linear expression.


AMC
AIME
trigonometry
AMC 8
USA(J)MO
USAMO
AMC 10

[quote="atm"]Im a sophomore and I plan to take AMC 12 this Febuary since I heard that its easier to get 100 than 120 in AMC 10, especially after the 1.5 reduction change this yr. So how can I register to take AMC 12 to my teacher as a sophomore? Do I have to tell her in advance or in the test day, I just raise my hand when she asks who needs the AMC 12 test? Is this legal (I mean is there any rules in the AMC rules against this)?

Thanks in advance.[/quote]
You should just tell her in advance.

Yes, let your teacher know.  I'm pretty sure that you don't have to choose which one you're taking until you are actually sitting there and your teacher hands you a test.  Basically, the schools register, order so many tests depending how many they're expecting to take the AMCs and then they administer them.  If your teacher has a lot of extra AMC 12's then it wont be a problem.  But it would be a problem if there are too many upper classmen for there to be extra tests.  Additionally, your teacher might not know that it's OK for you to take the AMC 12.  So you should tell her now and convince her (if she thinks that only 11th and 12th graders should take the AMC 12 or something like that...).

No.  I took the AMC 10 as a 7th grader, and the AMC 12 as a freshman (9th grader).  I'm 100% sure you can take the AMC 12 whenever you want, in high school.  You could even take it in middle school or earlier, if you wanted to.

My daughter has had this problem at her school.  We couldn't resolve it last year but here's what we are trying this year.

Find out who will be coordinating the AMC 10/12 at your school. It might not be your teacher. Coordinators typically register the school, purchase the exams, and proctor the contest.  That person  determines how many AMC 10's and AMC 12's to purchase before the deadline (Dec 15.)  It's possible that your coordinator will order the contests earlier than the deadline so talk to them soon.  Convince that person that you need to take the AMC 12 before he/she orders the exams so that there will be enough.  If you can succeed at this, you might suggest that the school offer the contest on both the A & B dates (i.e. AMC 10 on A and AMC 12 on B) so that you can take both contests.  This requires an extra day of contest administration and some extra work on the part of your coordinator.

Your coordinator will probably go to the AMC website to register for the contest (web registration is allowed this year.)  You might e-mail a link to the AMC 10/12 contest info on the MAA site ([url]http://www.unl.edu/amc/whatswhat.shtml[/url].)  It's clear from this site that there are no minimum age/grade level eligibility requirements for the AMC 12.[/url]

[quote="atm"]Thanks for the help
But then how do I convince her? Like is there any rules in the AMC set of rules that says AMC 12 is not only for juniors and seniors? I dont think shes gonna believe me if I tell her because she doesnt know much about the AMC rules thou.[/quote]

There's probably something in the rulebook.

Or you could show her the results on the AMC website, which shows the average score for each grade.  You can't really have an average score on the AMC 12 for 10th graders if 10th graders can't take the test.

oh thanks a lot you guys. All the information is really appreciated. :)

does that mean I could ask my teacher to take the AMC 12 if I wanted? I am in 8th grade.  :)

I think so.  I am in 7th grade and will probably take either the AMC 10 or 12.  :D

As I noted in another thread, there isn't a younger age limit for either the AMC 10 or the AMC 12. The member of AoPS known as A+MATH began taking the AMC 12 in seventh grade, after beginning to take the AMC 10 in sixth grade. He is not the only one. As long as you don't take two tests on one test date (that is, as long as you only take one "A" test and one "B" test), you can take both the AMC 10 and the AMC 12 each year until you run out of eligibility for the AMC 10.

In making this choice, the key question to ask is how much of the curriculum you've had. The AMC-12 has algebra and trigonometry that the AMC-10 doesn't; this material is taught in 10th grade for most students who take the AMC exams. If you've had that material already, you're probably better off taking the 12. If you haven't, you're probably better off taking the 10.

[quote="now a ranger"]should I ask my teacher to take the amc 10a and 10b? and is there a chance she might say no?

I already took the AMC 8, which has nothing to do with this.

I just need two chances to get a 120.[/quote]

You can't take both AMC. Atleast I don't think so.

[quote="now a ranger"]i heard you can.  :lol:[/quote]

I heard you can take the AMC 10a and then the AMC 12b or vice versa.

no, you can also take 10a and 10b

or 12a and 12b

I asked like several people.

Let's clear this up: as long as you are not too old, or too high in school grade placement, you are eligible for BOTH the AMC 10 and the AMC 12. There are two test dates each year for those tests. You may take exactly ONE test on each test date. You can take those tests in any combination, that is:

10A and 12B

or

10A and 10B

or

12A and 10B (this is what my son does each year)

or

12A and 12B.

After you reach a certain age, you are no longer eligible for the AMC 10 and can only take the AMC 12. That's it as to the rules of the AMC. Some schools have obnoxious local rules, but that is why I offer a testing group to homeschooled students and public school students able to get an excused absence from school each year in Minnesota.

When in doubt, read the rules in the teachers manuals posted on the AMC Web site. In general, read the AMC Web site a LOT; it is full of helpful information.

I took the AMC8, 10, 12, AIME, and USAMO last year.

-Ravi

A redundant answer:
YES, you can take the AMC 12 if you're a sophomore. I am a freshman, and I just took the AMC 12A, after a few minutes of convincing the entire math faculty.  :lol:

Another redundant answer, I took AMC 8/10/12 all in one year when I was in 5th grade.  Did it again this year 6th grade (well AMC8/10) as long as your teacher doesn't refuse to hand you a test, you can take either the AMC 10 or 12.

~lyra

[quote="AMCDirector"]On page 5, Section III of the 2007 AMC 10A/AMC 12 A Teachers' Manual it will say: (The 2007 Teachers' Manual is not yet printed, nor yet posted on the AMC website.  It will be in about 2 weeks.)

III. Eligibility

Any student who is officially enrolled in high school (or below) and is taking at least one course at the high school level, and has not graduated, is eligible to take the AMC 10 or AMC 12 or AIME (with qualifying score). Please note: students in grades 11 & 12 can not take the AMC 10. However, students in grades 9 & 10 may choose which contest they take.

This statement has been unchanged for as long as I have been here. You can find it on our Website, in the AMC 10/AMC 12 Archive in the 2006 Teachers' Manual, see
http://www.unl.edu/amc/d-publication/d1-pubarchive/2005-6pub/2006AMC1012/html/06tm-04-text.html

Steve Dunbar
MAA Direcotr for the AMC[/quote]

We had a really good Math Department head at our school who left (to be a principal at the middle school) last summer. A teacher that was just hired is now the math department head. However, she did a very good job in going over the rules and read this one in particular aloud, and in fact encouraged people to take the harder test either for the first date or to come to the second date, in addition.


function
complex analysis
algebra
polynomial
complex numbers
complex analysis open

Hey all, my flat-mate left this question on the fridge for me (it's a game we play), and it's been driving me nuts!  Any help will be greatly appreciated.

Let f be an entire function such that Re(f(z))+Im(f(z)) < 1
for all z. Show that f is constant.

So far, I've got that f is holomorphic and entire...

Thanks for any help.

what do you mean, you've got that f is holomorphic and entire? doesnt entire mean f is holomorphic everywhere on the complex plane?

Since any entire function which is not a polynomial has an essential singularity at $\infty$, we must have that $f$ is a polynomial by Casorati-Weierstrass. I am not going to write out a detailed argument, but it's fairly easy to realize that only a constant polynomial will work.

Of course the problem is killed immediately by Picard's theorem.

A third way would perhaps be to prove a weak version of Picard's theorem using Liouville: The range of a non-constant entire function is dense in $\mathbb{C}$.

Fourth way: The described region is simply connected, so by Riemann's mapping theorem there is a function $g$ mapping the region bijectively and biconformally onto the unit disk. Then $g(f(z))$ is a bounded entire function, so it is constant. Therefore $f$ is constant. We don't need Riemann's mapping theorem here. The described region is not dense in $\mathbb{C}$ so it is easy to construct an analytic univalent map that maps the region onto a bounded region. I suppose this way is actually the same as the third way when I think about it.

This is definitely not an open question, it's a very weak problem.

Another appoach: Re(f(z))+Im(f(z)) is harmonic function. If it's <1 then it's constant. Then the image $f(\mathbb{C})$ cannot be open. Hence f(z)=const

[quote="Kalle"][quote="RIP"]Another appoach: Re(f(z))+Im(f(z)) is harmonic function. If it's <1 then it's constant. [/quote]
Is this because it is the real part of an entire function $h$ and then $e^{h}$ is bounded so $h$ is constant, or is there some more general reason? I don't know a lot about harmonic function theory.[/quote]
There's Liouville theorem for harmonic functions:
If $u(x_{1},x_{2},\ldots,x_{n})$ is harmonic in $\mathbb{R}^{n}$ and $u(x)< const$, then $u=const$ :D


calculus
integration
function
parameterization
derivative
complex analysis
calculus computations

How to integrate:
$\int_{-\infty}^{\infty}{\frac{cos(x)}{1+x^x}}$

$x^x$ doesn't make any sense for $x<0$. Do you want it to be $\int_0^\infty$ instead?

If a solution is reasonable, it isn't going to come from finding an antiderivative.

I guess one has to use complex function to solve it. One may need build a path that avoid zero point.  Without reviewing complex analysis, I am not sure how to do this.

Complex analysis isn't the only trick for solving problems like this; series expansions or introducing parameters to allow differentiation also sometimes work. Since I haven't done anything with this problem yet, I don't know whether any of those tricks works or not.


calculus
integration
algebra
polynomial
calculus computations

Hello Everyone,

     I'm new to the board.  I had some problems while doing my homework, and was wondering if someone here could help me out.  I'd really appreciate it.

Okay, here are the three problems, I have the answers but I need to know how to solve them.

1.   Problem:  Integral [(sqrt x) / ((sqrt x) - 3)]dx    It is indefinate.
      Answer:   x + 6sqrt x + 18 ln(absvalue of sqrt (x-3)) + C

2.   Problem:  Integral [(x(x-2)) / (x-1)^3]dx    It is indefinate.
      Answer:   ln(absvalue of (x-1)) + [1/(2(x-1)^2)] + C

3.   Problem:  Find volume of the solid by revolving around the region.
                     xy = 1, y = 0, x = 1, x = 4 about the line y = 4
      Answer:   PI/[4(32 ln 4  - 3)]

Thank you very much. =)

Hi. Pretty standard stuff indeed.

1. "Integration by substitution". t=sqrt(x)=u(x). dt = u'(x)dx = u'(u^-1(t))dx. You get Integral[2*t^2/(t-3)]dt. Integrate using standard techniques for "rational fractions": decompose into simple fractions over R.

2. Also a rational fraction. If you see no quicker/simpler way, you can always integrate these by dividing the two polynomials and decomposing the remainder fraction into simple fractions. In this case, of course, the degree of the numerator is lower, so the decomposition will have the form  a/(x-1) + b/(x-1)^2 + c/(x-1)^3.

3. Compute Integral[1/x]dx from 1 to 4, use formulae to find the volume of the "revolution solid".

Hope it helps.

Hello Phils,

      Last night after I posted this, I kept playing around and managed to figure out 1 and 2, but not 3.  However I think I should be able to do three now.

Thank you! :lol:

Hi  :D

The solutions are included in the attachment.


The problem is -4 x 2/3 - 7/8.  i keep getting -101/24 but it says the answer is -85/24.  I think the book is wrong.

Thanks.. for some reason that was so hard to get.