Hi, I have a writing question for everyone, and I'm not sure why the answer is what it is.

29. The woman [u]whom[/u] the board picked [u]to design[/u] the new building is a [u]renowned[/u] architect and [u]has received[/u] many awards. [u]No error[/u]

[hide="I think the answer is..."]
HAS RECEIVED because should the sentence have parallel structure? Wouldn't it be better if the sentence read "...is a renowned architect and the recipient of many awards."?[/hide]

[hide="But the answer really is..."]
NO ERROR[/hide]

All help appreciated!

The problem is that it is grammatically correct, but isn't that stylish. Don't fix style; fix grammar.

You can even say: I am tall, ate potatoes yesterday, will be hanging on AoPS for a few hours tomorrow, and had not finished reading my book by the time I had to return it to the library. It is still grammatically correct as far as I can tell. Your version is fine too (except it should be "...is [color=red]a[/color] recipient...")

Okay, but then why do all my english teachers say to be consistent with parallel structure? Isn't that one of the grammar rules you should know?

I understand the grammar, not style point, but I guess I'm not sure whether parallel structure is [i]completely[/i] considered style, not grammar.

Well, give an example of a sentence that you teachers will consider unacceptable and we'll try to see what exactly the differences are. Mixing tenses in moderate amounts is fine: "I got this book yesterday and will read it tomorrow" is something that nobody will object. The monstrous sentence I concocted suffers not from tense mixing but from very weak connection between different statements, so it may be unacceptable, but only for that reason, which is very hard to formalize.

"The man was a basketball star and a football star"

is better than

"The man was a basketball star and he played football"

right?

Your two examples convey two different things semantically.  The first says that he was a star in both basketball and football; the second says that he was a star in basketball but may or may not have been a football star.  So, since they don't contain the same information, it's not valid to compare them stylistically.

Also, as a linguist, I would urge you not to take "school grammar" too seriously.  Many of the unyielding grammatical "rules" you are taught are no more than stylistic conventions (like mixing tenses in parallel structure, which you bring up), many are hundreds of years out-of-date (such as the rule about not splitting an infinitive), and many are also flat-out wrong (such as the rules about "whom" and singular "they").  If you want to understand how English grammar really works, check out the Merriam-Webster Dictionary of English Usage ([url]http://www.amazon.com/Merriam-Websters-Dictionary-English-Usage-Merriam-Webster/dp/0877791325[/url]) or A Student's Introduction to English Grammar ([url]http://www.amazon.com/Students-Introduction-English-Grammar/dp/0521848377/ref=sr_1_3?ie=UTF8&s=books&qid=1229965438&sr=1-3[/url]).  The bottom line is that, if you're a native speaker of English, it's not grammar you need help with so much as style, so just practice reading and writing with an ear to what is clear and reads well.

We should also note that "The man was a basketball star" is a full sentence as is "He played football," so to be picky you should have a comma before the conjunction: "The man was a basketball star, and he played football."

(You know, since we're dissecting whether things are correct grammatically).

Yeah, that was a bad example I posted above, but does anyone else know what I'm talking about? I'm not concerned about English in general, I'm more concerned about why the answer was NO ERROR. Because many times, in a similarly structured sentence, the book will say that the sentence is [i]wrong[/i] because of a lack of parallel structure.

And I agree with Osud's point about not taking "school grammar" too seriously. I know the nit-picky rules don't really matter in the long run. But when I asked this question, I was really only referring to this one problem that was in the book. (In other words, I don't mean to take this question and try to apply it to the rest of my life).  :)

[quote="buzzer11"]many times, in a similarly structured sentence, the book will say that the sentence is [i]wrong[/i] because of a lack of parallel structure.[/quote]
Could you give an example?  This one is quite forgivable; being a renowned architect and receiving awards aren't actually being paralleled in this sentence, just put together as two things that happened.  There is [b]nothing[/b] grammatically wrong with that construction.

If you want the standard prescriptive ("school") grammar treatment of parallel structure, a good place to start would be the classic of the field, Strunk & White's [i]Elements of Style[/i].

Here's a link to the book, and use your web browser to find the word "parallel" and you'll be at the right section:  [url]http://www.bartleby.com/141/strunk5.html[/url]

Here's Strunk & White's example:

Bad:
[quote]Formerly, science was taught by the textbook method, while now the laboratory method is employed.[/quote]

Good:
[quote]Formerly, science was taught by the textbook method; now it is taught by the laboratory method.[/quote]

Strunk & White's explanation of the examples is, "The left-hand version gives the impression that the writer is undecided or timid; he seems unable or afraid to choose one form of expression and hold to it. The right-hand version shows that the writer has at least made his choice and abided by it."

I find that reasoning for favoring one style over another pretty weak; I would have preferred reasoning along the lines of "parallel construction makes it easier for your reader to see exactly what you are comparing or contrasting in the two clauses."  But, there you go.

Edited to add:  Also, I wouldn't knock myself out on trying to understand the SAT's reasoning; it may not even be correct.  For an example of faulty grammatical reasoning on an SAT writing exam, see [url]http://itre.cis.upenn.edu/~myl/languagelog/archives/001863.html[/url]  As Liberman notes, coming up with the "right" answer can boil down to deciding how sophisticated the test designers were.

[quote="Osud"]
Also, as a linguist, I would urge you not to take "school grammar" too seriously.  Many of the unyielding grammatical "rules" you are taught are no more than stylistic conventions (like mixing tenses in parallel structure, which you bring up), many are hundreds of years out-of-date (such as the rule about not splitting an infinitive), and many are also flat-out wrong (such as the rules about "whom" and singular "they").[/quote]

[quote]Edited to add: Also, I wouldn't knock myself out on trying to understand the SAT's reasoning; it may not even be correct.[/quote]

Agreed... but the problem is, no matter how ridiculous "school grammar" may be, the lives of today's students depends on it. So we still have to deal with it.

In answer to your question, buzzer11, I would say that a good rule of thumb is that those test-makers only test parallelism if (1) there are three or more parallel phrases (term?) or if (2) the two parallel phrases actually have [i]several words in common[/i] besides just having the same grammatical structure (WARNING: I just came up with that off the top of my head). With your example, if

[quote]
The woman whom the board picked to design the new building is a renowned architect and has received many awards.[/quote]

were changed to

[quote]
The woman whom the board picked to design the new building is a renowned architect and the recipient of many awards.[/quote]

then the two parallel phrases would be "a renowned architect" and "the recipient of many awards". The only thing in common about these two phrases is that they are both noun phrases. No words are shared. If it were changed to

[quote]
The woman whom the board picked to design the new building is a renowned architect and is the recipient of many awards.[/quote]

instead, then the two parallel phrases would be "is a renowned architect" and "is the recipient of many awards". Now they have [i]one[/i] word in common - "is". But that's still just one, so if I were sitting there taking the test I wouldn't mark it as wrong.

Now with Osud's example,

[quote]Formerly, science was taught by the textbook method; now it is taught by the laboratory method. [/quote]

the two parallel phrases are "taught by the textbook method" and "taught by the laboratory method". Not only are they the same grammatical structure, they share the words "taught by the ___ method" in common.

I guess it's usually easy to spot that the test-makers want you to use parallelism when there are three phrases that are supposed to be parallel.


Prove that the set of integers of the form 2^k-3(k=2,3,......) contains an infinite subset in which every two members are relatively prime.

We can construct the sequence $2^{a_{n}}-3$ by induction.
Take $a_{1}=3$.
Now, suppose that the construction is done until $a_{n}$.
Let $N=\prod_{k=1}^{n}(2^{a_{k}}-3)$, and take
$a_{n+1}=\phi(N)$
Since $N$ is odd, we have
$2^{a_{n+1}}-3\cong-2 mod (N)$
So, $2^{a_{n+1}}-3$ is coprime with $2^{a_{k}}-3$ for all $1\leq k\leq n$


geometry
3D geometry
LaTeX
inequalities open
inequalities

Sorry everyone but I don't know Tex.
Let x,y,z be positive real numbers so that x + 2r(xy) + 3r (xyz) =3/4
notice that nr(x) mean n root of number x (ex. 2r(x) mean square root of x, 3r(x) mean cube root of x).find the maximum of (x+y+z).this is from a mathematical magazines for [b]Secondary school students in Vietnam.
I'm very grateful to s.one who change this one to Tex.thank you[/b]

[quote="Yutzpracachia"]Let $x,y,z$ be positive real numbers so that $x+\sqrt{xy}+\sqrt[3]{xyz}= \frac{3}{4}$

find the maximum of $(x+y+z)$.[/quote]

$\LaTeX$'d.


function
real analysis
real analysis unsolved

Hi, I'm studyng the identity function in a special space.

Let $ X\equal{}\left\{\left(x_n\right)_{n\equal{}1}^\infty\subseteq \mathbb{R}: \sum_{n\equal{}1}^\infty|x_n|<\infty\right\}$. Consider $ X_p\equal{}\left(X,\parallel{}\cdot \parallel{}_p\right)$ with $ \parallel{}(x_n)\parallel{}p\equal{}\left(\sum_{n\equal{}1}^\infty|x_n|^p\right)^\frac{1}{p}$.

I'd like to prove that $ I: X_p\to X_1,\left(x_n\right)\to\left(x_n\right),p> 1$ isn't continuous.

I tried to prove that $ \sup\left\{\frac{\parallel{}\left(x_n\right)\parallel{}_1}{\parallel{}\left(x_n\right)\parallel{}_p}\right\}\equal{}\infty$, but had no success.

PD: My apologies if it has been solved in the past :maybe:

Presumably $ p>1$. Consider sequences that look like $ 1, 1, ..., 1, 0, 0, ...$

I like geometric sequences. For $ 0<r<1,$ let $ x_n\equal{}r^{n\minus{}1}.$ Then $ \|x\|_p\equal{}\frac1{(1\minus{}r^p)^{1/p}}$ and $ \frac{\|x\|_1}{\|x\|_p}\equal{}\frac{(1\minus{}r^p)^{1/p}}{1\minus{}r}.$

Now try taking the limit of that as $ r\to 1.$


calculus
geometry
induction
vector
MIT
college
linear algebra

Hi,
I've been out of the loop for a few years now.  I took Calculus BC sophomore year, and haven't taken any math since then (I'm going to college this year).  What should I study to prepare for the rigors of college mathematics courses?

Depends what you're taking.  If you are going to take calculus again, run through your old notes and possibly a textbook if you have one.  If you are going to calc 3, linear algebra, or some other higher math, try to take a look at a textbook so you can at least see what it looks like and get a feel for it.

Seems like JRav's advice is good.  Get the textbook early and get a head start---work through the first few chapters, do a lot of problems.

If you're going to major in math in college, try to get good at understanding proofs and coming up with them yourself.  If you're going to do any kind of technical work in life, practice problem solving by hanging out at AoPS and doing problems.

If you're taking Honors math (i.e. directed at mathematicians, not engineers or physicists) then you should learn also how to write a proof.

Thanks for the tips.
I've had some limited experience in the area of proofs.  I've done a few direct proofs and a few induction proofs but that's about it.  I think I have a lot to do...

If you want to get started with proofs, start looking at a good linear algebra book.  I used Friedberg, Insel, and Spence and I can say it is a good book with hundreds of proofs (some easy, some not).  I also used Colley's [u]Vector Calculus[/u] for my multivariable calculus class.  While we didn't focus on the proofs that much, the book is still filled with them, including in the HW problems.

Also, you may want to look at a more rigorous calculus book, based on proofs. You'll have familiarity with the concepts and computations and get a far better understanding of the math than Calc BC would give you from the proofs. It doesn't hurt to get a head start. You can also look at course materials from other college honors courses. MIT posts them at OCW, and I'm sure other colleges do this also.

To follow up Anaxerzia's advice, if you want a rigorous, proof-based calc book, you have to go with Spivak's [u]Calculus[/u].  It's essentially an introduction to analysis but if you have taken calc before, you should have no trouble understanding the first 24 chapters or so.


Is there a nice formula (in closed form) for
$\sum_{i=1}^{n}i^{k}$

I know the answers when $k=1$, $k=2$, and $k=3$, but what about the general case?

Prove your results.

vishalarul, many of the questions you post (including this one) have appeared on the forum before.  This one, for instance, has appeared many, many times.  Try a search, or just look around the older posts on the forums.

Sorry.  :oops:
What should be the key word when I search for it? (I didn't know what to put, so I just made a new topic)

try dr. math.  there is an article about the first m nth powers.


More of a philosophical than a mathematical question. Either way, feel free to discuss. :)

The broader question, and one which must be addressed first, is what [i]is[/i] truth in the physical/meta-physical world? Can natural or mathematical language really embody physical/meta-physical truth? And does the truth of statements in classical logic accurately correspond to truth in the "physical/meta-physical" world?

I would say - The laws of logic are very much necessary for understanding this complex world, to comprehend and solve abstruse problems in science. But I feel it is not mandatory to apply the laws of logic everywhere, I feel some of [i] the greatest discoveries in science [/i] were made through intuitive and irrational leaps of thought rather than adhering to logical principles/laws.

Perhaps I shall explain the terms used in my question.

[b]necessary[/b] means it must be true, no matter what the circumstances are or could be (it must be true in every possible world, even one not containing humans or the laws of physics).
[b]contingent[/b] means it is not necessarily true and not necessarily false either (it could have been otherwise).


ratio
geometric series
geometry proposed
geometry

The angles of a triangle  form a geometric series with ratio 2.

Prove that the midpoints of its sides and the feet of its altitudes are vertices of a regular 7-sided  polygon

A',B',C' are the midpoints of BC, CA, AB; A'', B'', C'' are the feet of the corresponding altitudes; R = circumradius.
We have allways :
B'C' = C''A' = A'B'' = R sin(A)
A''B'' = R |sin(2C)|
C'C'' = R |sin(A-B)|
Now, if A =Pi/7, B= 2Pi/7, C=4Pi/7, the five lengths above are R sin(Pi/7) thus
B', C', C'', A', B'', A'' are 6 consecutive vertices of a regular heptagon.


limit
real analysis
real analysis unsolved

Let $ (a_n)_{n\ge{1}},(b_n)_{n\ge{1}}$ 2 sequences of real numbers ($ a_n\ge{0},b_n\ge{0}$),with the property that exist $ q\in{(0,1)}$ so that $ a_{n \plus{} 1} \le{qa_n \plus{} b_n}$,$ n \equal{} 1,2,3,.....$
 Prove that if $ \lim_{n\to\infty}{b_n} \equal{} 0$ than  $ \lim_{n\to\infty}{a_n} \equal{} 0$

$ \lim_{n\rightarrow \infty}b_n\equal{}0$ then $ \forall \alpha>0 \quad\exists N_1>0 \quad\forall n\geq N_1: b_n<\alpha(1\minus{}q)$
Then $ \forall n> N_1$ we have
$ a_n\minus{}\alpha<q(a_{n\minus{}1}\minus{}\alpha)<...<q^{n\minus{}N_1}(a_{N_1}\minus{}\alpha)$
-) If $ a_{N_1}\leq\alpha$ then $ a_n<\alpha$ for all $ n>N_1$. OK!
-) If $ a_{N_1}>\alpha$
Because $ \lim_{n\rightarrow \infty} q^{n\minus{}N_1}\equal{}0$ then $ \forall \beta>0 \quad\exists N_2>N_1>0 \quad\forall n\geq N_2: q^{n\minus{}N_1}<\frac{\beta}{a_{N_1}\minus{}\alpha}$
Then $ \forall n>N_2$ we have $ a_n<\alpha\plus{}\beta$. OK!

Or just use the fact that $ \limsup (x_n\plus{}y_n)\le\limsup x_n\plus{}\limsup y_n$, which immedeiately results in $ \limsup a_n\le q\limsup a_n\plus{}\limsup b_n\equal{}q\limsup a_n$ and, finally, in $ \limsup a_n\le 0$


Let $ f(x)=x^4-x^3+8ax^2-ax+a^2$. Find all real number $ a$ such that $f(x)=0$ has four different positive solutions.

[quote="math10"]Let $ f(x) \equal{} x^4 \minus{} x^3 \plus{} 8ax^2 \minus{} ax \plus{} a^2$ .Find all real number $ a$ ,suth that: $ f(x) \equal{} 0$ has four different positive solution.[/quote]

Consider $ a\neq 0$ (since $ a \equal{} 0$ gives $ 3$ similar roots $ 0$).
We have $ f(x) \equal{} \frac {x^4}{a^2}f(\frac ax)$ $ \forall x\neq 0$ and so $ z\neq 0$ root $ \implies$ $ \frac az$ root too.

So we can try to write $ f(x) \equal{} (x^2 \minus{} ux \plus{} a)(x^2 \minus{} vx \plus{} a)$ which implies $ u \plus{} v \equal{} 1$ and $ uv \equal{} 6a$
In order to have $ 4$ distinct real roots, we need $ u,v$ be different real numbers such that $ u^2 > 4a$ and $ v^2 > 4a$  and so :
1) $ x^2 \minus{} x \plus{} 6a$ must have two real roots (distinct) and so $ a < \frac {1}{24}$ and so $ u \equal{} \frac {1 \minus{} \sqrt {1 \minus{} 24a}}{2}$ and $ v \equal{} \frac {1 \plus{} \sqrt {1 \minus{} 24a}}{2}$

2) $ u^2 > 4a$ and $ v^2 > 4a$ and so $ \frac {1 \minus{} 12a \minus{} \sqrt {1 \minus{} 24a}}{2} > 4a$ and $ \frac {1 \minus{} 12a \plus{} \sqrt {1 \minus{} 24a}}{2} > 4a$ and so :
$ 1 \minus{} 20a > \sqrt {1 \minus{} 24a}$ and so (since $ a < \frac {1}{24}$ implies $ 1 \minus{} 20a > 0$) : $ 1 \minus{} 40a \plus{} 400a^2 > 1 \minus{} 24a$ and so $ 25a^2 > a$.

And so $ f(x)$ has $ 4$ (maybe not distinct) real roots $ \iff$ $ a\in( \minus{} \infty,0]\cup[\frac {1}{25},\frac {1}{24}]$
We know that $ a\in\{0,\frac {1}{25},\frac {1}{24}\}$ implies multiple roots. It remains to check if out of these values, roots are distinct.

And this is easy to check :
If $ a\notin\{0,\frac {1}{25},\frac {1}{24}\}$, $ u\neq v$ and both polynomial $ x^2 \minus{} ux \plus{} a$ and $ x^2 \minus{} vx \plus{} a$ have two distinct roots. It remains to check if some root $ r$ may be root of both polynomials. But in such a case, both $ r$ and $ \frac ar$ would be root of the two polynomials and these two polynomials would be identical, and so $ u \equal{} v$.

Hence the result : $ f(x)$ has $ 4$ distinct real roots $ \iff$ $ a\in( \minus{} \infty,0)\cup(\frac {1}{25},\frac {1}{24})$

More, we have the $ 4$ roots :

$ \frac {1 \minus{} \sqrt {1 \minus{} 24a} \minus{} \sqrt {2(1 \minus{} 20a \minus{} \sqrt {1 \minus{} 24a})}}{4}$

$ \frac {1 \minus{} \sqrt {1 \minus{} 24a} \plus{} \sqrt {2(1 \minus{} 20a \minus{} \sqrt {1 \minus{} 24a})}}{4}$

$ \frac {1 \plus{} \sqrt {1 \minus{} 24a} \minus{} \sqrt {2(1 \minus{} 20a \plus{} \sqrt {1 \minus{} 24a})}}{4}$

$ \frac {1 \plus{} \sqrt {1 \minus{} 24a} \plus{} \sqrt {2(1 \minus{} 20a \plus{} \sqrt {1 \minus{} 24a})}}{4}$

Btw, I forgot that the problem required that roots must be positive. So we need $ u,v$ and $ a$ all positive, and so $ a > 0$ (since $ a > 0$ implies $ u \plus{} v > 0$ and $ uv > 0$). And so :
$ f(x)$ has $ 4$ distinct positive real roots $ \iff$ $ a\in(\frac {1}{25},\frac {1}{24})$


Two times the smallest of three consecutive odd integers is nine more than three times the middle integer. Find the largest of the three integers.

You have this equation:

$ 2x \minus{} 9 \equal{} 3(x \plus{} 2)$

Distribute.

$ 2x \minus{} 9 \equal{} 3x \plus{} 6$

Subtract $ 2x$ from both sides.

$ \minus{} 9 \equal{} x \plus{} 6$

Subtract $ 6$ from both sides.

$ \minus{}15 \equal{} x$

Thus, the largest integer must be $ x \plus{} 4$, or $ \boxed{\minus{}11}$.


n is a constant and k is a variable

let 1*a1+2*a2+...+k*ak=n, such that a1,a2,...,a(k-1)>1 and ak>=1

prove
Sum(all possible sequences of ai for all possible values of k)
(a_1 - 1)(a_2 - 1)...(a_(k-1) - 1)(a_k)
= number of non-negative solutions to (b1*1+b2*2+...+bn*n=n)

***
it is a_k and NOT a_k - 1.  this is *NOT* a typo
***

for example
for n=6
we have:
1*(6)        ====>      6
1*(4)+2*(1)  ====>  (4-1)(1)=3
1*(2)+2*(2)  =====>   (2-1)(2)=2

6+3+2=11, you can verify this is the parition of 6

I can reduce this problem to the following identity

\sum x k^2*1/(1-x)2*1/(1-x2)2*...*1/(1-xk)2=1/(1-x)*1/(1-x2)*1/(1-x3)*..., k=0,1,2,...

But I can't proof this identity.

I am just looking at LHS.  and the coefficient infront of x^6
the left hand is given by:

x(1+x+xx+...)+xxxx(1+x+xx+...)(1+xx+xxxx+...)+...
and hence the coefficient for x^6 is only

1+2=3

but the partiton of 6 is 11...

if i understand things correctly, you counted the number of partitions satisfying the property satisfying the equation (1*a1+2*a2+...+k*ak=n), but you did not calculate the sum of (a1 - 1)(a2 - 1)...(a_(k-1) - 1)ak

bugzpodder, PLEASE look carefully.

LHS is 1+x(1+2x+3xx+...)+xxxx(1+2x+3xx+...)(1+2xx+3xxxx+...)+...

sorry about that....  yes you are right
how did you come up with the g.f. on the LHS?

anyways i think they are true because LHS is sum of paritions of n with at most k parts (squared or something, which the g.f. can be found http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A008284), and the exact same identity can be found here: http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000041

i am still confused as to how you derived LHS from the given info, please explain?  thx

I simply reformulated problem:
if 1*a1+2*a2+...+k*ak=n, a1,a2,...,a(k-1)>1 and ak>=1 then
1*(a1-2)+2*(a2-2)+...+k*(ak-1)=n-k2 ==>
1*b1+2*b2+...+k*bk=n-k2, b1,b2,...,bk \geq 0 and (a_1 - 1)(a_2 - 1)...(a_(k-1) - 1)(a_k)=(b1+1)...(bk+1).

The sum 1*b1+2*b2+...+k*bk appears in product of polynoms of form s0+s1*xl+s22l+... as a power of x. Final step is finding of proper values for s0,s1,... Each item k*bk correspond to factor sk and in the product (b1+1)...(bk+1) this item correspond to factor (bk+1) ==> sk=bk+1, i.e. polynoms have form 1+2*xl+3x2l+... After all we obtain required identity.


Sorry for posting the previous hyperfactorial problem twice by accident. This one is different.

Define the hyperfactorial $n^{!}$ for integers $n\geq1$ as follows:

$n^{!}=\displaystyle\prod_{k=1}^n k!$

find the smallest n such that 10^k divides $n^{!}$ ?

what is $k$??? (because the answer varies...). Or do we express answer in terms of $k$?

If we assume that $k$ is any positive integer, then the solution is obviously 5... do you want an answer in terms of $k$? (For each $k$, find the smallest $n$ that satisfies the requirements?)

I guess you probably want us to find smallest $n$ such that

$10^n|n^{!}$

When $n=5$ to $9$ , there are each one factor of $5$ . When $n=10$ to $14$ , there are each two factor of $5$ .

So by trial and error , we found $n=13$ is the smallest  :)

That's not what I intended. Let me change the problem notation to avoid confusion. Suppose $j$ is a constant. What is the smallest value of $n$ such that $10^{j}|n^{!}$


[b]1) [/b]

α). Αν έχεις μόνο ένα ευρώπουλο, δεν είσαι πλούσιος
β). Αν δεν είσαι πλούσιος, το να σου δώσουμε άλλο ένα ευρώ, δεν σε κάνει πλούσιο.

[b]2) [/b]

Η καμπυλότητα της περιφέρειας ενός κύκλου είναι αντιστρόφως ανάλογη της ακτίνας του. Δηλαδή όσο μεγαλώνει η ακτίνα ενός κύκλου (ή και σφαίρας) η καμπυλότητά του μειώνεται. Πάρτε παράδειγμα τη γη πχ.. Εχει τόσο μεγάλη ακτίνα που η καμπυλότητά της είναι αμελητέα, άρα και τη θεωρούμε επίπεδη.
Το όριο στη μείωση της καμπυλότητας, είναι η ευθεία.
Άρα ένας άπειρος κύκλος είναι μια ευθεία..

Πάρτε π.χ. και το νούμερο 2 - σύμφωνα με τον Ζήνωνα, δεν μπορείτε να ολοκληρώσετε κανένα απολύτως ταξίδι. Δηλαδή για να φτάσετε εκεί που πάτε, πρέπει πρώτα να καλύψετε το μισό της απόστασης. Και μετά από αυτό το μισό του υπόλοιπου μισού, και μετά το μισό του μισού του υπόλοιπου μισού κ.ό.κ. Δηλαδή όσο και να προχωρήσετε, πάντα θα υπάρχει μια απόσταση μέχρι το τέλος του προορισμού.Και φυσικά πριν από όλα αυτά θα πρέπει να έχετε κάνει το πρώτο μισό του πρώτου μισού, και πριν απο αυτό το μισό αυτού, και το μισό του μισού κ.ο.κ επ άπειρον..
Άρα μην κουράζεστε να πάτε πουθενά, αφού ποτέ δεν θα φτάσετε :)

[b]3)[/b]

Σκεφτείτε το σύνολο S όλων των συνόλων που δεν περιέχουν τον εαυτό τους! Το S περιέχει τον εαυτό του? Αν ναι, τότε εξ'ορισμού δεν πρέπει, αλλά τότε εξ'ορισμού πρέπει να τον περιέχει...

[b]4)[/b]

Έστω πως μας πιάνει μια τρέλλα και θέλουμε να δούμε αν όλα τα κοράκια είναι μαύρα. Τι θα κάνουμε? Θα αρχίσουμε να παρατηρούμε κοράκια, μέχρι να δούμε κάποιο που δεν είναι μαύρο, οπότε και ο ισχυρισμός "όλα τα κοράκια είναι μαύρα" θα καταρριφθεί. Ναι αλλά σιγά μην βρούμε όλα τα κοράκια του κόσμου. Απλά κάθε φορά που θα βλέπουμε άλλο ένα μαύρο κοράκι, ο ισχυρισμός μας θα ενισχύεται. Θα σκεφτούμε όμως: Η πρόταση "όλα τα κοράκια είναι μαύρα" είναι λογικά ισοδύναμη με την πρόταση "όλα τα μη μαύρα αντικείμενα είναι μη κοράκια" . Άρα απλά παρατηρώντας το οτιδήποτε γύρω μας, ο ισχυρισμός μας ενισχύεται κάθε φορά και από λίγο. Δηλαδή, μια και τα μη-μαύρα μη-κοράκια αντικείμενα στο σύμπαν είναι περισσότερα απο τα μαύρα κοράκια, κοντεύουμε να φτάσουμε στο συμπέρασμά μας, αλλά αυτό είναι μια κακή ιδέα, αν και λογικά στέκει. Δυστυχώς όμως πέσαμε σε μια μικρή αλλά σημαντική συνάμα παγίδα: Το ότι βλέπετε ένα κόκκινο μήλο δεν είναι μόνο παράδειγμα ενός μη-μαύρου μη-κορακιού, αλλά και ενός μη-άσπρου μη-κορακιού. Άρα, ένα κόκκινο μήλο ενισχύει τόσο την πρόταση "όλα τα κοράκια είναι μαύρα" όσο και την "όλα τα κοράκια είναι άσπρα".

Στο 3 νομίζω ότι οδηγούμαστε σε άτοπο αφού όταν κάνουμε την υπόθεση ότι το S περιέχει τον ευατό του και μετά εξορισμού βγάζουμε ότι είναι άτοπο τότε δεν χρειάζεται παραπάνω ανάλυση...

[quote="mostel"]Δηλαδή για να φτάσετε εκεί που πάτε, πρέπει πρώτα να καλύψετε το μισό της απόστασης. Και μετά από αυτό το μισό του υπόλοιπου μισού, και μετά το μισό του μισού του υπόλοιπου μισού κ.ό.κ. Δηλαδή όσο και να προχωρήσετε, πάντα θα υπάρχει μια απόσταση μέχρι το τέλος του προορισμού.Και φυσικά πριν από όλα αυτά θα πρέπει να έχετε κάνει το πρώτο μισό του πρώτου μισού, και πριν απο αυτό το μισό αυτού, και το μισό του μισού κ.ο.κ επ άπειρον..
Άρα μην κουράζεστε να πάτε πουθενά, αφού ποτέ δεν θα φτάσετε :)

[/quote]

1/2+1/4+....=1

(απόδειξη:
1+1/2+1/4+....=2<=>1/2+1/4+....=1 οεδ :P )

Ωραίος thatin..  :lol:

Έχουν πλάκα αυτά όμως  :)

thanx!
α και για το 3, πως ξέρεις ότι μπορείς να ορίσεις ένα τέτοιο σύνολο? κάθε πρόταση δεν ορίζει σύνολο!

Μα, γι' αυτό αποκαλούνται παράδοξα  :)

lolol ναι!
ξέρετε το παράδοξο του απροειδοποίητου διαγωνίσματος?

νομιζω οτι το ξερω αλλα πες το


topology
advanced fields
advanced fields unsolved

If we consider the space $ (\mathbb{R}, \tau_{discrete} )$ .

Let $ p \notin \mathbb{R}$ .

Consider $ X \equal{} \mathbb{R} \cup \{p\}$ with open sets of $ \tau$ as follows :

$ \forall x \in \mathbb R$, the open set will be of the form  $ \{x\}$.

The neighborhood base at the point $ p$ will be of the form $ \{p\} \cup A$  s.t $ \mathbb{R} \minus{} A$ is finite set. where $ A \subseteq \mathbb{R}$.

I know $ (X,\tau)$ is not separable ( How we can prove that  ?) .

how could I prove that space is Lindelöf Space.? :huh:

Thank you in advanced.

Not separable:
Note that for any basis $ B$, for any $ x \in R$, there exists a basis element $ B_x$, so that $ x \in B_x \subset \{x\}$, i.e. each $ \{x\}$ must be a basis element. $ \mathbb{R}$ is uncountable means that we don't have a countable basis.

Lindelof:
Note that for any open cover, there exists one open set that covers $ p$. This open set, by definition, would have covered everything but a finite set. For each element in that finite set, choose an open set from the open cover to cover it. Then you have your finite open cover (which is of course countable)


Seventeen people are at a party.  It turns out that for each pair of people present, exactly one of the following statements is always true: They haven't met," "They are good friends," or "they hate each other."  Prove that there must be a trio (3) of people, all of whom are either mutual strangers, mutual good friends, or mutual enemies.

Are you sure this is a getting started problem? It's an interesting problem, no doubt, and not terribly hard, but if the problem statement says "Prove . . ." I would take it to be ipso facto at the intermediate level, at least. People who are getting started are more used to finding numerical answers than they are familiar with writing proofs.

What does everyone else think?

Yeah, I think this isn't getting started level.
On the other hand, I think 99% of problem solving is writing proofs, no matter what level you are at..

I'd be inclined to say it wasn't Getting Started on notational grounds.

I think that "trio" was supposed to be the word t-r-i-o.  But I just realized that if you type the word t-r-i-o without hyphens, the forum changes it to a triangle followed by the letter o.  The same thing happens with t-r-i itself, and t-r-i followed by a single letter.  I think it's a bug.  I'll report it on the Support board.

[quote="Ravi B"]I think that "trio" was supposed to be the word t-r-i-o.[/quote]

I was wondering where that notation came from. That's funny. I'll try it here:

tri

t-ri

t-r-i

trio

tripod

triple

triceratops

This was fun to look at in preview.

tokenadult, you're a :tri:p, but you're going to have to stop playing such :tri:x.  Meanwhile, I need to brush up on some :tri:g, after my hair gets its :tri:-weekly :tri:m. :)

[Edited to make the joke still work.]


Suppose x is a rational number. Can you prove that x^x is irrational unless x is an integer?

x is a rational number. Then x = p/q, where p is an integer and q is a positive integer, and p and q are in lowest terms.

Then x^x = (p/q)^(p/q) = qth root of (p^p/q^p).

This is irrational, unless (p^p/q^p) is a perfect qth power.

But this is impossible, since we required p and q to be in lowest terms.

What steps did i miss here?

If $ \frac{p}{q}$ is in lowest terms, they share no common factors, so if $ \frac{p}{q}$ is a perfect $ q^{th}$ power of a rational then $ p, q$ are both perfect $ q^{th}$ powers.

Here is the step you missed:  for $ q \ge 1$ we have $ 2^q > q$, so $ q$ is a perfect $ q^{th}$ power if and only if $ q \equal{} 1$.


We can use the "Code" menu item to enclose some code in a box.  At the top of the box it says "code" . . . any way I could have it say "Mathematica Code"?   I mean wouldn't it look a little more professional if the code block explicitly stated what type of code it was?  Here's an example:

[code]In[152]:=
n = 5;
w = Exp[Pi*(I/2)];
roots = z /. Solve[z^n == w];
Table[Simplify[Times @@ (Abs[roots[[num]] - #1] & ) /@
     Drop[roots, {num}]], {num, 1, 5}]

Out[156]=
{5, 5, 5, 5, 5}[/code]

You are asking for a new (customisable) tag so it isn't a specific LaTeX issue but a suggestion for the AoPS forum; I am moving it to the Questions, Suggestions, & Announcements forum where it seems more relevant.

The code that is used to show code that is not being parsed by BBCode in the site, not a particular programming language code. That's all there is to it, and that's not going to change.


Oh!As i thought, nobody among 82 members can solve this problem. :huh:  :huh:  :huh:


geometry
geometric transformation
reflection

If you have a flying bird in a bottle (sealed), would it register on a scale? If you have a flying bird in a cage (open), would it register on a scale? If you have a flying bird in the sky, you could get it to stay still, and there wasn't any wind or anything, would it register on a scale that was placed directly beneath it? What if you did the same experiments with an airplane (i.e. it doesn't flap wings so there's no downward force of air)?

Please do not attempt to perform any of these experiments.

If the bird were in the bottle, then yes it would register.  Something's got to be keeping that bird up; by flapping its wings, it sends air downwards, which has got to push on the scale as much as the bird weighs.

If the cage were open, then some of the air could go out, and it wouldn't register as much on the scale.  But if it were a large enough cage, then I don't suppose the effect would be noticeable.

Same thing when it's flying in the sky.  The air molecules are going to be moving around so much that the momentum they gain from the bird's wings wouldn't deflect them enough to hit a scale directly below.

I'm sorry that my explanation is very fuzzy.

I fully agree with daveed. I would only like to add that although airplane doesn't flap its wings, it too creates a downward force - there has to be something keeping it in the air.

Also, take a look at this topic:

http://www.mathlinks.ro/Forum/viewtopic.php?t=52680


find the center of a circle only by compass.

this probelem has disscused many times and times by:

$yetti,arne,michal,stergiu,jensen,amir2,ashegh,...$

its too hard for this forum...

i want the moderator to move or lock this topic...


Consider the following equation
$ x^n \equal{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} a_{n \minus{} 2}x^{n \minus{} 2} \plus{} \cdots \plus{} a_1x \plus{} a_0$
where $ a_i \in(0; \plus{} \infty)$ for $ i \equal{} 1,2,\cdots,n \minus{} 1.$
Show that this equation has only one positive root.

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 , but unlock!

Now, with your problem, use theorem on sign of $ a_i's$, called Decart's theorem.

Yes but this problem more general :wink:
$ f(x)\equal{}x^n\minus{}a_{n\minus{}1}x^{n\minus{}1}\minus{}...\minus{}a_0$
$ f(0)<0,f(\infty)>0$ so it has a positive solution.
$ n\equal{}\text{deg}P(x)$
We prove by induction on $ n$
$ n\equal{}1$ it true.
Suppose it true for $ n$
We will prove for $ n\plus{}1$
$ f'(x)\equal{}nx^{n\minus{}1}\minus{}...\minus{}a_1$ has unique solution on $ (0,\plus{}\infty)$
Call it is $ x_{n\minus{}1}$
Now consider sign of $ f(x),f'(x)$ on $ (0,\plus{}\infty)$ easy to has proof.

:lol: . On $ (0,\infty)$ that equation is equivalent to $ 1\equal{}\frac{a_{n\minus{}1}}{x}\plus{}\cdots\plus{}\frac{a_0}{x^n}\equal{}h(x)$ and $ h(x)$ is decreasing on $ (0,\infty)$.


Prove that $ x^{8} \minus{} x^{5} \plus{}x^{2}\minus{}x\plus{}1>0$ for all real numbers $ x$.

[hide]Hi,

For $ x \ge1$

$ x^8 \ge x^5$
$ x^2 \ge x$

The inequality is true.

For $ x \le o$, all terms are non-negative. The inequality is true.

For $ o < x < 1$

Let $ y \equal{} \frac {1}{x}$

$ y > 1$

The inequality becomes

$ 1 \minus{} y^3 \plus{} y^6 \minus{} y^7 \plus{} y^8 \ge 0$

As $ y^8 > y^7$ and $ y^6 > y^3$, the inequality is true.[/hide]

[quote="AndrewTom"]Prove that $ x^{8} \minus{} x^{5} \plus{} x^{2} \minus{} x \plus{} 1 > 0$ for all real numbers $ x$.[/quote]
[hide="solution"]Denote $ f(x) \equal{} x^{8} \minus{} x^{5} \plus{} x^{2} \minus{} x \plus{} 1$

If $ x \leq 0$ we have $ f(x) \equal{} x^8 \plus{} x^2(1 \minus{} x^3) \plus{} (1 \minus{} x) > 0$

If $ 0 < x\leq1$ and $ x > 1$ we have: $ f(x) \equal{} x^5(x^3 \minus{} 1) \plus{} x(x \minus{} 1) \plus{} 1 > 0$[/hide]


does the solution manual for each text in the intro series include solutions and answer key or is the answer key in the textbook?

As far as I know, the the textbook does work out some problems but for all the problems in the back, you need to get the manual.

The Solution Manuals go over problems in the section (like 4.2 or 9.1), but the textbook does [b]not[/b] go over Review and Challenge problems.
Hope it helps!
:)

The solutions manuals contain complete solutions to all exercises, review problems, and challenge problems in the texts.  There are no answers to any of these in the textbooks themselves.


linear algebra
matrix
algebra
polynomial
vector
search
linear algebra unsolved

If C is the companion matrix of p(x), how can I proof that p(C)=0, and C has minimal polynomial p(x)?

1) $ det(C-xI_n)=\pm{p(x)}$, thus by Cayley-Hamilton $ p(C)=0$.
2) $ C$ is cyclic because $ e_1,C(e_1),\cdots,C^{n-1}(e_1)$ is a basis ($ e_1$ is the first vector of the canonical basis).
Let $ q$ be a non zero polynomial of degree $ m<n$ s.t. $ q(C)=0$. Then for all vector $ y$ the system $ y,C(y),\cdots,C^m(y)$ is not free, that is a contradiction.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=49967623&t=127175]here[/url] for a proof that does not invoke Cayley-Hamilton.


show that if p(x) is an element of [b]R[/b][x] is a non-negative polynomial, that is p(x)>=0 for all x contained in [b]R[/b], then p(x)=f(x)^2+g(x)^2 for some polynomials f and g.


Find minimum and maximum value of:
\[ f(x)=\frac{2+x+x^3+2x^4}{(1+x)^4}  \]
if $x \ge 0$.
[hide="Read it"]Don't use calculus.[/hide]


inequalities
function
inequalities theorems

[quote="alain-connes"]hello
can any one explain me very simpely:
*what does it mean homogenisation?
*what does it mean a homogeneous inequality?
*why; if the inequality is homogeneous;we can suppose that a+b+c=1?
please don't use polynomels to explaine me  :blush:
thanx and sory for my bad english[/quote]

* Function  F(x, y, ..., z) is called homogenus (of degree k) if for all  t  and for all x, y, ..., z we have
               F(tx, ty, ..., tz) = t^kF(x, y, ..., z)
* Inequality of the form F(x, y, ..., z) >= 0 where F be homogenus function is called homogenus.
* Homogenization: The process to make non-homogenus inequality become homogenus one
* If we have f(x, y, ..., z) >= 0 for x + y + ... + z = 1 then for arbitrary (x, y, ..., z) we have
              f(x, y, ..., z) = t^kf(x/t, y/t, ..., z/t) >= 0 where t = x + y + ...+ z, and so x/t + y/t + ... +z/t = 1.

Namdung

thanks a lot
please can you give me an application to illustrate that?

View this ebook
[url]http://ultrametric.googlepages.com/tin2006new.pdf[/url]


geometry
circumcircle
3D geometry
prism
geometry proposed

Let $ l$ be a tangent of the incircle of triangle $ ABC.$ Let $ M$ be a moving point on the circumcircle of the triangle. Let tangents from $ M$ to the incircle intersect $ l$ at $ X_{1},X_{2},$ respectively. Prove that the circumcircle of triangle $ MX_{1}X_{2}$ always passes through the Miquel point of the complete quadilateral determined by four lines $ l,AB,BC,CA.$
P.S Original problem as follows
Let $ l$ be a tangent of the incircle of triangle $ ABC.$ Let $ M$ be the Miquel point of the complete quadilateral determined by four lines $ l,AB,BC,CA.$ Let tangents from $ M$ to the incircle cut another moving tangent of the incircle at $ P,Q$ respectively. Denote by $ X$ the second intersection of  the circumcircle of triangle $ MPQ$ with of triangle $ ABC.$ $ l$ cuts the circumcircle of triangle $ MPQ$ at $ Y,Z,$ respectively. Show that lines $ XY,XZ$ are tangent to the incircle of triangle $ ABC.$

Is not this a variation of [b]pohoatza[/b] problem ? :idea: Let $ l$ cut the circumcircle $ (O)$ at $ B', C'.$ By Poncelet porism, tangents to the incircle $ (I)$ from $ B', C'$ other than $ l$ intersect at $ A' \in (O).$ Circumcircle $ \odot (MX_1X_2)$ goes through the tangency point $ P \in (O)$ of the mixtilinear incircle of the $ \triangle A'B'C'$ in the angle $ \angle C'A'B',$ which does not depend on $ M \in (O).$ Let $ MX_1, MX_2$ cut $ (O)$ again at $ Y_1, Y_2.$ By Poncelet porism, $ (I)$ is incircle of the $ \triangle MY_1Y_2.$ $ P$ is fixed Miquel point of the complete quadrilateral $ (l, Y_1Y_2, Y_2M, MY_1)$ for any $ M \in (O),$ including the case $ M \equiv A,$ when $ (l, Y_1Y_2, Y_2M, MY_1) \equiv (l, BC, CA, AB).$


geometry
3D geometry
algebra
system of equations

Let $ a,b,c\in \mathbb{Q}$ such that $ a\sqrt[3]{3}\plus{}b\sqrt[3]{4}\plus{}c\sqrt[3]{5}\equal{}0$ . Prove that $ a\equal{}b\equal{}c\equal{}0$ .

There is an advanced theory (where  this means that the three cubic roots are linearly independent over $ \mathbb{Q}$) that gives immediate answer to such questions. Otherwise just separate into $ a\sqrt[3]{3} \plus{} b\sqrt[3]{4} \equal{} \minus{}c\sqrt[3]{5}$, cube it, group again with all rationals on one side, cube it, ..., until you get some linear system of equations with rational coefficients in $ \sqrt[3]{6}$ and $ \sqrt[3]{6^2}$. Solve this, and use the fact that these two cubic roots are irrational to get conditions on the (rational) coefficients, ..., until you finally should reach $ a\equal{}b\equal{}c\equal{}0$. Boooooring.

wow, the dude above me said exactly what i was going to. right on :!:


geometry
geometric transformation
reflection
arithmetic sequence

ok
once again, i have 5 problems
PLEASE PLEASE PLEASE answer fully and explain:

1) In the U.S.A., the symbol 5/2 means the 5th month, 2nd day, or May2. But in England, 5/2 means the fifth day, 2nd month, or February 5. How many days of the year each have the same symbol in both the U.S.A. and England?

2)The product of 2 numbers is 504 and each of the numbers is divisible by 6. However, neighter of teh two numbers is 6. What is the larger of the two numbers?

3) A rectangular garden is 14 ft by 21 ft and is bordered by a concrete walk 3 ft wide. How many sq. ft. are in the surface area of just the concrete walk?

4) 4 numbers are arranged in order of size and the difference between any 2 adjacent numbers is the same. suppose 1/3 is the first and 1/2 is the fourth of these numbers. What are the two numbers between 1/3 and 1/2?

5) Each of the 3 diagrams shows a balance of weights using different letters. How many ::'s will balance a O?

NOTE: The symbol :: is used as one unit.

O           = A :: ::
O ::        = A V
A           =  V ::

[hide="1"]since the days are the same, the dates must be same, so
1/1;2/2;3/3;...11/11;12/12
so 12 days[/hide]
[hide="2"]let 6a be one number, and 6y be the other
$6a \times 6y=36(ay)=504$
$a\times y=14$, since neither a=1 nor y=1 because 6a or 6y would then equal 6. so (x,y)=(2,7) or (7,2) the larger one is 7. $7\times6=42$the larger one is 42. [/hide]
[hide="3"]$14 \times 3 \times 2=84$feet
now subtract a foot from 21 because u just used it above.
$20 \times 3 \times 2=120$feet. 120 + 84=204 feet [color=blue]204 feet [/color][/hide]

[hide]We can note the numbers $a-r$, $a$, $a+r$ and $a+2r$.
We have $a-r=\frac{1}{3}$ and $a+2r=\frac{1}{2}$ By substracting we find $r=\frac{1}{18}$
and then $a=\frac{7}{18}$ and $a+r=\frac{8}{18}=\frac{2}9$. :roll: [/hide]

[hide="Answer to fifth question"]
1. O = A + 2
2. O + 1 = A + V
3. A = V + 1
Substituting the O in 1 into 2
A + 3 = A + V
3 = V
Substituting the V in 2 into 3
A = 4
Substituting the A in 3 into 1
[b]O = 6[/b]
[/hide]

Answer to question 4:

The two numbers are 7/18 and 4/9.

[hide="SOLUTION to question 4"]
It is an arithmetic sequence because the difference of two adjacent numbers is what they go up by. The difference of $\frac{1}{2}$ and $\frac{1}{3}$ is $\frac{1}{6}$, and there are three differences (first to second, second to third, third to fourth). $\frac{1}{6} \div 3 = \frac{1}{18}$ So the difference is 1/18, $second: \frac{7}{18}, third: \frac{4}{9}$[/hide]

polarpan,

You can help the community out by following a few rules of etiquette:

Don't post multiple problems at once.
Don't post homework.
Don't title posts "Help!!!!" -- the title should reflect the topic or the problem.

Thanks.

For questions number 1, the answe is 12. This is true because  is u have the date 9/2 in America and In europe it would be September 2nd in America and Feburary 9 th in Europe. If the numbers are the same such as 5/5 Then in both cases it would be May 5th. The month and the day are in different positions in Europe and in America so in order for them to be the same both numbers msut be the same.  :)

[hide="Solution 1"] The possible days are: 1/1, 2/2, 3/3, 4/4, ..., 12/12, so the answer is $\boxed {12}$ [/hide]


Help me with this problem:
  Given a set of n<101 domino pieces where each side is marked with two digits from 0 to 6. Your task is to arrange pieces in a line such way, that they touch through equal marked sides. It is possible to rotate pieces changing left and right side.
Time limit: 0.25sec.

Just construct a (non-simple) graph on 7 vertices (0-6), connecting i and j if a domino has those two numbers on it. Then the problem is just to find a Eulerian path, which can be done very quickly.


algebra polynomial induction inequalities

Is it possible to find a set $A$ of eleven positive integers such that no six elements of $A$ have a sum which is divisible by $6$?

The theorem of Erdös-Ginzburg-Ziv says:

Let $ S$ be a set of $ 2n \minus{} 1$ integers (we can allow them to occure more than once in the set). Then there is some $ n$ element subset of $ S$ having sum divisible by $ n$.

The whole theorem is proven by showing that if it holds for $ n \equal{} a,b$, then it holds for $ n \equal{} ab$. By this it's reduced to primes, where one can show it on various ways, e.g. using the theorem of Chevalley-Warning or alternatively using that if Cauchy-Davenport.
If you want, I can post a full proof.

This one is $ n \equal{} 6$:
There is no problem to check it for $ n \equal{} 2,3$ by hand, then one only needs to care about the proof of multiplicativity for this case here.

I remember that Arne (from ML) had a nice direct solution for this (short and no casework or anything), but I can't remember it. It was on Kalva but disappeared when the British MO was taken off.

This was Arne's proof.

[quote="Arne"]Lemma 1: among any three integers, two integers can be found having an even sum.

Lemma 2: among any five integers, two integers can be found having a sum which is divisible by 3.

(Proof: almost trivial, pigeonhole principe.)

Now, given $ a_{1},a_{2},a_{3},\,\cdots,a_{11}$, we can assume without loss of generality that $ b_{1}\equal{} a_{1}\plus{}a_{2}$ is even, by Lemma 1. Also, we can suppose that $ b_{2}\equal{} a_{3}\plus{}a_{4}$ is even, ..., and that $ b_{5}\equal{} a_{9}\plus{}a_{10}$ is even. Among the five integers $ b_{1},b_{2},\,\cdots,b_{5}$, three can be found having a sum which is a multiple of 3, by Lemma 2. But the sum of these $ b_{i}$ is equal to the sum of six of the numbers $ a_{i}$, so we have found six numbers among the given ones which have their sum divisible by 6.

We can conclude that such a set cannot exist.[/quote]

Maybe this proof can be generalized. :-)

@ZeTaX: Yes, it would be nice if you could post a full proof. :D

Sorry, haven't seen your response.
In fact, Arne's proof is the one I mentioned above.

Here a proof of the general thing (essentially consisting of two posts I made on MathLinks a longer time ago):

[b]
[u]Erdös-Ginzburg-Ziv:[/u][/b] From any $ 2n \minus{} 1$ integers, someone can find exactly $ n$ of them, such that their sum is divisible by $ n$.

At first, some requisites:

A [i]monomial[/i] in $ n$ variables is a term of the type $ c\cdot x_{1}^{v_{1}}x_{2}^{v_{2}}...x_{n}^{v_{n}}$ and it's [i]degree[/i] is defined as $ v_{1} \plus{} v_{2} \plus{} ... \plus{} v_{n}$.
Then for a [i]polynomial[/i] $ p(x_{1},x_{2},..,x_{n})$ in $ n$ variables (which is a sum of such monoms by definition), the [i]degree[/i] $ \mathrm{deg}(p)$ is defined as the highest degree occuring in it's monomials.

Now it's possible to state the important [b]theorem of Chevalley-Warning[/b]:
Let $ p$ be a prime and $ f_{1}(x_{1},x_{2},..,x_{n}) ,\ f_{2}(x_{1},x_{2},..,x_{n}) ,\ ... ,\ f_{k}(x_{1},x_{2},..,x_{n})$ be $ k$ polynomials in $ n$ variables $ x \equal{} (x_{1},x_{2},...,x_{n})$.
If we have $ \mathrm{deg}(f_{1}) \plus{} \mathrm{deg}(f_{2}) \plus{} ... \plus{} \mathrm{deg}(f_{k}) < n$, then the number of $ \mod p$ different solutions to the system
\[ \\
f_{1}(x_{1},x_{2},..,x_{n})\equiv 0\mod p \\
f_{2}(x_{1},x_{2},..,x_{n})\equiv 0\mod p \\
... \\
f_{k}(x_{1},x_{2},..,x_{n})\equiv 0\mod p
\]
is divisible by $ p$.

[u]Proof:[/u]

Lets work in the field $ \mathbb{F}_{p}$ (because this prevents a lot of writing and means nothing else than everything is $ \mod p$).

We need a
[b]Lemma[/b]:
Let $ k\in\mathbb{Z}, 0\leq k < p \minus{} 1$, then $ \sum_{x\in\mathbb{F}_{p}}x^{k} \equal{} 0$ (here $ x^{k}$ is seen as polynomial, thus $ x^{0}$ means the constant polynomial $ 1$)
[u]Proof [/u]of Lemma:
$ z^{k} \minus{} 1 \equal{} 0$ has at most $ k$ solutions $ z$ in our field. As $ k < p \minus{} 1$, there is a nonzero $ y\in\mathbb{F}_{p}$ such that $ y^{k} \minus{} 1\neq 0$. Now we get
\[ \sum_{x\in\mathbb{F}_{p}}x^{k} \equal{} \sum_{xy^{ \minus{} 1}\in\mathbb{F}_{p}}(xy)^{k} \equal{} y^{k}\sum_{x\in\mathbb{F}_{p}}x^{k}
\]
giving $ (y^{k} \minus{} 1)\cdot\sum_{x\in\mathbb{F}_{p}}x^{k} \equal{} 0$, resulting in$ \sum_{x\in\mathbb{F}_{p}}x^{k} \equal{} 0$ as $ y^{k} \minus{} 1\neq 0$.

Back to the [u]proof[/u] of the theorem:

Let be $ s$ the number of solutions of the system of equations. We want to prove $ s \equal{} 0$ in $ \mathbb{F}_{p}$ (be aware of that this just means that the number of solutions is divisible by $ p$).

Lets consider the polynomial $ R(x) : \equal{} \prod_{i \equal{} 1}^{m}\left(1 \minus{} \left(f_{i}\left(x_{1},...,x_{n}\right)\right)^{p \minus{} 1}\right)$.
Then $ s \equal{} \sum_{x\in\mathbb{F}_{p}^{n}}R(x)$, because $ R(x) \equal{} 1$ iff $ x$ is a solution and $ \equal{} 0$ otherwise.

Now lets look at the monoms $ g(x) \equal{} c\cdot\prod_{i \equal{} 1}^{n}x_{i}^{v_{i}}$ that occure in $ R$. For any single one of them, there is an $ i$, w.l.o.g. $ i \equal{} 1$,  with $ v_{i} < p \minus{} 1$, because otherwise the condition on the degrees would not be fulfilled. But then
\[ \sum_{x\in\mathbb{F}_{p}^{n}}g(x) \equal{} \sum_{x_{1}\in\mathbb{F}_{p}}\sum_{x_{2}\in\mathbb{F}_{p}}...\sum_{x_{n}\in\mathbb{F}_{p}}g(x) \equal{} c\cdot\sum_{x_{1}\in\mathbb{F}_{p}}x_{1}^{v_{1}}\cdot\sum_{x_{2}\in\mathbb{F}_{p}}...\sum_{x_{n}\in\mathbb{F}_{p}}x_{2}^{v_{1}}x_{3}^{v_{1}}... x_{n}^{v_{1}}
\]
and the last one is $ 0$ as $ \sum_{x_{1}\in\mathbb{F}_{p}}x_{1}^{v_{1}} \equal{} 0$ by the lemma. So $ s \equal{} 0$ because we are summing up monoms summing up to $ 0$. This proves the theorem.

Now let's start a [u]proof[/u] of the large goal, the theorem of [u]Erdös-Ginzburg-Ziv:[/u]

Let $ EGZ_{n}$ mean the theorem for this special $ n$, so stating that we can choose from any set of $ 2n \minus{} 1$ (or more) integers $ n$ of them such that $ n$ divides their sum.

Lets split it up:

[b]Lemma 1[/b]: If we can prove $ EGZ_{a}$ and $ EGZ_{b}$, we can show that $ EGZ_{ab}$ is true, too.
[u]Proof[/u]:
Let $ 2ab \minus{} 1$ numbers be given. Since we assume $ EGZ_{a}$ to be true, we can choose $ a$ numbers of them such that their sum $ s_{1}$ is divisible by $ a$. Then $ 2a(b \minus{} 1) \minus{} 1$ numbers are left, and we can again choose $ a$ of them such that their sum $ s_{2}$ is divisible by $ a$. Go on and on and on till there are just $ a \minus{} 1$ numbers left. We calculate to see that we took $ 2b \minus{} 1$ times $ a$ numbers this way and call their sums $ s_{1},s_{2},...,s_{2b \minus{} 1}$.
Now it's nice to have exactly $ 2b \minus{} 1$ such sums, so that we can apply $ EGZ_{b}$ and find $ b$ of these $ s_{i}$ such that their sum $ S$ is divisible by $ b$.
But looking back at how we got the $ s_{i}$, we see that $ S$ is the sum of exactly $ ab$ of the original numbers. And looking further, their sum is divisible by $ ab$, which we wanted to show.

So applying this lemma often enough, we are left with proving the whole theorem for primes only. Thus it suffices to prove the following:

[b]Lemma 2[/b]: $ EGZ_{p}$ is true for primes $ p$.
[u]Proof[/u]:
Let any $ 2p \minus{} 1$ integers $ a_{1},a_{2},...,a_{2p \minus{} 1}$ be given.
Look at the system
\[ x_{1}^{p \minus{} 1} \plus{} x_{2}^{p \minus{} 1} \plus{} ... \plus{} x_{2p \minus{} 1}^{p \minus{} 1}\equiv 0\mod p \\
a_{1}x_{1}^{p \minus{} 1} \plus{} a_{2}x_{2}^{p \minus{} 1} \plus{} ... \plus{} a_{2p \minus{} 1}x_{2p \minus{} 1}^{p \minus{} 1}\equiv 0\mod p
\]
of equations $ \mod p$ in $ 2p \minus{} 1$ variables. Since the sum of their degrees is $ (p \minus{} 1) \plus{} (p \minus{} 1) \equal{} 2p \minus{} 2$, lower than $ 2p \minus{} 1$, the number of variables, we can use the theorem of Chevalley-Warning.
There is one solution $ x_{1} \equal{} x_{2} \equal{} ... \equal{} x_{2p \minus{} 1} \equal{} 0$, and since the number of solutions is divisible by $ p > 1$, there is another solution.
For that other solution, set $ y_{i} \equal{} x_{i}^{p \minus{} 1}$. By Fermat's theorem, $ y_{i}\equiv 0\mod p\iff x_{i}\equiv 0\mod p$ and $ y_{i}\equiv 1\mod p$ otherwise.
We conclude that all $ y_{i}$ are $ \equiv 0,1\mod p$. By definition, we have that not all $ y_{i}$ are $ 0$, but the first of the two equations says
\[ y_{1} \plus{} y_{2} \plus{} .. \plus{} y_{2p \minus{} 1}\equiv 0\mod p
\]
giving that exactly $ p$ of the $ y_{i}$ are $ \equiv 1\mod p$ (and the others $ 0$).
Setting this into the second one
\[ a_{1}y_{1} \plus{} a_{2}y_{2} \plus{} .. \plus{} a_{2p \minus{} 1}y_{2p \minus{} 1}\equiv 0\mod p
\]
shows us that a sum of exactly $ p$ of the $ a_{i}$ is $ \equiv 0\mod p$, in other words divisible by $ p$, finishing the proof

[b]q.e.tee.[/b]

I just wanted to note that there is needed a very little correction to ZetaX's proof.

[quote="ZetaX"]
Now it's nice to have exactly $ 2b \minus{} 1$ such sums, so that we can apply $ EGZ_{b}$ and find $ b$ of these $ s_{i}$ such that their sum $ S$ is divisible by $ b$.
But looking back at how we got the $ s_{i}$, we see that $ S$ is the sum of exactly $ ab$ of the original numbers. And looking further, their sum is divisible by $ ab$, which we wanted to show.
[/quote]

From the fact that, for example, $ b|s_1\plus{}s_2...\plus{}s_b$ it doesn't follow that $ ab|s_1\plus{}...\plus{}s_b$. But of course that's not a problem, since we can write $ s_i\equal{}ak_i$ and from the numbers $ k_1, k_2, ..., k_{2b\minus{}1}$ we can choose $ b$ of them with the sum divisible by $ b$. And then everything is allright.

As[b] ZetaX[/b] mentioned [b]Erdös-Ginzburg-Zif's theorem[/b] follows from [b]Chevalley-Warning[/b] or [b]Cauchy-Davenport [/b]theorem.And the last theorem has very elementary and short proof by induction,that I've found out at TST,as the first theorem was given just like one of our problems.And this proof doesn't use any polynomials just some beautiful combinatorics ideas...
I didn't post it because maybe this proof was already discussed here...But if not,I can post the solution.

[quote="Erken"]I didn't post it because maybe this proof was already discussed here...But if not,I can post the solution.[/quote]Well, I don't see it in this topic, so go ahead.

Let me at first prove [b]particular case of [/b]

[b]Cauchy-Davenport Theorem[/b](let me call it [b]T.E[/b]'s lemma :P  :

Let $ A_1,A_2,\dots,A_n$ be two-element subsets of $ \mathbb{Z}_p$,where $ p$ is prime number.Then the following inequality holds:

$ |A_1 \plus{} A_2 \plus{} \dots \plus{} A_n|\geq\min\{p,n \plus{} 1\}$.
[b]
Proof:[/b]
We will proceed by induction on $ n$:

It is not hard to check that theorem holds for $ n \equal{} 2$.

Suppose that we've already proved the theorem for $ n\leq k \minus{} 1$.

[b]First[/b] case:$ k\leq p \minus{} 1$:

Let $ A_i \equal{} \{x_i,y_i\}$,for all $ i \equal{} 1,2,\dots,k$.

Suppose to the contrary that $ |A_1 \plus{} A_2 \plus{} \dots \plus{} A_k| < k \plus{} 1$,it follows that $ |A_1 \plus{} \dots \plus{} A_{k \minus{} 1}| \equal{} |A_1 \plus{} A_2 \plus{} \dots \plus{} A_k| \equal{} k$.

And let $ S \equal{} A_1 \plus{} \dots \plus{} A_{k \minus{} 1} \equal{} \{S_1,S_2,\dots,S_k\}$ and $ S' \equal{} A_1 \plus{} A_2 \plus{} \dots \plus{} A_k$.

Also let's consider two sets $ X \equal{} \{S_1 \plus{} x_k,S_2 \plus{} x_{k},\dots,S_k \plus{} x_k \}$ and $ Y \equal{} \{S_1 \plus{} y_k,S_2 \plus{} y_k,\dots,S_k \plus{} y_k\}$.

It is clear that $ |X| \equal{} |Y| \equal{} k$ and $ \boxed{X \equal{} Y\subset S'}$,otherwise we would obtain a contradiction.

Therefore,sum of elements in $ X$ and $ Y$ are equal,so:

$ S_1 \plus{} S_2 \plus{} \dots \plus{} S_k \plus{} k\cdot x_k \equal{} S_1 \plus{} S_2 \plus{} \dots \plus{} S_k \plus{} k\cdot y_k$,which is equivalent to

$ p$ divides $ k(x_k \minus{} y_k)$ but $ k < p$ and $ x_k\neq y_k$.Contradiction.

[b]Second[/b] case: $ k\geq p$ is obvious. :lol:

Now let's prove our theorem:

[b]Erdös-Ginzburg-Zif[/b]:

From any $ 2n \minus{} 1$ integers,someone can find exactly $ n$ of them,such that their sum is divisible by $ n$.

[b]Proof:[/b]
As it was mentioned by [b]ZetaX[/b] it is enough to consider the case when $ n$ is prime.

Let $ a_{2p \minus{} 1}\geq a_{2p \minus{} 2}\geq \dots\geq a_1$ be given integers.As we did it before,we will consider two cases and again one of them is obvious :lol: :

[b]First[/b] case:

When there exist $ i\in\mathbb{N}$,such that $ a_i \equal{} a_{i \plus{} p \minus{} 1}$,then we can take these $ p$ equal integers.

[b]Second[/b] case:

When $ a_i\neq a_{i \plus{} p \minus{} 1}$,for all $ i\in\mathbb{N}$.

Let $ A_i \equal{} \{a_i,a_{i \plus{} p \minus{} 1}\}$,where $ i \equal{} 1,2,\dots,p \minus{} 1$ and $ a_{2p \minus{} 1}$ doesn't belong to any set $ A_i$.

Applying [b]T.E[/b]'s lemma:

$ \boxed{|A_1 \plus{} A_2 \plus{} \dots \plus{} A_{p \minus{} 1}|\geq p}$,therefore, $ A_1 \plus{} A_2 \plus{} \dots \plus{} A_{p \minus{} 1} \equal{} \mathbb{Z}_p$.

But it means that:

$ \boxed{\minus{} a_{2p \minus{} 1}\in A_1 \plus{} A_2 \plus{} \dots \plus{} A_{p \minus{} 1}}$.

So their sum is divisible by $ p$.
[b]
Q.E.D[/b]

[b]Note:[/b]I hope you will enjoy my proof,any comments would be appreciated.


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Can anybody minimize $ x^2 \plus{} y^2 \plus{} z^2$ given that $ (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2 \equal{} 1$ using lagrange multipliers? That is to say, give a full & rigorous solution.

(of course it is easy using homogenization...)

Let $ f(x,y,z)\equal{}x^2\plus{}y^2\plus{}z^2$
Let $ g(x,y,z)\equal{}x^2\plus{}y^2\plus{}z^2\minus{}2xy\minus{}2yz\minus{}2xz\minus{}1$
We remark that $ f$ and $ g$ are both smooth (infinitely differentiable) functions on the entire plane.

We want to find the minimum value of $ f$ on $ g^{\minus{}1}(0)$.

Now, the gradient of $ g$ is $ (2x\minus{}2y,2y\minus{}2z,2z\minus{}2x)$.  This is zero only when $ x\equal{}y\equal{}z$ and in particular is not $ 0$ anywhere on $ g^{\minus{}1}(0)$.  Thus we can apply Lagrange multipliers and conclude there exists some real number $ \lambda$ such that the gradient of $ f$ is $ \lambda$ times the gradient of $ g$ at the minimum of $ f$ on $ g^{\minus{}1}(0)$.  This gives us the following system of equations:

$ 2x\equal{}\lambda \cdot (2x\minus{}2y)$
$ 2y\equal{}\lambda \cdot (2y\minus{}2z)$
$ 2z\equal{}\lambda \cdot (2z\minus{}2x)$
and
$ (x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(z\minus{}x)^2\equal{}1$

Solving this is something of a pain, but anyways the above verifications suffice to prove that any local minimum of $ f$ on $ g^{\minus{}1}(0)$ occurs at a triplet $ (x,y,z)$ such that there exists some $ \lambda \in \mathbb{R}$ for which $ (x,y,z,\lambda)$ is a solution to the above system.

In general if you want to find an extreme value of $ f(x_1,x_2,...,x_n)\equal{}0$ subject to $ g(x_1,x_2,...,x_n)\equal{}0$ via Lagrange multipliers you must check that:
1) f and g have continuous partial derivatives in an open neighborhood of $ g^{\minus{}1}(0)$
2) The gradient of $ g$ does not vanish on $ g^{\minus{}1}(0)$

This is not enough. An argument guaranteeing a minimum is also necessary.

[quote="n^4+4"]Let $ g(x,y,z) \equal{} x^2 \plus{} y^2 \plus{} z^2 \minus{} 2xy \minus{} 2yz \minus{} 2xz \minus{} 1$[/quote]
Shouldn't $ x^2$ be $ 2x^2$ (and same for $ y$ and $ z$)?  Does this change affect the rest of your argument?

the main point is that grad g is zero at x=y=z

if i replaced 'minimum' with 'maximum' your argument would not change...

since there is obviously not a maximum, then thats another problem

An interesting approach is via space geometry.

Consider triangle $ ABC$ with $ A\equal{}(x,y,z), B\equal{}(y,z,x), C\equal{}(z,x,y)$.
Then $ ABC$ is equilateral with sidelength 1.

Note that $ ABC$ lie on the sphere with the center $ O$ and radius $ \sqrt{x^2\plus{}y^2\plus{}z^2}$.
Clearly, the minimum of this radius is when $ O$ coincides with circumradius of triangle $ ABC$.
Thus the minumum of $ x^2\plus{}y^2\plus{}z^2\equal{}R_{ABC}^2\equal{}\frac{1}{3}$.

The condition holds iff $ A,B,C, O$ are coplanar. Hence
$ \begin{vmatrix}x&y&z&1\\y&z&x&1\\z&x&y&1\\0&0&0&1\end{vmatrix}\equal{}0$
or $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}0 \text{ } \iff \text{ } (x\plus{}y\plus{}z)\equal{}0$.

The same thing you get using Lagrange Multiplyers:
Consider $ f(x,y,z,\alpha)\equal{}x^2\plus{}y^2\plus{}z^2\minus{}\alpha((x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(x\minus{}z)^2\minus{}1).$
The minumum holds when

$ 2x\minus{}4\alpha x\plus{}2\alpha y\plus{}2\alpha z\equal{}0$
$ 2y\minus{}4\alpha y\plus{}2\alpha x\plus{}2\alpha z\equal{}0$
$ 2z\minus{}4\alpha z\plus{}2\alpha x\plus{}2\alpha y\equal{}0$
$ (x\minus{}y)^2\plus{}(y\minus{}z)^2\plus{}(x\minus{}z)^2\minus{}1\equal{}0$.

So $ x(3\alpha\minus{}1)\equal{}y(3\alpha\minus{}1)\equal{}z(3\alpha\minus{}1)\equal{}\alpha(x\plus{}y\plus{}z)$
if $ (3\alpha\minus{}1)\ne 0,$ then $ x\equal{}y\equal{}z$ - false. Thus $ \alpha\equal{}\frac{1}{3}$ and $ x\plus{}y\plus{}z\equal{}0$.
From here we immediatelly deduce that this is the case when minimum of $ x^2\plus{}y^2\plus{}z^2$ occurs
and it is $ \frac{1}{3}$.

Here is my solution. It is my problem by the way :D

Let $ \sigma_1\equal{}x\plus{}y\plus{}z$ and $ \sigma_2\equal{}xy \plus{}yz\plus{}zx$. Suppose that $ \sigma_1^2\minus{}2\sigma_2\equal{}k$ and $ 2\sigma_1^2\minus{}6\sigma_2\equal{}1$. By rolle's theorem and blah blah a clever application, there are real a,b such that $ \sigma_1\equal{}\frac{3}{2}\cdot (a\plus{}b)$ and $ \sigma_2\equal{}3ab$.

WLOG $ a\ge b$
Then the restriction becomes $ a\minus{}b\equal{}\sqrt{\frac{2}{9}}$ and the thing that we have to minimize is \[ \frac{3}{4}\cdot (3a^2\minus{}2ab\plus{}b^2)\equal{}\frac{1}{2}\cdot (6b^2\plus{}2\sqrt{2} b\plus{}1)\ge \frac{1}{3}\] easily. It is easy to check equality.


For $ a,b,c$ are positive real numbers, prove that:
$ \frac {a^4}{b^3 \plus{} c^3} \plus{} \frac {b^4}{c^3 \plus{} a^3}\plus{} \frac {c^4}{a^3 \plus{} b^3} \ge \frac {3}{2}$

[quote="AngelHolder"]For $ a,b,c$ are positive real numbers, prove that:
$ \frac {a^4}{b^3 \plus{} c^3} \plus{} \frac {b^4}{c^3 \plus{} a^3} \plus{} \frac {c^4}{a^3 \plus{} b^3} \ge \frac {3}{2}$[/quote]

It's not true for  $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$

With $ a\plus{}b\plus{}c\equal{}3$
=>True
Chebyshev can slove it


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Show that $ \frac{d \tan x}{dx} \equal{} 1 \plus{} \tan^{2} x$.

$ F(x) \equal{} \int_{0}^{\infty} \frac {1}{x^{2} \plus{} y^{2}}\ dy$.

Find $ F(x)$ in terms of $ x$ and sketch the graph of $ F(x)$ for $ x > 0$.

Hence sketch the graph of $ G(x) \equal{} \int_{0}^{\infty} \frac {1}{x^{4} \plus{} y^{4}}\ dy$.

Can you at least do the first two parts?  Show us what you've been able to do.

Yes, atomicwedgie, I can: I haven't posted the question because, as you wrongly assume, I can't do it, but because I think it's a little unusual and may be of interest to some people out there.

Are you suggesting that you can do the third part (if I do the first two)? If so:

$ \frac{d(tan x)}{dx} \equal{} \sec^{2}x$, (by the quotient rule if necessary),

$ \equal{} 1\plus{}\tan^{2}x$

From the standard result, $ \int \frac{1}{a^{2}\plus{}x^{2}}\ dx \equal{} \frac{1}{a} \tan^{\minus{}1} (\frac{x}{a}) \plus{} c$,

$ F(x) \equal{} \frac{1}{x} [\tan^{\minus{}1}(\frac{y}{x})]_{0}^{\infty} \equal{} \frac{\pi}{2x}$.

hello, by the quotient rule we get
$ (\tan(x))^'=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=1+\tan^2(x)$ by the definition of $ \tan(x)$.
Sonnhard.

hello, i have got
$ \int_{0}^{\infty}\frac{1}{x^4\plus{}y^4}\,dy\equal{}\frac{\pi\left(\frac{1}{x^4}\right)^{\frac{3}{4}}}{2\sqrt{2}}$.
Sonnhard.

Hi Sonnhard (and Happy Christmas), I've also got $ \frac {\pi}{2\sqrt {2}x^{3}}$ but the question is to sketch the graph of $ F(x)$ and hence sketch the graph of $ G(x)$. I've got to go out. I'm still eagerly awaiting atomicwedgie's solution to this part.

So you're singling me out because I made an incorrect assumption?  Sorry.

No, atomicwedgie, (you've made another incorrect assumption), I'm not singling you out: I thought you might have a neat solution. Sorry if I offended you.

For the last part, I integrated, as Sonnhard did, to check my ideas. It seemed that if we had cubes in the denominator, instead of squares as in the second part, we'd have squares in the denominator when we integrate; and similarly for quartics as in $ G(x)$, we'd have cubes in the denominator when we integrate, which checked out. As the question seemed to be to deduce the graph of $ G(x)$ from that of $ F(x)$, I looked at what happens as $ y$ approaches zero and as it approaches infinity. (Sorry, my CTRL key is faulty again (and my backslash key never works) so I can't LaTeX at the moment, making it too wordy.) It seems that there will be a larger gap on the left and a smaller one underneath but that otherwise it will look like the positive branch of the rectangular hyperbola of the second part.

Perhaps someone has a better idea?


Prove that the number $ (\underbrace{9999 \dots 99}_{2005}) ^{2009}$ can be obtained by erasing some digits of $ (\underbrace{9999 \dots 99}_{2008}) ^{2009}$ (both in decimal representation).
[i]Yudi Satria, Jakarta[/i]

Outline: Write $ 999...9\equal{}10^n\minus{}1$. So $ 999...9^{2009}\equal{}(10^n\minus{}1)^{2009}\equal{}\sum\binom{n}{k}10^{nk}\minus{}\binom{n}{k\minus{}1}10^{n(k\minus{}1)}$. Therefore $ 999...9^{2009}$ can be written as blocks of numbers. It is easy to prove the each block of $ (\underbrace{9999\dots 99}_{2008})^{2009}$ can be erased to get a block of $ (\underbrace{9999\dots 99}_{2005})^{2009}$


I have this science homework but I just can't seem to be sure.

is this:

-cyte (cyto) means cell or hollow vessel

neuro- means brain or nerve

hydrology means study of water???

necrology means study of death???

arachnophobia is fear of spider right? Or is it fear of Arachnid wholeself?

cytology means study of cells and structures???

osteocyte means bone with cell surrounding it???

neurocyte is what??? I can't seem to find what it is....

hydrophilic mens love of water???

osteoporosis is kind of weird.. What I found was disease in which bones became extremely porous?

arachnology means study of arachnid?

acrophobia means fear of high places right?

What's phobophobia???

nuerology is study of brain or study of nerves?

necrosis is process of death?? It sounds kind of stupid.. So is it then process of cells through injury or disease?? I'm not kind of sure because necro actually meant death (or that's what I know)

THanks for big help.. I haven't in the site for while but anyway, great thanks!  :lol:

Whoa,  man.  Doesn't your book give you definitions for these?  These are things that you could very easily find on an online dictionary or you could search for these.  It would be much better if you used google, or something like dictionary.com... doing this would be much faster for you and would teach you how to do research on your own which is very important.

We don't mind you posting questions from homework as long as the info. for doing it can't be found on a reference site or it is something that is truly confusing.

I got some from yahoo dictionary but some of them I couldn't find.

So, help for those...

And neuro, too.. Nerve or brain?

dictionary.com actually has a vast majority of these words... check that.. :?

I'll just answer phobophobia, cause that one rocks.

Fear of fears.

http://www.miriamwebster.com

great site

[quote]Phobophobia - Fear of fears[/quote]

Thanks a lot!

I'll just summarize. Just scam throu and if you see something that makes you say "HEY!!!" Point out. THANKX

Some might not be ones I posted but here it goes:

1. -ology - study
2. -phobia - fear
3. -philic - love
4. -osteo - bone
5. -sis - process of
6. -cyte (cyto-) - cell
7. hyper - too much
8. hypo - not enough, less
9. necro- : death
10. hydro- : water
11. zoo- : animal
12. thermo- : heat
13. techno- : new
14. neuro- : nerves
15. biology - study of living things
16. hydrology - study of water
17. necrology - study of death
18. hydrophobia - fear of water
19. arachnophobia - fear of spiders
20. cytology - study of cells and structure
21. osteocyte - A branched cell embedded in the matrix of bone tissue
22. zoology - study of animals
23. hydrophilic - loves water
24. osteoporosis - A disease in which bones became extremely porous
25. arachnology - study of Anarchnids
26. hydrosis - process of water
27. thermophilic - loves heat
28. acrophobia - Fear of high places
29. phobophobia - fear of fear
30. neurology - study of nervous system
31. technophilic - loves technology
32. necrosis - process of death
33. zoophilic - loves animals

THANKS>>>>>

That's it.  :maybe:

From what I know "hyper" and "hypo" is more correctly thought of as "over" and "under" so you can think of hypothermia as "under normal" instead of "not enough". Not very importent, but if you ever study vocab the distinction is important.

My personal favorite: Hippopotomonstruosesquipedaliophobia=fear of long words.


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real analysis
real analysis unsolved

Prove that the function $ f(z)=\sum_{n\geq 1}{\frac{z^{2^n}}{n!}}$ can be extented to a smooth function on all the complex plane. Note that it is pretty obvious that it cannot be extended analytically beyond the unit disc.

What is smooth function here?  :?: Somewhat $ C^\infty[x,y]$?

Yes, it means it is differentiable with continuous derivatives as a function of two variables and this happens at any order of differentiation.

Seems like it is time to post a solution to this one (not to make harazi wait for one year again :P). Of course, there is nothing sacred about the particular series given; it could be any series $ \sum_k c_k z^k$ where $ c_k$ decay faster than any power. Let $ F: \mathbb C\to\mathbb C$ be any compactly supported $ C^\infty$ function such that $ F(z)\equal{}z$ for $ |z|\le 1$. Then $ \sum_k c_k F(z^k)$ is a desired extension. Indeed, the $ k$-th term is non-zero only if $ |z|^k$ is bounded by some fixed constant and when differentiating it $ j$ times, you get only something of order $ k^j (1\plus{}|z|^{kj})$ as a factor in front of some derivative of $ F$, so the formal series defining any derivative converges absolutely and uniformly in $ \mathbb C$, which is more than enough to ensure smoothness.

Nice, as usual, Fedja. I used instead Seeley's lemma in order to get the desired extension. Thank you!


geometry calculus rotation rectangle integration logarithms limit

Prove without calculus that the area of the region in Quadrant I surrounded by the graphs of the equations $x=0$, $y=0$, and $y=\frac{1}{x}$ is infinite.

It is possible to do this with absolutely no calculus whatsoever.

Wait isn't it finite?

I thought the area under this part of the curve to the right of x=1 was e, then just rotate the plane 90 degrees counter clockwise to find another region with area e, then we're left with a region that is closed, so clearly it is finite.

Unless I'm missing someting...

It's finite. As X goes from 1 to $\infty$, we see that the area is $e$. Also, as Y goes from 1 to $\infty$, we see the area is also $e$.
We are left with a unit square, so we can see the area should be $2e+1$

Is this right?

Here's what I have:
[hide]Draw rectangles that have lengths of 1.  The bottoms of the rectangles will be on the x-axis.  The top-right point of each rectangle will touch the graph of $y=\frac{1}{x}$.  The first rectangle has an area of 1, the second 1/2, the third one 1/3, and so on.
The area of the $n^\text{th}$ rectangle has an area of $1/n$.
Since we can clearly draw an infinite number of rectangles, the area is at least $\sum_{i=1}^\infty \frac{1}{i}$.
A proof that $\sum_{i=1}^\infty \frac{1}{i}=\infty$:
Break it up into groups of 1, 2, 4, 8, 16...
$1+\frac12+\left( \frac13+\frac14\right)+\left( \frac15+\frac16+\frac17+\frac18\right)+\ldots$
$>1+\frac12+\frac12+\frac12+\ldots$
So it is obvious that $\sum_{i=1}^\infty \frac{1}{i}=\infty$.

Since we can inscribe an infinite area within that region, the area of the region itself must be infinite.

$\mathbb{QED}$[/hide]

I'm not sure where you guys are getting an area of $e$ from.

[hide="The area to which you refer is..."] $\int_{1}^{\infty}\frac{1}{x}\, dx = \ln x |_{1}^{\infty}= \lim_{x \to \infty}\left( \ln x-\ln 1 \right)$ which is clearly infinite. [/hide]

I know we're not supposed to use calculus but if you do, you'll find that the area is infinite because the antiderivative of $1/x$ is $\ln x$ and we are going from 1 to $\infty$ so we will end up with $\ln \infty$ which is $\infty$.  So the area is infinite.

Also, using the way lingomaniac used with the rectangles, the lengths will stay the same but the heights diverge since they are in a harmonic sequence.

[quote="t0rajir0u"][hide="And the final connection, ladies and gentlemen"] $\lim_{n \to \infty}\left( \sum_{k=1}^{n}\frac{1}{k}-\ln n \right) = \gamma$ [/hide][/quote]

For those of us without massive knowledge of strange constants...

[url]http://mathworld.wolfram.com/Euler-MascheroniConstant.html[/url]

[quote="DiscreetFourierTransform"]Uggh. I realize our mistake:
The area from $x = 1$ to $x = e$ under the curve is $1$, not from $x=1$ to $x=\infty$.[/quote]

Ohh I guess that's where I screwed up.

I wasn't trying to make a topic that was related to calculus.  Yes, it can be solved using calc, but it's possible to use none whatsoever, and I wanted to see if anyone could figure it out.

Well, you already showed the solution.

[hide="In particular..."] The area to which we refer is bounded below by the area of the rectangles of the LRAM, which is

$\sum_{k=1}^{\infty}\frac{1}{k}$

There are several popular methods of showing that this diverges; here is one that I think is fairly simple.  Assume it converges to $L$.  Let

$E = \sum_{k=1}^{\infty}\frac{1}{2k}$
$O = \sum_{k=1}^{\infty}\frac{1}{2k-1}$

If $L$ exists, then $E, O$ also exist, and $L = E+O$.  However,

$E = \frac{1}{2}L = \frac{1}{2}(E+O) \Leftrightarrow E = O$

Which is impossible, since $O-E$ is clearly positive.  Hence $L$ does not exist. [/hide]

Do they ever ask to solve a problem without using calculus at USAMO or other big competitions? And I think I may have to throw a party since I did this the same way as t0rajir0u [quote]$\int_{1}^{\infty}\frac{1}{x}\, dx = \ln x |_{1}^{\infty}= \lim_{x \to \infty}\left( \ln x-\ln 1 \right)$ which is clearly infinite.[/quote]...even if it was a really easy problem, lol.

As far as the USAMO is concerned, calculus is strongly discouraged.  If your solution is even slightly off, I hear they're not very forgiving about points.

You're allowed to test for convexity (for inequalities) without too much punishment though. There aren't that many problems that can be more easily solved by calculus than non-calculus means anyway.

it is pretty obvious if you look at the left hand riemann sum, one value is undefined

[hide="solution"]We want the region bounded by y=1/x when x is positive. As x gets arbitrarily close to ∞, y gets arbitrarily close to 0, and vice-versa. Since y goes to infinity when x apporaches 0, and x neer reaches 0, the area of the region in mind is ∞.[/hide]

[quote="Altheman"][quote="paladin8"]To Altheman: The left hand Riemann sum is greater than the area under the curve though...

To nutz_for2.718281828: Your argument is flawed, the function $-\ln{x}$ also goes to $\infty$ as $x \rightarrow 0$, but the area under the curve from $x = 0$ to $x = 1$ is finite.[/quote]

$\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}f(x)*\frac{1}{n}=\int_{0}^{1}f(x) \, dx$

regardless, the left side is undefined, so the area is undefined, since it is obvious that there is some area, then it is clear that there is infinite area in that interval, (it does not matter that it is an over approximation, because as the limit is taken to infinity, they are equal (in a limit sense))[/quote]

But that runs into the same problem with the case $f(x) =-\ln{x}$; you would say that the LHS is "undefined" but the integral converges.

[quote="paladin8"][quote="Altheman"]$\int_{0}^{1}-\ln x$ is undefined... recall the definition of an integral,

$f$ must be continuous on $[0,1]$ which it is clearly not...(at $x=0$ )[/quote]

What... then $\int_{0}^{1}\frac{1}{x}dx$ is no more defined... we're talking about improper integrals here, or

$\lim_{a \rightarrow 0}\int_{a}^{1}-\ln{x}$.[/quote]

alright...i concede you are correct, the definition i gave was the only definition that i have seen in my (incomplete) calculus study...after reading the section on improper integrals, i see your point,

but i still have trouble with that definition...how can you speak of an area when there is no point bounding at x=0,
suppose
$f(x)=$\begin{eqnarray*}1 &\text{for}&x\ne 1\\ \text{undefined}&\text{for}&x=1\\ \end{eqnarray*}

then $\int_{0}^{2}f(x)\, dx$ would be $2$ regardless of the fact that the area is not bounded at $x=1$ endpoint

[quote="Altheman"]
but i still have trouble with that definition...how can you speak of an area when there is no point bounding at x=0,
suppose
$f(x)=$\begin{eqnarray*}1 &\text{for}&x\ne 1\\ \text{undefined}&\text{for}&x=1\\ \end{eqnarray*}

then $\int_{0}^{2}f(x)\, dx$ would be $2$ regardless of the fact that the area is not bounded at $x=1$ endpoint[/quote]

Well since the integral is improper, it's not saying the area is exactly $2$. It's saying that, as you get arbitrarily close to $1$ from both sides, the area goes toward (or, in this case, stays constant at) $2$. Limits are a funny thing.

The Riemannian integral is defined in a lot of interesting situations.  For example,

[quote="Kent Merryfield"]Example (famous):

Let $f(x)=\begin{cases}\frac1q&\text{if }x=\frac{p}{q}\text{ in lowest terms}\\0&\text{if }x\text{ is irrational}\end{cases}$

There is no interval on which this function is continuous, so it cannot be piecewise continuous. It is continuous at every irrational number and discontinuous at every rational number. Nonetheless it it Riemann integrable on any interval, and has integral zero.

--

It is true that a piecewise continuous function is Riemann integrable. This would follow from the following:

(1) A continous function on a bounded interval is Riemann integrable. The proof uses the notion of uniform continuity.

(2) A piecewise integrable function is integrable, just by partitioning appropriately.

--

There is a theorem which characterizes the Riemann integrable functions: A bounded function on a bounded interval is Riemann integrable if and only if the set of its discontinuities has Lebesgue measure zero.[/quote]


[quote="digitalboy1"]Find the minimum value of $|x-p|+|x-15|+|x-p-15|$, if $x$ satisfy $p\leq x \leq 15$ and $0<p<15$[/quote]
Was this is an old AIME problem?
Anyways, $p\leq x$ so $|x-p|=x-p$.
$x\leq15$ so $|x-15|=15-x$ and $|x-p-15|=15+p-x$.
Then $|x-p|+|x-15|+|x-p-15|=x-p+15-x+15+p-x=30-x$.
Since $0<x\leq15$, the minimum value occurs at $x=15$, and it will be $15$.

EDIT: Never mind. All of these are AIME problems.

people need to start using search, i think everyone should watch that DarkKnight's SPAM video. ok, even though this is not a SPAM topic, it can really reduce the amount of posts because the old posts are not here just to take cyber-dusts and be put into a corner, they are here to reference it and be rediscussed, if you like, if the previous discussions are not clear enough. below is a topic with many AIME problems discussed already more than a year ago

[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x-p+x-15&t=9297]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x-p+x-15&t=9297[/url]


A set of seven integers has a median of 73, a mode of 79, and a mean of 75. What is the least possible difference between the maximum and minimum values in the set?

The sum of the integers is $ 75 \times 7$, or $ 525$.  Obviously, $ 73$ is one of these numbers, so the sum left is $ 452$.

Let's make the digit right after $ 73$ as small as possible, so let's use $ 74$.  Since $ 79$ is the mode, we must have $ 2$ of those.  Subtract $ 74 \plus{} 2 \times 79$ from $ 452$, to receive $ 220$.

But then, you discover that this doesn't work, because $ 220 \div 3 > 73$...

Instead, we must use a number which leaves a sum of $ 213$ (so we can use $ 70$, $ 71$, and $ 72$).  Let's use $ 81$.  Then, your sum left will be $ 213$.

Thus, your set is $ {70,71,72,73,79,79,81}$.  Your answer must then be $ \boxed{11}$.

Nice solution, but there's a small mistake.  Notice that if we make 79 appear 3 times, the minimum value is increased, and the maximum value is decreased, reducing the difference.  Thus, the actual set is $ 71, 72, 72, 73, 79, 79, 79$ and the answer is $ \boxed{8}$

Hello, I don't understand the solution(s) to this problem. Anyone have anything?


AMC USA(J)MO USAMO trigonometry function ratio AoPSwiki

Prove that the average of the numbers $ n \sin n^{\circ} \; (n \equal{} 2,4,6,\ldots,180)$ is $ \cot 1^{\circ}$

[hide="my solution"]We have to prove that $ \frac{2sin2\plus{} 4sin4 \plus{} \dots \plus{} 180sin180}{90} \equal{} cot1$

Since $ sin(a) \equal{} sin(180\minus{}a)$, we can simplify the left side of the above equation by adding terms with degree values that add to 180:

$ \frac{180(sin2 \plus{} sin4 \plus{} sin6 \plus{} \dots \plus{} sin88)\plus{} 90sin90}{90}$

$ \frac{180(sin2 \plus{} sin4 \plus{} sin6 \plus{} \dots \plus{} sin88) \plus{} 90sin90}{90}$

We divide the numerator by the denominator, yielding

$ 2(sin2 \plus{} sin4 \plus{} sin6 \plus{} \dots \plus{} sin88) \plus{} sin90$

Now, we know that $ sin(a) \equal{} cos(90 \minus{} a)$, so we can change our expression to

$ 2(cos2\plus{} cos4 \plus{} cos6 \plus{} \dots \plus{} cos88) \plus{} cos0$

Expanding, we can rewrite this as

$ cos0 \plus{} cos2 \plus{} cos2 \plus{} cos4 \plus{} \dots \plus{} cos88 \plus{} cos90$

(We can add $ cos90$ because it is equal to zero).

We know from angle sum formulas that $ cosAcosB \equal{} \frac{cos(A \plus{} B) \plus{} cos(A \minus{} B)}{2}$.

So, grouping every two terms together, and setting $ B\equal{}1$, we simplify into

$ 2cos1(cos1 \plus{} cos3 \plus{} cos5 \plus{} \dots \plus{} cos89)$

We notice that $ cot1 \equal{} \frac{cos1}{sin1}$. It may be easier to simplify the expression into $ cot1$ if we have it in the expression, and make the rest of the expression cancel out. So, we rewrite our expression as

$ 2sin1cot1(cos1 \plus{} cos3 \plus{} cos 5 \plus{} \dots \plus{} cos89)$

$ cot1(2sin1cos1 \plus{} 2sin1cos3 \plus{} \dots \plus{} 2sin1cos89)$

$ cot1(sin2 \plus{} sin6 \plus{} sin10 \plus{} \dots \plus{} sin178)$

We can apply the same angle sum logic from above here:

$ cot1(sin2 \plus{} sin4 \minus{} sin 2 \plus{} sin6 \minus{} sin4 \plus{} \dots \plus{} sin90 \minus{} sin88)$

This is a telescoping sum that leaves us with

$ cot1(sin90)$

which is just

$ cot1$

QED [/hide]

[quote="Arvind_sn"]

$ cot1(2sin1cos1 \plus{} 2sin1cos3 \plus{} \dots \plus{} 2sin1cos89)$

$ cot1(sin2 \plus{} sin6 \plus{} sin10 \plus{} \dots \plus{} sin178)$

We can apply the same angle sum logic from above here:

$ cot1(sin2 \plus{} sin4 \minus{} sin 2 \plus{} sin6 \minus{} sin4 \plus{} \dots \plus{} sin90 \minus{} sin88)$

QED [/hide][/quote]

... OK... so we have the following $ \sin (A \plus{} B) \plus{} \sin(A \minus{} B) \equal{} 2\sin A\cos B$.

So.... $ 2\sin 1\cos 1 \equal{} \sin 2 \plus{} \sin 0$

$ 2\sin 1\cos 3 \equal{} \sin 4 \plus{} \sin ( \minus{} 2) \equal{} \sin 4 \minus{} \sin 2$
and in general, $ 2\sin 1\cos (2k \plus{} 1) \equal{} \sin (2k \plus{} 2) \minus{} \sin ( \minus{} 2k)$

This gives us the telescoping series... Not what you wrote $ cot1(sin2 \plus{} sin6 \plus{} sin10 \plus{} \dots \plus{} sin178)$... Actually i don't even know what this is because the next line is correct.

BTW use [code]$\sin x$ and $\cos x$[/code]

Dude calm down,

What the heck is a sin of n,

I only know up to logs.

Hmm... My instinct when I saw this problem was to evaluate the sum as $ \frac {1}{45}\;\text{Im}\left(\sum_{k \equal{} 1}^{90}ke^{2ik}\right)$.

It maybe more straightforward that way.

[b]@saszs:[/b] Don't most schools teach trig before logs? Hm. Maybe not. $ \sin$ is a function that maps an angle value $ \theta$ to the ratio of the opposite side to the hypotenuse of a right triangle with angles $ \theta$, $ 90^\circ\minus{}\theta$, and $ 90^\circ$. AoPSWiki has a lot on trigonometry, I believe. Anyway, if you don't know trig, you're probably looking at the wrong kinds of problems (i.e. USAMO).

I'll just ask for help you log base 2 to 8=3

OK so I just looked at this problem again... Another solution as follows (basic idea is similar)

Claim: $ 2\sin 2\plus{}4\sin 4\plus{}6\sin 6\plus{}\ldots \plus{}178\sin 178\equal{} 90 \cot 1$ (don't need $ \sin 180\equal{}0$)

This is the same as showing $ 2\sin 1\sin 2\plus{}2\cdot 2\sin 1\sin 4\plus{}3\cdot 2\sin 1\sin 6 \plus{}\ldots \plus{}89\cdot2\sin 1\sin 178$

Then simply use the formula $ 2\sin a\sin b\equal{} \cos (a\minus{}b)\minus{}\cos (a\plus{}b)$

So we have $ \cos 1\minus{}\cos 3\plus{}2\cos 3\minus{}2\cos 5 \plus{}\ldots\plus{}89 \cos 177\minus{}89\cos 179$

Which gives us $ \cos 1\plus{}\cos 3\plus{}\cos 5\plus{}\ldots \plus{}\cos 177 \minus{}89\cos 179\equal{}\cos 1\plus{}(\cos 3\plus{}\cos 177)\plus{}\ldots\minus{}89\cos 179\equal{} \cos 1\plus{} 89\cos 1\equal{}90\cos 1$ and so we are done.

Another way is using complex numbers...

[quote="grn_trtle"]Hmm... My instinct when I saw this problem was to evaluate the sum as $ \frac {1}{45}\;\text{Im}\left(\sum_{k \equal{} 1}^{90}ke^{2ik}\right)$.

It maybe more straightforward that way.

[b]@saszs:[/b] Don't most schools teach trig before logs? Hm. Maybe not. $ \sin$ is a function that maps an angle value $ \theta$ to the ratio of the opposite side to the hypotenuse of a right triangle with angles $ \theta$, $ 90^\circ \minus{} \theta$, and $ 90^\circ$. AoPSWiki has a lot on trigonometry, I believe. Anyway, if you don't know trig, you're probably looking at the wrong kinds of problems (i.e. USAMO).[/quote]

What do you mean by the lm? This is a sign that I haven't come across before; could you please explain? :)

Im is the imaginary part of a complex number; i.e. $ bi$ for $ a\plus{}bi$. Do you understand the trigonometric/exponential form of complex numbers and what they have to do with this problem?

$ \text{Im}(a\plus{}bi)\equal{}b$, not $ bi$. Careful!

[quote="resurrection"]

Another way is using complex numbers...[/quote]

[quote="grn_trtle"]Hmm... My instinct when I saw this problem was to evaluate the sum as $ \frac {1}{45}\;\text{Im}\left(\sum_{k \equal{} 1}^{90}ke^{2ik}\right)$.
[/quote]

I'm not very experienced in solving higher level problems with complex numbers. However, I do have a fairly well knowledge of them. Could you please explain this solution?

[quote="worthawholebean"]Next time, please post these in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=353049#353049]original thread[/url]. That helps us keep discussion for each problem in one place.[/quote]

Thank you, I'll be sure and do this in the future. ^__^

The Im function is the imaginary part - $ \text{Im}(a\plus{}bi)\equal{}b$. The general idea with these sort of solutions is that we can easily calculate sums of exponentials as geometric series and roots of various polynomials and then separate the sums into real and complex parts afterwards. Here, instead of finding the sum of $ 2k\sin\left(\frac{2k\pi}{180}\right)$, we find the sum of $ 2ke^{\frac{2k\pi i}{180}}$, which is easier.

Euler proved that $ \cos\theta\plus{}i\sin\theta\equal{}e^{i\theta}$. Therefore, $ \text{Im}(e^{i\theta})\equal{}\sin\theta$.

We must prove that $ \text{Im}(z\plus{}w)\equal{}\text{Im}(z)\plus{}\text{Im}(w)$. Letting $ z\equal{}a\plus{}bi$ and $ w\equal{}c\plus{}di$, we find that $ \text{Im}(z\plus{}w)\equal{}b\plus{}d$, and $ \text{Im}(z)\plus{}\text{Im}(w)\equal{}b\plus{}d$ as well, so we've proved this fact.

Then a sum like $ \sum_{k\equal{}0}^{n}\sin k\equal{}\sin 0\plus{}\sin 1\plus{}\cdots\plus{}\sin n$. But each $ \sin k\equal{}\text{Im}(e^{ik})$, so this sum is equivalent to $ \text{Im}(e^{0i})\plus{}\cdots\plus{}\text{Im}(e^{ni})\equal{}\text{Im}(e^{0i}\plus{}\cdots\plus{}e^{ni})\equal{}\text{Im}\left(\sum_{k\equal{}0}^{n}e^{ik}\right)$.

This is significantly easier to solve because it is a geometric series.


http://www.cbsnews.com/stories/2008/03/19/national/main3949353.shtml?source=RSSattr=HOME_3949353

"A Senate candidate has legally changed his name to Pro-Life and will appear on the ballot that way this year, state election officials say.

As Marvin Pro-Life Richardson, the organic strawberry farmer from Letha, 30 miles northwest of Boise, was denied the use of his middle name when he ran unsuccessfully for governor in 2006 because the state's policy bars the use of slogans on the ballot.

Now, though, officials in the Idaho secretary of state's office say they have no choice because Pro-Life is his full and only name. He says he will run for the highest state office on the ballot every two years for the rest of his life, advocating murder charges for doctors who perform abortions and for women who obtain the procedure. "


In triangle $ABC$ the medians $AM$ and $CN$ to sides $BC$ and $AB$, respectively, intersect in point $O$.  $P$ is the midpoint of side $AC$, and $MP$ intersects $CN$ in $Q$.  If the area of triangle $OMQ$ is $n$, then the area of triangle $ABC$ is:
\[ \text{(A)}\ 16n\ \ \ \ \text{(B)}\ 18n\ \ \ \ \text{(C)}\ 21n\ \ \ \ \text{(D)}\ 24n\ \ \ \ \text{(E)}\ 27n \]

[hide] To solve this problem, it is crucial to know the relationship between the ratio of sides of triangles and the areas. E.g. a) the area ratio of 2 similar triangles is the square the side ratio. b) the ratio of the area of two triangles with same height is the ratio of their bases. Knowing these properties, it is easy to solve this problem.

First of all, triangle OMQ is similar to triangle OAN (since PM//AB) and the ratio of the sides OM/OA=1/2 (properties of medians of triangles). therefore the ratio of the area is (1/2)^2=1/4. Thus, are of OAN=4n. Since OAN and OBN have the same height and their bases are equal, their areas are equal too. therefore OAB=2*4n=8n. [edited] Similarly, since OAB and CAB have the same base and heights in a 1/3 ratio, therefore their areas are in the same ratio as well. CAB=3*8n=24n.

Answer is D

P.S: I initially made the error of confusing OAB with AMB which gave the answer of A.[/hide]

[hide]
Since $\triangle{ABC}=3\cdot\triangle{AOB}=6\cdot\triangle{AON}=24\cdot\triangle{OMQ}.$
So the answer is D[/hide]


An airplane loses altitude at the rate of 400 miles/hour. What is the rate of decrease of the visible portion of the surface of the earth when the plane is one mile high? (Assume the earth is a sphere)

[quote="1=2"]Isn't this physics? :huh:[/quote]

No, but it is calculus and should be moved. :wink:


algorithm
number theory
relatively prime
greatest common divisor

Let a, b and c be integers whose greatest common divisor is 1. Show that there exist integers m and n such that a + mc and b + nc are relatively prime (i.e. have greatest common divisor 1.)

Should $ m$ and $ n$ be positive?

And... we could use Euclidean algorithm on this


How many pairs of real numbers $ \left(x,y\right)$ are there such that $ x^{4} \minus{} 2^{ \minus{} y^{2} } x^{2} \minus{} \left\| x^{2} \right\| \plus{} 1 \equal{} 0$, where $ \left\| a\right\|$ denotes the greatest integer not exceeding $ a$.

$\textbf{(A)}\ 0   \qquad\textbf{(B)}\ 1   \qquad\textbf{(C)}\ 2   \qquad\textbf{(D)}\ 4   \qquad\textbf{(E)}\ \text{Infinitely many}$

Rewrite the equation as

$ (x^2\minus{}1)^2\plus{}\{x^2\}\plus{}x^2\left(1\minus{}2^{\minus{}y^2}\right)\equal{}0$

Every term on the RHS is non-negative, hence we must have

$ x^2\equal{}1\land x\in\mathbb{Z}\land 1\minus{}2^{\minus{}y^2}\equal{}0$, from where we get $ (x,y)\equal{}(\pm 1,0)$, hence the answer is $ C$.


Given a rectangle ABCD ( anti-clockwise from bottom left), AD = BC = 1, DC = DG + GF + FC, where DG = tan 50, GF = tan 60 and FC = tan 70. Prove that angle ACB = 80 degrees.

Why are these questions different? http://www.artofproblemsolving.com/Forum/viewtopic.php?t=266566

(Topic locked for this reason)

They're not. I've just seen the sheet that the question came from and it's got a diagram which makes it nice.

In this question, you asked to show $ m\angle ACB \equal{} 80^\circ$, where $ \tan\angle ACB \equal{} \frac{\textrm{opposite}}{\textrm{adjacent}} \equal{} \tan 50^\circ \plus{} \tan 60^\circ \plus{} \tan 70^\circ$.
In the other thread, you asked to show $ \tan 50^\circ \plus{} \tan 60^\circ \plus{} \tan 70^\circ \equal{} \tan 80^\circ$.

Do you see how the second statement almost immediately implies the first? The diagram doesn't matter; these two questions are essentially the same.

I realise that; I just thought the diagram was neat.


Given a string of numbers $a_1, a_2, \cdots, a_n$ with minimum value of $a_x =1$ and maximum value $a_x = 10$, find the number of such string when either a $+$ or a $-$ is placed between any two terms, the end result of such a string is $10$.
A) For $n=5$
B) For $n=10$
The $+$ and $-$ can be placed in any order in a given string, all that matters is that the end result of the string is $10$. Does this make sense or is it too convoluted? An example of a satisfying string for part A is $4,2,1,5,2 \rightarrow +4+2+1+5-2 = 10$. Note, a $+$ or $-$ [b]does[/b] precede the first term.

[hide="Hint"]In other words, the question basically asks,
A) In how many ways can 5 numbers between -10 and 10 be chosen such that their average is 2?
B) In how many ways can 10 numbers between -10 and 10 be chosen such that their average is 1?[/hide][/hide]


How many different sets of three numbers, each number chosen from the integers 1-16, could be the side length of a right triangle?

So we have to try all numbers and you have learned them


Let $ a,b,c>0$ such that$ : a\plus{}b\plus{}c\equal{}1$.Prove that
$ : \sum\frac{a^2\plus{}b^2c}{b\plus{}c}\geq\frac{2}{3}$

[quote="DCTPKTCSPHN"]Let $ a,b,c > 0$ such that$ : a \plus{} b \plus{} c \equal{} 1$.Prove that
$ : \sum\frac {a^2 \plus{} b^2c}{b \plus{} c}\geq\frac {2}{3}$[/quote]

Setting:$ q\equal{}ab\plus{}bc\plus{}ca \le \frac{1}{3}$
We have:
$ \sum \frac{a^2}{b\plus{}c} \ge \frac{3(a^2\plus{}b^2\plus{}c^2}{2}\equal{}\frac{3(1\minus{}2q)}{2}$
$ \sum \frac{b^2c}{b\plus{}c} \ge \frac{(ab\plus{}bc\plus{}ca)^2}{a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca}\equal{}\frac{q^2}{1\minus{}q}$
We need prove that:
$ \frac{3(1\minus{}2q)}{2}\plus{}\frac{q^2}{1\minus{}q} \ge \frac{2}{3}$
$ <\equal{}>(3q\minus{}1)(8q\minus{}5) \ge 0$
Done! :P


What is the sum of the 100th number in each of the columns A, B and C in the   array shown?

                          Column A                                 Column B                                 Column C

Row 1                     1                                              200                                       1

Row 2                     10                                             195                                       4

Row 3                     19                                             190                                       9

Row 4                     28                                             185                                       16

Row 5                     37                                             180                                       25

   .                          .                                                .                                          .

   .                          .                                                .                                          .

   .                          .                                                 .                                         .

[hide]In Column A

the numbers are going up by 9 starting at 1so the 1--th term is 1+(99*9)
which is 892.

In Column B the hundreth term is 200+(99*-5)
which is -295.

In column c the hundreth term is $ 100^2$=10000

so the sum is 892-295+10000=10,597[/hide]