calculus
inequalities
three variable inequality

[quote="orl"]Let be given $a,b,c>0$. Prove that: $\sum \frac{a^3}{(a+b)^3} \geq \frac{3}{8}$.[/quote]
.$\sum \frac{a^3}{(a+b)^3}=\sum\frac{a^4}{a(a+b)^3}$. By Cauchy's inequlity .

[quote="xiaozg"].$\sum \frac {a^3}{(a+b)^3}=\sum\frac{a^4}{a*(a+b)^3}$. By Cauchy's inequlitity and A-G inequlitity.[/quote]
Ehhh... sorry but how can you prove that with cauchy ??  :?  :?

Well, Cauchy (in engel form) gives us

$\sum \frac {a^4}{a(a+b)^3}\geq \frac {(a^2+b^2+c^2)^2}{\sum a(a+b)^3}$

but I believe this is too weak.

it is not necesary bad...
in fact you can use better inequalities.. for example, use that $\sum{a^4}+\sum{a^2bc} \geq \sum{a^3b}+\sum{ab^3}$ and it will sufice to prove that $2\sum{a^4}+4\sum{a^2b^2} \geq 6\sum{a^3b}$ which doesn't look so obviously wrong...

 apart from that, you can not cancel terms when you have cyclic sums..

firstly, my cancelling was correct.

--

at the second step, we have only expanded and cancelled.  So lets go there.

working with:  $\displaystyle \sum 5a^4 + 7a^2b^2 \geq \sum 9a^3b + 3ab^3$...

how did you find a a^2bc term?

^

I got the same thing when I expanded as singular did, the sums $\sum a^4$ and $\sum a^2b^2$ are not really cyclic, its only the $\sum a^3b$ terms.

Does anyone has a better solution for this problem ?

can you read the other thread of this problem?
a solution is posted already.

[quote="erdos"]yes I know but I just asked for a better solution, not one which is using calculus[/quote]

I think the one I saw is not using calculus?
Power mean + an old ineq on the forum.

Yes, in ML circles, the inequality

$a,b,c,d\geq0$, $abcd=1$,

$\sum \frac {1}{(1+a)^2}\geq 1$

is about as famous as it gets

It's really true Singular
$\sum 5a^4+\sum 7a^2b^2\ge \sum 9a^3b+\sum 3ab^3$
We have $4a^4+b^4+7a^2b^2-9a^3b-3ab^3=(a-b)^2(4a^2-ab+b^2)\ge 0$
And the same with the 2 other.

My student suggested following proof:

$\sum a^3/(a+b)^3 * \sum c^3(a+b)^3 \ge (\sum ac^{3/2})^2$

Then, this enough to show that

$8(\sum ac^{3/2})^2 \ge 3(\sum c^3(a+b)^3) (1) $

Put $\sqrt(ab) = x, \sqrt(cb) = y, \sqrt(ac) = z$ (1) becomes

$8(\sum x^3)^2 \ge 3(\sum (x^2+y^2)^3) $

which leads to an trivial inequality.

But I think, the nicest proof it's the one that uses lemma

$ 1/(1+x)^2 + 1/(1+y)^2 \ge 1/(1+xy) $

utkarshgupta I posted a solution regarding this problem:

Prove that for all $a,b,c$ positive real numbers the following statement is true, $ \frac{a^{3}}{\left(a+b\right)^{3}}+\frac{b^{3}}{\left(b+c\right)^{3}}+\frac{c^{3}}{\left(c+a\right)^{3}}\geq\frac{3}{8} $. What is wrong?

nkalosidhs, you are trying to prove another inequality. :wink:

Oh...excuse me...I saw the first inequality on page 1 and then tried to post my solution without realizing that it is posted on third page! Well, I would like to just check whether my solution is correct!  :wink:

Read your post please.
[quote="nkalosidhs"][quote="orl"]Let be given positive reals $a$, $b$, $c$. Prove that: $\frac{a^{3}}{\left(a+b\right)^{3}}+\frac{b^{3}}{\left(b+c\right)^{3}}+\frac{c^{3}}{\left(c+a\right)^{3}}\geq \frac{3}{8}$.[/quote]

Let denote $A=\frac{a^3}{(b+c)^3}+\frac{b^3}{(a+c)^3}+\frac{c^3}{(a+b)^3}$
then we should prove that $A\geq\frac{3}{8}$
[/quote]
We don't need to prove that $A\geq\frac{3}{8}$. :wink:

Well, I don't think that the logic in my solution changes dramatically now...

I think I would agree but rechecking would be helpful !

[quote=nkalosidhs][quote="orl"]Let be given positive reals $a$, $b$, $c$. Prove that: $\frac{a^{3}}{\left(a+b\right)^{3}}+\frac{b^{3}}{\left(b+c\right)^{3}}+\frac{c^{3}}{\left(c+a\right)^{3}}\geq \frac{3}{8}$.[/quote]

Let denote $A=\frac{a^3}{(a+b)^3}+\frac{b^3}{(b+c)^3}+\frac{c^3}{(c+a)^3}$
then we should prove that $A\geq\frac{3}{8}$
We have by holder's inequality, $A[(a+b)+(b+c)+(c+a)][(a+b)^2+(b+c)^2+(c+a)^2]\geq(a+b+c)^3$
So, $2A(a+b+c)[(a+b)^2+(b+c)^2+(c+a)^2]\geq(a+b+c)^3$
which means that,
$2A[(a+b)^2+(b+c)^2+(c+a)^2]\geq(a+b+c)^2$
we multiply both sides with $3(a^2+b^2+c^2)$
So we have,

$2A3(a^2+b^2+c^2)[(a+b)^2+(b+c)^2+(c+a)^2]\geq3(a^2+b^2+c^2)(a+b+c)^2$      $(1)$
So due to the fact that, $3(a^2+b^2+c^2)\geq(a+b+c)^2$
which means that from $(1)$ we get that

$2A[(a+b)^2+(b+c)^2+(c+a)^2]\geq3(a^2+b^2+c^2)$

[/quote]

I think  this is wrong!

How can you deduce from $(1)$  to

$2A[(a+b)^2+(b+c)^2+(c+a)^2]\geq3(a^2+b^2+c^2)$?

Clearing the denominator, then we use Muirhead inequality(Bunching),

$a^5bc^3+a^3b^5c+ab^3c^5 \geq a^4b^3c^2+a^2b^4c^3+a^3b^2c^4$

Obviously this holds by AM-GM inequality. :)

Let $a+b+c=1$, by Rearrangement,

$\sum_{cyc}\frac{a^3}{(a+b)^3}\geq \sum_{cyc}\frac{a^3}{(b+c)^3}=\sum_{cyc}\frac{a^3}{(1-a)^3}$

We're going to using Tangent Line method.

Let $f(x)=\frac{x^3}{(1-x)^3}(0<x<1)$, then

$f(x)\geq \frac{27x-7}{16}$
$\iff$
$\left(x-\frac{1}{3}\right)^2\left(3(x-1)^2+4\right)\geq 0$

This holds when $0<x<1$.Thus

$\sum_{cyc}f(a)\geq \frac{27-21}{16}=\frac{3}{8}$ :D

[quote=mudok][quote=Takeya.O]
$\sum_{cyc}\frac{a^3}{(a+b)^3}\geq \sum_{cyc}\frac{a^3}{(b+c)^3}$
[/quote]

This step is wrong.[/quote]

I see, I missed :ewpu: Thanks for teaching me, Mr.mudok :D

Let $a,b,c$ be  positive reals . Prove that $$\frac{a^3}{(b+c)^3}+\frac{b^3}{(c+a)^3}+\frac{c^3}{(a+b)^3}+\frac{3abc}{8(a^3+b^3+c^3)}\ge\frac{1}{2}$$

Old?  Where?
[url=https://artofproblemsolving.com/community/c6h1198209p5879029]h[/url] [url=https://artofproblemsolving.com/community/c6h299899p2590936]h[/url]


Hello everyone,

Shortly I'll be posting all of the 2006 AIME I problems for the [url=http://www.artofproblemsolving.com/Forum/resources.php]Olympiad Resources/Contests[/url] section of AoPS-Mathlinks.  Some of you are probably familiar with the Resources section and know most of the details.  However, I would like to go over some things if you want to post a solution.

If you want to post a solution and you want it to go in the Resources section, here's what you have to do:

1.  Post a full, complete, and correct solution.  Please use proper english including proper grammar, spelling, and punctuation.
2.  Put [i]nothing[/i] but your solution in your post.  Use of hide tags is fine and strongly encouraged.  But putting extraneous comments such as "this problem was the most interesting on the whole test" will decrease your solution's chances of being selected.
3.  Use $\LaTeX$.

Hopefully that covers everything.

Thanks

To expand on Joe's second point, in this instance if you have a comment we recommend you double posting and putting anything that is not part of the solution in the second post. Also, please do not "quote" the question or make any markings to indicate that you're posting a solution. We'll (or rather, he'll) recognize them when he sees them.

do you want us to post solutions here or in another forum?

Each problem will have a thread dedicated solely to the discussion of that problem. You should post solutions in the appropriate threads.


Given that  2/3  of the committee use  3/4  of the chairs in the room, what is the least number of members on the committee?

(A)8 (B) 9 (C) 12 (D) 16 (E) 18

You can easily guess check this, starting with A. A obviously is not the right answer since it isn't divisible by 3. If B is right, then there are 6 committee members sitting in chairs, and thus 8 chairs. B works, so our answer is $ \boxed{(B)}$.


geometry
geometric transformation
rotation

i am not sure but,
 for both a and b answer is 240 hours.:?:
solution:
now the relative gain of one watch is 3 minutes(1/20 hours) over the other.
 the faster watch must gain 12 hours(complete rotation) to be at the same position as the slower one.
to gain 1/20 hours,it takes 1 hour,therefore to gain 12 hours the time taken will be (12)/(1/20)=240 hours.

in b,as they both are required to come back to 8,the same(above) solution applies. :D

[quote="Arvind_sn"]OK, my cousin has holiday homework and his school starts tomorrow!!! Fortunately he only needs the answer to this one---

You have two twelve hour watches set correctly at 8:00 AM. One gains two minutes per hour,the other loses one minute per hour.

a. When will both watches again show the same time?
b. When will they both show 8 o clock at the correct time?

For both, answer in hours. (ex. 10 hours later, 235873058273 hours later)

Thanks!!! :D[/quote]

Hmm.

a.
[hide]Hard to explain, but when one watch moves 8 hours forward, the other would move 4 hours backward. This would make each show 4:00.
So 4 hours at one minute per hour, that's 60*4 = 240 hours.[/hide]

b.
[hide]Every 240 hours, the watches will both be at a time 4 hours earlier (8 o clock, 4 o clock, 12 o clock, 8 o clock, etc)
So in 720 hours, they will meet at 8:00.[/hide]

oops!made a DUMB mistake :blush: .the answer is 720 hours like klebian has posted. :)


I've decided to take the AP Physics C exams next year, but I realize I do actually need to [b]learn [/b]the stuff.. I was just wondering if anyone knew where it would be best to take an online course? I've found several places, like EPGY and UCCP, but I'm not sure which is better.

Or is the course simply not worth it and should I try to teach myself the information from a textbook? I have already taken both the Calc BC and Physics B classesand exams (though the scores won't arrive until July... egh), so I know the basics, sort of, for the Physics C-track.

Any suggestions would be greatly appreciated!

Thanks!

At my high school Physics AP was not offered because there were no students who were advanced enough to take it. In fact their was only one physics class period, with about 25 students, it was at the honors level. I decided, like you, that I wanted to take the physics C exam anyway, and I opted to just get a textbook and study on my own (it was the only option in my case). I cannot tell you which online class is better, I can only tell you my personal experience. I read and learned the material on my own and studied for the exam on my own, using Physics: Calculus by Hetch. Looking back, I am not sure if this was the best book, but it's what I had. In any event, I had no problems learning the material or preparing for the exam. I went through the Princeton Review and Barron's and both helped (I think). Understanding the proofs is most important (IMO), if you do that, you'll be fine no matter what. If you are motivated, no matter which course you decided to take, you should succeed.

in my opinion, EPGY is definitely the way to go if money isn't an issue.

I plan on taking AP PHysics C w/ EPGY next year because my school doesn't offer it.

I plan on doing both Mech and EM Physics C from EPGY in the fall as well because my school doesn't offer AP Physics C, only B. (Our district's math placement has students taking Calculus senior year, and BC is invite only, so they figure to only offer B) which is evil, and IMHO, a sign of a lazy administration and a jacked up school.

(I hope those comments are allowed on this forum -- I didn't specify which school I go to)

at least they offered Physics B, and Calc BC... My school had neither.


A combination lock uses three integers in the combination,
and the dial is numbered with the integers 0, 1, 2 and 3. If
consecutive numbers in the combination cannot be the same,
how many possible combinations are there?

For the first number, we have $ 4$ choices. For the second, we have $ 3$ choices. And for the third, we have $ 3$ choices again. (Do you see why?)

$ 4 \times 3 \times 3 \implies \boxed{36}$


Let $ a,b,c$ be real numbers with $ a\plus{}b\plus{}c\equal{}1$. Show that
\[ \frac{a^2\plus{}b^2\plus{}c^2}{4}\plus{}\frac{ab\plus{}bc\plus{}ca}{5}\ \geq\ \frac{3}{20}
\]
(NB: $ a,b,c$ can be negative/zero as well as positive.)

Let $ A\equal{}a^2\plus{}b^2\plus{}c^2$ and $ B\equal{}ab\plus{}ac\plus{}bc$. We have:

$ (a\plus{}b\plus{}c)^2\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2ac\plus{}2bc\equal{}A\plus{}2B\equal{}1$

So, $ B\equal{}\frac{1\minus{}A}{2}$. Now:

$ \frac{a^2\plus{}b^2\plus{}c^2}{4}\plus{}\frac{ab\plus{}ac\plus{}bc}{5}\equal{}\frac{A}{5}\plus{}\frac{1\minus{}A}{10}\equal{}\frac{3A\plus{}2}{20}$
But we know that: $ 1\equal{}a\plus{}b\plus{}c\equal{}a\cdot 1\plus{}b\cdot 1\plus{}c\cdot 1\leq \sqrt{1^2\plus{}1^2\plus{}1^2}\cdot \sqrt{a^2\plus{}b^2\plus{}c^2}$. And: $ A\equal{}a^2\plus{}b^2\plus{}c^2\geq \frac{1}{3}$
Finally: $ \frac{3A\plus{}2}{20}\geq \frac{3\cdot\frac{1}{3}\plus{}2}{20}\equal{}\frac{3}{20}$  :surf:

Multiplying both sides by 20 and homogenizing, the inequality becomes
\[ 5a^2 \plus{} 5b^2 \plus{} 5c^2 \plus{} 4ab \plus{} 4ac \plus{} 4bc \ge 3 \equal{} 3(a \plus{} b \plus{} c)^2 \equal{} 3a^2 \plus{} 3b^2 \plus{} 3c^2 \plus{} 6ab \plus{} 6ac \plus{} 6bc,\]
which is equivalent to
\[ 2a^2 \plus{} 2b^2 \plus{} 2c^2 \ge 2ab \plus{} 2ac \plus{} 2bc,\]
which in turn is equivalent to
\[ (a \minus{} b)^2 \plus{} (a \minus{} c)^2 \plus{} (b \minus{} c)^2 \ge 0.\]


inequalities
search
function
vector
inequalities theorems

hello,everyone:
     I see the $ karamate's$ $ inequality$ first time the day before yesterday(if my memory works well) in $ myth's$,I see it also in the forum.but there no proof for  $ karamate's$ $ inequality$ :( --the great and useful inequality :) .can anybody post it for me??thank you in advance.

no, darij proved it before, just search for it.

thank you first.but you are wrong,I think :D
I see the post by darij.
but there is no proof.
I ask darij personly for the proof,
but he tell me no proof now :P

Just Have a look here : [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=14975[/url]

zhaobin:
what and where is '[b][color=red]myth's[/color][/b]'

[quote="ondrob"]Yes, I proved it another way (but not so elegant) in other thread (ineq. solved problems: convex function).[/quote]

I posted in the topic "convex function" a link to a taiwan site, which has an interesting article on majorization. It found a constraint for X > Y if X and Y have at most 2 different terms. It also showed that for any two sequences that X > Y, we can find X > Z1 > Z2 > ... >Zn > Y where (X,Z1), (Z1,Z2),...,(Zn,Y) differ by at most two terms. This shows that we only need to prove karamata for two variables, which is simple.

[quote="suweijie"]Now I have found a proof in a book,let me post it here.Maybe my English is poor. :blush:
Suppose [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/be86156d54f526b5b670467980553c24.gif[/img]
then for arbitrary convex function f the ineq(below) is true.
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/4994a74cc03f1a13857dd38211c94364.gif[/img]
defined [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/ea70490be4bc30dde73e57c84467bde3.gif[/img]
since f is convex,hence [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f4bbc08bcddb4e0b8282adc7f5b41517.gif[/img]

[/quote]

why is it true?

[quote="ondrob"]Yes, I know, but I am unable to read it ;) My solution in other thread can be remade to showing that you can make one vector to another by finite number of compressions (I saw Darij calling it compressions), it's the same what you are mentioning Siuhochung. With these so-called compressions you can also prove Muirhead inequality easily ;)[/quote]

i'm eager to see the proof of the much used muirhead

It's true because of the convexity of functions. It is somehow a property of convex functions. I remember blahblahblah posted a proof somewhere but i cannot find it.

For muirhead, Peter Scholze and darij proved it.
http://www.mathlinks.ro/Forum/viewtopic.php?p=140365#140365
A generalized version.

[quote="siuhochung"]It's true because of the convexity of functions. It is somehow a property of convex functions. I remember blahblahblah posted a proof somewhere but i cannot find it.[/quote]

i found in kedlaya's notes that we must have $a \geq b \geq c \geq d$ for such an inequality to hold. here i think this is not always the case

[quote="Rafal"]I wonder why anyone has mentioned: [url=http://www.mathlinks.ro/Forum/topic-41966.html]www.mathlinks.ro/Forum/topic-41966.html[/url] with an inductive proof....as far as I know there are few versions of these majorization inequalities.[/quote]
This version is weaker than the known version on the forum (the one mahbub posted).

[quote="suweijie"]Now I have found a proof in a book,let me post it here.Maybe my English is poor. :blush:
Suppose [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/be86156d54f526b5b670467980553c24.gif[/img]
then for arbitrary convex function f the ineq(below) is true.
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/4994a74cc03f1a13857dd38211c94364.gif[/img]
defined [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/ea70490be4bc30dde73e57c84467bde3.gif[/img]
since f is convex,hence [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f4bbc08bcddb4e0b8282adc7f5b41517.gif[/img]
then we have [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/1e846364bd39b5598116910af06f77f5.gif[/img]
where [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/2863342235680af68717d091cf076f24.gif[/img]
hence [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/3432a6589d995c7ca4933f484d51c0ce.gif[/img]
so we obtain [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/6f4fccd694bfd367db11a59cd31fb263.gif[/img]
since [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/c20f21e98dd8c86966e9208f21c4fe13.gif[/img]
so [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/163a50a2be565707cb62c3fb273e6547.gif[/img]
now we have proved it.Cheers! :D[/quote]

BAH...

Do you have the proof stored on your computer?
Please don't link to LaTeXrender graphics, they are not kept online for long!

  Darij

A very straight forward and elegant proof can be found here:

[url]http://www.imocompendium.com/tekstkut/ineq_im.pdf[/url]

A lot of goodies in this article and on the website too.


algebra
polynomial
induction
modular arithmetic
algebra proposed

Let $n,\ k$ be positive integers and let $P(x)$ be polynomial with degree $\geq n.$ If all the coefficients of the term with degree $\leq n$ of the polynomial $(1+x)^{k}P(x)$ are integers,  then prove that all the coefficients of the term with degree $\leq n$ of $P(x)$ are integers.Note that you may regard constant term itself as coefficient.

The case $k=1$ is easy, and the rest follows trivially by induction on $k$.

EDIT another solution would be
$(1+x)^{k}P(x)\equiv Q(x)\pmod{x^{n+1}}$ for some $Q\in\mathbb{Z}[x]$.
Then
$\begin{array}{rcll}P(x)&\equiv & (1-x^{2n+2})^{k}P(x)&\pmod{x^{n+1}}\\ &=&(1+x^{2}+\cdots+x^{2n})^{k}(1-x)^{k}(1+x)^{k}P(x)&\\ &\equiv & (1+x^{2}+\cdots+x^{2n})^{k}(1-x)^{k}Q(x) & \pmod{x^{n+1}}\end{array}$
with $(1+x^{2}+\cdots+x^{2n})^{k}(1-x)^{k}Q(x)\in\mathbb{Z}[x]$.


inequalities
trigonometry
function
AMC
USA(J)MO
USAMO
inequalities unsolved

Prove that in triangle $ABC$ we have the following inequality:

$1+\frac{r}{R}\leq \sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\leq \sqrt{2+\frac{r}{2R}}$

i'll prove just the LHS inequality...

recall that
$1+\frac{r}{R}=\cos A+\cos B+\cos C=\sin (90-A)+\sin (90-B)+sin(90-C)$

then, it's easy to verify that $[90-A,90-B,90-C]$ majorizes $[A/2, B/2, C/2]$, so the result follows from the karamata inequality applied to the function $\sin x$.  :D

minutes later i came with some ideas for the RHS inequality

again, let's use the identity

$1+\frac{r}{R}=\cos A+\cos B+\cos C=2(\cos^{2}A/2+\cos^{2}B/2+\cos^{2}C/2)-3$

then,

$\sqrt{2+\frac{r}{2R}}=\sqrt{\cos^{2}A/2+\cos^{2}B/2+\cos^{2}C/2}$.

we could try squaring it, and then the inequality is equivalent to

$(\sin A/2+\sin B/2)^{2}+(\sin B/2+\sin C/2)^{2}+(\sin C/2+\sin A/2)^{2}\leq 3$

this looks similar to the inequality $\sum_{cyc}(\cos A+\cos B)^{2}\leq 3$, which turned to be very famous in the forum some time ago and which can be proven to be equivalent with the usamo 2001 inequality...

hope somebody can continue this  :lol:

just to finish this out... i didn't wonder how close I was  :P

according to [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=14567[/url] we have that

$\sum_{cyc}(\cos A+\cos B)^{2}\leq 3$, for $A,B,C$ acute angles.

then, make the $A,B,C=90-X/2, 90-Y/2, 90-Z/2$, with $X,Y,Z$ angles of a triangle, and then

$\sum_{cyc}(\sin X/2+\sin Y/2)\leq 3$, and we're done.


quadratics
modular arithmetic
number theory solved
number theory

can anyone show me how to prove the four identities in the end of this page: http://mathworld.wolfram.com/LegendreSymbol.html

I guess(3) follows from the following:

(a/p)=a^((p-1)/2) (mod p)

[b]Solution:[/b]

if p|a then a^((p-1)/2)=0=(a/p) (mod p)

Lets consider that not ( p|a) then because of the fact that:

(a^((p-1)/2)-1)(a^((p-1)/2)+1)=a^(p-1)-1=0(mod p) we get that
a^((p-1)/2)= \pm1 (mod p).
if a is a quadratic residue mod p then we have that:
a^((p-1)/2)=1=(a\p)(mod p)
if it's not a quadratic residue mod p then
a^((p-1)/2)<>1(mod p) so
a^((p-1)/2)=-1=(a/p) (mod p)

cheers! :D :D

(3) comes if u take in the above a=-1;

(2) (ab/p)=(a/p)(b/p)
comes from the relation (a/p)=a^((p-1)/2) (mod p)

we have that (a/p)(b/p)=a^((p-1)/2)b^((p-1)/2)=(ab)^((p-1)/2)=
(ab/p)(mod p) so
(a/p)(b/p)=(ab/p) (mod p)
but because the only values of (a/p);(b/p);(ab/p) can only be 0 or  \pm1 and p>=3 then we have the relation above! :D

cheers!

(4)

(2/p)=(-1)^((p^2-1)/8)

Let e be a root of order 8. so that e^8=1 and e^4<>1
so we can take e=cos(2pi/8)+isin(2pi/8)=sqrt(2)/2(1+i)
we have that e^4=-1 and e^2+e^(-2)=0.
lets denote t=e+e^(-1) then t^2=2.
we can see that e,t are algebric integers and we shall use congruence in the algebric integer rings omega.
we have that: t^(p-1)=(t^2)^((p-1)/2)=2^((p-1)/2)=(2/p)(mod p)
so: t^p=(2/p)*t(mod p)
and also:
t^p=e^p+e^(-p)(modp)
because e^8=1 we have for p= \pm1(mod 8); e^p+e^(-p)=e+e^(-1)
and for p= \pm3(mod p) we have e^p+e^(-p)=-(e+e^(-1)).
so
e^p+e^(-p)=
t if p= \pm1 (mod 8 )
-t if p= \pm3 (mod 8 )
which means e^p+e^(-p)=(-1)^(eps(p))*t
where:
eps(p)=
1 if p= \pm1 (mod 8 )
-1 if p= \pm3 (mod 8 )

so we get that
(2/p)*t=t^p=e^p+e^(-p)=(-1)^(eps(p))*t(mod p)
now in the last relation we lumtiply it with t and we know that t^2=2 we get that:
(2/p)*2=(-1)^(eps(p))*2(mod p)
so (2/p)=(-1)^(eps(p))
now it's easy to see that eps(p)=(p^2-1)/8 (mod 2)
so we got the indetity!!!

cheers! :D :D

Nice proof for (4), Lagrangia. I have sen this before in a book somewhere. You can also find a proof here:

[url]http://www.mathlinks.ro/phpBB/viewtopic.php?t=2154[/url]

It's somewhat messier than what Lagrangia posted.


analytic geometry
function
real analysis
real analysis unsolved

f:[0,1]->R^n is a continuous path of finite length. (by the way, what is a path? Each coordinate is a function of t, t in [0,1]?). Show that if n>1 then the n-dimentional Leb measure of the path is zero. (what about n=1, still measure zero?) Thanks.

Using your definition of path and assuming $n>1$:

For each $\epsilon>0$ you can find a partition of $[0,1]$ by points $t_{i}$ such that the length of the arc between $f(t_{i-1})$ and $f(t_{i-1})$ is less than $\epsilon$ for all $i$. The balls of radius $|f(t_{i})-f(t_{i-1})|$ centered at $f(t_{i})$ cover the path. Since $\sum_{i}|f(t_{i})-f(t_{i-1})|\le L$ ($L$ is the length), we have $\sum_{i}|f(t_{i})-f(t_{i-1})|^{n}\le L\epsilon^{n-1}$, which is good.


The United States Mint produced 13.5 billion pennies in 1995. If the value of these pennies is exchanged for the same amount in dimes, how many dimes would be needed?

There are 13.5 billion pennies.

10 pennies equals 1 dime.

So divide 13.5 by 10 to get the answer.

The answer is 1.35 billion dimes.


inequalities
function
inequalities unsolved

[b]This is not true![/b]

Let $a=b$ and $c$ goes to 0, then both $a,b$ go to 1. So LHS goes to 2 and RHS goes to 3.

Maverick, I don't think your inequality is true. For if it would be true, then we could "de-homogenize" it and obtain the inequality

[tex]\frac{a^{2}}{b^{2}+bc+c^{2}}+\frac{b^{2}}{c^{2}+ca+a^{2}}+\frac{c^{2}}{a^{2}+ab+b^{2}}\geq \frac{3\left( a^{3}+b^{3}+c^{3}\right) /2}{\left( a^{3}+b^{3}+c^{3}\right) /2+3abc}[/tex]

for all positive numbers a, b, c.

But taking a = 1, b = 1 and $c=\sqrt{2}-1$, we get

[tex]\frac{a^{2}}{b^{2}+bc+c^{2}}+\frac{b^{2}}{c^{2}+ca+a^{2}}+\frac{c^{2}}{a^{2}+ab+b^{2}}-\frac{3\left( a^{3}+b^{3}+c^{3}\right) /2}{\left( a^{3}+b^{3}+c^{3}\right) /2+3abc}\approx -4.\,5\cdot 10^{-2}<0[/tex].

On the other hand, the inequality with reversed sign is not true either, since a = 1, b = 1 and c = 2 gives

[tex]\frac{a^{2}}{b^{2}+bc+c^{2}}+\frac{b^{2}}{c^{2}+ca+a^{2}}+\frac{c^{2}}{a^{2}+ab+b^{2}}-\frac{3\left( a^{3}+b^{3}+c^{3}\right) /2}{\left( a^{3}+b^{3}+c^{3}\right) /2+3abc}\approx 0.255>0[/tex].

What, I think, is valid would be the inequality

[tex]\frac{a^{2}}{b^{2}+bc+c^{2}}+\frac{b^{2}}{c^{2}+ca+a^{2}}+\frac{c^{2}}{a^{2}+ab+b^{2}}\geq \frac{\left( a^{3}+b^{3}+c^{3}\right) /2}{\left( a^{3}+b^{3}+c^{3}\right) /2+3abc}+\frac{2}{3}[/tex].

But I have no idea how to prove it.

  Darij

[quote="darij grinberg"]What, I think, is valid would be the inequality

[tex]\frac{a^{2}}{b^{2}+bc+c^{2}}+\frac{b^{2}}{c^{2}+ca+a^{2}}+\frac{c^{2}}{a^{2}+ab+b^{2}}\geq \frac{\left( a^{3}+b^{3}+c^{3}\right) /2}{\left( a^{3}+b^{3}+c^{3}\right) /2+3abc}+\frac{2}{3}[/tex].

But I have no idea how to prove it.

  Darij[/quote]

Post it in the [b]Open Questions[/b] section, see whether somebody can prove or disprove it!

[quote="fuzzylogic"]Post it in the [b]Open Questions[/b] section, see whether somebody can prove or disprove it![/quote]

Well, I have not spent any time with trying to prove it - I have not much time now -, so it may be very easy and I don't want to spam the Open Questions section with such problems...

  Darij

Darij, your examples are wrong: $(1,1,\sqrt{2}-1)\,\ (1,1,2)$  do not satisfy $a^3+b^3+c^3=2$ and also fuzzylogic $c$ cannot be 0.
I'm not sure if the problem is wrong or not, but here is the way I thought it:

$\sum{\frac{a^2}{b^2+bc+c^2}}= \sum{\frac{a^3}{ab^2+abc+ac^2}}\\
\frac{a^3}{ab^2+abc+ac^2}+\frac{b^3}{bc^2+bca+ba^2}+\frac{c^3}{ca^2+cab+cb^2}\geq \frac{a^3+b^3+c^3}{3} \cdot \sum{\frac{1}{ab^2+abc+ac^2}}\\
\frac{2}{3} \cdot \sum{\frac{1}{ab^2+abc+ac^2}}\geq \frac{6}{\sum{ab(a+b)}+3abc}\geq \frac{6}{a^3+b^3+c^3+6abc}=\\
\frac{6}{2+6abc}=\frac{3}{1+3abc}$

Fuzzylogic is right.
First inequality written by Maverick in his last message is wrong, since
for [i]a[/i] goes to 0 obtain $\displaystyle finite\geq \infty$.

[quote="Maverick"]Darij, your examples are wrong: $(1,1,\sqrt{2}-1)\,\ (1,1,2)$  do not satisfy $a^3+b^3+c^3=2$[/quote]

Maverick, you probably skipped the first line of my message. I wrote that I "de-homogenized the inequality". If you really want to have a counterexample with $a^3+b^3+c^3=2$, then take $\displaystyle a=\root{3}\of{\frac{2}{1^{3}+1^{3}+\left( \sqrt{2}-1\right) ^{3}}}\cdot 1$, $\displaystyle b=\root{3}\of{\frac{2}{1^{3}+1^{3}+\left( \sqrt{2}-1\right) ^{3}}}\cdot 1$ and $\displaystyle c=\root{3}\of{\frac{2}{1^{3}+1^{3}+\left( \sqrt{2}-1\right) ^{3}}}\cdot \left( \sqrt{2}-1\right) $, then you have $a^{3}+b^{3}+c^{3}=2$ but still

[tex]\frac{a^{2}}{b^{2}+bc+c^{2}}+\frac{b^{2}}{c^{2}+ca+a^{2}}+\frac{c^{2}}{a^{2}+ab+b^{2}}-\frac{3}{1+3abc}\approx -4.5\cdot 10^{-2}<0[/tex].

[quote="Maverick"]and also fuzzylogic $c$ cannot be 0.[/quote]

But it can get arbitrarily close to 0 and the function is continuous, so...

[quote="Maverick"]$\frac{a^3}{ab^2+abc+ac^2}+\frac{b^3}{bc^2+bca+ba^2}+\frac{c^3}{ca^2+cab+cb^2}\geq \frac{a^3+b^3+c^3}{3} \cdot \sum{\frac{1}{ab^2+abc+ac^2}}$[/quote]

If you applied Chebyshev here, then it's wrong: The number arrays $\left( a^{3},\;b^{3},\;c^{3}\right) $ and $\displaystyle \left( \frac{1}{ab^{2}+abc+ac^{2}},\;\frac{1}{bc^{2}+bca+ba^{2}},\;\frac{1}{ca^{2}+cab+cb^{2}}\right) $ are (in general) not equally sorted.

  Darij


integration
calculus
derivative
probability
algebra
polynomial
conditional probability

a funny question given to a friend during an interview for a job:

A man picks a number $ x$ (uniformly) in $ [0;1]$, and writes it down on a pice of paper. On another piece of paper, he writes $ x^2$. Then you enter the room and choose one of the two pieces of paper. You have the choice between:
1/ keep this piece of paper, and the man will give you $ y$ dollars ($ y$ is what is written on the piece of paper you chose).
2/ decide to take the other piece of paper, and the man will give you what is written on the other paper.

What is your strategy ?

[hide="Something I do not believe any more."]I get the following optimal strategy: examine the number on the first slip. If it is greater than $ \frac14,$ keep it. If it is less than $ \frac14,$ switch to the other slip.[/hide]

My method is probably unnecessarily complicated; there's probably an easier argument.

Let $ X$ be the original number that is uniform on $ [0,1].$ Let $ Y$ be the number written on the slip you choose first. Let $ L$ be the event that $ Y \equal{} X.$

$ P(Y\in [y,y \plus{} \Delta y]\text{ and }L) \equal{} P(X\in [y,y \plus{} \Delta y]\text{ and }L) \equal{} \frac {\Delta y}2.$

$ P(Y\in [y,y \plus{} \Delta y]\text{ and }L^C) \equal{} P(X\in \left[\sqrt {y},\sqrt {y \plus{} \Delta y}\right]\text{ and }L^C)\approx \frac {\Delta y}{4\sqrt {y}}.$

From that I get that

$ P(L\mid Y \equal{} y) \equal{} \frac {\frac12}{\frac12 \plus{} \frac1{4\sqrt {y}}} \equal{} \frac {2\sqrt {y}}{1 \plus{} 2\sqrt {y}}.$

Now, let us explore a strategy of switching for $ Y < b$ and keeping for $ Y > b$ for some $ b$ to be determined. If you switch, you get $ Y^2$ if $ L$ and $ Y$ if not $ L.$ If you keep, the payoffs reverse. I get that the expected payoff for this strategy is

[hide="An argument not to trust."]$ g(b) \equal{} \int_0^b\frac {y^22\sqrt {y}}{1 \plus{} 2\sqrt {y}} \plus{} \frac {y}{1 \plus{} 2\sqrt {y}}\,dy \plus{} \int_b^1\frac {y^2\sqrt {y}}{1 \plus{} 2\sqrt {y}} \plus{} \frac {y^2}{1 \plus{} 2\sqrt {y}}\,dy$

We can differentiate this by the fundamental theorem of calculus. The derivative simplifies to

$ g'(b) \equal{} \frac {(b \minus{} b^2)(1 \minus{} 2\sqrt {b})}{1 \plus{} 2\sqrt {b}}.$

From this we see that we have a maximum at $ b \equal{} \frac14,$ hence my answer.[/hide]

I just hid some parts of the last post because I realized that they depended on the assumption that $ Y$ is uniformly distributed - and it's not.

The probability of $ L,$ which is $ \frac12,$ should be the expectation of the conditional probability of $ L.$ But

$ \int_0^1\frac{2\sqrt{y}}{1\plus{}2\sqrt{y}}\,dy\equal{}\frac12\ln3,$ which is not $ \frac12.$

Some parts of the work I did might have uses, but otherwise, it's time to try again.

There's a borderline number $ b$ which is approximately $ 0.31944846.$

I get the following optimal strategy: examine the number on the first slip. If it is greater than $ b,$ keep it. If it is less than $ b,$ switch to the other slip.

Let $ X$ be the original number that is uniform on $ [0,1].$ Let $ Y$ be the number written on the slip you choose first. Let $ L$ be the event that $ Y = X.$

$ P(Y\in [y,y + \Delta y]\text{ and }L) = P(X\in [y,y + \Delta y]\text{ and }L) = \frac {\Delta y}2.$

$ P(Y\in [y,y + \Delta y]\text{ and }L^C) = P(X\in \left[\sqrt {y},\sqrt {y + \Delta y}\right]\text{ and }L^C)\approx \frac {\Delta y}{4\sqrt {y}}.$

From that I get that

$ P(L\mid Y = y) = \frac {\frac12}{\frac12 + \frac1{4\sqrt {y}}} = \frac {2\sqrt {y}}{1 + 2\sqrt {y}}.$

Note also that the probability density of $ y$ is simply $ \frac12 + \frac1{4\sqrt {y}}.$ As a consequence, we get that

$ P(L\mid Y = y)\cdot f_Y(y) = \frac12$ and $ P(L^C\mid Y = y)\cdot f_Y(y) = \frac1{4\sqrt {y}}.$

Now, let us explore a strategy of switching for $ Y < b$ and keeping for $ Y > b$ for some $ b$ to be determined. If you switch, you get $ Y^2$ if $ L$ and $ \sqrt{Y}$} if not $ L.$ If you keep, you get $ Y.$ I get that the expected payoff for this strategy is

$ g(b) = \int_0^by^2\cdot \frac12 + \sqrt{y}\cdot\frac1{4\sqrt {y}}\,dy + \int_b^1y\left(\frac12 + \frac1{4\sqrt {y}}\right)\,dy$

We can differentiate this by the fundamental theorem of calculus. The derivative becomes

$ g'(b) = \frac{b^2}2+\frac14-\frac{b}2-\frac{\sqrt{b}}{4}.$

Setting this equal to zero results in a quartic polynomial equation for $ \sqrt{b}.$ $ b=1$ is one root; factoring this out gives a cubic equation. The number given at the top is a numerical approximation to the root of this equation.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147596]Two[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32063]threads[/url] on a related "paradox".

another way to write the same proof as Kent Merryfield:

Let $ X$ and $ \epsilon$ be two independant random variables and suppose that $ X$ is uniformly distributed in $ [0;1]$, and that $ \epsilon$ is uniformly distributed in $ \{0,1\}$. One can modelize the number written on the slip you choose first by $ Y_1 \equal{} \epsilon X \plus{} (1\minus{}\epsilon) X^2$, and the second number by $ Y_2 \equal{} (1\minus{}\epsilon) X \plus{} \epsilon X^2$. Then the good strategy is to choose the second number when $ E[Y2 | Y1] \geq Y1$. But standart calculations show that $ E[Y2 | Y1] \equal{} \frac{Y_1^2 \plus{} \frac{1}{2}}{1 \plus{} \frac{1}{2 \sqrt{Y1}}}$. Hence one has to find the interval where $ \frac{x^2 \plus{} \frac{1}{2}}{1 \plus{} \frac{1}{2 \sqrt{x}}} \geq x$.


function
integration
limit
real analysis
calculus
derivative
search

Let $ \phi (x) , x\in\mathbb{R}^n$ be a bounded measurable function, such that $ \phi (x) \equal{} 0$ for $ \parallel{}x\parallel{}\geq 1$, and $ \int \phi \equal{} 1$. For $ \epsilon > 0$ let $ \phi _{\epsilon}(x): \equal{} \epsilon ^{ \minus{} n}\phi (\frac {x}{\epsilon})$ (approximation to the identity) .
If $ f\in L(\mathbb{R}^n)$ show that $ \lim_{\epsilon \rightarrow 0}(f\ast \phi_{\epsilon})(x) \equal{} f(x)$  in the Lebesgue set of $ f$.

Here's a start. Note that $ \int\phi_{\epsilon} \equal{} 1$ and that $ |\phi_{\epsilon}(x)|\le\epsilon^{ \minus{} n}M.$

$ (f*\phi_{\epsilon})(x) \minus{} f(x) \equal{} \int_{\mathbb{R}^n}\phi_{\epsilon}(x \minus{} y)f(y)\,dy \minus{} \int_{\mathbb{R}^n}\phi_{\epsilon}(x \minus{} y)f(x)\,dy$

$ |(f*\phi_{\epsilon})(x) \minus{} f(x)|\le\int_{B(x;\epsilon)} |\phi_{\epsilon}(x \minus{} y)|\,|f(y) \minus{} f(x)|\,dy$

$ \le \epsilon^{ \minus{} n}M\int_{B(x;\epsilon)}|f(y) \minus{} f(x)|\,dy$

Can you finish it from there?

Next up: weakening the hypotheses on $ \phi$ from bounded and supported on a ball to majorized by a decreasing radial integrable function.

Thanks, I have two question: why is this function ($ \phi_{\epsilon}$) called "approximation to the identity" (I just realized I wrote it wrong the last post  :blush: )? My opinion is,the problem is put there out of context ( it's an intro to harmonic analysis book).
Next one, what do you mean by [b]radial[/b]? I'm lost on terminology.

"radial" means "radially symmetric- the value depends only on the distance from the origin.
As for why this is an "approximation to the identity"- functions form a ring, with convolution as the multiplication. This sequence converges weakly to the multiplicative identity in that ring, although we need to extend to distributions for that identity ($ \delta$) to actually exist.

For a textbook or course that studies Euclidean harmonic analysis (that is, harmonic analysis on $ \mathbb{R},\mathbb{T},\mathbb{Z},\mathbb{Z}/N\mathbb{Z}$ and cartesian products of those) and which assumes a knowledge of the Lebesgue integral, this exercise is absolutely and totally [b]in[/b] context. This result and its relatives are things that any student of the subject should be instantly and intensely familiar with.

This problem is not out of Harmonic Analysis. It's wellknown that there's connection between results on maximal functions and results on differentiability and pointwise convergence, say [url=http://planetmath.org/encyclopedia/LebesgueDifferentiationTheorem.html]Lebesgue's Differentiation theorem[/url]. As you see the first step in hints which  Kent Merryfield showed, you could finish it by using this theorem. The pattern of the above comments were reproduced to obtain other differentiability results, from estimates on maximal functions. This motivates the search for extensions of the orginal [url=http://planetmath.org/encyclopedia/HardyLittlewoodMaximalTheorem.html]maximal theorem of Hardy Littlewood[/url], namely $ f\to \sup_{\epsilon}|\phi_{\epsilon}*f(x)|$ is bounded in $ L^p$ for $ 1 < p < \infty$.

Let $ \omega\equal{}e^{2\pi i/N}$ - a primitive $ N$th root of unity. Suppose $ f$ is a complex-valued function defined on $ \mathbb{Z}_N$ (another notation for $ \mathbb{Z}/N\mathbb{Z}$). Then it has a Fourier transform $ \widehat{f},$ also a function on $ \mathbb{Z}/N,$ defined by
\[ \widehat{f}(k)\equal{}\frac1N\sum_{n\equal{}0}^{N\minus{}1}\omega^{\minus{}kn}f(n)\]
from which we can recover $ f$ by
\[ f(n)\equal{}\sum_{k\equal{}0}^{N\minus{}1}\omega^{kn}\widehat{f}(k).\]
(There are other possible conventions for handing the factor of $ N.$)

Alternately, this can be seen as a matrix operation. In all of our vectors and matrices, let the indices run from $ 0$ to $ N\minus{}1.$ Consider the $ N\times N$ matrix $ F$ whose $ (k,n)$th entry is $ \omega^{kn}.$ Then $ \widehat{f}\equal{}\frac1NF^*f,$ and $ f\equal{}F\widehat{f}.$ Here, by $ F^*,$ we mean the Hermitian adjoint, or conjugate transpose. $ F$ is within a constant of being a unitary matrix: $ F^*F\equal{}NI.$

We can now build all of the usual structures of harmonic analysis: convolution (with the idea that convolution on one side is pointwise multiplication on the other), norms and norm inequalities, including a 2-norm equality (a version of Plancherel's Theorem), diagonalization of translation-invariant linear operators, and so on. We don't need much in the way of heavy theorems of analysis, since for the most part we are working with finite sums.

The four domains $ \mathbb{R},\mathbb{T},\mathbb{Z},$ and $ \mathbb{Z}_N$ can be related to each other. Start with a function $ f$ on $ \mathbb{R}.$ We can create a function on $ \mathbb{Z}$ by periodically sampling it: $ \varphi(n)\equal{}f(nh)$ for some $ h>0.$ Or we can create a function on $ \mathbb{T}$ by summing it in a particular way: $ g(x)\equal{}\sum_{m\equal{}\minus{}\infty}^{\infty}f(x\plus{}mP),$ where $ P$ is the period. Sample and sum both (in either order) and you get a function on $ \mathbb{Z}_N.$ These sampling and summing operations have predictable and useful consequences for the Fourier transforms - this is called the Poisson summation formula, along with a number of related theorems.

The operation that takes $ f$ to $ \widehat{f}$ over $ \mathbb{Z}_N$ is called the [i]discrete Fourier transform[/i] (DFT). By the raw definition of matrix multiplication, it requires $ O(N^2)$ arithmetic operations to execute. However, if $ N$ is a highly composite integer (nothing but small factors), this can be speeded up to $ O(N\log N)$ arithmetic operations. Such an algorithm is a [i]fast Fourier transform[/i] (FFT).

[quote="pluricomplex"]Kent Merryfield, would you please tell me something about Harmonic Analysis on $ {\mathbb Z}/\mathbb NZ$ ?[/quote]

I refer you to the book entitled "An Introduction to Harmonic Analysis (Cambridge Mathematical Library) " by Katznelson.

I don't recall that much about the finite case in Katznelson. But, returning to Mr. Doe's original question, Katznelson would be an excellent textbook or resource for the course of study that his question suggests. For something with a little more emphasis on the $ \mathbb{R}^n$ case, also consider the Stein and Weiss book (from the Princeton series).

I don't see much speacial DFT appear in book of Katznelson but FT on $ \mathbb T,R$ or on LCA.
Thanks Kent for excellent post.  :)


I have written my USAMTS solutions in pencil.  Is pencil fine, or do we have to use pen?

Pencil is fine, but I'd strongly suggest mailing it instead of faxing it (pencil doesn't fax too well).  And I'd suggest photocopying it and keeping a copy for yourself.


Hello! :)
Let $\mu: 2^N\to [0,+\infty]$ and $\mu(E)=card E$. Let $f: N\to R$. Proof that $f$ is $\mu$ - integrable if and only if series $\sum_{n=1}^{+\infty}f(n)$ is absolutely summable. Thanks.

isn't it straightfoward ? what's the definition of $\mu$-integrable ..


What's a good book for learning amc/aime algebra?

Any AOPS book. This question is in the wrong forum by the way.


Which natural numbers $t$ can be represented as \[t = \sum^n_{k=1} x_k\] where $ x_k  \in \mathbb{N}_0$ with fixed $n,$ $n = 4 \cdot z, z \in \mathbb{N}$ and $x_i \cdot x_{i+1} = x_{i+2}  \cdot x_{i+3}$ where $i = 4 \cdot q + 1, q \in \mathbb{N}_0$ and $i = 1, \ldots, n-3.$

Ugh? :? What a complex wording! Did I understand it correct if I'm rewriting it like this? :

$\displaystyle t = \sum^{4n}_{k=1}x_k \ (x_k \in \mathbb{N}_0)$

and $\displaystyle x_{4m+1}\cdot x_{4m+2} = x_{4m+3} \cdot x_{4m+4}$ for  $m<n$, with $m,n \in \mathbb{N}_0$.


function
search
calculus
calculus computations

If I am asked find the infinite sum of:

sin(x) + sin(x/2) + sin( x/4) + sin(x/8) .... to infinity;

is there a way i can find the answer? and does the answer exist?

Thanks

[quote="madness"]note that $ \sin \theta \plus{} \sin 2\theta \plus{} \cdots \plus{} \sin n\theta \equal{} \frac {\sin 2^n \theta}{2^n \sin \theta}$
From here employ the limit :)[/quote]

wow there madness, i think somebody's getting confused a bit. could you please derive this formula?  : :wink:

sorry :wallbash_red: im totally confused :fool: thanks hash_include  :)

hey everyone, im confused now too, so there isnt a solution to this question or that wasnt the right approach?

I'm afraid that the function that arises this way is not elementary (i.e., there is no reasonable formula for it), though I'm very far from having a formal proof of that. Anyway, if I were to give a shot at the problem, I would start with something like what I did [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=31408439&t=36673]here[/url]


Let ABCD be a convex quadrilateral and all sides of it are constant.AC and BD diagonals intersect at P point.Prove that the sum of AC and BD diagonals are maximum iff 1/AP +1/PC =1/BP+1/PD.
Find a formula for the maximum and minumum value of the sum of AC and BD diagonals.

Interesting problem.  I've tried some numerical examples, and it appears to be true.

However, even the simple case of a kite (AB = DA, BC = CD) seems intractable.  Any thoughts?  ssupermath, does this have a nice solution?

Unfortunately,I don't have any solution.Perhaphs somebody can solve it...


quadratics
number theory proposed
number theory

Let p be a prime number and let K={1;2;3;...;p-1}. Evaluate the sum of all the quadratic residues
mod p of K

Do $ p\equal{}2,3$ manually (you get $ 1 \mod p$). For $ p>3$, the sum of squares of all residues is $ 0 \mod p$ by http://www.mathlinks.ro/Forum/viewtopic.php?t=40171 . Here we counted every quadratic residue twice, thus the sum of all quadratic residues is $ 0 \mod p$.

Oh, I know that the sum is ≡ 0 mod p. I want to know the sum as a number ( in Z not in Zp)

For $ p \equiv 1 \mod 4$ it's not hard:
$ x$ is a quadratic residue iff $ p \minus{} x$ is one. Thus we have $ \frac {p \minus{} 1}4$ such pairs of quadratic residues, and every pair sums to $ p$. Thus the sum is $ \frac {p(p \minus{} 1)}4$.
For $ p \equiv \minus{} 1 \mod 4$, it's more difficult:

Let in general be $ s(p)$ be the desired sum. Then $ s(p) \equal{} \sum_{i \equal{} 1}^{p \minus{} 1} \frac 12 (\left( \frac ip \right) \plus{} 1)i$, where $ \left( \frac ip \right)$ is the Legendre-symbol (it is $ 1$ for quadratic residues and $ \minus{} 1$ for nonresidues). So in fact $ s(p) \equal{} \frac 12 \cdot \left( \sum_{i \equal{} 1}^{p \minus{} 1} \left( \frac ip \right) i \plus{} \sum_{i \equal{} 1}^{p \minus{} 1} i \right) \equal{} \frac {p(p \minus{} 1)}4 \plus{} \frac 12 \cdot \sum_{i \equal{} 1}^{p \minus{} 1} \left( \frac ip \right) i$.
So we are interested in $ h(p) \equal{} \sum_{i \equal{} 1}^{p \minus{} 1} \left( \frac ip \right) i$.
For $ p \equiv 1 \mod 4$, this is well known to be $ 0$, giving the result from above. But if $ p \equiv \minus{} 1 \mod 4$ and $ p \neq 3$, this is known to be $ \minus{} p$ times the class number $ h_{ \minus{} p}$ of $ \mathbb Q [\sqrt { \minus{} p}]$. Thus we have $ s(p) \equal{} p \cdot \left( \frac {p \minus{} 1}4 \minus{} \frac {h_{ \minus{} p}}2 \right)$.

ZetaX :
I think that this sum is p(p-3)/4 when p≡3 (mod 4). It is posisble?

That would mean $ h_{ \minus{} p} \equal{} 1$, but this is only true for $ p \equal{} 3, 7, 11, 19, 43, 67, 163$. In fact, Gauß' class number conjecture (which is proven) says that for every $ n$ there are only finitely many primes $ p$ such that $ h_{ \minus{} p} \equal{} n$.

My mistake, sorry

Initialy I try to solve this problem:
(American Mathematical Monthly, 1999) Let p be a prime number with p ≡7 (mod 8), and let Lp =
{1, 2, . . . , (p − 1)/2}. Prove that the sum of the quadratic residues (i.e. the squares) modulo p in Lp
equals the sum of the quadratic nonresidues modulo p in Lp. (The sums are taken in Z)

And I make a mistake that make necessay that the sum of the quadratics was p(p-3)/4
javascript:emoticon(':blush:')
Blush
However.
You can help me to solve that (the AMM)?


todos  sabemos  que las lineas  que  unen  los vertices respectivos  de 2 triangulos  homotéticos  concurren  en  el  centro de  homotcia.  pero.. :

 como  lo  pruebo  si  no  supiera  nada de  homotecia  utilizando  ceva??

Bueno por lo que a mi respecta :

Se llama homotecia de centro O y razón k (distinto de cero) a la transformación geometrica que hace corresponder a un punto A otro A´, alineado con A y O, tal que: OA´=k·OA. Si k>0 se llama homotecia directa y si k<0 se llama homotecia inversa. ( o sea no habria nada que probar )

Ahora la pregunta seria la siguiente :

Dados dos triangulo semejantes ABC y A'B'C' y ademas se cumple : AB // A'B'  ; BC // B'C' y CA // C'A' . Probar que : AA', BB' , CC' son concurrentes .

Bueno eso creo que seria asi :

sea O el punto de interseccion de AA' y BB' luego como los triangulo ABC y A'B'C' son semejantes entonces : AB/A'B' = BC/B'C' = CA/C'A' = k . entonces por semejanza de ABO con A'B'O se cumple que : OB/OB' = k y el angulo entre OB y BC es congruente angulo entre  OB'/B'C' o sea los triangulos OBC y OB'C' son semejantes . con ello los angulos BOC y B'OC' son congruentes y esto garantiza que C , O y C' sean colineales .

Nota : la condicion de que AB // A'B'  ; BC // B'C'  y CA // C'A' implica que los triangulo ABC y A'B'C' sean semejantes . asi que se podria replantear la situacion .

Bueno ..
podemos encontrar algo de informacion adiconal sobre [color=darkblue]CENTRO DE HOMOTECIA[/color]....
aqui pongo un link desde http://mathworld.wolfram

[url=http://mathworld.wolfram.com/HomotheticCenter.html]Homotecia - english version![/url]

y su version traducida:

[url=http://translate.google.com/translate?u=http%3A%2F%2Fmathworld.wolfram.com%2FHomotheticCenter.html&langpair=en%7Ces&hl=es&ie=UTF-8&oe=UTF-8&prev=%2Flanguage_tools]Centro de Homotecia (español)[/url]

Carlos Bravo :)
Lima -PERU


Suppose $ X$ is a metric space such that every open cover has a finite subcover. I'm proving that this implies that every sequence has a convergent subsequence.

So take a sequence $ x_n$ of points in $ X$. Assume that it has no convergent subsequence. Thus for every $ x \in X$, we choose an open ball $ U_x$ such that $ U_x$ contains finitely many terms of the sequence. But we can cover $ X$ with finitely many of these balls, and since each contains only finitely many terms of the sequence, it follows that the sequence has finitely many terms, a contradiction.

Yet this proof appears to work for any topological space, which I believe is not true (just replace $ U_x$ with an arbitrary open neighborhood around $ x$ containing finitely many terms of the sequence). Where did I go wrong?

In a space with the right separation properties, i.e. one that lets you pick $ U_x$ to be small enough, this result should still hold.  I'm not very clear on the details, though, so I couldn't tell you exactly which separation axiom(s) you need.  (For example, the argument fails if there exists a point $ x$ such that the smallest open set containing $ x$ is the whole space.)

Edit:  See [url=http://en.wikipedia.org/wiki/Compact_space#Other_forms_of_compactness]the Wiki article[/url].

I actually believe I see the issue. It doesn't always hold that if every open set around a point contains infinitely many terms of the sequence, then there exists a convergent subsequence. For a metric space, we could consider balls of radius $ 2^{\minus{}n}$ and arbitrarily choose an element of the sequence there. In fact, this argument can be applied to all first countable spaces (and I was previously aware that the implication is supposed to hold for first countable spaces).


Show that $ x^7\minus{}2x^5\plus{}10x^2\minus{}1$ has no roots greater than 1.

I have tried making the substitution $ x\equal{}y\plus{}1$ and showing that the new equation has no positive roots but don't seem to be getting anywhere.
Any suggestions?

Just a thought

[hide]$ f(x)\equal{}x^7 \minus{} 2x^5 \plus{} 10x^2 \minus{} 1$
$ f(1)\equal{}8$

Then show $ |f(x)|>|f(x\minus{}1)|$[/hide]

[quote="SimonM"]Just a thought

[hide]$ f(x) \equal{} x^7 \minus{} 2x^5 \plus{} 10x^2 \minus{} 1$
$ f(1) \equal{} 8$

Then show $ |f(x)| > |f(x \minus{} 1)|$[/hide][/quote]

This would only prove that no [i]integer[/i] roots greater than $ 1$ exist.  To prove for all reals greater than $ 1$, you would have to prove that $ f(x) > f(x\minus{}\epsilon)$, which is essentially proving that $ f'(x) \ge 0$ (i.e. calculus).

:!:  :!:  :!:  This may be one of the rare instances where [hide] Descartes' Rule of Signs [/hide] is useful, although you would have to check that the coefficients are indeed all positive. Of course, you don't need Descartes' to see that a polynomial with all positive coefficients can have only negative roots.

[quote="SnipedYou"][hide]Try factoring it like this

$ (x^{5} \plus{} 10)(x^{2} \minus{} 2) \equal{} \minus{} 19$

and every $ x \ge 2$ would make the left side positive.[/hide][/quote]
sorry but i do not see how this helps.
also I thought about using Descartes but realized that I would have to expand after the substitution, but i guess that is what i will have to do.

[quote="t0rajir0u"][quote="SimonM"]Just a thought

[hide]$ f(x) \equal{} x^7 \minus{} 2x^5 \plus{} 10x^2 \minus{} 1$
$ f(1) \equal{} 8$

Then show $ |f(x)| > |f(x \minus{} 1)|$[/hide][/quote]

This would only prove that no [i]integer[/i] roots greater than $ 1$ exist.  To prove for all reals greater than $ 1$, you would have to prove that $ f(x) > f(x \minus{} \epsilon)$, which is essentially proving that $ f'(x) \ge 0$ (i.e. calculus).[/quote]

Yeah, I should have written

$ f(x)>f(y), x>y$

Here is another seemingly unmotivated solution... $ x^7\minus{}2x^5\plus{}10x^2\minus{}1\equal{}8\plus{}(x\minus{}1)(17\plus{}(x\minus{}1)(11\plus{}(x\minus{}1)(x^4\plus{}3x^3\plus{}4x^2\plus{}4x\plus{}3) ) )$


Suppose $p(x) \in \mathbb{Z}[x]$ and $P(a)P(b)=-(a-b)^2$ for some distinct $a, b \in \mathbb{Z}$. Prove that $P(a)+P(b)=0$.

[quote="Peter"][quote="Tiks"]we suppose that $P(a)P(b)$ is not equal $0$[/quote]In fact, we are even given that (since $a,b$ are distinct).[/quote]
Ooops,I forgot that :oops: .

$P(a)-P(b)$ is divided $a-b$
the rest is easy. :thumbup:


algebra
function
domain
articles
number theory

Two questions:

1. Do PDFs on the web written by reliable people like [url=http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf]Naoki Sato's Number Theory[/url] count as reliable sources to cite without proof?

2. Suppose you find a completely correct proof (not just a statement, a full proof) on an "unreliable" source like AoPS forums or Wikipedia. Can you use the proof?

2. I think citing the proof would be a bad idea, but copy-pasting never hurt anyone... (I'm sure wiki is in the public domain, but I'm not sure about AoPS.)

is this for USAMT's? wats the point of using someone elses proof!?

[quote="Wikipedian1337"]1. Do PDFs on the web written by reliable people like [url=http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf]Naoki Sato's Number Theory[/url] count as reliable sources to cite without proof?[/quote]No.  Although Naoki is reliable, it doesn't fit the criteria of rule #8:
[quote="USAMTS Rule #8"]Students may cite (without proof) results from a reputable source, such as a widely-accepted book or peer-reviewed journal, or from a professionally-maintained reference website such as Wolfram MathWorld.[/quote]

[quote]2. Suppose you find a completely correct proof (not just a statement, a full proof) on an "unreliable" source like AoPS forums or Wikipedia. Can you use the proof?[/quote]Yes, a proof is a proof.  If it's valid, then it's legitimate.  Of course, if you find such a proof, you need to cite when you found it (otherwise you're stealing someone else's work without credit).  However, it pretty much defeats the purpose of participating.  The point of the contest is not to do internet research to try to find a proof--the point is to try to solve the problem yourself.

[quote]2. Suppose you find a completely correct proof (not just a statement, a full proof) on an "unreliable" source like AoPS forums or Wikipedia. Can you use the proof?[/quote]Yes, a proof is a proof.  If it's valid, then it's legitimate.  Of course, if you find such a proof, you need to cite when you found it (otherwise you're stealing someone else's work without credit).  However, it pretty much defeats the purpose of participating.  [b]The point of the contest is not to do internet research to try to find a proof[/b]--the point is to try to solve the problem yourself.[/quote]

Thats what I thought!

[quote="USAMTS Rule #8"]Students may cite (without proof) results from a reputable source, such as a widely-accepted book or peer-reviewed journal, or from a professionally-maintained reference website such as Wolfram MathWorld.[/quote]

Really?  I cited Wolfram functions last year for the fact $ \sum_{n\equal{}1}^{\infty} \frac{\phi(n)}{n^4} \equal{} \frac{\displaystyle \sum_{n\equal{}1}^\infty \frac{1}{n^3}}{\displaystyle \sum_{n\equal{}1}^\infty \frac{1}{n^4}}$ yet still lost a point for it.  Is this rule new?  Because I thought there was a rule like this when I cited it.

[quote="facis"][quote="USAMTS Rule #8"]Students may cite (without proof) results from a reputable source, such as a widely-accepted book or peer-reviewed journal, or from a professionally-maintained reference website such as Wolfram MathWorld.[/quote]

Really?  I cited Wolfram functions last year for the fact $ \sum_{n \equal{} 1}^{\infty} \frac {\phi(n)}{n^4} \equal{} \frac {\displaystyle \sum_{n \equal{} 1}^\infty \frac {1}{n^3}}{\displaystyle \sum_{n \equal{} 1}^\infty \frac {1}{n^4}}$ yet still lost a point for it.  Is this rule new?  Because I thought there was a rule like this when I cited it.[/quote]

Well, I was able to get to $ \sum_{n \equal{} 1}^{\infty} \frac {\phi(n)}{n^4}$ in like 1 line so proving that this is equal to $ \frac {\displaystyle \sum_{n \equal{} 1}^\infty \frac {1}{n^3}}{\displaystyle \sum_{n \equal{} 1}^\infty \frac {1}{n^4}}$ is the only non-trivial part of the proof. Citing it makes the entire part of the proof trivial... I think. That should be a guideline to citing stuff..

[quote="facis"]Xaero, I slightly agree with you, but as I am completely clueless about how to manipulate Dirilecht series... Anyway, for me that was not "line 1" but one of my first steps once I got past proving that the circles cannot overlap and started the second part of the problem.

So I'd like to learn more about Dirilecht series, but I've had very little luck... any ideas would be appreciated.[/quote]

learning Dirichlet series need results from some somewhat difficult to understand number theory, but for that particular problem, we can look at the wikipedia article and try to understand the proof.

[url]http://en.wikipedia.org/wiki/Phi_function#Generating_functions[/url]
You can sort of intuitively check that the proof is correct by writing out the series and multiplying them instead of using convolution. Oh, I think it's necessary to prove
$ \sum_{d|n} \varphi(d) \equal{} n$ to get full credit.

Yeah, I just looked back at that and it makes a lot more sense now.  I'm still not sure what $ *$ is supposed to mean but the key step seems to be that the product of the two dirilecht series can be written as a single dirilecht series: $ \sum_{g \equal{} 1}^\infty \frac {1}{g^s} \cdot \sum_{f \equal{} 1}^\infty \frac {\phi(f)}{f^s} \equal{} \sum_{h \equal{} 1}^\infty \frac{\displaystyle\sum_{f|h} \phi(f)}{h^s}$ if we let $ fg \equal{} h \implies f|h$ and then the sum of the totients of the divisors...

If I had had the inspiration to look in this direction, I might have been able to conjecture that $ \sum_{d|n} \phi(d) \equal{} n$ but I probably wouldn't have figured out how to prove that... anyway, I'll know better for next time ;-)

[quote="123456789"]Last year, for the same problem, I cited that fact, and also provided the proof for it, and still got a point off.[/quote]

How did you get marked off if you provided the proof? I think you should have been able to protest that (but there's nothing you can do about it now..)


algebra
polynomial
number theory proposed
number theory

let $x_1,x_2,\dots,x_m\in \mathbb{Z}$ be distinct numbers and $P(x)$ be polynomials with $m-1$ degree. prove that at least one of the numbers $|P(x_1)|,\dots,|P(x_m)|$ is not less than $\frac{(m-1)!}{2^{m-1}}$

You can get easy proof, using Lagrang's interpolational polynom.


On the AIME answer form, there is a space where one bubbles which previous AMC exam was taken. If someone, such as myself, took both the AMC 10, and the AMC 12, which bubble(s) should we fill in.

I'm not positive, but I think it would be the one that you did better on (in your case AMC 12). But if my memory is really accurate, it actually doesn't matter, as they will check both scores anyway. Maybe bubbling in the better one will reduce the likelihood of error though.

The test-taker I coach here will just put in his AMC 12 score, which is lower NUMERICALLY than his AMC 10 score (as one would expect) but probably more significant for someone of the age to take either test.

>  the one that you did better on (in your case AMC 12). But if my memory is really accurate, it actually doesn't matter, as they > will check both scores anyway.

This is correct.  Fill in the higher of the two scores.  We check both.

Steve Dunbar
Director, American Mathematics Competitions

[quote="AMCDirector"]Fill in the higher of the two scores.  We check both.[/quote]

That's clear. The higher score it is then.

[quote="randomdragoon"]What should you get on the AMC 12 to be considered "better" than a passing AMC 10 score?[/quote]

I guess you could weight it so that 120 (on the 10) = 100 (on the 12) and then go up proportionally there to 150. Too much work though  ;) .

[quote="dogseatcheese"]It would be too much work to proportionatly see which one is a better score. Just put down the one you like the best and well they check both anyway.[/quote]

Acutally I realized not really:

Take your AMC 10 score. subtract 120. (Only if you got above 120) Then multiply by 3/5. Then add 100. And there you go.

KK that's not that accurate but hey it's not bad.

Someone should make a converter for AMC 10 and 12 scores....what if someone.....*me* cough got less than 120 on one of the AMC 10s and over on the other.....then I would have a negative score after using your operation.


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Now that you have all seen the "lets make a deal" or monty hall problem, here is one that is of a similar confusing nature. Please vote in the poll what you personally think the answer is, don't go search for it on the web or something.

We plan to put Beauty to sleep by chemical means, and then well flip a fair coin. If the coin lands Heads, we will awaken Beauty on Monday afternoon and interview her. If it lands Tails, we will awaken her Monday afternoon, interview her, put her back to sleep, and then awaken her again on Tuesday afternoon and interview her again. The (each?) interview is to consist of the one question : what is your credence now for the proposition that our coin landed Heads? When awakened (and during the interview) Beauty will not be able to tell which day it is, nor will she remember whether she has been awakened before. She knows about the above details of our experiment. What credence should she state in answer to our question?

Uhh...[hide]1/3, since an interview can be of three different natures:

- it could be on the Monday, and we flipped a heads

- it could be on the Monday, and we flipped a tails

- it could be on the Tuesday, and we flipped a tails.

Sleeping Beauty doesn't know which one of these it is.  But she does know that if you selected a random interview that may have taken place, 1 in 3 times it will be because we flipped a heads and 2 in three times it will be because we flipped a tails.  So the chance that we did flip a heads is 1/3.[/hide]

Oops, I'm supposed to vote.

I think its [hide]1/3 [/hide] because the probability that she [hide]wakes up and its heads [/hide]is[hide] 1/3[/hide]... and the probability [hide]that she wakes up is 1[/hide]... so [hide](1/3)/1 = 1/3[/hide]

MysticTerminator wrote:gah. i'm tempted to say [hide]1/3[/hide] because [hide]there's a 1/3 chance that it's the interview for which the coin landed heads, and 2/3 chance it's one of the tail interviews[/hide] but it seems like she should just say [hide]1/2[/hide] because, [hide]well, there's a 1/2 chance the coin will land heads.[/hide]. gah. you have ruined my perfect life, stephen. i think i'll vote [hide]1/3[/hide].

But [hide]it doesn't matter what the coin's chances are.  She knows she is in an interview, and she knows that the chances that any given interview is a "heads" interview is 1/3.[/hide]

Ok, here is an argument for each case. Don't read these until you've made up your own mind.. and then see if you can disprove either.

[hide]

1) For the halfers: The probability the coin toss came up heads was 1/2. SB gains absolutely no new information from being interviewed, as this was always going to happen. Therefore the probability must be 1/2.

2) For the thirders: 2/3 of the interviews take place when the coin has come up tails. Therefore the probability must be 1/3.

[/hide]

There are various other arguments too..

[quote="TripleM"]Theres two types of people in this world - halfers, who think the answer is obviously 1/2, thirders who think its obviously 1/3. Both are convinced they are right.. and I have never seen a convincing proof that either is wrong.[/quote]
look at my post, and you'll see i'm not either. i think the question itself maybe slightly ambigous, but i can't quite pin it down.

i'll give a reason for my last comment: for the monty hall problem, we could actually enact it with dice "choosing" random curtains, etc, and find the probability is indeed close to 2/3 if we switch (or w/e, i can't remember exactly). for this problem, we can't do any such thing. i think it depends on how we define probability again.

Okay, here is my confusing take on it:
That probability doesn't exist.  Probability measures the expected number of times an event will occur when a trial is repeated indefinitely.  This is always the definition of probability.  The problem is that the way this problem is stated, it [i]explicitly forbids[/i] doing an unlimited number of trials -- she has to deduce probability from a single trial.  This is not possible.  In essence, the question she is being asked (I believe) is meta-mathematical, and not truly mathematical.  The question she can answer is "What do you think the coin flip landed?" to which it is in her best interest to answer "tails."  But the way the question is phrased, it doesn't really have an answer. [/confusion]

Ok, so how about we repeat this experiment as many times as you want to get this probability. Each time, she still gains no information from waking up and being interviewed, and the past experiments can have no effect on later tosses of the coin. So we get the same thing..

Exactly -- each repitition is completely blind.  She [i]cannot[/i] compute a probability because, as a person [i]in[/i] the problem, [i]she[/i] cannot collect any repeated data.  And she's the one who needs the data, because of the way the problem is phrased, not us.

My answer is she should answer 1/2 the first time and 0 if she is waken up the second time; I believe even if you are put to sleep by chemical means your memory will not degenerate in such a short period of time. Of course this is not the case here so the answer is 1/2.

1/3 is incorrect because the three times she wake up does not have anything to do with the previously flipped coin, those two events are completely independant. No matter what the coin was flipped whether she's awakened, sleeping, or dead. According to this logic she could have been woken 5 times when the coin showed tails and the probability would've been 1/5, and so on until you get 0, which is obviously not possible. Therefore the answer is 1/2.

By the way if the problem asked for her credence on whether or not this interview was [b]because [/b]the coin showed heads then the answer is 1/3, see the difference?

Joel, where did you get the idea that we can only define probability if we can repeat an experiment?  If I just bought a fair coin, the probability of getting heads on my very first toss is 1/2, even though I can't repeat the experiment (because later tosses won't be the first).

I believe the answer to the original question is [hide]1/3[/hide], using conditional probability.  I will leave out my proof until more people have thought about the question.

[quote]If I just bought a fair coin . . .[/quote]

Ravi -- how can you buy a fair coin?  What does that mean?  It means that someone, at some point, did something which either (a) flipped the coin some infinite number of times to determine that it does in fact land heads half the time or (b) carried out some experiment or calculation that determined the same thing.  Either way, someone had to do something, before you ever bought the coin, that determined that it was fair.  There had to have been such a determination for you to buy a fair coin in the first place.

I still think the question is fundamentally unsolvable.

Ok, let us look at the begging, chance of A, B and c is each 1/3, we chose b, thus the chance of a or c being right is 2/3, no we have been told c isn't right, so a's chance is 2/3? am i wrong somewhere?

ml2, if you're thinking of the "let's make a deal" game, this one is different.  In that situation, the game show host purposefully reveals which answer is incorrect based on which one you picked (i.e. if you had picked the correct one, he will reveal either of the two wrong ones, but if you had picked an incorrect one, he will reveal the OTHER incorrect one).  In this situation, the teacher is unaware of your current response, so his saying that choice C is incorrect is completely random.

So I'd say it doesn't matter if you switch or not.

Yeah, I also get that it makes no difference whether you swap or not.

So now where does the paradox come from?

Mathfanatic's explaination is exactly right. The way the question is worded, you have to infers that the teacher's announcement was independent of your answer choice. So in theory they [i]could[/i] have announced the same answer choice that you had put, making the sample space of possibile outcomes larger than in the Monty Hall problem.

The paradox comes from the fact that this problem sounds so similar to the Monty Hall Problem, but a small inference makes such a big difference in the outcome. The physical situation of the two problems are the same, but from psychological inference the two problems become different. The psychology of infering that the teacher was not making the announcment just to you is what changes the problem!

Keep in mind that a paradox isn't a problem without an answer. A paradox is a problem that seemingly has different reasonble interpretation leading to different answers. The fact that I didn't explicitly [b]say[/b] whether the teacher's announcement was independent of my answer leaves that determination up to an inference. That ambiguity of interpretation is where the paradox comes from.

i still don't see the difference, if you had siad that you picked [i]a door[/i] and the teacher told you that c isn't right then maybe, but since you did pick B and the teacher did (indepndently or not) do the same thing that the game host does.

Yes but in one case you *know* what Monty Halls strategy is, but not the teachers.

could you explain that more clearly, i can't quite understand what you mean

OK.

In Monty hall problem, he says he will open a wrong door, but not the one you have chosen.
Suppose you chose door A.

The probabilities are:
A correct, open B: 1/3 x 1/2 = 1/6
A correct, open C: 1/3 x 1/2 = 1/6
B correct, open C: 1/3 x 1 = 1/3
C correct, open B: 1/3 x 1 = 1/3.

So given that door C was opened, the probability A is correct is 1/6 / (1/6 + 1/3) = 1/3.

Now for the teachers problem. She doesn't know you have chosen B and will tell you one of the incorrect answers.
The probabilities are:
A correct, told B: 1/3 x 1/2 = 1/6
A correct, told C: 1/3 x 1/2 = 1/6
B correct, told A: 1/3 x 1/2 = 1/6
B correct, told C: 1/6
C correct, told A: 1/6
C correct, told B: 1/6.

So the probability that B is correct, given that you were told C is incorrect, is 1/6 / (1/6 + 1/6) = 1/2.

I am still undecided as to whether I think the best answer is 1/2 or 2/3, and I keep arguing myself from one side to the other.  Here are my answers to Ravi's questions that I would make if i were defending the answer of 1/2:
1.  1/2
2.  3/4  -  This is probably wrong, but this is my reasoning: there is a half of a chance to get a heads - and if this is the case there is a 100% chance of being woken up on monday.  (I think this question may be just as much a source of debate as the original one.)
3.  1/2  -  Given that today is monday there is a 2/3 probability that the coin landed heads, hence the answer is 3/4*2/3.
4.  1/4  -  1/3 probability given that today is monday.
5.  0
6.  1/4  -  The probability of a tails is 1 if we know that today is tuesday.

I think that I used these last four to answer the first ones (is that a valid method?).  Anyways, I will answer as best I can for now and possibly change them later.
7.  2/3
8.  0
9.  1
10.  1/2

This problem is one of my favorite in all of mathematics because it demonstrates how difficult it is to even define probability.

First, let us consider what probability means.

The frequentist interpretation of probability is based on repeated experiments.  I personally find this interpretation superficial because it doesn't relate to reality.  We can never reset the initial conditions of an experiment.  It also doesn't jive with events that we can examine such as sporting events.  How is it that bookies can set odds that seem to work so well in the long run [i]before each event is played out[/i]?

I personally prefer the subjectivist interpretation of probability.  All probabilities are subject to the information that I have about the events being considered.

This interpretation has the benefit of making problems with information in the form of frequencies solvable directly through mathematical reasoning.

However, this interpretation also allows me to make statements such as "I think the probability that the Cardinals will beat the Padres when next they play is 3/4 because I am [i]indifferent[/i] toward betting for or against the Cardinals as 3:1 money favorites."

This may seem a little goofy at first, but we can extend this seemingly blurry idea of subjective probability to problems of quantum physics -- what do we mean when we say that an electron has a 1/3 probability of being in some region of its orbital around an atomic nucleus?  We can never reset initial conditions because we can never even determine what those initial conditions were (due to the Heisenberg Uncertainty Principle).  We can however make a statement such as "I am indifferent toward the proposition of betting for or against the event given 2:1 money odds against the proposition".

As the subjectivist interpretation relates to the Sleeping Beauty problem, she should answer 1/3, not 1/2.  If she were to make a bet, she would want 2:1 odds because she'll be asked the question twice following one result and only once following the other.  She is not answering the question "what was the probability that the coin heads?"  She is answering the question "given the circumstance that we are interviewing you, what odds would you need us to lay such that you are indifferent toward betting on heads (vs. tails)?"

This jives nicely with the Bayesian (dependent probability) arguments put forth numerous times in this thread.

The only problem with the subjectivist interpretation is that is doesn't take into account utility concerns over volatility.  Volatility destroys utility and every time a person makes a wager they are subjecting themselves to volatility (this is a good reason not to play games of [i]chance[/i]).  For instance, who makes more money, a person who earns 100% on some years and loses 50% on other years with approximately equal frequency (for an expected yearly return of 25%) or the person who earns 10% every year without fail?

This problem again might be analogous to real world physics.  Why is it that given all other information about an electron, we can only pinpoint its location to within an interval?

Why is it that if I take volatility into consideration that I can only set a range on what odds I would take for the Yankees to win the World Series?

MCrawford, those are interesting points, and I basically agree with you.

Ian, you are the first halfer to at least give self-consistent answers to my set of 10 questions.  Congratulations.  But that doesn't mean I agree with those answers. :)  Let me focus on question 7: "Now what is the conditional probability that the coin landed heads given that today is Monday?".  Your answer was 2/3.  Do other halfers agree with that answer?  Let me argue that the answer should be 1/2.  Basically, here we are telling Sleeping Beauty the day, and then asking her the probability of heads.  It happens to be Monday.  But Sleeping Beauty can reason that, whether the coin was heads or tails, she would be interviewed on Monday.  Thus her being woken on a Monday has given her no information on whether the coin is heads or tails.  So she should answer 1/2.

In other words, if  she knows today is Monday, it is irrelevant that she sometimes gets interviewed on Tuesdays.  She should answer the same as if someone had flipped a coin in front of her eyes: 1/2.

Hmmm..... I think I need something clarified about conditional probability.

Your question was "What is the probability that the coin landed heads given that today is monday?"

Im wondering -- which of the following situations is that analagous to:

1.  The sleeping beauty is told beforehand that she would be told what day it is when she is awoken.

2.  The sleeping beauty is not told what day she will be awoken but happens to notice a calendar with the day marked on it as Monday.

These are different situations, I believe -- in the first I would say the probability that the coin landed on heads would be 1/2; whereas in the second I maintain the probability that the coin landed heads would be 2/3.

Well, I had been thinking of the first situation.  But the second situation seems equivalent to me, for the purpose of calculating probability.  Unless Sleeping Beauty knew something like "the interviewers make calendars available only when the coin came out heads", both situations just filter out Tuesday outcomes.  A third situation to think about conditional probability is that no one tells Sleeping Beauty what day it is; instead she just makes a conditional statement such as "If [or given that] today is Monday, then the probability of heads is ...".  All three situations are equivalent to me.  Whichever situation you prefer, it should make the formula Pr(Monday and heads) = Pr(Monday) Pr(heads | Monday) work out, i.e., let you go from Question 7 to Question 3.

Basically, conditional probability means to filter out those outcomes that do not meet the condition, and then scale up the probabilities of the remaining outcomes so that they sum to 1.

To hopefully clarify our differences, let me broaden the discussion.  To solve any question about "applied" probability, there is usually a two-step process:
(1) Decide on a probability space that models the question.  A probability space just means a sample space (set of outcomes) and a probability distribution on those outcomes.
(2) With the probability space in Step 1 and the basic definitions of probability, calculate the probability of the event in question.  In other words, add up the probabilities of those outcomes that satisfy the event.

It seems to me that we disagree on the appropriate probability space in step 1.  To make our differences explicit, I will state here the probability space that I am using.

Had there been no restriction against heads interviews on Tuesdays, the sample space would consist of 4 outcomes {Mh, Mt, Th, Tt}, where Mh stands for "an interview on Monday and the coin came up heads", and similarly for the other three outcomes.  The probability distribution would be 1/4 for each.  But in fact there is a restriction against heads interviews on Tuesdays; that is, the original problem is itself a conditional probability.  Following the definition of conditional probability, our sample space becomes {Mh, Mt, Tt}, and the probability of each outcome scales up to 1/3 for each.  That is the sample space I use.  That explains why I think the answer to the original question is 1/3.  It also explains why my answer to say Question 7 (the probability of heads given Monday) is 1/2.  In fact, this sample space explains all my answers.

So, Ian (and other halfers), let me ask you: What is the probability space that you are using?  The one you use to answer the original question and all the other questions.  Thanks.

Sorry that it took me so long to respond to this thread.

First to clear up the difference (that I think exists, but may be mistaken in doing so) between the two cases that I gave:

[quote]1. The sleeping beauty is told beforehand that she would be told what day it is when she is awoken.

2. The sleeping beauty is not told what day she will be awoken but happens to notice a calendar with the day marked on it as Monday. [/quote]

The difference that I see between these two situations is the same as the difference between the Monty-Hall problem and the one gauss202 mentioned earlier in this thread:

[quote]Suppose that you're taking a multiple choice test with 3 equally likely answer choices, A, B, and C. Towards the end of the test, your teacher announces to the class that the answer to problem 1 is NOT C. You had previously marked answer choice B (but not having any idea what the question was even asking you really just chose it at random). Should you stick with your answer choice B, or should you switch your answer to A? [/quote]

The answer to gauss's problem is that it doesnt matter whether you switch or not -- whereas in the Monty-Hall problem you have a higher chance of winning if you switch.  The difference between these two problems is that while you are taking the test, the teacher announces the answer is not C [i]without knowing which one you have picked[/i], but in the Monty Hall problem the person who shows you a door that doesnt contain the prize [i]purposely avoids[/i] picking the door that you chose.

I think that, for the same reason that gauss's problem is not equivalent to the Monty-Hall problem, the knowledge of whether she will be told what day it is [i]does[/i] change the conditional probability that the coin landed heads.

Next, my probability space is the same as yours {Mh, Mt, Th, Tt} except with probabilities as follows:
Mh: 1/2
Mt: 0
Th: 1/4
Tt: 1/4
(The different outcomes are permitted to have different wights, right?)

I agree that if the interviewers strategically determine when to make calendars available, then they can fool Sleeping Beauty.  For that matter, they can fool Sleeping Beauty even more easily by not waking her according to the rules.  I was assuming that the interviewers were honest.  That's why I added the caveat about "unless Sleeping Beauty knew something like the interviewers make calendars available only when the coin came out heads".

To avoid such complications, we can stipulate that for calculating conditional probability, Sleeping Beauty knows in advance that she will be told the date.  Your answer for conditional probability in that case was 1/2, which I agree with.  What about the third possibility I mentioned, where Sleeping Beauty just makes conditional statements to herself, such as "If  today is Monday, then the probability of heads is ...".  How would she complete such a sentence?  (Again, I think she would say 1/2.)

Thanks also for listing your probability distribution.  I don't see where your numbers come from, but at least I know where you stand.  By the way, I think you made some typos in your probability distribution.  For example, you wrote that the probability of Mt is 0.  But I think you meant that the probability of Th is 0.

Let me try another tack to explain why I think the answer to the original Sleeping Beauty problem can't be 1/2.  When we say that the probability of an event is 1/2, we generally mean that we are indifferent (as MCrawford would say) to betting for or against the event.  For example, for halfers, Sleeping Beauty would be indifferent to betting that the coin was heads.  In other words, she would be indifferent to gaining a dollar whenever she was awoken because of heads, and losing a dollar whenever she was awoken because of tails.

Now consider the following revision of the original problem.  Here she is guaranteed to be interviewed on both Monday and Tuesday, regardless of the coin outcome.  (She knows about the new rules.)  Here I think we would all agree that the probability of heads is 1/2.  That is, Sleeping Beauty would be indifferent to gaining a dollar whenever she was awoken because of heads, and losing a dollar whenever she was awoken because of tails.

Now compare the two games.  The only difference between them is what happens on Tuesday heads.  In the original problem, she neither gains nor wins a dollar on such days.  In the revised problem, she gains a dollar.  So how can she be indifferent to both games?  Wouldn't she much prefer the second game to the first?


Solve in positive integers:

$x^{2}+tx=y^{2}+ty$

where $t$ is given positive integer.

[hide="some notes"]related to http://www.mathlinks.ro/Forum/viewtopic.php?t=103637[/hide]

[quote="TomciO"]Err... $(x-y)(x+y+t)=0$?[/quote]

 :oops:

so all the solutions are:

$(x,y)\in\{(s,s),(r,t-r);s,r\in N,r<t\}$

maybe I should start think about some harder ones..


Thanks, see the next post.

And, just wondering:

Is this summation very well known?
And, are these types of proofs generally possible?

[quote="Altheman"]wahoo...this solution has no motivation

$\frac{x}{\sqrt{1-4x^{2}}}=\sum_{k=0}^{\infty}\binom{2k}{k}x^{2k+1}$

now, this is a trivial exercise in maclurian series, use the binomial series and y=4x^2[/quote]

How did you get your expression?

I remember seeing that expansion on my friend's Combinatorics homework at PROMYS.  It's probably derived from expanding $(1-4x^{2})^{-\frac{1}{2}}$ by the Generalized Binomial Theorem.

Edit:  Notice that ${2n \choose n}= \frac{4n-2}{n}{2n-2 \choose n-1}$.  The binomial expansion fits this exactly, I think.

i got it from plugging in a few values instead of 3 (i.e. 4,5,6,...and squared that, and made them into fractions)...

[quote="bloodysloth"][quote="scorpius119"]well, it follows from
\[\frac{1}{\sqrt{1-4t}}=\sum_{k=0}^\infty \binom{2k}{k}t^{k}\]
which can be attacked using the binomial series, or can be done if there's a nice counting proof of
\[\sum_{a+b=s}\binom{2a}{a}\binom{2b}{b}=4^{s}\]
[/quote]

hmm... i see... how do you prove it with binomial series then?

or the counting proof or whatever.[/quote]

I actually don't know of any counting proof, but it appears like there should be one.

[hide="for the binomial series"]
The binomial coefficient $\binom{r}{k}=\frac{r(r-1)(r-2)\cdots (r-k+1)}{k!}$ can be defined for any $r$ and nonnegative integer $k$, and it turns out that
\[(1+u)^{r}=\sum_{k=0}^\infty \binom{r}{k}u^{k}\]
for all $u$ such that the sum on the right converges.

For the problem, set $u=-4x$ and $r=-1/2$. Now consider the coefficient of $x^{k}$: it is
\[(-4)^{k}\binom{-1/2}{k}=(-4)^{k}\cdot \frac{(-1/2)(-3/2)\cdots (-(2k-1)/2)}{k!}=2^{k}\cdot \frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2k-1)}{k!}\]
\[=2^{k}\cdot \frac{1\cdot 2\cdot 3\cdot \cdots \cdot 2k}{k!\cdot 2\cdot 4\cdot \cdots \cdot 2k}=\frac{(2k)!}{(k!)^{2}}=\binom{2k}{k}\]
proving the identity.
[/hide]

note the extended binomial theorem is just an application of taylor series


trigonometry
geometry
angle bisector
geometry proposed

Point $A$ is on the angle bisector of $\angle{XOY}$. Through $A$ construct a line that meets OX and OY at P and Q. Prove $\frac{1}{OP}+\frac{1}{OQ}$ is a constant.

$ OA=\frac{2OP\cdot OQ}{OP+OQ}\cos{\frac{O}{2}}$ so $\frac{1}{OP}+\frac{1}{OQ}=\frac{2\cos{\frac{O}{2}}}{OA}$ so iti is constant. :)


inequalities
LaTeX
inequalities proposed

Let $ x;y;z\ge 0$;and $ x\plus{}y\plus{}z\equal{}1$
Prove that
$ 27(x^3\plus{}yz)(y^3\plus{}xz)(z^3\plus{}xy)\ge 64x^2y^2z^2$

It's the following ineq of a well-know ineq :
$ (a^2 \plus{} 3)(b^2 \plus{} 3)(c^2 \plus{} 3) \ge 64 \forall a\plus{}b\plus{}c\equal{}3$

[quote="mitdac123"]It's the following ineq of a well-know ineq :
$ (a^2 \plus{} 3)(b^2 \plus{} 3)(c^2 \plus{} 3) \ge 64 \forall a \plus{} b \plus{} c \equal{} 3$[/quote]
I can prove your inequality but I can't prove the inequality of tranvanluan  :wink:

[quote="tranvanluan"]Let $ x;y;z\ge 0$;and $ x \plus{} y \plus{} z \equal{} 1$
Prove that
$ 27(x^3 \plus{} yz)(y^3 \plus{} xz)(z^3 \plus{} xy)\ge 64x^2y^2z^2$[/quote]
I proven the problem of tranvanluan but I didn't use your the problem   :wink:
this my proof
Lemma: $ r\le \frac {q^2(1 \minus{} q)}{2(2 \minus{} 3q)}$

and $ r\ge \frac {(4q \minus{} 1)(1 \minus{} q)}{6}$

$ 27(x^3 \plus{} yz)(y^3 \plus{} xz)(z^3 \plus{} xy)\ge 64x^2y^2z^2$

$ \Leftrightarrow 27r^3 \plus{} 27q^4 \minus{} 54q^2r \plus{} 125r^2 \plus{} 108qr^2 \plus{} 27r \minus{} 108rq\ge 0$

setting $ f(r) \equal{} 27r^3 \plus{} 27q^4 \minus{} 54q^2r \plus{} 125r^2 \plus{} 108qr^2 \plus{} 27r \minus{} 108rq$

the first case: $ 81r^2 \minus{} 54q^2 \plus{} 250r \plus{} 216rq \plus{} 27 \minus{} 108q\ge 0$

$ f'(r) \equal{} 81r^2 \minus{} 54q^2 \plus{} 250r \plus{} 216rq \plus{} 27 \minus{} 108q\ge 0$

$ \Rightarrow f(r)\ge f(\frac {(4q \minus{} 1)(1 \minus{} q)}{6}) \equal{} \frac { \minus{} 1}{72}(3q \minus{} 1)(192q^5 \minus{} 1808q^4 \plus{} 476q^3 \minus{} 267q^2 \plus{} 518q \minus{} 83)\ge 0$

the second case: $ 81r^2 \minus{} 54q^2 \plus{} 250r \plus{} 216rq \plus{} 27 \minus{} 108q\le 0$

$ f'(r) \equal{} 81r^2 \minus{} 54q^2 \plus{} 250r \plus{} 216rq \plus{} 27 \minus{} 108q\le 0$

$ \Rightarrow f(r)\ge f(\frac {q^2(1 \minus{} q)}{2(2 \minus{} 3q)}) \equal{} \frac {q^2(3q \minus{} 1)(9q^6 \plus{} 192q^5 \plus{} 1061q^4 \minus{} 3490q^3 \plus{} 4064q^2 \minus{} 2160q \plus{} 432)}{8( \minus{} 2 \plus{} 3q)^3}\ge 0$

we have done  :wink:

Like crazy :P
Tranvanluan's ineq is equivalent to :
$ \prod ( \frac{3x^4}{xyz}\plus{}3) \ge 64$ (1)
Setting :$ 3x\equal{}a,3y\equal{}b,3z\equal{}c$
By am-gm , we have :
$ LHS (1) \ge ( a^4\plus{}3)(b^4\plus{}3)(c^4\plus{}3) \ge \frac{1}{64}(  (a^2\plus{}3)(b^2\plus{}3)(c^2\plus{}3) )^2  \ge 64$
;)
Note :  It's better if you think more about classical ineq before use modern tech ;)

[quote="mitdac123"]$ LHS (1) \ge ( a^4\plus{}3)(b^4\plus{}3)(c^4\plus{}3) \geq \frac{1}{64}( (a^2\plus{}3)(b^2\plus{}3)(c^2\plus{}3) )^2 \geq 64$[/quote]
I think this step may not be true always, given $ x\plus{}y\plus{}z\equal{}1$. :wink:

[quote="CoBa_c_Kacka"]Sorry but why is that true (a^2 + 3)(b^2 + 3)(c^2 + 3) \ge 64 \forall a+b+c=3?[/quote]
Actually I pointed out the same thing. Use $ LaTeX$ to make your post readable. :wink:


algebra
polynomial
linear algebra
linear algebra theorems

[quote="bchui"]If the resultant polynomial ${\rm Res}(f,g)$ of two polynomials $f$ and $g$ equals to zero then we know that $f$ and $g$ has a common root. How about the meanings of ${\rm Res}(f,g)>0$ and ${\rm Res}(f,g)<0$? Can we conclude anything from ${\rm Res}(f,g)>0$?[/quote]

Taking into account the sign and degree of the highest term in each polynomial, the sign of Res indicates the number of crossings between roots of $f$ and roots of $g$ when slides one set of roots across the other  (pulling $f$ in the positive direction and $g$ to the negative).

[quote="bchui"]So, ${\rm Res}(f,f')=0$ means $f(x)=0$ has a repeated root. What does it mean for ${\rm Res}(f,f')>0$ and ${\rm Res}(f,f')<0$?[/quote]

In that case the relative position of $f$ and $f'$ on the real line is fixed by the number of real roots of $f$, because the roots interlace (Rolle's theorem).  The sign of Res is something like $(-1)^{k(k-1)/2}$  where $k$ is the number of real roots of $f$.  This is after taking into account the sign and degree of the highest degree term of $f$.

By the way, do we have a definition of "resultant" for two multinomials $f(x_{1},.\ldots,x_{n})$ and $g(x_{1},\ldots,x_{n})$?


parameterization
function
probability
probability and stats

A continuous random variable is said to have an exponential distribution with parameter $ \lambda$ if its density function is $ f(t) \equal{} \lambda e^{\minus{}\lambda t}$  ($ 0 \leq t < \infty$). If $ X_1$ and $ X_2$, which are independent random variables, have exponential distributions with parameters $ \lambda_1$ and $ \lambda_2$ respectively, find an expression for the probability that either $ X_1$ or $ X_2$ (or both) is less than $ x$. Prove that if $ X$ is the random variable whose value is the lesser of the values of $ X_1$ and $ X_2$, then $ X$ also has an exponential distribution.

Route A and Route B buses run from my house to my college. The time between buses on each route has an exponential distribution and the mean time between busses is 15 minutes for Route A and 30 minutes for Route B. The timings of buses on the two routes are independent. If I emerge from my house one day to see a Route A bus and a Route B bus just leaving the stop, show that the median wait for the next bus to my college will be approximately 7 minutes.

I have only looked at the first part.

If X = min(X1,X2) then
P[X<x] = 1-P[X>x] = 1-P[x1>x;X2>x].
Because of independence of X1 and X2, we have this equals to:
1 - P[X1>x]P[X1>x]

From here , you can easily deduce the distribution function of X and differentiating this function will give you the density function of X which is of a Exponential distribution as expected.

ok! Thanks! But what I do not understand is the second part:

density function of Route A distribution: $ \lambda_1 e^{\minus{}\lambda_1 t}$ where $ \lambda_1 \equal{} \frac{1}{15}$

density function of Route B distribution: $ \lambda_2 e^{\minus{}\lambda_2 t}$ where $ \lambda_2 \equal{} \frac{1}{30}$

Hence the density function of the minimum time by part one is:

$ f(t) \equal{} (\lambda_1 \plus{} \lambda_2)e^{\minus{}(\lambda_1 \plus{} \lambda_2)t}
       \equal{} \frac{1}{10} e^{\minus{}\frac{1}{10}t}$

But the mean of this distribution is 10 minutes and not 7 as stated in the problem. Where's my mistake?

Well, it says that you have to find the median, not the mean..


calculus
integration
trigonometry
limit
search
real analysis
real analysis unsolved

Prove that
$ \int_{0}^{\infty} \frac{\sin x}{x} \mathrm{d}x \equal{} \frac{\pi}{2}$
using real analytic methods.

I know a proof using Residue Theorem but I want to see a straightforward real analytic approach.

Thanks.

[hide]$ \mathcal{L} \left(\frac{\sin x}{x}\right) \equal{} \int_0^{\plus{} \infty} \frac{\sin x}{x} e^{\minus{}sx}\,dx \equal{} \arctan \left(\frac{1}{s}\right)$.

Therefore,

$ \int_0^{\plus{} \infty} \frac{\sin x}{x}\,dx \equal{} \lim_{s \to 0^\plus{}} \int_0^{\plus{} \infty} \frac{\sin x}{x} e^{\minus{}sx}\,dx \equal{} \lim_{s \to 0^\plus{}} \arctan \left(\frac{1}{s}\right) \equal{} \frac{\pi}{2}$.[/hide]

hello, this may help you
http://www.physicsforums.com/archive/index.php/t-22531.html
Sonnhard.

Carcul's proof requires an analogue for integrals of Abel's theorem for series - see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1409951753&t=19202]here[/url]

I've posted my own favorite version of this here many times, but am not sure how to find it. Here's the outline:

Step 1: Integrate by parts to show $ \int_0^{\infty}\frac{\sin x}x\,dx\equal{}\int_0^{\infty}\frac{1\minus{}\cos x}{x^2}\,dx.$

(The significance of step 1 is that we now have an absolutely convergent integral with a nonnegative integrand.)

Step 2: Replace $ \frac1{x^2}$ with $ \int_0^{\infty}te^{\minus{}tx}\,dt.$

Step 3: Interchange the order of integration (justified by Tonelli's theorem - nonnegative integrands).

Step 4: Do some calculus and algebra.

Thanks.

When I was reading some Fourier Analysis today, I came across a proof using the following:
$ \frac {1}{2} a_{0} \plus{} \sum_{n \equal{} 1}^{\infty}a_n \equal{} \frac {f(0 \plus{} ) \plus{} f(0 \minus{} )}{2} \int_{0}^{\infty} \frac {\sin x}{x} \mathrm{d}x$
where $ a_i$'s are the Fouriers coefficients involving the $ \cos$ terms, and $ f$ satisfies certain conditions like boundedness and integrability in $ [ \minus{} \pi,\pi]$, monotonicity in $ [\minus{}\delta,0]$ and $ [0,\delta], 0 < \delta  < \pi$.

Using that for $ f(x) \equal{} 1, \forall x$ yields our required result.

Lemma: If $ \lim_{x \to \infty} f(x) \equal{} 0$, then $ \int_{0}^{\infty} \sin x f(x) dx \equal{} \lim_{h \to 0} h \sum_{n \equal{} 1}^{\infty} f(nh) \sin nh$*.

Applying the above lemma and noting that $ \sum_{n \equal{} 1}^{\infty} \frac {\sin \theta n}{n} \equal{} \frac {\pi \minus{} \theta}{2}$ for $ \theta \in (0, 2\pi)$ give the desired result for the integral in question.

* sine can be replaced with cosine too.

romanovodka: my preferred pathway is to reverse the logic on your post #5. That is, first compute the integral in question by some other means, and then use that to establish the truth of the theorem you quote (or something similar to it).

Another proof I came across:
$ \lim I_n \equal{} \lim_{n \rightarrow \infty} \int_{0}^{\delta} \frac {\sin nx}{\sin x} \mathrm{d}x \equal{} \lim_{n \rightarrow \infty} \int_{0}^{\delta}\frac {\sin nx}{x} \mathrm{d}x$ where $ \delta > 0$
(which follows by applying the Riemann Lebesgue lemma on $ \frac {1}{\sin x} \minus{} \frac {1}{x}$ defining as $ 0$ at $ 0$).

Its easy to see that with a change of variable the RHS becomes,
$ \lim_{n \rightarrow \infty} \int_{0}^{n \delta }\frac {\sin x}{x} \mathrm{d}x \equal{} \int_{0}^{\infty} \frac {\sin x}{x} \mathrm{d}x$

We can prove easily that the limit in the LHS exists i.e. $ \lim_{n \rightarrow \infty} I_n$ exists for any $ \delta$.(use Cauchy's principle of convergence after change of variable as before)
Prove that:
$ \lim_{n \rightarrow \infty} I_n \equal{} \lim_{n \rightarrow \infty} \int_{0}^{\frac {\pi}{2}}\frac {\sin nx}{\sin x} \mathrm{d}x$
Proof:
$ \lim_{n \rightarrow \infty} \int_{0}^{\frac {\pi}{2}}\frac {\sin nx}{\sin x} \mathrm{d}x \equal{} \lim_{n \rightarrow \infty} \int_{0}^{\delta}\frac {\sin nx}{\sin x} \mathrm{d}x \plus{} \lim_{n \rightarrow \infty} \int_{\delta}^{\frac {\pi}{2}}\frac {\sin nx}{\sin x} \mathrm{d}x$
The 2nd part of the RHS sum goes to $ 0$ by the Riemann-lebesgue lemma.
So done.

We take n=2m+1 for ease in evaluation.
$ \frac {\sin nx}{\sin x} \equal{} 2(\frac {1}{2} \plus{} \cos 2x \plus{} \cos 4x \cdots \plus{} \cos 2mx)$
$ \Rightarrow \int_{0}^{\frac {\pi}{2}} \frac {\sin nx}{\sin x} \mathrm{d}x \equal{} \frac {\pi}{2}, \forall m \in \mathbb{N}$

QED.

@Kent Merryfield
I made an error in mentioning the theorem. The denominator should have $ \pi$ instead of $ 2$.
I think the proof I referred to is not circular logic. You can derive it without using value of the integral. A more standard result that is usually mentioned, I guess, is
$ \frac {1}{2}a_{0} \plus{} \sum_{n \equal{} 1}^{\infty}a_{n} \equal{} \frac {f(0 \plus{} ) \plus{} f(0 \minus{} )}{2}$ which can be derived from the theorem by substituting the value of the integral. But nonetheless, it does not make it circular.