How many square inches are in the area of a square inscribed in a circle of radius 6 inches?

Diagonal measures 12, so 2s^2=144

 s^2=72, which is our answer.

We know that if the square is inscribed in a circle then the diameter of the circle is equal to the diagonal of the square. Since the radius of the circle is 6, the diameter is 12, which means the diagonal of the square is also 12. There is a trick to solving this problem or anyone similar. If you know the diagonal of a square and you need to find the area, just square the diagonal and divide by 2. So 12(12) = 144. 144 / 2 equals 72, which is the area of the square.


geometry
projective geometry
geometry unsolved

A convex quadrilateral ABCD with AC ≠ BD is inscribed in a circle with center O. Let E be the intersection of diagonals AC and BD.  If P is a point inside ABCD such that $ \angle PCB\plus{}\angle PAB\equal{}\angle PDC\plus{}\angle PBC\equal{}90^{o}$. Prove that O, P and E are collinear.

this is an old problem,maybe it has been posted before.
anyway,u can solve it using PASCAL's Theorem,or another way is to use the properties of orthogonal circles.


A particle has velocity vector(V) and there is a force vector(F) acting upon it.
Fiind acceleration vector(a).

Wrong answer.and you must fiind Vector(a) as a expression of other 2 vectors.i repeat,a relation between vectors

Just a tiny modification of [b]Namiqfizik[/b]'s answer:
\[\vec a = \frac{\vec F}{m}\left(1-\frac{v^{2}}{c^{2}}\cos^{2}\phi\right)^{3/2}, \]
where $\phi = \measuredangle(\vec v, \vec F)$.

[b]edit[/b]: lapsus calami :(

you must have vector v too.the final answer must contain vector F and V ,m (moovement velocity)and c

$\vec v$ is in the formula as well, since $\phi$ depends on its direction.

What you should reach is vec(a)=vec(F)/m-[vec(F)*vec(v)]*vec(v)/(m*sqr(c))

I don't get your second line. There's a mistake there, it should be $\vec v \cdot \frac{\text d\vec p}{\text dt}$, but I don't get the following step. Could you explain how you got it?

yes there was a mistake.sorry.here is more explicitely.

Um, could you attach that ppt file again? There seems to be a bug of some sort, I can't download it.


You're adding $ a_2$ to itself $ n$ times?

Do you mean:

$ a_1^2 \plus{} a_2^2 \plus{} \dots a_n^2$?

Lol yeah, I remember being confused about that before. Its just
$ a_1^2 \plus{} a_2^2 \plus{} \dots a_n^2 \equal{} (a_1)^2 \plus{} (a_2)^2 \plus{} \dots \plus{} (a_n)^2$

We're just two lazy to use parenthesis. The $ _i^j$ isnt a new notation or anything. $ a_i$ is its own term, and its being raised to the second power.

tHanks thunder thats what u meant exceptit didnt come ou properly.

Q:A piece of equipment cost a certain factory Rs600,000.If it depriciates in value 15% the first,13.5% the next year,12% the third year,and so on.What will be its value at the end of 10 years ,all percentages applying to the original cost?


[hide]Treating $a$ as $x$ and $b$ as $y$, the solution can be written as 4 ordered pairs$: (19, -18) , (19, 18) , (-19, 18) , (-19, -18)$

$a^2 - b^2 = 37$ can be rewritten as $(a+b)(a-b) = 37$
Since $37$ is prime, it only has $2$ factors. Thus a and b both being integers must produce $2$ integers which are $1$ and $37$ (the two factors of $37$) as their sum or difference.

$a + b = 37$
$a - b = 1$

$a = 19, b = 18$

$a + b = 1$
$a -b = 37$

$a = 19, b = -18$

Switch the signs of the two pairs to make the sum and differences negatives so the equation becomes $2$ negative factors of $37$, which still yields $37$.
[/hide]

I don't understand the difference between the two. Should the first one be "Find [b]one[/b] such pair"?

[hide="Answer"]Second part: $a^2-b^2=(a+b)(a-b)=37$, so one is $\pm 37$ and the other is $\pm 1$. If $a+b=37$, then $a-b=1$, so we have $(a,b)=(19,18)$. If $a+b=1$, then $a-b=37$, so we have $(a,b)=(19,-18)$. We can simply switch the signs of these for the other two answers, so we have $(-19,-18)$ and $(-19,18)$.[/hide]

The first one is positive integers and the second one is integers.

[quote="mna851"]Silverfalcon, do you live in Nassau county and participate in this competition, I do, tell me if you do so I can look for you on the top 50.[/quote]

Nah.  :D

I don't even live there.. But I just know the ways to solve them.  :lol:

[hide]We factor to get that (a+b)(a-b) = 37. Because a and b are both integers, a+b and a-b must also be integers.

Since 37 is prime, one of the factors have to be $\pm37$ and the other factor has to be $\pm1$. If the factors are 37 and 1:

If a+b=37, then a-b=1, and a=19, b=18.
If a-b=37, a+b=1, a=19, b=-18.

Now let the factors be -37 and -1:

If a+b=-37, then a-b=-1, and a=-19, b=-18.
If a+b=-1, a-b=-37, then a=-19, b=18.

The positive pair is (19,18).
All pairs are (19,18), (19,-18), (-19,-18), (-19,18). [/hide]

[quote="IntrepidMath"][quote="Silverfalcon"]This is a challenge problem so if you are above this level, leave this to Getting Started.

Find the ordered pair of positive integers $(a,b)$ for which $a^2-b^2 = 37$.

Further challenge: Find ALL ordered pair of integers $(a,b)$ for which $a^2-b^2 = 37$.[/quote]
[hide]
Factor the eqn. into $(a+b)(a-b)=37$.
37=1*37.
2 cases, $a+b=37$ and $a-b=1$ OR $a-b=37$ and $a+b=37$
CASE 1:
$a+b=37$
$a-b=1$
$2a=38$
$a=19$
$19-b=1$
$b=18$
$(19,18)$
CASE 2:
$a-b=37$
$a+b=1$
$2a=38$
$a=19$
$19-b=37$
$-b=18$
$b=-18$
$(19,-18)$
----END HARD WORK----
These are the only 2 integer solutions.
The answer to number 1 is (19,18), and the answer to number 2 is (19,-18) and (19,18).
thank you and have a nice day.
[/hide][/quote]
You forgot some other possible cases...
Did you read what JesusFreak197 and 236factorial wrote?


If it is true it can surely be proved by newton/maclaurin inequalities... No time for this now

Let us put x=abc,y=bcd,z=cda,w=dab
and homogenize to a+b+c+d=1
(a+b)(b+c)(c+d)(d+a)=(ac-bd)^2+(a+b+c+d)(x+y+z+w)
and xyz+yzw+zwx+wxy=(abcd)^2(a+b+c+d)

therefore enough to prove
(x+y+z+w)^3>= 16(xyz+yzw+zwx+wxy)
but this is trivial by AM-GM. :wink:

so then this problem doesnt belong to this section Alisher,this section is for open questions:[b]An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not.[/b]
you should have posted this problem into [b]Inequalities Unsolved Problems[/b]`s section: [url]http://www.mathlinks.ro/index.php?f=51[/url] which is for [b]Problems you couldn't solve and to which you know that there is a solution (i.e. a problem from a contest, etc.) but you don't know it.[/b]...


calculus
integration
inequalities
trigonometry
Cauchy Inequality
real analysis
real analysis unsolved

Let $f:\mathbb{R}\to\mathbb{R}$ be an integrable function.Prove the following inequality:

         $\left(\int_{a}^{b} f(x)sinx dx\right)^2$ $+$ $\left(\int_{a}^{b} f(x)cosx dx\right)^2\leq\left(\int_{a}^{b} |f(x)|dx \right)^2$

I am sorry to announce you, but you should change the source and say "classical". Actually no, you should rather write "wrong inequality".

Is it wrong now? well if you say it is classical can you give me some refferences? :what?:

Making the face  :what?:  really does not impress me. And please don't ask me to give refference to Cauchy inequality (I choosed this one since it solves this problem) for example, since all of us know it very well. Of course, one can insist and call it new and even more spectacular "cool", but it remains classical.

[quote="harazi"]And please don't ask me to give refference to Cauchy inequality (I choosed this one since it solves this problem) for example, since all of us know it very well.[/quote]
Hehe, it was one of my homework problems. Apply Cauchy-Schwarz intequality,
$
\left(\int_{a}^{b} f(x)\sin x dx\right)^2 + \left(\int_{a}^{b} f(x)\cos x dx\right)^2
\leq
\left(
\int_a^b \sqrt{|f(x)|}\cdot \sqrt{|f(x)|}|\sin x|dx
\right)^2+
\left(
\int_a^b \sqrt{|f(x)|}\cdot \sqrt{|f(x)|}|\cos x|dx
\right)^2
\leq \int_a^b |f(x)|dx \int_a^b |f(x)|\sin^2 x dx + \int_a^b |f(x)|dx \int_a^b |f(x)|\cos^2 x dx
= \left(\int_a^b |f(x)|dx\right)^2
$

p.s. Is it Schwarz or Schwartz? I know they are both mathematicians, which one is the correct name here?


Hi, :blush:
I registered today
I can't write English well, sorry
...... $x=x$ oh It's cool,,
Anybody teach me english with E-mail or MSN messenger??
I'm looking for pen pal.

cool, you already know a bit of latex. Where are you really from? :)

Do you know Kimchi?
Im from Republic of Korea.
Im a middle school student, $9^{0}.5$grade.

I've got it! Are you from North Korea?

[edit] I'm lovin' kimchi! My mom makes it. Even though we are in USA.

oh no :lol:  I'm from South Korea
I don't like Kim Jeong Ill.

[quote="SplashD"][quote="lotrgreengrapes7926"][quote="nutz_for2.718281828"]I've got it! Are you from North Korea?[/quote]
I don't think they have computers in North Korea. :|  :fool:[/quote]

Is North Korea communist?[/quote]
I don't know about communist, but I know it's dictated by Kim Jeong Ill, who's caused a lot of problems last week. :noo:  :help:

[quote="Inspired By Nature"]cool, you already know a bit of latex. Where are you [b]really[/b] from? :)[/quote]
[quote="SplashD"]He's from Earth, The Galaxy
Can't you tell?[/quote]
I think you should read your signature splashD. :ninja:
click the smilie in case you're still confused.--->
[hide=" :D "] joking [/hide]

Yes NK is communist.

It sucks to be so close to them.

My parents occasionally eat kimchi but I don't wanna try. I'm cantonese btw.

I think I've seen North Korea before.  :o I was 8 or something and I went on a boat ride in that gulf thing between China and North Korea.

There was a front cover on the economist making fun of Kimmy.

HOW DO YOU NOT LIKE KIM CHI????


This isn't BS. When I started focusing on math, everything else besides maybe science just wasn't important to me anymore. If I didn't care so much about my grades (for my college application) I would've completely blown off all my other subjects to focus on my math.

[quote="moogra"]all the math people[/quote]
Do you mean current students or historical mathematical figures?  There are certainly some anecdotes about the latter, but as to the former I have a suspicion that most competent math students come from backgrounds that predispose them to care about their grades  :P

i didn't care about my grades until AFTER i started doing math

[quote="archimedes1"]I recently discovered math and now I'm finding it hard to cope with all the Honors classes while being active here.  It's hard to do homework at the same time as AoPS.  AoPS is not like MySpace/Facebook/Youtube/...; you have to actually concentrate on what you're doing on AoPS.[/quote]

Wow... that is like the same thing I was thinking.

Actually, this year, I've done far less math than I did last year as an 8th grader. Instead, I am beginning to focus a lot more on school subjects, but maybe that is due to the fact that I'm not completely adjusted to high school yet? It's impossible for me to do as much of each (school subjects + extra math) as you would like to do on a weekly basis.

[quote="moogra"]I've been hearing a lot about all the math people, or at least some having extremely low GPAs. I heard that all they do well in is math, and since they spend all their time on math, somehow their GPAs drop. I don't believe it (since good at math and good at school should be proportional), but many people say so. So is this bs?[/quote]

This is largely a myth.  There are certainly counterexamples, but many of your USAMO winners are also valedictorians.  Sure, there are some examples of math-only types who do math to the exclusion of all else and don't have any interest in grades, but most math superstars are at or near the top of their school in nearly everything (particularly science and computer science, as you might expect).

That said, to become really good at something you love, you do have to make some trade-offs.  The one I recommend is noting that maybe it's not so important to take 13 AP exams, including 9 in subjects you don't care at all about.  And if you hate playing <whatever instrument or sport here> at age 13 or 14, then maybe it's time to reasses whether or not it's worth putting 10 hours a week into it.  You have to make choices at some point. . .   If important work is suffering because of your dedication to something you enjoy, then you have to reasses: is that important work so important after all (maybe, maybe not), or is the thing you enjoy really worth spending that much time on (maybe, maybe not).

A lot of the young people who are prominently good at math (i.e. do very well on contests) are so because they can think very quickly. This is a skill that's applicable in almost every field of study, although it is often most useful in math and science (since it is usually best expressed by the ability to do calculations and come to logical conclusions quickly). But when someone is taking a large number of difficult classes in high school, it helps a lot to be able to plow through the homework quickly (often during other classes) so that there's time at home for doing more math.

That's certainly how my friends and I in high school maintained a high GPA, took many diverse AP tests, and still managed to do decently in a number of different math and science competitions.

Another possible trait is the ability to devote all free time to working. I certainly can't do this, but some people are able to come home and just spend every waking minute doing homework or studying math and science (or whatever else interests them). Even if you don't work very quickly, if you spend all of your time working, you will surely get a lot done. Personally, I don't see how someone could do this and still be happy, but I'm not going to make judgments about what's the "'right" way to go about studying and doing homework.

[quote]Personally, I don't see how someone could do this and still be happy[/quote]

... so true.


求通解公式的表达式

算法比拼中

imi00网友的直接for循环法——

public   class   test   {
public   static   void   main(String[]   args)   {
int   count   =   0   ;
for(int   i   =   1;   i   <100;   i++   )
for(int   j   =   i   +   1;   j   <   100;   j++   )
for(   int   k   =   j   +   1;   k   <   100;   k++   )
for   (int   m   =   k   +   1;   m   <   100;   m++   )
for(int   n   =   m   +   1   ;   n   <100;   n++   )
if(i   +   j   +   k   +   m   +   n   <   100){
count++;
// System.out.println(   i   +   "   "   +   j   +   "   "
// +   k   +   "   "   +   m   +   "   "   +   n   +   "   sum: "   +   (i   +   j   +   k   +   m   +   n)   );
}
System.out.println(count);
}
}
最直接最原始的算法：答案为455175，循环次数为71523114

wqs1106网友的改良方案
——如果没有数学公式可以直接计算,我想我这中方法应该算是循环次数比较少的了,有多少种组合就循环多少次,一次不多一次不少.

public       class       Test3       {
public       static       void       main(String[]       args)       {
int       count       =       0       ;
for(int   i=1;i <86;i++)
{
for(int       j=i+1;j <(88-i);       j++       )
{
for(       int   k=   j+1;k <(91-i-j);       k++       )
{
for       (int       m       =k+   1;m <(95-i-j-k);       m++       )
{
for(int       n       =   m+     1;n <(100-i-j-k-m);       n++       )
{
if(i       +       j       +       k       +       m       +       n       <       100)
{
count++;
}
}
}
}
}
}
System.out.println(count);
}
}
答案为455175，循环次数为455175。

jseasidej网友的试错法

public   class   less100
{
public   static   void   main(String[]   args)
{
int   a1,a2,a3,a4,a5;
int   sum=0;
int   time=0;
for(a1=1;a1 <=17;a1++)
{
for(a2=a1+1;a2 <=23;a2++)
{
if((a2*4+6+a1)> =100)   break;
else
for(a3=a2+1;a3 <=31;a3++)
{
if((a3*3+3+a1+a2)> =100)   break;
else
for(a4=a3+1;a4 <=46;a4++)
{
if((a4*2+1+a1+a2+a3)> =100)   break;
else
{
time++;
                                                          sum=sum+(99-a1-a2-a3-a4)-a4;
}
}
}
}
}
System.out.println( "结果是 "+sum+ "       循环次数： "+time);
}
}
运行结果：       结果是455175       循环次数：   24260

jacaranda网友的递归法——比较全面的分析啊

C#   code
using   System;
using   System.Collections.Generic;
using   System.Text;
namespace   Test1
{
        class   Program
        {
                static   void   Main(string[]   args)
                {
                        int   total=0;
                        int   num   =   0;
                        for   (int   i   =   1;   i   <   18;   i++)   //第一个数最大只能取到17
                                for   (int   j   =   i   +   1;   j   <   24;   j++)   //第二个数最大只能取到23
                                        for   (int   k   =   j   +   1;   k   <   32;   k++)   //第三个数最大只能取到31
                                                for   (int   l   =   k   +   1;   l   <   47;   l++)   //第四个数最大只能取到46
                                                {
                                                        //num++;   放到if   里面
                                                        int   m   =   99   -   i   -   j   -   k   -   l;
                                                        if   (m   >   l)
                                                        {
                                                                num++;//这里
                                                                total   =   total   +   (m   -   l);
                                                        }
                                                }
                        System.Console.WriteLine(total);
                        System.Console.WriteLine(num);
                        System.Console.ReadLine();
                }
        }
}
输出：
455175               //答案
24260                   //循环次数


While traveling home from school, Alyssa fell asleep halfway through the journey. When she awoke, she still needed to travel one-fourth of the distance that she had traveled while sleeping. For what part of the journey was she awake?

[hide]she was awake for the first $\frac 12$, so if she was awake in the last half for $x$ time, then the total amount she was asleep was $4x$ so $5x=\frac 12$ and $x=\frac{1}{10}$. She was asleep for $\frac{2}{5}$, so she was awake for ${\boxed{\frac 35}}$[/hide]

[hide]She was awake for half of the journey.  Set the part when she was asleep $x$.  If she was asleep for $x$ of the journey, she was awake $\frac{1}{2} + \frac{1}{4}x$ of the journey.  Also, $x+\frac{1}{4}x=\frac{1}{2}$ because that was the second half of the journey. Solving gives $x=\frac{2}{5}$. Therefore, she was awake $\frac{1}{2}+\frac{1}{10}$ or [size=200]$\frac{3}{5}$[/size].[/hide]

[hide]1/4x + x = 1/2, 5/4 x = 1/2, so she was asleep for 2/5 of the journey x. Therefore she was awake for 3/5.[/hide]

[quote="JesusFreak197"]You answered the wrong thing. [/quote] :fool:  :ewpu:
Edited


trigonometry
inequalities
limit
geometric sequence

find all real x, such that
cos x, cos 2x, cos 4x, cos 8x, ...
are all negative.

Did you keep in mind that cos x < 0 for 90 + k.360 < x < 270 + k.360?
[k is an integer]

By the way, cos 2x = 2 cos2 x - 1.

I really hope that you have a solution for a beginner or that you joke. As far as I remember, this is a russian problem and from what I've seen it's not at all easy. I would like to see a solution for a beginer!

the problem is asked by one of my classmates.. and my teacher didn't solve it. But i don't think it is difficult.

If the story I heard about it is true, then it is really difficult. I heard that when it was proposed on the russian olympiad, it was marked as a very difficult problem.

Because of the geometric progression of the number in front of x, wouldn't it just keep cycling around and around?  I don't think you could find a value (which would need to be between -pi/2 and pi/2, where cosine is negative) such that multiplying it by those numbers would keep the cosine negative

It isn't that simple, and the set in question is not empty. An example of such a number: $\frac\pi3.$

[hide]I think its just a matter of solving these two inequalities (if |cosx| is more than 1/2 then cos2x > 0):

[tex] \displaystyle 0 > \cos{x} \ge -\frac{1}{2} \\
|2\cos^2{x}-1| \le \frac{1}{2} \\ [/tex]

Solving the second inequality, we have [tex] \displaystyle \frac{\pi}{6} \le x \le \frac{\pi}{3} [/tex]. However, including the first inequality we know that [tex] \displaystyle \frac{\pi}{2} < x \le \frac{2\pi}{3} [/tex] or [tex] \displaystyle \frac{4\pi}{3} \le x < \frac{3\pi}{2} [/tex]. So then there are no solutions, right? [/hide]

btw, [tex] \displaystyle \cos{\frac{\pi}{3}} = \frac{1}{2} > 0 [/tex] so that doesn't work.

EDIT: nvm, i thought about it, my method is not right...

[hide="Hint"]Look at the base 2 representation of $\frac{x}{2\pi}$.[/hide]

I think jmerry meant $\frac{2\pi}{3}$.

[quote="Ravi B"][hide="Hint"]Look at the base 2 representation of $\frac{x}{2\pi}$.[/hide]
[/quote]
Cute! (I read the hint!)

I think I have a solution.but I'm not sure about it.can somebody check it for me :) ?
we can know $ \cos x,\cos 2x, \cdots ,cos 2^{n}x,\cdots $ are all negetive is equal to $ \cos x,\cos \frac{x}{2}, \cdots ,\cos \frac{x}{2^n},\cdots $ are all negetive.
so we can get $\cos x<0$
$-\frac{\sqrt{2}}{2}<cos \frac{x}{2}<0$
$\cdots $
let ${a_0=0,a_{n+1}=-\sqrt {\frac {a_{n}+1}{2}}}$
we can get $a_n<cos \frac{x}{2^n}<a_{n-1}$
because $\lim_{n\rightarrow \infty } a_n= -\frac{1}{2}$
then we get $cos \frac{x}{2^n}= -\frac{1}{2}$
so $\cos x= -\frac{1}{2}$
am I right?

This is where I posted my solution: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=343568#p343568[/url]
I think the proof is deeply related to the base 2 representation of $\alpha/2\pi$.


What are the last two digits of $2^{66}-4$?

[hide]I think he meant $2^{66}-4$.

You should try to realize people mistakes and then solve the question.

We see that there is a pattern every 20.  I won't type them all out, so this means we are just looking at $2^{6}-4=62$.

$62$[/hide]

[quote="SplashD"]do you mean $2^{66}-4$? :huh:[/quote] yes I did.
and mathgenius is wrong.

[hide]Yes, I am wrong.

I menat to say 64-4=60, not 62.

So yes, $60$[/hide]

make a table of this and you can see tat there is a pattern among the answers

(no, i will not do latex)

s= square
v= value

s v
0 1
1 2
2 4
3 8
4 16
5 32
6 64
...

after all tat, u can see tat the pattern tat 2^66-4=...60

[quote="bpms"]There is a way to do this without the pattern ;)[/quote]

yeah, all-out multiplication

I made it so that this is easily factored...
What is Euler's totient theorem...

It is just a fancy name for Euler-Fermat, I believe.

It just uses the $\phi$ function, in it.

omg i dont kno any of these stuff

wat r these theorems u speak of

[quote="mathgeniuse^ln(x)"]I think you mean Fermat's Little Theorem.

Last would be useless in this problem.[/quote]

oh right, lol

[quote="bpms"][quote="slamchess"][quote="mathgeniuse^ln(x)"]I think you mean Fermat's Little Theorem.

Last would be useless in this problem.[/quote]

what are those two fromulas?[/quote]
Fermat's last $a^{x}+b^{x}$isn't equal to$c^{x}$, for any natural numbers a, b, c, and x if x>2.
Fermat's little $a^{p}=a(mod\ p)$ if a is prime.[/quote]

fermat's little sounds important

thnx bpms

[quote]fermat's little sounds important [/quote]

but worth barely anything in mc

[quote="akalra1"][/quote][quote="bpms"][quote="slamchess"][quote="mathgeniuse^ln(x)"]I think you mean Fermat's Little Theorem.

Last would be useless in this problem.[/quote]

what are those two fromulas?[/quote]
Fermat's last $a^{x}+b^{x}$isn't equal to$c^{x}$, for any natural numbers a, b, c, and x if x>2.
Fermat's little $a^{p}=a(mod\ p)$ if a is prime.[/quote][quote="akalra1"]

fermat's little sounds important

thnx bpms[/quote]
Please, please, please do not use chatspeak. I find it very hard to read some your posts because of the chatspeak. Thanks.  :wink:

[quote="moogra"]1:2
2:4
3:8
4:16
5:32
6:64
7:28
8:56
9:12
10:24
11:48
12:96
13:92
14:84

ok yo uget the point (it does go to 20)
64-4=60[/quote]
My solution has alot less computations and doesn't require knowledge of the pattern.

$2^{66}-4$
First take the 4th root of $(2^{6}6)$ because $2^{x}$ is a 4peat.

Then you get $2^{2}$.

Finally the units digit will be $2^{2}-4=0$

So 0 is the units digit.

Next we have to figure out the tens digit.

Hmm, gimme some time im not sure if there is a pattern to find or not.

I found out that the tens digit of the number has a pattern of every 20.

So then it wouuld be $2^{6}$ which is 64 so its tens digit is 6 and units digit is 0. So 60.

PS i didnt read it right i thought it was the units digit not the last 2 digits


calculus
integration
trigonometry
limit
real analysis
real analysis unsolved

Or can you show

$ \int_0^\infty \frac {2x\cos \pi x^2}{e^{2\pi x} \minus{} 1}\, dx \equal{} \frac {2 \minus{} \sqrt {2}}{8}$

$ \int_0^\infty \frac {2x\sin \pi x^2}{e^{2\pi x} \minus{} 1}\, dx \equal{} \frac {\sqrt {2}}{8} \minus{} \frac {1}{2\pi}$

$ \int_0^\infty \frac {2x\cos 2\pi x^2}{e^{2\pi x} \minus{} 1}\, dx \equal{} \frac {1}{16}$

$ \lim_{N\in\Bbb{N}} \int_0^N 4x\, \cos \pi x^2\, \sin(2\pi ux)\, dx\equal{}\sqrt{2}\, u\, \left(\sin\pi u^2\minus{}\cos \pi u^2\right)$

$ \lim_{N\in\Bbb{N}}\int_0^{2N} \frac {2x\sin \pi x^2}{1 \minus{} e^{ \minus{} 2\pi x}}\, dx \equal{} \frac {\sqrt {2}}{8} \minus{} \frac {1}{2\pi}$

$ \lim_{N\in\Bbb{N}}\int_0^{2N \plus{} 1} \frac {2x\sin \pi x^2}{1 \minus{} e^{ \minus{} 2\pi x}}\, dx \equal{} \frac {\sqrt {2}}{8} \minus{} \frac {3}{2\pi}$

$ \lim_{N\in\Bbb{N}}\int_0^{N \plus{} \frac12} \frac {2x\sin \pi x^2}{1 \minus{} e^{ \minus{} 2\pi x}}\, dx \equal{} \frac {\sqrt {2}}{8} \minus{} \frac {1 \minus{} \sqrt {2}}{2\pi}$

$ \lim_{N\in\Bbb{N}}\int_0^{N} \frac {2x\cos \pi x^2}{1 \minus{} e^{ \minus{} 2\pi x}}\, dx \equal{} \frac {2 \minus{} \sqrt {2}}{8}$ ?

And what is $ \lim_{N\in\Bbb{N}}\int_0^{N \plus{} \frac12} \frac {2x\cos \pi x^2}{1 \minus{} e^{ \minus{} 2\pi x}}\, dx$ ?

[quote=Doe John]Can you show

$ \int_{\minus{}\infty}^\infty \frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\, dx\equal{}\frac{2\minus{}\sqrt{2}}{2}$ ?[/quote]

https://math.stackexchange.com/questions/61605/tough-integral-involving-sinx2-and-sinh2-x

Let's evaluate $ \int_0^\infty \frac {2x\sin \pi x^2}{e^{2\pi x} - 1}\, dx $
$$I=\int_0^\infty \frac {2x\sin \pi x^2}{e^{2\pi x} - 1}\, dx =\int_0^\infty \frac {4x\sin (\pi x^2/2)\cos (\pi x^2/2)}{e^{2\pi x} - 1}\, dx =\frac{2}{\pi}\int_0^\infty \frac {d\Bigl(\sin^2(\pi x^2/2)\Bigr)}{e^{2\pi x} - 1}\, =4\int_0^\infty \frac {\sin^2(\pi x^2/2)e^{2\pi x}}{(e^{2\pi x} - 1)^2}\, dx$$
$$=\int_0^\infty \frac {\sin^2(\pi x^2/2)}{\sinh^2\pi x}\, dx=\frac{1}{4}\int_{-\infty}^\infty \frac {1-\cos\pi x^2}{\sinh^2\pi x}\, dx=\frac{1}{4}\Re \,J\,, \,\,\text {where} \,J=\int_{-\infty}^\infty \frac {1-e^{i\pi x^2}}{\sinh^2\pi x}\, dx$$
$x=0$ is a removable singular point, therefore we can consider both terms in the principal value sense
$$J_1=\int_{-\infty}^{-r} \frac {1}{\sinh^2\pi x}\, dx+\int_{r}^{\infty} \frac {1}{\sinh^2\pi x}\, dx=-\frac{1}{\pi}\Bigl(\coth |_{-\infty}^{-r}+\coth |_{r}^{\infty}\Bigr)=-\frac{2}{\pi}\,+\,\frac{2}{\pi^2 r}$$
Following the approach [url=https://math.stackexchange.com/questions/2717244/is-there-a-rapider-or-more-elegant-way-to-evaluate-int-0-infty-frac-cos?noredirect=1](here)[/url] we introduce function $f(x)=\frac{e^{\pi i x^2}e^{\pi x}}{\sinh(2\pi x)\sinh(\pi x)}$. It is easy to check that $f(x)-f(x+i)=\frac{e^{\pi i x^2}}{\sinh^2(\pi x)}$
Next, we integrate along rectangle in the comples plane: from $-R$ to $R$, from $R$ to $R+i$, from $R+i$ to $-R+i$ and from $-R+i$ to $-R$ ($R\to\infty$). We also add two small semi-circles of radius  $r\to0$ (going clockwise) above $x=0$ and below $x=i$ - to close the contour.
$$J_2=\int_{-\infty}^\infty \frac {e^{i\pi x^2}}{\sinh^2\pi x}\, dx=\int_{-\infty}^\infty \bigl(f(x)-f(x+i)\bigr) dx=\oint\frac{e^{\pi i x^2}e^{\pi x}}{\sinh(2\pi x)\sinh(\pi x)}dx -\int_{C_1}-\int_{C_2}-(1)-(2)$$
where $\int_{C_1}$ and $\int_{C_2}$ are integrals along small semi-circles, and (1) and (2) - integrals along right and left path (from $R$ to $R+i$ and from $-R+i$ to $-R$; (1) and (2) $\to0$ as $R\to\infty$).
We have a single simple pole inside the contour at $x=\frac{i}{2}$,  therefore
$$\oint=2\pi i Res_{x=i/2}\frac{e^{\pi i x^2}e^{\pi x}}{\sinh(2\pi x)\sinh(\pi x)}=2\pi i\frac{e^{-\pi i/4} e^{\pi i/2}}{-2\pi i}=-\frac{1+i}{\sqrt2}$$
$\int_{C_1}=-\frac{1}{\pi^2r}-\frac{i}{2}\,$ and $\int_{C_2}=-\frac{1}{\pi^2r}-\frac{i}{2}$
$$J_2=-\frac{1+i}{\sqrt2}+\frac{2}{\pi^2r}+i$$
Taking all together
$$I=\frac{1}{4}\Re\,J=\frac{1}{4}\Re\,(J_1-J_2)=-\frac{1}{2\pi}+\frac{1}{4\sqrt2}$$

As a bonus,
$$\int_{-\infty}^\infty \frac {\sin{\pi x^2}}{\sinh^2\pi x}\, dx=\Im \int_{-\infty}^\infty \frac {e^{i\pi x^2}}{\sinh^2\pi x}\, dx= -\Im J=\Im J_2=1-\frac{1}{\sqrt2}$$


logarithms
calculus
calculus computations

If exist lim and equal a, then $ e^a\equal{}\frac{a^2}{1\plus{}ln n}\plus{}o(1)\to a\equal{}\minus{}\infty$.

[quote="Rust"]If exist lim and equal a, then $ e^a \equal{} \frac {a^2}{1 \plus{} ln n} \plus{} o(1)\to a \equal{} \minus{} \infty$.[/quote]

Why  :wink:

[quote="Rust"]If exist lim and equal a, then $ e^a \equal{} \frac {a^2}{1 \plus{} ln n} \plus{} o(1)\to a \equal{} \minus{} \infty$.[/quote]

Would you care to explain if you let $ n \rightarrow \infty$, how come $ \frac{1}{1\plus{}\ln n}$ sticks? Thank you! :)


calculus
integration
trigonometry
limit
logarithms
calculus computations

Calculate the integral:

\[ \int\limits_{0}^{\infty}\int\limits_{0}^{\infty}\frac{\sin^{3}(x\plus{}y)}{(x\plus{}y)^{3}}dxdy.\]

My first thought is to apply the change of variables $ x\equal{}uv,y\equal{}u(1\minus{}v),$ so $ 0<u<\infty,$ $ 0<v<1,$ $ dx\,dy\equal{}u\,du\,dv,$ and $ x\plus{}y\equal{}u.$ This leads to

$ \int_0^{\infty}\frac{\sin^3 u}{u^2}\,du.$

Well, it's a first step. I don't see the next step yet. At least it's clear that the integral converges absolutely.

A change of variables $ (u, v) = (x + y, x - y)$ would give the same result $ \int_{0}^{\infty} \frac {\sin^3 u}{u^2} \, du$. Then we can do like this:

\begin{eqnarray*} \int_{0}^{\infty} \frac {\sin^3 u}{u^2} \, du & = & \int_{0}^{\infty} \sin^3 u \left( \int_{0}^{\infty} t \, e^{ - ut} \, dt \right) \, du \\
& = & \int_{0}^{\infty} t \left( \int_{0}^{\infty} \sin^3 u \, e^{ - tu} \, du \right) \, dt \\
& = & \int_{0}^{\infty} t \left( \int_{0}^{\infty} \frac {3\sin u - \sin 3u}{4} \, e^{ - tu} \, du \right) \, dt \\
& = & \int_{0}^{\infty} \frac {t}{4} \left( \frac {3}{1 + t^2} - \frac {3}{9 + t^2} \right) \, dt \\
& = & \lim_{R\to\infty} \int_{0}^{R} \frac {3}{8} \left( \frac {2t}{1 + t^2} - \frac {2t}{9 + t^2} \right) \, dt \\
& = & \lim_{R\to\infty} \frac {3}{8} \left[ \log\left( \frac {1 + t^2}{9 + t^2} \right) \right]_{0}^{R} \\
& = & \frac {3}{4} \log 3 \end{eqnarray*}


Hi guys. I will try to keep this as simple as possible, so that it is more or less comprehensible, but I am afraid this is not going to work very well. I will appreciate a lot if you could read this till the end and advise me what to do.

For past year-year and half or so I have been having some advances in 'logistics' (if that's the correct word). I gradually improved my analytical and logical skills by questioning the reasons behind various happenings and things in the world around me. I managed to discover some interesting things, for example:

-That people do pretty much everything to achieve some sense of comfort, on conscious or sub-conscious level.

-Believing that there is no God (basically being an atheist) is not 'smarter' than believing in one since both sides lack evidence to support their opinions (hence making the debate pointless).

-That people who oppose racism, sexism, and religionism based on an ideology of treating people fairly regardless of their 'predefined' characteristics (or 'predefined' in some cases, such as religion) should be opposing the idea of nationalism (putting one nationality/ethnicity above/below other(s), stereotyping based on their nationality/ethnicity etc.), or they are illogical - it's quite easy to find a person with 'Western moral values' who is anti-racist/sexist/etc. but is yet patriot of his own country and 'own people', as opposed to being patriotic of the whole planet Earth.

-Moral values vary from society to society not because one is 'wrong' and the other one is 'correct', but because different groups of people have different aims. If one aims to build a society with mutual respect and peaceful coexistence between its members, so that this society can spend time in more meaningful way (exploring unknown, i.e. space) that it is now (wars?) then it perfectly makes sense to advocate Unity in Diversity, respect to people regardless of their skin color and similar attributes, and so on.

These are just several things I picked, and they may sound not as convincing as they originally did (I will mention later why exactly, and that's actually the reason I created this topic). But in a nutshell, I progressed a lot over the last year or two. If 5 years ago I would call a "you give money to poor people only to make yourself feel better by helping people on a sub-conscious level" absurd, I would give it a full consideration and analyze it now. If 5 years ago I would argue just to defend my point regardless of it being logical or not, now I would use any argument as a source of learning, and I would change my opinion if presented some logical evidence against it. I came to understand that people are not equal, they are different, but that's what makes us so unique, and what we should embrace, and that people should get fair treatment regardless of their pre-defined attributes/relation to some group that may cause stereotypes, and should rather be judged (if ever) on their own views and opinions, as opposed to the things mentioned earlier. I have learned to prevent my own nationality/previous opinions/race/gender/etc. from affecting my thinking, as I tried to look at things from a perspective; now I can understand virtually any opinion/action (by that I mean understanding the reasons that led to it), without necessarily supporting it. Some people called it an emotionless 'computer' approach, but I enjoyed it because this 'neutrality of opinion' led me to explore and understand A LOT. I have explored the 'logistics' further, and was very disappointed to find out that logic is not absolute either - it relies on basic axioms used to build a 'tree'. For example, for most people 'human beings are born with a right of being treated fairly' sounds logical, but there is no proof that it actually is logical, since it's pretty much a fundamental statement/axiom we use to build more complex 'trees' (there probably is a way to prove it using even more fundamental statement(s), but I am not going into there because I only wanted to show a basic example). So, based on  the aforementioned statement, one can say 'gay people are born with a right of being treated fairly, since they are human beings as well.'

I have just re-read what I wrote, and it has a lot of 'I's. Oh well.

Long story short, after waking up about three days ago, I realized that my 'skills' were gone. While I enjoyed logical debates before this incident, now I started to encounter a 'mental block' each time I tried to have a similar conversation.

[quote]Before:

Person x: Abortion is wrong because it is against human rights.
Me: The pregnant person is also a human being, and since the baby is in her body, she has rights what to do with her body too
Person x: Killing fetus is immoral because you by doing so you prevent birth of a potential human being
Me: Then, would you consider using condoms immoral, since they do the very exact thing?

Now:

Person x: Abortion is wrong because it is against human rights.
Me: Uhh... Oh... *DEAD END*.[/quote]

Since there are a lot of math-oriented people, I will try to explain it this way - before 'waking up' on that day, I could foresee several possible solutions/answers to the debate question - my mind was automatically analyzing everything faster that I spoke, so I could reply instantly, at the same time predicting several replies I could get back from the person I am talking to, and prepare answers for them too, etc. Kind of like in a chess game - if you succeed in 'brute forcing' your opponent and foreseeing his combinations farther than he does yours, you win. Or at a math olympiade, you know the theory needed to solve the problem, but since you need to work fast, you need to 'load' the problem in your head and figure out where to approach the problem from, sometimes trying a certain solution intuitively, noticing something specific, or so. Up until this time, this worked well - I could 'load' pretty much anything, and then analyze and discuss it immediately. Now, it's the exact opposite - when I try to debate, I open my mouth expecting the ideas to flow as usual, but it's not happening.

As mentioned above, I became anti-patriotic/anti-nationalistic after analyzing and deciding that nation/ethnicity are not more important than race/gender in any way, and should not be used to discriminate people. I started to strongly dislike pretty much any nationalistic symbols after this. I couldn't stand patriotic brainwashy ads on TV or flags and national anthems sung in solemnity (why saying I am proud to be white is OhMyGod, but saying I am proud to be from [insert country's name] is great?) After waking up on [b]that morning[/b], I didn't care. I turned on the tv, found a nationalistic ad, and was surprised to notice that I didn't feel any disgust. Same reaction followed when encountering several pretentious people (I prefer, or should I say used to prefer, simple/down-to-earth people, and dislike pretentiousness and pompousness). Now, I didn't care at all - some kind of apathy. I asked several friends to debate with me on various topics - I am still able to argue 'logically' about the topics we have covered before, mainly because I remember what I used to say to respond to their claims, but if we debate about something new - I am TOTALLY lost. It took a while for some of my friends to realize I am not kidding or pretending. I have tried some iq tests and math problems -- my performance on those didn't change. I have tried to read several pages, and it looks like while my reading speed didn't change, my comprehension is lagging - I can't immediately grasp the hidden points as I used to.

Recalling Leonardo Da Vinci's quote 'There are three types of people, those who see, those who see when shown, and those who do not see', I can say that I moved from the first category to the second one, for no apparent reason. And that's what terrifies me.  I can still understand the previous logical debates I had, debates of other people, but I can no longer 'see' or analyze as fast as I used to. If I take significantly more time to think, I might get it, but might encounter a mental block.

Rereading the whole thing I wrote, I now recall Einstein's 'If you cannot explain it simply then you do not understand it well enough.' What I wrote is not simple for sure, and several days ago I would have been able to fit this in 3 times less symbols.

Do any of you guys have any idea what I could do?  :maybe: Try brain food such as fish oil?  :lol: I am feeling both weird and funny at the same time - I have always thought this happens only in movies. I feel like a basketball player who woke up and realized that he can't dribble for his life, or like a math major who has realized he can't solve anything beyond arithmetic. And as I said before - I didn't do anything different that could have possibly lead to a change in me, it just suddenly came on me, and now I can't even find logical solutions to everyday arguments such as 'omg my boyfriend doesn't spend enough time with me' which would be extremely easy for me ~3-4 days ago. Should I treat this as if I was 1-2 years ago, and try to begin everything from level 0? Thanks for reading this until the end. :)

:lol:  I didn't have the patience to read your entire post, so I can't label it (it is unfair too) is as being less "productive". I think you should be prepared to face some "snark" and brickbats from our distinguished moderator [b] JBL [/b].  :D

[quote="Gen8"] I think you should be prepared to face some "snark" and brickbats from our distinguished moderator [b] JBL [/b].  :D[/quote]
I think you should stop being bitter and passive-aggressive and think about why you get negative responses to your posts.  (Were you trying to say "bric-a-brac"?)

As for the OP's question, your brain may just be tired out.  If rest doesn't work, you might want to see a doctor; this might be a neurological symptom of some disorder.

[quote] I think you should stop being bitter and passive-aggressive and think about why you get negative responses to your posts. (Were you trying to say "bric-a-brac"?) [/quote]

[b] @ t0rajir0u [/b] : I never expected such a knee jerk reaction from you! Well, I was not bitter or passive-aggressive as you may think. I didn't understand what you meant by -- "get negative response to your posts". (I acknowledge that some of my queries in the Round Table were highly hypothetical or sometimes silly in nature, so, I might not have got fruitful/elaborate replies)

I was saying "brickbat". You may want to refer to the link below:

[url]http://www.thefreedictionary.com/brickbat[/url]

I was saying that "posts" that exhibited the [b] OP [/b] "post pattern" (lengthy, etc)were sometimes mocked.

[quote="Gen8"]I was saying that "posts" that exhibited the [b] OP [/b] "post pattern" (lengthy, etc)were sometimes mocked.[/quote]
I'm not sure what posts you're referring to that didn't have other problems.

Maybe a little of topic?

I am confused too. I can't think of a possible reason. You probably could have a week ago  :lol: .

After thinking about it, i have to ask how did it start? Did it start overnight or gradually?

Ignore Gen8.

This skill is incredible. To tell you the truth I envy you. :)
Did you do anything different on that fateful night? Change of diet, living conditions, etc.?

[b] @ 3100 : [/b] Hey! My comment was not meant to offend you (or anything like that), I was just cautioning you. If I hurted you, I apologize.  :(

[b] @ 1=2: [/b] I haven't done anything "offensive", so it would be better if you desisted from giving unwarranted suggestions (The first line of your post) :ninja:

Let's put an end to this meta-discussion, shall we?  Those of you brave enough to read through the entire treatise (I am not among your number) and come out believing there's something worth saying in response, feel free to carry on with actual discussion.

It's ok, I am not mad or anything. I am afraid there is nothing to be jealous of -- the 'skill' is gone :(. I woke up in the morning, did my daily routines, then when talking to my friend sometime after noon I realized that I couldn't find a simple solution for 'omg my bf doesn't want to spend enough time with me' or something along the lines. I then started to hectically recall some logical puzzles to see if something is wrong with my mind, and I encountered 'mental blocks' when trying to solve them at some point. By 'mental block' I mean some feeling of laziness coupled with unease of thinking about any of the problems. I can't think of doing something differently, I tried to recall all the 'different' things I have done, and the only thing that comes to my mind is physical workouts which I began to do. Those were rapid, but not really tiring, and I stopped doing them as soon as I got these 'mental blocks' to check if there was any relation - no improvements so far. I am not sure it's because my brain is tired out either - I wasn't really loading it over in last year, and all the analysis I have done up until this point was more like a hobby - it was very easy to do, and I didn't even need to concentrate much. I probably would have to concentrate and tire myself if I would go in-depth trying to find the very fundamentals of logic and how they are related to human psychology, but I wasn't going there just yet (and I can't imagine it being that tiring either). The fact that it happened rapidly as opposed to gradually is the one that frightens me the most - if it was something gradual, like me getting 'dumber' for past 2 months, then I might have been able to think of a reason. But this way it just left me totally puzzled. The irony is that if it was a physical disability of some sort then I could have asked a doctor to help me, I can't imagine psychologist listening to me seriously and advising me what to do... I am actually more [b]normal[/b] person now than I was several days ago. The only places I can ask for ANY type of advice is probably some high iq society board or this forum, and hope for a miracle. :maybe: Losing my mind has been one of the things I am actually afraid of, and I guess it happened now. I will try consuming some 'brain food' such as fish oil, nuts, etc. and try to treat this situation as if I was 2 years back from now, and had to build all of my 'skills' from zero.

I can tell you that if you waste money seeing a shrink, now would be the moment to stop doing so.

I highly doubt your problem is a 'simple' physical deal...I would not venture to say that it is purely neurological, or a direct reprecussion of eating patterns.

I think you are lost. But not in the way you think you are. Lemme explain..

Have you ever looked back at a time in your childhood, i.e., watched home videos, read something you wrote back then with all ernesty, but now you realize that it was just a simple thought from a simpler time? When you refocus things, put things into a new perspective, your point of view is bound to change.  Oftentimes, this takes a rather long duration of time, a new accumulation of life experiences, etc. This is one possibility....I would guess that you've come to look back at your opinions, no matter how recently formed, and you've started to question.

I'm guessing that you're questioning your logic, your insight, your clear-mindedness that was 'second-nature' until that faithful morning when, *sound effect*, everything went KABLOOEY!

Questioning is good if it doesn't completely rip to shreds what little sanity human beings have to start with.

---

But at this time of the post, I think you should take some deep breaths, and take a break for a while, and not freak out about this. That was a run-on sentence, but don't let that distract you from the relatively sound advice.  You can't hear it, so, assume it's sound advice.  Try not to think about this as a crisis of sorts, just a 'phase' that you're going through.  Try to relax and do other things besides well reasoned arguments.

---

So now, perhaps you [i]should[/i] start back at 'level 0'.

I guess, in essence, what I'm trying to say is that you requestioning everything you believe in doesn't necessarily indicate something is wrong with you.  It's part of this inevitable little process called change, and it's perfectly normal.  I think you should think about your current confidence level, and re-evaluate priorities and experiences that led to your 'breakthrough' conclusions in the first place.  I think you would find that helpful, and although it would not entirely stop your mind from playing tricks on you, it's where I would start.

Alright. Ending my ramblings. Lemme know how that goes for you. :)

Thanks, The Scintillator. You might be correct, although I would expect a change in my views to occur much less rapidly. I have tried to pull my stuff together and analyze all the possible ways to get back on track (I don't worry about the cause anymore, hoping it doesn't happen again), and it looks like (surprisingly) SAT prep back in Nov-Dec-Jan has helped me quite some bit. I am talking about reading passages as a 'robot' and neglecting the common emotional reactions you get in favor of more analytical, 'computer', critical reading. Seems to be helping me a little by little -- who would ever think standardized testing can be of such use. :)

Here's a lossless and much more readable paraphrase of the original post. I would advise the OP to replace his/her original post with this so that people might actually care to read the thread. Note that I used hide tags for a quasi-relevant part of the summary.

[quote]For past year-year and half or, I’ve gradually improved my analytical and logical skills by questioning/pondering about the world around me. [hide="Click here to see some of my conclusions so far"]- Almost everything people do is in pursuit of physical, emotional/moral*, or spiritual comfort, whether or not they are aware of it. That is, people are selfish.
- Atheists should shut up since they are equally lacking in evidence.
- Nationalism is a form of discrimination, just like racial/gender/religious/orientation discrimination
- Morality is only relevant within a particular societal/cultural context. We can’t say anyone’s morals are right or wrong; that would be discrimination.[/hide]

Three days ago, I started getting a mental block every time I get into a debate. When I try to debate, I open my mouth expecting the ideas to flow as usual, but it's not happening. I can hold familiar debates, since I remember the points I’ve used in the past, but I can’t think of new points on the fly.

As mentioned above, I became anti-patriotic/anti-nationalistic after analyzing and deciding that nation/ethnicity are not more important than race/gender in any way, and should not be used to discriminate people. I became very anti-nationalist. Now, I feel apathy – I don’t care anymore.

I’ve tried IQ tests and math problems, and my performance hasn’t changed. But when I read, it takes longer for me to grasp the underlying meaning of the author.

What do I do?[/quote]

[quote]
As mentioned above, I became anti-patriotic/anti-nationalistic after analyzing and deciding that nation/ethnicity are not more important than race/gender in any way, and should not be used to discriminate people. I became very anti-nationalist. Now, I feel apathy – I don’t care anymore.
[/quote]

I'm antipatriotic and antinationalist too....
I am Greek but i'm feeling that my homeland is the earth and not Greece....I love Greeks the same as Albanians,Indians,Italians etc...
I hate when people believe that they are above all the others because of their nationality and i strongly believe that ''when your sleeping the night patriot you wake up fascist''.

BUT i cant understand what do you  mean that you feel apathy.You feel apathy against...?

On the other hand when i read about these ideas and make them my own i became more sensitive about :
social inequalities, refugees etc and i'm 'fighting' to pretend their rights....
So just make to yourself clear what you are believe and why you believe this and you will feel better :wink:

I cant expreess myself as i would like because of my mediocre English but i believe that you can understand what i want say...
Dimitris

About how time consuming was this hobby of yours?
It could be overload: you have become obsessed with this certain hobby and tried to learn as much as possible so that you burnt yourself out.

What types of discussions are you having with your friends now? Are they more complex than before? Perhaps you don't know where to go with your "guideline" for thinking that you outlined in your testimony. Maybe this is an indication of necessary adaptation: experience/read more of life so that you can improve your former assumptions, because maybe you have outgrown your conclusions as scintillator pointed out.

Good luck!

[quote="3100"]
-Believing that there is no God (basically being an atheist) is not 'smarter' than believing in one since both sides lack evidence to support their opinions (hence making the debate pointless).
[/quote]
the average atheist, at least among the people i know, does not specifically believe that thier is not a god, the just don't specifically believe in any god.

also the existence of no gods is more likely then the existence of gods by occam's razor since it is a simpler system

Play tennis or something? No? Well, let's pretend you do.

Match point, ad-out, your serve, what's the worst thing that could happen? You realize what situation you're in. - oh crap if i miss the first serve which i've been doing at a frequency of about 60% this game (goddam first serve's not working) then i'll have to rely on my second serve which he's been constantly pounding down the alley but what if he takes it cross-court and catches me off guard but crap i haven't even served yet i should do that but etc.

I don't have some secret window into your life, every situation is different, and so I may be off. In the end, only you can properly diagnose the problem because only you are completely immersed in its context. Which is the problem. I think you are being too self-conscious about your lack of logic. Instead of tuning your mind into solving the problem, you tune your mind into why you aren't solving the problem. You realize the situation and then your mind pops up a level and focuses at a paradigm it shouldn't be focusing on. You may not realize you're doing this - certainly the test problem you give yourself after reading this post will be all too self-aware about your self-awareness - but it's happening on some level which is fundamentally inhibiting the process. I can only see it be more affected when that is the unconscious level, which is usually reserved for grabbing different associations between ideas but in your case is too focused on how it should be grabbing ideas. Started off simply enough - you tried to solve a problem, noticed your ability, and from then on you always noticed your ability (and now lack thereof).

How to fix this, if this is indeed the case? Other than putting yourself into a Skinner box, I don't know. I'm afraid that a diagnosis probably only worsens the symptoms. You can try convincing yourself that you don't have an ability. This is why so many people are so humble concerning themselves (e.g. omg i failed OR ugly solution, above is better, how did you think of that?) By consciously deflecting their thoughts away from how strong their abilities are - or used to be, in your case - they don't ever fall into the trap of self-awareness. The mindset is: how good you are is trivial, what matters is the solution. So convince yourself you never had anything, and it might just come back.

I think you'll figure yourself out with time and with experience living life. If that's not fast enough for you, then go out and try something new. Maybe you'll figure yourself out better. I feel like you're too focused on tapping this one, hyper-analytical part of yourself, and have a very narrow idea of how your analytical mind can contribute to your life.

I remember my mathematical abilities sort of stagnated after 10th grade, when I stopped having time to invest into reading math books and such. A huge boost in my competition math abilities came, to my surprise, from doing mock trial. Going back and connecting the dots, it all makes sense. Another example -- I have yet to have a technical job. I remember, in particular, working at a courthouse the summer after my senior year, where I helped people file claims/motions/cases, which involved some customer service type stuff and giving out some legal advice. Because this involved a lot of dealing with lots of people and communicating analytical thoughts to less analytical people, I developed these skills, which I now find invaluable when dealing with professors, looking for research, etc.

While neither of these examples have directly to do with your debating ability and only tangentially to your analytical mind, they have the relevant message -- branch out, and then go back and connect the dots. The dots [i]will[/i] connect, and you'll be surprised at what you'll be able to do.

Interestingly, I heard Steve Jobs say something similar at his speech at Stanford's commencement. While I won't guarantee that dropping out of college and taking calligraphy classes will lead to you becoming like Steve Jobs, I will guarantee that if you explore a bit, open your mind, and let your gut guide you, the dots will connect.

The serious reply here is that you're obsessing way too much over this. Judging by the length of your posts and points raised, it would seem you're quite concerned. If you actively pick up on every fault you make, every time you're not quick enough (by some imaginary standards), every time you can't remember something then you will find many things indeed. It sounds to me a lot like fairly ordinary anxiety/obsession. A similar example would be a person who feels depersonalised and is constantly thinking about this, judging every experience by this.

Of course I am probably reading too much into your posts, given you've only felt this way for a few days, but you can rest assured that you don't need to punish yourself by pondering this kind of stuff. Nothing has happened. Your abilities and skills are intact, a "neurological symptom" is ridiculously unlikely. That your "ability of comprehension" could have changed in 3 days is just such total nonsense if you break it down; I am 100 % you are just worrying ("imagining things").

So 3100, care to give us a status report?

Many others, (too lazy to list everyone) have been suggesting lots of wonderful ideas, such as going out and trying new things, taking a break, and just in general not being too concerned with your problem.

I think stuck said it best:
[quote="stuck"]In the end, only you can properly diagnose the problem because only you are completely immersed in its context. Which is the problem. I think you are being too self-conscious about your lack of logic.[/quote]

So the best way, in my opinion? Take some more time off and let us know how you feel. :)

[quote="dreambold22"]@ Latios315: The definition of "atheism" is "the assertion that dieties do not exist". In other words, it's saying there is no god.

Not being sure of the existence of a god is agnostism (did I spell that right?).

Uh... I forgot the word that means "believing in more than one god"..... (sweatdrops)

Simply not really asserting/believing anything is being "non-religious". It is not the same thing as "atheist".

At least, that's what I think. Other than that, I hav absolutely nothing to contribute to the discussion. Carry on.  :lol:[/quote]
I know that, which is why when someone asks my religion i say i am agnostic. however many people use the term "atheist" to describe anyone that does not believe in god

@OP that's a very strange situation, If i were you i would go to the doctor just to make sure you don't have a brain tumor or other disease as the suddenness of this is very wierd

That would seem more like a power than a burden, right? I mean, its a pretty cool thing that your so insightfull now! :lol:

I agree with The_Scintillator. We need a status report.

TZF, thanks for shortening it, and the message after that one. It is quite helpful.

[quote]I'm antipatriotic and antinationalist too....
I am Greek but i'm feeling that my homeland is the earth and not Greece....I love Greeks the same as Albanians,Indians,Italians etc...
I hate when people believe that they are above all the others because of their nationality and i strongly believe that ''when your sleeping the night patriot you wake up fascist''.

BUT i cant understand what do you mean that you feel apathy.You feel apathy against...?

On the other hand when i read about these ideas and make them my own i became more sensitive about :
social inequalities, refugees etc and i'm 'fighting' to pretend their rights....
So just make to yourself clear what you are believe and why you believe this and you will feel better Wink

I cant expreess myself as i would like because of my mediocre English but i believe that you can understand what i want say...
Dimitris[/quote]

Nice to see a fellow anti-nationalist, Dimirtis. I was trying to say that if a day before I was all sure about my position (since I could logically back it), a day after I did not feel 'allergic' to patriotic ads and other brainwashing techniques which would make me 'vomit' several days before this accident. I became apathetic, and that was frightening. :) Keep it up, and your English is very good, don't worry about it.

LightsAbove, it was pretty time consuming -- I was applying to everything I did in my life, but I think it's more that I can't get the conclusions with the same ease as I used to - I can't 'see' as fast right now.

latios315, if I do not mistake, occam's razor principle states that one should make as less assumptions as possible, so assuming that there is(are)/is(are) no God(s) is equally 'wrong' according to this principle. Not to assume anything, you would have to state that you 'don't know anything on this issue, since there is no solid evidence to support either of the opinions.'

stuck, thanks, your way to solve this problem might actually work. I will try it out (actually, I have already been trying it out, albeit unconsciously.)

Kalle, I wish I was imagining things :) I know you probably feel that I am just going crazy -- if I had read something like that myself I would be like 'what the hell is he talking about?' The situation is so absurd and unbelievable, sometimes I wonder if I am dreaming or not. But, drawing an analogy, I went from solving multi-variable calc problems to not being able to factor the (a+b)^2. I used to 'think logically' with such an inspiration, that I couldn't find something hard enough to keep me engaged for a long while, and I can't think of something that could prevent me from solving something trivial I had stumbled upon.

I have thought of several things, and well, it might have been one specific national dish that got me in this 'mental block' position -- I am recalling feeling sleepy and kind of zoned out. The other possible reason might have been that some building right across my room's window was being renewed, and I had all this plaster powder and like flying in the direction of my room, and I assume it's not all that good for one's health. Can't think of any other reasons  :blush: I have stopped eating that dish and changed my room, and it seems that the 'mental block' is gone -- I haven't regained my abilities, but now I feel like I can do it if I try, that might happen (previously, it was like mental block -- you can't do it.) This was an interesting occurrence though, like a chance from god (metaphorically speaking) to understand people who are not trying to be 100% logical. Earlier, when I told people that crying when someone dies is pointless, because it doesn't change anything and it is [b]normal[/b] to die, I was surprised that majority of them didn't agree with me (despite not coming up with counter-arguments.) Now I kinda understand them, yes it is logical, but looks like people are not all about logic, and nature/animal instincts still play big part in our responses to the things around us. I am kind of back to the level I was at a year or two ago; I have reluctantly found out that now I care what others people say, what I wear, where I leave, what my social status is, and so on. Two months ago, I wouldn't mind leaving in a tub as Diogenes did (for example), but now I am kind of worried that this will not be viewed 'correctly' by the society in the scope of [b]current[/b] moral values and established views. I hope I will manage to prescind from the everyday routines once again and focus on more 'global' issues, like before. I want to thank everyone who tried to help me -- looks like I have achieved some progress by following several advices suggested on here.

[quote="3100"]
latios315, if I do not mistake, occam's razor principle states that one should make as less assumptions as possible, so assuming that there is(are)/is(are) no God(s) is equally 'wrong' according to this principle. Not to assume anything, you would have to state that you 'don't know anything on this issue, since there is no solid evidence to support either of the opinions.'
[/quote]
occam's razor says that the system making the least assumptions is most likely, since you make less assumptions in believing that there is no god, then believing in god, the existence of no god is more likely then the existence of a god, that is all i was saying. I agree with you that it is best to make no assumptions and be agnostic

This conversation is so deep, I think that 3100 should just carry on and start from scratch.

You know what they say, have fun in life, nobody ever makes it out alive.

dreambold, the word for believing in more than one god is:

polytheism.

The noun form is polytheist.

But isn't this whole conversation derived from the words:

why

how

who

what

where

when

After reading through all that, I can only agree with the poster who told you to take a break. Maybe you're trying too hard to see something/think of something when before your power lay in your ability to just think and go with the flow of thought.

I'd also like to address "atheists should shut up because they have just as little evidence". Although technically the definition of atheism is belief that no god exists, many choose it as the only logical choice after pointing out that it makes no sense for "my god" to be better than "your god". Atheism tends to be more "anti-religious-ism" than the actual technical definition itself (referring to the common person, at least in my experience).

Sorry for the off topicness -- feel free to not respond.

@tcjy: Thank you for telling me the word for "believeing in more than one god". And I know my post was extremely off-topic... That's why I added the last two sentences. ("Other than that, I hav absolutely nothing to contribute to the discussion. Carry on.") I'm not really sure what you're trying to tell me, but I'm inferring you want me to either shut up or get on topic. I think I'll shut up now. (unless someone brings this particular subject about religions up again.)

Okay. I am now officially shutting up. Carry on.

Off topic: I envy dreambold, she is lost in a sushi.

On topic: I am an anti-nationalist too [hide="(conversation)"]Someone else: July the 4th. This country was born.
Me: Whatever.[/hide], and read my response to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=286761]this[/url] topic for my religion. I think you should just relax about the loss of ability, let it go for a few days, and maybe it will come back.

Oh--and one more thing.
I was reading this one book, and it was kind of interesting....it might explain why people aren't agreeing with you:

Pure logic won't always win the competition--you need to be persuasive. Which sounds weird and manipulative, but most people don't respond well to cold logic.  Even people on this site. Just a thought.

I am quite pleased to hear you are doing better. Be happy! :lol:

[quote="The_Scintillator"]
Pure logic won't always win the competition--you need to be persuasive. [/quote]
Actually, even more than that. If you want to convince somebody, you should play on his feelings, desires, and other subconscious things much more than on his reasoning. That's why the TV commercials (most of which are perfect examples of ultimate idiocy as far as the pure logic is concerned) are so effective. Besides, beyond some elementary cases, correct application of pure logic requires complete information about the subject, which is almost always absent in the real life arguments.

As to "nationalism" you dislike so much, there are many faces of it too. I do not mean here the existence of some mild forms for which the word "patriotism" is more often used. Rather I mean the state of Israel. In under 20 years the Jews built a thriving country in what is technically a mountainous desert with a narrow coastal strip and without any significant natural resources. That was quite a feat by all standards and most people who were in the leading positions were extreme nationalists. They just were clever enough not to equate equate the preservation of cultural identity with the world domination. I do not see anything logically wrong with the former, it is just the latter that is questionable (though even here there is a huge difference between the Victorian British empire and the Nazi Germany). Most ironically, the abhorrence of all kinds of nationalism often comes from the standard logical mistake, which is "If A is B and C is B, then A is C" with B being "nationalism", C being "German nazism" and A everything else.

you should be passive and nice and friendly

I think the Scintillator is right, and you should get some rest :wink:

with a very modesty i may think that :
1. the deductive proof scheme we use is natural(mechanical) like breathing or say walking or running
2. you are using this proof scheme consciously. So it gives up functionning mecanically since you took it in your hand(your mental hand).
3. If you concentrate on you breath or on the moment you fall asleep, or on your walking you will make a disorder on them.(many poeple had breathing and sexual desorders by practicing tantra yoga and concentrationg on the prana)
3.But i dont think what happened to you is   irreversible. It seems to me that if you stop doing so(forcing the deductive proof scheme) and let him fall from your hand, it will take its normal function.
4. I think also that the deductive proof scheme is the last think to trust, because the fundamental think to have is big datta and information ground because fast deduction can be damaging. So it's better for us if we concentrate on checking informations "pertinece" according to the different situations and take time to say that way be we are wrong instead of fast deductions that kills the audience virtually only.
In your case you have erased virtually all your backing by experiencing fast deductin that made you see  that you  can have arguments against anything, even if they are contradictory, so there's no reason for backing for you so when you sollicitated it, it doesnt emerge,(but be sur that it is in your memory).

Sorry to revive, but when reading this, I did come up with an interesting analogy:

Consider your situation to when you start noticing your breathing (which you should after reading this). If you think about it, it doesn't come naturally anymore and you are suddenly cognizant of it for the next few minutes right?. In time, hopefully, your mind will drift so that you breath more naturally and don't have to worry about breathing, but I think this is the case with your thinking.
Perhaps that day was when you became more cognizant of your ability to think rapidly? Hopefully the same principle applies: that you will regain your natural thought structure back. But in this case, there is something that I hope you learned:

[quote="3100"]
Now I kinda understand them, yes it is logical, but looks like people are not all about logic, and nature/animal instincts still play big part in our responses to the things around us.
[/quote]

that logic cannot run everything. Yes, logic is the purest form of reason, but that is not all our thoughts and actions are limited to (hopefully, as long as you don't believe in determinism). That additional factor of soul and emotions must also play a factor in any conclusion.

Also, since its been so long, what's happened since then 3100?


if (x²+x+2)(x+3)= ax^3+bx^2+cx+d, for all values of x, what is a+b+c+d?

I just need an answer for this one

if x and y are positive integers, x^2-y^2= 35 and x^2+2xy+y^2= 49, what is x-y?

explain this one please

[quote="Forte"][hide] For number one, the answer is 16. You multiply, and corresponde to the second equation, and then add them up.[/hide][/quote]

and can you do the 2nd one too?
also

y=x^2-6x+9 what is the least value of y?

thanks in advance

[quote="ernie"][hide="The First One"]Do a manuever which is almost like Foil-ing.

$ x^2 \times x + 3x^2 + x \times x + 3x + 2x + 2 \times 3 = x^3 + 4x^2 + 5x + 6$  Thus, the answer is $ 1 + 4 + 5 + 6$, or $ \boxed{16}$.[/hide]

[hide="The Second One"]Factor both equations:

$ (x - y)(x + y) = 35$

$ (x + y)(x + y) = 49$

Take the square root of the second equation.

$ x + y = 7$

Plug this into the first.

$ 7(x - y) = 35$

$ x - y = \boxed{5}$[/hide][/quote]

Look at the way you factored y= x²-6x+9.

Correct answer- The smallest value occurs when x-3= 0. Therefore the answer is 0.

[quote="PiLuvah"]im confused from ernie's post  :([/quote]

Don't be, he just messed up.  :P

If you're confused by the third question, I edited it out.


Very nice Erken . :) .

My solution :

The inequality is equivalent to :

$ \sqrt{(m\plus{}n\plus{}p)\left(\frac{1}{m}\plus{}\frac{1}{n}\plus{}\frac{1}{p}\right)}\ge 1\plus{}\sqrt [3]{(\frac{m}{n}\plus{}1)(\frac{n}{p}\plus{}1)(\frac{p}{m}\plus{}1)}(1)$.

($ m \equal{}\frac{a}{b},n \equal{}\frac{b}{c},p \equal{}\frac{c}{a}$ then $ mnp \equal{} 1$ )

Let $ t \equal{}\frac{m}{n}\plus{}\frac{n}{m}\plus{}\frac{m}{p}\plus{}\frac{p}{m}\plus{}\frac{n}{p}\plus{}\frac{p}{n}$

So $ t\ge 6$.

$ (1)\Leftrightarrow\sqrt{t\plus{}3}\ge 1\plus{}\sqrt [3]{t\plus{}2}(1)$

Let $ t \equal{} x^{2}\minus{}3,x\ge 3$.

$ (2)\Leftrightarrow x\minus{}1\ge\sqrt [3]{x^{2}\minus{}1}\Leftrightarrow x(x\minus{}1)(x\minus{}3)\ge 0$

Done . :)

[quote="quangpbc"][quote="Erken"]I think that you mean $ t \equal{}\frac{m}{n}\plus{}\frac{n}{p}\plus{}\frac{p}{m}\plus{}\frac{m}{p}\plus{}\frac{p}{n}\plus{}\frac{n}{m}$.
But anyway your solution is very nice.My solution is longer and isn't so nice,i solved it using ABC theorem.[/quote]

Sorry, i have edited it . :D .

By ABC -theorem , we must consider only case $ a \equal{} b$, but i think , it is long solution . Please post it , Erken  :wink:[/quote]
Do you see my post about ABC theorem?I used 4th Consequences from there.But if $ a\equal{}b$ then the inequality is obvious,isn't it?
Anyway i'll post my solution later. :)
Is there any other solutions to this inequality?

WOW !Very very nice short and beautiful solution!Thank you very much! :lol:

[quote="Erken"]Prove that:
$ \sqrt{(a^{2}b\plus{}b^{2}c\plus{}c^{2}a)(ab^{2}\plus{}bc^{2}\plus{}ca^{2})}\geq abc\plus{}\sqrt [3]{(a^{3}\plus{}abc)(b^{3}\plus{}abc)(c^{3}\plus{}abc)}$,for all $ a,b,c > 0$.[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90015

Just want to provide an alternate proof for the inequality $ a,b,c > 0$, $ abc = 1$, then
$ \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\ge 1+\sqrt[3]{(a+b)(b+c)(c+a)}$

Note that
$ 1+\sqrt[3]{(a+b)(b+c)(c+a)}$
$ =\sqrt{\left(1+\sqrt[3]{(a+b)(b+c)(c+a)}\right)\left(1+\sqrt[3]{\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{c}+\frac{1}{a}\right)}\right)}$
$ \leq\sqrt{1+\frac{2(a+b+c)}{3}}\sqrt{1+\frac{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}{3}}$
$ \leq\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$


For $ n \in \mathbb{N}$, let $ f(n)\equal{}1^n\plus{}2^{n\minus{}1}\plus{}3^{n\minus{}2}\plus{}...\plus{}n^1$. Determine the minimum value of: $ \frac{f(n\plus{}1)}{f(n)}.$

we [color=#f00][b]Claim:-[/b][/color]$\frac{f(n+1)}{f(n)}>3$ $\forall$ $n>4$\\

[color=#00f][b]Proof:-[/b][/color] we need to show that $1^{n+1}+2^{n}+\cdots n+1>3(1+2^{n-1}+\cdots+n)$\\

now what we observe is the series of these inequalities:\\

$1^{n+1}+2^{n}+\cdots+(n-1)^3+n^2+n+1>1^{n+1}+2^{n}+3^{n-1}+(4^{n-2}+5^{n-3}+\cdots+(n-1)^3+n^2)$\\

now $(4^{n-2}+5^{n-3}+\cdots+(n-1)^3+n^2)>3(4^{n-3}+5^{n-4}+\cdots+(n-1)^2+n)$\\

so our goal is to prove that $1^{n+1}-3\cdot 1^{n}+2^{n}-3\cdot 2^{n-1}+3^{n-1}-3^{n-1}+4^{n-2}-3\cdot 4^{n-3}+5^{n-3}-3\cdot 5^{n-4}+\cdots+n^2-3\cdot n >0$ which is true for $\forall$ $n>4$ , hence we get\\

$f(n+1)>3f(n)$ $\forall$ $n>4$ , claim proved $\blacksquare$\\

so we check for $n=1,2,3,4$ where the minimum is getting attained, clearly by inspection we observe for $n=2,  \left(\frac{f(n+1)}{f(n)} \right)_{\text{min}}=\boxed{\frac{8}{3}}$ $\blacksquare$

Excellent idea for n>4! Congratulations!


Ok, I'm trying to solve this problem and would appreciate some help.

Here it is:
[b]A Circular Massacre[/b]
a)  100,000 sailors are arranged in a circle on board a ship.  Every other one is shoved overboard and the last one standing gets to live.  Where should you stand to survive (ex. 4th position).
b)Answer the same question if every third person is thrown overboard.
c)Do the problem with every 4th person being eliminated.

Ok, so ya, I have to find a pattern in each one.  I have already completed a and b, but am not totally sure on the answers.  If you people know a formula/solution/link/strategy/whatever It would be greatly appreciated.
[img]http://instagiber.net/smiliesdotcom/games/diablo/KNTWALK.GIF[/img]

a. Maybe... [hide] We can see that each time the people gone are 2^n + 1 (mod 2^(n+1)), such as 2 (mod 2), 3 (mod 4), 5 (mod 8), etc. So we want the maximum n such that 2^n - 1 < 100,000 which is 17. So the last person living is 2^17 + 1 = 65537 ? Sounds a little too simple... [/hide]

Ok guys I really need help on this.  So plz don't just blow me off.  Anyway I think I got an answer for a b and c.

a)68929
b)89644
c)89689

Is that what you guys think is right?  I really need someone to post a formula/answer/link to a site that has this.

Can you at least tell me what a math problem that requires elimination is called?  If I know that I can plug it into a search enine and see if I can find a formula.  Plz some1 help.

[quote="marcos"]I think this is what you're looking for:
http://mathworld.wolfram.com/JosephusProblem.html

Marcos[/quote]

wow thank you soooo much.  Anyway could someone help me know with solving the equation?  K here is what I got:
L(n,m)=1+2n-2^(1+[lgn])

lg is log base 2, n is survivor is every mth man is killed.

What I really don't get is this [quote]Given a group of n men arranged in a circle under the edict that every mth man will be executed going around the circle until only one remains, find the position L(n,m) in which you should stand in order to be the last survivor[/quote]

position L(n,m)????   I understand logs etc, but could you help me with this part, or maybe even solve it for me?  I need it with:

every other person, every third, and every 4th.  They are all for 100,000 people.  Tnx again


Let $ ABC$ be an equilateral triangle, and $ P$ an arbitrary point within the triangle. Perpendiculars PD, PE, PF are drawn to the three sides of the triangle. Show that, no matter where $ P$ is chosen,

$ \frac {PD \plus{} PE \plus{} PF}{AB \plus{} BC \plus{} CA} \equal{} \frac {1}{2\sqrt {3}}$.

[hide="triv solution"]Recall our formula for the area of an equilateral triangle in terms of one of its side lengths. Draw a diagram and see that PD, PE, and PF are the altitudes of three triangles whose area sums to the area of ABC. Equate![/hide]

[hide]Let the side length of $ \triangle ABC$ be $ s$.  We connect $ PA$, $ PB$, and $ PC$, forming three triangles in $ \triangle ABC$ with these three semgents as altitudes.  Finding $ [ABC]$ in two ways, we get: $ \frac {s^2\sqrt {3}} {4}\equal{}\frac {s(PD\plus{}PE\plus{}PF)} {2}.$
Solving for $ PD\plus{}PE\plus{}PF$, we get that $ PD\plus{}PE\plus{}PF\equal{}\frac {s\sqrt {3}} {2}$.  Thus, $ \frac {PD\plus{}PE\plus{}PF} {AB\plus{}BC\plus{}AC}\equal{}\frac {1} {2\sqrt {3}}$.[/hide]
Isn't this kinda, as you say, trivial/as I say, uber-easy? :P

[quote="Math Geek"][hide]Let the side length of $ \triangle ABC$ be $ s$.  We connect $ PA$, $ PB$, and $ PC$, forming three triangles in $ \triangle ABC$ with these three semgents as altitudes.  Finding $ [ABC]$ in two ways, we get: $ \frac {s^2\sqrt {3}} {4} \equal{} \frac {s(PD \plus{} PE \plus{} PF)} {2}.$
Solving for $ PD \plus{} PE \plus{} PF$, we get that $ PD \plus{} PE \plus{} PF \equal{} \frac {s\sqrt {3}} {2}$.  Thus, $ \frac {PD \plus{} PE \plus{} PF} {AB \plus{} BC \plus{} AC} \equal{} \frac {1} {2\sqrt {3}}$.[/hide]
Isn't this kinda, as you say, trivial/as I say, uber-easy? :P[/quote]

indeed, however it was on the 1969 Canadian National Olympiad, so go-figure.


What is the minimum number of equilateral triangles, of side length 1 unit, needed to cover an equilateral triangle of side length 10 units?

We're spreading an area (square unit) with a given measurement of side length (linear unit). Thus the answer is $ 10^2\equal{}\fbox{100}$.


Two bugs begin their journey at a vertex of an equilateral triangle with side length $3$, and travel a distance of $1$ towards a second vertex. Bug $A$ continues on its way, turning $60^\circ$ towards the third vertex and traveling $\frac{2}{3}$ of the distance it just traveled. It continues turning in the same direction, with the same ratio, until it approaches one location and gets really dizzy :P . Bug $B$, on the other hand, travels $\frac{2}{3}$ of its previous distance [u]directly towards[/u] the third vertex, then the first, then the second, and so on until it too approaches one location and gets quite dizzy.

Now, what is the distance from their originating vertex to each of the bugs? Which is farther away?

merr...whats the point of two of them...once you have one, the other is annoying

OMG I THINK WE NEED TO BUST OUT SOME VECTORS

This problem becomes trivial once you use third roots of unity.


[quote=chien than]Let $a,b,c$  be positive real number such that: $abc=1$. Prove that:
$\sqrt[4]{\frac{a}{15b+1}}+\sqrt[4]{\frac{b}{15c+1}}+\sqrt[4]{\frac{c}{15a+1}}\geq \frac{3}{2}$[/quote]
[url]https://artofproblemsolving.com/community/c6h554702p3222734[/url]
[url]https://artofproblemsolving.com/community/c6h451457p2538878[/url]


Columbia
probability
geometry
perimeter
floor function

The problems are here [url]http://www.columbiacollegesc.edu/academics/math/sample_level_II_questions.pdf[/url] but there are no answers as far as I could find.  It'd be amazing if someone could check to see if the ones I answered are right and help me with the couple that I haven't gotten.

1. [hide][b]C.[/b] Dimensions are 6*4*8[/hide]

2. Is this one even a question?  If it is, it has eluded me. @_@

3. [hide][b]D. [/b] 4 odd and 1 even, or 2 odd and 3 even, which leads to 5*4 + 10*4.[/hide]

4. [hide][b]H.[/b] There are 210 total paths from A to B, 20 from A to P, and 4 from P to B.  So 20*4/210.[/hide]

5. [hide][b]A.[/b] If the first card is an ace, the probability of getting a 2nd ace is (4/52)(3/51).  If the first card wasn't an ace, it's (48/52)(4/51).  Adding those together yields 1/13.[/hide]

6. Problematic Problem (aka, I don't know.  Help, please?)

7. [hide][b]B.[/b]  r^2 = 1.1(xr)^2 --> x=.95 [/hide]

8. I knew how to do this once upon a time...but now I think I'm over-compensating for repeated multiples. @_@

9. [hide][b]B.[/b] a-b=5, ab=24.  a^2 - 2ab + b^2 = 25 --> a^2 + b^2= 25+48=73.
a^2 + 2ab + b^2 = x^2, x=a+b
a^2 + b^2 = x^2-48 = 73.
x^2=121, x=11 [/hide]

10. [hide][b]H.[/b] [/hide]

11. [hide][b]B.[/b] xyz=125, wxyz=625*4, w=(wxyz)/(xyz)=20 [/hide]

12. [hide][b]F.[/b] Side of the square = s. Perimeter of square=4s, as does circumference of the circle.
Radius of circle: 4s/(2pi).
Area of square: s^2
Area of circle: pi(2s/pi)^2= (4s^2)/pi
Square:Circle :: pi:4 [/hide]

13. [hide][b]F.[/b] 2/x=a.
a^2 - 2a + 1 = 0 --> (a-1)^2=0 --> a=1 [/hide]

14. [hide][b]B.[/b] (1000/25)6=240 classes.  240/5=48 teachers. [/hide]

[quote="avcamethyst"]
6. Problematic Problem (aka, I don't know.  Help, please?)[/quote]
One brute force method is: you can determine $ y$ and $ z$ in terms of $ x$ from the first two equations, and if you insert $ y$ and $ z$ into the third, you'll get a value for $ x$. From that you can get the values of $ y$ and $ z$.

[quote]8. I knew how to do this once upon a time...but now I think I'm over-compensating for repeated multiples. @_@[/quote]
Use the inclusion-exclusion principle.
[hide="Solution."]
I will denote the first 1000 positive integers with [1000].

$ A_2 \equal{} \{ n \in [1000] \colon 2|n\}$, $ A_5 \equal{} \{ n \in [1000] \colon 5|n\}$, $ A_7 \equal{} \{ n \in [1000] \colon 7|n\}$. Therefore, $ |A_2| \equal{} \lfloor 1000/2 \rfloor$, $ |A_5| \equal{} \lfloor 1000/5 \rfloor$, $ |A_7| \equal{} \lfloor 1000/7 \rfloor$, $ |A_2 \cap A_5| \equal{} \lfloor \frac {1000}{2\cdot 5} \rfloor$, $ |A_2 \cap A_7| \equal{} \lfloor \frac {1000}{2\cdot 7} \rfloor$, $ |A_5 \cap A_7| \equal{} \lfloor \frac {1000}{5\cdot 7} \rfloor$, $ |A_2 \cap A_5 \cap A_7| \equal{} \lfloor \frac {1000}{2\cdot 5\cdot 7} \rfloor$, where $ \lfloor\cdot\rfloor$ is the floor function.

$ |A_2\cup A_5 \cup A_7| \equal{} |A_2| \plus{} |A_5| \plus{} |A_7| \minus{} |A_2 \cap A_5| \minus{} |A_2 \cap A_7| \minus{} |A_5 \cap A_7| \plus{} |A_2 \cap A_5 \cap A_7|\equal{}$ $ 657$.[/hide]

[hide="One approach for Number 6:"]
Take the reciprocal of each equation to get
$ \frac {x \plus{} y}{xy} \equal{} \frac {1}{x} \plus{} \frac {1}{y} \equal{} 7$

$ \frac {x \plus{} z}{xz} \equal{} \frac {1}{x} \plus{} \frac {1}{z} \equal{} 15$

$ \frac {y \plus{} z}{yz} \equal{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} 16$

Subtracting the first equation from the second gives
$ \frac {1}{z} \minus{} \frac {1}{y} \equal{} 8$

Adding this to the third equation we get
$ \frac {2}{z} \equal{} 24$. So $ \frac {1}{z} \equal{} 12$, and back substituting leads to $ \frac {1}{y} \equal{} 4$ and $ \frac {1}{x} \equal{} 3$.

So $ x \equal{} \frac {1}{3}$, $ y \equal{} \frac {1}{4}$, and $ z \equal{} \frac {1}{12}$; and $ x \plus{} y \plus{} z \equal{} \frac {2}{3}$.[/hide]


superior algebra
superior algebra solved

Let $A \in M_2(C) $ with two distincts eigenvalues

B=
(A  A  A  A)
(A  0  0   A)
(A  0  0   A)
(A  A  A   A)

Prove that B is diagonalisable

$ou \beta ino \lambda $

Let $T=\left(\begin{array}{cccc} I_2& 0&0&0 \\ 0&0&0&I_2 \\ 0&0&I_2&0\\ 0&I_2&0&0 \end{array}\right)$.

Then $T^{-1}=T$ and $B_1=T^{-1}BT=\left(\begin{array}{cccc} A&A&A&A \\ A&A&A&A \\ A&A&0&0\\ A&A&0&0 \end{array}\right)$

Let $U^{-1}AU=D$ such that D is diagonal. This is possible because the eigenvalues of A are distinct, so A is diagonalizable.

Let $R=\left(\begin{array}{cccc} U&0&0&0 \\ 0&U&0&0 \\ 0&0&U&0\\ 0&0&0&U \end{array}\right)$. Then $R^{-1}B_1R=\left(\begin{array}{cccc} D&D&D&D \\ D&D&D&D \\ D&D&0&0\\ D&D&0&0 \end{array}\right)$

The last is similar to $\left(\begin{array}{cccc}  xD&0&0&0 \\ 0&xD&0&0 \\ 0&0&yD&0\\ 0&0&0&yD \end{array}\right)$ because $\left(\begin{array}{cc} 1&1\\1&0\end{array}\right)$ is similar to $\left(\begin{array}{cc} x&0\\0&y\end{array}\right)$, where $(X-x)(X-y)=X^2-X-1$.


AMC
USA(J)MO
USAMO
algebra proposed
algebra

Let $a,b,c,d,e,f$ be six real numbers with sum 10, such that \[ (a-1)^2+(b-1)^2+(c-1)^2+(d-1)^2+(e-1)^2+(f-1)^2 = 6. \] Find the maximum possible value of $f$.

[i]Cyprus[/i]

Let the a,b,c,d, and e expressions be all treated as (x-1)^2
because they're not being asked to be maximized.

5(x-1)^2 + (f-1)^2 = 6

And because we have the variables adding up to 10, then we have:

5x + f = 10

Then f = 10 - 5x

Substituting,

5(x-1)^2 +(9 - 5x)^2 = 6

>
>
>  Steps left out here
>
>

30x^2 - 100x + 86 = 6

30x^2 - 100x + 80 = 0

3x^2 - 10x + 8 = 0

(3x - 4)(x - 2) = 0

x = 4/3,  x = 2

f = 10/3, f = 0

The maximum f = 10/3.

You can do it the same way as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47256]1978 USAMO, #1[/url], but the numbers do not seem to come out as nice.

Sorry to revive, but here is a solution, similar to what nsato was probably describing.

[hide="Solution"]Let $A=a-1$, $B=b-1$, $C=c-1$, $D=d-1$, $E=e-1$, $F=f-1$. Then $A^2+B^2+C^2+D^2+E^2+F^2=6$ and $A+B+C+D+E+F=4$.

Then $A+B+C+D+E=4-F$ AND $A^2+B^2+C^2+D^2+E^2=6-F^2$. By Cauchy-Schwarz, we have $5(A^2+B^2+C^2+D^2+E^2) \ge (A+B+C+D+E)^2 \Rightarrow 5(6-F^2) \ge (4-F)^2$. Expanding gives $30-5F^2 \ge F^2-8F+16$ so $6F^2+8F-14 \le 0 \Rightarrow 3F^2+4F-7=(F-1)(3F+7) \le 0$. Therefore $-\frac{7}{3} \le F \le 1$ so $F$ is at most $1$. Therefore, $f$ is at most $1+1=\boxed{2}$.[/hide]

[quote="AwesomeToad"]... Expanding gives $30-5F^2 \ge F^2-8F+16$ so $6F^2+8F-14 \le 0 \Rightarrow 3F^2+4F-7=(F-1)(3F+7) \le 0$. Therefore $-\frac{7}{3} \le F \le 1$ so $F$ is at most $1$. Therefore, $f$ is at most $1+1=\boxed{2}$.[/quote]
The error stems from the fact that from $30-5F^2 \ge F^2-8F+16$ follows $6F^2-8F-14 \le 0$, i.e. $3F^2-4F-7=(F+1)(3F-7) \le 0$. Therefore $-1 \le F \le \frac{7}{3}$.

Equality occurs when $A=B=C=D=E = \frac {1} {3}$ and $F = \frac {7} {3}$, thus for $a=b=c=d=e=\frac {4} {3}$ and $f = \frac {1} {3}$.


Suppose that $ ABC$ is an isosceles triangle with $ AB \equal{} AC$. Let $ P$ be the point on side $ AC$ so that $ AP \equal{} 2CP$. Given that $ BP \equal{} 1$, determine the maximum possible area of $ ABC$.

[hide="Solution by Mass Points"]
Draw a symmetric trisector from $ C$ to side $ AB$ to form point $ Q$, and let $ O \equal{} \overline{BP} \cap \overline{CQ}$. Now, draw in ray $ \vec{AO}$ and let $ R \equal{} \overline{BC} \cap \vec{AO}$.

Now, apply a mass of $ 1$ to point $ A$. Then, $ B$ and $ C$ have masses of $ 2$ because of trisection. Thus, the masses of $ P,Q,R$ are $ 3,3,4,$ respectively.

From our masses we see that $ BO \equal{} OC \equal{} \frac {3}{5} BP \equal{} \frac {3}{5}$ and that $ RO \equal{} \frac {1}{5} AR \implies [ABC] \equal{} 5[AOC]$.

By the sine formula for area, $ [ABC] \equal{} 5 \cdot \frac {1}{2} \cdot AO \cdot OC \cdot \sin \angle AOB \equal{} \frac {5}{2} \left( \frac {3}{5} \right) ^2 \sin \angle AOB \equal{} \frac {9}{10} \sin \angle AOB$

This is clearly maximized when $ \sin \angle AOB \equal{} 1 \implies [ABC] \equal{} \frac {9}{10}$[/hide]

I thought this was both a nice contest problem and teaching problem. However, I think it should've been placed earlier in the round  :)


calculus integration function algebra domain logarithms trigonometry

Hello... Good morning!

I'm trying to solve these 4 integrals using the calculus of residues.
Please help.   :huh:

(1) $\int_{0}^{\infty}\frac{\ln(1+x)}{1+x^{2}}dx$
PS: I know how to solve
$\int_{0}^{\infty}\frac{\ln(x)}{1+x^{2}}dx$
which equals 0, where ln(z) is a multiple-valued function
in the complex domain with branch point z=0.
But I didn't know what contour to use for (1) since
the branch point of ln(1+z) is at z=-1. If I indent it
at z=-1 and use a similar shaped contour to that of
ln(z), I get the contribution from -1 to 0 in addition to
the contribution from 0 to $\infty$, which
throws me off.

(2) $\int_{0}^{\infty}\frac{\ln(1+x+x^{2})}{1+x^{2}}dx$
PS: I know how to solve
$\int_{0}^{\infty}\frac{\ln(1+x^{2})}{1+x^{2}}dx$
which equals $\pi \ln 2$, but the presence of "x"
within $\ln(1+x+x^{2})$ in (2) is giving me a hard time.

(3) $\int_{0}^{\infty}\frac{\ln^{3}(1+x^{2})}{1+x^{2}}dx$

(4) $\int_{0}^{\infty}\frac{\ln(1+x^{3})}{1+x^{3}}dx$

Thank you.
---Nasim

$ I(\alpha) \equal{} \int_0^\infty \frac {\ln(1 \plus{} 2 \sin\alpha\,\, x \plus{} x^2)}{1 \plus{} x^2}\, dx$ with $ 0<\alpha<\pi$ is after

substitution $ x\mapsto \tan x \qquad \int_0^{\frac {\pi}{2}} \ln\left(\frac {1}{\cos^2 x} \plus{} 2\sin\alpha\, \tan x\right) dx$

$ \equal{} \underbrace{\int_0^{\frac {\pi}{2}} \ln\left(1 \plus{} 2\sin\alpha\, \sin x\, \cos x\right) dx}_{ \equal{} : A} \plus{} \underbrace{\int_0^{\frac {\pi}{2}} \minus{} \ln\left(\cos^2 x\right) dx}_{\pi\ln 2}$

$ A \equal{} \int_0^{\frac {\pi}{2}} \ln\left(1 \plus{} \sin \alpha\, \sin 2x\right) dx \equal{} \frac12 \int_0^\pi \ln\left(1 \plus{} \sin\alpha \, \sin x\right) dx$

$ \equal{} \frac12 \sum_{k \equal{} 1}^\infty \frac {( \minus{} 1)^{k \minus{} 1}}{k}\, (\sin \alpha)^k \int_0^\pi (\sin x)^k \, dx$

$ \equal{} \frac12 \sum_{k \equal{} 1}^\infty \frac { \minus{} 1}{2k}\, (\sin \alpha)^{2k} \int_0^\pi (\sin x)^{2k} dx \plus{} \frac12 \sum_{k \equal{} 0}^\infty \frac {1}{2k \plus{} 1}\, (\sin \alpha)^{2k \plus{} 1} \int_0^\pi (\sin x)^{2k \plus{} 1} dx$

whereat $ \int_0^\pi (\sin x)^{2k}\, dx \equal{} \frac {\pi}{2^{2k}} {2k\choose k}$ and $ \int_0^\pi (\sin x)^{2k \plus{} 1}\, dx \equal{} \frac {2^{2k \plus{} 1}}{2k \plus{} 1} \frac {1}{{2k\choose k}}$ .

So $ A \equal{} \frac {\pi}{2} \underbrace{\sum_{k \equal{} 1}^\infty \frac { \minus{} 1}{2k} \frac {(\sin \alpha)^{2k}}{2^{2k}} {2k\choose k}}_{ \equal{} \ln \left(\cos^2 \frac {\alpha}{2}\right)} \plus{} \underbrace{\frac12 \sum_{k \equal{} 0}^\infty \frac {(2\sin \alpha)^{2k \plus{} 1}}{(2k \plus{} 1)^2} \frac {1}{{2k\choose k}}}_{ \equal{} : B}$

To calculate $ B$ first differentiate $ (\arcsin z)^2 \equal{} \frac12 \sum_{k \equal{} 1}^\infty \frac {(2z)^{2k}}{k^2\, {2k\choose k}}$ to obtain

$ \frac {\arcsin z}{\sqrt {1 \minus{} z^2}} \equal{} \sum_{k \equal{} 1}^\infty \frac {(2z)^{2k \minus{} 1}}{k\, {2k\choose k}} \equal{} \sum_{k \equal{} 0}^\infty \frac {(2z)^{2k \plus{} 1}}{(k \plus{} 1) {2k \plus{} 2 \choose k \plus{} 1}} \equal{} \sum_{k \equal{} 0}^\infty \frac {(2z)^{2k}\, z}{(2k \plus{} 1) {2k\choose k}}$

Hence $ B \equal{} \int_0^{\sin \alpha} \frac {\arcsin z}{z\, \sqrt {1 \minus{} z^2}}\, dz \equal{} \int_0^\alpha \frac {z}{\sin z}\, dz \equal{} \left[z\, \ln\left(\tan \frac {z}{2}\right)\right]_0^\alpha \minus{} \int_0^\alpha \ln\left(\tan \frac {z}{2}\right) dz$

For the last integral use the fourier expansion $ \minus{} \ln\left(\tan \frac {z}{2}\right) \equal{} 2\sum_{k \equal{} 0}^\infty \frac {\cos (2k \plus{} 1) z}{2k \plus{} 1}$ .

$ \minus{} \int_0^\alpha \ln\left(\tan \frac {z}{2}\right) dz \equal{} 2\sum_{k \equal{} 0}^\infty \frac {\sin (2k \plus{} 1) \alpha}{(2k \plus{} 1)^2}$

With that $ I(\alpha)$ can be written as $ \frac {\pi}{2}\ln\left[\left(2\cos \frac {\alpha}{2}\right)^2\right] \plus{} \alpha \ln\left(\tan \frac {\alpha}{2}\right) \plus{} 2\, \sum_{k \equal{} 0}^\infty \frac {\sin (2k \plus{} 1) \alpha}{(2k \plus{} 1)^2}$

In particular for $ \alpha \equal{} \frac {\pi}{6}$

$ 2\sum_{k \equal{} 0}^{3N} \frac {\sin\left((2k \plus{} 1)\frac {\pi}{6}\right)}{(2k \plus{} 1)^2} \equal{} \sum_{k \equal{} 0}^{3N} \frac {( \minus{} 1)^k}{(2k \plus{} 1)^2} \plus{} 3\sum_{k \equal{} 0}^{N \minus{} 1} \frac {( \minus{} 1)^k}{(6k \plus{} 3)^2}\to K \plus{} \frac {K}{3}$ .

Therefore $ I\left(\frac {\pi}{6}\right) \equal{} \int_0^\infty \frac {\ln(1 \plus{} x \plus{} x^2)}{1 \plus{} x^2} \, dx$

$ \equal{} \frac {\pi}{2} \ln\Big(2 \plus{} \sqrt {3}\Big) \plus{} \frac {\pi}{6} \ln\Big(2 \minus{} \sqrt {3}\Big) \plus{} \frac {4K}{3} \equal{} \frac13 \left(\pi \ln\Big(2 \plus{} \sqrt {3}\Big) \plus{} 4K\right)$ .

[quote="misan"]Hello... Good morning!

I'm trying to solve these 4 integrals using the calculus of residues.
Please help.   :huh:

[/quote]

@Dr. Doe: I didn't really read your entire solution, but it didn't appear that you used residues.

I'm curious as well how to define a contour to do these with residues. Does anyone know how to do it this way?


How many ordered pairs of integers (a,b) satisfy the inequalities:

$ a^2\plus{}b^2<16$
$ a^2\plus{}b^2<8a$
$ a^2\plus{}b^2<8b$

?

[hide="Graphical solution"]
If you graph the three inequalities they are three circles with different center, and the ordered pairs that satisfy are their intersection. This gives 6
[/hide]
Problems with an algebraic solution?
If you subtract the second inequality from the first, you get $ 0<16\minus{}8a \rightarrow a<2$. However (2,2) is clearly a solution. Where's the error?

From $ a<b$ and $ c<d$ it doesn't necessarily follow that $ a\minus{}c<b\minus{}d$.  For example, take $ (a,b,c,d)\equal{}(0,1,\minus{}666666666,1)$.

[hide]$ a$ and $ b$ are positive from the 2nd and 3rd inequalities.

$ b < 4$ and $ a < 4$
A little case work

and we have $ (1,1),(1,2),(2,1),(2,2),(2,3),(3,2)$[/hide]


function
number theory proposed
number theory

Find all functions $f:\mathbb N\to\mathbb Z$ for which $f(1)=1$, $f(2)=0$, and
\[f(m)^2-f(n)^2=f(m+n)f(m-n),\quad \forall m,n \in \mathbb N, m>n.\]

[quote="TTsphn"]Find all function such that :
$ f: N\to Z$
$ f(m)^2 \minus{} f(n)^2 \equal{} f(m \plus{} n)f(m \minus{} n),\forall m > n$
$ f(1) \equal{} 1 ,f(2) \equal{} 0$[/quote]

Let $ P(m,n)$ the property : $ f(m)^2 \minus{} f(n)^2 \equal{} f(m \plus{} n)f(m \minus{} n)$

$ P(n\plus{}2,n)$ gives $ f(n\plus{}2)^2\equal{}f(n)^2$ which implies $ f(2p)\equal{}0$ and $ f(2p\plus{}1)^2\equal{}1$
$ P(2p\plus{}1,2p)$ gives then $ 1\equal{}f(4p\plus{}1)$
$ P(2p\plus{}2,2p\plus{}1)$ gives $ \minus{}1\equal{}f(4p\plus{}3)$

And so we have a full definition of $ f(n)$ :
$ f(2p)\equal{}0$
$ f(4p\plus{}1)\equal{}1$
$ f(4p\plus{}3)\equal{}\minus{}1$

And it is easy to verify that these necessary conditions are also sufficient.

More than find all function satisfy conditon :
$ f: N\to Z$ such that :
$ f(m)^2\minus{}f(n)^2\equal{}f(m\plus{}n)f(m\minus{}n)$


limit
trigonometry
calculus
calculus computations

Compute the following limit:

$ \lim_{x\rightarrow 0}{\dfrac{(x\minus{}\sin{x})^2}{\left(e^{x^{2}}\minus{}1\right)x^4}}$

Since there is $ x^4$ in the denominator, we substitute

$ sinx \equal{} x\minus{}\frac{x^4}{4!}$

Taking $ x^2\equal{}k$, and applying L'Hospital, we arrive to the limit $ \frac{1}{12}$.

Near zero, we have:

$ \sin x \sim x \minus{} \frac {x^3}{6}$

$ e^{x^2} \sim 1 \plus{} x^2$

This gives us:

$ \lim_{x\rightarrow 0}{\frac {(x \minus{} \sin{x})^{2}}{\left(e^{x^{2}} \minus{} 1\right)x^{4}}} \equal{} \lim_{x\rightarrow 0}{\frac {(x \minus{} x \plus{} \frac {x^3}{6})^{2}}{\left(1 \plus{} x^2 \minus{} 1\right)x^{4}}} \equal{} \frac {1}{36}$

EDIT:

[quote="soumik"]Since there is $ x^4$ in the denominator, we substitute

$ sinx \equal{} x \minus{} \frac {x^4}{4!}$ [/quote]

This makes no sense... You can approximate $ \sin x$ only in ways it is possible, not in ways it would be convenient in a given situation. What was your reasoning that let you derive this?

My one comment is that "L'Hôpital rule application" is precisely the title that I [i]wouldn't[/i] give to this problem. I spend a significant amount of effort telling my students [i]not[/i] to use L'Hôpital most of the time, and to instead use either general knowledge of size - what grows or shrinks faster than what else - and power series. This very problem would be a very good problem for me to use in this campaign - a problem for which L'Hôpital's rule will be nightmarishly messy but for which power series methods will be straightforward and easily followed.

Which is to say that my actual solution would be functionally identical to the first part of milin's post, with insignificant differences in notation and intermediate steps.

hello, the right answer is $ \frac{1}{36}$.
Sonnhard.