analytic geometry
calculus
calculus computations

Find the coordinates of the turning points of the curve $ y\equal{}27x^{3} \minus{}27x^{2} \plus{}4$. Sketch the curve and deduce that $ x^{2}(1\minus{}x) \le \frac{4}{27}$ for all $ x \ge 0.$

Given that each of the numbers $ a, b$ and $ c$ lies between $ 0$ and $ 1$, prove by contradiction that at least one of the numbers $ bc(1\minus{}a), ca(1\minus{}b)$ and $ ab(1\minus{}c)$ is less than or equal to $ \frac{4}{27}$.

Given that each of the numbers $ p$ and $ q$ lies between $ 0$ and $ 1$, prove that at least one of the numbers $ p(1\minus{}q)$ and $ q(1\minus{}p)$ is less than or equal to $ \frac{1}{4}$.

$ y'\equal{}81x^2\minus{}54x\equal{}27x(3x\minus{}2)$, whose roots are $ 0, \frac{2}{3}$. $ y'(x)\leq 0$ for $ x\in [0,\frac{2}{3}]$. moreover, $ \frac{2}{3}$ is a minimal turning point and $ y(\frac{2}{3})\equal{}0$. hence for $ x\geq 0$ we have $ 27x^3\minus{}27x^2\plus{}4\geq 0$, i.e. $ x^2(1\minus{}x)\leq \frac{4}{27}$.

we may assume WLOG that $ b, c \leq a$. then $ bc(1\minus{}a)\leq a^2(1\minus{}a)\leq \frac{4}{27}$.

the last problem is similar: prove that $ x(1\minus{}x)\leq \frac{1}{4}$ for $ x\geq 0$.


trigonometry
function
calculus
derivative
Support
calculus computations

If you observe the graph of the relation $ y = \cos [({{\frac {1}{x})^{( \frac {1}{y})}}]}$, you will observe a strange plot, as given in the first image. Zooming in further will reveal strange but repeating intricate patterns of "designs" as shown in the second image. Why does this happen?

The patterns are moiré artifacts that are a result of the method used to plot the relation.  It is not a true property of the graph.

You can convince yourself that the relation is in fact smooth by solving explicitly for x:

$ x \equal{} (\cos^{\minus{}1} y)^{\minus{}y}$,

which is a function of y.  Note that since the cosine is a periodic function of period $ 2\pi$, we can reconstruct the relation by considering curves of the form

$ f(m,k;x) \equal{} ((\minus{}1)^m \cos^{\minus{}1} y \plus{} 2\pi k)^{\minus{}y}$,

where the inverse cosine is restricted to the range $ [0,\pi]$, for $ m \in {0,1}, k \in Z$, and $ y \in [\minus{}1,1]$, whenever such a function is defined.  The union of all such f over m and k then yields the same graph, but it is clear that the function is smooth for each choice of m and k except when |y| = 1.  At these points, the derivative is undefined, but then if we go back to the original relation, that simply means dy/dx = 0.

[quote="atomicwedgie"]
You can convince yourself that the relation is in fact smooth by solving explicitly for x:

$ x \equal{} (\cos^{ \minus{} 1} y)^{ \minus{} y}$,

which is a function of y.  Note that since the cosine is a periodic function of period $ 2\pi$, we can reconstruct the relation by considering curves of the form

$ f(m,k;x) \equal{} (( \minus{} 1)^m \cos^{ \minus{} 1} y \plus{} 2\pi k)^{ \minus{} y}$,

where the inverse cosine is restricted to the range $ [0,\pi]$, for $ m \in {0,1}, k \in Z$, and $ y \in [ \minus{} 1,1]$, whenever such a function is defined.  The union of all such f over m and k then yields the same graph, but it is clear that the function is smooth for each choice of m and k except when |y| = 1.  At these points, the derivative is undefined, but then if we go back to the original relation, that simply means dy/dx = 0.[/quote]
Convincing. The graph plot looks thus:

[quote="atomicwedgie"]You only plotted one portion of the relation, corresponding to k = 0 and m = 1.  If you plot f(m,k;y) for all m, k for which the function is defined, you recover the original relation.[/quote]
my application doesn't support that, but I do agree with what you say.


real analysis
function
real analysis unsolved

Let $ A\subseteq\mathbb{R}$ Lebesgue measureable such that $ \lambda(A)\equal{}1$.
Prove that there exists a $ B\subseteq A$ Lebesgue measureable such that $ \lambda(B)\equal{}\frac{1}{2}$.

Consider $ \lambda(A\cap ( \minus{} \infty,x])$ as a function of $ x$.
[url=http://en.wikipedia.org/wiki/Atom_%28measure_theory%29#Non-atomic_measures]More general result[/url].


a, b, c are positive numbers, such that a2+b2+c2=1. Find the smallest possible value of the sum
      S= $ {\frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b}}$

If
$\displaystyle  \frac{ab}{c}=x$

$\frac{bc}{a}=y$

${ \frac{ca}{b}}=z$

then it is easy to prove that $x+y+z\geq\sqrt{3(xy+yz+zx)}$.

so $\displaystyle  {\frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b}} \geq\sqrt{3(a^2+b^2+c^2)}=\sqrt{3}$

Equality holds iff $x=y=z$

                     iff $a=b=c=\frac{\sqrt{3}}{3}$

finally $minS=\sqrt{3}$

This is a known problem that was given in many contests, too.
It is quite good because it has this not so easy-to-see point ($x+y+z\geq\sqrt{3(xy+yz+zx)}$.....)

 :)  :lol:


Is there a specific reason why no one has posted here recently, or is it just too much of an out-of-the-way corner?

I don't think we have that many active members from Ohio.


One morning each member of Angela’s family drank an $ 8$-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

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

[hide="this is the long way"]If Angela drank c cups of coffee and m cups of milk (very creative I know), then

c + m = 8 (this accounts for Angela)
and 6c + 4m = 8n  (this represents the family, where n is the number of people in the family)

Solving this system, c = 4n - 16, and m = 24 - 4n.

The only integer n (since the number of people is usually a whole number) for which both c and m are positive is n=5. [/hide]

[hide="Solution"]
Let $ c$ be the amount of coffee and $ m$ be the amount of milk, both in ounces. Let $ f$ be the number of family members.

[b]Equations 1a,1b,1c:[/b] [i]The total coffee and milk being consumed adds to 8 ounces times the number of family members[/i]
$ c\plus{}m \equal{} 8f \qquad 2c\plus{}2m \equal{} 16f \qquad 3c\plus{}3m \equal{} 24f$

[b]Equations 2a,2b:[/b] [i]Angela drinks 1/6 of the coffee and 1/4 of the milk, which adds to 8 ounces[/i]
$ \frac{1}{6} c \plus{} \frac{1}{4} m \equal{} 8 \qquad 2c\plus{}3m \equal{} 96$

[2b]-[1b] gives $ m \equal{} 96\minus{}16f$. Since $ m>0$, we have $ 96\minus{}16f > 0 \implies 96 > 16f \implies 6>f$
[1c]-[2b] gives $ c \equal{} 24f \minus{} 96$. Since $ c>0$, we have $ 24f\minus{}96 > 0 \implies 24f > 96 \implies f > 4$

Combining the two inequalities gives $ 6>f>4 \implies f\equal{}5 \implies \boxed{C. \; 5}$[/hide]

[b]Edit:[/b] [i]See above -- it's not that long, really...[/i]

[quote="The Zuton Force"]
[b]Edit:[/b] [i]See above -- it's not that long, really...[/i][/quote]

I remember that there was a much shorter way of doing this problem, so I did some searching and found that on #13 of this page:

http://www.goodlymaths.0catch.com/contests/amc/amc12_2000s.pdf

there is a much shorter solution.

Quickest solution: Let m+c equal to the total amount of fluid drunk. If Angela drank 1/4m+1/6c=k(the 8 ounces part is unnecessary), then 4 people drunk 4k ounces of fluid, which is equivalent to m + 4/6c (4k=4(1/4m+1/6c)) which is not enough. Let's say 6 people are in her family, then we have 6k, or 6/4m + c, which is over the amount of fluid drunk in total(m+c). Therefore there must have been 5 people, or (c).

[hide=sol]Intuitively she drank more milk then the average person and less coffee that the average person $\boxed{{C}}$[/hide]


Using exactly 5 "i"'s**, 1 pair of parentheses (for grouping and/or multiplication), 1 "-" (for negation or subtraction),
2 "^" symbols (for exponentiation), and 1 "/" (for division), each time, but no other symbols, make (an) expression(s) that has a value of 0.

** where i = the square root of -1

For this puzzle, I am not allowing ii, iii, iiii, iiiii (or their negatives) to be counted.
Otherwise, all eleven symbols must be used each time.  Also, as a check, the expression should work correctly in a TI-83 Plus calculator.

Okay... so far.  Now it is incomplete.

You answered before an edit was put in the "subject description."

All of the (remaining) valid expressions are wanted typed out here.

Another expression (not saying it's the last one remaining) is
>
>
.
.
.
.
.
.
.
.
.
.
>
>

  (i - i)/i^i^i


Find all positive integers $ n$ such that $ \sqrt{\sqrt{n}\plus{}\sqrt{n\plus{}2009}}$ is an integer.

$ \sqrt{n}\plus{}\sqrt{n\plus{}2009}\equal{}k^2$
$ k^4\minus{}2k^2 \sqrt{n}\equal{}2009$
$ k^2(k^2\minus{}2\sqrt{n})\equal{}2009$, obviously $ \sqrt{n}$ is an integer, and $ 2009\equal{}7^2 \times 41$.

So $ k\equal{}1$ or $ 7$, after checking gives $ n\equal{}16$ as the only solution.


geometry
trigonometry
angle bisector
geometry proposed

Let a quadrilateral $ABCD$ have an incircle with center $I$. Then show that

$\frac{AI^{2}}{DA\cdot AB}+\frac{BI^{2}}{AB\cdot BC}+\frac{CI^{2}}{BC\cdot CD}+\frac{DI^{2}}{CD\cdot DA}=2$.

  Darij

The problem turned out to be quite simple, I just didn't see the forest for the trees, made some errors in the calculations and followed a misleading track.

We have

  < BIC + < AID = (180° - < IBC - < ICB) + (180° - < IDA - < IAD)
        = 360° - < IBC - < ICB - < IDA - < IAD
        = 360° - B/2 - C/2 - D/2 - A/2
        = 360° - (A + B + C + D) / 2
        = 360° - 360° / 2 = 180°.

Now, let R be a point on the side AB such that < BRI = < BIC. Then,

  < ARI = 180° - < BRI = 180° - < BIC = < AID

(since < BIC + < AID = 180°). Hence, the point R could have been defined by the equation < ARI = < AID as well.

Now, < BRI = < BIC, and < RBI = < IBC (since I lies on the angle bisector of the angle ABC). Thus, the triangles BRI and BIC are similar, so that $\frac{BI}{BR}=\frac{BC}{BI}$, and $\frac{BI^{2}}{BC}=BR$, hence

  $\frac{BI^{2}}{AB\cdot BC}=\frac{BI^{2}/BC}{AB}=\frac{BR}{AB}$.

Similarly,

  $\frac{AI^{2}}{DA\cdot AB}=\frac{AR}{AB}$.

Therefore,

  $\frac{AI^{2}}{DA\cdot AB}+\frac{BI^{2}}{AB\cdot BC}=\frac{AR}{AB}+\frac{BR}{AB}=\frac{AR+BR}{AB}=\frac{AB}{AB}=1$.

Similarly,

  $\frac{BI^{2}}{AB\cdot BC}+\frac{CI^{2}}{BC\cdot CD}=1$;
  $\frac{CI^{2}}{BC\cdot CD}+\frac{DI^{2}}{CD\cdot DA}=1$;
  $\frac{DI^{2}}{CD\cdot DA}+\frac{AI^{2}}{DA\cdot AB}=1$.

This not only entails

  $\frac{AI^{2}}{DA\cdot AB}+\frac{BI^{2}}{AB\cdot BC}+\frac{CI^{2}}{BC\cdot CD}+\frac{DI^{2}}{CD\cdot DA}=2$,

but also gives

  $\frac{AI^{2}}{DA\cdot AB}=\frac{CI^{2}}{BC\cdot CD}$;
  $\frac{BI^{2}}{AB\cdot BC}=\frac{DI^{2}}{CD\cdot DA}$.

As an easy exercise, you may want to prove

  $\frac{AB}{CD}=\frac{AI\cdot BI}{CI\cdot DI}$.

(14th German (GDR) NMO 1974/75, 4th Round, Class 10).

  Darij

Nice property !
My solution :
In the triangle $AIB$ :
$\frac{AI}{AB}=\frac{\sin{AIB}}{\sin{AIB}}=\frac{\sin{\frac{B}{2}}}{\sin{\frac{C+D}{2}}}$
Similarly for the triangle $AID$
$\frac{AI}{AD}=\frac{\sin{\frac{D}{2}}}{\sin{\frac{B+C}{2}}}$
So $\frac{AI^{2}}{AD\cdot AB}=\frac{\sin{\frac{B}{2}}\cdot\sin{\frac{D}{2}}}{\sin{\frac{C+D}{2}}\cdot\sin{\frac{B+C}{2}}}$
Let $\sin{\frac{B+C}{2}}=\sin{\frac{A+D}{2}}=\sin{\alpha}$
and $\sin{\frac{A+B}{2}}=\sin{\frac{C+D}{2}}=\sin{\beta}$
Then $\sum{\frac{AI^{2}}{AD\cdot AB}}=2\frac{\sin{\frac{B}{2}}.\cdot \sin{\frac{D}{2}}+\sin{\frac{A}{2}}\cdot\sin{\frac{C}{2}}}{\sin{\alpha}\cdot\sin{\beta}}$
Now all we have to prove is
$\sin{\frac{B}{2}}\cdot \sin{\frac{D}{2}}+\sin{\frac{A}{2}}\cdot\sin{\frac{C}{2}}=\sin{\alpha}\cdot\sin{\beta}$
$\Leftrightarrow \sin{\frac{B}{2}}\cdot \sin{\frac{D}{2}}+\sin{\frac{A}{2}}\cdot\sin{\frac{C}{2}}=\frac{1}{2}(\sin{\frac{A+D}{2}}\cdot\sin{\frac{A+B}{2}}+\sin{\frac{B+C}{2}}\cdot\sin{\frac{C+D}{2}})$
$\Leftrightarrow \cos{\frac{A-C}{2}}+\cos{\frac{B-D}{2}}-\cos{\frac{A+C}{2}}-\cos{\frac{B+D}{2}}=\frac{1}{2}(2\cos{\frac{B-D}{2}}-\cos{\frac{2A+B+D}{2}}-\cos{\frac{2C+B+D}{2}})$
$\Leftrightarrow \cos{\frac{A-C}{2}}=\cos{\frac{A-C}{2}}$
which is trivial

[b] Remark.[/b] From this result we can prove that:
$\frac{DI.AI.BI}{CI.DA.AB}+\frac{AI.BI.CI}{DI.AB.BC}+\frac{BI.CI.DI}{AI.BC.CD}+\frac{CI.DI.AI}{BI.CD.DA}=2$


members of a chess club hold a round robin tournament,in which each player plays every other player. All the players get the same points, except for 4 junior players,whose net points are 17.5.how many players were in the chess club? given-you get 1 point for a win,0.5 for draw and 0 for a loss.

OKay. Set it so the n-4 non juniors draw every game against one another and constant outcome against the 4 juniors.
Then the equation:
$4(17.5)+(n-4)(n-5)/2=n(n-1)/2$ must hold, solve it now.

im sorry but i dont get it. how can you just set it it to a specific outcome?isnt that manipulating the problem to get a solution? besides you misunderstood the question. each of the juniors doesnt get 17.5,all of them together get 17.5, hence the word "net".

There are ${ n \choose 2}$ games. Each game nets 1 point no matter what the outcome. Hence, always ${ n \choose 2 }$ points.

Yes of course, but that was not the question :wink:

You can in fact assume a certain outcome because scores and such don't affect how many people there are.

Select all (Ctrl+A) to reveal hidden content.
[color=#d2f8f8]9 and 11 both work.  Say the non-juniors all draw with each other.  The number of non-juniors (x) has to be odd.  4x=17.5+x(plus or minus).5x; 3x=17.5(plus or minus).5x; x=5 or x=7 (I actually did deduce those formulas, but I don't know how I did so).  I probably botched that.[/color]

uhhh i am really sorry if the problem is still unclear.  :blush:

each of the juniors did not get 17.5 points

that is to say if we name the juniors to be A,B,C,D then;

17.5 = points of A + points of B + points of C + points of D

and recursive acronym, 9 and 11 are not the correct answers

Ok lets do this problem right guys

x seniors.

Juniors vs themselves net 4C2 = 6 points.
Juniors vs seniors net 17.5 - 6 = 11.5 points.

Total points in seniors vs juniors is 4x.
So seniors net 4x-11.5 points in senior vs. junior games.
They net xC2 points in senior vs. senior games.

So total senior points is 4x-11.5+xC2. When x divides this total it must divide evenly or have a decimal of .5. In other words, x divides 2*(4x-11.5+xC2). So x divides x^2+7x-23. x can be any multiple of 23.

(For x=23, points per senior is 14.5)


logarithms
limit
calculus
calculus computations

investigate the absolute or conditional convergence (or divergence)
of the following 8 series.....
[ $ \ln$ refers to the logarithm to the base $ e$, i.e. to the base $ 2.718281828\cdots$ ]

[ and where in case of convergence, is it possible to compute the sum ?  :| ]

$ \textbf{(i)}\;\sum_{n=2}^\infty\;\frac{\ln n}{n}$

$ \textbf{(ii)}\;\sum_{n=2}^\infty\;\frac{(-1)^{n}\;\ln n}{n}$

$ \textbf{(iii)}\;\sum_{n=3}^\infty\;\frac{\ln \ln n}{\ln n}$

$ \textbf{(iv)}\;\sum_{n=3}^\infty\;\frac{(-1)^{n-1}\;\ln \ln n}{\ln n}$

$ \textbf{(v)}\;\sum_{n=2}^\infty\;\frac{1}{\ln n}$

$ \textbf{(vi)}\;\sum_{n=2}^\infty\;\frac{(-1)^{n}}{\ln n}$

$ \textbf{(vii)}\;\sum_{n=3}^\infty\;\frac{1}{\ln \ln n}$

$ \textbf{(viii)}\;\sum_{n=3}^\infty\;\frac{(-1)^{n-1}}{\ln \ln n}$

(i) $ \frac{\ln n}{n}> \frac{1}{n}$, and as $ \sum_{n=1}^{\infty}\frac{1}{n}$ diverges, then also our series diverges.

(ii) We have seen in (i) that the series of the modulus diverges. Let's see if we are in the conditions of the Leibniz test:

- $ \lim_{n\to\infty}\frac{\ln n}{n}= 0$
- $ a_{n}= \frac{\ln n}{n}$ is monotone decreasing for n > 2, because $ \left(\frac{\ln x}{x}\right)' = \frac{1-\ln x}{x^{2}}$ is streactly negative for x > e.

So, our series converges conditionally, and its sum equals $ \ln 2 \left(\gamma-\frac{\ln 2}{2}\right)$.

(v) As $ \ln n < n$ then $ \frac{1}{\ln n}> \frac{1}{n}$ and the series diverges.

(iii) $ \frac{\ln \ln n}{\ln n}> \frac{1}{\ln n}$ and so this series also diverges.

(iv) $ \lim_{n\to\infty}\frac{\ln \ln n}{\ln n}= 0$, and $ a_{n}= \frac{\ln \ln n}{\ln n}$ is monotone decreasing (for n > 16) because $ \left(\frac{\ln \ln }{\ln x}\right)' = \frac{1-\ln \ln x}{x(\ln x)^{2}}< 0$ for $ x > e^{e}$. So, the series converges but only conditionally.

(vi) $ a_{n}= \frac{1}{\ln n}$ decreases to 0, and so this series is also conditionally convergent.

(vii) This series diverges because ln(ln n) < n.

(viii) The series converges conditionally, because $ \frac{1}{\ln \ln n}$ decreases to 0.


calculus
integration
trigonometry
logarithms
calculus computations

hello, do you mean really
$ \int\frac{\sec^2(x)}{\tan(x)\cdot\tan(x)}\,dx$?
Sonnhard.

The code is \int\frac {sec^2(x)}{tan(x)*(tan(x)\ + 1)} which is almost certainly meant to be $ \int\frac {\sec^{2}(x)}{\tan(x)\cdot(\tan(x) \plus{} 1)}\,dx$.  Since the denominator is already factored for us, this makes it much easier to take the first step and write
\[ \int\frac {\sec^{2}(x)}{\tan(x)\cdot(\tan(x) \plus{} 1)}\,dx \equal{} \int \frac{\sec^{2} x}{\tan x} \minus{} \frac {\sec^{2} x}{1 \plus{} \tan x}\,dx.\]

hello, simplifying the integrand we get $ \frac{1}{\sin(x)(\sin(x)\plus{}\cos(x))}$ integrating this term with respect to $ x$ we get $ \ln\left|\frac{\tan(x)}{\tan(x)\plus{}1}\right|\plus{}C$
Sonnhard.

hello, i don't know, but why is my solution not so straight as the solution of JBL?
Sonnhard.

Because JBL's solution leads to the immediate substitution $ u \equal{} \tan x$, $ du \equal{} \sec^2 x \, dx$, as well as $ v \equal{} 1 \plus{} \tan x$, $ dv \equal{} \sec^2 x \, dx$.  I don't know of a substitution that immediately yields the antiderivative for $ 1/(\sin x (\sin x \plus{} \cos x))$.  It is not obvious.


Let a,b,c are non-negative numbers and setting $ \frac {(a\minus{}b)(b\minus{}c)(c\minus{}a)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\equal{}t$. Prove that
$ \frac ab\plus{}\frac bc\plus{}\frac ca\minus{}3\ge 6\sqrt[3]{t^2}\plus{}2t$
P/s:
This inequality is stronger than
$ \frac ab\plus{}\frac bc\plus{}\frac ca\plus{}8t\ge 3$
____________________
Nguyen The Cuong

We can prove it by apply a nice indentity
$ \sum_{cyc}{\frac ab}-3=\sum_{cyc}{\frac{(a-b)^2}{b(a+b)}+2\frac{(a-b)(b-c)(c-a}{(a+b)(b+c)(c+a)}}$
And apply AM_GM inequality we have the inquality is true.
It is very nice solution!


When reading a book, Charlie made a list by writing down the
page number of the last page he finished reading at the end of
each day.  (He always finished reading a page that he started.)
His mom thought his list indicated the amount of pages he
had read on each day.  At the end of the 8$ ^{th}$ day of reading, she
added the numbers on his list and thought Charlie had read
432 pages.  If Charlie started reading the book on page one,
and he read the same amount of pages each day of this eight-
day period, how many pages did he actually read by the end of
the 8$ ^{th}$ day?

First, we write an equation for what Charlie's mom thought it was. Let's call the number of pages he reads each day $ x$. We have:

$ x \plus{} 2x \plus{} 3x \plus{} 4x \plus{} 5x \plus{} 6x \plus{} 7x \plus{} 8x \equal{} 432$

This simplifies to:

$ 36x \equal{} 432$

$ x \equal{} 12$

So Charlie reads $ 12$ pages a day, and he reads for $ 8$ days, so our answer is $ 12 \times 8 \equal{} \boxed{96}$.

My brain is off again: why did you multiply 12 by 8

He read 12 pages a day, as we figured out from our equation, and he read for 8 days, so $ 12 \times 8$.


Given a circle with center $O$ and radius $r$ inscribed in triangle $ABC$. The line joining $O$ and the midpoint of side $BC$ intersects the altitude from vertex $A$ at $I$. Prove that $AI = r$.

Take $J$ the touch point of $(O)$ at $BC$, $M$ the midpoint of $AJ$ and $N$ the midpoint of $BC$. Then $NO$ pass through $M$ and $I$, hence the triangles $AIM$ and $MOJ$ are congruent, hence $AI=OJ=r$.

[quote="M4RI0"]Given a circle with center $I$ and radius $r$ inscribed in triangle $ABC$.
The line joining $I$ and the midpoint $M$ of side $BC$ intersects the altitude from vertex $A$ at $N$. Prove that $AN = r$.[/quote][color=darkblue]Suppose w.l.o.g. $b>c\ .$ Denote the projections $D,E$ of the points $A,I$ on the line $BC\ .$ Prove easily that $MD=\frac{b^{2}-c^{2}}{2a}$ and $ME=\frac{b-c}{2}\ .$ Therefore, $\frac{ND}{IE}=\frac{MD}{ME}=\frac{b+c}{a}$ $\Longrightarrow$ $ND=\frac{r(b+c)}{a}$ $\Longrightarrow$ $NA=AD-ND=h_{a}-\frac{r(b+c)}{a}=\frac{ah_{a}-r(b+c)}{a}=\frac{2pr-r(b+c)}{a}=r$, i.e. $NA=r\ .$[/color]

[quote="juancarlos"]Then $NO$ pass through $M$ and $I$.[/quote]

juancarlos could you prove it, in this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=112509  :?:


limit
integration
logarithms
real analysis
real analysis unsolved

$ \lim_{n \to \plus{}\infty}\sum_{k\equal{}1}^n \frac{k^3}{n^4} \cdot e^{(\frac{k}{n})^2}$

Awesome, thanks :)

Let $ f: (0,2)\to \mathbb{R}$ and $ f(1) \equal{} 4 \equal{} f`(1)$
Find:
$ \lim_{h \to 0^ \plus{} }(\frac {f(1 \plus{} h)}{f(1 \plus{} 3h)})^{1/h}$

Edit: Error fixed.

L'Hopital quickly kills that one.

$ \lim_{h\to 0^ \plus{} } \left ( \frac {f(1 \plus{} h)}{f(1 \plus{} 3h)} \right )^{\frac {1}{h}}$

$ \lim_{h\to 0^ \plus{} } e^{\frac {\ln \frac {f(1 \plus{} h)}{f(1 \plus{} 3h)}}{h} }$

Apply L'hopital to the fraction.

$ \lim_{h\to 0^ \plus{} } e^{\frac {f(1 \plus{} 3h)}{f(1 \plus{} h)}\cdot \frac {f'(1 \plus{} h)f(1 \plus{} 3h) \minus{} 3f(1 \plus{} h)f'(1 \plus{} 3h)}{[f(1 \plus{} 3h)]^2} }$

$ \lim_{h\to 0^ \plus{} } e^{\frac {f'(1 \plus{} h)f(1 \plus{} 3h) \minus{} 3f(1 \plus{} h)f'(1 \plus{} 3h)}{f(1 \plus{} h)f(1 \plus{} 3h)} }$

$ e^{\minus{}2}$

Tell me if you're not allowed to use L'hopital.

Ow yes thanks, I did it without L`Hospital and get $ e^{\minus{}2}$ now was looking who did mistake nice that you fixed it such quick. Thanks for help :)


I would think this is covered in multivariable, but I'm not sure. I was reading a bit about general relativity, and I came across the equation
\[d\tau^2=dt^2-dx^2-dy^2-dz^2\]
How does one integrate that? I'd understand what to do if the terms weren't all squared, but... they are.

I kind of feel weird telling you this isn't where it belongs, but shouldn't you put this in the calculus forum?

(Number 2 on "using this forum" that Julie posted specifically says that you can ask Calculus questions on this forum.)

[quote="Magnara"]I don't think you can just integrate it....I imagine if you're given four of the functions, you could solve for the last d(variable), and then integrate.[/quote]
They're not functions; t, x, y, and z are all independent variables. $\tau$ is the only function, and it's a function of t, x, y, and z.

HSPotter - Yea, that still holds because it's not about problem solving, plus it's a physics-related question.  See #2 of my list.  :)

[quote="confuted"][quote="h_s_potter2002"]Yeah, but that was before the merge.  Tha caluclus forum used to be part of MAthLinks, not AoPS.[/quote]
It's also not exactly a "problem solving" level inquiry - just a simple question of method.[/quote] The calculus section is open to any questions about calculus. Of course this being more of a physics question, it's place is here, but other calculus related questions should be placed in this forum http://www.mathlinks.ro/Forum/viewforum.php?f=66 (only questions not problems, the problems should go into the sub-forums there). :)


Calculate the determinant of the matrix with the following entries:

$ a_{i,i\plus{}2}\equal{}\minus{}2$ for $ i \in \left\{1,\ldots, n\minus{}2\right\}$
$ a_{i,i\plus{}1}\equal{}\minus{}1$ for $ i \in \left\{1,\ldots, n\minus{}1\right\}$
$ a_{i,i}\equal{}2$ for $ i \in \left\{1,\ldots, n\right\}$
$ a_{i\plus{}1,i}\equal{}1$ for $ i \in \left\{1,\ldots, n\minus{}1\right\}$

All the other entries are $ 0$.


geometry
geometric transformation
reflection
geometry proposed

Let $ABC$ be a triangle. Let $C^{\prime}$ and $A^{\prime}$ be the reflections of its vertices $C$ and $A$, respectively, in the altitude of triangle $ABC$ issuing from $B$. The perpendicular to the line $BA^{\prime}$ through the point $C^{\prime}$ intersects the line $BC$ at $U$; the perpendicular to the line $BC^{\prime}$ through the point $A^{\prime}$ intersects the line $BA$ at $V$. Prove that $UV \parallel CA$.

  Darij

let $X = {BA'}\cap{C'U}$ and $Y = {BC'}\cap{A'V}$.

suppose that $a\ge{c}$.

by menelaus, we have $\frac{A'C'}{CC'} . \frac{CU}{UB} . \frac{BX}{XA'} = 1$. (*)

again by menelaus we have $\frac{AA'}{A'C'} . \frac{C'Y}{YB} . \frac{BV}{VA} = 1$. (**)

we have to prove that $\frac{BV}{VA} = \frac{BU}{UC}$,

but this is the same that prove that $AA'/CC' . C'Y/YB . BX/XA' = 1$ (multiplying (*) and (**).

$AA'/CC' = AD/DC = tgC/tgA$.

$C'Y/YB = tg(YBA')/tg(YC'A') = tgB/tgC$.

$BX/XA' = tg(C'A'B)/tg(C'BA') = tgA/tgB$.

(because $\triangle A'C'B$ is congruent to $\triangle ABC$)

E.L


group theory
abstract algebra
linear algebra
matrix
linear algebra unsolved

Let $ G$ be an additive subgroup of $ \mathbb{R}$ and let $ a,b,c,d$ be elements of $ G$. Suppose $ a\plus{}b \equal{} c\plus{}d$. Let

$ A \equal{}\left(\begin{matrix}a & b\\ c & d\\ \end{matrix}\right)$

Find the eigenvalues of $ A$ and show that each eigenvalue is an element of $ G$.

Clearly we have $ a\plus{}b$ or $ c\plus{}d$, whichever you prefer, as eigenvalues since $ A\left(\begin{matrix}1\\ 1\end{matrix}\right) \equal{} (a\plus{}b)\left(\begin{matrix}1\\ 1\end{matrix}\right)$.

Also, since $ G$ is an additive group it is clear that $ a\plus{}b\in G$. I don't see any other obvious e-vals. There should be another right?

The sum of the eigenvalues is the trace, and the trace is $ a\plus{}d.$ Hence if we know the trace and one eigenvalue, we can find the other. The other eigenvalue would be $ (a\plus{}d)\minus{}(a\plus{}b)\equal{}d\minus{}b.$ You can also write that as $ a\minus{}c.$

Pursuing that, we see that this has the eigenvector $ \begin{pmatrix}b\\\minus{}c\end{pmatrix}.$

If $ b\equal{}\minus{}c\ne0,$ then then $ A$ would fail to be diagonalizable.


Hope you guys can help.  I purchased over 300 songs from itunes.  Now I just switched over to Sony MP3 player.  Sonic Stage (sony's software for this MP3 player) cannot get songs that were downloaded from Itune.  Any idea?  Thanks so much.

How do I convert these to MP3 files so Sonic Stage can read?  Again, thanks so much.

Its one of the bad things about buying songs online. Any song you buy on an online store will have Digital Rights Management protection, or DRM for short. In the case of iTunes, because Apple is so selfish, this DRM says that the only things that can play these songs are the computer that bought it and an ipod.

I'm sure there are programs out there that allows you to somehow crack open the songs and convert them to mp3, but you have to be internet savvy to find those and I'm pretty sure they are illegal.

My brother said to burn the songs to cds, and then rip it to Sonic Stage.  I dunno if that would work or not though.

You could try ripping it from it burned on a cd.. I seriously doubt that would do anything, it would still mean that Sonic Stage could convert the file, which isnt true.

As to the program to crack it open - no there are none, I've looked. DRM has been cracked a few times in different ways but every time it has, Apple has updated its software (automatically) to override it. So no, theres no crack for DRM.

[quote="Iversonfan2005"]audio convertors....the sony probably does not play mp4s, which are what itunes usually converts the music into when downloaded. However, an audio convertor which you could easily download will let you convert them into mp3 format.[/quote]

I dont think you got the point - the music you download from iTunes is impossible to convert to mp3 because a DRM is put on the song when you download it. Also, iTunes does not convert the music after you download it, it gives it to you in AAC, not mp4 format. Regardless, online-store-bought music cant be converted or played on a cd player.


calculus
integration
trigonometry
calculus computations

$ \int {\sqrt{\frac{1\minus{}\sqrt{x}}{1\plus{}\sqrt{x}}}} dx \equal{} ?$

I am able to get an answer in terms of $ sine$ (not sure of the answer) but I want the answer in terms of $ cos^{\minus{}1}$.
Thanks for any help.

hello, setting $ t\equal{}\sqrt{\frac{1\minus{}\sqrt{x}}{1\plus{}\sqrt{x}}}$ then we get $ x\equal{}\left(\frac{1\minus{}t^2}{1\plus{}t^2}\right)^2$ and $ dx\equal{}\frac{8t(t^2\minus{}1)}{(t^2\plus{}1)^3}\,dt$, and our integral is $ 8\int\frac{t^2(t^2\minus{}1)}{(t^2\plus{}1)^3}\,dt$
Sonnhard.

$ \begin{gathered}
  x \equal{} t^2  \Leftrightarrow dx \equal{} 2tdt \hfill \\
  I \equal{} \int {\sqrt {\frac{{1 \minus{} \sqrt x }}
{{1 \plus{} \sqrt x }}} } dx \equal{} \int {\sqrt {\frac{{1 \minus{} t}}
{{1 \plus{} t}}} .2tdt \equal{} } \int {\frac{{1 \minus{} t}}
{{\sqrt {1 \minus{} t^2 } }}} 2tdt \hfill \\
   \equal{} \int {\frac{{2tdt}}
{{\sqrt {1 \minus{} t^2 } }}}  \minus{} \int {\frac{{2t^2 dt}}
{{\sqrt {1 \minus{} t^2 } }}}  \hfill \\
  I_1  \equal{} \int {\frac{{2tdt}}
{{\sqrt {1 \minus{} t^2 } }}}  \hfill \\
  t \equal{} \cos \theta  \Leftrightarrow dt \equal{}  \minus{} \sin \theta d\theta  \hfill \\
  I_1  \equal{} \int {\frac{{2\cos \theta dt}}
{{\sqrt {1 \minus{} \cos ^2 \theta } }}}  \equal{}  \minus{} \int {\frac{{2\cos \theta \sin \theta d\theta }}
{{\sin \theta }}}  \equal{}  \minus{} 2\sin \theta  \plus{} C_1  \hfill \\
   \equal{}  \minus{} 2\sqrt {1 \minus{} t^2 }  \plus{} C_1  \equal{}  \minus{} 2\sqrt {1 \minus{} x}  \plus{} C_1  \hfill \\
  I_2  \equal{} \int {\frac{{2t^2 dt}}
{{\sqrt {1 \minus{} t^2 } }}}  \hfill \\
  t \equal{} \cos \theta  \Leftrightarrow dt \equal{}  \minus{} \sin \theta d\theta  \hfill \\
  I_2  \equal{}  \minus{} \int {\frac{{2\cos ^2 \theta \sin \theta d\theta }}
{{\sin \theta }}}  \equal{}  \minus{} \int {2\cos ^2 \theta d\theta }  \equal{}  \minus{} \int {\cos 2\theta d\theta  \minus{} \int {d\theta } }  \hfill \\
   \equal{}  \minus{} \sin 2\theta  \minus{} \theta  \plus{} C_2  \equal{}  \minus{} 2t\sqrt {1 \minus{} t^2 }  \minus{} \cos ^{ \minus{} 1} t \plus{} C_2  \hfill \\
   \equal{}  \minus{} 2\sqrt x \sqrt {1 \minus{} x}  \minus{} \cos ^{ \minus{} 1} \sqrt x  \plus{} C_2  \hfill \\
  I \equal{} I_1  \minus{} I_2  \equal{} \left( { \minus{} 2\sqrt {1 \minus{} x}  \plus{} C_1 } \right) \minus{} \left( { \minus{} 2\sqrt x \sqrt {1 \minus{} x}  \minus{} \cos ^{ \minus{} 1} \sqrt x  \plus{} C_2 } \right) \hfill \\
   \equal{}  \minus{} 2\sqrt {1 \minus{} x}  \plus{} 2\sqrt x \sqrt {1 \minus{} x}  \plus{} \cos ^{ \minus{} 1} \sqrt x  \plus{} C_1  \minus{} C_2  \hfill \\
   \equal{} 2(\sqrt x  \minus{} 1)(\sqrt {1 \minus{} x} ) \plus{} \cos ^{ \minus{} 1} \sqrt x  \plus{} C \hfill \\
\end{gathered}$


when i had 2 minutes left for the test, i quickly summed up all the problems that i knew for sure are correct. here's what i know:

1. at least 1 question that will be thrown out (the one about the human brain)
2. a misprint that excluded the score count on the bird/minnow problem in B (presumably worth 2 points)

i estimate around 153 points on the exam, of which 3 extra points will be awarded for free.

i calculated my score to be at least 101, which is only a 2-point improvement over last year. for those of you who made it, do you agree that this was harder than last year's, and that the cutoff, instead of 104, is probably lower than that? perhaps by 2 points?

Wow, 104 seems like a really low cutoff, yet after taking the test it seems pretty hard. I thought it used to be like 124 pts? Is this cause the test was worth less points this year?

Anyway, does anyone remember the pts values for section C (just the main ones)?

124 was the highest score anyone in the US got on the semifinal exam last year, not the cut-off.

it got significantly harder the last two years, and last year's cutoff was 103.5/150+. my bio teacher said this year was even more difficult than last, and i along with most of my friends zeroed the last page of test C. so i guess that means the cutoff will fall again this year.

yeah. the fact that the last page wasnt really in campbell was certainly not fun. if its 103, i think im going to barely miss it

Anyone remember what the 2 compounds did for the plant respiration? It was the 3rd FR.

which two compounds? meanwhile i'm freaking out over the two questions in part 1 that refer to interleukin 2.

wait omg you got your results already?!  :o

apparently for the west coast the results are postponed to friday...

btw congrats  :)

i am from the west coast: they call the finalists. apparently i'm the first one to be called, so that means they start from the bottom 20 and work to the top because my friend hasn't gotten a call yet and i'm 99% certain he scored higher.

Hey everyone, I got the call today, apparently I got a 101 and I made finals! but the trouble is that I made RSI as well this year, do you know if I can somehow attend both?
I doubt I'll make IBO but I'm ok with the USABO camp
The other thing I was wondering is that (I'm a junior btw) if I end up not going to USABO camp can I still put on my resume for college apps that I was a USABO finalist but I just went to RSI?

I guess I just mean will colleges, especially the Ivys, buy that?

Thanks

Ok looks like results are out for non-finalists.

Well I certainly failed...

2 people at my school were extremely close though. Assuming the cutoff is 101, both of them are well within 10 points of becoming finalists. If enough finalists drop from USABO camp, then they'll probably get a chance to go.

[quote="loneph03nixx"]don't worry about it. the IVYs will know that you made USABO top 20, it's just a schedule conflict. go to the RSI thing. it's a lot more prestigious than USABO anyway. you can always make USABO next year if you made it this year already.[/quote]

i think what phoenix is saying is that u can go 2 rsi this year and go 2 usabo next but u cant go the other way e.g. u cant say i want to go to usabo this year and rsi next, cuz i think they only take sniors 2 be or whatever

so ya i would say go to rsi

the cutoff is between 99 and 101. 101 i believe, which makes me #20 ;P

In the case you go to RSI this summer and not make USABO next year, yes, you can put you down that you are a USABO finalist on your college apps. I had a friend who did just that (skip out on USABO for something else), and I'm sure USABO finalist was one of the several factors that got her accepted to the top schools.

Congrats to all the finalists! (Were there really 21 of us this year?) See you in June! :D

Never mind about the 21 question; yes, there will be that many of us this year, unless some of them choose to drop USABO. :lol:

I'm trying to remember Ms. Reedy's exact words.. something about this year's test being an assault/massacre/something equally unpleasant compared to last year's. So I would say YEAH it was waaaaay more difficult this year.

the more difficult = the better for those who prepared.


Find all functions $f: \mathbb N \to \mathbb N$ such that
\[f(x^{2}\plus{}f(y))\equal{}xf(x)\plus{}y\]
for all $x, y \in \mathbb N$.

[quote="cnyd"]$ f: \mathbb{N} \mapsto \mathbb{N}$  (not $ \mathbb{N}_{0})$

$ f(x^{2} \plus{} f(y)) \equal{} xf(x) \plus{} y$[/quote]

Let $ P(x,y)$ be the assertion $ f(x^2\plus{}f(y))\equal{}xf(x)\plus{}y$
Let $ a\equal{}f(1)$

$ P(1,x)$ $ \implies$ $ f(f(x)\plus{}1)\equal{}x\plus{}a$
$ P(1,f(x)\plus{}1)$ $ \implies$ $ f(f(f(x)\plus{}1)\plus{}1)\equal{}f(x)\plus{}a\plus{}1$, and, using previous line : $ f(x\plus{}a\plus{}1)\equal{}f(x)\plus{}a\plus{}1$ and so $ f(x\plus{}p(a\plus{}1))\equal{}f(x)\plus{}p(a\plus{}1)$ $ \forall p\in\mathbb N_0,x\in\mathbb N$

Then $ P(x\plus{}a\plus{}1,y)$ $ \implies$ $ f(x^2\plus{}f(y)\plus{}(a\plus{}1)(a\plus{}1\plus{}2x))\equal{}(x\plus{}a\plus{}1)f(x\plus{}a\plus{}1)\plus{}f(y)$. Using then the property established in the previous line :

$ f(x^2\plus{}f(y))\plus{}(a\plus{}1)(a\plus{}1\plus{}2x)\equal{}(x\plus{}a\plus{}1)(f(x)\plus{}a\plus{}1)\plus{}f(y)$
$ \iff$ $ f(x^2\plus{}f(y))\plus{}(a\plus{}1)^2\plus{}2x(a\plus{}1)\equal{}xf(x)\plus{}(a\plus{}1)^2\plus{}(a\plus{}1)f(x)\plus{}x(a\plus{}1)\plus{}f(y)$
$ \iff$ $ f(x^2\plus{}f(y))\plus{}x(a\plus{}1)\equal{}xf(x)\plus{}(a\plus{}1)f(x)\plus{}f(y)$

Subtracting $ P(x,y)$ from the above line, we get $ x(a\plus{}1)\equal{}(a\plus{}1)f(x)$ and so $ f(x)\equal{}x$ since $ a\plus{}1>1$

It's immediate to check back that this function indeed is a solution. Hence the unique solution : $ \boxed{f(x)\equal{}x}$ $ \forall x\in\mathbb N$


function
trigonometry
algebra
polynomial
real analysis
abstract algebra
advanced fields

i didnt find the proof of this fact that if f(x) is a function of period 2pi then it can be written as a linear combination of sines & cosines. pls give this to me.

That's because there is no such proof (hopefully) as this proposition is clearly false.

What is true though is that every continuous function can be uniformly approximated by such sums with any given precision (Stone-Weierstrass theorem) and every, say, $ C^1$-function can be represented as a uniformly convergent Fourier series. The proofs can be found in any decent textbook in Fourier analysis, so I do not see any point in reproducing them here. On the other hand, if you have some trouble with understanding some particular proof, you are always welcome to post your questions. If you do not have access to printed textbooks, google "Fourier analysis notes" or something similar and spend half an hour finding the most relevant links. Start with looking at [url=http://www.csun.edu/~ac53971/courses/math650/fourier.pdf]this[/url]. It is a very short introduction into Fourier series with a couple of nice applications.

I've posted a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=249955]link[/url] to my own lecture notes.

I will say that I'm less concerned there with proofs of validity (although I do make a few nods in that direction) then I am with properties and manipulations. I also use complex exponentials everywhere rather than sines and cosines - that's really the same thing, it's just a notational change.

well i think the westress thorem gives us approximation of a series by polynomials. couldnt find it for sines & cosines. i really didnt find this approximation proof also.

Could you please make at least some attempt to spell proper names correctly? "westress" is one of the worst versions of "Weierstrass" that I've seen.

fedja actually said "Stone-Weierstrass." That's a powerful and general theorem that contains as special cases both the original Weierstrass theorem about ordinary polynomials and the "trigonometric Weierstrass theorem" about uniform approximation of continuous functions by trigonometric polynomials.

There's also a very satisfying direct proof of the trigonometric Weierstrass theorem: the Cesàro means of the Fourier series of a given function are given by the convolution of that function with the Fejér kernel. The Fejér kernel is well-behaved approximation of the identity, which gives us uniform convergence for $ f$ continuous (and both norm convergence and pointwise [i]a.e.[/i] convergence for $ f$ in $ L^p,1\le p<\infty.$)

aradhita chattopadhyaya: if you said you didn't find anything, you really weren't looking that hard.

What do you mean aradhita chattopadhyaya? :? The first six pages in the notes I linked to contain the full proof of the convergence result for Lipschitz functions. Did you bother to read them? Of course, this is not the most general result and the presentation may be not the best possible one (though I find it very good) but it gives what you asked for and, together with Kent's notes, provides quite a good overview of the subject at the beginner's level.

However,you could find the coefs of the series.But no one could say the series is convergent,even it is,it may not converge to the former function...


group theory
abstract algebra
superior algebra
superior algebra unsolved

I feel like there must be an easy example of this. I was thinking about it, and if you take a group $ G$ and quotient by one of its normal subgroups, then you can't always get one of its subgroups back, can you?

Ex: $ \frac{S_3}{<(123)>}\cong <(12)>$ a subgroup of $ S_3$. Is every quotient group a subgroup of the original group (up to isomorphism)?

It seems that it can't be too small or be Abelian, just from experimentation.

It's easy to construct cases in which the quotient $ G/N$ isn't isomorphic to a complementary subgroup $ H$ ($ HN\equal{}G$). We need more size to make it not isomorphic to any subgroup. The simplest case: $ G\equal{}\mathbb{Z}$, $ N\equal{}2\mathbb{Z}$.

We can also find subgroups that aren't isomorphic to any quotient; that's quite a bit easier.


Find the sum of all the real numbers x such that

$ (2^x\minus{}4)^3 \plus{} (4^x\minus{}2)^3 \equal{} (2^x\plus{}4^x\minus{}6)^3$

Let $ a\equal{}2^x\minus{}4$ and $ b\equal{}4^x\minus{}2$. We have

$ a^3\plus{}b^3\equal{}(a\plus{}b)^3$
$ 3ab\cdot (a\plus{}b)\equal{}0$.

$ (a\plus{}4)^2\equal{}b\plus{}2$.

we can easily solve from here...if we want to be tricky, we can note that

$ 2^{x_1}\cdot 2^{x_2}\equal{}2^{x_1\plus{}x_2}$...


geometry
does a circle exist
3D geometry
pyramid
conics
ellipse
induction

Does a circle exist? Yes, I can draw one... but I'm only placing graphite-atoms on a sheet of paper, you know? The question is a metaphysical one, about the ontological status of mathematical entities.

I don't expect anyone to give "the answer"... I was only wondering what the points of view are within the mathematical community.

[size=75]Suggested [url=http://en.wikipedia.org/wiki/Philosophy_of_mathematics#Contemporary_schools_of_thought]reading[/url]before making redundant comments about the beauty of mathematics, and how it can describe the world truthfully.[/size]

The definition of a circle is the set of points equidistant from a fixed point. Points are 0-dimensional, so you could technically say that it doesn't exist.

In my opinion, it exists in almost its own dimension.

I think like this: a circle is practically a symbol of a unit of distance. The only way to make a "perfect" circle is to find the points a certain distance from a point. When drawing, treating a circle like a polygon will be extremely futile, as a circle has the equivalence of infinite sides. Thus, in my opinion, the existence of the circle in human definition is confirmed, but a new dimension of special curves need definition. Thus, in nature, if any circle or anything related to a circle in curvature is found, it is confirmed if and only if a meaning of distance exists in the circumstance.

A perfect circle doesn't exist in real life. It only exists as an ideal form in our minds.

[quote="Jacobson"]Does a circle exist?[/quote]

Such a question requires careful scrutiny.
What do you mean by 'circle'? A tangible circle? A...?
And, the more difficult question: what do you mean by 'exist'?
Is the question meaningful?

Look at the Platonism part of that link: [url]http://en.wikipedia.org/wiki/Philosophy_of_mathematics#Platonism[/url]
Plato's belief in a "World of Ideas" is that "the everyday world can only imperfectly approximate an unchanging, ultimate reality." Mathematics is based on this concept. Drawn circles can only imperfectly approximate a true circle.

[quote="Mij"]Mathematics is based on this concept.[/quote]  This is certainly not true: many mathematicians are Platonists (about mathematics) and Platonism validates the study of mathematics (and these two things may be related), but mathematical validity is independent of whether or not mathematical Platonism is true.

Slightly more fundamental question: Do "points" exist?

THE underlying fundamental question: What do you mean by "existence"?

Just to give it some more twist: does matter exist? I mean, if we believe the quantum mechanics, all those things we observe are just undounded self-adjoint operators in some Hilbert space :P. We can view it as a "description", of course, but where is the borderline between "a description of reality" and "reality" itself?

This kind of questions has led mathematician to create non euclidean geometry.
Omar kheyyami a muslim mathematician frow nishapor,has tried to prove the fifth postulate of euclide elements, and was faced by the proof of the existence of "rectangle" .

But we can see that the boundary of the full moon is a circle, so it exists visually then sensibly. so is it for the rising sun.
we can also deduce that the intersecting point of the cone of light made by the sun on earth is a circle.We know for example that the path of the shadow cast by the apex of the pyramid from some time before noon to some time after noon is a hyperbola.
we can obtain ellipses and circles from the perspective of a sphere.
If we have istrument to locate object in any space, we can say that any nonlinear three objects  determine uniquely a triangle....
This trend  of determining the shadow cast of sensible  object may lead us to say by induction that the geometrical objects have astronomical origin.

Circles, points, lines - they are abstract, imaginary. We can only try to sketch them, but we can't draw perfect circles...

But from where does a definition of a cirle fall?
The origin of all what we know is related to our senses, or discovered by analysis of what we already know. Imagination cannot do anything but imitating sensitive world or producing thing from the image of sensitive things.
If we loose our memory we loose all the science we acquired before, and memory is fulled by informations extracted from the sensitive world or from things produced from the image of the sensitive world.
This means that what ever you imagine, its origin and all what it is composed is in the empirical world.

So I imagine a purple unicorn with 8 legs. Does it exist?
Things imagined don't necessarily have physical existence.

[quote="Ihatepie"]It exists in your mind.[/quote]
That's almost precisely what gauss202 had implied with the statement preceding yours.

I'm pretty sure that's who's post he was responding to...

The problem is that the purple color, the corn, the legs,... exists in the physical world, and the unicorn you imagine is constructed from those thinks. You can't do it otherwise and this is an example to show that imaginatin imitate the physical world. So the definition of a circle cannot fall from nowhere so it must emerge from empirical experiences.
And this shows how models are constructed to explain the world.

[quote="gauss202"]So I imagine a purple unicorn with 8 legs. [/quote]
Well, if you go a little further and play a bit with some genetic engineering, you may quite well present it at the meeting of the Zoological society in, say, 2135 AD. At that point there will be no denial that it does exist as a physical entity. Now, where exactly is the point at which it came to "existence"?

If you do not like this genetic engineering story, consider an architect who conceives a completely new building in his head today, checks its feasibility using abstract math. within the next month, draws the blueprints a month after that, passes them to the construction firm a year later and suddenly dies on the next day. Due to financial troubles, the construction starts only 10 years later and is successfully accomplished (with minor modifications) in 3 years. In 20 years after that, a powerful earthquake destroys the entire town and the building is never rebuilt but the blueprints still lie in the library on some remote shelf. Question: at which periods of time in this story does the building "exist"?


Find all natural $n\in N$ that:
$\sigma(\sigma(n))=2n$

Let $d_{1}, d_{2}, \cdots , d_{k}$ be all the divisors ( including $1$ and $n$ ) of $n$. It's easy to prove that if $A_{n}= \frac{d_{1}+d_{2}+\cdots+d_{k}}{k}$ and $H_{n}= \frac{k}{\frac{1}{d_{1}}+\frac{1}{d_{2}}+\cdots+\frac{1}{d_{k}}}$ , then: $n = A_{n}.H_{n}$.
( just remember that if $d$ is a divisor of $n$ then $\frac{n}{d}$ is a divisor of $n$ too ).
now we have : $\sigma(\sigma(n)) = 2 \times \frac{\sigma(n)}{\frac{1}{d_{1}}+\frac{1}{d_{2}}+\cdots+\frac{1}{d_{k}}}$
Let $D_{1}, D_{2}, \cdots , D_{l}$ be all the divisors ( including $1$ and $\sigma(n)$ ) of $\sigma(n)$.
 Thus : $\sigma(\sigma(n)) = \frac{2 \sigma(\sigma(n))}{( \frac{1}{d_{1}}+\frac{1}{d_{2}}+\cdots+\frac{1}{d_{k}})( \frac{1}{D_{1}}+\frac{1}{D_{2}}+\cdots+\frac{1}{D_{l}})}$

As a consequence : $( \frac{1}{d_{1}}+\frac{1}{d_{2}}+\cdots+\frac{1}{d_{k}})( \frac{1}{D_{1}}+\frac{1}{D_{2}}+\cdots+\frac{1}{D_{l}}) = 2$

Can anyone continue? :D

If $n=2^{k}$ then $\sigma(n)=2n-1$, therefore $\sigma (\sigma (n))=2n$ if and only if $2n-1=2^{k+1}-1$ is Mersene prime. It is easy to check, that for another cases $\sigma\sigma (n))\not =2n$.
It give $n=(M_{p}+1)/2$, were $M_{p}$ Mersenes primes, known 44 Mersens primes, last $M_{p}=2^{32582657}-1$.

What are Mersenne primes ?? N do u have more details about that ?  :maybe:

Mersene guessed that there are infinitely many primes in the form $2^{k}-1$  , where $k$ is a positive integer.
No one has proved it yet, but as [b]Rust[/b] said the biggest Mersene prime that has been found is $2^{32582657}-1$.
Nice solution Rust  :lol:

So by my solution we can also say that if $n = \frac{M_{p}+1}{2}$  where $M_{p}$ is a Mersene prime and if $D_{1}, D_{2}, \cdots , D_{l}$ are all the divisors of $\sigma(n)$ then:
$\frac{1}{D_{1}}+\frac{1}{D_{2}}+\cdots+\frac{1}{D_{l}}= \frac{1}{1-\frac{1}{2^{k}}}$
nice formula, isn't it ? :D


real analysis
calculus
integration
limit
inequalities
triangle inequality
real analysis unsolved

Hello, I attempted a reductio ad absurdum proof of such theorem which goes:

Let $ E \subseteq \mathbb R^N$ measurable and $ f: E \rightarrow [0,\plus{}\infty]$ a summable function. Then for all $ \varepsilon >0$ exists $ \delta > 0$ such that

 if $ F \subseteq E$ with $ m(F) < \delta$ then $ \int_F |f|dx < \varepsilon$
proof:

for purpose of contradiction suppose that exists $ \varepsilon >0$ such that for all $ \delta >0$ every subset $ F$ of $ E$ with $ m(F) < \delta$ then $ \int_F |f|dx \geq \varepsilon$.

Let $ F_n \equal{} E \cap Q \left (0,\frac 1 n \right)$ (where Q(0,r) is the square of center 0 and half-side r, sides oriented where you like)

We have $ \int_{F_1} |f|dx < \infty$ because f is summable.

The sequence $ \{F_n\}_n$ is decreasing, $ \bigcap_{n\equal{}1}^\infty F_n \equal{} \{0\}$ and $ m(F_n) \leq \frac 1 n$.

We have

$ \int_{F_n} |f|dx \geq \varepsilon$
we let $ n \rightarrow \infty$ and we get

$ \lim_{n \rightarrow \infty} \int_{F_n} |f|dx \equal{} \int_{ \{0\}} |f|dx \equal{} 0 \geq \varepsilon$
Contradiction.

Is that correct?

You negated the statement incorrectly. The negation of
"for all $ \epsilon > 0$ there exists $ \delta > 0$ such that for every subset $ F\subset E$ with measure $ m(F) < \delta$ we have $ \int_F |f|\,dx < \epsilon$"
is
"there exists $ \epsilon > 0$ such that for all $ \delta > 0$ there exists a subset $ F\subset E$ with measure $ m(F) < \delta$ for which $ \int_F |f|\,dx\ge \epsilon$"

I would prove this statement by splitting the set $ E$ into two parts: where $ |f| > M$ and where $ |f|\le M$. You can choose $ M$ large enough so that the integral over the first part is less than $ \epsilon/2$ [of course, this itself requires a proof]. Then deal with the second part.

If $ E_1$ and $ E_2$ are the parts of $ E$ defined above, then for any set $ F\subset E$ we have
\[ \int_F |f| \equal{} \int_{F\cap E_1}|f| \plus{} \int_{F\cap E_2}|f|
\]
The first integral on the right does not exceed $ \int_{E_1}|f| < \epsilon/2$. The second does not exceed $ M m(F\cap E_2)\le M\delta$. This should tell you how to choose $ \delta$.

and how can you say that $ \int_F |f|fx \geq \epsilon$ ?

Following what mlok did, if you pick $ M$ large enough then you get (by the triangle inequality) that the integral of $ |f|$ over $ F$ is at most $ \varepsilon/2\plus{}M\delta$.  You want this to be $ \leq \varepsilon$.  So you want $ M\delta$ to be less than $ \varepsilon/2$, which means that picking $ \delta<\varepsilon/(2M)$ works.  There's no need to try for a contradiction, everything goes through directly.

[quote="blahblahblah"] You want this to be $ \leq \varepsilon$.[/quote]

The negation of the claim wants it to be $ \geq \varepsilon$.

I think you're reading mlok's post (#2) incorrectly.  The first part tells you what the negation of what you want to prove is.  The second part has nothing to do with negations or proof by contradiction, and gives an indication of how the desired result may be proved directly.

ok, now I get it. Btw do you have any clues for a proof by contradiction?


my \begin{enumerate} was working just fine until i got to number 17. It won't label the \item after that. can anyone help me with this? My document's shown below right where it got messed up.

\item
\flushleft{\ce{FeO_{(s)} + CO_{(g)} -> Fe_{(s)} + CO2_{(g)}}}\hfill{$\bigtriangleup$\ce{H^{\circ}}= -11kJ}
\medskip \\

\item For the following reaction ...

And one more problem. These two items are different \items, but they always seem to end up together when i build it.

\item Calculate the following...
\medskip \\
\underline{Solution} \\

\medskip \\

\item Calculate the ...

[quote="NeverOddOrEven"]And one more problem. These two items are different \items, but they always seem to end up together when i build it. [/quote]Should be items not the command \items (which is shown in TeXnicCenter in blue and in the errors as Undefined control sequence)

yea i know i used \item for each thing, but it seems to be combining two different \item

when i type this:

\item
We use the formula $E=h\nu$ to get\\
\medskip \\
$E=(6.63\times 10^{-34}\mbox{J\cdot s})(2.89\times 10^{10}\mbox{s^{-1}})=1.92\times 10^{-23}\mbox{J/photon}$
\newpage
\item Then whatever i type here...

it combines the two \item. Can anyone help me with this?

It doesn't help that your code is in a mess where you have used \mbox (which means text) and then tried to use math mode without dollar signs. This gives you lots of errors on compiling which you [i]must[/i] sort out.

This works fine:[code]\begin{enumerate}
\item We use the formula $E=h\nu$ to get
\[E=(6.63\times 10^{-34}\text{J}\cdot \text{s})(2.89\times 10^{10}\text{s}^{-1})=1.92\times 10^{-23}\text{J/photon}\]
\item Then whatever I type here...
\end{enumerate}[/code]I have removed \newpage as it has no place here - LaTeX will go to a new page when necessary and you generally get better results if you rely on it. It means you can add or delete stuff before and know that the new page won't be in the wrong place.

You should also use display for your equation not use \\ medskip \\. LaTeX will then lay it out properly. [code]\[...\][/code]is for displaying equations.

The result looks like this:
$\par\begin{enumerate}\item We use the formula \(E=h\nu\) to get\begin{displaymath}E=(6.63\times 10^{-34}\text{J}\cdot \text{s})(2.89\times 10^{10}\text{s}^{-1})=1.92\times 10^{-23}\text{J/photon}\end{displaymath}\item Then whatever I type here... \end{enumerate}$


Yann and Camille go to a restaurant.  If there are 10 items on the menu, and each orders one dish, how many different combinations of meals can Yann and Camille order (note that they are allowed to both order the same dish, and that two combinations are considered different if either Yann or Camille orders a different meal)?

Yann can have 10 different meals, and Camille can have 10 different meals. We multiply 10*10 to get [i]100[/i].

But isn't there stuff overcounted, like dish A and dish B, lets say Yann orders A and Camille orders B, this is the same as Yann ordering B and Camille ordering A, so don't we have to divide by 2?

[quote="DerekDW"]Yann orders A and Camille orders B, this is the same as Yann ordering B and Camille ordering A[/quote]

How are these the same?

Technically, the question should be asking how many [i]permutations[/i] can be made, not combinations.

In permutations, AB <> BA.

My solution:
[hide]Just find out how many items Yann and Camille can pick from, then multiply them together: 10 choices for Yann and 10 choices for Camille, so the equation becomes $10\cdot10=\boxed{100\text{ possible ways of eating}}$[/hide]

I agree with DerekDW. 'It does not matter who orders what' implies that (for example) Yann ordering #1 & Camille ordering #2 is a combination of dish #1 & dish #2, but Camille ordering #1 & Yann ordering #2 is the same combination: a combination of dish #1 & dish #2. So with the given wording, it should not be counted twice. Still, that's a vague clause. Better to add a clarification like 'Yann ordering a certain dish & Camille ordering another certain dish should be counted as a combination separate from Camille ordering the first dish and Yann ordering the second.'

6stars is actually correct i believe, his reasoning makes the most sense i the the people who got 100 made a mistake somewhere

[quote]
[b]Problem:[/b]
Yann and Camille go to a restaurant. If there are 10 items on the menu, and each orders one dish, how many different combinations of meals can Yann and Camille order? (Note that they are allowed to order the same dish, and that it does matter who orders what.)

[b]Solution:[/b]
Yann can order 10 different dishes. After he has chosen a dish, Camille can also choose 10 different dishes. Thus there are a total of $10\cdot 10 = \boxed{100}$ different possible combinations of meals.

[b]Your Report:[/b]
combinations are unordered groups. if it matter who orders what, then this problem is asking about permutations, not combinations. permutations are ordered arrangements. it should ask "..., how many different permutations of meals..." instead.

[b]Response:[/b]
We believe that the word "combinations" here is correct. The way we assign the meals is indeed unordered, in the sense that Yann ordering meat and Camille ordering fish is the same as Camille ordering fish and Yann ordering meat. The word "combination" doesn't by itself imply that we will need a binomial coefficient.
[/quote]

The word "combination" has a specific meaning of "unordered selection".  The problem states explicitly that "it does matter who orders what".  We know that order matters because it matters [i]who[/i] orders [i]what[/i]: we must assign each dish to a specific person.  The order MF (meat, fish) is only different from FM when we define the first position to correspond to Yann and the second to correspond to Camille.  If one were to list out all 100 possible meals in this way, one would enumerate the 2-permutations with replacement of 10 objects.  The use of the word "combination" is, therefore,  inappropriate.  Instead of "combination", it should say "permutations", or even just "ways" (e.g. "how many ways can they order their meals").  [url=https://xkcd.com/386/]Relevant xkcd #386[/url].

[quote=Mrdavid445][quote="6stars"]It's clear:
[hide]Like seriously, There are 10 choices for Yann, and 9 choices for Camille, so we have $10*9=\fbox{90}$[/hide][/quote]
I don't mean to revive, but you forgot that they could order the same dish. So we have to add $10$ more giving us $\boxed{100}$[/quote]

Um... actually Mrdavid445 is correct. It doesn’t matter who ordered what, and they are allowed to pick the same thing, so then it would be [hide=this]$10*10=$$\boxed{100}$[/hide]

[hide = My Solution]
Yann and Camille both can choose $1$ of $10$ meals, since we have no restrictions we can just do $10 \cdot 10$(they both have 10 ways to choose a meal), therefore we get $\boxed{100}$.[/hide]

[quote=6stars]It's clear:
[hide]Like seriously, There are 10 choices for Yann, and 9 choices for Camille, so we have $10*9=\fbox{90}$[/hide][/quote]

First of all, don't say stuff like "seriously", and "its clear". We all know you are implying that the answer is obvious and easy, which may not be the same for everyone.

Secondly, you got the problem WRONG, which gives you no right to say stuff that implies that you "easily" got the correct answer.

[quote=darkdragon2019]Don’t be unkind about saying that their answer is wrong. Please respect people.[/quote]

yes guys let's all tiptoe around we dont want to hurt someone.

like does anyone actually think that jerryli's comment actually offended someone

also 6stars posted this years ago he's not gonna see it to be offended

[hide=My solution]Yann can order 10 dishes. Camille can order 10 dishes. So they can order 10 $\cdot$ 10 = $\boxed{100}$ combinations of dishes.[/hide] [hide=Note]If Yann orders dish A and Camille orders dish B, then that is different than vice versa. The combanation AB is different than the combination BA.[/hide]


[quote="JesusFreak197"]#4 was even easier than I thought... think about this. If every third student stayed until the next round, then the answer is $3^n$ for all integer values of $n$. So, since it's the first rather than the third, subtract $2$, and then just a little tweak for even cases. :roll:[/quote]

thats kinda what i said but i have a feeling im not getting a 5
i said they had to be -2 mod 3^n

[quote="TheAmazingOne"]What do you mean, "A little tweak for even cases"?[/quote]

Well, you multiply by $2$ for even cases. We can get this by noticing that switching the survivor of each triplet to the first student allows the $n^{th}$ student to be the second student in the smallest group as well as the third, so there are two possibilities for each $3^n$.

I said it had to be $3^x-2$ where x is some positive integer greater than 1.  Small careless mistake.  I hope they don't take off too much for it.


Εστω ότι χωρίζουμε το σύνολο των θετικών ακεραίων σε $ n$ αριθμητικές προόδους, με κοινές διαφορές $ r_1,r_2,... r_n$ ( για την καθεμία αντίστοιχα). Να δείξετε ότι:
$ \frac {1}{r_1} \plus{} \frac {1}{r_2} \plus{} ... \plus{} \frac {1}{r_n} \equal{} 1$

Ίσως χάνω κάτι...
[hide]αλλά μου φαίνεται ότι αρκεί να δεις τί γίνεται στους αριθμούς $ \{1\ldots \prod\nolimits_{i \equal{} 1}^n r_i)\}$.[/hide]
Cheerio,

Durandal 1707

Edit: nevermind, νόμιζα ότι ήθελες να μην αλληλεπικαλύπτονται κιόλας οι πρόοδοι. Βασικά είναι λίγο πρόβλημα αυτό στη διατύπωση γιατί μπορώ απλώς να πάρω πολλές φορές την ίδια πρόοδο και να ξεπεράσω το $ 1$ (π.χ. να πάρω την πρόοδο με κοινή διαφορά 1 και την πρόοδο με κοινή διαφορά 2...)

Edit2: naah, λες "χωρίζουμε", aok :)

Edit3: Η παραπάνω συνθήκη είναι αναγκαία. Είναι όμως και ικανή?

[quote="Durandal"]
[hide]αλλά μου φαίνεται ότι αρκεί να δεις τί γίνεται στους αριθμούς $ \{1\ldots \prod\nolimits_{i \equal{} 1}^n r_i)\}$.[/hide][/quote]
 Δεν πολυκαταλαβαίνω πως βγαίνει αυτό... θα μπορούσες να το εξηγήσεις λίγο?

[quote="Durandal"]
Edit: nevermind, νόμιζα ότι ήθελες να μην αλληλεπικαλύπτονται κιόλας οι πρόοδοι. Βασικά είναι λίγο πρόβλημα αυτό στη διατύπωση γιατί μπορώ απλώς να πάρω πολλές φορές την ίδια πρόοδο και να ξεπεράσω το $ 1$ (π.χ. να πάρω την πρόοδο με κοινή διαφορά 1 και την πρόοδο με κοινή διαφορά 2...)

Edit2: naah, λες "χωρίζουμε", aok :)
[/quote]

Για να γίνω πιο ξεκάθαρος (με τη λέξη χωρίζουμε: set of positive integers is partitioned) είναι σαν να λέμε ότι παίρνουμε τα σύνολα $ A_1,A_2,...A_n$ ώστε η ένωση τους να κάνει το σύνολο των φυσικών, να είναι διαφορετικά όλα μεταξύ τους και να αποτελούνται το καθένα από όρους μιας αριθμητικής προόδου.

Ελπίζω να έγινα ξεκάθαρος...

[quote="bilot4"][quote="Durandal"]
[hide]αλλά μου φαίνεται ότι αρκεί να δεις τί γίνεται στους αριθμούς $ \{1\ldots \prod\nolimits_{i \equal{} 1}^n r_i)\}$.[/hide][/quote]
 Δεν πολυκαταλαβαίνω πως βγαίνει αυτό... θα μπορούσες να το εξηγήσεις λίγο?
[/quote]

Έστω $ R \equal{} \prod_{i\equal{}1}^n r_i$. Στο σύνολο $ \{1,2,\ldots,R\}$, έχουμε ακριβώς $ R/r_1$ όρους από την ακολουθία $ A_1$, $ R/r_2$ όρους από την ακολουθία $ A_2$, ... , $ R/r_n$ όρους από την ακολουθία $ A_n$.

[quote="bilot4"][quote="Durandal"]
[hide]αλλά μου φαίνεται ότι αρκεί να δεις τί γίνεται στους αριθμούς $ \{1\ldots \prod\nolimits_{i \equal{} 1}^n r_i)\}$.[/hide][/quote]
 Δεν πολυκαταλαβαίνω πως βγαίνει αυτό... θα μπορούσες να το εξηγήσεις λίγο?[/quote]

Sure! Απλώς μέτρα πόσα στοιχεία από κάθε πρόοδο βρίσκονται στο σύνολο $ \{1\ldots, r_1\cdots r_n\}$. Με λίγη προσοχή βρίσκεις ότι η $ i$ πρόοδος έχει $ r_1\cdots r_{i \minus{} 1}\cdot r_{i \plus{} 1}\cdots r_n$ στοιχεία και, αφού το σύνολο είναι διαμερισμένο από τις προόδους θα ισχύει:
$ r_2 \cdots r_n \plus{} \cdots \plus{} r_1\cdots r_{n \minus{} 1} \equal{} r_1\cdots r_n$
Διαρείς και... :)

Πρακτικά η συνθήκη που έχεις γράψει λέει ότι η πιθανότητα να ανήκει ένας αριθμός στην αριθμητική πρόοδο που έχει βήμα $ r_i$ είναι $ \frac {1}{r_i}$... όμορφο αποτέλεσμα :)

Cheerio,

Durandal 1707

Edit: Ninja'ed από τον Dem :ninja:


Two carts are accelerated separately by different constant forces along a horizontal track with negligible friction. Cart A is pulled by one spring stretched by a certain amount, while cart B is pulled by two identical springs stretched by the same amount. The mass of cart A is one-fifth of the mass of cart B. What is the ratio of the net forces acting on the carts?
What is the ratio of their accelerations?
What is the ratio of the time intervals the carts take to travel the length of the track, assuming they both start with zero velocity?

Before practicing his routine on the rings, a 64.1 kg gymnast hangs motionless, with one hand grasping each ring and his feet touching the ground. Both arms make an angle of 25.9° with the vertical. If the force exerted by the rings on each arm has a magnitude of 284 N, what is the magnitude of the force exerted by the floor on his feet?


Let $n$ be a positive integer, and let $a>0$ be a real number. Consider the equation:
\[ \sum_{i=1}^{n}(x_i^2+(a-x_i)^2)= na^2 \]
How many solutions ($x_1, x_2 \cdots , x_n$) does this equation have, such that:
\[ 0 \leq x_i \leq a, i \in N^+ \]

[hide]
Expanding LHS, we get
$\sum_{i=1}^n (2x_i^2-2ax_i + a^2) = na^2$
$2\sum x_i^2 - 2a\sum x_i + na^2 = na^2$
$\sum_{i=1}^n x_i^2 = a\sum_{i=1}^n x_i$
Dang... can't get any further.  I think I'm missing something easy...
[/hide]

Let's continue....

$\sum_{i=1}^n x_i^2 =a\sum_{i=1}^n x_i$ or $\sum_{i=1}^n x_i^2 -\sum_{i=1}^n ax_i=\sum_{i=1}^nx_i(x_i-a)\leq0$
because $0\leq x_i \leq a$

so $x_i=a$ $\forall i=1,2,3, \cdots$

An other solution :

$f(x) = x^2 + (x-a)^2 < a^2$ for $0 \leq x \leq a$ except when $x = 0$ or $x = a$

So there is equality iff $x_i \in \{ 0,a \}$ and there are $2^n$ solutions


function
superior algebra
superior algebra unsolved

How can I prove the following:

if the cardinality of A1 = the cardinality of A2, the cardinality of B1 = the cardinality of B2, B1 is a subset of A1, and B2 is a subset of A2,
then the cardinality of (A1 / B1) = the cardinality of (A2 / B2) where / is the set difference operation

I know there exists a bijective function f: A1 to A2 and a bijective function g: B1 to B2. I need to find a bijective function h: (A1 / B1) to (A2 / B2). Can someone fill in the details?

Thanks for your help.

The statement is false. Let $ A_1 \equal{} A_2 \equal{} \mathbb Z$, $ B_1 \equal{} \mathbb Z$, $ B_2 \equal{} 2{\mathbb Z}\equiv\{2i: i\in Z\}$. Clearly, $ |A_1| \equal{} |A_2| \equal{} |B_1| \equal{} |B_2|$. But $ A_1\setminus B_1 \equal{} \emptyset$, $ A_2\setminus B_2\neq\emptyset$.

However, the statement holds if $ |B_1| < |A_1|$ since then
\[ |A_1\setminus B_1| \equal{} |A_1| \equal{} |A_2| \equal{} |A_2\setminus B_2|
\]
(unless $ A_1$ is finite, in which case the statement holds trivially).


You are in the 2nd seat for a 3-person relay and presented with the question below.  T=TNYWR values are passed to you under the hide tags at the elapsed time indicated.  What do you pass to the person in the 3rd seat and when?

Question R-2:

Let T=TNYWR and let K = T - 5.
The positive integer n is even, and all its divisors (except n itself) divide n/2.  Compute the largest K-digit number n with this property.

0:20 [hide="1st pass"]16[/hide]
2:40 [hide="2nd pass"]4[/hide]
4:50 [hide="3rd pass"]8[/hide]

When you're finished discussing this one, I'll post another (if someone else doesn't beat me to it.)

Great idea. Thanks!

[hide]Using my awesome relay strategy*, I would immediately recognize that we're dealing with powers of two and pass 8, 64, 512, 8192 all in a row. It would be pretty incredible if none of these were right. I would ignore the first two answers, and strongsend 512 after the third answer. Then if #3 used 512 for their first turn-in, they would leave it alone, and if they didn't, they would go again, since I have just confirmed that the answer is definitely 512.

Of course, if we were using my awesome relay strategy, that last pass would only occur if I had been strongsent 8. Otherwise, I would just think, "this guy doesn't know what he's doing" and not send anything up for fear of reducing our pointage (or maybe I would just weaksend 8192).

*My awesome relay strategy has been cut to preserve NC secrets. I will say that it involves weakpassing, strongpassing, and longpassing.[/hide]

Edited to take into account the surprising fact that 8-5=3.

Very nice, jb05, though we don't want to give away too much strategy  :D

My dad & I were talking about it while waiting for a haircut; using that same basic strategy, there was a problem I did that did not require an answer from the first person.

BTW, $8-5=3$  :P

dude jb05 take your awesome relay strategy off of aops somebody might steal it.  share it at practice.  because seriously that's a good strategy.

AND IF SOMEBODY MARKS THIS AS SPAM YOU ARE A TRAITOR TO THE NORTH CAROLINA TEAM.  just don't rate this post (you could rate it a 6 so my poster rating would increase - that is the one exception)

thanks for reading this public service announcement.

[hide="my strategy"]
mutter under my breath 'dude wtf man' until the person in the #1 position realizes oh s* that's wrong and passes me the right answer.  at which point i hope that the #3 is amazing at guessing and already got the answer.
[/hide]

hmm yeah my strategy if I were passed stuff like that would be to start punching the guy in front of me

this is a pretty effective strategy

[quote="Teki-Teki"]

[hide="my strategy"]
mutter under my breath 'dude wtf man' until the person in the #1 position realizes oh s* that's wrong and passes me the right answer.  at which point i hope that the #3 is amazing at guessing and already got the answer.
[/hide][/quote]

no, you should have some code phrase like "the moose is too far from the barn"

hmm? how are we stacking? plus the relays today were hilarious at #1 spot

(FIND THE LEAST PERFECT SQUARE GREATER THAN 2007!!)

[quote="MysticTerminator"]guys guys do we really want to stack the teams that much[/quote]

No! That was a dreadful idea, and I have no idea who came up with it. And I refuse to be a #1. I have ALWAYS been a #3... everywhere. #1 is just like "what is the least square greater than 2007"... umm, what?

well we always stack teams

it's just that when we did that this time, it sort of ended up as ... OH MAN YOU ME AND JEREMY FINISHED IN LIKE ONE MINUTE AND THEN WE JUST SAT AROUND BEING BORED

and that's kinda not that helpful

and uh now I'm kinda confused because I always have jeremy passing to me so then who usually passes to jeremy

but in any case stacking that much is maybe sort of stupid

also

#1 probs are like usually not triv

like maybe they were during the practice

I'm not sure why

but often they're like urgh counting things and then you just like miss one or something

like I would often be a pretty bad #1 because I would be kinda like "eh screw it"


calculus
derivative
function
trigonometry
calculus computations

How do you find $ d^2x/dt^2$ as a function of $ x$ when all I know is that $ dx/dt\equal{}x\sin{x}$?

$ x'\equal{}x\sin x$

$ x''\equal{}x'\sin x\plus{}x(\sin x)'\equal{}x\sin^2 x\plus{}x x'\cos x\equal{}x\sin^2 x\plus{}\frac{x^2\sin(2x)}{2}$


There were 2 guys walk in the bar. The 1st guy said:"How many sons/daugters do you have?". The 2nd guy said:"I have 3". Then, the 1st guy asked:"Well, then how old are they?". The 2nd guy hesitated and said:"I can't tell you but when you multiply the ages, you get 72". The 1st guy said:"I don't have enough information". Then, the 2nd guy said:" Yeah, I think so too." After that, the 1st guy called the check. And then the 2nd guy said:"When you add the ages, it would be your bar tab (the price)". The 1st guy now said:"I don't have enough information yet". After that, the 2nd guy said:"Ok, well. My oldest daughter can play the piano". Eventually, the 1st guy said:"Ok, now I have enough information".
  Can any of you solve this by using indirect proof? I need help with this problem. Thanks. :lol:

First, factor 72 into every possible combination of 3 factors. This gives
1 2 36
1 3 24
1 4 18
1 6 12
1 8 9
2 2 18
2 3 12
2 4 9
2 6 6
3 3 8
3 4 6
And when you add the terms together, the sums you get are all different except for 2, 6, 6, and 3, 3, 8 which each give 14. Since the 2, 6, 6 combination has no [b]older[/b] daughter, then the only combination is 3, 3, 8, and thus those are their ages.

^yah. I remember a problem like this.... older is the key word, the piano skills don't matter :P

I don't understand how you can come to the conclusion that the check is for 14 dollars...

-BLt

If the check was anything other than $14, then he could have deduced their ages without having heard the "oldest daughter" statement.

Well, 9 of the 11 possible combinations of ages results in a unique sum. However, two of them, namely 2,6,6 and 3,3,8 result in the [i]same[/i] sum of 14. If their ages had been any other set out of the eleven, he could have deduced it, however this is not the case as he said: "I don't have enough information yet." Ergo, the sum of their ages [i]must[/i] be 14. Then the piano statement comes in, implying that one of the children was older than the others, ruling out 2,6,6 as that would result in no child being oldest.

Think of it as you are in the shoes of the man trying to guess their ages. For the scenario to progress the way it does, the man's daughters ages must sum to a number that is not unique among the triples of the divisors of 72. The only sum that is not unique is 14, which is why the man progresses by saying "I still don't have enough information". The key to these is looking at it from the person's point of view. There was one on the USAMTS a few weeks ago (more complicated though).

Nice proof. But isn't there 12 combinations. I think you miss 1,1, and 72. :)

Ah, yes, I did miss 1,1,72. Doesn't change anything though, as it's sum is unique.


function
inequalities
limit
logarithms
calculus
derivative
inequalities proposed

Consider a function which is : $f(x)=(\frac{a^{x}+b^{x}}{2})^\frac{1}{x}$.
Now we know that for x=1 we will get Arithmetic Mean (AM) and for x=-1 we will Harmonic Mean (HM) ... then prove that if x=0, we will have a Geometric Mean (GM)!

*you can solve by every way you want i mean not only inequality way is acceptable

thank you very much

davron

Altough it is not the appropiate place to solve it we have $\lim_{x\rightarrow 0}(\frac{a^{x}+b^{x}}{2})^\frac{1}{x}=e^{\lim_{x\rightarrow 0}\frac{a^{x}-1}{2x}+\frac{b^{x}-1}{2x}}$.Now we have that $\lim_{x\rightarrow 0}\frac{a^{x}-1}{x}=ln a$.We are done. :lol:

Please can you explain the proof I am new starting working with limits etc

davron

Let $g(x) = \left( \frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}\right)^\frac{1}{x}$.

We have that $\lim_{x \to 0}g(x) = \lim_{x \to 0}\exp \left( \frac{\ln \frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}}{x}\right) = \exp \left( \lim_{x \to 0}\frac{\ln \frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}}{x}\right)$.

We write $\lim_{x \to 0}\frac{\ln \frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}}{x}$ as $\lim_{x \to 0}\frac{\ln \frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}-\ln \frac{a_{1}^{0}+a_{2}^{0}+\ldots+a_{n}^{0}}{n}}{x-0}$, which is by definition equal to the derivative of $\ln \frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}$ evaluated at $x=0$.

Using the following rules $\left( f(x)+g(x) \right)^\prime = f^\prime (x)+g^\prime (x), \, \left( c f(x) \right)^\prime = c f^\prime (x), \, \left( f(g(x)) \right)^\prime = f^\prime (g(x)) \cdot g^\prime (x)$ we get that \[\left( \ln \frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}\right)^\prime = \frac{\left( \frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}\right)^\prime}{\frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}}\] - note that $\left( \ln x \right)^\prime = \frac1{x}$ - \[= \frac{\frac{a_{1}^{x}\ln a_{1}+a_{2}^{x}\ln a_{2}+\ldots+a_{n}^{x}\ln a_{n}}{n}}{\frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}}\] - note that $\left( a_{i}^{x}\right)^\prime = a_{i}^{x}\ln a_{i}$ -

Evaluating this at zero: \[\left. \frac{\frac{a_{1}^{x}\ln a_{1}+a_{2}^{x}\ln a_{2}+\ldots+a_{n}^{x}\ln a_{n}}{n}}{\frac{a_{1}^{x}+a_{2}^{x}+\ldots+a_{n}^{x}}{n}}\right|_{x=0}= \frac{\frac{\ln a_{1}+\ldots+\ln a_{n}}{n}}{\frac{n}{n}}= \ln \sqrt[n]{a_{1}\ldots a_{n}}.\] Now take the $\exp$ of this and you're done.

The difference between mateivld's solution and the above is that I've taken some derivatives as "known", whereas he computed them.

I strongly recommend reading a calculus book (for beginners). It's gonna be a lot easier for you  :)

PS: I saw this solution in someone else's post, but I can find that post now, so I rewrote it.

ps : thank you very much I will do so ...
also thank you for the solution ~

davron


How would you use latex to express the $n$ root of $x$ Example:  Third root of 27.  I only know how to make square roots $\sqrt{2}$  Thanks!

\sqrt[n]{x}

gives $\sqrt[n]{x}$

put n in brackets right after \sqrt


algebra
polynomial
MATHCOUNTS
geometry
Gauss
quadratics
AMC 8

Solve for all x.  x^(x^5-13x^3+36x)=1
EDITED

Scrambled wrote:Solve for all x.  x^(x^5-13x^3-36x)=1

[hide]

This seems to me to be a rather difficult problem! There are two trivial real solutions, but I don't see how you expect anyone to find the rest (there is at least one more) without a calculator.[/hide]

nr1337 wrote:Hmm.. this is hard for a getting started problem.  1 and -1 work since 1^anything = 1 and (-1)^(some even number) = 1, but the fifth degree polynomial doesn't have any rational roots.  (I'm trying to do this without a calculator)

[hide]0 also works, depending on your definition of 0^0. But there is at least one which, as far as I could tell, was irrational[/hide]

It's definitely solvable for it -- perhaps you are doing the wrong thing with that quintic.  Hint: [hide]You should be setting either x = :pm: 1, which you've done, or setting the quintic = 0.  One root of that quintic is 0; after that, it takes the form of a solvable quartic.[/hide]

However, this is clearly not a getting started question.  Scrambled, the you seem to have a serious misconception about what a getting started question is supposed to be.  It is not a question that you are able to solve, because you are not a getting-started-level problem solver.  It should be a question that is covered by a middle-school math curriculum or AMC-8 or Mathcounts level geometry, combinatorics, etc.

This question is at least 2 steps harder than easy but tricky questions I've seen on some high school contests.  I get the sense that you make your own questions up, and that is fine (and more than fine, it is commendable), but you are going to have to develop a better sense of what your questions actual difficultly level is.  As a rule of thumb, perhaps, just post everything on Intermediate -- it seems that that is where most of it really belongs.

[quote="Scrambled"]I made a typo.[/quote]
It's much better now.

It's not very hard now that he edited it.  Before it was a lot more difficult.

[quote]However, this is clearly not a getting started question. Scrambled, the you seem to have a serious misconception about what a getting started question is supposed to be. It is not a question that you are able to solve, because you are not a getting-started-level problem solver. It should be a question that is covered by a middle-school math curriculum or AMC-8 or Mathcounts level geometry, combinatorics, etc. [/quote]

He does post stuff that is really hard in "getting started," but this question is not really hard at all - just tests exponent properties and factoring.  Everyone seems to be complaining about the level of problems everywhere, don't give up!

[quote]This question is at least 2 steps harder than easy but tricky questions I've seen on some high school contests. I get the sense that you make your own questions up, and that is fine (and more than fine, it is commendable)[/quote]

This question has been asked before (actually you helped me solve one of these problems), in a thread in getting started (?) called "algebra problems" which gauss posted.  Here's the link if you want to try similar problems, scrambled.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11063&highlight=

EDIT: Oh okay, sorry, thanks joml and nr.

[quote="JS1527"][quote]However, this is clearly not a getting started question. Scrambled, the you seem to have a serious misconception about what a getting started question is supposed to be. It is not a question that you are able to solve, because you are not a getting-started-level problem solver. It should be a question that is covered by a middle-school math curriculum or AMC-8 or Mathcounts level geometry, combinatorics, etc. [/quote]

He does post stuff that is really hard in "getting started," but this question is not really hard at all - just tests exponent properties and factoring.  Everyone seems to be complaining about the level of problems everywhere, don't give up![/quote]
IT WAS EDITED!!!!
i made a typo and that screwed everything up

Not to get off the subject of math, but I was sortof sticking up for you (as in "please try the harder problems instead of saying they are too difficult.")

Sorry if I have insulted you.  PM me if there's a problem.

First of all, all your typo did was change nice answers into messy answers.  The process for solving was [i]exactly[/i] the same before, and answers could be found [i]exactly[/i] as easily by applying the quadratic formula and following it up with a square root.  So, for those of you who are saying the typo made it harder, I strenuously disagree -- it just made the answers look nasty.

It would take a lot of explaining for someone to show me why questions with answers that are integers belong in an easy forum while those that have irrational answers, but are in all other respects identical, are somehow miraculously harder.  I stand by my claim that the problem, edited or otherwise, should not be in getting started.

I would further point out that the question, when posted in getting started again, STILL has typos.

While I again commend your efforts to put interesting problems on these boards, I am left wondering why you have difficulty checking them for typos and, further, why you are so insistent that your problems end up in the easier forums to begin with.

[quote="JBL"I stand by my claim that the problem, edited or otherwise, should not be in getting started.[/quote]
Why does a quadratic equation not belong in getting started?

Look I don't see what the big deal is that people post more difficult problems in getting started... Challenges are good, just ask questions if you need help.  I don't want to get off the subject of math, which is what we do by arguing about this.

[quote="nr1337"][quote="JS1527"]
Guys this really does not seem hard:
[/quote]

That's because Scrambled fixed a typo.  Now all of the roots of the polynomial are rational.  Before, they could not easily be found without a calculator.[/quote]

It still could before.
It would just be more tedious

[quote="Scrambled"][quote="Ravi B"]While we are discussing which topics are appropriate for which forums, may I also recommend choosing titles that are more meaningful than "Something"?[/quote]i didnt have a better name[/quote]

Usually its helpful just to know a general idea of what the topic is about. So saying "algebra problem" or "solve for x" would be a lot more useful, say, if people were searching through and only wanting to look at a particular type of problem.

Taking care to come up with an informative title is part of the art of problem-posing too, and is enormously helpful for people who want to come back and find the same thread the next time they are at the computer. Good challenge problems deserve good titles.


Find the maximum value of the expression :
 $ A = x^2y + y^2z + z^2t + t^2x - xy^2 - yz^2 - zt^2 - tx^2 $ , where $ x, y , z, t \in [0,1] $

I guess the maximum will be $\frac{1}{4}$ :)

yes.I am wrong ;)
but I think the solution is coming up :) .as following:
let $f(x,y,z,t)=A=(x-y)xy+(y-z)yz+(z-t)zt+(t-x)tx$
and $f(1,y,z,t)-f(x,y,z,t)=y(1-x)[(1+x-y)y-(1+x-t)t]$
we can assume $(1+x-y)y \geq (1+x-t)t$ so that
$f(1,y,z,t) \geq f(x,y,z,t)$
now we have to calculate the maximum of $f(1,y,z,t) $
because $(z-t)zt+(t-1)t \leq 0 $
so $f(1,y,z,t)  \leq (1-y+yz-z^2)y \leq y(1-y+\frac{y^2}{4})$
let $g(y)=y(1-y+\frac{y^2}{4})$
we can get $g'(y)=\frac{(3y-2)(y-2)}{4}$
so $g(y)$ is the maximun when $y=\frac{2}{3}$
so we get the maximum of A is $\frac{8}{27}$
 when $x=1,y=\frac{2}{3},z=\frac{1}{3},t=0$
am I right? :)


I'm wondering if there's any significance in looking for patterns in $ (p\minus{}1)!$ in mods other than $ p$. Specifically, I'm looking for a generalization mod $ p^2$... I looked at the primes up to 37, and there was no obvious pattern..

Thanks in advance.

The thing about Wilson's Theorem is that if an integer $ p$ satisfies the conditions, then it is a prime (if and only if).  I haven't played around with other mods but mod $ p$ guarantees that $ p$ is (or is not) a prime.

$ (p\minus{}1)! \equiv \minus{}p$ modulo $ p^2$ if and only if $ p$ is prime is my conjecture. I don't have a proof of it yet.

JRav: Yes, I know. I wanted to know if this applied to p^2 in any way (not -1, of course, but...).

darkdieuguerre: I don't think this is true..

I've brute-forced primes up to $ 300000$ without a counterexample  :|

I still don't have a proof, but I'll bug my number theory professor about it tomorrow.

What about $ p \equal{} 5$?

[url]http://en.wikipedia.org/wiki/Wilson's_theorem[/url]

You should look at the bottom of the page for the actual generalization to arbitrary mods.:ninja:

Cheers,

Rofler

His conjecture works for p=5, but how about 7?

$ 6! \equal{} 720 \ne \minus{} 7 (\mod{7^2})$

EDIT: the generalization on wikipedia says $ (p^2)! \equiv \minus{} 1$ mod p^2, but i'm looking at $ (p\minus{}1)!$.

I screwed up, I got this trivial result that follows directly from Wilson's:

$ p! \equiv \minus{} p$ modulo $ p^2$.


combinatorical geometry:

The football's surface is covered with with black 5-gons and white 6-gons. At the edges (sides) of every 5-gon there only 6-gons. But at the edges of every 6-gon there are 5-gons and 6-gons alternately.

Based on this information determine the number of 5-gons and 6-gons of the football !

just for knowledge,this is the structure of the fullerene molecule CARBON-60. :D  :D .not sure though
sorry abt introducing chemistry :D
cheers

in every polyhedron composed only of hexagons and pentagons there must be exactly 12 pentagons (proof: angle deficit :)).  then you can work out the hexagons easily enough...

Suppose that all 5-gons and 6-gons are regular so one will grasp from the problem's statement that in every vertex a 5-gon and 6-gon meet. Define the angle deficit of a polyhedron's vertex as the difference from 360 degrees and and the sum of all the polygon's inner angle who are connected with the respective vertex. Hence the angle deficit of each vertex is 12 degrees.

Euler's polyhedron theorem for this special case:

e+f+s = k+2

f denotes the number of pentagons, g denotes the number of 6-gons, k the edges and e the vertices.

With the help of this theorem one prooves that the angle defect of all polyhedron's vertices is 720 degrees. Hence e = 720 : 12 = 60 degrees. Adding up all the 5-gons and 6-gons which meet at every polyhedron's vertex we get 60 5-gons and 120 6-gons. The 5-gons are counted f times each and the 6-gons six times each. Hence we get 12 5-gons and 20 6-gons !


1/2+(1/2+1/3)+(2/3+1/4)+(3/4+1/5)+(4/5+1/6)+(5/6+1/7)+(6/7+1/8)+(7/8+1/9)+(8/9+1/10)

This is a very simple fraction problem that can be calculated each fraction by each fraction but I want to know if there's a way to calculate it quickly.

Thanks.

I found this problem in a past Cayley contest.[/list]

Yes, it can be easily done quickly.

Hint: [hide]Try pairing up terms so that they add up nicely.[/hide]

Oh!!! I see now...I'm really bad with looking at questions in a different direction...

Thanks to all.


Euler percent calculus derivative Support

Leonhard Euler had proposed a proof of the existence of God, '$(a+b^{n})/n=x$,hence God exists', but I don't understand how this can prove the existence of God, can someone please help.
Thank you. :lol:

1. This story is apocryphal -- it most likely did not actually happen.

2. The "lesson" of the story has nothing to do with the existence of any deity, it has to do with innumeracy.  The "point" is that because Diderot did not know algebra (one aspect of the story that leads one to believe it is false), he was unable to tell that Euler's argument was nonsensical.

Tell me you're being sarcastic.

(Miss, not JBL)

[quote="mathnerd314"]Tell me you're being sarcastic.[/quote]

I am not being sarcastic at all, :mad:
I just wonder how Euler's proof related to the existence of God.

Don't you get it? He said some stuff that obviously doesn't make sense for anyone knowing algebra. For a regular person that's a valid equation.

So if someone tells you he spoke to God, it's just like Euler showing you that formula that you don't know what it means. It can or it cannot be true.

i think that defining god (or a god) the way most people do, it's impossible to prove or disprove his existence.

[quote="drunner2007"]i think that defining god (or a god) the way most people do[/quote] I think most people don't bother to define god at all, because most people don't actually spend time thinking hard enough about their religious beliefs.  (If they did, probably fewer people would claim to believe.)

Quite true, very nearly any argument that a religious philosopher puts forth can be broken by a careful analytical attack.
Consider the following revived by Descartes-
'God is the greatest being that can be, yet it is greater to exist than to not exist, therefore God exists.'

Sounds quite solid on the surface doesn't it? Yet 'existence' is not a valid predicate for there cannot be values attached to it.

Returning to Euler I believe that he gave this argument as a deterrant to those who refute God's existence, in that a believer may always resort to tautology in that the rationality of belief extends only so far as a believer can stump the other into silence.

Say we fix integers a, b and rational x. Our system S (a, b, x) is said to contain a God if we can find a positive integer n such that $a+b^{n}=nx$. Under such axioms, you can see that Euler's argument makes sense.

I like Descartes's reasoning better:

The only way a being can imagine a superior being is if the superior being places the thought of a superior being into the minds of the regular beings.

That's just a massive example of begging the question.

Not only God but most of the things which have not yet been defined properly because that's we fall short of words ...our language is not developed ...we have the inability to express ourselves

[hide="click after you have read"]See i mysely am not satisfied with what i have written i am unable to express what i want completely something u understand is not the way it is or it  is defined or is defined with ambiguity.in short our logic is incomplete because of our inability to express ourselves or may be even the converse :maybe: [/hide]

I don't quite agree with pardesi, language is one of the most adaptable entities known to man, somewhat perplexing for it is a 'creation' of man.
If we wish to, we can define God with arbitrary perfection, but people will always be looking for more. What we have is the inability to agree. And logic isn't always the only means towards an end, that's why mathematics isn't always logical or intuitive. Godel's incompleteness theorem is a fine example.

I don't quite agree with what Yasiru89 has to say about logic and intution ..i feel after a certain level it is the i[i]ntution that lays the  fundamental block to everything ...and logic completely defines what we thought of or even what we are thinking[/i].we are probabaly having Godels incompleteness theorem because we are unable to complete our theory with th'e knowledge language expressions we have .see we humans can only use '[i]if' ,'or' 'and' and few other words to  complete a logical scenario[/i] .See if you propose any theory no matter how accurate it is there probably will exist a person who will disagre to the fact but that doesn't matter here we are not speaking about relative agreements or betterness of our definition but rather depending on absolute understanding of the fact no i am in no way going against Einstein ,everything is relative but at the same time we should stress ohn definitions axioms which are the fundamental blocks of our logic ,why is that now and then many people come up with proofs like 1=2 ,farces ,proving that all triangles are isosceles so and so which are atleast 80 to 90 percent logically correct that's because the building blocks on which we stand are highly intutive and are at some stage though not always Practical.

Godel's incompletenes theorem in it'e essence tells us this.[i]See before writing the proof which depend on definitions and axioms it is virtuially assumed that the words or rather symbols with which you write down are known and their meanings is predfined and understood[/i].lbut the big question is how?if these are the building blocks then how to understand or state what i am stating without knowing anything.So where does it all start ..

P.S.-Just when i posted this post there was a topic above me which read "13 times 7 = 28" :wink:

What utter rubbish. I tried not to dignify pardesi with a response but I am compelled to.
First of all 'intuition' as you call it, shares more with 'logic' than one would care to admit. Yet to believe that something underlies 'everything' is as radical and unjustified a leap of faith as believing in God, which I believe what this discussion was about.
Yet I choose to entertain you for a while. Why must a theory of everything be such that it shall depend on entirely mathematical 'truths'? And why if so and keeping with your narrow view, is mathematics as pointed out not always logical or 'intuitive'? To resort to fundamental axioms and fixed definitions in this regard is foolhardy at best. Hence my reference to the incompleteness theorems, allow me to clarify.
The basic implication I am concerned with is that a theory that expresses certain 'truths' cannot be both complete and consistent at once. Simply put it is impossible to find axioms that justify the entirety of a consistent system, say mathematics. With a definite regard to axioms a truth is 'unprovable'!

[quote]I tried not to dignify pardesi with a response but I am compelled to.[/quote]

Thank you. :P

Yes whatever i wrote was indeed utter rubbish and will be to those who think theorems and proofs are magical derivatives of some kinda potion which the god had given to mathematicians when they stepped down to earth .These theorems are nothing put spurts of intution which come to those who are always thinking,living,dreaming of maths.The best example i could state of is that of [i]Ramanujan [/i]and also of all other mathematician i could  remember.

For the rest part i couldn't understand what you have understood of what i tried to make you understand in my previous post.What i am stressing is and was when i say we have 'few' tools of logic to prove our ideas.I don't understand what you meant when you wrote
[quote]With a definite regard to axioms a truth is 'unprovable'![/quote]
if something is unprovable how can you with assurance you tell that it is truth.

I have surely nothing to do with Godel's completness theorem what i am stressing is the most 'axioms' we learn and are there are the most intutive one's without which nothing would have been possible.
You may not build acomplete consistent theory but whatevere you do are because of certain a[i]xioms and definitions[/i].And finally why are there some discrepancie?that's because our logical tool is finite  :rotfl:

[quote]Yet to believe that something underlies 'everything' is as radical and unjustified a leap of faith as believing in God, which I believe what this discussion was about. [/quote]

Just remember that to believe that the world exists in reality is also a pretty big jump of faith, and also without any real logical justification. For all you know, you might just be a brain in a vat and your entire life an illusion. Everybody in this world has to have plenty of faith, it's just what they put their faith in that makes the difference between one person and another.

[quote="Caradhras"]
Just remember that to believe that the world exists in reality is also a pretty big jump of faith, and also without any real logical justification. For all you know, you might just be a brain in a vat and your entire life an illusion. Everybody in this world has to have plenty of faith, it's just what they put their faith in that makes the difference between one person and another.[/quote]

Faith, sadly, has the attribute of often being misplaced. Reality is not some grand, purely objective basis for everything, it is what I deem it to be, and while reality is a concept that coincides for most people it isn't always perfectly coincident. In essence you may be correct to some degree, but why not put faith in a complete detachment, most of us choose not to, and if the aforementioned attribute holds true could that not be where the readily if reluctantly sought truth lie?

Let's also not forget that the truth is not influenced by what we believe in. You can think anything you want, but the truth remains the truth, even if that truth can't be proven or definitely discovered.

Yes, but there is a [i]convincing basis[/i] for assuming that some things are truth and others aren't.  In particular, the possibility that I [i]might[/i] be a brain in a vat of fluid and the entire history of the universe is a hallucination does not mean its a possibility worth considering.  An hypothesis which has no evidence to support it and which is unfalsifiable [i]might[/i] be true, but it's not interesting, philosophically or otherwise.

hm, i think i now the background story of this "proof": euler was on a dinner with some important people and one philosopher told that god doesn't exist. euler told him that he has found the proof that god truly exists and showed him that what miss wrote. the philosopher didn't say anything any more (he didn't know math  :lol: )

See my first comment on this thread.  The philosopher in question was Diderot, and the story is most likely fictional, so providing additional details is more an exercise in story telling (or story re-telling) than it is in history.