calculus
integration
logarithms
function
algebra
partial fractions
linear algebra

Evaluate $ \int  \frac{x^2\minus{}1}{x^4\plus{}x^2\plus{}1} d\ x$

$ P \equal{} \int \frac{x^2\minus{}1}{x^4\plus{}x^2\plus{}1}\,dx \equal{} \int \frac{1 \minus{} \frac{1}{x^2}}{x^2 \plus{} 1 \plus{} \frac{1}{x^2}}\,dx \equal{} \int\frac{1 \minus{} \frac{1}{x^2}}{\left(x \plus{} \frac{1}{x}\right)^2 \minus{} 1}\,dx$

and now if we consider $ x \plus{} \frac{1}{x} \equal{} t$ then we get

$ P \equal{} \int\frac{1}{t^2\minus{}1}\,dt \equal{} \frac12 \int \left(\frac{1}{t\minus{}1} \minus{} \frac{1}{t\plus{}1}\right)\,dt \equal{} \frac12 \ln \left|\frac{t\minus{}1}{t\plus{}1}\right| \plus{} C$

and finally

$ P \equal{} \frac12 \ln \left|\frac{x^2 \minus{} x \plus{} 1}{x^2 \plus{} x \plus{} 1}\right| \plus{} C$.

For a somewhat more "calculus conventional" approach:

Note that $ x^4\plus{}x^2\plus{}1$ is a factor of $ x^6\minus{}1,$ so its complex roots are the four of the roots of $ x^6\minus{}1$ that are neither $ 1$ nor $ \minus{}1.$ Collecting these into conjugate pairs, we get the factorization
\[ x^4\plus{}x^2\plus{}1\equal{}(x^2\minus{}x\plus{}1)(x^2\plus{}x\plus{}1)\]
(Or we just trial-and-error our way into this factorization directly.) This leads us to the partial fractions problem:

$ \frac{x^2\minus{}1}{(x^2\minus{}x\plus{}1)(x^2\plus{}x\plus{}1)}\equal{}\frac{Ax\plus{}B}{x^2\minus{}x\plus{}1}\plus{}\frac{Cx\plus{}D}{x^2\plus{}x\plus{}1}.$

This may look intimidating, but once you get into it, the linear algebra isn't that hard to do by hand. We get

$ \int \frac{x^2\minus{}1}{(x^2\minus{}x\plus{}1)(x^2\plus{}x\plus{}1)}\,dx\equal{}\int \frac{x\minus{}\frac12}{x^2\minus{}x\plus{}1}\,dx \minus{}\int\frac{x\plus{}\frac12}{x^2\plus{}x\plus{}1}\,dx$

In each of these two integrals, the "natural substitution" of letting $ u$ equal the denominator works, leaving us with an answer of

$ \frac12\ln(x^2\minus{}x\plus{}1)\minus{}\frac12\ln(x^2\plus{}x\plus{}1)\plus{}C$

which agrees with Carcul's answer, of course.

what about $ \int{\frac{dx}{x^4\plus{}x^2\plus{}1}}$[/hide]

Partial fractions result:

$ \frac {1}{x^4 \plus{} x^2 \plus{} 1} \equal{} \frac12\left(\frac {x \plus{} 1}{x^2 \plus{} x \plus{} 1} \minus{} \frac {x \minus{} 1}{x^2 \minus{} x \plus{} 1}\right)$

There's a routine finish to the problem after this. The answer will contain both logarithms and arctangents.

Note:

$ \int_{\minus{}\infty}^{\infty}\frac{x^2\minus{}1}{x^4\plus{}x^2\plus{}1}\,dx\equal{}0.$

However, $ \int_{\minus{}\infty}^{\infty}\frac{1}{x^4\plus{}x^2\plus{}1}\,dx$ is the integral of a positive function and must be positive. Compute it.

$ \int \frac {1}{x^4 \plus{} x^2 \plus{} 1}\,dx \equal{} \int \frac {\frac {1}{x^2}}{x^2 \plus{} 1 \plus{} \frac {1}{x^2}}\,dx \equal{} \frac12 \int\left(\frac {1 \plus{} \frac {1}{x^2}}{x^2 \plus{} 1 \plus{} \frac {1}{x^2}} \minus{} \frac {1 \minus{} \frac {1}{x^2}}{x^2 \plus{} 1 \plus{} \frac {1}{x^2}}\right)\,dx$

$ \equal{} \frac12 \int\left[\frac {1 \plus{} \frac {1}{x^2}}{\left(x \minus{} \frac {1}{x}\right)^2 \plus{} 3} \minus{} \frac {1 \minus{} \frac {1}{x^2}}{\left(x \plus{} \frac {1}{x}\right)^2 \minus{} 1}\right]\,dx \equal{} P$

Considering now $ x \minus{} \frac {1}{x} \equal{} t$ and $ x \plus{} \frac {1}{x} \equal{} u$ we have

$ P \equal{} \frac12 \int\left(\frac {dt}{t^2 \plus{} 3} \minus{} \frac {du}{u^2 \minus{} 1}\right) \equal{} \frac {1}{2 \sqrt {3}} \arctan\left(\frac {t}{\sqrt {3}}\right) \minus{} \frac14 \ln \left|\frac {u \minus{} 1}{u \plus{} 1}\right| \plus{} C$, and switching back to the original variable x,

$ P \equal{} \frac {1}{2 \sqrt {3}} \arctan\left(\frac {x^2 \minus{} 1}{x \sqrt {3}}\right) \minus{} \frac14 \ln \left(\frac {x^2 \minus{} x \plus{} 1}{x^2 \plus{} x \plus{} 1}\right) \plus{} C$.


function
factorial
ratio
limit
calculus
calculus computations

What is a good approach for evaluating a limit containing factorials?. For instance,

lim(((n+1)^3*(3n)!)/(3(n+1))!),n->infinity. I know it converges to 1/27 by using Maple, but how does one evaluate the old-fashioned way. I have several of these to do and never really encountered limits with factorials before and I'd like to know an approach.

Can someone lend a hand?.

Thank you,
Cody

Almost all such problems can be killed with Stirling's formulae:
\[n!\approx n^ne^{-n}\sqrt{2\pi n},\]
where $\approx$ means that ration $LHS/RHS$ tends to 1 when $n\to \infty$.
Additional useful tool is $\lim_{n\to\infty}\left(1+\frac{a}{n}\right)^n=e^a$.

Oh, indeed, it works for most [i]simple[/i] problems. ;)


function
algebra
polynomial
real analysis
real analysis unsolved

I Have a function $ f_n(x)$ and the first $ 2n$ values from $ x\equal{}0$ to $ x\equal{}2n\minus{}1$.
The series is definied by $ f_n(x)\equal{}\sum _{j\equal{}0}^{n\minus{}1} e^{\frac{2 i \pi  j x}{n}} \left(A_j\plus{}x B_j\right)$

How Can I find $ A,B$?

Look at $ f_n(x\plus{}n)\minus{}f_n(x)$ to find $ B$. Once you know $ B$'s, to find $ A$'s shouldn't be a problem.

Actually it is a problem!
$ B_j\equal{}\sum _{x\equal{}0}^{n\minus{}1} \frac{f(n\plus{}x)\minus{}f(x)}{n^2 e^{\frac{2 i \pi  j x}{n}}}$
$ A_j\equal{}\sum _{x\equal{}0}^{n\minus{}1} \frac{f(x)\minus{}\sum _{k\equal{}0}^{n\minus{}1} x B_k e^{\frac{2 i \pi  k x}{n}}}{n e^{\frac{2 i \pi  j x}{n}}}\equal{}\left(\sum _{x\equal{}0}^{n\minus{}1} \frac{f(x)}{n e^{\frac{2 i \pi  j x}{n}}}\minus{}\sum _{x\equal{}0}^{n\minus{}1} \sum _{k\equal{}0}^{n\minus{}1} \sum _{z\equal{}0}^{n\minus{}1} \frac{x (f(n\plus{}z)\minus{}f(z))}{n^3 e^{\frac{2 i \pi  (j x\plus{}k (z\minus{}x))}{n}}}\right)$

It is too complex.
But thanks anyway. I didn't see that.

Too complex=too slow on a computer?

Compute the [b]values[/b] of the $ B$-part at $ 0,\dots,n\minus{}1$ first and $ A$'s afterwords. If $ n$ is up to you to choose, take a power of $ 2$ and use FFT everywhere.

you didn't get. I don't know a priori $ n$.
So FFT is useless. the formula were intended to be used for manipulation only. not numertically

How can you not know $ n$ if you're given the first $ 2n$ values of the function?  Anyway, $ f_n$ has characteristic polynomial $ (x^n \minus{} 1)^2$, and the first $ 2n$ values are enough in principle for you to compute the numerator of the generating function.  I don't see the problem.

[quote="EulerIntegral"]the formula were intended to be used for manipulation only. not numertically[/quote]
OK, then why don't you tell us what the actual problem you are trying to solve is. :)


inequalities
triangle inequality
advanced fields
advanced fields unsolved

How do you show that if $ \{X_n\}$ $ (n\in\mathbb{N})$ is a sequence
in some normed space X such that:

$ \parallel{} X_n \minus{} X_{n\plus{}1} \parallel{} \leq2^{\minus{}n},$

for every n then $ \{X_n\}$ is a Cauchy Sequence.

Use this trick:
If $ p$ is a positive integer we have, based on triangle inequality, that
$ \parallel{}X_{n}\minus{}X_{n\plus{}p}\parallel{}\leq \parallel{}X_{n}\minus{}X_{n\plus{}1}\parallel{}\plus{}\parallel{}X_{n\plus{}1}\minus{}X_{n\plus{}2}\parallel{}\plus{}\cdots\plus{}\parallel{}X_{n\plus{}p\minus{}1}\minus{}X_{n\plus{}p}\parallel{}$

$ <2^{\minus{}n}\plus{}2^{\minus{}(n\plus{}1)}\plus{}\cdots\plus{}2^{\minus{}(n\plus{}p\minus{}1)}<\frac{1}{2^{n\minus{}1}}<\epsilon$ for $ n\geq N(\epsilon)$.


1. A microbe either divides into two identical copies
of itself or else dies. If the probability of division is p = 1/3,
what is the probability that the descendants of the microbe (the
microbe colony) will eventually become extinct?

2. Player A throws 3 dice and player B throws 2
dice. The trials go on simultaneously and successively until a
6 appears on at least one of the dice. Find the probabilities of
the following events:
1. Player A, and not B, first obtains a 6;
2. Player B, and not A, first obtains a 6;
3. Players A and B obtain a 6 simultaneously.

I reckon part b) can be calculated as by working out the probability of each of the situations happening on any given turn:
1. $\frac{3.5^{4}+3.5^{3}+5^{2}}{6^{5}}=\frac{2275}{6^{5}}$ by considering how many of the dice thrown by A are 6s.
2. $\frac{2.5^{4}+5^{3}}{6^{5}}=\frac{1375}{6^{5}}$
3. $\frac{(3.5^{2}+3.5+1)(2.5+1)}{6^{5}}=\frac{1001}{6^{5}}$

I suspect that this is right, since the sum of these is equal to $1-(\frac{5}{6})^{5}$ which should be the probability that a six occurs at all.

To obtain the final probability that each happens on the first occurence of a six, you must divide by the probability that a six appears at all, which is $\frac{6^{5}-5^{5}}{6^{5}}$ Therefore the overall probabilities are:
$\frac{2275}{6^{5}-5^{5}}$, $\frac{1375}{6^{5}-5^{5}}$, $\frac{1001}{6^{5}-5^{5}}$

[quote="Inquisitor"]Hmm, thinking about the first question:
Suppose the probability that the descendants of a given microbe become extinct is 'x'. But the probability that the descendants of a given microbe can also be given by the following:
There is a 2/3 prob that it dies.
If it doesn't the chance that its two descendants both die = $x^{2}$. Therefore, the total probability = $\frac{2}{3}+\frac{x^{2}}{3}=x$.
Therefore x=1! So it always becomes extinct. I expect that there is a mistake somewhere though, since this does not seem, intuitively, to work...[/quote]

That's correct. The population will [i]almost surely[/i] die. (look that up on Wiki.)

[hide="Another way of looking at it:"]
It has a $\frac{2}{3}$ chance of dying at the first event, then $\frac{1}{3}$ chance of making it to the second event, at which point the cycle repeats. So, \[P= \frac{2}{3}+\left( \frac{1}{3}\right) \frac{2}{3}+\left( \frac{1}{3}\right)^{2}\frac{2}{3}+\cdots = \frac{ \frac{2}{3}}{ 1-\frac{1}{3}}= \boxed{1}\] [/hide]

If the population is allowed to go on indefinitely, it will die with probability 1 because the expected increase in population at each step is negative...


trigonometry
limit
calculus
derivative
logarithms
function
calculus computations

Evaluate $\displaystyle\lim_{x\to0}x^{tanx}$

[hide="here is what i am getting"]
$e^{-1}$
[/hide]
im not sure if this is correct could someone please tell me what the answer is.

This isn´t my answer. We have $x^{tan(x)}=e^{tan(x)ln(x)}$. I´m considerating $x>0$. And note that we can write $tan(x)ln(x)=ln(x)\frac{sin(x)}{cos(x)}$ $\to$ $tan(x)ln(x)=\frac{1}{cos(x)}\frac{ln(x)}{\frac{1}{sin(x)}}$. And now we can aplly Lhopital´s Rule to obtain that $\lim_{x\to 0+}\frac{ln(x)}{\frac{1}{sin(x)}} = \lim_{x\to 0+}-\frac{sin(x)^2}{xcos(x)}$. And we can note that this last limit is $0$, because $sin(x)$ is equivalent to $x$ when $x\to 0+$. And so we have that our limit is $e^0=1$.

o i realized what i did wrong. i took the derivative totally wrong. lol. :rotfl:  stupid mistake.

[hide="here is what i did"]
$\displaystyle\lim_{x\to0}x^{tanx}$
$f(x)=x^{tanx}$ so $\displaystyle\lim_{x\to0}e^{lnf(x)}$ is the same as the original limit.
so $\displaystyle\lim_{x\to0}lnf(x)=\frac{lnx}{\frac{1}{tanx}}=\frac{lnx}{cotx}$
from here we can use LHopital's rule. and get
$\displaystyle\lim_{x\to0}\frac{\frac{1}{x}}{-csc^2x}=\frac{-sin^2x}{x}=(-sinx)(1)=0$
so $\displaystyle\lim_{x\to0}lnf(x)=0$
so the finaly answer is $e^0=1$
when i did this the first time i said that the derivative of $cotx$ was $-cscx$ not $-csc^2x$
[/hide]

In my dislike of L'Hopital, here is an alternative:

\[ x^{\tan x} = e^{\ln x\cdot tan x} = e^{\ln x\cdot(x + O(x^3))} = e^{x\ln x(1 + O(x^2))} \to e^0 = 1, x\to0^+  \]

Here we've used the standard limit $x\ln x \to 0, x\to 0^+$ and that $exp$ is continuous.

If you have defined the power function for negative x, I don't know how to show that the left limit is also 1. Maybe someone else can do that.

I believe that the left limit doesn´t exist.


Let ABC be a triangle, M ,M' inside it. X,X' on segment BC . Y,Y' on
segment CA. Z,Z' on segment AB and

M'X//AM , M'Y//BM, M'Z//CM; MX'//AM'; MY'//BM'; MZ'//CM' .

Prove that AX,BY,CZ are concurrent <=> AX',BY',
CZ' are concurrent.

My sol:
MX' cuts M'X at A' , MY' cuts M'Y at B' , MZ' cuts M'Z at C'.
easily to see that : AC'//CA' ,AB'//BA' , B'C//C'B
Applying "ceva sin" we have conclusion.

I don't see how you use the points A', B', C' in your solution, but maybe this is just because I don't really understand your solution. Here is my solution of an extended version of your problem (the points M and M' need not lie [i]inside[/i] triangle ABC; similarly for the points X, X', Y, Y', Z and Z'):

[b]Extended problem:[/b] Let ABC be a triangle, and M and M' two points in its plane. Let X and X' be two points on the line BC, let Y and Y' be two points on the line CA, and let Z and Z' be two points on the line AB. Assume that

M'X || AM; M'Y || BM; M'Z || CM; MX' || AM'; MY' || BM'; MZ' || CM'.

Prove that the lines AX, BY, CZ concur if and only if the lines AX', BY', CZ' concur.

[i]Solution.[/i] Applying the sine law to triangles BM'X and CM'X, we get

$BX=BM^{\prime}\cdot\frac{\sin\measuredangle BM^{\prime}X}{\sin\measuredangle BXM^{\prime}}$;
$XC=CM^{\prime}\cdot\frac{\sin\measuredangle CM^{\prime}X}{\sin\measuredangle CXM^{\prime}}$.

Hence,

$\frac{BX}{XC}=\frac{BM^{\prime}\cdot\frac{\sin\measuredangle BM^{\prime}X}{\sin\measuredangle BXM^{\prime}}}{CM^{\prime}\cdot\frac{\sin\measuredangle CM^{\prime}X}{\sin\measuredangle CXM^{\prime}}} = \frac{BM^{\prime}}{CM^{\prime}}\cdot\frac{\sin\measuredangle BM^{\prime}X}{\sin\measuredangle CM^{\prime}X}\cdot\frac{\sin\measuredangle CXM^{\prime}}{\sin\measuredangle BXM^{\prime}}$.

The angles < BXM' and < CXM' are either equal or complementary (depending on whether the point X lies inside or outside the segment BC); in both cases, sin < BXM' = sin < CXM', and thus the above simplifies to

$\frac{BX}{XC}= \frac{BM^{\prime}}{CM^{\prime}}\cdot\frac{\sin\measuredangle BM^{\prime}X}{\sin\measuredangle CM^{\prime}X}$.

Similarly,

$\frac{CY}{YA}= \frac{CM^{\prime}}{AM^{\prime}}\cdot\frac{\sin\measuredangle CM^{\prime}Y}{\sin\measuredangle AM^{\prime}Y}$;
$\frac{AZ}{ZB}= \frac{AM^{\prime}}{BM^{\prime}}\cdot\frac{\sin\measuredangle AM^{\prime}Z}{\sin\measuredangle BM^{\prime}Z}$.

Thus,

$\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}$
$=\left(\frac{BM^{\prime}}{CM^{\prime}}\cdot\frac{\sin\measuredangle BM^{\prime}X}{\sin\measuredangle CM^{\prime}X}\right)\cdot\left(\frac{CM^{\prime}}{AM^{\prime}}\cdot\frac{\sin\measuredangle CM^{\prime}Y}{\sin\measuredangle AM^{\prime}Y}\right)\cdot\left(\frac{AM^{\prime}}{BM^{\prime}}\cdot\frac{\sin\measuredangle AM^{\prime}Z}{\sin\measuredangle BM^{\prime}Z}\right)$
$=\frac{\sin\measuredangle BM^{\prime}X}{\sin\measuredangle CM^{\prime}X}\cdot\frac{\sin\measuredangle CM^{\prime}Y}{\sin\measuredangle AM^{\prime}Y}\cdot\frac{\sin\measuredangle AM^{\prime}Z}{\sin\measuredangle BM^{\prime}Z}$.

But after the Ceva theorem, applied to triangle ABC, the lines AX, BY, CZ concur if and only if

$\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=1$.

Hence, the lines AX, BY, CZ concur if and only if

$\frac{\sin\measuredangle BM^{\prime}X}{\sin\measuredangle CM^{\prime}X}\cdot\frac{\sin\measuredangle CM^{\prime}Y}{\sin\measuredangle AM^{\prime}Y}\cdot\frac{\sin\measuredangle AM^{\prime}Z}{\sin\measuredangle BM^{\prime}Z}=1$.

Similarly, the lines AX', BY', CZ' concur if and only if

$\frac{\sin\measuredangle BMX^{\prime}}{\sin\measuredangle CMX^{\prime}}\cdot\frac{\sin\measuredangle CMY^{\prime}}{\sin\measuredangle AMY^{\prime}}\cdot\frac{\sin\measuredangle AMZ^{\prime}}{\sin\measuredangle BMZ^{\prime}}=1$.

But since M'X || AM and MY' || BM', the angles < BM'X and < AMY' are either equal or complementary to each other, and thus sin < BM'X = sin < AMY'. Similarly, sin < CM'X = sin < AMZ', sin < CM'Y = sin < BMZ', sin < AM'Y = sin < BMX', sin < AM'Z = sin < CMX' and sin < BM'Z = sin < CMY'. Hence, the equation

$\frac{\sin\measuredangle BM^{\prime}X}{\sin\measuredangle CM^{\prime}X}\cdot\frac{\sin\measuredangle CM^{\prime}Y}{\sin\measuredangle AM^{\prime}Y}\cdot\frac{\sin\measuredangle AM^{\prime}Z}{\sin\measuredangle BM^{\prime}Z}=1$

is equivalent to

$\frac{\sin\measuredangle AMY^{\prime}}{\sin\measuredangle AMZ^{\prime}}\cdot\frac{\sin\measuredangle BMZ^{\prime}}{\sin\measuredangle BMX^{\prime}}\cdot\frac{\sin\measuredangle CMX^{\prime}}{\sin\measuredangle CMY^{\prime}}=1$,

what is, in turn, equivalent to

$\frac{\sin\measuredangle BMX^{\prime}}{\sin\measuredangle CMX^{\prime}}\cdot\frac{\sin\measuredangle CMY^{\prime}}{\sin\measuredangle AMY^{\prime}}\cdot\frac{\sin\measuredangle AMZ^{\prime}}{\sin\measuredangle BMZ^{\prime}}=1$.

But the first of these equations is the criterion for the lines AX, BY, CZ to concur, and the last of these equations is the criterion for the lines AX', BY', CZ' to concur. Hence, the lines AX, BY, CZ concur if and only if the lines AX', BY', CZ' concur. $\blacksquare$

  Darij


Sam is writing at-least-2-digit positive whole numbers on a piece of

paper. After how many numbers written on the paper can we be

sure that there are three of them that start with the same

digit and also end wih an identical digit. (The first and last digits do not

have to be the same.)

[hide="Assuming that..."]
... $0$ cannot be a first digit, there are $9\times10=90$ possibilities for the combination of the first and last digit. If there are 180 numbers, each combination could have been used twice- but if there are 181 numbers, there must be a combination of first and last digit that has been used three times, by the pigeonhole principle.[/hide]


trigonometry
geometry
circumcircle
power of a point
radical axis
geometry solved

Why need inversion? Trigo just works fine for this problem.

Let $X$ be the midpoint of $AC$.
$\frac{\sin\angle ABX}{\sin\angle BCX}=\frac{\sin A}{\sin C}=\frac{\sin\angle MKB}{\sin\angle MLB}=\frac{\frac{BM}{\sin\angle MLB}}{\frac{BM}{\sin\angle MKB}}=\frac{\sin\angle KBM}{\sin\angle LBM}$. Therefore B,X,M are collinear, that is, BM bisects AC.

[color=blue][b]Problem.[/b] Let $ABC$ be a triangle and $BD$ be one of its altitudes. The circle with diameter $BD$ intersects the sides $AB$ and $BC$ at the points $K$ and $L$, respectively (apart from $B$). The tangents to this circle at the points $K$ and $L$ intersect each other at a point $M$.
Prove that the line $BM$ bisects the segment $AC$.[/color]

I don't know whether you are acquainted with the theory of antiparallels and symmedians in a triangle, but in case you are, here is what I believe to be the most explanatory solution of this problem:

We will prove:

[b]Lemma 1.[/b] The line AC is antiparallel to the side KL of the triangle KBL. (Here and in the following, the notion of antiparallelism is always considered with respect to the triangle KBL.)

[b]Lemma 2.[/b] The line BM is the symmedian of the triangle KBL issuing from its vertex B.

As soon as these two lemmas will be verified, the problem will readily follow: It is known that in every triangle, the symmedian from any vertex bisects any segment antiparallel to the opposite side. Now, apply this fact to the triangle KBL, with the vertex B, the corresponding symmedian BM and the antiparallel AC to the opposite side KL, and you see that the line BM bisects the segment AC. This solves the problem.

So it remains to prove the lemmas 1 and 2.

[i]Proof of Lemma 1.[/i] We will use directed angles modulo 180°.

Since the points K and L lie on the circle with diameter BD, we have < KLB = < KDB. But

  < KDB = < (DK; BD) = < (DK; AB) + < (AB; CA) + < (CA; BD) = - < (AB; DK) + < (AB; CA) + < (CA; BD).

Since the point K lies on the circle with diameter BD, we have < BKD = 90°, or, equivalently, < (AB; DK) = 90°. Also, since the line BD is an altitude in triangle ABC, it is perpendicular to the side CA, and thus < (CA; BD) = 90°. Thus,

  < KDB = - 90° + < (AB; CA) + 90° = < (AB; CA) = < BAC = - < CAB.

Since < KLB = < KDB, this becomes < KLB = - < CAB. In other words, < CAB = - < KLB. And therefore, the line AC is antiparallel to the side KL of the triangle KBL. This proves Lemma 1.

[i]Proof of Lemma 2.[/i] Since the points K and L lie on the circle with diameter BD, this circle with diameter BD is the circumcircle of triangle KBL. Hence, the point M is the point of intersection of the tangents to the circumcircle of triangle KBL at the points K and L. But it is well-known that in a triangle, the point of intersection of the tangents to the circumcircle at two vertices lies on the symmedian from the third vertex (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=564645#564645]http://www.mathlinks.ro/Forum/viewtopic.php?t=99571 post #2[/url] Theorem 1 for a proof). Hence, the point M must lie on the symmedian of triangle KBL from the vertex B. In other words, the line BM is the symmedian of triangle KBL issuing from the vertex B. This proves Lemma 2.

And thus, the solution of the problem is complete.

  darij

Both of your solutions are true.
But it has a very nice solution with $Harmonic$  $Conjugates$.

Let $P$ be the intersection of the tangents in $K,L$ to the circle, and let $T$ be the intersection between the circle and $BP$. $BLTK$ is a harmonic quadrilateral, so, if $d$ is the tangent in $B$ to the circle, then $(d,BT;BL,BK)=-1\ (*)$. However, $d$ is parallel to $AC$, meaning that in order for $(*)$ to hold, $BT$ must bisect $AC$, Q.E.D.

Ofcourse Amir 2 cant solve it even with or without inversion :D . I will solve it by this power full tool.

Consider an inversion with center$B$ and an arbitrary radius.

We have a circle $BA'C'$, which is tangent to line $K'L'$ at $D'$.

 a circle which is tangent to $K'L'$ at $K'$, and goes through $B$.

and acircle which is tangent to $K'L'$ at $L'$, and goes through $B$.

and the intersection of these two circles is $M'$.

it is clear that $BM'$ is the radical axis of these two circles.

And also it is clear that $BM'$ , bisects segment $K'L'$. because $K'L'$, is tangent to these two circles.

…………………………………………………………………………………………………………….

$\angle BKD=\angle BDA=90$, and it means that $BKD$ and $BAD$ are similar to each other ,

and : $BK.BA=BD^2$.

$\angle BKD=\angle BDC=90$, and it means that $BLD$ and $BCD$ are similar to each other ,

and: $BL.BC=BD^2$.

It is concluded that: $BK.BA=BL.BC$ ,  and it means that $AKLC$ is cyclic.

A good fact which we should consider is:

 $KL$ is parallel to $A'C'$ .and $AC$ is parallel to $K'L'$.

………………………………………………………………………………………………………………..

then because of $BM'$ bisects $K'L'$. and because of  $AC$ is parallel to $K'L'$,

$BM'$ bisects $AC$ too.

It means that, $BM$ bisects $AC$.

Is it ok, my dear.

Ask me each part which u couldn’t understand.

[quote="silouan"][quote="grobber"] $BLTK$ is a harmonic quadrilateral, so, if $d$ is the tangent in $B$ to the circle, then $(d,BT;BL,BK)=-1\ (*)$. .[/quote]
Why $BTLK$  is harmonic and what means  $(d,BT;BL,BK)=-1$  ??[/quote]

because if u intersect $KL$and $MB$,  and name it $N$,

then: $(BTNM)=-1$, and then the quadrilateral is harmonic.

grobber said that: $BB=d$, means that, he named the tangent from

$B$ to the circle, $d$.

then lines $BB,BK,BT,BL$, are harmonic and we have:

$(d,BK,BT,BL)=-1$

[quote="kvmedya"][i]Amir2 solve it with an [u]inversion[/u][/i],pleaze :roll:  :roll:[/quote]

another thing is that, there not exist any $amir 2$.

because Now he is $Amir.s$


Δυο τριγωνα με πλευρες $ a,b,c$ και $ 1\minus{}a,1\minus{}b,1\minus{}c$ ειναι ισεμβαδικα και εγγεγραμενα και τα δυο σε κυκλο ακτινας $ R$.
Αν $ E$ και $ P$ ειναι το εβαδο και η περιμετρος αντιστοιχα του πρωτου τριγωνου να βρεθει το $ (P\plus{}16RE)_{max}$

Θα ισχύει $ \frac {abc}{4R} \equal{} \frac {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}{4R}$. Οπότε $ abc \equal{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)$ .
 Άρα $ \frac {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}{abc} \equal{} 1$.
Θέτω $ \frac {1 \minus{} a}{a} \equal{} \frac {x}{y}$ , $ \frac {1 \minus{} b}{b} \equal{} \frac {y}{z}$ και $ \frac {1 \minus{} c}{c} \equal{} \frac {z}{x}$ .

 Άρα $ a \equal{} \frac {y}{y \plus{} x}$,   $ b \equal{} \frac {z}{y \plus{} z}$ και $ c \equal{} \frac {x}{x \plus{} z}$ .
Είναι
$ A \equal{} P \plus{} 16RE \equal{} a \plus{} b \plus{} c \plus{} 4abc$ ή
$ A\equal{} \frac {x}{x \plus{} z} \plus{} \frac {y}{x \plus{} y} \plus{} \frac {z}{z \plus{} y} \plus{} \frac {4xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}$ (1).
Θα δείξω ότι $ A\leq 2$ . Άρα $ \frac {x}{x \plus{} z} \plus{} \frac {y}{x \plus{} y} \plus{} \frac {z}{z \plus{} y} \plus{} \frac {4xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\leq 2$   (2).
Μετά τις πράξεις στην (2) έχω ότι $ (x^2)y \plus{} (y^2)z \plus{} (z^2)x\geq 3xyz$ που είναι προφανής από την ανισότητα A.M.-G.M. Άρα (P+16RE)max=2, για α=b=c=1/2.


Which of the following describes the graph of the equation $ (x \plus{} y)^2 \equal{} x^2 \plus{} y^2$?

$ \textbf{(A)}\text{ the empty set}\qquad \textbf{(B)}\text{ one point}\qquad \textbf{(C)}\text{ two lines}$

$\textbf{(D)}\text{ a circle}\qquad \textbf{(E)}\text{ the entire plane}$

[hide]

Square it out and you get:

$x^2+2xy+y^2=x^2+y^2$

Then, we have

$2xy=0$ and $xy=0$  So, either x=0 and/or y=0, which is a horizontal and a vertical line, both through the origin.

C[/hide]

[quote=ishanvannadil2008]@above
Was this post necessary? Not to be rude, but it shouldn't bother you that people are bumping. In all of their posts (in bumped threads), rft hasnt postfarmed[/quote]

yea he made a good totient solution on one
not really sure about @above tho
he's been bumping the really early end problems from 13 years ago, which is usually solved in one step

[quote=sugar_rush][hide=solution]Similar to 2021 12A #2, however that problem has to have $x+y\geq 0$ but this one doesn't :)
[quote]Under what conditions does $x^{2}+y^{2}=(x+y)^{2}$ hold, where $x$ and $y$ are real numbers?[/quote]
\begin{align*}
x^{2}+y^{2}=x^{2}+2xy+y^{2} \\
2xy=0 \\
xy=0 \\
x=0\ \text{or}\ y=0
\end{align*}
two lines :)
hope this helps
It's funny how an old AMC 12 #11 becomes a #2[/quote]

The answer to 2021 AMC 12A P2 is B.

[quote=sugar_rush][hide=solution]Similar to 2021 12A #2, however that problem has to have $x+y\geq 0$ but this one doesn't :)
[quote]Under what conditions does $x^{2}+y^{2}=(x+y)^{2}$ hold, where $x$ and $y$ are real numbers?[/quote]
\begin{align*}
x^{2}+y^{2}=x^{2}+2xy+y^{2} \\
2xy=0 \\
xy=0 \\
x=0\ \text{or}\ y=0
\end{align*}
two lines :)
hope this helps
It's funny how an old AMC 12 #11 becomes a #2[/quote]

this one didnt have a square root. so domian is different. totally diferent problemos

The equation simplifies to $2xy=0$. This includes the $\boxed{\textbf{(C)}\ \text{two lines}}$ $x=0$ and $y=0$.


[quote="charikaar"]anyone know how to do this questin.
please help.

sugar cubes come in boxes of 400 cubes.the cubes are piled 10 high and 5 deep.how many cubes wide are that?

charikaar[/quote]

Is it 8 because 10*5*8=400.

[quote="charikaar"]you mean the answer is 250[/quote]

Uhh... Why would it be 250?

So... We are trying to find the width and we already have two dimensions and the total number of cubes right?

Because if that's the question then the answer is 8.

10*5*x=400
5*x=40
x=8

[quote="charikaar"]this question is tilted as mixed homework problem(i don't have the answer to check whether you are right or wrong).

Charikaar[/quote]

Oh. Ok.

Thank you very much for the prompt respones.

Charikaar

The answer is 8.... 400/(5*10) is 8, for the width..


hello, we get
$ \frac{9!}{5!\cdot 4!}\equal{}\frac{5!\cdot6\cdot7\cdot8\cdot9}{5!\cdot2\cdot3\cdot4}\equal{}14\cdot9 \equal{}126$
Sonnhard.


[quote="Fermat -Euler"]Find
$P=\displaystyle{\prod_{k=1}^{+\infty} (1-6^{-k})}$[/quote]
Several asymptotic results are known about the general product
\[(q)_{\infty}=\prod_{k=1}^{\infty}(1-q^{-k})\]
(see [url]http://mathworld.wolfram.com/q-PochhammerSymbol.html[/url])
but I think that the exact value of $(6)_{\infty}$ is not known.


calculus
integration
function
inequalities
algebra unsolved
algebra

Prove that for every continuous function $f:[-1,1]\rightarrow \mathbb{R}$ the inequality holds, $\int_{-1}^1 f^2(x)dx\geq \frac 12 (\int _{-1}^1 f(x)dx)^2+\frac 32 (\int _{-1}^1 xf(x)dx)^2$

I have a solution, which consider the infinite-dimensional inner product space of all real-valued continuous function on [-1,1], where $<f, g>=\displaystyle \int_{-1}^1 f(x)g(x)dx$.

The idea is this: Let $a(x)=1$ and $b(x)=\sqrt{3}x^2$. Then $|a|^2=|b|^2=2, <a, b>=0$ and what we have to prove is
$|f|^2>=\frac{<f, a>^2+<f, b>^2}{2}$

Hint: prove the inequality for an even function and for an odd function (this can be done by using the integral form of Cauchy-Schwartz). Then, for an arbitrary $f$, write $f=f_1+f_2$, where $f_1(x)=\frac{f(x)+f(-x)}{2}$ and $f_2(x)=\frac{f(x)-f(-x)}{2}$.
Finally, you will find that equality occurs iff $f$ is a linear function.
P.S. The problem was proposed by Dorian Popa.


[i]Moderator's Note:  This was cut out of a discussion about who should be elected President of the United States because, as you see, it has very little to do with that. --JBL[/i]

The only reason I still am a strong Democrat is because of the few white racists that still exist in the Republican Party. If one day the Republicans annihilate those racists I might support the Republican Party. [b]But there is still some hope, [/b]George Wallace to the best of my knowledge, Helms has never done what the late George Wallace did well before his death -- recant and apologize for his use of racial issues. And that use was blatant.  Many of the accounts of Helms's retirement linked him with another prospective retiree, Sen. Strom Thurmond of South Carolina. Both these Senate veterans switched from the Democratic to the Republican Party when the Democrats began pressing for civil rights legislation in the 1960s. But there is a great difference between them. [b]Thurmond, who holds the record for the longest anti-civil rights filibuster, accepted change. For three decades he has treated African Americans and black institutions as respectfully as he treats all his other constituents.[/b]

For years now the vitriol that  Senator Jesse Helms has directed against anyone [b]who is not a white, heterosexual, small town southerner[/b], has made him the "Great Satan" of liberal politics. Recently, even conservatives have turned against Helms.

He is the only politician to continuously have radical racist commericals and threatening mail to minorities till this new century. Jeese Helms was a liar and cheater. He started his career as a talk show host and was a campaign supporter for a person. He cut pictures of a black man and the incumbent's wife and pieced it together and had it publichsed in the newspaper.

In 1953, Helms returned to North Carolina as executive director of the state's banking association, spending the next seven years fighting to enrich his bosses. He won a seat on the Raleigh City Council and, in 1960, took a job as a TV commentator. He spent the decade railing against [b]King, "Negro hoodlums," the media, "sex perverts," and anyone on welfare.[/b] As he explained in one of his nightly five-minute broadcasts, "A lot of human beings have been born bums."

Since Helms won election to the Senate, no "bums" have felt his rage as fiercely as citizens of poor nations. Over the years, the senator has proposed hundreds of measures to slash foreign aid, overthrow governments he doesn't like, and block administration policies. As the new chair of the Foreign Relations Committee, he has made it clear that his first priority is to enact deeper cuts to a foreign aid budget already slashed nearly 30 percent in the past decade.

"The fact is that the American people are sick and tired of this whole foreign aid concept anyhow," Helms said last year. "I find myself wishing that somehow we could put it on a national ballot and say: 'What do you think of this?'"

Those cuts will hit hardest in the Third World, [b]where Helms has long been a staunch ally of right-wing military dictators[/b] like Augusto Pinochet in Chile, Raoul Cedras in Haiti, and Roberto D'Aubuisson in El Salvador. Confronted with evidence that D'Aubuisson directed death squads to murder civilians, Helms made it clear that some things are more important than human life. "All I know," he replied, "is that D'Aubuisson is a free enterprise man and deeply religious."

North Carolina AFL-CIO President Christopher Scott, who has followed Helms' Senate career closely. "He's never won an election by more than 55 percent, and when he's been seriously challenged it's been closer than that. He's been able to artfully use racism and negative character assassination to pull just enough votes over the line to win."

But respect only goes so far--so the Helms campaign hedges its bets by cheating. In 1986, the Federal Election Commission penalized the North Carolina Congressional Club $10,000 and ordered it to reorganize, saying it had illegally subsidized Helms' 1984 campaign. Last year, a decade after the race, the FEC penalized the Helms for Senate committee $25,000 for [b]accepting $700,000 in illegal contributions[/b]. And in 1992, the Helms campaign and the Congressional Club settled a Justice Department complaint over a [b]pre-election mailing of postcards falsely threatening 125,000 black voters with jail if they went to the polls.[/b]

 A year before the election, when public polls showed Helms trailing by 20 points, he launched a Senate filibuster against the bill making the birthday of Martin Luther King Jr. a national holiday. Thurmond and the Senate majority were on the other side, but the next poll showed Helms had halved his deficit.

All year, Peterson reported, "Helms campaign literature sounded a drumbeat of warnings about black voter-registration drives. . . . On election eve, he accused Hunt of being supported by 'homosexuals, the labor union bosses and the crooks' and said he feared a large 'bloc vote.' What did he mean? 'The black vote,' Helms said." He won, 52 percent to 48 percent.

In 1990, locked in a tight race with an African American Democrat, former Charlotte mayor Harvey Gantt, [b]Helms aired a final-week TV ad that showed a pair of white hands crumpling a rejection letter[/b], while an announcer said, "You needed that job and you were the best qualified. But they had to give it to a minority because of a racial quota." Once again, he pulled through.

[quote="churchilljrhigh"]The only reason I still am a strong Democrat is because of the few white racists that still exist in the Republican Party. [/quote]
And there exists no racism in the Democratic Party?

Senator Robert Byrd of West Virginia was an ex-Klu-Klux-Klansman. He said that joining the Klu-Klux-Klan "offered excitement" and because the Klan was an "effective force" in "promoting traditional American values." Nothing like the thrill of gathering around a midnight bonfire, roasting s'mores, tying nooses, and promoting white supremacy with a bunch of your hooded friends.
The ex-Klansman allegedly ended his ties with the group in 1943. He may have stopped paying dues, but he continued to pay homage to the KKK. Republicans in West Virginia discovered a letter Senator Byrd had written to the Imperial Wizard of the KKK three years after he says he abandoned the group.
He wrote: "The Klan is needed today as never before and I am anxious to see its rebirth here in West Virginia" and "in every state in the Union."
The ex-Klansman later filibustered the landmark 1964 Civil Rights Act -- supported by a majority of those "mean-spirited" Republicans -- for more than 14 hours. He also opposed the nominations of the Supreme Court's two black justices, liberal Thurgood Marshall and conservative Clarence Thomas. In fact, the ex-Klansman had the gall to accuse Justice Thomas of "injecting racism" into the Senate hearings. Meanwhile, author Graham Smith recently discovered another letter Sen. Byrd wrote after he quit the KKK, this time attacking desegregation of the armed forces.
The ex-Klansman vowed never to fight "with a Negro by my side. Rather I should die a thousand times, and see Old Glory trampled in the dirt never to rise again, than to see this beloved land of ours become degraded by race mongrels, a throwback to the blackest specimen from the wilds."

True, lets check out Jeesie Helms' quotes.

"No, I do not. And neither do the people in the armed forces. Mr. Clinton better watch out if he comes down here. He'd better have a bodyguard."
-- When asked in 1994 on CNN if he thought President Clinton was "up to the job" of serving as Commander-in-Chief

"Let me adjust my hearing aid. It could not accommodate the decibels of the Senator from Massachusetts. I can't match him in decibels or Jezebels."

[b]-- After Ted Kennedy made an emotional speech to let foreigners with HIV become US citizens, 1993.
"The New York Times and Washington Post are both infested with homosexuals themselves. Just about every person down there is a homosexual or lesbian."
-- 1995[/b]

"The University of Negroes and Communists"
-- Reference to the University of North Carolina devised by Mr. Helms when he worked for Willis Smith's 1950 U.S. Senate campaign.

"All I know is that D'Aubuisson is a free enterprise man and deeply religious."
-- Responding to evidence that Roberto D'Aubuisson directed Salvadoran death squads that murdered thousands of civilians.

"Your tax dollars are being used to pay for grade-school classes that teach our children that CANNIBALISM, WIFE-SWAPPING and MURDER of infants and the elderly are acceptable behavior."
-- Fund raising mailer, 1996

[b]"All Latins are volatile people. Hence, I was not surprised at the volatile reaction."
-- After Mexicans protested his visit in 1986[/b]

"It's their deliberate, disgusting, revolting conduct that is responsible for the disease."
-- Justifying his refusal to give financial support to families of AIDS victims.

"Homosexuals are weak, morally sick wretches."
-- 1995 radio broadcast

[b]"She's a damn lesbian. I am not going to put a lesbian in a position like that. If you want to call me a bigot, fine."
-- Explaining why he was opposing the appointment of a woman for a cabinet post.[/b]

"They should ask their parents if it would be all right for their son or daughter to marry a Negro."
-- In response to Duke University students holding a vigil after Martin Luther King, Jr. was assassinated, 1968

Helms remonstrated ten female members of the House of Representatives to "act like ladies" when they interrupted a Senate Foreign Relations Committee hearing to demand support of a U.N. treaty against gender discrimination, and subsequently had them removed from the hearing by Capitol police. (St. Louis Post-Dispatch, 10/28/99)

[quote="churchilljrhigh"]True, lets check out Jeesie Helms' quotes.[/quote]
I'm not going to argue with you over who is more racist. But what I'm trying to say here is that you claim the Republican Party has a handful of racist politicians, but the Democratic Party does too. Therefore, your support of the Democratic Party based on the fact that racism exists in the Republican Party is very hypocritical.

I know I'm on leave here...lol...just a single comment:

What party's candidate described southerners as white guys with confederate flags on their pickups?

Also, look back at the reason why the Republican Party was formed in the first place. Sure, you might be able to pick out a handful of racist Republicans...but I'm sure you could easily do the same for any political party, and the democrats are no exception. A party's beliefs and status and whatnot should not be determined by the beliefs of certain individuals. (Even if they are in office or are running for office)

Listen, if you want to fight the history of racism in each party at a national level, you are going to lose, badly.  The reason the South is a Republican stronghold nowadays is attributable almost entirely to racial politics during and after the civil rights movement.  That's just life.  I recommend that you back off of that and try and defend on issues where history is less clear.

By the way, it's worth reading the Dean quote in full and actually understanding what he was saying.  He wasn't calling [i]everyone[/i] in the south a "guy with a confederate flag on his pickup."  He was saying that he wanted to be the president for those guys -- part of a larger point that, in voting Republican, poor rural southerners have consistently been voting in favor of their own economic [i]dis[/i]advantage.

Sample Republican Campaign Picture

[img]http://www.evilgopbastards.com/klan2004sm.jpg[/img]

America was one of the most racist countries in the world. Blacks, Irish, Italians, Chinese, Japanese, Jews,Native Americans.... also catholics, atheists, homosexuals... Muslims could be the next

No difference between Republicans and Democrats

Beta, you have a point, but it's not a very good one.  To the extent that both parties include racists, it's good.  To the extent that the core goals of each party are equally reflective of racist beliefs or ideals, or to the extent that it implies that historically, the two parties are equivalent, it is not a good point.

The history of all this really is interesting, by the way -- when Republicans began fostering racist sentiments to win elections in the South and so on.  It's a good history.

All Im gonna say is the Republican party is more racist. This being the perdominatly "rich party" as my history teacher (whose a democrat) puts it, I would expect this but I know that there are more racists in the Republican party coming from a Republican family and community.

I'm confused, I never created a thread called this. Is this one of JBL's ingenious moderator skills?

Yep, sorry, forgot to put a moderator's note at the top -- I'll go do that now.

There are very good arguments that affirmative actions is inherently racist, since it is choosing and therefore discrimination by race, and therefore those who support it are in effect racist.

Not that I'm going to go that far and say that all democrats are racist, but it is a loose term and we should avoid throwing it around without very good support.

And the republican party does have its share of racists. So do the democrats. How many black people in the democratic party absolutely HATE white people? This does count as racism.

I do wonder why people continually avoid actually confronting the point I'm making.

Affirmative action is a *really* different debate than this.  We've had it here at least twice.  We could have it again, but I think the arguments for and against affirmative action, and the people who will be making them here, are pretty clear.

As for your last point, it seems unlikely to me that a particularly larger percentage of African-Americans would "hate" white people than vice-versa.  I've been the target of anti-white discrimination in my neighborhood, and I don't like it, but it's also a much less significant problem, from a societal point of view, than anti-black and anti-hispanic racism.  Furthermore, there are hardly enough black politicians to make the claim that that would at all influence the racist tendancies of the respective parties.

There might be more anti-black and anti-hispanic racism simply because there are more white people. A lot of the black/hispanic population near here is pretty racist, while not too many whites around here are...

I would like to see some study done regarding percentages of black, hispanic, white that are racist though... i suppose it would be difficult to conduct, yet the results might be interesting.

[b]The Truth Behind Nukular's Comment [/b]
[i]
*Moderator's Note* The text of what this post used to be can be found by looking at Nukular's most recent comment and by visiting the following website:[/i]

http://davidgrenier.com/000098.html

[i]--JBL[/i]

One of the weirdest things about what's happened to racism in America is that it's impossible to find anyone who thinks that they are a racist.  In the 1960's, you could go around asking survey questions like "If your daughter told you she had a Negro friend she wanted to invite over, would you (a) tell her it was all right?  (b) tell her she could play with the girl at school but not bring her home or go to her home?  or (c) forbid her from playing with the girl?"  and people would give you honest answers.  Nowadays, the only ways to test racism are very roundabout, and usually involve tests like showing a series of pairs of faces and asking individuals which face they associate with a positive adjective, or some such thing.  Even as someone who believes that racism is (unfortunately) alive and well, I think they're rather silly methods.

As for your point about there being more white people, it's also quite relevant to consider that whites hold a disproportionate share of power, which is what makes that racism more significant, from a political point of view.

again churchill i cant even read past your first paragraph... the guy that came up with the "white" scholarship (thats like $350 or something like that... it was intended as political protest) was <i>hispanic</i>, and got a hispanic scholarship to go to school himself.

And secondly, I do believe in financially based affirmative action, which takes care of people in urban areas and is fairly objective; what I do have a problem with is racially based affirmative action, unless you would like to assume that all black people are poor and need a boost and all white people are rich and deserve to be forced to work harder to get into college...


function
limit
real analysis
real analysis unsolved

Let $f_n$ be a sequence of non-negative continuous functions on $\mathbb R$. Let $A$ be the set of all $x$ such that $\sum_{n\ge 1}f_n(x)<+\infty$. It is more or less clear that $A$ always is an $\mathcal F_{\sigma}$-set (i.e., a countable union of closed sets). Can every $\mathcal F_{\sigma}$-set arise this way?

We claim that $A_m=\{\sum_{n\ge 1}f_n(x)\le m\}$ - is closed. Now we prove that $CA_m$ - is open.
Let $x_0\in CA_m$. Then $\sum_{n\ge 1}f_n(x_0)> m.$ Thus $\lim_{N\to \infty} \sum_{n=1}^{N}f_n(x_0)> m$ and $\exists N,\varepsilon>0$ suh that $\sum_{n=1}^{N}f_n(x_0)> m+\varepsilon.$ Now $\forall 1\le n\le N,\ \exists \delta_n$ such that $|x-x_0|< \delta_n \Rightarrow |f_n(x)-f_n(x_0)|<\varepsilon/2^{n+2}.$ Let $\delta = \min_{1\le n\le N} \delta_n$. Then \[ \forall x\in (x_0-\delta;x_0+\delta) \ \sum_{n=1}^{\infty}f_n(x)\ge \sum_{n=1}^{N}f_n(x)>\sum_{n=1}^{N}f_n(x_0) - \sum_{n=1}^{N}\varepsilon/2^{n+2} > m+\varepsilon -\sum_{n=1}^{\infty}\varepsilon/2^{n+2}=m+\varepsilon-\varepsilon/4>m.  \] It follows that  $(x_0-\delta;x_0+\delta) \subset CA_m$ and $CA_m$ is open. It follows that $A_m-$ is closed, and \[ A=\bigcup_{m\ge 1} A_m.  \]

anik_andrew, what you just did was to add details to fedja's incidental sentence, "It is more or less clear that $A$ always is an $\mathcal F_{\sigma}$-set (i.e., a countable union of closed sets)."

What you didn't do is answer the question fedja actually asked, "Can every $\mathcal F_{\sigma}$-set arise this way?"


1. Consider the following circuit, composed by one dc source, four resistors, and two capacitors [i]initially discharged[/i].

(a) What is the current intensity in the circuit in the initial instant?

(b) What is the voltage across the terminals of $ R_{3}$ and $ R_{4}$ in that same instant?

2. Consider the following circuit, composed by one dc source, five resistors, and two capacitors [i]initially charged[/i].

(a)  What is the current intensity in the circuit in the initial instant?

(b) What is the voltage across the terminals of both capacitors in that same instant?

Can you show your reasoning for both problems?

I used the result that for an [b]uncharged[/b] capacitor connected across a circuit, it [b]instantaneously[/b] behaves as a [b]conductor[/b], while if the capacitor is [b]initially charged[/b], it [b]instantaneously[/b] behaves as an [b]insulator[/b]. You can get these results from equation showing the variation of charge with time. (There are also such results for equilibrium conditions.) Using this, the problems reduce to simple resistor systems.

Ok. So, for problem 1 how do you proceed and calculate I for example? Specifically, how do you calculate the equivalent resistance?

In problem 1, replace capacitors with wires. We see that $ R_{1}$, $ R_{2}$,$ R_{3}$,$ R_{4}$ are in parellel, hence, $ R_{eq}\equal{}\frac{R_{1}R_{2}R_{3}R_{4}}{R_{1}R_{2}R_{3}\plus{}R_{2}R_{3}R_{4}\plus{}R_{3}R_{4}R_{1}\plus{}R_{4}R_{1}R_{2}}$
In problem 2, remove the capacitors, We see that $ R_{1}$, $ R_{2}$ are in parellel and together in series with $ R_{3}$, $ R_{4}$ and $ R_{5}$
Hence, $ R_{eq}\equal{}\frac{R_{1}R_{2}}{R_{1}\plus{}R_{2}}\plus{}R_{3}\plus{}R_{4}\plus{}R_{5}$

Ok, that's right. Thanks.

By the way, how did you get $ V_{1}$ and $ V_{2}$ for question 2.b)?


We can write a closed formula to this expression.

I'm not so sure about that...  :blush:  How?

I will show a general way to find formulas for such things on this example (for this special one there is an easier way):

Use $1+x+x^2+...+x^n=\frac{x^{n+1}-1}{x-1}$, derivate and multiply it with $x$ to get $1x+2x^2+3x^3+... +n x^n = x\frac{(n+1)x^n(x-1)-x^{n+1}+1}{(x-1)^2}=\frac{nx^{n+2}-(n+1)x^{n+1}+x}{(x-1)^2}$.
Set $x=2$ and you get $1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + .. + n \cdot 2^n =2n2^{n+1}-(n+1)2^{n+1}+2=(n-1)2^{n+1}+2$.


function
number theory proposed
number theory

It is known  x^19+y^19=x^99+y^99   such,    x,y∈R,|x┤>1,|y├>1  It is he true x+y=0   ???? :wink:  :lol:

Function $ f(x) \equal{} x^{99} \minus{} x^{19}$ is strictly increasing on $ (\minus{}\infty,\minus{}1) \cup (1, \plus{}\infty)$. The given equation writes $ f(x) \equal{} f(\minus{}y)$, hence $ x\equal{} \minus{}y$, or $ x\plus{}y\equal{}0$.


How many digits are in the decimal representation of the product of $ 2^{12}$ and $ 5^7$?

[quote=math1108]how do u do it without the calculator?[/quote]

Just see the order of $2$ and $5$. And see which limits the product, in this case $5^7$ limits. So, we get $5^7.2^7=(8 \text{digits}$ then we know $2^5=32$. So, $5^7.2^{12}=9\text{digits}$

ok... im still kinda confused about what limits the product

That's my language! Just see no more 10 can be produced since 5 is finished up! :D

ok thanks so much you have to friend me tho


I need to find the numbers 1-20 using only four fives. YOU can use can use adding, subtracting,multiplication,dividing,order of operations,parentheses,grouping,decimals.

i basically just cant use exponets.

Please help

[quote="ch1n353ch3s54a1l"]$17=\displaystyle\left\lceil\frac{5!}{5.5}\right\rceil - 5$
Can you use that function?[/quote]
[quote="abcdefg"]YOU can use can use adding, subtracting,multiplication,dividing,order of operations,parentheses,grouping,decimals. [/quote]
So I don't think factorials, floor functions or ceiling functions are allowed.


This thing is troubling me since the late February that is when I heard the news that 7 HODs of Bansal left and opened a new institute named Vibrant.
I wanted to go to Bansal because of its best faculty and highly competitive atmosphere. Now, the Bansal faculties have left but still most of the students want to remain in Bansal only. So, the advantages of Bansal (for me) have split.
Now, what should I do? Go for Bansal or Vibrant?
Also, if you know when and where will Vibrant conduct its entrance exam for students of X (going to XI), then please tell me?

well well this would put many in doubts!!!!

Please answer as soon as possible as seats are limited in Vibrant Academy.

If anybody of you has studied under Nitin Jain sir or Narendra Avasthi sir, then please tell your experiences under them.

well yes this is true that top 7 faculty teaching p1-p2-p3 batches has left bansal but still it has tried to almost compensate the loss.........they have bougth many good teachers........

a d gupta of physics..........
best of carrer point as maths.........
and others........

I don't know much but some of my acquaintances have left bansals and they were in gud enuf batches.

bansals needs to fix it quickly

from my vantage point now, I can say coaching institutes suck, they only help eat up your precious plus two's, but of course by the time you understand that, it's always too late.
I wish I could convey to you youngsters in some way the utter contempt we hold these 'Bansal sir's in now.

I strongly disagree with going to Kota for coaching. But if you have some truly awesome coaching in your hometown like me, which finihes course by september-october and doesn't screw your boards, I'm more than sattisfied. Besides our coaching teachers are so nice, they hold extra classes at our request to teach us the school topics. :)

And i wanted to screw up my boards somehow!!! :D  :)

I also want to screw up my boards and I want to study under great teachers................but for Vibrant one thing that is troubling me is that there's always a chance that Vibrant teachers are not able to manage and fight and break away. In that way, what shall we do?

I again repeat my request, please if anybody has studied under Nitin Jain or Narendra Avasthi sir, please tell your experiences.

1 of my ex-classmates in P2 is leaving for Vibrant with her friends coz those were all their fav teachers. And another good friend is leaving bansal for Resonance in 12th.
I'd strongly suggest neither... Just join a good coaching in your hometown and do more self study.

Thats i wanted to say!!
being a nerd wont help ur coz!! :D

@nuclear.............
u mean shivangi with ex classmeate???????

Ermmm firstly, don't mention this to anyone I said about Shivangi. She didn't want to. Thats why I never took her name and I thought you'd not remember lol. Anyways, the girl in P2 was Shivangi and the other guy- Shantanu. Forgot Shantanu's batch lol. He's in Lucknow giving school exams these days, but we ambush his house every other day for multiplayer games :P


function
number theory unsolved
number theory

[quote="hien"]Let a function $f(x) = 5x^{13}+13x^{5}+9ax$. Find minimum positive integer $a$ such that $f(x)$ is divisble by 65 for all integer numbers $x$[/quote]
65=5*13. $5|13x^{5}-13x \ \forall x$. Therefore must be $5|(9a+13)x \ \forall x$. $13|(5x^{13}-5x \ \forall x$, therefore must be $13|(9a+5) \ \forall x.$ It give $9a=2(mod \ 5), \ 9a=8(mod \ 13)$ or $a=3(mod 5), a=11(mod 13)$ or $a=63(mod \ 65)$.


Hi! I like this puzzle but I don´t know if any competitions exits anyway.....

[quote="laughinghead505"]What do you mean by that? Post the puzzle. I am eager for a good puzzle.  :lol:  (oh, and welcome to aops!) :)[/quote]
Here's a link to an online version of the puzzle:
[url=http://www.holotronix.com/samlloyd15.php]Online Puzzle[/url]
Here's a link to information about the puzzle (and how it's related to math  :D ):
[url=http://mathworld.wolfram.com/15Puzzle.html]Puzzle Info[/url]

Oh, I see. That is a pretty cool puzzle! :lol:

I have known this puzzle since I was a little kid.
Has anyone seen or knows a 3D version of this?(ie 4*4*4 cube with 63 uint cubes numbered from 1 to 63) I don't even know if this 3 version exists, this is just a totally random thought. It would be nice if it exists though

My records:

http://www.speedcubing.com/

Unofficial World records-Other Puzzles-Samuel lloyd´s...

I am David and I think I am good, unfortunately there are no competitions I think...


trigonometry
inequalities unsolved
inequalities

$ x$ , $ y$ ,$ z$ $ \in \mathbb R$  such that          $ \sin(x)\plus{}\sin(y)\plus{}\sin(z) \geq \frac{3}{2}$           , prove that :

$ \sin\bigg(x\minus{}\frac{\pi}{6}\bigg)\plus{}\sin\bigg(y\minus{}\frac{\pi}{6}\bigg)\plus{}\sin\bigg(z\minus{}\frac{\pi}{6}\bigg) \geq 0$

Presume $ \sum {\sin(x \minus{} \frac {\pi}{6})} < 0$ . Then we have  $ \frac {\sqrt {3}}{2}(\sum {\sin x}) < \frac {1}{2}(\sum {\cos x}).$

  But $ \sum {\sin x}\geq \frac {3}{2}$ , result $ \sum {\cos x} > \frac {3\sqrt {3}}{2}$ and then $ \sum {\cos^2 x} \plus{} 2\sum {\cos x\cos y} > \frac {27}{4}$

 Because $ \sum {\sin x}\geq\frac {3}{2}$  result $ \sum {\sin^2 x} \plus{} 2\sum {\sin x\sin y}\geq\frac {9}{4}$

 Then  $ 3 \plus{} 2\sum {\cos (x \minus{} y)} > 9$ or $ \sum {\cos (x \minus{} y)} > 3$ false.

thank's but why $ \sum cos(x\minus{}y) \geq 3$ is false

It is false because the maximum value $ cos(x)$ can take is 1. Forget what $ x\minus{}y$ equals, since 3 $ cos(x)$ are being added together its maximum is 3, and since he has shown it would be strictly $ >3$,  it is false. So the assumption in his first line is false. Hence he proved.. bla bla


algebra polynomial calculus integration function

:D  :D  :D

Let $ P(x)$ be a polynomial with integral coefficients such that $ P^{n}(x) \equal{} x$ for some $ n > 2$ and  and some integer  $ x$  then prove that $ P^{2}(x) \equal{} x$ [b]for all integral $ x$[/b] where $ P^{i}(x) \equal{} P(P(P(....P(x))...)$
where the $ P$'s are $ i$ times

[quote="pardesi"]:D  :D  :D

Let $ P(x)$ be a polynomial with integral coefficients such that $ P^{n}(x) \equal{} x$ for some $ n > 2$ then prove that $ P^{2}(x) \equal{} x$ where $ P^{i}(x) \equal{} P(P(P(....P(x))...)$
where the $ P$'s are $ i$ times[/quote]

I have an idea but am not sure how to proceed.

Anyways we have to prove that $ f(x)$=$ f^{\minus{}1}(x)$

Consider f(x) to be a polynomial of degree k

$ f(x) \equal{} akx^k \plus{} ....... \plus{} a0$

Consider $ f(f(x)) \equal{} ak (f(x)^k) \plus{} ... \plus{} a0$

So $ f^2(x)$ has degree $ k^2$....

And so going on $ f^n(x)$ has degree $ k^n$. And by plain logic k = 1. And constant term should be zero again by stupid logic.

So f(x) = cx
and f(x) = x as well because c=1 unless you can say for sure that n is even.

So, f(x) = x.
Hence Proved.

[quote="pardesi"][quote="Cyber-Shreyas"][quote="pardesi"]sorry see the edited question :([/quote]

My answer remains the same.[/quote]
no ur proof won't hold because u can't compare coefficients since u don't know how it behaves for reals :)[/quote]

Poda Dog. Anyways Have to go. Will be only be back on Sun. Cya!

@Pardesi: Here you can compare the coefficients coz shre is comparing the constant only, and it is obvious that the coefficient is 0.
And the other comparison is of the degree, where the degree is $ k^n$ always..

no u can't because if $ P(x)\equal{}R(x)$ for all $ x \in \mathbb{R}$ then  and only then u have the coefficients but here $ P^{k}(x)\equal{}x$ holds only for integral $ x$

I feel the soln there now is perfect. I donno if he's edited stuff..

see my edited reply da ...i mis read u :)

Hmm .... The function (P^n)(x) - x has infinitely many  roots, doesnt it ?


geometry
conics
ellipse
incenter
geometry unsolved

$ G$ is the centroid of △$ ABC$, $ PQ$ pass through $ G$, and $ QG \equal{} 2PG$, △$ DEF$ is the ceva triangle of $ Q$ aboat △$ ABC$. Prove that：$ S(PEAF) \equal{} S(PBC)$ .
When $ P \equal{} K$($ K$ is the symmedian point of △$ ABC$), then $ Q \equal{} H'$($ H'$ is Isometric conjugate point of $ H$)

[hide]Need use [b]Newton Line[/b][/hide]

[quote="behdad.math.math"][quote="lym"]
[hide]Need use [b]Newton Line[/b][/hide][/quote]

What is Newton line?[/quote]

[b]Newton line [/b]is a line which is formed by three midpoints of three diagonals of a complete quadrilateral

Nobody？ there are some ways to prove, let me give one：

Let $ M, T, N, S$ respectively be the midpoints of $ BC, PQ, EF, AQ$ . First we can easy get $ AP//MT$, then $ M, T, S$ are on a line .
Becase of $ MNS$ is a newton line, so $ N$ is on $ MT$, hence $ S(NAP)\equal{}S(MAP)$,  and as known $ 2S(NAP)\equal{}S(FAP)\minus{}S(EAP), 2S(MAP)\equal{}S(BAP)\minus{}S(CAP)$ .
So we get $ S(FAP)\minus{}S(EAP)\equal{}S(BAP)\minus{}S(CAP)$, hence $ S(PBF)\equal{}S(PCE)$, Similarly, $ S(PAF)\equal{}S(PCD), S(PAE)\equal{}S(PBD)$
Then we get $ S(APEF)\equal{}S(PBC)$ .

[b]Remark[/b]：
Let $ CX\equal{}BD$, Similarly delimit $ Y, Z$, then $ S(PAZ)\equal{}S(PAY)$, so $ AP$ equally $ YZ$, Similarly, $ BP, CP$ equally $ XZ, XY$ .
[color=red]So we can get a very important conclusion：A ellipse which's center is $ P$ tangent $ BC, AC, AB$ at $ X, Y, Z$ .[/color]
Let $ P\equal{}K$, $ K$ is symmedian point of △$ ABC$, then we get $ X, Y, Z$ are the Pedals, $ MK$ equally $ AX$ .
Let $ P\equal{}I$, $ I$ is incenter, then we found $ Q\equal{}Na$, $ Na$ is Nagel point .


What is the sum of all integers $ x$ for which $ |x\minus{}2| <4$?

This is 2 different equations:
$ x\minus{}2<4$ and $ x\minus{}2>\minus{}4$
$ x<6$ and $ x>\minus{}2$
The integers that satisfy this are -1 thru 5, inclusive, which is $ \boxed{7}$

And the sum is $ \minus{}1\plus{}0\plus{}1\plus{}2\plus{}3\plus{}4\plus{}5\equal{}\boxed{14}$


Let a, b be integers such that 47/151<a/b<34/109, what is the minimum of b?
(A)58 (B)61 (C)77 (D)93

[hide]Well, 47/151 is about 0.31125, and 34/109 is about 0.31193. The smallest fraction that works is 19/61, which is about 0.31148.

I did this by brute force; I don't know how you would systematically solve it.[/hide]

[hide="Discussion"] Consider the lattice points $ A(0, 0), B(151, 47), C(a, b), D(109, 34)$.  The problem condition says that $ AC$ is between $ AB$ and $ AD$, therefore that $ ABCD$ is a (concave) quadrilateral.  In particular, $ C$ is the lattice point inside triangle $ ABD$ that is furthest to the left.  This should motivate a solution by [url=http://www.cut-the-knot.org/ctk/Pick.shtml]Pick's theorem[/url].  :)[/hide]


Let $x,y,z \in R$ and satisfying$x^{2}+y^{2}=1$ and $y^{2}+2y(x+z)=6$
Prove that:$y(z-x) \leq 4$


modular arithmetic
algebra
polynomial
Vieta
linear algebra
matrix
number theory

In sequence $\{ b_n \}$, $b_1=0$, $b_2=2$, $b_3=3$ and $b_{n+1}=b_{n-1}+b_{n-2}$. If $p$ is a prime,
Prove: $p|b_p$.

And also, if $p|b_p$, can we get $p$ is a prime from it?.

Actually, it is the Perrin sequence: [url=http://mathworld.wolfram.com/PerrinSequence.html]mathworld.wolfram.com/PerrinSequence.html[/url]

It is clear how to construct other similar sequences $\{b_n\}$ with the property $p\mid b_p$ for every prime $p$.

Take any polynomial (of any degree) with integer coefficient and zero second but leading coefficient. For example, $f(x)=x^4 + 3x^2 - 7x - 1$.
Using [url=http://mathworld.wolfram.com/SymmetricPolynomial.html]Newton-Girard formulas[/url] compute power sums $S_k$ of the roots of $f(x)$ for $k=0,...,\deg(f)-1$ and let them be initial terms of the sequence $\{b_n\}$. For the specified $f(x)$ we have
$b_0 = S_0 = \deg(f) = 4$
$b_1 = S_1 = 0$
$b_2 = S_2 = -6$
$b_3 = S_3 = 21$
the other terms are defined by $f(x)$ as a characteristic polynomial.
In our example $b_n = -3 b_{n-2} + 7b_{n-3} + b_{n-4}$. Hence, the sequence starts with
$4, 0, -6, 21, 22, -105, 75, 490, -938, -1050, 6319, -2926, -27245, 51961, \ldots$
where we observe (as expected)
$2\mid b_2=-6,$
$3\mid b_3=21,$
$5\mid b_5=-105,$
$7\mid b_7=490,$
$11\mid b_{11}=-2926,$
$13\mid b_{13}=51961$
and so on.

[quote="Rushil"]this is Lucas series right??[/quote]
A Lucas series is one whereby the first $2$ terms of the series is fixed, and every term is the sum of the previous two terms. For example, the Fibonacci sequence is an example of a Lucas series/sequence...

[quote="maxal"]If \[x^3-x-1\equiv (x-q_1)(x-q_2)(x-q_3)\pmod p\] then for $n=0,1,\ldots,$
\[b_n \equiv q_1^n + q_2^n + q_3^n\pmod p\]
and \[b_p \equiv q_1^p + q_2^p + q_3^p \equiv q_1 + q_2 + q_3 \equiv 0\pmod p\]
implying $p\mid b_p$.

If $x^3-x-1$ is not a product of linear factors over the field $F_p$ we can cosider an algebraic closure $c(F_p)$ and factorization $x^3-x-1=(x-q_1)(x-q_2)(x-q_3)$ over it. Then note that $q_1^p + q_2^p + q_3^p=0$ since $(x^3-x-1)^p = x^{3p}-x^p-1$ over $c(F_p)$.[/quote]
\[b_p \equiv q_1^p + q_2^p + q_3^p \equiv q_1 + q_2 + q_3 \equiv 0\pmod p\]

why?Is that Fermat's Little Theorem?But in Fermat's Little Theorem the base should be a natural number.

[quote="dingdongdog"]\[b_p \equiv q_1^p + q_2^p + q_3^p \equiv q_1 + q_2 + q_3 \equiv 0\pmod p\]

why?Is that Fermat's Little Theorem?But in Fermat's Little Theorem the base should be a natural number.[/quote]
Here $q_1, q_2, q_3$ are some residues modulo $p$. Fermat's Little Theorem implies $q_1^p + q_2^p + q_3^p \equiv q_1 + q_2 + q_3\pmod p$ and Vieta's formula implies $q_1 + q_2 + q_3 \equiv 0\pmod p$.

In Fermat's Little Theorem the base should be a natural number，isn't it?

Well, they are integers here (or more exactly: residue classes) ;-)
But in the general case you only know that $q_1^p + q_2^p + q_3^p \equiv (q_1 + q_2 + q_3)^p \equiv 0 \mod p$ which is also enough to prove the result.

[quote="maxal"]If \[ x^3-x-1\equiv (x-q_1)(x-q_2)(x-q_3)\pmod p \] then for $n=0,1,\ldots,$
\[ b_n \equiv q_1^n + q_2^n + q_3^n\pmod p \]
and \[ b_p \equiv q_1^p + q_2^p + q_3^p \equiv q_1 + q_2 + q_3 \equiv 0\pmod p \]
implying $p\mid b_p$.

If $x^3-x-1$ is not a product of linear factors over the field $F_p$ we can cosider an algebraic closure $c(F_p)$ and factorization $x^3-x-1=(x-q_1)(x-q_2)(x-q_3)$ over it. Then note that $q_1^p + q_2^p + q_3^p=0$ since $(x^3-x-1)^p = x^{3p}-x^p-1$ over $c(F_p)$.[/quote]
Guys I can't undestande,why we can finde tree numbers $q_1,q_2,q_3$ that
\[ x^3-x-1\equiv (x-q_1)(x-q_2)(x-q_3)\pmod p \] ? :?

[quote="ZetaX"]Read the last two lines of the cited part ;)[/quote]
[b]ZetaX[/b] can you explain me that part? :?

Here it would also be enough to consider a finite extension:

Let $q_1,q_2,q_3$ be any residues $\mod p$ that are all (not even distinct) roots of that polynomial (the difference is: we define them to be, we don't say they are 'normal' residues itself; this is like defining $i$ as a solution of $x^2+1=0$ over the reals just to be able to handle such equations).
Now consider all numbers we can 'reach' by addition, subtraction, multiplication and division (like all numbers of the type $a+bi$ at the complex case). One can show that we can calculate with them as used (they form a field, so there are inverses and everything else).

Also one well known property holds again: $p=0$, since that also hold before considering that numbers (note: we work $\mod p$!).
Thus $(a+b)^p = a^p + b^p$ can be proven the same way as usuall, where $a,b$ can now be any numbers of that new ones too.

The author of the original post used even the (more brutal) algebraic closure:

A field $A$ is called algebraically closed when every polynomial with coefficients in $A$ has a root in $A$ (one example for such a field are the complex numbers).
Now one can show that for every field $F$, there exists a (minimal) field $A$ being an extension of $F$ (thus also contains $F$) and algebraically closed itself. This means also that every polynomial over $F$ has a root in $A$. This $A$ is called the algebraic closure of $F$.
Now apply that to $\mathbb{F}_p$, the field of the residues $\mod p$, to be able to work somewhere where every polynomial can be written as product of linear terms.

Again it holds $p=0$ then, giving the same property $(a+b)^p = a^p + b^p$ for all $a,b$ from that $A$ (and the $q_i$ exist be definition).

It may worth to notice that $b_n = trace(B^n)$, where
\[ B=\left(\begin{matrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 1 & 0\end{matrix}\right), \]
and then to prove that for prime $p$, characteristic polynomial of the matrix $B^p$ is $x^3 - x - 1$ modulo $p$.


Hello Everyone,

When playing AoPS, I came across quadratic equations, series of numbers (2+4+6+8+...+48, 1+3+...+79, etc.) and infinite series (1/2 + 1/4 + 1/8 + ... + 1/2^n where n is infinity, etc.).
Please explain how to solve these topics because I always miss them.

akhil0422

Quadratic equations: Use the quadratic formula. If ax^2+bx+c=0, then x=-b+-sqrt(b^2-4ac)/2a

For the finite series like 2+4+6+8+...+84, all you have to remember is the formula for the sum of the first n counting numbers...

       n(n+1)/2

For example.. the sum 1+2+3+...+10 = 10(11)/2 = 55

So for 2+4+...+84, we will have 2(1+2+...+42) = 2(42)(43)/2 = 42(43) = 1806...

Actually, all you have to do is play with the formula.. and you'll find the answer...

Try doing the 1+3+...+79, by first finding a way to express it as 1+2+...+n.

For the infinite series... express the series in the form x + x^2+x^3+... and use the formula
            x/(1-x).
Note: This formula only applies if |x|<1.

For example... 1/2 + 1/4 + 1/8 + ... = 1/2 + (1/2)^2 + (1/2)^3 +... = (1/2)/(1-1/2) = 1...

in some cases it is hard to express it to the form x + x^2+..., but if you pay close attention to th form of the series.. i guess you'll manage to do it...

Note: There is actually a general formula, but sadly i can't remember it... and I'm already satisifed with the process I showed you... ^_^

Quadratic equations: If the quadratic is of the form $ ax^2\plus{}bx\plus{}c\equal{}0$, the roots are always $ x\equal{}\frac{\minus{}b\pm\sqrt{b^2\minus{}4ac}}{2a}$.

Series of numbers: When we evaluate series, we try to find the average of all the terms and then multiply by the amount of terms.  For example, take $ 2\plus{}4\plus{}6\plus{}\ldots\plus{}48$.  The average will be the sum of the first and last divided by $ 2$, or $ \frac{48\plus{}2}{2}\equal{}25$.  To find the number of terms, try dividing the entire sequence by $ 2$.  Then we have a list of consecutive integers: $ 1,2,3,\ldots,24$.  There are obviously $ 24$ numbers, and multiplying by the average, we get $ \boxed{600}$.


I have a round bar with a radius of 6 cm.How do I find the area of the round cross section of this bar ?

If you're asking for the area of the circle, then it's...

[hide]$6^{2}\pi=36\pi$[/hide]

[quote="guess"]here is the question exactly as stated in my book.
 A round bar has a radius of 6 cm. The area of the round cross section of this bar is
 A. 7.1 cm2
 B. 18.8 cm2
 C. 28.26 cm2
 D. 113.04 cm2

  I really don't know how to come up with the answer. I would appreciate any help in explaining to me how to figure it out. I want to find the answer out but I want to know how to figure it out.[/quote]

The book wants you to assume that $\pi=3.14$

[quote="guess"]here is the question exactly as stated in my book.
 A round bar has a radius of 6 cm. The area of the round cross section of this bar is
 A. 7.1 cm2
 B. 18.8 cm2
 C. 28.26 cm2
 D. 113.04 cm2

  I really don't know how to come up with the answer. I would appreciate any help in explaining to me how to figure it out. I want to find the answer out but I want to know how to figure it out.[/quote]
[hide]$36\pi\approx36\cdot3.14=113.04\Rightarrow D$[/hide]


What is the unit digit of the expansion of

"7 to the power 7 to the power 7"

I know how to do it for "7 to the power 7" but how to go about this one?

will the final answer be the same as ".... 3 to the power 7"

[hide = Ans] 3 is the units digits
I have used cyclicity funda here.
First, we find remainder when exponent 7^7 divided by 4 (since cyclicity of 7 is 4)
(-1)^7 = -1 implies remainder is 3
now it boils down to finding units digit of 7^3, which is [b] 3 [/b]
[/hide][/b]

If you meant $ 7^{(7^7)}$, as the above poster did:
[hide]In case you don't know what cyclicity is, here is the same thing rephrased.

Check out that:
$ 7^7 \equiv ( \minus{} 1)^7 \equiv \minus{} 1 \equiv 3 \pmod{4}$ and that $ 7^4 \equal{} 49^2 \equiv 9 ^2 \equiv 81 \equiv 1 \pmod{10}$

We then have for some $ k \in \mathbb{N}$, that $ 7^{7^7} \equal{} 7^{4k \plus{} 3} \equal{} (7^4)^k \cdot (7^3) \equiv 1^k \cdot 7^3 \equiv 7^3 \pmod{10}$

So, it is the same as the units digit of $ 7^3$.

$ 7^3 \equiv 49 \cdot 7 \equiv 9 \cdot 7 \equiv 63 \equiv 3 \pmod{10}$ so the unit's digit is $ \boxed{3}$

In case you don't know what "funda" is, it roughly equates to "dealy"  :D[/hide]

If you meant $ (7^7)^7$, then it is just:[hide]
$ \equiv ((\minus{}3)^7)^7 \equiv ((\minus{}3)^6 (\minus{}3))^7 \equiv ((9^3 (\minus{}3))^7 \equiv ((\minus{}1)^3 (\minus{}3))^7 \equiv 3^7 \pmod{10}$, as you stated. From here, you can now do $ 3^7 \equiv (3^6)(3) \equiv (9^3)(3) \equiv ((\minus{}1)^3)(3) \equiv \minus{}3 \equiv \boxed{7} \pmod{10}$[/hide]

This topic brings up a question I've been thinking about recently. Is $ 7^{7^{7}}$ equal to $ \left(7^{7}\right)^{7}$ or $ 7^{\left(7^{7}\right)}$?

The second. The first can be easily rewritten as $ 7^{7 \cdot 7}$, but the second can't be simplified in a similar manner, so the notation $ 7^{7^7}$ implies $ 7^{(7^7)}$, and similarly, $ a^{b^{c^d}} \equal{} a^{(b^{(c^d)})}$.

The ambiguity in this case is because it is written out in words.

Thanks for the replies. i figured it out... its same as what ganwar as used.


inequalities
algebra
polynomial
inequalities proposed

For any three positive reals a, b, c, we have

$\displaystyle \frac{a^3+abc}{(b+c)^3}+\frac{b^3+abc}{(c+a)^3}+\frac{c^3+abc}{(a+b)^3} \ge \frac{3}{4}$

WLOG suppose $a \ge b \ge c$. Then $a^3+abc \ge b^3+abc \ge c^3+abc$ and $\frac{1}{(b+c)^3} \ge \frac{1}{(c+a)^3} \ge \frac{1}{(a+b)^3}$.
By Chebysev's inequality,
\[\displaystyle\sum \frac{a^3+abc}{(b+c)^3} \ge \frac{1}{3}(\displaystyle\sum a^3+abc)(\displaystyle\sum \frac{1}{(b+c)^3}).\]
By Schur's inequality,
\[\displaystyle\sum a^3+abc \ge a^2(b+c)+b^2(c+a)+c^2(a+b).\]
Hence
\[\displaystyle\sum \frac{a^3+abc}{(b+c)^3} \ge \frac{1}{3}(\displaystyle\sum a^3+abc)(\displaystyle\sum \frac{1}{(b+c)^3}) \ge \frac{1}{3}(\displaystyle\sum a^2(b+c))(\displaystyle\sum \frac{1}{(b+c)^3}).\]
By Cauchy-Schwarz inequality,
\[(\displaystyle\sum a^2(b+c))(\displaystyle\sum \frac{1}{(b+c)^3}) \ge (\displaystyle\sum \frac{a}{b+c})^2.\]
Therefore finally by Nesbitt's inequality,
\[\displaystyle\sum \frac{a^3+abc}{(b+c)^3} \ge \frac{1}{3}(\displaystyle\sum a^2(b+c))(\displaystyle\sum \frac{1}{(b+c)^3}) \ge \frac{1}{3}(\displaystyle\sum \frac{a}{b+c})^2 \ge \frac{1}{3} \cdot (\frac{3}{2})^2 = \frac{3}{4}.\]

I think the following stronger ineq holds

$\displaystyle \frac{a^3+2abc}{(b+c)^3}+\frac{b^3+2abc}{(c+a)^3}+\frac{c^3+2abc}{(a+b)^3} \ge \frac{9}{8}.$ ;)

And maybe,

$\displaystyle \frac{a^3+3abc}{(b+c)^3}+\frac{b^3+3abc}{(c+a)^3}+\frac{c^3+3abc}{(a+b)^3} \ge \frac{3}{2}$ .

[quote="Vasc"]$\displaystyle \frac{a^3+2abc}{(b+c)^3}+\frac{b^3+2abc}{(c+a)^3}+\frac{c^3+2abc}{(a+b)^3} \ge \frac{9}{8}.$ ;)[/quote]
By Chebysev's inequality,
\[\displaystyle\sum \frac{a^3+2abc}{(b+c)^3} \ge \frac{1}{3}[\displaystyle\sum (a^3+2abc)](\displaystyle\sum \frac{1}{(b+c)^3}).\]
By Schur's inequality,
\[\displaystyle\sum (a^3+2abc) \ge (ab+bc+ca)(a+b+c),\]
so
\[\displaystyle\sum \frac{a^3+2abc}{(b+c)^3} \ge \frac{1}{3}(ab+bc+ca)(a+b+c)(\displaystyle\sum \frac{1}{(b+c)^3}).\]
By Chebysev's inequality again,
\[\displaystyle\sum \frac{1}{(b+c)^3} \ge \frac{1}{3}(\displaystyle\sum \frac{1}{(b+c)^2})(\displaystyle\sum \frac{1}{(b+c)}),\]
so
\[\displaystyle\sum \frac{a^3+2abc}{(b+c)^3} \ge \frac{1}{9}(ab+bc+ca)(\displaystyle\sum \frac{1}{(b+c)^2})(a+b+c)(\displaystyle\sum \frac{1}{a+b}).\]
By Cauchy-Schwarz inequality,
\[(a+b+c)(\displaystyle\sum \frac{1}{a+b}) \ge \frac{9}{2},\]
so
\[\displaystyle\sum \frac{a^3+2abc}{(b+c)^3} \ge \frac{1}{2}(ab+bc+ca)(\displaystyle\sum \frac{1}{(b+c)^2}).\]
Now by Iran MO 1996 question 1  ;) ,
\[\displaystyle\sum \frac{a^3+2abc}{(b+c)^3} \ge \frac{1}{2}(ab+bc+ca)(\displaystyle\sum \frac{1}{(b+c)^2}) \ge \frac{1}{2} \cdot \frac{9}{4} = \frac{9}{8}.\]

yptsoi , you are faster than me :)

but this is my proof:

use lemma:$\sum\frac{a(a-b)(a-c)}{(b+c)^3}\geq0$

so our ineq is translated :

$(ab+bc+ca)(\sum\frac{a}{(b+c)^3})\geq\frac{9}{8}$

the rest like yptsoi's way

by tsebysev and iran 1996 ineq :D

[quote="Vasc"]

And maybe,

$\displaystyle \frac{a^3+3abc}{(b+c)^3}+\frac{b^3+3abc}{(c+a)^3}+\frac{c^3+3abc}{(a+b)^3} \ge \frac{3}{2}$ .[/quote]

WLOG suppose $a\ge\ b\ge\ c$ Then $ \frac{1}{(b+c)^2} \ge \frac{1}{(c+a)^2} \ge \frac{1}{(a+b)^2} $ and
$\frac{a^3+3abc}{b+c} \ge \frac{b^3+3abc}{c+a} \ge \frac{c^3+3abc}{a+b}$
By Chebysev's inequality,

$\sum \frac{a^3+3abc}{(b+c)^3} \ge \frac{1}{3}(\sum \frac{a^3+3abc}{b+c})(\sum \frac{1}{(b+c)^2})\ge \frac{1}{3}2(ab+bc+ca) (\sum \frac{1}{(b+c)^2}) \ge \frac{2}{3} .\frac{9}{4}= \frac{3}{2}$
By inequality
$\frac{a^3+3abc}{b+c}+\frac{b^3+3abc}{c+a}+\frac{c^3+3abc}{a+b} \ge \ 2(ab+bc+ca)$
http://www.mathlinks.ro/Forum/topic-32537.html
and Iran 1996 ineq

With the same conditions, the generalized inequality

$ \frac {a^3 \plus{} kabc}{(b \plus{} c)^{3}} \plus{} \frac {b^3 \plus{} kabc}{(c \plus{} a)^{3}} \plus{} \frac {c^3 \plus{} kabc}{(a \plus{} b)^{3}}\geq\frac {3(1 \plus{} k)}{8}$

holds if and only if $ \minus{} 29.6505\cdots \equal{} k_3\leq k\leq \frac {13}{3},$

where $ k_3$ is the least real root of the following irreducible polynomial over $ \mathbb{Q}:$

$ 4k^5 \plus{} 171k^4 \plus{} 1476k^3 \minus{} 2250k^2 \plus{} 1512k \minus{} 513;$

with equality if $ k \equal{} k_3,b \equal{} c \equal{} 1,a \equal{} 2.25245\cdots$ is the greatest real root of the following irreducible polynomial

$ 2a^5 \plus{} 11a^4 \plus{} 8a^3 \minus{} 38a^2 \minus{} 98a \minus{} 77.$

I'm expecting nice solutions for $ k \equal{} \frac {13}{3}:$

$ \frac {3a^3 \plus{} 13abc}{(b \plus{} c)^{3}} \plus{} \frac {3b^3 \plus{} 13abc}{(c \plus{} a)^{3}} \plus{} \frac {3c^3 \plus{} 13abc}{(a \plus{} b)^{3}}\geq 6$

and $ k \equal{} \frac {3647}{ \minus{} 123}:$

$ \frac {123a^3 \minus{} 3647abc}{(b \plus{} c)^{3}} \plus{} \frac {123b^3 \minus{} 3647abc}{(c \plus{} a)^{3}} \plus{} \frac {123c^3 \minus{} 3647abc}{(a \plus{} b)^{3}}\geq \frac {2643}{ \minus{} 2}.$


Joan and Kate are both standing at the beginning of a straight pathway and are both facing the same direction. One of Joan's steps is the same length as one of Kate's steps. If Joan takes two steps forward and one step back, and Kate takes three steps forward and five steps back, how many steps will Joan be ahead of Kate?

We can picture this on a number line.

Joan is two steps forward minus one step back, so one step from the starting place. $ J \equal{} 2 \minus{} 1 \equal{} 1$

Kate is three steps forward and five steps back, so two steps behind the starting place. $ K \equal{} 3 \minus{} 5 \equal{} \minus{}2$

$ 1 \minus{} (\minus{} 2) \equal{} \boxed{3}$

[quote=GameBot]Joan and Kate are both standing at the beginning of a straight pathway and are both facing the same direction. One of Joan's steps is the same length as one of Kate's steps. If Joan takes two steps forward and one step back, and Kate takes three steps forward and [size=100]$\boxed{\bold{give}}$[/size] steps back, how many steps will Joan be ahead of Kate?[/quote]

"give" should be replaced by "five".


1) Prove that for any integer $ k>3$ then the equation $ x^2\minus{}(k^2\minus{}4)y^2\equal{}\minus{}1$ has no positive integer solution.

2) Find all positive integer $ k$ so that $ x^2\minus{}(k^2\minus{}4)y^2\equal{}\minus{}24$ has integer solutions

My teacher give a proof using something not elementary, does anyone has a simple solutions for these?   :maybe:

[quote="Boatman"]1) Prove that for any integer $ k > 3$ then the equation $ x^2 \minus{} (k^2 \minus{} 4)y^2 \equal{} \minus{} 1$ has no positive integer solution.

[/quote]

here is a rather basic demonstration :

$ x^2 \minus{} (k^2 \minus{} 4)y^2 \equal{} \minus{} 1$ $ \implies$ $ (ky \minus{} x)^2 \plus{} (2y)^2 \plus{} 1 \equal{} k(ky \minus{} x)(2y)$ $ \implies$ $ u^2 \plus{} v^2 \plus{} w^2\equal{} kuvw$ for some positive integers $ u,v,w$

But the equation $ u^2 \plus{} v^2 \plus{} w^2 \equal{} kuvw$ has solutions only for $ k\in\{1,3\}$ (see http://www.mathlinks.ro/viewtopic.php?p=1500981#1500981) and so no solution exists for $ k > 3$


let an acute triangle ABC with H is an orthocentre , O is an midpoint of the segment BC. A line d via H cut AB, AC respectively at M, N . IF OM = On prove HM = HN.

[hide="Hint."] $ \left\{\begin{array}{ccc}
U\in BH & , & MU\perp BH\\\
V\in CH & , & NV\perp CH\end{array}\right\|\implies UV\parallel BC$ .[/hide]

if u draw a prependicular line $ l$,from point $ H$,to line $ d$,and suppose $ l$,cuts $ AB,AC$,at

$ M,N$,i claim that $ HM\equal{}HN$,and it will proved accordingly that $ OMH$ is isoceles and $ OM\equal{}ON$

ok,if u draw a circle with diameter $ BC$,u can see the theorem of $ BUTTERFLY$,exactly!!!

then,its obvious,and we dont need to continue... :wink:

[quote="o0o_nhocemma_o0o"][color=darkred] Let $ ABC$ be an acute triangle with the orthocenter $ H$ and the  middlepoint $ O$ of the side $ [BC]$ .
A line $ d$ via $ H$ cut the sides $ [AB]$ , $ [AC]$ at $ M$ , $ N$ respectively . Prove that $ OM = ON\Longleftrightarrow HM = HN$ .[/color] [/quote]
[color=darkblue][b][u]Proof I.[/u][/b] Denote the points $ \|\begin{array}{ccc} E\in AC & , & BE\perp AC \\
\ F\in AB & , & CF\perp AB \\
\ U\in BE & , & UM\perp MN \\
\ V\in CF & , & VN\perp MN\end{array}$ . Observe that the quadrilaterals $ BCEF$ , $ UMEN$ , $ VNFM$ are cyclically : the first is inscribed in the circle with the diameter $ [BC]$ , the second is inscribed in the circle with the diameter $ [UN]$ and the third is inscribed in the circle with the diameter $ [VM]$ . Thus, $ HU\cdot HE = HM\cdot HN = HV\cdot HF$ , i.e. $ HU\cdot HE = HV\cdot HF$ what means that the quadrilateral $ UFEV$ is cyclically. Therefore, $ \widehat {HBV}\equiv$ $ \widehat {EBV}\equiv$ $ \widehat {EFV}\equiv$ $ \widehat {EFC}\equiv$ $ \widehat {EBC}\equiv\widehat {HBC}$ , i.e. $ \widehat {HBV}\equiv\widehat {HBC}$ . Hence $ UV\parallel BC$ . Denote the middlepoint $ W$ of the segment $ [UV]$ . Observe that $ W\in [HO]$ because $ UV\parallel BC$ . Thus, $ HM = HN$ $ \Longleftrightarrow$ In the right trapezoid $ UMNV$ the middle line $ HW\parallel UM$ $ \Longleftrightarrow$ $ HW\perp MN$ $ \Longleftrightarrow$ $ OH\perp MN$ $ \Longleftrightarrow$ the triangle $ MON$ is isosceles $ \Longleftrightarrow$ $ OM = ON$ .

[b][u]Proof II.[/u][/b] Apply the theorem of the butterfly to the point $ H$ in the circle with the diameter $ [BC]$ a.s.o. See :

http://www.mathlinks.ro/viewtopic.php?t=46358
http://www.mathlinks.ro/viewtopic.php?t=181632
http://www.mathlinks.ro/viewtopic.php?t=153907
http://www.mathlinks.ro/viewtopic.php?t=46146
http://www.mathlinks.ro/viewtopic.php?t=28062
http://www.mathlinks.ro/viewtopic.php?t=72094
http://www.mathlinks.ro/viewtopic.php?t=73560
http://www.mathlinks.ro/viewtopic.php?t=13061

 [/color]

[quote="highly_effective_teens"]I think 2 people had made wrong answer . the second ppl , can u see carefully the Provement???????.....theoreom BUTTERFLY only apply in yr case and my problem is opposite ............, the first ppl , I draw as yr hint......but may be u had made somt mistake.......[/quote]

dont be angry!!! :)

infact i meant ure comment :wink:

the opposite form of BUtterfly is as hard as the butterfly theorem,

in steps of the proof u should just put $ \Longleftrightarrow$ betwin each step!

in steps of the proof u should just put  betwin each step!... can u see carefully .....somt steps dont put  <=>


Let $ a_i$ (for $ i \equal{} 1,2,\cdots ,n$) be real numbers. Prove the following inequality.

$ \frac {2^{a_1} \plus{} 2^{a_2} \plus{} \cdots \plus{} 2^{a_n}}{n}\geqq 2^{\frac {a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n}{n}}$.

This is AM-GM, applied to the numbers $2^{a_1}$, $2^{a_2}$, ..., $2^{a_n}$. Basta!

  dg


Given that $ 64^m \equal{} 8$ and $ m^n \equal{} 8$, evaluate $ mn$. Express your answer as a common fraction.

Since $ \sqrt{64}\equal{}8$, $ m\equal{}\frac{1}{2}$.

$ 2^{\minus{}n}\equal{}8$, so $ n\equal{}\minus{}3$.

$ mn\equal{}\boxed{\minus{}\frac{3}{2}}$


LaTeX
function
combinatorics proposed
combinatorics

[color=blue][size=150]PERU TST IMO - 2006[/size]
Saturday, may 20.[/color]

[b]Question 03[/b]
In each square of a board drawn into squares of $2^n$   rows and
$n$ columns $(n\geq 1)$ are written a 1 or a -1, in such a way
that the rows of the board constitute all the possible sequences
of length $n$ that they are possible to be formed with numbers 1
and -1.
Next, some of the numbers are replaced by zeros.
Prove that it is possible to choose some of the rows of the board
(It could be a row only) so that in the chosen rows, is fulfilled that the
sum of the numbers in each column is zero.

----
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88511]Spanish version[/url]

$\text{\LaTeX}{}$ed by carlosbr

It's also problem 7 of the 3rd QEDMO.
Here is my (=the "official") solution:

A binary row $(b_{1}, b_{2}, ..., b_{n})$ is a sequence of $0$'s and $1$'s.
An original row is one of those rows consisting of $n$ numbers from $\{1,-1\}$ that were there at the beginning.
A h4xx0red row is one of those $2^{n}$ rows (with $0$'s, $1$'s and $-1$'s), that are archieved after replacing some numbers by $0$.

Lemma: For every binary row $b=(b_{1}, b_{2}, ..., b_{n})$ we can find a h4xx0red row $p=(p_{1}, p_{2}, ..., p_{n})$ such that $b+p$ (addition at each of the $n$ entries) is again a binary row. This $p$ can be choosen injectively if seen as function of $b$ (thus no two different binary rows force us to choose the same h4xx0red row).

Proof: take that original row $o=(o_{1}, o_{2}, ..., o_{n})$ with $o_{k}=1$ iff $b_{k}=0$ and $o_{k}=-1$ iff $b_{k}=1$. It's easy to see that, independent if $o_{k}$ was changed to $0$ afterwards, always $p_{k}+b_{k}\in \{0,1\}$. This choice was injective.

Construction: We start with the binary row $0=(0, 0, ..., 0)$. As done in the lemma, we add h4xx0red rows (and always get binary rows again), until we first reach some binary row for the second time (there are only $2^{n}$, so finitely, many binary rows, thus this happens for sure).
Between this two occurences of the same two binary rows, we added exactly a sum of $0$. To finish the proof, we are left to show the following:

Lemma: no h4xx0red row was added twice.

Proof: If some h4xx0red row would have been added twice, say to the binary lines $b$ and $c$, then by injectivity in the first lemma $b=c$. But then our construction would have already stopped at $c$, contradiction.

It's originally from [b]Tournament of towns, Spring 1996, Senior A-Level P6[/b].


what is moosing b/c i don't know i might "moose"...

It's like mooing but with more moose. Am I right? Am I right?

birds can fap their wings if they try very very [b]hard[/b]

But what is it? If treething is serious then I suspect you guys smack your computers so please tell me what it is!!?!

Why would I say moose even daily?

NEVER

I STILL DON'T GET WHAT MOOSING IS.
AND WHAT'S FAPPING?

YEAH LET'S KILL IVERSONFAN! We NEED to KNOW!

[hide="I DONT KNOW DUDE WHATS MOOSING"][url]http://www.urbandictionary.com/define.php?term=moose[/url][/hide]

hahaha i play that but i don't yell moose..instead i yell the person's name

...I play that game daily except that I do it with "MOOOOOOO" instead

no! he was asked if you [i]moose[/i].  please stay on topic.   :spam:

this isn't a game guys  :mad:

my moosing amount is inversely related to my fapping amount

that being said i never moose

joml we weren't trying to spam...now i moose and have gotten smacked by people for wrecking their hair many times.

I have never moosed, but I will soon begin. :moose:

def 44 of moose:

My sweet husband, who is cute as a bugs ear. Juicy moosie..

[quote="akalra1"]i do moosing

its tat i jus smack them really hard and call them a cheesedoodle[/quote]

excuse you

[quote="akalra1"]EXCUSE YOU *****[/quote]

(in scolding tone)
*click*

[quote="akalra1"]EXCUSE YOU *****[/quote]

um ok...your ***** does not offend me you know...

just take your moose and go moosing somewhere else


I saw this problem in a forum and I was like totally confused when I see someones solution perhaps someone can solve and explain me his solution:

The function $ f\left(n\right)$ is a function on positive real numbers such as: $ f\left(1\right)\equal{}2001$ and $ \sum_{i\equal{}1}^n f\left(i\right)\equal{}n^2 f\left(i\right)$,    $ \forall n >1$ Find $ f\left(2001\right).$

Thanking you in advance  :)

$ f(1) \plus{} f(2) \plus{} .... \plus{} f(n) \equal{} n^2 \cdot f(n)$
or $ f(1) \plus{} f(2) \plus{} .... \plus{} f(n \minus{} 1) \equal{} f(n) \cdot (n^2 \minus{} 1)$
therfore $ f(n)\cdot (n^2 \minus{} 1) \equal{} f(n \minus{} 1)\cdot (n \minus{} 1)^2$
therefore $ f(n) \equal{} f(n \minus{} 1)\cdot \frac {(n \minus{} 1)^2}{n^2 \minus{} 1} \equal{} f(n \minus{} 1)\cdot \frac {n \minus{} 1}{n \plus{} 1} \equal{} f(1)\frac {(n \minus{} 1)!}{(n \plus{} 1)!} \equal{} \frac {1}{2002}$

Well before I ask you about some parts in your solution. The final answer of the solution I saw was different from yours if you are still sure about your answer I will put the solution I saw.

Nice solution different from the one I saw.
but I want to ask does function have special way to treat them or something???
And if you please guys can someone till me any source to learn more about basic functions (book or site etc.)?

Thanking you

OK RoFlLoLcOpT this is the solution I saw before

[quote]
$ \mbox{First we have: }\\
n^2 f(n) \equal{} \sum_{i\equal{}1} ^ n f(i) \equal{} \sum_{i\equal{}1} ^ {n\minus{}1} f(i)\ \plus{} f(n) \\
\Rightarrow\ f(n) \equal{} \frac{ \sum_{i\equal{}1}^{n\minus{}1} f(i)}{n^2\minus{}1}\ ...(1)$

$ \mbox{ Now we have from (1): }\\
 \sum_{i\equal{}1} ^ n f(i) \equal{} \sum_{i\equal{}1} ^ {n\minus{}1} f(i)\ \plus{} f(n) \equal{} \sum_{i\equal{}1}^{n\minus{}1} f(i)\ \plus{}\frac{ \sum_{i\equal{}1}^{n\minus{}1} f(i)}{n^2\minus{}1} \equal{}\sum_{i\equal{}1}^{n\minus{}1} f(i)\ (1\plus{}\frac{1}{n^2\minus{}1}) \ ...(2)$

By repeating (2) we get:

$ \sum_{i\equal{}1}^n f(i) \equal{} f(1) \prod_{k\equal{}2}^n (1\plus{}\frac{1}{k^2\minus{}1})\\
\equal{}f(1) \prod_{k\equal{}2}^n (\frac{k^2}{k^2\minus{}1} )\\
\equal{}f(1) (\prod_{k\equal{}2}^n  \frac{k}{k\minus{}1} )
\prod_{k\equal{}2}^n \frac{k}{k\plus{}1} \equal{} \frac{2001 \times 2n}{n\plus{}1}\\
\therefore n^2 f(n) \equal{} \frac{2001 \times 2n}{n\plus{}1}\\
\therefore 2001^2 f(2001)\equal{}\frac{2001^2 \times 2}{2002} \Rightarrow\ f(2001) \equal{}\frac{1}{1001}$[/quote]

[quote="hasan4444"]$ \equal{} f(1) (\prod_{k \equal{} 2}^n \frac {k}{k \minus{} 1} ) \prod_{k \equal{} 2}^n \frac {k}{k \plus{} 1} \equal{} \frac {2001 \times 2n}{n \plus{} 1} \\
\therefore n^2 f(n) \equal{} \frac {2001 \times 2n}{n \plus{} 1} \\
\therefore 2001^2 f(2001) \equal{} \frac {2001^2 \times 2}{2002} \Rightarrow\ f(2001) \equal{} \frac {1}{1001}$[/quote]
There's a typo in the first line. The $ 2001$ comes from nowhere and should be another $ n$.
Maybe that's why you were confused. It's essentially the same proof as the RoFl's.

Well I'm not familiar with functions at all and those notaions make worse for me. I think some people like to use some fancy notaions just to make people say WOW (like the product one) that can be better further more explained in a situation when people don't understand like me  :D (proud of it)

Anyway thanks a lot for your replies and concerning  :)


linear algebra
matrix
superior algebra
superior algebra unsolved

Given a field K and a commutative family S of nxn matrices with elements in K, is it true that there exists T in $GL_n(K)$   such that for all A in S we have $Trans(A)=S^{-1}AS$ , where trans(A) is the traspose matrix?  Please, the one who has a solution to post it. Thanks

Yes, the idea is that the basis changement basis is the same for all members of the family.


inequalities
number theory
greatest common divisor
algebra unsolved
algebra

Hello, I've got a problem with the following task:
For $ {d}$, $ {e}$,$ {f}$ $ \in N$, $ {d}$ $ \geq 2$, $ {e}$ $ \geq 2$, $ {f}$ $ \geq 2$ and $ {x},{y},{z} \in N, {x}\geq 1, {y}\geq 1, {z}\geq 1$ it is:
$ \frac{1}{x} \equal{}\frac{1}{e} \plus{}\frac{1}{f}$
$ \frac{1}{y} \equal{}\frac{1}{d} \plus{}\frac{1}{f}$
$ \frac{1}{z} \equal{}\frac{1}{d} \plus{}\frac{1}{e}$

Proove, that: $ gcd(d,e,f)\geq 2$

I'm sorry, but I have no real idea, only that it is obvious, that there has to be:
$ gcd(d, e)\geq 2$
$ gcd(d, f)\geq 2$
$ gcd(e, f)\geq 2$

But how to show, that: $ gcd(d,e,f)\geq 2$

Conditional is  equivalent :
$ e\plus{}f|ef,d\plus{}f|df,e\plus{}d|de$
Let :
$ c_1\equal{}\gcd(e,f),c_2\equal{}\gcd(d,f),c_3\equal{}\gcd(e,d)$
We only to prove that at least on of three $ \gcd(c_1,c_2),\gcd(c_2,c_3),\gcd(c_1,c_3)>1$
Suppose that each of three number equal 1 .
From condition we can resume that $ c_1<c_2<c_3$
$ d\equal{}c_2d_2\equal{}c_3d_3$
Because $ \gcd(c_2,c_3)\equal{}1$ so $ d_3\equal{}md_2$
It gives that $ d\equal{}mc_2c_3$
Similar $ e\equal{}nc_1c_3,f\equal{}pc_1c_2$
From the condition we have : $ nc_3\plus{}m.c_2|c_1$
It gives contradiction .
So $ \gcd(e,f,d)>1$

[quote="Atan"]First of all: Thanks a lot for your instant endeavor.

But excuse me please, I'm a newcomer here, so that not everything is easy to follow for me:
Why can you state, that:
$ c_1 < c_2 < c_3$

And could you explain to me the following steps a little bit more detailed? You also forgot to say, what are $ d_2$ and $ d_3$, so unfortunately, I cannot get your messages completely.[/quote]
Be cause $ \gcd(c_i,c_j)\equal{}1$ and $ c_i>1$
If exist $ c_i\equal{}c_j$ it gives contradiction.
$ c_2|d$ so exist an integer $ d_2$ such that $ d\equal{}c_2d_2$

Okay, these steps now are clearly to follow. Excuse me please, but I rather ask once more than live with an uncertainty.

(1) Your inequality is only "without loss of generality", isn't it? That means, it is only a presumption, that $ c_1 < c_2 < c_3$. It also could be $ c_3 < c_2 < c_1$ ?

(2) How can you follow from  $ gcd(c_2, c_3)\equal{}1$ that $ d_3\equal{}md_2$ ? If you presume that $ 1\equal{} \frac {c_3*d_3}{c_2*d_2}$ than $ d_3$ cannot be greater than $ d_2$ because $ c_3$ is already bigger than $ c_2$, so that the result 1 could not be reached.

(3) You show, that $ d\equal{}m*c_2*c_3 , e\equal{}n*c_1*c_3 , f\equal{}p*c_1*c_2$. Which of the three conditions do you use to build $ nc_3\plus{}mc_2 | c_1$ ? Or should the m rather be the letter p?

[quote="Atan"]Okay, these steps now are clearly to follow. Excuse me please, but I rather ask once more than live with an uncertainty.

(1) Your inequality is only "without loss of generality", isn't it? That means, it is only a presumption, that $ c_1 < c_2 < c_3$. It also could be $ c_3 < c_2 < c_1$ ?

(2) How can you follow from  $ gcd(c_2, c_3) \equal{} 1$ that $ d_3 \equal{} md_2$ ? If you presume that $ 1 \equal{} \frac {c_3*d_3}{c_2*d_2}$ than $ d_3$ cannot be greater than $ d_2$ because $ c_3$ is already bigger than $ c_2$, so that the result 1 could not be reached.

(3) You show, that $ d \equal{} m*c_2*c_3 , e \equal{} n*c_1*c_3 , f \equal{} p*c_1*c_2$. Which of the three conditions do you use to build $ nc_3 \plus{} mc_2 | c_1$ ? Or should the m rather be the letter p?[/quote]
(1) You can see that it depend on $ e,f,d$ and we can resume that .
(2) $ c_2d_2\equal{}c_3d_3$
So $ c_3|c_2d_2$ but from $ \gcd(c_3,c_2)\equal{}1$ so $ c_3|d_2$
Oh I have a typo here .
But the next is true. Exist an integer m such that $ d_2\equal{}mc_3$ and so $ d\equal{}mc_2c_3$
(3) It from condition $ f\plus{}e|\gcd(f,e)\equal{}c_1$

Let me ask a final question - let's hope so :-)

A given condition is:
$ e\plus{}f|ef$ or in other terms $ \frac {ef}{e\plus{}f}\equal{}x$. But that's not the same like your point $ e\plus{}f|gcd(e,f)$, because: How can (e+f) be a divisor of its gcd?

I'm sorry, but your final step is totally unclear for me. Perhaps, I cannot get the message, then I would like you to explain this last step more in-depth.

[quote="Atan"]Let me ask a final question - let's hope so :-)

A given condition is:
$ e \plus{} f|ef$ or in other terms $ \frac {ef}{e \plus{} f} \equal{} x$. But that's not the same like your point $ e \plus{} f|gcd(e,f)$, because: How can (e+f) be a divisor of its gcd?

I'm sorry, but your final step is totally unclear for me. Perhaps, I cannot get the message, then I would like you to explain this last step more in-depth.[/quote]
It must be $ {e \plus{} f}|\gcd(e,f)^2$

Wow, the more we discuss, the more I get confused. Wouldn't it be possible, that you explain your final step more detailed? I cannot understand, from which condition follows your last post and why it follows from this condition and how to argue with this idea from your last post?

Why is $ e\plus{}f|gcd(e,f)^2$? I don't understand this? Could you explain it to me, because it is essentiel to understand your final step and likewisely the whole task.

$ \gcd(e,f)\equal{}m$ then exist $ (x,y)$ such that
$ e\equal{}mx,f\equal{}my$ and $ \gcd(x,y)\equal{}1$
Because $ e\plus{}f|ef$ so $ m(x\plus{}y)|m^2xy$
$ \Leftrightarrow x\plus{}y|mxy$
Because $ \gcd(x\plus{}y,xy)\equal{}1$ so $ x\plus{}y|m$

Okay, now I can follow this idea:
Inserting x and y gives us:
$ (nc_3\plus{}pc_2)|c_1$
But this might be not possible, because you said, that $ c_1<c_2<c_3$, or isn't it so?

And otherwise: Why should this be a contradiction to $ gcd(c_1,c_2)\equal{}1$? If we ignore your condition $ c_1<c_2<c_3$, because that cannot work here, then imagine the numbers: n=1, p=1, $ c_1\equal{}5, c_2\equal{}3, c_3\equal{}2$. The gcd is still 1, but both numerator and denominator are divisible as well, so that there is no contradiction.

If $ \gcd(c_1,c_2)\equal{}k>1$ then
$ k|e,f,d$ so $ k|\gcd(d,e,f)$

But we have still yout presumption, that $ gcd(c_1,c_2)\equal{}1$. Your idea directs us to the point, that:
$ (nc_3\plus{}pc_2)|c_1$

But why this is a contradiction to our presumption? If we take the example of $ n\equal{}1, p\equal{}1, c_1\equal{}5, c_2\equal{}2, c_3\equal{}3$, then the presumption is fullfilled and also $ gcd(c_1,c_2)\equal{}1$, becauso $ c_1\equal{}1$ and $ c_2\equal{}1$.

Something with your final explanations cannot work. Although the inequality of $ c_1<c_2<c_3$ seems to be odd.

Is there anyone who is able to help me in order to understand the final step(s) of this task?