probability integration limit calculus function parameterization algebra

Hi, I read some demonstrations of the central limit theorem and wrote some of my own, but I always end up requiring a certain hypothesis.
Let f(x) bè a probability distribution:
\[ \begin{array}{l} f(x) \ge 0 \\ \int\limits_{ - \infty }^\infty {f(x)dx} = 1 \\ E[x] = \int\limits_{ - \infty }^\infty {\left[ {f(x)x} \right]dx} = \mu \\ x_c = x - \mu \\ E[x_c ] = 0 \\ E[x_c^2 ] = \int\limits_{ - \infty }^\infty {\left[ {f(x_c )x_c^2 } \right]dx_c } = \sigma ^2 \\ \end{array}  \]

Let \[ x_1 ,x_2 ,...x_n  \] be n different samples of the variable x. Let z be \[ z = \sum\limits_{i = 1}^k {x_i }  \]
Then the probability distribution of z as the sum of n variables, called \[ f_n (z)  \], can be defined recursively as:\[ f_n (z) = \int\limits_{ - \infty }^{ + \infty } {f(x)f_{n - 1} (z - x)dx}  \]

Where \[ f_1 (z) = f(z)  \]

Now we can use this relationship to prove that for every integer n:
\[ \begin{array}{l} \int\limits_{ - \infty }^{ + \infty } {f_n (z)dz = } 1 \\ E_n [z] = \int\limits_{ - \infty }^{ + \infty } {f_n (z)zdz = } n\mu \\ E_n [(z - n\mu )^2 ] = n\sigma ^2 \\ \end{array}  \]

Now, the central limit theorem states that:
\[ \mathop {\lim }\limits_{n \to \infty } f_n \left( {\frac{{z - n\mu }}{{\sigma \sqrt n }}} \right) = \frac{1}{{\sqrt{2\pi } }}\exp \left[ {\frac{{ - w^2 }}{2}} \right]  \]

In all the central limit theorem demonstrations I have seen it is necessary that for every integer k \[ \left| {E[x_c^k ]} \right| = \left| {\int\limits_{ - \infty }^\infty {\left[ {f(x_c )x_c^k } \right]dx_c } } \right| < L < + \infty  \]
.

Is it necessary for f(x) to be a probability distribution to satisfy this condition. Because otherwise, I can define a f(x) such that has overall integral 1, finite mean and variance, but for certain integers k E[x^k] diverges.
For instance, consider this function:
\[ \begin{array}{l} \left| x \right| < = L \to f(x) = a \\ \left| x \right| > L \to f(x) = \frac{{aL^4 }}{{x^4 }} \\ \end{array}  \]

We can see that E[x]=0, E[x^2] is finite, E[x^(2k+1)]=0, but E[x^2k] diverges for k>1.

\[ \begin{array}{l} f(x) = f( - x) \to \int\limits_{ - \infty }^{ + \infty } {f(x)x^{2k + 1} } dx = 0 \\ \int\limits_{ - \infty }^{ + \infty } {f(x)dx = 2\left[ {\int\limits_0^{ + \infty } {f(x)dx} } \right]} = 2\left[ {\int\limits_0^L {adx + \int\limits_L^{ + \infty } {\frac{{aL^4 }}{{x^4 }}dx} } } \right] = \frac{8}{3}aL \\ aL = \frac{3}{8} \\ \int\limits_{ - \infty }^{ + \infty } {f(x)x^2 dx = \frac{2}{3}} aL^3 + 2aL^3 = \left( {\frac{8}{3}aL} \right)L^2 = L^2 \\ L = \sigma \\ a = \frac{3}{8\sigma} \end{array}  \]

But as we said:
\[ k > 1 \to \int\limits_{ - \infty }^{ + \infty } {f(x)x^{2k} dx = + \infty }  \]

Now my question is: is the central limit theorem valid for this particular f(x) ?

[quote="andrea_roxbury"]Now, the central limit theorem states that:

\[ \mathop {\lim }\limits_{n \to \infty } f_n \left( {\frac{{z - n\mu }}{{\sigma \sqrt n }}} \right) = \frac{1}{{\sqrt{2\pi } }}\exp \left[ {\frac{{ - w^2 }}{2}} \right] \] [/quote]
That statement seems a little technically garbled (what's $w$?), but that's not the point I'm going to make. What you have there is pointwise convergence of the density functions to the density of the standard normal distribution. That's far too strong a notion of convergence, and with it leaves you with a far too narrow and restricted a theorem. For instance: wouldn't you like it to be true that the binomial $(n,p)$ distribution, properly scaled, tends to the standard normal as $n\to\infty$? But for each $n,$ the scaled binomial distribution doesn't even have a density, since it is a discrete distribution.

The appropriate standard for a useful version of the Central Limit Theorem is [b]convergence in distribution.[/b] Here's a sample version:

Suppose $X$ is a random variable with a finite second moment. $(E(X^2)<\infty.)$ Let $\mu=E(X)$ and $\sigma=\sqrt{\text{Var}(X)}.$ Suppose $(X_k)$ is a sequence if independent identically distributed random variables with the same distribution as $X.$

Let $Z_n=\frac1{\sqrt{n}\,\sigma}\left(\sum_{k=1}^nX_k-n\mu\right).$

Let $F_n(z)=P(Z_n\le z).$ This is the cumulative distribution function.

Then, $\forall\,z,\,\lim_{n\to\infty}F_n(z)= \int_{-\infty}^z\frac1{\sqrt{2\pi}}e^{-x^2/2}\,dx.$

That's convergence in distribution. It's certainly possible for a sequence of discrete random variables to converge in distribution to a continuous random variable.

The proofs I know involve Fourier-analytic methods (or as the probabilists would say, the "characteristic function" $\chi_X(t)=E(e^{itX}).)$ But in answer to your question: the proofs I know do not require that moments higher than the second moment be finite. Your example (after we tweak the parameters $a$ and $L$ to make sure it's a random variable) would be included in this theory; its second moment is finite and that's all we need.

First of all, the case of discrete distributions can be easily treated using Dirac's delta functions.
The operation of calculating the probability ${\rm{P(z) = P(Z}}_{\rm{n}} \le z)$ where $Z_n=\frac1{\sqrt{n}\,\sigma}\left(\sum_{k=1}^nX_k-n\mu\right)$, is "regularizing": this means that even if we start with an irregular probability distribution (such as with the discrete case), we end up at the limit with a very regular function: in fact, the probability of $Z_n$ being a precise value is 0, since a single point has zero measure.
Anyway, I wondered how you could use the characteristic function (which is an inverse fourier transform) without requiring that moments of greater than second order must be finite: for instance, the page [url]http://mathworld.wolfram.com/CentralLimitTheorem.html[/url]. It is used the fact that the distribution of the variable $s = \sum\limits_{i = 1}^N {x_i }$, $f_N (s)$, has as its characteristic function $E_N \left[ {e^{2\pi ifx} } \right] = \left[ {E\left[ {e^{2\pi ifx} } \right]} \right]^N$. Then a substitution of variables, taylor series etc are used to reach the limit of the characteristic of a normal distribution. However, we suppose that ${E\left[ {e^{2\pi ifx} } \right]}$ exists (at least for certain values of f). But we know that:
$E\left[ {e^{2\pi ifx} } \right] = \int\limits_{ - \infty }^{ + \infty } {f(x)e^{2\pi ifx} dx = } \int\limits_{ - \infty }^{ + \infty } {f(x)\left( {\sum\limits_{k = 0}^\infty {\frac{{\left( {2\pi if} \right)^k }}{{k!}}x^k } } \right)dx = \sum\limits_{k = 0}^\infty {\frac{{\left( {2\pi if} \right)^k }}{{k!}}} } \int\limits_{ - \infty }^{ + \infty } {f(x)x^k dx = } \sum\limits_{k = 0}^\infty {\frac{{\left( {2\pi if} \right)^k }}{{k!}}} E\left[ {x^k } \right]$

Now I can't see how it can be possible that the characteristic of f exists when f has moments of finite order that are not finite. I suspect I might be making a basic calculus mistake in my question, then could you please show me where I'm wrong?

I can see anyway that:

$\left\| {\int\limits_{ - \infty }^{ + \infty } {f(x)e^{2\pi ifx} dx} } \right\| \le \int\limits_{ - \infty }^{ + \infty } {\left\| {f(x)e^{2\pi ifx} } \right\|dx} = \int\limits_{ - \infty }^{ + \infty } {\left| {f(x)} \right|dx} = 1$

Anyway it is possible that, although finite in modulus, the limit might not exist because of the oscillating properties of the imaginary exponential.

I tried to see if we could find a characteristic function of the cumulative probability function, but it turned useless, since erf(z) has no characteristic function. In fact, we can see that given a probability distribution function f(x) and the cumulative probability function $g(x) = \int\limits_{ - \infty }^x {f(\tau )d\tau }$, if f has a characteristic function then g hasn't, and if g has a characteristic function then f hasn't. I'll prove it in the following lines, but notice the consequence that since we know that a normal distribution has a well known characteristic function, then erf(z) hasn't.
$\int {g(x)e^{sx} dx = \frac{1}{s}} g(x)e^{sx} - \frac{1}{s}\int {\frac{{dg}}{{dx}}e^{sx} dx} = \frac{1}{s}g(x)e^{sx} - \frac{1}{s}\int {f(x)e^{sx} dx}$

$\int\limits_{ - \infty }^{ + \infty } {g(x)e^{sx} } dx = \frac{1}{s}\mathop {\lim }\limits_{x \to + \infty } \left[ {g(x)e^{sx} - g( - x)e^{ - sx} } \right] - \frac{1}{s}\int\limits_{ - \infty }^{ + \infty } {f(x)e^{sx} dx} = \frac{1}{s}\mathop {\lim }\limits_{x \to + \infty } \left[ {g(x)e^{sx} - g( - x)e^{ - sx} } \right] - \frac{1}{s}E\left[ {e^{sx} } \right]$

Now, if we know that $E\left[ {e^{sx} } \right]$ exists, in order for the characteristic of g(x) to exist, it is necessary that the limit $\mathop {\lim }\limits_{x \to + \infty } \left[ {g(x)e^{sx} - g( - x)e^{ - sx} } \right]$ exists.

$s = 2\pi if$

$\mathop {\lim }\limits_{x \to + \infty } \left[ {g(x)e^{sx} - g( - x)e^{ - sx} } \right] = \mathop {\lim }\limits_{x \to + \infty } e^{ - sx} \left[ {g(x)e^{2sx} - g( - x)} \right] = \mathop {\lim }\limits_{x \to + \infty } e^{ - sx} \mathop {\lim }\limits_{x \to + \infty } \left[ {g(x)e^{2sx} - g( - x)} \right]$

Therefore, since
$\begin{array}{l} \mathop {\lim }\limits_{x \to + \infty } e^{ - 2\pi ifx} = ? \\ \mathop {\lim }\limits_{x \to + \infty } \left\| {e^{ - 2\pi ifx} } \right\| = 1 \\ \end{array}$

It is necessary that $\mathop {\lim }\limits_{x \to + \infty } \left[ {g(x)e^{2sx} - g( - x)} \right] = 0$

But we can see that this limit doesn't exist:
$\mathop {\lim }\limits_{x \to + \infty } g(x)e^{2sx} - \mathop {\lim g( - x) = }\limits_{x \to + \infty } \mathop {\lim }\limits_{x \to + \infty } g(x)e^{2sx} = \left[ {\mathop {\lim }\limits_{x \to + \infty } g(x)} \right]\mathop {\lim }\limits_{x \to + \infty } e^{2sx} = \mathop {\lim }\limits_{x \to + \infty } e^{2sx} = ?$

since $s = 2\pi if$

I don't have time right now to reply to all of this, and it may be a few days until I get back to it. There are some other users who could comment on this, and I hope one of them does.

A couple of quick points:

[quote]First of all, the case of discrete distributions can be easily treated using Dirac's delta functions.[/quote]
But the point wasn't about being able to express discrete distributions (which is easy enough); the point was about the mode of convergence. When you say this sequence of distributions converges to the standard normal, exactly what sense of convergence do you mean?

[quote]Now I can't see how it can be possible that the characteristic function of f exists when f has moments of finite order that are not finite.[/quote]
The characteristic function does not have to be infinitely differentiable and does not have to be represented by a power series. For instance, we could take the distribution whose density is $\frac1{\pi(x^2+1)}.$ This distribution doesn't even have a first moment - its expectation is undefined. But it has a perfectly well-defined characteristic function of $e^{-|t|}.$ (Or, putting $2\pi$ in the exponent as you seem to be doing, $e^{-2\pi |t|}.)$ Of course that characteristic function isn't even differentiable at zero.

The manipulations you refer to in the proof of the CLT do not require the characteristic function to have a Taylor series - just a 2nd order Taylor polynomial with some kind of control of the error. And if the distribution has a finite 2nd moment, its characteristic function will be $C^2.$ That's the standard Fourier transform tradoff between smoothness on one side and rate of decay on the other.

I spent some time on the problem, I tried to put it in a different way, but ended up using Laplace/Fourier transforms again.
Given a probability distribution $f(v)$ of finite mean $\mu$ and variance $\sigma ^2$, the problem of determining the distribution of the variable $x = \sum\limits_{i = 1}^N {\frac{{\left( {v_i - \mu } \right)}}{{\sqrt n \sigma }}}$, $f_n (x)$, is similar of determining the probability distribution of the position of a particle at a certain time, when it changes velocity at regular intervals $\Delta t_n = \frac{1}{{\sqrt n \sigma }}$,  and where the velocity is a random variable of distribution $f(v)$, and finally given the fact that the particle is in x=0 for t=0.
To simplify everything, let us suppose that $\mu=0$.

The relationship we're interested in is:
$g\left( {(n + 1)\Delta t_n ,x} \right) = \int\limits_{ - \infty }^{ + \infty } {g(n\Delta t_n ,x - v\Delta t_n )} f(v)dv$

Now it is quite delicate to treat this problem. Let $g\left( {n\Delta t_n ,x} \right)$ be the solution to this equation, with a certain dependance from $f(v)$. We have to show that for n that approaches infinity, the solution depends only from the first and second moment of f(v), and is a normal distribution. Maybe we could take out a differential equation that has as its solution the normal distribution. However it would be easy to make mistakes in calculating the limits, therefore transforms are a very practical solution.
\[ \begin{array}{l} g\left( {(n + 1)\Delta t_n ,x} \right) = \int\limits_{ - \infty }^{ + \infty } {g(n\Delta t_n ,x - v\Delta t_n )} f(v)dv \\ g\left( {(n + 1)\Delta t_n ,x} \right) = \frac{1}{{\Delta t_n }}\left[ {\int\limits_{ - \infty }^{ + \infty } {g(n\Delta t_n ,x - z)} f\left( {\frac{z}{{\Delta t_n }}} \right)dz} \right] \\ \\ T[g\left( {n\Delta t_n ,x} \right)] = G(n\Delta t_n ,s) \\ T[f(z)] = F(s) \\ T[f(\lambda z)] = \frac{1}{\lambda }F\left( {\frac{s}{\lambda }} \right) \\ G\left( {(n + 1)\Delta t_n ,s} \right) = G\left( {n\Delta t_n ,s} \right)\left( {\frac{1}{{\Delta t_n }}T\left[ {f\left( {\frac{z}{{\Delta t_n }}} \right)} \right]} \right) \\ G\left( {(n + 1)\Delta t_n ,s} \right) = G\left( {n\Delta t_n ,s} \right)F\left( {\Delta t_n s} \right) \\ \end{array}  \]

And finally we obtain: $G(n\Delta t_n ,s) = G(0,s)\left[ {F\left( {\Delta t_n s} \right)} \right]^n$

Now we can find the transform of our final solution if we calculate the limit:
$\mathop {\lim }\limits_{n \to \infty } \left[ {F\left( {\Delta t_n s} \right)} \right]^n$

$\begin{array}{l} \mathop {\lim }\limits_{n \to \infty } s\Delta t_n = 0 \\ \bar s = s\Delta t_n \\ \mathop {\lim }\limits_{\bar s \to 0} F(\bar s) = \int\limits_{ - \infty }^{ + \infty } {f(v)dv = 1} \\ \\ \frac{{d^k F(\bar s)}}{{d\bar s^k }} = \int\limits_{ - \infty }^{ + \infty } {f(v)v^k e^{\bar sv} dv} \\ \\ \mathop {\lim }\limits_{\bar s \to 0} \frac{{d^k F(\bar s)}}{{d\bar s^k }} = \int\limits_{ - \infty }^{ + \infty } {f(v)v^k dv} = M_k \\ \end{array}$

You probably know how we end the demonstration, developing F in taylor series etc, so my real question is: until now we assumed that f(v) had a transform F(s). Now when we calculate the limit for n that approaches infinity, we are using the transform in a region where s approaches 0. If we know that f(v) has overall integral 1, we know that the transform is defined for s=0, and if we know that has finite first and second moment, we know that the first and second derivatives of F(s) in 0 would be finite. Is this enough to know that F(s) is defined for at least a small region around s=0 ?

Another thing: there's sort of a weak point, about the convergence as you said. It is quite difficult to speak of convergence of a irregular function, since we could have dirac's deltas in the discrete probability case. Anyway we have that the transforms of such irregular function are much more regular, so another question of mine is: once we have a very good convergence of the transforms, what kind of convergence we have for the antitransforms?
I'm sorry I couldn't manage to put the problem in terms of the cumulative probability functions (that are in fact much more regular than the probabilty distributions).


Solution:

The smallest triangles: 8
Group two of the smallest triangles together: 4
Group two of the smallest triangles together combined with a square: 4
Group four of the smallest triangles together with two squares: 4

$ 8 + 4 + 4 + 4 = \boxed{20}$

Let the smallest possible triangle in the picture have area/size $t$, and the smallest possible square have area/size $s$. Now, we count:

Area $t$ $\implies$ $8\triangle$s
Area $2t$ $\implies$ $4\triangle$s
Area $2t+s$ $\implies$ $4\triangle$s
Area $4t+s$ $\implies$ $4\triangle$s

$$8+4\cdot3 = \boxed{20}$$


Find the area of triangle ABC if a=40ft b=25ft and <C=30 degrees


function
integration
real analysis
real analysis unsolved

Let $ f$ be a twice differentiable real valued function satisfying:

$ f(x)\plus{}f''(x)\equal{}\minus{}xg(x)f'(x)$

where $ g(x)\geq 0$ for all real $ x$. Prove that $ |f(x)|$ is bounded

Multiplying both sides by $ f'$ gives
\[ ff' \plus{} f''f' \equal{} \minus{} xg(x)(f')^2 \\
\Rightarrow \frac {d}{dx}\left(f^2 \plus{} (f')^2\right) \equal{} \minus{} 2xg(x)(f')^2 \\
\Rightarrow f^2 \plus{} (f')^2 \equal{} E_0 \minus{} 2\int_0^xtg(t)(f')^2dt \\
\Rightarrow f^2 \plus{} (f')^2\leq E_0 \\
\Rightarrow |f|\leq \sqrt{E_0}
\]
where $ E_0 \equal{} f(0)^2 \plus{} f'(0)^2$ is some positive constant.  [hide="Remark"]This is a simple kind of 'energy method', and is motivated by the fact that the given ODE describes damped harmonic motion.  In this interpretation, the $ f^2$ term is proportional to the potential energy, $ (f')^2$ to the kinetic energy, and $ 2\int_0^xtg(t)(f')^2dt$ the total energy lost to damping up to time $ x$.[/hide]


inequalities
calculus
calculus computations

If $a,b$ be positive real numbers such that $ab=1$, then the following inequlaity
\[(m+a)^{2}+(m+b)^{2}\geq(m+1)^{2}(a^{p}+b^{p}) \]
holds for any nonnegative real numbers $m$ if and only if $0\leq p\leq\frac{\sqrt{2m+4}}{m+1}$.

If
\[(m+a)^{2}+(m+b)^{2}\geq(m+1)^{2}(a^{p}+b^{p}) \]
holds for any nonnegative real numbers $m$ then when $m\to\infty$ we have $2\geq a^{p}+b^{p}$, but $a^{p}+b^{p}\geq 2(ab)^{\frac{p}{2}}=2$.  :huh:

[quote="Ji Chen"]When $m\to+\infty$ we have  $p\to0$.[/quote]
Your problem isn't good  :lol:


geometry
rotation
integration
LaTeX
calculus
calculus computations

The designers of a bowl use an area rotated through $2\pi$ radians around the $x$ axis as the basis for their design.

The area used is between the curve $y=4\sqrt{x+2}$, the $x$ axis and the lines $x=0$ and $x=a$.

If the capacity of the bowl is to be $1000cm^3$, find $a$ correct to one decimal place.

The volume of this bowl is

\[ \textrm{V}=16\pi\int_{0}^{a}\left(x+2\right)\,dx=8\pi a^{2}+32\pi a.  \]
Set this equal to $1000$ and you get $125 =\pi a^{2}+4\pi a$.  Therefore you are looking for $a\approx 4.6$.

The first stage is correct, but an error has occurred in the rest. If you try the value for $a$ you have calculated, $1000$ is not the capacity of the bowl :oops:

Sorry about that I made a mistake in my arithmetic.  It should be $a\approx 4.6$.

Yes, thats correct.  :D
btw, using latex how do you get integrals with limits on them?

This should do it:

$\$$ \int_{a}^{b}f\left(x\right)\,dx $\$$ = $\int_{a}^{b}f\left(x\right)\,dx$

Just to let you know:  If you ever want to see how a formula is coded in $\LaTeX$ you can either click on the image or drag your mouse over it and the code will appear.


inequalities
Cauchy Inequality
algebra unsolved
algebra

We know that $ 1 \plus{} x^2\ge2x$ so $ \frac {2x^2}{1 \plus{} x^2} \le x$
so $ x \ge y$,we repeat the same thing for others we have $ x \equal{} y \equal{} z$
and we solve the equation $ x^3 \minus{} 2x^2 \plus{} x \equal{} 0$
so $ x \equal{} y \equal{} z \equal{} 1$

Oh! No. You are wrong. If $ x<0$ then $ \frac{2x^2}{1\plus{}x^2} \le x$ is not true.

[quote="thanhnam2902"]$ \begin {cases} \frac {2x^2}{1 \plus{} x^2} \equal{} y \\
\frac {2y^2}{1 \plus{} y^2} \equal{} z \\
\frac {2z^2}{1 \plus{} z^2} \equal{} x \end {cases}$[/quote]
$ x,y,z \geq 0$
If a among them is equal to 0 then that two numbers is equal to 0
=> $ x \equal{} y \equal{} z \equal{} 0$
If $ x,y,z \neq0$, we multiplicates ever all together thigh is three equation
$ (1 \plus{} x^2)(1 \plus{} y^2)(1 \plus{} z^2) \equal{} 8xyz$
Take Cauchy inequality We has equal bar happen
.............

$ (x;y;z) \equal{} (0;0;0); (1;1;1)$

It is a very well known problem similar problem is given at Canadian Math Olympiad - 1996.
The same problem is given at Bulgarian Math Olympiad 1977 or 1978 as well as Moskow Math Olympiad.


The table below represents the male population of students attending
Walker Middle School. One boy is selected at random from the male
students. What is the probability that he is in seventh grade? Express
your answer as a common fraction.
\begin{tabular}{|l|c|c|c|} \hline
& 7th Grade & 8th Grade & 9th Grade \\ \hline
Males & 25 & 22 & 23 \\ \hline
\end{tabular}

The total number of boys is 25+22+23=70. Only 25 of these are 7th grade boys, so the answer is 25/70 or 5/14.

$ \begin{tabular}{|l|c|c|c|} \hline
& 7th Grade & 8th Grade & 9th Grade \\ \hline
Males & 25 & 22 & 23 \\ \hline
\end{tabular}$


function
inequalities
inequalities proposed

Let $x_1,x_2,\dots , x_n$ be nonnegative reals such that
\[x_1x_2\cdots x_n+(n-1) = x_1+x_2+\cdots + x_n.\]
Prove that $\text{max}\{x_1,x_2,\dots , x_n\} \geq 1$.

The function $x_1+x_2+\ldots+x_n-(n-1)-x_1x_2\ldots x_n$ is linear in $x_1,\ldots,x_n$.

Hence it takes its maximum over $[0,1]^n$ at the vertices of $[0,1]^n$ (include $0$ as an admissible value for now)

If $x_1=1$, the function becomes $x_2+\ldots+x_n-(n-2)-x_2\ldots x_n$, and we can apply induction.

If $x_1=0$, the function becomes $x_2+\ldots+x_n-(n-1)$, which means that all of the remaining $x_i$s must be $1$.  In either case, the inequality has a maximum value of $0$, and since $(0,1]^n\subseteq [0,1]^n$, we're done.

It remains to show that equality is only achieved over $(0,1]^n$ if all of the variables are $1$.  If one of the variables $x_j$ is not $1$, we may increase the value of the function by changing $x_j$ to one, since

$x_j\left(1-\prod_{i\not=j}x_i\right)\geq 0$

Cool... I was thinking more along the lines of... say $y_i = x_1x_2\cdots x_i$ so assume $x_i < 1$ and hence $y_i <1$.  We have

$x_1x_2\cdots x_n + (n-1) - (x_1+x_2+\cdots +x_n) =  \sum_{i=1}^{n-1} (1+y_{i+1}-y_i-x_{i+1})$
$ = \sum_{i=1}^{n-1}(1-y_i)(1-x_{i+1}) > 0$,

contradiction!


2. You are given a piece of paper. You can cut the paper into 8 or 12 pieces. Then you can do so for any of the new pieces or let them uncut and so on.
Can you get exactly 60 pieces? Show that you can get every number of pieces greater than 60.

[quote="cosinator"][hide]Whenever you cut up a piece of paper, you add either $12$ or $8$ new pieces of paper, but lose the one you originally had (the one that's now cut up).  So, you can only add $11$ or $7$ pieces of paper every time.  Since you start with $1$ piece, all numbers that can be reached are of the form $1+7x+11y$.  By the chicken mcnugget theorem, the number $1+7\cdot11-7-11=60$ is the largest number that cannot be reached.  The answer follows.[/hide][/quote]

I don't think you should directly cite the chicken mcnugget theorem since it trivializes the problem.

it was pretty trivial to begin with

and you can be like 1 - 7 - 11 + 7*11 = (7 - 1)(11 - 1) = 60 and cite simon's favorite factoring trick

fun fun fun


How would I proof the following:
suppose  f: X--->Y is a function and A, B are subsets of Y
a) show that f inverse(A U B) = f inverse (A) U f inverse (B)
b) show that f inverse (A ∩ B)= f inverse (A) ∩ f inverse (B).

another one:
 suppose  f: X--->Y , A C X, and B C Y
a) show that A C f inverse(f(A)). give an example in which the containment is proper.
b) show that f(f inverse(B)) C B. give an example in which the conatainment is proper.

These aren't linear algebra questions, of course - they're part of what I sometimes informally call "functionspeak," which is the elementary background of many different parts of mathematics. I'd move this if I could think of a better place to put it - but i don't know where that would be.

You just have to consider the definitions, and keep asking what it would mean for a particular point to belong to a particular set.

For instance, $x\in f^{-1}(A)$ if and only if $f(x)\in A.$

Thus $x\in f^{-1}(A\cup B)$ iff $f(x)\in A\cup B$ iff $f(x)\in A$ or $f(x)\in B.$

Can you finish that thought?

As for the examples for your second pair of questions:

[hide="Hint for (a):"]Consider an example of a function that is not one-to-one.[/hide]
[hide="Hint for (b):"]Consider an example of a function that is not onto (does not map $X$ onto $Y$).[/hide]


Find all non-decreasing functions $ f : \mathbb{R} \rightarrow \mathbb{R}$ such that $ f(x\plus{}f(y))\equal{}f(f(x))\plus{}f(y)$ for all reals $ x$ and $ y$.

Let x,y=0. Then, f(f(0)) = f(f(0)) + f(0) or f(0) = 0.

Now let y = 0. We have f(x + f(0)) = f(f(x)) + f(0) or f(x) = f(f(x)).

f(x) = x is the only function that satisfies this. [/code]

... or $ f(x) \equal{} \lfloor x \rfloor$, or many other functions.  This solution is not complete by any means.

Is there like a way of proving that functions with ceilings or floors work

Like that seems kind of hard.

I think beezees has established that f(f(x)) = f(x). Here is a demonstration of how varied the solutions to that equation can be:

[hide]

Let S be the finite set of points in the range of f. f(f(x))=f(x) means that we have fixed points at (z,z) for every z in S.... these points lie along the line y = x.

Consider two consecutive elements of S, say z1 and z2. If there lies an x between these two points, then f(x) must be either z1 or z2 to preserve the non-decreasing criterion. Hence, we may have one of four schemes for each such interval:

$ \forall x : \; z_1 < x < z_2, \; f(x) \equal{} z_1$
$ \forall x : \; z_1 < x < z_2, \; f(x) \equal{} z_2$
$ \exists m : \; z_1 < m < z_2 \; : \; \forall x : \; z_1 < x < m, \; f(x) \equal{} z_1, \; \forall x : \; m < x \leq z_2, \; f(x) \equal{} z_2$
$ \exists m : \; z_1 < m < z_2 \; : \; \forall x : \; z_1 < x \leq m, \; f(x) \equal{} z_1, \; \forall x : \; m < x < z_2, \; f(x) \equal{} z_2$

Let's label these schemes 1, 2, 3, 4, and apply them to each finite-length interval between fixed-points (there are a countable number).

We can thus consider the set S of fixed points, an ordered list of m values, and an ordered list of scheme numbers, and it will uniquely define a function that satisfies f(f(x)) = f(x).

[/hide]

Now, you can look at this set of functions and determine which solve the given equation


In a circle of 20 cm of diameter, the distance from the center to a chord AB is 6 cm.  Find the length of the chord AB.

This problem appears in the test to enter in the university here in Chile (equivalent to SAT in USA)
So, by the Chord theorem we have :  16*4 = X*Y
Hence, we have one equation and we need another.
So, the answer is trivial if we “assume”  that  “the line from the center of the circle to the chord is ‘perpendicular’ to the chord“.
In this case that segment also bisects the chord. So X = Y ==> 16*4 = X*X
Then, X= 8 => chord AB = 16

However, my opinion is that this problem is wrong because there isn’t enough information.
Why ? in which part of the “announcement” of the problem said “explicitly” that the chord is ‘perpendicular’ to the line (in this case the radius) ?
So, this is my question, I means: Who, where? in the Math community defined (as a theorem) that each time that we read : “The distance  of a point (p) to a line, by definition is  always   the length of the  perpendicular segment   from the point ‘p’ to that line.“  Who defined that?
Could some one please give me a reference book or valid link in the internet that accept this definition as the “only convention accepted in the math world”. So, this is the only true assumption that we always need to do in this type of problem.
Thanks,

I don't have a good link, but here's the basic idea: The distance between a point and a line is defined as the length of the [i]shortest[/i] segment that connects the point and the line, which is perpendicular to the line. Otherwise, there would be infinitely many possible values for the distance. That would be very inconvenient. I hope this makes sense...

[quote="gva"]You mention: “…The distance between a point and a line is defined as the length of the shortest segment that connects the point and the line, which is perpendicular to the line.”

Well, I already know that, so you repeat the same. My question is : where is a valid reference, link etc that stablish that “DEFINITION” as a valid theorem or the only valid assumption in the math community.

Also you mention : “.. Otherwise, there would be infinitely many possible values for the distance. That would be very inconvenient….”

If I say : The area of a rectangle is 80. Find the “length” of the rectangle.
Using your  logic or resoning, you will conclude the same “…. would be infinitely many possible values for the Length. That would be very inconvenient….”
Obviously, The problem is NOT full determinate ( have another degree of freedom ) so you need to establish another equation. (see the attached document)[/quote]

wow, you wrote a lot in the pdf...but as frt said, the distance from a point to a line is defined as the shortest distance...which happens to be the perpendicular distance

the proof is pretty simple, it is a consequence of the pythagorean theorem...here is a proof anyway

say we have line $m$, point $P$ not on that line, $P'$ on line $m$ such that $P'P$ is perpendicular to $m$, and point $Q$ on $m$, distinct from $P'$,
then we have $P'M^2+P'P^2=MP^2$, but $P'M>0$ since those are distinct points, so we get
$PM>PP'$
so basically, the minimum distance occurs when you do the perpendicular distance...

as for a direct reference, look in a geometry textbook...this is kind of a standard knowledge thing...here is a link...

[url]http://softsurfer.com/Archive/algorithm_0102/algorithm_0102.htm#Distance%20to%20an%20Infinite%20Line[/url]

as for your length of a rectangle, that is a different problem...this problem is fully determined since we have that the perpendicular distance is 6, and a perpendicular line from the center of a circle bisects the chord, the rest is just pythagorean theorem

[quote="gva"]If I say : The area of a rectangle is 80. Find the “length” of the rectangle.[/quote]
We would never see this problem, unless they're asking for $0<l$. :?

the chord is 16 cm
it forms a right triangle - the midpt of AB, point A, and the center.


find all
\[ x\in R\]
such that
\[ \left[ x\right]\left\{ x\right\}\ge 3\]

Easy to check $ x> 4$
$ 1>\{x\}\geq\frac{3}{[x]}$
Have many $ x$ satisfy.

[quote="TTsphn"]Easy to check $ x > 4$
$ 1 >\{x\}\geq\frac{3}{[x]}$
Have many $ x$ satisfy.[/quote]

Your answer is : $ \forall x>4$, TTsphn ?  :( . Try $ x\equal{}4,0001$

All $ x>4$ satisfy
$ \{x\}\geq\frac{3}{[x]}$
I don't understand what is problem.
We can find out many number as above.


inequalities
SFFT
special factorizations

These were on a NEML last week:

1. 20 people are lined up in a row. Every minute, the people in positions 10 and 20 move up to positions 1 and 2, respectively, and the whole line shifts to make room for them. After how many minutes will the two people originally in positions 1 and 2 return to this position?

2. Solve for (a, b, c, d, e):

abcde = a + b + c + d + e

Umm...the first one, I got 20 minutes.  And on the second one, I got (0,0,0,0,0) Correct?  Wait, are answers allowed to be trivial?

For the second one(2, 2, 2, 1, 1) would work also (I'm guessing abcde is a*b*c*d*e rather than just one five-digit number because then there'll be 10 answers or more)

Yeah, 20 minutes for 1. For 2, there is more than one solution.

for 2, I got 8, 9 , and 10 as possible solutions. I broke it up according to the number of ones, i think. There can't be 4 ones, if there are 3 ones, we use Simon's favorite factoring trick, and if there are 2 ones, there is only one possible solution (which I think tare gave). This is a close to my original thinking as I can remember.

For the second one, there are definitely infinitely many solutions, unless you give me a restriction on a, b, c, d, and e.

i think you could solve the second one with AM-GM, knowing that the product is greater than 5 and less than whatever. by the way, the second one wanted all values for the product if a,b,c,d,e are pos integers JBL.

Eh, yeah, don't have the test in front of me, #2 might be for positive integers only.

Well,  I don't think AM-GM really tells you very much in that case.  If it's restricted to positive integers, try setting first 4 of them equal 1, then 3, then 2, then 1, then none, and see what happens in each of the cases.

yah... that's what i did, except it is along the lines of guess and check... i got them both... but missed two easy ones... taught me to always read the questions carefully...

[quote]If it's restricted to positive integers, try setting first 4 of them equal 1, then 3, then 2, then 1, then none, and see what happens in each of the cases.[/quote]

But why would you do that?

And for #1, does anyone have a nice solution that does not involve listing out the positions for each of the first 20 min (which is more or less what I did not the actual test)?

well just find the pattern of the changing of the positions. i beleive it changes by 1 until like the 11th time then by 2 or something. i worked it out some time back, so i dont really remember the exact changes

Ok... at first one clang moves the person back 2 spots... so after the first clang, he is in the 3rd spot... and he goes like that until after the 5th clang, he is in the 11th position. Now after 1 clang, he only moves back one, because the 10th is already ahead of him, and only 20th moves in front, so after the 14th clang, he is in the 20th spot. On the 15th clang, he moves back to the 2nd spot. Again, after each clang, he moves back 2 spots like at the beginning. so after the 19th clang, he'll be in spot 10. So the 20th clang will bring him back to 1st position.

[quote="Fierytycoon"][quote]If it's restricted to positive integers, try setting first 4 of them equal 1, then 3, then 2, then 1, then none, and see what happens in each of the cases.[/quote]

But why would you do that?[/quote]

Because 1 is clearly a special number for this problem -- it only changes one of the two sides.

For #1, ok, I guess I sorta did the same thing you all did.

For #2,

[quote]Because 1 is clearly a special number for this problem -- it only changes one of the two sides.[/quote]

So how can you prove that there can't be a solution w/o at least one of a,b,c,d,e being equal to 1?

try 2*2*2*2*2... its the smallest product that doesn't contain a 1 in it, but its sum isn't big enough... so there's not a combination w/0 1.

yeah Chinaboi got it. For each case (according to number of ones) try the rest all equal to 2 (that's the smallest product) and see whether there are any solutions.

did anyone read the solution that they gave for #2?  I didnt get to read it fully, but i saw they had some proof, im not exactly sure what it was though.

no, i just glanced at the answer page, and never saw it again...

for number 2, I think it meant all the numbers must be different or else it would probably be written like this aaaaa=a+a+a+a+a

[quote]2. Solve for (a, b, c, d, e):

abcde = a + b + c + d + e[/quote]
 I think a,b,c,d,e are integers. We can assume that e >= a,b,c,d. Then abcde=a+b+c+d+e <= 5e.
So abcd <= 5. From the last inequality we will find  solutions. :)

e[quote]2. Solve for (a, b, c, d, e):

abcde = a + b + c + d + e[/quote]
Ooops, it is not given that a,b,c,d,e are positive integers. So my first solution it is not full.
  1) if a,b,c,d,e are positive it is easy to find solutions.
  2) if a,b,c,d,e are negative there are no any problem.
  3) If some of a,b,c,d,e are negative and some are positive.
 In the third case : We can assume that a >= b,c,d >= e. So 5a >= a+b+c+d+e=abcde >= 5e.
Then bcde <= 5 <= abcd. If a is negative then b,c,d,e are negative. So a is nonnegative and bcd is nonnegative , too.
 Since bcd >= 0.  If e >= 0 then a,b,c,d  >= e >= 0. So e  <= 0. Let's denote e=-k then k >= 0.
We have a+b+c+d-k+abcdk=a+b+c+d+k(abcd-1)=0. If a,b,c,d  >= 0 there are no problem. Since   bcd >= 0
only two of them can be negative, say b and c. Let b=-m, c=-n. Then a+d-m-n-k+abmnk=0
For a >= 3   a+d+abmnk >= 3+d+ 3bnmk >= 3+d+ bnmk+bnmk+bnmk>m+n+k. We have to check a=2,a=1,a=0
I think this solution is true, but I don't like it myself :blush: . Is there another solution?


Maybe Alcumus can provide hints to problems? I currently do not own Intro. to Counting and Probability (although I ordered it over a week ago) so this would be very helpful with most of the problems.

me agrees. This way the woreser ppl( like me) don't have to give up as much.

I volunteer to help write hints, if such is decided a good idea.

I volunteer too, to hint on the C&P problems. And watch the videos currently available if you want help. Or if you are in the diagnostic stage, then you'll have to wait.


FInd the relationship between arcsin (cos (arcsin x)) and arccos (sin (arccos x))

in other words we have to link them in some way....ne ideas?

Let $\theta=\arcsin x$.  So then

\begin{eqnarray*} \arcsin(\cos(\arcsin x)) &=& \arcsin(\cos\theta)\\ &=& \arcsin(\sin(90^\circ-\theta))\\ &=& 90^\circ-\theta \end{eqnarray*}

and

\begin{eqnarray*} \arccos(\sin(\arccos x)) &=& \arccos(\sin(90-\theta))\\ &=& \arccos(\cos \theta)\\ &=& \theta \end{eqnarray*}

So they add to $90^\circ.$

Try with complementar angles: i suppose the sum of the two terms is $\frac{\pi}{2}$


trigonometry
function
Support
geometry
geometric transformation
homothety
geometry proposed

Let $ABCD$ be a convex quadrilateral with its sides tangent to a circle $(O)$. Let $(O_1, r_1)$ and $(O_3, r_3)$ be circles that are different from $(O)$ and respectively tangent to segment $AB$, ray $CB$, ray $DA$ and sement $CD$, ray $BC$, ray $AD$. Similarly, we define $(O_2, r_2)$ and $(O_4, r_4)$ be circles that are different from $(O)$ and respectively tangent to segment $DA$, ray $BA$, ray $CD$ and sement $BC$, ray $AB$, ray $CD$. Prove that $r_1.r_3 = r_2.r_4$.

i just did copy/paste so try inderestandig serbian. It was in some journal and it's well known Sangaku problem

4. Ako su poluprecnici spolja pripisanih kruznica tangentnog
cetvorougla $ABCD$ nad strancima $AB,BC,CD,DA$ redom
$r_1,r_2,r_3,r_4$, dokazati da je $r_1 \cdot r_3=r_2 \cdot r_4$\\

Resenje:\\

Neka je tacka $O_1$ centar pripisane kruznice nad stranicom $AB$ i
neka je podnozje normale iz $O_1$ na $AB$ tacka $M$. Onda je
$r_1=O_1M$. Neka su unutrasnji uglovi cetvorougla $ABCD$ redom
$\alpha, \beta, \gamma, \delta$, a duzine stranica redom
$a,b,c,d$. Onda iz trougla $AO_1M$ imamo ${r_1 \over \cos{{\alpha \over 2}}}=AO_1$. Slicno je i $O_1B={r_1 \over \cos{{\beta \over 2}}}$. Izjednacavanjem povrsine trougla $ABO_1$ dobijamo
$ar_1={AO_1\cdot BO_1 \cdot \sin {{\alpha + \beta \over 2}} \over 2}$. Iz ovoga je $r_1={a \cos{{\alpha \over 2 }}\cdot \cos{{\beta \over 2 }} \over \sin{{\alpha+\beta \over 2}}}$.\\

Sada se slicno dobijaju i formule $r_2={b\cdot \sin{{\beta+\gamma \over 2}} \over \cos{{\beta \over 2}}\cdot \cos{{\gamma \over 2}}}$, $r_3={c\cdot \sin{{\gamma+\delta \over 2}} \over \cos{{\gamma \over 2}}\cdot \cos{{\delta \over 2}}}$ i
$r_4={d\cdot \sin{{\delta+\alpha \over 2}} \over \cos{{\delta \over 2}}\cdot \cos{{\alpha \over 2}}}$, pa se zadatak svodi na to
da se dokaze
\begin{center}
$ac \cdot \sin{{\alpha+\beta \over 2}}\cdot \sin{{\gamma+\delta \over 2}}=bd \cdot \sin{{\beta+\gamma \over 2}} \cdot \sin{{\alpha+\delta \over 2}}$    ($\ast$)
\end{center}\\

Neka je centar upisane kruznice u cetvorougao $ABCD$ tacka $O$ i
neka upisana kruznica dodiruje stranice $AB,BC,CD,AD$ u tackama
$M,N,P,Q$ redom. Neka je i poluprecnik upisane kruznice $r$. Onda
je $a=AM+BM=r \cot{{\alpha \over 2}}+r \cot{{\beta \over 2}}=r{\sin{\alpha \over 2}\cdot \cos{{\beta \over 2}}+\sin{{\beta \over 2}}\cdot \cos{{\alpha \over 2}} \over \sin{{\alpha \over 2}}\cdot \sin{{\beta \over 2}}}=r{\sin{{\alpha+\beta \over 2}}\over \sin{{\alpha \over 2}}\cdot \sin{{\beta \over 2}}}$.
Slicno se dobija i $b=r{\sin{{\beta+\gamma \over 2}}\over \sin{{\beta \over 2}}\cdot \sin{{\gamma \over 2}}}$,
$c=r{\sin{{\gamma+\delta \over 2}}\over \sin{{\gamma \over 2}}\cdot \sin{{\delta \over 2}}}$ i $d=r{\sin{{\delta+\alpha \over 2}}\over \sin{{\delta \over 2}}\cdot \sin{{\alpha \over 2}}}$.

Sada jednostavnom proverom ( zamenjujuci ove izraze u ($\ast$) )
dobijamo da je ($\ast$) tacno pa je tacna i njegova ekvivalencija,
odnosno polazna pretpostacka $r_1\cdot r_3=r_2 \cdot r_4$, sto je
i trebalo dokazati.

Dule-00,

  treegoner gave us a hint that it was a silly problem,

  so there was no need to get so serious and post solution in serbo-horvathian.

  So, next time please try to post in some silly language, like ancient babylonian.

   I've just found out that trigonometric functions were not available in 17th century

   Japan, the time of Sangaku(s). They struggled with geometric problems using

   medieval methods.

   Thank you.

   M.T.

Dule_00, although Armpist uses a callous language (with good reason), he is right. You must understand here can use only the English language and/or the U.M.L. with the its support LaTeX.
I will use the universal mathematical language ([b]U.M.L.[/b]). Denote:
$\blacktriangleright AB=a,\ BC=b,\ CD=c,\ DA=d;\ A=2x,\ B=2y,\ C=2z,\ D=2t$.
$\blacktriangleright$ the incircle $C(I,r)$ is tangent to the sides $AB$, $BC$, $CD$, $DA$ in the points $M$, $N$, $P$, $R$ respectively;
$\blacktriangleright$ the exincircles $C(I_1,r_1)$, $C(I_2,r_2)$, $C(I_3,r_3)$, $C(I_4,r_4)$ are tangent to the sides $AB$, $BC$, $CD$, $DA$ in the points $M'$, $N'$, $P'$, $R'$ respectively.
From the well-known property result: $AM'=BM$, $BN'=CN$, $CP'=DP$, $DR'=AR$.
Thus, $AM'=r_1\tan x,\ BM=r\cot y\Longrightarrow r_1=r\cot x\cot y$.
Similarly, $r_2=r\cot y\cot z$, $r_3=r\cot z\cot t$ and $r_4=r\cot t\cot x$.
Therefore, $r_1r_3=r_2r_4=r^2\cot x\cot y\cot z\cot t$.

The only thing armpist is right about is that there is no need to post a solution in Serbian, but that's only because there is no need to post a solution of any problem in any language whatsoever.  I remember him quite well posting in Russian on an English forum ([url=http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=205&forum=DCForumID3#4]here[/url], message #6, Golland = armpist). As for the rest of his rattle, you would think that carving the solution in wood would make him content.

Let $P \equiv AB \cap CD$, $Q \equiv BC \cap DA$. WLOG, let P lie on the opposite side of BC than A and let Q lie on the opposite side of CD than A. Let $T_1, T_2, T_3, T_4$ be the tangency points of $(O_1), (O_2), (O_3), (O_4)$ with AB, BC, CD, DA. The vertex A is the internal homothety center of $(O_4), (O_1)$ and the vertex C is the internal homothety center of $(O_2), (O_3)$. Hence, $\frac{r_4}{r_1} = \frac{AT_4}{AT_1},\ \frac{r_2}{r_3} = \frac{CT_2}{CT_3}$. The convex quadrilateral ABCD and the concave quadrilateral APCQ are both tangential with an incircle, hence,

$AB + CD = BC + DA,\ \ AP + CQ = QA + PC$

Starting from $BC + CQ = BQ$, add $AP = AB + BP$ to both sides:

$AP + BC + CQ = AB + BP + BQ$

Substitute $AP + CQ = QA + PC$ on the left:

$QA + PC + BC = AB + BP + BQ$

Rearrange:

$BC + CP - PB = AB + BQ - QA$

$s_2 - PB = s_1 - QA$

where $s_2, s_1$ are semiperimeters of the triangles $\triangle BCP, \triangle ABQ$. Since $(O_2)$ is an incircle of the $\triangle BCP$ and $(O_1)$ is an excircle of the $\triangle ABQ$,

$CT_2 = AT_1$

Similarly, we obtain $CT_3 = AT_4$. Consequently,

$\frac{r_4}{r_1} = \frac{AT_4}{AT_1} = \frac{CT_3}{CT_2} = \frac{r_3}{r_2}$.

Dear treegoner,

  What was the reason to call this problem  'very silly' ?

  What is your proof?

   T.Y.

   M.T.

I have no idea why I put it "very silly"   :ninja: My proof is similar and based on the following lemma:

[i]Lemma[/i] Suppose $ABC$ is a triangle. Then $\frac{r_a}{r} = tg{\frac{B}{2}}tg{\frac{C}{2}}$.


LaTeX
induction
strong induction
number theory unsolved
number theory

i don't use the latex like you  :) Writing like that is easier ...but i hope that you can understand it!

Since x_i are increasing, there is a unique i>= for which x_i<= y < x_i + 1 . Using  strong induction on i...if y=xi<b, we let a=y-x_i and we're done. Otherwise x_1=b <= y-x_i <x_(i+1)-x_i= x_(i-1). Thus there is a unique j>=1 such that x_i<= y-x_i < x_(j+1), and j<i,so we finish by applying the inductive hypothesis yo y-x_i.

Thank you fatih, but i couldn't understand :blush:

[quote="snowflake"]Please! prove quickly....
Every integer y>=b can be expressed in the form a + x_i_1 +x_i_2+...+x_i_a,
where 0<=a<b, i_1,i_2,...,i_a, and  x_i_a<= y < x_i_a + 1[/quote]
I think if I understood the problem, this has NO sense let me write in a good $\LaTeX$ way:
Every integer $y\ge b$ can be expressed in the form
\[
a+x_{i_1}+x_{i_2}+\cdots+x_{i_a},
\]
where
\[
0\le a <b,\, i_1,\, i_2,\, \cdots , i_a, \quad {\rm and } \quad x_{i_a}\le y <x_{i_a}+1.
\]
If I am right, are $x_{i_j}$ integers? what is the order between them?  Please, write in a better way if you can.


abstract algebra
function
group theory
superior algebra
superior algebra unsolved

let G be abelian group
let $H \leq G$ so that $G/H$ be infinite and cyclic.
proove that $G \simeq H\times G/H$

this is the problem...
its taken from Ion D. Ion - "Probleme de Algebra"
im now writing some parts of the solution given in the book that i dont understand
they say that if G/H is cyclic then $\exists \hat{a}$ so that $<\hat{a}\geq G/H$
ok...this i understand now...
they say that if $G/H$ is infinite it means that $\forall g\in G/H$ $\exists ! k\in N$ so that
$g=(\hat{a})^k=\hat{a}_i^k$
and they say that this means that $\exists !$ $h \in H$ so that $g=ha^k$
i dont understand this
now they say that we can consider
$\phi : G\to G/H$ canonical epimorphism and also $\psi : G\to H$ with $\psi (g)=h$
so that the isomorphism we searched for is $f: G\to H\times G/H$ with
$f(g)=(\phi(g),\psi(g))$
they say that it is easy to verify that $\phi,\psi$ are both morfisms.
Also that does imply that f should be isomorphism...that is not very clear to me but seems intuitive.

[quote="spx2"]let G be abelian group
let $H \leq G$ so that $G/H$ be infinite and cyclic.
proove that $G \simeq H\times G/H$

this is the problem...
its taken from Ion D. Ion - "Probleme de Algebra"
im now writing some parts of the solution given in the book that i dont understand
they say that if G/H is cyclic then $\exists \hat{a}$ so that $<\hat{a}\geq G/H$
ok...this i understand now...
they say that if $G/H$ is infinite it means that $\forall g\in G/H$ $\exists ! k\in N$ so that
$g=(\hat{a})^k=\hat{a}_i^k$
and they say that this means that $\exists !$ $h \in H$ so that $g=ha^k$
i dont understand this
[/quote]
Ok, the fact that $\forall g\in G/H$ $\exists ! k\in N$ so that
$g=(\hat{a})^k=\hat{a}_i^k$ means in other words that $gH=(aH)^k=a^kH$, so $(a^k)^{-1}g \in H$, thus there exists $h \in H$ such that $g=ha^k$.(The uniqueness of this $h$ is easy by supposing the contrary)

[quote="spx2"]
now they say that we can consider
$\phi : G\to G/H$ canonical epimorphism and also $\psi : G\to H$ with $\psi (g)=h$
so that the isomorphism we searched for is $f: G\to H\times G/H$ with
$f(g)=(\phi(g),\psi(g))$
they say that it is easy to verify that $\phi,\psi$ are both morfisms.
Also that does imply that f should be isomorphism...that is not very clear to me but seems intuitive.[/quote]
No, exactly the fact that they are morphisms doesn't imply that it's isomorphism, but it follows from the written above:
it's injective(due to uniqueness of $k$ and $h$ above), and surjective, because for every pair $(h,\hat g)$, there exist a pair $(h,k)$ where $k$ is the power in $\hat g = (\hat a)^k$, so we can put $g=ha^k$ and then $f(g)=(g, \hat g)$
Injectivity with surjectivity together with the fact that it is morphism gives us the isomorphism.

but from what you say $h=1$.
what exactly is that function $\psi$ and how does it
interact with $G$ and $H$ ?
i mean...how can we take an element from G and associate it to
an element in H wich is smaller and at the same time avoiding
colissions wich would compromise injectivity(probably if some
collision like that would occur then probably $\phi$ would change
so injectivity of $f$ is not compromised but i dont see exactly how) ?

now there is something i noticed $|H\times G/H|=|H|\cdot |G/H|=|H|\cdot \frac{|G|}{|H|}=|G|$
so they should really be isomorphic.

[quote="spx2"]but from what you say $h=1$.
[/quote]
Sorry, it was my typo, instead of "unicity" i meant $uniqueness$.

[quote="spx2"]
what exactly is that function $\psi$ and how does it
interact with $G$ and $H$ ?
i mean...how can we take an element from G and associate it to
an element in H wich is smaller and at the same time avoiding
colissions wich would compromise injectivity(probably if some
collision like that would occur then probably $\phi$ would change
so injectivity of $f$ is not compromised but i dont see exactly how) ?
[/quote]
You see, $\psi$ is not actually injective, for example you may have for some $g=ha^k$ and $g_1=ha^m$. It is true that for every $g \in G$, the pair $(h,k)$ is really unique. So, there shouldn't be confusion with that $G$ is bigger than $H$ and we map $G$ on $H$ (this map is not injective).
And as $f(g)$  is surjective together with you observation
[quote="spx2"]
now there is something i noticed $|H\times G/H|=|H|\cdot |G/H|=|H|\cdot \frac{|G|}{|H|}=|G|$
so they should really be isomorphic.[/quote]
we have isomorphism.


here is a science problem/math problem.

a person last year got it in 10 seconds but chess64 took 4 minutes then gave up  :lol:

balance the equation:

$H_{3}PO_{4}+(N H_{4})_{2}MoO_{4}+HNO_{3}\to N_{3}H_{1}2 PO_{4}Mo_{12}O_{36}+NH_{4}NO_{3}+H_{2}O$

if you used a method other than counting (which i'm sure you did) then please show work

uhh...if you cannot just see the answer, all you do is assign variables and solve a linear system

[quote="me@home"]What element is $Mo$?[/quote]

Molybdenum. (pronounced mo-[b]lib[/b]-de-num)

never mind turns out we solved it during class using calculators and matrices...

[quote="me@home"]How else would you pronounce molybdenum?

I kind of guessed it was something like that (I'm in AP Chem right now but I guess I am still inept in chemistry :oops: ). I forget by know but now that I think about it there was something like a P that stood for something else...[/quote]

 :)  I've just heard people trying to pronounce it as moly-bden-um and attempting a "bd" sound, which just comes out weird.


hello, from the equation $ xy=\left(\frac{x}{3}\right)^{\frac{y}{4}}$ we get $ x=3e^\frac{3\ln(3y)}{y-4}$ inserting this in $ xy=\left(\frac{y}{2}\right)^{\frac{x}{6}}$ we get the equation
${ 3e^{\frac{4\ln(3y)}{y-4}}y-\left(\frac{y}{2}\right)^{\frac{1}{2}e^{\frac{4\ln(3y)}{y-4}}}}=0$,solving this by a numerical method we get $ y\approx 0.7447582853156541580$
Sonnhard.


geometry
3D geometry
probability
MATHCOUNTS

[quote="aidan"]A cube made of yellow-coloured rubber with side length $8$ cm is painted blue. It is cut with planes parallel to the sides of the cube into cubes of side lengths $2$ cm.

(i) How may (side length $2$ cm) cubes have no blue paint on any of their faces?

(ii) How many (side length $2$ cm) cubes are painted blue on $1$ face?

(iii) How many (side length $2$ cm) cubes are painted blue on $2$ faces?

(iv) Peter picks up a (side length $2$ cm) at random and tosses it like a die so that it lands on the floor. What is the probability that the (side length $2$ cm) cube has no blue paint on any of its 5 faces showing?

Source: Singapore ASEAN Scholarship Math Entrance Examination[/quote]

[hide="i"]All the cubes inside the top layer exposed to paint have no blue paint. This is $2^3=8$. [/hide]

[hide="ii"]All the cubes not edges or corners have one face painted blue. That is $6\times 2\times 2=24$
[/hide]

[hide="iii"]This is all the edges of the cube, or $2\times12=24$[/hide]
[hide="iv"]
This is the probability he either rolls a one color cube and it is face down (1/6) or a no color cube. There is an 8/64 chance of rolling a cube with no blue, and a (1/6)(24/64) chance he does the thing with the one blue face cube. Add them together to get $\frac{3}{16}$.[/hide]

I got the exact same answers, which is a good sign.  :)

I did too, but these weren't on Mathcounts, darn it.

[hide="i"]They are the cubes in the center, 2x2x2=8.[/hide]
[hide="ii"]There are 4 for each of the 6 faces, 4x6=24.[/hide]
[hide="iii"]There are 2 for each of the 12 edges, 2x12=24.[/hide]
[hide="iv"]There is a 8/64 chance of picking a cube with no blue paint, and a 24/64 chance of picking a cube with 1 blue face, but for the cubes with 1 blue face, there;s only a 1/6 chance that the blue face is not showing.
$\frac{8}{64}+\frac{24}{64\times 6}=\frac{3}{16}$.[/hide]


[tex]\[
\{ (x_1 ,x_2 ) \in \mathbb{R}^2 :0 \leqslant x_2  \leqslant 1 - x_1 ^2 \}
\]
[/tex]

Show if the set is bounded,open,closed,compact.

[hide]I would say, not open, closed and bounded (so compact, we are in finite dimension).[/hide]

[hide]Definitely bounded, because if x_1 is less than -1, then (x_1) :^2:  is greater than one, which the rules for x_2 don't allow, and x_1 can't be greater than one for the same reason.  Thus -1 :le: x_1 :le: 1 and 0 :le: x_2 :le: 1[/hide]


A metallic chloride A,when treated with NaOH/H2O2 gives yellow solution due to  the formation of B.The colour of the solutionchanges to Orange when dilut H2SO4 is added.It is due to the formation of compound C.When a compound D is heated with C in presence of conc.H2SO4,a red volatile liquid E is formed.E when absorbed on NaOH solution,gives a yellow solution of B which when treated with (CH3COO)2Pb soln gives a yellow ppt.C when treated with NH4Cl
forms compund F,which decomposes on heating giving a colourless gas,water and green residue G.D gives golden yellow flame.

Identify A to G

Befor clicking post your answer

@pardesi-click the link
[hide]That's correct[/hide]


calculus
integration
function
real analysis
real analysis solved

Show that for all t>=0, equation x^3 + tx - 8 has the only positive real solution x(t). Calculate \int_{0..7}[x(t)]^2dt.

<i>Source: VUMC 1997 </i>

Of course, the equation has at least one real root and it is obvious that it is positive. Of course, we cannot have two positive roots for a certain t. For x>0 define f(x)=-x^2+8/x. Then x(t)=f^(-1)(t) for all t>0 and the integral is  integral from 0 to 7 from (f^(-1)(t))^2. Make the change of variable t=f(u) and it becomes integral from 2 to 1 from (u^2*f'(u))du and this one is 15,5.


This time of year always reminds me of Tolkien's The Lord of the Rings. It was november when I read the book for a first time and couple years later I was waiting for the movie.
Right now I can't decide what to read:Hiperion by Den Simons Shards of Honor by Lois McMaster Bujold.
What books you like to read? Do you have any recomendation for reading?

Have you ever read Enders Game?  If not, then you are a crazy person!!  It's by Oson Scott Card, and is extremely riviting.  10 out of 10! :10:

I agree. Then read Enders Shadow, which is just as good or getter in my opinion!

[quote="K81o7"]There's a little book I recommend called The Art and Craft of Problem Solving, by Paul Zeitz.  Good choice.

A good one: Fyodor Dostoevsky's Crime and Punishment, if you're into somewhat dense books which provoke thought about psychology.  Dostoevsky's books are generally good for that purpose.

And then there's that classic, How to Eat Fried Worms (the author's name escapes me at this point).  I have to remember to cite that book in an SAT essay sometime...[/quote]

lol  :rotfl:

the sorrows of young werther by j. w. goethe
the stranger by a. camus

Lord of the Flies- William Golding, pretty good book.

The Alex Rider Series (Book 1,2,3,4,5,6)- Anthony Horowitz.

[quote="neelnal"]LORD OF THE RINGS ALL THE WAY!!!

I reccomend reading Gulliver's Travels if you have not yet.[/quote]

Lord of the Rings is a good book to read. Fun and Entertaining!

Thanks for suggestions. :)

I read LOTR 4 times and planning to read it again. :jump:

Ignite168-who is the writer?

K81o7-I read Dostoevsky and he's really fantastic. Please remember who is the writer of How to Eat Fried Worms .

deimos-I managed to skip The sorrows of young werther when I was in high scool, but maybe I should try it now.  :maybe:

[quote="Ignite168"]The Books of Genesis, Exodus, Deuteronomy, Judges, Ruth, I Samuel, and II Samuel are all very good.[/quote]

[quote="Nienna"]Ignite168-who is the writer?[/quote]

according to the bible and church, the real author is god.

[quote="Nienna"]deimos-I managed to skip The sorrows of young werther when I was in high scool, but maybe I should try it now. :maybe: [/quote]

well, it's a good book. you should read it. have you read the stranger by camus? any of camus's works? his philosphy is quite interesting.
schiller is ... oh well, never mind.

[quote="Nienna"]Please remember who is the writer of How to Eat Fried Worms.[/quote]

You do realize that's a book for..maybe 7-8 year olds?

Nevertheless, it is a classic.

http://en.wikipedia.org/wiki/How_to_Eat_Fried_Worms
"How to Eat Fried Worms is the title of a children's novel written by Thomas Rockwell [...]"
 :wink:


linear algebra
matrix
algebra
polynomial

How can I compute the eigenvalues for a four by four matrix, all of whose entries are ones? Also, how can I compute the eigenvalues of a four by four matrix with (0 1 0 1) in row 1, (1 0 1 0) in row 2, (0 1 0 1) in row 3, (1 0 1 0) in row 4?

Is there a shortcut?

Thanks

Yes, there's a shortcut. If $ J$ is the $ n\times n$ matrix with all entries equal to $ 1$, $ J^2\equal{}nJ$. It has rank 1, so you get the multiplicities as well: $ n$ with multiplicity $ 1$ and $ 0$ with multiplicity $ n\minus{}1$.
Try something similar with your other matrix.

[quote="bythecliff"]How can I compute the eigenvalues for a four by four matrix, all of whose entries are ones?[/quote]
Let's make this even more general: the $ n\times n$ matrix of all $ 1$s, conventionally known as $ J.$ Multiply that by itself, and you find that $ J^2\equal{}nJ.$

[Side note: I am assuming that we are working over a field of characteristic zero. bythecliff is probably assuming the field is $ \mathbb{R}.$]

The matrix satisfies the polynomial equation $ x^2\minus{}nx\equal{}0.$ If a matrix satisfies a polynomial equation, then the only possible eigenvalues are the roots of that equation. In this case, that means that the only possible eigenvalues are $ 0$ and $ n.$

We further note that $ J$ has rank $ 1.$ By the rank-nullity theorem, that means that its null space has dimension $ n\minus{}1.$ That means that $ 0$ is an eigenvalue of geometric multiplicity $ n\minus{}1,$ hence algebraic multiplicity at least $ n\minus{}1.$

On the other hand, $ \text{trace }J\equal{}n,$ and the trace is the sum of the eigenvalues. Hence the eigenvalues can't all be zero. We must in fact have $ n\minus{}1$ copy of the eigenvalue $ 0$ and one copy of the eigenvalue $ n.$

It's also not hard to see that $ [1\;1\;1\;\dots\;1]^T$ is an eigenvector. The structure of $ J$ is just begging for us to try this out. It turns out to be the eigenvector associated with the eigenvalue $ n.$

Ah, jmerry beat me to that one. We both had this way of thinking drilled into us by the same individual, a fellow professor of mine at CSULB named Robert Mena.

Let's look at your second question: $ A\equal{}\begin{bmatrix}0&1&0&1\\ 1&0&1&0\\ 0&1&0&1\\ 1&0&1&0\end{bmatrix}.$

This is a symmetric matrix, hence diagonalizable (with real eigenvalues), hence the algebraic multiplicity of each eigenvalue equals its geometric multiplicity.

$ A$ fairly clearly has rank $ 2,$ hence nullity $ 2,$ hence $ 0$ as an eigenvalue of multiplicity $ 2.$ But the trace is zero, so the other two eigenvalues must be a balanced $ \pm$ pair.

It would be nice to find a polynomial that $ A$ satisfies. Let me use some block structuring to get at that. Write $ B\equal{}\begin{bmatrix}0&1\\1&0\end{bmatrix}.$ Note that $ B^2\equal{}I.$

Then, $ A\equal{}\begin{pmatrix}B&B\\B&B\end{pmatrix}.$

Square that, using that $ B^2\equal{}I,$ to get;

$ A^2\equal{}\begin{pmatrix}2I&2I\\2I&2I\end{pmatrix}.$

Multiply that through again:

$ A^3\equal{}A^2A\equal{}\begin{pmatrix}4B&4B\\4B&4B\end{pmatrix}\equal{}4A.$

Hence $ A$ satisfies $ x^3\minus{}4x\equal{}0,$ which has roots $ 0$ and $ \pm2.$

We can put together what we've found to say that the eigenvalues of $ A$ are $ 2,\minus{}2,0,0.$


function
algebra
polynomial
blogs
algebra unsolved

Define $ f_n(x)$ as $ \underbrace{f(f(...(f(x))...))}_{n\ times}$ and let $ f: \mathbb{Z}^{\plus{}}\leftarrow\mathbb{Z}^{\plus{}}$.
Does this sentence is true?
There exist a function $ f$ such that $ f_n(x)\equal{}kx$ iff $ \sqrt[n]{k}\in\mathbb{Z}^{\plus{}}$ and $ f$ can be represented in one case (i.e. "$ f(x)\equal{}x$ when $ x$ is odd and $ \equal{}2x\plus{}3$ when $ x$ is even" is prohibited.).
And how do you call a function that can be represented in one case?

You have to give a precise definition of "representation in one case", for example, $ f$ is a polynomial.

What happens if $ f$ is not a polynomial?

[quote="MellowMelon"]Let's say this "representation in one case" means a polynomial. Then it is fairly easy to prove that $ f_n(x) \equal{} kx$ exists iff $ k$ is an $ n$th power. Just show that $ 1$ is an upper bound for $ f$'s degree.

Let's omit this representation condition entirely. Then there always exists a function $ f$ such that $ f_n(x) \equal{} kx$ if $ n$ and $ k$ are positive integers. See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=185272]my blog[/url] for a generalization.[/quote]
i have see you blog $ MellowMelon$.
i will searsh for $ c)$ and for the cas where $ k\not|S|$

$ a)$ we suppose that $ S\equal{}\{b_i: \ i\in\mathbb{N}\}$ where $ b_{i\plus{}1}>b_i$
we have $ \forall n\in\mathbb{N}:$ exists exacly one $ (m,p,h)\in\mathbb{N}\times \mathbb{N}\times \{0,1,2,..,k\minus{}1\}$ such that $ n\equal{}f^m(b_{pk\plus{}h})$
if $ h\in\{0,1,2,...,k\minus{}2\}: \ g(n)\equal{}f^{m}(b_{pk\plus{}h\plus{}1})$ and if $ h\equal{}k\minus{}1: \ g(n)\equal{}f^{m\plus{}1}(b_{pk})$
so if easy to chek that $ \forall n\in\mathbb{N}: \ g^k(n)\equal{}f(n)$

$ b)$ we suppose that $ |S|\equal{}dk$ and $ S\equal{}\{b_0,b_1,b_2...,b_{dk\minus{}1}\}$ where $ b_{i\plus{}1}>b_i$
we have $ \forall n\in\mathbb{N}:$ exists exacly one $ (m,p,h)\in\mathbb{N}\times \{0,1,2,..,d\minus{}1\}\times \{0,1,2,..,k\minus{}1\}$ such that $ n\equal{}f^m(b_{pk\plus{}h})$
if $ h\in\{0,1,2,...,k\minus{}2\}: \ g(n)\equal{}f^{m}(b_{pk\plus{}h\plus{}1})$ and if $ h\equal{}k\minus{}1: \ g(n)\equal{}f^{m\plus{}1}(b_{pk})$
so if easy to chek that $ \forall n\in\mathbb{N}: \ g^k(n)\equal{}f(n)$

$ f: \mathbb{N}\to\mathbb{N},n\mapsto\{\begin{array}{cl}2n &\text{ for }n \text{ odd}\\
4 &\text{ for }n=2\\
2 &\text{ for }n=4 \\
n &\text{ for }n>4 \text{ even}\end{array}$
Then $ \mathbb{N}\setminus f(\mathbb{N})$ is infinite and $ f\not =g^2$ for any $ g: \mathbb{N}\to\mathbb{N}$.

[quote="olorin"]$ f: \mathbb{N}\to\mathbb{N},n\mapsto\{\begin{array}{cl}2n & \text{ for }n \text{ odd} \\
4 & \text{ for }n = 2 \\
2 & \text{ for }n = 4 \\
n & \text{ for }n > 4 \text{ even}\end{array}$
Then $ \mathbb{N}\setminus f(\mathbb{N})$ is infinite and $ f\not = g^2$ for any $ g: \mathbb{N}\to\mathbb{N}$.[/quote]
but i have prove that $ g$ exists forall $ k\in \mathbb{N}^*$ if $ \mathbb{N}\setminus f(\mathbb{N})$ is infinite and $ f$ injective.
in your exemple $ f$ is not injective $ f(4) = f(1)$.

 :arrow: however that in your exemple that $ f\neq g^2$

Sorry, forgot about $ f$ injective.

$ f: \mathbb{N}\to\mathbb{N},n\mapsto\{\begin{array}{cl}
2 & \text{ for }n = 1 \\
1 & \text{ for }n = 2 \\
n^2 & \text{ for }n > 2\end{array}$
Then $ f: \mathbb{N}\to\mathbb{N}$ injective and $ \mathbb{N}\setminus f(\mathbb{N})$ is infinite
and $ f\not = g^2$ for any $ g: \mathbb{N}\to\mathbb{N}$.

If $ g(1)=1\Rightarrow 2=f(1)=g^2(1)=1$.
If $ g(1)=2\Rightarrow 2=f(1)=g^2(1)=g(2)\Rightarrow 1=f(2)=g^2(2)=2$.
If $ g(1)=n>2\Rightarrow n^4=f^2(g(1))=g(f^2(1))=g(1)=n$.

[quote="olorin"]Sorry, forgot about $ f$ injective.

$ f: \mathbb{N}\to\mathbb{N},n\mapsto\{\begin{array}{cl} 2 & \text{ for }n = 1 \\
1 & \text{ for }n = 2 \\
n^2 & \text{ for }n > 2\end{array}$
Then $ f: \mathbb{N}\to\mathbb{N}$ injective and $ \mathbb{N}\setminus f(\mathbb{N})$ is infinite
and $ f\not = g^2$ for any $ g: \mathbb{N}\to\mathbb{N}$.

If $ g(1) = 1\Rightarrow 2 = f(1) = g^2(1) = 1$.
If $ g(1) = 2\Rightarrow 2 = f(1) = g^2(1) = g(2)\Rightarrow 1 = f(2) = g^2(2) = 2$.
If $ g(1) = n > 2\Rightarrow n^4 = f^2(g(1)) = g(f^2(1)) = g(1) = n$.[/quote]
yes $ good$ :D ,
my fault is that $ Imf=\bigcup_{m\in\mathbb{N}}f^m(S)$ where $ S=\mathbb{N}-Im(f)$
in your exemple we have $ 1,2\in Imf$ but $ 1,2\not\in \bigcup_{m\in\mathbb{N}}f^m(S)$ :wink:


Find the lenght of each edge of a cube that has a volume twise that of a cube with and adge of 10cm . give your answer to the nearest tenth of a cuntimeter

The volume of the original cube is 1000 cm^3. Thus, the new cube has a volume of 2000 cm^3. To find the side length of the new cube, compute the cube root of its volume. The cube root of 2000 is 2^(1/3) x 10. Thus, the side length of the new cube is 2^(1/3) x 10 cm.


symmetry
number theory open
number theory

Find all pairs $(x,y)$ of natural  numbers $x>1$, $y>1$, such that $3x+1$ is divisible by $y$ and simultaneously $3y+1$ is divisible by $x$.

Because of the symmetry, if $(x_0,y_0)$ is a solution, then $(y_0,x_0)$ is also a solution.

Put

$(S): =\begin{cases}3x+1=ay & (1) \\3y+1=bx & (2)\end{cases}$

where $a,b\in\mathbb{Z}^{+}$. We have

\begin{eqnarray*}xy|(3x+1)(3y+1)\implies xy|3x+3y+1\implies xy\leqslant 3x+3y+1\implies (x-3)(y-3)\leqslant 10\end{eqnarray*}

Therefore, $2\leqslant x\leqslant 13$ and $3\nmid x$ because of $(2)$, giving $x\in\{2,4,5,7,8,10,11,13\}.$

(Note: for $x>13$ we get either $y=3$, which can't satisfy $(1)$, or $y=2$, which is covered below)

Casework:

For $x=2$ $(S)$ becomes $7=ay\land 3y+1=2b$. The only possibility is $y=7$, which gives an integer solution for $b$. Hence, one pair is $(2,7)$, which also covers $(7,2)$.

For $x=4$ $(S)$ becomes $13=ay\land 3y+1=4b$. The only possibility is $y=13$, which gives an integer solution for $b$. Hence, another pair is $(4,13)$, which also covers $(13,4)$.

For $x=5$ $(S)$ becomes $16=ay\land 3y+1=5b.$ Since $y-3\leq\frac{10}{x-3}=5$, the possibilities are $y=8,4,2$, but only $y=8$ gives an integer $b$. Hence, another pair is $(5,8)$, which also covers $(8,5)$

For $x=10$ $(S)$ becomes $31=ay\land 3y+1=10b.$ The only possibility is $y=31$, but it contradicts $(x-3)(y-3)\leqslant 10.$

For $x=11$ $(S)$ becomes $34=ay\land 3y+1=11b$. The possibilities are $y=2,17,34$, but the first one gives no integer $b$, and the other two contradict $(x-3)(y-3)\leqslant 10.$

It's easy to check that $x=7,8,13$ give no new solutions.

Summary: $\{x,y\}\in\{\{2,7\},\{4,13\},\{5,8\}\}$


[quote="spiders_hate_chestnut"]i need help on this question

after collecting candy for the candy drive, Sally had eight 1-dollar bills, five 5-dollar bills and three 10-dollar bills. She ramdomly selected 3 bills without replacement. What is the probability that she chose one of each type of bill? express answer in common fraction.

please have the solutions too.

also, how do u make a dollar sign on Latex?[/quote]

Dollar sign:
[code]
$\$$
[/code]

[hide]
So if we have $ 8\times 5\times 3\equal{}120$ ways to choose one of each type of bill.  THe total number of choices is $ 16\times 15\times 14\equal{}3360$ (?)  So it's $ \frac{120}{3360}$ which I believe simplifies to $ \frac{1}{28}$.
[/hide]

Anyone from Jenks who could write a detailed explanation for what alanchou did wrong?

Don't you have to multipy the 1/28 by 6?
I don't remember why.

Because there are 3!=6 ways to arrange the bills.  In the "solution", the order didn't matter in the numerator and did in the denominator.  That=Bab  :mad:

[quote="spiders_hate_chestnut"]so permutation of how much u need to pick?[/quote]

No, more like Combination, I think.

alanchou did permutation I think... but if you do that, you obviously over count.


Hello,
I have a server for Enemy Territory, anyone wants to play?

It runs ET 2.60b, Jaymod 2.1.7 (XP save forever, unlimited shotguns and molotov) and another mod which puts bots. Usually there are 2-3 of humans on it, myself and a couple of friends of mine.

Unfortunately we only use default maps because they are the only I can have the bots "know" what to do in them, but I am working on it.

If interested please add/contact me on MSN to check out the ip address (it's dynamical) and to know what files you need (direct download from server is veeery slow).

Bye

making the thread come up
("up" was too short :) )


I am a student from Romania and this year our school will be participating at the AMC 8 . My problem is that my parents sent me to school when I was 8 years old . This means that now I am in the 8th grade but I have the age of 15 since September the 28th . In the AMC 8 Teacher's Manual says "Any student 14.5 years of age or younger on the day of the contest and not enrolled in grades 9, 10, 11, or 12 or equivalent is eligible to participate." Does this mean that I can't participate at the AMC 8 ? Should I participate directly at the AMC 10/12 ?

Since you are 15 years old, I don't think you're eligible for AMC 8.

Given the eligibility rule that you quoted from the Teacher's Manual, you should be looking at the AMC 10.

Steve Dunbar
MAA Director, American Mathematics Competitions

why school so late??? ur like my age... That definately has more advantages that setbacks tho (being old)


trigonometry
induction
algebra
polynomial
function
complex numbers
binomial theorem

Prove that for all natural n:

$ \tan{nA} \equiv \frac {\binom{n}{1}\tan{A} \minus{} \binom{n}{3}\tan^3A \plus{} \binom{n}{5}\tan^5A \ldots} {\binom{n}{0} \minus{} \binom{n}{2}\tan^2A \plus{} \binom{n}{4}\tan^4A \ldots}$

My advice

[hide] Try to use induction using the angle sum formula for tangents.  You will need to put terms under common denominators and I think you need Pascal's Identity.[/hide]

Or, try using complex numbers, De'Moivre's Theorem, and the Binomial Theorem.
[hide]
Let $ a \equal{} \tan(A)$. Note that $ 1 \plus{} ai \equal{} r \cdot cis(A)$ for some $ r \in \mathbb{R}^ \plus{}$.

Using De'Moivre's Theorem, and the Binomial Theorem:

$ r^n \cdot cis(nA) \equal{} (1 \plus{} ai)^n \equal{}$ $ \left(\dbinom{n}{0} \minus{} \dbinom{n}{2}a^2 \plus{} \dbinom{n}{4}a^4\ldots\right) \plus{} \left(\dbinom{n}{1}a \minus{} \dbinom{n}{3}a^3 \plus{} \dbinom{n}{5}a^5\ldots\right)i$

Therefore: $ \tan(nA) \equal{} \dfrac{Im(cis(nA))}{Re(cis(nA))} \equal{} \dfrac{\dbinom{n}{1}a \minus{} \dbinom{n}{3}a^3 \plus{} \dbinom{n}{5}a^5\ldots}{\dbinom{n}{0} \minus{} \dbinom{n}{2}a^2 \plus{} \dbinom{n}{4}a^4\ldots}$ as desired.[/hide]

Related problem:

1991 Tokyo University entrance exam/Science

(1) For natural number $ n\equal{}1,\ 2,\ 3,\ \cdots$, show that there exist some polynomials $ p(x),\ q(x)$ and we can write $ \sin n\theta \equal{}p_n(\tan \theta)\cos ^ n \theta ,\ \cos n\theta \equal{}q_n(\tan \theta)\cos ^ n \theta .$

(2) If the relation in (1) holds, then for $ n>1$, prove that $ p_n'(x)\equal{}nq_{n\minus{}1}(x),\ q_n'(x)\equal{}\minus{}np_{n\minus{}1}(x).$

Thanks for your help guys, this problem was driving me insane last night. I've never used complex numbers in trig before, but after I looked up the definitions of trig functions using $ r \cdot cis(\theta)$ I understood your proof.


limit
calculus
integration
function
real analysis
calculus computations

Let $x_{j}= \frac{j}{k}$ and $\Delta x = \frac{1}{k}$. By MVT,

$\sqrt{1-x_{j}^{2}}-\sqrt{1-x_{j-1}^{2}}=-\frac{c_{j}\Delta x}{\sqrt{1-c_{j}^{2}}}$

for some $x_{j-1}< c_{j}< x_{j}$. Then this Riemann sum [u]converges[/u](really?) to the integral

$\int_{0}^{1}\sqrt{1+\left(\frac{x}{\sqrt{1-x^{2}}}\right)^{2}}\; dx = \int_{0}^{1}\frac{1}{\sqrt{1-x^{2}}}\; dx = \frac{\pi}{2}$

I'm not sure if my solution is correct...

You're right to be cautious. Riemann sums don't automatically converge when the function is unbounded.
You'll need another argument to show the part for $x$ near 1 doesn't blow up.

A simple version of that argument:
For each $j<k$, the $j$th term of the sum is less than $\frac1k\cdot\frac1{\sqrt{1-\left(\frac{j}{k}\right)^{2}}}$, which is less than $\int_{j/k}^{(j+1)/k}\frac1{\sqrt{1-x^{2}}}\,dx$. The sum of the first $k-1$ terms is thus less than $\int_{1/k}^{1}\frac1{\sqrt{1-x^{2}}}\,dx$, which tends to $\frac{\pi}2$.

That leaves the last term, $j=k$. This term evaluates to $\sqrt{\frac1{k^{2}}+1-\frac{(k-1)^{2}}{k^{2}}}=\sqrt{\frac{2k}{k^{2}}}$, which goes to zero as $k\to\infty$. We therefore have that $\limsup S_{k}\le\frac{\pi}2$, where $S_{k}$ is our sum.
For a lower bound, we use the estimate that the $j$th term is greater than $\frac1k\cdot\frac1{\sqrt{1-\left(\frac{(j-1)^{2}}{k^{2}}\right)}}$, which is greater than $\int_{(j-2)/k}^{(j-1)/k}\frac1{\sqrt{1-x^{2}}}\,dx$ for $k\ge 2$. Adding these up, the sum of all terms is greater than $\int_{0}^{(k-1)/k}\frac1{\sqrt{1-x^{2}}}\,dx$, which tends to $\frac{\pi}2$, and the $\liminf$ is at least $\frac{\pi}2$. These two results bracket the sum well enough, and the limit is indeed $\frac{\pi}2$.


trigonometry
analytic geometry
geometry
circumcircle
geometry proposed

Given $\triangle ABC$ inscribed $(O)$ with any $M\in (O)$ prove that
$a\cos A.MA^{2}+b\cos B.MB^{2}+c\cos C.MC^{2}=abc$
A special case $ABC$ is equaliteral $\cos A=\cos B=\cos C=\frac{1}{2}\Rightarrow MA^{2}+MB^{2}+MC^{2}=2a^{2}$  :D

[color=darkblue][b][u]Gemath[/u][/b], you begun use the barycentrical coordinates ![/color]

[color=darkred][b]Lemma.[/b] If the point $P$ has the barycentrical coordinates $(x,y,z)$ w.r.t. the triangle $ABC$, then for any point $M$ there is the following relation :
$(*)\ \ \ x\cdot MA^{2}+y\cdot MB^{2}+z\cdot MC^{2}=MP^{2}-p_{w}(P)\ ,$ where $p_{w}(X)$ is the power of the point $X$ w.r.t. the circumcircle $w=C(O,R)$ of the triangle $ABC\ .$[/color]

[color=darkblue][u]Particularly[/u], for $P: =O$ with the barycentrical coordinates $\frac{x}{\sin 2A}=\frac{y}{\sin 2B}=\frac{z}{\sin 2C}=\frac{R^{2}}{2S}$ , $p_{w}(O)=-R^{2}$ and for any point $M\in w$, i.e. $MO=R$,

relation $(*)$ $\implies$ $\frac{R^{2}}{2S}\cdot \sum\sin 2A\cdot MA^{2}=2R^{2}$ $\implies$ $\sum \sin 2A\cdot MA^{2}=4S$ $\implies$ $\sum a\cos A\cdot MA^{2}=abc$ because $\sin 2A=\frac{a\cos A}{R}$ and $4RS=abc\ .$

[b]Exercise 1[/b]. Let $ABC$ be a triangle and for a interior point $P$ denote the measures $x,y,z$ of the angles $\widehat{BPC}$ , $\widehat{CPA}$ , $\widehat{APB}$ respectively.

Prove that the relations $\cos x+\cos y+\cos z\ge-\frac{3}{2}$ and $\sum a^{2}\cdot PA^{2}=2\sum bc\cos (x-A)\cdot PB\cdot PC\ .$ Study some particular cases.

[b]Exercise 2.[/b] Let $ABC$ be a triangle and let $x$ be a real number. Prove that the identity $4R\cdot\prod \sin\frac{A+x}{2}=2R\cdot\sin x\sin \frac{x}{2}+r\cdot\cos\frac{x}{2}+p\cdot\sin\frac{x}{2}\ \ (1)$

and study some particular cases. Solve the equation $p=2R\cdot \sin \phi+r\cdot \cot \frac{\phi}{2}$, where $\phi \in (0,\pi )\ .$

[b]Remark.[/b] For $x: =0$ and $x: =\pi$ in the relation $(1)$ obtain the well-known identities $\prod\sin\frac{A}{2}=\frac{r}{4R}$ and $\prod\cos\frac{A}{2}=\frac{p}{4R}\ .$

For $x: =-x$ the identity $(1)$ becomes $4R\cdot\prod\sin\frac{A-x}{2}=\sin\frac{x}{2}\cdot\left(2R\cdot\sin x+r\cdot\cot\frac{x}{2}-p\right)\ .$[/color]

Hi thank Virgil for the good idea, I found it intricately, but solution by barycentric is very nice, maybe in in your lemma need $x+y+z=1$ but in your solution it isn't ? :)

Evidently. When I say "barycentrical coordinates" I understand "normal", i.e. $x+y+z=1\ .$
You can joke with the point $M$ which belong another fixed circle and not by all means to the circumcircle. Maybe you"ll obtain a more general identity. Success !


algebra
polynomial
number theory unsolved
number theory

Prove that, for infinitely many positive integers $ n$, there exists a polynomial $ p$ of degree $ n$ with real coefficients such that $ p(1), p(2), ..., p(n\plus{}2)$ are different whole powers of 2.

You surely don't mean "with real coefficients", do you?

http://www.mathlinks.ro/Forum/viewtopic.php?t=1366

Wait, maybe you mean http://www.mathlinks.ro/viewtopic.php?t=121070 ?

  darij


geometry
angle bisector
geometry unsolved

Can you help me with these:
1. If ABC is an equilateral triangle and O its incircle. D and E are points on AB and AC respectively and DE is a tangent to O. Is it neccessary that angle DOE = 60 degree?
2. In ABC, P and Q are on BC and AP and AQ are trisector of angle A and BQ = QC and $AC=AQ\cdot\sqrt2$. Find angle A.
3. In ABC, angleB = angleC = 80 and P is on AB with AP = BC. Find angle BPC.
4. Angle bisectors of AF, BG and CH in ABC are drawn. angle A = 120, find angle GFH.
5. Points $A_2$, $B_2$ and $C_2$ are the midpoints of the altitudes $AA_1$, $BB_1$, $CC_1$ of acute triangle ABC. Find the sum of angles $\measuredangle B_2A_1C_2$, $\measuredangle C_2B_1A_2$ and $\measuredangle A_2C_1B_2$.

[b]Problem 1[/b] is almost trivial: I assume that you mean that the line DE is tangent to the incircle of triangle ABC, while O is the center of this incircle. Well, indeed the angle < DOE = 60°. In fact, since the lines BD and DE both touch the incircle of triangle ABC, the center O of this incircle lies on the angle bisector of the angle BDE, so that < ODE = < BDE / 2; similarly, < OED = < CED / 2, and thus,

  < DOE = 180° - < ODE - < OED
        = 180° - < BDE / 2 - < CED / 2
        = (360° - < BDE - < CED) / 2
        = (< DBC + < ECB) / 2

(since the sum of the angles in the quadrilateral BCED is 360°), so that

  < DOE = (< ABC + < ACB) / 2 = (60° + 60°) / 2 = 60°,

qed.

[b]Problem 5[/b] is simple, too: We have $\measuredangle B_2A_1C_2 + \measuredangle C_2B_1A_2 + \measuredangle A_2C_1B_2 = 180^{\circ}$.

[i]Proof.[/i] Since $\measuredangle BB_1C = 90^{\circ}$ and $\measuredangle BC_1C = 90^{\circ}$, the points $B_1$ and $C_1$ lie on the circle with diameter BC. Hence, the segments $BB_1$ and $CC_1$ are two chords in the circle with diameter BC. Therefore, the center of this circle, i. e. the midpoint M of the segment BC, lies on the perpendicular bisectors of these chords $BB_1$ and $CC_1$. Since the midpoints of these chords are $B_2$ and $C_2$, respectively, we have $MB_2 \perp BB_1$ and $MC_2 \perp CC_1$. In other words, $\measuredangle MB_2H = 90^{\circ}$ and $\measuredangle MC_2H = 90^{\circ}$, where H is the orthocenter of triangle ABC. Also, clearly, $\measuredangle MA_1H = 90^{\circ}$. Hence, the points $B_2$, $C_2$ and $A_1$ lie on the circle with diameter MH. Thus, the points $B_2$, $C_2$, $A_1$ and M are concylic, and thus $\measuredangle B_2A_1C_2 = \measuredangle B_2MC_2$. Now, since $MB_2 \perp BB_1$ and $BB_1 \perp CA$, we have $MB_2 \parallel CA$, and similarly, $MC_2 \parallel AB$, so that the angles $\measuredangle B_2MC_2$ and $\measuredangle CAB$ are equal since their corresponding sides are parallel. In other words, we have proven $\measuredangle B_2A_1C_2 = \measuredangle B_2MC_2 = \measuredangle CAB$. The same procedure, with possible minor modifications, yields $\measuredangle C_2B_1A_2 = \measuredangle ABC$ and $\measuredangle A_2C_1B_2 = \measuredangle BCA$, and thus,

  $\measuredangle B_2A_1C_2 + \measuredangle C_2B_1A_2 + \measuredangle A_2C_1B_2 = \measuredangle CAB + \measuredangle ABC + \measuredangle BCA = 180^{\circ}$.

Proof complete.

As for [b]Problem 4[/b], the angle < GFH is 90°. See http://www.mathlinks.ro/Forum/viewtopic.php?t=1336 .

[b]Problem 3[/b] was solved on http://www.mathlinks.ro/Forum/viewtopic.php?t=19719 .

  Darij


I'm looking for the proof of associativity of multiplication.
For example: ((5*6)*7)*8 = 5*(6*(7*8)). I know from elementary school that multiplication is associative but I don't know why...
Can you show any proof except a proof by induction??

[quote="cosinator"]I'm pretty sure that associativity is a given in the field of complex (and real) numbers closed under multiplication and addition. The hard part is proving commutativity.[/quote]

Associativity and commutativity are both axioms as far as I know.


calculus
derivative
trigonometry
real analysis
real analysis unsolved

y=ax+bcos(x). What values of a guarantee the nonexistence of an inverse of y(x)?

$ f(x) \equal{} ax \plus{} b\cos x$, with $ a,b \neq 0$

$ f^\prime(x) \equal{} a \minus{} b\sin x$, which is continuous

$ f^\prime(x) \equal{} 0\Leftrightarrow \sin x \equal{} \frac {a}{b} \ \ (1)$

$ \bullet$ If $ |a| > |b|$ then $ (1)$ never holds, so $ f'$ has fixed sign, so $ f$ is either increasing or decreasing

$ \bullet$ Even if $ |a| \equal{} |b|$, then $ f'$ has a discrete set of roots, so $ f$ is still increasing or decreasing

$ \bullet$ Only if $ |a| < |b|$ we have intervals at which $ f'$ is positive and other intervals at which $ f'$ is negative

For example, $ f'\left(\frac {\pi}{2}\right)\cdot f'\left( \minus{} \frac {\pi}{2}\right) \equal{} (a \plus{} b)(a \minus{} b) \equal{} a^2 \minus{} b^2 < 0$

So, at this final case there is no inverse


I am new to this forum.  How to copy mathequaitons from math-equation editor to the thread?
Regards
Sridhar[/code]

Do you mean like this? $ \frac{\pi}{\sqrt{2}}$

If so, this would probably be the wrong forum. In the forums, you place dollar signs (\$) around your LaTeX equation. Here are a few basic commands:

\frac{}{} : the two curly braces take in two arguments for the numerator and denominator of the fraction
\sqrt{} : the braces puts your argument under a square root sign
\pi : the symbol pi

There are more in the AoPSWiki Guide.

By math-equation editor, do you mean the one on word?  If so, copying is not possible.  You must learn $ \text{\LaTeX}$ and rewrite them (see post above).

A more thorough FAQ can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=700733#700733]here[/url].


number theory
relatively prime
algebra unsolved
algebra

Determine all integer m for which all solutions of the equation

$ 3x^{3}\minus{}3x^{2}\plus{}m\equal{}0$

are rational

For root's to be rational, the discriminant must be a perfect square. In other words:

$ 9 \minus{} 12m\equal{}k^2$

where $ k \in \mathbb{N}$. And since $ m \in \mathbb{N}$ we can say that there is no solution since $ k^2$ must be positive. But, if we take $ 0$ to be an element of $ \mathbb{N}$ then the solution $ m\equal{}0$ would give the appropriate result.

But its not quadratic. Not necessarily $ m\in \mathbb{N}$ I think there's something wrong.
Converting rational to relatively prime integers $ (a,b)$:
$ 3a^3\minus{}3a^2b\plus{}mb^3\equal{}0$. Thus $ b|3$. For $ b\equal{}3$, $ a^3\minus{}3a^2\plus{}9m\equal{}0$, meaning $ 3|a$, impossible. Thus $ b\equal{}1$ and $ m\equal{}3t^2\minus{}3t^3$ for all integers $ t$.

P.S.I really have the feeling I did something wrong.

I apologize.. I misread the post. My solution is WRONG!!


calculus
derivative
analytic geometry
graphing lines
slope
limit

what the heck is it? I mean i know that its the natural base, but why can't ten be the natural base or i don't know 42 be the natural base?  :huh:  :huh:

E is the limit of $ (1 \plus{} (1/x))^x$ as x approaches infinity. This number has many special properties though. For example, think of the formula $ (1 \plus{} (1/x))^x$ as a small amount of interest (1/x) compounded a large (x) amount of times. Thus, it is useful in computing things that compound continuously, such as money or bacteria.
Also, e is important in calculus. The derivative of $ e^x$ is itself and the derivative of $ ln x$ is $ 1/x$. This means if you look at the graph of $ e^x$, the slope of the tangent line at any point will be the y coordinate! And if you look at the graph of $ ln x$, the slope of the tangent line at any point will be the recipriocal of the x coordinate.

EDIT:(cleaner)

$ e\equal{}\lim_{x\to\infty}(1\plus{}\frac{1}{x})^x$

$ e\equal{}\lim_{x\to0}(1\plus{}x)^{\frac{1}{x}}$

i suppose you'll not find $ e$ as important till you start calculus.

Once you'll start calculus,you'll tend to forget base $ 10$ or anything,then only $ e$ remains.It is equally important for calculus as $ \pi$ is for geometry(circles)

For more details check [url]http://mathworld.wolfram.com/e.html[/url]

well, i am doing derivative calculus and i just read that $ \frac{\text{d}e}{dx} \equal{} e$. That's pretty cool!

Well, in fact, $ \frac {de}{dx} \equal{} 0$, as $ e$ is just a real constant. What you probably wanted to write is $ \frac {de^x}{dx} \equal{} e^x$.


I am using the itemize environment to list some stuff but the spacing between the items of the list is too great for my liking. How can I decrease the this spacing between the items to something like the normal line spacing in a paragraph of text?

Inside the itemize environment you can change two vertical distances, \parsep and \itemsep. \parsep sets the vertical spacing between paragraphs and itemsep adds extra spacing between items. Try using \setlength{\itemsep}{0pt}to remove the extra spacing (if that's what you are thinking of)
[code]\begin{itemize}
\setlength{\itemsep}{0pt}
\item Item one
\item Item two
\end{itemize}[/code]


analytic geometry
Ramsey Theory
combinatorics open
combinatorics

Color all points in  plane with RED or GREEN .
Let $ABC$ is a triangle .
Prove:
We can choose $M,N,P$ are the same color ,such that :
triangle$MNP$ ~$ABC$.
[i](PS: It 's from a easy question:ABC is a regular-triangle)[/i]

In fact the result follows more directly from Gallai's theorem than from Ramsey's one.
Gallai's theorem states that :
For any finite subset $E$ of $\mathbb{R}^{n}$ and any finite coloring of $\mathbb{R}^{n}$ then we may find a monochromatic set $F$ which is homothetic to $E$.

Pierre.

For triangles, you can even work only on a suitably chosen lattice, and imitate [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=waerden&t=31161]this[/url] proof. More precisely, we can perform an affine transformation on the plane that turns $ABC$ into a right isosceles triangle, say, and then apply the cited problem directly.

Of course, that's not such a big deal, since it uses Van der Waerden's Theorem, and once you've proved that, Gallai's Theorem shouldn't be too hard to prove either; the idea is the same :).

But why do the affine transformation would preserve coloration?

Pierre.

I don't really know what that means :?.

The affine transformation $T$ sends each point $x$ to $Tx$. In this new setting, color the point $y$ with the same color that $T^{-1}y$ used to have in the old setting. You know that you can find a monochromatic triangle $\Delta$ homothetic to $T(ABC)$ (by that problem that I posted a link to), so in the old coloring, $T^{-1}(\Delta)$ must have been a monochromatic triangle homothetic to $ABC$.

Yes, sorry, I understood that you transport the coloration too a little late.

Pierre.

Here's a proof which doesn't need the results mentioned. We will first give the proof assuming the following (a very very simple case of Van der Waerden's theorem):
  [b](1)[/b] For any coloring of a line with two colors there exist three points of the same color on the line with one of them being a midpoint of the other two.

       Let $P, Q,R$ be such points all colored, let's say, red with $Q$ as midpoint of $PR$. Let $PMR$ be a triangle similar to $ABC$. Let $R_{1}, P_{1}$ be the midpoints of $PM, RM$ respectively. Note that triangles $PR_{1}Q$, $QP_{1}R$, and $PMR$ are all similar to triangle $ABC$. Therefore if one of $R_{1}, P_{1}, M$ is red we are done and if not we have $R_{1}, P_{1}, M$ to be of the same color and $R_{1}, P_{1}, M$ is similar to $ABC$. Thus proved.

[b]Proof of (1):[/b] As I can't draw a diagram I'm going to use a coordinate system. Pick two points $P,Q$ of the same color, say red,  and set up a coordinate system with $P$ as origin and $PQ$ as x-axis. Let $Q=(o,a)$. Let $Q_{1}=(0,-a), Q_{2}=(0,a/2), Q_{3}=(0,2a)$. Now, if $Q_{1}$ is red we are done ($Q_{1}, P, Q$ are the desired triple). Similarly if $Q_{2}, Q_{3}$ are red we are done as well. And if not $Q_{1}, Q_{2}, Q_{3}$ are all of the same color and $Q_{2}$ is midpoint of the other two.

This would been simpler to describe if I could draw pictures.


calculus
derivative
function
integration
Support
limit
real analysis

Suppose that $f: \mathbb{R}\to \mathbb{R}$ is a continuous function such that $\int_{\mathbb{R}}f(x) \phi^{''}(x) dx \geq 0$ for every smooth function $\phi$ positive with compact support. Prove that $f$ is convex.

Let $\psi$ be a smooth "bump" function: nonnegative, supported on $[-1,1],$ and $\int\psi=1.$ Define $\psi_{\epsilon}(x)= \frac1{\epsilon}\psi\left(\frac{x}{\epsilon}\right).$

Fix $a<b$ and $t\in(0,1).$ Let

$\phi''(x)=(1-t)\psi_{\epsilon}(x-a)-\psi_{\epsilon}(x-(1-t)a-tb)+t\psi_{\epsilon}(x-b).$

Make $\epsilon$ small enough that the support of $\phi''$ consists of three separated intervals.

Then $\phi(x)=\int_{-\infty}^{x}\int_{-\infty}^{u}\phi''(v)\,dv\,du$

We need to argue two things about $\phi.$ First, that is equal to zero for all sufficiently large $x$ and hence compactly supported, and second, that it is nonnegative. I'll leave both of those claims as exercises for the reader.

We can then show that $\lim_{\epsilon\to0}\int_{\mathbb{R}}f(x)\phi''(x)\,dx = (1-t)f(a)-f((1-t)a+tb)+tf(b).$

By hypothesis, this is nonnegative. Since we can do this for all $a,b,$ and $t,$ we conclude that $f$ is convex.

Suppose $f$ is a distribution whose (weak) second derivative is nonnegative. That is, $f$ satisfies precisely the condition stated in alekk's post. Can $f$ be identified with a continuous function, which by the argument above would be convex?

I think that a positive distribution $T$ is actually a postive measure. To see this, take some positive smooth functions $f_{n}$ with support included in $[n-1; n+1]$ and such that $\sum f_{n}= 1$. Then, if $\phi$ is a test function (smooth with compact support), then
$(T, \phi) = \sum (T; f_{n}\phi) = \sum (f_{n}T; \phi)$

Because each distribution $f_{n}T$ has compact support, it suffices to show that every positive distribution $D$ with compact support $K$ is a measure. To see that, one can take another compact $K \subset G$ and some positive test functions $F_{K}$ and $F_{G}$ such that $F_{K}=1$ on $K$ and $F_{G}=1$ on $G$. Then if $\phi$ is a test function:
$(D; \| \phi \|_{\infty}F_{G}-F_{K}\phi) \geq 0$
so
$(D; \phi)=(D; F_{K}\phi) \leq \| \phi \|_{\infty}(D; F_{G}) = C \| \phi \|_{\infty}$
Similarly one could show that
$-(D; \phi)=(D;-F_{K}\phi) \leq \| \phi \|_{\infty}(D; F_{G}) = C \| \phi \|_{\infty}$
Hence $D$ is a continuous functional on $C^{\infty}_{0}(\mathbb{R})$ : this is a measure. Positivity shows that this is a positive measure.

Now, suppose that $T''$ is a positive distribution, hence a positive measure $\mu$. That is $(T'', \phi) = \int_{\mathbb{R}}\phi d\mu$. Then $S(x)=\int_{(0;x)}d \mu$ is a left continuous increasing function vanishing at $x=0$ and $f(x)=\int_{(0;x)}S(x) dx$ is a continuous function whose weak second derivative is $T''=[\mu]$. Because a distribution with null derivative is constant, one easily conclude that $T$ is given by the function $x \to f(x)+(ax+b)$. Is it OK ?


I think I've found a way to get any piece of $\LaTeX$ to run on the AoPSWiki!  Just throw in the \displaystyle command right after <math> tag.  I was able to use that to fix three things that just refused to render.

The only downside is that they all rendered forum-style.  Is this discouraged?

They are meant to be rendered forumstyle... eum... wait:
- do they work with 2 different caches? (what a waste?)
- are they letting the cache being polluted again by non-displaystyled posts? :fool:


vector
group theory
abstract algebra
induction
superior algebra
superior algebra solved

This a famous problem, i just want to see different elementary solutions for it :
$G$ is a groupe such that $\forall x \in G{\text{ }}: {\text{ }}x^2 = e$
show that $\left| G \right|$ is a power of 2.

because it can be seen as a vector space of finite dimension over $\frac{Z}{2Z}$.
But I bet it is your solution  :)

that's why i talked about "elementary solution" , more precisely no linear algebra and without cauchy theorem :)

First, since $x \mapsto x^2$ is a homomorphism, we see that it is abelian.
Now take any nontrivial element, and let $H$ be the subgroup generated by it, so $|H|=2$.
When we have already proved the theorem for smaller groups, we get that $|G/H|=2^s$ for some $s$, so $|G| = |G/H| |H| = 2^{s+1}$.
Thus, it's simple induction


Prove the following Theorem:

Let W be the set of all positive integers whose decimal expansion only contains the digit 9 (i.e. W = {9, 99, 999, 9999, ...}).  For and odd prime p>5, p divides infinitely many distinct elements of the set W.

$ W \equal{} \{10^n\minus{}1\: : n\in N\}$

If $ p>5$, then $ p \nmid 10$ and by Fermat's little theorem

$ 10^{p\minus{}1}\minus{}1 \equiv 0 \mod p$
$ 10^{2(p\minus{}1)}\minus{}1 \equiv 0 \mod p$
$ 10^{3(p\minus{}1)}\minus{}1 \equiv 0 \mod p$
...


geometry
geometric transformation
rotation

How many noncongruent triangles have vertices at the vertices of a regular 101-gon?

Note that there is a 1-1 Correspondence to this and the number of solutions to the equation $ a + b + c = 101$ such that $ a\geq b\geq c\geq1$ (do you see why?).

Since $ a,b,c\geq1$, there are $ \dbinom{(101 - 3) + 3 - 1}{3 - 1} = \dbinom{100}2$ choices with only the restriction $ a,b,c\geq1$.

For every $ (a,b,c)$ such that $ a,b,c$ are all different, there are $ 6$ permutations of that, and only one which satisfies $ a > b > c\geq1$.

It is not possible to have an equilateral triangle, since $ 3\not|101$. However, it is possible to have isosceles triangles, which there are only $ 3$ permutations of (and only one that satisfies $ a\geq b > c\geq1$).

Thus, we must break up $ \dbinom{100}2$ into two cases: isosceles and scalene triangles.

[hide="Isosceles Triangles"]
Let the sides be $ a,a,c$. Then we want $ 2a + c = 101$, which has a unique solution for each of $ 1\leq a\leq50$, or $ 50$ solutions (not counting order).

With order, there are $ 3\cdot50 = 150$ ways to arrange this, but $ 50$ isosceles triangles.
[/hide]

Because there are $ \dbinom{100}2 = 4950$ triangles total (counting order of $ a,b,c$), $ 4950 - 150 = 4800$ scalene triangles (counting the rotations of $ a,b,c$).

Finally, we see that there are $ \frac {4800}6 + 50 = \boxed{850}$ triangles that satisfy the given condition.

Isn't it just (101 C 3)/101/2 for rotations and flips of the same triangle.  That would give 825, not 850, right?

[quote="math154"]Note that there is a 1-1 Correspondence to this and the number of solutions to the equation $ a \plus{} b \plus{} c \equal{} 101$ such that $ a\geq b\geq c\geq1$ (do you see why?).
[/quote]
I don't see why this is so. Could someone explain?

The 101-gon has 101 sides and 101 vertices, of which you pick three for the triangle. Let's call these vertices $x$, $y$, and $z$. A certain number $a$ of the 101-gon's sides lie between vertices $x$ and $y$, $b$ between $y$ and $z$, and $c$ between $x$ and $z$.

The values of $a$, $b$, and $c$ determine the side-lengths of the triangles, and whenever one of them changes you get a different triangle. Therefore to solve the problem you count the different values for $a$, $b$, and $c$. Because the 101-gon has 101 sides, $a+b+c=101$.


Another question:

I also need to do get these mini-recommendations from various teachers so that my counselor has something to write about me when he does his recommendation for me. He suggests that I choose five teachers who I will NOT be asking for my teacher recommendation for college. I have narrowed it down to exactly five teachers who I might ask for my college recommendation, and I am not sure who to pick.

Teacher A

subject: English

- taught me for several years (9th, 10th, 11th) for Honors English
- advisor of a club for which I hold a leadership position
- might exaggerate how smart I am more so than other teachers
- on friendly terms with me

Teacher B

subject: Social Studies

- taught me for AP US History and AP US Gov & Politics
- went to Harvard (which is where I plan to apply)
- on friendly terms with me

Teacher C

subject: Science

- taught me for Honors Biology
- might exaggerate how smart I am more so than other teachers
- on friendly terms with me

Teacher D

subject: Math

- taught me for AP Calculus AB
- might teach me again for AP Statistics next year
- does not know me as well, but I might get to know him better near the end of the year (when all the seniors leave, and I am the only junior in the class)
- i'm doing better than all the seniors in my class (A+)

Teacher E

subject: Science

- taught me for Accelerated Physics, AP Physics
- thinks that I am: hardworking, intelligent
- if he says something negative it would be that I am a bit too grade conscious
- on friendly terms with me

I know I should have one math/science teacher recommendation and one humanities recommendation, but I cannot decide which two to choose for my teacher recommendation and which three to leave for the "mini" recommendation.

Can anyone give me suggestions?

Also, would it be better to choose an AP teacher rather than an honors teacher?

[quote="JBL"]My vote is to ask the English teacher for a college recommendation and the social studies teacher for a mini-rec.[/quote]

I agree; the English teacher seems to know you better and probably has a pretty high opinion of you.

Good luck =)

What about my math/science recommendation? Who should I go with?

[quote="Mr. T"]I would suggest using Teacher B for your college app.  I'm pretty sure that Harvard weighs advice from its previous students pretty well.[/quote]

They won't know it's their own student.  This is not a good reason to use him for applying to Harvard.

I'd say E for your math/science, just b/c the other two options don't look so hot.

I return! Anyways, it might be too late but I would recommend the Harvard teacher because a good rec letter for Harvard coming from a teacher who went to Harvard would definitely state that he is a Harvard alumnus.  This would look VERY good for you...