calculus
integration
vector
function
algebra
domain
analytic geometry

Let $ C_r$ denote the circle or radius $ r$ about hte origin in the $ xz$ plane. oriented counterclockwise as viewed from the positive y-axis.  Suppose $ \textbf{F}$ is a $ C^1$ vector field on the complement of the y-axis in $ \mathbb{R}^3$ such that $ \int_{C_1}\text{F}\cdot d\textbf{x}\equal{}5$ and $ \text{curl }\textbf{F}(x,y,z)\equal{}3\textbf{j}\plus{}\frac{z\textbf{i}\minus{}x\textbf{k}}{(x^2\plus{}z^2)^2}.$
compute $ \int_{C_r}\textbf{F}\cdot d\textbf{x}$ for any $ r$.

[quote="e^pi"]Let $ C_r$ denote the circle or radius $ r$ about hte origin in the $ xz$ plane. oriented counterclockwise as viewed from the positive y-axis.  Suppose $ \textbf{F}$ is a $ C^1$ vector field on the complement of the y-axis in $ \mathbb{R}^3$ such that $ \int_{C_1}\text{F}\cdot d\textbf{x} \equal{} 5$ and $ \text{curl }\textbf{F}(x,y,z) \equal{} 3\textbf{j} \plus{} \frac {z\textbf{i} \minus{} x\textbf{k}}{(x^2 \plus{} z^2)^2}.$
compute $ \int_{C_r}\textbf{F}\cdot d\textbf{x}$ for any $ r$.[/quote]

Stoke's theorem states that:

$ \int_{C_r}\textbf{F}\cdot d\textbf{x} \equal{} \iint\nabla\times\textbf{F}\cdot \hat{\bold{n}}\, da$.

In this case, $ \hat{\bold{n}} \equal{}\textbf{j}$ and $ da \equal{} r dr d\theta$.

Therefore, $ \iint\nabla\times\textbf{F}\cdot \hat{\bold{n}}\, da \equal{} \iint 3 r dr d\theta \equal{} 3\pi r^2$.

Hence $ \int_{C_r}\textbf{F}\cdot d\textbf{x}\equal{} 3\pi r^2$.

There's a problem with that. Stokes' theorem does require that the function and its curl satisfy some reasonable regularity or continuity conditions on the surface in question. But in this case $ \nabla\times\mathbf{F}$ has a glaring singularity all along the $ y$ axis, and we must reasonably assume that $ \mathbf{F}$ is similarly singular. Hence we cannot apply Stokes' theorem directly.

Let's split $ \mathbf{F}$ into $ \mathbf{G} \plus{} \mathbf{H}$ where $ \nabla\times G \equal{} 3\mathbf{j}$ and $ \nabla\times\mathbf{H} \equal{} \frac {z\mathbf{i} \minus{} x\mathbf{k}}{(x^2 \plus{} z^2)^2}$

We have (as in Dr. No's computation) that $ \int_{C_r}\mathbf{G}\cdot d\mathbf{r} \equal{} 3\pi r^2.$ But note the hypothesis that $ \int_{C_1}\mathbf{F}\cdot d\mathbf{r} \equal{} 5.$ It then follows that

$ \int_{C_1}\mathbf{H}\cdot d\mathbf{r} \equal{} 5 \minus{} 3\pi.$

We cannot apply Stokes' theorem directly to the integral of $ \mathbf{H}$ on a circle around the $ y$ axis, but we can apply it to a curve made up of two concentric such circles parameterized in opposite directions, surrounding an annular surface. That gives us an integral of zero, meaning that the two circles gave the same result.

Hence, $ \int_{C_r}\mathbf{H}\cdot d\mathbf{r} \equal{} 5 \minus{} 3\pi$ for any positive $ r.$

Our final answer is that

$ \int_{C_r}\mathbf{F}\cdot d\mathbf{r} \equal{} 5 \minus{} 3\pi \plus{} 3\pi r^2.$

As a side question: does such a vector field $ \mathbf{F}$ exist? That doesn't seem obvious to me.

A choice of $ F$: $ \frac{\minus{}1}{x^2\plus{}z^2}\mathbf{j}\plus{}3z\mathbf{k}$.

I'm not sure I've got the right signs, but the essence is there.

[quote="Kent Merryfield"]

$ \int_{C_r}\mathbf{G}\cdot d\mathbf{r} \equal{} 5 \minus{} 3\pi \plus{} 3\pi r^2.$
[/quote]

this supposed to be F correct?

[quote="Kent Merryfield"]...Hence, $ \int_{C_r}\mathbf{H}\cdot d\mathbf{r} \equal{} 5 \minus{} 3\pi$ for any positive $ r.$

Our final answer is that

$ \int_{C_r}\mathbf{G}\cdot d\mathbf{r} \equal{} 5 \minus{} 3\pi \plus{} 3\pi r^2.$

As a side question: does such a vector field $ \mathbf{F}$ exist? That doesn't seem obvious to me.[/quote]

I've been pondering this since I posted my reply earlier this evening. I didn't notice the inconsistency until after I left but then it was too late to address it. My own opinion is that there is no such vector field. Stokes clearly shows that $ \int_{C_1}\mathbf{F}\cdot d\mathbf{x} \equal{} 3\pi$; yet, the problem statement says $ \int_{C_1}\mathbf{F}\cdot d\mathbf{x} \equal{} 5$.

If both are true then $ 3\pi \equal{} 5$.  Conclude: $ \neg\exists\mathbf{F}$

[quote="Dr. No"]Stokes clearly shows that $ \int_{C_1}\mathbf{F}\cdot d\mathbf{x} \equal{} 3\pi$[/quote]
No, it does not do that. Stokes' theorem cannot be directly applied to that curve, since $ \nabla\times\mathbf{F}$ is not defined over all of any set in the $ xz$ plane that contains $ x \equal{} 0,z \equal{} 0,$ and furthermore, $ \nabla\times\mathbf{F}$ is not integrable over any such set, as it goes to infinity like $ r^{ \minus{} 3}$ as you approach the singularity.

e^pi: yes, that's a typo. I'll go back and fix it.

jmerry: try doing the line integrals for that $ \mathbf{F}.$ It doesn't look right to me.

Here's an illustration of the problem, starting over with a new vector field.

Let $ \mathbf{W}\equal{}\frac{z\mathbf{i}\minus{}x\mathbf{k}}{x^2\plus{}z^2}$ everywhere except the $ y$ axis, where it is undefined.

(Physically, this gives the magnetic field caused by a steady current in an infinite straight wire along the $ y$ axis.)

We can calculate the curl: $ \nabla\times W\equal{}\mathbf{0}$ everywhere that it is defined - which is not everywhere.

But we cannot use Stokes' theorem to compute $ \int_{C_r}\mathbf{W}\cdot d\mathbf{r}.$

If we could use Stokes, since the curl is zero, we would get an integral of zero. But that is not correct. By direct computation,

$ \int_{C_r}\mathbf{W}\cdot d\mathbf{r}\equal{}2\pi.$

Now, we can use Stokes to show that the integral on any curve that winds once around the $ y$ axis is the exact same $ 2\pi.$ This is similar to my talk of an annular region in one of my previous posts.

jmerry: should you be adding a constant times this $ \mathbf{W}$ to your proposed construction for $ \mathbf{F}?$

Yes, that's it. I was focused on the curl piece and forgot the other condition. I was thinking of the magnetic field for a wire with current anyway, and didn't realize we needed that to make the integrals around circles right.

[quote="Kent Merryfield"][quote="Dr. No"]Stokes clearly shows that $ \int_{C_1}\mathbf{F}\cdot d\mathbf{x} \equal{} 3\pi$[/quote]
No, it does not do that. Stokes' theorem cannot be directly applied to that curve, since $ \nabla\times\mathbf{F}$ is not defined over all of any set in the $ xz$ plane that contains $ x \equal{} 0,z \equal{} 0,$ and furthermore, $ \nabla\times\mathbf{F}$ is not integrable over any such set, as it goes to infinity like $ r^{ \minus{} 3}$ as you approach the singularity.[/quote]

Kent, I respect your opinion and I have no desire to get into a food fight with you over this; nevertheless, what you've said here is gibberish.

First, the integrand of the Stokes' integral is $ \nabla\times\mathbf{F}\cdot\hat{\mathbf{n}}$, [b]not[/b] $ \nabla\times\mathbf{F}$. Second, the line integral $ \int_{C_1}\mathbf{F}\cdot d\mathbf{x}$ is calculated along a path of constant radius, $ r \equal{} 1$.  It doesn't even approach the singularity at $ x \equal{} z \equal{} 0$. In other words, the integrand is well defined over the domain of integration, and the result is easily calcuable: $ \int_{C_1}\mathbf{F}\cdot d\mathbf{x} \equal{} 3\pi$

Finally, and most importantly, Stokes' theorem is a theorem; as such it is a logically self-consistent statement whose validity does not depend upon any assumed premises. Positing a counterexample to it is futile and foolish.

[quote="Kent Merryfield"]Please. Try to apply the exact same reasoning you just gave to the field $ \mathbf{W}$ in my most recent post. Use a circle for simplicity of computation.[/quote]

I'd be happy to do so.

First, $ \mathbf{W} = \frac {z\mathbf{i} - x\mathbf{k}}{x^2 + z^2}$.

Doing a little work to recast this in polar coordinates, much easier to use if we are considering a circular region:

$ \frac {z}{x^2 + z^2} = \frac {\sin\theta}{r}$.

$ \frac { - x}{x^2 + z^2} = \frac { - \cos\theta}{r}$.

$ \mathbf{i} = \hat{e_r}\cos\theta - \hat{e_{\theta}}\sin\theta$.

$ \mathbf{k} = \hat{e_r}\sin\theta + \hat{e_{\theta}}\cos\theta$.

Putting all together ... $ \mathbf{W} = - \frac {\hat{e_{\theta}}}{r}$.

$ \int_{C_r} \mathbf{W}\cdot d\mathbf{x} = \int_{0}^{2\pi} - \frac {\hat{e_{\theta}}}{r}\cdot \hat{e_{\theta}} r d\theta$

[this line was edited by blahblahblah to fix a $ \text{\LaTeX}$ typo]

(Thanks, blahblahblah. I caught it but not in time :))

$ = \int_0^{2\pi} - d\theta = - 2\pi$.

Even in this case, the integrand is well defined over the domain of integration.

I believe the thing that's bothering you is the stipulation that:

$ \int_{C_r} \mathbf{W}\cdot d\mathbf{x} = \iint \nabla\times\mathbf{W}\cdot\hat{\mathbf{n}} da.$

Substituting $ \int - d\theta$ into the RHS,

$ \iint \nabla\times\mathbf{W}\cdot\hat{\mathbf{n}} r dr d\theta = \int - d\theta$.

Differentiating once we obtain:

$ \int \nabla\times\mathbf{W}\cdot\hat{\mathbf{n}} r dr = - 1$.

Differentiating again, $ \nabla\times\mathbf{W}\cdot\hat{\mathbf{n}} r = 0$.

From the above it is observed that either $ r = 0$, $ \nabla\times\mathbf{W}$ and $ \hat{\mathbf{n}}$ are mutually orthogonal, or $ \nabla\times\mathbf{W} = 0$.

We know that it is the third case that is true. However, the more important point is that one can integrate over a zero function. It's kind of silly to bother, but one can do it nonetheless; the result is always a constant.

I realize that the argument of my last post was psychologically unsatisfying.

Let me therefore consider the general case of circular symmetry.

Suppose we have a field $ \mathbf{G} = g(r,\theta) \hat{\mathbf{e}_{\theta}}$.

The function $ g(r,\theta)$ may be rewritten by separation of variables as: $ g(r,\theta) = r^n f(\theta)$.

It is observed that $ \nabla \times \mathbf{G} = \hat{\mathbf{e}_z}\frac{1}{r}\frac{\partial}{\partial r}(r g(r,\theta)) = \hat{\mathbf{e}_z}\frac{1}{r}\frac{\partial}{\partial r}(r^{n+1} f(\theta))$.

The first thing to notice is that when $ n = -1$ then $ \nabla \times \mathbf{G} = 0$, everywhere; the case where $ r = 0$ notwithstanding.

The line integral $ \int_{C_r} \mathbf{G}\cdot\, d\mathbf{x} = \int_{\theta_0}^{\theta_0 + 2\pi} r^{n + 1} f(\theta)\, d\theta = r^{n+1}[ F(\theta_0 + 2\pi) - F(\theta_0)]$.

Whereas, the Stokes' integral is $ \int_{\theta_0}^{\theta_0 + 2\pi}\int_{\epsilon}^{\epsilon + R} \frac{1}{r}\frac{\partial}{\partial r}(r^{n+1} f(\theta)) r \,dr \, d\theta$
$ = [ F(\theta_0 + 2\pi) - F(\theta_0)] \int_{\epsilon}^{\epsilon + R} \frac{\partial}{\partial r}(r^{n+1} ) \,dr$
$ = [ F(\theta_0 + 2\pi) - F(\theta_0)][(\epsilon + R)^{n + 1} - \epsilon^{n + 1}]$.

If we now replace $ r$ in the line integral result with $ \epsilon + R$, the difference between the line integral and the Stokes' integral is:

$ \int_{C_r} \mathbf{G}\cdot\, d\mathbf{x} -\iint_S \nabla \times \mathbf{G}\cdot \hat{\mathbf{n}}\, da = [ F(\theta_0 + 2\pi) - F(\theta_0)]\epsilon^{n + 1}$.

In order to complete the analysis I'm going to let above difference equal $ J$.

$ J = [ F(\theta_0 + 2\pi) - F(\theta_0)]\epsilon^{n + 1}$.

$ \lim_{\epsilon\to 0} J = \{ \begin {array}{rcl} 0 & \mbox{for} & n > -1 \\
 [ F(\theta_0 + 2\pi) - F(\theta_0)] & \mbox{for} & n = -1 \\ \text{indeterminant} & \mbox{for} & n < -1 \end {array}$.

$ F(\theta)$ is typically periodic with period $ 2\pi$, in which case $ \lim_{\epsilon\to 0} J = 0$ identically when $ n = -1$, and has the form $ 0\cdot\infty$ when $ n < -1$ (as such it is subject to being defined as identically equal to zero).

If $ F(\theta)$ is aperiodic on $ 2\pi$ then there is a non-trivial difference between the line integral and the Stokes' integral when $ n\leq -1$.

Thus Kent's original point is preserved: Stokes' theorem doesn't hold in all cases.

From a pragmatic, i.e. applied, standpoint Stokes' theorem is a tool useful when calculating a surface integral is simpler than calculating an "equivalent" line integral. Thus when powers of $ r$ less than or equal to minus 1 are present in the original field then one simply avoids calculating the Stokes' integral.

Considering the curl operation in isolation (apart from the Stokes' integral), that the curl of a particular field may be undefined at $ r = 0$ is physically superfluous because curl measures the circulation of a field about an axis. How much the field is circulating [b] on the axis [/b] itself is a vapid idea. How dizzy would you get if you were reduced to a line and then spun about an axis coincident with yourself?


algebra
polynomial
number theory proposed
number theory

Let $ p(\cdot)$ a non-costant polynomial with integer coefficients such that $ p(0) \neq 0$. Let $ a_1,a_2,\ldots,a_k$ be some integers fixed such that $ \min\{|a_1|,|a_2|,\ldots,|a_k|\} \ge 2$.

Show that exist infinitely many positive integer $ n$ such that $ \min\{\omega(p(a_1^n)),\omega(p(a_2^n)),\ldots,\omega(p(a_k^n))\} \ge 2$.

[i](Salvatore Tringali)[/i]

$ \omega(x)$ is the number of prime divisor of $ x$ (see [url=http://mathworld.wolfram.com/DistinctPrimeFactors.html]here[/url])


trigonometry
calculus
calculus computations

Yes, it's bounded. It's essentially a geometric sum.

$\sum_{k=1}^{n}\sin kx=\text{Im}\left(\sum_{k=1}^{\infty}e^{ikx}\right)$

$=\text{Im}\left(\frac{e^{ix}-e^{i(n+1)x}}{1-e^{ix}}\right)$

$=\text{Im}\left(\frac{e^{ix/2}-e^{i(n+1/2)x}}{e^{-ix/2}-e^{ix/2}}\right)$

Which leads to the forumula RIP just posted.

In general, this is bounded for all $n$ by $\frac1{2\sin\left(\frac{x}{2}\right)}.$ In your particular case, let $x=1$ in that formula and that bound.


Circles $ S$ and $ T$ intersect at $ P$ and $ Q$, with $ S$ passing through the centre of $ T$. Distinct points $ A$ and $ B$ lie on $ S$, inside $ T$, and are equidistant from the centre of $ T$. The line $ PA$ meets $ T$ again at $ D$. Prove that $ |AD| \equal{} |PB|$.

[hide="Solution"]Let $ O$ be the center of $ T$. Clearly $ PABQ$ is an isosceles trapezoid and so $ PB \equal{} AQ$.
$ \angle ADO \equal{} \angle APO \equal{} \angle AQO$.
$ \angle DAO \equal{} 180^{\circ} \minus{} \angle PAO \equal{} \angle PQO \equal{} \angle QPO \equal{} \angle QAO$.
Then $ \angle AOD \equal{} \angle AOQ$ and by SAS triangles $ AOD$ and $ AOQ$ are congruent.[/hide]

As seen above, PB = AQ, so we shall prove that AD and AB are symmetrical w.r.t. AO (O being the center of circle T), which is obvious, <OAD being half of the arc OP, while <OAQ is half of the arc OQ, equal arcs of the circle S.

Best regards,
sunken rock


function
real analysis
real analysis unsolved

$f_n$ a sequence of convex function defined on $]a,b[$, that simply converge towards $f$.
Is the convergence uniform on any compact set of $]a;b[$.
Rq: if you know the official sln, please do not post it.

I knew this problem beforehand, but I swear I've never tried to solve it :).

$f$, the pointwise limit of $(f_n)_n$, is convex on an open interval, and thus continuous. Assume we have restricted our attention to the compact subinterval $[c,d]$ of $(a,b)$. Take a $c'\in(a,c)$, consider the functions $\tilde f_n(x)=\frac{f_n(x)-f_n(c')}{x-c'}$, which are increasing and converge pointwisely to $\tilde f(x)=\frac{f(x)-f(c')}{x-c'}$, and then use Polya's Theorem stating that if a sequence of monotonic functions converges pointwisely to a continuous function defined on a compact interval, then the convergence is uniform (it's easy to see that the convergence $f_n\to f$ is uniform iff the convergence $\tilde f_n\to\tilde f$ is).


Hey,guys.I have 2 nice inequalities.May be it's old ...
For $a,b,c$ be positive numbers .Prove that:
$1/\sum\frac{b+c}{a^2+bc}\leq \sum\frac1a$
$2/\sum\frac{ab+c^2}{a+b}\geq\sum a$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=43872 for the first one; in this thread, the second inequality is also mentioned, but without a proof (and I could not find out where it was posted before), so let's prove it here:

We have to show that if a, b, c are any three positive numbers, then the inequality

$\frac{bc+a^2}{b+c}+\frac{ca+b^2}{c+a}+\frac{ab+c^2}{a+b}\geq a+b+c$

holds.

Well, by subtracting a + b + c on both sides, this inequality becomes

$\left(\frac{bc+a^2}{b+c}+\frac{ca+b^2}{c+a}+\frac{ab+c^2}{a+b}\right)-\left(a+b+c\right)\geq 0$
$\Longleftrightarrow\ \ \ \ \ \left(\frac{bc+a^2}{b+c}-a\right)+\left(\frac{ca+b^2}{c+a}-b\right)+\left(\frac{ab+c^2}{a+b}-c\right)\geq 0$
$\Longleftrightarrow\ \ \ \ \ \frac{1}{b+c}\left(a-b\right)\left(a-c\right)+\frac{1}{c+a}\left(b-c\right)\left(b-a\right)+\frac{1}{a+b}\left(c-a\right)\left(c-b\right)\geq 0$.

But this follows from the Vornicu-Schur inequality ( http://www.mathlinks.ro/Forum/viewtopic.php?t=24555 post #9 Lemma 1) for $x=\frac{1}{b+c}$, $y=\frac{1}{c+a}$ and $z=\frac{1}{a+b}$ (in fact, if we assume that $a\geq b\geq c$, then $\left(a+b+c\right)-a\leq\left(a+b+c\right)-b\leq\left(a+b+c\right)-c$, i. e. $b+c\leq c+a\leq a+b$, so that $\frac{1}{b+c}\geq\frac{1}{c+a}\geq\frac{1}{a+b}$, what is equivalent to $x\geq y\geq z$, and thus the Vornicu-Schur inequality can be applied). This completes the proof.

  Darij

Just want to make a remark.
we can obtain a similar solution $(\sum (a^2-b^2)(\frac{1}{b+c} - \frac{1}{a+c}) \ge 0$ if we do the subtraction $\sum \frac{bc+a^2}{b+c} - \frac{b+c}{2} \ge 0$

[i]Another simple and nice aproach.[/i]

  If we add $a$  to the first fraction , [tex ] b [/tex]  to second and $c$  to the third , the first nominator becomes
                                        $(a +b)(a+c)$.
But now it is clear that we must put
                      $a+b = x , a+c = y$  and  $b+c= z$ .
             We solve for $a , b , c$ ( too easy) and the inequality becomes
                          $\sum {\frac{xy}{z}} \geq \sum {x} = x + y +z$.

  But
                              $\sum {\frac{xy}{z}} = ... \geq xyz \frac{ {x+y+z}}{xyz}$.

according to the basic inequality

                             $\sum{(xy)^2} \geq xyz \sum{x}$

 We have equality only if a = b = c.

 [i]Babis[/i]


geometry
3D geometry
advanced fields
advanced fields unsolved

Is there any infinite  subset of natural numbers  like $ A$ such that the sum of elements of each finite subset of $ A$ is not pefect square?

It's not too hard to find such a set $ A$.  We will construct its elements $ a_1$, $ a_2$, and so on, one at a time.  Each $ a_i$ will be of the form $ b_i^2 \plus{} 1$, one more than a square.  We will insist that for each $ i$,
\[ b_i > a_1 \plus{} a_2 \plus{} \dots \plus{} a_{i \minus{} 1}.
\]
We claim that each finite sum is not a perfect square.  Suppose we have a finite sum whose biggest term is $ a_k$.  On the one hand, the sum must be bigger than $ b_k^2$.  On the other hand, the sum must be less than $ (b_k \plus{} 1)^2$.  The sum is between two consecutive squares, and so is not a square itself.

Or you can take the set of numbers of the form $ 2\cdot 6^n$, then in this case the sum is never a perfect square, cube, etc.

General problem : Prove that there exist an infinite sequence of natural number such that $ \sum_{x \in X} x$ is not a perfect power where X is a subset of this suquence . :wink:

Well, look closer at my example and you'll see the answer to your general problem.  :wink:

Or you can construct such a set inductively :Choose the successive element such that the sum of it and each finite subset of the preceding elements  is divisible by p but not p^2 for some p.


vector
calculus
integration
calculus computations

do i have to draw the vector field first, then i can compute the volume?

how do i do it?

[img]http://farm4.static.flickr.com/3230/3117153037_19e1177ca5_o.png[/img]

For the second one, break apart your integral into its three components and use the multivariable version of the fundamental theorem of calculus?  But first check whether or not the vector field is conservative.  If not, you can't use my suggestion.

Problem 4, (ii): you can calculate the curl. I got $ \nabla\times \mathbf{F} \equal{} (0,4z,3x^2)$ which in general does not equal zero, so $ \mathbf{F}$ is not conservative: you will not be able to find $ f$ such that $ \mathbf{F} \equal{} \nabla f$

Problem 1:
$ \frac{d}{dt}(\mathbf{\dot{c}}(t)\cdot \mathbf{\dot{c}}(t))\equal{}2(\mathbf{\dot{c}}(t)\cdot \mathbf{\ddot{c}}(t))\equal{}2\equal{}\frac{d}{dt}|\mathbf{\dot{c}}(t)|^2$

Hence $ |\mathbf{\dot{c}}(t)|^2\equal{}2t\plus{}K$. Begin condition: $ |\mathbf{\dot{c}}(0)|^2\equal{}K\equal{}3^2\equal{}9$

So the answer is $ |\mathbf{\dot{c}}(t)|\equal{}\sqrt{2t\plus{}9}$

Please, don't ever change the opening post. Now nobody knows what the question was. It's very annoying.

Was my above answer helpful?


geometry
cyclic quadrilateral
perpendicular bisector
geometry unsolved

Let $X$and$Y$ be the mid-points of $SA$and$SB$.

Because$PX=SA/2=QY$,$RY=SB/2=QX$and$<QXP$=$<QXS+<PXS$=$<QXS+2<CAD$=$<QYS+2<CBD$=$<QYS+<RYS$=$<RYQ$,we have △$QXP$≌△$RYQ$,so $PQ=RQ$.

[quote="cancer"]Let $Q$ be the midpoint of the side $AB$ of a cyclic quadrilateral $ABCD$ and let $S$ be the intersection of its diagonals. Denote by $P,R$ the orthogonal projections of $S$ on $AD,BC$ respectively. Prove that $PQ=QR$.[/quote]

This is equivalent to claiming that the midpoint $Q$ of the side $AB$ lies on the perpendicular bisector of the segment $PR$. This was proven at http://www.mathlinks.ro/Forum/viewtopic.php?t=39107 .

  Darij


If two triangles are similar, are all of their respective interior segments (ie altitudes, angle bisectors, medians) also similar in the same ratio as their sides are similar? If so, is there a way to prove this result?

Thanks.

if you draw them out, it shouldnt be too hard to prove which ones are....um i think all 4 special ones are.

Oh, ok, yeah, its not too hard to prove for say altitudes and angle bisectors. But its not true for all cevians?

yeah, if in the two similar triangles the cevians cut the opposite sides in the same ratio, because of sas similarity.

You can be really sneaky on contests when you want to say something like that. You can say that they're similar by homothety.


trigonometry function algebra polynomial induction

Is there any formula for $\tan(x_{1}+x_{2}+...+x_{n})$?

$\tan(x+y) = \frac{\tan x+\tan y}{1-\tan x \tan y}$ (*)

Let $T(x)$ be the tangent function, and $a' = T(a)$ for all a.

We have the familiar $T(a+b+c) = \frac{a'+b'+c' -a'b'c'}{1-a'b'-b'c'-c'a'}$.

Using (*), we have $T(a+b+c+d) = \frac{ \frac{a'+b'+c' -a'b'c'}{1-a'b'-b'c'-c'a'} + d' }{1 -  \frac{a'+b'+c' -a'b'c'}{1-a'b'-b'c'-c'a'} d'}$

$= \frac{a'+b'+c' -a'b'c' + d' - ab'd' - b'c'd' - a'c'd' }{1-a'b'-b'c'-c'a' -a'd' -b'd' -c'd' + a'b'c'd'}$

This _strongly_ suggests the following.  Suppose $a_1, a_2, \dots a_n$ have symmetric polynomials $s_1, s_2, \dots s_n$  (in example, $s_1 = \sum a_i$, $s_2 = \sum_{i<j} a_ia_j$, ..., $s_n = \prod a_i$).  Define $s_0 = 1$.

We claim $T(a_1, a_2, \dots, a_n) = \frac{s_1 - s_3 + s_5 - \dots}{s_0 - s_2 + s_4 -\dots}$.   $(\square)$ We prove this by induction.

We use (*) to complete the induction step.  Let $s'_k$ be the kth symmetric polynomial of the terms $a_1 \dots a_{n+1}$.

----

Then $T(a_1, \dots, a_n, a_{n+1}) = \frac{(s_1 - s_3 + s_5 - \dots) + a_{n+1}(s_0 - s_2 + s_4 -\dots)}{(s_0 - s_2 + s_4 -\dots) - (s_1 - s_3 + s_5 - \dots)a_{n+1}}$

where it remains to verify $(s'_1 - s'_3 + s'_5 - \dots) = (s_1 - s_3 + s_5 - \dots) + a_{n+1}(s_0 - s_2 + s_4 -\dots)$ (%)
and $(s'_0 - s'_2 + s'_4 -\dots) = (s_0 - s_2 + s_4 -\dots) + (s_1 - s_3 + s_5 - \dots)a_{n+1}$ (%%)

We verify % first.  Notice that $s_i$ contains all terms in $s'_i$ except the $a_{n+1}$ terms.  Notice $a_{n+1}s_{i-1}$ are these terms.

We verify %% in the same way.

Thus, $(\square)$ is proved.


hello, your equation can be simplified to
$ \minus{}\frac{(x\minus{}5)x(12x^4\minus{}110x^3\plus{}345x^2\minus{}423x\plus{}158)}{12(x\minus{}4)(x\minus{}3)(x\minus{}2)(x\minus{}1)}\equal{}0$
Sonnhard.

Please, hoangvn.fix, could you give us your own clever solution to find the four non trivial real roots of this equation, instead of my ugly numeric calculus one ?

Thanks

beautiful graph, wonderfully contrived . . . .  :roll:

dear pco for this problem i think after finding $ x\equal{}0 and x\equal{}5$ be two roots you can use cardano to find 4 non trivial roots

hello,no but then you have to solve
$ 12x^4\minus{}110x^3\plus{}345x^2\minus{}423x\plus{}158\equal{}0$
Sonnhard.


Bonjour amis français :)
Je suis en DEUG MIAS 2ème année et je dois faire un stage d'environ 120h en entreprise ou enseignement. Je crois (enfin je pense) que Pierre est chercheur et je m'adresse donc à toi Pierre ou a d'autres plus ou moins proche du domaine de la recherche. Je voudrais savoir si il est possible de trouver un stage dans un labo car j'aimerais avoir un maximum d'information sur le travail de chercheur, j'hésite actuellement entre 3 options: continuer en licence de maths, tenter le second concours de l'ENS Lyon, passer le concours de l'armée de l'air pour éventuellement devenir pilote.
Merci

Heu...non,non. Moi, je suis prof en lycée.

Pierre, désolé.

Euh vu que tu es en DEUG, la recherche c'est pas pour tout de suite....
Si tu veux vraiment faire ça, il faut que ton projet professionnel murisse, et que tu en parles avec tes profs de fac.
Parce que même en passant pas l'ENS, le mieux c'est de faire une thèse par la suite, quand même.
Donc tu as le temps d'y penser....

Voilà...

Merci du conseil mais tu aurais du regarder la date du post :D
Cette année je suis à l'ENS Lyon donc j'ai déjà fait mon stage l'année dernière en lycée et c'était moyennement interessant. Cette année c'est stage en recherche donc ça devrait être interessant


If $m+n>50$ and $m-n \leqslant 10$, what is the smallest possible integer value of n?

Is it [hide]1[/hide]?
Because after a bit of inspection, we find it can't be negative or 0.

Again we find n can't be negative or 0, so...

[hide]Answer 21?[/hide]

[hide]mean of 10 and 50 is 30, so the smallest n is 21 ?[/hide]

What is that weird symbol $\leqslant$? is it $\leq$?

[hide]
It should be 21.
The greatest the difference can be is 10.
The sum has to be greater that 50.
start with 1,11 the sum is too low, by 51-12=39. So you have to add half that to both m and n. You can't have fractions, so add 20 to get 21,31.


You meant this, right? $\frac{n+2-\frac{n}{25}}{2}= \frac{8n+24}{25}$
What did you mean by
[quote]for evry n(=IN.[/quote]
[hide]
Solving for $n$
$\frac{n+2-\frac{n}{25}}{2}= \frac{8n+24}{25}$
$25n+50-n = 16n+48$
$8 n =-2$
$\boxed{n=\frac{1}{4}}$

That is probably not what was meant from the problem.
[/hide]

I think n(=IN is supposed to be $n\in\mathbb{N}$.

[quote="boukharfane"]i mean by [] la partie entiere and (= appartient.[/quote]
Not to be rude, but that's not English, right?

:blush: Yeah, I thought the same thing, but I was waiting for someone who knows what it meant to answer

i m very sorry because my inglish is very bad.
(= mean belong to
[x] mean floor(x).

[quote="boukharfane"][b][color=darkred][size=150]prove that:
[(n+2-[n/25])/2]=[(8n+24)/25][/size][/color][/b] for evry n(=IN.[/quote]

So, corrected problem:

Prove that
\[\left\lfloor \frac{n+2-\lfloor \frac{n}{25}\rfloor}{2}\right\rfloor = \left\lfloor \frac{8n+24}{25}\right\rfloor \]
for all $n \in \mathbb{N}$ (or do you mean $n \in \mathbb{Z}$?)

Suggestion at boukharfane:

Use LaTeX to type math.  People can confuse the $[\text{ }]$ in something like $[(x+3)^{3}-10x]^{2}$ to be brackets, and not the floor function.  That's why $\lfloor\text{ }\rfloor$ is used instead.

[quote="lena_siz"][quote="AstroPhys"]Since the previous solution I posted yeilded an $n$, where $n \not\in \mathbb N$, it is proven false.[/quote]

But this equation is of degree 1, so it should have only one solution. I got -0.25 . How come n can belong to any natural number?[/quote]

-.25 is correct without floors, but the edited equation is no longer has a unique solution because of the floors...

Let $n=25k+m$. We must show \[\left\lfloor \frac{n+2-\lfloor \frac{n}{25}\rfloor}{2}\right\rfloor = \left\lfloor \frac{8n+24}{25}\right\rfloor \]
which is equivalent to \[\left\lfloor \frac{25k+m+2-k}{2}\right\rfloor = \left\lfloor \frac{200k+8m+24}{25}\right\rfloor \]
Now \[\left\lfloor \frac{24k+m+2}{2}\right\rfloor = 12k+1+\left\lfloor \frac{m}{2}\right\rfloor\]
Also \[\left\lfloor \frac{200k+8m+24}{25}\right\rfloor = 8k+\left\lfloor \frac{8m+24}{25}\right\rfloor =8k+1+\left\lfloor \frac{8m-1}{25}\right\rfloor \] since $8m>1$
Now it becomes clear that this was either a joke or a misunderstanding, because \[8k+1+\left\lfloor \frac{8m-1}{25}\right\rfloor =12k+1+\left\lfloor \frac{m}{2}\right\rfloor\]
can't be true for all reasonable $k,m$ since if $m$ is held constant, this is equivalent to $12k=8k$ for all $k$ which is absurd.


Suppose that $f : [0,1]\to [0,1]$ has the property that  equation
\[ f\left(f(x)\right)-x=0 \; \; ,\; \; \; x \in [0,1] ,  \]
 has a unique solution.  [b] Prove that equation $f$ has a  unique fixed point . [/b]

[quote="Gauß"]Maybe $f$ must be continuous?[/quote]
No !

Since for any fixed point $x_0$ of $f\circ f$, both $x_0$ and $f(x_0)$ turn out to be fixed points of $f\circ f$, it means that $x_0=f(x_0)$, so it's also a fixed point of $f$. If there were at least two, then we would have at least two fixed points for $f\circ f$, so it must also be unique.

If you read the message above carefully, you'll see why in this case, this is true.

[quote="Gauß"]To grobber.

I had the same idea, but there is a gap, I think. If the function $f(f(x))$ has a fixed point, why does $f(x)$ have a fixed point too?[/quote]
if not,f(f(x)) will have two fixed point at least.


Let $ABC$ be a triangle with its circumcircle $\mathcal W$.Let $\ell_1,\ell_2$ be two lines tangent to $\mathcal W$ at $B$ and $C$.
Let the bisector of $AC$ meets $AB$ at $k$.Let $\mathcal F$ be a line Prependicular to the bisector of $AC$ at $k$.
Prove that $\mathcal F$ and $\ell_1$ and $\ell_2$ are congurrent.

Let's take the parallel $d$ to $AC$ through $P=\ell_1\cap\ell_2$ and show that it passes through $K$.

Let $K'=d\cap AB$. We get $\angle K'BC=\angle B=\pi-\angle ACP=\angle K'PC$, so $K'BPC$ is cyclic. This means that $\angle PK'C=\angle PBC=\angle A$, and, at the same time, since $PK'\|AC$, we have $\angle PK'C=\angle ACK'$, so we have $\angle ACK'=\angle A=\angle CAK'$, meaning that $ACK'$ is isosceles, i.e. $K'=K$, as desired.


limit
topology
advanced fields
advanced fields unsolved

Let M be a compact metric space. Show that a sequence in M is convergent $ \Longleftrightarrow$
 has only one aderent point.

Let $ (x_n)$ a sequence in $ M$

An "adherent" point is a limit of subsequence of any sequence of M ( i think you call it "closure" )

( $ \implies$ ) If there are more than one point adherent$ \implies (x_n)$ doesn't converges

( $ \Longleftarrow$ )

Suposse that $ (x_n)$ doesn't converges. This implies that $ \exists \epsilon > 0, \forall n\in \mathbb{N}: d(x_n,a)\geq \epsilon$, $ forall a\in M$

Because $ M$ is compact, we have that $ (x_n)$ have a convergent subsequence in $ M$, put $ \lim x_{n_{k}} \equal{} m\in M$.

For all $ \epsilon > 0$ we have that in $ B(m;\epsilon)$ there are infinty points of $ x_{n_{k}}$, and outside of $ B(m;e)$ we have infinity points of $ (x_n)$ . We may consider a new sequence, take the terms of $ (x_n)$ outside $ B(m;e)$. This new sequence has a subsequence (note that this new subsequence is also subsequence of $ (x_n)$ ) that converges in a point outside of $ B(m;e)$, then , $ (x_n)$ has two different adherent points, and this is a contradiction

[quote="jorgeston"]An "adherent" point is a limit of subsequence of any sequence of M ( i think you call it "closure" )[/quote]
I've never seen that term.  For me, the "closure" of a set $ E$ means smallest closed set that completely encloses $ E$, so those sound like two different things.  I have heard of a limit point, which I think is pretty self-explanatory.  Maybe that's what you're referring to?

i'll write the definition of " adherent point" , from "espaços metricos- elon langes lima" ( this text is in portuguese)

[quote]Um ponto $ a$ diz-se aderente a um subconjunto $ X$ de um espaço metrico $ M$ quando $ d(a,X) \equal{} 0$ . Isto significa que existem pontos de $ X$ arbitrariamente próximos de a, ou seja , para cada $ \epsilon > 0$ , podemos encontrar $ x\in X$ tal que $ d(a,x) < \epsilon$[/quote]

[b][u]"Mini"-Traduction:[/u]
[/b]
[quote]
A point is "adherent" to  $ X\subset M$, whit $ M$ a metric space, if $ d(a,X) \equal{} 0$. [/quote](bla bla bla...)

This definition is equivalent to say that an "adherent" point value is a limit of a subsequence of any sequence of $ M$. If $ M$ has an only adherent point, all subsequences converges to that value.


In rectangle $ RECT$, $ M$ is the midpoint of $ \overline{RE}$, $ N$ is the midpoint of $ \overline{CT}$ and $ P$ and $ Q$ trisect $ \overline{RT}$. What percent of the area of $ RECT$ is in the area of $ \triangle ROQ$? Express your answer as a mixed number.
[asy]draw((0,0)--(3,0)--(3,2)--(0,2)--cycle);
draw((0,1)--(3,1));
draw((0,0)--(2/3,1)--(2,0));
dot((1,0));
label("$R$",(0,0),SW);
label("$M$",(0,1),W);
label("$E$",(0,2),NW);
label("$O$",(2/3,1),N);
label("$P$",(1,0),S);
label("$Q$",(2,0),S);
label("$T$",(3,0),SE);
label("$N$",(3,1),E);
label("$C$",(3,2),NE);[/asy]


calculus integration trigonometry logarithms function limit algebra

Just let everything be real.  This one is tough I think and I've not yet been able to show it:
Given $ \displaystyle \psi(x)=\sum_{n=1}^{\infty} e^{-n^2 \pi x}$ and the relation $ \displaystyle  4\psi'(1)=\psi(1)=-1/2$ show:
\begin{align*}
\Xi(t)&=1/2-(t^2+1/4)\int_{1}^{\infty} \psi(x)x^{-3/4}\cos\big[t/2\ln(x)\big]dx\\
&=4\int_1^{\infty}\frac{d}{dx}\left\{x^{3/2}\psi'(x)\right\}x^{-1/4}\cos\left[t/2\ln(x)\right]dx
\end{align*}

I've already spent some time on it.  I'm pretty sure I can get it eventually.  I'm just posting it here because I think it's beautiful and maybe others may be interested in it if it's a challenge for you.   If you wish, I'll give you the same meager hint Titchmarsh gave me:

[hide]Use integration by parts[/hide]

shawtoos, i am failing to see what is the interest or motivation behind this integral...  :maybe: (and the other one which you posted prior to this ) can you elaborate  as to why it may be of consequence ? i don't see it yielding any closed form that could  involve  constants like $ \pi,\quad e,\quad \gamma,\quad \zeta(3),\quad \beta(2),\quad  \text{etc.}$

or, r u saying the [i]final[/i] integral, i.e.

$ 4\;\int_1^{\infty}\;\frac{d}{dx}\;\left\{x^{3/2}\;\psi'(x)\right\}\;x^{\minus{}1/4}\;\cos\;\left[t/2\; \ln\;(x)\right]\;\;dx$

is [i]actually[/i] doable ? 'cause then i would find some interest to try and tackle it, otherwise not.

The interest is in using integral representations to derive properties of the functions. Expressing constants in terms of other constants is an abandoned field, and not the point here.

Hello Misan.  Sorry about that.  I should have said:  Express this:
\[ 1/2\minus{}(t^2\plus{}1/4)\int_{1}^{\infty} \psi(x)x^{\minus{}3/4}\cos\big[t/2\ln(x)\big]dx\]
as this:
\[ 4\int_1^{\infty}\frac{d}{dx}\left\{x^{3/2}\psi'(x)\right\}x^{\minus{}1/4}\cos\left[t/2\ln(x)\right]dx\]
and I'm finding it really tough and I'd disappointed in myself because I'm not having much success and I'm pretty sure I've given all that's needed to do it but then I remember Riemann was a master at integration, kinda' like some of you I think (really).  The other illustrates the first appearance of one of the most elegant concepts in mathematics:``und es ist schr wahrscheinlich, dass alle Wurzeln reele sind'' (it is very likely that all the roots are real).

you should take it easy on yourself.  :(  by the way, i tried to understand what it is that you are trying to do and i think i know now exactly what it is you are trying to get at. you are essentially trying to recreate some of what riemann did in his paper.

 start with  euler's gamma function  :

$ \Gamma (z) \equal{} \int_0^\infty e^{ \minus{} x} \; x^{z \minus{} 1} \; dx \qquad ( \mathbb Re(z) > 0)$

replace $ x\quad with \quad n^2\pi x\quad\text{and}\quad z\quad with \quad \frac {z}{2}$  to get

$ \Gamma \left(\frac {z}{2}\right) \equal{} \int_0^\infty e^{ \minus{} n^2\pi x} \; \left( n^2\pi x\right) ^{z/2 \minus{} 1} \;\; d \left( n^2\pi x\right)$

$ \Rightarrow \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \;\frac {1} {n^z} \; \equal{} \; \int_0^\infty e^{ \minus{} n^2\pi x} \;x^{z/2 \minus{} 1}\; dx$

now take $ \sum_{n \equal{} 1}^\infty$ of both sides to get

$ \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \;\sum_{n \equal{} 1}^\infty\; \frac {1} {n^z} \; \equal{} \;  \int_0^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \;x^{z/2 \minus{} 1}\; dx$

$ \Rightarrow \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \; \equal{} \; \int_0^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \;x^{z/2 \minus{} 1}\; dx$

now break $ \int_0^\infty \qquad into \qquad \int_0^1 \plus{} \int_1^\infty$

$ \equal{} \int_0^1 \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \;x^{z/2 \minus{} 1}\; dx\; \plus{} \; \int_1^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \;x^{z/2 \minus{} 1}\; dx$

substitute $ x \equal{} \frac {1}{x}$ in the 1st integral,

$ \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \; \equal{} \; \int_\infty^1 \left( \sum_{n \equal{} 1}^\infty e^{\frac { \minus{} n^2\pi}{ x}} \right) \;x^{1 \minus{} z/2}\;\; d\left(\frac {1}{x}\right) \; \plus{} \; \int_1^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \;x^{z/2 \minus{} 1}\; dx$

$ \equal{} \int_1^\infty \left( \sum_{n \equal{} 1}^\infty e^{\frac { \minus{} n^2\pi}{ x}} \right) \;x^{ \minus{} 1 \minus{} z/2}\;\; dx \; \plus{} \; \int_1^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \;x^{z/2 \minus{} 1}\; dx$

 notice the following ( i am leaving  it as an exercise for you to show this, but if you can't,  i'll provide hints .... )

$ 1 \plus{} 2 \sum_{n \equal{} 1}^\infty e^{\frac { \minus{} n^2\pi}{ x}} \equal{} \sqrt x \left( 1 \plus{} 2 \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right)$

hence, we have

$ \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \; \equal{} \; \frac {1}{2} \int_1^\infty \left[ \sqrt x \left( 1 \plus{} 2 \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \minus{} 1 \right]\; x^{ \minus{} 1 \minus{} z/2}\; dx \plus{} \int_1^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) x^{z/2 \minus{} 1}\; dx$

collect the terms involving $ \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x}$  within a single integral,

$ \equal{} \int_1^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \left[ x^{z/2 \minus{} 1}\; \plus{} \; x^{ \minus{} (1 \plus{} z)/2} \right] \; dx \plus{} \frac {1}{2} \int_1^\infty \left[ x^{ \minus{} (1 \plus{} z)/2} \minus{} x^{ \minus{} 1 \minus{} z/2} \right] \; dx$

$ \equal{} \int_1^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \left[ x^{z/2}\; \plus{} \; x^{(1 \minus{} z)/2} \right] \; \frac {dx}{x} \plus{} \frac {1}{2} \int_1^\infty \left[ x^{(1 \minus{} z)/2} \minus{} x^{ \minus{} z/2} \right] \;\frac { dx}{x}$

recall that

$ \int_1^\infty \; x^{ \minus{} m} \;\frac { dx}{x} \equal{} \int_1^\infty \;\frac { dx}{x^{m \plus{} 1}} \equal{} \frac {1}{m}\qquad ( where\quad m > 0 )$

and so,

$ \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \; \equal{} \; \int_1^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \left[ x^{z/2}\; \plus{} \; x^{(1 \minus{} z)/2} \right] \; \frac {dx}{x} \plus{} \frac {1}{2} \left[ \frac {2}{z \minus{} 1} \minus{} \frac {2}{z} \right]$

$ \Rightarrow \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \; \equal{} \; \int_1^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \left[ x^{z/2}\; \plus{} \; x^{(1 \minus{} z)/2} \right] \; \frac {dx}{x} \minus{} \frac {1}{z\; (1 \minus{} z) } \qquad \ldots \qquad \ldots \qquad (1)$

this is  called a "functional equation", i.e. if you substitute  $ (1 \minus{} z)$ in place of $ z$, the equation remains [b]unchanged.[/b]

it is a very important relationship, one that you will see   often in functional analysis (e.g. when studying whittaker and watson's text )......

[  look at equation $ (14)$ of http://mathworld.wolfram.com/RiemannZetaFunction.html   ]

in any event, equation $ (1)$ above is an important  one to arrive at. now we proceed to show the relationship that  you are interested in.

(continued on next post)....

from my previous post, i had

$ \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \; \equal{} \; \int_1^\infty \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \left[ x^{z/2 \minus{} 1}\; \plus{} \; x^{ (1 \minus{} z)/2 \minus{}1} \right] \; dx\; \minus{} \; \frac {1}{z\; (1 \minus{} z) }$

integrating by parts gives

$ \equal{} \left\{\left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right) \left[ \frac { x^{z/2} } { z/2} \plus{} \frac { x^{(1 \minus{} z)/2} } { (1 \minus{} z)/2} \right]\right\} \quad \bigg|_1^\infty \minus{} \int_1^\infty \; \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right)' \; \left[ \frac { x^{z/2}} { z/2} \plus{} \frac { x^{(1 \minus{} z)/2}} { (1 \minus{} z)/2} \right] \; \; dx \; \minus{} \; \frac {1}{z\; (1 \minus{} z) }$

$ \equal{} \left\{ 0 \minus{} \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi} \right) \left[ \frac {2}{z} \plus{} \frac {2}{1 \minus{} z} \right] \right\} \minus{} \int_1^\infty \; \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right)' \; \left[ \frac { x^{z/2}} { z/2} \plus{} \frac { x^{(1 \minus{} z)/2}} { (1 \minus{} z)/2} \right] \; \; dx \; \minus{} \; \frac {1}{z\; (1 \minus{} z) }$

$ \equal{} \; \minus{} \; \frac {1}{z\; (1 \minus{} z) } \left( \; 1 \plus{} 2\; \psi(1) \; \right) \minus{} \int_1^\infty \; \psi'(x)\; \left[ \frac { x^{z/2}} { z/2} \plus{} \frac { x^{(1 \minus{} z)/2}} { (1 \minus{} z)/2} \right] \; \; dx$

$ \left( \text{where} \qquad \psi(1) \equal{} \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi} \qquad \text{and} \qquad \psi' (x) \equal{} \left( \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right)' \;\;\; \right)$

 [ i am  using shorthands here... ]

factor out a $ x^{3/2}$ inside the integral to get,

$ \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \; \equal{} \; \minus{} \; \frac {1}{z\; (1 \minus{} z) } \left( \; 1 \plus{} 2\; \psi(1) \; \right) \minus{} \int_1^\infty \;\left( x^{3/2}\; \psi'(x)\;\right) \; \left[ \frac { x^{ \minus{} (1 \minus{} z)/2 \minus{} 1}} { z/2} \plus{} \frac { x^{ \minus{} z/2 \minus{} 1}} {(1 \minus{} z)/2 } \right] \; \; dx$

now integrate by parts once again

$ \equal{} \; \minus{} \; \frac {1}{z\; (1 \minus{} z) } \left( \; 1 \plus{} 2\; \psi(1) \; \right) \minus{} \left\{\;\left( x^{3/2}\; \psi'(x)\;\right) \; \left[ \minus{} \frac { x^{ \minus{} (1 \minus{} z)/2}} {z/2\;\;\; (1 \minus{} z)/2 } \minus{} \frac { x^{ \minus{} z/2}} {(1 \minus{} z)/2 \;\; \;z/2} \right] \;\right\} \; \;\; \bigg|_1^\infty \; \plus{} \; \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \; \left[ \minus{} \frac { x^{ \minus{} (1 \minus{} z)/2}} {z/2\;\;\; (1 \minus{} z)/2 } \minus{} \frac { x^{ \minus{} z/2}} {(1 \minus{} z)/2 \;\; \;z/2} \right] \;\; dx$

$ \equal{} \; \minus{} \; \frac {1}{z\; (1 \minus{} z) } \left( \; 1 \plus{} 2\; \psi(1) \; \right) \minus{} \left\{ 0 \; \plus{} \;\psi'(1)\; \left[ \frac {4} {z\;\;(1 \minus{} z)} \plus{}   \frac {4} {z\;\;(1 \minus{} z)} \right] \;\right\} \; \minus{} \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \; \frac {4}{z\;\;(1 \minus{} z)} \; \left[ x^{ \minus{} z/2} \plus{} x^{ \minus{} (1 \minus{} z)/2} \right] \; dx \qquad \ldots \qquad \ldots \qquad (2)$

  remember i used (in the earlier post)

$ 1 \plus{} 2 \sum_{n \equal{} 1}^\infty e^{\frac { \minus{} n^2\pi}{ x}} \equal{} \sqrt x \left( 1 \plus{} 2 \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right)$

i.e.

$ 1 \plus{} 2\; \psi \left( \frac {1}{x}\right) \equal{} \sqrt x \left( 1 \plus{} 2\; \psi (x) \right)$

 differentiate both sides w.r.t. $ x$ to get

$ 2 \;\psi' \left( \frac {1}{x}\right)\;\left( \minus{} \frac {1}{x^2}\right) \equal{} \sqrt x \left( 2\; \psi' (x) \right) \plus{} \frac {1}{2\sqrt x} \; \left( 1 \plus{} 2\; \psi (x) \right)$

which gives, (after you put $ x \equal{} 1$ )

$ \minus{}2 \; \psi'\; (1) \equal{} 2 \; \psi'\; (1)\plus{}  \frac {1}{2}  \left( 1 \plus{} 2\; \psi (1) \right)$

$ \Rightarrow \frac {1}{2} \plus{} \psi\;(1) \plus{} 4\; \psi'\; (1) \equal{} 0$

[ note that the relation  you have in the original question  has a  typo: you have an $ " \equal{} "$  instead of a $ " \plus{} "$ ]

and so,

$ \minus{} \;\psi'\; (1) \; \equal{} \; \frac {1}{4} \left( \frac {1}{2} \plus{} \psi\;(1)\; \right)$

now put this in $ (2)$ to get

$ \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \; \equal{} \; \minus{} \; \frac {1}{z\; (1 \minus{} z) } \left( \; 1 \plus{} 2\; \psi(1) \; \right) \plus{} \frac {1}{4} \left(\frac {1}{2} \plus{} \psi\;(1) \right) \left[ \frac {8} {z\;\;(1 \minus{} z)} \right]  \; \minus{} \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \; \frac {4}{z\;\;(1 \minus{} z)} \; \left[ x^{ \minus{} z/2} \plus{} x^{ \minus{} (1 \minus{} z)/2} \right] \; dx$

(continued on next post)....

simplify to get (you can see the first $ 2$ terms on the $ RHS$ cancel each other out)

$ \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \; \equal{} \; \minus{} \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \; \frac {4}{z\;\;(1 \minus{} z)} \; \left[ x^{ \minus{} z/2} \plus{} x^{ \minus{} (1 \minus{} z)/2} \right] \; dx$

factor out a $ x^{ \minus{} \frac {1}{4}}$ to get

$ \equal{} \; \minus{} \frac {8}{z\;\;(1 \minus{} z)} \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \;x^{ \minus{} 1/4} \left[ \frac {1}{2} \left( x^{ \minus{} z/2 \plus{} 1/4} \plus{} x^{z/2 \minus{} 1/4 } \right) \right] \; dx$

$ \equal{} \; \minus{} \frac {8}{z\;\;(1 \minus{} z)} \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \;x^{ \minus{} 1/4} \left[ \frac {1}{2} \left( x^{ \minus{} z/2 \plus{} 1/4} \plus{} x^{\minus{} ( \minus{}z/2 \plus{} 1/4) } \right) \right] \; dx$

$ \equal{} \; \minus{} \frac {8}{z\;\;(1 \minus{} z)} \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \;x^{ \minus{} 1/4} \left[ \frac {1}{2} \left( e^{\left( \minus{} z/2 \plus{} 1/4\right) \; \ln \; x } \plus{} e ^ { \minus{} \left( \minus{} z/2 \plus{} 1/4\right) \; \ln \; x } \right) \right] \; dx$

$ ( \quad \because x^ a \equal{} e ^ {a\; \ln \; x} )$

now since
 $ \cosh\theta \equal{} \frac {1}{2} \left( e^ \theta \plus{} e ^ { \minus{} \theta } \right)$

 we have

$ \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \; \equal{} \; \minus{} \frac {8}{z\;\;(1 \minus{} z)} \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \;x^{ \minus{} 1/4} \; \cosh \; \left( \left( \minus{} \frac{z}{2} \plus{} \frac{1}{4}\right) \; \ln \; x\right) \; dx$

it seems to me  riemann must have expressed the whole thing only after multiplying both sides by $ \left( \minus{} \; \frac {z\;\;(1 \minus{} z)}{2} \right)$

in other words,

$ \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \cdot \left( \minus{} \; \frac {z\;\;(1 \minus{} z)}{2} \right) \; \equal{} \;$

$ 4 \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \;x^{ \minus{} 1/4} \; \cosh \; \left( \left( \minus{} \frac{z}{2} \plus{} \frac{1}{4} \right) \; \ln \; x\right) \; dx$

you can express the entire thing on the $ LHS$ as

$ \Xi \; (z) \equal{} \Gamma \left(\frac {z}{2}\right) \; \pi ^{ \minus{} z/2} \; \zeta(z) \cdot \left( \minus{} \; \frac {z\;\;(1 \minus{} z)}{2} \right)$

so that

$ \Xi \; (z) \equal{} 4 \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \;x^{ \minus{} 1/4} \; \cosh \; \left( \left( \minus{} \frac{z}{2} \plus{} \frac{1}{4}\right) \; \ln \; x\right) \; dx$

now take a  look at my avatar, tells you what kind of an argument riemann must have picked for $ z$

in other words, put

$ z \equal{} \frac {1}{2} \plus{} \imath \; t$

$ \Rightarrow \minus{} \frac {z}{2} \plus{} \frac {1}{4} \;\equal{}\; \minus{} \imath\; \frac {  t}{2 }$

to finally get

$ \Xi \;\left( \frac {1}{2} \plus{} \imath \; t\right) \; \equal{} \; 4 \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \;x^{ \minus{} 1/4} \; \cosh \; \left( \left(\minus{} \imath\; \frac { t}{2 } \right) \; \ln \; x\right) \; dx$

recall that

$ \cosh \; ( \minus{} \imath\; \theta ) \equal{} \cos \; \theta$

and so

$ \Xi \;\left( \frac {1}{2} \plus{} \imath \; t\right) \; \equal{} \; 4 \int_1^\infty \; \left( \; x^{3/2}\; \psi'(x)\; \right)' \;x^{ \minus{} 1/4} \; \cos \; \left( \left( \frac {t}{2 } \right) \; \ln \; x\right) \; dx$

which, i believe, is essentially what you were going after !  :|

:wow:

Can I get some of whatever you're taking?

Hey Misan,
I delayed logging onto the forum until just now so I could try one more time and I just figured it out!  I appreciate your help and would have relied on your solution above had I not solved it today.  Somebody here, I don't know, I don't remember who  :wink: , said, "they have to learn how to do it themselves".  I made a mistake above.  The relation is
\[ 4\psi'(1) \plus{} \psi(1) \equal{} \minus{} 1/2
\]
That's a nice explanation you gave of the first part of Riemann's paper leading up to the equations.  I'll spend time working on it as well as your method for the second step above.  I see you used an approach different than what I used.  I'll post my method later.

[quote="misan"]

 notice the following ( i am leaving  it as an exercise for you to show this, but if you can't,  i'll provide hints .... )

$ 1 \plus{} 2 \sum_{n \equal{} 1}^\infty e^{\frac { \minus{} n^2\pi}{ x}} \equal{} \sqrt x \left( 1 \plus{} 2 \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi x} \right)$
[/quote]

. . . took a while . . . guess I should have already known this:

The relation comes out of the Poisson Summation Formula for the Fourier Transform of the function $ f(x) \equal{} e^{ \minus{} \pi t x^2}$:
\[ \mathcal{F}_x\left\{e^{ \minus{} \pi t x^2}\right\} \equal{} \frac {1}{\sqrt {t}}e^{ \minus{} \pi \xi^2/t}
\]
Then by Poisson's Summation Formula:
\[ \sum_{n\in\mathbb{Z}}e^{ \minus{} \pi t n^2} \equal{} \frac {1}{\sqrt {t}}\sum_{n\in\mathbb{Z}}e^{ \minus{} \pi n^2/t}
\]
Splitting out the positive integers:
\[ \left(1 \plus{} 2\sum_{n \equal{} 1}^{\infty}e^{ \minus{} \pi t n^2}\right) \equal{} \frac {1}{\sqrt {t}}\left(1 \plus{} 2\sum_{n \equal{} 1}^{\infty}e^{ \minus{} \pi n^2/t}\right)
\]
I'm working through the rest of it.  Noticed you spotted my error.  Sharp. :)

sounds like u r having fun.  :)

wanna try something ?  show

$ \psi(1) \;\equal{}\; \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi} \;\equal{}\; \frac{1}{2}\left(\frac{\Gamma\left(\frac{1}{4} \right)}{\sqrt[4]{\;4\pi^3 \;}} \minus{}1\right)$

my solution is different from the one ramanujan used.
(by the way, this is one of the sums in ramanujan's letter that impressed hardy)

beautiful isn't it ?

and if you want to have even more fun,  show

$ \psi(\sqrt 2) \;\equal{}\; \sum_{n \equal{} 1}^\infty e^{ \minus{} n^2\pi\sqrt 2}  \;\equal{}\; \frac{1}{2}\left( \frac{\Gamma\left(\frac{1}{8} \right)} {\sqrt { \; 2^{9/4}\;\;\pi\;\; \Gamma\left(\frac{1}{4} \right)\;}} \minus{}1\right)$

to this date, i haven't been able to show this.  :(


calculus integration limit logarithms algebra polynomial derivative

Let $a,b$ be real numbers such that $a<b$.
Evaluate

\[\lim_{b\rightarrow a} \frac{\displaystyle\int_a^b \ln |1+(x-a)(b-x)|dx}{(b-a)^3}\].

[hide="Hint."]It should be possible to do this without actually computing the primitive. Use Taylor polynomials, see the "big-O, little-o" handout.

Personally I like making things symmetric, and I would make the substitution $y=x-\frac{a+b}2.$[/hide]

We use l'Hopital's rule and differentiation (w.r.t. b) under the integral sign to obtain:

\[\lim_{b\rightarrow a} \frac{\displaystyle\int_a^b \ln |1+(x-a)(b-x)|dx}{(b-a)^3}=\lim_{b\rightarrow a} \frac{\displaystyle\int_a^b \frac{x-a}{1+(x-a)(b-x)}dx}{3(b-a)^2}\]\[=\lim_{b\rightarrow a} \frac{b-a+\displaystyle\int_a^b \frac{-(x-a)^2}{(1+(x-a)(b-x))^2}dx}{6(b-a)}=\frac{1}{6}\]

I hope this is correct.

P.S. I enjoy your problems, kunny!

Thank you for your reply, rookworm.

Your answer is correct. :)

kunny


[quote="Twins"]Prove that $ \sum{\frac{a(a-b)(a-c)}{a^{2}+2bc}}\geq 0$[/quote]

I think $ a,b,c$ is non-negative.
Assume that $ c=min(a,b,c)$.
We have:
\[ LHS=M(a-b)^{2}+N(a-c)(b-c)\]
In which:
\[ M=\frac{c(2a^{2}+2b^{2}+3ab-2ac-2bc)}{(a^{2}+2bc)(b^{2}+2ca)}\ge 0\]
\[ N=\frac{c}{c^{2}+2ab}\ge 0\]
DONE.

lasha was mistaken, he just hurried, Schur does not directly work here, but as the inequality makes us remember about Shur it is natural to try to prove the inequality as we prove Schur itself, taking a>=b>=c without loss of generality ,

then clearly the only negative is $ b(b-c)(b-a)/b^{2}+ac$
and $ a(a-b)(a-c)/a^{2}+bc$>=$ b(b-c)(a-b)/b^{2}+ac$ (just multiply ;) )
so inequality only when a=b=c :D

it is just so natural...

and I bet as in Schur we don't need that a,b,c are non-negative, because then $ c(a-c)(b-c)/c^{2}+ab$>=$ b(b-c)(a-b)/b^{2}+ac$ should work

[quote="Twins"]Prove that $ \sum{\frac{a(a-b)(a-c)}{a^{2}+2bc}}\geq 0$[/quote]
For non-negative numbers we assume $ a\geq b\geq c.$
Then $ \frac{c(c-a)(c-b)}{c^{2}+2ab}\geq0$ and $ \frac{a(a-b)(a-c)}{a^{2}+2bc}+\frac{b(b-a)(b-c)}{b^{2}+2ac}=$
$ =\frac{(a-b)^{2}(2a^{2}+3ab+2b^{2}-2ac-2bc)c}{(a^{2}+2bc)(b^{2}+2ac)}\geq0.$


We should really have a trivia contest. It will be like group jeopardy, or maybe it will be like the weakest link. That would be really fun!!! I was wondering if anyone else was interested in the idea - I just was thinking and it sounded logical, since were all so bright ;) . LoL.

Sounds like a good idea to me.  Any suggestions for contest parameters?

Perhaps, there should be teams of 3. The contest will be held through AOL Instant Messenger. Each team will meet the coordinator, and they will be given a series of lets say, 30 questions? I have tons at my disposal. They will have 30 seconds per question? They can discuss it in a separate window. I think a double elimination will be good. There can be 2 people or even one person on a team depending on how many people are interested. The trivia will be the following subjects, all non-math since this is the non-math forum. I think the questions will be slightly difficult, since ties would be hard to break. The subjects that I have most questions on are Global View, Sound and Screen, News, The Written Word, Innovations, and Game Time.

how about a political section group? i wouldn't mind, but im sure my mom would be against me doing it.  specially cause its random non-math stuff.  but if we really are doing it, count me in.

suenified

[quote="starrynightskies"]how about a political section group? i wouldn't mind, but im sure my mom would be against me doing it.  specially cause its random non-math stuff.  but if we really are doing it, count me in.

[/quote]

I think Global View and News both involve lots of politics, so that should be Ok. By the way, not all the questions are current.

Me and Danny (SingTheSorrow) play a game like this.  It's really fun.  We just make up 10 questions to ask each other (on all different subjects).  If you want to be serious you could do something like this, and make a bracket like joml88 did for the math contest.  I'll do it though I have forgot all my random knowledge and don't know anything about sports! =)

Yeah, I'm in!  This will definitely be interesting.  (btw, AIM s/n = KillingInSleep)

I think I will create a bracket like joml88, just to see how things go. Meanwhile, what do you guys think about the topics so far?

I'd like to join too.

An about the categories, you should try to have some more academic ones, instead of just all just fun stuff.  One or two would be okay.

I'll include a science one, but I don't know what level everyone is at in science. If not science, what else? Perhaps there should be history?

History is always good.  Do we like music?

If we are to have sports, I think I will need help, because I don't have that many sports related questions currently. Making all of them up from scratch will be difficult.

Yeah! I'll join. I'm not too great at trivia, outside of my areas of knowledge, television (only basic cable though), movies, and sports.

If you need a place for trivia questions, try [url]funtrivia.com[/url].  They're supposed to have the most trivia questions on the internet.

I checked it out - questions will be unlimited, so I'm not worried about that anymore! Anyone else, please tell me if you are interested by next week. I will post a full detailed version of the competition once we have enough people. There are 7 people interested so far.

I'd be interested. we did a thing like this in school which was a blast. there are all sorts of interesting little facts out there that could be asked.

Music, History, Science, TV...

I think the fun part about it is preparing questions for your opponent.  Could it be arranged this way?

The problem with making questions for your opponent is that your questions might be almost impossible compared to your opponnent's questions.

But if you want to make trivia questions for other people to play, that web-site I mentioned ([url]funtrivia.com[/url]) has a way.

Yeah. One person might give extremely difficult questions while another person may give easy questions - it would be better if there were only one or two people administering the questions. What format would you guys like to do this competition in. Team or Individual? Immediate Response(IM) or Rounds(Posts)? What times are good? How many rounds?

What dates and times are available to everyone?

I can't do it from the 14th to the 24th.

I woudl prefer a team thing with posts, but AIM is also okay (If I get it by then).

I join, but not sure about the times, right now....

Count me in.  I'd personally prefer a post-based system, since my schedule's rather hackneyed, but I'll try and participate however I can.

sometime on the weekdays either 7:00 am-10:00am or 7:00 pm-9:00pm. sorry about the weird times.

I will choose the evening time becuase in accordance to eastern time, the morning time you suggested, means that people on the west coast would have to wake up at 4:00 AM and no thanks to that.

Can people do it on the 11th or the 12th probably at 7pm?

12ths no good for me (and danny too)  ;)

I have to work at night at mcdonalds  :( , so if we could do it as a post based thing that would be great

I'm in

My AIM is hezgotzbeenz, just send an invite when you are ready.

I can do it both dates, but I'd like a team method

Sorry everyone - my internet was down for a whole week, so I couldn't make it. My sincere apologies :blush:  :blush:  :blush: .

[quote=ObjectZ]16 year bump[/quote]

But why? Why would you do this?


calculus
integration
limit
function
real analysis
real analysis unsolved

Evaluate $ \lim_{n\to\infty}\int^n_0 (1\plus{}\frac{x}{n})^n e^{\minus{}2x} dx$

$ \int_0^{\infty}e^{\minus{}x}\,dx \equal{} 1$.

[quote="Carcul"]$ \int_0^{\infty}e^{ \minus{} x}\,dx \equal{} 1$.[/quote]

Why can we "swap" the limit and the integral? Besides, $ n$ is in both the upper bound and in the integrand. How to deal with that?

Hi :) !!

we can also prove that:

$ \int_0^n \left( 1 \plus{} \frac {x}{n} \right)^n e^{ \minus{} 2x}dx \equal{} \left( \frac {e^{ \minus{} 2}}{2}\right)^n \sum_{k \equal{} 1}^{n}2^{2k \minus{} 1} \minus{} \left( 2^{ \minus{} n} \plus{} \frac {1}{4n}(1 \minus{} e^{ \minus{} 2n}) \right) \plus{} 1$

thans the limit are $ 1$

derekhohk: two things.

First, deal with the $ n$ in the limit of integration by saying it is
\[ \lim_{n\to\infty}\int_0^{\infty}\chi_{[0,n]}(x)\left(1 \plus{} \frac xn\right)^ne^{ \minus{} 2x}\,dx
\]
Second, we justify the the swap of limit and integral by either the DCT or the MCT. Can you either prove that the sequence is monotone, or find an appropriate dominating function?

[quote="Kent Merryfield"]
\[ \lim_{n\to\infty}\int_0^{\infty}\chi_{[0,n]}(x)\left(1 \plus{} \frac xn\right)^ne^{ \minus{} 2x}\,dx
\]
What is $ \chi_{[0,n]}$?

It's the function defined by $ \chi_{[0, n]}(x) \equal{} 1$ if $ x \in [0, n]$ and $ 0$ otherwise.  The greek letter chi for "characteristic function" (of the interval $ [0, n]$).

Completely unrelatedly, this is a terribly-named thread  :wink:

I go back and forth between using $ \chi$ and calling it the "characteristic function" versus using $ \mathbf{1}_{[0,n]}(x)$ and callling in the "indicator function."

Since this idea - a function that is $ 1$ on some set and zero off of it, or $ 1$ when some condition is true and zero when that condition is false - is incredibly useful in many branches of mathematics, including algebra and combinatorics, what name/notation do algebraists and combinatorists like?

As a (former) combinatorialist, I tend to use the notation promoted by Knuth, Graham, and Patashnik in [i]Concrete Mathematics[/i].  Namely, if $ b$ is a boolean (true or false), then $ [b]$ is 1 if $ b$ is true and $ 0$ otherwise.  One downside is that brackets have other meanings, but no notation is perfect.

So then the analyst's $ \chi_A(x)$ could be written as $ [x\in A]?$ Or $ \chi_{[0,1]}(x)=[0\le x\le 1]?$

I also have a specifically $ \text{\TeX}$ problem with $ \chi_A:$ the letter chi is too low, and the subscript doesn't really look like a subscript - that's one reason for using $ \mathbf{1}_A.$ Although I'm also not sure that \mathbf{1} is the right way to render that symbol.

[quote="Kent Merryfield"]So then the analyst's $ \chi_A(x)$ could be written as $ [x\in A]?$ Or $ \chi_{[0,1]}(x) \equal{} [0\le x\le 1]?$[/quote]
Yes.

In addition to the two notations you mentioned, I've also seen versions using delta (though maybe only used for the equality of several expressions, e.g., $ \delta_{m, n, k}$ as the indicator for $ m \equal{} n \equal{} k$) and square brackets $ [0 \leq x \leq n]$, though this last one I think is just atrocious for several reasons.  I personally tend to use $ \chi$, though I guess this might lead to notational collision with the representation-theoretic aspects of algebraic combinatorics.  Which, uh, does almost nothing to answer your question :).

Edit: and look at what happens when you go have dinner in the middle of posting a comment!  Since Ravi's already introduced the bracket notation, I might as well say what I don't like about it: first of all, nothing about it suggests what it actually is, while $ \chi$ and $ \delta$ at least have similar meanings in other contexts.  (The $ \bf 1$ notation clearly has a lot going for it here.)  Second of all, because square brackets have many other uses, it's not immediately clear when you run across it in an equation what's going on (though maybe if you go all the way through a text like [i]Concrete Mathematics[/i] or a course based on it you get fully used to it by the time you reach the end).  Third of all, I really like square brackets as a notation for coefficient-extraction from a generating function.  While this suffers from the first problem and somewhat from the second, there aren't any really good replacement notations that I've seen for it, while there are (as we've noted) several decent alternatives for indicator functions.  I think that this thread has now convinced me to use the bolded 1 in the future.

I will show that $ \lim_{n\to\infty}(\int_0^ne^{ \minus{} x}dx \minus{} \int^{n}_{0}(1 \plus{} \frac {x}{n})^{n}e^{ \minus{} 2x}dx) \equal{} 0$
i.e.
 $ \lim_{n\to\infty}(\int^{n}_{0}e^{ \minus{} x}[1 \minus{} (1 \plus{} \frac {x}{n})^{n}e^{ \minus{} x}]dx \equal{} 0$
Now, $ \frac {d}{dx}(1 \minus{} (1 \plus{} \frac {x}{n})^{n}e^{ \minus{} x}) \equal{} \frac {x}{n}e^{ \minus{} x}\left(1 \plus{} \frac {x}{n}\right)^{n \minus{} 1}$
In the closed interval $ [0,n]$ the function
$ f: x\mapsto e^{ \minus{} x}\left(1 \plus{} \frac {x}{n}\right)^{n \minus{} 1}$ is increasing, so it has a maximum for $ x \equal{} n$.

So $ 0 \le 1 \minus{} (1 \plus{} \frac {x}{n})^{n}e^{ \minus{} x}\le \int_0^x \frac {u}{n}e^{ \minus{} n}(1 \plus{} \frac {n}{n})^{n \minus{} 1}\ du \le \frac {x^22^{n \minus{} 1}}{2ne^n}$.
Now $ |\int^{n}_{0}e^{ \minus{} x}[1 \minus{} (1 \plus{} \frac {x}{n})^{n}e^{ \minus{} x}]dx| \le \int^{n}_{0}e^{ \minus{} x}\frac {x^22^{n \minus{} 1}}{2ne^n}\ dx \equal{} \frac {2^{n \minus{} 1}}{2ne^n}\int_0^nx^2e^{ \minus{} x}\ dx$
and this has limit 0 for $ n$ to $ \infty$.


A book has 500 pages numbered 1, 2, 3, and so on.  How many times does the digit 1 appear in the page numbers?

[hide="da answer"]
between...
1-100 there are 20
101-200 there are 117
201-300 there are 18
301-500 there are 36
that adds up to $191$
i probalby made a careless error though.[/hide]

[hide]there is 10-19, or 11 ones there (11 has 2 ones) and 1, 21, 31, ...91 has 9 ones, for a total of 20 ones from 1 to 99
from 100-199, there is another one in front of all the number from 1 to 99, so there are 100 extra ones, or 120 ones.
from 200 to 299, there are still only 20 ones, as with each hundred from 300 to 500.
120 + 4(20) = [color=blue]200 ones[/color].

I think that I made an error too for some reason.[/hide]


if we are bouncing a basket ball, what is the probability that it will bounce on the same point twice in a row.

i got some strange stuff because there are an infinite number of points and i guess the surface area might matter, im not sure.

but this is what i said;

[hide]there are an infinite number of points, that means we have an infinite number of choices for a point.  next, the probability of choosing that same point on the secon bounce would be 1/(infinity)

so that would mean the probability would be (infinity)[1/(infinity)]
that would either be 0, or some other number maybe 1 but im not sure if infinity is a number so its kind of hard to decide.[/hide]

Infinity is definitely not a number. If it were a number, then what is 1+infinity? However, I am not sure about your proof. You treat infinity like a number, and I'm not sure you can do that for certain problems to get the answer.

-rts2007

If there are an infinite number of points, and each is equally likely to be bounced on, then the probability of bouncing on any one is $ 1/\infty \equal{} 0$, and the probability of bouncing on a point once you bounced on it (the same thing as the probability that you bounce on a specific point, since the first bounce "specifies" the point) is also $ 0$.

This seems sortof strange, right?  Well the issue is that you are comparing a point to an infinite set of points - a whole surface.  You could choose points forever off this surface and never choose them all - in fact, you could never make anything surface with a non-zero area without an infinite number of points.

Now, if you were to change your question to say, what is the probability that if we pick two random points within $ 1$ unit of each other?  This would have a finite answer if the points are constrained to be in a specific polygon/circle/figure.

[quote="iYOA"]
[hide]there are an infinite number of points, that means we have an infinite number of choices for a point.  next, the probability of choosing that same point on the secon bounce would be 1/(infinity)

so that would mean the probability would be (infinity)[1/(infinity)]
[/hide][/quote]

Lets try this problem for 6 points.

There are 6 points, that means we have 6 choices for a point. next, the probability of choosing that same point on the second bounce would be 1/6

so that would mean the probability would be $ 6\frac{1}{6}\equal{}1$

Clearly there was an error in our methodology.

The proper way to do this problem would be to
A.) Choose any point, the probability of choosing that point again is 1/6, therefore the probability is 1 out of 6
or
B.) There are 6 choices for the first bounce, and 6 choices for the second bounce. There are 6 times when you choose both points at the same time. The probability is $ \frac {6}{6^2}\equal{}\frac{1}{6}$

In either case your answer should be 1/infinity, which I say is equal to 0.

The answer to this question depends on how you define "bounce" and "point."  In an idealized model where the ball moves perfectly up and down, it will of course bounce on the same point every time.  Similarly, in an idealized model where this doesn't occur and the surface on which the ball bounces is modeled by, say, the unit interval $ [0, 1]$, the probability that it bounces at [b]any[/b] given point is zero.  That's because this isn't a meaningful question to ask in this situation; the meaningful question is the probability that it bounces in, say, $ [4.99, 5.01]$.

There's a second subtlety here, which is that the "infinity" you're used to thinking of (say, the integers) is not the "infinity" of the unit interval $ [0, 1]$; in fact, the unit interval is much, much bigger, in a certain [url=http://en.wikipedia.org/wiki/Countable]precise sense[/url].

Here's a precise version:

[i]A basketball is called "$ \epsilon$-bouncing" if its location $ p(t) \in \mathbb{R}$ at time $ t$ changes according to the following rule:  at time $ t \plus{} 1$ the ball is at a point chosen uniformly in $ [p(t) \minus{} \epsilon, p(t) \plus{} \epsilon]$ for some given $ \epsilon > 0$.  Given another positive constant $ \delta > 0$, what is the probability at time $ t$ that the ball has remained in the interval $ [p(0) \minus{} \delta, p(0) \plus{} \delta]$ as a function of $ t, \epsilon, \delta$? [/i]

In the limit as $ t \to \infty$ this approaches a [url=http://en.wikipedia.org/wiki/Central_limit_theorem]normal distribution[/url], i.e. a Bell curve.  Here's a "discrete" version:

[i]A basketball bounces to a sequence of locations $ p(t) \in \mathbb{Z}$ according to the following rule:  at time $ t \plus{} 1$ the ball is either at $ p(t) \plus{} 1$ or at $ p(t) \minus{} 1$ with equal probability.  What is the probability at time $ t$ that the ball has returned to $ p(0)$?[/i]

This is an example of a [url=http://mathworld.wolfram.com/RandomWalk1-Dimensional.html]one-dimensional random walk[/url].


Let n>1 be an integer and let A1...An be real numbers such that A1 \leq A2 \leq .... \leq An

Prove that
                    (4n-4)A1An \leq (A1+....+An) 2

If  all Ai are negative
A1 \leq A2 \leq .... \leq An \leq 0
0 \geq A1+...+An \geq (n-1)A1+An in this case
((n-1)A1+An)^2 \geq (A1+...An)^2
Myth what is the temperatur in winter in winter ?

But, in that case, will the inequality manlio posted be always true? Maybe he meant to say positive numbers?

Moubinool, it is obvious we can WLOG suppose Ai \geq 0.

But how do you solve the problem if Ai <0? I don't think you can assume them to be >=0. Did you see what happens when they are <0? You get ((n-1)A1+An)^2 >= (A1+...An)^2, so you can't solve the problem like this: (A1+...+An)^2 >= ((n-1)A1+An)^2 >= 4(n-1)A1An  :? I'm getting a bit confused here... I'm sure you'll enlighten me, Myth.. :D  :D

My solution is like Myth's solution:

If 0 \leq A1 \leq A2 \leq ...

(An-(n-1)A1)^2  \geq 0 so (An-(n-1)A1)^2  \geq 4(n-1)A1An  so

(A1+...+An)^2 \geq (An+(n-1)A1)^2 \geq 4(n-1)A1An

If A1 \leq ... \leq An \leq 0

(A1-(n-1)An)^2 \geq 0   so  (A1+(n-1)An)^2 \geq 4(n-1)A1An  so

(A1+...+An)^2  \geq (A1+(n-1)An)^2  \geq 4(n-1)A1An

If A1<0 and An>0 the LHS<0 and RHS>0 and the result follows.

Oh, now I get it! I didn't see that at the beginning. So if they're negative you use A1+(n-1)An instead of (n-1)A1+An...Kind of stupid of me not to notice that :D  :D


Hello.  I am starting to think of majors that I would like to do in college, and I am unsure if I should focus on finance or engineering.  Could someone post the various carreers available for each option, and what kind of education is neccessary (i.e. computer-based, physics-based, 4-yr, 5-yr.).

Sorry if this is very broad.  Any help is appreciated.

Whatever you decide, do know that it's not final.  I majored in chemical engineering, but a couple internships convinced me that it wasn't for me.  So, I ended up in trading for four years.  And decided it wasn't for me.  The rest, I guess is history.

That said, it's much easier to get training in engineering (as long as it includes a decent programming background) and switch into finance and trading than it is to study finance and switch into engineering!  A great many of my Princeton classmates ultimately stapled an MBA to their engineering degrees (MIT has a very good program for this), and have been able to put their backgrounds in finance and engineering together in their careers.


This is a simple concept:

$1, 2, 4, 8, 16 .....$
This is a geometric sequence with the first term, $a =1, r =2$

so, if u have n number of squares on the board, the sum of all the rice is : $\frac{a(r^{n-1}-1)}{2-1} = 2^{n-1} - 1$

for any nth square, the number of rice is: $2^{n-1}$
 and since the sum of the rice on all the squares before (1,2,....n-1) is $2^{n-1} - 1$,
adding $1$ will give you number of rice on the nth square ($2^{n-1}$)

What about just the 1 2 4 = 7. 7+1=8 the concept works all the way through. Why? is there a name to concept. I'm sorry math basic here.
Any way to explain it without formulas?

This is a question that really belongs in the Getting Started forum.  If you happen to know how (in general) to add up a sequence like that, violetcraze's solution should be good enough.  If you don't, the nicest way to do this is with a proof by induction.

Here's a trickier way I just thought up:

Consider adding $1 + \frac12 + \frac14 + \ldots + \frac 1{2^{n-1}}$.  Each new term halves the distance from where we were to 2.  At first, the distance is 1, but we add $\frac12$ and so the distance becomes only $\frac12$.  Then we add $\frac 14$, so the distance is onyl $\frac 14$.  Then we add $\frac18$, and the distance shrinks to $\frac18$.  So each time we cut the distance in half, and the remaining distance is always equal to the last term we added.  Thus $(1 + \frac12) + \frac12 = 2$ and $(1 + \frac12 + \frac14) + \frac14 = 1$ and so on.  In the general case, that says $(1 + \frac12 + \frac14 + \ldots + \frac 1{2^{n-1}}) + \frac1{2^{n-1}} = 2$.  Now, if you multiply both sides of this equation through by $2^{n-1}$, you get on the right-hand side just $2^n$, while on the left-hand side you get $(2^{n - 1} + 2^{n - 2} + \ldots + 1) + 1$, and that's just your sum, plus one more.

[quote="meghanchev"]Thanks for the Help so far. Ok, with same stuff if the answer is 1024.  How would I in the simplest manner explain, to lets say a child, how I got to 1024 without using a formula or n power or square. I know this is below you folks but your help has been very appreciated.[/quote]

i don't understand what you mean by $1024$. If you're referring to the chess board question, the answer will never be even since the first term is $1$, and the successive terms are all even.

[quote="meghanchev"]I need help understanding.  The question posed to me was you have a chess board add a grain of rice to 1st square then double amout of rice to the rest of the squares how much rice is on board? 1 2 4 8 16 ect ect.  I got the answer using a calulator.  The interesting fact that was then told to me is if you add the rice on the previous squares then +1 you get the amount on the next square.  1 2 4 = 7. 7+1 =8 the amount on the 4th square. This pattern continues.  Why does this work? Why +1? Is this some type of math concept or theory? What is it called?[/quote]

Here's a really interesting extension of this pattern.

$9(1 + 10) + 1 = 100$
$9(1 + 10 + 100) + 1 = 1000$
$9(1 + 10 + 100 + 1000) + 1 = 10000$

Hopefully the reason why [b]this[/b] pattern works should be clear.

You can then apply this reasoning to powers of $2$ instead of powers of $10$, looking at it in binary!  :)

For an algebraic explanation,

$\frac{a^{n} - 1}{a - 1} = 1 + a + a^2 + ... + a^{n-1}$

(You can prove this using sum of a geometric series or you can just take it on faith.  :P)  So, setting $a = 2$, we get

$2^{n} - 1 = 1 + 2 + 4 + 8 + ... + 2^{n-1}$

:)

[b]Edit:[/b]  I am now going to attempt a completely intuitive explanation of this pattern!  :D

Specifically, let's work at it in reverse.  We have the number of grains of sand on the $n^{th}$ square, $2^n$, and we want to [b]subtract[/b] (rather than add) the previous squares.  Now, taking each square starting from the end,

$2^n - 2^{n-1} = 2^{n-1}$
$2^{n-1} - 2^{n-2} = 2^{n-2}$
...

And so forth, until

$2 - 1 = 1$

:)

That last one is actually the same thing I did, but slightly simpler.

The sum you're summing, where each term is a constant multiple of the previous term, is known as a geometric series.

[quote="emosbagels"]

Since there are 64 squares on a chess board, and as someone mentioned earlier, the number has to be odd, isn't the answer just

$2^{64}+1$[/quote]

The answer to what?

1)  The number of grains on the $64^{th}$ square is $2^{64}$.
2)  The total number of grains on the chess board is $2^{65} - 1$.


Asymptote
linear algebra
matrix
LaTeX
articles
trigonometry
vector

Hi.

I have asymptote figure that needs to go in a specific place on a page. The source looks like this:

[code]\noindent \#problem

(a) [blahblah]

Asymptote Figure 1

Explanation

(b) [blahblahblah]

Asymptote Figure 2

Explanation[/code]

But the output looks like this:

\[ \noindent \#problem \linebreak
\linebreak
(a) [blahblah]\linebreak
\linebreak
A. Fig. 1\linebreak
\linebreak
Exp. 1\linebreak
\linebreak
(b) [blahblah]\linebreak
\linebreak
Exp. 2\linebreak
\linebreak
\noindent \#nextProblem\linebreak
\linebreak
A. Fig. 2\linebreak
\linebreak\]

Which of course makes the assignment hard to read. I have tried both [code]\begin{figure}[h][/code] and [code]\begin{figure}[h!][/code] but I keep getting the same result. Any tips?

Thanks.

Floats can be difficult to deal with so could you give us a complete minimal example which would include the preamble (the stuff before \begin{document}) and some text and actual asymptote figures so that we can see where the problem lies?

No problem. Sorry. First Post.

Here is a trimmed down version of my LaTeX/Asymptote source up until the point where the problem occurs. I cut out a bunch of general text and all of my page setup and custom commands. For that reason I need to point out that \bfig and \efig are just \begin and \end figure as typical. '\basy'='\begin{asy}'. The \bsmat, \esmat, \bmat and \emat are shortcuts for matrix drawing commands. You can see I have two Asymptote drawings embedded in my document. I'm using the latest version of TeXnicCenter and have configured Asymptote to work in TC as specified in the AoPSWiki.

The problem I am having is precisely this: I want the figures to appear before the computation and explanation in parts (a) and (b) of #18. The drawing for (a) appears right where it's supposed to. However, (b) appears in the output PDF right after the prompt for #20 (the end of the document in this trimmed version). I have tried the options [h] and [h!] to prevent LaTeX from moving the second figure, but it just gives me a warning and says that [h] has been changed to [ht] (or [h!] to [ht!]). Here's the source:

[code]
\documentclass[11pt]{article}

\pagestyle{plain}

\usepackage[pdftex]{graphicx}
\usepackage{amsmath}
\usepackage{asymptote}
\usepackage{amsfonts}
\usepackage{amssymb}

\begin{document}

\noindent \#10: Is the projection matrix $P$ invertible? Why or why not?

[A paragraph of answer and computation]

\noindent \#12: Find the matrix that projects every point in the plane onto the line $x + 2y = 0$.

[Computation and answer]

\noindent \#18: Draw the projection of $b$ onto $a$ and also compute it from $p = \hat{x} a$.

(a) $b = \left[\bsmat \cos{\theta}\\\sin{\theta}\\ \esmat \right]$ and $a = \left[ \bsmat 1\\0\\ \esmat \right]$.

\bfig[h]
\basy
import math;
include graph;
size(3cm);
pair O=(0,0);
pair a=(1,0);
pair b=dir(50);
draw(O--a,EndArrow);
label("$\vec{a}$",a,S);
draw(O--b,EndArrow);
label("$\vec{b}$",b,NW);
pair P=extension(b,b+S,O,a);
draw(b--P,dotted);
draw("$\mbox{proj}_{\vec{a}} \vec{b}$",O--P,linewidth(0.75mm),Arrow(1.5mm));
\end{asy}
\efig

$$
p = \hat{x}a = \frac{a^T b}{a^T a}a = \frac{ \left[ \bsmat 1 & 0 & \\ \esmat \right] \left[ \bsmat \cos\theta \\ \sin\theta \\ \esmat \right] }{\left[ \bsmat 1 & 0 & \\ \esmat \right] \left[ \bsmat 1 \\ 0 \\ \esmat \right] } \bmat 1 \\ 0 \\ \emat = \bmat \cos\theta \\ 0 \\ \emat
$$

(b) $b = \left[ \bsmat 1 \\ 1 \\ \esmat \right]$ and $\left[ \bsmat 1 \\ -1 \\ \esmat \right]$.

\bfig[h]
\basy
import math;
include graph;
size(3cm);
pair b = (1,1);
pair a = (1,-1);
pair O = (0,0);
draw(O--a,EndArrow);
label("$\vec{a}$",a,S);
draw(O--b,EndArrow);
label("$\vec{b}$",b,N);
dot(O);
label("$\mbox{proj}_{\vec{a}} \vec{b}$",O,W);
\end{asy}
\efig

$$
p = \hat{x}a = \frac{ \left[ \bsmat 1 & -1 & \esmat \right] \left[ \bsmat 1 \\ 1 \\ \esmat \right] }{\left[ \bsmat 1 & -1 & \esmat \right] \left[ \bsmat 1 \\ -1 \\ \esmat \right] } \bmat 1 \\ -1 \\ \emat = \bmat 0 \\ 0 \\ \emat = \vec{0}
$$

The vectors are orthogonal, so the projection is $\vec{0}$.

\noindent \#20: Construct the projection matrices $P_1$ and $P_2$ onto the lines through the $a$'s in Problem 18. Is it true that $(P_1 + P_2)^2 = P_1 + P_2$? This \emph{would} be true if $P_1 P_2 = 0$.

\end{document}
[/code]

If it's no trouble, how would I center the figure after I get it on the right page? Right now all of my figures are left-justified.

I don't get a problem but that could be that I guessed at what \bsmat etc might mean. This is what I used
[code]\documentclass[11pt]{article}

\pagestyle{plain}

\usepackage[pdftex]{graphicx}
\usepackage{amsmath}
\usepackage{asymptote}
\usepackage{amsfonts}
\usepackage{amssymb}

\begin{document}
\newcommand{\esmat}{\end{pmatrix}}
\newcommand{\bsmat}{\begin{pmatrix}}
\newcommand{\emat}{\end{pmatrix}}
\newcommand{\bmat}{\begin{pmatrix}}

\noindent \#10: Is the projection matrix $P$ invertible? Why or why not?

[A paragraph of answer and computation]

\noindent \#12: Find the matrix that projects every point in the plane onto the line $x + 2y = 0$.

[Computation and answer]

\noindent \#18: Draw the projection of $b$ onto $a$ and also compute it from $p = \hat{x} a$.

(a) $b = \left[\bsmat \cos{\theta}\\\sin{\theta}\\ \esmat \right]$ and $a = \left[ \bsmat 1\\0\\ \esmat \right]$.

\begin{figure}[h]
\begin{center}
\begin{asy}
import math;
include graph;
size(3cm);
pair O=(0,0);
pair a=(1,0);
pair b=dir(50);
draw(O--a,EndArrow);
label("$\vec{a}$",a,S);
draw(O--b,EndArrow);
label("$\vec{b}$",b,NW);
pair P=extension(b,b+S,O,a);
draw(b--P,dotted);
draw("$\mbox{proj}_{\vec{a}} \vec{b}$",O--P,linewidth(0.75mm),Arrow(1.5mm));
\end{asy}
\end{center}
\end{figure}

$$
p = \hat{x}a = \frac{a^T b}{a^T a}a = \frac{ \left[ \bsmat 1 & 0 & \\ \esmat \right] \left[ \bsmat \cos\theta \\ \sin\theta \\ \esmat \right] }{\left[ \bsmat 1 & 0 & \\ \esmat \right] \left[ \bsmat 1 \\ 0 \\ \esmat \right] } \bmat 1 \\ 0 \\ \emat = \bmat \cos\theta \\ 0 \\ \emat
$$

(b) $b = \left[ \bsmat 1 \\ 1 \\ \esmat \right]$ and $\left[ \bsmat 1 \\ -1 \\ \esmat \right]$.

\begin{figure}[h]
\begin{center}
\begin{asy}
import math;
include graph;
size(3cm);
pair b = (1,1);
pair a = (1,-1);
pair O = (0,0);
draw(O--a,EndArrow);
label("$\vec{a}$",a,S);
draw(O--b,EndArrow);
label("$\vec{b}$",b,N);
dot(O);
label("$\mbox{proj}_{\vec{a}} \vec{b}$",O,W);
\end{asy}
\end{center}
\end{figure}

$$
p = \hat{x}a = \frac{ \left[ \bsmat 1 & -1 & \esmat \right] \left[ \bsmat 1 \\ 1 \\ \esmat \right] }{\left[ \bsmat 1 & -1 & \esmat \right] \left[ \bsmat 1 \\ -1 \\ \esmat \right] } \bmat 1 \\ -1 \\ \emat = \bmat 0 \\ 0 \\ \emat = \vec{0}
$$

The vectors are orthogonal, so the projection is $\vec{0}$.

\noindent \#20: Construct the projection matrices $P_1$ and $P_2$ onto the lines through the $a$'s in Problem 18. Is it true that $(P_1 + P_2)^2 = P_1 + P_2$? This \emph{would} be true if $P_1 P_2 = 0$.

\end{document} [/code]
The figures seem to be in the correct place and are centred. See attached pdf output. If your code is producing a different result perhaps you could supply code which I can compile immediately without having to guess at what some commands mean, as they could be causing the problem in your case.

Thank you very much. I can center my figures now. Unfortunately, TeXnicCenter and I had a disagreement about automatic saving, and so my source is lost. But I'll use the skeleton from this topic and remake it. I'll be back if it still doesn't work. Again, thank you.


I don't know how necessary this is but it does give some sense of context for where I am coming from sorry its so long but it's pasted from an old post I made in the college playground and I didn't have time to read through and find what could be eliminated.

Personality/Story/ Projected Major or Career Path

Alright so here is the thing...I am a junior in high school and I absolutely love math and science (specifically physics) I mean what can I say it is amazing, so beautiful,so elegant and its everywhere and everything. There is nothing quite like working on a really interesting and engrossing math problem. In school though I always found math class rather easy still I pressed forward accelerating through classes and what not thinking that once I was taking an advanced enough class the math would become more interesting. It didn't it was just pretty much the same I mean I learned stuff I didn't know before but it wasn't all that deep if you know what I mean, just regular old school math. Then I happened across math contests! Here were problems that were actually interesting and fun, problems that dealt not only with stuff like algebra but with lots of other really cool math, stuff like number theory and combinatorics etc. There was just one problem I sucked at contest problems. Even though I had always found math class easy, these new problems I encountered were unlike anything I had ever seen before. Had I not liked math I suppose I probably wouldn't have cared much, but as it was this really sucked for me being terrible at something that I enjoyed so much. So I decided to work on it and try and get better. So ya that's how I started doing the AMC and joined AOPS, the math team and what not. Ive learned a lot of cool new math from AOPS and gotten better. However, I am still bad at contests (one of my biggest problems being I always have trouble figuring how I should begin to approach a problem, of course now that I think about I am actually getting better at that).

As for what I wanna do in the future I know people change and everything so of course nothing is set in stone but I really want to study theoretical physics and mathematics and become a professor/researcher. I am also hoping to get good at math so that unlike the AMC and other my other high school contest experiences I can do well on the Putnam and other contests when I go to college.  Idk though I mean I try not to really let this worry me much but sometimes the thought does creep up on me and it's rather depressing. I mean I really love math and want to like be a mathematician (I mean seriously what could possibly be cooler than that) but sometimes the thought does creep up on me that based on my current performance on the AMC etc. it would seem I am not very good at math. Especially when I consider that many of the great mathematicians that I read about and admire apparently were quite good at this sort of thing. It's a problem cause I like math a lot it really is a very big part of who I am and how I see the world around me. A very very small part of me very rarely but sometimes wonders whether I would be better off not liking math then I wouldn't care or need to worry, but then again I can't imagine myself actually disliking something so beautiful.I just don't know how bad of a sign this is. Like I said it's quite depressing so I make it a point not to think about it but every once in awhile it does make me wonder just a little bit whether or not I can actually be successful in what I want to do. So what I am wondering is how much concern do you guys think this warrants?

Well first of all, if you wanted to shorten it, you could get rid of about half of the first paragraph.  Honestly, that is nothing to worry about.  Just because you don't make USAMO doesn't mean you can't go to a good school.  The thing with contests is that they can only help you.  If you do well in contests, it will certainly help, although it won't guarantee anything.  If you don't do well, so what.  If you express interest in math and do well in courses, that's what matters.

[quote]In school though I always found math class rather easy still I pressed forward accelerating through classes and what not thinking that once I was taking an advanced enough class the math would become more interesting. It didn't it was just pretty much the same I mean I learned stuff I didn't know before but it wasn't all that deep if you know what I mean, just regular old school math.[/quote]

I agree that high school courses, even more advanced high school courses, tend to be totally the same. But when it actually gets good is college! So if you just wait until college, you will probably find that it actually will become more interesting  :wink:

Ya I agree I think college will definitely be a lot better.

P.S I decided that this was indeed very long and also a lot easier to shorten than I had originally expected. Unfortunately it would no longer let me edit the post so I just cut it down and posted it anew.


[quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ \frac {a^2 \plus{} bc}{b \plus{} ca} \plus{} \frac {b^2 \plus{} ca}{c \plus{} ab} \plus{} \frac {c^2 \plus{} ab}{a \plus{} bc} \geq\ 3$
 :)[/quote]
It 's easily by AM-GM .

we have:

$ \sum\frac {3(a^2 \plus{} bc)}{b^2 \plus{} ab \plus{} bc \plus{} ca \plus{} 2ac}\geq\frac {3\sum(a^2 \plus{} bc)}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca} \equal{} 3$ :blush:


Ok, after I failed miserably in a relatively not too hard math contest, I came up with an idea to help me. Before executing that idea, I want your opinion to whether or not it would MAKE SENSE if I do it...

I planned to send e-mails to a local university's teachers, planning to ask them if they know any good math teacher who would be willing to help me through and through with math competitions... working with me and strengthen my weak points...unconditional. (My mom is having financial problems and I don't want to bother her further.)

I have been working independently for quite a long time and I found out recently that I did not progress. I know that most of you have no problem working on your own, but I do. Is it so wrong to be looking for a great math coach?

I have math teachers at my school but I'm not used to how they look at the solution and explain the solution to the problem without giving me any approaching thoughts that I found addictive. I think I got so used to wdragyn and joejia teaching me math one-on-one in the IS... I mean they are just wonderfully spendid perfect teachers... what are the chance of me finding a good teacher like them in my community?

thanks for your input!

Well certainly it would be great if you could find such a person, but these "personal trainers" are scarce and expensive. It is perfectly normal to want someone to coach you, you'll make much more progress that way. Best of luck!!! :)

Erm, mostly it's difficult to find a good teacher in person. My friend's dad is a math professor at the University of Delaware, but he's always busy. He's a good teacher at Olympiad stuff though. He did some of that when he was younger in China... but time is a major problem. If I have questions, I can call him, but that's about it. I think AOPS is my best tutor by FAR. :idea:

You know, you can do stuff online as well.  There are plenty of good olympiad students who would possibly be willing to help other students out.  However, I think your idea is great; just know that there is a good chance you won't get any useful feedback.  But depending on your community, it may be worth a try as you never know and the payoff would be good.


algebra
polynomial
floor function
modular arithmetic
number theory
relatively prime
number theory proposed

Let P(n) is a polynomial with integer coefficient and not constant  such that $ P(0) \equal{} 0$ . Let $ r$ be a positive real  such that for each pairs of integers $ (m,n)$ satisfy $ m|n$ then $ [mr]$ devides $ {P}([nr])$ where $ [x]$ denote the largest integer not great then $ x$ .Prove that $ r$ is an integer .

[quote="TTsphn"]Let P(n) is a polynomial with integer coefficient and not constant  such that $ P(0) \equal{} 0$ . Let $ r$ be a positive integer such that for each pairs of integers $ (m,n)$ satisfy $ m|n$ then $ [mr]$ devides $ {P}([nr])$ where $ [x]$ denote the largest integer not great then $ x$ .Prove that $ r$ is an integer .[/quote]
" Let $ r$ be a positive integer  ... Prove that $ r$ is an integer " :P :P :P

Are you sure $ P(0)\equal{}0$ is necessary?
[hide]Suppose $ r$ is not an integer.  Let $ n\equal{}mk$ for integer $ k$.  Also let $ \lfloor mr\rfloor\equal{}a$ so that $ mr\equal{}a\plus{}t$.  Choose $ m$ so that $ t$ is nonzero.  Notice that there are infinitely such $ m$ since $ r$ is not an integer.  Then $ \lfloor nr\rfloor\equal{}\lfloor mkr\rfloor\equal{}\lfloor ka \plus{} kt\rfloor\equal{}ka\plus{}\lfloor kt\rfloor$.  Choose $ k$ so that $ \lfloor kt\rfloor\equal{}1$.  Notice that this is always possible since $ 0<t<1$.  Then we have $ \lfloor nk\rfloor\equal{}ka\plus{}1$.

We are given that $ P(x)\equal{}x(c_px^p\plus{}c_{p\minus{}1}x^{p\minus{}1}\plus{}\cdots\plus{}c_0)$.  Since $ P(\lfloor mr\rfloor)|P(\lfloor nr\rfloor)$ we have $ P(a)|P(ka\plus{}1)$.  Obviously, $ a\not| ka\plus{}1$ so $ a|P(ka\plus{}1)/(ka\plus{}1)$.  Since $ ka\plus{}1\equiv1\pmod{a}$, this implies that $ c_p\plus{}c_{p\minus{}1}\plus{}\cdots\plus{}c_0\equiv0\pmod{a}$.  Since there are infinitely many $ m$, there are infinitely many $ a$, and thus there is not such polynomial.[/hide]

[quote="cosinator"]Are you sure $ P(0) \equal{} 0$ is necessary?
Suppose $ r$ is not an integer.  Let $ n \equal{} mk$ for integer $ k$.  Also let $ \lfloor mr\rfloor \equal{} a$ so that $ mr \equal{} a \plus{} t$.  Choose $ m$ so that $ t$ is nonzero.  Notice that there are infinitely such $ m$ since $ r$ is not an integer.  Then $ \lfloor nr\rfloor \equal{} \lfloor mkr\rfloor \equal{} \lfloor ka \plus{} kt\rfloor \equal{} ka \plus{} \lfloor kt\rfloor$.  Choose $ k$ so that $ \lfloor kt\rfloor \equal{} 1$.  Notice that this is always possible since $ 0 < t < 1$.  Then we have $ \lfloor nk\rfloor \equal{} ka \plus{} 1$.

We are given that $ P(x) \equal{} x(c_px^p \plus{} c_{p \minus{} 1}x^{p \minus{} 1} \plus{} \cdots \plus{} c_0)$.  Since $ P(\lfloor mr\rfloor)|P(\lfloor nr\rfloor)$ we have $ P(a)|P(ka \plus{} 1)$.  Obviously, $ a\not| ka \plus{} 1$ so $ a|P(ka \plus{} 1)/(ka \plus{} 1)$.  Since $ ka \plus{} 1\equiv1\pmod{a}$, this implies that $ c_p \plus{} c_{p \minus{} 1} \plus{} \cdots \plus{} c_0\equiv0\pmod{a}$.  Since there are infinitely many $ m$, there are infinitely many $ a$, and thus there is not such polynomial.[/quote]
Your solution doesn't complete because $ c_1\plus{}...\plus{}c_p$ may be equal 0 . Condition $ P(0)\equal{}0$ may be moved but it makes the converse problem still be true .

[quote="TTsphn"]Your solution doesn't complete because $ c_1 \plus{} ... \plus{} c_p$ may be equal 0 . Condition $ P(0) \equal{} 0$ may be moved but it makes the converse problem still be true .[/quote]

Ah, sorry.  To complete the proof, simply notice that we can make $ \lfloor nk\rfloor \equal{} ka \plus{} 2,ka \plus{} 3,...,ka \plus{} i,...$.  Choosing $ i$ so that infinitely many $ a$ are relatively prime to $ ka \plus{} i$ reveals that $ Q(x) \equal{} c_px^p \plus{} c_{p \minus{} 1}x^{p \minus{} 1} \plus{} \cdots \plus{} c_0$ must have an infinite amount of roots which is impossible.

My solution is quite same with your solution ,Cosinator . For more clearly ,we can divide in two case : the first when $ r$ is irrational ,and the other $ r$ is a positive rational but not an integer . This problem was taken from my school contest  ,in case $ P(x)\equal{}x^2\plus{}x$ .


real analysis
function
logarithms
real analysis unsolved

Suppose $ f_n$ are measurable functions on some measure space $ (X,A,µ)$, and

$ |\left\{ |f_n(x)| \geq \lambda \right\}| \leq C e^{\frac{-\lambda^2}{n}}$ for all $ \lambda > 0$.

Prove that

$ \limsup_n \frac{f_{2^n}}{\sqrt{2^n \log(\log(2^n))}} \leq 1, \ a.e.$


geometry
inequalities proposed
inequalities

let a,b,c are bases of triagle ABC,S is the area

prove that:

$ {x}^{2}$ $ \times$ a+$ {y}^{2}$ $ \times$ b+$ {z}^{2}$ $ \times$ c$ \ge$ 4$ \times$ S$ \times$ $ \sqrt{xy\plus{}yz\plus{}zx}$

[quote="trannguyenquyND_VN"]let a,b,c are bases of triagle ABC,S is the area

prove that:

$ {x}^{2}$ $ \times$ a+$ {y}^{2}$ $ \times$ b+$ {z}^{2}$ $ \times$ c$ \ge$ 4$ \times$ S$ \times$ $ \sqrt {xy \plus{} yz \plus{} zx}$[/quote]
must:$ xa^2\plus{}yb^2\plus{}zc^2\geq\ 4\sqrt{xy\plus{}yz\plus{}zx}S$
We have:
$ xa^2\plus{}yb^2\plus{}zc^2
 \equal{}x(a^2\minus{}b^2\minus{}c^2)\plus{}(x\plus{}y)b^2\plus{}(x\plus{}z)b^2
  \geq\ \minus{}2xbccosA\plus{}2\sqrt{(x\plus{}y)(x\plus{}z)}bc$
on the other hand
$ (xcos A\plus{}\sqrt{xy\plus{}yz\plus{}zx}sinA)^2\le\ (cos^2A\plus{}sin^2A)(x^2\plus{}xy\plus{}yz\plus{}zx)\equal{}(x\plus{}y)(x\plus{}z)$
THERE FORE $ xa^2\plus{}yb^2\plus{}zc^2\geq\ 2\sqrt{xy\plus{}yz\plus{}zx} bcsinA$
=>dpcm