trigonometry
geometry
geometric transformation
rotation
analytic geometry
3D geometry
sphere

How come current is induced? There is no flux change when it is in equilibrium so you have to run a current through a battery in the coils.

This is nice and interesting problem, I will let you try this till your next post and if you couldnt solve, i will reply.

[quote="Rajiv"]Click and zoom into the image to read your question.

[hide="Right Approach?"]
A priori, the induced current causes a magnetic dipole which is ultimately responsible for the torque and further calculations are simple.

But, do [b]we[/b] have to introduce a current or is the current induced by the field $ \vec B$? A friend of mine argues it is the field $ \vec B$. I argued that if the ball is in equilibrium, the current will stop. He feels the ball should momentarily stop, thats all that is required!

Please express your views. I'd also like to know if there's any other way to crack this problem and, of course, if this approach is correct.
[/hide][/quote]

[hide="My thoughts"]

To obtain static equilibrium:

\[ F_m \minus{} mg\sin\theta \equal{} 0\]

\[ \vec{F}_m \equal{} I\vec{\ell}\times\vec{B} \\ \Rightarrow F_m \equal{} I\cdot (N 2\pi R) B_z\sin\theta\]

*The factor of $ \sin\theta$ in the above comes from rotation of the system of coordinates, not the cross product.

\[ \boxed{\therefore I \equal{} \frac{mg}{N 2\pi R B_z}}\][/hide]

[quote="Immanuel Bonfils"]The force due an uniform magnetic field in a coil is null. It can cause onlly a torque.[/quote]

No, it isn't null. The current flows along the azimuthal direction. The B-field is pointed in the z-direction.

Therefore, the Lorentz force is given by $ d\vec{F} = dQ\, \vec{v}\times \vec{B}$

In a cylindrical coordinate frame that implies $ dF_r = dQ\, v_{\phi}\, B_z$.

Let $ \vec{v}=\frac{d\vec{\ell}}{d t}$.

Further let $ dQ = \frac{\partial Q}{\partial \vec{\ell}} d\vec{\ell}$ then ${ dQ\,\vec{v} =\frac{\partial Q}{\partial \vec{\ell}}\frac{d\vec{\ell}}{d t} d\vec{\ell} = I d\vec{\ell}}$

Thus one can also say $ d\vec{F} = I d\vec{\ell}\times \vec{B}$.

I second Immanuel Bonfils. We have to consider friction and balance its torque by the coil's torque. Also this friction balances mg along the slope.

[quote="Conjurer"]Want to recheck your above post  :P[/quote]
Oops! I have edited it! Thanks for pointing it out!

[quote="Rajiv"][quote="Conjurer"]We have to consider friction and balance its torque by the coil's torque. Also this friction balances mg along the slope.[/quote]

I'm of the same opinion. We should consider the torque caused by the magnetic dipole resulting out of the current through the coil. The torques due to $ mg$ and friction should cancel with the torque due to $ \vec B$ for rotational equilibrium. If the ball does not rotate, why would it move down the plane? Hence, complete equilibrium-[b]rotational[/b] as well as [b]translational[/b]![/quote]

There is a torque on the sphere resulting from the Lorentz force. I did not insinuate that there isn't. Nevertheless, I disagree with Immanuel where he says that the magnetic force on the coil (due to a uniform B-field) is "null" -- he's wrong. Furthermore, if that were the case then there would be no torque on the coil either. Since the torque (that it is produced by the magnetic dipole notwithstanding) is given by: $ \vec{\tau} \equal{}\vec{r}\times \vec{F}_m$.

One certainly could consider the complete set of equilibrium equations, translational and rotational, however I am not certain that it is necessary to do so in this case. The objective of the problem is to find the current through the coil necessary to stop the sphere from rolling down the plane. Said current is a single unknown quantity, thus if we write more than one equation our solutions of that system will either be redundant or inconsistent, i.e. the system will be over specified.

Also, you point out that if the ball does not rotate then it will not roll down the plane. However, it is also true that if the ball's center of mass does not translate, neither will it roll down the plane.

Finally, you are concerned about friction, but the problem does not mention it; the problem does say to show that the result does not depend on $ \theta$. Therefore, if you do consider friction then it seems to me that you would have to assume $ \mu \equal{}\tan\theta$, and that is going to introduce a dependence upon $ \theta$ in your solution, violating the problem statement.

[quote="Conjurer"]Want to recheck it again?  :P[/quote]

Now what? :mad:

[quote="Conjurer"]No torque due to mg. :P[/quote]

Ok...!

[quote="Immanuel Bonfils"]Dr.No

Did you ever heard of couple of forces, or magnetic dipole associated to a coil?[/quote]

Yes, I am. The torque on a coil is given by

\[ \vec{\tau} \equal{} \vec{\mu}\times\vec{B}\]

but it is also true that $ \vec{\tau} \equal{}\vec{r}_{\perp}\times\vec{F}_m$.

I have no problem with your calculating the current from the torque if that's the way you want to tackle this. My objection to your argument is based on your claim that the magnetic force on a coil due to a uniform B-field is null. That is clearly untrue.

The magnetic force on a wire loop in the presence of a z-directed uniform B-field is given by:

\[ \vec{F}_m \equal{}\int_{0}^{2\pi} I r_c\mathrm{d}\vec{\phi}\times B\,\hat{z} \equal{} I(2\pi r_c)B\,\hat{r}_c\]

The radius, $ r_c$ , is in cylindrical coordinates and $ \vec{r}_c \neq \vec{r}_s$, where $ r_s$ is the radius in spherical coordinates, and therefore $ \hat{r}_s\times\hat{r}_c \neq 0$. There will be a torque about the $ \phi$ axis, which is precisely what we expect.

[quote="Immanuel Bonfils"]
Did you ever heard of couple of forces, or magnetic dipole associated to a coil?

[/quote]
[color=green]Reference:[/color] [b]Section 30-5: A Current Carrying Coil as a Magnetic Dipole[/b] in [b]The Fundamentals of PHYSICS, Extended[/b] [i](Sixth Edition)[/i] by [b]Halliday / Resnick / Walker.[/b]

It says:
The torque $ \vec{\tau}$ acts on a current carrying coil placed in a magnetic field $ \vec B$ is given by:

$ \vec{\tau} \equal{} \vec{\mu} \times \vec B$

Where $ \vec{\mu}$ is the magnetic dipole moment of the coil and its magnitude is given by $ NiA$.

The direction of $ \vec{\mu}$ is given by the Right-Hand Rule.

There is no net force on the coil due to the magnetic field.

The PE of coil and magnetic field, $ \minus{}\mu.B$ is constant hence its derivative is also 0.

if the ques is not done yet,
just use $ \tau \equal{} NiBAsin\theta \equal{} \vec{r}  \times  \vec{F}$

$ F\equal{}mg$
$ r\equal{}2r$(diameter)
$ A\equal{}\pi r^2$

Clearly, $ sin \theta$ will cancel out and u'll get i :)

I have reflected upon this a great deal since yesterday. Having done so, I concede Immanuel's point. (It's not that the magnetic force on the coil is null but that the effect of that force is null. [I wasn't seeing that part.  :oops: ])

Considering a circular wire loop, there is a radial force on said loop due to the presence a z-directed B-field. However, in as much as the force is radial, the force pointed along any given radius is completely offset by the force pointed along the opposite radius. Hence there is no net force on the center of mass.

Notwithstanding, there is still a torque about the $ \phi$ axis of the sphere due to this radial force. Given that is the case, the most direct way of calculating that torque is
\[ \vec{\tau} =\vec{\mu}\times\vec{B}\]

I also agree with Rajiv. There is a torque on the sphere due to its weight. When this torque balances the torque due to the magnetic field then the sphere will no longer accelerate down the plane. (If it is already in motion then it will continue to roll at constant speed according to Newton's first law.)

The torque due to the sphere's weight is given by $ \tau = r_s mg\sin\theta$. (There is a component of the weight directed parallel to the face of the plane equal to $ mg\sin\theta$. The moment arm is the radius of the sphere.)

\[ \vec{\tau} = \vec{\mu}\times\vec{B}\implies NI\pi r_{s}^{2}B\hat{n}\times\hat{z}\]

When the plane of the coil is parallel to the face of the plane that the sphere is sitting upon then the angle between the normal vector and the z-direction is itself $ \theta$ and thus $ |\hat{n}\times\hat{z}| = |\sin\theta|$.

Therefore,

\begin{align*} NIB\pi r_{s}^{2}\sin\theta &= r_s mg\sin\theta \\ \implies I &=\frac{r_s mg}{NB\pi r_{s}^{2}} \\ \implies I &=\frac{mg}{NB\pi r_s}\end{align*}

\[ \boxed{I =\frac{mg}{NB\pi r_s}}\]

[quote="Conjurer"]No there is no torque due to mg. How can it ever be, since the sphere is rolling about its CoM.[/quote]

No, the sphere is rotating about its point of contact with the inclined plane. The tangential component of force $ mg\sin\theta$ acts through the center of mass giving this force a moment arm equal to $ r_s$.

Consider this, the only force other than the magnetic force acting on the sphere is gravity. If there were no torque on the sphere due to gravity, would there be any reason for it to roll? Would it not rather slide?

@Dr no-- I posted the exact thing just before u :)

@Conjurer -- i understand ur point-but there is no other way to do it :P

[quote="Conjurer"]
No there is no torque due to mg.

[/quote]

He meant no [b]effect due to[/b] torque by $ mg$.

[quote="Dr. No"]
No, the sphere is rotating about its point of contact with the inclined plane. The tangential component of force $ mg\sin\theta$ acts through the center of mass giving this force a moment arm equal to $ r_s$.

Consider this, the only force other than the magnetic force acting on the sphere is gravity. If there were no torque on the sphere due to gravity, would there be any reason for it to roll? Would it not rather slide?
[/quote]

There's no effect of the torque due to $ \mu mg\cos\theta$ since it is applied at the IAR and the torque due to the magnetic dipole tends to rotate it about the [b]geometric centre[/b] of the body.

[quote="Conjurer"]
I dont understand why you dont people dont consider friction here. Friction force will equal = mgsin@.

[/quote]

[i]We people[/i] (thta's the way [b]you[/b] put it!  :rotfl: )do consider friction, the only point I'd like to stress is that the friction causes no torque since it passes through the IAR having a moment arm = $ 0$.

The only torque, besides the one by the coil, is that due to $ mg\sin\theta$.

[quote="Conjurer"]
@Dr No. : Its rolling about its center of mass and doing pure rotation about point of contact.

[/quote]

Just as I said!

[quote="I"]
There's no effect of the torque due to friction since it is applied at the IAR and the torque due to the magnetic dipole tends to rotate it about the geometric centre of the body.

[/quote]

[quote="Conjurer"]I dont understand why you dont people dont consider friction here. Friction force will equal = mgsin@. So that the block is in rotational as well as translational eq.[/quote]
Nothing about friction is given in the problem statement. Therefore it is safe to assume that the author meant for us to neglect it. Incidentally, the force of friction is equal to $ mg\sin\theta$ only if the coefficient of friction is equal to $ \tan\theta$, and that is the case only if $ \theta$ is the [b]angle of repose[/b]. There is no indication in the problem statement that $ \theta$ is anything other than arbitrary.

[quote="Conjurer"]@Dr No. : Its rolling about its center of mass and doing pure rotation about point of contact.[/quote]
It is unclear how you're making this distinction. The center of rotation is the point of contact with the inclined plane, period.

[quote="Conjurer"]I dont understand why you dont people dont consider friction here. Friction force will equal = mgsin@. So that the block is in rotational as well as translational eq.

@Dr No. : Its rolling about its center of mass and doing pure rotation about point of contact.

@ritu: I guess this post would have sufficed to show there is a correct way.[/quote]

Well this a resnick ques(odd numbered :P )
i solved the same way s i did it hear and my answer matched with the ans given at back :)

Dr. No

If there isn't friction force, why doesn't the sphere slide down along the inclined plane ?

Rajid, Cojurer, rituraj007, etc.

Remember that the torque depends on the pole you choose; isn't that the origin of all this arguing ?

[quote="Immanuel Bonfils"]Dr. No

If there isn't friction force, why doesn't the sphere slide down along the inclined plane ?

[/quote]

It very well might do that, but if so, it can only do so under the influence of gravity.

Recall Newton's 1st law: [i]An object in motion tends to remain in motion and an object at rest tends to remain at rest until acted upon by an external force.[/i]

In this case, gravity is the only external force there is and it is gravity that we are attempting to counteract by applying a current to the coil wrapped around the sphere.

[quote="Immanuel Bonfils"][quote="Dr. No"]I have reflected upon this a great deal since yesterday. Having done so, I concede Immanuel's point. (It's not that the magnetic force on the coil is null but that the effect of that force is null. [I wasn't seeing that part.  :oops: ])

Considering a circular wire loop, there is a radial force on said loop due to the presence a z-directed B-field. However, in as much as the force is radial, the force pointed along any given radius is completely offset by the force pointed along the opposite radius. Hence there is no net force on the center of mass.

[/quote]

[quote="Dr. No"]It very well might do that, but if so, it can only do so under the influence of gravity.

Recall Newton's 1st law: An object in motion tends to remain in motion and an object at rest tends to remain at rest until acted upon by an external force.

In this case, gravity is the only external force there is and it is gravity that we are attempting to counteract by applying a current to the coil wrapped around the sphere.

[/quote]

I think I'll give up[/quote]

Why? This isn't that difficult. The external force causing the sphere to roll (or slide) down the plane is $ F \equal{} mg\sin\theta$.

The torque on the sphere created by said force is $ r F$.

To ensure rotational equilibrium $ r F \equal{} NIB \pi r^2\sin\theta$.

Therefore, $ F \equal{} NIB\pi r \sin\theta$. (And God said, "Let there be translational equilibrium.")

Hence, $ mg\sin\theta \equal{} NIB\pi r\sin\theta$. (And there was translational equilibrium, and He saw that it was good.)

It simply isn't necessary to consider friction.

As for the magnetic force, it is entirely radial. Therefore when the plane of the coil is parallel to the face of the inclined plane, there is a component of force pointed up the face of the inclined plane equal to $ \plus{}F_m$ and a component of force pointed down the face of the inclined plane equal to $ \minus{}F_m$. These cancel each other. Thus [b]gravity is the only external force acting on the sphere.[/b]

[quote="Immanuel Bonfils"]
Dr. No

If there isn't friction force, why doesn't the sphere slide down along the inclined plane ?

[/quote]

Friction is a force that opposes motion.

Here $ \mu mg\cos\theta$ is the limiting force, and friction force, $ f \le \mu mg\cos\theta$. Here, first of all, since the friction is applied at the Instantaneous Axis of Rotation, it produces no torque. That doesn't mean it has no effect on the sphere. It may cause linear acceleration although no angular acceleration. However, since the ball is in equilibrium, there is no fricition force acting on it.

For example, consider a block of mass $ m$ resting on a horizontal surface with co-efficient of friction $ \mu$. The block is at rest, hence no friction force acts on it. If any external force $ F$ acts on the block, it can move the block only if $ F\ge \mu mg$.

However, the friction force,

$ f \equal{} F; \; \; F<\mu mg
\\
\; \; \;\equal{}\mu mg\cos\theta ; \; \; F\ge\mu mg\cos\theta$

[quote="And you"]
[b]Rajid[/b], Cojurer, rituraj007, etc.

Remember that the torque depends on the pole you choose; isn't that the origin of all this arguing ?

[/quote]

My name is Rajiv. :mad:  :furious:

Rajiv, sorry for the mistake on the name: also Conjurer went wrong (something failed int he copy - paste operation).

Back to the friction force.
[quote="Rajiv"]
  However, since the ball is in equilibrium, there is no fricition force acting on it.
[/quote]

The ball is in equilibrium  just because the friction force is  [b]acting[/b] on it and balancing the effect of the tangential component of the weight!
Maybe your intention is to say that the friction force is not "working", i.e. doing a work ?...

[quote="Immanuel Bonfils"]
Maybe your intention is to say that the friction force is not "working", i.e. doing a work ?...

[/quote]
Yes, the friction does no work here-I meant that, sorry!

I just want to stress that friction does not contribute to rotational equilibrium.

[quote="Rajiv"]Friction is a force that opposes motion.[/quote]

and then (in the same post),

[quote="Rajiv"]... It may cause linear acceleration although no angular acceleration. ...[/quote]

Which is it, my friend, does friction oppose motion or cause it?

I'm not trying to be contrary here, but I would like to remind everyone: [b]The problem statement does not mention friction; therefore, it is safe to neglect it.[/b]

The objective is to determine the coil current necessary to nullify the effect of gravity -- that question has already been answered unambiguously.

[quote="Rajiv"]... It may cause linear acceleration although no angular acceleration. ...[/quote]

Hehe... :D  :oops: I meant retardation.

[quote="Dr. No"]
I'm not trying to be contrary here, but I would like to remind everyone: [b]The problem statement does not mention friction; therefore, it is safe to neglect it.[/b]

The objective is to determine the coil current necessary to nullify the effect of gravity -- that question has already been answered unambiguously.

[/quote]

I agree with Dr. No that we must neglect friction here since the problem makes no mention of it.

[quote="Immanuel Bonfils"]Absurd!
Wake up people!
I said already that I would give up, an Dr.No answered that I shouldn't, that it wasn't so difficult. Sure he misunderstood me; what is difficult is to argue on so contradictory statements. But I'll insist with all my patience, because it's impossibly you can't understand.

[b]The electromagnetic resultant force is null.

The Plane is inclined.

If there is no friction, what is the force that holds the sphere, against the tangential component of weight that would accelerate it downward the slope???   [/b][/quote]

Immanuel,

You're missing something fundamental.

The sum of tangential forces, those directed parallel to the plane, is given by:

\[ \sum F_t\colon F \minus{}mg\sin\theta \equal{} 0\]

Likewise the sum of (clockwise) torques is given by:
\[ \sum \tau_{cw}\colon r_sF \minus{} mgr_s\sin\theta \equal{} 0\]

But we know that    $ r_sF \equal{} NIB\pi r_s^2\sin\theta$.

$ \implies F \equal{} NIB\pi r_s\sin\theta$.

The ball neither translates nor rotates. Both equilibria are produced by the same current.

What is F ???!!! Sure it's not the electromagnetic resultant force (that you have already agreed that its zero).

And what is $ r_s$ ? Is it the sphere radius? Does it represent the moment arm of which forces wrt to cw ???

[quote="Immanuel Bonfils"]What is F ???!!! Sure it's not the electromagnetic resultant force (that you have already agreed that its zero).[/quote]

The [b]resultant[/b] magnetic force is zero parallel to the coil radius. (The force itself however is non-zero -- it is entirely radial in cylindrical coordinates.)

[quote="Immanuel Bonfils"]And what is $ r_s$ ? Is it the sphere radius? Does it represent the moment arm of which forces wrt to cw ???[/quote]

Yes, $ r_s$ is the radius of the sphere; $ r_c$ is the radius of the coil. In general, $ r_c \equal{} r_s\sin\beta$. I use $ \beta$ here to represent the vertical angle because $ \theta$ is already used for the angle of incline of the inclined plane.

\[ r_c \equal{} r_s\sin\beta \implies \hat{r}_c \equal{} \sin\beta \hat{r}_s \plus{} \cos\beta \hat{\beta}\]

It follows that $ \vec{F}_{r_c} \equal{}\vec{F}_{r_s}\sin{\beta} \plus{}\vec{F}_{\beta}\cos\beta$.

Then the torque due to the current in the coil in the presence of the B-field is given by:

\[ \vec{r}_s\times\vec{F}_{r_c} \equal{} r_sF_{\beta}\cos\beta \hat{\phi}\]

Intially $ \beta \equal{}\frac{\pi}{2} \minus{}\theta$ ; therefore, for rotational equilibrium:

\[ r_sF_{\beta}\cos(\frac{\pi}{2} \minus{}\theta) \minus{} mgr_s\sin\theta \equal{} 0 \\ \implies r_sF_{\beta}\sin\theta \equal{} mgr_s\sin\theta \\ \implies F_{\beta} \equal{} mg\]

When the sphere is on a level surface $ \beta \equal{}0$ and torque is given by $ r_sF_{\beta} \equal{} NIB\pi r_s^2 \implies F_{\beta} \equal{} NIB\pi r_s$.

Ultimately we have $ NIB\pi r_s \equal{} mg \implies I \equal{}\frac{mg}{NB\pi r_s}$.

Therefore provided there is a current $ I$, as defined above, the sphere will be in simultaneous translational and rotational equilibrium.

[quote="Immanuel Bonfils"][quote="Dr. No"]
The resultant magnetic force is zero parallel to the coil radius. (The force itself however is non-zero -- it is entirely radial in cylindrical coordinates.)[/quote]

Sorry but after that one I really give up![/quote]

I'm sorry this is so confusing for you.

It really should not be that difficult. Considering just the coil, in isolation apart from the sphere, in standard position, there is a force on the coil in the radial direction.

\[ \vec{F} = NI\vec{\ell}\times \vec{B} = NI(2\pi r_c)\hat{\phi}\times B\hat{z} = 2\pi r_c BNI\hat{r_c}\]

The question is does this force have any effect tangent to the plane?

\[ \left[\begin{array}{c}F_{x'} \\ F_{y'}\end{array}\right]=\left[\begin{array}{cc}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{array}\right]\left[\begin{array}{c}F_{x} \\ F_{y}\end{array}\right]\]

\[ \implies \left[\begin{array}{c}F_{x'} \\ F_{y'}\end{array}\right]=\left[\begin{array}{cc}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{array}\right]\left[\begin{array}{c}F_{r_c}\cos\phi \\ F_{r_c}\sin\phi\end{array}\right]\]

\[ \implies F_{x'} = F_{r_c}\left(\cos\phi \cos\theta + \sin\phi \sin\theta\right)\]

For any angle $ \phi$ you can choose you can also choose an angle $ \phi + \pi$.

Therefore you obtain both $ F_{x'}$ and $ -F_{x'}$; summing these, the total is zero.

The Lorentz force contributes nothing to keep the sphere from sliding or rolling.

Nonetheless, there is a torque on the sphere due to the magnetic moment.

\[ \mathbf{\tau} =\mathbf{\mu}\times\mathbf{B}\]

From which we find $ \tau = |\mu| |B| |\sin\beta|$.

The work-energy theorem gives us $ W = \int \vec{\tau}\cdot\mathrm{d}\beta = \mp |\mu\parallel{}B|\cos\beta + C = U_0 -U(\beta)$.

Letting $ C = U_0 \implies U(\beta) = \pm |\mu| |B|\cos\beta$

By definition the maximum potential occurs where $ \beta = \pi$ and that implies $ U(\beta) = -|\mu| |B|\cos\beta \implies U(\beta)\stackrel{\textrm{def}}= -\vec{\mu}\cdot\vec{B}$.

We now have, $ U(r_c,\beta) = -NI(\pi r_c^2)B\cos\beta$.

As you might know, minus the gradient of a potential function is a force.

The gradient of $ U(r_c,\beta)$ in polar coordinates:

\begin{align*}\nabla U(r_c,\beta) &= \frac{\partial U}{\partial r_c}\hat{r_c} +\frac{1}{r_c}\frac{\partial U}{\partial \beta}\hat{\beta} \\ \\ &= -NI(2\pi r_c)B\cos\beta\hat{r_c} + NI(\pi r_c)B\sin\beta \hat{\beta}\end{align*}

The total magnetic force vector is then
\[ \vec{F} = NI(2\pi r_c)B\cos\beta\hat{r_c} - NI(\pi r_c)B\sin\beta \hat{\beta}\]

Note that the $ r_c$ component is the Lorentz force multiplied by $ \cos\beta$.

Considering the $ \beta$ component of force,

\[ \hat{\beta} = \hat{x}\cos\phi\cos\beta +\hat{y}\sin\phi\cos\beta -\hat{z}\sin\beta\]

\begin{align*}\hat{x'}\cdot\hat{\beta} &= (\hat{r_c}\cos\theta +\hat{z}\sin\theta)\cdot(\hat{x}\cos\phi\cos\beta +\hat{y}\sin\phi\cos\beta -\hat{z}\sin\beta)\\ &=(\cos^2\phi\cos\beta\cos\theta +\sin^2\phi\cos\beta\cos\theta -\sin\beta\sin\theta)\\ &= (\cos\beta\cos\theta -\sin\beta\sin\theta) \\ &= \cos(\beta + \theta)\end{align*}

From the above it is evident that when $ \beta = -\theta$ then $ \hat{x'}\cdot\hat{\beta} = 1$ implying that $ \beta$ and $ x'$ -- the direction tangent to the surface of the plane -- are parallel.

This implies $ F_{\beta} - mg\sin\theta = 0$; the sphere is in translational equilibrium.

Multiplying through by $ r_s =r_c = r$ we obtain the equation for rotational equilibrium.

In any case, the current required to obtain this equilibrium is $ I =\frac{mg\sin\theta}{-NB\pi r\sin\beta} = \frac{mg\sin\theta}{NB\pi r\sin\theta} =\frac{mg}{NB\pi r}$.

\[ \boxed{I =\frac{mg}{NB\pi r}}\]

As has been demonstrated several times by now!


A 15-foot by 32-foot rectangular pool is surrounded by a 3-foot wide  walkway so that the pool and walkway make a larger rectangle.  How many 6-inch by 6-inch square tiles are required to pave this walkway?

A.318	B.636	C.954	D.1024	E. 1272

[hide=" click to reveal hidden content"]

$ 38\times21 \minus{}15 \times 32 \equal{} 318$

$ 318$ divided by $ \frac{1}{4}$ $ \equal{} 1272$.

[/hide]


vector
analytic geometry
Functional Analysis
advanced fields
advanced fields unsolved

Is it possible that infinite dimensional vector space be isomorphic to its proper subspace?

What about topological isomorphism (assuming our vector space to be normed)?

Sure. $\ell^{2}$ is isomorphic (actually, isometric) to its proper subspace given by $x_{1}=0$ (the isometry is just the coordinate shift). You can do this trick with many other standard spaces. Actually, it is much more interesting to ask if there exists an infinite dimensional Banach space that [b]is not[/b] isomorphic to any its proper subspace.

Of Course $\mathbb R[x] \simeq x\mathbb R[x]$ and also they have norm make them topologically isomorph.

Yes, just take your favorite \ell^p space, and then the shift operator gives you the desired isomorphism.

So what's an \ell^p space?  It's the space of sequences of real numbers with finite \ell^p norm.  The \ell^p norm of a sequence is the infinite sum of the p-th powers of the numbers in the sequence.

Oops, someone else already posted the same answer.

The answer turned out to be very easy. Thank you.


geometry
geometric transformation
reflection
trigonometry
circumcircle
geometry proposed

$ABC$ is a triangle. Reflect each vertex in the opposite side to get the triangle $A'B'C'$. Find a necessary and sufficient condition on $ABC$ for $A'B'C'$ to be equilateral.

[quote="N.T.TUAN"]$ ABC$ is a triangle. Reflect each vertex in the opposite side to get the triangle $ A'B'C'$. Find a necessary and sufficient condition on $ ABC$ for $ A'B'C'$ to be equilateral.[/quote]
If I understand "reflect" correct, it is just the same as letting $ G \equal{} O$, where $ G$ is the intersection of the medians, and $ O$ is origo. And then multiply every vertex with $ \minus{} 2$. (e.g. $ (2,\minus{}3)$ becomes $ (\minus{}4,6)$)  So the nescessary and sufficient condition is that $ ABC$ is equilateral.
Did I understand the problem correctly?

He meant $ A'$ is the image of $ A$ after the reflection about $ BC$, $ B'$ is the image of $ B$ after the reflection about $ CA$, $ C'$ is the image of $ C$ after the reflection about $ AB$.

There are other triangles except equilateral ones, for example, see [url=http://www.mathlinks.ro/viewtopic.php?t=229877]here[/url]

I think trig works nicely:

[color=blue]$ \Delta BAC\cong \Delta B'AC$ and similarly.
So $ \angle B'AC' \equal{} 3\angle A$
By cosine rule $ B'C'^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc\cos 3A$
$ \Delta A'B'C' \text{ equilateral}\iff$\[ b^2 \plus{} c^2 \minus{} 2bc\cos 3A \equal{} c^2 \plus{} a^2 \minus{} 2ca\cos 3B \equal{} a^2 \plus{} b^2 \minus{} 2ab\cos 3C\]\[ \iff a^2 \plus{} 2bc\cos 3A \equal{} b^2 \plus{} 2ca\cos 3B \equal{} c^2 \plus{} 2ab\cos 3C
\]$ 2bc\cos 3A \equal{} 2bc\cos A(1 \minus{} 4\sin^2 A) \equal{} 2bc \left(\frac {b^2 \plus{} c^2 \minus{} a^2}{2bc}\right)(1 \minus{} 4\sin^2 A) \equal{}$
$ \equal{} (b^2 \plus{} c^2 \minus{} a^2)(1 \minus{} 4\sin^2 A)$

Let $ \sin^2 A \equal{} x,\sin^2 B \equal{} y,\sin^2 C \equal{} z,R\equal{}\text{circumradius}$
By sine rule: $ a^2 \equal{} 4R^2x,b^2 \equal{} 4R^2y,c^2 \equal{} 4R^2z$
$ a^2 \plus{} 2bc\cos 3A \equal{} a^2 \plus{} (b^2 \plus{} c^2 \minus{} a^2)(1 \minus{} 4\sin^2 A) \equal{} 4R^2(x \plus{} (y \plus{} z \minus{} x)(1 \minus{} 4x))$
So we have now:\[ x \plus{} (y \plus{} z \minus{} x)(1 \minus{} 4x) \equal{} y \plus{} (z \plus{} x \minus{} y)(1 \minus{} 4y) \equal{} z \plus{} (x \plus{} y \minus{} z)(1 \minus{} 4z)\]$ x \plus{} (y \plus{} z \minus{} x)(1 \minus{} 4x) \equal{} y \plus{} (z \plus{} x \minus{} y)(1 \minus{} 4y)$
$ \iff (x \minus{} y)(1 \plus{} 4z \minus{} 4x \minus{} 4y) \equal{} 0$. So:\[ (x \minus{} y)(1 \plus{} 4z \minus{} 4x \minus{} 4y) \equal{} (y \minus{} z)(1 \plus{} 4x \minus{} 4y \minus{} 4z) \equal{} (z \minus{} x)(1 \plus{} 4y \minus{} 4z \minus{} 4x) \equal{} 0\]
If at least two of $ (x \minus{} y),(y \minus{} z),(z \minus{} x)$ are $ 0$ then $ \Delta ABC$ must be equilateral.
If none of $ (x \minus{} y),(y \minus{} z),(z \minus{} x)$ are $ 0$ then we have the linear system on $ x,y,z$:\[ 1 \plus{} 4z \minus{} 4x \minus{} 4y \equal{} 1 \plus{} 4x \minus{} 4y \minus{} 4z \equal{} 1 \plus{} 4y \minus{} 4z \minus{} 4x \equal{} 0\]This system implies $ x \equal{} y \equal{} z$. Contradiction.
If exactly one of $ (x \minus{} y),(y \minus{} z),(z \minus{} x)$ is $ 0$ then WLOG $ x \equal{} y$. We also have:\[ 1 \plus{} 4x \minus{} 4y \minus{} 4z \equal{} 1 \plus{} 4y \minus{} 4z \minus{} 4x \equal{} 0\]Which implies $ z \equal{} \frac {1}{4}\equal{}\sin^2 C\implies \sin C\equal{}\frac{1}{2}\implies \angle C\equal{}30^0\text{ or }150^0$. Also $ x\equal{}y\implies \angle A\equal{}\angle B$\[ \boxed{\Delta A'B'C'\text{ equilateral}\iff \Delta ABC\text{ is isosceles with angle between equal sides }30^0\text{ or }60^0\text{ or }150^0}\][/color]

[quote="gauravpatil"]it need not be isoscales at all
infact this is true for any triangle ABC
for instance cnsider a triangle A'B'C' (equalateral)
you consider angles 180-a,180-b, 180-c and (directed angles)
and then let the three arcs intersect at P draw perpendicular bisectors of A'P,B'P,C'P
this gives yo triangle ABC[/quote]
I don't understand a single thing you said :blush: Could you please explain it more for me? :blush:
I drew an arbitrary triangle ABC and found A'B'C' not equilateral, see figure below.
where should I consider the angles 180-a,180-b, 180-c? and what are a, b, c?
What do you mean by the three arcs? where are they?
How will giving me triangle ABC help? how should I know if it is arbitrary or conditioned? Wouldn't we need to consider an arbitrary triangle and then show A'B'C' equilateral to prove that ABC can be arbitrary?

Edit: It just occurred to me that I and you might not be interpreting the problem in the same way. I read the problem statement again several times and couldn't find an alternate interpretation :( .

Edit2: I again read your post thoroughly.

I think you thought the problem was this:
The image of a point P after reflection on the sides of a triangle ABC is A'B'C'. Find a necessary and sufficient condition on triangle ABC such that there exists a point P such that A'B'C' is equilateral.

But I don't think the problem said that. There's no such point P. It says reflect each [b]vertex[/b] in the opposite side to get ...
A' is the reflection of A itself on the side BC and similarly.

Angle BAC(i.e. the vertical angle of triangle ABC) = 30

And Yes The Triangle HAS to be isosceles with AB = AC.

i am really sorry my comment is a complete mess


function
induction
algebra proposed
algebra

Find all functions f defined in the Field of positive racional numbers such that:
1)       f(1/x)=f(x)
2)       xf(x)=(x+1)f(x-1)
3)       f(1)=1; ;)

I think there's no such function. Because the positive rational numbers form no field.  ;)

No just kidding, I've also got a proof: With 2) we get  by induction
\[ \forall x \in \mathbb{Q}_{>0} , n \in \mathbb{N}_{<x} : f(x)=\frac{x+1}{x-n+1} f(x-n)  \]
Especially for $n = x-1$ with 3)
\[ \forall x \in \mathbb{N}_{>0} : f(x)=\frac{x+1}{2}  \]
With these two formulas and 1), we can calculate each image. For example:
\begin{eqnarray*} f\left(\frac{7}{13}\right)&=&f\left(\frac{13}{7}\right)=\frac{20}{13} * f\left(\frac{6}{7}\right) = \frac{20}{13} * f\left(\frac{7}{6}\right) = \frac{20}{13} * \frac{13}{7}*f\left(\frac{1}{6}\right)\\ &=&\frac{20}{13} * \frac{13}{7}*f\left(6\right) = \frac{20}{13} * \frac{13}{7} * \frac{7}{2} = \frac{20}{2} \end{eqnarray*}
In general, it can be shown (by induction on residues)
\[ f\left(\frac{p}{q}\right)=\frac{p+q}{2}  \]
But that's the problem: The fraction is independant from representatives, whereas the image not. This is a contradiction. For example we have
\[ \frac{6}{2} = f\left(\frac{2}{4}\right)=f\left(\frac{1}{2}\right)=\frac{3}{2}  \]
This disproof shows even more: There's a unique function $f : \mathbb{N}_{>0} \times \mathbb{N}_{>0} \to \mathbb{Q}_{>0}$  satisfying the properties 1),2),3), or rather analogues ones. The operations on $\mathbb{N}_{>0} \times \mathbb{N}_{>0}$ are like the ones on $\mathbb{Q}_{>0}$.


trigonometry
limit
integration
function
real analysis
real analysis unsolved

I have a question my friend gave me; it's hard and I am stuck too:

Is the set {sinx, sin 2x, sin3x .... } precompact or compact in L(0, 1)? Precompact means
that this set has a subset which is Cauchy; compact means it is also closed (i.e. the
limit is in the above set. My guess is no. We were able to prove that the set is NOT precompact in C[0,1];
here we used the sup norm. (hence it is not compact either)

If we consider $ A = \{ \sin n x \, | \, n = 1, 2, 3, \cdots \}$ as a subset of $ L^1 (0, 2\pi)$, then $ A$ is not precompact. To show this, fix a positive integer $ m$ and consider $ \|\sin mx - \sin (m+n) x \|_{1}$. We claim that this values are eventually bounded below by a positive number independent of both $ m$ and $ n$. Before the proof, we introduce a lemma:

[b]Lemma)[/b] If $ f \in C[a, b]$, $ g \in L^1[a, b]$ and $ g(x+b-a) = g(x)$ a.e., then $ \lim_{n\to\infty} \int_{a}^{b} f(x)g(nx) \, dx = \left( \frac{1}{b-a} \int_{a}^{b} g(x) \, dx \right) \left( \int_{a}^{b} f \, dx \right)$.

The proof of this statement is dealt with for many times, so I will skip the proof. Returning to the proof, addition formula for sine yields

$ \|\sin mx - \sin (m+n) x \|_{1}$ $ = \int_{0}^{2\pi} |\sin mx - \sin (m+n) x | \, dx  \geq \int_{0}^{2\pi} |\sin mx| (1 - \cos nx) - |\cos mx| |\sin nx| \, dx$.

Taking $ \liminf_{n\to\infty}$ to both sides and applying the lemma, we have

\begin{eqnarray*}\liminf_{n\to\infty} \|\sin mx - \sin (m+n) x \|_{1}
& \geq & \int_{0}^{2\pi} |\sin mx| \, dx - \frac{2}{\pi} \int_{0}^{2\pi} |\cos mx| \, dx\\
& = & \frac{4(\pi - 2)}{\pi},
\end{eqnarray*}

which confirms our claim and hence completes the proof.

But I'm not sure if $ A$ is also not precompact in $ L^1 (0, 1)$.

That whole argument runs more simply by noting that the sequence is discrete in $ L^2$ ($ \int_0^{2\pi} (\sin(mx)\minus{}\sin(nx))^2\,dx\equal{}2\pi$), and that if $ |f(x)|\le M$ for all $ x$, $ \|f\|_1\equal{}\frac1{|I|}\int_I |f|\ge \frac1{|I|}\int_I \frac{|f|^2}{M}\equal{}\frac1{M|I|}\|f\|_2^2$. For uniformly bounded functions on a finite interval, all $ L^p$ norms, $ 1\le p<\infty$, induce the same topology.

On the smaller interval, just consider a subsequence $ \sin(knx)$ for some fixed $ k$ such that $ \sin(kx)$ has a full period inside the interval. This subsequence is discrete for the same reasons as before.


can you tell me what is the basic properties of the real numbers.(like :if $x,y,z$ are real numbers then $x+(y+z)=y+(x+z)$).
thanks. :D

Your example (the associative property of addition) is a field property.

Axioms for the real numbers can be split into three groups.

I. Field properties. I'll just mention these by name; you can probably write all of them out yourself just from hearing the names.

A field (such as the real numbers as one example) is closed under each of two arithmetic operations, called addition and subtraction. Addition is associative and commutative, there is an additive identity $(0)$, and each element has an additive inverse. Multiplication is associative, commutative, and distributes over addition. There is an multiplicative identity $(1)$, and every nonzero element has a multiplicative inverse. And $0\ne1.$

II. Order properties. There are several ways to present these; here's one possibility. There is a set $P$ (intuitively, the positive elements) with these properties:

For each $x$, exactly one of the following is true: $x\in P,\,-x\in P,$ or $x=0.$

If $x,y\in P$ then $x+y\in P$ and $xy\in P.$

Any set satisfying I and II is an ordered field. Every ordered field must at least contain a copy of the rational numbers, but there are many possiblitities, some of them fairly exotic.

III. Completeness.

There are several equivalent ways to express this. One of the more popular is the least upper bound axiom: every nonempty subset with an upper bound has a least upper boound.

There is (up to isomorphism of ordered field) only one complete ordered field. That is the real numbers.

If you are interested in such things, I would suggest you read [i]Intermediate Real Analysis[/i], published by Springer.  It's a great book on many great things.

Masoud Zargar


let $ x_1;x_2 ... x_n$ be n real numbers satisfying $ max(x_1;x_2;...;x_n) \minus{} min(x_1;x_2;...;x_n)\equal{}1$
and let $ y_i\equal{}\frac{x_1\plus{}x_2\plus{}...\plus{}x_i}{i}$ ,
find $ max(y_1;y_2;...;y_n)\minus{}min(y_1;y_2;...;y_n)$

Do you mean you want the largest value of $ \text{max}(y_i) \minus{} \text{min}(y_i)$?

in the sequence $ y_1;y_2;...y_n$
first you chose the largest value , let it be $ y_k$
and then you chose the smallest value and let it be $ y_t$
and we have to find out the value of $ y_k\minus{}y_t$


calculus
integration
trigonometry
algebra
polynomial
calculus computations

Could I get some assistance on how to compute the following integral:

$ \int tan^5xsec^3x dx$

Thanks in advance.

Well the answer is $ \frac{\sec^{7}x}{7}\minus{}\frac{2\sec^{5}x}{5}\plus{}\frac{\sec^{3}x}{3}\plus{}C$.  Now you can try to get there.  Some possible suggestions (some may help, some may not): u-substitutions, integration by parts, trig identities.  And just for future reference, when you use TeX, you can clean up your trig expressions by adding \ in front of the expression.  So instead of $ sin$, you get $ \sin$.

Thanks very much.  I was hoping to get a push in the right direction however.
I am fairly certain that a u-substitution is required.
Do I let $ u\equal{}\tan x$ or let $ u\equal{}\sec x$?

Thank you kindly.

As it is, u-substitution won't work because you need the difference in powers to be 1, which you don't have.

An even power of $ \tan$ could be turned algebraically into an even power of $ \sec$ and vice versa, from the Pythagorean identity $ \sec^2x\equal{}\tan^2x\plus{}1.$ But trying to do that with odd powers would introduce square roots, which we'd rather not do.

Instead of asking what to make $ u,$ let's ask what to make $ du.$ If we let $ du\equal{}\sec x\tan x\,dx,$ then we'd pull that out and leave behind even powers of both $ \sec x$ and $ \tan x.$ That would be the substitution $ u\equal{}\sec x,$ we we try to turn all the even powers into secants. So we try that:

$ \int\tan^5x\sec^3x\,dx\equal{}\int\tan^4x\sec^2x\sec x\tan x\,dx$

$ \equal{}\int(\sec^2x\minus{}1)^2\sec^2x\sec x\tan x\,dx$

$ \equal{}\int(u^2\minus{}1)^2u^2\,du$

Now we've got polynomials; it should be clear how to get from here to the result JRav quoted.

I really appreciate your help.  I get the overall picture of how to do these questions, but just to get started and to know to use $ \sec^2 x \equal{} \tan^2 \plus{}1$ in solving, I have a hard time.
To continue where Kent left off:

$ \equal{} \int(u^4\minus{}2u^2\plus{}1)u^2 du$

$ \equal{} \int(u^6\minus{}3u^4\plus{}u^2) du$

$ \equal{} \frac {u^7}{7}\minus{}\frac{2u^5}{5}\plus{}\frac{u^3}{3}\plus{}C$

$ \equal{} \frac{\sec^7 x}{7}\minus{}\frac{2\sec^5x}{5}\plus{}\frac{\sec^3 x}{3}\plus{}C$

Thanks again.


I don't know if it was posted before or well-known for somebody.
Prove that number of ordered pairs (x,y) which are solutions to the equation
x2 + y2 = n is r(n) = 4(d1 - d2).
where d1, d2 reprezents the numbers of divisors of the form 4k+1 and 4k+3,
respectively. Is it right ?

I think we can consider that there are no divisors of the form $4k+3$ since we can eliminate them and the right factor doesn't change, while the left one doesn't gain or loose any solutions because it is well known that if $4k+3=p|x^2+y^2$ and p is prime then $p|x$ and $p|y$. Now, if n is a prime number of the form $4k+1$, your equation has 2 solutions, $(x,y)$ and $(y,x)$. Maybe the factor $4$ should be $2$?

No, I think this is it if we also consider negative solutions. For example, if $n=5$, the solutions we consider are $(1,2),(-1,2),(1,-2),(-1,-2),(2,1),(-2,1),(2,-1),(-2,-1)$ (the divisors of the form $4k+1$ are $1,5$, so the formula gives $4(2-0)=8$).


If you don't get accepted through early application an test A, can you do test B and apply again through normal application?

You certainly may, but we'd encourage you to put a complete effort into your first application so that this is unnecessary. If, for some reason, you believe that your second application will be considerably better than your first (for example, if you were struggling to meet the deadline and had very little time to work on Test A or your essays), then we invite a resubmission. However, a second application is subject to a second application fee, so it's better to only have to send one.

You're also welcome to retake the Admission Test (visit [url]http://www.awesomemath.org/application.shtml[/url] for more information about that.)

But it's in everyone's best interest to submit the best application you can the first time.


Find the value of $ a$ such that the planes $ ax\plus{}y\plus{}z\equal{}0$, $ x\plus{}3z\equal{}0$, and $ 5y\plus{}6z\equal{}0$ have a line in common.

hello
$ x\plus{}3z\plus{}\alpha(5y\plus{}6z)\equal{}0$ will become $ ax\plus{}y\plus{}z\equal{}0$ when
$ \frac{5\alpha}{1}\equal{}\frac{3\plus{}6\alpha}{1}$
hence $ \alpha\equal{}\minus{}3$
now
$ \frac{a}{1}\equal{}\frac{1}{5\alpha}$
or $ a\equal{}\frac{\minus{}1}{15}$
thank u


Is there a better way to  solve this problem other than trying all the combinations?

Use grouping symbols to make each equation true.

5.6 + 8 ÷ 4 • 2 = 7

Wait, is the 5. just the number of the problem? In other words, is the real problem 6+8/2*4=7?
Well, I don't know what other way for complex ones, but this seems pretty easy...
[hide]$ 6+\frac{8}{(2*4)}$[/hide]

Lol.. that might have been my problem. I thought that the five was part of the question.  :blush: If it isn't though, it would be quite easy. Thanks!

Dohhh lol

anyways i gots the answer:
[hide="answer to da problam"]
\[ 6+[8/(2*4)]=7\]
[/hide]


[quote="GameBot"]140 teachers in a school system are having a meeting at which they will divide into discussion groups. The number of teachers in each group is equal, with no more than 10 teachers to a group. Given that each group must have more than one member, how many possible group sizes are there?[/quote]

Factors of 140 less than or equal to 10:

1,2,4,5,7,10. so 6 possibilities.

Sorry if this is "reviving", but I thought I would point something out.
[quote]Given that each group must have [b]more[/b] than one member[/quote]
This means that the answer is actually $ 5$.


I need help on this problem. I got 2 of the 5 answers they wanted but I'm not sure how to find the other 3.

[b]A triangle is called Heronian if its sides and area are integers. Determine all five Heronian triangles whose perimeter is numerically the same as its area.[/b]

[hide="My_Answer"]
5,12,13
6,8,10

But that's found by guess and check on right triangles.[/hide]

[hide="My_Algebraic_Attempt"]
Since this problem never said about right triangle, I decided to apply the Heron's Formula.

So, the question is saying:

$a+b+c = \sqrt {s(s-a)(s-b)(s-c)}$ Then:

$2s = \sqrt {s(s-a)(s-b)(s-c)}\\
4s^2 = s(s-a)(s-b)(s-c)\\
4s^2 = (s^2-as)(s^2-bs-cs+bc)$

But as you can see, this works out quite nasty. I know that it can't be huge number of a,b,c because then area would be a lot bigger than the perimeter for sure.

Is there any slick way to do this or solution or anything? The site that I got the problem from doesn't have answer nor solution with it.[/hide]

Thanks.

Clearly, any triangle (with integer side lengths) with area equal to perimeter will be Heronian. Now, with $s, a, b, c$ standing for the semiperimeter and the three side lengths, make the following substitutions: $s-a = x, s-b = y, s-c = z$. We know $x, y, z$ are positive from the triangle inequality. Now, the constraint that the area is equal to the perimeter can be written (through Hero's formula) as follows: $2s = \sqrt{sxyz} \Rightarrow 4s = xyz$. Now, at the moment, we don't know if $s,x,y,z$ are integers. It is possible that $s$ is half an integer, in which case $x,y,z$ would all be half-integers. However, in the equation $4s=xyz$ we know the left-hand side must be an integer as $4s = 2(a+b+c)$, so if $s$ were not an integer, neither would $x,y,z$ be, so the right-hand side could not be an integer. Thus $s,x,y,z$ are all integers. Now, note that $s = x+y+z$, so we substitute that in to find that we must have $4(x+y+z) = xyz, x,y,z \in \mathbb{Z}^+$. This is symmetric in $x,y,z$ so let us assume without loss of generality that $x \ge y \ge z$. Then $xyz = 4(x+y+z) \le 4(x+x+x) = 12x \Rightarrow yz \le 12$. We could do some further analysis, such as noting that at least one of $x,y,z$ must be even, but at this point, it's easy enough just to try all possible pairs $(y,z)$ and see which of them yield integral $x = \frac{4(y+z)}{yz-4}$. We find the triples $(x,y,z) = (24,5,1), (14,6,1), (9,8,1), (10,3,2), (6,4,2)$. Now, to find $(a,b,c)$ from these, we have the formulas $a = y + z, b = x + z, c = x + y$. Thus these five triples of $(x,y,z)$ yield the following triangles: $(29,25,6), (20,15,7), (17,10,9), (13,12,5), (10,8,6)$. Incidentally, it appears that only the last two are right triangles.


function logarithms algebra polynomial calculus derivative integration

For every complex number $z$ different from 0 and 1 we define the following function
\[  f(z) := \sum \frac 1 {  \log^4 z } \]
where the sum is over all branches of the complex logarithm.

a) Prove that there are two polynomials $P$ and $Q$ such that $f(z) = \displaystyle \frac {P(z)}{Q(z)} $ for all $z\in\mathbb{C}-\{0,1\}$.

b) Prove that for all $z\in \mathbb{C}-\{0,1\}$ we have

\[ f(z) = \frac { z^3+4z^2+z}{6(z-1)^4}. \]

[quote="Valentin Vornicu"]where the sum is over all branches of the complex logarithm. [/quote]
Maybe a stupid question, but what does that mean?

I guess you need university vocabulary & knowledge to understand these questions :D

it just means all the values the logarithm takes. actually, since the exponential function is periodic with period $2\pi i$, the value of the logarithm is only defined up to a natural multiple of $2\pi i$. now, the sum is taken over all possible values of the logarithm. this explains why the value of the resulting function is actually something quite natural.

Peter

yeah...beautiful problem. having read some real good maths books it is "straightforward" to proof, but anyway, it is so nice how one can derive b) directly from a)...

Peter

Hi guys,

Having read a little bit of complex analysis myself, I will atempt to do something fairly reasonable .  consider first the function g(z)=f(z)*(z-1)^4.  now, as was pointed out before, the valuse of ln(z) is periodic with period 2pi*i.    Therefoe, we can easily prove convergence for g(z) when z is not equal to 0 by comparing to 1/z^4, for example. consider now (z-1)^4/ln(z)^4 as z-->1, in the branch where
ln(1)=0. Using l'hpoitals rule 4 times, we see that the limit is 1. Since the value at all other branches clearly converge as does their sum, we have that limg(z) as
 z-->1 exists. Therefore, g(z) is analytic over the entire complex plane. Also, consider |g(z)| as z-->infinity.
re(ln(z))=ln(|z|))-->inf as z-->infinity, and since ln(z) values are periodic with period 2pi, we can check that |f(z)|~|SUM(-inf<x<inf){1/(ln(|z|+2pi*i*x)^4}|
<SUM(inf<x<inf)1/|(1+2pi*i*x)|^4=k for some fixed constant k, and we can easily check that the final sum converges. in particular, f(z) is bounded as z-->inf. so g(z) grows asymptotically with at most order z^4 as z-->inf. Hence, since g(z) is also an entire function, g(z) must be a polynomial. This proves part (a). YAY! now, lets move on to part (b). We have that g(z) is a polynomial with order at most 4. Now, we have firstly proved that g(1)=1 earlier in the argument.    now, consider g(0) we know that ln(z)-->infinity faster than z times any constant.(in any branch)using, for instance, l'hopitals rule. Hence, we have easily that g(0)=0. We can calculate g(-1) using the infinite sum of -4th powers(well known fact, easily derivable from the taylor series of sin(x)). Now, we can notice that f(i) and f(-i) is just 16 times f(-1), and so we can get g(i) and g(-i) as well(we can actually calculate them, and i would show the details if i wasnt so lazy). So we have 5 values now, determinintg uniquely our polynomial of degree 4. There. That wasnt so bad, now was it?

hi, sorry for the double post, firstly.

Second, i just forgot to mention that to establiush the analyticity of g(z) for z=0, we have to show that g(z)=0 as z-->0, which is trivial cause (z-->0)ln(z)-->-inf faster than any power of z times any constant.

yeah that's about the same i wanted to do. i love proofs like that one :D

Peter

[quote="Sasha"][quote="Jacob Tsimerman"]Hence, since g(z) is also an entire function, g(z) must be a polynomial. [/quote]
?????[/quote]When (and where) did Tsimerman write this? :)

Hi guys,

I wrote that sice g(z) is an entire function that grows asymptotically no faster than z^4, it must be a polynomial.  This is actually a well known fact in complex analysis. To see this, consider the 5'th derivative of g(z). We have th weel known formula \[ g''''(z)=\frac {5!}{2\pi}\int \frac {g(w)}{(w-z)^6}\ dw\]  where the integral is taken over any contour around the point z. However, g(w)<w^5 for |w| big enougg, so by taking the contour to be side enough, we canbound the integral as small as we desire. so g'''''(z)=0. Hence,g(z) is a polynomial of degree at most 4.

Can you give more information about this and related formulas, please :).

Try googling "Cauchy integral formula". [url=http://mathworld.wolfram.com/CauchyIntegralFormula.html]Here[/url]'s a link.


I'm not sure what you did ducky, but I think typo is correct.  Here's the explanation on why it is.  We split it up into two cases:
CASE 1
They're both red
The probability is $ \frac {4}{10}\cdot\frac {3}{9} \equal{} \frac {12}{90}$
CASE 2
They're both blue
The probability is $ \frac {6}{10}\cdot\frac {5}{9} \equal{} \frac {30}{90}$
Now we add then together.  We get $ \frac {42}{90} \equal{} \boxed{\frac {7}{15}}$

I think ducky did only the red marbles.

Anyways, $ \frac{7}{15}$ should be the correct answer.

For some reason, I thought you had to multiply the two cases, not add them.

I just added the red and the blue. We have $\frac{30}{90}+\frac{12}{90}=\frac{7}{15}$

Wow... $11$ years old.

[quote=myyellowducky82][quote="GameBot"]A bag has 4 red and 6 blue marbles.  A marble is selected and not replaced, then a second is selected.  What is the probability that both are the same color?[/quote]
I got $ \frac{2}{15}$...[/quote]

The official answer is 7/15

[hide=Solution]
We have $2$ draws from $4+6=10$ marbles. Since the marbles chosen aren't put back into the bag, we have $10\cdot9=90$ ways for the total.

If we have both marbles red, we have $4\cdot3=12$ successes.

If we have both marbles blue, we have $6\cdot5=30$ successes.

So, our probability is $\frac{12+30}{90}=\frac{42}{90}=\frac{7}{15}.$

[/hide]

[hide=wrong sol]so the number of ways to pick two different colors is $\frac{4\times6}{10\times9}=\frac{24}{90}=\frac{4}{15}.$ there are two ways to do this, so we multiply by $2$ to get $\frac{8}{15}.$ we can only pick two different colors or the same color, so the probability is $1-\frac{8}{15}=\boxed{\frac{7}{15}}$[/hide]

thx fruitmonster

[quote=Lamboreghini][hide=wrong sol]so the number of ways to pick two different colors is $\frac{4\times6}{10\times9}=\frac{24}{90}=\frac{4}{15}.$ we can only pick two different colors or the same color, so the probability is $1-\frac{4}{15}=\boxed{\frac{11}{15}}$[/hide]

what did i do wrong

halp[/quote]
There are $2$ ways to pick different marbles: Red, Blue or Blue, Red.


[quote="albertfermat"]Does anyone know how to type in PDF forms. I am applying to a bunch of competitions, and they all have PDF forms. They mention that you can print them and fill them out by hand or type. However, I do not know how to type on PDFs. Is this possible??[/quote]
It's possible to make PDF forms that have "clickable" fields for text, but I've only ever seen one and it didn't work properly. Needless to say, I don't know how to do it. However, what they mean when they say "type" is "print this page and use a typewriter" or "print this page and figure out how to make everything line up when you run it through the printer again (ah, the nightmares!)". They don't mean that you can actually type it right on that form from Adobe Reader.

You might also be able to paste it onto Word and then position text boxes where you need to type.

PDF forms can be filled out with both Adobe Reader and Adobe Acrobat. However, only Acrobat can make and save PDF forms. If you use Reader, you must fill out the form in one session; the filled-out information is lost when you reopen the document.


inequalities
calculus
inequalities solved

Let x_2,x_3,..,x_n be positive real numbers and s=x_2+x_3+...+x_n. Show that

sum(k=2,..,n)x_k^(1-1/k) <s + 2 \sqrt s

I think this is a very hard inequality.

It has been proposed in our french 'Concours gnral' in 1989.

Pierre.

Yes, but I have a non-strict inequality. I did not check if we can manage to have a strict one.

Pierre.

Ok, so let's post...

Let x_1 be any positive real number.
Let S = s + x_1.
Let s' denotes the LHS of the desired inequality, and S' = s' + (x_1) 1-1/1  = s' + 1.

First, we prove the following lemma :
For each   t > 1, we have S' <  t*S +  t/(t - 1).

Proof.
Let  t > 1. Note that  t /(t - 1) =  \sum (k = 0 to +oo) 1/t  k  >  \sum (k = 0 to n-1) 1/ t k .
Then, the lemma will be proved if (x_k) 1-1/k  \leq  t*x_k + 1/t k-1  for x_k > 0, and k = 1,...,n.
For k = 1, the inequality is trivial. So let k > 1.
Let f(x) = x 1-1/k  - t*x. It is easy, using calculus, to verify that f reaches its maximum at x = x_0 = ((k-1)/k) k /t k-1 , and that f(x_0) = (k-1) k-1 /(k k t k-1 ).
Since, for k  \geq 2, we have (k-1) k-1  < k k , we deduce that f(x)  \leq 1/t k-1  for all x > 0, and we are done.

Now, let's solve the problem :
Let g(t) = t*S + t/(t-1). Clearly, for each t > 1, we have g'(t) = S - 1/(t-1) 2 . It follows that g reaches its minimum for t = 1 + 1/ \sqrt s, and that this minimum is  (\sqrt S + 1) 2.
Fromm the lemma, we deduce that
S' < g(1 + 1/ \sqrt S ) = ( \sqrt S + 1) 2 .
That is :
s' + 1 < s + x_1 + 2 \sqrt (s + x_1) + 1
then : s' < s + 2 \sqrt (s + x_1) + x_1.
This holds for all x_1 > 0. So taking the limit as x_1 tends to 0, we get :
s'  \leq  s + 2 \sqrt s , as desired.

Pierre.


search
number theory proposed
number theory

Let $ a_0,a_1,\dots,a_{n \plus{} 1}$ be natural numbers such that $ a_0 \equal{} a_{n \plus{} 1} \equal{} 1$, $ a_i>1$ for all $ 1\leq i \leq n$, and for each $ 1\leq j\leq n$, $ a_i|a_{i \minus{} 1} \plus{} a_{i \plus{} 1}$. Prove that there exist one $ 2$ in the sequence.

Let $ a_{i\minus{}1}\plus{}a_{i\plus{}1}\equal{}(k_i\plus{}1)a_i, k_i\ge 0$, then $ 2\equal{}\sum_{i\equal{}1}^n k_ia_i$.
Therefore we had one of 3 cases:
1. $ a_i\equal{}2, k_i\equal{}1, k_j\equal{}0 \ i\not \equal{}j$ in these case $ i\equal{}1, n\equal{}1,a_1\equal{}2.$
2. $ a_i\equal{}1,k_i\equal{}1,a_j\equal{}1,k_j\equal{}1, \ \forall m\not\equal{}i,j  \ \ k_m\equal{}0$.
In these case $ a_m\equal{}a_{m\minus{}1}\plus{}a_{m\plus{}1},m\not \equal{}i,j \ 2a_i\equal{}a_{i\minus{}1}\plus{}a_{i\plus{}1},2a_j\equal{}a_{j\minus{}1}\plus{}a_{j\plus{}1}$.
It give $ i\equal{}1$ (else $ a_2\equal{}0$), and $ j\equal{}n$ (else $ a_{n\minus{}1}\equal{}0$), and $ n\le 3$ (else $ a_4<0$).
$ n\equal{}2$ solution $ a_0\equal{}a_1\equal{}a_2\equal{}a_3\equal{}1$.
$ n\equal{}3$ give contradition.
3. $ a_i\equal{}1,k_i\equal{}2, \ k_j\equal{}0 \ \forall j\not \equal{}i$. In these case not solutions.
Therefore only 2 solutions $ n\equal{}1$ $ a_0\equal{}1,a_1\equal{}2,a_2\equal{}1$
and $ n\equal{}2$ $ a_0\equal{}a_1\equal{}a_2\equal{}a_3\equal{}1.$

[quote="Rust"]Let $ a_{i \minus{} 1} \plus{} a_{i \plus{} 1} \equal{} (k_i \plus{} 1)a_i, k_i\ge 0$, then $ 2 \equal{} \sum_{i \equal{} 1}^n k_ia_i$.
[...] [/quote]
we have $ \sum_{i\equal{}0}^{n\minus{}1}a_{i } \plus{} \sum_{i\equal{}2}^{n\plus{}1}a_{i } \minus{}\sum_{i\equal{}1}^{n}a_i\equal{}\sum_{i\equal{}1}^{n} k_i a_i$
then $ 2\equal{}a_1\plus{}a_n\plus{}\sum_{i\equal{}1}^{n} (k_i\minus{}1) a_i$  :maybe:

And this solution does not work. This does not solve the problem.

I can post my solution.

How about $ n\equal{}1$ $ a_0\equal{}a_1\equal{}a_2\equal{}1$ ?

I think the numbers cannot all be 1. Otherwise the problem is senseless.

Yes, we must add a condition that $ a_i>1$ for $ 1\leq i\leq n$. Can the moderators add this condition to the problem?

Dear Omid,

I would like to see a hint or two for solving this problem, If not the whole solution. Would this be possible?

Thanks in advance

[hide="Hint:"]Prove that for some $ 1\leq i\leq n$:
\[ a_i\equal{}a_{i\minus{}1}\plus{}a_{i\plus{}1}\] [/hide]

[quote="Omid Hatami"][hide="Hint:"]Prove that for some $ 1\leq i\leq n$:
\[ a_i \equal{} a_{i \minus{} 1} \plus{} a_{i \plus{} 1}\]
[/hide][/quote]

I had gotten that far. But not much further

Thanks again.

Perhaps the much harder problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1593884703&t=270561]here[/url] might inspire somebody.

[hide="further hint?"]What happens if you delete $ a_i$ that has the property $ a_i\equal{}a_{i\minus{}1}\plus{}a_{i\plus{}1}$ from the sequence?  :) [/hide]

Wow, Brain's turning rusty. Thank you guys

amazing!
take $j$ in such a way that $a_{j}$ be the maximum number among $a_{i}$'s and $j$ is the least with this property
then we have:
$a_{j}$|$a_{j-1}$+$a_{j+1}$ but $a_{j-1}+a_{j+1}$ < $2a_{j}$ so $a_{j}=a_{j-1}+a_{j+1}$
now for the rest:
omit the $a_{j}$ mentioned above from the sequence then we would gain another sequence which satisfies the conditions of the problem and has a length less than the mother sequence,continuing this process we will obtain a sequence like this:
$1,a,1$ and so the $a$ is exactly equal to $2$.


First off, let me say I am a math dummy, so please give any responses as though you were talking to a 3 year old :-) Here is my problem.

I have a 4" diameter circle and I have to drill out 1" diameter cores. I need to know how many 1" cores I can fit and how much space will be left over.

THEN, let's say I have to drill out 2" cores - how many can I drill, and will there be room left to drill out smaller 1" cores in the remaining space?

This can go on and on, but I think you get my problem. Here is one for all the people who ever said "When would I ever use THAT in real life?"

Many, many thanks to anyone who can help me with this issue.

[hide]Try drawing a diagram. For the first one I believe it is 7. (I set them up with one in the middle and 6 surrounding it)

Well the big one has radius 2 and the little ones have radius $ \frac {1}{2}$

So the area of the large one is $ 4\pi$ and the area of the smaller ones are $ \frac {\pi}{4}$

So the amount of space left over would be $ 4\pi \minus{} \frac {7\pi}{4}$ So that leaves us with $ \frac{9}{4}$

For the second I was only able to fit 2 2" circles and 2 1" circles. [/hide]

Can someone check this?


What is $ ( \minus{} 1)^1 \plus{} ( \minus{} 1)^2 \plus{} \cdots \plus{} ( \minus{} 1)^{2006}$?

$ \textbf{(A) } \minus{} 2006 \qquad \textbf{(B) } \minus{} 1 \qquad \textbf{(C) } 0 \qquad \textbf{(D) } 1 \qquad \textbf{(E) } 2006$

[hide="Solution"]
$(C)$ All of the terms with odd exponents equal $-1$ whereas those with even exponents equal 1.  So pairing up each two terms, we see that the answer is 0. [/hide]

[quote=megarnie]Let $x=-1$. We have $1+x^1+x^2+\cdots+x^{2006}=\frac{x^{2007}-1}{x-1}$.

So $x^1+x^2+\cdots+x^{2006}=\frac{x^{2007}-1}{x-1}-1=\frac{-2}{-2}-1=\boxed{\textbf{(C)}\ 0}$.[/quote]

This is kind of overkill but I guess it gets the job done.

Observe that there are even number of terms, so half of the 1s will be negative and vice versa, Thus, after cancellation answer would be 0

Note that $-1$ is a primitive 2nd root of unity. Thus, if $x=-1$, then $x+x^2=0$. Also, $x^2=1$, so the answer is $(0)+(0)+(0)...=0$


I need some help with a computer problem.

I recently set my computer up to dual boot Windows XP / Ubuntu Linux. I created a FAT32 partition for my documents so that they can be accessed from either operating system. It is called E: in Windows and /docs in Linux. I typed up some homework in Linux and saved it in /docs. It told me that it couldn't save a backup, but I figured that was okay. I think that may have been an issue with permissions, since you can't change the permissions of a FAT32 partition normally. The document was in LaTeX, so I then built the DVI file and viewed it to see if it was right. But when I went to print it, I didn't have the printer driver configured, and I didn't have much time, so I switched to Windows. I had to force the computer to shut down. When I booted Windows (it was hibernating), the files were gone. The TeX file, the DVI file, etc. were all gone. However, a different TeX file that I had worked on earlier in Linux was still there. When I switched back to Linux, the files were still gone. I don't know what happened. I saved the file, and I know it was saved because otherwise I wouldn't be able to build the DVI.

It might have been in a temporary folder.  :(

How do I tell if it was a temporary folder? I don't think that was the problem, as the folder I saved it in was created in Windows, and already had documents in it, none of which disappeared.

Mmmmmmm, that's what tends to happen when you use fat32 and you don't cleanly unmount. Sorry :(, but it is gone.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13841[/img]

 :(

From R_1 (radius of C_1) and d is easy to find the radius of required circle, knowing it's centered on C_1. Draw then a circle O centered P with the found radius; it will usually cut C_1 at 2 points, the positions of the center of C, so basically there are 2 solutions. When the circles O and C_1 are tangent it will be a single solution, and none if the circles do not cut each other.

Best regards,
sunken rock

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=13908[/img]

 :thumbup:


function
integration
calculus
complex analysis
complex analysis unsolved

Consider the following theorem from "Complex variables and applications" By Brown & Churchill page185:
"If a function $f$ is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour $C$, then
$\int_C f(z)dz=2\pi Res_{z=0}\left [ \frac{1}{z^2}f(\frac{1}{z})\right]$."

My question is, why to restrict ourselves to a [b]finite[/b] number of singular points? I went through the proof and it did'nt really appear to me.

If there are infinitely many singularities, there are non-isolated singularities.

That said, the hypothesis is far stronger than we need. We only need $f$ to be analytic outside $C$ and continuous on $C$; the nature of $f$ inside $C$ is irrelevant.

The reason for the stronger hypothesis comes when we apply the ordinary residue theorem to this integral; the residues inside $C$ exactly cancel the resdue at $\infty$.


[quote="GameBot"]In the diagram, $ \Delta ABC$ is an isosceles right triangle that overlaps square ADEF. $ EF \equal{} 4$ and $ AC \equal{} 8$. What is the ratio of the area of quadrilateral EDBC to the area of pentagon AFECB? Express your answer as a common fraction.

[asy]pair A,B,C,D,E,F;
A=(4,4); B=(4,0); C=(0,0); D=(4,1); E=(1,1); F=(1,4);
draw(A--C--B--A--F--E--D);
label("A",A,E);label("B",B,E);label("C",C,W);
label("D",D,E);label("E",E,W);label("F",F,W);[/asy][/quote]

ADEF has area 16. Also, ADE has area 8. since AC=8, AB=4sqrt2. Thus, EDBC has area 32-8=24. The pentagon has area 16+24=40. $ \frac{24}{40}\equal{}\frac{3}{5}$.

I think I failed...

Actually, I believe that this is the solution.  Square $ ADEF$ has an area of 16, as you mentioned.  Since $ \Delta ABC$ is isosceles, and has a hypotenuse of 8, then it's legs are each $ 4\sqrt2$, making the area $ 16$.  $ \Delta ADE$ is half of the area of square $ ADEF$, so it's area is $ 8$, making the area of quadrilateral $ EDBC \equal{} 8$.  The area of the pentagon is square $ ADEF$+quadrilateral $ EDBC$=$ 8 \plus{} 16 \equal{} 24$.  Therefore, the ratio that we are looking for is $ \frac {8}{24} \equal{} \boxed{\frac {1}{3}}$


function
limit
integration
group theory
abstract algebra
linear algebra
matrix

[b]The 12th International Scientific Olympiad on Mathematics for University Students
10–13 July 2007
Tehran, Iran[/b]

[b]I. Mathematical Analysis (Pure and Applied Mathematics)[/b]

1. (25 points) Suppose that $ f: \mathbb{R}^{n}\to\mathbb{R}^{n}$ is a function satisfying the following two conditions:
i) $ f(K)$ is compact whenever $ K$ is a compact subset of $ \mathbb{R}^{n}$;
ii) $ f\left(\bigcap_{n=1}^{\infty}K_{n}\right)=\bigcap_{n=1}^{\infty}f(K_{n})$ whenever $ \{K_{n}\}_{n=1}^{\infty}$ is a decreasing sequence of compact subsets of $ \mathbb{R}^{n}$.
Prove that $ f$ is continuous.

2. (25 points) Suppose $ f$ is a real valued twice differentiable function defined on $ [a,b]$. Show that there are $ \xi$, $ \eta$ in $ [a,b]$ such that
\[ f(\eta)-f(a)\frac{b-\eta}{b-a}-f(b)\frac{\eta-a}{b-a}-\frac{1}{2}(\eta-a)(\eta-b)f''(\xi)=0.\]

3. (25 points) Let $ f_{n}: \mathbb{R}\to\mathbb{R}$ be a sequence of functions with the following properties:
i) each $ f_{n}$ ($ n=1,2,\ldots$) is a periodic function with period $ T$;
ii) each $ f_{n}$ ($ n=1,2,\ldots$) is continuous on $ \mathbb{R}$;
iii) the sequence $ \{f_{n}\}_{n=1}^{\infty}$ is uniformly bounded on $ \mathbb{R}$.
Then $ \{f_{n}\}_{n=1}^{\infty}$ is an equicontinuous sequence of functions on $ [0,T]$.
Prove or disprove this statement.

4. (25 points) Suppose $ f: [0,\infty)\to\mathbb{R}$ is continuous and $ \lim_{x\to\infty}f(x)=L$. Show that for each $ a>0$
\[ \lim_{n\to+\infty}\int_{0}^{a}f(nx)\,\mathrm{d}x=aL.\]

[b]II. Numerical Analysis (Pure and Applied Mathematics)[/b]

1. (25 points) Let $ g$ be a function defined on $ [a,b]$ with properties:
i) if $ x\in[a,b]$ then $ g(x)\in[a,b]$;
ii) $ g'$ is continuous on $ [a,b]$;
iii) for all $ x\in[a,b]$, $ \lvert g'(x)\rvert<1$.
1) Prove that the equation $ x=g(x)$ has exactly one root in $ [a,b]$.
2) Show that if condition i) is not satisfied then the equation $ x=g(x)$ may have no roots in $ [a,b]$.

2. (25 points) Let $ f$ be continuous on $ [a,b]$, $ h=\frac{b-a}{n+1}$ and $ x_{i+1}-x_{i}=h$, $ i=0,1,\ldots,n$, $ a=x_{0}<x_{1}<\ldots<x_{n}<x_{n+1}=b$.
1) Find weights of the open Newton-Cotes quadrature rule:
\[ \int_{a}^{b}f(x)\,\mathrm{d}x\approx\sum_{i=1}^{n}w_{i}f(x_{i}).\]
2) Find 3 points open Newton-Cotes formula for $ \int_{0}^{4h}f(x)\mathrm{d}x$.
3) Why open Newton-Cotes quadrature rules are not commonly used?

3. (25 points) Let $ \alpha$ be a fixed positive real number and define
\[ x_{n+1}=\frac{x_{n}^{3}+3\alpha x_{n}}{3x_{n}^{2}+\alpha},\quad n=0,1,\ldots\]
Assume $ x_{0}$ is so that the sequence $ \{x_{n}\}$ is convergent.
1) Find $ \lim_{n\to\infty}x_{n}$.
2) Determine the order of convergence of the sequence $ \{x_{n}\}$.

4. (25 points) Find coefficients $ a$, $ b$, $ c$, $ d$, $ m$, $ n$ and $ p$ in order that for differential equation $ y'=f(x,y)$ the Runge-Kutta formulas given as
$ k_{1}=hf(x,y)$,   $ k_{2}=hf(x+mh,y+mk_{1})$,
$ k_{3}=hf(x+nh,y+nk_{2})$,
$ k_{4}=hf(x+ph,y+pk_{3})$,
$ y(x+h)-y(x)\approx ak_{1}+bk_{2}+ck_{3}+dk_{4}$
conform with the Taylor series method of order $ h^{4}$.

[b]III. Algebra (Pure Mathematics)[/b]

1. (25 points) Let $ R$ be a ring with unity.
Prove that:
1) If every invertible element is central then every nilpotent element is central.
2) If every nilpotent element is central then every idempotent element is central.

2. (25 points) Let $ G$ be a non-cyclic group of order $ p^{n}$ where $ p$ is a prime number. Prove that $ G$ has at least $ p+3$ subgroups.

3. (25 points) Let $ G$ be a finite group with exactly 50 Sylow 7-subgroups. Let $ P\in Syl_{7}(G)$ and $ N=N_{G}(P)$.
1) Prove that $ N$ is a maximal subgroup of $ G$.
2) If $ N$ has a Sylow 5-subgroup $ Q$ and $ Q\lhd N$ then prove that $ Q\lhd G$.

4. (25 points) Let $ R$ be a ring such that $ x^{3}=0$ implies that $ x=0$. Let for each $ a,b\in\mathbb{R}$, $ (ab)^{2}=a^{2}b^{2}$. Prove that $ R$ is commutative.

[b]Operations Research (Applied Mathematics)[/b]

1. (25 points) A matrix $ A$, $ m\times n$, $ m\le n$, with integer components is said to be totally unimodular if every submatrix $ B$ composed of a set of $ m$ distinct columns of $ A$ is so that $ \lvert\det(B)\rvert=1$ (every such $ B$ is also said to be unimodular).

1) Prove that if an integer matrix $ B$, $ m\times m$, is unimodular then $ B^{-1}$ is also an (integer) unimodular matrix.
2) Consider the (LP) problem below:
$ \min z=c^{T}x$
$ \text{s.\,t.\ }Ax=b$ (LP)
$ x\ge0$
where $ A$ is an integer $ m\times n$ matrix, $ m\le n$ and vector $ b$ has integer components. Prove that if $ A$ is totally unimodular and (LP) has an optimal solution then the simplex method for solving (LP) will find an optimal integer solution (assuning that (LP) is nondegenerate).

2. (25 points) Consider the linear programming problem below:
$ \min_{x\in\mathbb{R}^{n}}q^{T}x$
$ \text{s.\,t.\ }Mx\ge-q$ (LP)
$ x\ge0$
where $ M=-M^{T}$ and $ q$ are given. Prove that:
1) $ x\ge0$ is an optimal solution of (LP) if and only if we have
\[ s(x)=Mx+q\ge0,\quad x^{T}s(x)=0.\]
2) Two feasible points $ x$ and $ y$ for (LP) are optimal points of (LP) if and only if we have
\[ x_{i}s_{i}(y)=y_{i}s_{i}(x),\quad i=1,\ldots,n.\]
3) If $ q\ge0$ then the problem (LP) has an optimal solution.

3. (25 points) Prove 1) and 2) using elementary definitions of convex sets and convex functions.
1) Show that the set
\[ S=\{x\in\mathbb{R}^{n}\mid Ax=b,\ x\ge0\}\]
is convex.
2) Show that if $ x^{*}$ is an optimal (local) solution for the problem
$ \min z=c^{T}x$
$ \text{s.\,t.\ }Ax=b$ (LP)
$ x\ge0$
then $ x^{*}$ is a global solution.
Prove or disprove 3) and 4).
3) Problem (LP) can be infeasible.
4) Problem (LP) can be unbounded.

4. (25 points) Assume that $ S$ is a nonempty open convex set in $ \mathbb{R}^{n}$ and $ f: S\to\mathbb{R}$ is differentiable on $ S$. Prove that if $ f$ is convex on $ S$ then we have
$ \bigl(\nabla f(x_{2})-\nabla f(x_{1})\bigr)^{T}(x_{2}-x_{1})\ge0$ for all $ x_{1},x_{2}\in S$.

[b]IV. Linear Algebra (Pure and Applied Mathematics)[/b]

1. (33.3 points) Let $ A\in M_{n}(\mathbb{C})$ and $ \mathbb{C}[A]=\{f(A)\mid f(x)\in\mathbb{C}[x]\}$. Prove that the ring $ \mathbb{C}[A]$ has no non-zero nilpotent element if and only if $ A$ is diagonalizable.

2. (33.3 points) Prove that if $ n$ is a positive integer then there exists a $ 2\times2$ invertible matrix $ A=(a_{ij})$ such that $ a_{ij}\in\mathbb{R}$, $ A^{n}=I$ and $ A^{k}\ne I$ for every $ 1\le k<n$.

3. (33.3 points) Let $ X$ be an invertible matrix with columns $ X_{1},X_{2},\ldots,X_{n}$. Let $ Y$ be a matrix with columns $ X_{2},X_{3},\ldots,X_{n},0$. Show that the matrices $ A=YX^{-1}$ and $ B=X^{-1}Y$ have rank $ n-1$ and have only 0's for eigenvalues.

Here is my solution to the Mathematical Analysis problem 4):

Using the substitution $ nx=y$ we get that

$ l=\int_{0}^{a}f(nx)dx=\frac{1}{n}\int_{0}^{na}f(y)dy$

and an application of Cesaro Stoltz lemma  shows that

$ =\int_{na}^{(n+1)a}f(y)dy=af(c_{n}))\rightarrow aL$,

where $ c_{n}\in (na, (n+1)a)$.

Note that $ \lim f(c_{n})=L$ since $ \lim f(x)=L$.


geometry
3D geometry
number theory unsolved
number theory

Show that there is a set of 2002 distinct positive integers such that the sum of one or more elements of the set is never a square, cube, or higher power.

Posted and solved before...
Look at the Resources section.

We can easily contruct an infinite set with this property :

Let $p_{1},p_{2},\cdots$ be the sequence of the primes in increasing order.
Let $x_{1}= p_{1},x_{2}= p_{1}^{2}p_{2}, \cdots, x_{n}= p_{1}^{2}\cdots p_{n-1}^{2}p_{n},\cdots$.
The set $S$ of all the $x_{i}$'s has the desired property, because for the sum $s(A)$ of the elements of any finite $A$ subset of $S$, if $x_{k}$ is the least element of $A$, then $s(A)$ is divisible by $p_{k}$ but not by $p_{k}^{2}$, so it cannot be a perfect power.

Pierre.

Oops...too late...
Well, it avoids you a search.

Pierre.

It is obviosly. Let $x_{n}=2^{y_{n}}-1, \ y_{n}=\prod_{k=1}^{n}p_{k}^{n-k}(p_{k}-1), \ n=1,2,...$, were $p_{k}$ is k-th prime.
It is easy to show, that $x_{n}$ satisfyed your condition.


calculus
integration
function
calculus computations

$ F(x)$ is a differentiable function such that

$ F'(a\minus{}x)\equal{}F'(x)$

for all $ x$ satisfying $ 0\leq x\leq a$.

Evaluate

$ \int_{0}^aF(x)dx$

and give an example of such a function.

[quote="pankajsinha"]$ F(x)$ is a differentiable function such that

$ F'(a \minus{} x) \equal{} F'(x)$

for all $ x$ satisfying $ 0\leq x\leq a$.

Evaluate

$ \int_{0}^aF(x)dx$

and give an example of such a function.[/quote]

$ \int_{\frac{a}{2}}^t (F'(x)\minus{}F'(a\minus{}x))dx\equal{}0$
$ \Leftrightarrow \int_{\frac{a}{2}}^t F'(x)dx\minus{}\int_{a\minus{}t}^{\frac{a}{2}}F'(x)dx\equal{}0$
$ \Leftrightarrow F(t)\plus{}F(a\minus{}t)\equal{}2F(\frac{a}{2})$

$ 2\int_{0}^{a}F(\frac{a}{2})dx\equal{}\int_{0}^{a}F(x)dx\plus{}\int_{0}^{a}F(a\minus{}x)dx\equal{}2\int_{0}^{a}F(x)dx$

So $ \int_{0}^{a}F(x)dx\equal{}\int_{0}^{a}F(\frac{a}{2})dx\equal{}aF(\frac{a}{2})$

Example: $ a\equal{}2 \pi$, $ F(x)\equal{}sin x$


A wooden cube with edge length $ n$ units (where $ n$ is an integer $ >2$) is painted black all over.  By slices parallel to its faces, the cube is cut into $ n^3$ smaller cubes each of unit length.  If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is $ n$?

$ \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ \text{none of these}$

[hide="Solution"]$ 6(n \minus{} 2)^2$ is the number of cubes with one side painted.
$ (n \minus{} 2)^3$ is the number of cubes that are unpainted.
$ (n \minus{} 2)^3 \equal{} 6(n \minus{} 2)^2\implies (n \minus{} 2) \equal{} 6\implies n \equal{} \boxed{\textbf{(D)}\ 8}$.[/hide]

[quote="grn_trtle"][hide="Solution"]$ 6(n \minus{} 1)^2$ is the number of cubes with one side painted.
$ (n \minus{} 1)^3$ is the number of cubes that are unpainted.
$ (n \minus{} 1)^3 \equal{} 6(n \minus{} 1)^2\implies (n \minus{} 1) \equal{} 6\implies n \equal{} \boxed{\textbf{(C)}\ 7}$.[/hide][/quote]

Are you sure its (n-1)^3?

This is my solution:

Let the side length be x. On each face, there will be (x-2)^2 cubes that have exactly one face (Try drawing a diagram).

Also, the cubes that have no faces colored are inside the large cube. This is (x-2)^3.

Thus, $ 6(x\minus{}2)^2\equal{}(x\minus{}2)^3$.

Since x>2, we can divide by $ (x\minus{}2)^2$.

Thus, $ 6\equal{}x\minus{}2$. Therefore, $ x\equal{}8$.

Yeah, sorry, I meant $ (n\minus{}2)^3$.
I even wrote $ (n\minus{}2)^3$ on paper but when I was typing...


geometry
MATHCOUNTS
calculus
inequalities
rectangle
perimeter
3D geometry

[quote="Ignite168"]What is the radius of a cylinder with volume 12 cubic centimeters and the least possible surface area? Express your answer as a common fraction. However, it does not have to be in simplest radical form.[/quote]

[hide]
Given $V = \pi r^{2}h = 12$ minimize

$A = 2\pi r h+2\pi r^{2}$

Using the Lagrange multiplier technique one finds:

$\lambda = 2/r = 1/r+2/h$

$\implies r^{3}= \frac{6}{\pi}$[/hide]

[quote="Kent Merryfield"]There is no way this problem belongs anywhere close to the Mathcounts forum. Lagrange multipliers is a very nice technique, but it's a calculus technique.

Actually, the Lagrange multipliers lead directly to the condition $2r=h.$ That is, the optimal cylinder has height equal to its diameter - when seen from the side, it has a square cross-section.

It should be possible to solve this using some variant of AM-GM, Cauchy-Schwarz, or other well-known inequality, but even that should be at some high school level. The problem should never have been posted in this forum, and should be moved.[/quote]

Through manipulation of the formula for surface area of cylinder in terms of r and h, you can get a fraction that has $r^{2}$ on the numerator and $r+h$ on the denominator. It basically follows how to maximize same way you maximize area of a rectangle when given a perimeter.

Can you post a complete solution? I'm very interested in how to do this, as it had me stumped for a while.

[hide]
So you have $r^{2}h\pi=12$ and you want to minimize $2r^{2}\pi+2rh\pi$.

Let's simplify the surface area formula a bit.

First, we factor out $2$ to get $2(r^{2}\pi+rh\pi)$.

Then, we multiply that by $\frac{rh}{rh}$ to get $\frac{2rh(r^{2}\pi+rh\pi)}{rh}$.

We distribute to get $\frac{2(r^{2}h\pi*r+r^{2}h\pi*h)}{rh}$.

We substitute $12$ for $r^{2}h\pi$ to get $\frac{2(12r+12h)}{rh}$.

Then, we factor to get $\frac{24(r+h)}{rh}$.

We want to make this fraction as small as possible, and we also have $r^{2}h\pi=12$.

The fraction is basically like perimeter of a rectangle over the area. Now, $2r=h$ follows from the fact that this is basically like a rectangle has a certain amount of area, minimize the perimeter as you obviously want a square so in your cylinder, diameter should be equal to height.

So now, we substitute $h$ out for $2r$ to get $2r^{3}\pi=12$.

Dividing gets $r^{3}=\frac{6}{\pi}$.

We then take the cube root of both sides to get $\boxed{r=\frac{\sqrt[3]{6}}{\sqrt[3]{\pi}}}$.

[/hide]

Excellent. Makes sense. Great Presentation too. Thanks.
Still, seems a bit hard for math counts..

[quote="Ignite168"]Now, $2r=h$ follows from the fact that this is basically like a rectangle has a certain amount of area, minimize the perimeter as you obviously want a square so in your cylinder, diameter should be equal to height.[/quote]

Not even remotely rigorous.  Not a valid argument at all without the underlying calculus.

It's just a bit of easy differential calculus (now that I see the outline for the solution), but because calculus shouldn't be needed for mathcounts, I don't think this should be in here.

[hide="Elementary solution"]$V=r^{2}\pi h\implies h=\frac{V}{r^{2}\pi}$

$S=2r^{2}\pi+2r\pi h=2r^{2}\pi+2r\pi{V\over r^{2}\pi}=2r^{2}\pi+{2V\over r}$

Using AM-GM

$2r^{2}\pi+{2V\over r}=2r^{2}\pi+{V\over r}+{V\over r}\geqslant 3\sqrt[3]{2r^{2}\pi\cdot{V\over r}\cdot{V\over r}}=3\sqrt[3]{2\pi V^{2}}$

Equality is satisfied iff $2r^{2}\pi={V\over r}\iff r=\sqrt[3]{{V\over 2\pi}}=\sqrt[3]{{6\over\pi}}$

That gives $h=\sqrt[3]{{4V\over \pi}}=2r$, hence the axial section of the cylinder must be a square.[/hide]

[quote="t0rajir0u"][quote="Ignite168"]this question was a [b]proof[/b] so I don't see how it has to be [b]rigorous[/b][/quote]

I don't follow.  Of course a proof has to be rigorous.

There's a difference between [b]rigor[/b] and [b]completeness.[/b]  You don't need to prove every theorem you use in a proof, for example, as long as they're well-known, but the conclusion you drew was

1.  Not well known (i.e. did not have a name), and
2.  Not obvious

Any statement of this sort that you use needs to be proven.  Your intuitive argument is not valid because it would fail for [b]many[/b] other optimization problems.[/quote]

That obviously should've said wasn't looking at the original problem.

As for your next point, please show me a problem where you have an optimization with something like $\frac{x+y}{xy}$ that needs to be right and uses some other shape than a square.

[quote="t0rajir0u"]What is the radius of a "pill" (a cylinder with two hemispheres of the same radius on each end) with volume $12$ cubic centimeters and the least possible surface area?[/quote]

[hide="Solution"]$S=2r\pi h+4r^{2}\pi=2r\pi(h+2r)$

$V={4\over 3}r^{3}\pi+r^{2}\pi h=r^{2}\pi({4\over 3}r+h)\iff h={V\over r^{2}\pi}-{4r\over 3}$

$S=2r\pi({V\over r^{2}\pi}-{4r\over 3}+2r)=2r\pi({V\over r^{2}\pi}+{2r\over 3})={2V\over r}+{4r^{2}\pi\over 3}$

AM-GM:

$S={V\over r}+{V\over r}+{4r^{2}\pi\over 3}\geqslant 3\sqrt[3]{{4V^{2}\pi\over 3}}=12\sqrt[3]{3\pi}$

The equality holds iff ${V\over r}={4r^{2}\pi\over 3}\iff r=\sqrt[3]{{3V\over 4\pi}}=\sqrt[3]{{9\over\pi}}$

But if ${V\over r}={4r^{2}\pi\over 3}$, then ${V\over r^{2}\pi}={4r\over 3}\iff h=0$. Therefore, the desired "pill" is actually a pure sphere - which is all but expected.[/hide]

[quote="Ignite168"]I haven't even taken Pre-Calc yet, I really don't see how I accidentally stumbled onto calculus  :|  :huh:

However, this question was a proof so I don't see how it has to be rigorous at all. While what I used won't stand up logically, it's good enough for solving this problem, isn't it?[/quote]
You "stumbled into" the calculus because of the type of problem this is.  Calculus is the easiest way to solve problems where you want the min or max of some function. In fact it didn't occur to me that the problem could be solved without the calculus and that's why I hid my solution and didn't explicitly show how I arrived at the intermediate result $h = 2 r.$  Given that this is high school problem you were obviously supposed to use the AM-GM inequality to solve it. The AM-GM inequality is sufficiently well known such that results from it can be considered proved. The method was used by the ancient Greeks who did not have the calculus.

[quote="Ignite168"]Now, $2r=h$ follows from the fact that this is basically like a rectangle has a certain amount of area, minimize the perimeter as you obviously want a square so in your cylinder, diameter should be equal to height.[/quote]

Alright, so I understand what you were saying better now.  You're actually making two mistakes that cancel out.

First, you claim that $\frac{24(r+h)}{rh}$ is the multiple of a function that gives the ratio of the perimeter of a rectangle to the area.  This would imply that $r, h$ are the sides of the rectangle.

Then you claim that the "rectangle" you're looking at is the rectangle formed by a given diameter and by the height, in which case $2r, h$ are the sides of the rectangle.

In both cases you ignore the condition $r^{2}h \pi = 12$, which is the real reason $2r = h$ is the correct equality condition.

Just as a little side note, could someone tell me what AM-GM is?

Arithmetic Mean of a set of numbers is always greater than or equal to the Geometric Mean of the same set.
This can be handy for finding max/mins.
The other way to do this is with a bit of calc.

Yeah, with calc it's a pretty easy optimization problem...


logarithms
inequalities unsolved
inequalities

hello, this is equivalent with $ e^t>1\plus{}t$ and this is well-known.
Sonnhard.

$ t> \log (1\plus{}t) \implies t> t\minus{} \frac{t^2}{2}\plus{}\frac{t^3}{3}\minus{}\frac{t^4}{4}\plus{}...$
$ 0>\left (\minus{}\frac{t^2}{2}\plus{}\frac{t^3}{3}\right )\plus{}\left (\minus{}\frac{t^4}{4}\plus{}\frac{t^5}{5}\right )\plus{}...$
ok?

Why does $ \log (1 \plus{} t) \equal{}t \minus{} \frac {t^2}{2} \plus{} \frac {t^3}{3} \minus{} \frac {t^4}{4} \plus{} ...$ hold?

thats supposed to be maclaurin series for $ \log (1 \plus{} x)$.
...anything wrong..?
it is clearly visible that for large $ t$, $ e^t > 1 \plus{} t$ but above makes it clear to understand even for small $ t$.

That doesn't make sense. :P

[quote="airrafer"]Proof 1/x > ln(1+1/x)  x>0[/quote]

[url=http://www.mathlinks.ro/viewtopic.php?t=310769]See here[/url]


calculus
integration
logarithms
function
induction
real analysis
real analysis solved

Prove that
\[ \sum_{n=1}^\infty {1\over n^n} = \int_0^1 x^{-x}\,dx. \]

$\int_0^1x^{-x}dx=\int_0^1e^{-x\ln x}dx=\int_0^1\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}x^k\ln^kx\,dx$

Interchange sum and integral to get $\sum_{k=0}^{\infty}\frac1{k!}\int_0^1 (-1)^kx^k\ln^kx\,dx$

In the integral $\int_0^1(-1)^kx^k\ln^kx\,dx,$ make the substitution $u=-\ln x,$ which can be written as $x=e^{-u},\,dx=-e^{-u}du.$ The integral becomes

$\int_0^{\infty}u^ke^{-ku}e^{-u}\,du=\int_0^{\infty}u^ke^{-(k+1)u\,du}$

Now make the substitution $v=(k+1)u,\,u=\frac{v}{k+1},\,du=\frac{dv}{k+1}.$ The inetegral becomes:

$\frac1{(k+1)^{k+1}}\int_0^{\infty}v^ke^{-v}\,dv=\frac{k!}{(k+1)^{k+1}}$

Put that back into the original sum, getting

$\int_0^1x^{-x}dx=\sum_{k=0}^{\infty}\frac1{k!}\cdot\frac{k!}{(k+1)^{k+1}} =\sum_{n=1}^{\infty}\frac1{n^n}$

where we made the substitution $n=k+1.$

I am sure that it has been posted before in this forum.
It is an old problem, as it appears in my analysis problems book.

Yes I agree
Can you find other fuctions $f$ s.t:$\sum_{i=1}^{\infty}f(n)=\int_0^1f(x)dx?$

For example $ f(x)\equal{}a$ if $ 0\leq x\leq 1$ and $ f(x)\equal{}0$ as $ x>1$.

Can anyone give a complete characterization of such functions?

Well, not that much. The simple reason is that the integral depends on the values of the function from 0 to 1, while the sum depends on the discreet values beyond that, so we are looking at the same object, the function, from so different perspectives that we can't easily say we are looking at the same object enough to find its characterization. For example imagine you have a function whose integral from 0 to 1 is [i]a[/i], then you have infinite number of extensions of the same function that would satisfy given equation. All that is needed is that sum over N, f(1)+f(2)+... is again [i]a[/i]. Except in the interval 0,1 other real non-integer values are totally irrelevant.

So from this:
1. The function should be summable over N* (N without 0)
2. The function should be integrable over the interval 0,1
3. Its infinite sum over N* should be equal to its integral over the interval 0,1

we might not go further than this:

Imagine a function F(x) satisfying 1. and 2. where integral is equal to [i]a[/i] and the sum to [i]b[/i]. Suppose we create two scaled function G(x) = a/b F(x) and H(x) = b/a F(x). We can combine these three functions to get two solutions of original problem:
1. create a new function taking F(x) for x from 0 to 1, excluding 1, and G(x) for the remaining values
2. create a new function taking H(x) for x from 0 to 1, excluding 1, and F(x) for the remaining values

The used procedure does not affect integrability and summability: we used scaling and the integral value does not depend on the possibly changed value at 1. So the process is reversible. If we have a function that is a solution to the problem we can easily find another function we could start our procedure from and eventually find that solution. If a function is not a solution we can create one. It means that the procedure describes the entire space of the solutions. Since we can arrive to a solution from EVERY function satisfying 1. and 2. and the steps involve nothing more than scaling I would say we didn't get much of characterization.

So nothing fancy, I guess, some scaling applied does not make it anything special.


Hi. Basically, I will be moving to the Chicago area in the fall and I need information regarding the area. You can just read this if you don't feel like reading more details I am listing below.

I am a current 8th grader and my father very recently got a job with Northwestern University. I would like information about which High Schools in the Northern Chicago suburbs are good, which cities you suggest we move to and any information that may help us. My father would like to stay close to Evanston. My friend recently moved to Buffalo Grove and he says Adlai Stevenson High School is a good one. I have already contacted Northside High School and I have missed their application deadline. As of right now, we are open to any suggestions. I know of Illinois Math and Science Academy and will apply there for 10th grade. Thanks and we need as much advice as we can get.

Also, is this reliable? http://www.suntimes.com/images/cds/pdf/top50high-2008.pdf

The Evanston area has Evanston Township High School and New Trier, which are both pretty good public schools, although NT often gets a bad rap for being a rather elitist school   :maybe: . They are both huge schools in terms of geographical size and population.

New Trier also has a very strong athletics program, and a very strong Science Olympiad program, if you are interested in those.

Stevenson is also a good school, but it is not as close as ETHS and NTHS. It also has a good SciOly program.

Given that you are on this forum (and thus are a math enthusiast), and that you wish to settle close to Evanston, those three are probably your top bets for public schools.

The top 5 schools are usually Naperville North, IMSA, Naperville Central, Stevenson, and Walter-Payton. ETHS and NTHS do not have as strong math teams, but you'll still get a strong education, and the opportunity to participate in math team.

I would not necessarily trust a newspaper's ranking of high schools, because high schools can be judged in so many ways. What might matter to the general public (the % of students who graduate, the % of students who continue on to college, etc.) might not be so important to you.

I don't know if this is possible, but it might be advantageous for you to shadow a friend or sit in some classes at the schools just so you get a feel for the people and the classes (remember - the people you are around in high school may matter just as much to your future as the classes you are in).

Other than the Evanston area, the Naperville schools are known to be strong, although I do have a bias and inclination. But our math teams traditionally do well (top 5), and Naperville Community Unit School District 203 invests a lot of money per student if you look at the statistics (well above average in IL). Naperville is mainly known for its super-competitive, broad-curriculum schools, as well as its livability (it's safe, and people are nice, etc.). It is about an hour from Northwestern, however, and about 30 to 45 minutes from downtown Chicago.

Good luck with your search! You can pm me or email me (10thsymphony@gmail.com) if you have any questions, and I'll try my best to answer them.

Thanks for the reply. I have heard of Naperville before because of their large Chinese community (or so my mom says) but it is a little too far. I have been looking at New Trier High School but it seems like a very large school (are all Chicago high schools 4000+?) and it seems to be comprised of mainly rich white and asian kids. I am not too concerned about whether or not the math team is good or not but as long as they are offered.

Between Stevenson and New Trier which would you think has better students, teachers (you probably wouldn't know), community, athletics... It looks to me that they have similar test scores, similar size and similar ethnicity breakdown. What things have you heard about the two?

Another good choice (though I may be biased) would be the Glenbrook school district in Northbrook and Glenview. These suburbs are a little bit further from Evanston than Wilmette (about 25-30 minutes from downtown Evanston), but Glenbrook South and Glenbrook North High Schools are great schools. They are large public high schools, but not quite on the scale of New Trier and Stevenson (graduating class around 650-700).

From personal experience (I am an alum of Glenbrook South, and I've had many interactions with Glenbrook North), I can tell you that both of these schools are excellent. Both have math teams, as well as many other extracurricular activities. You should contact me directly if you have more questions about either.

Well, in my opinion, New Trier is probably the best school for anything, and I don't even go there. The arts, sports, academics. They have a hugely talented/competitve student body and enough funds to start their own bank. They can produce stars in every field, which you can pretty much tell by looking at their alumni list.

Be careful though; correlation does not imply causation. The success of students at New Trier and other schools could very well be in large part due to the backgrounds of the students who go there.

It's much easier to succeed when you're well-off to start with.  :wink:

It is funny to hear New Trier bashing and then realize that I went there.

Like I suppose that there are people who go to the school that you describe, but you probably won't be interacting with them at all. The classes are split based on levels and these people you describe aren't really in those classes. But I suppose that it is your choice to subscribe to those stereotypes.

And to just comment in general, it really doesn't matter much which of these schools you go to. They are all great and I think the minor differences between schools balance out. It is really on you to make the best of your time in high school. You choose who your friends are and how you choose to be a part of the community. You choose what opportunities to seek and what teachers to get to know.

<heckle>[quote="Altheman"]It is funny to hear New Trier bashing and then realize that I went there. [/quote]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=798911#798911]New Trier bashing[/url]?  Hmm...I remember it like it was yesteryear...</heckle>

[quote="Altheman"]And to just comment in general, it really doesn't matter much which of these schools you go to. They are all great and I think the minor differences between schools balance out. It is really on you to make the best of your time in high school. You choose who your friends are and how you choose to be a part of the community. You choose what opportunities to seek and what teachers to get to know.[/quote]
Altheman is completely correct.  All of those schools are very good.  You really cannot go wrong with any of them.

Good luck with the move.

coach fritz

[quote="coachfritz"]<heckle>[quote="Altheman"]It is funny to hear New Trier bashing and then realize that I went there. [/quote]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=798911#798911]New Trier bashing[/url]?  Hmm...I remember it like it was yesteryear...</heckle>
[/quote]

Haha, owned.

But yes, I'll reiterate that all of these schools are excellent schools and you can't go wrong with any of them.

Thanks everyone. I have more than enough information.

I'm glad you solved your problem. I'm sure you'll enjoy Illinois no matter where you live. Maybe we'll run into each other without knowing it  :lol:


Prove that there is no function $f:\mathbb {Z}\rightarrow \mathbb {Z}$ such that $f(f(x))=x+1, \forall x\in \mathbb {Z}.$
Prove that there are infinitely many functions $g:\mathbb {Z}\rightarrow \mathbb {Z}$ such that $g(g(x))=-x, \forall x\in \mathbb {Z}.$

1. $f(0)=k, f(k)=f(f(0))=1, f(1)=f(f(k))=k+1,$ and then if $f(n)=k+n, f(k+n)=f(f(n))=n+1, f(n+1)=f(f(k+n))=k+n+1$, then $f(a)=k+a$ by induction for non-negative $a$. But then either $f(k)=2k \neq f(f(0))=1$ for $k$ positive, or $f(k+n)=n+1=2k+n$, so $2k=1$, again impossible.

Mr enescu, you should probably post the extremely beautiful problem that the two teachers proposed in the TST in the same year, related to iterated functions that give linear functions. That problem is a real challenge: nobody solved it. Unfortunately, I do not remember very well the details of the statement, so please post it.

I will do that. In fact, it was problem 4, 2nd selection test 1991.

Thanks and sorry for missing the right year.  :)


geometry
perimeter
incenter
trigonometry
geometry unsolved

For the maximum perimeter, from my solution at http://www.mathlinks.ro/Forum/viewtopic.php?t=163107, it follows that $ P$ must be the incenter of triangle $ ABC$, and then you can just use trigonometry to finish this part of the problem.