Country Boy

Or, think of the electrons in the wire as being right next to one another so the form an "electron rod". Move an electron at one end slightly and the electron at the other end moves (almost) instantly!

Then I'll have to ask again what definition of "perpendicular" you are using.

I've never heard of points being "perpendicular". Yes, we can say that planes are perpendicular if their normal lines are perpendicular. Whether or not you can use the term "perpendicular" in higher dimension depends upon the precise definition of "perpendicular" you are using.

Let [math]u_n= \frac{p}{q}[/math] in your first formula:

[math]\frac{p}{q}-\frac{\frac{p^2}{q^2}-2}{\frac{2p}{q}}[/math]

Multiply numerator and denominator of the second fraction by q

[math]\frac{p}{q}- \frac{\frac{p^2}{q}-2q}{2p}[/math]

Get common denominator 2pq

[math]\frac{2p^2}{2pq}- \frac{p^2- 2q^2}{2pq}[/math]

and combine fractions

[math]\frac{p^2+2q^2}{2pq}[/math]

The denominator of that is exactly the same the other form

[math]\frac{2p^2-1}{2pq}[/math]

so they will be the same as long as the starting values satisfy

[math]p^2+ 2q^2= 2p^2- 1[/math]

That is the same as [math]2q^2= p^2-1[/math] which happens to be true for p/q= 3/2.

First thing you will have to do is DEFINE "perpendicular" for cubes. As far as I know "perpendicular" is only defined for lines (or, more generally, curves, but always one dimensional) in any dimension space.

Since you don't mention squares, what would it mean for two squares to be "perpendicular"?

In other words, what message are you talking about???

A first obvious step is to multiply through by H!

You now have [math]H"+ \frac{1}{r}H'= 0[/math]. Let Y= H' and now

[math]Y'+ \frac{1}{r}Y= 0[/math] or [math]Y'= -\frac{1}{r}Y[/math] which is a simple separable equation. Solve for Y, then solve H'= Y. Remember that the solution is only valid for H not equal to 0.

Hello everybody !

This is my question:

Suppose you have a mass m0 inside an ellipse (at rest).

Suposse someone see it from another frame, from this frame he will see the ellipse contracted.

I know there's no black hole, but, how can explain our observer this result?

I'm trying to see only the frame of our observer, I know he knows about relativity and he can calculate our m0, and our r0 (ellipse's radius at rest), and then conclude there's no black body. But what is seeing at really our observer? I mean, he can explain it seeing our viewpoint and noting that there's no black hole, but how can explain it from his own frame?

 

Thanks !

I don't understand why you even mention "black holes"! Assuming you have mass m0 inside an ellipse, traveling at high speed relative to an observer, yes, he will see the ellipse contracted. He will NOT see a 'black hole'.

 

PS: I have read many times articles like:

If you go too fast do you become a black hole?

http://www.math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html

If you were under the impression that the ellipse, at high speed, will contract to a "black hole", this very link explains why that is NOT true!

 

and again, I don't think there's a paradox here, I'm just trying to see what is seeing our observer.

Maybe is analogous to:

When we see a moving frame, we see their atoms contracted in the direction of motion. Then the orbitals don't follow the expected symmetry, how can be stable these "deformed" atoms?

What orbitals are you talking about? Contracted orbitals are still orbitals.

Take [math]\mathbb{R}[/math] and adjoin two points to it, namely [math]+\infty[/math] and [math]-\infty[/math], denote this new set by [math]\left[-\infty, +\infty\right][/math].

 

Define addition as follows :

 

[math] a + b = \begin{cases} a + b & \text{ if } a, b \in \mathbb{R} \\ +\infty & \text{ if } a = +\infty \text{ or } \left( a \in \mathbb{R} \text{ and } b = +\infty \right) \\ -\infty & \text{ if } a = -\infty\ \text{or } \left( a \in \mathbb{R} \text{ and } b = -\infty \right) \end{cases}[/math]

 

This operation is closed and associative so we have a semigroup with both [math]+\infty[/math] and [math]-\infty[/math] acting as zeros on [math]\mathbb{R}[/math] under addition and is an extension of the normal addition on [math]\mathbb{R}[/math]. In fact both added points are left-zeros of the entire space

 

The trick here is to not demand that addition be commutative for all possible values, only for real values (else define a local semigroup and work in the two point compactification of [math]\mathbb{R}[/math] if you want to keep the commutativity.) For our original problem though we only actually need to add one point to [math]\mathbb{R}[/math] so the issue is actually a moot point.

You HAVEN'T defined [math]\infty+ (-\infty)[/infty][/math]so this is not a well defined operation.

First this should go in "General Math" not "Linear Algebra and Group Theory".

Second, there is no one answer. Given any finite number of points, there exist an infinite number of functions that will give those points.

However, the tree's suggestion is very good. There exist only one LINEAR function through those points. Do you know the general form for a linear function?

It's hard to understand either of these two posts. Newtonian theory does not give any hypotheses as to how gravity "works" but the general theory of relativity certainly does.

As for the "hard drive"- it wasn't necessary to disassemble it, exactly the same effect can be seen with a bicycle wheel. In fact it's a standard example in high school physics classes- one student stands on a small platform that is free to rotate, holding a bicycle wheel by an extended axis. Get the bicycle wheel rotating at high speed. When the student tries to tilt the wheel to one side or the other, the entire platform rotates. That's because the "torque" vector of the wheel is at right angle to the wheel itself. It's the basis of the "gyroscope" effect. I don't see why you would think it had anything to do with gravity.

Tom, I would interpret this to mean that A is the ring and a some member of A. aA= {ab| b contained in A}.

If it is true that aA= A for all a, then x-> ax is a one-to-one mapping from A onto aA. If a were a zero-divisor, then there exist b such that ab= 0= a0, so that the mapping is not one-to-one.