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[quote="EquinoX"]What is the flux through a cylinder of radius R and height h from a point charge at the center[/quote]

The flux through any closed figure by a point charge $ q$ is $ \left(\frac{q}{\epsilon_0}\right)$ if all the field due to it crosses the surface of the figure.

For example, the flux through a cube with a charge $ Q$ placed at a corner is $ \frac{Q}{8\epsilon_0}$.

[quote="EquinoX"]What is the flux through a cylinder of radius R and height h from a point charge at the center[/quote][hide]
\[ \newtheorem*{GL}\begin{GL}\textbf{Gauss's Law: }\textrm{The total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.}\end{GL}\]

Written as an integral equation, the above theorem is $ \iint_S \mathbf{E}\cdot\mathrm{d}\mathbf{A} =\frac{Q}{\varepsilon_{0}}$.

One would imagine that the author intended for you to calculate the flux and show that:
\[ E_r(2\pi R h) =\frac{q}{\varepsilon_0}\]

However according to the problem statement, the author asks, "What is the flux ...?" to which the answer is: $ \Phi_E = \frac{q}{\varepsilon_0}$.[/hide]

[quote="Immanuel Bonfils"]
The point charge is in [b] the center [/b].

[/quote]

[quote="I"]
[b]For example[/b], the flux through a cube with a charge $ q$ placed at a corner is $ \frac{Q}{8\epsilon_0}$.

[/quote]

That was an example so as to help the author understand better what I meant.

The answer to the original question is $ \frac{q}{\epsilon_0}$, as everyone has pointed out.

[quote="Conjurer"][quote="Dr. No"]

One would imagine that the author intended for you to calculate the flux and show that:
\[ E_r(2\pi R h) \equal{} \frac {q}{\varepsilon_0}
\]
[/quote]

I dont see how this is true.[/quote]

\[ \iint \vec{E}\cdot \mathrm{d}\vec{A} \equal{} E_r A_r \plus{} E_z A_z \minus{} E_z A_z \equal{}\frac{q}{\varepsilon_0}\]

The "direction" of any given section of surface is the direction of its outward normal unit vector.

$ A_z$ is the area of an end plate, $ A_z \equal{} \pi R^2$. There are two end plates, one whose outward normal is $ \plus{}\hat{z}$, and the outward normal of the other is $ \minus{}\hat{z}$.

In order to calculate $ A_r$ imagine cutting the cylinder down its height, and laying it out like a flat sheet. $ A_r$ is the area of this sheet. It has height $ h$ and width equal to the circumference of the cylinder, $ 2\pi R$.

\[ A_r \equal{} \textrm{height }\times\textrm{ width } \equal{} h(2\pi R) \equal{} 2\pi R h\]

[quote="Conjurer"]Its a point charge, obviously $ \hat{E}.\hat{dA} \neq EA$  since the field lines are not perpendicular to all the elements.[/quote]

What?

Its a point charge: its E-field is isotropic. That means the field lines [b]are[/b] perpendicular to all the (surface) elements. However even if that weren't the case, Gauss's law guarantees that
\[ \iint \vec{E}\cdot \mathrm{d}\vec{A} \equal{} \frac{q}{\varepsilon_0}\]

In other words, if there is a charge present, point charge or not, then $ \vec{E}\cdot\vec{A} \neq 0$. Furthermore, from the definition of the dot product $ \vec{E}\cdot\vec{A} \equal{} E_1A_1 \plus{} E_2A_2 \plus{} E_3A_3 \plus{}\ldots \plus{} E_nA_n$.

In this particular case $ n \equal{} 3$, and $ A_2 \equal{} A_3 \equal{} \pi R^2$

$ A_1 \equal{} 2\pi R h$ and $ E_1 \equal{} E_2 \equal{} E_3$ since the field is isotropic.

\[ \implies \vec{E}\cdot\vec{A} \equal{} E_1(2\pi R^2 h) \plus{} E_2(\pi R^2) \plus{} E_3(\pi R^2) \implies E(2\pi [R h \plus{} R^2]) \equal{}\frac{q}{\varepsilon_0}\]

(I messed up in my previous post. I broke down the field into only two components but three area elements. Since the field is isotropic all the components add!)

[quote="Dr. No"]

In other words, if there is a charge present, point charge or not, then $ \vec{E}\cdot\vec{A} \neq 0$. Furthermore, from the definition of the dot product $ \vec{E}\cdot\vec{A} \equal{} E_1A_1 \plus{} E_2A_2 \plus{} E_3A_3 \plus{} \ldots \plus{} E_nA_n$.[/quote]

That is not how you define the dot product of 2 vectors or even the circulation integral of $ \vec E \cdot d\vec A$

[quote="Dr. No"]

$ A_1 \equal{} 2\pi R h$ and $ E_1 \equal{} E_2 \equal{} E_3$ since the [b]field is isotropic[/b].
\[ \implies \vec{E}\cdot\vec{A} \equal{} E_1(2\pi R^2 h) \plus{} E_2(\pi R^2) \plus{} E_3(\pi R^2) \implies E(2\pi [R h \plus{} R^2]) \equal{} \frac {q}{\varepsilon_0}
\]
(I messed up in my previous post. I broke down the field into only two components but three area elements. [b]Since the field is isotropic[/b] all the components add!)[/quote]

You conveniently forget to mention everywhere that the field is [b]radially isotropic[/b]. This seems to be the very basis of why you seem to be getting the wrong result.

As Conjurer pointed out, the field and the element vector are not along the same direction at every point on the cylinder, so that their dot product is not necessarily the product of their magnitudes.

Dr. No
At least you fixed on the sign of the flux in the basis ( by the way there is no [b]plate[/b]; the whole cylinder is an abstract closed surface).
The field being isotropic doesn't means it's field lines are perependicular to all the surface elements, unless they belong to a spherical surface centered in the source (the point charge).
Actually the flux in each basis would be

\[ {q\over{2\epsilon_0}} ( {1 - {h \over {\sqrt{4R^2+ h^2}}})}\]

and in the lateral surface

\[ {{qh \over{\epsilon_0 \sqrt{4R^2+ h^2}}}\;\;}\]

[quote="hell_ever"][quote="Dr. No"]

In other words, if there is a charge present, point charge or not, then $ \vec{E}\cdot\vec{A} \neq 0$. Furthermore, from the definition of the dot product $ \vec{E}\cdot\vec{A} = E_1A_1 + E_2A_2 + E_3A_3 + \ldots + E_nA_n$.[/quote]

That is not how you define the dot product of 2 vectors...[/quote]

Sure it is. Look it up.  :furious:

Let $ \mathbf{a} =\left[\begin{array}{c}a_1 \\ a_2 \\ a_3 \\ \vdots \\ a_n \end{array}\right]$

Likewise let $ \mathbf{b} = \left[\begin{array}{c}b_1 \\ b_2 \\ b_3 \\ \vdots \\ b_n \end{array}\right]$

Then

\begin{align*} \mathbf{a}\cdot\mathbf{b}\, &\stackrel{\triangle}=\, \mathbf{a}^{\tt T}\mathbf{b} \\ &=\left[\begin{array}{ccccc}a_1 & a_2 & a_3 & \ldots & a_n \end{array}\right]\left[\begin{array}{c}b_1 \\ b_2 \\ b_3 \\ \vdots \\ b_n \end{array}\right] \\ &= a_1b_1 + a_2b_2 + a_3b_3 +\ldots + a_nb_n\end{align*}

[quote="hell_ever"]...or even the circulation integral of $ \vec E \cdot d\vec A$[/quote]

$ \iint_S \vec{E}\cdot\mathrm{d}\vec{A}$ [b]is not a circulation integral.[/b] It's a surface integral over a closed region $ S$. The circle that often appears around the double integral symbol (something I can't represent in $ \text{\LaTeX}$) means that $ S$ is closed.

If you're going to be condescending toward others, especially toward people whom you don't know, it behooves you to know what you're talking about!

[quote="Dr. No"]

$ A_1 = 2\pi R h$ and $ E_1 = E_2 = E_3$ since the [b]field is isotropic[/b].
\[ \implies \vec{E}\cdot\vec{A} = E_1(2\pi R^2 h) + E_2(\pi R^2) + E_3(\pi R^2) \implies E(2\pi [R h + R^2]) = \frac {q}{\varepsilon_0}
\]

(I messed up in my previous post. I broke down the field into only two components but three area elements. [b]Since the field is isotropic[/b] all the components add!)[/quote]

[quote="hell_ever"]You conveniently forget to mention everywhere that the field is [b]radially isotropic[/b]. This seems to be the very basis of why you seem to be getting the wrong result.[/quote]

No, I explained why I got the wrong result at first. The result that you're criticizing in this post is actually correct.

There are three regions for the cylinder, two end plates, and the circular tube. In my first post I assumed that there was a single axial component of E-field pointed in one direction. If true that would mean that since the outward normal vectors of each end plate are pointed in opposite directions, these when dotted with the E-field imply that the flux contributions of the end plates would cancel each other. However, since the E-field is (radially) isotropic -- meaning that it has the same value in every direction one can look when standing on the point charge -- my assumption of a single axial component of E-field was incorrect. There is a component of E-field pointed along the direction of each outward normal vector, and as such the contributions of the end plates don't cancel, they add.

[quote="hell_ever"]As Conjurer pointed out, the field and the element vector are not along the same direction at every point on the cylinder, so that their dot product is not necessarily the product of their magnitudes.[/quote]

As I said, please know what you're talking about before launching into a critique. The E-field of a point charge is isotropic: it has the same magnitude in every direction. Therefore necessarily $ \vec{E}\cdot (2\pi R h)\hat{r} = E(2\pi R h)$, and $ \vec{E}\cdot (\pi R^2)\hat{z_+} = E(\pi R^2)$, and likewise $ \vec{E}\cdot (\pi R^2)\hat{z_-} = E(\pi R^2)$.

To obtain the total flux over the entire cylinder you merely have to add each of these flux contributions. (I don't know if you know much calculus, but one can evaluate integrals in a piecewise fashion, and that is precisely what I've done here.) The result being

\[ E(2\pi [R h + R^2]) =\frac{q}{\varepsilon_0}\]

[quote="Immanuel Bonfils"]Dr. No
At least you fixed on the sign of the flux in the basis ( by the way there is no [b]plate[/b]; the whole cylinder is an abstract closed surface).[/quote]

I understand that the cylinder is an abstraction, as is my use of the term "plate." I used the word to distinguish that particular region of the surface of cylinder. There is no actual no material object capping the cylinder on either end. Nonetheless, if there were an actual material plate it would make no difference to the mathematics -- this is a physics problem after all. We do this so we can discuss real objects.

[quote="Immanuel Bonfils"]The field being isotropic doesn't means it's field lines are perependicular to all the surface elements, unless they belong to a spherical surface centered in the source (the point charge).[/quote]

You might want to re-think that comment. [url=http://www.thefreedictionary.com/isotropic]Isotropic[/url]

[quote="Immanuel Bonfils"]Actually the flux in each basis would be
\[ {q\over{2\epsilon_0}} ( {1 - {h \over {\sqrt {4R^2 + h^2}}})
}\]
and in the lateral surface
\[ {{qh \over{\epsilon_0 \sqrt {4R^2 + h^2}}}\;\;
}\]
[/quote]

I'm sorry if you went to the trouble to calculate some grossly complicated integral, my friend.

You are incorrect.

It's a point charge. As such, the z-component of E-field at "top" end plate is:

\[ E_z =\frac{q\frac{h}{2}}{4\pi \varepsilon_0 (R^2 + \frac{h^2}{4})^{\frac{3}{2}}}\]

At the "bottom" plate the z-component of the E-field is has equal magnitude but is pointed in the opposite direction.

The flux through either of these regions is then $ E_z(\pi R^2)$.

Notwithstanding, none of that is necessary to solve this problem. The point of Gauss's law is to simplify electric field calculations. The electric flux across the surface of the cylinder is $ \frac{q}{\varepsilon_0}$, period.

If you then apply the mean value theorem to the Gauss integral you obtain $ E_{ave} = \frac{q}{\varepsilon_0 (2\pi [Rh + R^2])}$.

[quote="Immanuel Bonfils"]Fantastic; the magnitude of [b]your[/b] Coulombian Field, $ \; E \;$ is constant over all the surface of the cylinder !!![/quote]

You guys are missing the point of Gauss's law.

It is Gauss's law that tells you that the E-field of a point charge is spherically symmetric in the first place.

Consider a point charge enclosed in a shell of some arbitrary shape.

By Gauss's law $ \iint_S \mathbf{E}\cdot \mathbf{\hat{n}}\, \mathrm{d}A \equal{} \frac {q}{\varepsilon_0}$.

From the integral we immediately obtain, $ E_n \equal{} \frac {q}{\varepsilon_0 A}$.

We can now make the substitution $ A \equal{} \Omega R^2$, where $ \Omega$ is a solid angle that defines the shape of the surface enclosing $ q$.
\[ E_n \equal{} \frac {q}{\Omega R^2 \varepsilon_0}
\]
Let $ R$ be fixed. If you now probe the field at a $ R$ from the charge, you find said field is constant no matter where you insert your probe. That implies $ \Omega \equal{} 4\pi$ (you didn't find any voids, attenuation or gain.).

Thus $ E_n \equal{} \frac {q}{4 \pi R^2\varepsilon_0}$. Now you know that $ 4\pi R^2$ is the surface area of a sphere; therefore, the field of a point charge is spherically symmetric.

Dr. No

Please don't misunderstand me, but I'm trying to help you. Seems you forgot our dialog some months ago; it should be about June, last year.The subject was Induction or Magnetic Induction, something of this sort... and you where kind of asking my help or explanation. I did it and didn't get any answer from you.

Then, you stayed away from this Forum's section for some months, and now you're back and I can see you improved a lot your Physics, but still you have more to learn.

  [quote="Dr. No"]I'm sorry if you went to the trouble to calculate some grossly complicated integral, my friend. [/quote]

It wasn't [b]grossly complicate[/b] at all, they where quite wright, and may be  [b]you[/b] could do it as exercises, and improve more in your knowledges.

[quote="BlueVelvet"][quote="Dr. No"]As I said, please know what you're talking about before launching into a critique. The E-field of a point charge is isotropic: it has the same magnitude in every direction. Therefore necessarily $ \vec{E}\cdot (2\pi R h)\hat{r} \equal{} E(2\pi R h)$
[/quote]This doesn't make any sense. The E-field of a point charge is spherical symmetric, i.e. in the radial direction. The surface of a cylinder is obviously not spherical, hence the dot product $ \vec{E}\cdot d\vec{A}$ is not simply $ E\cdot dA$. This simplification would only be correct in the case of a sphere instead of a cylinder.

Here's a crappy drawing I made:[/quote]

I appreciate your drawing. It's actually quite good.

Gauss's law does not assume that the field produced by a given charge has a particular symmetry.

I showed in a previous post how Coulomb's law is derivable from Gauss's law.

That is because a free charge not contained within any particular volume of space produces an isotropic E-field, as a consequence said field has spherical symmetry.

However in this case, the charge [b]is[/b] contained within a particular volume of space.

Admittedly, my calculation for $ E \equal{} \frac{q}{\varepsilon_0 2\pi (Rh \plus{} R^2)}$ represents little more than the charge distribution -- surface charge density -- on the cylindrical surface. However, if your objective is to use Gauss's law to compute the E-field at the surface of the cylinder then it is best to do so piecewise.

We have already discovered that the fields at the end plates are equal in magnitude but opposite in direction thus they cancel one another. The radial component of field however survives.

\[ E_R \equal{} \frac{q}{\varepsilon_0 2\pi R h}\]

This is useful if there is an actual cylinder containing the charge $ q$.

Otherwise, it is more precise to model $ q$ as a free charge then the field has spherical symmetry and we can model it as:

\[ \vec{E}(r, z) \equal{}\frac{q}{4\pi \varepsilon_0} \frac{(r \minus{} r_0)\hat{r} \plus{} (z \minus{} z_0)\hat{z}}{(\sqrt{(r \minus{} r_0)^2 \plus{}  (z \minus{} z_0)^2})^3}\]

where $ (r_0, z_0)$ is the position of $ q$.

Rather than trying to reconcile these two models for the E-field it is easier to recognize that $ E_R$ as computed from Gauss's law represents an average E-field, and that from the mean value theorem of calculus it is the actual E-field at some point on the surface of the cylinder.

[quote="Immanuel Bonfils"]Dr. No

Please don't misunderstand me, but I'm trying to help you. Seems you forgot our dialog some months ago; it should be about June, last year.The subject was Induction or Magnetic Induction, something of this sort... and you where kind of asking my help or explanation. I did it and didn't get any answer from you.

Then, you stayed away from this Forum's section for some months, and now you're back and I can see you improved a lot your Physics, but still you have more to learn.[/quote]

Immanuel, I appreciate that you're trying to help me; but in all seriousness, there are probably others here who can use your help more than I. As you can probably tell from my writings I am not an undergrad. I too come here to help others, and in no small measure because I enjoy solving problems.

I do admit to my share of errors, but I also correct most of them in subsequent posts, and where I do make errors I make them mostly from a lack of intuition (or perhaps a surfeit of it), not because of a lack of understanding of the physics. I have a Master's degree in the subject -- my screen name is something of a pun. I call myself Dr. No because I didn't finish my doctorate. (Also because "No" is a diminutive of my real name.)

If you felt that I abandoned our previous discussion prematurely, I'm apologize. I only come here when time permits and I seldom stick around a thread if it gets off topic or starts a flame war.

  [quote="Dr. No"]I'm sorry if you went to the trouble to calculate some grossly complicated integral, my friend. [/quote]

[quote="Immanuel Bonfils"]It wasn't [b]grossly complicate[/b] at all, they where quite wright, and may be  [b]you[/b] could do it as exercises, and improve more in your knowledges.[/quote]

No worries. Right or wrong, the point of this problem was not to find the flux or the field anyhow.

I would remind everyone that if appealing to Gauss's law does not make your calculation of electric field easier than it would be otherwise you're probably missing the point.

Master's degree or not, you are and have been making [b]a lot [/b]of elementary mistakes.
Everyone agrees that the answer to the original question (the total flux through a cylinder) is just $ \frac {q}{\varepsilon_0}$, like the flux through any closed surface. We all know Gauss' Law (hopefully).

But if you want to calculate the flux explicitly, i.e. work out the integral $ \int \vec{E}\cdot d\vec{A}$ explicitly, you obviously can't use Gauss law. You can use Coulom's law to determine the E-field at every point, and calculate the integral (separately for the three parts of the surface and then add hem) (and yes I know how to derive Coulomb from Gauss).

Let's calculate the flux through one of the top circular plates. So we have a point charge and some circular disc of radius R, the distance between the charge and the center of the disc being h/2, half of the height of the cylinder. Using spherical coordinates, we can look at a constant radial vector to obtain for a surface element $ d\mathbf{a} \equal{} r^2\sin\theta d\theta d\phi \hat{r}$.

$ \mbox{Flux} \equal{} \phi_E \equal{} \int \mathbf{E}\cdot d\mathbf{a} \equal{} \int \left(\frac {q}{4\pi\epsilon_0}\frac {1}{r^2}\hat{r}\right)\cdot(r^2\sin\theta d\theta d\phi \hat{r}) \equal{} \frac {q}{4\pi\epsilon_0}\int\int \sin\theta d\theta d\phi$

$ \phi$ goes all around, $ \phi\in[0,2\pi]$, and $ \theta\in[0,\theta_0]$. We can determine $ \theta_0$ with elementary trigonometry, $ \cos\theta_0 \equal{} \frac {h/2}{\sqrt {R^2 \plus{} (h/2)^2}}$.
Then we get $ \phi_E \equal{} \frac {q}{2\pi\epsilon_0}\left(1 \minus{} \frac {h}{2\sqrt {R^2 \plus{} h^2/4}}\right) \equal{} \frac {q}{2\pi\epsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)$, just like Immanuel Bonfils got.

It's ok to be confident, but a critical look at your own calculations won't hurt. So please, read all messages in this thread (again) carefully, because several mistakes have been pointed out already. These parts for example, are all plain incorrect:

[quote="Dr. No"]from the definition of the dot product $ \vec{E}\cdot\vec{A} \equal{} E_1A_1 \plus{} E_2A_2 \plus{} E_3A_3 \plus{} \ldots \plus{} E_nA_n$.

In this particular case $ n \equal{} 3$, and $ A_2 \equal{} A_3 \equal{} \pi R^2$

$ A_1 \equal{} 2\pi R h$ and $ E_1 \equal{} E_2 \equal{} E_3$ since the field is isotropic.
\[ \implies \vec{E}\cdot\vec{A} \equal{} E_1(2\pi R^2 h) \plus{} E_2(\pi R^2) \plus{} E_3(\pi R^2) \implies E(2\pi [R h \plus{} R^2]) \equal{} \frac {q}{\varepsilon_0}
\]
[/quote][quote="Dr. No"]
As I said, please know what you're talking about before launching into a critique. The E-field of a point charge is isotropic: it has the same magnitude in every direction. Therefore necessarily $ \vec{E}\cdot (2\pi R h)\hat{r} \equal{} E(2\pi R h)$, and $ \vec{E}\cdot (\pi R^2)\hat{z_ \plus{} } \equal{} E(\pi R^2)$, and likewise $ \vec{E}\cdot (\pi R^2)\hat{z_ \minus{} } \equal{} E(\pi R^2)$.

To obtain the total flux over the entire cylinder you merely have to add each of these flux contributions. The result being
\[ E(2\pi [R h \plus{} R^2]) \equal{} \frac {q}{\varepsilon_0}
\]
[/quote]

[quote="BlueVelvet"]Master's degree or not, you are and have been making [b]a lot [/b]of elementary mistakes.[/quote]

In your opinion.

[quote="BlueVelvet"]Everyone agrees that the answer to the original question (the total flux through a cylinder) is just $ \frac {q}{\varepsilon_0}$, like the flux through any closed surface. We all know Gauss' Law (hopefully).[/quote]

Agreed. Everyone should know that, but some don't; hence the question.

[quote="BlueVelvet"]But if you want to calculate the flux explicitly, i.e. work out the integral $ \int \vec{E}\cdot d\vec{A}$ explicitly, you obviously can't use Gauss law.[/quote]

Why would anyone want to calculate the flux explicitly? This is an electrostatic case with a point source. You already know the expression for the E-field of a point charge. The flux tells you nothing -- it's an exercise in calculation for the sake of calculation.

[quote="BlueVelvet"]Let's calculate the flux through one of the top circular plates. So we have a point charge and some circular disc of radius R, the distance between the charge and the center of the disc being h/2, half of the height of the cylinder. Using spherical coordinates, we can look at a constant radial vector to obtain for a surface element $ d\mathbf{a} \equal{} r^2\sin\theta d\theta d\phi \hat{r}$.

$ \mbox{Flux} \equal{} \phi_E \equal{} \int \mathbf{E}\cdot d\mathbf{a} \equal{} \int \left(\frac {q}{4\pi\epsilon_0}\frac {1}{r^2}\hat{r}\right)\cdot(r^2\sin\theta d\theta d\phi \hat{r}) \equal{} \frac {q}{4\pi\epsilon_0}\int\int \sin\theta d\theta d\phi$

$ \phi$ goes all around, $ \phi\in[0,2\pi]$, and $ \theta\in[0,\theta_0]$. We can determine $ \theta_0$ with elementary trigonometry, $ \cos\theta_0 \equal{} \frac {h/2}{\sqrt {R^2 \plus{} (h/2)^2}}$.
Then we get $ \phi_E \equal{} \frac {q}{2\pi\epsilon_0}\left(1 \minus{} \frac {h}{2\sqrt {R^2 \plus{} h^2/4}}\right) \equal{} \frac {q}{2\pi\epsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)$, just like Immanuel Bonfils got.[/quote]

Not that I mind, but this is not the right forum to demonstrate your skill at integral calculus. Beside which, after all that, you got the answer wrong. Immanuel got it right, you didn't. (It's not a glaring mistake, but it is a mistake nonetheless. Given your penchant for pedantry, I would be remiss not to point it out.)

[quote="BlueVelvet"]It's ok to be confident, but a critical look at your own calculations won't hurt. So please, read all messages in this thread (again) carefully, because several mistakes have been pointed out already. These parts for example, are all plain incorrect:

[quote="Dr. No"]from the definition of the dot product $ \vec{E}\cdot\vec{A} \equal{} E_1A_1 \plus{} E_2A_2 \plus{} E_3A_3 \plus{} \ldots \plus{} E_nA_n$.

In this particular case $ n \equal{} 3$, and $ A_2 \equal{} A_3 \equal{} \pi R^2$

$ A_1 \equal{} 2\pi R h$ and $ E_1 \equal{} E_2 \equal{} E_3$ since the field is isotropic.
\[ \implies \vec{E}\cdot\vec{A} \equal{} E_1(2\pi R^2 h) \plus{} E_2(\pi R^2) \plus{} E_3(\pi R^2) \implies E(2\pi [R h \plus{} R^2]) \equal{} \frac {q}{\varepsilon_0}
\]
[/quote][/quote]

As I have pointed out to others previously, this is not wrong. $ E$ in the above calculation is the average electric field at the surface of a cylindrical body that is charged with a charge $ q$. Furthermore, the mean value theorem implies that there must exist at least one point on the surface where the E-field is actually equal to $ E$.

[quote="Dr. No"]
As I said, please know what you're talking about before launching into a critique. The E-field of a point charge is isotropic: it has the same magnitude in every direction. Therefore necessarily $ \vec{E}\cdot (2\pi R h)\hat{r} \equal{} E(2\pi R h)$, and $ \vec{E}\cdot (\pi R^2)\hat{z_ \plus{} } \equal{} E(\pi R^2)$, and likewise $ \vec{E}\cdot (\pi R^2)\hat{z_ \minus{} } \equal{} E(\pi R^2)$.

To obtain the total flux over the entire cylinder you merely have to add each of these flux contributions. The result being
\[ E(2\pi [R h \plus{} R^2]) \equal{} \frac {q}{\varepsilon_0}
\]
[/quote][/quote]

This snippet is essentially the same as the one above.

Dr. No, when you make a mistake very obvious, and many have pointed it out, then you should at least try to see your work with a critical eye, just as blue velvet said.

Actually your mistake is a mathematical one, not a physical mistake. You just calculated, the surface integral over a cylinder wrongly. That's all.

And I'm sure you have noticed your mistake already, If you have the education you described above.
Just my question is then, why dont you admit, "yes, I was wrong", and that would be it....

In your last post you wrote:

[quote]Immanuel got it right, you didn't. (It's not a glaring mistake, but it is a mistake nonetheless...[/quote]

So my above assumptions are true, you do see that what you did is wrong, since, here you admit, that Immanuel's work is correct..., but you still you dont say that "I was wrong" and go on trying to justify yourself.

Here is another thing I dont understand, what it has to do with the given problem:

[quote]...the above calculation is the average electric field at the surface of a cylindrical body that is charged with a charge...[/quote]

A cylinder "charged" with a charge? The cylinder here is just an "imaginary" surface, we are integrating the electric field of a SINGLE charge on, it doesnt have any charge..

[quote="Dr. No"]
$ \iint_S \vec{E}\cdot\mathrm{d}\vec{A}$ [b]is not a circulation integral.[/b] It's a surface integral over a closed region $ S$. The circle that often appears around the double integral symbol (something I can't represent in $ \text{\LaTeX}$) means that $ S$ is closed.

[/quote]

Is it this $ \oint$?

If $ \frac{q}{\epsilon_0}$ is the correct answer, what is everyone arguing for?

What if the charge were placed on the circumference of any one circular surface? What would the flux be in that case?

[quote="Dr. No"]

As I have pointed out to others previously, this is not wrong. $ E$ in the above calculation is the average electric field at the surface of a cylindrical body that is charged with a charge $ q$. Furthermore, the mean value theorem implies that there must exist at least one point on the surface where the E-field is actually equal to $ E$.
[/quote]

I think you should give up now.

[quote="Rajiv"][quote="Dr. No"]
$ \iint_S \vec{E}\cdot\mathrm{d}\vec{A}$ [b]is not a circulation integral.[/b] It's a surface integral over a closed region $ S$. The circle that often appears around the double integral symbol (something I can't represent in $ \text{\LaTeX}$) means that $ S$ is closed.

[/quote]

Is it this $ \oint$?

If $ \frac {q}{\epsilon_0}$ is the correct answer, what is everyone arguing for?[/quote]

$ \oint$ is the symbol for a closed path integral. I believe that is what hell_ever might have had in mind when he wrote "circulation integral." Indeed a circulation integral, something that is properly called a circulation integral, arises from Stokes's theorem and is the surface integral, $ \iint_S \nabla \times \vec{F}\cdot \mathrm{d}\vec{A}$, where $ \vec{F}$ is some vector field. However, the double integral, with a circle around both integral symbols means the integral over a closed surface just as $ \oint_C$ means the integral over a closed path $ C$. It does not necessarily mean a circulation integral. In any case the closed surface integral, double integral with a circle around both S's, does not have a built-in $ \text{\LaTeX}$ command.

I haven't got a clue why every one is arguing except perhaps that we have a room full of folks who all very much want to win the "smartest man in the world" contest, often at the expense of others.  :wink:

[quote="Rajiv"]What if the charge were placed on the circumference of any one circular surface? What would the flux be in that case?[/quote]

I'll try to answer this question in a further post. Last night I actually performed the integration that both Immanuel and BlueVelvet insisted was necessary and came to some interesting conclusions.

[quote="Conjurer"]I think you should give up now.[/quote]

Me too.

[quote="Thaakisfox"]Dr. No, when you make a mistake very obvious, and many have pointed it out, then you should at least try to see your work with a critical eye, just as blue velvet said.[/quote]

I am my own biggest critic, and though it might not seem so, I do seriously consider the criticisms of others. If you've read a good portion of my other posts here at AoPS you know that I am as prone to error as anyone here (perhaps more so) and when I am wrong I admit it. Notwithstanding, in this case I am firmly convicted that I'm correct (at least in terms of my approach to the problem even if not in its final result). It's not a matter of a lack of humility on my part, it is a matter of whether I will concede a falsehood. I won't, because that is a matter of integrity.

[quote="Thaakisfox"]Actually your mistake is a mathematical one, not a physical mistake. You just calculated, the surface integral over a cylinder wrongly. That's all.[/quote]

Perhaps, but then so have my critics as I shall show shortly, if you all will permit me.

[quote="Thaakisfox"]So my above assumptions are true, you do see that what you did is wrong, since, here you admit, that Immanuel's work is correct..., but you still you dont say that "I was wrong" and go on trying to justify yourself.[/quote]

Not necessarily. What I do admit is that when it comes to manually performing the operation of integration I'm lazy. I find it tedious and I will avoid it if at all possible. (It is in that spirit that I have approached this problem. From an electrostatic standpoint, Gauss's law is a tool to avoid tedious integrations. If one can't avoid them by it then it is utterly useless. Notwithstanding, GL has electrodynamic implications that make it interesting in and of itself but this is not one of those problems.) It is a profound misjudgment, that I find highly annoying, to presume that my preference for avoiding integration stems from an inability on my part to integrate a given function. That, and arrogance mixed with ignorance is a terrible combination in my opinion that I shall continue to eschew wherever I find it.

[quote="Thaakisfox"]Here is another thing I dont understand, what it has to do with the given problem:

[quote]
...the above calculation is the average electric field at the surface of a cylindrical body that is charged with a charge...[/quote]

A cylinder "charged" with a charge? The cylinder here is just an "imaginary" surface, we are integrating the electric field of a SINGLE charge on, it doesnt have any charge..[/quote]

The situation is a mental contrivance certainly, but it represents an actual physical phenomenon that we are supposedly trying to understand by doing the math. Gauss's law essentially says that electric charges are sources of electric fields, and further that if the volume of space containing a given charge is electrically conductive then the charge contained in said volume will migrate to its surface. At that point, the surface itself is said to be charged. Conductivity of a given region notwithstanding, when a volume of space contains a charge it is said to be "charged." This is what I meant when I wrote "charged with a charge $ q$."

[quote]
Immanuel got it right, you didn't. (It's not a glaring mistake, but it is a mistake nonetheless...

[quote="Thaakisfox"]So my above assumptions are true, you do see that what you did is wrong, since, here you admit, that Immanuel's work is correct..., but you still you dont say that "I was wrong" and go on trying to justify yourself.[/quote][/quote]

I defended myself against the accusations of a hypocrite. It is the definition of hypocrisy to accuse others of being in error when one is in error one's self; nevertheless, with respect to the members of AoPS, I prefer to think that this behavior stems from ignorance rather than malice.

As for Immanuel's work being correct, his computation of the integral that he integrated was correct. BlueVelvet's, assuming that BlueVelvet's integral was the same one that Immanuel had computed, was not. At this point, having performed the integration for myself, I cannot say that either of them selected the right boundary conditions for their computations however.

Rather than continue all of this finger pointing though, I would really rather just do the integral, properly, and have everyone come to their own conclusions. Before I do that I'd like to comment specifically on what, in my opinion, is wrong with BlueVelvet's calculation.

[quote="BlueVelvet"]Let's calculate the flux through one of the top circular plates. So we have a point charge and some circular disc of radius R, the distance between the charge and the center of the disc being h/2, half of the height of the cylinder. Using spherical coordinates, we can look at a constant radial vector to obtain for a surface element $ d\mathbf{a} \equal{} r^2\sin\theta d\theta d\phi \hat{r}.$[/quote]

If you use spherical coordinates for this, you end up integrating over a dome shape (like the top of an ice cream cone) rather than a flat circular plate. To my mind, cylindrical coordinates are most appropriate since the body in question is in fact a cylinder.

[quote="BlueVelvet"]$ \mbox{Flux} \equal{} \phi_E \equal{} \int \mathbf{E}\cdot d\mathbf{a} \equal{} \int \left(\frac {q}{4\pi\epsilon_0}\frac {1}{r^2}\hat{r}\right)\cdot(r^2\sin\theta d\theta d\phi \hat{r}) \equal{} \frac {q}{4\pi\epsilon_0}\int\int \sin\theta d\theta d\phi$

$ \phi$ goes all around, $ \phi \in [0, 2\pi]$ , and $ \theta\in [0,\theta_0]$ . We can determine $ \theta_0$ with elementary trigonometry, $ \cos\theta_0 \equal{} \frac {h/2}{\sqrt {R^2 \plus{} (h/2)^2}}.$
Then we get $ \phi_E \equal{} \frac {q}{2\pi\epsilon_0}\left(1 \minus{} \frac {h}{2\sqrt {R^2 \plus{} h^2/4}}\right) \equal{} \frac {q}{2\pi\epsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)$, just like Immanuel Bonfils got.[/quote]

Almost, you failed to cancel the factor of $ \pi$ in the denominator of the leading coefficient upon integration over $ \phi$. Other than that, as I said, you integrated over the top of an "ice cream cone."

Setting aside the work of others for the moment, the E-field of a point charge located at (0, 0, 0) in cylindrical coordinates is given by:
\[ \vec{E}(r, z) \equal{} \frac {q}{4\pi \varepsilon_0}\frac {r\,\hat{r} \plus{} z\,\hat{z}}{(r^2 \plus{} z^2)^{\frac {3}{2}}}
\]
Considering the differential flux across the top end plate:
\[ \vec{E}(r, \frac {h}{2})\cdot r\mathrm{dr}\mathrm{d}\phi\, \hat{z} \equal{} \frac {q}{4\pi \varepsilon_0}\frac {\frac {h}{2} r\mathrm{d}r\mathrm{d}\phi}{(r^2 \plus{} \frac {h^2}{4})^{\frac {3}{2}}} \implies \int_{0}^{R}\int_{0}^{2\pi} \vec{E}(r, \frac {h}{2})\cdot r\mathrm{dr}\mathrm{d}\phi\, \hat{z} \equal{} \left. \minus{} \frac {q}{2\varepsilon}\frac {h}{\sqrt {4r^2 \plus{} h^2}}\right|_{0}^{R} \equal{} \frac {q}{2\varepsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)
\]
The integral over the bottom plate is identical because though the bottom plate is at $ z \equal{} \minus{} \frac {h}{2}$, the differential area it is dotted with is pointed in the minus z-direction, the minus sign of the E-field cancels the minus sign of the unit vector of the differential area giving us an integral identical to the one just computed.

Thus the combined flux across both end plates is:
\[ \frac {q}{\varepsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)
\]
Now, for the tube portion:
\[ \vec{E}(R, z)\cdot R\mathrm{dz}\mathrm{d}\phi\, \hat{r} \equal{} \frac {q}{4\pi \varepsilon_0}\frac {R^2\mathrm{d}z\mathrm{d}\phi}{(R^2 \plus{} z^2)^{\frac {3}{2}}} \implies \int_{\frac { \minus{} h}{2}}^{\frac {h}{2}}\int_{0}^{2\pi} \vec{E}(R, z)\cdot R\mathrm{dz}\mathrm{d}\phi\, \hat{r} \equal{} \left. \frac {q}{2\varepsilon_0}\frac {z}{\sqrt {R^2 \plus{} z^2}}\right|_{{ \minus{} h}{2}}^{\frac {h}{2}} \equal{} \frac {q}{\varepsilon_0}\frac {h}{\sqrt {4R^2 \plus{} h^2}}
\]
This last integral is easily solved using trigonometric substitution, but I'm loath to show it -- that tedium business again -- however to check it I plugged it into [url=http://integrals.wolfram.com/index.jsp?]The Integrator(tm)[/url]. I got the same result.

Finally, adding this result to the one for the end plates given above we obtain [i]drum roll, please ...[/i]
\[ \mbox{Flux} \equal{} \frac {q}{\varepsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right) \plus{} \frac {q}{\varepsilon_0}\frac {h}{\sqrt {4R^2 \plus{} h^2}} \equal{} \boxed{\frac {q}{\varepsilon_0}}
\]
Gauss's law works, all is well with the world. So enough of the :cursing: ad hominem nonsense, everyone, please.

[quote="Dr. No"]Why would anyone want to calculate the flux explicitly? This is an electrostatic case with a point source. You already know the expression for the E-field of a point charge. The flux tells you nothing -- it's an exercise in calculation for the sake of calculation.[/quote]
Exactly my thoughts, but I thought it was you who brought the whole computation up in the first place?  :maybe:
It is not necessary to compute [i]anything[/i] because we have Gauss, but if people are posting wrong calculations I think it's a good thing to correct that.
Besides, this calculation can be useful to some, otherwise we wouldn't even have this discussion at all.
[quote="BlueVelvet"]Let's calculate the flux through one of the top circular plates. So we have a point charge and some circular disc of radius R, the distance between the charge and the center of the disc being h/2, half of the height of the cylinder. Using spherical coordinates, we can look at a constant radial vector to obtain for a surface element $ d\mathbf{a} \equal{} r^2\sin\theta d\theta d\phi \hat{r}$.

$ \mbox{Flux} \equal{} \phi_E \equal{} \int \mathbf{E}\cdot d\mathbf{a} \equal{} \int \left(\frac {q}{4\pi\epsilon_0}\frac {1}{r^2}\hat{r}\right)\cdot(r^2\sin\theta d\theta d\phi \hat{r}) \equal{} \frac {q}{4\pi\epsilon_0}\int\int \sin\theta d\theta d\phi$

$ \phi$ goes all around, $ \phi\in[0,2\pi]$, and $ \theta\in[0,\theta_0]$. We can determine $ \theta_0$ with elementary trigonometry, $ \cos\theta_0 \equal{} \frac {h/2}{\sqrt {R^2 \plus{} (h/2)^2}}$.
Then we get $ \phi_E \equal{} \frac {q}{2\pi\epsilon_0}\left(1 \minus{} \frac {h}{2\sqrt {R^2 \plus{} h^2/4}}\right) \equal{} \frac {q}{2\pi\epsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)$, just like Immanuel Bonfils got.[/quote]
[quote]Not that I mind, but this is not the right forum to demonstrate your skill at integral calculus. [/quote]
I just can't believe this. These are precisely the remarks that make you sound rather arrogant, even childish. We are discussing some calculation, and I present the calculation which I believe is the right one. You repeatedly said Immanuel's answer is incorrect. The only sensible thing to do then is writing out the calculation, so you can point out where the mistake would be. How can you possibly interpret that as "demonstrating my skill at integral calulus"?
[quote]Beside which, after all that, you got the answer wrong. Immanuel got it right, you didn't.[/quote]
Obviously the $ \pi$ shouldn't be there.
[quote]As I have pointed out to others previously, this is not wrong. $ E$ in the above calculation is the average electric field at the surface of a cylindrical body that is charged with a charge $ q$. Furthermore, the mean value theorem implies that there must exist at least one point on the surface where the E-field is actually equal to $ E$.
(...)
This snippet is essentially the same as the one above.[/quote]
It's a mystery to me how you still are able to say this is true.

[quote="Dr. No"]As for Immanuel's work being correct, his computation of the integral that he integrated was correct. BlueVelvet's, assuming that BlueVelvet's integral was the same one that Immanuel had computed, was not. [/quote]
[quote]Almost, you failed to cancel the factor of $ \pi$ in the denominator of the leading coefficient upon integration over $ \phi$. [/quote]
Yes, as said before, my $ \pi$ was misplaced. I just copied the Latex-code and forgot to delete it (that's why I said 'the same as Immanuel's answer'). Pointing it out three times is more than enough  :)
[quote]Rather than continue all of this finger pointing though, I would really rather just do the integral, properly, and have everyone come to their own conclusions. [/quote]
Thank you....finally  :D
[quote]
Before I do that I'd like to comment specifically on what, in my opinion, is wrong with BlueVelvet's calculation.

If you use spherical coordinates for this, you end up integrating over a dome shape (like the top of an ice cream cone) rather than a flat circular plate. To my mind, cylindrical coordinates are most appropriate since the body in question is in fact a cylinder.

as I said, you integrated over the top of an "ice cream cone."
[/quote]
You are right, I am integrating over a dome shape instead of a flat circular plate. But ask yourself this: does that matter?  :wink:
[quote]
Setting aside the work of others for the moment $ \frac {q}{2\varepsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)$
(...)
Thus the combined flux across both end plates is:
$ \frac {q}{\varepsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)$
[/quote]
He, and what do we have here. The answer that was give already three times? I am glad you finally agree with us, although it took quite some time. Hopefully you will now appreciate the comments given before a bit more.
[quote]
This last integral is easily solved using trigonometric substitution, but I'm loath to show it -- that tedium business again -- however to check it I plugged it into [url=http://integrals.wolfram.com/index.jsp?]The Integrator(tm)[/url]. I got the same result.[/quote]
I'd love to show how it's done, but I realise this is not the place to show my integral calculus skills  :lol:

[quote="BlueVelvet"][quote="Dr. No"]Why would anyone want to calculate the flux explicitly? This is an electrostatic case with a point source. You already know the expression for the E-field of a point charge. The flux tells you nothing -- it's an exercise in calculation for the sake of calculation.[/quote]
Exactly my thoughts, but I thought it was you who brought the whole computation up in the first place?  :maybe:[/quote]

Yes, it was I who brought it up, a decision I now regret.  :(

[quote="BlueVelvet"]It is not necessary to compute [i]anything[/i] because we have Gauss, but if people are posting wrong calculations I think it's a good thing to correct that.[/quote]

Usually so, except when the calculations are not wrong.

[quote="BlueVelvet"]Besides, this calculation can be useful to some, otherwise we wouldn't even have this discussion at all.[/quote]

Agreed. I have nothing against seeing whatever kind of solutions someone wishes to post. But Introductory Physics is strictly 'plug and chug' when it comes to acceptable solutions.

[quote="BlueVelvet"]Let's calculate the flux through one of the top circular plates. So we have a point charge and some circular disc of radius R, the distance between the charge and the center of the disc being h/2, half of the height of the cylinder. Using spherical coordinates, we can look at a constant radial vector to obtain for a surface element $ d\mathbf{a} \equal{} r^2\sin\theta d\theta d\phi \hat{r}$.

$ \mbox{Flux} \equal{} \phi_E \equal{} \int \mathbf{E}\cdot d\mathbf{a} \equal{} \int \left(\frac {q}{4\pi\epsilon_0}\frac {1}{r^2}\hat{r}\right)\cdot(r^2\sin\theta d\theta d\phi \hat{r}) \equal{} \frac {q}{4\pi\epsilon_0}\int\int \sin\theta d\theta d\phi$

$ \phi$ goes all around, $ \phi\in[0,2\pi]$, and $ \theta\in[0,\theta_0]$. We can determine $ \theta_0$ with elementary trigonometry, $ \cos\theta_0 \equal{} \frac {h/2}{\sqrt {R^2 \plus{} (h/2)^2}}$.
Then we get $ \phi_E \equal{} \frac {q}{2\pi\epsilon_0}\left(1 \minus{} \frac {h}{2\sqrt {R^2 \plus{} h^2/4}}\right) \equal{} \frac {q}{2\pi\epsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)$, just like Immanuel Bonfils got.[/quote]
[quote]Not that I mind, but this is not the right forum to demonstrate your skill at integral calculus. [/quote]
[quote="BlueVelvet"]I just can't believe this. These are precisely the remarks that make you sound rather arrogant, even childish. We are discussing some calculation, and I present the calculation which I believe is the right one. You repeatedly said Immanuel's answer is incorrect.[/quote]

I owe him an apology then. Where I said Immanuel was wrong I based my opinion on the immediate assessment that it was beside the point to actually evaluate the integral -- something I still believe -- that, and it appeared to me at the time as if Immanuel might have chosen the wrong boundary conditions where he did evaluate the integral. In retrospect, on that point, I was self-evidently mistaken.

[quote="BlueVelvet"]The only sensible thing to do then is writing out the calculation, so you can point out where the mistake would be. How can you possibly interpret that as "demonstrating my skill at integral calulus"?[/quote]

No calculus in here is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84055]Rule #1[/url].

[quote="BlueVelvet"][quote]Beside which, after all that, you got the answer wrong. Immanuel got it right, you didn't.[/quote]
Obviously the $ \pi$ shouldn't be there.[/quote]

As you said, "Obviously."

[quote="BlueVelvet"][quote]As I have pointed out to others previously, this is not wrong. $ E$ in the above calculation is the average electric field at the surface of a cylindrical body that is charged with a charge $ q$. Furthermore, the mean value theorem implies that there must exist at least one point on the surface where the E-field is actually equal to $ E$.
(...)
This snippet is essentially the same as the one above.[/quote]
It's a mystery to me how you still are able to say this is true.[/quote]

[color=red][b]Warning![/b][/color][i] The following sidebar contains graphic scenes of differential and integral calculus that may not be suitable for younger viewers. Parental discretion is strongly advised.[/i]

The use of Gauss to solve electrostatic problems comes primarily from the fact that Gauss's law allows one to obtain a gauge of the electric field for a body of arbitrary shape (or one for which obtaining an analytic closed form solution is impossible -- and in nature such things are far more plentiful than ones for which you can calculate a closed form solution.).

With that idea in mind, it is certainly true that when $ \vec{E}$ is constant, $ {E} \equal{}\frac{Q_{enc}}{\varepsilon A}$, where $ A$ is the surface area of the body in question, and I have specified $ {\varepsilon}$ as opposed to $ \varepsilon_0$ to account for the possibility that the body is a dielectric.

However, Gauss's law is more generally applicable than merely for cases of constant E-field.

The mean value theorem of calculus states:

\[ \textrm{If } f(x)\textrm{ is a function continuous and differentiable for }x\in [a, b]\textrm{ then }\\ \\ \exists c\in [a, b]\textrm{ such that } f^{\prime}(c) \equal{}\frac{f(b) \minus{} f(a)}{b \minus{} a}\].

Applying this principle to the Gauss integral:

\[ \mbox{For some }r^2\in [0, A],\,\frac{d \Phi_E(r^2)}{d A} \equal{}\frac{\Phi_E(A) \minus{}\Phi_E(0)}{A \minus{} 0} \equal{}\frac{Q_{enc}}{\varepsilon A}.\]

Since $ \Phi_E \equal{}\iint_S \vec{E}\cdot\mathrm{d}\vec{A}$, then $ \frac{d \Phi_E(r^2)}{d A} \equal{}\frac{d}{dA}\iint_S \vec{E}\cdot\mathrm{d}\vec{A} \equal{} E(r^2)$.

Given the above result, I am supremely confident with regard to my original calculation: $ E\equal{}\frac{q}{\varepsilon_0 2\pi R h}$, though not so much of my final result (with the area of the end plates included) because the surface of the cylinder ceases to be differentiable at the ends.

Furthermore, since the use of calculus [i]ist verboten hier[/i] my thought was if calculation was intended then this was how it was to be done. As I have said many times this is the principal use of Gauss's law in electrostatics.

[quote="BlueVelvet"][quote="Dr. No"]As for Immanuel's work being correct, his computation of the integral that he integrated was correct. BlueVelvet's, assuming that BlueVelvet's integral was the same one that Immanuel had computed, was not. [/quote]
[quote]Almost, you failed to cancel the factor of $ \pi$ in the denominator of the leading coefficient upon integration over $ \phi$. [/quote]
Yes, as said before, my $ \pi$ was misplaced. I just copied the Latex-code and forgot to delete it (that's why I said 'the same as Immanuel's answer'). Pointing it out three times is more than enough  :)[/quote]

Yes, and one persnickety observation of what is in essence bad grammar deserves another.

[quote="BlueVelvet"][quote]Rather than continue all of this finger pointing though, I would really rather just do the integral, properly, and have everyone come to their own conclusions. [/quote]
Thank you....finally  :D
[/quote]

I'm happy to oblige. :)

[quote="BlueVelvet"][quote]Before I do that I'd like to comment specifically on what, in my opinion, is wrong with BlueVelvet's calculation.

If you use spherical coordinates for this, you end up integrating over a dome shape (like the top of an ice cream cone) rather than a flat circular plate. To my mind, cylindrical coordinates are most appropriate since the body in question is in fact a cylinder.

as I said, you integrated over the top of an "ice cream cone."
[/quote]
You are right, I am integrating over a dome shape instead of a flat circular plate. But ask yourself this: does that matter?  :wink:[/quote]

What?!? You are the one insisting upon pointless exactitude here. Since the area of a circle is what you were supposed integrate over, any curvature that departs from absolutely flat is an error.  The fact that in this case the limits of integration were such that the error didn't show up in the final result is God's grace to you.

[quote="BlueVelvet"][quote]
Setting aside the work of others for the moment $ \frac {q}{2\varepsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)$
(...)
Thus the combined flux across both end plates is:
$ \frac {q}{\varepsilon_0}\left(1 \minus{} \frac {h}{\sqrt {4R^2 \plus{} h^2}}\right)$
[/quote]
He, and what do we have here. The answer that was give already three times? I am glad you finally agree with us, although it took quite some time. Hopefully you will now appreciate the comments given before a bit more.
[quote]
This last integral is easily solved using trigonometric substitution, but I'm loath to show it -- that tedium business again -- however to check it I plugged it into [url=http://integrals.wolfram.com/index.jsp?]The Integrator(tm)[/url]. I got the same result.[/quote]
I'd love to show how it's done, but I realise this is not the place to show my integral calculus skills  :lol:[/quote]

I bet you would. ;)

[quote="Dr. No"]But Introductory Physics is strictly 'plug and chug' when it comes to acceptable solutions.[/quote]
[quote]No calculus in here is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84055]Rule #1[/url][/quote]
[quote]Furthermore, since the use of calculus [i]ist verboten hier[/i] [/quote]
I am sorry, I was not aware of that rule. Sometimes 'simple' problems can lead to a discussion on a more fundamental/advanced level. In my opinion it's up to the moderators to split the topic up* if it gets too advanced.
[quote]Yes, and one persnickety observation of what is in essence bad grammar deserves another.[/quote]
* If my English is bad, I apologize. It's not my first language.

Shall we leave it at this? I think it's clear to everyone how it's done, and the discussion is getting a bit tedious. Thanks for your contributions.

I hope I could leave this discussion, but can't waste the opportunity, commenting that Gauss Law has many important uses besides simplifying the field calculation.   [quote="Dr. No"]From an electrostatic standpoint, Gauss's law is a tool to avoid tedious integrations. If one can't avoid them by it then it is utterly useless.[/quote]
  Here we have an happy coincidence to use Gauss Law for an important subject:
[quote="Dr. No"]
What?!? You are the one insisting upon pointless exactitude here. Since the area of a circle is what you were supposed integrate over, any curvature that departs from absolutely flat is an error. The fact that in this case the limits of integration were such that the error didn't show up in the final result is God's grace to you.[/quote]
 It's not an error if it simplifies the calculus. God's grace was to Gauss understand that this wasn't a simple coincidence. Do you know why?

If Gauss' Law gives you an answer in a jiffy, why go for the all-around-the-world-and-back-to-the-same-place method :P .

If the cylinder had had a billion spikes all over it, each a kilometer long, I think computing it with the above method will surely take you a lifetime, and given the number of debate-loving folks here, maybe two more.

[quote="Once upon a time, I"]
What if the charge were placed on the circumference of any one circular surface? What would the flux be in that case?

[/quote]

[quote="Rajiv"]If Gauss' Law gives you an answer in a jiffy, why go for the all-around-the-world-and-back-to-the-same-place method :P .[/quote]

I have no idea. :surrender:

[quote="Rajiv"]If the cylinder had had a billion spikes all over it, each a kilometer long, I think computing it with the above method will surely take you a lifetime, and given the number of debate-loving folks here, maybe two more.[/quote]

Evidently, since the surface of such an object is not differentiable.

[quote="Once upon a time, I"]
What if the charge were placed on the circumference of any one circular surface? What would the flux be in that case?

[/quote]

That's easy!
\[ \boxed{\Phi_E \equal{}\frac{q}{\varepsilon_0}}\]

[quote="Immanuel Bonfils"]I hope I could leave this discussion, but can't waste the opportunity, commenting that Gauss Law has many important uses besides simplifying the field calculation.[/quote]

I agree that Gauss's divergence theorem, of which Gauss's law is a special case, was [i]a stroke of genius[/i] if you'll pardon the pun, but Gauss's law is a relation of three variables, $ (\vec{E},\vec{A}, Q_{enc})$, i.e. given any two find the third. What kind of other uses did you have in mind?

[quote="Dr. No"]From an electrostatic standpoint, Gauss's law is a tool to avoid tedious integrations. If one can't avoid them by it then it is utterly useless.[/quote]
You omitted my statement immediately following ...
[quote="where I"]Notwithstanding, GL [Gauss's law] has electrodynamic implications that make it interesting in and of itself but this is not one of those problems.[/quote]
[quote="Immanuel Bonfils"]Here we have an happy coincidence to use Gauss Law for an important subject:
[quote="Dr. No"]
What?!? You are the one insisting upon pointless exactitude here. Since the area of a circle is what you were supposed integrate over, any curvature that departs from absolutely flat is an error. The fact that in this case the limits of integration were such that the error didn't show up in the final result is God's grace to you.[/quote]
It's not an error if it simplifies the calculus. God's grace was to Gauss understand that this wasn't a simple coincidence. Do you know why?[/quote]

[i]"Make everything as simple as possible, but not simpler."[/i] -- [b]Albert Einstein[/b]

I cannot presume to know what Gauss understood, but I do know this:

The transformation between $ z$ in cylindrical coordinates, and a vertical vector in spherical coordinates is $ z \equal{} r_s\cos\theta$. If one's intention is to integrate over a circular region at a fixed height then $ z \equal{} r_s\cos\theta \implies dz \equal{} \cos\theta dr_s \minus{} r_s\sin\theta d\theta$ and $ dz \equal{} 0$, implying that both $ dr_s$ and $ d\theta$ are also equal to zero. However, if one chooses integrate using spherical coordinates, holding $ r_s$ fixed, then $ dz \equal{} \minus{}r_s\sin\theta d\theta$, which is an error.

The redeeming factor in this particular case is that we are dealing with a point charge, and the E-field of a point charge has spherical symmetry, so everything turns out alright in the end. However, in general there is no way to know what kind of symmetry a given charge distribution has and that is why one invokes Gauss's law in the first place!

[quote="BlueVelvet"]* If my English is bad, I apologize. It's not my first language.[/quote]

No need to apologize. Your English is almost flawless.  :)

[quote="BlueVelvet"]Shall we leave it at this? I think it's clear to everyone how it's done, and the discussion is getting a bit tedious. Thanks for your contributions.[/quote]

I think perhaps we should. Those of us still debating the issue, have begun to do so in circles.

You're welcome. As always I come here to help.

[quote="Dr. No, about the flux in case the point charge is on a circumference"]
That's easy!
\[ \boxed{\Phi_E \equal{} \frac {q}{\varepsilon_0}}
\]

[/quote]

[quote="I"]
Had this been a charge $ Q$ placed at one corner of a cube, the flux would be $ \frac{Q}{8\varepsilon_0}$

[/quote]

If the charge is placed on the [b]boundary[/b] of the figure, will the flux not change?

I'm confused....

Gauss Law doesn't apply when there is electric charge in the Gaussian Surface.  Unless you consider, classically, that charge is a matter property so its dimensions can be neglected (to be considered as point charge) only for points far enough; in the case you are considering the charge is [b]in the point[/b]. The Gaussian Surface, that is a [b]geometric surface[/b] with no "thickness" (that is why I don't like "plates",  "tubes", "dome", etc.) would "cut" the charge in two parts : one that is inside the Gaussian, and contributes to the flux, and other outside tha doesn't.

In your former example the charge would be kind of spherically uniform volumetric distribution centered in the corner, so only one eighth of the charge contributes. For an mathematical model, [b] let be a point with a property charge[/b], one has to complete with seven adjacent  alike cubics surfaces to get to one of double edge, with the charge in it's center. Found if it's possible something of this kind in your second proposition.

Since the charge is a point charge, we can utilize this to our benefit and twist the picture according to our convenience (an approach inspired by Indian politics :wink: ).

So in case of the cube (the example I was talking about before all the argument began), we may consider the charge to be a cube with side $ a\to0$ and the charge uniformly distributed throughout its volume. So, the corner of the cube coincides with the geometric center of the charge.

The flux $ \Phi$ through any solid body's surface by Gauss' Law is $ \Phi_{\text{net}} \equal{} \frac {q_{\text{enclosed}}}{\varepsilon_0}$. So, here, the $ q_{\text{enclosed}} \equal{} \frac {Q}{8}$.

Similarly, we could do it with the cylinder.

3-D geometry is limited to planes and lines in the IIT-JEE syllabus so I don't know how to find the common volume between two "intersecting" three dimensional bodies. Further, I'm not even sure if this approach is correct. Hence I'm waiting for a solution from the others (you, Immanuel, included).

Very good

It's equivalent to the "spheric trick" . But you already did the intersection of two solids, right now.

Oh, thank you for including me (he, he, he).

[quote="Rajiv"][quote="Dr. No, about the flux in case the point charge is on a circumference"]
That's easy!
\[ \boxed{\Phi_E \equal{} \frac {q}{\varepsilon_0}}
\]
[/quote]

[quote="I"]
Had this been a charge $ Q$ placed at one corner of a cube, the flux would be $ \frac {Q}{8\varepsilon_0}$

[/quote]

If the charge is placed on the [b]boundary[/b] of the figure, will the flux not change?

I'm confused....[/quote]

Rajiv,

I'm sorry, I sort of glossed over that when you originally posted it.

The answer is no, the flux does not change when the charge is on the surface as opposed to being contained in the volume bounded by it. If you ponder this for a bit, I'm sure you'll see that it makes sense. I understand that we have been talking about a fictitious surface, but you might want to consider what happens when the surface is [url=http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/gausur.html#c3]conductive.[/url]

[quote="Immanuel Bonfils"]Gauss Law doesn't apply when there is electric charge in the Gaussian Surface.[/quote]
Immanuel, this bit is not true.

[quote="Immanuel Bonfils"]Unless you consider, classically, that charge is a matter property so its dimensions can be neglected (to be considered as point charge) only for points far enough; in the case you are considering the charge is [b]in the point[/b]. The Gaussian Surface, that is a [b]geometric surface[/b] with no "thickness" (that is why I don't like "plates",  "tubes", "dome", etc.) would "cut" the charge in two parts : one that is inside the Gaussian, and contributes to the flux, and other outside tha doesn't.[/quote]
Plates, tubes, domes, etc. are only metaphors for the purpose of visualization. Such terms are used even in calculus where it is perfectly obvious that the objects one refers to are only mathematical abstractions.

[quote="Immanuel Bonfils"]In your former example the charge would be kind of spherically uniform volumetric distribution centered in the corner, so only one eighth of the charge contributes. For an mathematical model, [b] let be a point with a property charge[/b], one has to complete with seven adjacent  alike cubics surfaces to get to one of double edge, with the charge in it's center. Found if it's possible something of this kind in your second proposition.[/quote]
This is not true, either.

Consider a cube with a point charge $ q$ in each corner.

\[ \Phi_E \equal{}\iiint_{\Omega} \frac{\rho}{\varepsilon_0}\mathrm{d}V\]

Now since we have a configuration of point charges,

\[ \rho \equal{} q\delta (x) \delta (y)\delta (z) \plus{} q\delta (x)\delta (y \minus{} y_1)\delta (z)\\ \plus{} q\delta (x \minus{}x_1)\delta (y \minus{}y_1)\delta (z) \plus{} q\delta (x \minus{} x_1)\delta(y)\delta(z)\\ \plus{} q\delta (x \minus{} x_1)\delta (y)\delta (z\minus{}z_1) \plus{} q\delta (x \minus{}x_1)\delta (y \minus{}y_1)\delta (z\minus{}z_1)\\  \plus{} q\delta (x)\delta (y \minus{} y_1)\delta (z \minus{} z_1) \plus{} q\delta (x)\delta (y)\delta (z \minus{} z_1)\]

Thus,

\[ \Phi_E \equal{} \int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{\rho}{\varepsilon_0} \mathrm{d}x \mathrm{d}y \mathrm{d}z \equal{} \frac{8 q}{\varepsilon_0}\]

You can't just split up the flux by dividing it by 8. The physics doesn't work like that.

[quote="Dr. No"]
The answer is no, the flux does not change when the charge is on the surface as opposed to being contained in the volume bounded by it. If you ponder this for a bit, I'm sure you'll see that it makes sense. I understand that we have been talking about a fictitious surface, but you might want to consider what happens when the surface is conductive.

[/quote]

We may consider the charge is [b]completely inside[/b] the volume (yes, imaginary) since it is a [b]point[/b] charge. But the boundary of the surface will be equally "thin" and we can say some of the charge is [b]outside[/b] the body. And a charge placed outside a Gaussian surface does not affect the flux $ \Phi$ through it.

[quote="Rajiv"][quote="Dr. No"]
The answer is no, the flux does not change when the charge is on the surface as opposed to being contained in the volume bounded by it. If you ponder this for a bit, I'm sure you'll see that it makes sense. I understand that we have been talking about a fictitious surface, but you might want to consider what happens when the surface is conductive.

[/quote]

... a charge placed outside a Gaussian surface does not affect the flux $ \Phi$ through it.[/quote]
That's absolutely true. If a charge is outside a Gaussian surface, it does not contribute to the flux across said surface.

[quote="Rajiv"]We may consider the charge is [b]completely inside[/b] the volume (yes, imaginary) since it is a [b]point[/b] charge. But the boundary of the surface will be equally "thin" and we can say some of the charge is [b]outside[/b] the body.[/quote]

The boundary layer may be considered infinitesimally thin, yes. Nevertheless, it is a mistake to view a point charge [b]on[/b] a Gaussian surface as if it were half inside and half outside that surface.

Point charges are zero dimensional; hence, the name [b]point[/b] charge. By virtue of being zero dimensional, point charges on a Gaussian surface are considered to be bounded by said surface, from which it follows that point charges on a Gaussian surface do contribute fully to the flux across it.


geometry
3D geometry
function
complex numbers

[hide]take the 125 across, find the cube root of both sides and multiply through by (2 - cube root x), solve for the cube root and cube both sides to get x=512/27.[/hide]

[hide]I'm stupid  :wallbash_red: [/hide]

Yes, when I hitted to delete the post I got "you cant delete messages that have been replied to"  :blush:

i don't think that you should be looking for complex solutions because the cube root is not a function for complex numbers (it is a triple valued "function")

Why isnt it a function for complex values? The problem said 'solve', not mentioning anything about reals, so i think we should be looking for complex soluti0ons aswell.

I agree with seamusoboyle. The cube root (and any odd root, except 1, which is undefined) does have complex solutions sometimes.

Even $\sqrt[3]{1}$ has complex roots: $w$ and $w^2$,
$w=\frac{-1-i\sqrt{3}}{2}$

I think its all the time, just looking at it in terms of the complex plane, to get the cubed (nth) root you just divide the angle by 3, and go out the nth root of the lenght, (this time taking it it be the real, positive value.) Its judt De Moivre.


The area of the largest rectangle inscribed in a circle of radius
5 cm is

a)25 cm^2   b)50 cm^2    c) 100 cm^2    d)10 square root(2)cm^2      e) 20 square root(2)cm^2

[hide]

Since the largest rectangle that can be inscribed is a square and the diameter = the diagonal of the square = 10 cm, then the side length of the square is 5sqrt(2) cm. Square this to get the area, and you will get 50 cm^2 = B.

[/hide]

just saying biblioman, your method works great, but a kinda quick shortcut (for me, i depends on how your mind works), is that for any quadrilateral with diagonals that are perpendicular, and this is only for quadrilaterals that ALWAYS have perpendicular diagonals, the product of the diagonals divided by 2 is the area of the square. For example, a square inscribed in a circle radius 4 would have an area of 32, because the diameter is 8, which is also the diagonal of the square. So, $ 8*8\equal{}64$, then $ 64/2$ equals 32. I hope this might help those of you who don't really like to work with radicans. :)


hello..i was wondering (just to clarify) what do the following terms mean:

injective

surjective

bijective

i looked on mathworld.wolfram.com, which gave me kind of a vauge understanding, but didn't help that much (im not that good w/transformations or maps) From my understanding, these terms play somewhat of a large role in IMO (i may be wrong, but it seems most training problems leading to IMO contain at least one problem involving one of the above terms)

Injective means one-to-one.

Surjective means onto.

Bijective means both one-to-one and onto.

oh okay, i see how it works...somewhat :?

Let's give examples so that you can see clearly the difference between these types of mappings:

Example of bijection: all mappings f:R->R with inverses g=f-1 like f(x) = x.

Example of surjection but not injection: the mapping f:R->R+ defined by f(x) = x2 because for any element y of R+, there exists an element x of R such that y=f(x), which is x= \pm  \sqrt y. Notice that f is not injective because f(a)=f(b) or a2=b2 does not imply a=b.

Example of injection but not surjection: the mapping f:{1,2,...,n}->N defined by f(n)=n. Easily f(i) = f(j) implies i = j and yet this mapping is not surjective because there is no element from {1,2,...,n} that maps to n+1\inN

A rule of thumb that worked out for me while I was in high school was this, so this is my opinion and use it at your own risk: if f:A->B
1-1 means one element gives another element, no duplicates, [b]usually[/b] AB
bijective means inverse exists, [b]always[/b] A=B by Cantor-Schroeder-Bernstein Theorem (I hate these long name theorems  :D )
This is just an intuitive sense so if anyone disagree, it's your own problem.

okay, yeah..this is much better (not to say that your "short" defintions were bad or anything gauss :D, they were somewhat okay), but now that examples are shown, its more clear

[quote] if f:A->B  1-1 means one element gives another element, no duplicates, usually A<B [/quote]

i guess its okay :?, oh well i will just have to see how it works out when im solving problems :D. How did you come by this anyway?


I am trying to make a table with 5 rows and 2 columns. I want the table to expand to fit the entire page even if I don't type anything into the cells. I also want every row to be the same height and every column to be the same width. Is this possible?

Hi,
Use the p option for fixed column width: [code]\begin{tabular}[c]{|p{1.0 in}|p{1.0 in}|}[/code] will give two 1.0 in wide columns. The rows should be of the same height by default.
SSN.


Find all pairs $(x;y)$ of positive integers satisfying the equation:

$(x+y)^2-2(xy)^2=1$

Note that for positive integers x and y:
$x+y\leq xy$ unless $x = 1$ or $y=1$ or $1\leq x,y\leq 2$
Direct checking shows that $(0,1), ( 2,1), (1,0)$ and $(1,2)$ are all the possible solutions.
The solutions to the problem are $(1,2)$ and $(2,1)$ only.


Ring Theory
superior algebra
superior algebra unsolved

By considering factorization of Z[root (-2)] (similar to Gaussian Integers which is a UFD), find all integer solutions to the equation X^2+2=Y^3. :)

Thanks.

Classic problem.

Let $ a\equal{}x\minus{}\sqrt{\minus{}2}$, $ b\equal{}x\plus{}\sqrt{\minus{}2}$, so $ y^3 \equal{} ab$.  Since $ b \minus{} a \equal{} 2\sqrt{\minus{}2}\equal{}\minus{}(\sqrt{\minus{}2})^3$, and $ \sqrt{\minus{}2}$ is prime, we either have two cases:

Case I: $ \sqrt{\minus{}2}$ divides $ a$.  Then $ \sqrt{\minus{}2}$ divides $ x$, so we have $ \sqrt{\minus{}2}(c\plus{}d\sqrt{\minus{}2})\equal{}x$ for some $ c,d\in\mathbb{Z}$.  So $ c\equal{}0$ and $ \minus{}2d\equal{}1$, contradiction.

Case II: $ (a,b)\equal{}1$.  Then each of $ a$ and $ b$  must be a cube.  So there exist some $ c,d\in\mathbb{Z}$ with $ (c\plus{}d\sqrt{\minus{}2})^3 \equal{} x \minus{} \sqrt{\minus{}2}$.  Writing it all out...

...we see that $ c^3 \minus{}6cd^2 \equal{} x$ and $ 3c^2 d \minus{} 2 d^3 \equal{} \minus{}1$, so $ d\equal{}\pm 1$, hence $ 3c^2 \minus{} 2  \equal{} \pm 1$.  It is easy to find all solutions to this equation: we must have $ c\equal{}\pm 1$, so $ x\equal{}\pm 5$.

Correct answer!

But I would expect something involving irreducibles and mulplicative norms (or not?) rather than number theoretic method.

You are hereby invited to explain to me what the difference is.

[quote="LydianRain"]So $ c \equal{} 0$ and $ \minus{} 2d \equal{} 1$, contradiction.[/quote]
This part appears to be wrong.  It should read: $ \minus{}2d\equal{}x$, so $ y^3$ is $ 2$ mod $ 4$, contradiction.


When will the grand statistics for AMC 10 and 12 be ready?

[quote]Scanning for the 2004 AMC 10 A and 2004 AMC 12 A ended yesterday. That
means we have all data entered, and we have some raw statistics. More
complete statistics should be available later this week.

[/quote]

The grand statistic for AMC A should be up about a week ago, according to the AMC Director.

I meant - the grand statistics that show the average scores by grade and gender.   Following this we get information on the states' statistics - highest scores for different grades, school's team scores, etc.


ARML
induction
algebra
polynomial
symmetry
Rational Root Theorem

Not a solution but a story this reminds me of:
First day of precalc with math (as opposed to this is what the course will be like) my teacher "taught" about the whole idea of closedness, like how the integers are closed over addition because if you add any two integers together, you'll get another integer.  So she said that the irrationals were closed over addition.  I told her no.  She said yes.  I told her  :rt2: and (1- :rt2: ) as an example.  She sat there for 5 minutes before finally admitting I was right.

And the moral of the story is: that's why I had enough time to learn calc and switch to BC second semester.

When I was in middle school, I had a teacher who told me that irrationals were closed under addition. Eventually I convinced him that that was complete nonsense. It took me more like two weeks though rather than five minutes.

In sixth grade I had to correct a teacher.  The question was something along the lines of put these in increasing order:
-1 -1/2 -3/2

She said it was -1/2 -1 -3/2.  Booyah.

I wonder if anyone will ever solve this problem, or if we'll just keep going on like this.

[hide]I think you should be able to show that the sum of two square roots of integers that aren't integers is irrational and then do the rest by induction. There was a problem like that on an ARML power round one year (2001 maybe?), and I think we did it like that, and it wasn't too hard.[/hide]

[quote="Nukular"]when I was in middle school, I spent a week convincing my math teacher that -(a-b) = b-a...

I think thats a bit more pathetic than anything someone else is gonna post about this. (I win :) )[/quote]

At least your teacher probably believed in the fact that using 3.14 for pi and the actual pi value might give you different answers when calculating nice stuff like a circle with radius 970 in. (Answers to be expressed to nearest hundredth of a square inch)

Maybe I'm being idiotic, but I don't see how to prove that say :rt2: + :rt3: + :rt5: + :sqrt:(7) is irrational.  Can someone enlighten me?

Here we go:

[hide]

k, n, m are rational numbers.

k - (n)^(1/2) = (m)^1/2 ?

if that is true then the sum could be a rational number.

but if we square the equation:

k^2 - 2k(n)^(1/2) + n = m

we get a contradiction.

then apply the induction method.

[/hide]

You have shown that the sum of any two terms is irrational.  But how does that prove that the entire sum is irrational?

You can't induct.  Just because the sum of two is irrational doesn't mean that the sum of three will be.

Well, you have to do a bit of work to induct, but it can be done.

sums of sqrts of integers can be dealt with nicely through the use of polynomials and the rational root theorem.  Let's say that you have A =  :sqrt: (a1) + :sqrt: (a2) + ... + :sqrt: (an) with ai a positive integer.

consider f(x) := product of (x :pm: :sqrt: (a1) :pm: (a2)  :pm: ... :pm: :sqrt: (an)) which has 2^n roots, and you know that f(A) = 0.  Take some ai, and for each term in the f(x), group everything except for the ai, what we will notice, is that lets say we pick a term, and call the part not including ai, G.  In our product of f(x) we will have (G - :sqrt: (ai))(G + :sqrt: (ai)) = G^2 - ai, and this will be true of all the terms, so f(x) will not contain :sqrt: (ai).  By symmetry, this will be true for all of the ai's so f(x) will only have integer coefficients, and it will be monic.  Then by the rational root theorem, its roots can only be integers, so all we need to do is show that A is not an integer, and it would prove that it is irrational.

I hope that made sense, its not very easy to type out.

Okay, here is my outline of a solution to the original problem.

(Oops, I see that zscool has just posted something similar.)

[hide]I will first show that any sum of square roots of nonnegative integers must be the root of a monic polynomial with integer coefficients.  The proof is by induction on the number of terms in the sum.  Suppose we have a sum whose first term is :sqrt:(a).  By induction, the sum of the remaining terms is a root of some monic polynomial p with integer coefficients.  But then the sum of all the terms is a root of the polynomial p(x - :sqrt:(a)) p(x + :sqrt:(a)).  It's not too hard to see that this product polynomial is itself a monic polynomial with integer coefficients.  Thus we have finished the inductive proof.

In particular, by the Rational Root Theorem, if a sum of square roots of integers is rational, then it must be an integer.

So all we have to do is prove that the original sum is not an integer.  But that's pretty easy, because each term is just a shade more than an integer.  Namely, if n is an integer, then

:sqrt:(n2 + 1) - n = 1 / (:sqrt:(n2 + 1) + n) < 1 / (2n).  Adding up for all integers n from 1001 to 2000, we see that the sum of these terms is less than 1, but more than 0.  Thus it can't be an integer.[/hide]

This is the proof that I found...however, I don't understand one part of it...perhaps you guys can explain it.

The proof:

[hide]We must do two things: 1) show that the sum in question is not an integer and 2) show that it is a zero of a monic polynomial.

For the first step, observe that if n > 1, then n <  :sqrt:(n :^2: + 1) < n + 1/n.  Let us call the sum in question S.  Then

S = 1001 + x_1 + 1002 + x_2 + ... + 2000 + x_1000

where each x_i lies between 0 and 1/1001.  Consequently,

0 < x_1 + x_2 + ... + x_1000 < 1

so S is not an integer.

Next, we will show that S is a zero of a monic polynomial.  More generally, we shall prove that for all positive integers n, the quantity

 :sqrt: a_1 +  :sqrt: a_2 + ... +  :sqrt: a_n

is a zero of a monic polynomial if each a_i is an integer that is not a perfect square.  We proceed by induction.  If n = 1, the assertion is true because  :sqrt: a_1 is a zero of the monic polynomial x :^2: - a_1.   Now assume that

y =  :sqrt: a_1 +  :sqrt: a_2 + ... +  :sqrt: a_n

is a zero of the monic polynomial P(x) = x^r + (c_r-1)x^(r-1) + ... + c_0.  We will produce a monic polynomial that has x = y + :sqrt: a_n+1 as a zero.  We have

0 = P(y) = P(x - :sqrt: a_n+1) = (x - :sqrt: a_n+1)^r + c_r-1(x - :sqrt: a_n+1)^(r - 1) + ... + c_0

Notice that the expansion of each (x -  :sqrt: a_n+1)^k term can be separated into terms that have integer coefficients and terms with coefficients equal to an integer times :sqrt: a_n+1.  Thus we have

0 = P(x -  :sqrt: a_n+1) = x^r + Q(x) + (:sqrt: a_n+1)R(x)

where Q(x) and R(x) are polynomials with integer coefficients, each of degree at most r - 1.  Putting the radicals on one side of the equation yields

x^r + Q(x) = -(:sqrt: a_n+1)R(x),

and squaring both sides leads to

x^(2r) + 2x^rQ(x) + (Q(x)) :^2: - a_n+1(R(x)) :^2: = 0

The term with the highest degree is x^(2r).  Since all coefficients are now integers, we have produced a monic polynomial with x = y + :sqrt: a_n+1 as a zero, as desired.  (This is the part I don't get: why is x = y + :sqrt: a_n+1 a zero of the polynomial with degree x^(2r)??)[/hide]

Surely there's a geometric proof of this problem?


function
calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

1)Find the gradient of the function $y: |\rightarrow x^{3}+x^{2}-2x$ at the point(s) where it crosses the x-axis and where it cuts the y-axis.

2)If $y=f(x)g(x)h(x)$,show that $\frac{dy}{dx}=f'(x)g(x)(h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x).$

3)Find the equation of the tangent to teh curve $y=x^{2}-2x$ that is parallel to teh line with equation $y=4x+2$

1)the function crosses the x axis at x = -2,0,1
the derivative of the function is $3x^{2}+2x-2$
so the derivatives at -2,0,1 are 6,-2, and 3, respectively
it cuts the y axis at x = 0, which we already showed was -2

2)use the product rule with one function being f(x) the other g(x)h(x)...
$f'(x)g(x)h(x)+f(x)[g(x)h(x)]' = f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$

3)the slope of the given line is 4, set the derivative( 2x - 2) equal to 4, one gets x = 3, y = 3 at x = 3, so point slope formula gives:
$y-3 = 4(x-3)$
$y = 4x-9$


[quote="funcia"]Mine is [color=yellow]GOLD GLITTER!![/color]
I understand not everyone is a girl so why not make one up?
I used to have red under my glitter!!!
EDIT: I am serious. Very serious. If you are not wearing nailpolish pretend you are.[/quote]

I'm sorry, but I just can't do that.

[quote="funcia"]Mine is [color=yellow]GOLD GLITTER!![/color]
I understand not everyone is a girl so why not make one up?
I used to have red under my glitter!!!
EDIT: I am serious. Very serious. If you are not wearing nailpolish pretend you are.[/quote]

No. Comment.

NOW I AM GOING TO PUT ON [color=red]RED[/color] this person asked me to buy gold nail polish for them and everyone touches my hands saying BLING BLING so i decided it is a bit distracting

[quote="mtms5467"][quote="funcia"]Mine is [color=yellow]GOLD GLITTER!![/color]
I understand not everyone is a girl so why not make one up?
I used to have red under my glitter!!!
EDIT: I am serious. Very serious. If you are not wearing nailpolish pretend you are.[/quote]
Uhhh...I have never worn nail polish in my life and I never will. This is a weird topic. :roll:[/quote]
and crazy too

[quote="Treething"]i wear pink nail polish all the time

ALL THE TIME[/quote]
no wonder hoe you won mc....

blue sparkles still on from homecoming. Which was a long time ago... i neded to redo them

the last time i had any on (which was probably about a year ago!) it was blue chrome. it matched my swim cap and i was known as the girl whose toenails matched her swim cap!!! :rotfl: :rotfl:  :rotfl:

Honestly, I don't know what to say.

But seriously, 12 posts?  ...let's see just how long a topic like this can go.

my swimcap is yellow and my team is green and gold so i guess i sort of used to match!

my swimcap was always black (I don't swim anymore) and no way was I going to paint my toenails black!! but to tell the truth it never even occurred to me to match colors.

i didn't think of it either until my friend said your nails are the same color as your cap! i said oooooooooookay and it was the exact same random shade! :thumbup:

THIS TOPIC MUST KEEP GOING AND GOING AND GOING

Huh.  Barf purple.

yeah i gonna buy red cap JK Bout that grammar but...

i took off my nailpolish but I believe i will paint my nails blue next time to remember old times

The movie Brother Bear says nothing about my nailpolish ...  :maybe:

but perhaps it does  :wink:
go watch it again
and right the book Brother Bear: What Will Happen In The Next Movie  :lol: That's what I think other brother bear fans would like.

yeah i guess so though i only watched brother bear 1  :blush:

now i have really chipped red nail polish on my left hand :(

Nail polish is not equal to nailpolish which is what we are REALLY chatting about.


The Badgers play the Cougars in a series of seven basketball games. Each team has an equal chance of winning each game. What is the probability that the Badgers will win at least four games? Express your answer as a common fraction.

i'm not sure i quite understand this problem, like why it's 1/2.  explanation, please?

A team has to win at least four games to win the series. Since both teams are equally likely to win the series, they are equally likely to win at least four games; the probability is $ \boxed{\frac12}$.

Symmetry; this is kind of trollololololololololololololol :D

Pretty much one team must win at least four games. Since the teams are equally matched, this team will be the Badgers with probability $\boxed{\frac{1}{2}}$.

[hide=...]Everything is syemetry :D 1/2[/hide]

[hide=Solution]
We can think of this as almost any professional sports playoff series that is best of 7 and divide it into cases based on how many games it takes the Badgers to win.

Let's start with the simplest case: Badgers win in 4 games. The probability is $\left(\frac{1}{2}\right)^4=\frac{1}{16}$.

Now, suppose they win in exactly 5 games. For a particular sequence of wins in losses ending after 5 games, the probability of it occurring is $\left(\frac{1}{2}\right)^5=\frac{1}{32}$. We must find how many sequences of wins and losses are possible that result in the Badgers winning in exactly 5 games. Clearly, they must win the final game (or otherwise they would have won in four games), but they can then choose any of the first four games to be a loss. This results in $\binom{4}{1} = 4$ possible sequences, so the probability of the Badgers winning in 5 games is $4\cdot\frac{1}{32}=\frac{1}{8}$.

If the Badgers win in 6 games, the probability of a particular sequence of wins and losses occurring is $\left(\frac{1}{2}\right)^6=\frac{1}{64}$. Again, they must win the final game, but we will have to choose two of the first five games to be games that the Badgers lose. There are $\binom{5}{2}=10$ ways of doing this, so the probability of the Badgers winning in 6 games is $10 \cdot \frac{1}{64}=\frac{5}{32}$.

And finally, we consider the possibility that the Badgers win in 7 games. The probability of a specific sequence of wins and losses occurring is $\left(\frac{1}{2}\right)^7=\frac{1}{128}$. The Badgers must win the final game, but the Badgers must lose three of the first six games. There are $\binom{6}{3}=20$ ways this can happen, so the probability of the Badgers winning in 7 games is $20 \cdot \frac{1}{128}=\frac{5}{32}$. (Same probability of winning in 6 games. Coincidence or not?)

Thus the overall probability of the Badgers winning 4 games in the series is $\frac{1}{16}+\frac{1}{8}+\frac{5}{32}+\frac{5}{32}=\boxed{\frac{1}{2}}$. $\boxed{}$

[/hide]

I'm guessing this is from the Counting and Probability "think about it" topic.

who else did [hide=this]$\sum_{k=4}^{7} \binom{7}{k}\left(\frac{1}{2^7}\right)=\frac12$[/hide] because they didn't see the obvious answer :P

I thought I was clever because I did this:
[hide= a sol] The probability is equivalent to $\left( \frac{1}{2^7}\right)\left(\binom{7}{0} + \binom{7}{1} + \binom{7}{2} + \binom{7}{3} \right)$, and then I notice that

$\binom{7}{0} + \binom{7}{1} + \binom{7}{2} + \binom{7}{3} + \binom{7}{4} + \binom{7}{5} + \binom{7}{6} + \binom{7}{7} = 2^7$

And $\binom{7}{4} + \binom{7}{5} + \binom{7}{6} + \binom{7}{7}  = \binom{7}{3} + \binom{7}{2} + \binom{7}{1} + \binom{7}{0}$

So $\binom{7}{0} + \binom{7}{1} + \binom{7}{2} + \binom{7}{3} = 2^6$, and the given probability is precisely $\frac{2^6}{2^7} = \frac{1}{2}$

Then I go click submit and then realize how easy it could’ve been.
[/hide]


Heather-M, just a suggestion: maybe you should post problems like this one and those other two equations you posted in the middle school forum?  ;)


function
algebra
polynomial
analytic geometry
superior algebra
superior algebra unsolved

Let $ V \subset \mathbb{A}^n(k)$ be affine variety.
Show there is a one-to-one correspondence between morphims $ V \to \mathbb{A}^1(k)$ and functions $ f \in k[V]$

I'm assuming in this post that your definition for a morphism is that it is a polynomial in all its coordinates (a regular map).

First we make the identification $ \mathbb{A}^1_k$ with $ k$. Then we see that polynomials $ f \in k[x_1,...x_n]$ are really regular maps to the affine line. Also notice two functions are the same if they take the same value on $ V$ so this means that they are elements of the quotient ring $ k[x_1,...x_n]/I(V)$. This is just the coordinate ring $ k[V]$


abstract algebra
superior algebra
superior algebra unsolved

Let $ A$ be a noetherian ring. A proper ideal $ I$ of $ A$ is called a perfect ideal if $ \textrm{grade}(I) \equal{} \text{proj dim}(A/I)$. Prove that for a regular local ring $ A$ and $ P \in \textrm{Spec}(A)$ we have that $ P$ is perfect $ \Leftrightarrow$ $ A/P$ is Cohen-Macaulay.

According to the author this follows directly from the following:
i) Auslander and Buchsbaum's theorem: $ \textrm{proj dim M} \plus{} \textrm{depth M} \equal{} \textrm{depth A}$.
ii) Serre's theorem: $ A$ is regular $ \Leftrightarrow \textrm{gl dim A} \equal{} \textrm{dim A} \Leftrightarrow \textrm{gl dim A} < \infty$.

Unfortunately I don't quiet see  how... :(

I think $ \text{grade}(I)$ means depth. Then use that a regular local ring is Cohen-Macaulay ($ \text{dim} \equal{} \text{depht}$) and universally catenary.

Because $ \textrm{gl dim}_A A \leq \textrm{gl dim}_A A/P < \infty$ we have by (ii) that $ A/P$ is regular, and a regular local ring is CM, hence both $ A/P$ and $ A$ are CM.

But from (i) we see that $ \textrm{proj dim}_A A/P \equal{} \textrm{depth}_A A \minus{} \textrm{depth}_A A/P \equal{} \textrm{dim }A \minus{} \textrm{dim }A/P \equal{} \textrm{ht }P$, where the last equality follows from the fact that $ A$ is catenary. Furthermore is $ \textrm{ht }P \equal{} \textrm{depth}(P,A) \equal{} \textrm{grade } P$ in a CM ring, so the conclusion follows.

Is this what you meant or is my reasoning somewhere flawed?


logarithms
inequalities
inequalities unsolved

Denote:
$ n$ an integer such that $ n > 1$
$ x_i$ $ (1\leq i \leq n)$ some real positive numbers
with: $ \sum \frac {1}{1 \plus{} x_i} \equal{} 1$

prove: $ \sqrt [n]{\prod xi}\geq n \minus{} 1$

let $ \frac{1}{1\plus{}x_{i}}\equal{}a_{i},a_{i} <1$
$ \Rightarrow \sum a_{i}\equal{}1$
to prove $ \prod \frac{(1\minus{}a_{i})}{a_{i}} \geq (n\minus{}1)^{n}$
which follows from the convexity of $ f(x)\equal{}\ln \frac{1\minus{}x}{x},0<x<1$

[quote="pardesi"]l
$ \prod \frac {(1 \minus{} a_{i})}{a_{i}} \geq (n \minus{} 1)^{n}$
which follows from the convexity of $ f(x) \equal{} \ln \frac {1 \minus{} x}{x},0 < x < 1$[/quote]
How?

$ \prod \frac {(1 \minus{} a_{i})}{a_{i}} \geq (n \minus{} 1)^{n}$
Equiv to
$ \sum ln(\frac {(1 \minus{} a_{i})}{a_{i}}) \geq nln((n \minus{} 1))$
From jensen:
$ LHS \geq n$ $ ln(\frac {1}{n}\sum \frac {1 \minus{} ai}{ai}))$

Then??

u have taken the logs wrongly
$ \ln a^{x}\equal{}x\ln |a|$ and not $ \ln x|a|$
rest follows

[quote="pardesi"]u have taken the logs wrongly
$ \ln a^{x} \equal{} x\ln |a|$ and not $ \ln x|a|$
rest follows[/quote]

Just a typo, Ive edited them.

[quote="pardesi"]so is everything clesr now?[/quote]

 :D yes
Thanks  :wink:


I registered for the Precalculus class and my dad says he got a message saying the book has been shipped and that it would arrive in two to three weeks. If the book is finished, why doesn't AoPS make it available on the bookstore, rather than having to register for the class to get the book?

That email is in error.  The book is not yet in print.


[quote="symmetry"]I do not understand how to set up a truth table for this question:

If x=y, then 2x = 2y. It is false that 2x = 2y and y is not equal to z. Therefore if y is not equal to z, then x is not equal to y. PLEASE ASSIST. Thanks

MY WORK:  (May be wrong)

I've set up some variables that I think are ok.

x:y

2x:2y

x AND y IMPLIES 2x AND 2y
NOT (2x AND 2Y)

THEREFORE: y OR NOT z IMPLIES x OR NOT y.

Is this correct?  If not, can someone help me step by step?[/quote]

you say that $2x=2y$, then you say that $2x\ne2y$
why? clearly a contradiction, also the
$y\ne z$ implying $y \ne x$ is false because we know nothing about the relationship of x and z

$2x = 2y \thickspace \wedge \thickspace y \neq z = F$

Remember the equivalence: $p \wedge q \thickspace \equiv \thickspace \neg ( \neg p \vee \neg q)$

$\neg(2x \neq 2y \thickspace \vee \thickspace y = z) = F$

Remember the equivalence: $p \rightarrow q \thickspace \equiv \thickspace \neg p \vee q$

$\neg(2x = 2y \thickspace \rightarrow \thickspace y = z) = F$

Then...

$2x = 2y \thickspace \rightarrow \thickspace y = z$

Therefore...

$x = y \thickspace \rightarrow \thickspace y = z$

...

Bruno,

Thank you.  What is the meaning of the symbol (--\) that you have infront of THREE equal signs?  What is the meaning of 3 equal signs?

$\neg$ means NOT operator.
$\equiv$ means equivalence/identity.


algebra
polynomial
superior algebra
superior algebra solved

show that:
  P(x)  =  x^((n-1)*p^n) + x^((n-2)*p^n) + x^((n-3)*p^n) + x^((n-4)*p^n) +  ...  x^(p^n)   +   1
is irreducible over Q?  [p is prime, n natural number]

thanks! :)

Write it as $\frac{x^{n \cdot p^{n}}-1}{x^{p^{n}}-1}$.
But this is not always irreducible, e.g. take $p=2,n=3$.

This is only irreducible when $n=p$:
It is the $n \cdot p^{n}$-th cyclotomic polynomial iff it is irreducible, thus iff $\varphi(n \cdot p^{n}) = (n-1) p^{n}$ iff $\varphi(pn) = p(n-1)$.
The latter one happens iff $n=p$.


Given a triangle with sides $a,b,c$ and angles $A,B,C$, prove that:

$a \sin(A/2) + b\sin(B/2) + c\sin(C/2) \ge \frac{a+b+c}{2}$

An equivalent inequality was posted yesterday [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80834]here[/url].


integration
logarithms
calculus
calculus computations

During the black plague, the number of people that were infected at any time increases at a rate proportional to the number of people that were infected at that time. If $ 1000$ people were infected when the plague was first discovered, $ 1200$ people were infected $ 7$ days later, how many people were infected $ 12$ days after the plague was first discovered?


Νίκο, είσαι σίγουρος πως είναι έτσι ;

Δε το δοκίμασα στο χέρι, αλλά το mathematica δε μου το υπολογίζει !  :|

Στέλιος

[quote="NickNafplio"]Να υπολογισετε το γενικευμενο ολοκληρωμα:

$ \int_{0}^{\frac {\pi}{2}}{\frac {dx}{1 + (\tan{x})^{\sqrt [5]{2009}}}}$[/quote]

Όντως, very cute! :) Αν δεν κάνω λάθος, το ολ/μα είναι $ \pi/4$.

[hide="Hint"]Θεωρείστε τη συνάρτηση $ f(k) = \int_0^\frac {\pi}{2} \frac {dx}{1 + \tan^k x}$.
Τότε,  $ f'(k) = - \int_0^\frac {\pi}{2} \frac {\log(\tan x) \tan^k x}{(1 + \tan^k x)^2}dx$ και, αν δείξετε ότι το τελευταίο ολ/μα είναι 0, τελειώσατε (αφού $ f(0) = \frac {\pi}{4}$) :D[/hide]

Cheerio,

Durandal 1707

Ναι τοσο ειναι. Ωραιος. Στη Σεμφε στο μαθημα της Αναλυσης ο καθηγητης μας το ελυσε με εναν αλλο καταπληκτικο τροπο στον οποιο δεν χρησημοποιει καθολου μεθοδους ολοκληρωσης

[hide]Στο καρτεσιανο επιπεδο το γραφημα της συναρτησης που ειναι μεσα στο ολοκληρωμα στο διαστημα [0, π/2) ειναι συμμετρικο ως προς το σημειο (π/4, 1/2), οποτε το εμβαδον του χωριου ισουτε με το μισο του εμβαδου του ορθογωνιου με κορυφες τα Α(0,0), Β(1,0), Γ(1,π/2), Δ(0,π/2) :wink: [/hide]

ΥΓ: Το ολοκληρωμα ειναι ανεξαρτητο απο τη δυναμη στην οποια υψωνουμε το $ tanx$

Κανονικά το ολοκλήρωμα είναι γενικευμένο αλλα αν γράψουμε
$ \frac {1}{1 \plus{} tan^kx} \equal{} \frac {sin^kx}{sin^kx \plus{} cos^kx}$ η συνάρτηση γίνεται συνεχής στο [0,π/2] οπότε αν Ι είναι το ολοκλήρωμα με την αλλαγή μεταβλητής $ u \equal{} \pi /2 \minus{} x$ βγαίνει 2Ι=π/2


modular arithmetic
number theory unsolved
number theory

Prove that $ 3^n \minus{} 2$ is a perfect square only for $ n \equal{} 1, 3$.

We assume that there exist $ n>3$ so that $ 3^n\minus{}2$ is a square. Let $ u^2\equal{}3^n\minus{}2$, where $ u\in\mathbb Z_\plus{}$. Take modulo $ 8$ then $ n$ must be odd. Put $ v\equal{}3^{(n\minus{}1)/2}$ then $ u^2\minus{}3v^2\equal{}\minus{}2$. This is a Pell equation. Let $ u\equal{}x\plus{}3y$ and $ v\equal{}x\plus{}y$ this lead us to $ x^2\minus{}3y^2\equal{}1$. From this it is not hard to see that $ v$ must be $ v_n$ for some $ n$, where the sequence $ \{v_n\}$ is defined as the following: $ v_0\equal{}1$, $ v_1\equal{}3$ and $ v_{n\plus{}2}\equal{}4v_{n\plus{}1}\minus{}v_n$. Because $ \frac{n\minus{}1}2>1$ so $ 9\mid v$. Take on modulo $ 9$ we have $ 9\mid v_n$ iff $ n\equiv 4\pmod 9$. But take on modulo $ 17$ we see that $ n\equiv4\pmod 9$ iff $ v_n\equiv0\pmod{17}$. So $ v_n$ can not be a power of $ 3$.

So $ n\leq 3$. Take $ n\equal{}1,2,3$ a quick calculation shows that $ n\equal{}1,3$ are all possible values.

[quote="Zuza"]Nice, thanks :)
Just, how did you exactly pinpoint the 17 as the correct modulus to watch?[/quote]

Well, you can compute $ v_4$ and will see that $ v_4 \equal{} 153 \equal{} 9\times 17$ that's why I try with modulo $ 17$.

Btw, in the first table, it should be $ u_n\mod 9$, not $ u_n\mod 19$.

Hello,it was posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=368694]here[/url]
not much ago.

[quote="sri3"]I did not see this topic before i posted....
And i like this solution...
Anything wrong? :?:[/quote]
No,I later understood this.