TITLE: Chebotarev Density Theorem answers to factorization of polynomials
QUESTION [4 upvotes]: I've read of Chebotarev Density Theorem. The statement over $\mathbb{Q}$ is:
Let $K$ be Galois over $\mathbb{Q}$ with Galois group $G$. Let $C$ be a conjugacy class of $G$. Let $S$ be the set of (rational) primes $p$ such that the set of $\{\text{Frob}_{} |  \ \text{above } p\}$ is the conjugacy class $C$. The set $S$ has density $\frac{|C|}{|G|}$.
I understand the statement, and my question is not about that, but the relation of this statement with facts about polynomials. There is the following fact, which tells that there is a correspondence:
Let $K = \mathbb{Q}(\alpha)$, where $\alpha$ is an algebraic integer. Let $f$ be the minimal polynomial of $\alpha$, and let $L$ be the Galois closure of $K$ with Galois group $G$. Let $p$ be a (rational) prime which does not ramify. The factorization of $f$ modulo $p$, that is, $f \equiv f_1f_2 \ldots f_k \pmod{p}$, is the same as the cycle type $(f_1, f_2, \ldots, f_k)$ of the $Frob_ \in G$ permuting $f$'s roots, where $ \in L$ is a prime ideal above $p$.
(I would also be happy to see a reference/proof to this theorem, but that's not the main issue. I'm also interested to know if there is some other way to motivate the connection between Frobenius maps and polynomials)
Now, for the questions.

Can we apply the Chebotarev Density Theorem for determining the density of a given factorization type for a given polynomial? For example, if we have the polynomial $x^4 + x + 1$, can we calculate how often we have the factorization type $(1, 3)$. This seems possible to me using the other fact I wrote, but I am not able to fill in the details.
Moreover, can this factorization type be reached from more than one conjugacy class? That is, can there be two (or more) conjugacy classes $C_1, C_2$ such that the $p$ with $\sigma_p = \{\text{Frob}_ |  \text{ above } p\} \in C_1, C_2$ have the cycle type $(1, 3)$?
Does the situation in 1. change for reducible polynomials - is the calculation still possible?

I also have a loosely-related question about applying the density theorem to obtain some other results:
4.
In http://websites.math.leidenuniv.nl/algebra/Lenstra-Chebotarev.pdf there is an example application of the Chebotarev density theorem for proving the Dirichlet's theorem on arithmetical progressions at page 4 in the paragraph starting with "If you apply this theorem in the abelian case, ..."
In general, I'm very confused about the proof. For me the most intuitive way to approach the problem would be looking at the cycle types of $\sigma_p$, but the proof seems to "cheat" by looking at $\sigma_p$ as integers modulo $m$. There seems not to be a polynomial whose factorization we would inspect, which confuses me - how does this other method precisely work, what happens in the step $\sigma_p \longleftrightarrow (p \text{ mod } m)$? I kind of feel like I've answered my own question already, but if someone has some clarifying ideas, please share.

REPLY [2 votes]: Now that I have learnt more about algebraic number theory and such, I decided to answer my own question. We work with $\mathbb{Q}$ all the way, though the general case is not very different.
First a remark on the Frobenius symbol. Let $P \in \mathbb{Z}[x]$. For each prime $p$ there is a field automorphism of the algebraic closure of $F_p$ [one could work with $\mathbb{F}_{p^k}$
for $k = \deg(P)!$ -- the point is that we take some field which contains the roots of $P$ modulo $p$], called the Frobenius endomorphism, which maps $x$ to $x^p$. Denote this by $\sigma_p$.
It is an easy exercise to prove that the only elements fixed by this map are those in $F_p$. We see, in particular, that the Frobenius endomorphism $\sigma_p$ fixes a root of $P$ in the closure $F_{p^{\infty}}$ if and only if this root is in $F_p$, i.e. this root is indeed a root modulo $p$. In general one can prove that if $\alpha$ is a root of $P$ in $F_{p^{\infty}}$ with $[F_p(\alpha) : F_p] = d$, then the sequence
$$\alpha, \sigma_p(\alpha), \sigma_p(\sigma_p(\alpha)), \ldots$$
is periodic with period equal to the degree $d$.
For example, if $P(x) = x^2 + 1$ and $p \equiv 3 \pmod{4}$, then the roots of $P$ in $F_{p^{\infty}}$ are not in $F_p$. One may think of the roots of $P$ as $i$ and $-i$, and now $i$ maps to $-i$ and $-i$ maps to $i$ (but keep in mind that here $i$ and $-i$ are not complex numbers, even though they have the property that $i^2 = (-i)^2 = -1$) . The cycle length in this case is $2$, corresponding to the fact that we "picked a root out of a quadratic irreducible factor of $P$ modulo $p$".
We thus see that the orbits of the roots under the Frobenius symbol contain all of the information on the factorization of $P$ modulo $p$.
Recall that the Frobenius symbol (also called the Artin symbol - I'll use this term to not mix it up with Frobenius endomorphism) of a prime $p$ with respect to the splitting field $F_P$ of $P$ is a conjugacy class $C \subset \text{Gal}(F_P/\mathbb{Q})$ which "mimics" the behaviour of the Frobenius endomorphism $\sigma_p$. (Remember that we have to pick a conjugacy class instead of a single element, but elements in the same conjugacy class are quite similar, so this doesn't really bother us.) More precisely, when we view the Artin symbol as a (conjugacy class of) permutation(s) of the roots of $P$ in $F_P$, it has the same cycle pattern as $\sigma_P$ on the roots of $P$ in $\mathbb{F}_{p^{\infty}}$.
We are now ready to answer the questions.

This indeed is possible. Consider the case $P(x) = x^4 + x + 1$. There is some set of conjugacy classes in $\text{Gal}(F_P/\mathbb{Q})$ such that the elements in those conjugacy classes permute the roots of $P$ with cycle pattern $(1, 3)$. Now those $p$ for which the Artin symbol of $p$ with respect to $F_P$ belongs to some of these conjugacy classes are exactly the primes for which $P$ splits into factors of degrees $1$ and $3$ modulo $p$ due to the connection with the Frobenius endomorphism, Artin symbol and the degrees of the irreducible factors of $P$ modulo $p$. Note, however, that it is not easy to determine the Galois group of a given polynomial - thus, it could be that the number of desired conjugacy classes in this case is $0$. (For the polynomial $x^4 + x + 1$ this is not the case - see the answer of Jyrki Lahtonen for the calculation of the Galois group.)
In general, yes. If there was a bijection between factorization types and conjugacy classes, it would be that the Frobenius density theorem was no weaker than Chebotarev's. This, however, is not the case. For example, we know that Chebotarev implies Dirichlet's theorem while Frobenius does not. In the exposition linked in the question there is more about this aplication.
Reducible polynomials are no problem. It is convenient to formulate the Frobenius and Chebotarev density theorems for irreducible polynomials (if one wants to formulate them in terms of polynomials), as in the irreducible case the Galois group of the polynomial is transitive. The general case follows follows from the irreducible case. Consider, for example, the case of $P(x) = P_1(x)P_2(x)$, where $P_1$ and $P_2$ are irreducible. One may now take the compositum of the splitting fields of $P_1$ and $P_2$ and work in there. The extra ingredient one needs is the following natural property of the Artin symbol: Let $\mathbb{Q} \subset K \subset L$ is a chain of fields. The restriction of the Artin symbol of $p$ w.r.t. $L$ to $K$ is the Artin symbol of $p$ w.r.t. $K$. So, for example, if $p$ is such that the Artin symbol of $p$ w.r.t. the compositum $F_{P_1}F_{P_2}$ is the identity, then the Artin symbol of $p$ w.r.t. $F_{P_1}$ is the identity, and we then know that $P_1$ splits completely modulo $p$.
As commented by reuns, it turns out that the Artin symbol of a prime $p$ with respect to the cyclotomic field $\mathbb{Q}(\zeta_m)$ depends on what $p$ is modulo $m$ and only on that. This is non-trivial. Here's one reference (Lemma 2.4.1). This is precisely the reason why cyclotomic fields correspond to congruence conditions. (In many cases the converse is true too: congruence conditions imply cyclotomic fields. This, apparently, is one of the main philosophies in class field theory.) Application: It is true that $\mathbb{Q}(\sqrt{n})$ is a subfield of a cyclotomic extension. (For an overkill proof, use the Kronecker-Weber theorem.) It does not come as a surprise that the set of primes $p$ for which a given integer $n$ is a quadratic residue is determined by congruence conditions.


TITLE: Convolution of ring Delta function
QUESTION [0 upvotes]: Assume $f(r)=\delta(r-R)$ where $\delta(\cdot)$ is a ring delta function. In other word, $f$ is a circular delta function on a circle with radius $R$. I want to do the convolution of $f$ with itself ($f*f$).
How can I do convolution in polar(spherical) coordinates. I read that I can use Hankel and inverse Fourier transforms to obtain the convolution, however I am not into the subject and I will appreciate if some could help me with this.

REPLY [1 votes]: Although convolutions of distributions are not defined in general, you can do it in this case.  Let
$$ f_\epsilon(x) = \epsilon^{-1} I_{|x| \in [R, R+\epsilon]} .$$
Then compute $f_\epsilon*f_\epsilon$, and then let $\epsilon \to 0$.
What you will see is that $f*f(x)$ is the "area" of the intersection of the boundaries of the circles of radius $R$, one centered at $0$, and the other centered at $x$.  Obviously this area isn't properly defined, but you could think of thickening the boundaries of the circles so that their thicknesses are $\epsilon$, computing the area of this intersection, dividing by $\epsilon^2$, and then letting $\epsilon \to 0$.  You will realize that the answer depends only on the angle at which the two circles intersect each other.


TITLE: Work to pump water out of a tank with radius $10$
QUESTION [1 upvotes]: Water density is 1000, tank is a half sphere with radius $10$.
9.8 for gravity, 1000 for density, 2pi to find the volume and all of the rest gives me work.
$$ 9.8(1000)(2\pi) \int r^2  dy$$
To find the radius I just use $(10 - y)$
This is wrong, but why?

REPLY [2 votes]: The mass of an infinitely thin layer of water at depth $y$ is $1000\pi r^2dy$, so the force needed to lift it is $9.8\cdot1000\pi r^2dy$, and the work done in lifting it is $9.8\cdot1000\pi r^2ydy$. Your integral should be
$$9800\pi\int_0^{10}r^2ydy\;.$$
Moreover, $r$ is not $10-y$: $r^2+y^2=10^2$, so $r^2=100-y^2$.


TITLE: question about zero of $p$-adic $L$-function
QUESTION [3 upvotes]: For simplicity, let $p$ be an odd prime. Let $P(X) \in \mathbf{Z}_p[X]$ be a distinguished polynomial.
Suppose $\mathrm{ord}_p(P(u^k-1))$ is bounded as $k$ varies in the whole integer. Suppose there exist positive integers $k, k^{\prime}$ such that $\mathrm{ord}_p(P(u^k-1)) \neq \mathrm{ord}_p(P(u^{k^{\prime}}-1))$, can we prove that $P(X)$ has zero in $\mathbf{Z}_p$ and make the assumption of bounded failed? Where $u=1+p$. And what about $P(X)$ is the Iwasawa polynomial defining $p$-adic $L$-function, does it has zero in $\mathbf{Z}_p$?

REPLY [2 votes]: I have no insight concerning your first question. Why do you think that the condition $ord_p P(u^k - 1) \neq ord_p P(u^{k'} - 1)$ should imply the existence of a zero of $P(X)$ ? Or is it just wishful thinking ? Given a power series $f \in \mathbf Z_p[[X]]$, there exist general considerations about its zeroes and the $p$-adic maximum principle (see e.g. the report by Hugo Castillo on « Kubota-Leopoldt’s $p$-adic $L$-functions », thm. 5.1.2 of chap. 5 on « The compact-open topology »), but they are too general to be of use here.
A little more is known about the Iwasawa power series $f_\chi (X)$ which enters the definition of the Kubota-Leopoldt function $L_p (s, \chi)$. By construction, $L_p (s, \chi)$ is identically null if  the Dirichlet character $\chi$ is even. If $\chi$ is odd, it can be shown as an exercise that  $L_p (s, \chi)=0$ iff $\chi\omega^{-1}(p) = 1$, where $\omega$ is the so called Teichmüller character, which gives the Galois action on the $p$-th roots of unity. But note that in general, a zero of the series $f_\chi (X)$ will not yield a corresponding zero of the function $L_p (s, \chi)$ (numerical counterexamples computed by Wagstaff , circa 1982). Yet a precise relationship can be established by modifying suitably $s$ and $\chi$ in the $L_p$-function (Childress & Gold,  Acta Arithmetica, 1987).
The zeroes of the Iwasawa power series are more readily computed because, thanks to the Main Conjecture ( = the theorem of Mazur-Wiles), $f_\chi (X)$ has an algebraic meaning : roughly speaking, it is the « characteristic  series » (in the sense of linear algebra) which describes the Galois action on the inverse limit of adequate  $p$-class groups along the cyclotomic extension obtained by adjoining  to $\mathbf Q$ all the $p^n$-th roots of unity. This does not mean that all its zeroes are known. But « special non-zeroes » have a particular arithmetic meaning. Example : replacing $\mathbf Q$ by a totally real number field $K$, the fact that $0$ is not a zero of an adequate generalization of the « characteristic series » above is equivalent to the validity of the celebrated Leopoldt conjecture for $K$.


TITLE: find this limit $\lim_{x\to0^{+}}\frac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$
QUESTION [7 upvotes]: find the limit.

$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$

my try:
$$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$
so
$$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\tan^3{x})=x+\dfrac{1}{3}x^3+\dfrac{1}{3}(x+\dfrac{1}{3}x^3)^3+o(x^3)$$
$$\tan{(\sin{x})}=\sin{x}+\dfrac{1}{3}(\sin{x})^3+o(\sin^3{x})=x-\dfrac{1}{6}x^3+\dfrac{1}{3}(x-\dfrac{1}{6}x^3)^3+o(x^3)$$
so
$$\tan{(\tan{x})}-\tan{(\sin{x})}=\dfrac{1}{2}x^3+o(x^3)$$
$$\tan{x}-\sin{x}=\dfrac{1}{2}x^3+o(x^3)$$
Have other metods? Thank you

REPLY [1 votes]: Factor a $\tan(x)$ from the denominator:
$$\frac{\tan\tan(x) - \tan\sin(x)}{\tan(x)-\sin(x)} = \frac{\frac{\tan\tan(x)}{\tan(x)} - \cos(x)\frac{\tan\sin(x)}{\sin(x)}}{1-\cos(x)}.$$
The quotients in the numerator go to $1$, so the numerator goes to $1-\cos(x)$. This suggests cancellation.
After two derivatives, the denominator of this quotient will go to $1$, so one strategy is L'Hopital's rule. (I don't have time to compute it but maybe I'll have time later today.)


TITLE: Advice for choosing a book in Complex Analysis.
QUESTION [2 upvotes]: I am a 4th year mathematics undergraduate student. I took a "Complex Variables" (complex analysis made less rigorous) course in second year, though I feel at the time I wasn't able to properly appreciate the subject since I was in second year. Now, I am in 4th year and I have a background in real analysis, including two introductory course in the subject, a course in measure theory, a course in topology, and (currently) a course in functional analysis. I am looking for a complex analysis book that I will enjoy in my spare time.
What I am looking for:

Most importantly, lots of exercises that have (at least some) selected solutions
A book that explains well and emphasizes the theory

Something under $200 for a physical copy would be nice (though beggers can't be choosers).
Thanks in advance for the advice!

REPLY [0 votes]: Look at this page and download "basic complex analysis" and "selected topics in complex analysis". They are really good notes.


TITLE: Is this sequence bounded ? (An open problem between my schoolmates !)
QUESTION [11 upvotes]: Let $f$ be a smooth function (say $\mathcal{C}^{\infty}$) in its two real variables ($t$ and $T$). I consider the following sequence defined by
  $$A_n:=\lim_{T \to \infty} \int_{0}^{1} e^{-n t} f(t,T)dt \quad \quad (n\geq 1)$$
  where we suppose that this limit exists and is finite for all integer $n$.
  I would like to conclude that the sequence $(A_n)$ is bounded. Is that true ?

This is not an obvious problem : assuming certain conditions on $f$ then the exercise becomes classic. However, very little can be said about $(A_n)$ in the general case. In fact I deeply think that the answer is no and that we can find a $f$ such that $A_n$ tends to infinity.
If one takes $f(t,T)=2\sin(tT)/\pi t$, then $(A_n)$ is the constant sequence equal to $1$. This shows that $f$ does not necessarily tend to $0$.
Thanks for your help !

REPLY [6 votes]: The sequence $\{A_n\}$ need not to be bounded. To see this, one could for example as $f(t,T)$ choose something that approximates a derivative of a delta distribution as $T\to+\infty$. I wish to give credits to my colleague Tomas Persson who came up with that idea.
I will give such an approximating example. My example is non-smooth, but that is just to make the calculations more transparent.
Let
$$
g(t,T)=
\begin{cases}
\frac{T}{2} & |t|\leq\frac{1}{T}\\
0 & |t|>\frac{1}{T}.
\end{cases}
$$
This is an approximation of the delta distribution as $T\to+\infty$. We then let $f$ be the following difference quotient:
$$
f(t,T)=\frac{g(t-1/T,T)-g(t-2/T,T)}{1/T}
$$
It is then a simple matter to calculate the integral
$$
\int_0^1 e^{-nt}f(t,T)\,dt=\frac{T^2}{2n}\Bigl(1+e^{-3n/T}-e^{-2n/T}-e^{-n/T}\Bigr)
$$
Hence,
$$
A_n=\lim_{T\to+\infty}\int_0^1 e^{-nt}f(t,T)\,dt = n,
$$
which of course is unbounded.
Update
Let me, for completeness, add a smooth function $f$ that also gives $A_n=n$:
$$
f(t,T)=(T^2-T^3t)e^{-Tt}.
$$
The argument is the same, it approximates a derivative of the delta distribution.


TITLE: Prove that $|u(x,t)|\leq\sup\limits_{y\in R^n}|g(y)|$
QUESTION [2 upvotes]: Let
  $$u(x,t)=\int\limits_{\mathbb{R}^n}\Phi(x-y,t)g(y)dy$$
  where
  $$\Phi(x-y,t)=\frac{1}{(4\pi t)^{n/2}}e^{\frac{-|x-y|}{4t}}$$
  (this is the solution for the homogeneous heat equation), where $g$ is a continuous function with $\lim\limits_{|x|\to\infty}g(x)=0$. Prove that $$|u(x,t)|\leq\sup\limits_{y\in \mathbb{R}^n}|g(y)|$$
  for all $x\in \mathbb{R}^n, t>0$.

My Attempt:
For a fixed $x,t$, we have
$$e^{\frac{-|x-y|}{4t}}\leq 1$$
Hence
\begin{aligned}
\lvert u(x,t)\rvert &\leq\int\limits_{\mathbb{R}^n} \lvert \Phi(x-y)g(y) \rvert dy \\
        &=\int\limits_{\mathbb{R}^n}|\Phi(x-y)||g(y)|dy\\
        &\leq\frac{1}{(4\pi t)^{n/2}} \int\limits_{\mathbb{R}^n}|g(y)|dy\\
        &\leq \sup\limits_{y\in \mathbb{R}^n}|g(y)|\frac{1}{(4\pi t)^{n/2}}\int\limits_{\mathbb{R}^n}dy
\end{aligned}
But after that how can I get rid of the integral?

REPLY [2 votes]: Oh.. I have missed an important lemma.

This is from the book:
Partial differential equations
Book by Lawrence C. Evans second edition page 46
So with that It will solve the problem.


TITLE: Nagell-Lutz theorem
QUESTION [1 upvotes]: I'm a question about Nagell-Nutz theorem.
For example, the point $P'=( \frac{31073}{2704},-\frac{5491823}{140608})$ on the curve $$y^2=x^3+8$$ does not meet the theorem Nagell-Nutz. So I can say what the point $P=(2,4)$ has infinite order by the Nagell-Lutz theorem? Since $P'=4P$.
I am not fluent in English. Sorry for any error.
Thank you very much!

REPLY [1 votes]: Yes, that conclusion is correct. By Nagell-Lutz the points of finite order must have integer coordinates, thus $P'$ has infinite order and by the relation $4P = P'$ we conclude that $P$ has infinite order too.


TITLE: Determining the refractive index of a glass slab using a travelling microscope
QUESTION [0 upvotes]: In this practical video of finding the refractive index, the reading is being taken by placing the travelling microscope perpendicular to the slab. But I've learned in another video(Frame Shot) if you are watching directly from the above (incident angle is 0°) then the Real depth = Apparent depth.And  I am sure that In the above situation the below refractive index formula won't work.

My questions are :

How did he calculate the Apparent depth and the refractive index?

If he actually tilted the microscope in (1.) How was he able to calculate SP' and PP'?
3.Why aren't there much resources on this topic in the internet? (Most of the resources i've seen are
from Indian sites. Do western countries use different method or instrument)

Any help or referrals to online resources is welcomed. Kindly ask in the comments, if you have any confusions regarding my question.

REPLY [1 votes]: if you are watching directly from the above (incident angle is 0°) then the Real depth = Apparent depth

To form the final image the objective of the travelling microscope collects a cone of rays with a range of small angles of incidence (as assumed in the derivation of the formula for refractive index) from a point on the object and not just one ray with an incident angle of zero degrees.
Update as a result of a comment from @SivaManasan
The derivation of the formula for the refractive index is done in many books and in many Internet sites but they all make an approximation of the type if angle $\theta$ is small $\sin \theta \approx \theta,\: \sin \theta \approx \tan \theta $ etc and here is one of those derivations.
My point about the cone of rays I have tried to illustrate below.

The first thing you need to know is that you only move the travelling microscope you make no adjustments to it so the distance between the object and the final image is fixed.
These are distances $CC'$ and $DD'$ in my diagram.
The other important property of the microscope is that its depth of field (distance between the nearest and the furthest objects that are in acceptably sharp focus in an image) is small.
Without the glass block (grey rays) the cone of rays $ACB$ originating from a point object are focussed by the microscope lens $AB$ to form an image in the plane of the cross wires of the microscope at $C'$.
Putting the glass block in the way effectively moves the observed position of the object for $C$ to $D$ and now $D$ acts as the object for the microscope lens.
To focus the cone of rays apparently originating from point $D$ the microscope lens has to be moved to position $A'B'$ which is distance $h$ in my diagram.
On moving the microscope (lens) the cone of rays $A'DC'$ are now focused by the lens at $D'$.
Note that all the rays (except the zero incidence ray) undergo refraction (bending of a light ray as it travels from one medium to another) and so the derived formula for the refractive index is valid particularly as the angles involved are small because of the small diameter of the microscope lens.

Most of the resources I've seen are from Indian sites. Do western
  countries use different method or instrument?

There are other methods with pins etc but one of the main reasons why Indian sites have a lot of reference to to this method of measuring refractive index might be that the topic is in the syllabuses of Indian examination boards whereas, for example in the UK, this method has been omitted in favour of other methods which mostly rely on the measurement of angles.


TITLE: Magnetic field from different reference frame
QUESTION [1 upvotes]: We know if a particle moves under a magnetic field then the force applied on it is given by
F=q(V×B)
Where V is the velocity vector and B is the magnetic field intensity.
Now consider a hypothetical situation that I throw a charged ball in a uniform magnetic field and very naturally it will get deflected under the magnetic force.
Now again this time,say,I am sitting in a car and have the ball in my hand so in my frame the magnetic field should be the same but the velocity of the ball is zero so there should be no magnetic force at all and the ball should not get deflected.
A physical phenomena if I am observing from an outside static frame or from a  car frame should not be different,at least upto the extent of viewing a deflection in ball's path.
This is an absurd unrelaistic situation,so the only solution is that the magnetic field should change by some means.
So,does it really happen in practice?

REPLY [2 votes]: Yes, the electric and magnetic fields change.  When you move to a different reference frame with relative speed $\vec v$, the components of the fields which are perpendicular to your new velocity become:
$$ \vec E'_\perp = \gamma\left(\vec E_\perp +\vec v \times \vec B\right)$$
$$ \vec B'_\perp = \gamma\left(\vec B_\perp -\frac{1}{c^2}\vec v \times \vec E\right)$$
where
$$ \gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$
The components of the fields which are parallel to your new velocity are unchanged.
As you can see, the electric field gains a component perpendicular to your new velocity which precisely compensates for the fact that the magnetic field does not act on the charged ball which is stationary in your new reference frame.
The result is that both observers ("stationary" and co-moving) would agree$^\dagger$ on the forces acting on the ball, but would disagree on whether those forces were due to the electric or the magnetic fields.  In this sense, the electric and magnetic fields aren't separate entities, but are part of a unified electromagnetic field; how much of the force on a particle is due to the electric or magnetic fields depends on one's reference frame.
$^\dagger$Clarification: The observers would not actually agree on the magnititude of the forces because of the factor of $\gamma$ which is present in the new reference frame.  This reflects the fact that force and acceleration are not 4-vector quantities - without going into too much detail, the difference can be chalked up to time dilation, and can be neglected if $v/c <<1$.