TITLE: What is the permeability of a permanent magnet?
QUESTION [5 upvotes]: The following is a diagram that shows the B- and H-fields of a permanent magnet.
Inside the magnet, the H-field is in the opposite direction to the B-field, because of the magnetisation M-field and the relationship
$$ \vec{B} = \mu_0 (\vec{H} + \vec{M})$$
(in SI units).
It is also commonly stated that the relationship between $\vec{B}$ and $\vec{H}$ can be written as $\vec{B} = \mu \vec{H}$, where in general, $\mu$ can be a tensor that depends on the H-field, position, time, temperature etc. (i.e. not necessarily restricted to linear media).
Yet when I look up the relative permeability for magnets (for example here for Neodymium magnets), it is given as a positive scalar $>1$. But surely if the B-field and H-field are in opposite directions then $\mu$ is negative?
What is the source of my confusion?

REPLY [0 votes]: The relation between $B$ and $H$ (or for the electric case between $E$ and $D$) are completely material dependent and have no closed form expression. They are called constitutive relations. There are only models for these relations. The permeability model is just the simplest one.
Even when one would say $μ$ is tensorial and allowed to depend on space, time, $H$ and $B$, this (extended permeability) model is still not fully general. This is because there may be dynamic effects ($M$ may depend on past values of $H$) and/or dependencies on spatially neighboring points of $H$. Dynamic effects are very common (a.k.a. frequency dependence)
If one really wants to write down an equation capturing all this, one has probably something like this (this is from the Wikipedia link above):
$M = \frac{1}{\mu_0} \iiint d^3r' dt' \chi_m(r',t',r,t,H) H(r',t')$
However, I think this equation is not very useful. In my opinion, it is equally helpful to say $M$ is formed by applying a causal operator to $H$ (potentially nonlinear and time variant). But even that is not most general:
Magnetization can also be caused by other physical quantities such as temperature (Seebeck effect), electric field (Hall Effect), stress (Piezomagnetic effect), radio frequency waves (Nuclear magnetic resonance -> e.g. Bloch equaitions) with all these effects having different equations to describe magnetization.
Concluding, it is probably best to see magnetization (and its electric equivalent polarization) as a quantity which is given by many physical effects and has to be determined by the equations specific for the effects to be taken into account. A permeability based model is only valid for isotropic material, low fields and slow temporal variations.
For permanent magnet the first breaks down (fields are constantly there and are not "low") and the second very quickly with increasing frequency.
So for a magnetized magnet a permeability based model makes no sense.
One can specify a permeability for the (unmagnetized) material the magnet is made of, though. This is probably what you saw.
A good starting point is probably an advanced book on electromagnetism, such as Jacksons. More specific topics are likely handled in special book on magnetics, superconduction, spectroscopy (NMR or else methods), nonlinear optics or optoelectronics.


TITLE: modular arithmetic commutative group problem.
QUESTION [1 upvotes]: Im trying to do an exercise in Roger Godements algebra  which has me stumped.
Let $m,n$ be rational integers show there exists an integer $r$ such that $r=0\mod n$ and $r=1\mod n$ if and only if $m,n$ are coprime. Deduce from this that if $G$ is a commutative group and $x,y$ are elements of $G$ of orders $m,n$ respectively, with $m,n$ coprime then $z=xy$ has order $mn$ and the subgroup of $G$ generated by $z$ contains $x$ and $y$.
My first confusion comes from the lack of a variable $m$ in the equations for $r$.
I know that I can rewrite $r=0+nm$ and $r=1+nq$ which implies that $nm =1 +nq$ but from the two modulo equations but I dont see any relation between $q,m$ that must hold (I want to say that $m-q =1 =n$ but i feel that this is wrong). Am I missing something fundamental?
For the second part I think I could relate the powers of $z$ which generate $G$ to the exponents via a substitution using the equations from part a, but i dont think i can proceed without understanding the first part.
any hints are appreciated, thanks!

REPLY [2 votes]: If $(m,n)=1,$ then there exist $a,b\in\Bbb Z$ such that $am+bn=1.$ Let $r=am.$ Then $r\equiv 0\pmod m$ and $r\equiv 1\pmod n.$
Let $G$ be abelian, with $|x|=m,|y|=n,$ where $|\cdot|$ denotes the order of elements $x,y\in G.$ Since $G$ is abelian, we have $(xy)^{mn}=x^{mn}y^{mn}=(x^m)^n(y^n)^m=1^n\cdot 1^m=1.$ Thus, we find that $|xy|$ divides $mn.$
On the other hand, $(xy)^r=x^ry^r=1\cdot y=y$ by the choice of $r$ as above. This shows that the subgroup generated by $xy$ contains $y.$ By setting $r'=bn$ and using the same method, we see that this subgroup also contains $x.$
In particular, since $x\in\langle xy\rangle,$ we must have $|x|=m$ divides the order of $xy,$ because $|xy|=|\langle xy\rangle|.$ Similarly, since $y\in\langle xy\rangle,$ we must have $|y|=n$ divides $|xy|.$ Hence, using that $(m,n)=1,$ we get $mn$ divides $|xy|,$ and so we have $mn=|xy|.$


text:
{\bf Problem.} Ten families have an average of 2 children per family. If exactly two of these families are childless, what is the average number of children in the families with children? Express your answer as a decimal to the nearest tenth.
{\bf Level.} Level 3
{\bf Type.} Prealgebra
{\bf Solution.} There are $10(2)=20$ children total.  If $2$ families are childless, $8$ have children.  So the average number of children for a family with children is

$$\frac{20}{8}=\boxed{2.5}$$

set_name: MATH


TITLE: Find the values of r at which bifurcation occus and classify those as saddle node, transcritical, or pitchfork bifurcation
QUESTION [1 upvotes]: $$f(x)= rx-\frac{x}{1+x}$$
Find the values of $r$ at which bifurcation occus and classify those as saddle node, transcritical, or pitchfork bifurcation.
I found the fixed points as $x^*=0,\frac{1}{r}-1$.
I am not sure how to identify what TYPE of bifurcation it is out of the three types. Is there a way to test which one?

REPLY [2 votes]: $f(x)=rx-{x\over1+x}$. Fixed points where $f(x)=x$, so $$x=rx-{x\over1+x}$$ $x^*=0$ or $$1=r-{1\over1+x},\quad {1\over1+x}=r-1,\quad 1+x={1\over r-1},$$ $$x^*={1\over r-1}-1={2-r\over r-1}$$ I don't know how you got $x^*={1\over r}-1$.
Bifurcations occur when $f'(x^*)=\pm1$. When $f'(x^*)=-1$, you may get a period-doubling bifurcation, but that's not among the ones you listed, so I'll ignore that. $$f'(x)=r-{1\over(x+1)^2}$$ so we need $r-{1\over(x+1)^2}=1$. When $x^*=0$, this gives $r=2$. The other fixed point, $x^*={2-r\over r-1}$, gives nothing (but check my algebra here).
So, bifurcation occurs at $$(r^*,x^*)=(2,0)$$ Just from drawing the graphs of $x=0$ and $x={2-r\over r-1}$ and seeing how they cross at $(2,0)$ is, I think, enough to see that it's a transcritical bifurcation. But you can always calculate $f'(x)$ for fixed points nearby $(2,0)$ to see where you have stable fixed points and where unstable, and analyze the bifurcation that way. Or, you can use the formulas in Elaydi, for example, if $${\partial f\over\partial x}(r^*,x^*)=1,\quad {\partial f\over\partial r}(r^*,x^*)=0,\quad {\rm and}\quad {\partial^2 f\over\partial x^2}(r^*,x^*)\ne0$$ it's transcritical.


TITLE: Trying to find rotation transformation matrix using basis
QUESTION [3 upvotes]: I am studying linear transformations on my own and I have met the following theorem.

Suppose the vectors $v_1,\dots,v_n$ are the basis for the space $\mathbf{V}$, and vectors $w_1,\dots,w_m$ are basis for $\mathbf{W}$. Each linear transformation $T$ from $\mathbf{V}$ to $\mathbf{W}$ is represented by a matrix $\mathbf{A}$ whose $j$th column is found by applying $T$ to the $j$th basis vector $v_j$, and writing $T(v_j)$ as a combination of the $w$'s:
$$\text{Column}~~~\mathbf{j}~~~\text{of}~~~\mathbf{A}~~~T(v_j)=Av_j=a_{1j}w_1 + a_{2j}w_2 + \dots + a_{mj}w_m$$

Given this theorem we can compute the transformation matrix $\mathbf{A}$ of differentiation of polynomials. $\frac{d}{t}:\mathbf{P}_n \rightarrow \mathbf{P}_{n-1}$ where $\mathbf{P}_n$ are the polynomial of degree $n+1$ (consider constant term). Thus given the natural basis of third and second degree polynomials by $\mathbf{V} = \{1,t,t^2,t^3\}$ and $\mathbf{W} = \{1,t,t^2\}$ respectively, we may compute transformation matrix $\mathbf{A_{diff}}$. More specifically,
$$\frac{d}{dx}v_1 = 0~~~\text{or}~~~\mathbf{A_{diff}}v_1 =0w_1+0w_2+0w_3+0w_4$$, the coefficients $[0,0,0,0]^T$ form the first column of $\mathbf{A_{diff}}$.
$$\frac{d}{dx}v_2= 1~~~\text{or}~~~\mathbf{A_{diff}}v_2 =1w_1+0w_2+0w_3+0w_4$$, the coefficients $[1,0,0,0]^T$ form the second column of $\mathbf{A_{diff}}$. With similar way, we can compute the rest of the columns to obtain
$$\mathbf{A_{diff}}=\begin{bmatrix}
0&1&0&0\\
0&0&2&0\\
0&0&0&3\\
\end{bmatrix}$$
Similarly I have tried to compute the rotation transformation matrix $\mathbf{A_{rot}}$. Given the basis of the initial space $\mathbf{V}=\{[1, 0]^T, [0, 1]^T\}$, the basis of the rotated space is $\mathbf{W}=\{[cos\theta, sin\theta]^T, [-sin\theta, cos\theta]^T\}$
Could you please help to find the $\mathbf{A_{rot}}$ using the same steps as with $\mathbf{A_{diff}}$?
Thanks!

REPLY [2 votes]: Let $T:V \to W$ be rotation by $\theta$ in $\mathbb{R}^2$. Pick the standard basis for $V = \mathbb{R}^2$. We have that $T([1,0]) = [\cos \theta,\sin \theta]$ and $T([0,1]) = [-\sin\theta, \cos\theta]$.
If you choose the standard basis $([1,0],[0,1])$ for $W = \mathbb{R}^2$ then the matrix that you get is $$A_\text{rot} = \begin{pmatrix} \cos \theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.$$
If on the other hand you choose the basis $([\cos \theta,\sin \theta],[-\sin\theta, \cos\theta])$ for $W = \mathbb{R}^2$ then the matrix that you get is the identity $$A_\text{rot} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$


TITLE: Solving $\sqrt{(\log_2x)^2 - 5|\log_2x| + 6} \le 2\sqrt{5}$
QUESTION [1 upvotes]: My attempt:
Given,
$$\sqrt{(\log_2x)^2 - 5|\log_2x| + 6} \le 2\sqrt{5}$$
Taking $t=|\log_2x|$:
$$\sqrt{t^2-5t+6} \le 2\sqrt{5}$$
I know that squaring both sides will lead to extraneous solutions. In fact, I solved the rest of the problem by squaring sides and got $x \in \left[\dfrac{1}{128},128\right]$ whereas the correct answer according to Wolfram Alpha is $x \in \left[\dfrac{1}{128},\dfrac{1}{8}\right] \cup \left[\dfrac{1}{4}, 4\right] \cup [8,128]$.
What is the correct way of solving this problem that avoids extraneous solutions?

REPLY [2 votes]: This inequality is quite fiddly, but with care we can get to the solution. First, note that any inequality of the form $\sqrt{a}\le b$ is equivalent to $a\le b^2$ and $a\ge0$, meaning
\begin{align}
\sqrt{t^2-5t+6}\le2\sqrt{5} &\iff t^2-5t+6\le20 \quad\text{and}\quad t^2-5t+6\ge0 \\[5pt]
&\iff (t+2)(t-7)\le0\quad\text{and}\quad(t-2)(t-3)\ge0 \\[5pt]
&\iff t\in[-2,7]\quad\text{and}\quad t\not\in(2,3) \\[5pt]
&\iff t\in\left([-2,7]\setminus(2,3)\right)\\[5pt]
&\iff t\in[-2,2]\cup[3,7] \, .
\end{align}
Hence, $-2\le|\log_2(x)|\le 2$ or $3\le|\log_2(x)|\le 7$.
If $x\ge1$, then these inequalities become $-2\le\log_2(x)\le2$ (which has the solution $\frac{1}{4}<x\le4$) and $3\le\log_2(x)\le7$ (which has the solution $8\le x\le 128$). So if $x\ge1$, then $\frac{1}{4}\le x\le 4$ or $8\le x\le 128$. This boils down to $1\le x\le 4$ or $8\le x\le128$.
If $0<x<1$, then the inequalities become $-2\le-\log_2(x)\le 2$ (which has solution $\frac{1}{4}\le x\le2$) and $3\le -\log_2(x) \le 7$ (which has solution $\frac{1}{128}\le x\le\frac{1}{8}$). This boils down to $\frac{1}{128}\le x\le\frac{1}{8}$ or $\frac{1}{4}\le x\le1$.
Combining all of these solutions, we find that $\frac{1}{128}\le x\le\frac{1}{8}$ or $\frac{1}{4}\le x\le 4$ or $8\le x\le 128$.
I will give an example of how I solved one of the above inequalities. The rest of them are left as exercises:
\begin{align}
3\le -\log_2(x)\le 7 &\iff -3 \ge \log_2(x) \ge - 7 \\[5pt]
&\iff -7 \le \log_2(x) \le -3 \\[5pt]
&\iff 2^{-7} \le x \le 2^{-3} \\[5pt]
&\iff \frac{1}{128} \le x \le \frac{1}{8} \, .
\end{align}
Finally, there is a simpler way to solve $-2\le|\log_2(x)|\le 2$ or $3\le|\log_2(x)|\le 7$, but it is not obvious. A number $y$ satisfies these inequalities if and only if $\frac{1}{y}$ satisfies them. So once we have the solutions $1\le x \le 4$ or $8 \le x \le 128$, we can easily find the others. Admittedly, I didn't realise this until I solved this inequality the long-winded way.


TITLE: Inequality for positive definite matrices of specific type
QUESTION [0 upvotes]: We call a matrix good if it is symmetric positive definite with all its eigenvalues below $1$ .
Assume that we have a good matrix $B$.
How to show that for $n\in \mathbb{N}$:
$$\left(\mathbb{I} - (\mathbb{I}-B)^n\right)^2\prec nB$$
I know that, given a good matrix $B$, the matrices $(\mathbb{I}-B)$ and $(\mathbb{I} - (\mathbb{I}-B)^n)$ are good as well. So, it is easy to show that
$$\left(\mathbb{I} - (\mathbb{I}-B)^n\right)^2\prec I $$
But $\left(\mathbb{I} - (\mathbb{I}-B)^n\right)^2\prec nB$ does seem to be more difficult to prove.

REPLY [3 votes]: Suppose that $Q$ is orthogonal and diagonalizes $B$. Then
$$
Q'BQ = D
$$
is diagonal (where the prime means 'transpose'), with all diagonal entries strictly between 0 and 1, and each column of $v$ is a unit eigenvector of $B$.
It turns out that such eigenvectors are also eigenvectors of $I - B$ (with eigenvalues $1-c$, where $c$ is the corresponding eigenvalue in $B$), hence of $(I - B)^n$, and hence of $(I - (I-B)^n)$, and hence of $(I - (I- B)^n)^2$. In short: $Q$ also diagonalizes the matrix you're looking at.
I believe (but I might be wrong -- kinda early in the morning) that in the $Q$ basis, your matrix looks like
$$
(I - (I - D)^n)^2
$$
Now all you have to do is check that for each diagonal entry $c$ of $D$, you have
$$
(1 - (1-c)^n)^2 \le nc
$$
which is presumably either false or a not-too-hard calculus problem. I leave that part to you.
I'm a little uncertain about the less-than-or-equal vs less-than matter; I suspect that $A \preceq 0$ really means that $A$ is positive semidefinite, but I can't be sure of that. Anyhow, you can decide which one it is, and then you can do the appropriate calculus problem.


TITLE: For a prime $p$, determine the number of positive integers whose greatest proper divisor is $p$
QUESTION [3 upvotes]: I'm having a bit of difficulty writing a graceful proof for the following problem:

For a prime $p$, determine the number of positive integers whose greatest proper divisor is $p$.

Let $A$ be the set of positive integers whose greatest proper divisor is $p$. I will show that $A=\{\alpha p\,|\,\alpha \;\text{prime},\; \alpha \leq p\}$ so that $|A|=\pi(p)$, the number of primes less than or equal to $p$.
Assume that $n=\alpha p$ for some prime $\alpha \leq p$. The factors of $n$ are $1, \alpha, p, \alpha p$. (In the case that $\alpha =p$, the factors of $n$ are $1,p, p^2$.) The greatest proper divisor of $n$ therefore is $p$ and so $n\in A$.
Conversely, for any number in $A$, clearly we have that $p$ is the greatest prime dividing it. Moreover, any three primes dividing it would contradict $p$ being the greatest divisor (this can be seen in the following way: if $\alpha_1\neq \alpha_2$ are two primes dividing the number, neither of which is $p$, then we have $\alpha_1<p<\alpha_1p<\alpha_1\alpha_2p$ from which we infer that $p$ is not the greatest proper divisor.) Therefore, a number in $A$ must have at most two prime factors, one being $p$. (We can rule out that a number in $A$ has exactly the one prime factor, $p$, since it has no proper divisors.) Hence, $A\subset \{\alpha p\,|\,\alpha \;\text{prime},\; \alpha \leq p\}$.
Perhaps there's a more elegant proof, but I'm concerned with my writing style.  Can anyone help me rephrase my argument in a smoother way or critique it?
edit: incorporated the changes suggested by Henning.

REPLY [4 votes]: Looks sound and direct to me. There is only minor copy-editing to do, such as:

The condition $2\le\alpha$ is redundant; it is already implied by $\alpha$ being prime.
The part "Let $n\in\{\alpha p\,|\,\alpha \;\text{prime},\; 2\leq \alpha \leq p\}$. Then
$n=\alpha p$ for some prime $\alpha$, $2\leq \alpha \leq p$" is probably more verbose than you need to be. I would just write, "Assume that $n=\alpha p$ for some prime $\alpha \leq p$."
To make the structure of the proof more explicit, you could write "Conversely," rather than "Now," at the beginning of the third paragraph. This tells the reader that you have now finished something and is proceeding to the opposite direction of what you just proved.
Calling the other prime $q$ would be somewhat more conventional than $\alpha$.


TITLE: Find all natural numbers for which $3\binom{2n}{n+1}=2\binom{2n+1}{n-1}$ holds true
QUESTION [1 upvotes]: I end up getting a quadratic equasion with no natural answers, so I am probably wrong.
(Dont know if the tag is right, its part of the combinatorics section in my book)

REPLY [1 votes]: You had the right idea;
$$ 3 \times {2n \choose n+1} = 2 \times { 2n+1 \choose n-1}$$
$$ 3 \times \frac{(2n)!}{(n+1)!(n-1)!} = 2 \times \frac{(2n+1)!}{(n-1)!(n+2)!}$$
Cancelling gives:
$$ 3  = 2 \times \frac{(2n+1)}{(n+2)}$$
and thus $$3n+ 6 = 4n+2$$
and so $n=4$.
We can check:
$$ 3 \times {8 \choose 5} = 3\times 56 = 168$$
and
$$ 2 \times { 9 \choose 3} = 2\times 84 = 168$$


TITLE: continuity and open subset
QUESTION [1 upvotes]: let say  E is a subset of R and f be the real number function o E then
  f is continuou on E if an only if $f^{-1}(V)$ is open in E for every
  open subset $V$ of $R$

what is opn subset V mean? in a continuous function?
I thought open set mean the set contain all its interier points
but what open set mean in he conticuose function
and also what is the closed set mean in continuous function?

REPLY [1 votes]: Let me rephrase it to (hopefully) make it more clear.

Let's say $E\subseteq\Bbb R$ and $f:E\to\Bbb R.$ The following are equivalent:

The function $f$ is continuous on $E$.
For every $V\subseteq\Bbb R$ with $V$ open, we have that $f^{-1}(V)$ is (relatively) open in $E$.

When we talk about $V$ being open, we mean exactly that every point of $V$ is an interior point. (Every set contains all its interior points, but open sets don't have any non-interior points.) When we talk about $f^{-1}(V)$ being relatively open in $E,$ we mean that there is a set $U$ that is open (in the usual sense) such that $f^{-1}(V)=U\cap E.$
The notation $f^{-1}(V)$ denotes the preimage or inverse image of the set $V$, and is defined by $$f^{-1}(V):=\{x\in E\mid f(x)\in V\}.$$ Put another way, it is the set of elements of $E$ that $f$ sends to $V$.


TITLE: Proof about set theory and distance function. Compact set and closed set.
QUESTION [2 upvotes]: $A$ : closed set in $\mathbb{R}^n$
$B$ : compact set in $\mathbb{R}^n$
$A \cap B=\phi.$
$d$ : distance function
Then, prove that
\begin{equation}
\text{For all } x \in B, \text{ there exists } a_x \in A \text{  such that } d(x, A)=d(x, a_x).
\end{equation}
$\bigg(d(x,A)=\inf\{d(x,y) | y \in A\} \bigg)$
I think that I should use the continuity of distance function.
For all $x\in B$,
define
$f(x):=d(x,A) \ (x\in B).$
Because $B$ is compact and $f$ is continuous, $f(B)$ is also compact. Thus there exists maximum value and minimum value of $f(B)$.
But this idea didn't work.
I would like you to give me some ideas.

REPLY [3 votes]: Compactness of $B$ is not needed!
$d(x,A)$ can be written as $\lim d(x,a_n)$ for some squence $(a_n) \subset A$. Now boundedness of $d(x,a_n)$ implies boundedness of $(a_n)$. Hence there is a subsequence converging to some $a_x$. Since $A$ is closed we see that $a_x \in A$.  Can you see that $d(x,A)=d(x,a_x)$?


TITLE: When will the ball hit the ground.
QUESTION [2 upvotes]: A girl is throwing a ball.
If I have an equation $h= -4.9*t^2+20t+1.2$,
 where $t$ is number of second after the girl has thrown the ball, and $h$ is the height above ground.
I want to find out when will the ball hit the ground. My solution is
set $-4.9*t^2+20t+1.2=0$, and then I get $t=4.1$ seconds. Is it correct?
An it is wrong to say that we solve $-4.9*t^2+20t+1.2=4.1$ and then solve for $t$. Am I right?

REPLY [0 votes]: So that the question will not remain unanswered, here is an answer.
You calculated the (positive) value of $t$ such that $-4.9t^2+20t+1.2=0$ correctly, and this was the right thing to do.
I get that $t\approx 4.14$. That kind of precision (to the nearest hundredth of a second), is physically unreasonable, since the quadratic model assumes that there is no air resistance.
As to the final question, setting $-4.9t^2+20t+1.2=4.1$ has no connection to solving the problem of finding the time when the ball hits the ground.


TITLE: Prove that f satisfies Cauchy-Riemann Equations but is not differentiable
QUESTION [0 upvotes]: Prove that $f$ defined by $f(z) = \frac{z^5}{|z|^4} | (z\neq 0), f(0)=0$ satisfies the Cauchy-Riemann equations at $z=0$ but is not differentiable there.
So I know that $u_x = v_x$.
Therefore let $u(x,y) = |z|^4 = (x^2+y^2)^2$ and let $v(x,y)=
z^5 = (x+iy)^5$.
These are not equal. Where have I made my mistake?
Thank you for any guidance.

REPLY [7 votes]: There is no reason to take $u(x,y)=\lvert x+yi\rvert^4$ and $v(x,y)=(x+yi)^5$. Note that it is not even true that $v$ is a function from $\mathbb{C}$ into $\mathbb R$.
Note that, if $x\in\mathbb{R}\setminus\{0\}$, then $f(x)=\frac{x^5}{x^4}=x$ and that therefore $u(x,0)=x$ and $v(x,0)=0$. So, $u_x(0,0)=1$ and $v_x(0,0)=0$. On the other hand, $y\in\mathbb{R}\setminus\{0\}\implies f(yi)=\frac{(yi)^5}{y^4}=yi$ . Therefore, $u(0,y)=0$ and $v(0,y)=y$. So, $u_y(0,0)=0$ and $v_y(0,0)=1$. So, $(0,0)$ is a solution of the Cauchy-Riemann equations.
However, $f$ is not differentiable at $0$. The limit$$\lim_{z\to0}\frac{z^5}{z|z|^4}=\lim_{z\to0}\frac{z^2}{\overline z^2}$$does not exist. Just see what happens when $z\in\mathbb R$ and when $z$ s of the form $(1+i)t$, for a real $t$.


TITLE: How is the spin of hurricanes explained from an inertial frame?
QUESTION [1 upvotes]: I have read that hurricanes spin because of the Coriolis effect. Since the Coriolis force is a ficticious force, which is only present in a frame that is rotating w.r.t. to an inertial one, I am wondering if an inertial observer would have an explanation which does not rely on ficticious forces.

REPLY [2 votes]: Ok, the explanation that doesn't use the concept of fictitious force.
I need to mention first that explanation of physics taking place is independent of how you represent it.
Example: a wind blows over a tree. When you see the fallen tree you know the cause: the wind blew it over. Whether you represent that in an inertial frame or a rotating frame has no bearing on identifying the cause.

The diagram below shows a schematic representation of the formation of cyclonic flow.

This representation is not to scale of course, the size is exaggerated for clarity.
The most interesting part of the diagram is the the air flow from east to west that is contributing to the formation of cyclonic flow. That east-to-west flow is curved so much that it moves inside of the latitude line that it is starting from.
To explain why east-to-west flowing air mass will move inside the latitude line it is starting from I need to discuss the way the rotating Earth is in an equilibrium shape.

the diagram shows a celestial body with a far larger equatorial bulge than the Earth.
Any buoyant object is subject to a normal force (red arrow).
When a celestial body has an equatorial bulge the gravitational force is not exactly opposite in direction to the normal force. So there is a resultant force, which in Geophysical Fluid Dynamics is referred to as the poleward force.
In the case of the Earth: polar radius is 6357 kilometer, equatorial radius is 6378 kilometer. The difference is 21 kilometer.
This means that from the equator to the pole is a downhill slope. If that slope would not be there then the water of the Earth would flow to the Equator. The Earth's equatorial bulge prevents that.
Example: at 45 degrees latitude the downward slope is 0.1 degree. That slope provides the required centripetal force to remain co-rotating with the Earth. At 45 degrees, to find the amount of required centripetal force you divide by 580, that is the ratio.
To get a feel for how much centripetal force is required (at 45 degrees latitude), divide your own weight by 580. If you have a weighing utensil you can push: that will give you a feel for it.

So: when you are at rest with respect to the Earth you are co-rotating with the Earth.
Now take the case of buoyant mass, flowing from west-to-east. Moving west-to-east the mass is circumnavigating the Earth's axis faster than the Earth itself. The mass is on the slope (like a car on a banked circuit), but the mass is speeding, so it will swing wide.
So: buoyant mass that is flowing from west-to-east will depart from the latitude that it is starting from, veering off towards the equator.
Now buoyant mass that is flowing from east-to-west.
Moving east-to-west this mass is circumnavigating the Earth's axis slower than the Earth itself. But the slope is the slope voor co-rotating mass. Comparison: if a car is on a banked circuit, and the car is going too slow the car will slump down
So that is what happens to air mass that is flowing from east-to-west: the poleward force pulls that air mass closer to the nearest pole.
For completeness:
When air mass is north-south, flowing away from the (nearest) pole the poleward force is doing negative work, slowing down the angular velocity of the air mass.
When air mass is south-north, flowing towards the (nearest) pole the poleward force is reeling in that mass, causing the angular velocity of the air mass to increase.

That is, highly schematicy, the process of formation of cyclonic flow.

Of course, that raises the question: why do many authors attribute the formation of cyclonic flow to a fictitious force?
To explain that let me discuss a simpler case: motion of a single object over the surface of a rotating celestial body.
The weight of the object is supported by the surface so the object is subject to the poleward force. At every latitude the magnitude of the poleward force is precisely the required centripetal for to keep co-rotating with the Earth, at the same distance to the axis of rotation.
(I'm also simplifying to planar motion, instead of three spatial dimenstions)
The formula for required centripetal force when circumnavigating at angular velocity $\Omega$
$$ F = -m \Omega^2 r $$
With the minus sign to indicate that this force acts back to the center, not outward.
The equation of motion for motion with respect to the rotating coordinate system has three terms:
the term for the poleward force
the coriolis term
the centrifugal term
$$ -m \Omega^2 r + 2m \Omega v + m \Omega^2 r  $$
The fact that the Earth's equatorial bulge is an equilibrium state is crucial here. The Earth's shape is such that at every latitude the poleward force provides the required centripetal force for co-rotating with the Earth. That is why the term for the poleward force has the angular velocity value $\Omega$ of the Earth rotation.
That means that mathematically the term for the poleward force and the centrifugal term drop away against each other.
The equations of Geophysical Fluid Dynamics do incorporate the poleward force. If they would not incorporate the poleward force then the centrifugal term makes everything fly outward. The centrifugal term is far larger than the coriolis term. What happens is that the poleward force is incorporated and then it is the case that mathematically the term for the poleward force and the centrifugal term drop away against each other.


TITLE: Find the number of local extrema of that function without calculus.
QUESTION [0 upvotes]: I need to find the number of local extrema of that function without derivate or using calculus.
I know that in $x = 1$ and $x = 3$ $f(x) = 0$ ... in which way I can affirm that this function has at least $4$ different local extrema?
$$
f(x) = (x-1)^2(x-3)^4(x+5)^{1/3}
$$
Thanks a lot!

REPLY [1 votes]: Looking at $g = f^3$ instead will not alter the number (or existence) of extrema since $x \mapsto x^3$ is bijective.
$$ g(x) = (x-1)^6(x-3)^{12}(x+5)$$
From this, since $g$ has degree 19, we know $g(x) \to -\infty$ with $x \to -\infty$ and $g(x) \to \infty$ with $x \to \infty$. We also know it has an odd zero at $x=-5$, followed by an even zero at $x = 1$, and another even zero at $x = 3$.
We pick up an extremum between $x=-5$ and $1$, another at $x=1$ (even zero), another between $x=1$ and $3$, and a final one at $x=3$. That's four extrema. These should coincide with the extrema of $f$.

REPLY [0 votes]: By using continuity of $f$ and by using its graph we see that between $-5$ and $1$, there is a local extrema because there is a point where the tangent line is horizontal (Rolle's theorem), and between $1$ and $3$ there is another one, and two more at $1$ and $3$ themselves are local min each. Thus at least we can see $4$ already.


TITLE: For how many $n\in \{1,...,500\}$ is $n$ a multiple of one or more of $5,6$ or $7$
QUESTION [1 upvotes]: For how many $n\in \{1,...,500\}$ is $n$ a multiple of one or more of $5,6$ or $7$

So I believe I want to find the number of integers in the list divisible by $5,6,$ or $7$ and then subtract the times I double counted the ones which are divisible by more then one.
But I'm not sure how I know how many are divisible by $5,6,7$?

REPLY [2 votes]: The number of integers in range $[1,n]$ divisible by some integer k is equal to:
$$\lfloor{\frac{n}{k}}\rfloor$$
(Those weird brackets are the floor function, which is essentially rounding down to an integer. $\lfloor{1.9}\rfloor = 1$, $\lfloor{87.2}\rfloor=87$.
This is because there is multiple of k every k numbers, which is practically the definition of division. You take the floor of it to avoid counting zero.
So to do this problem you would get
$$\lfloor{\frac{500}{5}}\rfloor + \lfloor{\frac{500}{6}}\rfloor + \lfloor{\frac{500}{7}}\rfloor - $$
$$(\lfloor{\frac{500}{lcm(5,7)}}\rfloor + \lfloor{\frac{500}{lcm((5,6)}}\rfloor + \lfloor{\frac{500}{lcm(6,7)}}\rfloor) +$$
$$(\lfloor{\frac{500}{lcm(5,6,7)}}\rfloor)$$
Which evaluates to $217$


TITLE: Solve recurrence formula
QUESTION [3 upvotes]: Thanks! That helps a lot. I think the substituting is the way to go

REPLY [1 votes]: There is a linear recurrence formula hidden not far.
By the change of varibale $n=2^k$ and setting $x_k:=T(2^k)$, the problem amounts to solving
$$
x_k=2^kx_{k-1}^2\qquad x_0=\frac{1}{3}.
$$
Of course, $x_k>0$ for all $k$ by induction, so we consider the sequence $y_k=\log_2x_k$, which satisfies
$$
y_k=2y_{k-1}+k\qquad y_0=-\log_23.
$$
The general solution of the homogeneous part is $C2^k$. And by the method of undetermined coefficients, we look for a particular solution of the form $y_k=Ak+B$. Plugging this is the reccurrence formula yields $A=-1$ and $B=-2$.
So the general solution is:
$$
y_k=C2^k-k-2.
$$
Given $y_0=C-2=-\log_23$, we get
$$
y_k=(2-\log_23)2^k-k-2
$$
hence
$$
x_k=2^{(2-\log_23)2^k-k-2}\quad\Rightarrow\quad x_k=\frac{2^{2^{k+1}}}{2^{k+2}3^{2^k}}.
$$
Going back to $T(n)$, this gives
$$
T(n)=\frac{4^{n-1}}{n3^n}.
$$


TITLE: Inner Product on space of piecewise continuous functions
QUESTION [0 upvotes]: Let $PC_2[0,1]$ denote the vector space of piecewise continuous functions $f$ on $[0,1]$ such that $$\int_0^1|f(t)|^2dt < \infty.$$ Then I wish to show that $$\langle f, g\rangle =\int^1_0 f(t)\overline{g(t)} dt$$ defines an inner product on $PC_2[0, 1]$.
My question is how to show conjugate symmetry $\langle f,g\rangle=\langle g,f\rangle$? Is it just $$\int_0^1f(t)\overline{g(t)}dt=\int_0^1\overline{\overline{f(t)}g(t)}=\overline{\langle g,f\rangle}\;?$$
For positive definiteness, do I have to use epsilon delta definition of continuity to show it?

REPLY [1 votes]: Your proof that inner product is conjugate symmetric is correct. $ \langle f, f \rangle =\int_0^{1}|f(x)|^{2} \, d x \geq 0$ and equality holds iff $f =0$. [ If $f(x) \neq 0$ for some $x$ then there is an interval $(a,b) \subset (0,1)$ and a positive number $\delta$ such that $|f(y)| >\delta$ for all $y \in (a,b)$. But then $\int_0^{1}|f(x)|^{2} \, d x \geq \int_a^{b}|f(x)|^{2} \, d x \geq \delta^{2} (b-a)>0$, a contradiction].


TITLE: How to prove $x\ln^2(x)\leq(x-1)^2$ for every real number $x$
QUESTION [1 upvotes]: I really need help with this proof for my analysis class. I have to prove the following inequality :
$$(\forall x \in \mathbb R^*_+) \qquad \quad x\cdot \ln^2(x)\leq(x-1)^2$$
I tried, with a simple calculus approach, to compute $f'(x)=[(x-1)^2-x \cdot \ln^2(x)]'$, yet it is too long and I guess it is not the best way to prove it. ( I have to compute $f''(x)$ and $f'''(x)$). Any help ?

REPLY [0 votes]: If
$$
g(x) = \frac{(x - 1)^2}{x} \quad (x > 0),
$$
then
$$
g\left(\frac1x\right) = g(x) \quad (x > 0).
$$
But also
$$
\ln\left(\frac1x\right)^2 = \ln(x)^2 \quad (x > 0),
$$
therefore it is enough to prove
$$
\ln(x)^2 < \frac{(x - 1)^2}{x} \quad (x > 1).
$$
Equivalently:

$$
\ln(x) < \frac{x - 1}{\sqrt{x}} \quad (x > 1).
$$

This follows from the convexity of the logarithm function.
Proof. Write $x = t^2,$ where $t > 1.$ Then, because the graph of a convex function lies below any of its chords,
$$
\ln(x) = 2\ln(t) = 2\int_1^t\frac{du}u < 2(t - 1) \cdot \frac12\left(1 + \frac1t\right)
= \frac{t^2 - 1}t = \frac{x - 1}{\sqrt{x}}.
$$


TITLE: Interesting mathematical artifact: Equality sign wrong for exponential function.
QUESTION [3 upvotes]: I found an interesting case where it seems like an equality sign works wrong.
Let's consider the following construction:
$\frac{1+\Lambda}{2} e^{i\Lambda \phi}$ where $\Lambda = \pm1$, so $\Lambda^2=1$.
Then I apply Euler formula:
$\frac{1+\Lambda}{2} e^{i\Lambda \phi} = \frac{1+\Lambda}{2} (\cos \phi + i\Lambda \sin \phi)= \frac{1}{2} \cos \phi + \frac{\Lambda}{2} \cos \phi + \frac{i\Lambda}{2} \sin \phi + \frac{i}{2} \sin \phi = \frac{1+\Lambda}{2} e^{i\phi}$
where I have used $\sin(\Lambda \phi) = \Lambda \sin \phi$ and $\cos (\Lambda \phi) = \cos \phi$.
However, this is just wrong! $e^{i\Lambda \phi}\neq e^{i \phi}$ even though the equality sign was not broken anywhere in between (at least it doesn't seem to be broken to me). What am I doing wrong?

REPLY [0 votes]: You are correct that $e^{i\Lambda \phi}\neq e^{i \phi}$.
But your computation doesn't show that $e^{i\Lambda \phi}= e^{i \phi}$; it shows that $\frac{1+\Lambda}{2}e^{i\Lambda \phi}= \frac{1+\Lambda}{2}e^{i \phi}$, which is true for $\Lambda=\pm 1$.