In the $3D$ plane, how is $z$ always the "vertical" axis? In all the graphs I see of the 3d cartesian system, $x$ and $y$ are "horizontal", although they can change positions, but $z$ is always the vertical one.

Even if we had a choice to graph the $yz$ plane, the $z$ will always be the vertical axis, in almost every textbook and example. Why is this?

Is this convention?

I even see in a few youtube videos that $z$ is always the vertical axis, no matter what. How come this is the case?

Besides as you said it is convention, the direction of the $z$-axis can be determined by using the "Right Hand Rule". When we take unit vectors $\hat{i}$ and $\hat{j}$ from $x$-axis and $y$-axis respectively, we have $\hat{i} \times \hat{j} = \hat{k}$, where $\hat{k}$ is the unit vector on $z$-axis. Since the direction of cross-product will be determined by the Right Hand Rule, whether it is vertical or not is all about how the directions for $x$-axis and $y$-axis are determined (And since they generally are lying on a horizontal plane, your result follows).

It is the same reason why we call some cars as righthand drive, one could have equally well have called it rigthfoot drive (we use a our foot a lot to drive)


Prove that the equation $xyzt=2\, xy +2xz +2xt +2yz +2yt +2zt +4\,x+4\,y+4\,z+4\,t+4$ has finite numbers of integer solutions. Prove that the equation $$xyzt=2\, xy +2xz +2xt +2yz +2yt +2zt +4\,x+4\,y+4\,z+4\,t+4$$  has finite numbers of integer solutions.

My attempt. Suppose that $x \leq y \leq z \leq t$ and $t=\max(t,4)$ ( the case $t<4$ is trivial). Then
$$
xyzt\leq 2\, xt +2xt +2xt +2yt +2yt +2zt +4\,t+4\,t+4\,t+4\,t+t=6xt+4yt+2 zt+21t,
$$
and we obtain the inequality $xyz\leq 6x+4y+2z+21$.

Then by the same idea, assuming $z>21,$ I get that $$xyz\leq 33z \implies xy\leq 33,$$
so, there are only finite integer solutions for $x,y.$ Now, for any fixed $x,y$ there are only finite numbers integer solutions for $z$ of the inequality $xyz\leq 6x+4y+2z+21$. Then the same for $t.$

Is it ok? Is there any more elegant solution?

Equation above shown below:

$xyzt=2\, xy +2xz +2xt +2yz +2yt +2zt +4\,x+4\,y+4\,z+4\,t+4$

Above has numerical solution:

$(x,y,z,t)=(2,4, (-1),(-2))$


How is $(\mathbb{S}^2,\pi)$ a covering of $\mathbb{RP}^2$? My lecture notes state that a basic example of covering space is $(\mathbb{S}^n,\pi)$ where $\pi$ is the projection $\pi(p) = [p]$ to $\mathbb{RP}^n$ the projective space considered as the sphere with opposite points identified.

Next, I wonder what is the intuition behind this result. How does one visualize this covering? Let me explain myself: in the case of $\mathbb{R}$ and $\mathbb{S}^1$ it is easy to see that each loop of the fundamental group can be unrolled into a segment $[k,k+n]$. Is there a similar view in this case?

Response to first thing: No, it is the quotient map
$$\pi: S^n \to S^n/\sim (i.e., \mathbb{R}P^n)$$
$$x \mapsto [x].$$
Response to second thing: I don't see how your example is an "intuition for being a covering map". But the fact analogous to your example is that every loop in the fundamental group of $\mathbb{R}P^n$ can be corresponded to half a great circle connecting two antipodal points and its concatenations.

However, I'd rather say that "the intuition"  for being a covering map coincides with its proof: it is due to the fact that it is a quotient of a properly discontinuous action of a discrete group (namely, $\mathbb{Z}/2\mathbb{Z})$.

$\mathbb{R}P^2$ is the quotient of $S^2$ endowed with the equivalent relation $xRy$ if and only if $x=y$, or$ x=-y$. So it is this quotient map. It is a cover since the restriction of the canonical projection $\pi:S^n\rightarrow \mathbb{R}^n$ to an open hemisphere is a diffeomorphism onto its image. It like so you can visualize this covering, since $x$ and $-x$ are in two different open hemispheres, so $\pi$  is a double covering map.


Is there a predefined mathematical function for $ f(x) = \left\{\begin{aligned} &x &&: x < n\\ &x - n &&: x \ge n \end{aligned} \right.$ if n is a constant, is there a predefined/shorter mathematical function that can replace this function :
$$
f(x) = \left\{\begin{aligned}
&x &&: x < n\\
&x - n &&: x \ge n
\end{aligned}
\right.$$

where x >= 0 and x < 2n

You can build this function from the Heaviside step:

$$f(x):= x-nH(x-n).$$

Make sure to use the convention $H(0):=1.$

actually there IS a function that can do the same, which is the modulus

the above function is the same as

x mod n

although it isn't true for all values of x, it's what I have been looking for


Trying to show a set is a subgroup of $\mathbb{Q}$ I'm trying to show that for a fixed $n\in\mathbb{Z}^+$ the set of rational numbers whose denominators divide $n$ (call the set $H$) is a subgroup of $\mathbb{Q}$. If I let $a/b,c/d\in H$, then adding them gives $(ad+bc)/bd$. However, does $bd$ necessarily divide $n$? If I let $n=bk$ and $n=dl$ for appropriate $k,l\in\mathbb{Z}$, then I get $n^2=bdkl$. Not sure why this shows $bd$ divides $n$.

Don't use the formula
$$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$$
Instead, using the hypothesis, just express each fraction with a denominator of exactly $n$.

Explicitly, suppose $x \in H$. By definition of $H$, we can write $x = {\large{\frac{a}{b}}}$, where $a,b \in \mathbb{Z}$, and $b|n$.

Then $b|n$ implies $n=bc$, for some $c \in \mathbb{Z}$, hence
$$x = \frac{a}{b} = \frac{ac}{bc} = \frac{ac}{n}$$
Thus, each element of $H$ can be expressed with a denominator of $n$.

It follows that $H$ is the set of rational numbers which can be expressed with a denominator of $n$.

From there, verifying the group axioms is easy.

Consider the homomorphism $\phi: \mathbb Q \to \mathbb Q$ given by $q \mapsto nq$.

Then $H = \phi^{-1}(\mathbb Z)$ and so is a subgroup.


Differentiating the sum of increasing nonnegative functions Let $f_n:[0,1] \to \mathbb R_{+}$ be a sequence of nondecreasing functions such that $f_n(0) = 0$ and $\sum_{n=1}^{\infty} f_n(1) \leq \infty$. Show that the sum can be differentiated term by term almost everywhere, i.e. $f' = \sum^{\infty}_{n=1} f'$.

It appears that the set up of the problem is hinting Dominated Convergence Theorem. So I want to view the summation as the integration against the counting measure. In order for the theorem to apply, I need a bound on the derivative of $f_n$. Is it possible to derive a bound from the information given?

Since $f_n$ is non-decreasing and $f_n(0)=0$, we have $f_n \ge 0$. Moreover $\sum_{n=1}^\infty f_n(1) < \infty$ (the series should convergence?) implies that the function $g(x):= \sum_{n=1}^\infty f_n(x)$ converges on $[0,1]$ and is also non-decreasing. Any monotonic function is (by a Theorem of Lebesgue) almost everywhere differentiable.

Now, let $g_m(x):= \sum_{n=1}^m f_n(x)$. Since $f_n$ is non-decreasing, we have $f_n' \ge 0$ almost everywhere. This implies that $$g_m' \leq g_{m+1}' \leq g'$$
almost everywhere. Thus $g'_m(x) = \sum_{n=1}^m f_n'(x)$ converges almost everywhere. In order to get $g_m' \rightarrow g'$ it is enough to show this for a subsequence.

For this, choose $m_k$ such that $g(1) - g_{m_k}(1) \le 2^{-k}$. Because $H_k(x) := g(x) - g_{m_k}(x)$ is non-negative, non-decreasing with $H_k \le 2^{-k}$, we can apply the same argument as above for $H_k$. Thus $L:=\sum_{k=1}^\infty H_k$ converges almost everywhere and is differentiable with $\sum_{j=1}^k H_j' \le L'$ for any $k \ge 1$. Hence $\sum_{j=1}^k H_j'$ converges almost everywhere. In particular $H_j' \rightarrow 0$ almost everywhere.

This statement was proven by Fubini and sometimes stated as "Fubini's Theorem on termwise differentiabilty of series with monoic coefficients".

I have myself a different solution to the same problem.
Since each $F_n$ is increasing, $F$ is increasing (on $[0, 1]$). Furthermore, this means $F'$ exists on $(0, 1)$. Since we're only asked for a derivative formula to hold a.e. I'm going to discard the endpoints $0$ and $1$ now to avoid technicalities with one-sided derivatives/limits.
Now, fix $x \in (0, 1)$. By our given formula for $F$, we have the following formula for the symmetric difference quotient where $h > 0$ is sufficiently small so that $x + h, x - h \in (0, 1)$. Such $h$ exists since $(0, 1)$ is an open set. Then
$$\frac{F\left(x + \frac{h}{2}\right) - F\left(x - \frac{h}{2}\right)}{h} = \int_{\mathbb{N}}\frac{F_n\left(x + \frac{h}{2}\right) - F_n\left(x - \frac{h}{2}\right)}{h}d\mu(n),$$
where $d\mu(n)$ denotes the counting measure on $\mathbb{N}$. Now, by applying the Fundamental Theorem of Calculus on the right-hand side,
$$\frac{F\left(x + \frac{h}{2}\right) - F\left(x - \frac{h}{2}\right)}{h} = \int_{\mathbb{N}}\int^{x + \frac{h}{2}}_{x - \frac{h}{2}}\frac{F_n'(y)}{h}dyd\mu(n)$$
Since the functions $F_n$ are all increasing, the integrand on the right-hand side is nonnegative, and measurable with respect to the product measure (which I'll denote $d(y \times n)$). Then by the Fubini Theorem,
$$\frac{F\left(x + \frac{h}{2}\right) - F\left(x - \frac{h}{2}\right)}{h} = \int^{x + \frac{h}{2}}_{x - \frac{h}{2}}\frac{1}{h}\int_{\mathbb{N}}F_n'(y)dyd\mu(n)$$
The limit on the left-hand side as $h \rightarrow 0$ exists since $F$ is differentiable at $x$, so
$$F'(x) = \lim_{h \rightarrow 0}\int^{x + \frac{h}{2}}_{x - \frac{h}{2}}\frac{1}{h}\int_{\mathbb{N}}F_n'(y)dyd\mu(n)$$
The limit on the right-hand side actually exists as well. The inner integral is independent of $h$, so we only concern with the outer integral. The length of the interval is $h$, and we are dividing by $h$, so this is a mean-value integral in one dimension. Thus, as $h \rightarrow 0$, the integrand approaches the value at the center of the interval. Thus
$$F'(x) = \int_{\mathbb{N}}F_n'(x)d\mu(n),$$
completing the proof. $\square$


Defining a metric space I'm studying for actuarial exams, but I always pick up mathematics books because I like to challenge myself and try to learn new branches.  Recently I've bought Topology by D. Kahn and am finding it difficult.  Here is a problem that I think I'm am answering sufficiently but any help would be great if I am off.

If $d$ is a metric on a set $S$, show that
$$d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ is a metric on $S$.

The conditions for being a metric are $d(X,Y)\ge{0}, d(X,Y)=0$ iff $X=Y$, $d(X,Y)=d(Y,X)$, and $d(X,Y)\le{d(X,Z)+d(Z,Y)}$.  Thus, we simply go axiom by axiom.

1)  Since both $d(x,y)\ge{0}$ and $1+d(x,y)\ge{0},$ it is clear that $d_1(x,y)\ge{0}$.  (Is this a sufficient analysis?)
2) $d_1(x,x)=\frac{d(x,x)}{1+d(x,x)}=\frac{0}{1+0}=0$.
3) $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}=\frac{d(y,x)}{1+d(y,x)}=d_1(y,x).$
4) $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}\le{\frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}}=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}\lt\frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}=d_1(x,z)+d_1(z,y).$

However, #4 is strictly less, not less than or equal to, according to my analysis, so where did I go wrong?

There is someing wrong in 4), just as Brian comments. Here I offered a proof for you:

  Proof: Notice that $f(x)=\frac{x} {1+x}$ is increasing on $\mathbb R^+$: to see this, let $g(x)=\frac{1}{x+1}$. It is easily to see that $g(x)$ is descreasing on $\mathbb R^+$. And note that $f(x)+g(x)=1$. Therefore, $f(x)$ is a increasing function on $\mathbb R^+$.

Since $d(x,y) \le d(x,z)+d(z,y)$, we have $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}\le{\frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}}=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}\lt\frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}=d_1(x,z)+d_1(z,y).$

Hope this be helpful for you.

ADDed: $d(z,y)$ and $d(x,z)$ could be zero.

Yeah, you forgot that z=y. Remember, no where is it stated that each of these points are different. This will give the in equality that you are looking for.


Possible permutations of a grid I hope this is the correct place to post this, as I don’t study maths. But I do need help calculating the possible permutations of a grid based game I’m currently programming. This isn’t to help out with the game logic, but rather to help me understand how many different combinations of each puzzle I can get by randomising the placement of the tiles.

I think that to someone with a good understanding of maths, this should be a pretty basic problem to solve. I would give it a shot myself, but I didn’t study further maths and I really want a correct answer/algorithm.

Okay so this is the scenario:

I have a 5x5 grid of coloured tiles. There are 5 different colours of tiles. By default, the tiles are ordered into rows of 5. My question is, if every tile can go to any position in the 5x5 grid, how many possible permutations of the grid are there? (also taking into account that tiles of the same colour count as the same tile, so 'red tile A’ is exactly the same as ‘red tile B’ as far as the game mechanics are concerned)

It would be useful to know what you think the permutations figure is for this specific grid but also the calculation you used to arrive at that answer. The reason for this is that the size of the grid may change in the future and I’d like to be able to use the same algorithm to calculate its permutations. It's also likely that I will have other layouts which aren't arranged into rows, and may have different quantities of each block, so if anyone could point me in the right direction with this, it would be greatly appreciated!

Thanks for any advice!!

Below is an example screenshot of what I mean, you can ignore the numbers as they are just for debugging purposes.

If the tiles were distinguishable, we could permute them in $25! $ ways, since for the first tile there would be $25 $ places, for the second one $24 $ etc... But now we can still interchange tiles of a given color with themselves (and leave the permutation "the same"). For instance the red tiles can be permuted between eachother in $5!$ ways. Therefore the final answer should be (divide by $5!$ for each color):

$$\frac {25!}{5! 5! 5! 5! 5!}=\frac{(5^2)!}{(5!)^5}=623360743125120$$

Usind the same reasoning, for an $n \times n$ grid we get:

$$\frac {(n^2)!}{(n!)^n}$$

Now note that the total number of permutations has nothing to do with the grid being a square, so we can generalise even further, to a grid with $k$ tiles and number of tiles per colour $n_1,\cdots,n_m$ (with $n_1+\cdots+n_m=k$). Then we get:

$$\frac{k!}{n_1!*\cdots *n_m!}$$
different permutations.

Let's start with a blank grid of 25 possible places. At this point there are twenty five possible permutations of any given block. We'll say that this starting point is block 1, after this we have to fill up twenty four more so multiply twenty five by twenty four and get six hundred. This means that there is a one in six hundred chance that a randomly generated sequence would have put this as the space filled. Then multiply six hundred by twenty three because there are twenty three possible places left. Hopefully you understand the process  so just repeat this until you finish filling up the grid and take the chance that you got that sequence and that is how many possible permutations there are of that grid with one by one blocks.
I see that uncountable has already answered but I thought it might be useful.


How could we define the factorial of a matrix? Suppose I have a square matrix $\mathsf{A}$ with $\det \mathsf{A}\neq 0$.

How could we define the following operation? $$\mathsf{A}!$$

Maybe we could make some simple example, admitted it makes any sense, with

$$\mathsf{A} =
\left(\begin{matrix}
1 & 3 \\
2 & 1
\end{matrix}
\right)
$$

For any holomorphic function $G$, we can define a corresponding matrix function $\tilde{G}$ via (a formal version of) the Cauchy Integral Formula: We set
$$\tilde{G}(B) := \frac{1}{2 \pi i} \oint_C G(z) (z I - B)^{-1} \, dz ,$$
where $C$ is an (arbitrary) anticlockwise curve that encloses (once each) the eigenvalues of the (square) matrix $B$. Note that the condition on $C$ means that restrictions on the domain of $G$ determine restrictions on the domain of $\tilde{G}$.
So, we could make sense of the factorial of a matrix if we had a holomorphic function that restricted to the factorial function $n \mapsto n!$ on nonnegative integers. Fortunately, there is such a function: The function $$F: z \mapsto \Gamma(z + 1),$$ where $\Gamma$ denotes the Gamma function, satisfies $F(n) = n!$ for nonnegative integers $n$. (There is a sense in which $F$ is the best possible function extending the factorial function, but notice the target of that link really just discusses the real Gamma function, which our $\Gamma$ preferentially extends.) Thus, we may define factorial of a (square) matrix $B$ by substituting the second display equation above into the first:
$$\color{#df0000}{\boxed{B! := \tilde{F}(B) = \frac{1}{2 \pi i} \oint_C \Gamma(z + 1) (z I - B)^{-1} \, dz}} .$$
The (scalar) Cauchy Integral Formula shows that this formulation has the obviously desirable property that for scalar matrices it recovers the usual factorial, or more precisely, that $\pmatrix{n}! = \pmatrix{n!}$ (for nonnegative integers $n$).
Alternatively, one could define a matrix function $\tilde G$ (and in particular define $B!$) by evaluating formally the power series $\sum_{i = 0}^{\infty} a_k (z - z_0)^k$ for $G$ about some point $z_0$, that is, declaring $\tilde G(B) := \sum_{i = 0}^{\infty} a_k (B - z_0 I)^k$, but in general this definition is more restrictive than the Cauchy Integral Formula definition, simply because the power series need not converge everywhere (where it does converge, it converges to the value given by the integral formula). Indeed, we cannot use a power series for $F$ to evaluate $A!$ directly for our particular $A$: The function $F$ has a pole on the line segment in $\Bbb C$ with endpoints the eigenvalues of $A$, so there is no open disk in the domain of $F$ containing all of the eigenvalues of $A$, and hence there is no basepoint $z_0$ for which the series for $\tilde F$ converges at $A$.
We can define $\tilde G$ in yet another way, which coincides appropriately with the above definitions but which is more amenable to explicit computation: If $B$ is diagonalizable, so that we can decompose $$B = P \pmatrix{\lambda_1 & & \\ & \ddots & \\ & & \lambda_n} P^{-1} ,$$
for eigenvalues $\lambda_a$ of $B$ and some matrix $P$, we define
$$\tilde{G}(B) := P \pmatrix{G(\lambda_1) & & \\ & \ddots & \\ & & G(\lambda_n)} P^{-1} .$$ Indeed, by substituting and rearranging, we can see that this coincides, at least formally, with the power series characterization. There is a similar but more complicated formula for nondiagonalizable $B$ that I won't write out here but which is given in the Wikipedia article Matrix function.
Example The given matrix $A$ has distinct eigenvalues $\lambda_{\pm} = 1 \pm \sqrt{6}$, and so can be diagonalized as $$P \pmatrix{1 - \sqrt{6} & 0 \\ 0 & 1 + \sqrt{6}} P^{-1} ;$$ indeed, we can take $$P = \pmatrix{\tfrac{1}{2} & \tfrac{1}{2} \\ -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}}}.$$
Now, $F(\lambda_{\pm}) = \Gamma(\lambda_{\pm} + 1) = \Gamma (2 {\pm} \sqrt{6}),$
and putting this all together gives that
\begin{align*}\pmatrix{1 & 3 \\ 2 & 1} ! = \bar{F}(A) &= P \pmatrix{F(\lambda_-) & 0 \\ 0 & F(\lambda_+)} P^{-1} \\ &=  \pmatrix{\tfrac{1}{2} & \tfrac{1}{2} \\ -\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}}} \pmatrix{\Gamma (2 - \sqrt{6}) & 0 \\ 0 & \Gamma (2 + \sqrt{6})} \pmatrix{1 & -\frac{\sqrt{3}}{\sqrt{2}} \\ 1 & \frac{\sqrt{3}}{\sqrt{2}}} .\end{align*}
Multiplying this out gives
$$\color{#df0000}{\boxed{\pmatrix{1 & 3 \\ 2 & 1} ! = \pmatrix{\frac{1}{2} \alpha_+ & \frac{\sqrt{3}}{2 \sqrt{2}} \alpha_- \\ \frac{1}{\sqrt{6}} \alpha_- & \frac{1}{2} \alpha_+}}} ,$$
where $$\color{#df0000}{\alpha_{\pm} = \Gamma(2 + \sqrt{6}) \pm \Gamma(2 - \sqrt{6})}. $$
It's perhaps not very illuminating, but $A!$ has numerical value
$$
\pmatrix{1 & 3 \\ 2 & 1}! \approx \pmatrix{3.62744 & 8.84231 \\ 5.89488 & 3.62744} .
$$
To carry out these computations, one can use Maple's built-in MatrixFunction routine (it requires the LinearAlgebra package) to write a function that computes the factorial of any matrix:
MatrixFactorial := X -> LinearAlgebra:-MatrixFunction(X, GAMMA(z + 1), z);

To evaluate, for example, $A!$, we then need only run the following:
A := Matrix([[1, 3], [2, 1]]);
MatrixFactorial(A);

(NB executing this code returns an expression for $A!$ different from the one above: Their values can be seen to coincide using the the reflection formula $-z \Gamma(z) \Gamma(-z) = \frac{\pi}{\sin \pi z} .$ We can further simplify the expression using the identity $\Gamma(z + 1) = z \Gamma(z)$ extending the factorial identity $(n + 1)! = (n + 1) \cdot n!$ to write $\Gamma(2 \pm \sqrt{6}) = (6 \pm \sqrt{6}) \Gamma(\pm \sqrt{6})$ and so write the entries as expressions algebraic in $\pi$, $\sin(\pi \sqrt{6})$, and $\Gamma(\sqrt{6})$ alone. One can compel Maple to carry out these substitutions by executing simplify(map(expand, %)); immediately after executing the previous code.) To compute the numerical value, we need only execute evalf(%); immediately after the previous code.
By the way, we need not have that $\det B \neq 0$ in order to define $B!$. In fact, proceeding as above we find that the factorial of the (already diagonal) zero matrix is the identity matrix: $$0! = \pmatrix{\Gamma(1) & 0 \\ 0 & \Gamma(1)} = I .$$ Likewise using the formula for nondiagonalizable matrices referenced above together with a special identity gives that the factorial of the $2 \times 2$ Jordan block of eigenvalue $0$ is, somewhat amusingly,
$$\pmatrix{0 & 1\\0 & 0} ! = \pmatrix{1 & -\gamma \\ 0 & 1} ,$$ where $\gamma$ is the Euler-Mascheroni constant.

Travis' answer is very nice.

It would be good to mention that (almost) any matrix function can be made into a power-series expansion, which eventually involves the values of the function on the eigenvalues of the matrix multiplied by the eigenvectors.

In other words the matrix function is completely characterised by the values it takes on the eigenvalues of the matrix (even if a power-series expansion may be needed).

The above hold for matrices which are diagonalisable (i.e. the number of linearly independent eigenvectors is equal to the matrix dimension). There are ways to expand an arbitrary matrix into what is referred to as generalised eigenvectors, but this will not be pursued further here.

Furthermore, since any square, finite-dimensional, matrix satisfies its characteristic polynomial equation (if seen as a matrix function), aka Cayley-Hamilton theorem, the powers of $A^k$ for $k \ge n$ ($n$ is the dimension) can be expressed as a function of the powers of $A$ up to $n$. So eventually the matrix function power-series expansion collapses to polynomial expansion (for square matrices). Finally, this polynomial expansion, for a given function, can be found more easily by methods such as variation of parameters or polynomial modeling.


If $A$ is a subobject of $B$, and $B$ a subobject of $A$, are they isomorphic? In category theory, a subobject of $X$ is defined as an object $Y$ with a monomorphism, from $Y$ to $X$. If $A$ is a subobject of $B$, and $B$ a subobject of $A$, are they isomorphic? It is not true in general that having monomorphisms going both ways between two objects is sufficient for isomorphy, so it would seem the answer is no.

I ask because I'm working through the exercises in Geroch's Mathematical Physics, and one of them asks you to prove that the relation "is a subobject of" is reflexive, transitive and antisymmetric. But it can't be antisymmetric if I'm right...

I don't think this can be true in general. What if we just take the category consisting of two objects $A$, $B$ and take morphisms $f:A\to B$, $g:B\to A$ with no relations between the morphisms, but forcing associativity. Then certainly $f$ and $g$ are monomorphisms but $A$ and $B$ are not isomorphic (since there are no relations between the morphisms).

In Top (topological speces) take the subspaces of the real line:  $B=[0,1]$ and $A=[0,1]\cup \{4/3\}$ considering the inclusion $B \subset A$ and imbedding map $f: A\to B: x \mapsto x/2$. But $A$ isn't isomorphic  (i.e. homeomorphic) to $B$


Probability generating function for negative values of random variables? What if we have negative integral values for a random variable?Then is it possible to write a probability generating function for it? All definitions I have seen so far is for non negative integer values. I hope someone could assist me. Thanks

Using probability generating functions $s\mapsto E(s^X)$ for $s$ in $[0,1]$, for negative valued random variables $X$ would be taking the risk of manipulating divergent series. This is the reason why random variables with negative values (integer or not) are best dealt with using characteristic functions $t\mapsto E(\mathrm e^{\mathrm i tX})$ for every real number $t$, always well defined and allowing every manipulation that probability generating functions would.

Yes, you could define a possion distribution on negative integers as follows. Let support of $X$ be $\{0,-1,-2,...\}$ and let pdf of $X$ be
$$f(x)=\frac{e^{-\lambda}\lambda^{-x}}{(-x)!}$$
compared to $Y$ with support $\{0,1,2,...\}$ with possion has pdf
$$f(y)=\frac{e^{-\lambda}\lambda^{y}}{(y)!}$$


Balls & Bins has N Numbers with Capacity of 1000 each I am new to this Community. Actually I am myself a Software Engineer and our company has stuck to a problem and has not come up with a solution for many days. So, I just remembered an old school Math Chapter of Probability and a King to pass 8 doors etc. I thought Mathematicians can solve this problem.

So, here is the problem, we have an increasing numbers of Email Addresses and Buckets. And each Bucket can hold 1000 Emails

Just for a side note, if it is hard to work on alphabats, we can convert each email to numbers first like a=1, b=2, c=3 ... z=26 with 0 (Zero) appended for making it unique and +s for @ and . characters. For Example

 abcd@gmail.com  ->  10203040+901301090+3015013

So, now we have numbers which make it easier to apply any calculations. Once we get the solution, we will make a Hashing Library to solve problems in Software Industry

Each email can be given a unique primary key. This can be given relative to their positions when you sort them alphabetically, or convert strings to numbers or whatever way.

Now we essentially need to find which bucket holds that key.

If each bucket can hold $1000$ keys,
then key $1-1000$ are in first bucket, keys $1001-2000$ are in second bucket and so on.

Then given a key $K$ it lies in the $\lceil \frac{K}{1000}\rceil$ bucket.

Similarly if each bucket can hold $n$ emails, then you can replace $1000$ with $n$.


Graph Theory question? Assuming that friendship is always mutual, prove that in any group of n  2 persons, there are at least 2 persons with the same number of friends in the group.

How do I answer this question with a graph ?

You can view this as a question about simple graphs, but it will boil down to the pigeon hole principle. Suppose everyone has a different number of friends, so listing those numbers in order yields $\{ 1,...,n-1\}$ or $\{ 0,1,...,n-2\}$ (depending on what kind of assumptions you want to make). Regardless, you have $n-1$ possibilities with $n$ people.

This is already on the internet here:

http://puzzlesland.com/mathprob/graph-samedeg.html


Indicator variable of the sign of a difference in a math program I am interested in a mathematical program with objective:

$\max \sum_{i \in I} x_i $

where $x_i$ is a binary defined variable as follows:

$x_i = 1$ if $y_i - A \geq 0$

$x_i = 0$ if $y_i - A < 0$

where $y_i$ is a real or integer variable, and $A$ is a constant. I am not able to express such a constraint.

My first thought was a constraint like:

$x_i \leq y_i - A +1$

but it works only if $y_i \geq A$. If the difference is negative it does not work.

I hope this can be expressed as a linear program. Thanks

You're on the right track, just have to keep going.

$x_i = 1$ if $y_i - A \leq 0$

$x_i = 0$ if $y_i - A > 0$

Which is the same as

$x_i = 1$ if $y_i \leq A$

$x_i = 0$ if $y_i > A$

Now, I'll follow up what you were doing. We can try to find another solution later, but I want to show a way of using what you've discovered to get to find the answer.

You've find this constraint, which you say it works only when $y_i \geq A$

That's fine, but remember we need it to work for $y_i > A$

$x_i \leq y_i - A - 1$

If $y_i > A$, I'll say $y_i = A + surplus$ then

$x_i \leq A + surplus - A -1$

$x_i \leq surplus - 1$

This doesn't help us in that case, because this inequation would only work if the surplus is between 1 and 2. Remembering that we want to force a 0 on $x_i$ on this case.

Then again if we could just at least force the 0 on this case, we would be closer to achieving what we want. How can you force the 0 ? There are probably many ways. I'll write one that I've just thinked, it's probably not the best, but it'll work, we can think later how to improve how it looks.

I'll start easy, with something that looks like what you were doing.
We will focus on setting the 0 when $y_i \leq A$

Then this is my first attempt:
$$x_i \leq y_i - A$$

If $Yi \leq A$ then it should do nothing. We check that, and it's not true when the difference is less than 1. Then we correct that.

$$x_i \leq y_i - A + 1$$

Now we have something which does nothing when $yi > A$, but in the case we want to force the 0, we can reach something like $xi \leq -1$ which is not nice.

So i' say, that's fine, that equation is valid, but if that one is telling you to set that variable to something negative, let it be a 0 because if not, this won't have a solution.

Let M be a big enough number, in this case, M could be 2. Let U be a bivalent variable-

$$x_i \leq y_i - A+ 1 + M*U$$
$$x_i \leq 0 +  (1-U)* M$$

Now since $x_i$ can't be less than 0, on a negative number on the first equation, the second is forced to be active, and the first is forced to be inactive.

In other cases you have an interval where it's the same which one is active. And another one where both can be. When both can be, the important thing is that the less restrictive constraint can be picked. We will use that to force the 1 on the other case.

Setting a 1 can always be a little bit easier than a 0. We can do that with only one constraint. As always, there are other ways, this is one which is really easy, but has some drawbacks. In theory every M big enough can work. Because of the limitations of a computer, you have to avoid having a variable too close to 0 because the computer could just say it's 0. In a real work, you have to pay attention with what M to pick. If it's too big the computer gives you trouble, if it's too little the math won't work.

$x_i > (A - y_i)/M_2$

When $y_i > A$ it sets a 1 on $x_i$ and in the other case it does nothing.

Now the only thing left is to "fuse" those equations. In this case, it's only putting all of them together.

$$x_i \leq y_i - A+ 1 + M*U$$
$$x_i \leq 0 +  (1-U)* M$$
$$x_i \geq (y_i - A)/M_2$$

Now, when we have all the equations, it's good to check what happens on the limit between both sets. In this case, when $y_i = A$.

We can see that when $y_i = A$ the first set does nothing, and the other one does nothing. So we have a little trouble. Usually to fix this you can use a little constant $\epsilon$

$$x_i \leq y_i - A+ 1 + M*U$$
$$x_i \leq 0 +  (1-U)* M$$
$$x_i \geq (y_i + \epsilon - A)/M_2$$

That'd be all. If you need another approach, there are some general solutions to this kind of problems on some book. The second equation I used is not usually the most common approach, feel free to use another one you know. Or ask again if you need other one.

Something extra: To simplify this kind of problems, you can also take a look on what you're optimizing. Let's say that in your problem it's always better to have a 0 than a 1. Then you can just forget about setting the 0, and just use an equation to set the 1.


Image of the unit square under linear transformation Let $\mathbf{B} = \begin{pmatrix}1&-1 \\ 3 & -2\end{pmatrix}.$ The
  image of the unit square in $\mathbb{R}^2$ under the linear mapping
  $T_{\mathbf{B}}: \mathbb{R}^2 \to \mathbb{R}^2$ defined by
  $T_\mathbf{B}{\mathbf{x}} = \mathbf{Bx}$ is

  (A) a square; (B) a non-square rectangle; (C) a non-rectangular
  parallelogram; (D) a line segment; (E) the point $(1, −2)$; (F) none of
  these.

I think it takes $(0,0) \mapsto (0,0)$, $(1,0) \mapsto (1,3)$, $(0,1) \mapsto (-1,-2)$, and $(1,1) \mapsto (0,1)$. Is this right? I can't figure out what shape it should be.

Here is an image of the unit square image:

If it’s a rectangle adjacent sides will be perpendicular. Take the dot product of adjacent sides. If $0$ it’s a rectangle or a square. If not $0$ it’s a parallelogram. Btw if two adjacent sides point in the same direction it’s a line. If the determinant is zero it’s a point. No matrix is a translation so if determinant is zero, it’s the point $(0,0)$.


Can you determine the average second derivative from a set of points? Let us say we have a smooth function $f$. We can find the exact average of $f'$ on the interval $[a,b]$ via

$$\bar{f'}=\frac{f(b)-f(a)}{b-a}$$

My question is, can you find the exact average of the second derivative with only a finite number of points? If not, can we determine a range of values the average second derivative must be in? What methods are there to approximate the average of the second derivative from a (finite) set of points?

If you can generalize your answer to nth derivatives, that would be even better.

You can't compute the exact average of the second derivative with just the function's values at the points.  You need the value of the derivatives at those points too.  Then the value is given by
$$\bar{f}''=\frac{f'(b)-f'(a)}{b-a}$$
The functions's value at a f finite set of points give you an average of the derivative, but no estimate of the value of the derivative at those points, so there is not enough information here.


basis and dimension of the splitting field of $x^4+5x^2+6$ Please help me finding the basis and dimension of the splitting field of the polynomial $x^4+5x^2+6$ in the rational field.

Thanks

Hint: You can factor this as $(x^2+3)(x^2+2)$ so the splitting field is $\mathbb{Q}(\sqrt{-2},\sqrt{-3})$. You should then apply the tower rule to get that $[\mathbb{Q}(\sqrt{-2},\sqrt{-3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{-2},\sqrt{-3}][\mathbb{Q}(\sqrt{-2}):\mathbb{Q}]$. To find each of the degrees in the product try and find the degree of the minimal polynomial. The hint for this is that a quadratic annihilates things and so you'r each degree 1 or 2 at each step--then recall that you're degree one only if you were in the field all along!

Here, x=±√2 i,±√3 i. So splitting field is Q(√2 i,√3 i) whose elements are of the form Q((√2 i) )(√3 i)=Q_1 (√3 i), in particular, {q_0+q_1 (√2 i);q_i ϵQ_1} that is {(a_0+a_1 √2 i)+(a_2+a_3 √2 i)√3 i;a_i ϵQ}. Therefore, basis is {1,√2 i,√3 i,-√6} and dimension is 4


Probability of collecting 3 balls of a kind There are a lot of balls, each color has more than or equal to 3 balls.
Now pick up the ball randomly one by one without replacement.
What is the probability that the first color collected 3 of a kind is red.
For example 6 red balls, 7 blue balls, 8 yellow balls and 9 green balls.
I can solve this question by Markov chain, but it turns out to be too complicated, so I would like to seek for alternative method, thank you.

Let us consider the general problem: there are $N_1,N_2,\dots,N_k$ balls of $k$ colors. What is the probability of the event that the first color witn $n $ collected balls is the color "1"?
Obviously the above event is realized if the sequence of balls drawn before the $n$-th ball of color "1" contains besides $n-1$ balls of this color at most $n-1$ balls of any other color. We shall refer to such sequences as "good".
With this observation the probability in question reads:
$$\small
\frac{\sum\limits_{0\le n_2,n_3,\dots n_k<n}\color{red}{\binom{N_1}{n-1}\binom{N_2}{n_2}\cdots\binom{N_{k}}{n_{k}}(n-1+n_2+\cdots+n_k)!}\color{magenta}{(N_1-n+1)}\color{blue}{(N_1-n+N_2-n_2+\cdots+N_k-n_k)!}}{(N_1+N_2+\cdots+N_k)!}\\
=\frac{n}{N_1+N_2+\cdots+N_k}\sum\limits_{0\le n_2,n_3,\dots n_k<n}\frac{\binom{N_1}{n}\binom{N_2}{n_2}\cdots\binom{N_k}{n_k}}{\binom{N_1+N_2+\cdots+N_k-1}{n+n_2+\cdots+n_k-1}},
$$
where the $\color{red}{\text{red}}$ term counts the number of good sequences, the $\color{magenta}{\text{magenta}}$ term counts the number of ways to choose the $n$-th ball of the 1st color, and $\color{blue}{\text{blue}}$ term counts the number of ways to arrange the rest balls. The second line is a trivial simplification of the first one. I did not check if the expression can be furter simplified.

Here's another way to do it, but it's still a lot of work.  I'd want to write a computer program.  Consider the situation just before the third red ball is drawn.  We've previously drawn $2$ red balls, and from $0$ to $2$ of each of the other $3$ colors, so there are $27$ possibilities.  For each of these possibilities, compute the probability that the situation arises and multiply by the probability that the next draw is red.  Add up the results for all $27$ cases.
One example will sufficiently indicate the computations.  Suppose we have chosen $2$ red balls, $2$ green balls, $1$ yellow ball and one blue ball.  There are $\frac{6!}{2!2!}=180$ permutations of the string "RRGGYB".  For each of these, we have $6\cdot5$ ways to choose the red balls, $9\cdot8$ ways to choose the green balls, $8$ ways to choose the yellow ball, and $7$ ways to choose the blue ball, giving $6\cdot5\cdot9\cdot8\cdot8\cdot7$ ways  in all.  We have chosen $6$ balls from $30$ so there were $30\cdot29\cdot28\cdot27\cdot26\cdot25$ possible choices.  Now there are $24$ balls left, and $4$ of them are red, so the probability that the next draw is red is $\frac4{24}.$  The probability for this case is $$\frac{180\cdot6\cdot5\cdot9\cdot8\cdot8\cdot7}{30\cdot29\cdot28\cdot27\cdot26\cdot25}\cdot\frac4{24}$$
I wrote a python script to do the indicated calculations for each color, and it seems to have worked correctly, in that the computed probabilities sum to $1$.
Here's the script:
from itertools import product
from math import factorial

def probabilities(balls, goal):
    '''
    balls is the list a the number of balls of each color
    goal is the number of balls of one color to be drawn
    returns a list of the probabilities that each color wins
    '''
    if min(balls)<goal:
        raise ValueError('Each color must have at least goal balls')
    answer = []
    for idx, b in enumerate(balls):
        answer.append(winner([b]+balls[:idx]+balls[idx+1:],  goal))
    return answer

def winner(balls, goal):
    '''
    Compute probability that the winner is the first color
    in the list.
    '''
    n = sum(balls)
    others = balls[1:]
    answer = 0
    for p in product(range(goal), repeat = len(others)):
        total = ((goal-1),) + p
        num = factorial(sum(total))
        for idx, t in enumerate(total):
            num //= factorial(t)
            b = balls[idx]
            num *= factorial(b)/factorial(b-t)
        left = n-sum(total)
        den = factorial(n)//factorial(left)
        draw = (balls[0]-goal+1)
        answer += (num/den)*(draw/left)
    return answer

probs= probabilities([6,7,8,9], 3)
print(probs)
print(sum(probs))

and the output:
[0.13806360223151828, 0.2058007126972644,
0.28417096812898923, 0.37196471694222827]
1.0000000000000002


Cholesky factors of covariance and precision matrix Let $\Sigma$ be a covariance matrix (symmetric positive-definite), and $\Omega = \Sigma^{-1}$ the corresponding precision matrix, which is also SPD (the quotients of positive eigenvalues are positive).
Let $T$ and $U$ be the upper-triangular Cholesky factors respectively of $\Sigma$ and $\Omega$, such that:
$$
\Sigma = T^t T \qquad\text{and}\qquad \Omega = U^t U
$$
writing down the inverse of $\Sigma$ explicitly, we also have:
$$
\Sigma^{-1} = T^{-1} T^{-t} = \Omega = U^t U
$$
and similarly the other way around, writing down the inverse of $\Omega$.
In this equality, we have a product of upper-times-lower triangular matrices on the left $T^{-1} T^{-t}$, and a product of lower-times-upper on the right $U^t U$. I get confused with such equations, and am not sure what to do next; considering products like $\Sigma\Omega$ do not lead anywhere (that I can see). Can we say anything about how $T$ and $U$ are related?

$\Sigma^{-1} = T^{-1} (T^{*})^{-1} = R^*R\implies I= TR^*RT^*=\big(RT^*\big)^*\big(RT^*\big)\implies Q^* = RT^*$
for some unitary $Q$
Multiply each side on the left by $Q$ and on the right by $(T^*)^{-1}$ to get
$(T^*)^{-1} = QR$
That is, they are related by the unique QR factorization of $(T^*)^{-1}$ such that $R$ has positive diagonals (uniqueness follows since the Cholesky factorization is unique for PD matrices).

(Could not edit the answer of @user8675309, so here is a cleaner derivation.)
The main equality relating $T$ and $U$ is:
$$ T^{-1} T^{-t} = U^t U $$
Multiplying by $T$ on the left, and $T^t$ on the right, the left-hand side becomes the identity:
$$ I = T U^t\ U T^t = (U T^t)^t\ UT^t $$
so the matrix $Q=UT^t$ is orthogonal. We can then write:
$$ T^{-t} = Q^t\ Q\ T^{-t} = Q^t\ UT^t\ T^{-t} = Q^t\ U $$
where we "recognize" the QR decomposition of $T^{-t}$. Uniqueness follows since the Cholesky factorization is unique for PD matrices.


Another flavour of Hilbert's Hotel Many of you have probably heard about Hilbert's Hotel problem. Mr Hilbert owns a hotel with countably infinite amount of one-bed rooms. All the rooms are, of course, taken.

A (finite or infinite) group of k people walks in and wishes for accommodation. However, here comes the tricky part. The current guests are quite tired and Mr Hilbert does not wish to make them move from one room to another. How does he achieve accommodating the new batch of people without moving the already accommodated ones?

This problem, although it's quite known and documented is nowhere to be found here, or anywhere else on the Internet (which is kind of strange).

My understanding of the Hilbert Hotel is that the hotel has $n$ rooms. Since the number of rooms is infinite a new guest will get the room $n_{+1}$. So if $k$ people arrive the will stay in the rooms $n_{+1},...,n_{+k}$, and there is no need for the other guests to move.

The usual specification is that the hotel is already full when the new guests arrive.  If that is true, you cannot accommodate them without moving the guests that are already there.  If we just have to have an infinite number of guests in the hotel and accommodate a number more without moving the existing ones, just put the existing guests in the odd numbered rooms and put the new guests in the even numbered rooms.  The question is not clearly stated.


Find a simplified form for $n!+(n-1)!+(n-2)!+(n-3)!+ \dots +1!$ . Find a simplified form for $n!+(n−1)!+(n−2)!+(n−3)!+\dots+1!$ .

By simplified, I mean that there should not be "..." in the equation.

If there isn't one, prove it. Thanks!

Edit 1: Is it possible not to have $\sum$ in it?

This number is just
$$1111\cdots1111$$
with $n+1$ digits in factorial number system.

∑(2-k)!

This is a simplified form of n!+(n−1)!+(n−2)!+(n−3)!+...+1!


How many ways are there to arrive at point $E$ in $10$ minutes or less? Otto is traveling along octagon $ABCDEFGH$. He starts at $A$ and every minute, he will move to one of the adjacent vertices. He will stop once he reaches $E$. How many ways are there to arrive at point $E$ in $10$ minutes or less?

This question seems hard as time is the constraint here. How do I account for the time constraint here? my first approach would be to do case work with the amount of minutes: in $4$ minutes there are two possibilities, it is impossible for $5$ minutes, and then the casework gets more tricky.

You could draw something similar to Pascal's triangle where each cell is the sum of the two cells diagonally above it (except in columns D and F since you cannot come from E).

It might look something like this, and you then have to add up the numbers in the two E columns.

E   F   G   H   A   B   C   D   E
0   0   0   0   1   0   0   0   0
0   0   0   1   0   1   0   0   0
0   0   1   0   2   0   1   0   0
0   1   0   3   0   3   0   1   0
1   0   4   0   6   0   4   0   1
0   4   0   10  0   10  0   4   0
4   0   14  0   20  0   14  0   4
0   14  0   34  0   34  0   14  0
14  0   48  0   68  0   48  0   14
0   48  0   116 0   116 0   48  0
48  0   164 0   232 0   164 0   48

Apparently I calculated this in 2001

To go forward 4 places he must have 7 steps forward and 3 backward. He must make 10 steps so the number of ways is ${10 \choose 7} = 120$. He could also get there by 7 steps backward and 3 forward so double this to get $240$.


Series of square of factorial I recently found this series and the result:
$$
  \sum_{k=1}^{\infty}{\frac{((k-1)!)^2}{(2k)!}} = \frac{\pi^2}{18}
$$
However I'm not able to justify why. Is there a special function (zeta, beta) that I don't know and that may help?
Thank you

for an even $k$, $(2k)!=2^kk!(2k-1)!!$ which may help, also notice that:
$$\sum_{k=1}^\infty\frac{((k-1)!)^2}{(2k)!}=\sum_{k=1}^\infty\frac{(k!)^2}{k^2(2k)!}$$
also since:
$$k!=\prod_{n=1}^kn,(2k)!=\prod_{n=1}^{2k}n\Rightarrow\frac{k!}{(2k)!}=\prod_{n=k+1}^{2k}\frac 1n=\prod_{n=1}^k\frac{1}{n+k}$$
and so:
$$\frac{(k!)^2}{(2k)!}=\prod_{n=1}^k\frac{n}{n+k}$$
so we can rewrite our expression as:
$$S=\sum_{k=1}^\infty\prod_{n=1}^k\frac{n}{k^2(n+k)}$$
now we should try and do something with this product:
$$\frac{n}{n+k}=1-\frac{k}{n+k}$$

$$\ln\left(\frac{n}{n+k}\right)=\sum_{j=1}^\infty\frac{(-1)^j n^j}{j(n+k)^j}$$
although I'm not sure where to go from here


Weird ordinary differential equation I am searching for the functions $f,g:[0,\infty) \to [0,\infty)$ which are both increasing, $f'$ is strictly increasing, $g$ is the inverse of $f$ and

$$ g(x)^2=f(x)^2\cdot g'(x)$$

My approach was to get rid of $f$ by plugging $x=g(x)$ and obtain something which contains only $g$:

$$ (g\circ g)^2(x) = x^2 \cdot g' \circ g(x)$$

but this seems to lead nowhere. Do you have some ideas?

I can't even find an example, since $f(x)=x$ doesn't work because $f'$ should be strictly increasing...

$\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$$
\totald{}{x}\bracks{1 \over {\rm g}\pars{x}} = -\,{1 \over \fermi^{2}\pars{x}}
\quad\imp\quad
{1 \over {\rm g}\pars{x}} = -\int{\dd x \over \fermi^{2}\pars{x}} + \mbox{a constant}.
$$


Elementary number theory proof without contradiction Question: suppose $a, b, c$ are integers and $\gcd(a,b) = 1$. Prove that if $c|(a^2+b^2)$ then $\gcd(a,c) = gcd(b,c) = 1$.

Prove by contradiction (naive approach) : assume $p \in PRIMES$ and $\ne1$ such that: $p|a$ and $p|c$. Since $c|(a^2+b^2)$ then $p|(a^2+b^2) \Rightarrow p|b^2$ (because $p|a$ which means $p|a^2$). Thus, as $p|b^2$ then $p|b$. It contradicts with the fact that: $\gcd(a, b) = 1$

My question is that is there another way to prove it? (preferably without contradiction).

Any help would be appreciated.

Suppose $a^2+b^2=kc$ and, as implied by $\text{gcd}(a,b)=1$, we have $ma+nb=1$ for some integers $m$ and $n$. Then
$$
1=m^2a^2+n^2b^2+2mnab.
$$
Consider the RHS above:
\begin{aligned}
1=\text{RHS}&=m^2(a^2+b^2)+(n^2-m^2)b^2+2mnab=c(km^2)+b[(n^2-m^2)b+2mna];\\
1=\text{RHS}&=n^2(a^2+b^2)+(m^2-n^2)a^2+2mnab=c(kn^2)+a[(m^2-n^2)a+2mnb].
\end{aligned}
Then, respectively, the two lines above prove $\text{gcd}(c,b)=1$ and $\text{gcd}(c,a)=1$.

The proof for this is simple:

$\gcd(a, b)$ $=$ $1$, therefore $\gcd(a + b, a, b) $=$ $1, similarity $\gcd(a^n + b^n, a, b) $=$ $1. Any prime $c$ dividing $a^n + b^n$, the $\gcd($c$, $b$, $a$) $=$ $1. Your question is when $n$ $=$ $2$.


Proving sine of sum identity for all angles Could anyone present a proof of sine of sum identity for any pair of angles $a$, $b$?

$$\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$

Most proofs are based on geometric approach (angles are $<90$ in this case). But please note the formula is supposed to work for any pair of angles.

The other derivation I know is using Euler's formula, namely this one.

There's one thing I don't feel comfortable with - we know that we add angles when multiplying two complex numbers. This is proven with sine of sum identity. So first we prove how multiplication of two complex exponentials works using sine of sum identity, and then use multiplication of complex exponentials to prove sine of sum identity. Can you tell me how it's not a circular argument?

here is a geometric proof i saw in an old american mathematics monthly which uses the unit circle. first show that the square of the chord connecting $(1,0)$ and $(\cos t, \sin t)$ is $2(1-\cos t)$ using the distance formula. now reinterpret
$$\text{ length of chord making an angle $t$ at the center is } 2 - 2\cos t $$

now compute the length squared between $\cos t, \sin t), (\cos s, \sin s)$ in two different ways:

(i) distance formula gives you $2 - \cos t \cos s - \sin t \sin s$

(ii) chord making an angle $t - s$ is $2 - \cos(t-s)$

equating the two gives you $$\cos (t-s) = \cos t \cos s + \sin t \sin s \tag 1$$

now use the fact $\cos \pi/2$ to derive  $\cos (\pi/2 - s) = \sin s$ by putting $t = \pi/2$ in $(1)$

put $t=0,$ to derive $\cos$ is an even function. put $t = -\pi/2,$ to show $\sin$ is an odd function. after all these you derive
$$\sin(t-s) = \sin t \cos t - \cos t \sin s $$ and two for the sums.

The statement:

$ e^{ix} = \sin(x) +i\cos(x) $

is derived using Taylor series which can be proven like so

Just sum $\cos(x)$ and $\sin(x)$ for $f(x)$ to convince yourself. If you sum up those two Taylor series, it will give you the Taylor series for $e^{ix}$.

We never aknowledged the existence of sum angle identity in this proof, hence there is not circular reasoning.


Show the limit holds true. I have to show that: lim$_{x\rightarrow 2} \displaystyle\frac{1}{x} = \frac{1}{2}$

So far I have:

$|f(x)-L|<\epsilon \leftrightarrow |\displaystyle\frac{1}{x}-\frac{1}{2}|<\epsilon \leftrightarrow |\displaystyle\frac{x-2}{2x}|< \epsilon \leftrightarrow \frac{1}{2x} \cdot |x-2| < \epsilon \leftrightarrow |x-2| < \frac{2x}{\epsilon}$. So $\delta=\displaystyle\frac{2x}{\epsilon}$.

My question is: Is it okay for $\delta$ to depend on $x$?

Remember that by definition

$$\lim_{x\rightarrow a} f(x) = L \iff \forall \varepsilon >0\, \exists \delta > 0: 0<\vert x-a\vert <\delta \implies \vert f(x)-L\vert <\varepsilon$$

To find the optimal $\delta$ you can proceed as follow

$$\left|\frac1x-\frac12 \right|<\epsilon \iff -\epsilon<\frac1x-\frac12 <\epsilon\iff -\epsilon+\frac12<\frac1x<\epsilon+\frac12$$

$$\iff \frac{1-2\epsilon}2<\frac1x<\frac{1+2\epsilon}2\iff \frac2{1+2\epsilon}<x<\frac2{1-2\epsilon}$$

$$\iff \frac2{1+2\epsilon}-2<x-2<\frac2{1-2\epsilon}-2$$
$$\iff \frac{-4\epsilon}{1+2\epsilon}<x-2<\frac{4\epsilon}{1-2\epsilon}$$

thus

$$|x-2|<min\{\frac{4\epsilon}{1-2\epsilon},\frac{4\epsilon}{1+2\epsilon} \}=\frac{4\epsilon}{1+2\epsilon}=\delta$$

No, it is not okay if your $\delta$ depends on $x$. It can depend on $x_0$, in your case $x_0 = \dfrac{1}{2}$.

But why do you need a $\delta$ or do you want to proof continuity?


Why does $\frac {1}{a}-\frac{1}{b}=\frac {b-a}{ab}$? I really don't understand the expression.
$$\frac {1}{a}-\frac{1}{b}=\frac {b-a}{ab}$$
I generally have a hard time understanding non-intuitive things in math and this is one of them. Normally when I don't understand something I use an app , photomath, to help explain expressions/equations I don't understand however, I still need help with this expression.
I'm told that to get to $\frac {b-a}{ab}$ you need to expand the fraction to the least common denominator:
$$\frac {1}{a}-\frac{1}{b} \to \frac {\pmb b\times 1}{\pmb b a} - \frac {\pmb a \times 1}{\pmb a b} \to \frac {b}{ab} - \frac {a}{ab} \to \frac {b-a}{ab}$$
What I don't understand is this
$$ \frac {\pmb b \times 1}{\pmb b a} - \frac{\pmb a \times 1}{\pmb a b}$$
I don't understand how exactly the $a$ and $b$ seemingly 'appear' in the expression.

It is easier to grasp if you start with the RHS.
Note that $$\frac {b-a}{ab} = \frac {b}{ab} -\frac {a}{ab}=$$
$$ \frac {1}{a} -\frac {1}{b}$$

Original Equation
$\frac{1}{a} - \frac{1}{b} \stackrel{?}{=} \frac{b - a}{ab}$
Multiply both sides by $ab$
$ab \cdot \big(\frac{1}{a} - \frac{1}{b}\big) \stackrel{?}{=} ab \cdot \big(\frac{b - a}{ab}\big)$
Distribute the $ab$
$\frac{ab}{a} - \frac{ab}{b} \stackrel{?}{=} \frac{ab \cdot (b - a)}{ab}$
Cancel out variables and both sides have identical expressions
$b - a \stackrel{\checkmark}{=} b - a$


Proving Sylow's first theorem Let $p$ be prime and $G$ a group such that $|G| = p^n k$, where $p \nmid k$. We want to show that $G$ has at least one Sylow $p$-subgroup.

  Let $\mathcal{S} := \{ S \subset G | \, |S| = p^n\}, |\mathcal{S}| := N$. Then, $N = \binom{p^n k}{p^n} \equiv k \text{ (mod } p)$. Let $G$ act on $\mathcal{S}$ by $(g, S) \mapsto gS = \{gs| \, s \in S\}$. Let $S_1, \dots, S_r$ represent the disjoint orbits of $\mathcal{S}$. Then, $p$ cannot be a divisor of the cardinality of all orbits, because $N \equiv k \text{ (mod } p)$. Choose $S_i$ such that $p \nmid |GS_i|$. Then, $\operatorname{Stab}_G(S_i) = S_i$ and thus $S_i$ is a Sylow $p$-subgroup.

Why is $\operatorname{Stab}_G(S_i) = S$? We defined the stabilizer as follows:

$$\operatorname{Stab}_G(x) := \{g \in G| \, gx = x\}$$

Maybe I'm just confusing things, but if we take $g \in G \setminus S_i$, isn't still $gS_i = S_i$?

You are confusing yourself by the notation. Also observe that the proof requires the notion of stabilizers of sets not points. But let's use your notation. Let $T \in S_i$, where the cardinality of orbit $S_i$ is not divisible by $p$. Put $H=Stab_G(T) = \{g \in G | gT=T\}$. We are going to argue that $H$ is a Sylow $p$-subgroup. First, since $p$ does not divide $|S_i| = [G:H]$, it follows that $p^n | |H|$, hence $p^n \leq |H|$. Since $H$ stabilizes $T$ by left multiplication, for any $t \in T$ we have $Ht \subseteq T$. Hence $|H| = |Ht| \leq |T| = p^n$. We conclude that $|H| = p^n$ and we are done.

I surmise that you took the proof in the wrong direction: we do not show that the stabiliser equality, but show that the stabiliser is a sub-group with the correct order.
I think also you meant to say that $p$ cannot divide the cardinality of the orbit of $S_i$, not the order of $S_i$, where $S_i$ is a subgroup in that orbit. Then let the orbit containing $S_i$ be $\mathscr O$.
So $|\mathscr O|=\mid G\mid/\mid H\mid$, where $H$ is the stabiliser of $S_i$.
Thus all the powers of $p$ appearing in the factorisation of $\mid G\mid$ must divide the order of $H$. This is equivalent with saying that $p\mid |H|$.
If the proof is fine in the direction described, tell me, for I cannot see how your proof is going to end. Regards.
Edit The above proof is not complete: we also have to show that the cardinality of $H$ is exactly $p^n$, not just divisible by $p^n$.
Since $H$ stabilises $S_i$, we know that, for every $s\in S_i$, $Hs\subset S$, and hence $\mid Hs\mid=\mid H\mid\le p^n$.


Power series with square summable coefficients An exercise in my textbook asks to prove the following:

  Suppose $\sum |a_n|^2<\infty$. Show that the power series $\sum
 a_nz^n$ has radius of convergence at least equal to $1$.

My reasoning is as follows:

Since $\sum |a_n|^2<\infty$ we should have $|a_n|^2\to 0$ and hence $|a_n|\to 0$.

Then $\limsup |a_n|^{1/n}\le 1$ and hence the radius of convergence is at least one.

Although I believe my solution is correct, I still don't understand the reason for assuming square summability. We could have assumed that $\sum |a_n|^r<\infty$ for any $r>0$, couldn't we? Does $r=2$ have any particular application or significance?

Your arguments are correct and you are correct that it would work for any $r > 0$. The main point is probably that square-summability is a weaker condition than (absolute) summability, yet still sufficient to guarantee the result. It might not be quite as obvious without Hadamard's formula.

According to your  radius of convergence is at most one no


method to solve Cubic Equation solve the cubic eq with respect to $w$?

$$8p^3(1-c)w^3-4p^2 w^2-pw-1=0 $$

I try to solve it by take the factor of constant term


I shall assume that we want $f,g\in \Bbb Z[X]$ (or else $\gcd(f(3),g(6))$ make little sense).

If $g$ is constant, we need $f(30)=30$, but then $f(30)-f(0)=54$ is not a multiple of $30$, contradiction.
Similarly, if $f$ is constant, we need $g(-24)=-24$, but then $g(-24)-g(0)=-54$ is not a multiple of $-24$, contradictiom.
We conclude that neither $f$nor $g$ is constant.

Write  $f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots$ and $g(x)=b_mx^m+b_{m-1}x^{m-1}+\ldots$ with $a_n, b_m$ non-zero and $n\ge 1$, $m\ge 1$.
Then
$$f(g(x))=a_nb_m^nx^{nm}+\ldots \qquad g(f(x))=b_ma_n^mx^{nm}+\ldots. $$
Equality is possible only in the following cases:

$n=m=1$, $a_n,b_m$ arbitrary
$n=1$, $a_n=1$, $b_m,m$ arbitrary
$m=1$, $b_n=1$, $a_n,n$ arbitrary
$a_n=b_m=1$, $n,m$ arbitrary.

In the first case, $f(X)=aX+30$, $g(X)=bX-24$, $f(g(X))=abX-24a+30$, $g(f(X))=abX+30b-24$, so in this case a necessary and sufficient condition is that $5b+4a=9$.
We can parametrize this as $a=5k+1$, $b=-4k+1$.
Then $f(3)=3a+30=15k+33$, $g(6)=6b-24=-24k-18$.
From $8f(3)+5g(6)=174$, we conclude $\gcd(f(3),g(6))\mid 174=2\cdot 3\cdot 29$.
On the other hand, clearly $3\mid \gcd(f(3),g(6))$.
We have $2\mid \gcd(f(3),g(6))$ iff $k$ is odd.
We have $29\mid f(3)\iff k\equiv 21\pmod {29}$ and also $29\mid g(6)\iff k\equiv 21\pmod{29}$.
In summary, the case of linear $f,g$ allows the following values for $\gcd(f(3),g(6))$:

$3$, for example with $f(X)=X+30$, $g(X)=X-24$
$6$, for example with $f(X)=6X+30$, $g(X)=-3X-24$
$87$, for example with $f(X)=251X+30$, $g(X)=-199X-24$
$174$, for example with $f(X)=106X+30$, $g(X)=-83X-24$

In the second case, $f(X)=X+30$, so $g(X+30)=g(X)+30$; it follows that $g(X)-X$ has period $30$, hence must be constant. This takes us back to the first case.

In the third case, $g(X)=X-24$, so $f(X-24)=f(X)-24$, and again we end up in the first case.

In the fourth case, we may assume $n\ge 2$ and $m\ge 2$.
This might be treated by consideraing the intermediate coefficients of $f(g(X))=g(f(X))$. (But at least it is still clear that $3\mid f(3)-f(0)$ and $6\mid g(6)-g(0)$, so $3\mid\gcd(f(3),g(6))$).

$f(x) = x-24$ and $g(x) = x+30$ satisfy the conditions. These give $f(3) = -21$ and $g(6) = 36$.


Finding the $n$th term for the sequence $1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, \dots$ I have tried using a negative exponent.  I need one statement not two, the pattern is
$$1, \; \frac{1}{2}, \; 3, \; \frac{1}{4}, \; 5, \; \frac{1}{6}, \dots$$

Hint: You can express it by $n^{a_n}$, where the $a_n$ give the appropriate power needed. For $n$ even, you want $a_n=1$ and for $n$ odd, you want $n=-1$ (recall $x^{-y}=1/x^y$).

So what formula  gives $1,-1,1,-1,\ldots$?

The answer is simple!
$(2n+1),\frac{n}{2}$ where $n$EZ.


What are the most overpowered theorems in mathematics? What are the most overpowered theorems in mathematics?

By "overpowered," I mean theorems that allow disproportionately strong conclusions to be drawn from minimal / relatively simple assumptions. I'm looking for the biggest guns a research mathematician can wield.
This is different from "proof nukes" (that is, applying advanced results to solve much simpler problems).  It is also not the same as requesting very important theorems, since while those are very beautiful and high level and motivate the creation of entire new disciplines of mathematics, they aren't always commonly used to prove other things (e.g. FLT), and if they are they tend to have more elaborate conditions that are more proportional to the conclusions (e.g. classification of finite simple groups).
Answers should contain the name and statement of the theorem (if it is named), the discipline(s) it comes from, and a brief discussion of why the theorem is so good.  I'll start with an example.

The Feit-Thompson Theorem. All finite groups of odd order are solvable.

Solvability is an incredibly strong condition in that it immediately eliminates all the bizarre, chaotic things that can happen in finite nonabelian simple groups.  Solvable groups have terminating central and derived series, a composition series made of cyclic groups of prime order, a full set of Hall subgroups, and so on. The fact that we can read off all that structure simply by looking at the parity of a group's order is amazing.

I would include the pigeonhole principle in such a list. This principle is one that is easy to explain to a 4 year old, and at the same time it is used throughout all areas of mathematical research. As an example see Proof of Van der Waerden's theorem (in a special case).

there are no theorems whose conclusions contain more that their fully-spelled-out premises.

all purported examples of such just use, e.g., definitions that allow a short syntactic statement of the theorem while shoving all the actual work of it under the rug.

well, unless math is inconsistent at any rate - in which case we have bigger problems.


$f:\mathbb R \to \mathbb R$ be differentiable such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ ; then is $f(x)>0,\forall x>0$? Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ ; then is it true that $f(x)>0,\forall x>0$ ?

Because $f'(x)>f(x)$ we have that $ e^{-x} f(x)$ is strictly increasing.That is the conclusion.

Let $y(x)=e^{-x}f(x)$. Then $ f$ (strictly) positive $ \iff  y$ (striclty) positive.

$\forall x $,  $   y'(x)=e^{-x}(f'(x)-f(x)) \ge 0$ and if $x> 0, y'(x)>0$.

Therefore $y$ is positive for $x \ge 0$.

Now lets suppose it exists $t>0$ such that $y(t)=0$. Then for  $\epsilon >0 $ small enough, because $y'(t) >0$, for $x \in ]t-\epsilon,t[$ , $y(x) < y(t) $, which is in contradiction with the previous point.


Prove directly from the definition that $\lim: \lim_{x\to -2} x^2 + x - 5 = -3$ So I am a bit stuck on applying the limit definition for this problem. This is what I have so far:

$\lvert x - c \rvert < \delta $ then $\lvert f(x) - l \rvert < \epsilon$$ $$\lvert x + 2 \rvert < \delta$ $$\lvert x^2 + x -2 \rvert < \epsilon $$

which means $\lvert x -1 \rvert \lvert x + 2 \rvert < \epsilon $

$\lvert x + 2 \rvert < {\epsilon\over \lvert x -1 \rvert} $

let $\delta = 1 $ then: $\lvert x + 2 \rvert < \delta < 1 $ which means
$-1 < x -1 < 1 $
$-4 < x -1 < -2 $
Thus the upper bound of $\lvert x -1 \rvert < -2 $ is $-2$

let $$\delta = \min(1, {\epsilon\over -2}) $$

But what is confusing me here is the fact that, how is $ \lvert x + 2 \rvert < {\epsilon\over -2} $

Because if $\epsilon$ is any number more than $0$, then ${\epsilon\over -2} $ will be negative and it can't be greater than $ \lvert x + 2 \rvert $.

Note that\begin{align}x^2+x-5-(-3)&=x^2-4+x+2\\&=(x-2)(x+2)+x+2\\&=(x-1)(x+2).\end{align}So, if $\lvert x+2\rvert<1$, then $\lvert x-1\rvert\leqslant\lvert x+2\rvert+3<4$. So, take $\delta=\min\left\{1,\frac\varepsilon4\right\}$. Then, if $\lvert x+2\rvert<\delta$, then$$\lvert x^2+x-5-(-3)\rvert<4\frac\varepsilon4=\varepsilon.$$

If $0<ε<2+$$\frac{1}4$$,then:
|f(x)-l|<ε$ $\Rightarrow$ $|x^2+x-5+3|<ε$ $\Rightarrow$$-ε<x^2+x-2<ε$ $\Rightarrow$ $(0<x^2+x-(2-ε)) \land (x^2+x-(2+ε)<0)$ $\Rightarrow$ $(x<\frac{-1-\sqrt{1+4(2-ε)}}2,\frac{-1+\sqrt{1+4(2-ε)}}2<x ) \land (\frac{-1-\sqrt{1+4(2+ε)}}2<x<\frac{-1+\sqrt{1+4(2+ε)}}2)$

By combining the inequalities we acquire two intervals containig 1 and -2 respectively.We are interested in the interval that for every 0<ε<2+1/4 contains -2, and then you are done.You can fill in the details.
If $ε>2+$$\frac{1}4$$,then:
-ε<x^2+x-2 $ holds true for every ε, and the other inequality simplifies to an interval that contains both 1 and -2.So you do basically the same work with the first case but you only have one interval to work with. I hope you found this helpful. I thank Martin Sleziak for the help with the edit.


Is $f(2x)/f(x)$ nonincreasing for concave functions with $f(0)=0$? I have a question about concave functions.

Let $f:R_+\rightarrow R_+$ be any nonidentically zero, nondecreasing, continuous, concave function with $f(0)=0$. Do we have that the ratio function $f(2x)/f(x)$ is nonincreasing on $(0,+\infty)$?

No. Let $f$ be the function
$$
f(x) \;=\; \begin{cases}2x & \text{if }0\leq x \leq 1 \\ x+1 & \text{if }1 \leq x.\end{cases}
$$
Then $f$ satisfies the hypotheses you have given, but
$$
\frac{f(2)}{f(1)} = \frac{3}{2} \qquad\text{and}\qquad \frac{f(4)}{f(2)} = \frac{5}{3} > \frac{3}{2}\text{,}
$$
so $f(2x)/f(x)$ increases from $x=1$ to $x=2$.

Edit: There are also differentiable counterexamples, e.g. $f(x) = (x^2+10x)/(x+1)$.

Edit: This proof has an error. See if you can spot it without looking at the comments!

Suppose $f$ is differentiable. Then $f' \geq 0$ is non-increasing, and your ratio is
$$ \frac{\int_0^{2x} f'(y) dy}{\int_0^x f'(y) dy}. $$
The derivative of the fraction is
$$\frac{f'(2x) f(x) - f'(x) f(2x)}{f(x)^2}. $$
Now $0 \leq f(x) \leq f(2x)$ and $0 \leq f'(2x) \leq f'(x)$ so the derivative is non-negative. Hence in this case, the ratio function is non-increasing.

This proof can probably be extended to the non-differentiable case.


Let $N$ be an abelian minimal normal subgroup of $G$. Then $N$ has a complement in $G$ iff $N\not\leq \Phi(G)$. Let $N$ be an abelian minimal normal subgroup of a finite group $G$. Then $N$ has a complement in $G$ iff $N\not\leq \Phi(G)$, where $\Phi(G)$ is the Frattini subgrup of $G$.
The Schur-Zasenhaus theorem says that if in addition $(N,G/N)=1$ then $N$ has a complement but I don't think I can use it here.
(a) Let $N\not\leq \Phi(G)$. Then there exists $M$ maximal in $G$ such that $N\not\leq M$ (1). As $M\leq NM$ either $NM=M$ or $NM=G$. If the former then $N\leq M$, absurd. Therefor $G=NM$ (2). Now suppose $M$ normal in $G$. Then $N\cap M$ is an abelian normal subgroup of $G$ and hence $N\cap M=1$ due to the minimality of $N$ and the thesis follows from (2). However $M$ could well be not normal. On the other hand $\Phi(G)$ is not a candidate for the complement because in $D_6=\langle \sigma,\tau\rangle$, $|\sigma|=3$, $\langle\sigma\rangle$ is abelian minimal normal but $\Phi(D_6)=1$.
(b) Let $H$ be a complement of $N$ in $G$. If $H$ maximal in $G$ then as $N\not\leq H$ I have $N\not\leq\Phi(G)$ as desired. But again why is $H$ going to be maximal? Could somebody give me a hint to this problem?
EDIT: From scratch: Let $N\not\leq\Phi(G)$. Then there exists $M$ maximal in $G$ such that $N\not\leq M$ (1). As $G$ finite, it has an abelian minimal normal subgroup $N_1$. What is more $N_1\leq N\cap M$ necessarily. To simplify assume $N_1=N\cap M$. To apply the inductive hypothesis to $M$, whose order is less than $|G|$, I still need to prove $N_1\not\leq\Phi(M)$. Assume $N_1\leq\Phi(M)$. And here I hope a contradiction will arise but can't find it. Can you?
Another thing. The author of this book (Kurzweil and Stellmacher, The theory of finite groups, an introduction, New York, 2004) has just proved two theorems about complements: the Schur-Zassenhaus and the Gaschutz theorems. Could I not use one of them to solve the problem?

If $NA=G$ and $N\cap A=1$ then $NH=G$ for every maximal subgroup $H$ containing $A$. Hence $N\not\subseteq H$ and $N\not\subseteq \Phi(G)$.
Conversely if $N\not\subseteq \Phi(G)$ then $N\not\subseteq H$ for some maximal $H$. Then  $NH=G$. Then $N\cap H\ne N$ is normal in $G$ (because the intersection is normal in $H$ and in $N$ as $N$ is abelian), so this intersection is 1 and $H$ is a complement.

Hi: Let $N\not\leq\Phi(G)$. Then there exists $M$ maximal in $G$ such that $N\not\leq M$ (1). But $M\leq NM$ and therefor either $M=NM$ or $G=NM$. If the former then $N\leq M$ contradicting (1). So $G=NM$ (2).
Let $A=N\cap M$. I'll show $A$ is normal in $G$. Let $a\in A$ and $nm\in G$. Then $b:=(nm)^{-1}anm= m^{-1}am$ (because both $a$ and $n$ are in $N$ which is abelian). But $A$ normal in $M$. So $b\in A$ and $A$ normal in $G$.
That is $A$ is an abelian normal subgroup of $G$ included in $N$ which means either $A=1$ or $A=N$. The latter contradicts (1). So $A=1$ wich together with (2) proves one of the problem statement inferences. How can I prove the other one?


Cancel partial derivative in fraction The question is related to this one.

Let $f(x,y) = xy$ and $g(x,y) = f(x,y)^2 = x^2y^2$.
Now consider the fraction of partial derivatives
\begin{align}
\frac{\frac{\partial g}{\partial x}}{\frac{\partial f}{\partial x}} = \frac{2xy^2}{y} = 2xy
\end{align}

I was wondering if I can cancel $\partial x$ in the fraction, i.e.,
\begin{align}
\frac{\frac{\partial g}{\partial x}}{\frac{\partial f}{\partial x}} = \frac{\partial g}{\partial f} = \frac{\partial f^2}{\partial f} = 2f = 2xy.
\end{align}

Is this some general principle or coincidence?

This is not by accident. As N74 said, this is chain rule. So $$\frac{\partial g}{\partial f}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial f}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial f}=\frac{\frac{\partial g}{\partial x}}{\frac{\partial f}{\partial x}}+\frac{\frac{\partial g}{\partial y}}{\frac{\partial f}{\partial y}}$$


Give the interval for which $\frac{\ln(t)}{1+t^{x}}$ is differentiable We are given the function: $$f(x) = \frac{\ln(t)}{1+t^{x}}$$ and are asked to show for which interval $I \subset \Bbb{R}$, where I can be unbounded, $f(x)$ is derivable.

I'm not sure how to start working this out, so any help would be appreciated.
Thank you in advance.

I assume you mean differentiable.

Since it's a function of $x$, the factor $\ln(t)$ can be viewed as a constant, just use the linearity of the derivative in the end. Since nothing is mentioned I assumed that $f: \mathbb{R} \to \mathbb{R}$ and then $t>0$,

$$
\lim_{h \to 0}\frac{\frac{1}{1+t^{x+h}} - \frac{1}{1+t^x}}{h} = \lim_{h \to 0} \frac{1+t^x - (1 + t^{x+h})}{h(1+t^x)(1+t^{x+h})} = \\ \lim_{h \to 0} \frac{t^x(1-t^{h})}{h(1+t^x)(1+t^{x+h})} =  \frac{t^x}{1+t^x} \cdot \lim_{h \to 0}\frac{1-t^h}{h(1+t^{x+h})} =\\ / \text{ h'opitals rule }/ = \frac{t^x}{1+t^x} \cdot \lim_{h \to 0}\frac{-t^h \ln(t)}{t^{h+x} + 1 + h\ln(t)t^{h+x}} \\ = \frac{-t^x \ln(t)}{(1+t^x)^2}
$$

Thus $f'(x) = \frac{-t^x \ln(t)^2}{(1+t^x)^2}$


Irrational number $\sqrt2$ proof I have seen the proof that  $\sqrt2$ is irrational
However I'm a bit confused. The proof goes something like this. first assume $m$ and $n$ are integers such that $$ \frac{m}{n}  = \sqrt{2} $$ $$ \frac{m^{2}}{n^{2}}  = 2 $$ $$m^{2}  = 2 n^{2}$$  $\implies m^{2} $ is divisible by $2 \implies m$ is divisible by 2. Let $2q = m$. Then $$m^{2}  = 4 q^{2} = 2n^{2}$$ $$ n^{2}  = 2 q^{2}$$ This means that $n^{2}$ is divisible by $2 \implies n$ is divisible by $2$...... Hence $\sqrt{2}$ is irrational. $$$$If $ m^{2} $ is divisible by $2$ how do we know that $m$ is also divisible by $2$? If you say it's a prime number then I can argue that I didn't need to do all these steps. I could have just said that its factors don't multiply to $2$ hence $\sqrt{2}$ is irrational

Suppose that $m$ is odd, then $m=2k+1$, and thus
$$m^2=(2k+1)^2=2(2k^2+k)+1$$
and thus $m^2$ is odd too. That prove that if $m^2$ is even then $m$ is even too.

You can generalize this result : If a prime number $p\mid m^s$ for $s\in\mathbb N$ then $p\mid m$. Indeed, let $m=p_1^{\alpha_1}...p_n^{\alpha_n}$ the decomposition of $m$ in prime factor. Then $$m^s=p_1^{s\alpha_1}...p_n^{s\alpha_n}$$
and thus all prime that divide $m^s$ also divide $m$.

Your entire proof is based on the assumption that m and n are integers. Till this assumption holds true. There is no fallacy in the argument.

But, Lets say m and n can be anything. That is, we don't assume anything for m andn. Then, your statement

  m^2 is divisible by 2 => m is divisible by 2

no longer holds true. for eg: m=k*sqrt(2).
 Then m is not divisible by 2. But m^2=2*k^2 is divisible by m.

Hence, by contradiction sqrt(2) is irrartional.


How to find the equation of a line inside a circle given the center point of the line? I have a circle with center (0,0) and radius 1. I have calculated a random point inside a circle by generating a random angle $a=random()\cdot 2\pi $ and a random distance smaller than or equal to the radius $b=random()\cdot r$. The center of the line is $(cos(a)\cdot b,sin(a)\cdot b)$. The line has to be inside the circle like this:

Knowing this information how can I calculate the equation of the line in the most efficient way for a computer? Thanks.

The random function generates a random Real Number between 0 and 1.

In polar coordinates,

$$\rho\cos(\theta-a)=b,$$

hence
$$x\cos a+y\sin a=b.$$

you can just do this with basic coordinate geometry. To find the line equation you need the slope and a point. Fortunately, the point is given. You can simply find the slope for the line since the product of two perpendicular lines' slope is -1.


Prove $\gcd(nn!, n!+1)=1$ For any $n \in \mathbb{N}$, find $\gcd(n!+1,(n+1)!+1)$. First come up with a conjecture, then prove it.

By testing some values, it seems like $\gcd(n!+1,(n+1)!+1) = 1$

I can simplify what's given to me to $\gcd(nn!, n!+1)=1$ but I can't find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance?

$\gcd(n!+1,(n+1)!+1) = 1 \implies \gcd(n!+1,(n+1)n!+1) = 1 \implies \gcd(n!+1,nn!+n!+1) = 1 \implies \gcd(nn!, n!+1) = 1$

You don’t need to use induction; you just need to prove the statement in the title. Suppose that $p$ is a prime factor of $nn!$; can $p$ divide $n!+1$?

I dont know i am right or wrong but i can do this example in following way,\
Let $$\gcd(n\cdot n!,n!+1)=d$$
    $$\therefore d\mid n\cdot n!\ ,\ d\mid n!+1$$
    $$\Rightarrow d\mid n,\ d\mid n!,\ d\mid n!+1$$
    $$ \Rightarrow d\mid n!+1-n!$$
Thus $d\mid 1\ \Rightarrow d=1.$


How to find the multi variable limit Here is the multi variable function:
$$\lim_{(x,y)\to (0,0)} \frac{x^2-2|x|-2|y|}{|x|+|y|}$$
I have no idea to solve the limit

Hint: Use the Squeeze theorem as follows:
$$\color{red}{-2} \le \frac{x^2}{|x|+|y|} - 2 \le \frac{|x|^2}{|x|} -2 = \color{red}{|x| - 2}$$

Since all variables return non-negative results, you can assume $x,y>0$.
Rewriting limit: $\lim_{(x,y)\to (0,0)} \frac{x^2}{x+y} - 2$
Then using polar coordinates $x=rcos\theta$ and $y=rsin\theta$ as $r\to 0$, we get limit is $-2$


Calculating integrals with asymptotes? Find $\displaystyle\int^2_0 \dfrac{1}{(1-x)^2} dx$.

Is there a way of doing this without considering the asymptote at $x=1$? What if you didn't know at first that there was indeed an asymptote at this point?

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
One alternative is the Cauchy Principal Value $\ds{\color{#f00}{\mrm{P.V.}}}$:
\begin{align}
\color{#f00}{\mrm{P.V.}\int_{0}^{2}{\dd x \over \pars{x - 1}^{2}}} &
\,\,\,\stackrel{\mbox{def.}}{=}\,\,\,
\lim_{\epsilon \to 0^{+}}\bracks{%
\int_{0}^{1 - \epsilon}{\dd x \over \pars{x - 1}^{2}} +
\int_{1 + \epsilon}^{2}{\dd x \over \pars{x - 1}^{2}}}
\\[5mm] & =
\lim_{\epsilon \to 0^{+}}\bracks{%
{1 \over 1 - \pars{1 -\epsilon}} - {1 \over 1 - 0} + {1 \over 1 -2} -
{1 \over 1 - \pars{1 + \epsilon}}} =
\lim_{\epsilon \to 0^{+}}\pars{{2 \over \epsilon} - 2}
\\[5mm] & = \color{#f00}{\infty}
\end{align}

  Even the Cauchy Principal Value $\underline{diverges}$.

I upvoted the other answer because I agree. However there are a few Nice identities

https://en.m.wikipedia.org/wiki/Sokhotski%E2%80%93Plemelj_theorem

https://en.m.wikipedia.org/wiki/Hadamard_finite_part_integral

Or on the complex plane

https://en.m.wikipedia.org/wiki/Kramers-Kronig_relations


Stuck on the formula of spectral radius I am having two questions on spectral radius:

We denote spectral radius of a matrix $A$ by $\rho(A)$

Question 1: Is it true that for any two matrices $A,B$,

$\rho(A+B)\le \rho(A)+\rho(B)$

When does it hold?

Question 2:

Is it true that $\rho(A)=\max_{||x||=1} x^TAx$?

If it holds then is it  not true that $\rho(A+B)= \rho(A)+\rho(B)$?

because  $\rho(A+B)=\max_{||x||=1} x^T(A+B)x=\max_{||x||=1}
x^T(A)x+\max_{||x||=1} x^T(B)x=\rho(A)+\rho(B)$

I am stuck on these two questions.
Can someone please help me out?
If some more information is needed please reach me.

Question 1 holds when;

$A$ and $B$ are normal. Because in this case $\rho(A) = \lVert A \rVert_2$, which means
$$ \rho(A+B) \leq \lVert A + B \rVert_2 \leq \lVert A \rVert_2 + \lVert B \rVert_2 = \rho(A) + \rho(B) $$
Note: $A+B$ may not be normal, but the first inequality holds for all matrices.
$A$ and $B$ are simultaneously upper-triangulizable, i.e. there exists an invertible $T$ such that $U_A:=T^{-1}AT$ and $U_B:=T^{-1}BT$ are upper triangular, in which case equality occurs.
$$ \rho(A+B) = \rho(T^{-1} (A+B) T) = \rho(U_A + U_B) = \rho(A) + \rho(B)$$
$AB = BA$ (special case of 2)

For Question 2, both claims are not true. However, for a symmetric matrix $A$
$$\lambda_\max(A)=\max_{\lVert x \rVert=1} x^T A x$$
holds. Note that this is not equivalent to $\rho(A)$, which is the maximum absolute value of the eigenvalues. So, for symmetric matrices
$$\lambda_\max(A+B)=\max_{\lVert x \rVert=1} x^T (A+B) x \leq \max_{\lVert x \rVert=1} x^T A x + \max_{\lVert x \rVert=1} x^T B x = \lambda_\max(A)+\lambda_\max(B)$$
holds.

For example, try $A = \pmatrix{0 & 1 \cr 0 & 0\cr}$ and $B = A^\top$.


How many powerball tickets will guarantee a 3/5 match? Trying it figure this out, but looks like the obvious answer might not be right. According to the probabilities there is a 1 in 579.76 chance of matching 3 of 5 numbers in a Powerball ticket. So how many tickets will you need to guarantee a 3/5 match? Will you need 580 tickets?

Doesn't seem right when considering that to guarantee at least 1 number match  you will need to buy 14 tickets as each ticket has 5 numbers. So 14 tickets can give you all possible 69 numbers.

However, probability of matching exactly 1 number would be around 0.28 using the formula ( 5 choose 1 * 64 choose 4 / 69 choose 5  ). Probability of matching at least 1 will be higher. So based on that you should need only 4 tickets or less to guarantee that at least one number matches.

Note: Powerball allows you to pick 5 numbers from 1 to 69. Probabilities given above may or may not consider bonus but the problem stays the same.

A crude upper bound for the number of tickets needed to guarantee a win is $18106$, as explained below . . .

Suppose a lottery is defined as a triple $(n,c,w)$ of integers, with $1\le w\le c\le n$, where

$S=\{1,...,n\}$ is the set of numbers from which $c$ distinct numbers are chosen.$\\[4pt]$
A ticket is an arbitrary choice of $c$ distinct numbers from $S$.$\\[4pt]$
To win, you need to match only $w$ of the $c$ chosen numbers.

In the context of the given question, we want to find an upper bound for the number of tickets required to guarantee a win for the lottery $(n,c,w)=(69,5,3)$.

Claim:

An upper bound for the general $(n,c,w)$ lottery is $f(n,c,w)$ where $f$ is defined recursively by
$$
f(n,c,w)=
\begin{cases}
\text{if}\;n=c,\;\text{then}\\[4pt]
\;\;\;\;\;1\\[4pt]
\text{else if}\;w=1,\;\text{then}\\[4pt]
\;\;\;\;\;\left\lceil{\large{\frac{n}{c}}}\right\rceil\\[4pt]
\text{else}\\[4pt]
\;\;\;\;\;f(n-1,c,w)+f(n-1,c-1,w-1)\\[4pt]
\end{cases}
$$
Implemented in Maple, we get $f(69,5,3)=18106$.

To justify the upper bound, the following ticket buying strategy, though not claimed to be optimal, suffices to guarantee a win, and requires buying exactly $f(n,c,w)$ tickets  . . .

If $n=c$, there is only one possible draw, and only one legal ticket, which is therefore guaranteed to win. Thus, for this case, we get $f(n,c,w)=1$.

If $w=1$, only one number needs to be matched, so an optimal covering can be had by buying tickets corresponding to $\left\lfloor{\large{\frac{n}{c}}}\right\rfloor$ pairwise disjoint $c$-element subsets of $S$, plus one more ticket if $n$ is not a mutilple of $c$ (in order to cover the remainder). Thus, for this case, we get $f(n,c,w)=\left\lceil{\large{\frac{n}{c}}}\right\rceil$.

Otherwise, recursively, use the same strategy to:

Buy $f(n-1,c,w)$ blank tickets for the $(n,c,w)$ lottery, and mark them as if the goal was to guarantee a win for an $(n-1,c,w)$ lottery. Note: Those tickets will guarantee a win for the $(n,c,w)$ lottery if none of the drawn numbers is equal to $n$.$\\[4pt]$
Buy $f(n-1,c-1,w-1)$ blank tickets for the $(n,c,w)$ lottery, choosing $n$ for one of the $c$ numbers, and for the other $c-1$ choices, mark them as if the goal was to guarantee a win for an $(n-1,c-1,w-1)$ lottery. Note: Those tickets will guarantee a win for the $(n,c,w)$ lottery if one of the drawn numbers is equal to $n$.

Guaranteed means $p = 1$. Like I said in my comment, if you buy one more than all possible $0, 1$ and $2$ number matches you are guaranteed at least a $3$ number match.

For $0$ number matches, there are $^{64}C_5 = 7624512$ ways to get that.

For $1$ number matches, there are $5$ ways to match that one number and for every one of those there $^{64}C_4 = 635376$ ways to not match the other $4$ making a total of $3176880$.

For $2$ number matches, there are $10$ ways to match $2$ numbers from $5$ and for every one of those there are $^{64}C_3 = 41664$ ways not to match the other $3$ making a total of $416640$.

The number of tickets to purchase to guarantee a $3$ number match is therefore:

$7624512 + 3176880 + 416640 + 1 = 11218033$

For an interpretation of this problem whereby guaranteed = likely, as in the same birthday problem, that is $23$ random people are required for a likely match of two people with the same birthday where $p > 0.5$.

So, the minimum number of tickets purchased for a probability $>0.5$ of $3$ out of $5$ matching numbers is the number of iterations of $1 - \frac{11218032}{11238512}\cdot \frac{11218031}{11238512}\cdot \frac{11218030}{11238511}...............\frac{11217652}{11238133} = .500001$. Hence $\approx 380$ tickets are needed to have a minisculely better than even chance of getting $3$ out of $5$ matching numbers.


How prove this $\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+e}+\frac{d-e}{d+2e+a}+\frac{e-a}{e+2a+b}\ge 0$ Let $a,b,c,d,e$ are postive real numbers,show that
$$\dfrac{a-b}{a+2b+c}+\dfrac{b-c}{b+2c+d}+\dfrac{c-d}{c+2d+e}+\dfrac{d-e}{d+2e+a}+\dfrac{e-a}{e+2a+b}\ge 0$$

My try: since
$$\Longleftrightarrow\sum_{sym}\left(\dfrac{a-b}{a+2b+c}+\dfrac{1}{2}\right)\ge\dfrac{5}{2}$$
$$\Longleftrightarrow \sum_{sym}\left(\dfrac{3a+c}{a+2b+c}\right)\ge 5$$
use Cauchy-Schwarz inequality,we have
$$\sum_{sym}\dfrac{3a+c}{a+2b+c}\sum_{sym}((3a+c)(a+2b+c))\ge\left(\sum_{sym}(3a+c)\right)^2$$

$$\Longleftrightarrow 16(\sum_{sym}a)^2\ge5\sum_{sym}(3a+c)(a+2b+c)$$

becasue this is not hold
$$\Longleftrightarrow 16(\sum_{sym}a)^2\ge5\sum_{sym}(3a+c)(a+2b+c)$$for $a,b,c,d,e>0$
and this method  is from this simaler inequality :
How prove this inequality $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$
then I can't,Thank you

You need to try a different fraction to add, so that you get better terms on applying Cauchy-Schwarz.  For e.g. here we have the equivalent inequality:

$$\sum_{cyc} \left(\frac{a-b}{a+2b+c}+\frac15\right)\ge 1$$

$$\iff \sum_{cyc} \frac{6a-3b+c}{a+2b+c} \ge 5$$

By Cauchy-Schwarz, we have:
$$ \sum_{cyc} \frac{6a-3b+c}{a+2b+c} \ge \frac{\left(\sum_{cyc} (6a-3b+c)\right)^2}{\sum_{cyc}\left((a+2b+c)(6a-3b+c)\right)} = \frac{16(a+b+c+d+e)^2}{\sum_{cyc}(a^2+8ab+7ac)}$$

So it is sufficient if we can show that:
$$16(a+b+c+d+e)^2 \ge 5 \sum_{cyc}(a^2+8ab+7ac) $$

But we can express the above $LHS - RHS$ as:
$$4\sum_{cyc}(a-b)^2 + \frac32 \sum_{cyc}(a-c)^2 \ge 0$$

Added based on a comment in linked post - this application of Cauchy Schwarz requires the numerator of the fraction $6a-3b+c$ to be non-negative, and hence covers only the cases with a condition like $a,b,c,d,e \in [\frac1k,k]$ where $k^2=\frac73$.

I'll solve the three variable case (the $a,b,c,d,e$ case is similar).

Assume without loss of generality $a \geq b \geq c$. Multiplying by the denominators we obtain

$$
a^3+a^2 c+a b^2-6 a b c+b^3+b c^2+c^3 \geq 0
$$

Now use the rearrangment inequality twice. We can split the inequality in two parts:
$$
a^2c+b^2a+c^2b \geq 3abc
$$
and
$$
a^3+b^3+c^3 \geq 3abc
$$
For the first one, use the rearrangment inequality to the sequences $ab, ac, bc$ and $a,b,c$ (which are listed in increasing order).
As for the second one, you can use the extension of the rearrangement inequality (also explained in the link) with three equal sequences $a,b,c$. Sum up the two inequalities and you are done.


4 normals can be drawn from any point to an ellipse proof doubt: How do we know the key quartic has 4 distinct real roots? Theorem: 4 normals can be drawn from any point to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. $( a > b > 0)$

I had a doubt in the proof of the above statement given in my book.

Proof given in Book.
The normal at the point with eccentric angle $\phi$ is $ax\sec\phi - by \csc\phi = a^2 - b^2 $. Let P$(h,k)$ be any point in $\mathbb{R^2}$. If the normal passes through P:
$$ah\sec\phi - bk \csc\phi = a^2 - b^2  = a^2e^2 $$
Let $ t = \tan \frac{\phi}{2}$, then $\cos \phi = \frac{1 - t^2}{1 + t^2}$ and $\sin \phi = \frac{2t}{1 + t^2}$. Substituting and rearranging, gives:

  $$bkt^4 + 2t^3(ah + a^2e^2) + 2t(ah - a^2e^2) - bk = 0 $$

This is a quartic in $t$, hence has 4 roots, corresponding to 4 points on the ellipse which are conormal.

  Doubt: My question is how do we know that the roots of that quartic equation are real and distinct? Because if not, then the theorem is not yet proved. I looked up the discriminant of a quartic on Wikipedia (https://en.wikipedia.org/wiki/Quartic_function), but the inequalities are too difficult to handle. How can I show that the roots are all real and distinct?

I think the theorem is simply false. Here is a rendering of the set of lines normal to an ellipse $(a=1, b=1/2)$:

There is a region (if I had to guess, bounded by the evolute of the ellipse: it would be interesting to prove this) where every point is indeed intersected by four normals to the ellipse. Outside this region, only two normals can be drawn.

Here is an example of such 4 normal:

Hope this helps!


Arrangement of colored cars in traffic Seeing helicoptor captures of car packed streets, and the sort order of colors can be observed, if for simplicity we limit the scope of the pictures to say 9 or 10 cars, I wondered whether it is possible to count all the unique arrangements of the 9 colored cars in a line (assuming we have 4 blue and 5 green ones) where the order g..g..b..g..b can always be found in the arrangement. That is, an order involving only 5 cars is sought for out of the 9.

I have a feeling it is very similar to introductory problems in probability involving a bowl of colored balls, where we know how the bowl is composed, and we are asked how many arrangements of this and that color can be observed in order, if we pick a number of balls at a time.

The regular expression that describes the valid sequences of cars is:

b*g b*g g*b b*g g*b (g|b)*

being b "blue car", g "green car", "*" zero or more, "|" alternative and blanks are used only for explicitness.

That means we must initially solve two problems of integer partition with fixed number of parts (see here), one for blues, one for greens.

For blue cars, we must find 4 parts that adds 2 cars (the remainder after subtract the two in mandatory positions to the four available) . The result is:
(0,0,0,2)
(0,0,1,1)
(0,0,2,0)
(0,1,0,1)
(0,1,1,1)
(0,2,0,0)
(1,0,0,1)
(1,0,1,0)
(1,1,0,0)
(2,0,0,0)

and 2 green cars in 3 parts:

(0,0,2)
(0,1,1)
(0,2,0)
(1,0,1)
(1,1,0)
(2,0,0)

Each pair of these two list of solutions defines a start of the sequence and the amount of green and blue cars that are available in the queue part "(b|g)*". This part has variants defined by the permutations with repetition $\frac{(m+n)!}{m!n!}$. By example:

(0,0,0,2) and (0,0,2) : ggbgb + pv{b,b,g,g}, with 4!/(2!2!)=6 alternatives (bbgg, bggb, bgbg, ....)
for the last item.
(0,0,1,1) and (0,1,1) : ggbbggb + pv{b,g}, with
2!/(1!1!)=2 alternatives for the last item (bg and gb).
...

** Addendum **

If the sequence has fixed start (g) and end (b), just discard these two cars. The the problem is now:

b*g g*b b*g (g|b)*

with 3 b and 4 g available in total. Using same method:

Partition for b:
(0,0,2)
(0,1,1)
(0,2,0)
(1,0,1)
(1,1,0)
(2,0,0)

partition for g:
(0,2)
(1,1)
(2,0)

that means 18 starts where each one has ending variants :

(0,0,2) - (0,2) : 4!/(2!2!) = 6

(0,1,1) - (0,2) : 3!/(2!1!) = 3

(0,2,0) - (0,2) : ... = 1

(1,0,1) - (0,2) : ... = 3

(1,1,0) - (0,2) : ... = 1

(2,0,0) - (0,2) : ... = 1

(0,0,2) - (1,1) : ... = 3

(0,1,1) - (1,1) : ... = 2

(0,2,0) - (1,1) : ... = 1

(1,0,1) - (1,1) : ... = 2

(1,1,0) - (1,1) : ... = 1

(2,0,0) - (1,1) : ... = 1

(0,0,2) - (2,0) : ... = 1

(0,1,1) - (2,0) : ... = 1

(0,2,0) - (2,0) : ... = 1

(1,0,1) - (2,0) : ... = 1

(1,1,0) - (2,0) : ... = 1

(2,0,0) - (2,0) : ... = 1

total is 31.

Hint

Notice, that there are only 4 cases, when this sequence can not be found in line:

ABC, where A contains $a$ greens, B contains 4 blues, C contains $c$ greens ($a+c=5$)
ABCDE, where A contains $a$ greens, B contains b blues, C contains $c$ greens, D contains $d$ blues and E contains e greens($a+c+e=5$, $b+d=4$, $a\leq 1$, $b,c,d,e >0$)
ABCDEF, where A contains $a$ blues, B contains 1 green, C contains $c$ blues, D contains $d$ greens and E contains e blues, F contains f greens($a+c+e=4$, $d+f=4$, $a,c,d,e,f>0$)
ABCDEFG, where A, C, E, G contains 1 blues, B contains 1 green, D contains $d$ greens, F contains f greens( $d+f=4$, $d,f>0$)

All these cases are disjoint and easy to calculate the number of possibilities.


Name the property $f(x) \ge x$ It's a really one of the simplest properties you could imagine for a function. But I haven't been able to find a name for it. What do you call a function $f$ with the following property:

$$f(x) \ge x$$

Note, this is not monotonicity. Monotonicity is the following property:

$$x \ge y \implies f(x) \ge f(y)$$

As mentioned, this property is called expansion, as the element $x$ can be expanded, or stretched-out and thus contained in, the element $f(x)$.

However, this property is also termed reflexivity.

Why? Well, recall that a relation is reflexive if it relates identical elements.
More precicely,

$\;\;\; R \text{ reflexive}
\\\equiv \forall x:: \; x \, R \, x
\\\equiv \forall x,y:: \; x = y \implies x \, R \, y
\\\equiv \forall x,y:: \; x \, I \, y \implies x \, R \, y \text{, where $I$ is the identity relation: $I=\{(x,y)\,|\,x=y\}$}
\\\equiv I \subseteq R
$

Even more so, recall that an ordering $\leq$ on elements can be lifted to functions by $$f \overset{.}{\leq} g \,:\equiv\, (\forall x :: f \ x \leq g \ x)$$
With this, and letting $1$ be the identity function $x \mapsto x$, we have
$$f \text{ reflexive } \ \equiv \ 1 \overset{.}{\leq} f$$

Hope that helps!

I call it the opposite of a contraction; ie the concave variant; it would follow from star concavity at $x_0=0$ where $F(x_0=0)\geq 0$with
now as $$F(1)=1 \land \text{dom(F)}=[0,1]$$.

star concavity at 0 $$\forall (t\in [0,1]):\forall (x \in [0,1]):F((1-t)\times x_0+t\times y )\geq t\times F(y)+(1-t)\times F(x_0)$$.

ie
$$\to F(t\times y)=F( (1-t)\times 0+ t\times y)\geq (1-t)\times F(0)+t\times f(y)$$

As $(t,1-t)\in [0,1]\land F(x_0=0)\geq 0$

$$\to F(t\times y)\geq (1-t)\times F(0)+t\times f(y)\geq  t \times F(y)$$

star concavity  at $0$ with $F(0)\geq 0(B)$
$$\to (B)F(t\times y)\geq t\times F(y)$$

(2) now as $$F(1)=1 \land \text{dom(F)}=[0,1]$$.

$$t=x \to F(x)=F(t)$$
$$F(x)=F(t)= F(1 \times t)\to F(x)=F(1\times t)$$
concavity

$$\to F(x)=F(1\times t)\geq t\times F(1)$$

substitution of $F(1)=1$
$$\to F(x)\geq t\times F(1)=1\times t=t$$
$$\to F(x)\geq t $$=

where t=x so:
$$\to F(x)\geq x$$


Equations to select pseudorandom number in a set I asked a similar question here and it was recommended I check out group theory.
Basically, I want to generate a string of numbers from 5 to 100 characters in length. The sets might be "every 5-number sequence" or "every 15 number sequence" or "every 100 number sequence". Every n number sequence.
Actually, I don't want to generate it. I want to most optimally find the next random (or pseudo-random) value using something like Group Theory! I can sort of vaguely imagine a way to use modulo to iterate through numbers in a set in a seemingly random way, and then at the same time never encounter the same number twice. (Or know if you are, and have some optimized solution to finding a nearby number or something).
Is there any way to accomplish this in Group Theory or otherwise? If not, that would be good to know to, why you can't accomplish this. But maybe there is an equation to do this.
For example, one thing I'm thinking that is better than a straight sequence is to go like a dampening wave. Say the set is from 100-999 (all 3-letter sequences). Then you would go:
100
999
101
998
102
997
...

That still doesn't appear random though, you can figure out that pattern in your head (even though I don't know what the equation is).
I would like something like that, where I can simply increment and index and have an equation to spit out the value.

If the set is large it's easy to just pick a random number and repeat if you get the same one, so that you get unique ones.
If the set is very small it might be best to randomly shuffle the list and pick the first n of that.
In the middle, perhaps the best way would be to store a list of indices that have already been taken and increment the randomly generated index for each time it is >= a taken index.


Related Rates: Two sides of a triangle are 6 m and 8 m in length and the angle between them is increasing at a rate of 0.06 rad/s. Two sides of a triangle are 6 m and 8 m in length and the angle between them is increasing at a rate of 0.06 rad/s. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is π/3 rad

Also, I have been using the following template below; however, I do not understand how they computed 63/25? I have used this solved calculation (below) as a template; yet with my problem that I want to solve in the "title", I keep getting 14.4/20?

Help welcomed.

$42 \times 0.06 = 42 \times \frac {6}{100} = 42 \times \frac {3}{50}=\frac {126}{50}=\frac {63}{25}$

You should get $24 \times 0.06 = 24 \times \frac {6}{100} = 24 \times \frac {3}{50}=\frac {72}{50}=\frac {36}{25}$

This is what you should have instead of the $\frac {63}{25}$ in their example. You then need to go on to multiply by $\frac 1 2$.

You have $\frac {14.4}{20}$ as your final answer or the shaded part?

If the final answer, then all is well. $\frac {14.4}{20} = \frac {144}{200}= \frac {18}{25}$