When to use the modulus symbol and when not to use the modulus symbol in integration and differentiation? I am facing this conceptual doubt for quite some time now.

We know $$\frac{d}{dx}{(\sec^{-1}{x})}=\frac{1}{|x|\sqrt{x^2-1}}$$ whereas $$\frac{d}{dx}{(\csc^{-1}{x})}=\frac{-1}{|x|\sqrt{x^2-1}}$$

Now suppose I need to find the integral
$$\int\frac{1}{x\sqrt{x^2-1}}dx$$ then will the answer be $\sec^{-1}{x}$ or $\csc^{-1}{x}$ in case the modulus function is not used for $x$ in the denominator? Why?

Another similar doubt I have is that $$\int{\frac{1}{x^2-a^2}} \, dx$$ equals $$\frac{1}{2a}\ln\left(\frac{x-a}{x+a}\right)+C$$ or $$\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C \text{?}$$

Some books use the former formula and some use the latter.Which one is correct and why?

Pardon me if you find this question too trivial.But really I'm confused with this thing from the past few months!

The functions $sec^{-1}$ and $-cosec^{-1}$ only differ by a constant in the points where they are both defined. The same is true for $sin^{-1}$ and $-cos^{-1}$. So, when they are both defined, you can choose one or another as you please, as long as you add an additive constant.

This happens because of the formula $\cos(x) = \sin(\pi/2-x)$ which says that $\sin(x)$ and $\cos(-x)$ are obtained one from the other with an horizontal translation, so their inverse functions correspond one to the other with a vertical translation.

To be more precise, (and to discuss the case of $\log x$ versus $\log |x|$) if the domain of the integrand function is not an interval, it is not enough to add a constant. You need to add a combination of different constant on each different connected component of your domain.

Example. Many books say that
$$
\int \frac{1}{x}\, dx = \log |x| + C.
$$
This is not completely true. All the antiderivatives of $1/x$ are given by a family of functions with two parameters, not one:
$$
  F(X) = \begin{cases} \log x + C_1 & \text{if $x>0$,}\\
  \log (-x) + C_2 & \text{if $x < 0$.}
\end{cases}
$$

It is true that in most applications you only need to find an antiderivative in one fixed interval (for example if you are solving differential equations). So, given the interval, there can be a preferred choice for the solution. If you are integrating $1/x$ for $x<0$, the solution can be written as $log(-x)+C$. If you are integrating $\frac{1}{x\sqrt{x^2-1}}$ on $(-1,0)$ the solution can be written as $cosec^{-1}(x) + C$.

The modulus expressions are absolutely fine.

So if you differentiate $\csc^{-1}x$ w.r.t. $x$ where $|x|>1$ from definition, you get after simplification,

$\large\frac{d}{dx}(\csc^{-1}x)=-\frac{1}{\csc y.\cot y}$ (where $y=\csc^{-1}x$)

$=\large-\frac{1}{\pm \csc y\sqrt {\csc^2y-1}}=-\frac{1}{|x|\sqrt {x^2-1}}$;$\quad (|x|>1)$

$[\because$ for $-\frac{\pi}{2}<y<\frac{\pi}{2}$, $\csc y\cot y>0 ]$

Similarly if $z=\sec^{-1}x$,

then $\large\frac{d}{dx}(\sec^{-1}x)=\frac{1}{\sec z\tan z}$

$\large=\frac{1}{\pm \left[\sec z\sqrt {\sec^2z-1}\right]}=\frac{1}{|x|\sqrt {x^2-1}}$; $\quad (|x|>1)$

$[\because$ for $0<z<\pi \quad (z\neq \frac{\pi}{2}), \sec z\tan z>0]$

Hence, $\displaystyle\int \frac{dx}{x\sqrt {x^2-1}}=\sec^{-1}|x|+c$ $\quad$or$\quad$ $-\csc^{-1}|x|+c;$ $($both for $|x|>1)$.

Note that whenever differentiations of inverse trigonometric functions are considered, only the principal values of the functions are taken into account, and the respective ranges against each of the formulas are to be taken care of. For instance, we see that the formula $\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt {1-x^2}}$ does not remain valid for $\frac{\pi}{2}<\sin^{-1}x<\frac{3\pi}{2}$.

As for your last integral, it makes sense to include the modulus sign if you accept the fact that $\displaystyle\int \frac{dx}{x}=\ln {|x|}+c; \quad (x\neq 0)$, which is the right thing to say after combining the two cases of $x>0$ and $x<0$.

So, for completeness, you should use the modulus symbol in these cases.


Pedro's barn and how to find constant "In Pedro's barn, the number of mice is inversely proportional to the number of cats. When he owned 5 cats, there were 48 mice in the barn. He increased the number of cats to 8. Based on the increased number of cats, how many mice are in the barn? Assuming that the relationship between the number of nice and the number of cats does not change, write an equation that Pedro could to to determine the number of mice (m), for any number if cats (c)."

I know what inversely proportional means. I know that in this case, as the number of cats increases, the number of mice decreases. I also know that the equation would be: m=k/c, k=constant. I need to know how to find the constant, and the steps leading up to it. Thanks.

Edit: I figured it out, thanks for your help anyway.

You find the constant by substituting the known values into the equation $m=\dfrac{k}{c}$:

$$48=\dfrac{k}{5}\implies k=240.$$

We have that
$$5=\frac k{48} \implies k= 240$$
and
$$m=\frac{240}c$$
with $c=8$. The result is then $m=30$.


Can the gradient operator $\nabla_\mathbf{x}$ be treated as a standalone vector? Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a scalar function. Then, the gradient of $f(\mathbf{x})$ is defined as:
$$
\nabla_\mathbf{x} f(\mathbf{x}) =
\begin{bmatrix} \frac{\partial f(\mathbf{x})}{\partial x_1} \\
\frac{\partial f(\mathbf{x})}{\partial x_2} \\
\vdots \\
\frac{\partial f(\mathbf{x})}{\partial x_n}
\end{bmatrix}
$$
However, if $f(\mathbf{x})$ is a scalar, then wouldn't the following also be valid?
$$
(\nabla_\mathbf{x}) (f(\mathbf{x})) =
\begin{bmatrix} \frac{\partial}{\partial x_1} \\
\frac{\partial}{\partial x_2} \\
\vdots \\
\frac{\partial}{\partial x_n}
\end{bmatrix} f(\mathbf{x}) =
\begin{bmatrix} \frac{\partial f(\mathbf{x})}{\partial x_1} \\
\frac{\partial f(\mathbf{x})}{\partial x_2} \\
\vdots \\
\frac{\partial f(\mathbf{x})}{\partial x_n}
\end{bmatrix}
$$
In other words, the gradient $\nabla_\mathbf{x}$ is treated as a standalone vector, and then it is multiplied by the scalar $f(\mathbf{x})$. Interestingly, if this is indeed true, then I can build the Hessian matrix of $f(\mathbf{x})$ using the following outer product:
$$
(\nabla_x \nabla_x^T) (f(\mathbf{x}))
$$
In other words, I left-multiply the row vector $\nabla_x^T$ by the column vector $\nabla_x$ to get an $n \times n$ matrix of second-order partial derivative operators. I then multiply this matrix by the scalar $f(\mathbf{x})$ to get the Hessian matrix. However, I am wondering if this is just a coincidence.

This notation is often formally used, notably in the definition of the Laplace operator. We can make it explicit though.
For convenience, lets only work with functions that are $C^\infty$, such that all their partial derivatives exist and can again be partially differentiated and so on (think about the issues that happen when we don't do this).
A partial derivative is just a function (operator) that sends functions to functions:
$$\frac{d}{dx_i}: C^\infty(\mathbb{R}^n, \mathbb{R}) \to C^\infty(\mathbb{R}^n, \mathbb{R})$$
We can regard the gradient as a $1 \times n$ matrix with such operators as coefficients, so the gradient is then an element of the set $C^\infty(\mathbb{R}^n, \mathbb{R})^{1\times n}$. This gradient, and any other matrix of operators, is then itself an operator that returns a (real) matrix, by applying the operators pointwise to a function.
To be able to multiply matrices with coefficients in $C^\infty(\mathbb{R}^n, \mathbb{R})$, we need to be able to multiply the coefficients with each other. The multiplication of operators can be defined as composition. In this setting we indeed have
$$ \nabla^T \nabla = L$$
$$ \nabla \nabla^T = H$$
where $L$ is the Laplace operator and $H$ is the Hessian operator.

$\nabla \nabla^T$ is not a common notation.
Although sometimes the notation $(\nabla \cdot \nabla) f = \nabla^2 f$ is used to denote $\Delta f$, where $\Delta$ is the Laplace operator, it's important to note that
$$(\nabla \cdot \nabla) f = \Delta f = \operatorname{tr} H_f$$
results in the trace of the Hessian $H_f$ of $f$. Some consider this notation strongly misleading: The square of first order derivatives, as $(\nabla \cdot \nabla) f$ might suggest, does not equate to the second order derivative. After all, with $f = x^2$, the second order derivative is $f'' = 2$ and not $\left(f'\right)^2 = (2x)^2 = 4x^2$.
Your observation however has a common application in optimization. Second-order optimization methods use the Hessian to improve an initial guess $x_0$ of a minimum of $f$. Instead of calculating the exact Hessian at $x_0$, the Gauss–Newton algorithm approximates the Hessian using
$$H_f(x_0) \approx J^T(x_0)\,J(x_0),$$
where $J$ is the Jacobian. This question discusses this approximation.


If $\Gamma \cup \{ \neg \varphi \}$ is inconsistent, then $\Gamma \vdash \varphi$ If $\Gamma \cup \{ \neg \varphi \}$ is inconsistent, then $\Gamma \vdash \varphi$

Here, a set of formulas is inconsistent means they syntactically imply some formula as well as its negation. Syntactic implication here consists of (1) given formulas, (2) all first-order logical axioms, (3) modus ponens.

I have tried to explicitly construct a syntactic proof for the conclusion but failed to come up with one.

You need the axiom (or tautology) :

  $\vdash (¬φ→¬ψ)→((¬φ→ψ)→φ)$.

Having already proved :

  if $Γ∪{¬φ}$ is inconsistent, then $Γ⊢¬φ→ψ$ and $Γ⊢¬φ→¬ψ$,

it is enough to aplly modus ponens twice to conclude with :

    $Γ⊢φ$.

Suppose $\Gamma\cup\{\neg\alpha\}$ is inconsistent.  Then for some $\beta$ we have $\Gamma\vdash\beta$ and $\Gamma\vdash\neg\beta$.  It remains only to show that $\{\beta,\neg\beta\}\vdash\alpha$.


How many ways can you put $20$ cards in $12$ envelopes if the cards are (a)Distinct (b)identical (c)identical and no envelope is left empty You have $20$ cards and $12$ envelopes(labelled $1.....12$). In how many ways can you put $20$ cards in envelopes if

a) the cards are distinct

b) the cards are identical

c) the cards are identical and no envelope can be left empty.

Please provide me with some briefing on how to solve this problem. Thank you

a) for each of the $20$ cards there are $12$ possibilities, so...

b) To be found is the number of sums $c_1+\cdots+c_{12}=20$ where the $c_i$ are nonnegative integers. For this you can apply stars and bars.

c) To be found is the number of sums $c_1+\cdots+c_{12}=20$ where the $c_i$ are positive integers.

Define $c'_i:=c_i-1$ and find the number of sums $c'_1+\cdots+c'_{12}=20-12=8$ where the $c'_i$ are nonnegative integers, again with stars and bars.

a) $$ \frac{31!}{11!} $$

A example:

$$ 1, 3, 5 | 2 | 4 | | | 6, 7, 8 | | | | | | 9, 10, 11, 12, ..., 20 $$

31 Symbols (20 numbers + 11 separators). The 11 separators repeating. The separators, separating the envelopes.

b) $$ \frac{31!}{11! \cdot 20!} $$

A example:

$$ A, A, A | A | A | | | A, A, A | | | | | | A, A, A, A, ..., A $$

31 Symbols (20 'A' + 11 separators). The 11 separators repeating. The separators, separating the envelopes.

c) $$ \frac{19!}{11! \cdot 8!} $$

A example:

$$ A, A, A | A | A | A | A | A, A, A | A | A | A | A | A | A, A, A, A, A $$

31 Symbols (20 'A' + 11 separators). The 11 separators repeating. The separators, separating the envelopes.

In this example we take an A from each envelope and it remains:

$$ A, A | |  |  |  | A, A |  |  |  |  |  | A, A, A, A $$

19 Symbols (8 'A' + 11 separators). The 11 separators repeating. The separators, separating the envelopes.


How can I find $x$ from this logarithms equation? Solve $\log_9x + \log_{81}3x = 1$


Khan Academy calculus problem solution differentiation at a point: algebraic, incorrect. I am practicing calculus on Khan Academy but appear to be having trouble with Differentiability at a point: algebraic. I'll provide an example problem outlining my approach to a solution and how it compares to the solution on Khan Academy.

  Is the function given below continuous/differentiable at, $x = 3$?
    $$f(x) =
  \begin{cases}
    x^2, & x \le 3\\
    6x - 9 & x \geq 3\\
  \end{cases}.$$

Finding the limit from both sides show that the function is continuous.
$$
   \lim \limits_{x \to 3^-} f(x) = 9, \quad
   \lim \limits_{x \to 3^+} f(x) = 9.
$$

To see if they are differentiable at $x = 3$ I use the following equation provided to me.

$$
   \lim \limits_{x \to 3} f(x) = \frac{f(x)-f(3)}{x-3} = \frac{f(x)-9}{x-3}.
  $$

$$
   \lim \limits_{x \to 3^-} = \frac{x^2-9}{x-3} = \frac{(x+3)(x-3)}{x-3} = x+3
  $$

$$
   \lim \limits_{x \to 3^+} = \frac{6x-9-9}{x-3} = \frac{6x-18}{x-3} =
   \frac{6(x-3)}{x-3} = 6
  $$

Now you would think that the function $f(x)$ is continuous and not differentiable but the actual answer is, in fact, both continuous and differentiable. My results for calculating the differentiation are incorrect based on their answers. Mind you they don't show how they reached that answer.

$$
   \lim \limits_{x \to 3^-} = \frac{x^2-9}{x-3} = 6
  $$

$$
   \lim \limits_{x \to 3^+} = \frac{6x-9-9}{x-3} = 6
  $$

I've looked at this problem from different angles and I am perplexed by how they reached that answer. Where did I happen to go wrong?

TL;DR I got a math problem wrong on Khan Academy. I want to know where I went wrong.

You should write

\begin{equation*}
\lim_{x\rightarrow3^{-}}\frac{x^2-9}{x-3}=\lim_{x\rightarrow3^{-}}\frac{(x+3)(x-3)}{x-3}=\lim_{x\rightarrow3^{-}}x+3=3+3=6
\end{equation*}

(don't simply write "$\text{lim}$" at the start of the line). This will help you avoid getting $x+3$ at the end of your limit (your variable should not be there after you impose $x\rightarrow 3$).

In fact, we never write:

\begin{equation*}
\lim_{x\rightarrow a}=\text{something}
\end{equation*}

It's always the limit of a function, as $x$ approaches some value $a$ that makes sense symbolically. Hence you always get:

\begin{equation*}
\lim_{x\rightarrow a}f(x)=\text{a real number or}\pm\infty
\end{equation*}

(if the limit exists).

Your solutions seem to directly indicate that the function is differentiable!

The derivative approaches $x+3$ from the left and $6$ from the right.

$x+3$ approaches $6$ as $x$ approaches $3$ from the left.

Because the derivative approaches $6$ from both sides, the function is differentiable at that point (given that you proved it is continuous already)


Simulink model from a nonlinear State Space I have the nonlinear state space already constructed in MuPAD as shown:

u is the input and y is the output.

What is the best way for me to take this to Simulink?

I would not use Simulink for simulating the system. But just use plain matlab code and the ODE solver...

Read about it: http://nl.mathworks.com/help/matlab/math/ordinary-differential-equations.html

It would look something like...

time       = 0:0.01:5;
x0         = [0 0 0 0];
odeOptions = odeset('RelTol',1e-6,'AbsTol',[1e-8 1e-8]);
u          = sin(time);
[t,y]      = ode45(@system,time,x0,odeOptions,u);

And then in your ODE function system you define your (non)linear system.

function dxdt = system(t,x,u)
dxdt = [x(2); ...
        -7.01*sin(x(3)) + (0.163*x(4) - 0.00816*u - 4.0e-6*sin(x(3)) + ...;
        x(4); ...
        -(400/7*x(4) - 20/7*u - ...];
end

Exactly for this case we published a small tool that enables to generate simulink models: ODESCA TOOL (Github)


$v(x,y), u(x,y)$ are differentiable. $v_x=-u_y, u_x=v_y$. Given $x=r\cos\theta, y=r\sin\theta$ $v(x,y), u(x,y)$ are differentiable. $v_x=-u_y, u_x=v_y$. Given $x=r\cos\theta, y=r\sin\theta$, Prove that for every $(x,y)\neq(0,0) : v_r=\frac{-1}{r}u_\theta, u_r=\frac{1}{r}v_\theta$.

I'm clueless in this case. How should I approach this type of question?

by the chain rule :

$ \begin{align} u_r = u_x x_r+u_yy_r = u_x \cos \theta +u_y \sin \theta \\
   v_r = v_x x_r+v_yy_r = v_x \cos \theta +v_y \sin \theta \\
u_{\theta} = u_x x_{\theta}+u_yy_{\theta} = -rv_y\sin \theta -rv_x \cos\theta = -rv_r \implies v_r=\frac{-1}{r}u_{\theta} \\
v_{\theta} = v_x x_{\theta}+v_yy_{\theta} = ru_y\sin \theta +ru_x \cos\theta = ru_r \implies u_r=\frac{1}{r}v_{\theta}
\end{align} $

we have, (dv/dx)=−(du/dy) and (du/dx)=(dv/dy)
'derivative of v with respect to x= product of derivative of r with respect to x and derivative of v with respect to r' as well as in y and θ

given, x=rcosθ,y=rsinθ => r=x/cosθ=y/sinθ and cosθ=(x/r)and sinθ=(y/r)
dr/dx=1/cosθ, dr/dy=1/sinθ and sinθ.dθ=-(1/r).dx, cosθ.dθ= (1/r).dy

(dv/dx)=−(du/dy) =>(dr/dx)(dv/dr)=−(dθ/dy)(du/dθ) =>(1/cosθ)(dv/dr)=-(1/(r.cosθ)).(du/dθ)=>(dv/dr)=-(1/r).(du/dθ)

(du/dx)=(dv/dy)=>(dr/dx)(du/dr)=(dθ/dy)(dv/dθ) =>(1/sinθ)(dv/dr)=(1/(r.sinθ)).(du/dθ)=>(du/dr)=(1/r).(dv/dθ)


How to find function for the following input output table? I have the following input/output table:

$$
\begin{matrix}
\textbf{In} & \textbf{Out} \\
-13 & 15 \\
 -4 &   2 \\
 -1 &   5 \\
  0 &  -9 \\
 11 &  -1 \\
 17 &   5 \\
 20 &  11
\end{matrix}
$$

Q1) I would like to find a function for the table.

Q2) Can someone recommend some tips how to approach similar problems ? E.g. different input values same output value the function uses an absolute value operator.

Thank you for reading.

The general answer is there are many functions that you might use to model the input and output. Here are two simple ways:

Lagrange interpolation

Let's assume that we have a collection of points $\{(x_k,y_k)\}_{k=1}^{n}$ and want a minimal degree polynomial which exactly goes through each point. Then we can define the Lagrange interpolation polynomial to be the function $L : \mathbb{R} \to \mathbb{R}$ by
$$
L(x) = \sum_{k=1}^{n} y_k \prod_{j \ne k} \frac{x-x_j}{x_k-x_j}.
$$
This has a nice closed-form result, but in general the polynomial you get out of this has degree $n$. It means you have 100% accuracy for the data you've provided, but a lot of variation around it.

Here's how your data looks when fit with a Lagrange interpolation polynomial.

Linear regression

If we require that only a low-degree (say $n=1$) polynomial is allowed, we avoid the greater variation from before, but we have to sacrifice fidelity to the original data. This is the linear regression technique, which looks for the line of best "fit" (according to some measure) for the data. In particular, the linear regression is simply $L : \mathbb{R} \to \mathbb{R}$ by
$$
L(x) = \alpha x + \beta,
$$
where $\alpha$ and $\beta$ minimize the sum of squared differences:
$$
S(\alpha, \beta) = \sum_{k=1}^{n} (y_k - \alpha x_k - \beta)^2.
$$
Here's how your data looks when fit with a linear regression.

As you can see, there's a big trade-off here between capturing the input-output data and "simplicity" of the model. Of course, these are only two common techniques. There are plenty of others as well.


Picard Iteration Error I have a question concerning Picard's iterative method. I have created an algorithm which solves (first, second order and system of DEs) initial value problems (IVPs) using the Symbolic toolbox in MATLAB.

Upon running the algorithm the iterations produce a Taylor series which approximates the exact solution of the IVP.  My question is what error would I use to show that function converges closer and closer to the exact solution?

I understand that the relative error can be used, but are there any other errors I can use for the same purpose?

Thank you for the help everyone!

Assume $y^n(t)$ is the $n$th Picard iteration. If it happens also that $y^n(t)$ is the same $n$th Taylor polynomial of the exact solution, say $y(t)$, then from Taylor's theorem the error is:
$$y^n(t)-y(t)=\frac{y^{(n+1)}(t_0)}{(n+1)!}(t-t_0).$$


Is there a Möbius torus? Does the concept of a Möbius torus make sense: taking a cylinder (instead of a rectangle as in the case of the Möbius strip) and twisting it before joining its ends? Or will the resulting twisted torus be indistinguishable from the normal torus in any relevant respect?

[This equivalent to the well-known Möbius strip should be called Möbius cylinder but it would have so much in common with a torus that I preferred to call it a Möbius torus.]

Embedded in Euclidean space the twisted and untwisted torus "look" the same - opposed to Möbius strip and cylinder -, the difference would be only in their intrinsic properties. But can there be such differences? And how do I specify them?

PS: I posted a follow-up question here.

As remarked above: What you get is the Klein bottle. Put differently: The result is what you get when you take two Möbius strips (which both have one boundary) and glue both boundaries together (which does not work when embedded in 3d space but works in theory). See this image from http://im-possible.info/english/articles/klein-bottle/:

The question is one of boundary between two and three dimensional space.  For example, bisect cut a bicycle tube which lays flat, draw a green line down one side of the tube, draw a red line down the other side of the tube.  Then rotate one end 180-degrees and cement it to the other end.  The green line meets the red line on both sides of the tube.     It looks nothing like a Klein bottle.  If you now inflate the tube from a two dimensional flat plane to a three dimensional torus, in effect you are creating an infinite loop where say, an ant travelling in one direction along the inside of the tube has twice the distance to walk to return to his original starting point.  This topography is not found in nature, at least not on the surface of a sphere.   If the tube is blown up enough, say to the size of a universe, you have finite space WITHIN the tube, although it is quite possible to have infinite space on the outside of the tube, surrounding it.  We simply don't know at this point since all we can comprehend is our interior space.  All that we do know is that we exist in a two-dimensional closed (finite) space within a three-dimensional open (infinite) space.  To say we live in a three dimensional world (height, length, depth) is naive at best.


I have 3 pieces of candy to place in 4 lunch boxes. In how many ways can I do this if... I am a middle school student, so it would be super helpful if anyone could explain this problem with simple words. I was able to do (a) and (b), but I got stuck on (c) and onwards.

For (c), I think there are three general cases, 3-0-0-0, 2-1-0-0, and 1-1-1-0. For 3-0-0-0, I know there would be 4 ways, because the 3 candies could go into any of the 4 boxes. I don't know how to tackle the other two cases, though.

I have 3 pieces of candy to place in 4 lunch boxes. In how many ways can I do this if:

(a) The candies are all different and the lunch boxes are all different?

(b) The candies are all the same and the lunch boxes are all the same?

(c) The candies are all the same and the lunch boxes are all different?

(d) The candies are all different and the lunch boxes are all the same?

(e) Exactly two of the candies are the same (but the third is different) and all of the lunch boxes are different?

a) If the candies are all different, and so are the lunch boxes, then placing each candy in each box affects which outcome we get. Each candy has four lunch boxes to choose from, and there are three candies, making the number of possible choices
$4 \times 4 \times 4 = 64.$

b) If the candies are the same and the lunch boxes are the same, then the possibilities are threefold: all the candy is in one box, two candies are in one and one is in another, or all three candies are in different boxes. The total number of possibilities is
$$3.$$

c) If the candies are the same and the lunch boxes are all different, then  there are cases, which are the three possibilities from b):

i) One box holds all three candies. In this case, there are $4$ boxes to choose from.

ii) One box holds two, another holds one. In this case, we have $4$ boxes to choose from putting the two in, and $3$ remaining boxes to choose to put the remaining one in. So, in total, this gives us $12$ possibilities.

iii) Each candy lies in a different box. In that case, since the candies are all identical, this means we are just choosing a box to not contain anything. We can choose our empty boxes in $4$ ways.

In total, the number of choices are
$$4 + 12 + 4 = 20.$$

d) If the candies are different and the lunch boxes are the same, then again, we tackle this with cases:

i) The first case, with $1$ possibility, is that all three candies are in the same box.

ii) Two candies could be in one box, and one in the other. Since the candies are now different, there are $3$ ways to do this, as we have three ways to choose the lone candy.

iii) All three candies are in separate boxes. Since the boxes are indistinct, there is $1$ way to do this.

Thus, the number of ways to do this is
$$1 + 3 + 1 = 5.$$

e) We do this in two steps: choose where the different candy goes, then choose where the same candies go. Note that these processes are totally independent, in that doing one first does not change the number of ways we can do the other. By choosing the different candy to go in one of the $4$ possible boxes, the boxes become no more or less distinct than they did previously.

Then, the question becomes, how many ways can we put two identical candies in four similar boxes? The answer becomes similar to c): we use cases.

i) Both candies are in the same box. There are $4$ boxes to choose from.

ii) The candies line in different boxes. We can choose one of the $4$ boxes first to host one candy, then one of the remaining $3$ to host the other.

This produces a total of $4 + 4 \times 3 = 16$. Putting in our initial choice of box for our distinct candy, we get
$$4 \times 16 = 64$$
possibilities.

Exactly two of the candies are the same (but the third is different) and all of the lunch boxes are different?

All in 1                               4ways

2 the same  in 1 box and the odd one in another   4*3=12

2 different  in 1 box and the other in another          4*3=12

all in different boxes          4C2*2 = 6*2=12ways

4+12+12+12 = 40ways

From web 2.0 calc
BTW the answer in your last part is wrong I think so because in ii) you have to also consider the different candy.


Expected value of a sure win game Suppose that there are 2 players playing the following game:

Given 10 coins, each player takes turn to collect either 1, 2 or 3 coins. The winner is the one who collects the last coin.

Assume that both players are smart. Then in this game, the player who start first by picking 2 will always wins as he can always picks 2, 6, 10 to collect the last coin.

If the winner gets 100 dollars, then what is the fair price of this game?

My attempt is that the winner has expected value 100 since he will always wins.

Is my reasoning above correct?

You have indeed found the correct strategy to win this game, which is to always present your opponent with a multiple of four coins. Taking two coins in the first round will leave your opponent with eight coins in their first turn, after which you will be able to leave four coins in their second turn and take the last coin yourself.

When it comes to determining the fair price of this game, it rather depends on who gets to start. If a player is certain of getting the first turn, then yes, the game has an expected return of \$100. If the starting player is assigned at random, however, it has an expected return of \$50 only.

Let's say that you have a probability $p$ with $p \in[0,1]$ to be the the starting player, then the fair price of the game should be $p\cdot 100$ .


Is $u \frac{dv}{du} + v (1+u) = u e^{u}$ a linear or nonlinear ODE? I am having a hard time deciding if the following differential equation is linear or not because of the last term on the left handside.

$$u \frac{dv}{du} + v (1+u) = u e^{u}$$

Can $v$ be multiplied by a linear function of $u$ like $(1+u)$ and still make the differential equation linear?

Thank you

Any equation first order of this form is linear:
$$y'+\alpha(x)y=\beta(x)$$
You DE is of that form
$$v'+\frac {u+1}{u}v=e^u
\implies v'+\alpha ({u})v=\beta(u)$$
So it's linear.

The solution is given by $$v(u)=\frac{c_1 e^{-u}}{u}+\frac{e^u \left(\frac{u}{2}-\frac{1}{4}\right)}{u}$$


The area under the curve 1/x in the first quadrant (integral) I've got a problem that I can't remember how to solve.
I'm meant to calculate the area under a curve 1/x in the first quadrant. Or, in other words, the integral of 1/x within these limits.

How do I go about doing this? I know that the primitive function (is that the correct term in English?) is: * ln(x) + C *.
What do I do now to get the area?

Thanks in advance!

You have an improper integral.  You get $\lim_{x\to\infty}\ln(x)-\lim_{x\to0}\ln x=\infty$...


How to prove that ${1,x,x^2}$ is a basis of a real polynomial functions space. Let $V$ be the real vector space of all polynomial functions from $\mathbb{R}$ to $\mathbb{R}$ at most second degree. That is, the space of all functions with form $f(x)=c_0+c_1x+c_2x^2$ with $c_i\in\mathbb{R}$

I need to prove that $\{1,x,x^2\}$ is a basis of $V$. But for that, first I need to prove the linear independence of those vectors, right? Well, here is where I get stuck. The only way I can think to do it, is defining an isomorphism $ T:V \to \mathbb{R^3} $ where each coeficient $ T(f)=(c_1,c_2,c_3) $ and proof the linear independece on $\mathbb{R^3}$.

Despite it work, probably it would be better if the proof is made without using transformations. Any suggestions?

hint: Suppose that $a_1\cdot 1 + a_2\cdot x + a_3\cdot x^2 = 0, \forall x \in \mathbb{R}$, then substitute $x = 0,1,2$ into the equation and solve for $a_1,a_2,a_3$. What do you get?

First, $V$ is the space of polynomials of degree $\le 2,$ not the space of all polynomials. Suppose $c_0+c_1x+ c_2x^2$ is the zero function. Doesn't that imply all $c_j=0$?


Is $4^{x+6}-3$ an algebraic expression? $$4^{x+6}-3$$

Can we say this is an algebraic expression? I believe it is an exponential expression, but is it also an algebraic expression?

"In mathematics, an algebraic expression is an expression built up from integer constants, variables, and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number)." [Wikipedia: Algebraic expression]

According to this definition, $4^{x+6}-3$ is not an algebraic expression - because the integer constants ($3,4,6$) and the variables ($x$) are connected in the expression with help of non-algebraic operations. $()^x$ is not an algebraic operation, because transcendental (= non-algebraic) operations are needed, e.g.: $a^x=e^{x\ln(a)}$.

But a completely different question is, if the value of the expression is algebraic (an algebraic number) or not.

An exponential expression is a mathematical expression that contains an exponent that is a non-rational expression of the considered variables.

According to this definition, $4^{x+6}-3$ is an exponential expression.

Remember that exponentiation by an exponent that is a rational number is an algebraic operation, but exponentiation by an exponent that is not a rational number is a non-algebraic operation.


Is the set of all invertible $n \times n$ matrices a vector space? I'm studying Algebra and I'm asked to prove or disprove "Is the set of all invertible $n \times n$ matrices a vector space?" I assume with respect to the usual matrix-sum and scalar multiplication.  I found that is true, but I'm not sure how to prove it.

My problem here is that this statement is too "broad", i.e. I cannot create a matrix with arbitrary values a,b,c,d,(...) considering that I don't know the size of the matrix.

My idea was to prove in a first time that the set of all invertible 2 x 2 matrices of real number is a vector space and then to show that this property could be extended to bigger matrices. However I don't know how to do it and it's precisely here that I need a little help ; how can I show that we can extend our statement?

Is it enough to say that the matrix's size doesn't affect in any way properties we need to check?
Is my overall strategy wrong?
Any form of help would be very appreciated on this dubious answer.

The set of all invertible $n\times n$ matrices of real numbers is NOT a vector space.

Let for example $I$, the unit matrix is invertible and so is $-I$. But their sum $I+(-I)=0$ is definitely not invertible!

The set of all invertible $n \times n$ matrices is not a vector space with respect to the typical matrix addition and scalar multiplication operations and the typical matrix zero.

However,

The special orthogonal group (rotation matrices) is a vector space if you use matrix multiplication for the addition operator and the identity matrix as the zero matrix.

You also have to make scalar multiplication exponentials.

$ s \otimes A = A^s$

You can also change the zero matrix to be a matrix $C$ with:

$A\oplus B = A B C^{-1}$

and

$ s \otimes A = A^s C^{-s}$.

Then,

$ A \oplus C = A$.

And invertible $n \times n$ matrices are just rotation matrices with a scaling part and so are also vector spaces.


What is the probability that balls left are white? From a bag containing $b$ black balls and $a$ white balls, balls are successively drawn without replacement until only those of the same colour are left. What is the probability that balls left are white?

the favourable event occurs when the balls drawn consist of

$0$ white and $b$ black balls
$1$ white and $b$ black balls
$2$ white and $b$ black balls

$\space\space\space$:
$\space\space\space$$a$. $a-1$ white and $b$ black balls

the last ball drawn cannot be white. it has to be black.
the probability of drawing $i$ white and $b$ black balls, successively without replacement from a bag containing $b$ black balls and $a$ white balls, such that the last ball drawn is  black is $$p_i = \frac{\binom{a}{i}\binom{b}{1}\binom{b-1}{b-1}(b-1+i)!}{\binom{a+b}{b+i}(b+i)!} = \frac{\frac{a!}{(a-i)!i!} \cdot b \cdot 1 \cdot (b-1+i)!}{\frac{(a+b)!}{(b+i)!(a-i)!}(b+i)!} = \frac{a! \cdot b \cdot (b-1+i)!}{(a+b)! \cdot i!}$$
$$\text{the probability that the balls left are white } = \sum_{i=0}^{a-1}p_i = \frac{a! \cdot b!}{(a+b)!}\sum_{i=0}^{a-1}\frac{(b-1+i)!}{i!(b-1!)}$$
$$ =  \frac{a! \cdot b!}{(a+b)!}\sum_{i=0}^{a-1} {b-1+i \choose i} = \frac{a! \cdot b!}{(a+b)!}\sum_{i=0}^{a-1} {b-1+i \choose i}{1 \choose 0}{a-i \choose a-i} $$

the given answer is  $\frac{a}{a+b}$

my question is:
is what I have done correct? if yes, how do I proceed further? if no, then what is wrong in my approach?
while I know that there is a very elegant proof for this question, I am curious as to why a very simple approach to solving this question is not working.
Note to myself:
for the summation $\sum_{i=0}^{a-1} {b-1+i \choose i}{1 \choose 0}{a-i \choose a-i}$

The vandermonde identity unfortunately does not apply
since, the summation is from $0$ to $a-1$, the last slot is always chosen. i.e. the last ball chosen is always white.

It seems that your calculation is correct. Using your final formula Mathematica obtained for all $(a,b)\in[100]^2$ the official value ${a\over a+b}$.
In reality it is expected that you solve this problem without any calculation. Just don't stop drawing until you have drawn all $a+b$ balls. The color of the last ball is the "remaining" color you are interested in. If you agree that the probability for the first ball being white is ${a\over a+b}$ then you should also  agree that the probability for the last ball being white is ${a\over a+b}$.

Lets take two cases-

where all ball left are white
where all balls left are black

Case 1:
You can take out all the black balls at one go and then successively pick out a white ball one by one. The number of ways in which you can do this such that the bag doesn't get emptied will be 'a'(This works since all the white balls are identical.).
Case 2:
Similarly, you can take out all the white balls at one go and then successively pick out a black ball one by one. The number of ways in which you can do this such that the bag doesn't get emptied will be 'b'(This works since all the black balls are identical.).
So the probablity will be simply  $a/(a+b)$


How to find the measure of an angle? I have a triangle with the following angles:

84 degrees,
x + 59,
and x + 51.

x + 51 is angle A

I have to find angle A's measure.

Since the interior angles of a triangle add up to be 180 degrees, I set up the equation like so:

x + 59 + x + 51 + 84 = 180 My equation

2x + 194 = 180 add like terms

2x = -19 subtract 194 from both sides of the equation

x = -7 divided both by 2x to get x alone

It has occurred to me that I am finding x here (it still makes no sense to get a negative number) but this is the only equation I know to set up.

Notes:

This is for geometry homework and I realize just asking you to solve my problems is academically dishonest off topic if I have not put genuine effort into the question, which I have (put effort in, that is). I have put a lot of mental effort into figuring this out, though, so I will tell you what I did and hopefully you can tell me where I went wrong and how to correct my mistake, because I think it is obvious I made one. For the record I accidentally deleted my math notes from my iPad and Googling the question this morning produces hard to understand results. My geometry teacher was unavailable while I was working last night due to my lack of internet connection.

84+x+59+x+51 = 180 so x=-7 therefore angles are -7+51=44 which is A and -7+59=52 Thus angles are 84,44,52. NOTE x isn't a angle so dont get confused all angles are positive here. Thanks!

enter image description here

Just ı want to learnd what the x and how did you do


Prove that $I+\lambda A$,$I+\lambda B$,$I+\lambda C$ are invertible and A=B=C Given the matrices $A,B,C\in M_n(\mathbb R)$ which satisfy the condiditions:

$A+B+C+\lambda ΑΒ=0 \quad(1)$

$Α+Β+C+\lambda BC=0 \quad(2)$

$A+B+C+\lambda CA=0 \quad(3)$

for some $\lambda\neq0$.

(a) Prove that $I+\lambda Α$, $Ι+\lambda Β$, $I+\lambda C$ are invertible and $AB=BC=CA$.

(b) Prove that $A=B=C$.

For (a) i understand that the matrices are invertible, because we have non zero eigenvalue. In addition we have AB=BC=CA, because the (1)-(2) and the (2)-(3) give us these equalities.
I have been stucked in (b). I tried to use the equality $AB=BC=CA$ and the fact that $I+\lambda Α,Ι+\lambda Β,Ι+\lambda C$ are invertible, but i can't prove that $A=B=C$.
I thought that A, B and C are null matrix, but i don't find a way to prove that.
I want help for (b).

For (a), the equations are equivalent to $-\lambda AB=-\lambda BC=-\lambda CA=A+B+C$. Since $\lambda$ is nonzero, we have $AB=BC=CA$.

Now suppose $(I+\lambda A)x=0$. Then $x=-\lambda Ax$. Hence $Cx=C(-\lambda Ax)=-\lambda CAx=(A+B+C)x$. Subtract $Cx$ from both sides and rearrange the remaining terms, we get $Bx=-Ax$. It follows that $Cx=-Ax$ too, because
$$
C\left(-\frac1\lambda x\right)=CAx=
ABx=-A^2x=-\frac1{\lambda^2}x.
$$
But then from $AB=BC$, we obtain $-\frac1{\lambda^2}x=-A^2x=ABx=BCx=A^2x=\frac1{\lambda^2}x$. Therefore $x=0$ and $I+\lambda A$ is invertible. By similar arguments, $I+\lambda B$ and $I+\lambda C$ are invertible too.

The statement in (b) is not true. For a counterexample, pick any three different scalars $a,b,c$ such that $a+b+c=0$ and consider
$$
A=\pmatrix{0&a\\ 0&0},\ B=\pmatrix{0&b\\ 0&0},\ C=\pmatrix{0&c\\ 0&0},\ \lambda=1.
$$
However, it is always true that $A^2=B^2=C^2$. This can be shown by considering $(A+B+C+\lambda AB)C=A(A+B+C+\lambda BC)$ and similar equalities.

This is not an answer, these are just ideas that came up into my head and I am not sure they'll help (I don't have much time now to try it) but try for example multiplying the equations by $I+\lambda A$ you'll get $A^2+AB+AC+\lambda AB=0$

Or maybe find the inverse of $I+\lambda A$, maybe it has a form that'll help.


Evaluate $\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx$ Evaluate $$\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx$$ by any method. In short I am interested in any method that overcomes the lack of convergence of this integral and gives an "number" to it.

EDIT

As I'm getting answers regarding convergence test, this should clear the question.
This is integral is divergent. Do other integration theories (Lebesgue, Henstock-Kurweil,..) overcome this problem and actually help assign a non-infinite value to this. (P.S. I'm engineering guy, need incentive to go beyond Riemann and this is a start)

Here is a heuristic reasoning why one should not expect to assign a number to this integral, even if we speak about a different type of "convergence/divergence".

Lets assume we can get, in some sense, a number $I$ so that

$$\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx =I \,.$$

After the substitution $x=\frac{u}{a}$ with $a>1$ we get

$$I= \int \limits_{0}^{\infty} \frac{\frac{u}{a}}{1+\frac{u^2}{a^2}} \frac{1}{a} du  = \int \limits_{0}^{\infty} \frac{u}{a^2+u^2} du< \int \limits_{0}^{\infty} \frac{u}{1+u^2} du =I$$

Thus $I <I$.

Of course, the "convergence" could be weak enough so that standard properties of integrals are not true anymore, but then it is unprobable that that type of convergence would be helpful.

Note that for all $R >0$

$$  \int \limits_{0}^{R} \frac{x}{1+x^2} dx =\int \limits_{0}^{aR} \frac{x}{a^2+x^2} dx  \,.$$

So any type of convergence, must either make those two limits different, or fail the following property:

$f <g , f,g$ continuous implies $\int_a^ \infty f < \int_a^ \infty g$.

[and in our example $f(x) < g(x)$ for all $x$, which means that $\int_a^\infty f =\int_a^b f +\int_b^\infty f$ should probably also fail..]

This is not difficult:

$$\operatorname{st}\int \limits_{0}^{\infty} \frac{x}{1+x^2} dx=\frac{\pi ^2}{2}-4+\psi\left(\frac{3}{2}\right)=0.971292...$$


Bayesian belief network A child inherits a gene X with probability 50%. A disease will develop if child inherited gene from both parents. The disease will not develop if child got gene from just one of parents.
Jain and Max have 0.25 probability of having that gene. They are parents of two children Mark and Eva. What is probability that Eva is healthy? What probability that Eva is healthy given Mark is healthy?

Let's start with the first question.
$$P(\text{Eva healthy}) = 1 - P(\text{Eva unhealthy})$$

Here we'll use law of total probability

$$P(\text{Eva unhealthy}) = P(\text{Both parents pass gene onto Eva}|\text{Both parents have the gene}).P(\text{Both parents have the gene})$$
$$P(\text{Eva healthy}) =1- 0.5^{2}0.25^{2}=\frac{63}{64}$$

Now for the second question we have more information.
The conditional will remain unchanged
$$P(\text{Both parents pass gene onto Eva}|\text{Both parents have the gene}) = 0.5^{2}$$
However, now that we know that Mark is healthy, our belief of the parents having the gene decreases. So now,

$$P(\text{Eva unhealthy}) = P(\text{Both parents pass gene onto Eva}|\text{Both parents have the gene}).P(\text{Both parents have the gene}|\text{Mark is healthy})$$

$$P(\text{Both parents have the gene}|\text{Mark is healthy}) = \frac{3*0.25^{2}*0.5^{2}}{3*0.25^{2}*0.5^{2} + 2*0.25*0.75 + 0.75^{2}}$$

$$P(\text{Both parents have the gene}|\text{Mark is healthy}) = \frac{1}{21}$$

Consequently,
$$P(\text{Eva healthy}|\text{Mark healthy}) =1- 0.5^{2}\frac{1}{21}=\frac{83}{84}$$


Chain rule for matrix exponentials I need help in proving the following theorem:

  If $M(t)$ is an $n \times n$ matrix of differentiable functions, then
  $$ \frac{d}{dt}\left( \exp(M(t))\right) = \frac{d}{dt}M(t) \exp(M(t)) = \exp(M(t)) M'(t) $$
  if and only if $M(t)$ and $\frac{d}{dt} M(t)$ commute.

Please this is not homework. I am reading about the chain rule for matrix exponentials and I came across it.

Clearly Avitus proves nothing at all !

The key is the derivative of $f(t)=M^k$. Indeed $f'(t)=\sum_{i=1}^k M^{i-1}M'M^{k-i}$. If $M'$ and $M$ commute, then $f'(t)=kM'M^{k-1}=kM^{k-1}M'$. $\exp$ is an entire function, then we can derive the associated series term by term.
 $(\exp(M))'=\sum_{i=1}^{\infty} (M^i/i!)'=\sum_{i=1}^{\infty}M'M^{i-1}/(i-1)!=\sum_{i=1}^{\infty}M^{i-1}/(i-1)!M'$ and we are done.

About the converse: assume that for every $t$, $(\exp(M))'=M'\exp(M)=\exp(M)M'$. Does this imply $M'M=MM'$ ? I don't know.

EDIT 1. Assume that if $\lambda,\mu$ are distinct eigenvalues of $M$, then $\exp(\lambda)\not=\exp(\mu)$ ; then $B$ and $M$ commute iff $B$ and $\exp(M)$ commute. Thus if $M$ is $2i\pi$ congruence free (that is, for every $\lambda,\mu\in spectrum(M)$, $\lambda-\mu\notin2i\pi\mathbb{Z}\setminus\{0\})$, then $M'M=MM'$. It remains to study the values of $t$ such that $M(t)$ is not $2i\pi$ congruence free. For instance take $M(t)=\begin{pmatrix}t^2&-2\pi\\2\pi&t\end{pmatrix}$. $\exp(M(0))=I_2$ commute with $M'(0)=\begin{pmatrix}0&0\\0&1\end{pmatrix}$ but $M'(0)$ and $M(0)$ don't commute. I think that, in general, there's not a lot of such values of t (here $t=0$); thus if we assume that $M'$ is continuous, then the result seems to be true. Moreover, note that $(\exp(M))'_{t=0}=1/2I_2$, that is the equality $(\exp(M))'=\exp(M)M'$ is not satisfied when $t=0$.

EDIT 2. Let $ad(M):H\in\mathcal{M}_n:H\rightarrow MH-HM$. Then the derivative of $\exp$ is $D(\exp(M))'(H)=\exp(M)\phi(M)H$ where $\phi(M)=\dfrac{1-\exp(-ad(M))}{ad(M)}$. Here $(\exp(M))'=\exp(M)\phi(M)M'=\exp(M)M'$, that implies $\phi(M)(M')=M'$, that is $M'$ is in the eigenspace of $\phi(M)$ associated to the eigenvalue $1$ (assume $M'\not=0$). Let $(\lambda_i)_i$ be the spectrum of $M$. Then the spectrum of $ad(M)$ is $(\mu)_{i,j}=(\lambda_i-\lambda_j)_{i,j}$ and the spectrum of $\phi(M)$ is $(\rho_{i,j})_{i,j}=(\dfrac{1-\exp(-\mu_{i,j})}{\mu_{i,j}})_{i,j}$. Note that $\rho_{i,j}=1$ implies $\mu_{i,j}=0$ that is the case if $i=j$ OR $\mu_{i,j}\not=0$ and $\exp(-\mu_{i,j})=1-\mu_{i,j}$, equation that has solutions in $\mathbb{C}$.

EDIT 3. OK , the solution is given in W. Ma ,B. Shekhtman. Do the chain rules for matrix functions hold without commutativity ? Linear and multilinear algebra.Vol 58, 1,79-87,2010.

$(A\exp(A)-\exp(A)A)'=0=A'\exp(A)+AA'\exp(A)-\exp(A)A'A-\exp(A)A'=[A,A']\exp(A)$ implies that $[A,A']=0$

Moreover $(\exp(A))'=A'\exp(A)$ does not imply $[A,A']=0$. Take $A=\begin{pmatrix}c&F(t)\\0&0\end{pmatrix}$ where $c\not=0,\exp(c)=1+c$. Note that $A'\exp(A)\not=\exp(A)A'$. The previous counter-example is in $M_2(\mathbb{C})$ ; yet one can deduce a counter-example in $M_2(\mathbb{R})$.

I show the first equality, to start with.

Using the definition

$$\frac{d\exp(M(t))}{dt}=\lim_{h\rightarrow 0}\frac{\exp(M(t+h))-\exp(M(t))}{h}=\lim_{h\rightarrow 0}\frac{\exp(M(t+h)-M(t)+M(t))-\exp(M(t))}{h}=\\
\left(\lim_{h\rightarrow 0}\frac{\exp(M(t+h)-M(t))-1}{h}\right)\exp(M(t));$$

Using the definition of derivative and exponential function

$$M(t+h)-M(t))=\frac{dM}{dt}h+O(h^2) $$
$$\exp(M(t+h)-M(t))=1+\frac{dM}{dt}h+O(h^2)$$

we arrive at

$$\frac{d\exp(M(t))}{dt}=\left(\lim_{h\rightarrow 0}\frac{\exp(M(t+h)-M(t))-1}{h}\right)
\exp(M(t))=\\
\left(\lim_{h\rightarrow 0}\frac{dM}{dt}+O(h)\right) \exp(M(t))=\frac{dM}{dt}\exp(M(t)).  $$


Prove that statements are equivalent Let $m,n \in \mathbb{N}$. Prove that following statements are equivalent:
(1) $m, n$ - mutually prime numbers;
(2) for every group $G$, any subgroup $A \subseteq G$ of order $m$ and any subgroup $B \subseteq G$ of order $n$ the following is true: $A \cap B =\{e\}$.

It is important to notice that $G$ is not necessarily a finite group.
I have following ideas on my mind:
$1 \rightarrow 2:$ if $\gcd(m, n)=1$ then as of Lagrange's theorem given that $Q=A\cap B$, $|Q|$ should be a divisor of $|A|$ and $|B|$, which is true only for $|Q|=1$, given that and the fact that every subgroup should contain unitary element we have: $A\cap B=Q=\{e\}$.
$2 \rightarrow 1:$ let's prove that by contradiction: if $\gcd(m, n)\neq 1$, then lets take group  $G=(\mathbb{Z}_{m\cdot n}, +)$, $A=(\mathbb{Z}_{m}, +)$ and $B=(\mathbb{Z}_{n}, +)$. Then $A\cap B=(\mathbb{Z}_{gcd(m, n)}, +)$, so if $\gcd(m, n)\neq 1$, then $A \cap B \neq \{e\}$, what lead us to the contradiction itself. We may say that, as in the statements there are words: 'every' and 'any'.
Are my ideas correct? I'm asking as my teacher said, that mentioned solution wouldn't work for infinite G and some other specific cases. Can you help me with fixing it? And may anyone explain me, why $A\cap B=(\mathbb{Z}_{\textbf{gcd(m, n)}}, +)$, we were given that with no proof, however I don't want to use it with no ideas why does it work.

Your "$1\to 2$" implication is fine.

The "$2\to 1$" implication is not precise enough for my taste. First, what exactly is $\mathbb{Z}_k$? For the sake of this post I will assume that $\mathbb{Z}_k=\{0,1,\ldots,k-1\}$ with "modulo $k$" addition.

Now you consider $\mathbb{Z}_{mn}$, that is fine. But neither $\mathbb{Z}_m$ nor $\mathbb{Z}_n$ is a subgroup of $\mathbb{Z}_{mn}$. The $\mathbb{Z}_{mn}$ group contains cyclic subgroups of order $m$ and $n$, but these are only isomorphic to $\mathbb{Z}_m$ and $\mathbb{Z}_n$, not literally equal. For example $\mathbb{Z}_6=\{0,1,2,3,4,5\}$ and $\mathbb{Z}_2=\{0,1\}$ but $\{0,1\}$ is not even a subgroup of $\mathbb{Z}_6$. The cyclic subgroup of order $2$ in $\mathbb{Z}_6$ is actually $\langle 3\rangle=\{0,3\}$. It is indeed, isomorphic to $\mathbb{Z}_2$, but not literally equal.

Now take again $G=\mathbb{Z}_{mn}$. The subgroups you are looking for are $A=\langle n\rangle$ and $B=\langle m\rangle$, which are of order $m$ and $n$ respectively. So how to show that $A\cap B\neq \{0\}$? This follows from the fact that if $G$ is a cyclic group and $d$ is a divisor of $|G|$ then $G$ contains a unique cyclic subgroup $H\subseteq G$ of order $d$. See this: Each finite cyclic group of order $n$ contains unique subgroup of order $d$ where $d\mid n$. And so both $A$ and $B$ contain cyclic subgroups of order $\gcd(n,m)\neq 1$. By the uniqueness this is one and the same subgroup. And so it is a subgroup of $A\cap B$, which thus is nontrivial.

For $\ 2\rightarrow 1$, we again use Lagrange's theorem.

$\ | A\cap B |$ divides $\ |A|$ and $\ |B|$ as the $\ A\cap B $ is subgroup of both $\ A$ and $\ B$. Now if $\ gcd(m,n)\ne 1$ then $\ |A\cap B| \ne 1\implies  A\cap B \ne \{e\}$. Contrapositively you can say if $\ A\cap B =\{e\}$ then $\ gcd(m,n)= 1$


Numerical solution to ${d\over dt} \nabla^2 p = {dp\over dx}$ I need help numerically solving
$$\frac{\partial \nabla^2 p} {\partial t} = \frac{\partial p} {\partial x}$$

I know that to solve

\begin{equation}
     \frac{\partial p} {\partial t} = \frac{\partial p} {\partial x}
\end{equation}

the finite difference formula is
\begin{equation}
     \frac{p_{i,j}^{n+1} - p_{i,j}^n} {\Delta t} = \frac{p_{i+1,j}^n - p_{i-1,j}^n} {2 \Delta x}
\end{equation}
This simplifies to
\begin{equation}
     p_{i,j}^{n+1} = r \left( p_{i+1,j}^n - p_{i-1,j}^n \right) + p_{i,j}^n
\end{equation}
where $r = \frac{\Delta t} {2 \Delta x}$, which allows me to explicitly calculate $p$ at the next time step from $p$ at the current time step.

Following the same procedure for $\frac{\partial \nabla^2 p} {\partial t} = \frac{\partial p} {\partial x}$, I get

\begin{multline}
     p_{i-1,j}^{n+1} + p_{i+1,j}^{n+1} + p_{i,j-1}^{n+1} + p_{i,j+1}^{n+1} - 4p_{i,j}^{n+1} = \\
     (1-r) p_{i-1,j}^n + (1+r) p_{i+1,j}^n + p_{i,j-1}^n + p_{i,j+1}^n - 4p_{i,j}^n
\end{multline}
where $r = \frac{\Delta \Delta t} {2}$ and $\Delta x = \Delta y = \Delta$. But now I must solve $p$ at five points simultaneously. I cannot simply march through the grid, updating each point as I go. How can I solve this equation cheaply and avoid inverting a matrix?

Here is a fast way to solve it. Take the 2D Fourier transform of the equation to get

$$\frac{\partial \hat{p}(\vec{k},t)}{\partial t} = -\frac{i k_x}{k^2}\hat{p}(\vec{k},t)$$

where $\hat{p}(\vec{k},t) = \mathcal{F}(p)$ is the Fourier transform of $p$. The solution of this equation is simply

$$\hat{p}(\vec{k},t) = u_0(\vec{k})e^{-\frac{i k_xt}{k^2}}$$

where I assume you have $u_0(\vec{k}) \equiv \mathcal{F}(p(x,y,0))$ as an initial condition at $t=0$. The solution in real space, at any time $t\geq 0$, can be found from

$$p(x,y,t) = \mathcal{F}^{-1}\left( u_0(\vec{k})e^{-\frac{i k_xt}{k^2}}\right)$$

where $\mathcal{F}^{-1}$ is the inverse Fourier transform. Numerically calculating the transforms is really fast and also easy to implement if you use, for example, the FFTW library. Note that if you only want the solution at one particular time-step then this method only requires two FFTs - one to get $u_0$ and one to get back to $p$ - no matter what time you are interested in (which is for example done in milliseconds for a $50\times 50$ grid).

Presumably you are using some boundary conditions such that $u$ uniquely determines $p$ (otherwise the problem is not well-posed).
Since you didn't specify that, I'll assume homogeneous Dirichlet boundary conditions.  Solve (in a discretized version) with these boundary conditions
$\nabla^2 p = e_{ij}$ for each grid point $(i,j)$, where $e_{ij}$ is $1$ at
$(i,j)$ and $0$ everywhere else, obtaining the Green's functions $g(x,y,i,j)$.
Then given $u$, the solution of $\nabla^2 p = u$ is $p(x,y) = \sum_{i,j} u(i,j) g(x,y,i,j)$.  Now given $p^n$, let $u^n = \nabla^2 p^n$, $u^{n+1} = u^n + (\Delta t) \partial_x p^n$, and $p^{n+1} = \sum_{i,j} u^{n+1}(i,j) g(\cdot, \cdot, i, j)$.


Can any Real number be typed in a computer? Suppose that we have a computer program, my question is whether a human can type in any real number - say in $[0,10]$ - that she would like to type in a finite amount of time? Suppose that the program allows typing in any basic expressions like sum, multiplication, square-root, limits etc.

in a more formal way:
Choose one real number $x\in [0,10]$, is it possible to express $x$ in a way that a computer can understand*?

*can understand = if $y$ is another such number, computer can tell whether $x>y, y>x$ or $x=y$.

EDIT: For example, square-root 2 can be typed as $\sqrt{2}$ and that is okay. As asked in comments, computer does not have an infinite memory, What is imporant is that the program can distinguish any two numbers that are typed in. For example, even though it is not possible to represent $\sqrt{2}$ in a computer, it can understand that $\sqrt{3}$ is bigger than $\sqrt{2}$.

When you have finite or countable many symbols:

No, when you only allow symbols from a finite (or countable) set of possible symbols like $\lim$, $\sqrt{}$, digits and so on, the set of all possible terms (with finte length) someone can type is countable. But there are uncountable many reals in $[0,10]$. Thus there are reals which are "untypable"...

This is related to definable numbers: There are only countable many expressions to define a number in first order logic but there are uncountable many reals -> There are undefinable numbers. See also the answers and comments to the question Is there an example for an undefinable number? for examples of undefinable/uncomputable numbers.

When you have uncountable many symbols:

Then you can assign to each real number a different symbol which stands for it (like the symbol "$\pi$" stands for the number $\pi$) and voila: Every real can be typed by using its symbol ;-)

(Let's assume that the continuum hypothesis is true as a axiom for this answer: There is no uncountable set with cardinality less then the cardinality of $\mathbb R$)

Update: There are computable numbers $x$ for which you cannot decide whether $x=0$ or $x > 0$. To cite a comment of Robert Israel to this answer:

  Consider a predicate $S(n)$ that can be computed for each natural number $n$. Define $x=\sum_{n=1}^\infty 2^{−n}d(n)$, where $d(n)=1$ if $S(n)$ is true and 0 otherwise. Since each $S(n)$ can be computed, $x$ is computable in the sense that arbitrarily good rational approximations of $x$ can be computed. Now it might be that all $S(n)$ happen to be false, but there is no proof (in your favourite consistent formal system) of this fact. Then it is impossible to decide (in that system) whether $x=0$ [or $x > 0$].

Any real number $\xi$ you, or a more experienced mathematician, can think of can be defined and described in terms of text and universally accepted formulas, whence: be expressed in TeX-code on less than two ASCI-pages or so. Any computer will accept this as input. Such an input determines $\xi$ as an element of ${\mathbb R}$ once and for all, in other words: to  any desired number of decimal places. Of course it may be the case that your pocket calculator will not be sufficient to compute these digits one after the other.


How find the number of $z$,such that$ |a^2-b^2-b+1|\le 10$ let $f:C\longrightarrow C$ and $f(z)=z^2+zi+1$,find the number of $z$,such $Im(z)>0$
and $|Im(f(z))|\le 10,|Re(f(z))|\le 10$,and $ Im(f(z)) ,Re(f(z)) $are integer numbers.
where $i=\sqrt{-1}$,

$A:399,   B:401,  C:413,   D:431,   E:441$

my idea:let $z=a+bi,b>0$, then
$$f(z)=z^2+iz+1=(a+bi)^2+i(a+bi)+1=(a^2-b^2-b+1)+(2ab+a)i$$

so:
$|a^2-b^2-b+1|\le 10,|2ab+a|\le 10$,and
$a^2-b^2-b+1\in Z,2ab+a\in Z,b>0$
then this problem equalitly find the number pairs$(a,b)$? But I can't find .

Thank you everyone,

If this is a competition or homework or exam and you will use anything below, you should clearly indicate it in your answer. It is a matter of honor. Or stop reading.

We are told that the real and imaginary parts of $f(z)$ are integers not greater than 10 in the absolute value and the imaginary part of $z$ is positive. So the imaginary part of $f(z)$ is between $-10$ and $10$ (21 possibilities). The real part is between $-10$ and $+10$ (21 possibilities). When multiplied, we have $21\times 21=441$ possibilities for what $f(z)$ may be. But we're not finished yet.

Now, we want to  count the possibilities for $z$. Here, $z^2+zi+1=f(z)$ for a fixed $f(z)$ is a quadratic equation that typically has two distinct solutions for $z$; the map is mostly two-to-one and about one of the two solutions has $f(z)\gt 0$. There are a few subtleties, however. In some cases, the two solutions may coincide. And we must take care of the special case ${\rm Im}(z)\sim 0$; some of those values of $z$ have to be eliminated (see below).

First, the quadratic equation can't have double (degenerate) roots because the discriminant $b^2-4ac$ only vanishes for $f(z)=5/4$ which hasn't an integral real part so this is not a problem. Also, the pair of solutions for one choice of $f(z)$ is always different from any element of any pair for another value of $f(z)$ because the solutions are given as $1/2(-i\pm \sqrt{4f(z)-5})$. It is self-evident that different values of $f(z)$ produce different values of $z$ symmetrically distributed from $-i/2$.

So the only problem is to look at the values of $f(z)$ for which the imaginary part of $z$ isn't positive exactly for one of the two solutions. But that may only occur if $|{\rm Im}(\sqrt{4f(z)-5})|\leq 1$ (the factors of $1/2$ cancel). That may occur if the square root is close enough to the real axis. It means that $4f(z)-5$ must be close enough to the positive real semiaxis.

One has to go through the few dozens of numbers near the positive real semiaxis. We eliminate all the roots for $f(z)$ equal to $1,2,3,\dots 10$ (ten values of $f(z)$), and $2\pm i,3\pm i,4\pm i,\dots 10\pm i$ (eighteen values of $f(z)$ in total here), and $5\pm 2i,6\pm 2i,\dots 10\pm 2i$ (twelve values), $10\pm 3i$ (two values). Note that all the initial numbers $1;2\pm i; 5\pm 2i; 10\pm 3i$ were eliminated "marginally" and would stay there is one allowed all non-negative ${\rm Im}(z)$ rather than the strictly positive ones.

So for $10+18+12+2=42$ values of $f(z)$, there are no acceptable solutions $z$ because of the condition on the imaginary part of $z$. For the remaining $441-42=399$ of $f(z)$ values, there is a unique solution per point and they're all distinct. So the right answer is A:399.

Your approach was counterproductive. There's no reason to divide the complex number $z$ to its real and imaginary part; on the contrary, it's right to view this number as a unified complex number for as long as possible a part of the calculation. The simple grid-like integrality condition affected $f(z)$ and not $z$ which is why we looked at the grid in the $f(z)$ plane and not the $z$ plane as you proposed.

From these: $b>0$ and $|2ab+a|\leq 10$ it follows:
$$-10\leq 2ab+a\leq 10 \iff \frac{-10-a}{2}\leq ab\leq\frac{(10-a)}{2}. $$
Then you should take three cases: when $a>0$, $a=0$ or $a<0$. Just for an example of a needed pair let's take $a=0$. Then everything above holds and $|a^2-b^2-b+1|=|-b^2-b+1|$, so $|-b^2-b+1|\leq 10$,  for example $b=1,2$ and pairs $(a,b)=(0,1), (0,2)$. But you can find more pairs by evaluating inequalities.


In M/M/1 Markov process, why must entering and leaving the zero state be equal? According to the image below, which I snipped from this article, the rate of leaving State 0 and the rate of arriving into State 0 are equal. The article simple asserts this but does not explain why.

Conceptually, this makes no sense to me, as there are many cases where $\lambda$ is greater or less than $\mu$.

Is there something I am missing? Why must they be equal?

Define:

(i) $P_{{\bf a},t}{}:={}P(1\,{\bf a}\mbox{rrival in }(t,t+\delta t]\,\vert \, \mbox{system in state 0 at time }t )$;

(ii) $P_{{\bf c},t}{}:={}P(1\,{\bf c}\mbox{ompletion in }(t,t+\delta t]\,\vert \, \mbox{system in state 1 at time }t )\,;$

(iii) $\pi_{i,t}{}:={}P(\mbox{ system in state }i\mbox{ at time }t)\,.$

At equillibrium, for each system state, the probability of the system being in the given state does not change with time. Therefore,
$$
\lim\limits_{\delta t\to 0}\dfrac{\pi_{0,t+\delta t}{}-{}\pi_{0,t}}{\delta t}{}={}0\,.
$$

This means that,

$$
\begin{eqnarray*}
&\overbrace{\pi_{0,t+\delta t}}^{\begin{array}{c}probability\,system\\is\,in\,state\,0\,at\,time\\t+\delta t\end{array}}{}={}\overbrace{\pi_{0,t}\left(1{}-{}P_{{\bf a},t}\right)}^{\begin{array}{c}probability\,system\\undergoes\,no\,change\\from\,state\,0\,in\,(t,t+\delta t]\end{array}}{}+{}\overbrace{\pi_{1,t}P_{{\bf c},t}}^{\begin{array}{c}probability\,system\\undergoes\,change\\from\,state\,1\,to\,0\,in\,(t,t+\delta t]\end{array}}&\newline
&&\newline
{}\implies{}&\lim\limits_{\delta t\to 0}\dfrac{\pi_{0,t+\delta t}{}-{}\pi_{0,t}}{\delta t}{}={}-\pi_{0,t}\lim\limits_{\delta t\to 0}\dfrac{P_{{\bf a},t}}{\delta t}{}+{}\pi_{1,t}\lim\limits_{\delta t\to 0}\dfrac{P_{{\bf c},t}}{\delta t}\newline
&&\newline
{}\implies{}&0{}={}-\pi_{0,t}\lambda{}+{}\pi_{1,t}\mu\,,&
\end{eqnarray*}
$$

from which the assertion follows.

I figured out the answer to my confusion.

To recap, I was wondering why it must be that $$P_0\lambda = P_1\mu$$

It seemed strange to me, since I knew that $\lambda$ and $\mu$ could be different from one another and $P_0$, $P_1$ could be anything else. But I asked, why should the outflow of State $0$ be equal to the inflow to $0$ via State $0$.

Solution

What I realized is that $State \ 1$ is directly dependent on the outflow from $State \ 0$. Therefore the probability of $State \ 1$ even ever existing is built in to the expression $P_0\lambda$. So it makes sense to say that the flow back from $1$ - given that we are in State 1 - to $0$ is be equal to the flow from $0$ to $1$

To put it into words:

The probability of being in $State \ 0$ and moving to $State \ 1$ is equal to the probability of being in $State \ 1$ and moving to $State \ 0$


Is there a notion of similarity for shapes on the sphere? There are similar shapes in the plane, such as similar triangles. So is there a similar shape on the sphere? For example, is a great circle similar to a small circle on a sphere? I think great circles and small circles are similar, so there are similar shapes on the sphere. So as shown in the figure, are there similar two sides on the sphere?

Here is my opinion:

Can two rectangles of different sizes be similar on a plane? Obviously, rectangles can be similar. Under what conditions are they similar?

The proportion of their corresponding sides is equal;
Their corresponding angles are equal.

These two conditions must be met at the same time.

Squares are also similar because they meet both conditions 1 and 2.

The conditions for the similarity of rectangle and triangle are not exactly the same. Although, we can think of each rectangle as consisting of two triangles.

In the plane, two regular polygons with the same number of sides are similar because they meet the requirements of conditions 1 and 2 at the same time.

We can think that any two circles are regular polygons with the same number of sides, so any two circles are similar. Because they meet both conditions 1 and 2. (their corresponding angles are all π).

If we take two points on each circle, we get some shapes with two sides. Can these shapes be similar? Obviously they can be similar. As long as they meet the requirements of condition 1 (their corresponding angles are π).

Can two shapes formed by closed curves be similar? I think they can be similar, and the conditions of judgment are similar, but the judgment is more complicated.

Are two circles similar on a sphere? Obviously they are similar. Because they meet both conditions 1 and 2.

On a sphere, are the two shapes shown in the figure similar (they are composed of arcs)? Obviously they can be similar as long as they meet the requirements of conditions 1 and 2 at the same time.

All is exploration. I'm not sure what I said is right. I hope you can talk about your knowledge.

Similarity can be tricky business.

Instead of the broad category of "shapes", first consider triangles.

In Euclidean geometry, "similar" triangles have the "same shape", a notion we clarify with the properties

  Corresponding angles are congruent.
  Corresponding sides are proportional.

Conveniently, property (1) implies (2), and vice-versa. Perhaps because angle-congruence is easier to check, "angles are congruent" is sometimes taken as the defining characteristic of similar triangles, with "sides are proportional" being a useful consequence. It doesn't seem to matter. Of course, we when consider "similar polygons", the logical equivalence breaks: squares and, say, golden rectangles have congruent angles, but their sides aren't proportional; conversely, squares and non-square rhombi have proportional sides, but their angles aren't congruent. "Same shape"-ness for polygons in general requires both (1) and (2).

Defining "same shape"-ness for curves is a little more nuanced, since we can't compare angles or sides directly. (We can get technical and establish a "similarity transformation" involving a dilation.) Nevertheless, it's "obvious" that all circles have the "same shape", so if the descriptor "similar" applies to any figures, it certainly applies to them.

On the sphere, triangles already drive a wedge between (1) and (2). If corresponding sides are proportional, but not congruent, the corresponding angles are not congruent. (Specifically, larger sides make for larger angles. For example, consider a tiny-tiny equilateral triangle, whose angles are close to $60^\circ$ each, and the equilateral triangle joining the North, "East", and "West" Poles, whose angles are $90^\circ$.) Contrariwise, if corresponding angles are congruent, then the corresponding sides are congruent; "Angle-Angle-Angle" is a congruence pattern! This phenomenon can be summarized as

  "There are no similar triangles in spherical geometry."

which is shorthand for "There are no similar triangles ---in the sense of (1) and (2)--- that aren't fully congruent triangles, so we have no use for the term 'similar'".

But what about circles?

In the vague sense of "same shape"-ness, then all circles on the sphere should be considered "similar". In the technical sense of "similarity transformation"-ability, too: simply align centers with an isometry and apply an appropriate dilation. Case closed!

And yet ...

On the (unit-radius) sphere, a spherical circle with radius $\theta$ as measured along the surface of the sphere is a plane circle with radius $\sin\theta$ as measured on the plane containing the circle; such a circle has circumference $C = 2\pi\sin\theta$. But, then, a circle with (surface) radius $2\theta$ has circumference $2\pi\sin 2\theta \neq 2C$: doubling the radius does not double the circumference; indeed, the circumference could get smaller! Likewise for other scale factors (apart from $1$).

This isn't the kind of behavior we've come to expect from "same shape" figures in the Euclidean plane. In light of this situation, we recognize that our (1) and (2) are actually insufficient to capture the entirety of our expectations. We find that a refinement is in order:

  2'. Corresponding lengths are proportional.

Here, "length" isn't limited to just sides. It applies to medians, altitudes, arbitrary cevians, midpoint segments, etc, for triangles; diagonals of quadrilaterals and polygons; perimeters for everything; radii and circumferences for circles; and on and on and on. Here again, Euclidean geometry spoils us: figures that satisfy (1) and (2) ---and/or admit a similarity transformation--- automatically satisfy (2'); it's a freebie. Spherical geometry, however, reveals (2') to be a stronger condition of "same shape"-ness; a condition that spherical circles do not satisfy unless they are fully congruent.

In the context of $(2')$, we find that, apart from segments,

  "There are no [non-congruent] similar figures at all in spherical geometry."

If you choose to reject $(2')$ as a requirement, then arbitrary circles might reasonably be called "similar", but in a way that's almost-no help to further investigation and that's almost-certain to cause confusion. It would be better to apply a different descriptor ("quasi-similar"?).

It should not be said that the great circle is geodesic, so we think that the great circle and the small circle are not similar. Because we can also ask: are two small circles with different diameters similar? When the diameter increases, the small circle can coincide with the great circle, and the small circle can coincide with the small circle. So I think: on the sphere, any two circles are similar, whether they are great circles and small circles, or small circles and small circles. Only in this way can our geometric theory be unified, otherwise our geometric theory will be split.


Sequences are Cauchy depending on the norm? Assume that we have two Banach spaces $B_1, B_2$ with their underlying sets being identical. Is it possible that a Cauchy sequence in one of the spaces would fail to be Cauchy in the other? I am assuming that it is possible since the Cauchy test directly depends on the norm. However, this seems kind of unnatural, will someone please clarify.

Here are some  details to add to Jonas' answer in the link of my comment above.

Let $V$ be a vector space of Hamel dimension $c$. Let $T$ be a linear bijection from $V$ to $\ell_1$ and let $S$ be a linear bijection from $V$ to $\ell_2$.  This can be done since any two separable infinite dimensional
 Banach spaces have  Hamel dimension $c$.

Define two norms on $V$ via
$\Vert x\Vert_T=\Vert Tx\Vert_{\ell_1}$ and
$\Vert x\Vert_S=\Vert S
x\Vert_{\ell_2}$. Then $(V,\Vert\cdot\Vert_S)$ is isomorphic to $\ell_2$  and  $(V,\Vert\cdot\Vert_T)$ is isomorphic to $\ell_1$. In particular, these two norms are complete.

Since $\ell_1$ is not isomorphic to $\ell_2$, it follows that these two norms are not equivalent. Per Julien's answer, these two norms have different Cauchy sequences.

It's definitely possible. Consider the real numbers with (1) the usual metric and (2) the discrete metric. Both are complete, yet the only Cauchy sequences in (2) are the eventually constant sequences, while (1) has many more Cauchy sequences.


Reduce power modulo a semiprime quickly by hand I worked through problem 7 in this GRE prep material (https://math.uchicago.edu/~min/GRE/files/week1.pdf), and I got the right answer, but I'm wondering if there is a more efficient way to do this, particularly considering that the GRE math subject test doesn't permit using a calculator. The question is to find the remainder upon dividing $ 13 ^{2019} $ by $95$. So we must solve for $x$ in $ 13^{2019} \equiv_{95} x.$ Since 95 factors into 5*19, I first found each of those residues.

$ 13^{2019} \equiv_{5} {{(13 \pmod 5})}^{2019 \pmod{\phi(5)}}\equiv_{5} 3^{2019 \pmod{4}}\equiv_{5} 3^3=27\equiv_{5}2.$
$ 13^{2019} \equiv_{19} {13}^{2019 \pmod{\phi(19)}}\equiv_{19} {13}^{2019 \pmod{18}}= 13^3\equiv_{19}(-6)^3=36*(-6)\equiv_{19}(-2)*(-6)\equiv_{19}12.$

Since $2 \equiv_5 12$, that means $13^{2019}$ is equivalent 12 with respect to each modulus, which means that $13^{2019} \equiv_{95} 12.$
Are there any other techniques I could have used to get the answer more quickly?


Convergence of the series $\sum\limits_n \left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}} \right)^n$ for $\epsilon>0$ This is not a homework problem. It has come up in my research. I am trying to show that $$\sum_n \left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}} \right)^n$$ converges for  $\epsilon>0$. I have no idea how to approach the problem.

Start by an asymptotic expansion of $\log\left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}}\right)$, that is
$$\log\left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}}\right)=-\frac K{n^{1-\varepsilon}\sqrt{\log n}}-\frac{K^2}{2n^{2-2\varepsilon}\log n}(1+\delta_n),$$
where $\lim_{n\to\infty}\delta_n=0$. This gives
$$\exp\left(n\log\left(1-\frac{K}{n^{1-\epsilon}\sqrt{\log n}}\right)\right)=
\exp\left(-\frac{n^\varepsilon}{\sqrt{\log n}}-\frac{K^2}{2n^{1-2\varepsilon}\log n}(1+\delta_n)\right).$$

The main term is of order
$$a_n=\left(1-\frac{1}{n^{1-\epsilon}\sqrt{\ln n}}\right)^n\sim e^{-\dfrac{n^\varepsilon}{\sqrt{\ln n}}}\leq e^{-n^\varepsilon}\leq \frac{1}{n^2}$$
for arbitrary large $n$, so the series converges.


Could the Monty-Hall Problem be applied to multiple choice tests? Given a multiple choice test where each question contains 4 possible answers, what would happen if before beginning the test (before reading the questions), someone were to make a random selection for each question?

At this point it seems logical that for a given question the student has a 1/4 chance of their choice being correct and a 3/4 chance of one of the other choices being correct.

Let's say that they now begin to read the questions and in some cases they can deduce that one of the provided answers which was not the one that they picked is not correct (let's assume that there is no error in this deduction). In the scenario with the Monty Hall Problem, the probabilities did not change once the door was opened, they just shifted.

By applying the same logic, the original selected answer has a 1/4 chance of being correct and the other three have a 3/4 chance of being correct, except that since one was deduced to be incorrect, the two remaining options have a 3/4 chance of being correct and so switching answers would increase the odds of being correct to $\frac{1}{2} * \frac{3}{4}$.

Is this an accurate assumption or are there pitfalls in doing this?

If this is the case, then what happens if another deduction is made such that their original answer was determined to be incorrect? It seems that there would be no change in the odds, but that seems unlikely.

Instinct tells me no - switching shouldn't make a difference, but instinct is what gets most people the incorrect answer for Monty Hall.

However, I'll try to explain it:

In the Monty hall problem, the host chooses which doors to show you (or to show you which answers are incorrect).

In your exam, the person who can "show you" incorrect answers is you. You do not, unlike the host, have knowledge of what the incorrect/ correct answers are which I think is the fundamental difference between the situations. Therefore, finding out whether an answer is incorrect is random. You have an equal chance of finding out that the answer you selected was incorrect as finding out one of the other answers was incorrect. In Monty Hall, the host will never tell you that the door you selected was a goat.

You are assuming you read the question and answer before picking a answer. This makes the probability void.
FIRST you pick an answer randomly, THEN you check if it may or may not be correct.
4 possible answers, P(Correct)= 1/4 P(FALSE)= 3/4
meaning there is 75% chance that the answer is in the other 3 answers (you have not looked at the content as of yet!)
Now you read. Assuming that you KNOW that 1 of the other 3 answers is wrong, then there is only 2 answers left to split 75% -> (3/4)*(1/2)=3/8 37.5%
Assuming you can tell the answer YOU randomly picked is wrong, you can just pick one of the other 3 anyway, 33% chance of correct answer

This only works when you picked an answer under BLIND circumstances, and that the answers left have NO info what so ever, that they are indistinguishable from each other, which in MCQs are almost never correct.

NOTE*** if you can not gleam ANY info from the question, stick to your original answer, with gives P(C)= 25%, since picking erratically is known to lower the your score


Prove that $ \ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x}$? How can one prove that
$$\ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x} \, ?$$

Considering $f(t) = \log t$ and applying Lagrange's theorem in the interval $[x,x+1], x > 0$ you get
$$
f(x+1) - f(x) = f'(\xi_x) (x+1-x) = \frac{1}{\xi_x}, \quad \xi_x \in ]x,x+1[
$$

But, since $f(x+1)-f(x)= \log (x+1)-\log x = \log \left(\frac{x+1}{x}\right)$ and $\frac{1}{\xi_x} > \frac{1}{x+1}$ you can conclude that

$$
\log \left(\frac{x+1}{x}\right) > \frac{1}{x+1}.
$$

This is a heuristic:

Assuming $x>0,$ we have $$x\log\left(\frac{1+x}{x}\right)>\left(\frac{1+x}{x}\right)^{-1}.$$ Letting $A=1+1/x,$ we see that we need to show that $$xA\log A>1.$$ Now, since $A>1$ for all $x>0,$ but so that $A\to 1$ as $x\to \infty,$ and that $\log A$ therefore goes to $0$ in this limit, we need only show that $x\log A$ stays more than $1$ as $x\to \infty.$ That is, the limiting value must not fall below $1.$ But $$x\log A=\log\left(1+1/x\right)^x,$$ which indeed goes to $1$ near infinity, so that the inequality always holds for $x>0.$


Is the set of all definable subsets of the natural numbers recursively enumerable? I asked myself similar questions before, for example "Are the definable real numbers countable"? It seemed to me that the set of all explicitly and unambiguously definable objects is "countable", because we could write down the text defining a specific object in the English language. (Or use the English language to define a formal language appropriate to write down the definition.) But English isn't unambiguous, and the meaning of "countable" depends on the used set theory.

Right now I'm thinking about second order logic. Assuming that the natural numbers exist and are well defined seems unproblematic to me. However, the question which subsets of the natural numbers (should we assume to) exist has no good answer. If we say that "all" subsets exist, we run into all sorts of set theoretic questions. If we resort to axiomatic set theory for these questions, we end up with the fact that the natural numbers can't be defined up to isomorphism. The alternative seems to be to assume that only the definable subsets of the natural numbers exist. But my guess is that this won't work out either, i.e. that the "set of all definable subsets of the natural numbers" turns out to be a very strange beast. But if it is a strange beast, I guess it won't be recursively enumerable.

Question
Assuming the Church-Turing thesis (or a suitable analogue for definability, or some other sufficiently solid foundation), what can be said about the "set of all definable subsets of the natural numbers"? Is it well defined and recursively enumerable?

Edit It has been pointed out that my use of "recursively enumerable" is ambiguous without further clarifications, especially that I have to specify how a definable subset of the natural numbers is expected to be encoded as a natural number. Given a natural number, take its binary representation and interpret it as a document in the "OpenDocument" format (ODF). Assume that we can efficiently decide whether the document is valid ODF and whether its content is written sufficiently clear and tries to define a subset of the natural numbers. Let's call this a well formed definition. Assume further that each definable subset has at least one well formed definition which can be recognized to succeed in defining a subset of the natural numbers. Now the question is whether there exists a recursively enumerable subset $S$ of the natural numbers such that each number in $S$ corresponds to a well formed definition which succeeds, and that for each definable subset the set $S$ contains a corresponding definition.

The question is somewhat ambiguous about exactly how the sets would be enumerated. The natural way to read it in computability theory is the following, which is the usual sense in which a sequence of sets can be r.e.:

  Is there a computable double enumeration $\{ n_{i,j} : i,j \in \omega\}$ such that for each definable set $S \subseteq \omega$ there is an $i_0$ such that $S = \{ n_{i_0,j} : j \in \omega\}$.

The answer to that is certainly "no", because if it was true then every definable set would be r.e., which is not the case.

As long as every "definable" number may be defined using a finite string in some finite alphabet you may sort those strings by length and enumerate numbers using this sort.


Minimal model of ZFC without power set axiom We know that $L$ is the minimal standard model of ZFC.

The question is, what is the minimal "standard" model of ZFC$^-$ (meaning ZFC without the Power Set axiom)?

This is really two questions:

Is there such a minimal model (this is non-trivial, but sound intuitively true to me)? If so, let's call it $L^-$.
Assuming $L^-$ exists, is it equal to, or is it a proper subset of $L$ (obviously $L$ itself is a model of ZFC$^-$ so $L^- \subseteq L$)?

Both possible answers for (2) seem puzzling to me. If $L^-=L$ then somehow the Power Set axiom is sufficiently embedded in ZFC$^-$, that even if we don't assume it, it kind of crops up by itself in the minimal model (kind of the way AC and GCH do in $L$).

If $L^-\subsetneq L$ then a whole host of new questions pop up - like whether we have AC if we only assume ZF$^-$ to begin with, and what happens to all the other nice properties $L$ has. Furthermore, if $L^-$ doesn't have the exact same cardinals as $L$, then it appears to me we have some ordinal in $L$ that the power set axiom was necessary to show it has the same cardinality (which sounds peculiar because the bijection has the same cardinality, so why did we "need" anything bigger to construct it? I would have thought the axiom of replacement would be enough).

This isn't an answer, but a bit long for a comment; hope it helps.

First, there is nothing 'obvious' about the existence of a minimal model.
In practice, you could very well be able to prove the existence of some set $X$ such that you can't prove what is exactly its structure, such that there are two models of $X$, one being 'large' and the other 'deep' (in Hamkins words, one fat and one tall, more on that below), so that there is no embedding of one in the other, and both are minimal. So there's some (nontrivial) work to do to show that.

You could generalize this question of non-embedability to "are there two models that can't be embedded into each others?". Turns out that this question was answered negatively for ZFC by Joel D. Hamkins [0]. Now, since ZFC minus powerset is a strictly weaker axiomatic system, the result could not apply. It turns out this specific axiomatic system was also studied by Hamkins and co-authors [1], and results in the paper include reasons to be worried.
Now your best bets are to either work the answer as a corollary of these papers, or contact him directly.

[0] J. D. Hamkins, “Every countable model of set theory embeds into its own constructible universe,” J. Math. Log., vol. 13, iss. 2, p. 1350006, 27, 2013.

[1] Gitman, Victoria, Joel David Hamkins, and Thomas A. Johnstone. "What is the theory ZFC without power set?." arXiv preprint arXiv:1110.2430 (2011).

In ZFC you can not prove the existence of a minimal model, even assuming ZFC is consistent. Since you would have a minimal model of ZFC if you had your $L^-$, by taking closure under the operation of power set, you can not have $L^-$ either (as a consequence of ZFC).

So, basically, we can't know the answer to your question without assuming consistency of a theory not strictly contained in ZFC. I'd be weary of any such proof.


$f_{X,Y}(x,y)=\frac{1}{2}xyI_{(0,x)}(y)I_{(0,2)}(x)$ Find the marginal distribution of $X\mbox{ and }Y$ Consider two random variables $X$ and $Y$ having a joint probability density function

$f_{X,Y}(x,y)=\frac{1}{2}xyI_{(0,x)}(y)I_{(0,2)}(x)$

(a) Find the marginal distribution of $X\mbox{ and }Y$

$f_y(y)=(y) $

http://www.wolframalpha.com/input/?i=+%28integral+%28xy%29%2F2+dx++from+0+to+2+%29

$f_x(x)=(x^3/4) $

http://www.wolframalpha.com/input/?i=+%28integral+%28xy%29%2F2+dy++from+0+to+x+%29

(b) Are $X$ and $Y$ independent?

Given that $f_x(x) * f_y(y) = (x^3/4)* (y) \not=  (xy)/2 = f_{xy}(x,y) $ then $X$ and $Y$ aren't independents

Am I right???

(c)What is $E(Y)$?

Thanks for your time

I try to fix my previous answer, which was obviously wrong:

In order to calculate $E(y)$, one can use directly the joint distribution:

$$E(y) = \int_0^2 \int_0^x y f(x,y) dy dx = \int_0^2 \int_0^x \frac{xy^2}{2} dy dx = \int_0^2\frac{x^4}{6}dx = \frac{2^5}{30}=\frac{16}{15}$$


Explaining Horizontal Shifting and Scaling I always find myself wanting for a clear explanation (to a college algebra student) for the fact that horizontal transformations of graphs work in the opposite way that one might expect.

For example, $f(x+1)$ is a horizontal shift to the left (a shift toward the negative side of the $x$-axis), whereas a cursory glance would cause one to suspect that adding a positive amount should shift in the positive direction. Similarly, $f(2x)$ causes the graph to shrink horizontally, not expand.

I generally explain this by saying $x$ is getting a "head start". For example, suppose $f(x)$ has a root at $x = 5$. The graph of $f(x+1)$ is getting a unit for free, and so we only need $x = 4$ to get the same output before as before (i.e. a root). Thus, the root that used to be at $x=5$ is now at $x=4$, which is a shift to the left.

My explanation seems to help some students and mystify others. I was hoping someone else in the community had an enlightening way to explain these phenomena. Again, I emphasize that the purpose is to strengthen the student's intuition; a rigorous algebraic approach is not what I'm looking for.

Here is a setting where the students will have experience, having nothing to do with skills in algebra: daylight savings time (since you're in the USA).  When we advance clocks ahead by an hour we lose an hour of sleep.  Thus replacing $t$ with $t+1$ shifts your schedule for the day back by one hour.

Here, $t$ is the old "function" and $t + 1$ is the new "function". $t$ is one hour behind $t + 1$, but after this transformation, $t + 1$ shortens the day by one hour. So in the end, we lose an hour.  Point out to the students that it also works the other way when we turn the clocks back ($t$ is replaced with $t-1$) and then gain an hour in our schedule.

Just as an extension to the KCd's answer, we can think of this not only as daylight saving problem, but also as a timezone problem. For example, if you leave New York for Singapore you have to shift time on your watch $+12$ hours in order to keep up with local Singapore time. But if you want to express Singapore time in terms of New York time (due to the jet lag, perhaps) you could say "it's $x - 12$ hours in Singapore now in terms of New York time" or "it's $x - 12$ hours in New York now".


Proof that spanning set is greater than linearly independent subset Let G be a set that generates V containing n vectors and H be a linearly independent subset of V containing m vectors. Prove that $m \le n$.
My attempt:
We will use a proof by induction with base case $m=0$.
$m=0$ implies that $H =∅$ and since m cannot be less than $0$, $m \le n$.

Our induction hypothesis is that the theorem holds for some $n>0$. We need to show that it holds for $m+1$, ie. $m+1 \le n$.

Suppose the opposite, that $m+1 \ge n$ $\implies m>n$ which cannot be true but I am not sure how to prove it.
Any help is appreciated!

If $G = \{ v_1, \cdots, v_n\}$ generates $V$, then $n \geq\dim V $ by the definitions of basis and dimension. If $H = \{w_1, \cdots, w_m \}$ has $m$ linearly independent vectors, again by the definitions of basis and dimension we have $m \leq \dim V$. So $m \leq \dim V \leq n$, as we wanted to prove.


How does $ x^{2} -xz- y^{2}+yz $ become $(x-y)(x+y-z)$? I need some help in simplifying this expression. I want sums of terms into products instead.
$$ x^{2} -xz- y^{2}+yz $$
The answer is supposed to be:
$$(x-y)(x+y-z)$$
I just don't understand how, I tried doing completing the square with no success. I can't find anything to factor out that's common amongst all the terms.

Factorize $x^2-y^2$ and $yz - xz$ separately to find some like terms. $x^2-y^2$ becomes $(x-y)(x+y)$ and $yz-xz$ becomes $z(y-x)$. You can convert $z(y-x)$ to $-z(x-y)$. Now we can take $(x-y)$ common, so we get $(x-y)(x+y-z)$. Hence, we get our answer. Hope it helps

Writing your expression $(x^2-y^2)+(yz-xz)$ make things clearer. Indeed, we have:
$$x^2-y^2=(x-y)(x+y)$$
$$yz-xz=(y-x)z=(x-y)(-z)$$
Therefore, your expression is equal to $(x-y)(x+y)+(x-y)(-z)$. Can you figure out the last step?


Convergence of Series Whose Terms are Defined Recursively My recursively defined sequence $(a_n)_{n\in\mathbb{N}}$ is given trough

$$a_1 = 1, \quad a_2=\frac{1}{2}\quad a_{n+2}=a_{n}a_{n+1}\quad \text{for } n\geq1$$

and I have to show that the series

$$\sum_{n=1}^{\infty}a_n$$

converges.

I have 4 questions to my approach(es) I did to deal with that:

I noticed that $a_n$ can be represented as $$a_n = 2^{-f_{n-1}}$$
whereas $f_n$ is the $n$-th Fibonacci number. For this claim I did this proof by induction:

$$ n=1: a_1 = 1 = 2^0 = 2^{f_0} = 2^{-f_0} = 2^{-f_{1-1}} \\
 n=2: a_2 = \frac{1}{2} = 2^{-1} = 2^{-f_1} =  2^{-f_{2-1}}$$
and by multiplying the premises $A(n)$ and $A(n+1)$ in order to get $A(n+2)$
$$"n+1\to n+2":\quad a_n \cdot a_{n+1}  = 2^{-f_{n-1}} \cdot 2^{-f_n} \quad\Longleftrightarrow\\  a_{n+2} = a_n \cdot a_{n+1} = 2^{-f_{n-1}} \cdot 2^{-f_n} = 2^{-(f_n + f_{n-1})} = 2^{-f_{n+1}}.$$
...is that valid?
Given that $a_n = 2^{-f_{n-1}}$ is true I proceeded with the ratio test
$$q := \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left|\frac{2^{-f_{n}}}{2^{-f_{n-1}}}\right| = \lim_{n\to\infty} \frac{2^{f_{n-1}}}{2^{f_{n}}} =\\
 = \lim_{n\to\infty} \frac{1}{2^{f_{n} - f_{n-1}}} = \lim_{n\to\infty}\frac{1}{2^{(f_{n-2} + f_{n-1}) - f_{n-1}}} = \lim_{n\to\infty}\frac{1}{2^{f_{n-2}}} < 1.$$
Is that sufficient to show the convergence? Follow-up question: given the $a_n = 2^{-f_{n-1}}$ term - is there a better way than this ratio test to show convergence of $\sum a_n$ now that I have this representation?
Another approach would be to find a sequence $(b_n)_{n\in\mathbb{N}}$ with
$$\sum_{n=1}^{\infty}b_n < \infty \quad\wedge\quad |b_n| > |a_n|\quad\forall n\geq1.$$
For what it's worth I struggle to find such a series - am I missing something? Is there something immediate that should come to mind? $a_n$ and $b_n$ have to be zero sequences, how can I do the estimate in its recursively defined form?
Another "please correct me if I'm wrong"-question: I did an approach where I proved that $a_{n+1}< a_{n}$ for all $n\geq1$ (which is similar to question 1:

$$n=1:\quad a_1 = 1 \geq \frac{1}{2} = a_2\\
n=2: a_3 = a_1a_2 = 1 \cdot \frac{1}{2} = \frac{1}{2},\quad\therefore a_2 = \frac{1}{2} \geq \frac{1}{2} = a_3$$
and
$$"n+1 \to n+2":\quad a_n \cdot a_{n+1} \geq a_{n+1} \cdot a_{n+2} \Longleftrightarrow\\
a_{n+2} = a_n \cdot a_{n+1} \geq a_{n+1} \cdot a_{n+2} =  a_{(n+1)+2} = a_{n+3}$$
respectively) and after that I did the ratio test
$$q := \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left|\frac{a_{n+2}}{a_{n+1}}\right| = \lim_{n\to\infty} \left|\frac{a_n a_{n+1}}{a_{n+1}}\right| = \lim_{n\to\infty}\left|a_{n}\right| \stackrel{*}{<} 1.$$
(with (*) denoting that $a_{n+1}<a_{n}$ and $a_2 <1$). Is that a sufficient answer to my problem?

...thank you so much, guys!

I would greatly prefer your approach no.3 because such series are actually easy to find. One example is the geometric series with ratio 1/2 (up to an initial "stutter" term which is easy to discard).


A space is Hausdorff iff the diagonal is closed Let X be a topological space. Prove that X is a T2-space if and only if
$A = \{ (x,x) : x \in X \}$ is closed in $X \times X$.

I've been able to prove first part of the result but not the converse.

For the first part,

We have X is T2. hence X x X is also T2. Hence X x X is T1.

Then for any z belonging to A and y not belonging A, there exist an open set U in X x X such that y belongs to U but z doesn't belong to U.

Hence for every z belonging to A and for a fixed y not belonging to A , there exists an open set V(y) in X x X such that "V(y) intersection A" is empty.

hence (X x X) - A = Union of all V(y)'s such that y doesn't belong to A.

since arbitrary union of open sets is open, hence (X x X) - A is also open in
X x X , Implying that A is closed in X x X. ( Is this proof valid ? )

Now how to prove the converse part ?

your prove is not correct at the part "hence (X x X) - A = Union of all V(y)'s such that y doesn't belong to A.".
This question can be solved directly by using the fact that  every points have neighbourhood  $U_1 \times U_2$ with$U_1,U_2$ is open subset. try to use thit :)


Two doors, 2 guards logic problem ("is XOR a hypothetical statement?") You're trapped in a room with 2 doors, one on your left and one on your right. One door leads to heaven, one door leads to hell. There are two birds in the room. One is a bad bird, which always lies, one is a good bird, which always tells the truth. You don't know which door is which, nor do you know which bird is which. The birds know which door is which.

  You get to ask one question to one bird. How do you figure out which door leads to heaven?

This puzzle has some typical "expansions" to make it more difficult, like

  You may not ask about the other bird

or equivalently (I think?)

  There is only one bird which either always tells the truth or always lies.

From this expansion on, there is a post often going around online, considered a "troll post" (http://i.imgur.com/huBYtN5.jpg), which adds the following:

  You may not use IF statements or any hypothetical statements in your question

I believe adding the restriction that forces a question to have a reply of either "true" or "false" does not change the difficulty, so let's work with that.

A solution I came up with the help from a stranger online, was asking the following question:
"(Does the door on the left lead to heaven) XOR (Do you always speak the truth)"

My doubts about this are:

1) Is XOR not a hypothetical statement?

2) Would a lying bird lie about the entire question, or about the subquestions that are being XOR'd, or about all? (This may just be "definition"?)

In answer to 2): either type of bird could, and therefore presumably would, respond to the entire question. A lying bird would say "false" if the XOR statement evaluates to true, and vice versa. There's no issue here.

Question 1) is a little less straightforward. On the one hand, in logical terms XOR is a perfectly ordinary two-place truth-functional connective, just like AND and OR. And like those other two, it's symmetric: A XOR B means the same as B XOR A. Why would one think there's anything hypothetical about it?

On the other hand, A XOR B can be read as "A or B, but if A, then not B, and if B, then not A." Which really brings us back to the old problem of how to read "if-then." If we "If A, then B" as equivalent to "Either B, or else not-A" (this interpretation is called the material conditional), then we have something that's very easy to work with, but we have made it impossible to distinguish if-then statements from OR statements, because either form can be converted into the other. And since OR statements are interconvertible with AND statements, ... you see where this leads. No wonder the ban on IF statements is considered a trollish move.

what would the other bird say if you asked if the left door was heaven


The difference between $\Delta x$, $\delta x$ and $dx$ $\Delta x$, $\delta x$ and $dx$ are used when talking about slopes and derivatives. But I don't know what the exact difference is between them.

$\Delta x$ is about a secant line, a line between two points representing the rate of change between those two points. That's a "differential" (between the two points).
$dx$ is about a tangent line to one point, representing an instantaneous rate of change. That makes it a "derivative."
$\delta x$ is about a tangent line to a partial derivative. That's a rate of change or derivative in one direction, holding a number of other directions constant.

$∆x~$ is small change in $~x~$.

$~dx~$ is small part of $~x~$ but represents independent change.
& $~\frac{dy}{dx}~$ means slope of tangent at a point where it touches to the curve
$~\frac{∆y}{∆x}~$ s the slope through two points.

We say $~∆x~$ tends to zero .

It becomes $~\frac{dy}{dx}~$  which is slope of tangent at a point (reason of why $~∆x~$ tends to zero).

That is it


Proof writing involving propositional logic: (x ∨ y) ≡ ( x ∧ y ) → x ≡ y Prove by using propositional logic:

(x ∨ y) ≡ ( x ∧ y ) → x ≡ y

I'm a bit lost here proving by propositional logic that the statement is valid. I don't know how to start this problem. Any help? I know the statement is true since x ≡ y, thus the premise (x ∨ y) ≡ ( x ∧ y ) does not matter, it will be still true according to the → operation. Any ideas? Any help will be greatly appreciated, thanks.

Edit:

Apart from true tables.

You want to show that $x \equiv y$ from the premise $(x\vee y) \equiv (x \wedge y)$. I assume it's enough to derive $(x\to y) \wedge (y \to x)$.

$(x\vee y) \equiv (x \wedge y) \qquad \text{(premise)}$
Assume $x \qquad\qquad \text{(assumption)}$
$x \vee y \qquad\qquad\qquad \text{by $p \to (p \vee q)$}$
$x \wedge y\qquad\qquad\qquad \text{by 1.}$
$y \quad\qquad\qquad\qquad \text{by $(p\wedge q) \to q$}$
$x \to y \quad\qquad\qquad \text{by 2. and 5., discharging 2.}$
-- 11. similarly derive $y \to x$.

$$a\equiv b=ab+\overline a\overline b$$

Then

$$(x+y)(xy)+\overline{(x+y)}\overline{xy}=xy+\overline x\overline y(\overline x+\overline y)=xy+\overline x\overline y.$$


New symbol for an exact sequence A sequence $A\to B \to C$ is exact if $\operatorname{im} f = \ker g$, where $ f:A\to B$ and $g:B\to C$.

Why is there not a symbol to denote such a sequence?
Which one would you suggest?

@egreg's comment is totally right — because there's no need in such symbol.

Also, there are at least two reasons to not introduce it. First one is that heavily symbolized mathematical writing is usually unreadable. Second is high probability of  ambiguity in case when diagram is somewhat complicated.

So I would suggest to just write «this sequence is exact at $B$», or that «pair of morphisms $f, g$ is exact» (which is better, in my opinion).

Here are three ideas. I prefer the last one.