Meaning behind differentials So I think I understand what differentials are, but let me know if I'm wrong.

So let's take $y=f(x)$ such that $f: [a,b] \subset \Bbb R \to \Bbb R$.  Instead of defining the derivative of $f$ in terms of the differentials $\text{dy}$ and $\text{dx}$, we take the derivative $f'(x)$ as our "primitive".  Then to define the differentials we do as follows:

We find some $x_0 \in [a,b]$ where there is some neighborhood of $x_0$, $N(x_0)$, such that all $f(x)$ in $\{f(x) \in \Bbb R \mid x \in N(x_0)\}$ are differentiable.  Then we choose another point in $N(x_0)$, let's call it $x_1$, such that $x_1 \ne x_0$.  Then let $dx = \Delta x = x_1 - x_0$.  Now this $\Delta x$ doesn't actually have to be very small like we're taught in Calculus 1 (in particular it's not infinitesimal, it's finite).  In fact, as long as $f(x)$ is differentiable for all $x \in [-10^{10}, 10^{10}]$ we could choose $x_0 = -10^{10}$ and $x_1 = 10^{10}$.

Then we know that $\Delta y = f'(x_0) \Delta x + \epsilon(\Delta x)$, where $\epsilon(\Delta x)$ is some nonlinear function of $\Delta x$.  If $f(x)$ is smooth, we know that $\epsilon(\Delta x)$ is equal to the sum of powers of $\Delta x$ with some coefficients, by Taylor's theorem.  But of course, $\epsilon(\Delta x)$ won't be so easy to describe if $f(x)$ is only once differentiable.  So we define $dy$ as $dy = f'(x_0) dx$: that is, $dy$ is the linear part of $\Delta y$.  This has the very useful property that $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx} = f'(x_0)$.  This is then not a definition of the derivative, but a consequence of our definitions.

It can be seen from this $dy$ really depends on what we choose as $dx$, but $f'$ is independent of both.

This definition can be extended to functions of multiple variables, like $z = f(x, y)$ as well, by letting $\Delta x = dx,\ \Delta y=dy$ and defining $dz$ as $dz = \frac{\partial f(x_0, y_0)}{\partial x}dx + \frac{\partial f(x_0, y_0)}{\partial y} dy$.  So $dz$ is the linear part of $\Delta z$.  Does all of the above look correct?

If so, then where I'm having a problem is: 1) how then do we define the derivative of $f(x)$ if not by $f'(x_0) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$? 2) how do we apply this definition of $dx$ to $\int_a^b f(x)dx$?  It seems like the inherit arbitrariness of $dx$ is really going to get in the way of a good definition of the integral.

$\mathrm{d}y$ depends only on $y$: it doesn't depend on any choice of $x$ or anything else: that's one of the big advantages to differentials (as opposed to, say, partial derivatives).

A differential is a gadget that expresses how something varies. There are three main things you can do with such a gadget:

You can compare two differentials: e.g. if $x$ and $y$ are dependent on one another in a differentiable way, then they are multiples of each other. e.g. if $y = f(x)$, then $\mathrm{d}y = f'(x) \mathrm{d}x$.
Given a differential, you can ask if it has an antiderivative: e.g. $2x \mathrm{d}x$ is the differential (often called the "exterior derivative") of $x^2$.
You can compute a (path) integral to 'add up' along a path all of the variations the differential expresses. e.g. $\int_0^1 2x \mathrm{d}x$ means we 'accumulate' all of the variations $2x \mathrm{d}x$ as we go from $x=0$ to $x=1$. And as we know $2x \mathrm{d}x = \mathrm{d}(x^2)$, our intuition is satisfied in the sense that accumulating how $x^2$ varies from $x=0$ to $x=1$ works out to $1^2 - 0^2$.

You can also ask the differential to give you an ordinary number expressing a variation along a (tangent) vector. A common notation for this is, e.g. in $(x,y)$ coordinates, to let the symbol $\partial/\partial x$ and $\partial/\partial y$ denote vectors, and for a differential $\omega$, the notation $\frac{\partial}{\partial x} \omega$ means ordinary number that $\omega$ yields for a variation by the vector $\partial/\partial x$.

e.g. we have
$$ \frac{\partial}{\partial x} \mathrm{d}x = 1
\qquad \qquad \frac{\partial}{\partial x} \mathrm{d}y = 0
\qquad \qquad \frac{\partial}{\partial y} \mathrm{d}x = 0
\qquad \qquad \frac{\partial}{\partial y} \mathrm{d}y = 1$$

This is consistent with the notation for partial derivatives you've learned, in that, e.g.,

$$ \frac{\partial}{\partial x} f = \frac{\partial}{\partial x}
\mathrm{d} f $$

where the left hand side is the meaning taken from introductory multivariable calculus, and the right hand side is the meaning I describe above. (usually first introduced in differential geometry)

Incidentally, I think partial derivative notation is absolutely terrible, and I avoid using it whenever possible. I also think differentials are more intuitive than partial derivatives as well, and I prefer to do all of my calculus in terms of differentials these days. A convenient analog to $f'$ for multivariable functions is to let, e.g., $f_1$ denote the derivative of $f$ in its first argument, $f_2$ denote the derivative in the second argument, and so forth. So I would prefer to write

$$ \mathrm{d}f(x,y) = f_1(x,y) \mathrm{d}x + f_2(x,y) \mathrm{d}y $$

rather than anything resembling the traditional notion of partial derivatives. If I want derivatives in the direction where $y$ is held constant, I express that as setting $\mathrm{d}y = 0$ rather than resorting to partial derivatives.

This use of combining vectors with differentials is related to the (unfortunately common) mistake / abuse of notation that you often see, where the notation $\mathrm{d}x$ is treated an actual change in $x$, rather than as a gadget that can tell you what the change in $x$ is.

Differentials are infinitely small changes in x or y. For instance, the concept of the integral is the sum of the areas of an infinite number of rectangles under a curve. The height of each is f(x) and the width is dx.


Find constants so that the directional derivative of $f(x,y,z) = axy^2+byz+cx^3z^2$ has maximum value $32$ in point $P$ given the direction I am asked to find $a$, $b$ and $c$ so that the directional derivative of $$f(x,y,z) = axy^2+byz+cx^3z^2$$ has maximum value of $32$ in the point $P(1,2,-1)$ and in the direction $\overrightarrow{u} = (0,0,1)$.

What I have so far is

$$
\frac{\partial f}{\partial x} = ay^2 + 3cx^2z^2\\
\frac{\partial f}{\partial y} = 2axy+bz\\
\frac{\partial f}{\partial z} = by+2cx^3z\\
\nabla f(1,2,-1) = (4a+3c,4a-b,2b-2c)$$

$$
D_{u} f(1,2,-1) = (4a+3c, 4a-b, 2b-2c) \cdot (0,0,1) = 32 \therefore b-c = 16
$$

and I'm not sure how to proceed.

Answer:
$$a = 3\\
b = 12\\
c = -4$$

Find values of the constants $a, b, c$ such that the directional derivative of $$f(x, y, z) = axy^2 +
byz +cz^2x^3$$ at the point $(1, 2, −1)$ has a maximum value of $32$ in the direction parallel to the $z$-axis.
First compute the gradient:
$$\nabla f = (ay^2 + 3cz^2x^2)i + (2axy + bz)j + (by + 2czx^3)k$$
and
$$\nabla f(1, 2, −1) = (4a + 3c)i + (4a − b)j + (2b − 2c)k$$
We know that the gradient vector points in the direction of the greatest rate of change. So, if $f$
attains a maximum value in a direction parallel to the $z$-axis, this means that $\nabla f(1, 2, −1)$ points
in a direction parallel to the $z$-axis (but we don’t know if it points up or down). This gives the
equations
$$4a + 3c = 0, \quad
 4a − b = 0, \quad
 2b − 2c = 32$$
where the final equation follows from the fact that we know that the maximum value attained by $f$
is $64$, and the fact that the maximum value of the gradient vector equals the length of the gradient
vector. Solving this, we get
$$(a, b, c) = (3, 12, −4) \text{ or } (a, b, c) = (−3,12,-4)$$


The meaning of the symbol $\infty$ in Spivak's calculus book Spivak in "Calculus" writes

  ... symbols of $\infty$ and $- \infty$ are purely suggestive: there is no number $``\infty"$ which satisfies $\infty \geq a$ for all numbers $a$.

What is the meaning of being purely suggestive?

Source:

$\infty$ and $-\infty$ are not real numbers. So when ever we use them in real analysis we need to define their use carefully, as he has done in the extract you quote.

For instance the interval $(a,b)$ is defined as $\{ x \in \mathbb R \ : \ a < x < b \}$ whenever $a, b$ are both real numbers.

This definition is in trouble if the symbols $a$ or $b$ are $-\infty$ or $\infty$. In those cases, we need a different definition, as Spivak gives. This is to say, up to that point of Calculus, we know what $$x < y$$ means for real numbers $x$ and $y$. We have not yet defined what for instance $-\infty < x < b$ would mean.

This might seem pedantic, especially in the example of interval notation, but mathematical results can often turn on carefully delineated definitions and expressions. So it's important to get even these basic definitions right in one's early training in rigorous mathematics, which is what Spivak's Calculus is all about.

In addition to the use of $\pm\infty$ in interval notation, Spivak will later give careful definitions  of expressions such as

$$\lim_{x\to\infty} f(x),$$ $$\lim_{n\to\infty} a_n,$$

$$ \lim_{x\to a} f(x) = \infty \quad\text{ where $a$ is a real number,}$$

and

$$\int_a^\infty f(x) \ dx \quad\text{ where $a$ is a real number}.$$

Those definitions will not treat $\infty$ as a real number. Instead, the definitions are expressed using real numbers in such a way that the suggestion of $\infty$ is formalized.

You may have heard the adage, "infinity is not a number, it's a concept" (or at least you have now). This is in fact true, since $\infty$ is not a real number (if it were, terrible things would happen).

Infinity $-$ $\infty$ $-$ is the object such that no number is greater than it:
$$ \forall x\in \mathbb R : |\infty| > |x|$$

We use this object to talk about the "extreme" ends of the real numbers (e.g., when talking about taking limits), because it is easier to say "as $x \to \infty$, foobar is true" than it is to say "for any [possibly very large] number $M \in \mathbb R$ there exists a number $N>M$ such that foobar is still true".


Chessboard, Tiling, Maths A square-shaped garden is divided into an n x n square grid by footpaths (like a chessboard).

The owner wishes to have all the sections covered by grass; however, the grass grows in a peculiar way. If at least two of the neighbouring squares (i.e. squares that share an edge) of a given square become fully covered with grass, then so will the given square.

What is the minimum number of squares that need to be planted initially so that the grass will eventually extend to the whole garden?

The solution in the book (see my comment on the original question) is terrific: just note that the perimeter of the grassy region never increases.

Take $2*2$ case, and plant grass in diagonal squares initially this will cover complete 4 squares. Now add 5 squares along adjacent edges of given square to make it a $3*3$ square. No squares of the newly added squares grow grass on its own. So you need to plant one square, any square of the newly added squares with grass and you are done. This logic can be extended to higher $n$. Thus minimum square need to be planted with grass initially is $n$.


Can anyone help me solve this differential equation? $y''+9y=0, y(\frac{\pi}{3})=3, y'(\frac{\pi}{3})=3$. I have got a general solution of $y(x)= C\cos(3x)+ D\sin(3x)$ but I am having trouble solving for the initial value portion

Your first initial condition gives you

$$y(\pi/3)=3=C \cos(3 \pi/3) + D \sin(3 \pi/3) = C \cos(\pi) + D \sin(\pi)=-C$$

so $C=-3$. Now differentiate and do the same thing for $y'(\pi/3)$.

In general you would need to solve a system of linear equations to deal with the initial conditions. Here we are fortunate that the equations nicely decouple, so we don't have to do that.


Find the range of |z| when $|z−\frac{4}{z}|=2$ I have tried using the triangle inequality but got two answers.
$z≤1+\sqrt5$ and one imaginary quadratic. I think here is one more equation which I am not getting. Please help.

Extended hint:

$$
\begin{align}
4=\left|z−\frac{4}{z}\right|^2=\left(z−\frac{4}{z}\right)\left(\bar z−\frac{4}{\bar z}\right) & = |z|^2+\frac{16}{|z|^2}-4\left(\frac{z}{\bar z}+\frac{\bar z}{z}\right) \\ & = \left(|z|+\frac{4}{|z|}\right)^2 - 4\left( 2+ 2 \operatorname{Re}\frac{z}{\bar z}\right)
\end{align}
$$

Since $\left|\frac{z}{\bar z}\right| = 1$ it follows that $-1 \le \operatorname{Re}\frac{z}{\bar z} \le 1$ so:

$$
\left(|z|+\frac{4}{|z|}\right)^2 = 4\left( 3+ 2 \operatorname{Re}\frac{z}{\bar z}\right) \in [\,4,\,20\,]
$$

$$
|z|+\frac{4}{|z|} \in [\,2,\,2\sqrt{5}\,]
$$

From there on it's left to determine the range of the positive real roots of the quadratics in $|z|\,$:

$$
\begin{cases}
\begin{align}
|z|+\frac{4}{|z|} & \ge 2 &&\iff |z|^2 - 2|z| + 4 \ge 0 \tag{1}\\
|z|+\frac{4}{|z|} & \le 2 \sqrt{5} && \iff |z|^2 - 2 \sqrt{5} \,|z| + 4 \le 0 \tag{2}\\
\end{align}
\end{cases}
$$

$(1)$ is satisfied for $\forall |z| \in \mathbb{R^+}$ while $(2)$ gives the solution $|z| \in [\sqrt{5}-1,\sqrt{5}+1]$.

As suggested using  Inequalities Among Complex Numbers,

$$\left|z-\dfrac4z\right|\ge|z|-\dfrac4{|z|}\implies|z|-\dfrac4{|z|}\le2\ \  \  \ (1)$$

and   $$\left|z-\dfrac4z\right|=\left|z+\left(-\dfrac4z\right)\right|\le|z|+\left|\dfrac{-4}z\right|=|z|+\dfrac4{|z|}\implies|z|+\dfrac4{|z|}\ge2\ \  \  \ (2)$$

Now use the followings on $(1),(2)$

$(x-a)(x-b)\le0;a\le b\implies a\le x\le b$

and $(x-a)(x-b)\ge0;a\le b\implies$ either $x\le a$ or $x\ge b$


Sample: don't confuse measurements with actual values? In Wikipedia's article on Sample there is the following remark:

''Note that a sample of random variables (i.e. a set of measurable functions) must not be confused with the realizations of these variables (which are the values that these random variables take). In other words, $X_i$ is a function representing the measurement at the $i$-th experiment and $x_i = X_i(ω)$ is the value we actually get when making the measurement.''

I'm afraid I don't understand this passage, can anybody please explain the point?

Let's assume that there are infinitely many humans and that their IQ has a normal distribution with mean 100 and variance 15. Here is a possible experiment:

  grab 10 people downtown and measure their IQ

As a mathematical model, we accept the setting that there is a probability space $\Omega$ and "grab human number i" is modelled by a "random variable" $X_i$ which is a measurable function
$$
X_i: \; \Omega \to \mathbb{R}
$$
such that the $X_i$ are independent and have a gaussian aka normal distribution.

When you go out and acutally perfom this experiment, you'll get ten humans and ten values (real numbers) for theiy IQ's. Let's call them $IQ_i$. With regard to our mathematical model, this means that for every human you convinced to do the IQ-Test there is a corresponding $\omega \in \Omega$, each of which is called an event, such that
$$
IQ_i = X_i(\omega)
$$

When you get out again and perform this experiment again, you can accept the same random variables $X_i$ as a model for your experiment, but you'll encounter different humans, get different results from the IQ-Tests, which corresponds to a different $\omega ' \in \Omega$.

Or, to put it shortly as Didier Piau did: The description of the experiment entails as a mathematical model the random variables  aka measurable functions $X_i$; but everytime when you actually perform the experiment, it will result in a tuple of values of these functions.

When a random variable takes a value, you cannot always explicitly know this exact value, you can only measure it, i.e. use a function of the result which outputs you the measuring. For instance, would you be sampling real numbers in $[0,1]$ and measuring them only up to 3 digits of precision, if your first random variable $X_1$ gets value $\pi/4$ for instance, you could only say $X_1 = 0.785$, since the measurable function $X_1$ maps any $\omega$ in the interval $(0.7845,0.7855]$ to $0.785$.

In other words, $\omega$ is the "actual value", if we could say this, of the random variable and $X_1$ is a way to measure the actual value by letting it become a measurable function.

Hope that helps,


Column Stochastic Matrix Let $$A =\begin{pmatrix}0&0 &0.5&0.5&0\\1/3&0&0&0&0\\1/3&1/2&0&1/2&1\\1/3&1/2&0&0&0\\0&0&1/2&0&0\end{pmatrix}$$
and $$S=\dfrac{1}{5}\begin{pmatrix}1&1 &1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\end{pmatrix}.$$

$M=0.85A+0.15S$ is column stochastic and absolute value of second largest eigenvalue of $M \approx 0.6113.$ I have calculated the eigenvector of $M$ corresponding to eigenvalue $1$ to be $q=(0.2371, 0.0972,0.3489, 0.1385,0.1783)^T.$

If $z=(0.2, 0.3, 0.4, 0.05, 0.05)^T,$ it is known that $\text{lim}_{k\to \infty}\dfrac{\lVert M^{k}z-q \rVert_1 }{ \lVert M^{k-1}z-q \rVert_1 }= 0.6113,$ where $\lVert . \rVert_1 =$ sum of modulus of the entries of a column vector.

For $k=50,\dfrac{\lVert M^{k}z-q \rVert_1}{ \lVert M^{k-1}z-q \rVert_1}=\dfrac{7.8781\times 10^{-12}}{1.2888\times 10^{-11}} \approx 0.615 $

But for $k=70, \dfrac{\lVert M^{k}z-q \rVert_1 }{ \lVert M^{k-1}z-q \rVert_1}= \dfrac{5.8287\times 10^{-16}}{4.4409\times 10^{-16}} =1.3125$ I have also tested larger values of $k$ and the ratio seems to converge to $1$ instead.

Could anyone please advise? Thank you.

I feel it is the accuracy. You already got to $10^{-16}$ range in one of your calculations, but your $q$ is likely a double-precision floating number, which is accurate only up to 15 decimal points.


A recurrence relation question - transforming $a_{n+1}+{a_n}^2-2a_n=0$

I guess it is solved by transforming $a_n$ to some form of $b_n$. But I could not see the way. Would you explain the solution in details? Thanks.

Note that
$$\begin{align}
a_{n+1}+{a_n}^2-2a_n=0 & \iff a_{n+1} = -a_n^2 + 2a_n \\
& \iff a_{n+1}=-(a_n^2-2a_n+1)+1 \\
& \iff a_{n+1}=-(a_n-1)^2+1 \\
\end{align}$$
and a simple heurist reasoning suggests that
$$a_n \approx -( \ldots ((a_0^2)^2)^2 \ldots )^2 = -a_1 ^ {2^n}$$

Now, since solving quadratic recurrence relation is quite hopeless  because most quadratic maps are not solvable (however there are some tricks you can try) the first thing to try is to give some values, and one can see that
$$\begin{cases}
a_1 = 1 \Rightarrow a_n = 1 \qquad \forall n \in \mathbb{N}\\
a_1 = 2 \Rightarrow a_n = 0 \qquad \forall n \in \mathbb{N}, n \ge 2\\
a_1 = 3 \Rightarrow a_2 = -3, a_3 = -15, a_4 = -255, a_5 = 65535 \ldots
\end{cases}$$
and it's clearly visible a pattern $\ldots$
$$a_{n} = -(a_1 -1)^{2^n}+1$$


Each digit of $\dfrac{n(n+1)}{2}$ equals $a$ For which digits $a$ does there exist an integer $n \geq 4$ such that each digit of $\dfrac{n(n+1)}{2}$ equals $a$?

I was first thinking of looking at $\dfrac{n(n+1)}{2} \pmod{100}$, but it doesn't look like it has a small period so it may be hard to arrive at a conclusion. Is there an easier way to solve this question?

Note that $n(n+1)/2 = z$  has an integer solution iff $1+8z$ is a square.
So the question is whether $1 + \dfrac{8 a}{9} (10^k-1)$ can be a square, or equivalently whether $9 - 8 a + 8 a \cdot 10^k$ can be a square.
You can rule out $a = 2, 3, 4, 7, 8, 9$ this way because $9-8a$ is not a square $\mod (8 a \cdot 10^k)$ for $k=1$ or $2$.
That leaves $a=1$.

EDIT: For $a = 1$, you want $1 + 8 \cdot 10^k$ to be a square with $k \ge 2$.  Suppose it is $x^2$.  Since $x^2 \equiv 1 \mod 5^k$, we may assume  $x \equiv 1 \mod 5^k$.
So let $x = 1 + 5^k y$.  Then we have $1 + 2 y \cdot 5^k + y^2 \cdot 5^{2k} = 1 + 8 \cdot 10^k$, or $2 y + y^2 \cdot 5^k = 2^{k+3}$.  In particular, $y \mid 2^{k+3}$, so $y$ is a power of $2$.  If $y = 2^j$, the equation becomes
$2^{j+1} + 2^{2j} \cdot 5^k = 2^{k+3}$.  Considering factors of $2$, two of $j+1, 2j, k+3$ must be equal.  But it's easy to rule out all these possibilities.

See also OEIS sequence A045914.

EDIT: It's also interesting to consider other bases than $10$.  Here there are some infinite families of solutions, as well as some (such as $52 \dfrac{98^4-1}{98-1} = \dfrac{9944 \cdot 9945}{2}$ or $105 \dfrac{159^6-1}{159-1} = \dfrac{4634175 \cdot 4634176}{2}$) for which I don't see an obvious pattern.

In base $8j+1$, where $j$ is an integer, $j \dfrac{(8j+1)^{2k}-1}{8j+1-1} = \dfrac{n(n+1)}{2}$ where $n = \dfrac{(8j+1)^k-1}2$.

Use induction to show that $\sum_{i=0}^n i = \frac {n(n+1)}{2} $

Then the number with digit $a$, $aaa...a = \sum_{i=0}^n i$ for some $n $

These are the triangular numbers

Now to start the problem we can look at lists of Triangular numbers to see that both 55 and 66 are candidates.

Bellew and Weger showed that 666 is the largest such number.

Then the complete list of repdigit triangular numbers is 55, 66, and 666.


How to find dimension of vector space In $\mathbb{R}^5$ there is given vector space $V$. Its dimension is 3. In $\mathbb{R}^{6,5}$ consider the subset $X = \{A \in \mathbb{R}^{6,5} : V \subset \ker A\}$. I have to show that $X$ is a vector space in $\mathbb{R}^{6,5}$ and find its dimension. To show that $X$ is vector space consider $x_1, x_2 \in X$ and $v \in V$. We know that $x_1 v = 0$ and $x_2 v = 0$ so $(\alpha x_1 + \beta x_2) v = \alpha (x_1 v) + \beta (x_2 v) = 0$ so $V \subset \ker (\alpha x_1 + \beta x_2)$. But I don't know how to find $X$'s dimension. Any ideas?

I understand $V \subseteq \mathbb{R}^5$ is a subspace, $\dim V = 3$

$X = \{A \in \mathbb{R}^{6 \times 5} : V \subseteq \ker A \}$

To show that $X$ is a vector space, it suffices to show it is a subspace of $\mathbb{R}^{6 \times 5}$.

$0 \in X$, clearly because $V \subseteq \mathbb{R}^5 = \ker 0$
For $\alpha_i \in \mathbb{R}$ and $A_i \in X$, if $v \in V$, $A_i v = 0$, and so $ (\sum_i \alpha_i A_i) v = 0$.

So it is a subspace.

Let $B = \{v_1, v_2, v_3 \}$ be a basis of $V$, and extend it to a basis $\{v_1, v_2, v_3, v_4, v_5 \}$ of $\mathbb{R}^5$.  As $Av_i = 0$ for $1 \leq i \leq 3$, you only have to say where goes $A v_i$ for $i=4$ and $5$.

So you have 5-3=2 degrees of freedom in the domain and 6 in the codomain, that gives $2 \cdot 6 = 12$.  I suspect the dimension is 12. Hope that helps.

I add this: you can think of $A$ as the matrix representation of some linear transformation $f: \mathbb{R}^5 \to \mathbb{R}^6$ with respect to the bases the extended version of $B$ above for $\mathbb{R}^5$ and and the standard basis for $\mathbb{R}^6$, so $f(v_i) = 0$ for $1 \leq i \leq 3$, and you can decide where goes $f(v_i)$ for $i=4,5$.

For the dimension of $X$, if $A\in X$, $A$ has to nullify $V$ and can do anything on the $2$-dimensional orthogonal complement of $V$. So the dimension of $X$ is the same as the dimension of the space of linear transformations from a $2$-dimensional space to a $6$-dimensional space, which is $12$.

EDIT: A commenter is unfamiliar with the direct bijection between $6\times 5$ matrices and linear transformations from $\mathbb{R}^5$ to $\mathbb{R}^6$;  we can reinvent like four or five wheels and say the same thing but longer:

Let $V$ have basis $\{v_1,v_2,v_3\}$. Use Gram Schmidt in $\mathbb{R}^5$ so that $\{v_4,v_5\}$ is a basis for $V^{\perp}$. Let $\{e_1,e_2,e_3,e_4,e_5,e_6\}$ be the standard basis for $\mathbb{R}^6$.

Consider the matrices $A_{ij}$ defined as the matrix that takes $v_j$ to $e_i$ and all other $v_k$ to the zero vector in $\mathbb{R}^6$. Explicitly, $A_{ij}=C_{ij}P$, where $C_{ij}$ is a $6\times5$ matrix with all zeros except a $1$ in the $i,j$ position, and $P$ is the change of basis matrix from $\{v_1,v_2,v_3,v_4,v_5\}$ to the standard basis for $\mathbb{R}^5$. Explicitly, $P=\begin{bmatrix}|&|&|&|&|\\v_1&v_2&v_3&v_4&v_5\\|&|&|&|&|\end{bmatrix}^{-1}$. Check: $A_{ij}v_j=C_{ij}Pv_j=C_{ij}\epsilon_j=e_i$ where $\epsilon_j$ are the standard basis vectors in $\mathbb{R}^5$.

There are $30$ of these $A_{ij}$, which are matrices in $M_{6\times 5}$. They are linearly independent because if $\sum_{i,j} c_{ij}A_{ij}=[0]$, then multiplying against each of the $v_j$ one at a time gives $\sum_i c_{ij}e_i=[0]$, and since the $e_i$ are independent, the $c_{ij}$ must equal $0$.

But 18 of them are not in $X$, because they have one of $Av_1,Av_2,Av_3$ nonzero. Only the twelve matrices $A_{i4}$ and $A_{i5}$, which all have $Av_1=Av_2=Av_3=0$ are in $X$.

$M_{6\times5}$ is $30$-dimensional. We have established 12 linearly independent vectors that are in the space $X$. So $\dim V\geq12$. We have established 18 more linearly independent vectors that are not in $X$. So $\dim V\leq30-18-12$. So $\dim V=12$.


Why are even primes notable? There are much-discussed theorems like Fermat's theorem on sums of two squares which make statements about odd primes only. This makes $2$ seem to be a "special" prime. In their book The book of numbers, Conway and Guy accordingly state that "Two is celebrated as the only even prime, which in some sense makes it the oddest prime of all."

On the other hand, the fact that $2$ is the only even prime is completely trivial, because the term "even" means the same thing as "divisible by $2$" and every prime number has the property that it is the only prime which is divisiable by itself.

So my question is: Is there really something special about even primes and if yes, what is it? Does aesthetics with regard to the theorems we are looking for play a role or is there a mathematical reason? Do we have theorems about primes which aren't divisible by $3, 5, ... $ or are there only results which don't apply to even primes?

Edit: As the user A.G has mentioned in a comment below, in many cases where we have a regular pattern, the fact that $2$ is too small for the pattern to kick in yet seems to be the decisive thing. So in these cases, the notable thing is not that $2$ is the only even prime but that it is the smallest prime.

The quip about "2 being the only even prime" is a bit silly, as you say, since 3 is the only divisible-by-3 prime, etc. For that quip, it's just that parity (odd-or-even) exists in the ambient language.

For $p$ prime, the $p$th roots of unity are in $\mathbb Q$ for $p=1$. Similarly, the $p$th roots of unity lie in all finite fields (of characteristic not $p$...) only for $p=2$.

Quadratic forms and bilinear forms behave differently in characteristic two.

Groups $SL(n,\mathbb F_q)$ do not assume their general pattern yet for small $n$ and $q=2$.

The index of alternating groups in symmetric groups is $2$.

Subgroups of index $2$ are normal.

The canonical anti-involution on a non-commutative ring, that reverse the order of multiplication, is of order $2$.

Another reason why 2 is so notable is because it is the most important part in finding a square root without a calculator as not all numbers can be divided by an odd prime

However, what is distinct is that since 2 is so small, (tell me if I'm wrong) in engineering, it is impossible to balance something sturdily with two beams supporting an item. The best numbers for supporting are primes like 3, 5, 7 etc. and although any other number larger than two can support, I'm pretty sure that the best numbers are prime numbers other than two.


Derivative of a reciprocal function If I want to prove that a partial derivative $\frac{\partial D}{\partial x}$ is either positive or negative, and $D=\frac{1}{E}$, then how can I use the partial derivative $\frac{\partial E}{\partial x}$ (which is a simplier derivative) to prove this?

Since $\;\cfrac{\partial D}{\partial x}=\cfrac{\partial}{\partial x}\left(\cfrac{1}{E}\right)=-\cfrac{1}{E^2} \cfrac{\partial E}{\partial x}\;,\;$ then
$\;\cfrac{\partial D}{\partial x}>0\;\iff\;\cfrac{\partial E}{\partial x}<0\;,$
$\;\cfrac{\partial D}{\partial x}<0\;\iff\;\cfrac{\partial E}{\partial x}>0\;.$
So, if you want to prove that $\;\cfrac{\partial D}{\partial x}\;$ is positive, it is sufficient to prove that $\;\cfrac{\partial E}{\partial x}\;$ is negative, and vice versa.

I don't think that is possible because continuity of the function itself might change.
For example
If your function is $xy$, then upon taking the reciprocal, $1/xy$ isn't continuous at $x=0$ and $y=0$. So ideally being able to reciprocate to differentiate will not give you an answer.
Also, let us assume the domain isn't an issue for us. Still, assume a function y=x which is a straight line with a positive slope. If you rewrite it as $y=1/x$, it is a  rectangular hyperbola that has a negative slope. So, this as per my knowledge doesn't seem possible


If $(X,d)$ is compact (why?) and $(E,d')$ is complete then $\mathscr{C}(X,E)$ is complete (equipped with the topology of unif. convergence). A theorem in a book that I am looking into states that if $(X,d)$ is compact and $(E,d')$ is complete then $\mathscr{C}(X,E)$ is complete (equipped with the metric $d_u$ of uniform convergence).

Now the thing is that I do not see why the compactness on $(X,d)$ is necessary. Is it maybe so that $d_u(f,g)$ would be finite for $f, g \in \mathscr{C}(X,E)$?

Can we "overcome" this problem if we replace $d'$ by $d''= \frac{d'}{1+ d'}$ as a metric on $E$?

The condition that $X$ is compact is essential, because:

Every continuous function on a compact set is bounded.  The uniform  metric is given by $d(f,g)=\sup_{x\in X}|f(x)-g(x)|$.  Since the function is bounded the set $\{f(x):x\in X\}$ is bounded and so its supremum exits.


Why is $SU(2)$ diffeomorphic to $S^3$? As the title,my question is why $SU(2)$ is diffeomorphic to $S^3$ ? How can I structure a diffeomorphic map between $SU(2)$ and $S^3$?

Please help me,and give me more details!

Thank you very much!

By definition, we have that  $\displaystyle SU(2)=\left\{\begin{pmatrix}a &-\overline{b}\\  b &\overline{a}\end{pmatrix}: a, b\in\mathbb{C}, |a|^2+|b|^2=1\right\}.$

Since $\mathbb{R}^4\cong\mathbb{C}^2 $, we may think of $S^3$ as $\displaystyle S^3=\{(a,b)\in\mathbb{C}^2: |a|^2+|b|^2=1\}$.

Define a map from $S^3$to $SU(2)$ as:
\begin{aligned}F: S^3&\to SU(2)\\  F(a,b)&=\begin{pmatrix}a &-\overline{b}\\  b &\overline{a}\end{pmatrix}.  \end{aligned}

Can you show that $F$ is well defined, $F$ is injective, and $F$ is surjective (i.e. an isomorphism)? After that, you should show that $F$ and $F^{-1}$ are smooth (or $C^{\infty}$). (Hint: Show $\displaystyle SU(2)$ is a submanifold of of the set of 2 x 2 matrices with complex entries, or better yet $\mathbb{R}^8$. What can you say about $F$ and its inverse restricted to these submanifolds?)

Every SU2 matrix is a linear combination of Pauli matrices p1, p2, p3 + unity matrix I which is a base of the associated Lie algebra of the SU2 group.
So every SU2 matrix  X can be written a X=x1.p1 +x2.p2 +x3.p3 +x4.I and the condition that detX = 1 gives x1power2 + x2power2 +x3power2 +x4power2 = 1 which is the S3 sphere. and every point on that sphere represents an SU2 matrix


Fixed point of $\cos(\sin(x))$ I can show that $\cos(\sin(x))$ is a contraction on $\mathbb{R}$ and hence by the Contraction Mapping Theorem it will have a unique fixed point. But what is the process for finding this fixed point? This is in the context of metric spaces, I know in numerical analysis it can be done trivially with fixed point iteration. Is there a method of finding it analytically?

The Jacobi-Anger expansion gives an expression for your formula as:

$\cos(\sin(x)) = J_0(1)+2 \sum_{n=1}^{\infty} J_{2n}(1) \cos(2nx)$.

Since the "harmonics" in the sum rapidly damp to zero, to second order the equation for the fixed point can be represented as:

$x= J_0(1) + 2[J_2(1)(\cos(2x)) + J_4(1)(\cos(4x))]$.

Using Wolfram Alpha to solve this I get $x\approx 0.76868..$

Since it's a contraction, just iterate. Pick an arbitrary starting point and keep applying the operator until you achieve the desired accuracy.


Dartboard paradox and understanding independence By definition, events $A$ and $B$ are independent if $$P(A \cap
    B) = P(A)\:P(B).$$ Thus if an event $A$ happens almost never
$\left(P(A)=0\right),$ then
$A$ is independent of all events, including itself. So, hitting the
exact centre of a dartboard (happens almost never) and hitting within its inner ring are independent events.

On the other hand, it is standard to characterise
independence as follows:

Two events are independent if the occurrence of one does not
affect the probability of occurrence of the other.

Since hitting the exact centre of a dartboard guarantees hitting
within its inner ring (which is otherwise not guaranteed), the
two events are dependent.

Isn't this a contradiction?? If yes, then do we accept that the definition of independence is not meant to fully correspond to its verbal/intuitive characterisation?

Is there any semantic difference between the following two versions?

$(i)$ Two events are independent if the occurrence of one does not
affect the probability of occurrence of the other.

$(ii)$ Two events are independent if the occurrence of one does not
affect the occurrence of the other.

ADDENDUM
On further pondering, I have resolved question 2:

Consider this experiment: flip two fair coins, letting $H_1$ be the
event that the first coin lands on Heads, and $X$ be the event that
the coins land on different sides. Then $$ P\left(H_1 \cap
X\right)=\frac14=P(H_1)\:P(X);$$ i.e., $H_1$ and $X$ are independent
events.
Knowledge that $H_1$ has occurred reduces the known number of ways that $X$ can eventuate—from $2$ (HT, TH) to $1$ (HT)—but does not alter the probability $(\frac12)$ assigned to $X.$

The following revision characterises pairwise independence more clearly and accurately:

Events $A$ and $B$ are independent if $A$ having occurred does not
change the probability of $B.$

It still doesn't apply when $A$ is an impossible event though; and remains problematic when $A$ is an almost-never event. (What does it even mean to say that an almost-never event has occurred: in what sense have I hit the exact centre of a dartboard?)

Events $A$ and $B$ are indeed defined to be independent if $P[AB] = P[A]P[B]$.  This definition holds for all cases, including cases when $P[A]=0$ and/or $P[B]=0$. The definition leads to your desired interpretation "knowing if $B$ occurred does not affect the likelihood of $A$"  in the special case $P[B]>0$, since it means $P[A|B] = P[A]$ in that case. It does not allow your interpretation in the case when $P[B]=0$, since we cannot condition on a zero-probability event.  So indeed, there are peculiarities when $P[B]=0$ since intuition fails in that case. Nevertheless, your intuition is still useful since it applies when $P[B]>0$.
It may also be helpful to observe that if $P[B]=0$ then $P[A|B^c] = P[A]$, which is intuitively what we would want when $A$ and $B^c$ are independent.

A situation about darts similar to the one you posed can be addressed by the definitions of conditioning with respect to a random variable. For a dartboard, we can imagine $(X,Y)$ uniformly distributed over a 2-d ball of radius 1.  Then define $R = \sqrt{X^2+Y^2}$ as the distance from the center.  Consider the events:
$$ A = \{R \leq 1/10\}, B = \{R = 1/10\}$$
Now $P[B]=0$ so $A$ and $B$ are independent.  Nevertheless this is similar to the situation you describe because $B \subseteq A$, so "intuitively" we would want $P[A|B]=1$. This holds if we use:
$$ P[A|R=r] = \frac{f_{R|A}(r)P[A]}{f_R(r)} $$
which shall be applied when $f_R$ and $f_{R|A}$ are continuous at point $r$ and $f_R(r)\neq 0$,
and we observe:
\begin{align}
f_R(r) &= \left\{\begin{array}{cc}
2r  &  \mbox{ if $r \in [0,1]$}\\
0 & \mbox{else}
\end{array}\right.\\
f_{R|A}(r) &= \left\{\begin{array}{cc}
\frac{2r}{P[A]} & \mbox{ if $r \in [0,1/10]$} \\
0 & \mbox{ else}
\end{array}\right.
\end{align}
so
$$ P[A|R=r] = \left\{\begin{array}{cc}
0 & \mbox{if $1/10 < r<1$} \\
1 & \mbox{ if $0<r\leq 1/10$}
\end{array}\right.$$
Our expression $P[A|R=r]$ cannot be viewed as an instance of the definition of conditional probability $P[A|B]$ for events $A$ and $B$, since the event $\{R=r\}$ has probability 0.  Rather, it is an expression related to the definition of conditioning on a random variable $R$. Formally, you can say the random variable $Z$ defined by
$$ Z = \left\{\begin{array}{cc}
0 & \mbox{if $R>1/10$} \\
1 & \mbox{ if $0\leq R \leq 1/10$}
\end{array}\right.$$
is a "version" of the conditional expectation $E[1_A|R]$.  As such, it is almost surely the same as any other version. Thus, while this version takes the liberty to define $Z=1$ when $R=0$, and $Z=0$ when $R\geq 1$, this is fine because it only affects a set of measure 0.

When two events A and B are disjoint it means that if one of them occurs, the other one cannot occur, i.e., $A\cap B= \phi$. Suppose $A=${hitting the exact centre of a dartboard} is an impossible event, then based on the above argument non-happening of A implies the happening of other events, and hence there is no independence between A and the other events. Further, the dartboard is an example of continuous sample space and an event like A may be a rare event, but one claim that it is an almost impossible event.


Evaluating $\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi n t)}{t^z} dt$ I need help evaluating any of:
$$\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi n t)}{t^z} dt$$
$$\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi (2n - 1) t)}{t^z} dt$$
Where $z \in \Bbb{C}$

Also I would like to ask if this is true:
$$\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac {\cos(2 \pi n t)}{t^z} dt = \int_{1}^{\infty} \frac {\sum_{n=1}^{\infty} \cos(2 \pi n t)}{t^z} dt$$
If yes then how to in clever way calculate:
$$\sum_{n=1}^{\infty} \cos(2 \pi n t)$$

I will solve the first sum only, since the second one may be tackled in a similar way. Assume that $\operatorname{Re}(z) > 0$. Then
\begin{align*}
\sum_{n=1}^{N} \int_{1}^{\infty} \frac{\cos(2\pi n t)}{t^z} \, \mathrm{d}t
&= \int_{1}^{\infty} \left( \sum_{n=1}^{N} \cos(2\pi n t) \right) \frac{\mathrm{d}t}{t^z} \\
&= z \int_{1}^{\infty} \left( \sum_{n=1}^{N} \frac{\sin(2\pi n t)}{2\pi n} \right) \frac{\mathrm{d}t}{t^{z+1}}
\end{align*}
Now, from the knowledge on Fourier series, we know that
$$ S_N(t) := \sum_{n=1}^{N} \frac{\sin(2\pi n t)}{2\pi n} $$
is bounded uniformly in $N$ and $t$ and that
$$ \lim_{N\to\infty} S_N(t) = \frac{1}{4} - \frac{t - \lfloor t \rfloor}{2} $$
for each $ t \in \mathbb{R}\setminus\mathbb{Z}$. So by the dominated convergence theorem,
\begin{align*}
\sum_{n=1}^{\infty} \int_{1}^{\infty} \frac{\cos(2\pi n t)}{t^z} \, \mathrm{d}t
&= z \int_{1}^{\infty} \left( \frac{1}{4} - \frac{t - \lfloor t \rfloor}{2} \right) \frac{\mathrm{d}t}{t^{z+1}}.
\end{align*}
The last integral can be computed by decomposing it over the subintervals of the form $[k, k+1]$ for $k = 1,2, \dots$, and the result is
$$ \frac{\zeta(z)}{2} - \frac{1}{4}\left(\frac{z+1}{z-1}\right). $$
This function is analytic on all of $\operatorname{Re}(z) > 0$.

In a very general manner, the sum and integration operations can be switch iff they both converge in both cases.
For the evaluation of the integral I would do the following:
$$
\sum_{n=1}^\infty \int_0^\infty \frac{\cos(2\pi nt)}{t^z}dt=\sum_{n=1}^\infty \int_0^\infty \frac{\frac{e^{i2\pi nt}-e^{-i2\pi nt}}{2}}{t^z}dt\\
\frac{1}{2}\sum_{n=1}^\infty \int_0^\infty \frac{e^{i2\pi nt}-e^{-i2\pi nt}}{t^z}dt
$$
Now try interchanging the sum and integral, I think this would be a good way to approach the problem


is nthroot of a negative number imaginary if n is not whole number I wanted to find the domain of a function which is defined as $y=x^{1/x}$
For negative $x$, if $x$ is even then the result will be imaginary, if $x$ is odd then the result is real. However, what about $x=-2.5$ or $-e$?

The n-th root of a number is that number raised to the power  1/n. So the 1/3-th root of a number is just that number cubed. Thus, by these definitions, the 1/3-th root of -5 is -125, which is not imaginary. Or did I misunderstand the question?


Evaluating limit without L'Hospital's I would like to evaluate the limit $$\lim_{x\to0}\frac{2x\sin(x)}{1-\cos(x)}$$ It can be done with L'Hospital, but I would like to know if it is possible to do so without L'Hospital. I tried multiplying through by the conjugate of the denominator, which leads me to $$\lim_{x\to0}\frac{2x(1+\cos(x))}{\sin(x)}$$ which is not very helpful, because it is still $0/0$. Any help would be much appreciated, thanks!

From the later form you can use $\lim_{x\to 0} \frac{\sin x}{x} = 1$ directly with the product rule to show that the limit is $4$. This identity can be proven without L'Hopital so it's valid to use it. A lot of limits involving trigonometric functions can be solved this way


Example of an infinite dimensional Hilbert space that is not an RKHS I have been studying Reproducing Kernel Hilbert Spaces (RKHS). The definition I am using is as follows: An RKHS is a Hilbert space $\mathcal{H}$ of real-valued functions on a set $X$ such that for all $x \in X$, the evaluation functional $E_x : \mathcal{H} \to \mathbb{R}$ defined by $E_x(f) = f(x)$ is continuous. (Recall that continuity in this context means $\sup_{\|f\|\leq 1}|f(x)| < \infty$ for all $x$.)

As one of the first steps in my study I want to see some positive and negative examples of these spaces. So I want to find a non-trivial (meaning infinite dimensional) space of functions which is a Hilbert space but not an RKHS, meaning the evaluation functionals are not continuous.

The problem is that in all the references I find, the only example they give is the space of functions $L_{2}$. However, this example is trivial since $L_2$ fails for the simple reason that it is not a space of functions, rather it contains equivalence classes of functions.

Can someone please help me find an infinite dimensional Hilbert space of functions that is not an RKHS?

There are no explicit examples of this kind. Indeed, if one could show you such a space $\mathcal H$, then point evaluation $f\mapsto f(x_0)$  would give an explicit example of a discontinuous linear functional on a Hilbert space. And there can be no explicit examples of this kind, because it's consistent with ZF that they do not exist. Only with the Axiom of Choice can one give a (non-constructive) proof of their existence. This is discussed in many places, such as

Discontinuous linear functional
Unbounded linear operator defined on $l^2$

So, any time someone gives you a concrete Hilbert space of functions, you can be sure that it's either RKHS, or is not really a space of functions.

Consider $L_2$. It is a Hilbert space. Then, consider the following function:
$$
f_{\epsilon}(x)=\max\{0,1-\frac{|x-\frac{1}{2}|}{\epsilon}\}.
$$
You can see that this function belongs to $L_2$. However, the function evaluation at $x=\frac{1}{2}$ is always equal to 1. Therefore you cannot find $M$ such that
$$
f_{\epsilon}(\frac{1}{2})\leq M \|f_{\epsilon}\|_{\mathcal{H}}
$$
where $\mathcal{H}=L_2$. The reason is that  $\|f_{\epsilon}\|_{\mathcal{H}}$ goes to zero.


Show that $Ha=Hb$ if and only if $ab^{-1} \in H$ where $a,b\in G$ and $H$ is a subgroup of $G$ Suppose $G$ is a finite group,  $a,b\in G$ and $H$ is a subgroup of $G$.

Show that $Ha=Hb$ if and only if $ab^{-1} \in H$.
  (Notice that $Ha$ and $Hb$ are both right cosets of $H$ by $G$).

This is what I have so far:

(-->) Suppose $Ha=Hb$. We know that $H$ is a subgroup and therefore $1_G\in H$ and from this we can derive that $a\in H$ and $b\in H$.

For a certain $h\in H$, we have:

$b\in Ha \implies b=ha \implies 1_G=h(ab^{-1})$

Is it enough to derive that $ab^{-1}\in H$ ?(because we know $1_G\in H$)

(<--) Suppose $ab^{-1} \in H$. I'm not quite sure what direction should I take from here in order to show that $Ha=Hb$. Can you please give me an hint how to start this direction?

Thank you very much,

You have some mistakes. If $1_G\in H$ then we cannot derive that $a\in H$ and $b\in H$.

We have that either $Hg_1=Hg_2$ or $Hg_1\cap Hg_2=\emptyset$

$(->)$

If $Ha=Hb\Rightarrow a=hb$ for some $h\in H$.

Thus $ab^{-1}=h\in H$.

$(<-)$

Let $ab^{-1}\in H$. Then $ab^{-1}=h\in H\Rightarrow a=hb=\rightarrow a\in Hb\Rightarrow Ha=Hb$.

For the converse, $ab^{-1}\in H$ implies:

$ab^{-1} = h_0$ for some $h_0\in H\implies a = h_0b\implies a\in Hb\implies ha \in Hb$ for every $h\in H$.

$\qquad\therefore x\in Ha\implies x\in HB$.

$(ab^{-1})^{-1} \in H$, by invertibility of group $H\iff ba^{-1} \in H$.

$\qquad\therefore x\in Hb\implies x\in Ha$, by applying similar argument in (1).

Hence $Ha=Hb$.

$\Box$


Find a point $X$, in the plane of regular pentagon $ABCDE$, that minimizes $\frac{XA+XB}{XC+XD+XE}$. Find such a point $X$, in the plane of the regular pentagon $ABCDE$, that the value of expression $$\frac{XA+XB}{XC+XD+XE}$$ is the lowest.

I tried using Ptolemy's theorem but don't know how to make use of inequalities it gives.

I'd be really grateful for any help :)

We use Ptolemy inequality : For an arbitrary quadrilateral $PRQS$ we have: $$PQ\cdot RS \leq PR\cdot QS+PS\cdot RQ$$

Let $AB = s$ and $AC=m$ and $XA =a$, $XB =b$... Now we use this inequality on $3$ different quadrilaterals:

$$AXBE:\;\;\;\;\; es\leq bs+am$$

$$ AXBC:\;\;\;\;\; cs\leq as+bm$$

$$ AXBD:\;\;\;\;\; ds\leq am+bm$$

Adding these three we get $$ s(c+d+e)\leq (a+b)(2m+s)$$so we have $$ {s\over 2m+s}\leq {a+b\over c+d+e}$$

So your expression is always $\geq  {s\over 2m+s}$ and this value is achieved if we put $X=A$ or $X=B$ or at any point on arc between $AB$ on circumcircle for $ABCDE$

(Remember that equality is true iff all inequalities become equalities and that is if $AXBC$, $AXBD$ and $AXBE$ are cyclic with that order of points on circle).

If $X$ is on $AB$, $AX+BX$ is minimum.  And if $X=A$ , $CX+DX+EX$ is maximum. So cesareo's answer follows.


Recurrence - Master Theorem - Asymptotic Question Sorry if this question has been asked before, but I am trying to figure this out. I am using the CLRS text, Introduction to Algorithms. In the Recurrences chapter, in the Master Theorem section, the following example is given with the solution:

$$T(n) = 3T\left(\frac{n}4\right) + n\log n $$

Here,

$$ a = 3, b =4, f(n) = n\log n $$

Using the Master method, it says

$$f(n) = \Omega(n^{\log_4 3 + \epsilon})  $$

I don't understand how $f(n)$ was "assumed" to have this
lower bound ($\Omega$). I would be extremely greatful to anyone who points out to me why this is so.

If more details are needed to understand the problem, please let me know. I will add more details to this question.

TIA,

Jake Clawson

EDIT: I thank all those who have responded below. I have replied to each below individually. But my question still stands. The book directly goes on to say that $f(n)$ is $\Omega$. I don't understand this and this is the key to applying the Master Theorem because once $f(n)$ is known that we know which case to apply.

So you want to understand why $n\log n=\Omega(n^{\log_4 3+\epsilon})$

The easiest way is to look at the limit

$\lim_{n->\infty}\frac{n\log n}{n^c}$

which is evaluated to

$\lim_{n->\infty}\frac{\log n+1}{cn^{c-1}}$ where $c=log_4 3+\epsilon$

since $\log_4 3\lt1$, there exists $\epsilon\gt0$ such that $\log_4 3+\epsilon=c\lt1$. Therefore, the limit would tends to $\infty$ and hence $nlogn=\Omega(n^{\log_43+\epsilon})$

Take $c=\lg\left(n_{0}\right)$ in the following definition for the
$\Omega$-notation:

$$
\Omega\left(g\left(n\right)\right)=\left\{ f\left(n\right):\exists\left(c>0\wedge n_{0}>0\right)\backepsilon0\leq cg\left(n\right)\leq f\left(n\right)\forall n\geq n_{0}\right\}
$$

which appears on page 48 of the third edition of the book. Thus, with
$g\left(n\right)=n$, for $n\geq n_{0}$ we have that

$$
n\lg\left(n_{0}\right)\leq n\lg\left(n\right)
$$

This shows that $f\left(n\right)=\Omega\left(n\right)$, which answers
the question.


Predictive distribution with divergent integral I am having some trouble understanding the basic definition of a posterior predictive distribution and how to apply it to simple examples.

I have that if $$y \sim g(y\mid \theta, x)$$

then the predictive posterior distribution is

$$g(y\mid x)=\int g(y\mid\theta,x)\pi(\theta\mid x) \,d\theta$$

So I am trying to work through an example of this where

$x \sim \operatorname{Normal}(\theta, \sigma^2)$ and $y \sim \operatorname{Normal}(ex,\sigma^2)$ and $\pi(\theta,\sigma^2)=\frac{1}{\sigma^2}$

And I am not sure I understand all the notations correctly. I don't have any examples either of actually solving for this, but I would be interested in seeing any.

So I know that $$\pi(\theta\mid x)  \propto \pi(x\mid\theta) \pi(\theta)$$

so $$\pi(\theta,\sigma^2\mid x) \propto \exp\left(\frac{-1}{2\sigma^2}{(x-\theta)^2}\right) \frac{1}{\sigma^2}$$

But then I am not sure how to proceed.

I assume I am looking for $\pi(Y\mid X=x)$

Well  $P(Y\mid X=x,\theta , \sigma^2) = \exp(\frac{-1}{2\sigma^2}(y-ex)^2)$

so,

$$P(Y\mid X=x,\theta , \sigma^2) \pi(\theta,\sigma^2 \mid x)= \frac{1}{\sigma^2} \exp\left(\frac{-1}{2\sigma^2}[(y-ex)^2+(x-\theta)^2]\right)$$

Note that if we try to double integral to normalise over $\pi(x\mid\theta , \sigma^{2})\pi(\theta,\sigma^2)$ we get a divergent integral. So we cannot make equality.

To sum up, you should know the prior distribution of $\sigma^2$. And, yes, you need double integral to integrate out both $\sigma^2, \theta$, unless $\sigma^2$ has been assumed to be a given constant.

Let's clarify what you are looking for. Your are looking for the posterior predictive, i.e. the distribution of $Y | X=x$. Your model is as follows. Given constant $e$,

$$
\begin{aligned}
X \mid \theta, \sigma^2 &\sim N(\theta,\sigma^2) \\[7pt]
Y \mid x, \theta, \sigma^2 &\sim N(ex,\sigma^2) \\[7pt]
π(θ \mid \sigma^2) &= 1/\sigma^2 I_\Theta \\[7pt]
\sigma^2 &\sim f_{\sigma^2}
\end{aligned}
$$

So you have

$$
\begin{aligned}
\pi(\theta, \sigma^2 \mid x ) &\propto \pi(x \mid \theta, \sigma^2) \pi(\theta \mid \sigma^2) \pi(\sigma^2)\\[7pt]
&\propto \exp(-\frac{1}{2\sigma^2} (x-\theta)^2) 1/\sigma^2 \cdot f_{\sigma^2}
\end{aligned}
$$

Thus

$$
\begin{aligned}
P(Y \mid X=x) &= \int_{\sigma^2} \int_\theta P( Y, \theta, \sigma^2 \mid X=x) \\[7pt]
&= \int_{\sigma^2} \int_\theta \pi(Y \mid x, \theta, \sigma^2) \pi (\theta \mid \sigma^2) f_{\sigma^2} \\[7pt]
&= \int_{\sigma^2} \int_\theta N(ex,\sigma^2) 1/\sigma^2 f_{\sigma^2} \\[7pt]
\end{aligned}
$$


What K value minimizes this function I got a function $$ nK^{\frac{-1}{\beta}}\log{K} + ZK\log{n} $$ where $K$ and $n$ are integers and $Z > 0$.
I followed the typical way of differentiating the function and equating to zero.  But I couldn't find a closed form.  I got the following result
$$
\frac{Z\log{n}}{n} = \frac {\frac{\log{K}}{\beta}-1}{K^{\frac{1+\beta}{\beta}}}
$$
I am not sure how can I go from here.

The last equation you wrote gave me an idea for simplifications.
For simplifications, let $$K=x^{\frac{\beta }{\beta +1}}$$ which makes the function to be
$$f(x)=x^{-\frac{1}{\beta +1}} \left(\frac{\beta  n}{\beta +1} \log (x)+x Z \log(n)\right)$$
$$f'(x)=\frac{\beta  x^{-\frac{\beta +2}{\beta +1}} ((\beta +1) (x Z \log (n)+n)-n \log(x))}{(\beta +1)^2}$$
So, after simplifications, we look for the zero of
$$g(x)=(\beta +1) (x Z \log (n)+n)-n \log (x)$$ The only analytical solution of this equation express in terms of Lambert function
$$x=-\frac{n\,\,W(t)}{(\beta +1) Z \log (n)} \quad \text{where} \qquad t=-\frac{e^{\beta +1} (\beta +1) Z \log (n)}{n}$$ Then $K$ from $x$.
It remains to prove that this solution is a minimum. So,
$$f''(x)=\frac{x^{-\frac{1}{\beta +1}-2} (\beta  (\beta +2) n \log (x)-\beta  (\beta +1)
   ((\beta +3) n+x Z \log (n)))}{(\beta +1)^3}$$ I have not been able to prove it.
Let us try using $n=56789$, $Z=\pi$  and $\beta=\log(10)$. It would give $K=10.4097$. But, this is a maximum.
So, if you look for a minimum value of the function other than $K=0$, we need to use the second branch of Lambert function, that is to say
$$x=-\frac{n\,\,W_{-1}(t)}{(\beta +1) Z \log (n)} \quad \text{where} \qquad t=-\frac{e^{\beta +1} (\beta +1) Z \log (n)}{n}$$ For the worked example, this will give $K=213.69$
For the evaluation of $W(t)$, assuming that $\beta >-1$ (wich makes $t<0$) you can use the series expansion
$$W(t)\approx L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(-t)$ and $L_2=\log(-L_1)$ .
If you just look for bounds, in the linked page, you will notice that in $2013$ was established the double inequality
$$-1 - \sqrt{2u} - u < W_{-1}\left(-e^{-(u+1)}\right) < -1 - \sqrt{2u} - \tfrac{2}{3}u $$
For the worked example this gives
$208.178 < K <229.578$

K and n are constant integers. $\beta$ is constant real. Being in form a + b Z as a linear relation.. minimum when Z=0.


How can a piece of A4 paper be folded in exactly three equal parts? This is something that always annoys me when putting an A4 letter in a oblong envelope: one has to estimate where to put the creases when folding the letter. I normally start from the bottom and on eye estimate where to fold. Then I turn the letter over and fold bottom to top. Most of the time ending up with three different areas. There must be a way to do this exactly, without using any tools (ruler, etc.).

Fold twice to obtain quarter markings at the paper bottom.
Fold along the line through the top corner and the third of these marks.
The vertical lines through the first two marks intersect this inclined line at thirds, which allows the final foldings.

(Photo by Ross Millikan below - if the image helped you, you can up-vote his too...)

This is a slight variation of the method of Egor Skriptunoff that works for paper of any aspect ratio, not just A4, and so is of use for our non-Continental readers.  It uses the solution to the crossed ladders problem, and requires one more fold than Egor's method but fewer than the method of Hagen von Eitzen which also works for any aspect ratio.

Fold the paper in half along the long side (constructing segment $EF$ in the diagram below, connecting the midpoints of $AD$ and $BC$).
Fold the paper along the segment $BE$.
Fold the paper along segment $AC$.

One can show that $GH = \frac{1}{3}AD$, so folding the paper along segment $IJ$ (and then either folding again at the crease, or folding along the segment $DF$ to get another intersection point) will produce the desired fold into thirds.
If the ratio $AB/AD = \sqrt{2}$ as in A4 paper, then we have $AC \bot BE$ (the converse is also true) allowing the omission of steps 1 and 2 that construct $BE$, and we can proceed as in Egor's solution.


Confused about the Definition of an Isolated Point For any point $x \in \mathbb{R}$ and a set $S$, $x$ is called an isolated point if $x \in S$ and $x \notin S'$, where the set $S'$ denotes the set of accumulation points. But what confuses me is that, if $x \in S$, then every neighborhood of $x$ has a nonempty intersection with $S$, but if $x \notin S'$, then for some $\epsilon > 0$, the deleted neighborhood $N^*(x;\epsilon)$ satisfies $N^*(x;\epsilon) \cap S = \varnothing$. To me, this seems like a contradiction, because the first claims that every neighborhood of $x$ contains at least one point in $S$, but the second one states that there is at least one deleted neighborhood of $x$ whose members do not belong in $S$.

What am I overlooking?

The more insightful definition of an isolated point of $S$ is:

There is some $\varepsilon>0$ such that $N(x; \varepsilon) \cap S = \{x\}$.

This implies that $N^\ast(x; \varepsilon) \cap S = \emptyset$, indeed, so $x \notin S'$. But clearly the definition implies $x \in S$ as well.

Indeed every neighbourhood of $x$ contains a point of $S$ (namely $x$ itself!), and for small neighbourhoods (smaller radius than $\varepsilon$) this is the only point this neighbourhood has in common with $S$. In the deleted neighbourhood we delete this $x$ itself, so then there are no more points in common with $S$ in the deleted neighbourhood (for those small ones).

The first claim is wrong as $\{x\}$ may be a neighbourhood of $x$.


Confusion in understanding continuous function Okay so I've started to study derivatives and there is this idea of continuity. The book says "a real valued function is considered continuous at a point iff the graph of a function has no break at the point of consideration, which is so iff the values of the function at the neighbouring points are close enough to the value of the the function at the given point"

So what I dont understand is that why is it that values of the function at the neighbouring points should be close enough to the value of the function at the given point, isn't it enough if they are defined why do they have to be close enough the value of the function at the given point?

Intuitively, the idea is that the closer and closer you get to $x$, the closer and closer $y$ gets to $f(x)$. A small neighborhood of $x$ must correspond to a small neighborhood of $f(x)$.

When there is a discontinuity, the image of a neighborhood of $x$ always includes that discontinuity and that blocks it from being as small as you want.

If the values of the function in points $x$ close enough to a specific point $x_0$ are not arbitrarily close to $f(x_0)$ the graph will be (usually) broken.

This definition is not completely rigorous, as it assumes we have intuitive understanding what a "graph with no break is", and it can create some misunderstandings. For example, it can be proven that the graph of function $$ f(x) = \left\{\begin{array}{l} \sin\frac{1}{x} & \text{for }x\neq 0 \\ 0 &\text{for }x=0 \end{array}\right. $$
is connected, but the function is still non-continuous.

It's better to consider "iff the values of the function at the neighbouring points are close enough to the value of the the function at the given point" as the definition, or, rigorously speaking

  Function $F : X \rightarrow Y$ is called continuous at point $x_0\in X$, iff $$ \forall \epsilon>0 \exists \delta>0 \forall x\in X: \Big((|x-x_0|<\delta) \Rightarrow \big(|f(x)-f(x_0)|< \epsilon\big)\Big)$$

which means: "For arbitrary distance $\epsilon$ I can choose a neighbourhood of $x_0$ small enough for the values of function to be within distance of $\epsilon$ from the value of function at point $x_0$".


Using integration by parts show that: $$\int_0^{+\infty} e^{-x} \cos x\, dx = \int_0^{+\infty} e^{-x} \sin x\, dx$$

So far I have just expanded out, and I don't see an end, is there a trick I am missing?

A rather simple and detailed approach :

$$\int_0^\infty e^{-x}\cos(x)\mathrm{d}x = \int_0^\infty e^{-x}(\sin(x))'\mathrm{d}x = \big[ e^{-x}\sin(x)\big]_0^\infty - \int_0^\infty(e^{-x})'\sin(x)\mathrm{d}x$$

$$=$$

$$\lim_{x\to \infty} \big[ e^{-x}\sin(x)\big] - 0 - \int_0^\infty-e^{-x}\sin(x)\mathrm{d}x $$

$$=$$

$$\boxed{\int_0^\infty e^{-x}\sin(x)\mathrm{d}x}$$

apply integration by part twice then you'll get the same integral with extra terms then call integral I then it is a first order linear equation so, solve it for I.
Other approach add
$$\int e^{-x} i \cos(x)$$ to original integral. So desired result is real part of  $$\int_{0}^{\infty} e^{-x}(\sin(x)+ i \cos(x))dx = \int e^{(-1+i)x} dx$$.


Subtleties of "unknown" vs. "variable" I'm trying to pin down the difference between "unknown" and "variable". I have always understood that in the equations $2x+1=10$ or $x^2+5x+6=0$, $x$ is an unknown (short for "unknown constant"), since its value can be determined. In the expression $2x+1$, however, $x$ can take any value, therefore it is a variable.

What about in the equation $2x+3y=10$? $x$ and $y$ can both take infinitely many values, but once one is fixed, the other becomes fixed. Does this mean they are both variables? Does it mean that one (say $x$) is a variable, but the other is an unknown (since its value is determined by the variable)?

I'd appreciate some insight. Thanks.

It depends what your question is. While they are kind of interchangeable, these two terms are used in different contexts.  Unknown is usually employed in equations,so for example you could ask how to solve the equation $2x=1$,where $x$ is the unknown. On the other hand the term variable is more used in case of functions. You could ask for example what is the second derivative according to the $x$ variable of the function $f(x,y)=x^2+y+2$. Hope that clears things up a bit.

I would say...a variable is an unknown but an unknown doesn't necessarily have to be a variable. A variable means it could be any number, it is not fixed but a unknown means it is a specific number that we do not know as yet. Therefore a variable is an unknown because it could be any number but an unknown doesn't have to be a variable because it is a fixed number that we do not know. Hope that makes sense.


Lipschitz-type estimate... True or false? I have two parameters $\alpha,\varepsilon>0$ and the following difference:
$$D:=\left|\,\varphi\left(\frac{x-\alpha^2-\varepsilon}{\alpha}\right)-\varphi\left(\frac{x-\alpha^2+\varepsilon}{\alpha}\right)\,\right|,$$
where $\varphi(z)=\frac{1}{2\pi}e^{-z^2/2}$. I have the impression that $D\leq C\varepsilon$ for some constant $C>0$ independent of $x$ and $\alpha$. Nevertheless, applying the Lipschitz property "brutaly" we get
$$D\leq \max_{z}\left|\varphi'(z)\right| \frac{2\varepsilon}{\alpha} \leq C \frac{\varepsilon}{\alpha}.$$

Observe that when we send $\alpha\to 0$ then the RHS above explodes, while in $D$ it seems to be bounded. Where is the error? or how can I show that the above estimate holds true? Any idea?

Thanks a lot!

Given any $\varepsilon\gt0$, let $x=\alpha^2+\epsilon$. Then $\frac{x-\alpha^2-\varepsilon}{\alpha}=0$ and $\frac{x-\alpha^2+\varepsilon}{\alpha}=\frac{2\varepsilon}\alpha$. Then
$$
\varphi\left(\frac{x-\alpha^2-\varepsilon}{\alpha}\right)-\varphi\left(\frac{x-\alpha^2+\varepsilon}{\alpha}\right)=\frac1{2\pi}-\varphi\left(\frac{2\varepsilon}{\alpha}\right)\tag{1}
$$
And by letting $\alpha\to0$, we get that the difference in $(1)$ tends to $\frac1{2\pi}$ no matter how small $\varepsilon$ is. Therefore, it is not possible to get
$$
\left|\,\varphi\left(\frac{x-\alpha^2-\varepsilon}{\alpha}\right)-\varphi\left(\frac{x-\alpha^2+\varepsilon}{\alpha}\right)\,\right|\le C\varepsilon\tag{2}
$$
independently of $x$ and $\alpha$.

It's an inequality. You may have given something away on the right hand side, but that doesn't imply an error.

A better estimate would be

$$|D| \le \sup \{|\varphi '(z)|: (x-\alpha^2 -\epsilon)/\alpha \le z \le (x-\alpha^2 +\epsilon)/\alpha\}\cdot (2\epsilon/\alpha).$$

Since $|\varphi '(z)| \to 0$ fast as $|z|\to \infty,$ the above should help.


$A_5$ is generated by all 3 cycles There are some similiar questions, but it seems that there is some different equal definition to $A_5$.
The definition I'm using for $A_5$ is that this is the kernel of the sign homomorphism- i.e all the permutations with even number of transpositions.
I would like to get help with the equivalence of this definition , to the definition of subgroup of $S_5$ , generated by all cycles of order 3.
EDIT- THERE IS AN ANSWER ON THE LINK IN COMMENTS

A5 is the group [3,3,3]+ or [3,5]+ in Coxeter notation.
You could make it, I suppose, out of (AB)³ = (BC)³ = (CD)³ = I, and (AC)² = (BD)² = (DA)² = I, which is what the first symbol means.
The second one means (AB)³ = (BC)^5 = (AC)² = I, which is what the second definition.
A5 is a relatively primitive group, but you can indeed generate A5 from purely cycles of three.


Why can you turn clothing right-side-out? My nephew was folding laundry, and turning the occasional shirt right-side-out. I showed him a  "trick" where I turned it right-side-out by pulling the whole thing through a sleeve instead of the bottom or collar of the shirt. He thought it was really cool (kids are easily amused, and so am I).

So he learned that you can turn a shirt or pants right-side-out by pulling the material through any hole, not just certain ones. I told him that even if there was a rip in the shirt, you could use that to turn it inside-out or right-side-out, and he was fascinated by this and asked "why?"

I don't really know the answer to this. Why is this the case? What if the sleeves of a long-sleeve shirt were sewn together at the cuff, creating a continuous tube from one sleeve to the other? Would you still be able to turn it right-side-out? Why? What properties must a garment have so that it can be turned inside-out and right-side-out?

Sorry if this is a lame question, but I've always wondered. I wouldn't even know what to google for, so that is why I am asking here.

If you know the answer to this, could you please put it into layman's terms?

Update: Wow, I really appreciate all the participation. This is a really pleasant community and I have learned a lot here. It seems that the answer is that you need at least one puncture in the garment through which to push or pull the fabric. It appears that you can have certain handles, although it's not usually practical with clothing due to necessary stretching.

Accepted (a while ago actually -- sorry for not updating sooner) Dan's answer because among the answers that I understand, it is the highest ranked by this community.

First, a warning.  I suspect this response is likely not going to be immediately comprehensible.  There is a formal set-up for your question, there are tools available to understand what's going on.  They're not particularly light tools, but they exist and they're worthy of being mentioned.  Before I write down the main theorem, let me set-up some terminology.  The tools belong to a subject called manifold theory and algebraic topology.  The names of the tools I'm going to use are called things like: the isotopy extension theorem, fibre-bundles, fibrations and homotopy-groups.

You have a surface $\Sigma$, it's your shirt or whatever else you're interested in, some surface in 3-dimensional space.  Surfaces have automorphism groups, let me call it $\operatorname{Aut}(\Sigma)$.  These are, say, all the self-homeomorphisms or diffeomorphisms of the surface.   And surfaces can sit in space.  A way of putting a surface in space is called an embedding.  Let's call all the embeddings of the surface $\operatorname{Emb}(\Sigma, \mathbb R^3)$.  $\operatorname{Emb}(\Sigma, \mathbb R^3)$ is a set, but in the subject of topology these sets have a natural topology as well.  We think of them as a space where "nearby" embeddings are almost the same, except for maybe a little wiggle here or there.  The topology on the set of embeddings is called the compact-open topology (see Wikipedia, for details on most of these definitions).

Okay, so now there's some formal nonsense.  Look at the quotient space $\operatorname{Emb}(\Sigma, \mathbb R^3)/\operatorname{Aut}(\Sigma)$.  You can think of this as all ways $\Sigma$ can sit in space, but without any labelling -- the surface has no parametrization.  So it's the space of all subspaces of $\mathbb R^3$ that just happen to be homeomorphic to your surface.

Richard Palais has a really nice theorem that puts this all into a pleasant context.  The preamble is we need to think of everything as living in the world of smooth manifolds -- smooth embeddings, $\operatorname{Aut}(\Sigma)$ is the diffeomorphism group of the surface, etc.

There are two locally-trivial fibre bundles (or something more easy to prove -- Serre fibrations), this is the "global" isotopy-extension theorem:

$$\operatorname{Diff}(\mathbb R^3, \Sigma) \to \operatorname{Diff}(\mathbb R^3) \to \operatorname{Emb}(\Sigma, \mathbb R^3)/\operatorname{Aut}(\Sigma)$$

$$\operatorname{Diff}(\mathbb R^3 \operatorname{fix} \Sigma) \to \operatorname{Diff}(\mathbb R^3, \Sigma) \to \operatorname{Aut}(\Sigma)$$
here $\operatorname{Diff}(\mathbb R^3)$ indicates diffeomorphisms of $\mathbb R^3$ that are the identity outside of a sufficiently large ball, say.

So the Palais theorem, together with the homotopy long exact sequence of a fibration, is giving you a language that allows you to translate between automorphisms of your surface, and motions of the surface in space.

It's a theorem of Jean Cerf's that $\operatorname{Diff}(\mathbb R^3)$ is connected. A little diagram chase says that an automorphism of a surface can be realized by a motion of that surface in 3-space if and only if that automorphism of the surface extends to an automorphism of 3-space.  For closed surfaces, the Jordan-Brouwer separation theorem gives you an obstruction to turning your surface inside-out.  But for non-closed surfaces you're out of tools.

To figure out if you can realize an automorphism as a motion, you literally have to try to extend it "by hands".  This is a very general phenomena -- you have one manifold sitting in another, but rarely does an automorphism of the submanifold extend to the ambient manifold.  You see this phenomena happening in various other branches of mathematics as well -- an automorphism of a subgroup does not always extend to the ambient group, etc.

So you try your luck and try to build the extension yourself.  In some vague sense that's a formal analogy between the visceral mystery of turning the surface inside-out and a kind of formalized mathematical problem, but of a fundamentally analogous feel.

We're looking for automorphisms that reverse orientation. For an arbitrary surface with boundary in 3-space, it's not clear if you can turn the surface inside out.  This is because the surface might be knotted. Unknotted surfaces are examples like your t-shirt.  Let's try to cook up something that can't be turned inside-out.

The automorphism group of a 3-times punctured sphere has 12 path-components (12 elements up to isotopy).  There are 6 elements that preserve orientation, and 6 that reverse.  In particular the orientation-reversing automorphisms reverse the orientation of all the boundary circles.  So if you could come up with a knotted pair-of-pants (3-times punctured surface) so that its boundary circles did not admit a symmetry that reversed the orientations of all three circles simultaneously, you'd be done.

Maybe this doesn't seem like a reduction to you, but it is.

For example, there are things called non-invertible knots:

http://en.wikipedia.org/wiki/Invertible_knot

So how do we cook up a knotted pair-of-pants from that?

Here's the idea.  The non-invertible knot in the link above is sometimes called $8_{17}$.  Here is another picture of it:

http://katlas.org/wiki/8_17

Here is a variant on that.

Interpret this image as a ribbon of paper that has three boundary circles.  One boundary circle is unknotted.  One is $8_{17}$.  The other is some other knot.

It turns out that other knot isn't trivial, nor is it $8_{17}$.

So why can't this knotted pair of pants be turned inside-out?  Well, the three knots are distinct, and $8_{17}$ can't be reversed.

The reason why I know the other knot isn't $8_{17}$?  It's a hyperbolic knot and it has a different ($4.40083...$) hyperbolic volume than $8_{17}$ ($10.9859...$).

FYI: in some sense this is one of the simplest surfaces with non-trivial boundary that can't be turned inside-out.  All discs can be turned inside-out.  Similarly, all annuli (regardless of how they're knotted) can be turned inside-out. So for genus zero surfaces, 3 boundary components is the least you can have if you're looking for a surface that can't be turned inside-out.

edited to correct for Jason's comment.

comment added later: I suggest if you purchase a garment of this form you return it to the manufacturer.

A shirt with no hole, or all holes sealed is homeomorphic to a sphere.

A shirt with holes is homeomorphic to a sphere with punctures.

Now you intuitively see that a rubber sphere with punctures can be turned inside out by continuous motion, whereas a perfect sphere cannot.

To understand why, let's concentrate on 2-dimentional plane, taking a circle instead of a sphere. It can be proven that any continuous deformation that inverts the circle will make it cross itself. However if the circle has 'holes', or gaps, then it topologically is equivalent to one or more line segments, which can all be inverted in place without touching each other (appropriate scaling may be necessary).

Similar feature holds for a sphere too. Hence you cannot turn a shirt inside out if all the holes are sealed, but you can do it even if you have one hole (and provided that the shirt is flexible enough to go through that hole if the hole is small).


Proving that a sequence is periodic Given the following sequence,

  $$1,1,2,3,5,8,3,1,4,5,9,4,3,7....$$
  $$a_{n+2}=(a_n+a_{n+1}) \mod{10}\;\;\;\; \forall\;\;{n\geq{1}}$$
  Prove that it is periodic?

My Attempt:

There can be atmost $10 \times 10=100$ unique pairs of integers $(a,b)$ with $0\leq{a,b}\leq9$.

So pairs will certainly start repeating after $100$ pairs.

Let the first pair repeating after $100$ is $(x,y)$.

The problem I am facing is that how can prove that $(x,y)$ will have unique predecessors and successors, i.e., if $$...a,b,x,y,c,d...\in\text{sequence}$$

also,

$$...e,f,x,y,g,h... \in \text{sequence}$$

then , $a=e,b=f,c=g,d=h...$

If I can prove this, I can easily prove that $(x,y)$ will trace back to $(1,1)$ and the sequence is periodic.

Any hints on how I can show this.

Also let me know if I can make the question more clearer.

First of all, when you say that the sequence follows the rule that you specify just by looking at its first entries, you are making a guess ... a reasonable guess, of course, but a guess nevertheless.

Anyway, let's just say that the rule you indicate does indeed specify the sequence we are interested in. Is it periodic?

Your observation that given that the number of pairs of numbers is finite, and hence that there has to be a repeat of some a pair of numbers at some point is correct.

However, there is no need to worry about the predecessors: once you have a repeat of $x,y$ in your sequence, then, given the rule, from then on everything will be the same again, and that is all you need to show it is cyclic.

In fact, there are many sequences that are cyclic, but where the cycle is preceded by a little 'start-up' that does not get repeated ... i.e. where its graph looks like a '6', not a '0'. For such sequences, you would not be able to prove that the predecessors are always the same as well, because that would not be true. But, you still have that cycle, and that makes it periodic.

In sum, the sequence will be periodic, though it may not go back to $(1,1)$

That said, it is actually easy to show that the sequence will go back to $(1,1)$, for all the predecessors will be the same: if you have

$$b,x,y$$

and

$$f,x,y$$

then it must be the case that

$$b+ x =y \mod 10$$

and

$$f+x=y \mod 10$$

and from that it immediately follows that $b=x$ given that $0 \le b,f \le 9$

Once you have established that $b=f$ it likewise follows that $a=e$ ... but you really don't need to show that anymore: once you have shown the immediate predecessor of two numbers is fixed, then all predecessors are fixed. In short, from any $a_n$ and $a_{n+1}$, both $a_{n+2}$ and $a_{n-1}$ are fully determined.

As an element of the sequence only depends on its two predecessors, the argument "$100$ unique pairs" holds and there is a period, not longer than $100$.

From a given pair, say $a,b$, you will see new pairs $b,c$, then $c,d$... But sooner or later you will see a pair for the second time, and this delimits a period.


Evaluate $\int_0^1(x\ln(x))^{50}dx$ Evaluate $$\int_0^1(x\ln(x))^{50} dx.$$

Here are my steps so far using differentiation under the integral sign:

$$I(t) = \int_0^1(x\ln(x))^t dx$$
$$I'(t) = \frac d{dt}\int_0^1(x\ln(x))^t dx = \int_0^1\frac \partial{\partial t}(x\ln(x))^t dx = \int_0^1(x\ln(x))^t\ln(x\ln(x)) dx$$

I can't find a way to continue so hints are appreciated.

Hint. Following your approach by differentiation under the integral sign, note that
$$\int_{0}^{1} x^t (\ln(x))^n \, dx=\frac{d^n}{dt^n}\left( \int_0^1 x^t dx\right)=\frac{d^n}{dt^n} \left((t+1)^{-1}\right).$$


Analytic solution to Poisson equation I need to find the analytic solution to this equation, in order to compare it with solution I get from using a numerical solution. However, I have not been able to find the solution.

I think I can't use separation of variables since the R.H.S cannot be separated. I tried solving it using Green's theorem but to no luck. Any ideas how to approach this?

$\bigtriangleup u = 2x(y −1)(y −2x + x y +2)e^{x−y}$, $(x, y) ∈ (0, 1)×(0, 1)$

Boundary conditions

$u(x, 0) = u(x, 1) = 0$, $x ∈ [0, 1]$

$u(0, y) = u(1, y) = 0$, $y ∈ [0, 1]$

I found the solution:

Since

$\bigtriangleup u = 2x(y −1)(y −2x + x y +2)e^{x−y}$, $(x, y) ∈ (0, 1)×(0, 1)$

The function u(x,t) must have $e^{x−y}$

From the boundary conditions:

$u(x, 0) = 0$ then u(x,t) must be 0 for x=0, so we assume u(x,t) must
have $e^{x−y}*x$
$u(x, 1) = 0$ then u(x,t) must be 0 for x=1, so we assume u(x,t) must
have $e^{x−y}*x*(1-x)$
$u(0, y) = 0$ then u(x,t) must be 0 for y=0, so we assume u(x,t) must
have $e^{x−y}*x*(1-x)*y$
$u(1, y) = 0$ then u(x,t) must be 0 for y=1, so we assume u(x,t) must
have $e^{x−y}*x*(1-x)*y*(1-y)$

The solution should look like this: $u(x,t)=e^{x−y}*x*(1-x)*y*(1-y)$

Doing the calculations leads to:

$\bigtriangleup u = 2x(y −1)(y −2x + x y +2)e^{x−y}$

Which due to the uniqueness of the solution, is the solution.

Calculations can be seen below:


Find the area of the region determined by the system: \begin{align} y & \ge |x| \\ y & \le -|x+1| +4 \\ \end{align} Find the area of the region determined by the system: \begin{align} y
 & \ge |x| \\ y & \le -|x+1| +4 \\ \end{align}

My attempt

Assuming $x>0$

I have the system
$$\begin{cases}\begin{align}
y &\ge x \\
y & \le -x+3 \\
\end{align}\end{cases}$$

Assuming $x<0$

$$\begin{cases}\begin{align}
y &\ge -x \\
y & \le x+5 \\
\end{align}\end{cases}$$

A relevant interval is $-1<x<0$ because in this interval $y \le -|x+1| +4 $ is still positive while $y \ge |x|$ is negative.

How do I combine now this information to solve the problem ?

I don't see how I can  get something of the form $y \ge a$ and $y \le b$ which give clear bounds about  area... in this problem it seems more complicated.

If you can use graphs, then draw the graphs of the functions. The graph of $y=|x|$ is a pretty basic one. You can use transformation to get the graph of the other one. The following is the resulting picture:

You can see that it forms a rectangle. Can you proceed from here?

By the way, if you use algebraic way, you have to discuss three cases:
$$x\leq-1\\
-1\leq x \leq 0\\
x\geq 0$$

Hint:You have 4 equations of form $y=mx+c$ so find $4$ points of intersection $2$ each with opposite slopes also its a polygon with $2$ pair of lines parallel . Can you proceed further.


Are there mini-mandelbrots inside the julia set? I've seen a julia set zoom but it is not nearly as interesting as a mandelbrot zoom. I also have not seen corresponding julia sets for zooms in the mandelbrot deeper than the original image. I'm wondering if it is possible to locate some form of a mini mandelbrot inside the julia set.

This answer has two parts: a partial negative and a partial positive.

Why Julia sets are typically much simpler than the Mandelbrot set

Julia sets of rational functions can be computed using an inverse iteration technique that shows them to be something close to self-similar.  This helps explain the extreme regularity displayed when zooming into most Julia sets.  For example, here we zoom in to the Julia set for $f(z)=z^2-1$ increasing the magnification by a factor of the Golden ratio with each step.

Unlike this, the Mandelbrot set gets more complicated with each step in magnification, rather than staying at the same basic level of complexity.

Julia sets that contain part of a Mandelbrot set

Nonetheless, these inverse images are not strict self-similarities (just not so far off for many functions) so more complicated behavior is possible.  In fact, the Julia sets of some rational functions do indeed contain some Mandelbrot type behavior.  Here's the Julia set of

$$f(z) = \frac{z^3-z}{1+4 z-z^2}$$

Note the appearance of something that looks like the period two bulb of the Mandelbrot set.  I believe that there is, in fact, a genuine correspondence here but that the true nature is not yet fully understood.  It appears to be related to the fact that the period two bulb is missing from the parameter space for a related family of functions:

$$f_a(z) = \frac{z^3-z}{1+az-z^2}$$

The study of this family is ongoing by Jane Hawkins and her students at UNC Chapel Hill and you can find more in her papers, particularly the ones that have the term "rational maps" in the title.

Due to the Julia set being much simpler than the Mandelbrot (it never actually repeats smaller copies of itself), there’s no Mandelbrots in there.


Defining a uniform distribution of points in the plane I saw this question, and was wondering what is the best way to describe a random distribution of points in the plane such that the expected number of points in any region of unit area is $c$, where $c$ is a given positive constant, and such that the number of points in every pair of disjoint regions are independent.
$\def\lfrac#1#2{{\large\frac{#1}{#2}}}$

I can mimic the derivation of the Poisson distribution in the following manner:

Take any $n \in \mathbb{N}_{>c}\,\,.$ Partition each unit square of the plane into an $n \times n$ grid of square cells.
With probability $\lfrac{c}{n^2}$ put a point into each cell, distributed uniformly within the cell, independently of other cells.
Then every region comprising $n^2$ cells has expected number of points $c$.
Also every pair of disjoint regions have independent numbers of points.
Now if $n \to \infty$, I presume that this will more and more accurately resemble the desired distribution of points.

But I have no idea how to take such a limit, nor what the limit distribution is!. I suspect it might be important that the cells are getting smaller in both dimensions, but I don't know. (If each square is partitioned into $n$ horizontal rectangles instead, each rectangle comprising $1 \times n$ cells, and the probability changed to $\lfrac{c}{n}$, the number of points in two vertical regions each comprising $n^2 \times 1$ cells will not be independent. This dependence might not affect the final limiting distribution, if such a limit exists at all, but a priori it is not clear to me.)

So how can we rigorously define such a distribution, and is it really the limit of the above concept, in which case how do we rigorously define this limit?

There is one notable example of such a random collection of points in $\Bbb R^d$.

This example is the Poisson Point Process, also called the Spatial Poisson Process. This random set $N$ of points has the property that for any Borel set $B \subset \Bbb R^d$, we have that $N\cap B$ is distributed as a Poisson random variable with parameter $\lambda(B)$ where $\lambda$ is Lebesgue measure, and such that disjoint regions have independent numbers of points intersecting $N$.

If you're interested and want to know more about random point measures and Poisson Point Processes, you can see the Wiki article and its sources therein.

The construction of such a random cloud of points can be done in great generality, for instance on abstract measure spaces (not just $\Bbb R^n$ with Lebesgue measure). However, such a construction uses the Kolmogorov Extension Theorem and I won't be bothered to write out the details here (although you may want to try it as an exercise). In $\Bbb R^2$ there might be a much simpler construction without using such powerful abstract machinery.

See this PDF for a simple construction of the Poisson random measure using a method similar to yours. And see here for another (possibly unnecessarily convoluted, and rather poorly written) abstract construction using Kolmogorov Extension.

Say the expected number of sites in a given region is $c$; divide the region into $m$ parts.  Approximate the desired distribution by saying you put a site into each part with probability $c/m$.  This forbids the improbable event of more than one site in one of those $m$ parts, so it's not exact.  We need a limit as $m\to\infty$.

The probability that the number (capital) $X$ of sites in the region is (lower-case) $x$ is
\begin{align}
\Pr(X=x) & = \binom m x \left( \frac c m \right)^x \left( 1 - \frac c m \right)^{m-x} \\[10pt]
& = \frac{m(m-1)(m-2)(m-3)\cdots(m-x+1)}{x!} \cdot \frac{c^x}{m^x} \cdot \left( 1-\frac c m \right)^m \left( 1- \frac c m \right)^{-x} \\[10pt]
& = \frac{m(m-1)(m-2)(m-3)\cdots(m-x+1)}{m^x} \cdot \frac{c^x}{x!} \cdot \left( 1 - \frac c m \right)^m \left( 1 - \frac c m \right)^{-x} \\[10pt]
& = \underbrace{\frac{m(m-1)(m-2)(m-3)\cdots(m-x+1)}{m^x}}_{\text{This } \rightarrow \, 1} \cdot \frac{c^x}{x!} \cdot \underbrace{\left( 1 - \frac c m \right)^m}_{\text{This }\rightarrow\, e^{-c}}\  \underbrace{\left( 1 - \frac c m \right)^{-x}}_{\text{This }\rightarrow \,1} \\[15pt]
& \to \frac{c^x e^{-c}}{x!} \qquad \text{as } m\to\infty.
\end{align}

(Notice that the factor $\dfrac{c^x}{x!}$ does not change as $m\to\infty$, so its limit is not problematic (to say the least).)

In other words in the limit, $X$ has a Poisson distribution with expected value $c$.


Let $A,B$ be $m\times m$ matrices such that $AB$ is invertible. Show $A,B$ invertible. Let $A,B$ be $m\times m$ matrices such that $AB$ is invertible. Show $A,B$ invertible.

Here is what I have attempted:

We have $AB$ invertible so there exists an $m\times m$ matrix $C$ such that $C(AB)=(AB)C=I_{m}$. Since matrix multiplication is associative we immediately have a right and left inverse respectively: $(CA)B=A(BC)=I_{m}$. I know the fact that I have square matrices needs to play a roll and gaurentees me that the product will be, in general, defined. I have tried to multiply $CA$ and $BC$ on other sides but that didn't get me anywhere.

I've asked myself what would happen if they weren't square and its obvious, by definition, that they would not be invertible but I am not seeing a way to use the fact that they are to gaurentee ourselves a general inverse.

Okay, so I figured this out.

We have a left and right inverse for our matrices $B$ and $A$ respectively. Since $A,B$ both correspond to linear transformations from a space of dimension n to another of dimension n we know that they are functions which possess a right and left inverse respectively, and thus are onto and one-to-one respectively. Therefore they are both one-to-one and onto since having one property gives the other when we are dealing with vector spaces of finite and equal dimension. Since $A,B$ both correspond to invertible transformations we know they are invertible. Q.E.D.

If $AB$ is invertible, then there exists $D$ s.t. $ABD=I$

Then $A$ is invertible with inverse $BD$.

It remains to show that $B$ is invertible.

Now we have

$$ABD=I$$
$$\to BD=A^{-1}$$
$$\to B=A^{-1}D^{-1}$$

Then $B$ is invertible as it is the product of invertible matrices...wait, is that given?

QED if yes. Idk o/w.


Simplest or nicest proof that $1+x \le e^x$ The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is accessible to students with limited calculus experience, and does not involve too much analysis of different cases.

Another way (not sure if its "simple" though!):  $y = x+1$ is the tangent line to $y = e^x$ when $x= 0$.  Since $e^x$ is convex, it always remains above its tangent lines.

$1+x \le e^x$

Take the ln of both sides

$\ln(1+x) \le x$

differentiate both sides w.r.t $x$

$\frac{1}{1+x} \le 1$

which holds $\forall x \in \mathbb{R}, x \neq -1$.


Prove a functions inverse is differentiable on a domain if it is always increasing or decreasing My calculus textbook gave this theorem without a formal proof:

theorem Suppose that the domain of a function f is an open interval on which $f' (x) > 0$ or on which $f '(x) < 0$. Then $f$ is one-to-one, $f^{−1} (x)$ is differentiable at all values of $x$ in the range of $f$, and the derivative of $f^{−1} (x)$ is given by Formula (2). $\dfrac{dx}{dy}= \dfrac{1}{\dfrac{dy}{dx}}$
Anton, Howard. Calculus, 10th Edition (Page 428). Wiley Higher Ed. Kindle Edition.

For my class I do not need to know the proof but an understanding of what would a formal proof would look like would be helpful.
Furthermore is there a name to this proof.
I conceptually understand this proof as an increasing or decreasing function's inverse would be a function. Therefore, since the function is differentiable and it's inverse is also a function then the inverse would be differentiable. Correct me if I am wrong or if my wording makes no sense.

The fact that the inverse is a function doesn't mean it's differentiable.  For example, $f(x) = x^3$ is an increasing differentiable function on $(-\infty, \infty)$, its inverse is a function, but the inverse is not differentiable at $0$.  You need the added information that $f'$ is never $0$.  See inverse function theorem.


A simple inequality: $(x+y)^p>x^p+y^p?$ Let $x,y>0$ and $p>1$. Is it always true that $$(x+y)^p>x^p+y^p?$$

I think it is true. But can not prove it.

Since the inequality is homogeneous, we may assume without loss of generality that $y=1$ and $x\in(0,1]$, then prove:
$$ \forall x\in(0,1],\qquad (1+x)^p > 1+x^p.\tag{1}$$
On the other hand, if we set $f(x)=(1+x)^p-x^p$, it is straightforward to check that $f'(x) = (p-1)\cdot\left((1+x)^{p-1}-x^{p-1}\right)>0$, hence $f$ is increasing on $(0,1]$. Since $f(0)=1$, $(1)$ is proved.

Let $x,y>0$ and $p>1$ so we must prove $(x+y)^p > x^p + y^p$. We can prove this by saying that $(x+y)^p = x^p + y^p + c(x,y)$ where $c(x,y)$ are terms of $x,y$ that are combined. If $x,y>0$ then so must $c(x,y)$ because it will be the sum of products of positive numbers. Example for $c(x,y)$ $(x+y)^2 = x^2 + y^2 + 2xy$. Thus as we know $c(x,y)>0$ we can say $(x+y)^p = x^p + y^p + c(x,y) > x^p + y^p$ and if we subtract $x^p + y^p$ from both sides we get $c(x,y)>0$ which is true.


Entropy of isometric extension A similar question to mine was asked before at the address below but it was not answered there so I am asking it again. Also there is a more specific case I am interested in.

Topological entropy of isometric extension

This is Exercise 2.5.6 in 'Introduction to Dynamical Systems' by Michael Brin and Garrett Stuck. Let $Y,Z$ be compact metric spaces and let $X = Y \times Z$. Let $d$ be the metric on $X$. Write $\pi:X \to Y$ be the projection onto the $Y$-coordinate. Suppose $f: X \to X$ is an isometric extension of $g: Y \to Y$; that is to say $\pi \circ f = g \circ \pi$ and $d(f(x_1),f(x_2)) = d(x_1,x_2)$ for all $x_1,x_2 \in X$ with $\pi(x_1) = \pi(x_2)$. Prove that $h(f) = h(g)$, where $h$ is the topological entropy of a map.

I understand that points in the same fiber do not diverge, since $f$ acts isometrically there. However, in order to show entropy does not increase we need to control the divergence of points in nearby fibers. I don't see how to do this.

I've been focusing on trying to understand the following specific case. Let $Y = Z = S^1$ so that $X$ is the standard two-dimensional torus. Let $g$ be the identity map on $Y$, so clearly $h(g) = 0$. Take any continuous function $\phi:S^1 \to S^1$ and define a homeomorphism $f_\phi: X \to X$ by letting $f_\phi$ leave every fiber setwise invariant and rotating the fiber over $y \in Y$ by an angle of $\phi(y)$. Then clearly $f_\phi$ is an isometric extension of $g$. However in the case $\phi(y)$ is not Lipschitz I am having a hard time proving that $h(f_\phi) = 0$.

Here is a hint for the proof: We can write $f=\pi(f)\times\pi_Z(f)$ where $\pi,\pi_Z$ are the natural projections onto $Y$ and $Z$ respectively. Now, $\pi_Z(f)$ acts as an isometry on each fiber $\pi^{-1}(y)$, $y\in Y$ and the entropy of an isometry is $0$.


Verifying Serret-Frenet equations I need to verify the Serret-Frenet equations for $ \gamma(t) = (4/5 \cos t, 1-\ \sin t, -3/5 \cos t)$ That is I need to verify $\dot t = \kappa n, \dot n = -\kappa t+ \tau b, \dot b = -\tau n$
Here from the given $\gamma(t)$ I can find $t$ and $\dot t$. But how to find $n$? If I use the relation $n=\dot t/{\kappa}$ won’t it be a circular argument since that is what I need to verify?

For clarity, we need to use different notation for $t$ vector and $t$ as a parameter.
By definition we have that
$$T=\frac{\dot{\gamma(t)}}{\left|\dot{\gamma(t)}\right|}$$
and
$$N=\frac{\dot{T}}{\left|\dot{T}\right|}=\frac{\ddot{\gamma(t)}}{\left|\ddot{\gamma(t)}\right|}$$
with $N\cdot T=0$, indeed
$$\frac d{dt}(T\cdot T)=2T\cdot \dot T=0$$
Refer also to the related

Tangent, Normal and Binormal vectors
How to find unit tangent, normal, and binormal vectors?


Show that $d/dx (a^x) = a^x\ln a$. Show that
$$
\frac{d}{dx} a^x = a^x \ln a.
$$

How would I do a proof for this.  I can't seem to get it to work anyway I try.

I know that
$$
\frac{d}{dx} e^x = e^x.
$$
Does that help me here?

Let $y = a^x$

Then taking log on both side to the base $e$

We have;

$ \ln y = \ln a^x$

$ \ln y = x\cdot  \ln a$

Taking derivative with respect to $x$;

$\frac d{dx} \ln y = \frac d{dx} x\cdot \ln a$

(I have applied chain rule here>>)

$ \frac 1y \frac{dy}{dx} = \ln a + 0$

$ \frac {dy}{dx} = y\cdot  \ln a$

We know $y= a^x$

So, $\frac {dy}{dx} = a^x \ln a$

Hope it was easier than other methodologies used to derive it!


A problem with "Crossed Ladders Theorem" In the following diagram, $AY \parallel BZ$, $AB$ is base. $M$ is $5$ above $N$ and $N$ is $4$ above $O$. What is the height of the triangle $\Delta AOB$.

My Work
There is a theorem named Crossed Ladder Theorem.
In the following Diagram $\dfrac{1}{h} = \dfrac{1}{H_1} + \dfrac{1}{H_2}$

I think this problem can be transformed into this. But I don't know how to. Any hint will be helpful.

Note that there is no statement about the angle between $AY$ and $AB$. The only constraint is that $AY$ and $BZ$ are parallel. It is therefore permissible to move $Y$ to give (amongst many others) the different arrangements shown below:

In all of these $|MC|=5$ and $|EO|=4$ and you will see that $|OF|=h$ is independent of the angle between $YA$ and $AB$.

Of particular interest is the diagram on the right, where $YA$ is perpendicular to $AB$.

This is essentially the same as your diagram for the crossed ladders:

$AM=H_2=h+4+5$

$BN=H_1=h+4$

Using the relationship you described, you now have:

$$\frac 1 h = \frac 1 {h+9} + \frac 1 {h+4}$$


Physical meaning of the null space of a matrix What is an intuitive meaning of the null space of a matrix? Why is it useful?

I'm not looking for textbook definitions. My textbook gives me the definition, but I just don't "get" it.

E.g.: I think of the rank $r$ of a matrix as the minimum number of dimensions that a linear combination of its columns would have; it tells me that, if I combined the vectors in its columns in some order, I'd get a set of coordinates for an $r$-dimensional space, where $r$ is minimum (please correct me if I'm wrong). So that means I can relate rank (and also dimension) to actual coordinate systems, and so it makes sense to me. But I can't think of any physical meaning for a null space... could someone explain what its meaning would be, for example, in a coordinate system?

Thanks!

If $A$ is your matrix, the null-space is simply put, the set of all vectors $v$ such that $A \cdot v = 0$. It's good to think of the matrix as a linear transformation; if you let $h(v) = A \cdot v$, then the null-space is again the set of all vectors that are sent to the zero vector by $h$. Think of this as the set of vectors that lose their identity as $h$ is applied to them.

Note that the null-space is equivalently the set of solutions to the homogeneous equation $A \cdot v = 0$.

Nullity is the complement to the rank of a matrix. They are both really important; here is a similar question on the rank of a matrix, you can find some nice answers why there.

Think of an observer and n number of speakers at different distance and in directions. Now make a matrix of equations for sound from each speaker, based on contribution of their amplitude, frequencies and phase. Null space will be formed of all possible combination that you can set in a way that, the total/superimposed sound at observer location will be zero. Means, observer will not hear anything even if the speakers are playing.


Lagrangian Torus of a symplectic manifold Suppose $(M,\omega)$ is a $2n$-dimensional symplectic manifold. Consider the $n$-torus $\mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n.$ Then $\mathbb{T}^n$ is a Lagrangian submanifold of the symplectic $2n$-torus $\mathbb{T}^{2n}$, equipped with the unique symplectic form that pulls back to the canonical symplectic form on $\mathbb{R} ^{2n}$. Now, how does it follow that there exists a Lagrangian submanifold of $M$ that is diffeomorphic to $\mathbb{T}^{n}$?

The fact that $\mathbb T^n$ is a Lagrangian submanifold of $\mathbb T^{2n}$ plays no role. Indeed, by Darboux's theorem, for any symplectic manifold $(M^{2n}, \omega)$ and for any $x\in M$, there is $U_0 \subset \mathbb R^{2n}$ and a local coordinates chart $\phi : U_0 \to U$ so that $\phi^* \omega = \omega_{std}$ and $x\in U$, where $\omega_{std}$ is the standard symplectic structure on $\mathbb R^{2n}$. Since it is easy to find Lagnrangian in $(U_0, \omega_{std})$ diffeomorphic to $\mathbb T^n$: choose $p =(p_1, \cdots, p_n)\in U_0$ and $r_1, \cdots, r_n >0$ small enough, then
$$ \mathbb S^1 (p_1) \times \cdots \times \mathbb S^1 (p_n) $$
is such an example.
Of course this also shows the existence of infinitely many Lagrangian tori on any symplectic manifold.


When does the integral $\iiint_{\Bbb R^3}\frac{1}{(1+x^2+y^2+z^2)^p}\,dxdydz$ converge? Consider $$\iiint_{\Bbb R^3}\frac{1}{(1+x^2+y^2+z^2)^p}\,dxdydz$$

I believe it converges when $p<\frac{1}{2}$, but I couldn't show it. Any hints?

Using spherical coordinates,

$$I=4\pi\int_{r=0}^\infty\frac{r^2}{(1+r^2)^p}dr.$$

The integrand is asymptotic to $r^{2-2p}$ so that the integral converges for $p>3/2.$

Consider the spherical coordinates $r ∈ [0, \infty), \theta ∈ [0, \pi], \phi ∈ [0, 2π),$ with
$$x = r\sin\theta \cos \phi, y = r\sin \theta \sin \phi, z= r\cos \theta,$$
then you get $$4 \pi \int_O ^\infty \frac{r^2}{(1+r^2)^p} dr,$$
which converges iff $2p -2 >1.$


The irrationals, the Cantor Set, and $F_\sigma$ sets I know that $\mathbb{R}\setminus\mathbb{Q}$ is not an $F_\sigma$ set.  However, what about $\mathbb{R}\setminus\mathbb{Q}\cap$ Cantor set? Is that an $F_\sigma$?

HINT: Let $C$ be the Cantor set. $(\mathbb{R}\setminus\mathbb{Q})\cap C=C\setminus\mathbb{Q}$, which in some ways is a lot like $\mathbb{R}\setminus\mathbb{Q}$. How do you prove that $\mathbb{R}\setminus\mathbb{Q}$ is not an $F_\sigma$? Can you use the same basic idea on $C\setminus\mathbb{Q}$?

Do you know the properties of the Cantor set, i.e., is it closed, open, etc? Find out, and that will answer your question.


Can you suggest some challenging calculus questions? I am a calculus 2 student, just passing high school. I would really appreciate it if you can suggest some cool challenging questions for a high school student with solutions or hints too!

Here are two that I've seen on this network.
Problem 1. Show that
$$
\sum_{n=1}^{\infty}\big(\frac{1}{4n+1}-\frac{1}{4n}\big)=\frac{1}{8}(\pi-8+6\ln 2)
$$
Problem 2. Evaluate the indefinite integral
$$
\int e^{\sin x}(x\cos x-\sec x\tan x)dx
$$