What is the meaning of "Hermitian"? Google search-bar gives the definition of Hermitian as:

  Hermitian: denoting or relating to a matrix in which those pairs of elements that are symmetrically placed with respect to the principal diagonal are complex conjugates

I have thought that Hermitian was synonymous with "real", meaning, if the matrix (A, for example) is Hermitian then that means there are no complex values in the matrix. I also believe it means the complex conjugate of the matrix is equal to the matrix like so:
$$A = A^\dagger.$$

However, there also exist Hermitian functions (which are complex?!) and the Hermitian operator (does not have to be real). Could  someone please tell me what does the word "Hermitian" mean and what are the differences between the three: Hermitian matrix, Hermitian function, and Hermitian Operator? I am confused.

(PS: Please feel free to correct me if I have tagged this question incorrectly.)

A Hermitian matrix is a matrix that is equal to its conjugate transpose. This generalizes the concept of a "symmetric matrix", since every real symmetric matrix is Hermitian. However, there are certainly complex matrices that are Hermitian, such as

$$ \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix} $$

By the Spectral theorem, every Hermitian matrix is unitarily diagonalizable with all real eigenvalues. Therefore, while a Hermitian matrix can have complex entries, in an appropriate basis it behaves like a real matrix.

When we describe a linear operator (or linear function) as being Hermitian, we are using a basis-free approach to Hermitian-ness. A linear operator $T: V \rightarrow V$ on an inner product space $V$ is said to be Hermitian if $T = T^*$, where $T^*$ is the unique operator such that $\langle Tx, y \rangle = \langle x, T^* y \rangle$ for every $x,y \in V$.

The connection between the two concepts is that an $n \times n$ Hermitian matrix is the matrix representation of a Hermitian operator on $\mathbb{C}^n$ equipped with the standard dot product.

I have never heard about a Hermitian function but Wikipedia defines it to be a function such that $f(-x) = \overline{f(x)}$. A Hermitian matrix is a matrix that is equal to its tranconjugate, that is to the complex-conjugate of its transpose matrix. In order to speak about a Hermitian operator, one has to be in a complex vector space $E$ with a Hermitian inner product $\langle \cdot,\cdot \rangle$ on it. Then a linear map $f$ from $E$ to itself is Hermitian if it is equal to its adjoint, that is, if we have the equality
$$\langle f(x),y\rangle = \langle x, f(y)\rangle.$$

A matrix with real coefficients has no particular name, except real matrix. Nevertheless, remark that a Hermitian matrix has real values on the diagonal. Moreover, the spectral theorem says that any Hermitian matrix is conjugate by a unitary matrix, to a diagonal matrix with real entries.


Fisher information for exponential family: Regularity conditions for the Fisher-Information to be defined certain regularity conditions have to be fulfilled (like in Lemma 5.3. in Theory of Point Estimation by E.L. Lehmann or on slide 2 here: http://www.stat.nus.edu.sg/~stachenz/40-530Statistics-6.pdf).

I know that any exponential family satisifies these conditions. But how does one formally show property (iv) on slide 2 in the last link for exponential families, i.e. that one can interchange derivative and integral/expectation? Should one use the dominated convergence theorem? And how does one prove the second assumption (ii) on slide 5 of the link? If you know literature containing the proof I am interested in reading it.

Looking forward to your answers!

Thanks!

I think that this follows from Theorem 5.8 in Theory of Point Estimation by Lehmann and Casella (1998, p.27):

you can choose $\eta_i = \theta_i$ (the parameters of a distribution) in the abovementioned formula.

To show both of the properties hold, you can use the Leibniz Integral Rule which lets you pull the derivative inside the integral. Since, for the exponential family $p(x,\theta)$, the partial derivatives of $p(x,\theta)$ with respect to t exist, and is continuous.


Limit of n-th root of $1/n$ I'm struggling to calculate the limit

$$\lim_{n\rightarrow\infty}\left(\frac{1}{n}\right)^{\frac{1}{n}}$$

I have tried taking log,

$\log(\frac{1}{n})^{\frac{1}{n}}=\frac{1}{n}\log({\frac{1}{n}})$
and setting $t=\frac{1}{n}$ and rewite the desired limit as
$e^{\lim_{t\rightarrow 0}t\log(t)}$,

and I'm stuck here because $\log$ is not defined at $0$.

Did I miss something?

Thank you for reading.

$\left(\dfrac{1}{n}\right)^{\frac{1}{n}} = \dfrac{1}{\sqrt[n]{n}} \to 1$ because:

$1 < \sqrt[n]{n} = \sqrt[n]{1\cdot 1\cdot 1...\cdot \sqrt{n}\cdot \sqrt{n}} < \dfrac{1+1+...+1 + 2\sqrt{n}}{n} = \dfrac{n-2+2\sqrt{n}}{n} = 1 -\dfrac{2}{n} + \dfrac{2}{\sqrt{n}}$. Apply squeeze theorem the answer follows.

this is a standard matter.  I hope you've worked out a conjecture first by using a computer programme or just plugging in some numbers on the formula. (You should, as it is a healthy thing).

Either you continue as you started (continuous variable $t$), and write your limit as

$$\frac{log(t)}{1/t},$$ (i.e. a quotient of infinities) and apply whichever result you please to your newly-found indeterminate limit (L'Hopital), or you can use a standard result on sequences.

I suggest e.g. D'Alembert's Criterion or the Quotient Criterion (look up Apostol's Calculus, for instance, its chapter on sequences). (Stoltz's theorem also works).

A nicer way, which can accompany you for the rest of your life, is as follows.

Consider $1+x_n=n^{1/n},$ and write the identity
$$n=(1+x_n)^n.$$

You can use Newton's binomial theorem here, and establish (choosing your favourite term, I'll choose the degree-2 term) that, for instance,
$$n> \frac{n(n-1)}{2} x_n^2 \geq 0.$$

After isolating, you'll see immediately what the expected result is.

Your sequence is very much related to this one.

There are many resources available online, e.g.
http://planetmath.org/proofoflimitofnthrootofn

REGARDING POLYNOMIALS AND EXPONENTIALS:  Actually, there are recipes such as this one: the exponential always beats a polynomial, or a power of $x$, as $x\to +\infty.$  The argument would be very similar.  Say, you want to show that, for $a>1$,

$$\frac{a^x}{x^p}$$ tends to infinity; you may restrict to $x=n$, and then compare (write $a=1+b$, for $b>0$)

$$(1+b)^n= 1 + nb + \cdots + \frac{n(n-1)\ldots (n-p+1)}{(p+1)!} b^{p+1} + \cdots,$$ which clearly is bigger than $n^p$, for $n \to \infty$.

I leave it as an exercise for you to see what happens as $x\to +\infty$ (take the integer part $n=n(x)$ of $x$, and bound numerator and denominator appropriately comparing $f(x)$ to $f(n)$).


Integrate $\displaystyle\int{\frac{1}{2+3x}dx}$ Integrate $$\int{\frac{1}{2+3x}dx}$$

What should I do for this integral?  How do I evaluate it?

Thanks!

HINT:

Put $2+3x=u$ so that $3dx=du$

and we know $$\int \frac{du}u=\ln |u|+C$$ where $C$ is an arbitrary constant for indefinite integration

Integrate 1/2+3xdx............ let u=2+3x, so that 3dx=du, .'. dx=du/3u.......and we know........int:du/u=In\u/+C, where c is the arbitrary constant for indifinite integration........ Therefore, int:du/3u=1/3 In u+c..........final answer is...........1/3In(2+3x)+C. Hope this help........ By O'john


$(P\implies Q) \implies [(R ∨ P)\implies (R ∨ Q)]$ is a tautology I'm currently trying to work on the proof for this tautology.
But every time I derive the right side, I end up with a lone $R$ that will never cancel out.

Like I always end up with

$$(P\implies Q) \implies  ¬P∨Q∨R$$

And the $R$ never goes away at this point.

What can I do to cancel out the $R$?

HINT: $(P\to Q)\to\lnot P\lor Q\lor R\equiv(P\to Q)\to((P\to Q)\lor R)$.

Here is a proof using natural deduction in a Fitch-style proof checker. The proof is similar to what Joshua Taylor provided.

The inference rules used are disjunction introduction ($\lor$I), disjunction elimination ($\lor$E), implication introduction or conditional proof ($\to$I) and implication elimination or modus ponens ($\to$E).

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/


Converting my C# code into a mathematical formula How can I convert my C# code into a mathematical formula? I would like to use discrete mathematics and convert my code into a mathematical formula.
Here is the example C# code that I have written. The aim is to calculate a rank based on the positive and negative weight that is assigned to each variable. The positive weight is added to the rank if the variable is higher than 0. The negative weight is taken away from the rank if the variable is a negative number. Please keep in mind that each variable can have any negative or positive weight that can be configured differently.
var aPosWeight = 3;
var aNegativeWeight = -3;

var bPosWeight = 2;
var bNegativeWeight = -2;

var cPosWeight = 1;
var cNegativeWeight = -1;

var a = 1;
var b = 0;
var c = -1;

var rank = 0;

if (a > 0)
{
    rank += aPosWeight;
}
else if (a < 0)
{
    rank += aNegativeWeight;
}

if (b > 0)
{
    rank += bPosWeight;
}
else if (b < 0)
{
    rank += bNegativeWeight;
}

if (c > 0)
{
    rank += cPosWeight;
}
else if (c < 0)
{
    rank += cNegativeWeight;
}

Console.WriteLine(string.Format("Your final rank is {0}", rank));

Here is the mathematical formula that I have come up with that I think represents the C# code written above. My main concern is that the mathematical formula is completely wrong and can't be used to interpret the C# code that I've written above.
$$
f(x)=\begin{cases}\text{PosWeight},\quad\text{if }a>0\\
\text{NegWeight},\quad\text{if }a<0\\
0\hphantom{egw.eight},\quad\text{if }a=0\end{cases}
$$
$$
\text{sum}=\sum_{i=1}^3f(x_i)
$$

Let $p,\,n$ denote the positive and negative weights, so $f(x)$ is $p[x>0]+n[x<0]$ in terms of Iverson brackets.

If you're trying to do this automatically, you could write a parser and input the specific 'final value' and then determine how it was composed by building a tree of derivation steps and then look at how it was used, the existence of datasets etc and tie these activities to mathematical symbols.
eg.
float inputa, inputb, finalvalue;
..
..
for (int x=0; x < 5; x++)
{
inputa = inputb *x;
finalvalue+=inputb;
}
where finalvalue is the derived value.
in this case the for loop would look like a sum (sigma) from 0 to 5  with  bx as its term.
anyway, after doing this you could use a latex renderer to create a graphic of your equation, like they're doing above with their answers.


Show that the closure of $A$ is the intersection of all closed sets containing $A$, topology proof needed I want to show that given $(X, \mathcal{T})$, we define $\overline A = \{x \in X| \forall U \in \mathcal{T}, x \in U \implies U \cap A \neq \varnothing\}$ (definition of closure from Munkres), then

  Show that $\overline A = \bigcap\{C \subseteq X | C \text{ is closed }, A
\subseteq C\}$

I find this really hard to tackle because some unnaturalness in that $\overline A$ is specified with respect to open sets, but then it is alternatively defined as intersection of closed sets..how to juggle between open and closed?

Several other posts also doesn't help...

The closure of A is the smalled closed set containing A is proved in terms of accumulation points and limit points which I do not define
Proving that the closure of a subset is the intersection of the closed subsets containing it is defined wrt of metric spaces

I am stuck on both inclusions and need some help

Attempt:

$(\overline A \subseteq \bigcap\{C \subseteq X | C \text{ is closed }, A
\subseteq C\})$

Let $x \in \overline A$, then $\forall U \in \tau, x \in U \implies A
   \cap U \neq \varnothing$. We want to show that $x \in \bigcap\{C
   \subseteq X | C \text{ is closed }, A \subseteq C\}$
So we know that $x$ is contained in some $U' \in \mathcal{T}$ that
has non-empty intersection with $A$, $x$ not necessarily in $A$.
Let $C_1$ be a closed set containing $A$, then $U' \cap C_1 \neq
   \varnothing$.  Let $C_2$ be a closed set containing $A$, then $U'
   \cap C_2 \neq \varnothing$. Assuming $C_1 \subseteq C_2$, then   $U'
   \cap C_1 \cap C_2 \neq \varnothing$.
Continue this way, $U' \cap \bigcap\limits_{\alpha \in I} C_\alpha
   \neq \varnothing$, where $\bigcap\limits_{\alpha \in I} C_\alpha$ is
the intersection of all closed sets containing $A$
We know already that $x \in U'$, but from above how can we see that $x
   \in \bigcap\limits_{\alpha \in I} C_\alpha$? From figure below, it
seems that $x$ will not be in $\bigcap\limits_{\alpha \in I}
   C_\alpha$

$( \bigcap\{C \subseteq X | C \text{ is closed }, A
\subseteq C\} \subseteq \overline A)$

Let $x \in  \bigcap\{C \subseteq X | C \text{ is closed }, A
   \subseteq C\}$, we want to show that $x \in \overline A$. It suffices
to show that $\forall U \in \mathcal{T}, x \in U \implies U \in A
   \neq \varnothing$.
Since $x \in \bigcap \{C\}$, then there exists some closed set $C'
   \subseteq X$ such that $x \in C'$. Let $U \in \mathcal{T}$ containing
$x$, then we will show that $U \cap A \neq \varnothing$
We know that $x \in C' \cap U$, then $x \in \bigcap{C} \cap U$. At this point however I still don't know whether $U \cap A \neq \varnothing$. Couldn't we have a case in figure below where $x \in \bigcap \{C\}$ and $x \in U$, but $U \cap A = \varnothing$?

You’re getting bogged down in the details of the definitions and thereby making it much harder than it really is.

For the first inclusion, start, as you did, with an arbitrary $x\in\operatorname{cl}A$. Let $C$ be any closed set such that $A\subseteq C$. Suppose that $x\notin C$: then $x\in X\setminus C$, and $X\setminus C$ is open, so $(X\setminus C)\cap A\ne\varnothing$. But on the other hand we know that $A\subseteq C$, so $A\cap(X\setminus C)=\varnothing$. This contradiction shows that $x\in C$, and since $C$ was an arbitrary closed set containing $A$, we conclude that

$$\operatorname{cl}A\subseteq\bigcap\{C\subseteq X:A\subseteq C\text{ and }C\text{ is closed}\}\;.$$

For the opposite inclusion just observe that $\operatorname{cl}A$ is one of the closed sets containing $A$, so if $x\in\bigcap\{C\subseteq X:A\subseteq C\text{ and }C\text{ is closed}\}$, then automatically $x\in\operatorname{cl}A$. It follows that

$$\bigcap\{C\subseteq X:A\subseteq C\text{ and }C\text{ is closed}\}\subseteq\operatorname{cl}A$$

and hence that

$$\bigcap\{C\subseteq X:A\subseteq C\text{ and }C\text{ is closed}\}=\operatorname{cl}A\;.$$

Let $x$ in $\bar{A}$. Note that $\bar{A}$ is closed, so $x$ is in a closed set. Suppose for some closed $C$ containing $A$ that $x$ is not in $C$. That means $x$ is in $C$'s complement, which is open. But that contradicts the fact that $x$ was in a closed set, so $x \in C$. Since $x$ was arbitrary in $\bar{A}$, and $C$ was an arbitrary member of all $C$ containing $A$, we have proved $\bar{A} \subset \mathscr{C}$, where the fancy "C" is the set of all closed sets containing $A$. Keep in mind that if a set exists in every one of a collection of sets, then it must also be in their intersection.

Now, for equality we need to show $\bar{A} \supset \bigcap{C}$. Let $x$ in $\bigcap C$. Suppose $x$ not in $\bar{A}$, then $x$ is in $\bar{A}^c$ which is open. But we just said $x$ is in an intersection of closed sets, so it must be in a closed set. Contradiction again.

The one gap in this proof is the fact that these sets could have been $\mathbb{R}^d$. But in that case equality is automatic.


Why defining tangent bundle? I'm learning a bit about smooth manifolds, and currently I'm learning about tangent bundles (just definitions mainly) and vector field.

This is my reference : Tu's Introduction to Manifolds.
I was also watching a video about tangent bundles (because I was struggling with the concept).

Once I understood the definition however I realized I have an issue understanding why we need the notion of tangent bundle. I can't remember where I read this but am I right when I say that tangent bundles are necessary if we want to generalize the notion of function on a manifold?

Consider a smooth manifold $M$, if we wanted to define what a vector field is to me the definition should reflect the fact that for each $p \in M$ we have a $v \in T_p M$, therefore it should be a map.

This is probably the key why such association isn't good as definition because a map needs both domain (in this case $M$) and an image space, however my naive definition involves for each $p$ a different space $T_p M$ and this is why we need the notion of tangent bundle.

Is this observation correct?

Remember that the tangent bundle is the disjoint union of the tangent spaces:
$$TM = \coprod_{P \in M} T_P M.$$
It has the topology of a smooth manifold in the following manner. Let $(U_\alpha, \phi_\alpha)$ be an atlas for $M$, and let $\pi: TM \longrightarrow M$ be the natural projection, i.e. if $(P, v) \in T_P M \subset TM$, then $\pi(P, v) = P$.

Edit: On the Why? bit

The point of this is just to establish a mathematical framework in which we can talk about points on a base manifold, and all the possible curves through any point on the manifold. We can talk about the base points $x$, together with possible directions at $x$.

Actually we know how to do calculus (differentiation, integration...) for functions on $R^n$. And if,
$f:M\rightarrow N$
is smooth map then its differential
$df:TM\rightarrow TN$
Will be map between tangent spaces so overall we will need knowledge of tangent spaces if we want to study smooth maps between manifolds.


Does graph theory have application in pure mathematics? I have read some part of graph theory,the  question pop up in my mind, does
it have application in pure mathematics?
I mean like number theory,analysis,algebra,toplogy .... etc
Can it be used to prove some problem in pure mathematics ?
Anyone have any idea ?

Graph Theory should be Pure Mathematics because we can study its elements in the way we would study any other geometric object in Geometry. Things get out of control when people call Inequalities Applied Mathematics or Analysis of Convex Functions Convex Analysis, but they do that... In my point of view, Inequalities and Convex Analysis are a horrendous  mistake. An inequality involving Real Analysis should be studied inside of Real Analysis: if we call that inequalities, then it is a branch of Analysis, not Applied Mathematics. If we study a real convex function, that could be inside of Real Analysis. In the same way, Graph Theory should be inside of Geometry if what we are doing there is dealing with geometric objects in the traditional way: studying the shape, producing reasoning that leads to calculations, and so on.


Minimum of an approximated function I have the following function
$$ f(x,y) = 100(y-x^2)^2 + (1-x)^2$$
which has minimum at point $(1,1)$. I was asked to find it's quadratic approximation by using Taylor's theorem. By using the derivatives
$$f_x=400(x^3-xy)+2x-2$$
$$f_y=200(y-x^2)$$
$$f_{xx}=400(3x^2-y)+2$$
$$f_{yy}=200$$
$$f_{xy}=-400x$$

I found Taylor's expansion around $(0,0)$ as follows:
$$Q(x, y) = f(0,0)+f_x (0,0)x +f_y(0,0)y+\frac{1}{2}f_{xx}(0,0)x^2+f_{xy}xy + \frac{1}{2} f_{yy}(0,0)y^2 $$

$$=1 -2x +0y +\frac{1}{2}2x^2+0xy+\frac{1}{2}200y^2$$
$$=x^2-2x+100y^2+1$$

However, this new function has a minimum at point $(1,0)$. My profesor told me that this can't be right and that the new approximated function $Q$ should have min at the same point $(1,1)$.

Im not sure why is this true and I cant find any mistakes in my solution. Am I wrong in thinking that it makes sense that the new fucntion $Q$ doesnt have the same minima with the old?

Does anyone have any insight?

Thanks in advance !


Supremum over all simple closed curves of integral This is a question from an old qualifying exam at my university. I expect there's some trick but I couldn't spot it.

The question is to find the supremum over all simple closed curves $C$ of $$\int_C (y^3 - y) \  dx - 2x^3 dy$$ and then determine if the supremum is actually obtained or not.

I thought that perhaps if this integral was actually independent of path, then it would follow that over simple closed curves, this integral would always be $0$ and that would make the supremum easy - just $0$. But this isn't true, as just integrating around the regular unit circle gives $-5\pi/4$.

My next thought was maybe Green's Theorem. I can write this as $$\int_{S_C} (-6x^2 - 3y^2 + 1) \  dx \ dy$$ where $S_C$ is the interior of the simple closed curve $C$ Then by linearity of the integral it appears that there is a term which just keeps track of the area of the region, which I can make as large as I like. I don't know how the other terms might effect things in this situation though, and if the supremum isn't infinity, I don't know if I'm any closer to working it out like this than I was the first way.

All the problems on the exam are straight forward, and none require any particularly high powered weaponry. So probably there is an easy way to do it. Can anyone point me in the right direction?

If you take, say, a rectangle oriented clockwise with vertices $[0,0]$, $[0,1]$, $[X,1]$, $[X,0]$ you can make the integral arbitrarily large.
Or was there an assumption on the orientation of the curve?


integral of absolute value of a function. The problem is as following:
straight lines are drawn joining every point on $y=\sqrt{3}x\sin{x}$ to the origin. what fraction of the x-y plane is occupied by these lines.

My solution is:

I observe that the limit of $y=\sqrt{3}x\sin{x}$ is oscillating between positive and negative. So the graph of the function cross x-axis infinitely many times as $x$ goes to infinity. What the problem ask is actually to compute the fraction of the integral of absolute value of the funciton to the area of the $xy$-plane. Since $y=\sqrt{3}x\sin{x}$ is even function, it is enough to consider right half of $xy$-plane. I compute the integral of absolute value of $y=\sqrt{3}x\sin{x}$ on the right half plane by spliting the right half $xy$-plane by $k\pi$'s, where $k$ represents an integer range from $0$ to infinity.

$\frac{\sum_{k=0}^{\infty}|\int_{k\pi}^{(k+1)\pi}\sqrt{3}x\sin{x} dx|}{2\sum_{k=0}^{\infty}\pi\lim_{x\rightarrow \infty}{x}}$

After computation, the result on the top of the fraction is $\sqrt{3}\sum_{k=0}^{\infty}(2k+1)\pi$ and the result on the bottom of the fraction is $\lim_{n\rightarrow\infty}(2n\pi\lim_{x\rightarrow \infty}{x})$.

Hence the fraction is result of the fraction is $\lim_{n\rightarrow\infty}\frac{\sqrt{3}(n+1)^2}{2n{\lim_{x\rightarrow \infty}{x}}}$, which is just to compute the limit of the expression $\lim_{x\rightarrow \infty}\frac{\sqrt{3}(x+1)^2}{2x^2}=\frac{\sqrt{3}}{2}$.

But the place I find this question, it is with an answer that is $\frac{2}{3}$.

So I am wondering if these are any mistakes of my solution? or I am correct and the answer I found is wrong?

I think you complicated your life.

Here is the graph of $\sqrt{3}x\sin(x)$.

Since $|\sin(x)|<1$ then $-\sqrt{3}x\le f(x)\le\sqrt{3}x$ and the set of lines joining every point of the curve to the origin is just the part of the plane between these two straight lines (imagine the blue part, all filled with blue).

Since this region is infinite as well as the plane, I guess the ratio is calculated by evaluating the angular filling.

The angle of $y=\pm\sqrt{3}x$ are $\pm 60°$ so the overall angular load factor is $\dfrac{4\times60°}{360°}=\dfrac 23$

Let me explain myself in another way by answering a question: What fraction of $xy$-plane occupied by the area bounded by $|\sqrt{3}x\sin{x}|$ and the $x$-axis?

After computing the integral of $|\sqrt{3}x\sin{x}|$ as what I did in my first post, the solution leads to compute the limit of
$\frac{\sqrt{3}\lim_{M\rightarrow\infty}(M+1)^2}{2\lim_{N\rightarrow \infty}N^2}$
 where both top and bottom are discrete functions.

For this question, I think we can choose any function, to compute the area of the $xy$-plane ( as long as this function is able to get the area. e.g we can choose either discrete or continuous function). But for this specific problem, we'd better to use discrete function, because the top of the fraction is discrete. In order to use the quotient rule for limit, we require that both top and bottom of the quotient to have the same property (both discrete or both continuous).

How can we use the quotient rule to calculate limit of a function with multiple variables? This arises to same problem as to compute the limit:
$\lim_{x\rightarrow 0, y\rightarrow 0}\frac{xy}{x^2+y^2}$
We can solve this question by change the Cartesian coordinate to polar coordinate. But for the question here, this technique is not applicable.

I compute the limit of this quotient with different relation between $M$ and $N$. For example use linear relation $N=\sqrt{3}M$ or $N=M^2$. We get different answer.

Conclusion: The limit of this quotient do not exist. the theorem I use is that a function can only have one limit if its limit exists.

Hence, the solution of this question does not exist.

We should not change different variables directly to the same variables. As this changes, we narrow our computation to a fix relation between two variables. Actually, if not specifically mentioned, there will be no relation between variables. in this question, there is no extra conditions mentioned to restrict the variable for the functions on the top of the quotient and on the bottom of the quotient to only one relation: $M=\sqrt{3}N$. We should not add a relation ourselves, which may cause incorrect judgement for the existence of the limit.


How to prove $\lim_{x\to0}\sin x\log x = 0$ How to prove $\lim_{x\to0}\sin x\log x = 0$

My try:

I tried using expansions of sin and log but that does not seem to work.

you can't use any taylor expansion for $\log(x)$

But you can say that $\sin x \sim x$, then you have $$\lim_{x \to 0} x\log x = 0$$

Because $x$ goes to $0$ so much faster than $\log x$ goes to infinity.

Or if you have to justify that,  rewrite the limit as

$$\frac{\log(x)}{\frac{1}{x}}$$ then use l'Hopital

lim x tends to 0 log sin x/1/x
=Limx tends to 0cosx/sinx/1/x^(-2)
=limx tends to 0 cotx/-1/x^2
=Limx tends to 0 -x^2/tanx
=Limx tends to 0 -2x/sec^2x
=0


Subgradients of non-convex functions In these notes (section 2.3), it is stated that:

  A point $x^*$ is a minimizer of a function $f$ (not necessarily convex) if and only if $f$ is subdifferentiable at $x^*$ and $0 \in\partial f(x^*).$

Could anybody provide me with references for a proof of the above statement?

Is there a reference where we can learn more about subgradients of non-convex functions?
In Section 3. Calculus of subgradients of the above notes,many properties of subgradients are presented for convex functions. I would like to know which ones among these properties still hold for non-convex functions.

Thank you in advance!

$$x^* \text{ minimizes } f(x) \Longleftrightarrow 0\in\partial f(x^*)$$
is trivial by definition of subgradients: $$f(x) - f(x^*) \ge \partial f(x^*)^T (x-x^*)\quad \forall x.$$

Thanks to Michael Grant for his comments.

The answer above is true only when $f$ is convex.


How to solve $\int\ x^{\ln x} dx$? How to solve this integral

  $$\int\ x^{\ln x} dx$$

step by step?

We assume $x^{\ln x}$ is well defined.

Then we make the change of variable $x=e^{u}$, $dx=e^{u}\:du$, obtaining
$$
\begin{align}
\int x^{\ln x}\:dx&=\int e^{(\ln x)^2}dx
\\&=\int e^{u^2+u}du
\\&=e^{-1/4}\int e^{(u+1/2)^2}du
\\&=\frac{\sqrt{\pi }}2e^{-1/4}\: \text{erfi}\left(u+\tfrac12\right)
\end{align}
$$ where we have used the special function $\text{erfi}(\cdot)$.

Finally,

  $$
\int x^{\ln x}\:dx=\frac{\sqrt{\pi }}2e^{-1/4}\: \text{erfi}\left(\ln x+1/2\right)+C.
$$

One may notice that
$$
\operatorname{erfi}(z)= \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{z^{2n+1}}{n! (2n+1)},\quad z \in \mathbb{C}.
$$

get $$u=Lnx$$
so $$du=\frac{1}{x}dx$$
the main integral became
$$\int x^{u}xdu=\int x^{u+1}du\overset{x=e^{u}}{\rightarrow}\int (e^{u})^{u+1}du=\int e^{u^2+u}=\int e^{(u+\frac{1}{2})^{2}-\frac{1}{4}}du\overset{p=u+\frac{1}{2}}{\rightarrow}e^{\frac{-1}{4}}\int e^{p^2}dp$$
We can calculate the last integral by double integration.
$$I=\int e^{p^2}dp=\int e^{q^2}dq$$
so $$I^2=\int \int e^{p^2+q^2}dpdq\overset{r^{2}=p^2+q^2}{\rightarrow}\int \int re^{r^2}drd\theta =2\pi\frac{1}{2}e^{r^2}$$
finally $$I=\sqrt{\pi}e^{r}$$
and your answer is $$ans=\sqrt{\pi}e^{r-\frac{1}{4}}$$


Proof that Category Product is a Category let A and B be categories, and let's define category product as:

$A \times B \;\;$ where:

$(1)$ the objects are pairs $(a, b)$ such that $a \in A$ and $b \in B$.

$(2)$ the arrow work like $(f,g) : (A, B) \rightarrow (C, D)$ where:

$f \in A, \;\;f:A \rightarrow C$
$g \in B, \;\;f:B \rightarrow D$

$(3)$ the identity is defined as $\;1_{(A,B)} = (1_{A}, 1_{B})$

$(4)$ the composition is defined as $(f, g)\circ (f',g')$ = $(f\circ f',\;g\circ g')$

I think I should prove it like this:

1 - Prove that the identity is identity for composition operation

2 - Prove that the composition is associative.

[1] let $(f,g) = fg $

$fg = fg \circ 1_{(A,B)}$

so

$fg( (a,b) ) = fg \times 1_{(A,B)}(a,b)$

$fg( (a,b) ) = fg ((a,b))$

my questions are:

do I have to prove what I said I should?
is the identity proof correct?
how can I prove associativity over composition?

Why would you set $(f,g)=fg$? That looks like you're confusing a morphism $(f,g)$ in the product with the composition $fg$ of two morphisms in the same category. Due to this confusion, none of the rest of your argument makes sense. You need to prove, for any $(f,g):(A,B)\to (A',B'),$ that $(f,g)=(f,g)\circ 1_{A,B}=1_{A',B'}\circ (f,g)$.

The proof should look something like this:
$(f,g)\circ 1_{A,B}=(f,g)\circ (1_A,1_B)=(f\circ 1_A,g\circ 1_B)=(f,g)$, and similarly for sticking the identity on the other side. Then you should do the same kind of thing for associativity.

Associativity is a statement about three morphisms, so it doesn't look like you've checked that here.  The statements are true and the definition is standard; see https://ncatlab.org/nlab/show/product+category


Area of a circle segment on sphere, given radius (meters) and central angle (degrees) Situation

I have a circle segment and some information about the circle it belongs to.

Given Information:

radius of the circle in meters
central angle in degrees
lat/long of all three points on the image below.

I need to calculate the area of the circle segment on sphere. It should depend on the radius of the Earth (like great-circle distance).

Problem

I have no idea how to do that and couldn't find any algorithms.

Thanks for help

Compute first the area of a full circle, then scale the result by $\theta/(2\pi)$ to get the area of the sector corresponding to central angle $\theta$. Finally, subtract from that the area of spherical triangle $XYZ$.


A Question on the Application of Liouville's Theorem Let $f(z)$ be an entire function that satisfies $|f(z)|\leq 1+|z|, \ \forall z\in\mathbb{C}$. Show that $f(z)=az+b$ for fixed complex numbers $a$ and $b$.

My attempt:

Consider a circle, $\Gamma$, centered at the origin with radius $R$. Now, by Cauchy's Integral Formula:
$$f(z)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n \ \ \forall z,$$
where, $$f^{(n)}(0)=\frac{n!}{2\pi i}\int_\Gamma \frac{f(z)}{z^{n+1}} \ dz.$$
Now suppose $\Gamma=\Gamma_R=Re^{it} \ \ t\in[0,2\pi]$. Then if $|z|=R$, we have:
$$\left|\frac{f(z)}{z^{n+1}}\right|\leq\frac{1+R}{R^{n+1}}.$$
So, $$|f^{(n)}(0)|=\frac{n!}{2\pi}\left|\int_{\Gamma_R} \frac{f(z)}{z^{n+1}} \ dz\right|\leq\frac{n!}{2\pi}\frac{1+R}{R^{n+1}}2\pi R=\frac{n!(1+R)}{R^n}.$$
Now $|f^{(n)}(0)|\rightarrow 0$ as $R\rightarrow\infty$ for $n\geq 2$. But what is the answer (what is $a$ and $b$)?

What you have proved is  $f^{(n)}(0)=0$ for $n \geq 2$ because $f^{(n)}(0)$ does not depend on $R$. From the power series expansion you get $f(z)=f(0)+f'(0)z$ so $a=f'(0), b=f(0)$.

Each pair $(a,b) \in \mathbb C^2$ with $|a|,|b| \le 1$ will do the job !


Show that the determinant of $A$ is equal to the product of its eigenvalues Show that the determinant of a matrix $A$ is equal to the product of its eigenvalues $\lambda_i$.

So I'm having a tough time figuring this one out. I know that I have to work with the characteristic polynomial of the matrix $\det(A-\lambda I)$. But, when considering an $n \times n$ matrix, I do not know how to work out the proof. Should I just use the determinant formula for any $n \times n$ matrix? I'm guessing not, because that is quite complicated. Any insights would be great.

Suppose that $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$. Then the $\lambda$s are also the roots of the characteristic polynomial, i.e.
$$\begin{array}{rcl} \det (A-\lambda I)=p(\lambda)&=&(-1)^n (\lambda - \lambda_1 )(\lambda - \lambda_2)\cdots (\lambda - \lambda_n) \\ &=&(-1) (\lambda - \lambda_1 )(-1)(\lambda - \lambda_2)\cdots (-1)(\lambda - \lambda_n) \\ &=&(\lambda_1 - \lambda )(\lambda_2 - \lambda)\cdots (\lambda_n - \lambda)
\end{array}$$
The first equality follows from the factorization of a polynomial given its roots; the leading (highest degree) coefficient $(-1)^n$ can be obtained by expanding the determinant along the diagonal.
Now, by setting $\lambda$ to zero (simply because it is a variable) we get on the left side $\det(A)$, and on the right side $\lambda_1 \lambda_2\cdots\lambda_n$, that is, we indeed obtain the desired result
$$ \det(A) = \lambda_1 \lambda_2\cdots\lambda_n$$
So the determinant of the matrix is equal to the product of its eigenvalues.

I think this is right...

Write A =
$$\begin{pmatrix}a_{11} & a_{12}&\cdots&a_{1n}\\\ \vdots & \vdots&&\vdots\\\ a_{n1}& a_{n2}&\cdots& a_{nn} \end{pmatrix}$$

Let the n eigenvalues of A be $\lambda_1, \cdots , \lambda_n$. Finally, denote the characteristic polynomial of A by
$p(\lambda) = |\lambda I − A| = \lambda_n + c_{n−1}\lambda{n−1} + \cdots + c_1λ + c_0$.

Note that since the eigenvalues of A are the zeros
of $p(\lambda)$, this implies that $p(\lambda)$ can be factorised as $p(\lambda) = (\lambda − \lambda_1)\cdots(\lambda − \lambda_n)$.
Consider the constant term of $p(λ), c_0$. The constant term of $p(\lambda)$ is given by $p(0)$, which can be
calculated in two ways:

Firstly, $p(0) = (0 − λ_1)\cdots(0 − λ_n) = (−1)^nλ_1 \cdots λ_n$.

Secondly, $p(0) = |0I − A| = | − A| = (−1)^n
|A|$.

Therefore $c_0 = (−1)^nλ_1 \cdots λ_n = (−1)^n
|A|$, and so  $λ_1 \cdots λ_n = |A|$.

That is, the product of the n
eigenvalues of A is the determinant of A.


How different is Beta computation using Covariance and Linear Regression? I wanted to compute Beta for a Stock against an Index (Say Stock X against S&P 500).

I computed the daily returns for over one year applied the following logic :

Beta =  COVAR(X, S&P 500)/VARP(S&P 500)

Where:

COVAR :  Returns Covariance, the average of the products of deviations for each data point pair.
VARP : Variance of the entire population.

The problem I run into is, X has few missing data points, and the daily returns has lot of NAN, hence I seem to get some bad COVAR.

Linear Regression was suggested here, I would like to know how Linear Regression can solve the bad data issue here, also how different is Beta computation using COVAR and Linear Regression.

You will get the same answer using linear regression or using the covariance formula. This is because the covariance formula is derived from a linear regression.

In more details, if $X_t$ is the return of the stock on day $t$ and $S_t$ is the return of the index, and $\epsilon_t$ is the error, then you have a model

$$X_t = \alpha + \beta S_t + \epsilon_t$$

Performing a linear regression of $X_t$ against $S_t$ will return the parameters $\alpha$ and $\beta$. You can show that the returned value for $\beta$ will be

$$\beta = \frac{E(XS) - E(X)E(S)}{E(S^2)-E(S)^2} = \frac{\mathrm{Cov}(X,S)}{\mathrm{Var}(S)}$$

which is the same as the formula you have. Unfortunately there's not a lot you can do except get better data.

@Chris Taylor:

This formula is only valid for regressions with only one explanatory variable. Adding regressors makes the link disappear as regressions give you the conditional correlation/covariance when the cov(x,y) gives you the unconditional covariance.


Probability that for two random points in unit ball, one is closer to the center than to the other point If I have two points $p_1, p_2$ uniformly randomly selected in the unit ball, how can I calculate the probability that one of them is closer to the center of the ball than the distance between the two points?

I know how to calculate the distribution of the distance between two random points in the ball, same for one point from the center, but I'm not sure how to use the two distributions to get what I'm after.

$d(p_1,O)\lt d(p_1,p_2)$ or $d(p_2,O)\lt d(p_1,p_2)$ if and only if one of $d(p_1,O)$ and $d(p_2,O)$ is the least of the three distances. By symmetry, the probability for this is twice the probability that $d(p_2,O)$ is the least of the three distances.

Fix $p_1$ at $(0,r)$. Then $d(p_2,O)$ is the least of the three distances if $p_2$ lies within the circle of radius $r$ around the origin and below $y=\frac r2$, and thus in a circular segment with radius $r$ and angle $\frac{4\pi}3$, so the desired probability is

$$
2\cdot2\cdot\frac1\pi\int_0^1r\mathrm dr\,\frac{r^2}2\left(\frac{4\pi}3+\frac{\sqrt3}2\right)=\frac23+\frac{\sqrt3}{4\pi}\approx80\%\;,
$$

where one $2$ is the symmetry factor above, another $2$ normalises for $p_1$ and $\frac1\pi$ normalises for $p_2$.

P.S.: I just realised that I simply assumed that you meant the unit ball in $2$ dimensions but you hadn't actually specified the number of dimensions. In case you meant the unit ball in three dimensions, we need to use a spherical cap of height $\frac32r$ and adjust the radial density and the normalisation; the probability in this case is

$$
2\cdot 3\cdot\frac3{4\pi}\int_0^1r^2\mathrm dr\,\frac{\pi\left(\frac32r\right)^2}3\left(3r-\frac32r\right)=\frac{81}{16}\int_0^1\mathrm dr\,r^5=\frac{27}{32}\approx84\%\;.
$$

In case you wanted the result in arbitrary dimensions, you'll find some information about the volumes of the resulting hyperspherical caps with height $\frac32r$ at Stars in the universe - probability of mutual nearest neighbors.

For one of the points to be closer to the center than the other point, both points should lie outside the region of sphere which subtends a solid angle of
2*pi*(1-cos(a)).

Where a= 1 radian (180/pi).

and solid angle subtended at the center by the entire sphere is 4pi.
Hence, required probability is 1-[{2*2*pi(1-cos(a))}/{4*pi}]. Which gives the value 0.5403.


A is an open set and B is a closed set. Prove that A\B is open and B\A is closed. The problem consists of using the epsilon neighborhood definition to prove that A\B is open and B\A is closed.

I'm thinking of using a proof by contradiction and assuming that A\B is not open and vice versa for B\A. Any help would be greatly appreciated.

Sorry, forgot to mention that we can't use the fact that a set is open iff its compliment is closed and vice versa.

$A$ is open and $B$ is closed. Let $x \in A \setminus B$. Then there is some $r>0$ such that $B(x,r) \subseteq A$ as $A$ is open and $x \in A$.

$x \notin B$ so $x$ is not a limit point of $B$ (as $B$ is closed it would contain all its limit points), so there is soem $s>0$ such that $B(x,s) \cap B =\emptyset$.

But then $B(x, \min(r,s)) \subseteq A\setminus B$ showing $x$ to be an interior point of $A \setminus B$, so $A \setminus B$ is open.

Now suppose $x$ is a limit point of $B\setminus A$. Then $x$ is also a limit point of the larger set $B$ and as $B$ is closed, $x \in B$. Suppose that $x \in A$, then we'd have $r>0$ such that $B(x,r) \subseteq A$ and then $B(x,r) \cap (B \setminus A)=\emptyset$, and $x$ wouldn't be a limit point of $B \setminus A$. So, in fact, we must have that $x \notin A$, which together with $x \in B$ implies $x \in B\setminus A$, which thus contains all its limit points and is thus closed.

If we assume A/B is not open, then there must be a point $x$ in $A \setminus B$ for which all neighbourhoods intersect $(A \setminus B)^c$. $(A \setminus B)^c$ is the set of all points which either lie in $B$, or lie not in $A$ (or both).

Pick a point in $A \setminus B$. It lies in $A$, so at least one of its neighbourhoods lies completely in $A$. It lies in $B^c$. There has to be at least one of its neighbourhoods which lies completely in $B^c$ (i.e. does not intersect $B$), as $B$ is its own closure.

Take the intersection of it with the first neighbourhood to provide a counterexample for our assumption that $A \setminus B$ is not open.

A similar idea could be used for the $B \setminus A$ case.


Find eigenvectors of an infinite dimensional vectorspace Pn(R)? Define a linear operator T:Pn(R)--->Pn(R) by T(p(x))=p(x)+p(2)x.

(a) How many distinct eigenvalues does T have?

(b) What are the dimensions of their corresponding eigenspaces?

At first I started with writing the matrix for T in terms of the standard bases and I got a big matrix that I think it will be impossible to find its characteristic polynomial. the matrix I got is in the link : http://imgur.com/GQlmTqN

so that didn't help.
Then, I tried to use the definition of an eigenvalue. T(x)=Lambda x
so I let x= an(x^n)+.....+a0           .  with (an) doesn't equal 0 (because x has to be non zero)
then: since T(x)=lambda x
so, [an(x^n)+an(2^n)x]+.......[a0+a0(x)] = lambda an(x^n)+......lambda a0

My, intention was to cancel things out but I cannot see how this is going to work.

any suggestions on how to approach this problem?
remember that the question is How many distinct eigenvalues and what are the dimensions of their eigenspaces, so if there is way to get that without finding the eigvecs and the eigvalues then thats probably easier.
Thank you in advance!

Edit:
all of my a0's should have been a1's. because if it was a0 the dimension of the vectorspace would be n+1; however its just n

Hint: you should be able to show that there is a eigenspace of dimension $n$ for eigenvalue $\lambda = 1$  (think about what has to happen for $T(p(x)) = p(x)$ -- try to describe all such $p(x)$).  Then find some other eigenvector for a different eigenvalue.

Expanding to a more complete solution:

$T(p(x)) = p(x)$ if and only if $p(2) = 0$.  Notice that the polynomials $p_k(x) = x^k - 2^k$ are linearly independent for different $k$ and have $p_k(2) = 0$, so this gives us an eigenspace of dimension $n$ (since $p_k(x)$ for $1\leq k\leq n$ gives $n$ linearly independent eigenvectors) for the eigenvalue $\lambda = 1$.  On the other hand, for $p(x) = x$ we have $T(p(x)) = x + 2x = 3p(x)$; this gives us an eigenspace of dimension $1$ for the eigenvalue $\lambda = 3$.  Since this exhausts the whole space ($P_n(\mathbb{R})$ has dimension $n+1$) we are done, and these are the full eigenspaces.

By the way, this actually works for the infinite dimensional space of all polynomials, though one does have to check that the polynomials above in fact span the whole space rather than just relying on a dimension argument, but this isn't hard.

(Edited to reflect the clarification that $P_n(\mathbb{R})$ is indeed supposed to be the polynomials over $\mathbb{R}$ of degree at most $n$.)


$\lVert u +v \rVert =\lVert u \rVert +\lVert v\rVert$ only when $u$ and $v$ have the same direction How can I prove that $\lVert u +v \rVert =\lVert u \rVert +\lVert v\rVert$ only when $u$ and $v$ have the same direction?

I started by raising the two parts of the equation to the second power, but I got stuck at $u \cdot v = \lVert u \rVert \, \lVert v \rVert$.

Use the fact that $u \cdot v = \|u\| \|v\| \cos \theta$, where $\theta$ is the angle between $u$ and $v$. You can show that $$\|u+v\|^2  = \left( \|u\| + \|v\| \right)^2$$ if and only if $\cos \theta = 1$, that is, if and only if $\theta = 0^\circ$.
If this condition holds then a quick calculation verifies $u = \alpha v$ for $\alpha = \frac{u \cdot v}{v \cdot v} > 0$ because in this case $\|u - \alpha v\|^2 = 0$.

As stated in the comments earlier, start with $(u+v)^2$:
\begin{align}
(u + v)^2 &= u^2 + 2 u\cdot v + v^2  \iff \\
\lVert u + v \rVert^2 &= \lVert u \rVert^2 + \lVert v \rVert^2 +
2 \lVert u \rVert \lVert v \rVert \cos \angle(u, v) \\
&= (\lVert u \rVert + \lVert v \rVert)^2 +
2 \lVert u \rVert \lVert v \rVert (\cos \angle(u, v) - 1) \iff \\
\lVert u + v \rVert&=
\sqrt{(\lVert u \rVert + \lVert v \rVert)^2 + 2 \lVert u \rVert \lVert v \rVert (\cos \angle(u, v) - 1)}
\end{align}
where I assumed $\cdot$ to be the standard scalar product and $\lVert \cdot \rVert$ to be the Euclidean $\lVert \cdot \rVert_2$ norm.

The claimed result holds if
$$
\cos \angle(u, v) - 1 = 0
$$
which is true if $u$ and $v$ have angle $0$ between them, i.e. show in the same direction.


Does $f_n(x)=\frac{x}{n}\log(\frac{x}{n})$ converge uniformly in $(0,1)$? $f_n(x)=\frac{x}{n}\log\frac{x}{n}$

I tried to expand the bounds to $[0,1]$ and prove that the hypotheses of Dini's uniform convergence criterion are satisfied, but I'm not even sure I can expand the bounds since $f_n(0)$ isn't defined.

First, let us compute the limit of this sequence of functions:

$$\lim \limits _{n \to \infty} \frac x n \log \frac x n = \lim \limits _{t \to 0, \ t > 0} t \log t = \lim \limits _{u \to \infty} \frac 1 u \log \frac 1 u = \lim \limits _{u \to \infty} - \frac {\log u} u = 0 .$$

So, if $f_n (x) = \frac x n \log \frac x n$, we have obtained $f_n \to 0$.

To see whether the convergence is uniform, all we have to do is study whether

$$\sup \limits _{x \in (0,1)} |f_n (x) - 0| = 0 .$$

Since $f_n \le 0 \ \forall n \ge 1$, we must check whether $\sup \limits _{x \in (0,1)} g_n = 0$ where

$$g_n(x) = |f_n (x)| = - \frac x n \log \frac x n .$$

Let us explicitly find the $\sup$ of $g_n$ on $(0,1)$. We have that $$g_n ' = - \frac 1 n \log \frac x n - \frac x n \frac 1 {\frac x n} \frac 1 n = - \frac 1 n (\log \frac x n + 1) .$$

The critical point are the roots of $\log \frac x n + 1 = 0$, which means $\frac x n = \frac 1 {\rm e}$, which gives $x = \frac n {\rm e}$. Note that for $n \ge 3$ this point lives outside of $(0,1)$, so for $n \ge 3$ we have that $g_n ' \ge 0$, so $g_n$ is non-decreasing, therefore its $\sup$ must be reached in $1$ and be $- \frac 1 n \log \frac 1 n$.

Finally, note that $\lim \limits _{n \to \infty} - \frac 1 n \log \frac 1 n = 0$ (with the same technique as in the beginning of this answer), so $\sup \limits _{x \in (0,1)} g_n = 0$ as desired, so yes, the convergence is uniform.

There is an alternative aproach to this problem, along the lines of your unachieved (but essentially correct) proof.

First, extend $f_n$ by continuity to a function $F_n$ defined on $[0,1]$: noting that $\lim \limits _{x \to 0, \ x > 0} f_n (x) = 0$ allows us to define $F_n (x) = \left\{ \begin{eqnarray} 0 &,& \quad x = 0 \\ \frac x n \log \frac x n &,& \quad x \in (0,1) \\ \frac 1 n \log \frac 1 n &,& \quad x = 1 \end{eqnarray} \right.$.

It is easy to see that $F_n$ is continuous, increasing ($F_n ' \ge 0$), and that $F_n \to 0$ pointwisely. This means that we may apply Dini's theorem and get that $F_n \to 0$ uniformly on $[0,1]$.

Finally, to show that $f_n \to 0$ uniformly on $(0,1)$ note that

$$0 \le \sup \limits _{x \in (0,1)} |f_n (x) - 0| \le \sup \limits _{x \in (0,1)} |F_n (x) - 0| \le \sup \limits _{x \in [0,1]} |F_n (x) - 0| \to 0$$

(where the last convergence encodes the fact that $F_n \to 0$ uniformly), so $f_n \to 0$ uniformly on $(0,1)$.

As you see, you were very close to the correct proof, your intuition was good but there were a few technical details missing.

Hint:
$$
f_n(x)=\frac{x\log x}{n}-\frac{\log n}{n}\,x
$$
and $\lim_{x\to0^+}x\log x=0$.


Finding 4 closest points So I am trying to use bilinear interpolation to extrapolate data for certain values of a 3-dimensional array. But, the issue is that the bilinear interpolation requires an input of the 4 closest numbers by distance in order for the equations to be solved. Does anyone have an idea of how I can obtain these 4 values. I am going to put some context below for better understanding.
62  0   63  0
0   0   0   0
63  0   61  0
0   0   0   0

I am going to refer to the coortinates of the elements in the form of (row,column).
So for example, consider this array above. For context, I need a way to find the 4 closest non-zero points to (1,2). If I solved this manually, the points would be (1,1),(1,3),(3,1),(3,3).
This is only a small part of a very large array that I have given for context. If anyone can help me with the coding, it would be appreciated.

EDIT:  If you are looking for the four points in 3space with odd coordinates that are closest to $(x,y,z)$ they are the nearest point and three of its neighbours.
$$X=2floor(x/2)+1\\Y=2floor(y/2)+1\\Z=2floor(z/2)+1$$
The nearest point is $(X,Y,Z)$.
If $x\gt X$, another point is $(X+2,Y,Z)$, otherwise $(X-2,Y,Z)$
If $y \gt Y$, the third point is $(X,Y+2,Z)$ otherwise $(X,Y-2,Z)$
If $z\gt Z$, the fourth point is $(X,Y,Z+2)$ otherwise $(X,Y,Z-2)$
If you are looking for the coordinates of the square that contains a point $(x,y)$, where those coordinates are odd, let
$$X=2×floor\left(\frac{x+1}2\right)\\
Y=2×floor\left(\frac{y+1}2\right)$$
Then the coordinates are
$$(X+1,Y+1),(X+1,Y-1),(X-1,Y+1),(X-1,Y-1)$$
If you are close enough to one of the lattice points, then the four nearest don't form a square, but form a T instead.  Is that important ?
If the points are irregular, you might run through the list, keeping just the four nearest so far.  Then you need only run through the list once.


Prove inequality $2e^x>x^3+x^2$ If $x \in \Bbb R$, show that
$$2e^x>x^3+x^2$$
This inequality is right, see

Own ideas: If $x\in \Bbb R$,

$$f(x)=2e^x-x^3-x^2$$
$$f'(x)=2e^x-3x^2-2x$$
$$f''(x)=2e^x-6x-2$$
$$f'''(x)=2e^x-6$$
$$f''''(x)=2e^x>0$$

Like the symbol cannot judge $f$ sign.

So how can we show this $$f(x)>0 \text{ for } x \in \Bbb R?$$

For $x \leq 1$, notice that
$$ x^3 + x^2 \leq x + 1 \leq e^x \leq 2e^x. $$
For $x \geq 1$, it suffices to prove that $f(x) := \log(2e^x) - \log(x^3 + x^2) > 0$. Differentiating twice,
$$ f'(x) = 1 - \frac{2}{x} - \frac{1}{x+1}, \qquad f''(x) = \frac{2}{x^2} + \frac{1}{(x+1)^2} $$
This shows that $f$ is strictly convex and attains global minimum on $[1, \infty)$ at $x = 1+\sqrt{3}$. Now the conclusion follows from
$$ f(1+\sqrt{3}) = 1+\sqrt{3} - 2\log(2+\sqrt{3}) \approx 0.098135 > 0. $$

Consider that for all x in $\mathbb R$, $e^{x}$ > 0, so  $2e^{x}$ > 0 for all $\mathbb R$.So clearly for $x\leq 0$ in $\mathbb R$, the statement is true since $x^2 \geq 0$ and  $x^3 \leq 0$.  Since $|x^3| \geq x^2$ for x in $\mathbb R$, $x^3 + x^2 \leq 0$ for $x\leq 0$ in $\mathbb R$.

Now consider where $x>0$ in $\mathbb R$. Let's use the exponential power series since we know it converges for all real numbers. Assuming x > 0,rearranging and substituting gives $e^x > \frac{1}{2} (x^3 + x^2)$ =  $\sum_{n=1}^{\infty} \frac{x^n}{n!} > \frac{1}{2} (x^3 + x^2)$

The question now is which side grows larger as $x\rightarrow +\infty$. Clearly both go to infinity in this case. So consider the following:

$\frac{e^x} {\frac{1}{2} (x^3 + x^2)}$

Now use L'Hospital's rule:

$\frac{e^x} {\frac{3}{2} (x^2 + x)}$

Apply again:

$\frac{e^x} {3x + 1}$

One last time.

$\frac{e^x} {3}$

Clearly as $x\rightarrow +\infty$ then  e^x > 3. That completes the proof. Whew!


From [a,b] to [$0$,$1$] I was checking an exercise and it's proof and I couldn't understand why the interval [a,b] can be 'replaced' by [0,1].

This is the exercise:

Prove that every continuous function f:[a,b]$\rightarrow \mathbb R$ is the uniform limit of a sequence of even polynomials if and only if (a,b) doesn't contain the origin.

Proof:

Suppose (a,b) doesn't contain the origin.
And for simplicity consider $[0,1]$ instead of [a,b].

.

. (and the proof continue)

.

So why is it equivalent to work with $[0,1]$ ?

Is there a proof to show that $[0,1] \Leftarrow \Rightarrow [a,b]$ ?

Suppose we have the result for $[0,1].$ If now $f\in C([0,b]),$ define $g:[0,1] \to [0,b]$ by setting $g(x) = bx.$ Then $f\circ g \in C([0,1]).$ So there exists a sequence $P_n$ of even polynomials such that $ P_n \to f\circ g$ uniformly on $[0,1].$ This implies $P_n\circ g^{-1} \to f$ uniformly on $[0,b].$ Since each $P_n\circ g^{-1}$ is an even polynomial, we have the result for $[0,b].$

If now $0<a<b$ and $f\in C([a,b]),$ we can extend $f$ to a continuous function $F$ on $[0,b],$ simply by setting $F(x) = f(a)$ for $x\in [0,a).$ By the above, there is a sequence $P_n$ of even polynomials such that $ P_n \to F$ uniformly on $[0,b].$ This of course implies $ P_n \to F$ uniformly on $[a,b].$ Since $F=f$ on $[a,b],$ we have the desired result for $[a,b].$

The result for intervals $[a,b]$ with $a<b\le 0$ follows easily from this.

Yes, actually you can map $[a,b]$ to $[0,1]$ using the following function: $x \mapsto \frac{x - a}{b - a}$. This is a bijection between the two intervals, and it is also a continuous function as you can check.


Is there a term for two polygons with the same angles but different side lengths? Suppose polygons $A$ and $B$ have the same number of sides, and there is a correspondence between the vertices of $A$ and $B$, in consecutive order around both polygons, so that the angles at corresponding vertices are equal.  However, the lengths of the sides need not be equal.

For example:

any two similar polygons are related in this way
any two rectangles are related in this way
these four hexagons share this relationship (modulo my drawing errors; all angles are supposed to be $90^\circ$ or $270^\circ$):

Is there a standard term for this relationship or property?

"Angular congruence", perhaps?  I see it used occasionally, most clearly in http://www.gutenberg.ca/ebooks/dantzig-poincare/dantzig-poincare-00-h-dir/dantzig-poincare-00-h.html:

Thus, two rectangles may be dissimilar, although the corresponding angles are certainly congruent in this case; again, the sides of any rhombus are certainly proportional to the sides of any square, and yet the two figures are generally dissimilar. Angular congruence does by no means entail proportionality of lines.
The case of two similar triangles is an important exception. Here the congruence of corresponding angles does entail the proportionality of corresponding sides and, consequently, the similarity of the two figures. This property of similar triangles enabled Euclid to eliminate allusion to proportion and reduce the criteria of similarity of two polygons to congruence tests.

(Italics in original; bold face mine.)

I'd call such polygons (as those you depicted before your edit) parallel, and have heard them being called that way in personal conversations with other mathematicians, but I don't know whether this is in some way established. And it doesn't exactly fit if you only consider relative angles, i.e. if you rotate one of the figures as you did in your edit. You might want to call that a rotated parallel polygon, but that term sounds enough like a contradiction that I wouldn't use it without an accompanying definition.

I'm posting this as an answer, not a comment, so that people can express their (dis-)agreement using votes.


Norm of ideals in the ring of integers Let $\alpha \in \mathbb{Z}_k=\mathcal{O_k}$ (Ring of integers) be non-zero.

Then $$ N_{K/\mathbb{Q}}((\alpha))=N_{K/\mathbb{Q}}(\alpha) $$

$(\alpha)$ denotes the principal ideal generated by $\alpha$. $N$ is the norm.

Proof. Let $\mathbb{Z}_k=\mathbb{Z}\omega_1+\mathbb{Z}\omega_2+\dots+\mathbb{Z}\omega_n$.
Then $(\alpha)=\mathbb{Z}\alpha\omega_1+\mathbb{Z}\alpha\omega_2+\dots+\mathbb{Z}\alpha\omega_n$. Then $N_{K/\mathbb{Q}}((\alpha))=|\mathbb{Z}_k/(\alpha)| $ ; if we write $\alpha \omega_i=\sum_{j=1}^n a_{ji} \omega_j$ , then the index of $\alpha$ in $\mathbb{Z}_k$ is just $\det(a_{ij})$ . But we know that $ N_{K/\mathbb{Q}}(\alpha)=\det(a_{ij})$  , and so the result follows .

I do not understand why the index of ($\alpha$) in $\mathbb{Z}_k$ is $\det(a_{ij})$ .

I know that the index gives the number of the cosets and each element is of the form $\beta+(\alpha)$ but this does not bring me further .

Thanks for the help .

You have certainly seen before that the ring of integers $O_k$ is a free $\mathbf Z$-module of rank $n$ equal to the degree of $k/\mathbf Q$. The $\mathbf Z$-submodule $\alpha O_k$ has also rank $n$ because the multiplication by $\alpha$  is a bijection of $O_k$ onto $\alpha O_k$. But the classical structure theorem of finitely generated free modules over a PID ($\mathbf Z$ here) shows the existence of a basis $(e_1,..., e_n)$ of $O_k$ and natural non null integers $a_1,..., a_n$ s.t. $a_i$ divides $a_{i+1}$ for $i\le n-1$ and $(a_1e_1,...,a_ne_n)$ is a basis of $\alpha O_k$ (the bases are called $adapted$). Hence $O_k/\alpha O_k\cong \mathbf Z/a_1\mathbf Z \times ...\times \mathbf Z/a_n\mathbf Z$ and has order $a_1...a_n$. If $f$ denote the $\mathbf Z$-linear application of $O_k$ onto $\alpha O_k$ defined by $f(e_i)=a_ie_i$, then det($f)=a_1...a_n$.

Besides, since $(\alpha e_1,..., \alpha e_n)$ is a basis of $\alpha O_k$, there is an automorphism $g$ of the $\mathbf Z$-module $\alpha O_k$ s.t. $g(a_ie_i)=\alpha e_i$. Then det($g)$ is invertible in $\mathbf Z$, i.e. det($g)=\pm 1$. But by definition  $g.f$ is the multiplication by $\alpha$ and its determinant is N$(\alpha)$ (your argument). As det($g.f$)=det($g$).det($f$), it follows that N$(\alpha)=\pm a_1...a_n=\pm$ card $(O_k/ \alpha O_k$) ./.


Name of the highest power of 2 smaller than or equal to a given number For a number $x$, I would like to know whether there is a common name for the number $2^n$ such as $2^n \leq x < 2^{n+1}$ (e.g. If $x = 7$, then $2^n = 4$, $n = 2$).

I have some computer science related article where I extensively use such a number and I need a name to give it in order to explain how an algorithm works without having to repeat the number definition over and over every time I need to use it. I currently call it a "base $2$", saying for example that "$4$ is the base $2$ of $7$" (see example above), and that we need to "compute the base $2$ of the number", but this name feels wrong. Do you know whether a common name exists for such a number?

Note: actually, the article I am talking about deals with Gray codes. I am looking for a term that looks like it comes from math and not from computer science since many terms from computer science that deal with powers of two tend to be references to the usual binary representations of numbers. As an example, with Gray codes $2^3$ is 0b1100 and not 0b1000 so I am trying to avoid names that would literally mean the $n$th set bit, hence the question on Math.SE.

Note 2: as it has been highlighted in the many answers and comments, the goal of this question, once clearly reformulated, is to find a terse, pronounceable name for the function $2^{\lfloor \log_2(x) \rfloor}$ so that it is possible to say that "some number is the [insert name here] of $x$".

In the context of data structures - specifically, the van Emde Boas layout - I've heard this referred to as the hyperfloor of $x$. See “Cache-Oblivious B-Trees (Wayback Machine) by Bender, Demaine, and Farach-Colton for details - it defines the hyperfloor of $x$, denoted $\lfloor \lfloor x \rfloor \rfloor$, to be $2^{\lfloor \log_2 x \rfloor}$.

Hope this helps!

You could maybe use the fact that the upper end-point is twice as large as the lower end-point. Something like even interval or double interval ..


Insert variable into function and solve for another variable I have the following production function:

$$y=x_1^{0.4}\cdot x_2^{0.4}$$

and I know that $x_1 = \dfrac{x_2}2$.

I need to plug $x_1$ into the production function and solve it for $x_2$.

I've tried it this way:

$$y = \left(\dfrac{x_2}2\right)^{0.4}\cdot x_2^{0.4}$$

Unfortunately, I'm not sure how I can solve this for $x_2$.

I would be grateful for any help!

Hint: I would simplify your expression by replacing an awkward term like $x_1^{0.4}$ by the term $\sqrt[5]{x_1^2}$. Now you can proceed by using more algebraic techniques.


Derivation of 2$\pi$ in the period of a simple pendulum Most of us know that the period of a simple pendulum is given by

$$2 \pi \sqrt{\dfrac{\ell}{g}}.$$

But how did the $2\pi$ term get into that argument. From dimensional analysis, we can find the period but not the constant.

Is there a calculus argument as some limit is taken? Is it based on an energy equation? Surely there is a way to derive that $2 \pi$.

The pendulum movement equation is given by
\begin{align*}
\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + \frac{g}{\ell}\sin(\theta) = 0
\end{align*}

For small values of $\theta$, we can make the approximation $\sin(\theta) \approx \theta$, from whence we obtain the equation
\begin{align*}
\ddot{\theta} + \frac{g}{\ell}\theta = 0
\end{align*}

whose associated characteristic equation is
\begin{align*}
x^{2} + \frac{g}{\ell} = 0 \Longrightarrow x = \pm i\sqrt{\frac{g}{\ell}}
\end{align*}

Thus the solutions are $\theta(t) = c_{1}\sin\left(\displaystyle t\sqrt{\frac{g}{\ell}}\right) + c_{2}\cos\left(\displaystyle t\sqrt{\frac{g}{\ell}}\right)$. Finally, for $T = 2\pi\sqrt{\displaystyle\frac{\ell}{g}}$, one has
\begin{align*}
\theta\left(t + 2\pi\sqrt{\frac{\ell}{g}}\right) = c_{1}\sin\left(t\sqrt{\frac{g}{\ell}} + 2\pi\right) + c_{2}\cos\left(t\sqrt{\frac{g}{\ell}} + 2\pi\right) = \theta(t)
\end{align*}
Therefore we conclude that $T$ is the period indeed.

The expression $\sqrt{\frac{g}{l}}$ represents the frequency in radians per second. Divide by $2\pi$ radians per cycle to get $\frac{1}{2\pi}\sqrt{\frac{g}{l}}$ cycles per second.

Now take the reciprocal to get $2\pi\sqrt{\frac{l}{g}}$ seconds per cycle which, by definition, is the period.


$A^T A$ eigenvalue bounding Let $A$ be a square matrix with real entries.

Is there anything like any eigenvalue of $A^tA \leq \max({1,\lambda^2})$ where
$\lambda$ is an eigenvalue of $A$ and max is taken over all eigenvalues?

Consider $A=\begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix}$

We know that $A$ and $A^t$ have the same set of eigenvaluse, and we also know that the matrix products $AB$ and $BA$ have the same sets of eigenvalues. These two bits of information may be of help. Good luck!


A construction of sigma-algebras - surely not new, right? I know no descriptive set theory. I've stumbled on something that must be well known, being so simple. But it contradicts something I've been told by smart people; the question is whether it's well known (or at least known).

A complete version of this post would be a little long. It seems clear to me that anyone who knows the answer is going to understand what I mean by the standard this and the usual that; if not I can clarify.

Context. Say $G$ is a group and $S\subset G$. There are at least three ways to say what $H$, the subgroup of $G$ generated by $S$, is:

Let $S_0=S$, and let $S_{n+1}$ be $S_n$ together with all the $xy$ and $x^{-1}$ for $x,y\in S_n$. Then $H=\bigcup_{n=0}^\infty S_n$. (I call this the bottom-up construction.)
$H$ is the set of all words in the elements of $S$.
$H$ is the intersection of all the subgroups of $G$ containing $S$ (this I tend to think of as "top-down".)

Not that anyone would ever use (1) in this context. But now let's say $S\subset P(X)$, the powerset of $X$, and we want to describe $A$, the sigma-algebra generated by $S$. The analogue of (3) is what you see in every book on measure theory; some books include the transfinite-recursion analogue of (1) when they want to show that the Borel sets on the line have cardinality $c$. For some time I've speculated on whether there was something like (2) here.

There is.

(2') The elements of $A$ are exactly the roots of the trees $T$ such that $T$ is well-founded, every node of $T$ is a subset of $X$, every terminal node is an element of $S$, and every other node has either one or countably many subnodes; if a node $E$ has one subsnode $F$ then $E$ is the comlement of $F$, while if $E$ has countably many subnodes it is the union of the subnodes. (Taking advantage of the arity to avoid coloring the nodes to say which operation we're using is probably in poor taste, sorry.)

The proof that this works is totally elementary and straightforward from the definition of $A$ as the intersection of all sigma-algebras containing $S$ - no need to confuse the poor students with ordinals. And one can use it to show, for example, that the Borel sets on the line have cardinality $c$.

I'd be certain that this was perfectly well known except that smart people have told be that the only way to show the Borel sets have cardinality $c$ is transfinite induction.  Hence the question: This actually is standard, those guys were just unaware of it as I was until yesterday, right?

First, a small technical point: in order to count them it's best to look at trees which are subsets of $\omega^{<\omega}$, explicitly. This actually matters if the axiom of choice fails: just because each successor set is countable, and the tree is well-founded, doesn't mean the tree is countable if the countable union of countable sets isn't countable! In fact, there are models of ZF in which every set of reals is Borel. See also Arnold Miller's wonderful paper "Long Borel Hierarchies" (http://arxiv.org/abs/0704.3998)

Your trees are Borel codes. Since a lot of descriptive set theory really does use transfinite induction in an essential way, we often don't talk about Borel codes; however, they are intuitively fundamental, even if often just implicit, and there are also many times when explicitly talking about codes is invaluable, such as effective descriptive set theory (https://en.wikipedia.org/wiki/Borel_hierarchy#Lightface_hierarchy), reverse mathematics (as in https://math.berkeley.edu/~antonio/papers/Delta4Det.pdf), and algebras of sets beyond the Borel (analytic sets - which contain the Borel sets - are defined in terms of tree representations, finding tree representations for classes of sets is a big tool in descriptive set theory, and als see https://en.wikipedia.org/wiki/Infinity-Borel_set).

As far as showing that Borel codes really do code the Borel sets, I think there is a bit of a subtlety actually: how do you show that there is no proper subset of $A$ which is the class of Borel sets? That is, you need to show that every countable well-founded tree yields a set which must be Borel; to do this it seems to me you need to use induction on the rank of the well-founded tree.

This can be done in a way which does not seem to use transfinite induction: given a well-founded tree, if the root is not Borel then we can find some node which is not Borel, but all of whose successors are Borel (otherwise we could build a path through the tree). This is of course a contradiction.

I would argue, though, that this is transfinite induction in disguise.

For a set of reals $X$, define $ord(X)$ to be the length of Borel hierarchy on $X$. What are the possible values of $ord(X)$? Arnold Miller proved a lot of interesting results on this in his thesis. In particular, he showed that for each $\alpha < \omega_1$ it is consistent to have a set of reals $X$ with $ord(X) = \alpha$. To construct this model, he made use of what he called $\alpha$-forcing to add a generic $\mathbf{\Pi}^0_{\alpha}$-set. See his book especially Section 7.


Is diffirentiability in finite dimensional space is equivalent to the existence of partial derivatives Let $f$ be a continuous mapping, defined on $\mathbb{R}^2$. I want to prove that $f$ is differentiable at a point $(a,b)\in \mathbb{R}^2$ : Is it sufficient to prove that $f$ has first order derivative at $a$ and $b$ i.e $f_x(a,b)$ and $f_y(a,b)$ exists (since we have a finite dimensional space)?
Is there a reference book I could look at to find such result ?

Define $f(x) = \min(x_1,x_2)$ for $x\ge 0$ and zero otherwise.
$f$ is not differentiable at zero but the partials exist at $x=0$.

$f$ might even fail to be continuous at $(a,b)=(0,0)$ :
$$
f(x,y)=\left\{
\matrix{
0, & \text{ if } x=0 \text{ or } y=0,\hfill\cr
1, & \text{ if } x\neq0 \text{ and } y\neq0.
}
\right.
$$


How can I find a linearly independent subset of a set of vectors in $\mathbb{Z}^n$? I have a set of $M$ vectors in the module $\mathbb{Z}^n$ ($M>n$) over $\mathbb{Z}$.

Question 1: How can I find a linearly independent subset of these vectors? (so that others can be written as a linear combination of this subset with integer coefficients)

Question 2: Let's assume that this linearly independent subset forms the basis for an $n$-dimensional lattice (or equivalently there are $n$ of them). Is there a way that without finding the subset, I can find the determinant of the matrix whose columns are these basis?

To clarify, let's say for example I have these $M=3$ vectors in $\mathbb{Z}^2$:
$$
v_1 = (1,1) \\
v_2 = (2,0) \\
v_3 = (0,2)
$$

What I want for the linearly independent subset is either the subset $\{v_1,v_3\}$ or $\{v_1,v_2\}$ but not $\{v_2,v_3\}$. Because for example in the first case I can write $v_2 = 2v_1-v_3$.

For question 1, write the extended matrix
$$\begin{pmatrix}
1&2&0\\
1&0&2\\
\end{pmatrix}$$
and then bring it to Hermite Normal Form (HNF). The answer will always be of the form
$$\begin{pmatrix}
c | 0
\end{pmatrix}$$
ie a square lattice c that gives the basis you are asking for, followed by zero columns. You might want to search for "Hermite Decomposition" algorithm if you are not sure how to perform the transformation. Since you are working in $\mathbb{Z}^n$ all the operations of the algorithm can be performed mod n.

For question 2 I am not familiar with any possible way of calculating the determinant without going through step 1 first. The determinant will not be $\pm1$ as written in another answer. Even for the example mentioned here, where
$$c=\begin{pmatrix}
1&0\\
1&2\\
\end{pmatrix}$$
the determinant is 2. Once $c$ is known calculating det$(c)$ is straightforward.

Problem 2: It's not quite as straightforward as for a field but you can still get a very concrete answer. A square matrix over a commutative ring with unit $R$ defines an injective map $R^n \rightarrow R^n$ if and only if its determinant is not a zero divisor; and it defines a surjective map $R^n \rightarrow R^n$ if and only if its determinant is invertible.

(In particular, surjective maps $R^n \rightarrow R^n$ are always injective. This isn't really trivial.)

A set of $n$ column vectors is a basis of $R^n$ is a basis if and only if the matrix they form is a bijective map. Over $\mathbb{Z},$ this means the matrix has determinant $\pm 1$, since these are the only integers whose multiplicative inverse is also an integer.

Problem 1) To find a linearly independent subset, just take the first vector that isn't zero. Maximal linearly independent subset is not really a good concept over $\mathbb{Z}$: for example, in $\mathbb{Z}^1$, the set $8 \mathbb{Z}$ should be considered smaller than $2 \mathbb{Z}$ although $\{8\}$ and $\{2\}$ are both sets of size $1$.


Are infinitesimals, i.e. $dx = ...$, rigorous and correct notation? In many fields of physics and engineering, when we want to describe an infinitesimal, for example, the electric field, we could say $dE(x_i,y_i) = e^{jkr}...dx_0dy_0$

Since derivatives are not fractions, does it still make sense, in a mathematically rigorous way to talk about infinitesimals and would this be the correct notation of describing it? What would be the rigorous way of dealing with infinitesimals?

--Edit--

The history of this question went from "duplicate to an irrelevant question" but once reopened was closed for being "missing context" despite proper english, a clear yes or no question with already great answers (which would be impossible if the question was indecipherable). It seems like those with points have been abusing their powers an incorrectly closing questions that they simply don't like.

There are basically two rigorous ways to deal with differentials. One is to treat them as differential forms. This is kind of an algebraic way of doing things, it sets rules for how you can manipulate differentials without trying to describe them as, say, "limits of small differences".

The other way is nonstandard analysis, of which there are at least two incompatible types. One of those is the one from which that name originated, which originally used the idea of a nonstandard model (from model theory) to construct a self-consistent theory containing infinite and infinitesimal "hyperreal" numbers. This originated with Robinson. A different formalism with the same semantics (which is probably easier to understand for non-logicians) was made later by Nelson.

An entirely different semantics arises in smooth infinitesimal analysis. SIA is somewhat alien to "mainstream" mathematicians, because it works in a field which has nonzero nilpotent elements (e.g. $dx \neq 0$ but $(dx)^2=0$). Such a thing is  a contradiction in terms in classical logic, so this subject requires a weaker logic called intuitionistic logic in order to function (and even then $dx \neq 0$ is really "it cannot be proven that $dx=0$", a weaker statement).

Honestly, most mathematicians, scientists, and engineers don't need either one. It is better to learn methods for manipulating differentials in formal (i.e. "regarding only form", which is sort of like "non-rigorous") calculations. Optionally you can also learn proofs in standard analysis (which use finite but arbitrarily small numbers). These never explicitly use differentials.

One can deal with infinitesimals properly, although it is far from trivial.

But, more importantly, equalities of the kind $$\tag1 dy=f(x)\,dx$$ are just notation. What I mean is that you never do any algebra nor any other operation with $(1)$: you don't add it, you don't multiply it, you don't take its square root.

So, whether $dy$ and $dx$ in $(1)$ have meaning as objects is irrelevant. One uses an expression as $(1)$ to indicate  a substitution in an integral; and, in an integral, $dx$ and $dy$ are notation and not mathematical objects per se.


If $0<x<1$ then prove that $x^a \leq x < 1$ for all $a\in \mathbb{R}, a\geq 1$. If this is true then can somebody please help me to get the proof. Thanks!
I am trying to see the proof of this for all $a\in \mathbb{R}, a\geq 1$.I saw that this is easy inequality to prove if $a\in \mathbb{N}, a\geq 1$.  It follows easily by induction.
 I was trying to do this for $a\in \mathbb{R}, a\geq 1$. Now, I have figured out the proof which I have given below in this thread.

I have a proof for this inequality for all $a \geq 1, a\in \mathbb{R}$. Fix $0<x<1$ and
define a function $f(a) = x^a - x$. Then $f^{\prime}(a) = x^a \ln x-0<0$. Note that $\ln x <0$ for $0<x<1$. So the function is decreasing for $a \geq 1$. That is $f(a) \leq f(1)$ as desired.
This requires no induction and more powerful than induction.

If $a > 0$ and $b < c$, then $ab < ac$.

So, for any integer $n\geq 1$: if $x^{n-1} < x^n$, we must also have
$$
x\cdot x^{n-1} < x \cdot x^{n}
$$
which leads (by induction) to the desired result.


Slope field of $y'=x^2 - y^2$ I don't know how I am supposed to go about creating a table with slope values for the graph so that I can sketch them. I knew how to do it when $y'$ equations had $y$ only or $x$ only, but not when both are in the equation.

At each point, make a tiny line segment with slope $x^2-y^2$.  For instance, at the point $(1,2)$, make a tiny line segment of slope $1^2-2^2=-3$.  A full graph is here:  http://www.wolframalpha.com/input/?i=slope+field+of+y%27%3Dx%5E2-y%5E2

You haven't provided much information but here are two possible ways of solving it.

Solution one:
This is assuming you know at least one point on the integrated graph.

$\frac{dy}{dx} = x^2 - y^2$

$y = \int x^2 - y^2 dx$

$ y = \frac{(x^3)}{3} + y^2x + c$

The following is where you should solve for c if you have your point, I am just going to carry it through

$xy^2 - y + (\frac{x^3}{3} + c) = 0$

You can either substitute into the quadratic formula here for y in terms of x.

$y = \frac{ 1 \pm\sqrt{1 - 4x(\frac{x^3}{3} + c})}{2x}$

Because your original equation ($ \frac{dy}{dy} = x^2 - y^2$) requires $y^2, $square both sides.

$y^2 = (\frac{ 1 \pm \sqrt{1-4x(\frac{x^3}{3} + c)}}{2x})^2$

Now you can substitute that in.

Solution two: This is assuming you know at least one point on the graph ($y'$).
I'm going to use $(a,b)$ as my point.

$b = a^2 + y^2$

$b - a^2 = y^2$

If you don't have either of the requirements for the solutions then I don't think you can graph that. (If I have made a mistake please point it out)


Difference between a proposition and an assertion It may be a silly doubt, but let me ask this.

  What is the difference between a proposition and an assertion?

I know there's a very thin line between  the two terminologies, but I'm unable to get that.

There is a subtle difference between the two:

A proposition is a statement in either a natural or a formal language, for which it makes sense to ask whether it is either true or false.
An assertion is a statement which one claims to be true.

As such, assertions are more restrictive than propositions. For example:

In a logic textbook, we read "It's raining" as an example of a proposition. This is not an assertion.
I look out of the window and say: "It's raining." From the context, it is clear that I am asserting this proposition to be true. Hence, it is an assertion.

The distinction is fine, because everything that is a proposition can also be an assertion, while all assertions are necessarily propositions. So the distinction between the two depends entirely on the context.

Both are statements.
For a Proposition, we can say whether it is true or false.
But for an assertion, no chance to be false and definitely it is true.
So you can identify that all assertions are prepositions also.
But all propositions should not be assertions.


Show that $\exists x \in (0, \pi/2)$ such that $\cos(x)=x$ Let $f(x) = \cos(x)$, $g(x)=x$, both functions are continuous. $f(0)=1$, $f(\pi/2)=0$, so, by the Intermediate Value Theorem, for any $z\in [0,1]$, there exists $c\in [0,\pi/2]$ such that $f(x)=z$.

This should be simple to prove, but for some reason I have a problem with IVT, don't know why. Would appreciate some help.

I think that the fact that $g(x)=x$ is onto $[0,1]\subset [0, \pi/2]$ should be used here somehow.

Consider $h(x)=\cos(x)-x$. Then, $h(0)=1$ and $h\left( \displaystyle\frac{\pi}{2}\right)=-\displaystyle\frac{\pi}{2}$. Hence, $h(0)>h\left( \displaystyle\frac{\pi}{2}\right)$ and, clearly, $h(x)$ is continuous. For the IVT, there exist a $z\in\left( 0,\displaystyle\frac{\pi}{2}\right)$ such that $h(z)=0$.

$h(z)=0$ if and only if $\cos(z)-z=0$ if and only if $\cos(z)=z$

The proof is immediate by the use of graphs of $y=\cos x$ and $y=x$ in the given interval. See this Wolfram Alpha link for plots


Any diffeomorphism between the Minkowski indicatrix and the Euclidean sphere? Consider the Finsler Minkowski space $(R^n,F)$ and the Euclidean space $(R^n,||.||)$. Consider the Finsler Minkowski indicatrix of radius $r$, that is $$\Sigma(r)=\{x\in R^n\ :\ F(x)=r\}$$ FYI. The indicatrix of a Monkowski Finsler metric is (topologically) a spherical fiber bundle over $R^n$. Furthermore consider the Euclidean sphere of radius $r$, that is $$S^n(r)=\{x\in R^n\ :\ ||x||=r\}.$$ Now if there exists any diffeomorphism (an injective and surjective map whose inverse is also injective and surjective) $$f: (R^n,F)\to (R^n,||.||)$$ that takes the indicatrix to the sphere?

I mean I am asking about the indicatrix of a Monkowski Finsler metric which is (topologically) a spherical fiber bundle over $R^n$ is diffeomorphic to the sphere in $R^n$?

The only things that I found is the diffeomorphism between the punctured spaces as $$f:(R^n\backslash\{0\},F)\to (R^n\backslash\{0\},||.||)$$ s.t. $f(x)= \frac{||x||}{F(x)} x$ and its inverse $h(x)= \frac{F(x)}{||x||}x$.

I am going to assume that $F(x)$ is infinitely smooth (or smooth
enough) on $\mathbb{R}^n \setminus \{0\}$. It is continuous everywhere because it is a norm. Per your notations, $\Sigma(r) = \{ x \in \mathbb{R}^n \,
: \, F(x) = r\}$ and $S(r) = \{ x \in \mathbb{R}^n \, : \, \|x\| =
r \}$. Fix $\Sigma(1)$. Then there exists a large enough $R>0$
such that $\Sigma(1) \, \subset \, D_E(R) = \{ x \in \mathbb{R}^n
\, : \, \|x\| < R \}$. Consider an infinitely smooth
(rotationally symmetric) bump function
$$\lambda \, : \, \mathbb{R}^n \, \to \,\,\mathbb{R}$$ such that
\begin{align}
&\lambda(x) \equiv 0 \,\,\,\text{ for all } \, x \in \mathbb{R}^n \setminus D_E(2R)\\
0 \leq &\lambda(x) \leq 1 \,\,\,\text{ for all } \, x \in D_E(2R)\setminus D_E(R)\\
&\lambda(x) \equiv 1 \,\,\,\text{ for all } \, x \in D_E(R)\\
\end{align}
Rotationally symmetric means it has the form $\lambda(x) =
\kappa\big(\|x\|\big)$ for a smooth function $\kappa(s), \,\, s
\in [0,\infty)$ (actually, one can be very specific about the way
$\kappa$ looks). Define another similar bump function by setting
$$\lambda_{\epsilon}(x)  = \lambda\left(\frac{R}{\epsilon}x\right) = \kappa\left(\frac{R}{\epsilon}\|x\|\right) $$ which has the properties
\begin{align}
&\lambda_{\epsilon}(x) \equiv 0 \,\,\,\text{ for all } \, x \in \mathbb{R}^n \setminus D_E(2\epsilon)\\
0 \leq &\lambda_{\epsilon}(x) \leq 1 \,\,\,\text{ for all } \, x \in D_E(2\epsilon)\setminus D_E(\epsilon)\\
&\lambda_{\epsilon}(x) \equiv 1 \,\,\,\text{ for all } \, x \in D_E(\epsilon)\\
\end{align}
Take $\epsilon > 0$ very small, so small that $D_E(3\epsilon)$ is
entirely contained in the interior of $\Sigma(1)$.

Consider the expression  $H(x) = \big(1-\lambda(x)\big)\,\|x\| +
\lambda(x) \, F(x)$ which by construction is an infinitely smooth function for all points outside the disk $D_E(\epsilon)$. Define the infinitely smooth vector field on
$\mathbb{R}^n \setminus D_E(\epsilon)$
$$Y(x)  = - \, \frac{\nabla H(x)}{\|\nabla H(x)\|^2}$$
where $\nabla$ is the ordinary gradient in $\mathbb{R}^n$ (i.e.
roughly speaking it is the one defined via $\| \cdot\|\,$ ). One
can verify that, due to the way $H$ is constructed,
$\nabla \, H(x) \neq 0$ for all $x \in \mathbb{R}^n \setminus D_E(\epsilon)$ (recall that $\lambda$ was chosen to be rotationally symmetric and the two
norms are convex functions).

Furthermore, define the infinitely smooth vector field $$X(x) =
\big(1-\lambda_{\epsilon}(x)\big)\, Y(x)$$ for all $x \in
\mathbb{R}^n\setminus D_E(\epsilon)$ and $X(x) = 0$ for $x \in D_E(\epsilon)$. Then $X(x)$ is an
infinitely smooth vector field on the whole space $\mathbb{R}^n$.

Let $\phi^t(x)$ be the flow of the vector field $X$, i.e.
$$\frac{d}{dt} \, \phi^t(x) = X\big(\phi^t(x)\big)$$ Moreover, due
to the fact that the vector filed $X$ is bounded everywhere on
$\mathbb{R}^n$, i.e. there exists a large enough cpnstant $M>0$ such that
$$\|X(x)\| = \big|\big(1-\lambda_{\epsilon}(x)\big)\big| \, \|Y(x)\| \leq M$$
the flow extends for all $t \in \mathbb{R}$. By construction, given $x \in \mathbb{R}^n \setminus D_E(3\epsilon)$ the following relation holds
$$H\big(\phi^t(x)\big) = H(x) - t$$ for $t \in (-\infty, H(x) - 2\epsilon)$.
Indeed, for any point $x \in \mathbb{R}^n \setminus D_E(3\epsilon)$  we have that $X(x) = Y(x)$ so whenever $t \in (-\infty, H(x) - 2\epsilon)$
\begin{align}
H\big(\phi^t(x)\big) - H(x) &= \int_{0}^t \, \frac{d}{d\tau} \, H\big(\phi^{\tau}(x)\big) \, d\tau = \int_{0}^t \, \Big( \, \nabla H\big(\phi^{\tau}(x)\big) \, \cdot \, \frac{d}{d\tau} \, \phi^{\tau}(x) \, \Big) \, d\tau\\
&= \int_{0}^t \, \Big( \, \nabla H\big(\phi^{\tau}(x)\big) \, \cdot \, X\big(\phi^{\tau}(x)\big) \, \Big) \, d\tau\\
&= \int_{0}^t \, \Big( \, \nabla H\big(\phi^{\tau}(x)\big)
\, \cdot \, Y\big(\phi^{\tau}(x)\big) \, \Big) \, d\tau\\
& = - \, \int_{0}^t \,\frac{ \Big( \, \nabla H\big(\phi^{\tau}(x)\big) \, \cdot \,  \nabla H\big(\phi^{\tau}(x)\big) \, \Big)}{\|\nabla H\big(\phi^{\tau}(x)\big) \|^2} \, d\tau\\
& = - \, \int_{0}^t \,\frac{\|\nabla H\big(\phi^{\tau}(x)\big) \|^2}{\| \nabla H\big(\phi^{\tau}(x)\big) \|^2} \, d\tau =  - \, \int_{0}^t  \, d\tau\\
& = - t
\end{align}
The phase flow $\phi^t(x)$ is a diffeomorphism for any fixed $t \in
\mathbb{R}$ and can move any level hyper-surface $H(x) = r_1$ to any level
hyper-surface $H(x) = r_2$ as long as $r_1 > 2\epsilon$ and $r_1 >
2\epsilon$.

Then, by construction $$\{x \in \mathbb{R}^n \, : \, H(x) = 3R\,\}
= S(3R) \,\,\,\,\ \text{ and } \,\,\,\, \{x \in \mathbb{R}^n \, :
\, H(x) = 1\,\} = \Sigma(1)$$ Set $t_0 = 3R - 1$. Then, again by
construction, $\phi^{t_0}\big(S(3R)\big) = \Sigma(1)$. Then the
map
$$\phi = \phi^{t_0}|_{S(3R)} \, : \, S(3R) \, \to \, \Sigma(1)$$
is a smooth diffeomorphism between the two submanifolds $S(3R)$
and  $\Sigma(1)$ and in fact $$\phi^t(x) \, : \, t \in [0, 3R-1]$$
is a smooth isotopy of $\mathbb{R}^n$ which isotopes
$S(3R)$ to $\Sigma(1)$. To get an ambient isotopy from $S(1)$ to
$\Sigma(1)$, simply precompose $\phi^t(x)$ with a linear
homothetic isotopy that stretches homogeneously $\mathbb{R}^n$,
taking $S(1)$ to $S(3R)$.

Yes, the indicatrix of a Minkowski space is diffemorphic to a sphere.

Let $I$ and $S$ be the indicatrix of a Minkowski (V,F) and the sphere of a Euclidian space, resp. The aplication $f:S\to I$ give by $$v\mapsto \frac{v}{F(v)}$$
is a diffomorphism. For more details vide the Proposition 2.6 on [Javaloyes and Sànches, On the definiton and examples of Finsler metrics].


Gradient of a line using 2 points Why is that the gradient of the line? Why is it 1/ (OP/OQ)?

The definition of gradient in this case is the same as the derivative: $\frac{\Delta y}{\Delta x}$. you don't even need to put the limit $\Delta x\to 0$, since we are talking about a line.
$$\frac{\Delta y}{\Delta x}=\frac{y_P-y_Q}{x_P-x_Q}=\frac{y_O-y_P}{x_P-x_O}=\frac{-|y_O-y_P|}{|x_P-x_O|}=-\frac{OQ}{OP}=-\cot\theta$$

The gradient of a line is the same as the slope of the line. The slope of a line is defined to be $$\frac{\text{change in }y}{\text{change in }x },$$

or if we want to we can write this as $$\frac{1}{\left(\frac{\text{change in }x }{\text{change in }y}\right)}.$$

Now, in this problem the change in $x$ is given by $OP$ and the change in $y$ is given by $-OQ$, because $OQ$ only gives the length of that line, but as $x$ increases $y$ decreases, so the change is negative.  Putting it together, then, we get that the slope is $$\frac{1}{\left(\frac{\text{change in }x }{\text{change in }y}\right)} = \frac{1}{\left(\frac{OP}{-OQ}\right)} = -\frac{1}{\left(\frac{OP}{OQ}\right)}.$$


complex number : exam exercise I don't realize how to solve this:

$z^4\cdot{\displaystyle {\bar {z}}} + 243 = 0$

I just know this, but it doesn't help me at all.

$z\cdot{\displaystyle {\bar {z}}}=|z|^2$

You know that $z\bar{z}=|z|^2$, so then $z^3 \cdot z\cdot \bar{z}=z^3|z^2|=-243$.

Now, $||z^3|z^2|=|z^5|=|-243|$, so $|z|=3$.

Going back to $z^3|z^2|=-243$, we have $z^3=-27$.

  That means the three roots are $z=3e^{\frac{\pi i}{3}},3e^{\frac{3\pi i}{3}},3e^{\frac{5\pi i}{3}} $. Do you see how those three roots come from $-27$?

Let $z=r(\cos\theta+i\sin\theta),$ where $r>0$ and $0\leq \theta<360^{\circ}.$

Thus, $$r^4(\cos4\theta+i\sin4\theta)r(\cos(-\theta)+i\sin(-\theta))=-243$$ or
$$r^5(\cos3\theta+i\sin3\theta)=-243,$$ which gives
$$\left|r^5(\cos3\theta+i\sin3\theta)\right|=|-243|$$ or $$r^5=243$$ or
$$r=3$$ and it remains to solve:
$$\cos3\theta+i\sin3\theta=-1.$$
Can you end it now?


draw an Ellipsoid given its quaderatic equation in Matlab How can I draw an ellipsoid in Matlab with following equation:

$aX^2 + bY^2 + cZ^2 + d XY+e XZ +f YZ +gX +hY+iZ=R^2$

with (for example)
$a=1;
b=2;
c=3;
d=4;
e=5;
f=6;
g=7;
h=8;
i=9;
R^2=10.$

Since the question's been bumped by the Community bot yet again, I'm posting my comment as an answer:

You can just use isosurface.

[X, Y, Z] = meshgrid(-5:5, -5:5, -5:5);
isosurface(X, Y, Z, a*X.^2 + b*Y.^2 + c*Z.^2 + d*X.*Y + e*X.*Z + f*Y.*Z + g*X + h*Y + i*Z, r^2)

You can treat the ellipsoid as a transformed sphere, with points $v$ given by an equation of the form $v^t M v = 1$. Because the matrix $M$ will turn out to be symmetric with nonnegative eigenvalues, it has a square root, say $S$, so that $S^t S = M$. Then the equation becomes $v^t S^t S v = 1$, so $Sv$ is a vector $u$ satisfying $u^t u = 1$, i.e., a point of the unit sphere. Letting $v = S^{-1}u$, you're done. here's the code:

function ellipsoid(a, b, c, d, e, f, g, h, j, R)
M = (1/R^2) * [
    a d e g;
    0 b f h;
    0 0 c j;
    0 0 0 1];
M = (M + M')/2;
S = M^0.5

n = 30;
[th, ph] = meshgrid(linspace(0, 2*pi, n), linspace(0, pi, n));
spherepoints = zeros(4, n, n);
spherepoints(1, :,:) = cos(th) .* sin(ph);
spherepoints(2, :,:) = sin(th) .* sin(ph);
spherepoints(3, :,:) = cos(ph);
spherepoints(4, :, :) = 1;
sp = reshape(spherepoints, 4, []);
epoints = S \ sp;
epoints = reshape(epoints, 4, n, n);

surf(squeeze(epoints(1, :,:)),...
    squeeze(epoints(2,:,:)),...
    squeeze(epoints(3,:,:)));
set(gca, 'DataAspectRatio', [1,1,1]);
figure(gcf);

Thanks to the anonymous reviewer who noted that I'd written epoints = S * sp instead of S \ sp.


How to calculate the total number of possible triangles where the perimeter is given, and all sides are integers? How would I go about calculating the total number of possible triangles where the perimeter is given, and all sides are integers?

Exempli gratia:

$a + b + c = 15$
$d + e + f = 16$
$x + y + z = 100$

This is given in A005044.  The generating function is $\frac{x^3}{(1-x^2)(1-x^3)(1-x^4)}$  Some other formulas and references are given as well

An efficient approach can be found with little algebraic manipulation :

Since $a + c > b$ or, $p – b > b$ or, $b < p/2$. Thus iterating $b$ from $1$ to $p/2$, calculating $a$ and storing only the whole number $a$ would give all solutions for a given $p$. There are no right triangles are possible for odd $p$.


Is this definition of splitting field pleonastic? I found this definition of a splitting field but I am wondering if the second condition does not implies the first one.

If $L$ is generated over $K$ by the zeros of the polynomials of the family, does not it follows that such zeros belong to $L$ and so every $f_i$ splits into linear factors?

If (2) does not hold, there would be infinitely many "splitting fields". The condition (2) guarantees that L is the smallest field in which the function family factors into linear factors.


The Integral that Stumped Feynman? In "Surely You're Joking, Mr. Feynman!," Nobel-prize winning Physicist Richard Feynman said that he challenged his colleagues to give him an integral that they could evaluate with only complex methods that he could not do with real methods:

One time I boasted, "I can do by other methods any integral anybody else needs contour integration to do."
So Paul [Olum] puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration! He was always deflating me like that. He was a very smart fellow.

Does anyone happen to know what this integral was?

I doubt that we will ever know the exact integral that vexed Feynman.
Here is something similar to what he describes.

Suppose $f(z)$ is an analytic function on the unit disk.
Then, by Cauchy's integral formula,
$$\oint_\gamma \frac{f(z)}{z}dz = 2\pi i f(0),$$
where $\gamma$ traces out the unit circle in a counterclockwise manner.
Let $z=e^{i\phi}$.
Then
$\int_0^{2\pi}f(e^{i\phi}) d\phi = 2\pi f(0).$
Taking the real part of each side we find
$$\begin{equation*}
\int_0^{2\pi} \mathrm{Re}(f(e^{i\phi}))d\phi = 2\pi \mathrm{Re}(f(0)).\tag{1}
\end{equation*}$$
(We could just as well take the imaginary part.)
Clearly we can build some terrible integrals by choosing $f$ appropriately.

Example 1.
Let $\displaystyle f(z) = \exp\frac{2+z}{3+z}$.
This is a mild choice compared to what could be done ...
In any case, $f$ is analytic on the disk.
Applying (1), and after some manipulations of the integrand, we find
$$\int_0^{2\pi}
\exp\left(\frac{7+5 \cos\phi}{10+6\cos\phi}\right)
\cos \left(
\frac{\sin\phi}{10+6 \cos\phi}
\right)
d\phi = 2\pi e^{2/3}.$$

Example 2.
Let $\displaystyle f(z) = \exp \exp \frac{2+z}{3+z}$.
Then
\begin{align*}\int_0^{2\pi} &
\exp\left(
 \exp\left(
  \frac{7+5 \cos \phi}{10+6 \cos \phi}
 \right)
 \cos\left(
  \frac{\sin \phi}{10+6 \cos \phi}
 \right)
\right) \\
& \times\cos\left(
 \exp\left(
     \frac{7+5 \cos \phi}{10+6 \cos \phi}
    \right)
 \sin\left(
     \frac{\sin \phi}{10+6 \cos \phi}
    \right)
\right) = 2\pi e^{e^{2/3}}.
\end{align*}

It was something related to calculating a definite integral that required Feynman to calculate some digits of $\pi$ after the decimal point. An exact reference to this integral would be very difficult to find if it exists in literature.


Prove $\sum\limits_{r=0}^{2n} r \binom{2n}{r}^2 = 2n \binom{4n-1}{2n}$ I expanded

$(1+x)^{2n}$ = $\sum\limits_{r=0}^{2n} \binom{2n}{r} x^r $

Differentiating both sides, we get

$2n(1+x)^{2n-1}$ = $0$ + $\binom{2n}{1}$ + $2\binom{2n}{2}x$ + $3\binom{2n}{3}x^2$ .....

Put $x=1$.

As you see, I'm not able to get the 'squares'.Any idea on how to go about it?

Out of $4n$ people, $2n$ men and $2n$ women, select a committee of $2n$ people and choose a woman to be the chair of the committee. If $r$ women are chosen, there are $\binom{2n}{r} \binom{2n}{2n-r} = \binom{2n}{r}^2$ ways to choose the committee and $r$ ways to choose the chair, so the left hand side gives the number of ways to do so. Alternatively, first choose one of the $2n$ women as chair and then select the remaining members of the committee in $\binom{4n-1}{2n-1} = \binom{4n-1}{2n}$ ways.

$\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\sum_{r = 0}^{2n}r{2n \choose r}^{2} = 2n{4n - 1 \choose 2n}:\ {\large ?}}$

  Hereafter we'll use the identity:
  $$\color{#c00000}{%
{m \choose \ell} = \int_{\verts{z} = 1}{\pars{1 + z}^{m} \over z^{\ell + 1}}
\,{\dd z \over 2\pi\ic}\,,\qquad m, \ell \in {\mathbb N}\,,\quad m \geq \ell}
$$

\begin{align}
\color{#00f}{\large\sum_{r = 0}^{2n}r{2n \choose r}^{2}}&
=\sum_{r = 0}^{2n}r{2n \choose r}
\int_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z^{r + 1}}\,{\dd z \over 2\pi\ic}
=\int_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z}
\sum_{r = 0}^{2n}{2n \choose r}{r \over z^{r}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\int_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z}
\lim_{\mu \to 1/z}\bracks{\mu\,\partiald{}{\mu}
\sum_{r = 0}^{2n}{2n \choose r}\mu^{r}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\int_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z}
\lim_{\mu \to 1/z}\bracks{\mu\,\partiald{\pars{1 + \mu}^{2n}}{\mu}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\int_{\verts{z} = 1}{\pars{1 + z}^{2n} \over z}
\bracks{{1 \over z}\,2n\pars{1 + {1 \over z}}^{2n - 1}}\,{\dd z \over 2\pi\ic}
=2n\int_{\verts{z} = 1}{\pars{1 + z}^{4n - 1} \over z^{2n + 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\color{#00f}{\large 2n{4n - 1 \choose 2n}}
\end{align}


Functions images and inverse images The objective of this question is to find if the function is a bijective function or not and if it is a bijective find its images and inverse images.

$$ f:\mathbb{Z^2} \to \mathbb{Z}$$
$$ f(n,k) = n^2k $$

We have to find inverses of   $ f^{-1}(\{0\}) $,  $ f^{-1}(\mathbb{N}) $  and  $  f(\mathbb{Z} \times \{1\}) $

But I fail to understand the approach to this problem, I do understand that they need to have unique mappings and co-domains must be matched, but could anyone help me make it analogous to this situation?

questions such $$y = x^2 $$ is not bijective since they have multiple images and are not bijective. Their inverse will be a sqaure root with + and - and hence its an invalid case. Could someone please  correct my approach?

We have

$$ f:\mathbb{Z^2} \to \mathbb{Z}$$
$$ f(n,k) = n^2k $$

This function is obviously not bijective. Many elements get mapped to zero. For example, $f(1,0) = f(-1,0) = 0$ (not injective). The function is surjective though.

So this means that $f$ does not have a well defined inverse function, as bijectivity is required for that.

However, we can consider the inverse image, even when the inverse function does not exist. You are asked to find $$f^{-1}(\{0\})$$

or in other words:

Find the pairs $(n,k)$ such that $f(n,k) = 0$.

So $f(n,k) = 0 \iff n^2k = 0 \iff n = 0 \lor k = 0$

Hence:

$$f^{-1}(\{0\}) = \{(n,k)|n = 0 \quad \mathrm{or} \quad k = 0\} = \{(n,0)|n \in \mathbb{Z}\} \cup \{(0,k)|k \in \mathbb{Z}\}$$

Can you proceed now?

We have
$$f:\mathbb{Z^2} \to \mathbb{Z}$$
$$f(n,k) = n^2k$$

For Surjection,
$$n^2k = n.n.k$$
example
$$2 . 2. 2 = 16$$ $$and$$ $$1.1.16 = 16$$
both map to the same number in the codomain,
Since surjective function is a function whose image is equal to it's codomain it is a surjective function.
For injection,  let's assume
$$n^2k = p^2q$$
example $$2.2.9 \neq 3.3.4$$
Hence, the function is not injective
$f^{-1}(\mathbb{N})$
$$n^2k \in \mathbb{N}$$
$$= \mathbb{Z} \times \mathbb{N}$$
$f^{-1}(0)$
$$n^2k = 0$$
$${n = 0} \cup {k = 0}$$
$${ {0}\times\mathbb{Z} }\cup{ \mathbb{Z} \times {0} }$$
$f^{-1}({-1})$
$$n^2k = -1$$
$${n = 1} \cup{n = -1}$$
Since n is squared, it can either be positive or negative, will always give +1.
$$k = -1$$
$${ {1} \times {-1} }\cup{ {-1} \times {-1} }$$
$f(\mathbb{Z}\times{1})$
$$n^2k = integer^2 \times 1 = natural$$


Why is 8 so special? I have been reading about multi-dimensional numbers, and found out that it's been proven that the Octonions are the composition algebra of the largest dimension. I was wondering why, despite having infinitely many different dimensions of numbers, the only composition algebras are of 1, 2, 4, and 8 dimensions. What's so special about 8?

You might be interested in Hurwitz's proof of his theorem (which is not as strong as Wikipedia's statement). Here is the original German and an English translation. The maximal $n$ turns out to be the solution of $2^{n-2} = n^2$.

Your information is wrong: tessarines of any $2^n$ dimension (for instance, 16-dimensional tessarines) is a commutative, associative, composition algebra.