prove or disprove an inequality on bounds of derivatives for radial functions Suppose $f$ is a radial function, i.e., $f(x)=f(|x|)$,
and $f \in C^\infty(\bar{B})$, where $\bar{B}$ is the closure of the unit ball in $\mathbb{R}^n$.
Prove or disprove the following.

Given any positive integer $k$, $$\sup_{|\alpha|=k,x\in B} |D^\alpha f(x)| \leq \sup_{r < 1} \lvert f^{(k)}(r) \rvert,$$
where $\alpha$ is a multi-index and $D^\alpha f$ is the corresponding derivative of $f$.
By $f^{(k)}(r)$, we mean the $k^{th}$ derivative of $f$ as a function of $r=|x|.$ The norm on the left is in fact $|\cdot|_{W^{k,\infty}(B)}.$

I try some functions and the inequality hold for all of them.
The case where $k=1$ is easy to prove but I can't prove for a general $k$.

Update: Instead of a general smooth $f$,  can we prove the assertion for polynomials in which
$f(r) = \sum_{j=0}^m c_j r^{2j}$ ?

This question has been posted on Math Overflow. The answer can be found here:Math Overflow answer


probability of rolling a sum of $14$ when rolling a $20$-sided die twice or a $20$-sided die with a $4$-sided die depending on outcome of first roll A $20$-sided die is rolled. If the result is $10$ or less, the $20$-sided die is rolled again. If the result is $11$ or more, a $4$-sided die is rolled. In either case, the results are summed after the second die roll. What is the probability that the sum will be $14$?

My Approach:
there are 10 ways we can get a sum of 14 if the first dice is  $<= 10$ {$(10,4),(9,5),(8,6),(7,7),(6,8),(5,9),(4,10),(3,11),(2,12),(1,13)$} and $3$ ways we can get sum of $14$ if the first dice has value  $>= 11$ {$(11,3),(12,2),(13,1)$} so the probability is: $(10/400)$ + $(3/80) = 1/16$

Hints:

$Pr(A\cap B) = Pr(A)Pr(B\mid A)$

$Pr(B) = Pr(A_1\cap B)+Pr(A_2\cap B)+\dots+Pr(A_n\cap B)$ where $A_1,\dots,A_n$ form a partition of the sample space

Let $B$ be the event the sum is $14$.

Let $A_1$ be the event the first die rolls a number $1-10$.  Let $A_2$ be the event the first die rolls a number $11-13$.  Let $A_3$ be the event the first die rolls a number $14+$.

Additional Hint:

   Regardless what was rolled on the first die, so long as it was less than $14$, there will be exactly one outcome on the second roll that would let the total sum to $14$.

The probability that you obtain $14$ using the $20$-sided die on the second role is

$$\frac{10}{20} \cdot \frac{1}{20}$$

The probability you obtain $14$ using the $4$-sided die on the second role is

$$\frac{3}{20} \cdot \frac{1}{4}$$

Add these.


Deeper understand of $\cos$ and $\sin$ I was hoping to understand $\sin$ and $\cos$ a bit more in detail regarding how they came about, not so much about the history but rather and the relationships of these functions with nature and other math topics. In school and college (depending on what your major is), we focus just on the mechanical aspects of $\sin$ and $\cos$. But I'm still having a hard time wrapping my head around how these functions came about.
On the unit circle, why does $\sin$ correspond to the $y$ coordinate on a point on the circle and $\cos$ correspond to the $x$ coordinate(was this just chosen by someone a long time ago)? Or why the vertical distance from the center of the circle to the tip of the line gives us the amplitude of the sine wave as shown on this site? Or why is there a relationship between these two trig functions and a circle? Lastly, are $\sin$ and $\cos$ found in cyclic parts of nature, or are these functions used as good approximations regarding cyclic behaviors in nature?
Sorry if the questions are open-ended, it's just that I've read some wiki articles and visited some other random websites, but I still wasn't able to find answers to these questions. Thanks for your time and help!

Let's look at the most basic definition of $\sin x$ and $\cos x$, using a right-angled triangle.

$\sin x=\frac{\text{Opposite}}{\text{Hypotenuse}}$ and $\cos x=\frac{\text{Adjacent}}{\text{Hypotenuse}}$
This gives us a clear geometric interpretation of what we mean by $\sin x$ and $\cos x$. If we want to compute the sides of a triangle (or the angles of a triangle), then these definitions of $\sin x$ and $\cos x$ will suffice. $\sin x$ and $\cos x$ are useful because if we know what the value of the angle $x$, then the ratios stays the same regardless of how big or small the right-angled triangle is.
At this point, it is unclear why we would need to know what $\sin(100°)$ is. Indeed, using the above definition, $\sin(100°)$ doesn't even have a defined value. It is only when we start using a unit circle that $\sin(100°)$ is meaningful. Why would even try to define $\sin(100°)$? Well, it's not immediately clear. At some point in history, mathematicians were scratching their heads and wondering what the use of defining the $\sqrt{-1} \text{ as } i$ was (if this makes your head spin, don't worry!). Again, it's not immediately clear. But often, when we extend the definition of something, its uses becomes apparent later on. (Incidentally, $i$ is now used routinely in physics and engineering.)
$\sin(100°)$ is similar to $i$ in that it now has many real-world applications. According to Wikipedia, 'the sine function can be used to model phenomena such as sound and light waves'. It is much easier to extend the arguments of $\sin x$ and $\cos x$ to all real numbers, than to stick rigidly to the definition provided by the right-angled triangle. Therefore, while $\sin(100°)$ is certainly an abstract concept at first, ultimately it is very useful.
Now that we have established the motivation for extending the arguments of $\sin x$ and $\cos x$, all that is left to answer is exactly how. Fortunately, there are many sources that explain this. One of your questions was about why the unit circle definition is used, and why $\sin x$ is simply the $y-$coordinate. In this Wikipedia article, all is explained. Let me know if you have any questions.
And, we can keep on going. Remember when I mentioned $i$, the $\sqrt{-1}$ earlier? That's exactly the kind of argument that modern defintions of $\sin x$ and $\cos x$ can accommodate. Again, more abstract, but more useful. I'll spare you the details, but just remember, that there is always a rationale behind what we do in mathematics. No one ever just decided that $\sin x$ was something without good reason for it.


Integration by parts. Which is the simplest way for this integral? Which is the easiest way to compute the following integral by parts?
$$\displaystyle\int\limits_{x}^{\infty}\dfrac{2}{\sqrt{\pi}}e^{-s^2}ds$$?

Given that $\int f'(x)\cdot g(x)=f(x)\cdot g(x)-\int f'(x)\cdot g(x) +c$

could I choose $f(x)=e^{-s^2}$ and $g'(x)=1$?

Which is the simplest way to the solution of the integral by parts?

To nitpick a few comments, it is impossible to find the antiderivative of this function in terms of elementary functions. However, it is perfectly possible to compute this integral.

First, recall that the integral is a linear operator, which is a fancy term for saying that $\int af(x)dx = a \int f(x)dx$ and $\int (f(x)+g(x))dx =  \int f(x)dx+\int g(x)dx$.

So to start, we can express the integral as:

$$\frac{2}{\sqrt{\pi}}\int_x^\infty e^{-s^2}ds$$

If you wanted to find the antiderivative of this function, you would need a series expansion, and to express the result in terms of a series. However, it is perfectly possible to find this specific integral with a little trick.

Let's set $I = \int_x^\infty e^{-s^2}ds$ (we'll multiply by the constant fraction in the end). Let's also restrict this answer to $x \geq 0$: it's easy to observe that $e^{-x^2}$ is even, so all we would need to do to find the answer for $x<0$ is compute $2*\int_0^\infty e^{-s^2}ds-\int_{|x|}^\infty e^{-s^2}ds$. Observe that $I = \int_x^\infty e^{-t^2}dt$ as well, so:

$$I^2 = \int_x^\infty e^{-s^2}ds*\int_x^\infty e^{-t^2}dt$$

$t$ and $s$ are constant realative to each other, so:

$$I^2 = \int_x^\infty \int_x^\infty e^{-s^2}e^{-t^2}dsdt = \int_x^\infty \int_x^\infty e^{-(s^2+t^2)}dsdt$$

Observe that, if we were to plot the region of integration with respect to s and t, we would find that we are integrating from the quarter circle with radius $x$ in quadrant I to the quarter circle with radius $c$ in quadrant I, where $c$ increases without bound.

So how does this help us? Well, we can integrate much more easily if we use polar coordinates, because while the differential in linear coordinates is $ds*dt$ (the area of a small rectangle in those coordinates), if we integrate with variables $r = \sqrt{s^2+t^2}$ and $\theta  = arctan(t/s)$, the differential is $r*dr*ds$. So:

$$I^2 = \int_x^\infty \int_0^{\pi/2} e^{-r^2}*r*d\theta dr = \frac{\pi}{2}\int_x^\infty e^{-r^2}*rdr = -\frac{\pi}{4}[e^{-r^2}]_x^\infty = \frac{\pi}{4} e^{-x^2}$$

So, for $x \geq 0$

$$I = \frac{\sqrt{\pi}}{2} e^{-\frac{x^2}{2}}$$

Which means that for $x \geq 0$:

$$\frac{2}{\sqrt{\pi}}\int_x^\infty e^{-s^2}ds = e^{-\frac{x^2}{2}}$$

And for $x < 0$:

$$\frac{2}{\sqrt{\pi}}\int_x^\infty e^{-s^2}ds = 2*e^{-\frac{0^2}{2}}-e^{-\frac{x^2}{2}} = 2-e^{-\frac{x^2}{2}}$$

Note that this does not contradict the idea that $e^{-x^2}$ does not have an elementary antiderivative. You cannot differentiate the result here to get back the original function. This can only be used to calculate actual numerical areas given that one of your upper bound is infinite.

calculus is not sufficient to solve your integral. You need to look at complex analysis books which includes error function.


Find the intersection of the images of two matrices I am studying for a big test and have this problem:

I am given two matrices A and B

$$A = \left(\begin{array}{crc}
 1 & 0\\
 2 & 1\\
 1 & -1\\
\end{array}\right)$$

$$B = \left(\begin{array}{crc}
 0 & 1\\
 1 & 0\\
 0 & 1\\
\end{array}\right)$$

1) Decide if the im(A) and the im(B) intersect and if yes find all the vectors of the intersection.Explain why the intersection is a subspace of $R^3$ and find its dimension.

2) Find a vector in the im(A) that is not in the im(B)

I have tried to solve it and from my understanding they do not intersect which is prob wrong.

Any help appreciated. Thanks!

Here are some hints.

First, you write "from my understanding they do not intersect". That's definitely wrong, because $im(A)$ and $im(B)$ are subspaces of $\mathbb R^3,$ and so both contain $0,$ the zero vector. So, we have at least $\{0\} \subseteq im(A) \cap im(B).$

Second, we can use dimesions. By "visual inspection", we see that the columns of the matrix $A$ are linearly independent. So, since $A$ has two columns, we must have $\dim\ im(A) = 2.$ In a completely similar manner, we find $\dim\ im(B) = 2.$ Now, there is a "well known" formula in linear algebra which states that for subspaces $U$ and $V,$ we have
$$
\dim U + \dim V = \dim(U+V) + \dim(U \cap V).
$$
We can apply this to the subspaces $im(A)$ and $im(B)$ of $\mathbb R^3.$ Note that $im(A) + im(B) \subseteq \mathbb R^3,$ and so $\dim(im(A) + im(B)) \leq \dim \mathbb R^3 = 3.$ Plugging everything we know about $im(A)$ and $im(B)$ in the formula above, we get
$$
\begin{align}
4 & = 2 + 2 \\
& = \dim\ im(A) + \dim\ im(B) \\
& = \dim(im(A) + im(B)) + \dim(im(A)\cap im(B)) \\
& \leq 3 + \dim(im(A)\cap im(B)),
\end{align}
$$
which leads us to
$$
\dim(im(A)\cap im(B)) \geq 1.
$$
So there must be a nonzero vector in $im(A)\cap im(B).$ In order to find it, you can use the fact that $im(A)$ and $im(B)$ are planes in $\mathbb R^3,$ and thus described by their normal vectors. The normal vector of $im(A)$ is the cross product of the columns of $A$, and similar for $im(B)$. Now, find a vector that is orthogonal to both these normal vectors. This is a homogeneous system $X$ of two linear equations in three variables. One solution to this system $X$ is
$$
v = \begin{pmatrix}1 \\ 2 \\ 1 \end{pmatrix}.
$$
We also can see that $X$ has rank 1, and thus we must have $im(A) \cap im(B) = \mathbb R v.$

Finally, note that every element of $im(B)$ is of the form
$$
w = \begin{pmatrix}\lambda \\ \mu \\ \lambda \end{pmatrix}
$$
for arbitrary $\lambda,\mu \in \mathbb R.$ From this, we see that for every element of $im(B),$ the first and last coordinate must be equal. From this in turn, we see that the second column of $A$ is contained in $im(A),$ but not in $im(B).$

Notice that the first column of $A$ is in the image of $B$.  In fact the image of $B$ consists of vectors of the form $(a,b,a)$.  What would happen if you added a nonzero multiple of the second column of $A$ to a multiple of the first column?  Would that still be in the image of $B$?


How many $3$-Letter Strings from come from $ABRACADABRA$? How many three-letter strings are there that come from the word $ABRACADABRA$?

I know I can split this in to cases, but I was hoping that there would be a better way of solving it.

Here is an approach using exponential generating functions.  Whether this is an improvement over case by case analysis depends on how you feel about algebra and maybe how you feel about using a computer algebra system.  Personally, I use a CAS and feel no guilt.

More generally, let's see if we can find how many $r$-letter words can be formed; call this number $a_r$.  Define the exponential generating function of $a_r$ by
$$f(x) = \sum_{r=0}^{\infty} \frac{1}{r!} a_r x^r$$
Then it's "easy to see" (ahem) that
$$f(x)=(1+x+\frac{1}{2!}x^2 +\frac{1}{3!}x^3 +\frac{1}{4!}x^4 +\frac{1}{5!}x^5) \cdot (1+x+\frac{1}{2!}x^2)^2 \cdot (1+x)^2$$
If we expand this (probably using a CAS), we will have the solution to the problem for all values of $r = 0, 1, 2, \dots , 11$.  But for your problem, all we really need is the coefficient of $\frac{1}{3!} x^3$.  It's possible to find this coefficient by pencil and paper without a lot of work, but doing so is very similar to case by case analysis, so I'm not sure that is an improvement over the case by case approach.

Using a CAS, it turns out that the coefficient of $x^3$ is 97/6, so the coefficient of $\frac{1}{3!} x^3$ is 97.  That's your answer.

I think that it is like that:

$$\binom{11}{3} \cdot \frac{11!}{5! \cdot 2! \cdot 2!}$$

You want a string of $3$ letters,so from the 11 possible letters you choose three ones.
But you have to consider that there are $5$ $A$s, $2$ $B$s and $2$ $R$s.


Help with integral with parameter. I have this integral with parameter:
$$g(y)=\int_{0}^{\frac{\pi}{y}}\frac{\tan^2(xy)}{2x}dx$$

Also, $y>0$
So i want to calculate $g'$.
How can i do this?
Any help would be really helpful.

Hint. Use the Leibniz integral rule. The first step is
$$g'(y)=\frac{\tan^2((\pi/y)\cdot y)}{2(\pi/y)}\cdot \frac{d}{dy}\left(\pi/y\right)
+\int_{0}^{\pi/y}\frac{d}{dy}\left(\frac{\tan^2(xy)}{2x}\right)dx.$$

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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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  Lets consider a 'more' general case: $\ds{\,\mrm{G}\pars{y} \equiv \int_{0}^{\pi/y}{\mrm{f}\pars{xy} \over 2x}\,\dd x}$ for 'some'
  $\ds{\,\mrm{f}}$ such that $\ds{\mrm{G}\pars{y}}$ converges. Indeed,
  $\ds{\,\mrm{G}\pars{y}}$ is $\ds{y}$-independent:

\begin{align}
\mrm{G}\pars{y} & =
\int_{0}^{\pi/y}{\mrm{f}\pars{xy} \over 2x}\,\dd x
\,\,\,\stackrel{x\ \equiv\ t/y}{=}\,\,\,
\int_{0}^{\pi}{\mrm{f}\pars{t} \over 2t}\,\dd t\quad\imp\quad
\color{#f00}{\mrm{G}'\pars{y}} = \color{#f00}{0}
\end{align}

  As pointed out by $\texttt{@Jack Lam}$, the "OP-original integral" diverges because
  $\ds{\tan^{2}\pars{t} = {1 \over \tan^{2}\pars{\pi/2 - t}} \sim
{1 \over \pars{t - \pi/2}^{2}}}$
  as $\ds{t \to \pi/2}$.


Existance of an (in)finite theory having infinite model Please help me to study the following simple cases:

Let $P$ be a binary predicate symbol. I am trying to find out, if there exists a satisfiable $T$ having infinite models only, for the following cases:

1) T is under {P} language with "="

2) T is under {P} language with "=" and T is finite

3) T is under {P} language without "="

4) T is under {P} language without "=" and T is finite

Where I am:

1)
$\varphi_i \equiv \forall x_1 \forall x_2 ... \forall x_i \exists y \land ^i_{j=1} \lnot( x_j=y)$

$T = \{\varphi_i | i>0 \}$ so there exists a T having only infinite models

2) 4) can probably have use of Löwenheim-Skolem theorem, but I don't see how to apply it properly.

3) No adequate idea so far

Hint: it is possible to axiomatize that $P$ is the comparison relation for a linear order (i.e., it plays the role of "$\leq$") without endpoints. Think about how those axioms depend on the presence of the equality relation, and how you might modify them to eliminate that dependency while still ensuring your models are infinite.

To make Mike's answer just a little more explicit:

When $P$ is the inequality symbol of a partial order, it interprets equality in the sense that the definable set $\{m \in M \operatorname{ | } P(m,m)\}$ is exactly the diagonal relation $\{m \in M \operatorname{ | } m = m\}$.

In particular, we know that such a $P$ must satisfy: $$\forall a \forall b, P(a,b) \land P(b,a) \rightarrow P(a,a),$$
$$\forall a \forall b \forall c, P(a,b) \land P(b,c) \rightarrow P(a,c),$$
$$\forall a, P(a,a).$$

After writing these out, can you think of just one more axiom we can add to $T$ such that any chain in our partial order must be infinite? (Hint: given an element $a$ in a partial order $M$, the set of all elements $b$ less than $a$ is $a$-definable.)


Permutations with Repetition (example) Assume a box that contains $20$ colored balls.

\begin{align}
&-Red: 2 \\
&-Blue: 1 \\
&-White: 4 \\
&-Orange: 3 \\
&-Black: 5 \\
&-Yellow: 2 \\
&-Green: 3
\end{align}
The possible ways to put all those $20$ in order are:
$$
N_{20}=\frac{20!}{2!1!4!3!5!2!3!}
$$
However, what if we want to pick only $5$ and put them in order? I tried picking ordered groups of $5$ out of $20$ but ended up double-counting. Any hints?

Strategy

Notice that $5$ can be partitioned in the following ways.
\begin{align*}
5 & = 5\\
  & = 4 + 1\\
  & = 3 + 2\\
  & = 3 + 1 + 1\\
  & = 2 + 2 + 1\\
  & = 2 + 1 + 1 + 1\\
  & = 1 + 1 + 1 + 1 + 1
\end{align*}
This allows us to consider cases.

Five balls of one color:  We must select the black balls, then fill all five positions with them.

Four balls of one color and one ball of a different color:  We have two choices for which color to use for the four balls of the same color.  Choose four of the five positions for that color.  Choose one of the other six colors to fill the remaining position.

Three balls of one color and two balls of a different color:  We can have four choices for which color to use for the three balls of the same color.  Choose three of the five positions for that color.  No matter which color we have used, we are left with five choices for the color to use for the two balls of the same color.  Those balls must fill the remaining two positions.

Three balls of one color and one ball each from two of the other colors:  We have four choices for which color to use for the three balls of the same color.  Choose three of the five positions for that color.  That leaves us with six colors from which to choose the remaining two colors.  Arrange those two distinct balls in the remaining two positions.

Two balls each of two colors and one ball from one of the other colors:  We have six choices for which colors to use for the two pairs of balls of the same color.  Choose two of the five positions for the selected color that appears first in an alphabetical list.  Choose of the the remaining three positions for the other selected color.  Choose one of the other five colors to fill the remaining position.

Two balls of one color and one ball each from three other colors:  We have six choices for which color to use for the pair.  Choose two of the five positions for that color.  Choose three of the remaining six colors.  Arrange those three distinct balls in the remaining three positions.

One ball from each of five colors:  Choose five of the seven colors.  Arrange the five distinct balls in the five positions.

Fine, here we have 20 balls they aren't identical (by means of color) and you want to pick only 5 from them, order them and find out in how many ways we can do that if yes . It is indeed a very big problem when you pick 5 from 20 we are also considering the probability of picking up balls (each of different colors). So we need to consider the probability of picking a red, blue, white, orange, black, yellow, green which will get quite complex and will split up eventually into many cases. We need to order them after this in order to find out answer for your question !!

[edit]: or as the comment suggests you can also use exponential generating functions to solve this in a much more easier way


Must probability density be continuous? From other materials that I've read, the probability density of a continuous random variable must itself be continuous. Is this correct? If it is, I don't understand why that would be so, why can't the probability change abruptly?

Michael Chernick asks for an example of a probability distribution with a density that is everywhere discontinuous.

As discussed in this question, there exists a measurable set $A \subset \mathbb{R}$ such that for every interval $I$, we have $0 < m(A \cap I) < m(I)$, and moreover $m(A) < \infty$.  Then $f(x) = \frac{1}{m(A)} 1_A(x)$ is a nonnegative measurable function with $\int_\mathbb{R} f(x)\,dx = 1$, so it can be taken as the density of a continuous probability distribution.  $f$ is nowhere continuous because every interval contains points of $A$ and $A^C$.  Moreover, any function $g$ with $f=g$ a.e. is also nowhere continuous.

No, need not be. However, the cumulative distribution function (CDF), is always continuous (mayn't be differentiable though) for a continuous random variable.
For discrete random variables, CDF is discontinuous.


Example: Continuous function having a limit along every arithmetic sequence but having no limit over the reals The following question seems to be very elementary and must be a folklore, but we are not able to find an answer.

Let $f: [0,\infty)\to \mathbb R$ be a continuous function such that for every $a>0$,
$$
\lim_{n\to+\infty} f(na) = 0,
$$
where $n$ is restricted to be integer.

Does this imply that
$$
\lim_{t\to+\infty} f(t) = 0,
$$
where $t$ is allowed to be real?

Yes. The important thing here is that $f$ is continuous so if limit on discrete variable is zero, limit on continuous variable is also zero.


Books on classical geometry I'm curious to whether you guys have any tips on book concerning classical euclidean geometry. I'd like somewhat of an advanced treatment, around the same level as Coxeter's "Geometry revisited". I'd however want something more comprehensive, therefore I'm asking you for tips!

Hopeful for answers ^^

For a careful and very comprehensive treatment:

Robin Hartshorne: Geometry: Euclid and Beyond; Springer, 2000.

Here is one at advanced high school level: Plain Plane Geometry by Amol Sasane: http://www.worldscientific.com/worldscibooks/10.1142/9907


Why definition of Out(G) does actually make sense? I would like to ask - why does $\operatorname{Out}(G)$ := ${\operatorname{Aut}(G) / \operatorname{Inn}(G)}$ make sense? As $\operatorname{Out}(G)$ is outer automorphism group, I would expect it to include outer automorphisms, but for nontrivial $\operatorname{Inn}(G)$ the quotient is a group of cosets which are sets of functions, not functions. Isomorphism would be totally intuitive for me but equality I do not understand.

Thank you in advance!

The set of outer automorphisms (taking an "outer automorphism" of a group to be an automorphism that is not inner) does not give a group. The easiest way to see this is to consider that the identity automorphism is an inner automorphism. But there can also be other cases where the composition of two outer automorphisms gives a nontrivial inner automorphism.

Thus it makes sense to consider equivalence classes of outer automorphisms.  Looking at $\mathrm{Aut}(G)/\mathrm{Inn}(G)$, you have 1: an actual group, and 2: every nonzero equivalence class consists entirely of outer automorphisms.

Since the definition is everything that survives modding out by Inn, the definition is precisely as those equivalence classes of automorphisms which are not inner.


Are half of all numbers odd? Plato puts the following words in Socrates' mouth in the Phaedo dialogue:

  I mean, for instance, the number three, and there are many other examples. Take the case of three; do you not think it may always be called by its own name and also be called odd, which is not the same as three? Yet the number three and the number five and half of numbers in general are so constituted, that each of them is odd though not identified with the idea of odd. And in the same way two and four and all the other series of numbers are even, each of them, though not identical with evenness. (104a-b)

The philosophical point is that there exists an Idea (or Form) called Odd and odd numbers are merely specific instances of Odd.  The are not, themselves, identical to the concept of Oddness.

But I bolded a throwaway phrase that prompts my question: Are half of all integers odd?

Plato likely did not consider one to be odd nor did he likely consider either zero or negative integers among his set of "numbers in general".  I don't see his proof (if he had or was aware of one), but I would imagine it be something along the lines that for every even number N there is an odd number N+1.  Therefore half of all numbers greater than 1 are odd.

I'm aware that zero is even, which makes me think there is one extra even number than odd numbers.  My thinking is that if half of all positive numbers are odd and half of all negative numbers are odd, than leaving out zero, half of all integers are odd.  But when you add in the only non-positive, non-negative number, which is even, you have an extra even number.  (The programmer in me wants to add the concepts of -0 and +0 for symmetry.  The rest of me thinks that's nuts!)

However, I think my imagined platonic proof still works as long as zero is definitely even and won't work if it's either odd or neither parity.

Are either of these proofs valid?

The basic problem with your argument is that the concept of half of an infinite set is not well-defined. It’s possible to pair up the even integers with the odd integers with none left over in either set, and if we were talking about finite sets, that would be a demonstration that each was half of the whole set of integers. However, it’s also possible to pair up the multiples of $100$, say, with all the rest of the integers with none left over in either set, and the multiples of $100$ are obviously only part of the set of even numbers. Clearly, then, this kind of pairing argument cannot lead to any very useful notion of half of the set of integers.

There is a notion of asymptotic density of a set of positive integers that does a pretty good job of capturing many people’s intuitive sense of what half (or any other fraction) of the set of positive integers should mean. Let $A\subseteq\mathbb{Z}^+$; for each $n\in\mathbb{Z}^+$, $$\frac{|A\cap\{1,2,\dots,n\}|}n$$ is the fraction of the first $n$ positive integers that are in the set $A$. If this fraction approaches a limit $\alpha$ as $n$ increases, i.e., if $$\lim_{n\to\infty}\frac{|A\cap\{1,2,\dots,n\}|}n=\alpha\;,$$ the set $A$ is said to have asymptotic density $\alpha$. (Note that the limit need not exist: not all sets of positive integers have asymptotic densities. It’s not hard to show that if $A$ is the set of even integers or the set of odd integers, its asymptotic density is $1/2$, so it’s meaningful to say that each of these sets comprises half of the positive integers in terms of asymptotic density. Similarly, for each positive integer $n$, the set of multiples of $n$ can easily be shown to have density $1/n$, exactly as one would wish.

Added: In the comments Srivatsan points out an excellent example: the perfect squares can be paired up one-for-one with the integers, but their asymptotic density is $0$: they get sparser and sparser as you get further away from $0$. Thus, in one sense there are just as many perfect squares as there are integers, and yet in another sense almost every positive integers is a non-square.

Your question brings up your confusion around the problem of "0" and suggests that this might lead one to think that there is one extra even number.

While this is erroneous and can be proven mathematically as others have above, there is also a simpler explanation which stems from a philosophical question: Why do you assume that the "center" of the natural numberline is "0"?

The concept of "center" is something that works with finite groups. But the numberline is infinite. (What is the midpoint of an infinite ordered set?)

If we had randomly placed the number "-1" as the "center" (ie. there are infinite numbers less than it and larger than it on either side), would that have led you to think that there is one extra odd number?

The point is to make you realize that, in reality, mathematicians' choice of "0" being at the center of every numberline we draw is just a tradition and is just an agreed upon matter. It has no "scientific" basis and should not affect the way you look at things

Now, from a mathematical perspective: $\lim_{x\to\infty}\frac{x}{x+1} =\lim_{x\to\infty}\frac{x}{x} =1$

So even if you have one extra value, it wont affect the overall result because we're dealing with non-finite sets!

So I guess the summary of what I'm trying to tell you is that your initial problem of how to deal with the extra "0" is not actually a problem after all :). You can safely (mathematically) say that half of all numbers are odd.


Measure theory question about sum of sequence of functions Let $\left(f_n\right)_{n\ge1}$ be a sequence of measurable real valued functions. Prove that there exist a sequence of constants $c_n$ $>0$ such that $\sum_{n=1}^{\infty} c_nf_n $ converges for almost every x $\in \mathbb R$.

Any hints is appreciated.
Thanks

This work when the involved measure space $(X,\mathcal A,\mu)$ is $\sigma$-finite. Let $\left(A_n\right)_{n\geqslant 1}$ be a non-decreasing sequence of measurable subsets of finite measure such that $\bigcup_{n\geqslant 1}A_n=X$. Notice that for all fixed $n$, the following convergence holds:
$$
\lim_{R\to +\infty}\mu\left(A_n\cap\left\{x\in X\mid \left\lvert f_n(x)\right\rvert\gt 2^{-n} R\right\}\right)=0
$$
hence using the definition of the limit with $\varepsilon=2^{-n}$, we see that we can choose $R_n\gt 0$ such that
$$
\mu\left(A_n\cap \left\{x\in X\mid \left\lvert f_n(x)\right\rvert\gt 2^{-n} R_n\right\}\right)\lt 2^{-n}.
$$
Choose $c_n=1/R_n$. Then for all $n$, the following inequality holds
$$
\mu\left(A_n\cap \left\{x\in X\mid c_n\left\lvert f_n(x)\right\rvert\gt 2^{-n} \right\}\right)\lt 2^{-n}.
$$
Fix a $N\geqslant 1$. Since $A_N\subset A_n$ for $n\geqslant N$, it follows that for such $n$'s,
$$
\mu\left(A_N\cap \left\{x\in X\mid c_n\left\lvert f_n(x)\right\rvert\gt 2^{-n} \right\}\right)\lt 2^{-n}.
$$
hence exploiting the convergence of the series $\sum_n \mu\left(A_N\cap \left\{x\in X\mid c_n\left\lvert f_n(x)\right\rvert\gt 2^{-n} \right\}\right) $, we derive that there exists $E_N\subset A_N$ of measure $0$ such that for all $x\in A_N\setminus E_N$, there exists $m(x)$ such that for all $n\geqslant m(x)$,  $c_n\left\lvert f_n(x)\right\rvert\leqslant 2^{-n}$. This proves that for all $x\in A_N\setminus E_N$, the series $\sum_n c_n\left\lvert f_n(x)\right\rvert$ converges. Finally, let $E:=\bigcup_{N\geqslant 1}E_N$. For all $x\in X\setminus E$, there exists a $N$ such that $x\in A_N\setminus E_N$ and $E$ has measure zero.

I think this is too general a statement.
If f(n)=n^m and c(n)=1/n^(m+2) , the sum will be =1/1^2+1/2^2+1/3^2+...=pi^2/6 .
If c(n)=1/n^(m+s) , then the sum will be the Zeta function Z(s)
--B.Sahu


Proving a quantified statement wrong The domain is the natural numbers set.
I want to prove this statement wrong, however, I'm not sure how should I go about that since, theoretically, I would have to go through all y values in order to find at least one that would be fit for every single x value. Can anyone give me a tip?
Here's the statement:
∃y∀x(2x - y = 0)

Asserting that$$\exists y\forall x(2x-y)=0$$is false is the same thing as asserting that$$\forall y\exists x(2x-y\ne0)$$is true. And it is indeed true: for any $y$, just take $x=y$ if $y\ne0$ and $x=1$ if $y=0$.


slowness of growth of polynomials How to show that for any $d\in\mathbb{N}$ and $\delta>0$ there exists $\varepsilon=\varepsilon(d,\delta)>0$ such that is $f$ is a polynomial of degree $d$, and
$$|f(t)|\leq 1,\quad t\in[0,1],$$ then $$|f(t)|<1+\delta,\quad t\in[0,1+\varepsilon].$$
 (This is an exercise from "Introduction to Ratner Theorems..." by D.W. Morris). It is immediate for $d=0,1,2$, but I am probably lacking some trick to approach the general case. Any help is greatly appreciated.

Claim: For every $d\in\mathbb{N}$, there exists $M_d>0$, such that for $f(x)=\sum_{i=0}^da_ix^i$, if $|f(t)|\le 1$ when $t\in[0,1]$, then $|a_i|\le M_d$, $i=0,\dots d$.

Proof of Claim:
Given $f(x)=\sum_{i=0}^da_ix^i$ and given $b_0,\dots,b_d\in[-1,1]$, consider the following system of linear equations with unknowns $a_0,\dots,a_d$:
$$f\big(\frac{i}{d}\big)=b_i,\quad i=0,\dots,d.\tag{1}$$
Note that the coefficient matrix of $(1)$ is a non-singular Vandermonde matrix whose entries are independent of $b_0,\dots,b_d\in[-1,1]$. It follows that $(1)$ has a unique solution $(a_0,a_1,\dots,a_d)$ and there exists  $M_d>0$, only dependent on $d$, such that $|a_i|\le M_d$, $i=0,\dots d$, which proves the claim. $\quad\square$

The statement follows from the claim immediately. If $f(x)=\sum_{i=0}^da_ix^i$ satisfies that $|f(t)|\le 1$ when $t\in[0,1]$, then $|a_i|\le M_d$, $i=0,\dots d$, and for every $\delta>0$, there exists $\varepsilon=\min(1,\frac{\delta}{2^dM_d})$, such that when $t\in[1,1+\epsilon]$,

$$|f(t)|\le |f(1)|+|f(t)-f(1)|\le 1+\sum_{i=0}^d|a_i|(t^i-1)\le M_d\varepsilon\sum_{i=1}^d 2^{i-1}\le 1+2^dM_d\varepsilon\le 1+\delta.$$

Edit: An alternative way(indirect but a little more general) to prove the claim is to use the fact that all norms on finite-dimensional vector spaces are equivalent. The claim follows from comparing the two norms of $\mathbb{R}^{d+1}$ below:
$$\|(a_0,\dots,a_d)\|:=\max_{0\le i\le d}|a_i|~,\quad \|(a_0,\dots,a_d)\|':=\max_{0\le t\le 1}|\sum_{i=0}^d a_it^i|.$$

Clearly $f $ is uniformly continuous on closed intervals, So given $ \delta >0 $ there exists $ \epsilon'(d,\delta)>0 $ such that if $ |y-t| \leq \epsilon' $ then for all such $y$ we have
$$ |f(y)-f(t)|< \delta $$
Take $ \epsilon = \min\{ \epsilon',1\} $, then for all $ t \in [1,1+\epsilon] $ we have $ 0 \leq 1-\epsilon \leq t-\epsilon \leq 1 $ and thus $ t-\epsilon \in [0,1] $. So we can choose $y \in [t-\epsilon,1]\subseteq [0,1] $ which clearly satisfies $ |y-t|\leq \epsilon $. Hence we have for all $ t \in [1,1+\epsilon] $
$$ |f(t)|\leq |f(y)|+|f(y)-f(t)| < 1 + \delta $$
For $ t\in [0,1] $ we trivially have $ |f(t)| \leq 1 < 1 + \delta $. So $ |f(t)| < 1+ \delta $ for all $ t \in [0,1+\epsilon ] $


Calculating probability in a Bayesian Network $$\require{enclose}\newcommand{\circletext}[1]{\enclose{circle}{\bbox[0.75ex]{\rm #1}}} \begin{array}{c} && \circletext{Influenza} &&&& \circletext{Smokes} \\ & \swarrow & \downarrow & \searrow &  &\swarrow \\ \circletext{Sore\, throat} && \circletext{Fever} & & \circletext{Brochitis}\\ &&& \swarrow & & \searrow \\ && \circletext{Coughing}&&&&\circletext{Wheezing}\end{array}$$

Suppose I have a network defined as above.
What would be the formula for calculating
$\mathsf P(\text{Wheezing})$?

Do you mean calculate it from the root probabilities and branch conditionals?

Letting $W$ stand for "Wheezing", $B$ for "Bronchitis", $I$ for "Influenza", $S$ for "Smoker", et cetera.

$$\def\P{\operatorname{\mathsf P}} \P(W) ~{=~ \sum\limits_{b\in\{B,B^\complement\}} \P(W\mid b)\P(b) \\ =~\sum\limits_{b\in\{B,B^\complement\}} \P(W\mid b)\Big(\sum\limits_{i\in\{I,I^\complement\}}\underline{\phantom{ \P(i)}}\big(\sum\limits_{s\in\{S,S^\complement\}}\underline{\phantom{\P(s)\P(b\mid s, i)}}\big)\Big) }$$

You have that:

$$P(I, SH, F, B, S, C, W) = P(I)P(SH|I)P(F|I)P(B|I,S)P(S)P(C|B)P(W|B)$$

Which will give you:

$$P(W) = \sum_{I,SH,F,B,S,C} P(I)P(SH|I)P(F|I)P(B|I,S)P(S)P(C|B)P(W|B)$$
$$P(W) = \sum_{I} P(I) \sum_{SH}P(SH|I) \sum_{F}P(F|I)\sum_{B} P(B|I,S) \sum_{S} P(S) \sum_{C} P(C|B)P(W|B)$$
$$P(W) = P(W|B) \sum_{C} P(C|B)$$
$$P(W) = P(W|B)$$


What is the correct quantifier order for 'almost everywhere' statements? For an expression like $P[X=Y]=1$ what is the correct way to unpack the expression into quantifier-like language?

Supposing X and Y are random variables with domain $\Omega$, is this correct?
For almost every $\omega\in\Omega$, $X(\omega)=Y(\omega)$.

Also, is it correct to think of expressions like 'almost every' as quantifiers?

We work on the space $(\Omega,\mathcal{F},P)$.

$P(X=Y)=1$ means $X$ and $Y$ are equal a.s. which can be written that way : $\exists A\in \mathcal{F},P(A)=0, \forall \omega \in A, X(\omega)\neq Y(\omega) $. (Here, just chose $A=(X\neq Y)$).


Find two numbers $a,b$ such that $a^a=b^b$ Playing around on Desmos with the equation $y=x^x$, I noticed that the function, for numbers $0<x<1$, has two $x$ values for every $y$ value. For instance, $\frac12^{\frac12}=\frac14^{\frac14}$, which we can rewrite as $\sqrt[2]\frac12=\sqrt[4]\frac14$.

What I'm trying to find is: Given some number $a$ such that $0<a<1$, find $b$ such that $\sqrt[a]\frac1a=\sqrt[b]\frac1b$.

Is there a general formula for this? I've tried using Newton's Method, but that just approximates the result; it doesn't give an exact answer.

I'm struggling with tagging this question properly, so I'd appreciate some help with that as well.

Let $t=a/b$ be a positive number other than $1$.  Then

$(bt)^{bt}=b^b$

$(bt)^t=b$

$bt=b^{1/t}$

$t=b^{(1-t)/t}$

$\color{blue}{b=t^{t/(1-t)}}$

From this we then get

$\color{blue}{a=bt=t^{1/(1-t)}}$

For instance, if $t=2$ then $b=2^{2/(1-2)}=1/4, a=2^{1/(1-2)}=1/2$.


Show that linear functional $L(f) = \int_0^1 f(x) dx$ is continuous Let $(C[0,1], d_1)$ be a metric space of all continuous functions $f:[0,1] \to \mathbb{R}$, $d_1$ is the $L_1$ metric
$$d_1(f,g) = \int\limits_0^1 |f(x) - g(x)| dx$$

  Show that linear functional $L(f) = \int\limits_0^1 f(x) dx$ is
  continuous

I'm not sure what would be the easiest way to prove this claim, I feel like I am on the right track

Proof Attempt:

Recall $L$ is continuous if for all convergent sequence $(f_n), f_n
   \to f$ as $n \to \infty \implies L(f_n) \to L(f)$ as $n \to \infty$

Let $(f_n)$ be a convergent sequence in $(C[0,1], d_1)$, we wish to
show that $L(f_n) \to L(f)$ as $n \to \infty$
$f_n \to f$ if $\forall \epsilon >0, \exists N \in \mathbb{N}$, such
that $\forall x \in [0,1], \forall n \geq N, d_1(f_n(x), f(x)) <
   \epsilon$

By definition, $d_1(f_n(x), f(x)) < \epsilon \implies \int\limits_0^1
   |f_n(x) - f(x)| dx < \epsilon$
If $L(f_n) \to L(f)$ as $n \to \infty$ then $|L(f_n) - L(f)| = |\int\limits_0^1 f_n(x) - f(x) dx| < \epsilon$
But $|\int\limits_0^1 f_n(x) - f(x) dx|  \leq \int\limits_0^1 |f_n(x) - f(x)| dx $ by triangle inequality, and $\int\limits_0^1 |f_n(x) - f(x)| dx  < \epsilon$
Therefore,  $L(f_n) \to L(f)$ as $n \to \infty$

Does this look correct?

Your proof is correct. An easier way to prove the claim is to notice that $d_1$ is the metric induced by the norm:

$$\|f\|_1 = \int_0^1 |f(t)|dt$$

Note that $L:(C([0,1]), d_1) \to \Bbb R$ is continuous iff $L: (C([0,1]), \| \cdot \|_1) \to \Bbb R$ is continuous.

Hence, to prove $L$ is continuous, it is sufficient to find a constant $C$ such that $|Lf| \le C \|f\|_1$ for all $f \in C([0,1])$. But this is obvious:

$$\left| L(f) \right| \le \int_0^1 |f(t)|dt = \|f\|_1$$

Let's apply the following result:

Theorem
For the linear operator $T: X \rightarrow X$ the following properties are equivalient:

(1): $T$ is bounded

(2): $T$ is continuous

(3): $T$ is continuous at zero

So in order to show that $L(f) = \int_{0}^{1}{f(x)dx}$ is a continuous linear operator $(C[0,1], d) \rightarrow (C(0,1), d)$ it is sufficient to check that it is bounded. Indeed, $$||L|| = \sup_{f: ||f|| \
\leq 1}{|L(f)|} = \sup_{f: ||f|| \leq 1}{\int_{0}^{1}{f(x)dx}} = \sup_{f: ||f|| \leq 1}{||f||}=1$$


An element $\alpha\in L/K$ is integral over A if its minimal polynomial over $K$ has coefficients in $A$. I suspect it is a very trivial question, it is marked as such in algebraic number theory books by Neukirch and Lee, and also in lectures I'm attending. But I can't wrap my head around it.
So, $L/K$ is an algebraic field extension, $A\subseteq K$ - an integrally closed integral domain, such that $Quot(A) = K$.
In the books, "iff" version of this statement is proved, but actually only the opposite direction is proved, and the one I'm asking about is stated as trivial.

If $f = a_n X^n + \dots + a_0$, $a_i\in A,$ is the minimal polynomial of $\alpha$, how do we get rid of $a_n$?

Edit: I forgot that minimal polynomial is defined to be monic. Can be closed.

One direction is trivial. I'll show the non-trivial one, that is, if $\alpha \in L$ is integral over $A$, then the minimal polynomial of $\alpha$ over $K$ has its coefficients in $A$.

Since $\alpha$ is integral over $A$, there is $p\in A[x]$, a  monic polynomial, for which $p(\alpha)=0$.

Of course $\alpha $ is algebraic over $K$, so let $q\in K[x]$ be the minimal polynomial of $\alpha$ over $K$. Then $q$ divides $p$ in $K[x]$. Hence the roots of $q$ are all integral over $A$ (because these roots are just some of the roots of $p\in A[x]$).

Write $$q(x)=x^n+a_{n-1}x^{n-1}+\cdots +a_0=\prod_i (x-\alpha_i) \in K[x].$$
But you know we can express the coefficients $a_i$ in terms of symmetric polyomials in the roots $\alpha_i$'s. Hence, it follows that each $a_i$ is integral over $A$. But as $A$ is integrally closed, it follows $a_i \in A$ for all $i$.

"$\alpha$ integral over $A$" : $\alpha$ is a root of some polynomial $\in A[x]_{monic}$ , equivalently $A[\alpha]$ is a finitely generated $A$-module, equivalently $\alpha \in End(M)$ where $M$ is a finitely generated free $A$-module (since then $\alpha$ is a root of the characteristic polynomial of the endomorphism $\in End(M)$ of the multiplication by $\alpha$).
If $A$ is integrally closed in $Frac(A)$ then this is the same as "$f\in K[x]_{monic}$, the minimal polynomial of $\alpha$, is in $A[x]$". One direction is obvious, the other is that $\alpha$ is a root of $g=\det(\alpha-xI)\in A[x]_{monic}$, since every root of $f$ is a root of $g$ we have $f^m=g\in A[x]$ which implies that the coefficients of $f$ are integral over $A$. Since $A$ is integrally closed in $K$ it means $f \in A[x]$.
Thus in that case it is easy to see that if $\alpha,\beta$ are integral then $A[\alpha,\beta]$ is a finitely generated $A$-module so all its elements are integral and  hence the integral closure of $A$ is a ring.


Inverse of a Fourier Transform is $L^1 \cap L^{\infty}$ I was reading a Navier-stokes paper by Kato, and he affirm that if $k\geq0$ than $$\mathcal{F}^{-1}(| \xi |^ke^{-|\xi|^2})\in L^1(\mathbb{R}^2)\cap L^\infty(\mathbb{R}^2)$$

(here $\mathcal{F}^{-1}$ denotes the inverse Fourier Transform).

Does anyone knows how to do it or where I can find something to help me with this?

Index.

Introduction.
A solution based on integration by parts.
A solution based on Sobolev embeddings (not entirely sure).
A failed naïve attempt.

0: Introduction.

Denoting $G(x):=e^{-|x|^2}$ and $f:=|\nabla|^k G$, the non-trivial part is to show that $f\in L^1(\mathbb R^d)$. Here
$$
|\nabla|^k h(x):= \mathcal F^{-1} |\xi|^k \hat{h}(\xi).$$

1: Integrating by parts.
Here we assume $d=1$. Since $f\in L^\infty$, if we can show that $f(x)=O(|x|^{-2})$ as $|x|\to \infty$ then we can conclude that $f\in L^1$. Now integrating by parts we have
$$\begin{split}
f(x)&=\int_{-\infty}^\infty |\xi|^ke^{-|\xi|^2} e^{ix\xi}\, d\xi \\ & = \frac{-1}{ix}\left[k\int_{-\infty}^\infty |\xi|^{k-2}\xi e^{-|\xi|^2}e^{i x \xi}\, d \xi - 2 \int_{-\infty}^\infty |\xi|^k\xi  e^{-|\xi|^2} e^{i x \xi}\, d\xi\right].\end{split}$$
We claim that both summands in the square bracket are $O(|x|^{-1})$. If $k>1$, this is immediately proven by redoing the integration by parts as before. However, if $0<k< 1$, the first summand poses a problem, because
$$\tag{1}\int_{-\infty}^\infty |\xi|^{k-2}\xi e^{-|\xi|^2}e^{i x \xi}\, d \xi = \frac{-(k-1)}{ix}\text{pv}\int_{-\infty}^\infty |\xi|^{k-2} e^{-|\xi|^2}e^{i x \xi}\, d\xi + O(|x|^{-1}),
$$
where we have a nonintegrable singularity in the integrand and that's why the integral has to be taken in principal value sense.

At first sight, it is unclear whether the integral is convergent. However we know that the Fourier transform of $|\xi|^{k-2}$ is, up to multiplicative constants, $|x|^{1-k}$ (see here). On the other hand, the Fourier transform of $e^{-|\xi|^2}$ is $e^{-|x|^2}$ up to irrelevant constants. So, rewriting the pv-integral in (1) as a convolution, we find (up to multiplicative constants)
$$
|x|^{1-k} \ast e^{-|x|^2} =\int_{-\infty}^\infty |x-y|^{1-k}e^{-|y|^2}\, dy, $$
and here it is clear that the integral on the right-hand side is convergent, thanks to the strong decay of $e^{-|y|^2}$. This proves our claim.

The case $k=1$ is still untreated. In this case, integration by parts produces a Dirac delta:
$$\int_{-\infty}^\infty |\xi|^{-1}\xi e^{-|\xi|^2}e^{i x \xi}\, d \xi = \frac{-2}{ix}\int_{\infty}^\infty \delta(\xi)e^{-|\xi|^2}e^{i x \xi}\, d\xi + O(|x|^{-1}), $$
and the integral on the right-hand side evaluates to $1$. The proof is complete.

2: Sobolev embedding.

This is a conditional answer, since it relies on the Sobolev embedding (!) (see below), and I am not entirely sure it is correct for a reason that I will explain below.

Now, if the following inequality is true:
$$\tag{!}
\| f\|_{L^1(\mathbb R^d)}\le C_d \| |\nabla|^{-d/2} f \|_{L^2(\mathbb R^d)}, $$
then we can argue that
$$
\|f\|_{L^1(\mathbb R^d)}^2 \le C_d^2\||\nabla|^{k-d/2} G\|_{L^2(\mathbb R^d)}^2 =C_d^2 \int_{\mathbb R^d} |\xi|^{2k-d}e^{-2|\xi|^2}\, d\xi, $$
and the right hand side is a convergent integral for all $k > 0$.

WHY AM I NOT SURE? The Sobolev embedding (!) is never stated with a negative power of $|\nabla|$ in the right-hand side. I should see if the proof extends to the case at hand.

3: Failed attempt. Here follows a naïve attempt of mine, which fails at the label (WRONG). I am not deleting it because I am convinced that we learn from errors at least as much as we learn from correct answers.

In comments, amsmath suggests that this follows from $|\xi|^k e^{-|\xi|^2}$ being a Schwartz function, which is not entirely correct because the function fails to be smooth at $\xi=0$ when $k$ is not an even integer. However, there is a simple workaround. Consider a smooth cutoff function $\zeta$ that equals $1$ in a neighborhood of $0$ and is compactly supported. Write
$$
f= \mathcal F^{-1}(\zeta|\cdot|^k e^{-|\cdot|^2}) + \mathcal F^{-1}((1-\zeta)|\cdot|^k e^{-|\cdot|^2})=f_0 + f_1.$$
Now $f_1$ is the inverse Fourier transform of a genuine Schwartz function and so it is itself a Schwartz function. On the other hand,
$$
\begin{split}
\lVert f_0\rVert_1 &= \int_{\mathbb R^d} \left\lvert \int_{\mathbb R^d} \zeta(\xi)|\xi|^k e^{-|\xi|^2}e^{i x\xi}\, d\xi\right\rvert\,dx \\
&\le R^k \int_{\mathbb R^d} \left\lvert \int_{\mathbb R^d} \zeta(\xi) e^{-|\xi|^2}e^{i x\xi}\, d\xi\right\rvert\,dx \ \text{(WRONG)}\\
&= R^k\lVert \mathcal F^{-1} (\zeta e^{-|\cdot|^2)}\rVert_1
\end{split}
$$
where $R$ is a constant that depends only on the support of $\zeta$. Since $\mathcal F^{-1} (\zeta e^{-|\cdot|^2})$ is again a genuine Schwartz function, its $L^1$ norm is finite and the proof is complete.

If what @amsmath said on the comments is true,i.e. $|\xi|^ke^{|\xi|^2}\in \mathcal{S}$ (here $\mathcal{S}$ denotes the Schwarz space). Then:

Note that $\mathcal{F}$  is an isomorphism of $\mathcal{S}$ into itself (you can see this result in corollary 8.28 from Folland's Real Analysis). Hence $\mathcal{F}^{-1}(|\xi|^ke^{-|\xi|^2})\in\mathcal{S}$.

Using the fact that  $\mathcal{S}\subset L^1\cap L^\infty$, you conclude the result.


Finding the actual area represented on a map This question is similar to finding the actual distance represented on a map. The question is:

  Find the actual area, in m², represented by 1cm² on the map. The scale of map is 1: 180,000

A 1 cm by 1 cm square on the map represents a 180,000 cm by 180,000 cm region of land. That is, a 1,800 m by 1,800 m region of land. Its area is 3,240,000 square meters (1,800 times 1,800 is 3,240,000).

Calculate an actual area in Km2 represented by 40 cm2 on a map, the scale of the map given as 1:50,000.
Solution:
As 1 cm in the map represents 50,000 cm on ground
therefore    1 cm x 1 cm = 50,000cm x 50,000 cm
          1 cm2 (on map) =2,500,000,000  cm2 (on ground)
                         = 2,500,000,000/(100,000x100,000)
         1 cm2 (on map)  = 0.25 km2    (on ground)
        40 cm2 (on map)  =40 x 0.25 km2 = 10 km2   (on ground)


What structure is this? I had hoped that this would be easier.

I have power series that I'm using.  I want to know what simple structure describes the operations that I'm performing on them.  It would be better if I could find something that non-mathematicians (actually computer scientists) would understand, but I'd like to know the technical terminology too.

First, let me define three power series, $f(x)$, $g(x)$, and $h(x)$:

$f(x) = \sum_{i=m_1}^{n_1}{c_i x^i}$

$g(x) = \sum_{i=m_1}^{n_1}{d_i x^i}$

$h(y) = \sum_{i=m_2}^{n_2}{e_i y^i}$

Then the two operations that I perform are as follows:

(1) Subtraction $r(x) = f(x) - g(x)$

$r(x) = \sum_{i=m_1}^{n_1}{c_i - d_i}$

(2) Multiplication $r(x,y) = f(x)h(y)$

$r(x,y) = \sum_{i=m_1}^{n_1}{\sum_{j=m_2}^{n_2}{c_i \cdot e_j}}$

All of the coefficients and variables are integers.

I'm writing a paper, and I'm searching for a more "definite" description than what I have.  It is somewhat ackward to describe, and I want to convey this information exactly and concisely.  Plus, I don't want the reader to have any doubts.

I've been considering that I should just write out the math, since it seems to leave less doubt.


Proving convergence by proving a sequence does not diverge Is it possible to prove convergence by proving that a sequence does not diverge?

Especially I don't know how to deal with periodic sequences such as $\lim \limits_{n \to \infty } \sin(n)$.

By definition any sequence that does not converge, diverges. Thus, by contraposition, if a sequence does not diverge, it must converge. So technically, yes, you could do that. I am not sure how useful this would actually be.

In this particular case your limit diverges, so it does not really apply here.

your limit diverges? i think it must be the given sequence, the limit doesn't exist


Can the archimedian postulate prove the limit of the series $1/n$ is $1$? So I know that the sequence $a_n = 1/n$ will converge to $0$ as $n$ approaches infinity.
By the definition of a limit this means that for all $\epsilon>0$ there exists $N$ in the naturals such that for all $n\geq N$, $|1/n - 0|<\epsilon$ and this can be proved using the Archimedian postulate, since a $N$ can be found s.t $N>1/\epsilon$ this is all well and good and makes sense.

But the problem comes up if I were to guess the limit wrong and say tried to prove that $1/n\to 1$ as $n$ approached infinity. This would mean showing $|1/n - 1|<\epsilon$ which can also be done using the Archimedian postulate as an $N$ can be found s.t $N>1/(\epsilon+1)$. I don't understand how they can both be proved as the second is clearly untrue.

This is the sequence, not the series. (These are two different things).
Let's do as you suggest: fix any $\varepsilon > 0$. Indeed, as you point out there exists an integer $N$ such that, for all integers $n\geq N$, $$
n \geq N > \frac{1}{\varepsilon+1}\,.
$$
This is true. But what does that give? It is equivalent to
$$
\forall n \geq N, \qquad \frac{1}{n} < \varepsilon+1
$$
itself equivalent to
$$
\forall n \geq N, \qquad \frac{1}{n} - 1  < \varepsilon
$$
This is true, but is not the same as
$$
\forall n \geq N, \qquad \left\lvert \frac{1}{n} - 1\right\rvert  < \varepsilon
$$
since $\frac{1}{n} - 1$ can (and will) be negative...

In short: you forgot to worry about the absolute values.

You appear to have used the supposed identity $|1/n-1|<\epsilon\iff|1/n|<\epsilon+1$

But this is not a correct identity because the absolute value of $1/n-1$ measures that $1/n$ moves away from $1$ as $n$ increases, despite $1/n-1$ itself being negative for all $n>1$.

Therefore pulling the $-1$ out of the absolute value function exchanges the sign of the part within, so for the identity to hold you would need to exchange the signs elsewhere.


Solving a recurrence relation with special cases I need help solving the following problem with a recurrence relation.

  A miner is trapped in a mine with three doors. The first door will lead him to safety
  in two hours. The second door leads him back to the same place in three hours. The third door
  leads him to a maze which takes him back to the same place in 5 hours. However, if the miner
  ever travels through the third door again, it takes him only 1 hour to go back to the same place
  as he remembers how to get out of the maze. The three doors look the same and so the miner
  always chooses each of them with equal probability, independent of which door he has taken
  before. What is the expected time until the miner reaches safety?

I am trying to solve it by using Markov chain with 2 states, one which indicates the miner has gone through door 3 already and the other that he has not but I'm somewhat confused. Any help will be appreciated.

P.S. This is not homework but practice for an upcoming midterm.

The state space is on the right track but there are 3 states to consider.

$$S_0\text{ - the initial state}$$
$$S_1\text{ - the original room after passing through the maze}$$
$$S_2\text{ - the exit}$$

With these the transition probabilities and times are straightforward:

$$\begin{align}
P_{00}&=\frac{1}{3}&t=3\\
P_{01}&=\frac{1}{3}&t=2\\
P_{02}&=\frac{1}{3}&t=5\\
P_{10}&=0&t=N/A\\
P_{11}&=\frac{1}{3}&t=2\\
P_{12}&=\frac{2}{3}&t=1\\
P_{20}&=0&t=N/A\\
P_{21}&=0&t=N/A\\
P_{22}&=1&t=0\\
\end{align}$$


How to solve for the expectation of the Ito Integral: $\int_0^4 B_t^2 dB_t$? I would like to find the expectation of the Ito Integral: $\int_0^4 B_t^2 dB_t$. My strategy is to use Ito's general formula with:

$$
f(t, B_t) = f(0,0) + \int_0^t \frac{df}{dx}(s, B_s) dB_s + \int_0^t \frac{df}{dt}(s, B_s) ds + \frac{1}{2}\int_0^t \frac{d^2f}{dx^2}(s, B_s) ds
$$

Then, my candidate function is $f(t, B_t) = \frac{B_t^3}{3}$. However, the problem I am having is in the derivative $\frac{df}{dt}(s, B_s)$. Since I want to have my integration go up to $4$, then I necessarily have:

$$
\int_0^4 \frac{df}{d4}(s, B_s) ds = \int_0^4 \frac{\frac{B_t^3}{3}}{d4} ds
$$

I know that there is probably a bad/stupid mistake here, and that I should just take the derivative with respect to $s$, but I cannot figure out why exactly we are having the $dt$ MATCH the upper limit of the integral, $t$. Can anyone shed some light here? Thanks!

Let $f(x) = \frac{x^3}{3}$ and for reference, note that $f'(x) = x^2$ and $f''(x) = 2x$. It\^o's formula gives us that \begin{eqnarray*}
f(B_t) - f(0) &=& \int_0^t f'(B_s) dB_s + \frac{1}{2} \int_0^t f''(B_s) ds.
\end{eqnarray*} Therefore, we see that \begin{eqnarray*}
\frac{1}{3} B_t^3 &=& \int_0^t B_s^2 dB_s + \int_0^t B_s ds \\
\therefore \ \mathbb{E} \left( \frac{1}{3} B_t^3 \right) &=& \mathbb{E} \left( \int_0^t B_s^2 dB_s + \int_0^t B_s ds \right) \\
\therefore \ \frac{1}{3} \mathbb{E} (B_t^3) &=& \mathbb{E} \left( \int_0^t B_s^2 dB_s \right) + \mathbb{E} \left( \int_0^t B_s ds \right) \ \ \ \text{(linearity of expectation)}, \\
0 &=& \mathbb{E} \left( \int_0^t B_s^2 dB_s \right) + \mathbb{E} \left( \int_0^t B_s ds \right) \\
\therefore \ \mathbb{E} \left( \int_0^t B_s^2 dB_s \right) &=& - \mathbb{E} \left( \int_0^t B_s ds \right) =  - \int_0^t \mathbb{E}(B_s) ds \ \ \ \text{(Fubini)} \\
&=& 0,
\end{eqnarray*} where the last line follows from the fact that the Brownian motion is normally distributed with mean zero.


What is the probability that less than four of the balls are numbered 2 I'm a high school student and I'm struggling with one problem we've been given. Since we are allowed to use calculators, TI84's I am kinda confused If I can use in this problem binomcdf or is there any other way how can I calculate it?

Six balls numbered 1,2,2,3,3,3 are placed in a bag. Balls are taken one at a time from the bag at random and the number noted. Throughout the question a ball is always replaced before the next ball is taken.

-- The question is: Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.

the probability is (the number of events that you want/the number of all events):
the number of all events is 6^10
events that you want:
1)3 balls are numbered 2:
-Suppose that the first 3 ball are numbered 2 and The another 7 ball would not be numbered 2, like this: 2,2,2,X,X,X,X,X,X,X. For each X there are 4 possibilities, then in this case there would have $4^{7}$ possibilities. There are $C_{10,3}$
ways to choose the positions of balls numbered 2, each of way with $4^{7}$ possibilities, so the total of cases with 3 ball  numbered 2 is $4^{7}*C_{10,3}$

2)2 balls are numbered 2:
In a similar way: $4^{8}*C_{10,2}$

1)1 ball are numbered 2:
In a similar way: $4^{9}*C_{10,1}$

4)0 ball are numbered 2:
$4^{10}$

the result is $\frac{4^{7}*C_{10,3}+4^{8}*C_{10,2}+4^{9}*C_{10,1}+4^{10}}{6^{10}}$


Prove distributive law of sets Let $A,B,C$ be sets. Prove the distributive law

$$A \cap\left(B\cup C\right) = \left(A\cap B\right)\cup \left(A\cap C\right)$$

proof

First we'll show that $A \cap\left(B\cup C\right) \subset \left(A\cap B\right)\cup \left(A\cap C\right)$, and then the converse.

If $x$ is in $A \cap\left(B\cup C\right)$, then $x$ must be in $A$ and $x$ must be in $B$ or $C$. An element $x$ can satisfy this membership by being in either $A$ and $B$, or $A$ and $C$. In symbols,

$$x \in \left[\left(A\cap B\right)\cup \left(A\cap C\right)\right]$$

I'm not going to write the other direction. Because I know once I have this way down I can go the other way. What I'm looking for is -- in addition to correctness -- whether my argument is clunky or ambiguous. This "or" and "and" wording throws me for a loop when I'm trying to use it as a math operation and a word. Ideally, I'd like to know if there's a simpler way to write these types of proofs.

Your proof of the one direction looks perfectly fine. It is very common to use "and" and "or" written in a meta-level proof. However, if you feel this getting into the way of displaying an argument, you may benefit by some more formalism, e.g.

\begin{align}x\in A\cap(B\cup C) &\Rightarrow x\in A\land x\in (B\cup C)\\ &\Rightarrow x\in A\land (x\in B\lor x\in C)\\&\Rightarrow (x\in A\land x\in B)\lor (x\in A\land x\in C)\\&\Rightarrow x\in A\cap B\lor x\in A\cap C\\ &\Rightarrow x\in(A\cap B)\cup (A\cap C)\end{align}

I'm using $\land,\lor$ here as symbols of the "and","or" respectively.

I this specific case, the proof relies on using the distributive property of $\land,\lor$ as logical connectives to prove the corresponding property of unions and intersections via set membership. (this connection of course comes as no surprise through the common connection to boolean algebra(s)).

So, having this translation of very similar connectives/operations into one another as the essence of a proof can seems a little ambiguous, although it is the heart of the argument.

In general, I think the key to a clear exposition of course always depends on the context, but it is almost always a good mix of formatting, formalism and non-formalism.

E.g. no one wants to write precise first-order set theory statements for every mathematical concept, although it would be formally strict. In cases like the above, or in general with chains of implications or bi-implications, a structured formatting of the steps, like the aligned presentation above, can already improve the exposition a lot.

EDIT: Note, that you can turn every implication in the above chain into a bi-implication, i.e. by the argument above, you may prove directly that $x\in A\cap (B\cup C)\Leftrightarrow x\in(A\cap B)\cup (A\cap C)$. Thus, by extensionality of sets, you straightforwardly have $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$.

EDIT2: The distributive property of the logical connectives $\land,\lor$ may be verified by corresponding truth tables. It is important to note, that $\cap,\cup$ are defined operations in the theory of sets while the underlying logic(where you proceed with your reasoning with $\land,\lor$) is the first-order logic(of set theory). The distributive property of the logical connectives is a theorem of first-order logic which can then be used in your proof to apply it to propositions about the set-membership relation.

The reasoning is less circular as it is referential. In the end, you derive a property of your defined connectives by using a similar property of the connectives used inside their formal definition.

When proving "$P$ or $Q$" you can instead prove "If not $P$, then $Q$."

I cannot think of a way at the moment to simplify "$P$ and $Q$".


Finding a good difficult example function to minimize I am comparing some code for non-linear function minimization in multiple variables, like quasi-Newton methods etc. I am looking for a nice function to use as a test case.

So far I have been using $f(\bf x) = \frac{1}{2} {\bf x}^T A {\bf x}$ where $\bf A$ is a random symmetric positive definite matrix. This is nice because I can easily compute the gradient and Hessian and condition number. However it is too easy.

I'm looking for a function of two or more variables where one can analytically compute the gradient and Hessian. It does not need to be convex, so long as I can start the minimization with an initial point so that simple gradient descent will find a well known minimum. Ideally I'd like the function to have a narrow valley that is curved and not a straight shot from the starting point to the minimum, so I can test how well algorithms deal with following such narrow valleys without excessive zig zagging function evaluations.

Peter Spellucci's test cases for DONLP2. The documents on that page provide both text examples and the results from DONLP2, so you can check your answers against his. Many of his examples are constrained optimization problems, but quite are few are not, so they should be usable.

One interesting example is the Rosenbrock "banana" function
$$
f(x,y) = (1 - x)^2 + 100 (y - x^2)^2
$$
Oh -- I see that this function was already mentioned in the comment above. I'll leave it here in my answer, anyway, because I don't think the site's search function looks at comments.

You can compare your results with Mathematica here.

Here, a list of functions, given the shape, you can see, it's difficult to optimize.....

Griewank Function: http://i.stack.imgur.com/8dLq0.png

Kursawe Function: http://i.stack.imgur.com/qvrBD.png


How to solve a Second order and Second Degree Differential Equation? I have came across a second order as well as second degree differential equation in study of cosmic inflation. The full equations is quite messy so let me give a prototype:
        $$A(x) y''(x)^2 + B(x) y(x) + C(x) = 0 $$ $$y'(0) = a$$ $$y(0) = b$$
What I usually do in this type of cases is to ask Mathematica to solve it. And she usually solve it. But, sometimes mathematica show some error, for instance "singularity or stiffness at x=d". Now, I am trying to solve them numerically in some basic language ( read python ). But, the way we solve 2nd order differential equation is not applicable here, i.e., writing it as two first order differential equations.

Can any one provide me with an algorithm for solving the problem numerically and/or direct me to the literature where I can learn more about the methods.

Thanks in advance.

N.B. I have asked this question in stackoverflow, now asking here as suggested in a helpful comment.

But, the way we solve 2nd order differential equation is not applicable here, i.e., writing it as two first order differential equations.

Why not?  Let $z = y'$;  then we have
$$
y'(x) = z(x) \\
z'(x) = \pm \sqrt{- \frac{B(x) y(x) + C(x)}{A(x)}} \\
z(0) = a \\
y(0) = b
$$
This will provide you with a local solution for $y(x)$ so long as $y''(x)$ does not cross zero (i.e., $z'(x)$ never switches sign.)  In fact, the solution in a neighborhood of 0 will not be unique (if it exists—the quantity under the square root could be negative);  there will be two solutions, one with $z'(0) > 0$ and one with $z'(0) < 0$.

I would try using the Laplace Transform, isolate the function and then apply the inverse transform.


Number of positive integer solution, n=2x+y n is arbitnary positive integer.
then Cosider this equation   n=2x+y.
Where x and y are positive integer.
Finally, if gcd (x,y)=1, what's numer of pair of (x,y)?

Without the condition $\gcd(x,y)=1$ then the number of solutions is $\lfloor\frac{n-1}2\rfloor$, because $x$ can only range from $1 \text{ to }\lfloor\frac{n-1}2\rfloor$. One is the low value because $x$ must be positive. The high value of $x$ comes from $y$'s minimum values: $y=1=n-2\lfloor\frac{n-1}2\rfloor$ for odd $n$ and $y=2=n-2\lfloor\frac{n-1}2\rfloor$ for even $n$.

Remembering that $\gcd(x,y)=1$ is a condition: consider the case where $n$ is prime. Let $g=gcd(x,y), u=\frac{x}g, v=\frac{y}g$, then $n=2gu+gv$. Because $n$ is prime $g=1$. Therefore $\lfloor\frac{n-1}2\rfloor$ is the number of solutions when $n$ is prime.

If $n$ is composite the number of solutions to discard because $g>1$ must be accounted for.

Consider $n$ even, then: $1 \leq x \leq \frac{n}2-1$. Each even $x$ in this range must be discarded. Similarly if $3|n$ then each $x$ divisible by 3 must be discarded. This is true for all prime factors of $n$ with the exception of $\frac{n}2$ being prime, because $n\over 2$ isn't in the range.

Let $P_n = \{\text{distinct prime factors of n not including }\frac{n}2\}$ and $m=\frac{n}2-1$ then the solution for even n is: $$m - \sum_{i \in P_n} \lfloor\frac{m}i\rfloor$$

For odd $n$, with $P_n = \{\text{distinct prime factors of n}\}$ and $m=\frac{n-1}2$:
$$m - \sum_{i \in P_n} \lfloor\frac{m}i\rfloor$$.

For an arbitrary natural number x, there are infinitely many values for y which satisfy gcd(x,y)=1. This can be shown by the fact that there are infinitely many prime numbers, each of which, by definition, can only have a gcd of 1 or itself with another number, and the latter case only occurs when the number is a multiple of that prime; therefore, every prime which x is not a multiple of will have a gcd of 1 with x, and since the prime factorization for a given natural number must be finite, there will be an infinite set of primes for which this is true. It follows that there are infinitely many pairs (x,y)  that satisfy these conditions.


Balancing Coursework with original research I am a first-year graduate student in a US university pursuing a PhD in mathematics.

I am a bit frustrated in trying to balance coursework with original research. I saw students who spend most of their time with classes, but only produce one or two mediocre papers by the end of their PhD careers (I am not saying that not publishing a lot is a bad thing, as that depends on the research group in which the student is involved).

In my opinion, one should learn by reading papers rather than taking courses at the PhD level. If one wants to study spectral theory, for example, one should read the classic papers by Weyl alongside modern treatments, and should talk with his advisor (hopefully he or she is at an institution with strong research group) with a view towards trying to contribute something new to the field. One should also try to keep up with the new contributions to the field (sometimes a graduate program is blessed the resources that enable it to bring people from the US and/or world to give talks on active topics).

Unfortunately, with teaching 4 times a week and taking 3-4 math courses with long homework assignments, the idealized scenario that I just described is very hard to follow. My life has turned into finishing homework assignments, and I am not really convinced that spending my time in this way is going to help me. Perhaps, there is a reason why some graduate math programs (such as Princeton, for example) do not require the students to take courses and get grades.

In doing research, if one encounters some new concepts he or she is not familiar with (perhaps, due to not taking enough courses), then he or she should develop the skills necessary to efficiently learn that field/ concept to move forward with his or her research; this can come through with having conversations with fellow graduate students well versed in the field or self-study. Unfortunately, a lot of the graduate programs do not stress this enough in the early stages (and one cannot always a read a book or paper(s) front-to-back); however, Harvard's math department does require the student to write a minor thesis ("The minor thesis is complementary to the qualifying exam. In the course of mathematical research, the student will inevitably encounter areas in which s/he is ignorant. The minor thesis is an exercise in confronting gaps of knowledge and learning what is necessary efficiently."), which is crucial.

Has anyone else felt this frustration? Obviously, complaining about this is not going to do anything, and I want to find some way to cope, work around or come to terms with this frustration.

You are in a mediocre program to produce mediocre students; don't let it bring you down...learn to play the game even if it is boring so that you do not sacrifice original research in the long run. I am sorry you have to face this bureaucratic BS... if you are PHD student in mathematics, ideally you will be learning by yourself and building relationships with good researchers in your area of research (hopefully in your institution).

Taking and learning from generic courses like abstract algebra, complex analysis, etc. will not do you much good at any stage of your graduate career (unless you were underprepared as an undergrad). From the very beginning, you ought to see live, breathing mathematics with a beating heart. The rest- if you intensely dedicate yourself to yoru research- will fall into place. If you only want to work 9-5 with small-talk chitchat with your fellow cubicle mates who lack passion...then just go the mediocre route and take a lot of courses, do homework all day and just get a feel for the mathematical literature...

And you will have a a strong publication record by the time you graduate, whereas most graduate students from mediocre (most are...) institutions will maybe write one or two with their name slapped onto the paper with their adivsor.


Using Squeeze Theorem, find limit $(5^{n} + 6^{n})^{\frac{1}{n}}$. Find limit of $(a_{n}) = (5^{n} + 6^{n})^{\frac{1}{n}}$ given $b^{\frac{1}{n}} \to 1, b \in R^{+}$

Is it too cheeky to say let $b = (5^{n} + 6^{n})$ and say by definition $(5^{n} + 6^{n})^{\frac{1}{n}} \to 1$?

As pointed out in the comments, we have $6^n\leq5^n+6^n\leq2\cdot6^n$ so the desired limit is inbetween $\lim_{n\to\infty}\left(6^n\right)^{1/n}=6$ and $\lim_{n\to\infty}\left(2\cdot6^n\right)^{1/n}=6\cdot\lim_{n\to\infty}2^{1/n}=6\cdot1=6$. Both limits equal $6$ so the final answer is $6$ as well.

$$(5^n+6^n)^{1/n}=6\left(1+\left(\frac56\right)^n\right)^{1/n}.$$

Now as

$$1<1+\left(\frac56\right)^n<2,$$

$$1<\left(1+\left(\frac56\right)^n\right)^{1/n}<2^{1/n},$$
you can use the hint given.


A non trivial system verifying some statements For every abelian group $(V,\oplus)$  and an unitary field $\left( {{\Bbb F},{\text{ }} + ,{\text{  }}{\text{.}}} \right)$ acting on $(V,\oplus)$ by the external operation $\odot$ .
Question  : Is there a non trivial system such that  :

$\forall x, y \in V, \  \ $  $\forall \alpha \in \mathbb{F}, \  \ $ $\alpha \odot \left( x \oplus y \right) = \left(\alpha \odot x \right)\oplus\left( \alpha \odot y\right)$ .

$\forall x \in V, \  \ $  $\forall \alpha, \beta \in \mathbb{F}, \  \ $ $\left( \alpha + \beta \right) \odot x =  \left(\alpha \odot x \right)\oplus\left( \beta \odot x\right)$ .

$\forall x \in V, \  \ $  $\forall \alpha, \beta \in \mathbb{F}, \  \ \alpha \odot \left( \beta \odot x \right) = \left( \alpha\cdot \beta \right) \odot x$ .

$\exists u \in V , \  \  u \ne {0_V} \Rightarrow {{\text{1}}_{\Bbb F}} \odot {\text{ }}u  \ne u $ .
And if yes, is there a general theorem of a such system?

Here's a recipe for a more general class of counterexamples. Start with a vector space $V$ over $\Bbb{F}$ (with operations $+, \cdot$). Fix a linear projection operator $\phi$ on $V$ (i.e. $\phi^2 = \phi$). Define a new scalar multiplication
$$\alpha \odot v = \alpha \cdot \phi(v).$$
Then $\odot$ satisfies all the axioms except possibly the final axiom.

Note that this counterexample covers both of the counterexamples presented, plus others.

The first four axioms can be summarized to a single axiom:

  $(V,+)$ is an abelian group

The next three can be summarized as:

  $\Bbb F$, viewed as non-unitary ring, acts on $(V,+)$.

With this, let $U=\{\,1\cdot v\mid v\in V\,\}$.
Then $1\cdot u=u$ for all $u\in U$ and in fact $U=\{\,v\in V\mid 1\cdot v=v\,\}$. One wuickly check that $U$ is in fact a subgroup of $V$ and closed under scalar multiplication, hence is a vector  space.

Now let $W=\{\,v-1\cdot v\mid v\in V\,\}$. Then for $w\in W$,
$1\cdot w=1\cdot v-1\cdot 1\cdot v=0 $ and in fact $W=\{\,v\in V\mid 1\cdot w=0\,\}$.
One quickly checks that $W$ is in fact a subgroup of $V$.

Finally one sees that $V=U\oplus W$ as abelian groups. Thus the most general counterexample is of the form $V=U\oplus W$ where $U$ is a vector space and $W$ an abelian group (non.trivial, to make it actually a counterexample) and
$\alpha\cdot (u,w):=(\alpha u,0)$.


How to construct the subring generated by a set, T? I'm trying to find a constructive way of describing the subring generated by some subset, T, of a ring R.  I think I could describe it as all finite sums of finite products of elements of T, but I have no idea how to write that as a set or how to prove that it is equal to the subring generated by T.

Let $S$ be a subset of a commutative ring $R$. By the universal property of polynomial rings, there exists a (unique) morphism of $\mathbb{Z}-$algebras $\phi:\mathbb{Z}[x_{s}: s\in S]\rightarrow R$ such that $\phi(x_{s})=s$ for all $s\in S$. Then $\mathbb{Z}[S]$, the image of this morphism, is the subring of $R$ which is generated by $S$. In fact, it is easy to see that $\mathbb{Z}[S]$ is the intersection of all subrings of $R$ that are contain $S$. One can observe that $a\in\mathbb{Z}[S]$ if and only if $a=f(s_{1},...,s_{n})$ where $f(x_{s_{1}},...,x_{s_{n}})\in\mathbb{Z}[x_{s}: s\in S]$ and $s_{i}\in S$ for all $i$.

Your guess is correct: the subring of $R$ generated by $T \subseteq R$ is the set $S$ of all finite sums of finite products of elements of $T$, i.e. $S = \{ \sum_{\text{finite}} \left( \prod_{\text{finite}} t_i \right) \mid t_i \in T \}$. To see this, check that $S$ is a subring (closure under multiplication follows from distributivity, the empty product is $1$ if $R$ has a $1$, and the empty sum is $0$), and also that any other subring $S'$ of $R$ containing $T$ also contains $S$.

If $R$ is commutative with unity, another way to think about $S$ is by giving it a presentation as a $\mathbb{Z}$-algebra. If $x_t$ is a set of indeterminates, one for each element of $T$, then $\mathbb{Z}[T] := \mathbb{Z}[x_t \mid t \in T]$ is the free commutative ring on the set $T$. There is a universal map $\mathbb{Z}[T] \to R$ sending $x_t \mapsto t$, and if $I$ is the kernel, then $\mathbb{Z}[T]/I$ is isomorphic to the subring $S$ defined above. In this view, the subring $S$ is often written as $\mathbb{Z}[t \mid t \in T]$

If $R$ is not commutative but still has a $1$, then one would need to take the free noncommutative algebra over $\mathbb{Z}$ instead. Finally, although the universal construction may seem rather abstract, note that the kernel $I$ above can be easily described as the set of relations, under operations of $R$, satisfied by elements of $t$.


Is saying each pt of a topo. manifold has nbhd homeomorphic to R^n the same thing as saying there is a local coordinate system at each point? Is saying each point of a topological manifold has a neighborhood homeomorphic to $\Bbb{R}^{n}$ the same thing as saying there is a local coordinate system at each point?

I'm not really sure what I mean by coordinate system. It's just a term I have heard used. If someone could define that at the same time that would be awesome.

Yes. This is true by definition:

A topological space $M$ is said to be locally Euclidean of dimension $n$ if each point $x \in M$ has a neighbourhood $U$ homeomorphic to an open subset $V$ of $\mathbb{R}^n$. A homeomorphism of this sort is by definition a coordinate chart.

generally when imposing coordinate systems there is the additional condition that they are compatible. Thus if two neighbourhood's intersect then the intersection the associated two homeomorphisms yield a map between two subsets of $R^n.$ This map must also be a homeomorphism for a coordinate system.


Prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$ How can I prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$?

This is an exercise in my textbook associated with orthogonal projections and Gram-Schmidt process, but I am unsure how they are relevant.

Let $\mathbf{x} \in N(A)$ where $N(A)$ is the null space of $A$.

So, $$\begin{align} A\mathbf{x} &=\mathbf{0} \\\implies A^TA\mathbf{x} &=\mathbf{0}  \\\implies \mathbf{x} &\in N(A^TA) \end{align}$$ Hence $N(A) \subseteq N(A^TA)$.

Again let $\mathbf{x} \in N(A^TA)$

So, $$\begin{align} A^TA\mathbf{x} &=\mathbf{0} \\\implies \mathbf{x}^TA^TA\mathbf{x} &=\mathbf{0} \\\implies (A\mathbf{x})^T(A\mathbf{x})&=\mathbf{0} \\\implies A\mathbf{x}&=\mathbf{0}\\\implies \mathbf{x} &\in N(A) \end{align}$$ Hence $N(A^TA) \subseteq N(A)$.

Therefore $$\begin{align} N(A^TA) &= N(A)\\ \implies \dim(N(A^TA)) &= \dim(N(A))\\ \implies \text{rank}(A^TA) &= \text{rank}(A)\end{align}$$

Since elementary operations do not change the rank of a matrix we have $\text{rank}(A^TA) = \text{rank}(E^TA^TAE)$, where $E$ is a multiplication of several elementary operations which make $AE = [A_1, A_2]$, where $A_1$ is a column full rank matrix with $\text{rank}(A_1) = \text{rank}(A)$.

Thus we can find a matrix $P$ such that $A_1P= A_2$ and $AE = [A_1, A_1P] = A_1[I, P]$.

Thus $\text{rank}(E^TA^TAE) = \text{rank}(A_1[I, P])^T(A_1[I, P])$. In this equation,  the matrices are all of full rank and the rank equals $\text{rank}(A)$, so on a real space $\text{rank}(A^TA) = \text{rank}(A)$, completing the proof.


Discrete Sets in Certain Metric Spaces are Countable I've recently started my first real analysis course, and we're studying metric spaces from Rudin's Principles of Mathematical Analysis (Baby Rudin).

We have the following definition:

Let $(X,d)$ be a metric space, and $A \subset X$. A point $p \in A$ is said to be isolated if $p$ is not a limit point of $A$. Furthermore, $A$ is said to be discrete if all its points are isolated.

Now, consider $\mathbb{R}^{n}$ with the usual metric $d(x,y)=|x-y|$. Prove that if $A \subset \mathbb{R}^{n}$ is discrete, then it is at most countable.

What I had in mind was something along the lines of:

Since we know that for every $p \in A$ there exist some $r \in \mathbb{R}$ such that $B_{r}(p)  \cap A = \left \{ p \right \}$. We can find some rational $q$ with $0<q<r$, and isolate $p$ with the ball of radius $q$. Then we define a function which assings to each $p$ said $q$. If said function were injective, we would be done, but I don't know how to do that, since all points in $A$ could very well be isolated by finitely many different reals $r$.

Any ideas?
Thanks in advance!

PS: I've previously shown that a discrete set $A \subset \mathbb{R}^{n}$ is compact if and only if it is finite. But it could not think of any useful ways to apply this to the problem at hand.

Hint. Show that for each point $p\in A$ there is an open ball $B$ with rational center and radius, such that $B\cap A=\{p\}$. (I.e., the radius is a rational number, the center is a point in $\mathbb R^n$ all of whose coordinates are rational numbers.) The set of all such balls is countable.

In reference to the hint,we do the following.
Choose  r positive real such that B:= B(p,r)'s closure, has no point of A other than p.let (r(n)) be a sequence of rational coordinates in B(p,r)such that it converges to p.Let s(n)=d(r(n),bdB) and let p(n)=d(r(n),p) for all n. p(n) converges to 0 and s(n) converges to r.
Let  v(n)=s(n)-2p(n) for all n.Thus v(n) converges to r.Since r is positive,there exists N a natural number,
Such that v(N)=s(N)-2p(N)is positive.Hence 2p(N) is strictly less than s(N).
B(r(N),2p(N)) is contained in B(p,r) since Eucledian balls are convex and straight lines in Eucledian spaces are connected.
Now there exists a positive rational t such that t is in between p(N) and 2p(N). Also r(N) is rational in all coordinates.Hence B(r(N),t) is inside B thus B(r(N),t) has p as the only point of A.
(Note p is in B(r(N),t)). Hence is the proof.


Different sums in a $3\times 3$ grid We write the numbers $1, 2, 3$ in the cells of grid $3 \times 3$ and we calculate the sums in rows and columns. What is the maximal number of different sums?

I am learning combinatorics, and this is a problem I got stuck with. I cannot show my work, since I am new here and am not fluent in mathjax.

The possible sums range from $3$ to $9$. This gives $4$ odd numbers and $3$ even numbers in range.
Consider two cases - 1) we have both $3$ and $9$ as sums, using $(1,1,1)$ and $(3,3,3)$. Then these are either both on rows or both on columns, since they cannot intersect. Say they are two rows; then the third row can be all different ($1,2,3$), in which cases we have two totals of $6$, or not all different, in which we have a matching pair of column totals. In either case the maximum number of different values is $5$.
Case 2) - we don't have both $3$ and $9$. Now there are $3$ odd totals and $3$ even totals to aim for, but we cannot have all of them, since then the total of all rows and columns would be odd, which is not possible since that is twice the total grid sum and must be even. So again we cannot have $6$ different row and column sums, and it's easy to demonstrate grids with $5$ different sums.
So the maximal number of different row and column sums from such a grid is $5$.

For each row/column the minimum sum which is reachable is [1,1,1] => 3
and going the same way the maximum sum which can be reached is [3,3,3] => 9
So All sums of range [3,9] are achievable.. I hope you can find the answer now ..


Find the matrix $\mathbf{A}$ if $A\binom{7}{-1} = \binom{6}{2}.$ Find the $2\times2$ matrix $A$ where $A^2=A$ and
$$A\begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$$

I tried plugging in: $A= \begin{pmatrix}a&b\\c&d\end{pmatrix}$ but that became messy very quickly. I got the equations:

$7a-b = 6$

$7c-d = 2$

$a^2+bc = a$

$ab+bd = b$

$ac + cd = c$

$bc + d^2 = d$ from trying that method. What should I do?

As $A^2=A$, you also know that $A\begin{pmatrix}6\\2\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$. Since $\begin{pmatrix}7\\-1\end{pmatrix},\begin{pmatrix}6\\2\end{pmatrix}$ are linearly independant, this determines $A$ completely. We conclude that both columns of $A$ are multiples of $\begin{pmatrix}6\\2\end{pmatrix}$, so
$$ A=\begin{pmatrix}6x&6y\\2x&2y\end{pmatrix}$$
for certain $x,y$. Now $A\begin{pmatrix}6\\2\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$ just means that $6x+2y=1$, and $A\begin{pmatrix}7\\-1\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$ means that $7x-y=1$. Can you find $x,y$ now?

Write out the equations and use a simple linear solver to find:

${1 \over 10}\left(
\begin{array}{cc}
 9 & 3 \\
 3 & 1 \\
\end{array}
\right)
$


Does pi have any combination of two numbers that the distance between them equal some natural number Let's assume that $\pi = \sum_{i=0}^{\infty}\frac{a_i}{10^i}.$ Also, let $n$ be an arbitrary natural number.

Is there $j,k,m \in \mathbb{N}, (j < k < m)$ such that $|b_1 - b_{2}| = n$ when $b_1 = \sum_{i=j}^{k}a_i10^{k-i}, b_2 = \sum_{i=k+1}^{m}a_i10^{m-i}$?

for example:   0.14(15)(92)... so for |92-15|=77 there are j,k,m as wanted.

the original question was:

can a number $x = \sum_{i=0}^{\infty}\frac{x_i}{10^i}$ such that $\forall n \in \Bbb N\ \exists j,k,m \in \mathbb{N}, (j < k < m)$ such that $|b_1 - b_{2}| = n$ when $b_1 = \sum_{i=j}^{k}a_i10^{k-i}, b_2 = \sum_{i=k+1}^{m}a_i10^{m-i}$ is a rational number?

i would like for now a solution only for the first part but any solution will be welcomed.

It is widely believed that the decimal expansion of $\pi$ contains every string of digits, see for example Does Pi contain all possible number combinations?

If one accepts this as true, your statement follows readily.
However, to show this is considered as very difficult.

While your problem is somewhat simpler, I am afraid it is sufficiently similar that it is too difficult to be answered definitely at this point in time (and likely quite a bit of time in the future).


Prime gaps distribution It is well-known that gaps between successive primes have i.e. multimodal distribution
(with peaks at $6 k$):

I'm interested to know: what is the most suitable approximation for
such weird distributions? The envelope of peaks looks like $\chi^2$  ??

Asymptotically it depends on the radical of k: basically take
$$
f(k)=\prod_{p|k,\ p>2}\frac{p-1}{p-2}
$$
and compare for different k. So f(6) = 2 > 1 = f(8), so gaps of length 6 are asymptotically twice as common as gaps of length 8. (Obviously k needs to be even.)

You asked (in a comment) for a smooth envelope. Using Mertens' theorem and the Prime Number Theorem it can be shown that
$$
f(k)=O(\log\log k)
$$
and this bound is tight in the sense that there is some $\alpha$ with $f(k)>\alpha\log\log k$ for infinitely many $k$. (This can be computed without too much difficulty, if desired.)

You might notice that this does not resemble the curve you have drawn. This is the result of a number of separate factors:

The envelope necessarily ignores low points, so only 2, 6, 30, ... are relevant.
There is a discretization error such that the small members do not fit the curve very nicely. It works better for large k, say $k\ge 2\cdot3\cdot5\cdot7$. The error diminishes approximately as $O(1/\log\log k)$.
Your graph uses a small number of prime gaps. This error decreases approximately as $O(k/\log x)$ where x is the number of prime gaps used. In particular, every time you double k you need to square the number of prime gaps used to keep the error roughly constant.

EMPIRICAL RESULTS:

We are talking about primes in some interval between xbar-dx to xbar+dx where xbar is up to around 10^40, and dx is up to around sqrt(xbar)

The number of gaps G between successive primes p = 6*k+1 in this interval can be approximated by a log-linear distribution:
log( G(6*k) ) = A + B*k, for k = 1, 2, ... N small enough.

Similarly, The number of gaps G between successive primes p = 6*k-1 in this interval can be approximated by a log-linear distribution:
log( G(6*k) ) = C + D*k, for k = 1, 2, ... N small enough.

The results for (6*k+1) and (6*k-1) need to be combined to get the approximation for gaps between all the primes.


Integral of the square of the Dirac delta What is the following integral?

$$\int_{-\infty}^{\infty} \delta^2(x) {\,\rm d} x$$

I think it should be one, by Parseval's Theorem.

There is no such thing as $\delta^2(x)$ and $\int_{-\infty}^{\infty} \delta^2(x) {\,\rm d} x$.

Nevertheless, there is a natural extension of the $L^2$ norm to the distributions, which of course assigns $\infty $ to $\delta$ and many others.

Take any $\phi \in C^\infty_c(-1,1),\phi \ge 0,\|\phi\|_{L^1}=1$, $\phi_n(x)=n\phi(nx)$ then $$\|T\|_{L^2} = \lim \sup_{n\to \infty} (\int_{-\infty}^\infty |T\ast \phi_n(x)|^2dx)^{1/2},\qquad T\ast \phi_n(x)= \langle T,\phi_n(x-.)\rangle$$

It is an interesting exercice to show (using the Fourier series of $ (T\ast \phi_n)\psi$) that $\|T\|_{L^2}$ is finite iff the distribution $T$ is (represented by) a function in $L^2$. Moreover $\|T\|_{L^2}=\|\hat{T}\|_{L^2}$.

Note that the same idea gets more complicated for $L^1$ : $\delta$ is a finite measure but it is not in $L^1$.

$$\int_{-\infty}^{\infty} \delta^2(x) dx=2\int_{-\infty}^\infty dx$$


What does it mean to write a linear operator in a particular basis? I am working on a past Algebra exam paper and have come across a problem which requires me to write the linear operator associated to a given matrix $M$ in the standard basis of $\mathbb{R}^4$.

What does it mean to 'write a linear operator IN a given basis'?

Thank you.

NB: I have looked at this link, but I am not sure if the concept of writing a matrix in a given basis is synonymous to the concept of writing a linear operator in a given basis.

EDIT:

I think I should be more specific.

The statement of the problem I am working on is as follows:

Let $M$ be the matrix:
$$\begin{bmatrix}
-2 & 3 & 7 & -3 \\
-6 & 1 & 16 & 1 \\
-2 & 1 & 6 & -1 \\
-2 & -1 & 6 & 3 \\
\end{bmatrix}$$

Write the linear operator associated to $M$ in the standard basis of $\mathbb{R}^4$.

The model solution is as follows:

The linear operator associated to $M$ is the linear map given by

$$M
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
\end{bmatrix} =
\begin{bmatrix}
-2x_1 + 3x_2 + 7x_3 -3x_4 \\
-6x_1 + x_2 + 16x_3 + x_4 \\
-2x_1 + x_2 + 6x_3 - x_4 \\
-2x_1 - x_2 + 6x_3 + 3x_4
\end{bmatrix} $$

My current issue is that I don't fully understand how the model solution is an example of the linear operator written in the standard basis of $\mathbb{R}^4$.
In particular, I'm still trying to grasp the concept of writing a linear operator IN a given basis.  What does this really mean?

A linear operator can be written as a matrix in a given basis.

For example, suppose we have the linear operator, T, from $R^2$ to $R^2$ that maps (x, y) to T(x, y)= (x- y, 2y).  Since that is from $R^2$ to $R^2$, in can be written as a 2 by 2 matrix: $\begin{bmatrix}a & b \\ c & d \end{bmatrix}$.  If we use the "standard basis" for $R^2$, (1, 0) and (0, 1), then (x, y)= x(1,0)+ y(0, 1) so $\begin{bmatrix} x \\ y\end{bmatrix}$ is the representation in the standard basis.  The operation, in matrix form is $\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} ax+ by \\ cx+ dy\end{bmatrix}$.

We want that to be $T(x, y)= \begin{bmatrix} x- y \\ 2y\end{bmatrix}$ so we must have ax+ by= x- y and cx+ dy= 2y.  That is two equations for the four unknowns, a, b, c, and d, but remember this must be true for all x and y.  In particular, taking x= 1, y= 0 we get a(1)+ b(0)= a= 1- 0, so a= 1, and c(1)+ d(0)= 2(0), so c= 0.  Taking x= 0, y= 1, we get a(0)+ b(1)= 0- 1, so b= -1, and c(0)+ d(1)= 2(1) so d= 2.  The matrix representing linear operator, T, in this particular basis, is $\begin{bmatrix} 1 & -1 \\ 0 & 2\end{bmatrix}$.  This particular choice of "x= 0, y= 1" and "x= 1, y= 0" makes the calculations especially easy since (0, 1) and (1, 0) are the basis vectors.
 Notice that $\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}$ and $\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}b \\ d \end{bmatrix}$.  That is, applying the linear operator to each basis vector in turn, then writing the result as a linear combination of the basis vectors gives us the columns of the matrices as those coefficients.

For another example, let the vector space be the set of all polynomials of degree at most 2 and the linear operator, D, be the differentiation operator.  That is, any such "vector" can be written as $P= ax^2+ bx+ c$ and $DP= 2ax+ b$.
If we take $\{x^2,x, 1\}$ as basis $ax^2+ bx+ c$ will be written as $\begin{bmatrix} a \\ b \\ c\end{bmatrix}$.  Applying the derivative operator to the first "basis vector", $x^2= 1x^2+ 0x+ 0$, gives $2x= 0x^2+ 2x+ 0$ so the first column of the matrix representation is $\begin{bmatrix}0 \\ 2 \\ 0\end{bmatrix}$. Applying the derivative operator to the second "basis vector", $x= 0x^2+ 1x+ 0$, gives $1= 0x^2+ 0x+ 1$  so the second column of the matrix representation is $\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$.  Finally, applying the derivative operator to the third "basis vector", $1= 0x^2+ 0x+ 1$, gives $0= 0x^2+ 0x+ 0$  so the third column of the matrix representation is $\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$.   The matrix representing the derivative operator in this basis is $\begin{bmatrix}0 & 0 & 0 \\ 2 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}$

Assume that you have a result for a given basis.

T(e1) = [a1,a2....an]

T(e2) = [b1,b2....bn]

and so on. For this result and the given basis you can find the transformation matrix (by arranging the results in columns). So that is with respect to the given basis.

Now if we change the basis and write the result for each basis vectors, we get a new set of results non similar to the initial set of results. These can be written in columns to form the new transformation matrix which is definitely different from the initial transformation matrix which can also be called as linear operator.


Is there such a thing as the number of axioms? This question was inspired by this question.

Does it really make sense to say that a formal system has some number of axioms, say three, or ten, etc? E.g., take a formal system that admits conjuction elimination and has axioms $ A_1, A_2,..., A_n $. But a formal system that has the single axiom $ A_1\wedge A_2\wedge ...\wedge A_n $ is essentially the same formal system!

So, is the concept of number of axioms in a formal system subjective, just as there is no objective difference between a lemma and a theorem?

P.S. I'm indeed talking about finitely axiomatizable formal systems only. I.e. when I say "axioms", I mean it - no axiom schemata allowed.

Certainly if I give you a formal system with an explicit list of axioms, say 42 of them, then it makes sense to say that the system has 42 axioms. It's hard to dispute that.

But you make a good observation, that there many equivalent formal systems with all sorts of different numbers of axioms. To be precise, say two formal systems (using the same symbols) are equivalent if they prove exactly the same theorems. It's often useful consider formal systems only up to equivalence, when what we care about is the theorems they prove, not their presentations.

So the question is really about whether by formal system you mean "explicitly defined formal system" or "formal system up to equivalence". The former has a well-defined number of axioms, the latter does not.

First off, let's make things clear and point out that not all theorems are the same even from a syntactic point of view.  Some theorems can serve as a single axiom for some logical system in the sense that under some rule of inference, some theorem can get generated from that axiom which is not a substitution instance of the axiom.  Other theorems can't generate anything under that same rule.

For instance, in some systems with detachment as the rule of inference, CCpqCCqrCpr is a theorem.  If the system has countably infinite variables, the system {CCpqCCqrCpr} under detachment and substitution has countably infinite theorem which are not substitution instance of CCpqCCqrCpr.  On the other hand, in some systems CCCpqpp is a theorem.  But, the system {CCCpqpp} only has substitution instance of CCCpqpp as theorems.

Now one part of a formal system, according to Wikipedia consists of its deductive apparatus.  The deductive apparatus consists of the axioms and/or rules of inference of the formal system.  Thus, if the axioms differ, we have a different deductive apparatus.  Consequently, even though the consequences of two deductive apparatuses may end up the same, the deductive apparatuses differ and consequently the formal systems differ.  Axioms also differ from theorems in that they belong to the deductive apparatus, while theorems do not (at least according to the Wikipedia definition).  Additionally, in terms of proving certain theorems, or what can/has gotten called the development of a formal system, the systems can differ in that some theorem might come as provable in fewer steps in system A than in system B.

So, no, the concept of the number of axioms in a formal system is not subjective.  And yes it makes sense to say that a formal system has a certain natural number of axioms.