Need help with the general formula for the Taylor series of $e^x+\sin(x)$ A while ago, I asked about ways to derive the sigma notation for the infinite series of $e^x+sin(x)$.
$$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$$
In this thread,
Clive Newstead brilliantly gave me two formulas.
There, he said that the pattern of the series, regarding the coefficient of the numerators, is as followed: 1,2,1,0,1,2,1,0
He said that the pattern is "is periodic with period 4"
This comment still puzzles me. I have tried to validate his formula and it is true, they are all correct, but I don't know why he can come up with succinct representation of the pattern:
$$\sum_{k=0}^{\infty} \left( \dfrac{x^{4k}}{(4k)!} + \dfrac{2x^{4k+1}}{(4k+1)!} + \dfrac{x^{4k+2}}{(4k+2)!} \right)$$
How does one realize that the power is related to $4k, 4k+1$ and $4k+2$ and $4k+3$?
This seems to relate to finding the general formula for a sequence or series of number, for example, trivially
$1, 3, 5, 7, ...$ can be represented as $2k+1$
$0, 2, 4, 6, ...$ can be represented as $2k$
with $k$ as the position of the term in the sequence.
So my thought is you to count the terms that have the numerator's coefficient as $1$, $2$ and $0$
The ones that have $1$ is $1^{st}$, $3^{sd}$, $5^{th}$, $7^{th}$, etc.
The ones that have $2$ is $2^{sd}$, $6^{th}$, $10^{th}$, $14^{th}$, etc.
The ones that have $0$ is $4^{th}$, $8^{th}$, $12^{th}$, $16^{th}$, etc.
So the difference between the positions of the terms that contain the coefficient $1$ is $2$
So the difference between the positions of the terms that contain the coefficient $2$ is $4$
So the difference between the positions of the terms that contain the coefficient $0$ is $4$
I find that the general term for terms that contain $1$ is $2k+1$
I find that the general term for terms that contain $2$ is $4k-2$
I find that the general term for terms that contain $0$ is $4k$
But I don't know how to derive these into $4k$, $4k+1$, $4k+2$, $4k+3$
I realize that this is similar to finding patterns of a sequence on an IQ test, but I don't know how to derive it. One my friend said I should look into Lagrange's interpolation, but I don't why this technique has anything to do with writing down the general terms for sequences and series.
Could you help me on this?
Methinks, this is a simple question. I am not good at writing down general term of a sequence yet.

$e^x = 1 + x + \frac {x^2}{2} + \frac {x^3}{3!} + \cdots$
It seem pretty natural to write this as
$e^x = \sum_\limits{n=0}^\infty \frac {x^n}{n!}$
But we could look at pairs of terms.
$e^x = \sum_\limits{n=0}^\infty \left(\frac {x^{2n}}{2n!}+\frac {x^{2n+1}}{(2n+1)!}\right)$
Or, even triples of terms.
$e^x = \sum_\limits{n=0}^\infty \left(\frac {x^{3n}}{3n!}+\frac {x^{3n+1}}{(3n+1)!}+\frac {x^{3n+2}}{(3n+3)!}\right)$
And do the same thing with $\sin x$
$\sin x = \sum_\limits{n=0}^\infty (-1)^n \frac {x^{2n+1}}{(2n+1)!} = \sum_\limits{n=0}^\infty \left(\frac {x^{4n+1}}{(4n+1)!} - \frac {x^{4n+3}}{(4n+3)!}\right)$
Now find a representation of $e^x$ that plays nicely with our representation of $\sin x$ add them together and you have what you show above.

The mod $4$ periodicity can perhaps be most easily understood using the formula $e^{ix}=\cos x+i\sin x$, so that $\sin x={e^{ix}-e^{-ix}\over2i}$, which gives
$$e^x+\sin x={1\over2i}\sum_{n=0}^\infty{(2i+i^n-(-i)^n)x^n\over n!}$$
Since $i^4=1$, the coefficients $2i+i^n-(-i)^n$ cycle through the values
$$\begin{align}
2i+i^0-(-i)^0&=2i+1-1=2i\\
2i+i^1-(-i)^1&=2i+i-(-i)=4i\\
2i+i^2-(-i)^2&=2i-1-(-1)=2i\\
2i+i^3-(-i)^3&=2i-i-i=0
\end{align}$$


What is the integral of $e^x a^x$ Can you confirm that my answer below is correct?

$$\int (a^x e^x)dx $$

My attempt:

$$\int a^x e^x \, dx = \int (ae)^x \, dx $$
$$\int a^x e^x \, dx = \frac{(ae)^x}{\ln(ae)} + C $$
$$\int a^x e^x \, dx = \frac{a^xe^x}{\ln(ae)} + C $$

Yes, I can confirm that your answer above is correct.


What will be the domain of the function $f(x)=\sqrt{2\{x\}^2-3\{x\}+1}$? I need to find the domain of the function

$$f(x)=\sqrt{2\{x\}^2-3\{x\}+1}$$

where $x \in [-1,1]$ and $\{.\}$ represents the fractional part of $x$

So here's what I tried:

Clearly the part inside the square root has to be greater than or equal to zero for it to exist, and by factoring the quadratic in terms of $\{x\}$ I get,

$$(\{x\}-1)(2\{x\}-1)\geq0$$

So, from here I got $$\{x\} \in \bigg(-\infty,\frac{1}{2}\bigg] \cup \bigg[1,\infty\bigg)$$

But we know, that the fractional part of x can only vary between 0 and 1, ie,

$\{x\} \in[0,1)$

So finally, I get $\{x\} \in \bigg(-\infty,\frac{1}{2}\bigg]$, which further reduces to the following

$$\{x\} \in \bigg[0,\frac{1}{2}\bigg]$$

And, from the question, $x \in[-1,1]$

But how do I finally get the domain for $x$ here? I'm seemingly stuck at the last step

Hint:
Write $X = \left \lfloor X \right \rfloor + \left \{ X \right \}$
$\\ And  \left \lfloor X \right \rfloor = 0 \  or \ -1$


Is collapsing considered a legitimate proof? For example if I want to prove that $2^n - 1 = 1 + 2 + 4 + 8 +...+ 2^{n-1}$ I can obviously use induction and that is accepted. But I can also collapse it like:

To Prove $2^n = S(n)$:

$S(n) = (1 + 1) + 2 +...+ 2^{n-1}$
$S(n) = (2 + 2) + 4 + 8 +...+2^{n-1}$
$S(n) = (4 + 4) + 8 +...+2^{n-1}$

and so on until
$S(n) = 2^{n-1} + 2^{n-1} = 2^n$

Is this method of collapsing considered a legitimate and presentable proof?

Well, sort of, but in fact, writing proofs like the one you want to write is why induction exists. Whenever people say something like "and so on until", they're expressing your intuition that it's possible to continue the argument by induction. The whole point of the method of induction is to make intuitions like this one precise.

Let $S(k)=2^k + 2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1}$. Then what we want to show is that $S(0) = 2^n$. Your proof basically amounts to saying $S(0) = S(1) = S(2) = ...$ "and so on", until we get $S(0) = S(n-1)$. Notice that $S(n-1) = 2^{n-1} + 2^{n-1}$, which obviously equals $2^n$. So we get $S(0) = 2^n$. To phrase this as a proof by induction, we're going to prove by induction that $S(0) = S(k)$ for all $k<n$, thus we'll obtain $S(0)=S(n-1)$ at the end.

Obviously, $S(0) = S(0)$. Now suppose $S(0) = S(k)$. Then:

$$\begin{align}S(0) &= S(k) \\&= 2^k + 2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1} \\&= 2\cdot2^k + 2^{k+1} + 2^{k+2} + ... + 2^{n-1} \\&= 2^{k+1} + 2^{k+1} + 2^{k+2} + ... + 2^{n-1}\\&=S(k+1)\end{align}$$

the consensus here seems to be "yeah, the proof is ok but rather informal since it invokes induction 'coded' in the 'and-so-on' statement". I don't agree. Imho this is no proof at all and there is no induction argument given.

The Argument starts with the claim (short of subtracting one from both sides of the equation). Why bother and continue?
The modifications made in order to get to the next line are completely unclear to me. Are all the elements in the series doubled? Then the result (left side) should be doubled as well but it isn't?


Finding determinant of following matrix I need to find determinant of following matrix.

$$\begin{bmatrix}1&0&0&0&0&2\\0&1&0&2&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{bmatrix}$$

Original Image

I did it by simply doing $R_5$ - $R_1$. and then evaluating the determinant. But its a lengthy process but answer came out..

But another thing i have noticed afterwards is that it is symmetric matrix. So finding determinant my be easy, but i don't know how it helps which is my main concern. Thanks

Might be easiest to row-reduce just enough so that the matrix becomes a triangular matrix. Then the determinant is the product of the diagonal entries. In this case, we can row reduce to get $$\begin{bmatrix} 1 & 0 & 0 & 0 &0 & 2\\0& 1&0& 2 & 2 &0\\0 & 0 & 1&2&0&0\\0&0&0&-3&0&0\\0&0&0&0&-3&0\\0&0&0&0&0&-3\end{bmatrix},$$ and so the determinant is $-27$.


Find the value of $(a^3 + b^3 + c^3)/(abc)$ if $a/b + b/c + c/a = 1$. Find the value of $$\frac{a^3+b^3+c^3}{abc}\qquad\text{ if }\quad \frac ab + \frac bc + \frac ca = 1.$$

I tried using Cauchy's inequality but it was of no help. Please guide me.

$a, b, c$ are real.

There is not enough information to solve this problem. Clearing out denominators, your hypothesis is
$$a^2 c + a b^2 + b c^2 = abc \quad (1)$$
and your desired conclusion is
$$a^3+b^3+c^3=kabc \quad (2)$$
for some constant $k$.

Suppose, for the sake of contradiction, there were a $k$ such that $(1)$ implied $(2)$. Since the polynomial $a^3+b^3+c^3-abc$ is irreducible, this would mean that  $a^2 c+a b^2+b c^2-abc$ would divide $a^3+b^3+c^3-kabc$. But the two polynomials are both cubics, so the only way for the first to divide the second is the first is a scalar multiple of the second, and it isn't.

It's also easy to generate points on $(1)$ and see that the ratio $(a^3+b^3+c^3)/(abc)$ isn't constant. Just choose random values for $a$ and $b$ and equation $(1)$ turns into a quadratic; solving that quadratic gives you some points to try. You'll see very quickly that nothing like this is true.

Take $a=1$ and $b=-\epsilon$. Then $c=O(1/\epsilon)$ ($\epsilon\rightarrow 0$). By taking $\epsilon$ suitably small, we can make the expression to be evaluated, of order $1/\epsilon^3$, big or much bigger.


Simultaneous equation $x^2 +y=10050$ and $y^2 + x= 2600$ solution Here's the problem:
Jim and Tim are sharing money. If I square Jim’s money and add on Tim’s, I get £10,050. If I square Tim’s money and add on Jim’s, I get £2,600. How much do they each have?
From this the equations are as follows  $x^2 +y=10050$ and $y^2 + x= 2600$
This is a grade 10 math problem and want to solve it w/o using differentiation, integration etc.

$\text{Jim} = x$, $\text{Tim} = y$
$$\begin{cases} x^2 +y=10050\\
y^2 + x= 2600\end{cases}$$
Subtracting the equations,
$$x^2 – y^2 +y – x = 10050-2600=   7450
\\( x-y) (x+y-1)= 7450 = 149 \times 50$$
Since $149$ is prime and $x\ge y$,
$$\begin{cases} x- y = 50 \\
x + y -1 = 149\end{cases}$$
After solving $x= 100$ and $y = 50$.


How to identify rules of inference that establishes validity? I've been trying to determine an explanation for the falsity of a logical statement for some time now and I've had no luck in figuring out exactly how to go about it. The two part question goes as follows:

Consider the arguments below. If the argument is valid, identify the rule of
inference that establishes its validity. If not, explain why.

a. If Robert understands the concepts correctly, he will be able to finish his
assignment in two hours. Robert finished his assignment in more than two
hours. Therefore, Robert did not understand the concepts correctly.

b. If taxes increase, the housing market will decrease. Taxes are not
increasing. Therefore, the housing market will not decrease.

Perhaps I'm misunderstanding the way to determine the falsity of a logical statement? Any help is appreciated.

The "logical form" of the premise of a) is :

  if $p$, then $q$.

The following is a valid argument :

  $p \rightarrow q \vDash \lnot q \rightarrow \lnot p$,

thus, from the above premise we can correctly conclude with :

  if not $q$, then not $p$

which is exactly the conclusion of a).

The argument in b) is not valid, because :

  $p \rightarrow q \nvDash \lnot p \rightarrow \lnot q$;

thus, from the premise : if $p$, then $q$, we cannot conclude with : if not $p$, then not $q$.

a) is valid by modus tollens. Because recall, modus tollens is,

P → Q, ~Q  |- ~P

The argument is identical


There exists an integer k such that $n = 3k+1$. Then $n^2 = (3k+1)^2 =9k^2 + 6k + 1 = 3 (3k2 +2k)+1$. Consider the following proof fragment.

There exists an integer $k$ such that $n = 3k+1$.

Then $n^2 = (3k+1)^2 =9k^2 + 6k + 1 = 3 (3k^2 +2k)+1$.

For each of the statements, $(a), (b), (c)$, below, answer the following.

Does the fragment provide a proof of the statement? If yes, explain why. If no, explain why not. The letter n denotes an integer.

(a) If $n$ is odd, then $n^2$ is odd.

(b) If $n^2$ is divisible by $3$, then $n$ is divisible by $3$.

(c) If $n$ leaves remainder $1$ on division by $3$, then so does $n^2$.

I am having some trouble understanding the question.
I have started a) by assuming $n$ is odd and I took $n=3k+1$ and then I showed that $n^2=(3k+1)^2=3(3k^2+2k)+1$
and I said that if $n=3k+1$ is odd $3k$ is even so $k$ is even and can be writen as $k=2m$ where $m$ is an integer. After that I sub. $k=2m$ into the $n^2=3(3k^2+2k)+1$ and I find that $n^2=2$(somthing)$+1$ so it is odd.

I have been told by my Professor that is I should not use the statment given to me to answer $a,b$, and $c$. I was wondering if for $a)$ I am just supposed to take $n=2k+1$ and then show that $n^2$ is odd.
Can someone help me out?

Generally, what does it mean for a number to be odd? Well mathematically we say:

$(2k)$ is even, and $(2k+1)$ is odd, for $k\in \mathbb{N}$.

Let's consider part (a):

Prove if $n$ is odd, $n^2$ must also be odd:

Let n be an odd number, such that $n=2k+1$ for $k\in \mathbb{N}$.

Then, $n^2=(2k+1)^2=(4k^2+4k+1)=2(2k^2+2k)+1\Rightarrow n^2$ is odd.

Why? Well any natural number multiplied by an even number is even, so what happens when we add 1 to an even number? It turns odd!

Consider part (b):

Prove if $n^2$ is divisible by $3$, then $n$ is divisible by $3$.

If $n^2$ is divisible by $3\Rightarrow n^2=3k$ for $k\in \mathbb{Z}$

Well, this comes directly from the fundamental theorem of arithmetic, that is, every number is either prime, or the product of prime numbers, such that each composite number has a unique sequence of products. (excluding the order in which primes are multiplied of course).

Well, this implies that $n^2$ is the product of a sequence of prime numbers, that is, $n^2$ is a number multiplied by itself. So intuitively it makes sense that $n$ must also be divisible by $3$, and that's the proof!

Consider part (c):

Can you express $n^2=3M+R$ such that $M\in \mathbb{N}$ and $R$ is your remainder, in this case $R=1$?

Question has been answered in the comments by JMoravitz.


Relationship between the column space of a matrix $A$ and its non-free (pivot) columns Given an $m\times n$ matrix $A$ with $m\leq n$, with the rank of $A$ being less than $n$, is it necessarily true that the columns in $A$ representing the free variables are linear combinations of the pivot columns? If I am to figure out the column space of $A$, without having to calculate which of the columns are redundant (i.e. linear combinations of other columns), can I reliably say that $C(A)$ is the span of all (and only) the pivot columns in $A$? I was watching a video by Khan Academy where it seemed that this was the case, at least for the example given... but I don't know if it generalizes for all matrices $A$ where the null space does not equal $\{\vec{0}\}$

Example:

$$A=\left[\begin{array}{rrrr}1 & 1 & 1 & 1\\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \end{array}\right]$$

Its column space is the span of the two vectors $\left[\begin{array}{r}1\\2\\3 \end{array}\right]$ and $\left[\begin{array}{r}1\\1\\4\end{array}\right]$, which just so happen to be the only two pivot columns. The other two are free variable columns.

This is true for all matrices. Elementary row operations preserve linear relationships between the columns of a matrix. Suppose we have a matrix $A$ with columns $\mathbf{a}_i$ along with the Reduced Row Echelon Form $R$ with columns $\mathbf{r}_i$. Then for any set of coefficients $c_i$, we have
$$\sum_{i=1}^nc_i\mathbf{a}_i = \mathbf{0}\iff \sum_{i=1}^nc_i\mathbf{r}_i = \mathbf{0}$$
The pivot columns in $R$ correspond to a basis for the columnspace of $R$, it follows that the same columns in $A$ form a basis for the columnspace of $A$.

It is not necesary true. We have a result which says

  If the matrices $A$ and $B$ are related by a elementary row operation, then their row spaces are equal. Hence, row-equivalent matrices have the same row space, and hence also, the same row rank.

But with column space is different, another related result is

  Row operations do not change the column rank.

which do not say nothing about column space, only column rank. For example, consider the effect on the column space of this row reduction

$$\left( {\begin{array}{*{20}{c}}
   1 & 2  \\
   3 & 4  \\
\end{array}} \right)\mathop  \to \limits^{ - 2{\rho _1} + {\rho _2}} \left( {\begin{array}{*{20}{c}}
   1 & 2  \\
   0 & 0  \\
\end{array}} \right).
$$

The column space of the left-hand matrix contains vectors with second component that is nonzero. But the column space of the right-hand matriz is different because it contains only vectors whose second component is zero.

Above comments says that we can not, always, express the column space of a initial matrix as span of the pivot columns of the echelon form of this matrix because, as above example, it can change with row operations. The only thing that do not change is the column rank.


Conditional probability - sum of dice is even given that at least one is a five Question:

  Calculate the conditional probability that the sum of two dice tosses is even given that at least one of the tosses gives a five.

I'm a bit confused by this. Shouldn't the probability just be 1/2, since we know that at least one of the dice tosses gave us a five, thus the other must give us an odd number?

A = event when one of the tosses gives a five. (Sample space for the conditional probability)
Let (n1, n2) represent the outcomes of die1 and die2
A = { (1,5), (2,5), (3,5), (4,5), (5,5), (6,5),
      (5,1), (5,2), (5,3), (5,4), (5,6) } // (5,5) must be counted once only

Thus n(A) = 11

B = sum of two dice tosses is even
n(B|A) = { (1,5), (3,5), (5,5), (5,1), (5,3)|

P(B|A) = n(B|A)/n(A) = 5/11

Examine this sample space

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),6,6)}

If you did not know anything about the two die, then probability of having even sums when you have at least a 5 is 5/11. Your intuition for 1/2 however requires you to have some information about which die has turned out a 5. If you knew for sure everything about column 5 (or row 5) of this matrix, i.e. {(...,5), (...,5), (...,5), (...,5), (...,5), (...,5)}, this means you know which die has a five (first or second die). Then you only have row 5 (or column 5) to deal with. And in that case you probability is 1/2, i.e. your outcomes are:

$\{(5,1), (5,3), (5,5)\}$

In a sample space of $\{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$.


Does a closed point of a scheme have an affine open environment with the same dimension? Consider a scheme $X$, and an a closed point $x\in X$. I am wondering whether there is an affine open neighborhood $x\in U\subseteq X$ such that
$$\dim \mathcal O_{X,x}=\dim U.$$

I tried the following approach: Let $V=\mathrm{spec}(R)$, then $\mathcal O_{X,x}=R_{\mathfrak m_x}$, where $\mathfrak m_x$ is the maximal ideal corresponding to the closed point $x$. Let $f\in R$ with $f\notin \mathfrak m_x$, then $U=\mathrm{spec}(R_f)$ is an open subset of $V$ that contains $x$. As every chain of prime ideals in $R_{\mathfrak m_x}$ is a chain in $R_f$ as well, we have $\dim\mathcal O_{X,x}\leq \dim U$. Unfortunately, I do not understand if we can choose an $f$ for which we obtain equality.

If this is not possible in general, can we prove it for a variety, i.e. an integral, separated scheme of finite type over $\mathbb C$?

The answer to your first question is no.

Let $k$ be a field and $m_i$ a sequence of positive integers such that $m_{i+1}-m_i >m_i-m_{i-1}$. For each positive integer $i$, let $\mathfrak{p}_i=(x_{m_i+1},\dots,x_{m_{i+1}})$, which is a prime ideal of $B=k\left[x_1,\dots,x_n,\dots\right]$. Let $S$ be the union of the $\mathfrak{p}_i$'s and $A$ the localization of $B$ at $S$ (the ring $B$ has been introduced by Nagata). Then, the $S^{-1}\mathfrak{p}_i$'s are maximal ideals of $A$ (see Remark 1 of Tomo's answer in this MSE question).

Now, let $X=\text{Spec}(A)$ and chose $x=S^{-1}\mathfrak{p}_i$ a closed point of $X$: note that $\dim\mathscr{O}_{X,x}=\text{ht}(S^{-1}\mathfrak{p}_i)=m_{i+1}-m_i$. Now let $D(f)$ be a principal open set containing $x$ for some $f\in A$. One may assume that $f$ comes from an element $g$ of $B$ (take $g$ to be the numerator of a representative of $f$) and let $l>i+1$ such that $l$ is greater than all the indices of the indeterminates occurring in $f$: then, for any $j\geq l$, you may check that $S^{-1}\mathfrak{p}_j\in D(f)$ (this comes from the choice of $l$: none of the $x_n$'s occur in $f$ for $n\geq l$). This shows that $\dim D(f)=\infty>\dim\mathscr{O}_{X,x}$.

Nonetheless, the answer is positive for schemes locally of finite type over a field.

Indeed, if $X$ is locally of finite type over a field $k$, then one may show that for any point $x\in X$ one has that $$\dim_xX=\dim\mathscr{O}_{X,x}+\text{degtr}(\kappa(x)/k)$$ (where $\dim_xX=\min_{U\ni x}\dim U$). When $x$ is closed, one gets that $\dim_xX=\dim\mathscr{O}_{X,x}$ which answers your question.


Fit exponential with constant I have data whic would fit to an exponential function with a constant.

$$y=a\cdot\exp(b\cdot t) + c.$$

Now I can solve an exponential without a constant using least square by taking log of y and making the whole equation linear. Is it possible to use least square to solve it with a constant too ( i can't seem to convert the above to linear form, maybe i am missing something here) or do I have to use a non linear fitting function like nlm in R ?

A direct method of fitting (no guessed initial values required, no iterative process) is shown below.
For the theory, see the paper (pp.16-17) : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales

I tried to deduce the formula for 'b' in case of equdistant ti-s, but I have got a different one. There is a reciprocal under the logarithm in my result. (Please check it. Maybe I am not right.)

Sorry. I see ... The t3-t1 minus sign makes the log content right.


Is there a name for this special matrix? $$C_{n\times n}={\begin{bmatrix}
    1 & x_{12} & \dots  & x_{1n} \\
    {x_{12}}^{-1} & 1 & \dots  & x_{2n} \\
    \vdots & \vdots & \ddots & \vdots \\
    {x_{1n}}^{-1} & {x_{2n}}^{-1} & \dots  & 1
\end{bmatrix}}$$

Criteria

$$x_{ab} = \frac{1}{x_{ba}}, \qquad x_{ab}>0, \qquad x_{aa} = 1$$

Alternatively

$$C \cdot C^{T} =  \underline1$$

I know that this matrix models currency exchange (without commission or fluctuation). So I'm guessing it's called a currency matrix or a trade matrix. I'm just after a name so I can search it's properties.

I couldn't spot it in the Matrix Reference Manual. http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/special.html

I'm curious about some of it's algebraic properties, for example it has only 1 non-zero eigenvalue. What does this eigenvalue signify?

$$\text{Eigenvalues of} \begin{bmatrix}
    1 & 6 & 30  & 210 \\
    1/6 & 1 & 5  & 35 \\
    1/30 & 1/5 & 1 & 7 \\
    1/210 & 1/35 & 1/7  & 1
\end{bmatrix} = 0,0,0,4$$

Any help, ideas, advice greatly welcomed. Thanks

$C$ is a positive reciprocal matrix

to be an exchange matrix the additional condition that $a_{ij} \times a_{jk} = a_{ik}$ is required

$C$ is then a Positive Saaty-consistent Reciprocal matrix (PS-cR)

the eigenvalues of an $n\times n$ PS-cR matrix are $n,0,..,0$

Thank you @Algebraic Pavel for the term reciprocal.

It is laplacian matrix which is normally I-A, where A is adjacency matrix.
https://en.m.wikipedia.org/wiki/Laplacian_matrix


Proof or disprove that the following statement is true I have a problem I need help with:

Prove or disprove that P(M $\cup$ N) $\neq$ P(M) $\cup$ P(N).

What I have tried so  far

Consider: M = $\{ a,b,c\}$ and N = $\{\text{red},\text{blue},\text{yellow}\}$

$\Rightarrow$ P(M) ${}= 3^3 = 27$ and P(N) ${} = 3^3$

Combine P(M)$\cup$P(N)

$\Rightarrow \bigcup \{M,N\} = \{x\mid(x\in M) \text{ and } (x\in N)\}$

$\Rightarrow$ P(M ${}\cup{}$ N) ${}= \{a,b,c,\text{red}, \text{blue}, \text{yellow} \}$

$\Rightarrow$ P(M,N) ${}= 6^6 = 46656$

$\Rightarrow$ P(M $\cup$ N) ${}\neq{}$ P(M) ${}\cup{}$ P(N)

Is my answer correct?

Your answer has a good idea but there are some technical mistakes in it.

For example, you say $P(M)=3^3$, which is not true.

First of all, $P(M)$ is a set, while $3^3$ is a number. The two are not equal.
What you probably want to say $|P(M)|=3^3$, i.e. that the size of the set $P(M)$ is $3^3$. But even then, no, that is not true.

To try the proof again, I suggest you try to construct a smaller counterexample, and ditch the idea of set sizes. Simply look at the power sets directly.

For an added challenge, try proving this statement:

  $P(A\cup B)=P(A)\cup P(B)$ if and only if $A=\emptyset$ or $B=\emptyset$.

In your example,  $\vert A\vert=\vert B\vert = 3$ and $\vert A \cup B \vert=6$ so
$\vert P (A)\vert =\vert P (B)\vert =2^3$ and
$\vert P (A \cup B) \vert =2^6$. ..
Now $\vert P (A) \cup P (B)\vert= \vert P (A)\vert +\vert P (B)\vert - \vert P (A)\cap P (B)\vert=2^3+2^3-1=2^4-1= 15$,
 so they are not equal. ..


If P(n) divides Q(n) for all integers n then does P(x) divides Q(x)? Let $S = \{ f \in \mathbb{Q}[x]\mid f(n) \in \mathbb{Z}, \forall n \in \mathbb{Z} \}$.
Let  $P(x), Q(x) \in S $ be polynomials such that for every $ n \in \mathbb{Z}$ either $\frac{Q(n)}{P(n)} \in \mathbb{Z}$  or both $P(n), Q(n)$ are $0$ . Can we conclude $\frac{Q(x)}{P(x)} \in S$ ?, or do we have a counter example?.
What if we work in $\mathbb{Z}[x]$?
[I know that a $\mathbb{Z}$-basis for $S$ is $x \choose i$ where $i \in \mathbb{Z}_{\geq 0}$.]

Let $dQ(x)=P(x)A(x)+r(x)$ for some integer $d$, with $A$ and $r$ having integer coefficients, and degree of $r$ less than degree of $P$. Then $P(n)$ divides $r(n)$ for all $n$. But this is impossible unless $r$ is identically zero, since $P(n)$ grows faster than $r(n)$.

If there exist some $n\in \mathbb{Z}$ such that $Q(n)$ & $P(n)$ are both zero then $\frac{Q(n)}{P(n)}$ is not exist. Hence $\frac{Q(x)}{P(x)}$ doesn't belongs to $S$. So for all $n\in \mathbb{Z}$ $\frac{Q(x)}{P(x)}\in\mathbb{Z}$. so we can't concluded that $\frac{Q(x)}{P(x)}\in S$ until the second either case not omitted.


The Gamma function and factorial satisfy $\Gamma(n+1) = n!$ This was a question from a mock paper, for my upcoming exam however my teacher unhelpfully did not post any solutions.

  Prove that $\Gamma(n+1)=n!$.

Can anyone check if my proof is correct.

Thank you for reading.

-Alexis

Note that he property $$G(n + 1) = n G(n)$$ you establish also holds for any constant multiple of $\Gamma$, including the zero function.

Since the proof you give is basically an inductive argument (it might be useful to say a little more in your solution about how this goes), it suffices to add a base case, that is, show that the identity holds for the lowest applicable value of $n$. Since $0!$ makes sense (but $(-1)!$ is not defined), one should show that $\Gamma(0 + 1) = 0!$.


Pure mathematics in our society Is there some book or essay which deals with the sociological and economical justification of doing and funding pure mathematics? I'm looking for a modern version of Hardy's A Mathematician's Apology, but if possible from a non-mathematician's point of view. For example, the book should answer the following type of questions:

  Why is it (and should it be?) possible in our society to earn a living by calculating, say, K-theory groups of spectra? Why does the society support a subculture of mathematicians who solve abstract problems which "obviously" don't have any connection to the rest of the world?

I'm not asking for personal opinions from math.SE users - this is a reference request.

Not an essay or book, but there is a rather informative and clear video address by Timothy Gowers (a Fields Medalist) in 2000 on the "Importance of Mathematics" for an audience containing non-mathematicians in Paris.

He specifically addresses the issue of the (enormous) benefits that mathematics yields on a (very small) investment, its intrinsic cultural value and the interconnectedness of mathematics that make the "useful" areas inseparable from the "useless" ones.

It is available for download from the Clay Math Institute's website here, you'll have to scroll down to the bottom of that page for the download link.

There doesn’t seem to be published very much:

Mathematics in Society and History: Sociological Inquiries
  S. Restivo                  (Springer)

On the Sociology of Mathematics
D. J. Struik
(Science & Society)   (JSTOR)

http://www.amazon.com/What-Mathematics-Really-Reuben-Hersh/dp/0195130871/ref=sr_1_1?s=books&ie=UTF8&qid=1404379719&sr=1-1&keywords=mathematics+hersh
(And other books by same author)


Divergence of the sequence $\sin(n!)$ Does the sequence $\sin(n!)$ diverge(converge)?

It seems the sequence diverges. I tried for a contradiction but with no success.
Thanks for your cooperation.

Depends on whether the parameter of $\sin$ is in radians or degrees. If in degree, $n!$ becomes multiple of 360 and after that function value will be zero, for all value of $n$.

In radians this will not happen as $\pi$ is irrational.

Hint: Take a subsequence in which $a_{n_i} \approx \pi ( 4i+1)/2 $ and another where $b_{n_j} \approx \pi ( 4i +3)/2$.


Show that if $\int_0^x f(y)dy \sim Ax^\alpha$ then $f(x)\sim \alpha Ax^{\alpha -1}$ Let $f$ be a real, continuous function defined on $[0,\infty)$ such that $xf(x)$ is increasing for all sufficiently large values of $x$. Show that if

$$\int_0^x f(y)\,dy \sim Ax^\alpha \quad \left(\,x\to \infty\right)$$

for some positive constants $A$ and $\alpha$, then

$$f(x)\sim \alpha Ax^{\alpha -1} \quad \left(\,x\to \infty\right).$$

Clearly, I have to use differentiation somewhere, but I don't know how to manipulate $\lim_{x\to \infty}\frac{\int_0^x f(y)\,dy}{Ax^\alpha}=1$ to get the desired result.

From suggestions given, I know that by L'hospital's rule, it's enough to show that the limit $\lim \frac{f(x)}{\alpha Ax^{\alpha -1}}$ exists, and this limit equals $
\lim \frac{xf(x)}{\alpha Ax^{\alpha}}$. From here, I'll need to use the given assumption that $xf(x)$ is eventually increasing. But how can I show the existence of the limit based on these facts?

I would greatly appreciate any solutions, hints or suggestions.

Ooh, a tauberian theorem with a simple elementary proof, cool.

Assume $x$ is always so large below that $xf(x)$ is increasing. Fix $\delta>0$. It follows that $$\int_x^{(1+\delta)x}f(y)\,dy\sim A((1+\delta)^\alpha-1)x^\alpha.$$In particular, if $x$ is large enough we have $$\int_x^{(1+\delta)x}f(y)\,dy\le(1+\delta)A((1+\delta)^\alpha-1)x^\alpha.$$But $yf(y)$ increasing shows that $$\int_x^{(1+\delta)x}f(y)\,dy\ge\int_x^{(1+\delta)x}\frac{xf(x)}{y}\,dy=xf(x)\log(1+\delta),$$and combining this with the previous inequality shows that if $x$ is large enough then$$\frac{f(x)}{x^{\alpha-1}}
\le\frac{(1+\delta)A((1+\delta)^\alpha-1)}{\log(1+\delta)}.$$Letting $\delta\to0$ now shows that $$\limsup\frac{f(x)}{x^{\alpha-1}}\le\alpha A.$$The inequality $\liminf f(x)/x^{\alpha-1}\ge\alpha A$ is proved similarly, starting with $$\int_{(1-\delta)x}^xf(y)\,dy\sim A(1-(1-\delta)^\alpha)x^\alpha.$$

First, obviously, $Ax^\alpha\to+\infty$

Now, we can see that

$\lim\frac{\int_0^x f(y)dy}{Ax^\alpha}=1$ which means that

$\lim \int_0^x f(y)dy = \lim Ax^\alpha=+\infty$

Now, you can use L'hopital's rule.

Even though we already know the limit is $1$, L'hopital's statement is more than that.

It states that $\lim \frac{f}{g}=\lim\frac{f'}{g'}$ which technically means you get what you want.


How to use prove this $p^4\equiv p\pmod {13}$ let prime number $p$,and $n$ postive integer, such $$p|n^4+n^3+2n^2-4n+3$$
show that
$$p^4\equiv p\pmod {13}$$

A friend of mine suggested that I might be able to use the results problem

I highly respect  subtle mathematics  but here, at MSE, I give priority to the elementary.  I think mainly of beginners who, for obvious reasons, do not understand anything if the reasoning is of medium high level.

We have to prove that the polynomial $f(x)=x^4+x^3+2x^2-4x+3$ (which, say it, is always divisible by the prime $3$ because
$f(n)=n(n-1)(n+1)^2+3(n^2-n+1)$)  is such that putting $$f(n)=\prod_{i=1}^{i=r}p_i^{\alpha_i}$$  where $n$ is an arbitrary natural,  it is verified  for all $p_i$  the congruence
$$p_i^4\equiv p_i\pmod{13}$$ so it is clear that the primes $p_i$ belong to a certain class excluding a lot of other primes.

We can write $$4f(n)=(2n^2+n+5)^2-13(n+1)^2$$  from which $$4f(n)\equiv(2n^2+n+5)^2\pmod{13}$$
A straightforward calculation gives for $g(x)=(2x^2+x+5)^2$
$$g(\mathbb F_{13})=\{0,4,10,12\}$$ so we have (adding details) in $\mathbb F_{13}$  (where $0=13k;1=13k+1$, etc)
$$\begin{cases}4f(n)=0 \hspace{10mm}\text {for }  n=0\space  \text {and 3}\\4f(n)=4\hspace{10mm}\text {for }  n=2,4,8,11\\4f(n)=10\hspace{8mm}\text {for }  n=7,9,10,12\\4f(n)=12\hspace{8mm}\text {for }  n=1,5,6\end{cases}$$
On the other hand the inverse of $4$ modulo $13$ is $10$ so we have

$$\begin{cases}f(n)= 0\hspace{10mm}\text {for }  n=0\space  \text {and 3}
\\f(n)=1\hspace{10mm}\text {for }  n=2,4,8,11\\f(n)=9\hspace{10mm}\text {for }  n=7,9,10,12\\f(n)=3\hspace{10mm}\text {for }  n=1,5,6\end{cases}$$
Now the primes $p_i$ above can be only $13$ and those of the form $13k+1,13k+9$ and $13k+3$.

This property can be easily verified because if $p^4\equiv p\pmod{13}\iff p(p^3-1)=13$k then $p^3-1$ is divisible by $13$ when   $p\ne13$. In fact
$$1^3-1=0=13\cdot0\\\hspace{5mm}9^3-1=728=13\cdot56\\3^3-1=26=13\cdot2$$
But none of the nine integers below are divisible by $13$
$$2^3-1\\4^3-1\\5^3-1\\6^3-1\\7^3-1\\8^3-1\\10^3-1\\11^3-1\\12^3-1$$

We will have modulo $13$ all the time. Notice that $$N:= n^4+n^3+2n^2-4n+3 \equiv (n^2-6n-4)^2 \equiv (n-3)^4$$

Clearly $13\mid N\iff 13\mid n-3$, so we assume $p\ne 13$ and thus $13$ does not divide $n-3$. Say prime divisor $p$ of $N$ is good if $p^4{\equiv}p$.

Suppose the statment is not true, so there exists a prime $q$ which is not good.

Also we can assume that there is no good prime divisor of $N$: if $p\ne 13$ is good than we can divide $N$ by $p\equiv p^4$ modulo $13$ and then observe $N' = N/p^4$ modulo $13$.  Also we can reduce $N$ with all divisors of $N$ of form $d^4$. So if we make now a prime factorisation for $N$ modulo $13$ we have $$N\equiv  2^a4^b5^c6^d7^e8^f10^g11^h12^i$$
(clearly there is no $0,1,3$ and $9$) where all the exponents are nonnegative and less than $4$ and at least one is positive. Of course, we can reduce this even more:
$$N\equiv  2^{a+2b+d+3f+g+2i} 3^{d+i} 5^{c+g}7^e11^h$$
or $$N\equiv  2^x 3^y 5^z(-6)^e(-2)^h$$
where or exponents are nonegative and less than $4.$ Again we can assume if $y=0$ (else we divide $N$ by $3$ modulo $13$). So we can write:$$ N\equiv  (-1)^{3z+e+h} 2^{x+3z+e+h} = (-1)^t2^{x+t}$$
where $t= 3z+e+h$. So $$N\equiv   -1\pm2,\pm4,\pm8$$
Now this can not be of the form $(...)^4$ and we have a contradiction.


Ordinary generating function of powers of 2 Is there a good closed form expression for the generating function of the formal power series
$$
A(z) := \sum_{n=0}^\infty z^{2^n} = z + z^2 + z^4 + z^8 + z^{16} + \cdots.
$$
Is there a tractable way to retrieve the coefficient of $z^m$ in powers of $A(z)$, say in $A(z)^k$ for $k \geq 1$? Thanks.

The value $A(1/2)=\kappa$ is known as the Kempner number, and was proven transcendental in 1916. The paper "The Many Faces of the Kempner Number", by Adamczewski, may provide some insight for you.

How about $A(z)=\frac{z^2}{1-z^2}$? This works if $|z|<1$.

I got this idea from expanding $\frac{1}{1-z}$.


Functions of the form $\int_a^x f(t) dt$ that are commonly used. I am a graduate student and teaching assistant, and I am teaching Calc 1 for the first time. In a few weeks I will be covering the Fundamental Theorem of Calculus.  I'm using James Stewart's Calculus textbook, and I was hoping to give students several "real world" examples of functions of the form $g(x)=\int_a^x f(t) dt$ to make the first part of the FTC more accessible.  Stewart gives one example, the Fresnel function

$$S(x)=\int_0^x \sin (\pi t^2/2)$$

which is (apparently) used in the theory of the diffraction of light waves.  But I was hoping for more examples from physics, chemistry, etc.  Any thoughts or ideas?

$$
\Phi(z) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-z^2/2}\,dz
$$

This is the cumulative distribution function of the standard normal distribution, seen in every course on statistics.  Physicists often talk about the "error function", in which $z^2$ appears instead of $z^2/2$ (and then the normalizing constant is different).  That's trivially equivalent to this one, but with this version, the standard deviation is $1$.

Abraham de Moivre considered this function in the 18th century while thinking about a coin-tossing problem.  Suppose you toss a coin 1800 times.  What is the probability that the number of heads is in the range (say) $\{886,\ldots, 908\}$?  An exact answer is computationally far too expensive, but the integral above does it.

The derivative of $\Phi$ is conventionally denoted $\varphi$.

The one seen tabulated in the back of every statistics textbook is $\Phi$.

Why do you need any examples?

In my opinion, FTC is the most important part in Calculus because how it unites the other two parts (dir/int).

The derivative-form of FTC implies that the overall properties of a function (integral) is decided by the domestic properties (derivative).

The integral-form of FTC of FTC indicates that the domestic properties are regulated by the overall properties.

With FTC, it's much easier to do integral with anti-derivatives. It's an example to tell the students how the advancements in mathematics make it easier to solve problems.

I don't think it's needed to make the "first part" of FTC accessible. Actually FTC is FTC and the two so-called parts are essentially the same. They are just represented in two different forms. Similarly, differentiation and integration are essentially the same. So do the two mean value theorems.


Time Complexity of DFS My understanding is that:
1) given a graph G with n vertices and m edges, DFS is O(n + m)
2) DFS can be used to produce a list of all simple paths between 2 vertices u and v

This would mean that DFS can produce a list of all simple paths between u and v in polynomial time.

However, if we could could list all simple paths between u and v in polynomial time, we should be able to decide if G has a Hamilton Path between u and v in polynomial time.

Since determining if G has a Hamilton Path is NP-Complete, my understanding must be incorrect. I'm hoping someone can clarify what I'm missing?

What you're missing is that there could be an exponential number of paths  that stop short of connecting $u$ and $v$.  Discovering these dead ends using depth-first search and backtracking is what makes DFS require worst-case exponential time to find a Hamiltonian path.

For traversing all the nodes, you will only visit each point for one time, thus linear time for traversing.

For finding all the paths, you might need to visit each node for many times(e.g. you need to backtrack to a node many times), and the number of possible paths is exponential thus exponential time for finding all paths.


How to show $\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$? I am able to evaluate the limit $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$ for a given $n$ using l'Hôspital's (Bernoulli's) rule.

The problem is I don't quite like the solution, as it depends on such a heavy weaponry. A limit this simple, should easily be evaluable using some clever idea. Here is a list of what I tried:

Substitute $y = x - 1$. This leads nowhere, I think.
Find the Taylor polynomial. Makes no sense, it is a polynomial.
Divide by major term. Dividing by $x$ got me nowhere.
Find the value $f(x)$ at $x = 1$ directly. I cannot as the function is not defined at $x = 1$.
Simplify the expression. I do not see how I could.
Using l'Hôspital's (Bernoulli's) rule. Works, but I do not quite like it.

If somebody sees a simple way, please do let me know.

Added later: The approach proposed by Sami Ben Romdhane is universal as asmeurer pointed out. Examples of another limits that can be easily solved this way:

$\lim_{x \to 0} \frac{\sqrt[m]{1 + ax} - \sqrt[n]{1 + bx}}{x}$ where $m, n \in \mathbb{N}$ and $a, b \in \mathbb{R}$ are given, or
$\lim_{x \to 0} \frac{\arctan(1 + x) - \arctan(1 - x)}{x}$.

It sems that all limits in the form $\lim_{x \to a} \frac{f(x)}{x - a}$ where $a \in \mathbb{R}$, $f(a) = 0$ and for which $\exists f'(a)$, can be evaluated this way, which is as fast as finding $f'$ and calculating $f'(a)$.

This adds a very useful tool into my calculus toolbox: Some limits can be evaluated easily using derivatives if one looks for $f(a) = 0$, without the l'Hôspital's rule. I have not seen this in widespread use; I propose we call this Sami's rule :).

Let
$$f(x)=x+x^2+\cdots+x^n-n$$
then by the definition of the derivative we have

$$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1}=  \lim_{x \to 1}\frac{f(x)-f(1)}{x - 1}=f'(1)\\[10pt] = \left[ \vphantom{\frac11} 1 + 2x + 3x^2 + \cdots + nx^{n-1} \right]_{x=1} = \frac{n(n + 1)}{2}$$

Your objection to using l'Hospital's rule is on the basis that it feels like it's "too powerful" a tool for the problem, right?

Lets go all the way to the other end, then, and prove the limit using just the definition.

$$\lim_{x\to a} f(x) = L \iff \forall \varepsilon \gt 0 \ \exists \delta  \gt 0 \ni \left| x-a \right| \le \delta \Rightarrow  \left| f(x)-L \right| \le \varepsilon$$

Now, all we need to do is figure out a way that we can always pick a $\delta$ small enough to keep the function within $\varepsilon$ of $L$.

The particular limit in question:
$$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$

From that, we can take:
$$\begin{align*}
a&=1\\
f(x)&=\frac{x + x^2 + \dots + x^n - n}{x - 1}\\
L&=\frac{n(n + 1)}{2}
\end{align*}$$

So we need to solve
$$\left| \frac{x + x^2 + \dots + x^n - n}{x - 1} - \frac{n(n + 1)}{2}\right| < \varepsilon$$
for $\left|x-1\right|$.

I really don't feel like doing any of that crunchwork solving that equation.  If someone out there does feel like it, please do so, and edit it into my answer.  For now, though, I am going to skip ahead bunch of steps, assume we solved it, and have our solution of
$$\left|x-1\right| \ge g(\varepsilon)$$

Now, we know from the definition that $\left| x-1 \right| \le \delta$, so we can conclude that if we pick $\delta = g(\varepsilon)$, we ensure that the value of $f(x)$ is within $\varepsilon$ of $L$, satisfying our definition of the limit, and proving that $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$


Evaluate $3\int_{0}^{2\pi} \sin(t) \cos(t) \,{\rm d}t$ $$3\int_{0}^{2\pi} \sin(t) \cos(t) \,{\rm d}t$$
My calculus is a bit rusty and I can not find where I get it wrong. Setting $u= \sin(t)$, I get ${\rm d} u=\cos(t) \,{\rm d} t$ and, thus,
$$3\int_{u=0}^{u=0}u \,{\rm d}u=0$$

The substitution is correct. If $f$ is a function with an antiderivative $F$, one has by the fundamental theorem of calculus
\begin{equation}
\int_a^b f(u(t))u'(t)dt = \int_a^b(F(u(t)))' dt = F(u(b)) - F(u(a)) = \int_{u(a)}^{u(b)}f(x) d x
\end{equation}
A sufficient hypothesis is that $u'$ is continuous in $[a,b]$ and that $f$ is continuous on an interval that contains $u([a,b])$.
In your case $u(t) = \sin(t)$ and $f(t)=t$.
I'm amazed that so many people are puzzled by this simple application of the fundamental theorem of calculus.
Let us take the case of the integral
\begin{equation}
I = \int_0^\pi \sin(x) d x
\end{equation}
and let $u(x) = - \cos(x)$ as suggested in the comments, with $f(x)= 1$. The above substitution formula gives
\begin{equation}
I = \int_{-\cos(0)}^{-\cos(\pi)} dx = \int_{-1}^1 d x = 2
\end{equation}
which is the correct result.
There is no contradiction with this counterexample because in the counterexample, the invalid substitution is $u(x) = \sin(x)$. It would imply $f(u) = \pm\frac{u}{\sqrt{1-u^2}}$ and the original integral must be split at $\pi/2$ to choose between $+$ and $-$. It does not invalidate the above formula with the continuity condition on $f$.

The result is correct, because the function $x \mapsto \sin(x) \cos(x)$ is $\pi$-periodic.
However, this was more of a lucky strike here, because the sub is not appropriate.
Another (bad) example would be the integral $\displaystyle\int_0^\pi \sin(x) \mathrm{d}x$. If you set $u = \sin(x)$, then you get $\mathrm{d}x = \dfrac{\mathrm{d}u}{\sqrt{1 - u^2}}$, and after substitution, you end up with $\displaystyle\int_0^0 \dfrac{\mathrm{d}u}{\sqrt{1 - u^2}} = 0$, which is clearly false. So something went wrong somewhere.
The main problem, in both cases, is that the inverse sine function is only defined to give you outputs in the "rightmost" part of the trigonometric circle (in other words, the image of $\sin^{-1}$ is $[-\pi/2, \pi, 2]$).
So, since in your example, your range is $[0, 2\pi]$, it messes up as soon as $x$ goes beyond $\pi/2$.
That's why you need to be careful when making trig subs. To do it properly, you'd have to split your intervals in parts where $\sin$ is injective, so that you can get a well defined inverse.


Find the smallest number $\alpha$, such that for all $x,y,z$ $\alpha(x^2-x+1)(y^2-y+1)(z^2-z+1)\ge(xyz)^2-xyz+1$ Find a smallest number $\alpha$, such that for all $x,y,z$ (not all of which are positive) inequality
$$\alpha(x^2-x+1)(y^2-y+1)(z^2-z+1)\ge(xyz)^2-xyz+1$$

My work so far:
Let $f(t)=t^2-t+1$. Then $f(t) \ge \frac34$.
If $x=0, y=z=\frac12$, then $$\alpha\ge \frac{16}9$$

In the starting formulation the answer is $\frac{16}{9}$.

If $x$, $y$ and $z$ are non-positives so after replacing $x\rightarrow-x$...
we need to prove that
$$\frac{16}{9}(x^2+x+1)(y^2+y+1)(z^2+z+1)\geq x^2y^2z^2-xyz+1$$

which is true for all non-negatives $x$, $y$ and $z$ because we'll prove now that even
$$\frac{16}{9}(x^2+x+1)(y^2-y+1)(z^2-z+1)\geq x^2y^2z^2+xyz+1$$.

If $x\leq0$, $y\leq0$ and $z\geq0$ we need to prove that
$$\frac{16}{9}(x^2+x+1)(y^2+y+1)(z^2-z+1)\geq x^2y^2z^2-xyz+1$$
for non-negatives $x$, $y$ and $z$, which follows from
$$\frac{16}{9}(x^2+x+1)(y^2-y+1)(z^2-z+1)\geq x^2y^2z^2+xyz+1$$
again.

Now we'll prove it:

$$\begin{align}
&16(x^2+x+1)(y^2-y+1)(z^2-z+1)-9(x^2y^2z^2+xyz+1)
\\&\phantom{aaa}=(16(y^2-y+1)(z^2-z+1)-9y^2z^2)x^2+(16(y^2-y+1)(z^2-z+1)-9yz)x
\\
&\phantom{aaaaa}+16(y^2-y+1)(z^2-z+1)-9
\\&\phantom{aaa}
\geq\left(16\left(\frac{3}{4}y^2+\left(\frac{y}{2}-1\right)^2\right)\left(\frac{3}{4}z^2+\left(\frac{z}{2}-1\right)^2\right)-9y^2z^2\right)x^2+
+(3y\cdot3z-9yz)x
\\&\phantom{aaaaa}+
16\left(\frac{3}{4}+\left(\frac{1}{2}-y\right)^2\right)\left(\frac{3}{4}+\left(\frac{1}{2}-z\right)^2\right)-9
\\&\phantom{aaa}\geq0
\end{align}$$


Prove $f|_{U_0}$ is $m$-to-$1$ except at $z_0$. Let $f$ be analytic on a domain $U$, $z_0\in U$, and $w_0=f(z_0)$. Suppose that $\mbox{ord}_{z_0}(f-w_0)=m\in\mathbb N$. Prove that there is an open set $U_0$ with $z_0\in U_0\subset U$ such that $f^{-1}(w_0)\cap U_0=\{z_0\}$ and $f^{-1}(w)\cap U_0$ contains exactly $m$ elements (without repetition) for $w\in f(U_0)\setminus\{w_0\}$. This means that $f|_{U_0}$ is $m$-to-$1$ except at $z_0$.

Proof: First, since $\mbox{ord}_{z_0}(f-w_0)=m$, when writing $f-w_0$ as its Laurent series, then $a_m \ne 0$, and $a_n=0$ for all $n<m$ (this is the definition of $\mbox{ord}_{z_0}(f-w_0)=m$, which essentially says $m$ is the minimum power actually appearing in the Laurent series with a nonzero coefficient.) Then we can write
\begin{align*}
f(z)-w_0&=\sum_{n=m}^\infty a_n(z-z_0)^n \\
&=a_m(z-z_0)^m+a_{m+1}(z-z_0)^{m+1}+\cdots
\end{align*}
This also means $z_0$ is either a removable singularity or a pole of $f-w_0$. Since $m\in \mathbb{N}$, we know $z_0$ is removable and after removing the singularity, we say that $z_0$ is a zero of $f-w_0$ (of order $m$). Also, it means there exists a holomorphic function $g$ in the disk $D(z_0,r)$ with $g(z_0) \ne 0$ such that $f(z)-w_0=(z-z_0)^m g(z)$. Would this be correct so far?

We have just finished talking about the Open Mapping Theorem and Inverse Mapping Theorem, so it's could be possible this question is an application of one of those.

Actually this claim is included in some proof of the open mapping theorem, for example, see W. Rudin's Real and Complex Analysis, Thm 10.23.

Your discussion is correct. Let's start from $f(z)-w_0=(z-z_0)^m g(z)$, where $g(z)$ is holomorphic in $D(z_0,r)$ and never vanishes.

Claim: There exists a holomorphic function $h(z)$, such that $g(z)=(h(z))^m$ in $D(z_0,r)$.
Proof: Consider $g'/g$ which is holomorphic in $D(z_0,r)$. Hence we can find some $G$ holomorphic in $D(z_0,r)$, such that $G'=g'/g$. Hence, we can choose one $G$ such that $\exp G=g$. Thus, $h(z)=G(z)/m$ is what we want.

Now we have $f(z)-w_0=((z-z_0)h(z))^m$. Choose a small neighborhood $V$ of $z_0$ such that $(z-z_0)h(z)$ is one-to-one in $V$, by the open mapping theorem, since the derivative of that is not zero at $z_0$. Now the conclusion follows from the fact that $z\mapsto z^m$ is $m$-to-$1$.

As you said, you can write $f(z)-f(c)=(z-c)^ng(z)$ where $g(c)\neq 0$ is holomorphic where $f$ is. Since $f$ is holomorphic, we can restrict ourselves to a nbhd $N$ where $f$ attains the value $f(c)$ only once, by the identity theorem.

Since $g(c)\neq 0$ there is a neighborhood $V\subseteq N$ of $c$ where $g$ is nonzero. In such neighborhood you can pick an $n$-th root $q$ of $g$; i.e. $q$ holomorphic with $q^n=g$.

If you let $h=(z-c)q$ then $f(z)-f(c)=q(z)^n$, and $h'(c)=q(c)$ and since $q(c)^n=g(c)\neq 0$; $h'(c)\neq 0$. This means there is further a nbhd $U$ where $h:U\to h(U)$ is biholomorphic, i.e. we have written $f(z)=f(c)+h(z)^n$ with $h$ biholomorphic, all this in $U$. Since $h(c)=0$, the open mapping theorem guarantees there is an open ball $B=B(0,r)$ inside $h(U)$. Let $W=h^{-1}(B)$, so that $\tilde h=r^{-1}h:W\to B(0,1)$. Note then that we can write $f$ has a series of compositions $t\circ p \circ \tilde h$ where $p(z)=z^n$ and $t(z)=r^nz+f(c)$.

The end result is we have factored $f$ into a biholomorphism $\tilde h:W\simeq B(0,1)$, a linear biholomorphism $t:B(0,1)\to B(f(c),r^n)$ and the usual $n$-fold cover of the unit disk (minus the origin!) by itself $z\mapsto z^n$.

This is known as the local normal form of a holomorphic mapping.


Probablity of picking up 3 vowels from the word "MATHEMATICS" Each of the letters in the word "MATHEMATICS" is on a letter tile in a
  bag. Foool picks three without replacement. what is the probability
  that he will get all vowels?

My approach,

The number of ways to the letters of the words "MATHEMATICS" can be arranged 3 at a time is coefficient of $x^3 $ $$ 3! \times  (1+x)^5 \times \left(1+x+\frac{x^2}{2}\right)^3$$ which is $399$

The number of arrangements of 2A's, 1I and 1E taken 3 at at a time is coefficient of $x^3$ in  $$ 3! \times  (1+x)^2 \times  \left(1+x+\frac{x^2}{2}\right)$$ which is $12$.

Then the required probability is given by $\frac {12}{399} $ but apparently this is not the right answer. What exactly I am missing here?

How many ways are there to pick 3 letters? How many ways are there to pick 3 vowels? By my count, the right answer should be $$\binom{4}{3}/\binom{11}{3}=\frac{4}{165}.$$

We obtained  vowels in a word =5 (a ,e i,  o ,u )
    And numbers of word = 11

A/Q        the probability of picking up a vowel = 5/ 11  ans


See my solution from 2006  here:

https://artofproblemsolving.com/community/c6h22937p427220

I have a direction which may be fruitful but didn't work out all cases. Rewrite the inequality as:
$$
\frac{x^3}{2x^2+y^2}-\frac{x}{3}+\frac{y^3}{2y^2+z^2}-\frac{y}{3}+\frac{z^3}{2z^2+x^2}-\frac{z}{3}\ge0
$$

Let $f(t)=\frac{1-t^2}{2+t^2}$, then the inequality is equivalent to: $xf(a)+yf(b)+zf(c)\ge0$, where $a=y/x, b=z/y $ and $c=x/z$.

Case 1. Assume that $a,b,c \in (0,2]$. Note that $f(t)\ge \frac{1-t}{2}$ for each $t \in (0,2]$. Then we have $$xf(a)+yf(b)+zf(c) \ge \frac{x}{2}(1-\frac{y}{x})+\frac{y}{2}(1-\frac{z}{y})+\frac{z}{2}(1-\frac{x}{z})=0$$

Case 1 is settled.

Case 2 $x\ge y\ge z$ and $x\ge 2z$ remains to be settled.

Case 3 $z\ge y\ge x$ and $z\ge 2y$ OR $y\ge 2x$ remains to be settled.


Find $\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{1}{2n}\right)\ldots\left(1+\dfrac{1}{2^{n-1}n}\right)$ Find $\displaystyle\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{1}{2n}\right)\left(1+\dfrac{1}{4n}\right)\ldots\left(1+\dfrac{1}{2^{n-1}n}\right)$.

(This is not my homework. One of my friends gave this to me.)

Apply the GM-AM inquality to get:

$$\begin{align}
\prod_{j=1}^n\left(1+\frac1{2^{j-1}n}\right)&\leq\left(1+\frac1{n^2}\frac{1+\cdots+2^{n-1}}{2^{n-1}}\right)^n\\
&\leq\left(1+\frac2{n^2}\right)^n
\end{align}$$

Since last expression tends to $1$, the limit is $\leq1$

Since each factor is greater than $1$, the limit must be $1$.

Isn't it $1$? The fractional expressions inside the parentheses are zero as $n\to \infty$.

$$\lim\limits_{n\to\infty}\frac{1}{n}=0$$
$$\lim\limits_{n\to\infty}\frac{1}{2n}=0$$
$$\vdots$$
$$\lim\limits_{n\to\infty}\frac{1}{2^{n-1}n}=0$$

Consequently the answer is $1$


Fixed point in one-to-one function on $\mathbb{R}$ Given that $f:[a,b]\to[a,b]$ is a real continuous one-to-one function, and that neither $a$ or $b$ are fixed points of $f$, show that there exists a fixed point in $(a,b)$.

  Proof:

Since $f(a)\ne a$ and $f(b)\ne b$, and $f$ is one-to-one on $[a,b]$, by IVT, $\exists c_1, c_2 \in (a,b)$ such that $f(c_1)=a$ and $f(c_2)=b$. Since $a$ and $b$ are boundary points, $\exists d_1, d_2\in (a,b)$, with $d_1\ne d_2$ (WLOG, let $d_1 < d_2$), such that $f'(d_1)=f'(d_2)=0$. Because of this and since $f$ is continuous, $\exists k\in (d_1,d_2)$ such that $\left| f'(k)\right|=1$. This implies that $f(x)$ coincides with $g(x)=x$ at least once, which implies that a fixed point exists in $(a,b)$.

I think that rigour suffers someplace in this proof. Would appreciate some feedback and suggestions.

I don't really understand your point about $d_1, d_2$, but I'd say $f(x) = a+b-x$ is a counterexample. Plus, you don't even know if $f$ is differentiable.

Consider now $g(x) =f(x) - x$. $f$ is a continuous bijection, so it's monotone, and you can then easily check that $f(a) =b$ and consequently $f(b) =a$, so $g(a) > 0, g(b) < 0$ and you just need to apply Bolzano.


Solve $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{(1+\log x+\log y)^2}$ Solve $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{(1+\log x+\log y)^2}$

I tried solving this differential equation, I got stuck after a few steps
I got
$(a+\log(xy))^2d(xy) = xdx$
I don't know what to do after this

If you write $$\frac{dy}{dx}+\frac{y}{x} = \frac{1}{(1+\log(xy))^2}$$ and substitute $v=xy$, then wonderful things happen.  Then $$\frac{dv}{dx} = x\frac{dy}{dx} +y$$ Plug stuff in and the equation becomes $$\frac{1}{x}\frac{dv}{dx}=\frac{1}{(1+\log v)^2},$$ separable.

With $u = xy$, the differential equation becomes
separable.


$1+2+3... =-\frac{1}{12}$ - Question regarding this So, I'm not a big expert in this subject but I know $1+2+3...=-\dfrac{1}{12}$ isn't to do with 'real' maths but it's all to do with the zeta function; however I was watching a maths video and the equation:

$$
\frac{x(x+1)}{2}
$$

... is actually a perfect equation for the series $1+2+3...$ etc. where $x$ represents $n$ in a series and $y$ is the sum of the series up to $n$. So, you can conclude that:

$$
\sum^{n}_{i=1}1+2+3...=\frac{x(x+1)}{2}
$$

However, this is where it gets weird; as you have probably guessed, the roots of the equation is $x=0,-1$  but if I want to find the integral of the roots from $-1$ to $0$ which is under the $x$ axis, I get the following:

$$
\int_{-1}^{0} \frac{x(x+1)}{2}\:dx=-\frac{1}{12}
$$

So, my question is why is this the case; what connection is there between the value of the integral under the $x$ axis that this graph has compared to the summation of the series?

Link to Desmos graph for more clarity

Your result for this integral is just a consequence of a general formula wich gives the value of the zeta function at negative integer. Integrating Faulhabert* (or maybe bernoulli's, i'm not sure) formula between -1 and 0 leads to the formula.

*The formula is the one that gives the n-th partial sum of k-th power of integers in term of polynomials of degree k+1.


Why isn't this valid? If I have an equation, say $x^2 + 9x-93=0$, why can't I just take the derivative on both sides to get $2x = -9$ and therefore $x = -\dfrac{9}{2}$. What is the reason behind why what I did is wrong?

Your equation only holds for certain values of $x$. If you have two functions, places where they are equal are not necessarily places where their derivatives are equal.

If you have an identity, though, you can take the derivative of both sides and keep it valid. For instance: $$\frac{d}{dx}(\sin^2(x) + \cos^2(x)) = \frac{d}{dx}1$$

$$2\sin x \cos x - 2 \cos x \sin x = 0$$

$$0 = 0$$

Differentiation needs two variables that have a mutual dependence.  You have only one variable.

Note the more complete way to describe differentiation is differentiation with respect to a variable.

That last bit is what you cannot do here.  $x$ is related to no other variable.

The expression $x^2+9x-93=0$ is equivalent to stating that x has two possible values ( the roots of that expression.

So it's equivalent to saying something like :

$$x=constant$$

Now it's a trap I fell into myself to say that we can differentiate this and get $1=0$, but in fact there is nothing to differentiate with respect to.  All we have is a statement of equality and, again, no dependence.

Why fall into this trap ?  Because we start out with a mindset that we should be differentiating something and because that sexy $x$ looks beguilingly like a variable.  But it's just a no-good cheap constant under that variable make-up.

Be wary of such disguises.


$U$ bisected by all hyperplanes $\implies$ $U$ symmetric? Let $U \subset \mathbb{R}^n$ be a bounded open convex set, such that every hyperplane passing through the origin divides $U$ into two sets of equal volume ($n$-dimensional Lebesgue measure?).

  Does it follow that $U$ is symmetric with respect to the origin, i.e. $U = -U$?

I'm fairly sure this holds for $n=2$: If two distinct hyperplanes (lines) through the origin divide $U$ into four sets $A, B, C, D$, then we have
\begin{align*}
\text{vol}(U)/2 &= \text{vol}(A)+\text{vol}(B) = \text{vol}(C)+\text{vol}(D) \\
                &= \text{vol}(B)+\text{vol}(C) = \text{vol}(D)+\text{vol}(A)
\end{align*}
Hence $\text{vol}(A) = \text{vol}(C)$, $\text{vol}(B) = \text{vol}(D)$. The boundary of $U$ is a simple closed curve, and a limiting argument using this equality of volumes implies that opposite points on this curve ("opposite" meaning the line between them passes through the origin) have the same distance from the origin. Therefore the boundary of $U$ is symmetric, so $U$ is symmetric.

For $n > 2$, if $U$ is divided by $n$ hyperplanes, then opposite volumes are not necessarily equal, so the above approach does not apply.

(Leaving as CW since it does not resolve the question, but just indicates some possible points of view.)

Walsh functions

The fact that the argument that the OP gave works in 2 but higher dimensions is related to properties of the Walsh functions. In $n$ dimensions with $n$ hyperplanes we divide the body into $2^n$ regions. The regions can be numbered by binary numbers as follows: the $k$th bit of the number is $1$ if it the region is "above" the $k$th hyperplane, and $0$ if the region is "below" the $k$th hyperplane.

Now let $V$ be the vector of length $2^n$ whose entries are the volumes of the corresponding regions, ordered according to the numbering above. The reflection condition gives $n$ equations of the form
$$ A_k \cdot V = 0 \tag{W1}$$
where $A_k$ is a vector of length $2^n$, whose entries take value $\pm1$ corresponding to whether the corresponding region lies "above" or "below" the $k$th hyperplane.

It is clear that each $A_k$ is one of the Walsh functions for $2^n$ dimension.

The method that works in 2 dimensions as outlined in the OP is the claim that the equation (W1) implies
$$ (1,0\ldots, 0, -1) \cdot V = 0 \tag{W2}$$

And this is true in 2 dimension since of the four Walsh functions, namely
$$ (1,1,1,1) \\ (1,1,-1,-1) \\ (1,-1,-1,1) \\ (1,-1,1,-1) $$
the first and third are orthogonal to $(1,0,0,-1)$ and the second and fourth are exactly the ones corresponding to $A_1$ and $A_2$ in (W1). Using that the Walsh functions form an orthogonal bases for $\mathbb{R}^{2^n}$ we see that this allows us to reconstruct $(1,0,0,-1)$ as a superposition of $A_1$ and $A_2$.

In $>2$ dimensions this is no longer true, simply from a counting argument. $(1,0,0,\ldots,0,-1)$ is orthogonal to exactly half of the Walsh functions, so you need knowledge of exactly half of the Walsh functions to reconstruct it. However, for $n > 2$ we have $2^{n-1} > n$ so (W2) cannot be derived as a consequence of (W1).

A related problem

Instead of considering the volume of a convex body, we can consider the following set-up:

  Problem 2 Let $f:\mathbb{S}^{n-1} \to \mathbb{R}$ be a Lebesgue-integrable function defined on the sphere. Given a hyperplane $\Pi$ through the origin, we denote by $H_\pm$ the two hemispheres. Suppose that for every $\Pi$ it is true that $\int_{H_+} f \mathrm{d}\sigma = \int_{H_-} f \mathrm{d}\sigma$, is it true that $f$ is an even function? ($f(-x) = f(x)$)

A positive answer to the above would give a positive answer to the question in the OP, since the volume for the convex body can be computed by performing the radial integral first and then integrating over the sphere, which reduces it to a question like  the above.

Of course, for Lebesgue integrable functions the claim is not true, at least not without replacing "$f$ is even" by "$f$ is even almost everywhere."

We note then the proof given in the OP also solves the $n = 2$ case of Problem 2 if we assume $f$ is continuous; this extends the result from "convex" bodies to "star-shaped bodies" such that the distance from the boundary to the center is a continuous function on the sphere.

Harmonic analysis

One can replace the Walsh functions approach by a harmonic analysis approach, considering Problem 2. Suppose $f$ is $L^2$ integrable on the sphere, we know that it can be decomposed into spherical harmonics. Using the symmetry properties of the spherical harmonics, Problem 2 would be resolved if we can show that the spherical harmonic coefficients for $f$ vanishes whenever angular momentum parameter $\ell$ is odd.

One method to approach this is to show that each spherical harmonic can be reconstructed from the hemispherical integrals by superposition. More precisely, let $v\in \mathbb{S}^{n-1}$ and let $\chi_v(x) = \mathrm{sgn}(x\cdot v)$, this function is positive on one hemisphere and negative on the other. Our goal then is to reconstruct the spherical harmonics by finding for each harmonic $Y$ a corresponding $\rho$ such that
$$ Y(x) = \int \rho(v) \chi_v(x) \mathrm{d}v \tag{H1}$$
This is the continuous version of what happened in the Walsh functions section.

It turns out that in 2 dimensions solving (H1) is pretty easy: to reconstruct the harmonic $\sin(k \theta)$ for $\theta\in [0,2\pi)$ we can let $\rho$ be $\frac1{4k} \cos (k\theta)$. But as to whether this can be done in higher dimensions: I have no idea.


Why can you find a $k$-clique in polynomial time, but determining if there is a $k$-clique is NP-complete? You can find a $k$-clique in $n^k$ time by examining all possible sets of vertices of size $k$.  So why is it NP-complete to determine if there is a clique of size greater than $k$?  It looks like you can solve it in polynomial time if $k$ is a constant.

Indeed we could resolve the $k$-clique problem for fixed $k$ by inspecting $\binom{n}{k}$ subgraphs.  However, the number $k$ is considered an input to the problem too.

  In the $k$-clique problem, the input is an undirected graph and a number $k$, and the output is a clique of size $k$ if one exists (or, sometimes, all cliques of size $k$). -- Wikipedia

If e.g. $k=\lfloor n/2 \rfloor$, then $\binom{n}{k}$ increases exponentially (not polynomially) with the input size.

I think finding a k-clique is simply O(m+n). If we know k, we can store a list of all the nodes which have a degree of atleast k-1. Then for every node in that list, we check if all of its neighbors or atleast n-1 of its neighbors (depending on whether degree of the node is n-1 or > n-1) have degree n-1 or not. Using adjacency lists, this step takes O(m+n). So total complexity is O(m+n). It's linear.


Prove inequality and when does equality hold? Prove that for any real positive numbers $a$ and $b$
$$\left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right) \geq 4.$$
When does equality hold?

Hint:

$$ \;\; \text{For any two real numbers $a$ and $b$, } \;\;(ab - 1)^2 \ge 0 $$

Simplify the above expression and try to use the positivity of $a$ and $b$ to arrive at your inequality. And if you do get your inequality then the answer to your follow-up question is pretty much embedded in the hint.

Let's just multiply it out.

$ab + \frac{1}{ab} \ge 2$

Intuïtively we can see that the left expression has minimal value when a=b=1. Its value then is 4.


How to fastest approximate definite integrals I know that a definite integral is a limit of Riemann sums. So if one wanted to estimate a definite integral (because one might not be able to find an antiderivative), then one can just take enough subintervals.

I have recently learnt that one can also write the function considered as a power series (Taylor series or Maclaurin series). Then it is easy to find an antiderivative power series that one can then use to estimate the definite integral.

My question is: Which of these two methods is in general fastest?

As an example I considered $f(x) = x^3\arctan(x)$ and the integral
$$
\int_0^{1/2} f(x) dx.
$$
I get that
$$
\int_0^{1/2} f(x) dx = \sum_{n=0}^{\infty} (-1)^n\frac{(1/2)^{2n+5}}{(2n+1)(2n+5)}.
$$
The first three terms alone give an estimate of $0.0059$.

Using three left rectangles, I get $0.1211$ and the real answer is $0.00591592$. It looks like it is faster to use the Taylor series approach. Is this in general true?

Taylor series don't always converge.  The Taylor series method can be very good if the interval of integration is inside the radius of convergence of the series, otherwise it may not converge at all.

But as Yves Daoust mentioned, you should be comparing this to more sophisticated numerical methods, not to Riemann sums.  In practice, Riemann sums are almost never used for numerical integration.

Take a look at the wiki page for Monte-Carlo Integration:

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

This is how mathematica and other systems approximate definite integrals.

And it's devilishly clever.


How to evaluate these complex integrals? I am trying to learn string theory in AdS. Following this review:
 https://arxiv.org/abs/1104.2604, I have come across the following integrals when trying to calculate some OPEs:

$$
I = \int_{\mathbb{C}} \frac{ d^{2} u }{(u - w)^{2} (u - z)}.
$$

The author claims on page 137, equation (D.4), that

$$
I = - 2 \pi \frac{\bar{z} - \bar{w}}{(z - w)^{2}}.
$$

How is this possible? I have no idea how to compute this, or even how to
search for it in the literature. Any help is appreciated.

I have found a way. It is likely to be incorrect, but it reproduces all integrals of the paper.

Define

$$
I  = \int \frac{d^{2} u}{(u - w)^{2} (u -z)}
$$

and take

$$
\frac{\partial I}{ \partial \bar{z}} = - 2 \pi\int \frac{d^{2} u}{(u-w)^{2}} \delta^{2} (u-z) = - 2 \pi \frac{1}{(z - w)^{2}}.
$$

The same way, compute

$$
\frac{\partial I}{\partial \bar{w}} = -2 \pi \frac{\partial}{\partial w}\int \frac{d^{2}u}{(u-z)} \delta^{2} (u - w) =   2 \pi \frac{1}{(z - w)^{2}}.$$

It follows then

$$
I = - 2 \pi \frac{\bar{z} - \bar{w}}{(z-w)^{2}}.
$$

Notice that

$$
\frac{\partial}{\partial \bar{z}} \left[ \frac{\bar{z}}{(z-w)^{2}} \right]
=  \frac{1}{(z-w)^{2}} - \bar{z} \frac{\partial}{\partial z}\, \delta^{2} (z-w) = \frac{1}{(z-w)^{2}}
$$

because the delta multiplies the odd function $\bar{z}$.

If someone knows a way to obtain this result in any other way, please post it.


Number of ways of cutting a stick such that the longest portion is of length n We are given a stick of length $L$ (say). We make cuts such that the longest piece is of length $n$ (say) at most.
What are the minimum number of pieces we will get and in how many ways this can be done? e.g

We have a stick of length 7. We want the longest piece of the stick to be of length 3 at most.

Soln. :The minimum number of pieces is 3 and there are 6 ways to make 2 cuts :

positions: 1 4 (length of portions will be 1(0-1), 3(1-4), 3(4-7))
positions: 3 4 (length of portions will be 3,1,3)
positions: 3 6 (length of portions will be 3,3,1)
positions: 2 4 (length of portions will be 2,2,3)
positions: 2 5 (length of portions will be 2,3,2)
positions: 3 5 (length of portions will be 3,2,3)

How can this be solved for large values of $L$ and $n$?

Try L=8, n=2; that means k=4 and applying the above formula, its 5, which i think is incorrect. I cant be making 4 cuts in such a way that i get only 4 pieces of size 2 or less.

the correct answer is one (as i can do this in only one way)

The minimum number of pieces would be $k=\lceil \frac{L}{n} \rceil$. Because you want to put as much as you can in each piece to minimise the number of pieces. Therefore, you want to have each piece to have the length $n$. There can be $\lfloor \frac{L}{n} \rfloor$ piece of length $n$, plus one piece of a length less than $n$.

Having the number of pieces $k$, you may use stars and bars method to deal with the rest. First you compute the number of possible positions, while having no upper limit for any of the pieces. So, the problem is as below

$$x_1+x_2+\ldots+x_k=L-k$$

and the solution is

$$\binom{L-1}{k-1}$$

Then, you need to take some possibilities, that have lengths greater than $n$, out. To do so, you take one of $x_i$ at a time and choose its length to be $n$. There are $\binom{k}{1}=1$ ways to choose one. Then, for each $x_i$, solve the following problem.

$$x_1+x_2+\ldots+x_k=L-(k+n)$$

$$\binom{L-n-1}{k-1}$$

The final solution would be

$$\binom{L-1}{k-1}-k\binom{L-n-1}{k-1}$$


When does the normality of the family of derivatives imply that the family itself is normal? Let $\mathcal{F}$ be a family of analytic functions in a domain $G$. Denote by $\mathcal{F}':=\{ f' \mid f \in \mathcal{F}\}$. It does not suffice to know that $\mathcal{F}'$ is normal to ensure that $\mathcal{F}$ is normal (namely take $\mathcal{F} = \{ n \mid n \in \mathbb{N}\}$ as a counterexample) but someone assured me that if you assume at some fixed point $z_0 \in G$, if you know that $f(z_0) = 0$ for all $f \in \mathcal{F}$ and that $\mathcal{F}'$ is normal, then $\mathcal{F}$ is normal. But I can't see how to do it.

A friend tried something with line integrals to show that the $f$ are locally bounded on compact subsets, but he only considered compact sets that were the image of a path, which I don't think is sufficient. Which characterization of normality should I try (local boundedness? convergence in the space of functions?)? Does anyone have an idea?

Let $\mathcal{F}$ be a family of analytic functions in a (connected) domain $G$
and assume that

$\mathcal{F}':=\{ f' \mid f \in \mathcal{F}\}$ is normal
and
$\{ f(a)  \mid f \in \mathcal{F}\}$ is bounded for some
$a \in G$.

Then $ \mathcal{F}$ is normal.

Part 1:
First we prove this for the special case that $G = D$ is an open disk:
Let $(f_n)$ be a sequence in $\mathcal{F}$. $\mathcal{F}'$ is normal,
so there exists a subsequence $f_{n_k}$ such that $f_{n_k}'$ is locally uniformly convergent
in $D$. $\{ f_{n_k}(a)  \mid f \in \mathcal{F}\}$ is bounded and has
a convergent subsequence, so by taking
another subsequence we can assume that
$$
 f_{n_k}' \to g \text{ in } D \, , \quad f_{n_k}(a) \to w
$$
Now define for $z \in D$
$$
 G(z) = w + \int_{[a, z]} g(t) \, dt
$$
where $[a, z]$ is the straight line from $a$ to $z$. $G$ is holomorphic in
$D$ and $G' = g$, therefore
$$
  f_{n_k}(z) - G(z) = f_{n_k}(a) - w + \int_{[a, z]} (f_{n_k}'(t) - g(t)) \, dt
$$
If $K \subset D$ is compact then choose a closed disc $B$ such that
$a \in B$ and $K \subset B \subset D$. Then $[a, z] \subset B$ for all $z \in K$ and therefore
$$
| f_{n_k}(z) - G(z)| \le | f_{n_k}(a) - w | + \text{diam}(D) \sup_{t \in B} | f_{n_k}'(t) - g(t)|
$$
and that converges to zero uniformly on $K$.

Part 2:
In order to generalize this to an arbitrary domain $G$ we define
$$
A := \{ z \in G \mid \mathcal{F} \text{ is normal in a neighborhood of } z  \} \, .
$$
The aim is to show that  $A = G$ by proving

$A$ is open in $G$.
$G \setminus A$ is open in $G$.
$A$ is not empty.

Now (1) follows directly from the definition of $A$. Part 1 shows that $a \in A$, this is (3).

To prove (2) assume that $b \in G \setminus A$. Let $D$ be a disk
with center $b$ which is contained in $G$.
We'll show that $D \subset G \setminus A$.

Assume that there exists a $c \in D \cap A$. Then $\mathcal{F}$
is normal in a neighborhood of $c$ and that implies that
$\{ f(c)  \mid f \in \mathcal{F}\}$ is bounded.
(Otherwise $|f_n(c)| \to \infty$ for some sequence in $\mathcal{F}$,
but $(f_n)$ has a subsequence which converges in particular at
the point $c$.)
From part 1 it follows that $\mathcal{F}$
is normal in $D$ in contradiction to the assumption that $b \in G \setminus A$.

This proves that $ G \setminus A$ is open.

Part 3:
$G$ is connected, therefore from (1, 2, 3) above it follows
that $A = G$, i.e. every $z \in G$ has a neighborhood $U(z)$
such that $\mathcal{F}$ is normal in $U(z)$.

Since $G$ is the countable union of compact subsets, and each
compact subset is covered by finitely many $U(z)$, a classical
diagonal argument shows that $ \mathcal{F}$ is normal in $G$.
(Compare If a family of meromorphic functions is normal near each point in a region, then it's normal in the region.)

I may have worked it out. If anyone finds an issue, let me know!

Let $\{f_n\}$ be a sequence in $\mathcal F$. It suffices to show that $\{ f_n\}$ has a subsequence which converges uniformly on compact subsets of the domain $G$. Let $\{f_n'\}$ be the corresponding sequence of derivatives. Since $\mathcal F'$ is a normal family, we have that $\{ f_n'\}$ has a convergent subsequence, $\{f_{n_k}'\}$ which converges uniformly on compact subsets to a function $g \in \mathcal F'$. By the definition of $\mathcal F'$, we must have that $g = f'$ for some $f \in \mathcal F$.

Now we'd like to show that $\{f_{n_k}\}$ converges uniformly on compact sets to $f$. But since the $f_{n_k}' \rightarrow f'$, applying the Fundamental Theorem of Calculus and continuity of the integral we have $\int f_{n_k}' \rightarrow \int f'$ so $f_{n_k}(z) + C_{n_k} \rightarrow f(z) + C$. But since each $f_{n_k}$ and $f$ is in $\mathcal F$, we have that $f(0) = f_{n_k}(0) = 0$ and so the $C_{n_k} \rightarrow C$. Thus $f_{n_k} \rightarrow f$ uniformly on compact subsets, and so since the sequence $\{f_n\}$ was arbitrary, $\mathcal F$ is a normal family.

I'm worried about playing with the constants of integration (basically the FTC step in general) but I think it should work.


For $\lim_{x→1}(2−1/x)=1$, finding δ, such that if $0<|x−1|<δ, then |f(x)−1|<0.1$ I don't know what I did wrong. Can anyone point out my mistake? The problem is:

  For $\lim\limits_{x\to1}(2-1/x)=1$, finding $\delta$, such that if $0<|x-1|<\delta$, then $|f(x)-1|<0.1$

Here is what I did:

Since $|f(x)-1|=|2-1/x-1|=|1-1/x|=|x-1|/x<0.1$, then $|x-1|<0.1x$.
We know that  $0<|x−1|<δ$, so $\delta=0.1x=x/10$

The answer sheet says $\delta=1/11$. I don't understand it.

$\delta$ must be a constant determined by $.1$ and not dependent on the value of $x$.

(How could you chose an $x$ so that $|x -1| <\frac x{10}$ is we don't know what $x$ is in the first place?)

So you need to find a constant $\delta$ so that $|x - 1| < \delta \implies |2-\frac 1x-1|=|1-\frac 1x| < .1$.

To find that $|1 - \frac 1x| < .1$ so

$.9 < \frac 1x < 1.1$ so

$\frac {10}{11} < x < \frac {10}{9}$

$1 - \frac 1{11} < x <  1 + \frac 1{9}$

$-\frac 1{11} < x -1 < \frac 1{9}$

$|x - 1| < \min(\frac 1{11},\frac 1{9}) = \frac 1{11}$.

So that is a (and the maximum possible) delta.

NOW it's easy to chose and $x$ so that $|x-1| < \frac 1{11}$.  And when we do, it will logically (by following all those steps backwards) must be that $|2 - \frac 1x - 1| < .1$


How do people perform mental arithmetic for complicated expressions? This is the famous picture "Mental Arithmetic. In the Public School of S. Rachinsky." by the Russian artist Nikolay Bogdanov-Belsky.

The problem on the blackboard is:
$$
\dfrac{10^{2} + 11^{2} + 12^{2} + 13^{2} + 14^{2}}{365}
$$

The answer is easy using paper and pencil: $2$.
However, as the name of the picture implies, the expression ought be simplified only mentally.

My questions:

Are there general mental calculation techniques useful for performing basic arithmetic and exponents?
Or is there some trick which works in this case?
If so, what is the class of problems this trick can be applied to?

I wasn't smart enough to group them the way Barry Cipra did, so I just did the arithmetic: $169 + 121 = 290$; adding $144$ makes $434$; adding $196=200-4$ makes $630$, adding $100$ makes $730$; wow, that's twice $365$, so the answer is $2$.


Computing parametrizations for a differentiable $2$-manifold with boundary Consider the following subset of $\mathbb{R}^{3}$
\begin{equation}
C=\{(x,y,z)\in\mathbb{R}^{3}\:|\:0\leq x\leq 1,\:0\leq y\leq 1,\:z=x^{2}+y^{2} \}.
\end{equation}
Intuitively, this looks like a differentiable $2$-manifold with boundary, the boundary being the points at $z=1$.

I am looking to parametrize $C$ using the least possible charts. I think that I'll need two.

I have tried over and over to find such charts, but since I have to get the boundary, the problem got a little hard. I could come up with something like (for most of the manifold, missing a point)
\begin{equation}
f_{1}\colon[0,1)\times[0,1)\to\mathbb{R}^{3}\qquad\qquad f_{1}(u,v)=(1-u,\:1-v,\:1-u^{2}-v^{2})
\end{equation}

But this one has the issue that $[0,1)\times[0,1)$ is not an open set of $\mathbb{H}^{2}=\{(x,y)\in\mathbb{R}^{2}\:y\geq 0\}.$

Any tips? Thank you all.

EDIT

I feel like my failure could be in assuming that I could do that with only two charts. Since the boundary is kind of circular, it doesn't really look like $\mathbb{H}^{2}$.

One chart is enough. Take $I=[0,1]$ and $$f:I^2\rightarrow\mathbb{R}^3:(u,v)\mapsto (u,v,u^2+v^2)$$.


Loops when drawing constantly changing angles and lines. I start by drawing a line of 1 unit on the $x$ axis. I turn left (from the perspective of an ant on the line) by an angle of $\alpha$ and I draw a second segment of length $u$ from my endpoint of the first segment. I then turn another angle $\alpha$ and then draw a $3^{rd}$ segment of length 1, etc. alternating between segments of length $1,u,1,u,...$ and rotating by a fixed angle each time. I am interested in when does this process returns to the origin and how long does it take for it to return when it does.

We work in the complex plane. Let $z=e^{i\alpha}$. Then the path is:
$$1+uz+z^2+uz^3+z^4+uz^5+\cdots$$
If $1+uz=0$, meaning $u=1$ and $\alpha=\pi$, then we return to $0$ after $2$ steps. Otherwise, assume that $1+uz\neq0$.

Suppose the path returns to $0$ after $2N$ steps for $N>1$. Then we can solve for this:
$$0=(1+uz)\sum_{k=0}^{N-1}z^{2k}$$
Divide out $(1+uz)$:
$$0=\sum_{k=0}^{N-1}z^{2k}$$
This is true if and only if $z^{2N}=1$ and $z^2\neq1$. In other words, $\alpha$ is a multiple of $\pi/N$, but not $0$ or $\pi$. Note that $u$ is irrelevant!

Tackling an odd number is harder. Suppose the path returns to $0$ after $2N+1$ steps. Then:
$$0=z^{2N+1}+(1+uz)\sum_{k=0}^{N-1}z^{2k}=z^{2N+1}+(1+uz)\frac{1-z^{2N}}{1-z^2}$$
$$0=z^{2N+1}-z^{2N+3}+1+uz-z^{2N}-uz^{2N+1}$$
That's a really messy equation, and I'm not sure if there's a simple class of solutions.

The named system of equations solving for $n=2k,\alpha,u$ should be:
$$
u \sum_{i=0}^k \cos((2k+1)\alpha)+\sum_{i=0}^k \cos(2k\alpha)=0\\
u \sum_{i=0}^k \sin((2k+1)\alpha)+\sum_{i=0}^k \sin(2k\alpha)=0
$$
Hence the solution for a given $k$, if exists, is given by:
$$
u = -\frac{\sum_{i=0}^k \cos(2k\alpha)}{\sum_{i=0}^k \cos((2k+1)\alpha)}\\
 \frac{\sum_{i=0}^k \cos(2k\alpha)}{\sum_{i=0}^k \cos((2k+1)\alpha)} = \frac{\sum_{i=0}^k \sin(2k\alpha)}{\sum_{i=0}^k \sin((2k+1)\alpha)}
$$
Which as you see, would have a closed form for some circular symmetry values for $\alpha$....


Groups reluctant to have infinite subgroup Is there a group with only one infinite subgroup?

Yes.

If you want a group with no proper infinite subgroups then take a Tarski monster group.
If you want a group with precisely one infinite proper subgroup then take a Tarski monster group $T$ and consider the direct product with a cyclic group of prime order, for example, $G=T\times C_2$.

A Tarski monster group is a finitely generated, infinite simple group where every proper, non-trivial subgroup is cyclic of order a fixed prime $p$. Hence, when you form $T\times C_2$ the only infinite subgroup you obtain is the copy of $T$. The proof that Tarski monster groups exist is highly non-trivial.

The circle group $\mathbb{T}$ has only one infinite subgroup. The torsion subgroup of $\mathbb{T}$ is the $n$th roots for unity for all $n$.


Path connected spaces with same homotopy type have isomorphic fundamental groups I was try to understand the following theorem:-

Let $X,Y$ be two path connected spaces which are of the same homotopy type.Then their fundamental groups are isomorphic.

Proof: The fundamental groups of both the spaces $X$ and $Y$ are independent on the base points since they are path connected. Since $X$ and $Y$ are of the same homotopy type, there exist continuous maps $f:X\to Y $ and $g:Y\to X$ such that $g\circ f\sim I_X$ by a homotopy, say, $F$ and $f\circ g \sim I_Y$ by some homotopy, say $G$. Let $x_0\in X$ be a base point. Let
$$f_\#:\pi_1(X,x_0)\to \pi_1(Y,f(x_0))$$ and $$g_\#:\pi_1(Y,f(x_0))\to \pi_1(X,g(f(x_0)))$$ be the induced homomorphisms. Let $\sigma$ be the path joining $x_0$ to $gf(x_0)$ defined by the homotopy $F$.

After that the author says that $\sigma_\#$ is a isomorphism. obviously $\sigma_\#$ is a homomorphism but I could not understand how it becomes a isomorphism.

Can someone explain me please. thanks for your kind help and time.

I suppose $σ$ is a path from $x_0$ the a point $x_1$ and the map $\sigma_\#$ is defined as:
$$σ_\#:\pi_1(X,x_0)\to π_0(X,x_1)\\σ_\#([p])=[σ\cdot p\cdot\barσ]$$ where $σ⋅p$ is the path which first traverses the loop $p$ and then $σ$, and $\bar σ$ is the reversed path.

This is a bijection because it has the inverse $\barσ_♯$ which we see by calculating $$\barσ_\#σ_\#([p])=[\barσ⋅σ⋅p⋅\barσ⋅σ]=[\barσ⋅σ]⋅[p]⋅[\barσ⋅σ]=[p]$$ as the homotopy class of $[\barσ⋅σ]$ is trivial.  Similarly, we get $σ_\#\barσ_\#([p])=[p]$

$f_{\#}:\pi_1(X,x_0) \to \pi_1(Y,f(x_0))$ is an isomorphism since $f$ is an homotopy equivalence; moreover $\pi_1(Y,f(x_0)) \simeq \pi_1(Y,y_0)$ since $Y$ is path-connected.


Finding a bijective function from $\mathbb R \setminus \{0\}$ to $\mathbb R$ Basically I wanna show that $|\mathbb R \setminus \{0\}|=|\mathbb R|$, so I need to find a bijective function $f: \mathbb R \setminus \{0\} \rightarrow \mathbb R$.

What I tried:

$\mathbb R \setminus \{0\} = (-\infty,0) \cup (0, \infty)$

And playing with that. I showed $\mathbb R \setminus \{0\}  \sim (-1,0) \cup (0,1) $. What can I do to continue?

Thanks!

Try this: $$f(x)=\begin{cases} x & x\notin \mathbb{N}\\ x+1 & x\in \mathbb{N}\end{cases}$$

Note: for this function, the domain is $\mathbb{R}$, and $\mathbb{N}=\{0,1,2,3,\ldots\}$.  This function is the inverse of the one sought by OP.

Each has the same cardinality. Thus, each is in bijection with the corresponding cardinal number. The composition of bijection is a bijection, so the composition bijects your two sets.


Proving unsatisfiability with propositional resolution I'm having trouble understanding how to use the resolution rule to prove if a statement is satisfiable or unsatisfiable.

I watched this course lecture on propositional resolution and unsatisfiability (linked with a time stamp to skip to the part in question):
"5.4 - Resolution Method - Propositional Resolution - [Introduction To Logic] By Michael Genesereth" https://www.youtube.com/watch?v=IpixwNupBIs#t=2m21s

For those that don't want to watch the video, my understanding is that he is saying that for a statement in conjunctive normal form (CNF) we can test whether the statement is satisfiable by repeatedly applying the simplifying resolution rule:
If there is a pair of clauses of the form
$$(a \vee X)$$
$$(\neg a \vee Y)$$
this can be replaced by a single clause
$$(X \vee Y)$$

The claim is that a statement is unsatisfiable if and only if repeated application of this rule leads to an empty clause.

But this doesn't make sense to me. I think there are some important details that are left out. For instance in the example given in the video, consider the statement with the clauses

$(p \vee q)$
$(p \vee \neg q)$
$(\neg p \vee q)$
$(\neg p \vee \neg q)$

By inspection, this is unsatisfiable.  But let's find this using the resolution rule, following along with the video...

Applying the resolution rule on pair 1,2 gives: $p$

Then applying the resolution rule on pair 3,4 gives $\neg p$

So now we just have the clauses $p, \neg p$, which after applying the resolution rule one more time gives the empty clause.

That is how the example goes in the video. But look what happens if we apply the rule in another order:

Applying the resolution rule on pair 1,4 gives: $q \vee \neg q$

Then applying the resolution rule on pair 2,3 gives $\neg q \vee q$

So now we are just left with two clauses which are tautologies.  So we simplified the statement to just $\top$.  So by this order, it claims the statement is satisfiable.

Clearly something is wrong.  What am I misunderstanding?  Or is the course lecture leaving out something crucial?

You can see :

Melvin Fitting, First-Order Logic and Automated Theorem Proving (1990), Ch.3.3 Propositional Resolution, page 45-on

for a complete treatment of the proof system based on the Resolution Rule.

The proof system "expands" the set of cluases, applying the Resoultion Rule in order to add e new clause (a disjunction) to the original set of clauses. The rule has the property that "preserves stisfiability", i.e. the new set of clauses is satisfiable iff the original one is.

Thus, if after iterated applications of the rule the empty clause is produced, being it unsatisfiable (it is another way of writing $\bot$, i.e. a contradiction) this is enough to conclude that also the original set of clauses is unsatisfiable.

I think this algorithm should help you . You need to add the resolved clause to your set of clauses everytime you resolve and so your set of clauses grow. Also before adding, you need to make sure that when you resolve you do not end up generating the clause which you might have already generated before in the process of resolution or( which already exists in your set of clauses). Look the algorithm pictures

Unsatisfiability algo pic

Saturation Algo pic


$S_n$ acting transitively on $\{1, 2, \dots, n\}$ I am reading Dummit and Foote, and in Section 4.1: Group Actions and Permutation Representations they give the following example of a group action:

  The symmetric group $G = S_n$ acts transitively in its usual action as permutations on $A = \{1, 2, \dots, n\}$. Note that the stabilizer in $G$ of any point $i$ has index $n = | A |$ in $S_n$ (My italics)

I am having trouble seeing why the stabilizer has index $n$ in $S_n$. Is this due to the fact that we have $n$ elements of $A$ and thus $n$ distinct (left) cosets of the stabilizer $G_i$?

And how would I see this if I were to work with the permutation representation of this action? The action is not very clear to me.

We have that $\mbox{Stab}_{S_n}(i)\cong S_{n-1}$ for every $i\in \{1,\ldots,n\}$.  (Why?)

   $S_n$ contains all possible permutations on $n$ vertices.  The set of those permutations which fix $i$ are simply the ones that permute the set $\{1,\ldots,n\}\setminus i$ in every possible way.  Thus  $\mbox{Stab}_{S_n}(i)\cong S_{n-1}$.

Now, $|S_{n-1}|=(n-1)!$, so $[S_n:\mbox{Stab}_{S_n}(i)]=\ldots?$

   $$[S_n:\mbox{Stab}_{S_n}(i)]=\frac{n!}{(n-1)!}=n.$$

And how big is $A$?

$\sigma \in \textrm{ Stab } (i)$ for $1\le i \le n$ iff $\sigma $ fixes $i.$ You just have to count the number of permutations that fix $i$ - working in the usual order, there are $n-1$ choices for the image of the first element in the domain, $n-2$ for the second, and so on, so that $|\textrm{ Stab } (i) | = (n-1)!.$ Apply Lagrange's Theorem.


Definition of the convolution with tempered distributions and Schwartz function In the book where I'm studying there is the following exercise.

If $x \in \mathbb{R}^n$, $\varphi \in \mathcal{S}(\mathbb{R}^n)$ and $u \in \mathcal{S}'(\mathbb{R}^n)$ we define $(u \ast \varphi)(x)=\langle \tau_x \widetilde{\varphi} , u \rangle$, where we place $(\tau_x \widetilde{\varphi})(y):=\widetilde{\varphi}(y-x):=\varphi(x-y)$. Then

(a) $(u \ast \varphi)(x)$ is continuous with respect to $u \in \mathcal{S}'(\mathbb{R}^n)$, with respect to $\varphi \in \mathcal{S}(\mathbb{R}^n)$, and with respect to $x \in \mathbb{R}^n$

(b) $u \ast \varphi$ is a tempered distribution.

(c) If $\psi \in \mathcal{D}(\mathbb{R}^n)$, we have
\begin{align*}
\langle \psi, u \ast \varphi \rangle = u \left ( \int_{\mathbb{R}^n} \psi(x) (\tau_x \widetilde{\varphi})(\cdot) dx \right )
\end{align*}
and extend this identity to case $\psi \in \mathcal{S}(\mathbb{R}^n)$.

HINT: To prove (b), check that $|(u \ast \varphi)(x)| \leq C(1+|x|^2)^N$
by proving an estimate $q_N(\tau_x \varphi) \leq 2^N(1+|x|^2)^N q_N(\varphi)$.

Note that for me there are these definitions. Let $\mathcal{S}'(\mathbb{R}^n)$ the topological dual space of $\mathcal{S}(\mathbb{R}^n)$.
We have that the mapping $u \in \mathcal{S}'(\mathbb{R}^n) \longmapsto v=u_{|\mathcal{D}(\mathbb{R}^n)} \in \mathcal{D}'(\mathbb{R}^n)$ is linear and one-to-one  because convergence in $\mathcal{D}(\mathbb{R}^n)$ implies convergence in $\mathcal{S}(\mathbb{R}^n)$, and $u_{|\mathcal{D}(\mathbb{R}^n)} \in \mathcal{D}'(\mathbb{R}^n)$ determines uniquely $u \in \mathcal{S}'(\mathbb{R}^n)$. Then a distribution $v \in \mathcal{D}'(\mathbb{R}^n)$ is the restriction of an element $u \in \mathcal{S}'(\mathbb{R}^n)$ if and only if there exist $N \in \mathbb{N}$ and a constant $C_N>0$ such that

\begin{align*}
|u(\varphi)| \leq C_N q_N(\varphi)=C_N \sup_{x \in \mathbb{R}^n; |\alpha| \leq N} (1+|x|^2)^N |D^\alpha \varphi(x)| , \forall  \varphi \in \mathcal{D}(\mathbb{R}^n)
\end{align*}

where $q_N(\varphi)$ are seminorm that make $\mathcal{S}(\mathbb{R}^n)$  a Fréchet space. The elements of $\mathcal{S}'(\mathbb{R}^n)$ or their restriction to  $\mathcal{D}(\mathbb{R}^n)$ are called tempered distributions.

To prove (a). I thought can be done with an application of the closed graph theorem, proving that

$\tau_a \cdot : \varphi \in \mathcal{S}(\mathbb{R}^n) \longmapsto \tau_a(\varphi)=\varphi(x-a) \in \mathcal{S}(\mathbb{R}^n)$

$\widetilde{\varphi} \cdot : \varphi \in \mathcal{S}(\mathbb{R}^n) \longmapsto \widetilde{\varphi}=\varphi(-x) \in \mathcal{S}(\mathbb{R}^n)$

are continuous with respect to convergence in $\mathcal{S}(\mathbb{R}^n)$. Is it correct, no?

Do you have any idea to the point (b) and (c)?

Note that (b) and (c) I tried to show in a different way, as here Tempered distributions and convolution

Thanks for any help

Let us start by proving the inequality in the hint:
For fixed $x\in\mathbb{R}^{n}$, we have
$$
\partial_y^\alpha(\tau_x\tilde{\psi})(y) = \partial_y^\alpha(\psi(x-y)) = (-1)^{|\alpha|}(\partial^\alpha\psi)(x-y)
$$
and
$$
(1+|x-z|^2)^N \leq (1+2|x|^2+2|z|^2)^N \leq 2^N(1+|x|^2+|z|^2)^N \leq 2^N (1+|x|^2)^N(1+|z|^2)^N.
$$
Hence
$$
\sup_{|\alpha|\leq N,y\in\mathbb{R}^{n}} (1+|y|^2)^{N} |\partial^\alpha (\tau_x\tilde{\psi})(y)| = \sup_{|\alpha|\leq N,z\in\mathbb{R}^{n}} (1+|x-z|^2)^{N}|\partial^\alpha \psi(z)|
\leq 2^N(1+|x|^2)^N q_N(\psi).
$$
Now observe that since $\tau_x\tilde{\psi}\in\mathcal{S}(\mathbb{R}^{n})$, for all $u\in\mathcal{S}'(\mathbb{R}^{n})$ and all $N\in\mathbb{N}$, there is $C_N>0$ such that
$$
|\psi*u(x)| = |\langle \tau_x\tilde{\psi},u\rangle |\leq C_n q_N(\tau_x\tilde{\psi}) \leq 2^N (1+|x|^2)^N C_N q_N(\psi).
$$
Since $u\ast\psi$ is continuous and therefore locally integrable, we get that $u\ast\psi$ is a distribution since, for $\varphi\in\mathcal{D}(\mathbb{R}^{n})$
$$
\langle \varphi, u\ast\psi\rangle = \int_{\mathbb{R}^{n}} \varphi(x)(u\ast\psi)(x)\;{\rm d}x.
$$
Using the above estimate for $N=1$, we get
$$
|\langle \varphi, u\ast\psi\rangle| \leq \int_{\mathbb{R}^{n}} |\varphi(x)||(u\ast\psi)(x)|\;{\rm d}x \leq
2 C_1 \int_{\mathbb{R}^{n}} (1+|x|^2) \frac{(1+|x|^2)^{n+2}}{(1+|x|^2)^{n+2}} |\varphi(x)|\;{\rm d}x
\leq 2 C_1 q_1(\psi) q_{n+2}(\varphi) \int_{\mathbb{R}^{n}} \frac{{\rm d}x}{(1+|x|^2)^{n+1}} \leq C q_{n+2}(\varphi),
$$
where $C:=2 C_1 q_1(\psi)\int_{\mathbb{R}^{n}}\frac{{\rm d}x}{(1+|x|^2)^{n+1}}$.

Now the claim should follow from the fact that $\mathcal{D}(\mathbb{R}^{n})\subset\mathcal{S}(\mathbb{R}^{n})$ is dense.

I am also learning this stuff at the moment so I'll let you know what I think. I don't see how you would use closed graph theorem, in fact, you would need either $\mathcal{S}=\mathcal{S}(\mathbb{R}^d)$ to be a Banach space or compact Hausdorff. I do not believe either conditions are satisfied ($\mathcal{S}$ is normable or compact, at least I don't think so). Instead, it might just be easier to prove (a) directly. Say you want to prove (a) with respect to $x$. Just take a sequence $x_n$ in $\mathbb R^n$ that converges to $x$, say. Then

$$\lim_{n\to \infty} \langle\tau_{x_n} \tilde{\varphi},u\rangle= \langle\lim_{n\to\to \infty}\tau_{x_n} \tilde{\varphi},u\rangle=\langle \tau_x \tilde{\varphi},u\rangle$$
where the last inequality holds once you prove $\tau_{x_n}\varphi\to \tau_{x}\varphi$ in $\mathcal{S}$, which shouldn't be too hard to prove. For continuity with respect to $u$, you want to prove that the function $u\mapsto u*\varphi(x)$ is continuous for each pair $(x,\varphi)$. This follows by definition of convergence of tempered distribution since $u_n\to u$ as tempered distribution implies that
$$\lim_{n\to \infty} \langle\tau_{x} \tilde{\varphi},u_n\rangle= \langle\tau_{x} \tilde{\varphi},u\rangle.$$
I'm too lazy to do (a) w.r.t $\varphi$. But I would imagine it's the same sort of argument.

For (b) and (c), there is a proof in Classical Fourier Analysis by Grafakos, and it's Theorem 2.3.20. in the 3rd edition. Unfortunately I don't completely understand his proof, especially the Riemann sum part.


Can all surfaces be turned inside out? I saw this YouTube video claiming that spheres, double torues, triple toruses, etc. can all be turned inside out, but what about other surfaces? Are there any surfaces that can't be turned inside out?

In mathematical terms, your question may be restated as "does every closed orientable manifold of dimension two admit an orientation reversing diffeomorphism?" As noted by others, the question does not make sense for non orientable surfaces, and so we have to limit ourselves to orientable spaces.

A key phrase here is the classification of surfaces, which puts surfaces into three categories: the sphere, tori with several holes (surfaces of genus $g$), and several copies of the projective plane glued together. It can be shown that any closed $2$-manifold must be of this type. So in short yes, since the orientable surfaces are precisely the tori with several holes, all orientable surfaces admit an orientation reversing diffeomorphism.

It should be noted that the answer is negative for higher dimensional orientable spaces. For example, even dimensional complex projective spaces do not admit orientation reversing diffeomorphisms.

Edit: I am limiting myself here to closed orientable surfaces. Non-closed surfaces are not as nice and I'm not sure there's a clear answer, perhaps someone else can fill in those details.

For the class of surfaces called non-orientable, there's no such thing as inside or out.

This class includes the famous Moebius Strip and Klein Bottle, for instance.


Measures of expanding maps of the circle and their coding It is well known that the dynamics of linear examples $f(x)=mx(mod1)$ for natural $m\geq 2$ is semi conjugated to the full shift on the space of one-sided sequences of digits $\{0,…,m−1\}.$ Is it true that the semi conjugated map measures theoretical isomorphism for all measures? If not, under what (minimum assumption on maps) that is.
Thanks in advance.

You can start by noting the obvious:
The semiconjugacy is not a conjugacy because there are points with more than one coding.
For example, for $m=2$ these are the points in $[0,1]$ having a base-$2$ representation with infinitely many consecutive $1$'s. Note that all of them are of the form $k/2^n$ for some integers $k$ and $n$. They form a Borel set.
So you obtain an isomorphism if and only if this Borel set has measure zero.
Continuing the example, since the set is countable, it has zero Lebesgue measure. So Lebesgue measure induces a measure on the shift (in this case the Bernoulli measure with probabilities $1/2$, $1/2$) for which we have an isomorphism.
You can make a similar reasoning for any $m$.

To talk about a measure theoretic isomorphism you first need measures on both spaces.  The most classic example is if you give $[0,1]$ Lebesgue measure and give sequence space $\{0,1,\dots,m-1\}^{\mathbb{N}}$ the infinite product of the uniform measure on $\{0,1,\dots,m-1\}$.  In this case the semiconjugacy map is a measure theoretic isomorphism between the two systems.  There are also many other choices of measures you can do this with.