Groups with $|G| = p^2q$. Prove that if $p$ and $q$ are primes, then there are no simple groups of order $p^2q$. Groups with $|G| = p^2q$.

Prove that if $p$ and $q$ are primes, then there are no simple groups of order $p^2q$.

Also another question, do p and q have to be distinct for this to hold?
On top of that, I know we have to use the Sylow Theorems but I do not understand them as they are currently.

First, suppose $p<q$. Then $n_q\equiv 1\pmod{q}$, and $n_q\mid p^2$. So the only possibilities are that $n_q=1,p,p^2$. If $n_q=1$, the group is not simple. But $n_q=p$ is impossible, else $q\mid p-1$. So $n_q=p^2$. Then we have $p^2$ groups of prime order $q$, so they have pairwise trivial intersection, and this forces the $p$-Sylow subgroup to be unique.

The $q$-Sylow subgroups give a total of $p^2(q-1)$ elements of order $q$. This accounts for all but $p^2$ elements of the group, so any $p$-Sylow subgroup must be these remaining elements, hence unique.

The case where $q<p$ is quite quick. Any $p$-Sylow subgroup has index $q$. This is smallest prime dividing $|G|$, so your $p$-sylow subgroup is normal.

Either $p^2 < q$ or $q < p^2$.  Assume WLOG the former.

By Sylow I, there must be at least one Sylow $q$-subgroup.  Now by Sylow III, the total number of $q$-subgroups must be congruent to $1 \pmod{q}$, and further, the total number of such subgroups must divide $|G| = p^2q$.

The only number that satisfies these criteria is $1$, so there must be $1$ Sylow $q$-subgroup.  By Sylow II, all Sylow $p$-subgroups are conjugate to one another for a specific $p$ and there's only $1$ Sylow $q$-subgroup, then it is necessarily normal in $G$.

$\Longrightarrow G$ is not simple.


Factorization in prime elements of $\mathbb{Z}\left[\sqrt{p}\right]$ for a prime number $p$ I'm having troubles with the following problem:

Let $p$ be a prime number in $\mathbb{Z}$, and $\alpha\in\mathbb{Z}\left[\sqrt{p}\right]$ which is not a unit. Prove that $\alpha$ have a factorization in irreducible elements of  $\mathbb{Z}\left[\sqrt{p}\right]$.

At the beginning I though that that ring was a Euclidean Domain, but that fails for $\mathbb{Z}\left[\sqrt{5}\right]$. So, I don't know where to start now.

Thanks in advance.

Hint: Consider the norm $N(a+b\sqrt{p})=|a^2-b^2p|$. Prove that if $\delta$ divides $\alpha$, then $N(\delta) \le N(\alpha)$. Then use induction on $N(\alpha)$.

Or, more sophisticatedly, argue that $\mathbb{Z}\left[\sqrt{p}\right]$ is a Noetherian ring and so all ascending chains of principal ideals eventually stop.

Suppose $\alpha$ is not irreducible. Then there exist elements $a_1$ and $a_2$ that are not units such that
$$\alpha=a_1a_2$$
If $a_1$ and $a_2$ are irreducible, then we are done. Otherwise, $a_1$ is a product of two nonunit elements $a_3$ and $a_4$, so
$$\alpha = a_3a_4a_2$$
In general there is an increasing chain of ideals
$$(a_1)\subseteq (a_3)\subseteq \cdots$$
Since $\mathbb{Z}[\sqrt{p}]$ is Noetherian, this must stabilize at some element $a_{2n+1}$, which is irreducible. Thus, reindexing,
$$\alpha = a_1a_2$$
with $a_1$ irreducible. Thus you can iterate this to write
$$\alpha = a_1\cdots a_n$$
with $a_1,\ldots, a_{n-1}$ irreducible. This gives you another increasing sequence of ideals
$$(a_{n,1})\subseteq (a_{n+1,2})\subseteq \cdots$$
This must also stabilize, until you end up with a factorization into irreducible elements.


Between Groups Mean Square expectation Show that in the Normal one-way classification the Between Groups Mean Square has expectation $E(\frac {B}{k-1})=\sigma^2$.

$E(\frac {B}{k-1})=\frac {1}{k-1}E(\sum_1^k$$n_i$$(\bar{x_i}-\bar{x})^2)$=$\frac {1}{k-1}E(\sum_1^k$$n_i$$(\bar{x_i}^2-2\bar{x}\bar{x_i}+\bar{x}^2))$. And then I thought I could use the fact that $E\bar{x_i}^2=\sigma^2+\mu^2$ and $E\bar{x}^2=\mu^2+ \sigma^2/k$ and also $\sum_1^k\bar{x_i}=\bar{x}k$. But I am not sure how to deal with the $n_i$ in the sum and I can't get to $\sigma^2$ because of it. Do I need to use more definitions or it is possible to get the answer from this information?

Excuse me if I misunderstand your question, but, I think the expectation is not what you state.  If you take the simple case in which all $n_i$’s =n, then, using your terminology, $E(\frac {B}{k-1})=\sigma^2 + n\sigma_b^2$, where $\sigma_b^2$ is the between-means variance (subscript $_b$ for between-means).  This derives from a model in which an observation within a class is the sum of a between-means component and a within-class component and the variance of the latter is assumed to be $\sigma^2$.


How is the set of all programs countable? I'm having a hard time seeing how the number of programs is not uncountable, since for every real number, you can create a program that's prints out that number.  Doesn't that immediately establish uncountably many programs?

I don't know your definition of 'program', but I'm fairly sure that any program will be a finite length string of characters over some finite alphabet.  For any finite set $X$, the set $X^*$ of all finite length strings over $X$ is countable (by the same sort of argument you would use to show the rationals are countable).

Well ,why don't we define a program as a circuit rather to give us an output of one's and zero's?

That is i am saying that rather than looking at the program at the highest level only. Lets look at what a program does at the physical level. All it does its take a set of 1's and 0's as input and then give us another set of 1's and 0's as output. So, the question will now change to whether the number of sets of 1's and 0's is finite or not?

Obviously this will be again be infinite. as the cardinality of this set of 1's and 0's will be an integer and the number of integers will be infinite so, again we can say that the number of programs is indeed infinite.

Though i don't understand what is the definition of uncountable.

We normally say that the grains of rice in a pack are uncountable which is blantantly wrong as if i had the time and motivation i could definitely(very easily too i might add) count the exact number of grains in say 1 kg of rice.

Thus the concept of uncountable is in itself flawed and gives a wrong perception about things. According to me only one which is infinite can be considered as uncountable.


Does a "cubic" matrix exist? Well, I've heard that a "cubic" matrix would exist and I thought: would it be like a magic cube? And more: does it even have a determinant - and other properties? I'm a young student, so... please don't get mad at me if I'm talking something stupid.

Thank you.

P.S. I'm 14 years old. I don't know that much about mathematics, but I swear I'll try to understand your answers. I just know the basics about PreCalculus.

If we're working with three-dimensional vectors, a matrix is a $3\times 3$ array of 9 numbers. If I'm understanding your question right, you're asking whether there is something like a $3\times 3\times 3$ array of 27 numbers with interesting properties.

Yes, there is such a thing; it is called a tensor. Tensors are a generalization of both vectors and matrices:

A number is a "rank-0 tensor".
A vector is a "rank-1 tensor"; it contains $D$ numbers when we're working in $D$ dimensions.
A matrix is a "rank-2 tensor", containing $D\times D$ numbers.
Your "cubic" thing is a "rank-3 tensor", containing $D\times D\times D$ numbers.

... and so forth.

One use for a rank-3 tensor is if you want to express a function that takes two vectors and produces a third vector, with the property that if you keep any one of the arguments constant, the output is a linear function of the other input. (That is, a bilinear mapping from two vectors to one). One familiar example of such a function is the cross product. In order to completely specify such a thing you need 27 numbers, namely the 3 coordinates of each of $f(e_1,e_1)$, $f(e_1,e_2)$, $f(e_1,e_3)$, $f(e_2,e_1)$, etc. Using linearity to the left and right, this is enough to determine the output for any two input vectors.

I haven't heard of any generalization of determinants to higher-rank tensors, but I cannot offhand think of a principled reason why one couldn't exist.

The study of tensors belongs in the field of multilinear algebra. It's quite possible to get at least an undergraduate degree in mathematics without ever hearing about them. If you take physics, you'll see lots and lots of them, though.

I write to you a cubic matrix (and with thr side of lengh 2), so it has 8 elements; You can arrange them in a column A; take a second cubic matrix B with side-lengh 2; you get
A= a1, a2, ..., a8
B= b1, b2, ..., b8
a1 could be baby girl shoes, a2 = socks, ... a8 = t-shirt;
b1 could be price a a single pair of baby shoes, b2 = price of a single pair of socks,..., b8 = price o a t-shirt.
c1 = N° of baby girl shoes sold, ..., c8 = N° of t-shirts sold.
You may do the product B.C = D, where D is a cubic matrix, this way: d1 = b1 . c1, ..., d8 = b8 . c8.
In the same way you can do it with even higher "dimension". When things look complicated, make them simpler!
Enzo


Solutions to $ax^2 + by^2 = cz^2$ The integer solutions to the equation $x^2 + y^2 = z^2$ are very well studied. I'm wondering if there's any literature about the integer solutions to the equation $ax^2 + by^2 = cz^2$ where a,b,c are non-zero integers. Here are some of my questions:

Given a,b,c, is there an explicit way of writing all the solutions?
If not, could you generate all the solutions given one of the solutions?
If this is the case, are there any algorithms that are designed to look for one of these solutions?
Are there any additional conditions that these solutions must satisfy?
What happens when you're only given a and b?

Proof of Legendre's theorem on the ternary quadratic form

I like the discussion of Legendre's theorem in Cassels, Rational Quadratic Forms, Theorem 4.1, pages 80-82, as he is careful about pointing out that certain products must be quadratic residues for each prime dividing each entry $(a,b,c).$

Next, if $a,b,c$ are positive, and there is at least one nontrivial solution $(x_0,y_0,z_0)$ then we may find all rational points on the ellipse $a \alpha^2 + b \beta^2 = c$ by drawing all lines of rational slope through $(x_0/z_0, y_0/z_0).$ For each such line, there is a second rational point on the ellipse, with the slope $m$ as a parameter. Multiplying back through by the denominators gives a new integral solution to $a x^2 + b y^2 = c z^2.$ It is exactly this method that gives primitive Pythagorean triples $(2rs, s^2 - r^2, s^2 + r^2).$   I've done that a few times, i will try to find some of  my answers on this site.

Found one anyway, Generating Pythagorean triples for $a^2+b^2=5c^2$?

Found another: note that more than one (two, here) recipe may be required to get all solutions. Pythagorean triples with additional parameters

Oh, for this few variables it is good enough to take lines with rational slope, but in general, such as $a x^2 + b y^2 + c z^2 = d w^2,$ it is better to find some rational point on the ellipsoid $(x_0/w_0, y_0/w_0, z_0/w_0)$ and then use a parameter $t$ and integers, call them $p,q,r,$ and find the other ellipsoid point
$(x_0/w_0 + pt, y_0/w_0 + qt, z_0/w_0 +rt), $ where we can then eliminate $t.$

Above mentioned equation is shown below:

$ax^2 + by^2 = cz^2$

The above equation has parametric solution for $(a,b,c)=(35,10,3)$

$x=5k^2-20k-4$

$y=10k^2+20k-32$

$z=25k^2-20k+60$

For $k=1$ we get:

$35(19)^2+10(2)^2=3(65)^2$


Improper Integral $\int\limits_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx$ How can I find a closed form for the following integral
$$\int_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx?$$

I tried the partial integration, with

$\frac{d \log^2(2\sin(\pi x))}{dx} = \frac{1}{3}\log^3(2\sin(\pi x))\frac{1}{\frac{1}{2\sin(\pi x)}2cos(\pi x)\pi} = \frac{1}{3}\log^3(2\sin(\pi x))\frac{sin(\pi x)}{\pi cos(\pi x)} = \frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}$

$\int_0^{\frac{1}{2}}(2x-1)^2\log^2(2\sin(\pi x))\,dx =$

$[(2x-1)^2\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} - \int_0^{\frac{1}{2}}(4x-2)\log^2(2\sin(\pi x))\,dx =$

$[(2x-1)^2\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + [(2 - 4x)\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + 4\int_0^{\frac{1}{2}}\log^2(2\sin(\pi x))\,dx =$

$[(2x-1)^2\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + [(2 - 4x)\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + 4[\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}}$

of course it's possible I messed it up somewhere
with $\cos(\frac{\pi}{2})$ you have a $0$ in the denominator which makes it rather hard to evaluate the integral.


Shortcut to find remainder (mod) of a number to a negative power? How would one find the remainder to a question of the type $a^{-x} / b$ really fast?

Came across this situation while solving the remainder of $12^{107} / 37$

Using Fermat's little theorem we get
$12^{108} \mod 37 = 1$
which would mean the question is nothing, but the remainder of $(12^{-1} \times 12^{108} ) / 37 \implies 12^{-1} \times 1 / 37$ (Basic remainder theorem)

Is there an easy and fast method to solving this question from this step?

I did manage to solve it but with 4 additional steps, where I basically got  the numerator to a 3-digit value applying  Fermat's and basic remainder theorem  multiple times. But in a competitive time crunch exam situation, I do not think my method is ideal, there must be a way to easily solve $12^{-1} \mod 37$ fast.

It is trivial to invert a number that divides modulus $\pm 1,\,$ e.g. here $\,12\mid 37-1,\,$ i.e $\,3\cdot 12 = 37-1\,$ so $\,{\rm mod}\ 17\!:\,\ 3\cdot 12\equiv -1\,$ so $\,12^{-1}\equiv -3.\ $  Said in fraction language

$$\dfrac{1}{12}\equiv \dfrac{-36}{12} \equiv \dfrac{-3}{1}$$

This is essentially an optimzation of the extended Euclidean algorithm (when it terminates in one or two steps).  There are other ad-hoc methods that are best expressed in terms of fraction arithmetic (e.g. using factorizations, cancellation, etc), e.g. see some of my prior posts using modular fractions

What you want is the answer to
$$a\equiv 12^{-1}\pmod{37}$$
or equivalently, on multiplying through by $12$ to
$$12a\equiv 12\cdot12^{-1}\equiv1\pmod{37}$$
where $a$ is the inverse of $12$ modulo $37$.

Now by Fermat's Little Theorem we have:
$$12^{37-1}=12^{36}\equiv1\pmod{37}$$
On noting $12a\equiv 12^{36}\pmod{37}$ we have, on dividing through by $12$, that
$$a\equiv 12^{35}=(12^{5})^{7}=248832^7\equiv 7^{7}=823543\equiv34\equiv-3\pmod{37}$$
So $a\equiv-3$ is the answer.


Leibniz integration rule reference I'm looking for a nice reference for the "Leibniz integral rule".
Do you know one?

Thank you

Might be a bit late, but one reference that looks nice is

Differentiation Under the Integral Sign

Harley Flanders

The American Mathematical Monthly , Vol. 80, No. 6 (Jun. - Jul., 1973) , pp. 615-627

Published by: Mathematical Association of America

Article DOI: 10.2307/2319163

Article Stable URL: http://www.jstor.org/stable/2319163

Avaiable at jstor


Is there a way to determine exactly the difference between $N$th & $(N+1)$th prime number? So I was trying to find the time complexity of an algorithm to find the $N$th prime number (where $N$ could be any positive integer).

So is there any way to exactly determine how far $(N+1)$th prime number will be if $N$th prime number is already known ?

You're got what looks like two questions.

The time complexity of the nth prime.  In practice it is $O\big(\frac{n^{2/3}}{\log^2n}\big)$ using fast prime count algorithms.  In theory this could be lowered to something on the order of $O(n^{1/2})$.  See:

Most efficient algorithm for nth prime, deterministic and probabilistic?
https://cs.stackexchange.com/questions/10683/what-is-the-time-complexity-of-generating-n-th-prime-number
primesieve README

Does knowing P(n-1) allow one to quickly find P(n)?  Yes, in the sense that we have reduced the problem to a single nextprime call.  Once beyond trivial sizes, this is just a loop calling isprime.  One might want to use a wheel or a partial sieve to skip obvious composites.

See Cramér's conjecture for a thought on how long this would take.  In practice this is quite efficient, with inputs of thousands of digits being straightforward to calculate.  You can look up the concept of "merits" for gaps.  On average you'll find the next prime about $\log n$ away.  For large inputs, I found sieving to a distance of $30 \log n$ (30 merits) was plenty to get excellent performance with exceptionally few gaps that need to look farther.

Knowing the previous prime hasn't really given us any special information though, as Gerry pointed out.

A good method for the operation of a prime number program is to output only the primes between an upper-bound and the upper-bound - k . Then the program just appears to be jumping from one section of a list to the other on each computation. Of course required consecutive primes are obviously found within the range of the output.

Here is my previous post:

Odd numbers can be wheeled in multiplications to output only odd composite numbers. Then the odd numbers that are not output are the prime numbers.

Now each inner loop can stop at a multiplication that reaches the value of an upper-bound and the outer loop can stop at the square root of the upper-bound.

Furthermore the loops from the number 11 can increment with 2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,8,6,4,6,2,4,6,2,6,6,4,2,4,6,2,6,4,2,4,2,10,2,10 ... as continuing and repeating. That's a big performance gain because multiples of 3, 5, and 7 are removed from the sequence of odd numbers.

A prime number application really works best when outputting prime numbers between an upper bound and the upper bound - k. Then the application appears to be just scrolling sections (or jumping to sections) of a list on each computation. And in this case the loop increments are really only needed on the outer loop because a single division operation jumps over unneeded sections of the inner loop. Of course, array subscript locations can work with translations such that the same array can handle any of the segmented computations and then not use very much memory.


Advice for how to learn more advanced math for audio signal processing? I am very interested in learning about audio from a signal processing standpoint.  However, whenever I try to further my education by reading books, I get extremely frustrated because the books use all kinds of crazy math notation that makes absolutely no sense to me...

An example would be this book:

The first chapter (1.1) starts out completely clear...  A nice diagram showing analog -> digital -> analog..  Great.. got it..

Then 1.2 goes to show $X(\Omega) = \int_{-\infty}^\infty x(t)e^{-i\Omega t} dt\ldots$

Uhhh.. yeah.. sorry..  I am totally lost and frustrated at this point.

Can anyone guide me to some sort of book or online site that will help me make sense of this notation?

EDIT:  Some of your comments asked about my background.  Unfortunately, I've only got a background in algebra, and that itself is slightly shaky as I haven't done a lot with it since high school.

This is probably heretical for math.SE, but you don't need to understand that equation.  Just skim over it.  You aren't going to use it for anything anyway.

Signal processing isn't mathematically rigorous (see the intro of Dirac delta "function", for instance).  You don't actually work out integrals to find Fourier transforms.  Instead, you memorize the most common Fourier transform pairs, and learn how mathematical operations in the time domain translate to the frequency domain (multiplication ⇔ convolution, for instance), so you can represent complicated signals as a combination of simple signals that you can work with easily.

Engineering is all about applying mathematics to build practical things, and taking lots of shortcuts and simplifications in the process.  We transform to the Laplace domain and use phasors to avoid doing differential equations, converting them into polynomials and algebra.  We memorize tables of common Fourier transforms to avoid doing the integrals, etc.

Fourier transform pairs:

For instance, say you have a recording of a tuning fork at 440 Hz (a sine wave), and you want to send it over the radio at 1 MHz.  To do this, you multiply the 440 Hz sine wave with another sine wave at 1 MHz.  This is amplitude modulation.

$x(t) = \cos(2 \pi 440 t) \cdot \cos(2 \pi 1000000 t)$

You know the Fourier transform of each sinusoid is a Dirac spike (as in the above graphic), and you know that multiplication in the time domain is equivalent to convolution in the frequency domain, so you can convolve the spectra of the two sine waves to get the spectrum of the result.  Once you learn convolution, you'll know that this is just two spikes at the sum and difference frequencies: 1000000-440 and 1000000+440.  You don't actually go through the trouble of solving the integral

$X(\Omega) = \int_{-\infty}^\infty \cos(2 \pi 440 t) \cdot \cos(2 \pi 1000000 t)e^{-i\Omega t} dt$

Solving this is not trivial, but applying transform tables is.  It's more important to see in your head what's happening.

To demodulate at the other end, you multiply by 1 MHz again, producing frequency components at the sum and difference frequencies again, which are now 440 Hz, 2000440 Hz, and 1999560 Hz.  The latter two can be thrown away by filtering, which just means multiplying by 0 in the frequency domain using a rectangle function, and you're left with the original recording.  (And again, this is not mathematically rigorous; real filters are not rectangular, and calculating real filters'  actual effects mathematically can be very difficult.)

For the stuff you want to know about audio signal processing, this is sufficient.  When you get into more advanced stuff and need to know the details, you can go back and learn it in more depth.

  The relationship of formal mathematics to the real world is ambiguous. Apparently, in the early history of mathematics the mathematical abstractions of integers, fractions, points, lines, and planes were fairly directly based on experience in the physical world. However, much of modern mathematics seems to have its sources more in the internal needs of mathematics and in esthetics, rather than in the needs of the physical world.  Since we are interested mainly in using mathematics, we are obliged in our turn to be ambiguous with respect to mathematical rigor.  Those who believe that mathematical rigor justifies the use of mathematics in applications are referred to Lighthill and Papoulis for rigor; those who believe that it is the usefulness in practice that justifies the mathematics are referred to the rest of this book.  (Hamming, Digital Filters, 1998 Dover edition, page 72.)

You first need to have a fairly good background in undergraduate real variables. That's really going to be essential regardless of how deeply into the theoretical aspects you want to do. That means working your way through Rudin's Principles Of Mathematical Analysis or Charles Pugh's Real Mathematical Analysis. You need to do that before tackling anything specific in signal processing. (Of course, it goes without saying you need to be very good at basic calculus first!) You also need to be pretty good at linear algebra since so much of signal processing theory is decomposition by linear basis function approximation. An applied book like Strang will do very nicely for your purposes since it does everything carefully,but with an applied flavor.

The great book serious applied mathematicians learned Fourier analysis and integrals from for a generation was Dym and McKean's Fourier Series And Integrals. This is absolute must read if you're interested in the theory behind signal processing-it presents the basic theory very rigorously,but very concretely with a lot of applications. Another excellent book at the same level,but with more of an applied bent for physicists and engineers, is James Walker's Fourier Analysis. It's another terrific text with everything done carefully,but with lots of applications.

Lastly,one of the great unmentioned books on Fourier Analysis is the 2 book text by Tom Korner. Loads of applications,beautifully written and with lots of mathematical insight you won't find in other texts.

All those books will give you a great place to start and from there, you can begin reading texts specifically on signal processing. Good luck!


Entropy of sum is sum of entropies Having $X$ and $Y$ discrete random variables above finite set. Z is defined as $Z=X+Y$ when does the following happen:

$$H(Z)=H(X)+H(Y)$$

That is not true in general.

$$H(X,Z)=H(Z)+H(X \mid Z) = H(X)+H(Z \mid X) $$

But $H(Z \mid X)=H(X+Y \mid X)=H(Y \mid X)$.

Further, if we assume $X,Y$ are independent, then $H(Y \mid X)=H(Y)$ and

$$ H(Z)=H(X)+H(Y)-H(X \mid Z)$$

We still need $H(X \mid Z)=0$ (knowing the sum lets me know the summands). This is only true in the very special case in which $x_1+y_1 \ne x_2+y_2$ for any set of values with positive probability.
If both $X$ and $Y$ comes from the same set, then it's only true in the trivial case of a single element support.

I am afraid H(Z|X)  is not equal to H(Y|X), if X and Y are independent and Z = X+Y, because say X and Y are binary and independent, then X and Y say nothing about each other, but Z = 0 => X = 0, and Z = 2 => X = 1. So H(X|Z) is not equal to H(X|Y)=H(Y).


minimum value of $2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$ Let $\alpha,\beta$ be real numbers ; find the minimum value of

$2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$

What I tried :

$\bigg|4\cos \beta+(2\cos \alpha+3\sin \alpha)\sin \beta\bigg|^2 \leq 4^2+(2\cos \alpha+3\sin \alpha)^2$

How do I solve it ? Help me please

In
$$(2\cos\alpha+3\sin\alpha)\sin\beta+4\cos\beta$$ the parenthesised factor takes values in $[-\sqrt{2^2+3^2},\sqrt{2^2+3^2}]$. Using the largest value, the minimum of $$\sqrt13\sin\beta+4\cos\beta$$ is

$$-\sqrt{13+4^2}.$$

Justification:

$$a\cos t+b\sin t$$ is the scalar product of $(a,b)$ with the unit rotating vector $(\cos\theta,\sin\theta)$, which takes the extreme values $\pm\|(a,b)\|=\pm\sqrt{a^2+b^2}$.

We use this property twice.

Property to note : $a\cos x + b\sin x = \pm\sqrt{a^2 +b^2}$, So,

$$2\cos\alpha + 3\sin\alpha = \pm\sqrt{13}$$
Taking the minimum value of the expression,
$$-\sqrt{13}\sin\beta +4\cos\beta = \pm\sqrt{29}$$
Therefore, the minimum value of the expression is $-\sqrt{29}$.


How to prove that interval $(x-1,x]$ contains exactly one integer? The exersise says that "Show that for any real number x, the interval $(x-1,x]$ contains exactly one integer", though there's a hint which says that consider three separately cases : $x\in \mathbb{Z}$,$(x \in \mathbb{R\setminus Z})\wedge(x>0)$, $(x \in \mathbb{R\setminus Z})\wedge(x<0)$, I don't know how to write them explicitly.
My attempt are belows:
Step1: try to prove the uniqueness.
If there are two integers $t$,$s$ and $t$ is not equal to $s$. So, |s-t|>=1, but distance between every two real numbers in (x-1,x] should strictly less than 1, it's contradict, so at most one integer in (x-1,x].
Step2: try to prove the existance.
Think about set S={$s \in \mathbb{Z}$|s<=x}, due to archimedian's property, we can conclude that S is not empty.And S has an upper bound x, so due to "maximum exists in subset of $\mathbb{Z}$", S has a maximum $l \in \mathbb{Z}$. Now I try to prove that l is contained in (x-1,x]. From definition of S, we know that $l$<=$x$.Now we try to prove that $l$>$x-1$. If not, that is $l$<=$x-1$, so $l+1$>=$l$,and $l+1$<=$x$, so $l+1 \in S$, it contradicts that $l$ is the maximum.
Can anyone tell me if my proof is true? Or if there some place where need to improve?

I can not explicitly write the solution in Logic notation but I have the answer.
Existence of the integer: Suppose $x$ is not an integer. Take $[x]$ as the greatest integer function. Clearly $x-1<[x]<x$ so the existence is justified.
Uniqueness of the integer: Clearly there is no integer between $x$ and $[x]$ as then it would contradict the definition of greatest integer function and also there can not be any integer between $x-1$ and $[x]$ (why? Because if there is an integer $y$ such that $x-1<y<[x]<x$ then we have $0<[x]-y<x-y<1$, now the difference between two integers can never be non-integer! hence it is a contradiction).
This completes the proof.

For every $y,z \in (x-1,x]$ we have $|y-z|<1$ because size of this interval is $1$ and you can't take elements on it's edge. For every two integers $n,m$ we have $|n-m|\geq 1$ so it would be contradiction if they were both in this interval.


Determine whether a number is prime How do I determine if a number is prime?  I'm writing a program where a user inputs any integer and from that the program determines whether the number is prime, but how do I go about that?

For very small numbers (less than a million), trial division is the best way: divide by 2, 3, 5, and so on until the square root of the number.  If you find a factor, the number is composite; otherwise, the number is prime.

For larger numbers there are better methods, but choosing which one depends on how much work you're willing to put into the program.  It is now known that there are no BPSW-pseudoprimes below $2^{64}$, so if you can write that test (see here for details) then you have a very quick test for primality.

If you only need to test up to $2^{32}$, you can simply check if the number is a 2-strong pseudoprime.  If so, test if it's one of 2314 exceptions (this can be done in 12 or 13 steps with a binary search); if the test fails or it's an exception, the number is composite, otherwise prime.  (You can go higher than $2^{32}$ if you're willing to build an appropriate table of exceptions.)

For larger numbers, the work is usually split into two parts: determining with high probability (say, 99.99999999%) that the number is prime, then actually proving that it is.  What type of proof depends on the form and size of the number.

Construct a list of primes up to the maximum input $n$. (Construct using prime sieves.) Then, trial division for every such prime $\le \sqrt{n}$. This should work.


Weakly closed implies sequentially closed Another problem involving the weak topology:

Let $X$ be a normed space and $A \subset X$ weakly closed. Then $A$ is sequentially closed, that is: If $(x_n) \subset A$ and $x_n \xrightarrow{w}x$, then $x \in A$.

I know this characterisation is also used as definition of weakly closed. So I guess it should be easy to prove. Yet I have trouble doing so. ;(

I tried proving it directly and towards a contradiction without success. I'm afraid my problem is a lack of understanding of weak closedness. I know how weakly open sets are generated, but this doesn't give me a concrete representation them, or of a weakly closed subset.

However, I know that $A$ is also closed with respect to the norm of $X$. Weak convergence of $(x_n)$ to $x$ means $f(x_n) \rightarrow f(x)$ for every $f \in X^*$. But it doesn't give me any statement related to the norm convergence of $(x_n)$, at least not that I know of. So the norm-closedness of $A$ doesn't really help. I could also find out that norm closedness is not sufficient for closedness with respect to weak convergence - I think convexivity has to be added to make the implication valid, is that correct?

So of course I have to fail if I weak convergence with norm convergence alone.

I've tried to work with balls around $f(x)$, too (for a contradiction). But again, the continuousness of $f$ only gives me control over $|f(x_n)-f(x)|$ if I have some bound for $||x_n-x||$. And I want it the other way around. I have a feeling that this way is wrong because it would need some implication between weak and norm convergence that I know isn't there...

It makes me mad that the proof should be rather simple, yet I'm not able to do it. Some hint please!? :(

Instead of saying "$A$ is sequentially closed", the conclusion of the problem really should say that $A$ is "weakly sequentially closed" or "sequentially weakly closed."

The problem is asking to show that closed in the weak topology implies sequentially closed in the weak topology.

A solution can be given that does not use the specifics of the weak topology at all. For any topological space, closed implies sequentially closed.

Let $X$ be any topological space and let be $A$ any closed subset of $X$. We need to show that $A$ is sequentially closed. Let $x \in X$ and $(x_n) \subseteq A$ be such that $x_n \to x$. We need to show $x \in A$. Let $U$ be an arbitrary  open neighbourhood of $x$. Since $x_n \to x$, we have $x_n \in U$ for all but finitely many $n$, and so $U$ contains at least one point of $A$. Since $U$ was an arbitrary open neighbourhood of $x$, this proves $x$ is in the closure of $A$. Since $A$ is closed, $x$ is in $A$.

Can't you just say that since $A$ is weakly closed then $A$ is strongly closed, then $A$ is sequentially closed? Cause I think every weakly closed set is strongly closed and in metric spaces (we are in a normed space which means in a metric space) (strongly) being closed set is equivalent to being sequentially closed.


Calculating the lateral area of a cone On my book, it says

  ...thinking of the lateral surface as swept out by revolving a generator (i.e. slant height) about the axis: the lateral area equals the length of this generator multiplied by the distance traveled by its midpoint, $ s*2pi*(\frac{1}{2}r) = pi*r*s $ ...

However, I can't understand why midpoint is involved. I looked up two proofs online (one of them), but both of them are different from my book's proof, and actually they seem more reasonable to me. Is my book wrong?

Update: my question is that how do you get the left hand side of the equation $ s*2pi*(\frac{1}{2}r) = pi*r*s $. Namely, why do you use $ \frac{1}{2}r $ and what's the point of using midpoint?

I'm reading the same book and, no surprise, had the same question. I've given it some thought and here's my best guess:

Think about what's happening when we revolve the generator (or slant length, s) around the axis. At the very top of the cone sits a single point, which doesn't move at all. At the very bottom of the cone (on its edge, which is one of the endpoints of the generator) sits a point which rotates around the entire base of the cone. The distance traveled by this point is 2*pi*r.

Now, how far does the average point on the generator travel on its trip around the axis? Intuition suggests it's the average of the distance traveled by each endpoint. That equals (0 + 2*pi*r)/2, which equals (1/2)(2*pi*r).

So, the points composing the generator move an average distance of pi(r) as they "sweep out" the lateral area of the cone. This implies that the lateral area is equal to the length of the generator multiplied by pi(r), which as shown above is the distance traveled by the generator's midpoint.

One thing to note: the author says that "the lateral area equals the length of this generator multiplied by the distance traveled by its midpoint." He then asserts (without proof) that the midpoint of the generator lies at the point on the cone where the cross-sectional radius is equal to 1/2 the radius of the cone's base. See if you can convince yourself if this assertion is justified.

See cone is formed by joining the end of the sector of a circle.

Let us take a circle of radius small r and the angle between the two radius is theta(this is the sector) so the area of the sector is theta divided by 360 degrees multiplied by pi*r^2

Next we should eliminate the theta, for that I am taking the circle at the base of the cone of radius r'(this circle is formed by joining the ends of the sector). Now the circumference of the circle at the base of the cone is (2*pi*r')

This circumference will be equal to that sector's circumference that is theta/360 multiplied by (2*pi*r)

Now equate,

2*pi*r' = theta/360 multiplied by (2*pi*r)

Now we can get theta angle to be 360*r' divided by r .

Now substitute the theta value in the first equation

It will be 360*r'/r divided by 360 multiplied by pir^2
This will be equal to pir*r'

Here r is equal to the slant height (s) that implies pisr'

Hence proved


Difference between a set and a class I don't understand the difference between a set and a class. The definition which I studied is:
 A set $A$ is a class such that there exists a class $B$ such that $A \in B$.
 But isn't it always true as we can have a collection of all sub classes of $A$, which will again be a class.  Something like power set, but in this case I should call it a power class. Somebody please clarify.

Set theory is a mathematical theory, like any other mathematical theory it has a "universe", and the axioms and properties are required to hold within the universe.

In set theory, the objects in the universe are called sets.

But since we don't live inside that universe, but rather work from the outside (in one way or another) we are free to talk about collections of elements from that universe. We can talk about "all the rational numbers which are negative or their square is strictly smaller than $2$", and we can talk about the collection of all sets which have a certain property.

What is confusing in the case of set theory is that sets come to model, mathematically, the notion of a collection of mathematical objects. So if classes are also collections are mathematical objects, why aren't classes sets?

As it turns out, not every collection which we can define form a set. That's the essence of Russell's paradox.

So we limit ourselves to some collections of mathematical objects, and require that they satisfy certain axioms. Classes are collections which need not be in our universe, and therefore don't have to satisfy the axioms of set theory.

Let me reiterate this point. Sets are elements of the model of set theory, and they have to satisfy the axioms, e.g. the axiom of power set. Classes are collections of elements from a model of set theory, but they don't have to correspond to any element in the model, and they don't have to obey the axioms. Just like a real number can be seen as a set of rational numbers, but it doesn't mean that it can be written as a quotient of two integers.

Sets are strongly regulated by some axioms and only those objects that satisfies these axioms are called sets.

Classes are also used in mathematics. For example, the class of all sets includes any object having the property of being a set. Classes are used e.g. in category theory to make a mathematical object of a property.

But it's a good thing in general to be able to strictly define a set.


what is the fractal dimension of the henon map? I have some questions about the Henon map that are not clear for me. I have seen that the correlation dimension of the Henon map is approximately 1,21, is that measure similar to its fractal dimension?

Also, can the Henon map could be considered like a multifractal? I am using a program that measures the multifractal spectra and the maximum value of Q, sometimes it is 1,21, when I rerun the program it can be 1,27 and when I run it again it can be also 1,31, why is that change?

That multifractal program measures the multifractal spectra based on the figure of the henon map

Any help?

The maximum value of of the f(alpha) spectrum occurs at the value of alpha for which Q=0.  The alpha(Q) spectrum is numerically derived from the generalized dimension spectrum, D(Q).  The D(Q) spectrum is calculated from the slope of a log-log plot of experimental data (related to the sum over some probability distribution raised to the power Q vs boxsize).

The value of the maximum can change depending on which portion of data is linearly fit, as the slope of the line will be different.


Complex-number inequality $| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$ Let $z_1, z_2 \ldots z_m$ be complex numbers, $m \in \mathbb{N}$. Can anybody tell me how to prove the following inequality?

$| z_1 z_2 \ldots z_m - 1 | \leq e^{|z_1 - 1| + \ldots + |z_m - 1|} - 1$

In case you're wondering, this is asserted without proof in a paper by Von Neumann, about infinite tensor products of Hilbert spaces.

The inequality in question bounds how far you can get from $1$ by multiplying several complex numbers that may individually not be far from $1$. So it makes sense to try to derive a bound for the product of just two complex numbers, and then proceed by induction.

Lemma: Suppose $\lvert z_1 - 1\rvert = \alpha_1$ and $\lvert z_2 - 1\rvert = \alpha_2$. Then $\lvert z_1 z_2 - 1\rvert \le (1 + \alpha_1)(1 + \alpha_2) - 1$.

Proof: We know $\alpha_1\alpha_2 = \lvert z_1z_2 - z_1 - z_2 + 1\rvert$. By triangle inequality on the three points $z_1 z_2$, $z_1 + z_2 - 1$, and $1$, we have
$$\begin{align}
\lvert z_1 z_2 - 1\rvert &\le \lvert z_1 z_2 - z_1 - z_2 + 1\rvert + \lvert z_1 + z_2 - 2\rvert \\
&\le \alpha_1\alpha_2 + \alpha_1 + \alpha_2 \\
&= (1 + \alpha_1)(1 + \alpha_2) - 1.
\end{align}$$

Now, for several numbers,
$$\begin{align}
\lvert z_1 z_2 \cdots z_m - 1\rvert &\le (1 + \alpha_1)(1 + \alpha_{2,\ldots,m}) - 1 \\
&\le (1 + \alpha_1)(1 + \alpha_2)(1 + \alpha_{3,\ldots,m}) - 1 \\
&\vdots \\
&\le (1 + \alpha_1)(1 + \alpha_2)\cdots(1 + \alpha_m) - 1,
\end{align}$$ where $\alpha_{2,\ldots,m}$, for example, is my hopefully transparent abuse of notation to denote $\lvert z_2\cdots z_m - 1\rvert.$ Finally, since $1+x \le e^x$ for real $x$, the desired inequality follows.

EDIT: This answer is wrong.

It all boils down to the inequality $|xy-1| \leq |x-1|+|y-1|$, which I expect to be true.

Given this inequality, prove by induction that

$|z_1\cdots z_m - 1| \leq |z_1 - 1| + \cdots + |z_m - 1|$.

Now use $e^x \geq 1 + x$.


Let $z=(x)^y$, what is $\frac{ \partial ^2z}{\partial x \partial y}$? Let $z=x^y$, what is $\displaystyle \frac{\partial ^2z}{\partial x \partial y}$ ?

My answer is
$ \displaystyle \frac{ \partial z}{ \partial y} = x^y \ln x$
$ \displaystyle \frac{ \partial ^2z}{\partial x \partial y} = \frac{x^y}{x} + \ln x(yx^{y-1})$

The choices are listed below but I don't seem to get the right answer. I am not sure if I have done it wrongly.

(A) 1
(B) $x/y$
(C) $(x)^y/x + y^2{(x)^y}^{-2}$
(D) $(x)^y/x$
(E) $y(x)^y$

You are correct.
$$z=x^y$$

\begin{equation}
\frac{\partial^2z}{\partial x\partial y}=\frac{\partial}{\partial x}x^y\ln x=\left(\frac{\partial}{\partial x}x^y\right)\ln x+x^y\frac{\partial}{\partial x}\ln x=yx^{y-1}\ln x+x^{y-1}=(y\ln x+1)x^{y-1}
\end{equation}

Take logarithms:

$$\log z=y\log x$$

Then:

$$\frac{1}{z}\frac{\partial z}{\partial y}=\log x$$

From there you can compute the further derivatives.


Is linear map f * f = f injective? Let's say that $V$ is Vector Space and $f : V \to V$ is linear map with $f \circ f=f$

I have to define what is $ker(f)$ of this function.

Can we say that this function is also injective and so $ker(f) = {0} $?

Hint: $f$ injective and $f \circ f=f$ iff $f=\text{id}$.

If f•f=f and f is a linear transformation then there are two possibilities for f, either f is zero transformation or identity transformation.
If f is zero transformation then it is not injective and Ker(f)=V.
If f is an identity transformation then it is one one and ker(f)={0}.


Combination is odd or even I saw in other posts people asking how to know if a combination is odd or even, but in my case the formula for the combination is a little bit different:

$$C(n,k) = \frac{(n+k-1)!}{k!(n-1)!}$$

Can I check if the result is odd or even without calculate the value ?
I have the factorials pre-calculated with mod $10^9+7$, and if I use the values in the formula, results in a wrong answer.

Max value of N = 664579

Max value of K = 10000000

Actually what you are considering is $C(n+k-1,k)$ so you should just apply any formula you have for "usual combination" to this one.


Calculate the length of the sides of a triangle from the area and the angle I need to find the length of the sides of a triangle.  I have an angle and the area of the triangle.

I have the answer but I don't know how to figure it out so it doesn't help.

The area of the triangle is 18cm2 - the angle is 23 degrees.

Can someone explain how I can use the area and the angle to calculate the length of the sides?

Thanks

Not enough informations. Assume that $\widehat{ABC}=\widehat{A'B'C'}$ and $AB\cdot BC = A'B'\cdot B'C'$.

Then the triangles $ABC, A'B'C'$ have the same area and share an angle, but they are not necessarily congruent:

You obtain the 3 sides of a triangle by using this five equations (for a General triangle):

$a^2+b^2-2abcos(\alpha(a,b))=c^2$ (1)

$b^2+c^2-2bccos(\alpha(b,c))=a^2$ (2)

$a^2+c^2-2accos(\alpha(a,c))=b^2$ (3)

For the area: $A = \frac{1}{2}absin(\alpha(a,b))$ (4)

You also have the angle sum condition: $\alpha(a,b) + \alpha(b,c) +\alpha(a,c)= 180°$ (5)

From These you compute all other angles and the 3 sides of the triangle.


determining if a set is well ordered set consider following question

I am able to easily see that set of positive rationals is not well ordered set
but I have difficulty coming to conclusion with this

set of positive rationals with denominator less than 200

Is this well ordered set or not ?

I think it is well ordered set .If it is not, could someone give me non empty subset that is not bounded from below

another question regarding well ordered sets I read from wiki article that "the standard ordering ≤ of the integers is not a well ordering, since, for example, the set of negative integers does not contain a least element."I got that

Another relation for well ordering the integers is the following definition: x ≤z y iff (|x| < |y| or (|x| = |y| and x ≤ y)). This well order can be visualized as follows:

0 −1 1 −2 2 −3 3 −4 4 ...

how can this be well ordered set, when we use argument same as above

consider set of negative integers -1 -2 -3 -4 -5  but this is not bounded from below, so how can this be well ordered set

Hint:

Reduce all these rational numbers to denominator $200!$.

ThIs is the conclusion for the set of rational numbers  less than or equal to  200 not been well order..

Let $X= \{u: u\text{ is a rational number and }0 \ll u\ll200 \}$ and

$E=\{1/n : n \in \Bbb N\setminus 0\}$ which is a subset of $X$.

Hence there exist no first element --> not well ordered.


Probability: People sitting in a row (linear arrangement) Question: Ten persons are seated at random in a row. What is the probability that a particular couple will be seated together?
My attempt: 9! 2!/ 10! = $\dfrac{1}{5}$ , since there are 9! ways of sitting in pairs and 2! ways to arrange a couple.

The solution I'm given is $\dfrac{1}{63}$.

Can someone point out what I'm doing wrong?

You’re doing nothing wrong, assuming that your reasoning is similar to Alex Becker’s in his comment: the correct answer is indeed $\frac15$. Here’s another route to it.

There are $\binom{10}2=45$ pairs of seats, and the couple is equally likely to occupy any one of those $45$ pairs of seats. Nine of the $45$ pairs are adjacent, so the probability that they will occupy adjacent seats is $\frac9{45}=\frac15$, as you say.

And here is yet another. The man sits in an end seat with probability $\frac2{10}=\frac15$. If he’s in an end seat, only one of the remaining nine seats is adjacent to him, and his wife’s probability of getting that seat is $\frac19$. With probability $\frac45$ the man sits in one of the eight seats that have two neighbors, and in that case his wife’s probability of ending up next to him is $\frac29$. The overall probability that the end up sitting together is therefore

$$\frac15\cdot\frac19+\frac45\cdot\frac29=\frac9{45}=\frac15\;.$$

Added: And just for fun, here’s yet another. Imagine that the seats are arranged in a circle around a table. Wherever the wife is sitting, the husband’s probability of sitting next to her is $\frac29$. Then the table is taken away and the seats unwrapped into a straight line, with the breakpoint chosen at random: with probability $\frac1{10}$ it will fall between the husband and the wife, so with probability $\frac9{10}$ they will still end up together. Thus, they end up together with probability $$\frac9{10}\cdot\frac29=\frac15\;.$$

There are $10!$ ways how to put these $10$ persons in the row, the couple can seat this way
$12$, $23$, $34$, $45$, $56$, $67$, $78$, $89$, $10 \ 1$ or $21$, $32$, $43$, $54$, $65$, $76$, $87$, $98$, $1\ 10$ so $20$ possibilities but there are $8$ other persons, there are $8!$ possibilities to seat them so the probability is

$$P = 20 \cdot \frac{8!}{10!} = \frac{2}{9} = 22.2\%$$


Series $\displaystyle \sum\frac{1}{b_n}$ where $\displaystyle b_n=\sum_{k=1}^n(-1)^{k+1}\sqrt{k}.$ Let
$$b_n=\sum_{k=1}^n(-1)^{k+1}\sqrt{k}.$$
I look for the following series $\displaystyle \sum\frac{1}{b_n}$ is convergent or divergent ? Any suggestion ?

The series $b_n$ is not convergent, so $\sum_{k=1}^\infty \frac{1}{b_n}$ is probably convergent. Though you would have to use some sort of test, maybe ratio or integral or something of the sort.

EDIT: I interpreted $b_n$ with $\sqrt k$ and not $\sqrt n$, as I'm not totally sure what you're writing.


Well-conditioned problem vs Stable algorithm I was studying what it means for a problem to be well- and ill- conditioned and for an algorithm to be stable and unstable.

Of course the concepts are somehow related, but I always forget the difference and their meaning, maybe because I didn't actually understand their meaning fully? Yes.

According to this paper an algorithm is stable if small changes in the input produce only small changes in the output. To analize if a algorithm is stable we must analyze every step of the algorithm which could introduce errors, i.e. we must look at divisions, subtractions, square roots, etc. Then, at the end, it is stated that an algorithm is stable if it's well condition at each of this steps.

So what does it mean for a problem to be well-condition according to the same paper?

It starts by saying that every problem is based on an expression.  If a problem is well-conditioned, then small changes in the input to the expression, will lead to small changes in the results.

Honestly this seems to be the definition of a stable algorithm. Why wouldn't this be?

Moreover, I think it seems that in this last definition the author mixed what's the problem and what's the expression.

And by the way, what's exactly this expression? Isn't it also an algorithm?

The author gives an example. We want to evaluate the expression:

  $$y = \frac{x}{1 − x}$$

Is this the problem or the algorithm, or both?

Assuming this is an algorithm, to state if it is stable we should verify if all "atomic" (for simplicity, not in the sense of CPU atomic operations) operations are stable. Operations include $t = 1 - x$ and $\frac{1}{t}$ (are there any others?).

Assuming it's a problem, the author inserts some values in $x$ and states that "we may say that this algorithm is well condition or not around a certain $x_0$".

This means a problem can be well condition or ill condition depending on the input range? So why talking about well condition problem at all and in a general sense?

Maybe the only problem that I've is to distinguish between what's meant by a problem and an algorithm, what is the difference between the two.

Consider a linear problem $$Ax = b \tag{1}$$
This problem is well conditioned if difference between $\textbf{exact}$ solution to this problem and $\textbf{exact}$ solution of a slighty perturbed problem $(A+\delta A)y = b+\delta b$ is small for any small perturbations $\delta A$,  $\delta b$. If the problem is badly conditioned, then there exists some small perturbations $\delta A$, $\delta b$ such that difference between $x$ and $y$ is large.

Suppose, that the problem (1) is solved using some algorithm and denote the solution by $\hat{x}$. Usually $\hat{x}$ is not an exact solution to (1). A method used to solve (1) is called backward stable if for any $A$ and $b$ it computes a solution $\hat{x}$  with small backward error, that is $\hat{x}$ is an exact solution of some slightly perturbed problem $(A+\delta A)\hat{x} = b+\delta b$. For a problem (1), the backward error is closely related to norm of residuals $r = A\hat{x} - b$.

Forward error is defined as a distance between true solution $x$ to (1) and solution computed by given method $\hat{x}$. This forward error can be estimated as
$$\text{forward error} \leq \text{condition number} \times \text{backward error} \tag{2}$$

The backward error and condition number are completely unrelated. It is possible to have small backward error for badly conditioned problem or large backward error for well conditioned problem. For example consider a problem
$$\left[\begin{array}{cc}\epsilon & 1\\1 &\epsilon\end{array}\right]x = b$$
for some small $\epsilon$. This problem is well conditioned. However if this problem is solved using LU method without pivoting, then the backward error is large and computed solution is highly inaccurate.

If condition number is low and a stable algorithm is used (thus the backward error is also small), then we can be sure, that (1) is solved accurately. It is however possible, that an unstable algorithm applied to badly conditioned problem gives an accurate solution. It is always possible to estimate the forward error for a given problem (1) using aposteriori error analysis, which do not use (2).

It's not a good example, but it works. Usually the conditionning is defined for the stability of linear inversion problems. $$ y=Ax $$ We the conditioning of $A$ is bad, the algorithm won't matter, because the exact solution $A^{-1}y$ (if we could get it) differs a lot if the input $x$ varries a little. This means the problem is unlikley to be of practicle interest, because of inevitable noise in measurment/data.


Counting surjective functions We have $r$ objects and $n$ boxes. I have to count all the combinations possible if the objects and boxes could be both different. If that happens I can count the number of variations from $n$ to $r$ (number of functions between objects and boxes). I mean, $n^r$.

But if I have to count all the combinations possible if anyone of the boxes is empty ($r\ge n$). Now I have to count the surjective functions.

For example if I have $r$ objects and 1 box. The number of surjective functions is 1. Then, if I have $r$ objects and 2 boxes, the number of surjective functions is $2^r-2$. And finally if I have $r$ objects and 3 boxes, I will count $3^r-2\cdot[3\cdot2^r-3]$. How do I get the formula for $r$ objects and $n$ boxes?

We want the number of surjections from $r$ to $n$.

First, note that the Stirling number of the second kind
$$
\begin{Bmatrix}
x\\
y
\end{Bmatrix}
$$
is a quantity that solves most of the puzzle: this number equals the number of partitions of $x$ into $y$-many nonempty subsets.

Clearly, every surjection $f\colon [r]\to [n]$ determines such a partition: $\{f^{-1}(j)\mid 1\le j\le n\}$, and each such partition of $r$ corresponds to $n!$ different surjections (permutations of the range). (Here, $[m]$ denotes $\{1,\dotsc,m\}$.) So the total number of surjections $[r]\to [n]$ is:
$$
n!
\begin{Bmatrix}
r\\
n
\end{Bmatrix}
$$

The Stirling number of the second kind can be computed from the explicit formula
$$
\begin{equation*}
\begin{Bmatrix}
r\\
n
\end{Bmatrix}
=\frac{1}{n!}\sum_{j=0}^{n}(-1)^{n-j}\binom{n}{j}j^r
\end{equation*}
$$
As you can see, a kind of "inclusion/exclusion" principle is at work. Various answers to the following question (the special case $n=3$ of yours) provide some intuition about why that is so: Counting the number of surjections.

Notice the factor of $\frac 1 {n!}$, which we multiplied the Stirling number of 2nd kind by when counting surjections. So finally can simplify further: the number of surjections $[r]\to [n]$ is
$$
\sum_{j=0}^{n}(-1)^{n-j}\binom{n}{j}j^r
$$

Lets say you have $k$ empty boxes where $1\leq k \leq n-1$. Choose your empty boxes $\binom{n}{k}$ and then fill the other boxes with your objects $(n-k)^r$. Sum this up and you get $\sum_{k=1}^{n-1}\binom{n}{k}(n-k)^r$. The number of your surjective functions will be $n^r-\sum_{k=1}^{n-1}\binom{n}{k}(n-k)^r$.


Recurrence of a certain class of $2$-$d$ random walks As is well known, a symmetric random walk on $\mathbb{Z}^d$ (the lattice of $d$ dimensional vectors with integer components) is recurrent if and only if $d=1,2$. In particular it is transient for $d=3$.

What about a random walk on $\mathbb{Z}^2$, where the probability to move right is equal to the probability to move left, and the probability to move forward is equal to the probability to move backward, but they are not necessary all $0.25$? (They do sum to $1$ of course). Well it turns out that this is recurrent as well.

My question is this: Is there a non-computational proof of this? i.e. can one show in some rigorous sense that this is "the same" as a truly symmetric $2$-$d$ random walk and thus recurrent?

edit: perhaps I should replace "non computational" with "slick" or "short". I was given a problem on an exam that required as a lemma the fact that these walks are recurrent. When I went over the official solution, I was surprised that the lecturer stated without proof that in this type of walk the probability of returning to the origin at time n is of order 1/n, as though this was some immediate corollary of the symmetric case (where this holds). I guess he was just being sloppy...

In fact, there is a theorem in Durrett's book "probability theory and examples"(P186)：

When $d=2$, for all iid random variables $X_n, n=1,2,\dots$ if  $S_n/n^\frac{1}{2}\Rightarrow$ a normal distribution, then $S_n$ is recurrent

Since the 2d random walk is symmetric, by the central limit theorem you can see that $S_n/n^\frac{1}{2}\Rightarrow$ a normal distribution, then $S_n$ is recurrent


Constructing sigma algebras in countably many steps I'm learning about $\sigma$-algebras and was interested when my textbook briefly mentioned the impossibility of constructing the Borel $\sigma$-algebra of $\mathbb{R}$, $\mathcal{B}(\mathbb{R})$, from the open intervals in countably many steps. More precisely, let $\mathcal{L}_0$ be the collection of all open intervals $(a,b)\subseteq\mathbb{R}$. Given $\mathcal{L}_i$, we define $$\mathcal{L}_{i+1}=\bigg\{\bigcup_{k\in\mathbb{N}}A_k, \Big(\bigcup_{k\in\mathbb{N}}A_k\Big)^{c}\ :\ A_k\in \mathcal{L}_{i} \bigg\}\ \ \ \text{ and } \ \ \ \hat{\mathcal{L}}=\bigcup_{i\in\mathbb{N}}\mathcal{L}_i$$

Then my textbook says that $\hat{\mathcal{L}}\subsetneqq\mathcal{B}(\mathbb{R})$.

My question is this: can one (easily) exhibit a Borel set not in $\hat{\mathcal{L}}$? Is it necessary to use heavy machinery like AC or CH in order to construct such a set? I'm quite stuck.

Many thanks!

One can find proofs in many set theory texts for the stronger result that, for each countable ordinal level of the Borel hierarchy, there exist Borel sets not belonging to that level. Look in book indexes for "universal set". You can also google Borel + "universal set".

However, I've found very few published proofs that limit themselves to the finite levels situation that most measure theory texts state and which you have stated in your question. For this reason I’ve made note of such proofs when I’ve encountered them, and in case it could be of use to you or others, below are the only three such references that I currently know about.

[1] Patrick Paul Billingsley, Probability and Measure, 3rd edition, Wiley Series in Probability and Mathematical Statistics, John Wiley and Sons, 1995 [reprinted as “Anniversary Edition” in 2012], xiv + 593 pages.

  On pp. 30-32 there is a detailed construction of a Borel set that does not belong to any of the finite Borel classes.

[2] Mikls Laczkovich, Conjecture and Proof, Classroom Resource Materials, Mathematical Association of America, 2001, x + 118 pages.

  See pp. 98-101 and p. 105 and Exercise 17.7 on p. 106.

[3] Eric M. Vestrup, The Theory of Measures and Integration, Wiley Series in Probability and Statistics, Wiley-Interscience, 2003, xviii + 594 pages.

  See Section 1.7 (pp. 28-34). Vestrup’s construction is an expanded version of the construction given in the 3rd edition of Billingsley’s book [1]. For the stronger version involving transfinite Borel classes, see Exercise 2.23 on p. 36 (and Notes on the Problems for 2.23, on p. 555).

I published the proof given in (Billingsley, Probability and Measure, 3rd edition, pages 30-32) and people sent me to this thread. I have questions to this proof. I do not understand it. Somebody, please help if still interested in the topic.
Problem from the book Probability and Measure (by Patrick Billingsley), on constructing σ-Fields.


If the first 10 positive integer is placed in a circle(any order), 3 integer in consecutive locations around the circle that have a sum > 17? If the first 10 positive integer is placed around a circle, in any order, there exists 3 integer in consecutive locations around the circle that have a sum greater than or equal to 17?

This was from a textbook called "Discrete math and its application", however it does not provide solution for this question.

May I know how to tackle this question.

Edit: I relook at the actual question and realize it is sum greater or equal to 17. My apologies.

Gerry's answer shows that the average sum of the triples is $16.5$. If there's no sum above $17$, then at least five sums have to be $17$ for the average to be $16.5$. Since two successive sums can't be equal, at most five sums are $17$, and thus exactly five sums are $17$, and thus the other five sums are $16$ and they alternate. But that's impossible, because it implies that moving by three goes up or down by $1$ and moving another three goes down or up by $1$, respectively, leading to the same number again.

Solution: For any given the first 10 positive integers placed around a circle, in any order, there are exactly
10 choices of 3 consecutive numbers around the circle. And each number appears exactly 3 times among the
10 choices. Hence, the sum of all numbers in 10 choices of 3 adjacent number is
(1 + 2 + · · · + 10) × 3 = 165
This implies that at least one choice of 3 adjacent numbers has a sum greater than or equal to 17. Indeed,
if there is no such triple, the sum of all numbers in 10 triples will be strictly less than 160 because each one
is strictly less than 16. This contradicts to the sum of all triples is 165.


Difference between function and equation What is the precise difference between function and equation ?
In which case will it be wrong if used( common mistakes )?
Also will the Venn diagram overlap if I were to draw one ?
Any help and discussions will be appreciated .

A function is a transformation or mapping of one thing into another thing. It might be written as a rule (e.g. "Take the input and square it"), as a formula ("e.g. $f(x) = x^2$ or $x \mapsto x^2$), as a set of ordered pairs (e.g. $\left\{(1, 1), (2, 4), (3, 9), \ldots\right\}$, or any other way of showing how the output relates to the input. The function doesn't have to use numbers, either - a function could take two words and return their letters interlaced (so f(cat, dog) = cdaotg) or it could tell you what day of the week a given date falls on, or the post code/zip code of a given geographical location.

[In very formal terms, a function is a set of input-output pairs that follows a few particular rules.]

An equation is a declaration that two things are equal to each other. For example, $2^2 = 4$ is an equation stating that the square of 2 is 4. An equation may include variables of unknown value, and it may be true for all, some or none of the possible values of those variables. For example, $x^2 = 4$ is an equation that is true when $x = \pm 2$, and false for other values of $x$, while $x^2 = -4$ is an equation that is false for all real values of $x$.

What may be confusing you is that we often use equations to declare a relationship between two variables, often in the form of a function or formula. For example, $y = x^2$ is an equation stating that the value of $y$ is determined by the value of $x$ via the function $x^2$.

A function $f(x):D\to C$ must satisfy $\forall x\in C \ \text{where f is defined in this point $x$},\ \exists!\  f(x)\in D$; i.e. every point in the domain of $f$ for which $f$ is still defined may have only one image, which is a point on the line (a point in the codomain). Not to confuse with surjectivity. For example, $f:\mathbb{R}\to\mathbb{R}$ with $f(x)=\sin(x)$ is non-surjective for there is no $x\in \mathbb{R}$ such that $f(x)=2$. But the $\sin(x)$ is a function because there's no $x$-value with a multiple $f(x)$-value.
An equation can be every equalty: a function is an equality, a differential equation is an equality.
E.g. $x=y^2$ is an equation, but not a function if we view it with x in the domain and y in the codomain. For instance, $x=1$ has $y=1$ and $y=-1$ as solution (point in domain with two different images). Hence, not a function in the $(x,y)$-plane.


Find parameter a for which.. Find parameter $a$ for which $$\frac{ax^2+3x-4}{a+3x-4x^2}$$ takes all real values for $x \in  \mathbb{R}$

I have equated the function to a real value, say, k
which gets me a quadratic in x. I have then put $D\geq 0$ (since $x \in \mathbb{R}$) which gets me $a \geq -9/16$

How do I proceed further to get other parameter for $a $?

Find $a$ so that the equation

$$y= \frac{a x^2+3x -4}{-4x^2+3x+a}$$

has a root which is in the domain of the function

The discriminant must be $\geq 0$ for all $y$

$$(9+16a)y^2+\left(4a^2+46\right)y +(9+16a)\geq 0$$

This is the case if its discriminant is negative and $9+16a>0$

$$16 (-7+a) (-1+a) (4+a)^2<0$$

which says $a$ must be in the interval $(1,7)$

$ a \geq -9/16$ is not correct.  For $ a \leq -9/16$ the domain of the function $f(x)=\frac{ax^2+3x-4}{a+3x-4x^2}$ is $ \mathbb R.$


How to find the area of region $R$? Region $R$ contains all the points $(x,y)$ such that $x^2+y^2\leq100\;$  and $\:\sin(x+y)\geq0$. Find the area of region $R$.

$\:\:\:\:\:\:\:\:\:$$\sin(x+y)\geq0$
$\:\:\:\:\:\:\:\:\:$$\implies2n\pi\leq x+y\leq(2n+1)\pi$

Thus graph would be probably look like this:

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$

Now how to find the area?

For the points in the circle $(a,b)$ if $$\sin(a+b)\leq0$$  then for $(-a,-b)$ $$\sin((-a)+(-b))\geq0$$
Thus if our condition does not hold true for a single point we can take a symmetrically opposite point to make our conditions hold. Thus half the are of our circle will satisfy our condition

(note there will be lines that have $\sin(x+y)=0$ but as they are lines their contribution to the area is negligible, not affecting the probability, in reality it should be slightly greater than 1/2)

Due to innate symmetry, we can say that area of the common region is half of the area of circle. So required area is
$$A = \frac{1}{2} 100 \pi = 50 \pi$$

Explanation

If $P(x,y)$ satisfies $x^2 + y^2 \le 100$, then $P'(-x,-y)$ also satisfies. Since $P$ and $P'$ are diametrically opposite and have same modulus, there are as many $P$ as there are $P'$.

Now if $\sin(x+y) <0$, then $-\sin(x+y) > 0$ That means $\sin(-x-y) >0$.

So if $P$ lies in required region, $P'$ does not lie. Since they are equal in number, required area is half of the area of circle.


What character can replace word "let" in proofs? For example, suppose I have a line of a proof introducing new “variable” $x$:
$$\textrm{Let}\:\:x\in f(y)$$
I am looking for ways to express the word “let” in this context and I would like to avoid using natural language because Math is itself a universal language for expressing complex ideas.

I used to use a character like a right square bracket or similar to it. I once encountered it somewhere, but I don't know if it is a common practice. For example, I would write that line like this:
$$\sqsupset x\in f(y)$$

Now I failed to find this character anywhere, neither on the Web nor in Unicode symbol set. Instead, I discovered some similar symbols like $\buildrel \text{def}\over=$, or $:=$, or $\buildrel\triangle\over=$, or $:\Leftrightarrow$ in Wikipedia, but those are very limited and not so much useful in my case.

I would like to avoid using natural language because Math is itself a universal language for expressing complex ideas.

Honestly, I think this is a bad (by which I mean non-mathematical) reason to do anything. You will have great difficulty introducing any symbol into any widespread use, and consequently you will not be able to use such a symbol in any piece of work you wish to be taken seriously (because nobody will be able to read it!). I also don't recommend you teach people bad habits.

On the other hand, if these are just for personal notes, by all means invent your own symbol. I use lots of imprecise squiggly arrows, equals signs with quotation marks round them, equals signs decorated with question marks, and the like. I often use := to conjure a symbol into existence at the same time as defining it (because, unlike when programming, I don't need to declare my variables and I don't often redefine them in the same 'subroutine'), or simply =. I've seen people write an equals sign with "def" or "$\triangle$" above it too.

This is not unlike how real mathematicians work with each other. When two people collaborate, it's very convenient for there to be an implicit assumption along the lines of "every time I say X, until we solve this problem, I mean this particular object", or vague terminology like "nice" to describe classes of objects that you can't quite pin down. But of course, once it comes to a seminar or a paper, you start from scratch, (mostly) give everything real, sensible words, and don't force your audience to learn a page of jargon and squiggles when there's perfectly good English available for it.

Note that both × and · are relatively language-independent notations for multiplication.  One is "Cross Multiplication", the other is "Dot Multiplication"; both devolve to "simple linear multiplication" when being applied to degenerate (i.e. 1 by 1 (by 1 by ...)) matrices, and either will generally be recognized as such by any technical person in a locale that uses Arabic-derived mathematics.  In other words, Western countries and Near-Eastern countries, and most of their former colonies and subject-states - which pretty much covers everything but China (I don't know enough about the impact of the "modernization/westernization/digitization" of communications on their technical notations to comment on them).

Both mean entirely different things when the context is multidimensional matrices, however.  That is, in fact, WHY specific symbols are used.  If you just say "Matrix A times Matrix B", it is not clear whether you mean cross-multiplication or dot-multiplication - or even something else entirely.  Sure, there are conventions for ad-hoc interpretation; but for a precise and unambiguous statement, a narrowly-defined (and thus, of restricted use) symbolic language is required.


Relationship between tensor product and wedge product In Gravitation (Misner, Thorne & Wheeler), the wedge product over a vector space $V$ is introduced as just the antisymmetrised tensor product: for $u, v \in V$, they define
$$u \wedge v = u \otimes v - v \otimes u$$
I'm trying to work out how to fit this together with the characterisation of the exterior algebra $\Lambda V$ as the quotient of the tensor algebra $T(V)$ by the two-sided ideal $I$ generated by tensor products of the form $v \otimes v$. In that characterisation (at least in Wikipedia), the wedge product is defined by
$$[u] \wedge [v] = [u \otimes v]$$
where square brackets indicate equivalence classes, for the equivalence relation $\sim$ induced by the ideal (i.e., $u \sim v$ iff $v = u + i$, for some $i \in I$). At first, I took it that the way they fit together was that we would be able to show that

$$u \otimes v \sim u \otimes v - v \otimes u$$

But so far as I'm able to tell, this isn't the case: what we can show instead is that $u \otimes v \sim -v \otimes u$, and hence that
$$u \otimes v \sim \frac{1}{2} (u \otimes v - v \otimes u)$$
Having browsed a bit, this answer suggests to me that what's going on is just that there are two ways of relating the exterior algebra to the tensor algebra: the MTW characterisation amounts to treating $\Lambda V$ as a subalgebra of $T(V)$ (specifically, the subalgebra generated by antisymmetric tensor products), rather than as a quotient algebra of $T(V)$. Is that correct? If so, what are the advantages of a subalgebra vs a quotient algebra construction, and vice versa?

(Apologies if this is overly duplicative of the linked discussion - it was a bit too abstract for me to be sure I'd followed it correctly.)

Over a real vector space $V$ one can regard the second exterior
power $\Lambda^2 V$ as either a subspace of $V\otimes V$
(the space of antisymmetric tensors) or as a quotient space of $V\otimes V$
(by factoring out the subspace generated by the $v\otimes v$). Over $\Bbb R$ these definitions are equivalent; the equivalence class $[v\otimes w]$
corresponding to the tensor $\frac12(v\otimes w-w\otimes v)$.

But over a vector space of characteristic 2, or over a ring where $2$
is not invertible, these definitions can vary. I would say the right definition is as a quotient of $V\otimes V$. But let $V$ be a vector
space of dimension $n$ over the field $\Bbb F_2$. Here antisymmetry
is the same as symmetry, so the antisymmetric tensors in $V\otimes V$
have dimension $\frac12n(n+1)$. However the "right" definition as a quotient gives the dimension of $\Lambda^2(V)$ as $\frac12n(n-1)$.

Of course all these considerations extend to $\Lambda^k(V)$.
Over a field of characteristic zero, it matters not which approach we take.

From your computation, the two definition are equivalent apart from a factor of $2$.


The number of natural number $n$ in the interval of [$1005,2010$ ] for which the polynomial The number of natural number $n$ in the interval of [$1005,2010$ ] for which the polynomial $1$+$x$+$x^2$+ $x^3$ +...+$x^{n-1}$ divides the polynomial   $1$+$x^2$+$x^4$+$x^6$ ....+$x^{2010}$ is

Ans is the number of such numbers is 503.

My soln:
$$ (x^{503}-1)(x^{503} +1)( x^{1006}+1) $$ to be
Divided by $ x^n -1$ So n=503.

The polynomial $1+x+x^2+...+x^{n-1}$ divides $1+x^2+x^4+...+x^{2010}$ exactly when
$$
(x^n-1)=(x-1)(1+x+...+x^{n-1})|(x-1)(1+x^2+x^4+...)=\frac{x^2-1}{x+1}(1+x^2+...+x^{2010})=\frac{x^{2012}-1}{x+1}.
$$

If the polynomial $x^n-1$ divides the polynomial $\frac{x^{2012}-1}{x+1},$ then all the (complex) roots of $x^n-1$ must also be roots of $x^{2012}-1,$ but $x=-1$ cannot be a root of $x^n-1.$. Because $-1$ is a root of $x^n-1$ when $n$ is even, the value $n$ must be odd. If $n$ is odd, then it is sufficient that $x^n-1|x^{2012}-1.$ But the complex roots of $x^n-1$ are shared by $x^{2012}-1$ exactly when $n$ divides $2012$. The only value $n$ in the range $[1005, 2010]$ for which this happens is $1006$, but this even, so there are zero solutions for $n$ in the range $[1005,2010].$

If $f(x)=1+x+x^2+\cdots+x^{n-1}$ divides $g(x)=1+x^2+x^4+\cdots+x^{2010}$, then the quotient has integer coefficients. Therefore, $n=f(1)$ divides $1006=g(1)$. There is only one multiple of $1006$ in $[1005,2010]$, which is $1006$ itself. Thus, the only possible solution, if there is a solution, is $n=1006$.


How to constantly move position A "elastically" towards position B? I'm writing an advanced interface, but I don't yet have a concept of derivatives or integrals, and I don't have an easy way to construct infinite many functions (which could effectively delay or tween their frame's contributing distance [difference between B and A] over the next few frames).

I can store values for a frame, and I can also consume them or previous values and map into A.

For example, each frame could calculate the distance between B and A, then add that distance to A, but they would be perfectly in sync.

I can keep track of the last N frames' distances and constantly shift old distances off, but this would create a delay, not an elastic effect. Somehow, the function that's popping off past-frames' distances needs to adjust for how long it's been for those 10 frames.

Is there any function I can rewrite each frame, which picks up the progress from it's predecessor, and contributes it's correct delta amount, adjusting for the new total distance between B and A?

Does this question make sense? How can I achieve behavior where A is constantly catching up to B in a non-linear, exponential way?

You can truly simulate the physics of a dampened spring, which leads to a differential equation.

$$m\ddot x_A+d\dot x_A+k(x_A-x_B)=0$$

where $x$ is the position, $m$ the mass, $d$ a damping coefficient and $k$ the stiffness constant of the spring.

As you probably have enough with a qualitatively realistic solution, there is no need for a sophisticated ODE solver and the Euler method should be good enough.

Turn the equation in a system of two first order equations:

$$\begin{cases}\dot v_A=\dfrac{kx_B-kx_A-dv_A}m,\\\dot x_A=v_A.\end{cases}$$

Hence you will iterate

$$\begin{cases}v_{A,n+1}=v_{A,n}+\dfrac{kx_{B,n}-kx_{A,n}-dv_{A,n}}m\Delta t,\\x_{A,n+1}=x_{A,n}+ v_{A,n}\Delta t\end{cases}$$

for some time increment. This way you can compute a position from the previous, but you also need to remember the speed.

By varying the damping and stiffness coefficients, you can achieve more or less tight following of $B$, with an oscillatory behavior or not.

If you want a 2D simulation, $x$ will be a vector, and the shape of the equations remains.

Elastic motion obeys a time differential equation representing a dynamic system of order two or higher in which elasticity constants like $m,k$ are fixed. The simplest harmonic motion $ m \ddot x + k x=0 $ enforces distances and you have no further control except on the imposed boundary conditions.


Is random variable $X_i$ measurable on ${\mathcal F_{i+1}}$ or ${\mathcal F_{i-1}}$? Let say I have a filtration $\mathcal F_i$ with $\mathcal F_1$ contained in $\mathcal F_2$, $\mathcal F_2$ contained in $\mathcal F_3$ and so on...$\mathcal F_n$. $X_i$ is a stochastic process, $X_i$ is measurable with $\mathcal F_i$ (adaptable). Is random variable $X_i$ also measurable on $\mathcal {F_{i+1}}$ or ${\mathcal F_{i-1}}$?

Since $\sigma(X_i)\subset\mathcal F_i\subset\mathcal F_{i+1}$, $X_i$ is certainly $\mathcal F_{i+1}$-measurable.

But in general $X_i$ is not necessarily $\mathcal F_{i-1}$ measurable: take $\mathcal F_i:=\sigma(X_k,1\leqslant k\leqslant i)$ where the $X_k$ are independent integrable non degenerated random variables. Then $\mathbb E(X_i\mid\mathcal F_{i-1})=\mathbb E(X_i)\neq X_i$.

It justifies the notion of predictable sequence/process.

If $\mathcal{F}_i \subset \mathcal{F}_{i+1} \subset \mathcal{F}_{i+2}$ for $i = 1, 2, \ldots$, $X_i$ is $\mathcal{F}_{i+1}$-measurable, but not $\mathcal{F}_{i-1}$-measurable.

[ incorrect statements removed. ]


Can you make the circle into a vector space? I thought maybe use a set with elements of the form $e^{i\theta}$ but what field would the scalars be taken from? Multiplying $ke^{i\theta}$ with $k \in \mathbb{R}$ or $\mathbb{C}$ doesn't give you another point on the circle though, so what field is used?

For the unit circle in $\mathbb{R}^{2}$, call it $S = \{(x,y) \in \mathbb{R}^{2}:x^{2}+y^{2}=1\}$ is not a vector space. Indeed, the two vectors $(1,0)$ and $(-1,0)$ lie in $S$. We will show that the sum of these two vectors do not lie in $S$. From this, we can therefore conclude that $S$ is not a vetor space, as it will not be closed under addition. So since $(1,0)+(-1,0)=(0,0)$ and $(0,0) \notin S$, this finishes the proof.

Here is a question to contemplate. If one adjoins the zero vector $(0,0)$ to $S$, is $S \cup \{(0,0)\}$ a vector space?


Propositional logic problem: Sales, expenses and happiness of the boss Either sales will go up and the boss will be happy, or expenses will go
  up and the boss won’t be happy. Therefore, sales and expenses will not
  both go up.

I know the solution is that the conclusion is wrong, but I don't see why. I would say that the boss is either happy, or he is unhappy, but he can't be both. Because of this, it is impossible that the sales go up and the expenses go up, since this would entail a state of schizophrenia of the boss, and thus the conclusion is true.

Let $S$ be the proposition that sales will go up,
$E$ the proposition that expenses will go up,
$H$ the proposition that the boss will be happy.

Suppose we are given that $P$ is true, where
$P = (S \wedge H) \vee (E \wedge \neg H)$.

With the stated definitions, it then is sound to say it is impossible that the
boss will be both happy and unhappy; $\neg(H \wedge \neg H)$.
But it is not sound to say that the boss's future happiness
controls both sales and expenses simultaneously.

Suppose the boss will be happy ($H$ is true); then $(E \wedge \neg H)$
is false and so $(S \wedge H)$ must be true. It follows that $S$ is true.

Suppose the boss will not be happy ($H$ is false); then $(S \wedge H)$
is false and so $(E \wedge \neg H)$ must be true. It follows that $E$ is true.

But observe: knowing that $H$ is true tells us nothing about the
truth or untruth of $E$. It tells us only about the truth of $S$.
On the other hand, knowing that $H$ is false tells us nothing about
the truth or untruth of $S$; it tells us only about the truth of $E$.

In other words, by knowing the truth value of $H$ you could make a conclusion
about $S$ or a conclusion about $E$, but not both.

In fact, if both sales will go up and expenses will go up,
that is, if $S \wedge E$,
then $P$ is satisfied regardless of the boss's future happiness.
Either $H$ is true, so $S \wedge H$ is true and $P$ likewise
is true because its first clause is true;
or $H$ is false, so $(E \wedge \neg H)$ is true and $P$ likewise
is true because its second clause is true.

Surely you can have two different emotions about two different things at the same time? e.g. you could love dogs and hate cats simultaneously? This is the same kind of thing, because the event causes the emotion ABOUT THAT EVENT, but an overall emotion does NOT cause the event. You would be correct if the statement was 'either the boss will be happy so sales will go up, or the boss will be unhappy so expenses will go up. Therefore, sales and expenses will not both go up.'


Total combinations from number of unique combination? How can you find count of the total combinations possible if you have the count of unique combinations?
Suppose I need to form a sum of $3$ using $\{1,2\}$
Unique combn. $(1+1+1)$,
$(1+2)$
hence '$2$' unique
but total is '$3$' i.e,
$(1+1+1)$,
$(1+2)$,
$(2+1)$
I know the elements used to make the combination
Note: Number of unique combinations (ie. 2 here) is known.
How can I find the total number of combination given the unique combinations?

To find the number of restricted compositions (use the correct terminology) with a desired sum $n$ and available numbers $\{1,2\}$, what I will label $f(n)$, recognize the following:
$f(1)=1$ (as the only possibility is the sum with only one term: $(1)$)
$f(2)=2$ (as there are two possibilities: $(1+1)$ and $(2)$)

 alternatively, and preferably, recognize that $f(0)=1$ as there is in fact a sum whose sum is zero using terms $1$ and $2$... the "empty sum"... $(~)$.  This is somewhat abstract, but I strongly encourage coming to grips with this as using counts related to objects of size zero can greatly simplify arithmetic in many instances

Next, recognize that for each $n\geq 3$ you have $f(n) = f(n-1)+f(n-2)$ which is seen by noting that every such restricted composition either begins with a "$1+\dots$" with a sum of $n-1$ following which can be arranged in $f(n-1)$ ways, or it will begin with $2+\dots$ with a sum of $n-2$ following which can be arranged in $f(n-2)$ ways.
Now... you should recognize this... $f(1)=1, f(2)=2, f(n)=f(n-1)+f(n-2)$... this is precisely the fibonacci sequence.  So... here we found that $f(n)$ is precisely equal to $F_n$.
No knowledge of the related problem counting restricted partitions rather than restricted compositions was needed.
The technique can obviously be adapted... the number of restricted compositions with sum $n$ using terms $\{1,2,3\}$ for instance would be the tribonacci numbers satisfying the recurrence $t(1)=1, t(2)=2, t(3)=4, t(n)=t(n-1)+t(n-2)+t(n-3)$ or the number of restricted compositions with sum $n$ using terms $\{2,5\}$ for instance would be $a(1)=a(3)=0, a(2)=a(4)=a(5)=1$ and $a(n)=a(n-2)+a(n-5)$ for each $n\geq 6$ and so on...

 Coming to terms with the zero object can simplify a lot of the above for finding initial conditions.  We could instead have had $a(n)=0$ for all $n<0$ and $a(0)=1$, reducing the need to actually manually find initial conditions

Coming up with a closed form for these types of linear recurrences is already covered in great detail in related chapters in books, and otherwise for now the current presentation of the answer in terms of a recurrence relation is generally considered sufficient.

The two objects you are interested in here are commonly called "compositions" and "partitions" of integers, where we count two compositions as different if the order of the parts is different (i.e. $(1+2)$ is a different composition than $(2+1)$).
So when you say you want to count the unique combinations, that is the same as counting partitions of n.
The partition function is a very important function in number theory and is usualy denoted by $p(n)$ for the number of partitions of the natural number $n$.
There is no known exact closed-form expression for the partition function, but a way to compute it is via the recurrence
$$p(n) = \sum_{k=1}^{\infty}(-1)^k (p(n - \frac{k(3k-1)}{2}) + p(n - \frac{k(3k+1)}{2}))$$
known as the Euler pentagonal theorem.
There is plenty more to say about paritions of integers.


How does one search intelligently for solutions of a Diophantine equation? Before the proof of Fermat's last theorem, much evidence was accumulated in favor of the conjecture, by using computer searches to prove that a solution would need to have very large values. What are the ideas (if any) behind these computer searches which are not specific to the Fermat curve?

Suppose that we have an irreducible polynomial $f(x_1, \dots, x_n) \in \mathbf Z[x_1, \dots, x_n]$, of large degree, and we want to use a computer to efficiently list all integer solutions of $f(x_1, \dots, x_n) = 0$ in the region $\max(|x_1|, \dots, |x_n|)\leq N$. What is the best way to proceed?

You ask, what ideas of number theory, if any, can be implemented in order to facilitate the exhaustive computer search for solutions, regardless of any sort of special property of $f$? Good question, Bruno. Qiaochu Yuan is right and I would go a step further to answer in the negative. But if you are willing to expand your question a bit from no conditions on $f(x)$ to allowing $f(x)$ to have the property that the greatest integer of its slope is constant over large intervals, then there is something I can offer from my research that you may not be aware of. And if you are willing to expand your question a bit from ideas of number theory to ideas in complex analysis, then perhaps we can do even more. I expand on this latter approach in the conclusion.

There is a general approach we can use to design Euclidean-like algorithms for finding integer solutions to some types of equations. If over an interval, $f(x)$ is monotone and its domain exceeds its range, then we can narrow the interval for trial and error by finding integer solutions for $f^{−1}(x)$. And if we can do this repeatedly, then we have an algorithm.

Our objective is narrowing the interval for trial-and-error while still preserving the integer solutions. We meet this objective by performing successive iterations on $f(x)$ with the following four steps. First, we round the slope of $f(x)$ to the nearest integer. Depending on the nature of $f(x)$ and the interval, it may not be practical to use regular rounding; we may need to round up or down instead. Therefore, we round the slope of $f(x)$ to $[f'(x) + a]$ where brackets denote greatest integer and $a$ is a pre-determined constant with 0 $≤a<$ 1. Next, we subtract the integer-sloped linear equation, $y'=[f'(x) + a]x$, from $y=f(x)$. This step is only practical if $[f'(x) + a]$ is constant over a large interval. Then, we take the inverse of the difference. We repeat the process until the domain of $x$ is small enough for trial-and-error or this method becomes impractical. Finally, after an integer solution is found, we use back substitution to find the integer solution to the original equation, $y=f(x)$. We call an algorithm that uses this method a Diophantine algorithm because it works by eliminating non-integer solutions to Diophantine equations. Our method guarantees i. the preservation of all integer solutions and the non-preservation of all non-integer solutions and ii. the domain of $y-y'$ over any interval where $[f'(x) + a]$ is constant always exceeds its range.

So, our algorithm is practical for any function over any large interval where
$[f'(x) + a]$ is constant. Our algorithm obviously works for linear Diophantine equations because their slope is constant. Our algorithm obviously works for Pell-like equations of the form $x^2−ny^2=s$ where |$s$| is small because they are asymptotically linear. It is less obvious that we could engineer a Diophantine algorithm for the circle equation, $^2+^2=n$, but we did. Although the slope of the circle equation diverges as $x^2$ approaches $n$, $[f'(x) + a]$ is constant over large intervals. The bad news is that for large $n$, we have to break the interval into subintervals and those subintervals into smaller subintervals with each iteration. The good news is that with each iteration, we can cut the size of most of the intervals for $x$  not by half, but into sevenths and eighths! And since we end up with a semicircle equation after each reduction, we can reduce the size of the intervals again and again! So, our algorithm works considerably better than Cornacchia’s algorithm when $n$ is the product of large primes.

What I typed above is a summary of my results. If you are interested, you can learn more about it on my site, www.greatestintegerfunctionresearch.org. I am also looking to collaborate with someone to determine if we can solve some Diophantine equations using residues. This approach requires combining expressions for integer solutions (such as sums of expressions involving $f(x)$ and the greatest integer function) with an alternative mathematical expression for the greatest integer function to create a holomorphic function. See Chapter 5 in my pdf on the Full-Text page.


Is there such a thing as implicit integration? A problem I'm working on asks me at a certain point to integrate the following:

$$x=2\left(\sin\left(y\right)\right)^2.$$

I've never integrated anything in this form. I can't isolate for $y$ either. Is there some sort of implicit integration technique I should be aware of, to get the indefinite integral of this?

Thanks to the below comments, I was able to isolate y and get

$$\sin^{-1}\left(\sqrt{\left(\frac{x}{2}\right)}\right)=y$$

....which still looks insanely messy to integrate. I imagine I can use a u-substitution or something similar to solve this?

The question is from a very long time ago, but I can still attempt to provide an answer. Implicit integration is kind of like the topic in differential equations called exact differential equations. It’s pretty much tracing backwards from applying multivariable chain rule on a function of multiple variables. I’d say it’s the closest thing I’ve seen to a concept of “implicit integration”.

Make $y$ the subject. Then multiply $\sin x$ on both sides.


Is $\frac {x^4 +2y^2}{(x^2 +y^2)^{\lambda}}$ totally differentiable? Let $\lambda$ be a positive real number.
 When is the function
$$f(x,y) ＝ \begin{cases}0 & \text{if }(x, y) = (0, 0) \\
\dfrac{x^4 +2y^2}{(x^2 +y^2)^{\lambda}} &\text{if } (x, y) \neq (0, 0)\end{cases}$$
totally differentiable?

Find all $\lambda$ where f is totally differentiable.

This is a question from exam of Tokyo university.

https://www.ms.u-tokyo.ac.jp/kyoumu/a20170524.pdf
question No3（2）

I think there is no easy necessary and sufficient condition in this problem, so I tired to find some necessary conditions （and check that is sufficient）but I couldn't find it .

Thank you for your kind help.

Take limit with different cases

$$y= 0$$$$
\lim_{x\rightarrow 0} \frac{x^4}{x^{2\lambda}}$$

This limit is equal to 0 when $\lambda<2$

Take another case $ x=0$

$$\lim_{y\rightarrow 0} \frac{2y^2}{y^{2\lambda}}$$

Now here the limit goes to zero when $$\lambda<1$$

So we can summarise that when $$\lambda <1$$.We took $\lambda<1$ because in this range both cases limit to 0 but after $\lambda>1$ limit x goes to zero but y doesn't


Find a Fractional Linear Transformation that maps the region between $\{|z+1| = 1\}$ and $\{|z|=2\}$ to the region between $Im(z) = 1$ and $Im(z) = 2$ I'm trying to find a Fractional Linear Transformation (if one exists) that maps the region between the circles $\{|z+1| = 1\}$ and $\{|z|=2\}$ to the region between the horizontal lines $\operatorname {Im}(z) = 1$ and $\operatorname{Im}(z) = 2$.

I know that since $-2$ is a point of both circles, I need to find a transformation which has a pole at $-2$, so the denominator of the transformation should be $z+2$ so that both circles get mapped to parallel lines.

From here, I'm getting really stuck.  I know how to find a transformation between two chosen triples, but I'm having trouble figuring out what other two points I need for those triples besides $-2$ and $\infty$.

Any hint or link to a place where I could do some more reading would be really helpful!  Thank you!

Taking two triples $(-2, 0, 2)$ and $(\infty, \alpha , \beta )$, where $\alpha $ and $\beta $ should be on the lines $\operatorname{Im}\,w=1$ and $\operatorname{Im}\,w=2$ respectively(, or  $\operatorname{Im}\,w=2$ and $\operatorname{Im}\,w=1$ respectively), we try to determine $$f(z)=\frac{az+b}{z+2}$$ so that $f$ maps the region between the circles $|z+1|=1$ and $|z|=2$ to the region between the horizontal lines $\operatorname{Im}\,w=1$ and $\operatorname{Im}\,w=2$.

Example 1. Let $\alpha =2i, \beta =i$. Then $$f(z)=\frac{4i}{z+2}.$$ Succesful! See Fig.1 below.

Example 2. Let $\alpha =2+2i, \beta =1+i$. Then $f(z)=\frac{4(1+i)}{z+2}$.
Failure! See Fig.2 below. Why ?

Since the line $l$ from $z=0$ to $z=2$ is perpendicular to the circles $|z+1|=1$ and $|z|=2$ the image $f(l)$ should be perpendicular to the lines $\operatorname{Im}\,w=1$ and $\operatorname{Im}\,w=2$ by the conformality of $f$. Thus the image $f(l)$ should be a vertical line and hence we must take $\alpha $ and $\beta $ so that their real parts are identical. In example 2 we  mistaked this.

Conversely every pair of  $\alpha $ and $\beta $ having identical real parts produces good results. For example, $\alpha =3+i$ and $\beta =3+2i$ leads to $f(z)=\frac{(3+3i)z+6+2i}{z+2},$ which is an appropriate linear transformation.

Lets pause for the second after the sentence (if one exists). Go find the statement of Riemann mapping theorem (more general result would be Uniformization theorem theorem). Make sure you convince yourself such mapping exists.

Next step is to find the actual map. That is actually a harder problem. Linear fractional transformation are so interesting that you one can write a whole book on them. For example this is a good one

http://www.amazon.com/Fuchsian-Groups-Chicago-Lectures-Mathematics/dp/0226425835

You problem is kind easy and playing with composition of the generators of the Fuchsian it will be easy to solve.

But you are raising an interesting question. Can I find algorithmic way to find actual open mapping between to regions. The answer is yes and relies on the discreat version of Riemann mapping which is proved by Denis Sullivan but the place to start learning that business is from is Ken Stephens

http://www.math.utk.edu/~kens/

So a good semester long project would be to write computer program which will solve the above HW question for you.


Find in terms of $p$, $\tan(-\alpha)$, $\tan(\pi - \alpha)$ and $\tan(\frac{\pi}{2}-\alpha)$. Given that $\tan$ $\alpha = p$, where $\alpha$ is acute, find in terms of $p$, $\tan(-\alpha)$, $\tan(\pi - \alpha)$ and $\tan(\frac{\pi}{2}-\alpha)$.

Hint: Use the following facts:

$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$
$\sin(-\alpha) = -\sin(\alpha)$
$\cos(-\alpha) = \cos(\alpha)$
$\sin(\pi-\alpha) = \sin(\alpha)$
$\cos(\pi-\alpha) = -\cos(\alpha)$
$\sin(\frac{\pi}{2}-\alpha) = \cos(\alpha)$
$\cos(\frac{\pi}{2}-\alpha) = \sin(\alpha)$.

$$\cos\dfrac{\pi}{2+\alpha} = - \sin \alpha \\
\sin\dfrac{\pi}{2+\alpha} = \cos \alpha$$


Let $A$ and $B$ be complex $n\times n$ matrices such that $AB−BA$ is invertible and such that $A^2+B^2=c(AB−BA)$ for some rational number $c$. Let $A$ and $B$ be complex $n\times n$ matrices such that $AB−BA$ is invertible and such that $A^2+B^2=c(AB−BA)$ for some rational number $c$. Prove $c\in\{−1,0,1\}$ and show that $n$ is a multiple of $4$ when $c\neq0$.
Please give some hints for this problem. I could not find a way to proceed. I had taken determinant on both sides. But it does not helping much

This is user91684's repertoire. Since he/she has left this community, I will outline a possible approach using one of his/her favourite tricks (which for instance was applied in this answer to a similar question).
Let $A=X+iY$ and $B=iX+Y$ or equivalently, let $X=\frac12(A-iB)$ and $Y=\frac{1}{2i}(A+iB)$. The equality $A^2+B^2=c(AB-BA)$ can then be rewritten as
$$
(c-i)XY=(c+i)YX.
$$
Since $AB-BA=2(XY-YX)$ is invertible, neither $XY$ nor $YX$ is zero. Therefore $c\ne\pm i$. It follows that $XY=\omega YX$ where $\omega=\frac{c+i}{c-i}\ne 1$. Hence $AB-BA=2(XY-YX)=2(\omega-1)YX$ and $YX$ is invertible. Since $XY$ and $YX$ have identical spectra, the equality $XY=\omega YX$ implies that the spectrum of $XY$ is invariant when it is multiplied by $\omega$. It follows that $\omega$ is some primitive $m$-th root of unity and the spectrum of $XY$ is then a disjoint union of $m$-cycles of the form $\lambda\{1,\omega,\omega^2,\ldots,\omega^{m-1}\}$. Thus $m$ divides $n$. If you can show that the only values of $c$ such that $\omega=\frac{c+i}{c-i}$ is a primitive root of unity are $0,1$ and $-1$, then you are done because $m=4$ (and $\omega=\pm i$) when $c=\pm1$.
Remark. Two matrices $X$ and $Y$ that satisfies the equation $XY=\omega YX$ are said to be quasi-commutative or $\omega$-commutative. According to the aforementioned answer by user91684, there is an interesting result, discovered by Drazin, about quasi-commutativity. See section 5 of Holtz, Mehrmann and Schneider, Potter, Wielandt, and Drazin on the Matrix Equation $AB=\omega BA$: New Answers to Old Questions, Amer. Math. Monthly, 111:8 (2004), 655-667.

This can't be right. There are $n^2$ equations and $2 n^2$ unknown over complex numbers - there will be solution almost always.