What properties do similar matrices share? I'm wondering if special matrices are always similar only to other special matrices.

That is:
$\cdot$ Are all of the matrices similar to a symmetric matrix symmetric?
$\cdot$ Are all of the matrices similar to a skew-symmetric matrix skew-symmetric?
$\cdot$ Are all of the matrices similar to a normal matrix normal?
$\cdot$ Are all of the matrices similar to a self-adjoint matrix self-adjoint?
$\cdot$ Are all of the matrices similar to a Hermitian matrix Hermitian?
etc.

Let's try one:
A matrix $A$ is symmetric if $A=A^T$.  Then if $A=P^{-1}BP$, we have $A^T=P^TB^T(P^T)^{-1}$.  Therefore $A^T=A=P^TB^T(P^T)^{-1}=P^{-1}BP$.  Therefore $B=PP^TB^T(P^T)^{-1}P^{-1}$.  Because $P$ is not orthogonal in general, $B \ne B^T$?

NOTE: I could try this for all of them, but I'm pretty sure I'd reach a very similar conclusion in each of the above mentioned cases.

Assuming that this is the correct conclusion here, does this mean that in general the specialness (symmetry, Hermitianness, orthogonality, etc) of matrices is not preserved by a change of basis?  If not, then there's no way to discuss a symmetric (or otherwise special) transformation without first defining our basis?

Just curious.

All of these properties fail to hold for similar matrices in general.  However, they are true if we specify unitary similarity; that is, we only allow similarities such that $P^{-1} = P^*$ (or $P^{T}$, if $P$ is real).

Counterexamples:

$
\pmatrix{1\\&2}
$
is normal, symmetric, self-adjoint, and Hermitian.  But the similar matrix
$
\pmatrix{1&1\\0&2}
$
has none of those properties.

The matrix
$
\pmatrix{0&-2\\2&0}
$
is skew-symmetric, but the similar matrix
$
\pmatrix{0&-1\\4&0}
$
is not.

If $A$ is a special matrix other than a scalar multiple of the identity matrix and $A$ and $B$ are similar, but $B$ has a basis of eigenvectors such that there are two non-orthogonal eigenvectors $u$ and $v$ for distinct eigenvalues $\lambda$ and $\mu$, respectively, then $B$ will not have the special property.

(Note, by the way, a self-adjoint matrix is the same thing as a Hermitian matrix.)


Find the domain and the correspondence rule of $F$ $F=f ∘ f∘ f$, $f(x)=\frac{1}{1-x} ⇒ Dom f(x)=\mathbb{R}-\{1\} \land Ran f(x)=\mathbb{R}-\{0\}$, then:
\begin{align*}
 Dom(f∘ f)&=\{ x ∈ Dom(f) \land f(x) ∈ Dom(f)\} \\
           &=\{ x ∈ \mathbb{R}-\{1\} \land \frac{1}{1-x} ∈ \mathbb{R}-\{1\} \} \\
           &=\{ x ∈ \mathbb{R}-\{1\} \land x ∈ \mathbb{R}-\{0\}-\{1\}  \} \\
                &=\{ x ∈ \mathbb{R}-\{1\}  \} \\
  Dom(f∘ f∘ f)&=\{ x ∈ Dom(f∘ f) \land f(f(x)) ∈ Dom(f∘ f)\} \\
          &=\{ x ∈  x ∈ \mathbb{R}-\{1\} \land \frac{1}{1-\dfrac{1}{1-x}} x ∈ \mathbb{R}-\{1\}\} \\
            &=\{ x ∈ \mathbb{R}-\{1\} \land 1-x ∈
              \mathbb{R}-\{1\}\} \\
  &=\{ x ∈ \mathbb{R}-\{1\} \} \\
F&=x
\end{align*}
It's wrong?

From: Lumbreras Editors

Since $f(\cdot)$ only takes values $\neq 1$, $f(f(x))$ is defined for $x \neq 1$ (because we need that the inner $f(x)$ is defined) and all the values of $x$ for which $f(x) \neq 1$, that is, $f(f(x))$ is defined for $x \neq 0,1$. Similarly, $f(f(f(x)))$ is defined for $x \neq 0,1$ (because we need that $f(f(x))$ is defined) and all the values of $x$ for which $f(f(x)) \neq 1$, that is, $f(f(f(x)))$ is defined for $x \neq 0,1$ since $f(f(x))$ is always $\neq 1$.
Thus $\operatorname{dom}(F) = \mathbb R \setminus \{0,1\}$.


The Orthogonality of Hermite functions I was stuck in proving the orthogonality of Hermite functions.
Assume our Hermite functions is defined as $H_{n} = e^{-x^{2}/2}\frac{d^{n}}{dx^{n}}e^{-x^{2}/2}$. What I want to show is $$\int_{-\infty}^{\infty} H_{n}H_{m}e^{-x^{2}}dx = 0-(1).$$

If we modify the weight $e^{-x^{2}}$ in (1.) to be $e^{-x^{2}/2}$, in this case, we have
$$\int_{-\infty}^{\infty} H_{n}H_{m}e^{-x^{2}/2} dx = \int_{-\infty}^{\infty} H_{m} (\frac{d^{n}}{dx^{n}}e^{-x^{2}/2}) dx -(2)$$
To prove $\int_{-\infty}^{\infty} H_{m} (\frac{d^{n}}{dx^{n}}e^{-x^{2}/2}) dx=0$, it is equivalent to show that $\int_{-\infty}^{\infty} x^{m} (\frac{d^{n}}{dx^{n}}e^{-x^{2}/2}) dx=0$ for any m < n , which can be proved by using integration by parts m-times. However, this argument can not be applied to (1.).

Any hint or reference on how to prove (1.) would be greatly appreciated.
Thanks!

Forget about doing the integral. It can be done but it is not the most straight forward way.

Consider Hermite's equation;

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

Now if we consider the "Hermite differential operator":

$L_H \equiv \frac{d^2}{dx^2} - 2x\frac{d}{dx}$

and we can see that: $ L_H[y] = -\lambda y $

We can put it in Sturm-Liouville form (if you don't know what that is , you should look up Sturm Liouville theory) :

$ L_H = e^{x^2}[e^{-x^2} \frac{d^2}{dx^2} - 2xe^{-x^2}\frac{d}{dx}]$

So now we know that the Hermite polynomials solve the DE. Consider two different Hermite polynomials $H_n,H_m$  where:

$\frac{d^2H_n}{dx^2} -2x\frac{dH_n}{dx} = -\lambda_n H_n  \, \, \, (1)$

$\frac{d^2H_m}{dx^2} -2x\frac{dH_m}{dx} = -\lambda_m H_m \, \, \, (2) $

Now multiply $(1)$ by $H_m$ and $(2)$ by $H_n$:

$H_m\frac{d^2H_n}{dx^2} -2xH_m\frac{dH_n}{dx} = -\lambda_n H_nH_m  \, \, \, (3)$

$H_n\frac{d^2H_m}{dx^2} -2xH_n\frac{dH_m}{dx} = -\lambda_m H_mH_n \, \, \, (4) $

Now subtract $(4)$ from $(3)$ then divide by $e^{x^2}$ on the both sides and on the LHS you use the reverse product rule to get a derivative (I leave the details for you to workout):

$\frac{d}{dx}[e^{-x^2}(H_m\frac{dH_n}{dx} - H_n\frac{dH_m}{dx})] = (\lambda_m - \lambda_n)e^{-x^2}H_nH_m $

Then integrate both sides from $-\infty$ to $+\infty$

Using properties of the Hermite polynomials the LHS will be zero and you are left with:

$(\lambda_m - \lambda_n)\int_{-\infty}^{\infty}e^{-x^2} H_mH_n dx = 0$

and since $m \neq n$ then $\lambda_m - \lambda_n \neq 0$ so the integral must be equal zero.


Rotated Ellipse (Parametric) - Determining Semi-Major and Semi-Minor Axes Given the parametric equation
$$\big(\;a \cos(\alpha+\theta), \;\;b\sin(\beta+\theta)\;\big)$$
with parameter $\theta$, how can we determine the length of the semimajor and semiminor axes, as well as the angle of tilt of the ellipse?

By experimentation it can be shown that by varying $\alpha, \beta$, the resulting ellipses are tangential to the rectangle defined by $|x|=a, |y|=b$.

See Desmos implementation here.

The parameterization $\mathbf r(t)=\left(w\cos(t+\alpha),h\sin(t+\beta)\right)$ has the interesting property that $\dot{\mathbf r}(t)=\mathbf r(t+\pi/2)$, which means that for any $t$, the points $\mathbf r(t)$ and $\mathbf r(t+\pi/2)$ are endpoints of conjugate diameters. (I’ve changed the parameter names to $w$ and $h$ in order to use $a$ and $b$ for the ellipse’s semiaxis lengths. I’m also assuming that $w,h\gt0$ since the solutions for negative values of these parameters can be obtained by reflection.) The area of the triangle formed by the halves of a pair of conjugate diameters is constant, which gives us the identity $$a b = \det\begin{bmatrix}w\cos(t+\alpha) & h\sin(t+\beta)&1 \\ -w\sin(t+\alpha) & h\cos(t+\beta) & 1 \\ 0&0&1\end{bmatrix} = w h \cos(\alpha-\beta).$$ The sides of the bounding rectangle are perpendicular tangents to the ellipse, so the corners of the rectangle lie on the ellipse’s orthoptic, which in turn means that $$a^2+b^2=w^2+h^2.$$ We want the nonnegative solutions to this system of equations with $a\ge b$. With a bit of help from a symbolic algebra program, the semiaxis lengths of the ellipse can be found to be $$\left(\frac12\left(w^2+h^2\pm\sqrt{w^4+h^4-2w^2h^2\cos(2(\alpha-\beta))}\right)\right)^{1/2},$$ or equivalently, $${\sqrt2 w h \cos(\alpha-\beta) \over \left(w^2+h^2\mp\sqrt{w^4+h^4-2w^2h^2\cos(2(\alpha-\beta))}\right)^{1/2}}.$$ The linear eccentricity, a.k.a. distance from center to focus, is then $$\left(w^4+h^4-2w^2h^2\cos(2(\alpha-\beta))\right)^{1/4}.$$

There’s a very simple geometric construction for finding the axes of an ellipse: draw a circle with the same center as the ellipse that intersects it at four points. The sides of the rectangle thus formed are parallel to the axes of the ellipse. Unfortunately, in this case this construction doesn’t really translate well into an analytical solution, but we can turn to the polar equation of an ellipse relative to its center: $$r = {b\over\sqrt{1-(e\cos\theta)^2}}$$ from which $$\cos^2\theta = {r^2-b^2\over r^2e^2}.$$ We have $e^2=1-b^2/a^2$ and using $\mathbf r(t)\cdot\mathbf r(t)$ with a convenient value of $t$ for $r^2$, we get (again with the help of a program because I’m lazy) $$\cos^2\theta = \frac12 + {w^2-h^2 \over \sqrt{w^4+h^4-2w^2h^2\cos(2(\alpha-\beta))}}.$$ I’m not sure that there’s any good way to choose the correct signs for $\cos\theta$ and ultimately for $\theta$ itself automatically, but it’s easy enough to generate potential solutions from this and choose the correct one either by comparing to the graph of the curve or by trying out a few values.

There are other options for finding the axes of the ellipse, but they’re not as computationally attractive. One option is to solve $\lVert\mathbf r(t)\rVert^2=\mathbf r(t)\cdot\mathbf r(t)=a^2$ or $\mathbf r(t)\cdot\mathbf r(t)=b^2$ for $t$ and substitute back into $\mathbf r$. These equations can be solved analytically, although the solutions appear to be rather unpleasant-looking. Or, one might exploit symmetry: the reflection of $\mathbf r(t)$ in the line with slope $\tan\theta$ is $$x = h \sin (2 \theta ) \sin (\beta +t)+w \cos (2 \theta ) \cos (\alpha +t) \\ y = w \sin (2 \theta ) \cos (\alpha +t)-h \cos (2 \theta ) \sin (\beta +t).$$ Choose a convenient value of $t$ such as $\pi/2-\alpha$ or $-\beta$ and find values of $\theta$ for which the reflected point also lies on the ellipse. A general solution using this method doesn’t look promising, it might be useful for specific instances.

Yet another possibility is to find the values of $t$ for which the conjugate diameters are perpendicular: $$w^2\cos(t+\alpha)\sin(t+\alpha)=h^2\cos(t+\beta)\sin(t+\beta).$$ This equation also arises when trying to find the extrema of $\lVert\mathbf r(t)\rVert^2$. Since you know the linear eccentricity, it might also be possible to work up some other equations using the reflective property of ellipses, but I don’t think that they’re going to be any more tractable than these.

this solution isn't polar

we have $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ as our conic section equation which is rotated.

if $B^2-4AC<0$, then we have an ellipse (If $A=C$ and $B=0$, it is a circle).

if $B^2-4AC=0$, then we have a parabola.

if $B^2-4AC>0$, then we have a hyperbola (If $A+C=0$, then we have a rectangular hyperbola).

our goal is to find the canonical equation of the rotated ellipse

first, we should calculate the Eigenvalues of the following matrix:

$\alpha_1=\begin{bmatrix}A&\frac{B}{2}\\\frac{B}{2}&C\end{bmatrix}$.

for finding the Eigenvalues, we should rewrite the matrix as the following and then find its determinant; then find $\lambda$ (note that you'll find two lambdas) when $|\alpha_2|=0$ ($|\alpha_2|$ is the determinant of the matrix).

$\alpha_2=\begin{bmatrix}A-\lambda&\frac{B}{2}\\\frac{B}{2}&C-\lambda\end{bmatrix}$.

now, we should find the determinant of the following matrix:

$\beta=\begin{bmatrix}A&\frac{B}{2}&\frac{D}{2}\\\frac{B}{2}&C&\frac{E}{2}\\\frac{D}{2}&\frac{E}{2}&F\end{bmatrix}$.

now, find the canonical equation using the following formula:

$\frac{x^2}{-|\beta|/\lambda_1\lambda_2^2}$+$\frac{y^2}{-|\beta|/\lambda_2\lambda_1^2}=1$.

if you don't know how to find those determinants, please tell me in the comments.


What exactly is the paradox in Zeno's paradox? I have known about Zeno's paradox for some time now, but I have never really understood what exactly  the paradox is. People always seem to have different explainations.

From wikipedia:

In the paradox of Achilles and the Tortoise, Achilles is in a footrace with the tortoise. Achilles allows the tortoise a head start of 100 metres, for example. If we suppose that each racer starts running at some constant speed (one very fast and one very slow), then after some finite time, Achilles will have run 100 metres, bringing him to the tortoise's starting point. During this time, the tortoise has run a much shorter distance, say, 10 metres. It will then take Achilles some further time to run that distance, by which time the tortoise will have advanced farther; and then more time still to reach this third point, while the tortoise moves ahead. Thus, whenever Achilles reaches somewhere the tortoise has been, he still has farther to go. Therefore, because there are an infinite number of points Achilles must reach where the tortoise has already been, he can never overtake the tortoise.
__

And we then say that this is a paradox since he should be able to reach the tortoise in finite time? For me it seems like that in the paradox we are slowing down time proportionally. Aren't we then already using the fact that the sum of those "time sequences" make up finite time? I feel like there is some kind of circular logic involved here.

What exactly is the paradox?

It's only a paradox if you assume that the sum of (countable) infinitely many numbers cannot be finite. But modern mathematics has no problem with infinite sums that yield finite results - in the case of Zeno's paradox, the sum in question is $$
  \sum_{k=1}^\infty 2^{-k} = 1 \text{.}
$$

Not everything that is called a paradox is actually a logical inconsistency. Quite often, things only seem inconsistent because we inadvertedly make an additional assumption, which turns out to be wrong. In the case of Zeno's paradox, that is the assumption that infinite sums cannot yield finite results.

I think some people just can't believe that our universe works in such a way that there is an infinite descending chain of times even though the laws of Physics say that time is continuous and therefore there is an infinite descending sequence of times. They might be like "If the universe did work in that way, then some events at some times would have no original cause?" Although Calculus gives a solution to the problem, they might still be like "Of course calculus gives a solution in an ideal mathematical world but how is it possible that our universe really works in such a way that there is an infinite descending chain of times?" You can't mathematically prove that you won't some day go through a tunnel into an entirely different world like in the movie "Coraline", yet you strongly believe it will never happen and would consider it magic if it did happen. Maybe some people feel the same way about the universe working in such a way that time is continuous. Some people might try to come up with a possible solution that the fundamental laws are a Conway's game of life and when a small amount of time goes by in the emulated universe, a much larger amount of time goes by in the Conway's game of life.  There probably is a way for a Conway's game of life to simulate a universe where it's consistent with observations that time is completely continuous. In such a simulation, maybe we can never make a measurement to infinite precision after a finite amount of time but as more time goes by, we can keep making more accurate measurements and that's why we observe time as being continuous even though in the Conway's game of life, it's discrete.


Prove that there is a real number $a$ such that $\frac{1}{3} \leq \{ a^n \} \leq \frac{2}{3}$ for all $n=1,2,3,...$ Prove that there is a real number $a$ such that $\frac{1}{3} \leq \{ a^n \} \leq \frac{2}{3}$ for all $n=1,2,3,...$ Here, $\{ x \}$ denotes the fractional part of $x$.

My attempt: Clearly $a$ cannot be an integer because $\{ a^n \}=0$ for all $n \in \mathbb{N}$. Also $a$ cannot be a rational number because $\{ x \}$ is the same as $a$ modulo $1$. So the fractional part of $a$, at some point, will be less than $\frac{1}{3}$.

I try to find irrational $a$ which satisfies the inequality but I got no luck.

Can anyone give some hint to this question?

There exists $\alpha \in\left[\dfrac{16}{3},\dfrac{17}{3}\right]$ with the required property. To see this, we will construct an interval sequence
$$\left[\dfrac{16}{3},\dfrac{17}{3}\right]=[\alpha_{1},\beta_{1}]\supset [\alpha_{2},\beta_{2}]\supset\cdots\supset[\alpha_{n},\beta_{n}],$$
where $\alpha_{n}$ and $\beta_{n}$ are such that
$$\alpha^n_{n}-\dfrac{1}{3}=\beta^n_{n}-\dfrac{2}{3}=m_{n}\in \Bbb N^{+},$$
so that, for any $x\in [\alpha_{n},\beta_{n}]$, we have
$$\dfrac{1}{3}\le\{x^n\}\le\dfrac{2}{3}.$$

We construct the interval sequence by induction. Assume that we have $[\alpha_{n},\beta_{n}]$. Let
$$a=\alpha^{n+1}_{n},\quad\quad b=\beta^{n+1}_{n}.$$It follows that $$
b-a=(m_{n}+\dfrac{2}{3})\beta_{n}-(m_{n}+\dfrac{1}{3})\alpha_{n}>\dfrac{\alpha_{n}}{3}>\dfrac{5}{3}.$$
Then there exists $m_{n+1}\in \Bbb N^{+}$ such that
$$\left[m_{n+1}+\dfrac{1}{3},m_{n+1}+\dfrac{2}{3}\right]\subset[a,b].$$
We take
$$\alpha_{n+1}=\sqrt[n+1]{m_{n+1}+\dfrac{1}{3}},\qquad\beta_{n+1}=\sqrt[n+1]{m_{n+1}+\frac{2}{3}}.$$
Now $$\alpha^{n+1}_{n}=a<\alpha^{n+1}_{n+1}=m_{n+1}+\dfrac{1}{3}<\beta^{n+1}_{n+1}=m_{n+1}+\dfrac{2}{3}<b=\beta^{n+1}_{n},$$and hence $\alpha_{n}\le\alpha_{n+1}<\beta_{n+1}<\beta_{n},$ or
$$[\alpha_{n},\beta_{n}]\supset[\alpha_{n+1},\beta_{n+1}].$$

Obviously it depends how approximately a≈5.62144865272, but a quick check with windows calculator shows that a^14 ≈ 31468458796.31579.. or a^15 ≈ 176898325303.72424... for instance


remainder on division by 9 for a tricky number What will be the remainder when $ 32^{32^{32}} $ is divided by 9?

I was able to solve this by using the cyclicity of remainders when $2^{2^n}$ is divided by 9. For $n$ =even it gave remainder 7 and for $n$ =odd it gave remainder 4. So using this I worked out the above to get answer as 4.

However, can there be a more concise way possible using some theorem like Fermat's or something else?

Euler's theorem says that any number coprime to $9$, raised to the sixth power is congruent to $1$ mod $9$. We have that $32$ is coprime to $9$, so $32^6\equiv 1 \mod 9$.

Now, we need to know what the exponent $32^{32}$ is mod $6$. By the chinese remainder theorem, we can reduce to mod $2$ and mod $3$. The exponent is obviously even, so it's congruent to $0$ mod $2$. It then remains to check congruency modulo $3$. We have (working mod $3$):
$$
32^{32} \equiv 2^{32} \equiv (2^2)^{16} \equiv 1^{16} \equiv 1
$$
so that finally we get $32^{32}\equiv 4$ mod $6$.

Back to the original question, this means that for some natural number $n$
$$
32^{32^{32}} = 32^{6n + 4} \equiv 32^4 \equiv 5^4 \mod 9
$$
This is not too difficult to calculate, since $5^2 = 25 \equiv -2$ mod $9$, so $5^4 \equiv (-2)^2 \equiv 4$.


Is infinity a number? Is infinity a number? Why or why not?

Some commentary:
I've found that this is an incredibly simple question to ask — where I grew up, it was a popular argument starter in elementary school — but a difficult one to answer in an intelligent manner. I'm hoping to see a combination of strong citations and solid reasoning in the answers.

It comes down to the definition of "number," as well as the definition of "infinity." Personally I don't think it's worth having an opinion on this subject; there are more precise words than "number" and "infinity" in mathematics. Historically the word "number" has come to mean an increasingly general list of things:

a positive integer
an integer
a rational number
a constructible number (say, to the ancient Greeks)
a real number
a complex number
a cardinal number
an ordinal number
a surreal number...

The word "infinity" has also come to mean an increasingly general list of things: it might refer to

a countably infinite set
an uncountably infinite set
the point at infinity in projective geometry
one of the two new points in the two-point compactification of the real numbers
the new point in the one-point compactification of the complex numbers, also known as the Riemann sphere
an infinite cardinal number
an infinite ordinal number...

Some of these meanings are compatible, as the above list demonstrates. But again, there are more precise words than "number" and "infinity" in mathematics, and if you want to get anywhere you should learn what those words are instead.

Here are some of those more precise words.

A set is a formalization of the intuitive notion of a bag of objects, and we can talk about finite or infinite sets. For example, $\{ 1, 2, 3 \}$ is a finite set, whereas the set of natural numbers is an infinite set. One can do arithmetic with sets in a way that leads to the arithmetic of the natural numbers: for example, taking the disjoint union corresponds to addition, and taking the Cartesian product corresponds to multiplication. These ideas lead to the arithmetic of the cardinal numbers, and similar ideas lead to the arithmetic of the ordinal numbers.
A ring is a formalization of the intuitive notion of a set of things you can add and multiply, so in some sense one can regard elements of rings as "generalized numbers" (but note that not every generalization I listed above can be interpreted in this way). When certain people say that "infinity is not a number," what they mean is that you can't adjoin an element called $\infty$ to the ring $\mathbb{R}$ of real numbers such that addition and multiplication do what you want them to do, the basic problem being that $\infty - \infty$ can't be consistently defined to satisfy the other rules of arithmetic if you also want it to be true that $n + \infty = \infty$ for any finite $n$.
A field is a commutative ring in which it's also possible to divide by nonzero elements. Some people would like to say that $\frac{1}{0} = \infty$, but by mathematical convention the element $0$ never has a multiplicative inverse in a field, the basic problem being that $0 \cdot \infty$ can't be consistently defined to satisfy the other rules of arithmetic. However, one can make sense of the expression $\frac{1}{0}$ in projective geometry; it describes the point at infinity on the projective line.
A topological space is an abstract setting for ideas like nearness and taking limits. Sometimes we don't want to view $\mathbb{R}$ as a ring, but as a space (the real number line), and we can talk about embedding this space into a larger space where more limits exist: this is known as compactification, and is an extremely useful tool in mathematics and physics. For example, we would like to say that the sequence $1, 2, 3, ... $ has limit $\infty$ in some sense, and we can do this compactifying $\mathbb{R}$.

To me infinity is just a way to represent something which cannot be written or expressed with common mathematics writings.

I don't really understand infinity as a precise number, I just understand it as some kind of domain for huge numbers, some place beyond our imagination.

The Knuth arrow notation http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation is a very good way to represent big numbers and compare them, but it won't help you understand why some fractions with the exp and ln functions will behave: infinity is there as some kind of numerical evaluation, it has no precise use.

A pretty good way to understand what infinity really means in term of reality, so that you can understand infinity is just an abstract and flawed, unprecise way to represent a value, try to trace the sin(1/x) function with some tool or other, try to zoom and try to understand what is happenning.

http://www.math.tamu.edu/~sottile/teaching/03S/sin1x-movie.html

Try also to understand convergent and divergent sequences.


Expectation and sign of the Radon-Nikodym derivative I am new here. I have some questions about the Radon-Nikodym derivative. I hope someone is willing to help me with these. The questions are stated below. Also I added my attempts to the problem to below all questions.

Question 1)

Let $(\Omega, 2^{\Omega},P)$ be a probability space such that $\Omega$ has a finite number of elements, and let $P(\omega)>0, \forall \omega \in \Omega$. Let $Q$ be a probability measure, equivalent to $P$. Prove that there is a random variable $Z$ s.t. $Q(F)=E^{P}(1_{F}(\omega)Z), \forall F\in 2^{\Omega}$

Now it turns out that this random variable $Z$, denoted as $Z=\frac{dQ}{dP}$ is called the Radon-Nikodym derivative.

Question 2)

Prove that this random variable is nonnegative and has expectation 1 w.r.t. the probability measure $P$. Hint (to prove that this R.V. is non-negative almost sure): you could use $(-\infty,0)=\cup_{n \geq1, n \in \mathbb{N}}(-\infty,-\frac{1}{n}) $

My attempts:

Q1] $E^{P}(1_{F}Z)=\sum_{\omega \in \Omega}1_{F}(\omega)Z(\omega)P(\omega)=\sum_{\omega \in F}Z(\omega)P(\omega)=Q(F)=\sum_{\omega \in F}Q(\omega)$.

Hence $Z(\omega)=\frac{Q(\omega)}{P(\omega)}, \forall \omega \in F$, $\forall F \in 2^{\Omega}$. is a random variable s.t. $Q(F)=E^{P}(1_{F}(\omega)Z), \forall F\in 2^{\Omega}$.

Q2] I think I am able to prove that this random variable $Z$ has expectation 1. I would do this is follows, however I am not sure!
$E^{P}(Z)=E^{P}\frac{Q(\omega)}{P(\omega)}=\sum_{\omega \in \Omega} P(\omega)\frac{Q(\omega)}{P(\omega)} = \sum_{\omega \in \Omega}Q(\omega)=1 $

So actually my biggest problem is to prove that this random variable $Z$ is nonnegative almost sure. I really did think about it, but can't come up with a reasonable answer where I also use the given hint. I Hope you are willing to help me with this, since I have shown now that I did think about these questions. However, I'm really stuck with this last problem, that $Z$ is nonnegative almost sure.

Yes indeed, $Q(\omega)$ and $P(\omega)$ are bigger than zero by definition and therefore the ratio is bigger than zero. But My problem still is that I did not use the hint which is given which feels wrong to me. Also, I think that $\frac{Q(\omega)}{P(\omega)}$ is just an example of such a $Z$. So in that case, I've proven only for this special case that the sign is positive almost sure and that it's expectation is one. Or am I wrong with this?

Thanx in advance!

I think I might have found the answer for the nonnegativity.

Let $F_n = Z^{-1}[(−∞,−(1/n))] = \{\omega \in \Omega : Z(\omega) \in (-\infty,-(1/n))\}$.
Note: $F_n \in \mathcal{F}$, since $Z$ is a random variable.
Suppose $F_n \neq \emptyset$. Then we get:

$Q(F) = E^P[Z\chi_F]<0$. This contradicts with the fact that Q is a probability measure.
Hence, $F_n = \emptyset$. Now we find:

$Z^{-1}[(-\infty,0)] = Z^{-1}[\cup_{n\in\mathbb{N}}(-\infty,(1/n))] = \cup_{n\in\mathbb{N}}Z^{-1}[(-\infty,(1/n))] = \cup_{n\in\mathbb{N}}F_n = \cup_{n\in\mathbb{N}}\emptyset = \emptyset $.

Hence, there exists no $\omega \in \Omega$ such that $Z(\omega)<0$.

I always mess up with almost sure things, so maybe you can find how to include that, but I think this is most of the proof.


In a model category, is the full subcategory of fibrant objects a reflective subcategory? I apologize in advance if my question is utterly stupid, but I can't resist asking it. So...

Is it true that in a model category ( - for example $\mbox{Set}_\Delta$ with the Joyal model structure - ) the full subcategory of fibrant objects is a reflective subcategory? More concretely, is the fibrant replacement functor a left adjoint to the inclusion functor?

Thanks.

Although most model structures appearing "in practice" do not have a reflective subcategory of fibrant objects (unless every object is fibrant), the following result of A. Salch provides many examples that do.

Theorem. Let $\mathcal{C}$ be a complete and cocomplete category, and let $\mathcal{A}$ be a full subcategory which is reflective and replete (i.e., closed under isomorphisms). Then there exists a model structure on $\mathcal{C}$ such that $\mathcal{A}$ consists of the fibrant objects.

See Corollary 6.3 here:

http://arxiv.org/abs/1501.00508v1

The model structure is obtained as a (left) Bousfield localization of the discrete model structure on $\mathcal{C}$. The weak equivalences are the maps that become isomorphisms upon applying the reflector $\mathcal{C} \to \mathcal{A}$.

In the case that all isomorphisms in the model category are fibration, cofibration, and weak equivalences, (trivial cofibrations, fibrations) and (fibrations, trivial cofibrations) form factorization systems in the sense of Borceux:

  A factorization system in a category $\mathbf B$ is defined as a pair
  $(E, F)$ of classes of arrows in $\mathbf B$ such that

  every isomorphism belongs to both E and M
  both E and M are closed under composition
  $\forall e \in E \forall m \in M. e \perp m$ (i.e. e has the LLP wrt m)
  every arrow f in B can be factored as f = me with e in E and m in M

Then the following theorem in Borceux's Handbook of Categorical Algebra, volume 1, page 211, gives an affirmative answer:

  Suppose $\mathbf B$ is a category with a terminal object 1 and a factorization system (E, F). For each object $B \in \mathbf B$, factor the unique arrow $B \xrightarrow ! 1$ into $B \xrightarrow {e_B} r(B) \xrightarrow {f_B} 1$. Then $r$ is a reflection of $\mathbf B$ into the full subcategory spanned by all $r(B)$, and each $e_B$ is the unit of the reflection.


Discriminant of $x^n-1$ Question is to find discriminant of polynomial $x^n-1$

I consider $f(x)=x^n-1=(x-a_1)(x-a_2)(x-a_3)\cdots(x-a_n)$

Now, $$f'(x)=[(x-a_2)(x-a_3)\cdots(x-a_n)]+\cdots+[(x-a_1)(x-a_2)\cdots(x-a_{n-1})]$$

$f'(a_1)=(a_1-a_2)(a_1-a_3)\cdots(a_1-a_n)$

$f'(a_2)=(a_2-a_1)(a_2-a_3)\cdots(a_2-a_n)$

$f'(a_3)=(a_3-a_1)(a_3-a_2)\cdots (a_3-a_n)$

and so on.. Now i need to know how many sign changes do i need to get something which looks like discriminant

I would write this in a matrix form to get some idea...
$$\begin{bmatrix}12&13&14&15&\cdots&1n\\21&23&24&25&\cdots&2n\\31&32&34&35&\cdots&3n\\\\n1&n2&n3&n4&\cdots&n(n-1) \end{bmatrix}$$

See that i first row every element is in correct position i mean of the form $ij$ for $i<j$

In second row only one element is odd of the form $ij$ with $i>j$ but i want $i<j$ in discriminant so i would change this.. So my count starts... change sign 1

In third row there are two elements which are not behaving properly... So, I should change them also.. So, Now another  two changes... On the whole $1+2$ changes...

In fourth row there would be $3$ misbehaving children so my count is $1+2+3$

In last row every body is behaving badly so i have to make $n-1$ changes in that..

On the whole i have to make $1+2+3+\cdots +n-1=\dfrac{n(n-1)}{2}$ changes..

So, $f'(a_1)f'(a_2)\cdots f'(a_n)=(-1)^{\dfrac{n(n-1)}{2}}(a_1-a_2)^2(a_1-a_3)^2\cdots (a_{n-1}-a_n)^2=(-1)^{\dfrac{n(n-1)}{2}} Disc(f)$

But then $f'(x)=nx^{n-1}$

This tells me that $f'(a_i)=n(a_i)^{n-1}$

So, $$Disc (f)=(-1)^{\dfrac{n(n-1)}{2}}  n^n(a_1a_2\cdots a_n)^{n-1}=(-1)^{\dfrac{n(n-1)}{2}}  n^n(-1)^{(n-1)^2}=(-1)^{\dfrac{n(n-1)}{2}+(n-1)^2}$$

As $(-1)^{n-1}=(-1)^{(n-1)^2}$ i would replace my $(-1)^{(n-1)^2}$ with $(-1)^{n-1}$

$$Disc (f)=(-1)^{\dfrac{n(n-1)}{2}}  n^n(a_1a_2\cdots a_n)^{n-1}=(-1)^{\dfrac{n(n-1)}{2}}  n^n(-1)^{n-1}=(-1)^{\dfrac{n(n-1)}{2}+n-1}n^n$$

i.e., $$Disc(x^n-1)=(-1)^{\dfrac{(n-1)(n+2)}{2}}n^n$$

As nobody was giving answer to my question i was trying my best and started editing this when ever i thought i find some thing and this is its final stage... This is fully solved now...

Thank you..

Any other ways of approaches would be appreciated..

Thank you :)

Your method seems fine, you just have to be careful and count the number of sign changes you need to make in the formulas for $f'(a_i)$ in order to get it on the right form.

We know that

$$\text{disc}(f) = \prod_{i<j}(a_i-a_j)^2$$

Now

$$f'(a_1) = (a_1-a_2)(a_1-a_3)\ldots (a_1-a_n) = (-1)^{0}(a_1-a_2)(a_1-a_3)\ldots (a_1-a_n) $$
$$f'(a_2) = (a_2-a_1)(a_2-a_3)\ldots (a_2-a_n) = (-1)^1(a_1-a_2)(a_2-a_3)\ldots (a_2-a_n) $$
$$f'(a_3) = (a_3-a_1)(a_3-a_2)\ldots (a_3-a_n) = (-1)^2(a_1-a_3)(a_2-a_3)\ldots (a_3-a_n) $$
$$\ldots $$
$$f'(a_{n}) = (a_{n}-a_1)(a_{n}-a_2)\ldots (a_{n}-a_n) = (-1)^{n-1}(a_1-a_{n})(a_2-a_{n})\ldots (a_{n-1}-a_n) $$

Above I have switched the sign of the factors which has the wrong form: $(a_i - a_j)$ with $i>j$. This leads to a $(-1)^k$ factor.

Multiplying togeather the terms above and comparing with the definition of disc$(f)$ we see that:

$$\text{disc}(f) = \prod_{i<j}(a_i-a_j)^2 = \left(\prod_{i=1}^n f'(a_i)\right) \cdot (-1)^{1+2+\ldots + n-1} = (-1)^{n(n-1)/2} \prod_{i=1}^n n a_i^{n-1}$$

the last product is as you know just $\prod_{i=1}^n n a_i^{n-1} = \prod_{i=1}^n n a_i^{n}/a_i = \frac{n^n}{a_1a_2\ldots a_n} = (-1)^{1+n}n^n$ since $f(0) = (-a_1)(-a_2)\ldots (-a_n) = -1$. This gives

$$\text{disc}(f) = n^n(-1)^{n(n-1)/2}(-1)^{1 + n} = n^n(-1)^{n(n+1)/2 + 1}$$

Consider $f(x)=x^n-1$.

You want to find the discriminant.

One way of doing that is to find what are the roots of the equation.

The equation is $x^n=1$.

For $n$ odd, only one unique root, which is $x=1$.

For $n=2$, the roots are $1$ and $-1$.

For $n$ even and $n>2$, you can easily come back to the case where $n$ is odd and the case where $n=2$ since $n=2^pq$, with $p$ and $q$ integers, and $q$ odd.

And you can show very easily that except for $n>2$, there are at least 2 identical roots to the equation...

EDIT: I was placing myself in the case of real numbers. The other answer if the one in C.

EDIT 2: complete answer found here: Calculating $\prod (\omega^j - \omega^k)$ where $\omega^n=1$.


If I generate a random matrix what is the probability of it to be singular? Just a random question which came to my mind while watching a linear algebra lecture online.  The lecturer said that MATLAB always generates non-singular matrices. I wish to know that in the space of random matrices, what percentage are singular? Is there any work related to this?

A square matrix is singular if and only if the determinant is zero. The determinant is a polynomial in the entries of the matrix. So your question can be thought of as asking the probability that a matrix $A$, viewed as a point in $\mathbb{R}^{n^2}$, will land in the zero set of a polynomial.

The zero set of a polynomial has Lebesgue measure 0. For a proof, see for instance http://www1.uwindsor.ca/math/sites/uwindsor.ca.math/files/05-03.pdf.

Therefore its common to say that this probability is zero.

Note this isn't strictly correct because the Lebesgue measure on $\mathbb{R}^{n^2}$ is not a probability measure. If the entries of the matrix are, say, (multivariate) normally distributed, you are working with an actual probability measure, but the probability is still zero because this measure is absolutely continuous with respect to Lebesgue measure.

The claim is false.  MATLAB can be made to produce a matrix containing only zeroes.  For instance,

zeros(5,5)

produces a $5 \times 5$ matrix of zeroes.  This matrix is singular and MATLAB can determine this.

det(zeros(5,5))
(*  0  *)

Further, it is no challenge to direct MATLAB to create matrices which are just copies of a single nonzero vector, which will also be singular.  See, for instance, ones(), which can produce a matrix of all ones.


Use of inequality $1 - \cos (x) \leq x^2 /2$ In this answer the value of $1 - \cos(x)$ has to be evaluated in order to find its upper limit, if it exists.

In particular, $x = 2 \pi / n$. The answer is related to the length of a side of a regular $n$-gon inscribed into a unit-radius circumference; because the perimeter of the $n$-gon is always less than $2 \pi$, the single side must always be less than $2 \pi / n$.

The inequality

$$1 - \cos (x) \leq \displaystyle \frac{x^2}{2}$$
(1)

is used and the proof is completed with

$$2(1 - \cos(x)) \leq (2 \pi / n)^2$$

$$\sqrt{2(1 - \cos(x))} \leq 2 \pi / n$$

But it is well known that the cosine is a function $f(x) \in [-1;1]$, so $1 - \cos (x) \in [0,2]$. By using this information, we would obtain

$$1 - \cos (x) \leq 2$$
(2)

The proof would provide

$$2(1 - \cos(x)) \leq 4$$

$$\sqrt{2(1 - \cos(x))} \leq 2$$

which is a completely different result.

Why in that case it is preferable to use (1) instead of (2)?
How to choose when it is convenient to use (1) and when to use (2) in a proof?

The bound $-1\le \cos(x)\le 1$ is correct, but rather a crude one.

We can easily obtain tighter bounds using the inequality from elementary geometry

$$|\sin(\theta)|\le |\theta| \tag 1$$

for all $x$, and the half-angle formula for the sine function

$$1-\cos(\theta)=2\sin^2(\theta/2) \tag 2$$

Squaring both sides of $1$, substituting $\theta =x/2$, and using $(2)$ reveals

$$\begin{align}
\sin^2(x/2)&=\frac{1-\cos(x)}{2}\\\\
&\le x^2/4 \tag 3
\end{align}$$

whereupon we find for all $x$

$$\bbox[5px,border:2px solid #C0A000]{1-\cos(x)\le \frac12 x^2} \tag 4$$

For values of $x<2$, $\frac12 x^2<2$ and $(4)$ provides a tighter bound than $1-\cos(x)\le 2$.  For $x\ge 2$, it is still correct that $1-\cos(x)\le \frac12 x^2$, but the inequality $1-\cos(x)\le 2$ is obviously tighter.  Therefore, we can write

$$1-\cos(x)\le \begin{cases}\frac12 x^2&,x<2\\\\2&,x\ge 2\tag 5\end{cases}$$

So, $(5)$ provides a guideline for the appropriate use of the bounds for $1-\cos(x)$.

ncmathsadist's technique can be extended for a tighter bound as described below. It also extends the region over which the bound is better than $1 - \cos (x) \leq 2$ per Mark Viola's answer.
Cutting off the Taylor series for $\sin(t)$ before a negative term (the terms alternate sign), one gets a polynomial bound on $\sin(t)$ that can be integrated.
For example $\sin (t) \leq t-t^3/3!+t^5/5!$, therefore
$$1-\cos (x)=\int_0^x \sin (t)\ dt \leq \int_0^x t-\frac{t^3}{3!}+\frac{t^5}{5!}\ dt = \frac{t^2}{2!} - \frac{t^4}{4!} + \frac{t^6}{6!} $$
The cut-off approximation is an upper bound. The cut-off residual will always be negative. This can be shown by summing the terms pairwise. Since $\sin (t)$ is absolutely convergent, there isn't an issue with ordering of terms.
A bound which is better than $1 - \cos (x) \leq 2$ from $[0,2\,\pi]$ is:
$$1-\cos (x) \leq \frac{t^2}{2!} - \frac{t^4}{4!} + \frac{t^6}{6!} - \frac{t^{8}}{8!} + \frac{t^{10}}{10!} - \frac{t^{12}}{12!} + \frac{t^{14}}{14!} \leq 2$$
You can then see that all this really says is:
$$\cos (x) \geq 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \frac{t^6}{6!} + \frac{t^{8}}{8!} - \frac{t^{10}}{10!} + \frac{t^{12}}{12!} - \frac{t^{14}}{14!}$$
...or $\cos (x)$ is greater than its Taylor expansion with the cut-off starting at a positive term. Really not that impressive a result. :-(


Write each of the following functions as a composition of functions chosen from $f$, $g$, and $h$. Let $f(x)=|x|$, $g(x)=x-7$, and $h(x)=x^2$.

$A(x)=x^2-7$
$B(x)=x^2-14x+49$
$C(x)= |x-7|$
$D(x)=(x^2-7)^2$

For number one I tried using $(fog)(x)= f(g(x))$ but I got the answer as being the absolute value of $x-7$ which is wrong. Please help.
I understand how it works now. Thank you very much everyone!

$A = g\circ h$ since $g(h(x)) = h(x)-7 = x^2-7$
$B = h\circ g$ since $h(g(x)) = (x-7)^2 = x^2-2\cdot 7x + 49$
$C = f\circ g$ since $f(g(x)) = \lvert g(x)\rvert = \lvert x-7\rvert$
$D = h\circ g\circ h$ since $h(g(h(x))) = \left(g(x^2)\right)^2 = (x^2-7)^2$


$\epsilon-\delta$ argument for continuity for $f(x) = \sin(x)$ at $x=0$ Given
$$\lim_{x\to c} f(x) = L$$
We want to show that $\forall \epsilon>0, \exists \delta>0$ such that
$$|x-c| < \delta \implies |f(x) - L | < \epsilon$$

Our $f(x)=\sin(x)$ and our $c=0$, and $L=0$. So, we suppose
$|x|<\delta$ and need to show $|\sin(x)|<\epsilon$.

My teacher did this by drawing a picture and showing it holds, but I wanted to see how it would work strictly algebraically.

hint: consider $g(x)=x-sin(x)$. $g'(x)=1-cos(x) \ge 0$ which means that $g(x)$ is increasing and because $g(0)=0$ we conclude that $x - sin (x) \ge 0$

As was already written, it depends on definition of sine.
But if we suppose y-coordinate of point of unit circle as definition,
it is obvious that for $\varepsilon>1$ is any $\delta$ good
and if $0<\varepsilon<1$ then $\delta=\inf{\{t;t>0\wedge \sin{\left(t\right)}>\varepsilon\}}$


Integral solutions of $x^2+y^2+1=z^2$ I am interested in integral solutions of $$x^2+y^2+1=z^2.$$ Is there a complete theory comparable to the one for $x^2+y^2=z^2?$

The equation defines a conic with a rational point (0,0,1). The other rational solutions can be parametrized by lines through that point.  As with $x^2+y^2 = z^2$ the integer solutions can be deduced from the rational ones, and the quadratic form $x^2+y^2 - z^2$ has a large linear symmetry group allowing one to move between solutions.  In these respects the theory is the same as the one for Pythagorean triples.

There is a difference in the structure of the symmetry group.  For $a=0$ the hyperboloid $x^2+y^2 - z^2 = a$ becomes a cone, with additional scaling symmetries in addition to the linear transformations of the hyperboloid; in effect the $O(2,1)$ symmetry group of the rational solutions collapses into a product of scalings and circle isometries.  For integer solutions, with Pythagorean triples there is a reduction to the case of primitive triples, but when $a \neq 0$ there is a bound on the shared factors of $(x,y,z)$, and for $a = \pm 1$ all integer solutions are primitive.  The organization of Pythagorean triples using several 3x3 integer matrices as transformations connecting different solutions does use the $O(2,1)$ structure, and the solution set for the equation with $a \neq 0$ could be presented in the same way (possibly with more than one connected component when $|a| > 1$).

All of numbers can be any character.In Equation:  $qX^2+Y^2=Z^2+a$

If the ratio is factored so:  $a=(b-c)(b+c)$

Then we use the solutions of Pell's equation: $p^2-fs^2=\pm1$

where:  $f=(q+1)k^2-2kt-(q-1)t^2$

Then the solutions are of the form:

$X=2(ck-bt)ps+2(bk^2-(b+c)kt+ct^2)s^2$

$Y=bp^2+2c(k-t)ps-(b(q-1)k^2+2(b-qc)kt+b(q-1)t^2)s^2$

$Z=cp^2+2b(k-t)ps+(c(q+1)k^2-2(bq+c)kt+c(q+1)t^2)s^2$

All of numbers can be any character.


How to solve this trigonomectric equation $2^{\lvert\sin x\rvert} + 2^{\lvert\cos x\rvert} = 3$? I thought its solutions are $(\sin x;\cos x) = \{(\pm1;0),(0;\pm 1)\}$
But how can I solve it?
Please help :(

Give the periodicity and symmetries of the trig functions, it suffices to show that $f(u)=2^u+2^{\sqrt{1-u^2}}$ is increasing for $0\lt u\le1/\sqrt2$.  That will imply that the only solutions to $2^{|\sin x|}+2^{|\cos x|}=3$ occur at multiples of $\pi/2$.
We have
$$f'(u)=\ln2\left(2^u-{u\over\sqrt{1-u^2}}2^{\sqrt{1-u^2}}\right)$$
so it's enough to show that
$${2^u\over u}\gt{2^{\sqrt{1-u^2}}\over\sqrt{1-u^2}}$$
for $0\lt u\lt1/\sqrt2$. But notice that in this range we actually have $0\lt u\lt1/\sqrt2\lt\sqrt{1-u^2}\lt1$ (i.e., $0\lt\sin x\lt1/\sqrt2\lt\cos x\lt1$ for $0\lt x\lt\pi/4$). So we need only prove that the function $g(u)=2^u/u$ is decreasing for $0\lt u\lt1$. This is equivalent to showing that $h(u)=(\ln2)u-\ln u$ is decreasing on the same range, which is clear since
$$h'(u)=\ln2-{1\over u}\lt0\iff0\lt u\lt{1\over\ln 2}$$
and $\ln2\lt1$.

For simplicity we have to solve for $x\ge 0$ $$f(x)=2^x+2^{\sqrt{1-x^2}}=3$$.
Note that $$f'(x)=2^x\log 2-\frac{2^{\sqrt{1-x^2}}x\log 2}{\sqrt{1-x^2}}$$
if   $x\le \frac{1}{\sqrt{2}}$ then $x\le \sqrt{1-x^2}$
then we have
Lemma: $$\frac{2^x}{x}\ge \frac{2^{\sqrt{1-x^2}}}{\sqrt{1-x^2}}$$
proof: show that $g(x)=\frac{2^x}{x}$ is decreasing in given interval.
From this lemma it follows that $f'(x)\ge 0$.Thus for all $x\in (0,\frac{1}{\sqrt{2}}]$ :$f(x)>f(0)=3$
Similarly prove that $f(x)$ is decreasing for all $x\in [\frac{1}{\sqrt{2}},1]$ hence deduce that $f(x)>f(1)=3$ for all $x\in [\frac{1}{\sqrt{2}},1)$.
The conclusion is now obvious:i.e $x=0,x=1$ are the only solutions.


Is it technically incorrect to write proofs forward? A question on an assignment was similar to prove:
$$2a^2-7ab+2b^2 \geq -3ab.$$
and my proof was:
$$2a^2-4ab+2b^2\geq0$$
$$a^2-2ab+b^2\geq0$$
$$(a-b)^2\geq0$$
which is true.

However, my professor marked this as incorrect and the "correct" way to do it was:

Starting from $$(a-b)^2\geq0$$ we have:
$$a^2-2ab+b^2\geq0$$
$$2a^2-4ab+2b^2\geq0$$
$$2a^2-7ab+2b^2 \geq -3ab.$$

His point was that if we start with a false statement we can also reduce it to a true statement (like $-5 =5$ we can square for $25 = 25$). I argued however that you can go back and forth between my operations (if and only if, which doesn't work for the $-5 =5$ example). He still didn't give me the marks for it. Which leads me to my questions:

Is my proof equally valid?
Do real mathematicians all write one way or the other when writing in a paper?

There are two ways to interpret your argument (which hardly counts as a proof if this is exactly what you have written in your assignment).

First Way
You actually meant to write, \begin{align}2a^2-7ab+2b^2 \geq -3ab&\implies 2a^2-4ab+2b^2\geq0\\&\implies a^2-2ab+b^2\ge 0\\&\implies (a-b)^2\ge0\end{align}
Second Way
You actually meant to write, \begin{align}2a^2-7ab+2b^2 \geq -3ab&\iff 2a^2-4ab+2b^2\geq0\\&\iff a^2-2ab+b^2\ge 0\\&\iff (a-b)^2\ge0\end{align}

In the second case your argument is correct but in the first case it is not (as your professor has elaborated via an example).
Now the answer to your questions,

If you didn't say explicitly in your assignment in which way your argument is to be interpreted and then if your teacher interprets your argument in the first way, you can't blame him/her for not giving you the corresponding marks of the question simply because it was you who failed to be explicit )and the first way of interpreting your argument indeed shows a misunderstanding of the working of $\implies$). So, if I were in the position of your professor, I would give you no marks.

I don't know what you mean by "one way or the other" here. The best way to answer this question will be to read the papers of some real mathematicians. However, when real mathematicians write a paper it is in general clear what they are assuming to be true and how the steps lead to the conclusion (which is not the case in your argument as I have explained above).

The teacher is correct, but hasn't explained it well.

Working one way doesn't always prove the opposite way. This is less obvious in basic maths, but it's important to grasp.  Why it doesn't becomes more obvious when you consider the kinds of operations used. With operations like inequalities, complex numbers, set inclusion and undefined operations like division (by zero), it is often the case that what works one way just doesn't prove anything about what works the other way, unless you actually check it will work the other way. But you have to specifically check that.

Whichever way the answer comes to mind, you will have to work it out the right way to actually prove anything (rather than just suspect it)


An $N$-qubit system keeps $2^N$ pieces of information or should we say $2^{N+1}$? Say my quantum register is of size $3$ qubits. This means I'll need $2^3$ complex
numbers to describe all of its possible arrangements.  But each complex number requires $2$ real numbers, so maybe I could say I'll need $2^{3}\times2 = 2^4$ real numbers.  Is this correct?

Granted, a complex number is a number.  But when we implement a complex number on any classical computer, we always reserve two numbers for it.  A complex number is a number, but down to computer technology it really is a pair of numbers.  When it comes to storing information, the matter is more about technology than abstract science.

I know in complexity theory it doesn't matter if $2^{N}$ or $2^{N+1}$, but I haven't seen any mention of this little detail anywhere and I've been reading every introduction on quantum computing I can find.  Please do mention any source that you might know that has touched on this little technicality.  Thank you!

You are right to say that every complex number (that represents an quantum amplitude) is represented by a pair of real number on a classical computer.
But, in my opinion, you are confusing a simulation of a real life quantum system
with a classical simulation.

In a classical simulation of a real life quantum system (say a quantum circuit that has not been built) we would associate a pair of real number-for representing a complex number that is a quantum amplitude of that quantum circuit -as you pointed out.
But if we actually build the quantum circuit in real life (rather than simulating it on a classical computer) then those complex amplitudes are actually an intrinsic part of the qubit (which may be implemented by ions, or photons)-because quantum states occur naturally-this is just basic physics.

Another thing to know is that the "amplitude model" of reality is based on the postulates of quantum mechanics.


Does this graph contain $K_5$ or $K_{3,3}$ as subdivision or minor? Does this graph contain subdivision of $K_5$ or $K_{3,3}$?

Does this graph contain  $K_5$ or $K_{3,3}$ as minor?

I'm not sure if I'm correct, but I think the answer is yes for both questions. $K_5$ has 5 vertices of degree 4, this graphs has 6 of them, and $K_{3,3}$ has 6 vertices of degree 3, and this graph also has 5 of them. are these enough to conclude that this graph says $G$ contain both $K_5$ and $K_{3,3}$ as subdivision and minor?

Denote the six inner vertices as $A,B,C,D,E,F$ in anticlockwise order. Denote the outer vertices as $A',B',C',D',E',F'$ accordingly.

Let's try subdivision of $K_5$: To find such, we need five vertices of degree (at least) $4$. Since there are six such vertices, we have littel choice. In fact, by symmetry, all possible choices are the same, so we may pick $A,B,C,D,E$. Then to exhibit a subdivision, we need to have a path from each of these vertices to each other vertex - and these paths must be vertex-disjoint (apart from the end points)!
Since the vertices are only of degree $4$, we must use all edges originating from our chosen vertices. This already completes $6$ of our needed $10$ paths: $AC, AD, AE, BD, BE, CE$.
Of the paths from $C$ to $B$ and from $C$ to $D$, one must make use of $C'$ and one must ake use of $F$.
Then for the path from $A$ to $B$ we must take  $AA'B'B$ because $F$ is already used.  Similarly we have $EE'D'D$. But then the path from $C$ to $B$ or $D$ that goes via $C$ cannot be comppleted.  We conclude that this graph does not contain a subdivision of $K_5$ as subgraph.

the answer is yes. For more defails see following literature:
http://www.math.ucla.edu/~mwilliams/pdf/petersen.pdf


Limits (Three Variable function). We're given :

$f(x,y,z) = \dfrac{xyz}{x^{2}+y^{2}+z^{2}}$ ,

Also , it's given that $\lim_{(x,y,z) \to (0,0,0)} f(x,y,z)$ exists.
We need to prove that $\lim_{(x,y,z) \to (0,0,0)} f(x,y,z) = 0$.

What I figured :

First I approach $(0,0,0)$ along $x$ -axis , and thus the limit becomes :

$\lim_{(x,y,z) \to (0,0,0)\dfrac{x.0.0}{x^{2}+0^{2}+0^{2}}}$ which is $0$.

Similarly approaching along $y$ and $z$ axis gives the limit to be $0$.

Now , since the limit at $(0,0,0)$ exists , it should be unique throughout , thus ,

$\lim_{(x,y,z) \to (0,0,0)} f(x,y,z) = 0$

Could anyone tell , am I right in making the above statement ?

Yes. But what I don't understand is why you were told that the limit exists. It is a totally redundant assumption.

Let $r = \sqrt{x^2+y^2+z^2}$.

Then $|x| \le r$ and $|y| \le r$ and $|z| \le r$.

Thus $|xyz| \le r^3$ and hence $\left| \frac{xyz}{x^2+y^2+z^2} \right| \le r \to 0$ as $r \to 0$.

Therefore $\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2+y^2+z^2} = \lim_{r \to 0} \frac{xyz}{x^2+y^2+z^2} = 0$.

Little observation: in general (if you do not know whether the limit exists), if the limit exists along one path it does NOT mean that it exists, so your proof for $(x,0,0)$ is not sufficient.
For the general case, therefore, I would either use the demonstration of answer $1$ , or majorize at the numerator $|xy|$ by $\frac{x^2+y^2}{2}$, and minorize at the denominator $x^2+y^2+z^2$ by $x^2+y^2$, in order to get that $0\leq |f(x,y,z)|\leq \frac{|z|}{2}\leq\frac{\sqrt{x^2+y^2+z^2}}{2}$, which goes to zero for $(x,y,z)$ going to $(0,0,0)$.
For the sandwich theorem, the limit of your function at the origin is zero.


Does $b_n = a_{n+1} - a_n $ is a zero sequence implies $a_n$ is convergent? Let $(a_n)$ be a sequence in $\mathbb{R}$ and let $b_n = a_{n+1} - a_n $ be a zero sequence. According to my intuition, I would say that $a_n$ converges. But my solution says otherwise. How could that be? I just can't find a counterexample.

Another nice example.  $a_n = \sqrt{n}$.  Show that
$$
\sqrt{n+1} - \sqrt{n} \to 0\qquad\text{but}\qquad
\sqrt{n} \to +\infty
$$


Sequences and Asymptotic Similarity We define a sequence by $a_0 = 0$ and $a_n = \ln(n!) - \left(n+\frac{1}{2}\right)\ln(n) + n$ for positive integer $n$.

If we know $a_n$ converges, i.e.,

$$\ln(n!) - \left(n+\frac{1}{2}\right)\ln(n) +n \to l, \text{ for some $l$ when }n \to +\infty,$$

why do we know that $\ln(n!)$ is asymptotically equaivalent to $\left(n+\frac{1}{2}\right)\ln(n)-n$, or in other words, why is the following true: $$\lim_{n\to +\infty}\frac{\ln(n!)}{\left(n+\frac{1}{2}\right)\ln(n)-n}= L,\, \text{ for some finite } L?$$

It follows that
$$
\mathop {\lim }\limits_{n \to  + \infty } \frac{{\log n!}}{{\left( {n + \frac{1}{2}} \right)\log n - n}} = 1 + \mathop {\lim }\limits_{n \to  + \infty } \frac{{a_n }}{{\left( {n + \frac{1}{2}} \right)\log n - n}} \\ = 1 + \frac{l}{{\mathop {\lim }\limits_{n \to  + \infty } \left( {\left( {n + \frac{1}{2}} \right)\log n - n} \right)}} = 1.
$$


Galois group of $x^6+3$ over $\mathbb Q$ I'm having some difficulties finding the Galois group of the polynomial $g(x)=x^6+3$ over $\mathbb Q$.

Here's what I did :
I observed that the roots of the given polynomial are $\sqrt[6]3 \xi_{12}^{k}$ where $\xi_{12}$ is a primitive 12-th root of the unity and $k=1,3,5,7,9,11$. Called $\mathbb{K}$ the splitting field of $g(x)$ over $\mathbb{Q}$, is obvious that $\mathbb{Q}(\sqrt[6]3,\xi_{12})=\mathbb Q(\sqrt[6]{3},i)\supseteq\mathbb{K}$ so $6|[\mathbb K:\mathbb Q]\le12$. But from this point I'm not able to continue rigorously.
Seems to me that $[\mathbb K:\mathbb Q]=6$ but I'm not sure on how to proof that.
Can anyone please help me? Thanks in advance!

Let $\alpha = \sqrt[6]{-3}$ be a root of $g$.  I claim that the splitting field $\mathbb{K}$ of $g$ is $\mathbb{Q}(\alpha)$.  The roots of $g$ are $\alpha, \zeta \alpha, \cdots, \zeta^5 \alpha$ where $\zeta$ is a primitive $6^\text{th}$ root of unity, so to prove that $\mathbb{Q}(\alpha)$ is the splitting field, it suffices to show that $\mathbb{Q}(\alpha)$ contains a primitive $6^\text{th}$ root of unity.

Note that $\alpha^3$ is a root of $x^2 + 3$, so $\alpha^3 = \pm \sqrt{-3}$.  Then $\zeta = \frac{1 + \alpha^3}{2} = \frac{1 \pm \sqrt{-3}}{2}$ is a primitive $6^\text{th}$ root of unity and $\zeta \in \mathbb{Q}(\alpha)$, as desired.  Thus $\mathbb{K} = \mathbb{Q}(\alpha)$ is the splitting field of $g$, hence $[\mathbb{K} : \mathbb{Q}] = 6$.

(For computing the Galois group of $g$, see this post.)

First, it can't be true that $ [ \mathbb{K} : \mathbb{Q} ] = 6$ since $[ \mathbb{Q}(\sqrt[6]{3}) : \mathbb{Q} ] = 6$ ($x^6 + 3$ is irreducible by Eisenstein criterion) and $\xi_{12} = \exp (i \pi / 6) = \frac{\sqrt{3}}{2} + \frac{i}{2} \in \mathbb{C}$ and hence $\xi_{12} \notin \mathbb{Q}(\sqrt[6]{3}) \subset \mathbb{R}$. You need to find $[ \mathbb{Q}(\sqrt[6]{3},\xi_{12}) : \mathbb{Q}(\sqrt[6]{3})]$. Note that $$ x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1)$$ is reducible. Moreover, $\sqrt{3} = (\sqrt[6]{3})^3 \in \mathbb{Q}(\sqrt[6]{3})$ and $i = \sqrt{-1}$ is algebric of degree 2 over this field, thus $\mathbb{K} = \mathbb{Q}(\sqrt[6]{3},i)$ is of degree 12 over $\mathbb{Q}$ since $[ \mathbb{Q}(\sqrt[6]{3},i) : \mathbb{Q}(\sqrt[6]{3}) ] = 2$ ($x^2 + 1$ is still irreducible there).


Is an irrational number odd or even? My sister just asked this question to me: "Is an irrational number odd or even?" I told her that decimals are not odd or even and that does imply that not recurring and non repeating decimals will also not be odd or even. But I want a rigorous math proof for this. Can someone help me with this proof? Also, can I do this proof by contradiction?

Definition.

  Let $n\in \mathbb{Z}$ be an integer. $n$ is called even if ...

According to the definition only integers are even or odd. It is not something that you have to prove.

Closely related to other answers but might be good for the purpose. By contradiction:

For a given irrational a, assume it is odd.
a/2 is also irrational; call it b
2b is obviously even since odd + odd = even
so a is even.
So all irrationals are even,
which shows they cannot be split into two categories at least.
All rationals can be expressed as the product of two irrationals
(which is  just saying that a rational divided by an irrational is irrational)
so all rationals are even too.

∴ 1 is even


Extremely tough indefinite integral This integral does indeed use special functions, so do include them here. Evaluate:

$\int \frac{1}{\sqrt{x}\ln(x)} dx$

$x = {\sqrt{x}}^{2} \space \text{let} \space u = \sqrt{x}$

$= 2\int \frac{1}{\ln(u^2)} du$

Now, I am not very experienced with special functions such as $Ei, li, Li$, etc...

Help please?

Hint

Change variable $x=e^{2t}$ and arrive to $$I=\int \frac{1}{\sqrt{x}\ln(x)} dx=\int \frac{e^t}{t} dt=\text{Ei}\left(t\right)$$

Puiseux expansion along the positive real axis is given by $$\text{Ei}\left(t\right)=\gamma+\log(|t|)+\sum_{k=1}^{\infty}\frac{t^k}{k~~ k!}$$

For example, for $t=e^2$, the expansion gives an error of $0.1$% using $13$ terms and $0.01$% after $18$ terms.

You're looking for a primitive I guess:

Let F(t) = $= \int \frac{u^t}{\ln(u)} du$

Prove that you can differentiate under the integral sign, and you get:

F'(t) = $= \int u^t du$ , F' can be easily calculated. You can then integrate on the "t" parameter and take the value of F in 0. I did not do the calculation all the way but it should work .


What are the Frobenius groups of order $100$? Question. Which groups of order $100$ are Frobenius groups?

The OEIS says that there are two Frobenius groups of order $100$, but I am finding three of them and I'd be grateful if someone can point out where I am going wrong.

(I was not able to locate an independent source for the Frobenius groups of order $100$, and my results agree with the OEIS up to order $99$.)

The three groups that I am finding are Frobenius are those with small group IDs [[100, 3], [100, 11], [100, 12]].

In each of these cases, I get the following:

the Frobenius kernel $K$ is the (necessarily abelian) Sylow $5$-subgroup, which is also the Fitting subgroup. This is (uniquely, up to conjugacy) complemented, by Schur-Zassenhaus. (For SmallGroup(100,3), $K\simeq C_{25}$, for the other two $K\simeq C_5^2$.)
the Frobenius complement $H$ is the Sylow $2$-subgroup of order $4$
in all three cases, $H$ is core-free
in all three cases, the action of the group on $H$ is Frobenius

The first three items are easy to verify.  For the last, these are the permutation representations that I get on the cosets of the putative complement $H$.

For SmallGroup( 100, 3 ):

$$\langle (2,8,25,21)(3,6,15,10)(4,20,24,9)(5,18,23,12)(7,19,22,11)(13,16,17,14), (2,25)(3,15)(4,24)(5,23)(6,10)(7,22)(8,21)(9,20)
(11,19)(12,18)(13,17)(14,16)\rangle$$

For SmallGroup( 100, 11 ):
$$\langle (2,4,11,7)(3,6,15,10)(5,13,25,21)(8,20,24,14)(9,22,23,12)(16,17,19,18), (2,11)(3,15)(4,7)(5,25)(6,10)(8,24)(9,23)(12,22)(13,
    21)(14,20)(16,19)(17,18), (1,2,4,7,11)(3,5,8,12,16)(6,9,13,17,20)(10,14,18,21,23)(15,19,22,24,25),
  (1,3,6,10,15)(2,5,9,14,19)(4,8,13,18,22)(7,12,17,21,24)(11,16,20,23,25) \rangle$$

For SmallGroup( 100, 12 ):

$$\langle (2,7,11,4)(3,6,15,10)(5,17,25,18)(8,9,24,23)(12,20,22,14)(13,19,21,16), (2,11)(3,15)(4,7)(5,25)(6,10)(8,24)(9,23)(12,22)(13,
    21)(14,20)(16,19)(17,18), (1,2,4,7,11)(3,5,8,12,16)(6,9,13,17,20)(10,14,18,21,23)(15,19,22,24,25),
  (1,3,6,10,15)(2,5,9,14,19)(4,8,13,18,22)(7,12,17,21,24)(11,16,20,23,25)\rangle$$

In each case, I check that the group is transitive, that the stabiliser of $1$ is non-trivial (as it must be), and that the two-point stabilisers are all trivial.

I've also checked (redundantly) that $C_G(h)\leq H$, for all $h\in H\setminus 1$, and that $C_G(k)\leq K$, for all $k\in K\setminus 1$, in each case.

I have replicated these calculations in two computer algebra systems with the same results, so I suspect that, rather than a software bug, there is a bug somewhere in my understanding.

I agree with your calculation. A routine search through the small groups library shows that only the groups $\mathtt{SmallGroup}(100,i)$ for $i=3,11,12$ are Frobenius groups.

Or you could prove it theoretically. It is easy to see that the only possible order of a Frobenius kernel $K$ in such a group is $25$. A Frobenius complement must be cyclic of order $4$ and act fixed point freely by conjugation on $K$.

If $K$ is cyclic then ${\rm Aut}(K)$ is cyclic and there is only one possible action.

Otherwise $K$ is elementary abelian and ${\rm Aut}(K) \cong {\rm GL}(2,5)$. Elements of order $4$ are diagonalizable, and there are  two two conjugacy classes of subgroups of order $4$ with fixed point free action, with representatives $\left\langle \left(\begin{array}{cc}2&0\\0&2\end{array} \right)\right\rangle$ and $\left\langle \left(\begin{array}{cc}2&0\\0&3\end{array} \right)\right\rangle.$

There are three non-isomorphic Frobenius groups of order 100, two with kernel K = $C_{5}^2$. Let h be an element of order 4 in FG of order 100 with kernel K. If h is in the FG of order 600 with cyclic complement, the conjugate by h of every k in K is $k^2$ (or, isomorphically, $k^{-2}$). If h is in the FG of order 600 whose complement has normal subgroup Q8, we can label the six subgroups of order 5 as A, B, C, D, E, F use $k_A$ for a generator of A, k*h as the conjugate of k by H, so that $k_A$*h = $k_A^2$, $k_B$*h = $k_B^{-2}$, $k_C$*h = $k_D$, $k_D$*h = $k_C^{-1}$, $k_E$*h = $k_F$, $k_F$*h = $k_E^{-1}$.


Please explain uniform distribution to me I am confused about Uniform Distribution
why does $$P(v < 2b1)$$

equal 2b1 ?

I guess $v$ is a random variable of uniform distribution on the interval $[0,1]$. Then, if $0\le b_1\le 1/2$, we indeed have

  $P(v<2b_1)=$ length of $[0,2b_1)$ which is $2b_1$.


Symmetric C* property What can be said about elements $a$ of a C*-algebra which fulfil the 'symmetric C* property'
$$\| a^\ast a+ aa^\ast\|=2\| a\|^2$$

I'd guess that this is not a general property, but I don't have a good idea which elements (except for self-adjoint and normal ones, of course) should satisfy this.

If you assume that $a^*a$, $a a^\ast$ are compact (this is just to avoid working with continuous spectrum), the equality above holds iff there is an eigenvalue $\lambda$ with maximal norm $| \lambda | = \|a\|^2$ such that the eigenspaces of $\lambda$ for $a^\ast a$ and $a a^\ast$ have nontrivial intersection. That already give you that the condition is pretty rare.

Indeed, two projections $p,q \in \mathcal{K}(\ell^2)$ satisfy that there is a partial isometry $a$ such that $p = a^\ast a$ and $q = a a^\ast$ iff $\dim(p\ell^2) = \dim(q \ell^2)$. But you can take $p \perp q$ and in that case:
$$
\| a^\ast a + a a^\ast \| = \max \{ \| a^\ast a \|, \| a a^\ast\| \} = 1.
$$


Graph contains a cycle containing at least 4 nodes There is a finite undirected graph in which the degree of every node is at
least 3.

How to prove that this graph contains a cycle containing at least 4 nodes.

No need to prove that a cycle with a chord, only need to show is that it contains a cycle containing at least 4 nodes.

I am assuming that parallel edges are not allowed.
Firstly, the graph cannot be a tree. Because trees have at least two vertices of degree one (contradict with the assumption). So, as the graph is not a tree, it should contain a cycle.

Let's assume that all cycles in the graph is of length 3. It can be proved that each vertex is at least in one cycle. Using the two previous sentences, the desired graph should be built using cycles of length 3 (triangles). We need to connect triangles to raise the degree of each vertex of triangles. It can be done either by using an edge connecting two vertices of each of the two triangles or by simply attach two vertices from each triangle (one of the vertices should be eliminated). In either case, the two connected vertices would satisfy the required condition (each vertex should be of a degree higher than 3). However, after connecting two triangles, no other edges can be drawn between them, for it makes a cycle of 4. So, raising the degree of the other vertices should be done using more triangles. Now, Let's consider a triangle belonging to the graph. We connect it to three other triangles. The first triangle satisfies the degree condition. However, the other triangles don't. As we cannot connect the edges ,with the degree 2, to the other parts of the current graph (we make a cycle of length more than 3 this way), we need to introduce more triangles and connect them like before. Continuing the process shows that the graph can not be made, for we have vertices of degree two in each iteration. Consequently, the assumption that we made at the beginning of this paragraph, is wrong.
I hope it helps with the main idea.


For two vectors $a$ and $b$, why does $\cos(θ)$ equal the dot product of $a$ and $b$ divided by the product of the vectors' magnitudes? While watching a video about dot products (https://www.youtube.com/watch?v=WDdR5s0C4cY), the following formula is presented for finding the angle between two vectors:

  For vectors $a$, and $b$,
  $$\cos( \theta ) =  \frac{(a, b)}{  \|  a  \|   \| b \| }$$
  where $(a,b)$ is the dot product of $a$ and $b$.

How is this formula derived?

There are several derivations of this online. Here's where you can start.

Define two vectors $\textbf{a}$ and $\textbf{b}$. Then $ \textbf{a} - \textbf{b}$ is the vector that connects their endpoints and makes a triangle.

Therefore, we have a triangle with side lengths  $|\textbf{a}|$, $|\textbf{b}|$, and $|\textbf{a} - \textbf{b}|$. Let the angle between the two vectors be $\theta$. By the Law of Cosines, we have

$$|\textbf{a} - \textbf{b}|^2 = |\textbf{a}|^2 + |\textbf{b}|^2 - 2 |\textbf{a}|  |\textbf{b}| \cos (\theta)$$

Now, use the fact that
$$
\begin{align*}
|\textbf{a}- \textbf{b}|^2 &= (\textbf{a}- \textbf{b}) \cdot (\textbf{a}- \textbf{b})\\ &= \textbf{a} \cdot \textbf{a} - 2 (\textbf{a} \cdot   \textbf{b}) + \textbf{b} \cdot \textbf{b} \\ &=  |\textbf{a}|^2 - 2 (\textbf{a} \cdot   \textbf{b}) + |\textbf{b}|^2
\end{align*}
$$

Simplify this equation, and you will get the desired formula.

Dot product of two vectors is defined as the product of the magnitude of the two vectors together with the cosine of the angle between the two vectors. Mathematically,
a.b=|a||b|cos§


Number of distinct chess positions and games I was wondering how many total possible chess games/positions are there? I know Shannon calculated an upper bound of about $10^{120}$ legal chess games and $10^{43}$ legal chess positions. Have there been any more recent developments?

Also, will we ever be able to know exactly how many legal chess games/positions are there in total? It will require tons of computing power but by Moore's Law we should eventually have the necessary computing power. Based on current estimates, when will this happen?

To answer the second part of your question, the number of total possible chess positions will eventually be calculated, but the number of chess games is still unimaginably large. THIS - broken link site will have a slice of the answers:

The number of possible chess positions after White’s first ply move is 20 (16 pawn moves and 4 knight moves).  There are 400 possible chess positions after two ply moves (first ply move for White followed by first ply move for Black).

this is because there are $20$ moves for white's move 1, and $20^2$, or $400$ moves at the beginning of move 2. then, the number balloons. the least number of moves to checkmate is 3, called the fool's checkmate. as for the other numbers, look on the site! as for when the numbers will be calculated, i'm not sure---yet.


Prove the inequality involving integrals How would you show that (if $x\gt0$):
$$\int_x^1 \frac{dt}{1 + t^2} = \int_1^{1/x} \frac{dt}{1 + t^2}$$

Hint

Make a change of variable $t=\frac {1} {u}$ on the left hand side and see what happens for the integral over $u$.

*put*$$ z=\frac{1}{t} $$ then the integrals limit changes and then omitting negative sign you get the answer


A function that is not a derivative If this is possible, could somebody give me an example of a function $f:[a,b]\to \Bbb{R}$ continuous in $[a,b]$ but that THERE IS NOT a function, namely G, such that its derivative is the funciont f? i.e., such that
$$G'=f.$$

This isn't possible: Every such function is a derivative. For any continuous function $f: [a, b] \to \mathbb{R}$, we can define the function
$$G(x) := \int_a^x f(t) \,dt,$$
and by the Fundamental Theorem of Calculus,
$$G'(x) = f(x).$$
(This assumes that we define the derivative of a function at the endpoint of an interval to be the value of the appropriate one-sided limit.)

I'm not pretty sure but this could be what you are looking for:

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


Solving $y'' y-2(y')^2\geq 0$ Here is an attempt

Let $y(x)=F^{a}(x)$ then

$$y''y-2(y')^{2}=a(a-1)F^{a-2}(F')^{2}F^{a}+aF^{a-1}F''F^{a}-2(F^{a-1}F')^{2}$$

$$=(a(a-1)-2)F^{2(a-1)}(F')^{2}+aF^{2a-1}F''$$

so for $a(a-1)=2\Rightarrow a=-1,2$ we obtain the differential inequality $F^{2a-1}F''\geq 0$ and so the inequality is true for positive convex functions like $e^{2x}$.

Update

I apologize the previous attempt was wrong,as rightly pointed out by professor Israel.

That's wrong.  With $y(x) = e^{c g(x)}$ you should get

$$ y'' y - 2 (y')^2 = e^{2cg(x)} c (g''- c (g')^2) $$

BTW, with $g' = v$

$$ v' - c v^2 = q $$

is a Riccati D.E.  I don't think there is a known solution for general $q$.


If a right circular cone has three mutually perpendicular generators then find its semi-vertical angle. If a right circular cone has three mutually perpendicular generators
  then find its semi-vertical angle.

We see that if $ax^2+by^2+cz^2+2fyx+2gzx+2hxy=0$ has three mutually perpendicular generators, then $a+b+c=0$. But I don't know what will be the way to find the semi-vertical angle.

Added:

Let us consider three mutually perpendicular generators with direction cosines $l_i,m_i,n_i$ for $i=1,2,3$.  The direction cosines of the axis are $\frac{\sum l_1}{3},\frac{\sum m_1}{3},\frac{\sum n_1}{3}=l',m',n'$ (say)   Since these three generators are mutually perpendicular, we have
$l_il_j+m_im_j+n_in_j=0$ for $i\neq j$.   Also we can say that $l_i^2+m_i^2+n_i^2=1$ for $i=1,2,3$.

From the above relation,  we have
$l_1m_1+l_2m_2+l_3m_3=0$ etc.
$\cos\alpha=\frac{l_1l'+m_1m'+n_1n'}{\sqrt{l'^2+m'^2+n'^2}}=\frac{1}{\sqrt{3}}\implies \alpha=\tan^{-1}(\sqrt{2})$. Is my approach correct?

If the generators are in the directions of the usual coordinate
axes, then the axis of the cone is in the direction of the vector $(1,1,1)$.
So the semi-vertical angle is the angle between the vectors $(1,0,0)$ and $(1,1,1)$ etc.

Assume 3 mutually perpendicular generators as
 coordinate axes
The axis of circular cone makes equal angles with the axes
 if $(l,m,n )$ are DC's of axes then
$$l=m=n=\cos\theta=\left(\frac13\right)^\frac12$$
    $$\tanθ=(2)^\frac12\implies θ=\arctan(2^\frac12)$$


Continuous function of one variable Let $f(x)$ continuous function on $R$ wich can be in different signs. Prove, that there is exists an arithmetic progression $a, b, c   (a<b<c)$,  such that $f(a)+f(b)+f(c)=0$.

Try defining the following function on $R \times R^+$

$ g(x,h) = f(x-h) + f(x) + f(x+h) $

Argue that if $f$ can take positive and negative values, then so can $g$.  Then, use the intermediate value theorem.

HINT:

1) Think about the intermediate value theorem.

2) Think about some $x$ and some $y$ with $f(x)\gt0$ and $f(y)\lt0$.


Ways to solve a 2D Laplace equation I'm looking for a survey of methods to solve Laplace equation in two dimensions.

Is there a book describing them with hints regarding their applicability for various cases?

I mean analytical methods, not numerics.

Since Community bumped the question, I might as well put something relevant here. My reaction to seeing "two dimensions" and "Laplace equation" in the same sentence is "ooh, I can use conformal maps to do that". The idea is that since the Laplace equation is invariant under conformal maps, we can transform our boundary value problem into a BVP on a nicer domain, a disk or a halfplane. This works very well for the Dirichlet problem; for the Neumann problem the boundary conditions must be multiplied by the derivative of the conformal map, which makes life a bit harder.

Low-level explanation with lots of examples and pictures. Unfortunately it dates  back to 2003 when MathJax wasn't around. We are spoiled by MathJax now.
Conformal Mapping: Methods and Applications, Dover book by Roland Schinzinger and Patricio A. A. Laura. From the preview it looks very thorough and accessible.
Research-level explanation with emphasis on polygonal domains

Here, I am giving you a link. See the file. You will get some hint.
http://newton.ex.ac.uk/teaching/CDHW/EM/CW970317-2.pdf

Books:

Schaum's Series - Laplace Transformation
Laplace Transforms for Electronic Engineers by JAMES G. HOLBROOK


Composition of Taylor Series Suppose I have smooth functions $f,g,y_0$ and $y_1$ from $\mathbb{R}$ to $\mathbb{R}$, such that
$$y_1(x) = y_0(x) - \epsilon g(y_0(x))$$
Then I consider
$$f(y_0(x)) = f(y_1(x) + \epsilon g(y_0(x)))$$
Is there a closed form expression for the Taylor series in the small parameter $\epsilon$ in terms of derivatives of $f$ and $g$ and only the function $y_1$?

The first few terms are

$$f(y_0) = f(y_1) + \epsilon f'(y_1) g(y_0) +  \frac{1}{2}\epsilon^2 f''(y_1)g^2(y_0) +..$$
Where we interpret $f(y_0)$ as $f|_{y_0(x)}$ and treat $x$ fixed. Then we can again replace the $y_0$ in $g(y_0)$ with
$$g(y_0) = g(y_1)+ \epsilon g'(y_1)g(y_0) +...$$
giving
$$= f(y_1) + \epsilon f'(y_1) [g(y_1) + \epsilon g'(y_1)g(y_0) + ... ]$$
$$+  \frac{1}{2}\epsilon^2 f''(y_1)[g(y_1) + \epsilon g'(y_1)g(y_0) + ... ]^2 +...$$
Continuing to replace the $y_0$ with $g(y_0)$ like this and grouping terms gives
$$f(y_0) = f(y_1)+ \epsilon [f'g](y_1) + \epsilon^2[f'g'g + \frac{1}{2}f''g^2](y_1) + \epsilon^3[f'g'^2g + \frac{1}{2}f'g''g + \frac{1}{2}f''g'g + \frac{1}{6}f'''g^3](y_1) + O(\epsilon^4)$$
But is there some way to write this as a more compact sum like
$$f(y_0) \sim f(y_1) + g(y_1)\sum_{n=1}^\infty\sum_{m=0}^n \epsilon^n \alpha(n,m)f^{(n)}g^{(n-m)}(y_1)$$
I am having trouble identifying the pattern. I know there will be some product involved as well.

Edit:

Thinking about it some more it may suffice to just set $f=id$ and consider
$$y_0 = y_1 + \epsilon g(y_0)$$
$$y_0 = y_1 + \epsilon g(y_1 + \epsilon g(y_0))$$
$$y_0 = y_1 + \epsilon g(y_1 + \epsilon g(y_1 + \epsilon g(y_0)))$$
$$y_0 = y_1 + \epsilon g(y_1 + \epsilon g(y_1 + \epsilon g(y_1 + ...)))$$
and somehow use the chain rule
$$[f_1\circ f_2 \circ .... \circ f_n]' = \prod_{i=1}^n(f'_{i}\circ f_{i+1}\circ ...\circ f_n)$$

It looks like there is a common factor of εg(d/dy) being used as both a function and a multiplicand.

It's being used as a function through composition -repeatedly taking the derivative and multiplying by g N times-, and as a multiplicand -taking the derivative N times then multiplying by g to the N.

Each εⁿ term looks to be made up of the difference between these two operations. Not sure of the factor that each is scaled by before taking the difference. (Repeatedly taking the derivative of each term and then multiplying by εg to obtain the next term is off by a constant factor of N and an additional amount of (d^n/dy^n)f*(g^n) scaled by N!.)


How do i get a view frames axis directions given a camera's direction vector? Say I have some custom view orientation in which the global coordinate frame looks like the picture below with no axis aligned to the edges of the computer monitors screen:

Say the direction vector from the origin away from the camera (away from you) is given by this vector:

Vector (-0.3159932494163513, 0.5070327520370483, -0.8019139766693115)

(Note that i am working in orthographic projections if that matters.)

But this is only one direction vector perpendicular to the computers monitor.

How would I get the direction vector parallel to the top (or bottom) edge of the monitor (i.e, the horizontal direction)?
Likewise, how would i get the direction vector parallel to the left (or right) side edge of the monitor (i.e, the vertical direction)?

Do i have enough information to find the rest of this frames axis directions?

The short answer: Let $O$ be the screen image of the world-coordinate origin, so that we have $X-O=(x_h,x_v)$, $Y-O=(y_h,y_v)$ and $Z-O=(z_h,z_v)$ for the unit world-coordinate axis vectors in screen coordinates. Then the world coordinates of the direction vectors for the edges of the screen are simply $(x_h,y_h,z_h)$ and $(x_v,y_v,z_v)$.

Orthographic (a.k.a. orthogonal) projection makes this fairly easy to compute. This projects onto some plane perpendicular to the camera’s axis $\mathbf n$, which for the purposes of this problem we can take as passing through the origin. The projection can be represented as a change of basis from the global $xyz$ coordinate system to the camera’s $x'y'z'$ coordinate system, which is accomplished with a rotation, and then projecting orthogonally onto some plane perpendicular to the camera’s axis—the $z'$-axis. This projection amounts to deleting the $z'$ coordinate. There’s also a translation involved since the origin of the viewport coordinate system isn’t necessarily at the point that corresponds to the camera’s axis, but I’ll suppress this aspect of the projection since it has no effect on what you’re trying to compute.

Since orthographic projection is parallel to the camera’s axis, the actual location of the image plane is irrelevant—only its attitude, i.e., the camera’s orientation, matters. In terms of this orientation, which can be described via pitch, yaw and roll from some standard orientation, the unit vector $\mathbf n$ that gives the camera’s facing takes care of pitch and yaw, and we’re trying to recover the roll.

For simplicity, assume that the image plane passes through the origin. To avoid a change of orientation, the unit vector $\mathbf n$ that corresponds to the camera’s axis defines the negative $z'$-axis of the camera coordinate system. Let $\mathbf u$ and $\mathbf v$ be a pair of orthogonal unit vectors that define “right” and “up,” respectively for the camera. These two vectors are parallel to the edges of the viewport, so are the ones we’re trying to find. Together with $\mathbf n$ these vectors define the $x'y'z'$ camera coordinate system. They form an orthonormal basis for $\mathbb R^3$ and so are related to the world coordinate system by a rotation. (For a left-handed viewport coordinate system as is common in computer graphics, take $\mathbf n$ as defining the positive $z'$-axis instead.)

(The image plane is moved away from the world-coordinate origin for clarity.)

Recall that the columns of a transformation matrix are the images of the basis, so the matrix that maps from the camera coordinate system to the world coordinate system has $\mathbf u$, $\mathbf v$ and $-\mathbf n$ as its columns. The inverse of a rotation matrix is its transpose, so the matrix that maps from world coordinates to camera coordinates has these vectors as its rows. The projection is therefore given by the matrix product $$P=\left[\begin{array}{rrr}1&0&0\\0&1&0\end{array}\right]\begin{bmatrix}\mathbf u^T\\\mathbf v^T\\-\mathbf n^T\end{bmatrix}=\left[\begin{array}{r}\mathbf u^T\\\mathbf v^T\end{array}\right]=\left[\begin{array}{rrr}u_x&u_y&u_z\\v_x&v_y&v_z\end{array}\right]=\begin{bmatrix}P\mathbf e_x&P\mathbf e_y&P\mathbf e_z\end{bmatrix}.$$ The last equality highlights the fact that the columns of $P$ are the images of the world-coordinate basis vectors (expressed in viewport coordinates), so $\mathbf u$ and $\mathbf v$ can be read directly from the coordinates of these images.

Now, the image plane to screen mapping typically involves some scaling. If the horizontal and vertical scale factors are equal, then you can use the resulting vectors as is or normalize/scale them as needed. Otherwise (e.g., for non-square pixels), you’ll need to retrieve the scale factors from the camera’s intrinsic matrix.

For example, let’s say that with the unit normal in your question, you have for direction vectors of the images of unit world coordinate basis vectors the screen vectors $(18.5907, -3.80073)$, $(6.69494, 15.8854)$, and $(-3.09257, 11.5416)$, respectively. Taking the first coordinates of these vectors we have $(18.5907,6.69494,-3.09257)$, which when normalized is $\mathbf u=(0.929535, 0.334747, -0.154629)$. Similarly, the second coordinates produce $\mathbf v=(-0.190036,0.794268,0.577082)$. If you form the matrix $U=\begin{bmatrix}\mathbf u&\mathbf v&\mathbf n\end{bmatrix}$, you can verify that $U^TU=I$ so that this is an orthonormal set of vectors, and that $\det U=1$, so that it is a right-handed basis. This computation also verifies that for the projection matrix $P$ formed from these vectors, $P\mathbf u=(1,0)$ and $P\mathbf v=(0,1)$, which is what was wanted.

This works because of a reciprocity between orthonormal coordinate systems, which itself is a result of the symmetry of inner products. For any basis $\mathscr B=\{\mathbf e_1,\mathbf e_2,\mathbf e_3\}$ of $\mathbb R^3$, the coordinates of a vector $\mathbf v$ are the coefficients of the unique linear combination $\mathbf v=c_1\mathbf e_1+c_2\mathbf e_2+c_3\mathbf e_3$. If the basis is orthonormal, we have $\mathbf v=(\mathbf v\cdot\mathbf e_1)\mathbf e_1+(\mathbf v\cdot\mathbf e_2)\mathbf e_2+(\mathbf v\cdot\mathbf e_3)\mathbf e_3$, that is, the coordinates of $\mathbf v$ are the scalar projections of $\mathbf v$ onto the basis vectors. If we have another orthonormal basis $\mathscr B'=\{\mathbf e_1',\mathbf e_2',\mathbf e_3'\}$, we have $$\begin{align} \mathbf e_1 &= (\mathbf e_1\cdot\mathbf e_1')\mathbf e_1'+(\mathbf e_1\cdot\mathbf e_2')\mathbf e_2'+(\mathbf e_1\cdot\mathbf e_3')\mathbf e_3' \\ \mathbf e_2 &= (\mathbf e_2\cdot\mathbf e_1')\mathbf e_1'+(\mathbf e_2\cdot\mathbf e_2')\mathbf e_2'+(\mathbf e_2\cdot\mathbf e_3')\mathbf e_3' \\ \mathbf e_3 &= (\mathbf e_3\cdot\mathbf e_1')\mathbf e_1'+(\mathbf e_3\cdot\mathbf e_2')\mathbf e_2'+(\mathbf e_3\cdot\mathbf e_3')\mathbf e_3'.\end{align}$$ At the same time, we also have $$\begin{align} \mathbf e_1' &= (\mathbf e_1'\cdot\mathbf e_1)\mathbf e_1+(\mathbf e_1'\cdot\mathbf e_2)\mathbf e_2+(\mathbf e_1'\cdot\mathbf e_3)\mathbf e_3 \\ \mathbf e_2' &= (\mathbf e_2'\cdot\mathbf e_1)\mathbf e_1+(\mathbf e_2'\cdot\mathbf e_2)\mathbf e_2+(\mathbf e_2'\cdot\mathbf e_3)\mathbf e_3 \\ \mathbf e_3' &= (\mathbf e_3'\cdot\mathbf e_1)\mathbf e_1+(\mathbf e_3'\cdot\mathbf e_2)\mathbf e_2+(\mathbf e_3'\cdot\mathbf e_3)\mathbf e_3.\end{align}$$ Since $\mathbf a\cdot\mathbf b=\mathbf b\cdot\mathbf a$, we can see from the above that the $\mathscr B'$-coordinates of $\mathbf e_1$ are equal to the first $\mathscr B$-coordinates of the elements of $\mathscr B'$ and vice-versa, and similarly for the other basis vectors. This reciprocity is what allows us to read the world coordinates of $\mathbf u$ and $\mathbf v$ from the viewport/screen coordinates of $X$, $Y$ and $Z$, above.

The horizontal and vertical screen axes are gotten by a different method mentioned in the comments. They are black in color.

Amd posted a method and gave his values for u and v for the camera direction I provided. These are the blue lines.

I did the steps described by amd and projected my own points then read off the values. These gave the following red lines.

Something is off!

There is not enough information to find the remain two axis directions that are parallel to edges of the monitor. One cannot position the XYZ-frame to the monitor just by moving one of it's axis in a parallel direction to the view direction through the origin. The moved XYZ-frame can still take any orientation around it's view directed axis. More information is needed to fix it's orientation.


What is the probability I get at least one call each day if my telephone rings 12 times each week? This question has already been answered but I was curious as to why the answer I came up with is wrong. What is wrong with my logic here.

$$\frac{7! \binom{12}{7} 7^5}{7^{12}}$$

Where $\binom{12}{7}$ is the number of ways to choose the 7 calls that will be assigned to each day (guaranteeing each day has a minimum of one call) and $7!$ represents the number of ways to order those calls and $7^5$ represents the number of ways to distribute the five remaining calls. What am I getting wrong here?

You are counting each distribution of calls multiple times.  Say the calls are distributed as follows:

\begin{array}{c c c c c c c}
\text{Sunday} & \text{Monday} & \text{Tuesday} & \text{Wednesday} & \text{Thursday} & \text{Friday} & \text{Saturday}\\ \hline
c_1 & c_2, c_3 & c_4 & c_5, c_6, c_7 & c_8 & c_9, c_{10} & c_{11}, c_{12}
\end{array}
You count this scenario $1 \cdot 2 \cdot 1 \cdot 3 \cdot 1 \cdot 2 \cdot 2 = 24$ times, once for each way you could designate one of the calls you receive each day as the call you receive that day and the remaining calls as being among the additional five calls you receive that week.

You already have an explanation why the way to count does not work.

Here is one way you can count. First distribute 7 calls one on each day, then however you place the rest the condition will be fulfilled. You have 5 calls left to place. You can place them any way among 7 days. So you got two components right, but in wrong places:

$$\frac{\left(\begin{array}{c}12\\7\end{array}\right)}{7^{5}}$$

If I got it right it will be $\approx$ 4.7123 %. Simulating $10^7$ weeks in Octave give me 4.7031% of the time.


$\tan(z)$ with residue theorem Calculate
$$\oint_{|z|=2}\tan(z)\,dz$$
because $\tan(z)=\dfrac{\sin(z)}{\cos(z)}$ the poles are when $\cos(z)=0$

at $z=\pm\pi/2\pm n\pi, \;n\in\mathbb{Z}$

Poles inside $|z|=2$ are $\pm\pi/2$ and are of first order. Can I know use the theorem for simple poles and just calculate the residuals
$$\operatorname{Res} \limits_{z=\pi/2} = \lim_{z\to \pi/2}(z-\pi/2)\frac{\sin(z)}{\cos(z)}$$
But I get stuck here. Hints?

OR
$$\cos (z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}\cdots $$
How do I know which coefficient to choose for the residual. Is it $1$ because the coefficient when $n=0$ is $1$ in the series? So the residuals for $\pm \pi/2$ are $1$.

So the integral is $2\pi i (1+1)=4\pi i$ ?

Let $f(z) = \tan z=(\sin z)/(\cos z)$.
cos $z = o$  when $z= ±π/2 , ±3π/2 , ±5π/2,$...
These are the poles of the function $\tan z$
Among these  the two poles $z= ±π/2$ lie inside the circle $|z| = 2$.
Residue of $f(z)$ at  $π/2 = −1$ and Residue of $f(z)$ at $−π/2  = −1$.
By the Residue Theorem, we have
the answer to be $= 2πi \text {(sum of the residues )} = 2πi(-2)=−4πi.$


Prove that for $x\in\Bbb{R}$, $|x|\lt 3\implies |x^2-2x-15|\lt 8|x+3|$. The problem I have is:

Prove that for real numbers $x$, $|x|\lt 3\implies |x^2-2x-15|\lt 8|x+3|$.

Since there aren't really any similar examples in my book, I've been unsure how to first approach this problem and have been trying to use to find similar questions.

Do I try and divide the problem into regions, like at absolute value inequalities, using $x^2-2x-15=(x-5)(x+3)$

Or could I just simply have two cases, those being $x\ge 3$ and $x\lt3$ like at How to write an expression in an equivalent form without absolute values?

We know that $|x|<3$. Therefore $-3<x<3$, subtracting $5$ from each side gives:

$-8<x-5<-2$

And so:

$|x-5|<8$

This implies that:

$|x^2-2x-15|=|(x-5)(x+3)|=|x-5|\cdot |x+3|<8|x+3|$

As needed.

Implies means inclusion of sets of solutions.

Therefore, we could solve both inequalitites and show their sets of solutions are one inside the other.

For $|x|<3$, it is $(-3,3)$.

For $|x^2-2x-15|<8|x+3|$, we get $|(x-5)(x+3)|=|(x-5)(x+3)|<8|x+3|$. This is the same as $x\neq-3$ and $|x-5|<8$. This is $[5-8,5+8]=(-3,13]$.

Now we can simply check that $(-3,3)\subset(-3,13)$


Solving for $a$ in this equation $\sin^a(a) = b$ $$\sin^{a}a=b$$
Do you know how to solve for $a$ here algebraicaly? This is a little bit too abstract for my understanding. Thank you very much.

In the same spirit as amcalde's answer, and staying in the real domain which implies $a\in[2k \pi,(2k+1)\pi]$ (this as been pointed out by Quang Hoang in a comment)  and $b\in[0,1]$.

Limiting the analysis to the case of $k=0$,  considering  $f(x)= \sin^x(x) $, you could easily notice that, starting at $x=0$ $(f(0)=1)$, it decreases and reaches a minimum value close to $x_1\approx 0.398576$ $(f_1=f(x_1)\approx 0.68575)$, starts increasing up to $x_2=\frac \pi 2$ $(f(x_2)=1)$ and then decreases to $0$ for $x_3=\pi$ $(f(x_3)=0)$.

So, depending on the value of $b$, the equation $$\sin^x(x)=b$$ can have

two solutions if $b=1$ (namely $0$ and $x_2$)
three  solutions if $f_1<b<1$
two solutions if $b=f_1$
one solution if $b<f_1$

As said, only numerical methods (such as Newton) will solve the problem but the question is to have a reasonable starting value.

Just as amcalde answered, look at the plot, make a guess $x_0$ for the root(s) you want and start Newton iterations $$x_{n+1}=x_n-\frac{\sin ^{x_n}(x_n)-b}{\sin ^{x_n}(x_n) (x_n \cot (x_n)+\log (\sin (x_n)))}$$ to the required accuracy.

For illustration purposes, consider $b=0.8$. The plot shows roots close to $0.1$, $0.9$, and $2.0$.

Applying the method for the first root, the iterates will be $$x_1=0.0944109$$ $$x_2=0.0945459$$ $$x_3=0.0945460$$ which is the solution for six significant figures.

Applying the method for the second root, the iterates will be $$x_1=0.892865$$ $$x_2=0.892846$$

Applying the method for the third root, the iterates will be $$x_1=2.03211$$ $$x_2=2.03102$$

As you see, the solutions are obtained quite fast.

Edit

After Quang Hoang's good point, I had a look at the case $k>0$ which is quite interesting. For any value of $b\in[0,1]$, it seems that there are always two roots on each side of $(4k+1)\frac \pi 2$. The curves are sharper and sharper picks (they look like gaussian curves in spite of some skewness that numerical calculations can easily show - at mid height, corresponding to $b=0.5$, the distance between the two roots is more or less $2 \sqrt{\frac{\log (2)}{1+k\pi}}$ ) and the problem becomes simpler than for the case $k=0$ since, for $x\in[2k \pi,(2k+1)\pi]$, we can approximate very accurately $$\sin^x(x)\approx \exp \left(-(1+k\pi) \left(x-(4k+1)\frac{\pi}{2}  \right)^2\right)$$from which quite good estimates can be generated for each of the two roots. $$x_{\pm}=(4k+1)\frac{\pi}{2}\pm \sqrt{\frac{-\log(b)}{1+k\pi}}$$ For example, still using $b=0.8$ and $k=1$ the above approximation gives as estimates $x_-=7.62186$, $x_+=8.08610$ while the solutions are $7.61304$ and $8.08781$. To some extent, the formula also applies for the estimates of the largest solution when $k=0$ (for the case looked at, it gives as estimates $1.09842$ and $2.04318$).

On the other side, the cases of $k<0$ seem to be nightmares (it does not seem that solutions exist - According to Quang Hoang's comment, this is wrong : there are effectively two solutions if $b>1$).

Edit

For the case of $k=0$ and $0.68575<b<1$, the initial (very crude) estimates can be obtained

For the first root, $$x=x_1 \left(1-\sqrt{\frac{b-f_1}{1-f_1}}\right)$$ This was obtained after fitting the function by a quadratic polynomial going through points $(0,1)$, $(x_1,f_1)$ with an horizontal tangent at  $(x_1,f_1)$.
For the second root,$$x=\frac{\pi}{2}- \sqrt{-\log(b)}$$
For the third root,$$x=\frac{\pi}{2}+ \sqrt{-\log(b)}$$

For the case $b=0.8$, this wil generate the estimates $0.16$, $1.10$ and $2.04$ which seem to be quite reasonable; starting withe these values, Newton method converges in very  few iterations.

For the case where $k=0$ and $0<b<f_1$ the estimate of the solution is
$$x=\pi -(\pi -x_4) \sqrt{\frac{b}{f_1}}$$  This has been obtained after approximating the function by a quadratic polynomial going through points $( x_4,f_1)$, $( \pi ,0)$ with horizontal tangents at the end point ($x_4 \approx 2.14647$ corresponds to the largest solution of $\sin^x(x)=f_1$).

Let us try Newton method for $b=0.1$. The successive iterates are then $$x_0=2.76158$$ $$x_1=2.69242$$ $$x_2=2.70099$$ $$x_3=2.70112$$

Edit

The case to also consider is if $b$ is a very small number. The approximation given from the tails of the approximating gaussian curves is not very good. In such a case, what is prosed is to use as starting values

for the smallest root,  $x=2k\pi+b^{\frac 1{2k\pi}}$
for the largest  root,  $x=(2k+1)\pi-b^{\frac 1{2k\pi}}$

For illustration purposes, consider $k=1$ and $b=10^{-4}$. The estimates are then $6.51406$ and $9.19390$ while the solutions are $6.39888$ and $9.20013$.

Edit

When we generate an estimate $x_0$, it must be such that $x_0\in[2k \pi,(2k+1)\pi]$. With the gaussian approximation, the value of the function is not zero at the bounds. So, with a minor sacrifice for the value of the maximum of the curve, I suggest to use instead
$$\sin^x(x)\approx \exp \left(-(1+k\pi) \left(x-(4k+1)\frac{\pi}{2}  \right)^2\right)-\exp \left({-\frac{1}{4} \pi ^2 (1+k\pi )}\right)$$ which will give as estimates $$x_\pm=  \left(2 k+\frac{1}{2}\right)\pi\pm\sqrt{-\frac{\log \left(b+e^{-\frac{1}{4} \pi ^2
   (1+k\pi)}\right)}{1+k\pi  }}$$ which also works fine for small values of $b$.

Finally, in order to introduce some skewness in the gaussian, the function is defined independently for left and right sides according to $$\sin^x(x)\approx \exp \left(-\alpha \left(x-(4k+1)\frac{\pi}{2}  \right)^2\right)-\exp \left({-\frac{\alpha}{4} \pi ^2 }\right)$$ using $\alpha=\frac 34+k\pi$ for the left side and  $\alpha=\frac 54+k\pi$ for the right side of the curve.

For illustration purposes, consider again $k=1$ and $b=10^{-4}$. The estimates are then $6.35931$ and $9.19390$ while the solutions are $6.39888$ and $9.28798$.

You have to use power reducing trig identities. Wikipedia has a list of them: https://en.wikipedia.org/wiki/List_of_trigonometric_identities. Go to the power reducing section and use either the one for odd sine powers or the one for even sine powers. The formula is rather messy, which is why it isn't typically a problem discussed much. Even after reducing the power, you'll probably have to use a fair amount of algebra, to massage things into a nice format.


how to apply "$ \varepsilon$- $\delta$" definition of limit I am self studying calculus, and i came across a University quiz that required me to:

  Use the $ \varepsilon$- $\delta$ definition of limits to verify that the $\lim\limits_{x\to\pi} 3 = 3$?

how to apply such thing
(this definition have not been mentioned in the textbook I am refereeing to (calculus life saver))

The "$\varepsilon$ - $\delta$" definition of a limit is the following:

We say that $L$ is the limit of a function $f$ at $c$ (i.e., we say that $\lim_{x \to c} f(x) = L$) if, for every $\varepsilon > 0$ there exists some $\delta > 0$ such that, whenever $|x-c| < \delta$, we have $|f(x) - L| < \varepsilon$.

As stated in the comments, having a constant function makes things pretty easy, at least in terms of figuring out what $\delta$ needs to be. Following is the required proof:

Let $\varepsilon > 0$, and set $\delta = 1$. (In this specific case, I can choose any positive number I want for $\delta$, but I'm choosing 1 just to be concrete.) Our function $f$ is a constant function, meaning that for all $x$, we have $f(x) = 3$. Therefore, whenever $|x - \pi| < 1$, we see that $|f(x) - 3| = |3 - 3| = 0 < \varepsilon$ as required.

If you have any questions about this proof, please ask. If it's your first time seeing an $\varepsilon$ - $\delta$ proof (and it sounds like it would be), I would be surprised if it wasn't a bit confusing.

For any $\epsilon > 0$ let $\delta = 5.6784\times 10^{237}$.

If $|x - \pi| < \delta$ then $|3-3| = 0 < \epsilon$.

So $\lim_{x\to \pi} 3 = 3$

=====

This follows the definition precisely.

The definition is literally that

$\lim_{x\to a} f(x) = L$ if for any $\epsilon >0$ we can find a $\delta$ (based upon $\epsilon$) so that whenever $|x-a|< \delta$ it will follow that that $|f(x) -L| < \epsilon$.

What this means is that if we want to get $f(x)$ as close as we want to $L$ ($|f(x) - L|< \epsilon$) we can achieve that by take $x$ within a range of $a$ ($|x-a| < \delta$).  In general, the closer we want to get $f(x)$ to $L$ we will usually need to take a narrower range. (If we cant find such a range, that would mean $f(x)$ doesn't get that close to $L$ after all.  If we can always find a range no matter how small $\delta$ is then that means $f(x)$ gets close to $L$ as $x$ gets close to $a$.

But if $f(x) =3$ is a constant function then $|f(x) - 3| = 0$ always and we always have $f(x)$ as close to $3$ as possible.  SO to find the range $\delta$ that assures this is trivial.  $\delta$ could be anything.  Whatever $x$ we choose $|f(x) -3| = 0 < \epsilon$.

So I chose $\delta$ to be anything I wanted.  I like weird and ridiculous numbers, so I chose $5.6784\times 10^{237}$.


Solving an exponential equation, $ z^n = ...$ Assuming $n$ is a whole, positive number. Calculate:

  $$z^n = (z-1)^n.$$

Perhaps a hint on where to start will be appreciated.. I'm not looking for answers, but an explanation so that I can understand how to solve it!

Thanks

Hint: Note that $z = 0$ cannot hold. Dividing by $z$ gives $\bigl(\frac{z-1}z\bigr)^n = 1$. So $\frac{z-1}z$ has to be an $n$-th root of unity, denote it by $\zeta$, then
$$\frac{z-1}z = \zeta \iff z-\zeta z = -1 \iff z = -\frac1{1-\zeta} $$

The answer is correct!

Here is a solution

$$ \left(\frac{z-1}{z}\right)^n = 1 = e^{i2\pi k} \implies \frac{z-1}{z}= e^{\frac{i2\pi k}{n}},\,k=1,\dots, n-1  $$

  $$\implies z=\frac{1}{1-e^{\frac{i2\pi k}{n}}},\quad k=1,\dots,n-1. $$

Note: We did not consider the case $k=0$ because it corresponds to the case

$$z-1=z.$$