Greatest number of parts that n circles can divide the plane What is the greatest number of domains(or parts) that n circles could divide the plane?

From many small cases I get the feeling that intersecting circles would provide the greatest number of parts.
Is this recursion right C(n+1) = 2C(n) using the previous statement. Since the new circle intersects all the circles and doubles the parts. Here C(n) is the number of parts for n circles.

How could I prove formally? If I could get an inequality that I will know for sure that I have got the greatest number of parts.

http://oeis.org/A014206 says draw n + 1 circles in the plane; then $a(n)=n^2+n+2$ gives the maximal number of regions into which the plane is divided. There are links and references there.

The formula is'nt correct.

Looking at the problem as a planar garph:
(you can think of every circle as 2 vertices we never intercect & 2 edges)

When adding a circle, we can intercect each previous circle in only 2 point, adding 2 vertices and 3 edges (one from the old circle). by Euler's formula it means we added one face.

so, in each step we add at most $n+1$ faces.(one of the new circle, $n$ from the n other circles).

Still, the question of the exact number is interesting.


Intersection of cylinder and plane If a cylinder is $x^2+8y^2=1$ and a plane is $x+y+3z=0$, what's the form of the intersection?

I thought of substituting the $y$ variable from the plane's equation in the cylinder's equation.

Actually I think we could get better results (at least easier to handle) about the intersection passing through parametrization.
The cylinder can be parametrized in $(u, v)$ like this:

$$ x = \cos u \\\\
y = \frac{\sqrt 2}{4}\sin u \\\\
z = v$$

, with $u\in[0, 2\pi]$ and $v\in(-\infty,+\infty)$

Getting the $z$ from the plane equation we get:

$$ z = \frac{x+y}{3}$$

and replacing the parametrization we get the final curve, parametrized in $u$ only:

$$ x = \cos u \\\\
y = \frac{\sqrt 2}{4}\sin u \\\\
z = \frac{4\cos u+\sqrt2\sin u}{12}$$

, with $u\in[0, 2\pi]$

we have the equation $$x^2+8y^2=1$$ and $$x+y+3z=0$$ solving the second equation for $y$ we have
$$y=-3z-x$$ in the first equation we obtain
$$x^2+8(-3z-x)^2=1$$
simplifying we obtain
$$9x^2+72z^2+48xz=1$$
can you go further?


If ${X_n} < \infty$, then $\sup {X_n} < \infty$, True or False? This maybe very bad but please correct me. Given $X_n >0$.

Suppose ${X_n} < \infty$, then $\max{X_n} + 1 < \infty$.

This means $\max{X_n} + 1$ is a upper bound of ${X_n}$.

Thus, $\sup {X_n} \le \max{X_n} + 1 < \infty$

Why is this wrong? or right?

The notation $X_n < \infty$ is a bit sloppy, because if it is read literally, it holds for a specific value $X_n$, even if the use of subscript would suggest that $X_n$ is part of a sequence of values.  If we then talk about $\sup X_n$, then the context of a sequence is made clear, but it still doesn't imply that all of the terms of the sequence are bounded above.  For that, you need to explicitly state something like $$\forall n, \quad X_n < \infty,$$ or $$X_n < \infty, \quad n = 1, 2, \ldots.$$  If this is the intended meaning, then the counterexample $X_n = n$ suffices, since for each $n = 1, 2, \ldots$, it is true that $n < \infty$, yet $\sup n = \infty$.  The statement $$\forall n, \quad X_n < \infty$$ is not equivalent to saying $$\exists C < \infty \quad \text{s.t.} \quad \forall n, \quad X_n < C.$$  The reason why is because if one wanted to say the second statement, one would not have written $X_n < \infty$ in the first place.  One would have simply written $X_n < C$.

To say that $X_n < \infty$ means that there exists a (finite) number $C$ such that $$\forall\, n\ ,\quad X_{n}<C\ ;$$ in other words the number $C$ is an upper bound for the sequence $X_n\ .$ Since $\sup X_n$ is the smallest upper bound for the sequence, the number $\sup X_n$ cannot be greater than the upper bound $C\ ,$ hence $$\sup X_n \leq C\ .$$


Making an angle of 10 degree and its multiple using compass? We all know very well how to make an angle of 15 and 45 degree and its multiple using compass. Can anybody tell me how to make an angle of 10 degree and its multiple using compass?

It's not possible to construct an angle of $10$ degrees with straightedge and compass alone. See for example http://en.wikipedia.org/wiki/Angle_trisection for details.

yes we can, for that we first need to draw equilateral triangle & then divide opposite side of the triangle into 6 parts & then join them. In this way at the vertex of one of the opposite side you can find 10 degrees.


Show that from a group of seven people whose (integer) ages add up to 332 one can select three people with the total age at least 142. I need help with this problem, and I was thinking in this way:
$$ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} = 332 $$
and I need to find three of these which sum is at least 142.

But I don't know what next.
Any help would be appreciated

Suppose that any $3$ chosen people have at most a total age of $141$. Then, $x_1 + x_2 + x_3 \le 141$ and so is $x_4 + x_5 + x_6$. This indicates that $x_7 \ge 50$.

Repeating the same process exchanging the roles of $x_i$'s yields $x_i \ge 50$ for any $i$. Which leads to a contradiction.

The average integer age of all $7$ is $47$. $3 \times 47 = 141$. However, because $7 \times 47 = 329$, it follows that one person must have an age of $50$ in order for the total to be $332$. Take $2$ average people and the old guy and you have $144$.


Is it possible for a 2nd-degree monomial to have 3 variables? (What about $6p^0qr$?) Is it possible for a 2nd-degree monomial to have 3 variables?

I think it's not, and then I remembered this: $6p^0qr$. But the $p^0$ is $1$, so for me it's not an example of a 2nd-degree monomial with three variables.

What do you think?

A monomial in 3 variables (say $x,y,z$) is of the form $cx^py^qz^r$ where $c$ is a nonzero constant and $p,q,r$ are positive integers (so each at least $1$).

The degree of the monomial is $p+q+r$, so it must be at least $3$. This makes degree $2$ impossible.

Note: this assumes the minimal variable list contains three variables; that is, all three variables contribute to the value of the monomial. In other words, there are strictly three variables. But this is no more problematic than stating that $x^2+0$ is not a binomial.

A second degree monomial takes on the form of:

f(x) = ax^2

If you want the monomial to take on three variables then you could have:

f(x,y,z) = ax^2

Technically this would still be a function of three variables. Or you could have:

f(x,y,z) = a(xyz)^2

But I don't think this would be considered a polynomial anymore.

In your own example you have:

6p^oqr

Here p would be the variable and the oqr would have to be constants for the monomial to be degree 2. In other words:

oqr=2


Assuming $x>-1$,
$$ (x+1)^{\log(x+1)}=100(x+1) $$
implies:
$$ \log^2(x+1) = \log(100)+\log(x+1) $$
hence by setting $\log(x+1)=y$ we have:
$$ y^2-y-2\log(10) = 0 $$
or:
$$ y = \frac{1\pm\sqrt{1+8\log 10}}{2}$$
from which:
$$ x=\exp\left(\frac{1\pm\sqrt{1+8\log 10}}{2}\right)-1. $$
If $\log$ is intended to be in base-$10$, the same process leads to $x=99$ or $x=-\frac{9}{10}$.

Here is my solution.I have left solving of quadratic equation to you


infinitely descending natural numbers Show that there is no infinitely descending sequence of natural numbers.

I was thinking that there exists no infinite descending chain on the natural numbers, since every chain of natural numbers has a minimal element. And so it reaches a finite minimum.

P.s I am just looking for a well built solution, since i cannot express the proof very clearly

Suppose the first number in the sequence is $n$. Then there can be at most $n$ terms in the sequence, since the second term is at most $n-1$, the third $n-2$, etc.

An integer is finite, and a finite number cannot infinitely decent, by definition


Which is greater, |$\vec a + \vec b$| or |$\vec a$|+|$\vec b$|? Pardon me if this has been asked before, I have encountered this question and its many variations in many physics mock papers.
I want to know if there is some sort of way to logically attempt these types of questions, I am not in requirement of an answer but a method which will help in solving questions of this form...

\begin{align*}
  |\mathbf{a}+\mathbf{b}|^{2} &=
  (\mathbf{a}+\mathbf{b}) \cdot (\mathbf{a}+\mathbf{b}) \\
  &= \mathbf{a} \cdot \mathbf{a}+\mathbf{a} \cdot \mathbf{b}+
     \mathbf{b} \cdot \mathbf{a}+\mathbf{b} \cdot \mathbf{b} \\
  &= |\mathbf{a}|^{2}+2\, \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^{2} \\
  & \leq |\mathbf{a}|^{2}+2|\mathbf{a}||\mathbf{b}|+|\mathbf{b}|^{2}
         \quad (\because \: \mathbf{a} \cdot \mathbf{b} \leq
         |\mathbf{a}||\mathbf{b}|) \\
  & = (|\mathbf{a}|+|\mathbf{b}|)^{2} \\
  \therefore \quad   |\mathbf{a}+\mathbf{b}|
  & \leq |\mathbf{a}|+|\mathbf{b}|
         \quad (\because \: |\mathbf{a}+\mathbf{b}|, |\mathbf{a}|,
         |\mathbf{b}| \geq 0)
\end{align*}

|a|+|b| will have to be greater than or equal to |a+b| and we can see that without any math just using pure logic.
When you do a + b, b can do 2 things compliment a or destruct its value. Even in general math if a is positive and b is negative or vice versa, b would negate a, but on the contrary when you apply a || function first and then add, you are assuring that b always compliments a.
Hope it helps


Prove that a continuous inverse-transformation of $f: [0,1) \cup \{ 2 \} \to [0,1]$ exists I am having this transformation $f: [0,1) \cup \{ 2 \} \to [0,1]$

$$f(x) = \begin{cases}
x  & x \neq 2
\\1 & x = 2
\end{cases}$$

I've already proved that it is continuous.

Question: Is there a continuous inverse-transformation of this transformation?

My thoughts:
Because f is continous and strictly increasing, then f has an
inverse function $ f^{-1}: [0,1] \to [0,1)  \cup \{ 2 \} $

Your function is a bijection, so an inverse definitely exists. You could probably write that function down by hand if you needed to.

As for continuity: One theorem about continuous functions is that the image of a connected topological space under a continuous map is also connected. What does that tell you about the continuity of the inverse function?

Because f is continous and strictly increasing, then f has an
inverse function $ f^{-1}: [0,1] \to [0,1[  \cup \{ 2 \} $


Diagonalize the matrix using simultaneous row and column operations? I need to diagonalize the matrix $$\left[\matrix{1 & -1\\-1 &3}\right]$$ using simultaneous row and column operations.

My attempt is to do $R2+R1 \to R2$ and $C2+C1\to C2$, and this brings the matrix to $\left[\matrix{1 & 0\\0 &-4}\right]$. Doing the same row operations to the identity matrix, I get $\left[\matrix{1 & 0\\1 &1}\right]$.

From what I understand, in the form $P^TAP = D$, is that $P^T=\left[\matrix{1 & 0\\1 &1}\right]$ and $D=\left[\matrix{1 & 0\\0 &-4}\right]$.

But I have plugged this into Wolfram Alpha and I know that my answer is incorrect. I'm not sure where to go from here.

When I calculate $R2+R1 \to R2$ and $C2+C1\to C2$, I get
$$
\left[\matrix{1 & 0\\0 &2}\right].
$$
Check out your calculations once again. Note $\color{green}{2}=\color{red}{+}3-1$, $\color{green}{-4}=\color{red}{-}3-1$.

That matrix does not have eigenvalues 1 and 2 for sure. Also if you want to preserve similarity it is necessary that the simultaneous column operation is the inverse of the raw operation just done. Which is not in your case. You do get a diagonal matrix, but is not similar to the one you started with.


square root of a real matrix I want to compute the square root of a real symmetric positive definite matrix $S\in \mathcal{M}_{m,m}$ such that $S^{1/2}S^{1/2}=S$ and it's well known that this decomposition is unique.

My question is if I have $S$ a real matrix will its square root be real too or it could be a complex matrix. In case that it could be complex then $S$ could have infinitely many square roots.

Any comments?

A real matrix $S$ can possess infinitely many real or nonreal square roots. For example,
$$
S=I=\begin{pmatrix}1&t\\0&-1\end{pmatrix}^2
$$
for every $t\in\mathbb{C}$. Note that $S=I$ is real and positive definite.

This is a very late answer, but I think that especially Dominique's answer is not very satisfying. So, if $S$ is real, symmetric and positive definite, it is trivially self-adjoint (i.e., $A^* = A$). The square root of $S$ is then usually defined to be the unique positive definite self-adjoint matrix $S^{1/2}$ which satisfies $(S^{1/2})^2 = S$. However, it does not follow immediately that $S^{1/2}$ is real. But this is in fact the case:

We have $S = XDX^{*}$ with $D$ diagonal (eigenvalues on the diagonal) and the columns of $X$ being the particular normalized eigenvectors of $S$. These can be chosen to be real, since if $x$ is an eigenvector of $S$, then also $\overline x$ (each entry being the complex conjugate of the correspoding entry in $x$) is an eigenvector (corresponding to the same eigenvalue), so $x+\overline x$ is. Thus, $X$ can be chosen to be real*. Now, $S^{1/2} = XD^{1/2}X^T$ (square roots of the (positive) eigenvalues on the diagonal) and so $S^{1/2}$ is real.

*) In case of an eigenspace $\ker(S-\lambda)$ whose dimension is greater than one, one has to work a little more. Choose an eigenvector $u_1$ and define $x_1 := u_1 + \overline{u_1}$. Then $x_1=0$ iff $u_1$ is completely imaginary. In this case, also $iu_1$ is an eigenvector. Choose this one for $x_1$. Now, consider $x_1^\perp$ in $\ker(S-\lambda)$. Choose any $z$ in this space which is not purely imaginary and put $x_2 := z+\overline z$. Now, consider $\{x_1,x_2\}^\perp$ in $\ker(S-\lambda)$ and proceed as before. In the end, and after normalization of $x_1,x_2,\ldots$ we obtain an orthonormal basis of $\ker(S-\lambda)$ with real vectors only.


High School Geometry - If $BC$ is the greatest side of $\triangle ABC$, $D$ & $E$ are points on $BC, CA$... If $BC$ is the greatest side of $\triangle ABC$, and $D$ & $E$ are points on $BC$ & $CA$, respectively, prove that $BC \ge DE$.

Clearly, equality holds iff $D$ is on $B$ and $E$ is on $C$.

I have tried to prove the inequality comparing $\angle BAC$ and $\angle DEC$.

If $\angle DEC \ge \angle BAC,$ then $DC$ hence $BC$ is $\gt DE$.

Else, on similar grounds $DE\lt EC$ hence $\lt BC$.

Is there something I am missing out. And, most importantly, can someone suggest a more neat version?

We assume as known that in any triangle $PQR$, the largest side is opposite the largest angle, and conversely.

Lemma: Let $PQR$ be a triangle, and let $X$ be in the interval $PR$, and not at an endpoint. Then $QX \lt QP$ or $QX \lt QR$ (or both). More compactly, $QX\lt \max(QP,QR)$.

Proof: Since angles $QXP$ and $QXR$ add up to $180^\circ$, one at least is $\ge 90^\circ$. So $\angle QXP$ is the largest angle in $\triangle QXP$, or $\angle QXR$ is the largest angle in $\triangle QXR$, or both. If $\angle QXP$ is the largest angle in $\triangle QXP$, then $QP \gt QX$. Similarly, if $\angle QXR$ is the largest angle in $\triangle QXR$, then $QR\gt QX$.

Now we look at our triangle $ABC$.  Suppose that $D=B$. If $E=C$, then $DE=BE=BC$. If $E\ne C$, then by the Lemma $BE$ (that is, $DE$) is $\lt\max(BC,BA)=BC$.

Suppose now that $D\ne B$. The case $E=C$ is easy.

If $B\ne D$ and $E\ne C$, draw the line $BE$. By the Lemma, we have $BE\lt BC$. Again by the Lemma, $DE\lt \max(BE,CE)\lt BC$. This completes the proof.

I have a geometric solution:

By extending $ED$ and $AB$ will intersect at $F$, assuming they are not parallel. Thus in $D$ $\triangle AEF$ we observe $\angle FAE=\angle BAC<\angle DEC$, and since $\angle BAC$ is the greatest angle we have $DC>DE$.


Anna wrote down the following 3 statements, denoted P1, P2, P3, on a blank piece of paper Here's a problem presented in my proofs class that I cannot for the life of me figure out. Apparently the answer is P2 according to the answer book but I have no idea how Help is much appreciated, thanks in advance!

Anna wrote down the following 3 statements, denoted P1, P2, P3, on a blank piece of paper:

P1 : There is exactly one FALSE statement written on this piece of paper.

P2 : There are exactly two FALSE statements written on this piece of paper .

P3 : There are exactly three FALSE statements written on this piece of paper.

One of the above statements is TRUE. What statement is TRUE?

Hint. If exactly one of the above statements is TRUE, and there are exactly three statements in all, then exactly how many are FALSE?

You're told that one of these statement — which means precisely one of them — is TRUE. So you can simply consider all possible cases in a very straightforward manner:

maybe P1 is TRUE and the other two are FALSE;
or maybe P2 is TRUE and the other two are FALSE;
or maybe P3 is TRUE and the other two are FALSE.

When you consider these three possible cases, you'll see that only one of them works, while the other two lead to a contradiction.

EDIT. Note that we don't even need to be told that one of them is true. Even without this additional information there's only one non-contradictory way (the same way) to assign TRUE/FALSE values to these three statements.


Prove Two Functions are Simultaneously Continuous Let $f,g,h: \mathbb{R} \rightarrow \mathbb{R}$ so that $f$ is
differentiable, $g,h$ monotone and $f'=f+g+h$.
Prove that $g$ is continuous in $x_0$
iff $h$ continuous in $x_0$.

My attempt
Suppose $g$ is continuous in $x_0$ and $h$ is discontinuous in $x_0$.
Monotone functions have lateral limits everywhere, therefore $h$ has lateral limits in $x_0$.
It follows that $f'$ is also discontinuous in $x_0$ (otherwise the sum of continuous functions is also continuous).
So far I have no idea how to get a contradiction out of here.

Suppose $g$ is continuous in $t$ and $h$ is discontinuous in $t$. By definition of differentiability

$$\lim_{k\to 0^+} f'(t+k) = \lim_{k\to 0^-} f'(t+k)$$ So that:

$$\lim_{k\to 0^+} f(t+k)+g(t+k)+h(t+k)=\lim_{k\to 0^-} f(t+k)+g(t+k)+h(t+k)$$

However, we know:

$$\lim_{k\to 0^+} f(t+k) = \lim_{k\to 0^-} f(t+k) \\
 \lim_{k\to 0^+} g(t+k) = \lim_{k\to 0^-} g(t+k)$$

Which lives us with

$$\lim_{k\to 0^+} h(t+k) = \lim_{k\to 0^-} h(t+k)  $$

Which is a contraddiction


Second order approximation of first derivative gives odd results Assume I want to find the second order approximation of the first derivative in 0 of a function $f(x)$ (not defined for $x<0$). I will use the formulation for the forward derivative:

$f'(0)\approx \frac{-3f(0)+4f(h)-f(2h)}{2h}$

where $h$ is the step in the regular grid I am using. Now assume that the values of the function are:
$f(0)=0$, $f(h)=5$, $f(2h)=25$.
The function is clearly monotonically increasing and the first derivative in 0 is expected to be positive but, from the formula above, I get a negative value.

What am I missing? Does the issue all lie in the error (that I neglected)?

A difference quotient with three terms should be exact for quadratics. You can even use this fact to derive difference quotients. So, let's fit a quadratic to the points $\{(0,0),(h,5),(2h,25)\}$; we find that
$$
f(x) = \frac{15 x^2}{2 h^2}-\frac{5 x}{2 h}.
$$
It's easy enough to check that this quadratic passes through the desired points. It's also easy to check that
$$
f'(0)=\frac{4 f(h)-f(2 h)-3 f(0)}{2 h}=-\frac{5}{2 h}
$$
and, in particular, that the derivative is negative when $h>0$. In fact, it grows in absolute value as $h\searrow0$.


Probability an 6 sided die will be higher than a 8 sided die? Say one person rolls an 8 sided die and the other rolls a six, what is the probability that the six sided die is higher than the 8?

I know that the expected value of the eight is 4.5 and the six is 3.5 but am having trouble figuring out how to find the probability.

EDIT: Answer is 15/48 but still curious if there's a way of doing this without creating a grid.

Probability That The $\boldsymbol{6}$-Sided Die Will Be Higher (from the body of the question)
$$
\sum_{k=1}^6\overbrace{\frac16}^{k\text{ on d}6}\cdot\overbrace{\frac{k-1}8}^{\lt k\text{ on d}8}
=\frac{5}{16}
$$

Probability That The $\boldsymbol{8}$-Sided Die Will Be Higher (from the original title to the question)
$$
\sum_{k=1}^7\overbrace{\frac18}^{k\text{ on d}8}\cdot\overbrace{\frac{k-1}6}^{\lt k\text{ on d}6}
+\overbrace{\frac18}^{8\text{ on d}8}\cdot\overbrace{1\vphantom{\frac16}}^{\lt8\text{ on d}6}
=\frac{9}{16}
$$

If there is a die with $m$ sides and a die with $n$ sides, assuming that $m<n$, the probability that $m$ will be greater than $n$ assuming that those are the names given to the outcomes when both of them are rolled is

$$\frac{m-1}{2n}$$

Full disclaimer: I made this up with help from the grid. This is a formula very specific to your question.


Problem in applying $\log x \cdot\log y$ and $\log_a(a)$ in this question $\log_5 (10)\cdot\log_{10} (15)\cdot \log_{15} (20)\cdot\log_{20} (25)$

I thought I could apply $\log_a(a)$ but I am not able to understand how.

Hint

$$\log_{a}b=\frac{\log_{c}b}{\log_{c}a}$$

You can, for example, write

$$\log_{5}10=\frac{\log10}{\log5}$$

You can apply change of basis (for basis $10$), i.e., $log_ba=\frac{loga}{logb}$. Thus,
$$\log_5 (10)\cdot\log_{10} (15)\cdot \log_{15} (20)\cdot\log_{20} (25)=\frac{log10}{log5}\frac{log15}{log10}\frac{log20}{log15}\frac{log25}{log20}=\frac{log25}{log5}=log_525=2$$


Is there a size of rectangle that retains its ratio when it's folded in half? A hypothetical (and maybe practical) question has been nagging at me.

If you had a piece of paper with dimensions 4 and 3 (4:3), folding it in half along the long side (once) would result in 2 inches and 3 inches (2:3), which wouldn't retain its ratio. For example, here is a piece of paper that doesn't retain its ratio when folded:

Is retaining the ratio technically possible? If so, what is the side length and ratio that fulfills this requirement? Any help would be appreciated.

Update:

I added "once" because I got an answer saying that any recectangle would work, as any rectangle folded twice has the original ratio. Nice answer, but not quite what I was looking for. As for the other answers, I got 3x as much information as I needed! Thanks!

The $1:\sqrt{2}$ ratio ensures exactly that. That is the idea behind the ISO 216 standard for paper sizes, which was adopted from the German DIN 476 standard.

Its most common usage is the A series which especially in Europe is a collection of very common paper sizes. The base size, A0, has an area of a square meter, and every next smaller paper size is constructed by folding it in half.

Well, if you have paper that has a ratio equivalent to the square root of two with the length divided by the width, if you fold it in half hamburger style (from width to width), you will get another ratio that's equivalent to the square root of two of the same property, likewise if you fold it more in half that way.

How It Works:

$$\sqrt{2}/1=\sqrt{2}$$$$\sqrt{2}/2=1/\sqrt{2}=0.7071...$$

That's half the square root of two, and when one is divided by it, we get the square root of two again: $1/0.7071...=\sqrt{2}$ and it just keeps going on and on from number to number being divided by the square root of two to get that result.  Just divide the dividend by the quotient in those things.


Solving quadratic or higher degree congruence with very large modulus. Is there any general way to solve a polynomial congruence with a very large modulus?

An example could be

$$
x^2-377x+1\equiv 0 \pmod {8683317618811886495518194401279999999 }
$$

or
$$
x^2-29478x\equiv 13^{67} \pmod {17^{2001}}
$$

If you can factor the modulus into coprime factors, you can solve the congruence with respect to each factor and then use the Chinese Remainder Theorem.  On the other hand, being able to solve $x^2 = a \mod m$ would give you
the ability to factor $m$.

EDIT: For lots more on these questions, see "Basic algorithms in number theory" by Joe Buhler and Stan Wagon

Algorithms for this are included in Mathematica. E.g.:

Solve[x^2 - 29478*x == 13^67, x, Modulus -> 17^2001]

yields the two solutions in an eyeblink.


How to solve a congruence system? Here is a tricky congruence system to solve, I have tried to use the Chinese Remainder Theorem without success so far.

$$2x \equiv3\;(mod\;7)\\
x\equiv8\;(mod\;15)$$

Thank you very much
Li

You need to "get rid of" the $2$ from the right hand side of the first congruence in your system:

\begin{aligned}
2x & \equiv 3 \mod 7 & (1) \\
x & \equiv 8 \mod 15 & (2)
\end{aligned}

So here, you have to multiply both sides of equation $(1)$ by the inverse of $2$ modulo $7$, which is $4$. This means the first equation becomes $x \equiv 12 \equiv 5 \mod 7$ So this leaves us with the system:

$$\color{blue}{\begin{aligned}
x & \equiv 5 \mod 7 \\
x & \equiv 8 \mod 15
\end{aligned}}$$

Since $\gcd(7, 15) = 1$, there is a unique solution for $x$ modulo $(7)(15) = 105$. So we can use the Chinese remainder theorem on the system that is in $\color{blue}{\mathrm{blue}
 }$.

Now try using the Chinese remainder theorem. Recall for a system of two congruences:

$$x \equiv a_1 \mod n_1$$

$$x \equiv a_2 \mod n_2,$$
if $\gcd(n_1, n_2) = 1$, then the solution is given by:

$$x \equiv a_1 n_2 [n_2^{-1}]_{n_1} + a_2 n_1[n_1^{-1}]_{n_2},$$
where $[p^{-1}]_{q}$ means "the inverse of $p$ modulo $q$".

You will find this is the solution:
$$x \equiv 5\cdot15\cdot1 + 8\cdot7\cdot13 \equiv \color{green}{803} \mod 105$$
and
$$\color{green}{803} \equiv \color{red}{68} \mod 105,$$ so
$$\boxed{\color{red}{x=68}}.$$

$\color{red}{\mathrm{That}}$ is the least positive residue solution, modulo $105$. Since anything congruent to $68$ modulo $105$ will also be a solution, you could also say $\boxed{x=105n+68}$ for each integer $n$.


How can tell if there is a prime number between any two numbers? Is there a conjecture, statement, formula or anything that determines whether or not there is a prime between two numbers? I'm not looking for a computer algorithm. Any help would be appreciated.

I highly doubt this is possible. Otherwise you could simply check any number of being prime, just by considering the number itself and the number +1 or something along these lines. But checking numbers being prime is hard...

The best we can do is that for any $x \ge 89693$ there is always at least one prime between $x$ and $x + \dfrac{1}{\log^3 x}$. This is much stronger than Bertrand's postulate and was proved by Pierre Dusart.


Rate of change - Implicit differentiation A price $p$ (in dollars) and demand $x$ for a product are related by $$\left(2x^2\right)-2xp+50p^2 = 20600$$ If the price is increasing at a rate of $2$ dollars per month when the price is $20$ dollars, find the rate of change of the demand.

I was a little confused on how to proceed with this question. Am I supposed to use implicit differentiation (with the $x$ serving the same purpose as a $y$) and then find the derivative of $x$?

This is the implicit differentiation I tried:

$$4x\frac{dx}{dp}-2\frac{dx}{dp}+100p = 0$$

$$4x\frac{dx}{dp}-2\frac{dx}{dp} = -100p$$

$$\frac{dx}{dp}(4x-2) = -100p$$

$$\frac{dx}{dp} = \frac{-100p}{4x-2}$$

I believe this is the derivative I am looking for (thought not entirely sure) but I am not sure what values of $p$ and $x$ to input, as I am supposed to get a numerical final answer.

Any help?

Guide:

$$2x^2-2xp+50p^2=20600$$

Differentiate with respect to $t$.

We have information about $x, p, \frac{dp}{dt}$, and you are interested in finding $\frac{dx}{dt}$.

Note that when $p=20$, you can compute the corresponding $x$.

Blindly follow the rule that $\boxed{{\rm d}f = \frac{\partial f}{\partial x} {\rm d}x + \frac{\partial f}{\partial p} {\rm d}p}$.

In this case:

$$ 2x^2-2xp+50p^2 = 20600 $$

$$ {\rm d} \left( 2x^2-2xp+50p^2 = 20600  \right)  $$

$$ \frac{\partial}{\partial x} \left( 2x^2-2xp+50p^2 = 20600  \right) {\rm d}x + \frac{\partial}{\partial p} \left( 2x^2-2xp+50p^2 = 20600  \right) {\rm d}p$$

$$ \left( 4x -2p  \right) {\rm d}x + \left(100p - 2x  \right) {\rm d} p = 0 $$

$$ \frac{{\rm d}x}{{\rm d}p} = \frac{x-50 p}{2 x -p} $$


7 fishermen caught exactly 100 fish and no two had caught the same number of fish. Then there are three who have together captured at least 50 fish. $7$ fishermen caught exactly $100$ fish and no two had caught the same number of fish. Prove that there are three fishermen who have captured together at least $50$ fish.

Try: Suppose $k$th fisher caught $r_k$ fishes and that we have
$$r_1<r_2<r_3<r_4<r_5<r_6<r_7$$
and let $r(ijk) := r_i+r_j+r_k$.
Now suppose $r(ijk)<49$ for all triples $\{i,j,k\}$.
Then we have $$r(123)<r(124)<r(125)<r(345)<r(367)<r(467)<r(567)\leq 49$$
so  $$300\leq 3(r_1+\cdots+r_7)\leq 49+48+47+46+45+44+43= 322$$
and no contradiction. Any idea how to resolve this?
Edit: Actually we have from $r(5,6,7)\leq 49$ that $r(4,6,7)\leq 48$ and $r(3,6,7)\leq 47$ and then $r(3,4,5)\leq r(3,6,7) - 4 \leq 43$ and $r(1,2,5)\leq r(3,4,5)-4\leq 39$ and $r(1,2,4)\leq 38$ and  $r(1,2,3)\leq 37$ so we have:
$$300\leq 49+48+47+43+39+38+37= 301$$
but again no contradiction.

Let's work with the lowest four numbers instead of the other suggestions.

Supposing there is a counterexample, then the lowest four must add to at least $51$ (else the highest three add to $50$ or more).

Since $14+13+12+11=50$ the lowest four numbers would have to include one number at least equal to $15$ to get a total as big as $51$.

Then the greatest three numbers must be at least $16+17+18=51$, which is a contradiction to the assumption that there exists a counterexample.

The examples $18+17+15+14+13+12+11=100$ and $19+16+15+14+13+12+11=100$ show that the bound is tight.

a = 0
b = 1
c = 2
d = 3
e = 4
f = 6
g = 84
In other words no mathematical solution to this riddle since there are many possible answers


Operator on $L^2 (0,1)$ defined by convolution with $|x-y|^{-\alpha}$ Define $A: L^2 (0,1) \to L^2(0,1)$
$$Af(x) = \int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \quad , \quad \alpha \in (0,1)$$

For what values of $\alpha$ is it well defined? Bounded? Compact?

I tried doing the appropriate integrals but I really don't understand what am I supposed to do in this specific case.

HINT:

The Cauchy-Schwarz Inequality reveals that

$$\begin{align}
\left|Af(x)\right|^2 &= \left|\int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \right|^2\\\\
&\le \int_0^1\left|f(y)\right|^2\,dy\,\int_0^1\frac{1}{|x-y|^{2\alpha}} dy
\end{align}$$

And thus, the square of the operator norm is

$$\begin{align}
||A||_2^2&=\sup_{f\in \mathscr{L}^2} \left(\frac{\int_0^1\left|\int_0^1 f(y) \frac{1}{|x-y|^\alpha} dy \right|^2dx}{\int_0^1\left|f(y)\right|^2\,dy}\right)\\\\
&\le \int_0^1\int_0^1\frac{1}{|x-y|^{2\alpha}}dydx
\end{align}$$

Operator is well-defined and bounded for all $\alpha\in(0,1)$: $$\|Af\|^2=\int\limits_0^1\left|\int\limits_0^1\frac{f(y)}{|x-y|^\alpha}dy\right|^2dx=\int\limits_0^1\left|\int\limits_0^1\frac{1}{|x-y|^\frac{\alpha}{2}}\frac{f(y)}{|x-y|^\frac{\alpha}{2}}dy\right|^2dx\leq$$$$\leq\int\limits_0^1\int\limits_0^1\frac{dy}{|x-y|^\alpha}\int\limits_0^1\frac{|f(y)|^2dy}{|x-y|^\alpha}dx.$$$$\int\limits_0^1\frac{dy}{|x-y|^\alpha}=\int\limits_0^x\frac{dy}{(x-y)^\alpha}+\int\limits_x^1\frac{dy}{(y-x)^\alpha}=\frac{x^{1-\alpha}}{1-\alpha}+\frac{(1-x)^{1-\alpha}}{1-\alpha}\leq\frac{2}{1-\alpha}.$$$$\|Af\|^2\leq\frac{2}{1-\alpha}\int\limits_0^1\int\limits_0^1\frac{|f(y)|^2dy}{|x-y|^\alpha}dx=\frac{2}{1-\alpha}\int\limits_0^1|f(y)|^2\int\limits_0^1\frac{dx}{|x-y|^\alpha}dy\leq\frac{4}{(1-\alpha)^2}\|f\|^2.$$

Operator is compact for all $\alpha\in(0,1)$. If $0<\alpha<\frac{1}{2}$ this operator is a Hilbert-Schmidt operator and therefore compact. Consider the case $\frac{1}{2}\leq\alpha<1$.

Consider the operator $A^2f(t)=\displaystyle\int\limits_0^1\left(\int\limits_0^1\dfrac{ds}{|t-s|^\alpha|s-\tau|^\alpha}\right)f(\tau)d\tau$ and let $k(t,\tau)=\displaystyle\int\limits_0^1\dfrac{ds}{|t-s|^\alpha|s-\tau|^\alpha}$. Performing first a replacement $t-s=\xi$ and then a replacement $\xi=|t-\tau|\eta$, we obtain: $$k(t,\tau)=\displaystyle\int\limits_{t-1}^t\dfrac{d\xi}{|\xi|^\alpha|t-\tau-\xi|^\alpha}\leq\int\limits_{-1}^1\dfrac{d\xi}{|\xi|^\alpha|t-\tau-\xi|^\alpha}=|t-\tau|^{1-2\alpha}\int\limits_{-\frac{1}{|t-\tau|}}^{\frac{1}{|t-\tau|}}\dfrac{d\eta}{|\eta|^\alpha\left|\frac{t-\tau}{|t-\tau|}-\eta\right|^\alpha}\leq$$$$\leq|t-\tau|^{1-2\alpha}\int\limits_{-\frac{1}{|t-\tau|}}^{\frac{1}{|t-\tau|}}\dfrac{d\eta}{|\eta|^\alpha|1-|\eta||^\alpha}=\frac{2}{|t-\tau|^{2\alpha-1}}\int\limits_0^{\frac{1}{|t-\tau|}}\dfrac{d\eta}{\eta^\alpha|1-\eta|^\alpha}\leq\frac{2}{|t-\tau|^{2\alpha-1}}\int\limits_0^{\frac{2}{|t-\tau|}}\dfrac{d\eta}{\eta^\alpha|1-\eta|^\alpha}=$$$$=\frac{2}{|t-\tau|^{2\alpha-1}}\left(\int\limits_0^2\dfrac{d\eta}{\eta^\alpha|1-\eta|^\alpha}+\int\limits_2^{\frac{2}{|t-\tau|}}\dfrac{d\eta}{\eta^\alpha|1-\eta|^\alpha}\right).$$

The first integral in brackets converges to some positive number $c(\alpha)$. In second integral $\eta-1\geq\dfrac{\eta}{2}$, therefore, the second integral does not exceed $$2^\alpha\int\limits_2^{\frac{2}{|t-\tau|}}\dfrac{d\eta}{\eta^{2\alpha}}\leq2^\alpha\int\limits_1^{\frac{2}{|t-\tau|}}\dfrac{d\eta}{\eta^{2\alpha}}=\frac{2^\alpha}{2\alpha-1}\left(1-\left(\frac{|t-\tau|}{2}\right)^{2\alpha-1}\right)\leq\frac{2^\alpha}{2\alpha-1},\;\;\frac{1}{2}<\alpha<1.$$

Finally, $k(t,\tau)\leq\dfrac{d(\alpha)}{|t-\tau|^{2\alpha-1}}$. For $A^4$ similarly $k_1(t,\tau)\leq\dfrac{d_1(\alpha)}{|t-\tau|^{2(2\alpha-1)-1}}$. Then for sufficiently large $n$ for operator $A^{2^n}$ $k_n(t,\tau)\leq\dfrac{d_n(\alpha)}{|t-\tau|^{\beta}}$, where $\beta<\dfrac{1}{2}$. So $A^{2^n}$ is Hilbert-Schmidt operator, and it's compact. The compactness of the operator $A$ follows from the statement: if $A$ -- self-adjoint and $A^2$ -- compact in hilbert space, then $A$ also compact.

The case $\alpha=\dfrac{1}{2}$ remains. But since $\dfrac{1}{|t-s|^\frac{1}{2}}\leq\dfrac{1}{|t-s|^\frac{5}{8}}$, then this case reduces to the previous one.


closed form expression for $\sin 10^o$? As we know that $\sin 15^o$, $\sin 30^o$,$\sin 45^o$ have simple closed form expressions as these are multiples of 3, but i have never seen any simple closed form expression for $\sin 10^o$ or simply sine for any non-multiple of 3, if there exists a closed form expression, do  help me,

  PS. I know $\sin 10^o$ is solution of $8x^3-6x+1=0$ but i can't solve it as its too tedious.

Why is being multiple of 3 such a great thing for an angle???

Being a multiple of $3°$ isn't a great thing, as degrees are an arbitrary unit. It matters much more to be a small fraction of a full turn, as this leads to polynomial equations of a low degree.

E.g.

$$\sin 3x=-4\sin^3x+3\sin x=0$$ leads to the well-known $$\sin\frac\pi3=\frac{\sqrt3}2.$$

Some other fractions lead to closed-form expressions, but most others not, as explained by the Abel-Ruffini theorem. In particular, angle trisection (dividing by three) involves a cubic equation which is in general not solvable (the above case being an exception).

It is interesting to note that there are analytical formulas to solve cubics, but for some values of the coefficients (the so-called casus irreductibilis), the solution requires… trigonometric functions, with angle trisection, and you are circling in rounds.

I don't think that it's tedious.

Let $x=\cos\alpha.$

Thus, we need to solve $$4\cos^3\alpha-3\cos\alpha=-\frac{1}{2}$$ or$$\cos3\alpha=-\frac{1}{2},$$which gives $$3\alpha=120^{\circ}+360^{\circ}k,$$ where $k\in\mathbb Z$ or
$$3\alpha=-120^{\circ}+360^{\circ}k,$$ which for $k=0$ in the first sequence and for $k=1$ and $k=2$ in the second sequence gives:
$$x_1=\cos40^{\circ},$$
$$x_2=\cos80^{\circ}=\sin10^{\circ}$$ and
$$x_3=\cos200^{\circ}=-\cos20^{\circ}.$$


What is the probability of having a pentagon in 6 points If the probability that $5$ random points in the plane whose horizontal
coordinate and vertical coordinate are uniformly distributed on the
interval $\left(0,1\right)$ occur to be the vertices of a convex
pentagon is $\frac{49}{144}$, what is the probability that a subset
of $6$ random points in the plane whose horizontal coordinate and
vertical coordinate are uniformly distributed on the interval $\left(0,1\right)$
occurs to be the vertices of a convex pentagon? Thanks a lot.

This is the same as the probability that no two of the six points lie within the convex hull of the other four.
The number of cases for this complementary event to happen is
$${6\choose2}=15$$
and all these cases are (with 100% probability) mutually exclusive and equiprobable, except that we have triple-counted the cases when three of the points actually lie within the convex hull of the other three, of which there are
$${6\choose3}=20.$$
These cases are also all equiprobable, they are mutually exclusive except on a set of null measure, and they have to be re-included twice.  This completes the inclusion-exclusion.
Given this, it is easy to solve the original problem if we can calculate the probability distribution of the area of the convex hull of the first four randomly chosen points and the marginal probability q that the last two uniformly and independently chosen points both lie within this area, and likewise the marginal probability r that the last three points all lie within the convex hull of the first three.
Then the probability that any five of the six points chosen uniformly and independently on $(0,1)\times(0,1)$ form a convex pentagon is
$$p=1-15q+40r$$
So let $H_3(x)$ be the marginal cumulative distribution function of the area of the convex hull of the first three points, and $H_4(x)$ be the marginal cdf of the area of the convex hull of the first four points. Then
$$q=\int_0^1 x^2dH_4(x);$$ and
$$r=\int_0^1 x^3dH_3(x).$$
Computing $H_3(x)$ and $H_4(x)$ is tedious but straightforward calculus. Otherwise the problem is solved.
(Small hint: it may be easier to compute $H_4(x)$ after computing $H_3(x)$. The integrals are actually the second and third moments about the origin for these distributions, respectively. They are taken over the Kolmogorov space $[0,1]$ as the cdf ranges from 0 to 1.)
For $H_3(x)$ see The PDF of the area of a random triangle in a square.  For $H_4(x)$ see PDF of area of convex hull of four random points in square.

Your original set has cardinality 6, your target set has cardinality 5, set member order does not matter, so you have $C^6_5=6$ possible subsets. Each of these has a $\frac{49}{144}$ probability of forming a convex pentagon. What is the probability that at least one of these forms a convex pentagon? It is 1 minus the probability that none of them form a convex pentagon:

$1-(1-\frac{49}{144})^6\approx1-65.97\%^6=1-8.2445\%=91.76\% $


Is the sum of this series finite? Let $a\in\mathbb{R}_{+}$. I consider the series
$$
 f(q) = \left(\sum_{k=1}^q e^{-a(q-k)}\frac{1}{k}\right)^2
$$
and the sum
$$
   \sum_{q=1}^{\infty}f(q).
$$
My numerical simulations show that this sum is finite (at least, for $a=1$), but I do not know how to prove it analytically.

Here is the answer that I have found.

Let us start defining $$z(q) = \sum_{k=1}^q e^{-aT(q-k)}\frac{1}{k}.$$ Then $f(q) = \left(z(q)\right)^2$, and the sum that we are studying is $S(p) = \sum_{q=1}^p\left(z(q)\right)^2$. The goal is to show that this sum converges as $p\to \infty$.

Ok, the strategy is as follows. We know that for all $C>0$ and $\gamma>\frac{1}{2}$ the sum $$\sum_{q=1}^\infty\left(\frac{C}{q^\gamma}\right)^2$$ converges. Thus to show that $S$ converges we intend to show that there exists a series (having the converging sum) upper bounding the one we study. I.e., we want to show that there exist  $C>0$ and $\gamma>0$ such that for all $q\ge 1$ $$z(q) \le \frac{C}{q^\gamma}.$$

It can be seen that for $z(q)$ it holds $$z(q+1) = \frac{1}{e^{aT}}z(q) + \frac{1}{q+1}.$$ Suppose that for some $q$ it holds $z(q) \le \frac{C}{q^\gamma}$, and let us show that then $z(q+1)\le \frac{C}{(q+1)^\gamma}$. To this end, note that
$$z(q) \le \frac{C}{q^\gamma} \Rightarrow z(q+1) \le \frac{1}{e^{aT}}\frac{C}{q^\gamma} + \frac{1}{q+1}.$$
To simplify, choose $\gamma = 1$ and $C=2e^{aT}$. Then for $aT>0$ and $q\ge 1$
$$\begin{gathered}\left(2e^{aT}-3\right)q-2 >0 \\
\Rightarrow \frac{2}{q} + \frac{1}{q+1} \le \frac{2e^{aT}}{q+1} \\
\Rightarrow \frac{1}{e^{aT}}\frac{C}{q^\gamma} + \frac{1}{q+1} \le \frac{C}{\left(q+1\right)^\gamma} \\
\Rightarrow z(q+1)\le \frac{C}{\left(q+1\right)^\gamma},\end{gathered}$$
and $z(1)=1\le 2e^{aT}$. Thus, $f(q)$ is upperbounded by $\frac{4e^{2aT}}{q^2}$, and $S(p)$ converges.


Is there an explicit relationship between the eigenvalues of a matrix and its derivative? If we consider a matrix $A$ dependent of a variable $x$, the eigenvalues and eigenvectors satisfying the equation
$$
A \vec{v}=\lambda \vec{v}
$$

will also depend on $x$. If we consider the matrix $B$ such that
$$B_{ij}=\frac{ \mathrm{d}}{ \mathrm{d} x} A_{ij}$$
Then, could we express the eigenvalues of $B$ in terms of the eigenvalues of $A$?
I found the question very interesting and was not able to find a satisfying answer myself.

For example in the case for $2\times2$ matrices of the form
$$
A=\left (
\begin{matrix}
 a(x) & b(x) \\
 0 & c(x)
\end{matrix}
\right ),\implies B=\left (
\begin{matrix}
 a'(x) & b'(x) \\
 0 & c'(x)
\end{matrix}
\right )
$$
I noticed that $\lambda_B(x)= \lambda_A'(x)$. But I cannot generalise it to general $2\times 2$ matrices. Not even thinking about $n\times n$ matrices...

Thank you for your help and any idea!

It is not true in general that the eigenvalues of $B(x)$ are the derivatives of those of $A(x)$. And this even for some square matrices of dimension $2 \times 2$.

Consider the matrix

$$A(x) =
\begin{pmatrix}
1& -x^2\\
-x &1
\end{pmatrix}$$
It’s characteristic polynomial is $\chi_{A(x)}(t)=t^2-2t+1-x^3$, which has for roots $1\pm x ^{3/2}$ for $x>0$. Those are the eigenvalues of $A(x)$.

The derivative of $A(x)$ is
$$B(x) =
\begin{pmatrix}
0& -2x\\
-1 &0
\end{pmatrix}$$ and it’s characteristic polynomial is $\chi_{B(x)}(t)=t^2-2x$, whose roots are $\pm \sqrt{2} x^{1/2}$for $x>0$.

We get a counterexample as the derivative of $1+x^{3/2}$ is not $\sqrt{2}x^{1/2}$.

However in the special case of upper triangular matrices (that you consider in your original question) the eigenvalues of the matrix derivative are indeed the derivatives of the eigenvalues.

If $A(x)v= \lambda(x)v$, with v independent of x, then, differentiating on both sides of the equation by x, $A'(x)v= \lambda'(x)v$.  That is, the eigenvalues of A' are the derivatives of the eigenvalues of A, with same associated eigenvectors.


Is there a domain without unity in which every element is a product? In this answer, Fortuon Paendrag provides an example of a ring without unity such that every element is a product of some two elements. The example has zero divisors. Can a ring without a unity and without non-zero zero divisors satisfy this condition if it's

(a) commutative,

(b) non-commutative?

Added: A related question.

Here is a useful commutative example that one actually meets in the wild.  Let $M$ be a non-standard model of analysis, and let our ring $I$ be the collection of infinitesimals in $M$, together with $0$.

One can make the example sound more explicit by constructing $M$ via the ultrapower.

One can also construct many function ring examples. One type of example is the unitless ring of all finite sums $\sum a_i x^{e_i}$, where the $a_i$ range over the reals, and the $e_i$ range over the positive reals. Or else we can restrict the $e_i$ o positive rationals, or positive dyadic rationals.

For commutative (Abeliano) you can define a domain such as |R+(pairs, *) (positive pairs with multiplication) and it will be a a Domain without unity, is commutative and every element can be achieved as a product of other (except the first... or consider $2*2^0$... this is arguable of course)

For non-commutative the elements would have to gain some positional properties so commutativity do not work... for example :
$$
( A \ op \ B ) <> ( B \ op \ A )
$$
only if the position alter the meaning of the symbols (either a value or any other piece of information the simbols A and B represent)


Differential forms on vector spaces For a differential form on Euclidean space
$$dx_i:\mathbb{R}^n\rightarrow \mathbb{R}$$
$$dx_i:(\textbf{v})\rightarrow v_i$$
Looking more genereally at vector spaces, for a differential form on a vector space is it also true that
$$dx_i:V\rightarrow K$$
$$dx_i:(\textbf{v})\rightarrow v_i$$
Where $\textbf{v}\in V$, $V$ is a vector space over field $K$?

Sure, let $K$ be a field, $V$ a $K$-vector field and $(e_1,\ldots,e_n)$ be a basis for $V$, then $(\mathrm{d}x_1,\ldots,\mathrm{d}x_n)$ is defined to be the dual basis of $(e_1,\ldots,e_n)$, every linear form on $V$ is a linear combination of the $\mathrm{d}x_i$.


Definition of a peculiar quotient group of isometries I just wrote a text file to sum up my ideas about the Riemann Hypothesis. This text is a draft, and I don't expect people here to say if this approach is interesting or not (but if by chance you think so, please let me know). My question is about the definition of a peculiar quotient group, which might not be correct in definition 6.
Can someone tell me if it's ok or not?
Thank you in advance.

Here comes the text:

Main ideas towards a proof of GRH

Definition 1
Let $A$ be be a subclass of the Selberg class containing $1$, closed under products, and such that every element of $A$ can be factored in a unique fashion in a product of primitive elements of $S$, these primitive elements belonging to $A$. Such a subclass of $S$ will be called a Galois class of L-functions.

Definition 2
Let $A$ be a Galois class of L-functions. $T$ is an automorphim of $A$ iff the following properties simultaneously hold true:

1) $T$ is a bijective map from $A$ to itself
2) $T$ maps a primitive element of $A$ to a primitive element of $A$
3) for all $F$ in $A$, the degree of $F$ and the degree of $T(F)$ are the same
4) for all $F$, $G$ in $A$, $T(F.G)=T(F).T(G)$

Definition 3
$\tilde{\phi}:F\mapsto \phi\circ F\circ \phi^{-1}$ is said to be an isometric automorphism of $A$ if $\tilde{\phi}$ is an automorphism of $A$ and $\phi$ is an isometry of the complex plane preserving any rectangle centered in $1/2$.

Proposition 1
$\tilde{\phi}(F)=F$ implies $\phi(0)=0$.

Definition 4
A group of equivalence classes of isometries is said to be nice iff each equivalence class contains exactly one representant $\phi$ such that $\phi(0)=0$.

Let $F\in A$ and let $G_F$ be the group of all equivalence classes of isometries preserving the set of non trivial zeros of $F$, where the equivalence relation $R_F$ is defined by : $f R_F g$ iff for every non-trivial zero $s$ of $F$, $f(s)=g(s)$.

Let $G'_F$ be the group of isometric automorphisms of $A$ preserving $F$.

Proposition 2
$G_F$ and $G'_F$ are isomorphic iff $G_F$ is nice.

Definition 5

Let $A$ and $B$ two Galois classes of L-functions such that $A$ is a sub Galois class of L-functions of $B$. Let's define the L-type (Isometric) Structure group
of $B$ over $A$ as the group of (isometric) automorphisms of $B$ preserving $A$ pointwise. The L-type structure group of $B$ over $A$ will be denoted as LStr(B/A) and the L-type Isometric Structure Group of $B$ over $A$ LIStr(B/A).

Definition 6

Let $GZ(A)$ be the set $\bigcup_{F\in A}\{Z(F):=\{s,F(s)=0, 0<\Re(s)<1\}\}$. The group of all isometries preserving $GZ(A)$ will be denoted as $PG_A$.
Now let's consider the relation $R'_A$ defined by $f R'_A g,$, where $f$, $g$ are elements of $PG_A$, iff for all $F$, $H$ in $A$,
$f R_F g$ iff $f R_H g$. The quotient group $PG_A/R'_A$ will be denoted as $SG_A$.
Let's now consider the group of elements of $SG_B$ preserving $GZ(A)$ pointwise. This last group will be called "Z-type Isometric Structure Goup of $B$ over $A$" and denoted as ZIStr(B/A).

Conjecture 1
Let $A$ be a Galois class of L-functions, $F$ an element of $A$.
$G_F$ is nice iff for all $H$ in $A$, $G_H$ is nice.

Conjecture 2 (Main Conjecture)
Let $A$ and $B$ be two Galois classes of L-functions such that $A$ is a sub-Galois class of L-functions of $B$.
Then LIStr(B/A) and ZIStr(B/A) are isomorphic.

Idea of proof: prove that LIStr(B/A) and ZIStr(B/A) are both isomorphic to $Gal(F_{B}/F_{A})$, where $F_B$ is the field generated by the union of all images of $\mathbb{R}-\{1\}$ by the elements of $B$, and $F_{A}$ is defined in a similar way.
If $B$ is the maximal Galois class of L-functions and $A=\{\zeta^{n}, n\geq 0\}$, one can expect to have a proof of the Riemann Hypothesis.

$SG_{A}=PG_{A}/[Id]$, where $[Id]$ is the equivalence class of the neutral element of $PG_{A}$ for the relation $R'_{A}$. So that $SG_{A}$ is well defined.


Infinite length Composition series Let $G$ be a group (possibly infinite). Suppose $G$ has a composition series. I could show that any other composition series has the same length. But I cannot prove the following.

Let $G \triangleright G_2 \triangleright G_3 \triangleright \cdots$ be a series of normal groups (may or may not assume the successive quotients are simple). I want to show that this series should terminate, i.e., there is $N$ such that $G_i=G_N$ for any $i >N$.

This is not true, take $G=Z^N$ and $G_i$ the subgroup such that $x\in G_i$ if its first $i$ components are zero.


finding the probability of A happens before B in die rolling I'm thinking about the following old exam question:

A die is rolled infinitely many times, what is the probability that (6,6) (i.e. two consecutive sixes) happens before (1,2)?

Since the expectations of waiting time for (6,6) and (1,2) could be calculated, I was trying to get it from there but I haven't seen the connection between them. Is it possible to get $P(\xi_1<\xi_2)$ with only $E(\xi_1)$ and $E(\xi_2)$, where $\xi_1$ and $\xi_2$ are the waiting times for $(6,6)$ and $(1,2)$ respectively? I also tried writing down all the possible events and sum them up, but the summation seems to be hard to calculate. Thanks for any help.

Let $X$ be the random variable that represents the time of the first instance of $12$ and $Y$ be the random variable that represents the first time of the instance of $66$. Then $P(X=1)\neq P(Y=1)=0$ and $P(X=2)=P(Y=2)=1/36$ but $P(X=3)\neq P(Y=3)$.

Namely $P(X=3)=\frac{6}{216}$ and $P(Y=3)=\frac{5}{216}$. That's because any sequence $*12$ satisfies $X=3$, but $*66$ only satisfies $Y=3$ if the first roll is not $6$.

So it isn't as simple as the two being equal.

One tricky way to do this is to use a Markov process. Let $s_0$ be the start state. Let $s_1$ be the state of having a string ending in $1$ and not having a prior $66$ or $12$. Let $s_6$ be the state of ending in a $6$ with no prior $12$ or $66$. And let $s_{12}$ and $s_{66}$ be the state of having reached $12$ or $66$ first.

Then you are looking for:

$$(0,0,0,P(X<Y),P(Y>X))=\lim_{n\to\infty} (1,0,0,0,0) \begin{pmatrix}
\frac 46&\frac16&\frac 16&0&0\\
\frac36&\frac16&\frac16 &\frac16&0\\
\frac46&\frac16&0&0&\frac16\\
0&0&0&1&0\\
0&0&0&0&1
\end{pmatrix}^n$$

This can be done by expressing $(1,0,0,0,0)$ in terms of the left eigenvectors of the above matrix.

Wolfram Alpha gives me $P(X<Y)\approx 0.5538$ and $P(Y<X)\approx 0.4462$.That is $12$ has a better than 55% chance of coming before $66$.

Assuming (1,2) means 1 has to come before 2:
   Assuming the question means that both (6,6) and (1,2) need to be consecutive: The answer is 1/2. No fancy math here, just logic: Each roll of the dice doesn't depend on the previous roll. The probability of hitting 6 is 1/6. The probability of consecutive (6,6) is 1/36. The probability of 1 is 1/6; the probability of 2 is 1/6. Hence, the probability of consecutive (1,2) is 1/36. Therefore, both (6,6) and (1,2) have equal chance of happening.
   If only (6,6) needs to be consecutive: The answer is Sigma of (#rolls-1)/((#rolls-1)!+(#rolls-1)) where #rolls goes from 2 to infinity. First roll both 6 and 1 have equal probability of 1/6. Second roll has 5/6 probability of not being 6 and needing to restart 6&6 sequence over again; while not hitting 2 on the second roll does not necessitate restarting the 1&2. Therefore probability of getting non-consecutive (1,2) in 2 rolls=1/36, 3 rolls=1/36+1/36+1/36=3/36(first roll probability of 1/6 multiplied by second roll probability of 1/6 or first multiplied by third or second multiplied by third) , 4 rolls=6/36(1*2,1*3,1*4,2*3,2*4,3*4), 5 rolls=10/36, 6rolls=15/36, 7rolls=21/36, 8rolls=28/36, 9rolls=36/36; formula (#rolls-1)!/36. While the probability of consecutive (6,6) is always 1/36 independent of the number of rolls; probability of rolling consecutive (6,6) in 2 rolls = 1/36, 3 rolls = 2/36 (1+2 or 2+3), 4 rolls = 3/36 (1+2,2+3,3+4), 5 rolls = 4/36, 6 rolls = 5/36, 7 rolls = 6/36, 8 rolls = 7/36; formula (#rolls-1)/36. Therefore the probability of getting consecutive (6,6) before non-consecutive (1,2) is: 1/2 in 2 rolls, 2/5 in 3 rolls, 3/9 in 4 rolls, 4/14 in 5, 5/20 in 6, 6/27 in 7, 7/35 in 8, 8/44 in 9 rolls; formula (#rolls-1)/((#rolls-1)!+(#rolls-1)).


What is the ratio of the speed of the cow and speed of the train given the following conditions? There is a bridge of $40\,\text m$ length. A cow is standing $5\,\text m$ away from the middle of the bridge. A train is coming from the direction nearest to the cow. If the cow runs towards the opposite direction of the train then he escapes the bridge when the train is $2\,\text m$ away from the bridge and if the cow runs towards the same direction of the train then the train hits the cow $2\,\text m$ before the other end of the bridge. What is the ration of speed of the cow and the train?

My Approach

Let the train speed be $x$ and the cow speed be $y$

Therefore, $$\frac{d-2}x=\frac{15}y\tag1$$

Similarly $$\frac{d+38}x=\frac{d-23}y\tag2$$

From $(1)$ and $(2)$ I get $d=77$

Therefore, $$\frac{15}{d-2}=\frac{15}{75}=1:5$$

This is how I solved this problem.

A simple intuitive way with practically no algebra is to imagine two cows, simultaneously  running towards bridge ends $A$ and $B$ respectively.

train -----> $A$ .. $15m$ ..$\bullet$..... $25m$ ..... $B$

By the time one cow has reached $A$, the other is $(25-15) =10m$ from $B$, and by the time this cow has travelled another $8m,$ the train hits it, having travelled $2+40-2 = 40m.$

Thus ratio of speed of cow: speed of train $= 8:40 = 1:5$

f x is distance the train is from the bridge, c is cow speed and t is treen speed then from the first sentence we get:

13m / c = (2m + x) / t

from the second:

25m / c = (38 m + x) / t

Then t / c = (2m + x)  / 13m  = (38m + x) / 25m, x = 37m, t / c = 3, c / t = 1 / 3

Well, if I read this question now, it looks like my cow swapped the directions... I guess you right, true blue anil, to have 2 cows running in opposite directions, this way at least one will do the right thing... and I will never do questions while at work again, Karma and things like that.


Closed form of integral How does one show that, $\forall \ n \in \mathbb{N}$ $$\int \frac{\ln^n|\tan(x)|}{\sin(x)}dx = \frac{e^{i\pi n}(\ln|e^{ix}-1|-\ln|e^{ix}+1|)}{2^n}+C$$
Is this possible to do without using any complex analysis theory like the residue theorem or likewise? I'm looking for a single variable solution, but all the methods are welcome!

Aside from the wise method in the comment, if you did not know the answer, you might write
$$ \frac{\log^n|\tan x|}{\sin x} = \frac{1}{\tan x} \log^n \tan x \sec x.$$ Since the derivative of $\tan$ is $\sec^2,$ you are almost there...


Limit of series divided and outside factor What is the limit of $$\lim_{x \to \infty} \frac{\sum_{k=0}^{x} (k+1){e^k}}{x{e^x}}\quad \text{?}$$ I think that it may go to infinity but not sure and i don't know how to solve it.

For $x$ a natural.

$$
\sum_{k=0}^{x}\left(k+1\right)e^{k}=\frac{1}{\left(e-1\right)^2}\left(\left(x+1\right)e^{x+2}-\left(x+2\right)e^{x+1}+1\right)
$$
So
$$\frac{\displaystyle \sum_{k=0}^{x}\left(k+1\right)e^{k}}{xe^x}=\frac{1}{\left(e-1\right)^2}\left(\frac{x+1}{x}e^{2}-\left(x+2\right)\frac{e}{x}+\frac{1}{xe^{x}}\right)
$$
Hence, the limit exists and

  $$\frac{\displaystyle \sum_{k=0}^{x}\left(k+1\right)e^{k}}{xe^x} \underset{x \rightarrow +\infty}{\rightarrow}\frac{e^2-e}{\left(e-1\right)^2}=\frac{e}{e-1} \approx 1.58197$$


Find solutions to $\cot(x)+\csc(x)=\sqrt3$ in range $[0,2\pi]$ What is the best way to do the above? Are there any tricks I should be aware of.

I know how to simplify it to $\dfrac{\cos(x)}{\sin(x)} + \dfrac{1}{\sin(x)} = \sqrt{3}$

so multiplying both sides by $\sin(x)$, we get $\cos(x)+1=\sqrt{3}\sin(x)$.

But I'm stuck from there.

Let's exploit the half-angle identity

$$\csc x+\cot x=\cot (x/2)$$

Then, we have $\cot(x/2) =\sqrt{3}$ or $\tan(x/2)=\sqrt{3}/3$.

Solutions are

$$x=2\arctan(\sqrt{3}/3)+2n\pi =\pi/3+2n\pi$$

Rearranging, we obtain
$$\sqrt3 \sin(x) - \cos(x) = 1 \implies \dfrac{\sqrt3}2 \sin(x) - \dfrac12 \cos(x) = \dfrac12 \implies \sin\left(x-\dfrac{\pi}6\right) = \sin\left(\dfrac{\pi}6\right)$$
This gives us that
$$x-\dfrac{\pi}6 = n\pi + (-1)^n \dfrac{\pi}6 \implies x = n\pi +(-1)^n \dfrac{\pi}6 + \dfrac{\pi}6$$
Further, $\sin(x)\neq 0 \implies x =2n\pi + \pi/3$.


Regarding $\lim \limits_{x \to 0} \frac{e^x-1}{\sqrt{1-\cos x}}$ Question:

  Find the limit of: $$\lim_{x \to 0} \frac{e^x-1}{\sqrt{1-\cos x}}$$

My approach:
By rationalizing the denominator, we get:
$$\lim_{x \to 0} \frac{(e^x-1)(\sqrt{1+\cos x})}{\sin x}$$

Dividing both the numerator and the denominator by $x$:
$$\lim_{x \to 0} \frac{\frac{(e^x-1)}{x}(\sqrt{1+\cos x})}{\frac{\sin x}{x}}$$

It is known that:
$$\lim_{x \to 0}\frac{(e^x-1)}{x} = 1$$
$$\lim_{x \to 0}\frac{\sin x}{x} = 1$$

Thus, simplifying, we get the answer to be $\sqrt2$

However, it is given that the function is divergent at $x = 0$ and thus the limit is undefined. Why is this so?

What you did is wrong, remember that
$$|\sin x|=\begin{cases}\sin x&if\ x\in[0,\pi]\\ -\sin x&if\ x\in[0,-\pi]\end{cases}.$$

So,

$$\frac{e^x-1}{\sqrt{1-\cos x}}=\frac{(e^x-1)\sqrt{1+\cos x}}{\underbrace{\sqrt{1-\cos^2x}}_{=|\sin x|}}=\frac{(e^x-1)\sqrt{1+\cos x}}{|\sin x|}$$

and thus,
$$\lim_{x\to 0^+}\frac{e^x-1}{\sqrt{1-\cos x}}=\lim_{x\to 0^+}\frac{(e^x-1)\sqrt{1+\cos x}}{\sin x}=\sqrt 2$$
and
$$\lim_{x\to 0^-}\frac{e^x-1}{\sqrt{1-\cos x}}=\lim_{x\to 0^-}\frac{(e^x-1)\sqrt{1+\cos x}}{-\sin x}=-\sqrt 2.$$

Therefore $$\lim_{x\to 0}\frac{e^x-1}{\sqrt{1-\cos x}}$$ is not defined.

Using L'Hospitals rule directly:

$\frac{e^x-1}{\sqrt{1-cos x}}=\frac{(e^x-1)'}{(\sqrt{1-cos x})'}=\frac{2 e^x \sqrt{1-cos x}}{sin x}$. Now let $x \rightarrow 0$, what do you observe?


How to determine the outward normal vector of a face of a hexahedron if the orientation (CW or CCW) of vertices of the face is unknown Consider a single quadrilaterally-faced hexahedron. If given the co-ordinates of the vertices, $\mathbf{v}_i$, of a face in counter-clockwise orientation, I can compute the corresponding unit outward normal vector using a cross-product:

$\mathbf{n}=\frac{(\mathbf{v}_2-\mathbf{v}_1)\times(\mathbf{v}_4-\mathbf{v}_1)}{\parallel(\mathbf{v}_2-\mathbf{v}_1)\times(\mathbf{v}_4-\mathbf{v}_1)\parallel}$

However, consider now an arbitrarily shaped volume which is discretised into a large number of quadrilaterally-faced hexahedra. Consider in particular a face shared by two adjacent hexahedra. If now given the co-ordinates of the vertices of the face in a certain orientation, the orientation will be clockwise for one hexahedron and counter-clockwise for the other.

If I do not know the orientation beforehand, how can I ensure that the unit normal is pointing outwards (assuming you know the co-ordinates of all vertices of all faces of all hexehedra)?

Your hexahedron is convex, so you can do the dot product of the normal vector that you found, and a vector from one of the vertices to some point inside the hexahedron. If the dot product is negative, then your normal vector points out.
If you already know a point inside the hexahedron, then you're in luck. Otherwise you would have to find this, which would pose challenges of its own.

you can computee the dot product between the normal and the vector:

[center of element, center of the current face].

If it's > 0, the normal is outwards to the element.

If not, it's inward and you just have to do : normal = -normal, to change that


Number of binary trees with n unlabeled node I found a proof about the number in this site.

In that site the author says that the number of all possible binary trees with n labeled nodes is equal to the number of ways one can make n-1 edges from n nodes. Since each node can have 2 or less edges, number of all possible edges is 2n. So, the number of ways one can choose n-1 edges from 2n possible edges is 2n×(2n−1)×(2n−2)×....×(2n−(n−2))=(2n)!/(n+1)!. And finally merging all permutation trees into one tree gives the number of binary trees with n unlabeled nodes, (2n)!/((n+1)!n!)=2nCn/(n+1).

But, in the middle of the proof I don't get how one can find 1-1 corrospondence between the number of ways of choosing edges and the number of binary trees with n labeled nodes. For example, if n=3, and each node has name 1, 2 and 3, respectively, and I choose left edge and right edge from node 1 in that order, then there are two possible cases:

1
/\
2 3

1
/\
3 2

So, it seems that choosing 2 edges from 6 edges (3 nodes) does not correspond to a specific binary tree.

Can anyone provide me with the mapping rule between the sequence of chosen edges and the corresponding binary tree?

As per my understanding, an edge is defined by combination of nodes.
Like in the example you mentioned (1,2) and (1,3) are two edges
So if you are choosing (1,2) to be the left edge, you cannot switch the nodes.
So it is corresponding to one binary tree only.


I have 100 boxes. C of them have a gift. I can open up to 16 boxes. What is the number of C that will give me probability over 0.5 to find a gift? Details:
We start by opening a box. If nothing is in there, we open another one. Once we find a gift, we can stop. Each empty box that was opened is discarded (no revisit).
I can find the number of $C$ that will give probability over $0.5$ by writing a program to try for $C=1, C=2$ .. etc.. , but I can't solve the equation for $C$ to find a more "mathematical" and elegant answer.

My work until now is:

1) Found in 1st box: $P(1) = \frac{C}{N}$

2) Found in 2nd box: $P(2) = \frac{1-C}{N}\cdot\frac{C}{N-1}$

4) Found in 3rd box: $P(3) = \frac{1-C}{N}\cdot\frac{1-C/}{N-1}\cdot\frac{C}{N-2}$

Etc...

Adding them up makes things very complicated to solve for $C$.

Any ideas? Thank you in advance!

Start with a rough estimate: If the box contents were independent, the probability of losing would be $(1-C/100)^{16}$. Equating this to $0.5$ gives us $C\approx 4.2$.

Hence, we boldly check the cases $C=4$:
$C=4$ leads to a losing probability of $$\frac{96\choose 16}{100\choose 16}=\frac{96!84!}{80!100!}=\frac{84\cdot 83\cdot 82\cdot 81}{100\cdot 99\cdot 98\cdot 97}\approx 0.492$$
so a winning probability slightly above $\frac12$. A close look reveals that $C=3$ leads to a winning probability below $\frac12$, so the correct answer is $C=4$.

Note that the "true" breaking point is thus between $3$ and $4$, not between $4$ and $5$ as the rough estimate suggested - the box contents are not independent after all (namely, if you find a - rare - gift, the probability of finding a gift in another box falls dramatically).

An alternative approach is to work in base 10 logarithms.  Chance of failing on 16 tries is

$\displaystyle f(C) = \left(\frac{100-C}{100}\right) \times
\left(\frac{99-C}{99}\right) \times
\left(\frac{98-C}{98}\right) \times \cdots \times
\left(\frac{85-C}{85}\right).
$

Assume that you've written a computer program that calculates
$\;\log_{10}n\;$ for $n\in\{30, 31, \cdots, 100\}.$

Then it becomes a simple matter to calculate
$\;g(C) = \log_{10}f(C).$


How can i calculate first digit in $(13)^{139}$. How can i calculate first digit in  $(13)^{139}$.

$\bf{My\; Try}::$ Let $a$ be the first Digit of $13^{139}$.

So $10^{b}a < 13^{139}<10^{b}(a+1)$, where $b= \lfloor \log_{10}(13^{139})\rfloor$

Now taking $\log_{10}$ on both side

$b+\log_{10}(a)<139\cdot \log_{10}(13)<b+\log_{10}(a+1)$

Now I did not understand how can i solve after that

Help Required

Thanks

Take your favourite calculator, and calculate $139 \log_{10}13$. Now let $x$ be the fractional part of this, and calculate $10^x$. Deduce the result.

$13$ is congruent to $3  (\mod 10)$, so $13^4$ is congruent to $3^4$ or $81$ congruent to $1 (\mod 10)$, but $139$ is the product of 34 and 4 added to 3... So $13^{136}$ is congruent to $1 (\mod 10)$ so $13^{139}$ is congruent to $13^3$ or $7 (\mod 10)$ so the first digit is $7$.


Why does $\frac{\log_{e}k}{k}$ approach $0$ as $k$ approaches $\infty$? A interesting problem: why is it that as $k$ approach infinity, $$\frac{\log_{e} k}{k}$$ approaches $0$?

Another problem: why $e^n=(1+n/x)^x$ as $x$ approach infinity?

First.

You can also understand it by graph.

let

$$f(k)=k,\ g(k)=\ln k$$

then

$$f'(k)=1,\ g'(k)=\frac{1}{k}$$

If $k>1$,

$$f'(k)>g'(k),\ \lim_{k\to \infty}f'(k)=1>\lim_{k\to \infty}g'(k)=0$$

notice that

$$f(1)>g(1)$$

so

$$\lim_{k\to\infty}\frac{g(k)}{f(k)}=0$$

Second.

notice that

$$e=\lim_{x\to\infty}\left (1+\frac{1}{x}\right )^x$$

if $1/n$ is constant, we know

$$e=\lim_{x\to\infty}\left (1+\frac{n}{x}\right )^{\frac{x}{n}}$$

so

$$e^n=\lim_{x\to\infty}\left (1+\frac{n}{x}\right )^x$$

Important thing

$$e=\lim_{x\to\infty}\left (1+\frac{\bigstar}{x}\right )^\frac{x}{\bigstar}$$

$$\frac{\bigstar}{x}\times\frac{x}{\bigstar}=1$$

As $k\rightarrow 0^+$, $\ln(k)\rightarrow -\infty$ and $1/k\rightarrow +\infty$. So $\ln(k)/k\rightarrow-\infty$. The problem doesn't make sense for the two-sided limit since $\ln(x)$ is not defined for $x<0$.


Polynomials with complex coefficients I am reading a book called College Algebra and it defines polynomials as an expression of the form $a_nx^n + a_{n-1}x^{n-1} + ...+ a_2x^2 + a_1x + a_0$ where $a_j$ are real numbers ( I'll stop the definition there). Does polynomials only have real coefficients or it can be complex numbers?

I asked this on a wrong place (Cross Validated) my bad.

Polynomials can also have complex coefficients. There is actually quite a bit of work related to the topic. See, for example, here or here.

An example of such a polynomial would be

$$(42+42i)x^2+(7+i)x.$$

The roots of the above polynomial are at $$x=0 $$
and
$$ x = -\frac{2}{21} + \frac{i}{14}.$$

So, nobody hinders you to create polynomials with complex coefficients.

To be clear, it's better to state over which ring the polynomial is defined.

  A polynomial in $X$ over a ring $R$ is defined as $\sum\limits_{k = 0}^n = a_kX^K$, where $a_k \in R$ for $k \in \{0,\dots,n\}$.

In particular, a polynomial over $\Bbb{R}$ admits only real coefficients, whereas a polynomial over $\Bbb{C}$ admits complex coefficients.

Remarks: For pedagogical reasons, when one find roots (of polynomials) in algebra-precalculus, one usually focus on real roots and discard the complex ones, especially in exercises of quadratic-equation.  I guess it's a possible reason that your College Algebra book only consider polynomials with real coefficients.  Things will become clearer when you move on to abstract-algebra.


Non singular matrix $M$ such that $MAM^T=F$, normal form Show that there is a non-singular matrix $M$ such that $MAM^T = F$ for any antisymmetric matrix $A$ where the normal form $F$ is a matrix with $2 \times 2$ blocks on its principal diagonal which are either zero or $$\begin{pmatrix} 0 &1 \\ -1&0 \end{pmatrix}$$

To do so, consider the analogue of the Gram-Schmidt method for antisymmetric matrices using the antisymmetric inner product $\langle x,y \rangle = x^TA y$

I am just really looking for some hints to start. For vectors $x$ and $y$, $$x^T A y = (x_1 \dots x_n) \begin{pmatrix} A_{11} & A_{12} &..&A_{1n} \\ ..&..  \\ ..& ..&..&A_{nn} \end{pmatrix} \begin{pmatrix} y_1 \\ y_2\\ ..\\y_n \end{pmatrix}$$ and $A_{ii} = 0, A_{ij} = -A_{ji} $ but I am not seeing how the hint is useful.

Thanks!

We suppose that $A$ is written in the canonical basis. Consider the product defined by $\langle x,y\rangle=x^TAy$,  it is a bilinear antisymetric product. There exists a basis $e_1,...e_{2n}$ such that this product in this basis is $e_{n+1}\wedge e_{1}+...+e_{2n}\wedge e_{n}$. The matrix of the bilinear for in this basis is $\pmatrix{0 & 1 \cr -1 & 0}$.

Let $M$ be the transtion matrix from the  basis to $e_1,...,e_{2n}$ to the canonical basis $M^TAM =\pmatrix{ 0 &1\cr -1 & 0}$


Surface Area of Sphere as Stacked Circles I know the formula is $4 \pi r^2$. I think it makes sense to say that if I represent a sphere as a stack of circles, the surface area of the sphere should be equal to the sum of the circumferences of the circles or the average circumference of the circles multiplied by the height of the sphere. Since the height of the sphere is $2r$, that means that the average circumference must be $2 \pi r$ to achieve $4 \pi r^2$. $2\pi r$ is the maximum circumference of the circles, not the average. Why is multiplying the average circumference of stacked circles by the height of the sphere to get the surface area wrong?

Arc length should be used instead.

Let $(x,y)=r(\cos t, \sin t)$.  Then $(\dot{x},\dot{y})=r(-\sin t, \cos t)$.

\begin{align*}
  S &= \int_{-r}^{r} 2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^{2}} dx \\
    &= 2\pi\int_{-\pi}^{\pi} y\sqrt{\dot{x}^{2}+\dot{y}^{2}} dt \\
    &= 2\pi \int_{-\pi}^{\pi} r^{2} \sin t \, dt \\
    &= 4\pi r^{2}
\end{align*}

Your question:
Why is multiplying the average circumference of stacked circles by the height of the sphere to get the surface area wrong?

Answer:
You would get the area of a cone.  There is no curvature as you go up in height, just a 45º cone.


Derivative of a constant not making sense I'm studying calculus on my own and I came across something weird:

The derivative of $x^n$ is $n\cdot x^{n-1}$, and the derivative of any constant is $0$. Also, any constant $x = x^1$. However, the derivative of $x^1$ ($x$ being any integer constant) seems to be $1$: $n\cdot x^{n-1} = 1\cdot x^{1-1} = x^0 = 1$. What's failing in my reasoning?

Let $a$ be some number and $D$ be the differential operator (the derivative).  Just for fun we'll use the notation $x\mapsto f(x)$ to represent the function $f$.  This notation is supposed to be pretty intuitive: in it $x$ is an arbitrary element in the domain of $f$ and $f(x)$ is the number it maps to under $f$.

We know that $$D(x\mapsto x^n) = x\mapsto nx^{n-1}\tag{1}$$ and that $$D(x\mapsto a) = x\mapsto 0\tag{2}$$

So then does that mean that $D(x\mapsto a^n) \stackrel{?}= x\mapsto na^{n-1}$?  No.  Of course not.  $a^n$ is just another number so we follow rule $(2)$ to get $$D(x\mapsto a^n) = D(x\mapsto b) = x\mapsto 0$$ where $b=a^n$.

Please note that a constant is not a  constant function. We can try to derive functions, we can surely derive constant functions, but we certainly do not derive constants.  So, yes, anyway, derivative of a constant doesn't make any sense.
Another preliminary remark : the expression $x^n$ in the function $f : x \mapsto x^n $ is the value of the function on $x$, not the function. If you write the function $x^n$ (or worse $x^n$ alone) you should keep in mind that it's just a (bad) shorthand notation. For example you can't really derive $x^n$ and $c$ you derive the functions $f : x \mapsto x^n $  and $g : x \mapsto c$. If you forget that it can lead you to mistakes...

What's wrong in your reasoning is that you used the $x$ symbol with $2$ different meanings without noticing it.
When you write $x=x^1$ and say $x$ is a constant it's ok, it could be, even if we generally use $x$ as a variable name. But when you say you derive $x^1$ and its derivative is $1$, implicitly, you derive the identity function ($Id : x \mapsto x $) where $x$ represents no longer constant. Try to use different names for your constants and your variables in function definitions and it should be ok.  For example if you want a constant function whose value is $x$ write $f : t \mapsto x $ and if you don't really need to name $x$ the constant you could use this more common form : $f : x \mapsto c $ where $c$ is the constant and $x$ the variable.


Fill in quadratic function table. How would you fill in this quadratic table?

\begin{array}{|c|c|}\hline
 x& y \\ \hline
-3& 0 \\ \hline
-2& 2 \\ \hline
-1& ? \\ \hline
 0& ? \\ \hline
 1& 0 \\ \hline
\end{array}

Guide:

Step $1$: Find the quadratic equation that passeas through $(-3,0), (-2,2), (1,0)$
Step $2$: Evaluate that quadratic equation formula that you have foundat $-1$ and $0$.


finding two positive integers in an array I'm facing a problem these days, the question is saying the following:
" Imagine we are having an array of positive integers called (a) and a variable called (k). Among this integers, we are looking for two numbers such that the sum of these two would be the (k). we are in need of designing such algorithm with complexity O(n) that prints out the locations of this two number(if exist)."

for example:
a[6,3,2, 1, 8]  ----- n=5 ----- k=8 ------ numbers: 6,2

in this case what would be the designing of the algorithm?

Thank you all for your efforts, they worth a lot!

Assuming $k$ can be at least considered to be $O\left(n\right)$, but ideally that $k \ll n$, i.e., $k$ is much smaller than $n$, and there is sufficient memory available, here is a relatively simple method. Have a "list" (note I am calling this as a mathematical name, not necessarily as a computer construct, as things like single-linked lists have individual access times of $O\left(k^2\right)$ (so it'll require that $k \ll n$), but I am assuming will use a contiguous array, so each element can be accessed directly, making individual access times be a small constant), call it $L$, of $\left\lfloor \frac{k}{2} \right\rfloor$ ordered arrays of $2$ positive integers which add to $k$, i.e., $\left(1, k - 1\right), \left(2, k - 2\right), \ldots, \left(\left\lfloor \frac{k}{2} \right\rfloor, \left\lfloor \frac{k + 1}{2} \right\rfloor\right)$. For example, for $k = 3$, we have just $\left(1, 2\right)$ while for $k = 3$, we have $\left(1, 3\right), \left(2, 2\right)$. Next, have some sort of structure attached to each value to keep track of which, if any, element of the main array has been found for it, with special attention paid to the case where $k$ is even due to the repeated value of the last element in $L$, i.e., $\left(\frac{k}{2}, \frac{k}{2}\right)$.

Now, go through each element of the main array. If it is $\lt k$, check the appropriate element in $L$ to see whether or not it already has something associated with it. If so, ignore it. Otherwise, assign the value in $L$ to this main array element. If the other element in the array in $L$ already has something assigned to it also, then you are done. Note that this finds the first set of elements which add to $k$, but there may be other ones. Otherwise, if you finish checking the main array with no array in $L$ matching, then there is no such result. In either case, this is altogether an $O\left(n\right)$ algorithm, with a certain fixed amount for the initial set up of $L$, and a fairly small maximum # of steps for checking each element of the main array. Note the constant to use in the $O\left(n\right)$ function depends a fair bit on how large your $k$ can be in relation to $n$. Also, if there is no particular limit on $k$, especially if you can't have any assurance you can fit the list $L$ into memory, then I believe that both TonyK and Penguino are likely correct, i.e., there is no such algorithm.

Note I don't know what language you are using, so I didn't try to write it in program code. Also, it's been over $35$ years since I wrote any psuedocode, so I didn't want to try to do that either. However, I trust my explanation is sufficient enough for you to understand my method.

A naive solution would be to test each element i with each element j until the sum is achieved. That will be an O(n^2) algorithm

for i = 1 to n-1 do
  for j = 2 to n do
    if A[i]+A[j]=k then return (i,j) and halt
return "no solution"

A better solution would be to sort the array first then check for your equality. That will be O(nlogn)

sort A
i=1
j=n
while (i < j) and (A[i]+A[j] < k) do inc(i)
while (i < j) and (A[i]+A[j] > k) do dec(j)
if (i <> j) return (i,j) and halt else return "no solution"

I doubt that an O(n) solution exists for unsorted data. If the data is already sorted then the second algorithm, minus the first line, is O(n).


Is there a method to solve this congruence equation? Consider: $\qquad71\cdot 6^n\equiv a^2\pmod {239}$, where $a$ and $n$ are positive integers.
$a=1$ is a solution for $n=4$,  $a=6$ is another solution for $n=6$.
Is there a general method to find all the solutions to the above equation? I imagine that these are quadratic residues$\pmod {239}$, or not?

HINT.- $239$ is prime so you have to make calculations in the field $\mathbb F_{239}$. One has
$6^n=\dfrac{a^2}{71}$ in the field. Now $71^{-1}=203$ and $71=73^2$ and $71=166^2$ then
$$6^n=(203a)^2\\6^n=(166a)^2$$ What more? Is $6$ square in $\mathbb F_{239}$, for example? Is $n$ necessarily even?


How can $i^i = e^{-\pi/2}$ !! I was asked a homework question: find $i^i$. The solution provided was as follows:

Let $A = i^i$.

$\log A = i \log i$.

Now, $\log i = \log e^{i\pi/2} = \frac{i\pi}{2}$.

So, $\log A = -\frac{\pi}{2}$

Thus, $i^i = e^{-\pi/2}$.

I understood how the result was obtained, but it is illogical. I understand that multiplying by $i$ is equivalent to rotating the position vector of the complex number in Argand Plane by $90$ degrees anti-clockwise. How can rotating $i$ anti-clockwise $i$ number of times give $e^{-\pi/2}$?

So can somebody explain to me graphically or more intuitively, how $i^i = e^{-\pi/2}$ ?

By definition $$e^{ix} = \cos x+i\sin x = cisx$$

This definition can be proved by observing the Taylor expansions of both the RHS and LHS. You will find they are both identical.

Letting $x= \frac{\pi}{2}$ we get this:

$$i = e^{\frac{i\pi}{2}}$$

Then just play with the powers.

$$i^i = e^{\frac{i^2\pi}{2}}$$

$$i^2=-1$$
Hence,
$$i^i = e^{-\frac{\pi}{2}}$$

Of course $i^i$ attains an infinite elemental set of real values (due to the periodic nature of $cisx$) but since your question was to prove $i^i = e^{-\frac{\pi}{2}}$, I have done so accordingly.

If you would like, there is also another solution to this problem. Since we cannot count "i" times, the solution is to find a formula for doing exponents, namely, Euler's formula. When you plug in $e^{xi}$, the output will be $\cos(x)+i\sin(x)$ (in radians), also notated as the $\operatorname{cis}(x)$.  Now, we can define $i$ as $e^{π/2i}$. So therefore our goal is $(e^{π/2i})^i$. Luckily, with the way that exponents work, we can simplify this to $e^{π/2i\cdot i}$, which of course, by definition, is simplified to
$e^{-π/2}$.


Need help to prove power series I try to prove the following series:
$\frac{1}{(1-z)^{2}} = \sum_{n=0}^{\infty} (n+1)z^{n} \quad z \in \mathbb{C}, \quad |z| < 1$

What I have tries so far:

$\text{I tries to use }\frac{1}{1-z} = \sum_{n=0}^{\infty} z^{n}\quad |z| < 1 \text{ and } \frac{1}{1-2z + z^2} = \frac{1}{1-(2z-z^2)}$

$\Rightarrow \sum_{n=0}^{\infty} (2z-z^2)^{n} = \sum_{n=0}^{\infty} z^n(2-z)^n$

but I can't go anyway

You can see it in two different ways. As Michael Galuza has suggest you have:
$\frac{1}{(1-z)^2}=\big(\frac{1}{1-z}\big)'$ then

$$\Big(\sum_{\mathbb{N}}z^n\Big)'=\sum_{\mathbb{N}}(n+1)z^n.$$

Otherwise you can observe that:

$\frac{1}{(1-z)^2}=\big(\frac{1}{1-z}\big)^2$ so

$$\Big(\sum_{\mathbb{N}}z^n\Big)\Big(\sum_{\mathbb{N}}z^n\Big)=\sum_{n\in\mathbb{N}}\sum_{i=0}^n(z^i z^{n-i})=\sum_{n\in\mathbb{N}}(n+1)z^n$$

because $\sum_{i=0}^n(z^i z^{n-i})=(n+1)z^n$.


Evaluate $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\pi^{6k}}{6k!}$ Evaluate

$$\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\pi^{6k}}{(6k)!}$$

I was trying to find a closed form for this sum

$$\sum_{k=1}^{\infty} (-1)^{k+1}\frac{x^{6k}}{(6k)!}$$

I believe there is something to do with $$\cos(x)=\sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k}}{(2k)!}$$

I do not have a clear idea where to started. I am wondering if someone would be able help me out !

Let $\omega=\exp\frac{2\pi i}{3}$. Notice that $1+\omega^n+\omega^{2n}$ equals $3$ if $3\mid n$ and zero if $3\nmid n$. Deduce that

$$ \sum_{k\geq 0}\frac{(-1)^k x^{6k}}{(6k)!}=\sum_{3\mid k}\frac{(-1)^k x^{2k}}{(2k)!}=\frac{\cos(x)+\cos(\omega x)+\cos(\omega^2 x)}{3}.\tag{1}$$
Evaluate both sides at $x=\pi $ to get
$$ \sum_{k\geq 0}\frac{(-1)^k x^{6k}}{(6k)!}=-\frac{1}{3},\tag{2}$$
profit.

This series is not a simple function. Indeed the general series for $x$ gives:

$$\sum_{k = 1}^{+\infty} (-1)^{k+1}\frac{x^{6k}}{(6k)!} = 1-\, _0F_5\left(\frac{1}{6},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{5}{6};-\frac{x^6}{46656}\right)$$

Namely is an HyperGeometric Function.

In you case, though, the sum is simply

$$\frac{4}{3}$$


Prove that this is the solution to the given minimization problem I have the minimization problem

minimize $\displaystyle f_0  = \sum_{i=1}^{N} \mu_i \left( \left( 2^\frac{R_i}{\mu_i} - 1 \right) \right)$

with constraint

$\displaystyle\sum_{i=1}^{N} \mu_i = 1$

and believe that the solution is

$\displaystyle \mu_i = \frac{R_i}{\sum_{j=1}^{N} R_j}$.

The first derivative is
$\displaystyle f_0^\prime = \sum_{i=1}^{N}  \left( 2^\frac{{R}_i}{\mu_i} - \frac{{R}_i}{\mu_i} \rm{log}(2)2^\frac{R_i}{\mu_i} - 1 \right)$

$N$ and $R_i$ are given. I can show that for $i=2$, $f_0^\prime = 0$, by inserting $\mu_2 = 1 - \mu_1$ in $f_0$, taking the derivative and then inserting the solution.

But how can I show it for any $i$?

I couldn't just insert the solution into the first derivative, so I tried to argue that if it works for $i=2$ it would also work for higher $i$. Or that the problem can always be solved for $i=2$ and then subdivided further. But I am not sure.

Any hints are appreciated.

The usual Lagrange multiplier method works (if you are not familiar with the method, let me recommand reading the WP page before continuing, in particular you must realize that the quantity you denote by $f'_0$ simply does not enter the picture).

For every given $i$, the partial derivatives with respect to $\mu_i$ of the function $f_0$ to be minimized and of the constraint are
$$
2^{R_i/\mu_i}-1-(R_i/\mu_i)\log(2)2^{R_i/\mu_i}\quad\text{and}\quad 1\quad\text{respectively}.
$$
Hence the former must not depend on $i$, that is, for a given $C$,
$$
2^{R_i/\mu_i}(1-\log(2)R_i/\mu_i)=C.
$$
The function $x\mapsto 2^{x}(1-\log(2)x)$ is monotone on $x\ge0$ hence $R_i/\mu_i$ must not depend on $i$, that is, one must have $\mu_i=cR_i$ for a given $c$. Since $\sum\mu_i=1$, $1/c$ is the sum of $R_i$ over $i$ and you are done: the optimal vector $(\mu_i)$ is given by
$$
\mu_i=\frac{R_i}{R_1+\cdots+R_N}.
$$

Just trying to throw some insight on the abstract thinking for the accepted answer: let $g:\mathbb{R} \rightarrow \mathbb{R} $ be the function such that $g(x)=2^x(1-x\log2)$. Notice that $g|_{\mathbb{R}_{\geq 0}}$ is decreasing and hence injective, therefore $g(R_i/\mu_i)=C$ implies $R_i/\mu_i $ being constant.

Edit: The above statement is sufficient when $\mu_i,R_i$ are assumed positive. Not sure if it is implied in any sense or if it can be derived without such assumption.


Finding velocity in optimization problem Given $s=-16t^2+192t+144$, what is the velocity when $s=0$?

This is part of a larger optimization problem which I solved, except for this last part. The critical point occurs at $t=6$, so after $t=6$ the position function is decreasing since the slope ($s^{\prime}$) is negative.

To find the velocity when $s=0$, I tried $s=-16t^2+192t+144=0$, and quadratic formula gives the roots, which are $t=-12.71$ and $t=0.71$. Shouldn't we use the positive root, $t=0.71$, since $t$ must be positive, because $t$ is time and can't be negative?

The text answer says $-214.72$ but that comes from using the negative root, and I don't understand why negative time $t$ is used. Is $-214.72$ incorrect?

Thanks.

Multiply $-1$ on both sides of the equation and substitute $s=0$ to get $16t^2-192t-144=0$. Then, solve with result from quadratic variables. Let $a=16$, $b=-192$, $c=-144$ and solve with
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$$

well Dear kmitov
if we solve the quadratic equation we get to 2 roots
12.708 or -0.708

but this answer is not the answer quoted by user 437158
can you help me on this

further velocity and s has a realtion

ds/dt (differentail)

if we find velocity by differentiating

it is idependent of variable  s=0

and ds/dt = -32t+192

if we put velocity v = o  then we get the time when body was supposed to be staionery

at t= 192/32=0.1875 second , the body was sationery but where it was from its origin s=o

it means that we should find instantous velocity when t= 0.1875

so we may say that instanous velocity was vt= 192-6 units when s=0

thanks
Stormer wiki ghubaroo@gmail.com


How would you solve the inequality $\sin x \gt \cos x$? $$\sin x \gt \cos x, \qquad (-2\pi <x <2\pi)$$

I tried an approach saying that $\tan x\gt1$ but apparently the solution, which is $\frac{\pi}{4}<x<\frac{5\pi}{4}$ is not good.

It's a bit confusing.

First look at where $\sin x = \cos x$ (the unit circle will be of help).

Here is the unit circle available at Wikipedia. The ordered pairs along the circumference correspond to $\;(\cos \theta, \sin\theta)$

There are two such solutions $x_1, x_2 \in [0, 2\pi]$: $\;\pi/4, 5\pi/4$. $\sin (\pi/4) = \cos(\pi/4) = 1/2,\;$ and $\;\sin(5\pi/4) = \cos (5\pi/4) = -1/2$.

That will give you intervals to check for when $\sin x > \cos x$.  It looks like given your comment below your answer that you correctly found when $\sin x > \cos x$ on the interval $x\in [0, 2\pi]$!

for $$0<x<\frac{\pi}{4}, \qquad \sin x<\cos x$$ and  for  $$\frac{\pi}{4}<x<\frac{\pi}{2}, \qquad \sin x>\cos x$$


Trying to write the bundle structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$ I'm trying to find an atlas and in general the bundle structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$ respect to, for example, the local chart
$(U_{01},\psi_{01})$, where
$$U_{01}:=\{((x^0:x^1:x^2:x^3),(y^0:y^1:y^2:y^3))\in\mathbb P^3\times \mathbb P^3|x^0\ne 0,y^1\ne 0\}$$
and $\psi_{01}:U_{01}\subset\mathbb P^3\times \mathbb P^3\to\mathbb R^3\times \mathbb R^3$ is the standard chart that divides for the non-zero component.
In particular I was trying to build the bundle structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$ starting from $T^*(\mathbb P^3\times \mathbb P^3)$ and I did this:

Could you tell me if it's correct?
How could I "extend" this idea to build a smooth structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$?
Thank you in advance.

Question: "How could I "extend" this idea to build a smooth structure for Λ2(T∗(P3×P3))? Thank you in advance."
Answer: An "alternative" approach. Since your bundle is constructed using the pull back of bundles on projective 3-space there is another approach that is more indirect but simpler. It is easier to construct such a structure on $\mathbb{P}^3$ and then pull back this structure to your product $\mathbb{P}^3 \times \mathbb{P}^3$.
Let $X:=\mathbb{P}^3_{k}$ be real ($k:=\mathbb{R}$) projective 3-space and let $Y:=X \times_k X$. There is a formula for the cotagent bundle of a product:
$$\Omega^1_Y \cong p^*\Omega^1_X \oplus q^*\Omega^1_X .$$
There is an exact sequence
$$ 0 \rightarrow \Omega^1_X \rightarrow \mathcal{O}(-1)^{4} \rightarrow \mathcal{O}_X \rightarrow 0.$$
You seek a description of the exterior product
$$\wedge^2 (p^*\Omega_X \oplus q^*\Omega_X)$$
There is (Hartshorne, Ex.II.5.16) a filtration
$$0=F^3 \subseteq F^2 \subseteq F^1 \subseteq F^0:=\wedge^2 (\Omega^1_Y)$$
with
$$F^i/F^{i+1} \cong \wedge^i(p^*\Omega^1_X)\otimes \wedge^{2-i}(q^*\Omega^1_X).$$
This gives an isomorphism
$$C1.\text{  }\wedge^2(T^*_Y):=\wedge^2 (\Omega^1_Y) \cong  $$
$$ p^*(\wedge^2 \Omega^1_X)\oplus q^*(\wedge^2 \Omega^1_X) \oplus p^*\Omega^1_X \otimes q^*\Omega^1_X $$
since pull back commutes with wedge products. Hence if you can construct charts for $\Omega^1_X$, you get charts for $\wedge^2(T^*_Y)$ using formula $C1$ and the projection maps $p,q$.
Exmaple: If $x_0,x_1$ are homogeneous coordinates on $C:=\mathbb{P}^1_k$ you get local coordinates $t:=x_1/x_0$ on $U_0$  and $s:=x_0/x_1:=1/t$ on $U_1$.
The cotangent module $\Omega^1_C$ is on $U_0$ a free module (a trivial bundle) on the basis $dt$. On the open set $U_1$ it has the basis $ds:=d(1/t)=-1/t^2dt$. Hence on the intersection $U_{01}:=U_0 \cap U_1$ you get
$$ds=-\frac{1}{t^2} dt:=\psi_{10}(t,s)dt.$$
The function $\psi_{10}(t,s)$ is invertible on $U_{01}$ and defines the coordinate transformation on this open set. Since $\Omega^1_{C/k}$ is an invertible sheaf, the above calculation proves that
$$ \Omega^1_{C/k} \cong \mathcal{O}(-2).$$
You may work with the real rank one vector bundle on $C$ or the locally trivial sheaf of rank one. The above construction uses the language of "locally trivial sheaves". There are projection maps
$$p,q: C \times C \rightarrow C$$
and you get the formula
$$\Omega^1_{C\times C/k} \cong p^*\mathcal{O}(-2)\oplus q^*\mathcal{O}(-2)$$
and
$$\wedge^1(\Omega^1_{C\times C/k}) \cong p^*\mathcal{O}(-2)\otimes q^*\mathcal{O}(-2).$$
You get a similar calculation for $X$ and $Y$.


Translating statement into General Predicate Logic I have a question from Kleene's Mathematical Logic that I am struggling to answer.

'Every liberal advocates changes. Some conservatives favour no one who advocates changes. Therefore, some conservatives favour no liberal.' (Kleene, 1967, 147).

My solution is as follows.
Every $x$ is a liberal, and every $y$ is a conservative. Every x advocates change $(Cy)$ and some y favour no-one who advocates change $\neg (Cx)$.

$\forall x (Cx) , \exists y(\neg Cy) \therefore \exists y(\neg Cx)$

I think this is right?

EDIT:
Okay, so I know that the first premise is $\forall x(Lx \to Ax)$, or, for all $x$, such that $x$ is a (L)iberal, $x$ (A)dvocates change.

Hint

You have to translate every statement individually.

1st premise) Every liberal advocates changes.

  "for every $x$, if $x$ is Liberal, then $x$ Advocate changes".

2nd premise) Some conservatives favour no one who advocates changes.

  "there is some $y$ that is a Conservative and for every $z$, if $z$ Advocate changes, then $y$ does not Favour $z$".

Conclusion) Therefore, some conservatives favour no liberal.

  "there is some $w$ that is Conservative and for every $u$, if $u$ is a Liberal, then $w$ does not Favour $u$".

You can't just stipulate that $x$'s will be used for liberals, and $y$'s for conservatives. All variables are assumed to be objects of one and the same domain.

So, what you need to do is use predicates $L(x)$ for '$x$ is a liberal' and $C(x)$ for '$x$ is a conservative.  Of course, you just proposed to use $C(x)$ for '$x$ advocates change', but for that you could use $A(x)$ instead.

Also, for the favouring, you want to use a $2$-place predicate $F(x,y)$ that means '$x$ favours $y$'


What would be the roots of the derivative of this polynomial I got this $$ f=a(x-1)(x-2)...(x-101) $$ knowing a and that $1 ,2 \ldots ,101$ are it's roots ... I need the roots of $f'$ , how should I do it ? More specifc how can I derivate a polynomial like that, is there a formula or something ?

You can get the derivative thus.

Taking logs, we get $$\log f=\log a+\sum_{i=1}^{101}{\log(x-i)},$$ so that differentiating gives $$f'=f\sum_{i=1}^{101}{\frac{1}{x-i}}.$$


Solving recurrences with the Substitution Method I have several problems like this to do, but I am having a hard time understanding how they get from one step to the next in the example (see note in picture). If I can understand this example I think I'll be good.  Also want to point out that this is just an example, not the actual homework problem.

To give a direct answer to your question it comes from the guess in step $(a)$.

$T(n) = n \log_2 (n) + n$ is the guess you made in step $(a)$. Using that replace $n = \frac{n}{2}$, $T(\frac{n}{2}) = \frac{n}{2} \log_2 (\frac{n}{2}) + \frac{n}{2}$.

$T(\frac{n}{2}) = \frac{n}{2} (\log_2 (n) - \log_2(2)) + \frac{n}{2} = \frac{n}{2} \log_2 (n) - \frac{n}{2} + \frac{n}{2} = \frac{n}{2} \log_2 (n)$

Hence, $T(n) = 2 T(\frac{n}{2}) + n = 2 \frac{n}{2} \log_2 (n) + n = n \log_2 (n) + n$

While solving some recurrences it is good to recognize some nice things about the recurrence you are actually solving. For instance in this recurrence, notice that at each step you are dividing $n$ by $2$. So strictly speaking through the recurrence you can only obtain $T(2^m)$ where $m \in \mathbb{N}$. For instance, $T(3)$ cannot be obtained using this recurrence. However, such recurrences come up when you do certain things, when $n$ is really large, recursively using say a binary tree. When $n$ is really large you are interested mainly in how the solution behaves in an asymptotic sense and not the precise expression for the solution. In such cases, you assume that $n=2^m$ for the simplicity of calculations and the cost for all $n$ (even when $n$ is not a power of $2$) can be shown to obey the solution, obtained by the assuming $n=2^m$, in an asymptotic sense.

If you were to rewrite the recurrence in terms of $m$, and call $T(2^m) = S(m)$, then we would get

$S(m) = 2 S(m-1) + 2^m$ since $\frac{n}{2} = 2^{m-1}$.

$S(m-1) = 2 S(m-2) + 2^{m-1}$.

$S(m) = 4 S(m-2) + 2^m + 2^m$ and so on (hidden induction here) to get

$S(m) = 2^m S(0) + 2^m + 2^m + \cdots + 2^m + 2^m = m 2^m + 2^m$.

Hence $T(n) = n \log_2(n) + n$ since $n = 2^m$ i.e. $m = \log_2(n)$

Solve the following recurrence relation through substitution.

(i) $a_n = a_{n-1} + 5$, subject to initial condition $a_1 = 2$

(ii) $S_n = 5 S_{n-1}$, subject to initial condition $S_0 = 1$


Find the min value of $x+1/x$ on $(0, \infty)$ without using calculus. Using calculus, it is pretty obvious. I was trying to see if it can be done without calculus. May be some sort of inequalities.

With AM-GM we get $$x+\frac { 1 }{ x } \ge 2\sqrt { x\frac { 1 }{ x }  } =2$$

$(x-1)^2 \geq 0$ for $x>0$. This, in turn, implies that $x+1/x\geq 2$ for all $x>0$. I have left the details. So, the min value of $x+1/x$ is 2 on $x>0$.


Finding the complementary function of a second order linear differential equation I have a question stating

  Find the complementary function of  $\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 40y = x$

So I solved the auxillary equation, $ \lambda^2 + 4\lambda +40 = 0$, which gives $\lambda = -2 \pm 6i$. I believe I am correct in thinking this means the complementary solution is of the form $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, where $x \in \mathbb{R}$. I would have thought this is the final answer but the answer given is $e^{-2x}(A\cos{6x} +B\sin{6x})$. To try to get this I did
$$ \begin{align}
Ae^{(-2+6i)x} + Be^{(-2-6i)x} &= Ae^{-2x}(\cos{6x} +i\sin{6x}) +Be^{-2x}(\cos{6x}+i\sin{6x}) \\ &= e^{-2x}((A+B)\cos{6x} + (A-B)i\sin{6x})
\end{align}$$
How do you get to the given answer?

EDIT

To clarify. What I want to know is how to get from my answer, $Ae^{(-2+6i)x} + Be^{(-2-6i)x}$, to the textbook's(and wolfram alpha's) answer, $e^{-2x}(A\cos{6x} +B\sin{6x})$ ?

It is simply $(A+B)=C$ and $(A-B)= D$ to give $e^{-2x}(C\cos{6x}+D\sin{6x})$

It is easy..

Use $e^{ix}=\cos x+i \sin x$.

And in final step you can collect $\cos$ and $\sin$ values as constant (you can say $A+B=$ new constant let say $C$ and $(A-B)i=$ another constant let say $D$)


Algorithm for determining what order to perform a set of tasks in Consider a set of $n$ items. Each item has a date $d$ by which it must be completed. Each item also has a priority level of $p$ and takes a time $t$ to complete. Is there an algorithm for determining in what order the set of items should be completed?

I would suggest creating an ultimate weight for each task factoring d p and t, then executing the list according to weight (greatest to smallest).  You could scale each value to a range, for example date from 0-100, priority 0-60, and t 0-30.  Then add the values together for a total weight. An example implementation in java psuedocode:

class Task contains values:
 double d = 0-30 days due
 double p = 0-100 priority rating
 double t = 0-500 minutes to complete
 String n = task name

ArrayList<Task> tasks = new ArrayList<Task>(); // contains all tasks

//if you assume an initial range of likely values
//for example 0-30 days (d) 0-100 (p) 0-500 minutes (t)
//use Irritate's algorithm to get a new range with method scale

//for each value:

double dweight = scale(task.getD(),0,30,0,100);
double pweight = scale(task.getP(),0,100,0,60);
double tweight = scale(task.getT(),0,500,0,30);
double totalweight = dweight + pweight + tweight;

//now re-sort the arraylist based on totalweight.  TADA!

//rescale code
public double scale(final double valueIn, final double baseMin, final double baseMax, final double limitMin, final double limitMax) {
    return ((limitMax - limitMin) * (valueIn - baseMin) / (baseMax - baseMin)) + limitMin;
}