How do I calculate the integral $\int^{\infty}_{0}\frac{\cos(ax)-\cos(bx)}{x}dx$ without residues? Consider the integral:

$$\int^{\infty}_{0}\frac{\cos(ax)-\cos(bx)}{x}\,dx$$

I found this problem in a subject test problem set and I'm unsure how to solve it without using the residue theorem.

I've simplified it to $$\int^{\infty}_{0}-2\frac{\sin\left(\frac{(a+b)x}{2}\right)\sin\left(\frac{(a-b)x}{2}\right)}{x}\,dx$$ but still unsure where to go from here.

There's a hint in the problem that says: " Rewrite numerator as integral over sine, estimate, and exchange order of integration." but I don't see a helpful bound to use. Bounding using the fact that $|\sin|\le 1$ hasn't gotten me anywhere helpful.

Any ideas?

Notice that:
$$\frac{\cos (ax)-\cos (bx)}{x}=\int_{a}^{b}{\sin \left( xs \right)ds}$$
If you define
$$I\left( t \right)=\int_{0}^{\infty }{{{e}^{-tx}}\frac{\cos (ax)-\cos (bx)}{x}}dx$$
Then
$$I\left( t \right)=\int_{0}^{\infty }{{{e}^{-tx}}\left\{ \int_{a}^{b}{\sin \left( xs \right)ds} \right\}}dx$$
Reversing the order of integration gives:
$$I\left( t \right)=\int_{a}^{b}{\left\{ \int_{0}^{\infty }{{{e}^{-tx}}\sin \left( xs \right)dx} \right\}}ds$$
integration by parts twice gives
$$\int{{{e}^{-tx}}\sin \left( xs \right)dx=\frac{{{e}^{-tx}}\left( s\sin \left( xs \right)-x\cos \left( xs \right) \right)}{{{t}^{2}}+{{s}^{2}}}}$$
So
$$I\left( t \right)=\int_{a}^{b}{\left[ \frac{{{e}^{-tx}}\left( s\sin \left( xs \right)-x\cos \left( xs \right) \right)}{{{t}^{2}}+{{s}^{2}}} \right]}_{0}^{\infty }ds=\int_{a}^{b}{\frac{s}{{{t}^{2}}+{{s}^{2}}}ds}$$
Finally the change of variable $u={{t}^{2}}+{{s}^{2}}$ reduces the integral
$$I\left( t \right)=\frac{1}{2}\ln \left( \frac{{{t}^{2}}+{{b}^{2}}}{{{t}^{2}}+{{a}^{2}}} \right)$$
Can you guess what happens when $t\to 0$?

Split into sum of two terms, and then apply integral from table:

$$\int \frac{\cos ax}{x} dx = \ln x - \frac{(ax)^2}{2 \cdot 2!} + \frac{(ax)^4}{4 \cdot 4!}  - \frac{(ax)^6}{6 \cdot 6!} + ...
 $$


Solve the following system of equations - (4). Solve the following system of equations:
  $$\large
\left \{
\begin{align*}
xy^2 - 2y + 3x^2 = 0\\
x^2y + 2x + y^2 = 0
\end{align*}
\right.
$$

Here's what I did.

If $x = y = 0$ then the system of equations is solved.

If $x \ne 0$ and $y \ne 0$ then we have that

$\left \{
\begin{align*}
xy^2 - 2y + 3x^2 = 0\\
x^2y + 2x + y^2 = 0
\end{align*}
\right.
$ $
\implies
$ $
\left \{
\begin{aligned}
x^2y^2 = 2xy - 3x^3\\
x^2y^2 = -2xy - y^3
\end{aligned}
\right.
$ $
\implies
$ $
2xy - 3x^3 = -2xy - y^3
$

$
\implies
$ $
3x^3 - 4xy - y^3 = 0
$

And I am stuck.

One way we could go is to subtract $x$ times the second equation from the first, getting us: $$xy^2 - 2y + 3x^2-x(x^2y + 2x + y^2) = 0\\xy^2 - 2y + 3x^2-x^3y - 2x^2 -xy^2 = 0\\x^2-x^3y-2y=0\\x^2=x^3y+2y\\x^2=(x^3+2)y$$ Note that if we had $x^3+2=0,$ then the last equation above would become $x^2=0,$ so $x=0,$ contradicting our assumption that $x\ne 0.$ Thus, we don't have to worry about $x^3+2$ being zero, and so $$y=\frac{x^2}{x^3+2}.\tag{$\star$}$$ Substituting for $y$ in the second equation gets us $$x^2\cdot\frac{x^2}{x^3+2}+2x+\left(\frac{x^2}{x^3+2}\right)^2=0,$$ or $$x\left(\frac{x^3}{x^3+2}+\frac{x^3}{(x^3+2)^2}+2\right)=0,$$ which (since $x\ne 0$) is equivalent to $$\frac{x^3}{x^3+2}+\frac{x^3}{(x^3+2)^2}+2=0\\x^3(x^3+2)+x^3+2(x^3+2)^2=0\\x^6+3x^3+2(x^6+4x^3+4)=0\\3x^6+11x^3+8=0\\3\left(x^3\right)^2+11x^3+8=0\\x^3=\frac{-11\pm\sqrt{121-96}}6\\x^3=\frac{-11\pm \sqrt{25}}6\\x^3=\frac{-11\pm 5}6\\x^3=-1\textrm{ or }x^3=-\frac83$$

If we're looking for real solutions, then we have $x=-1$ or $x=-\frac{2}{\sqrt[3]3},$ but if we're interested in non-real solutions, too, then since $-\frac12\pm\frac{\sqrt3}2$ are cube roots of $-1,$ too, we also have $x=\frac12\pm\frac{\sqrt3}2$ and $x=-\frac{2}{\sqrt[3]3}\left(-\frac12\pm\frac{\sqrt3}2\right).$ (For brevity, I will denote $\omega=-\frac12+\frac{\sqrt3}2$ and $\overline\omega=-\frac12-\frac{\sqrt3}2.$)

When $x\in\left\{-1,\omega,\overline\omega\right\},$ we have $x^3=-1,$ so $x^3+2=1,$ and so $(\star)$ becomes $$y=x^2.$$ Thus, we obtain the solutions $(-1,1),$ $\left(\omega,\omega^2\right)=\left(\omega,-\overline\omega\right),$ and $\left(\overline\omega,\overline\omega^2\right)=\left(\overline\omega,-\omega\right).$

When $x\in\left\{-\frac{2}{\sqrt[3]3},-\frac{2\omega}{\sqrt[3]3},-\frac{2\overline\omega}{\sqrt[3]3}\right\},$ we have $x^3=-\frac83,$ so $x^3+2=-\frac23,$ and so $(\star)$ becomes $$y=-\frac32x^2.$$ Thus, we obtain the solutions $\left(-\frac{2}{\sqrt[3]3},-2\sqrt[3]3\right),$ $\left(-\frac{2\omega}{\sqrt[3]3},2\overline\omega\sqrt[3]3\right),$ and $\left(-\frac{2\overline\omega}{\sqrt[3]3},2\omega\sqrt[3]3\right).$

In summary, our solutions are: $$(0,0),(-1,1),\left(-\frac{2}{\sqrt[3]3},-2\sqrt[3]3\right),\left(\omega,-\overline\omega\right),\left(\overline\omega,-\omega\right),\left(-\frac{2\omega}{\sqrt[3]3},2\overline\omega\sqrt[3]3\right),\left(-\frac{2\overline\omega}{\sqrt[3]3},2\omega\sqrt[3]3\right),$$ with the first three being the real solutions.

HINT: Use the discriminant for the two you will find you answer


How can I find the vertex of a parabola using only $x$ intercepts. My teacher gave me this problem where I did a long jump and recorded the distance I went. He then asked us the height. My distance was 80 inches so the x-intercepts are $0,0$ and $80,0$. My question Is how can I make a quadratic Equation using only the $x$ intercepts to find height.

You need additional data to find the height.

For example, the parabolas $y=-x^2+80x$ and $y=-2x^2+160x$ both have the $x$-intercepts you specify, but vertices at $(40,1600)$ and $(40,3200)$, respectively.

Basically, you need to know three points on the parabola to determine it uniquely. The general form of a parabola is $ax^2 + bx + c$, which is three unknowns, which means we need three linear relations (given by three points on the curve) to determine it. Just knowing the x-intercepts does not give enough information to determine all three coefficients.


Show that if we exchange elements in 2 different basis will still give us a basis. I came across a proof of the following theorem:

Let V be a free module over a division ring D. Suppose X and Y are two finite basis of V, then |X| = |Y|.

that uses the fact that if we exchange elements in 2 different basis, we will still get a basis. Say $X = \{x_1, x_2, ..., x_n\}$ and $Y = \{y_1, ..., y_k\}$, then $\{x_1, x_2, ..., x_{n-1}, y_k\}$ is still a basis.

However, I am having trouble proving that, I can't show that if $0 = a_1 x_1 + ... a_n y_k$, then all the coefficients are 0. Can someone teach me how to solve this? Thanks

You have to be more precise.

First of all, it is convenient to assume that $k \le n$.

Then, $y_{k}$ will be a non-zero linear combination of the $x_{i}$. This means that when you write out $y_{k}$ as a linear combination of the $x_{i}$, at least one of the coefficients is non-zero. Permuting the $x_{i}$, one may assume that it is precisely the coefficient of $x_{n}$ that is non-zero.

Now you can proceed safely. That is, you have
$$
y_{k} = a_{1} x_{1} + \dots + a_{n} x_{n},
$$
with $a_{n} \ne 0$. So
$$
x_{n} = a_{n}^{-1} (y _{k} - a_{1} x_{1} - \dots - a_{n-1} x_{n-1}).
$$
This shows that $x_{1}, \dots, x_{n-1}, y_{k}$ is a system of generators, as
$$
\langle x_{1}, \dots, x_{n-1}, y_{k} \rangle \supseteq \langle x_{1}, \dots, x_{n-1}, x_{n} \rangle = V.
$$

If
$$
b_{1} x_{1} + \dots + b_{n-1} x_{n-1} + c y_{k} = 0,
$$
then
$$
(b_{1} + c a_{1}) x_{1} + \dots + (b_{n-1} + c a_{n-1}) x_{n-1} + c a_{n} x_{n} = 0.
$$
This implies $c = 0$, as $a_{n} \ne 0$, and thus all $b_{i}$ are zero. It follows that $x_{1}, \dots, x_{n-1}, y_{k}$ are also linearly independent.

Let $X=\{x_1, x_2\}$, $Y=\{y_1, y_2\}$ with $y_1=x_2$, $y_2=x_1$. Now $\{x_1, y_2\} = \{x_1\}$ is not a basis in general. Hence, your statement is wrong.


How do I determine if a set is open or closed?? I have a question about open and closed sets. As far as I know, a open set is a set that do not contains its boundary points. A closed set is a set that contains its boundary points. If we think of an interval on real line, such as $(0,1)$ and $[0,1]$, the first interval is open and the second one is closed. However, If I am given finite set such as $\{1 ,2 ,3 \}$ or $\{10, 19, -10\}$ in $\mathbb{R}$, how do I determine if the set is open or closed?? From those finite sets, how do I know what is its boundary points?? I am having real analysis class and having hard time. Can anyone give some explanation with example?? Thanks.

First and foremost, it is important to know that open and closed are not opposites; i.e, a set that is not closed is not necessarily open. Sometimes sets can be neither open nor closed. For example, $[0,1)$. Sometimes sets can be both open and closed. For example, the emptyset or $\Bbb{R}$. One way to define an open set on the real number line is as follows:

$S \subset \Bbb{R}$ is open iff for all $s \in S$, there exists an interval of the form $(a,b)$ such that $s\in(a,b) \subset S$.

Another way to tell if a set is open is if it is the complement of a closed set. If $C$ is a closed set, then $\Bbb{R} \setminus C$ is open. Let's consider the union of open sets $(-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty)$. This union is open (although you should prove that any union of open sets is open so you can know this). Now, the complement is $$\Bbb{R} \setminus [(-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty)]= \{1,2,3 \}$$ so we now see that the complement of $\{1,2,3 \}$ is open, allowing us to deduce that $\{1,2,3 \}$ is closed. Read the definitions carefully of open sets, closed sets, limit points and boundary points. A clear understanding of the differences and how they interact will take you far in real analysis and topology.

A finite set is equal to its set of boundary points.


How many solutions are there to a matrix over a finite field? Given a matrix of the form Ax=b of arbitrary dimensions, which is defined over a finite field Z with p elements (p being a prime number), how many solutions can the matrix have?

A matrix which is defined over R (all reals, infinite field) may have infinite solutions. This is impossible with a finite field.

(EDITED)
Let $A$ be an $m \times n$ matrix over your field $\mathbb F$.  In the case $A = 0$, $b=0$, all members of $\mathbb F^n$ are solutions, and there are $p^n$ of them.  If $A$ has rank $r$, the solutions (if any) form an affine subspace of dimension $n-r$, so there are $p^{n-r}$ solutions if there are any.

Apparently, such a matrix can have p to the power of n solutions, where n is the amount of free variables in the matrix.


Value of an element of a symmetric matrix with given eigenvalue. Let \begin{bmatrix}
    a & b &  c \\
    b & d & e \\
    c & e & f
\end{bmatrix} be a real matrix with eigenvalues $1$,$0$ and $3$. If the eigenvectors corresponding to $1$ and $0$ are $(1,1,1)^{T}$ and$(1,-1,0)^{T}$ respectively, then what is the value of $3f?$  How to find value of $3f?$

From the given conditions on eigenvalues we have $$a+b+c=1\\b+d+e=1\\c+e+f=1\\a-b=0\\b-d=0\\c-e=0.$$ Is there any short way to find value of $f?$ Please suggest me. Thanks a lot.

As Friedrich points out, we should make use of the assumption that the matrix is symmetric and therefore the eigenvectors are orthogonal each other. So the first task is to solve $$\left(\begin{array}{ccc}1 & 1 & 1\\1 & -1 & 0\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)=0$$ so that we obtain the remaining eigenvector (please check by yourself that the result is $(-1,-1,2)^T$).

Now we know the image of the transformation for one set of basis $\{e_1=(1,1,1)^T,e_2=(1,-1,0)^T,e_3=(-1,-1,2)^T\}$ for $\mathbb{R}^3$; namely $$Ae_1=\left(\begin{array}{c}1\\1\\1\end{array}\right),\quad Ae_2=\left(\begin{array}{c}0\\0\\0\end{array}\right),\quad Ae_3=\left(\begin{array}{c}-3\\-3\\6\end{array}\right).$$

To obtain a value of $f$, easiest way I believe is to evaluate $$A\left(\begin{array}{c}0\\0\\1\end{array}\right)(=\left(\begin{array}{c}c\\e\\f\end{array}\right)).$$
So we want to know the set of coefficients $(c_1,c_2,c_3)$ such that $c_1e_1+c_2e_2+c_3e_3=(0,0,1)^T.$ Nice thing about the matrix which consists of orthogonal (if not orthonormal) vectors is that we can obtain diagonal matrix by multiplying its transpose. Thanks to this, we can solve the system easily (check by yourself that $(c_1,c_2,c_3)=(1/3,0,1/3)$). After all, we obtain $f=7/3.$

You forgot to take into account that $3$ is another eigenvalue. Moreover, recall that eigenvectors corresponding to different eigenvalues are orthogonal to each other. So, you can obtain an eigenvector corresponding to $3$.


Solution conflict: Expected number of distinct birthdays for $100$ people I was given a homework question that is stated in the title. Although I have a conflict with the solution provided, and was wondering if you could help me understand why the solution is correct or if it is indeed incorrect.

Define $X$ to be number of distinct birthdays.

The answer given is to set up a RV $X_i$ which is $1$ if the ith day is a birthday or $0$ otherwise, where:

$P(X_i = 1) = P(\text{at least one person has birthday on day i}) = 1- P(\text{no one has birthday on this day}) = 1 - \frac{364}{365}^{100}$. And so $\mathrm{E}X_i = 1 - \frac{364}{365}^{100}$

Thus $\mathrm{E}X =\mathrm{E}[X_1 + X_2 \dots X_{365}] = 365\left (1 - \frac{364}{365}^{100} \right)$

I think this is incorrect, however. The reason being is that it seems like they are calculating the expected number of birthdays not the expected number of distinct birthdays.

The answer that I think is correct is to define $X_i$ as $1$ if the ith day is a distinct birthday and $0$ otherwise. Then:

$P(X_i = 1) = 100 \times \left(\frac{1}{365}\right)\left(\frac{364}{365}\right)^{99}$.

Thus $\mathrm{E}X =\mathrm{E}[X_1 + X_2 \dots X_{365}] = 365 \times 100 \times \left(\frac{1}{365}\right)\left(\frac{364}{365}\right)^{99} = 100 \times \left(\frac{364}{365}\right)^{99}$.

This has been bothering me for quite some time. Any help would be great.

The provided solution is correct.  When it computes the chance that somebody has a birthday on Jan 1, it doesn't care how many people share the birthday.  Then it says each day has the same chance of being somebody's birthday and uses the linearity of expectation.

We can see what is going on with smaller numbers.  Say we throw two dice and ask what is the expected number of different numbers seen.  We can do the problem directly by saying the first die is some number.  The second die has $\frac 56$ chance of adding a new number, so the expected number of distinct numbers seen is $\frac {11}6$.  This is less than $2$ because of the chance that the two numbers are the same.  The approach in the solution you quote is to say the chance $1$ does not appear is $(\frac 56)^2$, so the chance it does appear is $1-(\frac 56)^2=\frac {11}{36}$.  Then the expected number of numbers we see is $6 \cdot \frac {11}{36}=\frac {11}6$

I have the same answer as the given result. But I do think the idea of the given answer is not accurate. My way is: Let Xi be the ith person who has a distinct birthday. The probability of Xi is (1/365)*(364/365)^(i-1). The i has the range from 1 to 100. The sum of P(xi) is the probability of 100 persons all have distinct birthdays. And then time 365 days. We got the same result of 87.56.

Hope it's helpful.


Solve for $x$: $\frac1e = e^{2x}$ I tried  making it to $e^{-1} = e^{2x}$

and had the exponents equal each other $-1=2x$

and the I solved for $x$, making it $x=-1/2$, but that answer is wrong.

please help

I don't know why that answer is wrong.

Multiply both sides by $e$; then $e^{2x+1}=1$. The solutions to $e^z=1$ are given by $z=2\pi ni$ for $n\in\mathbb{Z}$, thus the solutions for $x$ are given by
$$
x=\frac{2\pi ni -1}{2},\qquad n\in\mathbb{Z}.
$$
If you only want real solutions, take $n=0$ to get $x=-1/2$ as you mentioned.

$\frac1e = e^{2x}$ is in fact $e^{-1} = e^{2x}$  which applying lans is $lane^{-1} = lane^{2x}$ but this is simply ${-1} = {2x}$ which can only mean $x=-0.5$, so you are , in fact, correct.

Peace and Chicken grease.


Proving $f(x) = x^2 + 1$ is surjective Let $f(x) = x^2 + 1$, where $x$ is a real number. Prove that $f$ maps $ \mathbb R$ onto $[1, \infty)$.

We must show that if $y \in Y$, then there exists an $x$ such that $f(x) = y$.

I am tempted to use the property $f(x) = y$ to replace $f(x)$ in  $f(x) = x^2 + 1$ with $y$ and solve $y = x^2 + 1$ for $x$. Would that be permissible what with the property $f(x) = y$ being in the conclusion  of the statement, rather than in the hypothesis and given conditions?

Let $y \in [1,\infty]$.  Then, choose $x=\sqrt{y-1}$, so that $f(x)=(\sqrt{y-1})^2+1=y-1+1=y$, and f is surjective.

Draw a rough sketch of f(x). It is an upward parabola with no real root. So, f(x) is an one-to-one function, but not onto, because co-domain of f(x)=R, whereas range of f(x)=Z(+). So, it is a surjective function.


Find $f''(x)$ if $f\circ f'(x) = 4x^2 + 3$ Can you tell me the solution of this question?

If: $f\circ f'(x)=4 x^2 +3$
then what is $f''(x)$?

This was a question in math test which I just took yesterday.
One function satisfying the equation above is $f(x)=x^2+3$, for which $f'(x)=2x$ and therefore $f''(x)=2$.
We can also see that $f'(x)$ is monotonic in $[0,+\infty)$, and in $(-\infty,0]$.
What other analytic solutions $f(x)$ exist ?
Can we express all analytic solutions $f(x)$ with a few parameters ?

we will show that the only smooth solution to the problem $f(f'(x)) = 4x^2 + 3 $ is $f(x) = x^2 + 3.$

the differentiable function $f$ is defined by
$$ f(f^\prime(x)) = 4x^2 + 3.\tag 1$$
differentiating $(1)$ gives
$$[f^\prime(f^\prime(x)) ] * f^{\prime \prime}(x)= 8x \tag 2  $$
differentiating $(2)$ gives
$$[f^\prime(f^\prime(x))]*f^{\prime \prime \prime}(x) +
\left( f^{\prime \prime} (x)\right)^2 f^{\prime \prime}(f^\prime(x))= 8 \tag 3$$
 we will note that the range of $f$ is contained in $[3, \infty)$ and in fact $3$ is in the range of $f$ because putting $x = 0$ in $(1)$ gives
$$f(f^\prime(0)) = 3, \ \ f^\prime(f^\prime(0)) = 0\tag 4$$
 the second equality follows  if we recognize that $3$ is a global minimum of $f.$

putting $x=0$ in $(3)$ and making use of  $(4)$ shows that
$f^{\prime \prime}(0) \neq 0.$
 more is true. in fact we will show that
$$ f^{\prime \prime}(x) \neq 0 \tag 5$$

suppose there is an $a$ such that $f^{\prime \prime}(a) = 0.$ putting $x = a$ in $(2)$ shows $a = 0.$ but we already showed that $f^{\prime \prime}(0) \neq 0$ and that establishes $(5).$  we can improve $(5)$ and will show that
$$ f^{\prime \prime}(0) = 2,\ \ f^{\prime \prime}(x) > 0 \text {  for all } x. \tag {5b} $$
this follows simply by putting $x = 0$ in $(3)$

we will now show that the only local extremum on the graph is $(0, 3)$

suppose $x=a$ is a local extremum. then by $(1)$ we have $a = \pm \sqrt{f(0) - 3}$. if these are distinct, then by rolle's thorem, there will be a point $c$ between them at which $f^{\prime \prime}(c) = 0$ contradicting $(5b).$
this tells us that
$$f(0) = 3, f^\prime (0) = 0, f^{\prime \prime }(0) = 2, xf^\prime(x) > 0 \text{ for } x \neq 0, f^{\prime \prime}(x) > 0 \text{ for all } x. \tag 6$$

differentiating $(3)$ we get

$$[f^\prime(f^\prime (x)]*f^{(4)}(x) +
f^{(3)}(x) f^{(2)}(f^\prime(x))f^{(2)}(x) +
2f^{(3)}(x) f^{(2)}(x)f^{(2)}(f^\prime(x) +
f^{3}(f^\prime(x))(f^{(2)}(x))^3 = 0
\tag 7$$

observe that in both $(3)$ and $(7)$ the highest order derivative is multiplied by $[f^\prime(f^\prime(x))]$ which is zero at $x = 0$ it can be seen that this persists for all orders.

now putting $x = 0$ in $(7)$ and making use of $(6)$ gives us
$4f^{(3)}(0) + 8 f^{(3)}(0) + 8f^{(3)}(0) = 0 $ which implies that $$f^{(3)}(0) = 0 $$

continuing this way i think i can show that $$f^{(n)}(0) = 0 \text{ for } n = 3, 4, 5, \cdots \tag 8$$
but it may be easier to expand
$$f(x) = 3 + x^2 + 2a_4x^4 + 2a_5x^5 +  \cdots  $$

$$f^\prime(x) = 2x\left(1 +  + 4a_4x^2 + 5a_5x^3 +  \cdots \right) $$

putting these in $(1)$ we get

$$3 + 4x^2 = 3 + 4x^2 \left( 1 + 4a_4x^2 + 5a_5x^3 +  \cdots \right)^2
+ 2a_4(2x)^4  \left( 1 +  4a_4x^2 + 5a_5x^3 +  \cdots \right)^4
+ 5a_5 (2x)^5 \left( 1 +  4a_4x^2 + 5a_5x^3 +  \cdots \right)^5+ \cdots $$

equating the coefficient of $x^4$ gives $a_4 = 0.$ now remove $a_4$ and show $a_5 = 0$ and by induction we should be able to show $(8)$.

p.s. thanks to the op for posting this problem.

the proof in the last stages can be streamlined using the following claim.

claim: if $f(x) = 3 + x^2 + 2ax^k, k \ge 3,$ then $a = 0.$

proof:

$\begin{align}
3 + 4x^2 &= f(f^\prime (x)) = f(2x(1 + akx^{k-2})) \\
 &= 3 + 4x^2(1 +kax^{k-2})^2 + 2a*(2x)^k(1 + kax^{k-2})^k\\
 &= 3 + 4x^2 + 8akx^k + 4k^2 a^2 x^{2k-2} + 2^{k+1}ax^k + \cdots \\
 &= 3 + 4x^2 + a(8k + 2^{k+1})x^k + \cdots
\end{align}$

which implies $ a = 0.$  this could be used to cut short the proof of $(8).$

Together with the 2 answers already given here , for completeness I add this relevant link

Can $f(g(x))$ be a polynomial?

That should make everything clear.


Prove that $f=x^4-4x^2+16\in\mathbb{Q}$ is irreducible Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible.

I am trying to prove it with Eisenstein's criterion but without success: for p=2, it divides -4 and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find $f(x\pm c)$ which is irreducible:

$f(x+1)=x^4+4x^3+2x^2-4x+13$, but 13 has the divisors: 1 and 13, so don't exist a prime number p such that to apply the first condition: $p|a_i, i\ne n$; the same problem for $f(x-1)=x^4+...+13$
For $f(x+2)=x^4+8x^3+20x^2+16x+16$ is the same problem from where we go, if we set p=2, that means $2|8, 2|20, 2|16$, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for x-2

Now I'll verify for $f(x\pm3)$, but I think it will be fall... I think if I verify all constant $f(x\pm c)$ it doesn't work with this method...  so have any idea how we can prove that $f$ is irreducible?

The associated quadratic polynomial $t^2-4t+16$ has negative discriminant, so there's no real root. Then the polynomial can be factorized over the reals as a product of degree two polynomial. You get them by a process similar to completing the square:
\begin{align}
x^4-4x^2+16
&=x^4+8x^2+16-12x^2\\
&=(x^2+4)^2-(\sqrt{12}\,x)^2\\
&=(x^2-\sqrt{12}\,x+4)(x^2+\sqrt{12}\,x+4)
\end{align}
These two polynomials have negative discriminant (no need to verify it) and so they're irreducible in $\mathbb{R}[x]$. If the given polynomial were reducible over the rationals, the two factorizations in $\mathbb{Q}[x]$ and $\mathbb{R}[x]$ would coincide.

Therefore the given polynomial is irreducible over the rationals.

What's the general rule? Suppose you have $x^4+px^2+q$, with $p,q$ integers and $p^2-4q<0$ (so $q>0$). Write $q=r^2$, with $r>0$ (it need not be integer), and
$$
x^4+px^2+q=x^4+2rx^2+r^2-(2r-p)x^2
$$
Note that $2r-p>0$: it's obvious if $p<0$; if $p\ge0$ it's the same as $4q>p^2$, which is true by hypothesis. Then
$$
x^4+px^2+q=(x^2-\sqrt{2r-p}\,x+r)(x^2+\sqrt{2r-p}\,x+r)
$$
is the decomposition of the polynomial in $\mathbb{R}[x]$. It is in $\mathbb{Q}[x]$ if and only if $\sqrt{q}$ and $\sqrt{2\sqrt{q}-p}$ are integers.

For example, $q=4$ and $p=0$ is a case. For $q=16$ we need $8-p$ to be a square, so $q=16$ and $p=4$ is another case.

f(x) = 0 leads to $x^4$ +16 = 4$x^2$ then for |x| = 0,1, 2 and for |x| ≥ 3 the equality is impossible.Then x can't be integer. On the other hand we have
($x^2$ - 2)^2 = -12 which gives $x^2$ = 2(1 +sqrt(-3)) and x can't be rational. What remains is to prove that f has not two quadratic factors of Z[x]  wich can be proved with undetermined coefficients. I stop here.


Why are mathematical proofs that rely on computers controversial? There are many theorems in mathematics that have been proved with the assistance of computers, take the famous four color theorem for example. Such proofs are often controversial among some mathematicians. Why is it so?

I my opinion, shifting from manual proofs to computer-assisted proofs is a giant leap forward for mathematics. Other fields of science rely on it heavily. Physics experiments are simulated in computers. Chemical reactions are simulated in supercomputers. Even evolution can be simulated in an advanced enough computer. All of this can help us understand these phenomena better.

But why are mathematicians so reluctant?

What is mathematics? One answer is that mathematics is a collection of definitions, theorems, and proofs of them. But the more realistic answer is that mathematics is what mathematicians do. (And partly, that's a social activity.) Progress in mathematics consists of advancing human understanding of mathematics.
What is a proof for? Often we pretend that the reason for a proof is so that we can be sure that the result is true. But actually what mathematicians are looking for is understanding.
I encourage everyone to read the article On Proof and Progress in Mathematics by the Fields Medalist William Thurston. He says (on page 2):

The rapid advance of computers has helped dramatize this point, because computers and people are very different. For instance, when Appel and Haken completed a proof of the 4-color map theorem using a massive automatic computation, it evoked much controversy. I interpret the controversy as having little to do with doubt people had as to the veracity of the theorem or the correctness of the proof. Rather, it reflected a continuing desire for human understanding of a proof, in addition to knowledge that the theorem is true.
On a more everyday level, it is common for people first starting to grapple
with computers to make large-scale computations of things they might have
done on a smaller scale by hand. They might print out a table of the first
10,000 primes, only to find that their printout isn’t something they really
wanted after all. They discover by this kind of experience that what they
really want is usually not some collection of “answers”—what they want is
understanding.

Some people may claim that there is doubt about a proof when it has been proved by a computer, but I think human proofs have more room for error. The real issue is that (long) computer proofs (as opposed to, something simple like checking a numerical value by calculator) are hard to keep in your head.
Compare these quotes from Gian-Carlo Rota's Indiscrete Thoughts, where he describes the mathematicians' quest for understanding:

“eventually every mathematical problem is proved trivial. The quest for ultimate triviality is characteristic of the mathematical enterprise.” (p.93)
“Every mathematical theorem is eventually proved trivial. The mathematician’s ideal of truth is triviality, and the community of mathematicians will not cease its beaver-like work on a newly discovered result until it has shown to everyone’s satisfaction that all difficulties in the early proofs were spurious, and only an analytic triviality is to be found at the end of the road.” (p. 118, in The Phenomenology of Mathematical Truth)
Are there definitive proofs?
It is an article of faith among mathematicians that after a new theorem is discovered, other simpler proofs of it will be given until a definitive one is found. A cursory inspection of the history of mathematics seems to confirm the mathematician’s faith. The first proof of a great many theorems is needlessly complicated. “Nobody blames a mathematician if the first proof of a new theorem is clumsy”, said Paul Erdős. It takes a long time, from a few decades to centuries, before the facts that are hidden in the first proof are understood, as mathematicians informally say. This gradual bringing out of the significance of a new discovery takes the appearance of a succession of proofs, each one simpler than the preceding. New and simpler versions of a theorem will stop appearing when the facts are finally understood. (p.146, in The Phenomenology of Mathematical Proof).

In my opinion, there is nothing wrong with, or doubtful about, a proof that relies on computer. However, such a proof is in the intermediate stage described above, that has not yet been rendered trivial enough to be held in a mathematician's head, and thus the theorem being proved is to be considered still work in progress.

"The part of a proof you let the computer do is extremely hard to check for correctness (except in simple cases). Software has a pronounced tendency to have bugs. You have to check the programme for correctness, and you have to check the compiler for correctness. And the hardware too, it is not unheard-of that hardware is buggy"

Today's Microsoft .NET software is maintained and updated constantly by thousands of people worldwide.  It is common practice for well designed code to build in testing facilities as you write the code, or Built In Self Test(BIST).  These facilities are built in to run on PCs, run on embedded micro-controller targets.  As you develop the code, side by side with the test code, numerical calculations and their results can use the BIST to ensure that i.e. windowed values, tolerances, floating point calculations, and other parameters are what they should be. Of course all of this is quantified based on the design specification as this drives the design of the application and electronic hardware itself.

Provided that the design has been implemented according to design specification, and design itself has been correctly defined to do the intended job, then expectation is that the developed system will work.  Of course there are differences between hardware and software design practices. As signals travel between hardware performing data acquisition and convert them into software quantities, there is 'loss' in the conversion.  not just quantization in the analog to digital conversion but also passive filtering, active filtering, DSP processing, firmware processing and software post processing to name a few.


What are properties, operators, and where do we get numbers from? I am a very confused person. My high school teachers told me division by 0 is undefined because "it just is. The mathematicians just did it that way." So when I found out for real why division by 0 is undefined, I realized that every basic thing I thought I learned couldn't be trusted.

I understand axioms as "something we make up to see what happens," postulates as "things we suspect are true, but haven't proven," and theorems as "things we've proven based on the axioms we've picked." Very tidy, I can dig it.

But I can't find anywhere in several textbooks up to Calc 1 that describes what a "property" is. Is it a particular kind of axiom? Or is it a simple theorem that's close to the axioms? I've found out about indicator functions, but it looks like they only describe properties, without being the properties themselves. How can I prove that $a \times 1 = a$?

Relations, I get them. Mappings from a set to a set, based on rules of any kind. And functions are just a kind of relation where we say each thing in the domain set only gets to have one mapping to the range set. And operators being functions being relations makes sense too, except, where do the operators come from? I'm used to functions being things we make up to study polynomials. But how do we "make up" addition? I can think of any number of algorithms to perform addition, and any number of plain English descriptions of addition, but I couldn't write the right-hand side of $(a, b) = ?$. All the functions I've ever seen are just compositions of addition and the other operators. And why do operators have to be functions, anyway? Couldn't we define division by 0 just by saying division is a relation but not a function? Then it could just return the set of all numbers for $a/0$.

And then numbers. There's an infinity of them, whatever kind you care to pick. How? I figure you could lay down the existence of 1 as an axiom, but how could you construct 2 from that without using addition? Another axiom? Then there'd have to be an infinity of axioms, and that can't be workable.

Yes. I am a very confused person. I hope I picked the right tags.

How many "big" questions... I'll try with a couple of them.

For natural numbers, see Peano axioms : two basic "notions" are assumed :

the existence of an "inatial" numebr : $0$
the existence of a "basic" operation : the successor function $S$.

The first fact is "codified" by the first axiom :

  $0$ is a natural number.

The second fact is established by the second axiom :

  For every natural number $n, S(n)$ is a natural number.

These two "simple" axioms are the rules for the "number game" : start from the beginning and go one step after the other, i.e. counting.

We start from $0$ and apply the successor function $S$ to it, getting a new number : $S(0)$. We call it $1$. Then apply $S$ to $1$, i.e. to $S(0)$ and we get $2=S(S(0))$; and so on...

In order to make this basic machinery to work ad infinitum, we need some further axioms; the third one :

  For all natural numbers $m, n, \ \ m = n$ if and only if $S(m) = S(n)$.

We want that every number has a unique successor.

Then we have :

  For every natural number $n, S(n) \ne 0$.

This axiom is needed in order to avoid that, after a certain "amount" of numbers, we find a "loop" going back to $0$.

Finally, we have the Induction axiom.

With these axioms, we can define the addition operation.

Regarding the "thorny" question of the division by $0$, the issue is quite simple.

We define subtraction as a "derived" operation starting from addition : if $a + b = c$, we want that $c - b = a$.

Thus, from $2 + 0 = 2$, we "derive" : $2 - 0 = 2$.

The same for division with respect to multiplication; from : $a \times b = c$, we "derive" $c/b=a$.

Unfortunately, we have $a \times 0 = 0$ for any $a$. Thus, what is the "expected value" of $n/0$ ? It must be a number $a$ such that, multiplied by $0$ will return $n$.

But no $a$, when multiplied by $0$ will gives us back the original $n$.

Thus, we are forced to agree with the awkward fact that : the division by $0$ is undefined.

A (binary) relation is a way to associate things to other things; the world is plenty of them : "$x$ is father of $y$" defines a relation.

The way mathematics formalizes it is :

  $Father = \{ (x,y) \mid x \ \ \text {is father of} \ \ y \}$.

Functions are relations that satisfy an additional condition, the "functionality" condition : for all $x$ there exists at most one $y$ such that ...

Thus, "father of" is not a function, because a father may have more than one son. The relation "son of" instead, is a function : every son has one father (and not two).

An operation is a function, and thus a relation. We can "describe" the sum as a relation in the following way :

  $Sum = \{ ((n,m),k) \mid k=n+m \}$.

Of course this is not the "recipe" to perform additions; we already have to know how to add $n$ and $m$, but it is a way to "decide", for any triple $n,m,k$ if it satisfy the relation or not, because :

  $((n,m),k) \in Sum$ iff $k=n+m$.

Thus, e.g. $((2,3),5) \in Sum$, while $((1,1),1) \notin Sum$.

In mathematical "parlace", a property is something expressed by a theorem regarding an object or a colelction of objects.

Consider e.g. Euclidean geometry : it is a theory regarding objects (the "geometrical" ones) like points, line, circles, triangles, ...

If we consider the Pythagorean theorem, it states that :

  the sum of the areas of the two squares on the legs of a right triangle equals the area of the square on the hypotenuse.

This theorem states that right triangles have the property that : "the sum of the areas of the two squares on the legs equals the area of the square on the hypotenuse".

But how do we "make up" addition? I can think of any number of algorithms to perform addition, and any number of plain English descriptions of addition, but I couldn't write the right-hand side of $(a, b) = ?$.

It's complicated.

You can start with Peano's Axioms that define the natural numbers. Here is the version I find most useful.

We define $\mathbb{N}, S, 0$ such that:

$0\in \mathbb{N}$
$S$ is a function mapping $\mathbb{N}$ to itself (the so-called successor function)
$S$ is injective.
For all $x\in \mathbb{N}$, we have $S(x)\ne 0$
For all subsets $P$ of $\mathbb{N}$, if $0\in P$ and for all $x\in P$, we also have $S(x)\in P$, then $P=\mathbb{N}$

Then, using the rules and axioms of logic and set theory, we can actually prove (with some difficulty) the existence of a unique addition function $+$ such that:

For all $x,y\in \mathbb{N}$, we have $x+y\in \mathbb{N}$
For all $x\in \mathbb{N}$, we have $x+1=S(x)$.
For all $x,y\in \mathbb{N}$, we have $x+S(y)=S(x+y)$

Then we can prove (also with some difficulty) the usual algebraic properties of addition: associativity, commutativity and cancelability.

The above development in its full detail would require hundreds of lines of formal proof. For obvious reasons, most authors don't bother with such a detailed development, and just "define" addition as above. Some simply include addition and multiplication in their definition of the natural numbers. So, you will have to forgive your high school teachers if they did not stop the class to explain all this.

I don't want to discourage your questions. Just be aware that a full explanation of some questions may be "beyond the scope" of the textbook or course.


If something is not false, is it true? For proving vacuous truths for empty sets, Halmos in his Naive Set Theory mentions that

  To prove that something is true about the empty set, prove that it cannot be false.

But now consider Russell's paradox where $A$ is the set of all sets not contained in it. $A \notin A$ being false there doesn't imply that $A\in A$.

So how does Halmos' method work?

I don't know what point you're trying to make about Russell's paradox, but all Halmos is doing is reminding you to try proof by contradiction. Answering your title question with "yes" is equivalent to asserting the law of the excluded middle, which is assumed in classical logic. If you had instead asked, "If something is not true, is it false?", an answer if "yes" would be equivalent to the law of non-contradiction. This law, which Halmos wants you to use, applies even in some non-classical logics (it fails in paraconsistent logics).

Russell's paradox is a pathology that arises because the definition of "set" is too vague in traditional set theory. That is, the problem is with the "definition" of $A$ as "the set of all sets that don't contain themselves" and furthermore assuming that the symbols "$A\in A$" and "$A \not\in A$" can be interpreted with the same semantics as usual for this specific object $A$.

This is different than the law of the excluded middle, which is the principle in classical logic that $p\vee \lnot p$ is a tautology for all propositions $p$. Hence we have the universally valid inference $(p \vee \lnot p) \wedge \lnot(\lnot p) \Rightarrow p$.

The point that Halmos is making is that due to the law of the excluded middle, you can do proofs by contradiction. For some of the facts about the empty set, this is  very useful.

Not all systems of logic accept the law of the excluded middle. Non-classical systems like intuitionist or paraconsistent logic do not accept it. In mathematics, there is a school called constructivism that rejects the law of the excluded middle, but most mathematicians accept it.


Determine Circle of Intersection of Plane and Sphere How can the equation of a circle be determined from the equations of a sphere and a plane which intersect to form the circle?  At a minimum, how can the radius and center of the circle be determined?

For example, given the plane equation $$x=\sqrt{3}*z$$ and the sphere given by $$x^2+y^2+z^2=4$$

What is the equation of the circle that results from their intersection?  I have used Grapher to visualize the sphere and plane, and know that the two shapes do intersect:

However, substituting $$x=\sqrt{3}*z$$ into $$x^2+y^2+z^2=4$$ yields the elliptical cylinder $$4x^2+y^2=4$$while substituting $$z=x/\sqrt{3}$$ into $$x^2+y^2+z^2=4$$ yields $$4x^2/3+y^2=4$$ Once again the equation of an elliptical cylinder, but in an orthogonal plane.

Why does this substitution not successfully determine the equation of the circle of intersection, and how is it possible to solve for the equation, center, and radius of that circle?

$\newcommand{\Vec}[1]{\mathbf{#1}}$Generalities: Let $S$ be the sphere in $\mathbf{R}^{3}$ with center $\Vec{c}_{0} = (x_{0}, y_{0}, z_{0})$ and radius $R > 0$, and let $P$ be the plane with equation $Ax + By + Cz = D$, so that $\Vec{n} = (A, B, C)$ is a normal vector of $P$.

If $\Vec{p}_{0}$ is an arbitrary point on $P$, the signed distance from the center of the sphere $\Vec{c}_{0}$ to the plane $P$ is
$$
\rho = \frac{(\Vec{c}_{0} - \Vec{p}_{0}) \cdot \Vec{n}}{\|\Vec{n}\|}
  = \frac{Ax_{0} + By_{0} + Cz_{0} - D}{\sqrt{A^{2} + B^{2} + C^{2}}}.
$$

The intersection $S \cap P$ is a circle if and only if $-R < \rho < R$, and in that case, the circle has radius $r = \sqrt{R^{2} - \rho^{2}}$ and center
$$
\Vec{c}
  = \Vec{c}_{0} + \rho\, \frac{\Vec{n}}{\|\Vec{n}\|}
  = (x_{0}, y_{0}, z_{0}) + \rho\, \frac{(A, B, C)}{\sqrt{A^{2} + B^{2} + C^{2}}}.
$$

Now consider the specific example
$$
S = \{(x, y, z) : x^{2} + y^{2} + z^{2} = 4\},\qquad
P = \{(x, y, z) : x - z\sqrt{3} = 0\}.
$$
The center of $S$ is the origin, which lies on $P$, so the intersection is a circle of radius $2$, the same radius as $S$.

When you substitute $x = z\sqrt{3}$ or $z = x/\sqrt{3}$ into the equation of $S$, you obtain the equation of a cylinder with elliptical cross section (as noted in the OP). However, you must also retain the equation of $P$ in your system. That is, each of the following pairs of equations defines the same circle in space:
\begin{align*}
x - z\sqrt{3} &= 0,       & x - z\sqrt{3} &= 0,             & x - z\sqrt{3} &= 0, \\
x^{2} + y^{2} + z^{2} &= 4; & \tfrac{4}{3} x^{2} + y^{2} &= 4; & y^{2} + 4z^{2} &= 4.
\end{align*}
These may not "look like" circles at first glance, but that's because the circle is not parallel to a coordinate plane; instead, it casts elliptical "shadows" in the $(x, y)$- and $(y, z)$-planes.

Note that a circle in space doesn't have a single equation in the sense you're asking.

You were very close.

y2 = 4 * (1 - x2/3)

y  = +/- 2 * (1 - x2/3)1/2 , which gives you two curves

z = x/(3)1/2  (you picked the positive one to plot)

So for a real y, x must be between -(3)1/2 and (3)1/2.


Prove $P \lor Q, P \rightarrow R \lor S, R \rightarrow W \land Z, Z \land \lnot S \rightarrow \lnot W \vdash \lnot S \rightarrow Q$ using ND The problem is to prove
$$P \lor Q, P \rightarrow R \lor S, R \rightarrow W \land Z, Z \land \lnot S \rightarrow \lnot W \vdash \lnot S \rightarrow Q$$
using natural deduction.

I'm studying natural deduction for my upcoming exam in Logic and I have come across this problem. Unfortunately, I cannot seem to find a path to its solution. What I have done is the following and I don't know how to proceed any further:

  $P \lor Q$
  $P \rightarrow R \lor S$
  $R \rightarrow W \land Z$
  $Z \land \lnot S \rightarrow \lnot W$
  $P \qquad H$

    $R \lor S \qquad E \rightarrow 2,5$
    $R \qquad H$

      $W \land Z \qquad E\rightarrow 3,7$
      $W \qquad E\rightarrow 8$
      $Z \qquad E\rightarrow 8$

My professor only uses the Fitch notation, so I would appreciate you using it in your answer.

EDIT: Using the suggestions in the answers provided and studying the rules some more, I have come up with this solution. Can you tell me whether it is acceptable?

$1 \quad P \lor Q
\\ 2 \quad P \rightarrow R \lor S
\\ 3 \quad R \rightarrow W \land Z
\\ 4 \quad Z \land \lnot S \rightarrow \lnot W
\\ 5 \qquad \lnot S \qquad assumption
\\ 6 \qquad \qquad P \qquad assumption
\\ 7 \qquad \qquad R \lor S \qquad implication \ elim \ 2,6
\\ 8 \qquad \qquad R \qquad disjunctive \ syll \ 7,5
\\ 9 \qquad \qquad W \land Z \qquad implication \ elim \ 3,8
\\ 10 \qquad \qquad Z \qquad conjunction \ elim \ 9
\\ 11 \qquad \qquad Z \land \lnot S \qquad conjunction \ intro \ 10,5
\\ 12 \qquad \qquad \lnot W \qquad implication \ elim \ 4,11
\\ 13 \qquad \qquad W \qquad conjunction \ elim \ 9
\\ 14 \qquad \qquad false \qquad negation \ elim \ 13,12
\\ 15 \qquad \lnot P \qquad negation \ intro \ 14
\\ 16 \qquad Q \qquad disjunctive \ syll \ 1,15
\\ 17 \quad \lnot S \rightarrow Q \qquad implication \ intro \ 5,16$

We have to start from $P \lor Q$ and use $∨$elim [see here for the ND rules used].

The branch starting with $Q$ will be :

$1_Q)$ $Q$ --- assumed for $\lor$-elim

    A) $\lnot S \to Q$ --- from 1) by $\to$-intro.

For the "branch" starting with $P$ we have :

$1_P)$ $P$ --- assumed for $\lor$-elim

2) $R \lor S$ --- from 1) and 2nd premise by $\to$-elim

3) $S$ --- assumed [a] from 2) for $\lor$-elim

4) $\lnot S$ --- assumed [b]

5) $\bot$ --- contradiction : from 3) and 4) by $\lnot$-elim

6) $Q$ --- from 5) by $\bot$-elim

  7) $\lnot S \to Q$ --- from 4) and 6) by $\to$-intro, discharging [b].

Now for :

8) $R$ --- assumed [c] for $\lor$-elim

9) $W \land Z$ --- from 8) and 3rd premises by $\to$-elim

10) $W$ --- from 9) by $\land$-elim

11) $Z$ --- from 9) by $\land$-elim

12) $\lnot S$ --- assumed [d]

13) $Z \land \lnot S$ --- from 11) and 12) by $\land$-intro

14) $\lnot W$ --- from 13) and 4th premise by $\to$-elim

15) $\bot$ --- contradiction : from 10) and 14) by $\lnot$-elim

16) $Q$ --- from 15) by $\bot$-elim

17) $\lnot S \to Q$ --- from 12 and 16) by $\to$-intro, discharging [d]

    B) $\lnot S \to Q$ --- from 2), 3)-7) and 8)-9) by $\lor$-elim, discharging [a] and [c].

Now from A), B) and the 1st premise: $P \lor Q$, we conclude by $\lor$-elim with:

  $\lnot S \to Q$,

discharging assumptions $1_Q)$ and $1_P)$.

After an edit the OP asks the following:

  Using the suggestions in the answers provided and studying the rules some more, I have come up with this solution. Can you tell me whether it is acceptable?

One way to tell if a proof is likely acceptable is to put it in a proof checker. Since the OP is using a similar Fitch-style natural deduction system as presented by forallx, the proof checker associated with that text could be used to see if the proof is acceptable.  Here is the result:

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/


What's the rule for solving nested sums? I have the following nested double sum :

$\sum_{t=1}^{T}\sum_{u=t}^{T} Z(u) \cdot (1+i)^{-u}$ with $0<i<1$ and $Z(u)$ being a non-specified function.

By working with an example of $T=3$  I suppose the solution should be

$\sum_{t=1}^T t \cdot Z(t) \cdot (1+i)^{-t}$.

Is that correct?

My problem is to figure out which calculation rule for sums is to apply here when you have the index of the outer sum as starting point for the second sum. The normal commutative and distributive rule don't seem to apply here. I can hardly argue "It works for $T=3$ so it must be right for all $T$." Could someone please tell me the general/universal calculation rule?

Does/how does the solution change when the sum is changed to

$\sum_{t=1}^{T}\sum_{u=t}^{T} Z(u) \cdot (1+i)^{-u+t}$ ?

In the first place not the rules of algebra (distributivity, $\ldots$) apply here, but the rules of describing sets. If you are given a nested sum
$$\Sigma:=\sum_{j=a}^b \ \sum_{k=c(j)}^{d(j)} z_{j,k}\ ,$$
each individual summand $z_{k,j}$ depends on the outer variable $j$ as well as on the inner variable $k$. Sometimes a factor depending only on the outer variable $j$ can be pulled out, but forget this at the moment.

The essential point is to get a clear picture of the set
$$S:=\{(j,k)\ |\ a\leq j\leq b,\ c(j)\leq k\leq d(j)\}$$
of lattice points in the $(j,k)$-plane over which the double sum $\Sigma$ ranges. To this end draw such a plane with a horizontal $j$-axis and a vertical $k$-axis and shade the "region" $S$.

In most cases this region will be convex. This facilitates the rearranging of the double sum $\Sigma$ when we want $k$ to become the outer variable and $j$ the inner variable. There will be a $k_\min:=\min\{k\ |\ \exists j:\ (j,k)\in S\}$ and similarly a $k_\max:=\max\{k\ |\ \exists j:\ (j,k)\in S\}$, and for each integer $k\in[k_\min\ .\, .\ k_\max]$ there are two well defined numbers $a(k)$, $b(k)$, such that
$$\{j\in{\mathbb Z}\ |\ (j,k)\in S\}=[a(k)\ .\, .\ b(k)]\ .$$
It follows that
$$\Sigma\ =\ \sum_{k=k_\min}^{k_\max}\ \sum_{j=a(k)}^{b_k} z_{j,k}\ .$$
Now let's turn to your example. Here the set $S$ is given by
$$S:=\{(t,u)\in {\mathbb Z}^2\ |\ 1\leq t\leq T,\ t\leq u\leq T\}$$
and is a "triangle" with vertices $(1,1)$, $(1,T)$, $(T,T)$. It follows that $u_\min=1$, $\ u_\max=T$, and for each given $u\in[u_\min\ .\,.\ u_\max]$ we have
$$\{t\in{\mathbb Z}\ |\ (t,u)\in S\}=[1\ .\,.\ u]\ .$$
Therefore your sum $\Sigma$ under the envisaged rearranging becomes
$$\eqalign{\Sigma&=\sum_{u=1}^T\ \sum_{t=1}^u Z(u)(1+i)^{-u}\cr &=\sum_{u=1}^T\left( Z(u)(1+i)^{-u}\sum_{t=1}^u 1\right)\cr &=\sum_{u=1}^T u Z(u)(1+i)^{-u}\ .\cr}$$
As $z_{t,u}:=Z(u)(1+i)^{-u}$ did not depend on the inner summation variable $t$ we could pull $z_{t,u}$ out of the inner sum which then became $\sum_{t=1}^u 1= u$.

Your expression

$\sum_{t=1}^T t \cdot Z(t) \cdot (1+i)^{-t}$

is correct for all $T$ and not just $T=3$.
There is a pattern that every $Z(t) \cdot (1+i)^{-t}$ is appearing $t$ times.

It depends on the terms in the expression. In

$\sum_{t=1}^{T}\sum_{u=t}^{T} Z(u) \cdot (1+i)^{-u+t}$,

you have both variable $u$ and $t$ in the same term so it
breaks that straightforward pattern to wrap it
all up in terms of $t$. You still can write it more compactly tough:

$\sum_{u=1}^T Z(u) \cdot (\sum_{t=u}^T (1+i)^{-u+t})$.

Just expand it with terms and see the pattern there is.


What are "tan" and "atan"? As the title says, I'm confused on what tan and atan are. I'm writing a program in Java and I came across these two mathematical functions. I know tan stands for tangent but if possible could someone please explain this to me. I have not taken triginomotry yet (I've taken up to Algebra 1) so I don't really need a very in depth explanation since i wouldnt understand but just a simple one so i could move on with my program would be great! Thanks in advanced. Also if possible could someone possibly give me a link to an image/example of a tangent and atan.

A quick google search of "java atan"  would tell you that it stands for "arctangent", which is the inverse of tangent.  Tangent is first understood as a ratio of non-hypotenuse sides of a right triangle. Given a non-right angle $x$ of a right triangle, $\tan(x)$ is the ratio $\frac{o}{a}$ where $o$ is the length of the leg of the triangle opposite $x$ and $a$ is the length of the leg adjacent to $x$. Arctangent takes the ratio as an input and returns the angle.

I don't think any single answer to your post will give you a complete understanding of tangent and arctangent. To get that, I recommend spending some time with sine, cosine, tangent and the unit circle. Here is a link to an image http://www.emanueleferonato.com/images/trigo.png   that uses an angle $A$ and triangle sides $a,o,h$. In the context of that photo, $\arctan\left(\frac{\text{opposite}}{\text{adjacent}}\right) = A$

atan stands for "arc tangent." It is the inverse function of tangent - this means it undoes the tangent function. So atan(tan(30)) = 30.


Functional equation with strange property about irrational numbers Let $f:\mathbb R \to \mathbb R$ be a function such that for any irrational number r, and any real number x we have $f(x)=f(x+r)$. Show that f is a constant function.

It's easy to see any constant function satisfies the original properti... But I don't see how to show this is the only solution.

If $y$ is irrational, the set $x = 0$, $r = y$ to get $f(y) = f(0)$.

If $y$ is rational, then set $x = y$, $r = \pi-y$ to get $f(y) = f(\pi)$

But by the first statement, $f(\pi) = f(0)$. Hence, $f$ is constant.

Wrong question.

The general solution of $f(x)=f(x+r)$ is $f(x)=\theta(x)$ , where $\theta(x)$ is an arbitrary periodic functions with period $r$ .

Adding the condition $f:\mathbb{R}\to\mathbb{R}$ the general solution will be modified to $f(x)=\theta(x)$ , where $\theta(x)$ is an arbitrary periodic $\mathbb{R}\to\mathbb{R}$ functions with period $r$ .

For example a particular solution $f(x)=\sin\dfrac{2\pi x}{r}$ , it is a $\mathbb{R}\to\mathbb{R}$ function, a periodic function with period $r$ and satisfies $f(x)=f(x+r)$ .


2 Trains and Fly Problem. Find the number of trips made by the fly back and forth. Question: A Train A is approaching at a speed of 10m/sec, another Train B moving in the opposite direction at a speed of 20m/sec. A fly whose absolute speed is 50m/sec goes repeatedly from A to B and back, without loosing any time at any of the trains. The repeated moving back and forth stops when the 2 trains crash into each other.
The initial distance between the 2 trains is 300m.

1>Find the distance traveled by the fly

2>Find the number of trips made by the fly back and forth

My Question: I get that the answer to the first question is 500m, since it takes 10 seconds for the 2 trains to crash, so in 10 seconds at a speed of 50m/sec the fly covers 50m/sec*10sec = 500m.

But how do I find the number of trips(back and forth) the fly makes until the trains crash?

But an engineering solution might be to say that the fly stops going back and forth when the distance between the trains is equal to the fly's length, say $1$ cm. After all the work he's done, the poor guy starts getting smashed after this time. The distance between the trains could be expressed as, $D(t)=300-30t$, where $D$ and $t$ are in units of meters and seconds, respectively. To avoid the infinite result, we can calculate the trips the fly makes up until he starts getting smashed, $T_{smash}=\frac{300-\frac{1}{100}}{30}=9.999667$ seconds. Seems like the number of round trips the fly makes ($N$) can be approximated as:

$$ N=\frac{1}{2}\int_{0}^{T_{smash}} \frac{dt}{\frac{D(t)}{v}} $$ where $v=50$ m/s is the fly's speed and the $1/2$ multiplier is to make $N$ equate to 'round trips'. This computes to $$N\dot = 8.59.$$

Cheers,

Paul Safier

The problem can be solved much more simpler by using graphs. A plot of d(t) vs t for the trains A and B will be two straight line who intersect at the time and position where the two trains crash. Furthermore, plotting the d(t) of the bird on the same plane will show clearly where the fly keeps going to and fro between the two train plots….The graphical solution is simple and easy to understand


Equation of a rectangle I need to graph a rectangle on the Cartesian coordinate system. Is there an equation for a rectangle? I can't find it anywhere.

Based on Raskolnikov's answer here, one can build an implicit Cartesian equation for a $2p \times 2q$ rectangle:

$$\left(\frac{x}{p}\right)^2+\left(\frac{y}{q}\right)^2=\sec\left(\arctan\left(\frac{x}{p},\frac{y}{q}\right)-\frac{\pi}{2}\left\lfloor\frac2{\pi}\arctan\left(\frac{x}{p},\frac{y}{q}\right)+\frac12\right\rfloor\right)^2$$

Another one is based on modifying the implicit equation of a Lamé curve:

$$\left|\frac{x}{p}+\frac{y}{q}\right|+\left|\frac{x}{p}-\frac{y}{q}\right|=2$$

For purposes of plotting with a computer, the implicit equation isn't terribly convenient to handle, so I'll throw in a set of parametric Cartesian equations for free, based on the parametric equations of the Lamé curve:

$$\begin{align*}x&=p\left(|\cos\,t|\cos\,t+|\sin\,t|\sin\,t\right)\\y&=q\left(|\cos\,t|\cos\,t-|\sin\,t|\sin\,t\right)\end{align*}$$

Here's another one, based on a special case of the parametric equations given in this answer:

$$\begin{align*}x&=p\left(\cos\left(\frac{\pi}{2}\lfloor u\rfloor\right)-(2u-2\lfloor u\rfloor-1)\sin\left(\frac{\pi}{2}\lfloor u\rfloor\right)\right)\\y&=q\left(\sin\left(\frac{\pi}{2}\lfloor u\rfloor\right)+(2u-2\lfloor u\rfloor-1)\cos \left(\frac{\pi}{2}\lfloor u\rfloor\right)\right)\end{align*}$$

...and another one:

$$\begin{align*}x&=p\max\left(-1,\min\left(\frac4{\pi}\arcsin\left(\sin\left(\frac{\pi u}{2}+\frac{\pi}{4}\right)\right),1\right)\right)\\y&=q\max\left(-1,\min\left(-\frac4{\pi}\arcsin\left(\cos\left(\frac{\pi  u}{2}+\frac{\pi}{4}\right)\right),1\right)\right)\end{align*}$$

...and I suppose I should stop here. ;)

Lets say the rectangle is centred at $c(h,k)$ and half the length is $a$ and half the width is $b$.

The equation for this rectangle would be:

$$(\frac{x-h}{a})^\infty+(\frac{y-k}{b})^\infty=1$$


Regularization of integral How does one regularize a divergent integral of the form,
$$
I = \int_0^\infty dx \, \cosh^4 x \hspace{0.5 in} ?
$$
Regularizing via simple subtraction of divergences (as we commonly do in physics) is not a viable option, since expanding $\cosh^4x$ in a power series reveals an infinite number of divergent contributions, with no finite piece to remain. However, I do know that there must be some regularization technique out there enabling us to evaluate this, since we have a similar result
$$
\int_0^\infty dx \, \cosh^4 x \sinh x = -{8 \pi \over 15}
$$
in common use. Any tips on how to obtain this sort of thing  (or even reference to a source where this is calculated) would be very helpful!

One may note that:

$$I=\int_0^\infty\cosh^4(x)~\mathrm dx=\frac1{16}\int_0^\infty e^{4x}+4e^{2x}+6+4e^{-2x}+e^{-4x}~\mathrm dx$$

Using $\displaystyle\int_0^\infty e^{-ax}~\mathrm dx=\frac1a$, we may regularize our integral to

$$I=\int_0^\infty\frac38~\mathrm dx$$

Which may be regularized to $\frac38\zeta(0)=-\frac3{16}$.

Same method for the second integral.

The answer by Simply Beautiful Art is totally wrong because of his last step.

$$I=\int_0^\infty\frac38~\mathrm dx=\frac38 \int_0^\infty 1 dx=\frac38\left(\int_0^{1/2}1dx+\sum_{k=1}^\infty 1\right)=\frac38(\frac12-\frac12)=0$$

The last integral can only be regularized to zero.


Square root of a square I'm little puzzled with square roots basics which says that square root of a squared number is absolute value of that number. I was in a impression that it would have both positive and negative roots.For example

The Basic says: $\sqrt{2^2} = |2| = 2$

While I think that: $\sqrt{2^2} = \sqrt{4} = 2$ or $-2$.

$\sqrt{a^2}=|a|=a$. This is because the square root function returns the principal (positive) root.

Note that both $a\times a$ and $-a\times-a$ equal $a^2$ though.

You may be wondering when do we use the above fact. An example is when solving a quadratic equation:

Solve for $a:$

$(a^2-2)=0\implies a^2=2\implies a=\sqrt{2}$ or $a= -\sqrt{2}$

So to summarize, when dealing with the square root, the principal root is returned, while solving for variables, you can consider all values for which the equation holds.

It is a function for the negative choice too. So the idea that the positive root is chosen to form a function is arbitrary.  The better approach is to respect, in any root problems, the possibility of multiple answers.  You can then apply your own preference to fit your application.  But that is your choice, not the math's.  Its like quantum mechanics, where more than one simultaneous state is possible, but the observation reduces to a specific instance.


Sum of reciprocals of squared and factoriels I want to find out the answer to sum of reciprocals of multiplication of $n!$ and $2^n$.

$$\sum_{n=1}^{\infty} \frac{1}{n!(2^n)}.$$

It is, hopefully, a well known result that
$$ \mathrm{e}^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$
Your series can be rewritten as
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n! 2^n}
&= \sum_{n=1}^{\infty} \frac{ \left(\frac{1}{2} \right)^n }{n!} \\
&= \left[
\frac{ \left(\frac{1}{2}\right)^0 }{0!} \color{red}{- \frac{ \left(\frac{1}{2}\right)^0 }{0!}}\right] + \sum_{n=1}^{\infty} \frac{ \left(\frac{1}{2} \right)^n }{n!} && (\text{add zero}) \\
&= \color{red}{-1} + \sum_{n=0}^{\infty} \frac{ \left(\frac{1}{2} \right)^n }{n!} \\
&= -1 + \mathrm{e}^{\frac{1}{2}} && (\text{use the well known result}) \\
&= \sqrt{\mathrm{e}} -1.
\end{align}

Maclauren exspansion of sqrt e is sum(1/(n! * 2n) , 0 < n < infinity. This is an identity. You specified starting at one, the sum of terms lack one.
hence, sqrt e -1 is sum(1/(n! * 2n) , 1 < n < infinity.
ref. and google  https://oeis.org/wiki/Sqrt(e) and Maclauren exspansion.


Lebesgue integration calculation problem? Let $f:[0,1]\to \Bbb R$ be a bounded, Lebesgue measurable function with satisfies $$\int_{[0,1]}f(x)x^kdx=\frac{1}{(k+2)(k+3)}=\frac{1}{k+2}-\frac{1}{k+3} $$ for each $k\in \Bbb N \cup{0}$. Show that $f(x)=x-x^2$ almost everywhere(with respect to Lebesgue measure) on $[0,1]$.

This problem looks obvious by calculus. I don't know where to start. I cannot thinks a theorem in Lebesgue integration I can use. Could someone kindly provide a hint? Thanks!

We can relax the boundedness assumption. Suppose $f\in L^1(0,1)$ and the integral condition holds for all $k=0,1,\dots $ Let $g(x) = f(x)-(x-x^2).$ Then $\int_0^1g(x)x^k\,dx = 0$ for all $k.$ Hence $\int_0^1g\cdot p = 0$ for all polynomials $p.$

If $h$ is continuous on $[0,1],$ then by Weierstrass there is a sequence of polynomials $p_n \to h$ uniformly. A simple DCT argument then shows $\int_0^1g\cdot h = 0.$

Define $s(x) = x/|x|, x \ne 0,s(0) = 0.$ Then $s(g(x))$ is bounded and measurable on $[0,1].$ Hence by a well-known corollary of Lusin's theorem, there is a sequence $h_n\in C([0,1]), |h_n(x)|\le 1$ for $x\in [0,1],$ such that $h_n(x) \to s(g(x))$ a.e. By the DCT, we get

$$\int_0^1|g| = \int_0^1 g\cdot s(g) = \lim_{n\to \infty}\int_0^1g\cdot h_n =0.$$

Thus $g=0$ a.e.

It is easy to prove. We know that if $\int_A (f-g)d\mu=0$ then $f=g\hspace 3mm a.e. on A$ then we have to prove that
$$\int_{[0,1]}(f(x)-(x-x^2))x^kdx=0$$
 But we have
$$\int_{[0,1]}((x-x^2)x^k)dx=\frac{1}{k+2}-\frac{1}{k+3}$$
So the problem is solved.


The inverse of bounded operator? Is the inverse of a bounded operator always bounded , if yes how to prove it ?

If You consider an invertible, i.e. bijective and bounded linear operator $A:X\rightarrow Y$, between two Banach-spaces (it´s important they are complete), then as a consequence of Baires category theorem A is open (open mapping theorem) and so $A^{-1}:Y \rightarrow X$ is continuous, i.e. bounded.

It depends on how you define "inverse" of an operator. In general, if you don't require it be bounded, then no. For example, consider the bijection from $\mathbb{R}$ onto $(-1,1)$. But sometimes we say an operator is invertable only if we can find a bounded operator such that the composition is identity.


Show $\forall \delta > 0, \exists n \in \mathbb{N}$ such that $\frac{1}{n} < \delta$ The question is in the title, but I have no idea how to solve it, so a few hints would be appreciated, thanks.

Suppose the statement is false. Then we have some $\delta>0$ such that $\delta^{-1}>n$ for all $n$. This contradicts the Archimedean property. Assuming $\mathbb{R}$ is defined axiomatically, this property is easily derived as a consequence of the least upper bound axiom.

Given $\delta$, let $k = \frac{1}{\delta}$

$$\delta = \frac{1}{\frac{1}{\delta}} = \frac{1}{k}$$

Consider $ \frac{1}{\lfloor k+1 \rfloor}$.


Am I going about testing for convergence correctly for this example? $$\sum_{n=1}^\infty \frac{(2+4^{-n})}{n^2}$$

Would I be correct in thinking that this is the comparison test and need to use something like $\cfrac{3}{n^2}$ ? This is a much more advanced fraction than in my notes and am cautious of the power of -n on the numerator

Your solution is good. You could even make it closer if you notice that, $\forall n \geq 1$, $4^{-n} \leq \frac 14$ making
$$\sum_{n=1}^\infty \frac{(2+4^{-n})}{n^2} \lt \frac 94\sum_{n=1}^\infty \frac{1}{n^2}$$

Still simpler: asymptotic equivalence. The numerator of the  general term is equivalent to $2$, so
$$\frac{(2+4^{-n})}{n^2}\sim_\infty\frac{2}{n^2},$$
which converges.


Extension field definition vs Kronecker's theorem Extension field definition:

  A field $E$ is an extension field of a field $F$ if $F\subseteq E$ and the operations of $F$ are those of $E$ restricted to $F$.

Kronecker's theorem:

  Let $F$ be a field and let $f(x)$ be a nonconstant polynomial in $F[x]$.
  Then there is an extension field $E$ of $F$ in which $f(x)$ has a zero.

My confusion comes from the proof of Kronecker's theorem:  For a nonconstant polynomial $f(x)$ in $F[x]$, you take one of its irreducible factors $p(x)$ and form the extension field out of $F[x]/ \langle p(x) \rangle$, but how does this satisfy the definition of an extension field?

I don't see how $F \subseteq F[x]/ \langle p(x) \rangle$, because elements of $F$ are just elements of the field, while  elements of $F[x]/ \langle p(x) \rangle$ are of the form $g(x) + \{h(x)p(x) : h(x) \in F[x]\}$ which is a coset?

For every $c\in F$, let $g(x)$ the constant polynomial $c$ in the expression $g(x) + \{h(x)p(x) : h(x) \in F[x]\}$, this enables to identify $F$ as a subfield of $F[x]/(p(x))$.


Is a set contained in its Cartesian product with itself? Given the Cartesian product $X = A \times A$, is $A \subset X$?

If $A=\emptyset$, then indeed $A\subset X$. Otherwise, there is a copy of $A$ inside $X$ for every element of $A$, but the set $A$ itself is not necessarily exactly a subset of $X$. For every $a\in A$ we have $A\times\{a\}\subset X$ and $\{a\}\times A\subset X$. There is an obvious bijection between $A$ and $A\times\{a\}$.

A simple example for which it does not hold is $A=\{a\}$ with $a\neq\emptyset$. Then $X=\{(a,a)\}$ and $a\neq(a,a)$, so $A\not\subset X$.

Unfortunately its not.
Let $m \in A$. Then , $(m,m) \in A \times A=X$.
Now, What is the definition of $A \subset B$? --> Its $x \in A \Longrightarrow x \in B$.

Let your assumption $A \subset X$ is true, then $x \in A \Longrightarrow x \in X$
But we know $m \in A$ and $(m,m) \in X$. So this becomes $m \in A \Longrightarrow m \in X$. But we know $(m,m) \in X$ so this is clear contradiction. Hence your assumption is wrong.
Here I don't used Quantifiers, because it needs a bit higher knowledge of logic, and you are saying I am not very good at set theory, so...


Integral of $dy/dx$ confusion Why is $\displaystyle \int \dfrac{dy}{dx} dx = y + c$, but for example $\displaystyle \int \dfrac{dy}{dx} y dx = \dfrac{1}{2} y^2 + c$ instead of $\dfrac{1}{2} y^3 + c$?

Using and abusing the mathematical notation as sometimes is done when dealing with differential equations, what you really have here is

$$\int\frac{dy}{\color{red}{dx}}\color{red}{dx}=\int 1\cdot dy=y+C\;,\;\;\text{since}\;\;\frac{d}{dy}(y+C)=1$$

OTOH,

$$\int\frac{dy}{\color{red}{dx}}y\,\color{red}{dx}=\int y\,dy=\frac{y^2}2+C\;,\;\;\text{since}\;\frac d{dy}\left(\frac12y^2+C\right)=y$$

I have another question and i hope someone can clarify it:
How do you find out what this is: $\int \frac{dy}{dx}$


$5^{th}$ degree polynomial expression $p(x)$ is a $5$ degree polynomial such that
$p(1)=1,p(2)=1,p(3)=2,p(4)=3,p(5)=5,p(6)=8,$ then $p(7)$

$\bf{My\; Try::}$ Here We can not write the given polynomial as $p(x)=x$
and $p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ for a very complex system of equation,
plz hel me how can i solve that question, Thanks

Like this problem, using difference of differences method,

Substitute $x$ for the appropriate values in your expression $$p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$$and you will have a system of linear equations in $6$ variables ($a,b,c,d,e,f)$ that can be dealt with your favorite method, including linear algebra tech.

For instance, $p(2)=1$ yields $32a+16b+8c+4d+2e+f=1$.

After solving said system, you will have an explicit formula for $p(x)$, and then all you have to do is plug $x=7$ into it to obtain $p(7)$.


Find a natural deduction proof to show ∃x∃y (S(x,y) ∨ S(y,x)) ⊢ ∃x∃y S(x,y) by predicate logic. I'm trying to prove $\exists x \exists y (S(x,y) \lor S(y,x)) \vdash \exists x \exists y S(x,y)$ in natural deduction, and I have already applied existential elimination to get $S(x_0,y_0)$, with $x_0$ and $y_0$ being the assumptions.

Yet I'm stuck on how to prove $S(x_0,y_0)$ from $S(x_0,y_0) \lor S(y_0,x_0)$ after getting $S(x_0,y_0) \lor S(y_0,x_0)$ from existential elimination.

Can someone help me out or is there other ways to approach this question?

From $S(x_0,y_0)$ you conclude $\exists x \exists y S(x,y)$ by using existential generalization twice, once to introduce $\exists y$ and once more to introduce $\exists x$. From $S(y_0,x_0)$ you also conclude $\exists x\exists y S(x,y)$. Therefore you can conclude $\exists x\exists y S(x,y)$ from $S(x_0,y_0) \lor S(y_0,x_0)$.

Here is a proof using a Fitch-style proof checker:

Note that I used universal elimination twice from the same line 3. This allowed me to ultimately derive a contradiction on line 13 from the two disjuncts on line 9.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/


Use Laplace transform to solve initial value prob. The problem is: $y" + 9y = e^t$, with the initial conditions $y(0) = 0, y'(0) = 0$.

I'm stuck at the inverse Laplace transform part.  Do I have to use partial fraction expansion or can I just split the equation $\frac{1}{(s-1)(s^2+9)}$?

You do not have to use partial fractions. I have no idea though about your meaning of split the equation. I think most people would infer splitting the equation as partial fractions decomposition. If you don't want to use partial fractions, we can use the inverse Laplace transform definition. That is,
$$
\mathcal{L}^{-1}\{F(s)\}(t) = \frac{1}{2i\pi}\int_{\gamma - i\infty}^{\gamma + i\infty}F(s)e^{st}ds = \sum_j\operatorname{Res}\{F(s);s_j\}\tag{1}
$$
where $F(s) = \frac{1}{(s-1)(s^2 + 9)}$. The poles of $F(s)$ occur when $s = 1, \pm 3i$; thus, we only have simple poles. Using equation $(1)$, we have
\begin{align}
\frac{1}{2i\pi}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{st}}{(s-1)(s^2+9)}ds & = \lim_{s\to 1}(s-1)\frac{e^{st}}{(s-1)(s^2+9)}+\lim_{s\to 3i}(s-3i)\frac{e^{st}}{(s-1)(s^2+9)}\\
&+\lim_{s\to -3i}(s+3i)\frac{e^{st}}{(s-1)(s^2+9)}\\
&= \lim_{s\to 1}\frac{e^{st}}{s^2+9}+\lim_{s\to 3i}\frac{e^{st}}{(s-1)(s+3i)}+\lim_{s\to -3i}\frac{e^{st}}{(s-1)(s-3i)}\\
&\tag{2}
\end{align}
After taking the limit in equation $(2)$ and simplifying, you should get
$$
\mathcal{L}^{-1}\{F(s)\}(t) = \frac{1}{30}\bigl[3e^t - 3\cos(3t)-\sin(3t)\bigr]
$$


Solve $33x≡9$ (mod 29) I'm having trouble solving congruences. I can't seem to find a method (a series of steps) to follow in order to find the solutions of a linear congruence. In particular, I have $33x≡9$ (mod 29). My thoughts: gcd (33,29) $=1$, so there is one solution. I don't know what to do next (I tried using Euclid's algorithm...). Thanks

Hint $\ {\rm\ mod}\ 29\!:\ x\equiv \dfrac{9}{33}\equiv \dfrac{-20}{4}\equiv -5\equiv 24$

Such reduction and ad-hoc sign twiddling works quite well for small numbers. For larger numbers one can use the Extended Euclidean Algorithm to invert the denominator. Or one can use Gauss's algorithm, since here the modulus is prime.

Beware $\ $ The use of fractions in modular arithmetic is valid only when the denominator is invertible, i.e. coprime to the modulus. Otherwise the quotient need not be unique, for example mod $\rm\:10,\:$ $\rm\:4\,x\equiv 2\:$ has solutions $\rm\:x\equiv 3,8,\:$ so the "fraction" $\rm\:x \equiv 2/4\pmod{10}\,$ cannot designate a unique solution of $\,4x\equiv 2.\,$ Indeed, the solution is $\rm\:x\equiv 1/2\equiv 3\pmod 5,\,$ which requires canceling $\,2\,$ from the modulus too, since $\rm\:10\:|\:4x-2\iff5\:|\:2x-1.\:$

Generally the grade-school rules of fraction arithmetic apply universally (i.e. in all rings) where the denominators are invertible. This fundamental property will be clarified conceptually when one learns in university algebra about the universal properties of fractions rings and localizations.


The integer part of $\sum_{k=0}^{44}\frac{1}{\cos(k^\circ)\cos((k+1)^\circ)}$ What is the integer part of the number $$\sum_{k=0}^{44}\frac{1}{\cos (k^\circ)\cos((k+1)^\circ)}$$

I tried to solve it using partial fractions but could not get a result. Please help me out.

Notice for any $a < b$, we have

$$\tan b - \tan a = \frac{\sin b\cos a - \sin a\cos b}{\cos b\cos a} = \frac{\sin(b-a)}{\cos b\cos a}$$
This implies
$$\begin{align}\sum_{k=0}^{44} \frac{1}{\cos(k^\circ)\cos((k+1)^\circ)}
&= \frac{1}{\sin 1^\circ}\sum_{k=0}^{44} \left(\tan((k+1)^\circ) - \tan(k^\circ)\right)\\
&= \frac{1}{\sin 1^\circ}\left(\tan(45^\circ) - \tan(0^\circ)\right)\\
&= \frac{1}{\sin\frac{\pi}{180}} \approx 57.29868849855018
\end{align}$$
The integer we seek is $57$.

Playing around.

Let
$s(n)
=\sum_{k=0}^{n}\frac{1}{\cos(k)\cos(k+1)}
$.

Looking at the sum in Wolfy,
it peaks at about 12
and has a period of about 22
with 7 peaks.
The extreme values
are about $\pm 10$.
This is probably due to
$\pi \approx \frac{22}{7}$.

In particular,
$s(44) \approx 1.9$
so the integer part is 1.

Also,

$\begin{array}\\
\frac{1}{\cos(k)}-\frac1{\cos(k+1)}
&=\frac{\cos(k+1)-\cos(k)}{\cos(k)\cos(k+1)}\\
&=-2\frac{\sin(k+\frac12)\sin(\frac12)}{\cos(k)\cos(k+1)}\\
\end{array}
$

but this doesn't seem to help much.

Can't think of
anything else.


How can I prove that $xy\leq x^2+y^2$? How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?

Let $y>=x. xy<=x^2+y^2$. Divide both sides by $y$, and you get $x<=(x^2/y)+y$. If $x$ and $y$ are both positive numbers, this is clearly true because we said from the beginning that $y>=x$.


Evaluation of $\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$ Evaluate
$$\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$$

My main concern is finding the indefinite integral as once i have that the rest is fairly straight forward. Please give a detailed answer with reference to why you made each substitution (what indicated that said substitution would work etc.)

My initial substitution was $x= \sin \theta$ which tidies it up a bit

$$\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \space dx$$

Also the answer from Wolfram Alpha is;

$$ = \frac{1}{4}\bigg( log(1-2 x^2)+2 \tanh^{-1}\bigg(\frac{x}{\sqrt{1-x^2}}\bigg)+2 \sin^{-1}x\bigg)+constant$$

I need to be able to answer questions like this in under 15 mins so any help is appreciated

There's a simple solution somewhere in M.SE which goes like this...

$$I = \int_0^{\pi/2} dt \frac{\cos{t}}{\cos{t}+\sin{t}} $$
$$J = \int_0^{\pi/2} dt \frac{\sin{t}}{\cos{t}+\sin{t}} $$

$$I + J = \frac{\pi}{2}$$
$$I - J = \int_0^{\pi/2} dt \frac{\cos{t}-\sin{t}}{\cos{t}+\sin{t}} = \int_0^{\pi/2} dt \frac{(\cos{t}+\sin{t})'}{\cos{t}+\sin{t}}=[\log{(\cos{t}+\sin{t})}]_0^{\pi/2} = 0$$

Thus $I = J = \frac{\pi}{4}$.

ADDENDUM

The above technique requires the easy computation of the antiderivative, which you should see is

$$\frac12 \left [t + \log{(\cos{t}+\sin{t})} \right ] + C = \frac12 \left [\arcsin{x} + \log{(x+\sqrt{1-x^2})} \right ] + C$$

Finding indefinite integral:

\begin{align}
&x\mapsto\sin u\\
&I=\int \frac{\cos u}{\sin u + \cos u}du\\
&=\frac12\int \frac{\cos u - \sin u + \cos u + \sin u}{\sin u + \cos u}du=\frac12 \int\frac{\cos u-\sin u}{\sin u + \cos u}du+\frac u2\\
&=\frac12 \ln (\cos u + \sin u) + \frac u2+C\\
&=\frac12 \ln (\sqrt{1-x^2} + x) + \frac {\sin^{-1}x}{2}+C\\
\end{align}


Direct product vs categorical product Is what we call direct products (of groups, vector spaces, modules,...) actually the same as what we call categorical products? Why the word "direct"?

no, direct sums are biproducts (coproducts and products). But what one usually writes as infinite "direct sum" is an infinite coproduct.


fourth powers as sums of squares Is it possible to have a fourth power that is the sum of two squares in four different ways, e.g., $w^4 = a^2 + b^2 = c^2 + d^2 = e^2 + f^2 = g^2 + h^2$ with the added restriction that $e = a+c$ and $g = a-c$ ?  What is the lowest example, or why is it impossible?

Since $e=a+c$ and $g=a-c$, then the OP wishes to solve the system,

$$a^2 + b^2= c^2 + d^2\tag1$$
$$a^2 + b^2= (a+c)^2 + f^2\tag2$$
$$a^2 + b^2= (a-c)^2 + h^2\tag3$$
$$a^2 + b^2= w^4\tag4$$

It turns out that, using an elliptic curve, there is an infinite number of primitive integer solutions to the first $3$ conditions. However, it is highly doubtful the $4$th can be satisfied as well.

The first was completely solved by Brahmagupta as,

$$a,\;b,\;c,\;d = p r + q s,\; p s - q r,\; p r - q s,\; p s + q r$$

The second and third become,

$$-(3 p^2 - q^2) r^2 + (p^2 + q^2) s^2 = f^2$$
$$(p^2 + q^2) r^2 + (p^2 - 3q^2) s^2 = h^2$$

For example, let $p=1,\;q=2$, then,

$$r^2+5s^2=f^2\tag5$$
$$5r^2-11s^2=h^2\tag6$$

with initial rational point $r=178,\; s=19$. Two quadratic polynomials to be made a square and that has a rational point can be transformed into an elliptic curve, so $(5),(6)$ has infinitely many primitive solutions.

Thus, one small solution to the first three conditions is then,

$$a,\;b,\;c,\;d = 216,\; 337,\; 140,\; 375$$
$$f,\;h =183, \;393 $$

The problem is the fourth condition which becomes,

$$a^2+b^2=(p^2+q^2)(r^2+s^2) = w^4\tag7$$

A third quadratic that is to be made not just a square but a fourth power, makes it highly unlikely that $(5),(6),(7)$ can be solved simultaneously.

Select a pythogorean triplet of your choice.square root the answer.multyply each x and y with the answer.say x is 3 , y is 4 .the square root is 5.then (3×5)squared plus(4×5)squared equals 5^4.two squares equal to a fourth power


Can someone formalize 'compactness of set' defined as follows? My text book says

A set K ⊆ R is compact if every sequence in K has a subsequence that converges to a limit that is also in K.

Wondering how to formalize this statement, my trial was this.

($\forall$$a_n$$\in K$) ($\forall$$\epsilon$>0)($\exists$M$\in$N)($\forall$k≥M) |$a_{n_k}$ - $\alpha$ | < $\epsilon$ & $\alpha$ $\in$ K

But, I'm not confident of quantifying sequence.

So, what is an adequate formalization?

Conventions. I am experimenting with writing my arrows backwards. Also, write $\omega$ for the least infinite ordinal number. So $\omega = \{0,1,2,3,4,\ldots\}.$ Furthermore, write $f \restriction I$ to denote the restriction of $f$ to $I$. With these conventions in place, a rigorous definition of compactness of subsets of $\mathbb{R}$ can be given as follows.

Definition. Consider $X \subseteq \mathbb{R}$. Then $X$ is compact iff for all functions $f : X \leftarrow \omega,$ there exists $I \subseteq \omega$ such that the following hold.

$I$ is cofinal.
There exists $x \in X$ such that the sequence $f \restriction I$ converges to $x$.

To rewrite the condition in text, for all sequences in K, there exists a subsequence, with limit in K. In symbols that's $\forall (a_n) \in K, \exists (i_n), \lim a_{i_n} \in K$ and that's all. You may also want to expand what $\lim \in K$ means. This is the usual $\exists \alpha \in K, \forall \varepsilon, \exists N, \forall n > N, |a_{i_n}-\alpha| < \varepsilon$.

The nice thing about maths is that it's all plug and play. You can start writing the definition of compactness using high level concepts such as a limit. You can later replace the limit with its definition just by renaming the variables it uses.


Indefinite Integral of Floor Function Integration by Substitution From knowing the anti-derivative of floor function to be x*floor(x), is it possible to find the derivative of a function contained within a floor function?

The particular question I had in mind is floor(y(x)/17) and I believe in the same way y in an equation can be treated as y function of x, I tried using integration by substitution.

Nevertheless, the integral appears to be too complex to integrate; as by using the rule for substitution, every integration opens up a new iteration of another integral, sometimes even equating to the L.H.S. but unable to be simplified.
It is a shame my rough work is too unorganized to be shown on here.

It is hoped that someone may shed some light how to integrate implicitly, an example of a similar iterated equation may be fine, but it would be best if the integral of the equation prior can be solved.

First, the antiderivative for $\;\lfloor x \rfloor$ is $x\lfloor x \rfloor - \frac12 \lfloor x \rfloor(\lfloor x \rfloor + 1)\;$, not $x \lfloor x\rfloor$.

For any continuous function $g(x)$ defined on $(0^{-}, L^{+})$ for some $L > 0$ which satisfies:

The points $\lambda_i \in [0,L]$ with integral value $g(\lambda_i)$ are all isolated.
For any such $\lambda_i$, $g(x)$ is strictly monotonic in some neighborhood of $\lambda_i$.
Define $\epsilon_i = \pm 1$ depends on whether $g(x)$ is increasing or decreasing there.

If one interpret all integrals involved as
Riemann Stieltjes integral, one can integrate by part the integrand $\lfloor g(t)\rfloor$ over an interval $[0,x ] \subset [0,L]$ and get:
$$\int_0^x \lfloor g(t) \rfloor dt
= \int_{0^{-}}^{x^{+}} \lfloor g(t) \rfloor dt
= x \lfloor g(x) \rfloor - \int_{0^{-}}^{x^{-}} t\,d \lfloor g(t)\rfloor
= x \lfloor g(x) \rfloor - \sum_{i : \lambda_i \in [0,x]} \epsilon_i \lambda_i
$$
The problem of computing the anti-derivative reduces to a summation over the position of discontinuity of $g(x)$.

Please look up the wiki entry of Riemann-Stieltjes integral, once you understand under what condition you can integration by part, a lot of integral over functions
with only jump discontinuous can be converted to and from corresponding sums easily.

$\lfloor f(t) \rfloor = \int_0^t  \sum_{i=0}^{t+\epsilon}\left(f(a_i+\epsilon) - f(a_i-\epsilon)\right) \delta(x-a_i) dx$ where the $a_i$'s are the points where $f(t) \in \mathbb{N}$. the integration by parts with Dirac works if :
it's an integral on a finite interval $[b,c]$, there is no Dirac at the limits, and the Dirac are integrated times a $C^1$ function on $[b,c]$. if the interval is $[b;\infty[$ then the limit $c\to \infty$ must be investigated. your problem is to reverse that integration by parts thus it works.
note that $\delta(x^2-a)$ is non-sense : only $\delta(x-a)$ is allowed and that every time you get a Dirac. So integrating a $\delta$ is nether a problem, and thus derivating the obtained function neither. this means don't try to make a change of variable when you integrate a Dirac ! (or at least respect the weird rules for it)


characterize all matrices $X$ such that $BA = X$ whenever $AB = X$ It is clear that if $A$ and $B$ are $n\times n$ matrices (over a field) with $AB = I$ then $BA = I$. I like to characterize all matrices $X$ such that $BA = X$ whenever $AB = X$.

If $AB=X$ then we have $BAB=BX$. So $BA=X$ can only follow if $BX=XB$ for all $B$. Therefore $X$ has to be in the center of the general linear group which is all multiples of the identity matrix $I$.

From mathworld.wolfram
"matrix multiplication is not, in general, commutative (although it is commutative if A and B are diagonal and of the same dimension)."

So if for example
$$A=\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{array}\right)$$
and
$$B=\left(\begin{array}{ccc} 4 & 0 & 0\\ 0 & 5 & 0\\ 0 & 0 & 6 \end{array}\right)$$
then
$$AB=BA=\left(\begin{array}{ccc} 4 & 0 & 0\\ 0 & 10 & 0\\ 0 & 0 & 18 \end{array}\right)$$


In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these

The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+\sin^2B+\sin^2C=\sin A\sin C+\sin A\sin B\sqrt3$

I am stuck here.

Since $0=a^2+b^2+c^2-ac-ab\sqrt{3}=\left(b-\frac{\sqrt{3}}2a\right)^2+\left(c-\frac12a\right)^2\,,$ we have $b=\frac{\sqrt{3}}2a$ and $c=\frac{1}{2}a$.  Thus, $\angle A=\frac{\pi}{2}$, $\angle B=\frac{\pi}{3}$, and $\angle C=\frac{\pi}{6}$.

It can be shown that, if $\alpha,\beta,\gamma\in(0,\pi)$ and $\alpha+\beta+\gamma=\pi$, then there is a unique triangle $ABC$, up to scaling, with $BC=a$, $CA=b$, and $AB=c$ such that $$a^2+b^2+c^2=2bc\cos(\alpha)+2ca\cos(\beta)+2ab\cos(\gamma)\,.$$
To show this, one observes that the matrix $$\textbf{X}:=\begin{bmatrix}1&-\cos(\gamma)&-\cos(\beta)\\-\cos(\gamma)&1&-\cos(\alpha)\\-\cos(\beta)&-\cos(\alpha)&1\end{bmatrix}$$ is positive-semidefinite and the eigenspace associated with the eigenvalue $0$ of $\textbf{X}$ is spanned by $\begin{bmatrix}\sin(\alpha)\\\sin(\beta)\\\sin(\gamma)\end{bmatrix}$.

If $\triangle ABC$ is equilateral, then $3x^2=x^2(\sqrt{3}+1)$, absurd.

Now, note that the condition is symmetric respect to $b$ and $c$, ie, if $(a,b,c)$ satisfies, $(a,c,b)$ too, and then $\triangle ABC$ is isosceles with $b=c$.


Chicken Mcnugget Theorem (Frobenius Coin) Problem If $a$ and $b$ must be nonnegative integers, what is the largest integer $n$ such that $13a + 18b = n$ has no solutions?

I have tried doing the postage stamp method, but this doesn't work with such large numbers. Any help?

It works with postage stamp method.

n= LCM(a,b)-a-b

=> LCM(13,18)-13-18 = 203


Expected averages of ordered draws from a uniform distribution. Suppose I take $n$ independent draws from a uniform distribution on $[0,1]$ and call the set of those draws $N$. Then define $X \subset N$ as the $K$ largest draws, and $Y \subset N$ as the $N-K$ smallest draws. What is the probability that the mean of $X$ is $k$ times larger than the mean of $Y$?

So I think, I can combine the order statistic of the uniform distribution and the Bates distribution as follows.

First, the order statistic of the $K^{th}$ largest draw is distributed via a Beta distribution with parameters $K, n-K+1$. So I can define a random variable $a$ based on that.

Then, because each draw is independent, I can use that a to bound intervals and apply the Bates distribution as follows:

$$pr(\mu(X) > k \mu(Y) ) =  pr(Bates\left( [a,1], K\right) > k \; Bates\left( [0,a], n-K \right)) $$

This implies:

$$\int_{0}^{1}Beta(K, N-K+1)\int_{0}^{1}Bates\left( [0,a], n-K \right) \int_{xt}^{1} Bates\left( [a,1], K\right) dx dt da.$$

Does that look right?


Finding the kernel of a homomorphism mapping polynomials to polynomial functions Consider the ring homomorphism $\phi: A[t]\to A^A$ where $A$ is an integral domain and $A^A$ is the ring of all functions on $A$ with values in $A$. Determine the kernel of $\phi$. Be as explicit as possible.

I reasoned that if $f$ has $n$ distinct roots, that is, if every element in $A$ is a root of $f$, then $f\in\ker\phi$. However, my professor disagreed with me and asked me to be more explicit. What am I missing?

Hint $\ $ I presume you mean that $\rm\:t\:$ maps to the identity function. Suppose $\rm\:f(t)\:$ maps to a zero function, i.e. $\rm\:f(A) = 0.\:$ If $\rm\:A\:$ is infinite then $\rm\:f = 0\:$ since a polynomial $\ne 0$ over a domain has no more roots than its degree. If $\rm\:A = \{a_1,\ldots,a_n\}$ is finite, then by the Factor Theorem

$$\rm f(A)=0\iff\forall\: i:\ t\!-\!a_i\ |\ f(t)\iff lcm(t\!-\!a_i) = (t\!-\!a_1)\cdots(t\!-\!a_n)\ |\ f(t),\:$$

Thus the kernel $\rm\:I\:$ is generated by $\rm\:(t-a_1)\cdots(t-a_n).\:$ But $\rm\:t^n - t\in I\:$ since $\rm\:a^{n-1} = 1\:$ for $\rm\:a\ne 0,\:$ by applying Lagrange's theorem to the multiplicative group of the finite field $\rm\:A.\:$ So both polynomials are equal (else their difference is a  polynomial $\ne 0$ of degree $\rm < n$ with $\rm\:n\:$ roots).

Your question isn't so clear so I'll try to answer in the more coplete way.

There are two cases to consider:

$A$ is a finite integral domain;
$A$ is an infinite integral domain.

In case 1. $A$ is a field, this can easily be proven observing that for every $a \in A$ because $A$ is an integral domain the map
$$x \in A \mapsto a x \in A$$
is a injective endo-function on a finite set, so bijective.
In this case clearly we know that for every polinomial $f \in A[x]$ is such that $\varphi(f)=0$ if and only if for every $a \in A$ we have $f(a)=0$ (or more correctely $\phi(f)(a)=0$). This, for the polynomial remainder theorem, imply that for each $a \in A$ have to be $(x-a) \mid f$ and so $\prod_{a \in A}(x-a) \mid f$ (because the polynomial $x-a$ are coprime).

Let's pass to the more interesting case, the case 2: $A$ is infinite.
Let $K$ be the field of fractions of $A$, clearly $K$ is infinite.
If $f \in A[x]$ is such that $\phi(f)=0$ this implies that for every $a \in A$ we have that $(x-a) \mid f$ as polynomial in $K[x]$, but from this would follow that $f$ should have infinite degree, that's possible if and only if $f=0$ in $K[x]$ and so in $A[x]$ too.


characterization of quotient rings In the ring $\mathbb{Q}{[x]}$ consider the following subset
$I = \{f(x) (x^2 + 3) | f(x) \in \mathbb{Q}[x]\}$
Prove that $I$ is an ideal in $\mathbb{Q}[x]$ and express the quotient ring $\mathbb{Q}[x]/I$.
First I'm not sure of all the conditions to verify to be sure in an ideal.
I verified $I$ is an additive subgroup since $f(x)(x^2 + 3) + g(x)(x^2 + 3) = h(x) (x^2 + 3)$ with $h = g + f$.
Then it is also a two-sided ideal $g(x) \cdot f(x)(x^2 + 3)\in I$ with $g \in \mathbb{Q}{[x]}, f(x)(x^2 + 3) \in I$ and $f(x)(x^2 + 3) \cdot g(x) \in I$ as well.
Is it enough, or are there other conditions to be satisfy?
Regarding the characterization of $\mathbb{Q}[x]/I$, I'm uncertain. This should be made of all congruence classes with regards to the equivalence relation ~ defined as:
$g \sim f \quad \iff f - g \in I \quad for \quad g,f \in \mathbb{Q}[x]$
my manual express it as $R/I := \{a+I | a \in \mathbb{Q}[x]\}$
with
$a+I := \{a + i | i \in I \} = \{x \in \mathbb{Q}[x] | x \sim a\}$

By the division algorithm, for every $p(x) \in \mathbb Q[x]$ there exists $q(x),r(x) \in \mathbb Q[x]$ such that $$p(x) = (x^2+3)q(x)+r(x)$$ and that $\deg r(x) < \deg(x^2+3) = 2$ whenever $r(x) \neq 0$; put in another way: for every $p(x) \in \mathbb Q[x]$ there exists $q(x) \in \mathbb Q[x]$ and $a,b \in \mathbb Q$ such that $p(x) = (x^2+3)q(x)+(a+bx)$, which is the same as saying $$p(x)+I = (a+bx)+I.$$ Thus $\mathbb Q[x]/I = \{(a+bx)+I : a,b \in \mathbb Q\}$.


$g_{ij}$ calculation of randers metric Let $F=\alpha + \beta $ where $\alpha=\sqrt{a_{ij}(x)y^iy^j}$ is a riemannian metric and $\beta =b_i(x)y^i$ is a one form.that is F is Randers metric on a manifold $M$.  I want to calculate $g_{ij}$ where
$g_{ij}= \frac{1}{2}[F^2]_{y^iy^j}$

Clearly $g_{ij}=F_{y^i}F_{y^i}+FF_{y^iy^j}$.
Now we will calculate $F_{y^i}$.
$F_{y^i}=\alpha_{y^i}+\beta_{y^i}$.
Now $\alpha^2=a_{ij}(x)y^iy^j$ gives $\alpha_{y^k}=\frac{a_{ik}y^i}{\alpha}$
so$F_{y^i}=\frac{a_{ij}y^i}{\alpha}+b_i$

Again for $F_{y^iy^j}$ after a short calculation we get it as $0$
So $g_{ij}=F_{y^i}F{y^j}$.
so $g_{ij}=(\frac{a_{ij}y^j}{\alpha}+b_i)(\frac{a_{ij}y^i}{\alpha}+b_j)$

Is it okay?

How can I simplify it so that I can calculate the inverse of $g_{ij}$ using the following formula?

If $g_{ij}=h_{ij}+\delta C_iC_j$ where $C_i$ is a column vector , then det $g=(1+\delta C^2)\det(h_{ij})$ and $C^i=h^{ij}C_j$ and $g^{ij}=h^{ij}-\frac{\delta C^iC^j}{1+\delta C^2}$ where $C^2=h^{ij}C_iC_j$.

Can someone help me please.

First of all, the second derivative of F is not zero. Try that part again and you will get a nonzero part you need to add to what you already have.

Second, Once you have that part, you can collect all the terms with $a_{ij}$ in them, "factor" it out, and write it in a form that can be used in your formula.

edit:

apparently someone doesn't like my answer, so let me be more explicit.

you should get $g_{jk}=(1+b\cdot \hat{y})a_{jk}+(\hat{y}+b)_j(\hat{y}+b)_k-(1+b \cdot \hat{y})\hat{y}_j\hat{y}_k$.

Next apply your formula to the first two terms. Finally, apply the formula to the whole thing. Here, y-hat means $\frac{y}{\alpha}$. Also, in that formula, the vector index is raised and lowered by the inverse of the matrix. When you apply the formula to the second and third parts, you will need the inverse of the matrices (which are not just $a^{jk}$).


Cycles of Specified Length in a Graph Let $G=(V,E)$ be a graph, and $A$ be its adjacency matrix. Define $n = |V|$.

Given $A$ and a natural number $m \le n$, I'm interested in the following problem:

  How many simple cycles of length $m$ exist in $G$?

By simple cycle, I mean no repeated vertices along the cycle is allowed (other than the starting and ending vertices, which coincide).

The problem is NP-hard. However, I'm not asking its complexity; I'm merely interested in whether there is a closed-form expression for computing it. (Thus, computing the expression can be NP-hard.)

An answer containing a recursive formula for the number of cycles can be found here.

I think the trace of $A^m$ gives the number of cycles of length $m$ of a graph $G$.


K-theory and usual (cohomoogical) Euler class Can someone explain the difference (and hence the sense of studying) between the usual and K-theoretical Euler class? Most famous and for now the only difference is showed during computations of tensor product of two line bundles.

Namely $e^K(L_1\otimes L_2)\neq e^H(L_1\otimes L_2)$, where $e^K$ and $e^H$ are K-theory and cohomology Euler classes.

Any intuition or maybe an article?

The Euler classes for ordinary cohomology and K-theory are examples of characteristic classes. Roughly speaking, characteristic classes take vector bundles over a space $X$ and give you classes in $E^*(X)$, where $E^*$ is some suitable generalized cohomology theory in the sense of Eilenberg-Steenrod.

Some people study these generally because they give us a homotopy theoretic handle on geometry. Others study them because they give us a geometric handle on homotopy theory. There's a lot to be gained from the interplay between these two areas.

Specializing now to the homotopy theory side, we study characteristic classes because it turns out that, quite often, a large amount of data about the cohomology theory is packed in the characteristic classes. This leads (eventually) to the chromatic picture of homotopy theory, which is beautiful and mysterious.


Can a ring have no zero divisors while being non-commutative and having no unity? I was wondering if, in a ring, the property of having no zero-divisors (except for zero itself) is independent from the ring being commutative or from having a unity (i.e.multiplicative identity) so I started looking for a ring with the following properties:

non-commutative
no unity (i.e. no multiplicative identity: a so-called "rng")
no zero-divisors

I came up with the set of 2 x 2 matrices with even entries: $M_2(2\Bbb Z)$  endowed with the usual matrix addition and matrix multiplication.
It is:

non-commutative: $$\begin{pmatrix}2&2\\2&0\end{pmatrix}\begin{pmatrix}0&2\\2&2\end{pmatrix}\neq\begin{pmatrix}2&2\\2&0\end{pmatrix}\begin{pmatrix}0&2\\2&2\end{pmatrix}$$
no unity: $$\begin{pmatrix}1&0\\0&1\end{pmatrix}\notin M_2(2\Bbb Z)$$

But unfortunately it does have zero divisors: $$\begin{pmatrix}2&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&2\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$

So, can you come up with a ring having those three properties? Or a proof that such a group cannot exist?

Let $R$ be the ring of polynomials with integer coefficients in two non-commuting variables $x,y$, and let $I$ be the ideal generated by $\{x,y\}$.

Then $I$, regarded as a ring, satisfies all your specified conditions.

Well, each semigroup $(S,\circ)$ with no unit element can be extended to a monoid $(S\cup\{e\},\circ,e)$ which has a unit element $e$ not belonging to $S$. This is a general construction.

So the question whether the ring (i.e., multiplicative semigroup) has a unit element or not is more or less pointless.


Prove the matrix is symmetric negative definite Given a curve of symmetric positive definite matrices $t\mapsto P(t)\in\mathbb{R}^{n\times n}$ and a constant symmetric negative definite matrix $A\in\mathbb{R}^{n\times n}$, how would you prove that the symmetric matrix $Q(t)=AP(t)+P(t)A$ is negative definite? Or, if it is not true, what property would you impose on $A$ so that this is true?
It is quite easy to prove that both $AP(t)$ and $P(t)A$ have negative eigenvalues, however, the sum of the two matrices I think needs additional assumptions on $A$ in order to have negative eigenvalues. I say this because for a given matrix $S$ with all negative eigenvalues, we can not say anything about the Eigenvalues of symmetric and antisimmetric part
It is clear that for some choice of $A$ it works, for example $A=-c^2Id$ but in general it does not hold for any $A$ negative definite.
An example where this does not work is $$A = \begin{bmatrix} -34 & 8 \\ 8 & -2\end{bmatrix}$$ and $$P=\begin{bmatrix} 26 & -4 \\ -4 & 2\end{bmatrix}$$ where the matrix $AP+PA$ has a one eigenvalue which is positive and one which is negative. I know this is not a nice example, but I've found it playing a little bit with some matrices.
Do you understand what could be an additional property to impose onto $A$ to guarantee that $AP+PA$ has negative eigenvalues?

Hints:
All the matrices here are real symmetric:

Any symmetric matrix is diagonalizable, so we can assume its form is
$$\;\begin{pmatrix}\lambda_1&0&\ldots&0\\
0&\lambda_2&\ldots&0\\
\ldots&\ldots&\ldots&\ldots\\
0&0&\ldots&\lambda_n\end{pmatrix}\;$$

Any symmetric matrix is positive definite (=PD) iff all its eigenvalues are positive, and it is negative definite (=ND) iff all its eigenvalues are negative.

The product of a PD matrix and a ND definite matrix is a ND matrix

The sum of two PD (ND) matrices is a PD (ND) matrix.

Now put the above together.