How can I calculate $\sum_{i=1}^j \sum_{j=1}^4 \left(2ij+1\right) $ by doing the second summation after the first one? Consider $$\sum_{i=1}^j \sum_{j=1}^4 \left(2ij+1\right) $$

  How can I calculate the second summation after the first one?

We assume OP's double sum is not correctly written and we have to consider instead
\begin{align*}
\sum_{j=1}^4 \sum_{i=1}^j \left(2ij+1\right) \tag{1}
\end{align*}
The representation (1) is valid. The outer sum varies $j$ from $1$ to $4$ and for each $j$ we consider the inner sum varying $i$ from $1$ to $j$.

  If we want to exchange inner and outer sum in (1) we can write
  \begin{align*}
\sum_{j=1}^4 \sum_{i=1}^j \left(2ij+1\right)
=\sum_{\color{blue}{1\leq i\leq j\leq 4}} \left(2ij+1\right) =\sum_{i=1}^4\sum_{j=i}^4\left(2ij+1\right)
\end{align*}

To avoid confusion where $j$ is both an upper limit of the outer summation and the indexing variable of the inner summation, I will refer to the $j$ on the outer summation instead as $J$.

Remember that nested summations are evaluated from the inside out.  That is to say:

$$\sum\limits_{i=1}^J\sum\limits_{j=1}^4 2ij+1 = \sum\limits_{i=1}^J\left(\sum\limits_{j=1}^4(2ij+1)\right)$$

Now, let us evaluate the innermost summation.  $$\sum\limits_{j=1}^4(2ij+1) = (2i+1)+(4i+1)+(6i+1)+(8i+1) = 20i+4$$

Notice, $j$ has disappeared from the result of the inner summation.

We have then $$\sum\limits_{i=1}^J\left(\sum\limits_{j=1}^4(2ij+1)\right) = \sum\limits_{i=1}^J\left(20i+4\right)$$

Notice, if it really bothers you that badly that I'm using $J$ instead of $j$, if you were to change $J$ back to $j$ you get effectively the same result so far... My point in changing $j$ to $J$ is to point out and really emphasize that the meaning and use of the $J$ on the outside is different and unrelated to the meaning and use of the $j$ on the inside.

Continuing, using the linearity of summation we have:

$$\sum\limits_{i=1}^J(20i+4) = 20\cdot \left(\sum\limits_{i=1}^J i\right) + 4\cdot \left(\sum\limits_{i=1}^J 1\right)$$

Both of these two summations should be well known.  $\sum\limits_{i=1}^J i = T(J) = \dfrac{J(J+1)}{2}$ is the $J$'th triangle number.  Search elsewhere on this site for a proof as to why the summation evaluates as it does.  The second summation $\sum\limits_{i=1}^J 1 = J$ because you are simply adding $1$ to itself a total of $J$ times.

We get as a result:

$$20\cdot \left(\sum\limits_{i=1}^J i\right) + 4\cdot \left(\sum\limits_{i=1}^J 1\right) = 20\cdot \dfrac{J(J+1)}{2} + 4\cdot J = 10J^2+14J$$

Replacing $J$ back with $j$ so that it matches the original problem, we have:

$$\sum\limits_{i=1}^j\left(\sum\limits_{j=1}^4(2ij+1)\right) = 10j^2+14j$$

Notice that this summation relies on the value of $j$.  It is bad practice to have a variable name be repeated as an indexing variable at the same time as it is in use elsewhere.

Compare this to where the summations were reversed.  The sum $\sum\limits_{j=1}^4\sum\limits_{i=1}^j (2ij+1)$ evaluates to

$((2+1))+((4+1)+(8+1))+((6+1)+(12+1)+(18+1))+((8+1)+(16+1)+(24+1)+(32+1))=140$


find general equation of $x''(t) + 5x'(t) + 4x(t) = 0$ Suppose $x_1(t)$ and $x_2(t)$ are two linearly independent solutions of the
equations:

$$x'_1(t) = 3x_1(t) + 2x_2(t)$$

and

$$x'_2(t) = x_1(t) + 2x_2(t)$$

where $x'_1(t)\text{ and }x'_2(t)$ denote the first derivative of functions $x_1(t)$ and $x_2(t)$
respectively with respect to $t$.

Find the general solution of

$x''(t) + 5x'(t) + 4x(t) = 0$

in terms of $x_1(t)$ and $x_2(t)$.

The general solution of the equation

$x''(t) + 5x'(t) + 4x(t) = 0$................................(1)

is

$x(t) = c _1 e^{-4t} + c _2e^{-t}$................................(2)

Now if i want to express equation (2) in term of $x_1(t)$ and $x_2(t)$ , what exactly i have to do ?

This question was asked on Indian Statistical Institute exam sheet some years ago.

Both $x_1(t)=e^{-4t}$ and $x_2(t)=e^{-t}$ are linearly independent solutions of the given differential equation.

Thus, the complete general solution is,
$x(t) = c_1 x_1(t)+c_2x_2(t)=c_1e^{-4t}+c_2e^{-t}$, where $c_1$ and $c_2$ are arbitrary constants.

Consider $a^2 + 5a + 4 = (a+1)(a+4)$
Now "substitute" the variable $a$ by "taking the derivative":
The system $y = x' + 4x$ and $y' + y = 0$ is equivalent to your second-order-equation. This is the important result that you can check, the thoughts above are just a help for guessing correctly.

You can rewrite the system as $x' = -4x+y$ and $y' = -y$ or $$\left(\matrix{x\\y}\right)' = \left(\matrix{-4 & 1 \\ 0 & -1}\right) \left(\matrix{x\\y}\right)$$

Now rewrite the given system of equations to the same form, too: $$\left(\matrix{x\\y}\right)' = \left(\matrix{3 & 2 \\ 1 & 2}\right) \left(\matrix{x\\y}\right)$$
Now find the transformation between these matrices and you get a basis for the vector space of the solutions of the first system as a linear combination of the solutions of the second, which are the given $x_{1/2}(t)$


Directional Derivative(DD) If Iam going to derive the DD of a function without using a unit vector   i.e.$u=<a,b>$ (not unit vector). I get the same result $Du=(Fx).a+(Fy).b$.So why are we considering unit vector always?

If  V = < a, b> is not a unit vector then afx + bfy is not the directional derivative of f at the direction of V. It is the directional derivative multiplied  by the norm of V. Therefore you do not get the same result unless you normalize V.

The formula stated here gives the derivative of f, a real valued function of two variables, with respect to a vector u whose components are 'a' and 'b' at a point where the partial derivatives(denoted in the question by fx and fy) of f are evaluated, in case the function f is differentiable there, as a function of two variables. If we put the additional condition that u is a unit vector we get what is usually referred to as the directional derivative of f,at the point, along u and which is equal to the component of the gradient vector of f along u.
Ref: Tom Apostol: Mathematical Analysis.


AX = b has solution iff $\langle A, Y\rangle = 0$ implies $\langle b,Y\rangle = 0$ Let $A$ be an $m\times n$ matrix of real numbers, $b \in \mathbb{R}^n$.
Show that the system of equations $AX = b$ has a solution if and only if $A^TY = 0$ always implies $b^TY = 0$.

I'm not sure what we can use to prove this.  For one, since $A$ is not square, it can't be invertible, so I don't think we can say anything about $A^TY = 0$ right?

I can try substituting $b = AX$ into $b^TY.$  This gives $A^TY = 0 \Rightarrow (AX)^TY \Rightarrow 0 = X^TA^TY = 0$.  I'm not sure if I can use this to make any progress.

Any help would be greatly appreciated.

Thanks.

$AX=b$ have solution means $b \in \operatorname{im} A$.

$A^TY=0$ implies $b^TY=0$ means $b\in \operatorname{Ann}\ker A^T $, where $\operatorname{Ann}$ is the annihilators.

So your statement is equivalent to show $\operatorname{im} A=\operatorname{Ann}\ker A^T$, the rest is yours. (Hint: show $\operatorname{im} A\subset \operatorname{Ann}\ker A^T$ and $\operatorname{Ann} \operatorname{im} A\subset \ker A^T$)


Why is the empty set bounded? Why is the empty set bounded below and bounded above? If it has no elements, how can you say that an upper or lower bound exists?

Recall that implication has the property that when the assumption is false, the implication is true. In other words, if $P$ is false, then $P\implies Q$ is true.

Let $S$ be a set of real numbers. Then $M$ is an upper bound for $S$ if the following implication holds, $$s\in S\implies s \leq M.$$ Now let us examine the case for the empty set, $\emptyset$.

Proposition Let $M$ be any real number. Then $M$ is an upper bound for the empty set (of real numbers).

Proof:

Since the statement, $s\in\emptyset$ is false, the implication, $$s\in\emptyset\implies s \leq M$$ is true.

Note: An almost identical proof works for the lower bound case.  A nice little slogan to remember here is

All things are true about the MEMBERS of the empty set.

We can say that the empty set is bounded if its not in R; that is if the empty set is the complement of R then we can say is bounded, because it lies in the extension real numbers.But if the empty set is a subset of R then it may be bounded or unbounded. If we assume it finite then its bounded set but if we look at its supremum and its infimum we see that they are not real numbers so the set is unbounded. I see that the empty set is both bounded and unbounded.


Show that a relation is equivalent if it is both reflexive and cyclic. A relation $R$ on set $X$ is called cyclic if whenever both $xRy$ and $yRz$ then $zRx$ where $x,y,z\in X$. Show that a relation on $X$ is an equivalence relation if and only if it is both reflexive and cyclic.

Since an equivalence relation is reflexive, symmetric and transitive and we already know the relation is reflexive I assume you prove that symmetric and transitive is the same as reflexive and cyclic. I honestly don't know where to start though or if my intuition is correct so any help would be great!

1) reflexive + cyclic => equivalence

a) reflexive => reflix

b) reflexive + cyclic => symmetry

xRy => xRy and yRy => yRx.

c) symmetry + cyclic => transitivity

xRy + yRz => zRx => xRz

====================

2)  Equivalence => reflixe + cyclic

a) reflexive => reflexive

b) transitive + symmetry => cyclic

xRy and yRz => xRz => zRx.


Proving convergence of series $\sum_{n=1}^{\infty}u_{n}$ I was asked to prove that the series; $\sum_{n=1}^{\infty}u_{n}$ converges, given that,

$\lim_{n \rightarrow \infty}n^{p}u_{n}=A<\infty$ and $p>1$.

The proof of this is:

If $u_{n}<\frac{A}{n^{p}}$ then applying the integral test shows convergence.

This may be a stupid question but why is it that we are able to make the assumption $u_{n}<\frac{A}{n^{p}}$ ?

Thanks.

Besides direct comparison, you have $ u_n \sim A / n^p $ hence $\sum u_n$ converges by limit comparison with $\sum A/ n^p.$


Solve: $f(x+\frac{1}{y}) + f(x-\frac{1}{y}) = 2f(x).f(\frac{1}{y})$ Here i have one functional equation:
If $$f(x+\frac{1}{y}) + f(x-\frac{1}{y}) = 2f(x)\cdot f(\tfrac{1}{y})\text{ for all x},y\in\mathbb{R}-{0}$$ and $f(0) = \frac{1}{2}$ , then find the value of $f(4)$.

As noted in the comments, $f(x)$ is symmetric.  Set $a=f(1), b=f(2), c=f(3), d=f(4)$.  Then we have the following four equations:

$$d+f(0)=2b^2$$
$$d+b=2ac$$
$$c+a=2ab$$
$$b+f(0)=2a^2$$

This has two solutions according to alpha; in both solutions $d=-\frac{1}{2}$.  Finding them by hand doesn't seem too difficult.

$f(x)=\frac{1}{2}\cos x$ satisfies the condition.


Interior etc. of $M = \left \{ \ \left(\begin{array}{c} x \\ y \end{array}\right) \in \mathbb{R^2}: x \geq y^2 \text{ and } y > 2x-3 \right \}$ How can one find out the interior $\mathring{M}$, the closure $\overline{M}$ and the boundary $\partial M$ of this set?

$$M = \left \{ \ \left(\begin{array}{c} x \\ y \end{array}\right) \in \mathbb{R^2}: x \geq y^2 \text{ and } y > 2x-3 \right \}$$

I looked at the definitions of the interior, closure and boundary, but wasn't able to find an "algorithm" as to how one can figure them out.
Can someone show me how it's done?

This is what I get when I put the functions on Desmos:

It's quite clear from your picture: the area is (topologically) just a "ball" in the plane with part of its boundary already included. The intersection area (with strict inequalities) is the interior, but the points on the edge are also topologically speaking the boundary: balls around them stick out in both parts: inside and outside.

So on inspection: the interior is

$$\{(x,y) \in \Bbb R^2: x> y^2 \text{ and } y > 2x-3\}$$

the closure includes the points that are on the edges (so have $=$ in the the inequalities):

$$\{(x,y) \in \Bbb R^2: x\ge y^2 \text{ and } y \ge 2x-3\}$$

while the boundary is a union of three curves/lines (the points of intersection of $x=y^2$ and $y=2x-3$ are easily found to be $(\frac94, \frac32)$ and $(1,-1)$):

$$\{(x,y) \in \Bbb R^2: y^2=x, 0\le x \le \frac{9}{4}\} \cup \{((x,y) \in \Bbb R^2: 0\le x \le 1, y=-\sqrt{x}\}\cup \{(x,y) \in \Bbb R^2: y=2x-3, 1\le x \le \frac94\}$$

HINT. Consider the functions  $f(x,y) = x-y^2$ and $g(x,y)=y-2x.$ Then $M=f^{-1}([0, \infty)) \cap g^{-1}((-3, \infty)).$ The rest should be really easy.


Formally proving that if $E[\sum_{i=1}^\infty {X_i}] < \infty$, then $\sum_{i=1}^\infty {X_i} < \infty$ almost surely Given a sequence of non-negative random variables $(X_i)_{i\in\mathbb{N}}$, I would like to show that

$$ \mathbb{E}[\sum_{i=1}^\infty {X_i}] < \infty$$

implies that

$$ \sum_{i=1}^\infty {X_i} < \infty $$

almost surely

The approach that I have in mind is to condition the expectation on the value of $\sum_{i=1}^\infty {X_i}$ as follows:

$\begin{align}
\mathbb{E}[\sum_{i=1}^\infty {X_i}]  =& \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} < \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} < \infty] + \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] \end{align}$

and then say that since

$\mathbb{E}[\sum_{i=1}^\infty {X_i}] < \infty$

I can then argue that the above conditions imply that

$\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} < \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} < \infty] < \infty$

$\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] < \infty$

Given that $\mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] = \infty $, this must mean that $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] = 0$

Although this explanation makes sense intuitively, it doesn't seem formal enough (in particular it relies on the notion that $0 \times \infty = 0$). Is there a more elegant or formal approach?

This argument is correct.  In measure theory, when you allow functions with value $\infty$, the convention is indeed that $0 \times \infty = 0$, while any positive number times $\infty$ is taken to be $\infty$.

Thus, if a random variable is infinite on a set of measure zero, this doesn't contribute to its expected value.  But if the function takes the value $\infty$ over a set of positive measure, then its expected value is infinite.  (This is just a rephrasing of your argument.)

As it turns out the approach that I outlined above is not entirely correct. The reason why is because it has conditioned the expectation on the event $\sum_{i=1}^\infty {X_i} = \infty$, which only makes sense if $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] > 0$.

A different approach which uses the same idea but avoids this annoying technicality is to use a proof by contradiction, where we assume that

$$\mathbb{E}[\sum_{i=1}^\infty {X_i}] < \infty$$

but there exists some $\omega \in \Omega$

$$\sum_{i=1}^\infty {X_i}(\omega) = \infty$$

so that $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] > 0$.

Accordingly, we can now condition the expectation on the value of $\sum_{i=1}^\infty {X_i}$ since $\mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] > 0$ by assumption.

As shown above, this argument then shows that:

$\begin{align}
\mathbb{E}[\sum_{i=1}^\infty {X_i}]  &= \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} < \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} < \infty] &+ \mathbb{E}[\sum_{i=1}^\infty {X_i} \ \Bigg | \sum_{i=1}^\infty {X_i} = \infty] \mathbb{P}[\sum_{i=1}^\infty {X_i} = \infty] \\
&= M + \infty \\
&= \infty \end{align}$

which yields a contradiction and proves the required result.


First-grader problem in arithmetic I found this problem in a text book on arithmetic for first graders (7 y.o.) of the former USSR* . The problem comes from the section that covers single-digit addition and subtraction. Here is the screenshot of the problem:

This is the entire problem: there is no textual description accompanying it, and of course no answer at the back of the book. Other problems in the section are of the 1 + 6 = and 8 - 7 = kind, so this should be an elementary problem as well. However, I cannot figure out what is being asked here: I do not remember this notation, because we used different text books.

Can anybody figure out what is being asked by this assignment?

*  A.S.Pchyolko, G.B.Polyak "Arithmetic" Fifth edition. Text book for the first grade of elementary school. Moscow, Printing house of the Department of Education, 1959

Possibly the student is to produce three numbers by adding the number in the centre of the circle to each of the three around it. (Of course by adding $-6$ I mean subtracting $6$, since I assume that the students had not yet been introduced to negative numbers.) I note that the indicated subtractions are all possible in the non-negative integers, a fact that can be viewed as some small evidence in favor of the interpretation.

I notice the first 2 numbers +6 and -6 can be produced by rearranging the given numbers with + and - signs:

2+4 = 6

10 - 9 - 7 = -6

But the last 2 aren't too clear


What is the minimum height of a binary tree with $n$ vertices? What is the minimum height of a binary tree with $n$ vertices?

Is it $$\lceil \log_2n\rceil$$,

$$\lfloor \log_2{n} \rfloor$$

or

$$\lceil \log_2{n+1} \rceil -1$$

So we can see that if we let $n$ and $x$ be a natural number then,

Height 0 - $2^0$

Height 1 - $2$ $\le$ n $\le $$3$

Height 2 - $4$ $\le$ n $\le $$7$

Height 3 - $8$ $\le$ n $\le $$15$

Height 4 - $16$ $\le$ n $\le $$31$

Using mathematical deduction we can conclude,

Height x - $2^x$ $\le$ n $\le $$2^{x+1}-1$

And since we want the least number of terms for the minimum height of the graph we take into consideration the term on the left of the inequality. Further, in order to get the term of the least number of vertices of a given height n must equal to $2^x$ and so,

$n=$ $2^x$ $\Longrightarrow$ $\log_2{n}$ $=x$

Therefore in order to get the least number in each interval we can conclude,

$\lfloor \log_2{n} \rfloor$ $=$ $x$

I think there is a mistake in your idea, which is about the range of $n$.

For example, with height $2$,  $3 \leq n$, not $4$, and of height $3$, it is $4 \leq n$.


Is The *Mona Lisa* in the complement of the Mandelbrot set. Here is a description of how to color pictures of the Mandelbrot set, more accurately the complement of the Mandelbrot set. Suppose we have a rectangular array of points. Say the array is $m$ by $n$. Suppose also we have a number of color names. Now suppose we assign the color name $j$ to a point in the array if the $j$-th iteration exceeds $2$. If the iterates do not exceed $2$ we color the point black. This process will yield a picture. By careful positioning the array of points can we get any picture we want? In particular can we get a digital representation of the Mona Lisa.

I do not know how to begin to prove or disprove this. My guess is that we can probably get any pictures.

Edit

A different way to color the array would be to use color $c$ if the first iterate to exceed $2$ is iterate $i_{1}$, $i_{2}$, $\cdots$, $i_{j_{c}}$. The iterates for different colors should be distinct. If someone wishes to use infinites lists for the number of iterates that are assigned to a color that would also be acceptable.

With this change the problem reduces to finding an $m$ by $n$ array where each point in the array has a different number of iterates before the value exceeds $2$.

I think yes

Consider the sequence of "westernmost" islands increasing in period:

Here are some examples, you can see them increasing in hairyness / spinyness.

Period 20:

Period 30:

Period 40:

Period 50:

Zooming in near to a sufficiently hairy high-period island, you can get very nearly parallel spines:

There is a natural grid with a 4:1 aspect ratio between successive escape time level sets in neighbouring spine gaps.  (Determined by measuring images: I have no proof of this seeming fact.  But it's not a strictly necessary detail.)

If you choose the angle of the spines carefully, you can create an NxN grid with square cells that satisfies "distinct level set at each pixel sampling point in an (m,n) image array".

There is a slight fuzziness due to the imperfect parallelism and the quantized angles available for any given period island, but increasing the period far enough and that won't matter - eventually it gets good enough for the finite width of the level sets to take care of it.

You are coloring the interval between the lemniscates http://mathworld.wolfram.com/MandelbrotSetLemniscate.html

When zooming into the outer parts of the mandelbot set (where z[k] tends to eventually diverge) , you will render the picture into the boundaries of a convex polygon (likely a square or a rectangle). By looking carefully at the different points where |z[j]| < 2 and z[j+1] > 2, you will notice that the minimal values for j lies on the border of the rendering surface.

This mean that mandelbrot rendering is no more than drawing the contour lines on the map of an area without lake or holes. It always looks like that http://en.wikipedia.org/wiki/Contour_line#/media/File:Contour2D.svg

This requires the model to look as pretty as an elevation map. Lisa Gherardini might not appreciate the comparison (nor my humour).

As a annoying side effect, this must prevent Leonardo to arbitrarily use color transitions, the artist must always follow the transitions color[j+min], color[j+min+1] color[j+min+2]...color[j+max], in that very order, everywhere in the paint. Isolated areas within the face (like the eyes) must follow the same color transitions. Look at the bottom of her eyes pupils, some colors are "missing" (blame Mandelbrot, not Leonardo).

The other approach could be to start numbering the colors from the masterwork, and verify (or enforce) the numbers to meet the topography constraints of the connected contours. THAT alone is a challenging problem.

Maybe, a realistic request would be to locate the topographic map of any location and scale you wish in the world.


What is the probability that at least one card is in the right box Four Christmas cards are randomly placed in their recipients' boxes
  (one in every box). What is the probability that at least one card is
  in the right box?

So, if we number the boxes with $1,2,3,4$ and let the first card should be placed in the box with number $1$, the second in the box with number $2$, etc. If only one card is in the right box, we have $4$ possibilities; if only two cards are in the right boxes, we have $6$ possibilities. I am having troubles with what happens when three cards are in the right boxes. Do we have only $1$ way or $4$ ways?

The probability that at least one card is in the right box is $1$ minus the probability that no cards are in the right boxes. The latter is a derangement.

There are $n!$ total arrangements, and you are interested in the probability of not getting a derangement, which is $$1-\displaystyle\frac{\left\lfloor\displaystyle\frac{n!}{e}+\frac{1}{2}\right\rfloor}{n!}$$
Your problem is the case $n=4$, for which we have

$$1-\displaystyle\frac{\left\lfloor\displaystyle\frac{4!}{e}+\frac{1}{2}\right\rfloor}{4!} =1 -\frac{9}{24} = \boxed{\,\frac{5}{8}\,\,} $$

If 3 cards are in the correct box, then the fourth card must be in the remaining box. There is only 1 way to do this.


Help formulating a conjecture about the parity of every cycle length in a bipartite graph and proving it I know a cycle in a graph $G=(V,E)$ is a sequence of vertices $$v_0, v_1,\ldots, v_k$$ such that $k\geq 3$, $v_k = v_0$, and
$G$ contains every edge between consecutive vertices: $(v_0, v_1)$, $(v_1, v_2), \ldots$, $(v_{k−1}, v_k)$.
I know eventually I want to prove any cycle in a bipartite graph has an even length as the conjecture.

A hint has been given to use simple induction, but I am having trouble proving a bipartite graph with length $k+1$ also can only contain cycles of even length.

If you want to prove that every cycle in a bipartite graph is even, you can reason by contradiction. In details :

Let $G$ be a bipartite graph with sets of vertices $V_0$ and $V_1$. Suppose now that you have an off cycle in $G$, of length $2k+1$ :
$$ v_0,v_1,\ldots,v_{2k}$$

Without loss of generality you can suppose that $v_0\in V_0$. Then $v_1$ must be in $V_1$, $V_2$ in $V_0$, etc. Formally $v_i\in V_0$ if and only if $v_{i+1}\in V_1$, or $v_i\in V_0$ if and only $i$ is even.

This implies that $v_{2k}$ is in $V_0$. Then we have
$$ \left\{ \begin{array}{l}
v_{0}\in V_0\\
v_{2k}\in V_0
\end{array}\right.\text{ and } (v_{2k},v_0)\in E(G)$$

A contradiction with the fact that $G$ is bipartite.

I don't know why we need to use an induction.
The problem: every cycle in a bipartite graph is even, is simply proved by a contradiction.

Assume that there is an odd cycle in a bipartite graph. However, this derives a contradiction from two facts: 1. any subgraph of a bipartite graph is bipartite, and 2. odd cycle is not a bipartite graph.

It is very well-known fact that

A graph $G$ is bipartite iff every cycle in $G$ is even.


Why don't we study algebraic objects with more than two operations? Undergraduates learn about algebraic objects with one operation, namely groups, and we learn about algebraic objects with two "compatible" operations, namely rings and fields. It seems natural to then look at algebraic objects with three or more operations that are compatible, but we don't learn about them. I asked one of my professors why this is so and she posed this question in response: we have left-modules and right-modules, but there are no top-modules or bottom-modules, or any other ways of combining two elements to produce a third. I can't think of any satisfactory answer to either of these questions. Can anyone shed any light on them?

Edit: Now that I know these objects are studied, what I meant by "we" is essentially "Why are these objects not introduced to undergraduates (at least in a standard curriculum) given how natural they seem?"

While it is true that algebraic structures based on binary operations are very common, other structures exist, are being studied, and are very important. Examples include:

$A_\infty $-spaces, where homotopy considerations mandate not just one binary operation, but an $n$-ary operation for all $n\in \mathbb N$.
Malcev operations, are examples of structures based on ternary operations.
operads also have $n$-ary operations.

For many of the structures above things like modules and actions make sense, and themselves involve $n$-ary operations.

So, mathematicians certainly do study such structures. Perhaps the reason they are not commonly introduced at the undergraduate level is that these structures are more complicated than ones based on binary operations.

As for the comment made by your professor, I can think of many ways elements can be combined to give a new element, so I really don't know what is meant by that.

And, since you are wondering if such $n$-ary based algebraic structures are too complicated for undergraduates, I'll just mention that of the three I mention above, $A_\infty $ - space are quite complicated but Malcev operations and operads are not. Operads come in many flavours, and if one considers what are known as coloured planar non-enriched operads (probably the simplest kind of operad) then this is a structure that can be understood by a first year student (and this is actually quite an important class of operads, so not just a toy algebraic structure). The reason why these structures are not introduced early on has more to do with the fact that university curricula and textbooks change and adapt very slowly. They very rarely reflect current trends. In 100 years it is likely that operads will make it to first year or second year textbooks, much like groups do today.

And while on the subject, one must also consider algebraic structures with operations of infinite arity. These too exist and provide some surprising examples. For instance, it is classical result that the category of compact Hausdorff spaces is algebraic, and that means that that category can be thought of as consisting of algebraic structures with operations of arity $\infty $. Other important examples include complete lattices.

What do you mean by "we" in your statement " ...we learn...". If your "we" refers to post graduates, then I think your first statement isn't quite true as when I was undergraduate, I learned these objects you are talking about in all the three sense- groups, rings and fields. And regarding what your prof says, I think she meant you are welcome to start the study of algebraic objects in a new sense that is not known before. So, it is not that we don't study algebraic objects in another operation, it's just that no one has done it yet perhaps. Abstract Algebra is among the old fields in mathematics with relatively least publications.


A frog sitting on the second of three stones jumps to left with probability $0.4$, what is the probability that it will reach the zeroth stone first? A blind frog sits on the second of three stones in a line. At each minute, it jumps either to left (with probability $0.4$) or right. To the right of third stone there is a lake in which an alligator is waiting for the frog. To the left of the first stone there is a stork waiting for the frog. Find the probability that the stork eats the frog.

I tried to solve this by recurrence relations, but couldn't establish a proper definition of $a_n$. I mean $n$ can denote the stone number that the frogs is sitting, but then I have find $a_0$, and I don't know whether that kind of configurations are allowed.

In another case, I can choose the second stone as my $n=0$, and $n$ can denote, again, the stone numbers, but then I need to include the negative $n$s, which, again, I don't know whether it is eligible, so I'm stuck.

Hint. After two jumps frog can either be eaten by either alligator (with which probability?) or stork (with which probability?), or return to the second stone and make at least two more jumps with the same set of choices with the same probabilities.

The probability is $p^2\frac{1}{1-2p(1-p)}$ for $p=0.4$, then the probability is $\frac{4}{13}$.

I call $P_N({n})$ the probability of being at stage n in step N. With 0 the initial stage and 2 the stork stage. It's easy yo check that $P_{N-2}({0})=2p(1-p)P_{N-4}({0})$ and the solution to that recurrence relation gives that $P_N({2})=p^2(p(1-p))^{\frac{N-2}{2}}$ and summing up all of those gives my result.


every compact normal operator can be represented by product of unitary operator and compact positive operator Could anyone show that every compact normal operator can be represented by commutative product of unitary operator and compact positive operator? I mean if $T$ is a compact normal operator, then there exists a compact positive operator $A$ and unitary operator $B$ such that $T=AB=BA$. Are these operators uniquely determined?

The key here that every compact normal operator $T$ on any complex Hilbert space $\mathcal H$ is diagonalizable (see e.g. "Introduction to Hilbert Space" by Berberian, 1976) so there exists an orthonormal system $(e_n)_{n\in\mathbb N}$ in $\mathcal H$ and a complex null sequence $(\tau_n)_{n\in\mathbb N}$ such that $|\tau_n|\geq |\tau_{n+1}|$ for all $n\in\mathbb N$ and

$$
T=\sum_{n=1}^\infty \tau_n\langle e_n,\cdot\rangle e_n.
$$

Now $A=\sum_{n=1}^\infty |\tau_n|\langle e_n,\cdot\rangle e_n=|T|$ is the positive-semidefinite (obvious) compact operator we are looking for. The compactness of $A$ can be shown e.g. by showing $A$ is the limit of the norm-convergent sequence of the finite-rank operators $(\sum_{n=1}^k|\tau_n|\langle e_n,\cdot\rangle e_n)_{k\in\mathbb N}$ or by using the following argumentation:

$$
|T|\text{ compact }\Leftrightarrow |T|^\dagger |T|=|T|^2=T^\dagger T\text{ compact }\Leftrightarrow T\text{ compact}
$$

The unitary operator $B$ now somehow has to transform the $\tau_n$ into their absolute value. As unitary operators are strongly connected to orthonormal bases, we extend $(e_n)_{n\in \mathbb N}$ to an orthonormal basis $(e_n)_{n\in J}$ of $\mathcal H$ with $\mathbb N\subseteq J$. Define $I:=\lbrace n\in\mathbb N\,|\,\tau_n\neq 0\rbrace$ and the operator $B$ via

$$
Be_n:=\begin{cases} \frac{\tau_n}{|\tau_n|}e_n & \text{if }n\in I \\ e_n & \text{if }n\in J\,\backslash \,I \end{cases}
$$

and its linear and continous extension onto all of $\mathcal H$. Observe that

$$
\Big\langle \frac{\tau_j}{|\tau_j|}e_j,\frac{\tau_k}{|\tau_k|}e_k\Big\rangle=\delta_{jk}
$$

for all $j,k\in I$ so $B$ maps $(e_n)_{n\in J}$ to an orthonormal basis of $\mathcal H$ and thus is unitary. With this it is easy to see that $T=BA$ and $BAB^\dagger =A$ (where the latter obviously equals $BA=AB$).

For your last question, note that $A$ and $B$ are not unique as the orthonormal system $(e_n)_{n\in\mathbb N}$ belonging to $T$ is not uniquely determined. For this one may consider $(e^{i\varphi}e_n)_{n\in\mathbb N}$ for any $\varphi\in(0,2\pi)$, this still is an orthonormal system and satisfies the above decomposition of $T$. For something more concrete one may choose $T=A=0$ so $B$ can be literally any unitary operator so satisfies the required properties.


Solve multivariable limit $$\lim_{(x,y) \to (0,0)} \frac{x^3 + y^4}{x^2 + y^2}$$

I am almost sure it is equal to $0$ but I can't prove it.
Please give me some hint.

A more Rigerous solution one could show, which would be expected in Real Analysis.

If the multivariate limit F(x,y) -> 0 for x and y approaching 0 ,then it must be true that {Xn} -> 0 and {Yn} -> 0 so that F({Xn}, {Yn}) -> 0 for all such sequences {Xn}, {Yn} that approach 0.  Examples of single variable sequences that approach 0 are {1/n}, {1/n^2}, etc. as n approaches infinity.

In other words if there exist two sequences that both approach 0 as n goes to infinity such that F({Xn}, {Yn}) = 0, but then there exists other sequences (one is sufficient) such that F({Xn}, {Yn}) = L where L is not equal to 0.  Then F must be divergent, because it is impossible for a converging sequence to have more than one limit point.

Let us have {Xn} = {1/n} and {Yn} = {1/n}, this implies

F({Xn}, {Yn}) =  F({1/n}, {1/n}) = lim n -> infinity ( (n+1)/2n^2 ) -> 0

So it was shown for two sequences, it would get very tiring to show all sequences approach 0.  The trick would be to assume that for two sequences approaching 0, that F({Xn}, {Yn}) -> L not equal to 0, and then show a contradiction arises if you do so.  By contradiction assume  there are two sequences that both approach 0 such that {Xn} -> 0 and {Ym} = {0} -> 0 for n and m approaching infinity.  (Yes {Ym} is constant), and

F({Xn}, {Ym}) = L not equal to 0, which implies

lim n, m -> infinity ( ({Xn}^3 + 0) / ({Xn}^2 + 0) ) = lim n,m -> infinity ( {Xn} ) = L not equal to 0.  But we asserted that {Xn} -> 0, so a contradiction has arisen.  This must mean that F({Xn}, {Ym}) -> 0.

Q.E.D.


Expectation of product of compositions Suppose $(\Omega,\mathcal{A},P)$ be a probability space, and let $X,Y$ be two random variables on $\Omega$. Also, consider the (measurable) indicator function $\chi_{(-\infty,x]}$ given any $x\in \mathbb{R}$. I know that
$$\int_\Omega (\chi_{(-\infty,x]}\circ X)(w) P(dw)=\int_{\mathbb{R}}\chi_{(-\infty,x]}(w')P_X(dw')=P_X((-\infty,x])=P(\{X\leq x\})$$
where $P_X$ is the probability distribution of $X$.

Now I'm struggling to show that

$$E[(\chi_{(-\infty,x]}\circ X)(\chi_{(-\infty,y]}\circ Y)]=\int_\Omega [(\chi_{(-\infty,x]}\circ X)(\chi_{(-\infty,y]}\circ Y)](w)P(dw)=P(\{X\leq x,Y\leq y\})$$

It looks simple, but I cannot progress. Can you help me?

*I noticed the identity $(\chi_{(-\infty,x]}\circ X)(\chi_{(-\infty,y]}\circ Y)=\chi_{(-\infty,x]\times (-\infty,y]}(X,Y)$. So I could use

$$\int_\Omega \chi_{(-\infty,x]\times (-\infty,y]}(X(w),Y(w))P(dw)=\int_{\mathbb{R}^2} \chi_{(-\infty,x]\times (-\infty,y]}(w_1,w_2)P_{XY}(dw_1\otimes d2_2)=P(\{X\leq x,Y\leq y\})$$

What do you think?
I'm not sure if the transformation theorem works for the bivariate case as I did...

Let $(\Omega, \mathcal F, \mathbb P)$ be probability space, let $X: \Omega \to \mathbb R^n$ be random variable. (We consider $\mathbb R^n$ with borel $\mathcal B(\mathbb R^n)$ sigma field.) Moreover, let $f:\mathbb R^n \to \mathbb R$ be arbitrary function and $\mu_X$ the distribution of $X$. Then we have:

$$\mathbb E[f\circ X] = \int_\Omega (f \circ X)(\omega)d\mathbb P(\omega) = \int_{\mathbb R^n} f(s)d\mu_X(s)$$

Proof:

1) case: Let $f = \chi_A$, where $A \in \mathcal B(\mathbb R^n)$.

Then $$\mathbb E[ f \circ X] = 1 \cdot \mathbb P(X \in A) + 0 \cdot \mathbb P(X \notin A) = \mu_X(A) = \int_{\mathbb R^n}f(s)d\mu_X(s)$$

We only need that one above in your problem, but let's just at least point out, how to prove the rest:

2) case: $f$ is a simple function, that is there are $A_1,...,A_m \in \mathcal B(\mathbb R^n)$ and $a_1,...,a_m \in \mathbb R$ such that $f = \sum_{i=1}^m a_i\chi_{A_i}$. Then we can use 1) case + linearity of integrals and we're done.

3) case: $f$ is any non-negative borel function. Then there exists an inscreasing sequence of non-negative simple functions $f_n$ converging to $f$. Use Lebesgue-Levy monotone convergence theorem and you're done.

4) case: any borel $f$. Then $f = f^+ - f^-$ and use your result from case 3) to $f^+$ and $f^-$ (note that at least one $\mathbb E[f^+ \circ X], \mathbb E[f^- \circ X]$ is finite).

Now in your problem, as our random variable $X$ take vector $(X,Y): \Omega \to \mathbb R^2$ and as $f$ take $\chi_A$, where $A = (-\infty, x] \times (-\infty,y]$.

Use case 1) here: $\mathbb E[ f \circ (X,Y) ] = \mu_{(X,Y)}(A) = \mathbb P( (X,Y) \in A) = \mathbb P(X \le x, Y \le y)$

The integrand is $1$ on $\{X\leq x, Y\leq y\}$ (which is measurable since $X,Y$ are random variables) and zero elsewhere, so the result immediately follows by the definition of the integral of a simple function.


Prove that in each year, the 13th day of some month occurs on a Friday Prove that in each year, the 13th day of some month occurs on a  Friday.

No clue... please help!

Let us ignore leap years for a moment.

And let us assume that some date (e.g., 13th) falls on Monday in January.

What happens in February? January has $31=4\cdot 7+3$ days, so the days move by 3 to Thursday.

So if we denote the days in week by numbers $\{0,1,2,3,4,5,6\}$, we only have to consider the number of days in a month, compute modulo 7 and see what happens:

0 January (31 days)
3 February (28 days)
3 March (31 days)
6 April (30 days)
1 May (31 days)
4 June (30 days)
6 July (31 days)
2 August (31 days)
5 September (30 days)
0 October (31 days)
3 November (30 days)
5 December (31 days)
1 January next year

(Computing day in January in the next year is irrelevant for this question; but comparing this with an actual calendar is a good sanity check.)

Notice, that all numbers 0,1,2,3,4,5,6 appear in the above table.

Now what happens if a year starts not on Monday (=0) but on Tuesday. You simply have to add $+1$ (computing modulo 7) in each row of the table. But since every number appeared at least one, we will get $6+1=0$ in some row. And the same is true for any other day.

So now you only have to create similar table for a leap year, check whether all numbers from 0 to 6 appear there, and you are done.

Somewhat related: Occurrence of Friday the 13th on Wikipedia.

I think this question has already good answers,

It is enough to show that there exist a month  starting on Sunday.

Let $x$ be the starting day of the year then next month's first day is $x+31\equiv x+3$ and next month's first day, $x+3+28\equiv x+3$...

At the and we have, $$x,x+3,x+3,x+6,x+1,x+4,x+6,x+2,x+5..$$ now we reached seven different days, $x,x+1...,x+6$ by pigeonhole principle we must have all days which conclude the result.


Can two solutions of a PDE with different initial conditions coincide after finite time? Consider an ODE
$$
\dot x=f(x),\quad x(0)=x_0,
$$
where $f$ is a smooth function. It is well-known that if $y$ is another solution to this ODE with different initial condition $y_0\neq x_0$, then the trajectories of $x(t)$ and $y(t)$ do not intersect - in particular, for any finite $t$ we have $x(t)\neq y(t)$.

I wonder if the same is true for PDEs. I'm interested in the following simple example:
$$
\partial_t u=\partial_{xx} u+f(t,(u(t,x)),
$$
with periodic boundary conditions. Here $x$ lives on a circle $[0,1]$, $t\ge0$ and $f$ is a nice function. Is it true that if $u,v$ are two solutions of this PDE with $u(0)\neq v(0)$, then $u(t)\neq v(t)$ for any $t$? (equivalent statement: if $u(1)=v(1)$ is it true that $u(0)=v(0)$?) How can we prove this?

UPD: the case $f\equiv0$ can be found inn Evans' book and it involves taking second derivative of the energy. Is it possible to extend this technique to the case $f\neq0$?

Tychonoff gives an infinite family of solutions of the heat equation on $(x,t) \in \Bbb{R} \times \Bbb{R}$, all of which are $0$ everywhere at time $t \leq 0$ and different subsequently.  These solutions are also discussed on MSE.

Such solutions are backwards from what you request -- they are nonunique at a prior time, and then subsequently become unique. However, by the choice $f(t, u(t,x)) = 2\partial_t u(t,x)$, time reversal is implemented.  Tychonoff's solutions are then distinct on $t <0$ and identically zero on $t \geq 0$.  It would be no great feat to adapt these to $(x,t) \in S^1 \times \Bbb{R}$.


Find function $f: U \subset \mathbb{R}^2 \to \mathbb{R}$ with $||Df(x,y)|| \leq 1$ such that $P,Q \in U$ exist with $|f(P) - f(Q)| > || P-Q||$ I am looking for a continuously-differentiable function $f: U \to \mathbb{R}, U \subset \mathbb{R}^2$ which satisfies the following requirements:

$U$ is open set
$U$ is connected (right word?), i.e. $\forall P,Q \in U\ \ \exists$ continuous $\gamma: [0,1] \to U$ with $\gamma(0)=P$ and $\gamma(1)=Q$
$||Df(x)|| \leq 1$ for all $x \in U$ ($Df(x)$ is Jacobi Matrix)
There are points $P,Q \in U$ with $|f(P)-f(Q)| > ||P-Q||$

Essentially I am looking for a counter-example for the mean value theorem if U is not convex.

I had been thinking about trying to construct an (open) spiral like this: http://www.mathematische-basteleien.de/spiral22.gif
, but can't really get it work.

Any ideas / hints / other examples?

Thanks a lot in advance!

Consider the domain $U \subset \mathbb{R}^2$ obtained from the annulus $1 <r <2$ by removing the real half axis $[0, \infty)$. Consider the function $f(x,y) = \text{arg}( x + i y)= \phi$ in polar coordinates. $f$ maps $U$ onto $(0,2 \pi)$. The gradient of $f$ is $\nabla \arctan(y/x) = (- \frac{y}{x^2 + y^2}, \frac{x}{x^2 + y^2})$ so its norm is $<1$. Check to see that points very close by in the distance from $\mathbb{R}^2$ have values of $f$ differing by about $2 \pi$.

The point is that the distance from $\mathbb{R}^2$ is in general not the intrinsic distance ( the infimum of lengths of paths ) if $U$ is not convex. However, for that intrinsic distance the inequality will still hold, in fact you can check on this example.

Answer was provided before edit required continuity.

F(x,y)=0 elsewhere

F(y,y^2)=1

Now consider the origin. All gateaux derivatives exist but the function isn't even continuous.


Transitive subgroup of symmetric group I'm working on the following question, and honestly have no idea how to begin.  Any hints would be greatly appreciated!

Let $H$ be a subgroup of $S_n$, the symmetry group of the set $\{1,2,\dots, n\}$. Show that if $H$ is transitive and if $H$ is generated by some set of transpositions, then $H=S_n$.

This might not be at the appropriate level for you, but I think this is a cute argument.

Suppose we wanted to transpose $1$ and $n$ in the list $1,\cdots,n$ but we were only allowed to swap adjacent numbers each step, how would we do it? Well, we could use adjacent transpositions to move $1$ all the way to the right, which would shift the numbers $2,\cdots,n$ each one left. Then we could move $n$ to the left with adjacent transpositions until we get the list $n,2,\cdots,n-1,1$.

In other words, we have

$$ (12)(23)\cdots(n-2\,n-1)\cdot(n\,n-1)\cdots(32)(21)=(n1). \tag{$\ast$} $$

(Remember, the rightmost permutations are applied to an input first. Interpret $(u\,v)$ as "swap the numbers that are $u$th and $v$th in the list.") We can use this...

Form a graph as follows: given a generating set $S$ of $H$ consisting of transpositions, let $\{1,\cdots,n\}$ be the vertex set and form an edge $(i,j)$ for every transposition $(ij)\in S$. The hypotheses that $H$ acts transitively and is generated by $S$ is equivalent to saying this graph is connected, so between any two values $a,b\in\{1,\cdots,n\}$ there is some sequence of transpositions

$$ (a_1\,a_2),(a_2\,a_3),\cdots,(a_{k-1}\,a_k)\in S $$

where $a_1=a$ and $a_k=b$. Then we can reuse the $(\ast)$ trick:

$$ (a_1\,a_2)(a_2\,a_3)\cdots(a_{k-2}\,a_{k-1})\cdot(a_k\,a_{k-1})\cdots(a_3\,a_2)(a_2\,a_1)=(a_k\,a_1)=(ab)\in H.$$

That is, $H$ contains every transposition $(ab)$, so it must be $S_n$.

So I am convinced that I am miss understanding the question and its probably due to the fact that I am miss understanding the definition of $H$ being a transitive subgroup of $S_n$

Since $H$ is a transitive subgroup of $S_n$ I believe that means for any $\sigma, \tau \in S_n$ there exists an $h \in H$ such that $\sigma h = \tau$.

If this definition is correct then we can consider an arbitrary $\sigma \in S_n$. Let $h_1 \in H$ since $H$ is a subgroup of $S_n$ we have $h_1 \in S_n$ and so since $H$ is a transitive subgroup of $S_n$ we get that there is an $h_2 \in H$ such that $h_2h_1 = \sigma$ but since $H$ is a group and hence closed under multiplication we have $\sigma = h_2h_1 \in H$ and since $\sigma \in S_n$ was arbitrary $S_n = H$

I am definitly not understanding something about this question because otherwise this problem was to easy and the fact that  $H$ is generated by some set of transpositions was completely irrelevant information.


Is there any way to find out $x! = n$ by hand? I've given myself this question but I can't see how to get it. I'm actually not taking calculus I'm in high school taking Algebra 2 but I've learned a bit or two from this site.

I did:

$x! = 10$

Trying to solve for $x$ but I had trouble doing so. Can anyone give me a hand here?

Wolfram alpha gives the roots of $\Gamma(x+1)=10$ as
$-4.995806,
-4.016334,
-2.947296,
-2.097191,
-1.095325,
3.390078.$

One root of $\Gamma(y)=x$ is approximated by $\frac{L(x)}{W(\frac{L(x)}e)}+\frac{1}2$ where $L(x)=\ln(\frac{x+c}{\sqrt{2\pi}})$, $c\approx0.036534$ and $W(x)$ is the principal branch of the inverse of $xe^x$. $W(x)$ can be approximated by selecting an intial $w_0$, and find succesive approximations $w_{j+1}=w_j-\frac{w_je^{w_j}-z}{e^{w_j}+w_je^{w_j}}$. $e^x$ and $\ln {1+x}$ can be found using $1+x+ \frac{x^2}/{2!}+\frac{x^3}{3!}...$ and $x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+...$ respectively. This can be done by hand but is tedious.
http://mathforum.org/kb/message.jspa?messageID=342551&tstart=0

EDIT: The expansion for $\ln(x+1)$ only works if $|x|<1$. Otherwise use $\ln x\approx \frac{\pi}{2M(1,\frac{4}s}-m\ln 2$ with $M(1,\frac{4}s)=$ thearithmetic geometric mean of $1$ and $\frac{4}s$ ,$s=x2^m$ and $m$ any chosen integer (larger $m$ give a closer approximation).

If you allow for approximate solutions, in other words, you interpret "=" as meaning "approximately equal to", then you might consider 3 as a solution, since 3!=6, and 4!=24.


Modular exponentiation by hand ($a^b\bmod c$) How do I efficiently compute $a^b\bmod c$:

When $b$ is huge, for instance $5^{844325}\bmod 21$?
When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, for instance $5^{69}\bmod 101$?
When $(a,c)\ne1$, for instance $6^{103}\bmod 14$?

Are there any other tricks for evaluating exponents in modular arithmetic?

This is being asked in an effort to cut down on duplicates, see here and here.

Wikipage on modular arithmetic is not bad.

When $b$ is huge, and $a$ and $c$ are coprime, Euler's theorem applies:
$$
    a^b \equiv a^{b \, \bmod  \, \phi(c)} \, \bmod c
$$
For the example at hand, $\phi(21) = \phi(3) \times \phi(7) = 2 \times 6 = 12$. $ 844325 \bmod 12 = 5$, so $5^5 = 5 \times 25^2 \equiv 5 \times 4^2 = 80 \equiv 17 \mod 21$.
When $a$ and $c$ are coprime, but $0<b<\phi(c)$, repeated squaring (or using other compositions of powers) is the fastest way to go (manually):
$$
\begin{eqnarray}
   5^4 \equiv 5 \times 5^3 \equiv 5 \times 24 \equiv 19 &\pmod{101}\\
   19^4 \equiv (19^2)^2 \equiv 58^2 \equiv (-43)^2 \equiv 1849 \equiv 31 &\pmod{101} \\
   31^4 \equiv (31^2)^2 \equiv (961)^2 \equiv 52^2 \equiv 2704 \equiv 78 &\pmod{101} \\
   5^{69} \equiv 5 \times 5^4 \times ((5^4)^4)^4 \equiv 5 \times 19 \times 78 \equiv 5 \times 19 \times (-23)\\
 \equiv 19 \times (-14) \equiv -266 \equiv  37 & \pmod{101}
\end{eqnarray}
$$
When $a$ and $c$ are not coprime, let $g = \gcd(a,c)$. Let $a = g \times d$ and $c = g \times f$, then, assuming $b > 1$:
$$
   a^b \bmod c = g^b \times d^b \bmod (g \times f) = ( g \times (g^{b-1} d^b \bmod f) ) \bmod c
$$
In the example given, $\gcd(6,14) = 2$. So $2^{102} \times 3^{103} \mod 7$, using Euler'r theorem, with $\phi(7) = 6$,  and $102 \equiv 0 \mod 6$, $2^{102} \times 3^{103} \equiv 3 \mod 7$, so $6^{103} \equiv (2 \times 3) \equiv 6 \mod 14 $.

Its not hard to show that the sequence

$$
x_n=a^n\mod{c}
$$

is periodic, with period $p$ (which is at most $c$). Evaluate the first few terms to get the period $\{x_0,x_1,\dots,x_{p-1}\}$. Then you can evaluate for any huge power $n$ as

$$
x_n=x_{n\mod{p}}
$$


Solve for $x$: $\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)$ $$\arctan 2x +\arctan 3x = \left(\frac{\pi}{4}\right)$$
$$\arctan \left(\frac{2x+3x}{1-2x*3x}\right)=\frac {\pi}{4}$$
$$\frac {5x}{1-6x^2}=\tan \frac{\pi}{4}=1$$
$$6x^2 + 5x -1 = 0$$
$$(6x-1)(x+1)=0$$
$$x=-1, \frac{1}{6}$$

The answer however rejects the solution $x=-1$ saying that it makes the L.H.S of the equation negative. I don't understand this, I don't see how $x=-1$ makes the L.H.S. negative.

Concep: Let $f$ be a function, which means  $$x_1=x_2\Rightarrow f(x_1)=f(x_2)$$
But note that if $f(x_1)=f(x_2)$, then $x_1=x_2$ is one of the possibility, among perhaps many possibilities if the function is not injective

Now Given $\tan^{-1}(2x)+\tan^{-1}(3x)=\frac{\pi}{4}$, taking $\tan$ of both sides(since $\tan$ is a function),we get
 $$\frac{2x+3x}{1-2x.3x}=1$$

which after simplification gives $x=\frac16,-1$. Taking $x=\frac16$, we want to find what is $\tan^{-1}\frac13+\tan^{-1}\frac12$. For that let us take $\tan^{-1}\frac13=\alpha,\tan^{-1}\frac12=\beta$, which is equal to $\tan\alpha=\frac13,and \tan\beta=\frac12$, where $\alpha,\beta$ being principal values belongs to $(-\frac{\pi}2,\frac{\pi}2)$, which means $\alpha+\beta\in(-\pi,\pi)$. Now we know $$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$$ which implies $$\tan(\alpha+\beta)=\frac{\frac13+\frac12}{1-\frac13\frac12}=1$$ Hence $$\alpha+\beta=\tan^{-1}1=\frac{\pi}4$$ or $$\tan^{-1}\frac13+tan^{-1}\frac12=\frac{\pi}4$$ and $x=\frac16$ becomes a root of the above equation

Now take $x=-1$, then the lhs of the equation becomes $$\tan^{-1}(-2)+\tan^{-1}(-3)$$ which is a negative angle $-\frac{3\pi}{4}$ , which can not be a positive number $\frac{\pi}{4}$.


Evaluating $\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx$ I want to evaluate $\displaystyle \int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx$ but it's quite difficult.
I have tried to rewrite the integral as
$$\int _0^{\frac{\pi }{2}}x\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx=\frac{\pi }{2}\int _0^{\frac{\pi }{2}}\tan \left(x\right)\ln ^2\left(\sin \left(x\right)\right)\:dx-\int _0^{\frac{\pi }{2}}x\tan \left(x\right)\ln ^2\left(\sin \left(x\right)\right)\:dx$$
I've also tried to integrate by parts in multiple ways yet I cant go forth with this integral, I also tried using the substitution $t=\tan{\frac{x}{2}}$ but cant get anything to work, I'll appreciate any sort of help.
I also tried using the classical expansion
$$\ln \left(\cos \left(x\right)\right)=-\ln \left(2\right)-\sum _{n=1}^{\infty }\frac{\left(-1\right)^n\cos \left(2nx\right)}{n},\:-\frac{\pi }{2}<x<\frac{\pi }{2}$$
But it only gets worse.

$$I=\frac{1}{2}\int_0^{\pi/2} x^2\frac{\ln^2\cos x}{\sin^2x} \, dx+\int_0^{\pi/2} x^2{\ln\cos x}\,dx$$ integrating by parts
$$\int_0^{\pi/2} x^2 \ln\cos x \, dx=\frac{\pi^3}{24}\ln2-\frac{\pi}{4}\zeta(3)$$ see Integral $\int_0^\pi \theta^2 \ln^2\big(2\cos\frac{\theta}{2}\big)d \theta$.
$$I=\int_0^{\pi/2} x^2\frac{\ln^2\cos x}{\sin^2x} \, dx = \frac{1}{4} \int_0^\infty\frac{(\arctan u)^2 \log^2(1+u^2)}{u^2} \, du$$  Put $$x=\arctan u$$
Closer form for $\int_0^\infty\frac{(\arctan{x})^2\log^2({1+x^2})}{x^2}dx$

I will be using the “make everything into a sum and then integrate” method. This means that we will call it I and manipulate the integral until it can be integrable with a gamma function:
$$I=\int_0^{π/2} x\cot(x)\ln^2\cos(x)\,dx=\int_0^{π/2}x\bigg(\sum_{n=0}^∞ \bigg[c_n=\frac{(-1)^n2^{2n}B_{2n}}{(2n)!}\bigg] x^{2n-1}\bigg)\bigg(-\sum_{k=1}^∞ \frac{(-1)^k\cos^k(x)}{k}\bigg)\bigg(-\sum_{m=1}^∞\frac{(-1)^m\cos^m(x)}{m}\bigg)\,dx$$ where $B_y$ are the Bernoulli Numbers.
Because all of the indices are constant from each other, they can be treated as such and moved around without needing any cauchy products are needed, using the complex definition of the cosine function and using the binomial theorem once more gets us our simplified integral to be integrated in the next step:
$$\int_0^{π/2}x\bigg(\sum_{n=0}^∞c_n=x^{2n-1}\bigg)\bigg(-\sum_{k=1}^∞ \frac{(-1)^k \cos^k(x)}{k}\bigg)\bigg(-\sum_{m=1}^∞\frac{(-1)^m\cos^m(x)}{m}\bigg)\,dx = \sum_{n=0}^∞ c_n\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{(-1)^{m+k}}{mk} \int_0^{π/2}x^{2n} \frac{(e^{\mathit ix}+ e^{-\mathit ix})^{m+k}}{2^{m+k}} \, dx = \sum_{n=0}^∞ c_n\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{(-1)^{m+k}}{mk} \int_0^{π/2}x^{2n}2^{-(m+k)}\sum_{j=0}^{m+k} \binom{m+k}{j}e^{-ijx} \, dx= \sum_{n=0}^∞ c_n\sum_{k=1}^∞ \sum_{m=1}^ ∞\frac{(-1)^{m+k}}{2^{m+k}mk}\sum_{j=0}^{m+k} \binom{m+k}{j} \int_0^{π/2} x^{2n} e^{-ijx}dx$$
By the way, here is proof this equals @pisco’s answer. From here on out, desmos cannot handle complex numbers. This means that if the following steps are correct, them we have a summation form of I because there are no restrictions on the binomial theorem that would give us conflicts and because this method is similar to the generalized Sophomore’s Dream integral with a sum and then a general powers of $x\ln(x)$ integration where here there is a general power of $x\cos(x)$ except with two more complicated powers.
Finally, we can integrate by using the generalized incomplete gamma function as the integral looks like the main definition of the gamma function and put it into a few forms including the summation form of the gamma function. As I am not completely as versed in substitution for finding the integral in terms of this function, a bit of machine help was used:
$$\sum_{n=0}^∞ c_n\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{(-1)^{m+k}}{2^{m+k}mk} \sum_{j=0}^{m+k} \binom{m+k}{j} \int_0^{π/2} x^{2n} e^{-ijx}\,dx= \sum_{n=0}^∞ c_n\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{(-1)^{m+k}}{2^{m+k}mk}\sum_{j=0}^{m+k} \binom{m+k}{j}(j^{-1}\mathit i・\mathit i^{-2n}j^{-2n}Γ(2n+1,ijx)|_0^{π/2})= \sum_{n=0}^∞ \frac{(-1)^n2^{2n}B_{2n}}{(2n)!}\sum_{k=1}^ ∞\sum_{m=1}^∞ \frac{(-1)^{m+k}}{2^{m+k}mk}\sum_{j=0}^{m+k} \binom{m+k}{j}\frac{(-1)^n\mathit iΓ(2n+1,\frac{π\mathit ij}{2},0)}{j^{2n+1}}= \sum_{n=0}^∞ \frac{B_{2n}}{(2n)!} \sum_{k=1}^ ∞\sum_{m=1}^∞ \frac{(-1)^{m+k}}{2^{k+m-2n}mk}\sum_{j=0}^{m+k} \binom{m+k}{j}\frac{\mathit iΓ(2n+1,\frac{π\mathit ij}{2},0)}{j^{2n+1}}= \sum_{n=0}^∞ \frac{2^{2n}B_{2n}}{(2n)!}\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{1}{(-2)^{m+k}mk} \sum_{j=0}^{m+k} \binom{m+k}{j}\frac{\mathit i(Γ(2n+1,\frac{π\mathit ij}{2})-(2n))!}{j^{2n+1}}= \sum_{n=1}^ ∞\frac{B_{2(n-1)}}{(2(n-1))!}\sum_{k=1}^∞ \sum_{m=1}^∞ \frac{(-1)^{m+k}}{mk}\sum_{j=0}^{m+k} \binom{m+k}{j}\frac{(-\mathit i)^{j-1}}{j^{2n-1}}\sum_{p=0}^{2(n-1)} \frac{(π\mathit i)^p}{2^{k+m+2(1-n)+p}p!}$$
I will try to evaluate this expression if I can. I apologize for the unwanted solution, but this took some effort to put together.:
$$I=\sum_{n=0}^∞ \sum_{k=1}^∞ \sum_{m=1}^∞ \sum_{j=0}^{m+k} \binom{m+k}{j} \frac{(-1)^{m+k} B_{2n} \mathit iΓ(2n+1,\frac{π\mathit ij}{2},0)}{2^{k+m-2n} j^{2n+1} mk (2n)!}=\sum_{n=0}^∞ \sum_{k=1}^∞ \sum_{m=1}^∞ \sum_{j=0}^{m+k} \binom{m+k}{j}\frac{(-1)^nc_n \mathit iΓ(2n+1,\frac{π\mathit ij}{2})-(2n)!}{(-2)^{k+m} j^{2n+1} mk}= \sum_{n=0}^∞ \sum_{k=1}^∞ \sum_{m=1}^∞ \sum_{j=0}^{m+k} \binom{m+k}{j}\frac{(-1)^nc_n \mathit iΓ(2n+1,\frac{π\mathit ij}{2})}{(-2)^{k+m} j^{2n+1} mk}-\sum_{n=0}^∞ \sum_{k=1}^∞ \sum_{m=1}^∞ \sum_{j=0}^{m+k} \binom{m+k}{j}\frac{2^{2n} B_{2n}\mathit i}{(-2)^{k+m} j^{2n+1} mk} $$
Let there be a focus on the second term:
$$S_1-I=S_2= \sum_{n=0}^∞ \sum_{k=1}^∞ \sum_{m=1}^∞ \sum_{j=0}^{m+k} \binom{m+k}{j} \frac{2^{2n} B_{2n}\mathit i}{(-2)^{k+m} j^{2n+1} mk} =  \mathit i\sum_{n=0}^∞ 2^{2n} B_{2n} \sum_{k=1}^∞ \frac{(-1)^{k}}{2^kk} \sum_{m=1}^∞ \frac{(-1)^{m}}{2^mm} \sum_{j=0}^{m+k} \binom{m+k}{j} j^{-(2n+1)}$$
I will fix this soon enough ...
As always, there is not much of a way to check the post integration answer, so please point out any errors here and give me feedback. As a side note, the index switch was needed to expand out the generalized incomplete gamma function with the 0 in the argument and exponential for this expansion for n>0 was already simplified.


Large deviation in relation with Wishard matrix I try to prove the following fact.

Let $A$ be an $m\times n$ matrix with iid standard normal random variable.
Then $B:=A^{t}A$ is a $Wishart$ matrix with $m$ degrees of freedom and covariane $Id_{m}$.

Then $B_{11}$ is distributed as an inverse chi-squared random variable with $m-n+1 $ degrees of freedom.

The author of a paper I read claims that one can verify easily by linear algebra the $1/B_{11}$ is equal to the squared distance between the first column of $A$ and the linear space spanned by the others. If this is true then it is clear that $B_{11}$ has the given distribution. However how to prove his claim?

I played with the matrix expression for orthogonal projections, but formulas get so complicated, there must e some easier way.

We first show that the squared distance of a vector $v\in\mathbb{R}^{m}$ to a subspace $L\subseteq \mathbb{R}^{m}$ spanned by the linearly independent vectors $v_{1},...,v_{n}$ is given by
$$d^2(v,L)=\frac{G(v_1,\ldots,v_n,v)}{G(v_1,\ldots,v_n),}$$
where
$$G(x_1,\ldots,x_k)=\det
\begin{pmatrix}
\langle x_1,x_1 \rangle & \ldots & \langle x_1,x_k \rangle\\
\ldots & \ldots & \ldots\\
\langle x_k,x_1 \rangle & \ldots & \langle x_k,x_k \rangle
\end{pmatrix}$$

is a Gram determinant. Note that $(x_{1},...,x_{k})$ is a linear dependent family if and only if $G(x_{1},...,x_{k})=0$. Indeed, if $(x_{1},...,x_{k})$ is a linear dependent family, then there is $(\lambda_{1},...,\lambda_{k})\neq 0$ such that $\sum_{i=1}^{k}\lambda_{i}x_{i}=0$. Then $\sum_{i=1}^{k}\lambda_{i}L_{i}=0,$ where $L_{1},...,L_{k}$ are the lines of Gram matrix of $x_{1},...,x_{k}$. Conversely if $G(x_{1},...,x_{k})=0$, then there is $(\lambda_{1},...,\lambda_{k})\neq 0$ such that $\sum_{i=1}^{k}\lambda_{i}L_{i}=0,$ where $L_{1},...,L_{k}$ are the lines of the Gram matrix of $x_{1},...,x_{k}$ as previously. Then we can see that $\sum_{i=1}^{k}\lambda_{i}x_{i}=0$ is orthogonal to all $x_{i}$, and is thus 0. Linear dependence follows. \

Now let $v=u+k$, where $u\in L$ and $k\in L^{\perp}$. By using the multi-linearity of the determinant, we find that

\begin{align*}
G(v_1,\ldots,v_n,v)&= G(v_1,\ldots,v_n,u)+G(v_1,\ldots,v_n,k)\\
                   &  =G(v_1,\ldots,v_n,u)+        \det \begin{pmatrix}
\langle v_1,v_1 \rangle  & \ldots & \langle v_1,v_n \rangle & 0\\
\ldots                   & \ldots & \ldots                  & \ldots\\
\langle v_n,v_1 \rangle  & \ldots & \langle v_n,v_n \rangle &0\\
*                        & *      & *                       & ||k||^{2}
\end{pmatrix} \\
                  & =G(v_1,\ldots,v_n,u)+ G(v_1,\ldots,v_n)||k||^{2}
\end{align*}
Now $G(v_1,\ldots,v_n,u)=0$, because $(v_1,\ldots,v_n,u)$ is a linear dependent family. So we get that
$$||u||^{2}=d^2(v,L)=\frac{G(v_1,\ldots,v_n,v)}{G(v_1,\ldots,v_n)}.$$\

Let $\Phi$ be an $m\times n$ matrix with independent and identically distributed standard normal random variables. Denote the columns of $\Phi$ by $C_{1}$ to $C_{n}$. Since $m\geq n$ it is clear that the family $(C_{1},...,C_{n})$ is linearly independent.  Then it is easy to see that $(\Phi^{*}\Phi)_{ij}=\left<C_{i},C_{j}\right>$. Let us denote the elements of the inverse Wishart matrix $(\Phi^{*}\Phi)^{-1}=:B$ by $B_{ij} \,(i,j=1,...,n)$. Using the basic formulas for computing the inverse of matrices we find that
\begin{align*}
B_{11}&=\frac{\det
\begin{pmatrix}
\langle C_2,C_2 \rangle & \ldots & \langle C_2,C_n \rangle\\
\ldots & \ldots & \ldots\\
\langle C_p,C_2 \rangle & \ldots & \langle C_p,C_n \rangle
\end{pmatrix}}{\det
\begin{pmatrix}
\langle C_1,C_1 \rangle & \ldots & \langle C_1,C_n \rangle\\
\ldots & \ldots & \ldots\\
\langle C_n,C_1 \rangle & \ldots & \langle C_n,C_n \rangle
\end{pmatrix}}\\
 &= \frac{G(C_{2},...,C_{n})}{G(C_{1},...,C_{n})}
\end{align*}
By what we have seen previously we see that
$$B_{11}=\frac{1}{d(C_{1},L)^2},$$
where $L$ is the vector space spanned by $C_{2},...,C_{n}$.
Now $$d(C_{1},L)^2=||\Pi_{L^{\perp}}(C_{1})||^2,$$ where $\Pi_{L^{\perp}}(C_{1})$ denotes the orthogonal projection of $C_{1}$ to $L^{\perp}$, a subspace of $\mathbb{R}^{n}$ of dimension $m-n-1$ (because $L$ is of dimension $n-1$). By Cochran's theorem  we have that
$$||\Pi_{L^{\perp}}(C_{1})||^{2} \sim \chi^{2}(m-n+1).$$

You need to apply the standard inner product on $\mathbb{R}^n$.

Edit: Do you know about $SO(3,1,\mathbb{F}_p )$ or $SO(3,1,\mathbb{C} )$?

Actul]ally the best way to solve this is via modular forms


Finding value of a quadratic The polynomial
\begin{equation*}
p(x)= ax^2+bx+c
\end{equation*}
has $1+\sqrt{3}$ as one of it's roots and also $p(2)=-2$. Is there any way to know the value of $a$, $b$ and $c$? I tried but I can form only $2$ equations how can I figure out value of $3$ unknown with $2$ equations so something must be missing.

Also $a$, $b$ and $c$ are rationals.

Hint: If one of the roots is $1 + \sqrt{3}$ and $a, b, c$ are rational then what must the other root be?

Solution 1:

The other root must then be $1 - \sqrt{3}$.

You have two options now: You can make a system of three linear equations in three variables and simply solve.

Or, you can simply expand $\alpha[x - (1 - \sqrt{3})][x - (1 + \sqrt{3})] = \alpha(x^2 - 2x - 2)$ and use $p(2) = -2$ to get $\alpha = 1$.

Solution 2:

We only need two equations if we know that $a, b, c$ are rational.

$$(1+\sqrt{3})^2 a + (1+\sqrt{3})b + c = 0$$

$$4a + 2\sqrt{3} a + b + \sqrt{3} b + c = 0$$

$a, b, c$ are rational, so we know that $-2a = b$ because the $\sqrt{3}$ terms must cancel each other.
So, $4a + 2b + c = -2$ gives us $c = -2$.

Substituting back into the above we get $4a + b = 2$, so $a = 1$ and $b = -2$.

Thus the quadratic is $x^2 - 2x - 2$.

Let $x = 1+\sqrt{3} \Rightarrow (x-1)^2 = 3 \Rightarrow x^2-2x-2 = 0 \Rightarrow a = k, b = -2k,c = -2k$, and $p(2) = -2 \Rightarrow 4a+2b+c=-2\Rightarrow 4k - 4k - 2k = -2 \Rightarrow k = 1 \Rightarrow (a,b,c) = (1,-2,-2)$.


Rank of a partitioned matrix Let $p$, $q$ and $n$ be positive integers such that $p+q \geq n$ and $p \leq n$. Denote by $\mathbf{I}$ the identity matrix of size $p$, by $\mathbf{0}$ the zero matrix of size $(p\times (n-p))$ and by $\mathbf{A}=(a_{ij})_{i=1,\ldots,q,j=1,\ldots,n}$ a matrix over $\mathbb{R}$.

Can one formulate simple conditions on $\mathbf{A}$, which guarantee that the partitioned matrix

\begin{equation}
\mathbf{C}:=
\left(
\begin{array}{cc}
\mathbf{I} & \mathbf{0}\\
\mathbf{A}
\end{array}
\right)
\end{equation}

has full rank (i.e. $\text{rank}(\mathbf{C}) =n$)?

CLARIFICATION: Note that $\mathbf{A}$ has $n$ columns, so the matrix $\mathbf{C}$ does not necessarily have any columns that are all zero.

Write, less confusingly, $A=(A_1~A_2)$, where $A_1$ is a $q\times p$ matrix and $A_2$ a $q\times(n-p)$ matrix, so that in block form
$$
  C=\begin{pmatrix}I&0\\A_1&A_2\end{pmatrix}.
$$
Then row operations will nullify $A_1$, so $\def\rk{\operatorname{rk}}\rk C=\rk I+\rk A_2=p+\rk A_2$, and this is maximal (i.e., equal to the number $n$ of columns of $C$) if and only if $\rk A_2$ is maximal (i.e., equal to the number $n-p$ of columns of $A_2$), in other words if and only if the columns of $A_2$ are linearly independent.

By the fact that these are equivalent conditions;

1)  Rank$(C)=n$

2)  The $n$ column vectors of C are linearly independent,

you can not achieve your goal.


How can I compute a presentation of the tangent bundle for a smooth manifold defined by a family of polynomials? Consider a smooth manifold $M$ given by a system of smooth functions
$$
\begin{align*}
f_1 = 0 \\
\cdots \\
f_k = 0
\end{align*}
$$
in $n$ variables. This has the algebraic description as the $\mathbb{R}$-algebra
$$
C^\infty(M) = \frac{C^\infty(\mathbb{R}^n)}{(f_1,\ldots,f_k)}
$$
How can I compute the tangent and cotangent bundle in terms of a similar presentation of algebras?

Here is a running example from the answers:

Consider the circle
$$
C^\infty(S^1) = \frac{C^\infty(\mathbb{R}^2)}{x^2 + y^2 - 1}
$$
Then, the normal vector fields to the manifold should be spanned by the vectors
$$
y\frac{\partial}{\partial x} - x \frac{\partial}{\partial y}
$$
since
$$
\left(y\frac{\partial}{\partial x} - x \frac{\partial}{\partial y}\right)(x^2 + y^2 - 1) = 2xy - 2xy = 0
$$
But, this confuses me since the tangent vector evaluated at (1,0) should be
$$
-\frac{\partial}{\partial y}|_{(1,0)}
$$
which looks like a tangent vector if you draw the picture.

For reference, this can be done by looking at the embedding of $M$ inside $\mathbb{R}^n$, restricting the euclidean tangent bundle to $M$ and cutting out the vector fields which vanish on $M$. This can be explicitly presented as the vanishing of $f_1,\ldots, f_k$ and $\nabla f_1, \ldots, \nabla f_k$ where
$$
\nabla f_i = \frac{\partial f_i}{\partial x_1} \partial_{x_1} + \cdots + \frac{\partial f_i}{\partial x_n} \partial_{x_n}
$$
where the $\partial_{x_j}$'s are the vertical coordinates of $T\mathbb{R}^n \to \mathbb{R}^n$

First of all, in order that  the subset $M\subset \mathbb R^n$ defined by the system $$
\begin{align*}
f_1 = 0 \\
\cdots \\
f_k = 0
\end{align*}
$$be a submanifold you have to assume that the Jacobian matrix $J(f)(x)=(\frac{\partial f_i}{\partial x_j}(x))$ has rank $k$ at every $x\in M$.
If this is the case the tangent space to $M$ at $x$  consists of the $n-k$ dimensional subspace $T_x(M)\subset T_x(\mathbb R^n)= \{x\}\times\mathbb R^n $  defined by: $$T_x(M)=  \{x\}\times \{v\in\mathbb R^n\vert d_x  f_i(v)=0   \quad i=1,\cdots ,k \} $$ where of course  $d_x  f_i(v)=\sum_{j=1}^n \frac{\partial f_i}{\partial x_j}(x)\cdot v_j$

Your example
In your example $n=2, k=1,M=S^1$ and $f_1=f=x^2+y^2-1$ .
Hence for $m=(a,b)$ with $a^2+b^2=1$ we have $d_mf(u,v)=2au+2bv$ and thus $$T_mS^1=\{m\}\times \{(u,v)\in \mathbb R^2\vert au+bv=0\}$$ which can (prudently!) be identified  with the line $au+bv=0$ in $\mathbb R^2$ .
In particular  $T_{(1,0)}S^1=\{(1,0)\}\times \{(u,v)\in \mathbb R^2\vert u=0\}$ which can (prudently!) be identified with $\{0\}\times \mathbb R\subset \mathbb R^2$

edit: I write sequence other way around, confused myself and gave wrong answer. Now it should be correct.

Let $M\subseteq \mathbb{R}^n$ be the manifold. We have the following exact sequence of vector bundles:
$$0\rightarrow TM\rightarrow T\mathbb{R}^n_{\mid M}\rightarrow\mathcal{N}_{M/\mathbb{R}^n}  \rightarrow 0$$

and now vector fields on $M$ correspond to restrictions of those vector fields $v$ from $\mathbb{R}^n$ that vanish on $f_1$,...,$f_k$ i.e:
$$v(f_1)=...=v(f_k)=0$$
So that the correct answer is:
$$\langle v_{\mid M} \mid v(f_1)=...=v(f_k)=0\rangle$$


Does the everywhere differentiability of $f$ imply it is absolutely continuous on a compact interval? Suppose $f$ is differentiable everywhere on $[0,1]$. Must $f$ be absolutely continuous on $[0,1]$? I know this is true if $f'$ is integrable but I'm not sure in this more general case.

$ f ( x ) = \frac 1 { x + \epsilon } $, $ \epsilon $ tiny positive value

is differentiable every where in $[0,1]$ but not uniformly continuous which implies it is not absolutely continuous either.


Stuck in expressing factor into a sum of three perfect squares. Two part question.

(i) Consider the function $f(x)=x^3-6kx+k^3+8$. Show that we can write $f(x)$ as $(x+k+2)P(x)$ where $P(x)$ is a quadratic function.

(ii) Show that $2P(x)$ can be written as the sum of three perfect squares and hence solve $f(x)=0$ for all values of $k$.

My attempt.

(i) By long division I have $P(x)=x^2-(k+2)x+k^2-2k+4$. I believe this part to be correct by trying a few cases of $k$ with Wolfram Alpha.

(ii) $2P(x)=2x^2-2kx-4x+2k^2-4k+8$. No idea how to continue.

You are correct on (i). For (ii), just split it up. First, notice the $-2kx$ term, that'd be the middle term in one square:

$$2x^2 - 2kx - 4x + 2k^2 - 4k + 8$$
$$= (x^2 - 2kx + k^2) + x^2 - 4x + k^2 - 4k + 8$$

Then $(x^2 - 4x)$ looks like the beginning of another square:

$$= (x^2 - 2kx + k^2) + (x^2 - 4x + 4) + k^2 - 4k + 4$$

And there's our third:

$$= (x-k)^2 + (x-2)^2 + (k-2)^2$$

The first zero is $x = -(k+2)$. The other zeros come from the fact that all three terms here are non-negative, the only possible zero arises when all three terms are themselves zero: $x = k = 2$ (which is a double root, or no real root).

For i: It's enough to show that $f(-k-2)=0$.

For ii: Since $P(x)=(x-k)^2+(x-2)^2+(k-2)^2$ then $f(x)=0$ when $x=-k-2$ or when $k=2$.


Prove that $X \in \mathfrak g.$ Let $E$ be a Banach space. Let $G \subseteq GL(E)$ be a closed subgroup i.e. $G$ is a linear Banach Lie group. Define $$\mathfrak g : = \left \{X \in \mathcal L(E)\ \big |\ \forall t \in \Bbb R, e^{tX} \in G \right \}.$$ Let $\gamma : (-\varepsilon, \varepsilon) \longrightarrow G$ be a $C^1$-map with $\gamma (0) = I.$ Show that $X = \dot {\gamma} (0) \in \mathfrak g.$

By Taylor's theorem I get for all $t \in (-\varepsilon, \varepsilon)$ $$\gamma (t) = I  + t X + o (t).$$
From here how do I conclude that $e^{tX} \in G,$ for all $t \in \Bbb R\ $? Would anybody please help me in this regard?
Thanks in advance.
EDIT $:$ If $G = GL(E)$ then we are through since $e^T \in GL(E)$ if $T \in \mathcal L (E).$

Lemma $:$ Let $\{A_n \}_{n \geq 1}$ be a sequence in $\mathcal L(E)$ with $\lim\limits_{n \to \infty} A_n = A.$ Then $$\lim\limits_{n \to \infty} \left (I + \frac {A_n} {n} \right )^n = e^A.$$
Let us prove the required result with the help of the above lemma.
Let $t \in \Bbb R.$ Then there exists $N \in \Bbb N$ such that for all $n \geq N$ we have $\left \lvert \frac {t} {n} \right \rvert \lt \varepsilon$ i.e. $\frac {t} {n} \in (-\varepsilon,\varepsilon).$ Hence it follows that $\gamma \left (\frac {t} {n} \right ) \in G,$ for all $n \geq N.$ Therefore for all $n \geq N$ we have $\gamma \left (\frac {t} {n} \right )^n \in G,$ since $G$ is a group. Now by Taylor's theorem we have $$\begin{align*} \gamma \left (\frac {t} {n} \right ) & = \gamma (0) + \frac {t} {n} \dot {\gamma} (0) + o \left (\frac {t} {n} \right ) \\ & = I + \frac {t} {n} X + o \left (\frac {t} {n} \right )\ \ (\because \gamma (0) = I\ \text {and}\ \dot{\gamma} (0) = X\ ) \end{align*}$$ Hence from the above lemma it follows that $$\lim\limits_{n \to \infty} \gamma \left (\frac {t} {n} \right )^n = \lim\limits_{n \to \infty} \left (I + \frac {tX + o(1)} {n} \right )^n = e^{tX}\ \ (\because o(1) \to 0\ \text {as}\ n \to \infty\ )$$
Since $\gamma \left (\frac {t} {n} \right ) \in G,$ for all $n \geq N$ and $G$ is closed it follows that $e^{tX} \in G.$ Since $t \in \Bbb R$ was arbitrarily taken it follows that $e^{tX} \in G,$ for all $t \in \Bbb R.$ Hence $X \in \mathfrak g,$ as required.
QED


If the columns of an $n \times n$ matrix are linearly independent, then the columns span $\mathbb{R}^{n}$ My textbook says that its true but I can't find a proof of this on the internet.

"If $A$ is an $n \times n$ matrix with linearly independent columns, then the columns of $A$ span $\mathbb{R}^{n}$."

The dimension of a finite dimensional vector space is the maximal number of linearly independent vectors, and it is also the minimal number of a system of generators. A maximal system of linearly independent vectors is a basis. A minimal system of generators is a basis.


Proof of $0 \cdot\vec{v}=\vec{0} $ for all $ \vec{v}$ in the vector space According to this webpage from this Lemma 1.17, which states:

  In any vector space  V , for any  $\vec{v}\in V  \text{ and }
r \in\mathbb{R} ,\text{ we have } $

  $$0\cdot\vec{v}=\vec{0} $$

Their Proof:

For 1, note that  $\vec{v}=(1+0)\cdot\vec{v}=\vec{v}+(0\cdot\vec{v})$ . Add to both sides the additive inverse of  $\vec{v}$ , the vector  $\vec{w} $ such that  $\vec{w}+\vec{v}=\vec{0}$ .

\begin{array}{rl}
\vec{w}+\vec{v}
&=\vec{w}+\vec{v}+0\cdot\vec{v} \\
\vec{0}
&=\vec{0}+0\cdot\vec{v} \\
\vec{0}
&=0\cdot\vec{v}
\end{array}

My Question:

Intuitively, I understand this proof. However, my concern is with its rigor. How do they know $\vec{v}=(1+0)\cdot\vec{v}=\vec{v}+(0\cdot\vec{v})$ based on the weblinks 10 definitions on the vector space? What definition(s) are they using to justify this?

To me , it seems a bit circular because they say hey we are going to show  $0\cdot\vec{v}=\vec{0} $, but we are going to use $\vec{v}+(0\cdot\vec{v})=\vec{v}$, implying $(0\cdot\vec{v})$ is a zero vector.

They actually don't use the fact that $0 \cdot \vec{v} = \vec{0}$ in the proof but I understand (firsthand) how easy it is to be confused by that.

What actually happens is they establish that $0 \cdot \vec{v} = \vec{0}$ by starting with this equation:

$$\vec{v} = (1 + 0) \cdot \vec{v}$$

This equation is completely fine because of course $1 + 0 = 1$, and of course $\vec{v} = 1 \cdot \vec{v}$ (this is condition #10 in Definition 1.1).

The next step is to use condition #7 in Definition 1.1.  This gives us:

$$(1 + 0) \cdot \vec{v} = 1 \cdot \vec{v} + 0 \cdot \vec{v}$$

But since $1 \cdot \vec{v} = \vec{v}$, then we can simplify this and say:

$$(1 + 0) \cdot \vec{v} = \vec{v} + 0 \cdot \vec{v}$$

Combining this with the first equation I wrote above, we have

$$\vec{v} = \vec{v} + 0 \cdot \vec{v}.$$

Next they add $\vec{w}$ to both sides, where $\vec{w}$ is the additive inverse of $\vec{v}$.  Such a $\vec{w}$ is guaranteed to exist by condition 5 in Definition 1.1.  This gives us:

$$\vec{v} + \vec{w} = \vec{v} + \vec{w} + 0 \cdot \vec{v}$$

Now, because $\vec{w}$ is the additive inverse of $\vec{v}$, then by definition of additive inverse we have $\vec{v} + \vec{w} = \vec{0}$.  So the equation simplifies to:

$$\vec{0} = \vec{0} + 0 \cdot \vec{v}$$

And by condition 4 in Definition 1.1, $\vec{0} + 0 \cdot \vec{v} = 0 \cdot \vec{v}$.  Therefore the equation simplifies to:

$$\vec{0} = 0 \cdot \vec{v}$$

Finally, we conclude that $0 \cdot \vec{v}$ must equal $\vec{0}$.

Personally, I prefer:

$0 = 0 + 0$ (field axiom)

$0v = (0 + 0)v$ (multiplying by the same scalar on both sides)

$0v = 0v + 0v$ (vector space axiom - distribution of scalar multiplcation over vector addition)

$0v - 0v = (0v + 0v) - 0v$ (subtracting $0v$ from both sides)

$0 = 0v + (0v - 0v)$ (associative law of vector addition - I probably should have added $-0v$ to both sides, but I get lazy)

$0 = 0v + 0$ (some identity something-something for vector spaces)

$0 = 0v$ (where we wanted to go)....


Probability that the majority of 3 classifiers are wrong 3 classifiers. Each classifier has $0.7$ accuracy and makes its error independently.

How do I calculate the probability that the majority of three classifiers are wrong?

how I'm trying to solve it:

(3 choose 2) * 0.3 * 0.3 * 0.7 * 0.3 * 0.3 * 0.3

A similar question has been asked/answered here, but with 17 classifiers. Simply use their solution and change from 17 to 3 to apply to here.
To improve the accuracy, a majority vote is taken. What is the probability of an incorrect classification?

Hint: Majority means it could potentially be either $3$ or $2$. The probability of majority of the classifiers being wrong is the probability that either $2$ are wrong or all $3$ are wrong.  Example: Probability that $k$ out of $n$ independent trials of a random experiment are successful, with the probability of success being $p$ is ${n \choose k}p^k(1-p)^{n-k}$.


Uniform convergence of a series For which $x \ge 0$ does the power series
$$
\sum_{n=0}^\infty \frac x{(1+x)^n}
$$
converge uniformly?

Okay, I see that for $n \ge 2$ we obtain from upon taking a derivative and setting to $0$ the critical value
$$
x=\frac 1{n-1}.
$$
At that critical value,
$$
\frac{x}{(1+x)^n} = \frac{(n-1)^{n-1}}{n^n}.
$$
That value is the maximum over all $x \ge 0$, which means
$$
\left|\frac{x}{(1+x)^n}\right| \le \frac{(n-1)^{n-1}}{n^n}.
$$
for all $x \ge 0$, $n \ge 2$.
Am I now left to establish that
$$
\sum_{n=0}^\infty \frac{(n-1)^{n-1}}{n^n} < \infty,
$$
so that, using the Weierstrass M-test, the power series converges uniformly for all $x \ge 0$?

You need to bound away from the origin as this series is geometric with common ratio $1/(1+x)$, (ignoring the finite x) so you need
$$
\frac{1}{1+x}<1\Rightarrow 1<1+x\Rightarrow x>0
$$
You can also see this by noting that if $x\in (\epsilon,\infty)$ you can bound
$$
\frac{1}{1+x}\leq\frac{1}{1+\epsilon}
$$
And then apply the Weirstrass M test.

$$
\sum_{n=0}^\infty \frac x{(1+x)^n} = x\sum_{n=0}^\infty \frac {1}{(1+x)^n}= x\sum_{n=0}^\infty (1+x)^{-n}= x(\frac{1}{x}+1)= (1+x)
$$

  $$
\lim\limits_{n \rightarrow \infty}{(1+x)} = (1+x)
$$

That is, that the constant sequence converges trivially to $(1 + x)$


Is it a vector space? \begin{align*}W=\left\{\left\{x_1,x_2,x_3\right\}|x_1-1=\frac{x_2-2}{2}=\frac{x_3-3}{3}\right\}\end{align*}

\begin{align*}\alpha _1=\left\{x_1,x_2,x_3\right\};\\\\&\\\alpha _2=\left\{y_1,y_2,y_3\right\};\end{align*}

Is $W$ a vector space? I know $\alpha _1+\alpha _2$ and to check whether fit the equation, but seems stuck, and have a little problem in the expression.

$x_1+y_1-2=\frac{x_2+y_2-4}{2}=\frac{x_3+y_3-6}{3}$

subtract 1 from both sides,

$x_1+y_1-1=\frac{x_2+y_2-2}{2}=\frac{x_3+y_3-3}{3}$

Now we are done.

At first, you should add $1$ to the equality constraint describing your set. Then, you get
$$
x_1 = \frac{x_2}2 = \frac{x_3}3
$$
Now, it should be easy to prove that it is a vector space.

It is not Stright line through origin in R3-space, that's why it cant be the subspace and hence vector space as it doesnot contain in this case the zero vector.


In $\triangle ABC$, if angle bisectors $AE$ and $CD$ meet at incenter $F$, and $|FE|=|FD|$, then the triangle is isosceles or $\angle B=60^\circ$ I was screwing around lately in GeoGebra and I realized something.

  Draw a $\triangle ABC$, and let the bisectors for $\angle A$ and $\angle C$ meet sides $BC$ and $AB$ at points $E$ and $D$, respectively. If the angle bisectors meet at the incenter, $F$, and if $FD \cong FE$, then either $\triangle ABC$ must be isosceles or $\angle B$ must be $60^\circ$.

However I was unable to prove why that is. Any help would be appreciated.

Let $a$ and $c$ be the half angles,

$$ a = \frac{∠A}{2}, \space  c = \frac{∠C}{2}.$$

Apply the sine rule to the triangles ΔBDF and ΔBEF, respectively,

$$ DF = BF \frac{\sin(∠B/2)}{\sin(2a+c)}\tag{1}$$
$$ EF = BF \frac{\sin(∠B/2)}{\sin(2c+a)}\tag{2}$$

where ∠BDC = $2a + c$ and ∠BEA = $2c + a$ are recognized.

Take the ratio of (1) and (2),

$$ \frac{DF}{EF} = \frac{\sin(2c+a)}{\sin(2a+c)}$$

If DF = EF, the equation yields,

$$\sin(2c+a)=\sin(2a+c).$$

which gives two solutions,

$$2c+a = 2a+c\tag{3}$$

$$2c+a = 180^\circ - (2a+c)\tag{4}$$

The solution (3) leads to $a = c$, which is the isosceles triangle case.
The solution (4) leads to $a + c = 60^\circ$ and in turn ∠B = $180^\circ -2(a+c)=60^\circ $, which is the 60-degree case.

Hence, both cases are proved.

In the standard notation by the angle bisector theorem we obtain:
$$\frac{FE}{AF}=\frac{CF}{AC}=\frac{\frac{ab}{b+c}}{b}=\frac{a}{b+c}.$$
Thus, $$\frac{FE}{AE-FE}=\frac{a}{b+c},$$ which gives
$$FE=\frac{a}{a+b+c}\cdot AE=\frac{a}{a+b+c}\cdot\frac{2bc\cos\frac{\alpha}{2}}{b+c}=\frac{2abc\cos\frac{\alpha}{2}}{(a+b+c)(b+c)}.$$
Similarly, $$FD=\frac{2abc\cos\frac{\gamma}{2}}{(a+b+c)(a+b)}.$$
Id est,
$$\frac{2abc\cos\frac{\alpha}{2}}{(a+b+c)(b+c)}=\frac{2abc\cos\frac{\gamma}{2}}{(a+b+c)(a+b)}$$ or
$$a(b+c-a)(a+b)^2=c(a+b-c)(b+c)^2$$ or
$$(a-c)(a+b+c)(a^2+c^2-ac-b^2)=0.$$
Can you end it now?


Proof by contradiction that $n!$ is not $O(2^n)$ I am having issues with this proof: Prove by contradiction that $n! \ne O(2^n)$. From what I understand, we are supposed to use a previous proof (which successfully proved that $2^n = O(n!)$) to find the contradiction.

Here is my working so far:

Assume $n! = O(2^n)$. There must exist $c$, $n_{0}$ such that $n! \le c \cdot 2^n$. From the previous proof, we know that $n! \le 2^n$ for $n \ge 4$.

We pick a value, $m$, which is gauranteed to be $\ge n_{0}$ and $\ne 0$. I have chosen $m = n_{0} + 10 + c$.

Since $m > n_0$:

$$m! \le c \cdot 2^m\qquad (m > n \ge n_0)$$

$$\dfrac{m!}{c} \le 2^m$$

$$\dfrac{1}{c} m! \le 2^m$$

$$\dfrac{1}{m} m! \le 2^m\qquad (\text{as }m > c)$$

$$(m - 1)! \le 2^m$$

That's where I get up to.. not sure which direction to head in to draw the contradiction.

In
Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$,
they have obtained
$$
n!\geq \left(\frac{n}{e}\right)^n.
$$
Hence
$$
\lim_{n\rightarrow\infty}\frac{n!}{2^n}\geq\lim_{n\rightarrow\infty}\left(\frac{n}{2e}\right)^n=\infty.
$$
Suppose that $n!=O(2^n)$. Then there exist $C>0$ and $N_0\in \mathbb{N}$ such that
$$\frac{
n!}{2^n}\leq C
$$
for all $n\geq N_0$. Lettting $n\rightarrow\infty$ in the aobve inequality we obtain
$$
\lim_{n\rightarrow\infty}\frac{n!}{2^n}\leq C,
$$
which is an absurd.

This is the cleanest proof I can think of:

Using the same techniques used to show $2^n=O(n!)$, we can show that $3^n = O(n!)$ as well.  If we had $n!=O(2^n)$, then this implies $3^n = O(2^n)$, contradiction.


Roots of a cubic equation When will the roots of a cubic polynomial $ax^3+bx^2+cx+d$ generate a cyclic group of order 3?
I literally mean: the cubic equation $ax^3+bx^2+cx+d$ will have three roots (possibly all three real, possibly one real and two complex), say $\zeta_1$, $\zeta_2$ and $\zeta_3$. Under what conditions is $\{\zeta_1, \zeta_2, \zeta_3\}$ isomorphic to $C_3$?

If a complex number has multiplicative order 3, then it is a primitive cubic root of unit. Hence, two of the roots are primitive cubic roots of unit and the other one is 1. In other words, the polynomial is $a(x^3-1)$.

One obvious condition to get C3 as the Galois group is that there are no simple permutations of the roots. So a pair of complex roots is out. You must have three real roots. Then there must be some algebraic condition between the roots which does not allow you to permutate them willy-nilly. An example of three roots with built-in structure would be cos(pi/7), cos(2pi/7), and cos(3pi/7). They are the roots of the cubic equation which comes from adding together conjucate roots of x^7-1=0. If they don't have a Galois group of C3 then I'll be very surprised.


Name of the class of graphs obtained by deleting $\mathcal{Q}_d$ from $\mathcal{Q}_n$ Let $\mathcal{Q}_n$ denote the $n$-cube graph. I would like to know if there is a name for the class of graphs obtained by deleting a ${\bf single}$ arbitrary copy of $\mathcal{Q}_d$ from $\mathcal{Q}_n$, where $1 \leq d < n$. For example, when $d = 1$, this would involve removing two vertices that are connected by an edge, when $d = 2$, this would involve removing a square (i.e $4$ vertices of $\mathcal{Q}_2$) from $Q_n$, and so forth. Do these types of graphs have a name/have been studied previously?

Edit: I know that they are partial cubes - https://en.wikipedia.org/wiki/Partial_cube - i.e an isometric subgraph of the hypercube, but I suspect something more specific can be said.

It look's like horn in simplicial sets / cubical sets (some explanations here). (N.B. It is not about graphs exactly. It is much wider and uses constructions of the nature @user340082710 asks about.)

Some recent works with them seem to be tightly connected with contemporary well-computated foundations of mathematics (homotopy type theory, cubical type theory). Look these references and notes to obtain more information.


Understanding a step in Stein's proof that the Schwartz space is closed under convolution. I am looking at Stein's proof that if $f,g \in \mathcal{S}(\mathbb{R})$, then so is $f*g$. The definition of the Schwartz space on $\mathbb{R}$ given is, the set of all indefinitely differentiable functions $f$ so that $f$ and all its derivatives are rapidly decreasing, in the sense that
$$\sup_{x\in \mathbb{R}}|x|^k |f^{(l)}(x)| < \infty \; \text{for every}\; k,l\ge 0.$$
 In the proof below, I have four questions.

First, how can I check the assertion that we have $\sup_x |x|^l |g(x-y)|\le A_l (1+|y|)^l$? I tried considering the two cases, but I cannot figure out how to get this bound. In fact, I don't really understand why we need to consider two separate cases. Shouldn't we be able to get $|x|^l |g(x-y)|\le |x-y|^l|g(x-y)|+|y|^l|g(x-y)|\le (1+|y|^l)A_l\le(1+|y|)^l A_l$, where $A_l=\sup_x |x|^l |g(x)|$?

Next, how can we guarantee that the integral $\int |f(y)|(1+|y|)^l dy$ is bounded for every $l\ge 0$?

Finally, in the last paragraph, how is the interchange of differentation and integration justified in the case by the rapid decrease of $dg/dx$? Also, why do we only take the differentiation on $g$ and not $f$? I am not aware of this kind of  theorem.

I would greatly appreciate any hints or solutions.

For your proof
$$|x|^l |g(x-y)|≤|x-y|^l |g(x-y)| + |y|^l |g(x-y)|$$
is incorrect, the triangle inequality doesn't hold for $|\cdot|^l$. Consider ie $x=2,y=1$, $l=2$ you have $|2|^2=4 \not≤|2-1|^2+|1|^2=2$.

I'm not sure how the hint should be used, but you can also do it like this:
$$|x-y|^l|g(x)|≤\sum_kC_k |x|^{l-k}|y|^k|g(x)|≤A_l(1+|y|)^l$$
Where the Schwartz property of $g$ was used to bound $|x|^{l-k}|g(x)|$ in each summand. Then
$$\sup_{x}|x|^l |g(x-y)|=\sup_{x} |x+y|^l|g(x)|$$
and the result follows.
The integral is bounded because
$$|f(y)|(1+|y|^l)(1+|y|^{2})≤C$$
follows from the Schwartz condition. Then you have since $|f|(1+|y|^l)$ is measurable and pointwise dominated by $\frac C{1+|y|^2}$ it is integrable.
$$\frac{d}{dx}\int f(y)g(x-y)dy=\frac{d}{dx}\lim_{n\to\infty}\int_{-n}^n f(y)g(x-y)dy$$
We want to swap the limit with the derivative, this is possible if $\int_{-n}^n f(y)g(x-y) dy$ and $\int_{-n}^n f(y) g'(x-y) dy$ converge uniformly on compacta (wrt $x$). Here:
$$\left|\int_{\mathbb R} f(y)g(x-y) dy -\int_{-n}^n f(y) g(x-y)dy\right|≤\left(\int_{-\infty}^{-n}+\int_n^\infty\right)|f(y)g(x-y)|dy$$
From the Schwartz condition bound $g$ by some constant and bound $f(y)$ by $\frac c{1+y^2}$ to see that the expression goes to zero uniformly. One does essentially the same for the derivative integral.
This gives that
$$\frac{d}{dx}\int f(y) g(x-y) dy = \lim_{n\to\infty}\frac{d}{dx}\int_{-n}^n f(y) g(x-y) dy = \lim_{n\to\infty}\int_{-n}^n f(y) g'(x-y) dy\\
=\int f(y) g'(x-y)dy$$
From the rule of derivation under the integral. Proceed via induction to get it for $\frac{d^k}{dx^k}$.
The expression $\frac{d^k}{dx^k}(f*g)=(f*g^{(k)})$ is useful because we know already that for any Schwartz $f,g$:
$$\sup_{x}|x|^l |f*g|(x)<\infty$$
Thus, since $g^{(k)}$ is Schwartz, you have:
$$\sup_{x}|x|^l|(f*g)^{(k)}|(x)=\sup_{x}|x|^l |f*g^{(k)}|(x)<\infty$$

Schwartz space is invariant under Fourier transform, and it is closed under pointwise multiplication. And Fourier transform takes convolution to pointwise multiplication.


Area enclosed by general equation of second degree What is the area of the figure enclosed by the curve: $5x^2+6xy+2y^2+7x+6y+6=0$.

My attempt: $2y^2+6(x+1)x+5x^2+7x+6=0$.

$\displaystyle y=\frac{-6(x+1)\pm\sqrt{36(x+1)^2-4(5x^2+7x+6)}}{4}$

$\displaystyle y =\frac{-3(x+1)\pm \sqrt{4x^2+11x+3}}{2}$

How do I solve it?

Factorize the given expression from the roots you derived and plot the equation for confirmation.

Study the graph. Recognize that the blue ellipse with axes parallel to axes has been shifted up by pushing up thin slices to form the given oblique ellipse leaving area unchanged. This follows geometrization of area/volume concepts.

Accordingly Area $ = \frac{\pi}{4} .d_1. d_2  = \frac{\pi}{4} .1 .2 = \pi/2$

I get that ellipse major semi-axis is
$$a=\frac{3+\sqrt5}{2\sqrt2}$$
minor semi-axis is
$$b=\frac{3-\sqrt5}{2\sqrt2}$$
Area is
$$S=\pi ab=\pi\cdot\frac{3+\sqrt5}{2\sqrt2}\cdot\frac{3-\sqrt5}{2\sqrt2}=\frac{\pi}{2}$$


If $\cos a = \cos a \cos b - \sin a \sin b$, then $b = 2n\pi$ The solutions to a tutorial question I am working on are as follows:
$$\cos\left(\frac{p}{\sqrt{\eta}}x\right) = \cos\left(\frac{p}{\sqrt{\eta}}x\right)\cos\left(\frac{p}{\sqrt{\eta}}2L\right)-\sin\left(\frac{p}{\sqrt{\eta}}x\right)\sin\left(\frac{p}{\sqrt{\eta}}2L\right)$$

Clearly if LHS is equal to RHS, then
$$\frac{2Lp}{\sqrt{\eta}} = 2n \pi$$

I am confused as to how this is clear. If anyone is able to provide more working that would be great.

Note: (From comment).  The question was originally about concluding $\cos\left(\frac p\eta x\right)$, given $$\cos\left(\frac p\eta x\right)= \cos\left(\frac p{\eta} x +\frac p{\eta}2L\right)$$  The solution then expanded by using the sum of angles formula for $\cos$.

Hint:  The RHS gives $$\cos\left(\frac p{\eta} x +\frac p{\eta}2L\right)^{(\dagger)}$$

So what can you conclude if $$\cos\left(\frac p\eta x\right) = \cos\left(\frac p{\eta} x +\color{blue}{\frac p{\eta}2L}\right)\;?$$

This equality only occurs when $\color{blue}{\dfrac {2Lp}{\eta}} = 2n\pi$, where $n$ is an integer.  (Note that $2\pi (n)$ is $n$ full revolutions of the unit circle).

$^{(\dagger)}$ If this is how the problem  originally appeared, I'd suggest NOT expanding using the sum of angles identity for $\cos$.)

Let $\frac{p}{\sqrt{\eta}}=l$
Then, $RHS=\cos(lx+2Ll)$
$\therefore \cos(lx+2Ll)=\cos(lx)$
$\therefore lx+2Ll=2n\pi \pm lx$
$\therefore 2Ll=2\eta\pi,2\eta\pi-2lx$
$\frac{2Lp}{\sqrt{\eta}}=2\eta\pi,2\eta\pi-2\frac{p}{\sqrt{\eta}}x$


Does a continuous group action induce an open map? I am currently reading the book "The Topology of Fibre Bundles" written by Norman Steenrod. The book says a sentence without a proof and I was trying to give a rigorous proof of the sentence.
The settings for the sentence are:
Let $B$ be a topological group which acts on a topological space $X$ continuously, that is, the map $B\times X \to X$ is continuous.
Let $e\in B$ be the identity element of $B$.
Fix an element $x_0\in X$ and define a map $p:B\to X$ by $p(b)=b\cdot x_0$ for $b\in B$.
The statement I wish to prove is:
Suppose that $p$ maps an open neighborhood of $e$ onto an open neighborhood of $x_0$.
Then $p$ is an open map.
It seems that the context of the book assumes the Hausdorff condition on the space $X$ but I guess that condition is irrelevant to this problem.

Without further assumptions, the claim is simply false. As an example, consider the group $G={\mathbb Q}$ equipped with discrete topology, $X=G$ equipped with the standard order topology. Assume that the action $G\times X\to X$ is the one given by the left multiplication. I will leave it to you to verify that the action is continuous. Take $x_0=0$. The corresponding orbit map $p: G\to X$ maps $G$ onto $X$. Thus, we found an open subset of $G$ (i.e. $G$ itself) whose image under $p$ is open. However, the map $p$ clearly is not open (just take the image of any singleton under the orbit map). So, what Tsemo proved for you is false: His mistake was to assume that every neighborhood of $e$ has open image under $p$.
However, the statement you are trying to prove holds under the following mild extra assumptions which hold in most natural applications of the fiber bundle  theory:

Assume that your group $G$  (I do not like using the letter $B$ for groups)
is locally compact and 2nd countable. (For instance, every Lie group is.)

Assume that $X$ is completely metrizable (i.e. admits a complete metric metrizing its topology). - Note that ${\mathbb Q}$ (with the order topology) in my example is not completely metrizable.

Then indeed the orbit map is open provided that one of the two equivalent conditions holds:
a. There exists an (open, as customary in the US literature on general topology) neighborhood $U$ of $e$ such that $p(U)$  is open in $X$.
b. $p(G)$ is an open subset of $X$.
A proof of openness of $p$ under these extra assumptions is an application of the Baire Category theorem (arguing as in the proof of Arens Lemma in the theory of transformation groups, see here).

Let $V$ be a non empty open subset of $B$, and $v\in V$, $W=v^{-1}V=\{v^{-1}w, w\in V\}$ is an open subset which contains $e$, we deduce that $p(W)$ is open. Since $e\in W$, $p(V)=v.p(W)$ is open since the map defined on $X$ by $f:y\rightarrow v.y$ is invertible, we deduce that $p(V)=f(p(W))$ is open.


Does $\leq$ imply $<$? If I know for two numbers $a$ and $b$ that
$${a < b }$$
Then is it correct to say that
$$ a \leq b $$
I know that the second statement is true as long as the first one is.
However is it true that
$$a \leq b  \implies a<b$$
That is does less than or equal imply less than?

$1\leq 1$ (since $1=1$) but does not hold that $1<1$.

You're correct in the first case.
But now consider the proposition $a\leq b.$ In this case if $a=b=4$, $a\leq b$ is true.
However given $a< b$, it is false that $4<4$.  So it is false that $a\leq b \to a\lt b$ is always true.


Define the relation $\mathscr{R}$ on $\Bbb Z$ by $x \mathscr{R} y$ iff $xy >0$. I wish to examine several properties of this relation. Considering on $\mathbb{Z}$ the following relation:

$$\mathscr{R} = \left \{ (x,z) \in \mathbb{Z} \times \mathbb{Z} \quad \mbox{ such that } \quad xz > 0  \right \}$$

i.e. $\forall x,z \in \mathbb{Z}, x \mathscr{R} z \iff \mbox{ the product } \, \, x \cdot z > 0 $

Check if the relation $\mathscr{R}$ is:

i) reflexive;
ii) symmetrical;
iii) transitive;
iv) anti-symmetrical;
v) of order;
vi) of equivalence.

What I have done is the following:
i) is it a reflexive relation?
$\forall x,x \in \mathbb{Z}, \quad x \mathscr{R} x \iff x\cdot x = x^2 > 0 $
since a square is always $> 0$ the relation is reflexive.

ii) is it a symmetrical relation?
$\forall x,z \in \mathbb{Z}, \quad x \mathscr{R} z \iff x \cdot z > 0$ by hypothesis,
since the operation of multiplication is commutative on $\mathbb{Z}$ the following is true:
$\forall x,z \in \mathbb{Z}, \quad z \mathscr{R} x \iff z \cdot x > 0$
hence, the relation is symmetrical.

iii) is it a transitive relation?
$\forall x,z \in \mathbb{Z}, \quad x \mathscr{R}z \iff x \cdot z > 0$ by hypothesys,
if we take another ordered pair
$\forall z,m \in \mathbb{Z}, \quad z \mathscr{R} m \iff z \cdot m > 0$
hence,
if $x \mathscr{R}z, \quad z \mathscr{R} m \Rightarrow x \cdot m > 0$
i.e. if $x \cdot z > 0, \quad z \cdot m > 0 \Rightarrow x \cdot m > 0$
so the relation is transitive.

iv) is it a anti-symmetrical relation?
No. Because it is not true that $z \mathscr{R}x \iff x=z$, as we have seen in the symmetrical relation.

v) is it a order relation?
No. Because it is not anti-symmetrical.

vi) is it a equivalence relation?
Yes, because it is reflexive, symmetrical and transitive.

What do you think about it?
Please, can you give me any suggestions? Many thanks!

$
\newcommand{\RR}{\mathscr{R}}
\newcommand{\Z}{\mathbb{Z}}
$
Note, intuitively, what we have. This relation $\RR$ on $\Z$ essentially means $x \RR y$ if and only if $x,y$ have the same sign and are nonzero. (Same sign ensures nonnegative, nonzero ensures positive.) Hopefully this makes the intuition clearer.
Onto the errors with your approach:

You were wrong on reflexivity. Yes, $x^2 > 0$ for any $x \in \Z$, but only if $x \ne 0$.
Your justification for transitivity is unclear, in my opinion.
Your justification for antisymmetry seems to amount to "it's symmetrical and thus not antisymmetrical." The properties are not necessarily opposed; a relation can be both, or neither. You're right in that this relation is not antisymmetrical, though, but even so. (In particular, a relation on $A$ of the form $\{(x,x) \mid x \in X\}$ for some $X \subseteq A$ is both antisymmetric and symmetric.)

Onto the overall right ideas:

Reflexivity: This does not hold, because $(0,0) \not \in \RR$. This is because $0 \cdot 0 = 0 \not > 0$.

Symmetry: This does hold under commutativity of integer multiplication. Assume $(x,z) \in \RR$. Then $xz > 0$. But then $xz = zx > 0$. Thus, $(z,x) \in \RR$ on the premise $(x,z) \in \RR$.

Transitivity: This does hold. It's easiest to refer back to the intuitive idea: $(a,b) \in \RR$ if and only if $a,b$ are nonzero and share the same sign. Suppose $(a,b),(b,c) \in \RR$. Then $a,b$ share the same sign, and are nonzero; likewise, the same is true of $b,c$. Thus, $a,b,c$ are nonzero and have the same sign. Thus, $a,c$ have the same sign and are nonzero, and thus $(a,c) \in \RR$.
Of course, this can be verified algebraically too. Since $(a,b),(b,c) \in \RR$, then $ab>0$ and $bc>0$. Multiply the inequalities and you have $ab^2c > 0$. Since $b \ne 0$ (otherwise neither inequality would hold) and $b^2 >0$ as a result, we may divide both sides by $b^2$ and see $ac > 0$. Thus, $(a,c) \in \RR$.

Antisymmetry: Anti-symmetry requires that, when $(a,b) \in \RR$ and $(b,a) \in \RR$, that $a=b$. However, commutativity of multiplication allows them to be distinct. More specifically: $(1,2),(2,1) \in \RR$ since $1 \cdot 2 = 2 = 2 \cdot 1 > 0$, but obviously $1 \ne 2$.

Order: An order relation requires antisymmetry, so we don't have this.

Equivalence: Since we don't have reflexivity, we don't have equivalence.

$x\cdot x\geq0$  (is always greater than zero), it is reflexive.

$x\cdot z\geq0,\,z\cdot x>0$, it is symmetry

$x\cdot z>0,\,z\cdot y>0,\,x\cdot y>0,\,x\cdot z\geq0$ therfore fore $x\cdot z=x\cdot y$ it is transitivity