Approximation argument for Poisson integral formula In the PDE book,

  for a harmonic function $u \in C^2(B_R(0))\cap C^1(\overline B_R(0)),$ we have the following Poisson integral formula
  $$
u(y)=\frac{R^2-|y|^2}{n w_n R}\int_{\partial B_R(y)}\frac{u(x)}{|x-y|^n}ds_x
$$
  An approximation argument shows that the Poisson integral formula continues to hold for $u \in C^2(B_R(0))\cap C(\overline B_R(0)).$

What is the approximation argument? For a harmonic function $u \in C(\overline B_R(0))$, is there a sequence of harmonic functions $u_k \in  C^1(\overline B_R(0))$ such that $u_k \to u$ in $C(\overline B_R(0))$? I have no idea for harmonic functions. May I use convolution?

Please let me know if you have any hint or comment for it. Thanks in advance!

A typical way to approximate a function defined on a ball $B_R(0)$ is
$$u_k(x) = u(\lambda_k x),\quad \lambda_k\nearrow 1$$
For any $u \in C(\overline B_R(0))$, this results in $u_k \to u$ in $C(\overline B_R(0))$ by virtue of the uniform continuity of $u$.  If $u$ is smooth (harmonic, holomorphic...) in the open ball $B_R(0)$, then $u_k$ has these properties  on a larger ball.

For $t\in(0,1)$, consider $\lambda: \overline{B_r(0)}\rightarrow B_r(0)$, $\lambda(y)=ty$, then $\lambda\in C^{\infty}(\overline{B_r(0)})$.  As $u\in C^2(B_r(0))$ follows that $\hat{u}(y)=u(ty)\in C^2(\overline{B_r(0)})$. Moreover
$$
\frac{u(ty)}{|x-y|^n}\rightarrow \frac{u(y)}{|x-y|^n}, \text{ as } t\rightarrow 1, \text{ and }
\frac{|u(ty)|}{|x-y|^n}
\leq \frac{\displaystyle\max_{\overline{B_r(0)}}|u|}{|x-y|^n}\in L^1(\partial B_r(0)).
$$
Therefore, from the dominated convergence theorem, for all $y\in B_r(0)$,
\begin{eqnarray*}
u(y)&=&\lim\limits_{t\rightarrow 1}u(ty)
=\lim\limits_{t\rightarrow 1}\hat{u}(y)\\
&=&\frac{r^2-|y|^2}{nw_nr} \lim\limits_{t\rightarrow 1}\int_{\partial B_r(0)}\frac{u(tx)}{|x-y|^n}ds_x
=\frac{r^2-|y|^2}{nw_nr} \int_{\partial B_r(0)}\lim\limits_{t\rightarrow 1}\frac{u(tx)}{|x-y|^n}ds_x\\
&=&\frac{r^2-|y|^2}{nw_nr} \int_{\partial B_r(0)}\frac{u(x)}{|x-y|^n}ds_x.
\end{eqnarray*}


Prove that square root of 2 is irrational using the principle of Mathematical Induction How do I prove that the square root of 2 is irrational using the principle of mathematical induction?

The usual proof starts out something like "suppose for a contradiction $\sqrt 2$ were rational, and write it as $p/q$ in lowest terms ...".

The "in lowest terms" hides an instance of induction, which you can unfold to get a proof in the shape of

  Theorem. For all positive integers $p$ and $q$, it holds that $(p/q)^2\ne 2$.

  Proof. By long induction on $q$. If $p$ and $q$ have a common factor $n>1$ then $(\frac pq)^2 = (\frac{p/n}{q/n})^2$, and since $q/n<q$, the induction hypothesis guarantees that $(\frac pq)^2\ne 2$. Now we consider the case that $p$ and $q$ are relatively prime ...

This is as simple as I could make it in my mind, working on the simply basis of multiple factors (similar to the above) but less algebraic manipulation:


Balls and Urn Paradox So, I came across the following paradox:

At $1$ minute before noon, put in balls $1 \sim 10$ and take out ball number $1$. At $1/2$ minute before noon, put in balls $11\sim20$ and take out ball number $2$ and so on. How many balls are there at noon?

None.

At $1$ minute before noon, put in balls $1 \sim 10$ and randomly take out a ball. At $1/2$ minute before noon, put in balls $11\sim20$ and randomly take out another ball and so on. How many balls are there at noon?

None.

Okay, so I understand the first paradox because one can describe the exact moment each ball was taken out. But, you can't apply a similar argument to the second paradox because we randomly take out a ball.
  I feel as if it's like infinitely summing $\frac{1}{n}$ and eventually there would be too many balls.

Can someone explain to me mathematically why this is the case?

Let $ n \in \mathbb{N} $

Let the ball labelled $ n $ be placed in the urn on step $ i_0 $.

The probability of removing the ball labelled $ n $ at step $ i $ is $ \dfrac {1} {9i+1} $.

The probability of it never being removed is therefore

$$ \prod_{i=i_0}^{\infty} 1 - \dfrac {1} {9i+1} $$

Which diverges to $ 0 $ since the sum

$$ \sum_{i=i_0}^{\infty} -\dfrac {1} {9i+1} $$

diverges to $ -\infty $.

So all balls will be removed at some point with probability $ 1 $.

It depends. How to the laws of the universe work in your hypothetical situation? If the laws of the universe work in such a way that number of balls in the urn is the limit as time approaches noon from the past direction of the number of balls in the urn at that time, then there will be infinitely many balls in the urn at noon, if you're careful what you really mean when you say that as I will describe later. If the laws of the universe work in such a way that each individual ball will be in the urn at noon if and only if it was in the urn at all times before noon that are sufficiently close to noon, then there will be none in the urn at noon.

Now I'll explain what I mean when I say the number of balls in the urn at noon is the limit as time approaches noon from the past direction of the number of balls in the urn at that time. In Calculus, there is only one kind of infinity. In set theory, there are multiple infinities. In set theory, a cardinal number is used to describe an equivalence class of sets that all of a bijection from one another, and it can be shown that there are multiple infinite cardinal numbers. When the limit as time approaches noon from the past side of number of balls in the urn at that time is the Calculus infinity, I declare it correct to say the number of balls in the urn at noon is the limit when the number of balls in the urn at noon is $\aleph_0$, the number of natural numbers there are, in the context of this answer.

Now I'll explain how it's possible that if each individual ball will be in the urn at noon if and only if it was in the urn at all times before noon that are sufficiently close to noon, there will be none in the urn at noon. Suppose each individual ball is in the urn at noon if and only if it was in the urn at all times before noon that are sufficiently close to noon. Consider the time when so far, you just put in balls 1-10 and took out ball one. At that time, there are just balls 2-10 in the urn. Eventually all the balls from 2-10 will be taken out when you put in balls 90-100 and take out ball 10. The balls currently in the urn at this time, 11-100 also will eventually be all taken out when you put in balls 990 to 1,000 and take out ball 100.

Now we can conclude that each individual ball some time before noon will be taken out and never put back in any time before noon. Given what the laws are, that means none of the balls will be in the urn at noon. However, as time approaches noon from the past direction, the number of balls in the urn approaches infinity. This shows that the number of balls in the urn at noon is not the limit as time approaches noon from the past direct of the number of balls in the urn at that time.


$p$ a prime and $p\equiv3\mod 4$, if $a^4=1_F$ where $F$ a field with characteristic $p$, then $x=a+a^{-1}\in F$ $p$ a prime and $p\equiv3\pmod 4$, if $a^4=1_F$ where $F$ a field with characteristic $p$, then $x=a+a^{-1}\in F$

I am not quite sure if the statement is true or not. Here is my approach

Clearly we have $x$ lies in extension field $F(a)$. And we have $x^p=(a+a^{-1})^p=a^p+a^{-p}=a^{4k+3}+a^{-4k-3}=a^3+a^{-3}=a^4a^{-1}+a^{-4}a=a^4(a^{-1}+a)=a^{-1}+a=x$. Notice that $a^2=-1_F\in F$, for any $y\in F(a)$, the only cases that $y\notin F$ is that $y$ generated by $1=1_F$, $a$ and $a^3$. Suppose $y=e_1+e_2a+e_3a^3$ with $e_1, e_2, e_3\in F$, then we have $y^p=(e_1+e_2a+e_3a^3)^p=(e_1+e_2a)^p+e_3^pa^{3p}=e_1^p+e_2^pa^p+e_3^pa^{3p}=e_1+e_2a^p+e_3a^{3p}$. However $a^p=a^3=a^{-1}\neq a$ and $a^{3p}=a\neq -a=aa^{2}=a^3$. Hence $y\neq y^p$. Hence $x\in F$.

Edit: I think I make a assumption that $a^2=-1$. If this is the case then above still hold. If $a^2=1$, then $a^3=a$, but still we have $a^p=a^{-1}\neq a$, so above still hold.

$a \in F$ and $a^{-1} \in F$ then by Closure property $x \in F$


Congruences- find the number of solutions Find the number of solutions of $ a_1x_1+ ... +a_nx_n =$ b (mod m).

I was thinking of denoting $ d = (a_1,a_2,...,a_n,m)$.
Then the congruence has no solutions if $d$ doesn't divide $b$. And it has solutions when $d$ divides $b$.
I was thinking of using induction, but I didn't have a clear idea.(I think it may have $dm^{n-1}$ solutions ??)
Could someone please help me show the solutions of these questions more concretely?

1. Find the number of solutions to the congruence $ x^3-2x^2-x+2 = 0$ (mod 143) without finding the actual solutions.

First Question:

  Theorem :
  The congruence $a_1x_1+\cdots+a_nx_n\equiv b\mod m$ has exactly $gcd(a_1,\cdots,a_n,m)m^{n-1}$ when $gcd(a_1,\cdots,a_n,m)$ divides $b$, and no solution otherwise.

Proof: by induction:

For $n=1$ let $d=gcd(a_1,m)$, $a=\frac{a_1}{d}$, $b'=\frac{b}{d}$ and $m'=\frac{m}{d}$ so the equation $ax \equiv b' \mod m'$ has exactly on solution $x_0$ in $\mathbb{Z_{m'}}$. which gives us $gcd(a,m)$ in $\mathbb{Z_{m}}$ ( the solutions are $x_0,x_0+m',\cdots,x_0+m'(d-1)$)
Assume that the congruence $a_1x_1+\cdots+a_nx_n\equiv b\mod m$ has exactly $gcd(a_1,\cdots,a_n,m)m^{n-1}$ when $gcd(a_1,\cdots,a_n,m)$ divides $b$ for every  integers $a_1,\cdots,a_n,m,b$. let's consider a congruence with $n+1$ variables:
$$a_1x_1+\cdots+a_nx_n+a_{n+1}x_{n+1}\equiv b\mod m \ \ \ \ (*)$$
with surely the constraint $gcd(a_1,\cdots,a_n,a_{n+1},m)$ divides $b$, let $d=gcd(a_1,\cdots,a_n,m)$ and consider that $x_{n+1}$ is fixed,the equation $$a_1x_1+\cdots+a_nx_n\equiv b-a_{n+1}x_{n+1}\mod m \ \ \ (**)$$ has exactly $dm^{n-1}$ solution for every fixed $x_{n+1}$ such that $d|b-a_{n+1}x_{n+1}$ or $a_{n+1}x_{n+1}=b \mod d$ and ther is no solution otherwise. Now let's count the number of possible values of $x_{n+1} \mod m$, every possible value of $x_{n+1}$ it's a solution of :
$$ \left(\frac{ma_{n+1}}{d} \right)x_{n+1}\equiv \frac{bm}{d} \mod m $$
and because as hypothesis, $gcd(a_{n+1},d)=gcd(a_1,\cdots,a_{n+1},m)$ divides $b$ so $k=gcd\left(\frac{ma_{n+1}}{d},m\right)=m\frac{gcd(d,a_{n+1})}{d}$ divides $\frac{bm}{d}$ hence this equation has exactly $k$ solutions, so there is exactly $k$ values possible of $x_{n+1}$ for which the equation $(**)$ has solutions and has for each value of $k$ $dm^{d-1}$ solutions . So the total number of solutions of $(*)$ is exactly:
$$kdm^{n-1}=gcd(a_1,\cdots,a_{n+1},m)m^{n} $$
which terminates the proof.

Second Question:

As pointed by @marty cohen, we have :
$$x^3-2x^2-x+2=(x-2)(x-1)(x+1) $$
this signifies that $x^3-2x^2-x+2\equiv 0 \mod 143$ if and only if $143$ divides $(x-1)(x+1)(x-2)$ so this is equivalent to $13$ divides one factor among $x-1$,$x+1$ and $x-2$ and $11$ also divides one factor among this factors. but every time we choose two factors we will deal with an equation of the form :
$$ x\equiv a_1 \mod 13\\x\equiv a_2 \mod 11$$
and because $gcd(11,13)=1$, the Chinese remainder theorem gives us the existence of an unique solution $\mod 13*11$ so the number of solutions is the number of ways of choosing two factors among $3$ which is $3^2=9$.

Verification by walframalpha.

  Generalization:Given $m$ primes $p_1\cdots p_m$ and $n$ different integers $a_1,\cdots,a_n\in \mathbb{Z_{p_1\cdots p_m}}$, the equation $(x-a_1)\cdots(x-a_n)\equiv 0 \mod p_1\cdots p_m$ has exactly $n^m$ solutions.

For 1
(I guess the first problem is 0),
$x^3-2x^2-x+2
=(x-2)(x^2-1)
=(x-2)(x-1)(x+1)
$
and
143 = 11*13.

Consider the possible values of $x$
that make
one or two terms in the product
divisible by 11 and 13.


Let a be a positive rational number. Let $A = \{ x\in\Bbb Q \mid x^2 < a \}$ . Show that $A$ is bounded in $\Bbb Q$ I need to prove the following:

  Let $a$ be a positive rational number.

  Let $A = \{ x\in\Bbb Q \mid x^2 < a \}$ .

  Show that $A$ is bounded in $\Bbb Q$

So I know I have to show that $A$ is bounded above and below in $\Bbb Q$. I have figured out that A is bounded above by $\sqrt(a)$ but I am unsure to prove that it is bounded below.

Because a square root can be either positive or negative, when you have $x^2<a$ and you take the square root of both sides, you actually get two inequalities out.

$$\begin{align}
x^2 &< a \\
\implies\quad x<\sqrt{a} \;\;\;&\text{and}\;\;\; x>-\sqrt{a}\\
\implies\quad -\sqrt{a}< \;& x < \sqrt{a}
\end{align}$$

That gives you a lower bound.


Differentiation of the determinant of the Jacobian I am working through  A Mathematical Introduction to Fluid Mechanics and I have come to a statement on showing what I am guessing is a corollary to Jacobi's Formula
https://en.wikipedia.org/wiki/Jacobi%27s_formula

I do not understand the step on the next page of this text where they use multilinearity of the determinant to apply the derivative in this way. It looks like that they use differentiation similar to a scalar, but I do not understand how  or why they are adding these Jacobians.

https://books.google.com/books?id=HBXSBwAAQBAJ&pg=PA8&lpg=PA8&dq=The+determinant+J+can+be+differentiated+by+recalling+that+the+determinant+of+a+matrix+is+multilinear+in+the+columns&source=bl&ots=2nBkkqc4sZ&sig=ACfU3U3MY4WFxZdLCITJUyywE7B_qaKcoA&hl=en&sa=X&ved=2ahUKEwiK5_2Fq4rgAhVNPN8KHZRNAsQQ6AEwAHoECAQQAQ#v=onepage&q=The%20determinant%20J%20can%20be%20differentiated%20by%20recalling%20that%20the%20determinant%20of%20a%20matrix%20is%20multilinear%20in%20the%20columns&f=false

The Jacobian is used in coordinate transformations. Picture the xy plane parallel to a slightly "curved plane" (let's say u,v). Imagine lines (vectors) taking all the points from x,y and putting them on a new plane, u,v. When doing so, we need to scale things from one plane to the other. That's my simplified understanding and over generalization of it.

Determinants are used to calculate area or volume, also.

You are right about thinking of it as a scalar.

As far as the adding three determinants when taking the derivative of the Jacobian,

You can multiply a row or column of a determinant by a scalar, or you can divide a row or column from a determinant: the operation only includes one row or one column. This does not change the value of the determinant.

Sorry trying to type on my phone.


Definition of a cube I was thinking today, what the best definition of a cube is. Google defines it as such:

  a symmetrical three-dimensional shape, either solid or hollow,
  contained by six equal squares.

And I was wondering if the "squares" part of the definition is necessary?

  a symmetrical three-dimensional shape, either solid or hollow,
  contained by six equal sides

Would this definition be equivalent?

Essentially the question is, is a 3 dimensional shape consisting of 6 equal sides, possible to be a non-cube? Is this provable?

No, it's not equivalent at all. You can for example fuse two triangular pyramids together to form a triangular bipyramid. This polyhedron has six equal faces which happens to be triangles instead (they could be equilateral). Since triangles are congruent if their side are the same and we can assign the edge lengths quite freely on such figure we don't need the triangles to be equilateral or even isosceles.

Another counter example is a rhombohedron. Here all the edges are the same, but the faces are rhobi. Given the side length of a rhomb it's determined up to congruence by the angle at one corner. In the same way the rhombohedron is determined by its edge length and the angles at one corner and the angles at the oposite corner will match due to parallellity making all faces congruent.

A more elaborate counter example could be to start with a rhombohedron and then
pick two opposite corners and then for each of the other we move them perpendicularily to or from the plane halfway the first corners (perpendicularily to the line connecting them) by equal amount.

Apart from this there are vaugenesses in your proposed definition. The concepts of "equal", "side" is not as precise as one might think. Normally one talks about congruence in geometry (or "similar"). In addition what a "side" is is vaugue: do you mean surface or edge? do you require it to be flat or may it be curved? in that case if it may be a curved surface wheres the border of the surface (you could have a sphere and state that it's made up of six congruent surfaces)?

Having that sorted out one could actually prove that there are basically only these two alternatives (the second is a special case of the third, and the cube is a special case of the second) and counter examples. Such a polyhedron would either have to be assembled the same way as a cube consisting of kite (quadrilateral where each edge has an adjacent edge of the same length) surfaces or assembled like a triangular bipyramid.

The proof is by considering a corner and then rule out the possibility that other than three faces meet there. Then one rules out that they be anything else than triangles or quadrilateral. Last one rules out that the quadrilateral can be anything else than kites (by ruling out anything else than pairs of edges with the same length and that they can't be parallellograms unless they are rhomboids and thereby kites).

After realizing the possibilities one could complete your proposal to apply only to cubes. Going along on the same theme it would be to require twelve equal edges and that all angles are the same.

Your argument is good but the problem is that the term $Side$ is defined for left or right position of a body (Even for mathematics), as google dictionary says:

  A side is an upright or sloping surface of a structure or object that is not the top or bottom and generally not the front or back.

OR

  A side is a position to the left or right of an object, place, or central point.

Another advantage of using the square is that $Square$ is a well defined mathematical shape, but the term $Side$ can have different  meanings under different circumstances.


find the least square solution for the best parabola find the least squares solution for the best parabola going through (1,1), (2,1), (3,2), (4,2)

so to solve this problem I have 4 equation set up

  a +  b + c = 1
 4a + 2b + c = 1
 9a + 3b + c = 2
16a + 4b + c = 2

I found my $A$ to be

1  1  1
4  2  1
9  3  1
16 4  1

and my $\mathbf{b}$ to be

1
1
2
2

then I apply the normal equation $x =(A^TA)^{-1} \, A^T b$ to solve for least square solutions

I got $\mathbf{x}$ =

0
0.4
0.5

which give me $a = 0, b = 0.4, c = 0.5$

my friend went through similar process and get $a = 0.5, b= 0.4, c = 0$

which should be the right solutions for this problem ?

was it me that getting the reverse answer? or was my answer correct?

thank in advance

Most likely, you set up the quadratic as $ax^2+bx+c$, and your buddy set it up as $a+bx+cx^2$. Either way is OK, so long as when you get to the end you both remember which way you set up your quadratic.

Just a small note, "the best parabola" may be ambiguous term.* If the linear regression is specified in the task, it's clear. However, the metric of squares would suit me more to take the Euclidean distance of the points from parabola. See the code:

%% given points
x = [1 2 3 4];
y = [1 1 2 2];
N = length(x);

%% optimisation function 1
S1 = @(a) 0;
for k1 = 1:N
 S1 = @(a) S1(a) + dist2(x(k1),y(k1),a);
end

%% optimisation function 2
S2 = @(a) 0;
for k1 = 1:N
 S2 = @(a) S2(a) + (y(k1) - fun(x(k1),a(1),a(2),a(3)) ).^2;
end

a1 = fminsearch(S1,[0 0 0]);
a2 = fminsearch(S2,[0 0 0]);
%% compare optimising functions
f1S1=S1(a1)
f1S2=S1(a2)
f1S2=S2(a1)
f2S2=S2(a2)

%% plot
xgrid = linspace(0,10,100);
plot(xgrid,fun(xgrid,a1(1),a1(2),a1(3))); hold on;
plot(xgrid,fun(xgrid,a2(1),a2(2),a2(3))); hold on;
scatter(x,y);

%% functions
function f=dist2(x,y,a)
 g = @(t) (x-t).^2 + (y-fun(t,a(1),a(2),a(3))).^2;
 f = fminsearch(g,0);
 f = g(f);
end

function f=fun(x,a,b,c)
 f = a*x.^2 + b*x +c;
end

In this case, I would even go further to use https://en.wikipedia.org/wiki/Parabola#General_case , but there are 5 parameters in a general parabola and only four points. So, there would be solutions including all the points on the parabola.

*And this nuance is definitely not just a mathematical prudence. Imagine for example, that you measured points in a 2D space and you know that your physical trajectory is a parabola. in this case the weight of both coordinates x and y is completely equivalent. I am almost sure that it's not OP's intent in this case. However in these kind of problems, where we usually analyse measured data, it may be good to keep in mind.


Why base of a logrithim function must be greater than one? I'm in domain section of my textbook. It says that for logarithm functions, the base must be greater than one.
I can understand why base shouldn't be one but what is problem with negative numbers?

for example, What's wrong with $\log_{-2} x$  ?

First off, there is an error in the textbook, as you can have the base of the logarithm less than 1 and greater than 0. The based of the real logarithm is generally $b\in(0,\infty),b\neq1$. As for negative bases, we experience some weird problems and inconsistencies. For example, it would be logical that
$$(-2)^2=4\implies\log_{-2}(4)=2$$
However, using the change of base identity,
$$2=\log_{-2}(4)=\frac{\log_2(4)}{\log_{2}(-2)}\implies\log_{2}(-2)=1\implies2^1=-2$$
which is obviously a contradiction.

Furthermore, we lose some pleasant properties such as continuity in the real plane. For example, it might be logical to say that
$$(-27)^{\frac13}=-3\implies\log_{-27}(-3)=\frac13$$
but an approximation to $\frac13$ to $n$ decimal places is $\frac{10^n-1}{3\cdot 10^n}$ (you should be able to see why), and
$$(-27)^{\frac{10^n-1}{3\cdot 10^n}}=\left((-27)^{10^n-1}\right)^{\frac1{3\cdot10^n-1}}\notin\mathbb{R}$$
since $10^n-1$ is odd, $3\cdot10^n$ is even, an odd power of a negative number is negative, and an even root of a negative number is not real. So while an exponent of $\frac13$ has a well-defined real root, the area close to it does not. It is similar with other powers and negative bases, and you end with a disjointed set of points on the real axis. Since the logarithm is the inverse of multiplication, we similarly get a mess of single points where it is defined.

So to preserve our log laws and maintain a smooth graph, it is ideal to restrict the base of the logarithm to avoid negative values.

In principle there is nothing wrong with this, it is an analytical function on the complex plane. However for real numbers you will run into difficulties, for example what is $\log_{-2}(2)$? Certainly not real.

A more interesting question may be why cant the base be in the interval $(0,1)$? You should find that if we extend our definition to include this interval then $\log_{1/b} = - \log_b$, so often we would just use the basis that is in the interval $(1,\infty)$. However, as  has been pointed out in the comments, this is no reason to remove $(0,1)$ from the definition and so it is ok to use if you genuinely feel it is best.


Why should $b$ groups of $a$ apples be the same as $a$ groups of $b$ apples? Why should $b$ groups of $a$ apples be the same as $a$ groups of $b$ apples?

We where taught this so it seems rather trivial but the more I think about it the more I feel that it is not.

I'm trying to avoid an argument that uses the fact that multiplication is commutative. Because I see that I am trying to PROVE that in $\mathbb{Z}^{+}-0$ multiplication is commutative if we define multiplication by repeated addition.

I would accept arguments using the fact that:

$a+b=b+a$ because if we define $+$ to be the operation combining to quantities then it should be rather trivial that $a$ apples and $b$ apples is the same as $b$ apples and $a$ apples.

Is it enough to draw $a$ groups of $b$ (1 by 1) squares and rotate this to show that it is the same as $b$ groups of $a$ (1 by 1) squares. It does not seem good enough for me because it uses a picture, and I was taught before that pictures in math do not prove anything.

The usual way of proving it is by induction.

First, define multiplication recursively: $$a\cdot 1 = a\\a\cdot(b+1)=a\cdot b + a$$

Next, show, by induction on $b$ that $1\cdot b=b\cdot 1$. That's relatively easy to do.

Next, prove by induction on $b$ that that $(a+1)\cdot b = a\cdot b + b$ by induction on $b$.

Finally, prove by induction on $a$ that for all $b$, $a\cdot b = b\cdot a$.

Multiplication is commutative because a collection of objects that has $a$ groups of $b$ objects in each group can be partitioned also into $b$ groups of $a$ objects. This is the equivalent to repeated addition:
$a+a+...+a$  ($b$ times) is the same as
$b+b+...+b$  ($a$ times).


Can A and B be independent if A is a subset of B? (except P(B)=1 and P(A)=0) I have a question about the independence of events.

Let's say I have positive integers from 1 to 10. Event A is integers below 2 and event B is integers below 3, then P(A)=0.2, P(B)=0.3 and A is a subset of B.

I am wondering whether A and B are independent. The probability that one event (P(A)) occurs in no way affects the probability of the other event occurring (P(B)) here. So from the definition of the independent events, A and B are independent. But it looks like there is some dependency between A and B because A is a subset of B. A always occurs where B occurs.

Any suggestions are appreciated.

Assuming independence and $A\subset B$ yields
$$p(A)p(B)=p(A\cap B)=p(A)$$
This implies $p(B)=1$ or $p(A)=0$.

If we have events $A, B$ s.t. $A \subseteq B$ they are independent iff

$$P(A)P(B) = P(A \cap B)$$

$$\iff P(A)P(B) = P(A)$$

$$\iff P(A) = 0 \ \color{red}{\text{or}} \ P(B) = 1$$

Hence, we need not have

$$P(A) = 0 \ \color{red}{\text{and}} \ P(B) = 1$$


What is $i^i$ and $(i^i)^i$ What is $i^i$ and $(i^i)^i$. I fond some answers online but is there a clearer explanation of this? Maybe something help me to understand.

In general, complex power of $a^b$ $(a\ne 0)$ is defined as
$$
a^b = e^{b\log a}.
$$
Thus
$$
i^i = e^{i\log i}=e^{i(\operatorname{Ln}1+i(\frac{\pi}{2}+2k\pi))}=e^{-\frac{\pi}{2} -2k\pi}
$$
for $k\in\mathbb{Z}$ and
$$
(i^i)^i =\left(e^{-\frac{\pi}{2} -2k\pi}\right)^i=e^{i\log\left(e^{-\frac{\pi}{2} -2k\pi}\right)}=e^{i(-\frac{\pi}{2}-2k\pi+i2k'\pi)}=-ie^{-2k'\pi}
$$
for $k'\in\mathbb{Z}$.

Start by representing $i$ in exponential form via the Euler relation:
$$i = e^{i\frac{\pi}{2}}$$
Then combine exponents to get the answer:
$$i^i = e^{i * i\frac{\pi}{2}} = e^{-\frac{\pi}{2}}$$
To get to your second term, combine exponents and then take advantage of the fact that powers of $i$ form a group with integer exponents modulo 4:
$$(i^i)^i = (e^{-\frac{\pi}{2}})^i = e^{-i\frac{\pi}{2}} = i^{-1} = i^{4-1} = i^3 = -i$$
I hope this helps.


$g \in L^q(F)$ if for all $f \in L^p(F)$, $f(\cdot)g(\cdot)$ integrable? Here's a problem I'm struggling to prove.

Suppose $g(x)$ a measurable function on a measurable set $F$.
And for all $f \in L^p(F)\ (1\le p < \infty)$, $f(\cdot)g(\cdot)$ is integrable. How can we prove that $g \in L^q(F)$ ? Here $\frac 1 p + \frac 1 q = 1$.

My thoughts are :

Let $g_n(x) = g(x)$ for all $x$ where $\vert g(x) \vert \le n$. Then we can show that $g_n(x) \in L^q(F)$, $g_n$ corresponds to a functional on $L_p(F)$, say $F_n$ , and $\Vert F_n \Vert$ is exactly $\Vert g_n \Vert_{L^q}$.

With  $F_n(f) = \int f\cdot g_n\ d\mu \rightarrow \int f\cdot g < \infty$ for all $f \in L^p(F)$, the Uniform-Bounded-Principle tells us that $F_n$ is uniformly bounded(and also $\Vert g_n \Vert$).

Here is the question in my proof. Can we conclude that $\int \vert g\vert ^q d\mu = \lim \int \vert g_n \vert ^q d\mu < \infty$ ? (I don't think we can use the Dominated Convergence Theorem here since we don't know yet if $\vert g \vert ^q$ is integrable.

(If the proof above is totally in the wrong direction, please feel free to point out.)

This is not true in general, we need some additional assumption like $\sigma$-finiteness for example.

For a counterexample in the general case consider $F = \{0,1\}$ equipped with the $\sigma$-algebra $\mathcal P(F)$ and a measure $\mu$ with $\mu(0) = 0, \mu(1) = \infty$. If we define $g = \chi_{\{1\}}$, where $\chi_A$ denotes the indicator function of $A$, then clearly $g\notin L^q$ if $q<\infty$.
However, if $f\in L^p$, then $f(1) = 0$ and therefore $fg = 0$ is integrable.

Your attempted proof for the case $p>1$ is almost good, it just needs a slight refinement.

If we consider $\sigma$-finiteness (although a weaker assumption might be enough aswell) we find a sequence $E_n$ of measurable sets with finite measure and $E_n\nearrow F$. Let us further assume that $g\geq 0$.1 Let now $g_n := \min\{n, g\}\cdot \chi_{E_n}$. For these $g_n$ it is clear that $g_n\in L^q$, since $$\int |g_n|^q \leq \int_{E_n} n^q = n^q\mu(E_n)<\infty.$$
For the ones defined in your proof this is not so clear. Now you conclude that $\int |g_n|^q$ are uniformly bounded as in your question. But also $g_n\to g$ monotonously(!), so the theorem of monotone convergence yields $\int|g_n|^q\to \int|g|^q$.

For the case $p=1$ let $T:L^1\to \mathbb C, f\mapsto \int fg$.
The uniform boundedness principle yields $\|T\|<\infty$ with the same proof as in the case $p>1$. Assume that $\mu\{g>\|T\|\} > 0$. By $\sigma-$finiteness, we find $A\subset \{g>\|T\|\}$ with $0<\mu(A)<\infty$.
But now set $f := \frac{1}{\mu(A)}\chi_A\in L^1$. Then $\|f\|_1 = 1$ and we conclude

$$\|T\| = \frac{1}{\mu(A)}\int \|T\|\chi_A < \frac{1}{\mu(A)}\int g\chi_A = \int gf \leq \|T\|\|f\| = \|T\|,$$ which is a contradiction and therefore shows $\|g\|_\infty \leq \|T\|<\infty$.

1Otherwise we pass to $|g|$ instead of $g$.

Since $L(f) = \int gf$ is a functional on $L^p$ by hypothesis, it must be that $L(f) = \int hf$ for some $h\in (L^p)^* = L^{q}$ where $p^{-1} + q^{-1} = 1$ by the Riesz representation theorem. Then you have $\int (g-h)f = 0$ for all $f\in L^p$, implying $g = h$ almost-everywhere.


Cover a polygon with polygons Besides right angled triangles, is there any polygon I could use to cover any given (regular or not) polygon? It's clear that given a triangle, square, hexagon or rectangle you would other options. Can this be proven, if right? That is, that the right angled triangle is the only polygon able to $100\%$ cover any other polygon.

You'll want to look at Laczkovich, M. "Tilings of Polygons with Similar Triangles." Combinatorica 10, 281-306, 1990.

46 triangles of type $45^\circ-60^\circ-75^\circ$ can perfectly cover a square.

In FEA (Finite Element Analysis), meshing (specially triangle meshing) is required to perform the analysis.

Perhaps some insights of your proof can be found in Wikipedia, Types of Mesh: https://en.wikipedia.org/wiki/Types_of_mesh


How can be 1 is equal to 2? It may be a silly question. But I don't know it. So I'm questioning. Recently I've got a proof that proves 1=2.
Is there any fault in the proof? If so then what is the fault??

The problem is here:

$a^2=b^2 \Rightarrow a=b \underbrace{\vee}_{\text{or}} a=-b$.

In this case, only the second one counts:

$(1-\frac32)^2=(2-\frac32)^2 \Rightarrow \require{cancel} \cancel{1-\frac32=2-\frac32} \vee 1-\frac32=-2+\frac32$, but the first one gives a contradiction, so it doesn't count.


If I can't use Alt. Series Conv. test, how do I determine if it converges or diverges? Stumped on this absolute convergence problem!
(Converge conditionally, absolutely, or diverges)

$$\sum_{n=1}^\infty \frac{(-1)^n n^2}{(n+3)^2} $$

First, I tested for absolute convergence, and the fraction reduced to $\frac{n^2}{n^2}=1$   .....So, it does not converge absolutely.

Next, to use the alternating series test on the original series, you need to make sure the magnitude of the terms are decreasing.  Well, they are not:

$$\sum_{n=1}^\infty \frac{(-1)^n n^2}{(n+3)^2} = -\frac{1}{16} + \frac{4}{25} - \frac{9}{36} + .....$$

So, without using the alternating series test, how do I determine if the orig. series converges or diverges?

Indeed, Clement's comment below is correct.  This sum is not convergent and has no well-defined value!

The method below is akin to resummation but still very problematic.  For this sum, the Cesaro sum converges to 0 and Abel's resummation method gives $\frac{1}{8} \left(-26+6 \pi ^2-48 \log (2)\right)\approx -0.00667$.  Both differ from the value computed below.

-- I'm leaving the equation below as a reference for Clement's comment, Beware --

One approach is to combine odd and even terms.  This gives a sequence with positive terms, for which convergence is easy to determine.  Thus, we have
$$ \sum_{n=1}^\infty \frac{(-1)^n n^2}{(n+3)^2} = \sum_{k=0}^\infty \left( -\frac{(2k+1)^2}{(2k+4)^2} + \frac{(2k+2)^2}{(2k+5)^2} \right) = \sum_{k=0}^\infty \frac{39+72k+24 k^2}{4(k+2)^2(2k+5)^2} = \frac{3 \pi^2 - 12 \ln (4)-11}{4} \approx 0.493 $$


How to determine the existence of all subsets of a set? Given

The definition of subset;
The axiom of power set: for any set $S$, there exists a set $\wp$ such that $X \in \wp$ if and only if $X\subseteq S$

we know what a subset is and what a power set contains.

In a simple case where a set $A$ is supposed to exist, with $A=\{a, b, c\}$, we know what is and what is not a subset of $A$:

$\{a\}, \{b\}, \{c\}, \{a, b\}, \{a,c\}, \emptyset$ and $A$ are subsets of $A$ and anything different is not.

  $\wp(A)=\big\{\{a\}, \{b\}, \{c\}, \{a, b\}, \{b,c\}, \{a,c\}, \{a, b, c\}, \emptyset\big\}$.

However the mere definition of something (and consequently it's recognition as such) does not guarantee its existence. $\emptyset$ and $A$ seem like the only subsets whose existence is immediate.

In other words, I know what a power set contains, but how do I know that the things it contains exist in the first place?

Because such a well-defined and existent set such as $\wp(A)$ should not contain nonexistent elements, to prove the existence of its elements is important. It seems that two alternatives arise:

Being a member of $\wp(A)$ automatically makes this thing to exist;

or

There should be an alternative to prove the existence of all subsets of $A$ without the axiom of power set.

The power set axiom just tells you what it says: for every $A$, there exists a set $\mathcal{P}(A)$ such that
$$
\text{for all $B$, $B\in\mathcal{P}(A)$ if and only if $B\subseteq A$}
$$
There is no claim of “existence” of any particular subset of $A$. In $\mathsf{ZFC}$ one can show that $|\mathcal{P}(A)|>|A|$, so there is plenty of subsets.

It should be noted that, if $A$ is infinite, there is no hope to find, for each subset of $A$, a formula “describing it”, because $\mathcal{P}(A)$ is uncountable. This is however not a problem: the axiom tells you that you have a “container” for all subsets of $A$; when you prove that a set $B$ is a subset of $A$, then you know it belongs to $\mathcal{P}(A)$; and conversely, if you pick $B\in\mathcal{P}(A)$, you know $B\subseteq A$.

The real purpose of the axiom is that the subsets of a set form a set. In particular, for instance, the equivalence relations on a set form a set that can be isolated from $\mathcal{P}(A\times A)$ using a suitable predicate and the axiom of separation.

I remember some good notes about this in Paul J. Cohen's “Set theory and the continuum hypothesis”.

The most relevant axiom is actually the one mentioned by MathematicsStudent1122 and MauroALLEGRANZA, namely the Axiom of Separation (sometimes called specification).  Whenever you have a property expressed for example by a formula, this axiom guarantees that there is a subset whose members are precisely those entities that have the property or satisfy the formula.

For example, you can write down the property of a natural number $n$ that $(\exists m\in\mathbb{N})\,\big(n=2m\big)$ meaning that $n$ is an even integer.  But how do you know that the set of even integers exists? That's the role of the separation axiom.

In certain interesting mathematical theories, you can have predicates that do not satisfy separation. One such theory is Nelson's.


Strictly increasing and strictly decreasing derivatives How do I prove this? We don't have that strictly increasing implies $f' > 0$, but we do have the converse; and, things like Rolle's Theorem and statements of the mean value theorem, and that $f'$ exists implies that $f$ is continuous:

  Suppose $f'$ exists on $(a, b)$ and $f'(x) \neq 0$ on $(a, b)$. Prove
  that either $f'(x) > 0$ for all $x \in (a, b)$ or $f'(x) < 0$.

My attempts don't seem rigorous.

I'm assuming the conclusion is supposed to be that $f'(x)$ is either always positive or always negative (rather than $f(x)$), since the conclusion is false for $f(x)$ (If it were true for a function $f$, then doing a suitable vertical translation you would be able to get a function which is sometimes positive and sometimes negative on $(a,b)$, without changing the derivative.

Assume there are points $d,e$ with$ a\lt d\lt e\lt b$ and $f'(d)\lt 0 \lt f'(e)$.

By the Extreme Value Theorem, $f(x)$ achieves a maximum and a minimum on $[d,e]$, and this minimum is achieved at a critical point or at an endpoint. By hypothesis, there are no critical points, so the minimum must be achieved at $d$ or at $e$.

Show that the fact that $f'(d)\lt 0$ implies that $f$ cannot achieve its minimum over $[d,e]$ at $d$. One can do this by using the definition of derivative as a limit.

Added. Here is how one can prove this without trying to show that $f$ is decreasing on a neighborhood of $d$: by definition,
$$\lim_{h\to 0}\frac{f(d+h)-f(d)}{h} = f'(d)\lt 0.$$
In particular,
$$\lim_{h\to 0^+}\frac{f(d+h)-f(d)}{h} = f'(d)\lt 0.$$
By definition of the limit, for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that if $0\lt h\lt \delta$, then $\left|\frac{f(d+h)-f(d)}{h} - f'(d)\right|\lt \epsilon$. Take $\epsilon= |f'(d)/2$, and let $\delta_0$ be the corresponding $\delta$. Then for any $h$, $0\lt h\lt \delta_0$ we have
$$\left|\frac{f(d+h)-f(d)}{h} - f'(d)\right| \lt \frac{|f'(d)|}{2}.$$
From this, we conclude that $\frac{f(d+h)-f(d)}{h}\lt 0$. Since $h\gt 0$, that means that $f(d+h)-f(d)\lt 0$, or $f(d+h)\lt f(d)$. This holds for all $h$ with $0\lt h\lt \delta_00$. That means that $f(d)$ is strictly larger than all values of $f$ on $(d,d+\delta_0)$, so $f(d)$ cannot be the minimum on $[d,d+\delta_0)$, and so cannot be the minimum on $[d,e]$.

Likewise, show that the fact that $f'(e)\gt 0$ implies that $f$ cannot achieve its minimum over $[d,e]$ at $e$. This contradiction arises from the assumption that we can find points $d$ and $e$ with $f'(d)\lt 0 \lt f'(e)$. Replace $f$ with $-f$ to show you cannot find points $d$ and $e$ with $f'(d)\gt 0 \gt f'(e)$ either.

Use  intermediate value property  for  derivative. If $f'$ is negative at  one point  and  positive  at  another,  then at  some  point  in between $f'$ will  take  the  value $0$ which  contradicts  the  hypothesis.


How to show an aymptotic expansion is uniformly valid? I have an equation
$$
  nt = u - \epsilon \sin(u)
$$
which asks for the first four terms in the asymptotic solution.
Hence if the solution is $u_0 + \epsilon u_1 + \cdots.$, expand $\sin(u)$ around $u_0$,
$$\sin(u) = \sin(u_0) + \cos(u_0)(\epsilon u_1 + \cdots)  - \frac{\sin(u_0)(\epsilon u_1 + \cdots)^2 }2- \frac{\cos(u_0)(\epsilon u_1 + \cdots)^3}6 + \cdots.$$
By collecting terms in different orders of $\epsilon$, I got the following equations
$$u_0 = nt$$
$$u_1 = \sin(u_0) = \sin(nt)$$
$$u_2 = \sin(nt)\cos(nt)$$
$$u_3 = \sin(nt)\cos^2(nt) - \frac{1}{2}\sin^3(nt)$$
The question is after getting these terms, how do I show "the four term expansion is uniformly valid for all $nt$".  I am thinking the calculation itself already shows, but there is a separate question asking for that.  I am not sure what needs to be done here.

A series $\sum_{m=0}^{\infty} x^m u_m(t)$ is asymptotic to $u(t)$ as $x\rightarrow 0$ uniformly in $t$ if for any $\epsilon>0$ there exists $\delta$ independent of $x$ such that
\begin{align}
|x|<& \delta& \Rightarrow&&
\left| u(t) - \sum_{m=0}^M x^m u_m(t)\right| <& \epsilon |x^M| &
\text{ for all } M=0,1,2,\dots.
\end{align}
This is probably equivalent to that
\begin{align}
\left|\frac{u_{m+1}(t)}{u_m(t)}\right| <& \infty &
\text{ for all } t, &&
\text{ for all } m=&0,1,2,\dots.
\end{align}
I think you can check it as you have the expressions for $\{u_m(t)\}$.


Showing that a group is lattice-ordered Say I have a set $S$ with a group operation $\cdot$ and lattice ordering $\leq$. Suppose further that:

$x\leq 1\implies xy\leq y$
$x\geq 1\implies xy\geq y$

For all $x,y$.

Does it follow that $(S,\cdot,\leq)$ is a lattice ordered group? i.e. does it hold that $(x\vee y)z=xz\vee yz$ and $(x\wedge y)z=xz\wedge yz$ for all $x,y,z$?

I think this is true because I can't think of any counter examples, but I'm having trouble proving it. Any hints?

EDIT: This was also asked on Math Overflow

If it is a partially ordered group then it is also a lattice ordered group. The usual way is to prove
$$a≤b⇒xay≤xby.$$

Edit: Your task is to prove that it is a partially ordered group.
Your implications are a consequence from the definition of right-ordered groups $$x≤y ⇒ xz≤yz$$.

You have either more information about your structure or you are stuck with a right-ordered group. So your question reads: Is every right-lattice-ordered group already a $\ell$-group?

So looking at this survey we know that such a group can be embedded into an $\ell$-group. As it doesn't tell anything about isomorphy neither as fact nor as open question nor as recently answered question, I guess the anwer to your question is: No.

Maybe someone who is familiar with right-ordered groups can help you.


Is the Alexandroff double circle separable? Is the Alexandroff double circle separable (i.e. has a countable dense subset)?

The Alexandroff double circle is the space with underlying set $C = C_1 \cup C_2$, where $C_i$ is the circle of radius $i$ and centre $0$ in the complex plane. The basic open sets are:

$\{z\}$ for every $z \in \mathbb{C}$ with $|z| = 2$, and
$U \cup \{ 2z : z \in U\} \setminus F$, where $U$ is open in the normal topology of $C_1$ (i.e., a union of open arcs) and $F$ is a finite subset of $C_2$.

So far, I was able to show it is compact, Hausdorff, first countable but not second countable. Is there a way to connect this information to determining if it is separable or not?

No, it's not separable. The points of $C_2$ form an uncountably infinite discrete subspace. If there were some countable dense subset $S$, there would have to be an element of $S$ in each open set $\lbrace z \rbrace$ for all $z \in C_2$. Therefore, $C_2 \subseteq S$, which contradicts $S$ being countable.


Nice problem Which is bigger $e^a$ and $a^3$ let $$a=\sum_{n=0}^{\infty}\dfrac{\left(\dfrac{n+1}{3}\right)^n}{(n+1)!}$$

My Question:

  Which is bigger $e^a$ and $a^3$

I guess

  $$e^a=a^3$$

but I can't prove it,and I think this is nice problem.Thank you evryone

This problem from this http://www.math.org.cn/forum.php?mod=viewthread&tid=3577&extra=&page=1

We can "avoid" the Lambert $W$ function by skipping directly to its series representation.  We'll call upon the Lagrange inversion formula, which can be found in my answer here.

Suppose we're looking for the smallest positive root of the equation $e^x = x^3$, which is also the smallest positive root of the equation

$$
\frac{x}{e^{x/3}} = 1.
$$

Let's call this root $a$.  The Lagrange inversion formula yields the desired series representation,

$$
\begin{align}
a &= \sum_{n=1}^{\infty} \frac{1}{n!} \left\{\left(\frac{d}{dx}\right)^{n-1} e^{nx/3} \right\}_{x=0} \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \left\{\left(\frac{n}{3}\right)^{n-1} e^{nx/3} \right\}_{x=0} \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \left(\frac{n}{3}\right)^{n-1} \\
&= \sum_{n=0}^{\infty} \frac{1}{(n+1)!} \left(\frac{n+1}{3}\right)^{n}.
\end{align}
$$

well if e^a and a^3 are graphed
the results are a=1.857 and a=4.5364
lim a=>-inf e^a is 0 and lim a=>-inf a^3 is -inf
 lim a=>inf e^a/a^3
lim a=>inf e^a/3a^2
lim a=>inf e^a/6a
lim a>inf e^a/6 is inf by L'hopital rule
so e^a will be greater than a^3 from interval (-inf,1.857)u(4.53864,inf)


Sine and cosine graph transformation I'm having some difficulties with this question

A bike is on a stand such that the highest point of the back wheel is 47 inches above the ground. If the pedal is turned counter clockwise, the back wheel turns CCW also. The tire valve on the back wheel is 13 inches from the wheel's center of rotation.

now I need to define a function that gives the height of the tire valve

1) above the floor, if both the pedal and the tire valve start at 3 o'clock, and one revolution of the pedal results in one revolution of the tire valve.

2) above the tire's horizontal diameter, if the tire valve starts 3π/4 radians clockwise from 3 o'clock, and one revolution of the pedal results in one revolution of the tire.

3) above the floor, if both the pedal and the tire valve start at 3 o'clock, and 2 revolutions of the pedal results in 5 revolutions of the tire valve.

May be you are referring to a Cycloid as valve locus. R=13.

$$ x = R( \theta-\sin \theta) , y= R ( 1- \cos \theta). $$

if the input is angle measure a, the number of radians swept by the pedal:

1) What I know so far: measured in inches, f(a)=13sin(a)+?
the floor to the tire bottom is 34"
the floor to the horizontal diameter is 34"+13/2"
I am unsure if ? is 34 or 40.5

2) measured in radians, g(a)=sin(a-3pi/4)
This answer is confirmed by my math homework

3) my guess is that the argument is 5/2*a


Prove $\ker(T) = \ker(T^2)$ Let $V$ be a vector space over R such that $\dim(V) =3$ and let $T : V → V$ be a
linear transformation which satisfies $T^3 = 0$ and $T^2 \ne 0$. Prove:

(i) $\dim(\ker(T)) = 1$

In my proof I claim that $$T^2x = 0, \forall x \in V$$

Does this imply that $$Tx = 0, \forall x \in V$$

Beacuse $$T^2 \neq 0$$

Your title and seems to ask a different question than your problem statement. In general, $ker(T) \neq Ker(T^2)$. This is easily shown with a counter-example.
Consider the nilpotent matrix $A$:

\begin{bmatrix}
        0 & 2 & 1 \\
        0 & 0 & 1 \\
        0 & 0 & 0 \\
        \end{bmatrix}

$A^2$ = \begin{bmatrix}
        0 & 0 & 2 \\
        0 & 0 & 0 \\
        0 & 0 & 0 \\
        \end{bmatrix}
$A^3 =$\begin{bmatrix}
        0 & 0 & 0 \\
        0 & 0 & 0 \\
        0 & 0 & 0 \\
        \end{bmatrix}
So all of the parameters of your problem are met. However, the kernel of $A$ is the span of:\begin{bmatrix}
        1 \\
        0 \\
        0 \\
        \end{bmatrix}
and the kernel $A^2$ is the span of:
\begin{bmatrix}
        1 \\
        0 \\
        0 \\
        \end{bmatrix} AND \begin{bmatrix}
        0 \\
        1 \\
        0 \\
        \end{bmatrix}

This is made more clear by considering the fact that the kernel is the same as the set of eigenvectors with associated eigenvalue $\lambda = 0$. The eigenvalues for a nilpotent matrix must be $0$ for all eigenvectors, therefore the kernel of a nilpotent matrix is comprised of all eigenvectors. This can be easily proven:

$Av = \lambda v$

$A^2v = \lambda^2 v$

$A^3v = \lambda^3 v$

but since $A$ is nilpotent of degree $3$, $A^3v = 0$, and so $\lambda^3v = 0$, which is only (non-trivially) zero if $\lambda = 0$. For the same reason, it is clear that if $v$ is an eigenvector of $A$, it is also an eigenvector of $A^2$. However, this argument does not run both ways. You cannot say that an eigenvector of $A^2$ is necessarily an eigenvector of $A$.

The statement that $T^2(x) = 0$ implies $T(x)=0$ is false, as nilpotent transformations are not isomorphic. I.e., there can be no inverse transformation for:

$T^{-1}(T^2(x)) = T^{-1}(0) = 0 = T(x)$.

To prove the question asked in your problem statement, you need to consider the fact that, for an $n$ dimensional nilpotent matrix, $dim(ker(T^k)) < dim(ker(T^{k+1}))$ for $k \le n$. In this problem, $dim(ker(T^3)) = 3$, as any $3$-dimensional vector mapped by $T^3$ is the zero vector. This consequently forces $dim(ker(T)) = 1$

In addition to @Bungo's observations in the comments,
$$  T = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}  $$
has the property that $T x \neq 0, \forall x$ and $T^2 x = 0, \forall x$, so the implication you ask about does not hold.


If $0^0 = 1$, then is it true that $0/0 = 1$? By Knuth, Concrete Mathematics (2nd ed.) page 162, it is convenient that

$$0^0 = 1$$

Then, is it true that

$$0^1/0^1 = 0^{1-1}= 0^0 = 1$$ and we are free to exclude indeterminate statement of $0/0$ ?

There's a good reason why it makes more sense to define $0^0$ as $1$ than it does to define $0/0$ as $1$. Notice that $0=\alpha\cdot 0$ for any real number $\alpha$. Substituting this in $\frac00=1$ would give $1=\frac00=\frac{\alpha\cdot 0}{0}=\alpha$, so this convention would not be consistent.

Conversely, the convention that $0^0=1$ is consistent, at least by this measure, because $0^{\alpha\cdot 0}=(0^0)^\alpha=1^\alpha=1$, and $(\alpha\cdot 0)^0=\alpha^0\cdot0^0=1\cdot 1=1$.

Assuming that $0^0=1$ is wrong because, if you take natural logarithms both sides, this will mean $$0=\ln (1) =\ln(0^0)=0 \cdot \ln (0)=0 \cdot \lim_{x \to 0^{+}} \ln (x)=0 \cdot (-\infty) $$ which is a contradiction, because $0 \cdot (-\infty)$ is an indeterminate (an operation that is meaningless.)

Similarly, suppose $\frac{0}{0}=1$ as evaluated by yourself, then that will mean
$$\infty -\infty = \ln \left( \frac{0}{0}\right)=\lim_{x \to 0} \ln (x) -\ln(x)=\ln(1)=0$$ which is also a contradiction.

Let me supply an example, we know that $0.0000001 \approx 0 $ and $0.01 \approx 0$, then $$\frac{0}{0} \approx \frac{0.01}{0.0000001}=100000>>1$$

Let me give another example, since the symbol $\infty$ means a biggest number, I can consider the number 120 to be my $\infty$ just like in a T-distribution table so that $$\infty-\infty=125-120=5 >0$$ This is the reason why these operations are called indeterminate.


What is the geometric interpretation of the constant c/||n||? (vectors) [See part (e) of the attache image]

Part (e) of the question attached: Find the geometric interpretation of the constant c/||n||, when both sides of the cartesian equation x•n=c is divided by the norm ||n|| of the vector n.

I understand how they obtained the equivalent Cartesian equation but am struggling to understand the geometric interpretation of the constant (on the RHS).

Question Part (e)

Hopefully you get to see this before your class, but try multiplying the dot product so you get to a Cartesian description of a line. From that, it should be easy to see what the constant means.


Convergence and Absolute Convergence of Arithmetic Mean of a sequence Suppose $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n |x_i|$ exists. Does $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i$ exist?

How about the converse?

My thoughts:

I guess for the sequence $\{x_1,-x_1,x_2,-x_2,\ldots\}$ the converse doesn't necessarily hold.
If I can show that the existence of $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i$ implies that $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n x_i 1\{x_i\geq 0\}$ and $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n |x_i| 1\{x_i\leq 0\}$ exist, then the first part would hold. But I'm not sure this is true, I need to find a counterexample.

A counterexample for the converse is $x_i = (-1)^i\sqrt i$.

A bit of an explanation of Cla's hint: Consider
$$
S^1_n = \frac{x_1 + \ldots x_n}{n} < \bigg| \frac{x_1 +\ldots x_n}{n} \bigg| <\frac{|x_1| +\ldots |x_n|}{n} = S^2_{n}
$$
by triangle inequality. Since we know that $S^2_n \to_n a< \infty$ and $S^1_n < S^2_n$, $S^1_n$ converges by comparison test.


Do coalgebras arise outside the study of bi/Hopf-algebras? Hopefully the title is fairly self explanatory.  I'm curious as to whether the coalgebra structure (that is, a vector space with a comultiplication and counit) comes up an any area of mathematics not specifically because it is the arrow reversal of a unital algebra.

If $X$ is a set, then the canonical duplication $X \to X \times X$ map gives a coalgebra on $\mathbb{K}[X]$, the vector space with basis $X$. This is because $\mathbb{K}[X \times X] \cong \mathbb{K}[X] \otimes \mathbb{K}[X]$.

The induced comultiplication $\mathbb{K}[X] \to \mathbb{K}[X] \otimes \mathbb{K}[X]$ can be used to define a Hopf-algebra structure on $\mathbb{K}[G]$ for $G$ a finite group.

There is also another coalgebra on $\mathbb{C}[G]$ the group ring.

$\delta : \mathbb{C}[G] \to \mathbb{C}[G] \otimes \mathbb{C}[G]$ "unmultiplies" an element:

$\delta (g) = \frac{1}{|G|}\sum_{h \in G} gh^{-1} \otimes h$

Here are two other examples where one is lead to consider coalgebras by themselves, that are not necessarily Hopf algebras.

The singular chain complex $\mathcal{S}_*(X)$ of a topological space gas a coalgebra structure, by the Eilenberg–Zilber theorem. This is also an instance of the codiagonal $X \to X \times X$ appearing.  If you work with field coefficients, this gives a coalgebra structure on the homology $H_*(X; \Bbbk)$ of the space. (The field coefficients are here to deal with the fact that the Künneth morphism goes the wrong way; see this MO Q&A for more details). If you want to get a Hopf algebra structure on homology you probably, need some kind of H-space structure on $X$.

In homological algebra, one is often interested in constructing resolution of algebras. Given an (augmented) algebra $A$, its bar construction $BA = (T^c(\Sigma A), b)$ is a coalgebra. If $A$ is not commutative then there is no (to my knowledge) natural Hopf algebra structure on $BA$ to consider.
Then the cobar construction $\Omega BA$ is an algebra which is a resolution of $A$: $\Omega BA \xrightarrow{\sim} A$ (see eg. the $n$Lab article for some pointers).

Given a quadratic algebra $A = T(V)/(S)$, one can first consider its Koszul dual algebra $A^! = T(V^*)/(S^\perp)$, but it's also interesting to consider its Koszul dual coalgebra $A^¡ = T^c(V, S)$, cogenerated(*) by $V$ with corelations $S$. There is a natural map of algebras $\Omega A^¡ \to A$, and this map is a resolution when the algebra is Koszul. In this case it gives a very interesting, "small" resolution of $A$.

More generally, you can consider quadratic operads (in some sense, operads are generalizations of unital algebras). Given such a quadratic operad $\mathtt{P}$, you can construct its Koszul dual cooperad $\mathtt{P}^¡$ (in the same sense a cooperad is a generalization of a coalgebra). When the operad is Koszul you get a resolution $\Omega \mathtt{P}^¡ \xrightarrow{\sim} \mathtt{P}$. See eg. the book Algebraic Operads of Loday and Vallette.
Even more generally, you can consider monogenic algebras over a quadratic operad, and this leads one to consider Koszul dual coalgebras over some quadratic cooperad. See Millès, "The Koszul complex is the cotangent complex". So for example you would see Lie coalgebras, cocommutative coalgebras, Poisson coalgebras... appear. In this case it might not even be necessarily clear what the correct notion of "Hopf algebra" would be.

(*) There are actually suspensions involved but I don't want to get bogged down in technical details. In fact most of the last paragraph is very imprecise.


Using Rules of Inference to imply conclusion This is Discrete Math.  I just learned how to use the rules of inference and I'm not sure if im doing this right at all.  The question is:

Use rules of inference to show that the hypotheses "If it does not rain or if it is not foggy, then the sailing race will be held and the lifesaving demonstration will go on."  "If the sailing race is held, then the trophy will be awarded," and "The trophy was not awarded" imply the conclusion "It rained."

Here is an outline.

"Trophy not awarded" and "If sailing race held, then trophy will be awarded" $\implies$ $A$

(What is $A$?)

$A$ and "If it does not rain or if it is not foggy, then the sailing race will be held" $\implies $B

(What is $B$?)

Let $p$ be the proposition "it rain", $q$ be "it is foggy", $r$ be "sailing race is held" , $s$ be "the lifesaving demonstration go on" and $w$ be "the trophy is awarded" .

Then the argument form will be:
\begin{align*}
&(\neg p\vee \neg q)\to (r\wedge s) \\
&s\to w\\
&\neg w\\
&\hline{\therefore\neg p}\\
&
\end{align*}
Now the first premises can be simplified using  De Morgan's, implication and simplification i.e.
\begin{equation*}
\begin{split}
(\neg p\vee \neg q)\to (r\wedge s) & \equiv \neg (p\wedge q)\to (r\wedge s) \quad \text{by De Morgan's rule}\\
& \equiv (p\wedge q)\vee (r\wedge s) \quad \text{by De Implication rule}\\
&  \equiv p\vee s \quad \text{by Simplification rule}
\end{split}
\end{equation*}
The second and third premises imply $\neg s$ by $\textit{disjunctive syllogism}$ rule. Then we have
\begin{align*}
&p\vee s \\
&\neg s\\
&\hline{\therefore\neg p}\\
&
\end{align*}
The conclusion also follows by disjunctive syllogism.


Is there a book on proofs with solutions? I am a biochemistry graduate student who works on cancer. I am interested in learning proofs as a personal interest. I use math as a tool, but would like to start building a deeper understanding on my own. I am not taking any course. Hence, I am looking for a book with theory, exercise, and solution manual, in case I am stuck. I find this forum extremely helpful, but I would still like to have a reference. Most books that I have started looking to buy do not have a solution manual. Can anyone recommend an author? Sorry for this general question. Thank you!

EDIT:

I watched the movie Good Will Hunting, so I feel confident! lol I

In my experience you won't tend to find a book with a solution manual and if you do you won't find one that does a lot of exercises as proofs can get pretty long and tedious in a hurry. Most authors, I think, assume that you will be interacting a fair amount with your professors/other students at that point so I don't think they get written very much.

For a basic text into abstract math I would suggest my undergraduate text "Bridge to Abstract Mathematics" it does a good job of making important remarks and breaking things down. There is, no surprise, not a solution manual that I am aware of but I have a set of YouTube videos going up that go through and lecture on each chapter and do a number of the examples/problems in detail. They will start going up in a few weeks under the username superphyz and I will link them in the comments later if you comment and let me me know if that will be useful to you.

Good luck!

Well, there are lots and lots of books with solutions.

An answer to your question depends on which part of mathematics you would like to explore. For instance, a series of great ones about real analysis come from Kaczor & Nowak


Evaluate the following infinite sum using a similar convergent one $$\sum_{n=1}^\infty \frac{3n^2-4n+2}{n!}$$

I have tried rewriting: $$\frac{3n^2-4n+2}{n!} = \frac{2}{n!} - \frac{1}{(n-1)!} + \frac{3}{(n-2)!}$$

And using: $$\sum_{n=0}^\infty \frac{1}{n!} = e$$
I am also aware of the Shift Rule, but I'm not sure how to apply it here where $N$ is a negative number. I can't seem to put these all together since all my calculations get me nowhere.

Will appreciate any help/hints calculating the second and third term of the sum!

Edit: I've also tried index shifting on the second and third sums, but I end up with the sum starting at $-1$ and I'm unsure how to proceed with a factorial.

Attention please
$$ S=\sum_{k=1}^{\infty}\frac{3n^2-4n+2}{n!}= \sum_{n=1}^{\infty} \frac{3n(n-1)-n+2}{n!}=$$
$$ \Rightarrow \sum_{n=1}^{\infty} \left( \frac{3}{(n-2)!}- \frac{1}{(n-1)!} +\frac{2}{n!} \right)=\sum_{p=0}^{\infty} \frac{3}{p!}-\sum_{q=0}^{\infty} \frac{1}{q!}+ \sum_{n=1}^{\infty} \frac{2}{n!}= (3e)-(e)+2(e-1)$$ $$\Rightarrow S =4e-2.$$Here we have used $p=n-2, q=n-1$,

Please note that if $N$ is zero, a natural number or non-negative integer, $(-N)!=\pm \infty$, so it does not matter if you get $1/(-2)!$ it will be 0. That is why $p=-1$ is not cosidered as $(-1)!= \pm \infty$ and its reciprocal contributes 0.


Find all functions for which $\ x \cdot f(xy)+f(-y)=x \cdot f(x)$ Does anyone have idea for solution (for all non-zero numbers)?

$$\ x ≠ 0,y ≠ 0$$

$$\ f: R \setminus \{0\} → R$$

$$\ x \cdot f(xy)+f(-y)=x \cdot f(x)$$

Thanks!

I can solve this completely given the additional assumption that $f$ is continuously differentiable on its domain. I suspect that it's possible to relax this to merely "continuous", but you'll probably need some sort of niceness assumption to get anywhere.

If we set $x=1$, the functional equation becomes
$$
f(y)+f(-y)=f(1) \, .
$$
So the equation is equivalent to
$$
x \, f(xy)-f(y)+f(1)=x \, f(x) \, ,
$$
which I will suggestively rewrite as
$$
x[f(xy)-f(x)]=f(y)-f(1) \, .
$$

Now, restrict to the case where $y$ is positive and set $y=1+\epsilon$. Since $f$ is differentiable, we can apply the mean value theorem to find $c_\epsilon \in [x, xy]$, $d_\epsilon \in [1, y]$ such that
$$
f(xy)-f(x)=x \epsilon \, f'(c_\epsilon)\\
f(y)-f(1)= \epsilon f'(d_\epsilon)
$$
(where we take the convention that interval endpoints are commutative: that is, if $a>b$, $[a,b]=[b,a]$).

So our equation becomes
$$
x^2\epsilon f'(c_\epsilon)=\epsilon f'(d_\epsilon) \, .
$$
But $f'$ is continuous by assumption, and we have $\lim_{\epsilon \to 0} c_\epsilon=x$, $\lim_{\epsilon \to 0} d_\epsilon=1$. So, passing to the limit,
$$
x^2 f'(x)=f'(1) \, ;
$$
that is, $f'(x)=\frac{C}{x^2}$ for all $x$ and some constant $C$, and so $f(x)=\frac{C}{x}+D$ for all $x$ and some constants $C,D$.

Now, we can plug this $f$ back into the original functional equation to find which values of $C,D$ actually work. After a little bit of algebra, we see that the left-hand side of the equation reduces to $D+Dx$, while the right-hand side reduces to $C+Dx$. So we must have $C=D$.

That is, the continuously differentiable $f$ which satisfy the equation are precisely those of the form $f(x)=C(1+\frac{1}{x})$ for some constant $C$.

I guess there doesn't lies any solution to this one .
Replacing all y's by x. We get x∗f(x^2)+f(−x)=x∗f(x)....(1)in this equation if we put x=1,we get
f(-1)=0 .So let us consider there lies a function Q(x) such that , f(x)=(x+1)Q(x). Replacing the f(x) in equation (1) we get one more equation, if we consider the lading coefficient of Q(x) be a1, and it be of degree n.....we see that n=0.
so there doesn't lie a function as Q(x). we get f(x)=(x+1);which actually does't satisfy equation 1.
so we conclude that f(x) is a constant function and is identically equal to zero.

ALTERNATIVELY
considering the given functional equation if we  differentiate both sides with respect to x  we get


How to use double angle identities to find length 'd'? Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.

A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:

Edit: Note that triangle ACB is not a right angled triangle. Should now read:
Note that triangle ACD is not a right angled triangle.

By using $$\cos 2x=2\cos^2 x-1$$
My working is:
$$\cos x=\frac d3$$
$$\cos 2x=\frac d3$$
therefore:
$$\frac d3=2(\frac d3)^2-1$$

Now becomes solvable as a quadratic equation

I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"

Using the law of cosines in triangles $ABC$ and $ACD$ we get
$$c^2=2^2+3^2-2\times 2\times 3 \cos x$$
$$b^2=d^2+2^2-4d\cos x$$
so that $$\cos x=\dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-\dfrac{4d(13-c^2)}{12}$$
The last equation is quadratic in $d$ and easy to solve.

$\cos(x)\neq \frac{2}{3}$ because the angle $\widehat{ACD} \neq \frac{\pi}{2}$.


About the onto functions If the range of a function belongs to all real numbers . Is the function necessarily onto?
My answer: No, we cannot say so until we know the co domain.

If the function is onto and the range of the function is every real number then we can conclude that domain of it is R. This is only valid if f : R -> R. Else in all other cases we can't guess the function domain.


Show that inf(E=1/p)=0 Let E be subset of Q (with the usual ordering) consisting of all fractions of the form 1/p, where p is a
prime number. Show that inf(E) = 0.

Here is my attempt to prove this. Suppose there exists a minimum of E min(e). 0

First, clearly $\;0\;$ is a lower bound of $\;E\;$ as for any prime $\;p\,,\;\;\frac1p>0\;$ .

Second, since for all $\;n\in\Bbb N\;$ there exists a prime $\;p\;$ s.t. $\;p>n\;$ (why?), we then get that

$$\forall\;n\in\Bbb N\;\;\exists\;\text{a prime}\;\;p\;\;s.t. \;\;\frac1p<\frac1n$$

Now, for any $\;\epsilon >0\;$ choose $\;n\in\Bbb N\;$ s.t. $\;\frac1n<\epsilon\;$ (why such a natural exists?), and then choose a prime $\;p\;$ s.t.

$$\frac1p<\frac1n<\epsilon$$

and you've shown $\;0\;$ is the highest lower bound, so it is the infimum of $\;E\;$ . It's easy now to see it can't be the minimum


infinity times infinitesimal - what happens? So what happens if we multiply infinite number by. Infinitesimal number? Like $dx \times \infty$ where $dx$ is treated as in one-dimensional integration.

Also, can we divide infinite number by infinite number and get a finite number?

In the ordinary calculus, there are no infinitesimals.

Abraham Robinson and others, from the $1950$'s on, developed non-standard analysis, which does have infinitesimals, and also "infinite" number-like objects, that one can work with in ways that are closely analogous to the way we deal with ordinary real numbers.

In non-standard analysis, an infinitesimal times an infinite number can have various values, depending on their relative sizes. The product can be an ordinary real number. But it can also be infinitesimal, or infinite. Similarly, the ratio of two "infinite" objects in a non-standard model of analysis can be an ordinary real number, but need not be.

The calculus can be developed rigorously using Robinson's infinitesimals. There are even some courses in calculus that are based on non-standard models of analysis. Some have argued that this captures the intuition of the founders of calculus better than the traditional limit-based approach.

For further reading, you may want to start with the Wikipedia article on Non-standard Analysis.

This is a question for surreal numbers. Surreal numbers are a really amazing thing invented by John Conway that include numbers like 0 and 3/4, but also things like "twice the square root of infinity, all plus an infinitesimal". This question depends on the values of the infinite and infinitesimal, but the way it works is this. The number ω is defined as the number of items in the set {0,1,2,3,4,5...}, so it's infinite. The number ε is defined as 1/ω. So ω*ε is obviously one. If you think about it a bit, it makes sense that 2ε^2 * ω is 2ε, and so on.

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


What is the maximum volume that can be contained by a sheet of paper? I was writing some exercises about the AM-GM inequality and I got carried away by the following (pretty nontrivial, I believe) question:

  Q: By properly folding a common $210mm\times 297mm$ sheet of paper, what
  is the maximum amount of water such a sheet is able to contain?

The volume of the optimal box (on the right) is about $1.128l$. But the volume of the butterfly (in my left hand) seems to be much bigger and I am not sure at all about the shape of the optimal folded sheet. Is is something boat-like?

Clarifications: we may assume to have a magical glue to prevent water from leaking through the cracks, or for glueing together points of the surface. Solutions where parts of the sheet are cut out, then glued back together deserve to be considered as separate cases. On the other hand these cases are trivial, as pointed by joriki in the comments below. The isoperimetric inequality gives that the maximum volume is $<2.072l$.

As pointed out by Rahul, here it is a way for realizing the optimal configuration: the maximum capacity of the following A4+A4 bag exceeds $2.8l$.

This problem reminds me of tension field theory and related problems in studying the shape of inflated inextensible membranes (like helium balloons). What follows is far from a solution, but some initial thoughts about the problem.

First, since you're allowing creasing and folding, by Nash-Kuiper it's enough to consider short immersions
$$\phi:P\subset\mathbb{R}^2\to\mathbb{R}^3,\qquad \|d\phi^Td\phi\|_2 \leq 1$$
of the piece of paper $P$ into $\mathbb{R}^3$, the intuition being that you can always "hide" area by adding wrinkling/corrugation, but cannot "create" area. It follows that we can assume, without loss of generality, that $\phi$ sends the paper boundary $\partial P$ to a curve $\gamma$ in the plane.

We can thus partition your problem into two pieces: (I) given a fixed curve $\gamma$, what is the volume of the volume-maximizing surface $M_{\gamma}$ with $\phi(\partial P) = \gamma$? (II) Can we characterize $\gamma$ for which $M_{\gamma}$ has maximum volume?

Let's consider the case where $\gamma$ is given. We can partition $M_{\gamma}$ into

1) regions of pure tension, where $d\phi^Td\phi = I$; in these regions $M_{\gamma}$ is, by definition, developable;

2) regions where one direction is in tension and one in compression, $\|d\phi^Td\phi\|_2 = 1$ but $\det d\phi^Td\phi < 1$.

We need not consider $\|d\phi^Td\phi\|_2 < 1$ as in such regions of pure compression, one could increase the volume while keeping $\phi$ a short map.

Let us look at the regions of type (2). We can trace on these regions a family of curves $\tau$ along which $\phi$ is an isometry. Since $M_{\gamma}$ maximizes volume, we can imagine the situation physically as follows: pressure inside $M_{\gamma}$ pushes against the surface, and is exactly balanced by stress along inextensible fibers $\tau$. In other words, for some stress $\sigma$ constant along each $\tau$, at all points $\tau(s)$ along $\tau$ we have
$$\hat{n} = \sigma \tau''(s)$$
where $\hat{n}$ the surface normal; it follows that (1) the $\tau$ follow geodesics on $M_{\gamma}$, (2) each $\tau$ has constant curvature.

The only thing I can say about problem (II) is that for the optimal $\gamma$, the surface $M_\gamma$ must meet the plane at a right angle. But there are many locally-optimal solutions that are not globally optimal (for example, consider a half-cylinder (type 1 region) with two quarter-spherical caps (type 2 region); it has volume $\approx 1.236$ liters, less than Joriki's solution).

I got curious so I implemented a quick-and-dirty tension field simulation that optimizes for $\gamma$ and $M_{\gamma}$. Source code is here (needs the header-only Eigen and Libigl libraries): https://github.com/evouga/DaurizioPaper

Here is a rendering of the numerical solution, from above and below (the volume is roughly 1.56 liters).

EDIT 2: A sketch of the orientation of $\tau$ on the surface:

This may not be the optimum. But an easy solution with larger volume than the box. 1.14228 l. (Of course we need to tape it at base to hold)


Solve the stochastic differential equation $dY_t = rdt + \alpha Y_tdB_t$ where $B_t$ is a Brownian motion. I need some help me with a stochastic differential equation that I am stuck on, since my professor loves to not give us any suggested solution.

I got a hint first to calculate the dynamics of the process $Y_{}tF_{t}$ where $$F_t = exp(-\alpha B_t+0.5\alpha^2t)$$

My solution is as follows:

$$d(F_tY_t) = F_t r dt  + \alpha F_t Y_t dB_t + 0.5 \alpha^2 F_t Y_t dt - \alpha F_t Y_t dB_t $$

$$ d(F_t Y_t) = 0.5 \alpha^2 F_t Y_t dt + F_t r dt$$

Take integral of both sides

$$F_t Y_t = \int ^{t}_{0}r F_s  ds + 0.5 \alpha^2 \int ^{t}_{0} F_s Y_s ds$$

So I am stuck at this point, any advice?

Since $F_t = \exp(-\alpha B_t + \alpha^2 t / 2)$, we can write $F_t = f(t,B_t)$, where $f(t,x) = \exp(-\alpha x + \alpha^2 t/2)$. Thus, by Itô's Lemma, $$ \mathrm d F_t = \frac 12 \alpha^2 F_t \,\mathrm d t -\alpha F_t \, \mathrm d B_t + \frac 12 \alpha^2 F_t \,\mathrm d t=\alpha^2 F_t \,\mathrm d t -\alpha F_t\, \mathrm d B_t.$$

Then, by the product rule, \begin{align*} \mathrm d(FY)_t &= F_t\,\mathrm d Y_t + Y_t\, \mathrm dF_t + \mathrm d [F,Y]_t \\
&= r F_t\,\mathrm d t + \alpha F_t Y_t \,\mathrm d B_t +\alpha^2 F_t Y_t \,\mathrm dt -\alpha F_t Y_t\,\mathrm d B_t -\alpha^2 F_t Y_t \,\mathrm d t \\
&=r F_t \,\mathrm dt.
\end{align*}

This means that \begin{align*}
F_tY_t &= F_0 Y_0 + r\int_{[0,t]} F_s \,\mathrm ds \\
&= Y_0 +r\int_{[0,t]} \exp\left( -\alpha B_s + \frac 12 \alpha^2 s \right) \,\mathrm d s.
\end{align*}

Since $F_t \neq 0$ for all $t\ge 0$, we have that $$ Y_t = \exp\left( \alpha B_t - \frac 12 \alpha^2 t \right)\left(Y_0 + r\int_{[0,t]} \exp\left( -\alpha B_s + \frac 12 \alpha^2 s \right) \,\mathrm d s\right). $$

Let's check that this satisfies the SDE $$ \mathrm d Y_t = r\,\mathrm d t + \alpha Y_t \,\mathrm d B_t.\label{*}\tag{*} $$

To this end, define the processes $$ X_t = \exp\left( \alpha B_t - \frac 12 \alpha^2 t \right) $$ and $$ Z_t = Y_0 + r\int_{[0,t]} \exp\left( -\alpha B_s + \frac 12 \alpha^2 s \right) \,\mathrm d s. $$ It's easy to check via Itô's lemma that $X$ satisfies  $$ \mathrm d X_t = \alpha X_t \,\mathrm d B_t. $$ Moreover, $$ \mathrm d Z_t = rF_t \,\mathrm dt. $$ Thus, as $Z$ has finite variation, and $FX = 1$, the product rule gives us that \begin{align*} \mathrm d (XZ)_t &= \alpha X_t Z_t \,\mathrm d B_t + r \,\mathrm d t \end{align*} which means that $Y = XZ$ satisfies the SDE \eqref{*}.

You had done the solution, there isn't any thing left to solve. It seemed wierd since unlike ODE, solution to SDE seldom have explicit form, after integration, you've got an adapted representation for $Y_t$, that's engouh.


Orthogonal projection and subspaces proof? Let's let $M$ be a subspace of $\mathbb{R}^n$ and let $N$ be a subspace of $M$.
Let $m$ and $n$ denote the orthogonal projection matrices onto $M$ and $N$. Show that $mn = nm = n$.

--

I'm thinking about inner products of $n$ and $m$, but they would only be zero in the area in which $N$ is a subspace of $M$. I'm a bit lost. Any advice/pointers?

Denoting matrices with small letters is confusing. I'll denote projections on $M$ and $N$ as $P_M$ and $P_N$ respectively. I will also denote $V = \mathbb{R}^n$, as this proof is good for any finite inner product space.

Note that we can write $V$ as a direct sum of three orthogonal spaces (see Wikipedia article on orthogonal complements):

$$V = M \oplus M^\perp = (V \cap M) \oplus M^\perp = ((N \oplus N^\perp) \cap M) \oplus M^\perp = N \oplus (N^\perp \cap M) \oplus M^\perp.$$

Let $v = x + y + z$, be any vector in $V$, where $x \in N$, $y \in N^\perp \cap M$, $z \in M^\perp$. Recall (Lemma 4.8 here) that for each $v \in V$ there exist (unique) $x,y,z$, so we can do this for any $v \in V$. Then

\begin{gather*}
P_N x = x, \quad P_M x = x, \\
P_N y = 0, \quad P_M y = y, \\
P_N z = 0, \quad P_M z = 0,
\end{gather*}

so

\begin{align*}
P_M P_N v &= P_M P_N (x + y + z) = x, \\
P_N P_M v &= P_N P_M (x + y + z) = x, \\
P_N v &= P_N (x + y + z) = x,
\end{align*}

which proves the statement.

Presumably the statement was meant to be: Show $mn =nm = n$.

Note that if $P$ is a projection, then $x \in {\cal R} P=$ iff $Px = x$. (One direction is trivial, the other follows from $x = Py = P(Py + (I-P) y) = P(x + (I-P)y) = P x$.)

Suppose $N = {\cal R} n \subset M = {\cal R} m$.

First show $mn=n$. Choose $x$. Then $nx \in {\cal R} n \subset{\cal R} m$, hence $nx = mnx$, from which the result follows. This is true for all projections, orthogonal or not.

Now show $nm = n$. Here we need $m$ and $n$ to be orthogonal. It is straightforward to show that a projection $m$ is orthogonal iff $m$ is self-adjoint.

Then we have $m^*(I-m) = m(I-m) = 0$. Since $mn=n$, we have $n^* m^* = n^*$, and so $n(I-m) = n^*(I-m) = n^* m^* (I-m) = n^* m (I-m) = 0$.

Now we have $nx = n(mx + (I-m)x) = nm x + n(I-m)x = nmx$, which shows $n = nm$.

(The result is true for any closed subsets $M,N$ of a Hilbert space.)


Combinations and Permutations in ice cream cones... what is the difference? The first two I am certain I have correct however c and d... I am struggling to understand the difference. I have done research on this site and have seen similar questions with explanations but I just want to double check my thinking.

A certain store sells 31 different flavors of ice cream.  How many different 3-scoop cones are possible if:

a. each flavor must be different and the order of the flavors is unimportant? $31 nCr 3 = 4495$

b. each flavor must be different and the order of the flavors is important? $31 nPr 3 = 26970$

c. Flavors need not be different and the order of the flavors is unimportant? (This is a nontrivial question.)
I think I need to use this formula $$\binom{n+m-1}{m}$$

where $n$ in the total numbers in the set, and m is how many numbers you want to choose.

So,  $$\binom{31+3-1}{3}=\binom{32}{3}=\frac{32\times 31\times 30}{1\times 2\times 3}=4960$$  Is this correct?

d. Flavors need not be different and the order of the flavors is important?
I am thinking this is $31 \cdot 31 \cdot 31 = 29791$

I think your (c) is wrong. You can count it by considering $3$ different cases: 1) $3$ flavors $X_3$; 2) $2$ flavors $X_2$ and 3) $1$ flavor $X_1$.

The case of $3$ flavors is easy:
$$
X_3 = {31 \choose 3}
$$
The case of $2$ flavors. You first choose $2$ flavors, then choose one of them to be the $2$ cones and another as the remaining cone.
$$
X_2 = {31 \choose 2}\cdot 2
$$
The case of $1$ flavor.
$$
X_1 = {31 \choose 1}
$$

(d) is correct.

For (c), you had the correct idea and formula, but you made an arithmetic error. $31 + 3 - 1$ is $33$, not $32$. Using $\binom{33}{3}$ would give you $5456$, which is the correct answer.


Method of Characteristics and Initial Value Problem $u_t + 3u_x = 2t$, $u(x,0)=\sin(x/2)$.

I used the method of characteristics to get the answer, $u(x,t)=t^2 + 2\sin^{-1}(x-3t)$.

Does this satisfy the initial condition? I checked for the first equation and it does; however I do not think it satisfies the intial value when $t=0.$ Am I correct in saying so?

answer for above question is $$u(x,t)=t^2+\sin(x−3t/2)$$


$a_n = {{2^n}\over n}$ diverges? How do I formally show that the sequence $a_n = {{2^n}\over n}$ diverges using a $\delta$-$\epsilon$ argument?

Here is a hint, since I don't know what your thoughts on the problem are.

Pick an arbitrarily large $M>0$. Can you show that
$$\frac{2^n}{n}\geq M$$
for every big enough $n$?

You can see the order of growth of $2^n > n$ for all $n=1,2,...$. So, since order of growth of nominator is larger than order of growth of denominator it is obvious that sequence diverges. So,

$$\lim_{n \to \infty}a_n=\lim_{n \to \infty} \frac{2^n}{n}= \infty$$


How can I solve $u_{xt} + uu_{xx} + \frac{1}{2}u_x^2 = 0$ with the method of characteristics. I am trying to solve the following PDE: $u_{xt} + uu_{xx} = -\frac{1}{2}u_x^2$, with initial condition: $u(x,0) = u_0(x) \in C^{\infty}$ using the method of characteristics.

I am a beginner with the method of characteristics and PDE in general. Here is what I have so far.

Define $\gamma(x,t)$ as the characteristic curves.

$\frac{\partial}{\partial t} u_x(\gamma(x,t),t) = u_{xt} + u_{xx}\gamma_t(x,t) = - \frac{1}{2}u_x^2$

Set $u_t = u_x$

$\Rightarrow \frac{\partial}{\partial t} u_x(\gamma(x,t),t)= (u_t)_x + u_{xx}\gamma_t(x,t)$

$ = u_{xx} + u_{xx}\gamma_t = - \frac{1}{2}u_x^2$

From this I get $\gamma_t = -\frac{1}{2}\frac{u_x^2}{u_{xx}} - 1$

However, I am not sure this is the right approach and do not fully understand how to use the method of characteristics when the solution $u(x,t)$ is constant on the characteristic curves.

Any help is much appreciated.

Edit: I made some progress by using $v=u_x$ and getting $\frac{dv}{dt} = \frac{-1}{2} v^2$ and $\frac{\partial x}{\partial{t}} = 1$. Then separating the first ODE, I get $\frac{2}{v} = t + c$. However, I am not sure if my solution after integrating with respect to $x$ and using the initial condition is correct. I end up with $u(x,t) = \frac{2}{t+c}x + c_1$, $u(x,0) = \frac{2}{c}x + c_1$.

Modifying the problem. Rather than consider the PDE $u_{xt}+u\ u_{xx}+\frac{1}{2}u_{x}^2=0$ with initial condition $u(x,0)=u_0(x)$ as asked above, I will consider the following variant.
$$
\text{Solve }u_{xy}+u\ u_{xx} + u_x^2=0\text{ subject to }u(x,0)=f(x).\qquad(\star)
$$
There are three differences between this question and that which was asked originally.

The coefficient of $u_x^2$ has changed from $\frac{1}{2}$ to $1$.
The variable $t$ has been renamed to $y$.
The initial function $u_0(x)$ has been renamed to $f(x)$.

Only (1) represents a significant modification of the problem. It makes the solution more tractable and enables it to be found using an elementary application of the method of characteristics. For these reasons, it is conceivable that this was the intended question.

Note: I will not delve into regularity of the solutions in this answer.

Reduction to a first order quasilinear PDE. Write the equation as
$$
\frac{\partial}{\partial x}\left(u_y+u\ u_x\right)=0.
$$
Thus $(\star)$ is equivalent to
$$
u_y+u\ u_x=g(y),\qquad u(x,0)=f(x),\qquad (\star\star)
$$
where $g(y)$ is an arbitrary function of $y$ (with sufficient regularity).

Method of characteristics.
Perhaps the simplest formulation of the method of characteristics is for quasilinear first order PDEs. These are PDEs of the form
$$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u).$$
To solve this equation, one regards the solution as a surface $z=u(x,y)$ in $xyz$-space. Let $s$ parametrize the initial curve $\bigl(s,0,f(s)\bigr)$  and let $t$ be a second parameter, which can be thought of as the distance flowed along a characteristic curve emanating from $\bigl(s,0,f(s)\bigr)$.

The characteristic equations are then
$$
\frac{dx}{dt}=a(x,y,z),\quad \frac{dy}{dt}=b(x,y,z),\quad \frac{dz}{dt}=c(x,y,z).
$$
Returning to our equation $(\star\star)$, this reduces to $a(x,y,u)=u$ and $b(x,y,u)=1$ and $c(x,y,u)=g(y)$.
Thus
$$
\frac{dx}{dt}=z,\quad \frac{dy}{dt}=1,\quad \frac{dz}{dt}=g(y)
$$
with initial conditions $x(0)=s$ and $y(0)=0$ and $z(0)=f(s)$.

The solution to this system is
$$
x=s+zt,\quad y=t,\quad z=f(s)+h(t),
$$
where $h(t)$ is the antiderivative of $g(t)$ satisfying $h(0)=0$. Since $g$ was arbitrary, so is $h$ given $h(0)=0$.

The solution. Now we eliminate all occurrences of $t$ by replacing them with $y$, then eliminate $s$ by writing $s=x-zy$. Finally, replace $z$ with $u$ to obtain the implicit equation
$$
\boxed{u=f(x-uy)+h(y)},
$$
where $h(y)$ is any sufficiently regular function satisfying $h(0)=0$. This is an implicit equation for the general solution of $(\star)$.

TL;DR. Change the $\frac{1}{2}$ in the original question to $1$ to obtain a PDE solvable by the method of characteristics.


Your system now reads

$$\begin{cases}x+az=b,\\y+cz=d.\end{cases}$$

Move $z$ to the RHS and you get a parametric solution

$$\begin{cases}x=b-az,\\y=d-cz.\end{cases}$$

You have obtained that one of your three equations is a linear combination of the other two, so it doesn’t give you any other information. So you have two equations in  three variables, so you have infinitely many solutions.

Maybe you didn’t use in the right way the Gaussian elimination: are you sure you have only linear variables? Can you show your system of equations (not matricial form)?


Solving Polynomial Equations and Inequalities The distance, in km, of a ship from its harbour is modeled by the function
  $d(t)= -3t^3 + 3t^2 + 18t$
  where $t$ is the time elapsed in hours since departure from the harbour.

a) When does the ship return to harbour?

b) Estimate the time that the ship begins its return trip back to the harbour.

c) When is the ship $24$ km from the harbour?

I believe I have figured the first part of this question out.  I substituted values for "$t$" until the equation ended with a value of zero.  This value was $d(3)$.  Meaning that it takes the ship 3 hours to reach the harbour.  The other questions I am just not sure how to approach.  c) I began by substituting $24$ as my distance, but I am not sure what to do after that.

As of your answer to a), I wouldn't recommend that, instead do:
$$d(t)=0\implies -3t^3+3t^2+18t=0\implies 3t(-t^2+t+6)=0\implies t=0\text{ or }-t^2+t+6=0\implies t=0\text{ or }t=3\text{ or }t=-2$$
Do you understand how we found these solutions? It should be clear that from the above we can conclude that the ship returns at $t=3$

As of b), when would you expect this?

As of c), we have to solve $d(t)=24$. Now, our method of a) doesn't work anymore, since we have to solve $$-3t^3+3t^2+18t-24=0$$
we can't factor out a term $t$. Substituting some standard values is the only solution (actually there are ways to solve this, but I don't wan't to bother you with that!). And then we find $t=1$ is a solution.


Birthday probability: Permutation or Combination? In a class of $30$ people, what is the probability that at least $2$ people share the same birthday?

I would think that I should use $^{365}C_{30}$ for the numerator since order shouldn't matter in this case (?) but I have come across $2$ resources that use permutations instead and I am curious as to why.

The resources mentioned above, if anyone needs to refer:
https://www.youtube.com/watch?v=9G0w61pZPig&feature=youtu.be
https://medium.com/i-math/the-birthday-problem-307f31a9ac6f

With permutation, order matters, and with combination, it doesn't.

Imagine you have $30$ people. How many ways are there to assign them unique birthdays?

The question you should ask yourself is: if we assigned Alice $\to$ Jan $10$ and Bob $\to$ Jun $5$, would we consider this a different assignment than if Alice $\to$ Jun $5$ and Bob $\to$ Jan $10$ (for example)? We would, and so order matters. Hence we use permutation.

Another way to see this is to write out your assignments as a $30$-tuple (of days, $1 \ldots 365$), where each spot in the tuple is a specific person. Then you see that $(x, y, \ldots)$ is a different assignment than $(y, x, \ldots)$.

The answer is $\approx70,63\%$. Look here and here

Assume we have $n$ people and we ask about the probability that two of the people have their birthdays in the same day.

If we ask each person when this person was born, then by the time we're done we would have two possible results.

The first result would be that they all have different birthdays. The second result would be that at least two persons (or at max all $n$ persons) have equal birthdays.

Let the probability for the first result be $p_1$ and for the second result be $p_2$. Because there are no other results to expect, we have $p_1+p_2=100\%$.

It is easier to calculate $p_1$, the probability that all birthdays are different from each other: In total we have $365$ days. The first person could have his/her birthday on any of the $365$ days, the second person on any of the remaining $364$ days, ..., the $n$th person on one of the $365-n$ days. Before asking the first person about her/his birthday, we can calculate the probability $p_1^{\text{person 1}}$ that the birthday of the first person is different to all others, this is $100\%$ sure, because we don't know any other birthdays yet. Before asking the second person about her/his birthday, we can calculate the probability for the second person having a different birthday than the others (than the first person). This probability is $p_1^{\text{person 2}}=1-1/365=364/365$, because all days have the same probability $1/365$ to be the birthday of the second person except for one day, except for the day, when person 1 has his birthday, if we want to know the probability of different birthdays for all persons.This goes on and on and on for all $n$ persons and we get finally

$$p_1=\underbrace{1\cdot 364/365 \cdot 363/365 \cdot \cdots \cdot (365-(n-1))/365}_{n \text{ times}}$$

which translates to

$$p_1 = (1\cdot 364\cdot 363 \cdot \cdots \cdot (365-n+1))/365^n = {}_{365}P_n/365^n$$

And the answer to the question is:

$$p_2= 1-{}_{365}P_n/365^n$$

For $n=30$ we get $p_2\approx70,63\%$


Is there a definition for antiderivative? So, We all know that the definition of the derivative is
$f'(x)=\lim_{h\to0}{f(x+h)-f(x)\over h}$
So does the antiderivative have a solid definition like this?

The antiderivative of a function $f$ on an interval $(a,b)$ is defined as the set of all functions $g$ such that $g'=f$ identically on $(a,b)$.

This definition is solid, despite your opinion. Since there is no unique antiderivative, there can be no formula whose result must be a single function. This is related to the fact that the operator $f \mapsto f'$ is not really invertible (on suitable spaces), since all constants have null derivative. To be consistent with this, we can only define the antiderivative as a set of functions.

Unfortunately, the antiderivative is the function F(x) such that F'(x)=f(x)...
Which explains why it is harder to integrate than to differentiate (just like for division VS multiplication)


Finding volume of a cone given density Let $C$ be the solid cone with the boundary surfaces $x^2 +y^2 = z^2$ and $z = 0$. The density of the solid at point $(x,y,z)$ is $z$.

Find the volume of the solid using the integrals in both the cylindrical coordinates and the spherical coordinates.

I really cant do this question.
I know that $V=\rho /m$ but what is $m$?

I also don't understand how you can even find the volume with integrals because there is no limit of $z$ given so it would just be infinity wouldn't it.

Can someone start me off, I obviously don't need anyone to show me how to integrate. Just struggling on what the density and stuff is relevant.

As a start cylindrical coordinates case is simpler perhaps by cylinders stacking summation

$$ dV = \rho \pi r^2 dz $$
$$ \rho = z,  r = z $$
$$ V = \pi \int z^3 dz  = z^4 \pi/4 $$ etc..

I don't think using the given density is necessary because the triple integral of 1 using the cone's boundaries will give the volume of the cone. Although I don't know why the density would be mentioned if it's not part of the solution.
Is this one of the questions from the UoB MVA module? If so, I'm taking the module too.. hi:D


Cantor's diagonal argument and alternate representations of numbers Cantor's diagonal argument works because it is based on a certain way of representing numbers. Is it obvious that it is not possible to represent real numbers in a different way, that would make it possible to count them?

Edit 1: Let me try to be clearer. When we read Cantor's argument, we can see that he represents a real number as an infinite sequence of binary digits. Using this representation, he shows that real numbers are uncountable. An intuitive counter argument could be that maybe there is another type (perhaps incredibly strange) way of representation that would make it possible to count the real numbers. A kind of trick, like the typical one used to show the countability of rational numbers. One could thus be tempted to think that when representing real numbers as infinite sequences of binary digits, it is those representations that are uncountable, but that some other representation could be countable. It seems to me that this can be summed up like this: Is a proof of the countability of a set dependent on the representation its members in the proof?

You are making the common mistake to confuse between a number and its decimal representation.

An easy way to see that $\Bbb R$ is uncountable, regardless to how we can or cannot represent real numbers, is to see there is an injection from $\mathcal P(\Bbb N)$ into $\Bbb R$, defined by $\displaystyle A\mapsto\sum_{n\in A}\frac1{3^{n+1}}$.

This function depends only on the fact that this sum is a convergent sum (as it is bounded by a convergent geometric sequence), and the fact that $\mathcal P(\Bbb N)$ is uncountable depends only on its property of being the power set of $\Bbb N$.

To prove $\mathbb{R}$ is uncountable, we assume that it isn't (i.e. that it is countable). We then make a list of all the real numbers, and state that this list is complete. However we can construct a new real from not the $1^{st}$ digit of list item $1$ (e.g. if the $1^{st}$ digit of item $1$ is $4$, pick any digit other than $4$ (and arguably not $0$ either)), and keep the rest the same ($2^{nd}$ digit from $2$, $3^{rd}$ digit from $3$, etc..), not the $2^{nd}$ from list item $2$ and the rest the same, and so on. This number is not in the list, and so $\mathbb{R}$ is uncountable.

This is Cantor's Diagonal argument.

As it is impossible to count $\mathbb{R}$, there is no way we can re-invent numbers in such a way as to make this possible.


Are there linear transformations from vector spaces over different fields? I'm in a matrix theory class, and today we started talking about linear transformations. My professor noted that the range and domain of a linear transformation must be vector spaces over the same field. This made sense to me at first because of the fields are different, then something like $T(cx)=cT(x)$ doesn't make any sense when c is only in one of the vector spaces.

But then I thought that maybe some fields might be compatible, like for example, $\mathbb{Z}_5$ and $\mathbb{Z}_7$

Are there any linear transformations for two spaces over different fields? or Are there functions that have all the same properties as linear transformations except that the spaces involved use different fields?

If the answer is no, can you prove it?

If you have an injective field endomorphism $\varphi$ from field $Fa$ to field $Fb$ - said differently:

$$\forall (x, y) \in Fa \times Fa, \varphi(a+b) = \varphi(a) + \varphi(b) \space \mathrm{and} \space \varphi(a.b) = \varphi(a).\varphi(b)$$

then $\varphi(Fa)$ is a sub field of $Fb$ isomorphic to $Fa$

If $Vb$ is a vector space over $Fb$ of finite dimension, you can exhibit a base from $Vb$. You can then build a sub set (not a sub vector space) of  $Vb$ from the base and the field $\varphi(Fa)$. This sub set (let us call it $Wb$) is a vector space over $\varphi(Fa)$ (*).

By identifying $Fa$ and $\varphi(Fa)$, you can indeed build a linear transformations (say $\Lambda$) from a vector space over $Fa$ (say $Va$) and $Wb$.

But you cannot call it a linear transformations from $Va$ to $Vb$ because $\Lambda(Va)$ is not a sub vector space of $Vb$ (except for the trivial $\Lambda(Va) = \{ 0_{Vb} \}$ see below). It is only a sub vector space of $Wb$.

Well to be honest, there is one trivial case where you could exhibit a pseudo linear transformation for $Va$ to $Vb$:

$$\begin{align}\Lambda: &Va \rightarrow Vb \\
&x \mapsto 0_{Vb}\end{align}$$

It respects all the properties for a linear transformation, and $\Lambda(Va) = \{ 0_{Vb} \}$ is indeed a sub vector space but it is not really interesting, and on a strict point of view, as the fields are different, it is not a true linear transformation.

(*) I suppose it could be extended to vector space of infinite dimensions, but I do not know how to prove it

Let's suppose $V$ is a vector space over a field $K$ (i.e., an abelian group onto which $K$ acts). If $L$ is another field, and $K$ can act on the addtive structure of $L$ (i.e., the additive abelian group of $L$ is a $K$ vector space), then we can extend $V$ to a vector space over $L$ using tensor products. This construction is known as the extension of scalars. We take the tensor product of $V$ and $L$ over $K$:

$$V \otimes_K L$$

It is naturally both a $K$ vector space (by the tensor product constrution) and a $L$ vector space (by multiplication on the $L$ component). There is a natural function from $\varphi: V \to V\otimes_K L$, which is additive, and compatible with multiplication by scalars on $K$:

$$\varphi(x + y) = (x+y)\otimes 1 = (x\otimes 1) + (y\otimes 1)$$
$$\varphi(kx) = (kx\otimes 1) = (x\otimes 1_k) = k(x\otimes 1)$$

Here, $1_k$ detones the image of the action of $k$ on the unit of $L$.


Find the equivalence class of 0 R is a relation defined on the integers by $(a,b) \in R$ is $a^2-b^2$ and is divisible by 3.

I set a or b to zero to get all the negative and positive values in the equivalence class. Although I want to say it is $(...,-9n,-6n,3n,0,3n,6n,9n,...)$ for some integer n. But I do not think this is correct because if n = 1, 6n does not belong to the relation. What am I doing wrong?

We have your equivalence relation R as $a \sim b$ if $3|(a^2 - b^2)$.

To find the class of elements equivalent to $0$, we need to set one of the elements to $0$ (just one because of reflexivity, symmetry, and transitivity, as this is an equivalence relation): suppose $a \sim 0$, i.e. $3|(a^2 - 0^2) \Longrightarrow 3|a^2$. Then by Euclid's Lemma, $3|a$. So the equivalence class of $0$ is the set of all integers that we can divide by $3$, i.e. that are multiples of $3: \{\ldots, -6,-3,0,3,6, \ldots\}$.

The set has the following equivalence relations.


Proof a graph is bipartite if and only if it contains no odd cycles How can we prove that a graph is bipartite if and only if all of its cycles have even order? Also, does this theorem have a common name? I found it in a maths Olympiad toolbox.

One direction is very easy: if $G$ is bipartite with vertex sets $V_1$ and $V_2$, every step along a walk takes you either from $V_1$ to $V_2$ or from $V_2$ to $V_1$. To end up where you started, therefore, you must take an even number of steps.

Conversely, suppose that every cycle of $G$ is even. Let $v_0$ be any vertex. For each vertex $v$ in the same component $C_0$ as $v_0$ let $d(v)$ be the length of the shortest path from $v_0$ to $v$. Color red every vertex in $C_0$ whose distance from $v_0$ is even, and color the other vertices of $C_0$ blue. Do the same for each component of $G$. Check that if $G$ had any edge between two red vertices or between two blue vertices, it would have an odd cycle. Thus, $G$ is bipartite, the red vertices and the blue vertices being the two parts.

One relatively simple way is to break the if and only if into its two parts:

Prove that if a graph $G$ is bipartite then it has no odd cycles, and
If $G$ has only even cycles, then you can partition the vertices into two independent sets.

I don't think this is a named theorem, it's too simple.