problems on topology involving basis and sub-basis Need some help guys, The question was: Can you define a basis generated by a sub-basis?

I tried to prove this today in class. My prof said that everything is correct except B3 (the 3rd basis). I didn't declare or define it properly and without that I claimed that there exists a basis B3 which follows the definition or 2nd criteria of a basis. How should I define that in my proof?

Any help will be greatly appreciated. Thanks in advance.

You did every thing correctly.

For the $B_3$ you looking for, note that elements of the basis $B$ are finite intersections  of subbasis $S$ and you reach to $B_2\cap B_2$ is in the basis $B$ because $B_1\cap B_2$ is a finite intersection of elements of $S$.

Now you looking for a $B_3$ such that $B_3\subseteq B_1\cap B_2$? Take $B_3=B_1\cap B_2$.

The base could also be written as:

$$\mathcal{B}=\{ \bigcap \mathcal{S}': \mathcal{S}' \subseteq \mathcal{S} \text{ finite }\}$$

with the convention (follows from void truth considerations) that $\bigcap \emptyset = X$, which then handles the first demand on candidate bases, that $\bigcup \mathcal{B}=X$. This way we don’t even have to ask for $\mathcal{S}$ to cover $X$, as some texts do, superfluously.

$\mathcal{B}$ is then by construction closed under finite (or dual) intersections, as the union of two (or finitely many) finite subfamilies of $\mathcal{S}$ is also finite, and this implies the second demand on bases (as e.g. Munkres defines them).

We can then just take $B_3= B_1 \cap B_2$ for all $x$ in that intersection. Maybe you should have made that fact more explicit.


proving it is Hausdorff Question:

Let $X$ be a topological space and let $A ⊂ X$. Define an equivalence relation $∼$ on
$X$ such that the equivalence classes are:

• $A$ itself, and,
• all singletons $\{x\}$ such that $x \notin A$.

Then define $X/A$ to be the quotient space $X/{∼}$.

Prove that if $X$ is regular and $A$ is closed then $X/A$ is Hausdorff.

My Answer:

The quotient mapping $q(x)$ is
$$q(x)=\begin{cases}
x,&\text{if }x∉A\\
A,&\text{if }x∈A\;.
\end{cases}$$

Pick two distinct points $x,y∈X/A$.  If neither point is $A$, then $x,y∈$$X$ $\backslash$ $A$.  With the given information in the problem, particularly that $A⊂X$ is closed, we may deduce that $X$ $\backslash$ $A$ is open in $X$.  Then there exist open neighborhoods $U_A$ of $x$ and $V_A$ of $y$ such that $U_A$ and $V_A$ are disjoint from the closed set A.  Given that $X$ is regular, we can deduce that $X$ is Hausdorff.  $x$ and $y$ are distinct and thus we can find open sets $U'$ and $V'$ disjoint from each other such that $x∈U'$ and $y∈V'$.  Now define $U=U_A∩U′$ and $V=V_A∩V′$.  Then $U$ and $V$ are open, disjoint neighborhoods such that $x∈U$ and $y∈V$, and $U$, $V$ are also disjoint from $A$.

If one of the points, say $y$, is $A$, then $x$, the other distinct point, is $∈$ $X$ $\backslash$ $A$.  Because $X$ is regular, every non-empty closed subset of $X$,including  $A ⊂ X$, and every point in $X$ contained in a closed subset, admit non-overlapping open neighborhoods.  Then the point $x$ $∈$ $X$ $\backslash$ $A$ has an open neighborhood and the closed set A has a non-overlapping open neighborhood.  If $y$ is $A$, then $x$ and $y$ are contained in disjoint open neighborhoods.

Therefore any two distinct points $x,y∈X/A$, whether or not they are $A$, have open neighborhoods $U$ of $x$ and $V$ of $y$ such that $U$ and $V$ are disjoint.
$X/A$ is Hausdorff. $□$

Is my proof even correct?
Please help me make my proof more rigorous and accurate.  I need everything to be absolutely clear and rigorous. Thank you.

Pick two distinct points $x,y∈X/A$.  If neither point is $A$, then
  $x,y∈$$X$ $\backslash$ $A$.  With the given information in the
  problem, particularly that $A⊂X$ is closed, we may deduce that $X$
  $\backslash$ $A$ is open in $X$.  Then there exist open neighborhoods
  $U_A$ of $x$ and $V_A$ of $y$ such that $U_A$ and $V_A$ are disjoint
  from the closed set A.  Given that $X$ is regular, we can deduce that
  $X$ is Hausdorff.  $x$ and $y$ are distinct and thus we can find open
  sets $U'$ and $V'$ disjoint from each other such that $x∈U'$ and
  $y∈V'$.  Now define $U=U_A∩U′$ and $V=V_A∩V′$.  Then $U$ and $V$ are
  open, disjoint neighborhoods such that $x∈U$ and $y∈V$, and $U$, $V$
  are also disjoint from $A$.

So you take disjoint open neighborhoods $U'\ni x$ and $V'\ni y$, and you want them to also be disjoint from $A$. (Why is that sufficient in order to have disjoint neighborhoods of $\{x\}$ and $\{y\}$ in $X/A$? Here you could explain a bit more.) What you do is correct, but one could shorten it a bit by simply intersecting $U'$ and $V'$ with the open set $X/A$.

  If one of the points, say $y$, is $A$, then $x$, the other distinct
  point, is $∈$ $X$ $\backslash$ $A$.  Because $X$ is regular, every
  non-empty closed subset of $X$, including  $A ⊂ X$, and every point in
  $X$ contained in a closed subset, admit non-overlapping open
  neighborhoods. Then the point $x$ $∈$ $X$ $\backslash$ $A$ has an
  open neighborhood and the closed set $A$ has a non-overlapping open
  neighborhood.  If $y$ is $A$, then $x$ and $y$ are contained in
  disjoint open neighborhoods.

What do you mean by "every point in $X$ contained in a closed subset"? The disjoint neighborhoods exist for every closed set $A$ and every point outside of $A$. Again, why does this show that there are disjoint open neighborhoods around $\{x\}$ and $A$ in $X/A$?
One more thing: It is bad style to write something like "then $x$, the other point, is $\in X/A$", as one should not mix text and symbols. You could write "then $x$, the other point, is in $X/A$" or "then $x\in X/A$"

  Therefore any two distinct points $x,y∈X/A$, whether or not they are
  $A$, have open neighborhoods $U$ of $x$ and $V$ of $y$ such that $U$
  and $V$ are disjoint. $X/A$ is Hausdorff. $□$


9 pirates have to divide 1000 coins... A band of 9 pirates have just finished their latest conquest - looting, killing and sinking a ship. The loot amounts to 1000 gold coins.

Arriving on a deserted island, they now have to split up the loot. You, as the captain of the band, have to propose a distribution plan (who gets what). What's your proposal?

Consider that this bunch is a democratic lot. If your proposal is accepted by half of the group, then everybody adheres to it. However, if folks feel you are getting greedy, and less than half of the band agrees to your proposal, then they kill you, and then your First Mate gets to make a proposal. And so it goes in decreasing order of hierarchy/seniority.

In order for pirates to make consistent decisions, we must start with two assumptions.  First, pirates are all completely logical, and that they are all completely logical is 'common knowledge' (see joriki's comment below).  Second, we must take a stance on the bloodthirsty vs. friendly issue (see Micah's comment above), and the problem isn't as interesting with friendly pirates, so we assume they are bloodthirsty.

Under these assumptions, it can be shown that with a loot of $G$ gold coins, and $n$ pirates, the first pirate has a proposal that will be accepted in which he receives $G-\lfloor\frac{n-1}{2}\rfloor$ gold coins (here we assume $G\ge\lfloor\frac{n-1}{2}\rfloor$).  Denoting a proposal as $(g_1,\ldots,g_n)$, where pirate $i$ receives $g_i$ coins, the first pirate should propose $(G-\lfloor\frac{n-1}{2}\rfloor,0,1,0,\ldots,\frac{1-(-1)^n}{2}).$

Immediately we note that in any proposal, exactly $\lfloor\frac{n}{2}\rfloor$ pirates receive no gold coins.

We prove that this proposal will be accepted by induction.  If $n=1$, the pirate will choose the $G$ gold coins over suicide, and if $n=2$, the first pirate will vote for his own proposal, thus obtaining a majority of the vote.

Next, suppose the said proposal is accepted for $n-1$, and assume there are $n$ pirates.  Since all pirates are completely logical, there are exactly $\lfloor\frac{n-1}{2}\rfloor$ pirates that know they will receive $0$ coins if they don't accept the proposal of the first pirate.  Since pirates are bloodthirsty, the first pirate must buy their votes with $1$ gold coin each.  This gives a proposal that passes with a majority $\lfloor\frac{n-1}{2}\rfloor+1$ votes, where the extra vote is from the first pirate himself.

In the case of $9$ pirates and $1000$ gold coins, the first pirate will receive $996$ gold coins.

The solution given in the textbooks is wrong if people have Muth Rational Expectations- i.e. they expect the prediction of the correct economic theory. If the correct economic theory is that share in the booty should be proportionate to the pirate's expected marginal product in contributing to a victorious fighting coalition then any other proposal gets voted down and its proposer is killed. The fact that pirates are rankable by ability means there is a Expected Marginal Product curve even if it is difficult to compute. Moreover expected values can be fractional, reflecting the fact that there is some non zero chance that a stronger pirate may be killed by a lucky blow from a weaker pirate.
As a matter of fact, only an irrational person or one under compulsion by an all powerful tyrant, would make the 'backward induction' offer. But if an all powerful tyrant exists then they can always devise rules such that whatever outcome they want is voted for- unless of course people are rational enough to defy such a tyrant.


Find a particular solution for this linear system $\mathbf{x'}=\begin{bmatrix}3 & 1 \\ 0 & 3\end{bmatrix}\mathbf{x}+\begin{bmatrix} e^{3t}\\e^{3t}\end{bmatrix}$

My attempt:

We first find a solution to the associated homogeneous system $\mathbf{x'}=\begin{bmatrix}3 & 1 \\ 0 & 3\end{bmatrix}\mathbf{x}$.

My answer is $\mathbf{x}_c(t)=a_0\begin{bmatrix}1 \\ 0\end{bmatrix}e^{3t}+b_0\left(\begin{bmatrix}1     \\ 0\end{bmatrix}t+\begin{bmatrix}0 \\ 1\end{bmatrix}\right)e^{3t}$

In order to eliminate duplication, let $\mathbf{x}_p(t)=\begin{bmatrix}a_1 \\ a_2\end{bmatrix}t^2e^{3t}+\begin{bmatrix}b_1 \\ b_2\end{bmatrix}te^{3t}+\begin{bmatrix}c_1 \\ c_2\end{bmatrix}e^{3t}$

When I substitute this into the given ODE, I get 6 equations in 6 variables but I do not manage to obtain values for all of the variables. How do
I proceed?

$$X'=AX+g(t)$$
Your particular solution is correct but since the matrix A is already in Jordan form.
$$\pmatrix {x \\y}'=\pmatrix{3 & 1 \\ 0 & 3}\pmatrix  {x \\ y}+\begin{bmatrix} e^{3t}\\e^{3t}\end{bmatrix}$$
$$y'=3y+e^{3t} \implies (ye^{-3t})'=1$$
$$y(t)=e^{3t}(t+K)$$
$$x'=3x+y+e^{3t}$$
$$x'=3x+e^{3t}(t+K+1) \implies (xe^{-3t})'=t+K+1$$
$$x(t)=(\frac {t^2}2+(K+1)t+C)e^{3t}$$

It appears that the variables that don't have values can take on arbitrary values that are consistent with the 6 equations. I checked one case using MATLAB and it works.


function that preserves compactness Let S be a compact topological space, let T be a topological space, and let f be a function from
S onto T . Which of the following conditions on f is the weakest sufficient condition to guarantee
compactness of T :
(a) f is a homeomorphism
(b) f is continuous and injective
(c) f is continuous
(d) f is injective
(e) f is bounded

I think (c) is the weakest condition that guarantees compactness of T.
I can construct a bounded function from a compact set to a non compact set, but I am not quite sure about (d). It means f is bijective (since it's given that f maps from S onto T), so I've been trying to find a bijective function that maps from a compact set, preferably [0,1], to a non compact set.

Take $S = [0, 1]$ with the usual topology and $T = [0, 1]$ with the discrete topology (all subsets are open) and take $f$ to be the identity function. $T$ is not compact: $\{\{x\} : x \in [0, 1]\}$ is an open cover with no finite subcover.

For anyone who is still reading this thread, the answer is actually A because continuity only guarantees that the image $f(S)\subseteq T$ is compact. However, we still need that image to be equal to the set $T$ to guarantee that $T$ will be compact, which we can satisfy with surjectivity, a condition only given by choice A.


How to find point in sphere surface? Let $S$ be a sphere with origin at $(0,0,0)$.

Let $P$ be a point in $S$ surface, whose $y$ coordinate is a given $y_0$.

Let $(x_0, 0, z_0)$ be any point of the line from $(0,0,0)$ to the "shadow" of $P$ (projection of $P$ in the plane $y=0$).

How can I find $P$ ?

By spherical coordinates, exchanging y and z with respect to the usual convention in the following image

$$R=y_0 \cos \phi\implies\cos \phi=\frac{R}{y_0}\implies\sin \phi=\pm\sqrt{1-\cos \phi}$$

Note: the sign depends upon the sign of $y_0$

From $x_0$ and $z_0$ we can obtain

$\cos \theta = \frac{x_0}{\sqrt{x_0^2+z_0^2}}$
$\sin \theta = \frac{z_0}{\sqrt{x_0^2+z_0^2}}$

then we can calculate

$x=R\sin \phi \cos \theta$
$z=R\sin \phi \sin \theta$

therefore we can determine $P(x,y_0,z)$.

Let $u = \frac{(x_0,0,z_0)}{\|(x_0,0,z_0)\|}$. Then, $\langle u, P\rangle = (p_1, 0, p_3)$ where $p_1$ is the x-coordinate of $P$ and $p_3$ is the z-coordinate of $P$. Since you know the $y$ coordinate, we have $P=(p_1, y_0, p_3) = u+(0,y_0,0)$.


Logic Puzzle with Eight Clues I am not sure where to start to solve this problem:

Smith, Jones, and Rodriguez are the engineer, brakeman, and fireman on a train, not necessarily in that order.
Riding on the train are three passengers with the same last names as the crewmembers, identified as passenger Smith, passenger Jones, and passenger Rodriguez.
The brakeman lives in Denver.
Passenger Rodriguez lives in San Francisco.
Passenger Jones long ago forgot all the algebra that he learned in high school.
The passenger with the same name as the brakeman lives in New York.
The brakeman and one of the passengers, a professor of mathematical physics, attend the same health club.
Smith beat the fireman in a game of tennis at a court near their homes.

Can you discover a theorem that tells the name of the engineer, the brakeman, and the fireman?
I am lost on where to start. Thank you

This kind of problem is often solved most easily by making a table whose rows are the names and whose columns are the occupations and working through the postulates, marking the combinations $1$ (correct) and $0$ (impossible). Here's the empty table.

$$\begin{array}{r|c}
&\text{Eng.}&\text{Bman.}&\text{Fman.}\\ \hline
\text{Smith}\\ \hline
\text{Jones}\\ \hline
\text{Rodriguez}
\end{array}$$

Now number the postulates that give specific information:

The brakeman lives in Denver.
Passenger Rodriguez lives in San Francisco.
Passenger Jones long ago forgot all the algebra that he learned in high school.
The passenger with the same name as the brakeman lives in New York.
The brakeman and one of the passengers, a professor of mathematical physics, attend the same health club.
Smith beat the fireman in a game of tennis at a court near their homes.

(6) clearly implies that Smith is not the fireman. From (2) and (4) you can infer that the brakeman is not Rodriguez: if he were, passenger Rodriguez would lie in New York, and he doesn't. At this point you can enter a couple of $0$'s in the table:

$$\begin{array}{r|c}
&\text{Eng.}&\text{Bman.}&\text{Fman.}\\ \hline
\text{Smith}&&&0\\ \hline
\text{Jones}\\ \hline
\text{Rodriguez}&&0
\end{array}$$

Now (1) and (5) imply that the mathematical physicist lives in or near Denver, so by (2) he can't be passenger Rodriguez. By (3) he also can't be passenger Jones, since he obviously hasn't forgotten high school algebra, so he must be passenger Smith. (4) tells us that the passenger with the same name as the brakeman lives in New York, so the brakeman isn't Smith, and we can enter another $0$ into our table. But Smith has to have one of the occupations, so he must be the engineer: it's the only remaining possibility. And of course nobody else can be the engineer, so we now have:

$$\begin{array}{r|c}
&\text{Eng.}&\text{Bman.}&\text{Fman.}\\ \hline
\text{Smith}&1&0&0\\ \hline
\text{Jones}&0\\ \hline
\text{Rodriguez}&0&0
\end{array}$$

And at this point it all falls into place; we must have

$$\begin{array}{r|c}
&\text{Eng.}&\text{Bman.}&\text{Fman.}\\ \hline
\text{Smith}&1&0&0\\ \hline
\text{Jones}&0&1&0\\ \hline
\text{Rodriguez}&0&0&1
\end{array}$$

Smith is the engineer, Jones is the brakeman, and Rodriguez is the fireman.

From (6) Smith is not Fireman
From (3) and (5) Passenger Jones can not be the Person with brakeman because he forgot his algebra and can not be Professor.
So passenger Jones live in New York and hence from (4) Jones is brakeman .
So Rodriguez is fireman because smith can not be Fireman.
And Smith is Engineer.


Are the following two integrals equivalent? I am new to Royden's Real Analysis and am wondering what the relationship between Lebesgue and Riemann integrals.

For example, are the two integrals $\int_{\mathbb R} 1_{[0,1]}dm$ and $\int _0^1 1 dx$ equivalent? (m is Lebesgue measure on R). If yes, why?

In general, how does one transform a Riemann integral into a Lebesgue integral (when possible) and vice versa?

Yes, they are equivalent. Every Riemann integrable function in the standard sence (not in the improper sense) is Lebesgue integrable. This is a standard result in measure theory.

Let $f: [a,b] \to \mathbb R$ a Riemann integrable function. Then you have

$$ \int_a^b f(x) dx = \int_{\mathbb R} f \cdot 1_{[a,b]} dm .$$

This is possible since every Riemann integrable function is Lebesgue integrable and the interval $[a,b]$ is Borel measurable and hence $1_{[a,b]}$ Lebesgue integrable.

Further there are functions that are Lebesgue integrable but not Riemann integrable (an example is the Dirichlet function). But there are also functions that are improper Riemann integrable but not Lebesgue integrable (an example is $\frac{\sin x}{x}$ on $(0, \infty)$, since it is not absolutely improper Riemann integrable).

Yes, the two integrals are equivalent.

The general theorem is:
if the Riemann integral exists, then so does the Lebesgue integral, and they are equal.

The relationship is, given $f$ measurable: $$\int_{a}^{b}{f(x)\,dx} = \int_{\mathbb{R}}{1_{[a,b]}(x)\,f(x)\,dL^1 (x)}$$

The vice versa is not guaranteed; you may take as an example $$x \mapsto 1_{[0,1]\cap \mathbb{Q}}(x)\,,$$ which has a finite Lebesgue integral whose value is 0 (because it has nonzero values in a numerable-sized set) but has no Riemann finite integral.

NB At the end of the day, you compute the integral in the same way. With the Lebesgue integral and measure theory though, you have access to several more results.


Is there an interpretation for this trigonometric identity? A while ago I came across the following identity in an online math forum (of which I don't remember the name):
$$\tan\left(\frac{\pi}{11}\right)+4\sin\left(\frac{3\pi}{11}\right)=\sqrt{11}.$$

It is not hard to give a proof by rewriting everything in terms of $\exp(i\pi/11)$ and applying a sequence of manipulations. I am wondering where this identity is coming from. Can somebody think of a geometric interpretation? Of an algebraic one?

Edit: Here's an example of what I mean by an algebraic interpretation: The identity $$\sin\left(\frac{\pi}{7}\right)\cdot\sin\left(\frac{2\pi}{7}\right)\cdot\sin\left(\frac{3\pi}{7}\right)=\frac{\sqrt{7}}{8}$$ expresses the fact that for the Chebyshev polynomial
$$T_7(x)=x(64x^6-112x^4+56x^2-7)$$
the product of the roots $\displaystyle \sin\left(\frac{k\pi}{7}\right)$, $1\leq k<7$, of the second factor is equal to the normalized constant term $\displaystyle \frac{7}{64}$.

I could only think of a direct trigonometric interpretation of the identity.

The radius of the circular sector is 1. The measures of the central angles and  the lengths of the line segments are:

The smaller angle: $\pi/11$ rad.
The bigger angle: $3\pi/11$ rad.
The red line segment: $\sqrt{11}$.
The vertical black line segment: $4\sin(3\pi/11)$.
The vertical light red segment: $\tan(3\pi/11)$.

The red line segment is the  hypotenuse of the right triangle  whose catheti are the line segment with length $\sqrt{10}$ and the orthogonal unit segment. The $\sqrt{10}$ line segment is the hypotenuse of the right triangle  whose catheti are the horizontal line segment with length 3 and the vertical line segment with length 1.

Edited: The angle $\pi/11=2\pi/22$ is not constructible with compass and straightedge (Wikipedia, Constructible polygon ). Therefore the figure is an impossible construction with compass and straightedge only.

Remember the algebra form is $a+bi$.
You should calculate the first:
$|z| = \sqrt{a^2+b^2}$,
and then consider the angle "$\Phi$" (there are about 4 cases)...
I will give you a hint: Remember complex numbers, and the end is $z=|z|(\cos{\Phi}+i \sin{\Phi})$.


Find the radix of quadratic equation with solutions given Recently, archaeologists unearthed a tablet of an ancient civilization in which the followingproblem was posed:
Solve the equation $3x^2 − 25x + 66.$
Farther down the tablet, the solutions $x = 4$ and $x = 9$ were offered.
What is the base for this civilization’s number system? You can assume that the symbols $2, 3,4, 5, 6,$ and $9$ have the same meaning for this civilization as for us, and that positional notation is used.

I am unable to solve this problem. I believe the answer is $17$, but that was found through guess-and-check. What is the algebraic and concrete way to proceed with this problem.

Hint:  let $\,b \gt 9\,$ be the ancients' base, then $\,25_{b}=2b+5\,$ and $\,66_{b}=6b+6\,$. Substituting $\,x=4\,$ in the equation then gives $\,3\cdot16-4\cdot(2b+5)+(6b+6) = 0 \iff b=17\,$. Note that the other root $\,x=9\,$ has been neither used nor verified thus far.


Matrix representation of linear transformation in different basis Given a linear transformation $$T:V\rightarrow V$$
And given 2 bases $B$ and $C$ of the subspace $V$

Is there any connection between $[T]_B$ and $[T]_C$ (the matrix representation of the transformation in the specified basis)

Does those 2 matrices have the same rank?signature?eigenvalues?
Are they congruence or similar?

Thanks :)

Update:

I'm leaning on thinking that they are congruence but no similar and there for have same rank and signature but not same eigenvaluess

Use that if $T_{B}$ and $T_{C}$ are matrices of the same linear map (or restriction of this map on some subspace of $V$), then exist the matrix $S:T_{B}=ST_{C}S^{-1}$

The expression you're looking for is $$B \circ [T]_B \circ B^{-1} = C \circ [T]_C \circ C^{-1}.$$ Let me explain.

The way I like to think of bases is:

  Definition 0. A basis of $V$ is an isomorphism $\varphi : V \leftarrow \mathbb{R}^{\mathrm{dim}(V)}$.

Now suppose we're given:

A vectorspace $V$
A natural number $n$ defined by $n = \mathrm{dim}(V)$
A linear transform $T : V \leftarrow V$.
Two bases $B$ and $C$ of $V$.

Then we get "matrices" $$B^{-1}\circ T \circ B : \mathbb{R}^n \leftarrow \mathbb{R}^n, \qquad C^{-1} \circ T \circ C : \mathbb{R}^n \leftarrow \mathbb{R}^n$$

which are really just linear transforms. So define:

$$[T]_B = B^{-1} \circ T \circ B, \qquad [T]_C = C^{-1} \circ T \circ C$$

Then obviously, these are related by:

$$B \circ [T]_B \circ B^{-1} = C \circ [T]_C \circ C^{-1}$$


What does algebraic number look like locally? Is there any theorem characterizing what algebraic number looks like locally (in completion)? For example, do all algebraic numbers live in some $\mathbb{Q}_p$? Does there exist algebraic number in extension of $\mathbb{Q}_p$ of degree p, say?

Let me formulate your question (or rather one of your questions) this way: if $K=\mathbb Q(\alpha)\supset\mathbb Q$ is a finite extension, is there an embedding $K\to\mathbb Q_p$ for some $p$? The answer is yes; supposing without loss of generality that $\alpha$ is an algebraic integer, if $f\in\mathbb Z[x]$ is the monic minimal polynomial of $\alpha$, there are infinitely many $p$'s such that the reduction of $f$ mod $p$ has a root in $\mathbb F_p$ (one can use the nuke called Chebotarev's density theorem to see it, but I guess there are simpler proofs). Take such a $p$ not dividing the discriminant of $f$, and by Hensel lemma $f$ will have a root in $\mathbb Q_p$.

$x^2 + 1 = 0$ does not have a solution in $\mathbb{Q}_3$ since $0^2 + 1 = 1$ and  $1^2 + 1 = 2^2 + 1 = 2 \text{ mod } 3$.


Dimension of the real plane without origin Is there an intuitive way to deduce the dimension of $\mathbb{R}^2 \setminus \{(0,0)\}$ (real plane without origin), considered as a manifold?

The space $\mathbb{R}^2 \setminus \{(0,0)\}$ is homeomorphic to $\mathbb{R}^2$.  This is analogous to the way that a line is homeomorphic to a circle that is missing one point.  (See "one-point compactification for details.")  Since $\mathbb{R}^2$ is a Euclidean plane with dimension $2$, your manifold has two dimensions.


Does $\sum^{\infty}_{n=1} \frac{1}{a_n+1}$ with $a_{n+1}=a_n(a_n+1)$ and $a_1=\frac12$ have a closed form? Mr Beast here. I've been working on a complicated math problem, and I obtained a recursive sequence $a_{n+1}=a_n(a_n+1)$ and that $a_1=\frac 12$. I've been thinking, is it possible to convert this to closed form, or otherwise evaluate $\sum^{\infty}_{n=1} \frac{1}{a_n+1}$?
The original question said that $a_{1}=\frac{1}{2}$ and that $a_{n+1}=a_{n}\left(a_{n}+1\right),$ asked me to prove that $\frac{1}{a_{1}+1}+\frac{1}{a_{2}+1}+\frac{1}{a_{3}+1}+\cdots+\frac{1}{a_{n}+1}<2$ for all $n$ larger than $1 .$ I wondered if it was possible to convert into closed form to evaluate the sum definitely. I originally thought to solve it through a telescoping series, but the denominator leaves me unsure of what to do.

Hint: show (using induction on $m$) that $$\sum_{n=1}^m\frac{1}{a_n+1}=\frac1{a_1}-\frac1{a_{m+1}}.$$

This is not an answer, but deals with the underlying problem disclosed by OP in a comment to the question.
Consider the terms $$b_i = \frac{1}{a_i + 1}$$where $a_1 = \frac{1}{2}$ and $a_{i+1} = a_i^2 + a_i$:
$$b_1 = \frac{2}{3}, \,
b_2 = \frac{4}{7}, \,
b_3 = \frac{16}{37}, \,
b_4 = \frac{256}{1033}, \,
b_5 = \frac{65536}{868177}, \, \dots$$
Your task is to find a sequence that does have a closed form expression, for example
$$\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6} \approx 1.64493407$$
where each term $c_k = \frac{1}{k^2}$.  The sequence must be such that
$$c_k \ge b_k \quad \text{for all} \quad k \ge n$$
Your proof is then based on sums
$$\begin{aligned}
B &= \sum_{k=1}^\infty b_k = B_1 + B_2, \quad
B_1 = \sum_{k=1}^{n-1} b_k, \quad
B_2 = \sum_{k=n}^\infty b_k \\
C &= \sum_{k=1}^\infty c_k = C_1 + C_2, \quad
C_1 = \sum_{k=1}^{n-1} c_k, \quad
C_2 = \sum_{k=n}^\infty c_k = C - C_1 \\
\end{aligned}$$
where $B$ and $B_2$ are not known exactly, only $B_1$ is known exactly; but $C$, $C_1$, and $C_2$ are known exactly.
Because
$$c_k \ge b_k \quad \text{ for } \quad k \ge n, \quad C_2 \ge B_2$$
and if you prove that
$$B_1 + C_2 \lt 2$$
(which is something you have closed expressions for, if the infinite sum $C$ has a known closed expression), you have proven that
$$B_1 + B_2 = B \lt 2$$

The series to focus on those related to the Riemann Zeta function at even values:
$$\zeta(n) = \sum_{k=1}^\infty \frac{1}{k^n}$$
The particular one I was thinking of is
$$\zeta(20) = \sum_{k=1}^\infty \frac{1}{k^{20}} = \frac{174611 \, \pi^{20}}{1531329465290625} \approx 1.000000953962$$
with $n = 10$.
You see,
$$c_k = \frac{1}{k^{20}} \ge b_k \quad \text{for all} \quad k \ge 10$$
In particular,
$$\begin{aligned}
b_{8} \approx 1.4255 \cdot 10^{-9} \\
b_{9} \approx 2.0321 \cdot 10^{-18} \\
b_{10} \approx 4.1292 \cdot 10^{-36} \\
b_{11} \approx 1.7051 \cdot 10^{-71} \\
c_{8} \approx 8.6736 \cdot 10^{-19} \lt b_{8} \\
c_{9} \approx 8.2253 \cdot 10^{-20} \lt b_{9} \\
c_{10} = 1.0000 \cdot 10^{-20} \gt b_{10} \\
c_{11} \approx 1.4864 \cdot 10^{-21} \gt b_{11} \\
\end{aligned}$$
so we know that
$$\sum_{k=10}^\infty b_k \lt \sum_{k=10}^\infty \frac{1}{k^{20}}$$
and because
$$\sum_{k=10}^\infty \frac{1}{k^{20}} = \frac{174611 \, \pi^{20}}{1531329465290625} - \frac{106662370149283833915501942267857492605696794711249007378798472708401}{106662268397529335923611516553737065783558248857600000000000000000000}$$
we can now calculate the exact sum,
$$\left( \sum_{k=1}^{9} b_k \right) + \left( \sum_{k=10}^\infty \frac{1}{k^{20}} \right) \lt 2$$
and therefore
$$\sum_{k=1}^\infty b_k \lt 2$$
Because floating-point numerical calculations do not have enough precision here, I used
$$C = \sum_{k=1}^\infty \frac{1}{k^{20}} = \frac{174611 \, \pi^{20}}{1531329465290625}$$
and
$$\begin{aligned}
C_1 &= \sum_{k=1}^9 \frac{1}{k^{20}} \\
~ &= \frac{106662370149283833915501942267857492605696794711249007378798472708401}{106662268397529335923611516553737065783558248857600000000000000000000} \\
\end{aligned}$$
and
$$\begin{aligned}
B_1 &= \sum_{k=1}^9 b_k \\
~ &= \frac{6494086337598191931522860064611183626609948977242275075491600909077850578669643051565257331009366963694648505112737011786954151363577920417599029974920037685716132002271729535802246547158018}{3247043168799095965761430032305591820008878453592436087532812953641848353074504436078825354366464203708206267593141994294414224765240674053815444080703044269735007407109149741117947776621057} \\
\end{aligned}$$
and verified that $B_1 + C - C_1 \lt 2$ via
$$C + B_1 - C_1 - 2 \lt 0$$
The left side, for evaluation in e.g. Maxima (a free CAS program, wxMaxima being its graphical version), is
174611*(%pi^20) / 1531329465290625 - 16492253350531080128239160104845001622128429492012917644483996917752120075662103069882162351389153630866311951764035014010230695125685435114364416532374072620244556261880185006228929268661858194735280962676368978555791749162077866217824355140382739573209517 / 16492237617562539365036586070920801101220100452147363781860837846145943893309011792682018865262830776584392846863644670314594880250899416573838450710644960492510494335952071191890802960534017689711520151393881057079583788816561717064499200000000000000000000

and Maxima gives it an approximate value $-6.6613\cdot10^{-16}$, which indeed is less than zero, proving the series sum $\sum_{k=1}^\infty b_k \lt 2$.


Proof by Contradiction (I think) Prove: If $x^2+x-6 \ge 0$, then  $x\le -3$ or $x \ge 2.$
This is my initial work, but I'm not sure if this is the way to go or if should do a direct proof.
Proof by contradiction: Assume $x^2+x-6 \ge 0$ and $-3<x<2$.  Is this the correct start?

$f(x)=x^2+x-6=(x+3)(x-2)$
From that one can see that $f(x)\geq 0$ when $x\geq 2$ of $x\leq -3$.

As you have assumed, let $-3<x<2$ which implies that $x+3>0$ and $x-2<0$, i.e. these are quantities of opposite signs, so their product must be negative. But, $(x+3)(x-2)=x^2+x-6$ is non-negative, as given, hence, contradiction!
An algorithmic way of looking at questions of this type is given the name wavy-curve method.
In the direct way, $x^2+x-6\ge 0 \implies (x+3)(x-2)\ge 0$, so both the factors must be of the same sign.
There are only really $3$ regions to check for signs. $(-\infty,-3],(-3,2),[2,\infty)$.
In $[2,\infty)$, both $x+3$ and $x-2$ are non-negative, so their product is also non-negative. As you go into $(-3,2)$, the sign of $x+3$ remains the same, i.e. it is still non-negative, but the sign of $x-2$ is now negative, so their product is now negative. Going in $(-\infty,-3)$, the sign of $x-2$ is still negative, as it was in the previous interval, but that of $x+3$ has changed to negative as well, so their product is now positive.
Thus, the given term $x^2+x-6$ changes signs at the junctions of crossover $(2,3)$, and it is non-negative in two of them.
You have to be careful with choosing the endpoints of the intervals, which depends on whether there is an equality sign or not in the given inequality.
Also, try this: For what $x$ is $(x-1)(x-2)^2(x-3)^3>0$? (specify regions, that is ...)


What is the point of countable vs. uncountable sets? I understand how to use these concepts and how to prove certain sets are countable or uncountable. However I don't get the point of it. What difference does it make whether a set is countable? People say that Cantor's proof that the real numbers are uncountable is a milestone of mathematics. Having read through the proof, I'm still struggling to understand what these ideas are important.

One important application is this: Consider the set of functions that take an integer argument and return an integer result.  It's not hard to show that this set is uncountable.  Now consider the set of computer programs.  This set is countable.  Therefore there are uncomputable functions—in fact most functions cannot be computed.

This is a pretty common type of argument in some contexts.

Sample a real number from $\mathbb{R}[0,1]$ as an infinite series of fair coin tosses.  Then, each sample is not only a refinement, also the start of a new sample.  The result being a rational may seem much less likely in the ergodic, but, sampling a rational sees sampling infinitely many copies of the terminus.  Sampling an irrational (in this manner) sees infinitely many different, thus that the irrational is represented less, of the balance that the rationals and irrationals are each dense in the reals.  It takes infinitely many fair coin tosses to sample a real number, that samples infinitely many, the sample is always plural.

Consider the function $\frac{n}{d}$ for naturals n,d with $0 \le n \le d$.  As $d \to \infty$, $lim_{n\to d} f = 1$.  Here $f$ ranges between zero and one and always includes zero and one.  Also, the differences $f(n+1)-f(n)$ are constant (uniform).  Then it is a CDF, uniform, with domain $\mathbb{N}$.

As that's in the countable it's a way to build notions of uniform probability from the natural integers, satisying various intuitions, and reasonably for application.  The point of the uncountable here is to see there are various ways to see what would be facts about these systems.

Matters of the infinite are relevant to mathematics, in the pure for what it is and the applied for what it is and does.

In terms of the applied and for practical purposes for real analysis:  additivity of measure is countable.  And, much work is done in formalizing analysis in the constructible and countable.

It's good to know because other people believe it, to understand.  But, except for its own sake, there aren't results solely due the uncountable.  Results not of the uncountable are available via asymptotics, methods of exhaustion, and concrete mathematics.  Here for example, point discontinuities can be discarded in solutions due their vanishing without appeal to their zero measure.  And, the computable can be described in the countable constructible, the concrete mathematics is quite heavy.  For example, given a wager whether a random integer is even or odd, the asymptotic density of even integers in integers is $\frac{1}{2}$, as an example of that the cardinalities of those two set being the same does not reflect their densities being different.

So, there is much intuition about the real numbers, not all satisifed with the uncountable.  However, it is quite standard and in the general curriculum not knowing the general features as instructed and thus shared would be ignorant, regardless their eventual disposition or lack thereof.

The ideas are important as an example of a way to consider the mathematically infinite.  It is important that their relevance is qualified, in their consideration.


Question about a proof in Rudin R&CA: $ \mu(\{x\in \mathbb R^k: d(x, A(B_1(\mathbf0))) \le \eta\}) < \varepsilon$ Suppose we have a linear function $A\colon \mathbb R^k \to \mathbb R^k$ which is not injective (hence not surjective). Clearly, we have $\mu(A(\mathbb R^k))=0$ where $\mu(\cdot)$ is the Lebesgue measure on $\mathbb R^k$.

  Then, for every $\varepsilon>0$ there exists a $\eta>0$ s.t.  $$
 \mu(\{x\in \mathbb R^k: d(x, A(B_1(\mathbf0))) \le \eta\}) <
 \varepsilon. $$ where $B_1(\mathbf 0)$ is the unit ball in $\mathbb
 R^k$.

Is this fact so obvious? I find it in the proof of a theorem on Rudin, R&CA (Chapter 7, thm. 7.24, pag 152) and I'm quite puzzled. I tried to justify this passage in the following manner: consider the sets
$$
E_n :=\left\{x\in \mathbb R^k: d(x, A(B_1(\mathbf0))) \le \frac{1}{n}\right\}
$$
Then
$$
\bigcap_{n\in \mathbb N}E_n=\{x\in \mathbb R^k: d(x, A(B_1(\mathbf0))) =0\}=\overline{A(B_1(\mathbf0))}
$$
(I'm not so sure about the last equality...). Then, by monotonicity of the measure, we get
$$
\mu\left(\bigcap_{n \in \mathbb N}E_n\right)=0
$$
which implies, if $\mu(E_1)<+\infty$
$$
\lim_{n\to+\infty} \mu(E_n)=0.
$$
From this, I think that Rudin's claim follows (just writing the $\varepsilon-\delta$ definition of limit). But I'm not sure: indeed, I can't show that $\mu(E_1)<+\infty$ and, secondly, I think there is something I'm missing, it can't be too difficult (otherwise Rudin would have explained it!).

Thanks in advance.

The set $A(\mathbb R^k)$ is contained in an hyperplane.

The set $A(B_1(0))$ is bounded.

So the intuitive idea is that the set of points which are at a distance less than $\eta$ away from $A(B_1(0))$ is a very flat set, with thickness less than $2\eta$. Its Lebesgue measure is thus small.

More precisely:

Up to an isometry (which does not change the Lebesgue measure of the sets you are considering), you can suppose that $A(B_1(0))$ is a bounded subset of $Span(e_1,\dots,e_{k-1})$, with $e_j:=(0,\dots,0,1,0,\dots,0)$, say $$A(B_1(0))\subset[-M,M]^{k-1}\times \{0\}$$
for some $M>0$.
Then observe that for any $b>0$ there exist an $\eta$ such that $$\{x\in\mathbb R^k\,|\,d(A(B_1(0)),x)\leq \eta)\}\subset[-(M+1),(M+1)]^{k-1}\times [-b,b]\,.$$
The measure of the set on the right hand side is $2b(2(M+1))^{k-1}$ it is enough to take $b$ small enough to get the result. ($b\leq\varepsilon/(2(2(M+1))^{k-1})$)


Estimating the next prime in a random sequence Is there a way (using statistics, or data analysis) to predict the nth term of a random sequence given the first k terms. For instance, what about predicting the $n$th in this sequence (random, appearing list of primes):

$p_n$ $=$ $2, 3, 11, 17, 23, 53, 79, 103, 197, 593, 971, 1021, 1663, 2099, 3571, 4447, 5419, 10781, 29347, 57901, 78517, 91129, 251893, 695407...$

(all terms less than $1M$ of this specific, random appearing sequence of primes)

Given $p_{24}$ $=$ $695407$ estimate $p_{70}$?

Is there a way to solve this sequence (find $p_n$ given the first 22 terms) with statistics and analyzing the behavior of it. In other words, there is an approximation formula for any single-set of sequcences to find any $n$th term greater than it. Thanks in advance.

Well I can't say I know much on the subject, but given the fact that your sequence is 'random' and strictly of prime numbers less than 1M, there's probably not any way to do what you are asking. Sorry.


Prove that if events A and B are independent, then the complement events of A and B are also independent. I know that:
$$\begin{gathered}P\left(A\cap B\right)=P\left(A\right)P\left(B\right)\\
P\left(A^{C}\right)=1-P\left(A\right)\\
P\left(B^{C}\right)=1-P\left(B\right)
\end{gathered}
$$

My proof so far:
$$\begin{gathered}P\left(A^{C}\cap B^{C}\right)=\left(1-P\left(A\right)\right)\left(1-P\left(B\right)\right)=\\
1-P\left(B\right)-P\left(A\right)+P\left(A\right)P\left(B\right)=1-P\left(B\right)-P\left(A\right)+P\left(A\cap B\right)
\end{gathered}
$$

After that, I'm stuck.  Any help would be appreciated.

Assume $A$ and $B$ are independent. Then
\begin{align}
P(A^c \cap B^c)
&= 1 - P(A \cup B) \\
&= 1 - P(A) - P(B) + P(A \cap B) \\
&= 1 - P(A) - P(B) + P(A)P(B) \\
&= (1-P(A))(1-P(B)) \\
&= P(A^c)P(B^c).
\end{align}

By definition if two events are independent then $P(A | B)=P(A)=P(A/B')$.

So, $P(A/B)=P(A \cap B)/P(B)=P(A)$ and hence by multiplying both sides by $P(B)$ we get $P(A \cap B)= P(A)P(B)$.

Hence proved.


State a system as a canonical minimum problem Suppose we have a linear system $$Ax=b,\,\,x\ge0,$$ where $A$ is a $m\times n$ matrix and $b$ is a given $n\times 1$ column vector.

Def: We say a problem is a canonical minimum problem if the problem requires us to find a $x\in\mathbb R^n $ s.t. $$Ax=b,\,\,x\ge0\text{ , minimize }c\cdot x \text{ for some given }c\in\mathbb R^n.$$

My problem: Consider the linear system $Ax=b,\,\,x\ge0$ with nothing to be optimized. Show how to state this system as a canonical minimum problem by the right choice for the vector $c$.

In fact, I am confused with what the problem says. And I am not sure what it wants me to do. Can someone help me understand it?

Thanks in advance.

LinAlg has almost answered this question.  I am posting this answer so as to clear this question from the unanswered question queue.

  Can someone help me understand it?

\begin{align}
S_1 &= \{ x \in \Bbb{R}^n \mid Ax=b, x \ge 0 \} \\
S_2 &= \arg\min_{x \in \Bbb{R}^n \\ Ax=b \\ x \ge 0} c^Tx
\end{align}

Determine $c \in \Bbb{R}^n$ such that $S_1 = S_2$.

The solution is trivial: take $c = 0 \in \Bbb{R}^n$, so that $c^Tx \equiv 0$.  In this case, any element in $S_1$ is a minimiser to the trivial "canonical LP", so $S_1 \subseteq S_2$.  The other inclusion should be clear, and I left the detailed explanation as an exercise.


Geometry of finitely generated convex sets After drawing some pictures, I guess the following is true: Let $y_{1},\ldots, y_{d}$ be points in a Euclidean space $\mathbb{R}^{N}$ (the Euclidean norm is denoted by $\left\Vert \, \cdot \, \right\Vert$) and let $C = {\operatorname{conv}}{\left\{y_{1},\ldots,y_{d} \right\}}$ be the convex hull of those points. Assume that $0 \in C$. Then there is $r >0$ such that $$\left\{ \lambda x : x \in C \text{ and } \lambda >0 \text{ such that } \left\Vert \lambda x \right\Vert \leq r \right\} \subset C.$$

Here is my question: Is the above property correct? If so, could you please give me some clues?

Any help would be appreciated.

Thank you very much for your help.

Given that $0\in C$, you can use convexity to show that $\{ \lambda x : x \in C \text{ and } \lambda \in [0,1]\} \subset C$. From here just take $r=\min\{|x|:x\in C\}$.


Examples of theorems in graph theory I'm looking for examples of uses of the incidence matrix in graphs for my class (apart from proving that for a graph $G=(V,E)$ we have $2|E|=\underset{v\in V}{\sum}d(v)$).
Can you think of anything interesting?
Thanks in advance for any help.

A classic example is the Friendship theorem of Erdos and Renyi.

There is also the classical result that the powers of the adjacency matrix describe the number of paths of a given length between two vertices.


Proving that a space is disconnected Show that a subspace $T$ of a topological space $S$ is disconnected iff there are nonempty sets $A,B \subset T$ such that $T= A\cup B$ and $\overline{A} \cap B = A \cap \overline{B} = \emptyset$. Where the closure is taken in $S$.

I've used this relatively simple proof for many of these slightly different types of questions so I was wondering if it's the right method. It seems pretty good, except for the 'where the closure is taken in $S$ part'.

$T$ is disconnected if and only if there exists a partition $A,B \subset T$ such that $T = A \cup B$ and $A \cap B = \emptyset$. Also, $A$ and $B$ are both open and closed therefore $\overline{A} = A$ and $\overline{B} = B$. The result follows.

It looks fine to me, in particular because as $\,A\subset \overline{A}\Longrightarrow A\cap B\subset \overline{A}\cap B\,$ , so if the rightmost intersection is empty then also the leftmost one is, which is the usual definition


Fixed point iteration contractive interval Consider the function $F(x) = x^2-2x+2$.

Find an interval in which the function is contractive and find the fixed point in this interval.

What is the convergence rate of the fixed point iteration: $x_{n+1} = F(x_n)$ in that interval?

I'm lost on fixed point iteration. I've watched a few youtube videos, but I'm still not understanding contractive intervals and methods for finding fixed points in functions.

You need an interval $[a,b]$ so that there's a $0\leq c<1$ such that $|f(x)-f(y)|\leq c|x-y|$ for all $x,y\in[a,b]$. In our case, we have
$$
|f(x)-f(y)|=|x^2-y^2-(2x-2y)|\leq c|x-y|.
$$
Notice that, in fact, $f(1)=1$, so you're going to want to choose an interval somewhere around $1$.

You need to consider the recurrence relation

$$ x_{n+1}= x_{n}^2-x_n+2 ,$$

which suggests considering the function

$$f(x) = x^2-x+2  \implies  f'(x) = 2x-1. $$

Now, to find the interval, use the condition $|f'(x)|<1$. See a related problem.


Synthetic division by $ax^2+bx+c$. I know that synthetic division can be used in order to find quotient $q(x)$ and remainder $r(x)$ of a polynomial $p(x)$ when it is divided by some linear polynomial like $x-c$. Now, does exist some procedure (another than long division) in order to find $q(x)$ and $r(x)$ when the divisor is $ax^2+bx+c$? Suppose $b^2-4ac<0$.

Assume your divisor is $x-d$ rather than $x-c$ since $c$ is already used in the formation of the original polynomial. You can write the dividend $y$ as:

$y = a(x-d)^2 + b(x-d) + 2adx - ad^2 + c + bd = a(x-d)^2 + b(x-d) + 2ad(x-d) + 2ad^2 -ad^2 + c + bd = a(x-d)^2  + (b+2ad)(x-d) + ad^2 + bd + c = (x-d)(a(x-d) + b+2ad) + ad^2+bd+c = (x-d)(ax + ad + b) + ad^2+bd+c \Rightarrow q(x) = ax+ad+b, r(x) = ad^2+bd+c$.

For $ax^2 + bx +c$ with $b^2 - 4ac $>0 find the roots of the polynomial and apply the syntectic division twice in a row .


Define infinitesimally small? Why do we ignore squares of infinitesimal quantities in differential calculus , doesn’t that cause error when we require very accurate measurements.?

when you find a derivative it is something like $\frac{dy}{dx}=\frac{a_0\ +\ a_1{dx}\ +\  a_2{dx}^2 .......}{dx}$. Here $a_0\ +\ a_1\ .....$can be a function of x. So if $a_0$ is not zero this thing will tend to infinity (so lets us assume it to be zero). Overall it is $\frac{dy}{dx}=a_1\ +\ a_2{dx}.... $ . Now obviously you can take dx to be very small which is tending to zero , which will not give error.(let us say u r calculating till one billionth decimal place , you can take dx to be even smaller than that)

In all empirical sciences there is error no matter how accurate the measurements are. They will always be rational numbers and the error term is always non-trivial. You have to justify ignoring infinitesimals and that typically involves showing the error term is smaller than the error of the measurement. in practice this happens routinely and so it's useful to study the case of negligible terms.


Show $\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$ Positive real numbers $a,b,c$ satisfy $abc=1$. Prove
  $$\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$$

I tried AM–GM, Cauchy–Schwarz and Jensen's but they all failed.

Following stewbasic's hint in the comments:
$$\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$$
$$\iff(b^2+b+1)(c^2+c+1)+(c^2+c+1)(a^2+a+1)+(a^2+a+1)(b^2+b+1)\ge(a^2+a+1)(b^2+b+1)(c^2+c+1)$$
The LHS expands as
$$a^2b^2+a^2b+ab^2+\color{blue}{a^2}+b^2+ab+a+b+1\\
+b^2c^2+b^2c+bc^2+\color{blue}{b^2}+c^2+bc+b+c+1\\
+c^2a^2+c^2a+ca^2+\color{blue}{c^2}+a^2+ca+c+a+1$$
while the RHS expands as
$$a^2b^2c^2+abc(ab+bc+ca)+\\
a^2b^2+b^2c^2+c^2a^2+abc(a+b+c)+\\
abc+a^2b+b^2a+b^2c+c^2b+c^2a+a^2c+\\
\color{green}{ab+bc+ca}+a^2+b^2+c^2+a+b+c+1$$
Cancelling and using the relation $abc=1$ we find that the inequality above is equivalent to
$$\color{blue}{a^2+b^2+c^2}\ge\color{green}{ab+bc+ca}$$
which can be seen to be true by applying the Cauchy–Schwarz inequality on $(a,b,c)$ and $(b,c,a)$. By following the biconditionals back we see that the original inequality is true.

Let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$, where $x$, $y$ and $z$ are positives.

Hence, by C-S and Rearrangement we obtain $$\sum\limits_{cyc}\frac{1}{a^2+a+1}=\sum\limits_{cyc}\frac{y^2}{x^2+xy+y^2}=\sum\limits_{cyc}\frac{y^2(y+z)^2}{(x^2+xy+y^2)(y+z)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2)(y^2+2yz+z^2)}=\frac{\sum\limits_{cyc}(x^4+2x^3y+2x^3z+3x^2y^2+4x^2yz)}{\sum\limits_{cyc}(x^4+2x^3y+x^3z+3x^2y^2+5x^2yz)}=$$
$$=\frac{\sum\limits_{cyc}(x^4+2x^3y+x^3z+3x^2y^2+5x^2yz)+xyz\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}-x-y-z\right)}{\sum\limits_{cyc}(x^4+2x^3y+x^3z+3x^2y^2+5x^2yz)}\geq1$$


A positive integer is divisible by $3$ iff 3 divides the sum of its digits I am having trouble proving the two following questions:

If $p|N$, $q|N$ and gcd(p,q)=1, then prove that $pq|N$
If $x$ is non zero positive integer number, then prove that $3|x$ if and only if 3 divides the sum of all digits of $x$.

For both questions I tried to use theorems of discrete mathematics, but I could not find the way to solve them.

(1)  The hypotheses $p|N$ and $q|N$ give us two integers $m,k$ so that $N=mp$ and $N=kq$.  This implies $mp=kq$ so that $q|mp$.  Now, since $gcd(p,q)=1$ and $q|mp$ we know that $q|m$ (think about why this is true if it's not clear).  Then  $m=sq$ for some integer $s$.  Putting this all together, we have $N=mp=sqp=s(pq)$ so $pq|N$.

(2) Here we have an if and only if statement so you'll have to prove two statements/directions(or both at once):

(i) if $3|x$ then $3$ divides the sum of the digits of $x$

(ii) if $3$ divides the sum of the digits of $x$ then $3$ divides $x$

The strategy is to write the number in base $10$, for example (not a proof):

$1356=1\cdot 10^3+3\cdot 10^2+5\cdot 10^1 + 6\cdot 10^0$

Now $10$ has remainder $1$ under division by $3$ so the remainder of $1356$ under division by $3$ is $1 \cdot 1^3 + 3\cdot 1^2 +5\cdot 1^1 +6\cdot 1^0=1+3+5+6$ which is exactly the sum of the digits.  Then the remainder under division by $3$ of $1356$ and $1+3+5+6$ are the same, and $3|1356$ if and only if $3|1+3+5+6$.

Try to do this in general for some number $n$ with logical steps following the idea of the example above to write a formal proof.

Hints.

(1) Use the definitions. What does it mean for an integer to divide another? What does it mean for two integers to be relatively prime? Do you know that the $\text{GCD}$ of two integers may be expressed as integral linear combinations of these integers?

(2) Just expand the number out in powers of $10$ in the usual decimal convention. Then note that any positive power of $10$ reduces to $1$ modulo $3.$ Can you see why the rule for divisibility for $3$ and $9$ are similar?


Differentiable $\mathbf{f}:\Bbb R \to \Bbb R^3$, $|\mathbf{f}(t)| = 1$ implies $\mathbf{f}(t)\cdot\mathbf{f}'(t) = 0$ This question comes from Rudin's PMA problem 9.13.

For a differentiable function $\mathbf{f} : \Bbb R \to \Bbb R^3$ and $|\mathbf{f}(t)|=1$ for all $t$, I want to prove that $\mathbf{f}(t)\cdot\mathbf{f}'(t) = 0$.

Notionally, this makes sense to me. $|\mathbf{f}(t)|=1$ means that $\mathbf{f}(t)$ describes a curve on the unit sphere in $\Bbb R^3$. Of course, the tangent to this curve at any point will lie in the tangent plane to that point on the unit sphere, which is normal to the vector from the origin to that point. Therefore, by orthogonality, the dot product will be zero.

However, I am struggling to prove it rigorously using the definition of derivatives.

I have

$$\mathbf{f}(t)\cdot\mathbf{f}'(t) = \lim_{h\to 0} \frac{1}{h} \left(\mathbf{f}(t) \cdot \left(\mathbf{f}(t+h)-\mathbf{f}(t)\right)\right)$$

by applying properties of limits, etc., but this seems to bring me right back in a circle (no pun intended). I also know that $|\mathbf{f}(t)| = 1$ means that $\mathbf{f}(t)\cdot\mathbf{f}(t) = 1$. I'm just not sure how to proceed from here. I want to argue that the numerator of the limit must be zero, but I can't; evaluation of the limit just brings me back to $\mathbf{f}\cdot\mathbf{f}'$.

This is a homework problem, so please just provide a nudge in the right direction.

Edit: I also don't "know" that $\mathbf{a \cdot b} = |\mathbf{a}||\mathbf{b}|\cos \theta$. I suppose I could prove it, but I feel as if that's avoiding the intent of the question.

Based on @bobobinks' comment, I can employ
$$ |\mathbf{f}(t)| = 1 \implies \mathbf{f}(t)\cdot \mathbf{f}(t) = 1$$

Differentiating this, applying the product rule, I would have
$$\mathbf{f}(t)\cdot\mathbf{f}'(t)+ \mathbf{f}'(t)\cdot\mathbf{f}(t) = 2\mathbf{f}(t)\cdot\mathbf{f}'(t) = 0$$
and this would finish the problem.

How would you answer if you defined $f'(t)*f(t)=0$  as $<f'(t),f(t)> = 0 $ instead? where f is defined as a differential mapping.


Show that a space $H$ is Hilbert. Let
\begin{equation}
H = \{f \in L^2(0,\infty):\int_0^\infty f^2(x)e^{-x}dx \lt\infty\}
\end{equation}

Show that $H$ is a Hilbert space with scalar product
\begin{equation}
\langle f,g \rangle = \int_0^\infty f(x)g(x)e^{-x}dx
\end{equation}

I was trying to show that it is complete about the norm coming from scalar product, but I think it's way too difficult. Instead, I was thinking to define a Unitary isomorphism between $H$ and $L^2(0,\infty)$ like this:
\begin{equation}
U:H\rightarrow L^2(0,\infty)
\end{equation}

such that $\langle U(f),U(g)\rangle_{L^2}$ = $\langle f,g\rangle_{H}$.

The idea is that the scala product on $H$ is really similar to $L^2$ 's one.
Any suggestions on how to define it properly?

I think this is false. $\{I_{(0,n)}\}$ is a Cauchy sequence in $H$ which is not convergent. If this sequence converges in $H$ then some sequence converges almost everywhere w.r.t. $e^{-x}dx$ hence w.r.t. Lebesgue measure and the limit has to be $1$ almost everywhere w.r.t. Lebesgue measure. But then the limit is not in $L^{2}(0,\infty)$, hence not in $H$.

Ok this is what I got in the end:

Again, we have

\begin{equation}
H = \{f \in L^2(0,\infty): \int_0^\infty f^2(x)e^{-x}dx < \infty\}
\end{equation}
and
\begin{equation}
\langle f,g \rangle_{H} = \int_0^\infty f(x)g(x)e^{-x}dx
\end{equation}

So, (I guess),
\begin{equation}
||f||_H = \sqrt{\langle f,f \rangle_{H}} = \left(\int_0^\infty f^2(x)e^{-x}dx \right)^{1/2}
\end{equation}

Taking $f_n = \chi_{(0,n)}$ we have $f_n \in L^2(0,\infty) \forall n$ and it is Cauchy in $H$ because, assuming $n >m$ we could consider
\begin{equation}
||\chi_{(0,n)}-\chi_{(0,m)}||_H = ||\chi_{(m,n)}||_H = \int_m^n e^{-x}dx = -e^{-n}+e^{-m}<\epsilon
\end{equation}
For $m,n$ sufficiently large. And also
\begin{equation}
||\chi_{(0,n)}-\chi_{(0,\infty)}||_H = \int_0^\infty (\chi_{(0,n)}-\chi_{(0,\infty)})^2e^{-x}dx < \infty
\end{equation}

Which step could be wrong?

I'm sorry for my mistakes, I'm here to learn, and also to find people to talk over with.

Thank you


Proving $e^{i\pi} = -1$ without proving $e^{ix} = \cos x + i\sin x$ Just as the title asks: I'm looking for a proof of Euler's identity without first proving the general case of $e^{ix} = \cos x + i\sin x$.

I outlined in a previous question a proof that $$\lim_{n\to\infty}\left(1+\frac{ix}{n}\right)^n = \cos x + i\sin x$$

That outline (about halfway through the answer) can be altered just to prove it for $x=\pi$ pretty directly. Basically:

$$\frac{\cos(\pi/n)+i\sin(\pi/n)}{1+i\pi/n}=1+g(n)$$

where $ng(n)\to 0$, so we can prove that:$$\left(\frac{\cos(\pi/n)+i\sin(\pi/n)}{1+i\pi/n}\right)^n\to 1$$

You can prove that $ng(n)\to 0$ geometrically.  You essentially only need to prove that:

$$\left|\,\cos(\pi/n) - 1\right| = O(1/n^2)$$ and $$\left|\,\sin(\pi/n)-\pi/n\right| = O(1/n^2)$$

The first is easy to prove. The distance from $(\cos x,\sin x)$ to $(1,0)$ is $\sqrt{2-2\cos x}$ and $x$ is the length of the arc of the circle between those two points.  So $$0\leq 1-\cos\pi/n\leq \frac{\pi^2}{2}\frac{1}{n^2}$$

The second requires more care. We can easily see $0\leq \sin\pi/n \leq \pi/n$ since $\sin \pi/n$ is the distance of $(\cos\pi/n,\sin\pi/n)$ to the real line, and $\pi/n$ is the length of a longer path to the real line.

Now, take the points $P=(1,0)$ and $Q=(\cos 2\pi/n,\sin 2\pi/n)$. Draw the tangents to the unit circles at these points and find their intersection, $R$. Then show that the length of
the path $PRQ$ is $2\tan \pi/n$. This is a path "outside" the circle, so it must be greater than the path of length $2\pi/n$ along the circle between $P$ and $Q$, so we get inequality: $$\tan\pi/n \geq \pi/n$$ This gives us the inequality:

$$\frac{\pi}{n}\cos \pi/n \leq \sin\pi/n \leq\frac{\pi}n$$

Which means that $$|\pi/n-\sin\pi/n|\leq \frac{\pi}{n}(1-\cos\pi/n) = O(1/n^3)$$

Just take the log of both sides:

$$\ln e^{i\pi}=\ln(-1)\Rightarrow i\pi\ln e=i\pi\Rightarrow i\pi=i\pi$$


Local implicit function theorem and Global implicit function theorem Suppose that $f=f(x,y)$ is $C^1$ in $x$ and $y$, where $x,y\in\mathbb{R}$ and $f(x_0,y_0)=0$. Moreover, $\frac{\partial f(x,y)}{\partial y}>0$ for all $x,y\in\mathbb{R}$. Could we conclude that $y$ can be solved implicitly from $f(x,y)=0$ in terms of $x$ as $y=y(x)$ for all $x\in\mathbb{R}$? If the answer is no and an additional assumption is imposed, i.e. we can obtain $y'(x)>0$ for all $x\in\mathbb{R}$ by implicitly differentiating $f(x,y(x))=0$ with respect to $x$, then these assumptions are enough to get the same conclusion?

I'm not sure what you want to know beyond what is shown in this question and it's answer, but that question shows already where you have to expect the result to break down. As an example, consider

$$F(x,y) = - e^{-y} +x$$
Clearly, $\frac{\partial F}{\partial y} = e^{-y}>0$ everywhere, but it is easy to see that  $F$ is strictly negative for $x\le0$ on the one hand side but admits a (obviously unique) solution of $F(x,y) = 0$ for each $x>0$. This shows that if the bounds on the derivates no longer hold, the solution may go off to infinity (Which is exactly what happens in this example).

For Global Implicit Function Theorem you need "somewhat" stronger assumption, namely,  $\frac{\partial f(x,y)}{\partial y}\geq d>0,$ for all $(x,y)\in\mathbb{R}^2,$ $d$ is constant. See, for example, Lemma 1 and its proof in:

Ge, S.S.; Wang, C., Adaptive NN control of uncertain nonlinear pure-feedback systems, Automatica 38, No.4, 671-682 (2002). ZBL0998.93025.


Solve a simple differential equation This is an equation I met while solving a probability theory problem:

$ dy = 2(m/x-y) \cdot dx/x $

m is a constant

Also it is known that $y(m) =1$.

I have the function which satisfies the equation and I verified that it works. But I haven't solved a single differential equation in many years so... I just got curious how one can derive this solution.

Is this equation of some particular type which is solvable through a well-known procedure? Any reference?

Sorry... I am typing on my phone... I will try to improve the equation outlook now.

$$\dfrac{dy}{dx}=\dfrac{2m}{x(x-y)}\\\dfrac{dx}{dy}=\dfrac{x^2-xy}{2m}\\ x'+\frac{y}{2m}x=\frac1{2m}x^2$$
It's Bernoulli form, make $z=x^{-1}$ then $\frac{dz}{dy}=-x^{-2}\frac{dx}{dy}$ isolate x' and substitute in the ode, it reduces into a Linear Form do integrating factor and some algebra and you get it in the form $x(y)= F(y) +C$


Prove that $N,R,F$ are collinear In a triangle $ABC$, let I be the incentre. Let $D$, $E$, $F$ be the intersections of $(ABC)$. with the lines through $I$ perpendicular to $BC$, $CA$, $AB$, respectively.
Define $O= BC \cap DE$ and $L= AC \cap DE$ . Define $IF\cap AB= R$ . Let $N=(BOF) \cap (LAF)$ .Prove that $N$,$R$,$F$ are collinear.
My progress: Since $F\in (ABC) $ , I thought of using simson points . So I took points $J$,$R$,$K$ as the simson points in $BC$,$BA$,$AC$ wrt point $F$ respectively. ( as shown in the diagram )
Then since $NBFO$ and $AFLN$ is cyclic, we get that $180- \angle ONF=\angle OBF=\angle CBF=180- \angle FAC=180 -\angle FAL = \angle FNL $ .
Hence points $O$,$N$,$L$ are collinear .
Now, I am stuck . I tried using phantom points but couldn't proceed . I am thinking of using Radical axis but still confused.
Here are some more observations which might be trivial but still, we have $BJFR$, $RFKA$,$CJFK$ concyclic . We also have $\Delta JFK \sim \Delta BFA $
Please post hints if possible.
Thanks in advance.
Ps: This is my own observation, so there is a very high chance that I might be wrong.
Below are a few diagrams for the problem.

This result is false! An IMOTCer said that when we use the relation tool, the geogebra shows that $N \not\in RF$ .
Thank you show much to everyone for trying this problem .


Orientation of multiplying integrals Consider,

$$I = \int_{-\infty}^{\infty} e^{-x^2} dx$$

The trick is to multiply by $I$ again to get $I^2$ But they often write:

$$I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dy\space dx$$

The issue is orientation.

$I$ is an integral, which is orientated in the $xy$ plane, and $x$ is the independent variable, y is the dependent variable.

When you consider the other one with $e^{-y^2}$, then $y$ is the independent variable and $x$ is the dependent variable.

How does this work?

You want to think of $x$ and $y$ as two independent variables, and something else, say $z$, would be the dependent variable.

In the first integral, if you're thinking of $y$ as the dependent variable, rename it as $z$, and then add $y$ as a second independent variable.

Does that help?

Just an other solution (if you are interested by it):

We know that $$\frac{1}{\sqrt{2\pi}}\int_{-\infty }^\infty e^{\frac{-x^2}{2}}dx=1$$
(it's the normal law), therefore, if you set $u=\sqrt{2}x$, you get

$$1=\frac{1}{\sqrt{2\pi}}\int_{-\infty }^\infty e^{\frac{-x^2}{2}}dx=\frac{\sqrt 2}{\sqrt{2\pi}}\int_{-\infty }^\infty e^{-u^2}du\implies \int_{-\infty }^\infty e^{-u^2}du=\sqrt{\pi}$$


Properties of logs. Why does the log of a power, does the exponent become a coefficient? I'm having trouble seeing number 3 clearly. Or rather, I'm having trouble seeing a proof. I've tested a series of numbers and it works, but I'm having trouble seeing why. Can anyone clarify?

Here is where I get confused:

I start with this log:

$$ \log_2 4^1 = 2$$

now if I raise both sides to the 2nd power then:

$$ \log_2 4^{1^2} = 2^2$$
$$ = \log_2 4^2 = 4$$
$$ = 2 \cdot \log_2 4 = 4$$

but if I raise both sides to the power of 1.5 then:
$$ \log_2 4^{2 \cdot 1.5} \ne 4^{1.5} \ne 8$$

By definition:
$$
a^{\log_a x}=x
$$

Now, if:
$$
\log_a (x^y)=\alpha
$$
we have
$$
a^\alpha=x^y =\left( a^{\log_a x}\right)^y = a^{y\log_a x}
$$
so
$$
\alpha=y\log_a x
$$

Logy=logx^a

e^y=x^a

e^(y/a)=x

y/a=logx

y=alogx

You have raised the power in wrong way for log one answer is same


Some basic terminology This question is related to a topic in Physics but i think that it is more of a Mathematical question.

BACKGROUND

I am trying to study Physics rigorously from the beginning.  I am trying to understand each and every word (of course along with the idea). The first topic that i have to study is Measurement. It involves words and phrases like "quantity" and "physical quantity" and i do not think i know what they really mean. After some critical thinking about these terms, i realised that i should know the meanings of four more terms: object, property, mathematical object and number. After analysing these all i ranked them according to the order in which i should study about each. Here are the ranks:

1 .Object

2 .Property

3 .Mathematical object

4 .Number

5 .Quantity

6 .Physical quantity

I looked up the terms in the dictionary, the internet and also thought about them and analysed them myself. This is my interpretation of them along with which part of speech they belong to:

1 .Object :(noun) An object is an entity which can be perceived by our senses. It has an identity, characteristics (properties or attributes) and behaviour.

Objects can be classified into two types:

(I)Abstract objects : Abstract objects are those objects which do not have physical existence.

(II)Concrete objects : Concrete objects are those objects which have physical existence.

2 .Property :(noun) Property is an abstraction of behaviour of an object. An object may have many properties.

Concrete objects basically have two types of property:

(I) Physical property : A physical property of an object is a property that can be observed without changing the composition of matter. Example : appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others, etc.

(II) Chemical property : A chemical property of an object is a property that can be observed during a chemical reaction and thus changing the objects composition (chemical composition). Example : heat of combustion, reactivity with other chemicals, flammability, and types of bonds, etc.

3 .Mathematical object :(noun) A Mathematical object is an object arising in Mathematics. In mathematical practice, an object is anything that has been (or could be) formally defined, and with which one may do deductive reasoning and mathematical proofs. e.g. :number, matrix, etc.

4 .Number :(noun) Number is a Mathematical object used to count, label and measure. There are many types of numbers - Natural numbers, Whole number, Real numbers, etc.. It is an idea. To represent numbers we use numerals and to write numerals we need digits.

5 .Quantity :(noun) The extent, size, or sum of countable or measurable discrete events, objects, or phenomenon, expressed as a numerical value.

6 .Physical quantity :(noun) A physical quantity is a quantity in used in physics to express the "size" of a physical property of an object.

MY QUESTIONS

Am i good to proceed?

How correct am i in describing the above terminologies accurately and rigorously?

Is there anything that is needed to added to make them more accurate and concise?

I'm not entirely sure this does belong on this site, since in essence you're asking:

"What basic terminology and definitions are the most fundamental starting point for a rigorous study of Physics?"

You've assumed that the answer includes many mathematical terms, which is why you've placed the question on this site.  However, I disagree with that assumption.  I'll answer your question in any case, in the above paraphrased form.

The most basic terms to learn to fully understand Physics* are as follows:

Physics
matter
energy
space
time
science
observation
scientific method
hypothesis
theory
experiment
object
property
measure

From there you can descend to the various properties of objects and how to measure them, and the theories which allow accurate prediction of the measured results of experiments.

It's true that understanding these terms may involve understanding other terms first—but that doesn't mean that the other terms are even physics terms, or that they belong in a list of key words for the subject of Physics.  The word "number" is an example; you likely can't fully clear "measure" without some understanding of "number," but "measure" is a key word for Physics and "number" is not.

*I won't say "rigorously" since you didn't define that term and the usual meaning applies specifically to mathematical proofs.


How to show that $\lim_{n \to \infty} \sum_{i=n}^\infty x_i = 0$ Let $\{ x_i \}$ be a decreasing sequence of non-negative real numbers such that $\sum_i x_i < \infty$.

How do I argue that

\begin{align}
\lim_{n \to \infty} \sum_{i=n}^\infty x_i = 0
\end{align}

Hint:

Let $$S=\sum_{i=1}^\infty x_i$$

Then, $$\sum_{i=n}^\infty x_i = S - \sum_{i=1}^{n-1} x_i$$

Now, take the limits of both sides.

The statement seems very much false.

$$x_i>0 \hspace{4pt} \forall \hspace{2pt} i$$
$$\Rightarrow \sum_{i=n}^\infty x_i>0$$


Continuity of piecewise functions Need help with this calculus worksheet. not exactly sure of what to do. I have an idea of how to do it but I don't know if it is right.

Suppose you have a definition of a piecewise function in the form
$$f(x)=\begin{cases}a(x) && \text{if }x\geq 0\\b(x) && \text{otherwise}\end{cases}$$
or something analogous, for continuous functions $a$ and $b$. If $f$ is continuous, then the limits $\lim_{x\rightarrow 0^+}f(x)$ and $\lim_{x\rightarrow 0^-}f(x)$ must agree. In particular, however,
$$\lim_{x\rightarrow 0^+}f(x)=\lim_{x\rightarrow 0^+}a(x)$$
since, as we approach $0$ from above, we will always be in the domain in which $f(x)=a(x)$. Similarly,
$$\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^-}b(x)$$
since, for $x$ less than $0$, it holds that $f(x)=b(x)$. Since $a$ and $b$ were assumed to be continuous, this is equivalent to saying that $a(0)=b(0)$.

This can be applied here, by considering, at each "transition" between one piece of the function to the next, whether the functions composing the part to the right and left of the boundary agree at the boundary.

![enter image description here](https://i.stack.imgur.com/bd63d.jpg.           X^3 if x <0.       E^x if x>0

$$
  g(x) =
\begin{cases}
  x^3 & x \le 0 \\
  e^{x} x > 0
\end{cases}
$$


Has there ever been an analysis of random variables with multiple separate probability spaces? Whenever we analyze random variables, we analyze them as functions from the same probability space to e.g. $\mathbb R$.

I am wondering if anyone has ever analyzed what kind of possible properties might arise from having multiple separate probability spaces, with separate probability measures.

Intuitively, I assume we cannot analyze such a thing except by treating them completely separately. Nevertheless, I'm wondering whether someone has come up with some new concepts that allows one to get interesting results from analyzing multiple separate probability spaces, and their interaction, in some way.

I know this is a vague question, and very exploratory, as I don't have any specific reason to believe that it would even make sense.

If we have a function $f:\Omega\to X$ then the measurable spaces $(\Omega,\mathscr A)$ and $(X,\mathscr B)$ are given. The pair of $\sigma$-algebras with respect to which $f$ is measurable may be different. More importantly, the measure $P_{\theta}$ over $(\Omega,\mathscr A)$ may vary too. Then, $f$ may have different distributions over $X$ as the parameter $\theta$ changes.

This is quite common in quantum mechanics. Consider, for example, the famous Stern-Gerlach experiment demonstrating the quatized nature of the magnetic momentum of certain objects, say, electrons. If an electron beam goes through a cascade of Stern-Gerlach apparatuses then the probability of observing "up" spin or "down" spin depends on the relative situation of the Stern-Gerlach magnets. So, the probability space modelling the behavior of the electrons is equipped with a probability which depends on the parameter related to the said relative situation (usually the angle between the vector of inhomogeneities). At the same time it is meaningless to talk about a common probability space unless one changes the angle randomly.


Reference suggestion: eigenvalues of tridiagonal matrices I would like to ask for a reference on the problem of computing the eigenvalues/eigenvectors of tridiagonal matrices (not necessarily with constant diagonals).

I have seen authors use continued fractions and generating functions. However, I have thus far been unable to really grasp the foundations of this idea. From what I can see, the idea is really to reduce it to a difference equations. Then, perhaps, my request is for a good book on difference equations.
Moreover, is there any technique which is really of a broad scope; ie applicable to a broad range of problems.

Thank you all in advance,

Gabieel

I tried looking up this subject on Google and found this article that seems extremely relevant to your problem: Eigenvalues of Several Tridiagonal Matrices. This article uses symbolic calculus to compute eigenvalues (which I barely know a thing about; I'm not one who works with linear algebra) of multiple tridiagonal matrices. I believe this one should help you, I've skimmed through it.

I strongly recommend the paper "Eigenvalue computation in the 20th century" by Golub & van der Vorst and the references therein.


Convert Hadamard product to Matrix product (simplified) So I have a matrix $Z \in \mathbb{R}^{m \times d}$ (which has repeated row vectors of size m) and $A \in \mathbb{R}^{m \times d}$ and I use hadamard product for them $Z \circ A$.
My goal is to somehow separate Z and A in a way A will stay in the right hand side of the brackets i.e. either $$Z \circ A = (Z\cdot X)A$$ or $$Z \circ A = (Z\circ X)A$$ .
Is there any way I can do this?

Matrix A is already on the right-hand side. I am not sure there is any operator between ) and matrix A on the right side in your question.
However, If exists $Z^T$ then, you could have $(Z*Z^T) * A$.


Let $F = \mathbb{Q}(\pi^3)$. Find a basis for $F(\pi)$ over $F$. Let $F = \mathbb{Q}(\pi^3)$. Find a basis for $F(\pi)$ over $F$.

How can I solve this. Can anyone help me please. Thanks.

Hint: The number $\pi$ is transcendental. So this is the same problem as finding a basis for $F(x)$ over $F(x^3)$.

Finding a basis for $F(x)$ over $F(x^3)$.

This means you need to find a basis $v_1(x),\cdots,v_n(x)$ such that $p_1(x)v_1(x)+p_2(x)v_2(x)+\cdots+p_n(x)v_n(x)$ generates any polynomial of degree $n\in\mathcal{N}$ in $F(x)$. But $p_i(x)\in F(x^3)$ for any $i$, then $p_i(x)=a_nx^{3n}+a_{n-1}x^{3n-3}+\cdots+a_0$ where $a_i\in F$.

It is clear that $\{v_1(x),v_2(x),v_3(x)\}=\{1,x,x^2\}$ is the easiest answer.


Show that the linear functional $f(x) = 3\xi_1 - 5\xi_2$ is bounded on $\ell^1$ and compute its norm How do I show that $f$ is bounded and determine the norm of $f$?

Starting with,

$|f(x)| = |3\xi_1 - 5\xi_2|$

I am not sure what the next move is to get to the form

$|f(x)| \leq c\|x\|$ where $\|x\| = \sum_{k=1}^{\infty} |\xi_k|$

I assume $x=(\xi_i)_i$ ?

Hints:

1) For all $a,b\in\mathbb C$ there holds $|a\pm b|\leq|a|+|b|$ and $|ab|=|a|\cdot|b|$. So, $|3\xi_1-5\xi_2|\leq \ldots$

2) $|\xi_1|\leq \|x\|$ and $|\xi_2|\leq \|x\|$

The dual space of $\ell^1$ is isometrically isomorphic to the space $\ell^\infty$. Furthermore, every bounded linear map $f : \ell^1 \to \mathbb{R}$ can be expressed as

$$f(x_n)_{n=1}^\infty = \sum_{n=1}^\infty \alpha_nx_n$$

for some sequence $(\alpha_n)_{n=1}^\infty \in \ell^\infty$, and vice-versa, every linear map of this form is in $(\ell^1)^*$. The isometrical isomorphism is given by $f \mapsto (\alpha_n)_{n=1}^\infty$, so in particular $\|f\| = \left\|(\alpha_n)_{n=1}^\infty\right\|_\infty$.

For your functional in particular we have that $f$ is represented by the vector $(3, -5, 0, 0, \ldots) \in \ell^\infty$ so we conclude that $f$ is bounded and $$\|f\| = \|(3, -5, 0, 0, \ldots)\|_\infty = \max\{|3|, |-5|\} = 5$$


Showing that a squence is not uniformly convergent Let $f_n(x)=\dfrac{\sin(nx)}{1+n^2x^2}$.

First note that $f_n(x) \to 0$ pointwise.  Also, if you fix $\epsilon = \dfrac{1}{4}$ and let $x=\dfrac{1}{n}$, then $$|f_n(\frac{1}{n})-f(x)|=|
\frac{\sin(1)}{2}-0|=|\frac{\sin(1)}{2}|\geq \frac{1}{4}.$$

My question is, how much more should I show?  That is, I'm not sure how to make this argument more clearer.  I was marked off for lack of details.

The sequence converges pointwise because
$\displaystyle \lim_{n\rightarrow +\infty}f_n(x_0)=\lim_{n\rightarrow +\infty}\frac{\sin(nx_0)}{1+n^2x_0^2}=0$.

To prove that the convergence is uniform, one must be show that $\displaystyle\lim_{n\rightarrow +\infty}\sup_{x\in[0,1]}\left|f_n(x)-f(x)\right|=0$, i.e.

$$\sup_{x\in[0,1]}\left|\frac{\sin(nx)}{1+n^2x^2}-0\right| \to 0 \quad(n \to \infty).$$


If $a^3 + b^3 +3ab = 1$, find $a+b$ Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.

I tried to factorize it but unable to do it.

There are a continuum of solutions to
$$
a^3+b^3+3ab=1
$$
Suppose that
$$
x=a+b
$$
then
$$
\begin{align}
1
&=a^3+(x-a)^3+3a(x-a)\\
&=x^3-3ax^2+3a^2x+3ax-3a^2
\end{align}
$$
which means
$$
\begin{align}
0
&=(x-1)\left(x^2+(1-3a)x+3a^2+1\right)
\end{align}
$$
So either $x=1$ irregardless of $a$, or
$$
x=\frac{3a-1\pm(a+1)\sqrt{-3}}{2}
$$
Thus, other than $x=1$, the only real $x$ is $-2$, which comes from $a=-1$.

That is, the only two real values of $a+b$ are $1$ and $-2$.


Is there a non-circular explanation for computing the probability of the intersection of two dependent events? The explanation with which I'm familiar goes like this. Define the conditional probability P(A|B) as the probability of their intersections P(A and B) divided by the probability P(B). To figure out the probability of the intersection P(A and B) then, simply rearrange the terms of equation algebraically to obtain P(A and B) = P(A|B) X P(B).

The problem: What I'm struggling to understand is how to apply this equation in practice. It seems like whenever we want to figure out P(A and B), we need to figure out the values of P(A|B) and P(B).But to figure out the value of P(A|B), we already must figure out the value of P(A and B). So how is the equation P(A and B) = P(A|B) X P(B) helpful for figuring out the value of P(A and B)? In other words, to solve for it, we must figure out the value of P(A|B), which we can't do without knowing the value of P(A and B).

In practice, a lot of time you can calculate $P(A|B)$ by assuming $B$ happened and see what is the probability of $A$. For example, let's say you draw 2 cards face down, then randomly choose 1 to flip up, and let $B$ is the event that exactly 1 red card and 1 blue card was drawn, and $A$ is the event that the flipped up card is blue. Then you know $P(A|B)$ without knowing $P(B)$ nor $P(A\bigcap B)$, because assuming you have 1 red card and 1 blue card faced down, randomly picking 1 of them to flip up give you blue card with probability $\frac{1}{2}$.

Or, … just compute $P(A\cap B)$ by whatever methods you may have at hand. Say, we are rolling a fair die and $A$ is "number is even" and $B$ is "number is prime", both having probability $\frac12$ - what could be more straightforward than to observe that "$A$ and $B$" means "number is $2$" and conclude that $P(A\cap B)=\frac16$?

There is nothing "inherent" that can avoid what you call circularity. In a way, working with  $P(A\mid B)$ (or $P(B\mid A)$) is just another way of expressing the fact that  $P(A\cap B)$ need not be the product of $P(A)$ and $P(B)$.


If I stretch a convex polygon, does the original fit into the streched version? Suppose you have a convex polygon $P=\mathrm{conv}(\{(x_1,y_1),\dots, (x_k,y_k)\})$ and you stretch it in one dimension, that is, we choose $\alpha>1$ and get a new polygon $P^\alpha=\mathrm{conv}(\{(\alpha x_1,y_1),\dots, (\alpha x_k,y_k)\})$.

Is it true that you can translate and rotate $P$ to make it fit into $P^\alpha$?

This seems true to me, as you are somehow making it bigger and "keeping the shape", but I have no additional insight in how one could prove such a thing.

Yes, this is always possible.
Sandwich $P$ between two lines of some variable slope $p$, touching $P$ at $A$ and $B$ respectively, where $AB$ has slope $q$.  Without loss of generality, assume $q > 0$ when $p = 0$.  We can vary this configuration continuously such that $p$ increases from $0$ while $q$ increases toward $\infty$, so $pq$ will increase continuously from $0$ toward $\infty$, while $p < q$ at all times.  (The one exception is if $AB$ becomes vertical while $p = 0$, but then we’re already done: we can fit $P$ into $P^\alpha$ without any rotation.)  Stop when $pq = \alpha$.
We claim that rotating $P$ clockwise about $A$ by $\theta = \tan^{-1} \frac{pq - 1}{p + q}$, then translating it by $\vec{AA^\alpha}$, sends it to a subset of $P^\alpha$.
To see this, it suffices to show that the linear transformation about origin $A$ composed of this rotation by $\theta$ followed by a stretching in the $x$ direction by ratio $\frac{1}{\alpha}$:
$$\begin{bmatrix}
  \frac1\alpha \cos \theta & \frac1\alpha \sin \theta \\
  -\sin \theta & \cos \theta
\end{bmatrix} = \frac{1}{\sqrt{(1 + p^2)(1 + q^2)}}\begin{bmatrix}
  \frac{p + q}{pq} & \frac{pq - 1}{pq} \\
  1 - pq & p + q
\end{bmatrix}$$
sends $P$ to a subset of itself.  You can verify that this transformation has eigenvectors
$$v_1 = \begin{bmatrix}1 \\ p\end{bmatrix}, v_2 = \begin{bmatrix}1 \\ q\end{bmatrix},$$
with eigenvalues
$$\lambda_1 = \frac1p \sqrt{\frac{1 + p^2}{1 + q^2}}, \lambda_2 = \frac1q \sqrt{\frac{1 + q^2}{1 + p^2}},$$
where $\lambda_1 < \lambda_2 < 1$ because
$$\begin{gather*}
\lambda_2 - \lambda_1 = \frac{(q - p)(pq - 1)}{pq\sqrt{(1 + p^2)(1 + q^2)}} > 0, \\
1 - \lambda_2^2 = \frac{(pq)^2 - 1}{(1 + p^2)q^2} > 0.
\end{gather*}$$
So we can reinterpret this transformation as the composition of a stretching along each line of slope $p$ about its intersection with $AB$ by ratio $\frac{\lambda_1}{\lambda_2}$, followed by a dilation about $A$ by ratio $\lambda_2$; both these component transformations send $P$ to a subset of itself because it’s convex.  This proves the claim.

I think this hipothesis not true.

With a simple transformation we can make the stretch direction as parallel to X-axis. So, I draw two consecutive edges $A,B,C$ of a convex polygon, let's say they are at "north", and do a X-scale to get $D,E,F$:

As you can see, the original $ABC$ must be translated to $UEV$, by an horizontal distance $BE=d=\alpha·B_x-B_x=(\alpha-1)B_x$. It must be also rotated so $U,V$ be inner to $DEF$

The problem arises when you want to apply the same translation to the rest of vertices.

$BE$ must be the greatest distance of all possible-applied translations, otherwise other vertices will fall outside of the stretched polygon. This itself does not seem a great issue, just choose the biggest one. But it may be that that chosen distance is "$d_{max}>>d$; in this case translating $B$ with $d_{max}$ will make it to fall at the very right of $E=\alpha·B$

To get things worst, each translation has an associated rotation. The likelihood of any $translation, rotation$ pair fitted for one vertex, to be valid for the entire polygon is very low, zero in general.

It's also obvious that there are cases where such a pair can be obtained. For example the triangle in the image above.


symmetric positive definite matrix question Let $A$ be an $n\times n$ symmetric positive definite matrix and let $B$ be an $m\times n$ matrix with $\mathrm{rank}(B)= m$. Show that $BAB'$ is symmetric positive definite.

The result is symmetric since $(BAB')'= BAB'$

Case  1: $m<n$

$$\begin{bmatrix}
.&. &. &.
\end{bmatrix}
\begin{bmatrix}
. &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&.
\end{bmatrix}
\begin{bmatrix}
.\\.\\.\\.
\end{bmatrix}
$$
This matrix is always positive definite since you can always denote the outer factors as $B'x = y$ and since $A$ is positive definite $y'Ay$ > 0

Case 2: $m=n$

$$
\begin{bmatrix}
. &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&.
\end{bmatrix}
\begin{bmatrix}
. &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&.
\end{bmatrix}
\begin{bmatrix}
. &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&.
\end{bmatrix}'
$$
We obtain $BAB'= C$. If $B$ is invertible then this is called congruence transformation. From Sylvester's Law of Inertia, the number of positive, zero and negative eigenvalues of C are equal to of $A$. Therefore, if $A$ is positive definite, so is $C$. Since matrix $B$ is full rank, it is invertible. If $B$ was not invertible, then there exists a nonzero $x\in\mathbb{R}^n$ such that $B'x=0$ Thus, $BAB'$ becomes only positive semi definite.

Case 3: $m>n$

Then $B$ cannot have a rank of $m$ but suppose the question was modified and only B is full rank is given.
$$
\begin{bmatrix}
. &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&.\\.&.&.&.\\.&.&.&.
\end{bmatrix}
\begin{bmatrix}
. &. &. &.\\. &.&.&.\\. &.&. &.\\.&.&.&.
\end{bmatrix}
\begin{bmatrix}
. &. &. &.&.&.\\. &.&.&.&.&.\\. &.&. &.&.&.\\.&.&.&.&.&.
\end{bmatrix}
$$
Similarly, this case leads only to a positive semidefinite product. The quickest way is to observe that The product is a matrix of $m\times m$ and the rank of this matrix is at max $n$ since we cannot arrive to a full rank matrix with a product of elements that have ranks less than the resulting matrix.You can think of the tensor(or outer) product of two vectors $a\otimes b = ab'$.

Take $\xi$ different from zero and show that $\xi' B A B' \xi$ is always positive. If not, then $\mathrm{rank}(B) < m$.


Did Einstein introduce anything new to mathematics? Newton introduced calculus, so I am wondering, did Einstein introduce anything important to mathematics?

Ten of Einstein's equations were known as the Einstein Field Equations, published in 1915. One of these equations describes how stress-energy causes curvature of space-time. Of course, since physics is so close to math, Einstein contributed many mathematical laws.


Prove that limit doesn't exists at all I have a function $f(x)=\frac{1}{x}+\sin(\frac{1}{x})$ which I have proved that $\lim_{x \to0^+}f(x)=\infty$, and I need to prove that $\lim_{x \to0^+}f'(x) \ne -\infty$.
I have got that $f'(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$
I am trying to prove that:
$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}$$
Does not equal to $-\infty$ (or to prove that it doesn't exists at all).
I have tried to assume by contradiction that
$$\lim_{x \to 0^+}-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\infty$$
So there exists $\delta_{1}>0$ such that for every $0<x<\delta_{1}$ we say that $$f(x)<\square $$
We choose $\delta=\delta_{1}$, and let $0<x<\delta$ so:
$$f(x)=-\frac{1}{x^2}-\frac{\cos(\frac{1}{x})}{x^2}=-\frac{1}{x^2}(1+\cos(\frac{1}{x}))\le -\frac{2}{x^2}<\frac{2}{\delta^2}=k$$
I do not know what to choose instead of $\square$, and if what I did is correct.
Appreciate your help!
I have also tried to think about the continuity of $f$, and saying that it is not bounded so it can't be continuous, so the limit doesn't exists.
Thanks a lot!

Because there exists a subsequence converging to $0$:
if you pick $\{x_k\}$ to be such that $x_k=\frac{1}{\pi + 2\pi k}$, then
$$f^\prime (x_k) = -(\pi + 2\pi k)^2 \Big(1+cos(\pi + 2\pi k)\Big) = 0,$$
since
$$cos(\pi + 2\pi k)=-1 \mbox{, } \forall k;$$
Also the sequence of $x_k$ converges to $0$ as you can observe letting $k\to +\infty$.

Assume by contradiction that for every $K<0$ there exists a $\delta>0$ such that for every $0<x<\delta$, we get $f'(x)<K$.
We choose $n \in \mathbb{N}$ such that $n>K$, and we choose $x_{n}= \frac{1}{\pi +2 \pi n}$.
Note that $0<\frac{1}{\pi +2 \pi n}<1$. (So in this case $\delta=1$)
And: $f'(x)=-(\pi +2\pi n)(\cos(\pi+2\pi n)+1)=0>K$
So $f'(x)>K$, in contradiction!
Is that correct?
Thanks!


How would I convert from frequency back to percent? I'm working on a web application that uses an oscillator. It has a parameter that goes from 0 to 100 percent. I translate that percent into hertz using this equation:
$$
\text{frequency} = 2^{(\lfloor  128*\text{percent} \rfloor - 69)/12}*440
$$

I am having some trouble solving the equation for percent so that given a frequency, I can get a percent value to send back to my app. Help please? :]

$$f = 2^{(x - 69)/12} \times 440$$
inverts into
$$x = 69 + 12 \log_2(f/440)$$
but in your case $x = \lfloor  128p \rfloor$, which is impossible to invert exactly because it maps intervals to single numbers (i.e. the function is not one-to-one and so does not have an inverse).

For example, $p = 10^{-9}$ and $p = 10^{-8}$ would yield the same $x=0$, so if I tell you $x=0$, there is no way to imply what the original $p$ was directly.

EDIT
You could bound it, however. $x$ must certainly be an integer and then, since you are rounding down, $x = \lfloor  128p \rfloor$ means
$$
x \leq 128p  < x+1 \Leftrightarrow \frac{x}{128} \leq p < \frac{x+1}{128}
$$

Since the floor function is not bijective, one can not invert this equation. Ignoring the floor function gives:
$$p=\frac{69}{128}+\frac{12\ln\left(\frac{f}{440}\right)}{128\ln 2}$$
where $f$ is the frequency and $p$ is the percentage.


Fourier transform of $t^2$ discrepancy I encountered a discrepancy when taking the fourier transform of $t^2$ that I don't understand. I would expect $\mathcal{F}[t^2]$,
$$\mathcal{F}[t^2]=\int_{-\infty}^\infty t^2 e^{-i \omega t}dt$$
 to diverge because $t^2$ isn't absolutely integrable. Typing this integral into Wolfram Alpha confirms that it does diverge.

However, typing "Fourier Transform of $t^2$" into Wolfrom Alpha produces the following result:
$$\mathcal{F}[t^2]=-\sqrt{2\pi}\delta''(\omega)$$

What is the reason for this discrepancy and what does this answer represent physically? I was taught that if a function isn't absolutely integrable, it was not possible to take its Fourier transform, but this seems to suggest otherwise.

EDIT:
I think I wasn't clear on my exact question. I'm more interested in the implications of this integral and why you can take the Fourier transform of $t^2$ even though it isn't absolutely integrable. However, that brings up another question - why does Wolfram Alpha tell me that the integral diverges when it is actually equal to the second derivative of the delta function?

We can build the theory of the Fourier transform up in several steps. The integral in the definition converges by an immediate estimate if the function is in $L^1 (\mathbb{R})$. Since $L^1 (\mathbb{R}) \cap L^2 (\mathbb{R})$ (the set of functions which are both absolutely integrable and square-integrable) is dense in $L^2 (\mathbb{R})$ (i.e. every function in $L^2$ can be approximated of functions in the former space which converges in the $L^2$ norm), and the Fourier transform is a continuous linear operator from $L^1 (\mathbb{R})$ to $L^2 (\mathbb{R})$, we can define the Fourier transform for f in $L^2 (\mathbb{R})$ by approximating $f$ with a sequence of functions $f_n \in L^1 (\mathbb{R}) \cap L^2 (\mathbb{R})$, and defining the Fourier transform of $f$ to be the $L^2$-limit of the Fourier transforms of $f_n$. This idea of taking a function defined on a dense subset and extending it by continuity to a larger set is very useful.

Of course, your example isn't in $L^2 (\mathbb{R})$ either. It turns out we can extend the Fourier transform to an even broader class of functions (actually, to a class that includes things that aren't functions in the traditional sense at all!). This broad class is the set of tempered distributions. We do this by first noting that the Fourier integral is well-defined on the Schwartz class, which is roughly speaking the set of functions which decay to zero at infinity faster than any polynomial (hence these functions are all in $L^1 (\mathbb{R})$. We give the Schwartz class a particular (locally convex) topology, which (in particular) determines which sequences in this space are convergent and which functions defined on the Schwartz class are continuous.

The set of tempered distributions is the dual space to the Schwartz class, which means it is the set of all continuous linear functions which map from the Schwartz class to the real (or complex numbers). An example is the Dirac delta, which satisfies $\delta(f) = f(0)$ for $f$ in the Schwartz class. Distributions are objects which act on functions. Operations on distributions may therefore be characterized by how they affect the way the distribution acts on functions. It turns out that for any tempered distribution $\phi$, we can define the Fourier transform $F \phi$ as the tempered distribution defined by
\begin{equation*}
F \phi (g) = \phi (Fg) \quad \text{ for all } g \text{ in the Schwartz class}.
\end{equation*}
So the Fourier transform of a tempered distribution is the distribution which first applies the Fourier transform to the test function, and then applies the original distribution to the result.

Roughly speaking, all functions with polynomial growth at infinity have corresponding tempered distributions, and hence we can define their Fourier transform. Given such a function $f$, we define the tempered distribution corresponding to $f$ by
\begin{equation*}
\Lambda_f (g) = \int_{-\infty}^\infty f(x) g(x) \, dx \text{ for any } g \text{ in the Schwartz class}.
\end{equation*}
In particular, the function in your example has a corresponding tempered distribution, and so we can define the Fourier transform of that distribution. Notice that the Fourier transform involves the Dirac delta, and hence is not a traditional function, but instead is defined in terms of how it acts on other functions, as we might expect from our definition of the Fourier transform on the Schwartz class.

edit: Responding to your question in another comment, it is correct that we cannot take the Fourier transform of $e^{-t}$, since $e^{-t}$ is not a tempered distribution (it "grows too fast"). There are several ways to prove this. Suppose that $f(t) = e^{-t}$ was a tempered distribution, and take for granted that the rules of derivatives and Fourier transforms work for tempered distributions more or less like you expect they should (they do). Then $e^{-t}$ would (as a distribution) solve the differential equation
\begin{equation*}
f' + f = 0.
\end{equation*}
Taking the Fourier transform, we obtain
\begin{equation*}
(2 \pi i \xi + + 1) \hat{f} (\xi) = 0 \text{ for every } \xi \in \mathbb{R}
\end{equation*}
where $\hat{f}$ is the Fourier transform of $f$. The only way for this equation to be satisfied is if $\hat{f}$ is identically zero. But (it turns out that) the Fourier transform on the set of tempered distributions is an isomorphism, and in particular is invertible, so $\hat{f} = 0$ implies $f = 0$, since the Fourier transform of $0$ is clearly zero. This is a contradiction, so $f$ cannot be a tempered distribution, so we can't define its Fourier transform.

You simply use a trick to bypass the question.

$$\int_{-\infty}^{+\infty}t^2 e^{-i\omega t}\ dt = -\int_{-\infty}^{+\infty} \frac{d^2}{d\omega^2} e^{-i\omega t} = -\frac{d^2}{d\omega^2} \int_{-\infty}^{+\infty} e^{-i\omega t} =  -\frac{d^2}{d\omega^2} \sqrt{2\pi}\delta(\omega) = -\sqrt{2\pi}\delta''(\omega)$$


Proof that $1$ is the least natural number using the Principle of Well Ordering I'm doing a question from a PDF on Abstract Algebra.(The pdf can be found here http://abstract.pugetsound.edu/ 2012 edition). I have to show that the Principle of Well-Ordering implies that $1$ is the smallest natural number; this is only part of the whole question.(Question 14 of Chapter 2).

This is what I got so far.

Proof.

Let $S$ be an arbitrary subset of the Natural Number excluding one. By the Principle of Well-Ordering there exist a least element $X \in S.$
Now add $1$ to $S.$ $1$ is now the least element. Since there is no $X < 1$ in any such set $S,$ $1$ is the smallest element of any set $S$ and therefore the smallest natural number.

By the Well-Ordering principle, there is a least element $a$ in $\mathbb{N}$. Suppose, to the contrary, that $a<1$. Multiplying both sides of the inequality by $a$, we have $a^2<a$. Since $\mathbb{N}$ is closed under multiplication, $a^2\in\mathbb{N}$, which contradicts the assumption that $a$ is the least element in $\mathbb{N}$.

let S=N{1},by well-ordering principle, S has a least element, namely x. We claim that x>1. Suppose x<1, and x is a natural number, then we have 1-1=0 is also a natural number, contradiction arises. Hence 1


Intuition of $ P( X = a) $ for a continuous random variable? Let $(\Omega, {\cal B}, P )$ be a probability space, $( \mathbb{R}, {\cal R} )$ the usual
 measurable space of reals and its Borel $\sigma$- algebra, and $X : \Omega \rightarrow \mathbb{R}$ a random variable.

The meaning of $ P( X = a) $ is intuitive when $X$ is a discrete random variable, because it's the definition of the probability mass function. I am not sure if my question makes sense, but how should I think of $ P( X = a) $ when $X$ is a continuous random variable?

This doesn't need any measure theory.

What's the probability of picking 0.5 from the interval [0,1] following a [insert continuous eg uniform] distribution?
Can you measure 100% precisely 1.1566672 cm with a ruler?
Open Excel/Google Sheets and in two cells simulate values from a continuous distribution eg normal. In a third cell, check whether the values in the two cells are equal. I don't think they will ever be equal.


Proving that $\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2}$ How would I prove the following trig identity?

  $$\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2} $$

My work thus far has been:
$$\dfrac{2\cos\dfrac{A+B}{2} \cos\dfrac{A-B}{2}}{-2\sin\dfrac{A+B}{2} \sin\dfrac{A-B}{2}} =-\cot\dfrac{A+B}{2} \cot\dfrac{A-B}{2} \ .$$

First, note that $\dfrac{\cos(A + B)}{\cos A - \cos B} \neq -\cot \left( \dfrac{A-B}{2} \right) \cot \left(\dfrac{A+B}{2} \right)$

In particular, if we use something like $A = \pi/6, B = 2\pi/6$, then the left is $0$ as $\cos(\pi/2) = 0$ and the right side is a product of two nonzero things.

I suspect instead that you would like to prove:

$$\dfrac{\cos A + \cos B}{\cos A - \cos B} = -\cot \left( \dfrac{A-B}{2} \right) \cot \left(\dfrac{A+B}{2} \right)$$

HINTS

And this looks to me like an exercise in the sum-to-product and product-to-sum trigonometric identities (wiki reference). In fact, if you just apply these identities to the top and the bottom, you'll get the result.

A couple of hints: use the $\cot$ identity on the RHS to put in terms of $\sin$ and $\cos$, use a product to sum identity, and pay attention to odd and even functions. This should simplify it to an easier expression.