Division by $0$ and its restrictions Consider the following expression:

$$\frac{1}{2} \div \frac{4}{x}$$

Over here, one would state the restriction as  $x \neq 0 $, as that would result in division by $0$.

But if we rearrange the expression, then:

$$\begin{align}
\frac12\div\frac4x &= \frac{1}{2} \times \frac{x}{4} \\
&= \frac{x}{8}
\end{align}$$

In this expression, there are no restrictions. If we substitute $x = 0$, then the answer is $\frac{0}{8} = 0$.

So, how come in the first unsimplified expression, when we substitute $ x =0$, we get undefined, whereas in the simplified expression we get $0$?

EDIT: I have made a small change in terminology based on the suggestion by @BPP in the comments below.

This is an excellent question!

The subtlety here is that in high-school mathematics we are taught to freely move between polynomials and polynomial functions. But if we do not make the distinction between the two carefully, then confusions like in your question will arise.

When I have the equality $$\frac{1/2}{4/x} = \frac{x}{8}$$ there is no problem when viewing this is an equality of rational fractions, because in this case, there is no question of evaluating $x$, and hence no question of what the domains are, simply because they are not functions.

On the other hand, when I look at the equality $$\frac{1/2}{4/x} = \frac{x}{8}$$ as an equality of rational functions, then it is necessary that the domain of both functions are the same. Since the function on the left-hand side is defined for all $x \neq 0$, I can take that to be the domain of the function, and so the domain of the function on the right-hand side is also $\mathbb{R} \setminus \{0\}$.

A lengthier explanation of this distinction follows below.

What is a polynomial?

Typically, it is taught at the high-school level that a polynomial (in one variable) is any expression of the form $$a_0 + a_1 x + \dots + a_n x^n$$ where $a_0,\dots,a_n$ are real numbers, and $n$ is a non-negative integer. We can denote this polynomial by $f(x)$. Here, $x$ is called a variable or indeterminate.

At the same time, we are also taught that we can substitute a real number, say $t$, for $x$ in order to evaluate the polynomial at $t$, and this is represented by $f(t)$. This is the point where the distinction between polynomials and polynomial functions should be made.

What is a polynomial function?

For our purposes, we will stick to functions from real numbers to real numbers. We define a polynomial function to be a function $f: \mathbb{R} \to \mathbb{R}$ such that there exist real numbers $a_0,\dots,a_n$ so that $f(t) = a_0 + a_1 t + \dots + a_n t^n$ for every $t \in \mathbb{R}$.

How are polynomials and polynomial functions the same?

At first glance, there may not appear to be much of a difference between polynomials and polynomial functions, for the following reason: given a polynomial $f(x) = a_0 + a_1 x + \dots + a_n x^n$ I can define a polynomial function $\tilde{f} : \mathbb{R} \to \mathbb{R}$ by $\tilde{f}(t) = a_0 + a_1 t + \dots + a_n t^n$ for each $t \in \mathbb{R}$. And, given a polynomial function $\tilde{g} : \mathbb{R} \to \mathbb{R}$ such that $\tilde{g}(t) = b_0 + b_1 t + \dots + b_m t^m$ for each $t \in \mathbb{R}$, I can define a polynomial $g$ by $g(x) = b_0 + b_1 x + \dots + b_m x^m$. This back and forth between polynomials and polynomial functions is compatible in the sense that the map $F$ from the space of polynomials (over $\mathbb{R}$ in one variable) to the space of polynomial functions (from $\mathbb{R}$ to $\mathbb{R}$) that sends the polynomial $f$ to the polynomial function $\tilde{f}$ by the above rule, preserves the structure of both spaces. More precisely, it satisfies the conditions:
$$
\begin{gather*}
f + g \mapsto \tilde{f} + \tilde{g}\\
cf \mapsto c\tilde{f}\\
fg \mapsto \tilde{f} \tilde{g}\\
1 \mapsto \operatorname{id}
\end{gather*}
$$
where $\operatorname{id}$ is the constant function $\operatorname{id}(t) = 1$ for all $t \in \mathbb{R}$. (As an aside, all this is often expressed by saying that the map $F$ given by $F(f) = \tilde{f}$ is an isomorphism of algebras.)

The fact that the function $F$ has these properties essentially means that polynomials and polynomial functions are not all that different.

How are polynomials and polynomial functions different?

Despite the previous section, it is important to make a distinction between polynomials and polynomial functions for the following reason. When we divide a polynomial $f(x)$ by another polynomial $g(x)$ we get the rational fraction $f(x)/g(x)$. (Of course, $g(x)$ could divide $f(x)$ without leaving a remainder, in which case $f(x)/g(x)$ is in fact a polynomial, so every polynomial can also be considered a rational fraction.) Two rational fractions $f_1(x)/g_1(x)$ and $f_2(x)/g_2(x)$ will be considered equal if, after cancelling their common factors, we end up with the same expression in both cases. Or equivalently, the two rational fractions will be considered equal if $f_1(x) g_2(x) = f_2(x) g_1(x)$. Note that this is completely analogous to the definitions we make when talking about rational numbers.

Note that a rational fraction is not a function (since polynomials are not 'really' functions). However, we can define a rational function $h$ to be a function such that there exist real numbers $a_0,\dots,a_n$ and $b_0,\dots,b_m$ so that for every $t$ in the domain of $h$, we have $$h(t) = \frac{a_0 + a_1 t + \dots + a_n t^n}{b_0 + b_1 t + \dots b_m t^m}.$$

Clearly, I can repeat the same process as earlier: to every rational fraction I can associate a rational function, and to every rational function I can associate a rational fraction, in the obvious way. So, where does the problem arise?

The problem arises in the fact that two different representations of the same rational fraction can give rise to two different rational functions. For example, let $f(x) = x^2 - 4$, $g(x) = x-2$, $h(x) = x+2$ and $k(x) = 1$. Clearly, $f(x)/g(x) = h(x)/k(x)$. However, the function $u$ defined by $u(t) = f(t)/g(t)$ and the function $v$ defined by $v(t) = h(t)/k(t)$ are not the same function, because their domains are different! (I am assuming that all the functions have their domain as the maximal set on which they are defined. So, the domain of $u$ is $\mathbb{R} \setminus \{ 2 \}$ and domain of $v$ is $\mathbb{R}$.)

So, let's say we started with the rational function $v$. We construct the 'natural' rational fraction using $v$, namely $h(x)/k(x)$. We write $h(x)/k(x) = f(x)/g(x)$. Then we construct the rational function $u$ from the rational fraction $f(x)/g(x)$. But we end up with a different function! So, the realisation is that the 'evaluation map', which is basically 'replace $x$ by the real number $t$', is not an 'isomorphism' between the set of all rational fractions and the set of rational functions. Hence, we cannot move back and forth freely between the two notions unless we sort out how we want to define the evaluation map compatibly.

How does all this answer the question?

When I have the equality $$\frac{1/2}{4/x} = \frac{x}{8},$$ this is certainly true when I view both sides as rational fractions. In this case, there is no question of evaluating $x$, and hence no question of what the domains are, simply because they are not functions.

On the other hand, when I look at the equality $$\frac{1/2}{4/x} = \frac{x}{8}$$ as an equality of rational functions, then it is necessary that the domain of both functions are the same. Since the function on the left hand side is defined for all $x \neq 0$, I can take that to be the domain of the function, and so the domain of the function on the right-hand side is also $\mathbb{R} \setminus \{0\}$.

The question, “Why am I able to make sense of the right-hand side at $x = 0$ and not able to make sense of the left-hand side at $x = 0$?” arises because we expect the evaluation map to be 'compatible', in some sense. That this is not so, is the content of the discussion above.

What have I glossed over?

I have certainly been handwaving a bit when talking about isomorphisms, but this can be made fully precise without much trouble. All these ideas are discussed in any senior undergraduate course in algebra.

Also, as @AlexProvost commented below, I have defined a rational fraction to be a ratio of polynomials, and yet I am calling $\frac{1/2}{4/x}$ a rational fraction, despite $4/x$ not being a polynomial. Here, I have glossed over the fact that dividing a rational fraction by a rational fraction gives one a rational fraction. This can be shown quite easily, but I had still skipped this point.

Explanation of the edit: Earlier I had called ‘rational fractions’ as ‘rational functions’, and ‘rational functions’ as ‘functions that are rational’. I think the new terminology is cleaner, and I am not sure whether there is any standard terminology.

Consider the function
f(x)=x^3/x^4
here if you apply limit and put x=0 then definitely one will get 1/0 form which is undefined.
So make use of the L'Hospitals rule here that is take derivative of numerator and denominator term separately that is,
Applying limit when x approaches to zero and then taking limit


proving golden ration with induction If $\displaystyle a=\frac{1+\sqrt{5}}{2}$ and $\displaystyle b=\frac{1-\sqrt{5}}{2}$, prove that $\displaystyle f_n=\frac{a^n-b^n}{\sqrt{5}}$ for all $n\in\mathbb{P}$

Would we start with a base case for proof of induction?

In that case, $n=0$ would be the base case

so how would you prove for $n$ and $n+1$?

Hint $\ $ Note $\,a^n\,$ is a solution of the fibonacci recurrence since

$$\ a^{n+2}-a^{n+1}-a^n = a^n(a^2-a-1) = a^n(0) = 0$$

Similarly $\,b^n\,$ satisfies the recurrence since it too is a root of $\,x^2-x-1.$ Hence by linearity any linear combination $\,c a^n + d b^n\,$ also satisfies the recurrence for any $\,c,d\,$ that are constants (i.e. independent of $\,n).$ In particular $\,g_n = (a^n-b^n)/\sqrt{5}\,$ satisfies the recurrence. Thus $\,f_n,\,g_n\,$ are both solutions of the recurrence and, as easily checked, they have the same initial conditions $\,f_0=0,\,f_1 =1.\ $ Now a simple induction proves that any two such solutions are equal (the uniqueness theorem). Equivalently, considering $\,f_n-g_n,\,$ it suffices to show that $\,0\,$ is the only solution with initial conditions both $= 0,\,$ an obvious induction.

Assume that
$$
f_0 = 0\\
f_1 = 1\\
f_{n+1} = f_n + f_{n-1}
$$
Check the relation for $n=0,1$ (easy).

The general solution of such an equation has the form
$$
f_n = Ar_1^n + Br_2^n
$$
with $r_{1,2}$ solutions of the characteristic equation:
$$
r^2 = r + 1
$$
Check that $$r_{1,2} = \frac{1\pm\sqrt{5}}2$$
and then you are done: as your given candidate gives the good values for $n=0,1$ this is the only solution.

Alternative: prove that if
$$
f_n = \frac{a^n + b^n}{\sqrt{5}}\\
f_{n+1} = \frac{a^{n+1} + b^{n+1}}{\sqrt{5}}
$$then
$$f_{n+2} = \frac{a^{n+2} + b^{n+2}}{\sqrt{5}}
$$


Prove for $x \in \mathbb{X}: x \in \bar{A} \Leftrightarrow d_A(x)=0$ For $A \subset \mathbb{X}$ non empty and $x \in \mathbb{X}$ define the distance of x to A by $$d_A(x)=inf_{a \in A} d(x,a)$$

I am trying to prove for $$x \in \mathbb{X}: x \in \bar{A} \Leftrightarrow d_A(x)=0$$

The proof starts of by saying if $x \in \bar{A}$ then there is some $y\in A$.

How can we say this?

How do we know there is some $y \in A$?

As others noted, you explicitly assumed that $A\neq \emptyset$.

Here is a relatively easy proof of the fact. We use a characterization of the closure of a subset $$\bar A = \{x\in X\,|\, \forall U\in\mathcal O_x: A\cap U\neq\emptyset\}$$ (this is just a restatement of the fact that $\bar A = A\cup A_0$ where $A_0$ is the set of limit points of $A$)

Now, for the proof itself:

\begin{align}
x\in \bar A & \iff (\forall U\in\mathcal O_x)\ A\cap U\neq\emptyset\\
& \iff (\forall\varepsilon>0)\ A\cap K_d(x,\varepsilon)\neq\emptyset\\
& \iff (\forall\varepsilon>0)(\exists a\in A)\ d(x,a)<\varepsilon \\
& \iff \inf_{a\in A}d(x,a) = 0
\end{align}

The closure of any nonempty set cannot be nonempty. This is easily seen from the definition of the closure of $A\subseteq X$ as the intersection of all closed subsets in X where A is a subset of each set: This intersection cannot be empty unless A = $\emptyset$.

As for the proof of the result, consider the metric as defined. Let $\bar A$= $\cap\mathbb F$ where $\mathbb F$= { $F\subset X$  | $F^c$ is open in X and $A\subset F$}. Let $x\in \bar A$. If $x\in A$, we're done. Let $x\notin A$. By definition, if a$\in A$, $d_A(x)=inf_{a \in A}$d(x,a).Since x$\in\bar A$ = $\cap^{\infty}_{i\in I}$ $F_i$, then  $d_A(x)$= min{$r_i\in\mathbb R > 0|d_A(y)= r_i $ where $y\neq x\in F_i\in \mathbb F$}.Therefore $d_A(x) < r^{*}_y$ = $inf_{a \in A}$ d(y,a) for every $y\in\bar A$. But since $r^{*}_y\in \mathbb R > 0$ is arbitrary, we're forced to conclude  $d_A(x)= 0$.

(Seem to have finally fixed the code myself. Never mind.Now I wish someone would tell me what's wrong with this corrected proof if anything.)


What question to ask to find right path Suppose a certain country is inhabited by truth teller and liars, who only respond to yes or no questions. A tourist comes to a fork in the road where one branch leads to the capital, and the other does not. There is a native standing at the fork.  What yes or no question should the tourist ask to determine which branch to take?

The tourist can ask "If I were to ask you if the left branch leads to the capital, would you say yes?"  By construction, if Yes, then left branch leads to the capital. If No, then it doesn't.

My book however, suggests letting $A$ be "you are a truth teller" and $B$ be "the left branch leads to the capital." Construct a formula from $A$ and $B$, call it $F$, so when asked "Is the statement F true, the native replies yes iff $B$ is true.

How can we construct $F$?

"If I ask the other kind of person, whether the right road leads to the capital , he will say YES, won't he? "

Explanation: ((( Assuming left road is the correct Path ... )))
                                Liar will say "Yes"
                                Truth teller -"Yes"

Liar knows that truth teller knows the truth and will say no when asked So liar will tell the opposite which is "yes".
Truth teller knows that the liar will lie and will say yes about right road being correct, So truth teller will also say "Yes" indicating that the liar will say yes.

So I will take the other road.

I think you're supposed to ask: "If you were a truth teller, would you tell me if the left branch leads to the capital?".
If the native is a truth teller, he will confirm that he would say yes if the left branch does indeed lead to the capital.
If the native is a liar (and the left branch does lead to the capital), he would tell you that the left branch doesn't lead to the capital. However, he would ALSO lie to you when you ask him the question above. He would lie about lying, so his answer would have to be "yes" as well.


Expected number of tosses until 3 heads in a row - via Martingale method (Quant job interviews - questions and  answers - Question 3.8)

For a fair coin, what is the expected number of tosses to get 3 heads in a row

The answer is stated as :

We gamble in such a way that we make money on heads but such that if we get a T on toss $n$, our position is $-n$.

We therefore gamble one unit on the first toss and on each toss after a $T$.

After one head, we gamble three. This guarantees that if we get a $T$ next then we go to $-n$.

After two heads we are therefore up 4, and so we gamble 7  to get us to $-n$ again. our gambling winnings is a martingale since we are making finite trades in a martingale (any bounded trading strategy in a martingale is a martingale).

After three heads our position is $11 - (n-3)=14-n$. the time taken to get three heads is a stopping time with finite expectation so if we stop at it we still have a martingale (Optional sampling theorem) thus $\mathbb E (14 -n) = 0 $ and we are done.

I realise there are a few answers to this already, however i am not sure the explanation above, which uses martingale theory, is entirely clear.

The author states that  any bounded trading strategy in a martingale is a martingale  but what is the underlying martingale here ?

Also I don't understand the underlying  motivation for gambling the way described, please could someone put it in more mathematical terms so i can understand the reason for the sizes of the bets ?

I believe that the strategy is that:

consider the payoff of the $i_{th}$ toss

(1) If $i == 1$ or $X_{i-1} = Tail$  , you bet 1 unit on Head. Which means that $Payoff_i(X_i=Tail) = -1$ $Payoff_i(X_i=Head) = 1$;

(2) If $X_{i-1} = Head$  , you bet 3 unit on Head. Which means that $Payoff_i(X_i=Tail) = -3$ $Payoff_i(X_i=Head) = 3$;

(3) If $X_{i-2} = Head$ and $X_{i-1} = Head$  , you bet 7 unit on Head. Which means that $Payoff_i(X_i=Tail) = -7$ $Payoff_i(X_i=Head) = 7$;

(4) stop tossing when you get HHH serial.

Conclusions about this strategy:

(1) If you get a Tail at $i_{th}$ toss, your total payoff is $-n$(you could test this conclusion for arbitrage simple series);

(2)For every toss, your expected payoff is zero. Your total payoff is a martingale.

$$ \text{
**any bounded trading strategy in a martingale is a martingale**
}
$$
I think that the first "martingale" in this quote is the total payoff $M_t = \sum_{s=0}^tpayoff_s$.
The trading strategy descripted above is also a maringale.

When this strategy stop, denote $n$ as the strategy stopping time. The expected payoff should be zero.which is

$E[-(n-4)+1+3+7]=14-E[n]=0$. $E[n]=14$.

I believe the author is using a great deal of words to define the wealth process : $$M_n = M_0 + \sum_{1}^{3} 2^{i}X_{i}\text{ with }M_0=1\text{ and  } X_i \text{ iid as }\{1,-1\}\text{  with p=}\frac{1}{2}$$

$\mathbb E(M_n|M_{n-1}) = \mathbb E(M_{n-1}+2^nX_n|M_{n-1})=M_{n-1}+2^n\mathbb E(X_{n-1})=M_{n-1}$ so $M_n$ is a martingale

then for the stopping time $\tau = \tau_{HHH}+1$ since we want the number of throws rather than the number of $M_i$s

$$\mathbb E (M_{\tau}) = 1 + 2^1 + 2^2 + 2^3 = \mathbb E(M_0)\mathbb E(\tau)=\mathbb E_{\tau}=1+E(\tau_{HHH})\text{ so } \mathbb E(\tau_{HHH})=14$$


How can adding an infinite number of rationals yield an irrational number? For example how come $\zeta(2)=\sum_{n=1}^{\infty}n^{-2}=\frac{\pi^2}{6}$. It seems counter intuitive that you can add numbers in $\mathbb{Q}$ and get an irrational number.

But for example $$\pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+\cdots$$ and that surely does not seem strange to you...

This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.

Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?


Structure group of tangent bundle of Riemannian manifold In the article "Spin structures on manifolds" by J Milnor, the author begins as follows -

  Let $M$ be an oriented, Riemannian manifold. Then the tangent bundle of $M$ has the rotation group $SO(n)$ as structural group.

I was wondering if anyone could tell me where I could find a proof of this fact (or give me hints to prove it myself). Google didn't help much.

EDIT : Definition of Oriented manifold being used -

A smooth manifold $M$ is oriented if it admits an orientable atlas. That is an atlas whose all transition functions have positive Jacobian determinant.

Thank you.

If $E\to M$ is a rank-$k$ vector bundle, a reduction of the structure group of $\boldsymbol E$ to a subgroup $G\subset GL(k,\mathbb R)$ is a covering of $M$ by local trivializations of $E$ such that the transition functions are of the form $(x,v) \mapsto (x,\tau(x)v)$, where $\tau$ takes its values in $G$.

Thus a reduction of the structure group of $TM$ to $SO(n)$ is a covering by local frames (NOT necessarily frames arising from coordinate charts) such that the change-of-frame matrix takes its values in $SO(n)$.

If $M$ is an oriented Riemannian manifold, you just take a covering by oriented orthonormal frames, and Bob's your uncle.

This is what I've gathered reading the book "Riemannian Geometry and Geometric Analysis, by Jurgen Jost" edition 6 page 45.

So we have coordinate charts on an orientable manifold $M$ such that ALL chart transitions have positive Jacobian. Any one of these charts (say $x:M\to \mathbb{R}^d$) induces a frame $\partial/\partial x^a|_{m}$ at a point $m\in M$. Consider another chart $y$ and the frame $\partial/\partial y^a|_{m}$. These frames are related by the Jacobian $\partial/\partial y^a=J^b_a\partial/\partial x^b$ (where $J^b_a:=\partial x^b/\partial y^a|_m$). This is precisely the Jacobian that has positive determinant by assumption of orientability. Then consider using the GRAM-SCHMIDT process on the frame $\partial/\partial x^a|_{m}$ to obtain a frame $e_a$ (where $a=1,...,dim(M)$) such that $e_a$ is orthonormal with respect to the metric -- that is $g(e_a,e_b)|_{m}=\delta_{ab}$. Going from the frame $\partial/\partial x^a|_{m}$ to $e_a$ can be written as $e_a=G^b_a\partial/\partial x^b|_{m}$, where $G\in GL(dim(M))$. Similarly, one can GRAM-SCHMIDT orthonormalize $\partial/\partial_y|_{m}$, i.e $\tilde{e}_a=\tilde{G}^b_a\partial/\partial_y|_{m}$ such that $g(\tilde{e}_a,\tilde{e}_b)=\delta_{ab}$. Then consider the matrix $M$ such that $\tilde{e}_a=M^b_ae_b$.

So to finally address the question: We want to show that this matrix $M$ lies in $SO(dim(M))$. To show that is is orthogonal we write $\delta_{ab}=g(\tilde{e}_a,\tilde{e}_b)=M^c_ag(e_c,e_d)M^d_b=M^c_a\delta_{cd}M^d_b=M^d_aM^d_b=(M^{T}M)_{ab}=\delta_{ab}$. Since it lies in the orthogonal group, $det M=\pm 1$. To show that it has $det M >0$ and therefore $detM=1$, first observe from all the previous definitions that $MG=\tilde{G}J$. Therefore $detM=det \tilde{G} det J/det{G}$. Then FINALLY -- and here is the crux of the matter -- the GRAM SCHMIDT matrix $G$ has $det G>0$ (see Determinant in Gram-Schmidt process for a proof). Then using the fact that $det J>0$ we immediately conclude that $det M>0$.

So I guess the moral of the story is that given an orientable manifold, one can always construct a tangent bundle with structure group SO(d). There always exists one because we can always use the GRAM-SCHMIDT process to construct one, but I don't think there is a unique tangent bundle with structure group SO(d), because you might have another orientation preserving diagonalization procedure.


Are A and B conditionally independent Are A and B conditionally independent given the class label?

I calculated that

$$P(A=1) = \frac{1}{2}$$
$$P(B=1) = \frac{2}{5}$$
$$P(A=1,B=1)=\frac{1}{5}$$

My answer is yes. I do it by anding $(A\text{ and }B)$
which shows that when $A = 1$ and $B = 1$ it doesn't imply Class will be +. How can I actually prove this without this pseudo prove I have come up with?

A and B are conditionaly independent of C iff for every value of C, $P(A\cap B|C) = P(A|C)P(B|C)$

Lets try class = +.

$P(A\cap B|+)=\frac{P(A\cap\ B \cap +)}{P(+)} = \frac{1}{5} \neq P(A|+)P(B|+) = \frac{3}{5}\frac{2}{5} = \frac{6}{25}$

So no, they are not.

Marginalize to get the mass functions $p_{A \mid CL}(a \mid cl),p_{B \mid CL}(b \mid cl)$ and $p_{A,B \mid CL}(a,b \mid cl).$
Check for every class label $CL=cl$ and $\forall a,b$ if the following holds (this is the definition of conditional independence of independence in general): $p_{A,B \mid CL}(A=a,B=b \mid CL=cl) = p_{A \mid CL}(A=a \mid CL=cl)p_{B \mid CL}(B=b \mid CL=cl)$ If it holds, then $A,B$ are conditionally independent of the class label $CL$


Adding random numbers I haven't taken statistics or calculus, but I'm curious: if you take $100$ perfectly random numbers from $-1$ to $1$ and add them, how close would it be to $0$? How about adding $10000$ perfectly random numbers from $-1$ to $1$? How about  infinity ($\infty$)?

Perfectly random should be defined.

To answer your question we can say that if you repeat selecting $100$ numbers randomly from $-1$ to $1$, many times, and record the sums, then these sums are approximately normally distributed with the mean of $0$

If by perfectly random you mean that each number is independent of the other number and have identical probabilities of being some value, there is a famous law called "The Law of Large Numbers" that states if you have infinite samples, the empirical average of all of those numbers will equal the true average.

To be more specific, Let $X_i$ be the $i^{th}$ sample that you pick between $[-1,1]$. The empirical average is $\frac{1}{n}\underset{i=1}{\overset{n}{\sum}} X_i$ which goes to $0$ exponentially as $n$ grows large. The exact rate is given by Hoeffding's Inequality, which when applied to this problem states that

$$|\underset{i=1}{\overset{n}{\sum}}X_i| \geq n \sqrt{2 \ln \frac{2}{\delta}}$$  with probability $1 - \delta$. So if you just plug in the amount of error you allow into $\delta$ you can get the minimal number of samples you need.


Evaluate $\int_{0}^\frac{1}{2}\frac{\sin^{-1}x}{\sqrt{1-x^2}} dx $ $$\int_0^\frac{1}{2} \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx $$

My question for this is can I take out the square root so it would be

$$\frac{1}{\sqrt{1-x^2}} \int_{0}^\frac{1}{2}{\sin^{-1}x} \,dx $$

So all i have to do is find the anti-derivative  of $\sin^{-1}x$ then multiply the $\dfrac{1}{\sqrt{1-x^2}}$ back in afterwards? Is there a simpler way?

Hint: let $u = \arcsin x$. Then $du = \frac{1}{\sqrt{1-x^2}} dx$. You're not allowed to move the $\frac{1}{\sqrt{1-x^2}}$ term out in front since it has an $x$ in it and the variable of integration is $x$--so the integral depends on that term too!

I may suggest that this integral must be like this Integral arc sin x d(arc sin x) from 0 to 1/2. thus the answer must be π^2/72


Why are eigenspaces associated with different eigenvalues orthogonal? My teacher stated the following proposition:

Let $f:V\to V$ be an orthogonal/unitary Endomorphism and $V$ a finite-dimensional vector space. Then, If $F\leq V$ and $f(F)\subseteq F$ we have: $$f(F^\perp) \subseteq F^ \perp$$

I understand this proposition and I was able to prove it myself, but then my teacher stated the following as a corollary:

Let $f:V\to V$ be an orthogonal/unitary Endomorphism and $V$ a finite-dimensional vector space. Then eigenspaces associated with different eigenvalues are orthogonal.

There are two things I don't understand here:

What does it mean for two subspaces $A,B$ to be orthogonal?
How can this be proven using the proposition above?

Thank you.

We say $A\perp B$ if $a\perp b$ for all $a\in A$, $b\in B$.
Note that if $A$, $B$ are eigenspaces of eigenvalues $\lambda_A$, $\lambda_B$, then for $a\in A$, $b\in B$, $$\lambda_A\langle a,b\rangle = \langle \lambda_A a,b\rangle=\langle f(a),b\rangle =\langle a,f(b)\rangle=\langle a,\lambda_Bb\rangle=\lambda_B\langle a,b\rangle $$


Is it possible to do modulo of a fraction I am trying to figure out how to take the modulo of a fraction.

For example:  1/2 mod 3.

When I type it in google calculator I get 1/2. Can anyone explain to me how to do the calculation?

One can perform arithmetic of fractions mod $\,m\,$ as long as all fractions $\,a/b\,$  have denominator $\,b\,$ coprime to $\,m,\,$ since then, by Bezout, $\,b\,$ is invertible mod $\,m\,$ so the fraction has the unique denotation $\,x = a/b = ab^{-1}$ (the unique solution of $\,bx = a).\,$  The usual rules of fraction arithmetic remain true (as long as one restricts to such fractions), e.g. from a prior answer:

Hint $\,\ {\rm  mod}\ 13\!:\  \dfrac{41}7 \equiv  \dfrac{28}7 \equiv 4\ \ $ by $\ \ 41\equiv 41\!-\!13 \equiv 28$

Alternatively $\,\ \dfrac{41}{7}\equiv\dfrac{(-2)(-1)}{-6}\equiv \dfrac{-2}{-2}\dfrac{12}3\equiv 4\ \ $ by $\ \ \begin{eqnarray}41&&\!\!\equiv\ \ 2\\ 7 &&\!\!\equiv -6\end{eqnarray}$

Alternatively $\,\ \dfrac{41}{7}\equiv \dfrac{2}7\equiv \dfrac{4}{14}\equiv \dfrac{4}1\ $ by Gauss's Algorithm.

Such twiddling (adding/subtracting the modulus from numerator or denominator till things divide or factor nicely) works quite well for small numbers. For larger numbers one can invert the denominator by the Extended Euclidean Algorithm, or Gauss's algorithm if the modulus is prime.

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

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

Finding the inverse of p^n mod q where p,q are relatively-prime primes using Fermat’s little equation,
p^(-n) mod q = p^x mod q where x = (-n) mod (q-1).
For example, 3^(-1) mod 5 = 3^[(5-1)-1] mod 5 = 3^3 mod 5 = 2.
Another example, 7^(-1) mod 13 = 7^11 mod 13 = 2 (done by calculator).
The formula is useful when one is working algebraically, rather than with specific numbers.


Probability, conditional on a zero probability event Is there a way to resolve probability of an event, given another event that never happens? Mathematically speaking the problem is:

Given that $P(B) = 0$,

$$P(A|B)=\frac{P(A \cap B)}{P(B)} = \frac{0}{0}$$

Is this probability vacuously $0$ of $1$? Can we show that it's one or the other?

The comment by Dilip Sarwate points to conditioning on the level of densities which can be interpreted as conditioning on a family of events of probability zero. It is a probabilistic version of Radon-Nikodym derivative.

One can also condition on an individual event of probability zero, if that event admits a natural approximation by events of positive probability. For example, begin with a  probability measure $\nu$ on the space $C[0,1]$ of  continuous functions such that $\nu(\{f(0)=0\})=0$.  Restrict it to the set of functions $f$ such that $|f(0)| \le \epsilon$. Normalize this restriction to a probability measure. If these normalized restrictions converge (in some sense) to a probability measure $\mu$ on $C[0,1]$, we can regard $\mu$ as conditional distribution on the zero-probability event $f(0)=0$.

This argument applies only to probability of events and not for probability with random variables.

Firstly the definition, $$P(A|B)=\dfrac{P(A \cap B)}{P(B)}$$ applies only when $P(B) \neq 0$.

What is it when $P(B)=0$?

Well, $P(A|B)$ is the probability of $A$ happening given that $B$ happened is ridiculous to even ask. How can you be given that an event that cannot happen has happened? So, it shall remain indeterminate in my opinion.

However, some texts, do give that $P(A|B)=0$ and some others assign $1$, or so my teacher has told us that he has seen one book doing the latter (asssigning $1$).

But, it just does not matter.


Show that $f$ is NOT Lebesgue integrable. Define $f:[0,1]\to \mathbb R$ by $f(x)=\begin{cases}(-1)^nn &\text{ if }\frac{1}{n+1}<x\le \frac{1}{n}\\0 &\text{ otherwise }\end{cases}$

Show that $f$ is NOT Lebesgue integrable in $[0,1]$.

I want to show that $\displaystyle \int_0^1|f(x)|\,dx=\infty$ , but I am unable to find it.

Notice that $f$ can also be written as
$$ f = 0\chi_{\{0\}} + \sum_{k=1}^\infty (-1)^k k \chi_{(1/(k+1),1/k]}
$$ For each $n \in \mathbb{N}$, define
$$ \phi_n = 0\chi_{[0,1/(n+1)]} + \sum_{k=1}^n k \chi_{(1/(k+1),1/k]}
$$
to be a simple function. (The $0$ term is there to ensure that each point is covered by a characteristic function.) It is easy to see that
$$\int \phi_n\,dx = \sum_{k=1}^n \frac{k}{k(k+1)} = H_{n+1} - 1$$
where $H_n$ is the $n$th Harmonic number and is known to diverge. Since $\phi_n \leq |f|$ for all $n$, $\int |f|\,dx = \infty$.


Almost sure convergence of equal weighted sum Let $Z_1, Z_2, ...$ be independent random variables in the same probability space defined as follows:

$$P(Z_n=n)=P(Z_n=-n)=\frac{1}{2n^2} \space \mathrm{and} \space P(Z_n=0)=1-\frac{1}{n^2}$$

Is it true that

$$\lim \frac{1}{\sqrt{n}}(Z_1+...+Z_n)  = 0 \ \text{a.s.}$$

?

I thought of using the Borel-Cantelli Lemmas, but I don't know how to calculate the probability that $$P(|Z_1 + \ldots + Z_n| \ge \epsilon)$$ where $\epsilon > 0$.

I already tried to apply Chebyshev's inequalities, but it didn't work.

Notice that the series $\sum_{n=1}^{+\infty}\mathbb P\left(Z_n\neq 0\right)$ converges, as $\mathbb P\left(Z_n\neq 0\right)=1/n^2$. This means that for almost every $\omega\in \Omega$, we can find an integer $N\left(\omega\right)$ such that if $n\geqslant N\left(\omega\right)$, then $Z_n\left(\omega\right)  = 0$. This implies that for $n\geqslant N\left(\omega\right)$,
$$\frac{1}{\sqrt{n}}\left(Z_1\left(\omega\right)+\dots+Z_n\left(\omega\right)\right)=\frac{Z_1\left(\omega\right) +\dots+Z_{N\left(\omega\right)}\left(\omega\right)}{\sqrt n}    ,$$
which gives the wanted result.

Two remarks:

We did not use independence of the sequence $\left(Z_n\right)_{n\geqslant 1}$.
The normalization $\sqrt n$ could be replaced by $a_n$ where $a_n\to +\infty$ as $n$ goes to infinity.

Hint:

Using BCL1, it looks like we have

$$P(\liminf(Z_n \ne n)) = 1$$

$$P(\liminf(Z_n \ne -n)) = 1$$

I think we can conclude that

$$P(\liminf(Z_n = 0)) = 1 \tag{*}$$

because:

$\exists m_1 \ge 1$ s.t. $Z_{m_1} \ne m_1$, $Z_{m_1+1} \ne m_1+1$, ...

$\exists m_2 \ge 1$ s.t. $Z_{m_2} \ne m_2$, $Z_{m_2+1} \ne m_2+1$, ...

So I think for $m := \max\{m_1, m_2\}$, we have

$$Z_m \ne m, -m$$

$$Z_{m+1} \ne m+1, -(m+1)$$

$$\vdots$$

which I think is $(*)$

Also not sure if relevant but by BCL2, it looks like

$$P(\limsup(Z_n = 0)) = 1$$

As for computing

$$p_n := P(|\sum_{i=1}^{n} Z_i| > 0)$$

If you're going to use BCL2 on $\sum p_n$, assuming it converges, it looks like you'll be able to conclude

$$P(\limsup(|\sum_{i=1}^{n} Z_i| > 0)) = 0$$

$$\to P(\liminf(|\sum_{i=1}^{n} Z_i| = 0)) = 1$$

$$\to P(\liminf(\sum_{i=1}^{n} Z_i = 0)) = 1$$

This looks pretty strong to me, but I guess it'll work assuming you compute $p_n$ and $\sum p_n$ converges.


Probability of 3 people winning 3 games out of 9 total. I have a question and I'm not sure if my answer is correct.
Suppose Al, Bill, Cara, and Bob play cards together. Al wins with probability $0.4$, Bill with $0.3$, Cara with $0.2$, and Bob with $0.1$. Assume they play 9 games total and the outcomes of each game are independent.
What is the probability Al, Bill, and Cara win $3$ games each and Bob does not win any?
My answer:
\begin{align*}
P(\text{Al wins $3$}) & = \binom{9}{3}(0.4^3)(0.6^6)\\
P(\text{Bill wins $3$}) & = \binom{9}{3}(0.3^3)(0.6^6)\\
P(\text{Cara wins $3$}) & = \binom{9}{3}(0.2^3)(0.8^6)
\end{align*}
I know they're independent events so I added them:
$$P(\text{Al wins $3$}) + P(\text{Bill wins $3$}) + P(\text{Cara wins $3$})$$

I don't have a solution, and I'm not sure if this is correct reasoning.

EDIT: The correct answer is to use the multinomial probability.

Using multinomial probability, the formula is the following:
$$p=\frac{n!}{n_1!n_2!n_3!\dots}p_1^{n_1}p_2^{n_2}p_3^{n_3}p_4^{n_4}\dots$$
You use this because each of the events (or games) are independent of each other and the probability of one doesn't affect the other. You also cannot use binomial because there are more than 2 factors (3 people playing).

$n$ is the number of trials. $n_i$ are the individual events, the number of times Al, Bill, Cara, and Bob win the game. Al, Bill, and Cara's events are 3, since they need to win 3 times. Bobs event is 0 since he is not winning. Then we have the answer as
$$\frac{9!}{3!3!3!0!}.4^3.3^3.2^3.1^0=0.02322432=2.32%$$


Determine the coefficient of $xy$ in the expansion of $(x+y+2)^7$ The problem: Determine the coefficient of $xy$ in the expansion of $(x+y+2)^7$

My approach:

We can rewrite the equation substituting $x+y =j$

$$(x+y+2)^7=(j+2)^7$$

This is simpler because we know the coefficients thanks to the formula:

$$(a+b)^{n}=\sum _{{k=0}}^{n}{n \choose k}a^{{n-k}}b^{{k}} $$

With $n=7$  we have $1,7,21,35,35,21,7,1$ as coefficients and the expansion looks like this:

$$j^7+14\cdot j^6+(21\cdot 2^2) j^5+(35\cdot 2^3) j^4+(35\cdot 2^4) j^3+(21\cdot 2^5) j^2+(7\cdot 26)j+2^7$$

Now the only time that $xy$ appears is in the expansion of $j^2$ therefore we have:
$$(21\cdot 2^5) j^2=(21\cdot 2^5) (x+y)^2= (21\cdot 2^5)(x^2+2xy+y^2)=21\cdot 2^5x^2+21\cdot 2^6xy+21\cdot 2^5y^2$$
The coefficient is $21\cdot 2^6$

Is this correct? Is there a simpler proof?

Yes to both questions.

We can write:

$$(x+y+2)^7=\sum_{p+q+r=7}\binom7{p,q,r}2^rx^py^q$$

To get $xy$ we make $p=q=1$, so the coefficient of $xy$ is:

$$\binom7{5,1,1}2^5=\frac{7!}{5!1!1!}\cdot 2^5=42\cdot 2^5=21\cdot2^6$$

(x+y+2)(x+y+2)(x+y+2)*(4 more times).
To get xy as a coefficient you  can choose x from any block and y from any other block in 7c2 ways and since xy can be permuted in two ways you get 2*7c2 ways and for the remaining you must choose 2 from each block so you get an extra 2^5 . so coefficient is 2* 7c2 *2^5 = 21*2^6.

so for any general case (ax+by+cz+dw)^n . Coefficient of x^r1*y*r2*z^r3*w^r4 . where
r1+r2+r3+r4 = n. can be written as >>

n!*(a^r1*b^r2*c^r3*d^r4)/(r1!*r2!*r3!*r4!)


Probability that a page has at least three figures A books with 120 pages has 6 figures. Every figure could be on a page with the same probability. What is the probability that a randomly selected page has at least three figures?

Expanding on lulu's comment, the formula for binomial probability is:

$$P(x) = {n \choose x}p^x q^{n-x}$$

where $n$ is the number of trials, $x$ is the number of successes, $p$ is the success probability of one trial, and $q = 1 - p$ or the failure probability of one trial.

In your case, $n = 6$, $x = 3$, and $p = \frac{1}{120}$, so the probability should be:

$${6 \choose 3} \left(\frac{1}{120}\right)^3 \left(1-\frac{1}{120} \right)^{3}$$

$$\approx 1.13 \cdot 10^{-5}$$

Probability of finding a figure on a page is $P(A) = 6/120 = 0.05$. So finding 3 such images in the same page would mean $P(A) * P(A) * P(A) = 0.000125$


How many odd days are there in 1600 years ? I calculated it but I guess the result is wrong? I had to calculate the number of odd days in 1600 years. I have read the answer to be equal to 0. But I don't get it to equal to 0.

This is the way I am calculating the number of odd days in  1600 years :

1600 years = 24 x 16 = 384 leap years   (100 years = 24 leap years)
(because 100 years have 24 leap years)
1 leap year = 2 odd days (52 weeks + 2 odd days)
384 leap years = 384 x 2 = 768 odd days --(A)
1600 years = 1600 - 384 = 1216 ordinary years
1 ordinary year = 1 odd day (52 weeks + 1 odd day)
1216 ordinary years = 1216 x 1 = 1216 odd days--(B)

Total number of odd days = (A) + (B) = 768 + 1216 = 1984 odd days in 1600 years

and 1984 is not divisible by 7 !

Am I making a mistake ? If yes,what is it ?

Remember that a year divisible by $400$ is a leap year.  Although $2100$ will not be a leap year. $2400$ will be.

So in $400$ years there are precisely $97$ leap years.

And yes, the calendar repeats every $400$ years, so the number of days in $1600$ years is divisible by $7$. For $400$ years, to the $(400)(364)$ days,  just add $400+100-3$ (ordinary advance by $1$ day, plus 100 for the leap years sort of, minus $3$ for the multiples of $100$ that are not multiples of $400$).  Then multiply by $4$. So for $1600$ we get $4(500-3)$ "additional" days. Your $1984$ was essentially computed correctly, except that we need $4$ additional days for the $4$ multiples of $400$.

Every 4th century is a leap year and no other century is a leap year. Therefore there will be 96 + 1=97 leap year and 303 ordinary years. So the equation becomes 303*1 +97*2 =497 odd days or 0 odd days.


Order of element in multiplicative group modulo n, where n is not prime I should calculate order of a givent element of multiplicative group modulo n. This n might, or might not be a prime.

I discovered that I can use algorithm 4.79 from Handbook of Applied Cryptography to do that:

The thing is, I need to know order of the group itself as input for this algorithm. For a group where n is prime, this is easy, the order is n-1. But what if n is a composite number? Do I really have to calculate the Euler totient function (as suggested here) to determine the order of the group and then use this algorithm to calculate the order of the element? Or am I somehow able to calculate the order of a given element based on just factorization of n (teacher suggested this should be possible)?

You link to the "Multiplicative group modulo n" page on Groupprops.  That page references the "Euler-phi function" with a link to the "Euler totient function" page on Groupprops.  That page has an "Explicit formula" section with $$ \varphi(n) = n \prod_{p \mid n} \left(1 - \frac{1}{p}  \right) $$ (although the cited page indicates "$p \mid n$" with the words "over all distinct prime factors of $n$").

Thus, the reference you give tells you how to compute $\varphi(n)$ from a prime factorization of $n$.  The method given is relatively fast.

Hint Use the Chinese Remainder Theorem.

By the CRT, if $$n=p_1^{e_1}...p_k^{e_k}$$
then the order of $a$ modulo $n$ is the least common multiple of the order of $a$ modulo $p_j^{e_j}$.


Evaluating a contour integral (don't understand my error) I'm trying to evaluate the integral

$$ I = \int_{\gamma} \frac{1}{(z-a)(z-b)} dz$$

over the the circle of radius $r$ at the origin (assume that $|a| < r < |b|)$. Please note I haven't yet reached Cauchy's theorem in my book.

Then we have

$$ I = \int_0^{2\pi} \frac{ire^{it}}{(re^{it}-a)(re^{it}-b)} dt$$

And then we note that:

$$
\frac{1}{(re^{it}-a)(re^{it}-b)} = \frac{a-b}{(a-b)(re^{it}-a)(re^{it}-b)} = \frac{(re^{it}-b)-(re^{it}-a)}{(a-b)(re^{it}-a)(re^{it}-b)} = \\
\frac{1}{(a-b)(re^{it}-a)} + \frac{1}{(b-a)(re^{it}-b)}
$$

So then we have

$$ I = \int_0^{2\pi} \frac{ire^{it}}{(a-b)(re^{it}-a)} dt + \int_0^{2\pi} \frac{ire^{it}}{(b-a)(re^{it}-b)} dt $$

When I try to make a substitution $u = r e^{it} - a$ in the first integral (and similarly for the second integral) I have an answer like

$$ I = \frac{ \log(r e^{it} - a) ]_0^{2\pi}}{a-b} + \frac{ \log(r e^{it} - b) ]_0^{2\pi}}{b-a}$$

which is $0$ and I do not think this is the correct answer; however I'm not sure what I did incorrectly along the way.

Added later: I seem to be getting downvoted because I did not answer the specific question asked. So be it. I think the below will be of help to others who have this question.

I would do this using series expansions: Write

$$\frac{1}{(z-a)(z-b)} = \frac{A}{z-a}
+ \frac{B}{z-b},$$

where the constants $A,B$ are easy enough to find. So we're left trying to figure out the integrals $\int_\gamma dz/(z-a),\int_\gamma dz/(z-b).$ The first integral equals

$$\tag 1\int_0^{2\pi}\frac{ire^{it}}{re^{it}-a}\,dt = i\int_0^{2\pi}\frac{1}{1-(a/r)e^{-it}}\,dt$$ $$ = i\int_0^{2\pi}\left (\sum_{n=0}^{\infty}(a/r)^ne^{-int}\right )\,dt = i\sum_{n=0}^{\infty}\int_0^{2\pi}(a/r)^ne^{-int}\,dt.$$

The only summand on the right that is nonzero is the $n=0$ term, so we see $(1)$ equals $2\pi i.$

The second integral can be handled similarly.


how to find two right-inverse functions of a function i am stuck in this problem. i need to find two right-inverse functions of this function:

$h: \Bbb N_0\times \Bbb N \to \Bbb N, (m,n)\mapsto m+n$.

i know that the function h' is a right inverse of a function h if and only if:

$h  \circ h' = m+n$.

how can i say this in mathematical way and as for the h function given above?  i did this:

$h': \Bbb N->\Bbb N_0\times\Bbb N, m+n\mapsto (m,n)$.
  $h \circ h' = h(h'(m+n))=m+n$

but now from here, i dont know how to finish the proof and how to show another right-inverse function of h

thanks for help in advance

As you point out, $h^{-1}$ maps naturals to pairs $(m, n)$. If you think about what $h^{-1}$ is doing, you'll see that you need to find a pair $(m, n)$, $m \in \mathbb{N}_0$, $n \in \mathbb{N}$ such that $m+n$ is the function $h^{-1}$'s input. That is, if $h^{-1}(k) = (m, n)$, then $m + n = k$. An easy approach is to fix some $m$ first, then let $n = k - m$. This will work provided $k > m$. We can then handle the remaning cases separately. This gives $h^{-1}(n) = (0, n)$ as a simple example, since $(0, n) \in \mathbb{N}_0 \times \mathbb{N}$. So that gives one right inverse. Another can be given by, for example,

$$h^{-1}(n) = \begin{cases}(1, n-1), & \mbox{if } n > 1\\(0, 1), & \mbox{if } n = 1\end{cases}$$

Can you devise a suitable function for $m = 2$? For $m = 3$?

Just to give a little more clarity. For each $m\in \mathbb{N}$, define $g_{m}:\mathbb{N}\rightarrow \mathbb{N}_{0}\times \mathbb{N}$ as
\begin{eqnarray*}
g_{m}(n)=\left \{
\begin{array}{cc}
(m,n-m)&\text{ if } n\geq m\\
(0,n)&\text{ if } n<m
\end{array}\right.
\end{eqnarray*}

Then each $g_{m}$ is a right inverse of $h$.


Formula for binary sequences of length m with no n consecutive 1s? Formula for binary sequences of length $m$ with no $n$ consecutive $1$s?

I know The number of binary strings of length $m$ without consecutive $1$s is the Fibonacci number $F_{m+2}$.

But how about no $n$ consecutive $1$s?

Any help will be appreciated! Thank you

The same idea will work to make a recurrence.  To make a sequence of length $m$, you can take one of length $m-1$ and add a zero, one of length $m-2$ and add $01$, one of length $m-3$ and add $011$ on up to $m-n$ adding a zero and $n-1\ \ 1$'s

This formula should work-

Total sequences possible $=2^m=s(let)$;
Sequences with no $n$ consecutive $1$'s $= \frac{(m-n+1)(m-n+2)}2=t(let)$

Answer$=s-t$


Solid Revolution question I'm trying to do the following problem:

Find the volume of the solid obtained by rotating the region bounded by the given curves about the $y=3$  axis specified by means of  the circular arrow drawn :

$$ y = 3 \sin x, y = 3 \cos x, 0 \le x \le π/4 $$

First off, am I drawing the picture correctly?

Second, am I setting up the integral correctly?

$$
\pi\int_0^{\pi/4} [(3-3\sin x)^2-(3-3\cos x)^2]dx
$$

Did I correctly identify $3-3 \sin x$ as the outer radius? Because it makes sense for it to be that way, but I saw this same question worked out on like yahoo answers and it instead says that $y=3-3 \cos x $ should be the outer radius?

Yahoo!Answers is notoriously unreliable on math and technical matters; my informal count is that the answer is alright only about half the time. (I suspect the poster overlooked the rotation axis being $ \ y = 3 \ , $ rather than the $ \ x-$ axis.) Your integrand looks fine and reduces to

$$ \ (9 - 18 \sin x + 9 \sin^2 x) \ - \ (9 - 18 \cos x + 9 \cos^2 x) $$

$$ = \ 18 \ (\cos x - \sin x) \ + \ 9 \ (\sin^2 x - \cos^2 x) \ = \ 18 \ (\cos x - \sin x) \ - \ 9 \ \cos 2x \ . $$

The evaluation of the volume is then

$$ \pi \ \left[ \ 18 \ (\sin x + \cos x) \ - \ \frac{9}{2}  \sin 2x \ \right]_0^{\pi/4} $$

$$ = \ \pi \ \left( \ [ \ 18 \ (\ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) \ - \ \frac{9}{2} \cdot 1  \ ] \ - \ [ \ 18 \ (0 + 1) \ - \ \frac{9}{2} \cdot 0 \ ] \ \right) \ $$

$$ = \ \pi \  ( \ 18 \sqrt{2}  \ - \ \frac{9}{2}   \ -  \ 18  \ ) \ = \ \pi \  ( \ 18 \sqrt{2}  \ - \ \frac{45}{2}    \ ) \ \ \text{or} \ \ \frac{9 \pi}{2} \ ( \ 4 \sqrt{2}  \ - \ 5    \ ) \ \ , $$

confirming JohnD's result in the comment above.

Drawing looks ok to me. If axis of rotation s $x$ axis, rotated volume comes from section area between two disks:

$$ \pi\int_0^{\pi/4} [(3\sin x)^2-(3\cos x)^2]dx $$

EDIT1/2

But if axis of rotation is $y=3$ , then

$$ \pi\int_0^{\pi/4} [(3-3\sin x)^2-(3-3\cos x)^2]dx=18 \pi\int_0^{\pi/4} (1-\sin x -\cos x) dx = \pi/4 - 1 $$


Assume $A \subset \mathbb{R}^n$ is connected such that $A^c$ is separated by $B,C$, then $A \cup B$ is connected. The following problem showed up on a previous qualifying exam:

Assume $A \subset \mathbb{R}^n$ is connected such that \begin{equation}A^c = B \cup C, \text{ such that  } \overline{B} \cap C = \overline{C} \cap B = \emptyset \end{equation} show that $A \cup B$ is connected. (We take $\mathbb{R}^n$ with the usual metric topology.)

This is what I have so far:

Assume for the sake of contradiction that $A \cup B$ is disconnected, then $A \cup B = D \cup E$ where $D,E$ are separated. Then as $A$ is connected, we must have $A \subset D$ or $A \subset E$; WLOG assume $A \subset D$.

I'm unsure how to proceed from here because I want to write $A$ as a union of two non-empty separated sets, but from $A \cup B = D \cup E$, I cannot see how to get a non-empty separated sets since intersecting them by $A$ or $B^c$ ends up with $E \cap A$ and $E \cap B^c$ being empty since $E \subset B$ as $A \subset D$.

Any help would be appreciated!

In fact the result is more general and true for any connected topological space $X$, which is the case for $\mathbb R^n$ endowed with the topology induced by the usual distance.

Let’s suppose that $A \neq \emptyset$. We’ll deal with the case $A = \emptyset$ at the end of the answer.

Suppose that $D,E$ are open sets of $X$ with $A \cup B \subseteq D \cup E$ and $D \cap E = \emptyset$. As $A$ is supposed to be connected, we can suppose without loss of generality that $A \subseteq D$. We’ll prove that $U=E \cap (A \cup B)$ is open and closed in $X$ hence empty as in a connected space the only subspaces that are both closed and open are the empty set and the space itself. This will provide the conclusion.

Let’s first notice that $U \subseteq B$ as $D \cap E= \emptyset$ and therefore $U= E \cap B$.

$U$ is open

Take $b \in U$. We have $b \notin \overline{A}$ as $b \in \overline{A}$ would imply the contradiction $E \cap A \neq \emptyset$ as $E$ is open. The hypothesis $B \cap \overline{C}= \emptyset$ allows to state $b \notin \overline{C}$. We also have $b \notin \overline{A \cup C}$ as $\overline{A \cup C} \subseteq \overline{A} \cup \overline{C}$. So $b$ belongs to the open subset $V =X \setminus \overline{A \cup C}$ that is included in $B$. And also to the open subset $E \cap V$ that is included in $U$. This proves that $b$ belongs to the interior of $U$. As this is true for all $b \in U$, $U$ is equal to its interior and is open.

$U$ is closed

Take $x \in \overline{U} = \overline{E \cap B}$. As $\overline{E \cap B} \subseteq \overline{E} \cap \overline{B}$, we have $x \in \overline{B}$ and $x \notin C$ according to the hypothesis $\overline{B} \cap C =\emptyset$. Hence $x \in A \cup B$. $x$ cannot belong to $A$: if that would be the case, $x$ would belong to the open set $D$ and $D$ would intersect $U$ and therefore $E$, a contradiction. We get $x \in B$ and as $x \notin D$, we have $x \in E$ and finally $x \in U$. So $U = \overline{U}$, proving that $U$ is closed.

The case $A = \emptyset$

In that case, the space $X$ is the union of $B$ and $C$. $B$ and $C$ are open and closed subsets of the connected space $X$ and are therefore connected. We’re done!

The result doesn’t hold if $X$ is not connected. Take for example $X=\{1,2,3\}$ endowed with the topology induced by the metric distance and $A=\{1\}$, $B=\{2\}$ and $C=\{3\}$.

Note: this answer is based on following one of the excellent French math site les-mathematiques.net.

If $A$ is connected then the complementary of $A$ is connected, because $\mathbb{R}^n$ is for $n>1$. Suppose $A\cup B$ disconnected. Then the not connected part of the union has to be $B$, leading to a contradiction: $B$ is a not connected subspace of a connected one. On the real line is true by taking $A$ the open connected unit sphere without boundary and $B$ one of the connected components of boundary.


Does the vector space spanned by a set of orthogonal basis contains the basis vectors themselves always? I used to think that in any Vector space the space spanned by a set of orthogonal
 basis vectors contains the basis vectors themselves. But when I consider the vector space $\mathcal{L}^2(\mathbb{R})$ and the Fourier basis which spans this vector space, the same is not true ! I'd like to get clarified on possible mistake in the above argument.

If by "the Fourier basis" you mean the functions $e^{2 \pi i n x}, n \in \mathbb{Z}$, then these functions do not lie in $L^2(\mathbb{R})$ as they are not square-integrable over $\mathbb{R}$, so in particular they can't span that space in any reasonable sense. The functions $e^{2 \pi i n x}$ do span $L^2(S^1)$ (in the Hilbert space sense).

Perhaps you are getting the Fourier transform for periodic functions mixed up with the Fourier transform on $\mathbb{R}$.

I want you guys to understand few things clearly:

1> The dirchlet conditions are "sufficient" not necessary  for F' transform'bility

2> The sinusoid is DEFINITELY in the span of {exp(j2PIft)} this follows from linear algebra!
However It does not mean that the fourier TRANSFORM exists ! .. however in case of the periodic signals with finite discontinuities , and INTEGRABILITY(ABSOLUTE) WITHIN A PERIOD, the fourier SERIES does exist !!!  ..

3> for a sine wave , Acos(wt) we (in some engineering contexts) agree that the fourier transform is the delta distribution along +/- w . However strictly we can only talk of the fourier series of such a function ! .

4> the fourier transform is the fourier series , approximated as the fundamental frequency being "epsilon" -->0 ; i.e the funda period being inf.

lastly ! do remember that fourier transformability is not necessary for SPANNING !!

This may not really be an answer to the above but is WORTH THINKING ABOUT:  The fourier "BASIS"  definitely does not span L2(r) , consider the simple rectangular pulse which is definitely square integrable over R. However If you do consider the infinite linear combination (of coefficients of a sinc(f) function whose inverse fourier transform is the pulse) , in the limit as we go on taking infinite (uncountable) sums , The result does NOT converge to the rect pulse in the point convergence sense , but only in the norm square sense. !

In essence , rect(t) is NOT FOURIER TRANSFORMABLE !!   if it were, then all the filter theory we study makes no sense !!!


Showing $Cor(X,Y) = 1$ if $a>0$ and $-1$ if $a<0$ Suppose X and Y are random variables such that $Y=aX+b$ and $a$ and $b$ are constants. Show that $Cor(X,Y) = \begin{cases} +1 &\mbox{if } a > 0 \\
-1 & \mbox{if } a < 0. \end{cases}$

For this, I am using the formula for correlation provided in my book:

$$Cor(X,Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}
= \frac{E(XY) - E(X)E(Y)}{\sqrt{(E(X^2) - E(X)^2)(E(Y^2) - E(Y)^2)}}$$.

I substitute in the values of X and Y=aX+b to get:

$$Cor(X,aX+b) = \frac{E(X(aX+b)) - E(X)E(aX+b)}{\sqrt{(E(X^2) - E(X)^2)(E((aX+b)^2) - E(aX+b)^2)}}$$

I know that I can pull out constants and use the sum rule, but using those just gets really messy and I don't think leads anywhere.

I know how to show this would be between -1 and 1, but I must show it is exactly -1 OR exactly 1. I assume that this would involve squaring this value, but I am lost with where to go from here. Anything I do just makes the equation monstrous. I have been told this is an "easy" proof.

Any tips on how to proceed?

Hint: Use the first part of your $Cor(X, Y)$ expression:
$$
Cor(X, Y)=\frac{Cov(X, Y)}{\sqrt{Var(X)Var(Y)}}
$$
and replace $Y$ by $Y=aX+b$. Use the rules for covariances and variances to simplify.


The intersection of two parallel lines My teacher told me that two parallel lines have a point of intersection, it is called Point at infinity. But I I can't understand how it can be true or how was it proved, can someone explain this to me?

Thank you.

This is not true in ordinary plane geometry, and so it cannot be proved.

It is true, sort of, in a different form of geometry known as projective geometry, however.

As a quick intuitive introduction to projective geometry, imagine that you're standing on the ordinary Euclidean plane. Your head is about 2 meters above the plane, so when you look down you see whatever is drawn on the plane, stretching out to the horizon. Details on the plane right where you stand look large to you; the same details a long distance away will look small to you and be seen very close to the horizon.

Now it's a common enough experience that if we draw to parallel infinite lines on a plane, when we look at them from a point above the plane, it will look as if they meet at the horizon. We can decide to consider the points on the horizon line "equally real" as points on the plane. The horizon then becomes the "line at infinity" and parallel lines in the plane actually do meet at a point on the line at infinity. Then any two lines always meet. Every line in the plane meets the line at infinity at a point determined by its direction; two lines in the plane with different directions intersect in the plane itself, and two lines in the plane with the same direction both meet the line at infinity at the same point, and therefore meet there.

So far there's a strange thing about points on the horizon: they are at the "edge of the world", with plane below them but nothing above them. This is sort of untidy, and there are two ways to fix that. One is simply to decide to draw stuff on the sky. That leads to spherical geometry, an ancient and venerated area of study that is the foundation of astronomy. But it's not what today's lecture is about.

In projective geometry, we're standing on the Euclidean plane, but we're in a virtual reality constructed by a careless and/or lazy programmer. When we look in any direction, our VR helmet computes the infinite line through our head in the direction we're looking, and figures out where that infinite line intersects the plane below us. Whatever is at the plane there is what we see. So when we look at the sky, what we see is the plane behind our head! Only when we look in a perfectly horizontal direction does this procedure not work, but we posit that there's nevertheless some points to look at there, which form a "line at infintity" as before. Because we're not distinguishing between points in front of us and points behind us, there are only 180° of horizon all in all; if we turn 180° we will be looking at the same points at infinity.

(A different way of looking at it is to imagine that we took a copy of the ordinary Euclidean plane and lifted it up to hover 2 meters above our head, and then turned it by 180° about a vertical axis. Note that since we're looking at the sky plane from below, things in the sky will be the mirror image of the same thing if the turn around and look down at them instead).

In projective geometry any two different lines have exactly one point in common. If they are non-parallel lines in the original plane they will cross once on the ground, and we will also see that crossing in the sky -- but these are just two images of the same point in the projective plane. Two parallel lines will cross exactly once at the line at infinity -- again we see two images of that crossing when we turn around, but they are by definition the same point. And any line in the plane will cross the line at infinity once.

The main thing that makes this cool is that the line at infintity now has no special properties. If we tilt our head and forget which way was up and down, there is no way we can deduce from the geometrical properties of what we see which direction the "true" horizon lies. Projective geometry is about properties of figures that are invariant of rotating our view of the world. It is also about properties of figures that don't change as we walk around on the plane, or move our head closer to (or farther from) the plane. So it has no concept of scale or distance either.

These movements both preserve lines (a line stays a line in our view when we tilt our head and/or walk around), so "line" is a projective concept. They don't preserve circles -- if we stand above the center of a circle, it will look circular, but when we walk away from it it starts to look like an ellipse instead. So "circle" and "ellipse" are not projective concepts. On the other hand, surprisingly, "non-degenerate conic section" is a projective concept, and all non-degenerate conics are equivalent. An ellipse is simply a circle seen from a distance. A parabola is a circle that is tangent to the line at infinity. A hyperbola is a circle or ellipse that intersects the line at infinity twice.

This is just an imaginary point. In practice, you can go on following the lines upto whatever distance you wish, I bet your generations will never find that intersection point. There is no point like that. In books they put it to relate it to algebra which can be expressed mathematically only.
But then the question is why do we say that ? We say that because for the human perception there is no infinity. Things have to end somewhere. Try to capture two parallel lines in a camera and they will converge in the image. You can see your own eye does the same.
So, theoretically they don't intersect, while we see them as intersecting somewhere and which shifts to somewhere else as we move closer to that point. People had to name this point. It seems like they intersect and still we can never stand upon that point.


What five odd integers have a sum of $30$? I've been asked the following question:

What five odd integers from the set
$\{1, 3, 5, 7, 9, 11, 13, 15\}$
that when summed together equals to $30$? Note that any integer can be used more than once.

If my limited knowledge of maths is correct, there should be no answer, as no odd number of odd integers summed together can give an even number.

As straightforward mathematics there is no answer.

As anyone who has ever placed hymn numbers in a hymn board will know, it is possible to turn $9$ upside down to get $6$, and if this is allowed by the wording you can get a sum of $30$.

Likewise if it is odd numbers which are chosen, but the digits rather than the numbers which are added, the set $3,5,7,9,15$ gives $3+5+7+9+1+5=30$ - again this depends on precisely how the question is worded.

I think this is a valid question. Normal addition base 10 you can't achieve this
since

(odd+odd)+(odd+odd)+odd=even+even+odd=even+odd= always odd
or
(2x+1)+(2y+1)+(2z+1)+(2u+1)+(2v+1)=2(x+y+z+u+v)+5 is odd

A good answer seems to be an addition with Base 5 arithmetic

3+3+3+3+3=30?


The image of a circle under $ z \mapsto \frac{z- \frac{1}{2}}{\frac{1}{2}z-1} $ Let $C_1 = \{ z: |z| = \frac{1}{7}  \}$ be a circle inside the unit circle $C_0 = \{ z: |z|=1\}$.  The fractional linear transformation:
$$ \phi: z \mapsto \frac{z- \frac{1}{2}}{\frac{1}{2}z-1} $$
is a map from the unit cirle to itself, $\phi: C_0 \to C_0$.  What is the image of $C_1$ ?  Obviously it is a circle, but what is the radius and center of $\phi(C_1)$?

I hope this figure not too confusing.  $\phi$ maps the light blue circle to the dark blue circle.  Similarly it maps light green to dark green.

In regard to 3 points... I know there is formula for 3 points $(x_1, y_1), (x_2, y_2),(x_3, y_3)$ going through a circle using a determinant but I have not used it:

$$ \left|\begin{array}{cccc} x^2 + y^2 & x & y & 1 \\
x_1^2 + y_1^2 & x_1 & y_1 & 1 \\
x_2^2 + y_2^2 & x_2 & y_2 & 1 \\
x_3^2 + y_3^2 & x_3 & y_3 & 1 \\ \end{array} \right|=0$$

Which 3 points should I pick anyway?  Even more alternatively, are there ways to solve this using power of a point from geometry?

I suspect @vvnitram's answer is wrong.  Here is a plot of the image circle (computed numerically) and the circle he proposes.  These are not quite the same.

Let $$w=u+iv=\frac{2z-1}{z-2}$$

Rearranging, we get $$z=\frac{2w-1}{w-2}$$

So the image of the circle $|z|=\frac 17$ is given by $$\left|\frac{2w-1}{w-2}\right|=\frac 17$$
$$\Rightarrow 7|2w-1|=|w-2|$$
$$\Rightarrow 49((2u-1)^2+4v^2)=(u-2)^2+v^2$$

This simplifies to $$195u^2+195v^2-192u+45=0$$

This is a circle, centre $(\frac{32}{65},0)$ and radius $\frac {7}{65}$

Note that $\phi(0)=1/2$ and $\phi(1/7)=5/13$.

Because is a moebious transform, send the center into center. Then, $\phi(C_1)=C$ where $C$ is the circunference with center $1/2$ and radious $1/2-5/13=3/26$


Finding the expected value of the length of a minimal spanning tree of n randomly generated nodes bound in a box with edge length a. Say we specify a number (n) of random points (x,y), bound within the axes and x=a, y=a.

Given the number of points and the constraints on the boundaries, how can you calculate the expected value of the weight or length of the minimum spanning tree defined by those points?

I've experimentally found the formula E = .6883*a*sqrt(x-2) to fit very well, but how can this be determined mathematically?

This is some data I got from running some tests. Contrained to a box with edge length 10, the blue dots represent the average weight of all MSTs in this box with number of points from 2 to 373. For each n, I randomly generated 100 sets of n points and then calculated the MST of each and took the average. The red dots show the graph of the formula I found experimentally.

Again, I'm wondering how I'd arrive at the expected value mathematically. Thanks for any help.

Ok, so this might be wrong, but here is a fuzzy argument for the growth rate of the function.

So lets assume that the square is of size $a$. The minimum spanning tree on $n$ points must be less than the sum of four minimum spanning trees on the square divided into 4 equal parts, each one connected.

So here is the fuzzy part, if we assume that the points are uniformly distributed, then might be able to assume that there are $n/4$ points in each of the four quadrants. Additionally, the pairwise distances of the points are scaled by half since we are halfing the $x$ and $y$ axis. So one gets the following recursion:

$$M(n) \leq \dfrac{4}{2}M(\frac{n}{4}) + O(a)$$,

Where $M(n)$ is the expected value of the MST of $n$ points. The recursion yields an $O(a\sqrt{n})$ solution. Of course, the expected value thing needs to be formalized but I think that is the general idea.

This is an upper bound, which seems to match your experimental results, so a lower bound might be possible. If you are careful, you might even find the constants! Although I think a lower bound will be hard, since adding an extra point to the rectangle can drastically change the MST.

Good luck!


$\{0,1\}^n$ and $[0,1]^n$ notations Can someone please help me clarify the notations/definitions below:

Does $\{0,1\}^n$ mean a $n$-length vector consisting of $0$s and/or $1$s?

Does $[0,1]^n$ ($(0,1)^n$) mean a $n$-length vector consisting of any number between $0$ and $1$ inclusive (exclusive)?

As a related question, is there a reference web page for all such definitions/notations? Or do we just need to take note of them individually as we learn. Thanks.

The notation $\{0,1\}^n$ refers to the space of all $n$-length vectors consisting of $0$s and $1$s, while the notation $[0,1]^n$ ($(0,1)^n$) refers to the space of all $n$-length vectors consisting of real numbers between $0$ and $1$ inclusive (exclusive).

Edit: I often find wikipedia's list of mathematical symbols useful for looking up the meaning of symbols, although I'm not sure it would help with this question.

The idea is simple… The abstract set of Topology

$$\displaystyle{\prod_{i\in I}X_i=\left\{x:I\rightarrow\cup_{i\in I}X_i \vert x(i)=x_i\in X_i,\;\forall i\in I \right\}}$$

where $x$ are continuous functions. (Also, you can see the continuous function as equivalence relation with $n$-length vector)

Examples:

If $X_i=\{0,1\}, \forall i\in I$ then
$$\displaystyle{\prod_{i\in I}X_i=\{0,1\}^I}$$
$$x=(x_i)_{i\in I}\in\{0,1\}^I$$
If $X_i=[0,1], \forall i\in I$ then
$$\displaystyle{\prod_{i\in I}X_i=[0,1]^I}$$
Important case is the Cantor's set.

P.D.: Excuse my English, please.


Show that $\frac {(\mathbb Z/p)[x]}{(x^2)}$ is not isomorphic to $\frac {(\mathbb Z/p)[x]}{(x)} \times \frac {(\mathbb Z/p)[x]}{(x)}$ I am given in the question that $\frac {(\mathbb Z/p)[x]}{(x^2)}$ has only one maximal ideal.

So my thinking for this is that   $(\mathbb Z/p)[x]$ is an integral domain.

Now my original thinking was to use Chinese Remainder Theorem as I thought the $\gcd(x,x) = x$ which means they are not coprime so they can't be isomorphic.

I have a secondary thought of showing that   $\frac {(\mathbb Z/p)[x]}{(x)}$ has one maximal ideal, therefore its product $\frac {(\mathbb Z/p)[x]}{(x)} \times \frac {(\mathbb Z/p)[x]}{(x)}$ will have two maximal ideals, but I am not sure whether this is appropriate too.

Hint:

$\mathbf Z/p\mathbf Z[x]/(x^2)$ has a nonzero nilpotent element, whereas

$\mathbf Z/p\mathbf Z[x]/(x)\times\mathbf Z/p\mathbf Z[x]/(x)\simeq\mathbf Z/p\mathbf Z\times\mathbf Z/p\mathbf Z$, a product of fields.

I would simply show you can't create a bijection that satisfies the properties to make the two isomorphic. $\Bbb{Z}_p [x] \big/ \langle x^2 \rangle$ will have polynomials of a greater order, so you cannot construct said bijection.


Give an example of an integral domain which has an infinite number of element yet its finite characteristics Give  an example  of an integral domain   which has an infinite  number  of element yet its  finite characteristics?

I thinks $\mathbb{Q}$

Is  it correct ??

Take $F := \mathbb{F}_p(x)$, the field of rational functions in one variable over the prime field $\mathbb{F}_p$.

X is a set of infinity elements . Then P(x) is also infinity . And P(x) with respect to symmetric difference & intersection forms a infinite integral domain with Cher 2


Name of the union of a set with its holes Given an arbitrary connected and compact set $S$ with holes in it, is there a name for the simply connected set formed by the union of $S$ and its holes?

For example, let $S = \{x\in \mathbb{R}^n\ |\ 0 < a \leq||x||^2_2 \leq b\}$, its hole is $H = \{x\in \mathbb{R}^n\ |\ ||x||^2_2 < a\}$. Is there a name for $S \cup H$?

If there is not a specific definition for it, can I just call this set "the union of $S$ and its holes", or is there a more precise way of describing it?

By a set with holes I refer to sets such as this, which is taken from the Wikipedia page on simply connected sets. Here is my attempt to properly define a hole (at least in $\mathbb{R}^n$), inspired by this answer to another question.

Given a connected and compact subset $S \subset \mathbb{R}^n$, its boundary $\partial S$ can be written as the finite union of connected and compact hypersurfaces $F_i \subset \mathbb{R}^{n-1}$, such that $\partial S = \bigcup_{i=1}^k F_i$. Each $F_i$ cuts $\mathbb{R}^n$ in exactly two pieces $A_i \subset \mathbb{R}^n$ and $B_i \subset \mathbb{R}^n$, such that the closure of $A_i$ is compact and the closure of $B_i$ is not compact. Then, we say that $A_i$ is inside $F_i$, and $B_i$ is outside. A hole $H_i$ of $S$ is any $A_i$ such that $A_i \cap S = \emptyset$. Then, the set of all holes is $H = \bigcup H_i$.

If $S \cup H = S$, then is said that $S$ has no holes. This condition is equivalent to $H=\emptyset$.

If $S \cup H \neq S$, then I want to know if $S \cup H$ has a proper name. This condition is equivalent to $H\neq\emptyset$, and $H$ is "filling the holes" of $S$.

I assume you mean a compact set in $\mathbb{R}^n$ (or some metric space) so that "bounded" makes sense. Also, that in your clarification you probably meant to say something like the following (but see the NOTE at the end)

  "take a COMPACT simply connected set $T$ and remove ( open) subsets of the interior to get a compact connected subset $S$."

The goal being to describe $T$ in relation to $S.$

sandwich gives a good answer and from it we see that the key is that "hole" is something like "(a subsequently removed) bounded component of the interior of $T$ " Which becomes "bounded component of the  complement of $S.$"

One could avoid double complements by saying "the union of $S$ and the bounded components of the complement of $S.$" That makes sense even in (some) situations which are not compact. For example

  "Let $T$ be a (finite union of) CLOSED simply connected set(s) and remove BOUNDED subsets of the interior.."

Of course that is the same as "the complement of (the union of) the unbounded component(s) of the complement of $S.$"

NOTE: Really I (and you) should say something like "remove a finite number of open subsets each of which has compact closure."  Although one could also allow the case of an infinite number of holes, perhaps with disjoint closures. It depends somewhat on the exact situation you wish to model.

LATER: In answer to your actual question: I am not aware of a name for what you want, although that does not at all imply there isn't one. You could call it the simply connected closure and define it as the intersection of all simply connected closed sets containing $S.$ That would definitely work provided that such an intersection is closed and simply connected. It does not immediately give one an intuitive picture of what that thing is.

A concise definition of the object you describe is "the complement of the unbounded component of the complement of $S$".

In two dimensions, the term simply connected hull is sometimes used: for example, in a MathOverflow post and in other places you can find with a search.

But in dimensions $\ge 3$ "simply connected" would be misleading: filling in a torus $\mathbb{T}^2$ creates a solid torus, which is still not simply connected.

Another term in use is filled-in set, for example Filled Julia set. This one can be used in any dimension.


What does $\rightarrow$ mean in $p \rightarrow q$ I was looking at an exercise where it asked the following:

$$\begin{array}{ccc}
p&q&p\rightarrow q \\
T&T&T \\
&\ldots
\end{array}$$

So, for the third column, I just put $T$ which was correct but I didn't understand what $\rightarrow$ meant. I have seen $\implies$ but I haven't the arrow. Are they the same thing?

Thanks a bunch!

The $\rightarrow$ symbol is a connective. It's a symbol which connects two propositions in the context of propositional logic (and its extensions, first-order logic, and so on).

The truth table of $\rightarrow$ is defined to be that $p\rightarrow q$ is false if and only if $p$ is true and $q$ is false.

Indeed this is the same meaning of $\implies$, but the difference is that $p\implies q$ is a statement about propositions, whereas $p\rightarrow q$ is a proposition. In some contexts, though, people don't make this distinction between material implication (the connective) and logical implication (the $\implies$ arrow). But they are not the same thing in every context of propositional logic.

Now, although I am only a rising 8th grader taking geometry, I can assure you that there is no difference to the arrows. I have seen a two sided arrow (p<-->q), but that is different. The single arrow just indicates a conditional statement.


Asymptotic expansion of a simple ODE I'm interested in the long-time behavior of the simple ODE
$$\dot{x} = -x^p + 1/t^q$$
with $p, q$ integers not necessarily equal and $x = x(t)$, $\dot{x} = \frac{d}{dt}x$. Furthermore, $x(0) \geq 0$ so we do not need to worry about instability with $x$ negative for $p$ even. To avoid the singularity at zero we can feel free to say $1/(t+\epsilon)^q$ for some $\epsilon > 0$, it does not matter to me. Is there a simple method to say that $x \sim 1/t^l$ for some $l$ in terms of $p$ and $q$?

hint: Consider two options for larger $t$:

$|x|^p<<t^{-q}$

$|x|^p >> t^{-q}$

solve for each case and check for consistency.


Why is this expression false? I have thought thoroughly over this expression $(\:∀x∃y\:(Q(x, y) ⇒ P(x))\:) ≡ (\:∀x\:(∃y\:Q(x, y) ⇒ P(x))\:) $ however I do not understand why they are not equivalent? I thought the "there exists" symbol distributes because P(x) does not use it? Similarly, why is $(\:∀x∃y\:(P(x) ⇒ Q(x,  y))\:) ≡ (\:∀x\:(P(x) ⇒ (∃y\:Q(x, y)))\:) $ true? Because the forall distributes across nonusers?

Sorry about the confusion, the expressions have been updated and finalized.

Remember that $Q(x,y)\to P(x)$ is equivalent to $\neg Q(x,y)\vee P(x)$

So when you distribute the existential over the implication into the antecedent, you have to convert to the dual (that is a universal).

$$\begin{align}\forall x\;\exists y\;\big(Q(x,y)\;\to\; P(x)\big) & \equiv \forall x\; \exists y\;\big(\neg Q(x,y)\;\vee\; P(x)\big) \\[1ex] ~ & \equiv \forall x \;\Big(\big(\exists y\;\neg Q(x,y)\big)\;\vee\; P(x)\Big) \\[1ex] ~ & \equiv \forall x\;\Big(\neg\big(\forall y\; Q(x,y)\big) \;\vee\; P(x)\Big) \\[1ex] ~ & \equiv \forall x\;\Big(\big(\forall y\;Q(x,y)\big)\;\to\; P(x)\Big)\end{align}$$

There is a parenthesis after P(x) in RHS. For all distributes over both sides entirely (both over P and Q) for the first expression.


Finding the $UL$ decomposition of a matrix. (Linear Algebra) Find the $A=UL$ decomposition of the following matrix.(Note the letters $UL$)

$$
A =
\left[ {\begin{array}{cc}
a & b+c & b \\
b & b+c& b \\
c & c & b
\end{array} } \right]
$$

To find $L$ i took the first row subtract the second row to get $
 =
\left[ {\begin{array}{cc}
a-b & 0 & 0 \\
b & b+c& b \\
c & c & b
\end{array} } \right]
$

Next, I took the second row and subtract the third row from it to get $
 =
\left[ {\begin{array}{cc}
a-b & 0 & 0 \\
b-c & b& 0 \\
c & c & b
\end{array} } \right]
$.   Hence  $$
 L=
\left[ {\begin{array}{cc}
a-b & 0 & 0 \\
b-c & b& 0 \\
c & c & b
\end{array} } \right]
$$ Form $L$ i found my $U$ to be
$$
U =
\left[ {\begin{array}{cc}
1 & 1 & 1 \\
0 & 1& 1 \\
0 & 0 & 1
\end{array} } \right]
$$

Is my working correct. Could anyone explain. Thanks

I see what you've done but I think the top right entry of your upper triangular matrix should be zero, since you're not really subtracting the third row from the first right?


How do I prove that there is a neighbourhood $U$ of the orign in $\mathbb{R}^2$ and, $|y_{2}-y_{1}|\geq \epsilon|x_{2}-x_{1}|$. Let V be a neighborhood of the origin in $\mathbb{R}^2$, and $f: V \rightarrow \mathbb{R}$ be continuously differentiable. Assume that $f(0,0)=0$ and $f(x,y)\geq -3x+4y$ for $(x,y) \in V$. Prove that there is a neighborhood U of the origin in $\mathbb{R}^2$ and a positive number $\epsilon$ such that if $(x_{1},y_{1}),(x_{2},y_{2}) \in U$ and $f(x_{1},y_{1})=f(x_{2},y_{2})=0$, then $|y_{2}-y_{1}|\geq \epsilon|x_{2}-x_{1}|$.
This is my idea to solve this question:
Since f is a continuously differentiable, and $f(0,0)=0$.
and $f_{y}\geq4$ at $(0,0)$. By implicit function theorem, there exists a neighborhood $W$ of $0$, and a neighborhood $U$ of $0$, and there a unique continuously differentiable map $g:W \rightarrow U$, such that $g(0)=0$, and $f(x,g(x))=0$ for $x \in W$.
for second part I am thinking to use Mean value theorem, I stuck at $|g(x_{2})-g(x_{1})|=|g^{'}(c)||x_{2}-x_{1}|$.
Can anyone please suggest me the direction of this solution?
Is my answer on a right way?

Well, the IFT also tells you how much $g'(0)$ is. Since $g'(0) \neq 0$ and $g$ is continuous at $0$, there is an $\epsilon >0$ such that $g'(c) \geq \epsilon$ for all $c$ in an open interval containing $0$. Also add to this the fact that $(x,g(x))$ are ALL of the zeros of the function $f$ in a neighborhood, that is, if $f(x,y)=0$, then $y=g(x)$.

You have a good idea. Just prove that $g'(0)\ne 0.$ Then $|g'| >a$ for some positve $a$ in a neighborhood of $0,$ and that implies what you want.
To show $g'(0)\ne 0,$ note that $f(x,g(x)) =  0$ for $x$ near $0.$ Differentiate this at $x=0$ to see
$$\tag 1 D_x f(0,0)\cdot 1 + D_y f(0,0)\cdot g'(0)=0.$$
Now because $f(0,0)=0$ and $f(x,y)\ge -3x+4y$ you can see that for $x<0,$ $f(x,0)\ge -3x = 3|x|.$ This implies $|D_x f(0,0)|\ge 3.$ Similarly $|D_y f(0,0)|\ge 4.$ Thus in $(1)$ we can solve for $g'(0)$ to see it is nonzero as desired.


How can I show that the conditional expectation $E(X\mid X)=X$? I tried to show that $E(X\mid X=x)=x$, which would lead me to get $E(X\mid X)=X$ but I am having trouble doing so. I know that the definition of conditional expectation (continuous case) is: $$E(X\mid Y=y)=\int_{-\infty}^{\infty}x f_{X\mid Y}(x\mid y)\,dx$$
This led me to write:
$$E(X\mid X=x)=\int_{-\infty}^{\infty}x f_{X\mid X}(x\mid x)\,dx$$ but I don't know what $f_{X\mid X}(x\mid x)$ is (if that's even a valid notation...)

I do note though that I don't know anything about $\sigma$-fields or Borel sets, as someone had tried to explain me once. Is it possible to prove this in an "elementary" way?

It is actually easier to define $E(X\mid Y)$ than $E(X\mid Y=y)$ hence let us recall how this is done. By definition, $E(X\mid Y)$ is any random variable $Z$ such that:

$\qquad$ (1.) $Z$ is a measurable function of $Y$.

$\qquad$ (2.) For every bounded random variable $v(Y)$, $E(Zv(Y))=E(Xv(Y))$.

It happens that when $X$ is integrable, such a random variable $Z$ exists and is almost surely unique, in the sense that if $Z$ and $Z'$ both satisfy conditions (1.) and (2.), then $P(Z=Z')=1$.

When $X=Y$, things are quite simple since $Z=X$ obviously satisfies conditions (1.) and (2.), thus, $E(X\mid X)=X$.

Note finally that $Z=X$ always satisfies (2.) but does not satisfy (1.) in general, and, likewise, that $Z=E(X)$ always satisfies (1.) but does not satisfy (2.) in general.

There's another way to see this.

You can define $ h(x) = E(X|X=x) $ in such a way that $ E(X|X) = h(X) $. Following this logic, the expectation of $X$ given that $X$ takes on the particular value $x$ is $x$, because the expectation of a constant function is the constant function itself, i.e., $$h(x) = E(X|X=x) = x,$$ and then we conclude that $$ E(X|X) = h(X) = X. $$


Solve $[\sin(\frac{\pi}{5})]^{x}+[\cos(\frac{\pi}{5})]^x$=1 Solve $[\sin(\frac{\pi}{5})]^{x}+[\cos(\frac{\pi}{5})]^x=1$

Here is another problem that I don't know how to solve. I know that $\sin^2(x)+\cos^2(x)=1$. How do I prove that here $x=2$?. Isn't this just a particular case of proving $\sin^2(kx)+\cos^2(kx)=1$ with $k$ being an integer?

Because both $sin\left(\frac\pi5\right)$ and $cos\left(\frac\pi5\right)$ are fractions, each will become smaller as you raise them to successively higher powers.  At $x=2$ they already add to exactly $1$.  Adding two smaller numbers will never add up to $1$.

http://www.wolframalpha.com/input/?i=sin+(pi%2F5)%5Ex+%2B+cos+(pi%2F5)%5Ex+%3D+1
This might help by equating LHS to RHS.

Here $(\frac {5}{8}-\frac {\sqrt 5}{8})^{\frac x2}+(\frac 14(1+\sqrt5))^x=1$.
Simplify and solve equation you will get x=2.
Also plotting graph will give the rational solution.


what is remainder when $(((3!)^{5!})^{7!})^{9!...}$ is divided by 11 $$(((3!)^{5!})^{7!})^{9!...}$$ when divided by 11 what will be the reminder?

Hint is appreciated

Sorry I do not know how to start this problem, so I have not shown my efforts!

Now we see that $(3!, 11)=(6,11)=1$. Hence by Fermat's theorem we have $(3!)^{10}\equiv 1[11]$ and hence $(3!)^{10m}\equiv 1[11]$. Moreover $10$ divides $5!$ so that $(((3!)^{5!})^{7!})^{...}\equiv 1[11]$.

Required reminder is 1.

see, the above number can be written like this

$(3!)^{5!*7!*9!*\dots}$

first look at the power of 3!, it will be a number ending with a lot of zeroes
let that number be n
hence $n(\mod 10)=0$
$\implies n=10*k$, $k$ is a positive integer
Now,
we have to find $6^n (\mod 11)=$?

by applying Fermat's little theorem $6^{10}(\mod 11)=1$
$=R[(6^{10})^k/11]$
$=R[1^k/11]$
$=1$.


Jointly distributed exponential random variables Attempting to solve the following problem I am confused about what to use as the probability density function

Problem

The time that it takes to service a car is an exponential
random variable with rate 1.

If A.J. brings his car in at time 0 and M.J.
brings her car in at time t, what is the probability
that M. J.'s car is ready before A. J.'s car?
(Assume that service times are independent and service
begins upon arrival of the car.)

Exponential Random Variable

A continuous random variable whose probability density
function is given, for some $\lambda > 0$, by

$
f(x) = \left\{
\begin{array}{l l}
\lambda e^{-\lambda x} &\text{if } x \ge 0\\
0 &\text{if } x < 0
\end{array} \right.
$

is said to be an exponential random variable (or, more simply,
is said to be exponentially distributed) with parameter $\lambda$.

How I'm approaching the problem

My thought is to let $\lambda = 1$, let $f( x, y) = \lambda^2 e^{-\lambda x} e^{-\lambda y}$, let $X$ denote A. J.'s car, let $Y$ denote M. J.'s car and solve for $P\{ X < Y\} = \int\limits_0^\infty \int\limits_t^y f( x, y) dx dy$  but immediately I am troubled with the though that this is not the proper joint distribution function?

I would approach the problem as follows.  Assume that $t \ge 0$.  First find the probability that the service time for A.J.'s car is $\le t$.  Of course if this happens, then A.J. will beat M.J.  If the first car is not serviced by time $t$, then by the memorylessness of the exponential, the probability that A.J. finishes before M.J. is $1/2$.  Now you can put the pieces together.

Or else more simply do the same calculation from M.J.'s point of view. For M.J. to "win," we need first of all to have A.J.'s  service time be $\ge t$. If that happens, the probability that M.J. wins is $1/2$.

The exponential distribution is memoryless. So by the time MJ arrives, AJs service starts out as a fresh new exponential. So $P=0.5$ since both have the same distribution


Learning by the Moore method - Books for self-study I recently read about the Moore method for learning mathematics (Moore method Wikipedia) and wanted to apply it to my own learning (undergraduate level). However, I am unable to find any books that follow such a method or something similar.

Specifically, I'm looking for books that introduce the reader to important mathematical concepts through problems/questions rather than blocks of text (for instance, instead of the book proving Lagrange's theorem in group theory, the reader is presented with a question which guides the reader by introducing the definition of cosets etc., so that the reader can prove it themself). I am not very picky about the topic, for I have not been able to find such books on any topic so far. Any recommendations? Thanks.

It’s long out of print, but John Greever’s Theory and Examples of Point-Set Topology is such a book. He does prove a few of the most difficult theorems and treat some of the more complicated examples in detail, but for the most part he gives you the definitions and the statements of the theorems, and you have to supply the proofs.
He wrote it because that’s how he taught the undergraduate topology course: at each class meeting he’d simply ask the next student on the roster to present the next theorem, and we were supposed to work far enough ahead to be prepared. (I took the course when the book was still in manuscript; it was one of the most enjoyable courses I’ve ever had, though I suspect that quite a few of my fellow students had a different opinion!)

I found the recently published book "Topology Through Inquiry" here: https://math.hmc.edu/su/topology-through-inquiry/
I actually don't know anything about this book (yet) but am posting this answer in case someone else like me finds this question. I continued my internet search and just stopped here to post what I've found.


Proving fixed points of a continuous function Prove that every continuous function $f:[0,1] \rightarrow [0,1]$ has a fixed point.

What does the statement $f:[0,1] \rightarrow [0,1]$ mean? I have never studied function rigorously so I am not well-versed with the notation.

Can somebody help me out or provide some resources that may prove beneficial?

It means that the domain of the function is $[0,1]$ and that the codomain of the function is $[0,1]$. An example of such a function would be $x^2$, for example, or maybe $\sin(x)$ or an infinitely many other functions.

To solve your problem, here's a couple guidelines:

Take a look at the function $g(x)=f(x)-x$.

What is the sign of $g(0)$?
What is the sign of $g(1)$?
What does the intermediate value theorem tell you?

Topology does assume that you know some basic set theory, among which the definition of functions.

If $X,Y$ are sets a function $f:X \to Y$ is just a set of pairs $f = \{x,y): x \in X. y \in Y\}$, such that whenever $(x,y) \in f$ and $(x,y') \in f$ we have $y=y'$ and for all $x \in x, \exists y \in Y: (x,y) \in f$ (every $x \in X$ has a unique image in $Y$, the unique $y$ for $x$ such that $(x,y) \in f$ is usually written $f(x) \in Y$.

Here we are in the situation that $X=Y=[0,1]$ and a fixed point, is an $x \in [0,1]$ such that $f(x) = x$.


Simple Calculus Function I am doing a course in calculus and I was given this problem :

Given that $f(x)=3x^4−6x^3+4x^2−7x+3$, evaluate $f(−2)$.

The answer is meant to be 129 according to the tutor, but no matter how many times I try and work it out I can't see how they got that answer.

Observe the magic of color:$$f(\color{#C00}{x}) = 3\color{#C00}{x}^4 - 6\color{#C00}{x}^3  + 4\color{#C00}{x}^2 - 7\color{#C00}{x} + 3$$Now, we essentially have to replace our colored expressions with numbers.$$f(-2) = 3(-2)^4 - 6(-2)^3 + 4(-2)^3 - 7(-2) + 3$$Now use the order of operations to simplify. For example, $3(-2)^4 = 3\times(-2)^4 = 3\times16 = {48}$.


Limit of the $n^{2}\left( \ln\left( 1+\frac{1}{n}\right) -\frac{1}{n+1}\right)$ How can one compute the following limit
$$
\lim_{n\rightarrow+\infty}n^{2}\left(  \ln\left(  1+\frac{1}{n}\right)
-\frac{1}{n+1}\right) \quad ?
$$
I wrote

$$
\lim_{n\rightarrow+\infty}n^{2}\left(  \ln\left(  1+\frac{1}{n}\right)
-\frac{1}{n+1}\right)  =\lim_{x\rightarrow0}\frac{\ln\left(  1+x\right)
-x}{x^{2}}\overset{l^{\prime}H}{=}\lim_{x\rightarrow0}\frac{\dfrac{1}{1+x}%
-1}{2x}\overset{l^{\prime}H}{=}\lim_{x\rightarrow0}\frac{-1}{2\left(
1+x\right)  ^{2}}=-\frac{1}{2}.
$$
But how can be a limit of a positive sequence be negative?

$$
\lim_{n\rightarrow+\infty}n^{2}\left(  \ln\left(  1+\frac{1}{n}\right)
-\frac{1}{n+1}\right)
$$
$$
=\lim_{n\rightarrow+\infty} \left( n^2 \ln\left(  1+\frac{1}{n}\right)
-n\frac{1}{1+\frac1n}\right)
$$

$$
=\lim_{n\rightarrow+\infty} \left(n^2\left[ \dfrac1n-\dfrac1{2n^2}+O\left(\dfrac1{n^3}\right)\right]-
n\left[1-\dfrac1n+O\left(\dfrac1n\right)\right]\right)=
$$

$$
=\lim_{n\rightarrow+\infty} \left( \left[ n-\dfrac1{2 }+O\left(\dfrac1{n}\right)\right]-
 \left[n-1+O\left(\dfrac1n\right)\right]\right)=\dfrac12
$$

Let $\dfrac1n=h,\dfrac1{1+n}=\dfrac1{1+\dfrac1h}=\dfrac h{h+1}$

$$\lim_{h\to0}\dfrac{\ln(1+h)-\dfrac h{h+1}}{h^2}$$

$$=\lim_{h\to0}\dfrac{\ln(1+h)-h}{h^2}+\lim_{h\to0}\dfrac{h-\dfrac h{h+1}}{h^2}$$

Use Are all limits solvable without L'Hôpital Rule or Series Expansion to find the first limit  $=-\dfrac12$

For the second, its $=1$ right?


Order of elements is lcm-closed in abelian groups How can I prove that if $G$ is an Abelian group with elements $a$ and $b$ with orders $m$ and $n$, respectively, then $G$ contains an element whose order is the least common multiple of $m$ and $n$?

It's an exercise from Hungerford's book, but it's not homework. I could not solve it when I take the course on groups and I think it should be easy. There is a hint that says to first prove when $m$ and $n$ are coprimes. I did this part. But I have no idea how to solve the general case.

Thanks.

Let $n = \prod_i p_i^{n_i}$ and $m = \prod_i p_i^{m_i}$, where the $p_i$s are the same. Thus $\mathrm{lcm}(n,m) = \prod_i p_i^{\max(n_i,m_i)}$. Let $N = \{ i : n_i \geq m_i \}$ and $M = \{ i : n_i < m_i \}$. Define $n' = \prod_{i \in N} p_i^{n_i}$ and $m' = \prod_{i \in M} p_i^{m_i}$. Now $a' = a^{n/n'}$ is of order $n'$, and $b' = b^{m/m'}$ is of order $m'$. Since $n',m'$ are coprime, we know that there is an element of order $n'm' = \mathrm{lcm}(n,m)$.

You can also prove it using the structure theorem for abelian groups, but that's an overkill.

Let $G$ be an abelian group. Let $a$ and $b$ belong to $G$ with order $m$ and $n$ respectively.

Three cases arise:

Case 1: If $m\mid n$ or $n\mid m$ then $\mathrm{lcm}(m,n)=m$ or $n$ in the two cases respectively. Hence our condition holds trivially.

Case 2: Suppose $m$ and $n$ are co-primes. Hence $\mathrm{lcm}(m,n) = mn$.
By abelian property of $G$ $(ab)^{mn} = a^{mn}b^{mn} =ee=e$.
If $o(ab)<mn$ then let $k$ be smallest value for which it holds.
Now $a^k$ and $b^k$ must not be $= e$ as if even one of them equals $e$ then $a^k=e$ implies $m\mid k$ as $m$ is the order of $a$.
Also then $b^k=e$ implies $n\mid k$ as $n$ is the order of $b$.
Hence $k$ is a multiple of $m$ and $n$ and is less than $mn$. Contradiction. Hence not possible.
Now if $a^k$ and $b^k$ are not equal to $e$ and $a^kb^k=e$
Then $a^k=(b^k)^{-1}$.
Hence $b^k$ belongs to $\langle a\rangle$.
Hence $o(b^k) \mid o(a)=m$,
but $o(b^k)\mid o(b)=n$,
hence $m$ and $n$ have a common factor.
Contradiction.
Hence $o(ab)=mn$

Case 3:
Let $m$ and $n$ not be coprime.
Let $d= \gcd(m,n)$.
Hence $n/d$ is coprime with $m$,
so $o(a)=m$ and $o(b^d)=n/d$.
But as $m$ and $n/d$ are coprimes, by Case 2 we get $o(a(b^d))=(mn)/d=mn/\gcd(m,n)=\mathrm{lcm}(m,n)$.
Hence we have an element which has order $\mathrm{lcm}(m,n)$.

Hence proved.


What is a suitable name for numbers like $a + b\sqrt{c}$ The motivation for this is to find a succinct name for a data type in a Python module.

Suppose I choose an integer $c$ and I want to talk about the set of numbers of the form $a + b\sqrt{c}$, where $a$ and $b$ are rational numbers. If $c = -1$, for example, then this set is called the set of Gaussian Rationals. Of course, we would usually think only of values of $c$ such that $|c|$ is non-square. If $|c|$ is a square, for example if $c = 4$, then we obtain either the set of rational numbers (if $c$ is nonnegative) or the set of Gaussian Rationals (otherwise).

Suppose, for example, that $c = 2$. What would the set of numbers of the form $a + b\sqrt{2}$ ($a$, $b$ rational) be called?

An obvious generalisation would be to cube and higher roots, for example numbers of the form $a + b \times 2^{1/3} + c \times 2^{2/3}$ where $a$, $b$ and $c$ are rational.

Since $\Bbb Q(\sqrt c)=\{a+b\sqrt c\colon a,b\in \Bbb Q\}$, you could call the elements of this set $\sqrt c\text{-rationals}$.

When c = -1 , we usually call that set as complex numbers.

When c = $\sqrt 2$, we can the set as surds (of type $\sqrt 2$) or simply surds because it is the simplest.


Is there a third dimension of numbers? Is there a third dimension of numbers like real numbers, imaginary numbers, [blank] numbers?

Alas, there are no algebraically coherent "triplexes". The next step in the construction as has been said already are "quaternions" with 4 dimensions.

Many young aspiring mathematicians have tried to find them since Hamilton in the 19th century. This impossibility links geometric dimensionality, fundamental properties of polynomial equations, algebraic systems and many other aspects of mathematics. It is really worth studying.

A quite recent book by modern mathematicians which details all this for advanced college undergraduates is Numbers by Ebbinghaus, Hermes, Hirzebruch, Koecher, Mainzer, Neukirch, Prestel, Remmert, and Ewing.

However, the set of quaternions with zero real part is an interesting system of dimension 3 with very interesting properties, linked to the composition of rotations in space.

There are of course, examples of R3, in the form X, Y, Z, which behave as three separate real numbers, with an invarient when X -> Y -> Z -> X applies.

Such numbers turn up in the study of the heptagon and enneagon.  The integer systems are discrete points forming a lattice in this space, the invariants cycle through the solutions to the heptagonal ($x^3-x^2+2x-1=0$) and enneagonal equations, which transforms eg {7} to {7/3} to {7/2}.

Such a projection converts an infinitely dense arrangemet of points onto a sparse lattice in three dimensions.  That is, in $x,y,z$ form, there is a sphere of size where there are no more than one legitimate value.  So if you evaluate a value even approximately in the three coordinates, it is possible to find the exact value.

One can, if one considers the span of numbers like $1.801937736$, $-1.2469796037$ and $0.445041867912$, which serve for the heptagon, the same role that $1.618033$ and $-0.618033$ do for the pentagon, then the cyclic rotation of this list will preserve multiplication relations, etc.  eg suppose that the order is $p, q, r$  Now $p^2=2-q$ and $q^2=2-r$, and $r^2=2-p$.  One notes that $p+q+r=p*q*r=1$.  These $p$, $q$, $r$ are the solutions to the cubic $x^3-x^2-2x+1=0$ in much the same way that 1.618 and -0,618 solve $x^2-x-1=0$.

The product, for example of $ap+bq+cr$ and its transforms ($aq+br+cp$ and $ar+bp+cq$), is always an integer, speciffically the product of any of these: 7, cubes, and of primes of the form $7n+1$, $7n-1$.

Since one can suppose that there is a real hypercomplex plane, of the form of numbers $a+bj$, where $j=\sqrt{1}$. The conjucate is $a-jb$. The pentagonal numbers belong here.  It features unit hyperbolae, and two 'real' axies, where the values which the value and its conjucates project onto.

And if one can do it meaningfully with two conjucates, it's possible with three, four, five, &c.  I use such systems extensively when wrangling with polygons.