Question on similar matrices and diagonalization. Let $A$ and $B$ be two $n\times n$ matrices with entries in a field $k$.

If $A$ and $B$ are similar, then can each one be obtained from the other via a combination of elementary row and column operations? Is the converse true?

I recall learning an equivalence relation on the set of $m \times n$ matrices with entries in $k$ that defined $Q$ and $P$ to be equivalent if they can be obtained from each other via combinations of elementary row and column operations, and I recall learning that this characterization is equivalent to saying there exists an invertible $m \times m$ matrix $U$ and an invertible $n\times n$ matrix $V$ such that $Q=UPV$. Is similarity the same relation restricted to the set of $n \times n$ matrices?

Any invertible matrix can be written as a product of elementary operations.  If you extend this arguement a bit, you'll see that, under the equivalence relation you've defined above, two matrices are equivalent if and only if they have the same rank.

If $A$ and $B$ are similar, then $A = M B M^{-1}$ for some invertible matrix $M$.  Thus, if two matrices are similar then they are equivalent under your relation.

However, most matrices that have the same rank are not similar.  From the Rational Canonical Form for Matrices, we know that every matrix is similar to a block diagonal matrix (called the Rational Canonical Form) where the blocks are the companion matrices of certian polynomials called invariant factors of the matrix.  Two matrices are similar if and only if they have the same rational canonical form up to reorderings of the blocks.  Equivalently, two matrices are similar if and only if they have the same multiset of invariant factors.  Thus, by choosing the invariant factors appropriately, we can construct matrices that have the same rank but are not similar.

If the field has more than two elements, then a simpler argument is possible since similar matrices must have the same multiset of eigenvalues.  However, we can choose diagonal matrices that have different sets of eigenvalues but still have the same rank.  (This corresponds to the case where the invariant factors are linear polynomials.)  The advantage of the above argument is that it works even for $\Bbb{F}_2$.

Not always. Note that if you could find similar matrices by simple row/column interchanges, then you could easily diagonalize a symmetric real matrix, which means you can easily find its eigenvalues. But finding eigenvalues is a strictly iterative method and cannot be done in polynomial number of steps, because if you could find all eigenvalues, it would essentially mean you could completely determine the roots of a polynomial of degree $n$, but by Abel-Ruffini theorem, there exists no closed form expression for the roots of a polynomial of degree greater than $4$.

But suppose $A$ is a scalar matrix then any row interchange and subsequent column interchange will give you only $A$, so $A$ is similar only to itself.


Minus sign in logarithm of integral's solution I want to solve the following integral:
$
\int \frac{dp}{a(1-p)u-bp}
$
where $a$, $b$ and $u$ are some constants.

After integration I get:
$
p = -\frac{\log(-apu+au-bp)}{au+b} + C.
$

According to WolframAlpha it is equivalent to (citing precisely: "Which is equivalent for restricted $p$, $a$, $b$ and $u$ values to"):
$
p = -\frac{log(a(p-1)u+bp)}{au+b} + C,
$
or transforming it a bit, it is equivalent to:
$
p = -\frac{\log(apu-au+bp)}{au+b} + C.
$
So for some reasons a minus is "removed" from the logarithm. Could you please explain to me why?

Hint: compute $D(\log(-x))$ and compare with $D(\log(x))$.

\begin{align}
I &= \int \frac{dp}{a(1-p)u - b p}  = \frac{1}{au} \, \int \frac{dp}{1 - \left( 1 + \frac{b}{au}\right) p } \\
&= \frac{1}{au+b} \, \int \frac{dx}{1-x} \hspace{10mm} \text{where} \quad  x = \left(1+\frac{b}{au}\right) \, p \\
&= - \frac{1}{au + b} \, \int \frac{dx}{x-1} \\
&= - \frac{1}{au+b} \, \ln(x-1) \\
&= - \frac{\ln(au\, (1-p) + bp) - \ln(au)}{au+b}.
\end{align}


Markov inequality with $>$ in place of $\geq$ The usual Markov inequality goes as follows: suppose $X$ is a random variable with $g \geq 0$. Then for any $r > 0$,
$$\mathbb{P}(g(X) \geq r) \leq \dfrac{\mathbb{E}[g(X)]}{r}\text{.}$$
Does the following statement still hold true?

$$\mathbb{P}(g(X) > r) < \dfrac{\mathbb{E}[g(X)]}{r}\text{.}$$

For context, suppose $\mathbb{E}[|X|] = 0$ and we want to find $\mathbb{P}(X > 1)$. A solution I read said that since $\mathbb{P}(X > 1) \leq \mathbb{P}(|X| > 1)$, we can show $\mathbb{P}(X > 1) = 0$ using the above.

I understand how the Markov inequality led to $\mathbb{P}(X > 1) = 0$, but I'm not comfortable using it in the $>$ case without justification.

The strict inequality will not be true in general, because it could be that $\mathbb{P}(g(X)>r)=\mathbb{P}(g(X)\geq r)=\frac{\mathbb{E}[g(X)]}{r}$.

But you can say that $\mathbb{P}(g(X)>r)\leq \mathbb{P}(g(X)\geq r)\leq \frac{\mathbb{E}[g(X)]}{r}$, which should be enough for most applications.

It does hold true iff $g(X) = 0$ does not hold almost surely. Here is a short proof:

$$
  \mathbb{E}[g(X)] \\= \mathbb{P}(g(X) \leq r) \mathbb{E}[g(X)|g(X) \leq r] + \mathbb{P}(g(X) > r) \mathbb{E}[g(X)|g(X) > r] \\> g(r) \mathbb{P}(g(X) > r)
$$

The last inequality holds since either $\mathbb{P}(g(X) > r) > 0$ or the first term (and the whole expression) will simply be $\mathbb{E}[g(X)] > 0$. Dividing by $g(r)$ yields $\mathbb{P}(g(X) > r) < \frac{\mathbb{E}[g(X)]}{g(r)}$.


Uses of the Logistic Growth Model My question concerns the logistic differential equation $$y'=y(b-ay),$$
where  $a\neq 0$ and $b\neq 0$, and its solution $$y(t) = \frac{b}{a+ae^{-bt}},$$ where $c$ is an arbitrary constant.

The logistic differential equation has well-documented uses in
population models, but can anyone provide a list of other scenarios that we can model with this equation? The more examples, the better.

I feel like this is a homework question.  Perhaps it will help to observe that this equation describes something growing (b*y) but competing with itself (-a*b*y^2) such that past a certain point (b/a) it cannot grow further due to competition with itself.  Given that, could you take a crack at answering the question and we can give you a sense of whether you're going the right direction?


Prime with highest power that divides $N$ Given a natural number $n$, how can I find the prime $p$ with highest power $e$ in the prime factorization of $n$.
E.g: $n=12=2^2.3$
$\therefore p=2 \ \ and\ \ e=2. $
E.g 2:
$n=450=5^2 * 3^2 * 2^1$ So the prime with highest power is 3 or 5. We can print any of them as a valid answer.
I can prime factorize a number in $O(logn)$ for $n<10^6$. How shall I solve this problem for $n$ upto $10^{12}$.

Since $n$ is bounded, this is a great way to use a look up table.

As there are approximately

$$\pi(n) \approx \frac{n}{\ln(n)}$$

which is your case equals only $\frac{10^6}{\ln(10^6)}\approx78498$
possible prime factors.
This should be the fastest approach.


If I roll two fair dice, the probability that I would get at least one 6 would be.... $11$ out of $36$? I got this by writing down the number of possible outcomes ($36$) and then counting how many of the pairs had a $6$ in them: $(1,6)$, $(2,6)$, $(3,6)$, $(4,6)$, $(5,6)$, $(6,6)$, $(6,5)$, $(6,4)$, $(6,3)$, $(6,2)$, $(6,1)$. Is this correct?

That's right. The easier approach would be to calculate the chance of not rolling a $6$ - that's just $\frac56$ for the first die, and $\frac56$ for the second die, so by the product rule (as the events are independent), the probability is $\frac56 \cdot \frac56 = \frac{25}{36}$.

Then the probability of rolling a $6$ is $1$ minus the probability of not rolling a $6$, which we just calculated: so it is $1-\frac{25}{36}=\frac{11}{36}$.

If calculating the probability of an event not occurring to calculate the probability of it occurring feels weird to you, you might want to read up on complementary events. The idea here is that the product rule can sometimes make probabilities smaller when that wouldn't make any sense --- in our example, if you had just multiplied $\frac16 \cdot \frac16 = \frac1{36}$, then that would've been obviously wrong. This is because when we work with probability, we're dealing with quantities in the $[0,1]$ interval, so multiplication usually makes things smaller, rather than larger.

The chance of getting a $6$ with the first dice is $1/6$ and the chance with the second dice is $1/6$ therfore the chance of getting a $6$ with either dice is $1/6 + 1/6 = 1/3$....not $11/36$!!!


Let $R=\mathbb{Z}\left[\sqrt{−5}\right]$. Let $I=\left(2,1+\sqrt{−5}\right)$. Show $I^2 = (2)$ in $R$. I don't know how to solve this problem. I tried to write out the sum of products in the definition of products of ideals, but that doesn't seem to lead anywhere.

The fact that $(2) \subseteq I^2$  follows immediatelly, as $2 = (1+\sqrt{-5})(1-\sqrt{-5}) - 2\cdot 2 \in I^2$

On the other hand we have that an element of $I^2$ is of the form $\sum_{finite} ab$, s.t. $a,b \in I$. Now we have $a = 2A_1 + (1+\sqrt{-5})A_2$ and $b = 2B_1 + (1+\sqrt{-5})B_2$ as $2$ and $1+\sqrt{-5}$ are generators of $I$. Now we have:

$$ab = 2^2A_1B_1 +  2(1+\sqrt{-5})(A_1B_2 + A_2B_1) + (1+\sqrt{-5})A_2B_2$$
$$ = 2\left(2A_1B_1 + (1+\sqrt{-5})(A_1B_2 + A_2B_1) + 3A_2B_2\right) \in (2)$$

So as each term in the sum is in $(2)$, so is their sum. Hence $I^2 \subseteq (2)$ and so $I^2 = (2)$

In general it holds that if $I = (a,b)$, then $I^2 = (a^2,ab,b^2)$. Using this lemma we have that:

$$I^2 = \left(4,2(1+\sqrt{-5}),2\sqrt{-5} - 4\right) = \left(4,2(1+\sqrt{-5}),2\sqrt{-5}\right) = \left(4,2,2\sqrt{-5}\right)= (2)$$

Observe that

$$2^2=4\;,\;\;(1+\sqrt{-5})^2=-4+2\sqrt{-5}=2(-2+\sqrt{-5})$$

so

$$I^2\subset\langle\,4,\,2(1+\sqrt{-5})\,\rangle=\langle2\rangle$$

since both generators on the left are contained in the right hand.

The other containment is also true ...can you do it?


Integrating with the indicator function of some random variables I have the following problem.

Suppose $X_1, X_2, \cdots, X_{n} $ are independent random variables which are all distributed in $U [ a, b]$. Define a new random variable $$N_{x, y} = \sum_{k=1}^y I\left\{X_k \le x \right\},\qquad x \in \mathbb{R}$$ so that $N_{x,y}$ is the number of sample variables less than or equal to $x$ with counting only first $y$ samples.

Then how to find the squared $L^2$ norm of $N_{x,y}-x$ over the probability space or integrate the following integral?
$$\left\| N_{x,y}-x \right\|_2^2 = \int \left( N_{x,y} - x \right)^2 dx$$

Can we simplify it by using some ordered random variables?

I assume that your integral is between $a$ and $b$?

Given any values of $X_1$, $X_2$, $X_3$, let us call them $x_1, x_2, x_3$, we can easily integrate some expression involving the indicator functions. Since $X_1$, $X_2$ and $X_3$ are all interchangeable, we can assume that $x_1 < x_2 < x_3$. So the integral is done between these values:

$$ \int_a^b (N_{x,y} - x)^2 dx = \int_a^{x_1} (0 - x)^2 dx + \int_{x_1}^{x_2} (1 - x)^2 dx + \int_{x_2}^{x_3} (2 - x)^2 dx + \int_{x_3}^{b} (3 - x)^2 dx $$

By multiplying out you can see that the term in $x^2$ doesn't change - so you can integrate this over $[a,b]$ and the other terms in the different sub-intervals.

When you have done this, you can take the expectation of this integral over the joint distribution of $X_1, X_2, X_3$. Or are you looking for the variance of this integral (the $L_2$ mean over the probability space?) In either case (since calculating variances involves taking expectations) you are simply going to do a triple integral, in each case over $[a,b]$, and then divide out by $(b-a)^3$.

When you take the expectation, you will be integrating over all cases for the relative size of $x_1, x_2, x_3$, not just the one where $x_1 < x_2 < x_3$. However, these cases are symmetric. If they weren't (eg if $N_{x,y}$ wasn't symmetric in the three random variables, you could just work out the integral above for all 6 different permutations of $1, 2, 3$.


Why is physical distance not modelled on rational numbers? Physicists use real numbers to model distance or physical line. My intuition is completely neutral on whether to use rational numbers or reals for physical quantities like distance, or temperature etc. Is there any historical or psychological reasons do prefer reals? Or there is some mathematical advantages working with real. Is calculus on rational numbers harder than calculus on reals?

Calculus over the rational numbers is really not so pleasant. For example, the Intermediate Value Theorem, The Maximum Value Theorem, and The Mean Value Theorem all fail for functions continuous on the rationals.

Edit: A better alternative for $\mathbb{Q}$ would be the completion of it with respect to a $p$-adic metric (or the real numbers, of course). Calculus over $\mathbb{Q}_p$ has been studied, see here. See also for $p$-adic quantum mechanics.

The very definition of "physical quantities" is that they are "measurable".  Irrational numbers cannot, by definition, be measurable.  So either our definition of physical quantities is wrong, or incomplete, or all operations on physical quantities must be represented by operations only on rational numbers which must reduce to operations on the reals in the limit of idealistic models of relationships between physical quantities, such as the use of calculus of variations to solve Lagrangians and Hamiltonians.


Proving that exponential growth at rate r equals exponential growth at rate 1 to the power of r This is a fairly basic result, but I could not find anything about it here.

How do you prove that the following relationship exists, and where does the basis for it originate from:

$$(1+r/∞)^∞=\left( \left( 1+\frac{1}{\infty}\right) ^{\infty}\right)^r$$

Improved by hjhjhj57:

$$\lim_{n\to\infty}\left(1+\frac{r}{n}\right)^n=\lim_{n\to\infty}\left(\left(1+\frac{1}{n}\right)^{n}\right)^r$$

That is to say, a constantly compounded interest at the period's interest rate $r$ is equivalent to a constantly compounded interest at a period's rate of $100$ % to the power of $r$.

The second formula is equivalent to the exponent function $e^r$. Can other functions with the same features (namely that the $f(dx) = f(x)$), be derived from the formula by changing the $r$ and maybe with some other alterations? I am asking this since using the $100$ % interest rate as a definition for $e$ seems somewhat arbitrary and I am wondering if other functions with the same features can be derived from the formula.

EDIT: The idea of this question is not necessarily to prove the equation above holds, but to explain where the relationship or the intuition comes from in the first place or how it was first discovered. In other words why does $e^r$ represent growth factor when the period's rate is $r$ and added constantly, given that $e^1$ equals period's rate of $100$%.

The second question was already answered here: Function is equal to its own derivative.

Consider the derivative of the differences with respect to r.

$\frac{1}{n}(1+\frac{r}{n})^{n-1}-nlog((1+\frac{1}{n})^{nr})(1+\frac{1}{n})^{nr-1}$

This expression tends towards 0 as n tends towards infinity so they are equal to each other.


Probability of flipped coin A bag has 3 coins a, b and c with probability of heads 0.9, 0.6 and 0.5.
I flip a randomly drawn coin and get heads.
What is the probability that it will come up heads again once re-flipped ?

probability of coin being 'a' is $\frac{0.9}{0.9+0.6+0.5}$ = 0.45

probability of heads in next turn = 0.9

.

probability of coin being 'b' is $\frac{0.6}{0.9+0.6+0.5}$ = 0.3

probability of heads in next turn = 0.6

.

probability of coin being 'c' is $\frac{0.5}{0.9+0.6+0.5}$ = 0.25

probability of heads in next turn = 0.5

Total probability = (0.45*0.9)+(0.3*0.6)+(0.25*0.5) = 0.71
\begin{align}
Ans = \bbox[yellow,5px,border:2px solid red]{0.71}
\end{align}

\begin{align}
P(\text{Next will be head}|\text{Current is head})
&= \frac{P(\text{Next will be head} \cap \text{Current is head}) }{P(\text{Current is head}) }
\\& = \frac{P(\text{Next will be head})P(\text{Current is head})}{P(\text{Current is head})}
\\&= P(\text{Next will be head})
\\&= P(a)P(a|a\text{ is head}) + P(b)P(b|b\text{ is head}) + P(c)P(c|c\text{ is head})
\\&= \bbox[yellow,5px,border:2px solid red]{\frac{1}{3}0.9 + \frac{1}{3}0.6 + \frac{1}{3}0.5}
\end{align}


Integer solutions to $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=4$ A serious challenge:

Can someone find 3 positive whole numbers that solve this equation?
$$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=4$$

The numbers must be whole!

An answer has been given by Michael Stoll at this MO-question, linked at this MSE-duplicate, using the elliptic curve

$$E_n \colon y^2 = x \bigl(x^2 + (4n(n+3)-3)x + 32(n+3)\bigr)
                  =: x(x^2 + Ax + B),$$

where in our case $n=4$. Then the curve is known to have rank $1$, and thus there is a solution in positive integers of "truly enormous size", see the article "An Unusual Cubic Representation Problem" by Andrew Bremner. It is given by
$$
x = 4373612677928697257861252602371390152816537558161613618621437993378423467772036;
$$

$$
 y = 36875131794129999827197811565225474825492979968971970996283137471637224634055579;
$$

$$
z = 154476802108746166441951315019919837485664325669565431700026634898253202035277999
$$
There are other solutions in positive integers, of course, but this is the smallest one.

$x=11$, $y=9$, $z=-5$

I brute forced it.


Inverse of $y = x^3 + x $? Can you help me find the inverse function for $y = x^3 + x$?  This question was posed at the beginning of AP Calculus, so we can't use any math beyond precalc.

Thanks!

Write $x^{3}+x-y=0$ and make $x=u+v$. You get $$u^{3}+v^{3}-y+\left(
u+v\right) \left( 3uv+1\right) =0.$$ Then solve the system

$$\left\{
\begin{array}{c}
u^{3}+v^{3}=y \\
u^{3}v^{3}=-\frac{1}{27},
\end{array}
\right. $$

which is equivalent to solving a quadratic equation in $u^3$ or $v^3$, because you know the sum of the two numbers $u^3,v^3$ and their product, e.g. the equation

$$\left( u^{3}\right) ^{2}-yu^{3}-\frac{1}{27}=0.$$

This technique is known as Cardano's method.

Reference: Sebastião e Silva and Silva Paulo, Compêndio de Álgebra, VII ano, pp.215-216, 1963

Its inverse function is $ x= y^3-y $

It is obtained by interchanging $x$ and $y$ or by mirroring graph with respect to line through origin $ x=y. $


Question to an exercise about relations which should be not reflexive but symmetric and transitive First I thought about a certain relation, but wasnt sure about the transitivity in the end, so I went a save route with:

For $x,y\in$ N :

$x\sim y \Longleftrightarrow x,y $: even

But the question was about my first thought! At first I thought I could get a not-reflexive, symmetrical, transitiv relation with the following:

For $ x \in M$:

$x\sim y \Longleftrightarrow | x-y|> 0$

But in this case I wasnt sure if it is transitive, since $x\sim y, y \sim z \Rightarrow x\sim z$. Can there be the case that $x = z$? So that we get the case of $x\sim x$, which isnt possible since the relation is not reflexive?

This is my first question, and as you can see I am just getting started with my math education.. I hope such questions dont bother the more experienced to much :)

Assume that $\sim$ is a relation that is symmetric and transitive.

Then, if $x\sim y$, we have $y\sim x$ by symmetry, so $x\sim x$ by transitivity. So, if $x$ is related to any element of the set, it has to be related to itself.

So, for the relation to not be reflexive, you need at least one element that is not related to any other element.

For example:

The empty relation on a nonempty set $X$.
Take any set $B$ with a symmetric, transitive (reflexive or not) relation and arbitrarily add one or more point with the property that the new points are not related to any point of the set.
A concrete example: the set of vertices $V$ of a nonoriented graph $G$ with the relation $\sim$ defined by $v\sim w$ if and only if there is a path between $v$ and $w$ in $G$. The relation $\sim$ is clearly symmetric and transitive, since the graph is nonoriented. If there are isolated vertices, then it is not reflexive.

Answering the title of ur problem ,"A relation which is symmetric and transitive but not reflrexive" is like this one.

On set $A=${$1,2,3$} consider the relation $R=${$(1,2),(2,1),(1,1),(2,2)$}. It is not reflexive as $(3,3)$ is missing.


Product of two Hölder continuous function on a bounded interval is also Hölder continuous. Prove that the product of two Hölder-1/2 continuous functions on a bounded interval is also Hölder-1/2 continuous.

Given a function on a bounded interval, $f:I\rightarrow \mathbb{R}$,  $f$ is Hölder-1/2 continuous if there exists some $M$ such that $|f(x_1)-f(x_2)|\leq M|x_1-x_2|^{1/2}$. But I'm having trouble how to link this to the product of two Hölder continuous functions, $fg:I\rightarrow\mathbb{R}$.


Does there exist something between topological space and metric space? As is well known, a metric induces a topology, therefore a metric space is always a topological space. However the reverse is not true, you can have topological spaces that do not correspond to a metric (that is, you cannot find a metric that induces that topology).

Now both topology and metrics have a certain concept of "nearness". In topological spaces it's concept of neighbourhood, but that's a relatively flexible concept (as all topological concepts are, by design). In particular, in topology it doesn't make sense to say how close a certain point is to another. On the other hand, metrics has the concept of distance (after all, that's what metric spaces are all about), which is about the most rigid concept of "nearness" you can think of: Every pair of points gets a number that tells you exactly how near or far they are to each other.

Now the question crossed my mind whether there is something in between. Say, a space $C$ based on the relation "$a$ is closer to $b$ than to $c$"; lets write that $b <_a c$. Clearly, a metric fixes that relation, by $b <_a c\iff d(a,b) < d(a,c)$. Also, that relation fixes a topology, by defining an open ball with center $c$ and border point $b$ as $B_b(c) = \{x\in C: (x <_c b)\}$ and then proceeding in the usual way. Clearly for a metric space, this is the open ball around $c$ with radius $d(b,c)$.

Note that this gives automatically a T1 space, as for any points $b,c\in C$, $b\notin B_b(c)$ and $c\notin B_c(b)$.

I think a reasonable definition for such a space would be:

A set $C$ with a ternary relation $(b <_a c)$ is a "closeness space" if the following axioms are fulfilled:

For fixed $a\in C$, the relation $x <_a y$ is a strict weak order.
For any $b\ne a$, $a <_a b$ (a is closer to itself than any other point).

However, does this really give anything new? That is, are there any "closeness spaces" that cannot be given a metric?

Assuming the answer to that question is "yes", are there any T1 spaces that cannot be obtained from a "closeness space"?

Edit: I misremembered the definition of Hausdorff; the condition I gave is not Hausdorff, but T1. I suspect a "closeness space" is also Hausdorff, but I'm not sure yet.

Edit 2: I just noticed that for a "closeness space" derived from a metric space the topology defined by the "comparison space" open balls is not always the same as that derived from the original metric space (the reason is that there can be metric open balls that are not closeness open balls).

For example, consider $C=\{0\}\cup (1,2)$ with the usual $d(x,y) = \left|x-y\right|$. Then in the topology derived from the metric, $\{0\}$ is an open set, but in the topology derived from the corresponding "closeness space", it is not. Indeed, the closeness space derived from that is the same as the closeness space derived from $[1,2)$ if identifying $0$ with $1$, and the topology derived from that "closeness space" is the topology derived from the distance on $[1,2)$.

Of course, in that example there still exists a metric that both induces the "closeness space" and the same topology as induced by the "closeness space", so the question whether there exists one not derivable from a metric is still open.

Answer: There is something like proximity spaces or "nearness spaces" which are a kind of topological spaces you are looking for. See the answer to Whatever Happened to Nearness Spaces? for more details and links. This paper proves some properties for those kind of spaces.

Site note: I wrote my bachelor thesis (it's in German) about a similar kind of relation. There I first defined the "distance relation" as $x \approx_{\epsilon} y \iff d(x,y) \le \epsilon$. With this relation I redefined the main concepts of analysis. For example continuity of a function $f:D\to\mathbb R$ at $x\in D$ can be expressed as

$$\forall \epsilon > 0\, \exists \delta > 0\, \forall y\in D: y\approx_\delta x \implies f(y) \approx_\epsilon f(x)$$

I also wrote a thesis in English about it, but I didn't publish it yet. I can send it to you via email, if you want (see this page for my email address).

Maybe the following is helpful to you:(I will just copy the beginning and give you the link)

The separation axioms attempt to answer the following:

Question: Given a topological space X, how far is it from being metrisable?

The link is:here


Properties of Two Inequalities If $a < b$ and $c < d$, then $a + c < b + d$ if $a, b, c, d$ are positive integers.

My hunch is that this is true, but I'm having difficulty proving it.  Is there a property or definition pertaining to inequalities that exists to support/disprove the above?

$a < b \implies a+c < b+c$

$c < d \implies b+c < b+d$

So $a < b$ and $c<d \implies a+c < b+c < b+d \implies a+c < b+d$


$\sigma$-algebra with cardinality $\aleph_0$ Can a $\sigma$-algebra in a set $X$ have cardinality $\aleph_0$, the cardinality of the naturals?

I do not have a clue on how to start with this? Can someone please give me a hint?

Pick a measurable. Then either it can be partitioned into two measurables or its complement can. If one of them doesn't can imagine looking at the one that can. Repeat inductively.

This gives you a sequence of disjoint measurables. But now any subsequence of them can be joined and get a different measurable. The number of subsequences is uncountable.


If $A$ and $B$ are subgroups of $G$ such that $[[A, B], B] = 1$, then $[A, B]$ is abelian. I am stuck on the following question:
If $A$ and $B$ are subgroups of $G$ such that $[[A, B], B] = 1$, then $[A, B]$ is abelian, where $[X, Y]$ denotes the commutator subgroup of $X$ and $Y$ as usual.
I have seen that $[A, B]$ is in the intersection of all the centralizer of each $b\in B$. And I am trying to show that $B$ is a normal subgroup of $G$, then it follows that $[A, B] \subseteq Z(B)$, where $Z(B)$ denotes for the centre of $B$. Then the result follows.
But I have no idea how to show that $B$ is normal in $G$ or maybe I am completely on the wrong way.
Anyone can help?

Lemma Let $A,B$ be subgroups a of group $G$, then both $A$ and $B$ normalize $[A,B]$.
Proof By symmetry we only prove this for $A$: let $a,a_1 \in A, b_1 \in B$, then
$$a^{-1}[a_1,b_1]a=(a_1a)^{-1}b_1^{-1}(a_1a)b_1.b_1^{-1}a^{-1}a_1^{-1}a_1b_1a=[a_1a,b_1][b_1,a] \in [A,B].$$
Proposition Let $A,B$ subgroups of a group $G$ and let $A$ or $B$ centralize $[A,B]$, then $[A,B]$ is abelian.
Proof Again, by symmetry, we only need to prove this if $[[A,B],B]=1$. Let $a_1,a_2 \in A$ and $b_1,b_2 \in B$, it is enough to show that the commutators $[a_1,b_1]$ and $[a_2,b_2]$ commute. Observe that by the Lemma $[a_2,b_2]^{a_1^{-1}} \in [A,B]$:
$$[a_1,b_1][a_2,b_2]=a_1^{-1}b_1^{-1}a_1b_1[a_2,b_2]\underset{b_1 \text { commutes with } [a_2,b_2]}=a_1^{-1}b_1^{-1}a_1[a_2,b_2]b_1=a_1^{-1}b_1^{-1}[a_2,b_2]^{a_1^{-1}}a_1b_1 =\underset{\text{ using the Lemma and } b_1^{-1} \text{ centralizes }[A,B]}=\\ a_1^{-1}[a_2,b_2]^{a_1^{-1}}b_1^{-1}a_1b_1=[a_2,b_2][a_1,b_1].$$
Note (added March 24th 2021) There is still another elegant way to show that $[A,B]$ is abelian, and that is by an appeal to the Three Subgroups Lemma:
Lemma Let $X,Y,Z$ be subgroups of a group $G$. If $[X,[Y,Z]]=1$ and $[Z,[X,Y]]=1$ then $[Y,[Z,X]]=1$.
Note that the subgroups/letters are shifted cyclically. For a proof and a slight generalization (replacing $1$ by containment in a normal subgroup $N$) see for example here. Proofs usually apply the so-called Hall-Witt Identity.
Now let's show that if $[A,B]$ is centralized by $B$, $[A,B]$ must be abelian. By the first Lemma, $A$ normalizes $[A,B]$, so $[A,[A,B]] \subseteq [A,B]$, implying $[B,[[A,B],A]]=1$. Further, $[A,[B,[A,B]]]=[A,1]=1$. The Three Subgroups Lemma now yields, $[[A,B],[A,B]]=1$, that is $[A,B]$ is abelian.


Finding the local extreme values of $f(x) = -x^2 + 2x + 9$ over $[-2,\infty)$. I'm tutoring a student, and we were trying to solve the following question:

  Find the local extreme values of $f(x) = -x^2 + 2x + 9$ over $[-2,\infty)$.

According to the textbook, the local extreme values are essentially the peaks and the valleys in the graph of the function $f$, so basically where $f'(x) = 0$. This is relatively easy to compute: $$f'(x) = -2x + 2,$$ of which the critical points are $x = 1$. Likewise, $f''(x) = -2 < 0$, which means $f$ is concave down everywhere, and thus $x = 1$ is where a maximum value occurs on the graph. The maximum is $f(1) = -1 + 2 + 9 = 10$.

Of course, the endpoint $x = -2$ yields $$f(-2) = -(-2)^2 +2(-2) + 9 = 1,$$ but since the graph is concave down everywhere, $\displaystyle \lim_{x\to\infty}f = -\infty$ implies there really is no minimum per se... right?

The online computer program tells us that $(-2,1)$ is a local minimum, and $(1,10)$ is a local maximum. But in accordance with the definition from the textbook, why is $(-2,1)$ where a local minimum of the graph occurs? It's neither a peak nor a valley in the graph. What exactly does local mean when the interval is infinite? It doesn't quite make logical sense, unless the definition is not as rigorous as it ought to be.

According to this Wikipedia page, the proper definition of local maxima and local minima is:

$f:X\longmapsto$Y is said to have a local (or relative) maximum point at the point $x^∗$ if there exists some $ε > 0$ such that $f(x^∗) ≥ f(x)$ for all $x\in$ X within distance $ε$ of $x^∗$. Similarly, the function has a local minimum point at $x^∗$ if $f(x^∗) ≤ f(x)$ for all $x\in$ X within distance $ε$ of $x^∗$.

Since we can fix a $\epsilon$-neighborhood with our choice of $\epsilon$
near $x=-2$ so that in this domain $(-2, 1)$ is a minimum point, it is indeed a local minimum.

For the purposes of local extrema we do not consider the endpoints of the interval in question. Thus the only local extremum (in this case a local maxiumum) occurs at $(1, 10)$. On the interval in question, $(1, 10)$ also happens to be an absolute maximum, and there are no local nor absolute minima on this interval.


How can we express "induction is the same as recursion", formally? Informally, the connection between induction and recursion is easy to see, especially when using induction to constructively prove the existence of something. For example, when proving that every natural number can be written as the sum of distinct powers of two, an inductive proof might go:

  The number $0$ can be written as the empty sum, which is vacuously a sum of powers of two. Now consider $n > 0$, and assume as inductive hypothesis that each $k < n$ may be written as a sum of distinct powers of two. If $n$ is even, then $n = 2k$, and $k$ can be written as a sum of distinct powers of two: therefore so can $n$, simply increase the exponents in the expression for $k$ by 1. If $n$ is odd, then $n-1$ is even, and therefore can be written as a sum of distinct powers of two not including $2^0$. Thus we add $2^0$ to the expression for $n-1$ to obtain a way of writing $n$ as a sum of distinct powers of two. This concludes the proof.

The "recursion" side of this proof goes:

def powers_of_two(n):
    if n == 0:
        return []
    if n % 2 == 0:
        return [2*x for x in powers_of_two(n // 2)]
    else:
        return powers_of_two(n - 1) + [1]

Informally this is all good and well. However, I am wondering if there is a formal version of the statement "induction and recursion are isomorphic", and in what context one might prove such a statement.

In a type system with propositions-as-types, such as Homotopy Type Theory (I will write what follows with that in mind) it is not something you would prove, but something that is built in.

Indeed in Homotopy Type Theory, the type of natural numbers, $\mathbb N$, is given ("freely") by $\mathbb N: \mathcal U, 0: \mathbb N, \mathrm{succ}: \mathbb{N\to N}$ and the following "recursion" principle (I'm using quotes here, because it is as relevant for recursion as it is for induction and since your question is specifically about the distinction between those two, one has to be careful) :

given a family of types $C: \mathbb N\to \mathcal U$, $p:C(0)$ and $f: \displaystyle\prod_{n:\mathbb N}(C(n)\to C(\mathrm{succ}(n))$, we have $rec(C,p,f): \displaystyle\prod_{n:\mathbb N} C(n)$, together with the following "rules" (defining equations) : $rec(C,p,f)(0) \equiv p$, $f(n)(rec(C,p,f)(n)) \equiv rec(C,p,f)(\mathrm{succ}(n))$

Now say you have a type $X$, $p:X$ and a map $h:\mathbb N\to X\to X$ and want to define a map $H:\mathbb N\to X$ such that $H(0)\equiv p$ and $H(\mathrm{succ}(n)) \equiv h(n)(H(n))$, then this recursion principle allows you to do that, by putting $C(n):\equiv X$ and $f(n):\equiv h(n)$, so it recovers what you seem to call recursion.

Now to recover induction say you have a proposition $P(n)$ that you want to prove by induction. But in this setting, $P(n)$ is a type a proving $P(n)$ is finding $p:P(n)$, so all in all you want an element of $\displaystyle\prod_{n:\mathbb N}P(n)$. Now if you have $p:P(0)$ (read "I have a proof of $P(0)$") and $f: \displaystyle\prod_{n:\mathbb N}(P(n)\to P(\mathrm{succ}(n))$ (read "I have a proof of $\forall n, P(n)\implies P(\mathrm{succ}(n))$"), then the induction principle gives you $q: \displaystyle\prod_{n:\mathbb N}P(n)$, that is, a proof of "$\forall n, P(n)$".

So in this context the two principles aren't just "formally similar" or "isomorphic", they're literally the same.

This can be seen category-theoretically as well. Say you have for simplicity a topos $T$ with a natural numbers object $N$, that is $N$ has a global element $0: 1\to N$, and a map $s: N\to N$ that have the obvious universal property. This universal property is literally what encodes recursion, but induction can be found from it too.

Indeed say you have a property $P$ on $N$, which really means (category-theoretically) a map $N\to \Omega$ ($\Omega$ being the subobject classifier). Then saying that "$P(0)$ holds" means essentially "the composite $1\overset{0}{\to} N \overset{P}{\to} \Omega$ is the $\mathrm{true}$ morphism".

Saying "for all $n$, $P(n)\implies P(s(n))$" can be interpreted as "the map $\{n:P\}\to N\overset{s}{\to} N$ factors through $\{n:P\}$" where $\{n:P\}\to N$ is the pullback of $\mathrm{true}: 1\to \Omega$ along $P:N\to \Omega$ (the subobject of $N$ defined by $P$).

This means that we have a commutative square $\require{AMScd} \begin{CD}\{n:P\} @>>> \{n: P\} \\
@VVV @VVV \\
N @>{s}>> N \end{CD}$ which can be extended with a triangle with $1\to N$ and $1\to \{n:P\}$

But now by the NNO property of $N$, this map $\{n:P\}\to N$ has a section (it is a section because the map to $N$ is unique) : however it was a monomorphism, therefore it having a section implies that it is an isomorphism, so $\{n:P\} = N$ and $N\overset{P}{\to} \Omega = N\overset{\mathrm{true}_N}\to \Omega$, in other words "for all $n:N, P(n)$", so we get back the induction principle as an easy corollary of the NNO-property, which is what defines recursion.

I don't know enough about topoi and specifically their internal language, but it is easy to see how the recursion principle can be found from the induction principle even set-theoretically, and I don't think one needs the law of excluded middle (although one may need some more assumptions on $N$, like that $0$ is not in the image of $\mathrm{succ}$, which doesn't follow from NNO-ness; these conditions would be needed to make sure for instance that $N$ is decidable even without the LEM) so there should be a way to do that internally to a topos, but someone with more knowledge should say something about that. If that's true (it definitely is in $\mathbf{Set}$ with LEM), then this shows that (in a topos) the two principles are again similar.

Unfortunately what I did was only give systems where the two principles are similar or can be derived from one another or are the same; I didn't answer the question of "how to state that they are somewhat isomorphic", so you'll have to tell me if that's fine by you !

Suppose we have $x_0 \in X$ and function $f: X\to X$. Then induction will hold on $(X,f,x_0)$ iff $X=\{ x_0, f(x_0), f(f(x_0)), \cdots \}$. Put another way, every element of $X$ but $x_0$ itself can be reached by repeated succession (using $f$ as the successor function) starting at $x_0$.

Example: Let $X=\{ a, b\}$, and let function $f:X \to X$ be such that $f(a)=b$ and $f(b)=a$. Notice that apart for $a$, there is only one other element of $X$, namely $b$. And $b$ can be reached by single application of the successor function $f$ starting at $a$. Therefore, induction holds on $(X,f,a)$, i.e. $\forall P \subset X: [a\in P \land \forall x\in P:[ f(x)\in P] \implies P = X]$


exp(X) has a normal distribution, what is the distribution of X? $\exp(X) $ has a normal distribution of mean $\mu$ and variance $\sigma^2$. What does it say of $X$ ? How do I sample $X$ from its distribution?

Can I simply apply the natural logarithm after having sampled $\exp(X)$?

Edit - Background:

In a paper (table 2, page 7 - 138)  I can read: $exp(k_{a\_int})$ has a normal distribution N(−2.372, 1.092).

You have $P(X<a)=P(\exp(X)<\exp(a))=\int_{-\infty}^{\exp(a)}$ $f_{\mu,\sigma}(x) dx$ with normal density $f_{\mu,\sigma}$.


How to find roots of $f(z)=(a+yg(z))^2+g(z)^2=0$? I want to find the roots of
$$f(z)=\left[a+zg(z)\right]^2+g(z)^2=0$$

Where $a$ is real number and:
$$
g(z)=\frac{1}{2\sqrt{z^2+1}}\ln\left(\frac{z+\sqrt{z^2+1}}{z-\sqrt{z^2+1}}\right)
$$

It said that $f(z)=0$ has double complex roots when $a\in(-\pi/2,\pi/2)$, no complex root when $a>\pi/2$ and four complex roots when $a<-\pi/2$.

It also said that the complex roots are pure imaginary, and come in conjugate pairs. The paper is published in a well known journal, here is the link, see page 3.

What I tried is that expand f(z) around $z=0$, I get:
$$
f(z)=(a^2-\frac{\pi^2}{4}) +i\pi(a+1)z +(1+2a)z^2+O(z^3)
$$

The constant term is $a^2-\pi^2/4$, does this constant term give the above claim? I don't know.

My question is: 1). how to determine the number of complex roots for different value of $a$? 2).If possible, can you express the roots as a function of $a$? 3).If it is very hard to express analytically, could you give a hint how can I find all of them numerically?

It is not hard, albeit numerically, to show all the features stated in this question. Because the roots come in pairs of pure imaginary numbers, I guess it is doable to show the features analytically as well.
The very first step is to derive
$$
a=\frac{\pm\mathrm{i}-z}{2\sqrt{z^2+1}}\ln\left(\frac{z+\sqrt{z^2+1}}{z-\sqrt{z^2+1}}\right).
$$
Then, here goes the plots (bird's eye-view & side view) of the real part of RHS (when the imaginary part is sufficiently small) for both $+\mathrm{i}$ and $-\mathrm{i}$. They are symmetric w.r.t. $x$-axis and $+\mathrm{i}$ and $-\mathrm{i}$ plots are reflections w.r.t. $y$-axis as to each other. The first two plots illustrate the imaginariness. A combination of the last two explains the root number properties. $+\mathrm{i}$ and $-\mathrm{i}$ cases together yield the even number of roots. And we observe one peak, one shallow trench and one deeper trench.
The peak: $0<a<\pi/2$ roots
The shallow trench: $-\pi/2<a<0$ roots
The shallow trench and the deeper trench: $a<-\pi/2$ roots
Note that the transition from 2 roots to 4 roots consists in the evident stepwise behavior from the peak to the deeper trench.

Please feel free to edit and elaborate!

Continuing from here:
$$
\frac{1}{2}(\frac{a}{q+\frac{\pi}{2}i}+\frac{q+\frac{\pi}{2}i}{a}) + \tanh q=0
$$
The power series representation of arctanh is
$$
arctanh(q)=x+\frac{x^3}{3}+\frac{x^5}{5}+\dots
$$
Using that approximation, we can solve with either the Cardano method to solve quartic equation, or with numerical methods (e.g. Newton-Raphson):
$$
\frac{-1}{2}(\frac{a}{q+\frac{\pi}{2}i}+\frac{q+\frac{\pi}{2}i}{a}) = \tanh q
$$
$$
arctanh(\frac{-1}{2}(\frac{a}{q+\frac{\pi}{2}i}+\frac{q+\frac{\pi}{2}i}{a})) = q
$$


Some Fundamental theory of calc questions I just wanted to sanity check these questions:

  Find the derivative of these functions

$$g(s) = \int_{5}^s (t - t^2) ^8 dt$$

$$ g'(s) = (t - t^2)^8$$

$$h(x) = \int_{1}^\sqrt{x} \frac{z^2}{z^4 +1} dz$$

$$h'(x) = \frac{1}{2\sqrt{x}} \cdot \frac{z^2}{z^4 + 1}$$

For the first one just replace $t$ with $s$ in the integrand.

For the second I suppose the chain rule gives:  $$h'(x)=\frac1{2\sqrt x}\cdot \frac x{x^2+1}$$.

The first should be $g'(s)=(s-s^2)^8$  and the second is  $h'(x)=\frac1{2\sqrt x}\cdot \frac x{x^2+1}$

This follows from the Leibniz integral rule of differentiation under the integral sign  , which is,

$\,\frac{d}{dx}\bigg(\huge \int_{\small a(x)}^{\small b(x)}\large f(x,t)\,dt\bigg) =\small f(x,b(x)).\frac{d(b(x))}{dx}-f(x,a(x)).\frac{d(a(x))}{dx}+\int_{\small{a(x)}}^{\small{b(x)}}\partial_xf(x,t)dt$.

In your case since the partial derivatives $\,0$  you can just omit them.

EDIT:

you are doing $\frac{d}{dx}(\int_1^{\sqrt x}\frac{z^2}{z^4+1})$. Thus  you sub in $z=\sqrt x$ after you differentiate(because of the above mentioned  rule) You get $\frac{(\sqrt x)^2}{(\sqrt x)^4+1}\cdot\frac1{2\sqrt x} = \frac{x}{x^2+1}\cdot\frac1{2\sqrt x}$. Also you dont get the term $\frac{1^2}{1^4+1}$ as $\frac{d}{dx}(1) =0$. Cancelling  that term.


let $c>0$ and $a_{1} = (c/2)$, $a_{n+1} = \frac{c + a_{n}^2}{2}$. Let $c>0$ and $a_{1} = \frac{c}2$, $a_{n+1} = \frac{c + a_{n}^2}{2}$. Determine all $c$ for which the sequence converges . For such $c$ find $\lim_{n\to\infty} a_{n}$.

Could anyone help me please in solving this?

First, $a_{2} = \frac{c + a_{1}^2}{2}>a_1=\frac{c}{2}$, and for $n\geq 2$
$$a_{n+1}-a_n=\frac{c + a_{n}^2}{2}-\frac{c + a_{n-1}^2}{2}=\frac{(a_n-a_{n-1})(a_n+a_{n-1})}{2}>0.$$
So the sequence$\{a_n\}$ is strictly increasing.
If the sequence is bounded, then the limit exists, suppose its limit is $a$, then it must satisfy $$a=\frac{c+a^2}{2}.$$
If this equation has solution($a=1\pm\sqrt{1-c}$), it must be $$0<c\leq 1.$$
When $c\in(0,1]$, we prove the sequence is bounded. By induction if $0<a_n<1-\sqrt{1-c}$ (you can check that $a_1<1-\sqrt{1-c}$), then
$$a_{n+1} = \frac{c + a_{n}^2}{2}<\frac{c+(1-\sqrt{1-c})^2}{2}=1-\sqrt{1-c}.$$
So when $c\in(0,1]$, the limit is $1-\sqrt{1-c}$.

Hint: Solve $$l=\frac{c+l^2}{2}$$

to get a list of possible values of the limit (I'll leave it to you to see why this is true)

Edit: Look at the comments for mroe hints


Expected value of $\sin^2(0.01n)$ for discrete $n$ Is it possible to calculate the expected value of $\sin^2(0.01n)$, with $n$ taking non-negative integer values?

Normally, when we wish to find the expected value of the sine function, we integrate (or add) its corresponding values within a single period (I think?), and then take its mean. Is such a method possible for the function under question?

Disclaimer: This is, so far, one of my most downvoted answers on the site. Since it is perfectly correct, and it answers the question, the downvotes might be due to some extra-mathematical reasons. In any case... happy reading!

The phrase "expected value" has no meaning when applied to a sequence $\mathbf x=(x_n)_{n\geqslant1}$, but one can look at the Cesàro means $(c_n(\mathbf x))_{n\geqslant1}$ and at their limit $\mathcal C(\mathbf x)$ defined, if it exists, as $$\mathcal C(\mathbf x)=\lim_{n\to\infty}c_n(\mathbf x)$$ where, for every $n\geqslant1$, $$c_n(\mathbf x)=\frac1n\sum_{k=1}^nx_k$$
In the present case, $x_n=\sin^2(an)$ with $a=0.01$ hence $2x_n=1-\cos(2an)=1-\Re(\mathrm e^{\mathrm ian})$ and $$2c_n(\mathbf x)=1-\frac1n\Re\left(\sum_{k=1}^n\mathrm e^{\mathrm iak}\right)$$
The sum in the RHS is geometric with ratio $z=\mathrm e^{\mathrm ia}$. Thus, the sums in the RHS are bounded since $|z|=1$ and $z\ne1$, hence the sequence $(c_n(\mathbf x))_{n\geqslant1}$ converges and its limit is $$\mathcal C(\mathbf x)=\frac12$$


Find scalar quantity linking two lengths The question is a GCSE maths question per the picture below. A friend asked me and I can't figure it out.. and I have a maths degree.. embarrassing! What would be a good approach to this question?

$|AP|=k\,|AB|$, $\angle NOB=\alpha$.

Let $[OBN]$ denote the area of $\triangle OBN$.

\begin{align}
[AOM] &=[AMB]= \tfrac12\,ab\sin\alpha
,\\
[OBN]&=\tfrac12\cdot3a\cdot2b\,\sin\alpha=3ab\sin\alpha
,\\
[ABN]&=[OBN]-2[AOM]=2\,ab\sin\alpha=4[AOM]
,\\
[OMN]&=\tfrac12[OBN]=\tfrac32 ab\sin\alpha
,\\
[AMP]&=k[AMB]
,\\
[APN]&=k[ABN]
,\\
[AMP]+[APN]&=
[OMN]-[AOM]=ab\sin\alpha
,\\
[AMP]+[APN]&=
k([AMB]+[ABN])=\tfrac52\,k\,ab\sin\alpha
,\\
k&=\tfrac25
.
\end{align}

![enter image description -
By putting X just in middle, I require this area in exact two equal area.
Means where Y fall ?
Means shaded area is equal to unshaded area.

](https://i.stack.imgur.com/c6fgP.jpg)


Real Analysis Archimedian Property If $x, y$ are real numbers, with $x < y$, show that $$x < ty + (1-t)x < y$$ for all $t$ such that $ 0 < t < 1$.

I know we need to use Archimedian principle, and I think it may be used twice, but I'm stuck on how to get the extra $x$ in there from $t(y-x) > 1$.

Any help would be great!

Since $x<y$ and $0<t<1$ so
$$x=tx+(1-t)x<ty+(1-t)x<ty+(1-t)y=y$$
Now let $x<z<y$, define the function $f:[0,1]\rightarrow [x,y]$ with
$$f(t)=ty+(1-t)x=t(x-y)+x$$
then $f(0)=x$ and $f(1)=y$ and also continuous injective, now mean value theorem implies that there is $t_0$ such that $f(t_0)=z$.


need to prepare for $5$ out of $6$ outcomes Four of six possible essay questions will be on the exam. If two questions are in part $A$ and two questions are in part $B$, and one question must be chosen from both part $A$ and $B$, how many essay questions must be studied for? I think it's $5$ but am not sure.

TBH honest the original directions were ambiguous but I think part A will have the same number as questions in it as part B, and one question must be chosen from part A and one question must be chosen from part B.

For completeness sake here is the original

  Four of  the questions  from  below  will  be  on  the  exam,  divided
  into  Part  A  and  Part  B.  You  will  have  to  answer  one
  question  from  part  A,  and  one question  from

Yes, that's right.

Assuming that the exam has two parts, A and B, both consisting of two essay questions, chosen from a collection of 6 possible questions, of which you must pick and answer one for each part, you will have to study for 5 of them. If you study for less, there is a chance that you haven't studied for both questions in at least one of the parts.

The number of parts to the exam doesn't matter, what matters is the smallest number of questions to a part, $k$ and the total number of questions, $n$. You will have to study for $n-k+1$ questions, otherwise there is a possibility that the part will feature exactly the $k$ questions you didn't study for.


Insights on coset in abstract algebra I am taking course on abstract algebra. I found the concept of coset is very important. However, I have not found any insights on my textbook (and by myself). I believe there must be great example on why coset is important in linear algebra(like solution to equations) or some other field of math. Could you provide any examples to connect coset to important theory or application?

I am NOT looking for what will appear on a standard textbook.

A coset can be thought about as a copy of a group, when interpreted as a subgroup of a larger group. So anywhere that groups occur, and their subgroups are interesting at all, we expect to see cosets.

(Note that the cosets of a normal subgroup are of course much more interesting than the cosets of a generic group, because of the group structure on the set of cosets. So these arise in applications much more often.)

Geometry provides many rich examples: here are two that come to my mind immediately: An affine space is a coset, for $G$ a vector space and $H$ a subspace. The orientation of a transformation is a coset where $G$ is $O(n)$ and $H$ is $SO(n)$.

Quotient groups and quotient spaces are enormously important constructions in mathematics.


Is 1 divided by 3 equal to 0.333...? I have been taught that $\frac{1}{3}$ is 0.333.... However, I believe that this is not true, as 1/3 cannot actually be represented in base ten; even if you had infinite threes, as 0.333... is supposed to represent, it would not be exactly equal to 1/3, as 10 is not divisible by 3.

0.333... = 3/10 + 3/100 + 3/1000...

This occured to me while I  discussion on one of Zeno's Paradoxes. We were talking about one potential solution to the race between Achilles and the Tortoise, one of Zeno's Paradoxes. The solution stated that it would take Achilles $11\frac{1}{3}$ seconds to pass the tortoise, as 0.111... = 1/9. However, this isn't that case, as, no matter how many ones you add, 0.111... will never equal precisely $\frac{1}{9}$.

Could you tell me if this is valid, and if not, why not? Thanks!

I'm not arguing that $0.333...$ isn't the closest that we can get in base 10; rather, I am arguing that, in base 10, we cannot accurately represent $\frac{1}{3}$


A fair coin is tossed $n$ times by two people. What is the probability that they get same number of heads? Say we have Tom and John, each tosses a fair coin $n$ times. What is the probability that they get same number of heads?

I tried to do it this way: individually, the probability of getting $k$ heads for each is equal to $$\binom{n}{k} \Big(\frac12\Big)^n.$$ So, we can do $$\sum^{n}_{k=0} \left( \binom{n}{k} \Big(\frac12\Big)^n \cdot \binom{n}{k}\Big(\frac12\Big)^n \right)$$ which results into something very ugly. This ugly thing is equal to the 'simple' answer in the back of the book: $\binom{2n}{n}\left(\frac12\right)^{2n},$ but the equality was verified by WolframAlpha -- it's not obvious when you look at it. So I think there's a much easier way to solve this, can someone point it out? Thanks.

The probability John gets $k$ heads is the same as the probability John gets $n-k$ heads since the coin is fair.

So the answer to the original question is equal to the probability that the sum of Tom's and John's heads is $n$.

That is the probability of $n$ heads from $2n$ tosses which is indeed $\frac{1}{2^{2n}}{2n \choose n}$.

The problem with this answer is that its not correct to begin with. Assume the problem pertains to two individuals throwing 2 coins then

the total outcomes is indeed $ \frac{1}{2^m} , m= 2n$ however the amount of possible outcomes is not $\binom{2n}{n}$ so in the case of each having 2 coins the resulting outcomes is:

1 head and two tails, this can be done 2 ways $2*2 = 4$

2 heads , this can be done only 1 way $1*1=1$

This is a total of $5$ and $\binom{2*2}{2} = 6$ this is of course not equal and difference of 1 holds as you continue.


English names for vector beginning and end I've done some research, but since English is not my native language, I'm struggling to find an answer to this:

Given a vector, what do you call its beginning and end points?

The best I've found so far is the word "base" for the beginning point of a vector, but I have no clue if that's correct. The best I've got for the end point is the "tip" of the vector, although the same applies as before.

A good question. Since an important property of a vector is its direction it is hard to talk about vectors without having words for where they start and end.

In my experience we have generally called the source or beginning of a vector its "tail" and the destination or end of the vector its "head".

Uh, you aren't actually talking about a vector here.  A vector is not defined by a beginning and end point.  It has a direction and magnitude (and equivalents in other coordinate systems) but no fixed origin.

The vector might have a location where the coordinates are taken.  This is particularly relevant in curvilinear coordinate systems.  If you draw a vector there, you typically draw the respective arrow with its base on the location of the vector, its shaft being straight and tangential to the coordinate directions at the origin of the vector, and the component sizes being in proportion.

Sometimes arrows are drawn for directed line segments.  Those do have a starting and an ending point.  In that case, the shafts of those arrows will follow any curvilinear coordinate system.  Visualizing a directed line segment specified by starting point and displacement vector is rarely done when more than one coordinate is actually changing since then the path along which the displacement is achieved has no obvious candidate: there is no obviously preferred route between starting and ending point any more.


Multiply the i-esim column of a matrix for the i-esim element of a vector Suppose you have a square matrix $A$ of order $n$ and a vector $\upsilon$ of dimension $n$ as well. Is there a way to combine elements in $\upsilon$ in the form of matrix (let's call this matrix $B$) such that a matrix operation between $B$ and $A$ allows me to get $C$.

Where $C$ is defined as follow:

$$C=\begin{bmatrix}
\upsilon_1a_{11} &  \upsilon_2a_{12} & \cdots  & \upsilon_na_{1n} \\
\upsilon_1a_{21} &  \upsilon_2a_{22} & \cdots & \upsilon_na_{2n}\\
\vdots & \vdots & \ddots & \vdots \\
\upsilon_1 a_{n1} &  \upsilon_2a_{n2} & \cdots  & \upsilon_ na_{nn}
\end{bmatrix}$$

Thanks for any help you can provide.

Yes, $C=AB$ if $B=\begin{bmatrix}v_1&0&\cdots & 0 \\
0& v_2 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots \\
0&0&\cdots & v_n
\end{bmatrix}$.


The name of the game (Hawk-Dove variant?) I stumbled upon the following symmetric two-person game. We have two objects $X,Y$ with positive value $x$ and $y$, and two persons that have to pick independently form each other simultaneous one of the objects. If a person is the only one that picks $X$, then she receives $x$ as payoff. If two persons pick $X$, then everybody gets $x/2$ as payoff. Some goes for $Y$.

This gives as payoff matrix
$$\begin{pmatrix}x/2,x/2 && y,x \\ x,y && y/2,y/2\end{pmatrix}.$$

Does this kind of game have a certain name? It looks similar to Hawk-Dove or chicken, but it is different. It seems to be a very natural instance of a game (also its generalization for more players), so I wonder if this is known as a classic example in game theory.

This game describes a conflicting situation and it is clearly an anti-coordination game. The (pure) Nash equilibrium payoffs are $(x,y)$ and $(y,x)$. We can safely say that this is an example of Hawk-Dove game. However, note that not every Hawk-Dove game must be of that type. That is, it is not necessary to put a restriction like $x/2$, it could be something else as long as it is strictly less than $x$. (to assure $(x,y)$ is an equilibrium outcome)

P.S. This game is definitely not prisoner's dilemma. First of all, there is no dominant strategy in this game.

I think this is a special case of the prisoners dilemma.

However, not in general, only if x>0 and y<0 (or vice versa).


construct a square without a ruler How can I construct a square using only a pencil and a compass, i.e. no ruler.

Given: a sheet of paper with $2$ points marked on it, a pencil and a compass.

Aim: plot $2$ vertices such that the $4$ of them form a square using only a compass.

P.S.: no cheap tricks involved.

On way to do this is to start by constructing the middle point of your segment like this:

Then you will easily have the middle of your square like this:

And you know can have a circonscrit circle for your square, and constructing the square is finally possible!

Edit.

Since you asked for it, I have made a few more drawings to illustrate how to construct the point $O$ from where we left it.

Where the last circle has for radius $CF$ and for center $A$.

Edit 2.

Since more details were requested, here is how to finish the proof once $O$ has been constructed.

The only difficulty is to create Perpendicular lines.
One possible way could be, if you are precise enough:

lets suppose the vertices are (P1,P2,P3,P4), where P1 and P2 are given two points.

Join the two points(P1,P2) and extend on both sides.

To draw a perpendicular at point(P1) on the line: Mark two points on the line equidistant from P1 on both sides, then place your compass at the marked points, there after create arcs using your compass such that they intersect.

joining the intersection of arcs and P1 would give you a perpendicular line on the given line.

Since you have created a perpendicular line, then it's not difficult to create a Square from it.


Finding a particular coefficient in a polynomial I'm trying to get the coefficient of $x^6$ of this polynomial product:

$$x^2(1+x+x^2+x^3+x^4+x^5)(1+x+x^2)(1+x^2+x^4).$$

I know with infinite series, you can use the closed form solution of the series to calculate the coefficient, but I haven't been able to figure out a way to calculate the coefficient for a finite series.

I hope there is a better way than doing the algebra.

Computing the coefficient directly doesn’t seem all that bad to me since you’re only interested in one term of the product. Its coefficient can be found without multiplying the whole mess out. Your function is $x^2$ times some polynomial in $x$, so you just need to count the number of ways that you can choose monomials from the other three terms in the product such that the sum of their exponents is $4$. The seven ways to do that are pretty easy to find by inspection.

There's a closed form for a finite series, too, e.g., $$1+x+\cdots+x^5={1-x^6\over1-x}$$ So your product is $$x^2{(1-x^6)^2(1-x^3)\over(1-x)^3(1+x)}$$ which is $$x^2(1-x^3-2x^6+2x^9+x^{12}-x^{15})(1+3x+6x^2+10x^3+\cdots)(1-x+x^2-x^3+\cdots)$$ Since you're only after the coefficient of $x^6$, you can simplify this a little to $$x^2(1-x^3)(1+3x+6x^2+10x^3+15x^4)(1-x+x^2-x^3+\cdots)$$ but this doesn't look any easier than what you call "doing the algebra".


Modal logic of contingency and necessity operators? Well let me say that this is a challenging question, I am stuck in it myself :(

I am fully aware of modal systems for necessity, i.e. being true in every accessible possible world, and possibility; i.e. being true in at least one accessible possible world. I am also aware of modal systems for contingency, i.e. being both possibly true and possibly false, and non-contingency, i.e. being necessarily true or being necessarily false. Both of these modal pairs are dual in normal sense. However I am looking for modal system between contingency and necessity, hereafter $C$ and $N$. notice that they are not dual like the two previous pairs. in fact only one direction of duality holds between them, namely $Np\rightarrow\lnot Cp$ but $\lnot Cp\rightarrow(Np\lor N\lnot p)$. I could not find any modal axiomatization of this two pairs.
Suppose we want the accessibility relation be partial order, non-total, non-dense, non-well-founded and non-convergent. So far I know that axioms $K$, $T$ and $4$ are held to this pairs and axiom $5$ is not held. So the logic should be at least as strong as $S4$ but weaker than $S5$. do you know any other interesting axiom which may be held for this pair? I think axiom $M$ and $G$ are not also held. I would appreciate if you have any suggestion about further axioms which are held for this pair. Thank you in advance!

Contingency can be somehow defined in terms of necessity:
$$\operatorname{C}p\leftrightarrow(\lnot\operatorname{N}p\,\land\,\lnot\operatorname{N}\lnot p)$$
You can add the above (defining) axiom to a system which you use to formalize necessity and have system for both necessity and contingency.

Note that the "meaning" of the above axiom is just "Contingency means being both possibly true and possibly false." .

You may want to look at the literature on contingency and non-contingency logic. E.g. Kuhn, S. T., “Minimal non-contingency logic,” Notre Dame Journal of Formal Logic, 1995, 36(2):230–234. Montgomery, H. and R. Routley, “Non-contingency axioms for S4 and S5,” Logique et analyse, 11 (1968), 422–424.

For the sort of substantial logic you are looking for you may want to look at the Sobocinski 'family of K systems', e.g. Sobocinski 1964. These are stronger than S4 and incompatible with S5.


Given an integer, how many ways can you divide into unequal portions I have been thinking about this problem for a while but have been unable to formalize a complete answer.

Given an integer $n \geq 3$, how many different sets can you sum the integer elements in the set to equal n ?

Constraints:

1) Each set must contain at least 2 elements.

2) All elements in a set must sum to $n$ (all elements must be greater than 0).

3) Normal set rules apply: No elements in a set can be repeated, and Order doesn't matter

Examples:

when n = 3 the only possible set is {1, 2} so the answer would be 1.

when n = 4 the only possible set is {1, 3} so the answer would be 1.

when n = 5 there are two sets {1, 4}, {2, 3} so the answer would be 2.

This is meant to be a programming question. I think the answer could be computed as a math equation, or at least a dynamic programming approach. My goal is for the run time to not exceed O(n), however O(1) would be great. If you want to provide code python like syntax would be fine.

Let $N_i(n)$ give the number sets of size $i$. You want $\sum_{i=2}^\infty N_i(n)$.

Clearly $N_2(n) = \lfloor\frac{n-1}{2}\rfloor$.

To compute $N_3(n)$, we're going to try to find the three-member set from least to greatest. If the least member is a $1$, then the other members are all at least $2$; we can subtract $1$ from the other $2$ members, and we need two unequal numbers adding to $n-3$; there are $N_2(n-3)$ such possibilities. Similarly, if the least member is $2$, there are $N_2(n-6)$ possibilities; etc. So $N_3(n) = \sum_{i=1}^{\lfloor\frac{n}{3}\rfloor} N_2(n-3i)$.

Similarly, $N_4(n) = \sum_{i=1}^{\lfloor\frac{n}{4}\rfloor} N_3(n-4i)$

You can try to devise some computation method based on these relations; however, I'm not sure it will be so computationally efficient.


Summation of the telescoping series $\sum_{n=1}^N \frac{x}{(1+(n-1)x)(1+nx)}$ $(i)$ Verify that $$\frac{1}{1+(n-1)x} - \frac{1}{1+nx} = \frac{x}{(1+(n-1)x)(1+nx)}$$
$(ii)$ Hence show that for $x  \ne 0$, $$\sum_{n=1}^N \frac{x}{(1+(n-1)x)(1+nx)}=\frac{N}{1+Nx}$$
Deduce that the infinite series $\frac{1}{1.\frac{3}{2}} +\frac{1}{2.\frac{3}{2}} +\frac{1}{2.\frac{5}{2}}.....$ is convergent and state the sum to infinity.

I need some help with the last part. I tried using the identity in the first part by letting $x=1$ and then varying $n$ but that did not produce the series that they have given. I also tried the same thing after setting $x=2$ but again could not get the series that they have given.

Hint

By $(i)$ we have
$$\frac{x}{(1+(n-1)x)(1+nx)}=\frac{1}{1+(n-1)x}-\frac{1}{1+nx}$$
and note that if $x>0$, then$$\sum_{n=1}^N\left(\frac{1}{1+(n-1)x}-\frac{1}{1+nx}\right)$$ is a telescoping sum.  Then, you need to show that $$\lim_{N\to\infty}\frac{N}{1+Nx}<\infty$$ for all $x\ne 0$ and find the limit.

From the LHS (The series), you can see that for:

At n=1, we have $\frac{x}{1+x}$

At n=2, we have $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}=\frac{2}{1+2x}$ (By simplifying)

At n=3, we have $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}=\frac{3}{1+3x}$ (Again by simplifying)

At n=4, we have  $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}+\frac{x}{(1+3x)(1+4x)}=\frac{4}{1+4x}$ (You see the partern?)

Therefore, By mathematical induction, we have......

At n=N, $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}+\frac{x}{(1+3x)(1+4x)}+.....+\frac{x}{(1+(N-1)x)(1+Nx)}=\frac{N}{1+Nx}$


Every Subset of a Finite Set is Finite (understanding Cain's proof) Firstly, I realize that there are many variations on this question already posted. My textbook, Cain's Topology gives a 2 line proof outline of Proposition 0.29 (see below) which I find quite confusing. I've tried to fill in the details, and would like someone to confirm if I understand it.

Definition. Two sets A,B
  are (set) equivalent if there is a bijection between them

Definition. Define $\mathbb{Z}_{n}=\{k\in\mathbb{Z}_{+}:k\leq n\}$
 . A set which equivalent with a subset of some $\mathbb{Z}_{n}$
  is said to be finite.

Proposition. 0.29a: (not in text): Let $g:A\rightarrow B$
  be a bijection. If $B\subseteq C$
 , then $f:A\rightarrow C$
  defined by $\forall a\in A(f(a)=g(a))$
  is an injection. (i.e, Enlarging the codomain does not affect the injectivity of a function).

Proof. Suppose $f(a_{1})=f(a_{2})$
 . By the definition of $f$
 , $f(a_{1})=g(a_{1})$
  and $f(a_{2})=g(a_{2})$
 . Thus $g(a_{1})=f(a_{1})=f(a_{2})=g(a_{2})$
 , so $g(a_{1})=g(a_{2})$
 . Since $g$
  is an injection, this implies $a_{1}=a_{2}$
 . Thus $f$
  is an injection. $\square$

Proposition. 0.29: Suppose A
  is finite and $B\subseteq A$
 . Then $B$
  is finite.

Proof. Since $A$
  is finite, there is a bijection between $A$
  and a subset of some $\mathbb{Z}_{n}$
 . By proposition 0.29a, this implies there is an injection $f:A\rightarrow\mathbb{Z}_{n}$. The restriction of $f$
  to $B$,
  $f|B:B\rightarrow\mathbb{Z}_{n}$, defined by $f|B(b)=f(b)$
  for each $b\in B$,
  is an injection into $\mathbb{Z}_{n}$
 , since if $b_{1},b_{2}\in B$
 , then:$f|B(b_{1}) = f|B(b_{2})\implies
f(b_{1}) = f(b_{2}) \implies b_{1} = b_{2}$.

Therefore, setting the codomain of $f|B$
  equal to $Im(f|B)$
  defines a bijection $g:B \rightarrow Im(f|B)$

Now, all we need to do is show $Im(f|B)\subseteq\mathbb{Z}_{n}$ Firstly, note $Im(f|B)$
  is a subset of $Im(f)$
 . Recall that $Im(f|B)=\{n\in\mathbb{Z}_{n}:\exists b\in B(n=f|B(b))\}$
 . Let $n\in Im(f|B)$
 . Then $\exists b\in B(f|B(b)=n)$
 . Since $f|B(b)=f(b)$
  for $b\in B$
 , we have, $\exists b\in B(f(b)=n)$
 . Since $B\subseteq A$
 , $b\in B$
  implies $b\in A$
 , and therefore $\exists b\in A(f(b)=n)$
 . Thus $n\in Im(f)$
 . Thus $Im(f|B)\subseteq Im(f)$
 , and $Im(f)\subseteq\mathbb{Z}_{n}$
  since the image is always a subset of the codomain, which in this case is $\mathbb{Z}_{n}$
 .

Thus $B$
  is equivalent with $Im(f|B)\subseteq\mathbb{Z}_{n}$
 , and is therefore finite.  $\square$

To demonstrate the idea of Proposition 0.29 I've attached the drawing below, with $A=\{a,b,c\},B =\{a,b\}$. In the first panel, we see that $A$ is finite, as there is an (unnamed) bijection between $A$ and the subset {1,2,3} of $\mathbb{Z}_4$. If we enlarge the codomain of the bijection to equal $\mathbb{Z}_4$,  the bijection becomes an injection only (proposition 0.29a), which we name $f:A\rightarrow \mathbb{Z}_4$.

In the second panel, we see that the restriction of $f$ to $B$, $f|B:B \rightarrow \mathbb{Z}_4$, is also an injection.

In the last panel, we let the codomain of $f|B$ equal the image $Im(f|B)$ thus implicitly defining the bijection $g:B \rightarrow Im(f|B)$

We observe that $Im(f|B) \subseteq \mathbb{Z}_4$. Thus $B$ is equivalent with the subset of $Im(f|B) \subseteq \mathbb{Z}_4$ and is therefore finite.

I think its quite straightforward:

If $A$ is finite, there is some subset $A' \subseteq \mathbb{Z}_n$ for some $n$ such that we have a bijection $f:A \to A'$. ($A$ is equivalent to a subset of $\mathbb{Z}_n$).

Now if $B \subseteq A$, then $f|B$ is still injective as you rightly claim and if we take as the codomain $B' = f[B] = \{f(b): b \in B\} \subseteq A' \subseteq \mathbb{Z}_n$, we see that $f|B$ is surjective onto $B'$ by definition and it is a bijection between $B$ and a subset $B'$ of $\mathbb{Z}_n$, so $B$ is finite too.

For contradiction let us suppose that some finite set have infinite subsetsLet set A be the subset of B
here B is finite and A is infinite
let us suppose X be the every possible Singleton subset of A
and hence X⊆A⊆B.
also ∪X⊆B
which suggest us that Set B is infinite which is not the case
hence our supposition was wrong hence every finite set has finite subset


why is the range of the function $\sqrt x$ the set of numbers $\geq 0$ I have tried looking for an explanation why the range of $$\sqrt x$$ is the set of non-negative real numbers, but I have found none.

I think the range should include negative real numbers because $$\sqrt 4= \pm 2 \ .$$

So why is it so?

Let me tell you first that
$\sqrt4$ is always $2$ not $\pm \,2$. Square root of a positive no. Is always positive.

Like $\sqrt{16}$ is just $4$, not $\pm \, 4$.

And we develop misconception from here.
In school we are taught to solve ques like
$$x^2 = 4$$
Then we take root both side,

$$x = \sqrt4$$
And then we write, $x = \pm\,2$
Although the above solution is correct but the way it is done is wrong.

The right method is
$$x^2 = 4$$
Taking root both side
$$\sqrt{x^2} = \sqrt4,$$
$$|\,X\,| = 2$$
Then we come to result that
$x= \pm\,2$
So in your $f(x) : \sqrt x$ ,
 x can't take negative values
And if it takes positive values then answer will be surely a positive number.

I hope you cleared your misconception today that $\sqrt4$ is just $2$.

A function is a relation from a domain to a codomain such that every element of the domain is related to exactly one element of the codomain.


Tricky trigonometry question Find all solutions of the equations $\tan(3\arccos{x})=11/2$

I got $x = \cos(\frac{1}{3}\arctan(11/2) + n\pi/3)$

Which gives $6$ unique values of $x$, but only $3$ of them seem to work when I sub them in using my calculator. How do I know which $3$ values of $n$ to pick?

HINT:

Let $\arccos x=\theta\implies 0\le \theta\le\pi$  and as the Principal value of $\cos$ lies in $\in[0,\pi]$

So, the problem becomes $$\tan3\theta=\frac{11}2$$ and as $$\tan3\theta=\frac{3t-t^3}{1-3t^2}$$

$$\implies 2t^3-33t^2-6t+11=0\  \  \  \  (1)$$ where $\displaystyle t=\tan\theta$

Solve $(1)$ for $\displaystyle t$

If $\displaystyle t=\tan\theta\ge0, \theta=\arccos x$ will lie in $\in[0,\frac\pi2]$

If $\displaystyle t=\tan\theta<0, \theta=\arccos x$ will lie in $\in(\frac\pi2,\pi)$

Have you covered the general solutions to trigonometric equations? For example, consider the general solution to $tan A = 11/12$. Tan is a positive number (in this case $+11/12$) in the first and third quadrants.

So, the general solution is $A = tan^-1(11/12) +k\pi = 0.74 radians + k\pi$. k is any integer.

But $A = 3cos^-1 (x)$.

So, $cos^-1(x) = 0.18 radians + k\pi/3$

Can you go on from here?


Find all integers n such that the quadratic $5x^2 + nx – 13$ can be expressed as the product of two linear factors with integer coefficients. I am unsure of how to approach this problem. I have thought about using the Rational root theorem, but I am unsure if this answers the question being asked.

Using the theorem, I get $\frac{p}{q} = \pm 1, \pm 13, \pm \frac{1}{5}$, and $\pm \frac{13}{5}$ as possible roots. Then I use synthetic division and Horner's method to get a remainder of $-(n+8)$. For this to be a solution, $-(n+8)=0$, so $n = -8$. Then I could do this for $+1, +13, -13,$ etc.

Is this the correct approach to answering the original question? Original question:

  Find all integers $n$ such that the quadratic $5x^2 + nx – 13$ can be expressed as the product of two linear factors with integer coefficients.

Why would I need to have a rational root to answer the problem? Couldn't I have complex solutions where I can express $5x^2 + nx - 13$ (where n is an integer) as a product of two linear factors with integer coefficients?

I greatly appreciate any insight you could provide on this. It's been about 2 years since I've done any mathematics (a brief foray into Neuroscience turned into a longer expedition than intended) and I am longing to return to the beautiful realm of mathematics. Thanks for your time in reading through this jumbled mathematical thought!

More simply, note that $5$ is prime, so the linear factors have to have coefficients of $x$ as $1$ and $5$ (or $-1$ and $-5$).

So, you have $(x-r)(5x-s) = 5 x^2 -nx -13$ for integers $r,s$. FOILING the left hand side gives $5x^2 - (5r+s) x + rs$, so $rs=-13$ and $5r+s=n$. Since $13$ is prime, we must have $r=-1,s=13$ or $r=1,s=-13$ to meet the constraint that $rs=-13$. You can then plug these in and find $n$.

You can do the same thing with $(-x -r) (-5x-s) = 5 x^2 -nx -13$ to get the other case.

As mentioned in the comments, the discriminant $\sqrt{n^2-4 \cdot 5 \cdot -13} = \sqrt{n^2+260}$ needs to be an integer (such as $k$) for both factors to have integer coefficients.
Let $n^2+260 = k^2$ where $k$ is an integer. Then $260 = k^2-n^2 = (k+n)(k-n)$, and factorising $260 = 2^2 \times 5 \times 13$, the factor pairs of $260$ are $(1, 260); (2, 130); (4, 65); (5, 52); (10, 26); (13, 20)$.
Discard all pairs where the numbers have opposite parity: (odd, even) or (even, odd), which leaves only $(2,130)$ and $(10,26)$. As to why this is, what happens to $k$ and $n$, which must be integers?
Therefore:
$$k + n = 130, k - n = 2 \Rightarrow n = 64 \tag{1}$$
$$k + n = 2, k - n = 130 \Rightarrow n = -64 \tag{2}$$
$$k+n=26, k-n=10 \Rightarrow n=8 \tag{3}$$
$$k+n=10, k-n=26 \Rightarrow n=-8 \tag{4}$$
which is no surprise, as substituting $n \to -n$ into $(1)$ gives $(2)$, and the same into $(3)$ gives $(4)$, and vice versa.


show $V = C_x$ iff there are no multiple eigenvalues $L: V\to V$ a diagonalizable linear operator on finite-dim vector space.

show that $V = C_x$ iff there are no multiple eigenvalues

here $C_x = \operatorname{span} \{x, L(x), L^2(x), \cdots\}$

basically it is a cyclic subspace generated by x that belongs to V.

my thought:

since L is diagonalizable Lx = ux for x is eigenvector and u is evalue

=> if V = C_x

V = C_x = span {x, L(x), L^2(x),.....}

= span {x, ux, (ux)^2,.....}

how does this show no multiple eigenvalues

<= ???

You first need to be clear about what the mystery $x$ is; if it is the zero vector for instance, you'll have a hard time getting $C_x$ to be $V$, no matter how simple the eigenvalues are. I'll assume you omitted a quantifier: there exists a vector $x$ with $C_x=V$ if and only if the characteristic polynomial has no multiple roots (which is a better thing to say than that there are no multiple eigenvalues, since different notion of multiplicity exist for them).

Now since $L$ is diagonalisable, you may suppose $x$ is expressed in a basis of eigenvectors for $L$. Then $L$, and consequently any polynomial in $L$, acts independently on each of the coordinates of$~x$. One thing that is useful, is that the dimension of $C_x$ equals the degree$~d$ of the minimal degree monic polynomial$~P$ such that $P[L](x)=0$: that equation allows $L^d(x)$ to be expressed as linear combination of $x,L(x),\ldots,L^{d-1}(x)$, and once this happens you can see by induction that the same is true for $L^i(x)$ whenever $i\geq d$. So you must show that that $d=n$ is obtained for some vector$~x$ if and only if the eigenvalues associated to each of the coordinates are all distinct.

For a given nonzero coordinate of $x$, whose position has an associated eigenvalue$~\lambda$, check that $P[L]$ makes that coordinate zero if and only if $\lambda$ is a root of$~P$. So in order to have $P[L](x)=0$, it is necessary and sufficient that $P[\lambda]=0$ whenever $\lambda$ is the eigenvalue associated to (the position of) a nonzero coordinate of$~x$.

Now I trust that you can show that the minimum degree a nonzero$~P$ must have in order to satisfy this is equal to the number of different such eigenvalues$~\lambda$. Then you can complete the proof.

Let $x=\sum_{i=1}^n α_i e_i$, where $e_i$ are the eigenvectors of $L$ and $α_i$ non-zero constants. Then

$$L^kx=\sum_{i=1}^nλ_i^kα_ie_i$$

and now you can argue about the representability of any other vector $y=\sum_{i=1}^n \beta_ie_i$ by considering the Vandermonde matrix of the eigenvalues in the system
$$
y=\sum_{i=0}^{n-1} c_iL^i(x)\iff
\pmatrix{β_1\\\vdots\\β_n}=
\pmatrix{α_1&&0\\&\ddots\\0&&α_n}
\pmatrix{1&λ_1&\dots&λ_1^{n-1}\\\vdots&&\vdots\\1&λ_n&\dots&λ_n^{n-1}}
\pmatrix{c_0\\\vdots\\c_{n-1}}
$$


How do I evaluate $e^{9\ln(x)(\frac{9}{x})}$ I am trying an FTC problem.

$$\frac{d}{dx}\int_0^{9\ln x}e^t \,dt=\ ?$$

How do I evaluate $$e^{9\ln(x)(9/x)}$$

EDIT:  Ok, I made a mistake above, and put the chain rule result in the exponent!
Wrong!  Here is the correct way, using FTC:

$$\huge{\frac{d}{dx}\int_0^{9\ln x}e^t \,dt=(e^{9lnx})\frac{9}{x}=(e^{lnx})^9(\frac{9}{x})=\frac{x^9}{1}\frac{9}{x}=9x^8}$$


Do I understand the concept of $x^{0.84}$ correctly? I'm trying to understand the concept of $x^{0.84}$ that Jeffrey Lagarias found for Collatz Conjecture. If I'm wrong, please correct me with an answer. I understand such that,

Suppose the interval $\left[1, 2^{1\,000\,000} \right]$ is given. In this interval we have at least $\lfloor{(2^{1\,000\,000}})^{0.84}\rfloor=2^{840\,000}$ natural numbers which is goes to $1$. In other words, we can choose at least such $2^{840\,000}$ natural numbers from this sequence $\left\{1,2,3,4,5,6,7,8,9\cdots 2^{1\,000\,000} \right\}$, which goes to $1$. Or, there are  at least $2^{840\,000}$ natural numbers  in this interval $\left[1, 2^{1\,000\,000} \right]$ (But, we don't know exactly what the numbers are ?), which gives result $1$. So, Collatz Conjecture is correct for at least $2^{840\,000}$ natural numbers.  Is my understanding correct?

(OP:)... So, Collatz Conjecture is correct for at least $2^{840\,000}$ natural numbers. Is my understanding correct?

As I've read Lagarias & Krasikov your take is correct. However in your sentence which I've cited here you should have included the clause:

  ... So, Collatz Conjecture is correct for at least $2^{840\,000}$ natural numbers below $2^{1\,000\,000}$. (...)

Otherwise you come in conflict with the property, that even infinitely many numbers are known to converge to $1$, for instance from the infinite sequence $\{1,2,4,8,\cdots,2^k, \cdots\}$ or from the infinite sequence $\{1,5,21,85, \cdots , {4^k-1 \over 3} , \cdots \}$ or from infinitely many infinite sequences generated in this style.

$x^{0.84} = x^\frac{21}{25} = (x^{21})^\frac{1}{25}$, which is the unique solution $y$ of $y^{25} = x^{21}$. For irrational $y$, $x^y = \sup_{z\in\mathbb{Q}\cap(-\infty,y)}x^z = \inf_{z\in\mathbb{Q}\cap(y,\infty)} x^z$.


Transition function for absorbed Brownian motion I need an help with the following exercise. I've already seen this question Prove that Brownian Motion absorbed at the origin is Markov but I don't understand the answer. Also I would like to prove the thing differently (at least the two ways seem different to me).

So let $\tau$ be the first time the Brownian motion arrives in $0$ and define

$W^x_t = \begin{cases} x+B_t &\mbox{if } t\leq \tau \\
0& \mbox{else }. \end{cases} $

Here we denote with $B_t$ the standard Brownian motion starting at $0$ and with $B_t^x$ the Brownian motion starting at $x$, so $B_t^x=x+B_t$.

I want to compute the transition function for the process, i.e. $p(t,x,A)$.

I've already got an hint: show that $\mathbb P(B_t^x\in A)= \mathbb P(B^x_t\in A,B_s^x>0 \,\forall s \in [0,t] )+\mathbb P(B_t^x\in -A)$, for each $x>0, A\subset (0+\infty).$

Now I see that if I prove the hint, I'm done, but I don't know how to prove the hint. I know that I should use reflection principle, but how?
Any help will be really appreciated.

$$\mathbb{P}(B^x_t\in A)=\mathbb{P}(B^x_t\in A,B^x_s>0,\forall s\in [0,t])+\mathbb{P}(B^x_\tau=0,B^x_t\in A)$$
where $$\tau=inf\{s,B^x_s=0,s\in[0,t]\}$$
And
$$\mathbb{P}(B^x_\tau=0,B^x_t\in A)=\mathbb{P}(B^x_\tau=0,B^x_t\in -A)=\mathbb{P}(B^x_t\in -A)$$


How to find the slope of curves at origin if the derivative becomes indeterminate What's the general method to find the slope of a curve at the origin if the derivative at the origin becomes indeterminate. For Eg--

What is the slope of the curve $x^3 + y^3= 3axy$ at origin and how to find it because after following the process of implicit differentiation and plugging in $x=0$ and $y=0$ in the derivative we get $0/0$.

Actually this question has been asked by me before and a sort of satisfactory answer that I got was

" For small $x$ and $y$, the values of $x^3$ and $y^3$ will be much smaller than $3axy$, so the zeroes of the function will be approximately where the zeroes of $0=3axy$ are -- that is, near the origin the curve will look like the solutions to that, which is just the two coordinate axes. So the curve will cross itself at the origin, passing through the origin once horizontally and once vertically.
(This is also why implicit differentiation can't work at the origin -- the solution set simply doesn't look like a straight line there under any magnification)."

Edit

If I approximate the function by saying that at (0,0) , the behavior is dominated be 3axy term as x^3 and y^3 are very small and then 3axy=0 and then tangents are x=0 and y=0 . Is doing so (saying x=0 and y=0) linear Approximation only. Because I am approximating the curve with a straight line at origin . But linear Approximation is  1st derivative (1st term of Taylor series). This cannot be right because Taylor series can't be formed where derivative doesn't exist*.

And if this is right then the function is approximately given by 3axy=0 at (0,0). But how does this give the tangent at (0,0).How shall I go about ?

Edit:

Is the answer give  right because the solpe does exist.

It's as simple of that: No (simple) curve, no derivative, no slope.

Use l'Hopital's rule.$ $ This rule, the proof of which is very complex, states that if you get 0/0 or infinity/infinity for some limit function $h(x)=f(x)/g(x)$, you can take the derivative of the top and bottom functions and recalculate. Thus, you would do $f'(x)/g'(x)$. If this result is 0/0 or infinity over infinity, try the next derative, and the next, and so forth. In your particular case, it can be confusing because the implicit function contains both y and x, so which do you take the derivative with respect to? The correct way is to take the derivative with respect to x because you are ultimately solving for dy/dx, even if you use implicit. When differentiating with resepect to x, remember to treat y as a constant. In this case, you should get  6x/constant after one application of l'Hops, with x being 0. Thus, the slope at x=0 is indeed 0. ps. the derivative is "technically" undefined even though it can be determined, so this is just how to get the slope.


Show that solution $(0,0)$ is unique Assume the system:
$$
\begin{align*}
& 2x-5y-x(x^2+2y^2)^2=0 \\
& 5x+2y-3y(2x^2+y^2)^2=0
\end{align*}
$$
Obviously, $(0,0)$ is a solution. The thing I find hard doing is showing that it's the only solution.

Attempt:

Suppose that there exist another solution $(\bar{x},\bar{y}) \neq (0,0)$. Then, for $x \neq0$ and $y \neq 0$:
$$
2x-5y=x(x^2+2y^2)^2 \iff 2-5\frac{y}{x}=(x^2+2y^2)^2=c(x,y)\geq 0 \tag{1}
$$
$$
5x+2y=3y(2x^2+y^2)^2 \iff 5 \frac{x}{y}+2=3(2x^2+y^2)^2=k(x,y) \geq 0\tag{2}
$$
From equations $(1)$ and $(2)$, we yield:
$$
2-5\frac{5}{k-2}=c\iff (2-c)(k-2)=25
$$
But I can't seem to spot a contradiction. Is there  another (more straightforward) way to prove this?


Rubik cube theory and commutative matrices I was reading the same paper on solving the Rubik cube. The 20 pages or so are mostly proofs and introduction to group theory which preface an algorithm for solving near the end. On pages 13-14 they introduce the commutator $PMP^{-1}M^{-1}$, then conjugation of $M$ by $P$, $PMP^{-1}$, then prove that conjugacy is an equivalence relation. $(PMP^{-1})M^{-1}$ looks so close to $PMP^{-1}$, and also to the Jordan decomposition of matrices $A = PJP^{-1}$, but they do not appear to discuss whether any of that is meaningful. The significance of the commutator and conjugacy as an equivalence relation also does not seem to be developed any further when they finally present the algorithm toward the end.

Is there some connection here between the conjugacy equivalence relation, the commutator, complete commutivity on a group, and the Jordan decomposition of matrices? Is conjugacy somehow useful to identify commutative groups, for example, a subset of matrices which form an abelian group?

(This answer only discusses under the context of the MIT paper OP referred to.)

I have done a research on "why the heck did they put commutator and conjugation in Rubik's Cube notes" (written in a more polite language of course) and read tons of materials related to Rubik's Cube.

Basically, there is no theoretical significance now. It is useful for practical reason and stays there for historic and educational reasons.

For people who has attempted solving the Cube, I'm sure it happens at some point of time that you say to yourself "let's put this cubie away using move $A$ so the next move $B$ (which usually does something helpful) does not affect it, and we can bring it back later with a reverse move $A^{-1}$". This is conjugation $ABA^{-1}$. Commutator has similar practical meaning.

They are useful when people want to figure out by themselves how to solve the Cube because they are simple and the effect is obvious to eyes, so they are used extensively in last century when people do not have the computational power to look at the mathematical essence of the Cube. I believe Singmaster's Notes on Rubik's Magic Cube and possibly one issue of Hofstadter's Metamagical Themas that introduces Rubik's Cube have significant influence on this.

Right now the academia is not really concerned about how to solve the Cube anymore since you can easily find some algorithms online. Rubik's Cube is mostly used for demonstrating or introducing concepts in group theory, and given that historically people often see commutator and conjugation in books about Rubik's Cube, it is natural to pick them as two of the concepts that are demonstrated by Rubik's Cube.


Can a convex combination of non-convex non-negative functions be convex function? It is well known that a convex combination of convex functions is still a convex functions. I am wondering is there possible that a convex function can be represented as a convex combination of both non-convex and convex and non-negative functions? Here, the concept of convex combination is relaxed to include an integration form, i.e. $\int_{a} f_a(x)d \mu(a)$ where $\mu$ is a probability measure satisfying  $\int_{a} d\mu(a) = 1$.

This is always possible. You can write the convex function $f$ as
$$
f = \frac23 (2f)  + \frac13(-f).
$$

Not always. Here is a counterexample: $f(x)=x^2$ (convex), $g(x)=-x^4$ (non-convex, concave) and the convex combination
$$
1/2\, f(x) + 1/2\, g(x)= \frac{x^2-x^4}2
$$
which is non-convex (and concave).


let $A$ be a 3 by 3 symmetric matrix over $\mathbb{R}$ such that $A^3+A^2-A-E=0$, then $|A+2E|=$? Since $A^3+A^2-A-E=(A-E)(A+E)^2=0$, so both 1 and -1 may be eigenvalues of $A$, what is the third eigenvalue of $A$, is it -1?
On the other hand $A=P^T\Lambda P$, so $\Lambda^3+\Lambda^2-\Lambda-E=0$, hence $(\Lambda-E)(\Lambda+E)^2=0$, therefore $\Lambda=E$ or $-E$, so $|A+2E|=27$ or $1$?

Diagonalize $A$ and we get $B=P^{-1} A P$, where $B=diag(a_1, a_2, a_3)$ and $a_i = 1$ or $a_i = -1$. Then $$\left\lvert A+2E \right\rvert = \left\lvert P^{-1} \right\rvert \left\lvert A+2E \right\rvert \left\lvert P \right\rvert = \left\lvert P^{-1}AP+2P^{-1}EP\right\rvert = \left\lvert B+2E \right\rvert$$
and you can calculate it. The answer is 1, 3, 9, 27.

As $A=P^T\Lambda P$, so $\Lambda^3+\Lambda^2-\Lambda-E=0$, hence $(\Lambda-E)(\Lambda+E)^2=0$, therefore $\Lambda=E$ or $-E$, now we have $|A+2E|=27$ or $1$, where $P$ is an orthogonal matrix over $\mathbb{R}$.

Note. This answer is wrong, As $\Lambda$ may not be neither $E$ nor $-E$.


Is a circle a multivalued function? I don't really understand multi-valued function. I hope one of you can make me understand it. What I've learned from google, I suppose that a multi-valued function is a binary relation that maps the domain more than once to a single point (i can't describe it well).
So, here, can I conveniently claim that a circle (on the real plane) is a multi-valued function since from one point on the domain we can get $2$ points as the range?
Thanks in advance.

The circle $S$ can be viewed as a binary relation on $\Bbb{R}$, simply because it is a subset of $\Bbb{R}^2$. And indeed if $(x,y)\in S$ then also $(x,-y)\in S$, so if $y\neq0$ then this shows that $S$ is multivalued. But $S$ is not a function on $\Bbb{R}$, because it is not defined on all of $\Bbb{R}$; there is no $y\in\Bbb{R}$ such that $(2,y)\in S$, for example. So the circle cannot be viewed as a multivalued function on $\Bbb{R}$.
There are two ways to adjust your example to valid example of a multivalued function:

We can adjust the domain; viewing the circle $S$ as a subset of $[-1,1]\times\Bbb{R}$, or even $[-1,1]^2$, we do get a multivalued function. This of course relates to the fact that every positive real number has precisely $2$ square roots, i.e. the function
$$f(x)=\sqrt{1-x^2},$$
can also be interpreted as a $2$-valued function on the interval $[-1,1]$.
We can consider a hyperbola instead, for example the set of all $(x,y)\in\Bbb{R}^2$ satisfying
$$x^2-y^2=-1.$$
I leave it to you to verify that this defines a $2$-valued function on all of $\Bbb{R}$.

$y=(R^2-x^2)^\frac{1}{2}$ is multivalue function because exponentiation to a negative even is multivalue by disjunction.
Yes, it is correct that any function which graphical representation has multiple y for single x would be multi-value by disjunction at that point.


Suppose $G$ is a group with exactly $8$ elements of order $10$. How many cyclic subgroups of order $10$ does $G$ have? In a cyclic subgroup of order $10$, there are $\phi(10)=4$ elements of order $10$.
Since, there are exactly $8$ elements of order $10$, we can choose $4$ elements out of the $8$ elements of order $10$ in $^8C_4=70$ ways and for each way, we have a cyclic subgroup of order $10 (4$ elements will of order 4 and the rest 4 will come from group G). Thus, $G$ has $70$ nos. of cyclic subgroups of order $10$.
But the answer to this question is $2$ since a cyclic subgroup of order $10$ can only have $4$ elements of order $10$ and hence there are $8/4$ cyclic subgroups only. But I don't understand why? There are $^8C_4=70$ ways to choose $4$ elements out of $8$ elements. So I think it should be $70$.
Please help me understand. Thanks in advance.
PS: I know that in a finite group, no. of subgroups should divide the order of the group.  However, in my case since order of $G$ is not given, I am confused.

I claim that there is no group with exactly $8$ elements of order $10$, which makes the whole question logically meaningless (except possibly as a lemma in the proof that there is no such group).

Such a group $G$ would have two exactly two distinct subgroups $A$ and $B$ of order $10$. If they were not normal in $G$, then they would be conjugate and their normalizer would have index $2$ and contain both $A$ and $B$, so we could replace $G$ by this subgroup to get a smaller example.

So we can assume that $G = \langle A,B \rangle = AB$ with $A,B \unlhd G$.

Since $|{\rm Aut}(A)| = 4$, the element $g$ of order $5$ in $B$ must centralize $A$. Then, if $g \not\in A$, then $\langle g, A \rangle \cong C_5 \times C_{10}$ has $6$ subgroups of order $15$, contrary to assumption.

So $g \in A$, $|A \cap B| = 5$, and $|G| = |AB| = 20$.

Now, of the five (isomorphism types of) groups of order $20$, $C_{20}$, $D_{20}$ and the dicyclic group have a unique subgroup of order $10$, $C_2 \times C_{10}$ has three such subgroups, and the final group, which is a semidirect product $C_5 \rtimes C_4$ with faithful action has none.

As you have correctly pointed out that each cyclic subgroup will have exactly 4 elements of order 10.

claim:- if ${H_k}\cap{H_l}$ contains an element (say) $a_i$ of order 10 then $H_k=H_l$

Since ${a_i}\in{H_k}\cap{H_l}$ and order of $a_i$ is 10 then,
$<a_i>=H_k$ and $<a_i>=H_l$
Hence $H_k=H_l$

Did you got why there will be only two cyclic subgroup of order 10?


Michael Spivak in "Calculus" asserts that $\sqrt2$ cannot be proven to exist, and that such a proof is impossible. What does he mean by "exist"? Michael Spivak in "Calculus" asserts that $\sqrt2$ cannot be proven to exist, and that such a proof is impossible. What does he mean by "exist"? How are you to prove that any number "exists"? Why can't we define $\sqrt2$ as a number that fits under some arbitrary definition of existence, while asserting that its most concise expression is with a functional root?

I'm sorry if these questions seem a bit sophomoric; in some ways it resembles an 8 year old repeatedly asking "why". But given that his prose is very concise and technical, his usage of "exist" was out of the ordinary.

(I used two tags representing the book's field of study; and one representing the actual relevant tag.)

edit

Oh, I'm sorry. I misquoted. My question still stands, though; how has he defined existence such that $\sqrt2$ might possible not be within it.

Direct quote: "We have not proved that any such number exists..." in reference to $\sqrt2$.

The point Spivak is making is that the properties of numbers that have been studied till that point in the book are not enough to prove that there is a number whose square is 2. This follows from the proof that no rational number will do for this task. Since the rational numbers satisfy all properties till that point, it is clear that some other property is needed (and that the rationals cannot have this property). That property is completeness and is a what characterizes real numbers. Using completeness, one can and does prove that there is a positive number whose square is 2; we call it $\sqrt2$.

I believe that the existence of √2 was proven in 300 B.C. give or take.

Just draw a right triangle, with the sides being exactly 1 unit in length, then the hypotenuse of the triangle - using the Pythagorean theorem - will be √2 in length.

I fully concede that I probably missed that whole point of this conversation, but that's how you prove the existence of √2 anyway :)


the order of the factor ring $\mathbb{Z}_{15}\left [ x \right ] / \left \langle 3x^{2}+5x \right \rangle$ What is the order of the factor ring $\mathbb{Z}_{15}\left [ x \right ] / \left \langle 3x^{2}+5x \right \rangle$ ?
Since $x$ and $3x+5$ are not relatively prime polynomial in $\mathbb{Z}_{15}$, so I can't use the Chinese Remainder Theorem.
Any help please. Thank you.

Let $f=3x^{2}+5x$ and $I=\langle f \rangle$.

$I \ni 3f = 9x^2+15x = 9x^2$.
$I \ni 5f = 15x^2+25x = 25x = 10x = -5x$.
$I \ni 5xf - 3f = 10x^2 - 9x^2 = x^2$.

Thus, $I=\langle 5x, x^2 \rangle$.

Therefore, for every $g \in \mathbb{Z}_{15}[x]$ we have $g \equiv ax+b \bmod I$, with $a \in \{0,1,2,3,4\}$.

Thus, $\mathbb{Z}_{15}[x] / I$ has $5 \cdot 15 = 75$ elements.

Hint: Here, $3x^2+5x=x(3x+5)$ both factors are irreducible and relatively prime.
Use Chinese reminder theorem for rings to find an isomorphism and to  SO get the order of ring.


Limit of integral with sinx I need to find this limit:

$$\lim\limits_{x\rightarrow 0}\int_{\sin x}^{x}\dfrac{1}{t^3(1+t^{1/2})}\,dt.$$

Thank you for any help.

$\int_{\sin x}^x\frac{1}{t^3(1+t^{1/2})}dt>\int_{\sin x}^x\frac{1}{x^3(1+x^{1/2})}dt=(x-\sin x)\frac{1}{x^3(1+x^{1/2})}$. Then using the taylor series of $\sin x$, you get the lower bound. The upper bound is in a similar way. Actually, you do not need to solve the integral.

Substitute $x^2=t$ and write the inner integral as $$2\int {1+x-x-x^2+x^2+x^3-x^3-x^4+x^4+x^5-x^5\over x^5(1+x)}$$$$=2\left(\int{1\over x^5}dx-\int{1\over x^4}dx+\int{1\over x^3}dx-\int{1\over x^2}dx+\int{1\over x}dx-\int{1\over 1+x}dx\right)$$

Now integrate and put the limits.See what you get..!


What's $P$ and what's $Q$ in this classic proof of the irrationality of $\sqrt 2$? In this proof extracted from the Wikipedia

  A classic proof by contradiction from mathematics is the proof that
  the square root of 2 is irrational. If it were rational, it could
  be expressed as a fraction $a/b$ in lowest terms, where a and b are
  integers, at least one of which is odd. But if $a/b = \sqrt 2$, then $a^2=$
  $2b^2$. Therefore $a^2$ must be even. Because the square of an odd number is
  odd, that in turn implies that $a$ is even. This means that $b$ must be
  odd because $a/b$ is in lowest terms. On the other hand, if $a$ is even,
  then $a^2$ is a multiple of $4$. If $a^2$ is a multiple of $4$ and $a^2=2b^2$,
  then $2b^2$ is a multiple of $4$, and therefore $b^2$ is even, and so is $b$. So
  $b$ is odd and even, a contradiction. Therefore the initial
  assumption—that $\sqrt 2$ can be expressed as a fraction—must be false.

Knowing that a proof by contradiction you assume P and Not(Q) what's P and what's not Q in this proof?

I'm assuming you are more accustomed to seeing proof by contradiction used largely with statements that are implications or conditionals. And indeed, when writing a proof by contradiction to prove statements of the form $$P \implies Q,$$ we typically assume $(P\land \lnot Q)$.

But in this particular case, we do not seem to have an implication to prove. Rather, we have the proposition:

  The square root of $2$ is irrational. $\quad$( $Q$).

There's no helpful "if, then", or "this implies that" to indicate any sort of implication being asserted. So we have an example of the use of a proof by contradiction where to prove a statement other than an implication.

What we can do is to think of the assertion to be proven as a simple "atomic" proposition: $\,Q.\,$ Then $\,\lnot Q\,$ is the statement to the effect:

  Suppose $\,\sqrt 2\,$ is not irrational. $\;$ Put differently, suppose $\,\sqrt 2\,$ is rational.$\quad(\lnot Q)$

The proof then proceeds, after having supposed $\,\lnot Q\,$ to invoke the definition of a rational number in order to arrive at a contradiction.

In a sense then, the proof amounts to a "bare-bones" proof-by-contradiction:

To prove that $\,Q,\,$ we assume $\,\lnot Q,\,$ and then we work to obtain a contradiction. Once we arrive at a contradiction, we can conclude that our assumption is false, and so we are justified in negating the false assumption: "therefore, $\lnot\lnot Q.$" $\;\;$ And this amounts to affirming the desired conclusion/assertion: therefore $Q$, since $\;\lnot \lnot  Q\equiv Q$.

The contradiction in this proof happens to come from our knowledge about the rational numbers, information which could be considered a premise: the "implicit" premise $P$ being the definition of a rational number.

P would be that $\sqrt{2}$ is rational, Q that it is irrational, and so not(Q) is that it is not irrational.