How is a system of axioms different from a system of beliefs? Other ways to put it: Is there any faith required in the adoption of a system of axioms? How is a given system of axioms accepted or rejected if not based on blind faith?

To paraphrase Robert Mastragostino's comment, a system of axioms doesn't make any assertions that you can accept or reject as true or false; it only specifies the rules of a certain kind of game to play.

It's worth clarifying that a modern mathematician's attitude towards mathematical words is very different from that of a non-mathematician's attitude towards ordinary words (and possibly also very different from a classical mathematician's attitude towards mathematical words). A mathematical word with a precise definition means precisely what it was defined to mean. It's not possible to claim that such a definition is wrong; at best, you can only claim that a definition doesn't capture what it was intended to capture.

Thus a modern interpretation of, say, Euclid's axioms is that they describe the rules of a certain kind of game. Some of the pieces that we play with are called points, some of the pieces are called lines, and so forth, and the pieces obey certain rules. Euclid's axioms are not, from this point of view, asserting anything about the geometry of the world in which we actually live, so one can't accept or reject them on that basis. One can, at best, claim that they don't capture the geometry of the world in which we actually live. But people play unrealistic games all the time.

I think this is an important point which is not communicated well to non-mathematicians about how mathematics works. For a non-mathematician it is easy to say things like "but $i$ can't possibly be a number" or "but $\infty$ can't possibly be a number," and to a mathematician what those statements actually mean is that $i$ and $\infty$ aren't parts of the game Real Numbers, but there are all sorts of other wonderful games we can play using these new pieces, like Complex Numbers and Projective Geometry...

I want to emphasize that I am not using the word "game" in support of a purely formalist viewpoint on mathematics, but I think some formalism is an appropriate answer to this question as a way of clarifying what exactly it is that a mathematical axiom is asserting. Some people use the word "game" in this context to emphasize that mathematics is "meaningless". The word "meaningless" here has to be interpreted carefully; it is not meant in the colloquial sense (or at least I would not mean it this way). It means that the syntax of mathematics can be separated from its semantics, and that it is often less confusing to do so. But anyone who believes that games are meaningless in the colloquial sense has clearly never played a game...

I am going to go ahead and claim that there is in many ways not much of a difference.

As others here have explained very well, mathematical axioms can be seen as a set of rules and statements describing a mathematical system. And in similar terms, religious beliefs can be seen as a set of rules and statements describing the world we live in (or, as some state, don't actually live in).

However, the term "faith" is mostly meaningless (and, I would claim, irrelevant) as far as mathematics are concerned. To have faith that something is true requires that there is uncertainty about that truth. A lack of proof. Axioms do not need proof, they are the basis of proofs. To have faith in an axiom is pointless.

As for the question of accepting or rejecting axioms, I would claim this to be a moot point. There is no need to accept or reject any particular axiom because there is nothing stopping two mathematical systems coexisting, one with the axiom in question and one without.

Of course, for practical purposes one cannot explore an infinite number of systems, so the question becomes one of usefulness. Does this system, with these axioms, allow me to make useful or interesting statements, or does it not? Most systems will not. But some, like for example that of Euclidean geometry, have resulted in a vast number of useful statements.

The accepting and rejecting of axioms is an issue for religions only because religions tend to make statements not about their own system, but about the world. This is, according to me, the one important way in which mathematical and religious axioms differ: mathematical axioms only make statements about their own systems,  while religious axioms may (and often do) make statements about anything and everything.

Where mathematical systems can coexist because they each create their own little hypothetical universe, religions often can not and require faith because we only have this one universe which we insist on making statements about.

To see what happens when we instead start making religious-style statements about another universe, have a look at any well written fantasy or science fiction book. Suddenly axioms which may appear very religious if taken out of context become more mathematical in nature; like mathematical systems they define their own universe, and having "faith" in them once again becomes meaningless.


The meaning of $dx$ in an indefinite integral This semester I'm taking integral calculus for the first time. We started with the differential (i.e. $dy=f'(x)\,dx$) and right after that with the indefinite integral. Since then, I've been trying to make sense of the $dx$ when it's part of an indefinite integral (i.e. $\int f(x) \, \boldsymbol{dx}$). I know there are already a bazillion answers regarding this question, but all of them refer to the definite integral and the ones that do touch on indefinite integrals only say things like "it's just a syntactical device to tell you the variable to differentiate with respect to or the integration variable" (Ihf, 2012, web). I don't like this answer, especially because I was taught that given two functions $f(x)$, $g(x)$ and the antiderivative of their product $\int f(x)g(x)\,dx$, if I were to assign $f(x)$ to $u$ and $g(x)\,dx$ to $dv$ in order to integrate by parts, then I would have to integrate $dv$ to find $v$. This only makes sense if $dx$ means something by itself and is not just a quirk of the notation, otherwise we would end up with something like the following: $$\int g(x)\,dx\space dx$$
After thinking about it a lot, I believe I finally found a way to make sense of the $dx$ as something that isn't purely and simply a notation device. My reasoning is as follows:

Given a function $f(x)$, let $y=f(x)$. Then $$\frac{dy}{dx}=f'(x)$$
Then from the fundamental theorem of calculus, we know that $$\int\frac{dy}{dx}\,dx=y$$
Let $dy=f'(x)\,dx$. Then the equation $$dy=\frac{dy}{dx}\,dx$$ holds and $$\int\frac{dy}{dx}\,dx=\int dy=y$$
Finally, integrate both sides of the equation $dy=f'(x)\,dx$ in order that $$\int dy=\int f'(x)\,dx\implies y=f(x)$$

From this I conclude that, by integrating a function, what we are really doing is integrating the differential of that function. This makes perfect sense to me, although I'm aware that seemingly logical things aren't necessarily logical. That's why I would appreciate it if someone could tell me whether the above is mathematically correct or pure gibberish.
PS, I had never written a mathematical proof before and I have taken no proofs courses yet, so any suggestions are welcome.
Reference: lhf. (2012, May 9). What does $dx$ mean?. Mathematics Stack Exchange. https://math.stackexchange.com/q/143262

An explanation is given in section 2.9 Fundamental Theorems of the Calculus in Introduction to Calculus and Analysis I by R. Courant and F. John.

It is quite customary to use a notation which is not perfectly clear without comment: we write
\begin{align*}
\color{blue}{F(x) = \int f(x)\,dx}
\end{align*}
when we mean that the function $F(x)$ is of the form
\begin{align*}
F(x) = c + \int_a^xf(u)\,du
\end{align*}
for suitable constants $c$ and $a$, that is, we omit the upper limit $x$, the lower limit $a$ and the additive constant $c$ and use the letter $x$ for the variable of integration.
Strictly speaking, of course, there is a slight inconsistency in using the same letter for the variable of integration and the upper limit $x$ which is the independent variable in $F(x)$. In using
the notation $\int f(x)\,dx$ we must never lose sight of the indeterminacy connected with it, that is, the fact that the symbol always denotes one of the primitive functions of $f$ only. The formula $F(x) = \int f(x)\,dx$ is
just a symbolic way of writing the relation
\begin{align*}
\color{blue}{\frac{d}{dx}F(x)=f(x)}.
\end{align*}

Remark
In mathematics it may be that "$\int$" and "$dx$" are considered to be brackets, inclosing the integrand:
$$
\int f(x)\;dx
$$
In physics it may be that "$\int dx$" is used as an operator, so it precedes the integrand:
$$
\int dx \;f(x)
$$


Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$ Evaluate the limit

  $$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$

My Attempt:

To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B = \left(\sqrt[3]{x^3-x^2+1}\right)$. Now

$$
\begin{align}
2x^2 &= A^3-B^3\\
x &= \sqrt{\frac{A^3-B^3}{2}}
\end{align}
$$

So the limit becomes

$$\lim_{x\rightarrow \infty}\left(\sqrt{\frac{A^3-B^3}{2}}-A-B\right)$$

How can I complete the solution from this point?

Note that $\sqrt[3]{x^3+x^2+1}=x\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^3}}$, with a similar expression for the other one. Make the substitution $t=1/x$. Our expression becomes
$$\frac{2-\sqrt[3]{1+t+t^3}-\sqrt[3]{1-t+t^3}}{t},$$
and we want to find the limit of this as $t$ approaches $0$ from the right. Note that the conditions for using L'Hospital's Rule are met. So our limit is
$$\lim_{t\to 0^+} -\frac{1}{3}(1+3t^2)(1+t+t^3)^{-2/3}-\frac{1}{3}(-1+3t^2)(1-t+t^3)^{-2/3}.$$
This one is easy.

As $x \to\infty$, $A\to x$ and $B\to x$. Therefore the expression would tend to $0$.


Question about complete orthonormal basis Let $V$ be an inner product space. Let $W$ be the Hilbert space obtained as the completion of $V$. Is there a complete orthonormal basis of $V$ which is still complete in $W$? This is true if we assume that $V$ is separable (Schumidt's method), but I don't know if this is true or not in general.

A complete orthonormal basis is one that cannot be extended to a larger orthonormal basis.  A complete orthonormal basis of an inner product space is usually not a Hamel basis (except in the finite-dimensional case), i.e. not every vector in the space is a linear combination of only finitely many members of the basis.  If one wants to speak of a linear combination of infinitely many vectors, one must understand what it means for such a series to converge, and for that the inner product is needed.  The distance between the $n$th partial sum and the limit is a sequence of real numbers that must approach $0$.  Without the inner product one cannot speak of distance.  And the idea of convergence is what brings us to the notion of "completion" of the space.  The completion contains only points that can be approached from within the space that is getting completed.  Therefore one adds no points that cannot be expressed as linear combinations of infinitely many points that were already there, in the space getting completed.

Bottom line: Every complete orthonoral basis of the space getting completed is a complete orthonormal basis of the completion.


Likelihood functions are same My approach: They both will not be same because if we try to find the maximum likelihood function, we will get different values for both the functions.

Is this correct?

No, that is a bit misleading.  Likelihood is a relative concept, and what is most relevant is not the value of the likelihood function, but the subset of the parameter space for which the likelihood is maximal (in the case of maximum likelihood estimation).

In other words, while the maximum likelihood of $\theta^2$ is $1$ in the first case and $100$ in the second, in both cases, the maximum is attained at $\theta = 1$, and it is this value that is of interest.

$$ \frac{\theta^2}{100\theta^2} = \frac 1 {100}, \text{ and this does not depend on } \theta. $$

Two likelihood functions are in many situations considered equivalent, or "the same", if one of them is a constant multiple of the other. "Constant" means not depending on $\theta.$

This means that not only will they yield the same maximum-likelihood estimators, but both will yield the same posterior distributions when given the same prior distributions.

For example, suppose $\Theta$ is uniformly distributed on $[0,1]$, so that $f_\Theta(\theta) = \begin{cases} 1 & \text{if } 0\le \theta\le 1, \\ 0 & \text{otherwise.} \end{cases}$

Then let $X$ be the number of successes in $8$ independent trials with probability $\Theta$ of success on each trial, and let $Y$ be the number of independent trials needed to get $5$ successes, again with probability $\Theta$ of success on each trial. If we observe $X=5,$ the likelihood is
$$
L(\theta) = \binom 8 5 \theta^5(1-\theta)^3 56\theta^5(1-\theta)^3.
$$
If we observe $Y=8$, then the likelihood is
$$
L(\theta) = \binom 7 4 \theta^4(1-\theta)^3\cdot\theta = 35\theta^5(1-\theta)^3.
$$
In either case the posterior density is
$$
f_{\Theta\,\mid\,X=5} (\theta) = 504 \theta^5(1-\theta)^3 = f_{\Theta\,\mid\,Y=8} (\theta),
$$
where the constant $504$ comes from the fact that
$$
\int_0^1 \theta^5(1-\theta)^3\,d\theta = \frac 1 {504}.
$$


Eigenvalues of a matrix $A$ and $e^{A}$ If I know the eigenvalues of $e^{A}$, what can I say about the eigenvalues of $A$ itself?

If $A$ is upper triangular, then it's easy: then $e^A$ is upper triangular too, and the diagonal elements of $e^A$ are the exponentials of the diagonal elements of $A$.

But this is always the case, because we can choose a basis in which $A$ is in Jordan canonical form (which is in particular upper triangular).

So the eigenvalues of $A$ are logarithms of the eigenvalues of $e^A$.

But beware that the eigenvalues of $A$ can be complex even if $A$ and $e^A$ are both real, and they are not necessarily the principal logarithms of $e^A$'s eigenvalues. For example, if $A=\pmatrix{0&2\pi\\-2\pi&0}$ then $e^A=I$, but the eigenvalues of $A$ are $\pm 2\pi i$.

This process May be right ,,,,
$$AX=\lambda  X$$
$$A^nX=\lambda^n X$$
 $$e^A=I +\frac{A}{1!}+\frac{A^2}{2!}+\ldots \tag{1}$$
Now multiplying $X$ vector  on both sides of equation (1), we get
$$e^A \cdot X=X+\frac{AX}{1!}+\frac{A^2X}{2!}+ \ldots$$
or,$$e^A\cdot X=X+\lambda\frac{X}{1!} + \lambda^2\frac{X}{2!}+\ldots$$
Or,$$e^A\cdot X=(1+\frac{\lambda}{1!}+\frac{\lambda^2}{2!}+ \ldots )X$$
or,
$$e^A\cdot X=e^{\lambda}X$$
Thus eigenvalues of $e^A$ are $e^{\lambda}$


Tangent line to a general conic at a point If I have a conic

$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$

and want to know the tangent line at $(x_0, y_0)$, I thought I would just find the derivative (implicit) and use the equation of a line $(y-y_0) = m (x-x_0)$. For the derivative I get:

$$2Ax + By + \frac{dy}{dx}Bx + 2Cy\frac{dy}{dx} + D + E \frac{dy}{dx} =0$$

Solving for $\frac{dy}{dx}$ I get

$$m = \frac{dy}{dx} = \frac{-2Ax - By - D}{Bx + 2Cy + E}$$

Substituting this into the equation for a line:

$$(y-y_0) = \frac{-2Ax - By - D}{Bx + 2Cy + E} (x-x_0) = $$

and grouping the $x$, $y$, and constant terms, I get

$$(2Ax_o+By_0+D)x + (Bx_0+2Cy_0+E)y -2Bx_0y_0 - 2Cy_0^2 - Ey_0 - 2Ax_0^2 - Dx_0 = 0 $$

I've seen things like this: http://demonstrations.wolfram.com/TangentLinesToAConicSection/ that show an $F$ in the expression. The $x$ and $y$ terms are the same, but the constant term is $-Ax_0^2 - Bx_0y_0 - Cy_0^2 + F$. In my method here, the $F$ term goes away in the differentiation immediately as it is a constant. Can anyone explain how these representations are equivalent?

$$-Ax_0^2 - Bx_0y_0 - Cy_0^2 + F = - 2Cy_0^2 - Ey_0 - 2Ax_0^2 - Dx_0$$

Theory: Equation of tangent at any point of the conic is given by $T = 0$.

$T$ is the transformed equation of the conic.

Following transformations are considered while transforming a conic at a point $(x_0, y_0)$

$$x^2 \implies x x_0$$

$$y^2 \implies y y_0$$

$$x \implies \frac{x+x_0}{2}$$

$$y \implies \frac{y+y_0}{2}$$

$$xy \implies x.y_0+y.x_0$$

Answer to your question: According to your question, the equation of conic is:

$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$

Tangent at the point: $(x_0,y_0)$ will be given by $T$= $0$.

$\implies$ $T$=$0$

$\implies$ $$A(x.x_0) + B(x.y_0+y.x_0) + C(y.y_0) + D.\frac{(x+x_0)}{2} + E.\frac{(y+y_0)}{2} + F = 0$$

This is required equation of tangent.

You have missed $-2Bx_0y_0$ at the Right hand side.

As $(x_0,y_0)$ lies on the curve, it will satisfy the equation of the given curve

Hence, $$Ax_0^2 + Bx_0y_0 + Cy_0^2 + Dx_0 + Ey_0+ F = 0$$

$$-Ax_0^2 - Bx_0y_0 - Cy_0^2 + F = - 2Cy_0^2 -2Bx_0y_0 - Ey_0 - 2Ax_0^2 - Dx_0$$
$$\iff Ax_0^2 + Bx_0y_0 + Cy_0^2 + Dx_0 + Ey_0+ F = 0$$


Taylor Series of $e^{-1/x^2}$ about $0$ So I worked out the Taylor series of $e^{-1/x^2}$ about $0$. However, all derivatives at $0$ came out to be $0$. So how can we write the Taylor series of the above function about $0$?

As you worked out, the Taylor series of this function is 0. It means that, around 0, this function and all of its derivatives evaluate to 0.

@user3650050 : I suppose that you expect a series which approximates the function $y(x)=\exp\left(-\frac{1}{x^2} \right)$ for $x$ close to $0$.

The pertinent comments and answer that you already received might disappoint you because whey don't give an approximate series as expected : As already pointed out, the Taylor series for $y(x)=\exp\left(-\frac{1}{x^2} \right)$ at $x=0$ doesn't fit the function for $x$ not exactly $0$.

I don't know if the next form of series can be useful for you, but I hope so :

Around $x=\epsilon\:\:$ close to $0$, the first terms of the series expansion are :
$$\exp\left(-\frac{1}{x^2} \right)=\exp\left(-\frac{1}{\epsilon^2} \right)\left(1+\frac{2}{\epsilon^3}x +\frac{2-3\epsilon^2}{\epsilon^6}x^2 +\frac{\frac43 -6\epsilon^2+4\epsilon^4}{\epsilon^9}x^3 +... \right)$$


Verifying a proof that if $x,y,z \geq 0$ and $x+y+z = 1$, then $0 \le xy + yz + zx - 2xyz \le \frac{7}{27}$ I was working some recreational problems from a book (The Art and Craft Of Problem Solving, Zeitz) and came across one from the '84 IMO:

  Suppose that $x, y, z$ are non-negative reals, with $x + y + z = 1$. Prove that
  $$0 \le xy + yz + zx - 2xyz \le \frac{7}{27}.$$

I'm afraid that I'm here today to ask for a proof check. I ask because I was able to prove the claim without using any sophisticated inequalities, and - although I have checked it myself - I can't help but feel a bit suspicious.

I have posted my solution as an answer below. Of course, I would love it if somebody would give it a careful read.

Shorter, more elegant solutions are also welcome.

Claim: (The above.)

Proof: First, notice that

\begin{eqnarray}
xy + yz + zx - 2xyz &=& xy(1 - z) + yz(1-x) + zx(1-y) + xyz\\
&=& xy(x+y) + yz(y+z) + zx(z+x) + xyz
\end{eqnarray}

since $x + y + z = 1$; since we can re-write our original expression as a sum of all positive terms, we plainly have

$$0 \le xy + yz + zx - 2xyz$$

which was the first part of the claim. For the second part, we first notice that

$$\frac{1}{3}(x + y + z)^3 = \frac{1}{3}(x^3 + y^3 + z^3) + (x + y + z)(xy + yz + yz) -xyz;$$

since $x + y + z = 1$, we get

\begin{eqnarray}
\frac{1}{3} = \frac{1}{3}(x^3 + y^3 + z^3) + [(xy + yz + yz) -2xyz] + xyz,
\end{eqnarray}

so we can re-write our original inequality as

$$\frac{1}{3} - \frac{1}{3}(x^3 + y^3 + z^3) - xyz \le \frac{7}{27}$$

or, just as well,

$$ \frac{1}{3}(x^3 + y^3 + z^3) + xyz \ge \frac{2}{27}.$$

Now, unfortunately, comes a wave of algebra, which I will not do out here. (I will just show you the results, which I checked carefully several times.)

We let $x = \frac{1}{3} + p, y = \frac{1}{3} + q, z = \frac{1}{3} + r$. Importantly, we have $p + q + r = 0$.

After a wave of algebra, our inequality can be re-written as

\begin{eqnarray} \frac{2}{27} + (p^3 + q^3 + r^3) + (p^2 + q^2 + r^2) + \frac{1}{3}(pq + pr + qr) + pqr &\ge& \frac{2}{27} \\ \iff (p^3 + q^3 + r^3) + (p^2 + q^2 + r^2) + \frac{1}{3}(pq + pr + qr) + pqr \ge 0. \end{eqnarray}

Recall that I earlier pointed out (with $x, y,$ and $z$ as the variables) that

$$\frac{1}{3}(p + q + r)^3 = \frac{1}{3}(p^3 + q^3 + r^3) + (p + q + r)(pq + pr + qr) -pqr;$$

we may thus conclude that $pqr = \frac{1}{3}(p^3 + q^3 + r^3)$. Similarly, expanding $(p+q+r)^2$ yields $pq + pr + qr = -\frac{1}{2}(p^2 + q^2 + r^2)$. Substituting, we see that

\begin{eqnarray} (p^3 + q^3 + r^3) + (p^2 + q^2 + r^2) + \frac{1}{3}(pq + pr + qr) + pqr \ge 0 \\ \iff \frac{4}{3}(p^3 + q^3 + r^3) + \frac{5}{6}(p^2 + q^2 + r^2) \ge 0 \end{eqnarray}

Finally, we obtain the equivalent inequality we will make a stand with:

$$ 5(p^2 + q^2 + r^2) + 8(p^3 + q^3 + r^3) \ge 0. \qquad(*) $$

Now, we're almost done. We need only consider the signs of $p, q,$ and $r$. WLOG, we must have

(1) $p, q, r > 0$,

(2) $p, q > 0$ and $r <0$,

(3) $p > 0$ and $q, r <0$, or

(4) $p, q, r < 0$.

(We ignore the cases where any of $p, q,$ or $r$ are zero because - since $p + q + r = 0$ - the inequality $(*)$ then becomes trivial.) Clearly, (1) and (4) are impossible (since $p + q + r$ must sum to something non-zero in those cases). Let us consider case (3): if $p$ and $q$ are both negative, then, in order that $p + q + r = 0$, we must have $0 < |p|, |q| < |r|$. But then $|p|^3 + |q|^3 < |r|^3$, and so $(p^3 + q^3 + r^3)$ is positive; the inequality $(*)$ then plainly holds.

So, the only case to worry about is (2). In that case, noting of course that $p + q = -r$, and also that $-\frac{1}{3} \le r$ (since $x,y $ and $z$ were positive) and therefore $p, q < p + q \le \frac{1}{3}$,

\begin{eqnarray} 5(p^2 + q^2 + (p+q)^2) + 8(p^3 + q^3 - (p+q)^3) &=& 10(p^2 + q^2 + pq) - 24(p^2q + q^2p)\\
 &=& 10pq + p^2(10 - 24q) + q^2(10 -24p)\\ &\ge& 10pq + p^2(10 -24(\frac{1}{3})) + q^2(10 -24(\frac{1}{3}) \\ &=& 10pq + 2p^2 + 2q^2 \ge 0. \end{eqnarray}

Thus, the inequality $(*)$ is proven in all cases; and so is the inequality $xy + yz + zx - 2xyz \le \frac{7}{27}$. The claim follows.

Here's what I think is a much simpler proof than all of the others.

By the Cauchy Schwarz Inequality, $$(xy+yz+xz)^2\leq (x^2+y^2+z^2)^2\\ \implies xy+yz+xz\leq x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz) \\ \implies3(xy+xz+yz)\leq 1 \implies xy+yz+xz\leq \frac{1}{3}$$

By AM-GM,
$$\sqrt[3]{xyz}\leq \frac{x+y+z}{3}\\\leq \frac{1}{3} \\ \implies xyz\leq \frac{1}{27}$$.

The original inequality is equivalent to $$xy+yz+xz\leq \frac{7}{27}+2xyz$$ which is clear given the bounds that we just proved. Equality occurs at $x=y=z=\frac{1}{3}$


If $x,y \in (0,\frac{\pi}{2})$ then expression $\sin x +\cos y +\tan^2y+\cot^2x+5>\ldots?$ Problem :

If $x,y \in (0,\frac{\pi}{2})$ then expression $\sin x +\cos y +\tan^2y+\cot^2x+5$ is always greater than :

(a)  $\ 7 $

(b)  $\ 8 $

(c)  $\ 9 $

(d)  $\ $none of these

Solution :

We can write the given expression $\sin x +\cos y +\tan^2y+\cot^2x+5$ as $\sin x +\cot^2x +\cos x +\tan^2y +5$

$\Rightarrow \sin x +\cot^2x +\cos y +\tan^2y +5  = \sin x + \csc^2x -1 +\cos x +\sec^2y-1+5$

$\Rightarrow \sin x + \csc^2x  +\cos x +\sec^2y+3$

Please guide me how to proceed further. Thanks.

I think you are doing a mistake here.
$ \sin(x)+\cos(y)+\tan^2(y)+\cot^2(x)+5 = (\sin(x)+ \csc^2(x)) + (\cos(y)+
\sec^2(y))+ 3$ . Not, what you have written.
Considering that to be a typo, Notice that the things in the bracket are independent, so we just find minimum value of each  separately.

Consider $f(x)=\sin(x)+\csc^2(x) \ge \sin(x)+\csc(x) \ge 2$.
(By AM-GM inequality) . Similarly consider $g(x)=\cos(y)+\sec^2(y)$

We get $7$ as the answer .

I don't think the simplifying is necessary.  We can just analyze the functions within the open interval $(0,\pi/2)$.  First, we can examine the minimum value of $\sin x+\cot^2x$.  When $x=\pi/2$, sine is $1$ and cotangent is $0$.  Cotangent increases without bound as we move towards $0$ from $\pi/2$.  So, we can assume $\sin x+\cot^2x$ is always greater than $1$ on the interval $(0,\pi/2)$.  The same reasoning applies to $\cos y+\tan^2 y$, however the minimum will be at $y=0$ now, and the minimum is again $1$.  So, just add these together with $5$, and we see that the entire function is always greater than $7$.


Is it possible to create a completely random integer between 1 and 13 using standard dice in a D&D dice kit? I am asking if it is possible to choose a random integer using a certain set of dice. Assume for these calculations that the dice are perfectly randomly distributed. There are 7 dice in a set, with the following possible outcomes for each dice:

  2, 4, 6, 8, 10, 12, and 20

You are free to interpret the outcomes as any value you like.

The dice can be rolled in any order, combination, or quantity desired.

The objective is to have 13 possible outcomes, perfectly random and uniformly distributed.

This is impossible if you require that there is a fixed upper bound on the number of possible rolls: in this case, the only probabilities you can get are of the form $\frac{n}{m}$ where $m$ is a product of the numbers $2, 4, 6, 8, 10, 12, 20$, and in particular cannot be divisible by $13$.

This is straightforward if you don't require such a fixed upper bound: as mentioned in the comments, you can discard and reroll. This strategy, like any successful strategy, necessarily has the property that it is possible that you'll have to reroll arbitrarily many times. But the probability is exponentially decaying so it's not really an issue.

Edit: An interesting follow-up question might be to find a strategy that minimizes the expected number of rolls. Suppose you have a strategy involving $n$ rolls per iteration, with probability $p$ of succeeding each iteration (and probability $1 - p$ of needing to reroll). Then the expected number of rolls turns out to be $E = \frac{n}{p}$. So, comparing the strategies that have been suggested so far:

Travis's $4$ and $10$ strategy has $n = 2, p = \frac{39}{40}$, so $E = \frac{80}{39} \approx 2.05$.
Amitai's $20$ strategy has $n = 1, p = \frac{13}{20}$, so $E = \frac{20}{13} \approx 1.54$.
M. Wind's "all the dice" strategy has $n = 7, p = \frac{921596}{921600}$, so $E = \frac{6451200}{921596} \approx 7.00$.

Amitai's strategy is in fact the only possible strategy with an expected number of rolls of less than $2$ (so I take back what I said about it requiring more rolls than Travis's!), although one might go on to calculate and compare the variance of the number of rolls as well...

Try rolling a D10 and a D6 with the D6 values being:

1-2 = 1
3-4 = 2
5-6 = 3


Find five positive integers whose reciprocals sum to $1$ Find a positive integer solution $(x,y,z,a,b)$ for which

$$\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} + \frac{1}{a} + \frac{1}{b} = 1\;.$$

Is your answer the only solution? If so, show why.

I was surprised that a teacher would assign this kind of problem to a 5th grade child. (I'm a college student tutor) This girl goes to a private school in a wealthy neighborhood.

Please avoid the trivial $x=y=z=a=b=5$. Try looking for a solution where $ x \neq y \neq z \neq a \neq b$ or if not, look for one where one variable equals to another, but explain your reasoning. The girl was covering "unit fractions" in her class.

The perfect number $28=1+2+4+7+14$ provides a solution:

$$\frac1{28}+\frac1{14}+\frac17+\frac14+\frac12=\frac{1+2+4+7+14}{28}=1\;.$$

If they’ve been doing unit (or ‘Egyptian’) fractions, I’d expect some to see that since $\frac16+\frac13=\frac12$, $$\frac16+\frac16+\frac16+\frac16+\frac13=1$$ is a solution, though not a much more interesting one than the trivial solution. The choice of letters might well suggest the solution

$$\frac16+\frac16+\frac16+\frac14+\frac14\;.$$

A little playing around would show that $\frac14+\frac15=\frac9{20}$, which differs from $\frac12$ by just $\frac1{20}$; that yields the solution

$$\frac1{20}+\frac15+\frac14+\frac14+\frac14\;.$$

If I were the teacher, I’d hope that some kids would realize that since the average of the fractions is $\frac15$, in any non-trivial solution at least one denominator must be less than $5$, and at least one must be greater than $5$. Say that $x\le y\le z\le a\le b$. Clearly $x\ge 2$, so let’s try $x=2$. Then we need to solve

$$\frac1y+\frac1z+\frac1a+\frac1b=\frac12\;.$$

Now $y\ge 3$. Suppose that $y=3$; then $$\frac1z+\frac1a+\frac1b=\frac16\;.$$

Now $1,2$, and $3$ all divide $36$, and $\frac16=\frac6{36}$, so we can write

$$\frac1{36}+\frac1{18}+\frac1{12}=\frac{1+2+3}{36}=\frac6{36}=\frac16\;,$$

and we get another ‘nice’ solution,

$$\frac12+\frac13+\frac1{12}+\frac1{18}+\frac1{36}\;.$$


$L^p$-space isomorphisms Let $G$ be a compact hausdorff abelian group.
There is a Haar integral $\int : C(G) \rightarrow \mathbb{R}$ on $C(G)$, the $C^*$-algebra of continuous functions from $G$ to $\mathbb{R}$. We can complete this $C^*$-algebra in various ways to get $L^p(G)$, the completion of $C(G)$ under the norm $||f||_p = \sqrt[p]{\int_G |f|^p}$.
We have a map
$$C(G) \rightarrow C(\hat{G})$$
sending $f : G \rightarrow \mathbb{R}$ to the map $\hat{f} : G \rightarrow \mathbb{R}$ sending $\chi$ to $\int_G f \chi$.
My question is about completing on one side or another by various norms. For which $p$ and $q$ do we have a map
$$L^p(G) \rightarrow L^q(\hat{G})$$
and for which $p$ and $q$ is this an isomorphism?
I suspect that for $p$ and $q$ such that $1/p + 1/q = 1$, we have such a map, and that it is an isomorphism. But a reference would be nice.

Hint: If $1 < p < \infty$, $L^p$ is reflexive. Otherwise not.


Calculate limit of Floor $\Rightarrow (n^2+n+1)*[\frac {2019}{n^2+n+1}]$ I can't figure out how to solve it.
The sequence is :
$(n^2+n+1)*[\frac {2019}{n^2+n+1}]$
I tried using the formula $x-1\lt [x] \le x$ and find the limit using the squeeze theorem.
so I get $\frac {2019}{n^2+n+1} -1 \lt [\frac {2019}{n^2+n+1}] \le \frac {2019}{n^2+n+1}$ , multyipling each side by ($n^2+n+1$).
$2019-(n^2+n+1) \lt (n^2+n+1)[\frac {2019}{n^2+n+1}] \le 2019$.
well , $\lim_{n\to \infty} (2019-(n^2+n+1)) = -\infty $.
$\lim_{n\to \infty} 2019 = 2019$
No common limit between them , so I can't use the squeeze theorem.
Thank you

You say ceiling, but you use the definition of floor. Ok, so:
if you meant floor
The floor of $[\frac{2019}{n^2+n+1}]=0$ is identically zero for $n$ big enough (say $n>2019$). So your sequence looks like $a_1, \dots, a_{2019}, 0, 0, 0, 0, 0, \dots$
In particular, then, its limit for $n \to \infty$ is $0$.
if you meant ceiling
Here things are more interesting. Note that $n \leq [n]$. Now you have an indeterminate form because $[\frac{2019}{n^2+n+1}] =1$ but $[\lim_{n \to \infty} \frac{2019}{n^2+n+1}] = [0] = 0$.
What you can do is consider the following:
$$ (n^2+n+1) \frac{2019}{n^2+n+1} \leq (n^2+n+1) \left[\frac{2019}{n^2+n+1}\right]$$
which tells you that the limit cannot be zero, as the sequence is uniformly bigger than $2019$.
Now note that for big enough $n$ (just as before, take $n>2019$),
$$(n^2+n+1) \left[\frac{2019}{n^2+n+1}\right] = n^2+n+1$$
since the quantity inside the brackets is nonzero, but certainly smaller than $1$, so its ceiling is $1$. In particular then your sequence gets arbitrarily big: for any number $M>0$, there is $N$ such that for all $n>N$ you have that $a_n > M$, so the limit is $+\infty$.

Answer is Zero.
start with inside of brackets.as $n$ goes to infinity $\frac{2019}{n^2+n+1}$ goes to zero and is positive so its floor is absolute zero and we know that multiply of each number by absolute zero is zero.


Measuring $\pi$ with alternate distance metrics (p-norm). How/why does $\pi$ vary with different metrics in p-norms? Full question is below.

Background

Long ago I did an investigation on Taxicab Geometry using basic geometry. One think I recall is that a circle (as defined by all points equal distance from a centre point) 'looks' like a diamond. The 'circumference' of this circle is 8. As an extension I looked at other metrics of the form:

$$D_n\left((x_1,y_1),(x_2,y_2)\right)=(|x_2-x_1|^n+|y_2-y_2|^n)^\frac{1}{n}$$

(My limited reading of wikipedia suggests I should call this a p-norm.)

More recently using differing values of $n$ I calculated the 'circumference' of unit circles in these metrics. I took the definition of a unit circle to be all points a distance of one unit from the origin. This gave me a formula for a semi-circle:

$$y=\left(1-|x|^n\right)^\frac{1}{n}$$

I took the normal arc length formula of:

$$\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}$$

and replaced all the powers of $2$ with powers of $n$ to get:

$$\int_a^b\left(1+\left|\frac{dy}{dx}\right|^n\right)^\frac{1}{n}$$

Combining the circle with the arc length formula (and take a quarter circle and times it by 4 gave) the following integral:

$$4\int_0^1\left(1+\left|\frac{d}{dx}\left(1-x^n\right)^\frac{1}{n}\right|^n\right)^\frac{1}{n}dx$$

Then $\pi(n)$ is found by dividing 'circumference' by two (twice the radius). Doing so led to this graph of $\pi(n)$ against $n$.

Interestingly $n=2$ is a minimum (both local and absolute) making our commonly thought of value of $\pi$ special.

Question

(EDIT) Math Question: Is my distance formula for a different metric correct? (Moishe Cohen's comment suggests it might not be).

Math Question: Assuming the math above is ok, is there a reason for $(2,\pi)$ to be a minimum?

Math/Philosophy Question: Assuming above ok, is this why we observe the metric $D_2$ in the real world?

Note

I have not formally studied metrics, tensors or vector spaces or related topics (but am happy to do some light reading if your answer requires it).

Math Question: Assuming the math above is ok, is there a reason for (2,π)
  (2,π)
   to be a minimum?

The $p=2$ is the only p-norm that has the $SO(3)$ Lie Group structure. In other words, it is rotation invariant. Try it, you can rotate the coordinate system without changing the length. You can't do that with the Taxicab metric, the length you get will change.

The deep answer to your question is that only $p=2$ has a continuous symmetry, rotation. All the other p-norms have a either a finite number of symmetries or symmetries. Here's the thing, a p-unit circle is something like every possible vector that can generated from the group glued together. Imagine starting with a small vector then copying and rotating it just a bit, attaching it to the original vector, then repeating with the copy. Eventually, you'll generate the circle. You generate any of the p-norm circles using a similar process using their symmetry, albeit a bit more opaque. So why is $p=2$ special? The circle is made with the shortest vectors, and only the shortest vectors available to construct it. Why? Because they're all the same length! This isn't true unless you're rotationally invariant and guess what? Only $p=2$ is!!

  Math/Philosophy Question: Assuming above ok, is this why we observe the metric D2
  D2
   in the real world?

Quick run through physics. In physics, you have a theory described by a lagrangains that describes the amount of action particles do over a length of time. It's a scalar function of distance, speed, and time. The symmetries of the lagrangian correspond to conserved quantities according to Noether's Theorem. In standard classical mechanics, we assume three things,

(1) The action of a particle moving through some path along space is unchanged if you shift the start of the motion forward or backward in time. (Energy is Conserved)

(2) The action is unchanged if you shift the translate the start of the motion to another location. (Momentum is Conserved)

(3) The action is unchanged if you rotate the motion of the particle. (Angular Momentum is Conserved)

The last one isn't talked about to much, however, it's very important and highly relevant to our discussion. Since, our lagrangian, the theory for our physical theory, is invariant under rotation, there can never be a preferred direction in our theory. This means that means the metric we are using to measure distance and speed must also be rotation invariant. The $p=2$ norm is the only p-norm that'll work.

Pi is the ratio of the circumference to the diameter. Lp geometries start from a point, becomes a coss, eventually becomes a circle at p=2, becomes more square, and eventually becomes a spiky ball. These happen as we go from dimension 0 to dimension n. This thing in its n>2 is a squircle.

The circumference gets larger and larger while the diameter does not change. This means the value of the p-norm or pi changes as n changes.

I read "π is the Minimum Value for Pi" and a portion of the paper it referenced, "π Goes Full Circle." The diagram in minimum value paper does not reflect this evolution. The shape is not a square at infinity. The subsequent paper asserted a smooth function. That is likewise false.

There a lot of mentions in Quora on the sphere packing problem. Spiky balls interlace so the max packing is much denser than researchers expected.

When using these distance functions, we ignore the square roots beyond the taxicab geometry. That causes problems. It changes the continuity of the metric. Spiky balls are not continuous.

Thanks for bringing your question back to my attention.


Solving a recurrence relation. Check my answer I have a recurrence relation:

T(n)=4 for n<=2
T(n)=3T(n/3)+5 for n>2

So I began solving it,

T(n/3)=3T(n/9)+5
then
T(n)=3(3T(n/9)+5)+5
= 3^k T(n/3^k)+5k
letting k=log_3_n
= 3^log_3_n T(n/3^log_3_n)+5log_3_n
= nT(1)+5log_3_n
= 4n+5log_3_n

Is this the correct answer?
I am confused on the part where there is +5k. Is that part correct?

This is a generalized Josephus recurrence, of which an in-depth solution is given in Concrete Mathematics (by Graham, Knuth, Patashnik).

  If $f$ is defined as:

  $$\begin{align}f(j) &= \alpha_j & \text{for }1\le j\lt d\\
f(dn+j) &= cf(n)+\beta_j & \text{for } 0 \le j \lt d\text{ and } n\ge 1
\end{align}$$

  Then $f\left((b_mb_{m-1}\dots b_1b_0)_d\right) = (\alpha_{b_m}\beta_{b_{m-1}}\beta_{b_{m-2}}\dots\beta_{b_1}\beta_{b_0})_c$
  (where $(b_mb_{m-1}\dots b_1b_0)_d$ is the radix-$d$ representation of $n$).

Applied to your problem, we have $d = 3$, $\alpha_j = 4$, $c = 3$, and $\beta_j = 5$.

Thus, your solution is:
$$\begin{align}
f\left((b_mb_{m-1}\dots b_1b_0)_d\right) &= (\alpha_{b_m}\beta_{b_{m-1}}\beta_{b_{m-2}}\dots\beta_{b_1}\beta_{b_0})_c\\
&= \underbrace{(4\;5\;5\dots5\;5)_3}_{m+1\text{ digits}}\\
&= \underbrace{(5\;5\;5\dots5\;5)_3}_{m+1\text{ digits}} - 3^{m} \\
&= 5\cdot \underbrace{(1\;1\dots1\;1)_3}_{m+1\text{ digits}} - 3^{m} \\
&= 5\left(\frac{3^{m+1}-1}{2}\right) - 3^{m}
\end{align}$$

(Where $m+1$ is the number of digits in the base-$3$ representation of $n$.)

Yes, that is correct. I don't see any problems with your answer.


If a random variable converge in distribution then scalar multiplication also converges in distribution. Let c be a constant and suppose that $X_n\xrightarrow{d}X$, prove that
i)  $cX_n\xrightarrow{d}cX$
ii) $X_n+c\xrightarrow{d}X+c$
I've tried to use that:
$lim_{n\rightarrow\infty} F_{cX_n}(x) = lim_{n\rightarrow\infty} P(cX_n\leq x) = lim_{n\rightarrow\infty} P\left(X_n\leq \dfrac{x}{c}\right) = lim_{n\rightarrow\infty} F_{X_n}\left(\dfrac{x}{c}\right) $
But that's not exactly what I was looking for.

The sequence $X_n$ converges in distribution to $X$ as $n \rightarrow \infty$ (denoted $X_n \Rightarrow X$) if and only if, for any bounded continuous function $f$ (denoted $f \in C_b$), we have
$$\lim_{n\rightarrow\infty} E[f(X_n)] = E[f(X)].$$
Let $c \in \mathbb{R}$ be given. The maps $g_c^\times(x) = cx$ and $g_c^+(x) = x + c$ are continuous, so for any $f \in C_b$, $f\circ g_c^\times$ and $f\circ g_c^+$ are also in $C_b$. Therefore, because $X_n \Rightarrow X$ we have
$$\lim_{n\rightarrow\infty} E[f(X_n^{c,\times})] = \lim_{n\rightarrow\infty} E[(f\circ g_c^\times)(X_n)] = E[(f\circ g_c^\times)(X)] = E[f(X_c^\times)],$$
and ditto for $X_n^{c,+}$.

Remember that $X_n \xrightarrow{d} X$ it's by definition:
$\lim\limits_{n \rightarrow \infty} F_{X_n} (x) = F_X(x)$
Hence:
a)
\begin{equation*}
\begin{split}
\lim\limits_{n \rightarrow \infty} F_{cX_n} (x) & = \lim\limits_{n \rightarrow \infty} \mathbb{P} (cX_n \leq x)  \\
 & = \lim\limits_{n \rightarrow \infty} \mathbb{P} \left( X_n \leq \dfrac{x}{c} \right) \\
 & = \lim\limits_{n \rightarrow \infty} F_{X_n} \left(\dfrac{x}{c}\right) \\
 & =  F_{X} \left(\dfrac{x}{c}\right) \\
 & =  \mathbb{P} \left( X \leq \dfrac{x}{c} \right) \\
 & =  \mathbb{P} \left( cX \leq x \right) \\
 & =  F_{cX} \left({x}\right) \\
\end{split}
\end{equation*}
\begin{equation*}
\therefore \lim\limits_{n \rightarrow \infty} F_{cX_n} (x) =  F_{cX} \left({x}\right) \end{equation*}
\begin{equation*}
\therefore cX_n \xrightarrow{d} cX
\end{equation*}
Analogously:
b)
\begin{equation*}
\begin{split}
\lim\limits_{n \rightarrow \infty} F_{X_{n}+c} (x) & = \lim\limits_{n \rightarrow \infty} \mathbb{P} (X_n + c \leq x)  \\
 & = \lim\limits_{n \rightarrow \infty} \mathbb{P} \left( X_n \leq x-c \right) \\
 & = \lim\limits_{n \rightarrow \infty} F_{X_n} \left( x-c \right) \\
 & =  F_{X} \left( x-c \right) \\
 & =  \mathbb{P} \left( X \leq x-c \right) \\
 & =  \mathbb{P} \left( X+c \leq x \right) \\
  & =  F_{X+c} \left( x \right) \\
\end{split}
\end{equation*}
\begin{equation*}
\therefore \lim\limits_{n \rightarrow \infty} F_{X_{n}+c} (x) =  F_{X+c} \left({x}\right)
\end{equation*}
\begin{equation*}
\therefore X_n+c \xrightarrow{d} X+c
\end{equation*}


Algebra: What allows us to do the same thing to both sides of an equation? I understand that the expressions on both sides of an equal sign are the same entity, and I know that when you modify one side, the other must be changed because it is referring to the same thing. What I do not understand is why making a new equation (adding or taking away from an expression) allows one to know what an unknown represents. What about equations lets us do this?

Equations
Here's one way of looking at it: if $a=b$, then $f(a)=f(b)$, no matter the function $f(x)$.
From this point of view, solving an equation amounts to applying a sequence of functions in order to generate an equation whose solutions are easy to read off. What this sort of argument shows is that the new equations are logical consequences of the original equation.
So, for example, suppose you're given that $$x+1=2$$
and you want to subtract one from both sides. Then just apply the function $f(s)=s-1$, yielding $$x=f(x+1)=f(2)=1$$
Aren't we done? We've isolated $x$, so what more is there to do? The problem is that high school trains us to stop thinking once we get to this point. (Actually it trains us not to think about what we're doing at all and instead to rely on the process to do the thinking for us. But manipulating symbols, by itself, doesn't constitute a mathematical argument. Experts often omit the details, but that's because they know how to fill them in if they needed to; beginners should be taught how to fill in the logical details above and beyond the symbol manipulation.)
Indeed there are some subtleties here. First of all, just because this last equation is a consequence of the original equation doesn't mean the last implies the original. (That would amount to the very common mistake of thinking a conditional and its converse are logically equivalent.) In other words, for an arbitrary function $f(a)=f(b)$ need not imply that $a=b$: the operation you perform on both sides might not be reversible. (It was in the example I just gave because the function I applied was linear, and all (non-constant) linear functions have inverses that don't require domain restrictions, which makes the transformation "reversible." Unfortunately in school almost all the examples we start out with are linear, so we have our intuition about equation-solving trained on a very special set of examples, which don't illustrate what can happen in general.)
The failure of $f(a)=f(b)$ to imply $a=b$ explains why certain operations—for example, squaring both sides—might generate "extraneous solutions." I put quotation marks around that phrase, because it's something of a misnomer: they aren't actually solutions to the (original) equation, precisely because they're extraneous. So, for example, if you apply the function $f(s)=s^2$ to the equation $$x=1$$ you deduce that $x^2=1$. You could then apply the function $g(s)=\sqrt{s}$ to deduce that $|x|=1$. At this stage you could analyze the problem into cases (depending on whether $x$ is positive or negative) using the definition of absolute value and deduce that either $x=1$ or $x=-1$. But this doesn't mean that either answer is a solution to the original equation. (Obviously $x=-1$ doesn't satisfy the original equation!) That's because the second step, of squaring both sides, isn't reversible. The chain of implication doesn't flow all the way backwards.
Another subtlety is that applying a certain transformation to both sides may require you to make an assumption without even realizing it. In other words, some operations, such as dividing by $x$, tacitly carry certain restrictions. The function $f(s)=\frac{s}{x}$, for example, requires that $x\neq0$; otherwise the value of the function doesn't make sense. So if you have $$x^2=x$$ and you apply the function $f(s)=\frac{s}{x}$ to both sides, you're tacitly assuming that $x\neq0$. That's why in other cases you might lose solutions rather than generate extraneous ones.
Of course, not all equations have solutions. For example, applying $f(s)=s-x$ to the equation $$x=x+1$$
yields $0=1$. What this argument shows is that $$(\exists x)(x=x+1)\implies0=1$$
By contraposition we conclude that $$\lnot(\exists x)(x=x+1)$$
or in other words that there is no value of $x$ that satisfies the equation $x=x+1$, because assuming there is such a value leads us into a contradiction. (As this example illustrates, if we're being totally rigorous we should really pay attention to quantifiers. But that's more than you're asking.)
And some equations are, in fact, true for all values of the variables. Such equations are called identities. A silly example is
$$x=x$$
but a slightly more interesting example is
$$x^2-1=(x+1)(x-1)$$
If you try to solve identities like this one, you'll produce a tautology like $0=0$.
Inequalities
For what it's worth, you can also extend this idea to the logic of inequalities. If you apply a function $f(x)$ to the statement $a<b$, you'd typically like to conclude something like $f(a)<f(b)$ or $f(a)>f(b)$. In other words, you want to know whether the process preserves the direction of the inequality or reverses it.
But to draw such a conclusion, you generally need to know whether $f(x)$ is

increasing, i.e., $a<b\iff f(a)<f(b)$ or
decreasing, i.e., $a<b\iff f(a)>f(b)$

on the interval from $a$ to $b$. So, for example, $f(x)=x+2$ is always increasing, and $g(x)=-x$ is always decreasing, so applying $f$ to $a<b$ yields
$$a+2=f(a)<f(b)=b+2$$
but applying $g$ yields
$$-a=g(a)>g(b)=-b$$
This second fact is just what we mean when we say "multiplying or dividing both sides of an inequality by a negative number reverses the direction of the inequality." It's simply a consequence of the fact that the function $f(x)=-x$ is decreasing.
How about squaring both sides of an inequality? In that case we're dealing with the function $h(x)=x^2$. Well, $h(x)$ is decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$, so you have to be careful about "squaring both sides" of an inequality. If $a<b$ and $b<0$, then $h$ is decreasing on $(a,b)$, so
$$a^2=h(a)>h(b)=b^2$$
but if $a>0$ then the inequality is reversed.
Summary

The moral of this story is:
when you do something to both sides of an equation or inequality, think carefully about what function you're applying. In particular, for equations, think about whether

It's invertible on the same domain you start out with.

Its application requires you to make any assumptions.

For inequalities, you also want to think about whether

the function is increasing or decreasing on the interval defined by the inequality you're starting from.

Instead of thinking of terms of numbers and operations and math garblygook, it may help to talk a real example.  Say you're a parent with a couple of twins. You want to treat them equally.

So if you give one a cupcake, you have to give the other a cupcake.

If you smash one kid's cupcake, now you have to smash the other one.

If you cut one in half, same with the other, etc.

As far as how it relates to solving for an unknown, there's of course the assumption that the two sides are in fact equal to begin with... hence the '=' sign.

So in the case of our kids, if they run off and steal a bunch of jolly ranchers, but one of the kid hides some in a shoe box under his bed, you can deduce from the amount the other kid has how many he's hiding (like by taking away all of the ones you can see, and counting how many the other kid still has left).  In other words -- by doing "like" operations to both sides, you can eventually narrow down the problem into a plain "unknown = known".

Hopefully this is just another way to view math that helps with any confusion.   Symplectomorphic's answer is also very good and starts to address some more mind-bending concepts.


calculate surface normal using light Imagine you have a sheet of glitter. It can be thought of as a thousands of tiny flat mirrors with varying surface normals. These varying surface normals are what allows for the shimmering effect of light as you move. I am attempting to figure out what those normals are.

Using the reflection of light, could I determine the surface normal of a planar mirror (aka, a single piece of glitter)?

My current thought is to have the sheet of glitter fixed and a light shines on a particular piece of glitter. Then, place a piece of paper or something underneath the single piece of glitter to catch the reflection of light. By moving that piece of paper closer to the single piece of glitter, the reflected light moves. Does that movement help in determining the angle of reflection? Does that even help us at all in determining the surface normal?

I know this is a strange question, and I will happily elaborate on it if you wish. I also welcome and suggestions on different ways of trying to obtain the surface normals of the sheet of glitter.

Gentle

Interesting.  I was contemplating a similar question while staring at the reflection of the sun of a choppy ocean: could I estimate the choppiness as a function of position by taking some sort of average of the intensity of the reflection.

The literature on optics is full of terminology concerning this sort of problem. In satellite imaging one wishes to analyze materials using spectral information, for example.  This requires a host of information so that you can determine how much of the light that you sense from a particular point was attenuated by the atmosphere or dust, how bright the source was, etc.  There are databases full of data collected in laboratories about how much light is reflected as a function of wavelength, angle of incidence, angle of reflection, and other parameters.

Generally speaking, the maximum amount of light will be reflected when the angle between the light source and the normal is the same as that between the sensor and the normal.  So if you can figure out a way to discover the orientation that produces this maximum you've got a start.

Of course doing this once only provides a plane of possible normals.  You need to repeat with a second orientation of the sensor (unless the light source is collocated with the sensor).

If you cannot control the orientation of the glitter you could try measuring the returned light for a number of points and assume that the maximum observed value represents those pieces of glitter with the optimal geometry and then try to infer how much orientation change is associated with various lesser values to guess the orientations of the rest of the glitter.  You would then have a graph that relates intensity to the angle between source and sensor -- roughly.

Don't know if you are still working on this problem, but I (with my students) wrote this paper that answers this question:

"SparkleGeometry: Glitter Imaging for 3D Point Tracking"

https://www.cv-foundation.org/openaccess/content_cvpr_2016_workshops/w16/papers/Stylianou_SparkleGeometry_Glitter_Imaging_CVPR_2016_paper.pdf

and there is related work that solves a similar calibration problem:

"SparkleVision: Seeing the world through random specular microfacets"

https://arxiv.org/abs/1412.7884

Good luck!


Understanding successor sets Is there an intuitive way to understand them? Let's use $ \mathbb N$ as an example.  $2$ is defined as $1 \,{\cup}\,\{1\}$, but I fail to understand why is it so and how it connects with the intuitive meaning of $2$.

If you're trying to understand why the definition has to be that way, don't - there's no reason it has to be that way. But the definition you're talking about works out very nicely - it's one of the reasons Von Neumann got the big money:

Don't think of it in terms of $S(n)=n\cup\{n\}$, think of it as $n=\{0,1,\dots,n-1\}$.

We want to give definitions of $0,1,2\dots$ in terms of sets.

It seems reasonable that whatever set we choose to represent the natural number $n$, it should be a set with exactly $n$ elements.

So what should $0$ be? The only choice is $\emptyset$, since that's the only set with no elements.

Now what should $1$ be? It should be a set with one element. What should that element be? We could say $1=\{\{\{\emptyset\}\}\}$,  since that set has exactly one element. But that seems needlessly complicated. Why not take that one element to be the simplest set possible? So we define $1=\{\emptyset\}$.

And now we notice that hmm, $1=\{0\}$ because of our previous definition.

Now $2$ should be a set with two elements. We already have the sets $0$ and $1$ lying around, and it might be cool or convenient to have $1$ be a subset of $2$, because after all $2$ is supposed to be "larger" than $1$. So why not say $2=\{0,1\}$?

And now we notice that $1=0\cup\{0\}$ and $2=1\cup\{1\}$...

It really does work out very nicely. Consider another possibility. We could have said $0=\emptyset$, $1=\{\emptyset\}$, $2=\{\{\emptyset\}\}$, etc. Let's say those are the ugly naturals. First, they're just not as pretty. For example, $0$ has no elements, while $1,2,\dots$ all have exactly one element.

More important, there's no ntaural way to extrapolate from the ugly naturals to infinite ordinals (saying $\omega=\{\{\dots\emptyset\dots\}\}$  or in other words $\omega=\{\omega\}$  messes up our set theory). But with the Von Neumann construction, as Akiva Weinberger pointed out, it's perfectly natural to say that the next thing after all the natural numbers is $\omega=\{0,1,2,\dots\}$. Looks just like a natural number except bigger. And then $\omega+1=\omega\cup\{\omega\}$ gives the next one.

As the $n$th successor of $0$: $1 = s(0), 2 = s^2(0), 3 = s^3(0)$, where $s^n(x) = \overbrace{s(\cdots s}^n(x))$.


Number theory!Polynomial modules From Fermats theorem we know that for every $a \in \mathbb{Z}$, $$a^p\equiv a \mod{p}$$. But the polynomial $x^p$ it is not equal to the polynomial $x$( as a Congruence ). Why?Also when you want to solve a polynomial equation with modules you use that fermats theorem to simplify the polynomial.Doesnt that contradicts that  $x^p$ is not equal to $x$. SInce for every $a \in \mathbb{Z}$ $$a^p\equiv a \mod{p}$$

You have to make a distinction between (more concrete) functions from $\Bbb Z / p \Bbb Z$ to $\Bbb Z/p \Bbb Z$ and (more abstract) polynomials with coefficients in $\Bbb Z/ p \Bbb Z$.

For each polynomial, there is an associated polynomial function. And as you have just discovered, it is NOT true that if the polynomial function is zero, then the polynomial is also zero.

On any finite ring $R$, you can make the polynomial $\prod_{x \in R} (X - x)$, whose polynomial function is obviously zero. But this polynomial is not zero !

Do you know the difference between Congruence Equation and Algebraic equation which obeys Fundamental theorem of algebra unlike the previous one?

For example $$x^2\equiv2\pmod3$$ is not solvable unlike $$x^2=2$$

Again, $$a^p\equiv a\pmod p$$ is basically Identity Congruence Equation i.e., any integer satisfies this equation   just like $$\prod_{r=0}^{p-1}(x-r)\equiv0\pmod p$$


What does "wedge" mean? In Allen Hatcher's Algebraic Topology, $X\vee Y$ means the "wedge sum" of two (topological) spaces $X$ and $Y$. However, in $\LaTeX$, \wedge is the notation for $\wedge$, while $\vee$ is wirtten \vee. Where does this inconsistency come from?

A wedge refers to the shape of this object:

It doesn't matter if it's facing up or down, it's still a wedge. But since there was a need for two commands to produce $\wedge$ and $\vee$, and since $\vee$ conveniently looks like the letter V, it was probably decided that one should get the command "vee" and the other "wedge" (instead of something like "upwedge" and "downwedge", which would be harder to remember). But both are called "wedge" and it's up to the context to disambiguate.

The thing is that 'wedge' may also refer to the wedge product in the exterior algebra (or differential forms), which is also writen as $\wedge$.

I'm unaware of the real reason, you may try on TeX.SE, but coming from a geometrical background I believe this is how it should be, even though some topologists may disagree.


At least one of $|f(x)|$ and $|g(x)|$ not less than $a+1$ Let $a\in(0,1),f(x)=ax^3+(1-4a)x^2+(5a-1)x-5a+3$,$g(x)=(1-a)x^3-x^2+(2-a)x-3a-1 $.
Prove that:

For any real number $x$ ,at least one of $|f(x)|$  and $|g(x)|$ not less than $a+1$

since
$$f(x)+g(x)=x^3-4ax^2+(1+4a)x-8a+2$$
if we can prove
$$|f(x)+g(x)|\ge 2(a+1)$$?but can not find any other and can not prove whether there is any other solution or not.

HINT: $max\{f,g\}=\frac{f+g+|f-g|}{2}$
and the problem is wrong. counterexample:
$f(1)=3-3a\geq a+1 \Rightarrow a \leq {1\over2}$ and
$g(1)=1-5a\geq a+1 \Rightarrow a \leq 0$ so if $a>1/2$ both f(1) and g(1) less than a+1


Which point of the graph of $y=\sqrt{x}$ is closest to the point $(1,0)$? This problem was assigned for an AP Calculus AB class and was not allowed a calculator:

Which point of the graph of $y=\sqrt{x}$ is closest to the point $(1,0)$?

We are not given answers and the teacher will be absent for $2$ weeks. I need to check my answer $(1/2,1/4)$ before she returns.

Here is my work:

$d=\sqrt{(x-1)^2+(y-0)^2}$

$d=\sqrt{(x-1)^2+x}$

$d^2=(x-1)^2+x$

$2dd'=2(x-1)+1$

$d'=(2(x-1)+1)/(2d)$

$0=2(x-1)+1$

$x=1/2$

Let $(x, \sqrt{x})$ a point on the graph. The distance to$(1,0)$ is then given by

$d(x)=((x-1)^2+\sqrt{x}^2)^{1/2}=((x-1)^2+x)^{1/2}$.

Let $f(x):=d(x)^2.$

It is now your turn to determine $x_0 \ge 0$ such that $f(x_0)= \min\{f(x): x \ge 0 \}.$

(we get $x_0 =1/2$).

We then have $d(x_0)= \min\{d(x): x \ge 0 \}.$


What languages to learn for maths? For people with "hands-on" experience in mathematical research, what languages are the most beneficial to learn? I know that many graduate programs require some degree of non-native lingual proficiency, so what would be the pros and cons of pursuing various languages?

(I realize that this question is a bit broad; please ask if anything needs to be specified.)

Most US-based PhD programs have, as a requirement for the degree, at least having reading proficiency in one of: French, German, Latin, Russian or, increasingly, Chinese. The first few for obvious historical reasons. In fact, except for Chinese, the impetus is predominantly historical nowadays; internationally, most cutting-edge mathematics is being published in English.

French for sure. Opens up a whole wide world. Much more opportunities. Non-francophone countries accept to do business in English.


First year french college. Is there a finite set of points of the plane A. Finite set of point A such as... Is there a finite set of at least 3 points in the plane A, such as every 3 points of A are not aligned, and such as every circle going through 3 points of more of A, it’s center is in A ?
I believe that there can’t be such a set. So I tried showing that for every n point of the plane you can’t arrange them to be set A.
Being based on a number n I also tried doing it by doing a « mathematical induction » to show that there can’t be a set A for n points.
I’m a french student in my first year of college, I’m sorry by advance if my english is a bit wacky.
edit: forgot that it's a set of at least 3 points sorry

Answer to be proved in details
Such a finite set of points $A$ can't exist.
Denote by $P_1, P_2, P_3$ three points for which the circumscribed circle $\mathcal C$ is achieving the minimum radius $R$ of all circumscribed circles. Let $O$ be its center. By hypothesis $O \in A$.
Assumption to be proven: one of the triangles $(O \ P_1 \ P_2)$, $(O \ P_2 \ P_3)$ or $(O \ P_1 \ P_3)$ has a circumscribed radius strictly less than $R$. A contradiction.

In the plane draw a circle, take  three points on the circle (I think you can take any  finite number of points but 3 will do). Call these points your set A. These 3 points are not aligned (not on the same line) and there is only one circle that goes through them and we know by construction that this circle is in the plane.


$A$ be a subset of $[0,1]$ with non-empty interior ; then is it true that $\mathbb Q+A=\mathbb R$? Let $A$ be  a subset of $[0,1]$ with non-empty interior ; then is it true that $\mathbb Q+A=\mathbb R$ ?

Sure. If $A$ has non-empty interior, that means it contains a ball around some point $x \in [0,1]$ of radius $2 r$. Since there's always a rational $q$ less than $r$ away from $y-x$ for any real number $y$, $y$ will be in the ball centered at $q+x$ of radius $2r$ in $q + A$, which is contained in $\mathbb{Q}+A$.

ℝ=ℚ +ℚ̅(irrational numbers),what ever we choose A other than irrational numbers,it is not possible to cover all ℝ,
so ℚ+A will be the proper subset of ℝ.


Inequality for probability distribution Suppose that $\sum_{I=1}^{\infty}p_i=1$ and $p_i\geq 0$. Show that $\sum_{i}p_i(1-p_i)^n$ goes to $0$ as $n \rightarrow \infty$.

My idea: I tried to use the inequality $(1-p_i)^n\geq 1-np_i$ but I think we need to use an inequality in opposite side.

What you should notice is that each term converges to zero: $\lim_{n\to\infty} p_i(1-p_i)^n = 0$. So the question is, how to harness this observation to produce the same conclusion for the infinite sum $\sum_{i=1}^{\infty} p_i(1-p_i)^n$. A natural idea will be to interchange the order of limit and summation. However, we also know that this is not always possible, so you will end up seeking a mathematical result that ensures this.

As many users pointed out, either monotone convergence theorem or dominated convergence theorem fits in your situation. Although not mentioned in comments, even Weierstrass $M$-test can be used to justify this. And the common reason why all these fancy techniques are available is that we have a summable dominating sequence $\{p_i\}$ for $\{p_i(1-p_i)^n\}$.

As a final blow, let me provide a low-tech solution (which of course critically rely on the above observation) : Fix $N$ and take limsup as $n\to\infty$ to

$$ \sum_{i=1}^{\infty} p_i (1-p_i)^n \leq \left( \sum_{i=1}^{N} p_i (1-p_i)^n \right) + \left( \sum_{i=N+1}^{\infty} p_i \right). $$

Then using the subadditivity of limsup, we have

$$ \limsup_{n\to\infty} \sum_{i=1}^{\infty} p_i (1-p_i)^n \leq \sum_{i=N+1}^{\infty} p_i. $$

Since the limsup in the left is a fixed number which is independent of $N$, we can let $N\to\infty$ to both sided to obtain $\limsup_{n\to\infty} \sum_{i=1}^{\infty} p_i (1-p_i)^n \leq 0$. Since the complementary inequality for liminf is trivial, it follows that the sum converges to $0$ a $n\to\infty$.

Been a while, but I think this does it.

Let a = $\sum_{i}p_i(1-p_i)^n$ and b=$\sum_{i}(1-p_i)^n$.

and c= $\sum_{i}(1-p_i)$^(n+1).

b-a=c, so b=a+c.

$0<p_i$<1, so $p_i^k$<$p_i^m$ if k>m. We also have $p_i(1-p_i)^n$<$(1-p_i)^n$ since (1-$p_i$ ) is also positive.

Note c and b have the same necessarily finite limit, so we avoid indeterminate terms. a being equal to b-c, with c and b being finite and determinate, lim a = lim (b-c)=0.


Losing at Spider Solitaire Spider Solitaire has the property that sometimes none of the cards in the final deal can "go" and so you lose, regardless of how much progress you have made beforehand. You would have known that you would lose had you seen the final ten cards before the game started.

I wonder if we can calculate the probability of this happening.

To be clear, I want to find the probability that the final ten cards out of two packs of well-shuffled cards comprise cards no two of which are exactly one away from each other numerically (only the values matter, not the suits).

Note: there are several variants of Spider solitaire. I'm primarily interested in the standard 104-card, four-suit game.

There can be $2$ to $7$ different card values, and $k$ card values can occur in $\binom{14-k}k$ different value combinations. The number of ways of choosing $10$ cards with exactly $k$ different card values can be found using inclusion–exclusion: There are $\binom{8k}{10}$ ways to choose $10$ cards with values in a given set of $k$ values, so there are

$$
\sum_{j=2}^k(-1)^{k-j}\binom kj\binom{8j}{10}
$$

ways to choose $10$ cards with values forming exactly a given set of $k$ values. Thus the number of combinations is

$$
\sum_{k=2}^7\binom{14-k}k\sum_{j=2}^k(-1)^{k-j}\binom kj\binom{8j}{10}=153020720928
$$

(computation) out of a total of $\binom{104}{10}=26100986351440$, so the probability for this to occur is

$$
\frac{153020720928}{26100986351440}=\frac{9563795058}{1631311646965}\approx0.00586264
$$

in agreement with Jonathan's simulations.

I don't believe your assumption is correct. If, in a final (or any other) deal, there is no single card that can be moved to another column, it may still be moved together with one or more cards that are beneath it. So the expert player will plan to leave one or more cards of the same suit and increasing order in one of the columns, to then move them together to some matching other column.


What is the link between homomorphisms and mutual information? Intuitively, there seems to be a link between the (kind of) homomorphism between two algebraic structures and the mutual information between two variables. However, since I'm not a mathematician, it's hard to see or formalize the exact connection. Anyone who can help me with this?

For example, if we'd take each variable to represent a set and its order relation, could we say that the very existence of a homomorphic map between the two would make their $ MI \gt 0 $ ?

And the other way around: would an $ MI = 0 $ between those variables make the map between them a random one?

Let's look at two examples.

In the case of vector spaces we have linear maps, or vector space homomorphisms. These preserve scalar products (1) and are linear(2).

Let $f: V \to W$ be a linear map.

$$f\big(\lambda x \in V \big) = \lambda f(x) \in W\quad (1)$$
$$f\big(x+y \in V\big) = f(x) + f(y) \in W \quad (2)$$

So we see that the operation of addition and scalar multiplication is preserved between $V$ and $W$.

In the case of groups, we have group homomorphisms. Similarly these preserve group operations.

Let $g: A \to B$ be a group homomorphism and $*$ the group operation.

$$g\big(a_1 * a_2 \in A\big) = g(a_1) * g(a_2) \in B.$$

So, in a sense, we can say the "mutual information" of the operations is preserved when a homomorphism maps an element from one object to another.


"Pathfinding" in a Category Suppose you are given the "action" of a morphism, some specification of what it does to elements; in $\textbf{Set}$ or related categories it would be the graph of a function. Is it possible to determine, in theory or practice, (whether there exists) a set of arrows whose composition matches the given action? If there are many such sets, is it possible to determine the one(s) requiring minimal compositions?
My intuition, stemming from the digraph axiomatization of a category, suggests that this is a simple example of pathfinding. However, most categories not only have infinite or large object sets, but infinite or large hom-sets. It's not clear how the standard algorithms would work on an infinite graph, let alone a graph with infinitely many edges between each vertex. I couldn't find any information on generalizations of pathfinding to such cases. However, it seems like this would be a common enough question in the subject to have been asked before. Is there something I'm missing?

There isn't a universal answer to this. Many problems (NP problems) could be easily written in such a formulation. For example : Hamiltonian cycle.

Your category is the set of nodes you've visited +
the node you're actually in. Your action is to move to a neighboring
vertex and mark it as visited. You have an Hamiltonian cycle iff the
shortest path from $\emptyset \times {x}$ to $V \times {x}$ is of
length exactly $|V|$)

So each answer will have to be designed specifically to the problem at hand, with the help of the geometry of the problem.
But this isn't a lost cause ! The edit distances (Levenshtein distance, tree edit distance, time-warp edit distance, etc) are generally defined as pathfinding distances on huge sets, but computed with dynamic algorithm that run in polynomial time of the input size.
If you can't find an "analytic" solution, maybe a pathfinding algorithm like A* can help you  find a solution faster. The idea is to have a lower-bound heuristic ("the path from X to Y is at least n"), and use this kind of bound to orient your search. This could help you explore the neighbors in an interesting order and not explore a huge part of the category.


Showing that meromorphic functions form a field (without using Laurent series) Thinking of meromorphic functions on a Riemann surface $X$ as holomorphic maps $X → \hat{ℂ}$, where $\hat{ℂ}$ is the Riemann sphere, how can one show that the set of meromorphic functions $\mathcal{M}(X)$ forms a field? I want to avoid Laurent series.

I tried to define addition and multiplication in the following way:
\begin{align}
f + g&\colon X → \hat{ℂ},\; z ↦ \lim_{w → z} (f(w) + g(w))\\
f · g&\colon X → \hat{ℂ},\; z ↦ \lim_{w → z} (f(w)·g(w))
\end{align}
I have to show that this is well-defined, then the assertion follows easily.
By the identity theorem, the poles of $f$ and $g$ are discrete and closed in $X$, so one can indeed look at the sums and products of $f$ and $g$ in a punctured neighbourhood of $z$.

So I have to establish that those limits exist.
I also know $\lvert h (w)\rvert → ∞ ⇔ h (w) → ∞$ which I then can apply to $h = f + g$ and $h = f · g$. This is where I get stuck.

I’m also open to different ways of proving that meromorphic functions form a field. For example one can prov it for the unit disk first and then lift it to $X$ by the use of charts. I’m also interested in plausible arguments to convince me, that Laurent series or lifting are the way to go.

One way is to start with a slightly different definition of a meromorphic function on $X$: namely, it is a holomorphic function $f$ on an open subset $Y$ of $X$ such that for each $x_0 \in X \setminus Y$ there is an open coordinate neighborhood* $U$ of $x_0$ such that $U \setminus \{x_0\} \subset Y$ and there is some positive integer $k$ such that
$x^k f$ extends to be holomorphic on $U$.  This gives a meromorphic function on your sense in an evident way, but it is easier to add, subtract, multiply and invert meromorphic functions as defined above.

*: Here I mean that there is a biholomorphic map from $U$ to the open unit disk in the  complex plane which carries $x_0$ to the origin.  We write $x$ for the pulback of the identity function $z \mapsto z$ under this map, so that $x$ has a simple zero at $x_0$.

Use another definition of meromorphic:

A function is meromorphic if it is analytical except for an isolated set of points.
This way things are quite easy now.

E.g. $f,g$ meromorphic with exceptional sets $F,G$ which consist of isolated points. $\Rightarrow f\cdot g$ meromorphic, since
$f\cdot g$ is analytical except on the set $F\cup G$ which also consists of isolated points, since we did no infinite union.
Etc. ...

So do not care so much about the function values/limits, work with the poles.


Transcendence degree for a $k$-algebra which is an integral domain Let $R$ be an integral domain over a field $k$.
Is it true, that $\deg.\mathrm{tr}_k \ \mathrm{Frac}(R)$ is the greatest
number of elements of $R$ algebraically independent over $k$?

Every transcendence basis of an algebra is a transcendence basis of its quotient field.
Because the set of algebraic elements over subfield is a subfield, and the lowest subfield containing algebra is its quotient field.


I need some clarification regarding functions. If $f(x) = 3 + 5$ and $() = 4^2 − 2$, find:

$(g(3))$
$g(3)$

Would you get the same answer for one and two? I think I read or heard somewhere that it doesn't matter whether you put a function within brackets or not.
I want to be absolutely sure because I'm not exactly a math genius.

It's a little ambiguous to write $fg(3)$, because it could either mean $$(f\circ g)(3) = f(g(3))$$ or it could be a sloppy way of writing $$(fg)(3)= f(3)\cdot g(3),$$ and these two things are not generally the same:
$$
(fg)(3) = f(g(3)) = f(30) = 95$$
whereas
$$f(3)\cdot g(3) = 14\cdot 30 = 420$$
If this is given in an assignment it's probably best to ask your instructor what they mean.  If you are looking for a shorter way of writing $f(g(3))$ you should go with $(f\circ g)(3)$


difference between implicit and explicit solutions? What is difference between implicit and explicit solution of an initial value problem?
Please explain with example both solutions(implicit and explicit)of same initial value problem? Or without example but in some way that is understandable.

thanks

As requested:

Let's use the example initial-value problem

$$y^\prime y=-x,\qquad y(0)=r, \qquad r\text{ constant}$$

One can derive both an implicit and explicit solution for this DE. The implicit solution to this DE is

$$x^2+y(x)^2=r^2$$

This solution implicitly defines $y(x)$; all we have here is an equation involving $y(x)$. On the other hand, the explicit solution looks like

$$y(x)=\pm\sqrt{r^2-x^2}$$

and in this case, $y(x)$ is explicitly defined: $y(x)$ is expressed here as an explicit function with $x$ as the only independent variable.

We aren't always this lucky when we solve differential equations that show up in practice. It often happens that we can only be content with an implicit solution (or a parametric solution, which is a somewhat better state of affairs than having just an implicit solution). One famous example is the differential equation that pops up in the brachistochrone problem:

$$(1+(y^\prime)^2)y=r^2$$

Let us consider a differential equation
$$
x +yy' =0 \label{1}\tag{1}
$$
Now, the relation $x²+y² -25 =0$ is an implicit solution of the above equation for all $x\in(-5,5)$. This relation define two real functions $f_1(x) =\sqrt{25 -x^2}$ and $f_2(x) = -\sqrt{25 -x²}$ in the interval $x\in(-5,5)$: here $f_1,f_2$ are explicit solutions of the differential equation.

Now is it quite easy to understand that what are implicit and explicit solutions?

  Implicit solution means a solution in which dependent variable is not separated and explicit means dependent variable is separated.

Now consider the relation
$$
x² +y² +25 =0
$$
Is it also an implicit solution of the differential equation \eqref{1}?
The answer is 'No': formally it is solution of the equation \eqref{1} but not implicitly because it won't be identically zero for any real values of $x,y$. Precisely If you write it in the form
$$
y  = -\sqrt{-25 -x^2}\:\text{}.
$$
you will get always an imaginary $y$ for all real $x$.


Can the axiom of choice be used to define the real numbers? I realize this is possibly a copy of another question (The Axiom of Choice and definability) but I would appreciate an explanation with less set theoretical explanations if that is at all possible.

I recently took the opportunity to look up the axiom of choice (AC). Wikipedia has both a formal and an informal definition.  Based on my shortcomings in set theory I have to rely on the informal version: “given any collection of bins, each containing at least one object, it is possible to make a selection of exactly one object from each bin”.

It seems that this can be used to define an endless series of decimals such as an arbitrary number x between 0 and 1, in term corresponding to an arbitrary positive real number through such a function as f(x)=1/x-1, i.e. it can be used to define any real number (assuming a real number is an infinite decimal series).

A beauty spot is the cases of 0.99999… and 1.00000… - discussed in a number of questions on the site. It points to several issues:

I argue, albeit in a roundabout way, that for a choice involving the AC to have any value then the number received through the choice must also allow for existing mathematical operations, i.e. 2*0.11111…=0.22222…  as discussed in another question. (“You're also assuming that you can multiply 0.9999... by 10 and get 9.9999... (or, rather, that arithmetic with infinite decimals works normally), which is not at all unreasonable to assume. – Isaac Jul 20 '10 at 20:59”)
i)  0.99999…must be equal to 1.00000… for the same reason.
ii) The fact that 0.99999…=1.00000…can challenge the one-to-one correlation between the real numbers, as we view as being without these doublets. However it can be disregarded as

I was assured in a recent question, in that removing a countable set such as 0,99999…, between the endless decimal series and 0.099999…and 0.0099999…  etc. from a uncountable set would not change the one-to-one correlation to numbers without this anomaly. (I was referred to i.a. the Cantor–Bernstein–Schroeder theorem).

Reasoning in this manner I perceive the AC to be identical of postulating - or at least allowing for the definition of – real numbers. My question is whether or not this is the case.

Edit much later: I should add that I primarily am looking at the problems of defing invividual real numbers, so called pointwise definition, rather than definition of the whole set of reals. I didn't realize at the time that these are two different issues.

The axiom of choice has nothing to do with the definition of the real numbers.

We define the real numbers either as Dedekind cuts over the rational numbers, which is simply a pair $(A,B)$ of subsets of the rational numbers which have some properties. Or we can define an equivalence relation on sequences of rational numbers and consider the equivalence classes as the real numbers themselves.

Then we can prove, again without any appeal to the axiom of choice, all sort of things about the real numbers. We can prove basic mathematical facts like convergence or divergence of some geometric series, the definition of $e$ and $\pi$, and $\sqrt2$. We can even prove that none of these are rational.

Without the axiom of choice we can even prove the Cantor-Bernstein theorem, and we can even prove that every set of real numbers whose complement is countable, has the same cardinality as the real numbers.

No, so far the axiom of choice isn't even playing a role. It does play a role if you want to prove other properties of the real numbers, such as "the real numbers are not the countable union of countable sets" (which is not to say they are countable!). But that doesn't seem to be what you're asking specifically.

It seems to me that you need the Axiom of Choice in order to get most of the Cauchy sequences whose equivalence classes constitute the real numbers.

Consider the set of all rational sequences. There are many which can be defined using a formula, i.e. an algorithm. For example $(1, 1, 1, 1, 1, \dots)$ has the formula $x_n = 1 \ \forall n$. Similarly you can define a rational sequence whose squares converge to $2$ (and thus you get a rational Cauchy sequence whose equivalence class will be named $\sqrt 2$) and thus you have defined that particular real number. It seems to me you can do that to construct all roots of polynomials (i.e. algebraic numbers), but what about all the transcendental numbers? $\Bbb e$ is easy enough. But what about just any old real number which lies somewhere on the number line. How do you define a sequence that converges to it? How can you specify the terms of such a sequence? It seems that you need the Axiom of Choice to do this. Because you're choosing an infinite number of elements from $\Bbb Q$, without any algorithm / formula.

The point is that to say that infinite sequences (other than ones that can be constructed algorithmically) exist requires the Axiom of Choice.

It's all very well to say you don't need AoC for constructing the reals because you just need cauchy sequences  - but the Cauchy sequences themselves (most of them - i.e. the ones that can't be specified by using some algorithm) need the Axiom of Choice.


Grade 8 simple algebra equation help I find this question hard, please help.
It is given that $x+\frac{1}{x}=3$
and $x^2+\frac{1}{x^2}=7$.

Please find the value of $x^3+\frac{1}{x^3}$.
Please show the steps.


Are there general methods that can be applied when using the Borel-Cantelli Lemma, to get a statement about a sequence of random variables? I hope the title in itself is clear, if not allow me to give an example.

In Class my Professor did the following:

  Given a sequence $(X_n)_{n \in \mathbb{N}}$ of non-negative i.i.d. RV $X_n \sim X$ with $E(X)< \infty$ then $\frac{X_n}{n} \to 0 $ almost surely

To prove this, it is easy to show that $\sum P (X_n/n > \epsilon) < \infty$ and thus conclude by Borel-Cantelli Lemma that $$P ( \limsup \lbrace X_n/n > \epsilon \rbrace ) =0 \iff P ( \lim \inf \lbrace X_n /n \leq \epsilon \rbrace )=1 $$
But one indeed looks for the answer that $\exists \Lambda \subset \Omega$ with $P( \Lambda)=1$ and for all realizations $\omega \in \Lambda$ we have that $X_n(\omega)/n \to 0$.

In the question Is it correct to say that ($\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) \supseteq \limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}$? Daniel Fischer gives an answer on how to get an inclusion of the form $\liminf \lbrace ... \rbrace \subset \lbrace \limsup ... \rbrace$.   I do believe that he makes use of the (complementary) statement which says that for $\epsilon >0$ $$ \lbrace \lim \sup | X_n - X| > \epsilon \rbrace \subset \lim \sup \lbrace | X_n -X | \geq \epsilon \rbrace \subset \lbrace \lim \sup |X_n - X | \geq \epsilon \rbrace \tag{*} $$
which gives $$ \lbrace \lim \sup |X_n -X | < \epsilon \rbrace \subset \lim \inf \lbrace | X_n - X | < \epsilon \rbrace \subset \lbrace \lim \sup |X_n -X | \leq \epsilon \rbrace $$

My question(s): 1) Are there more chains of inclusions as in (*) that I should know in order to give statements about the $\limsup$ or $\liminf$ of sequence of Random Variables rather than the $\limsup,  \liminf$ of the sequence of Events?

2) In the above chain of inclusions, does the $\epsilon >0$ actually depend on $k$? The Definition gives $$\limsup \lbrace X_n/n > \epsilon \rbrace = \bigcap_{n=0}^\infty \bigcup_{k=n} A_k \text{ where } A_k:= \lbrace X_k/k > \epsilon \rbrace $$ so intuitively I would say that $\epsilon=\epsilon(k)$ varies as $k$ increases.

3) Are there any general techniques you'd recommend to make a statement about the limit of a sequence of random variables using Borel-Cantelli Lemmas?

Q1,3

Williams - Probability with Martingales

Deduced similarly:

(iii) If $\liminf x_n > z$, then

$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)

(iv) If $\liminf x_n < z $, then

$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)

Q2

Please elaborate. $\varepsilon$ is supposed to be an arbitrary given. I think $k$ depends on $\epsilon$. For example, $k=5$ might work for $\epsilon = 0.5$ but $k=5$ might not work for $\epsilon = 0.01$ so $k$ would have to instead be say $10$


Existence of countable subgroup How can one show that there exists a countable subgroup $G$ of $SO(2)$?

(where $SO(2) =$ {$ M \in \mathbb R^{2 \times 2} | M^TM = Id, det(M) = 1$})

Consider a matrix in ${\rm SO}(2)$ it can be written (uniquely for some $0\leq \theta < 2\pi$) as

$$\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}.$$
This is of finite order if and only if $\theta$ is rational multiple of $2\pi$.

(1) You can choose $\theta$ which is not rational multiple of $2\pi$; order of above matrix will be infinite, so subgroup generated by it is one countable (infinite) subgroup.

(2) Or, you can collect all matrices of above type for all $\theta$ that are rational multiples of $2\pi$; this is certainly a countable subgroup.


Suppose that $P(X)=a$ and $P(Y)=b$, what is $P(X-Y)$? Suppose that we have:
 $$P(X)=a ~and ~ P(Y)=b$$
Now the question is : what is $P(X-Y)$? suppose that X and Y are independent.

$$\newcommand{\d}[1]{{\rm #1}}
\d P(\d A-\d B)=\d P(\d A)-\d P(\d A\d B)=\d P(\d A)-\d P(\d A)\d P(\d B)=a-ab$$
Since for independant events: $\d P(\d A\d B)=\d P(\d A)\d P(\d B)$

Assuming X and Y to be two sets of favourable events....

$P(X) = a$
$P(Y)= b$

We are given that X and Y are independent.

Now,

X-Y defines the set of favourable events which belongs to the set X but does not belong to the set Y.

So, we can write $X-Y = X\cap Y^c$

;where $Y^c$ defines the compliment of Y.

We already know from the question that $X$ and $Y$ are independent.

Also, without loss of generality $Y$ and $Y^c$ are also independent.

This leads us to conclude that $X$ and $Y^c$ are also independent.

So,

$P(X-Y) = P(X \cap Y^c) = P(X) * P(Y^c)= P(X)*(1- P(Y))=a(1-b) $

The result is just as everyone before me has concluded.

QED


Solve $\int_{-1}^1 f(x,t) \, dx=\log \left(\frac{t+1}{1-t}\right)$ Is it possible to find a function $f(x,t)$ so that
$$\int_{-1}^1 f(x,t) \, dx=\log \left(\frac{t+1}{1-t}\right)\text{?}$$
I tried the Laplace transform, but it did not work.

Easy!$$
\int_{-1}^1 \frac{1}{2}\log\left(\frac{t+1}{1-t}\right)dx = \log\left(\frac{t+1}{1-t}\right)
$$


Evaluating $\int_0^\infty e^{-x}\cos(x)dx$ By integrating over the contour around an appropriate sector, how does one solve  $$\int_0^\infty e^{-x}\cos(x)dx$$

Consider the integral
\begin{align}
\int_{C} e^{-|z|} \, e^{iz} dz
\end{align}
with a contour enclosing no poles.


Integrating powers of linear and quadratic functions How can I integrate function such as $(x+9)^3$? I obviously know that I can expand the function and integrate it normally. However, that is possible and feasible only as it is of third degree. What if the function is more like: $(x^2-9x+5)^7$? How could I integrate this function?

Setting $u = x+9 \implies du = \,dx$

That gives us $$\int (x+9)^3\,dx = \int u^3 \,du = \frac {u^4}4 + C = \frac 14(x+9)^4 + C$$

In your second example, we have a quadratic raised to a power, so the same method may not work. If you have $$\int x(x^2-9x+5)^7\,dx$$ then it will work since $x\,dx$ can be multiplied by a constant to give us $(x^2 - 9x + 5)' = 2x$.

With quadratics (raised to a power) for which you can complete the square, you can use trigonometric substitution. For example, $$(x^2 + 4x + 13) = (x^2 + 4x + 4 + 9) = (x+2)^2 + 3^2$$ Then the substitution $x+2 = 3 \tan \theta$ works well.

Note that for any function $f(x)$, $$\int f'(x) \cdot (f(x))^n \,dx = \dfrac{(f(x))^{n+1}}{n+1} + C$$


Expected number of rolls to get all sixes I'm struggling with the following problem:

  I have $N$ balanced 6-sided dice. I roll the dice simultaneously, and remove any sixes that occur. I roll the remaining dice again, and remove any more sixes. I repeat the process until there are no dice remaining. What is the expected number of rolls this will take?

So far I have calculated that by the $n^{th}$ dice roll there will be $N\left(\frac56\right)^n$ dice remaining: If we start with $N$ dice, on the first roll we expect $\frac{N}{6}$ sixes. Therefore, on the second roll we expect to have $N-\frac{N}{6}$ dice and hence expect $\frac{5N}{36}$ sixes, meaning after the second roll there are $\frac{25}{36}N$ dice remaining.  Repeating this calculation gives the sequence $N, \frac{25}{36}N, \frac{125}{216}N...$ which is equal to $N\left(\frac56\right)^n$ where $n$ is the roll number. I'm struggling with the next part: My thinking is we must find the expected number of rolls $n$ such that $N\left(\frac56\right)^n\lt0.5$, and therefore $n>\frac{\ln(\frac{0.5}{N})}{\ln(\frac56)}$. Taking $N$ to be $8$, the expected number of rolls is then about $15.21$. I used MATLAB to run the experiment $200,000$ times, and it gave me an average number of rolls of $15.4$. I seem to be close to the right answer, but I'm not sure what I've done wrong. What is the solution?

The probability a particular die has not shown a $6$ in the first $k$ rolls is $\left(\frac{5}{6}\right)^k$
The probability a particular die has shown a $6$ in the first $k$ rolls is $1- \left(\frac{5}{6}\right)^k$
The probability all $N$ dice have shown a $6$ in the first $k$ rolls is $\left(1- \left(\frac{5}{6}\right)^k\right)^N$
The probability all $N$ dice have shown a $6$ in $k$ rolls but not in $k-1$ rolls is $\left(1- \left(\frac{5}{6}\right)^k\right)^N - \left(1- \left(\frac{5}{6}\right)^{k-1}\right)^{N}$

This will give a mean of $$\sum_{k=0}^{\infty} \left(1- \left(1- \left(\frac{5}{6}\right)^k\right)^N\right)$$

For $N=8$ this seems to suggest a mean of about $15.40694347788$, close to your simulation result

The responses to this question seem to have a murkiness due to talking
about non-integer numbers of dice and defining when you have reached Zero
dice left.
  So I went for the results experimentally by having a (Pelles) C program do a million simulations, recording the number of total tosses it took for all N dice to show 6, and dividing by the number of trials.

The Code snippet, and the results for 1 to 100 dice is below. The results agree with the calculated results for small numbers of dice,  and shows the slow increase in tosses as the number of dice increases.

------------------------- C code for dice tossing experiment  -----------

NumTrials=100000;
for (N=1;N<=100;N++)
{

             printf(" # of Dice = %d\n ", N );

  for (i=0;i<10; i++) Counts[i]=0;

for (Trials=0;Trials < NumTrials; Trials++)
  {
   //Trial loop

  NewNum=N;    //start with N dice
    NewTrial=1;

  while (NewNum>0)  // trial is over when all N dice have hit 6.
  {
        next=NewNum;

         for(i=0;i<next;i++)
          {

           r=rand()%6;    // result of the toss is a random number mod 6, i.e.
                                  // a random number from 0 to 5
           if (r==0)NewNum--;   //reduce the number of remaining dice by 1whenever
                              // a dice comes up 6.  (i.e. 0 mod 6

           Counts[r+1]++;    // check the random numb gen

          }

           Counts[0]++;         // Record that another toss has happened.

    }//end of 'while next>0.  on to next trial

 }  // end of for Trial= .. loop

  Avg=(double)Counts[0]/NumTrials;

  fprintf(fp," N= %d, avg= %.2f \n", N, Avg);

-----------------------  RESULTS   -----------------------
C:\Pelles C Projects\rolling 6's.dat

trials = 200000

N= 1, avg= 6.01
 N= 2, avg= 8.73
 N= 3, avg= 10.56
 N= 4, avg= 11.93
 N= 5, avg= 13.02
 N= 6, avg= 13.94
 N= 7, avg= 14.72
 N= 8, avg= 15.39
 N= 9, avg= 16.01
 N= 10, avg= 16.57
 N= 11, avg= 17.07
 N= 12, avg= 17.52
 N= 13, avg= 17.95
 N= 14, avg= 18.35
 N= 15, avg= 18.70
 N= 16, avg= 19.06
 N= 17, avg= 19.37
 N= 18, avg= 19.66
 N= 19, avg= 19.95
 N= 20, avg= 20.23
 N= 21, avg= 20.50
 N= 22, avg= 20.76
 N= 23, avg= 20.98
 N= 24, avg= 21.21
 N= 25, avg= 21.42
 N= 26, avg= 21.66
 N= 27, avg= 21.88
 N= 28, avg= 22.01
 N= 29, avg= 22.22
 N= 30, avg= 22.41
 N= 31, avg= 22.59
 N= 32, avg= 22.72
 N= 33, avg= 22.90
 N= 34, avg= 23.09
 N= 35, avg= 23.23
 N= 36, avg= 23.39
 N= 37, avg= 23.56
 N= 38, avg= 23.67
 N= 39, avg= 23.83
 N= 40, avg= 24.00
 N= 41, avg= 24.11
 N= 42, avg= 24.22
 N= 43, avg= 24.37
 N= 44, avg= 24.49
 N= 45, avg= 24.62
 N= 46, avg= 24.72
 N= 47, avg= 24.86
 N= 48, avg= 24.93
 N= 49, avg= 25.07
 N= 50, avg= 25.17
 N= 51, avg= 25.30
 N= 52, avg= 25.37
 N= 53, avg= 25.50
 N= 54, avg= 25.60
 N= 55, avg= 25.67
 N= 56, avg= 25.79
 N= 57, avg= 25.89
 N= 58, avg= 25.98
 N= 59, avg= 26.08
 N= 60, avg= 26.17
 N= 61, avg= 26.26
 N= 62, avg= 26.35
 N= 63, avg= 26.47
 N= 64, avg= 26.53
 N= 65, avg= 26.61
 N= 66, avg= 26.68
 N= 67, avg= 26.76
 N= 68, avg= 26.86
 N= 69, avg= 26.94
 N= 70, avg= 27.02
 N= 71, avg= 27.09
 N= 72, avg= 27.14
 N= 73, avg= 27.26
 N= 74, avg= 27.31
 N= 75, avg= 27.41
 N= 76, avg= 27.46
 N= 77, avg= 27.53
 N= 78, avg= 27.60
 N= 79, avg= 27.68
 N= 80, avg= 27.71
 N= 81, avg= 27.80
 N= 82, avg= 27.84
 N= 83, avg= 27.97
 N= 84, avg= 28.03
 N= 85, avg= 28.08
 N= 86, avg= 28.13
 N= 87, avg= 28.21
 N= 88, avg= 28.25
 N= 89, avg= 28.32
 N= 90, avg= 28.39
 N= 91, avg= 28.43
 N= 92, avg= 28.51
 N= 93, avg= 28.55
 N= 94, avg= 28.62
 N= 95, avg= 28.68
 N= 96, avg= 28.72
 N= 97, avg= 28.79
 N= 98, avg= 28.85
 N= 99, avg= 28.88
 N= 100, avg= 28.93


Comparing sets of possible values Say $x$ is a real number, can one argue that the set of possible values for $x$ when $x < 2$ is less than when $x < 4$? Is this mathematically true?

One could argue that the set of possible values in the latter case contains that of the former case, plus other values. But, we are comparing infinite values.

It depends on your notion of size of a subset. If we partially order subsets of the real number using the subset relation then yes in a sense i.e. $\{x\in\mathbb{R}\mid x<2\}\subset \{x\in\mathbb{R}\mid x<4\}$.

If we compare the sets in terms of cardinality, then no. They have the same cardinality. Let $A=\{x\in\mathbb{R}\mid x<2\}$ and $B=\{x\in\mathbb{R}\mid x<4\}$. The map $f\colon A\to B$ given by $f(x)=x+2$ is a bijection.

If we compare using Lebesgue measure, then no. Both sets have measure equal to $\infty$.

mathematicaly a real line R is isomorphic with the interval $(0,1)$. I mean they have the same cardinality. can you imagine that??


It's given differential equation. How do I determine the solution curve of the begining values $x(0) = 0$,$p(0) = 1$ in the $(x, p)$ plane? It's given differential equation

$\dot{x}=p$

$\dot{p}=-x^3+x$

How do I determine the solution curve of the begining values $x(0) = 0$,$p(0) = 1$ in the $(x, p)$ plane?

As
$$
0=\dot x\dot p - \dot p\dot x=p\dot p+(x^3-x)\dot x
$$
solutions satisfy
$$
2p^2+(x^2-1)^2=C
$$
and for the given initial values $C=2·1^2+(0-1)^2=3$.


Problem with an Invertible matrix Good afternoon,

I need your help please in this question, and I'm sorry if my question is evident;

Let $X, Y\in M_n(\mathbb{R})$ be two real matrices and $I$ be the identity matrix. It is clear that
$$(X-I)(Y-I)=XY-X-Y+I$$

The question is: show that if $XY=X+Y$, then $I-X$ is invertible?

What I have done: if $XY=X+Y$ then one has $(X-I)(Y-I)=I$, how I continue?

PS: * I know that a matrix $X\in M_n(\mathbb{R})$ is invertible if and only if there exist $X'\in M_n(\mathbb{R})$ such that
$$XX'=X'X=I$$
** Also we know that a matrix $X\in M_n(\mathbb{R})$ is invertible if and only if $|X|\neq0$.

Thank you

If $ XY = X+Y, $ then $$ (X-I)(Y-I) = XY - X - Y + I = X + Y - X - Y + I = I $$
So $$(X-I)^{-1} = (Y-I)$$ and clearly if $(X-I)$ is invertible so is $(I-X) = - (X - I)$. We also have to show that $(Y-I)$ is a left inverse of $(X-I)$, but this follows quite easily.

If $PQ=I$ then $\det(PQ)=\det(I)$, and hence $\det(P)\cdot \det(Q) = 1$. It follows that $\det(P) \neq 0$ and $\det(Q) \neq 0$, i.e. both $P$ and $Q$ are invertible. Moreover, $P^{-1}=Q$ and $Q^{-1}=P$.

If $R$ is an $n$-by-$n$ matrix then $\det(-R)=(-1)^n\cdot \det(R)$ and hence $\det(R) \neq 0 \iff \det(-R) \neq 0$, i.e. $R$ is invertible if and only if $-R$ is invertible.

You have shown that $(X-I)(Y-I)=I$. If $P=X-I$ and $Q=Y-I$ then we can conclude that both $X-I$ and $Y-I$ are invertible. Moreover, it follows that both $I-X$ and $I-Y$ are invertible.


Is there active research in trigonometry? One learns trigonometry in high school/secondary school and either forgets it if one continues onto a career less mathematical or, possibly, uses it extensively in their work, as do engineers and physicists.

As a field of study in mathematics however, it seems that trigonometry is mostly "solved", at least it seems so for the familiar trigonometry in $\mathbb{R}^2$. Is this true, or are there still interesting questions that deal with trigonometry or, perhaps, generalizations of it?

Trigonometry on its own is for the most part no longer a very active area of research; though trigonometry remains a key tools in applied research today (for instance studying the rigidity of origami patterns). And the trig functions as eigenfunctions of the Laplacian have countless generalizations and applications.

One case I've seen of modern fundamental research on trigonometry is Wildberger's work on rational trigonometry, which seeks to reformulate trigonometry as purely algebraic relations of positions (see his web site). But NB that while as far as I can tell this work is sound, it is at the fringes of mainstream mathematical research.

In Āryabhaṭa's sine table the jya values or 'modern values' still have to be fully computed. So the Āryabhaṭa's computational method is still being researched.


$\lim\limits_{x\to -\infty}x^4 \arctan(x)$ problems! (easy I guess) I'm having some trouble calculating this limit:

$$\lim_{x\to -\infty}x^4 \arctan(x)$$

Because I get: $\infty \times (-\frac\pi2) = ?$

Now, I've learned you can't just treat infinity like a number, it's more like an idea or a concept, so you are not supposed to "multiply" $\infty$ with $-\frac\pi2$ and get $-\infty$.

But the answer to this is $-\infty$, so how do I show that in a valid manner?

$$\lim_{x\to -\infty}x^4\arctan(x)=-\infty$$

because
$$\lim_{x\to-\infty}x^4=+\infty$$
and

$$\lim_{x\to-\infty}\arctan(x)=\frac{-\pi}{2}<0$$